Solution manual for Mechanics of Materials, An Integrated Learning System 4th Edition by Philpot by Ace Test Banks

Solution manual for Mechanics of Materials, An Integrated Learning System 4th Edition by Philpot

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Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P1.1 A steel bar of rectangular cross section, 15 mm by 60 mm, is loaded by a compressive force of 110 kN that acts in the longitudinal direction of the bar. Compute the average normal stress in the bar.

Solution The cross-sectional area of the steel bar is A = (15 mm)( 60 mm) = 900 mm2 The normal stress in the bar is F (110 kN )(1,000 N/kN ) = = = 122.222 MPa = 122.2 MPa A 900 mm 2

Ans.

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P1.2 A circular pipe with outside diameter of 4.5 in. and wall thickness of 0.375 in. is subjected to an axial tensile force of 42,000 lb. Compute the average normal stress in the pipe.

Solution The outside diameter D, the inside diameter d, and the wall thickness t are related by D = d + 2t Therefore, the inside diameter of the pipe is d = D − 2t = 4.5 in. − 2 ( 0.375 in.) = 3.75 in. The cross-sectional area of the pipe is



 D − d ) = ( 4.5 in.) − ( 3.75 in.)  = 4.8597 in. (  4 4 2

The average normal stress in the pipe is F 42,000 lb = = = 8,642.6 psi = 8,640 psi A 4.8597 in.2

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P1.3 A circular pipe with an outside diameter of 80 mm is subjected to an axial compressive force of 420 kN. The average normal stress may not exceed 130 MPa. Compute the minimum wall thickness required for the pipe.

Solution From the definition of normal stress, solve for the minimum area required to support a 420 kN load without exceeding a normal stress of 130 MPa F F ( 420 kN )(1,000 N/kN ) =  Amin  = = 3, 230.77 mm 2 2 A  130 N/mm The cross-sectional area of the pipe is given by



(D2 − d 2 ) 4 Set this expression equal to the minimum area and solve for the maximum inside diameter d  2 (80 mm ) − d 2   3, 230.77 mm 2  4 4 2 (80 mm ) − d 2  ( 3, 230.77 mm2 )   d max  47.8169 mm The outside diameter D, the inside diameter d, and the wall thickness t are related by D = d + 2t Therefore, the minimum wall thickness required for the aluminum tube is D − d 80 mm − 47.8169 mm tmin  = = 16.092 mm = 16.09 mm 2 2

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P1.4 Three solid bars, each with square cross sections, make up the axial assembly shown in Figure P1.4/5. Two loads of P = 30 kN are applied to the assembly at flange B, two loads of Q = 18 kN are applied at C, and one load of R = 42 kN is applied at end D. The bar dimensions are b1 = 60 mm, b2 = 20 mm, and b3 = 40 mm. Determine the normal stress in each bar.

FIGURE P1.4/5

Solution Cut an FBD through bar (1). The FBD should include the free end of the assembly at D. We will assume that the internal force in bar (1) is tension. From equilibrium, the force in bar (1) is Fx = − F1 − 2 P + 2Q − R = 0

 F1 = −2 P + 2Q − R = −2 ( 30 kN ) + 2 (18 kN ) − 42 kN = −66 kN = 66 kN (C)

From the given width of bar (1), the cross-sectional area of bar (1) is 2 A1 = b12 = ( 60 mm ) = 3,600 mm2 and thus, the normal stress in bar (1) is F ( −66 kN )(1,000 N/kN ) 1 = 1 = = −18.333 MPa = 18.33 MPa (C) A1 3,600 mm2

Ans.

Cut an FBD through bar (2). The FBD should include the free end of the assembly at D. We will assume that the internal force in bar (2) is tension. From equilibrium, the force in bar (2) is Fx = − F2 + 2Q − R = 0

 F2 = 2Q − R = 2 (18 kN ) − 42 kN = −6 kN = 6 kN (C)

From the given width of bar (2), the cross-sectional area of bar (2) is 2 A2 = b22 = ( 20 mm ) = 400 mm2 The normal stress in bar (2) is Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

2 =

Timothy A. Philpot

F2 ( −6 kN )(1,000 N/kN ) = = −15.000 MPa = 15.00 MPa (C) A2 400 mm2

Ans.

Cut an FBD through bar (3). The FBD should include the free end of the assembly at D. We will assume that the internal force in bar (3) is tension. From equilibrium, the force in bar (3) is Fx = − F3 − R = 0

 F3 = − R = −42 kN = 42 kN (C)

The cross-sectional area of bar (3) is 2 A3 = b32 = ( 40 mm ) = 1,600 mm2 The normal stress in bar (3) is F ( −42 kN )(1, 000 N/kN ) 2 = 2 = = −26.250 MPa = 26.3 MPa (C) A2 1, 600 mm2

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P1.5 Three solid bars, each with square cross sections, make up the axial assembly shown in Figure P1.4/5. Two loads of P = 25 kN are applied to the assembly at flange B, two loads of Q = 15 kN are applied at C, and one load of R = 35 kN is applied at end D. Bar (1) has a width of b1 = 90 mm. Calculate the width b2 required for bar (2) if the normal stress magnitude in bar (2) must equal the normal stress magnitude in bar (1).

FIGURE P1.4/5

 F1 = −2 P + 2Q − R = −2 ( 25 kN ) + 2 (15 kN ) − 35 kN = −55 kN = 55 kN (C)

From the given width of bar (1), the cross-sectional area of bar (1) is 2 A1 = b12 = ( 90 mm ) = 8,100 mm2 and thus, the normal stress in bar (1) is F ( −55 kN )(1, 000 N/kN ) 1 = 1 = = −6.7901 MPa A1 8,100 mm2

 F2 = 2Q − R = 2 (15 kN ) − 35 kN = −5 kN

The normal stress in bar (2) must equal the normal stress in bar (1). Thus,  2 =  1 = −6.7901 MPa Solve for the required area of bar (2): Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

2 =

Timothy A. Philpot

F2 A2

F2 ( −5 kN )(1, 000 N/kN ) = 736.364 mm2 2 2 −6.7901 N/mm The width of bar (2) is therefore:  A2 =

b2 = 736.364 mm2 = 27.136 mm = 27.1 mm

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P1.6 Axial loads are applied with rigid bearing plates to the solid cylindrical rods shown in Figure P1.6/7. One load of P = 1,500 lb is applied to the assembly at A, two loads of Q = 900 lb are applied at B, and two loads of R = 1,300 lb are applied at C. The diameters of rods (1), (2), and (3) are d1 = 0.625 in., d2 = 0.500 in., and d3 = 0.875 in. Determine the axial normal stress in each of the three rods.

FIGURE P1.6/7

Solution Cut an FBD through rod (1). The FBD should include the free end of the assembly at A. We will assume that the internal force in rod (1) is tension. From equilibrium, the force in rod (1) is Fx = − P + F1 = 0

 F1 = P = 1,500 lb = 1,500 lb (T)

Use the given diameter to calculate the cross-sectional area of rod (1):

A1 =



d12 =



( 0.625 in.) = 0.3068 in.2 2

4 4 The normal stress in rod (1) is F 1,500 lb 1 = 1 = = 4,889.24 psi = 4,890 psi (T) A1 0.3068 in.2

Ans.

Cut an FBD through rod (2). The FBD should include the free end of the assembly at A. We will assume that the internal force in rod (2) is tension. From equilibrium, the force in rod (2) is Fx = − P + 2Q + F2 = 0

 F2 = P − 2Q = 1,500 lb − 2 ( 900 lb ) = −300 lb = 300 lb (C)

Use the given diameter to calculate the cross-sectional area of rod (2):

A2 =



d 22 =



( 0.500 in.) = 0.1963 in.2 2

4 4 The normal stress in rod (2) is F −300 lb 2 = 2 = = −1,527.89 psi = 1,528 psi (C) A2 0.1963 in.2

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Cut an FBD through rod (3). The FBD should include the free end of the assembly at A. We will assume that the internal force in rod (3) is tension. From equilibrium, the force in rod (3) is Fx = − P + 2Q − 2 R + F3 = 0

 F3 = P − 2Q + 2 R = 1,500 lb − 2 ( 900 lb ) + 2 (1,300 lb ) = 2,300 lb = 2,300 lb (T)

Use the given diameter to calculate the cross-sectional area of rod (3):

A3 =



d32 =



( 0.8750 in.) = 0.6013 in.2 2

4 4 The normal stress in rod (3) is F 2,300 lb 3 = 3 = = 3,824.92 psi = 3,820 psi (T) A3 0.6013 in.2

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P1.7 Axial loads are applied with rigid bearing plates to the solid cylindrical rods shown in Figure P1.6/7. One load of P = 30 kips is applied to the assembly at A, two loads of Q = 25 kips are applied at B, and two loads of R = 35 kips are applied at C. The normal stress magnitude in aluminum rod (1) must be limited to 20 ksi. The normal stress magnitude in steel rod (2) must be limited to 35 ksi. The normal stress magnitude in brass rod (3) must be limited to 25 ksi. Determine the minimum diameter required for each of the three rods.

FIGURE P1.6/7

Solution Cut an FBD through aluminum rod (1). The FBD should include the free end of the assembly at A. We will assume that the internal force in rod (1) is tension. From equilibrium, the force in rod (1) is Fx = − P + F1 = 0

 F1 = P = 30 kips = 30 kips (T)

The normal stress magnitude in aluminum rod (1) must be limited to 20 ksi. Therefore, the minimum cross-sectional area of rod (1) must be F 30 kips A1  1 = = 1.500 in.2 1 20 ksi The diameter must be

A1 



d12

 d1 



(1.500 in. ) = 1.382 in. 2

Ans.

Cut an FBD through steel rod (2). The FBD should include the free end of the assembly at A. We will assume that the internal force in rod (2) is tension. From equilibrium, the force in rod (2) is Fx = − P + 2Q + F2 = 0

 F2 = P − 2Q = 30 kips − 2 ( 25 kips ) = −20 kips = 20 kips (C)

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

The normal stress magnitude in steel rod (2) must be limited to 35 ksi. Therefore, the minimum crosssectional area of rod (2) must be F −20 kips A2  2 = = 0.5714 in.2 2 35 ksi The diameter of rod (2) must be

A2 



d 22

 d2 

( 0.5714 in. ) = 0.853 in.  4

Ans.

Cut an FBD through brass rod (3). The FBD should include the free end of the assembly at A. We will assume that the internal force in rod (3) is tension. From equilibrium, the force in rod (3) is Fx = − P + 2Q − 2 R + F3 = 0

 F3 = P − 2Q + 2 R = 30 kips − 2 ( 25 kips ) + 2 ( 35 kips ) = 50 kips = 50 kips (T)

The normal stress magnitude in brass rod (3) must be limited to 25 ksi. Therefore, the minimum cross-sectional area of rod (3) must be F 50 kips A3  3 = = 2.0000 in.2 3 25 ksi The diameter of rod (3) must be

A3 



d32

 d3 

( 2.0000 in. ) = 1.596 in.  4

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P1.8 Determine the normal stress in rod (1) for the mechanism shown in Figure P1.8. The diameter of rod (1) is 8 mm, and load P = 2,300 N. Use the following dimensions: a = 120 mm, b = 200 mm, c = 170 mm, and d = 90 mm.

FIGURE P1.8

Solution First, consider an FBD of the pulley to determine the reaction forces exerted on the pulley by the mechanism. Fx = Ax − P − P cos ( 60 ) = 0  Ax = ( 2,300 N ) + ( 2,300 N ) cos ( 60 ) = 3, 450.000 N Fy = Ay − P sin ( 60 ) = 0  Ay = ( 2,300 N ) sin ( 60 ) = 1,991.858 N

FBD of pulley

FBD of mechanism

Next, consider an FBD of the mechanism to determine the force in rod (1). Rod (1) is oriented at an angle of: c + d 170 mm + 90 mm tan  = = = 1.30 b 200 mm   = 52.431

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Rod (1) is a two-force member, and its axial force can be calculated from: M C = Ax c + Ay a − ( F1 cos  )( c + d ) = 0

 F1 =

( 3, 450.000 N )(170 mm ) + (1,991.858 N )(120 mm ) = 5, 207.523 N ( c + d ) cos  (170 mm + 90 mm ) cos ( 52.431) Ax c + Ay a

The area of rod (1) is

A1 =



d12 =



(8 mm ) = 50.265 mm2 2

The normal stress in the rod is thus F 5, 207.532 N 1 = 1 = = 103.601 MPa = 103.6 MPa A1 50.265 mm2

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P1.9 Determine the normal stress in bar (1) for the mechanism shown in Figure P1.9. The area of bar (1) is 2,600 mm2. The distributed load intensities are wC = 12 kN/m and wD = 30 kN/m. Use the following dimensions: a = 7.5 m and b = 3.0 m.

FIGURE P1.9

Solution Consider an FBD of the mechanism. Determine the angle  between rod (1) and the horizontal axis: a 7.5 m tan  = = = 2.5 b 3.0 m   = 68.199 Write an equilibrium equation for the sum of moments about C to compute the force in bar (1). Note: Bar (1) is a two-force member.

wC a a wD a 2a  −  =0 2 3 2 3 wC a 2 2wD a 2 2 + a 2 ( wC + 2wD ) ( 7.5 m ) 12 kN/m + 2 ( 30 kN/m )  6 6  F1 = = = = 242.332 kN b sin  6b sin  6 ( 3.0 m ) sin ( 68.199 )

M C = ( F1 sin  ) b −

The normal stress in bar (1) is thus: F (242.332 kN)(1,000 N/kN) 1 = 1 = = 93.205 N/mm2 = 93.2 MPa (T) 2 A1 2,600 mm

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P1.10 The rigid beam BC shown in Figure P1.10 is supported by rods (1) and (2) that have diameters of 0.875 in. and 1.125 in., respectively. For a uniformly distributed load of w = 4,200 lb/ft, determine the normal stress in each rod. Assume L = 14 ft and a = 9 ft.

FIGURE P1.10

Solution Equilibrium: Calculate the internal forces in rods (1) and (2).  9 ft  M C = − F1 (14 ft ) + ( 4, 200 lb/ft )( 9 ft )  =0  2   F1 = 12.150 kips 9 ft   M B = F2 (14 ft ) − ( 4, 200 lb/ft )( 9 ft ) 14 ft − =0 2    F2 = 25.650 kips

Areas: A1 = A2 =

 4

( 0.875 in.) = 0.601 in.2 2

(1.125 in.) = 0.994 in.2 2

Stresses:

1 =

F1 12.150 kips = = 20.206 ksi = 20.2 ksi A1 0.601 in.2

Ans.

2 =

F2 25.650 kips = = 25.804 ksi = 25.8 ksi A2 0.994 in.2

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P1.11 The rigid beam ABC shown in Figure P1.11 is supported by a pin connection at C and by steel rod (1), which has a diameter of 10 mm. If the normal stress in rod (1) must not exceed 225 MPa, what is the maximum uniformly distributed load w that may be applied to beam ABC? Use dimensions of a = 340 mm, b = 760 mm, and c = 550 mm.

FIGURE P1.11

Solution The cross-sectional area of rod (1) is  2 A1 = (10 mm ) = 78.540 mm 2 4 Since the normal stress in rod (1) must not exceed 225 MPa, the allowable force that can be applied to rod (1) is: F1,allow = 1 A1 = ( 225 N/mm2 )( 78.540 mm2 ) = 17,671.459 N Rod (1) is oriented at an angle of  with respect to the horizontal direction: c 550 mm tan  = = = 0.7237   = 35.893 b 760 mm Consider an FBD of rigid beam ABC. From the moment equilibrium equation about joint C, the relationship between the force in rod (1) and the distributed load w is:  a+b  M C = w ( a + b )   − ( F1 sin  ) b = 0  2  2b ( F1 sin  ) w = 2 (a + b)

Substitute the allowable force F1,allow into this relationship to obtain the maximum distributed load that may be applied to the structure: 2b ( F1 sin  ) w= 2 (a + b) =

2 ( 760 mm )(17, 671.459 N ) sin ( 35.893 )

( 340 mm + 760 mm )

= 13.014 N/mm = 13.01 kN/m

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P1.12 A simple pin-connected truss is loaded and supported as shown in Figure P1.12. The load P is 200 kN. All members of the truss are aluminum pipes that have an outside diameter of 115 mm and a wall thickness of 6 mm. Determine the normal stress in each truss member. Assume truss dimensions of a = 12.0 m, b = 7.5 m, and c = 6.0 m.

FIGURE P1.12

Solution Overall equilibrium: Begin the solution by determining the external reaction forces acting on the truss at supports B and D. Write equilibrium equations that include all external forces. Note that only the external forces (i.e., loads and reaction forces) are considered at this time. The internal forces acting in the truss members will be considered after the external reactions have been computed. The freebody diagram (FBD) of the entire truss is shown. The following equilibrium equations can be written for this structure: Fy = Dy − P = 0

 Dy = P = 200 kN

M D = Pa + Bx c = 0

P (12 m ) Pa =− = −2 P = −400 kN c 6m M B = Pa − Dx c = 0  Bx = −

 Dx =

Pa P (12 m ) = = 2 P = 400 kN c 6m

Method of joints: Before beginning the process of determining the internal forces in the axial members, the geometry of the truss will be used to determine the magnitude of the inclination angles of members AC and BC. Use the definition of the tangent function to determine AC and BC: c 6.0 m tan  AC = = = 1.3333  AC = 53.130 a − b 12.0 m − 7.5 m c 6.0 m tan  BC = = = 0.8  BC = 38.660 b 7.5 m

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Joint A: Begin the solution process by considering an FBD of joint A. Consider only those forces acting directly on joint A. In this instance, two axial members, AB and AC, are connected at joint A. Tension forces will be assumed in each truss member. Fx = FAB + FAC cos  AC = 0 (a) (b) Fy = FAC sin  AC − P = 0 Solve Eq. (b) for FAC: P 200 kN FAC = = = 250.0 kN sin  AC sin ( 53.130) and then compute FAB using Eq. (a): FAB = − FAC cos  AC

= − ( 250.0 kN ) cos ( 53.130 ) = −150.0 kN

Joint D: Next, consider an FBD of joint D. As before, tension forces will be assumed in each truss member. Fx = Dx − FCD = 0 (c) (d) Fy = Dy − FBD = 0 Solve Eq. (c) for FCD: FCD = Dx = 400.0 kN and solve Eq. (d) for FBD: FBD = Dy = 200.0 kN Joint C: Next, consider an FBD of joint C. As before, tension forces will be assumed in each truss member. Fx = FCD + FBC cos  BC − FAC cos  AC = 0 (e) (f) Fy = −FBC sin BC − FAC sin  AC = 0 Solve Eq. (e) for FBC: sin ( 53.130 ) sin  AC FBC = − FAC = − ( 250 kN ) = −320.1562 kN sin  BC sin ( 38.660 ) Eq. (f) can be used as a check on our calculations: Fy = − FBC sin  BC − FAC sin  AC

= − ( −320.1562 kN ) sin ( 38.660 ) − ( 250.0 kN ) sin ( 53.130 ) = 0

Checks!

Mechanics of Materials: An Integrated Learning System, 4th Ed. Section properties: For each of the five truss members:

d = 115 mm − 2 ( 6 mm ) = 103 mm

Timothy A. Philpot



2 2 (115 mm ) − (103 mm )  = 2, 054.602 mm2  4

Normal stress in each truss member: ( −150 kN )(1,000 N/kN ) = −73.007 MPa = 73.0 MPa (C) F  AB = AB = AAB 2, 054.602 mm2 ( 250.0 kN )(1,000 N/kN ) = 121.678 MPa = 121.7 MPa (T) F  AC = AC = AAC 2, 054.602 mm2 ( −320.156 kN )(1,000 N/kN ) = −155.824 MPa = 155.8 MPa (C) F  BC = BC = ABC 2, 054.602 mm2 ( 200.0 kN )(1,000 N/kN ) = 97.342 MPa = 97.3 MPa (T) F  BD = BD = ABD 2, 054.602 mm2 ( 400.0 kN )(1,000 N/kN ) = 194.685 MPa = 194.7 MPa (T) F  CD = CD = ACD 2, 054.602 mm2

Ans. Ans. Ans. Ans. Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P1.13 A horizontal load P is applied to an assembly consisting of two inclined bars, as shown in Figure 1.13. The cross-sectional area of bar (1) is 1.5 in.2, and the cross-sectional area of bar (2) is 1.8 in.2. The normal stress in either bar may not exceed 24 ksi. Determine the maximum load P that may be applied to this assembly. Assume dimensions of a = 16 ft, b = 8 ft, and c = 13 ft.

FIGURE P1.13

Solution Allowable member forces: Using the allowable stresses and the member areas, we can determine the allowable force for each member: F1,allow = 1,allow A1 = ( 24 ksi ) (1.5 in.2 ) = 36 kips

F2,allow =  2,allow A2 = ( 24 ksi ) (1.8 in.2 ) = 43.2 kips

(a) (b)

Equilibrium: The geometry of the two-bar assembly will be used to determine the magnitude of the inclination angles for members AB and BC. We can use the definition of the tangent function to determine AB and BC: a 16 ft tan  AB = = = 1.2308  AB = 50.906 c 13 ft b 8 ft tan  BC = = = 0.6154  BC = 31.608 c 13 ft Consider a free-body diagram (FBD) of joint B. The following equilibrium equations can be written for this joint: Fx = P − F1 cos  AB − F2 cos  BC = 0

Fy = F1 sin  AB − F2 sin BC = 0

Erroneous approach for finding maximum load P: Since we are trying to calculate P, the temptation at this point in the solution is to substitute the values from Equations (a) and (b) into Eq. (c) and simply solve for P: P = F1 cos  AB + F2 cos  BC = ( 36 kips ) cos ( 50.906 ) + ( 43.2 kips ) cos ( 31.608 )

= 59.493 kips (e) However, if we use the values from Equations (a) and (b) in Eq. (d), we find that equilibrium is not satisfied:

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Fy = F1 sin  AB − F2 sin  BC = ( 36 kips ) sin ( 50.906 ) − ( 43.2 kips ) sin ( 31.608 ) = 5.299 kips  0 Equilibrium must always be satisfied; therefore, we must conclude that F1 and F2 will not have the allowable values of Equations (a) and (b). The answer obtained in Eq. (e) is incorrect because equilibrium is not satisfied.

Correct method for calculating the capacity of the two-bar assembly: The allowable load that can be applied to this two-bar assembly will be the load P that produces the allowable load in either member (1) or member (2). Let’s return to Eq. (d), only this time, we are going to make an assumption. We will assume that the force in member (1) will control the capacity of the two-bar assembly. If this assumption is true, then the force in member (1) will equal its allowable force as given in Eq. (a), and the force in member (2) will be less than its allowable force as given in Eq. (b). sin ( 50.906 ) sin  AB F2 = F1 = F1 = 1.4809 F1 sin  BC sin ( 31.608 )

= 1.4809 ( 36 kips ) = 53.311 kips  F2,allow = 43.2 kips

N.G.

This calculation shows that the force in member (2) will exceed its allowable force when the force in member (1) equals its allowable force. Therefore, our assumption is proved incorrect. This result shows us that the force in member (2) will control the capacity of the two-bar assembly. We’ll return to Eq. (d), only this time, we know that member (2) will control. Set the force in member (2) to its allowable force from Eq. (b) and solve for the force in member (1) that is required to satisfy equilibrium. sin ( 31.608 ) sin  BC F1 = F2 = F2 = 0.6753F2 sin  AB sin ( 50.906 )

= 0.6753 ( 43.2 kips ) = 29.172 kips  F1,allow = 36 kips

O.K.

We now know the forces in members (1) and (2) that will satisfy the equilibrium equations without exceeding the allowable force in either member. Finally, we use these values to determine the load P from Eq. (c): P = F1 cos  AB + F2 cos  BC

= ( 29.172 kips ) cos ( 50.906 ) + ( 43.2 kips ) cos ( 31.608 ) = 55.188 kips

= 55.2 kips

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P1.14 The rectangular bar shown in Figure P1.14 is subjected to a uniformly distributed axial loading of w = 13 kN/m and a concentrated force of P = 9 kN at B. Determine the magnitude of the maximum normal stress in the bar and its location x. Assume a = 0.5 m, b = 0.7 m, c = 15 mm, and d = 40 mm. FIGURE P1.14

Solution Equilibrium: Draw an FBD for the interval between A and B where 0  x  a . Write the following equilibrium equation: +

⎯⎯ →Fx = (13 kN/m)(1.2 m − x) − (9 kN) − F = 0  F = (13 kN/m)(1.2 m − x) − (9 kN) The largest force in this interval occurs at x = 0 where F = 6.6 kN. In the interval between B and C where a  x  a + b , and write the following equilibrium equation: +

⎯⎯ →Fx = (13 kN/m)(1.2 m − x) − F = 0  F = (13 kN/m)(1.2 m − x) The largest force in this interval occurs at x = a where F = 9.1 kN. Maximum Normal Stress: (9.1 kN)(1,000 N/kN)  max = = 15.17 MPa at x = 0.5 m (15 mm)(40 mm)

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P1.15 The solid 1.25 in. diameter rod shown in Figure P1.15 is subjected to a uniform axial distributed loading along its length of w = 750 lb/ft. Two concentrated loads also act on the rod: P = 2,000 lb and Q = 1,000 lb. Assume a = 16 in. and b = 32 in. Determine the normal stress in the rod at the following locations: (a) x = 10 in. (b) x = 30 in. FIGURE P1.15

Solution (a) x = 10 in. Equilibrium: Draw an FBD for the interval between A and B where 0  x  a , and write the following equilibrium equation: +

⎯⎯ → Fx = (750 lb/ft)(1 ft/12 in.)(48 in. − x) + (2,000 lb) + (1,000 lb) − F = 0  F = (62.5 lb/in.)(48 in. − x) + 3,000 lb At x = 10 in., F = 5,375 lb. Stress: The normal stress at this location can be calculated as follows.

=



(1.25 in.)2 = 1.227185 in.2

5,375 lb = 4,379.944 psi = 4,380 psi 1.227185 in.2

Ans.

(b) x = 30 in. Equilibrium: Draw an FBD for the interval between B and C where a  x  a + b , and write the following equilibrium equation: +

⎯⎯ → Fx = (750 lb/ft)(1 ft/12 in.)(48 in. − x) + (1,000 lb) − F = 0  F = (62.5 lb/in.)(48 in. − x) + 1,000 lb At x = 30 in., F = 2,125 lb. Stress: The normal stress at this location can be calculated as follows. 2,125 lb = = 1,731.606 psi = 1,730 psi 1.227185 in.2

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P1.16 A block of wood is tested in direct shear using the test fixture shown below. The dimensions of the test specimen are a = 3.75 in., b = 1.25 in., c = 2.50 in., and d = 6.0 in. During the test, a load of P = 1,590 lb produces a shear failure in the wood specimen. What is the magnitude of the average shear stress in the wood specimen at failure?

FIGURE P1.16

Solution Visualize the surface that will be exposed when the specimen fails. The area of this surface will be AV = cd = ( 2.50 in.)( 6.0 in.) = 15.0 in.2 The average shear stress in the specimen at failure is thus P 1,590 lb  avg = = = 106.0 psi AV 15.0 in.2

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P1.17 A cylindrical rod of diameter d = 0.625 in. is attached to a plate by a cylindrical rubber grommet. The plate has a thickness of t = 0.875 in. If the axial load on the rod is P = 175 lb, what is the average shear stress on the cylindrical surface of contact between the rod and the grommet?

FIGURE P1.17

Solution Visualize the contact surface between the rod and the grommet. It will be a cylinder with a diameter of d and a height of t. The area of this cylinder will be AV =  dt =  ( 0.625 in.)( 0.875 in.) = 1.718 in.2 The average shear stress between the rod and the grommet is thus P 175 lb  avg = = = 101.9 psi AV 1.718 in.2

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P1.18 Two wood boards, each 19 mm thick, are joined by the glued finger joint shown in Figure P1.18. The finger joint will fail when the average shear stress in the glue reaches 940 kPa. Determine the shortest allowable length d of the cuts if the joint is to withstand an axial load of P = 5.5 kN. Use a = 23 mm and b = 184 mm. FIGURE P1.18

Solution We are considering the shear strength of the glued joint. The minimum shear area that is required for this connection can be determined from the load P and the shear strength of the glue. Consequently, we will need at least this much area P ( 5.5 kN )(1, 000 N/kN ) AV ,min = = = 5,851.064 mm 2  0.940 N/mm 2 to transmit the load P through the joint, based on the shear strength of the glue. For this particular joint, there are seven surfaces that will be glued. Each of these surfaces has a length of d and a thickness of 19 mm. Accordingly, the minimum length d required for each of the finger joints is 7 dt  5,851.064 mm 2 d 

5,851.064 mm 2 = 44.0 mm 7 (19 mm )

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P1.19 For the connection shown in Figure P1.19, determine the average shear stress produced in the 7/8 in. diameter bolts if the applied load is P = 32,000 lb.

FIGURE P1.19

Solution There are three bolts, and it is always assumed that each bolt supports an equal portion of the external load P. Therefore, the shear force V carried by each bolt is 32, 000 lb V= = 10, 666.667 lb 3 bolts The bolts in this connection act in single shear. The cross-sectional area of a single bolt is

Abolt =



2 d bolt =



(7 / 8 in.) 2 =



(0.875 in.)2 = 0.6013 in.2

4 4 4 Therefore, the average shear stress in each bolt is V 10,666.667 lb = = = 17,738.739 psi = 17,740 psi Abolt 0.6013 in.2

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P1.20 For the clevis connection shown in Figure P1.20, determine the maximum applied load P that can be supported by the 15 mm diameter pin if the average shear stress in the pin must not exceed 130 MPa. FIGURE P1.20

Solution Consider an FBD of the bar that is connected by the clevis, including a portion of the pin. If the shear force acting on each exposed surface of the pin is denoted by V, then the shear force on each pin surface is related to the load P by: Fx = P − V − V = 0  P = 2V The area of the pin surface exposed by the FBD is simply the cross-sectional area of the pin:

Apin =



2 d pin =



(15 mm) 2 = 176.715 mm 2

If the average shear stress in the pin must be limited to 130 MPa, the maximum shear force V on a single cross-sectional surface must be limited to V =  Abolt = (130 N/mm2 )(176.715 mm2 ) = 22,972.95 N Therefore, the maximum load P that may be applied to the connection is P = 2V = 2 ( 22,972.95 N ) = 45,945.9 N = 45.9 kN

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P1.21 The five-bolt connection shown in Figure P1.21 must support an applied load of P = 160 kips. If the average shear stress in the bolts must be limited to 30 ksi, what is the minimum bolt diameter that may be used for this connection?

FIGURE P1.21

Solution There are five bolts, and it is assumed that each bolt supports an equal portion of the external load P. Therefore, the shear force carried by each bolt is 160 kips V= = 32 kips 5 bolts Since the average shear stress must be limited to 30 ksi, each bolt must provide a shear area of at least: 32 kips/bolt AV = = 1.0667 in.2 /bolt 30 ksi Each bolt in this connection acts in double shear; therefore, two cross-sectional bolt surfaces are available to transmit shear stress in each bolt. AV 1.0667 in.2 /bolt Abolt = = = 0.5333 in.2 per bolt surface 2 surfaces per bolt 2 surfaces/bolt The minimum bolt diameter must be



2 d bolt  0.5333 in.2

 d bolt  0.824 in.

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P1.22 The handle shown in Figure P1.22 is attached to a 40 mm diameter shaft with a square shear key. The forces applied to the lever are P = 1,300 N. If the average shear stress in the key must not exceed 150 MPa, determine the minimum dimension a that must be used if the key is 25 mm long. The overall length of the handle is L = 0.70 m. FIGURE P1.22

Solution To determine the shear force V that must be resisted by the shear key, sum moments about the center of the shaft (which will be denoted O):  700 mm   700 mm   40 mm  M O = (1,300 N)  + (1,300 N)    − V = 0   2  2   2  V = 45,500 N Since the average shear stress in the key must not exceed 150 MPa, the shear area required is V 45,500 N AV  = = 303.3333 mm 2 2  150 N/mm The shear area in the key is given by the product of its length L (i.e., 25 mm) and its width a. Therefore, the minimum key width a is AV 303.3333 mm 2 Ans. a = = 12.1333 mm = 12.13 mm L 25 mm

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P1.23 An axial load P is supported by the short steel column shown in Figure P1.23. The column has a crosssectional area of 14,500 mm2. If the average normal stress in the steel column must not exceed 75 MPa, determine the minimum required dimension a so that the bearing stress between the base plate and the concrete slab does not exceed 8 MPa. Assume b = 420 mm.

FIGURE P1.23

Solution Since the normal stress in the steel column must not exceed 75 MPa, the maximum column load is Pmax =  A = (75 N/mm2 )(14,500 mm2 ) = 1,087,500 N The maximum column load must be distributed over a large enough area so that the bearing stress between the base plate and the concrete slab does not exceed 8 MPa; therefore, the minimum plate area is P 1,087,500 N Amin = = = 135,937.5 mm2 b 8 N/mm2 The area of the plate is a ×b. Since b = 420, the minimum length of a must be Amin = 135,937.5 mm2 = a  b

a 

135,937.5 mm2 = 324 mm 420 mm

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P1.24 The two wooden boards shown in Figure P1.24 are connected by a 0.5 in. diameter bolt. Washers are installed under the head of the bolt and under the nut. The washer dimensions are D = 2 in. and d = 5/8 in. The nut is tightened to cause a tensile stress of 9,000 psi in the bolt. Determine the bearing stress between the washer and the wood.

FIGURE P1.24

Solution The tensile stress in the bolt is 9,000 psi; therefore, the tension force that acts in the bolt is



Fbolt =  bolt Abolt = (9,000 psi) (0.5 in.) 2 = (9,000 psi)(0.196350 in.2 ) = 1,767.146 lb 4 The contact area between the washer and the wood is

Awasher =



(2 in.) 2 − (0.625 in.) 2  = 2.834796 in.2 4

Thus, the bearing stress between the washer and the wood is 1,767.146 lb b = = 623 psi 2.834796 in.2

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P1.25 For the beam shown in Figure P1.25, the allowable bearing stress for the material under the supports at A and B is b = 800 psi. Assume w = 2,100 lb/ft, P = 4,600 lb, a = 20 ft, and b = 8 ft. Determine the size of square bearing plates required to support the loading shown. Dimension the plates to the nearest ½ in. FIGURE P1.25

Solution Equilibrium: Using the FBD shown, calculate the beam reaction forces.

 20 ft  M A = By (20 ft) − (2,100 lb/ft)(20 ft)  − (4,600 lb)(28 ft) = 0  2   By = 27, 440 lb  20 ft  M B = − Ay (20 ft) + (2,100 lb/ft)(20 ft)  − (4,600 lb)(8 ft) = 0  2   Ay = 19,160 lb Bearing plate at A: The area of the bearing plate required for support A is 19,160 lb AA  = 23.950 in.2 800 psi Since the plate is to be square, its dimensions must be

width  23.950 in.2 = 4.894 in.

use 5 in.  5 in. bearing plate at A

Ans.

Bearing plate at B: The area of the bearing plate required for support B is 27,440 lb AB  = 34.300 in.2 800 psi Since the plate is to be square, its dimensions must be

width  34.300 in.2 = 5.857 in.

use 6 in.  6 in. bearing plate at B

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P1.26 A wood beam rests on a square post. The vertical reaction force of the beam at the post is P = 1,300 lb. The square post has cross-sectional dimensions of a = 6.25 in. The beam has a width of b = 1.50 in. and a depth of d = 7.50 in. What is the average bearing stress in the wood beam?

FIGURE P1.26

Solution Contact area: Visualize the contact area between the beam and the post. The contact area is Ab = ab = ( 6.25 in.)(1.50 in.) = 9.375 in.2 Bearing stress: The average bearing stress in the wood beam is P 1,300 lb b = = = 138.7 psi Ab 9.375 in.2

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P1.27 The pulley shown in Figure P1.27 is connected to a bracket with a circular pin of diameter d = 6 mm. Each vertical side of the bracket has a width of b = 25 mm and a thickness of t = 4 mm. If the pulley belt tension is P = 570 N, what is the average bearing stress produced in the bracket by the pin? FIGURE P1.27

Solution Pulley FBD: Consider an FBD of the pulley with the belt tensions. From equilibrium, the bracket exerts horizontal and vertical reaction forces Rx and Ry, respectively, on the pulley. Fx = Rx − P cos ( 30 ) = 0  Rx = P cos ( 30 ) = ( 570 N ) cos ( 30 ) = 493.634 N Fy = Ry + P + P sin ( 30 ) = 0  Ry = − P − P sin ( 30 ) = −570 N − ( 570 N ) sin ( 30 ) = −855.0 N

The resultant force exerted on the pulley by the bracket is thus

R = Rx2 + Ry2 =

( 493.634 N ) + ( −855.0 N ) 2

= 987.269 N Bearing stress in the bracket: From Newton’s Third Law, the pulley pin exerts an equal force R on the bracket. The bracket has two vertical pieces (i.e., a plate on each side of the pulley). The resultant force R is divided equally between these two vertical pieces. Therefore, the force exerted by the pin on one of the vertical bracket pieces is 493.635 N. The average bearing stress in the bracket is based on the projected area of the pin. Therefore, the average bearing stress produced in the bracket by the pin is R/2 R/2 493.635 N Ans. b = = = = 20.568 MPa = 20.6 MPa Ab dt ( 6 mm )( 4 mm )

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P1.28 The d = 15 mm diameter solid rod shown in Figure P1.28 passes through a D = 20 mm diameter hole in the support plate. When a load P is applied to the rod, the rod head rests on the support plate. The support plate has a thickness of b = 12 mm. The rod head has a diameter of a = 30 mm and the head has a thickness of t = 10 mm. If the normal stress produced in the rod by load P is 225 MPa, determine: (a) the average bearing stress acting between the support plate and the rod head. (b) the average shear stress produced in the rod head. (c) the punching shear stress produced in the support plate by the rod head.

FIGURE P1.28

Solution The cross-sectional area of the rod is:

Arod =



(15 mm)2 = 176.715 mm2

4 The tensile stress in the rod is 225 MPa; therefore, the tension force in the rod is Frod =  rod Arod = (225 N/mm2 )(176.715 mm2 ) = 39,760.782 N (a) The contact area between the support plate and the rod head is

Acontact =



(30 mm) 2 − (20 mm) 2  = 392.699 mm 2 4

Thus, the bearing stress between the support plate and the rod head is 39,760.782 N b = = 101.3 MPa 392.699 mm 2 (b) In the rod head, the area subjected to shear stress is equal to the perimeter of the rod times the thickness of the head. AV =  (15 mm)(10 mm) = 471.239 mm2 and therefore, the average shear stress in the rod head is 39,760.782 N = = 84.4 MPa 471.239 mm 2 (c) In the support plate, the area subjected to shear stress is equal to the product of the rod head perimeter and the thickness of the plate. AV =  (30 mm)(12 mm) = 1,130.973 mm2 and therefore, the average punching shear stress in the support plate is 39,760.782 N = = 35.2 MPa 1,130.973 mm2

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P1.29 A hollow box beam ABCD is supported at A by a pin that passes through the beam as shown in Figure P1.29. The box beam is also supported by a roller that is located at B. The beam dimensions are a = 2.5 ft, b = 5.5 ft, and c = 3.5 ft. Two equal concentrated loads of P = 2,750 lb are placed on the box beam at points C and D. The box beam has a wall thickness of t = 0.375 in., and the pin at A has a diameter of 0.750 in. Determine: (a) the average shear stress in the pin at A. (b) the average bearing stress in the box beam at A.

FIGURE P1.29

Solution Equilibrium: Determine the reaction force exerted on the beam by the pin at A. M B = − Ay a − Pb − P ( b + c ) = 0 b + (b + c ) 2b + c = −P a a 2 ( 5.5 ft ) + 3.5 ft = − ( 2, 750 lb ) 2.5 ft = −15,950 lb

Ay = − P

Average shear stress in the pin at A: The pin diameter is 0.750 in. The cross-sectional area of the pin is

Apin =



d2 =



( 0.750 in.) = 0.4418 in.2 2

4 4 From the support detail figure, we observe that this pin acts in double shear; therefore, the shear area of the pin is AV = 2 Apin = 2 ( 0.4418 in.2 ) = 0.8836 in.2 The average shear stress in the pin at A is thus Ay V 15,950 lb = = = = 18, 051.71 psi = 18, 050 psi AV AV 0.8836 in.2

Ans.

Average bearing stress in the box beam at A: The average bearing stress produced in the box beam by the pin is based on the projected area of the pin. The projected area is equal to the pin diameter times the wall thickness of the box beam, taking into account that there are two walls that contact the pin. Therefore, the average bearing stress in the box beam is Ay 15,950 lb b = = = 28,355.56 psi = 28, 400 psi Ans. 2dt 2 ( 0.750 in.)( 0.375 in.) Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P1.30 Rigid bar ABC shown in Figure P1.30 is supported by a pin at bracket A and by tie rod (1). Tie rod (1) has a diameter of 5 mm, and it is supported by double-shear pin connections at B and D. The pin at bracket A is a single-shear connection. All pins are 7 mm in diameter. Assume a = 600 mm, b = 300 mm, h = 450 mm, P = 900 N, and  = 55°. Determine the following: (a) the normal stress in rod (1) (b) the average shear stress in pin B (c) the average shear stress in pin A FIGURE P1.30

Solution Equilibrium: Using the FBD shown, calculate the reaction forces that act on rigid bar ABC. M A = F1 sin(36.87)(600 mm)

−(900 N)sin (55)(900 mm) = 0  F1 = 1,843.092 N

Fx = Ax − (1,843.092 N)cos(36.87) + (900 N)cos(55) = 0  Ax = 958.255 N Fy = Ay + (1,843.092 N)sin (36.87) − (900 N)sin (55) = 0  Ay = −368.618 N The resultant force at A is

A = (958.255 N)2 + (−368.618 N)2 = 1,026.709 N (a) Normal stress in rod (1).  Arod = (5 mm) 2 = 19.635 mm 2 4 1,843.092 N  rod = = 93.9 MPa 19.635 mm 2

Ans.

(b) Shear stress in pin B. The cross-sectional area of a 7 mm diameter pin is:

Apin =



(7 mm) 2 = 38.485 mm 2

4 Pin B is a double shear connection; therefore, its average shear stress is 1,843.092 N  pin B = = 23.9 MPa 2(38.485 mm2 ) (c) Shear stress in pin A. Pin A is a single shear connection; therefore, its average shear stress is 1,026.709 N  pin A = = 26.7 MPa 38.485 mm 2

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P1.31 The bell crank shown in Figure P1.31 is in equilibrium for the forces acting in rods (1) and (2). The bell crank is supported by a 10 mm diameter pin at B that acts in single shear. The thickness of the bell crank is 5 mm. Assume a = 65 mm, b = 150 mm, F1 = 1,100 N, and  = 50°. Determine the following: (a) the average shear stress in pin B (b) the average bearing stress in the bell crank at B

FIGURE P1.31

Solution Equilibrium: Using the FBD shown, calculate the reaction forces that act on the bell crank. M B = −(1,100 N)sin(50)(65 mm)

+ F2 (150 mm) = 0  F2 = 365.148 N Fx = Bx − (1,100 N) cos(50) +365.148 N = 0  Bx = 341.919 N

Fy = By + (1,100 N)sin(50) = 0  By = −842.649 N The resultant force at B is

B = (341.919 N)2 + (−842.649 N)2 = 909.376 N (a) Shear stress in pin B. The cross-sectional area of the 10 mm diameter pin is:

Apin =



(10 mm) 2 = 78.540 mm 2

4 Pin B is a single shear connection; therefore, its average shear stress is 909.376 N  pin B = = 11.58 MPa 78.540 mm 2

Ans.

(b) Bearing stress in the bell crank at B. The average bearing stress produced in the bell crank by the pin is based on the projected area of the pin. The projected area is equal to the pin diameter times the bell crank thickness. Therefore, the average bearing stress in the bell crank is 909.376 N Ans. b = = 18.19 MPa (10 mm)(5 mm)

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P1.32 The beam shown in Figure P1.32 is supported by a pin at C and by a short link AB. If w = 30 kN/m, determine the average shear stress in the pins at A and C. Each pin has a diameter of 25 mm. Assume L = 1.8 m and  = 35°.

FIGURE P1.32

Solution Equilibrium: Using the FBD shown, calculate the reaction forces that act on the beam.

 1.8 m  M C = − F1 sin(35)(1.8 m) + (30 kN/m)(1.8 m)  =0  2   F1 = 47.0731 kN

Fx = Cx − (47.0731 kN)cos(35) = 0  Cx = 38.5600 kN  1.8 m  M B = C y (1.8 m) − (30 kN/m)(1.8 m)  =0  2   C y = 27.0000 kN The resultant force at C is

C = (38.5600 kN)2 + (27.0000 kN) 2 = 47.0731 kN Shear stress in pin A. The cross-sectional area of a 25 mm diameter pin is:

Apin =



(25 mm) 2 = 490.8739 mm 2

4 Pin A is a single shear connection; therefore, its average shear stress is 47,073.1 N  pin A = = 95.9 MPa 490.8739 mm 2 Shear stress in pin C. Pin C is a double shear connection; therefore, its average shear stress is 47,073.1 N  pin C = = 47.9 MPa 2(490.8739 mm2 )

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P1.33 The bell-crank mechanism shown in Figure P1.33 is in equilibrium for an applied load of P = 7 kN applied at A. Assume a = 200 mm, b = 150 mm, and  = 65°. Determine the minimum diameter d required for pin B for each of the following conditions: (a) The average shear stress in the pin may not exceed 40 MPa. (b) The bearing stress in the bell crank may not exceed 100 MPa. (c) The bearing stress in the support bracket may not exceed 165 MPa. FIGURE P1.33

Solution Equilibrium: Using the FBD shown, calculate the reaction forces that act on the bell crank. M B = (7,000 N)sin(65)(200 mm)

− F1 (150 mm) = 0  F1 = 8, 458.873 N Fx = Bx + (7,000 N)cos(65) +8, 458.873 N = 0  Bx = −11, 417.201 N

Fy = By − (7,000 N)sin(65) = 0  By = 6,344.155 N The resultant force at B is

B = (−11,417.201 N)2 + (6,344.155 N)2 = 13,061.423 N (a) The average shear stress in the pin may not exceed 40 MPa. The shear area required for the pin at B is 13,061.423 N AV  = 326.536 mm 2 2 40 N/mm Since the pin at B is supported in a double shear connection, the required cross-sectional area for the pin is A Apin = V = 163.268 mm 2 2 and therefore, the pin must have a diameter of 4 Ans. d (163.268 mm2 ) = 14.42 mm 

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

(b) The bearing stress in the bell crank may not exceed 100 MPa. The projected area of pin B on the bell crank must equal or exceed 13,061.423 N Ab  = 130.614 mm 2 2 100 N/mm The bell crank thickness is 8 mm; therefore, the projected area of the pin is Ab = (8 mm)d. Calculate the required pin diameter d: 130.614 mm 2 Ans. d = 16.33 mm 8 mm (c) The bearing stress in the support bracket may not exceed 165 MPa. The pin at B bears on two 6 mm thick support brackets. Thus, the minimum pin diameter required to satisfy the bearing stress limit on the support bracket is 13,061.423 N Ab  = 79.160 mm2 2 165 N/mm d

79.160 mm2 = 6.60 mm 2(6 mm)

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P1.34 A structural steel bar with a 4.0 in. × 0.875 in. rectangular cross section is subjected to an axial load of 45 kips. Determine the maximum normal and shear stresses in the bar.

Solution The maximum normal stress in the steel bar is F 45 kips  max = = = 12.86 ksi A ( 4.0 in.)( 0.875 in.) The maximum shear stress is one-half of the maximum normal stress

 max =

 max 2

= 6.43 ksi

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P1.35 A stainless steel rod of circular cross section will be used to carry an axial load of 30 kN. The maximum stresses in the rod must be limited to 100 MPa in tension and 60 MPa in shear. Determine the required minimum diameter for the rod.

Solution Based on the allowable 100 MPa tension stress limit, the minimum cross-sectional area of the rod must equal or exceed ( 30 kN )(1, 000 N/kN ) = 300 mm2 F Amin  =  max 100 N/mm2 For the 60 MPa shear stress limit, the minimum cross-sectional area of the rod must be equal or exceed ( 30 kN )(1, 000 N/kN ) = 250 mm2 F Amin  = 2 max 2 ( 60 N/mm 2 ) Therefore, the rod must have a cross-sectional area of at least 300 mm2 to satisfy both the normal and shear stress limits. The minimum rod diameter D is therefore



2 d min  300 mm 2

 d min  19.54 mm

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P1.36 Two wooden members, each having a width of b = 1.50 in. and a depth of d = 0.5 in., are joined by the simple glued scarf joint shown in Figure P1.36/37. Assume  = 40°. If the allowable shear stress for the glue used in the joint is 90 psi, what is the largest axial load P that may be applied? FIGURE P1.36

Solution The angle  shown for the scarf joint is 40°. The normal force N perpendicular to the scarf joint can be expressed as N = P sin  and the shear force V parallel to the scarf joint can be expressed as V = P cos  The cross-sectional area of the bar is A = bd but the area along the inclined scarf joint is  d  A An =  b = sin   sin   Consequently, the shear stress nt parallel to the scarf joint can be expressed as V P cos  P  nt = = = sin  cos  An A / sin  A Given that the shear stress nt must be limited to 90 psi, solve for the maximum load P as: P  nt  sin  cos  A P 90 psi  sin 40 cos 40 bd ( 90 psi )(1.50 in.)( 0.5 in.) = 137.083 lb = 137.1 lb P sin 40 cos 40

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P1.37 Two wooden members, each having a width of b = 4.50 in. and a depth of d = 1.75 in., are joined by the simple glued scarf joint shown in Figure P1.36/37. Assume  = 35°. Given that the compressive axial load is P = 900 lb, what are the normal stress and shear stress magnitudes in the glued joint? FIGURE P1.37

Solution The angle  shown for the scarf joint is 35°. The normal force N perpendicular to the scarf joint can be expressed as N = P sin  and the shear force V parallel to the scarf joint can be expressed as V = P cos  The cross-sectional area of the bar is A = bd but the area along the inclined scarf joint is  d  A An =  b = sin   sin   Consequently, the normal stress n magnitude perpendicular to the inclined scarf joint can be expressed as N P sin  P n = = = sin 2  An A / sin  A 900 lb = sin 2 35 = 37.6 psi Ans. ( 4.50 in.)(1.75 in.) and the shear stress nt magnitude parallel to the scarf joint can be expressed as V P cos  P  nt = = = sin  cos  An A / sin  A =

900 lb sin 35 cos 35 = 53.7 psi ( 4.50 in.)(1.75 in.)

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P1.38 Two aluminum plates, each having a width of b = 7.0 in. and a thickness of t = 0.625 in., are welded together as shown in Figure P1.38/39. Assume a = 4.0 in. For a load of P = 115 kips, determine (a) the normal stress that acts perpendicular to the weld and (b) the shear stress that acts parallel to the weld.

FIGURE P1.38/39

Solution Begin by calculating the angle  for the weld joint. a 4.0 in. tan  = = = 0.5714 b 7.0 in.  = 29.745 The normal force N perpendicular to the weld joint can be expressed as N = P cos and the shear force V parallel to the weld joint can be expressed as V = P sin  The cross-sectional area of the bar is A = bt but the area along the inclined weld joint is A  b  An =  t = cos   cos   (a) Normal stress perpendicular to the weld: The normal stress n magnitude perpendicular to the inclined weld joint can be expressed as N P cos  P n = = = cos 2  An A / cos  A 115 kips = cos 2 29.745 = 19.82 ksi Ans. ( 7.0 in.)( 0.625 in.)

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

(b) Shear stress parallel to the weld: The shear stress nt magnitude parallel to the weld joint can be expressed as V P sin  P  nt = = = sin  cos  An A / cos  A =

115 kips sin 29.745 cos 29.745 = 11.32 ksi ( 7.0 in.)( 0.625 in.)

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P1.39 Two aluminum plates, each having a width of b = 5.0 in. and a thickness of t = 0.75 in., are welded together as shown in Figure P1.38/39. Assume a = 2.0 in. Specifications require that the normal and shear stress magnitudes acting in the weld material may not exceed 35 ksi and 24 ksi, respectively. Determine the largest axial load P that can be applied to the aluminum plates.

FIGURE P1.38/39

Solution Begin by calculating the angle  for the weld joint. a 2.0 in. tan  = = = 0.4 b 5.0 in.  = 21.801 The normal force N perpendicular to the weld joint can be expressed as N = P cos and the shear force V parallel to the weld joint can be expressed as V = P sin  The cross-sectional area of the bar is A = bt but the area along the inclined weld joint is A  b  An =  t = cos   cos   Normal stress perpendicular to the weld: The normal stress n magnitude perpendicular to the inclined weld joint can be expressed as N P cos  P n = = = cos2  An A / cos  A The normal stress perpendicular to the weld joint may not exceed 35 ksi. The allowable load P that satisfies this constraint is  A ( 35 ksi )( 5.0 in.)( 0.75 in.) P  n2 = = 152.25 kips cos  cos 2 ( 21.801 )

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Shear stress parallel to the weld: The shear stress nt magnitude parallel to the weld joint can be expressed as V P sin  P  nt = = = sin  cos  An A / cos  A The shear stress parallel to the weld joint may not exceed 24 ksi. The allowable load P that satisfies this requirement is ( 24 ksi )( 5.0 in.)( 0.75 in.) = 261.00 kips  nt A P = sin  cos  sin ( 21.801 ) cos ( 21.801 ) Allowable load P: The largest axial load P that can be applied to the aluminum plates is thus P  152.3 kips

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P1.40 Two wooden member are glued together as shown in Figure P1.40. Each member has a width of b = 1.50 in. and a depth of d = 3.50 in. Use  = 75°. Determine the average shear stress magnitude in the glue joint if P = 1,300 lb. FIGURE P1.40

Solution Using the notion of symmetry, we will consider an FBD for only the upper half of the left-hand wood piece. The central angle  for the joint is 75°. The shear force V parallel to the upper half joint can be expressed as P  V = cos 2 2

The cross-sectional area of the upper half member is d  A = b  2 but the area along the inclined upper half joint is bd An =  2sin 2 Consequently, the shear stress nt magnitude parallel to the joint can be calculated as P  cos V 2 = P sin  cos   nt = = 2 bd An bd 2 2 2sin



2 1,300 lb 75 75 = sin cos = 119.6 psi 2 2 (1.50 in.)( 3.50 in.)

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P1.41 Two bars are connected with a welded butt joint as shown in Figure P1.41. The bar dimensions are b = 200 mm and t = 50 mm, and the angle of the weld is  = 35°. The bars transmit a force of P = 250 kN. What is the magnitude of the average shear stress that acts on plane AB? FIGURE P1.41

Solution The angle  shown for the weld joint is 35°. The normal force N perpendicular to the weld joint can be expressed as N = P sin  and the shear force V parallel to the weld joint can be expressed as V = P cos  The cross-sectional area of the bar is A = bt but the area along the inclined weld joint is A  t  An =  b = sin   sin   Shear stress parallel to the weld: The shear stress nt magnitude parallel to the weld joint can be expressed as V P cos  P  nt = = = sin a cos  An A / sin  A

( 250 kN )(1, 000 N/kN ) sin 35 cos 35 = 11.75 MPa ( 200 mm )( 50 mm )

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P2.1 When an axial load is applied to the ends of the two-segment rod shown in Figure P2.1, the total elongation between joints A and C is 7.5 mm. The segment lengths are a = 1.2 m and b = 2.8 m. In segment (2), the normal strain is measured as 2,075 m/m. Determine: (a) the elongation of segment (2). (b) the normal strain in segment (1) of the rod. FIGURE P2.1

Solution (a) From the definition of normal strain, the elongation in segment (2) can be computed as  2 =  2 L2 = (2,075 10−6 mm/mm)(2,800 mm) = 5.81 mm

Ans.

(b) The combined elongations of segments (1) and (2) is given as 7.5 mm. Therefore, the elongation that occurs in segment (1) must be 1 =  total −  2 = 7.50 mm − 5.81 mm = 1.69 mm The strain in segment (1) can now be computed:  1.69 mm 1 = 1 = = 1, 408.33 10−6 mm/mm = 1, 408 μm/m L1 1, 200 mm

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P2.2 The two bars shown in Figure P2.2 are used to support load P. When unloaded, joint B has coordinates (0, 0). After load P is applied, joint B moves to the coordinate position (–0.55 in., –0.15 in.). Assume a = 15 ft, b = 27 ft, c = 11 ft, and d = 21 ft. Determine the normal strain in each bar.

FIGURE P2.2

Solution Given a = (15 ft )(12 in./ft ) = 180 in.

b = ( 27 ft )(12 in./ft ) = 324 in. c = (11 ft )(12 in./ft ) = 132 in. d = ( 21 ft )(12 in./ft ) = 252 in. The initial length of bar AB is LAB =

(180 in.) + (132 in.) = 223.2129 in. 2

and its length after deformation is LAB =

(180 in. − 0.55 in.) + (132 in. + 0.15 in.) = 222.8585 in. 2

The strain in bar AB is thus 222.8585 in. − 223.2129 in. −0.3544 in.  AB = = = −1,587.6  10−6 in./in. = −1,588 με 223.2129 in. 223.2129 in.

Ans.

The initial length of bar BC is LBC =

( 324 in.) + ( 252 in.) = 410.4632 in. 2

and its length after deformation is  = LBC

( 324 in. + 0.55 in.) + ( 252 in. + 0.15 in.) = 410.9895 in. 2

The strain in bar BC is thus 410.9895 in. − 410.4632 in. 0.5263 in.  BC = = = 1,282.2  10−6 in./in. = 1,282 με 410.4632 in. 410.4632 in.

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P2.3 Pin-connected rigid bars AB, BC, and CD are initially held in the positions shown in Figure P2.3 by taut wires (1) and (2). The bar lengths are a = 24 ft and b = 18 ft. Joint C is given a horizontal displacement of 5 in. to the right. (Note that this displacement causes both joints B and C to move to the right and slightly downward.) What is the change in the average normal strain in wire (1) after the displacement?

FIGURE P2.3

Solution Given a = ( 24 ft )(12 in./ft ) = 288 in. b = (18 ft )(12 in./ft ) = 216 in.

The initial length of wire (1) is the distance between joints A and C: L1 =

( 288 in.) + ( 216 in.) = 360 in. 2

Joint C displaces to the right by u = 5 in. Joint C also moves down. Use the geometry of the displaced structure to calculate a′ using the Pythagorean theorem: a = a 2 − u 2 =

( 288 in.) − ( 5.0 in.) = 287.9566 in. 2

After displacement, the coordinates of joint C, relative to joint A, are: C = ( b + u, a) = ( 216 in. + 5 in., 287.9566 in.) = ( 221 in., 287.9566 in.) The length of wire (1) after deformation is L1 =

( 221 in.) + ( 287.9566 in.) = 362.9876 in. 2

The change in strain in wire (1) is thus 362.9876 in. − 360 in. 2.9876 in.  AB = = = 8,298.9  10−6 in./in. = 8,300 με 360 in. 360 in.

Ans.

For completeness, note that the vertical displacement v is calculated as v = a − a = 287.9566 in. − 288 in. = −0.0434 in.

Mechanics of Materials: An Integrated Learning System, 4th Ed. P2.4 Bar (1) has a length of L1 = 2.50 m, and bar (2) has a length of L2 = 0.65 m. Initially, there is a gap of  = 3.5 mm between the rigid plate at B and bar (2). After application of the loads P to the rigid plate at B, the rigid plate moved to the right, stretching bar (1) and compressing bar (2). The normal strain in bar (1) was measured as 2,740 m/m after the loads P were applied. Determine the normal strain produced in bar (2).

Timothy A. Philpot

FIGURE P2.4

Solution From the measured strain, compute the deformation of bar (1) as: 1 = 1L1 = ( 2,740  10−6 mm/mm ) ( 2.5 m )(1,000 mm/m ) = 6.8500 mm Since it is attached to bar (1), rigid plate B displaces to the right by an equal amount. uB = 1 = 6.8500 mm → Plate B moves 3.5 mm before contacting bar (2). Therefore, the deformation produced in bar (2) is equal to the difference  2 = 1 −  = 6.8500 mm − 3.5 mm = 3.3500 mm However, bar (2) will be contracting; therefore, the deformation in bar (2) is negative.  2 = −3.3500 mm The normal strain produced in bar (2) is thus:  −3.3500 mm 2 = 2 = = −5,153.85  10−6 mm/mm = −5,150 με L2 650 mm

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P2.5 In Figure P2.5, rigid bar ABC is support by a pin at B and by post (1) at A. However, there is a gap of  = 10 mm between the rigid bar at A and post (1). After load P is applied to the rigid bar, point C moves left by 8 mm. If the length of post (1) is L1 = 1.6 m, what is the average normal strain that is produced in post (1)? Use dimensions of a = 1.25 m and b = 0.85 m.

FIGURE P2.5

Solution The deflected shape of the rigid bar is shown greatly exaggerated in the figure to the right. The rigid bar rotates about pin B. Consequently, a horizontal movement at C causes a downward deflection at A. Use similar triangles to determine the downward deflection at A: uC vA = b a a 1.25 m  vA = uC = (8 mm ) = 11.7647 mm  b 0.85 m

As the rigid bar deflects at A, it travels downward  = 10 mm before it contacts post (1). The amount of contraction in post (1) is equal to the difference between vA and . Therefore, 1 = vA −  = 11.7647 mm − 10 mm = 1.7647 mm The post becomes shorter after contact with the rigid bar; therefore, the deformation in the post is negative: 1 = −1.7647 mm The average normal strain produced in post (1) is thus  −1.7647 mm 1 = 1 = = −1,102.9 10−6 mm/mm = −1,103 με L1 1,600 mm

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P2.6 The rigid bar ABC is supported by three bars as shown in Figure P2.6. Bars (1) attached at A and C are identical, each having a length of L1 = 160 in. Bar (2) has a length of L2 = 110 in.; however, there is a clearance of c = 0.25 in. between bar (2) and the pin in the rigid bar at B. There is no strain in the bars before the load P is applied, and a = 50 in. After application of load P, the tensile normal strain in bar (2) is measured as 960 . What is the normal strain in bars (1)? FIGURE P2.6

Solution From the strain given for bar (2), we know that the elongation of bar (2) must be:  2 =  2 L2 = ( 960  10−6 in./in.) (110 in.)

= 0.1056 in. To stretch bar (2) by this amount, the rigid bar must move downward by 0.1056 in. However, there is a clearance in the pin at B of 0.25 in., and this clearance means that the rigid bar must deflect downward by 0.25 in. before it even begins to stretch bar (2). Consequently, the rigid bar will have to deflect downward by: vB = 0.1056 in. + 0.25 in.

= 0.3556 in. (downward) The deflection of the rigid bar at A is equal to the deflection at B. We know this because of the symmetry of the assembly…when the rigid bar deflects downward, it remains horizontal. Thus, v A = vB = 0.3556 in. (downward) As joint A deflects downward, it stretches bar (1). Since there are no gaps or clearances at A, any movement downward of joint A will elongate bar (1) by the same amount; thus, 1 = v A = 0.3556 in. From the definition of strain,  0.3556 in. 1 = 1 = = 2,222.5  10−6 in./in. = 2,220 με L1 160 in.

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed. P2.7 Rigid bar ABCD is supported by two bars as shown in Figure P2.7. There is no strain in the vertical bars before load P is applied. After load P is applied, the normal strain in bar (2) is measured as –3,300 m/m. Use the dimensions L1 = 1,600 mm, L2 = 1,200 mm, a = 240 mm, b = 420 mm, and c = 180 mm. Determine: (a) the normal strain in bar (1). (b) the normal strain in bar (1) if there is a 1 mm gap in the connection at pin C before the load is applied. (c) the normal strain in bar (1) if there is a 1 mm gap in the connection at pin B before the load is applied.

Timothy A. Philpot

FIGURE P2.7

Solution (a) The deformation of bar (2) can be calculated from the specified strain in bar (2)  2 =  2 L2

= ( −3,330  10−6 ) (1, 200 mm ) = −3.9600 mm

Since the connection at C is perfect (i.e., no gaps or clearances), the contraction of bar (2) corresponds to a downward deflection of the rigid bar at C of exactly the same magnitude. Thus, vC = 3.9600 mm (downward). From the deformation diagram of rigid bar ABCD vB v = C a a+b a 240 mm  vB = vC = ( −3.9600 mm ) = −1.4400 mm = 1.4400 mm  a+b 240 mm + 420 mm Since the connection at B is perfect (i.e., no gaps or clearances), the downward deflection of the rigid bar at B will create a deformation of the same magnitude in bar (1). Therefore, 1 = 1.4400 mm (elongation), and thus, from the definition of normal strain:  1.4400 mm Ans. 1 = 1 = = 900  10−6 mm/mm = 900 με L1 1,600 mm

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

(b) The deformation of bar (2) can be calculated from the specified strain in bar (2)  2 =  2 L2

= ( −3,330  10−6 ) (1, 200 mm ) = −3.9600 mm

Here, the connection at C is not perfect—there is a 1 mm gap in the connection at C. Therefore, the rigid bar must move downward by 1 mm before it contacts bar (2). We know that bar (2) has been contracted by −3.9600 mm, and we also know that the rigid bar moves 1 mm downward before even starting to contract bar (2). Thus, joint C on the rigid bar must have moved downward vC = 1 mm + 3.9600 mm = 4.9600 mm (downward). From the deformation diagram of rigid bar ABCD vB v = C a a+b a 240 mm  vB = vC = ( −4.9600 mm ) = −1.8036 mm = 1.8036 mm  a+b 240 mm + 420 mm The connection at B is perfect (i.e., no gaps or clearances); thus, the downward deflection of the rigid bar at B will create a deformation of the same magnitude in bar (1). Therefore, 1 = 1.8036 mm (elongation), and thus, from the definition of normal strain:  1.8036 mm Ans. 1 = 1 = = 1,127.3  10−6 mm/mm = 1,127 με L1 1,600 mm

= ( −3,330  10−6 ) (1, 200 mm ) = −3.9600 mm

Since the connection at C is perfect (i.e., no gaps or clearances), the contraction of bar (2) corresponds to a downward deflection of the rigid bar at C of exactly the same magnitude. Thus, vC = 3.9600 mm (downward). Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

From the deformation diagram of rigid bar ABCD vB v = C a a+b a 240 mm  vB = vC = ( −3.9600 mm ) = −1.4400 mm = 1.4400 mm  a+b 240 mm + 420 mm Here, the connection at B is not perfect—there is a 1 mm gap at joint B. The rigid bar must move downward 1 mm at B before it begins to elongate bar (1). Therefore, the elongation of bar (1) is 1 = vB − 1 mm = 1.4400 mm − 1 mm = 0.4400 mm

From the definition of normal strain:  0.4400 mm 1 = 1 = = 275  10−6 mm/mm = 275 με L1 1,600 mm

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed. P2.8 The sanding-drum mandrel shown in Figure P2.6 is made for use with a hand drill. The mandrel is made from a rubber-like material that expands when the nut is tightened to secure the sanding sleeve placed over the outside surface. If the diameter D of the mandrel increases from 2.00 in. to 2.15 in. as the nut is tightened, determine (a) the average normal strain along a diameter of the mandrel. (b) the circumferential strain at the outside surface of the mandrel.

Timothy A. Philpot

FIGURE P2.8

Solution (a) The change in strain along a diameter is found from D 2.15 in. − 2.00 in. D = = = 0.075 in./in. D 2.00 in. (b) Note that the circumference of a circle is given by D. The change in strain around the circumference of the mandrel is found from C  (2.15 in.) −  (2.00 in.) C = = = 0.075 in./in. C  (2.00 in.)

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P2.9 The normal strain in a suspended bar of material of varying cross section due to its own weight is given by the expression y/3E where  is the specific weight of the material, y is the distance from the free (i.e., bottom) end of the bar, and E is a material constant. Determine, in terms of , L, and E, (a) the change in length of the bar due to its own weight. (b) the average normal strain over the length L of the bar. (c) the maximum normal strain in the bar.

Solution (a) The strain of the suspended bar due to its own weight is given as y = 3E Consider a slice of the bar having length dy. In general,  = L. Applying this definition to the bar slice, the deformation of slice dy is given by y d  =  dy = dy 3E Since this strain expression varies with y, the total deformation of the bar must be found by integrating d over the bar length: L

= 0

y 3E

dy =

  y2 

 L2

  = 3E  2  0 6E

(b) The average normal strain is found by dividing the expression above for  by the bar length L  L2 / 6 E  L  avg = = L 6E

Ans.

(c) Since the given strain expression varies with y, the maximum normal strain occurs at the maximum value of y, that is, at y = L: y L  max =  y = L = = Ans. 3E y = L 3E

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P2.10 A steel cable is used to support an elevator cage at the bottom of a 2,000-ft deep mineshaft. A uniform normal strain of 250 in./in. is produced in the cable by the weight of the cage. At each point, the weight of the cable produces an additional normal strain that is proportional to the length of the cable below the point. If the total normal strain in the cable at the cable drum (upper end of the cable) is 700 in./in., determine (a) the strain in the cable at a depth of 500 ft. (b) the total elongation of the cable.

Solution Call the vertical coordinate y and establish the origin of the y axis at the lower end of the steel cable. The strain in the cable has a constant term (i.e.,  = 250 ) and a term (we will call it k) that varies with the vertical coordinate y.  = 250 10−6 + k y The problem states that the normal strain at the cable drum (i.e., y = 2,000 ft) is 700 in./in. Knowing this value, the constant k can be determined 700 10−6 = 250  10−6 + k (2, 000 ft) 700 10−6 − 250 10−6 = 0.225  10−6 /ft 2, 000 ft Substituting this value for k in the strain expression gives  = 250 10−6 + (0.225 10−6 /ft) y k =

(a) At a depth of 500 ft, the y coordinate is y = 1,500 ft. Therefore, the cable strain at a depth of 500 ft is Ans.  = 250  10−6 + (0.225  10−6 /ft)(1,500 ft) = 587.5  10−6 = 588 με (b) The total elongation is found by integrating the strain expression over the cable length; thus, 2000

 =   dy =   250  10−6 + (0.225  10−6 /ft) y  dy 0 2000

 0.225  10−6 2  −6 = (250  10 ) y + y  2  0 = 0.950 ft = 11.40 in.

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P2.11 A thin rectangular polymer plate PQRS of width b = 400 mm and height a = 180 mm is shown in Figure P2.11. The plate is deformed so that corner Q is displaced upward by c = 3.0 mm and corner R is displaced leftward by the same amount. Determine the shear strain at corner P after deformation. FIGURE P2.10

Solution Before deformation, the angle of the plate at P was 90° or /2 radians. We must now determine the plate angle at P′ after deformation. The difference between these angles is the shear strain at corner P. After point Q displaces upward by 3 mm, we calculate the angle  shown in the sketch to the right as c 3 mm tan  = = = 0.007500 b 400 mm   = 0.429710 After point R displaces 3 mm to the left, we calculate the angle  as c 3 mm tan  = = = 0.016667 a 180 mm  = 0.954841

After deformation, the angle of the plate at P is 90 +  −  = 90 + 0.954841 − 0.429710 = 90.525131 = 1.579962 rad The difference in the plate angle at P before and after deformation is the shear strain:  −  = 1.579962 rad 2   = −0.009165 rad = −9,170 μrad

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P2.12 A thin triangular plate PQR forms a right angle at point Q. During deformation, point Q moves to the right by u = 0.8 mm and upward by v = 1.3 mm to new position Q′, as shown in Figure P2.12. Determine the shear strain  at corner Q′ after deformation. Use a = 225 mm, b = 455 mm, and d = 319.96 mm.

FIGURE P2.12

Solution Before deformation, the angle of the plate at Q was 90° or /2 radians. We must now determine the plate angle at Q′ after the plate has been deformed. The difference between these angles is the shear strain. After deformation, the angle  is a+u 225 mm + 0.8 mm tan  = = = 0.70286 d + v 319.96 mm + 1.3 mm  = 35.10168 and the angle  is b−u 455 mm − 0.8 mm tan  = = = 1.41380 d + v 319.96 mm + 1.3 mm   = 54.72779 Therefore, after deformation, the angle of the plate at Q′ is  +  = 35.10168 + 54.72779 = 89.82946 = 1.56782 rad The difference in the angle at Q before and after deformation is the shear strain:  −  = 1.56782 rad 2   = 0.002976 rad = 2,980 μrad

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P2.13 A thin triangular plate PQR forms a right angle at point Q. During deformation, point Q moves to the left by u = 2.0 mm and upward by v = 5.0 mm to new position Q′, as shown in Figure P2.13. Determine the shear strain  at corner Q′ after deformation. Use c = 700 mm,  = 28°, and  = 62°.

FIGURE P2.13

Solution Let us first determine some dimensions of the triangular plate before deformation occurs. Based on the figure to the right and the specified values of c = 700 mm,  = 28°, and  = 62°, we calculate e sin  =  e = c sin  = ( 700 mm ) sin ( 28 ) = 328.6301 mm c f cos  =  f = c cos  = ( 700 mm ) cos ( 28 ) = 618.0633 mm c d sin  =  d = f sin  = ( 618.0633 mm ) sin ( 28 ) = 290.1632 mm f b cos  =  b = f cos  = ( 618.0633 mm ) cos ( 28 ) = 545.7175 mm f a = c − b = 700 mm − 545.7175 mm = 154.2825 mm

After deformation, corner Q moves to the left by u = 2.0 mm and upward by v = 5.0 mm. Before deformation, the interior angle at Q was a right angle. Our next task is to determine the interior angle at Q′ after deformation. Based on the figure to the right, we calculate a + v 154.2825 mm + 5.0 mm 159.2825 mm tan   = = = = 0.545183 d + u 290.1632 mm + 2.0 mm 292.1632 mm   = 28.59848 = 0.499138 rad b − v 545.7175 mm − 5.0 mm 540.7175 mm tan   = = = = 1.850738 d + u 290.1632 mm + 2.0 mm 292.1632 mm    = 61.61654 = 1.075412 rad Therefore, after deformation, the angle of the plate at Q′ is   +   = 0.499138 rad + 1.075412 rad = 1.574550 rad The difference in the angle at Q before and after deformation is the shear strain:  −  = 1.574550 rad 2   = −0.003753 rad = −3,750 μrad

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P2.14 A thin square polymer plate is deformed into the position shown by the dashed lines in Figure P2.14. Assume that a = 800 mm, b = 85 mm, and c = 960 mm. Determine the shear strain xy after deformation: (a) at corner P. (b) at corner Q.

FIGURE P2.14

Solution (a) Shear strain at corner P. After deformation, the angle that side PR makes with the vertical axis can be found from b 85 mm sin  = = = 0.088542 c 960 mm  = 0.088658 rad Before deformation, the interior angle at P was /2 radians. After deformation, the interior angle at P is /2 + 0.088658 radians. Refer to Figure 2.5a and 2.5b in the text. If we denote the interior angle at P after deformation as  −P 2 then the shear strain at corner P is:



−P =



+ 0.088658 rad

  P = −0.088658 rad = −0.0887 rad

Ans.

(b) Shear strain at corner Q. The change of angle at Q has the same magnitude as the angle change at P. After deformation, the interior angle at Q is therefore /2 – 0.088658 radians. If we denote the interior angle at Q after deformation as  −Q 2 then the shear strain at corner Q is:   −  Q = − 0.088658 rad 2 2   Q = 0.088658 rad = 0.0887 rad Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P2.15 A thin square plate PQRS is symmetrically deformed into the shape shown by the dashed lines in Figure P2.15. The initial length of diagonals PR and QS is d = 295 mm. After deformation, diagonal PR has a length of dPR = 295.3 mm, and diagonal QS has a length of dQS = 293.7 mm. For the deformed plate, determine: (a) the normal strain of diagonal QS. (b) the shear strain xy at corner P.

FIGURE P2.15

Solution (a) Normal strain of diagonal QS. The deformation of diagonal QS is QS = 293.7 mm − 295 mm = −1.3 mm Thus, the normal strain of diagonal QS is −1.3 mm  QS = = −0.004407 mm/mm = −4,410 με 295 mm

Ans.

(b) Shear strain at corner P. Before deformation, side PS makes an angle of /4 radians with the x axis. After deformation, the angle that side PS makes with the x axis can be found from ( 293.7 mm / 2 ) = 0.994582 rad tan  = ( 295.3 mm / 2 )

 = 0.782682 rad Before deformation, the interior angle at P (i.e., angle SPQ) was /2 radians. After deformation, the interior angle at P is 2(0.782682 radians) = 1.565364. Refer to Figure 2.5a and 2.5b in the text. If we denote the interior angle at P after deformation as  −P 2 then the shear strain at corner P is:



−  P = 1.565364 rad  P =

 2

− 1.565364 rad = 0.005433 rad = 5,430 μrad

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P2.16 An airplane has a half-wingspan of 96 ft. Determine the change in length of the aluminum alloy [ = 13.1×10–6/°F] wing spar if the plane leaves the ground at a temperature of 59°F and climbs to an altitude where the temperature is –70°F.

Solution The change in temperature between the ground and the altitude in flight is T = Tfinal − Tinitial = −70°F − 59°F = −129°F The thermal strain is given by T =  T = (13.110−6 /°F)(−129°F) = −1.6899 10−3 in./in. and thus the deformation in the 96 ft long wing is  = T L = (−1.6899 10−3 in./in)(96 ft) = −0.16223 ft = −1.947 in.

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P2.17 A square high-density polyethylene [ = 158×10–6/°C] plate has a width of 300 mm. A 180 mm diameter circular hole is located at the center of the plate. If the temperature of the plate increases by 40°C, determine: (a) the change in width of the plate. (b) the change in diameter of the hole.

Solution The thermal strain is given by T =  T = (158.0 10−6 /°C)(40°C) = 6,320 10−6 mm/mm (a) Change in width of the plate. The width w of the plate changes by w =  T w = ( 6,320 10−6 mm/mm ) ( 300 mm ) = 1.896 mm

Ans.

(b) Change in diameter of the hole. The diameter d of the hole changes by d = T d = ( 6,320 10−6 mm/mm ) (180 mm ) = 1.138 mm

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P2.18 A circular steel [ = 6.5×10−6/°F] band is to be mounted on a circular steel drum. The outside diameter of the drum is 50 in. The inside diameter of the circular band is 49.95 in. The band will be heated and then slipped over the drum. After the band cools, it will tightly grip the drum. This process is called shrink fitting. If the temperature of the band is 72°F before heating, compute the minimum temperature to which the band must be heated so that it can be slipped over the drum. Assume that an extra 0.05 in. in diameter is needed for clearance so that the band can be easily slipped over the drum. Assume that the drum diameter remains constant.

Solution To slip over the drum, the inside diameter of the steel band must be increased from 49.95 in. to 50.05 in. (including the 0.05 in. diameter tolerance). Therefore, the deformation required for the inside diameter is  d = 50.05 in. − 49.95 in. = 0.10 in. To produce this deformation, the temperature must be changed by at least  d =  T d =  Td = 0.10 in. T =

d 0.10 in. = = 308.00 F −6  d ( 6.5 10 / F ) ( 49.95 in.)

The initial temperature of the band is 72°F; therefore, the minimum final temperature to which the band must be heated is Ans. Tfinal  Tinitial + T = 72F + 308.0F = 380F

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P2.19 At a temperature of 60°F, a gap of a = 0.125 in. exists between the two polymer bars shown in Figure P2.19. Bar (1) has a length of L1 = 40 in. and a coefficient of thermal expansion of  = 47×10−6/°F. Bar (2) has a length of L2 = 24 in. and a coefficient of thermal expansion of  = 66×10−6/°F, respectively. The supports at A and D are rigid. What is the lowest temperature at which the gap is closed? FIGURE P2.19

Solution Write expressions for the temperature-induced deformations and set this equal to the gap a: 1T L1 +  2 T L2 = a T 1L1 +  2 L2  = a

T =

a 0.125 in. = = 36.085°F −6 1L1 +  2 L2 ( 47  10 / °F ) ( 40 in.) + ( 66  10−6 / °F ) ( 24 in.)

Since the initial temperature is 60°F, the temperature at which the gap is closed is 96.1°F.

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P2.20 An aluminum pipe has a length of 60 m at a temperature of 10°C. An adjacent steel pipe at the same temperature is 5 mm longer. At what temperature will the aluminum pipe be 15 mm longer than the steel pipe? Assume that the coefficient of thermal expansion for the aluminum is 22.5×10-6/°C and that the coefficient of thermal expansion for the steel is 12.5×10-6/°C.

Solution The length of the aluminum pipe after a change in temperature can be expressed as ( Lfinal ) A = ( Linitial ) A + ( L ) A = ( Linitial ) A +  A T ( Linitial ) A

(a)

Similarly, the length of the steel pipe after the same change in temperature is given by ( Lfinal )S = ( Linitial )S + ( L )S = ( Linitial ) S +  S T ( Linitial ) S

(b)

From the problem statement, we are trying to determine the temperature change that will cause the final length of the aluminum pipe to be 15 mm longer than the steel pipe. This requirement can be expressed as (c) ( Lfinal ) A = ( Lfinal )S +15 mm Substitute Eqs. (a) and (b) into Eq. (c) to obtain the following relationship ( Linitial ) A +  AT ( Linitial ) A = ( Linitial )S + S T ( Linitial )S + 15 mm Collect the terms with T on the left-hand side of the equation  AT ( Linitial ) A − S T ( Linitial )S = ( Linitial )S − ( Linitial ) A + 15 mm Factor out T T  A ( Linitial ) A −  S ( Linitial )S  = ( Linitial )S − ( Linitial ) A + 15 mm and thus, T can be expressed as ( Linitial )S − ( Linitial ) A + 15 mm T =  A ( Linitial ) A −  S ( Linitial ) S Convert all length dimensions to units of millimeters and solve 60, 005 mm − 60, 000 mm + 15 mm T = −6 (22.5 10 /°C)(60, 000 mm) − (12.5 10−6 /°C)(60, 005 mm) 20 mm 20 mm = = 1.35 mm/°C − 0.75 mm/°C 0.60 mm/°C = 33.3°C Initially, the pipes were at a temperature of 10°C. With the temperature change determined above, the temperature at which the aluminum pipe is 15 mm longer than the steel pipe is Ans. Tfinal = Tinitial + T = 10°C + 33.3°C = 43.3°C

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P2.21 The simple mechanism shown in Figure P2.21 can be calibrated to measure temperature change. Use dimensions of a = 25 mm, b = 90 mm, and L1 = 180 mm. The coefficient of thermal expansion for member (1) is 23.0×10–6/°C. Determine the horizontal displacement of pointer tip D of the mechanism shown in response to a temperature increase of 35°C. Assume that pointer BCD is not affected significantly by temperature change. FIGURE P2.21

Solution In response to the 35°C temperature increase, member (1) elongates by the amount 1 = T L1 = ( 23.0  10−6 /°C) ( 35°C)(180 mm ) = 0.1449 mm The horizontal displacement of joint B is equal to the deformation of member (1): u B = 1 = 0.1449 mm The scale reading can be determined from similar triangles uD uB = b a b 90 mm  uD = uB = ( 0.1449 mm ) = 0.522 mm  a 25 mm

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P2.22 For the assembly shown in Figure P2.22, high-density polyethylene bars (1) and (2) each have coefficients of thermal expansion of  = 88×10–6/°F. If the temperature of the assembly is decreased by 50°F from its initial temperature, determine the resulting displacement of pin B. Assume b = 32 in. and  = 55°.

FIGURE P2.22

Solution First, we will determine the initial length of member (1). Find the dimension a, which is shown in the figure to the right. b b 32 in. tan  = a = = = 22.40664 in. a tan  tan 55 The initial length of member (1) can be found from the Pythagorean theorem: L1 = a 2 + b 2 =

( 22.40664 in.) + ( 32 in.) = 39.06479 in. 2

In response to the 50°F temperature decrease, member (1) contracts by 1 = T L1 = (88  10−6 /°F) ( −50°F)( 39.06479 in.) = −0.17189 in. Therefore, the length of member (1) after the temperature decrease is: L1 = L1 + 1 = 39.06479 in. + ( −0.17189 in.) = 38.89290 in. By symmetry, the dimension a must remain unchanged. The dimension b′ is thus

L1 = a 2 + ( b )  b =

( L  ) − a = (38.89290 in.) − ( 22.40664 in.) = 31.78994 in. 2

The displacement of joint B from its original position B to its deflected position B′ is uB = b − b = 32 in. − 31.78994 in. = 0.210 in.

Ans.

P3.1 At the proportional limit, a 2 in. gage length of a 0.500 in. diameter alloy rod has elongated 0.0035in. and the diameter has been reduced by 0.0003 in. The total tension force on the rod was 5.45 kips. Determine the following properties of the material: (a) the proportional limit. (b) the modulus of elasticity. (c) Poisson’s ratio.

Solution (a) Proportional Limit: The bar cross-sectional area is   A = d 2 = (0.500 in.)2 = 0.196350 in.2 4 4 and thus, the normal stress corresponding to the 5.45 kip force is 5.45 kips = = 27.7566 ksi 0.196350 in.2 Based on the problem statement, the stress in the bar is equal to the proportional limit; therefore,  PL = 43.0 ksi (b) Modulus of Elasticity: The longitudinal strain in the bar at the proportional limit is  0.0035 in.  long = = = 0.001750 in./in. L 2 in. The modulus of elasticity is therefore  27.7566 ksi E= = = 15,860 ksi  long 0.001750 in./in. (c) Poisson’s ratio: The longitudinal strain in the bar was calculated previously as  long = 0.001750 in./in. The lateral strain can be determined from the reduction of the diameter: d −0.0003 in.  lat = = = −0.000600 in./in. d 0.500 in. Poisson’s ratio for this specimen is therefore  −0.000600 in./in.  = − lat = − = 0.343  long 0.001750 in./in.

Ans.

P3.2 A solid circular rod with a diameter of d = 16 mm is shown in Figure P3.2. The rod is made of an aluminum alloy that has an elastic modulus of E = 72 GPa and Poisson’s ratio of  = 0.33. When subjected to the axial load P, the diameter of the rod decreases by 0.024 mm. Determine the magnitude of load P.

FIGURE P3.2

Solution The lateral strain in the rod is d −0.024 mm  lat = = = −1,500  10−6 mm/mm d 16 mm Using Poisson’s ratio, compute the corresponding longitudinal strain:  −1,500  10 −6 mm/mm  long = − lat = − = 4,545.455  10−6 mm/mm  0.33 Use Hooke’s law to calculate the stress in the rod:  = Elong = (72,000 MPa)(4,545.455  10−6 mm/mm) = 327.273 MPa The cross-sectional area of the rod is:   A = d 2 = (16 mm) 2 = 201.062 mm 2 4 4 Consequently, the force P that acts on the rod must be P =  A = (327.273 MPa)(201.062 mm2 ) = 65,802.086 N = 65.8 kN

Ans.

P3.3 The polymer bar shown in Figure P3.3 has a width of b = 50 mm, a depth of d = 100 mm, and a height of h = 270 mm. At a compressive load of P = 135 kN, the bar height contracts by h = –2.50 mm, and the bar depth elongates by d = 0.38 mm. At this load, the stress in the polymer bar is less than its proportional limit. Determine: (a) the modulus of elasticity. (b) Poisson’s ratio (c) the change in the bar width b.

FIGURE P3.3

Solution (a) Modulus of elasticity: The bar cross-sectional area is

A = (50 mm)(100 mm) = 5,000 mm2

and thus, the normal stress corresponding to the 135 kN axial load is ( −135 kN )(1,000 N/kN ) = −27.0 MPa = 5,000 mm2 The longitudinal strain in the bar is  h −2.5 mm  long = = = = −9,259.259  10−6 mm/mm L h 270 mm The modulus of elasticity is therefore  −27.0 MPa E= = = 2,916 MPa = 2.92 GPa  long −9,259.259  10−6 mm/mm (b) Poisson’s ratio: The lateral strain can be determined from the elongation of the bar depth: d 0.380 mm  lat = = = 3,800  10−6 mm/mm d 100 mm Poisson’s ratio for this specimen is therefore  lat 3,800  10−6 mm/mm  =− =− = 0.410  long −9,259.259  10−6 mm/mm (c) Change in the bar width b: The change in bar width b can be found from the lateral strain: b  lat = b b =  lat b = ( 3,800  10−6 mm/mm ) ( 50 mm ) = 0.1900 mm

Ans.

P3.4 A 0.625 in. thick rectangular alloy bar is subjected to a tensile load P by pins at A and B as shown in Figure P3.4. The width of the bar is w = 2.00 in. Strain gages bonded to the specimen measure the following strains in the longitudinal (x) and transverse (y) directions: x = 1,140  and y = −315 . (a) Determine Poisson’s ratio for this specimen. (b) If the measured strains were produced by an axial load of P = 17.4 kips, what is the modulus of elasticity for this specimen? FIGURE P3.4

Solution (a) Poisson’s ratio for this specimen is   −315 με  = − lat = − y = − = 0.276  long x 1,140 με

Ans.

(b) The bar cross-sectional area is

A = ( 2.00 in.)( 0.625 in.) = 1.25 in.2

and so the normal stress for an axial load of P = 17.4 kips is 17.4 kips = = 13.920 ksi 1.25 in.2 The modulus of elasticity is thus  13.920 ksi E= = = 12,210.5 ksi = 12,200 ksi  long 1,140  10−6 in./in.

Ans.

P3.5 A 40 mm by 40 mm square ABCD (i.e., a = 40 mm) is drawn on a rectangular bar prior to loading (see Figure P3.5a). A uniform normal stress of  = 54 MPa is then applied to the ends of the rectangular bar, and square ABCD is deformed into the shape of a rhombus, as shown in the Figure P3.5b. The dimensions of the rhombus after loading are b = 56.88 mm and c = 55.61 mm. Determine the modulus of elasticity for the material. Assume that the material behaves elastically for the applied stress.

FIGURE P3.5a

FIGURE P3.5b

Solution The length of diagonal AC (and also diagonal BD) before deformation is LAC = LBD = a 2 + a 2 = a 2 = ( 40mm ) 2 = 56.569 mm

After the normal stress is applied, the longitudinal strain is c − a 2 55.61 mm − 56.569 mm  long = = = −16.9448  10−3 mm/mm 56.569 mm a 2 The modulus of elasticity is thus  −54 MPa E= = = 3,186.8 MPa = 3.19 GPa  long −16.9448  10−3 mm/mm

Ans.

P3.6 A nylon [E = 2,500 MPa;  = 0.4] bar is subjected to an axial load that produces a normal stress of . Before the load is applied, a line having a slope of 3:2 (i.e., 1.5) is marked on the bar as shown in Figure P3.6. Determine the slope of the line when  = 105 MPa.

FIGURE P3.6

Solution From the given stress and elastic modulus, compute the longitudinal strain in the bar:  105 MPa long = = = 0.04200 mm/mm E 2,500 MPa Use Poisson’s ratio to calculate the lateral strain: lat = −long = −(0.4)(0.04200 mm/mm) = −0.01680 mm/mm Before deformation, the slope of the line is 3 slope = = 1.500 2 After deformation, the slope of the line is 3(1 − 0.01680) 2.950 slope = = = 1.415 2(1 + 0.04200) 2.084

Ans.

P3.7 A nylon [E = 360 ksi;  = 0.4] rod (1) having a diameter of d1 = 2.50 in. is placed inside a steel [E = 29,000 ksi;  = 0.29] tube (2) as shown in Figure P3.7. The inside diameter of the steel tube is d2 = 2.52 in. An external load P is applied to the nylon rod, compressing it. At what value of P will the space between the nylon rod and the steel tube be closed?

FIGURE P3.7

Solution To close the space between the nylon rod and the steel tube, the lateral strain of the nylon rod must be: 2.52 in. − 2.50 in.  lat = = 0.0080 in./in. 2.50 in. From the given Poisson’s ratio, the longitudinal strain in the nylon rod must be  0.0080 in./in.  long = − lat = − = −0.0200 in./in.  0.4 Use Hooke’s law to calculate the corresponding normal stress:  = Elong = (360 ksi)( − 0.0200 in./in.) = −7.2000 ksi The cross-sectional area of the nylon rod is  A = (2.50 in.) 2 = 4.9087 in.2 4 Therefore, the force required to make the nylon rod touch the inner wall of the tube is Ans. P =  A = (−7.2000 ksi)(4.9087 in.2 ) = −35.3429 kips = 35.3 kips (C)

P3.8 A metal specimen with an original diameter of 0.500 in. and a gage length of 2.000 in. is tested in tension until fracture occurs. At the point of fracture, the diameter of the specimen is 0.260 in. and the fractured gage length is 3.08 in. Calculate the ductility in terms of percent elongation and percent reduction in area.

Solution Percent elongation is simply the longitudinal strain at fracture:  (3.08 in. − 2.000 in.) 1.08 in. = = = = 0.54 in./in. L 2.000 in. 2.000 in.  percent elongation = 54%

Ans.

The initial cross-sectional area of the specimen is  A0 = (0.500 in.) 2 = 0.196350 in.2 4 The final cross-sectional area at the fracture location is  Af = (0.260 in.) 2 = 0.053093 in.2 4 The percent reduction in area is A0 − Af percent reduction of area = (100%) A0

(0.196350 in.2 − 0.053093 in.2 ) (100%) = 73.0% 0.196350 in.2

Ans.

P3.9 A portion of the stress-strain curve for a stainless steel alloy is shown in Figure P3.9. A 350-mm-long bar is loaded in tension until it elongates 2.0 mm and then the load is removed. (a) What is the permanent set in the bar? (b) What is the length of the unloaded bar? (c) If the bar is reloaded, what will be the proportional limit?

FIGURE P3.9

Solution

(a) The normal strain in the specimen is  2.0 mm = = = 0.005714 mm/mm L 350 mm Construct a line parallel to the elastic modulus line that passes through the data curve at a strain of  = 0.005714 mm/mm. The strain value at which this modulus line intersects the strain axis is the permanent set: Ans. permanent set = 0.0035 mm/mm (b) The length of the unloaded bar is therefore:  =  L = (0.0035 mm/mm)(350 mm) = 1.225 mm

L f = 350 mm + 1.225 mm = 351.225 mm (c) From the stress-strain curve, the reload proportional limit is 444 MPa .

Ans. Ans.

P3.10 A plastic block is bonded to a fixed base and to a horizontal rigid plate as shown in Figure P3.10. The shear modulus of the plastic is G = 45,000 psi, and the block dimensions are a = 4.0 in., b = 2.0 in., and c = 1.50 in. A horizontal force of P = 8,500 lb is applied to the rigid plate. Determine the horizontal deflection of the rigid plate. FIGURE P3.10

Solution Determine the shear stress from the horizontal force P and the block area (i.e., the area of the y plane surface of the block). To determine the area, visualize the surface that is bonded to the fixed base. This area has dimensions of ab. The average shear stress acting on the block is therefore: V 8,500 lb = = = 1,062.5 psi A ( 4.0 in.)( 2.0 in.) Since the shear modulus G is given, the shear strain can be calculated from Hooke’s law for shear stress and shear strain:  1,062.5 psi  = G  = = = 0.023611 rad G 45,000 psi

Recall that shear strain is an angle. From the angle  and the block thickness c, the horizontal deflection u of the rigid plate can be determined from: u tan  =  u = c tan  = (1.5 in.) tan ( 0.023611 rad ) = 0.035423 in. = 0.0354 in. c

Ans.

P3.11 A 0.5 in. thick plastic panel is bonded to the pin-jointed steel frame shown in Figure P3.11. Determine the magnitude of the force P that would displace bar AB to the right by 0.8 in. Assume that a = 4.0 ft, b = 6.0 ft, and G = 70,000 psi for the plastic. Neglect the deformation of the steel frame.

FIGURE P3.11

Solution The horizontal displacement of bar AB is u = 0.8 in. This displacement creates a shear strain of

tan  =

u 0.8in. = = 0.016667 a ( 4.0 ft )(12 in./ft )

  = 0.016665 rad From Hooke’s law, determine the shear stress associated with this shear strain:  = G = ( 70,000 psi )( 0.016665 rad ) = 1,166.559 psi Determine the horizontal force P from the shear area of the plastic panel. This area has dimensions of bt. V P = = A bt

 P =  bt = (1,166.559 psi )( 6 ft )(12 in./ft )( 0.5 in.) = 42,000 lb

Ans.

P3.12 The complete stress–strain diagram for a particular stainless steel alloy is shown in Figure P3.12a/13a. This diagram has been enlarged in Figure P3.12b/13b to show in more detail the linear portion of the stress-strain diagram. A rod made from this material is initially 800 mm long at a temperature of 20°C. After a tension force is applied to the rod and the temperature is increased by 200°C, the length of the rod is 804 mm. Determine the stress in the rod, and state whether the elongation in the rod is elastic or inelastic. Assume the coefficient of thermal expansion for this material is 18×10−6/°C.

FIGURE P3.12a/13a

FIGURE P3.12b/13b

Solution The 4 mm total elongation of the rod is due to a combination of load and temperature increase. The 200°C temperature increase causes a normal strain of: T =  T = (1810−6 / C)(200C) = 0.003600 mm/mm which means that the rod elongates  T = T L = (0.003600 mm/mm)(800 mm) = 2.8800 mm The portion of the 4 mm total elongation due to load is therefore   =  −  T = 4 mm − 2.8800 mm = 1.1200 mm The strain corresponding to this elongation is  1.1200 mm  =  = = 0.001400 mm/mm L 800 mm By inspection of the stress-strain curve, this strain is clearly in the linear region. Therefore, the rod is elastic in this instance. For the linear region, the elastic modulus can be determined from the lower scale plot:  (400 MPa − 0) E= = = 200,000 MPa  (0.002 mm/mm − 0) Using Hooke’s law (or directly from the – diagram), the stress corresponding to the 0.001400 mm/mm strain is  = E = (200,000 MPa)(0.001400 mm/mm) = 280 MPa

Ans.

P3.13 A tensile test specimen of stainless steel alloy having a diameter of 12.6 mm and a gage length of 50 mm was tested to fracture. The complete stress–strain diagram for this specimen is shown in Figure P3.12a/13a. This diagram has been enlarged in Figure P3.12b/13b to show in more detail the linear portion of the stress-strain diagram. Determine: (a) the modulus of elasticity. (b) the proportional limit. (c) the ultimate strength. (d) the yield strength (0.20% offset). (e) the fracture stress. (f) the true fracture stress if the final diameter of the specimen at the location of the fracture was 8.89 mm.

FIGURE P3.12a/13a

FIGURE P3.12b/13b

Solution From the stress-strain curve, the proportional limit will be taken as  = 60 ksi at a strain of  = 0.002. (Obviously, there can be quite a bit of leeway in pulling numbers from such a limited plot.) (a) The modulus of elasticity is  60 ksi E= = = 30,000 ksi  0.002 in./in. (b) From the diagram, the proportional limit is taken as  PL = 60 ksi (c) The ultimate strength is U = 159 ksi (d) The yield strength is  Y = 80 ksi (e) The fracture stress is  fracture = 135 ksi

Ans.

Ans. Ans. Ans. Ans.

(f) The original cross-sectional area of the specimen is  A0 = (0.495 in.) 2 = 0.192442 in.2 4 The cross-sectional area of the specimen at the fracture location is Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Af =



(0.350 in.) 2 = 0.096211 in.2

4 The true fracture stress is therefore 0.192442 in.2 true  fracture = (135 ksi) = 270 ksi 0.096211 in.2

Ans.

P3.14 A 7075-T651 aluminum alloy specimen with a diameter of 0.500 in. and a 2.0-in. gage length was tested to fracture. Load and deformation data obtained during the test are given. Determine: (a) the modulus of elasticity. (b) the proportional limit. (c) the yield strength (0.20% offset). (d) the ultimate strength. (e) the fracture stress. (f) the true fracture stress if the final diameter of the specimen at the location of the fracture was 0.387 in.

(lb)

Change in Length (in.)

0 1,221 2,479 3,667 4,903 6,138 7,356 8,596 9,783 11,050 12,247 13,434

0 0.0012 0.0024 0.0035 0.0048 0.0060 0.0072 0.0085 0.0096 0.0110 0.0122 0.0134

Load

(lb)

Change in Length (in.)

14,690 14,744 15,119 15,490 15,710 16,032 16,295 16,456 16,585 16,601 16,601 16,489 16,480

0.0149 0.0150 0.0159 0.0202 0.0288 0.0581 0.0895 0.1214 0.1496 0.1817 0.2278 0.2605 fracture

Load

Solution The plot of the stress-strain data is shown below.

The same plot is shown below, enlarged to better show the linear portion of the stress-strain curve.

The initial cross-sectional area of the specimen is



( 0.500 in.) = 0.19635 in.2 2

(a) Using the data point for the 9,783 lb load and 0.0096 in. elongation, the modulus of elasticity can be calculated as 9, 783 lb = = 49,824.3 psi 0.19635 in.2 0.0096 in. = = 0.0048 in./in. 2 in.  49,824.3 psi E= = = 10,380, 000 psi Ans.  0.0048 in./in. (b) Using the data point for the 9,783 lb load and 0.0096 in. elongation, the proportional limit is calculated as  PL = 49,800 psi

Ans.

(c) The yield strength by the 0.20% offset method is  Y = 78,500 psi (d) The ultimate strength is 16, 601 lb U = = 84, 600 psi 0.19635 in.2 (e) The fracture stress is 16, 480 lb  fracture = = 83,900 psi 0.19635 in.2

Ans.

(f) The cross-sectional area of the specimen at the fracture location is

Af =



( 0.387 in.) = 0.11763 in.2 2

4 The true fracture stress is therefore 16, 480 lb true  fracture = = 140,100 psi 0.11763 in.2

Ans.

P3.15 A Grade 2 Titanium tension test specimen has a diameter of 12.60 mm and a gage length of 50 mm. In a test to fracture, the stress and strain data below were obtained. Determine: (a) the modulus of elasticity. (b) the proportional limit. (c) the yield strength (0.20% offset). (d) the ultimate strength. (e) the fracture stress. (f) the true fracture stress if the final diameter of the specimen at the location of the fracture was 9.77 mm.

(kN)

Change in Length (mm)

0.00 4.49 8.84 13.29 17.57 22.10 26.46 30.84 35.18 39.70 43.95 48.44

0.000 0.017 0.032 0.050 0.064 0.085 0.103 0.123 0.144 0.171 0.201 0.241

Load

(kN)

Change in Length (mm)

52.74 56.95 60.76 63.96 66.61 68.26 69.08 69.41 69.39 69.25 68.82 68.35 68.17

0.314 0.480 0.840 1.334 1.908 2.562 3.217 3.938 4.666 5.292 6.023 6.731 fracture

Load

Solution The plot of the stress-strain data is shown below.

The same plot is shown below, enlarged to better show the linear portion of the stress-strain curve.

The initial cross-sectional area of the specimen is



(12.60 mm ) = 124.690 mm2 2

(a) Using the data point for the 30.84 kN load and 0.123 mm elongation, the modulus of elasticity can be calculated as 30,840 N = = 247.333 MPa 124.690 mm 2 0.123 mm = = 0.002456 mm/mm 50 mm  247.333 MPa E= = = 100, 705.6 MPa = 100.7 GPa Ans.  0.002456 mm/mm (b) From the diagram, the proportional limit is taken as  PL = 247 MPa

Ans.

(d) The ultimate strength is 69, 410 N U = = 557 MPa 124.690 mm2 (e) The fracture stress is 68,170 N  fracture = = 547 MPa 124.690 mm 2

Ans.

(f) The cross-sectional area of the specimen at the fracture location is

Af =



( 9.77 mm ) = 74.969 mm2 2

4 The true fracture stress is therefore 68,170 N true  fracture = = 909 MPa 74.969 mm 2

Ans.

P3.16 Compound axial member ABC has a uniform diameter of d = 1.50 in. Segment (1) is an aluminum [E1 = 10,000 ksi] alloy rod with a length of L1 = 90 in. Segment (2) is a copper [E2 = 17,000 ksi] alloy rod with a length of L2 = 130 in. When axial force P is applied, a strain gage attached to copper segment (2) measures a normal strain of 2 = 2,100 in./in. in the longitudinal direction. What is the total elongation of member ABC?

FIGURE P3.16

Solution The strain in segment (2) is measured as 2,100 in./in. From Hooke’s law, the stress in segment (2) is:

 2 = E2 2 = (17,000 ksi ) ( 2,100 10−6 in./in.) = 35.70 ksi

The deformation in segment (2) is:  2 =  2 L2 = ( 2,100 10−6 in./in.) (130 in.) = 0.2730 in. The area of segment (2) is:   2 A2 = d 22 = (1.50 in.) = 1.7671 in.2 4 4 The diameters of segments (1) and (2) are equal; thus, A1 = A2 = 1.7671 in.2. The internal axial force in segments (1) and (2) is the same, and so are the normal stresses,  1 =  2 = 35.70 ksi From Hooke’s law, the strain in segment (1) must be  1 = E11

1

35.70 ksi = 3,570 10−6 in./in. E1 10, 000 ksi Consequently, the deformation of segment (1) is 1 = 1L1 = ( 3,570 10−6 in./in.) ( 90 in.) = 0.3213 in.  1 =

The total deformation of member ABC is equal to the sum of the deformations in segments (1) and (2):  ABC = 1 +  2 = 0.3213 in. + 0.2730 in. = 0.594 in.

Ans.

P3.17 An aluminum alloy [E = 70 GPa;  = 0.33;  = 23.0×10–6/°C] plate is subjected to a tensile load P as shown in Figure P3.17. The plate has a depth of d = 260 mm, a cross-sectional area of A = 6,500 mm2, and a length of L = 4.5 m. The initial longitudinal normal strain in the plate is zero. After load P is applied and the temperature of the plate has been increased by T = 56°C, the longitudinal normal strain in the plate is found to be 2,950 . Determine: (a) the magnitude of load P. (b) the change in plate depth d.

FIGURE P3.17

Solution (a) Magnitude of load P: The total strain in the plate is measured as 2,950  (elongation). Part of this strain is due to the stress acting in the bar and part of this strain is due to the temperature increase. The strain caused by the temperature change is T =  T = ( 23.0 10−6 / °C ) ( 56°C ) = 1, 288 10−6 mm/mm Since the total strain is  = 2,950  = 0.002950 mm/mm, the strain caused by the normal stress in the bar must be:

 =  − T = 2,950 10−6 mm/mm −1,288 10−6 mm/mm = 1,662 10−6 mm/mm From Hooke’s law, the stress in the bar is therefore:

 = E = ( 70,000 MPa ) (1,662 10−6 mm/mm ) = 116.34 MPa

and thus, the force in the bar is: P =  A = (116.34 N/mm2 )( 6,500 mm2 ) = 756, 210 N = 756 kN (b) Change in plate depth d: Due to the Poisson effect, the plate will contract in the lateral direction. However, the thermal strain will cause the width of the bar to increase. We will need to separately consider contract each of these strains. The strain in the lateral direction due to the Poisson effect is calculated from the longitudinal strain due solely to the normal stress. This portion of the overall lateral strain is:   = − lat  long   lat = − long = − ( 0.33) (1, 662 10−6 mm/mm ) = −548.46 10−6 mm/mm

The thermal strain is 1,288×10–6 mm/mm; and so, the total lateral strain is:  lat total =  lat Poisson +  T

= −548.46 10−6 mm/mm + 1, 288 10 −6 mm/mm = 739.54 10−6 mm/mm The change in plate width d is accordingly d =  lat total d = ( 739.54 10−6 mm/mm ) ( 260 mm ) = 0.1923 mm

Ans.

P3.18 The rigid plate in Figure P3.18 is supported by bar (1) and by a double shear pin connection at B. Bar (1) has a length of L1 = 60 in., a cross-sectional area of A1 = 0.47 in.2, an elastic modulus of E = 10,000 ksi, and a coefficient of thermal expansion of  = 13×10−6/°F. The pin at B has a diameter of 0.438 in. After load P has been applied and the temperature of the entire assembly has been decreased by 30°F, the total strain in bar (1) is measured as 570  (elongation). Assume dimensions of a = 12 in. and b = 20 in. Determine: (a) the magnitude of load P. (b) the average shear stress in pin B. FIGURE P3.18

Solution (a) Magnitude of load P: The total strain in bar (1) is caused partly by the axial force in the bar and partly by the decrease in temperature. The strain caused by the 30°F temperature decrease is: T =  T = (13 10−6 /F) ( −30F) = −0.000390 in./in. The strain caused by the axial force in the bar is thus: 1, = 1 − T = 0.000570 in./in. − ( −0.000390 in./in.) = 0.000960 in./in. The stress in bar (1) is 1 = E11, = (10,000 ksi )( 0.000960 in./in.) = 9.6 ksi and the force in bar (1) is F1 = 1 A1 = ( 9.6 ksi ) ( 0.47 in.2 ) = 4.512 kips Next, consider an FBD of the rigid plate. Use a moment equilibrium equation to compute load P: M B = F1b − Pa = 0 P =

b 20 in. F1 = ( 4.512 kips ) = 7.52 kips a 12 in.

Ans.

(b) Average shear stress in pin B: Use the following equilibrium equations for the rigid plate to calculate the reaction force at pin B, Fx = P + Bx = 0  Bx = − P = −7.520 kips Fy = By − F1 = 0  By = F1 = 4.512 kips B = Bx2 + By2 =

( −7.520 kips ) + ( 4.512 kips ) 2

= 8.770 kips

The cross-sectional area of pin B is



( 0.438 in.) = 0.15067 in.2 2

4 Since the pin is supported in a double-shear connection, the area that carries shear stress is AV = 2 0.15067 in.2 = 0.30134 in.2

(

)

and thus, the shear stress in pin B is B 8.770 kips = = = 29.1 ksi AV 0.30134 in.2

Ans.

P3.19 Member (1) is an aluminum bar that has a crosssectional area of A = 1.05 in.2, an elastic modulus of E = 10,000 ksi, and a coefficient of thermal expansion of  = 12.5 × 10−6/°F. After a load P of unknown magnitude is applied to the structure and the temperature is increased by 65°F, the normal strain in bar (1) is measured as –540 micros. Use dimensions of a = 24.6 ft, b = 11.7 ft, and c = 14.0 ft. Determine the magnitude of load P.

FIGURE P3.19

Solution The total strain in bar (1) consists of thermal strain as well as normal strain caused by normal stress:  =  + T The normal strain due to the increase in temperature is: T =  T = (12.5 10−6 /F ) ( 65F ) = 0.0008125 in./in. Therefore, the normal stress in bar (1) causes a normal strain of:   =  −  T = −0.000540 in./in. − 0.0008125 in./in. = −0.0013525 in./in. From Hooke’s law, the normal stress in bar (1) can be calculated as:

1 = E = (10,000 ksi )( −0.0013525 in./in.) = −13.525 ksi

and thus the axial force in bar (1) must be: F1 = 1 A1 = ( −13.525 ksi ) (1.05 in.2 ) = −14.201 kips Next, consider an FBD of the rigid bar. Use a moment equilibrium equation to compute load P: M A = − F1c − Pb = 0 c 14.0 ft  P = − F1 = − ( −14.201 kips ) = 16.99 kips b 11.7 ft

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P4.1 Seven bolts are used in the connection between the bar and the support. The ultimate shear strength of the bolts is 320 MPa, and a factor of safety of 2.5 is required with respect to fracture. Determine the minimum allowable bolt diameter required to support an applied load of P = 1,225 kN.

FIGURE P4.1

Solution Seven bolts support the load; therefore, the force acting on each bolt is 1, 225 kN Fbolt = = 175 kN/bolt 7 The allowable shear stress for the bolts is  320 MPa  allow = U = = 128 MPa FS 2.5 Each bolt acts in double shear; therefore, two cross-sectional surfaces on each bolt are subjected to shear stress. The shear force on each surface is thus Fbolt 175 kN/bolt Vbolt = = = 87.5 kN/surface 2 surfaces 2 surfaces/bolt The bolt cross-sectional area required to satisfy the allowable shear stress limit must equal or exceed: V 87,500 N Abolt  bolt = = 683.594 mm2 2  allow 128 N/mm The minimum bolt diameter is thus:  2 Ans. d bolt  683.594 mm 2  d bolt  29.5 mm 4

Mechanics of Materials: An Integrated Learning System, 4th Ed. P4.2 A thrust collar that rests on a 0.75 in. thick polymer plate supports the vertical nylon shaft shown in Figure P4.2. The shaft has a diameter of d = 0.50 in. The load acting on the shaft is P = 600 lb. (a) The ultimate shear strength of the nylon shaft is 2,700 psi, and a factor of safety of 2.5 with respect to shear is required. Determine the minimum thickness t required for the thrust collar. (b) The ultimate bearing strength of the polymer plate is 2,000 psi, and a factor of safety of 3.0 with respect to bearing is required. Determine the minimum outer diameter Dcollar required for the thrust collar.

Timothy A. Philpot

FIGURE P4.2

Solution (a) Minimum thrust collar thickness: The allowable shear stress of the nylon shaft is  2, 700 psi  allow = U = = 1, 080 psi FS 2.5 The area subjected to shear stress in the nylon shaft at the thrust collar is equal to the circumference of the shaft times the thickness of the collar: AV =  dt The load on the shaft is P = 600 lb. The minimum collar thickness required to limit the shear stress to 1,080 psi is P P  allow  = AV  d t

t 

 d allow

600 lb = 0.354 in.  ( 0.50 in.)(1, 080 psi )

Ans.

(b) Minimum outer diameter Dcollar for thrust collar: The allowable bearing stress of the polymer plate is  2, 000 psi  allow = U = = 666.667 psi FS 3.0 The contact area between the polymer plate and the thrust collar is expressed by:  2 Ab = ( Dcollar −d2) 4 The minimum outer diameter Dcollar required for the thrust collar can be determined as follows: P P  allow  = Ab  D 2 − d 2 collar 4

(

Dcollar 

 allow

)

+ d2 =

4 ( 600 lb )

 ( 666.667 lb/in. ) 2

+ ( 0.5 in.) = 1.181 in. 2

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P4.3 Two flat bars loaded in tension by force P are spliced using two rectangular splice plates and two 0.375 in. diameter bolts as shown in Figure P4.3. Away from the connection, the bars have a width of b = 0.625 in. and a thickness of t = 0.25 in. The bars are made of polyethylene having an ultimate tensile strength of 3,100 psi and an ultimate bearing strength of 2,000 psi. The bolts are made of nylon that has an ultimate shear strength of 4,000 psi. Determine the allowable load P for this connection if a safety factor of 3.5 is required. Consider tension and bearing in the bars and shear in the bolts. Disregard friction between the plates. FIGURE P4.3

Solution P based on normal stress in bar: The allowable normal stress is  3,100 psi  allow = Y = = 885.714 psi FS 3.5 The allowable load P based on the tensile strength of the bar is: Pallow   allow A =  allowbt = (885.714 psi )( 0.625 in.)( 0.25 in.) = 138.4 lb P based on bearing stress in bar: The allowable bearing stress is  2,000 psi  allow = U = = 571.43 psi FS 3.5 Use the projected area of a 0.375 in. diameter bolt bearing on the 0.25 in. thick bar to calculate the allowable bearing force for one bolt: Fb =  allow dt = ( 571.43 psi )( 0.375 in.)( 0.25 in.) = 53.57 lb/bolt Since each bar is connected with only one bolt, Pallow  (1 bolt ) Fb = (1 bolt )(53.57 lb/bolt ) = 53.57 lb P based on shear stress in bolts: The allowable shear stress of the bolts is  4,000 psi  allow = U = = 1,142.86 psi FS 3.5 Each bolt has a cross-sectional area of  2  2 Abolt = d bolt = ( 0.375 in.) = 0.11045 in.2 4 4 Each bolt is used in a double-shear configuration; therefore, one bolt can support Vbolt = 2 allow Abolt = 2 (1,142.86 psi ) 0.11045 in.2 = 252.45 lb/bolt

(

)

Since each bar is connected with only one bolt, Pallow  (1 bolt )Vbolt = (1 bolt )( 252.45 lb/bolt ) = 252.45 lb Allowable load P: Based on tension and bearing in the bars and shear in the bolts, the strength of this connection is Ans. Pallow  53.6 lb Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P4.4 A bar with a rectangular cross section is connected to a support bracket with a circular pin, as shown in Figure P4.4. The bar has a width of b = 40 mm and a thickness of 10 mm. The pin has a diameter of 12 mm. The yield strength of the bar is 415 MPa, and a factor of safety of 1.67 with respect to yield is required. The bearing strength of the bar is 380 MPa, and a factor of safety of 2.0 with respect to bearing is required. The shear strength of the pin is 550 MPa, and a factor of safety of 2.5 with respect to shear failure is required. Determine the magnitude of the maximum load P that may be applied to the bar.

FIGURE P4.4

Solution P based on normal stress in bar: The allowable normal stress is  415 MPa  allow = Y = = 248.503 MPa FS 1.67 The allowable load P based on the tensile strength of the bar is: Pallow   allow A =  allowbt = 248.503 N/mm2 ( 40mm )(10 mm ) = 99,401 N

(

)

P based on bearing stress in bar: The allowable bearing stress is  380 MPa  allow = U = = 190.000 MPa FS 2.0 Use the projected area of a 12 mm diameter pin bearing on the 10 mm thick bar to calculate the allowable load: Pallow   allow dt = 190.000 N/mm2 (12 mm )(10 mm ) = 22,800 N

(

)

P based on shear stress in pin: The allowable shear stress of the pin is  550 MPa  allow = U = = 220.000 MPa FS 2.5 The pin has a cross-sectional area of  2  2 Apin = d pin = (12 mm ) = 113.097 mm 2 4 4 The allowable load for the pin is Ppin   allow Apin = 220.000 N/mm2 113.097 mm2 = 24,881 N

(

)(

)

Allowable load P: Based on tension and bearing in the bar and shear in the pin, the maximum load that can be applied to the bar is Ans. Pallow  22.8 kN

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P4.5 In Figure P4.5, member (1) has a cross-sectional area of 0.7 in.2 and a yield strength of 50 ksi. Member (2) has a cross-sectional area of 1.8 in.2 and a yield strength of 36 ksi. A factor of safety of 1.67 with respect to yield is required for both members. Use dimensions of a = 60 in., b = 28 in., c = 54 in., and d = 72 in. Determine the maximum allowable load P that may be applied to the assembly. Report the factors of safety for both members at the allowable load.

FIGURE P4.5

Solution The allowable normal stress for member (1) is  50 ksi  allow,1 = Y ,1 = = 29.940 ksi FS 1.67 and the allowable force in member (1) is Fallow,1 =  allow,1 A1 = ( 29.940 ksi ) ( 0.7 in.2 ) = 20.958 kips

(a)

The allowable normal stress in member (2) is  36 ksi  allow,2 = Y ,2 = = 21.557 ksi FS 1.67 and the allowable force in member (2) is Fallow,2 =  allow,2 A2 = ( 21.557 ksi ) (1.8 in.2 ) = 38.802 kips

(b)

Consider an FBD of joint B. Compute the angles  and : a 60 in. tan  = = = 0.8333 d 72 in.  = 39.806

b 28 in. = = 0.5185 c 54 in.   = 27.408

tan  =

For joint B, write the following equilibrium equations: Fx = P − F1 cos  − F2 cos  = 0

Fy = F1 sin  − F2 sin  = 0

From Eq. (d): F2 = F1

sin  sin 

(e)

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Substitute the allowable force for member (1) into Eq. (e) and compute the force F2 that is required for equilibrium: sin ( 39.806 ) sin  (f) F2 = Fallow,1 = ( 20.958 kips ) = 29.147 kips sin  sin ( 27.408 ) Compare this result with Eq. (b) and note that F2  Fallow,2 Thus, we conclude that member (1) controls. The forces in members (1) and (2) that are required to satisfy equilibrium for joint B while also satisfying their respective allowable forces are now known: F1 = 20.958 kips

F2 = 29.147 kips Substitute these values into Eq. (c) and compute the allowable load P: P = F1 cos  + F2 cos  = ( 20.958 kips ) cos ( 39.806 ) + ( 29.147 kips ) cos ( 27.408 ) = 42.0 kips

The stress in member (1) is F 20.958 kips 1 = 1 = = 29.940 ksi A1 0.7 in.2 and its factor of safety is:  50 ksi FS1 = Y ,1 = = 1.670  1 29.940 ksi The stress in member (2) is F 29.147 kips 2 = 2 = = 16.193 ksi A2 1.8 in.2 and its factor of safety is:  36 ksi FS2 = Y ,2 = = 2.22  2 16.193 ksi

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P4.6 The bell-crank mechanism shown in Figure P4.6 is supported by a single-shear pin connection at B and a roller support at C. A load P acts an angle of  = 40° at joint A. The pin at B has a diameter of 0.375 in., and the bell crank has a thickness of 0.188 in. The ultimate shear strength of the pin material is 28 ksi, and the ultimate bearing strength of the crank material is 52 ksi. A minimum factor of safety of 3.0 with respect to shear and bearing is required. Assume a = 9.0 in. and b = 5.0 in. What is the maximum load P that may be applied at joint A of the bell crank? FIGURE P4.6

Solution Equilibrium: Based on an FBD of the bell crank, the following equilibrium equations can be written: Fx = P cos  + Bx = 0

Fy = P sin + By + Cy = 0

M B = − ( P cos ) a + Cyb = 0

(a) (b) (c)

From Eq. (c), express Cy in terms of the load P: ( 9.0 in.) cos 40 P = 1.378880 P a cos Cy = P= b 5.0 in.

(d)

From Eq. (a), express Bx in terms of P: Bx = − P cos = − P cos 40 = −0.766044 P From Eqs. (b) and (d), express By in terms of P: By = −P sin − Cy = −P sin 40 − 1.378880P = −2.021668P The resultant of the pin reaction at B can now be expressed in terms of P:

B = Bx2 + By2 =

( −0.766044 P ) + ( −2.021668P ) = 2.161935P 2

Consider pin shear: The allowable shear stress of the pin material is  28,000 psi  allow = U = = 9,333.33 psi FS 3.0 The cross-sectional area of the 0.375 in. diameter pin at B is:  2 Apin = ( 0.375 in.) = 0.11045 in.2 4 The shear force acting in the pin is equal to the resultant force at B. The pin acts in single shear and so the shear area is equal to the cross-sectional area of the pin. Solve for the allowable load P based on the strength of the pin at B:

Mechanics of Materials: An Integrated Learning System, 4th Ed.

 allow 

Timothy A. Philpot

B Apin

2.161935 Pallow 0.11045 in.2 ( 9,333.33 psi ) ( 0.11045 in.2 )  Pallow  = 476.8 lb (e) 2.161935 Consider bearing of pin on bell crank: The allowable bearing stress is  52,000 psi  allow = U = = 17,333.33 psi FS 3.0 Use the projected area of the 0.375 in. diameter pin bearing on the 0.188 in. thick bell crank to calculate the allowable load: (f) Pallow   allow dt = (17,333.33 psi )( 0.375 in.)( 0.188 in.) = 1,222.0 lb 9,333.33 psi 

Allowable load P: Compare the results in Eqs. (e) and (f) to find Pallow = 477 lb

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P4.7 The bell-crank mechanism shown in Figure P4.7 is in equilibrium for an applied load of F1 = 7 kN applied at A. Assume a = 150 mm, b = 100 mm, c = 80 mm, and  = 55°. The bell crank has a thickness of t = 10 mm, and the support bracket has a thickness of tb = 8 mm. The pin at B has a diameter of d = 12 mm and an ultimate shear strength of 290 MPa. The bell crank and the support bracket each have an ultimate bearing strength of 380 MPa. Determine: (a) the factor of safety in pin B with respect to the ultimate shear strength. (b) the factor of safety of the bell crank at pin B with respect to the ultimate bearing strength. (c) the factor of safety in the support bracket with respect to the ultimate bearing strength.

FIGURE P4.7

Solution Consider a free-body diagram of the bell crank. The equilibrium equations for the bell crank can be solved for Bx, By, and F2: Fx = − ( 7,000 N ) cos55 + Bx = 0  Bx = 4,015.035 N

M B = ( 7,000 N ) cos55 ( 80 mm ) + ( 7,000 N ) sin 55 (150 mm ) − F2 (100 mm ) = 0  F2 = 11,813.125 N

Fy = − ( 7,000 N ) sin 55 − F2 + By = − ( 7,000 N ) sin 55 − (11,813.125 N ) + By = 0  By = 17,547.189 N

The resultant force at pin B is: B = Bx2 + By2 =

( 4,015.035 N ) + (17,547.189 N ) = 18,000.676 N 2

(a) Pin B shear: The 12 mm diameter pin at B is supported in a double-shear connection; therefore, the shear force acting on one shear plane (which is simply equal to the cross-sectional area of the pin) is half of the resultant force at pin B: VB = 9,000.338 N. The cross-sectional area of the pin at B is:  Apin B = (12 mm) 2 = 113.097 mm 2 4 and therefore, the shear stress in pin B is: Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

B =

Timothy A. Philpot

VB 9,000.338 N = = 79.580 MPa Apin B 113.097 mm2

For pin B, the factor of safety is:  290 MPa FSB = U = = 3.64  B 79.580 MPa

Ans.

(b) Crank bearing stress at B: The bearing stress between the 12 mm diameter pin and the 10 mm thick bell crank at B is 18,000.676 N b = = 150.006 MPa (12 mm )(10 mm ) Therefore, the factor of safety is:  380 MPa Ans. FSB = U = = 2.53  b 150.006 MPa (c) Support bracket bearing stress at B: The reaction force carried by the pin at B is supported on two 8 mm thick brackets. Therefore, the bearing stress between the 12 mm diameter pin and two 8 mm thick bracket plates at B is 18,000.676 N b = = 93.754 MPa (12 mm )( 2  8 mm ) Therefore, the factor of safety is:  380 MPa Ans. FSB = U = = 4.05  b 93.754 MPa

Mechanics of Materials: An Integrated Learning System, 4th Ed. P4.8 Rigid bar ABC is supported by member (1) at B and by a pin connection at C, as shown in Figure P4.8. Member (1) has a cross sectional area of 1.25 in.2 and a yield strength of 40 ksi. The pins at B, C, and D each have a diameter of 0.5 in. and an ultimate shear strength of 72 ksi. Each pin connection is a double-shear connection. Specifications call for a minimum factor of safety in member (1) of 1.67 with respect to its yield strength. The minimum factor of safety required for pins B, C, and D is 3.0 with respect to the pin ultimate shear strength. Determine the allowable load P that may be applied to the rigid bar at A. Use overall dimensions of a = 54 in. and b = 110 in.

Timothy A. Philpot

FIGURE P4.8

Solution Equilibrium: Based on an FBD of the rigid bar, the following equilibrium equations can be written: Fx = C x = 0 (a) (b) Fy = F1 + Cy − P = 0

MC = P ( a + b ) − Fb 1 =0

(c)

From Eq. (c), express F1 in terms of the load P:

F1 =

a+b 54 in. + 110 in. P= P = 1.49091P b 110 in.

(d)

From Eq. (b), express Cy in terms of P:

Cy = P − F1 = P − 1.49091P = −0.49091P

The resultant of the pin reaction at C is simply:

C = Cx2 + C y2 = 0.49091P Consider member (1): The allowable normal stress for member (1) is  40 ksi  allow = Y = = 23.952 ksi FS 1.67 Solve for the allowable load P based on the strength of member (1): F  allow  1 A1

1.49091Pallow 1.25 in.2 ( 23.952 ksi ) (1.25 in.2 )  Pallow  = 20.082 kips 1.49091

23.952 ksi 

(e)

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Consider pin shear at B and D: The allowable shear stress of the pin material is  72 ksi  allow = U = = 24 ksi FS 3.0 The cross-sectional area of the 0.5 in. diameter pins is:  2 Apin = ( 0.5 in.) = 0.19635 in.2 4 The shear force acting in the pins at B and D is equal to the force F1 in member (1). The pins act in double shear and so the shear area is equal to twice the cross-sectional area of the pin. Solve for the allowable load P based on the strength of the pins at B and D: F  allow  1 2 Apin

24 ksi 

1.49091Pallow 2 ( 0.19635 in.2 )

 Pallow 

2 ( 24 ksi ) ( 0.19635 in.2 ) 1.49091

= 6.322 kips

(f)

Consider pin shear at C: Solve for the allowable load P based on the strength of the pin at C: C  allow  2 Apin

24 ksi 

0.49091Pallow 2 ( 0.19635 in.2 )

 Pallow 

2 ( 24 ksi ) ( 0.19635 in.2 ) 0.49091

= 19.199 kips

Allowable load P: Compare the results in Eqs. (e), (f), and (g) to find Pallow = 6.32 kips

(g)

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P4.9 In Figure P4.9, rigid beam AB is subjected to a distributed load that increases linearly from zero to a maximum intensity of wB = 17.5 kN/m. Beam AB is supported by a single-shear pin connection at joint A and by a double-shear connection to member (1) at joint B. Member (1) is connected to the support at C with a double-shear pin connection. Use dimensions of a = 2.8 m, b = 1.0 m, and c = 1.2 m. (a) The yield strength of member (1) is 340 MPa. A factor of safety of 1.67 with respect to the yield strength is required for the normal stress of member (1). Determine the minimum cross-sectional area required for member (1). (b) The ultimate shear strength of the material used for pins A, B, and C is 270 MPa. A factor of safety of 2.50 with respect to the ultimate shear strength is required for the pins, and all three pins are to have the same diameter. Determine the minimum pin diameter that may be used.

FIGURE P4.9

Solution Consider an FBD of rigid beam AB. Member (1) makes an angle of  with respect to the horizontal axis: c 1.2 m tan  = = = 1.2 b 1.0 m  = 50.1944 Note that member (1) is a two-force member. Equilibrium equations for horizontal member AB can be solved for the unknown reactions Ax and Ay and the force in member (1): w aa M B = B   − Ay a = 0 2 3 w a (17.5 kN/m )( 2.8 m )  Ay = B = = 8.1667 kN 6 6 wa Fy = Ay − F1 sin 50.1944 − B = 0 2 (17.5 kN/m )( 2.8 m ) wa Ay − B 8.1667 kN − 2 2  F1 = = = −21.2612 kN sin 50.1944 sin 50.1944 Fx = Ax + F1 cos50.1944 = 0  Ax = − F1 cos50.1944 = − ( −21.2612 kN ) cos50.1944 = 13.6111 kN

The resultant force at pin A is: A =

Ax2 + Ay2 =

(13.6111 kN ) + ( −21.2612 kN ) = 15.8731 kN 2

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

(a) Minimum area for member (1): The allowable normal stress for member (1) is  340 MPa  allow = Y = = 203.5928 MPa FS 1.67 The minimum area for member (1) is F 21,261.2 N A1  1 = = 104.4 mm2 2  allow 203.5928 N/mm

Ans.

(b) Minimum diameter for pins at A, B, and C: The allowable shear stress for the pins is computed from the ultimate shear strength and the factor of safety:  270 MPa  allow = U = = 108 MPa FS 2.5 Pin A: The pin at A is used in a single-shear connection. The shear force acting on this pin is simply the reaction force at A. The minimum area for pin A is A 15,873.1 N Apin A  = = 146.9736 mm2 2  allow 108 N/mm The minimum diameter of the pin at A is thus 4 dpin A  146.9736 mm2 = 13.6796 mm



(

)

Pins B and C: The force acting on both pin B and pin C is the force F1. The pins at B and C are used in double-shear connections. Consequently, the shear force acting on the cross-sectional area of each pin is F 21, 261.2 N Vpin B = 1 = = 10,630.6 N 2 2 The minimum area for pin B is V 10,630.6 N Apin B  B = = 98.4316 mm2 2  allow 108 N/mm The minimum diameter of the pin at B (and also C) is thus 4 dpin B  98.4316 mm2 = 11.1950 mm



(

)

Since all three pins are to have the same diameter, the minimum pin diameter that may be used is d  13.68 mm

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P4.10 Beam AB is supported as shown in Figure P4.10/11. Tie rod (1) is attached at B and C with double-shear pin connections while the pin at A is attached with a singleshear connection. The pins at A, B, and C each have an ultimate shear strength of 54 ksi, and tie rod (1) has a yield strength of 36 ksi. A uniformly distributed load of w = 1,800 lb/ft is applied to the beam as shown. Dimensions are a = 15 ft, b = 30 in., and c = 7 ft. A factor of safety of 2.5 is required for all components. Determine: (a) the minimum required diameter for tie rod (1). (b) the minimum required diameter for the double-shear pins at B and C. (c) the minimum required diameter for the single-shear pin at A. FIGURE P4.10/11

Solution Tie rod (1) is a two-force member that is oriented at  with respect to the horizontal axis: c 7 ft tan  = = = 0.4667  = 25.0169 a 15 ft From an FBD of beam AB, the following equilibrium equations can be written: Fx = Ax − F1 cos 25.0169 = 0 (a)

Fy = Ay + F1 sin25.0169 − wa = 0

(b)

M A = ( F1 cos 25.0169 ) b

a + ( F1 sin 25.0169 ) a − wa   = 0 2 From Eq. (c):

F1 =

( 0.5)(1,800 lb/ft )(15 ft )

(c)

= 23,522.6 lb  30 in.    ( cos 25.0169 ) + (15 ft )( sin 25.0169 )  12 in./ft  Substitute F1 into Eq. (a) to obtain Ax: Ax = F1 cos25.0169 = ( 23,522.6 lb) cos25.0169 = 21,315.8 lb

and substitute F1 into Eq. (b) to obtain Ay: Ay = wa − F1 sin 25.0169 = (1,800 lb/ft )(15 ft ) − ( 23,522.6 lb) sin 25.0169 = 17,052.6 lb The resultant pin force at A is found from Ax and Ay: A=

Ax2 + Ay2 =

( 21,315.8 lb ) + (17,052.6 lb ) = 27, 297.5 lb 2

Mechanics of Materials: An Integrated Learning System, 4th Ed. (a) Diameter for tie rod (1): The allowable normal stress for tie rod (1) is  36,000 psi  allow = Y = = 14, 400 psi FS 2.5 The minimum cross-sectional area required for the tie rod is F 23,522.6 lb Amin  1 = = 1.6335 in.2 2  allow 14,400 lb/in. Therefore, the minimum tie rod diameter is  2 d1  1.6335 in.2  d1  1.442 in. 4

Timothy A. Philpot

Ans.

(b) Diameter for double-shear pins at B and C: The allowable shear stress for the pins at A, B, and C is  54,000 psi  allow = U = = 21,600 psi FS 2.5 The double-shear pin connection at B and C must support a load of F1 = 23,522.6 lb . The shear area AV required for these pins is F 23,522.6 lb AV  1 = = 1.08901 in.2 2  allow 21,600 lb/in. The pin diameter can be computed from  2 Ans.  1.08901 in.2  d pin  0.833 in. ( 2 surfaces ) dpin 4 (c) Diameter for single-shear pin at A: The pin at A is a single-shear connection; therefore, AV = Apin. The shear area AV required for this pin is A 27,297.5 lb AV  = = 1.26377 in.2  allow 21,600 lb/in.2 The pin diameter can be computed from  2 Ans.  1.26377 in.2  d pin  1.269 in. (1 surface ) dpin 4

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P4.11 Beam AB is supported as shown in Figure P4.10/11. Tie rod (1) has a diameter of 50 mm, and it is attached at B and C with 24 mm diameter double-shear pin connections. The pin connection at A consists of a 40 mm diameter single-shear pin. The pins at A, B, and C each have an ultimate shear strength of 520 MPa, and tie rod (1) has a yield strength of 280 MPa. A uniformly distributed load of w is applied to the beam as shown. Dimensions are a = 4.0 m, b = 0.5 m, and c = 1.5 m. A minimum factor of safety of 2.5 is required for all components. What is the maximum loading w that may be applied to the structure?

FIGURE P4.10/11

Solution Tie rod (1) is a two-force member that is oriented at  with respect to the horizontal axis: c 1.5 m tan  = = = 0.3750  = 20.5560 a 4.0 m From an FBD of beam AB, the following equilibrium equations can be written: Fx = Ax − F1 cos 20.5560 = 0 (a)

Fy = Ay + F1 sin20.5560 − wa = 0

(b)

M A = ( F1 cos 20.5560 ) b

a + ( F1 sin 20.5560 ) a − wa   = 0 2 From Eq. (c), solve for w in terms of F1:  ( cos 20.5560 ) b + ( sin 20.5560 ) a  w = 2 F1   a2  

(c)

 ( cos 20.5560 )( 0.5 m ) + ( sin 20.5560 )( 4.0 m )  = 2 F1   2 ( 4.0 m )  

(d)

= ( 0.234082 m −1 ) F1

Alternatively, we may express F1 in terms of w as F1 = ( 4.272002 m) w Next, we will solve for the reaction forces at A in terms of w. From Eq. (a): Ax = F1 cos20.5560 = w( 4.272002 m) cos20.5560 = ( 4.00 m) w and substitute F1 into Eq. (b) to obtain Ay in terms of w: Ay = wa − F1 sin 20.5560 = w( 4.0 m) − w( 4.272002 m) sin 20.5560 = ( 2.5 m) w The resultant pin force at A is found from Ax and Ay: Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Ax2 + Ay2 = w ( 4.00 m ) + ( 2.5 m ) = ( 4.7170 m ) w

(e)

(a) Consider tie rod (1): The allowable normal stress for tie rod (1) is  280 MPa  allow = Y = = 112 MPa FS 2.5 The cross-sectional area of the tie rod is

A1 =



( 50 mm ) = 1,963.4954 mm2 2

4 and the allowable force in the tie rod is thus F1 =  allow A1 = (112 N/mm2 )(1,963.4954 mm2 ) = 219,911.5 N = 219.9115 kN From Eq. (d), we find that the allowable distributed load on the beam is w  ( 0.234082 m−1 ) F1 = ( 0.234082 m−1 ) ( 219.9115 kN ) = 51.4774 kN/m

(f)

(b) Consider double-shear pins at B and C: The allowable shear stress for the pins at A, B, and C is  520 MPa  allow = U = = 208 MPa FS 2.5 The cross-sectional area of a 24 mm diameter pin is  2  2 Apin = d pin = ( 24 mm ) = 452.3893 mm2 4 4 The pins at B and C are used in double-shear connections; therefore, the shear area for these pins is AV = 2 Apin = 2 452.3893 mm2 = 904.7786 mm 2

(

)

The strength of a 24 mm diameter double-shear pin connection is thus: V =  allow AV = ( 208 N/mm2 )( 904.7786 mm2 ) = 188,193.9 N = 188.1939 kN The shear force acting on pins B and C is equal to the force in tie rod (1). From Eq. (d), calculate the distributed load w that can be applied to the beam based on the strength of the pin connections at B and C: (g) w  ( 0.234082 m−1 ) F1 = ( 0.234082 m−1 ) (188.1939 kN ) = 44.0529 kN/m (c) Consider single-shear pin at A: The pin at A is a 40 mm diameter single-shear connection; therefore, AV = Apin. The shear area AV of this pin is

AV = Apin =



( 40 mm ) = 1,256.6371 mm2 2

4 The strength of a 40 mm diameter single-shear pin connection is thus: V =  allow AV = ( 208 N/mm2 )(1,256.6371 mm2 ) = 261,380.5 N = 261.3805 kN The shear force acting on pin A is equal to the reaction force at A. From Eq. (e), calculate the distributed load w that can be applied to the beam based on the strength of the pin connection at A: A 261.3805 kN w = = 55.4126 kN/m (h) 4.7170 m 4.7170 m Maximum loading w: Compare Eqs. (f), (g), and (h) to find w  44.1 kN/m

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P4.12 The idler pulley mechanism shown in Figure P4.12 must support belt tensions of P = l30 lb. Rigid bar BCD is supported by rod (1), which has a diameter of 0.188 in. and a yield strength of 35,000 psi. Rod (1) is connected at A and B with 0.25 in. diameter pins in double-shear connections. A 0.25 in. diameter pin in a single-shear connection holds rigid bar BCD at support C. Each pin has an ultimate shear strength of 42,000 psi. Overall dimensions of the mechanism are a = 12 in., b = 9 in., c = 7 in., and d = 3 in. Determine: (a) the factor of safety for rod (1) with respect to its yield strength. (b) the factor of safety for pin B with respect to its ultimate shear strength. (c) the factor of safety for pin C with respect to its ultimate shear strength.

FIGURE P4.12

Solution Equilibrium: First, consider an FBD of the pulley to determine the reaction forces exerted on the pulley by the mechanism. Fx = Dx − P − P cos ( 60 ) = 0  Dx = (130 lb ) + (130 lb ) cos ( 60 ) = 195 lb Fy = Dy − P sin ( 60 ) = 0  Dy = (130 lb ) sin ( 60 ) = 112.583 lb

FBD of pulley FBD of mechanism Next, consider an FBD of the mechanism to determine the force in rod (1) and the resultant force acting on the pin at C. Rod (1) is oriented at an angle of: c + d 7in. + 3 in. tan  = = = 0.8333 a 12 in.  = 39.8056

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Rod (1) is a two-force member, and its axial force can be calculated from: M C = Dxc + Dyb − ( F1 cos )( c + d ) = 0

 F1 =

(195 lb )( 7 in.) + (112.583 lb )(9 in.) = 309.579 lb ( c + d ) cos ( 7 in. + 3 in.) cos ( 39.8056) Dx c + Dyb

Determine the reaction forces at pin C from the following equilibrium equations: Fx = Cx − Dx + F1 cos ( 39.8056 ) = 0  Cx = 195 lb − ( 309.579 lb ) cos ( 39.8056 ) = −42.825 lb Fy = C y − Dy − F1 sin ( 39.8056 ) = 0  C y = 112.583 lb + ( 309.579 lb ) sin ( 39.8056 ) = 310.771 lb The resultant force at pin C is: C = C x2 + C y2 =

( −42.825 lb ) + ( 310.771 lb ) = 313.708 lb 2

(a) Factor of safety for rod (1) with respect to its yield strength: The area of rod (1) is

A1 =



d12 =



( 0.188 in.) = 0.0277591 in.2 2

4 4 The normal stress in the rod is: F 309.579 lb 1 = 1 = = 11,152.33 psi A1 0.0277591 in.2 Therefore, its factor of safety with respect to the 35,000 psi yield strength is

FS1 =

Y 35,000 psi = = 3.14 1 11,152.33 psi

Ans.

(b) Factor of safety for pin B with respect to its ultimate shear strength: The cross-sectional area of a 0.25 in. diameter pin is

Apin =



2 d pin =



( 0.25 in.) = 0.0490874 in.2 2

4 4 The pin at B is subjected to a shear force equal to the force in rod (1). Pin B is used in a double-shear connection; therefore, the shear area is twice the cross-sectional area of the pin. Calculate the shear stress in pin B as: V F 309.579 lb B = = 1 = = 3,153.34 psi AV 2 Apin 2 ( 0.0490874 in.2 ) Its factor of safety with respect to the 42,000 psi ultimate strength is

 B

FSB = U =

42,000 psi = 13.32 3,153.34 psi

Ans.

(c) Factor of safety for pin C with respect to its ultimate shear strength: Pin C is used in a singleshear connection. The shear stress in pin C is calculated as: C V 313.708 lb C = = = = 6,390.80 psi AV Apin 0.0490874 in.2 Its factor of safety with respect to the 42,000 psi ultimate strength is

 C

FSC = U =

42,000 psi = 6.57 6,390.80 psi

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P4.13 A 25 mm thick steel plate will be used as an axial member to support a dead load of 280 kN and a live load of 410 kN. The yield strength of the steel is 250 MPa. (a) Use the ASD method to determine the minimum plate width b required for the axial member if a factor of safety of 1.67 with respect to yielding is required. (b) Use the LRFD method to determine the minimum plate width b required for the axial member based on yielding of the gross section using the LRFD method. Use a resistance factor of t = 0.9 and load factors of 1.2 and 1.6 for the dead and live loads, respectively.

Solution (a) The service load on the axial member is P = D + L = 280 kN + 410 kN = 690 kN The allowable normal stress is  250 MPa  allow = Y = = 149.70 MPa FS 1.67 The minimum cross-sectional area required to support the service load is ( 690 kN )(1,000 N/kN ) = 4,609.20 mm2 P Amin  =  allow 149.70 N/mm2 Since the plate is 25 mm thick, the minimum plate width is therefore: A 4,609.20 mm 2 bmin  min = = 184.4 mm t 25 mm

Ans.

(b) The ultimate load for LRFD is U = 1.2D + 1.6L = 1.2 ( 280 kN ) + 1.6 ( 410 kN ) = 992 kN The design equation for an axial member subjected to tension can be written in LRFD as t Y A  U Consequently, the minimum cross-sectional area required for the tension member is ( 992 kN )(1,000 N/kN ) = 4,408.89 mm 2 U Amin  = t Y 0.9 250 N/mm 2

(

)

Since the plate is 20 mm thick, the minimum plate width is therefore: A 4, 408.89 mm 2 bmin  min = = 176.4 mm t 25 mm

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P4.14 A round steel tie rod is used as a tension member to support a dead load of 90 kN and a live load of 120 kN. The yield strength of the steel is 320 MPa. (a) Use the ASD method to determine the minimum diameter required for the tie rod if a factor of safety of 2.0 with respect to yielding is required. (b) Use the LRFD method to determine the minimum diameter required for the tie rod based on yielding of the gross section using the LRFD method. Use a resistance factor of t = 0.9 and load factors of 1.2 and 1.6 for the dead and live loads, respectively.

Solution (a) The service load on the axial member is P = D + L = 90 kN + 120 kN = 210 kN The allowable normal stress is  320 MPa  allow = Y = = 160 MPa FS 2.0 The minimum cross-sectional area required to support the service load is ( 210 kN )(1,000 N/kN ) = 1,312.50 mm2 P Amin  =  allow 160 N/mm2 and thus, the minimum tie rod diameter is  2 d min  1,312.50 mm 2  d min  40.9 mm 4

Ans.

(b) The ultimate load for LRFD is U = 1.2D + 1.6L = 1.2 (90 kN ) + 1.6 (120 kN ) = 300 kN The design equation for an axial member subjected to tension can be written in LRFD as t Y A  U Consequently, the minimum cross-sectional area required for the tension member is ( 300 kN )(1,000 N/kN ) = 1,041.67 mm2 U Amin  = t Y 0.9(320 N/mm2 ) and thus, the minimum tie rod diameter is  2 d min  1,041.67 mm 2  d min  36.4 mm 4

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P5.1 A steel [E = 200 GPa] rod with a circular cross section is 15 m long. Determine the minimum diameter required if the rod must transmit a tensile force of 300 kN without exceeding an allowable stress of 250 MPa or stretching more than 10 mm.

Solution If the normal stress in the rod cannot exceed 250 MPa, the cross-sectional area must equal or exceed P ( 300 kN )(1,000 N/kN ) A = = 1, 200 mm 2  250 N/mm 2 If the elongation must not exceed 10 mm, the cross-sectional area must equal or exceed PL ( 300 kN )(1,000 N/kN )(15,000 mm ) A = = 2, 250 mm 2 2 E ( 200,000 N/mm ) (10 mm ) Therefore, the minimum cross-sectional area that may be used for the rod is Amin = 2,250 mm2. The corresponding rod diameter is  2 Ans. d rod  2, 250 mm 2  d rod  53.524 mm = 53.5 mm 4

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P5.2 A rectangular bar of length L has a slot in the central half of its length, as shown in Figure P5.2. The bar has width b, thickness t, and elastic modulus E. The slot has width b/3. If L = 400 mm, b = 45 mm, t = 8 mm, and E = 72 GPa, determine the overall elongation of the bar for an axial force of P = 18 kN. FIGURE P5.2

Solution The internal force in the bar is equal to the external load P. The unslotted portions of the bar have an area of Aunslotted = (45 mm)(8 mm) = 360 mm2 The slotted portion of the bar has an area of 45 mm   2 Aslotted =  45 mm −  (8 mm) = 240 mm  3 The overall elongation of the bar is thus  P L L 18,000 N  200 mm 200 mm   =  unslotted + slotted  = + = 0.347 mm Ans. 2  2 E  Aunslotted Aslotted  72,000 N/mm  360 mm 240 mm2 

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P5.3 Compound axial member ABC shown in Figure P5.3 has a uniform diameter of d = 1.25 in. Segment (1) is an aluminum [E1 = 10,000 ksi] alloy and segment (2) is a copper [E2 = 17,000 ksi] alloy. The lengths of segments (1) and (2) are L1 = 84 in. and L2 = 130 in., respectively. Determine the force P required to stretch compound member ABC by a total of 0.25 in.

FIGURE P5.3

Solution The deformations in the two axial members are expressed by FL FL 1 = 1 1 and  2 = 2 2 A1E1 A2 E2 The total elongation of the assembly is equal to the displacement of end C relative to end A; which is simply equal to the sum of the deformations in the two axial members: FL F L uC / A = 1 +  2 = 1 1 + 2 2 A1E1 A2 E2 Since the internal forces F1 and F2 are each equal to external load P, this expression can be simplified to  L L  uC / A = P  1 + 2   A1E1 A2 E2  Both members have the same diameter and so the areas of the two members are each given by

A1 = A2 =



(1.25 in.) = 1.2272 in.2 2

The force P required to stretch the assembly by 0.25 in. is thus uC / A P= L1 L + 2 A1E1 A2 E2 =

0.25 in.

  84 in. 130 in. +   2 2 2 2  (1.2272 in. )(10,000 kips/in. ) (1.2272 in. )(17,000 kips/in. )  = 19.1185 kips

= 19.12 kips

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P5.4 Three solid cylindrical rods are welded together to form the compound axial member shown in Figure P5.4. The compound axial member is attached to a fixed support at A. Each rod has an elastic modulus of E = 40 GPa. Use the following values for the rod lengths and areas: L1 = 1,440 mm, L2 = 1,680 mm, L3 = 1,200 mm, A1 = 260 mm2, A2 = 130 mm2, and A3 = 65 mm2. What magnitude of external load P is needed to displace end D a distance of uD = 60 mm to the right?

FIGURE P5.4

Solution The deformations in the three axial members are expressed by FL FL FL 1 = 1 1 ,  2 = 2 2 , and 3 = 3 3 A1E1 A2 E2 A3 E3 The displacement of end D relative to fixed support A is simply equal to the sum of the deformations in the three axial members: FL FL F L uD = 1 +  2 + 3 = 1 1 + 2 2 + 3 3 A1E1 A2 E2 A3 E3 The internal forces F1, F2, and F3 are each equal to external load P. The elastic moduli are also the same for each of the three rods. Consequently, the expression for uD can be simplified to P L L L  uD =  1 + 2 + 3  E  A1 A2 A3  Solve for the force P required to stretch the assembly by 60 mm: EuD P= L1 L2 L3 + + A1 A2 A3

( 40, 000 N/mm ) ( 60 mm ) 2

1,440 mm 1,680 mm 1,200 mm   260 mm 2 + 130 mm 2 + 65 mm 2  = 65, 000 N

= 65.0 kN

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P5.5 The assembly shown in Figure P5.5 consists of a bronze rod (1) and an aluminum rod (2), each having a cross-sectional area of 0.15 in.2. The assembly is subjected to loads of P = 1,900 lb and Q = 3,200 lb, acting in the directions shown. Determine the horizontal displacement uB of coupling B with respect to rigid support A. Use L1 = 160 in., L2 = 40 in., E1 = 16 ×106 psi, and E2 = 10 ×106 psi.

FIGURE P5.5

Solution Draw an FBD that cuts through rod (1) and includes the free end of the assembly. From this FBD, the internal force in rod (1) is Fx = Q − P − F1 = 0

 F1 = Q − P = 3, 200 lb − 1,900 lb = 1,300 lb The deformation of rod (1) can be expressed as the difference between the displacements of joint A and coupling B: 1 = u B − u A However, joint A is fixed against translation, and so, we know that uA = 0. Consequently, the displacement of coupling B is equal to the deformation of rod (1): u B = 1 The deformation of rod (1) can be computed as (1,300 lb )(160 in.) = 0.08667 in. FL 1 = 1 1 = A1 E1 ( 0.15 in.2 )(16 106 lb/in.2 ) The displacement of coupling B relative to the support at A is thus:

uB = 1 = 0.08667 in. = 0.0867 in. = 0.0867 in. →

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P5.6 A compound steel (E = 30,000 ksi) bar is subjected to the loads shown in Figure P5.6. Bar (1) has a crosssectional area of A1 = 0.35 in.2 and a length of L1 = 105 in. Bar (2) has an area of A2 = 0.35 in.2 and a length of L2 = 85 in. Bar (3) has an area of A3 = 0.09 in.2 and a length of L3 = 40 in. The loads and angles are P = 40 kips,  = 36°, Q = 13 kips,  = 58°, and R = 9 kips. Determine the vertical deflection of end A.

FIGURE P5.6

Solution Draw an FBD that cuts through member (1) and includes the free end of the bar. From this FBD, the sum of forces in the vertical direction can be written as Fy = F1 + 2P sin  − 2Q sin  − R = 0 Calculate the internal axial force in member (1) as F1 = 2Q sin  + R − 2 P sin 

= 2 (13 kips ) sin ( 58 ) + 9 kips − 2 ( 40 kips ) sin ( 36 )

= −15.9736 kips The deformation that occurs in member (1) is. FL 1 = 1 1 A1 E1 =

( −15.9736 kips )(105 in.) = −0.1597 in. ( 0.35 in.2 ) ( 30, 000 ksi )

(contraction)

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Draw an FBD that cuts through member (2) and includes the free end of the bar. From this FBD, the equilibrium equation is Fy = F2 − 2Q sin  − R = 0 Therefore, the internal axial force in member (2) is F2 = 2Q sin  + R = 2 (13 kips ) sin ( 58 ) + 9 kips

= 31.0493 kips The deformation that occurs in member (2) is computed as FL 2 = 2 2 A2 E2 =

( 31.0493 kips )(85 in.) = 0.2514 in. ( 0.35 in.2 ) ( 30, 000 ksi )

(elongation)

Draw an FBD that cuts through member (3) and includes the free end of the bar. From this FBD, the equilibrium equation is Fy = F3 − R = 0 The internal axial force in member (3) is F3 = R = 9 kips

The deformation in member (3) is FL 3 = 3 3 A3 E3 =

( 9 kips )( 40 in.) = 0.1333 in. ( 0.09 in.2 ) ( 30, 000 ksi )

(elongation)

The deformations in each of the three members can be expressed as the difference between the upper and lower ends of each member as follows: 1 = vD − vC

 2 = vC − vB  3 = vB − v A We know that there is zero deflection at joint A of the bar. Use this fact to express the joint deflections in terms of the member deformations: 1 = vD − vC = −vC  vC = −1  2 = vC − vB  vB = vC −  2 = −1 −  2  3 = vB − v A  v A = vB −  3 = −1 −  2 −  3 The vertical deflection of end A is therefore: vA = −1 −  2 −  3 = − ( −0.1597 in.) − ( 0.2514 in.) − ( 0.1333 in.) = −0.2249 in. = 0.225 in. 

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P5.7 Rigid bar DB in Figure P5.7 is supported at pin B by axial member ABC, which has a crosssectional area of A1 = A2 = 75 mm2, an elastic modulus of E1 = E2 = 900 MPa, and lengths of L1 = 400 mm and L2 = 600 mm. A load of Q = 500 N is applied at end C of member (2), and a load of P = 1,500 N is applied to the rigid bar at a distance of a = 80 mm from pin support D and b = 50 mm from pin B. Determine the downward deflections of: (a) pin B. (b) pin C. FIGURE P5.7

Solution Equilibrium and deformations: Draw an FBD that exposes the internal axial force in member (2): Fy = F2 − Q = 0

 F2 = Q = 500 N The deformation of member (2) is computed as: ( 500 N )( 600 mm ) = 4.4444 mm FL 2 = 2 2 = A2 E2 ( 75 mm 2 )( 900 N/mm 2 ) Similarly, draw an FBD that exposes the internal axial force in member (1). Consider a moment equilibrium equation about D to determine the force in member (1): M D = − Pa − Q ( a + b ) + F1 ( a + b ) = 0 Pa + Q ( a + b ) (1,500 N )( 80 mm ) + ( 500 N )( 80 mm + 50 mm ) = a+b 80 mm + 50 mm = 1, 423.0769 N

 F1 =

The deformation of member (1) is computed as:

1 =

F1 L1 (1, 423.0769 N )( 400 mm ) = = 8.4330 mm A1 E1 ( 75 mm2 )( 900 N/mm2 )

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Deflections: The deformations in the two axial members can be expressed as the difference between the deflections of the upper and lower ends of each member as follows:  1 = v A − vB

 2 = vB − vC We know that there is zero deflection at joint A of the bar. Use this fact to express the joint deflections in terms of the member deformations as: 1 = vA − vB = −vB  vB = −1

 2 = vB − vC

 vC = vB −  2 = −1 −  2

(a) Deflection of pin B: The vertical deflection of pin B is: vB = −1 = − ( 8.4330 mm ) = 8.43 mm 

Ans.

(a) Deflection of pin C: The vertical deflection of pin C is: vC = −1 −  2 = − ( 8.4330 mm ) − ( 4.4444 mm ) = 12.88 mm 

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P5.8 A steel [E = 200 GPa] bar (1) supports beam AB, as shown in Figure P5.8. The cross-sectional area of the bar is 600 mm2. If the stress in the bar must not exceed 330 MPa and the maximum deformation in the bar must not exceed 15 mm, determine the maximum load P that may be supported by this structure. Use dimensions of a = 5.0 m, b = 3.0 m, and c = 6.5 m.

FIGURE P5.8

Solution (a) Draw an FBD of beam AB. Note that bar (1) is a twoforce member. Determine the angle  that bar (1) makes with the horizontal direction. c 6.5 m tan  = = = 0.8125 a + b 5.0 m + 3.0 m   = 39.094 From the sum of moments about pin B, express load P in terms of the force F1 in bar (1): M B = Pb − ( F1 sin  )( a + b ) = 0 P=

( a + b ) sin  F

b ( 5.0 m + 3.0 m ) sin ( 39.094 ) F = 1 3.0 m = 1.68158 F1

(a)

Consider stress: If the stress in bar (1) must not exceed 330 MPa, then the maximum force that can be applied to bar (1) is F1   allow,1 A1 = 330 N/mm2 600 mm2 = 198,000 N

(

)(

)

Consider deformation: The length of bar (1) is L1 =

( a + b ) + ( c ) = ( 5.0 m + 3.0 m ) + ( 6.5 m ) = 10.30776 m = 10,307.76 mm 2

If the maximum deformation in the rod must not exceed 15 mm, the maximum force in the bar must be limited to 2 2 1 A1 E1 (15 mm ) ( 600 mm )( 200, 000 N/mm ) F1  = = 174, 626 N L1 10,307.76 mm Comparison: Based on considerations of both stress and deformation, the maximum force permitted in bar (1) must not exceed 174,626 N. From Equation (a), the maximum load P that may be supported by this structure is: Ans. P = 1.68158F1 = 1.68158 (174,626 N ) = 293,647 N = 294 kN

Mechanics of Materials: An Integrated Learning System, 4th Ed. P5.9 The assembly shown in Figure P5.9 consists of rod (1) and tube (2), both made of steel (E = 30,000 ksi). The tube is attached to a fixed support at C. A rigid end plate is welded to the tube at B, and the rod is welded to the end plate. The rod is solid with a diameter of d1 = 0.375 in. The tube has an outside diameter of D2 = 2.25 in. and a wall thickness of t2 = 0.125 in. The dimensions of the assembly are a = 60 in. and b = 32 in. Determine the horizontal deflection at rod end A for an applied load of P = 4 kips.

Timothy A. Philpot

FIGURE P5.9

Solution Equilibrium: Before proceeding, it is convenient to determine the internal forces in each of the axial segments. Rod (1): Draw an FBD that cuts through rod (1) and includes the free end of the axial member. From this FBD, the sum of forces in the horizontal direction gives the force in segment (1): Fx = P − F1 = 0

 F1 = P = 4 kips Tube (2): Draw an FBD that cuts through tube (2) as well as rod (1). Include the end plate at B. From this FBD, the sum of forces in the horizontal direction gives the force in segment (2): Fx = F1 + F2 = 0  F2 = − F1 = −4 kips

Deformations Rod (1): The cross-sectional area of the rod is

A1 =



d12 =



( 0.375 in.) = 0.11045 in.2 2

4 4 The deformation in rod (1) can be computed as ( 4 kips )( 60 in. + 32 in.) = 0.11106 in. FL 1 = 1 1 = A1 E1 ( 0.11045 in.2 ) ( 30, 000 ksi ) Tube (2): The cross-sectional area of the hollow steel tube is

A2 =



 D − d ) = ( 2.25 in.) − ( 2.00 in.)  = 0.83449 in. (  4 4 2 2

2 2

The deformation in tube (2) is thus ( −4 kips )( 60 in.) FL 2 = 2 2 = = −0.00959 in. A2 E2 ( 0.83449 in.2 ) ( 30, 000 ksi )

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Deflections: The deformations in the two axial members can be expressed as the difference between the deflections at the right and left ends of each member as follows: 1 = u A − uB

 2 = uC − uB We know that there is zero deflection at fixed support C (i.e., uC = 0). Use this fact to express the joint deflections in terms of the member deformations as:  2 = uC − uB = 0 − uB  uB = − 2

1 = u A − uB

 u A = uB + 1 = − 2 + 1

Deflection of rod end A: The horizontal deflection of A is: u A = − 2 + 1 = − ( −0.00959 in.) + ( 0.11106 in.) = 0.1207 in. →

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P5.10 The system shown in Figure P5.10 consists of a rod, a spring, a tube, and a fixed base. Polymer rod (1) is attached to rod cap A at its upper end and it passes freely through the spring, tube cap B, tube (2), and the fixed base to rod end D, where a load P is applied. The load applied at rod end D is transmitted by the rod to rod cap A, causing the spring to compress. The force in the spring is then transmitted to the tube, creating compressive deformation in the tube. The polymer rod (1) has a diameter of d1 = 12 mm, an elastic modulus of E1 = 1,400 MPa, and a length of L1 = 640 mm. The linear compression spring has a spring constant of k = 850 N/mm. Tube (2) has an outside diameter of D2 = 50 mm, an inside diameter of d2 = 46 mm, an elastic modulus of E2 = 900 MPa, and a length of L2 = 350 mm. For an applied load of P = 750 N, determine the vertical displacement relative to C of (a) rod cap A, (b) tube cap B, and (c) rod end D. FIGURE P5.10

Solution Equilibrium: The force in rod (1) is equal to the external load P: F1 = P = 750 N Rod (1) is attached to rod cap A, which is supported by the spring, and in turn, tube (2). The tensile force in rod (1) creates compressive forces in both the spring and tube (2). Thus, Fs = F2 = − F1 = −750 N Deformations: The area of the 12 mm diameter polymer rod (1) is:

A1 =



d12 =



(12 mm ) = 113.097 mm2 2

4 4 The deformation in rod (1) is ( 750 N )( 640 mm ) FL 1 = 1 1 = = 3.0315 mm A1 E1 (113.097 mm 2 )(1, 400 N/mm 2 ) Tube (2): The cross-sectional area of tube (2) is

A2 =



 D − d ) = ( 50 mm ) − ( 46 mm )  = 301.593 mm (  4 4 2 2

2 2

The deformation in tube (2) is thus ( −750 N )( 350 mm ) FL 2 = 2 2 = = −0.9671 mm A2 E2 ( 301.593 mm 2 )( 900 N/mm 2 ) Spring s: The deformation of the spring is computed as F −750 N s = s = = −0.8824 mm k 850 N/mm Deflections: The deformations in the spring and the two axial members can each be expressed as the difference between their respective upper and lower ends. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

 1 = v A − vD  s = v A − vB  2 = vB − vC We know that the deflection at C is vC = 0. Use this fact to express the various deflections in terms of the member deformations.  2 = vB − vC = vB  vB =  2 = −0.9671 mm

 s = v A − vB

 v A = vB +  s =  2 +  s = −0.9671 mm + ( −0.8824 mm ) = −1.8495 mm

 1 = v A − vD

 vD = v A − 1 = −1.8495 mm − 3.0315 mm = −4.8810 mm

(a) Displacement of rod cap A

vA = 1.849 mm 

Ans.

(b) Displacement of tube cap B

vB = 0.967 mm 

Ans.

vD = 4.88 mm 

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P5.11 The mechanism shown in Figure P5.11 consists of rigid bar ABC pinned at C to nylon [E = 200 ksi] rod (1), which has a diameter of 0.25 in. and a length of L1 = 14 in. The linear compression spring, which is placed between the mechanism base at E and a rigid plate at D, has a spring constant of k = 940 lb/in. The mechanism base is rigid and fixed in position, and rigid bar ABC is horizontal before the load P is applied. Rod (1) is attached to plate D at the lower end of the spring. The nylon rod passes freely through the compression spring and through a hole in the base at E. Determine the value of load P that is needed to deflect end A downward by 2.0 in. Use dimensions of a = 9.0 in., b = 3.0 in., and c = 5.0 in.

FIGURE P5.11

Solution Equilibrium: From the sum of moments about pin B, express force F1 in terms of load P: M B = F1b − Pa = 0 a P b 9.0 in. = P = 3P 3.0 in.

F1 =

(a)

Rigid bar deflections: The rigid bar will rotate about pin B. The relationship between the deflections of the rigid bar at A and C can be determined by similar triangles: v vA =− C a b b  vC = − vA a 3.0 in. =− vA = −0.3333vA 9.0 in. Note that the negative sign is necessary because end C moves upward when end A moves downward. We are told that end A must deflect downward by 2.0 in. Therefore, end C must deflect upward by: vC = −0.3333vA = −0.3333( −2.0 in.) = 0.6667 in.  (b) Deformations of rod and spring: The amount that end C of the rigid bar deflects upward is dependent on two deformations: (a) the elongation of rod (1) due to the tensile force F1, and (2) the contraction of the spring due to the compressive force exerted on it by the rod. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Deformation of rod (1): The area of the 0.25 in. nylon rod:

A1 =



d12 =



( 0.25 in.) = 0.049087 in.2 2

4 4 The deformation in rod (1) can be expressed as F1 (14 in.) FL 1 = 1 1 = = (1.426028 10−3 in./lb ) F1 2 2 A1 E1 ( 0.049087 in. )( 200, 000 lb/in. ) Deformation of spring: The force in the spring is a compressive force equal in magnitude to the force in rod (1); thus, Fx = –F1. The deformation of the spring is expressed as F −F F1 s = s = 1 = − = − (1.063830  10−3 in./lb ) F1 k k 940 lb/in. Deflections: The deformations in the spring and the axial member can each be expressed as the difference between their respective upper and lower ends. 1 = vC − vD  s = v E − vD We know that the deflection at E is vE = 0 since E is on the base that supports the assembly. Use this fact to express the various deflections in terms of the member deformations.  s = vE − vD = − vD  vD = − s = (1.063830 10 −3 in./lb ) F1

1 = vC − vD  vC = vD + 1

= (1.063830 10−3 in./lb ) F1 + (1.426028 10 −3 in./lb ) F1 = ( 2.489858 10−3 in./lb ) F1

(c)

Substitute Equation (a) into Equation (c) to express the rigid bar deflection at C in terms of the load P: vC = ( 2.489858 10−3 in./lb ) F1 = ( 2.489858 10−3 in./lb ) ( 3P )

= ( 7.469574 10−3 in./lb ) P

(d)

Substitute the value for vC obtained in Equation (b) into the expression derived in Equation (d) and compute P: vC = ( 7.469574 10−3 in./lb ) P = 0.6667 in.

 P = 89.3 lb

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P5.12 The wooden pile shown in Figure P5.12 has a diameter of 100 mm and is subjected to a load of P = 75 kN. Along the length of the pile and around its perimeter, soil supplies a constant frictional resistance of w = 3.70 kN/m. The length of the pile is L = 5.0 m and its elastic modulus is E = 8.3 GPa. Calculate (a) the force FB needed at the base of the pile for equilibrium. (b) the magnitude of the downward displacement at A relative to B.

FIGURE P5.12

Solution (a) Force at base of pile: Consider the entire pile length L and sum forces in the vertical direction. Fy = − FB − wL + P = 0 The force FB required for equilibrium is thus Ans. FB = P − wL = 75 kN − (3.70 kN/m)(5.0 m) = 56.5 kN (b) Magnitude of the downward displacement at A: Consider the free-body diagram shown. The equilibrium equation for the pile is Fy = P − wy + F ( y) = 0 The internal force F(y) in the pile can now be expressed as F ( y) = wy − P = (3.70 kN/m) y − 75 kN The cross-sectional area of the pile is  A = (0.100 m) 2 = 7.8540  10−3 m 2 4 The total deformation of the pile can be calculated with the following integral L F ( y) 1 L  = dy = F ( y) dy 0 A( y ) E ( y ) AE 0 Thus, the total deformation of the pile is 5.0 1 L 1  3.70 kN/m 2  = (3.70 kN/m) y − 75 kN dy = y − (75 kN) y ( )  AE 0 AE  2 0

1  3.70 kN/m  (5.0 m) 2 − (75 kN)(5.0 m)  6 2  (7.8540  10 m )(8.3  10 kN/m )  2  −328.750 kN-m = = −5.04309  10 −3 m = −5.04 mm −3 2 6 2 (7.8540  10 m )(8.3  10 kN/m ) Since F(y) was assumed to be a tension force, a positive result here would indicate an elongation of the pile. The negative sign obtained for  indicates that the pile contracts in length. Therefore, the upper end of the pile moves downward by the same amount. =

−3

The magnitude of the downward displacement at A relative to B is thus

u A = 5.04 mm 

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P5.13 A 1 in. diameter steel [E = 29,000 ksi and  = 490 lb/ft3] rod hangs vertically while suspended from one end. The length of the rod is 320 ft. Determine the change in length of the bar due to its own weight.

Solution An incremental length dy of the bar has an incremental deformation given by F ( y) d = dy AE The force in the bar can be expressed as the product of the unit density of the steel (steel) and the volume of the bar below the incremental slice dy: F ( y ) =  steel A y Therefore, the incremental deformation can be expressed as  Ay  y d = steel dy = steel dy AE E Integrate this expression over the entire length L of the bar: L L L y   1  L2  =  steel dy = steel  y dy = steel  y 2  0 = steel 0 E E 0 E 2 2E The change in length of the bar due to its own weight is therefore: 3 2 3  1 ft  490 lb/ft )  320 ft 12 in./ft ) ( ( 2   L  12 in.   = steel = = 0.07209 in. = 0.0721 in. 2E 2 ( 29, 000, 000 lb/in.2 )

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P5.14 A homogenous rod of length L and elastic modulus E is a truncated cone with diameter that varies linearly from d0 at one end to 2d0 at the other end. A concentrated axial load P is applied to the ends of the rod, as shown in Figure P5.14. Assume that the taper of the cone is slight enough for the assumption of a uniform axial stress distribution over a cross section to be valid. (a) Determine an expression for the stress distribution on an arbitrary cross section at x. (b) Determine an expression for the elongation of the rod.

FIGURE P5.14

Solution (a) Determine an expression for the diameter of the truncated cone through the use of the similar triangles concept: 2d 0 − d 0 d ( x ) − d 0 = L x x  (a)  d ( x) = d 0 1 +   L Next, consider equilibrium of the truncated cone:  F ( x) = P  Fx = F ( x) − P = 0 The internal force F(x) can be expressed in terms of stress and area: F ( x) =   dA =  ( x) A( x) A

Equating this expression to the equilibrium equation gives: F ( x) =  ( x) A( x) = P P   ( x) = A( x) From Eq. (a), the area of the cone at any location x is expressed as 2 x 2  d 1 +  0    d ( x) 2 L A( x) = = 4 4 Therefore, the stress can be written as P 4P  ( x) = = 2 A( x) x 2  d 0 1 +   L

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

(b) The elongation in the axial member is found from Eq. (5.5): L L F ( x)  =  d =  dx 0 0 A( x ) E For the truncated cone axial member: L 4P 4 P L dx = dx = 2 2 0  Ed 02 0  x x   Ed 02 1 +  1+   L L 4P  − L  2 PL = = 2    Ed 0 1 + x  Ed 02    L 0 L

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P5.15 Determine the extension, due to its own weight, of the conical bar shown in Figure P5.15. The bar is made of aluminum alloy [E = 10,600 ksi and  = 0.100 lb/in.3]. The bar has a 2 in. radius at its upper end and a length of L = 20 ft. Assume the taper of the bar is slight enough for the assumption of a uniform axial stress distribution over a cross section to be valid.

FIGURE P5.15

Solution An incremental length dy of the bar has an incremental deformation given by F ( y) d = dy AE The force in the bar can be expressed as the product of the unit density of the aluminum (alum) and the volume of the bar below the incremental slice dy. The volume below the slice dy is a cone. At y, the cross-sectional area of the base of the cone is: 2

y  Ay =   r  L  and the volume of a cone is given by: 1 V = (area of base)(altitude) 3 The internal force at y can be expressed as 1 F ( y ) =  alum  Ay y 3 The incremental deformation can be expressed as  alum Ay y  y d = dy = alum dy 3 Ay E 3E Integrate this expression over the entire length L of the bar: L L L y    L2 1  =  alum dy = alum  y dx = alum  y 2  0 = alum 0 3E 3E 0 3E 2 6E The change in length of the bar due to its own weight is therefore:  alum L2 (0.100 lb/in.3 )(20 ft 12 in./ft) 2 = = = 9.0566 10−5 in. = 90.6 10−6 in. 6E 6(10,600,000 psi)

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P5.16 Rigid beam ABC shown in Figure P5.16 is supported by rods (1) and (2) that have identical lengths of L = 7.0 m. Rod (1) is made of steel [E = 200 GPa] and has a cross-sectional area of 175 mm2. Rod (2) is made of aluminum [E = 70 GPa] and has a cross-sectional area of 300 mm2. Assume dimensions of a = 1.5 m and b = 3.0 m. For a uniformly distributed load of w = 15 kN/m, determine the deflection of the rigid beam at point B. FIGURE P5.16

Solution Equilibrium: From an FBD of the rigid beam, write two equilibrium equations: Fy = F1 + F2 − wb = 0

(a)

b M C = − F1 ( a + b ) + wb   = 0 2 Solve Eq. (b) for F1:

(b)

(15 kN/m )( 3.0 m ) = 15 kN wb 2 F1 = = 2 ( a + b) 2 (1.5 m + 3.0 m ) 2

and backsubstitute into Eq. (a) to obtain F2: F2 = wb − F1 = (15 kN/m)( 3.0 m) −15 kN = 30 kN Deformations: Next, determine the deformations in rods (1) and (2): (15,000 N )( 7,000 mm ) = 3.000 mm FL 1 = 1 1 = A1E1 175 mm 2 200,000 N/mm 2

(

2 =

)(

)

( 30, 000 N )( 7, 000 mm ) = 10.000 mm F2 L2 = A2 E2 ( 300 mm 2 )( 70, 000 N/mm 2 )

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Rigid beam deflections: Sketch the deflected position of the rigid beam in response to the distributed load. The deformations of each rod can each be expressed as the difference between the deflections at their respective upper and lower ends.  1 = vD − v A  2 = vE − vC We know that the deflections at D and E are zero. Use this fact, and express the beam deflections in terms of the member deformations.  1 = vD − v A

 v A = −1 = −3.000 mm

 2 = vE − vC  vC = − 2 = −10.000 mm

The deflection at B can be determined from similar triangles: vC − vA vB − vA = L a Solve this expression for vB: a vB = (vC − vA ) + vA L 1.5 m = ( −10.000 mm ) − ( −3.000 mm )  + ( −3.000 mm ) 1.5 m + 3.0 m  = ( 0.3333)( −7.000 mm ) − 3.000 mm

= −5.33 mm = 5.33 mm 

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P5.17 In Figure P5.17, the horizontal rigid beam ABCD is supported by vertical bars (1) and (2) and is loaded at points A and D by vertical forces P = 30 kN and Q = 40 kN, respectively. Bar (1) is made of aluminum [E = 70 GPa] and has a crosssectional area of 225 mm2 and a length of L1 = 8.0 m. Bar (2) is made of steel [E = 200 GPa] and has a cross-sectional area of 375 mm2 and a length of L2 = 5.0 m. Assume dimensions a = 2.5 m, b = 2.0 m, and c = 1.50 m. Determine the deflection of the rigid beam (a) at point A and (b) at point D. FIGURE P5.17

Solution Equilibrium: Draw an FBD of the rigid beam ABCD. Note that member (1) is a two-force member; however, member (2) is not. Write two equilibrium equations and solve for the forces F1 and F2. (Clearly, Cx = 0 by inspection.) M B = − Pa − F2b + Q ( b + c ) = 0 F2 = =

− Pa + Q ( b + c ) b − ( 30 kN )( 2.5 m ) + ( 40 kN )  2.0 m + 1.50 m 

2.0 m M C = − P ( a + b ) + F1b + Qc = 0

= 32.5 kN

P ( a + b ) − Qc b ( 30 kN )( 2.5 m + 2.0 m ) − ( 40 kN )(1.50 m ) = 37.5 kN = 2.0 m

(a)

F1 =

(b)

Deformations of bars (1) and (2): Determine the deformations in bars (1) and (2): ( 37,500 N )( 8, 000 mm ) = 19.0476 mm FL 1 = 1 1 = A1 E1 225 mm 2 70, 000 N/mm 2

(

2 =

)(

)

( 32,500 N )( 5, 000 mm ) = 2.1667 mm F2 L2 = A2 E2 ( 375 mm 2 )( 200, 000 N/mm 2 )

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Rigid beam deflections: Sketch the deflected position of the rigid beam after loads P and Q have been applied. The deformations of each bar can be expressed as the difference between the deflection of its upper joint and the deflection of its lower joint.  1 = vB − vE

and

 2 = vC − vF We know that the deflection at E and F is zero. Thus, we can express the deflections at B and C in terms of the member elongations 1 and 2 as: 1 = vB  vB = 1 = 19.0476 mm

 2 = vC

 vC =  2 = 2.1667 mm

(a) Rigid beam deflection at A: The deflection of the rigid beam at A can be determined from the principal of similar triangles as: v A − vB vB − vC = a b a  v A = ( vB − vC ) + vB b 2.5 m = (19.0476 mm − 2.1667 mm ) + 19.0476 mm 2.0 m = 40.1488 mm = 40.1 mm 

Ans.

(b) Rigid beam deflection at D: Similarly, the deflection of the rigid beam at D is found from: vC − vD vB − vC = c b c  vD = vC − ( vB − vC ) b 1.5 m = 2.1667 mm − (19.0476 mm − 2.1667 mm ) 2.0 m = −10.4940 mm = 10.49 mm 

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P5.18 The truss shown in Figure P5.18 is constructed from three aluminum alloy members, each having a cross-sectional area of A = 850 mm2 and an elastic modulus of E = 70 GPa. Assume a = 7.0 m, b = 4.5 m, and c = 5.0 m. Calculate the horizontal displacement of roller A when the truss supports loads of P = 12 kN and Q = 30 kN acting in the directions shown. FIGURE P5.18

Solution To determine the horizontal displacement of roller B, we need only determine the deformation in member AB. This requires calculation of the internal force in member AB. Overall equilibrium: Begin the solution by determining the external reaction forces acting on the truss at support A. The free-body diagram (FBD) of the entire truss is shown. The following equilibrium equation can be written to directly obtain the reaction force at B. M B = Pb + Qc − Ay ( a + b ) = 0  Ay =

Pb + Qc (12 kN )( 4.5 m ) + ( 30 kN )( 5.0 m ) = = 17.7391 kN a+b 7.0 m + 4.5 m

Method of joints: The geometry of the truss will be used to determine the magnitude of the inclination angles of member AC. Use the definition of the tangent function to determine : c 5.0 m tan  = = = 0.714286  = 35.5377 a 7.0 m Joint A: Consider only those forces acting directly on joint A. In this instance, two axial members, AB and AC, are connected at joint A. Additionally, the reaction force Ay acts at joint A. Tension forces will be assumed in each truss member. Fx = FAB + FAC cos  = 0 (a) (b) Fy = Ay + FAC sin  = 0 Solve Eq. (b) for FAC: − Ay −17.7391 kN FAC = = = −30.5195 kN sin  sin 35.5377 and then compute FAB using Eq. (a): Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

FAB = − FAC cos 

= − ( −30.5195 kN ) cos 35.5377 = 24.8348 kN

Axial deformation of member AB: The axial deformation of member AB can be calculated from ( 24,834.8 N )(11,500 mm ) = 4.800 mm F L  AB = AB AB = AAB E AB 850 mm 2 70, 000 N/mm 2

(

)(

)

As member AB elongates, the roller support at A moves to the left; therefore uA = 4.80 mm 

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P5.19 Three bars, each with a cross-sectional area of 750 mm2 and a length of 5.0 m, are connected and loaded as shown in Figure P5.19. Bars (1) and (3) are made of steel [E = 200 GPa] and bar (2) is made of an aluminum alloy [E = 70 GPa]. At D, all three bars are connected to a slider block that is constrained to travel in a smooth, horizontal slot. When the load P is applied, the strain in bar (2) is found to be 0.0012 mm/mm. Assume that a = 2.4 m and b = 1.6 m. Determine: (a) the horizontal deflection of D. (b) the deformations in each of the three bars. (c) the magnitude of load P. FIGURE P5.19

Solution Initial Geometry: Each of the three bars has a length of L = 5.0 m. Let the angle between bar (1) and bar (2) be denoted , and let the angle between bars (2) and (3) be denoted . Calculate  and . a 2.4 m sin  = = = 0.48  = 28.6854 L 5.0 m b 1.6 m sin  = = = 0.32   = 18.6629 L 5.0 m (a) Horizontal deflection of D: From the strain given for bar (2), we can directly compute the deformation in bar (2):

2

 2 =  2 L = ( 0.0012 mm/mm )( 5, 000 mm ) = 6.00 mm L The deformation in bar (2) can be expressed as the difference in deflections between the joints at each end of the bar.  2 = uD − uB Since joint B is pinned, it does not deflect and thus, uB = 0. The horizontal deflection of joint D must equal the deformation of bar (2): Ans. uD =  2 = 6.00 mm →

2 =

(b) Deformations in each of the three bars: Sketch a diagram showing the deformations of the three bars. We have already found that joint D displaces to a new position D′, which is 6.00 mm to the right of its initial position. Because joint D is constrained in a smooth slot, we also know that joint D moves only in a horizontal direction. To understand the deformation produced in bar (1), we imagine that we temporarily unpin the undeformed bar (1) from joint D. We allow the released end of bar (1) to rotate slightly upwards, and then we stretch the bar enough so that it can be reconnected to pin D. The amount that we stretch bar (1) is the deformation 1. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Here, we make an approximation. When we allow bar (1) to rotate slightly upward, we assume that the angle  after deformation is essentially the same as our initial value of  because the deflections that occur are relatively small. The geometry involved in the deformation of bar (1) is illustrated in the sketch above. A similar argument can be made for bar (3). Since we know the distance between the initial position of D and the final position D′, we can determine the deformations of bars (1) and (3) from the geometry shown in the sketch above. Bar (1) deformation: From the sketch above, we find:

cos  =

1 2

1 =  2 cos  = ( 6.00 mm ) cos 28.6854 = 5.2636 mm = 5.26 mm

Bar (2) deformation: From the strain given for bar (2), we previously calculate:  2 = 6.00 mm Bar (3) deformation: From the sketch above, we find:

cos  =

3 2

 3 =  2 cos  = ( 6.00 mm ) cos18.6629 = 5.6845 mm = 5.68 mm

Ans.

Ans. Ans.

(c) Magnitude of load P: From the force-deformation relationship, we can determine the force in each bar from its known deformation: FL 1 = 1 1 A1 E1  F1 =

2 =

F2 L2 A2 E2

 F2 =

3 =

1 A1 E1

 2 A2 E2 L2

F3 L3 A3 E3

 F3 =

 3 A3 E3 L3

( 5.2636 mm ) ( 750 mm2 )( 200, 000 N/mm 2 ) 5, 000 mm

( 6.00 mm ) ( 750 mm2 )( 70, 000 N/mm 2 ) 5, 000 mm

= 63, 000 N

( 5.6845 mm ) ( 750 mm2 )( 200, 000 N/mm2 ) 5, 000 mm

= 157,908.3 N

= 170,535.2 N

Next, consider an FBD of joint D. Sum the forces in the x direction and solve for load P. Fx = P − F1 cos  − F2 − F3 cos  = 0

P = F1 cos  + F2 + F3 cos 

= (157,908.3 N ) cos 28.6854 + 63, 000 N + (170,535.2 N ) cos18.6629 = 363, 096.0 N = 363 kN

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P5.20 The pin-connected assembly shown in Figure P5.20 consists of solid aluminum [E = 70 GPa] rods (1) and (2) and solid steel [E = 200 GPa] rod (3). Each rod has a diameter of 16 mm. Assume a = 2.5 m, b = 1.6 m, and c = 0.8 m. If the normal stress in any rod may not exceed 150 MPa, determine: (a) the maximum load P that may be applied at A. (b) the magnitude of the resulting deflection at A. FIGURE P5.20

Solution The lengths of rods (1) and (2) are:

L1 = (1.6 m)2 + (0.8 m)2 = 1.78885 m = 1,788.8544 mm = L2 and the cross-sectional area of a 16 mm diameter rod is  A = (16 mm) 2 = 201.062 mm 2 4 Determine the inclination angles for rods (1) and (2): 0.8 m tan 1 = 1 = 26.5651 =  2 1.6 m (a) Maximum load P: From an FBD of joint A, the force in rod (3) is F3 = P Next, consider an FBD of joint B and write two equilibrium equations: Fx = − F3 + F1 cos 26.5651 + F2 cos 26.5651 = 0 Fy = F1 sin 26.5651 − F2 sin 26.5651 = 0 From the second equation, F1 = F2. The first equation gives F1 = 0.559017F3 = 0.559017P. The allowable normal stress for each rod is 150 MPa; therefore, the allowable force in any rod is Fallow =  allow A = (150 MPa)(201.062 mm2 ) = 30,159.3 N Setting this allowable force equal to F3, the force in rods (1) and (2) is F3 = Fallow = 30,159.3 N

F1 = F2 = 0.559017 F3 = (0.559017)(30,159.3 N) = 16,859.6 N  Fallow The maximum force P that can be applied to the assembly is thus: P = 30.2 kN

OK Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

(b) Deflection at A: The elongation of steel rod (3) for a load of P = 30,159.3 N is FL (30,159.3 N)(2,500 mm) 3 = 3 2 = = 1.8750 mm A2 E2 (201.062 mm2 )(200,000 N/mm2 ) The elongation of aluminum rod (1) [and rod (2)] is FL (16,859.6 N)(1,788.8544 mm) 1 = 1 1 = = 2.1429 mm A1E1 (201.062 mm2 )(70,000 N/mm2 ) Thus, the length of rod (1) [and rod (2)] after deformation is L1 = 1,788.8544 mm + 2.1429 mm = 1,790.9972 mm The distance c does not change with the deformation of rod (1), but the distance b will increase as the length of rod increases. The new dimension b' after the deformation of rod (1) is

b = (1,790.9972 mm)2 − (800 mm) 2 = 1,602.3954 mm This means that joint B moves to the left: uB = 1,602.3954 mm − 1,600 mm = 2.3954 mm  The deflection of joint A is the sum of this deflection and the elongation of rod (3): uA = 2.3954 mm + 1.8750 mm = 4.27 mm 

Ans.

Note: The deflection of joint B in the horizontal direction can be approximated from 1 2.1429 mm uB = = = 2.3958 mm cos 26.5651 cos 26.5651

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P5.21 A tie rod (1) and a pipe strut (2) are used to support a load of 25 kips as shown in Figure P5.21. Pipe strut (2) has an outside diameter of 6.625 in. and a wall thickness of 0.280 in. Both the tie rod and the pipe strut are made of structural steel with a modulus of elasticity of E = 29,000 ksi and a yield strength of Y = 36 ksi. For the tie rod, the minimum factor of safety with respect to yield is 1.5 and the maximum allowable axial elongation is 0.30 in. Assume that a = 21 ft, b = 9 ft, and c = 27 ft. (a) Determine the minimum diameter required to satisfy both constraints for tie rod (1). (b) Draw a deformation diagram showing the final position of joint B. FIGURE P5.21

Solution The angles of inclination for members (1) and (2) are: 21 ft tan 1 = 1 = 37.875 27 ft 9 ft tan 2 = 2 = 18.435 27 ft The member lengths are:

L1 = (21 ft)2 + (27 ft) 2 = 34.205 ft = 410.463 in. L2 = (9 ft)2 + (27 ft) 2 = 28.460 ft = 341.526 in. (a) Minimum diameter for tie rod (1): Consider an FBD of joint B and write two equilibrium equations: Fx = − F1 cos37.875 − F2 cos18.435 = 0 Fy = F1 sin 37.875 − F2 sin18.435 − 25 kips = 0 Solve these two equations simultaneously to obtain F1 = 28.5044 kips and F2 = −23.7171 kips.

The allowable normal stress for tie rod (1) is  36 ksi  allow = Y = = 24 ksi FS 1.5 The minimum cross-sectional area required to satisfy the normal stress requirement is F 28.5044 kips A 1 = = 1.1877 in.2  allow 24 ksi The maximum allowable axial deformation for the tie rod is 0.30 in. The minimum cross-sectional area required to satisfy the deformation requirement is F L (28.5044 kips)(410.463 in.) A 1 1 = = 1.3448 in.2 1E1 (0.30 in.)(29,000 ksi) Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

To satisfy both the stress and deformation requirements, the tie rod must have a minimum crosssectional area of A = 1.3448 in.2. The corresponding rod diameter is  2 d rod  1.3448 in.2 d rod  1.309 in. 4

Ans.

(b) The pipe has a cross-sectional area of A2 = 5.5814 in.2. The pipe deformation is FL (−23.7171 kips)(341.526 in.) 2 = 2 2 = = −0.0500 in. A2 E2 (5.5814 in.2 )(29,000 ksi) Rod (1) elongates 0.30 in. and pipe (2) contracts 0.0500 in. Therefore, the deformation diagram showing the final position of joint B is shown.

Mechanics of Materials: An Integrated Learning System, 4th Ed. P5.22 Two axial members are used to support a load of P = 72 kips as shown in Figure P5.22. Member (1) is 12 ft long, it has a cross-sectional area of A1 = 1.75 in.2, and it is made of structural steel [E = 29,000 ksi]. Member (2) is 16 ft long, it has a crosssectional area of A2 = 4.50 in.2, and it is made of an aluminum alloy [E = 10,000 ksi]. (a) Compute the normal stress in each axial member. (b) Compute the deformation of each axial member. (c) Draw a deformation diagram showing the final position of joint B. (d) Compute the horizontal and vertical displacements of joint B.

Timothy A. Philpot

FIGURE P5.22

Solution (a) Consider an FBD of joint B and write two equilibrium equations: Fx = − F1 + F2 cos55 = 0

Fy = F2 sin 55 − 72 kips = 0 Solve these two equations simultaneously to obtain F1 = 50.4149 kips and F2 = 87.8958 kips. The normal stress in member (1) is: F 50.4149 kips 1 = 1 = = 28.8 ksi A1 1.75 in.2 and the normal stress in member (2) is: F 87.8958 kips 2 = 2 = = 19.53 ksi A2 4.50 in.2 (b) The member deformations are F L (50.4149 kips)(12 ft)(12 in./ft) 1 = 1 1 = = 0.1430 in. A1E1 (1.75 in.2 )(29,000 ksi) FL (87.8958 kips)(16 ft)(12 in./ft) 2 = 2 2 = = 0.375 in. A2 E2 (4.50 in.2 )(10,000 ksi)

Ans.

Ans. Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

(c) Member (1) elongates 0.1430 in. and member (2) elongates 0.375 in. Therefore, the deformation diagram showing the final position of joint B is shown.

(d) The horizontal displacement of B is x = 0.1430 in. The vertical displacement is found from 0.375 in. + 0.1430cos55 sin 55 = y 0.375 in. + 0.1430cos55 0.375 in. + 0.0820 in.  y = = = 0.558 in.  sin 55 sin 55

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P5.23 A load P is supported by a structure consisting of rigid bar ABC, two identical solid aluminum alloy [E = 73 GPa] rods (1), and a solid bronze [E = 105 GPa] rod (2), as shown in Figure P5.23. The aluminum rods (1) each have a diameter of 12 mm and a length of L1 = 2.3 m. They are symmetrically positioned relative to middle rod (2) and the applied load P. Bronze rod (2) has a diameter of 18 mm and a length of L2 = 1.6 m. If all bars are unstressed before the load P is applied, determine the normal stresses in the aluminum and bronze rods after a load of P = 50 kN is applied to the rigid bar. FIGURE P5.23

Solution Equilibrium Consider an FBD of the rigid bar ABC. Sum forces in the horizontal direction to obtain: Fx = 2 F1 + F2 − P = 0

(a)

Geometry-of-deformation relationship For this configuration, the deformations of rods (1) and rod (2) will be equal; therefore, 1 =  2 (b) Force-Deformation Relationships The relationship between the internal force and the deformation of an axial member can be stated for members (1) and (2) as: FL FL (c) 1 = 1 1 2 = 2 2 A1E1 A2 E2 Compatibility Equation Substitute the force-deformation relationships (c) into the geometry-of-deformation relationship (b) to derive the compatibility equation: F1L1 F2 L2 = (d) A1E1 A2 E2 Solve the Equations Solve Eq. (d) for F2: L AE L A E F2 = F1 1 2 2 = F1 1 2 2 A1E1 L2 L2 A1 E1 Substitute this expression into equilibrium equation (a) and solve for F1:  L A E L A E  2 F1 + F2 = 2 F1 + F1 1 2 2 = F1  2 + 1 2 2  = P L2 A1 E1 L2 A1 E1  

(e)

(f)

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

For this structure, the areas and elastic moduli are given below: A1 = 113.097 mm2 A2 = 254.469 mm2

E1 = 73 GPa

E2 = 105 GPa

Substitute these values into Eq. (f). Use P = 50,000 N and calculate F1. Then, backsubstitute into Eq. (e) to calculate F2. F1 = 7,516.329 N

F2 = 34,967.341 N Normal stresses: The normal stresses in each axial member can now be calculated: F 7,516.329 N 1 = 1 = = 66.459 MPa = 66.5 MPa (T) A1 113.097 mm2

2 =

F2 34,967.341 N = = 137.413 MPa = 137.4 MPa (T) A2 254.469 mm2

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P5.24 A composite bar is fabricated by brazing aluminum alloy [E = 10,000 ksi] bars (1) to a center brass [E = 17,000 ksi] bar (2) as shown in Figure P5.24. Assume d = 0.75 in., a = 0.1875 in., L = 20 in. If the total axial force carried by the two aluminum bars must equal the axial force carried by the brass bar, calculate the thickness b required for brass bar (2). FIGURE P5.24

Solution Geometry-of-deformation relationship For this configuration, the deformations of all bars will be equal; therefore, 1 =  2

(a)

Force-Deformation Relationships The relationship between the internal force and the deformation of an axial member can be stated for members (1) and (2): FL FL (b) 1 = 1 1 2 = 2 2 A1E1 A2 E2 Compatibility Equation Substitute the force-deformation relationships (b) into the geometry-of-deformation relationship (a) to derive the compatibility equation: F1L1 F2 L2 = (c) A1E1 A2 E2 Constraint For this problem, the total axial force carried by the two aluminum bars must equal the axial force carried by the brass bar, or 2F1 = F2

(d)

Solve Substitute Eq. (d) into Eq. (c): F1L1 2F1L2 = A1E1 A2 E2 The lengths L1 and L2 are the same for the aluminum and brass bars; thus, E A2 = 2 A1 1 E2 which can be further simplified to E bd = 2ad 1 E2  b = 2a

E1  10, 000 ksi  = 2 ( 0.1875 in.)   = 0.221 in. E2  17, 000 ksi 

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P5.25 An aluminum alloy [E = 10,000 ksi; Y = 40 ksi] pipe (1) is connected to a bronze [E = 16,000 ksi; Y = 45 ksi] pipe at flange B, as shown in Figure P5.25. The pipes are attached to rigid supports at A and C. Pipe (1) has an outside diameter of 2.375 in., a wall thickness of 0.203 in., and a length of L1 = 6 ft. Pipe (2) has an outside diameter of 4.50 in., a wall thickness of 0.226 in., and a length of L2 = 10 ft. If a minimum factor of safety of 1.67 is required for each pipe, determine: (a) the maximum load P that may be applied at flange B. (b) the deflection of flange B at the load determined in part (a).

FIGURE P5.25

Solution Pipe section properties: The cross-sectional areas are:

A1 =



2.375 in.) − (1.969 in.)  = 1.3852 in.2 

A2 =



4.500 ) − ( 4.048 in.)  = 3.0345 in.2 

( 4

Allowable stresses: For a specified minimum factor of safety of 1.67, the allowable stresses for the aluminum and bronze pipes are:  40 ksi  1,allow = 1,Y = = 23.952 ksi FS 1.67  45 ksi  2,allow = 2,Y = = 26.946 ksi FS 1.67 Equilibrium: Consider an FBD of flange B. Sum forces in the horizontal direction to obtain: Fx = F2 − F1 − P = 0 (a) Geometry of Deformations: 1 +  2 = 0 Force-Deformation Relationships: FL FL 1 = 1 1 2 = 2 2 A1E1 A2 E2

(b)

(c)

Compatibility Equation: Substitute Eqs. (c) into Eq. (b) to derive the compatibility equation: F1 L1 F2 L2 (d) + =0 A1 E1 A2 E2 For this problem, it is convenient to express the compatibility equation in terms of the normal stresses 1 and 2:  1 L1  2 L2 (e) + =0 E1 E2 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Solve the Equations: The normal stress in aluminum pipe (1) is limited to 23.952 ksi. We will assume that the aluminum pipe controls, and then, we will calculate the corresponding stress in bronze pipe (2): L E 6 ft 16,000 ksi  2 = −1 1 2 = − ( 23.952 ksi ) = −22.994 ksi L2 E1 10 ft 10,000 ksi Here, we are interested in stress magnitudes, and so the magnitude of the normal stress in the bronze pipe is 22.994 ksi. This stress magnitude is less than the 26.946 ksi allowable stress for the bronze; therefore, this calculation shows that our assumption is correct—the aluminum pipe controls. (a) Maximum load P that may be applied at flange B: Now that the stresses are known, the allowable forces F1 and F2 can be computed: F1 =  1 A1 = ( 23.952 ksi ) (1.3852 in.2 ) = 33.178 kips F2 =  2 A2 = ( −22.994 ksi ) ( 3.0345 in.2 ) = −69.776 kips

Substitute these values into Eq. (a) to obtain the allowable load P: Pmax = F2 − F1 = −69.776 kips − 33.178 kips = −102.954 kips But wait, this doesn’t seem quite right because P came out as a negative value! Our result implies that the load P acts toward joint C. At this point in the calculations, we realize that we’ve got our forces turned around. By inspection, it seems likely that compression should exist in pipe (1) and tension should occur in pipe (2). Let’s correct this mistake by switching the signs on the allowable forces to find: Ans. Pmax = F2 − F1 = 69.776 kips − ( −33.178 kips ) = 102.954 kips = 103.0 kips

(b) Deflection of flange B: The deformation of pipe (1) can be expressed as the difference between the deflection of flange B and the deflection of flange A: 1 = u B − u A Since flange A is a fixed support, we know that uA = 0. Therefore, the deflection of flange B is simply equal to the deformation of pipe (1): F L ( −33.178 kips )( 6 ft )(12 in./ft ) u B = 1 = 1 1 = = −0.1725 in. = 0.1725 in.  Ans. A1 E1 1.3852 in.2 (10, 000 ksi )

(

)

Mechanics of Materials: An Integrated Learning System, 4th Ed. P5.26 A rectangular polypropylene [E = 6,200 MPa] bar (1) is connected to a rectangular nylon [E = 1,400 MPa] bar (2) at flange B. The assembly (shown in Figure P5.26) is connected to rigid supports at A and C. Bar (1) has a crosssectional area of A1 = 1,100 mm2 and a length of L1 = 1,450 mm. Bar (2) has a cross-sectional area of A2 = 2,800 mm2 and a length of L2 = 550 mm. After two loads of P = 4 kN are applied to flange B, determine: (a) the forces in bars (1) and (2). (b) the deflection of flange B.

Timothy A. Philpot

FIGURE P5.26

Solution Equilibrium: Consider an FBD of flange B. Sum forces in the horizontal direction to obtain: Fx = − F1 + F2 + 2 P = 0 (a) Geometry of Deformations: 1 +  2 = 0

(b)

Force-Deformation Relationships: FL FL 1 = 1 1 2 = 2 2 A1E1 A2 E2

(c)

Compatibility Equation: Substitute Eqs. (c) into Eq. (b) to derive the compatibility equation: F1 L1 F2 L2 + =0 A1 E1 A2 E2

(d)

Solve the Equations: Solve Eq. (d) for F2: L AE L A E F2 = − F1 1 2 2 = − F1 1 2 2 A1 E1 L2 L2 A1 E1 and substitute this expression into Eq. (a) to determine F1:   L A E  L A E  − F1 + F2 = − F1 +  − F1 1 2 2  = − F1 1 + 1 2 2  = −2 P L2 A1 E1    L2 A1 E1 

(e)

(f)

(a) Forces in bars (1) and (2): F1 can be computed as: 2P F1 = L A E 1+ 1 2 2 L2 A1 E1 =

2 ( 4, 000 N )

 1,450 mm   2,800 mm   1, 400 MPa  1+    2   550 mm   1,100 mm   6, 200 MPa 

= 3,180.5 N = 3,180 N (T)

8, 000 N 2.515329

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

and from Eq. (a), F2 has a value of F2 = F1 − 2P = 3,180.5 N − 2 ( 4, 000 N ) = −4,819.5 N = 4,820 N (C)

Timothy A. Philpot

Ans.

(b) Deflection of flange B: The deformation of bar (1) can be expressed as the difference between the deflection of flange B and the deflection of flange A: 1 = u B − u A Since flange A is a fixed support, we know that uA = 0. Therefore, the deflection of flange B is simply equal to the deformation of bar (1): ( 3,180.5 N )(1, 450 mm ) = 0.6762 mm = 0.676 mm → FL u B = 1 = 1 1 = Ans. A1 E1 1,100 mm 2 6, 200 N/mm 2

(

)(

)

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P5.27 In Figure P5.27, the two vertical steel [E = 200 GPa] rods that support rigid bar ABCD are initially free of stress. Rod (1) has an area of A1 = 450 mm2 and a length of L1 = 2.2 m. Rod (2) has an area of A2 = 325 mm2 and a length of L2 = 1.6 m. Assume dimensions of a = 3.0 m, b = 1.0 m, and c = 1.25 m. After a load of P = 80 kN is applied to the rigid bar at D, determine: (a) the normal stresses in rods (1) and (2). (b) the deflection of the rigid bar at D.

FIGURE P5.27

Solution Equilibrium Consider an FBD of the rigid bar. Assume tension forces in members (1) and (2). A moment equation about pin B gives the most useful information for this situation:

M B = Fa 1 + F2b − P ( b + c ) = 0

(a)

Geometry-of-deformation relationship Draw a diagram showing the deflection of the rigid bar. The deflections of the rigid bar are related by the principal of similar triangles: v v v − A= C = D (b) a b b+c Note that the slope of the deflected rigid bar, as drawn in the sketch, is negative. The negative sign on vA is necessary to be consistent; i.e., each term gives a negative slope.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

The deformations of each rod can be expressed as the difference between the deflection of its upper joint and the deflection of its lower joint.  1 = v A − vE and  2 = vF − vC We know that the deflection at joints E and F is zero. We will use this fact as we express the deflections at A and C in terms of the member elongations 1 and 2 as: 1 = vA  vA = 1

 2 = −vC

 vC = − 2

Equation (b) can be rewritten in terms of the member deformations as:

1  2 = a b

(c)

Force-Deformation Relationships The relationship between the internal force and the deformation of an axial member can be stated for members (1) and (2): FL FL (d) 1 = 1 1 2 = 2 2 A1E1 A2 E2 Compatibility Equation Substitute the force-deformation relationships (d) into the geometry-of-deformation relationship (c) to derive the compatibility equation: F1L1 a F2 L2 3.0 m F2 L2 3F2 L2 = = = (e) A1E1 b A2 E2 1.0 m A2 E2 A2 E2 Solve the Equations Solve Eq. (e) for F1: A E 3F L L A E 1.6 m 450 mm2 200 GPa F1 = 1 1 2 2 = 3F2 2 1 1 = 3F2 = 3.020979 F2 L1 A2 E2 L1 A2 E2 2.2 m 325 mm2 200 GPa and substitute into Eq. (a) to solve for F2: F1a + F2b = P ( b + c )

( 3.020979 F2 )( 3.0 m ) + F2 (1.0 m ) = (80 kN )(1.0 m + 1.25 m ) (80 kN )( 2.25 m ) F = 2

10.062937 m = 17.8874 kN

The value of F1 is:

F1 = 3.020979F2 = 3.020979 (17.8874 kN) = 54.0375 kN

(a) Normal stresses in rods (1) and (2): F 54,037.5 N 1 = 1 = = 120.1 MPa (T) A1 450 mm2

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

F2 17,887.4 N = = 55.0 MPa (T) A2 325 mm2 (b) Deflection of the rigid bar at D: Calculate the deformation of rod (1): ( 54, 037.5 N )( 2, 200 mm ) = 1.3209 mm FL 1 = 1 1 = A1 E1 ( 450 mm 2 )( 200, 000 N/mm 2 )

2 =

Previously, we found that vA = 1; thus, vA = 1.3209 mm. The relationship between the rigid bar deflection at D and the rigid bar deflection at A was expressed in Eq. (b). Therefore, the rigid bar deflection at D is: v v − A= D a b+c b+c 1.0 m + 1.25 m  vD = − vA = − (1.3209 mm ) = −0.9907 mm = 0.991 mm  a 3.0 m

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P5.28 The pin-connected structure shown in Figure P5.28 consists of a rigid bar ABCD and two bronze alloy [E = 100 GPa] bars, each with a cross-sectional area of 340 mm2. Bar (1) has a length of L1 = 810 mm and bar (2) has a length of L2 = 1,080 mm. Assume dimensions of a = 480 mm, b = 380 mm, and c = 600 mm. If the allowable normal stress in bars (1) and (2) must be limited to 165 MPa, determine: (a) the maximum load P that may be applied to the rigid bar at D. (b) the deflection at D for the load determined in part (a). FIGURE P5.40

Solution Equilibrium Consider an FBD of the rigid bar. Assume tension forces in bars (1) and (2). A moment equation about pin C gives the most useful information for this situation: M C = −F1 ( a + b ) − F2b + Pc = 0 (a)

Geometry-of-deformation relationship Draw a sketch of the deflected rigid bar. The deflections of the rigid bar are related by similar triangles. uA u v = B = D a+b b c

(b)

For simplicity, we will use absolute values for uA, uB, and vD. We will determine the directions of each deflection by inspection at the end of the calculation procedure.

Since there are no gaps or clearances at either pin A or pin B, the deformations of bars (1) and (2) will equal the deflections of the rigid bar at A and B, respectively.

1  = 2 a+b b

(c)

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Force-Deformation Relationships The relationship between the internal force and the deformation of an axial member can be stated for bars (1) and (2) as: FL FL (d) 1 = 1 1 2 = 2 2 A1E1 A2 E2 Compatibility Equation Substitute the force-deformation relationships (d) into the geometry-of-deformation relationship (c) to derive the compatibility equation: F1L1  a + b  F2 L2  480 mm + 380 mm  F2 L2 FL (e) = = = 2.263158 2 2   A1E1  b  A2 E2  380 mm A2 E2  A2 E2 Solve the Equations For this problem, it is convenient to express the compatibility equation in terms of stresses:  1L1  L = 2.263158 2 2 E1 E2  L  E   1, 080 mm  100 GPa  (f)  1 = 2.263158  2  1   2 = 2.263158     2 = 3.017544 2  810 mm  100 GPa   L1  E2  The allowable stress for each bar is 165 MPa. Substitute the allowable stress for bar (2) into Eq. (f) and solve for 1:

1 = 3.017544 (165 MPa ) = 497.895 MPa  165 MPa

This calculation shows that the stress in bar (1) controls. Substitute 1 = 165 MPa into Eq. (f) and solve for  2: 1 2 = OK (165 MPa ) = 54.680 MPa  165 MPa 3.017544 (a) Maximum load P that may be applied to the rigid bar at D. Now that the stress magnitudes are known, calculate the magnitudes of the allowable forces in the two bars F1 = (165 N/mm 2 )( 340 mm 2 ) = 56,100 N F2 = ( 54.6802 N/mm 2 )( 340 mm 2 ) = 18,591.3 N

Note that we know only the force magnitudes at this point in the calculation. From Eq. (a), calculate the load P: F ( a + b ) + F2b P= 1 c ( 56,100 N )( 480 mm + 380 mm ) + (18,591.3 N )( 380 mm ) = 600 mm Ans. = 92,184.5 N = 92.2 kN Accordingly, the maximum load P that can be applied to the rigid bar is:

P = 21.1 kN 

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

(b) Deflection at D: From Eq. (b), we can relate the rigid bar deflection at D to the rigid bar deflection at A or B. uA u v = B = D a+b b c We also know that the rigid bar deflection at A will equal the deformation in bar (1). Thus, vD u  = A = 1 c a+b a+b The deformation of bar (1) can be calculated as: F L  L (165 MPa )(810 mm ) 1 = 1 1 = 1 1 = = 1.3365 mm A1E1 E1 100,000 MPa And finally, the deflection of the rigid bar at D is: c 600 mm vD = 1 = (1.3365 mm ) = 0.9324 mm a+b 480 mm + 380 mm By inspection, we know that D must move downward; thus, vD = 0.932 mm 

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P5.29 In Figure P5.29, a load P is supported by a structure consisting of rigid bar BDF and three identical aluminum [E = 10,000 ksi] rods, each having a diameter of 0.625 in. Use dimensions of a = 72 in., b = 60 in., and L = 90 in. For a load of P = 16 kips, determine: (a) the tension force produced in each rod. (b) the vertical deflection of the rigid bar at B.

FIGURE P5.29

Solution Equilibrium: Consider an FBD of the rigid bar. Sum forces in the vertical direction to obtain: (a) Fy = F1 + F2 + F3 − P = 0 Sum moments about joint B to obtain: M B = F2 (a ) + F3 (2a) − P(b) = 0 (b)

Geometry of Deformations: When the load P is applied to the rigid bar, the bar will deflect downward but it will not remain horizontal. Since rods (1), (2), and (3) are equally spaced, we can readily observe that v + vF vD = B 2 For simplicity, we will use absolute values for vB, vD, and vF. We will determine the directions of each deflection by inspection at the end of the calculation procedure. The rods are attached to the rigid bar with perfect connections (i.e., no gaps or clearances), therefore:  1 = vB  2 = vD  3 = vF The geometry-of-deformation equation can be written in terms of the rod deformations as:  + 2 = 1 3 or 1 +  3 = 2 2 2 Force-Deformation Relationships: FL FL FL 1 = 1 1 2 = 2 2 3 = 3 3 A1E1 A2 E2 A3 E3

(c)

(d)

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Compatibility Equation Substitute the force-deformation relationships (d) into the geometry-of-deformation equation (c) to derive the compatibility equation: F1L1 F3 L3 FL + =2 2 2 A1E1 A3 E3 A2 E2 (a) Tension force in each rod: Since all three rods are identical, the compatibility equation can be reduced to: F1 + F3 = 2 F2 Solve Eqs. (a) and (f) simultaneously to find F2: ( F1 + F3 ) + F2 = P 2 F2 + F2 = P P F2 = 3 Substitute this result into Eq. (b) to find F3: F2 (a) + F3 (2a) = P(b)

(e)

(f)

 P   a + F3 (2a) = P(b) 3 P a  b 1 −   b −  = P  2a 3 2a 6  and therefore, F1 must equal:  P  b 1  2 b 1 5 b  F1 = 2   − P  −  = P  − +  = P −   3  2a 6   3 2a 6   6 2a  Given a = 72 in., b = 60 in., and P = 16 kips, the rod forces are: F1 = 6.67 kips F2 = 5.33 kips F3 = 4.00 kips F3 =

Ans.

(b) Rigid bar deflection at B: The rigid bar deflection at B will equal the deformation of rod (1):

 2 ( 0.625 in.) = 0.3068 in.2 4 ( 6.6667 kips )( 90 in.) FL 1 = 1 1 = = 0.19557 in. A1 E1 ( 0.3068 in.2 )(10,000 kips/in.2 ) A1 =

The displacement of joint B is thus: vB = 0.1956 in. 

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed. P5.30 A uniformly-distributed load w is supported by a structure consisting of rigid bar BDF and three rods, as shown in Figure P5.30. Rods (1) and (2) are 0.75 in. diameter stainless steel rods that have an elastic modulus of E = 28,000 ksi and a yield strength of Y = 36 ksi. Rod (3) is a 1.25 in. diameter bronze rod that has an elastic modulus of E = 16,000 ksi and a yield strength of Y = 48 ksi. Use a = 4 ft and L = 9 ft. If a minimum factor of safety of 1.8 is specified for the normal stress in each rod, calculate the maximum distributed load magnitude w that may be supported.

Timothy A. Philpot

FIGURE P5.30

Solution Equilibrium: Consider an FBD of the rigid bar. Sum forces in the vertical direction to obtain: (a) Fy = F1 + F2 + F3 − w(2a) = 0 Sum moments about joint F to obtain: M F = w(2a)(a) − F2 (2a) − F1 (3a) = 0

(b)

Geometry of Deformations: When the distributed load w is applied to the rigid bar, the bar will deflect downward but it will not remain horizontal. Rods (1), (2), and (3) are not equally spaced; however, we can use the principle of similar triangles to find that vB − vF vD − vF =  2vB + vF = 3vD 3a 2a For simplicity, we will use absolute values for vB, vD, and vF. We will determine the directions of each deflection by inspection at the end of the calculation procedure. The rods are attached to the rigid bar with perfect connections (i.e., no gaps or clearances), therefore:  1 = vB  2 = vD  3 = vF The geometry-of-deformation equation can be written in terms of the rod deformations as: 21 +  3 = 3 2 (c) Force-Deformation Relationships: FL FL FL 1 = 1 1 2 = 2 2 3 = 3 3 A1E1 A2 E2 A3 E3

(d)

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Compatibility Equation Substitute the force-deformation relationships (d) into the geometry-of-deformation equation (c) to derive the compatibility equation: FL FL FL 2 1 1 + 3 3 =3 2 2 A1E1 A3 E3 A2 E2

(e)

Section Properties For stainless steel rods (1) and (2),  2 A1 = A2 = ( 0.75 in.) = 0.44179 in.2 4 36 ksi  allow = = 20 ksi 1.8 F1,allow = F2,allow = ( 20 kips/in.2 )( 0.44179 in.2 ) = 8.8357 kips and for bronze rod (3),  2 A3 = (1.25 in.) = 1.22718 in.2 4 48 ksi  allow = = 26.6667 ksi 1.8 F3,allow = ( 26.6667 kips/in.2 )(1.22718 in.2 ) = 32.7249 kips Solve the Equations: Begin with the two equilibrium equations F1 + F2 + F3 = w(2a )

F1 (3a) + F2 (2a) = w(2a)(a) which can be simplified to 3F1 + 2 F2 = w(2a ) Combine these two equations to obtain: 2F1 + F2 = F3 Substitute this relationship into Eq. (e) to obtain a relationship between F1 and F2: F L (2 F1 + F2 ) L3 FL 2 1 1+ =3 2 2 A1E1 A3 E3 A2 E2  L  3L L  L  2 F1  1 + 3  = F2  2 − 3   A1E1 A3 E3   A2 E2 A3 E3  or, since all rods have the same length:  1  3 1  1  2F1  + −  = F2    A1E1 A3 E3   A2 E2 A3 E3  Solve for F1: 1  AE   3  − 3− 1 1     F AE A3 E3 F A3 E3 = 2  F1 = 2  2 2 2  1 + 1  2  1 + A1E1   A1E1 A3 E3   A3 E3 

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Enter values for area, elastic modulus, and the allowable force for rod (2). Also, note that A1E1 = A2E2. Solve for F1:   0.44179 in.2   28, 000 ksi   3 −   2  F2   1.22718 in.   16, 000 ksi   F2  3 − 0.63000  F1 = = = 0.72699 F2 2   0.44179 in.2   28, 000 ksi   2  1 + 0.63000  1+   2    1.22718 in.   16, 000 ksi   = 0.72699 ( 8.8357 kips ) = 6.4235 kips  F1,allow

Now, the forces F1 and F2 for the stainless steel rods are tentatively known. Next, check the force in rod (3), given the forces that have just been calculated.

F3 = 2F1 + F2 = 2 ( 6.4235 kips ) + (8.8357 kips ) = 21.6828 kips  F3,allow

The forces in all three rods have now been determined, with the strength of rod (2) controlling the capacity of the structure. Finally, the magnitude of the allowable distributed load w can be calculated from Eq. (a): F + F2 + F3 w= 1 2a 6.4235 kips + 8.8357 kips + 21.6828 kips = 2 ( 4 ft ) = 4.6178 kips/ft = 4.62 kips/ft

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P5.31 Links (1) and (2) support rigid bar ABCD shown in Figure P5.31. Link (1) is bronze [E = 15,200 ksi] with a cross-sectional area of A1 = 0.50 in.2 and a length of L1 = 24 in. Link (2) is cold-rolled steel [E = 30,000 ksi] with a cross-sectional area of A2 = 0.375 in.2 and a length of L2 = 32 in. Use dimensions of a = 14 in., b = 16 in., and c = 18 in. For an applied load of P = 9 kips, determine: (a) the normal stresses in links (1) and (2). (b) the deflection of end D of the rigid bar.

FIGURE P5.31

Solution Equilibrium Consider an FBD of the rigid bar. Assume tension forces in links (1) and (2). A moment equation about pin A gives the most helpful information for this situation: M A = F1 (a + b) − F2 a − P (a + b + c ) = 0 (a)

Geometry-of-deformations relationship Sketch the expected deflected position of the rigid bar. The deflections of the rigid bar are related by similar triangles: u uB uD = C = (b) a a+b a+b+c There are no gaps, clearances, or other misfits at pins B and C. The deformation of link (1) can be expressed as the difference between the deflection of pin C and pin E: 1 = uC − u E Since pin E is fixed against translation, we know that uE = 0. Therefore, the deflection of pin C is simply equal to the deformation of link (1): uC =  1 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Similarly, the deformation of link (2) can be expressed as the difference between the deflection of pin F and pin B:  2 = uF − uB Since pin F is fixed against translation, we know that uF = 0. Therefore, the deflection of pin B is: u B = − 2 Eq. (b) can now be rewritten in terms of the link deformations as: − 2  = 1 (c) a a+b Force-Deformation Relationships FL FL 1 = 1 1 2 = 2 2 A1E1 A2 E2

(d)

Compatibility Equation 1 F2 L2 1 F1L1 − = a A2 E2 a + b A1E1

(e)

Solve the Equations Solve Eq. (e) for F2: a F1 L1 A2 E2 a L1 A2 E2 F2 = − =− F1 a + b A1 E1 L2 a + b L2 A1 E1 14 in. 24 in. 0.375 in.2 30, 000 ksi F1 14 in. + 16 in. 32 in. 0.50 in.2 15, 200 ksi = −0.51809 F1

=−

and substitute this expression into Eq. (a) to solve for F1: F1 (a + b) − F2 a = P(a + b + c)

F1 (14 in. + 16 in.) − ( −0.51809 F1 )(14 in.) = ( 9 kips )(14 in. + 16 in. + 18 in.) F1 ( 37.25329 in.) = ( 9 kips )( 48 in.) F1 =

( 9 kips )( 48 in.)

37.25329 in. = 11.5963 kips

The value of F2 is:

F2 = −0.51809F1 = −0.51809 (11.5963 kips ) = −6.0079 kips

(a) Normal stresses in links (1) and (2): F 11.5963 kips 1 = 1 = = 23.2 ksi (T) A1 0.5 in.2 F −6.0079 kips 2 = 2 = = −16.02 ksi = 16.02 ksi (C) A2 0.375 in.2

Ans. Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

(b) Deflection of the rigid bar at D: Calculate the deformation of link (1): (11.5963 kips )( 24 in.) = 0.03662 in. FL 1 = 1 1 = A1 E1 0.5 in.2 15, 200 kips/in.2

(

)(

)

Previously, we found that uC = 1; thus, uC = 0.03662 in. The relationship between the rigid bar deflection at D and the rigid bar deflection at C was expressed in Eq. (b) as: uC uD = a+b a+b+c Solve this relationship for the rigid bar deflection at D: a+b+c uD = uC a+b 14 in. + 16 in. + 18 in. = ( 0.036620 in.) = 0.058592 in. = 0.0586 in. → 14 in. + 16 in.

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P5.32 In Figure P5.32, polycarbonate bars (1) and (2) are attached at their lower ends to a slider block at B that travels vertically inside of a fixed smooth slot. Load P is applied to the slider block. Each bar has a cross-sectional area of 85 mm2, a length of 1.25 m, and an elastic modulus of E = 8 GPa. Use dimensions of a = 1.15 m and b = 0.90 m. For an applied load of P = 1,700 N, determine: (a) the forces in bars (1) and (2). (b) the deflection of the slider block at B. FIGURE P5.32

Solution Initial Geometry: Each of the bars has a length of L = 1.25 m. Let the angle between bar (1) and the –x axis be denoted , and let the angle between bar (2) and the +x axis be denoted . Calculate  and . a 1.15 m cos  = = = 0.92  = 23.0739 L 1.25 m b 0.90 m cos  = = = 0.72   = 43.9455 L 1.25 m Equilibrium Consider an FBD of joint B. Assume tension forces in bars (1) and (2). The equilibrium equations for joint B are: Fx = − F1 cos  + F2 cos  + Bx = 0

Fy = F1 sin  + F2 sin  − P = 0

(a)

Geometry-of-deformation relationship Draw a deformation diagram. We imagine this process: (a) Temporarily removing the pin at B. (b) Permitting pin B to displace downward to position B′. (c) Slightly rotating bar (1) and stretching it enough so that it can be reconnected to the pin, which is now at its deflected position B′. The amount of stretch required to reconnect the bar is the deformation 1. (d) Slightly rotating bar (2) and stretching it enough so that it can be reconnected to the pin at B′. The amount of stretch required to reconnect the bar is the deformation 2. At this step of the calculation, we make an approximation. When we allow bar (1) to rotate slightly downward, we assume that the angle  after deformation is essentially the same as our initial value of  because the deflections that occur are relatively small. The geometry involved in the deformation of bar (1) is illustrated in the sketch above. A similar argument can be made for bar (2). From the geometry of the deformation diagram above, we can assert that:

sin  =

1

 vB =

1 sin 

and

sin  =

2 vB

 vB =

2 sin 

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

and thus, the geometry-of-deformation equation can be expressed as: 1  = 2 sin  sin 

(b)

Force-Deformation Relationships The relationship between the internal force and the deformation of an axial member can be stated for bars (1) and (2) as: FL FL (c) 1 = 1 1 2 = 2 2 A1E1 A2 E2 Compatibility Equation Substitute the force-deformation relationships (c) into the geometry-of-deformation relationship (b) to derive the compatibility equation: F1L1 sin  F2 L2 = (d) A1E1 sin  A2 E2 (a) Forces in bars (1) and (2): Solve Eq. (d) for F1: sin  L2 A1 E1 F1 = F2 sin  L1 A2 E2 From the problem statement, A1 = A2, L1 = L2, and E1 = E2. Use these facts and solve for F1: sin  L2 A1 E1 sin 23.0739 F1 = F2 = F2 = 0.56474F2 sin  L1 A2 E2 sin 43.9455 Substitute this result into Eq. (a) and solve for F2: F1 sin  + F2 sin  = P

( 0.56474 F2 ) sin 23.0739 + F2 sin 43.9455 = 1, 700 N 0.91531F2 = 1, 700 N F2 = 1,857.3 N = 1, 857 N

And then, solve for the force in bar (1): F1 = 0.56474F2 = 0.56474 (1,857.3 N ) = 1,048.9 N = 1,049 N

Ans. Ans.

(b) Deflection of the slider block at B: The downward displacement of pin B can be found from the deformation of either bar (1) or bar (2), as we found previously.   vB = 1 or vB = 2 sin  sin  Calculate the deformation of bar (1): (1,048.9 N )(1, 250 mm ) = 1.9281 mm FL 1 = 1 1 = A1E1 85 mm 2 8,000 N/mm 2

(

)(

)

Consequently, the displacement of the slider block at B is:  1.9281 mm vB = 1 = = 4.92 mm  sin  sin 23.0739

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P5.33 The pin-connected structure shown in Figure P5.33 consists of two cold-rolled steel [E = 207 GPa] bars (1) and a bronze [E = 105 GPa] bar (2) that are connected at pin D. All three bars have cross-sectional areas of 650 mm2. A load of P = 275 kN is applied to the structure at pin D. Using a = 2.4 m and b = 3.2 m, calculate: (a) the normal stresses in bars (1) and (2). (b) the downward displacement of pin D.

FIGURE P5.33

Solution Equilibrium Consider an FBD of joint D. Assume tension forces in bars (1) and (2). (Also note that the forces in the two steel bars must be the same because of symmetry.) a 2.4 m tan  = = = 0.75  = 36.870 b 3.2 m

Fx = − F1 cos  − F2 − F1 cos  + P = 0  2 F1 cos  + F2 = P

(a)

Fy = F1 sin  − F1 sin  = 0

Geometry-of-deformation relationship Draw a deformation diagram. We imagine temporarily removing the pin at D and allowing the members to elongate independently. Then, after deformation, bars (1) are rotated so that they intersect the lower end of bar (2), and the pin at D is reinserted. If we assume that the deformations will be small, then the angle after deformation is approximately the same as the angle before deformation. From the deformation diagram geometry, we can assert that:

1 = 2 cos 

(b)

Force-Deformation Relationships The relationship between the internal force and the deformation of an axial member can be stated for bars (1) and (2) as:

Mechanics of Materials: An Integrated Learning System, 4th Ed.

1 =

F1L1 A1E1

2 =

F2 L2 A2 E2

Timothy A. Philpot (c)

Compatibility Equation Substitute the force-deformation relationships (c) into the geometry-of-deformation relationship (b) to derive the compatibility equation: F1L1 FL = 2 2 (d) A1E1 cos  A2 E2 Solve the Equations From the problem statement, A1 = A2. The bar lengths are:

L1 =

( 2.4 m ) + ( 3.2 m ) = 4.0 m 2

L2 = 3.2 m

Use these facts and solve Eq. (d) for F2: 1  L1  E2  1  4.0 m  105 GPa  F2 =    F1 =    F1 = 0.79257 F1 cos   L2  E1  cos 36.870  3.2 m  207 GPa  Substitute this result in Eq. (a) and solve for F1: 2 F1 cos 36.870 + ( 0.79257 F1 ) = 275 kN  F1 =

275 kN = 114.9390 kN 2.39257

The force in bar (2) is therefore:

F2 = 0.79257 (114.9390 kN) = 91.0975 kN

(a) Normal stresses in bars (1) and (2). The normal stresses in bars (1) and (2) can now be calculated: F 114,939.0 N 1 = 1 = = 176.8 MPa Ans. A1 650 mm2 F 91,097.5 2 = 2 = = 140.2 MPa Ans. A2 650 mm2 (b) Displacement of pin D. The displacement of pin D is simply equal to the deformation of bar (2). ( 91, 097.5 N )( 3, 200 mm ) = 4.2712 mm FL 2 = 2 2 = A2 E2 ( 650 mm 2 )(105, 000 N/mm 2 ) Ans.  uD = 4.27 mm →

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P5.34 A pin-connected structure is loaded and supported as shown in Figure P5.34. Member ABCD is a rigid bar that is horizontal before a load of P = 6 kips is applied at end D. Bars (1) and (2) are made of steel [E = 30,000 ksi], and each bar has a cross-sectional area of 0.45 in.2. Dimensions of the structure are a = 220 in., b = 140 in., c = 100 in., and d = 220 in. Determine: (a) the normal stress in each bar. (b) the downward deflection of the rigid bar at end D. FIGURE P5.34

Solution Initial Geometry: Let the angle between bar (1) and the rigid bar be denoted , and let the angle between bar (2) and the rigid bar axis be denoted . Calculate  and . d 220 in. tan  = = = 1.0  = 45 a 220 in. d 220 in. tan  = = = 0.6111 a + b 220 in. + 140 in.   = 31.4296 The length of bar (1) is:

L1 = a 2 + d 2 =

( 220 in.) + ( 220 in.) = 311.1270 in. 2

The length of bar (2) is:

L2 =

( a + b ) + d 2 = ( 220 in. + 140 in.) + ( 220 in.) = 421.9005 in. 2

Equilibrium Consider an FBD of the rigid bar. Assume tension forces in bars (1) and (2). The equilibrium equation for the sum of moments about joint A is: M A = ( F1 sin  ) a + ( F2 sin  )( a + b ) − P ( a + b + c ) = 0 (a)

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Geometry-of-deformation relationship Draw a sketch of the deflected rigid bar. The deflections of the rigid bar are related by similar triangles. v vB vD = C = (b) a a+b a+b+c For simplicity, we will use absolute values for vB, vC, and vD. We will determine the directions of each deflection by inspection at the end of the calculation procedure.

Next, we must relate the deformations in bars (1) and (2) to the deflections of the rigid bar at B and C, respectively. We begin by sketching a deformation diagram at joint B. We imagine this process: (a) Temporarily removing the pin at B. (b) Permitting the rigid bar to deflect, causing pin B to displace downward to position B′. (c) Slightly rotating bar (1) and stretching it enough so that it can be reconnected to the pin, which is now at its deflected position B′. The amount of stretch required to reconnect the bar is the deformation 1. The deformation of bar (1) is related to the rigid bar deflection at B by:

vB =

1 sin 

(c)

At this step of the calculation, we make an approximation. When we allow bar (1) to rotate slightly downward, we assume that the angle  after deformation is essentially the same as our initial value of  because the deflections that occur are relatively small. The geometry involved in the deformation of bar (1) is illustrated in the sketch above. Next, sketch a deformation diagram at joint C. We imagine this process: (a) Temporarily removing the pin at C. (b) Permitting the rigid bar to deflect, causing pin C to displace downward to position C′. (c) Slightly rotating bar (2) and stretching it enough so that it can be reconnected to the pin, which is now at its deflected position C′. The amount of stretch required to reconnect the bar is the deformation 2. The deformation of bar (2) is related to the rigid bar deflection at C by:  (d) vC = 2 sin  As before, we assume that the angle  after deflection is essentially unchanged from its initial value. Substitute Eqs. (c) and (d) into Eq. (b) to obtain the geometry-of-deformation relationship:

1

a sin 

2

( a + b ) sin 

(e)

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Force-Deformation Relationships The relationship between the internal force and the deformation of an axial member can be stated for bars (1) and (2) as: FL FL (f) 1 = 1 1 2 = 2 2 A1E1 A2 E2 Compatibility Equation Substitute the force-deformation relationships (f) into the geometry-of-deformation relationship (d) to derive the compatibility equation: F1L1 a sin  F2 L2 = (g) A1E1 a + b sin  A2 E2 Forces in bars (1) and (2): Solve Eq. (g) for F1: a sin  L2 A1 E1 F1 = F2 a + b sin  L1 A2 E2 From the problem statement, A1 = A2 and E1 = E2. Use these facts and solve for F1: a sin  L2 A1 E1 F1 = F2 a + b sin  L1 A2 E2

220 in. sin 45 421.9005 in. F2 = 1.12374 F2 220 in. + 140 in. sin 31.4296 311.1270 in. Substitute this result into Eq. (a) and solve for F2: ( F1 sin  ) a + ( F2 sin  )( a + b ) = P ( a + b + c ) =

(1.12374 F2 )( 220 in.) sin 45 + F2 ( 220 in. + 140 in.) sin 31.4296 = ( 6 kips )( 220 in. + 140 in. + 100 in.) ( 362.53453 in.) F2 = ( 6 kips )( 460 in.) F2 = 7.6131 kips

And then, solve for the force in bar (1):

F1 = 1.12374F2 = 1.12374 ( 7.6131 kips ) = 8.5551 kips

(a) Normal stresses in bars (1) and (2): F 8.5551 kips 1 = 1 = = 19.01 ksi (T) A1 0.45 in.2 F 7.6131 kips 2 = 2 = = 16.92 ksi (T) A2 0.45 in.2

Ans. Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

(b) Deflection of the rigid bar at end D: The downward displacement of pin B can be found from the deformation of bar (1), as we found previously.

vB =

1 sin 

Calculate the deformation of bar (1): (8.5551 kips )( 311.1270 in.) = 0.197164 in. FL 1 = 1 1 = A1 E1 0.45 in.2 30, 000 kips/in.2

(

)(

)

Consequently, the deflection of pin B:  0.197164 in. vB = 1 = = 0.278833 in. sin  sin 45 and the deflection of end D can be found from Eq. (b): a+b+c vD = vB a 220 in. + 140 in. + 100 in. = ( 0.278833 in.) 220 in. = 0.583 in. 

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P5.35 A load of P = 100 kN is supported by an assembly consisting of rigid bar ABC, two identical solid bronze [E = 105 GPa] bars, and a solid aluminum alloy [E = 70 GPa] bar, as shown in Figure P5.35. The bronze bars (1) each have a crosssectional area of 125 mm2 and a length of L1 = 2,400 mm. Aluminum bar (2) has a cross-sectional area of 375 mm2 and a length of L2 = 1,500 mm. Assume a = 500 mm. All bars are unstressed before the load P is applied; however, there is a gap of  = 2 mm in the connection at B. Determine: (a) the axial forces in the bronze and aluminum bars. (b) the downward deflection of rigid bar ABC. FIGURE P5.35

Solution Equilibrium: By virtue of symmetry, the forces in the two bronze bars (1) are identical. Consider an FBD of the rigid bar. Sum forces in the vertical direction to obtain: (a) Fy = 2F1 + F2 − P = 0 Geometry of Deformations: For this configuration, the deflections of joints A, B, and C are equal: v A = vB = vC (b) The pin connections at A and C are ideal; therefore, the deflection of joints A and C will cause identical deformations of bars (1): v A = 1 (c) The pin at B has a 2 mm clearance; thus, the deformation of bar (2) is related to rigid bar deflection vB by: vB =  2 + 2 mm (d) Substitute Eqs. (c) and (d) into Eq. (b) to obtain the geometry of deformation equation: 1 =  2 + 2 mm (e) Force-Deformation Relationships: The relationship between the internal force and the deformation of an axial member can be stated for members (1) and (2): FL FL 1 = 1 1 2 = 2 2 (f) A1E1 A2 E2 Compatibility Equation: Substitute the force-deformation relationships (f) into the geometry-ofdeformation relationship (e) to derive the compatibility equation: F1L1 F2 L2 = + 2 mm A1E1 A2 E2

(g)

Solve the Equations: Solve Eq. (g) for F1: FL  AE L A E ( 2 mm ) A1E1 F1 =  2 2 + 2 mm  1 1 = F2 2 1 1 + L1 A2 E2 L1  A2 E2  L1 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Substitute this expression into Eq. (a)  L A E ( 2 mm ) A1E1  2 F1 + F2 = 2  F2 2 1 1 +  + F2 = P L1  L1 A2 E2  and derive an expression for F2:  L A E  ( 2 mm ) A1E1 F2  2 2 1 1 + 1 = P − 2 L1  L1 A2 E2  P−2

( 2 mm ) A1E1

L1 L A E 2 2 1 1 +1 L1 A2 E2

 F2 =

(h)

(a) Axial forces in the bronze and aluminum bars: For this structure, P = 100 kN = 100,000 N, and the lengths, areas, and elastic moduli are given below: L1 = 2, 400 mm L2 = 1,500 mm

A1 = 125 mm 2

A2 = 375 mm 2

E1 = 105, 000 MPa E2 = 70, 000 MPa Substitute these values into Eq. (h) and calculate F2 = 48, 076.9 N = 48.1 kN (T)

Ans.

Backsubstitute into Eq. (a) to calculate F1 = 25,961.5 N = 26.0 kN (T)

Ans.

(b) Downward deflection of rigid bar ABC: The deflection of the rigid bar can be determined from the deformation of bars (1): ( 25,961.5 N )( 2, 400 mm ) = 4.75 mm  FL vA = vB = vC = 1 = 1 1 = Ans. A1 E1 125 mm 2 105, 000 N/mm 2

(

)(

)

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P5.36 A bronze pipe (1) is to be connected to an aluminum alloy pipe (2) at flange B, as shown in Figure P5.36. When put in place, however, a gap of  = 0.25 in. exists between the two pipes. Bronze pipe (1) has an elastic modulus of E1 = 16,000 ksi, a cross-sectional area of A1 = 2.23 in.2, and a length of L1 = 5.0 ft. Aluminum alloy pipe (2) has an elastic modulus of E2 = 10,000 ksi, a cross-sectional area of A2 = 1.07 in. 2, and a length of L2 = 9.0 ft. If bolts are inserted in the flanges and tightened so that the gap at B is closed, determine: (a) the normal stresses produced in each of the members. (b) the final position of flange B with respect to support A.

FIGURE P5.36

Solution Equilibrium: Consider an FBD of flange B after the bolts have been tightened and the gap at B has been closed. Sum forces in the horizontal direction to obtain: Fx = F1 − F2 = 0  F1 = F2 (a)

Geometry of Deformations: 1 +  2 = 0.25 in.

(b)

Force-Deformation Relationships: FL FL 1 = 1 1 2 = 2 2 A1E1 A2 E2

(c)

Compatibility Equation: Substitute Eqs. (c) into Eq. (b) to derive the compatibility equation: F1L1 F2 L2 + = 0.25 in. A1E1 A2 E2

(d)

Solve the Equations: Substitute Eq. (a) into Eq. (d) and solve for F1:  L L  F1  1 + 2  = 0.25 in.  A1 E1 A2 E2      0.25 in.  F1 =  60 in. 108 in.    2.23 in.2 16, 000 ksi + 1.07 in.2 10, 000 ksi  ) ( ) )( )( ( = 21.2313 kips

(f)

Mechanics of Materials: An Integrated Learning System, 4th Ed. (a) Normal stresses: F 21.2313 kips 1 = 1 = = 9.5208 ksi = 9.52 ksi (T) A1 2.23 in.2 F 21.2313 kips 2 = 2 = = 19.8423 ksi = 19.84 ksi (T) A2 1.07 in.2 (b) The deflection of flange B is equal to the deformation of bronze pipe (1): ( 21.2313 kips )( 60 in.) = 0.0357 in. → FL 1 = 1 1 = A1 E1 2.23 in.2 16, 000 kips/in.2

(

)(

)

Timothy A. Philpot

Ans. Ans.

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed. P5.37 Rigid bar ABCD shown in Figure P5.37 is supported by a smooth pin at D and by vertical aluminum alloy [E = 10,000 ksi] bars attached at joints A and C. Bar (1) was fabricated to its intended length of L1 = 40 in. Bar (2) was intended to have a length of 60 in.; however, its actual length after fabrication was found to be L2 = 59.88 in. To connect it to the pin at C on the rigid bar, bar (2) will need to be manually stretched by  = 0.12 in. After bar (2) has been stretched and connected to the pin at C, a load of P = 20 kips is applied to the rigid bar at B. Use the following additional properties and dimensions: A1 = A2 = 0.75 in.2, a = 36 in., b = 54 in., and c = 48 in. Determine: (a) the normal stresses in bars (1) and (2). (b) the deflection of the rigid bar at A.

Timothy A. Philpot

FIGURE P5.37

Solution Equilibrium Consider an FBD of the rigid bar. Assume tension forces in bars (1) and (2). A moment equation about pin D gives the best information for this situation: M D = −F1 ( a + b + c ) − F2c + P (b + c ) = 0 (a)

Geometry-of-Deformations Relationship Sketch a diagram showing the expected deflection of the rigid bar. The deflections of the rigid bar are related by similar triangles: v vA = C (b) a+b+c c Since there is no clearance at pin A, the deformation of bar (1) will equal the deflection of the rigid bar at A. v A = 1 (c) Bar (2), however, has been made too short. Before it can be connected to joint C, and it must be stretched by  = 0.12 in. Once connected, bar (2) will be stretched more after load P is applied— stretched by an amount equal to the rigid bar deflection vC. We can express this situation by:  2 =  + vC  vC =  2 −  Note that absolute values will be used for vA and vC for simplicity.

(d)

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Substitute Eqs. (c) and (d) into Eq. (b) to derive the geometry-of-deformation relationship for this structure: 1  − = 2 a+b+c c a+b+c (e) 1 = ( 2 −  ) c Force-Deformation Relationships The relationship between the internal force and the deformation of an axial member can be stated for bars (1) and (2): FL FL 1 = 1 1 2 = 2 2 (f) A1E1 A2 E2 Compatibility Equation Substitute the force-deformation relationships (f) into the geometry of deformation relationship (e) to derive the compatibility equation:  36 in. + 54 in. + 48 in.  F2 L2  F1 L1 a + b + c  F2 L2 = −  = − 0.12 in.    A1 E1 c 48 in.  A2 E2   A2 E2 

FL  F1 L1 FL = 2.875  2 2 − 0.12 in.  = 2.875 2 2 − 0.345 in. A1 E1 A2 E2  A2 E2 

(g)

Solve the Equations: For this structure, P = 20 kips, and the lengths, areas, and elastic moduli are given below: L1 = 40 in. L2 = 59.88 in.

A1 = 0.75 in.2

A2 = 0.75 in.2

E1 = 10, 000 ksi E2 = 10, 000 ksi Substitute these values into Eq. (g) and determine an expression for F1: F L AE AE F1 = 2.875 2 2 1 1 − ( 0.345 in.) 1 1 A2 E2 L1 L1 = 2.875

L2 A1 E1 AE F2 − ( 0.345 in.) 1 1 L1 A2 E2 L1

( 0.75 in. )(10, 000 kips/in. ) 59.88 in. = 2.875 F2 − ( 0.345 in.) 40 in. 40 in. = 4.30388F2 − 64.68750 kips 2

(h)

Substitute Eq. (h) into Eq. (a) and solve for F2:

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

F1 ( a + b + c ) + F2c = P ( b + c ) F1 ( 36 in. + 54 in. + 48 in.) + F2 ( 48 in.) = P ( 54 in. + 48 in.) F1 (138 in.) + F2 ( 48 in.) = P (102 in.)

( 4.30388F2 − 64.68750 kips )(138 in.) + F2 ( 48 in.) = ( 20 kips )(102 in.) ( 641.93544 in.) F2 − 8,926.875 kip  in. = 2, 040 kip  in. F2 =

10,966.875 kip  in. = 17.0841 kips 641.93544 in.

Backsubstituting into Eq. (h) gives F1: F1 = 4.30388 F2 − 64.68750 kips

= ( 4.30388 )(17.0841 kips ) − 64.68750 kips = 8.8403 kips

(a) Normal stresses: F 8.8403 kips 1 = 1 = = 11.79 ksi (T) A1 0.75 in.2 F 17.0841 kips 2 = 2 = = 22.8 ksi (T) A2 0.75 in.2

Ans. Ans.

(b) Deflection of the rigid bar at A: Calculate the deformation of bar (1): (8.8403 kips )( 40 in.) = 0.047148 in. FL 1 = 1 1 = A1 E1 0.75 in.2 10, 000 kips/in.2

(

)(

)

The relationship between the rigid bar deflection at A and the deformation of bar (1) was expressed in Eq. (c). Therefore, the rigid bar deflection at A is: Ans. vA = 1 = 0.0471 in. 

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P5.38 A 0.625 in. diameter steel [E = 30,000 ksi] bolt (1) is placed in a copper tube (2), as shown in Figure P5.38. The copper [E = 16,000 ksi] tube has an outside diameter of 1.50 in., a wall thickness of 0.1875 in., and a length of L = 9.0 in. Rigid washers, each with a thickness of t = 0.125 in., cap the ends of the copper tube. The bolt has 20 threads per inch. This means that each time the nut is turned one complete revolution, the nut advances 0.05 in. (i.e., 1/20 in.). The nut is hand-tightened on the bolt until the bolt, nut, washers, and tube are just snug, meaning that all slack has been removed from the assembly but no stress has yet been induced. What stresses are produced in the bolt and in the tube if the nut is tightened an additional quarter turn past the snug-tight condition?

FIGURE P5.38

Solution Equilibrium: The force induced in the bolt by advancing the nut will be balanced by the force created in the tube; therefore: F1 = − F2 (a) Geometry-of-deformation relationship: When the nut is turned on the bolt, the bolt is shortened by that same amount. However, the resistance exerted by the tube causes the bolt to elongate and the tube to shorten. This relationship can be expressed by: 1 −  nut =  2 In words, this relationship says that the elongation of the bolt plus the shortening of the bolt caused by the nut advance (i.e., a negative deformation) must equal the deformation of the tube. When the nut is turned an additional quarter turn past the snug-tight condition, the nut has advanced 1 in. (0.25 turns) = 0.0125 in. 20 turns Therefore, the geometry-of-deformation equation can be written as 1 − 0.0125 in. =  2 (b) Force-Deformation Relationships: FL FL 1 = 1 1 2 = 2 2 A1E1 A2 E2 Compatibility Equation: Substitute Eqs. (c) in Eq. (b) to derive the compatibility equation. F1L1 FL − 0.0125 in. = 2 2 A1E1 A2 E2

(c)

(d)

Solve the equations: Substitute Eq. (a) into Eq. (b) to obtain:

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

F1L1 −F L − 0.0125 in. = 1 2 A1E1 A2 E2  L L  F1  1 + 2  = 0.0125 in.  A1E1 A2 E2  0.0125 in. F1 = L1 L + 2 A1E1 A2 E2 Numeric values:  2 A1 = ( 0.625 in.) = 0.306796 in.2 4 L1 = L + 2t = 9.25 in. E1 = 30, 000 ksi

(e)

A2 =



2 2 (1.5 in.) − (1.125 in.)  = 0.773126 in.2  4

L2 = 9.00 in. E2 = 16, 000 ksi

Substitute these values into Eq. (e) and solve for F1: 0.0125 in. 0.0125 in. F1 = = = 7.2147 kips L1 L2 9.25 in. 9.00 in. + + A1 E1 A2 E2 ( 0.306796 in.2 ) ( 30, 000 ksi ) ( 0.773126 in.2 ) (16, 000 ksi ) The force in the tube is the same magnitude; however, it is compression: F2 = −7.2147 kips Normal stresses: The normal stress in the bolt is: 7.2147 kips 1 = = 23.5 ksi (T) 0.306796 in.2 The normal stress in the tube is: −7.2147 kips 2 = = −9.3318 ksi = 9.33 ksi (C) 0.773126 in.2

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P5.39 A circular aluminum alloy [E = 70 GPa;  = 22.5 × 10–6/°C;  = 0.33] pipe has an outside diameter of 220 mm, a wall thickness of 8 mm, and a length of 5 m. The pipe supports a compressive load of 650 kN. After the temperature of the pipe drops 45°C, determine: (a) the axial deformation of the pipe. (b) the change in diameter of the pipe.

Solution Pipe area: The area of the pipe is: d = D − 2t = 220 mm − 2 (8 mm) = 204 mm

 2 2  2 2  D − d  = ( 220 mm ) − ( 204 mm )  = 5,328.14 mm2  4 4

(a) Axial deformation of the pipe: The relationship between the internal force, temperature change, and the deformation of an axial member is: FL = +  T L AE For this pipe: FL = +  T L AE ( −650, 000 N )( 5, 000 mm ) + 22.5 10−6 / °C −45C 5, 000 mm = )( ) ( )( ( 5,328.14 mm2 )( 70, 000 N/mm2 ) = −13.78 mm

Ans.

(b) Change in diameter of the pipe: The longitudinal direction of the pipe is the 5 m length. Thus, the diameter spans in the lateral direction. We must calculate the strain in the lateral direction for the pipe. Once we know the lateral strain, we can determine the change in diameter of the pipe. Recall that Poisson’s ratio defines the relationship between strains in the longitudinal and lateral directions:

 =−

 lat  long

The Poisson effect applies only to strains caused by stress—it does not apply to thermal strains. Consequently, we need to be careful to separate out the portion of the lateral strain caused by normal stress and the portion of the lateral strain caused by temperature change for this calculation. Lateral strain due to Poisson effect: The longitudinal normal strain produced in the pipe by the 650 kN load is: F −650, 000 N = = = −121.9938 MPa A 5,328.14 mm 2  −121.9938 MPa  = = = −1, 742.77  10−6 mm/mm E 70, 000 MPa Use Poisson’s ratio to calculate the corresponding lateral strain:

 lat = − long = − ( 0.33) ( −1,742.77 10−6 mm/mm ) = 575.1110−6 mm/mm

(a)

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Lateral strain due to temperature change: For thermal strain, it makes no difference whether we are talking about the longitudinal direction or the lateral direction. The thermal strain is the same in all directions (for an isotropic material). Thus, we can find the thermal strain in the direction of the pipe diameter from

T =  T = ( 22.5 10−6 /C) ( −45C ) = −1,012.50 10−6 mm/mm

(b)

Total lateral strain: The total normal strain in the direction of the pipe diameter is found from the sum of Eqs. (a) and (b):  total =   +  T = 575.1110−6 mm/mm + ( −1, 012.50 10−6 mm/mm ) = −437.39 10−6 mm/mm

The change in diameter of the pipe is thus:

D =  total D = ( −437.39 10−6 mm/mm ) ( 220 mm ) = −0.0962 mm

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P5.40 A solid 0.125 in. diameter steel [E = 30,000 ksi;  = 6.5 × 10–6/°F] wire is stretched between fixed supports so that it is under an initial tension force of 40 lb. If the temperature of the wire drops 75°F, what is the tensile stress in the wire?

Solution Initial stress: The initial stress in the wire is:



d2 =



( 0.125 in.) = 0.0122718 in.2 2

4 F 40 lb  initial = = = 3, 259.5 psi A 0.0122718 in.2

Force-Temperature-Deformation Relationship The relationship between internal force, temperature change, and deformation of an axial member is: FL = +  T L AE Since the bar is attached to rigid supports,  = 0. FL +  T L = 0 AE Let’s rewrite this expression in terms of the change in stress caused by the change in temperature: L  +  T L = 0 E Change in stress caused by 75°F temperature drop: Solve the preceding expression for the change in stress. E  = − T L = − TE = −6.5 10−6 /F ( −75F ) 30 106 psi = 14, 625 psi L

(

)

(

)

Final stress in the wire: Add the change in stress to the initial stress to calculate the final stress in the wire after the temperature drops:

 final =  initial +  = 3, 259.5 psi + 14,625 psi = 17,880 psi

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P5.41 An AISI 1040 hot-rolled steel [E = 207 GPa;  = 11.3×10–6/°C] bar is held between two rigid supports. The bar is stress free at a temperature of 25°C. The bar is then heated uniformly. If the yield strength of the steel is 414 MPa, determine the temperature at which yield first occurs.

Solution Force-Temperature-Deformation Relationship The relationship between internal force, temperature change, and deformation of an axial member is: FL = +  T L AE Since the bar is attached to rigid supports,  = 0. FL +  T L = 0 AE which can also be expressed in terms of the bar normal stress: L  +  T L = 0 E Temperature at which yield first occurs: Divide this expression by L and solve for T corresponding to a 414 MPa yield stress in the steel bar. Note that we are considering a compressive stress because we are told that the bar will be heated.

T = − =−

 E

−414 MPa = 176.99C (11.3  10 / C ) ( 207,000 MPa ) −6

Initially, the bar is at a temperature of 25°C. Therefore, the temperature at which yield first occurs is: Ans. Tfinal = Tinitial + T = 25C + 176.99C = 202C

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P5.42 A high-density polyethylene [E = 128 ksi;  = 88 × 10−6/°F;  = 0.4] bar is positioned between two rigid supports, as shown in Figure P5.42. The bar is square with cross-sectional dimensions of 2.50 in. by 2.50 in. and a length of L = 48 in. At room temperature, a gap of  = 0.25 in. exists between the block and the rigid support at B. After a temperature increase of 120°F, determine: (a) the normal stress in the bar. (b) the longitudinal strain in the bar. (c) the transverse (i.e., lateral) strain in the bar.

FIGURE P5.42

Solution If the polyethylene bar were completely free to elongate, a temperature change of 120°F would cause an elongation of

 =  T L = (88 10−6 /°F) (120F)( 48 in.) = 0.507 in.

Since this elongation is greater than the 0.25 in. gap, the temperature change will cause the polyethylene block to contact the support at B, which will create normal stress in the block. Force-Temperature-Deformation Relationship The relationship between the internal force, temperature change, and the deformation of an axial member is: FL = +  T L AE In this situation, the deformation of member (1) equals the 0.25 in. gap: FL +  T L = 0.25 in. AE This relationship can be stated in terms of normal stress as L  +  T L = 0.25 in. E (a) Normal stress: The normal stress in the block due to the 120°F temperature increase is: E  =  0.25 in. −  TL L 128,000 psi = 0.25 in. − ( 88  10−6 /°F ) (120F )( 48 in.) )  = −685.0 psi = 685 psi (C) 48 in. (b) Longitudinal strain: The longitudinal strain in the polyethylene block is:  0.25 in. = = = 0.0052083 in./in. = 5, 210 με L 48 in.

Ans.

(c) Transverse (i.e., lateral) strain: Recall that Poisson’s ratio defines the relationship between strains in the longitudinal and lateral directions:

 =−

 lat  long

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

The Poisson effect applies only to strains caused by stress—it does not apply to thermal strains. Consequently, we need to be careful to separate out the portion of the lateral strain caused by normal stress and the portion of the lateral strain caused by temperature change for this calculation. Lateral strain due to Poisson effect: The longitudinal normal strain produced by the stress in the bar is:  −685.2 psi  = = = −5,351.7  10−6 in./in. E 128, 000 psi Use Poisson’s ratio to calculate the corresponding lateral strain:

 lat = − long = − ( 0.4) ( −5,351.7 10−6 in./in.) = 2,140.7 10−6 in./in.

(a)

T =  T = (88 10−6 /F) (120F) = 10,560 10−6 in./in.

(b)

Total lateral strain: The total normal strain in the lateral direction of the bar is found from the sum of Eqs. (a) and (b):  total =   +  T

= 2,140.7 10−6 in./in. + 10,560 10−6 in./in. =12,700.7 10−6 in./in.

 lat = 12, 700 με

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P5.43 Rigid bar ABC is supported by bronze rod (1) and stainless steel rod (2) as shown in Figure P5.43. A concentrated load of P = 24 kips is applied to the free end of bronze rod (3). Determine the deflection of rod end D after the temperature of all rods has increased by 130°F. Use the following dimensions and properties: a = 3.5 ft, b = 6.5 ft. Rod (1): d1 = 1.00 in., L1 = 12 ft, E1 = 15,200 ksi,  = 12.20×10–6/°F. Rod (2): d2 = 0.75 in., L2 = 9 ft, E2 = 28,000 ksi,  = 9.60×10–6/°F. Rod (3): d3 = 1.25 in., L3 = 6 ft, E3 = 15,200 ksi,  = 12.20×10–6/°F. FIGURE P5.43

Solution Equilibrium Note that the force in rod (3) will equal the external load P. Therefore, F3 = P = 24 kips. Next, consider an FBD of the rigid bar. Assume tension forces in members (1) and (2). Solve for the force in rods (2) and (1):

M A = − F3a + F2 ( a + b ) = 0 a 3.5 ft F3 = ( 24 kips ) = 8.4 kips a+b 3.5 ft + 6.5 ft M C = − F1 ( a + b ) + F3b = 0  F2 =

 F1 =

b 6.5 ft F3 = ( 24 kips ) = 15.6 kips a+b 3.5 ft + 6.5 ft

Deformations: Next, determine the deformations in rods (1), (2), and (3). Include the thermal deformation in each calculation: Rod (1):



L1 = (12 ft )(12 in./ft ) = 144 in. 4 4 (15.6 kips )(144 in.) FL 1 = 1 1 + 1T L1 = + (12.2 10−6 /F ) (130F )(144 in.) 2 2 A1 E1 ( 0.78540 in. )(15, 200 kips/in. ) A1 =

d12 =

(1.0 in.) = 0.78540 in.2 2

= 0.41656 in.

Mechanics of Materials: An Integrated Learning System, 4th Ed. Rod (2):



Timothy A. Philpot

( 0.75 in.) = 0.44179 in.2

L2 = ( 9 ft )(12 in./ft ) = 108 in. 4 4 (8.4 kips )(108 in.) FL  2 = 2 2 +  2 T L2 = + ( 9.6 10−6 /F ) (130F )(108 in.) 2 2 A2 E2 ( 0.44179 in. )( 28, 000 kips/in. ) A2 =

d 22 =

= 0.20812 in. Rod (3):



(1.25 in.) = 1.22718 in.2

L3 = ( 6 ft )(12 in./ft ) = 72 in. 4 4 ( 24 kips )( 72 in.) FL  3 = 3 3 +  3T L3 = + (12.2 10−6 /F ) (130F )( 72 in.) 2 2 A3 E3 (1.22718 in. )(15, 200 kips/in. ) A3 =

d32 =

= 0.20683 in. Geometry of Deformations Relationship Draw a deformation diagram of the rigid bar. The deflections of the rigid bar are related by similar triangles: vB − vA vC − vA = (a) a a+b There are no gaps, clearances, or other misfits at pins A and C; therefore: v A = 1 (b) vC =  2 Substitute Eqs. (b) in Eq. (a) to obtain the rigid bar deflection at B: a a vB = ( vC − vA ) + vA = ( 2 − 1 ) + 1 a+b a+b 3.5 ft = ( 0.20812 in. − 0.41656 in.) + 0.41656 in. 3.5 ft + 6.5 ft = 0.34360 in. Determine the deflection of rod end D: The deformation of rod (3) can be expressed in terms of the deflections at the ends of the rod as:

 3 = vD − vB

Therefore, the deflection of D can be calculated as:

vD = vB + 3 = 0.34360 in. + 0.20683 in. = 0.55043 in. = 0.550 in. 

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P5.44 Two ductile cast iron [E = 24,400 ksi;  = 6.0×10– 6 /°F] bars are connected with a pin at B, as shown in Figure P5.44. Each bar has a cross-sectional area of 1.50 in.2. If the temperature of the bars is decreased by 75°F from their initial temperature, what force P would need to be applied at B so that the total displacement (caused both by the temperature change and by the applied load) of joint B is zero? Use a = 10 ft and  = 55°. FIGURE P5.64 Solution Equilibrium Consider an FBD of joint B. Assume tension forces in bars (1). Fx = − F1 cos  + F1 cos  = 0 Fy = F1 sin  + F1 sin  − P = 0  2 F1 sin  = P

(a)

Member Deformation From the problem statement, we are to determine the force P to be applied at B so that the total displacement of joint B is zero. If the displacement of joint B is zero, then the deformation of each bar must also be zero. Thus, we can state that:

1 =

F1L1 + 1T1L1 = 0 A1E1

Our task now is clear. We must determine the force F1 that makes 1 = 0. Once, we know that value for F1, we can substitute it into Eq. (a) and calculate P. Solve the Equations From the problem statement, A1 = 1.50 in.2. The length of bar (1) is: (10 ft )(12 in./ft ) = 146.493 in. a L1 = = sin  sin 55 Solve for F1: F1 L1 + 1T1 L1 = 0 A1 E1 AE F1 = −1T1 L1 1 1 = −1T1 A1 E1 L1

F1 = − ( 6.0  10 −6 / F ) ( −75F ) (1.50 in.2 ) ( 24, 400 ksi ) = 16.470 kips

Substitute this value into Eq. (a) to find force P:

P = 2F1 sin  = 2 (16.470 kips ) sin 55 = 26.983 kips = 27.0 kips

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P5.45 A solid aluminum alloy [E = 69 GPa;  = 23.6×10–6/°C] rod (1) is attached rigidly to a solid brass [E = 115 GPa;  = 18.7×10–6/°C] rod (2), as shown in Figure P5.45. The compound rod is subjected to a tensile load of P = 6 kN. The diameter of each rod is 10 mm. The rods lengths are L1 = 525 mm and L2 = 675 mm. Compute the change in temperature required to produce zero horizontal deflection at end C of the compound rod.

FIGURE P5.45

Solution The deformations in the two axial members are expressed by

1 =

F1L1 + 1T L1 A1E1

and

2 =

F2 L2 +  2 T L2 A2 E2

The total elongation of the assembly is equal to the displacement of end C relative to end A; which is simply equal to the sum of the deformations in the two axial members:

uC = 1 +  2 =

F1L1 FL + 1T L1 + 2 2 + 2 T L2 A1E1 A2 E2

The internal forces F1 and F2 are each equal to external load P. Also, the rods have equal diameters, and thus the areas of rods (1) and (2) are equal. The problem states that we are to find the change in temperature required to produce zero deflection at C. Using F1 = F2 = P and A1 = A2 = A, simply the expression and set the deflection at C equal to zero. PL L  uC =  1 + 2  + T 1L1 +  2 L2  = 0 (a) A  E1 E2  Both members have the same diameter, and so the areas of the two members are each equal to:

A1 = A2 = A =



(10 mm ) = 78.5398 mm2 2

Solve Eq. (a) for T: L1 L2 +  P  E1 E2 T = −    A  1 L1 +  2 L2 525 mm 675 mm + 2  6, 000 N  69, 000 N/mm 115, 000 N/mm 2 = − 2  −6 −6  78.5398 mm  ( 23.6 10 / C ) ( 525 mm ) + (18.7 10 / C ) ( 675 mm ) = − ( 76.3944 MPa )( 0.538861 C/MPa ) = −41.166C = −41.2C

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P5.46 At a temperature of 15°C, a gap of  = 5 mm exists between the two polymer bars shown in the figure below. Bar (1) has a length of L1 = 700 mm, a cross-sectional area of A1 = 1,250 mm2, a coefficient of thermal expansion of  = 130 × 10−6/°C, and an elastic modulus of E1 = 1,400 MPa. Bar (2) has a length of L2 = 500 mm, a cross-sectional area of A2 = 3,750 mm2, a coefficient of thermal expansion of  = 80 × 10−6/°C, and an elastic modulus of E1 = 3,700 MPa. The supports at A and D are rigid. Determine: (a) the lowest temperature at which the gap is closed. (b) the normal stress in the two bars at a temperature of 100°C. (c) the longitudinal normal strain in the two bars at 100°C.

FIGURE P5.46

Solution (a) Lowest Contact Temperature Before the gap is closed, only thermal strains and the associated axial elongations exist. Write expressions for the temperature-induced elongations and set this equal to the 5 mm gap: 1T L1 +  2 T L2 = 5 mm T 1 L1 +  2 L2  = 5 mm

T =

5 mm 5 mm = = 38.17°C −6 1 L1 +  2 L2 (130 10 /°C ) ( 700 mm ) + (80 10−6 /°C ) ( 500 mm )

Since the initial temperature is 15°C, the temperature at which the gap is closed is 53.2°C.

Ans.

(b) Equilibrium From the results obtained for part (a), we know that the gap will be closed at 100°C, making this a statically indeterminate axial configuration. Knowing this, consider an FBD at joint B. Assume that both internal axial forces will be tension, even though we know intuitively that both F1 and F2 will turn out to be compression. Note: We want a positive deformation caused by the internal force (i.e., elongation for a tension force) to be consistent with the thermal deformation (i.e., an elongation caused by a temperature increase). Fx = − F1 + F2 = 0 (a) Geometry of Deformations Relationship For this configuration, the sum of the deformations of members (1) and (2) must equal the initial gap:

1 +  2 = 5 mm

(b)

Force-Temperature-Deformation Relationships The relationship between the internal force, temperature change, and the deformation of an axial member can be stated for members (1) and (2):

1 =

F1L1 + 1T1L1 A1E1

2 =

F2 L2 +  2 T2 L2 A2 E2

(c)

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Compatibility Equation Substitute the force-temperature-deformation relationships (c) into the geometry of deformation relationship (b) to derive the compatibility equation:

F1L1 FL + 1T1L1 + 2 2 +  2 T2 L2 = 5 mm A1E1 A2 E2

(d)

Solve the Equations This part is no fun, but it must be done. From Eq. (a), F1 = F2. Substituting F1 = F2 into Eq. (d) gives: F1 L1 FL + 1T1 L1 + 1 2 +  2 T2 L2 = 5 mm A1 E1 A2 E2

F1 L1 F1 L2 + = 5 mm − 1T1 L1 −  2 T2 L2 A1 E1 A2 E2  L L  F1  1 + 2  = 5 mm − 1T1 L1 −  2 T2 L2  A1 E1 A2 E2  5 mm − 1T1 L1 −  2 T2 L2 F1 = L1 L + 2 A1 E1 A2 E2

(e)

For this structure, the lengths, areas, and elastic moduli are given below. The temperature change is the same for both members; therefore, T1 = T2 = T = 85°C: L1 = 700 mm L2 = 500 mm A1 = 1, 250 mm 2

A2 = 3, 750 mm 2

E1 = 1, 400 MPa

E2 = 3, 700 MPa

1 = 130 10 /°C

 2 = 80 10−6 /°C

−6

Substitute these values into Eq. (e) and calculate F1 = −14,069.9 N. Backsubstitute into Eq. (a) to calculate F2 = −14,069.9 N. Note that the internal forces are compression, as expected. Normal stresses The normal stresses in each axial member can now be calculated: F −14, 069.9 N 1 = 1 = = −11.2560 MPa = 11.26 MPa (C) A1 1,250 mm 2 F −14, 069.9 N 2 = 2 = = −3.7520 MPa = 3.75 MPa (C) A2 3,750 mm 2

Ans.

(c) Normal Strains The force-temperature-deformation relationships were expressed in Eq. (b). By definition,  = /L. Therefore, the normal strain for each axial member can be determined by dividing the relationships in Eq. (c) by the respective member lengths:

1 =

F1  + 1T1 = 1 + 1T A1E1 E1

2 =

F2  +  2T2 = 2 +  2T A2 E2 E2

Substitute the appropriate values to calculate the normal strains in each member:

Mechanics of Materials: An Integrated Learning System, 4th Ed.

1 = =

1 E1

+ 1T

−11.2560 MPa + (130  10−6 /°C ) ( 85C ) 1, 400 MPa

= 0.003010 mm/mm = 3, 010 με

2 = =

Timothy A. Philpot

2 E2

Ans.

+  2 T

−3.7520 MPa + ( 80  10−6 /°C ) ( 85C ) 3,700 MPa

= 0.005786 mm/mm = 5,790 με

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed. P5.47 A bronze [E = 15,200 ksi;  = 12.2×10–6/°F] rod (2) is placed within a steel [E = 30,000 ksi;  = 6.3×10– 6 /°F] tube (1). The two components are connected at each end by 0.5 in. diameter pins that pass through both the tube and the rod, as shown in Figure P5.47. The outside diameter of tube (1) is 3.00 in. and its wall thickness is 0.125 in. The diameter of rod (2) is 2.25 in., and the distance between pins is L = 21 in. What is the average shear stress in the pins if the temperature of the entire assembly is increased by 90°F?

Timothy A. Philpot

FIGURE P5.47

Solution Section properties: The cross-sectional areas of steel tube (1) and brass rod (2) are:

A1 =



( 4

3.0 in.) − ( 2.75 in.)  = 1.1290 in.2  2

A2 =



( 2.25 in.) = 3.9761 in.2 2

Equilibrium Consider an FBD cut through the assembly. Sum forces in the horizontal direction to obtain:

Fx = − F1 − F2 = 0

 F2 = − F1

(a) Geometry of Deformations Relationship For this configuration, the deformations of both members will be equal; therefore,

1 =  2

(b)

Force-Temperature-Deformation Relationships The relationship between internal force, temperature change, and deformation of an axial member can be stated for members (1) and (2): FL FL 1 = 1 1 + 1T L1  2 = 2 2 +  2 T L2 A1E1 A2 E2 (c) Compatibility Equation Substitute the force-deformation relationships (c) into the geometry of deformation relationship (b) to derive the compatibility equation:

F1L1 FL + 1T L1 = 2 2 +  2 T L2 A1E1 A2 E2

(d)

Solve the Equations Substitute Eq. (a) into Eq. (d), set L2 = L1, and solve for the force in the members:

Mechanics of Materials: An Integrated Learning System, 4th Ed. F1 =

Timothy A. Philpot

T ( 2 L1 − 1 L1 ) T ( 2 − 1 ) = L1 L1 1 1 + + A1 E1 A2 E2 A1 E1 A2 E2

( 90°F ) (12.2 10−6 / F ) − ( 6.3 10−6 / F )

1 1 + 2 2 (1.1290 in. )( 30, 000 kips/in. ) (3.9761 in. )(15, 200 kips/in.2 ) 2

= 11.5258 kips = 11.5258 kips (T)

and from Eq. (a), we know that the brass rod (2) has a compressive force of 11.5258 kips (C). Average shear stress in the pins: Now that we know the axial force produced in the assembly, we can compute the average shear stress produced in the pins. Consider an FBD of the pin passing through brass rod (2). From equilibrium, we find: Fx = − F2 − 2V = 0 F −11.5258 kips (e) V = − 2 = − = 5.7629 kips 2 2 The cross-sectional area of a 0.5 in. diameter pin is:

Apin =



( 0.5 in.) = 0.19635 in.2 2

4 And thus, the average shear stress produced in the pin is: V 5.7629 kips  pin = = = 29.4 ksi Apin 0.19635 in.2

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed. P5.48 A titanium [E = 16,500 ksi;  = 5.3×10–6/°F] bar (1) and a bronze [E = 15,200 ksi;  = 12.2×10–6/°F] bar (2), each restrained at one end, are fastened at their free ends by a pin of diameter d = 0.4375 in., as shown in Figure P5.48. The length of bar (1) is L1 = 19 in. and its cross-sectional area is A1 = 0.50 in.2. The length of bar (2) is L2 = 27 in. and its cross-sectional area is A2 = 0.65 in.2. What is the average shear stress in the pin at B if the temperature changes by 40°F?

Timothy A. Philpot

FIGURE P5.48

Solution Equilibrium Consider an FBD at joint B. Assume that both internal axial forces will be tension.

Fx = − F1 + F2 = 0

 F1 = F2

(a)

Geometry of Deformations Relationship

1 +  2 = 0

Force-Temperature-Deformation Relationships FL FL 1 = 1 1 + 1T1L1  2 = 2 2 +  2 T2 L2 A1E1 A2 E2

(b)

(c)

Compatibility Equation

F1L1 FL + 1T1L1 + 2 2 +  2 T2 L2 = 0 A1E1 A2 E2

(d)

Solve the Equations From Eq. (a), F1 = F2. The temperature change is the same for both members; therefore, T1 = T2 = T. Eq. (d) then can be written as:

F1L1 FL + 1T L1 + 1 2 +  2T L2 = 0 A1E1 A2 E2 Solving for F1: F1 L1 F1 L2 + = −1T L1 −  2 T L2 A1 E1 A2 E2  L L  F1  1 + 2  = −T 1 L1 +  2 L2   A1 E1 A2 E2  T 1 L1 +  2 L2  F1 = −  L1 L2  AE + A E   1 1 2 2

(e)

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

For this structure, the lengths, areas, coefficients of thermal expansion, and elastic moduli are given below. L1 = 19 in. L2 = 27 in. A1 = 0.50 in.2

A2 = 0.65 in.2

E1 = 16,500 ksi

E2 = 15, 200 ksi

1 = 5.3  10−6 /°F  2 = 12.2  10−6 /°F Substitute these values along with T = –40°F into Eq. (e) and calculate F1 = 3.4163 kips. From Eq. (a), F2 = F1 = 3.4163 kips. Note that if the temperature increased by this amount, the forces in bars (1) and (2) would be exactly the same magnitude, only in compression rather than tension. Only the force magnitude is important in determining the shear stress in pin B. Pin shear stress We now know that the pin at B will be subjected to a shear force of V = F1 = F2 = 3.4163 kips. Shear stress will exist on a single plane through the pin—the plane where bars (1) and (2) overlap. The shear area is equal to the cross-sectional area of the pin:

AV = Apin =

 2 ( 0.4375 in.) = 0.15033 in.2 4

The shear stress in pin B can be calculated as:

 pin =

V 3.4163 kips = = 22.7 ksi AV 0.15033 in.2

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P5.49 Rigid bar ABC is supported by two identical solid bronze [Y = 331 MPa; E = 105 GPa;  = 22.0 × 10−6/°C] rods and a single solid steel [Y = 250 MPa; E = 200 GPa;  = 11.7×10−6/°C] rod, as shown in Figure P5.49. Bronze rods (1) each have diameters of d1 = 25 mm and lengths of L1 = 3.0 m. Steel rod (2) has a length of L2 = 4.0 m. (a) Calculate the diameter d2 required for steel rod (2) so that the deflection of joint B is zero for any change in temperature. (b) Using the value of d2 determined in part (a), determine the maximum temperature decrease that is allowable for this assembly if a factor of safety of 2.0 is specified for the normal stress in each rod. FIGURE P5.49

Solution Equilibrium: Consider an FBD of bar BCD. Sum forces in the vertical direction to obtain: Fy = 2F1 − F2 = 0  F2 = 2F1 (a)

(a) Diameter d2 required for zero deflection at B: By symmetry, we know that rigid bar ABC will remain horizontal for any deflection that occurs. Therefore, if the deflection at B is to be zero, then we note that the deformations in bars (1) and (2) must each equal zero for any value of T. Additionally, from Eq. (a), we know that F2 = 2F1. Accordingly,

F1L1 + 1T L1 = 0 A1E1 FL  2 = 2 2 +  2T L2 = 0 A2 E2

1 =

F1 1 A1E1 F2 2F1 T = − =−  2 A2 E2  2 A2 E2 T = −

Equate these two expressions to find: F1 2 F1 = 1 A1 E1  2 A2 E2 21 E1 A2 d 22 = =  2 E2 A1 d12  22.0 10−6 /°C   105 GPa  1 E1  d 2 = d1 2 = d1 2    = 1.4051d1 −6  2 E2  11.7 10 /°C   200 GPa 

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

The diameter required for rod (2) is: d2 = 1.4051( 25 mm) = 35.1280 mm = 35.1 mm

Ans.

(b) Determine the maximum temperature decrease if FS = 2.0. Geometry-of-deformation Relationship:

1 +  2 = 0

(b)

Force-Temperature-Deformation Relationships:

F1L1 + 1T L1 A1E1

F2 L2 +  2 T L2 A2 E2

(c)

Compatibility Equation: Substitute Eqs. (c) into Eq. (b) to derive the compatibility equation: F1 L1 FL + 1T1 L1 + 2 2 +  2 T2 L2 = 0 A1 E1 A2 E2

(d)

1 =

2 =

Problem Data: For this structure, the areas, lengths, coefficients of thermal expansion, elastic moduli, and allowable stresses are given below:

A1 =



( 25 mm ) = 490.8739 mm2 2

A2 =



( 35.1280 mm ) = 969.1612 mm2

4 L1 = 3,000 mm

4 L2 = 4,000 mm

E1 = 105,000 MPa

E2 = 200,000 MPa

1 = 22.0  10−6 /°C

 2 = 11.7  10−6 /°C

331 MPa = 165.5 MPa 2.0 F1,allow = 81, 239.6 N

 2,allow =

 1,allow =

250 MPa = 125 MPa 2.0 F2,allow = 121,145.1 N

From Eq. (a), determine that the allowable force in rod (2) controls. F 121,145.1 N F1 = 2 = = 60,572.6 N  81, 239.6 N OK 2 2 F2 = 121,145.1 N

(e)

Maximum Temperature Decrease: Solve Eq. (d) for T, using the forces in Eq. (e):  FL F L  T 1L1 +  2 L2  = −  1 1 + 2 2   A1E1 A2 E2 

Mechanics of Materials: An Integrated Learning System, 4th Ed.

F1L1 F2 L2 + A1E1 A2 E2 T = − 1L1 +  2 L2

( 60,572.6 N )( 3,000 mm )

Timothy A. Philpot

(121,145.1 N )( 4,000 mm )

( 490.8739 mm )(105,000 N/mm ) (969.1612 mm )( 200,000 N/mm ) =− ( 22.0  10 /°C ) ( 3,000 mm ) + (11.7  10 /°C ) ( 4,000 mm ) 2

−6

3.5256 mm + 2.5000 mm 66.0  10−3 mm/°C + 46.8  10−3 mm/°C = −53.4188C

=−

= −53.4C

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P5.50 A polymer cylinder (2) is clamped between rigid heads by two steel bolts (1), as shown in Figure P5.50. The steel [E = 29,000 ksi;  = 6.5×10−6/°F] bolts have a diameter of 0.50 in. The polymer [E = 370 ksi;  = 39.0 × 10−6/°F] cylinder has an outside diameter of 6.625 in. and a wall thickness of 0.432 in. Assume a = 24 in. and b = 28 in. If the temperature of this assembly changes by T = 120°F, determine: (a) the normal stress in the polymer cylinder. (b) the normal strain in the polymer cylinder. (c) the normal strain in the steel bolts. FIGURE P5.50

Solution Equilibrium: By virtue of symmetry, the forces in the two steel bolts (1) are identical. Consider an FBD of the rigid head. Sum forces in the horizontal direction to obtain: (a) Fx = 2 F1 + F2 = 0 Geometry-of-deformation equation: For this configuration,

1 =  2

Force-Temperature-Deformation Relationships: The relationship between internal force, temperature change, and deformation can be stated for members (1) and (2): FL FL 1 = 1 1 + 1T1 L1  2 = 2 2 +  2T2 L2 A1E1 A2 E2

(b)

(c)

Compatibility Equation: Substitute the force-deformation relationships (c) into the geometry-ofdeformation relationship (b) to derive the compatibility equation:

F1L1 FL + 1T1L1 = 2 2 +  2 T2 L2 A1E1 A2 E2

(d)

Solve the Equations: For this situation, T1 = T2 = T. Solve Eq. (d) for F1: FL  AE L A E AE F1 =  2 2 +  2 T L2 − 1T L1  1 1 = F2 2 1 1 + T  2 L2 − 1L1  1 1 L1 A2 E2 L1  A2 E2  L1 Substitute this expression into Eq. (a) and solve for F2:

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

 L A E AE  2  F2 2 1 1 + T  2 L2 − 1L1  1 1  + F2 = 0 L1   L1 A2 E2  L A E  AE F2  2 2 1 1 + 1 = 2 T 1L1 −  2 L2  1 1 L1  L1 A2 E2  AE 2 T 1L1 −  2 L2  1 1 L1 F2 = L A E 2 2 1 1 +1 L1 A2 E2 For this structure, the areas, coefficients of thermal expansion, and elastic moduli are given below:

A1 =



( 0.50 in.) = 0.19635 in.2 2

A2 =



(e)

6.625 in.) − ( 5.761 in.)  = 8.40494 in.2 ( 

4 L1 = 28 in.

4 L2 = 24 in.

E1 = 29,000 ksi

E2 = 370 ksi

1 = 6.5  10−6 /°F

 2 = 39.0  10−6 /°F

Substitute these values along with T = 120°F into Eq. (e) and calculate F2:

F2 = −8,891.4 lb

Backsubstitute this result into Eq. (a) to calculate F1 = 4,445.7 lb. (a) Normal Stress in Polymer Cylinder: The normal stresses in the cylinder is computed as:

2 =

F2 −8,891.4 lb = = 1,058 psi (C) A2 8.40494 in.2

Ans.

(b) Normal Strain in Polymer Cylinder: The force-temperature-deformation relationships were expressed in Eq. (c). By definition,  = /L. Therefore, the normal strain for each axial member can be determined by dividing the relationships in Eq. (c) by the respective member lengths: F F 1 = 1 + 1T1  2 = 2 +  2 T2 (f) A1E1 A2 E2 Substitute the appropriate values to calculate the normal strains in the polymer cylinder: F  2 = 2 +  2 T2 A2 E2

−8,891.4 lb + ( 39.0  10−6 /°F ) (120F ) 2 2 (8.40494 in. )( 370,000 lb/in. )

= 0.0018209 in./in. = 1,821 με

Ans.

4, 445.7 lb + ( 6.5  10−6 /°F ) (120F ) 2 2 ( 0.19635 in. )( 29,000,000 lb/in. )

= 0.0015607 in./in. = 1,561 με

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P5.51 A load P will be supported by a structure consisting of a rigid bar ABCD, a steel [E = 29,000 ksi;  = 6.5 × 10−6/°F] bar (1) and an aluminum alloy [E = 10,000 ksi;  = 12.5 × 10−6/°F] bar (2), as shown in Figure P5.51. The bars have cross-sectional areas of A1 = 0.80 in.2 and A2 = 1.30 in.2, respectively. Dimensions of the structure are a = 3 ft, b = 3.75 ft, c = 5.0 ft, L1 = 12 ft, and L2 = 20 ft. The bars are unstressed when the structure is assembled. After a concentrated load of P = 35 kips is applied and the temperature has been increased by 60°F, determine: (a) the normal stresses in bars (1) and (2). (b) the vertical deflection of joint D. FIGURE P5.51

Solution Equilibrium Consider an FBD of the rigid bar. Assume tension forces in members (1) and (2). A moment equation about pin A gives the best information for this situation: M A = F1a + F2 ( a + b + c ) − P ( a + b ) = 0 F1 ( 3 ft ) + F2 ( 3 ft + 3.75 ft + 5.0 ft ) = P ( 3 ft + 3.75 ft ) F1 ( 3 ft ) + F2 (11.75 ft ) = P ( 6.75 ft )

(a)

Geometry of Deformations Relationship Draw a deformation diagram of the rigid bar. The deflections of the rigid bar are related by similar triangles: vB v vD = C = a a+b a+b+c vB vC vD (b) = = 3 ft 6.75 ft 11.75 ft There are no gaps, clearances, or other misfits at pins A and C; therefore, Eq. (c) can be rewritten in terms of the member deformations as:

1 2 = 3 ft 11.75 ft

(c)

Force-Temperature-Deformation Relationships FL FL 1 = 1 1 + 1T1L1  2 = 2 2 +  2 T2 L2 A1E1 A2 E2 Compatibility Equation   1  F1L1 1  F2 L2 + 1T1L1  = +  2 T2 L2    3 ft  A1E1  11.75 ft  A2 E2 

(d)

(e)

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Solve the Equations Note that the temperature change for both axial members is the same (i.e., T1 = T2 = T). Solve Eq. (e) for F1:  F1L1 3 ft  F2 L2 = +  2 T L2  − 1T L1  A1E1 11.75 ft  A2 E2  3 ft L2 A1 E1 3 ft AE +  2 T L2 1 1 − 1T A1E1 11.75 ft L1 A2 E2 11.75 ft L1 Substitute Eq. (f) into equilibrium equation (a): F1 = F2

(f)

  3 ft L2 A1 E1 3 ft AE +  2T L2 1 1 − 1T A1E1  ( 3 ft ) + F2 (11.75 ft ) = P ( 6.75 ft )  F2 L1  11.75 ft L1 A2 E2 11.75 ft  and solve for F2: P ( 6.75 ft ) − F2 =

( 3 ft )  T L A1E1 +  T A E 3 ft ) 2 2 1 1 1( 2

11.75 ft

( 3 ft )

(g)

L2 A1 E1 + 11.75 ft 11.75 ft L1 A2 E2

For this structure, P = 35 kips, and the lengths, areas, elastic moduli, and coefficients of thermal expansion are listed below: L1 = 12 ft L2 = 20 ft A1 = 0.80 in.2

A2 = 1.30 in.2

E1 = 29,000 ksi

E2 = 10,000 ksi

1 = 6.5  10 /°F  2 = 12.5  10−6 /°F Substitute these values along with T = 60°F into Eq. (g) and calculate F2 = 17.1926 kips. Backsubstitute into Eq. (f) to calculate F1 = 11.4125 kips. −6

(a) Normal Stresses: The normal stresses in each axial member can now be calculated:

F1 11.4125 kips = = 14.27 ksi (T) A1 0.80 in.2 F 17.1926 kips 2 = 2 = = 13.23 ksi (T) A2 1.30 in.2

1 =

Ans. Ans.

(b) Deflection of joint D: Calculate the deformation of member (2): FL  2 = 2 2 +  2 T2 L2 A2 E2

(17.1926 kips )( 20 ft )(12 in./ft ) + 12.5  10−6 /°F 60°F 20 ft 12 in./ft = 0.4974 in. ) ( ) ( )( )( (1.30 in.2 ) (10,000 ksi )

(h)

Since there are no gaps at pin D, the rigid bar deflection at D is equal to the deformation of member (2); therefore:

vD =  2 = 0.497 in. 

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P5.52 The pin-connected assembly shown in Figure P5.52 consists of two aluminum alloy [E = 69 GPa;  = 23.6 × 10−6/°C] bars (1) and a titanium alloy [E = 114 GPa;  = 9.5 × 10−6/°C] bar (2) that are connected at pin D. The aluminum bars have cross-sectional areas of A1 = 300 mm2, and the titanium bar has a cross-sectional area of A2 = 500 mm2. Dimensions for the assembly are a = 1.80 m and b = 2.50 m. All bars are unstressed at a temperature of 25°C. Calculate (a) the normal stresses in the aluminum and titanium bars and (b) the vertical displacement of pin D when the temperature of the assembly reaches 85°C. FIGURE P5.52

Solution Equilibrium Consider an FBD of joint D. Assume tension forces in bars (1) and (2). b 2.5 m tan  = = = 1.3889  = 54.2461 a 1.8 m Fx = − F1 cos + F1 cos = 0

Fy = − F1 sin  − F2 − F1 sin  = 0  F2 = −2 F1 sin 

(a)

Geometry-of-deformation relationship Sketch a deformation diagram that focuses on joint D. We imagine temporarily removing the pin at D and allowing the members to elongate independently. Then, after deformation, bars (1) are rotated so that they intersect the upper end of bar (2), and the pin at D is reinserted. If we assume that the deformations will be small, then the angles after deformation are approximately the same as the angles before deformation. From the deformation diagram geometry, we can assert that: 1 (b) = 2 sin  Force-Deformation Relationships The relationship between the internal force and the deformation of an axial member can be stated for bars (1) and (2) as: FL FL 1 = 1 1 + 1T1L1  2 = 2 2 +  2 T2 L2 (c) A1E1 A2 E2 Compatibility Equation Substitute the force-deformation relationships (c) into the geometry-of-deformation relationship (b) to derive the compatibility equation:  FL 1  F1L1 + 1T1L1  = 2 2 +  2 T2 L2 (d)  sin   A1E1  A2 E2 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Solve the Equations The bar lengths are: L1 =

(1.8 m ) + ( 2.5 m ) = 3.08058 m = 3,080.58 mm 2

L2 = 2.5 m = 2,500 mm

Additionally, the temperature change is the same for all members. Solve Eq. (d) for F2:  F2 L2 1  F1L1 = + 1T1L1  −  2 T2 L2  A2 E2 sin   A1E1 

F2 =

 A2 E2  F1L1 AE + 1T1L1  −  2 T2 L2 2 2  L2 sin   A1E1 L2 

 L1  A2 E2  1T1L1 A2 E2 AE = F1  −  2 T2 L2 2 2  + L2 sin  L2  L2 sin   A1 E1   L1  A2 E2   L1  = F1  1 −  2  A2 E2 T  +  L2 sin   A1 E1   L2 sin   2    500 mm 114 GPa  3.08058 m = F1    2  ( 2.5 m ) sin 54.2461   300 mm 69 GPa    3.08058 m + 23.6  10−6 / C ) − ( 9.5  10−6 /C )  ( 500 mm 2 )(114,000 N/mm 2 ) ( 60C ) (  ( 2.5 m ) sin 54.2461  = 4.18110 F1 + 90,063.1 N Substitute this result into Eq. (a) and solve for F1: F2 = −2 F1 sin  4.18110 F1 + 90,063.1 N = −2 F1 sin 54.2461 4.18110 F1 + 2 F1 sin 54.2461 = −90,063.1 N

F1 ( 4.18110 + 2sin 54.2461 ) = −90,063.1 N  F1 =

−90,063.1 N = −15,517.0 N 5.80417

The force in bar (2) is therefore: F2 = −2F1 sin = −2 ( −15,517.0 N ) sin54.2461 = 25,185.1 N (a) Normal stresses in bars (1) and (2). The normal stresses in bars (1) and (2) can now be calculated:

F1 −15,517.0 N = = 51.7 MPa (C) A1 300 mm2 F 25,185.1 N 2 = 2 = = 50.4 MPa (T) A2 500 mm2

1 =

Ans. Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

(b) Displacement of pin D. The displacement of pin D is simply equal to the deformation of bar (2). FL  2 = 2 2 +  2 T2 L2 A2 E2

( 25,185.1 N )( 2,500 mm )

( 500 mm )(114,000 N/mm ) 2

+ ( 9.5  10−6 / C ) ( 60C )( 2,500 mm )

= 2.5296 mm Therefore, the displacement of pin D is:

vD =  2 = 2.53 mm 

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P5.53 The pin-connected assembly shown in Figure P5.53 consists of a rigid bar ABC, a cast iron [E = 24,400 ksi;  = 6.0 × 10−6/°F] bar (1), and an aluminum alloy [E = 10,000 ksi;  = 13.1 × 10−6/°F] bar (2). Bar (1) has a cross-sectional area of A1 = 0.50 in.2 and a length of L1 = 18 in. Bar (2) has a crosssectional area of A2 = 1.60 in.2 and a length of L2 = 45 in. Dimensions of the assembly are a = 25 in. and b = 80 in. The bars are unstressed when the structure is assembled at 75°F. After assembly, the temperature of bar (2) is decreased by 90°F while the temperature of bar (1) remains constant at 75°F. Determine the normal strains in both bars for this condition. FIGURE P5.53

Solution Equilibrium Consider an FBD of the rigid bar. Assume tension forces in members (1) and (2). A moment equation about pin B gives the best information for this situation: a M B = − F1a + F2b = 0  F2 = F1 (a) b Geometry of Deformations Relationship Draw a deformation diagram of the rigid bar. The deflections of the rigid bar are related by similar triangles: −v A vC = a b

(b)

The deformation of bar (1) can be expressed in terms of the difference in deflections between its upper joint and its lower joint:

1 = vD − v A

Since the deflection of joint D is zero, the deflection of joint A can be expressed in terms of the deformation in bar (1) as:

v A = −1

Similarly, the deformation of bar (2) can be expressed in terms of the difference in deflections between its upper joint and its lower joint:

 2 = vE − vC

Since the deflection of joint E is zero, the deflection of joint C can be expressed in terms of the deformation in bar (2) as:

vC = − 2

There are no gaps, clearances, or other misfits at pins A and C; therefore, Eq. (b) can be rewritten in terms of the bar deformations as: 1 − 2 a (c) =  1 = −  2 a

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Force-Temperature-Deformation Relationships

1 =

F1L1 + 1T1L1 A1E1

2 =

F2 L2 +  2T2 L2 A2 E2

Compatibility Equation  F1L1 aFL + 1T1L1 = −  2 2 +  2 T2 L2  A1E1 b  A2 E2  For this assembly, T1 = 0. The compatibility equation can be simplified accordingly:  F1L1 aFL = −  2 2 +  2T2 L2  A1E1 b  A2 E2 

(d)

(e)

Solve the Equations: Substitute Eq. (a) into Eq. (e) and solve for F1:  a   F1  L2 2    F1L1 a b  a  a   L2  =−  +  2 T2 L2  = −     F1 −  2 T2 L2 A1E1 b  A2 E2 b  b   A2 E2    

L1 a a  L  F1 +    2  F1 = −  2 T 2 L2 A1E1 b  b   A2 E2  2

2  L1 a  a   L2   F1  +     = −  2 T2 L2 b  A1E1  b   A2 E2   a  2 T2 L2 b F1 = − 2 L1  a   L2  +    A1E1  b   A2 E2 

Thus,

25 in. (13.1  10−6 /°F) ( −90F)( 45 in.) 80 in. F1 = − 2 18 in. 45 in.  25 in.  +   2 2 2 ( 0.50 in. )( 24, 400 kips/in. )  80 in.  (1.60 in. )(10,000 kips/in.2 ) −0.016579688 in. 1.4754098  10 in./kips + 2.7465820  10−3 in./kips = 9.47374 kips

=−

−3

From Eq. (a), calculate F2: a 25 in. F2 = F1 = ( 9.47374 kips ) = 2.96054 kips b 80 in.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Strain in bar (1): Divide the force-deformation relationship for bar (1) by L1 to solve for 1:  F 9.47374 kips 1 = 1 = 1 = = 7.7654  10−4 in./in. = 777 με L1 A1E1 ( 0.50 in.2 )( 24,400 kips/in.2 )

Ans.

Strain in bar (2): Divide the force-deformation relationship for bar (2) by L2 to solve for 2:  F  2 = 2 = 2 +  2T2 L2 A2 E2

2.96054 kips + (13.1  10−6 /°F ) ( −90F ) 2 2 (1.60 in. )(10,000 kips/in. )

= 1.8503  10−4 in./in. − 1.1790  10−3 in./in. = −9.9397  10−4 in./in. = −994 με

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed. P5.54 A 100 mm wide by 8 mm thick steel bar is transmitting an axial tensile load of 3,000 N. After the load is applied, a 4 mm diameter hole is drilled through the bar, as shown in Figure P5.54. The hole is centered in the bar. (a) Determine the stress at point A (on the edge of the hole) in the bar before and after the hole is drilled. (b) Does the axial stress at point B on the edge of the bar increase or decrease as the hole is drilled? Explain.

Timothy A. Philpot

FIGURE P5.54

Solution (a) Stress at point A: Before hole is drilled: P 3,000 N A = = = 3.75 MPa A (100 mm)(8 mm) After hole is drilled: d 4 mm = = 0.04 D 100 mm

A =

Ans.

 K  2.89

PK (3,000 N)(2.89) = = 11.29 MPa Anet (100 mm − 4 mm)(8 mm)

Ans.

(b) Stress at point B: The axial stress at point B decreases. Since the average stress changes very little with the introduction of the small hole and since the stress at A is larger than the average stress, the axial stress far from the hole must be less than the average stress.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P5.55 The machine part shown in Figure P5.55 is 8 mm thick and is made of AISI 1020 cold-rolled steel (see Appendix D for properties). Determine the maximum safe load P if a factor of safety of 3 with respect to failure by yield is specified.

FIGURE P5.55

Solution  allow =

Y FS

427 MPa = 142.33 MPa 3

Fillets: D 120 mm = = 1.5 d 80 mm

Pallow =

 allow Amin K

r 16 mm = = 0.2 d 80 mm (142.33 MPa)(80 mm)(8 mm) = 49.8 kN 1.83

Hole: d 15 mm = = 0.125 D 120 mm

Pallow =

 allow Anet K

 allow Amin K

(a)

 K  2.67

(142.33 MPa)(120 mm − 15 mm)(8 mm) = 44.8 kN 2.67

Notches: r 10 mm = = 0.167 d 60 mm

Pallow =

 K  1.83

D 80 mm = = 1.333 d 60 mm

(142.33 MPa)(60 mm)(8 mm) = 30.0 kN 2.28

(b)

 K  2.28

(c)

Controlling Load: The notches control in this case; therefore,

Pallow = 30.0 kN

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P5.56 The 0.25 in. thick bar shown in Figure P5.56 is made of 2014-T4 aluminum (see Appendix D for properties) and will be subjected to an axial tensile load of P = 1,500 lbs. A 0.5625 in. diameter hole is located on the centerline of the bar. Determine the minimum safe width D for the bar if a factor of safety of 2.5 with respect to failure by yield must be maintained. FIGURE P5.56

Solution  allow =

Y FS

42 ksi = 16.8 ksi = 16,800 psi 2.5

By definition:

 max  nom

In this instance, the maximum stress equals the allowable stress:  16,800 psi K = max =  K nom = 16,800 psi

 nom

If we denote the net width of the bar at the hole location as w and the bar thickness at t, then w must satisfy: P P KP (a) K nom = K =K  16,800 psi w  Anet wt (16,800 psi) t And the bar width D is the sum of the net width w and the hole diameter d: (b) D = w+ d For a bar with a hole, the value of K is dependent on the ratio of d/D. Since K cannot be determined until D is known, and since the calculation of D from Eqs. (a) and (b) depends on K, a trial-and-error process can be established. One possible sequence is summarized in the table below.

d D

K [from Fig. 5.14]

w (in.) [from Eq. (a)]

D (in.) [from Eq. (b)]

0.25

2.43

0.868

1.430

0.393

0.39

2.25

0.804

1.366

0.412

0.41

2.23

0.796

1.359

0.414 (agrees with trial value)

Trial value of

Resulting value of

Therefore, the minimum bar width is:

Dmin = 1.359 in.

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P5.57 The machine part shown in Figure P5.57 is 10 mm thick, is made of AISI 1020 cold-rolled steel (see Appendix D for properties), and is subjected to a tensile load of P = 45 kN. Determine the minimum radius r that can be used between the two sections if a factor of safety of 2 with respect to failure by yield is specified. Round the minimum fillet radius up to the nearest 1mm multiple. FIGURE P5.57

Solution  allow = K=

Y FS

 allow Amin P

427 MPa = 213.5 MPa 2

(213.5 N/mm2 )(40 mm)(10 mm) = 1.90 45,000 N

D 80 mm = =2 d 40 mm Then, from Fig. 5.15 r = 0.214 d

 rmin = (0.214)d = (0.214)(40 mm) = 8.56 mm

 say rmin = 9 mm

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P5.58 The stepped bar with a circular hole, shown in Figure P5.58, is made of annealed 18-8 stainless steel. The bar is 12 mm thick and will be subjected to an axial tensile load of P = 70 kN. The normal stress in the bar is not to exceed 150 MPa. To the nearest millimeter, determine: (a) the maximum allowable hole diameter d. (b) the minimum allowable fillet radius r. FIGURE P5.58

Solution (a) Maximum hole diameter d: By definition:

 max  nom

In this instance, the maximum stress equals the 150-MPa allowable stress:  150 MPa K = max =  K nom = 150 MPa

 nom

If we denote the net width of the bar at the hole location as w and the bar thickness at t, then w can be expressed as w = D – d. The nominal stress at the hole location depends on Anet, which in turn can be expressed in terms of w: P P P KP K nom = K =K =K  150 N/mm2  (D − d )  Anet wt (D − d ) t (150 N/mm2 ) t From this relationship, the hole diameter d is: KP (a) d  D− (150 N/mm2 ) t For a bar with a hole, the value of K is dependent on the ratio of d/D. Since K cannot be determined until d is known, and since the calculation of d from Eq. (a) depends on K, a trial-and-error process can be established. One possible sequence is summarized in the table below.

d D

K [from Fig. 5.14]

d (mm) [from Eq. (a)]

0.10 0.18 0.23 0.26 0.28

2.73 2.56 2.47 2.42 2.39

23.83 30.44 33.94 35.89 37.06

0.29

2.37

37.83

Trial value of

Resulting value of

d D

0.183 0.234 0.261 0.276 0.285 0.291 (agrees with trial value)

Therefore, the maximum hole diameter is:

dmax = 37 mm

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

(b) Minimum fillet radius r: D 130 mm = = 1.3 d 100 mm Therefore, the curve for D/d = 1.30 on Fig. 5.15 will be used to determine the stress concentration factor.  max  allow 150 N/mm2 K= = = = 2.571  nom P / Amin (70,000 N) / (100 mm)(12 mm) From Fig. 5.15, r  0.047 d

 rmin = 0.047(100 mm) = 4.7 mm

say rmin = 5 mm

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P6.1 A solid circular steel shaft having an outside diameter of d = 60 mm is subjected to a pure torque of T = 2,400 N·m. What is the maximum shear stress in the shaft?

Solution The polar moment of inertia for the shaft is



( 60 mm ) = 1,272,345 mm4 4

32 The maximum shear stress in the steel shaft is found from the elastic torsion formula: Tc ( 2,400 N  m )( 60 mm/2 )  max = = = 56.6 MPa J 1,272,345 mm4

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P6.2 A hollow stainless steel shaft with an outside diameter of 42 mm and a wall thickness of 4 mm has an allowable shear stress of 100 MPa. Determine the maximum torque T that may be applied to the shaft.

Solution The polar moment of inertia for the shaft is



 4 4  D 4 − d 4  = ( 42 mm ) − ( 34 mm )  = 174,296 mm 4  32 32 

Rearrange the elastic torsion formula to determine the maximum torque T: 2 4  allow J 100 N/mm 174,296 mm T= = = 830 N  m c 42 mm / 2

(

)(

)

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P6.3 What is the minimum diameter required for a solid steel shaft to transmit a torque of 20,000 lb·ft if the maximum shear stress in the shaft must not exceed 8,000 psi?

Solution The polar moment of inertia for a solid shaft can be expressed as

 4 d 32

Rearrange the elastic torsion formula to group terms with d on the left-hand side:  d4 T = 32 (d / 2)  and simplify to d 3 T = 16  From this equation, the unknown diameter of the solid shaft can be expressed as 16T d=3



To support a torque of T = 20,000 lb·ft without exceeding the maximum shear stress, a solid shaft must have a diameter of

d3

16T



16 ( 20,000 lb  ft )(12 in./ft )

 (8,000 lb/in.2 )

= 5.35 in.

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P6.4 The compound shaft shown in Figure P6.4 consists of two steel [G = 80 GPa] pipe segments. The shaft is subjected to external torques TB and TC that act in the directions shown. Segment (1) has an outside diameter of 206 mm and a wall thickness of 10 mm. Segment (2) has an outside diameter of 138 mm and a wall thickness of 14 mm. The torque is known to have a magnitude of TC = 22 kN·m. The shear strain in segment (1) is measured as 500 rad. Determine the magnitude of external torque TB . FIGURE P6.4

Solution Equilibrium: From an FBD that cuts through shaft segment (1), write the equilibrium equation: M x = −T1 + TB − TC = 0 (a)

Section properties for segment (1):

J1 =



 4 4  D14 − d14  = ( 206 mm ) − (186 mm) )  = 59,290,775 mm4   32 32

Torque in segment (1): Use Hooke’s law to compute the shear stress in shaft segment (1): 1 = G1 1 = (80,000 N/mm2 )( 500  10−6 rad ) = 40 N/mm2 From this result, compute the torque in segment (1): Tc = J 2 4 1 J1 ( 40 N/mm )( 59,290,775 mm ) T1 = = = 23,025,544 N  mm = 23.026 kN  m c1 206 mm / 2

(b)

External torque at B: Substitute the results from Eq. (b) into Eq. (a) to find :

TB = TC + T1 = 22 kN  m + 23.026 kN  m = 45.0 kN  m

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P6.5 A solid constant-diameter circular shaft is subjected to the torques of TA = 420 lb·ft, TB = 1,040 lb·ft, TC = 850 lb·ft, and TD = 230 lb·ft, acting in the directions shown in Figure 6.5. The bearings shown allow the shaft to turn freely. (a) Plot a torque diagram showing the internal torque in segments (1), (2), and (3) of the shaft. Use the sign convention presented in Section 6.6. (b) If the allowable shear stress in the shaft is 6,000 psi, what is the minimum acceptable diameter for the shaft? FIGURE P6.9

Solution Equilibrium:

M x = −TD − T3 = 0 T3 = −TD = −230 lb  ft M x = −TD + TC − T2 = 0 T2 = TC − TD = 850 lb  ft − 230 lb  ft = 620 lb  ft M x = −TD + TC − TB − T1 = 0 T1 = −TB + TC − TD = −1,040 lb  ft + 850 lb  ft − 230 lb  ft = −420 lb  ft (a) Torque diagram: The maximum torque magnitude in the shaft occurs in segment (2): Tmax = 620 lb·ft.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

(b) Minimum acceptable diameter: The elastic torsion formula gives the relationship between shear stress and torque in a shaft. Tc = J In this instance, the torque and the allowable shear stress are known for the shaft. Rearrange the elastic torsion formula, putting the known terms on the right-hand side of the equation: J T = c  Express the left-hand side of this equation in terms of the shaft diameter D:  4 d  T 32 = d3 = d / 2 16  and solve for the minimum acceptable diameter: 16 T 16 ( 620 lb  ft )(12 in./ft ) d3 = = = 6.31527 in.3 2    (6,000 lb/in. )  d  1.848 in.

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P6.6 The nut on a bolt is tightened by applying a force of P = 16 lb to the end of a wrench at a distance of a = 6 in. from the axis of the bolt, as shown in Figure P6.6. The body of the bolt has an outside diameter of d = 0.375 in. What is the maximum torsional shear stress in the body of the bolt?

Solution The polar moment of inertia for the bolt is



( 0.375 in.) = 1.94144  10−3 in.4 4

The torque applied to the nut is T = Pa = (16 lb )( 6 in.) = 96 lb  in. The maximum shear stress in the bolt is found from the elastic torsion formula: Tc ( 96 lb  in.)( 0.375 in. / 2 )  max = = = 9, 270 psi J 1.94144  10−3 in.4

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P6.7 A solid circular bronze [G = 6,500 ksi] shaft, 5 ft long and 1.375 in. in diameter, carries a torque of 80 lb·ft. For this shaft, determine: (a) the maximum shear stress in the shaft. (b) the magnitude of the angle of twist.

Solution The polar moment of inertia for the shaft is



(1.375 in.) = 350.922  10−3 in.4 4

(a) The maximum shear stress in the steel shaft is found from the elastic torsion formula:

 max =

Tc ( 80 lb  ft )(12 in./ft )(1.375 in. / 2 ) = = 1,881 psi J 350.922  10−3 in.4

Ans.

(b) The magnitude of the angle of twist in a 5 ft long length of shaft is 2 80 lb  ft )( 5 ft )(12 in./ft ) ( TL = = JG ( 350.922  10−3 in.4 )( 6,500,000 lb/in.2 )

= 25.2522  10−3 rad = 0.0253 rad = 1.447

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P6.8 A solid titanium alloy [G = 114 GPa] shaft that is 900 mm long will be subjected to a pure torque of T = 160 N·m. Determine the minimum diameter required if the shear stress must not exceed 165 MPa and the angle of twist must not exceed 5°. Report both the maximum shear stress  and the angle of twist  at this minimum diameter.

Solution Consider shear stress: The polar moment of inertia for a solid shaft can be expressed as

 4 d 32

The elastic torsion formula can be rearranged to gather terms with d:  d4 d 3 T = = 32 (d / 2) 16  From this equation, the unknown diameter of the solid shaft can be expressed as 16T d=3



For the solid titanium alloy shaft, the minimum diameter that will satisfy the allowable shear stress is:

d3

16 (160 N  m )(1,000 mm/m )

 (165 N/mm2 )

= 17.02951 mm

Consider angle of twist: Rearrange the angle of twist equation: TL  TL = J = d4  JG 32 G and solve for the minimum diameter that will satisfy the angle of twist limitation:

d4

32 (160 N  m )(1,000 mm/m )( 900 mm ) 32TL =4 = 19.59533 mm G  ( 5 )( rad/180 ) (114,000 N/mm2 )

Therefore, the minimum diameter that could be used for the shaft is dmin = 19.60 mm

Ans.

The angle of twist for this shaft is  = 5° = 0.087266 rad. To compute the shear stress in a 19.60 mm diameter shaft, first compute the polar moment of inertia:



(19.59533 mm ) = 14,474.7 mm4 4

32 The shear stress in the shaft is thus: (160 N  m)(1,000 mm/m)(19.59533 mm / 2) = 108.3 MPa  max = 14,474.7 mm4

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P6.9 A compound shaft (Figure P6.9) consists of bronze segment (1) and aluminum alloy segment (2). Segment (1) is a solid bronze [G = 45 GPa] shaft with an outside diameter of 80 mm, a length of L1 = 780 mm, and an allowable shear stress of 65 MPa. Segment (2) is a solid aluminum alloy [G = 26 GPa ksi] shaft with an outside diameter of 95 mm, a length of L2 = 620 mm, and an allowable shear stress of 40 MPa. The maximum rotation angle at the free end of the compound shaft must be limited to A ≤ 2.5°. Determine the magnitude of the largest torque TA that may be applied at A. FIGURE P6.6/7

Solution From equilibrium, the internal torques in segments (1) and (2) are equal to the external torque TA. The elastic torsion formula gives the relationship between shear stress and torque in a shaft. Tc = J Consider shear stress: In this compound shaft, the diameters and allowable shear stresses in segments (1) and (2) are known. The elastic torsion formula can be rearranged to solve for the unknown torque. An expression can be written for each shaft segment: J  J T1 = 1 1 T2 = 2 2 c1 c2 For shaft segment (1), the polar moment of inertia is:

J1 =



(80 mm ) = 4,021,239 mm4 4

32 Use this value along with the 65 MPa allowable shear stress to determine the allowable torque T1: 2 4  1 J1 65 N/mm 4,021,239 mm T1  = = 6,534,513 N  mm c1 80 mm / 2

(

)(

)

(a)

For shaft segment (2), the polar moment of inertia is:

J2 =



( 95 mm ) = 7,996,396 mm4 4

32 Use this value along with the 40 MPa allowable shear stress to determine the allowable torque T2: 2 4  2 J 2 ( 40 N/mm )( 7,996,396 mm ) T2  = = 6,733,808 N  mm (b) c2 95 mm / 2 Consider angle of twist: The angles of twists in segments (1) and (2) can be expressed as: TL TL 1 = 1 1 2 = 2 2 J1G1 J 2G2 The rotation angle at A is the sum of these two angles of twist: TL TL A = 1 + 2 = 1 1 + 2 2 J1G1 J 2G2 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

and since T1 = T2 = TA:  L L   A = TA  1 + 2   J1G1 J 2G2  Solving for TA gives: A TA  L1 L + 2 J1G1 J 2G2 

( 2.5 ) 

 rad 

  180 

780 mm 620 mm + 4 2 ( 4,021, 239 mm )( 45,000 N/mm ) ( 7,996,396 mm 4 )( 26,000 N/mm2 )

 5,983, 254 N  mm

(c)

Compare the torque magnitudes in Eqs. (a), (b), and (c). The smallest torque controls; therefore, the maximum torque that can be applied to the compound shaft at A is Ans. TA = 5,983,254 N  mm = 5,980 N  m

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P6.10 The mechanism shown in Figure P6.10 is in equilibrium for an applied load of P = 20 kN. Specifications for the mechanism limit the shear stress in the steel [G = 80 GPa] shaft BC to 70 MPa, the shear stress in bolt A to 100 MPa, and the vertical deflection of joint D to a maximum value of 25 mm. Assume that the bearings allow the shaft to rotate freely. Using L = 1,200 mm, a = 110 mm, and b = 210 mm, calculate: (a) the minimum diameter required for shaft BC. (b) the minimum diameter required for bolt A.

FIGURE P6.10

Solution (a) Minimum diameter required for shaft BC. The torque applied to the shaft is T = Pb = (20,000 N)(210 mm) = 4.200  106 N  mm If the shear stress in the shaft is limited to 70 MPa, the minimum diameter required for the shaft is:  3 T 4.200  106 N  mm d  = = 60,000 mm3 16  70 N/mm2  d  67.356 mm If the vertical deflection of joint D is not to exceed 25 mm, then the rotation angle at C must not exceed 25 mm sin C  = 0.119048 C  0.1193306 rad 210 mm The rotation angle at C is equal to the angle of twist in the shaft; therefore, TL C =  0.1193306 rad JG The minimum polar moment of inertia required to satisfy this angular limit is TL (4.200  106 N  mm)(1,200 mm) J = = 527,944.944 mm4 2 G (0.1193306 rad)(80,000 N/mm ) Thus, the minimum required diameter to satisfy the deflection limit at D is



d 4  527,944.944 mm 4  d  48.156 mm

The minimum diameter required for the shaft BC is d = 67.4 mm

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

(b) Minimum diameter required for bolt A. Since the torque in the shaft is 4.200×106 N·mm, the force that acts on the bolt at A is 4.200  106 N  mm VA = = 38,181.818 N 110 mm The bolt acts in single shear. The area required to keep the average shear stress in the bolt to a value less than 100 MPa is 38,181.818 N Abolt  = 381.818 mm 2 2 100 N/mm Consequently, the bolt must have a minimum diameter of d  22.0 mm

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P6.11 A simple torsion-bar spring is shown in Figure P6.11. The shear stress in the steel [G = 11,500 ksi] shaft is not to exceed 10,000 psi, and the vertical deflection of joint D is not to exceed 0.5 in. when a load of P = 3,400 lb is applied. Neglect the bending of the shaft and assume that the bearing at C allows the shaft to rotate freely. Determine the minimum diameter required for the shaft. Use dimensions of a = 72 in., b = 30 in., and c = 18 in. FIGURE P6.11

Solution The torque applied to the shaft is

T = Pc = (3,400 lb)(18 in.) = 61,200 lb  in.

If the shear stress in the shaft is limited to 10,000 psi, the minimum diameter required for the shaft is:  3 T 61,200 lb  in. d  = = 6.120 in.3 2 16  10,000 lb/in.  d  3.147 in. If the vertical deflection of joint D is not to exceed 0.5 in., then the rotation angle at B must not exceed 0.5 in. sin B  = 0.027778 B  0.027781 rad 18 in. The rotation angle at B is equal to the angle of twist in the shaft; therefore, TL B =  0.027781 rad JG The minimum polar moment of inertia required to satisfy this angular limit is ( 61, 200 lb  in.)( 72 in.) TL J = = 13.792174 in.4 2  G ( 0.027781 rad ) (11,500,000 lb/in. ) Thus, the minimum required diameter to satisfy the deflection limit at D is



d 4  13.792174 in.4  d  3.443 in.

The minimum diameter required for the shaft is d = 3.44 in.

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P6.12 A rod specimen of ductile cast iron was tested in a torsion-testing machine. The rod diameter was 22 mm, and the rod length was 300 mm. When the applied torque reached 271.4 N·m, a shear strain of 2,015 microradians was measured in the specimen. What was the angle of twist in the specimen?

Solution The polar moment of inertia for the specimen is



( 22 mm ) = 22,998 mm4 4

The shear stress in the specimen at the specified torque is: Tc ( 271.4 N  m )( 22 mm / 2 )(1,000 mm/m ) = = = 129.81 MPa J 22,998 mm4 From Hooke’s law, calculate the shear modulus from the specified shear strain and the calculated shear stress:  129.81 N/mm2  = G G = = = 64,422.4 N/mm2 −6  2,015  10 rad The magnitude of the angle of twist in a 300 mm length of the specimen is TL ( 271.4 N  m )( 300 mm )(1,000 mm/m ) = = = 0.054955 rad = 0.0550 rad = 3.15 JG ( 22,998 mm4 )( 64, 422.4 N/mm2 )

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P6.13 A two-segment shaft is used to transmit power at constant speed through the pulleys shown in Figure P6.13. Power is input to the shaft at B through a torque of TB = 260 N·m. A torque of TA = 90 N·m is removed from the shaft at A, and a torque of TC = 170 N·m is removed from the shaft at C. The external torques act in the directions indicated in the figure. Both shaft segments are made of phosphor bronze [G = 42 GPa] and both segments are solid shafts. Segment (1) has a diameter of d1 = 25 mm and a length of L1 = 300 mm. Segment (2) has a diameter of d2 = 30 mm and a length of L2 = 900 mm. Determine the rotation angle of pulley C relative to pulley A. FIGURE P6.13

Solution Equilibrium

M x = TA + T1 = 0

 T1 = −TA = −90 N  m

M x = TC − T2 = 0

 T2 = TC = 170 N  m

Polar moments of inertia in the shaft segments will be needed for this calculation. Segments (1) and (2) are solid shafts. The polar moments of inertia for segments (1) and (2) are:

J1 = J2 =



d14 = d 24 =



( 25 mm ) = 38,349.52 mm4 4

( 30 mm ) = 79,521.56 mm4 4

Angles of twist: T L ( −90 N  m )( 300 mm )(1,000 mm/m ) 1 = 1 1 = = −0.016763 rad J1G1 38,349.52 mm4 42,000 N/mm2

(

2 =

)(

)

T2 L2 (170 N  m )( 900 mm )(1,000 mm/m ) = = 0.045810 rad J 2G2 ( 79,521.56 mm4 )( 42,000 N/mm2 )

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Rotation angles: The angles of twist can be defined in terms of the rotation angles at the ends of each segment: 1 = B −  A 2 = C − B The origin of the coordinate system is located at pulley A. We will arbitrarily define the rotation angle at pulley A to be zero (A = 0). The rotation angle at B can be calculated from the angle of twist in segment (1): 1 = B − A

B = A + 1 = 0 − 0.016763 rad = −0.016763 rad Similarly, the rotation angle at C is determined from the angle of twist in segment (2) and the rotation angle of pulley B: 2 = C − B C = B + 2 = −0.016763 rad + 0.045810 rad = 0.029047 rad Rotation angle of pulley C with respect to pulley A: Using the rotation angles determined for the system, the rotation angle of pulley C with respect to pulley A is simply: Ans. C = 0.029047 rad = 0.0290 rad

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P6.14 A compound shaft drives three gears, as shown in Figure P6.14. Segments (1) and (2) of the compound shaft are hollow bronze [G = 6,500 ksi] tubes, which have an outside diameter of 1.75 in. and a wall thickness of 0.1875 in. Segments (3) and (4) are solid 1.00-in.-diameter steel [G = 11,500 ksi] shafts. The shaft lengths are L1 = 60 in., L2 = 14 in., L3 = 20 in., and L4 = 26 in. The torques applied to the shafts have magnitudes of TB = 960 lb·ft, TD = 450 lb·ft, and TE = 130 lb·ft, acting in the directions shown. The bearings shown allow the shaft to turn freely. Calculate: (a) the maximum shear stress in the compound shaft. (b) the rotation angle of flange C with respect to flange A. (c) the rotation angle of gear E with respect to flange A.

FIGURE P6.14

Solution Equilibrium

M x = −TE − T4 = 0 T4 = −TE = −130 lb  ft M x = TD − TE − T3 = 0  T3 = TD − TE = 450 lb  ft − 130 lb  ft = 320 lb  ft Note: The internal torque in shaft segment (2) is the same as in segment (3); therefore, T2 = 320 lb·ft. M x = −TB + TD − TE − T1 = 0  T1 = −TB + TD − TE = −960 lb  ft + 450 lb  ft − 130 lb  ft = −640 lb  ft Polar moments of inertia in the shaft segments will be needed for this calculation. Segments (1) and (2) are hollow tubes with an outside diameter of 1.75 in. and an inside diameter of 1.75 in. – 2(0.1875 in.) = 1.375 in. The polar moment of inertia of segments (1) and (2) is:

J1 =



 4 4  D14 − d14  = (1.75 in.) − (1.375 in.)  = 0.569850 in.4 = J 2   32 32

Segments (3) and (4) are solid 1.00 in. diameter shafts, which have a polar moment of inertia of:

Mechanics of Materials: An Integrated Learning System, 4th Ed.

J3 =



d34 =



Timothy A. Philpot

(1.00 in.) = 0.098175 in.4 = J 4 4

32 32 (a) Shear stresses: The shear stresses in each segment can be calculated by the elastic torsion formula: T c ( −640 lb  ft )(1.75 in. / 2 )(12 in./ft ) 1 = 1 1 = = −11,793 psi J1 0.569850 in.4 T c ( 320 lb  ft )(1.75 in. / 2 )(12 in./ft ) 2 = 2 2 = = 5,896 psi J2 0.569850 in.4 T c ( 320 lb  ft )(1.00 in. / 2 )(12 in./ft ) 3 = 3 3 = = 19,557 psi J3 0.098175 in.4 T c ( −130 lb  ft )(1.00 in. / 2 )(12 in./ft ) 4 = 4 4 = = −7,945 psi J4 0.098175 in.4 The maximum shear stress in the compound shaft is  max =  3 = 19,560 psi

Ans.

Angles of twist: ( −640 lb  ft )( 60 in.)(12 in./ft ) = −0.124405 rad TL 1 = 1 1 = J1G1 ( 0.569850 in.4 )( 6,500,000 lb/in.2 )

2 =

( 320 lb  ft )(14 in.)(12 in./ft ) = 0.014514 rad T2 L2 = J 2G2 ( 0.569850 in.4 )( 6,500,000 lb/in.2 )

3 =

( 320 lb  ft )( 20 in.)(12 in./ft ) = 0.068024 rad T3 L3 = J 3G3 ( 0.098175 in.4 )(11,500,000 lb/in.2 )

4 =

( −130 lb  ft )( 26 in.)(12 in./ft ) = −0.035925 rad T4 L4 = J 4G4 ( 0.098175 in.4 )(11,500,000 lb/in.2 )

Rotation angles: The angles of twist can be defined in terms of the rotation angles at the ends of each segment: 1 = B −  A 2 = C − B 3 = D − C 4 = E − D The origin of the coordinate system is located at flange A. We will arbitrarily define the rotation angle at flange A to be zero (A = 0). The rotation angle at B can be calculated from the angle of twist in segment (1): 1 = B −  A

B =  A + 1 = 0 + ( −0.124405 rad ) = −0.124405 rad

Similarly, the rotation angle at C is determined from the angle of twist in segment (2) and the rotation angle of gear B: 2 = C − B

C = B + 2 = −0.124405 rad + 0.014514 rad = −0.109891 rad

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

The rotation angle at D is: 3 = D − C

D = C + 3 = −0.109891 rad + 0.068024 rad = −0.041867 rad and the rotation angle at E is: 4 = E − D

E = D + 4 = −0.041867 rad + ( −0.035925 rad ) = −0.077792 rad

(b) Rotation angle of flange C with respect to flange A: Using the rotation angles determined for the system, the rotation angle of flange C with respect to flange A is simply: Ans. C = −0.109891 rad = −0.1099 rad (c) Rotation angle of gear E with respect to flange A: Using the rotation angles determined for the system, the rotation angle of gear E with respect to flange A is simply: Ans. E = −0.077792 rad = −0.0778 rad

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P6.15 Figure P6.15 shows a cutaway view of an assembly in which a solid bronze [G = 6,500 ksi] rod (1) is fitted inside of an aluminum alloy [G = 4,000 ksi] tube (2). The tube is attached to a fixed plate at C, and both the rod and the tube are welded to a rigid end plate at A. The rod diameter is d1 = 1.75 in. The outside diameter of the tube is D2 = 3.00 in. and its wall thickness is t2 = 0.125 in. Overall dimensions of the assembly are a = 40 in. and b = 8 in. The allowable shear stresses for the bronze and aluminum materials are 10,000 psi and 8,000 psi, respectively. Further, the rotation angle of end B must be limited to a magnitude of 5°. Based on these constraints, what is the maximum torque TB that can be applied at end B? FIGURE P6.15

Solution Equilibrium: The torque T applied to the end of the rod at B creates a torque in rod (1) of T1 = TB (a)

In the second FBD (of the end plate), equilibrium requires that M x = T1 + T2 = 0

T2 = −T1 = −TB

(b)

Determine largest torque based on allowable shear stress: The polar moment of inertia for rod (1) is:

J1 =



(1.75 in.) = 0.920772 in.4 4

32 The largest torque that can be applied to the rod can be found from the specified allowable stress: 2 4  1,allow J1 (10,000 lb/in. )( 0.920772 in. ) T1  = = 10,523.1 lb  in. c1 1.75 in. / 2 From Eq. (a), we find that the magnitude of the largest torque that can be applied at B is TB  10,523.1 lb  in. (c)

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

The polar moment of inertia for the tube is

J2 =

 

3.00 in.) − ( 2.75 in.)  = 2.33740 in.4 (  4

32 The largest torque that can be applied to the tube can be found from: 2 4  2,allow J 2 ( 8,000 lb/in. )( 2.33740 in. ) T2 = = = 12,466.2 lb  in. c2 3.00 in. / 2 From Eq. (b), we find that the magnitude of the largest torque that can be applied at B is TB  12, 466.2 lb  in. (d) Determine largest torque based on specified rotation angle at B: The angles of twist can be defined in terms of the rotation angles at the ends of each segment: 1 = B −  A 2 = C −  A Locate origin of the coordinate system is located at fixed plate C, where the rotation angle is zero (C = 0). The rotation angle at A is determined from the angle of twist in tube (2): 2 = C −  A  A = −2 The rotation angle at B can be calculated from the angle of twist in rod (1) and the rotation of end plate A: 1 = B −  A B = 1 +  A = 1 − 2 Substitute torque-twist relationships for rod (1) and tube (2) into this expression to obtain: TL T L (e) B = 1 − 2 = 1 1 − 2 2 J1G1 J 2G2 Substitute Eqs. (a) and (b) into Eq. (e) and set this equal to the rotation angle limit specified for end B:  L TL  TL  L  B = B 1 −  − B 2  = TB  1 + 2   5 J1G1  J 2G2   J1G1 J 2G2  Solve for the largest torque that can be applied at B based on the specified rotation angle for B:  rad  ( 5 )    180  TB  = 7,095.8 lb  in. (f) 48 in. 40 in. + ( 0.920772 in.4 )( 6.5  106 lb/in.2 ) ( 2.33740 in.4 )( 4.0 106 lb/in.2 ) Maximum torque TB: Compare the results in Eqs. (c), (d), and (f) to find that the largest torque that can be applied at B is: Ans. TB  7,095.8 lb  in. = 591.32 lb  ft = 591 lb  ft

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P6.16 A compound shaft (Figure P6.16) consists of a titanium alloy [G = 6,200 ksi] tube (1) and a solid stainless steel [G = 11,500 ksi] shaft (2). Tube (1) has a length of L1 = 40 in., an outside diameter of D1 = 1.75 in., and a wall thickness of t1 = 0.125 in. Shaft (2) has a length of L2 = 50 in. and a diameter of d2 = 1.25 in. If an external torque of TB = 580 lb·ft acts at pulley B in the direction shown, calculate the torque TC required at pulley C so that the rotation angle of pulley C relative to A is zero.

FIGURE P6.16

Solution Equilibrium: From the two FBD’s shown, derive the following equilibrium equations: M x = TC − TB − T1 = 0

 T1 = TC − TB M x = TC − T2 = 0  T2 = TC

Section properties: The polar moments of inertia for the shaft segments will be needed for this calculation.

J1 = J2 =

 



1.75 in.) − (1.50 in.)  = 0.423762 in.4 (  4

(1.25 in.) = 0.239684 in.4 4

Geometry-of-deformation relationship: From the problem statement, the rotation angle of pulley C relative to A must equal zero. The rotation angle C is simply the sum of the two angles of twist. Therefore: C = 1 +  2 = 0 Torque-twist relationships: The torque-twist relationship for each shaft can be written as:

1 =

T1L1 J1G1

2 =

T2 L2 J 2G2

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Compatibility equation: Substitute the torque-twist relationships into the geometry relationship to derive: T1L1 T2 L2 + =0 J1G1 J 2G2 Solve: Substitute the equilibrium equations into the compatibility equation. (TC − TB ) L1 TC L2 + =0 J1G1 J 2G2 and solve for TC:  L L  T L TC  1 + 2  = B 1  J1G1 J 2G2  J1G1

 TC = TB

L1 J1G1 L1 L + 2 J1G1 J 2G2

= TB

L1  J G  L1 + L2  1   1   J 2   G2 

Calculate the torque required at C: TC = ( 580 lb  ft )

40 in.  0.423762 in.4   6, 200 ksi  40 in. + ( 50 in.)   4   0.239684 in.   11,500 ksi 

= ( 580 lb  ft )( 0.456313) = 264.662 lb  ft = 265 lb  ft

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P6.17 The copper pipe shown in Figure P6.17 has an outside diameter of 70 mm and a wall thickness of 5 mm. The pipe is subjected to a uniformly distributed torque of t = 500 N·m/m along its entire length. Using a = 1.0 m, b = 1.6 m, and c = 3.2 m, calculate: (a) the shear stress at A on the outer surface of the pipe. (b) the shear stress at B on the outer surface of the pipe. FIGURE P6.17

Solution Equilibrium: At A, the torque in the shaft is:

M x = t (b + c ) − T = 0

T = t (b + c )

and the torque at B in the shaft is:

M x = t ( c ) − T = 0

T = t ( c )

Section Properties: The polar moment of inertia for the hollow pipe is d = 70 mm − 2 ( 5 mm ) = 60 mm J=

 

( 32 

70 mm ) − ( 60 mm )  = 1,084,831 mm4  4

(a) Shear stress at A: Tc ( 500 N  m/m )(1.6 m + 3.2 m )( 70 mm / 2 )(1,000 mm/m ) = = J 1,084,831 mm 4

= 77.431 MPa = 77.4 MPa

Ans.

(b) Shear stress at B: Tc ( 500 N  m/m )( 3.2 m )( 70 mm / 2 )(1,000 mm/m ) = = J 1,084,831 mm 4

= 51.621 MPa = 51.6 MPa

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P6.18 The gear train system shown in Figure P6.18/19 includes shafts (1) and (2), which are solid 15 mm diameter steel shafts. The allowable shear stress of each shaft is 85 MPa. The diameter of gear B is DB = 200 mm, and the diameter of gear C is DC = 150 mm. The bearings shown allow the shafts to rotate freely. Determine the maximum torque TD that can be applied to the system without exceeding the allowable shear stress in either shaft. FIGURE P6.18/19

Solution Section properties:

J1 =

 4 (15 mm ) = 4,970.10 mm4 = J 2 32

Maximum torque in either shaft: 2 4  allow J1 ( 85 N/mm )( 4,970.10 mm ) Tmax = = = 56,327.8 N  mm c1 15 mm / 2 Torque relationship: T1 T D 200 mm = 2 or T1 = T2 B = T2 = 1.333T2 DB DC DC 150 mm

T1 controls

The torque in shaft (1) controls; therefore, T1 = 56,327.8 N·mm. Consequently, the maximum torque in shaft (2) must be limited to: D 150 mm Ans. T2 = T1 C = T1 = 0.75 ( 56,327.8 N  mm ) = 42, 245.8 N  mm = 42.2 N  m DB 200 mm

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P6.19 The gear train system shown in Figure P6.18/19 includes shafts (1) and (2), which are solid 0.75 in. diameter steel shafts. The diameter of gear B is DB = 8 in., and the diameter of gear C is DC = 5 in. The bearings shown allow the shafts to rotate freely. An external torque of TD = 45 lb·ft is applied at gear D. Determine the maximum shear stress produced in shafts (1) and (2). FIGURE P6.18/19

Solution Torque relationship: T1 T D 8 in. = 2 or T1 = T2 B = T2 = 1.6 T2 DB DC DC 5 in. Thus, T2 = TD = 45 lb  ft

T1 = 1.6T2 = 1.6 ( 45 lb  ft ) = 72 lb  ft Note: Absolute values used for torques since only stress magnitudes are asked for here. (b) Shear stress magnitudes for solid 0.75 in. diameter shafts:

 4  4 d = ( 0.75 in.) = 0.0310631 in.4 32 32 Tc J1

1 = 1 1 =

( 72 lb  ft )( 0.75 in. / 2 )(12 in./ft ) = 10, 430 psi

0.0310631 in.4 T c ( 45 lb  ft )( 0.75 in. / 2 )(12 in./ft ) 2 = 2 2 = = 6,520 psi J2 0.0310631 in.4

Ans. Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P6.20 In the system shown in Figure P6.20, the motor applies a torque of TA = 40 N·m to the pulley at A. Through a sequence of pulleys and belts, this torque is amplified to drive a gear at E. The pulleys have diameters of DA = 50 mm, DB = 150 mm, DC = 50 mm, and DD = 250 mm. (a) Calculate the torque TE that is produced at gear E. (b) Shaft (2) is to be a solid shaft, and the maximum shear stress must be limited to 60 MPa. What is the minimum diameter that may be used for shaft (2)? FIGURE P6.20

Solution (a) Calculate torque TE: The torque on pulley B is: D 150 mm TB = TA B = TA = ( 40 N  m )( 3) = 120 N  m DA 50 mm Shaft (1) has the same torque as pulley B; therefore, T1 = 120 N·m and TC = 120 N·m. The torque on pulley D is found from: D 250 mm TD = TC D = TC = (120 N  m )( 5) = 600 N  m DC 50 mm Shaft (2) has the same torque as pulley D; therefore, T2 = 600 N·m and Ans. TE = 600 N  m To summarize: T1 = 120 N  m

T2 = 600 N  m

(b) Determine minimum diameter for shaft (2): The elastic torsion formula can be rearranged and rewritten in terms of the unknown diameter d:



Tc = J

d4 J 32  T  = = d3 = c d / 2 16 

Using this expression, solve for the minimum acceptable diameter of shaft (2) for an allowable shear stress of 60 MPa: 16 ( 600 N  m )(1,000 mm/m ) 16 T2 d 23  = = 50,929.582 mm3 2   allow  60 N/mm

(

 d 2  37.067 mm = 37.1 mm

)

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed. P6.21 A motor provides a torque of 1,500 lb·ft to gear B of the system shown in Figure P6.21. Gear A takes off 900 lb·ft from shaft (1) and gear C takes off the remaining torque. Both shafts (1) and (2) are solid and made of steel [G = 11,500 ksi]. The shaft lengths are L1 = 15 ft and L2 = 9 ft, respectively. If the angle of twist in each shaft must not exceed 3.0°, calculate the minimum diameter required for each shaft.

Timothy A. Philpot

FIGURE P6.21

Solution The torque TA is 900 lb·ft. If the angle of twist in shaft (1) must be limited to 3.0°, then the minimum shaft diameter can be calculated as follows: 2 900 lb  ft )(15 ft )(12 in./ft ) ( T1 L1  4 J1 = = = 3.228490 in.4  d1 6 2 1G1 ( 3.0 )( rad /180 ) (11.5 10 lb/in. ) 32  d1  2.39 in.

Ans.

The torque remaining in the shaft at C is TC = 600 lb·ft. If the angle of twist in shaft (2) must be limited to 3.0°, then the minimum shaft diameter can be calculated as follows: 2 600 lb  ft )( 9 ft )(12 in./ft ) ( T2 L2  4 J2 = = = 1.291400 in.4  d2 −6 2 2G2 ( 3.0 )( rad /180 ) (11.5 10 lb/in. ) 32  d 2  1.904 in.

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed. P6.22 Two solid steel shafts are connected by the gears shown in Figure P6.22/23. The design requirements for the system require (1) that both shafts must have the same diameter, (2) that the maximum shear stress in each shaft must be less than 10,000 psi, and (3) that the rotation angle of gear D must not exceed 8°. Determine the minimum required diameter of the shafts if the torque applied at gear D is TD = 350 lb·ft. The shaft lengths are L1 = 78 in. and L2 = 60 in. The number of teeth on gears B and C are NB = 90 and NC = 52, respectively. Assume that the shear modulus of both shafts is G = 11,500 ksi and that the bearings shown allow free rotation of the shafts.

Timothy A. Philpot

FIGURE P6.22/23

Solution Torque relationships: The torque on gear D creates a torque in shaft (2) of: M x = TD + T2 = 0 T2 = −TD = −350 lb  ft = −4, 200 lb  in.

(a)

An FBD of gear C is shown to the right. The teeth of gear B exert a force F on gear C. Summing moments about the x′ axis gives: M x = −T2 − F  RC = 0

where RC denotes the radius of gear C. Let’s solve this equation to derive an expression for force F: T (b) F =− 2 RC An FBD of gear B is shown to the right. The teeth of gear C exert an equal magnitude force F on gear B, acting in the opposite direction. Summing moments about the x axis gives: (c) M x = −T1 − F  RB = 0 where RB denotes the radius of gear B.

Substitute Eq. (b) into Eq. (c) to relate torque T1 to torque T2:  T  R  T1 = − F  RB = −  − 2  RB = T2  B   RC   RC  The ratio of gear diameters is the same as the ratio of the number of teeth on the gears:

(d)

Mechanics of Materials: An Integrated Learning System, 4th Ed. RB N B = RC NC Substitute Eqs. (a) and (e) into Eq. (d) to determine the torque in shaft (1):

N   90 teeth  T1 = T2  B  = ( −4, 200 lb  in.)   = −7, 269.231 lb  in.  52 teeth   NC 

Timothy A. Philpot (e)

(f)

Shaft diameter based on allowable shear stresses: The elastic torsion formula gives the relationship between shear stress and torque in a shaft. Tc = J The torques in both shafts have been determined, and the allowable shear stress is specified. Rearrange the elastic torsion formula, putting the known terms on the right-hand side of the equation: J T = c  Express the left-hand side of this equation in terms of the shaft diameter D:



d4  T 32 = d3 = d / 2 16  Using the magnitude of the torque computed in Eq. (f), solve for the minimum acceptable diameter in shaft (1): 16 T1 16 ( 7, 269.231 lb  in.) d13  = = 3.702189 in.3 2    10, 000 lb/in.

(

)

 d1  1.547 in.

(g)

Using the magnitude of the torque computed in Eq. (a), solve for the minimum acceptable diameter in shaft (2): 16 T2 16 ( 4, 200 lb  in.) d 23  = = 2.139043 in.3 2    10, 000 lb/in.

(

)

 d 2  1.288 in.

(h)

Of these two values, d1 controls. Therefore, both shafts could have a diameter of 1.547 in. or more and the shear stress constraint would be satisfied. The angles of twists in shafts (1) and (2) can be expressed as: TL TL 1 = 1 1 2 = 2 2 J1G1 J 2G2 The rotation angles of gears B and C are related since the arclengths turned by the two gears have the same magnitude. The gears turn in opposite directions; therefore, a negative sign is introduced. RCC = − RBB

The angle of twist in shaft (1) can be expressed as the difference between the rotation angle at B (at the +x end of the shaft) and the rotation angle at A (at the –x end of the shaft): 1 = B −  A

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Since the rotation angle of flange A is zero, we can state that the rotation angle of gear B is B = 1. Consequently, the rotation angle of gear C can be expressed in terms of T1 as: R R R TL N TL C = − B B = − B 1 = − B 1 1 = − B 1 1 RC RC RC J1G1 NC J1G1 The angle of twist in shaft (2) can be expressed as the difference between the rotation angle at C (at the +x′ end of the shaft) and the rotation angle at D (at the –x′ end of the shaft): 2 = C − D

Thus, the rotation angle of gear D is: D = C − 2

and so the rotation angle of gear D can be expressed in terms of the torques T1 and T2: N TL T L D = − B 1 1 − 2 2  8 NC J1G1 J 2G2 Shafts having the same diameters and the same shear moduli are required for this system; therefore, J1 = J2 = J and G1 = G2 = G. Factor these terms out to obtain:   1  NB 1  NB T1 L1 − T2 L2   8  J  T1L1 − T2 L2  − − JG  NC G ( 8 )  NC   Express the polar moment of inertia in terms of diameter to obtain the following relationship:  32  N B d4  T1 L1 − T2 L2  −  G ( 8 )  NC  32  90 teeth  = − −7, 269.231 lb  in.)( 78 in.) − ( −4, 200 lb  in.)( 60 in.)  (  6 2   11.5 10 lb/in. (8 )  rad180  52 teeth

(

)

(

)

= 7.823839 in.4  d  1.672 in. Since this minimum diameter is greater than the diameter needed to satisfy the shear stress requirements [Eqs. (g) and (h)], the rotation angle constraint controls. Therefore, the minimum shaft diameter that satisfies all requirements is

d  1.672 in.

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P6.23 Two solid 120 mm diameter steel shafts are connected by the gears shown in Figure P6.22/23. The shaft lengths are L1 = 4 m and L2 = 3 m. The number of teeth on gears B and C are NB = 200 and NC = 115, respectively. Assume that the shear modulus of both shafts is G = 80 MPa and that the bearings shown allow free rotation of the shafts. If the torque applied at gear D is TD = 6,000 N·m, determine: (a) the internal torques T1 and T2 in the two shafts. (b) the angles of twist 1 and 2. (c) the rotation angles B and C of gears B and C. (d) the rotation angle of gear D. FIGURE P6.22/23

Solution

Torque relationships: The torque on gear D creates a torque in shaft (2) of: M x = TD + T2 = 0

T2 = −TD = −6,000 N  m

(a)

An FBD of gear C is shown to the right. The teeth of gear B exert a force F on gear C. Summing moments about the x′ axis gives: M x = −T2 − F  RC = 0

Substitute Eq. (b) into Eq. (c) to relate torque T1 to torque T2:

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

 T  R  T1 = − F  RB = −  − 2  RB = T2  B   RC   RC  The ratio of gear diameters is the same as the ratio of the number of teeth on the gears: RB N B = RC NC Substitute Eqs. (a) and (e) into Eq. (d) to determine the torque in shaft (1):

N   200 teeth  T1 = T2  B  = ( −6,000 N  m )   = −10,434.783 N  m N 115 teeth    C

(d)

(e)

(f)

(a) Internal torques T1 and T2 in the two shafts: From Eqs. (f) and (a):

T1 = −10,430 N  m

T2 = −6,000 N  m

Ans.

Section properties:

J1 =

 4 (120 mm ) = 20,357,520 mm4 = J 2 32

(b) Angles of twist in shafts (1) and (2):

T L ( −10,434.783 N  m )( 4 m )(1,000 mm/m ) 1 = 1 1 = = −0.025629 rad = −0.0256 rad J1G1 ( 20,357,520 mm4 )(80,000 N/mm2 )

Ans.

( −6,000 N  m )( 3 m )(1,000 mm/m ) = −0.011052 rad = −0.01105 rad TL 2 = 2 2 = J 2G2 ( 20,357,520 mm4 )(80,000 N/mm2 )

Ans.

(c) Rotation angles of gears B and C: The rotation angles of gears B and C are related since the arclengths turned by the two gears have the same magnitude. The gears turn in opposite directions; therefore, a negative sign is introduced. RCC = − RBB

Since the rotation angle of flange A is zero, we can state that the rotation angle of gear B is: B = 1 = −0.0256 rad

Ans.

The rotation angle of gear C can be expressed as: R N 200 teeth C = − B B = − B B = − ( −0.025629 rad ) RC NC 115 teeth

= 0.044572 rad = 0.0446 rad

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

(d) Rotation angle of gear D: The angle of twist in shaft (2) can be expressed as the difference between the rotation angle at C (at the +x′ end of the shaft) and the rotation angle at D (at the –x′ end of the shaft): 2 = C − D

Thus, the rotation angle of gear D is:

D = C − 2 = 0.044572 rad − ( −0.011052 rad ) = 0.055624 rad = 0.0556 rad

Ans.

P6.24 A solid 40 mm diameter shaft is rotating at a speed of 560 rpm with a maximum shear stress of 85 MPa. Determine the power in kW being delivered by the shaft.

Solution The torque in the shaft is: J=



J c

d4 =



( 40 mm ) = 251,327 mm4 4

32 (85 N/mm2 )( 251,327 mm4 )

40 mm / 2

= 1,068,142 N  mm = 1,068.142 N  m

The power being delivered by the shaft is:  1 min  P = T  = (1,068.142 N  m )( 560 rev/min )( 2 rad/rev )    60 s  Nm = 62,639 = 62,639 W = 62.6 kW s

Ans.

P6.25 A solid 1.50 in. diameter shaft is used to synchronize multiple mechanisms in a manufacturing plant. The torque delivered by the shaft is 3,100 lb·in. What horsepower must be delivered to the shaft to maintain a shaft speed of 600 rpm? What is the maximum shear stress developed in the shaft?

Solution Horsepower to maintain a shaft speed of 600 rpm:

 1 min  1 ft  P = T  = ( 3,100 lb  in.)( 600 rev/min )( 2 rad/rev )     60 s  12 in.     lb  ft  1 hp   = 16, 231.562  = 29.512 hp = 29.5 hp  s   550 lb  ft   s  

Ans.

Section properties: The polar moment of inertia of the shaft is:  4 J = (1.5 in.) = 0.497010 in.4 32 Maximum shear stress in the shaft: Tc ( 3,100 lb  in.)(1.50 in. / 2 ) = = = 6,677.977 psi = 6,680 psi J 0.497010 in.4

Ans.

P6.26 A solid steel [G = 11,500 ksi] shaft that has a diameter of 2.00 in. and a length of 30 ft transmits 14 hp while rotating at 90 rpm. What is the angle of twist in the shaft?

Solution The torque in the solid steel shaft is:  550 lb  ft/s  (14 hp )   P  1 hp  T= = = 816.995 lb  ft = 9,803.994 lb  in.   90 rev  1 min  2 rad       min  60 s  1 rev  The polar moment of inertia of the shaft is:  4 J = ( 2.00 in.) = 1.570796 in.4 32 The angle of twist in the shaft is T L ( 9,803.994 lb  in.)( 30 ft )(12 in./ft ) = = = 0.1954 rad JG (1.570796 in.4 )(11.5  106 lb/in.2 )

Ans.

P6.27 A solid stainless steel [G = 86 GPa] with a diameter of 30 mm and length of 900 mm transmits 42 kW from an electric motor to a compressor. If the allowable shear stress is 50 MPa and the allowable angle of twist is 1.5°, what is the minimum allowable speed of rotation?

Solution The polar moment of inertia of the shaft is:  4 J = ( 30 mm ) = 79,521.56 mm4 32 Torque based on shear stress: The maximum torque that can be transmitted by this shaft based on the allowable shear stress is: 2 4  J ( 50 N/mm )( 79,521.56 mm ) T = = 265,072 N  mm c 30 mm / 2 Torque based on angle of twist: The maximum torque that can be transmitted by this shaft based on the allowable angle of twist is:   rad  4 2 1.5 )  (  ( 79,521.56 mm )( 86,000 N/mm )  JG 180    T = = 198,934 N  mm L 900 mm Allowable torque: Based on these two constraints, the maximum torque that the shaft can transmit is T = 198,934 N  mm = 198.934 N  m Slowest allowable speed of rotation: From the power transmission equation, calculate: P 42,000 N  m/s  = = 211.125 rad/s T 198.934 N  m or  1 rev   = ( 211.125 rad/s )   = 33.602 rev/s = 33.6 Hz  2 rad 

Ans.

P6.28 A hollow aluminum alloy [G = 3,800 ksi] shaft having a length of 12 ft, an outside diameter of 4.50 in., and a wall thickness of 0.50 in. rotates at 3 Hz. The allowable shear stress is 6 ksi, and the allowable angle of twist is 5°. What horsepower may the shaft transmit?

Solution Section properties: The polar moment of inertia of the shaft is:   4 4 J= ( 4.50 in.) − ( 3.50 in.)  = 25.5254 in.4  32 Torque based on allowable shear stress: Compute the allowable torque if the shear stress must not exceed 6 ksi: 2 4  J ( 6 kips/in. )( 25.5254 in. ) (a) T = = 68.0678 kip  in. c 4.50 in. / 2 Torque based on angle of twist: Compute the allowable torque if the angle of twist must not exceed 5°:  rad ( 25.5254 in.4 )(3,800 kips/in.2 )  JG ( 5 ) 180  T = = 58.7816 kip  in. (b) L (12 ft )(12 in./ft )

(

)

Controlling torque: From comparison of Eqs. (a) and (b), the maximum torque allowed for this shaft is: Tmax = 58.7816 kip  in. = 4,898.47 lb  ft Power transmission at 3 Hz: The maximum power that can be transmitted at 3 Hz is:  3 rev  2 rad  P = T = ( 4,898.47 lb  ft )    = 92,333.9845 lb  ft/s  s  1 rev  or in units of horsepower, 92,333.9845 lb  ft/s P= = 167.9 hp 550 lb  ft/s 1 hp

Ans.

P6.29 The impeller shaft of a fluid agitator transmits 28 kW at 440 rpm. If the allowable shear stress in the impeller shaft must be limited to 80 MPa, determine: (a) the minimum diameter required for a solid impeller shaft. (b) the maximum inside diameter permitted for a hollow impeller shaft if the outside diameter is 40 mm. (c) the percent savings in weight realized if the hollow shaft is used instead of the solid shaft. (Hint: The weight of a shaft is proportional to its cross-sectional area.)

Solution Power transmission equation: The torque in the shaft is P 28,000 N-m/s T= = = 607.6825 N-m   440 rev  2 rad  1 min       min  1 rev  60 s  (a) Solid impeller shaft: The minimum diameter required for a solid shaft can be found from:  3 T (607.6825 N-m)(1,000 mm/m) d  = = 7,596.0313 mm3 16  allow 80 N/mm 2  d  33.8209 mm = 33.8 mm

Ans.

(b) Hollow driveshaft: From the elastic torsion formula: 4 4 Tc J   D − d  T   =  J c 32 D / 2  Solve this equation for the maximum inside diameter: 4 4  (40 mm) − d  (607.6825 N-m)(1,000 mm/m)  = 7,596.0313 mm3 2 32 40 mm / 2 80 N/mm 4 4 (40 mm) − d  1,547, 450.779 mm 4

d 4  2,560,000 mm 4 − 1,547, 450.779 mm 4 = 1,012,549.221 mm 4  d  31.7215 mm = 31.7 mm

Ans.

(c) Weight savings: The weights of the solid and hollow shafts are proportional to their respective crosssectional areas. The cross-sectional area of the solid shaft is  Asolid = (33.8209 mm)2 = 898.3803 mm 2 4 and the cross-sectional area of the hollow shaft is  Ahollow = (40 mm) 2 − (31.7215 mm) 2  = 466.3262 mm 2 4 The weight savings can be determined from ( A − Ahollow ) (898.3803 mm2 − 466.3262 mm2 ) weight savings (in percent) = solid = = 48.1% Ans. Asolid 898.3803 mm2

P6.30 A pulley with a diameter of D = 8 in. is mounted on a shaft with a diameter of d = 1.25 in. as shown in Figure P6.30. Around the pulley is a belt having tensions of F1 = 120 lb and F2 = 480 lb. If the shaft turns at 180 rpm, calculate: (a) the horsepower being transmitted by the shaft. (b) the maximum shear stress in the shaft. FIGURE P6.30

Solution Torque: From the belt tensions, calculate the torque exerted on the shaft:  8 in.  8 in. T = (480 lb)  − (120 lb)  = 1,440 lb-in.   2   2  (a) Shaft horsepower: The power of the shaft is  1 ft   180 rev   2 rad   1 min  P = T  = (1,440 lb-in.)  = 2,261.947 lb-ft/s  12 in.  min   1 rev   60 s  which is  1 hp  P = (2,261.947 lb-ft/s)  = 4.11 hp  550 lb-ft/s  (b) Maximum shear stress: The polar moment of inertia of the shaft is:  J= (1.25 in.) 4 = 0.239684 in.4 32 The maximum shear stress in the shaft is Tc (1, 440 lb-in.)(1.25 in./2) = = = 3,754.936 psi = 3,750 psi J 0.239684 in.4

Ans.

P6.31 A conveyor belt is driven by a 20 kW motor turning at 9 Hz. Through a series of gears that reduce the speed, the motor drives the belt drum shaft at a speed of 0.5 Hz. If the allowable shear stress is 70 MPa and both shafts are solid, calculate: (a) the required diameter of the motor shaft. (b) the required diameter of the belt drum shaft.

Solution (a) Motor Shaft: The torque in the motor shaft is: Nm 20,000 P s T= = = 353.678 N  m   9 rev  2 rad      s  1 rev  The minimum diameter required for the shaft can be found from: 16 ( 353.678 N  m )(1,000 mm/m ) 16T d3  = = 25,732.39 mm3 2  allow  ( 70 N/mm )  d  29.5 mm

Ans.

(b) Belt Drum Shaft: The torque in the belt drum shaft is: Nm 20,000 P s T= = = 6,366.200 N  m   0.5 rev  2 rad      s  1 rev  The minimum diameter required for the shaft can be found from: 16 ( 6,366.200 N  m )(1,000 mm/m ) 16T d3  = = 463,182.72 mm3 2  allow  ( 70 N/mm )  d  77.4 mm

Ans.

P6.32 A motor at A supplies 25 hp to the system shown in Figure P6.32/33. Sixty percent of the power supplied by the motor is taken off by gear C, and the remaining 40 percent of the power is taken off by gear D. Power shaft segments (1) and (2) are hollow steel tubes with an outside diameter of 1.50 in. and a wall thickness 0.1875 in. The diameters of pulleys A and B are DA = 4 in. and DB = 14 in., respectively. If the allowable shear stress for the steel tubes is 8,000 psi, what is the slowest permissible rotation speed for the motor? FIGURE P6.32/33

Solution Section properties: The polar moment of inertia of the steel tube is:    4 4 J1 = d14 = (1.50 in.) − (1.125 in.)  = 0.339753 in.4 = J 2  32 32 Allowable torque: Based on the 8,000 psi allowable shear stress, the shaft can transmit a torque of: 2 4  J (8,000 lb/in. )( 0.339753 in. ) T= = = 3,624.03 lb  in. = 302.002 lb  ft c 1.50 in. / 2 Power transmission and torques: Shaft (1) transmits 100% of the power from pulley B to gear C. This means that shaft (1) transmits 25 hp. From the problem statement, 60% of the 25 hp supplied to the system is removed by gear C. Consequently, shaft (2) transmits 10 hp. Shaft rotation speed: The slowest permissible rotation speed for shaft (1) is:  550 lb  ft/s  ( 25 hp )   P  1 hp  = 45.5295 rad/s  = T 302.002 lb  ft and the slowest permissible rotation speed for shaft (2) is:  550 lb  ft/s  (10 hp )   P  1 hp  = 18.2118 rad/s  = T 302.002 lb  ft Consequently, the slowest rotation speed for shafts (1) and (2) is: rad  1 rev  60 s      45.5295    = 434.774 rpm s  2 rad  1 min   Since pulley B is attached to the shaft, pulley B also rotates at this speed. Motor rotation speed: The rotation speed of pulley A and the motor is found with the use of the gear ratio:  A DA = B DB D 14 in. Ans.  A = B B = ( 434.774 rpm ) = 1,522 rpm = motor DA 4 in.

P6.33 A motor supplies sufficient power to the system shown in Figure P6.32/33 so that gears C and D provide torques of TC = 2.6 kN·m and TD = 1.2 kN·m, respectively, to machinery in a factory. Power shaft segments (1) and (2) consists of a hollow steel tube with an outside diameter of 90 mm and a wall thickness of 8 mm. The diameters of pulleys A and B are DA = 100 mm and DB = 360 mm, respectively. If the power shaft [i.e., segments (1) and (2)] rotates at 160 rpm, determine: (a) the maximum shear stress in power shaft segments (1) and (2). (b) the power (in kW) that must be provided by the motor as well as the rotation speed (in rpm). (c) the torque applied to pulley A by the motor. FIGURE P6.32/33

Solution Section properties: The polar moment of inertia of the steel tube is:  4   4 4 J1 = d1 = ( 90 mm ) − ( 74 mm )  = 3, 497,321 mm 4 = J 2  32 32 Torques: The torque magnitude in shaft (2) equals TD, and the torque magnitude in shaft (2) equals the sum of torques TC and TD. T1 = TC + TD = 2.6 kN  m + 1.2 kN  m = 3.8 kN  m = 3,800 N  m

T2 = TD = 1.2 kN  m = 1,200 N  m (a) Maximum shear stress in segments (1) and (2): T c ( 3,800 N  m )( 90 mm / 2 )(1,000 mm/m ) 1 = 1 1 = = 48.9 MPa J1 3,497,321 mm4

Tc J2

2 = 2 2 =

(1,200 N  m )( 90 mm / 2 )(1,000 mm/m ) = 15.44 MPa 3,497,321 mm4

Ans. Ans.

(b) Power transmission and rotation speed: The torque on pulley B is equal to T1. The corresponding power required to turn pulley B is found as: rev  2 rad  1 min   P = T11 = ( 3,800 N  m ) 160    min  rev  60 s   Nm = 63,669.61 = 63,669.61 W = 63.670 kW s The motor power is the same as the power required to turn pulley B: Ans. Pmotor = 63.7 kW

Motor rotation speed: The rotation speed of pulley A and the motor is found with the use of the gear ratio:  A DA = B DB D 360 mm Ans.  A = B B = (160 rpm ) = 576 rpm = motor DA 100 mm (c) Torque applied to pulley A by the motor: The torque applied to pulley A is found from the motor power and the motor rotation speed: Nm 63,669.61 P s T= = = 1,055.556 N  m = 1,056 N  m rev  2 rad  1 min     576    min  rev  60 s  

P6.34 In Figure P6.34/35, the motor at A produces 30 kW of power while turning at 8.0 Hz. The gears at A and B connecting the motor to line shaft BCDE have NA = 30 teeth and NB = 96 teeth, respectively. On the line shaft, gear C takes off PC = 50% of the power, gear D removes PD = 30%, and gear E removes the remainder. The line shaft will be solid, all segments will have the same diameter, and it will be made of steel [G = 80 GPa]. Segment lengths are L1 = 4.6 m, L2 = 2.3 m, and L3 = 2.3 m, and the bearings shown permit free rotation of the line shaft. Determine the minimum diameter required for line shaft BCDE for an allowable shear stress of 65 MPa and an allowable rotation angle of 10° of gear E relative to gear B.

FIGURE P6.34/35

Solution Power transmission and torques: The rotation speed of gear B is found from the motor rotation speed and the gear ratio:  A DA = B DB D N 30 teeth B =  A A =  A A = ( 8.0 Hz ) = 2.5 Hz DB NB 96 teeth The torque at gear B is P 30,000 N  m/s TB = = = 1,909,860 N  mm   2.5 rev  2 rad      s  1 rev  Accordingly, the torque in shaft segment (1) is T1 = 1,909,860 N·mm. The torque transferred by gear C is 50% of this value, leaving a torque in segment (2) of T2 = 954,930 N·mm. The torque transferred by gear D is 30% of TB, leaving a torque in segment (3) of T3 = 381,972 N·mm. Minimum diameter based on allowable stress: The minimum diameter required for the solid shaft based on the 65 MPa allowable stress can be found from:  3 T 1,909,860 N  mm d1  1 = = 29,382.462 mm3 16  1,allow 65 N/mm 2  d1  53.09 mm

(a)

Minimum diameter based on rotation angle of gear E: The rotation angle of gear E with respect to gear B is simply the sum of the angles of twist in segments (1), (2), and (3). Each segment has the same diameter and the same shear modulus. E / B = 1 + 2 + 3

T1L1 T2 L2 T3 L3 + + J1G1 J 2G2 J 3G3

1 T1L1 + T2 L2 + T3 L3   10 JG Solve this expression for the shaft diameter. =

 32

d4 

1 T L + T L + T L  (10) G 1 1 2 2 3 3

(1,909,860 N  mm )( 4,600 mm ) + ( 954,930 N  mm )( 2,300 mm ) + ( 381,972 N  mm )( 2,300 mm )  d4    rad  32 2 (10)   ( 80,000 N/mm )  180  32 11.860226  109 N  mm2 4 d  = 8,652,180 mm4 2  13,962.60 N/mm



(

 d  54.235 mm

)

(b)

Minimum diameter required for line shaft BCDE: The minimum diameter required for line shaft BCDE is found by comparing the results in Eqs. (a) and (b). Consequently, Ans. d  54.2 mm

P6.35 In Figure P6.34/35, the motor at A produces 25 hp while turning at 2,600 rpm. The gears at A and B connecting the motor to line shaft BCDE have NA = 30 teeth and NB = 140 teeth, respectively. On the line shaft, gear C takes off PC = 40% of the power, gear D removes PD = 40%, and gear E removes the remainder. Line shaft BCDE is a solid 1.25 in. diameter shaft made of an aluminum alloy that has a shear modulus of 3,800 ksi. Segment lengths are L1 = 80 in., L2 = 42 in., and L3 = 36 in. The bearings shown permit free rotation of the line shaft. Calculate: (a) the magnitude of the maximum shear stress in each shaft segment. (b) the rotation angle of gear E with respect to gear B.

FIGURE P6.34/35

Solution Power transmission and torques: The rotation speed of gear B is found from the motor rotation speed and the gear ratio:  A DA = B DB D N 30 teeth B =  A A =  A A = ( 2,600 rpm ) = 557.143 rpm DB NB 140 teeth The torque at gear B is  550 lb  ft/s  ( 25 hp )   P  1 hp  TB = = = 235.672 lb  ft   557.143 rev  2 rad  1 min      min   1 rev  60 s  Accordingly, the torque in shaft segment (1) is T1 = 235.672 lb·ft. The torque transferred by gear C is 40% of this value, leaving a torque in segment (2) of T2 = 141.403 lb·ft. The torque transferred by gear D is 40% of TB, leaving a torque in segment (3) of T3 = 47.134 lb·ft. Section properties: The polar moment of inertia of the line shaft is:   4 J1 = d14 = (1.25 in.) = 0.239684 in.4 = J 2 = J 3 32 32 (a) Maximum shear stresses: T c ( 235.672 lb  ft )(1.25 in. / 2 )(12 in./ft ) 1 = 1 1 = = 7,370 psi J1 0.239684 in.4

Tc J2

2 = 2 2 =

(141.403 lb  ft )(1.25 in. / 2 )(12 in./ft ) = 4, 420 psi

0.239684 in.4 T c ( 47.134 lb  ft )(1.25 in. / 2 )(12 in./ft ) 3 = 3 3 = = 1, 475 psi J3 0.239684 in.4

Ans. Ans. Ans.

(b) Rotation angle of gear E with respect to gear B: Note that the torques shown for gears C, D, and E all turn in the same direction. From the figure, we can determine that the internal torque in each segment is a positive value. The angles of twist in each segment are: T L ( 235.672 lb  ft )( 80 in.)(12 in./ft ) 1 = 1 1 = = 0.248402 rad J1G1 ( 0.239684 in.4 )( 3.8 106 lb/in.2 )

2 = 3 =

T2 L2 (141.403 lb  ft )( 42 in.)(12 in./ft ) = = 0.078247 rad J 2G2 ( 0.239684 in.4 )( 3.8 106 lb/in.2 )

T3 L3 ( 47.134 lb  ft )( 36 in.)(12 in./ft ) = = 0.022356 rad J 3G3 ( 0.239684 in.4 )( 3.8 106 lb/in.2 )

The rotation angle of gear E relative to gear B is simply the sum of these three angles of twist. E / B = 1 + 2 + 3 = 0.248402 rad + 0.078247 rad + 0.022356 rad = 0.349 rad

Ans.

P6.36 The motor at A is required to provide 40 hp of power to line shaft BCDE shown in Figure P6.36/37, turning gears B and E at 900 rpm. Gear B removes 70% of the power from the line shaft, and gear E removes 30%. Shafts (1) and (2) are solid aluminum alloy shafts with an outside diameter of 1.25 in. and a shear modulus of 3,800 ksi. Shaft (3) is a solid steel shaft that has an outside diameter of 0.75 in. and a shear modulus of 11,600 ksi. The shaft lengths are L1 = 8.25 ft, L2 = 4.25 ft, and L3 = 7.0 ft. The diameters of pulleys A and C are DA = 2 in. and DC = 10 in., respectively. The bearings shown allow free rotation of the shafts. Calculate: (a) the magnitude of the maximum shear stress in each shaft. (b) the rotation angle of gear E with respect to pulley C. (c) the torque magnitude and rotation speed required for motor A.

FIGURE P6.36/37

Solution Power transmission and torques: The torque applied to pulley C has a magnitude of  550 lb  ft/s  ( 40 hp )   P  1 hp  TC = = = 233.427 lb  ft   900 rev  2 rad  1 min       min  1 rev  60 s  The torque magnitude taken off at gear B is 70% of this value; thus, the magnitude of the torque at B is 163.399 lb·ft. The torque magnitude taken off at gear E is 30% of TC, leaving a torque magnitude of TE = 70.028 lb·ft. Since we are asked to determine rotation angles for this system, it is necessary that we determine the proper signs for each of these torque magnitudes. Consider an FBD that cuts through shaft (1) and includes gear B. M x = TB + T1 = 0 (a) T1 = −TB = −163.399 lb  ft Next, consider an FBD that cuts through shaft (3) and includes gear E. M x = TE − T3 = 0

T3 = TE = 70.028 lb  ft The torque in shaft (2) will be the same as the torque in shaft (3).

(b)

Now, we know the proper signs for each of the internal torques. Section properties: The polar moments of inertia of the solid aluminum and solid steel shafts are:  4  4 J1 = d1 = (1.25 in.) = 0.239684 in.4 = J 2 32 32 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

J3 =

 32

d34 =

 32

( 0.75 in.) = 0.031063 in.4 4

(a) Maximum shear stresses: T c (163.399 lb  ft )(1.25 in. / 2 )(12 in./ft ) 1 = 1 1 = = 5,110 psi J1 0.239684 in.4

Tc J2

2 = 2 2 =

( 70.028 lb  ft )(1.25 in. / 2)(12 in./ft ) = 2,190 psi

0.239684 in.4 T c ( 70.028 lb  ft )( 0.75 in. / 2 )(12 in./ft ) 3 = 3 3 = = 10,140 psi J3 0.031063 in.4

Ans. Ans. Ans.

(b) Rotation angle of gear E with respect to pulley C: The angles of twist in shafts (2) and (3) are:

( 70.028 lb  ft )( 4.25 ft )(12 in./ft ) = 0.047054 rad TL 2 = 2 2 = J 2G2 ( 0.239684 in.4 )( 3.8 106 lb/in.2 ) 2

( 70.028 lb  ft )( 7 ft )(12 in./ft ) = 0.195898 rad TL 3 = 3 3 = J 3G3 ( 0.031063 in.4 )(11.6 106 lb/in.2 ) 2

The rotation angle of gear E relative to pulley C is simply the sum of these two angles of twist. E / C = 2 + 3 = 0.047054 rad + 0.195898 rad = 0.243 rad

Ans.

(c) Torque magnitude and rotation speed required for motor A: The rotation speed of pulley A is found from the line shaft rotation speed and the gear ratio:  A DA = C DC D 10 in. Ans.  A = C C = ( 900 rpm ) = 4,500 rpm DA 2 in. The torque required from the motor is  550 lb  ft/s  ( 40 hp )   hp P   T= = = 46.7 lb  ft   4,500 rev  2 rad  1 min       min   1 rev   60 s 

Ans.

P6.37 The motor at A is required to provide 50 kW of power to line shaft BCDE shown in Figure P6.36/37, turning gears B and E at 8 Hz. Gear B removes 65% of the power from the line shaft, and gear E removes 35%. Shafts (1) and (2) are solid aluminum alloy [G = 26 GPa] shafts having an allowable shear stress of 45 MPa. Shaft (3) is a solid steel [G = 80 MPa] shaft that has an allowable shear stress of 60 MPa. The shaft lengths are L1 = 2.1 m, L2 = 1.2 m, and L3 = 1.8 m. The diameters of pulleys A and C are DA = 70 mm and DC = 300 mm, respectively. The bearings shown allow free rotation of the shafts. Calculate: (a) the minimum permissible diameter for aluminum shafts (1) and (2). (b) the minimum permissible diameter for steel shaft (3). (c) the rotation angle of gear E with respect to pulley C if the shafts have the minimum permissible diameters as determined in (a) and (b). (d) the torque magnitude and rotation speed (in Hz) required for motor A.

FIGURE P6.36/37

Solution Power transmission and torques: The torque applied to pulley C has a magnitude of P ( 50, 000 N  m/s )(1,000 mm/m ) TC = = = 994, 718 N  mm   8 rev  2 rad      s  1 rev  The torque magnitude taken off at gear B is 65% of this value; thus, the magnitude of the torque at B is 646,567 N·mm. The torque magnitude taken off at gear E is 35% of TC, leaving a torque magnitude of TE = 348,151 N·mm. Since we are asked to determine rotation angles for this system, it is necessary that we determine the proper signs for each of these torque magnitudes. Consider an FBD that cuts through shaft (1) and includes gear B. M x = TB + T1 = 0 (a) T1 = −TB = −646,567 N  mm Next, consider an FBD that cuts through shaft (3) and includes gear E. M x = TE − T3 = 0

T3 = TE = 348,151 N  mm The torque in shaft (2) will be the same as the torque in shaft (3).

(b)

Now, we know the proper signs for each of the internal torques. (a) Minimum diameter for shafts (1) and (2): The torque in shaft (1) is larger than the torque in shaft (2) so it will control the size of the aluminum shaft. The minimum diameter required for the solid aluminum shaft based on the 45 MPa allowable stress can be found from: Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

 16

d13 

1,allow

646,567 N  mm = 14,368.156 mm3 2 45 N/mm

 d1  41.827 mm = 41.8 mm

Ans.

(b) Minimum diameter for shaft (3): The minimum diameter required for the solid steel shaft based on the 60 MPa allowable stress can be found from: T  3 348,151 N  mm d3  3 = = 5,802.517 mm3 16  3,allow 60 N/mm2

 d3  30.917 mm = 30.9 mm

Ans.

(c) Rotation angle of gear E with respect to pulley C: The polar moments of inertia of the solid aluminum shafts (1) and (2) and the solid steel shaft (3) are:  4  4 J1 = d1 = ( 41.827 mm ) = 300, 489 mm 4 = J 2 32 32  4  4 J3 = d3 = ( 30.917 mm ) = 89, 698 mm 4 32 32 The angles of twist in shafts (2) and (3) are: ( 348,151 N  mm )(1.2 m )(1,000 mm/m ) = 0.053475 rad TL 2 = 2 2 = J 2G2 ( 300, 489 mm4 )( 26, 000 N/mm2 )

3 =

T3 L3 ( 348,151 N  mm )(1.8 m )(1,000 mm/m ) = = 0.087331 rad J 3G3 (89, 698 mm4 )(80, 000 N/mm2 )

The rotation angle of gear E relative to pulley C is simply the sum of these two angles of twist. E / C = 2 + 3 = 0.053475 rad + 0.087331 rad = 0.1408 rad

Ans.

(d) Torque magnitude and rotation speed required for motor A: The rotation speed of pulley A is found from the line shaft rotation speed and the gear ratio:  A DA = C DC D 300 mm Ans.  A = C C = ( 8 Hz ) = 34.286 Hz = 34.3 Hz DA 70 mm The torque required from the motor is P 50, 000 N  m/s TA = = = 232.101 N  m = 232 N  m A  34.286 rev  2 rad     s    1 rev 

Ans.

P6.38 The motor shown in Figure P6.38 supplies 15 kW at 1,700 rpm at A. Shafts (1) and (2) are each solid 30 mm diameter shafts. Shaft (1) is made of an aluminum alloy [G = 26 GPa], and shaft (2) is made of bronze [G = 45 GPa]. The shaft lengths are L1 = 3.4 m and L2 = 2.7 m, respectively. Gear B has 54 teeth, and gear C has 96 teeth. The bearings shown permit free rotation of the shafts. Determine: (a) the maximum shear stress produced in shafts (1) and (2). (b) the rotation angle of gear D with respect to flange A.

FIGURE P6.38

Solution Torques: The torque magnitude acting on flange A is: P 15, 000 N  m/s TA = = = 84.258 N  m   1,700 rev  2 rad  1 m      m   1 rev  60 s  This torque is transmitted through shaft (1) to gear B: T1 = TB = TA = 84.258 N·m. The torque applied to gear C is found using the gear ratio: N 96 teeth TC = TB C = (84.258 N  m ) = 149.793 N  m NB 54 teeth The torque magnitude on shaft segment (2) as well as gear D is equal to the torque applied to gear C. To determine the proper signs for T1 and T2, consider equilibrium. First, consider a FBD that cuts through shaft (2) and includes gear D. M x = TD − T2 = 0 T2 = TD = 149.793 N  m Since T2 is positive, the torque in shaft (1) must be negative; therefore, T1 = −84.258 N  m Section properties: The polar moment of inertia for both shaft (1) and shaft (2) is:  4  4 J1 = d1 = ( 30 mm ) = 79,521.6 mm4 = J 2 32 32 (a) Maximum shear stresses: T c ( 84.258 N  m )( 30 mm / 2 )(1,000 mm/m ) 1 = 1 1 = = 15.89 MPa J1 79,521.6 mm4

Tc J2

2 = 2 2 =

(149.793 N  m )(30 mm / 2 )(1,000 mm/m ) = 28.3 MPa 79,521.6 mm4

Ans. Ans.

Angles of twist in shafts (1) and (2): ( −84.258 N  m )( 3,400 mm )(1,000 mm/m ) = −0.138559 rad TL 1 = 1 1 = J1G1 ( 79,521.6 mm4 )( 26,000 N/mm2 )

2 =

T2 L2 (149.793 N  m )( 2,700 mm )(1,000 mm/m ) = = 0.113021 rad J 2G2 ( 79,521.6 mm4 )( 45,000 N/mm2 )

Rotation angles of gears B and C: The rotation of gear B is found from the angle of twist in shaft (1): 1 = B − A B = A + 1 = 0 rad + ( −0.138559 rad ) = −0.138559 rad As gear B rotates, gear C also rotates but it rotates in the opposite direction. The magnitude of the rotation is dictated by the gear ratio: N 54 teeth C = −B B = − ( −0.138559 rad ) = 0.077939 rad NC 96 teeth (b) Rotation angle of gear D: The rotation of gear D with respect to flange A is found from: 2 = D − C D = C + 2 = 0.077939 rad + 0.113021 rad = 0.1910 rad

Ans.

P6.39 A hollow circular cold-rolled stainless steel [G1 = 12,500 ksi] tube (1) with an outside diameter of 2.25 in. and an inside diameter of 2.00 in. is securely shrink fit to a solid 2.00 in. diameter cold-rolled bronze [G2 = 6,500 ksi] core (2) as shown in Figure P6.39/40. The length of the assembly is L = 20 in. The allowable shear stress of tube (1) is 60 ksi, and the allowable shear stress of core (2) is 25 ksi. Determine: (a) the allowable torque T that can be applied to the tube-and-core assembly. (b) the corresponding torques produced in tube (1) and core (2). (c) the angle of twist produced in the assembly by the allowable torque T. FIGURE P6.39/40

Solution Section Properties: For tube (1) and core (2), the polar moments of inertia are:   4 4 J1 = 2.25 in.) − ( 2.00 in.)  = 0.945316 in.4 (  32  J2 =

 32

( 2.00 in.) = 1.570800 in.4 4

Equilibrium: M x = T1 + T2 − T = 0

(a)

Geometry-of-Deformation Relationship: Since the tube and core are securely bonded together, the angles of twist in both members must be equal; therefore, 1 = 2 (b) Torque-Twist Relationships: TL TL 1 = 1 1 2 = 2 2 J1G1 J 2G2 Compatibility Equation: Substitute the torque-twist relationships [Eqs. (c)] into the geometry-ofdeformation relationship [Eq. (b)] to obtain the compatibility equation: T1L1 T2 L2 = J1G1 J 2G2

(c)

(d)

Solve Eq. (e) for  2: L G c  6,500 ksi   2.00 in. / 2   2 = 1 1 2 2 = 1  (f)  = 0.4622221  L2 G1 c1  12,500 ksi   2.25 in. / 2  Assume the tube controls: If the shear stress in tube (1) reaches its allowable value of 60 ksi (in other words, if the tube controls), then the shear stress in core (2) will be:  2 = 0.462222 ( 60 ksi ) = 27.73 ksi  25 ksi N.G. which is larger than the 25 ksi allowable shear stress for the core. Therefore, the shear stress in the core controls. Core shear stress actually controls: Rearrange Eq. (f) to solve for the shear stress in tube (1) given that the shear stress in core (2) is at its 25 ksi allowable value: 25 ksi 1 = = 54.087 ksi  60 ksi O.K. 0.462222 Now that the maximum shear stresses in the tube and the core are known, the torques in each component can be computed: 2 4  J ( 54.087 kips/in. )( 0.945316 in. ) T1 = 1 1 = = 45.448 kip  in. c1 2.25 in. / 2

T2 =

 2 J2 c2

( 25 kips/in. )(1.570800 in. ) = 39.270 kip  in. = 2

2.00 in. / 2

(a) Total Torque: From Eq. (a), the total torque acting on the assembly must not exceed: Tmax = T1 + T2 = 45.448 kip  in. + 39.270 kip  in. = 84.718 kip  in. = 84.7 kip  in.

Ans.

(b) Torques in each component: As computed previously, the torque in tube (1) is: T1 = 45.4 kip  in.

Ans.

and the torque in core (2) is: T2 = 39.3 kip  in.

Ans.

(c) Angle of Twist produced in a 20 in. length: Since both the tube and the core twist exactly the same amount [i.e., Eq. (b)], either torque-twist relationship can be used to compute the angle of twist of the entire assembly. ( 45.448 kip  in.)( 20 in.) = 0.076923 rad = 0.0769 rad TL 1 = 1 1 = Ans. J1G1 ( 0.945316 in.4 ) ( 6,500 ksi )

P6.40 The composite shaft shown in Figure P6.39/40 consists of an aluminum alloy tube (1) securely bonded to an inner brass core (2). The aluminum sleeve has an outside diameter of 60 mm, an inside diameter of 50 mm, and a shear modulus of G1 = 26 GPa. The solid brass core has a diameter of 50 mm and a shear modulus of G2 = 44 GPa. The length of the assembly is L = 600 mm. If the composite shaft is subjected to a torque of T = 5.0 kN·m, determine: (a) the maximum shear stresses in the aluminum tube and the brass core. (b) the rotation angle of end B relative to end A.

FIGURE P6.39/40

Solution Section Properties: For aluminum tube (1) and brass core (2), the polar moments of inertia are:   4 4 J1 = 60 mm ) − ( 50 mm )  = 658, 753 mm 4 (  32  J2 =



( 50 mm ) = 613,592 mm4 4

Equilibrium: M x = T1 + T2 − 5, 000 N  m = 0

(a)

Geometry-of-Deformation Relationship: Since the tube and core are perfectly bonded, the angles of twist in both members must be equal; therefore, 1 = 2 (b) Torque-Twist Relationships: TL TL 1 = 1 1 2 = 2 2 J1G1 J 2G2 Compatibility Equation: Substitute the torque-twist relationships [Eqs. (c)] into the geometry-ofdeformation relationship [Eq. (b)] to obtain the compatibility equation: T1L1 T2 L2 = J1G1 J 2G2

(c)

(d)

Solve the Equations: Solve Eq. (d) for T1:  658, 753 mm 4   26 GPa  L J G T1 = T2 2 1 1 = T2   = 0.634400 T2 4  L1 J 2 G2  613,592 mm   44 GPa  and substitute this result into Eq. (a) to compute the torque T2 in brass core (2): T1 + T2 = 0.634400 T2 + T2 = 1.634400 T2 = 5,000 N  m

T2 = 3,059.230 N  m

The torque in aluminum tube (1) is therefore: T1 = 5, 000 N  m − T2 = 5, 000 N  m − 3, 059.230 N  m = 1,940.770 N  m

(a) Maximum Shear Stress: The maximum shear stress in aluminum tube (1) is: T c (1,940.770 N  m )( 60 mm / 2 )(1,000 mm/m ) 1 = 1 1 = = 88.4 MPa J1 658, 753 mm4 The maximum shear stress in brass core (2) is: T c ( 3,059.230 N  m )( 50 mm / 2 )(1,000 mm/m ) 2 = 2 2 = = 124.6 MPa J2 613,592 mm4

Ans.

(b) Rotation Angle of B relative to A: Since both the tube and the core twist exactly the same amount [i.e., Eq. (b)], either torque-twist relationship can be used to compute the angle of twist of the entire assembly. T L (1,940.770 N  m )( 600 mm )(1,000 mm/m ) 1 = 1 1 = = 0.067988 rad = 0.0680 rad Ans. J1G1 ( 658, 753 mm4 )( 26, 000 N/mm2 )

P6.41 A solid 2.50 in. diameter cold-rolled brass [G = 6,400 ksi] shaft that is L1 + L2 = 115 in. long extends through and is completely bonded to a hollow aluminum [G = 3,800] tube, as shown in Figure P6.41. Aluminum tube (1) has an outside diameter of 3.50 in., an inside diameter of 2.50 in., and a length of L1 = 70 in. Both the brass shaft and the aluminum tube are securely attached to the wall support at A. External torques in the directions shown are applied at B and C. The torque magnitudes are TB = 6.0 kip·ft and TC = 2.5 kip·ft, respectively. After TB and TC are applied to the composite shaft, determine: (a) the maximum shear stress magnitude in aluminum tube (1). (b) the maximum shear stress magnitude in brass shaft segment (2). (c) the maximum shear stress magnitude in brass shaft segment (3). (d) the rotation angle of joint B. (e) the rotation angle of end C.

FIGURE P6.41

Solution Section Properties: The polar moments of inertia for tube (1) and brass shafts (2) and (3) are:

J1 =

 

( 32 

J 2 = J3 =

3.50 in.) − ( 2.50 in.)  = 10.8974 in.4  4

 32

( 2.50 in.) = 3.83495 in.4 4

Equilibrium: Consider a free-body diagram cut around end C through segment (3): M x = −T3 − TC = 0

T3 = −TC = −2.5 kip  ft

(a)

and also consider an FBD cut around joint B: M x = −T1 − T2 + T3 + TB = 0

T1 + T2 = T3 + TB = −2.5 kip  ft + 6.0 kip  ft = 3.5 kip  ft (b) Eq. (b) reveals that the portion of the shaft between A and B is statically indeterminate. The five-step solution procedure will be used to determine T1 and T2 in this portion of the shaft. Geometry-of-Deformation Relationship: Since the aluminum tube and the brass shaft are securely bonded together, the angles of twist in both members must be equal; therefore, 1 = 2 (c)

Torque-Twist Relationships: TL TL 1 = 1 1 2 = 2 2 J1G1 J 2G2 Compatibility Equation: Substitute the torque-twist relationships [Eqs. (d)] into the geometry-ofdeformation relationship [Eq. (c)] to obtain the compatibility equation: T1L1 T2 L2 = J1G1 J 2G2

(d)

(e)

Solve the Equations: Solve Eq. (e) for T1. Note that L1 = L2.  10.8974 in.4   3,800 ksi  L2 J1 G1 T1 = T2 = T2   = 1.68720 T2 4  L1 J 2 G2  3.83495 in.   6,400 ksi  and substitute this result into Eq. (b) to compute the torque T2 in brass shaft (2): T1 + T2 = 1.68720 T2 + T2 = 2.68720 T2 = 3,500 lb  ft

T2 = 1,302.471 lb  ft The torque in aluminum tube (1) is therefore: T1 = 3,500 lb  ft − T2 = 3,500 lb  ft − 1,302.471 lb  ft = 2,197.529 lb  ft

(a) Maximum Shear Stress in Aluminum Tube: The maximum shear stress in aluminum tube (1) is: T c ( 2,197.529 lb  ft )( 3.50 in. / 2 )(12 in./ft ) Ans. 1 = 1 1 = = 4, 230 psi J1 10.8974 in.4

(b) Maximum Shear Stress Magnitude in Brass Shaft Segment (2): T c (1,302.471 lb  ft )( 2.50 in. / 2 )(12 in./ft ) 2 = 2 2 = = 5, 090 psi J2 3.83495 in.4

(c) Maximum Shear Stress Magnitude in Brass Shaft Segment (3): T c ( 2,500 lb  ft )( 2.50 in. / 2 )(12 in./ft ) 3 = 3 3 = = 9, 780 psi J3 3.83495 in.4

Ans.

(d) Rotation Angle of Joint B: ( 2,197.529 lb  ft )( 70 in.)(12 in./ft ) = 0.044577 rad TL 1 = 1 1 = J1G1 (10.8974 in.4 )( 3.8 106 lb/in.2 ) or

2 =

T2 L2 (1,302.471 lb  ft )( 70 in.)(12 in./ft ) = = 0.044577 rad J 2G2 ( 3.83495 in.4 )( 6.4 106 lb/in.2 )

B = 0.0446 rad

Ans.

(e) Rotation Angle of End C: ( −2,500 lb  ft )( 45 in.)(12 in./ft ) = −0.055004 rad TL 3 = 3 3 = J 3G3 ( 3.83495 in.4 )( 6.4 106 lb/in.2 ) C = B + 3 = 0.044577 rad + ( −0.055004 rad ) = −0.01043 rad

Ans.

P6.42 The compound shaft shown in Figure P6.42/43 consists of two pipes that are connected at flange B and securely attached to rigid walls at A and C. Pipe (1) is made of an aluminum alloy [G = 26 GPa]. It has an outside diameter of 140 mm, a wall thickness of 7 mm, and a length of L1 = 9.0 m. Pipe (2) is made of steel [G = 80 GPa]. It has an outside diameter of 100 mm, a wall thickness of 6 mm, and a length of L2 = 7.5 m. If a concentrated torque of 15 kN·m is applied to flange B, determine: (a) the maximum shear stress magnitudes in pipes (1) and (2). (b) the rotation angle of flange B relative to support A.

FIGURE P6.42/43

Solution Section Properties: The polar moments of inertia for the two pipes are:   4 4 J1 = (140 mm ) − (126 mm )  = 12,970,130 mm4  32   4 4 J2 = (100 mm ) − (88 mm )  = 3,929,982 mm4  32 Equilibrium: M x = −T1 + T2 + TB = 0

−T1 + T2 = −TB

(a)

Geometry-of-Deformation Relationship: Since the two pipes are securely attached to fixed supports at A and C, the sum of the angles of twist in the two pipes must equal zero: 1 + 2 = 0 (b) Torque-Twist Relationships: TL TL 1 = 1 1 2 = 2 2 J1G1 J 2G2 Compatibility Equation: Substitute the torque-twist relationships [Eqs. (c)] into the geometry-ofdeformation relationship [Eq. (b)] to obtain the compatibility equation: T1L1 T2 L2 + =0 J1G1 J 2G2

(c)

(d)

Solve the Equations: Solve Eq. (d) for T1: L J G T1 = −T2 2 1 1 L1 J 2 G2  7.5 m   12,970,130 mm   26 GPa  = −T2    4   9.0 m   3,929,982 mm   80 GPa  = −0.893832 T2 4

and substitute this result into Eq. (a) to compute the torque T2: −T1 + T2 = − ( −0.893832 T2 ) + T2 = 2.893832 T2 = −15, 000 N  m T2 = −7,920.450 N  m The torque in member (1) is therefore: T1 = T2 + TB = −7,920.450 N  m + 15, 000 N  m = 7, 079.550 N  m

(a) Maximum Shear Stress: The maximum shear stress magnitude in member (1) is: T c ( 7,079.550 N  m )(140 mm / 2 )(1,000 mm/m ) 1 = 1 1 = = 38.2 MPa J1 12,970,130 mm4 The maximum shear stress magnitude in member (2) is: T c ( 7,920.450 N  m )(100 mm / 2 )(1,000 mm/m ) 2 = 2 2 = = 100.8 MPa J2 3,929,982 mm4

Ans.

(b) Rotation Angle of Flange B Relative to Support A: The angle of twist in member (1) can be defined by the difference in rotation angles at the two ends; hence, 1 = B −  A Since joint A is restrained from rotating, A = 0 and thus 1 = B The rotation angle at B can be determined by computing the angle of twist in member (1): TL 1 = 1 1 J1G1

( 7, 079.550 N  m )( 9, 000 mm )(1,000 mm/m )(1,000 N/kN )

(12,970,130 mm )( 26, 000 N/mm )

= 0.188943 rad = 0.1889 rad

Ans.

P6.43 The compound shaft shown in Figure P6.42/43 consists of a solid aluminum segment (1) and a hollow brass segment (2) that are connected at flange B and securely attached to rigid supports at A and C. Aluminum segment (1) has a diameter of 0.875 in., a length of L1 = 45 in., a shear modulus of 3,800 ksi, and an allowable shear stress of 6 ksi. Brass segment (2) has an outside diameter of 0.68 in., a wall thickness of 0.09 in., a length of L2 = 30 in., a shear modulus of 6,400 ksi, and an allowable shear stress of 8 ksi. Determine: (a) the allowable torque TB that can be applied to the compound shaft at flange B. (b) the magnitudes of the internal torques in segments (1) and (2). (c) the rotation angle of flange B that is produced by the allowable torque TB.

FIGURE P6.42/43

Solution Section Properties: The polar moments of inertia for the two shafts are:  4 J1 = ( 0.875 in.) = 0.057548 in.4 32   4 4 J2 = ( 0.68 in.) − ( 0.50 in.)  = 0.014855 in.4  32 Equilibrium: M x = −T1 + T2 + TB = 0

(a)

Geometry-of-Deformation Relationship: Since the two shafts are securely attached to fixed supports at A and C, the sum of the angles of twist in the two members must equal zero: 1 + 2 = 0 (b) Torque-Twist Relationships: TL TL 1 = 1 1 2 = 2 2 J1G1 J 2G2 Compatibility Equation: Substitute the torque-twist relationships [Eqs. (c)] into the geometry-ofdeformation relationship [Eq. (b)] to obtain the compatibility equation: T1L1 T2 L2 + =0 J1G1 J 2G2

(c)

(d)

Solve the Equations: Since the problem is expressed in terms of allowable stresses, it is convenient to rewrite Eq. (d) in terms of stresses. In general, the elastic torsion formula can be rearranged as: Tc T  =  = J J c which allows Eq. (d) to be rewritten as: 1L1  L (e) =− 2 2 G1c1 G2c2 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Solve Eq. (e) for  1: L G c  30 in.   3,800 ksi   0.875 in. / 2  1 = − 2 2 1 1 = − 2    = −0.509344 2  L1 G2 c2  45 in.   6,400 ksi   0.68 in. / 2 

(f)

Assume the shaft (2) controls: If the shear stress in shaft (2) reaches its allowable magnitude of 8,000 psi, then the shear stress magnitude in shaft (1) will be: 1 = −0.509344 (8,000 psi ) = 4,074.75 psi  6,000 psi O.K. This shear stress is less than the 6,000 psi allowable shear stress magnitude for shaft (1), and thus, our assumption that shaft (2) controls is confirmed. Now that the maximum shear stress magnitudes in the two shafts are known, the torque magnitudes in each component can be computed: 2 4  1 J1 ( 4, 074.75 lb/in. )( 0.057548 in. ) T1 = = = 535.989 lb  in. c1 0.875 in. / 2

T2 =

 2 J2

(8, 000 lb/in. )( 0.014855 in. ) = 349.534 lb  in. = 2

c2 0.68 in. / 2 Note that these are torque magnitudes. From inspection of the FBD of flange B and the associated equilibrium equation, it is apparent that T2 must act opposite to the direction assumed in the FBD, giving it a negative value. Therefore, by inspection T2 = −349.534 lb  in. (a) Total Torque: From Eq. (a), the total torque acting at flange B must not exceed: TB ,max = T1 − T2 = 535.989 lb  in. − ( −349.534 lb  in.)  1 ft  = ( 885.523 lb  in.)   = 73.8 lb  ft  12 in. 

(b) Torques Magnitudes: As computed previously, the torque magnitude in shaft (1) is:  1 ft  T1 = ( 535.989 lb  in.)   = 44.7 lb  ft  12 in.  and the torque magnitude in shaft (2) is:  1 ft  T2 = ( 349.534 lb  in.)   = 29.1 lb  ft  12 in. 

Ans.

(c) Rotation Angle of Flange B Relative to Support A: The angle of twist in member (1) can be defined by the difference in rotation angles at the two ends; hence, 1 = B −  A Since joint A is restrained from rotating, A = 0, and thus 1 = B The rotation angle at B can be determined by computing the angle of twist in member (1): ( 535.989 lb  in.)( 45 in.) TL B = 1 = 1 1 = = 0.110294 rad = 0.1103 rad J1G1 ( 0.057548 in.4 )( 3.5 106 lb/in.2 )

Ans.

P6.44 In Figure P6.44, shafts (1) and (2) are identical solid 1.25 in. diameter steel [G = 11,500 ksi] shafts. Shaft (1) is fixed to a rigid support at A and shaft (2) is fixed to a rigid support at C. Shaft (1) has a length of L1 = 4.25 ft, and shaft (2) has a length of L2 = 2.75 ft. The allowable shear stress of the steel is 8,000 psi. The number of teeth on gears B and D is NB = 72 teeth and ND = 54 teeth, respectively. Determine the allowable magnitude of torque TE.

FIGURE P6.44

Solution We will initially focus our attention on shafts (1) and (2). From the information given, we will determine the maximum torque that can be applied to gear B. Once TB is known, we will use the gear ratio to determine the corresponding torque on gear D. The torque TE will have the same magnitude as the torque on gear D. Equilibrium: M x = −T1 + T2 + TB = 0

(a)

(c)

(d)

Solve Eq. (e) for  1. Note that the diameters and the shear moduli are identical for shafts (1) and (2). L G c  2.75 ft  1 = − 2 2 1 1 = − 2  (f)  = −0.647059 2 L1 G2 c2  4.25 ft  Assume that shaft (2) controls: If the shear stress in shaft (2) reaches its allowable magnitude of 8,000 psi, then the shear stress magnitude in shaft (1) will be: 1 = −0.647059 (8,000 psi ) = 5,176.471 psi  8 ksi O.K. Since this stress is smaller than the 8,000 psi allowable shear stress magnitude for shaft (1), our assumption that shaft (2) controls is confirmed. Now that the maximum shear stress magnitudes in the two shafts are known, the torque magnitudes in each component can be computed: Section Properties: The polar moments of inertia for the two shafts are:  4 J1 = (1.25 in.) = 0.239684 in.4 = J 2 32 2 4  J ( 5,176.470 lb/in. )( 0.239684 in. ) T1 = 1 1 = = 1,985.150 lb  in. c1 1.25 in. / 2

T2 =

 2 J2

(8, 000 lb/in. )( 0.239684 in. ) = 3, 067.960 lb  in. = 2

c2 1.25 in. / 2 Note that these are torque magnitudes. From inspection of the FBD of flange B and the associated equilibrium equation, it is apparent that T2 must act opposite to the direction assumed in the FBD, giving it a negative value. Therefore, by inspection T2 = −3, 067.960 lb  in. Total Torque: From Eq. (a), the total torque acting at flange B must not exceed: TB ,max = T1 − T2 = 1,985.150 lb  in. − ( −3, 067.960 lb  in.) = 5, 053.110 lb  in. = 421.093 lb  ft

Torque relationship between gears B and D: TB TD D N 54 teeth = or TD = TB D = TB D = TB = 0.75TB DB DD DB NB 72 teeth From this relationship, the maximum torque that can be applied to gear B is limited to: TD = 0.75TB = 0.75 ( 421.093 lb  ft ) = 315.820 lb  ft The torque applied at E equals TD; therefore, the maximum allowable torque that can be applied to this system at E is: Ans. TE  316 lb  ft

P6.45 The torsional assembly shown in Figure P6.45 consists of solid 2.50 in. diameter aluminum [G = 4,000 ksi] segments (1) and (3) and a central solid 3.00 in. diameter bronze [G = 6,500 ksi] segment (2). Segment lengths are L1 = L3 = 36 in. and L2 = 54 in. Concentrated torques of TB = 2T0 and TC = T0 are applied to the assembly at B and C, respectively. If the rotation angle at joint B must not exceed 3°, determine: (a) the maximum magnitude of T0 that may be applied to the assembly. (b) the maximum shear stress magnitude in aluminum segments (1) and (3). (c) the maximum shear stress magnitude in bronze segment (2).

FIGURE P6.45

Solution Section Properties: The polar moments of inertia for the aluminum and bronze segments are:

J1 = J 3 = J2 =

 32



( 2.50 in.) = 3.83495 in.4 4

( 3.00 in.) = 7.95216 in.4 4

Equilibrium: Consider a free-body diagram cut around joint B, where the external torque 2T0 is applied: M x = −T1 + T2 + 2T0 = 0 (a) and also consider an FBD cut around joint C, where the external torque T0 is applied: M x = −T2 + T3 + T0 = 0 (b) Geometry-of-Deformation Relationship: Since the two ends of the torsion assembly are securely attached to fixed supports at A and D, the sum of the angles of twist in the three shafts must equal zero: 1 + 2 + 3 = 0 (c) Torque-Twist Relationships: TL TL 1 = 1 1 2 = 2 2 J1G1 J 2G2

3 =

T3 L3 J 3G3

(d)

Torque in Segment (1): From the problem statement, the rotation angle at B must not exceed 3°; consequently, the angle of twist in segment (1) must equal this value: 1 = 3 = 0.052360 rad From the torque-twist relationships, the internal torque in segment (1) can be computed as follows:

1 =

T1 L1 J1G1

T1 =

1 J1G1 L1

( 0.052360 rad ) ( 3.83495 in.4 )( 4,000 kips/in.2 ) 36 in.

= 22.311 kip  in.

Consider Rotation Angle of Flange B Relative to Support D: The rotation angle at B can be expressed in terms of the angles of twist in segments (2) and (3) as follows: 3 = D − C C = −3

2 = C − B

B = −2 + C = −2 − 3 The rotation angle at B must not exceed 3°. TL TL B = −2 − 3 = − 2 2 − 3 3  3 = 0.052360 rad J 2G2 J 3G3 from Eq. (a): T2 = T1 − 2T0 and from Eq. (b): T3 = T2 − T0 = (T1 − 2T0 ) − T0 = T1 − 3T0

(c)

(d) (e)

Substitute the expressions derived in Eqs. (d) and (e) into Eq. (c): (T − 2T0 ) L2 (T1 − 3T0 ) L3 − 1 −  0.052360 rad J 2G2 J 3G3 Simplify  L  2L L  3L3  −T1  2 + 3  + T0  2 +   0.052360 rad J G J G J G J G 3 3 3 3  2 2  2 2 Solve for T0:  L L  0.052360 rad + T1  2 + 3   J 2G2 J 3G3  T0  3L3 2 L2 + J 2G2 J 3G3 and compute:   54 in. 36 in. 0.052360 rad + ( 22.311 kip  in.)  +  4 4  ( 7.95216 in. ) ( 6,500 ksi ) ( 3.83495 in. ) ( 4, 000 ksi )  T0  2 ( 54 in.) 3 ( 36 in.) + 4 ( 7.95216 in. ) ( 6,500 ksi ) (3.83495 in.4 ) ( 4, 000 ksi )  14.023 kip  in. Backsubstitute T0 into Eq. (d) to compute T2: T2 = T1 − 2T0 = 22.311 kip  in. − 2 (14.023 kip  in.) = −5.735 kip  in.

and backsubstitute T0 into Eq. (e) to compute T3: T3 = T1 − 3T0 = 22.311 kip  in. − 3 (14.023 kip  in.) = −19.758 kip  in.

(a) Maximum magnitude allowed for T0: T0,allow = 14.02 kip  in.

Ans.

Maximum Shear Stress Magnitudes: The maximum shear stress magnitudes are: T c ( 22.311 kip  in.)( 2.50 in. / 2 ) 1 = 1 1 = = 7.272 ksi J1 3.83495 in.4

Tc J2

2 = 2 2 =

( 5.735 kip  in.)(3.00 in. / 2 ) = 1.082 ksi

7.95216 in.4 T c (19.758 kip  in.)( 2.50 in. / 2 ) 3 = 3 3 = = 6.440 ksi J3 3.83495 in.4

(b) Maximum Shear Stress Magnitude in Aluminum Segments: 1 = 7.27 ksi

Ans.

P6.46 The torsional assembly shown in Figure P6.46 consists of a hollow aluminum alloy [G = 28 GPa] segment (2) and two brass [G = 44 GPa] tube segments (1) and (3). The brass segments have an outside diameter of 114.3 mm, a wall thickness of 6.1 mm, and lengths of L1 = L3 = 2,400 mm. Aluminum segment (2) has an outside diameter of 63.5 mm, a wall thickness of 3.2 mm, and a length of L2 = 4,200 mm. If concentrated torques of TB = 4 kN-m and TC = 7 kN-m are applied in the directions shown, determine: (a) the maximum shear stress magnitude in brass tube segments (1) and (3). (b) the maximum shear stress magnitude in aluminum segment (2). (c) the rotation angle of joint C.

FIGURE P6.46

Solution Section Properties: The polar moments of inertia for the brass and aluminum segments are:

J1 = J 3 = J2 =

 

4 4 (114.3 mm ) − (102.1 mm )  = 6, 088, 070 mm4  32

( 32 

63.5 mm ) − ( 57.1 mm )  = 552, 603 mm4  4

Equilibrium: Consider a free-body diagram cut around joint B, where the external torque TB is applied: M x = −T1 + T2 + TB = 0 (a) and also consider an FBD cut around joint C, where the external torque TC is applied: M x = −T2 + T3 − TC = 0 (b) Geometry-of-Deformation Relationship: Since the two ends of the torsion structure are securely attached to fixed supports at A and D, the sum of the angles of twist in the three shafts must equal zero: 1 + 2 + 3 = 0 (c) Torque-Twist Relationships: TL TL 1 = 1 1 2 = 2 2 J1G1 J 2G2

3 =

T3 L3 J 3G3

(d)

Compatibility Equation: Substitute the torque-twist relationships [Eqs. (d)] into the Geometry-ofdeformation relationship [Eq. (c)] to obtain the compatibility equation: T1L1 T2 L2 T3 L3 (e) + + =0 J1G1 J 2G2 J 3G3 Solve the Equations: The approach used here will be to reduce the variables in Eq. (e) by replacing T1 and T3 with equivalent expressions involving T2. From Eq. (a): Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

T1 = T2 + TB and from Eq. (b): T3 = T2 + TC Substitute Eqs. (f) and (g) into Eq. (e) and simplify to derive an expression for T2: (T2 + TB ) L1 + T2 L2 + (T2 + TC ) L3 = 0 J1G1 J 2G2 J 3G3  L L  L T L T L T2  1 + 2 + 3  = − B 1 − C 3 J1G1 J 3G3  J1G1 J 2G2 J 3G3  TB L1 TC L3 + J1G1 J 3G3 T2 = − L L1 L + 2 + 3 J1G1 J 2G2 J 3G3 Since G1 = G3 and J1 = J3, Eq. (h) can be simplified further and T2 can be computed as: 1  (TB L1 + TC L3 )  T2 = −   J1G1  L1 + L3 + L2  J 2G2   J1G1

(f) (g)

(h)

  4  106 N  mm ) ( 2,400 mm ) + ( 7  106 N  mm ) ( 4,200 mm ) (   =− 2,400 mm + 2,400 mm 4, 200 mm  ( 6,088,070 mm4 )( 45,000 N/mm2 )  +  ( 6,088,070 mm 4 )( 45,000 N/mm 2 ) ( 552,603 mm 4 )( 28,000 N/mm 2 )    = −340,589 N  mm 1

Backsubstitute this result into Eqs. (f) and (g) to obtain T1 and T3: T1 = T2 + TB = −340,589 N  mm + 4,000,000 N  mm = 3,659, 410 N  mm

T3 = T2 + TC = −340,589 N  mm + 7,000,000 N  mm = 6,659, 410 N  mm Maximum Shear Stress: The maximum shear stress magnitudes in the three segments are: T c ( 3,659,410 N  mm )(114.3 mm / 2 ) 1 = 1 1 = = 34.352 MPa J1 6,088,070 mm4

Tc J2

2 = 2 2 = Tc J3

3 = 3 3 =

( 340,589 N  mm )( 63.5 mm / 2 ) = 19.569 MPa

552,603 mm4 ( 6,659,410 N  mm )(114.3 mm / 2 )

6,088,070 mm4

= 62.513 MPa

(a) Maximum Shear Stress Magnitude in Brass Segments:  3 = 62.5 MPa

Ans.

(b) Maximum Shear Stress Magnitude in Aluminum Segment:  2 = 19.57 MPa

Ans.

(c) Rotation Angle of Joint C Relative to Support A: The angle of twist in member (1) can be defined by the difference in rotation angles at the two ends: 1 = B −  A B =  A + 1

(i)

Similarly, the angle of twist in member (2) can be defined by: 2 = C − B C = B + 2

(j)

To derive an expression for C, substitute Eq. (i) into Eq. (j), and note that joint A is restrained from rotating; therefore, A = 0. C = 1 + 2 The angle of twist in member (1) is: ( 3,659, 410 N  mm )( 2,400 mm ) = 0.032786 rad TL 1 = 1 1 = J1G1 ( 6,088,070 mm 4 )( 44,000 N/mm 2 ) The angle of twist in member (2) is: ( −340,589 N  mm )( 4,200 mm ) = −0.092450 rad TL 2 = 2 2 = J 2G2 ( 552,603 mm 4 )( 28,000 N/mm 2 ) The rotation angle of flange C is thus: C = 1 + 2 = 0.032786 rad + ( −0.092450 rad ) = −0.059664 rad = −0.0597 rad

Ans.

P6.47 The gear assembly shown in Figure P6.47 is subjected to a torque of TC = 300 N·m. Shafts (1) and (2) are solid 30 mm diameter steel [G = 80 GPa] shafts, and shaft (3) is a solid 35 mm diameter steel shaft. The shaft lengths are lengths of L1 = L3 = 1,250 mm and L2 = 850 mm. The number of teeth on gears B and E is NB = 60 teeth and NE = 108 teeth, respectively. Determine: (a) the maximum shear stress magnitude in shaft (1). (b) the maximum shear stress magnitude in shaft segment (3). (c) the rotation angle of gear E. (d) the rotation angle of gear C.

FIGURE P6.47

Solution Section Properties: The polar moments of inertia for the shafts are:

J1 = J 2 = J3 =

 32



( 30 mm ) = 79,521.6 mm4 4

( 35 mm ) = 147,324 mm4 4

Equilibrium: Consider a free-body diagram cut through shaft (2) and around gear C: M x = TC − T2 = 0

T2 = TC = 300 N  m

(a) FBD Gear C

Next, consider an FBD cut around gear B through shafts (1) and (2). The teeth of gear E exert a force F on the teeth of gear B. This force F opposes the rotation of gear B. The radius of gear B will be denoted by RB for now (even though the gear radius is not given explicitly). M x = T2 − T1 − F  RB = 0 (b) FBD Gear B

Finally, consider an FBD cut around gear E through shaft (3). The teeth of gear B exert an equal magnitude force F on the teeth of gear E, acting opposite to the direction assumed in the previous FBD. The radius of gear E will be denoted by RE for now. T (c) M x = −T3 − F  RE = 0 F = − 3 RE FBD Gear E The results of Eqs. (a) and (c) can be substituted into Eq. (b) to give R T1 = 300 N  m + T3 B RE The ratio RB/RE is simply the gear ratio between gears B and E, which can also be expressed in terms of gear teeth NB and NE: N  60 teeth  (d) T1 = 300 N  m + T3 B = 300 N  m + T3   = 300 N  m + 0.555556T3 NE  108 teeth  Equation (d) summarizes the results of the equilibrium considerations, but there are still two unknowns in this equation: T1 and T3. Consequently, this problem is statically indeterminate. To solve the problem, an additional equation must be developed. This second equation will be derived from the relationship between the angles of twist in shafts (1) and (3). Geometry-of-Deformation Relationship: The rotation of gear B is equal to the angle of twist in shaft (1): B = 1 and the rotation of gear E is equal to the angle of twist in shaft (3): E = 3 However, since the gear teeth must interlock, the rotation angles for gears B and E are not independent. The arclengths associated with the respective rotations must be equal, but the gears turn in opposite directions. The relationship between gear rotations can be stated as: RBB = − REE where RB and RE are the radii of gears B and E, respectively. Since the gear rotation angles are related to the shaft angles of twist, this relationship can be expressed as: (e) RB1 = −RE3 Torque-Twist Relationships: TL TL 1 = 1 1 3 = 3 3 J1G1 J 3G3

(f)

Compatibility Equation: Substitute the torque-twist relationships [Eqs. (f)] into the geometry-of-deformation relationship [Eq. (e)] to obtain: TL TL RB 1 1 = − RE 3 3 J1G1 J 3G3 which can be rearranged and expressed in terms of the gear ratio NB /NE: Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

N B T1L1 TL =− 3 3 N E J1G1 J 3G3 Note that the compatibility equation has two unknowns: T1 and T3. This equation can be solved simultaneously with Eq. (d) to calculate the internal torques in shafts (1) and (3).

(g)

Solve the Equations: Solve for internal torque T3 in Eq. (g):  N   L   J  G  T3 = −T1  B   1   3  3   N E   L3   J1  G1  and substitute this result into Eq. (d): T1 = 300 N  m + 0.555556T3

  N   L   J  G   = 300 N  m + 0.555556  −T1  B   1   3  3     N E   L3   J1  G1     60 teeth   147,324 mm 4   = 300 N  m − 0.555556 T1   4    108 teeth   79,521.6 mm   = 300 N  m − 0.571797 T1 Collect the T1 terms to obtain: 300 N  m T1 = = 190.864 N  m 1.571797 Backsubstitute this result into Eq. (d) to find the internal torque in shaft (3): T1 = 300 N  m + 0.555556T3

T − 300 N  m 190.864 N  m − 300 N  m T3 = 1 = = −196.444 N  m 0.555556 0.555556 (a) Maximum Shear Stress Magnitude in Shaft (1): T c (190.864 N  m )( 30 mm / 2 )(1,000 mm/m ) 1 = 1 1 = = 36.0 MPa J1 79,521.6 mm4 (b) Maximum Shear Stress Magnitude in Shaft (3): T c (196.444 N  m )( 35 mm / 2 )(1,000 mm/m ) 3 = 3 3 = = 23.3 MPa J3 147,324 mm4

Ans.

(c) Rotation Angle of Gear E: ( −196.444 N  m )(1, 250 mm )(1,000 mm/m ) = −0.020835 rad TL 3 = 3 3 = J 3G3 (147,324 mm4 )(80, 000 N/mm2 ) E = 3 = −0.0208 rad

Ans.

(d) Rotation Angle of Gear C: T L (190.864 N  m )(1, 250 mm )(1,000 mm/m ) 1 = 1 1 = = 0.037502 rad J1G1 ( 79,521.6 mm4 )(80, 000 N/mm2 )

2 =

T2 L2 ( 300 N  m )( 850 mm )(1,000 mm/m ) = = 0.040083 rad J 2G2 ( 79,521.6 mm4 )(80, 000 N/mm2 )

C = B + 2 = 1 + 2 = 0.037502 rad + 0.040083 rad = 0.0776 rad

Ans.

P6.48 The aluminum alloy [G = 4,000 ksi] pipe shown in Figure P6.48 is fixed to the wall support at C. The bolt holes in the flange at A were supposed to align with mating holes in the wall support; however, an angular misalignment of 4° was found to exist. To connect the pipe to its supports, a temporary installation torque T ′B must be applied at B to align flange A with the mating holes in the wall support. The outside diameter of the pipe is 5.5625 in. and its wall thickness is 0.258 in. Segment lengths are L1 = 16 ft and L2 = 24 ft. (a) Determine the temporary installation torque T′B that must be applied at B to align the bolt holes at A. (b) Determine the maximum shear stress initial in the pipe after the bolts are connected and the temporary installation torque at B is removed. (c) Determine the magnitude of the maximum shear stress in segments (1) and (2) if an external torque of TB = 150 kip∙in. is applied at B after the bolts are connected. FIGURE P6.48

Solution Section Properties: The polar moment of inertia for the pipe is: J1 = J 2 =

 

( 32 

5.5625 in.) − ( 5.0465 in.)  = 30.3158 in.4  4

(a) Temporary Installation Torque T′B: The shaft, which is fixed to the support at C, must be rotated 4° so that the connection at A can be completed. Consequently, segment (2) must be twisted 4° (or 0.069813 rad) by the temporary torque T′B. From the torque-twist relationship for shaft (2) TL 2 = 2 2 J 2G2 the temporary torque T′B is: 4 2 2 J 2G2 ( 0.069813 rad ) ( 30.3158 in. )( 4,000 kips/in. )  TB = = = 29.4 kip  in. Ans. L2 ( 24 ft )(12 in./ft ) (b) Maximum Shear Stress after Temporary Torque is Removed: After T′B is removed, the 4° angle of twist now applies to the total pipe length, i.e., 40 ft. The internal torque magnitude in the pipe due to the 4° misfit is thus: 4 2 misfit JG ( 0.069813 rad ) ( 30.3158 in. )( 4,000 kips/in. ) Tinitial = = = 17.637 kip  in. ( L1 + L2 ) ( 40 ft )(12 in./ft ) and the maximum shear stress magnitude in the pipe due to this internal torque is: T c (17.637 kip  in.)( 5.5626 in. / 2 )  initial = initial = = 1.618 ksi J 30.3158 in.4

Ans.

After Bolts Installed: After the bolts are installed, the pipe is analyzed as a statically indeterminate torsion structure.

Equilibrium: M x = −T1 + T2 + TB = 0 (a) Note: By inspection, the internal torque T2 will end up having a negative value. Geometry-of-Deformation Relationship: The two shafts are securely attached to fixed supports at A and C; however, in this instance, the sum of the angles of twist in the two members must equal the misfit angle. Note: If the rotation angle at A is a positive 4° (as can be inferred from the problem sketch), then the pipe must twist in a negative sense to reach a zero rotation angle at C. 1 + 2 = −0.069813 rad (b) Torque-Twist Relationships: TL TL 1 = 1 1 2 = 2 2 J1G1 J 2G2

(c)

Compatibility Equation: Substitute the torque-twist relationships [Eqs. (c)] into the Geometry-ofdeformation relationship [Eq. (b)] to obtain the compatibility equation: T1L1 T2 L2 (d) + = −0.069813 rad J1G1 J 2G2 Solve the Equations: Solve Eq. (d) for T1: JG L J G T1 = ( −0.069813 rad ) 1 1 − T2 2 1 1 L1 L1 J 2 G2

( 30.3158 in. ) ( 4, 000 ksi ) − T  24 ft  = −0.069813 rad (

)

(16 ft )(12 in./ft )

   16 ft 

= −44.092551 kip  in. − 1.5 T2 and substitute this result into Eq. (a) to compute the torque T2: −T1 + T2 = − ( −44.092551 kip  in. − 1.5 T2 ) + T2 = 44.092551 kip  in. + 2.5 T2 = −150 kip  in. −194.092551 kip  in. = −77.637 kip  in. 2.5 The torque in member (1) is therefore: T1 = T2 + 150 kip  in. = −77.637 kip  in. + 150 kip  in. = 72.363 kip  in. T2 =

(c) Maximum Shear Stress: The maximum shear stress magnitude in member (1) is: T c ( 72.363 kip  in.)( 5.5625 in. / 2) 1 = 1 1 = = 6.64 ksi J1 30.3158 in.4 The maximum shear stress magnitude in member (2) is: T c ( 77.637 kip  in.)( 5.5625 in. / 2 ) 2 = 2 2 = = 7.12 ksi J2 30.3158 in.4

Ans.

P6.49 A stepped shaft with a major diameter of D = 20 mm and a minor diameter of d = 16 mm is subjected to a torque of 25 N·m. A full quarter-circular fillet having a radius of r = 2 mm is used to transition from the major diameter to the minor diameter. Determine the maximum shear stress in the shaft.

Solution Fillet: From Fig. 6.18: D 20 mm = = 1.25 d 16 mm

r 2 mm = = 0.125 d 16 mm

 K  1.30

Section Properties of Minor Diameter Section:  J= (16 mm) 4 = 6, 434.0 mm 4 32 Maximum Shear Stress: Tc (25 N  m)(16 mm / 2)(1,000 mm/m)  =K = (1.30) = 40.4 MPa J 6, 434.0 mm 4

Ans.

P6.50 A fillet with a radius of 1/2 in. is used at the junction of a stepped shaft where the diameter is reduced from 8.00 in. to 6.00 in. Determine the maximum torque that the shaft can transmit if the maximum shear stress in the fillet must be limited to 5 ksi.

Solution Fillet: From Fig. 6.18: D 8.0 in. = = 1.33 d 6.0 in.

r 0.50 in. = = 0.0833 d 6.0 in.

 K  1.41

Section Properties of Minor Diameter Section:  J= (6.00 in.) 4 = 127.2345 in.4 32 Maximum Torque: Tc  =K J  J (5 ksi)(127.2345 in.4 ) Tmax = allow = = 150.4 kip  in. Kc (1.41)(6.00 in. / 2)

Ans.

P6.51 A stepped shaft with a major diameter of D = 2.50 in. and a minor diameter of d = 1.25 in. is subjected to a torque of 1,200 lb·in. If the maximum shear stress must not exceed 4,000 psi, determine the minimum radius r that may be used for a fillet at the junction of the two shaft segments. The fillet radius must be chosen as a multiple of 0.05 in.

Solution Section Properties of Minor Diameter Section:  J= (1.25 in.) 4 = 0.239684 in.4 32 Fillet Requirement: From Fig. 6.18: Tc  J (4, 000 psi)(0.239684 in.4 )  =K K = = = 1.28 J Tc (1, 200 lb  in.)(1.25 in. / 2) D 2.5 in. = = 2.00 d 1.25 in. r   0.17 d Minimum Fillet Radius: r rmin =   d = (0.17)(1.25 in.) = 0.213 in. d 

say rmin = 0.25 in.

Ans.

P6.52 A stepped shaft has a major diameter of D = 100 mm and a minor diameter of d = 75 mm. A fillet with a 10 mm radius is used to transition between the two shaft segments. The maximum shear stress in the shaft must be limited to 60 MPa. If the shaft rotates at a constant angular speed of 500 rpm, determine the maximum power that may be delivered by the shaft.

Solution Fillet: From Fig. 6.18: D 100 mm = = 1.33 d 75 mm

r 10 mm = = 0.133 d 75 mm

 K  1.29

Section Properties of Minor Diameter Section:  J= (75 mm) 4 = 3,106,311.1 mm 4 32 Maximum Torque: Tc  allow = K J

Tmax =

 allow J Kc

(60 N/mm 2 )(3,106,311.1 mm 4 ) = 3,852, 789 N  mm = 3,852.789 N  m (1.29)(75 mm / 2)

Power transmission: The maximum power that can be transmitted at 500 rpm is:  500 rev  2 rad  1 min  Pmax = Tmax = (3,852.789 N  m)      min  1 rev  60 s  = 201, 731.6 N  m/s

= 202 kW

Ans.

P6.53 A semicircular groove with a 6 mm radius is required in a 50 mm diameter shaft. If the maximum allowable shear stress in the shaft must be limited to 40 MPa, determine the maximum torque that can be transmitted by the shaft.

Solution Groove: From Fig. 6.17: d = 50 mm − 2(6 mm) = 38 mm

D 50 mm = = 1.32 d 38 mm

r 6 mm = = 0.158 d 38 mm

 K  1.39

Section Properties at Minimum Diameter Section:  J= (38 mm) 4 = 204,707.75 mm 4 32 Maximum Torque: Tc  =K J  J (40 N/mm 2 )(204, 707.75 mm 4 ) Tmax = allow = = 310, 045.8 N  mm = 310 N  m Kc (1.39)(38 mm/2)

Ans.

P6.54 A 1.25 in. diameter shaft contains a 0.25 in. deep U-shaped groove that has a 1/8 in. radius at the bottom of the groove. The maximum shear stress in the shaft must be limited to 12,000 psi. If the shaft rotates at a constant angular speed of 6 Hz, determine the maximum power that may be delivered by the shaft.

Solution Groove: From Fig. 6.17: d = 1.25 in. − 2(0.25 in.) = 0.75 in.

D 1.25 in. = = 1.67 d 0.75 in.

r 0.125 in. = = 0.167 d 0.75 in.

 K  1.39

Section Properties at Minimum Diameter Section:  J= (0.75 in.) 4 = 0.031063 in.4 32 Maximum Torque: Tc  =K J  J (12, 000 psi)(0.031063 in.4 ) Tmax = allow = = 715.1219 lb  in. = 59.5935 lb  ft Kc (1.39)(0.75 in. / 2) Power transmission: The maximum power that can be transmitted at 6 Hz is:  6 rev  2 rad  Pmax = Tmax = (59.5935 lb  ft)     s  1 rev  = 2, 246.6219 lb  ft/s or in units of horsepower, 2, 246.6219 lb  ft/s Pmax = = 4.08 hp 550 lb  ft/s 1 hp

Ans.

P6.55 A solid 1.50 in. × 1.50 in. square shaft is made of steel [G = 11,500 ksi] that has an allowable shear stress of 16 ksi. The shaft is used to transmit power from a tractor to farm implements, and it has a length of 6 ft. Determine: (a) the largest torque T that may be transmitted by the shaft. (b) the angle of twist in the shaft at the torque found in part (a).

Solution Theoretical Basis: The maximum shear stress in a rectangular section is given by Eq. (6.22): T  max = 2 a b and the angle of twist is given by Eq. (6.23) TL = 3  a bG where a is the short side of the rectangle and b is the long side of the rectangle. Square Section 1.5 in. × 1.5 in.: From Table 6.1, aspect ratio = 1.0   = 0.208 Maximum torque: T =  max a 2b

 = 0.1406

= (16 kips/in.2 ) ( 0.208 )(1.50 in.) (1.50 in.) 2

= 11.232 kip  in. = 11.23 kip  in. Angle of twist: (11.232 kip  in.)( 6 ft )(12 in./ft ) TL = 3 = = 0.0988 rad  a bG ( 0.1406 )(1.50 in.)3 (1.50 in.) (11,500 kips/in.2 )

Ans.

P6.56 The allowable shear stress for each bar shown in Figure P6.56/57 is 50 ksi. Assume that a = 0.5 in. and determine: (a) the largest torque T that may be applied to each bar. (b) the corresponding rotation angle at the free end if each bar has a length of 48 in. Assume that G = 12,500 ksi.

FIGURE P6.56/57

Solution Theoretical Basis: The maximum shear stress in a rectangular section is given by Eq. (6.22): T  max = 2 a b and the angle of twist is given by Eq. (6.23) TL = 3  a bG where a is the short side of the rectangle and b is the long side of the rectangle. Rectangular Section (1) From linear interpolation of the values given in Table 6.1, 8a aspect ratio = 5.333   = 0.294  = 0.294 1.5a Maximum torque: 2 T =  max a 2b = ( 50 kips/in.2 ) ( 0.294 )( 0.75 in.) ( 4.00 in.) = 33, 080.9 lb  in. = 2, 750 lb  ft Angle of twist: ( 33, 080.9 lb  in.)( 48 in.) TL = 3 = = 0.256 rad  a bG ( 0.294 )( 0.75 in.)3 ( 4.00 in.) (12.5 106 lb/in.2 ) Rectangular Section (2) From Table 6.1, 4a aspect ratio = 1.333   = 0.225  = 0.180 3a Maximum torque: 2 T =  max a 2b = ( 50 kips/in.2 ) ( 0.225)(1.50 in.) ( 2.00 in.) = 50,563.5 lb  in. = 4, 210 lb  ft Angle of twist: ( 50,563.5 lb  in.)( 48 in.) TL = 3 = = 0.1594 rad  a bG ( 0.180 )(1.50 in.)3 ( 2.00 in.) (12.5 106 lb/in.2 )

Ans.

Rectangular Section (3) From Table 6.1, 6a aspect ratio = 3.0   = 0.267  = 0.263 2a Maximum torque: 2 T =  max a 2b = ( 50 kips/in.2 ) ( 0.267 )(1.00 in.) ( 3.00 in.) = 40, 081.2 lb  in. = 3,340 lb  ft Angle of twist: ( 40, 081.2 lb  in.)( 48 in.) TL = 3 = = 0.1948 rad  a bG ( 0.263)(1.00 in.)3 ( 3.00 in.) (12.5 106 lb/in.2 )

Ans.

P6.57 The bars shown in Figure P6.56/57 have equal cross-sectional areas, and they are each subjected to a torque of T = 550 N·m. Using a = 10 mm, determine: (a) the maximum shear stress in each bar. (b) the rotation angle at the free end if each bar has a length of 900 mm. Assume that G = 28 GPa. FIGURE P6.56/57

Solution Theoretical Basis: The maximum shear stress in a rectangular section is given by Eq. (6.22): T  max = 2 a b and the angle of twist is given by Eq. (6.23) TL = 3  a bG where a is the short side of the rectangle and b is the long side of the rectangle. Rectangular Section (1) From linear interpolation of the values given in Table 6.1, 8a aspect ratio = 5.333   = 0.294  = 0.294 1.5a Shear stress: ( 550 N  m )(1,000 mm/m ) = 103.9 MPa T  max = 2 =  a b ( 0.294 )(15 mm )2 (80 mm ) Angle of twist: ( 550 N  m )( 900 mm )(1,000 mm/m ) = 0.223 rad TL = 3 =  a bG ( 0.294 )(15 mm )3 ( 80 mm ) ( 28,000 N/mm 2 ) Rectangular Section (2) From Table 6.1, 4a aspect ratio = 1.333   = 0.225  = 0.180 3a Shear stress: ( 550 N  m )(1,000 mm/m ) = 68.0 MPa T  max = 2 =  a b ( 0.225 )( 30 mm )2 ( 40 mm ) Angle of twist: ( 550 N  m )( 900 mm )(1,000 mm/m ) = 0.0907 rad TL = 3 =  a bG ( 0.180 )( 30 mm )3 ( 40 mm ) ( 28,000 N/mm 2 )

Ans.

Rectangular Section (3) From Table 6.1, 6a aspect ratio = 3.0   = 0.267  = 0.263 2a Shear stress: ( 550 N  m )(1,000 mm/m ) = 85.8 MPa T  max = 2 =  a b ( 0.267 )( 20 mm )2 ( 60 mm ) Angle of twist: ( 550 N  m )( 900 mm )(1,000 mm/m ) = 0.1399 rad TL = 3 =  a bG ( 0.263)( 20 mm )3 ( 60 mm ) ( 28,000 N/mm 2 )

Ans.

P6.58 The steel [G = 11,200 ksi] angle shape shown in Figure P6.58 is subjected to twisting moments about its longitudinal axis (i.e., the x axis) of T = 400 lb·ft at its ends. The angle shape is 36 in. long, and its cross-sectional dimensions are b = 3.0 in., d = 5.0 in., and t = 0.375 in. Determine: (a) the angle of twist of one end relative to the other end. (b) the maximum shear stress in the angle. FIGURE P6.58

Solution Theoretical Basis: The maximum shear stress in a rectangular section is given by Eq. (6.22): T  max = 2 a b and the angle of twist is given by Eq. (6.23) TL = 3  a bG where a is the short side of the rectangle and b is the long side of the rectangle. We consider torsion of an angle shape as equivalent to the torsion of a simple rectangle. The short side of the rectangle is taken as the thickness t of the angle. The long side of the rectangle is taken as the length of the angle’s centerline. The equivalent rectangle dimensions a and b are:

short side dimension a = t = 0.375 in. long side dimension b = b + d − t = 3.0 in. + 5.0 in. − 0.375 in. = 7.625 in. From linear interpolation of the values given in Table 6.1, 7.625 in. aspect ratio = 20.333   = 0.333 0.375 in.

 = 0.333

(a) Angle of twist: ( 400 lb  ft )( 36 in.)(12 in./ft ) TL = 3 = = 0.1152 rad  a bG ( 0.333)( 0.375 in.)3 ( 7.625 in.) (11.2 106 lb/in.2 )

Ans.

(b) Maximum shear stress: ( 400 lb  ft )(12 in./ft ) T  max = 2 = = 13, 440 psi  a b ( 0.333)( 0.375 in.)2 ( 7.625 in.)

Ans.

P6.59 A torque of T = 9,000 lb·ft will be applied to the hollow, thin-walled stainless steel section shown in Figure P6.59. Dimensions of the cross section are a = 4.0 in., b = 6.0 in., c = 5.0 in., and d = 8.0 in. If the maximum shear stress must be limited to 12 ksi, determine the minimum thickness t required for the section. (Note: The dimensions shown are measured to the wall centerline.)

FIGURE P6.59

Solution The maximum shear stress for a thin-walled section is given by Eq. (6.28) T  max = 2 Amt For the stainless steel section shown, Am = a  d + b  c

= ( 4.0 in.)( 8.0 in.) + ( 6.0 in.)( 5.0 in.) = 62 in.2

Given that the maximum shear stress must be limited to 12 ksi, the minimum thickness required for the section is: ( 9, 000 lb  ft )(12 in./ft ) = 0.0726 in. T Ans. tmin = = 2 max Am 2 (12, 000 lb/in.2 )( 62 in.2 )

P6.60 A torque of T = 9.7 kN·m will be applied to the hollow, thinwalled aluminum alloy section shown in Figure P6.60. Dimensions of the cross section are a = 150 mm, t1 = 15 mm, and t2 = 6 mm. Determine the magnitude of the maximum shear stress developed in the section.

FIGURE P6.60

Solution The maximum shear stress for a thin-walled section is given by Eq. (6.28) T  max = 2 Amt For the aluminum alloy section,



t t  Am =  a − 1 − 2  4 2 2



15 mm 6 mm   2 = 150 mm − −  = (139.5 mm ) 4 2 2  4 2

= 15, 284.0 mm 2 The magnitude of the maximum shear stress developed in the section is thus: ( 9.7 kN  m )(1,000 N/kN )(1,000 mm/m ) = 52.9 MPa T  max = = 2 Amt 2 (15, 284.0 mm 2 ) ( 6 mm )

Ans.

P6.61 A torque of T = 4,200 kN·m is applied to a torsion member whose cross section is shown in Figure P6.61. Dimensions of the cross section are a = 600 mm, b = 1,080 mm, d = 1,660 mm, t1 = 18 mm, t2 = 9 mm, and t3 = 15 mm. The dimensions given for a, b, and d are measured to the wall centerline. Determine the average shear stress acting at points A, B, and C.

FIGURE P6.61

Solution The maximum shear stress for a thin-walled section is given by Eq. (6.28) T  max = 2 Amt For this section, ( 2a + b ) + b  d = a + b d Am = ( ) 2 = ( 600 mm + 1, 080 mm )(1, 660 mm )

= 2.7888 106 mm 2 Average shear stress at point A: T  max = 2 Amt1

( 4.2 kN  m )(1,000 N/kN )(1,000 mm/m ) = 4.2 106 N  mm 2 ( 2.7888 106 mm 2 ) (18 mm ) 2 ( 2.7888 106 mm 2 ) (18 mm )

= 41.8 MPa

Ans.

Average shear stress at point B: T 4.2 106 N  mm  max = = = 83.7 MPa 2 Amt2 2 ( 2.7888 106 mm 2 ) ( 9 mm )

Ans.

Average shear stress at point C: T 4.2 106 N  mm  max = = = 50.2 MPa 2 Amt3 2 ( 2.7888 106 mm 2 ) (15 mm )

Ans.

P6.62 A torque of T = 190 lb·ft is applied to a torsion member whose cross section is shown in Figure P6.62. Dimensions of the cross section are a = 0.75 in., b = 2.25 in., c = 1.00 in., d = 2.75 in., t1 = 0.065 in., and t2 = 0.120 in. Determine the maximum average shear stress developed in the section. (Note: The dimensions shown are measured to the wall centerline.)

FIGURE P6.62

Solution The maximum shear stress for a thin-walled section is given by Eq. (6.28) T  max = 2 Amt For this section,

Am = b  d −

( b − c )( d − a ) 2

= ( 2.25 in.)( 2.75 in.) −

( 2.25 in. − 1.00 in.)( 2.75 in. − 0.75 in.) 2

= 4.9375 in.

Maximum average shear stress developed in the section: T  max = 2 Amt1 =

(190 lb  ft )(12 in./ft ) 2 ( 4.9375 in.2 ) ( 0.065 in.)

= 3,550 psi

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P7.1 For the beam shown, (a) derive equations for the shear force V and the bending moment M for any location in the beam. Place the origin at point A.) (b) use the derived functions to plot the shear-force and bending-moment diagrams for the beam. Specify the values for key points on the diagrams. FIGURE P7.1

Solution Beam equilibrium: Fy = Ay − w0 L = 0  Ay = w0 L L M A = − M A − w0 L   = 0 2 w L2 MA = − 0 2

Section a–a: Fy = w0 L − w0 x − V = 0

V = w0 L − w0 x = w0 ( L − x) M a − a =

w0 L2  x − w0 Lx + w0 x   + M = 0  2 2

w0 L2 w0 x 2 w M = − − + w0 L x = − 0 ( L2 + x 2 ) + w0 Lx 2 2 2 (b) Shear-force and bending-moment diagrams

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P7.2 For the beam shown, (a) derive equations for the shear force V and the bending moment M for any location in the beam. Place the origin at point A.) (b) use the derived functions to plot the shear-force and bending-moment diagrams for the beam. Specify the values for key points on the diagrams. FIGURE P7.2

Solution Beam equilibrium: M A = − Pa − Q ( a + b + c ) + C y ( a + b ) = 0 Cy =

Pa + Q ( a + b + c )

a+b M C = Pb − Qc − Ay ( a + b ) = 0  Ay =

Pb − Qc a+b

Section a–a: For the interval 0 ≤ x < a:

Fy = Ay − V = 0 V =

Pb − Qc a+b

M a −a = − Ay x + M = 0 M =

Pb − Qc x a+b

Section b–b: For the interval a ≤ x < a + b:

Fy = Ay − P − V = 0 Pb − Qc −P a+b Pb − Qc P ( a + b ) Pa + Qc = − = − a+b a+b a+b

V =

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

M a − a = − Ay x + P ( x − a ) + M = 0 P ( x − a )( a + b ) Pb − Qc x− a+b a+b 2 Pbx − Qcx − Pax − Pbx + Pa + Pab −Qcx − Pax + Pa ( a + b ) = = a+b a+b

 M = Ay x − P ( x − a ) =

= Pa −

Pa + Qc x a+b

Section c–c: For the interval a + b ≤ x < a + b + c:

Fy = V − Q = 0 V = Q M a − a = − M − Q ( a + b + c ) − x  = 0  M = −Q ( a + b + c ) − x  = Qx − Q ( a + b + c )

(b) Shear-force and bending-moment diagrams

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P7.3 For the beam shown, (a) derive equations for the shear force V and the bending moment M for any location in the beam. Place the origin at point A.) (b) use the derived functions to plot the shearforce and bending-moment diagrams for the beam. Specify the values for key points on the diagrams. FIGURE P7.3

Solution Beam equilibrium: Fy = − wa a − wbb + C y = 0

 C y = wa a + wbb a  b M C = wa a  b +  + wbb   + M C = 0 2  2 a  b  M C = − wa a  b +  − wbb   2   2 Section a–a: For the interval 0 ≤ x < a: Fy = − wa x − V = 0

 x M a − a = wa x   + M = 0  2

 V = − wa x M =−

wa x 2 2

Section b–b: For the interval a ≤ x < b: Fy = − wa a − wb ( x − a ) − V = 0

 V = − wa a − wb ( x − a ) a   x − a M b − b = wa a  x −  + wb ( x − a )  +M =0   2  2 a  w ( x − a)   M = − wa a  x −  − b  2 2

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

(b) Shear-force and bending-moment diagrams

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P7.4 For the beam shown, (a) derive equations for the shear force V and the bending moment M for any location in the beam. Place the origin at point A.) (b) use the derived functions to plot the shear-force and bending-moment diagrams for the beam. Specify the values for key points on the diagrams. FIGURE P7.4

Solution Beam equilibrium: a  b M C = wa a  b +  + wbb   − Ay (a + b) = 0 2  2 w a (a + 2b) + wbb 2  Ay = a 2(a + b) b a  M A = − wa a   − wbb  a +  + C y (a + b) = 0 2 2  w a 2 + wbb(2a + b) Cy = a 2(a + b)

Section a–a: For the interval 0 ≤ x < a: Fy = Ay − wa x − V = 0

V = − wa x + Ay = − wa x +

wa a (a + 2b) + wbb 2 2(a + b)

 x M a − a = − Ay x + wa x   + M = 0  2 M = −

wa x 2 w x 2  w a(a + 2b) + wbb 2  + Ay x = − a +  a x 2 2  2(a + b) 

Section b–b: For the interval a ≤ x < b: Fy = Ay − wa a − wb ( x − a) − V = 0

V = Ay − wa a − wb ( x − a) =

wa a(a + 2b) wbb 2 + − wa a − wb ( x − a) 2(a + b) 2(a + b)

a   x − a M b − b = − Ay x + wa a  x −  + wb ( x − a )  +M =0   2  2 x − a)  wa a(a + 2b) a a wbb 2  ( ( x − a)    M = Ay x − wa a  x −  − wb =  x − wa a  x −  − wb +    2 2 2 2 2(a + b)   2(a + b) 2

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

(b) Shear-force and bending-moment diagrams

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P7.5 For the beam shown, (a) derive equations for the shear force V and the bending moment M for any location in the beam. Place the origin at point A.) (b) use the derived functions to plot the shearforce and bending-moment diagrams for the beam. Specify the values for key points on the diagrams. FIGURE P7.5

Solution Beam equilibrium: Fy = − P + C y = 0 Cy = P M C = P ( a + b ) − M B + M C = 0  MC = M B − P (a + b)

Section a–a: For the interval 0 ≤ x < a:

Fy = − P − V = 0

 V = −P

M a −a = Px + M = 0

 M = − Px

Section b–b: For the interval a ≤ x < b:

Fy = − P − V = 0

 V = −P

M b −b = Px − M B + M = 0

 M = M B − Px

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

(b) Shear-force and bending-moment diagrams If M B  Pa , then

If M B  P ( a + b ) , then

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

If M B  P ( a + b ) , then

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P7.6 For the beam shown, (a) derive equations for the shear force V and the bending moment M for any location in the beam. Place the origin at point A.) (b) use the derived functions to plot the shear-force and bending-moment diagrams for the beam. Specify the values for key points on the diagrams. FIGURE P7.6

Solution Beam equilibrium: w0 L  L   =0 2 3

 Ay =

w0 L 6

w0 L  2 L   =0 2  3 

 By =

w0 L 3

M B = − Ay L + M A = By L −

Section a–a:

Fy = Ay −

w0 x  x  w0 L w0 x 2 − V = − −V = 0   L  2 6 2L

w0 L w0 x 2 V = − 6 2L M a − a = − Ay x +

w0 x  x   x  w0 Lx w0 x 3 + M = − + +M =0    L  2  3 6 6L

w0 x 3 w0 Lx M =− + 6L 6

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

(b) Shear-force and bending-moment diagrams

P7.7 For the beam shown, (a) derive equations for the shear force V and the bending moment M for any location in the beam. Place the origin at point A.) (b) use the derived functions to plot the shear-force and bending-moment diagrams for the beam. Specify the values for key points on the diagrams. FIGURE P7.7

Solution Beam equilibrium:

w0 a  2a  b    − w0b  a +  + C y (a + b) = 0 2  3  2  w0  2a 2 + 6ab + 3b 2  Cy = 6 (a + b)

M A = −

a b w a M C = w0b   + 0  b +  − Ay (a + b) = 0 3 2 2  w0  a 2 + 3ab + 3b 2   Ay = 6 (a + b) Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Section a–a: For the interval 0 ≤ x < a: w x2 Fy = Ay − 0 − V = 0 2a

V = Ay −

w0 x 2 w0 w x2  a 2 + 3ab + 3b 2  − 0 = 2a 6 (a + b) 2a

w0 x 2  x  M a −a =   − Ay x + M = 0 2a  3   M = Ay x −

w0 x3 w0 x w x3  a 2 + 3ab + 3b 2  − 0 = 6a 6 ( a + b) 6a

Section b–b: For the interval a ≤ x < b: Fy = C y − w0 ( a + b − x ) + V = 0

V = −C y + w0 ( a + b − x ) = −

w0  2a 2 + 6ab + 3b 2  + w0 ( a + b − x ) 6 (a + b)

M b −b = C y ( a + b − x ) −  M = Cy ( a + b − x ) −

w0 2 (a + b − x) − M = 0 2

w (a + b − x) w0 w 2 2  2a 2 + 6ab + 3b 2  − 0 ( a + b − x ) (a + b − x) = 0 2 6 ( a + b) 2

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

(b) Shear-force and bending-moment diagrams

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P7.8 Use the graphical method to construct the shear-force and bending-moment diagrams for the beam shown. Label all significant points on each diagram and identify the maximum moments (both positive and negative) along with their respective locations. Clearly differentiate straight-line and curved portions of the diagrams. FIGURE P7.8

Solution Beam equilibrium:

M A = − ( 35 kN )( 4 m ) − ( 45 kN )( 8 m )

− (15 kN )(14 m ) + Dy (10 m ) = 0  Dy = 71 kN

Fy = Ay + Dy − 35 kN − 45 kN − 15 kN = Ay + 71 kN − 35 kN − 45 kN − 15 kN = 0  Ay = 24 kN

Shear-force and bending-moment diagrams:

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P7.9 Use the graphical method to construct the shear-force and bending-moment diagrams for the beam shown. Label all significant points on each diagram and identify the maximum moments (both positive and negative) along with their respective locations. Clearly differentiate straight-line and curved portions of the diagrams. FIGURE P7.9

Solution Beam equilibrium:

M A = − ( 4.5 kips/ft )(12 ft )( 9 ft ) + C y (12 ft ) = 0  C y = 40.50 kips Fy = Ay + C y − ( 4.5 kips/ft )(12 ft ) = Ay + 40.50 kips − ( 4.5 kips/ft )(12 ft ) = 0  Ay = 13.50 kips Shear-force and bending-moment diagrams:

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P7.10 Use the graphical method to construct the shear-force and bending-moment diagrams for the beam shown. Label all significant points on each diagram and identify the maximum moments (both positive and negative) along with their respective locations. Clearly differentiate straight-line and curved portions of the diagrams. FIGURE P7.10

Solution

Beam equilibrium: M A = − ( 300 lb/ft )( 6 ft )( 9 ft ) + ( 6, 000 lb  ft ) − (1, 200 lb )( 30 ft ) + E y ( 24 ft ) = 0

 E y = 1,925 lb M E = ( 300 lb/ft )( 6 ft )(15 ft ) + ( 6, 000 lb  ft ) − (1, 200 lb )( 6 ft ) − Ay ( 24 ft ) = 0  Ay = 1, 075 lb Shear-force and bending-moment diagrams:

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P7.11 Use the graphical method to construct the shear-force and bending-moment diagrams for the beam shown. Label all significant points on each diagram and identify the maximum moments (both positive and negative) along with their respective locations. Clearly differentiate straight-line and curved portions of the diagrams. FIGURE P7.11

Solution Beam equilibrium:

Fy = Ay − ( 40 kN/m )( 3 m ) + 50 kN = 0  Ay = 70 kN M A = − M A − ( 40 kN/m )( 3 m )(1.5 m ) + (50 kN)(3 m) − 60 kN-m = 0  M A = −90 kN-m

Shear-force and bending-moment diagrams:

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P7.12 Use the graphical method to construct the shear-force and bending-moment diagrams for the beam shown. Label all significant points on each diagram and identify the maximum moments (both positive and negative) along with their respective locations. Clearly differentiate straight-line and curved portions of the diagrams. FIGURE P7.12

Solution Beam equilibrium:

M A = (10 kips/ft )( 9 ft )( 4.5 ft + 6 ft ) + Dy ( 20 ft ) = 0  Dy = −47.25 kips Fy = Ay + Dy + (10 kips/ft )( 9 ft ) = 0  Ay = −42.75 kips

Shear-force and bending-moment diagrams:

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P7.13 Use the graphical method to construct the shear-force and bending-moment diagrams for the beam shown. Label all significant points on each diagram and identify the maximum moments (both positive and negative) along with their respective locations. Clearly differentiate straight-line and curved portions of the diagrams. FIGURE P7.13

Solution Beam equilibrium:

Fy = −10 kips + C y = 0  C y = 10 kips M C = (10 kips )(10 ft ) − 60 kip-ft + M C = 0  M C = −40 kip-ft

Shear-force and bending-moment diagrams:

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P7.14 Use the graphical method to construct the shear-force and bending-moment diagrams for the beam shown. Label all significant points on each diagram and identify the maximum moments (both positive and negative) along with their respective locations. Clearly differentiate straight-line and curved portions of the diagrams. FIGURE P7.14

Solution

Beam equilibrium:

M A = 74 kip  ft + 56 kip  ft + Dy ( 40 ft ) = 0

 Dy = −3.25 kips

M D = 74 kip  ft + 56 kip  ft − Ay ( 40 ft ) = 0

 Ay = 3.25 kips

Shear-force and bending-moment diagrams:

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P7.15 Use the graphical method to construct the shear-force and bending-moment diagrams for the beam shown. Label all significant points on each diagram and identify the maximum moments (both positive and negative) along with their respective locations. Clearly differentiate straight-line and curved portions of the diagrams. FIGURE P7.15

Solution Beam equilibrium:

M A = 80 kN-m − ( 25 kN/m )( 6 m )( 3 m ) − 50 kN-m + By ( 6 m ) = 0

 By = 70 kN Fy = Ay + By − ( 25 kN/m )( 6 m ) = Ay + 70 kN − ( 25 kN/m )( 6 m ) = 0  Ay = 80 kN Shear-force and bending-moment diagrams:

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P7.16 Use the graphical method to construct the shear-force and bending-moment diagrams for the beam shown. Label all significant points on each diagram and identify the maximum moments (both positive and negative) along with their respective locations. Clearly differentiate straight-line and curved portions of the diagrams. FIGURE P7.16

Solution

Beam equilibrium:

M E = 6, 000 lb  ft − ( 2, 400 lb )(15 ft ) + ( 600 lb/ft )( 9 ft )( 7.5 ft ) + M E = 0 Fy = 2, 400 lb − ( 600 lb/ft )( 9 ft ) + E y = 0

 M E = −10,500 lb  ft

 E y = 3, 000 lb

Shear-force and bending-moment diagrams:

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P7.17 Draw the shear-force and bending-moment diagram for the beam shown in Figure P7.17. Assume the upward reaction provided by the ground to be uniformly distributed. Label all significant points on each diagram. Determine the maximum value of (a) the internal shear force and (b) the internal bending moment. FIGURE P7.17

Solution Beam equilibrium: Fy = − ( 2 kips/ft )( 8 ft ) − 25 kips − 25 kips + w (16 ft ) = 0

 w = 4.125 kips/ft

Shear-force and bending-moment diagrams (a) Maximum value of internal shear force: V = 16.50 kips @ x = 4 ft

Ans.

V = −16.50 kips @ x = 12 ft

Ans.

(b) Maximum value of internal bending moment: M = 33 kip-ft @ x = 4 ft, 12 ft

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P7.18 Draw the shear-force and bending-moment diagram for the beam shown in Figure P7.18. Assume the upward reaction provided by the ground to be uniformly distributed. Label all significant points on each diagram. Determine the maximum value of (a) the internal shear force and (b) the internal bending moment. FIGURE P7.18

Solution Beam equilibrium: Fy = − ( 40 kN/m )(1 m ) − 50 kN − ( 40 kN/m )(1 m ) + w ( 4 m ) = 0  w = 32.5 kN/m

Shear-force and bending-moment diagrams (a) Maximum value of internal shear force: V = ±25 kN @ x = 2 m

Ans.

(b) Maximum value of internal bending moment: M = −4.62 kN-m @ x = 1.23 m M = −4.62 kN-m @ x = 2.77 m Mmax = 5.00 kN-m

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P7.19 Use the graphical method to construct the shear-force and bending-moment diagrams for the beam shown. Label all significant points on each diagram and identify the maximum moments along with their respective locations. Additionally: (a) Determine V and M in the beam at a point located 0.75 m to the right of B. (b) Determine V and M in the beam at a point located 1.25 m to the left of C. FIGURE P7.19

Solution Beam equilibrium:

M B = (15 kN )( 3 m ) − ( 40 kN/m )( 6 m )( 3 m ) − (18 kN )(10 m) + C y ( 6 m) = 0  C y = 142.50 kN Fy = By + C y − 15 kN − ( 40 kN/m )( 6 m ) − 18 kN = By + 142.5 kN − 15 kN − ( 40 kN/m )( 6 m ) − 18 kN = 0  By = 130.50 kN Shear-force and bending-moment diagrams Shear force V and bending moment M at specific locations: (a) At x = 3.75 m, V = 85.5 kN

Ans.

M = 30.4 kN-m

Ans.

(b) At x = 7.75 m, V = −74.5 kN

Ans.

M = 52.4 kN-m

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P7.20 Use the graphical method to construct the shearforce and bending-moment diagrams for the beam shown. Label all significant points on each diagram and identify the maximum moments along with their respective locations. Additionally: (a) Determine V and M in the beam at a point located 0.75 m to the right of B. (b) Determine V and M in the beam at a point located 1.25 m to the left of C. FIGURE P7.20

Solution Beam equilibrium: Fy = −75 kN − 60 kN + ( 35 kN/m )( 6 m ) + C y = 0  C y = −75.00 kN M C = ( 75 kN )( 6 m ) + ( 60 kN )( 3.5 m ) − 120 kN-m − ( 35 kN/m )( 6 m )( 3 m ) + M C = 0  M C = 90.00 kN-m

Shear-force and bending-moment diagrams Shear force V and bending moment M at specific locations: (a) At x = 3.25 m, V = −21.3 kN

Ans.

M = 16.09 kN-m

Ans.

(b) At x = 4.75 m, V = 31.25 kN

Ans.

M = 23.6 kN-m

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P7.21 Use the graphical method to construct the shear-force and bending-moment diagrams for the beam shown. Label all significant points on each diagram and identify the maximum moments along with their respective locations. Additionally: (a) Determine V and M in the beam at a point located 1.50 m to the right of B. (b) Determine V and M in the beam at a point located 1.25 m to the left of D. FIGURE P7.21

Solution Beam equilibrium:

M B = ( 52 kN )( 3 m ) − ( 35 kN/m )( 9 m )( 4.5 m )

− 150 kN-m − ( 36 kN )(12 m ) + Dy ( 9 m ) = 0  Dy = 204.83 kN

Fy = By + Dy − 52 kN − ( 35 kN/m )( 9 m ) − 36 kN = By + 204.83 kN − 52 kN − ( 35 kN/m )( 9 m ) − 36 kN = 0  By = 198.17 kN Shear-force and bending-moment diagrams Shear force V and bending moment M at specific locations: (a) At x = 4.5 m, V = 93.7 kN

Ans.

M = 23.9 kN-m

Ans.

(b) At x = 10.75 m, V = −125.1 kN

Ans.

M = 75.7 kN-m

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P7.22 Use the graphical method to construct the shearforce and bending-moment diagrams for the beam shown. Label all significant points on each diagram and identify the maximum moments along with their respective locations. Additionally: (a) Determine V and M in the beam at a point located 1.50 m to the right of B. (b) Determine V and M in the beam at a point located 1.25 m to the left of D. FIGURE P7.22

Solution Beam equilibrium:

M B = (160 kN )( 2 m ) + ( 50 kN/m )( 2 m )(1 m ) − ( 50 kN/m )( 2 m )(1 m ) − (120 kN/m )( 5 m )( 4.5 m ) + Dy ( 7 m ) = 0  Dy = 340 kN Fy = By + Dy − 160 kN − ( 50 kN/m )( 4 m ) − (120 kN/m )( 5 m ) = By + 340 kN − 160 kN − ( 50 kN/m )( 4 m ) − (120 kN/m )( 5 m ) = 0  By = 620 kN Shear-force and bending-moment diagrams Shear force V and bending moment M at specific locations: (a) At x = 3.50 m, V = 285 kN

Ans.

M = 63.8 kN-m

Ans.

(b) At x = 7.75 m, V = −190.0 kN

Ans.

M = 331 kN-m

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P7.23 Use the graphical method to construct the shear-force and bending-moment diagrams for the beam shown. Label all significant points on each diagram and identify the maximum moments along with their respective locations. Clearly differentiate straight-line and curved portions of the diagrams. FIGURE P7.23

Solution

Beam equilibrium: 1  2   M A = − ( 6 kips/ft )(12 ft )   (12 ft )  − ( 8 kips )( 30 ft ) + C y ( 22 ft ) = 0 2  3   1  1  ( 6 kips/ft )(12 ft )   (12 ft ) + 10 ft  − (8 kips )(8 ft ) − Ay ( 22 ft ) = 0 2  3   Shear-force and bending-moment diagrams M C =

 C y = 24 kips  Ay = 20 kips

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P7.24 Use the graphical method to construct the shear-force and bending-moment diagrams for the beam shown. Label all significant points on each diagram and identify the maximum moments along with their respective locations. Clearly differentiate straight-line and curved portions of the diagrams. FIGURE P7.24

Solution

Beam equilibrium:

 18 ft  M B = ( 50 kips )( 6 ft ) + 560 kip  ft − ( 20 kips/ft )(18 ft )  + 16 ft  + E y ( 28 ft ) = 0  2  18 ft   M E = ( 50 kips )( 34 ft ) + 560 kip  ft − ( 20 kips/ft )(18 ft )  6 ft −  − By ( 28 ft ) = 0 2   Shear-force and bending-moment diagrams

 E y = 290.71 kips  By = 119.29 kips

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P7.25 Use the graphical method to construct the shear-force and bending-moment diagrams for the beam shown. Label all significant points on each diagram and identify the maximum moments along with their respective locations. Clearly differentiate straight-line and curved portions of the diagrams. FIGURE P7.25

Solution Beam equilibrium:

M A = − M A − ( 50 kN/m )( 2 m )(1 m ) + ( 20 kN )( 2 m ) + ( 25 kN/m )( 3 m )( 3.5 m ) − ( 50 kN )( 5 m ) = 0  M A = −47.50 kN-m

Fy = Ay − ( 50 kN/m )( 2 m ) + 20 kN + ( 25 kN/m )( 3 m ) − 50 kN = 0  Ay = 55 kN Shear-force and bending-moment diagrams

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P7.26 Use the graphical method to construct the shear-force and bending-moment diagrams for the beam shown. Label all significant points on each diagram and identify the maximum moments along with their respective locations. Clearly differentiate straight-line and curved portions of the diagrams. FIGURE P7.26

Solution Beam equilibrium:

M C = ( 20 kips )(15 ft ) − ( 6 kips/ft )( 8 ft )(11 ft ) − (12 kips/ft )( 7 ft )( 3.5 ft ) + ( 70 kips )( 7 ft ) + M C = 0  M C = 32.00 kip-ft

Fy = −20 kips + ( 6 kips/ft )( 8 ft ) − 70 kips + (12 kips/ft )( 7 ft ) + C y = 0  C y = −42.00 kips

Shear-force and bending-moment diagrams

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P7.27 Use the graphical method to construct the shear-force and bending-moment diagrams for the beam shown. Label all significant points on each diagram and identify the maximum moments along with their respective locations. Clearly differentiate straight-line and curved portions of the diagrams. FIGURE P7.27

Solution Beam equilibrium:

Consider free-body diagram of DE:

M D = − ( 55 kN/m )( 3 m )(1.5 m ) + E y ( 3 m ) = 0  E y = 82.5 kN Fy = Dy + E y − ( 55 kN/m )( 3 m ) = Dy + 82.5 kN − ( 55 kN/m )( 3 m ) = 0  Dy = 82.5 kN

Consider free-body diagram of ABCD:

M A = 60 kN-m − ( 75 kN/m )( 5 m )( 2.5 m ) − (100 kN )( 2.5 m ) − Dy ( 5 m ) + C y ( 3.5 m ) = 60 kN-m − ( 75 kN/m )( 5 m )( 2.5 m ) − (100 kN )( 2.5 m ) − ( 82.5 kN )( 5 m ) + C y ( 3.5 m ) = 0  C y = 440 kN Fy = Ay + C y − ( 75 kN/m )( 5 m ) − 100 kN − Dy = Ay + 440 kN − ( 75 kN/m )( 5 m ) − 100 kN − 82.5 kN = 0  Ay = 117.5 kN

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Shear-force and bending-moment diagrams

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P7.28 Use the graphical method to construct the shear-force and bending-moment diagrams for the beam shown. Label all significant points on each diagram and identify the maximum moments along with their respective locations. Clearly differentiate straight-line and curved portions of the diagrams. FIGURE P7.28

Solution Beam equilibrium:

Consider free-body diagram of ABC: M A = − ( 500 lb/ft )(10 ft )( 5 ft ) + C y (15 ft ) = 0

 C y = 1,666.67 lb Fy = Ay + C y − ( 500 lb/ft )(10 ft ) = Ay + 1,666.67 lb − ( 500 lb/ft )(10 ft ) = 0  Ay = 3,333.33 lb Consider free-body diagram of CDE: Fy = −C y + E y − 1, 200 lb

= − (1, 666.67 lb ) + E y − 1, 200 lb = 0  E y = 2,866.67 lb M E = C y (10 ft ) + (1, 200 lb )( 8 ft ) + M E = (1, 666.67 lb )(10 ft ) + (1, 200 lb )( 8 ft ) + M E = 0  M E = −26, 266.67 lb-ft

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Shear-force and bending-moment diagrams

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P7.29 Use the graphical method to construct the shear-force and bending-moment diagrams for the beam shown. Label all significant points on each diagram and identify the maximum moments along with their respective locations. Clearly differentiate straight-line and curved portions of the diagrams. FIGURE P7.29

Solution Beam equilibrium:

M A = − ( 70 kN/m )( 7 m ) (3.5 m)

− 12 ( 70 kN/m )( 3 m )( 8 m )

− ( 55 kN )(10 m ) + By ( 7 m ) = 0

 By = 443.57 kN Fy = Ay + By − ( 70 kN/m )( 7 m ) − 12 ( 70 kN/m )( 3 m ) − 55 kN

= Ay + 443.57 kN − ( 70 kN/m )( 7 m ) − 12 ( 70 kN/m )( 3 m ) − 55 kN = 0  Ay = 206.43 kN Shear-force and bending-moment diagrams

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P7.30 Use the graphical method to construct the shear-force and bending-moment diagrams for the beam shown. Label all significant points on each diagram and identify the maximum moments along with their respective locations. Clearly differentiate straight-line and curved portions of the diagrams. FIGURE P7.30

Solution Beam equilibrium:

 8 ft  M B = 12 ( 6 kips/ft )( 8 ft )   3  − ( 4 kips/ft )(15 ft ) (14.5 ft) + Dy (17 ft ) = 0  Dy = 47.41 kips Fy = By + Dy − 12 ( 6 kips/ft )( 8 ft ) − ( 4 kips/ft )(15 ft ) = By + 47.41 kips − 12 ( 6 kips/ft )( 8 ft ) − ( 4 kips/ft )(15 ft ) = 0  By = 36.59 kips Shear-force and bending-moment diagrams

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P7.31 Use the graphical method to construct the shear-force and bending-moment diagrams for the beam shown. Label all significant points on each diagram and identify the maximum moments along with their respective locations. Clearly differentiate straight-line and curved portions of the diagrams. FIGURE P7.31

Solution Beam equilibrium:

M D = (25 kN)(6 m)

+ 12 ( 30 kN/m )( 3 m)( 3 m ) + M D = 0

 M D = −285.00 kN-m

Fy = Dy − 25 kN + 12 ( 30 kN/m )( 3 m) = 0  Dy = 70 kN

Shear-force and bending-moment diagrams

P7.32 For the beam and loading shown, (a) Use discontinuity functions to write the expression for w(x). Include the beam reactions in this expression. (b) Integrate w(x) twice to determine V(x) and M(x). (c) Use V(x) and M(x) to plot the shear-force and bending-moment diagrams. FIGURE P7.32

Solution Beam equilibrium: M A = − (180 lb)(2 ft) − (450 lb)(6 ft) + Dy (9 ft) = 0  Dy = 340 lb Fy = Ay + Dy − 180 lb − 450 lb = 0  Ay = 290 lb

Load function w(x): −1 −1 −1 −1 w( x) = 290 lb x − 0 ft − 180 lb x − 2 ft − 450 lb x − 6 ft + 340 lb x − 9 ft

Ans.

Shear-force function V(x) and bending-moment function M(x): 0 0 0 0 V ( x) = 290 lb x − 0 ft − 180 lb x − 2 ft − 450 lb x − 6 ft + 340 lb x − 9 ft

Ans.

M ( x) = 290 lb x − 0 ft − 180 lb x − 2 ft − 450 lb x − 6 ft + 340 lb x − 9 ft

Ans.

Shear-force and bending-moment diagrams:

P7.33 For the beam and loading shown, (a) Use discontinuity functions to write the expression for w(x). Include the beam reactions in this expression. (b) Integrate w(x) twice to determine V(x) and M(x). (c) Use V(x) and M(x) to plot the shear-force and bending-moment diagrams. FIGURE P7.33

Solution Beam equilibrium: M B = (10 kN)(2.5 m) − (35 kN)(3 m) + Dy (5 m) = 0  Dy = 16 kN Fy = By + Dy − 10 kN − 35 kN = 0  By = 29 kN

Load function w(x): −1 −1 −1 −1 w( x) = −10 kN x − 0 m + 29 kN x − 2.5 m − 35 kN x − 5.5 m + 16 kN x − 7.5 m

Ans.

Shear-force function V(x) and bending-moment function M(x): 0 0 0 0 V ( x) = −10 kN x − 0 m + 29 kN x − 2.5 m − 35 kN x − 5.5 m + 16 kN x − 7.5 m

Ans.

M ( x) = −10 kN x − 0 m + 29 kN x − 2.5 m − 35 kN x − 5.5 m + 16 kN x − 7.5 m

Ans.

Shear-force and bending-moment diagrams:

P7.34 For the beam and loading shown, (a) Use discontinuity functions to write the expression for w(x). Include the beam reactions in this expression. (b) Integrate w(x) twice to determine V(x) and M(x). (c) Use V(x) and M(x) to plot the shear-force and bending-moment diagrams. FIGURE P7.34

Solution Beam equilibrium: M A = − (30 kN)(3 m) − (20 kN)(7 m) −(15 kN)(15 m) + Dy (10 m) = 0  Dy = 45.5 kN Fy = Ay + Dy − 30 kN − 20 kN − 15 kN = 0  Ay = 19.5 kN

Load function w(x): −1 −1 −1 w( x) = 19.5 kN x − 0 m − 30 kN x − 3 m − 20 kN x − 7 m

+45.5 kN x − 10 m

−1

− 15 kN x − 15 m

Ans.

Shear-force function V(x) and bending-moment function M(x): 0 0 0 V ( x) = 19.5 kN x − 0 m − 30 kN x − 3 m − 20 kN x − 7 m

+45.5 kN x − 10 m − 15 kN x − 15 m 0

M ( x) = 19.5 kN x − 0 m − 30 kN x − 3 m − 20 kN x − 7 m 1

+45.5 kN x − 10 m − 15 kN x − 15 m 1

Ans. 1

Ans.

Shear-force and bending-moment diagrams:

P7.35 For the beam and loading shown, (a) Use discontinuity functions to write the expression for w(x). Include the beam reactions in this expression. (b) Integrate w(x) twice to determine V(x) and M(x). (c) Use V(x) and M(x) to plot the shear-force and bending-moment diagrams. FIGURE P7.35

Solution Beam equilibrium: Fy = C y − 5 kN = 0  C y = 5 kN M C = (5 kN)(6 m) − 20 kN-m + M C = 0  M C = −10 kN-m

Load function w(x): −1 −2 −1 −2 w( x) = −5 kN x − 0 m + 20 kN-m x − 3 m + 5 kN x − 6 m + 10 kN-m x − 6 m

Ans.

Shear-force function V(x) and bending-moment function M(x): 0 −1 0 −1 V ( x) = −5 kN x − 0 m + 20 kN-m x − 3 m + 5 kN x − 6 m + 10 kN-m x − 6 m

Ans.

M ( x) = −5 kN x − 0 m + 20 kN-m x − 3 m + 5 kN x − 6 m + 10 kN-m x − 6 m

Ans.

Shear-force and bending-moment diagrams:

P7.36 For the beam and loading shown, (a) Use discontinuity functions to write the expression for w(x). Include the beam reactions in this expression. (b) Integrate w(x) twice to determine V(x) and M(x). (c) Use V(x) and M(x) to plot the shear-force and bending-moment diagrams. FIGURE P7.36

Solution Beam equilibrium: Fy = Ay − (35 kN/m)(2 m) = 0  Ay = 70 kN M A = −(35 kN/m)(2 m)(4 m) − M A = 0  M A = −280 kN-m

Load function w(x): −1 −2 0 0 w( x) = 70 kN x − 0 m − 280 kN-m x − 0 m − 35 kN/m x − 3 m + 35 kN/m x − 5 m Shear-force function V(x) and bending-moment function M(x): 0 −1 1 1 V ( x) = 70 kN x − 0 m − 280 kN-m x − 0 m − 35 kN/m x − 3 m + 35 kN/m x − 5 m 35 kN/m 35 kN/m 1 0 2 2 M ( x) = 70 kN x − 0 m − 280 kN-m x − 0 m − x−3 m + x−5 m 2 2

Ans.

Ans. Ans.

Shear-force and bending-moment diagrams:

P7.37 For the beam and loading shown, (a) Use discontinuity functions to write the expression for w(x). Include the beam reactions in this expression. (b) Integrate w(x) twice to determine V(x) and M(x). (c) Use V(x) and M(x) to plot the shear-force and bending-moment diagrams. FIGURE P7.37

Solution Beam equilibrium:

M A = − (25 kN/m)(4 m)(2 m) − (32 kN)(6 m) + Dy (8 m) = 0  Dy = 49 kN Fy = Ay + Dy − (25 kN/m)(4 m) − 32 kN = 0  Ay = 83 kN

Load function w(x): −1 0 0 w( x) = 83 kN x − 0 m − 25 kN/m x − 0 m + 25 kN/m x − 4 m

−32 kN x − 6 m

−1

+ 49 kN x − 8 m

Ans.

Shear-force function V(x) and bending-moment function M(x): 0 1 1 V ( x) = 83 kN x − 0 m − 25 kN/m x − 0 m + 25 kN/m x − 4 m

−32 kN x − 6 m + 49 kN x − 8 m 25 kN/m 25 kN/m 1 2 2 M ( x) = 83 kN x − 0 m − x−0 m + x−4 m 2 2 0

−32 kN x − 6 m + 49 kN x − 8 m

Ans.

Shear-force and bending-moment diagrams:

P7.38 For the beam and loading shown, (a) Use discontinuity functions to write the expression for w(x). Include the beam reactions in this expression. (b) Integrate w(x) twice to determine V(x) and M(x). (c) Use V(x) and M(x) to plot the shear-force and bending-moment diagrams. FIGURE P7.38

Solution Beam equilibrium: M A = −(3,000 lb)(5 ft) + 8,000 lb-ft −(800 lb/ft)(7 ft)(12.5 ft) + E y (20 ft) = 0  E y = 3,850 lb Fy = Ay + E y − 3,000 lb − (800 lb/ft)(7 ft) = 0  Ay = 4,750 lb

Load function w(x): −1 −1 −2 w( x) = 4,750 lb x − 0 ft − 3,000 lb x − 5 ft − 8,000 lb-ft x − 5 ft

−800 lb/ft x − 9 ft + 800 lb/ft x − 16 ft + 3,850 lb x − 20 ft Shear-force function V(x) and bending-moment function M(x): 0 0 −1 V ( x) = 4,750 lb x − 0 ft − 3,000 lb x − 5 ft − 8,000 lb-ft x − 5 ft 0

−800 lb/ft x − 9 ft + 800 lb/ft x − 16 ft + 3,850 lb x − 20 ft 1

M ( x) = 4,750 lb x − 0 ft − 3,000 lb x − 5 ft − 8,000 lb-ft x − 5 ft 1

−1

Ans.

800 lb/ft 800 lb/ft 2 2 1 x − 9 ft + x − 16 ft + 3,850 lb x − 20 ft 2 2 Shear-force and bending-moment diagrams: −

Ans.

P7.39 For the beam and loading shown, (a) Use discontinuity functions to write the expression for w(x). Include the beam reactions in this expression. (b) Integrate w(x) twice to determine V(x) and M(x). (c) Use V(x) and M(x) to plot the shear-force and bending-moment diagrams. FIGURE P7.39

Solution Beam equilibrium: Fy = Ay − (800 lb/ft)(12 ft) − (800 lb/ft)(6 ft) = 0  Ay = 14, 400 lb M A = − (800 lb/ft)(12 ft)(6 ft) − (800 lb/ft)(6 ft)(21 ft) − M A = 0  M A = −158, 400 lb-ft

Load function w(x): −1 −2 0 w( x) = 14,400 lb x − 0 ft − 158,400 lb-ft x − 0 ft − 800 lb/ft x − 0 ft

+800 lb/ft x − 12 ft − 800 lb/ft x − 18 ft + 800 lb/ft x − 24 ft Shear-force function V(x) and bending-moment function M(x): 0 −1 1 V ( x) = 14,400 lb x − 0 ft − 158,400 lb-ft x − 0 ft − 800 lb/ft x − 0 ft 0

+800 lb/ft x − 12 ft − 800 lb/ft x − 18 ft + 800 lb/ft x − 24 ft 800 lb/ft 1 0 2 M ( x) = 14,400 lb x − 0 ft − 158,400 lb-ft x − 0 ft − x − 0 ft 2 800 lb/ft 800 lb/ft 800 lb/ft 2 2 2 + x − 12 ft − x − 18 ft + x − 24 ft 2 2 2 Shear-force and bending-moment diagrams: 1

Ans.

P7.40 For the beam and loading shown, (a) Use discontinuity functions to write the expression for w(x). Include the beam reactions in this expression. (b) Integrate w(x) twice to determine V(x) and M(x). (c) Use V(x) and M(x) to plot the shear-force and bending-moment diagrams. FIGURE P7.40

Solution Beam equilibrium: M A = −12 kN-m + (18 kN/m)(2 m)(2 m) + Dy (5 m) = 0  Dy = −12 kN Fy = Ay + Dy + (18 kN/m)(2 m) = 0  Ay = −24 kN

Reminder: The roller symbol implies that both upward and downward displacement is restrained.

Load function w(x): −1 −2 0 w( x) = −24 kN x − 0 m + 12 kN-m x − 0 m + 18 kN/m x − 1 m

−18 kN/m x − 3 m − 12 kN x − 5 m 0

−1

Ans.

Shear-force function V(x) and bending-moment function M(x): 0 −1 1 V ( x) = −24 kN x − 0 m + 12 kN-m x − 0 m + 18 kN/m x − 1 m

−18 kN/m x − 3 m − 12 kN x − 5 m 1

Ans.

M ( x) = −24 kN x − 0 m + 12 kN-m x − 0 m + 1

−

18 kN/m 2 1 x − 3 m − 12 kN x − 5 m 2

18 kN/m 2 x −1 m 2 Ans.

Shear-force and bending-moment diagrams:

P7.41 For the beam and loading shown, (a) Use discontinuity functions to write the expression for w(x). Include the beam reactions in this expression. (b) Integrate w(x) twice to determine V(x) and M(x). (c) Use V(x) and M(x) to plot the shear-force and bending-moment diagrams. FIGURE P7.41

Solution Beam equilibrium:

M A = −(6 kips/ft)(22 ft)(11 ft) 2(8 ft)   − 12 (9 kips/ft)(8 ft)  22 ft +  + By (22 ft) = 0  3   By = 110.73 kips Fy = Ay + By − (6 kips/ft)(22 ft) − 12 (9 kips/ft)(8 ft) = Ay + (110.73 kips) − (6 kips/ft)(22 ft) − 12 (9 kips/ft)(8 ft) = 0  Ay = 57.27 kips Load function w(x): −1 0 −1 0 w( x) = 57.27 kips x − 0 ft − 6 kips/ft x − 0 ft + 110.73 kips x − 22 ft + 6 kips/ft x − 22 ft −

9 kips/ft 9 kips/ft 1 1 0 x − 22 ft + x − 30 ft + 9 kips/ft x − 30 ft 8 ft 8 ft

Ans. Shear-force function V(x) and bending-moment function M(x): 0 1 0 1 V ( x) = 57.27 kips x − 0 ft − 6 kips/ft x − 0 ft + 110.73 kips x − 22 ft + 6 kips/ft x − 22 ft −

9 kips/ft 9 kips/ft 2 2 1 x − 22 ft + x − 30 ft + 9 kips/ft x − 30 ft 2(8 ft) 2(8 ft)

Ans.

6 kips/ft 6 kips/ft 2 1 2 x − 0 ft + 110.73 kips x − 22 ft + x − 22 ft 2 2 9 kips/ft 9 kips/ft 9 kips/ft 3 3 2 − x − 22 ft + x − 30 ft + x − 30 ft 6(8 ft) 6(8 ft) 2

M ( x) = 57.27 kips x − 0 ft − 1

Ans.

Shear-force and bending-moment diagrams

P7.42 For the beam and loading shown, (a) Use discontinuity functions to write the expression for w(x). Include the beam reactions in this expression. (b) Integrate w(x) twice to determine V(x) and M(x). (c) Use V(x) and M(x) to plot the shear-force and bending-moment diagrams. FIGURE P7.42

Solution Beam equilibrium: Fy = C y − (20 kN/m)(3 m) − 12 (30 kN/m)(3 m) = 0

 C y = 105 kN M C = (20 kN/m)(3 m)(2.5 m) + 12 (30 kN/m)(3 m)(2 m) + M C = 0  M C = −240 kN-m Load function w(x): w( x) = −20 kN/m x − 0 m − 0

30 kN/m 30 kN/m 1 0 1 x − 0 m + 20 kN/m x − 3 m + x−3 m 3m 3m −1

−2

+30 kN/m x − 3 m + 105 kN x − 4 m + 240 kN-m x − 4 m Shear-force function V(x) and bending-moment function M(x): 30 kN/m 30 kN/m 1 2 1 2 V ( x) = −20 kN/m x − 0 m − x − 0 m + 20 kN/m x − 3 m + x−3 m 2(3 m) 2(3 m) 0

+30 kN/m x − 3 m + 105 kN x − 4 m + 240 kN-m x − 4 m 1

−1

20 kN/m 30 kN/m 20 kN/m 30 kN/m 2 3 2 3 x−0 m − x−0 m + x−3 m + x−3 m 2 6(3 m) 2 6(3 m) 30 kN/m 2 1 0 + x − 3 m + 105 kN x − 4 m + 240 kN-m x − 4 m 2 Shear-force and bending-moment diagrams:

Ans.

M ( x) = −

Ans.

P7.43 For the beam and loading shown, (a) Use discontinuity functions to write the expression for w(x). Include the beam reactions in this expression. (b) Integrate w(x) twice to determine V(x) and M(x). (c) Determine the maximum bending moment in the beam between the two simple supports. FIGURE P7.43

Solution Beam equilibrium:

M B = 9 kN-m − 12 (18 kN/m)(3 m)(1 m) + C y (3 m) = 0  C y = 6 kN Fy = By + C y − 12 (18 kN/m)(3 m) = 0  By = 21 kN Load function w(x): −2 −1 0 w( x) = −9 kN-m x − 0 m + 21 kN x − 1 m − 18 kN/m x − 1 m +

18 kN/m 18 kN/m 1 1 −1 x −1 m − x − 4 m + 6 kN x − 4 m 3m 3m

Ans.

Shear-force function V(x) and bending-moment function M(x): −1 0 1 V ( x) = −9 kN-m x − 0 m + 21 kN x − 1 m − 18 kN/m x − 1 m 18 kN/m 18 kN/m 2 2 0 x −1 m − x − 4 m + 6 kN x − 4 m 2(3 m) 2(3 m) 18 kN/m 0 1 2 M ( x) = −9 kN-m x − 0 m + 21 kN x − 1 m − x −1 m 2 18 kN/m 18 kN/m 3 3 1 + x −1 m − x − 4 m + 6 kN x − 4 m 6(3 m) 6(3 m) +

Ans.

Maximum bending moment: Mmax = 5.66 kN-m at x = 2.59 m

Ans.

Shear-force and bending-moment diagrams:

P7.44 For the beam and loading shown, (a) Use discontinuity functions to write the expression for w(x). Include the beam reactions in this expression. (b) Integrate w(x) twice to determine V(x) and M(x). (c) Determine the maximum bending moment in the beam between the two simple supports. FIGURE P7.44

Solution Beam equilibrium:

M A = − 12 (5 kips/ft)(9 ft)(6 ft) + C y (14 ft) = 0  C y = 9.64 kips Fy = Ay + C y − 12 (5 kips/ft)(9 ft) = 0  Ay = 12.86 kips Load function w(x): w( x) = 12.86 kips x − 0 ft

−1

−

5 kips/ft 5 kips/ft 1 1 x − 0 ft + x − 9 ft 9 ft 9 ft

+5 kips/ft x − 9 ft + 9.64 kips x − 14 ft 0

−1

Ans.

Shear-force function V(x) and bending-moment function M(x): 5 kips/ft 5 kips/ft 0 2 2 V ( x) = 12.86 kips x − 0 ft − x − 0 ft + x − 9 ft 2(9 ft) 2(9 ft) +5 kips/ft x − 9 ft + 9.64 kips x − 14 ft 1

M ( x) = 12.86 kips x − 0 ft − 1

5 kips/ft 5 kips/ft 3 3 x − 0 ft + x − 9 ft 6(9 ft) 6(9 ft)

5 kips/ft 2 1 x − 9 ft + 9.64 kips x − 14 ft 2 Maximum bending moment: Mmax = 58.3 kip-ft at x = 6.80 ft +

Ans.

Ans. Ans.

Shear-force and bending-moment diagrams

P7.45 For the beam and loading shown, (a) Use discontinuity functions to write the expression for w(x). Include the beam reactions in this expression. (b) Integrate w(x) twice to determine V(x) and M(x). (c) Determine the maximum bending moment in the beam between the two simple supports. FIGURE P7.45

Solution Beam equilibrium:

M A = −(5 kips/ft)(6 ft)(3 ft) 21 ft   − 12 (9 kips/ft)(21 ft)  6 ft +  + C y (16 ft) = 0  3   C y = 82.41 kips Fy = Ay + C y − (5 kips/ft)(6 ft) − 12 (9 kips/ft)(21 ft) = 0  Ay = 42.09 kips Load function w(x): −1 0 0 0 w( x) = 42.09 kips x − 0 ft − 5 kips/ft x − 0 ft + 5 kips/ft x − 6 ft − 9 kips/ft x − 6 ft +

9 kips/ft 9 kips/ft 1 −1 1 x − 6 ft + 82.41 kips x − 16 ft − x − 27 ft 21 ft 21 ft

Ans.

Shear-force function V(x) and bending-moment function M(x): 0 1 1 1 V ( x) = 42.09 kips x − 0 ft − 5 kips/ft x − 0 ft + 5 kips/ft x − 6 ft − 9 kips/ft x − 6 ft 9 kips/ft 9 kips/ft 2 0 2 x − 6 ft + 82.41 kips x − 16 ft − x − 27 ft 2(21 ft) 2(21 ft) 5 kips/ft 5 kips/ft 9 kips/ft 1 2 2 2 M ( x) = 42.09 kips x − 0 ft − x − 0 ft + x − 6 ft − x − 6 ft 2 2 2 9 kips/ft 9 kips/ft 3 1 3 + x − 6 ft + 82.41 kips x − 16 ft − x − 27 ft 6(21 ft) 6(21 ft) +

Maximum bending moment: Mmax = 170.9 kip-ft at x = 7.39 ft

Ans.

Shear-force and bending-moment diagrams

P7.46 For the beam and loading shown, (a) Use discontinuity functions to write the expression for w(x). Include the beam reactions in this expression. (b) Integrate w(x) twice to determine V(x) and M(x). (c) Determine the maximum bending moment in the beam between the two simple supports. FIGURE P7.46

Solution Beam equilibrium:

M A = −(25 kN/m)(4.0 m)(4.5 m) 2(4.0 m)   − 12 (45 kN/m)(4.0 m)  2.5 m +  + Dy (8 m) = 0  3  Dy = 114.38 kN Fy = Ay + Dy − (25 kN/m)(4.0 m) − 12 (45 kN/m)(4.0 m) = 0  Ay = 75.63 kN Load function w(x):

w( x) = 75.63 kN x − 0 m +

−1

− 25 kN/m x − 2.5 m + 25 kN/m x − 6.5 m − 0

45 kN/m 1 x − 2.5 m 4.0 m

45 kN/m 1 0 −1 x − 6.5 m + 45 kN/m x − 6.5 m + 114.38 kN x − 8 m 4.0 m

Ans. Shear-force function V(x) and bending-moment function M(x): V ( x) = 75.63 kN x − 0 m − 25 kN/m x − 2.5 m + 25 kN/m x − 6.5 m − 0

45 kN/m 2 x − 2.5 m 2(4.0 m)

45 kN/m 2 1 0 x − 6.5 m + 45 kN/m x − 6.5 m + 114.38 kN x − 8 m 2(4.0 m)

Ans. 25 kN/m 25 kN/m 45 kN/m 2 2 3 x − 2.5 m + x − 6.5 m − x − 2.5 m 2 2 6(4.0 m) 45 kN/m 45 kN/m 3 2 1 + x − 6.5 m + x − 6.5 m + 114.38 kN x − 8 m 6(4.0 m) 2 Ans.

M ( x) = 75.63 kN x − 0 m − 1

Maximum bending moment: Mmax = 275 kN-m at x = 4.57 m

Ans.

Shear-force and bending-moment diagrams:

P7.47 For the beam and loading shown, (a) Use discontinuity functions to write the expression for w(x). Include the beam reactions in this expression. (b) Integrate w(x) twice to determine V(x) and M(x). (c) Determine the maximum bending moment in the beam between the two simple supports. FIGURE P7.47

Solution Beam equilibrium:

M C = (30 kN/m)(7.0 m)(3.5 m) + 12 (40 kN/m)(7.0 m)

2(7.0 m) 3

− (50 kN/m)(2.0 m)(1.0 m) − By (5.5 m) = 0  By = 234.24 kN Fy = By + C y − (30 kN/m)(7.0 m) − 12 (40 kN/m)(7.0 m) − (50 kN/m)(2.0 m) = 0  C y = 215.76 kN

Load function w(x): 40 kN/m 1 x−0 m 7.0 m 40 kN/m −1 0 1 +234.24 kN x − 1.5 m + 30 kN/m x − 7 m − x−7 m 7.0 m

w( x) = −30 kN/m x − 0 m − 40 kN/m x − 0 m + 0

+215.76 kN x − 7 m

−1

− 50 kN/m x − 7.0 m + 50 kN/m x − 9.0 m 0

Ans.

Shear-force function V(x) and bending-moment function M(x): 40 kN/m 1 1 2 V ( x) = −30 kN/m x − 0 m − 40 kN/m x − 0 m + x−0 m 2(7.0 m) 40 kN/m 0 1 2 +234.24 kN x − 1.5 m + 30 kN/m x − 7 m − x−7 m 2(7.0 m) +215.76 kN x − 7 m − 50 kN/m x − 7.0 m + 50 kN/m x − 9.0 m 0

Ans.

30 kN/m 40 kN/m 40 kN/m 2 2 3 x−0 m − x−0 m + x−0 m 2 2 6(7.0 m) 30 kN/m 40 kN/m 1 2 3 +234.24 kN x − 1.5 m + x−7 m − x−7 m 2 6(7.0 m) 50 kN/m 50 kN/m 1 2 2 +215.76 kN x − 7 m − x − 7.0 m + x − 9.0 m 2 2

M ( x) = −

Maximum bending moment: Mmax = 86.6 kN-m at x = 4.00 m

Ans.

Shear-force and bending-moment diagrams:

P7.48 For the beam and loading shown, (a) Use discontinuity functions to write the expression for w(x). Include the beam reactions in this expression. (b) Integrate w(x) twice to determine V(x) and M(x). (c) Determine the maximum bending moment in the beam between the two simple supports. FIGURE P7.48

Solution Beam equilibrium:

M B = (60 kN)(1.5 m) − 12 (90 kN/m)(4.5 m)

2(4.5 m) 3

+ Dy (6.5 m) = 0  Dy = 79.62 kN Fy = By + Dy − 60 kN − 12 (90 kN/m)(4.5 m) = 0  By = 182.88 kN Load function w(x):

w( x) = −60 kN x − 0 m +

−1

+ 182.88 kN x − 1.5 m

−1

−

90 kN/m 1 x − 1.5 m 4.5 m

90 kN/m 1 0 −1 x − 6 m + 90 kN/m x − 6 m + 79.62 kN x − 8 m 4.5 m

Shear-force function V(x) and bending-moment function M(x): 90 kN/m 0 0 2 V ( x) = −60 kN x − 0 m + 182.88 kN x − 1.5 m − x − 1.5 m 2(4.5 m) 90 kN/m 2 1 0 + x − 6 m + 90 kN/m x − 6 m + 79.62 kN x − 8 m 2(4.5 m) 90 kN/m 1 1 3 M ( x) = −60 kN x − 0 m + 182.88 kN x − 1.5 m − x − 1.5 m 6(4.5 m) 90 kN/m 90 kN/m 3 2 1 + x−6 m + x − 6 m + 79.62 kN x − 8 m 6(4.5 m) 2

Maximum bending moment: Mmax = 197.2 kN-m at x = 5.01 m

Ans.

Shear-force and bending-moment diagrams:

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.1 During fabrication of a laminated timber arch, one of the 10 in. wide by 1 in. thick Douglas fir [E = 1,900 ksi] planks is bent to a radius of curvature of 40 ft. Determine the maximum bending stress developed in the plank.

Solution From Eq. (8.3): E x = − y = −



1,900 ksi ( 0.5 in.) = 1.979 ksi = 1.979 ksi ( 40 ft )(12 in./ft )

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.2 A copper wire of diameter d = 2 mm is coiled around a spool of radius r. The elastic modulus of the copper is E = 117 GPa and its yield strength is 310 MPa. Determine the minimum spool radius r that may be used if the bending stress in the wire is not to exceed the yield strength.

Solution From Eq. (8.3): E 117,000 MPa  =− y=− ( 2 mm / 2 ) = 377.419 mm x 310 MPa

 =r+

d 2

r =  −

d 2 mm = 377.419 mm − = 376.419MPa = 376 mm 2 2

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.3 The boards for a concrete form are to be bent into a circular shape having an inside radius of 10 m. What maximum thickness can be used for the boards if the normal stress is not to exceed 7 MPa? Assume that the modulus of elasticity for the wood is 12 GPa.

Solution The radius of curvature of the concrete form is dependent on the board thickness:

 = 10,000 mm + From Eq. (8.3): E x = − y = −



t 2

t  7 MPa t   2   10,000 mm +  2 12,000 MPa

Solve for t: t t  12,000 MPa    7 MPa 10,000 mm +   2  2 6,000t  70,000 + 3.5t 5,996.5t  70,000  t  11.67 mm

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.4 A beam is subjected to equal bending moments of Mz = 3,200 N·m, as shown in Figure P8.4a. The cross-sectional dimensions (Figure P8.4b) are b = 150 mm, c = 30 mm, d = 70 mm, and t = 6 mm. Determine: (a) the centroid location, the moment of inertia about the z axis, and the controlling section modulus about the z axis. (b) the bending stress at point H. State whether the normal stress at H is tension or compression. (c) the bending stress at point K. State whether the normal stress at K is tension or compression. (d) the maximum bending stress produced in the cross section. State whether the stress is tension or compression.

FIGURE P8.4a

FIGURE P8.4b

Solution

(a) Centroid location in y direction: (reference axis at bottom of shape) yi Shape Area Ai (from bottom) yi Ai 2 (mm ) (mm) (mm3) (1) 828 3 2,484 (2) 840 35 29,400 (3) 288 67 19,296 1,956 51,180 3 yi Ai 51,180 mm y= = = 26.166 mm = 26.2 mm Ai 1,956 mm 2

Ans.

(measured upward from bottom edge of stem) Moment of inertia about the z axis: d = y − yi Shape IC (mm4) (mm) (1) 2,484 23.166 (2) 343,000 –8.834

d²A (mm4) 444,343.774 65,558.508

IC + d²A (mm4) 446,827.774 408,558.508

Mechanics of Materials: An Integrated Learning System, 4th Ed. 864 –40.834 480,224.049 Moment of inertia about the z axis (mm4) =

(3)

Timothy A. Philpot 481,088.049 1,336,474.331

I z = 1,336, 000 mm4

Ans.

Section moduli: I 1,336, 474.331 mm 4 S top = z = = 30, 489.197 mm3 ctop 70 mm − 26.166 mm I z 1,336, 474.331 mm 4 Sbot = = = 51, 077.448 mm3 cbot 26.166 mm  S = 30,500 mm3

Ans.

(b) Bending stress at point H: yH = d − t − y = 70 mm − 6 mm − 26.166 mm = 37.834 mm

x = − =−

My Iz

( 3, 200 N  m )( 37.834 mm )(1,000 mm/m ) = −90.589 MPa = 90.6 MPa (C) 1,336,474.331 mm 4

Ans.

x = − =−

My Iz

( 3, 200 N  m )( −20.166 mm )(1,000 mm/m ) = 48.284 MPa = 48.3 MPa (T) 1,336,474.331 mm 4

Ans.

(d) Maximum bending stress: The maximum bending stress occurs at either the top or the bottom surface of the beam. The top of the cross section is at y = 43.834 mm, and the bottom of the cross section is at y = −26.166 mm. The larger bending stress magnitude occurs at the larger magnitude of these two values; in this case, at the top of the cross section. My x = − Iz =−

( 3, 200 N  m )( 43.834 mm )(1,000 mm/m ) = −104.955 MPa = 105.0 MPa (C) 1,336,474.331 mm 4

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.5 A beam is subjected to equal bending moments of Mz = 45 kip·ft, as shown in Figure P8.5a. The cross-sectional dimensions (Figure P8.5b) are b1 = 7.5 in., d1 = 1.5 in., b2 = 0.75 in., d2 = 6.0 in., b3 = 3.0 in., and d3 = 2.0 in. Determine: (a) the centroid location, the moment of inertia about the z axis, and the controlling section modulus about the z axis. (b) the bending stress at point H. State whether the normal stress at H is tension or compression. (c) the bending stress at point K. State whether the normal stress at K is tension or compression. (d) the maximum bending stress produced in the cross section. State whether the stress is tension or compression.

FIGURE P8.5a

FIGURE P8.5b

Solution

(a) Centroid location in y direction: (reference axis at bottom of shape) yi Shape Area Ai (from bottom) yi Ai 2 (in. ) (in.) (in.3) (1) 11.25 8.75 98.437 (2) 4.5 5.0 22.5 (3) 6.0 1.0 6.0 21.75 126.937 yi Ai 126.937 in.3 y= = = 5.836 in. = 5.84 in. Ai 21.75 in.2

Ans.

(measured upward from bottom edge of section) Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Mechanics of Materials: An Integrated Learning System, 4th Ed. Moment of inertia about the z axis: d = y − yi Shape IC d²A 4 (in. ) (in.) (in.4) (1) 2.109 –2.914 95.515 (2) 13.500 0.836 3.147 (3) 2.000 4.836 140.333 Moment of inertia about the z axis (in.4) =

Timothy A. Philpot

IC + d²A (in.4) 97.624 16.647 142.333 256.604 I z = 256.6 in.4

Ans.

Section moduli: I 256.604 in.4 S top = z = = 70.038 in.3 ctop 9.5 in. − 5.836 in.

Sbot =

Iz 256.604 in.4 = = 43.968 in.3 cbot 5.836 in.

 S = 43.97 in.3

Ans.

(b) Bending stress at point H: yH = d3 − y = 2.0 in. − 5.836 in. = −3.836 in.

x = − =−

My Iz

( 45 kip  ft )( −3.836 in.)(12 in./ft ) = 8.073 ksi = 8.07 ksi (T) 256.604 in.4

Ans.

x = − =−

My Iz

( 45 kip  ft )( 2.164 in.)(12 in./ft ) = −4.554 ksi = 4.55 ksi (C) 256.604 in.4

Ans.

(d) Maximum bending stress: The maximum bending stress occurs at either the top or the bottom surface of the beam. The top of the cross section is at y = 3.664 in., and the bottom of the cross section is at y = −5.836 in. The larger bending stress magnitude occurs at the larger magnitude of these two values; in this case, at the bottom of the cross section. My x = − Iz =−

( 45 kip  ft )( −5.836 in.)(12 in./ft ) = 12.282 ksi = 12.28 ksi (T) 256.604 in.4

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.6 A beam is subjected to equal bending moments of Mz = 240 N·m, as shown in Figure P8.6a. The cross-sectional dimensions (Figure P8.6b) are a = 20 mm, b = 40 mm, d = 80 mm, and r = 12 mm. Determine: (a) the centroid location, the moment of inertia about the z axis, and the controlling section modulus about the z axis. (b) the bending stress at point H. State whether the normal stress at H is tension or compression. (c) the bending stress at point K. State whether the normal stress at K is tension or compression. (d) the maximum bending stress produced in the cross section. State whether the stress is tension or compression.

FIGURE P8.6a

FIGURE P8.6b

Solution (a) Centroid location in y direction: (reference axis at bottom of shape) yi Shape Area Ai (from bottom) yi Ai 2 (mm ) (mm) (mm3) rectangle 3,200 40.0 128,000.000 circle cutout –452.389 20.0 –9,047.787 2,747.611 118,952.213 3 yi Ai 118,952.213 mm y= = = 43.293 mm = 43.3 mm Ai 2,747.611 mm2

Ans.

(measured upward from bottom edge of section) Moment of inertia about the z axis: d = y − yi Shape IC d²A (mm4) (mm) (mm4) rectangle 1,706,666.667 3.293 34,699.588 circle cutout –16,286.016 23.293 –245,449.374 Moment of inertia about the z axis (mm4) =

IC + d²A (mm4) 1,741,366.254 –261,735.390 1,479,630.864 I z = 1,480,000 mm4

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Section moduli: I 1,479,630.864 mm 4 S top = z = = 40,309.2 mm3 ctop 80 mm − 43.293 mm S bot =

I z 1,479,630.864 mm 4 = = 34,177.2 mm3 cbot 43.293 mm

 S = 34, 200 mm3

Ans.

(b) Bending stress at point H: yH = d − a − y = 80 mm − 20 mm − 43.293 mm = 16.707 mm

x = − =−

My Iz

( −240 N  m )(16.707 mm )(1,000 mm/m ) 1,479,630.864 mm 4

= 2.710 MPa = 2.71 MPa (T)

Ans.

x = − =−

My Iz

( −240 N  m )( −11.293 mm )(1,000 mm/m ) 1,479,630.864 mm 4

= −1.832 MPa = 1.832 MPa (C)

Ans.

(d) Maximum bending stress: The maximum bending stress occurs at either the top or the bottom surface of the beam. The top of the cross section is at y = 36.707 mm, and the bottom of the cross section is at y = −43.293 mm. The larger bending stress magnitude occurs at the larger magnitude of these two values; in this case, at the bottom of the cross section. My x = − Iz

=−

( −240 N  m )( −43.293 mm )(1,000 mm/m ) 1,479,630.864 mm4

= −7.022 MPa = 7.02 MPa (C)

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.7 The dimensions of the beam cross section shown in Figure P8.7 are a = 7 mm and b = 45 mm. The internal bending moment about the z centroidal axis is Mz = 325 N·m. What is the magnitude of the maximum bending stress in the beam? FIGURE P8.7

Solution Moment of inertia about the z axis: Shape Rectangle Two cutouts on sides Moment of inertia about the z axis (mm4) =

IC (mm4) 277,830 –30,171.856 247,658.144

y = 3a = 3 ( 7 mm ) = 21 mm

x = − =−

My Iz

( 325 N  m )( 21 mm )(1,000 mm/m ) 247,658.144 mm 4

= 27.558 MPa = 27.6 MPa

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.8 The dimensions of the double-box beam cross section shown in Figure P8.8 are b = 150 mm, d = 50 mm, and t = 4 mm. If the maximum allowable bending stress is b = 17 MPa, determine the maximum internal bending moment Mz magnitude that can be applied to the beam. FIGURE P8.8

Solution Moment of inertia about the z axis: Shape

IC (mm4) 1,562,500 –852,012 710,488

Outer rectangle Two cutout rectangles Moment of inertia about the z axis (mm4) =

x =

M zc Iz

Mz =

 xIz

(17 N/mm )( 710, 488 mm ) = 2

c 50 mm / 2 = 483,132 N  mm = 483 N  m

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.9 An aluminum alloy [E = 10,300 ksi] tee-shaped bar is used as a beam. The cross section of the tee shape, shown in Figure P8.9, has a total depth of d = 7.5 in. The centroidal z axis of this cross section is located 2.36 in. above point H. The beam spans in the x direction and it bends about the z centroidal axis. A strain gage is affixed to the side of the tee stem at a distance of c = 1.25 in. below the top surface of the tee shape. After loads are applied to the beam, a normal strain of x = –810  is measured by the strain gage. What is the bending stress at point H? FIGURE P8.9

Solution The y coordinate at the strain gage is: ygage = d − c − y = 7.5 in. −1.25 in. − 2.36 in. = 3.890 in. The bending strain at H is thus:

 gage

ygage

H yH

 H =

yH −2.36 in.  gage = −810 10−6 in./in.) = 491.414  10−6 in./in. ( ygage 3.890 in.

The bending stress at H is:  x = E x = (10,300 ksi ) ( 491.414 10−6 in./in.) = 5.062 ksi = 5.06 ksi

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.10 The cross-sectional dimensions of the beam shown in Figure P8.10 are d = 17 in., bf = 10 in., tf = 1.0 in., tw = 0.60 in., and a = 3.5 in. (a) If the bending stress at point K is 5.4 ksi (C), what is the bending stress at point H? State whether the normal stress at H is tension or compression. (b) If the allowable bending stress is 30 ksi, what is the magnitude of the maximum bending moment Mz that can be supported by the beam? FIGURE P8.10

Solution The y coordinates at H and K are: d 17 in. yH = − t f = − 1.0 in. = 7.5 in. 2 2 d 17 in. yK = − + a = − + 3.5 in. = −5.0 in. 2 2 (a) Bending stress at H: Since the bending stress is linearly distributed over the depth of the section, we can calculate the bending stress at H from the principle of similar triangles:

H yH

K yK

 H =

yH 7.5 in. K = ( −5.4 ksi ) = 8.10 ksi (T) yK −5.0 in.

Ans.

Moment of inertia about z axis:

(10 in.)(17 in.) − (10 in. − 0.60 in.) 17 in. − 2 (1.0 in.) = 1, 450.417 in.4 I = 3

(b) Maximum bending moment Mz: M c x = z Iz M =

 xIz c

( 30 kips/in. )(1, 450.417 in. ) = 5,119.117 kip  in. = 427 kip  ft 2

17 in. / 2

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.11 The cross-sectional dimensions of the beam shown in Figure P8.11 are ro = 115 mm and ri = 95 mm. Given Mz = 16 kN·m,  = 30°, and  = 55°, what are the bending stresses at points H and K?

FIGURE P8.11

Solution Moment of inertia about the z axis: Iz =



 r − r ) = (115 mm ) − ( 95 mm )  = 73,395, 458 mm (  4 4 4 o

The y coordinates at H and K are: yH = ro sin  = (115 mm ) sin 30 = 57.5 mm yK = −ri sin  = − ( 95 mm ) sin 55 = −77.819 mm

Bending stress at H: My x = − Iz =−

(16 kN  m )( 57.5 mm )(1, 000 N/kN )(1,000 mm/m ) 73,395,458 mm4

= −12.535 MPa = 12.54 MPa (C)

Ans.

Bending stress at K: My x = − Iz =−

(16 kN  m )( −77.819 mm )(1, 000 N/kN )(1,000 mm/m ) 73,395,458 mm 4

= 16.964 MPa = 16.96 MPa (T)

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.12 The cross-sectional dimensions of the beam shown in Figure P8.12 are a = 5.0 in., b = 6.0 in., d = 4.0 in., and t = 0.5 in. The internal bending moment about the z centroidal axis is Mz = −4.25 kip-ft. Determine: (a) the maximum tension bending stress in the beam. (b) the maximum compression bending stress in the beam. FIGURE P8.12

Solution

(a) Centroid location in y direction: (reference axis at bottom of shape) yi Shape Area Ai (from bottom) yi Ai 2 (in. ) (in.) (in.3) (1) 2.50 3.75 9.375 (2) 4.00 2.00 8.00 (3) 4.50 0.25 1.125 11.00 18.50 3 yi Ai 18.50 in. y= = = 1.6818 in. (measured upward from bottom edge of section) Ai 11.00 in.2 Moment of inertia about the z axis: d = y − yi Shape IC 4 (in. ) (in.) (1) 0.05208 –2.0682 (2) 5.33333 –0.31820 (3) 0.09375 1.43180 Moment of inertia about the z axis (mm4) =

d²A (in.4) 10.69363 0.40500 9.22523

IC + d²A (in.4) 10.74571 5.73834 9.31898 25.80303

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

The y coordinates at the top and bottom surfaces of the section are: ytop = d − y = 4.0 in. − 1.6818 in. = 2.31820 in. ybottom = − y = −1.6818 in.

Bending stress at top surface: The maximum tensile bending stress occurs at the top surface of the beam for a negative bending moment: My x = − Iz =−

( −4.25 kip  ft )( 2.31820 in.)(12 in./ft ) = 4.582 ksi = 4.58 ksi (T) 25.80303 in.4

Ans.

Bending stress at bottom surface: The maximum compressive bending stress occurs at the bottom surface of the beam for a negative bending moment: My x = − Iz =−

( −4.25 kip  ft )( −1.6818 in.)(12 in./ft ) = −3.324 ksi = 3.32 ksi (C) 25.80303 in.4

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.13 Two uniformly distributed loads of w = 3,600 lb/ft act on the simply supported beam shown in Figure P8.13a. The beam spans are a = 8 ft and L = 16 ft. The beam cross section shown in Figure P8.13b has dimensions of b1 = 16 in., d1 = 6 in., b2 = 10 in., and d2 = 10 in. Calculate the maximum tensile and compressive bending stresses produced in segment BC of the beam.

FIGURE P8.13a

FIGURE P8.13b

Solution

Centroid location in y direction: (reference axis at bottom of shape) yi Shape Area Ai (from bottom) yi Ai 2 (in. ) (in.) (in.3) (1) 96.00 13.00 1,248.00 (2) 100.00 5.00 500.00 196.00 1,748.00 yi Ai 1, 748.00 in.3 (measured upward from bottom edge of section) y= = = 8.918 in. Ai 196.00 in.2 Moment of inertia about the z axis: d = y − yi Shape IC d²A 4 (in. ) (in.) (in.4) (1) 288.000 –4.082 1,599.334 (2) 833.333 3.918 1,535.360 Moment of inertia about the z axis (in.4) =

IC + d²A (in.4) 1,887.334 2,368.694 4,256.027

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Shear-force and bending-moment diagrams

The maximum bending moment in segment BC of the beam is –115,200 lb·ft. Bending stresses: The y coordinates at the top and bottom surfaces of the section are: ytop = d1 + d 2 − y = 6 in. + 10 in. − 8.918 in. = 7.082 in. ybottom = − y = −8.918 in.

Bending stress at top surface: The bending stress at the top surface of the beam in segment BC is: My x = − Iz =−

( −115, 200 lb  ft )( 7.082 in.)(12 in./ft ) = 2,300.185 psi 4,256.027 in.4

Bending stress at bottom surface: The bending stress at the bottom surface of the beam in segment BC is: My x = − Iz =−

( −115, 200 lb  ft )( −8.918 in.)(12 in./ft ) = −2,896.775 psi 4,256.027 in.4

Maximum bending stresses: The maximum tensile bending stress is thus:  x = 2,300 psi (T)

Ans.

and the maximum compressive bending stress is:  x = 2,900 psi (C)

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.14 An extruded polymer beam is subjected to a bending moment M as shown in Figure P8.14a. The length of the beam is L = 800 mm. The cross-sectional dimensions (Figure P8.14b) of the beam are b1 = 34 mm, d1 = 100 mm, b2 = 20 mm, d2 = 20 mm, and a = 7 mm. For this material, the allowable tensile bending stress is 16 MPa, and the allowable compressive bending stress is 12 MPa. Determine the largest moment M that can be applied as shown to the beam.

FIGURE P8.14a

FIGURE P8.14b

Solution Centroid location in y direction: Shape

yi (from bottom) (mm) 50 83

Area Ai yi Ai 2 (mm ) (mm3) Outer rectangle 3,400 170,000 Cutout square –400 –33,200 2 3,000 mm 136,800 yi Ai 136,800 mm3 y= = = 45.6 mm (measured upward from bottom edge of shape) Ai 3,000 mm 2 Moment of inertia about the z axis: d = y − yi Shape IC d²A 4 (mm ) (mm) (mm4) Outer rectangle 2,833,333.33 –4.4 65,824 Cutout square –13,333.33 –37.4 –559,504 Moment of inertia about the z axis (mm4) =

IC + d²A (mm4) 2,899,157.33 –572,837.33 2,326,320

Bending stresses: The y coordinates at the top and bottom surfaces of the cross section are: ytop = d1 − y = 100 mm − 45.6 mm = 54.4 mm ybottom = − y = −45.6 mm

Bending moment based on allowable tensile bending stress: For the indicated moment direction (i.e., counterclockwise), tensile bending stress will be produced above the neutral axis. The largest tensile

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

bending stress will be found at the top surface of the beam. Solve the flexure formula for M given that the bending stress at ytop must be less than or equal to 16 MPa: My −   allow Iz M  −

 allow I z ytop

(16 N/mm )( 2,326,320 mm ) = −684, 211.8 N  mm 2

=−

54.4 mm

Bending moment based on allowable compressive bending stress: For the indicated moment direction (i.e., counterclockwise), compressive bending stress will be produced below the neutral axis. The largest compressive bending stress will be found at the bottom surface of the beam. Solve the flexure formula for M given that the bending stress at ybottom must be less than or equal to 12 MPa: My −   allow Iz M  −

 allow I z ybottom

( −12 N/mm )( 2,326,320 mm ) = −612,189.5 N  mm =− 2

−45.6 mm

Largest moment M: The largest bending moment that can be applied as shown in Figure P8.14a is:

M max = 612,189.5 N  mm = 612 N  m

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.15 A 1.125 in. diameter solid steel shaft supports loads PA = 65 lb, PC = 150 lb, and PE = 55 lb as shown in Figure P8.15. Assume a = 10 in., b = 25 in., c = 13 in., and d = 16 in. The bearing at B can be idealized as a roller support and the bearing at D can be idealized as a pin support. Determine the magnitude and location of the maximum bending stress in the shaft. FIGURE P8.15

Solution Section properties Iz =



d4 =



(1.125 in.) = 78.6285 10−3 in.4 4

Shear-force and bending-moment diagrams Maximum bending moments positive M = 926.3 lb·in. negative M = −880 lb·in. Largest bending stress occurs at pulley C

x =

( 926.3 lb  in.)(1.125 in. / 2 )

78.6285 10−3 in.4 = 6, 626.653 psi

Maximum bending stress = 6,630 psi

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.16 A W18 × 40 standard steel shape is used to support the loads shown on the beam in Figure P8.16. Determine the magnitude and location of the maximum bending stress in the beam.

FIGURE P8.16

Solution Section properties From Appendix B:

d = 17.9 in.

I z = 612 in.4

Shear-force and bending-moment diagrams Maximum bending moments positive M = 132.03 kip·ft negative M = −100 kip·ft Since the shape is symmetric about the z axis, the largest bending stresses will occur at the location of the largest moment magnitude – either positive or negative. In this case, the largest bending stresses will occur where the moment magnitude is 132.03 kip·ft. Bending stresses at maximum moment (132.03 kip  ft )(17.9 in. / 2 )(12 in./ft ) x = 612 in.4 = 23.170 ksi (T) and (C)

Maximum bending stress = 23.2 ksi

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.17 A W410 × 60 standard steel shape is used to support the loads shown on the beam in Figure P8.17. Assume P = 125 kN, w = 30 kN/m, a = 3 m, and b = 2 m. Determine the magnitude and location of the maximum bending stress in the beam.

FIGURE P8.17

Solution Section properties From Appendix B:

d = 406 mm

I z = 216 106 mm4

Sx = 1,060 103 mm3

Shear-force and bending-moment diagrams Maximum bending moments positive M = 112.50 kN·m negative M = −150 kN·m Since the shape is symmetric about the z axis, the largest bending stresses will occur at the location of the largest moment magnitude – either positive or negative. In this case, the largest bending stresses will occur where the moment magnitude is 150 kN·m. Bending stresses at maximum moment 2 150 kN  m )( 406 mm / 2 )(1,000 ) ( x = 216 106 mm 4 = 140.972 MPa (T) and (C)

Maximum bending stress = 141.0 MPa

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.18 A W16 × 50 standard steel shape is used to support the loads shown on the beam in Figure P8.18. Assume P = 12 kips, w = 1.0 kips/ft, a = 30 ft, and b = 10 ft. Determine the magnitude and location of the maximum bending stress in the beam.

FIGURE P8.18

Solution Section properties From Appendix B:

d = 16.3 in.

I z = 659 in.4

Shear-force and bending-moment diagrams Maximum bending moments positive M = 64.22 kip·ft negative M = −110 kip·ft Since the shape is symmetric about the z axis, the largest bending stresses will occur at the location of the largest moment magnitude – either positive or negative. In this case, the largest bending stresses will occur where the moment magnitude is 110 kip·ft. Bending stresses at maximum moment (110 kip  ft )(16.3 in. / 2 )(12 in./ft ) x = 659 in.4 = 16.325 ksi (T) and (C)

Maximum bending stress = 16.33 ksi

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.19 A HSS12 × 8 × 1/2 standard steel shape is used to support the loads shown on the beam in Figure P8.19. The shape is oriented so that bending occurs about the strong axis. Determine the magnitude and location of the maximum bending stress in the beam. FIGURE P8.19

Solution Section properties From Appendix B:

I z = 333 in.4

d = 12 in.

Shear-force and bending-moment diagrams Maximum bending moments positive M = 124.59 kip·ft negative M = −72.00 kip·ft Since the shape is symmetric about the z axis, the largest bending stresses will occur at the location of the largest moment magnitude – either positive or negative. In this case, the largest bending stresses will occur at C, where the moment magnitude is 124.59 kip·ft.

Bending stresses at max moment magnitude (124.59 kip  ft )( 12 in. / 2 )(12 in./ft ) = 26.9 ksi x = − 333 in.4

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.20 A W360 × 72 standard steel shape is used to support the loads shown on the beam in Figure P8.20a. The shape is oriented so that bending occurs about the weak axis as shown in Figure P8.20b. Consider the entire 6 m length of the beam and determine: (a) the maximum tensile bending stress at any location along the beam, and (b) the maximum compressive bending stress at any location along the beam.

FIGURE P8.20a

FIGURE P8.20b

Solution Section properties From Appendix B:

I z = 21.4 106 mm4

bf = 204 mm

Shear-force and bending-moment diagrams Maximum bending moments positive M = 31.50 kN·m negative M = −25.87 kN·m Since the shape is symmetric about the z axis, the largest bending stresses will occur at the location of the largest moment magnitude – either positive or negative. In this case, the largest bending stresses will occur where the moment magnitude is 31.50 kN·m. Bending stresses at maximum moment (31.50 kN  m)(  204 mm/2)(1,000)2 x = − 21.4 106 mm4 = 150.1 MPa (T) and (C)

(a) Maximum tensile bending stress = 150.1 MPa (T)

Ans.

(b) Maximum compressive bending stress = 150.1 MPa (C)

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.21 A W310 × 38.7 standard steel shape is used to support the loads shown on the beam in Figure P8.21. What is the maximum bending stress magnitude at any location along the beam?

FIGURE P8.21

Solution Section properties From Appendix B:

d = 310 mm

I z = 84.9 106 mm4

Shear-force and bending-moment diagrams Maximum bending moments positive M = 78.14 kN·m negative M = −19.20 kN·m Since the shape is symmetric about the z axis, the largest bending stresses will occur at the location of the largest moment magnitude – either positive or negative. In this case, the largest bending stresses will occur where the moment magnitude is 78.14 kN·m. Bending stresses at maximum moment 2 78.14 kN  m )( 310 mm / 2 )(1,000 ) ( x = − 84.9 106 mm 4 = 142.658 MPa (T) and (C)

Maximum bending stress = 142.7 MPa

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.22 A WT305 × 41 standard steel shape is used to support the loads shown on the beam in Figure P8.22a. The dimensions from the top and bottom of the shape to the centroidal axis are shown on the sketch of the cross section (Figure P8.22b). Consider the entire 4.8 m length of the beam and determine: (a) the maximum tensile bending stress at any location along the beam, and (b) the maximum compressive bending stress at any location along the beam.

FIGURE P8.22a

FIGURE P8.22b

Solution Section properties From Appendix B:

I z = 48.7 106 mm4

Shear-force and bending-moment diagrams Maximum bending moments positive M = 45 kN·m negative M = −36 kN·m Bending stresses at max positive moment 2 45 kN  m )( 88.9 mm )(1, 000 ) ( x = − 48.7 106 mm 4 = 82.146 MPa (C)

( 45 kN  m )( −211.1 mm )(1, 000 )  =− x

48.7 106 mm 4 = 195.062 MPa (T)

Bending stresses at max negative moment 2 −36 kN  m )( 88.9 mm )(1, 000 ) ( x = − 48.7 106 mm 4 = 65.717 MPa (T)

( −36 kN  m )( −211.1 mm )(1, 000 )  =− x

48.7 106 mm 4 = 156.049 MPa (C)

(a) Maximum tensile bending stress = 195.1 MPa (T)

Ans.

(b) Maximum compressive bending stress = 156.0 MPa (C)

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.23 A flanged wooden shape is used to support the loads shown on the beam in Figure P8.23a. The dimensions of the shape are shown in Figure P8.23b. Consider the entire 18 ft length of the beam and determine: (a) the maximum tensile bending stress at any location along the beam, and (b) the maximum compressive bending stress at any location along the beam

FIGURE P8.23a

FIGURE P8.23b

Solution Centroid location in y direction: Shape

Area Ai (in.2) 20.0 16.0 12.0 48.0 in.2

top flange web bottom flange

yi A i A i

yi (from bottom) (in.) 11.0 6.0 1.0

yi Ai (in.3) 220.0 96.0 12.0 328.0 in.3

328.0 in.3 = 6.8333 in. (from bottom of shape to centroid) 48.0 in.2 = 5.1667 in.

(from top of shape to centroid)

Moment of inertia about the z axis: d = y − yi Shape IC d²A 4 (in. ) (in.) (in.4) top flange 6.667 –4.167 347.222 web 85.333 0.833 11.111 bottom flange 4.000 5.833 408.333 Moment of inertia about the z axis (in.4) =

IC + d²A (in.4) 353.889 96.444 412.333 862.667

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Shear-force and bending-moment diagrams Maximum bending moments positive M = 10,580 lb·ft negative M = −8,400 lb·ft Bending stresses at max positive moment (10,580 lb  ft)(5.1667 in.)(12 in./ft) x = − 862.667 in.4 = 760.4 psi (C)

(10,580 lb  ft)( − 6.8333 in.)(12 in./ft) 862.667 in.4 = 1, 005.6 psi (T)

x = −

Bending stresses at max negative moment (−8, 400 lb  ft)(5.1667 in.)(12 in./ft) x = − 862.667 in.4 = 603.7 psi (T)

(−8, 400 lb  ft)( − 6.8333 in.)(12 in./ft) 862.667 in.4 = 798.5 psi (C)

x = −

(a) Maximum tensile bending stress = 1,006 psi (T)

Ans.

(b) Maximum compressive bending stress = 799 psi (C)

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.24 The steel beam in Figure P8.24a/25a has the cross section shown in Figure P8.24b/25b. The beam length is L= 22 ft, and the cross-sectional dimensions are d = 16.3 in., bf = 10.0 in., tf = 0.665 in., and tw = 0.395 in. Calculate the maximum bending stress in the beam if w0 = 6 kips/ft.

FIGURE P8.24a/25a

FIGURE P8.24b/25b

Solution Moment of inertia about the z axis: Shape

Area Ai

top flange web bottom flange

(in.2) 6.6500 5.9131 6.6500

yi (from bottom) (in.) 15.9675 8.1500 0.3325

d = y − y

d²A

(in.4) (in.) (in.4) 0.2451 –7.8175 406.4035 110.4285 0.0000 0.0000 0.2451 7.8175 406.4035 Moment of inertia about the z axis (in.4) =

IC + d²A (in.4) 406.6486 110.4285 406.6486 923.7256

Moment in the simply supported beam: From a FBD of the beam, determine the reaction force at A: 1  L M C = w0 L   − Ay L = 0  2 2

 Ay =

w0 L 4

The maximum moment will occur in the center of the span at B. From the FBD shown, determine the bending moment M: w LL w L L M = 0   − 0   + M = 0 4 6 4 2 w L2 M = 0 12 (6 kips/ft)(22 ft) 2 = 12 = 242 kip  ft = 2,904 kip  in. Maximum Bending Stress: The maximum bending stress in the beam occurs at midspan: Mc ( 2,904 kip  in.)(16.3 in. / 2 ) x = = = 25.622 ksi = 25.6 ksi Iz 923.7256 in.4

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.25 The steel beam in Figure P8.24a/25a has the cross section shown in Figure P8.24b/25b. The beam length is L= 6.0 m, and the cross-sectional dimensions are d = 350 mm, bf = 205 mm, tf = 14 mm, and tw = 8 mm. Calculate the largest intensity of distributed load w0 that can be supported by this beam if the allowable bending stress is 200 MPa.

FIGURE P8.24a/25a

FIGURE P8.24b/25b

Solution Moment of inertia about the z axis: Shape

Area Ai

top flange web bottom flange

(mm2) 2,870 2,576 2,870

yi (from bottom) (mm) 343 175 7

d = y − y

d²A

(mm4) (mm) (mm4) 46,876.67 –168.00 81,002,880.00 22,257,498.67 0.00 0.00 46,876.67 168.00 81,002,880.00 Moment of inertia about the z axis (mm4) =

IC + d²A (mm4) 81,049,756.67 22,257,498.67 81,049,756.67 184,357,012.00

Allowable bending moment: Based on the allowable bending stress, the maximum moment that can be applied to this beam is  I (200 N/mm 2 )(184,357, 012 mm 4 ) M allow = allow z = = 210.694  106 N  mm c 350 mm / 2 Moment in the simply supported beam: From a FBD of the beam, determine the reaction force at A: 1  L M C = w0 L   − Ay L = 0  2 2

 Ay =

w0 L 4

The maximum moment will occur in the center of the span at B. From the FBD shown, determine the bending moment M: w L  L w L  L M = 0   − 0   + M = 0 4  6 4  2 w0 L2 M = 12

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Largest intensity of distributed load w0: Equate the allowable moment with the moment produced at midspan for this beam. w0 L2 = M allow 12 6 12M allow 12 ( 210.694 10 N  mm )  w0 = = = 70.231 N/mm = 70.2 kN/m Ans. 2 L2 ( 6, 000 mm )

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.26 A small aluminum alloy [E = 70 GPa] tee shape is used as a simply supported beam as shown in Figure P8.26a. For this beam, a = 160 mm. The cross-sectional dimensions (Figure P8.26b) of the tee shape are b = 20 mm, d = 30 mm, and t = 6 mm. After loads are applied to the beam at B, C, and D, a compressive normal strain of 440  is measured from a strain gage located at c = 8 mm below the topmost edge of the tee stem. What is the applied load P?

FIGURE P8.26a

FIGURE P8.26b

Solution Centroid location in y direction: (reference axis at bottom of shape) yi Shape Area Ai (from bottom) yi Ai 2 (mm ) (mm) (mm3) Lower flange 120.00 3.0 360.00 Stem 144.00 18.0 2,592.00 264.00 2,952.00 3 yi Ai 2,952 mm y= = = 11.182 mm (measured upward from bottom edge of section) Ai 264 mm 2 Moment of inertia about the z axis: d = y − yi Shape IC d²A 4 (mm ) (mm) (mm4) Lower flange 360.00 8.18 8,033.06 Stem 6,912 –6.82 6,694.21 Moment of inertia about the z axis (mm4) =

IC + d²A (mm4) 8,393.06 13,606.21 21,999.27

The y coordinate at the strain gage is: ygage = d − c − y = 30 mm − 8 mm −11.182 mm = 10.818 mm The bending strain at H is thus:

 gage

ygage

H yH

 H =

yH −11.182 mm  gage = −440 10−6 mm/mm ) = 454.805 10−6 mm/mm ( ygage 10.818 mm

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

The bending stress at H is:  x = E x = ( 70,000 MPa ) ( 454.805 10−6 mm/mm ) = 31.836 MPa From the flexure moment, determine the bending moment that produces this bending stress at H: My x = − Iz

M = −

 xIz y

( 31.836 N/mm )( 21,999.27 mm ) = 62, 633.6 N  mm 2

=−

−11.182 mm

Shear-force and bending-moment diagrams From the shear-force and bending-moment diagrams, we find that the bending moment at section 1–1 is: M = ( 480 mm) P Thus, the load P can be calculated as: ( 480 mm ) P = 62, 633.6 N  mm  P = 130.5 N

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.27 A solid steel shaft supports loads PA = 250 N and PC = 620 N as shown in Figure P8.27. Assume a = 500 mm, b = 700 mm, and c = 600 mm. The bearing at B can be idealized as a roller support and the bearing at D can be idealized as a pin support. If the allowable bending stress is 105 MPa, determine the minimum diameter that can be used for the shaft.

FIGURE P8.27

Solution Shear-force and bending-moment diagrams Maximum bending moment magnitude M = 142,615.4 N-mm Minimum required section modulus M x  S M S  x 

142,615.4 N-mm = 1,358.242 mm 3 2 105 N/mm

Section modulus for solid circular section d3 S= 32 Minimum shaft diameter d 3  1,358.242 mm 3 32  d  24.0 mm

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.28 A simply supported wood beam (Figure P8.28a/29a) with a span of L = 6 m supports a uniformly distributed load of w0. The beam width is b = 120 mm and the beam height is h = 300 mm (Figure P8.28b/29b). The allowable bending stress of the wood is 8.0 MPa. Calculate the magnitude of the maximum load w0 that may be carried by the beam.

FIGURE P8.28a/29a

FIGURE P8.28b/29b

Solution Beam equilibrium:  2 L  L  M A = − w0    + C y L = 0  3  3  2  C y = w0 L 9  2 L  2 L  M C = w0    − Ay L = 0  3  3  4  Ay = w0 L 9 Shear-force diagram Maximum bending moment from found from shear-force diagram 14  4  M max =  w0 L  L  29  9  8 = w0 L2 81 Allowable bending moment: I bh3 /12 bh 2 S= = = c h/2 6

(120 mm )( 300 mm ) = 6 = 1.8 10 mm3 6

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Maximum allowable bending moment M: M x  S  M   x S = ( 8 N/mm 2 )(1.8 106 mm3 ) = 14.4 106 N  mm = 14.4 kN  m

Maximum load w0 that may be carried by the beam: 8 w0 L2  14.4 kN  m 81 81  w0  14.4 kN  m ) = 4.05 kN/m 2 ( 8(6 m)

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.29 A simply supported wood beam (Figure P8.28a/29a) with a span of L = 24 ft supports a uniformly distributed load of w0 = 450 lb/ft. The allowable bending stress of the wood is 1,200 psi. If the aspect ratio of the solid rectangular wood beam is specified as h/b = 3.0 (Figure P8.28b/29b), calculate the minimum width b that can be used for the beam.

FIGURE P8.28a/29a

FIGURE P8.28b/29b

Solution Shear-force and bending-moment diagrams Maximum bending moment magnitude M = 25,600 lb·ft = 307,200 lb·in. Minimum required section modulus M x  S M S 

x



307,200 lb  in. = 256 in.3 1, 200 lb/in.2

Section modulus for solid rectangular section I bh3 /12 bh 2 S= = = c h/2 6

The aspect ratio of the solid rectangular wood beam is specified as h/b = 3.0; therefore, the section modulus can be expressed as: bh 2 b(3.0b) 2 S= = = 1.5b3 6 6 Minimum allowable beam width 1.5b3  256 in.3  b  5.547 in. = 5.55 in.

The corresponding beam height h is h / b = 3.0 h = 3.0b = 3.0 ( 5.547 ) = 16.64 in.

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.30 A cantilever timber beam (Figure P8.30a/31a) with a span of L = 4.25 m supports a linearly distributed load with maximum intensity of w0 = 5.5 kN/m. The allowable bending stress of the wood is 7.0 MPa. If the aspect ratio of the solid rectangular timber is specified as h/b = 0.67 (Figure P8.30b/31b), determine the minimum width b that can be used for the beam.

FIGURE P8.30a/31a

FIGURE P8.30b/31b

Solution Maximum moment magnitude: The maximum bending moment magnitude in the cantilever beam occurs at support A:  w L  L  w L ( 5.5 kN/m )( 4.25 m ) M max =  0    = 0 = = 16.5573 kN  m = 16,557.3 N  m 6 6  2  3  2

Minimum required section modulus M M (16,557.3 N  m )(1,000 mm/m ) x  S  = = 2,365,327.4 mm3 2 S x 7.0 N/mm Section modulus for solid rectangular section I bh3 /12 bh 2 S= = = c h/2 6 The aspect ratio of the solid rectangular wood beam is specified as h/b = 0.67; therefore, the section modulus can be expressed as:

bh2 b ( 0.67b ) 0.4489b3 S= = = = 0.074817b3 6 6 6 2

Minimum allowable beam width 0.074817b3  2,365,327.4 mm 3  b  316.202 mm = 316 mm

The corresponding beam height h is h / b = 0.67  h = 0.67b = 0.67 ( 316.202 mm ) = 212 mm

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.31 A cantilever timber beam (Figure P8.30a/31a) with a span of L = 14 ft supports a linearly distributed load with maximum intensity of w0. The beam width is b = 15 in. and the beam height is h = 10 in. (Figure P8.30b/31b). The allowable bending stress of the wood is 1,000 psi. Calculate the magnitude of the maximum load w0 that may be carried by the beam.

FIGURE P8.30a/31a

FIGURE P8.30b/31b

Solution Section modulus for rectangular cross section about horizontal centroidal axis

bh2 (15 in.)(10 in.) S= = = 250 in.3 6 6 2

Maximum allowable moment M x = S  M =  x S = (1, 000 lb/in.2 )( 250 in.3 ) = 250, 000 lb  in. = 20,833.33 lb  ft Maximum moment magnitude: The maximum bending moment magnitude in the cantilever beam occurs at support A: 2  w L  L w L M max =  0    = 0  2   3 6 Solve for the distributed load w0: 2 w0 L2 w0 (14 ft ) =  20,833.33 lb  ft 6 6 6 ( 20,833.33 lb  ft )  w0  = 637.755 lb/ft = 638 lb/ft 2 (14 ft )

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.32 The lever shown in Figure P8.32 must exert a force of P = 2,700 lbs at B as shown. The lever lengths are a = 10.5 in. and b = 46.0 in. The allowable bending stress for the lever is 12,000 psi. If the lever height h is to be three times the lever thickness t (i.e., h/t = 3), what is the minimum thickness t that can be used for the lever?

FIGURE P8.32

Solution Equilibrium M A = Pa − Qb = 0

Q =

a 10.5 in. P= ( 2, 700 lb ) = 616.304 lb b 46.0 in.

Maximum bending moment: The maximum bending moment magnitude in the lever occurs at B: M max = Q (b − a ) = ( 616.304 lb )( 46.0 in. −10.5 in.) = 21,878.8 lb  in. Minimum required section modulus M x  S M 21,878.8 lb  in. S  = = 1.8232 in.3 2  x 12, 000 lb/in. Section modulus for solid rectangular section I th3 /12 th 2 S= = = c h/2 6 The aspect ratio of the solid rectangular lever is specified as h/t = 3; therefore, the section modulus can be expressed as:

th2 t ( 3t ) S= = = 1.5t 3 6 6 2

Minimum allowable lever thickness 1.5t 3  1.8232 in.3  t  1.067 in.

The corresponding lever height h is h/t =3  h = 3t = 3 (1.067 in.) = 3.20 in.

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.33 The beam shown in Figure P8.33 will be constructed from a standard steel W-shape using an allowable bending stress of 30 ksi. The beam span is L = 24 ft, and the beam loads are w = 1.5 kips/ft and P = 20 kips. (a) Develop a list of five acceptable shapes that could be used for this beam. Include the most economical W14, W16, W18, W21, and W24 shapes on the list of possibilities. (b) Select the most economical W shape for this beam. FIGURE P8.33

Solution Shear-force and bending-moment diagrams Maximum bending moment magnitude M = 228 kip·ft Minimum required section modulus M x  S M S 

x



( 228 kip  ft )(12 in./ft ) = 91.2 in.3 30 kips/in.2

(a) Acceptable steel W-shapes W14  68, S = 103 in.3

W16  57,

S = 92.2 in.3

W18  55,

S = 98.3 in.3

W21  50,

S = 94.5 in.3

W24  55,

S = 114 in.3

(b) Most economical W-shape W21  50

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.34 The beam shown in Figure P8.34 will be constructed from a standard steel W-shape using an allowable bending stress of 30 ksi. The beam span is L = 24 ft, and the beam loads are w = 1,100 lb/ft and 2w = 2,200 lb/ft. (a) Develop a list of four acceptable shapes that could be used for this beam. Include the most economical W10, W12, W14, and W16 shapes on the list of possibilities. (b) Select the most economical W shape for this beam. FIGURE P8.34

Solution Shear-force and bending-moment diagrams Maximum bending moment magnitude M = 121.275 kip·ft Minimum required section modulus M x  S M S 

x



(121.275 kip  ft )(12 in./ft ) = 48.51 in.3 30 kips/in.2

(a) Acceptable steel W-shapes W10  45, S = 49.1 in.3 W12  40,

S = 51.5 in.3

W14  34,

S = 48.6 in.3

W16  40,

S = 64.7 in.3

(b) Most economical W-shape W14  34

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.35 The beam shown in Figure P8.35 will be constructed from a standard steel W-shape using an allowable bending stress of 165 MPa. (a) Develop a list of four acceptable shapes that could be used for this beam. Include the most economical W310, W360, W410, and W460 shapes on the list of possibilities. (b) Select the most economical W shape for this beam.

FIGURE P8.35

Solution Maximum moment magnitude: The maximum bending moment magnitude occurs at the base of the cantilever beam: 1 1 M max = (15 kN)(3.0 m) + (40 kN/m)(3.0 m) (3.0 m) 2 3 6 = 105.0 kN-m = 105.0 10 N-mm Minimum required section modulus M x  S M (105.0 kN-m)(1,000) 2 S  = = 636  103 mm3 2 x 165 N/mm (a) Acceptable steel W-shapes W310  60, S = 844 103 mm3 W360  44,

S = 688 103 mm3

W410  46.1, S = 773 103 mm3 W460  52,

S = 944 103 mm3

(b) Most economical W-shape W360  44

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.36 The beam shown in Figure P8.36 will be constructed from a standard steel HSS-shape using an allowable bending stress of 30 ksi. (a) Develop a list of three acceptable shapes that could be used for this beam. On this list, include the most economical HSS8, HSS10, and HSS12 shapes. (b) Select the most economical HSS-shape for this beam. FIGURE P8.36

Solution Shear-force and bending-moment diagrams Maximum bending moment magnitude M = 45.56 kip·ft Minimum required section modulus M x  S M S 

x



( 45.56 kip  ft )(12 in./ft ) = 18.22 in.3 30 ksi

(a) Acceptable steel HSS shapes HSS8 none are acceptable HSS10  4  3 / 8,

S = 20.8 in.3

HSS10  6  3 / 8,

S = 27.4 in.3

HSS12  6  3 / 8,

S = 35.9 in.3

HSS12  8  3 / 8,

S = 43.7 in.3

(b) Most economical HSS shape HSS10  4  3 / 8

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.37 A composite beam is fabricated by bolting two wooden boards (b = 3 in. and d = 12 in.) to the sides of a steel plate (t = 0.5 in. and d = 12 in.), as shown in Figure P8.37b. The moduli of elasticity of the wood and the steel are 1,800 ksi and 30,000 ksi, respectively. The simply supported beam spans a distance of L = 24 ft and carries a uniformly distributed load of w as shown in Figure P8.37a. (You may neglect the weight of the beam in your calculations.) (a) Determine the maximum bending stresses produced in the wooden boards and the steel plate if w = 400 lb/ft. (b) Assume that the allowable bending stresses of the wood and the steel are 1,200 psi and 24,000 psi, respectively. Determine the largest acceptable magnitude for distributed load w.

FIGURE P8.37a FIGURE P8.37b

Solution Let the wood be denoted as material (1) and the steel as material (2). The modular ratio is: E 30,000 ksi n= 2 = = 16.6667 E1 1,800 ksi Transform the steel (2) into an equivalent amount of wood (1) by multiplying its width by the modular ratio: b2, trans = 16.6667(0.50 in.) = 8.3333 in. Thus, for calculation purposes, the 12 in. × 0.50 in. steel plate is replaced by a wood board that is 12 in. deep and 8.3333 in. thick. Transformed moment of inertia about the horizontal centroidal axis d = y − yi Shape IC d²A (in.4) (in.) (in.4) wood (1) 864 0 0 transformed steel plate (2) 1,200 0 0 Moment of inertia about the z axis =

IC + d²A (in.4) 864 1,200 2,064 in.4

(a) Maximum bending stresses for w = 400 lb/ft: The maximum bending moment in the beam for a uniformly distributed load of w = 400 lb/ft can be found from the shear-force and bending-moment diagrams.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

The maximum bending moment in the simply supported beam with w = 400 lb/ft is: M max = 21, 600 lb  ft Bending stress in wooden boards (1) From the flexure formula, the maximum bending stress in wooden boards (1) is: Mc ( 21,600 lb  ft )( 6 in.)(12 in./ft ) 1 = = = 753.488 psi = 753 psi I 2,064 in.4

Ans.

Bending stress in steel plate (2) The bending stress in the transformed material must be multiplied by the modular ratio n. Therefore, the maximum bending stress in steel plate (2) is: ( 21,600 lb  ft )( 6 in.)(12 in./ft ) = 12,558.1 psi = 12,560 psi My Ans. 2 = n = (16.6667 ) I 2,064 in.4

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

(b) Determine maximum w Given that the allowable bending stress in the wood is 1,200 psi, the maximum bending moment that may be supported by the beam is: Mc 1 = I 2 4  1 I (1,200 lb/in. )( 2, 064 in. )  M max  = = 412,800 lb  in. = 34, 400 lb  ft c 6 in. Given that the allowable bending stress in the steel is 24,000 psi, the maximum bending moment that may be supported by the beam is: Mc 2 = n I 2 4  2 I ( 24, 000 lb/in. )( 2, 064 in. )  M max  = = 495,359 lb  in. = 41, 279.9 lb  ft nc (16.667 )( 6 in.) From these two results, the maximum moment that the beam can support is 412,800 lb·in. = 34,400 lb·ft. The maximum distributed load magnitude w can be found from the existing VM diagram by proportions: wmax 400 lb/ft = 34, 400 lb  ft 21, 600 lb  ft 34, 400 lb  ft Ans.  wmax = ( 400 lb/ft ) = 637 lb/ft 21, 600 lb  ft

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.38 A flitch beam can be fabricated by sandwiching two steel plates (2) between three wood boards (1), as shown in Figure P8.38. Dimensions for the cross section are b = 1.5 in., d = 7.25 in., and t = 0.5 in. The elastic moduli of the wood and steel are E1 = 1,800 ksi and E2 = 30,000 ksi, respectively. For a bending moment of Mz = +12 kip·ft, determine: (a) the normal stress in each material at point H, which is located at a distance of a = 1.75 in. above the z centroidal axis. (b) the normal strain in each material at point H. (c) the maximum bending stress in each material. FIGURE P8.38

Solution Let the wood be denoted as material (1) and the steel as material (2). The modular ratio is: E 30,000 ksi n= 2 = = 16.6667 E1 1,800 ksi Transform the steel (2) into an equivalent amount of wood (1) by multiplying its width by the modular ratio: b2, trans = 16.6667(2×0.50 in.) = 16.6667 in. Thus, for calculation purposes, the two 7.25 in. × 0.50 in. steel plates are replaced by a wood board that is 7.25 in. deep and 16.6667 in. wide. Transformed moment of inertia about the horizontal centroidal axis d = y − yi Shape IC d²A 4 (in. ) (in.) (in.4) wood (1) 142.9043 0 0 transformed steel plate (2) 529.2752 0 0 Moment of inertia about the z axis =

IC + d²A (in.4) 142.9043 529.2752 672.1795 in.4

(a) Bending stress in wood boards at H: From the flexure formula, the bending stress in the wood at H is: My 1 = − I (12 kip  ft )(1.75 in.)(12 in./ft )(1, 000 lb/kip ) =− 672.1795 in.4

= −374.8999 psi = 375 psi (C)

Ans.

Bending stress in steel plates at H: The bending stress in the transformed material must be multiplied by the modular ratio n. Therefore, the bending stress in the steel plates at H is: My  2 = −n I (12 kip  ft )(1.75 in.)(12 in./ft )(1, 000 lb/kip ) = − (16.6667 ) 672.1795 in.4

= −6, 248.3435 psi = 6, 240 psi (C)

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

(b) Normal strain in each material at point H: The normal strain at H is the same for both materials. To demonstrate, calculate the normal strain in the wood:  −374.8999 psi Ans. 1 = 1 = = −208.28 10−6 in./in. = −208 με E1 1,800,000 psi And, calculate the normal strain in the steel:  −6, 248.3435 psi Ans. 2 = 2 = = −208.28 10−6 in./in. = −208 με E2 30,000,000 psi (c) Maximum bending stress in wood boards: From the flexure formula, the maximum bending stress in the wood is: Mc 1 = I (12 kip  ft )( 7.25 in. / 2 )(12 in./ft )(1, 000 lb/kip ) = 777 psi Ans. = 672.1795 in.4 Maximum bending stress in steel plates: The maximum bending stress in the steel plates is: Mc 2 = n I (12 kip  ft )( 7.25 in. / 2 )(12 in./ft )(1, 000 lb/kip ) = 12,940 psi = (16.6667 ) 672.1795 in.4

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.39 A composite beam consists of a bronze [E = 105 GPa] bar (2) attached rigidly to an aluminum alloy [E = 70 GPa] bar (1) as shown in Figure P8.39. The dimensions of the cross section are b1 = 60 mm, b2 = 25 mm, and d = 40 mm. The allowable stress of the aluminum alloy is 165 MPa, and the allowable stress of the bronze is 210 MPa. What is the magnitude of the allowable bending moment Mz that may be applied to the composite cross section? FIGURE P8.39

Solution Denote the aluminum as material (1) and denote the bronze as material (2). The modular ratio is: E 105 GPa n= 2 = = 1.5 E1 70 GPa Transform the bronze bar (2) into an equivalent amount of aluminum (1) by multiplying its width by the modular ratio: b2, trans = 1.5(25 mm) = 37.5 mm. Thus, for calculation purposes, the 25 mm × 40 mm bronze bar is replaced by an aluminum bar that is 37.5 mm wide and 40 mm deep. Moment of inertia about the horizontal centroidal axis d = y − yi Shape IC (mm4) aluminum bar (1) 320,000 transformed bronze bar (2) 200,000 Moment of inertia about the z axis =

(mm) 0 0

d²A (mm4) 0 0

IC + d²A (mm4) 320,000 200,000 520,000 mm4

(a) Maximum bending moment magnitude based on allowable aluminum stress: Based on an allowable bending stress of 165 MPa for the aluminum, the maximum bending moment magnitude that be applied to the cross section is: Mc 1  I 2 4  I (165 N/mm )( 520, 000 mm ) (a) M  1 = = 4.29  106 N  mm = 4.29 kN  m c 40 mm / 2 Maximum bending moment magnitude based on allowable bronze stress: Based on an allowable bending stress of 210 MPa for the bronze, the maximum bending moment magnitude that be applied to the cross section is: Mc 2  n I 2 4  2 I ( 210 N/mm )( 520, 000 mm ) (b) M  = = 3.64 106 N  mm = 3.64 kN  m nc 1 .5 40 mm / 2 ( )( ) Maximum bending moment magnitude: From the values obtained in Eqs. (a) and (b), the maximum bending moment that can be applied to the cross section is M max = 3.64 kN  m Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.40 Two aluminum alloy plates (2) are attached to the sides a wood beam (1) as shown in Figure P8.40. Dimensions for the composite cross section are b1 = 80 mm, d1 = 240 mm, b2 = 10 mm, d2 = 120 mm, and a = 60 mm. Determine the maximum bending stresses produced in both the wood beam and the aluminum plates if a bending moment of Mz = +6,000 N·m is applied about the z axis. Assume E1 = 12.5 GPa and E2 = 70 GPa.

FIGURE P8.40

Solution Let the wood be denoted as material (1) and the aluminum as material (2). The modular ratio is: E 70 GPa n= 2 = = 5.6 E1 12.5 GPa Transform the aluminum plates (2) into an equivalent amount of wood (1) by multiplying the plate thicknesses by the modular ratio: b2, trans = 5.6(10 mm) = 56 mm (each). Thus, for calculation purposes, each 120 mm × 10 mm aluminum plate is replaced by a wood board that is 120 mm tall and 56 mm wide. Centroid location: Since the transformed section is doubly symmetric, the centroid location is found from symmetry. Moment of inertia about the z centroidal axis Shape wood beam (1) two transformed aluminum plates (2) Moment of inertia about the z axis =

IC (mm4) 92.16×106 2×8.06×106 108.29×106 mm4

Maximum bending stress in wood beam (1) From the flexure formula, the maximum bending stress in wood beam (1) is: M c ( 6, 000 N  m )( 240 mm / 2 )(1,000 mm/m ) 1 = z = = 6.65 MPa Iz 108.29 106 mm4

Ans.

Maximum bending stress in aluminum plates (2) The bending stress in the transformed material must be multiplied by the modular ratio n. Therefore, the maximum bending stress in the aluminum plates (2) is: ( 6,000 N  m )(120 mm / 2 )(1,000 mm/m ) = 18.62 MPa Mc  2 = n z = ( 5.6 ) Ans. Iz 108.29 106 mm4

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.41 A wooden beam (1) is reinforced with steel plates (2) rigidly attached to its top and bottom surfaces, as shown in Figure P8.40. Dimensions of the cross section are b1 = 6 in., d1 = 12 in., b2 = 4 in., and d2 = 0.5 in. The elastic moduli of the wood and steel are E1 = 1,250 ksi and E2 = 30,000 ksi, respectively. The allowable bending stresses of the wood and steel are 1,200 psi and 20,000 psi, respectively. Determine the largest concentrated vertical load that can be applied at midspan if a simply supported beam with this cross section spans 18 ft. FIGURE P8.41

Solution Denote the timber as material (1) and denote the steel as material (2). The modular ratio is: E 30,000 ksi n= 2 = = 24 E1 1,250 ksi Transform the steel plates into an equivalent amount of wood by multiplying their width by the modular ratio: b2, trans = 24(4 in.) = 96 in. Thus, for calculation purposes, the 4 in. × 0.5 in. steel plates can be replaced by wood plates that are 96 in. wide and 0.5 in. thick. Moment of inertia about the horizontal centroidal axis d = y − yi Shape IC 4

(in. ) transformed steel plate at top 1 wooden beam (1) 864 transformed steel plate at bottom 1 Moment of inertia about the z axis =

(in.) –6.25 0 6.25

d²A (in.4) 1,875 0 1,875

IC + d²A (in.4) 1,876 864 1,876 4,616 in.4

Maximum bending moment magnitude based on allowable stress for the wood: Based on an allowable bending stress of 1,200 psi for the wood, the maximum bending moment magnitude that be applied to the cross section is: 2 4 Mc1  1 I (1, 200 lb/in. )( 4,616 in. ) (a) 1  M  = = 923, 200 lb  in. I c1 12 in. / 2 Maximum bending moment magnitude based on allowable stress for the steel: Based on an allowable bending stress of 20,000 psi for the steel plates, the maximum bending moment magnitude that be applied to the cross section is: 2 4 Mc2  2 I ( 20, 000 lb/in. )( 4,616 in. ) (b) 2  n M  = = 591, 795 lb  in. I nc2 ( 24 )(12 in. / 2 + 0.5 in.) Maximum bending moment magnitude: From the values obtained in Eqs. (a) and (b), the maximum bending moment that can be applied to the cross section is: M max = 591, 795 lb  in. = 49,316.2 lb  ft

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Largest concentrated vertical load: For a simply supported beam with a concentrated load at midspan, the maximum bending moment is: PL M max = 4 For a beam spanning 18 ft, the maximum concentrated load is: PL  49,316.2 lb  ft 4 4 ( 49,316.2 lb  ft ) Ans. P  = 10,959.2 lb = 10,960 lb 18 ft

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.42 A bronze [E = 15,200 ksi] bar is rigidly attached to a stainless steel [E = 27,500 ksi] bar to form a composite beam (Figure P8.42b/43b). The dimensions of the cross section are b = 4.5 in., dB = 2.25 in., and dS = 1.00 in. The composite beam is subjected to a bending moment of M = 6,400 lb·ft about the z axis (Figure P8.42a/43a). Determine: (a) the maximum bending stresses in the bronze and stainless steel bars. (b) the normal stress in each bar at the surface where they contact each other.

FIGURE P8.42a/43a

FIGURE P8.42b/43b

Solution Denote the bronze as material (1) and denote the stainless steel as material (2). The modular ratio is: E 27,500 ksi n= 2 = = 1.8092 E1 15,200 ksi Transform the stainless steel bar (2) into an equivalent amount of bronze (1) by multiplying its width by the modular ratio: b2, trans = 1.8092(4.5 in.) = 8.1414 in. Thus, for calculation purposes, the 4.5 in. × 1.00 in. stainless steel bar is replaced by a bronze bar that is 8.1414 in. wide and 1.00 in. thick. Centroid location of the transformed section in the vertical direction Shape bronze bar (1) transformed stainless steel bar (2)

Width b (in.) 4.5

Height h (in.) 2.25

Area Ai (in.2) 10.1250

yi (from bottom) (in.) 2.125

8.1414

1.00

8.1414

0.50

18.2664

yi Ai (in.3) 21.5156 4.0707 25.5863

yi Ai

21.5156 in.3 y= = = 1.4007 in. (measured upward from bottom edge of section) Ai 18.2664 in.2

Moment of inertia about the horizontal centroidal axis d = y − yi Shape IC (in.4) 4.2715 0.6785

bronze bar (1) transformed stainless steel bar (2) Moment of inertia about the z axis =

(in.) –0.7243 0.9007

d²A (in.4) 5.3113 6.6053

IC + d²A (in.4) 9.5827 7.2837 16.8665 in.4

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

(a) Maximum bending stress in bronze bar (1): From the flexure formula, the maximum bending stress in the bronze bar is: y = d S + d B − y = 1.00 in. + 2.25 in. − 1.4007 in. = 1.8493 in.

1 = −

( 6, 400 lb  ft )(1.8493 in.)(12 in./ft ) = 8, 420 psi (C) My =− I 16.8665 in.4

Ans.

Maximum bending stress in stainless steel bar (2): The bending stress in the transformed material must be multiplied by the modular ratio n. Therefore, the maximum bending stress in the stainless steel bar is: y = − y = −1.4007 in.

 2 = −n

( 6, 400 lb  ft )( −1.4007 in.)(12 in./ft ) = 11,540 psi (T) My = − (1.8092 ) I 16.8665 in.4

Ans.

(b) Bending stress in bronze bar (1) at interface y = d S − y = 1.00 in. − 1.4007 in. = −0.4007 in.

1 = −

( 6, 400 lb  ft )( −0.4007 in.)(12 in./ft ) = 1,825 psi (T) My =− I 16.8665 in.4

Ans.

Bending stress in stainless steel bar (2) at interface y = d S − y = 1.00 in. − 1.4007 in. = −0.4007 in.

 2 = −n

( 6, 400 lb  ft )( −0.4007 in.)(12 in./ft ) = 3,300 psi (T) My = − (1.8092 ) I 16.8665 in.4

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.43 A bronze [E = 100 GPa] bar is rigidly attached to a stainless steel [E = 190 GPa] bar to form a composite beam (Figure P8.42b/43b). The dimensions of the cross section are b = 150 mm, dB = 85 mm, and dS = 40 mm. The allowable bending stresses for the bronze and stainless steel bars are 180 MPa and 225 MPa, respectively. Determine the allowable bending moment M (Figure P8.42a/43a) that can be applied to the composite beam.

FIGURE P8.42a/43a

FIGURE P8.42b/43b

Solution Denote the bronze as material (1) and denote the stainless steel as material (2). The modular ratio is: E 190 GPa n= 2 = = 1.9 E1 100 GPa Transform the stainless steel bar (2) into an equivalent amount of bronze (1) by multiplying its width by the modular ratio: b2, trans = 1.9(150 mm) = 285 mm. Thus, for calculation purposes, the 150 mm × 40 mm stainless steel bar is replaced by a bronze bar that is 285 mm wide and 40 mm thick. Centroid location of the transformed section in the vertical direction Shape bronze bar (1) transformed stainless steel bar (2)

Width b (mm) 150

Height h (mm) 85

Area Ai (mm2) 12,750

yi (from bottom) (mm) 82.50

yi Ai (mm3) 1,051,875

285

11,400

20.00

228,000

24,150 y=

yi Ai Ai

1,279,875

1,279,875 mm3 = 53.00 mm (measured upward from bottom edge of section) 24,150 mm 2

Moment of inertia about the horizontal centroidal axis d = y − yi Shape IC 4

(mm ) bronze bar (1) 7,676,562.5 transformed stainless steel 1,520,000.0 bar (2) Moment of inertia about the z axis =

(mm) –29.50 33.00

d²A (mm4) 11,095,687.5 12,414,600.0

IC + d²A (mm4) 18,772,250 13,934,600 32,706,850 mm4

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Maximum bending moment magnitude based on allowable stress for the bronze: Based on an allowable bending stress of 180 MPa for the aluminum, the maximum bending moment magnitude that be applied to the cross section is: y = d S + d B − y = 40 mm + 85 mm − 53.0 mm = 72.0 mm

1  −

My I

M  −

1I y

(180 N/mm )( 32, 706,850 mm ) = 81, 767,125 N  mm = 81.767 kN  m =− 2

72.0 mm

(a)

Maximum bending moment magnitude based on allowable stress for the stainless steel: Based on an allowable bending stress of 225 MPa for the stainless steel, the maximum bending moment magnitude that be applied to the cross section is: y = − y = −53.0 mm My  2  −n I M  −

 2I ny

( 225 N/mm )( 32, 706,850 mm ) = 73, 078,860 N  mm = 73.079 kN  m = − 2

(1.9 )( −53.0 mm )

(b)

Allowable bending moment M: From the values obtained in Eqs. (a) and (b), the maximum bending moment that can be applied to the cross section is M max = 73.079 kN  m = 73.1 kN  m Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.44 A wood beam (1) is reinforced on its lower surface by a steel plate (2) as shown in Figure P8.44. Dimensions of the cross section are b1 = 220 mm, d = 380 mm, b2 = 160 mm, and t = 20 mm. The elastic moduli of the wood and steel are E1 = 12.5 GPa and E2 = 200 GPa, respectively. The allowable bending stresses of the wood and steel are 7.5 MPa and 150 MPa, respectively. This cross section is used for a simply supported beam that spans 9 m. What is the largest uniformly distributed load that can be applied to the beam? FIGURE P8.44

Solution Denoted the wood as material (1) and denote the steel as material (2). The modular ratio is: E 200 GPa n= 2 = = 16 E1 12.5 GPa Transform the steel into an equivalent amount of wood by multiplying its width by the modular ratio: b2,trans = 16(160 mm) = 2,560 mm. Thus, for calculation purposes, the 160 mm × 20 mm steel plate is replaced by a wood board that is 2,560 mm wide and 20 mm thick. Centroid location of the transformed section in the vertical direction Shape wood beam (1) transformed steel plate (2) y=

yi Ai Ai

Width b (mm) 220 2,560

Height h (mm) 380 20

yi (from bottom) (mm) 210 10

Area Ai (mm2) 83,600 51,200 134,800

yi Ai (mm3) 17,556,000 512,000 18,068,000

18,068,000 mm3 = 134.04 mm (measured upward from bottom edge of section) 134,800 mm 2

Moment of inertia about the horizontal centroidal axis d = y − yi Shape IC 4

(mm ) wood beam (1) 1,005.99×106 transformed steel plate (2) 1.71×106 Moment of inertia about the z axis =

(mm) –75.96 124.04

d²A (mm4) 482.37×106 787.76×106

IC + d²A (mm4) 1,488.36×106 789.47×106 2,277.83×106 mm4

Determine maximum allowable moment Given that the allowable bending stress in the wood is 7.5 MPa, the maximum bending moment that may be supported by the beam is: c1 = t + d − y = 20 mm + 380 mm − 134.04 mm = 265.96 mm

1 =

Mc1 I

 M max 

 1I c1

( 7.5 N/mm )( 2,277.83  10 mm ) = 64.234  10 N  mm = 64.234 kN  m = 2

265.96 mm

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Given that the allowable bending stress in the steel is 150 MPa, the maximum bending moment that may be supported by the beam is: c2 = y = 134.04 mm

2 = n

Mc2 I

 M max 

 2I nc2

(150 N/mm )( 2, 277.83  10 mm ) = 159.316  10 N  mm = 159.316 kN  m = 2

(16 )(134.04 mm )

From these two results, the maximum moment that the beam can support is 64.234 kN·m. Largest uniformly distributed load: For a simply supported beam with a uniformly distributed load over the entire span, the maximum bending moment is: wL2 M max = 8 For a beam spanning 9 m, the maximum uniformly distributed load is: wL2  64.234 kN  m 8 8 ( 64.234 kN  m ) Ans. w  = 6.34 kN/m 2 (9 m)

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.45 A composite beam consists of concrete slab (1) that is rigidly attached to the top flange of a W16 × 40 standard steel shape (2). The cross section of this composite beam is shown in Figure P8.45. Concrete slab (1) has a width of b = 48 in., a thickness of t = 4 in., and an elastic modulus of 3,500 ksi. The elastic modulus of the steel shape is 30,000 ksi, and the cross-sectional dimensions are presented in Appendix B. For a bending moment of Mz = +150 kip·ft, determine: (a) the location of the centroid for the transformed section, measured upward from point K. (b) the normal stress at H in the concrete slab. (c) the normal stress at K in the steel shape.

FIGURE P8.45

Solution Denote the concrete as material (1) and denote the steel as material (2). For this cross section, it will be easier to transform the concrete into an equivalent amount of steel. Accordingly, we will express the modular ratio as: E 3,500 ksi n= 1 = = 0.1167 E2 30,000 ksi Transform the concrete slab (1) into an equivalent amount of steel (2) by multiplying its width by the modular ratio: b1,trans = 0.1167(48 in.) = 5.6016 in. Thus, for calculation purposes, the 48 in. × 4 in. concrete slab is replaced by a steel bar that is 5.6016 in. wide and 4 in. thick. From Appendix B, the cross-sectional dimensions and properties of the W16 × 40 standard steel shape are: d = 16.00 in.2 A = 11.8 in.2 I z = 518 in.4 (a) Centroid location of the transformed section in the vertical direction

Shape Transformed concrete slab (1) Steel wide-flange shape (2)

yi Ai Ai

Width b (in.) 5.6016

Height h (in.) 4 16.00

497.7152 in.3 = 14.5504 in. = 14.55 in. 34.2064 in.2

(measured upward from bottom edge of section)

Area Ai (in.2) 22.4064 11.8 34.2064

yi (from bottom) (in.) 18.0 8.0

yi Ai (in.3) 403.3152 94.4 497.7152 Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed. Moment of inertia about the horizontal centroidal axis d = y − yi Shape IC 4

(in. ) Transformed concrete slab (1) 29.867 Steel wide-flange shape (2) 518 Moment of inertia about the z axis =

(in.) –3.4496 6.5504

Timothy A. Philpot

d²A (in.4) 266.630 506.311

IC + d²A (in.4) 296.497 1,024.311 1,320.809 in.4

(b) Normal stress at H in the concrete slab: The bending stress in the transformed material must be multiplied by the modular ratio n. From the flexure formula, the normal stress at H is: yH = d + t − y = 16 in. + 4 in. − 14.5504 in. = 5.4496 in.

 H = −n

(150 kip  ft )( 5.4496 in.)(12 in./ft )(1,000 lb/kip ) = 867 psi (C) MyH = − ( 0.1167 ) I 1,320.809 in.4

Ans.

K = −

(150 kip  ft )( −14.5504 in.)(12 in./ft )(1,000 lb/kip ) = 19,830 psi (T) MyK =− I 1,320.809 in.4

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.46 A load of P = 1,200 N acts on a curved steel rod as shown in Figure P8.46. The rod is solid with a diameter of d = 15 mm. Assume b = 35 mm and a = 175 mm. Determine the normal stresses produced at points H and K.

FIGURE P8.46

Solution Section properties



Iz =

d2 =



d4 =

(15 mm ) = 176.715 mm2



(15 mm ) = 2,485.05 mm4 4

Internal forces and moments F = P = 1, 200 N

M z = Pb = (1, 200 N )( 35 mm ) = 42,000 N  mm Stresses F 1, 200 N = = 6.791 MPa (T) A 176.715 mm 2 M c ( 42, 000 N  mm )(15 mm / 2 )  bending = z = = 126.758 MPa Iz 2, 485.05 mm 4

 axial =

Normal stress at H By inspection, the bending stress at H will be compression; therefore, the normal stress at H is:  H = 6.791 MPa − 126.758 MPa = −119.967 MPa = 120.0 MPa (C)

Ans.

Normal stress at K By inspection, the bending stress at K will be tension; therefore, the normal stress at K is:  K = 6.791 MPa + 126.758 MPa = 133.549 MPa = 133.5 MPa (T)

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.47 A pulley is supported with the bar shown in Figure P8.47. The pulley has a diameter of D = 5 in., and the pulley belt tension is P = 45 lb. Assume a = 3 in. and c = 6 in. Determine the normal stresses produced at points H and K. The support bar cross-sectional dimensions at the section of interest are d = 1.5 in. by 0.25 in. thick. FIGURE P8.47

Solution Section properties A = ( 0.25 in.)(1.5 in.) = 0.3750 in.2

( 0.25 in.)(1.5 in.) = 70.3125 10−3 in.4 I = 3

Internal forces and moments F = 2 P = 2 ( 45 lb ) = 90 lb

d 1.5 in.    M z = 2 P  a +  = ( 90 lb )  3 in. +  = 337.5 lb  in. 2 2    Stresses

F 90 lb = = 240 psi (T) A 0.3750 in.2 M c ( 337.5 lb  in.)(1.50 in. / 2 )  bending = z = = 3, 600 psi Iz 70.3125  10−3 in.4

 axial =

Normal stress at H By inspection, the bending stress at H will be tension; therefore, the normal stress at H is:  H = 240 psi + 3,600 psi = 3, 840 psi = 3,840 psi (T)

Ans.

Normal stress at K By inspection, the bending stress at K will be compression; therefore, the normal stress at K is:  K = 240 psi − 3,600 psi = −3,360 psi = 3,360 psi (C)

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.48 A 30 mm diameter steel rod is formed into a machine part with the shape shown in Figure P8.48. A load of P = 2,500 N is applied to the ends of the part. If the allowable normal stress is limited to 40 MPa, what is the maximum eccentricity e that may be used for the part? FIGURE P8.48

Solution Section properties



A= I=

(30 mm)2 = 706.858 mm 2



(30 mm)4 = 39,760.782 mm4

Internal forces and moments F = 2,500 N

M = (2,500 N)e Stresses F 2,500 N = = 3.537 MPa (T) A 706.858 mm 2 Mc (2,500 N)(30 mm / 2) e  bending = = I 39,760.782 mm 4

 axial =

Maximum eccentricity Since the axial stress is tension, the largest combined stress will occur for the combination of the tension axial stress plus the tension bending stress. Therefore:  axial +  bending  40 MPa

3.537 MPa +

(2,500 N)(30 mm / 2) e  40 MPa 39,760.782 mm 4 (40 MPa − 3.537 MPa) e (39,760.782 mm 4 ) (2,500 N)(15 mm) e  38.662 mm = 38.7 mm

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.49 The bent bar component shown in Figure P8.49 has a width of b = 70 mm and a depth of d = 30 mm. At the free end of the bar, load P is applied at an eccentricity of e = 100 mm from the x centroidal axis. The yield strength of the bar is 310 MPa. If a factor of safety of 2.3 is required, what is the maximum load P that can be applied to the bar? FIGURE P8.49

Solution Section properties A = bd = ( 70 mm )( 30 mm ) = 2,100 mm 2 bd 3 ( 70 mm )( 30 mm ) = = 157,500 mm 4 12 12 3

Iz =

Stresses

 axial =

F −P = A 2,100 mm 2

P (1,500 mm 2 ) P (100 mm )( 30 mm / 2 ) M zc Pec P  bending = = = = = 4 4 Iz Iz 157,500 mm 157,500 mm 105 mm 2

 allow =

Y FS

310 MPa = 134.783 MPa 2.3

Normal stress on the upper surface of the bar The normal stress on the upper surface of the bar is equal to the sum of the compressive axial stress and the compressive bending stress. Since these stresses are expressed in terms of the unknown force P, the compressive normal stress is given by: −P P  upper surface = − 2 2,100 mm 105 mm 2

= − P(476.190 10−6 mm −2 + 9.5238110−3 mm −2 ) = − ( 0.010 mm −2 ) P The normal stress on the upper surface of the bar must be limited to –134.783 MPa; therefore, − ( 0.010 mm −2 ) P  −134.783 MPa P 

134.783 MPa = 13, 478.3 N = 13.48 kN 0.010 mm −2

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.50 A horizontal force of P = 1,200 lb is applied to the rectangular bar shown in Figure P8.50. Calculate the normal stress at points H and K. Use the following values: d = 1.50 in., b = 2.50 in., and e = 2.00 in.

FIGURE P8.50

Solution Section properties A = ( 2.50 in.)(1.50 in.) = 3.750 in.2

( 2.50 in.)(1.50 in.) = 703.125 10−3 in.4 I = 3

Internal forces and moments F = P = 1, 200 lb

M z = Pe = (1, 200 lb )( 2 in.) = 2, 400 lb  in. Stresses

F 1, 200 lb = = 320 psi (T) A 3.750 in.2 M c ( 2, 400 lb  in.)(1.50 in. / 2 )  bending = z = = 2,560 psi Iz 703.125 10−3 in.4

 axial =

Normal stress at H By inspection, the bending stress at H will be compressive; therefore, the normal stress at H is:  H = 320 psi − 2,560 psi = −2, 240 psi = 2, 240 psi (C)

Ans.

Normal stress at K By inspection, the bending stress at K will be tensile; therefore, the normal stress at K is:  K = 320 psi + 2,560 psi = 2, 880 psi = 2,880 psi (T)

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.51 The component shown has a width of b = 35 mm and a depth of d = 65 mm. At the free end of the bar, load P is applied at an eccentricity of e = 50 mm from the y centroidal axis. The yield strength of the bar is 340 MPa. If a factor of safety of 2.3 is required, what is the maximum load P that can be applied to the bar?

FIGURE P8.51

Solution Section properties A = bd = ( 35 mm )( 65 mm ) = 2, 275 mm 2 bd 3 ( 35 mm )( 65 mm ) Iz = = = 800,990 mm 4 12 12 3

Stresses

 axial =

F −P = A 2, 275 mm 2

P (1, 625 mm 2 ) P ( 50 mm )( 65 mm / 2 ) M zc Pec P  bending = = = = = 4 4 Iz Iz 800,990 mm 800,990 mm 492.9169 mm 2

 allow =

Y FS

340 MPa = 147.826 MPa 2.3

Normal stress on the upper surface of the bar The normal stress at point H on the bar is equal to the sum of the compressive axial stress and the compressive bending stress. Since these stresses are expressed in terms of the unknown force P, the compressive normal stress at H is given by: −P P H = − 2 2, 275 mm 492.9169 mm 2

= − P(439.560 10−6 mm −2 + 2.02874 10−3 mm −2 ) = − ( 2.46830 10−3 mm −2 ) P The normal stress at H must be limited to –147.826 MPa; therefore, − ( 2.46830 10−3 mm −2 ) P  −147.826 MPa P 

147.826 MPa = 59,889.8 N = 59.9 kN 2.46830 10−3 mm −2

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.52 A tubular steel column CD supports horizontal cantilever arm ABC, as shown in Figure P8.52. Column CD has an outside diameter of 8.625 in. and a wall thickness of 0.365 in. The loads are PA = 400 lb and PB = 600 lb. Dimensions of the structure are a = 6 ft, b = 8 ft, and c = 14 ft. Determine the maximum compression stress at the base of column CD.

FIGURE P8.57

Solution Section properties d = D − 2t = 8.625 in. − 2 ( 0.365 in.) = 7.895 in. A=



Iz =

 2 2  D 2 − d 2  = ( 8.625 in.) − ( 7.895 in.)  = 9.472 in.2   4 4 

  4 4  D 4 − d 4  = (8.625 in.) − ( 7.895 in.)  = 80.936 in.4  64 64

Internal forces and moments F = − PA − PB = −400 lb − 600 lb = −1, 000 lb

M = PA ( a + b ) + PBb = ( 400 lb )( 6 ft + 8 ft ) + ( 600 lb )(8 ft ) = 10, 400 lb  ft = 124,800 lb  in.

Stresses

F −1, 000 lb = = −105.58 psi A 9.472 in.2 M c (124,800 lb  in.)( 8.625 in. / 2 )  bending = = = 6, 649.72 psi I 80.936 in.4

 axial =

Maximum compression stress at base of column  compression = −105.58 psi − 6,649.72 psi = −6,755.30 psi = 6.76 ksi (C)

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.53 A load of P = 32 kN is applied parallel to the longitudinal axis of a rectangular structural tube as shown in Figure P8.53a. The cross-sectional dimensions of the structural tube (Figure P8.53b) are b = 100 mm, d = 150 mm, and t = 5 mm. Assume a = 500 mm and c = 40 mm. Determine the normal stresses produced at points H and K.

FIGURE P8.53b

FIGURE P8.53a

Solution Section properties A = (100 mm )(150 mm ) − ( 90 mm )(140 mm ) = 2, 400 mm 2

(100 mm )(150 mm ) − ( 90 mm )(140 mm ) = 7,545, 000 mm 4 I = 3

Internal forces and moments F = P = 32 kN = 32, 000 N

d 150 mm    6 M z = P  c +  = ( 32, 000 N )  40 mm +  = 3.68 10 N  mm 2 2     Stresses

 axial =

F 32, 000 N = = 13.333 MPa (T) A 2, 400 mm 2

M z ( d / 2 ) ( 3.68 10 N  mm ) (150 mm / 2 ) = = 36.581 MPa Iz 7.545 106 mm4 6

 bending =

Normal stress at H By inspection, the bending stress at H will be tensile; therefore, the normal stress at H is:  H = 13.333 MPa + 36.581 MPa = 49.914 MPa = 49.9 MPa (T)

Ans.

Normal stress at K By inspection, the bending stress at K will be compressive; therefore, the normal stress at K is:  K = 13.333 MPa − 36.581 MPa = −23.247 MPa = 23.2 MPa (T)

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.54 The bracket shown in Figure P8.54 is subjected to a load of P = 1,300 lb. The bracket has a rectangular cross section with a width of b = 3.00 in. and a thickness of t = 0.375 in. If the tension normal stress must be limited to 24,000 psi at section a–a, what is the maximum offset distance y that can be used?

Solution Section properties A = (3.00 in.)(0.375 in.) = 1.1250 in.2 (3.00 in.)(0.375 in.)3 I= = 0.013184 in.4 12

FIGURE P8.54 Internal forces and moments F = 1,300 lb

0.375 in.  M = (1,300 lb)  y +  = (1,300 lb) y + 243.75 lb-in.  2 Stresses

 axial =

F 1,300 lb = = 1,155.5556 psi A 1.1250 in.2

Mc  bending = = I

 (1,300 lb) y + 243.75 lb-in.  0.013184 in.4

0.375 in.  2

(243.75 lb-in.) y + 45.7031 lb-in.2 = 0.013184 in.4

(243.75 lb-in.) y + 45.7031 lb-in.2  24,000 psi 0.013184 in.4 15.2348 lb-in.2 + (243.75 lb-in.) y + 45.7031 lb-in.2  316.4160 lb-in.2

1,155.5556 psi +

Solve this equation for y: 316.4160 lb-in.2 − 45.7031 lb-in.2 − 15.2348 lb-in.2 255.4781 lb-in.2 y = 243.75 lb-in. 243.75 lb-in.  1.048 in.

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.55 A wall-mounted jib crane is shown in Figure P8.55a. Overall dimensions of the crane are a = 102 in., b = 52 in., c = 66 in., and e = 5.0 in. The cross-sectional dimensions of the crane rail (Figure P8.55b) are bf = 4.0 in., tf = 0.5 in., d = 6.0 in., and tw = 0.3 in. Assume that a hoist load of P = 2,000 lb is located as shown in Figure P8.55a. What is the maximum normal stress in the crane rail?

FIGURE P8.55a

FIGURE P8.55b

Solution Section properties A = 2 ( 4.0 in.)( 0.5 in.) + 6.0 in. − 2 ( 0.5 in.)  ( 0.3 in.) = 5.50 in.2

( 4.0 in.)( 6.0 in.) − ( 4.0 in. − 0.3 in.) 6.0 in. − 2 ( 0.5 in.) 3

Iz =

= 33.458 in.4

c 66 in. = = 0.42857 a + b 102 in. + 52 in.  = 23.199

tan  =

Equilibrium M A = ( F1 cos  ) e + ( F1 sin  )( a + b ) − Pa = 0

F1 e cos  + ( a + b ) sin   = Pa F1 = =

a P e cos  + ( a + b ) sin  102 in. ( 2, 000 lb ) ( 5 in.) cos 23.199 + (102 in. + 52 in.) sin 23.199

= 1.56300 ( 2, 000 lb ) = 3,125.991 lb Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Fx = Ax − F1 cos  = 0

 Ax = F1 cos  = ( 3,125.991 lb ) cos 23.199 = 2,873.239 lb

Fy = Ay + F1 sin  − P = 0  Ay = P − F1 sin  = 2, 000 lb − ( 3,125.991 lb ) sin 23.199 = 768.612 lb

Determine the axial force and the bending moment at the location of the hoist load, where the bending moment will be its maximum value. Fx = F + Ax = 0  F = − Ax = −2,873.239 lb Fy = Ay − V = 0 V = Ay = 768.612 lb M = M − Ay a = 0  M = Ay a = ( 768.612 lb )(102 in.) = 78,398.424 lb  in.

Stresses

F −2,873.239 lb = = −522.407 psi A 5.50 in.2 M c ( 78,398.390 lb  in.)( 6.0 in. / 2 )  bending = z = = 7, 029.495 psi Iz 33.458 in.4

 axial =

By inspection, the bending stress on top of the crane rail will be compressive. Thus, the compressive axial stress will combine with the compressive bending stress to produce the largest normal stress at the top of the crane rail at the location of the hoist load:  x = −522.407 psi − 7,029.495 psi = −7,551.902 psi = 7,550 psi (C) Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.56 The U-shaped aluminum bar shown in Figure P8.56 is used as a dynamometer to determine the magnitude of the applied load P. The aluminum [E = 70 GPa] bar has a square cross section with dimensions a = 30 mm and b = 65 mm. The strain on the inner surface of the bar was measured and found to be 955  What is the magnitude of load P? FIGURE P8.56

Solution Section properties A = (30 mm)2 = 900 mm2

(30 mm)4 I= = 67,500 mm4 12 Internal forces and moments F=P

M = P(65 mm + 30 mm/2) = (80 mm)P Stresses F P = A 900 mm 2 M c (80 mm)P(30 mm/2) P  bending = = =  (17.7778  10−3 mm −2 )P =  4 I 67,500 mm 56.25 mm2

 axial =

Normal stress on the inner surface By inspection, the bending stress on the inner surface will be tension; therefore, the normal stress on the inner surface can be expressed by:

=

P P P + = 2 2 900 mm 56.25 mm 52.94118 mm 2

Normal strain on the inner surface The normal strain on the inner surface was measured as 955  Use this value and the previous expression for  in Hooke’s Law to solve for P: P = (70,000 N/mm 2 )(955  10−6 mm/mm) 52.94118 mm2

 P = 3,539.118 N = 3.54 kN

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.57 The steel pipe shown in Figure P8.57 has an outside diameter of 195 mm, a wall thickness of 10 mm, an elastic modulus of E = 200 GPa, and a coefficient of thermal expansion of  = 11.7 × 10–6 /°C. Using a = 300 mm, b = 900 mm, and  = 70°, calculate the normal strains at H and K after a load of P = 40 kN has been applied and the temperature of the pipe has been increased by 25°C.

Solution Section properties d = D − 2t = 195 mm − 2(10 mm) = 175 mm   A =  D 2 − d 2  = (195 mm) 2 − (175 mm) 2  = 5,811.95 mm 2 4 4    D 4 − d 4  = (195 mm) 4 − (175 mm) 4  = 24,936,883 mm 4 I= 64 64

FIGURE P8.57

Internal forces and moments F = (40,000 N)sin 70 = 37,587.7 N

M = (40,000 N)sin 70(300 mm) + (40,000 N)cos70(900 mm) = 23,589,036.6 N-mm Stresses F 37,587.7 N = = 6.4673 MPa (C) A 5,811.95 mm 2 Mc (23,589,036.6 N-mm)(195 mm / 2)  bending = = = 92.2301 MPa I 24,936,883 mm 4

 axial =

Normal stresses at H and K  H = −6.4673 MPa − 92.2301 MPa = −98.6974 MPa

 K = −6.4673 MPa + 92.2301 MPa = 85.7628 MPa Normal strains at H and K due to load P −98.6974 MPa H = = −493.5  10−6 mm/mm 200,000 MPa 85.7628 MPa K = = 428.8  10−6 mm/mm 200,000 MPa Thermal strain T = T = (11.7  10−6 /C)(25C) = 292.5  10−6 mm/mm Total normal strains at H and K due to load P and temperature change  H = −493.5 με + 292.5 με = −201 με

 K = 428.8 με + 292.5 με = 721 με

Ans. Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.58 A short length of a rolled-steel [E = 29 × 103 ksi] column supports a rigid plate on which two loads P and Q are applied as shown in Figure P8.58a/59a. The column cross section (Figure P8.58b/59b) has a depth of d = 8.0 in., an area of A = 5.40 in.2, and a moment of inertia of Iz = 57.5 in.4. Normal strains are measured with strain gages H and K, which are attached on the centerline of the outer faces of the flanges. Load P is known to be 35 kips, and the strain in gage H is measured as H = +120 × 10–6 in./in. Using a = 6 in., determine: (a) the magnitude of load Q. (b) the expected strain reading for gage K.

FIGURE P8.58a/59a

FIGURE P8.58b/59b

Solution (a) Magnitude of load Q The axial normal stress caused at H by both P and Q will be compression. By inspection, the bending stress at H caused by load Q will be compression, and the bending stress at H caused by load P will be tension. The normal stress at gage H can be expressed by: Q P Q(6 in.)(8 in. / 2) P(6 in.)(8 in. / 2) H = − − − + 2 2 5.40 in. 5.40 in. 57.5 in.4 57.5 in.4 Q Q(24 in.2 ) P P(24 in.2 ) =− − − + 5.40 in.2 57.5 in.4 5.40 in.2 57.5 in.4 Q Q(24 in.2 ) 35 kips (35 kips)(24 in.2 ) =− − − + 5.40 in.2 57.5 in.4 5.40 in.2 57.5 in.4 Q Q(24 in.2 ) =− − + 8.1272 ksi 5.40 in.2 57.5 in.4 Normal strain at H The normal strain at H was measured as 120 × 10–6 in./in. Use this value and the previous expression for  in Hooke’s Law to solve for Q:

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

 H = E H −

Q Q(24 in. ) − + 8.1272 ksi = (29,000 ksi)(120  10 −6 in./in.) 2 4 5.40 in. 57.5 in.  1 24 in.2  Q + = 8.1272 ksi − (29,000 ksi)(120  10 −6 in./in.) 2 4  5.40 in. 57.5 in.  Q 0.6025765 in.−2  = 4.6472 ksi Q = 7.7122 kips = 7.71 kips

Ans.

(b) Expected strain reading for gage K The axial normal stress caused at K by both P and Q will be compression. By inspection, the bending stress at K caused by load Q will be tension, and the bending stress at K caused by load P will be compression. The normal stress at gage K for the value of Q determined in part (a) is: Q P Q(6 in.)(8 in. / 2) P(6 in.)(8 in. / 2) K = − − + − 2 2 5.40 in. 5.40 in. 57.5 in.4 57.5 in.4 7.7122 kips 35 kips (7.7122 kip)(6 in.)(8 in. / 2) (35 kips)(6 in.)(8 in. / 2) =− − + − 5.40 in.2 5.40 in.2 57.5 in.4 57.5 in.4 = −1.4282 ksi − 6.4815 ksi + 3.2190 ksi − 14.6087 ksi

= −19.2993 ksi Use Hooke’s Law to calculate the expected strain reading at gage K.  K = E K

K =

K E

−19.2993 ksi = −665.5  10−6 in./in. = −665 με 29,000 ksi

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.59 A short length of a rolled-steel [E = 29 × 103 ksi] column supports a rigid plate on which two loads P and Q are applied as shown in Figure P8.58a/59a. The column cross section (Figure P8.58b/59b) has a depth of d = 8.0 in., an area of A = 5.40 in.2, and a moment of inertia of Iz = 57.5 in.4. Normal strains are measured with strain gages H and K, which are attached on the centerline of the outer faces of the flanges. The strains measured in the two gages are H = –530 × 10–6 in./in. and K = –310 × 10–6 in./in. Using a = 6.0 in., determine the magnitudes of loads P and Q.

FIGURE P8.58a/59a

FIGURE P8.58b/59b

Solution The axial normal stress caused at H by both P and Q will be compression. By inspection, the bending stress at H caused by load Q will be compression, and the bending stress at H caused by load P will be tension. The normal stress at gage H can be expressed by: P Q P (6 in.)(8 in. / 2) Q(6 in.)(8 in. / 2) H = − − + − 2 2 5.40 in. 5.40 in. 57.5 in.4 57.5 in.4 P P (24 in.2 ) Q Q(24 in.2 ) =− + − − 5.40 in.2 57.5 in.4 5.40 in.2 57.5 in.4 = (0.2322061 in.−2 ) P − (0.6025765 in.−2 )Q (a) The axial normal stress caused at K by both P and Q will be compression. By inspection, the bending stress at K caused by load Q will be tension, and the bending stress at K caused by load P will be compression. Q P Q (6 in.)(8 in. / 2) P (6 in.)(8 in. / 2) K = − − + − 2 2 5.40 in. 5.40 in. 57.5 in.4 57.5 in.4 Q Q (24 in.2 ) P P(24 in.2 ) =− + − − 5.40 in.2 57.5 in.4 5.40 in.2 57.5 in.4 = ( −0.6025765 in.−2 ) P + (0.2322061 in.−2 )Q (b) Normal stresses from measured strains The normal strains at H and K were measured as H = –530 × 10–6 in./in. and K = –310 × 10–6 in./in. From these values, the stresses at H and K can be calculated from Hooke’s Law:  H = E H = (29,000 ksi)( −530  10−6 in./in.) = −15.3700 ksi

 K = E K = (29,000 ksi)( −310  10−6 in./in.) = −8.9900 ksi Substitute these values in Eqs. (a) and (b), respectively, to obtain the following equations. −15.3700 ksi = (0.2322061 in.−2 ) P − (0.6025765 in.−2 )Q

(c)

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

−8.9900 ksi = ( −0.6025765 in.−2 ) P + (0.2322061 in.−2 )Q

(d)

Solve Eqs. (c) and (d) simultaneously to obtain the values for P and Q: Load P −15.3700 −0.6025765 −8.9900 0.2322061 ( −15.3700)(0.2322061) − ( −8.9900)( −0.60257565) P= = 0.2322061 −0.6025765 (0.2322061) 2 − ( −0.6025765)2 −0.6025765 =

0.2322061

−8.9861629 = 29.0645 kips = 29.1 kips −0.3091788

Ans.

Load Q 0.2322061 −15.3700 −0.6025765 −8.9900 (0.2322061)( −8.9900) − ( −0.60257565)(−15.3700) Q= = 0.2322061 −0.6025765 (0.2322061)2 − ( −0.6025765)2 −0.6025765 =

0.2322061

−11.3491206 = 36.7074 kips = 36.7 kips −0.3091788

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.60 A wood beam with a rectangular cross section is subjected to bending moment magnitudes of Mz = 1,250 lb·ft and My = 460 lb·ft, acting in the directions shown in Figure P8.60. The cross-sectional dimensions are b = 4 in. and d = 7 in. Determine: (a) the maximum bending stress magnitude in the beam. (b) the angle  for the orientation of the neutral axis relative to the +z axis. Note that positive  angles rotate clockwise from the +z axis. FIGURE P8.60

Solution Section properties 3 7 in.)( 4 in.) ( Iy = = 37.3333 in.4 12

( 4 in.)( 7 in.) = 114.3333 in.4 I = 3

(a) Maximum bending stresses: For a shape having at least one axis of symmetry, Eq. (8.24) can be used to determine bending stresses. By inspection, it seems as though point A might have the largest bending stress. We will first compute the normal stress at point A where y = 3.5 in. and z = 2.0 in.: M z M y x = y − z Iy Iz =

( −460 lb  ft )( 2.0 in.)(12 in./ft ) − (1, 250 lb  ft )( 3.5 in.)(12 in./ft )

37.3333 in.4 = −295.7145 psi − 459.1838 psi

114.3333 in.4

= 755 psi (C) The normal stress at C will have the same magnitude, but it will be a tensile stress.

Next, let’s compute the normal stress at point B where y = 3.5 in., z = −2.0 in.: M z M y x = y − z Iy Iz =

( −460 lb  ft )( −2.0 in.)(12 in./ft ) − (1, 250 lb  ft )(3.5 in.)(12 in./ft )

37.3333 in.4 = 295.7145 psi − 459.1838 psi

114.3333 in.4

= 163.5 psi (C) The normal stress at D will have the same magnitude, but it will be a tensile stress.

Therefore, the maximum bending stress magnitude occurs at points A and C where  max = 755 psi

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

(b) Orientation of neutral axis For a shape having at least one axis of symmetry, Eq. (8.25) can be used to determine the orientation of the neutral axis: 4 M y I z ( −460 lb  ft ) (114.3333 in. ) tan  = = = −1.1270 M z I y (1, 250 lb  ft ) ( 37.3333 in.4 )

  = −48.4

( i.e., 48.4 CCW from + z axis to N.A.)

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.61 A hollow core concrete plank is subjected to bending moment magnitudes of Mz = 50 kip·ft and My = 20 kip·ft, acting in the directions shown in Figure P8.61. The cross-sectional dimensions are b = 24 in., h = 12 in., d = 7 in., and a = 5.5 in. Determine: (a) the bending stress at B. (b) the bending stress at C. (c) the angle  for the orientation of the neutral axis relative to the +z axis. Note that positive  angles rotate clockwise from the +z axis. FIGURE P8.61

Solution Area of a single core: Acore =



d2 =



( 7 in.) = 38.485 in.2 2

4 4 Moment of inertia about the z axis:

Shape

d = y − yi

d²A (in.) (in.4) 0 0 0 0 0 0 Moment of inertia about the z axis (in.4) =

IC + d²A (in.4) 3,456.000 –117.859 –117.859 3,220.282

Moment of inertia about the y axis: d = z − zi Shape IC d²A 4 (in. ) (in.) (in.4) Outer rectangle 13,824.000 0 0 Circular cutout –117.859 5.5 –1,164.171 Circular cutout –117.859 –5.5 –1,164.171 Moment of inertia about the y axis (in.4) =

IC + d²A (in.4) 13,824.000 –1,282.030 –1,282.030 11,259.970

Outer rectangle Circular cutout Circular cutout

IC (in.4) 3,456.000 –117.859 –117.859

(a) Bending stress at B: For a shape having at least one axis of symmetry, Eq. (8.24) can be used to determine bending stresses. At point B, where y = 6 in. and z = –12.0 in.: M z M y x = y − z Iy Iz =

( 20, 000 lb  ft )( −12.0 in.)(12 in./ft ) − ( 50, 000 lb  ft )( 6 in.)(12 in./ft )

11, 259.970 in.4 = −255.773 psi − 1,117.915 psi = 1,374 psi (C)

3, 220.282 in.4

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

(b) Bending stress at C: At point C, where y = –6 in. and z = –12.0 in.: M z M y x = y − z Iy Iz =

( 20, 000 lb  ft )( −12.0 in.)(12 in./ft ) − ( 50, 000 lb  ft )( −6 in.)(12 in./ft )

11, 259.970 in.4 = −255.773 psi + 1,117.915 psi

3, 220.282 in.4

= 862 psi (T)

Ans.

(c) Orientation of neutral axis For a shape having at least one axis of symmetry, Eq. (8.25) can be used to determine the orientation of the neutral axis: 4 M y I z ( 20 kip  ft ) ( 3, 220.282 in. ) tan  = = = 0.11440 M z I y ( 50 kip  ft ) (11, 259.970 in.4 )

  = 6.53

( i.e., 6.53 CW from + z axis to N.A.)

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.62 The moment M acting on the cross section of the tee beam is oriented at an angle of  = 55° as shown in Figure P8.62. Dimensions of the cross section are bf = 180 mm, tf = 16 mm, d = 200 mm, and tw = 10 mm. The allowable bending stress is 165 MPa. What is the largest bending moment M that can be applied as shown to this cross section?

FIGURE P8.62

Solution Section properties Centroid location in y direction: Shape

Width b (mm) 180 10

top flange stem

yi Ai Ai

Height h (mm) 16 184

722, 240 mm 3 = 153.017 mm 4,720 mm 2

Area Ai (mm2) 2,880 1,840 4,720

yi (from bottom) (mm) 192 92

yi Ai (mm3) 552,960 169,280 722,240

(i.e., from bottom of shape to centroid)

Note: Centroid is 46.983 mm below the top of the shape.

Moment of inertia about the z axis: d = y − yi Shape IC d²A (mm4) (mm) (mm4) top flange 61,440 –38.983 4,376,662 stem 5,191,253 61.017 6,850,457 Moment of inertia about the z axis (mm4) =

IC + d²A (mm4) 4,438,102 12,041,710 16,479,812

Moment of inertia about the y axis: 3 3 16 mm )(180 mm ) (184 mm )(10 mm ) ( Iy = + = 7, 791,333 mm4 12 12 Moment components M y = −M cos55

M z = M sin55

Maximum bending moment magnitude The maximum tensile bending stress should occur at point C, which has the (y, z) coordinates y = −153.017 mm and z = –5 mm: M z M y − M cos 55 ( −5 mm ) M sin 55 ( −153.017 mm ) x = y − z = −  165 MPa Iy Iz 7, 791,333 mm4 16, 479,812 mm4

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

M 368.0862 10−9 mm −3 + 7.605923 10−6 mm −3   165 N/mm 2  M  20.692 106 N  mm

(a)

The maximum compressive bending stress should occur at point A, which has the (y, z) coordinates y = 46.983 mm and z = 90 mm: M z M y − M cos 55 ( 90 mm ) M sin 55 ( 46.983 mm ) x = y − z = −  −165 MPa Iy Iz 7, 791,333 mm4 16, 479,812 mm4 − M 6.62555 10−6 mm −3 + 2.33536 10−6 mm −3   −165 N/mm 2  M  18.413 106 N  mm

Compare results from Eqs. (a) and (b) to find the largest bending moment M: M max = 18.41 kN  m

(b)

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.63 The box beam shown in Figure P8.63b has dimensions of b = 80 mm, d = 120 mm, and t = 8 mm. A downward concentrated load of P = 42 kN acts on wall BC of the box at a distance of a = 35 mm below the z centroidal axis as shown in Figure P8.63a. Determine the normal stress: (a) at corner A. (b) at corner D.

FIGURE P8.63a

FIGURE P8.63b

Solution Section properties A = (120 mm )( 80 mm ) − (104 mm )( 64 mm ) = 2,944 mm 2

(120 mm )(80 mm ) − (104 mm )( 64 mm ) = 2,848, 085 mm 4 I = 3

(80 mm )(120 mm ) − ( 64 mm )(104 mm ) = 5,520, 725 mm 4 I = 3

Moment components  b−t   80 mm − 8 mm  My = P  = ( 42, 000 N )   = 1,512, 000 N  mm 2  2    M z = − Pa = − ( 42, 000 N )( 35 mm ) = −1, 470, 000 N  mm (a) Normal stress at corner A For a shape having at least one axis of symmetry, Eq. (8.24) can be used to determine bending stresses. To compute the bending stress at A, use the (y, z) coordinates y = 60 mm and z = 40 mm: M z M y  bending = y − z Iy Iz =

(1,512, 000 N  mm )( 40 mm ) − ( −1, 470, 000 N  mm )( 60 mm )

2,848, 085 mm 4 5,520, 725 mm 4 = 21.235 MPa + 15.976 MPa = 37.211 MPa

In addition to bending stress, there will also be an axial stress acting over the entire cross section. This stress is calculated as: F −42, 000 N  axial = = = −14.266 MPa A 2,944 mm 2

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Thus, the combined normal stress at corner A is the sum of the axial and bending stresses:  x =  axial +  bending = −14.266 MPa + 37.211 MPa = 22.9 MPa (T)

Ans.

(b) Normal stress at corner D To compute the bending stress at D, use the (y, z) coordinates y = –60 mm and z = 40 mm: M z M y  bending = y − z Iy Iz =

(1,512, 000 N  mm )( 40 mm ) − ( −1, 470, 000 N  mm )( −60 mm )

2,848, 085 mm 4 5,520, 725 mm 4 = 21.235 MPa − 15.976 MPa = 5.259 MPa

The combined normal stress at corner D is:  x =  axial +  bending = −14.266 MPa + 5.259 MPa = 9.01 MPa (C)

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.64 The cantilever beam shown in Figure P8.64a is subjected to load magnitudes of Pz = 2.8 kips and Py = 8.3 kips. The flanged cross section shown in Figure P8.64b has dimensions of bf = 8.0 in., tf = 0.62 in., d = 10.0 in., and tw = 0.35 in. Using L = 84 in., determine: (a) the bending stress at point A. (b) the bending stress at point B. (c) the angle  for the orientation of the neutral axis relative to the +z axis. Note that positive  angles rotate clockwise from the +z axis.

FIGURE P8.64a

FIGURE P8.64b

Solution Moment of inertia about the z axis: 3 3 8.0 in.)(10.0 in.) ( 7.65 in.)(8.76 in.) ( Iz = − = 238.1255 in.4 12 12 Moment of inertia about the y axis: 3 3 0.62 in.)(8.0 in.) (8.76 in.)( 0.35 in.) ( Iy = 2 + = 52.9380 in.4 12 12 Moment components M y = Pz L = ( 2.8 kips )( 84 in.) = 235.2 kip  in. M z = − Py L = − ( 8.3 kips )( 84 in.) = −697.2 kip  in.

(a) Bending stress at A For a shape having at least one axis of symmetry, Eq. (8.24) can be used to determine bending stresses. To compute the normal stress at A, use the (y, z) coordinates y = 5 in. and z = 4 in.:

Mechanics of Materials: An Integrated Learning System, 4th Ed.

x = =

Myz Iy

−

Timothy A. Philpot

Mz y Iz

( 235.2 kip  in.)( 4 in.) − ( −697.2 kip  in.)( 5 in.)

52.9380 in.4 238.1255 in.4 = 17.7717 ksi + 14.6393 ksi = 32.4 ksi (T)

Ans.

(b) Bending stress at B To compute the normal stress at B, use the (y, z) coordinates y = 5 in. and z = –4 in.: M z M y x = y − z Iy Iz

( 235.2 kip  in.)( −4 in.) − ( −697.2 kip  in.)( 5 in.)

52.9380 in.4 238.1255 in.4 = −17.7717 ksi + 14.6393 ksi = 3.13 ksi (C)

Ans.

(c) Orientation of neutral axis For a shape having at least one axis of symmetry, Eq. (8.25) can be used to determine the orientation of the neutral axis: 4 M y I z ( 235.2 kip  in.) ( 238.1255 in. ) tan  = = = −1.5175 M z I y ( −697.2 kip  in.) ( 52.9380 in.4 )

  = −56.62

( i.e., 56.62 CCW from + z axis )

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.65 The U shaped cross section is oriented at an angle of  = 35° as shown in Figure P8.65. Dimensions of the cross section are b = 80 mm, d = 65 mm, and t = 5 mm. The allowable tensile bending stress is 150 MPa, and the allowable compressive bending stress is 80 MPa. What is the largest bending moment Mz that can be applied as shown to this cross section?

FIGURE P8.65

Solution Centroid location in y′ direction: Shape Left-side upright Right-side upright Bottom element

yi Ai Ai

Width b (mm) 5 5 70

Height h (mm) 65 65 5

Area Ai (mm2) 325 325 350 1,000

yi (from bottom) (mm) 32.5 32.5 2.5

yi Ai (mm3) 10,562.5 10,562.5 875.0 22,000

22, 000 mm3 = 22 mm (i.e., from bottom of shape to centroid) 1,000 mm 2

Note: Centroid is 43 mm below the top of the shape.

Moment of inertia about the z′ axis: Shape IC (mm4) Left-side upright 114,427.08 Right-side upright 114,427.08 Bottom element 729.17

Moment of inertia about the y′ axis: Shape IC (mm4) Left-side upright 677.08 Right-side upright 677.08 Bottom element 142,916.67

d = y − yi

IC + d²A (mm4) 150,258.33 150,258.33 133,816.67 434,333.33

d = z − zi

IC + d²A (mm4) 457,708.33 457,708.33 142,916.67 1,058,333.33

d²A (mm) (mm4) –10.5 35,831.25 –10.5 35,831.25 19.5 133,087.50 Moment of inertia about the z′ axis (mm4) =

d²A (mm) (mm4) –37.5 457,031.25 37.5 457,031.25 0 0 Moment of inertia about the y′ axis (mm4) =

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Moment components: Resolve Mz into components with respect to the y′ and z′ axes: M y = −M z sin35 M z = −M z cos35 Maximum bending moment magnitude The maximum tensile bending stress should occur at point B, which has the (y′, z′) coordinates of y′ = 43 mm and z′ = –40 mm. Set the stress at B equal to the 150 MPa allowable tensile bending stress and solve for Mz: M  z M y − M z sin 35 ( −40 mm ) − M z cos 35 ( 43 mm )  x = y − z = −  150 MPa I y I z 1, 058,333.33 mm4 434,333.33 mm4 M z  21.6785 10−6 mm−3 + 81.0979  10−6 mm−3   150 N/mm2  M  1.45948 106 N  mm = 1.45948 kN  m

(a)

The maximum compressive bending stress should occur at point D, which has the (y′, z′) coordinates of y′ = –22 mm and z′ = 40 mm. Set the stress at D equal to the –80 MPa allowable compressive bending stress and solve for Mz: M  z M y − M z sin 35 ( 40 mm ) − M z cos 35 ( −22 mm )  x = y − z = −  −80 MPa I y I z 1, 058,333.33 mm4 434,333.33 mm4 − M z  21.6785 10−6 mm −3 + 41.4920 10−6 mm −3   −80 N/mm 2  M  1.26641106 N  mm = 1.26641 kN  m

(b)

Compare the results in Eqs. (a) and (b) to find that the largest bending moment Mz that can be applied to the cross section is: Ans. M max = 1.266 kN  m

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.66 The moment acting on the cross section of the zee shape has a magnitude of Mz = 5,600 lb·ft and is oriented at an angle of  = 24° as shown in Figure P8.66. Dimensions of the cross section are bf = 3.0 in., tf = 0.35 in., d = 7.0 in., and tw = 0.20 in. Determine: (a) the moments of inertia Iy′ and Iz′. (b) the product of inertia Iy′z′. (c) the principal moments of inertia and the angle from the +z′ axis to the axis of the maximum moment of inertia. (d) the bending stress at points A and C. (e) the angle  for the orientation of the neutral axis relative to the +z axis. Note that positive  angles rotate clockwise from the +z axis. FIGURE P8.66

Solution Let y = 0 and z = 0. Moment of inertia about the z′ axis: d = y − yi Shape IC

Area Ai d²A (in.) (in.2) (in.4) –3.325 1.050 11.6084 0.000 1.260 0.0000 3.325 1.050 11.6084 Moment of inertia about the z′ axis (in.4) =

IC + d²A (in.4) 11.6191 4.1675 11.6191 27.4057

Moment of inertia about the y′ axis: d = z − zi Shape IC Area Ai d²A 4 2 (in. ) (in.) (in. ) (in.4) top flange 0.7875 –1.40 1.050 2.0580 web 0.0042 0.00 1.260 0.0000 bottom flange 0.7875 1.40 1.050 2.0580 Moment of inertia about the y′ axis (in.4) =

IC + d²A (in.4) 2.8455 0.0042 2.8455 5.6952

top flange web bottom flange

(in.4) 0.0107 4.1675 0.0107

Product of inertia about the centroidal axes: y − yi z − zi Shape (in.) (in.) top flange –3.325 –1.40 web 0.000 0.00 bottom flange 3.325 1.40

Area Ai (in.2) 1.050 1.260 1.050

( y − yi )( z − zi ) Ai 4

(in. ) 4.8878 0 4.8878 Product of inertia (in.4) =

Iyz (in.4) 4.8878 0 4.8878 9.7755

(a) Moments of inertia I y = 5.70 in.4

Ans.

I z = 27.4 in.4 (b) Product of inertia

Ans.

I yz = 9.78 in.4

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

I y + I z 2

 I  − I z  2   y  + I yz 2   2

 5.6952 in.4 − 27.4057 in.4  5.6952 in.4 + 27.4057 in.4 4 2 =    + ( 9.7755 in. ) 2 2   = 16.5504 in.4  14.6081 in.4

I p1 = 31.1586 in.4 = 31.2 in.4

Ans.

I p 2 = 1.9423 in.4 = 1.942 in.4

Ans.

 p = 21.0 CCW from +z axis to I p1 axis

Ans.

Moment components M y = − M z sin  = − ( 5, 600 lb  ft ) sin 24 = −2, 277.725 lb  ft M z = M z cos  = ( 5, 600 lb  ft ) cos 24 = 5,115.855 lb  ft

(d) Bending stress at points A and C Since the zee shape has no axis of symmetry, Eq. (8.21) or Eq. (8.22) must be used to determine the bending stresses. Equation (8.21) will be used here. Note: Take care to make the units consistent between the moment terms and the section properties. ( M z I y + M y I yz ) y + ( M y I z + M z I yz ) z x = − I y I z − I y2z I y I z − I y2z  ( 5,115.855 lb  ft ) ( 5.6952 in.4 ) + ( −2, 277.725 lb  ft ) ( 9.7755 in.4 )   (12 in./ft ) y = − 4 4 4 2   ( 5.6952 in. )( 27.4057 in. ) − (9.7755 in. )    ( −2, 277.725 lb  ft ) ( 27.4057 in.4 ) + ( 5,115.855 lb  ft ) ( 9.7755 in.4 )   (12 in./ft ) z  + 4 4 4 2   ( 5.6952 in. )( 27.4057 in. ) − (9.7755 in. )  

= ( −1,362.1656 lb/in.3 ) y + ( −2, 461.1691 lb/in.3 ) z 

To compute the bending stress at point A, use (y′, z′) coordinates of y′ = 3.5 in. and z′ = 2.9 in.:  x = ( −1,362.1656 lb/in.3 ) ( 3.5 in.) + ( −2, 461.1691 lb/in.3 ) ( 2.9 in.) = −11,904.97 psi = 11,900 psi (C)

Ans.

To compute the bending stress at point C, use (y′, z′) coordinates of y′ = –3.5 in. and z′ = –2.9 in.:  x = ( −1,362.1656 lb/in.3 ) ( −3.5 in.) + ( −2, 461.1691 lb/in.3 ) ( −2.9 in.) = 11,904.97 psi = 11,900 psi (T)

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

(e) Orientation of neutral axis Since the zee shape has no axis of symmetry, it is helpful to determine the orientation of the neutral axis from Eq. (8.23) to help identify points of maximum stress. M y I z + M z I yz tan  = M z I y + M y I yz

( −2, 277.725 lb  ft ) ( 27.4057 in.4 ) + ( 5,115.855 lb  ft ) (9.7755 in.4 ) = ( 5,115.855 lb  ft ) ( 5.6952 in.4 ) + ( −2, 277.725 lb  ft ) (9.7755 in.4 ) = −1.8068   = −61.04

(i.e., 61.04 CCW from + z axis)

The zee shape is rotated  = 24°. Accounting for the  angle means that the angle  relative to the +z axis is: Ans.  = 24 − 61.04 = −37.04 (i.e., 37.04 CCW from + z axis)

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.67 A beam cross section is fabricated by rigidly attaching two dimensions lumber boards together as shown in Figure P8.67. The dimensions of the cross section are b1 = 7.25 in., b2 = 9.25 in., and t = 1.50 in. The moment acting on the cross section is Mz = 930 lb·ft. Determine: (a) the moments of inertia Iy and Iz. (b) the product of inertia Iyz. (c) the principal moments of inertia and the angle from the +z axis to the axis of the maximum moment of inertia. (d) the maximum compressive bending stress in the cross section. FIGURE P8.67

Solution Section properties Centroid location in y direction: Shape

Width b (in.) 7.25 1.50

Horizontal leg Vertical leg y=

yi Ai Ai

Height h (in.) 1.50 9.25

172.922 in.3 = 6.987 in. 24.750 in.2

Area Ai (in.2) 10.875 13.875 24.750

yi (from bottom) (in.) 10 4.625

yi Ai (in.3) 108.750 64.172 172.922

(i.e., from bottom of shape to centroid)

Note: Centroid is 3.763 in. below the top of the shape.

Centroid location in z direction: zi Area Ai (from right edge) zi Ai 2 (in. ) (in.) (in.3) upright leg 10.875 3.625 39.422 bottom leg 13.875 0.750 10.406 24.750 49.828 3 zi Ai 49.828 in. z= = = 2.013 in. (i.e., from right edge of shape to centroid) Ai 24.750 in.2 Shape

Note: Centroid is 5.237 in. from the left edge of the shape.

Moment of inertia about the z axis: d = y − yi Shape IC d²A (in.4) (in.) (in.4) upright leg 2.039 –3.013 98.742 bottom leg 98.932 2.362 77.392 Moment of inertia about the z axis (in.4) =

IC + d²A (in.4) 100.781 176.324 277.105

Mechanics of Materials: An Integrated Learning System, 4th Ed. Moment of inertia about the y axis: d = z − zi Shape IC d²A 4 (in. ) (in.) (in.4) upright leg 47.635 –1.612 28.250 bottom leg 2.602 1.263 22.142 Moment of inertia about the y axis (in.4) = Product of inertia about the centroidal axes: y − yi z − zi Shape (in.) (in.) upright leg –3.013 –1.612 bottom leg 2.362 1.263

Area Ai (in.2) 10.875 13.875

Timothy A. Philpot

IC + d²A (in.4) 75.885 24.744 100.628

( y − yi )( z − zi ) Ai 4

(in. ) 52.819 41.392

Iyz (in.4) 52.819 41.392 94.211

(a) Moments of inertia I y = 100.63 in.4

Ans.

I z = 277.1 in.4 (b) Product of inertia

Ans.

I yz = 94.21 in.4

Ans.

I y + Iz 2

I −I    y z  + I yz2  2  2

 100.628 in.4 − 277.105 in.4  100.628 in.4 + 277.105 in.4 4 2 =    + ( 94.211 in. ) 2 2   = 188.867 in.4  129.081 in.4

I p1 = 317.947 in.4 = 318 in.4

Ans.

I p 2 = 59.786 in.4 = 59.8 in.4

Ans.

 p = 23.4 CCW from +z axis to I p1 axis

Ans.

(d) Maximum compressive bending stress Since the angle shape has no axis of symmetry, Eq. (8.21) or Eq. (8.22) must be used to determine the bending stresses. Equation (8.22) will be used here. Note that the bending moment component about the y axis is zero (i.e., My = 0); therefore, the first term in Eq. (8.22) is eliminated. By inspection, it seems possible that point B will have the largest compressive bending stress. Point B has (y, z) coordinates of y = 3.763 in. and z = –2.013 in.:

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

 − I y y + I yz z  Mz 2   I y I z − I yz 

x = 

 − (100.628 in.4 ) ( 3.763 in.) + ( 94.211 in.4 ) ( −2.013 in.)   ( 930 lb  ft )(12 in./ft ) = 4 4 4 2   100.628 in. 277.105 in. − 94.211 in. ( )( ) ( )    −568.310 in.5  = 11,160 lb  in.) 8 ( 19,008.809 in.   = −333.683 psi = 334 psi (C)

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.68 The extruded beam cross section shown in Figure P8.68 is subjected to a bending moment magnitude of Mz = 260 N·m acting in the indicated direction. The dimensions of the cross section are b1 = 40 mm, b2 = 70 mm, d = 90 mm, and t = 10 mm. Determine: (a) the moments of inertia Iy and Iz. (b) the product of inertia Iyz. (c) the principal moments of inertia and the angle from the +z axis to the axis of the maximum moment of inertia. (d) the maximum tensile bending stress in the cross section. FIGURE P8.68

Solution Section properties

Centroid location in y direction: Shape

Width b (mm) 40 10 70

(1) (2) (3) y=

yi Ai Ai

Height h (mm) 10 70 10

69, 000 mm 3 = 38.333 mm 1,800 mm 2

Area Ai (mm2) 400 700 700 1,800

yi (from bottom) (mm) 85 45 5

yi Ai (mm3) 34,000 31,500 3,500 69,000

(i.e., from bottom of shape to centroid)

Note: Centroid is 51.667 mm below the top of the shape.

Centroid location in z direction: zi Area Ai (from right edge) zi Ai 2 (mm ) (mm) (mm3) (1) 400 50 20,000 (2) 700 65 45,500 (3) 700 35 24,500 1,800 90,000 zi Ai 90, 000 mm3 z= = = 50.0 mm (i.e., from right edge of shape to centroid) Ai 1,800 mm 2 Shape

Note: Centroid is 20.0 mm from the left edge of the shape.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Moment of inertia about the z axis: Shape IC (mm4) (1) 3,333.333 (2) 285,833.333 (3) 5,833.333

Moment of inertia about the y axis: Shape IC (mm4) (1) 53,333.333 (2) 5,833.333 (3) 285,833.333

Timothy A. Philpot

d = y − yi

IC + d²A (mm4) 874,444.444 316,944.444 783,611.111 1,975,000

d = z − zi

IC + d²A (mm4) 53,333.333 163,333.333 443,333.333 660,000

d²A (mm) (mm4) –46.667 871,111.111 –6.667 31,111.111 33.333 777,777.778 Moment of inertia about the z axis (mm4) =

d²A (mm) (mm4) 0 0 –15 157,500 15 157,500 Moment of inertia about the y axis (mm4) =

Product of inertia about the centroidal axes: y − yi z − zi Shape (mm) (mm) (1) –46.667 0 (2) –6.667 –15 (3) 33.333 15

Area Ai (mm2) 400 700 700

( y − yi )( z − zi ) Ai

(mm4) 0 70,000 350,000

Iyz (mm4) 0 70,000 350,000 420,000

(a) Moments of inertia I y = 660, 000 mm4

Ans.

I z = 1,975, 000 mm4

Ans.

(b) Product of inertia I yz = 420, 000 mm 4

Ans.

I y + Iz 2

 I y − Iz  2    + I yz  2  2

 660, 000 mm 4 − 1,975, 000 mm 4  660, 000 mm 4 + 1,975, 000 mm 4 4 2 =    + ( 420, 000 mm ) 2 2   = 1,317,500 mm 4  780,196.289 mm 4

I p1 = 2, 097, 696.289 mm4 = 2, 098, 000 mm4

Ans.

I p 2 = 537,303.711 mm4 = 537, 000 mm4

Ans.

 p = 16.29 CCW from +z axis to I p1 axis

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

(d) Maximum tensile bending stress Orientation of neutral axis Since this shape has no axis of symmetry, it is helpful to determine the orientation of the neutral axis from Eq. (8.23) before beginning the stress calculations: M I + M z I yz I yz 420, 000 mm 4 tan  = y z = = = 0.63636 M z I y + M y I yz I y 660, 000 mm 4

  = 32.47

( i.e., 32.47 CW from + z axis )

Equation (8.22) will be used here to compute the bending stresses. Note that the bending moment component about the y axis is zero (i.e., My = 0); therefore, the first term in Eq. (8.22) is eliminated. By inspection, it seems likely that point B will have the largest tensile bending stress. Point B has (y, z) coordinates of y = 51.667 mm and z = –20 mm:  − I y + I yz z  x =  y Mz 2   I y I z − I yz 

 − ( 660, 000 mm 4 ) ( 51.667 mm ) + ( 420,000 mm 4 ) ( −20 mm )   ( −260 N  m )(1,000 mm/m ) = 4 4 4 2  ( 660, 000 mm )(1,975, 000 mm ) − ( 420,000 mm )    −42,500, 220 mm5  =  ( −260, 000 N  mm ) 12  1.127110  = 9.8040 MPa = 9.80 MPa (T)

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.69 A stainless-steel spring (shown in Figure P8.69/70) has a thickness of ¾ in. and a change in depth at section B from D = 1.50 in. to d = 1.25 in. The radius of the fillet between the two sections is r = 0.125 in. If the bending moment applied to the spring is M = 2,000 lb-in., determine the maximum normal stress in the spring. FIGURE P8.69/70

Solution From Figure 8.18 r 0.125 in. = = 0.10 d 1.25 in.

D 1.50 in. = = 1.20 d 1.25 in.

 K  1.69

Moment of inertia at minimum depth section: (0.75 in.)(1.25 in.)3 I= = 0.122070 in.4 12 Nominal bending stress at minimum depth section: My (2,000 lb-in.)(1.25 in./2)  nom = = = 10.2400 ksi I 0.122070 in.4 Maximum bending stress:  max = K nom = 1.69(10.2400 ksi) = 17.3056 ksi = 17.31 ksi

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.70 An alloy-steel spring (shown in Figure P8.69/70) has a thickness of 25 mm and a change in depth at section B from D = 75 mm to d = 50 mm. If the radius of the fillet between the two sections is r = 8 mm, determine the maximum moment that the spring can resist if the maximum bending stress in the spring must not exceed 120 MPa. FIGURE P8.69/70

Solution From Figure 8.18 r 8 mm = = 0.16 d 50 mm

D 75 mm = = 1.50 d 50 mm

 K  1.57

Determine maximum nominal bending stress:  120 MPa  nom = max = = 76.4331 MPa K 1.57 Moment of inertia at minimum depth section: (25 mm)(50 mm)3 I= = 260, 416.67 mm 4 12 Maximum bending moment:

M max 

 nom I y

(76.4331 N/mm2 )(260,416.67 mm4 ) = 796,178.3 N-mm = 796 N-m 50 mm/2

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.71 The notched bar shown in Figure P8.71/72 is subjected to a bending moment of M = 300 Nm. The major bar width is D = 75 mm, the minor bar width at the notches is d = 50 mm, and the radius of each notch is r = 10 mm. If the maximum bending stress in the bar must not exceed 90 MPa, determine the minimum required bar thickness b. FIGURE P8.71/72

Solution From Figure 8.17 r 10 mm = = 0.20 d 50 mm

D 75 mm = = 1.50 d 50 mm

 K  1.76

Determine maximum nominal bending stress:  90 MPa  nom = max = = 51.1364 MPa K 1.76 Minimum bar thickness b: M y M (d /2) 6M  nom  = = I bd 3 /12 bd 2 6M 6(300 N-m)(1,000 mm/m) b  = = 14.08 mm 2  nom d (51.1364 N/mm 2 )(50 mm)2

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.72 The machine part shown in Figure P8.71/72 is made of cold-rolled 18-8 stainless steel (see Appendix D for properties). The major bar width is D = 1.50 in., the minor bar width at the notches is d = 1.00 in., the radius of each notch is r = 0.125 in., and the bar thickness is b = 0.25 in. Determine the maximum safe moment M that may be applied to the bar if a factor of safety of 2.5 with respect to failure by yield is specified. FIGURE P8.71/72

Solution From Figure 8.17 r 0.125 in. = = 0.125 d 1.00 in.

D 1.50 in. = = 1.50 d 1.00 in.

 K  2.05

Moment of inertia at minimum depth section: (0.25 in.)(1.00 in.)3 I= = 0.020833 in.4 12 Maximum allowable bending moment: From the specified factor of safety and the yield stress of the material, the allowable bending stress is:  165 ksi  allow = Y = = 66 ksi FS 2.5 Thus, the maximum allowable bending moment can be determined from: My  allow  K I  I (66 ksi)(0.020833 in.4 )  M max  allow = = 1.3415 kip-in. = 111.8 lb-ft Ky (2.05)(1.00 in./2)

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.73 The C86100 bronze (see Appendix D for properties) shaft shown in Figure P8.73 is supported at each end by self-aligning bearings. The major shaft diameter is D = 40 mm, the minor shaft diameter is d = 25 mm, and the radius of the fillet between the major and minor diameter sections is r = 5 mm. The shaft length is L = 500 mm and the fillets are located at x = 150 mm and x = 350 mm. Determine the maximum load P that may be applied to the shaft if a factor of safety of 3.0 with respect to failure by yield is specified. FIGURE P8.73

Solution From Figure 8.20 r 5 mm = = 0.20 d 25 mm

D 40 mm = = 1.60 d 25 mm

 K  1.48

Moment of inertia at minimum diameter section:  I= (25 mm)4 = 19,174.76 mm 4 64 Maximum allowable bending moment:  331 MPa  allow = yield = = 110.33 MPa FS 3.0 My  allow  K I  I (110.33 N/mm 2 )(19,174.76 mm 4 )  M max  allow = = 114,357.58 N-mm Ky (1.48)(25 mm/2) Bending moment at x = 150 mm: P P M = x = (150 mm) = P(75 mm) 2 2 Maximum load P: P(75 mm)  114,357.58 N-mm  P  1,524.77 N = 1,525 N

Ans.

Check stress at midspan: PL (1,524.77 N)(500 mm) M midspan = = = 190,596.25 N-mm 4 4



(40 mm) 4 = 125,663.71 mm 4

64 M y (190,596.25 N-mm)(40 mm/2)  midspan = = = 30.33 MPa  110.33 MPa I 125,663.71 mm 4

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.74 The machine shaft shown in Figure P8.74/75 is made of 1020 cold-rolled steel (see Appendix D for properties). The major shaft diameter is D = 1.000 in., the minor shaft diameter is d = 0.625 in., and the radius of the fillet between the major and minor diameter sections is r = 0.0625 in. The fillet is located at x = 4 in. from C. If a load of P = 125 lb is applied at C, determine the factor of safety with respect to failure by yield in the fillet at B. FIGURE P8.74/75

Solution For 1020 cold-rolled steel:  Y = 62,000 psi From Figure 8.20 r 0.0625 in. = = 0.10 d 0.625 in.

D 1.000 in. = = 1.6 d 0.625 in.

 K  1.74

Moment of inertia at minimum diameter section:  I= (0.625 in.)4 = 0.0074901 in.4 64 Bending moment at x = 4 in.: M = Px = (125 lb)(4 in.) = 500 lb-in. Maximum bending stress: My (500 lb-in.)(0.625 in./2)  max = K = (1.74) = 36, 297.7 psi I 0.0074901 in.4 Factor of safety:

FS =

Y 62,000 psi = = 1.708  max 36,297.7 psi

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.75 The machine shaft shown in Figure P8.74/75 is made of 1020 cold-rolled steel (see Appendix D for properties). The major shaft diameter is D = 30 mm, the minor shaft diameter is d = 20 mm, and the radius of the fillet between the major and minor diameter sections is r = 3 mm. The fillet is located at x = 90 mm from C. Determine the maximum load P that can be applied to the shaft at C if a factor of safety of 1.5 with respect to failure by yield is specified for the fillet at B. FIGURE P8.74/75

Solution From Figure 8.20 r 3 mm = = 0.15 d 20 mm

D 30 mm = = 1.5 d 20 mm

 K  1.58

Moment of inertia at minimum diameter section:  I= (20 mm) 4 = 7,853.98 mm 4 64 Maximum allowable bending moment:  427 MPa  allow = Y = = 284.6667 MPa FS 1.5 My  allow  K I  allow I (284.6667 N/mm 2 )(7,853.98 mm 4 )  M max  = = 141,504.2261 N-mm Ky (1.58)(20 mm/2) Bending moment at x = 90 mm: M = Px = P(90 mm) Maximum load P: P(90 mm)  141,504.2261 N-mm  P  1,572.3 N = 1,572 N

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.76 The grooved shaft shown in Figure P8.76 is made of C86100 bronze (see Appendix D for properties). The major shaft diameter is D = 50 mm, the minor shaft diameter at the groove is d = 34 mm, and the radius of the groove is r = 4 mm. Determine the maximum allowable moment M that may be applied to the shaft if a factor of safety of 1.5 with respect to failure by yield is specified. FIGURE P8.76

Solution From Figure 8.19 r 4 mm = = 0.118 d 34 mm

D 50 mm = = 1.471 d 34 mm

 K  1.96

Moment of inertia at minimum diameter section:  I= (34 mm)4 = 65,597.24 mm 4 64 Maximum allowable bending moment:  331 MPa  allow = Y = = 220.6667 MPa FS 1.5 My  allow  K I  I (220.6667 N/mm 2 )(65,597.24 mm 4 )  M max  allow = Ky (1.96)(34 mm/2)

= 434,427.5 N-mm = 434 N-m

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.77 A curved bar with a rectangular cross section is subjected to a bending moment of M = 850 N·m as shown in Figure P8.77. Use b = 35 mm, d = 60 mm, and ri = 100 mm. Determine the normal stress at A and B.

FIGURE P8.77

Solution Radii: The inner radius of the curved bar is ri = 100 mm, and the outer radius is ro = ri + d = 100 mm + 60 mm = 160 mm. The radius from the center of curvature O to the centroid of the rectangular cross section is d 60 mm rc = ri + = 100 mm + = 130 mm (from center of curvature to centroid) 2 2 Next, we must determine the location of the neutral axis rn from the following formula: A rn = dA A r The solution of this integral can be found in Table 8.1. For a rectangular cross section, A rn = where A = b d ro b ln ri The distance from the center of curvature to the neutral axis is bd d 60 mm rn = = = = 127.6586 mm ro ro 160 ln b ln ln 100 ri ri Bending moment: The bending moment at A and B is M = 850 N·m. This moment is a positive moment since it creates compressive normal stress on the inner surface (i.e., at radius ri) of the curved bar. Bending stress: The bending stress is determined from the equation: M ( rn − r ) x = − r A ( rc − rn ) where r is the distance from the center of curvature to the location where the stress is to be determined.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

(a) Point A: At point A, r = ri = 100 mm. Thus, the bending stress at A is M ( rn − r ) ( 850 N  m )(127.6586 mm − 100 mm )(1,000 mm/m ) x = − =− r A ( rc − rn ) (100 mm ) ( 2,100 mm 2 ) (130 mm − 127.6586 mm ) = 47.8 MPa (C)

Ans.

(b) Point B: At point B, r = ro = 160 mm. Thus, the bending stress at B is M ( rn − r ) ( 850 N  m )(127.6586 mm − 160 mm )(1,000 mm/m ) x = − =− r A ( rc − rn ) (160 mm ) ( 2,100 mm 2 ) (130 mm − 127.6586 mm ) = 34.9 MPa (T)

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.78 The steel link shown in Figure P8.78 has a central angle of  = 120°. Dimensions of the link are b = 6 mm, d = 18 mm, and ri = 30 mm. If the allowable normal stress is 140 MPa, what is the largest load P that may be applied to the link?

FIGURE P8.78

Solution Radii: The inner radius of the curved bar is ri = 30 mm, and the outer radius is ro = ri + d = 30 mm + 18 mm = 48 mm. The radius from the center of curvature O to the centroid of the rectangular cross section is d 18 mm rc = ri + = 30 mm + = 39 mm (from center of curvature to centroid) 2 2 Next, we must determine the location of the neutral axis rn from the following formula given in Table 8.1. For a rectangular cross section, A rn = where A = b d ro b ln ri The distance from the center of curvature to the neutral axis is A = bd = ( 6 mm )(18 mm ) = 108 mm 2

rn =

A r b ln o ri

108 mm 2 = 38.2976 mm 48 ( 6 mm ) ln 30

Bending moment: The bending moment magnitude at section AB is computed as the product of P and the perpendicular distance e between the centroid of the cross section (i.e., the rectangle at section AB) and the line of action of applied load P.  120    M = Pe = P  rc 1 − cos  = P  ( 39 mm ) 1 − cos  = (19.5 mm ) P 2 2    This moment is a negative bending moment since it creates tensile normal stress on the inner surface (i.e., at radius ri) of the curved bar. Bending stress: The bending stress is determined from the equation: M ( rn − r ) x = − r A ( rc − rn ) where r is the distance from the center of curvature to the location where the stress is to be determined. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Combined stresses: Both axial and bending stress will be created at section AB. Load P is applied in such a way that it creates tensile normal stresses in the curved bar. The bending moment creates tensile bending stresses at A, and therefore, the combination of tensile axial and bending stresses at point A will control the magnitude of load P. The stresses at A can be expressed as:  x =  axial +  bending 140 MPa  

P M ( rn − r ) P − (19.5 mm ) P ( rn − r ) P  (19.5 mm )( rn − r )  − = − = 1 +  A r A ( rc − rn ) A r A ( rc − rn ) A r ( rc − rn )   (19.5 mm )( 38.2976 mm − 30 mm )  8.67859 P 1+ P = 2  108 mm  ( 30 mm )( 39 mm − 38.2976 mm )  108 mm 2

P  1, 742 N

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.79 The curved member shown in Figure P8.79 has a rectangular cross section with dimensions of b = 2 in. and d = 5 in. The inside radius of the curved bar is ri = 4 in. A load P is applied at a distance of a = 9 in. from the center of curvature O. If the tensile and compressive stresses in the member are not to exceed 24 ksi and 18 ksi, respectively, determine the value of load P that may be safely supported by the member.

FIGURE P8.79

Solution Radii: The inner radius of the curved bar is ri = 4 in., and the outer radius is ro = ri + d = 4 in. + 5 in. = 9 in. The radius from the center of curvature O to the centroid of the rectangular cross section is d 5 in. rc = ri + = 4 in. + = 6.5 in. (from center of curvature to centroid) 2 2 Next, we must determine the location of the neutral axis rn from the following formula: A rn = dA A r The solution of this integral can be found in Table 8.1. For a rectangular cross section, A rn = where A = b d ro b ln ri The distance from the center of curvature to the neutral axis is A = bd = ( 2 in.)( 5 in.) = 10 in.2

rn =

A r b ln o ri

10 in.2 9 ( 2 in.) ln 4

= 6.1658 in.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Bending moment: The bending moment magnitude at section AB is computed as the product of P and the perpendicular distance e between the centroid of the cross section (i.e., at section AB) and the line of action of applied load P. M = Pe = P ( a + rc ) = P ( 9 in. + 6.5 in.) = (15.5 in.) P This is a positive bending moment since it creates compressive normal stress on the inner surface (i.e., at radius ri) of the curved bar. Bending stress: The bending stress is determined from the equation: M ( rn − r ) x = − r A ( rc − rn ) where r is the distance from the center of curvature to the location where the stress is to be determined. Combined stresses: Both axial and bending stress will be created at section AB. Load P is applied in such a way that it creates compressive normal stresses in the curved bar at section AB. The bending moment creates compressive bending stresses at A and tensile bending stresses at B. The compressive stress is not to exceed 18 ksi. Consider the stresses at A, and determine the allowable value of load P:  x =  axial +  bending −18 ksi  − −

P M ( rn − r ) P (15.5 in.) P ( rn − r ) P  (15.5 in.)( rn − r )  − =− − = − 1 +  A r A ( rc − rn ) A r A ( rc − rn ) A r ( rc − rn )  P  (15.5 in.)( 6.1658 in. − 4 in.)  26.1121 1+ P =− 2  10 in.  10 in.2 ( 4 in.)( 6.5 in. − 6.1658 in.) 

P  6.89 kips

(a)

The tensile stress is not to exceed 24 ksi. Consider the stresses at B, and determine the allowable value of load P:  x =  axial +  bending 24 ksi  − −

P M ( rn − r ) P (15.5 in.) P ( rn − r ) P  (15.5 in.)( rn − r )  − =− − = − 1 +  A r A ( rc − rn ) A r A ( rc − rn ) A r ( rc − rn )  P  (15.5 in.)( 6.1658 in. − 9 in.)  −13.6054 1+ P =− 2  10 in.  10 in.2 ( 9 in.)( 6.5 in. − 6.1658 in.) 

P  17.64 kips

Compare the results in Eqs. (a) and (b) to find that the largest value of P is: Pmax = 6.89 kips

(b)

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.80 Determine the stresses acting at points A and B for the tube elbow shown in Figure P8.80 loaded by a moment of M = 2,500 lb·in. The outside and inside diameters of the tube are D = 1.75 in. and d = 1.50 in., respectively, and the inner radius of the elbow is ri = 1.50 in.

FIGURE P8.80

Solution Radii: The inner radius of the curved bar is ri = 1.50 in., and the outer radius is ro = ri + D = 1.50 in. + 1.75 in. = 3.25 in. The radius from the center of curvature O to the centroid of the circular cross section is D 1.75 in. rc = ri + = 1.50 in. + = 2.375 in. (from center of curvature to centroid) 2 2 Next, we must determine the location of the neutral axis rn from the following formula: A rn = dA A r The solution of this integral for a solid circular cross section can be found in Table 8.1: A rn = where A =  c 2 2 2 2 rc − rc − c

)

(

However, the tube elbow is a hollow circular cross section. For a hollow circular section, first we find the area of the tube as: A=



 D − d ) = (1.75 in.) − (1.50 in.)  = 0.6381 in. (  4 4 2

dA for a hollow circular section with an outer diameter of D and an inner r A diameter of d can be derived as: 2 2 2 2    dA D  d  D d      2 2 2 2 A r = 2 rc − rc −  2   − 2 rc − rc −  2   = 2 − rc −  2  + rc −  2        

The solution of the integral 

2 2  d D    2 2 = 2  rc −   − rc −     2  2   

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Consequently, the distance from the center of curvature to the neutral axis of the tube shape here is: A rn = 2 2  d D    2 2 = 2  rc −   − rc −    2  2    0.6381 in.2 = 2 2  1.50 in.  1.75 in.   2 2   2  ( 2.375 in.) −   − ( 2.375 in.) −     2   2    0.6381 in.2 = = 2.2307 in. 2 ( 2.2535 in. − 2.2079 in.) Bending moment: The bending moment is M = 2,500 lb·in. This moment is a negative bending moment since it creates tensile normal stress on the inner surface (i.e., at radius ri) of the curved tube. Bending stress: The bending stress is determined from the equation: M ( rn − r ) x = − r A ( rc − rn ) where r is the distance from the center of curvature to the location where the stress is to be determined. Stress at point A: At point A, r = ri = 1.50 in. Thus, the bending stress at A is M ( rn − r ) ( −2,500 lb  in.)( 2.2307 in. − 1.50 in.) x = − =− r A ( rc − rn ) (1.50 in.) ( 0.6381 in.2 ) ( 2.375 in. − 2.2307 in.) = 13, 225.9 psi = 13, 230 psi (T)

Ans.

Stress at point B: At point B, r = ro = 3.25 in. Thus, the bending stress at B is M ( rn − r ) ( −2,500 lb  in.)( 2.2307 in. − 3.25 in.) x = − =− r A ( rc − rn ) ( 3.25 in.) ( 0.6381 in.2 ) ( 2.375 in. − 2.2307 in.) = −8,515.1 psi = 8,520 psi (C)

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.81 Determine the stresses acting at points A and B for the curved solid rod shown in Figure P8.81. The rod has a diameter of d = 0.875 in. The inner radius of the bend is ri = 1.5 in. A load of P = 90 lb is applied at a distance of a = 5 in. from the center of curvature O.

FIGURE P8.81

Solution Radii: The inner radius of the curved bar is ri = 1.50 in., and the outer radius is ro = ri + d = 1.50 in. + 0.875 in. = 2.375 in. The radius from the center of curvature O to the centroid of the circular cross section is: d 0.875 in. rc = ri + = 1.50 in. + = 1.9375 in. (from center of curvature to centroid) 2 2 Next, we will determine the location of the neutral axis rn from the following formula found in Table 8.1: A rn = where A =  c 2 2 rc − rc2 − c 2

)

(

The area of the cross section is



( 0.875 in.) = 0.60132 in.2 2

4 The distance from the center of curvature to the neutral axis is A 0.60132 in.2 rn = = = 1.91248 in. 2   2 rc − rc2 − c 2 0.875 in. 2    2 1.9375 in. − (1.9375 in.) −     2    

(

)

Bending moment: The bending moment magnitude at section AB is computed as the product of P and the perpendicular distance e between the centroid of the cross section (i.e., at section AB) and the line of action of applied load P. M = Pe = P ( a + rc ) = P ( 5 in. + 1.9375 in.) = ( 6.9375 in.) P The sign of the bending moment will be positive since it creates compressive normal stress on the inner surface (i.e., at radius ri) of the curved bar.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Bending stress: The bending stress is determined from the equation: M ( rn − r ) x = − r A ( rc − rn ) where r is the distance from the center of curvature to the location where the stress is to be determined. Combined stresses: Both axial and bending stress will be created at section AB. Load P is applied in such a way that it creates compressive normal stresses in the curved bar at section AB. The bending moment creates compressive bending stresses at A and tensile bending stresses at B. Stress at point A: Calculate the stress at point A as follows:  x =  axial +  bending

=−

P M ( rn − r ) P ( 6.9375 in.) P ( rn − r ) P  ( 6.9375 in.)( rn − r )  − =− − = − 1 +  A r A ( rc − rn ) A r A ( rc − rn ) A r ( rc − rn ) 

=−

 ( 6.9375 in.)(1.91248 in. − 1.50 in.)  90 lb 1+  2  0.60132 in.  (1.50 in.)(1.9375 in. − 1.91248 in.) 

= −11,561.4 psi = 11,560 psi (C)

Ans.

Stress at point B: Calculate the stress at point B as follows:  x =  axial +  bending

=−

P M ( rn − r ) P ( 6.9375 in.) P ( rn − r ) P  ( 6.9375 in.)( rn − r )  − =− − = − 1 +  A r A ( rc − rn ) A r A ( rc − rn ) A r ( rc − rn ) 

=−

 ( 6.9375 in.)(1.91248 in. − 2.375 in.)  90 lb 1+  2  0.60132 in.  ( 2.375 in.)(1.9375 in. − 1.91248 in.) 

= 7,932.1 psi = 7, 930 psi (T)

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.82 A solid circular rod of diameter d is bent into a semicircle as shown in Figure P8.82. The radial distance to the centroid of the rod is to be rc = 3d, and the curved rod is to support a load of P = 5,000 N. If the allowable stress must be limited to 135 MPa, what is the smallest diameter d that may be used for the rod?

FIGURE P8.82

Solution Radii: The radius from the center of curvature O to the centroid of the circular cross section is to be rc = 3d. Consequently, the inner radius of the curved bar is ri = 2.5d, and the outer radius is ro = 3.5d. For a solid circular cross section, the neutral axis can be found from the formula given in Table 8.1: A rn = where A =  c 2 2 2 2 rc − rc − c

)

(

For this situation, we substitute rc = 3d to find:



rn =

 2  3d −  

2 2  d   3 d − ( )    2  



d 4 = = 2.97902d 2 ( 3d − 2.9580399d ) 8 ( 0.0419601)

Bending moment: The bending moment magnitude at section AB is computed as the product of P and the perpendicular distance e between the centroid of the cross section (i.e., at section AB) and the line of action of applied load P. M = Pe = Prc = P ( 3d ) This moment is a negative bending moment since it creates tensile normal stress on the inner surface (i.e., at radius ri) of the curved bar. Bending stress: The bending stress is determined from the equation: M ( rn − r ) x = − r A ( rc − rn ) where r is the distance from the center of curvature to the location where the stress is to be determined. Combined stresses: Both axial and bending stress will be created at section AB. Load P is applied in such a way that it creates tensile normal stresses in the curved bar at section AB. The bending moment Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

creates tensile bending stresses at A and compressive bending stresses at B. Since both the axial and bending stresses are tensile at point A, this point will experience the largest stress in the curved rod. Stress at point A: We can express the stress at point A as follows:  x =  axial +  bending

P M ( rn − r ) P − P ( 3d )( rn − r ) P  ( 3d )( rn − r )  − = − = 1 +  A r A ( rc − rn ) A r A ( rc − rn ) A r ( rc − rn ) 

4 ( 5, 000 N )  ( 3d )( 2.97902d − 2.5d )  1 +  d2  ( 2.5d )( 3d − 2.97902d ) 

4 ( 5, 000 N )  1.43706d 2  4 ( 5, 000 N ) 1 +  28.398665  0.052450d 2  = d2 d2  

Set this expression equal to the 135 MPa allowable stress and solve for d: 4 ( 5, 000 N ) 135 N/mm 2   28.398665 d2 4 ( 5, 000 N )( 28.398665 ) d2   (135 N/mm 2 )

d  36.6 mm

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.83 The curved tee shape shown in Figure P8.83 is subjected to a bending moment of M = 2,700 N·m. Dimensions of the cross section are b1 = 15 mm, d1 = 70 mm, b2 = 50 mm, and d2 = 20 mm. The radial distance from O to A is ri = 85 mm. Determine: (a) the radial distance from O to the neutral axis. (b) the stresses at points A and B.

FIGURE P8.83

Solution Radii: The inner radius of the curved bar is ri = 85 mm, and the outer radius is ro = ri + d1 + d2 = 85 mm + 70 mm + 20 mm = 175 mm. Centroid location for flanged shape: Next, the radial distance from the center of curvature O to the centroid and to the neutral axis will be calculated for the flanged shape as follows. Note that the flanged shape is subdivided into two rectangles. The inner and outer radii for each of the rectangles as well as the radial distance from the center of curvature O to the centroid of each rectangle will be used in the calculation. r  ro rci Ai rci ri b ln  o  Shape b h Ai  ri  2 3 (mm) (mm) (mm ) (mm) (mm) (mm) (mm ) (mm) flange 50 20 1,000 85 105 95 95,000 10.56545 stem 15 70 1,050 105 175 140 147,000 7.66238 2,050 242,000 18.22784 3 242, 000 mm rc = = 118.04878 mm (from center of curvature to centroid) 2, 050 mm 2 rn =

A  ro 

 b ln  r 

2, 050 mm 2 = 112.46533 mm 18.22784 mm

 i (a) Radial distance to neutral axis: rn = 112.47 mm

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Bending moment: The specified bending moment magnitude is M = 2,700 N·m. This moment is a positive bending moment since it creates compressive normal stress on the inner surface (i.e., at radius ri) of the curved bar. Bending stress: The bending stress is determined from the equation: M ( rn − r ) x = − r A ( rc − rn ) where r is the distance from the center of curvature to the location where the stress is to be determined. Point A: At point A, r = ri = 85 mm. Thus, the bending stress at A is M ( rn − r ) ( 2, 700 N  m )(112.46533 mm − 85 mm )(1,000 mm/m ) x = − =− r A ( rc − rn ) (85 mm ) ( 2, 050 mm 2 ) (118.04878 mm − 112.46533 mm ) = −76.22 MPa = 76.2 MPa (C)

Ans.

Point B: At point B, r = ro = 175 mm. Thus, the bending stress at B is M ( rn − r ) ( 2, 700 N  m )(112.46533 mm − 175 mm )(1,000 mm/m ) x = − =− r A ( rc − rn ) (175 mm ) ( 2, 050 mm 2 ) (118.04878 mm − 112.46533 mm ) = 84.29 MPa = 84.3 MPa (T)

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.84 The curved flanged shape shown in Figure P8.84/85 is subjected to a bending moment of M = 4,300 N·m. Dimensions of the cross section are b1 = 75 mm, d1 = 15 mm, b2 = 15 mm, d2 = 55 mm, b3 = 35 mm, and d3 = 15 mm. The radial distance from O to A is ri = 175 mm. Determine the stresses at points A and B.

FIGURE P8.84/85

Solution Radii: The inner radius of the curved bar is ri = 175 mm, and the outer radius is ro = ri + d1 + d2 + d3 = 175 mm + 15 mm + 55 mm + 15 mm = 260 mm. Centroid location for flanged shape: Next, the radial distance from the center of curvature O to the centroid and to the neutral axis will be calculated for the flanged shape as follows. Note that the flanged shape is subdivided into three rectangles. The inner and outer radii for each of the rectangles as well as the radial distance from the center of curvature O to the centroid of each rectangle will be used in the calculation. r  rci Ai ro rci ri b ln  o  Shape b h Ai  ri  (mm) (mm) (mm2) (mm) (mm) (mm) (mm3) (mm) Inner Flange (1) 75 15 1,125 175 190 182.5 205,312.5 6.16786 Web (2) 15 55 825 190 245 217.5 179,437.5 3.81351 Outer Flange (3) 35 15 525 245 260 252.5 132,562.5 2.07982 2,475 517,312.5 12.06119 3 517,312.5 mm rc = = 209.01515 mm (from center of curvature to centroid) 2, 475 mm 2 rn =

A  ro 

 b ln  r 

2, 475 mm 2 = 205.20365 mm 12.06119 mm

 i

Bending moment: The specified bending moment magnitude is M = 4,300 N·m. The moment is a negative bending moment since it creates tensile normal stress on the inner surface (i.e., at radius ri) of the curved bar.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Bending stress: The bending stress is determined from the equation: M ( rn − r ) x = − r A ( rc − rn ) where r is the distance from the center of curvature to the location where the stress is to be determined. Point A: At point A, r = ri = 175 mm. Thus, the bending stress at A is M ( rn − r ) ( −4,300 N  m )( 205.20365 mm − 175 mm )(1,000 mm/m ) x = − =− r A ( rc − rn ) (175 mm ) ( 2, 475 mm 2 ) ( 209.01515 mm − 205.20365 mm ) = 78.67 MPa = 78.7 MPa (T)

Ans.

Point B: At point B, r = ro = 260 mm. Thus, the bending stress at B is M ( rn − r ) ( −4,300 N  m )( 205.20365 mm − 260 mm )(1,000 mm/m ) x = − =− r A ( rc − rn ) ( 260 mm ) ( 2, 475 mm 2 ) ( 209.01515 mm − 205.20365 mm ) = −96.07 MPa = 96.1 MPa (C)

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.85 The curved flanged shape shown in Figure P8.84/85 has cross-sectional dimensions of b1 = 45 mm, d1 = 9 mm, b2 = 9 mm, d2 = 36 mm, b3 = 18 mm, and d3 = 9 mm. The radial distance from O to A is ri = 140 mm. Determine: (a) the radial distance rc – rn between the centroid and the neutral axis. (b) the largest value of M if the allowable normal stress is 250 MPa.

FIGURE P8.84/85

Solution Radii: The inner radius of the curved bar is ri = 140 mm, and the outer radius is ro = ri + d1 + d2 + d3 = 140 mm + 9 mm + 36 mm + 9 mm = 194 mm. Centroid location for flanged shape: Next, the radial distance from the center of curvature O to the centroid and to the neutral axis will be calculated for the flanged shape as follows. Note that the flanged shape is subdivided into three rectangles. The inner and outer radii for each of the rectangles as well as the radial distance from the center of curvature O to the centroid of each rectangle will be used in the calculation. r  rci Ai ro rci ri b ln  o  Shape b h Ai  ri  2 3 (mm) (mm) (mm ) (mm) (mm) (mm) (mm ) (mm) Inner Flange (1) 45 9 405 140 149 144.5 58,522.5 2.80367 Web (2) 9 36 324 149 185 167 54,108.0 1.94769 Outer Flange (3) 18 9 162 185 194 189.5 30,699.0 0.85504 891 143,329.5 5.60640 3 143,329.5 mm rc = = 160.86364 mm (from center of curvature to centroid) 891 mm 2 A 891 mm 2 rn = = = 158.92544 mm  ro  5.60640 mm  b ln  r   i Bending moment: The specified bending moment magnitude is M = 4,300 N·m. The moment is a negative bending moment since it creates tensile normal stress on the inner surface (i.e., at radius ri) of the curved bar. (a) Radial distance rc – rn between the centroid and the neutral axis: rc − rn = 160.86364 mm −158.92544 mm = 1.93819 mm = 1.938 mm

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

(b) Largest value of M if the allowable normal stress is 250 MPa. Bending stress: The bending stress is determined from the equation: M ( rn − r ) x = − r A ( rc − rn ) where r is the distance from the center of curvature to the location where the stress is to be determined. Point A: At point A, r = ri = 140 mm. The bending stress at A will be tensile. Solve for the allowable bending moment that will produce positive bending stresses less than or equal to 250 MPa: M ( rn − r ) x = − r A ( rc − rn ) 250 N/mm 2  −

− M (158.92544 mm − 140 mm )

(140 mm ) (891 mm 2 ) (160.86364 mm − 158.92544 mm )

18.92544 mm M 241, 771.0680 mm 4 M  3,193, 717 N  mm 

Ans.

Point B: At point B, r = ro = 194 mm. The bending stress at B will be compressive. Solve for the allowable bending moment that will produce negative bending stresses less than or equal to –250 MPa: M ( rn − r ) x = − r A ( rc − rn ) −250 N/mm 2  −

− M (158.92544 mm − 194 mm )

(194 mm ) (891 mm 2 ) (160.86364 mm − 158.92544 mm )

−35.07456 mm M 335, 025.6228 mm 4 M  2,387,953 N  mm 

Therefore, the largest value of M that can be applied is: M max = 2.387953106 N  mm = 2.39 kN  m

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.86 The curved tee shape shown in Figure P8.86 has cross-sectional dimensions of b1 = 1.25 in., d1 = 0.25 in., b2 = 0.25 in., and d2 = 1.00 in. The radial distance from O to A is ri = 2.00 in. Determine the stresses at A and B for an applied load of P = 560 lb.

FIGURE P8.86

Solution Radii: The inner radius of the curved bar is ri = 2.00 in., and the outer radius is ro = ri + d1 + d2 = 2.00 in. + 0.25 in. + 1.00 in. = 3.25 in.. Centroid location for flanged shape: Next, the radial distance from the center of curvature O to the centroid and to the neutral axis will be calculated for the flanged shape as follows. Note that the flanged shape is subdivided into two rectangles. The inner and outer radii for each of the rectangles as well as the radial distance from the center of curvature O to the centroid of each rectangle will be used in the calculation. r  rci Ai ro rci ri b ln  o  Shape b h Ai  ri  (in.) (in.) (in.2) (in.) (in.) (in.) (in.3) (in.) flange 1.25 0.25 0.3125 2.00 2.25 2.125 0.66406 0.14723 stem 0.25 1.00 0.2500 2.25 3.25 2.750 0.68750 0.09193 0.5625 1.35156 0.23916 3 1.25156 in. rc = = 2.40278 in. (from center of curvature to centroid) 0.56250 in.2 A 0.56250 in.2 rn = = = 2.35198 in.  ro  0.23916 in.  b ln  r   i Bending moment: The bending moment magnitude at section AB is computed as the product of P and the perpendicular distance e between the centroid of the cross section (i.e., at section AB) and the line of action of applied load P. M = Pe = Prc = ( 560 lb)( 2.40278 in.) = 1,345.56 lb  in. The sign of the bending moment will be positive since it creates compressive normal stress on the inner surface (i.e., at radius ri) of the curved bar.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Combined stresses: Both axial and bending stress will be created at section AB. Load P is applied in such a way that it creates compressive normal stresses in the curved bar at section AB. The bending moment creates compressive bending stresses at A and tensile bending stresses at B. Stress at point A: Calculate the stress at point A (r = 2.0 in.) as follows:  x =  axial +  bending

=−

P M ( rn − r ) − A r A ( rc − rn )

=−

(1,345.56 lb  in.)( 2.35198 in. − 2.00 in.) 560 lb − 2 0.56250 in. ( 2.00 in.) ( 0.56250 in.2 ) ( 2.40278 in. − 2.35198 in.)

= −995.56 psi − 8, 287.86 psi = −9, 283.4 psi = 9, 280 psi (C)

Ans.

Stress at point B: Calculate the stress at point B (r = 3.25 in.) as follows:  x =  axial +  bending

=−

P M ( rn − r ) − A r A ( rc − rn )

=−

(1,345.56 lb  in.)( 2.35198 in. − 3.25 in.) 560 lb − 0.56250 in.2 ( 3.25 in.) ( 0.56250 in.2 ) ( 2.40278 in. − 2.35198 in.)

= −995.56 psi + 13, 012.28 psi = 12, 016.7 psi = 12, 020 psi (T)

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.87 The curved bar shown in Figure P8.87 has a trapezoidal cross section with dimensions of b1 = 75 mm, b2 = 30 mm, and d = 100 mm. The radial distance from O to A is ri = 120 mm. Determine: (a) the radial distance rc – rn between the centroid and the neutral axis. (b) the largest value of M if the allowable normal stress is 200 MPa.

FIGURE P8.87

Solution Radii: The inner radius of the curved bar is ri = 120 mm, and the outer radius is ro = ri + d = 120 mm + 100 mm = 220 mm. Next, we must locate the centroid for the trapezoidal cross section.

Shape

rci

rci Ai

(mm ) (mm) (mm3) (1) ½(75)(100) = 3,750 120+⅓(100) = 153.333 575,000 (2) ½(30)(100) = 1,500 120+⅔(100) = 186.667 280,000 5,250 855,000 rci Ai 855, 000 mm3 rc = = = 162.85714 mm (from center of curvature to centroid) Ai 5,250 mm 2 From Table 8.1, the radial distance from the center of curvature to the neutral axis is calculated from: A rn =  r 1 ( b1ro − b2 ri ) ln o − d ( b1 − b2 )  d ri 

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Substitute the appropriate values and calculate the distance from the center of curvature to the neutral axis: 5, 250 mm 2 rn = 1 220   − (100 mm )( 75 mm − 30 mm )   ( 75 mm )( 220 mm ) − ( 30 mm )(120 mm )  ln 100 mm  120  3 3 525, 000 mm 525, 000 mm = = 220 3,319.15186 mm 2 2 2 12,900 mm ln − 4,500 mm ( ) 120 = 158.17294 mm (a) Radial distance rc – rn between the centroid and the neutral axis: rc − rn = 162.85714 mm −158.17294 mm = 4.68421 mm = 4.68 mm

Ans.

(b) Largest value of M if the allowable normal stress is 200 MPa. Bending moment: The moment is a negative bending moment since it creates tensile normal stress on the inner surface (i.e., at radius ri) of the curved bar. Bending stress: The bending stress is determined from the equation: M ( rn − r ) x = − r A ( rc − rn ) where r is the distance from the center of curvature to the location where the stress is to be determined. Point A: At point A, r = ri = 120 mm. The bending stress at A will be tensile. Solve for the allowable bending moment that will not produce positive bending stresses less than or equal to 200 MPa: M ( rn − r ) x = − r A ( rc − rn ) 200 N/mm 2  −

M (158.17294 mm − 120 mm )

(120 mm ) ( 5, 250 mm 2 ) (162.85714 mm − 158.17294 mm )

38.17294 mm M 2,951, 046 mm 4 M  −15, 461, 455 N  mm −

Point B: At point B, r = ro = 220 mm. The bending stress at B will be compressive. Solve for the allowable bending moment that will produce negative bending stresses less than or equal to –200 MPa: M (158.17294 mm − 220 mm ) −200 N/mm 2  − ( 220 mm ) ( 5, 250 mm 2 ) (162.85714 mm − 158.17294 mm ) −61.82706 mm M 5, 410, 251 mm 4 M  −17,501, 260 N  mm −

Therefore, the largest magnitude of M that can be applied as shown is: M max = 15.46147 106 N  mm = 15.46 kN  m

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.88 The curved bar shown in Figure P8.88 has a trapezoidal cross section with dimensions of b1 = 0.75 in., b2 = 1.50 in., and d = 2.00 in. The radial distance from O to A is ri = 1.50 in. A load of P = 1,400 lb acts on the bar. If the allowable stress is 18 ksi, what is the largest permissible distance a?

FIGURE P8.88

Solution Radii: The inner radius of the curved bar is ri = 1.50 in., and the outer radius is ro = ri + d = 1.50 in. + 2.00 in. = 3.50 in. Next, we must locate the centroid for the trapezoidal cross section.

Shape

Ai 2

(in. ) (1) ½(0.75)(2.00) = 0.75 (2) ½(1.50)(2.00) = 1.50 2.25 rci Ai 5.875 in.3 rc = = = 2.61111 in. Ai 2.25 in.2

rci

rci Ai

(in.) 1.50+⅓(2.00) = 2.16667 1.50+⅔(2.00) = 2.83333

(in.3) 1.625 4.250 5.875

(from center of curvature to centroid)

From Table 8.1, the radial distance from the center of curvature to the neutral axis is calculated from: A rn =  r 1 ( b1ro − b2 ri ) ln o − d ( b1 − b2 )  d ri  Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Substitute the appropriate values and calculate the distance from the center of curvature to the neutral axis: 2.25 in.2 rn = 1  3.5  − ( 2.00 in.)( 0.75 in. − 1.50 in.)   ( 0.75 in.)( 3.5 in.) − (1.50 in.)(1.5 in.)  ln 2.00 in.  1.5  3 3 4.50 in. 4.50 in. = = 2 3.5 + 1.50 in.2 1.81774 in. ( 0.375 in.2 ) ln 1.5 = 2.47561 in. Bending moment: The bending moment magnitude at section AB is computed as the product of P and the perpendicular distance e between the centroid of the cross section (i.e., at section AB) and the line of action of applied load P. M = Pe = P ( a + rc ) The sign of the bending moment will be positive since it creates compressive normal stress on the inner surface (i.e., at radius ri) of the curved bar. Bending stress: The bending stress is determined from the equation: M ( rn − r ) x = − r A ( rc − rn ) where r is the distance from the center of curvature to the location where the stress is to be determined. Combined stresses: Both axial and bending stress will be created at section AB. Load P is applied in such a way that it creates compressive normal stresses in the curved bar at section AB. The bending moment creates compressive bending stresses at A and tensile bending stresses at B. Stress at point A: Since point A will be subjected to compressive bending stress in addition to the compressive axial stress that will occur across the entire cross section, the stress at A will control the distance a that can be used. Set the combined stress at A equal to the allowable stress and solve for the allowable distance a:  x =  axial +  bending =−

−18, 000 psi  −

P M ( rn − r ) P P ( a + rc )( rn − r ) P  ( a + rc )( rn − r )  − =− − = − 1 +  A r A ( rc − rn ) A r A ( rc − rn ) A r ( rc − rn )  1, 400 lb  ( a + 2.61111 in.)( 2.47561 in. − 1.50 in.)  1 +  2.25 in.2  (1.50 in.)( 2.61111 in. − 2.47561 in.) 

 ( −622.22222 psi ) 1 + ( 4.8 in.−1 ) ( a + 2.61111 in.)  27.92857  ( 4.8 in.−1 ) ( a + 2.61111 in.) 5.81845 in.  a + 2.61111 in.  a  3.21 in.

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.89 The curved bar shown in Figure P8.89 has an elliptical cross section with dimensions of a = 0.9 in. and b = 0.4 in. The inner radius of the curved bar is ri = 1.50 in. For an applied moment of M = 1,700 lb·in., determine: (a) the radial distance rc – rn between the centroid and the neutral axis. (b) the maximum compressive stress. FIGURE P8.89

Solution Radii: The inner radius of the curved bar is ri = 1.50 in., and the outer radius is ro = ri + 2a = 1.50 in. + 2(0.9 in.) = 3.30 in. The radius from the center of curvature O to the centroid of the elliptical cross section is rc = ri + a = 1.50 in. + 0.9 in. = 2.40 in. Next, we will use the formula in Table 8.1 to determine the location of the neutral axis rn: A =  ab =  ( 0.9 in.)( 0.4 in.) = 1.13097 in.2 rn =

(

)

2 b rc − rc2 − a 2 a 1.13097 in.2 1.13097 in.2 = = 2 ( 0.4 in.)  0.48908 in. 2 2 2.40 in. − ( 2.4 in.) − ( 0.9 in.)     0.9 in.  = 2.31243 in.

(a) Radial distance rc – rn between the centroid and the neutral axis: rc − rn = 2.4 in. − 2.31243 in. = 0.087570 in. = 0.0876 in.

Ans.

(b) Maximum compressive stress: Bending moment: The bending moment is M = 1,700 lb·in. This moment is a positive bending moment since it creates compressive normal stress on the inner surface (i.e., at radius ri) of the curved bar. Bending stress: The bending stress is determined from the equation: M ( rn − r ) x = − r A ( rc − rn ) where r is the distance from the center of curvature to the location where the stress is to be determined. The maximum compressive stress will occur at ri = 1.50 in. At this location, the bending stress is M ( rn − r ) (1, 700 lb  in.)( 2.31243 in. − 1.50 in.) x = − =− r A ( rc − rn ) (1.50 in.) (1.13097 in.2 ) ( 2.4 in. − 2.31243 in.) = −9, 296.8 psi = 9,300 psi (C)

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P8.90 The curved bar shown in Figure P8.90 has a triangle cross section with dimensions of b = 1.5 in. and d = 1.2 in. The inner radius of the curved bar is ri = 3.8 in. For an applied load of P = 70 lb, determine the stresses at A and B.

FIGURE P8.90

Solution Radii: The inner radius of the curved bar is ri = 3.8 in., and the outer radius is ro = ri + d = 3.8 in. + 1.2 in. = 5.0 in. The radius from the center of curvature O to the centroid of the triangular cross section is rc = ri + d/3 = 3.8 in. + 1.2 in./3 = 4.2 in. Next, we will use the formula in Table 8.1 to determine the location of the neutral axis rn: 1 1 A = bd = (1.5 in.)(1.2 in.) = 0.9 in.2 2 2 A 0.9 in.2 0.9 in.2 rn = = = (1.5 in.)( 5.0 in.)  ln 5.0  − 1.5 in. 0.21523 in. b ro  ro     ln  − b 1.2 in. d  ri   3.8  = 4.18157 in.

Bending moment: The bending moment magnitude at section AB is computed as the product of P and the perpendicular distance e between the centroid of the cross section (i.e., at section AB) and the line of action of applied load P. M = Pe = Prc = ( 70 lb )( 4.2 in.) = 294 lb  in. The sign of the bending moment will be positive since it creates compressive normal stress on the inner surface (i.e., at radius ri) of the curved bar. Combined stresses: Both axial and bending stress will be created at section AB. Load P is applied in such a way that it creates compressive normal stresses in the curved bar at section AB. The bending moment creates compressive bending stresses at A and tensile bending stresses at B. Stress at point A: Calculate the stress at point A (r = 3.8 in.) as follows:  x =  axial +  bending

=−

P M ( rn − r ) − A r A ( rc − rn )

=−

( 294 lb  in.)( 4.18157 in. − 3.8 in.) 70 lb − 2 0.9 in. ( 3.8 in.) ( 0.9 in.2 ) ( 4.2 in. − 4.18157 in.)

= −77.78 psi − 1, 779.56 psi = −1,857.34 psi = 1,857 psi (C)

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Stress at point B: Calculate the stress at point B (r = 5.0 in.) as follows:  x =  axial +  bending

=−

P M ( rn − r ) − A r A ( rc − rn )

=−

( 294 lb  in.)( 4.18157 in. − 5.0 in.) 70 lb − 0.9 in.2 ( 5.0 in.) ( 0.9 in.2 ) ( 4.2 in. − 4.18157 in.)

= −77.78 psi + 2,900.93 psi = 2,823.15 psi = 2,820 psi (T)

Ans.

For problems P9.1 through P9.4, a beam segment subjected to internal bending moments at sections A and B is shown along with a sketch of the cross-sectional dimensions. For each problem: (a) sketch a side view of the beam segment and plot the distribution of bending stresses acting at sections A and B. Indicate the magnitudes of key bending stresses on the sketch. (b) determine the resultant forces acting in the x direction on the specified area at sections A and B, and show these resultant forces on the sketch. (c) determine the horizontal force required to satisfy equilibrium for the specified area and show the location and direction of this force on the sketch. P9.1 The beam segment shown in Figure P9.1a/2a is subjected to internal bending moments of MA = 1,640 lb·ft and MB = 760 lb·ft. The segment length is x = 7 in. Consider area (1) shown in Figure P9.1b/2b, and use cross-sectional dimensions of b = 3 in., d = 9 in., and d1 = 1.5 in.

FIGURE P9.1a/2a Beam segment.

FIGURE P9.1b/2b Cross section.

Solution Moment of inertia about the z axis:

bd 3 ( 3.0 in.)( 9.0 in.) Iz = = = 182.25 in.4 12 12 3

(a) Bending stress distribution

(b) Resultant forces acting on area (1) On section A, the resultant force on area (1) in the x direction is 1 FA = ( 485.93 psi + 323.95 psi )( 3 in.)(1.5 in.) = 1,822.22 lb = 1,822 lb (C) 2 and on section B, the horizontal resultant force on area (1) is

Ans.

FB =

1 ( 225.19 psi + 150.13 psi )( 3.0 in.)(1.5 in.) = 844.44 lb = 844 lb (C) 2

Ans.

 FH = 978 lb Ans. The horizontal shear force is directed from section B toward section A at the interface between area (1) and the web elements.

For problems P9.1 through P9.4, a beam segment subjected to internal bending moments at sections A and B is shown along with a sketch of the cross-sectional dimensions. For each problem: (a) sketch a side view of the beam segment and plot the distribution of bending stresses acting at sections A and B. Indicate the magnitudes of key bending stresses on the sketch. (b) determine the resultant forces acting in the x direction on the specified area at sections A and B, and show these resultant forces on the sketch. (c) determine the horizontal force required to satisfy equilibrium for the specified area and show the location and direction of this force on the sketch. P9.2 The beam segment shown in Figure P9.1a/2a is subjected to internal bending moments of MA = 1,430 N·m and MB = 3,140 N·m. The segment length is x = 180 mm. Consider area (2) shown in Figure P9.1b/2b, and use cross-sectional dimensions of b = 90 mm, d = 240 mm, and d2 = 70 mm.

FIGURE P9.1a/2a Beam segment.

FIGURE P9.1b/2b Cross section.

Solution Moment of inertia about the z axis:

bd 3 ( 90 mm )( 240 mm ) Iz = = = 103,680,000 mm4 12 12 3

(a) Bending stress distribution

(b) Resultant forces acting on area (2) On section A, the resultant force on area (2) in the x direction is 1 FA = (1.655 N/mm2 + 0.690 N/mm 2 ) ( 90 mm )( 70 mm ) = 7,385.85 N = 7.39 kN (T) 2 and on section B, the horizontal resultant force on area (1) is 1 FB = ( 3.634 N/mm 2 + 1.514 N/mm 2 ) ( 90 mm )( 70 mm ) = 16, 217.88 N = 16.22 kN (T) 2

Ans.

 FH = 8.83 kN Ans. The horizontal shear force is directed from section B toward section A at the interface between area (2) and the web elements.

For problems P9.1 through P9.4, a beam segment subjected to internal bending moments at sections A and B is shown along with a sketch of the cross-sectional dimensions. For each problem: (a) sketch a side view of the beam segment and plot the distribution of bending stresses acting at sections A and B. Indicate the magnitudes of key bending stresses on the sketch. (b) determine the resultant forces acting in the x direction on the specified area at sections A and B, and show these resultant forces on the sketch. (c) determine the horizontal force required to satisfy equilibrium for the specified area and show the location and direction of this force on the sketch. P9.3 The beam segment shown in Figure P9.3a/4a is subjected to internal bending moments of MA = −9.2 kN·m and MB = −34.6 kN·m. The segment length is x = 300 mm. Consider area (1) shown in Figure P9.3b/4b, and use cross-sectional dimensions of b1 = 250 mm, d1 = 15 mm, b2 = 10 mm, d2 = 370 mm, b3 = 150 mm, and d3 = 15 mm.

FIGURE P9.3a/4a Beam segment.

FIGURE P9.3b/4b Cross section.

Solution Centroid location in y direction: (reference axis at bottom of shape) Shape (1) (2) (3)

Width b (mm) 250 10 150

yi Ai

Depth d (mm) 15 370 15

2, 228, 750 mm3 y= = = 229.77 mm Ai 9,700 mm2 Moment of inertia about the z axis: Shape IC (mm4) (1) 70,312.50 (2) 42,210,833.33 (3) 42,187.50

Area Ai (mm2) 3,750 3,700 2,250 9,700

yi (from bottom) (mm) 392.5 200 7.5

yi Ai (mm3) 1,471,875 740,000 16,875 2,228,750

(measured upward from bottom edge of shape)

d = y − yi

d²A (mm) (mm4) –162.73 99,306,339.01 29.77 3,278,704.23 222.27 111,156,934.85 Moment of inertia about the z axis (mm4) =

IC + d²A (mm4) 99,376,651.51 45,489,537.57 111,199,122.35 256,065,311.43

(a) Bending stress distribution

(b) Resultant forces acting on area (1) On section A, the resultant force on area (1) in the x direction is 1 FA = ( 6.116 N/mm 2 + 5.577 N/mm2 ) ( 250 mm )(15 mm ) 2 = 21,925.1 N = 21.9 kN (T)

Ans.

and on section B, the horizontal resultant force on area (1) is 1 FB = ( 23.002 N/mm 2 + 20.975 N/mm 2 ) ( 250 mm )(15 mm ) 2 = 82, 457.4 N = 82.5 kN (T)

Ans.

(c) Equilibrium of area (1) Fx = −21,925.1 N + 82, 457.4 N = 60,532.3 N  0  FH = 60.5 kN Ans. The horizontal shear force is directed from section B toward section A at the interface between area (1) and the stem.

For problems P9.1 through P9.4, a beam segment subjected to internal bending moments at sections A and B is shown along with a sketch of the cross-sectional dimensions. For each problem: (a) sketch a side view of the beam segment and plot the distribution of bending stresses acting at sections A and B. Indicate the magnitudes of key bending stresses on the sketch. (b) determine the resultant forces acting in the x direction on the specified area at sections A and B, and show these resultant forces on the sketch. (c) determine the horizontal force required to satisfy equilibrium for the specified area and show the location and direction of this force on the sketch. P9.4 The beam segment shown in Figure P9.3a/4a is subjected to internal bending moments of MA = −45.0 kip·ft and MB = −26.0 kip·ft. The segment length is x = 9 in. Consider area (3) shown in Figure P9.3b/4b, and use cross-sectional dimensions of b1 = 6.6 in., d1 = 0.6 in., b2 = 0.4 in., d2 = 10.8 in., b3 = 4.2 in., and d3 = 0.6 in.

FIGURE P9.3b/4b Cross section.

FIGURE P9.3a/4a Beam segment.

Solution Centroid location in y direction: (reference axis at bottom of shape) Shape (1) (2) (3)

Width b (in.) 6.6 0.4 4.2

Depth d (in.) 0.6 10.8 0.6

Area Ai (in.2) 3.960 4.320 2.520 10.800

yi (from bottom) (in.) 11.700 6.000 0.300

yi Ai (in.3) 46.332 25.920 0.756 73.008

yi Ai

73.008 in.3 y= = = 6.76 in. (measured upward from bottom edge of shape) Ai 10.8 in.2 Moment of inertia about the z axis: d = y − yi Shape IC d²A 4 (in. ) (in.) (in.4) (1) 0.1188 4.94 96.6383 (2) 41.9904 -0.76 2.4952 (3) 0.0756 -6.46 105.1636 Moment of inertia about the z axis (in.4) =

IC + d²A (in.4) 96.7571 44.4856 105.2392 246.4819

(a) Bending stress distribution

(b) Resultant forces acting on area (3) On section A, the resultant force on area (3) in the x direction is 1 FA = (14.81 ksi + 13.50 ksi )( 4.2 in.)( 0.6 in.) = 35.67 kips = 35.7 kips (C) 2 and on section B, the horizontal resultant force on area (3) is 1 FB = ( 8.56 ksi + 7.80 ksi )( 4.2 in.)( 0.6 in.) = 20.61 kips = 20.6 kips (C) 2

Ans.

 FH = 15.06 kips

Ans.

The horizontal shear force is directed from section B toward section A at the interface between area (3) and the web elements.

P9.5 A cantilever wood beam (Figure P9.5a) consists of eight 2 in. thick planks glued together to form a cross section that is 16 in. deep, as shown in Figure P9.5b. Each plank has a width of b = 5.5 in. The cantilever beam has a length of L = 6 ft and it supports a concentrated load of P = 3,800 lb. (a) Calculate the horizontal shear stress at points A, B, C, and D. (b) From these results, plot a graph showing the distribution of shear stresses from top to bottom of the beam.

FIGURE P9.5a Cantilever beam.

FIGURE P9.5b Cross-sectional dimensions.

Solution Shear force in cantilever beam: V = 3,800 lb Shear stress formula: VQ = It Section properties: 3 5.5 in.)(16 in.) ( Iz = = 1,877.333 in.4 12 t = 5.5 in. Point



6 in.

Q QA = ( 5.5 in.)( 2 in.)( 7 in.) = 77 in.3

28.3 psi

4 in.

QB = ( 5.5 in.)( 4 in.)( 6 in.) = 132 in.3

48.6 psi

2 in.

QC = ( 5.5 in.)( 6 in.)( 5 in.) = 165 in.3

60.7 psi

QD = ( 5.5 in.)(8 in.)( 4 in.) = 176 in.

64.8 psi

P9.6 A 6 m long simply supported wood beam carries a 36 kN concentrated load at B, as shown in Figure P9.6a. The cross-sectional dimensions of the beam as shown in Figure P9.6b are b = 150 mm, d = 400 mm, a = 90 mm, and c = 30 mm. Section a–a is located at x = 0.8 m from B. (a) At section a–a, determine the magnitude of the shear stress in the beam at point H. (b) At section a–a, determine the magnitude of the shear stress in the beam at point K. (c) Determine the maximum horizontal shear stress that occurs in the beam at any location within the entire span. (d) Determine the maximum tensile bending stress that occurs in the beam at any location within the entire length.

FIGURE P9.6b Crosssectional dimensions.

FIGURE P9.6a Simply supported wood beam.

Solution Section properties: 3 150 mm )( 400 mm ) ( Iz = = 800 106 mm4 12

t = 150 mm

(a) Shear stress at H: QH = (150 mm )( 90 mm )(155 mm ) = 2.0925 106 mm 3

=

VQ It

( 24, 000 N ) ( 2.0925 106 mm 3 ) = (800 106 mm 4 ) (150 mm ) = 419 kPa

Ans.

(b) Shear stress at K: QK = (150 mm )( 30 mm )(185 mm ) = 832.5 103 mm 3

=

VQ It

( 24, 000 N ) (832.5 103 mm3 ) = (800 106 mm 4 ) (150 mm ) = 166.5 kPa

Ans.

(c) Maximum shear stress at any location: Qmax = (150 mm )( 200 mm )(100 mm ) = 3.0 106 mm3 6 3 V Q ( 24, 000 N ) ( 3.0 10 mm ) = = = 600 kPa It (800 106 mm4 ) (150 mm )

Ans.

(d) Maximum bending stress at any location: M max = 48 kN  m = 48 106 N  mm 6 M c ( 48 10 N  mm ) ( 200 mm ) x = = = 12.00 MPa I 800 106 mm 4

Ans.

P9.7 A 6 m long simply supported wood beam carries a uniformly distributed load of 14 kN/m, as shown in Figure P9.7a. The cross-sectional dimensions of the beam as shown in Figure P9.7b are b = 200 mm, d = 480 mm, yH = 80 mm, and yK = 160 mm. Section a–a is located at x = 1.2 m from B. (a) At section a–a, determine the magnitude of the shear stress in the beam at point H. (b) At section a–a, determine the magnitude of the shear stress in the beam at point K. (c) If the allowable shear stress for the wood is 950 kPa, what is the largest distributed load w that can be supported by the beam?

FIGURE P9.7a Simply supported wood beam.

FIGURE P9.7b Cross-sectional dimensions.

Solution Section properties: 3 200 mm )( 480 mm ) ( I= = 1.8432 109 mm4 12

t = 200 mm

(a) Shear stress magnitude at H: Q = ( 200 mm )(160 mm )(160 mm ) = 5.120 106 mm3

=

VQ It

( 25, 200 N ) ( 5.120 106 mm3 ) = (1.8432 109 mm4 ) ( 200 mm ) = 350 kPa

Ans.

(b) Shear stress magnitude at K: Q = ( 200 mm )( 80 mm )( 200 mm ) = 3.20 106 mm 3

= =

VQ It

( 25, 200 N ) ( 3.20 106 mm3 ) (1.8432 109 mm 4 ) ( 200 mm )

= 219 kPa

Ans.

(c) Largest distributed load w: The maximum shear force in the span is equal to V = wL/2. The maximum shear stress occurs at the neutral axis location for this rectangular cross section. The value of Q at the neutral axis is: Qmax = ( 200 mm)( 240 mm)(120 mm) = 5.76 106 mm3 Set the shear stress to 950 kPa and solve for the allowable shear force V. VQ = It  It Vallow  allow Qmax

( 0.950 N/mm )(1.8432 10 mm ) ( 200 mm )  ( 5.76 10 mm ) 2

 60,800 N

The largest distributed load is thus wL  60,800 N 2 2 ( 60,800 N ) w = 20, 267 N/m = 20.3 kN/m 6m

Ans.

P9.8 The simply supported wood beam shown in Figure P9.8a carries a uniformly distributed load of w = 780 lb/ft, as shown in Figure P9.8a, where LAB= 8 ft and LBC = 16 ft. The beam crosssectional dimensions as shown in Figure P9.8b are b = 6 in., d = 20 in., yH = 5 in., and yK = 7 in. Section a–a is located at x = 4 ft from A. (a) At section a–a, determine the magnitude of the shear stress in the beam at point H. (b) At section a–a, determine the magnitude of the shear stress in the beam at point K. (c) If the allowable shear stress for the wood is 150 psi, what is the largest distributed load w that can be supported by the beam?

FIGURE P9.8a Simply supported wood beam.

FIGURE P9.8b Cross-sectional dimensions.

Solution Section properties: 3 6 in.)( 20 in.) ( I= = 4, 000 in.4 12

t = 6 in.

(a) Shear stress magnitude at H: QH = ( 6 in.)( 5 in.)( 7.5 in.)

= 225 in.3 VQ = It =

( 4,160 lb ) ( 225 in.3 ) ( 4, 000 in.4 ) ( 6 in.)

= 39.0 psi

Ans.

(b) Shear stress magnitude at K: QK = ( 6 in.)( 3 in.)( 8.5 in.)

= 153 in.3 VQ = It

( 4,160 lb ) (153 in.3 ) = ( 4, 000 in.4 ) ( 6 in.) = 26.5 psi

Ans.

(c) Largest distributed load w: The maximum shear stress occurs at the neutral axis location for the rectangular cross section. The value of Q at the neutral axis is: Qmax = ( 6 in.)(10 in.)( 5 in.) = 300 in.3 Set the shear stress to 150 psi and solve for the allowable shear force V. VQ = It  It Vallow  allow Qmax 

(150 psi ) ( 4, 000 in.4 ) ( 6 in.)

( 300 in. ) 3

 12, 000 lb

The largest shear force will occur at support C, as can be seen from the shear-force and bending-moment diagrams. The largest V will equal the reaction force at C. If we consider the overall equilibrium of the beam, we can derive an expression for the reaction force at C in terms of the distributed load w.

Reaction force Cy. L   M A = − wLBC  LAB + BC  + C y ( LAB + LBC ) = 0 2   L  16 ft    wLBC  LAB + BC  w (16 ft )  8 ft +  2  2    Cy = = = (10.6667 ft ) w LAB + LBC 8 ft + 16 ft Set Cy equal to Vallow and solve for the largest allowable distributed load w: (10.6667 ft ) w  12, 000 lb w

12, 000 lb = 1,125 lb/ft 10.6667 ft

Ans.

P9.9 The wood beam shown in Figure P9.9a supports loads of w = 470 lb/ft and P = 1,000 lb on span lengths of LAB = 10 ft, LBC = 5 ft, and LCD= 5 ft. The beam cross section shown in Figure P9.9b has dimensions of b1 = 1.5 in., d1 = 6.0 in., b2 = 5.0 in., and d2 = 16.0 in. Determine the magnitude and location of: (a) the maximum horizontal shear stress in the beam. (b) the maximum tensile bending stress in the beam.

FIGURE P9.9a Simply supported wood beam.

FIGURE P9.9b Cross-sectional dimensions.

Solution Section properties: 3 3 5 in.)(16 in.) 1.5 in.)( 6 in.) ( ( I= +2 = 1, 760.67 in.4 12 12 (a) Maximum shear force: Vmax = 2,800 lb @ support A Check shear stress at neutral axis: Q = ( 5.0 in.)( 8.0 in.)( 4.0 in.) +2 (1.5 in.)( 3.0 in.)(1.5 in.) = 173.5 in.3 3 V Q ( 2,800 lb ) (173.5 in. ) = = = 34.49 psi It (1,760.67 in.4 ) (8 in.)

Check shear stress at top edge of cover plates: Q = ( 5.0 in.)( 5.0 in.)( 5.5 in.) = 137.5 in.3 3 V Q ( 2,800 lb ) (137.5 in. ) = = = 43.73 psi It (1,760.67 in.4 ) (5 in.)

Maximum shear stress in beam:  max = 43.7 psi

Ans.

(b) Maximum bending moment: Mmax = 8,340 lb·ft (between support A and point B) Maximum tensile bending stress: (8,340 lb  ft )( −8.0 in.)(12 in./ft ) My x = − =− I 1, 760.67 in.4

= 454.76 psi = 455 psi (T)

Ans.

P9.10 For the wood beam shown in Figure P9.10a, assume L = 1.2 m and P = 12 kN. The beam cross section shown in Figure P9.10b has dimensions of b = 50 mm and d = 250 mm. (a) Determine the maximum horizontal shear stress in the beam. (b) If the allowable shear stress for the wood is 850 kPa, what is the minimum width b that is acceptable for this beam?

FIGURE P9.10a Simply supported wood beam.

FIGURE P9.10b Cross-sectional dimensions.

Solution Section properties: 3 50 mm )( 250 mm ) ( Iz = = 65.104 106 mm4 12

t = 50 mm

(a) Maximum horizontal shear stress magnitude: Vmax = 18.00 kN = 18, 000 N

Q = ( 50 mm )(125 mm )( 62.5 mm ) = 390.625  103 mm3 VQ = It

(18, 000 N ) ( 390.625 103 mm3 ) = ( 65.104 106 mm4 ) ( 50 mm ) = 2,160 kPa

Ans.

(b) Minimum acceptable width b: Q = (125 mm )( 62.5 mm ) b = ( 7,812.5 mm 2 ) b I=

b ( 250 mm ) 12

=

= b (1.3021106 mm3 )

VQ  850 kPa It

(18, 000 N ) Q

b (1.302110 mm )  b 6

(18, 000 N ) ( 7,812.5 mm 2 ) b

(1.302110 mm ) b 6

 0.850 N/mm 2

108.0 N/mm  0.850 N/mm 2 b

 b  127.1 mm

Ans.

P9.11 A 0.375 in. diameter solid steel shaft supports two pulleys as shown in Figure P9.11. The shaft has a length of L = 9 in., and the load applied to each pulley is P = 60 lb. The bearing at A can be idealized as a pin support and the bearing at D can be idealized as a roller support. Determine the magnitude of the maximum horizontal shear stress in the shaft.

FIGURE P9.11

Solution Section properties: I=



d4 =



( 0.375 in.)

64 64 = 970.722 10−6 in.4

d 3 ( 0.375 in.) Q= = 12 12 −3 = 4.395 10 in.3

Maximum shear force magnitude: Vmax = 40 lb (between B and C)

Maximum bending moment magnitude: Mmax = 60 lb·in. (at B and C)

Maximum horizontal shear stress: ( 40 lb ) ( 4.395 10−3 in.3 ) VQ = = I t ( 970.722 10−6 in.4 ) ( 0.375 in.)

= 483 psi

(at neutral axis between B and C )

Ans.

P9.12 A 25 mm diameter solid steel shaft supports loads PA = 1,000 N, PC = 3,200 N, and PE = 800 N, as shown in Figure P9.11. Assume LAB = 80 mm, LBC = 200 mm, LCD = 100 mm, and LDE = 125 mm. The bearing at B can be idealized as a roller support and the bearing at D can be idealized as a pin support. Determine the magnitude and location of: (a) the maximum horizontal shear stress in the shaft. (b) the maximum tensile bending stress in the shaft.

FIGURE P9.12

Solution Section properties:  4  4 I= d = ( 25 mm ) 64 64 = 19,174.76 mm 4 d 3 ( 25 mm ) Q= = 12 12 = 1,302.08 mm 3

Maximum shear force magnitude: Vmax = 2,200 N (between C and D)

Maximum bending moment magnitude: Mmax = 120,000 N·mm (at C)

(a) Maximum horizontal shear stress: 3 VQ ( 2, 200 N ) (1,302.08 mm ) = = I t (19,174.76 mm 4 ) ( 25 mm ) = 5.9757 MPa = 5.98 MPa

(between B and C , at the neutral axis )

Ans.

(b) Maximum tensile bending stress: (120, 000 N  mm )( −25 mm / 2 ) My x = − =− I 19,174.76 mm4

= 78.228 MPa = 78.2 MPa (T)

(on bottom of shaft at C )

Ans.

P9.13 An aluminum alloy pipe with an outside diameter of 4.0 in. and a wall thickness of 3/16 in. cantilevers upward a distance of L = 18 ft, as shown in Figure P9.13. The pipe is subjected to a wind loading that increases from wA = 12 lb/ft at its base to wB = 20 lb/ft at its tip. (a) Compute the value of Q for the pipe. (b) What is the maximum vertical shear stress in the pipe?

FIGURE P9.13

Solution Section properties:



[D4 − d 4 ] =

 

( 64 

4.0 in.) − ( 3.625 in.)  = 4.09016 in.4  4

64 1 1 3 3 Q = [ D3 − d 3 ] = ( 4.0 in.) − ( 3.625 in.)  = 1.36377 in.3 = 1.364 in.3   12 12

Ans.

Maximum shear force magnitude: w + wA Vmax = B L 2 20 lb/ft + 12 lb/ft = (18 ft ) = 288.0 lb 2 (b) Maximum vertical shear stress: ( 288.0 lb ) (1.36377 in.3 ) VQ = = I t ( 4.09016 in.4 ) ( 4.0 in. − 3.625 in.)

= 256 psi

P9.14 The internal shear force at a certain section of a steel beam is V = 215 kN. The beam cross section shown in Figure P9.14 has dimensions of bf = 300 mm, tf = 19 mm, d = 390 mm, and tw = 11 mm. Determine: (a) the shear stress at point A, which is located at yA = 80 mm below the centroid of the wide-flange shape. (b) the maximum horizontal shear stress in the wide-flange shape.

FIGURE P9.14

Solution Moment of inertia about the z axis: d = y − yi Shape Width b Depth d IC d²A IC + d²A 4 4 (mm) (mm) (mm ) (mm) (mm ) (mm4) flange 300 19 171,475.0 –185.5 196,138,425 196,309,900.0 web 11 352 39,979,690.7 0.0 0 39,979,690.7 flange 300 19 171,475.0 185.5 196,138,425 196,309,900.0 Moment of inertia about the z axis (mm4) = 432,599,490.7 (a) Shear stress at A: 19 mm   QA = ( 300 mm − 11 mm )(19 mm ) 195 mm −  2    195 mm + 80 mm  3 + (11 mm )(195 mm − 80 mm )   = 1,192,518 mm 2  

( 215, 000 N ) (1,192,518 mm3 ) A = = 53.9 MPa ( 432,599, 490.7 mm4 ) (11 mm )

Ans.

(b) Maximum horizontal shear stress: 19 mm   Qmax = ( 300 mm − 11 mm )(19 mm ) 195 mm −  2    195 mm  3 + (11 mm )(195 mm )   = 1, 227, 718 mm 2  

( 215, 000 N ) (1, 227, 718 mm3 )  max = = 55.5 MPa ( 432,599, 490.7 mm4 ) (11 mm )

Ans.

P9.15 The extruded cross section shown in Figure P9.15 is subjected to a shear force of V = 420 N. Dimensions of the cross section are b = 40 mm, d = 60 mm, and t = 4 mm. Determine: (a) the value of Q associated with point H, located at a = 13 mm above the lower surface of the cross section. (b) the horizontal shear stress at point H. (c) the maximum horizontal shear stress in the cross section. FIGURE P9.15

Solution

Moment of inertia about the z axis: Shape Width b Depth d Area A (mm) (mm) (mm2) (1) 4 60 240 (1) 4 60 240 (2) 32 4 128 (2) 32 4 128 (3) 32 4 128 (3) 32 4 128

d = y − yi IC d²A 4 (mm ) (mm) (mm4) 72,000 0 0 72,000 0 0 170.667 –28 100,352 170.667 28 100,352 170.667 –10 12,800 170.667 10 12,800 Moment of inertia about the z axis (mm4) =

IC + d²A (mm4) 72,000 72,000 100,522.667 100,522.667 12,970.667 12,970.667 370,986.667

(a) Q for point H: 13 mm   QH = 2 ( 4 mm )(13 mm )  30 mm −  2   4 mm   3 + ( 32 mm )( 4 mm )  30 mm −  = 6, 028 mm 2  

(b) Shear stress at H: ( 420 N ) ( 6,028 mm3 ) H = = 0.85305 MPa = 853 kPa (370,986.667 mm4 ) ( 2  4 mm)

Ans.

(c) Maximum horizontal shear stress:  30 mm  Qmax = 2 ( 4 mm )( 30 mm )    2  4 mm   + ( 32 mm )( 4 mm )  30 mm −  2   + ( 32 mm )( 4 mm )(10 mm ) = 8, 464 mm3

( 420 N ) (8, 464 mm3 )  max = = 1.1978 MPa = 1,198 kPa (370,986.667 mm4 ) ( 2  4 mm)

Ans.

P9.16 The beam cross section shown in Figure P9.16 is subjected to a shear force of V = 215 kN. Dimensions of the cross section are b = 150 mm, d = 210 mm, and t = 6 mm. Determine: (a) the horizontal shear stress at point H, located at a = 60 mm below the upper surface of the cross section. (b) the maximum horizontal shear stress in the cross section. FIGURE P9.16

Solution

Moment of inertia about the z axis: Shape Width b Depth d Area A (mm) (mm) (mm2) (1) 6 210 1,260 (1) 6 210 1,260 (1) 6 210 1,260 (2) 66 6 396 (2) 66 6 396 (2) 66 6 396 (2) 66 6 396

d = y − yi IC d²A 4 (mm ) (mm) (mm4) 4,630,500 0 0 4,630,500 0 0 4,630,500 0 0 1,188 –102 4,119,984 1,188 –102 4,119,984 1,188 –102 4,119,984 1,188 –102 4,119,984 Moment of inertia about the z axis (mm4) =

IC + d²A (mm4) 4,630,500 4,630,500 4,630,500 4,121,172 4,121,172 4,121,172 4,121,172 30,376,188

(a) Shear stress at H: 6 mm   QH = 2 ( 66 mm )( 6 mm ) 105 mm −  2   60 mm   3 +3 ( 6 mm )( 60 mm ) 105 mm −  = 161, 784 mm 2  

( 215,000 N ) (161,784 mm3 ) H = = 63.616 MPa = 63.6 MPa (30,376,188 mm4 ) (3 6 mm)

Ans.

(c) Maximum horizontal shear stress: 6 mm   Qmax = 2 ( 66 mm )( 6 mm ) 105 mm −  2    105 mm  3 +3 ( 6 mm )(105 mm )   = 180, 009 mm 2   3 ( 215,000 N ) (180,009 mm )  max = = 70.783 MPa = 70.8 MPa (30,376,188 mm4 ) (3 6 mm)

Ans.

P9.17 The extruded plastic shape shown in Figure P9.17 will be subjected to an internal shear force of V = 300 lb. Dimensions of the cross section are b = 1.5 in., d = 3.0 in., c = 2.0 in., and t = 0.15 in. Determine: (a) the horizontal shear stress at points A and B. (b) the horizontal shear stress at point C, which is located at a distance of yC = 1.0 in. below the z centroidal axis. (c) the maximum horizontal shear stress in the extrusion.

FIGURE P9.17

Solution

Centroid location in y direction: Shape (1) (1) (2) (3)

yi Ai Ai

Width b

Depth d

(in.) 0.15 0.15 1.2 1.2

(in.) 3.0 3.0 0.15 0.15

1.710 in.3 = 1.3571 in. 1.26 in.2 = 1.6429 in.

Area Ai (in.2) 0.45 0.45 0.18 0.18 1.26

yi (from bottom) (in.) 1.5 1.5 0.075 1.925

yi Ai (in.3) 0.675 0.675 0.0135 0.3465 1.7100

(from bottom of shape to centroid) (from top of shape to centroid)

Moment of inertia about the z axis: Shape Width b Depth d (in.) (in.) (1) 0.15 3.0 (1) 0.15 3.0 (2) 1.2 0.15 (3) 1.2 0.15

Area A (in.2) 0.45 0.45 0.18 0.18

d = y − yi IC d²A IC + d²A 4 4 (in. ) (in.) (in. ) (in.4) 337.5×10–3 –0.1429 9.1892×10–3 346.6892×10–3 337.5×10–3 –0.1429 9.1892×10–3 346.6892×10–3 –6 –3 337.5×10 1.2821 295.8805×10 296.2180×10–3 337.5×10–6 –0.5679 58.0519×10–3 58.3894×10–3 Moment of inertia about the z axis (in.4) = 1.0480

(a) Shear stress at A and B:

1 in.   −3 3 QA = 2 ( 0.15 in.)( 3.0 in. − 2.0 in.) 1.6429 in. −  = 342.87 10 in. 2   −3 3 ( 300 lb ) (342.87 10 in. ) A = = 327 psi (1.0480 in.4 ) ( 2  0.15 in.)

Ans.

1.15 in.   QB = 2 ( 0.15 in.)(1.15 in.) 1.6429 in. −  2   + (1.2 in.)( 0.15 in.)(1.6429 in. − 1.075 in.) = 470.65  10−3 in.3

B =

( 300 lb ) ( 470.65 10−3 in.3 ) = 449 psi (1.0480 in.4 ) ( 2  0.15 in.)

Ans.

(b) Shear stress at C: 0.3571 in.   QC = 2 ( 0.15 in.)(1.3571 in. − 1.0 in.) 1.0 in. +  2   0.15 in.   −3 3 + (1.2 in.)( 0.15 in.) 1.3571 in. −  = 357.06 10 in. 2  

( 300 lb ) (357.06 10−3 in.3 ) C = = 341 psi (1.0480 in.4 ) ( 2  0.15 in.)

Ans.

(c) Maximum horizontal shear stress:  1.3571 in.  Qmax = 2 ( 0.15 in.)(1.3571 in.)   2   0.15 in.   −3 3 + (1.2 in.)( 0.15 in.) 1.3571 in. −  = 507.06 10 in. 2   −3 3 ( 300 lb ) 507.06 10 in.  max = = 484 psi 1.0480 in.4 ( 2  0.15 in.)

(

)

Ans.

P9.18 The extruded plastic shape shown in Figure P9.18 has dimensions of D = 30 mm, d = 24 mm, a = 85 mm, and t = 7 mm. (a) Determine the horizontal shear stress at point A for a shear force of V = 320 N. (b) If the allowable shear stress for the plastic material is 1,100 kPa, what is the maximum shear force V that can be applied to the extrusion? FIGURE P9.18

Solution

Section properties of circular portion



30 mm ) − ( 24 mm )  = 254.469 mm2 

 

30 mm ) − ( 24 mm )  = 23, 474.7657 mm4 

( 4

( 64 

Centroid location in y direction: Shape

Width b

Depth d

(mm)

(1) (2)

yi Ai Ai

38, 779.8654 mm3 = 52.091 mm 744.469 mm 2 = 47.909 mm

Area Ai (mm2) 254.4690 490 744.469

yi (from bottom) (mm) 85 35

yi Ai (mm3) 21,629.8654 17,150 38,779.8654

(from bottom of shape to centroid) (from top of shape to centroid)

Moment of inertia about the z axis: d = y − yi Shape Area A IC d²A 2 4 (mm ) (mm ) (mm) (mm4) (1) 254.469 23,474.7657 –32.909 275,590.5074 (2) 490 200,083.3333 17.091 143,130.1177 Moment of inertia about the z axis (mm4) =

IC + d²A (mm4) 299,065.2731 343,213.4510 642,279

(a) Shear stress at A: QA = ( 254.469 mm2 ) (85 mm − 52.091 mm ) = 8,374.4 mm3

( 320 N ) (8,374.4 mm3 ) A = = 0.5960 MPa = 596 kPa ( 642, 279 mm4 ) ( 7 mm )

Ans.

(b) Maximum shear force V: The maximum shear stress occurs at the neutral axis for this shape. The value of Q corresponding to the neutral axis location is:  52.091 mm  3 Qmax = ( 7 mm )( 52.091 mm )   = 9, 497.021 mm 2   Set the shear stress equal to the 1,100 kPa allowable shear stress and solve for V: VQ = It 1.100 N/mm  2

1.100 N/mm  2

V ( 9, 497.021 mm3 ) It V ( 9, 497.021 mm3 )

( 642, 279 mm ) ( 7 mm ) 4

V  520.747 N = 521 N

Ans.

P9.19 The internal shear force at a certain section of a steel beam is V = 9 kN. The beam cross section shown in Figure P9.19 has dimensions of b = 150 mm, c = 30 mm, d = 70 mm, and t = 6 mm. Determine: (a) the shear stress at point H. (b) the shear stress at point K. (c) the maximum horizontal shear stress in the cross section. FIGURE P9.19

Solution

Centroid location in y direction: Shape (1) (1) (2) (2) (3)

yi Ai Ai

Width b

Depth d

(mm) 6 6 24 24 138

(mm) 70 70 6 6 6

85, 740 mm3 = 43.834 mm 1,956 mm 2 = 26.166 mm

Moment of inertia about the z axis: Shape Width b Depth d Area A (mm) (mm) (mm2) (1) 6 70 420 (1) 6 70 420 (2) 24 6 144 (2) 24 6 144 (3) 138 6 828

Area Ai (mm2) 420 420 144 144 828 1,956

yi (from bottom) (mm) 35 35 3 3 67

yi Ai (mm3) 14,700 14,700 432 432 55,476 85,740

(from bottom of shape to centroid) (from top of shape to centroid)

d = y − yi IC d²A 4 (mm ) (mm) (mm4) 171,500 8.834 32,776.6 171,500 8.834 32,776.6 432 40.834 240,107.8 432 40.834 240,107.8 2,484 –23.166 444,357.4 Moment of inertia about the z axis (mm4) =

IC + d²A (mm4) 204,276.6 204,276.6 240,539.8 240,539.8 446,841.4 1,336,474

(a) Shear stress at H:

6 mm   3 QH = 2 ( 30 mm )( 6 mm )  43.834 mm −  = 14,700.4 mm 2   3 ( 9,000 N ) (14,700.4 mm ) H = = 8.25 MPa (1,336, 474 mm4 ) ( 2  6 mm)

Ans.

(b) Shear stress at K:

6 mm   3 QK = (150 mm )( 6 mm )  26.166 mm −  = 20,849.1 mm 2   3 ( 9,000 N ) ( 20,849.1 mm ) K = = 11.70 MPa (1,336, 474 mm4 ) ( 2  6 mm)

Ans.

(c) Maximum horizontal shear stress:  26.166 mm  Qmax = 2 ( 6 mm )( 26.166 mm )   2   6 mm   3 + (138 mm )( 6 mm )  26.166 mm −  = 23, 289.0 mm 2   3 ( 9,000 N ) 23, 289.0 mm  max = = 13.07 MPa 1,336, 474 mm4 ( 2  6 mm )

(

)

Ans.

P9.20 The beam cross section shown below in Figure P9.20a has been proposed for a short pedestrian bridge. The cross section will consist of two square tubes that are welded to a rectangular web plate. Dimensions of the cross section as shown in Figure P9.20b are d = 410 mm, tw = 12 mm, b = 120 mm, t = 8.0 mm. If the beam will be subjected to a shear force of V = 165 kN, determine: (a) the shear stress at point H, located at yH = 160 mm above the z centroidal axis. (b) the shear stress at point K, located at yK = 70 mm below the z centroidal axis.

FIGURE P9.20a Beam segment.

FIGURE P9.20b Cross-sectional dimensions.

Solution Square tube section properties 2 2 A = (120 mm ) − (104 mm ) = 3,584 mm 2

(120 mm ) − (104 mm ) = 7,531,178.7 mm4 I= 4

12 Moment of inertia about the z axis: d = y − yi Shape Width b Depth d IC d²A (mm) (mm) (mm4) (mm) (mm4) Tube 7,531,178.7 –265 251.6864×106 Web 12 410 68.9210×106 0.0 0.0 Tube 7,531,178.7 265 251.6864×106 Moment of inertia about the z axis (mm4) =

IC + d²A (mm4) 259.2176×106 68.9210×106 259.2176×106 587.3562×106

(a) Shear stress at H:  410 mm + 120 mm  QH = ( 3,584 mm 2 )   2   205 mm − 160 mm   3 + (12 mm )( 205 mm − 160 mm ) 160 mm +  = 1, 048,310 mm 2  

(165, 000 N ) (1, 048,310 mm3 ) H = = 24.5 MPa ( 587.3562 106 mm4 ) (12 mm )

Ans.

(b) Shear stress at K:  410 mm + 120 mm  QK = ( 3,584 mm 2 )   2   205 mm − 70 mm   3 + (12 mm )( 205 mm − 70 mm )  70 mm +  = 1,172,510 mm 2  

(165, 000 N ) (1,172,510 mm3 ) K = = 27.4 MPa ( 587.3562 106 mm4 ) (12 mm )

Ans.

P9.21 The cantilever beam shown in Figure P9.21a/22a is subjected to a concentrated load of P. The cross-sectional dimensions of the rectangular tube shape shown in Figure P9.21b/22b are b = 6 in., d = 10 in., and t = 0.375 in. (a) Compute the value of Q that is associated with point H, which is located to the right of the vertical centroidal axis at zH = 1.50 in. (b) If the allowable shear stress for the rectangular tube shape is 16 ksi, determine the maximum concentrated load P than can be applied as shown to the cantilever beam.

FIGURE P9.21a/22a Beam segment.

FIGURE P9.21b/22b Cross-sectional dimensions.

Solution Since the load P (and in turn the shear force V) acts in the z direction, we must determine the moment of inertia about the y axis. We will also calculate the first moment of area Q with respect to the y axis. Moment of inertia about the y axis:

(10 in.)( 6 in.) − ( 9.25 in.)( 5.25 in.) = 68.458 in.4 I = 3

(a) Q associated with point H:  6 in. 0.375 in.  QH = ( 9.25 in.)( 0.375 in.)  −  2  2  1.5 in.   6 in.  +2 ( 0.375 in.)  − 1.50 in.   1.50 in. +  2   2  = 12.2871 in.3 = 12.29 in.3

Ans.

(b) Maximum load P:

 6 in. 0.375 in.  Qmax = ( 9.25 in.)( 0.375 in.)  −  2  2   3 in.  3 +2 ( 0.375 in.)( 3 in.)   = 13.1309 in.  2  VQ  max = max  16 ksi It Vmax 

(16 ksi ) ( 68.458 in.4 ) ( 2  0.375 in.) 13.1309 in.3

= 62.5622 kips

For the cantilever beam shown here, V = P; therefore, Pmax = Vmax = 62.6 kips

Ans.

P9.22 The cantilever beam shown in Figure P9.21a/22a is subjected to a concentrated load of P = 25 kips. The cross-sectional dimensions of the rectangular tube shape shown in Figure P9.21b/22b are b = 5 in., d = 9 in., and t = 0.25 in. Determine: (a) the shear stress at point K, which is located to the left of the vertical centroidal axis at zK = 1.00 in. (b) the maximum horizontal shear stress in the rectangular tube shape.

FIGURE P9.21a/22a Beam segment.

FIGURE P9.21b/22b Cross-sectional dimensions.

( 9 in.)( 5 in.) − (8.5 in.)( 4.5 in.) = 29.203 in.4 I = 3

(a) Shear stress at point K Q associated with point K:  5 in. 0.25 in.  QK = ( 8.5 in.)( 0.25 in.)  −  2   2 1.5 in.   5 in.  +2 ( 0.25 in.)  − 1.00 in.  1.00 in. +  2   2  = 6.3594 in.3 = 6.36 in.3

( 25 kips ) ( 6.3594 in.3 ) K = = 10.89 ksi ( 29.203 in.4 ) ( 2  0.25 in.)

Ans.

(b) Maximum horizontal shear stress:  5 in. 0.25 in.   2.5 in.  3 Qmax = (8.5 in.)( 0.25 in.)  −  + 2 ( 0.25 in.)( 2.5 in.)   = 6.6094 in. 2 2 2    

( 25 kips ) ( 6.6094 in.3 )  max = = 11.32 ksi ( 29.203 in.4 ) ( 2  0.25 in.)

Ans.

P9.23 A W14 × 34 standard steel section is used for a simple span of 6 ft 6 in. Determine the maximum uniform load that it can support if the allowable bending stress is 30 ksi and the allowable shear stress is 18 ksi. Assume bending about the strong axis of the W shape.

Solution Section Properties for W14 × 34 from Appendix B b f = 6.75 in.

tf =

0.455 in.

14 in.

tw =

0.285 in.

Iz =

340 in.4 For a simply support beam with a uniformly distributed load, the maximum shear force in the span is: wL Vmax = 2 and the maximum bending moment is: wL2 M max = 8 Consider bending stress: The allowable bending moment for the W14 × 34 standard steel section is: Mc  allow  Iz

 M allow 

 allow I z

( 30 ksi ) ( 340 in.4 )

= c 14 in. / 2 = 1, 457.14 kip  in. = 121.43 kip  ft

The allowable distributed load for the 6.5 ft span is thus: 8 (121.43 kip  ft ) wL2  121.43 kip  ft  wallow  = 23.0 kips/ft 2 8 ( 6.5 ft ) Consider shear stress: For this shape, the maximum value of Q is calculated as: 0.455 in.   3 Qmax = ( 0.285 in.)( 7 in.)( 3.5 in.) + ( 6.75 in. − 0.285 in.)( 0.455 in.)  7 in. −  = 26.904 in. 2   The allowable shear force for the W14 × 34 standard steel section is: 4  allow I z tw (18 ksi ) ( 340 in. ) ( 0.285 in.) VQ  allow  Vallow  = = 64.83 kips It Qmax 26.904 in.3 The allowable distributed load for the 6.5 ft span is thus: 2 ( 64.83 kips ) wL  64.83 kips  wallow  = 19.95 kips/ft 2 6.5 ft Based on both bending and shear stress considerations, the maximum distributed load that can be supported by this beam is: Ans. wallow = 19.95 kips/ft Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

P9.24 A simply supported beam with spans of LAB = 9 ft and LBC = 27 ft supports loads of w = 5 kips/ft and P = 30 kips, as shown in Figure P9.24a. The cross-sectional dimensions of the wide-flange shape shown in Figure P9.24b are bf = 10 in., tf = 0.64 in., d = 27.7 in., and tw = 0.46 in. Determine the magnitude of: (a) the maximum horizontal shear stress in the beam. (b) the maximum bending stress in the beam.

FIGURE P9.24a

FIGURE P9.24b

Solution

Maximum shear force magnitude: Vmax = 85 kips (just to the right of B)

Maximum bending moment magnitude: Mmax = 472.50 kip·ft (at B)

Moment of inertia about the z axis:

(10 in.)( 27.7 in.) − (10 in. − 0.46 in.) 27.7 in. − 2 ( 0.64 in.) = 3, 050.5 in.4 I = 3

(a) Maximum horizontal shear stress: For this shape, the maximum value of Q is calculated as:  27.7 in.  27.7 in.   27.7 in. 0.64 in.  3 Qmax = ( 0.46 in.)  −   + (10 in. − 0.46 in.)( 0.64 in.)   = 126.73 in. 2   2  4   2 3 (85 kips ) 126.73 in.  max = = 7.68 ksi Ans. 3,050.5 in.4 ( 0.46 in.)

(

)

(b) Maximum bending stress: ( 472.50 kip  ft )( 27.7 in. / 2 )(12 in./ft ) = 25.7 ksi Mc x = =− I 3,050.5 in.4

Ans.

P9.25 Vertical beam AD supports a horizontal load of P = 120 kN as shown in Figure P9.25a. Dimensions of this structure are x1 = 220 mm, x2 = 5.78 m, y1 = 0.85 m, y2 = 3.15 m, and y3 = 1.60 m. The beam is a rectangular tube (Figure P9.25b) with dimensions of b = 100 mm, d = 180 mm, and t = 5 mm. At point H, which is located at a distance of a = 40 mm to the right of the z centroidal axis, determine: (a) the vertical shear stress magnitude. (b) the normal stress in the y direction produced by the bending moment. State whether it is a tensile or a compressive normal stress. (c) the normal stress in the y direction produced by the axial force. State whether it is a tensile or a compressive normal stress.

FIGURE P9.25b

FIGURE P9.25a

Solution

Equilibrium: y +y 0.85 m + 3.15 m tan  = 1 2 = = 0.69204 x2 5.78 m

 = 34.685 M A = ( FBC cos  )( y1 + y2 ) + ( FBC sin  ) x1 − P ( y1 + y2 + y3 ) = 0  FBC = =

P ( y1 + y2 + y3 ) ( y1 + y2 ) cos  + x1 sin 

(120 kN )( 5.6 m ) ( 4 m ) cos 34.685 + ( 0.22 m ) sin 34.685

= 196.815 kN

Fx = Ax + FBC cos  − P = 0

 Ax = P − FBC cos  = 120 kN − (196.815 kN ) cos 34.685 = −41.840 kN

Fy = Ay + FBC sin  = 0  Ay = − FBC sin  = − (196.815 kN ) sin 34.685 = −112.000 kN

Determine axial force F, shear force V, and bending moment M at point H: Fy = Ay − F = 0  F = Ay = −112.000 kN Fx = Ax + V = 0 V = − Ax = 41.840 kN M H = − Ax y1 − M = 0

 M = − Ax y1 = − ( −41.840 kN )( 0.85 m ) = 35.564 kN  m

Note that the bending moment acts about the z axis and that it creates tensile bending stresses at H. Section Properties: A = (100 mm )(180 mm ) − ( 90 mm )(170 mm ) = 2, 700 mm 2

(100 mm )(180 mm ) − ( 90 mm )(170 mm ) = 11.7525 106 mm 4 I = 3

Calculate Q for point H:

50 mm   QH = 2 ( 5 mm )( 50 mm )  40 mm +  2   5 mm   + ( 90 mm )( 5 mm )  90 mm −  2   = 71,875 mm3 (a) Vertical shear stress magnitude at H: ( 41,840 N ) 71,875 mm3 H = = 25.6 MPa 11.7525 106 mm4 ( 2  5 mm )

(

)

Ans.

(b) Bending stress at H: The sense of the bending stress (either tensile or compressive stress) is determined by inspection. 6 M z x ( 35.564 10 N  mm ) ( 40 mm ) y = = = 121.0 MPa (T) Ans. Iz 11.7525 106 mm 4 (c) Axial stress at H: F −112, 000 N  axial = = = 41.5 MPa (C) A 2, 700 mm 2

Ans.

P9.26 A rectangular beam is fabricated by nailing together three pieces of dimension lumber as shown in Figure P9.26. Each board has a width of b = 4 in. and a depth of d = 2 in. Pairs of nails are driven along the beam at a spacing of s, and each nail can safely transmit a force of 120 lb in direct shear. The beam is simply supported and carries a load of 800 lb at the center of a 9 ft span. Determine the maximum spacing s (along the length of the beam) permitted for each pair of nails.

FIGURE P9.26

Solution Maximum shear force For P = 800 lb, V = P/2 = 400 lb Maximum nail spacing s:

( 4 in.)( 6 in.) = 72 in.4 I = 3

12 Q = ( 4 in.)( 2 in.)( 2 in.) = 16 in.3

VQ ( 400 lb ) (16 in. ) q= = = 88.889 lb/in. I 72 in.4 Since two fasteners (i.e., two nails) connect the highlighted board to the rest of the beam, nf = 2 within the spacing interval s: qs  n f V f 3

s 

nf Vf q

( 2 nails )(120 lb/nail ) = 2.70 in. 88.889 lb/in.

P9.27 A wooden I-beam is fabricated from three pieces of dimension lumber as shown in Figure P9.27. The cross-sectional dimensions are bf = 90 mm, tf = 40 mm, dw = 240 mm, and tw = 40 mm. The I-beam will be used as a simply supported beam to carry a concentrated load P at the center of a 5 m span. The wood has an allowable bending stress of 8,300 kPa and an allowable shear stress of 620 kPa. The flanges of the beam are fastened to the web with screws that can safely transmit a force of 800 N in direct shear. (a) If the screws are uniformly spaced at an interval of s = 200 mm along the span, what is the maximum concentrated load P that can be supported by the beam based on the strength of the screw connections? Demonstrate that the maximum bending and shear stresses produced by P are acceptable. (b) Determine the magnitude of load P that produces the allowable bending stress in the span (i.e., b = 8,300 kPa). What screw spacing s is required to support this load magnitude? FIGURE P9.27

Solution Moment of inertia about the z axis: Shape Width b Depth d (mm) (mm) top flange 90 40 web 40 240 bottom flange 90 40

d = y − yi

IC (mm4) 480,000 46,080,000 480,000

(mm) –140 0 140

d²A (mm4) 70,560,000 0 70,560,000

IC + d²A (mm4) 71,040,000 46,080,000 71,040,000

Moment of inertia about the z axis (mm4) = 188.16×106 (a) Maximum concentrated load P:

Q = ( 90 mm )( 40 mm )(140 mm ) = 504, 000 mm 3 qs  n f V f q 

nf Vf s

(1 fastener )(800 N/fastener ) = 4.0 N/mm 200 mm

VQ I

6 4 qI ( 4 N/mm ) (188.16 10 mm ) V = = = 1, 493.333 N Q 504,000 mm 3

P 2

 Pmax = 2,986.667 N = 2,990 N

Check maximum bending and shear stresses:  ( 2,986.667 N )( 5, 000 mm )   320 mm     4 2 Mc ( PL / 4)c      max = = = = 3.175 MPa  8.3 MPa 6 4 Iz Iz 188.16 10 mm

Ans.

Qmax = ( 90 mm )( 40 mm )(140 mm ) + ( 40 mm )(120 mm )( 60 mm ) = 792, 000 mm3

3 VQmax (1, 493.333 N ) ( 792, 000 mm )  max = = = 157.14 kPa  620 kPa I zt (188.16 106 mm4 ) ( 40 mm )

(b) Magnitude of load P that produces the allowable bending stress in the span: Mc x =  8.3 MPa I 8.3 N/mm 2 )(188.16 106 mm 4 ) ( M  = 9.7608  106 N  mm = 9, 760.8 N  m 320 mm / 2 PL M max = 4 4 M max 4 ( 9, 760.8 N  m )  Pmax = = = 7,808.6 N = 7.81 kN L 5m

Ans.

Required nail spacing s: P 7,808.6 N Vmax = max = = 3,904.3 N 2 2 3 VQ ( 3,904.3 N ) ( 504, 000 mm ) q= = = 10.458 N/mm I 188.16 106 mm 4 qs  n f V f

s 

nfVf q

(1 fastener )(800 N/fastener ) = 76.5 mm

Ans.

10.458 N/mm

Check maximum shear stresses:

3 VQmax ( 3,904.3 N ) ( 792,000 mm )  max = = = 410.9 kPa  620 kPa I zt (188.16 106 mm4 ) ( 40 mm)

P9.28 A wooden box beam is fabricated from four boards, which are fastened together with nails, as shown in Figure P9.28b. The nails are installed at a spacing of s = 125 mm (Figure P9.28a), and each nail can provide a resistance of Vf = 500 N. In service, the box beam will be installed so that bending occurs about the z axis. Determine the maximum shear force V that can be supported by the box beam based on the shear capacity of the nailed connections.

FIGURE P9.28a

FIGURE P9.28b

Solution Moment of inertia Iz: (200 mm)(300 mm)3 (120 mm)(250 mm)3 Iz = − = 293,750,000 mm 4 12 12 First moment of area Q: Q = (200 mm)(25 mm)(137.5 mm) = 687,500 mm3 Shear flow q based on nail shear force: q s  nfVf

q 

nfVf s

(2 nails)(500 N/nail) = 8 N/mm 125 mm

Maximum shear force V:

VQ q= Iz

q I z (8 N/mm)(293,750,000 mm4 ) V = = = 3,418 N = 3.42 kN Q 687,500 mm3

Ans.

P9.29 A wooden box beam is fabricated from four boards, which are fastened together with screws, as shown in Figure P9.29b. Each screw can provide a resistance of 800 N. In service, the box beam will be installed so that bending occurs about the z axis, and the maximum shear force in the beam will be 9 kN. Determine the maximum permissible spacing interval s for the screws (see Figure P9.29a).

FIGURE P9.29a

FIGURE P9.29b

Solution Moment of inertia Iz: (190 mm)(250 mm)3 (140 mm)(150 mm)3 Iz = − = 208,020,833 mm 4 12 12 First moment of area Q: Q = (140 mm)(50 mm)(100 mm) = 700,000 mm3 Shear flow q based on beam shear force V:

VQ (9,000 N)(750,000 mm3 ) = = 30.285 N/mm Iz 208,020,833 mm4

Maximum spacing interval s: q s  nfVf

s 

nfVf q

(2 screws)(800 N/screw) = 52.8 mm 30.285 N/mm

Ans.

P9.30 A wooden beam is fabricated by bolting together three members, as shown in Figure P9.30a/31a. The crosssectional dimensions are shown in Figure P9.30b/31b. The 8 mm diameter bolts are spaced at intervals of s = 200 mm along the x axis of the beam. If the internal shear force in the beam is V = 7 kN, determine the shear stress in each bolt.

FIGURE P9.30a/31a

FIGURE P9.30b/31b

Solution Centroid location in y direction: Shape left board center board right board

Width b (mm) 40 40 40

Depth d (mm) 90 300 90

yi Ai

Area Ai (mm2) 3,600 12,000 3,600 19,200

3,636,000 mm3 y= = = 189.375 mm A i 19,200 mm 2 = 110.625 mm Moment of inertia about the z axis: Shape IC (mm4) left board 2,430,000 center board 90,000,000 right board 2,430,000

yi (from bottom) (mm) 255 150 255

yi Ai (mm3) 918,000 1,800,000 918,000 3,636,000

(from bottom of shape to centroid) (from top of shape to centroid)

d = y − yi

d²A IC + d²A (mm) (mm4) (mm4) –65.625 15,503,906.25 17,933,906.25 39.375 18,604,687.50 108,604,687.50 –65.625 15,503,906.25 17,933,906.25 4 Moment of inertia about the z axis (mm ) = 144,472,500.00

Shear force in each bolt Consider left board, which is held in place by the bolt: Q = (40 mm)(90 mm)(110.625 mm − 90 mm/2) = 236, 250 mm3 V Q (7,000 N)(236, 250 mm3 ) = = 11.4468 N/mm I 144,472,500 mm 4 Note that this value of q is the shear flow that must be transmitted by one surface of the bolt cross section. The cross-sectional area of the bolt is: q=

Abolt =



(8 mm)2 = 50.2655 mm2

4 Relate the shear flow and the bolt shear stress with Eq. (9.14): q s  n f  f Af

 f =

qs (11.4468 N/mm)(200 mm) = = 45.5 MPa n f Af (1 bolt surface)(50.2655 mm 2 )

Ans.

P9.31 A wooden beam is fabricated by bolting together three members, as shown in Figure P9.50a/31a. The cross-sectional dimensions are shown in Figure P9.50b/31b. The allowable shear stress of the wood is 850 kPa, and the allowable shear stress of the 10 mm diameter bolts is 40 MPa. Determine: (a) the maximum internal shear force V that the cross section can withstand based on the allowable shear stress in the wood. (b) the maximum bolt spacing s required to develop the internal shear force computed in part (a).

FIGURE P9.30a/31a

FIGURE P9.30b/31b

Solution Centroid location in y direction: Shape left board center board right board

Width b (mm) 40 40 40

Depth d (mm) 90 300 90

yi Ai

Area Ai (mm2) 3,600 12,000 3,600 19,200

3,636,000 mm3 y= = = 189.375 mm A i 19,200 mm 2 = 110.625 mm Moment of inertia about the z axis: Shape IC left board center board right board

(mm4) 2,430,000 90,000,000 2,430,000

yi (from bottom) (mm) 255 150 255

yi Ai (mm3) 918,000 1,800,000 918,000 3,636,000

(from bottom of shape to centroid) (from top of shape to centroid)

d = y − yi

d²A

IC + d²A

(mm) (mm4) (mm4) –65.625 15,503,906.25 17,933,906.25 39.375 18,604,687.50 108,604,687.50 –65.625 15,503,906.25 17,933,906.25 4 Moment of inertia about the z axis (mm ) = 144,472,500.00

Consider maximum horizontal shear stress: Q = (40 mm)(189.375 mm)(189.375 mm/2) = 717,257.813 mm3

=

VQ  850 kPa = 0.850 MPa It

(0.850 N/mm 2 )(144,472,500 mm 4 )(40 mm) Vmax  = 6,848.4 N = 6.85 kN 717,257.813 mm3

Ans.

Maximum bolt spacing Consider left board, which is held in place by the bolt: Q = (40 mm)(90 mm)(110.625 mm − 90 mm/2) = 236, 250 mm3 V Q (6,848.4 N)(236, 250 mm3 ) = = 11.1989 N/mm I 144,472,500 mm 4 Note that this value of q is the shear flow that must be transmitted by one surface of the bolt cross section. The cross-sectional area of the bolt is: q=

Abolt =



(10 mm) 2 = 78.5398 mm 2

4 Relate the shear flow and the bolt shear stress with Eq. (9.14): q s  n f  f Af

s 

n f  f Af q

(1 bolt surface)(40 N/mm 2 )(78.5398 mm 2 ) = 281 mm 11.1989 N/mm

Ans.

P9.32 The cantilever beam shown in Figure P9.32a is fabricated by connecting two hat shapes with rivets that are spaced longitudinally at intervals of s along each side of the built-up shape. The cross section of the built-up shape is shown in Figure P9.32b, and the section properties for a single hat shape are shown in Figure P9.32c. For an applied load of P = 6,800 N, determine the maximum spacing interval s that can be used if each rivet has a shear strength of 1,700 N.

FIGURE P9.32a Riveted beam.

FIGURE P9.32b Cross section.

FIGURE P9.32c Cross sectional Properties for a Single Hat Shape.

Solution Moment of inertia (both channels): 2 I z = 2 3.76 106 mm 4 + ( 60 mm ) ( 2, 400 mm 2 )  = 24.8 10 6 mm 4   Shear flow: Q = ( 60 mm ) ( 2,400 mm 2 ) = 144, 000 mm3 VQ ( 6,800 N ) (144, 000 mm ) q= = = 39.484 N/mm I 24.8 106 mm 4 3

Maximum spacing interval s: q s  nfVf

s 

nfVf q

( 2 rivets )(1, 700 N ) = 86.1 mm 39.484 N/mm

Ans.

P9.33 A steel wide-flange beam in an existing structure is to be strengthened by adding a cover plate to its lower flange, as shown in Figure P9.33. The wide-flange shape has dimensions of bf = 7.50 in., tf = 0.57 in., d = 18.0 in., and tw = 0.355 in. The cover plate has a width of b = 15 in. and t = 1.0 in. The cover plate is attached to the lower flange by pairs of 0.875 in. diameter bolts spaced at intervals of s along the beam span. Bending occurs about the z centroidal axis. (a) If the allowable bolt shear stress is 14 ksi, determine the maximum bolt spacing interval s required to support an internal shear force in the beam of V = 55 kips. (b) If the allowable bending stress is 24 ksi, determine the allowable bending moment for the existing wide-flange shape, the allowable bending moment for the wide-flange shape with the added cover plate, and the percentage increase in moment capacity that is gained by adding the cover plate.

FIGURE P9.33

Solution Wide-flange shape without cover plate: moment of inertia about the z axis: d = y − yi Shape IC d²A Width b Depth d (in.) (in.) (in.4) (in.) (in.4) Top flange 7.50 0.57 0.1157 –8.715 324.6915 Web 0.355 16.860 141.7816 0 0 Bottom flange 7.50 0.57 0.1157 8.715 324.6915 Moment of inertia about the z axis (in.4) = Wide-flange shape with cover plate attached: centroid location in y direction: yi Shape Width b Depth d Area Ai (from bottom) (in.) (in.) (in.2) (in.) Top flange 7.50 0.57 4.275 18.715 Web 0.355 16.860 5.985 10 Bottom flange 7.50 0.57 4.275 1.285 Cover plate 15 1 15 0.5 29.535 3 yi Ai 152.853 in. y= = = 5.175 in. (from bottom of shape to centroid) Ai 29.535 in.2

IC + d²A (in.4) 324.8072 141.7816 324.8072 791.3960

yi Ai (in.3) 80.007 59.853 5.493 7.500 152.853

= 13.825 in. (from top of shape to centroid) Wide-flange shape with cover plate attached: moment of inertia about the z axis: d = y − yi Shape IC d²A IC + d²A Width b Depth d (in.) (in.) (in.4) (in.) (in.4) (in.4) Top flange 7.50 0.57 0.1157 –13.54 783.7426 783.8583 Web 0.355 16.860 141.7816 –4.825 139.3345 281.1161 Bottom flange 7.50 0.57 0.1157 3.89 64.6897 64.8054 Cover plate 15 1 1.25 4.675 327.8344 329.0844 Moment of inertia about the z axis (in.4) = 1,458.8642 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

(a) Maximum bolt spacing for cover plate given that V = 55 kips: Consider the cover plate, which is connected to the wide-flange shape with two bolts: Q = (15 in.)(1 in.)( 5.175 in. − 0.5 in.) = 70.125 in.3 3 VQ ( 55 kips ) ( 70.125 in. ) q= = = 2.644 kips/in. I 1,458.8642 in.4 The cross-sectional area of a 0.875 in. diameter bolt is:

Abolt =



( 0.875 in.) = 0.6013 in.2 2

4 Relate the shear flow and the bolt shear stress with Eq. (9.14): qs  n f  f Af

s 

n f  f Af q

( 2 bolts )(14 ksi ) ( 0.6013 in.2 ) 2.644 kips/in.

= 6.37 in.

(b) Allowable bending moment for wide-flange shape without cover plate: My =  24 ksi I ( 24 ksi ) 791.3960 in.4  M allow  = 2,110.4 kip  in. = 175.9 kip  ft 18 in. / 2

(

)

Allowable bending moment for wide-flange shape with cover plate: My =  24 ksi I ( 24 ksi ) 1, 458.8642 in.4  M allow  = 2,532.6 kip  in. = 211 kip  ft 13.825 in.

(

)

Percentage increase in moment capacity: 2,532.6 kip  in. − 2,110.4 kip  in. % increase = (100%) = 20.0% 2,110.4 kip  in.

Ans.

P9.34 The tee shape shown in Figure P9.34 is constructed from two dimension lumber boards that are rigidly attached to each other. A vertical shear force of V = 1,350 lb acts on the cross section. Dimensions of the tee shape are bf = 7.25 in., tf = 1.50 in., d = 10.75 in., and tw = 1.50 in. Determine the shear flow magnitude and the shear stress magnitude that acts in the tee shape at section 1–1. FIGURE P9.34

Solution Centroid location in y direction: Shape

width (in.) 7.25 1.5

Top flange Stem

yi Ai Ai

depth (in.) 1.5 9.25

172.922 in.3 = 6.9867 in. 24.75 in.2

yi Ai (in.3) 108.750 64.172 172.922

(from bottom of shape to centroid)

= 3.7633 in.

Moment of inertia about the z axis: Shape IC (in.4) Top flange 2.039 Stem 98.932

Area Ai (in.2) 10.875 13.875 24.750

yi (from bottom) (in.) 10 4.625

(from top of shape to centroid)

d = y − yi

d²A (in.) (in.4) –3.013 98.742 2.362 77.392 Moment of inertia about the z axis (in.4) =

IC + d²A (in.4) 100.781 176.324 277.105

Shear flow at section 1–1: We will consider Q for the top flange, starting at section 1–1 and extending leftward to the free end of the flange. Q = ( 2.875 in.)(1.50 in.)(10 in. − 6.9867 in.) = 12.995 in.3

(1,350 lb ) (12.995 in.3 ) 277.105 in.4

= 63.3 lb/in.

Shear stress at section 1–1: (1,350 lb ) (12.995 in.3 ) = = 42.2 psi ( 277.105 in.4 ) (1.5 in.)

Ans.

P9.35 The wide-flange shape shown in Figure P9.35 has dimensions of bf = 7.25 in., tf = 0.8 in., d = 18.5 in., and tw = 0.5 in. A vertical shear force of V = 47,000 lb acts on the cross section. Determine the shear flow magnitude and the shear stress magnitude that acts in the shape: (a) at section 1–1. (b) at section 2–2. (c) at section 3–3 where a = 1.75 in.

Solution

FIGURE P9.35

Moment of inertia about the z axis: Shape Width b Depth d Top flange Web Bottom flange

(in.) 7.5 0.5 7.5

Area A

d = y − yi

d²A

IC + d²A

(in.2) 5.800 8.450 5.800

(in.4) 0.309 201.117 0.309

(in.) -8.850 0.000 8.850

(in.4) 454.271 0.000 454.271

(in.4) 454.580 201.117 454.580 1,110.277

(in.) 0.8 16.9 0.8

(a) Shear flow and shear stress at section 1–1: Calculate Q for the bottom flange, starting at section 1–1 and extending downward.  18.5 in. 0.8 in.  3 Q1−1 = ( 7.25 in.)( 0.8 in.)  −  = 51.33 in. 2 2  

q1−1 =

( 47, 000 lb ) ( 51.33 in.3 ) 1,110.277 in.4

= 2,170 lb/in.

( 47,000 lb ) (51.33 in.3 ) 1−1 = = 4,350 psi (1,110.277 in.4 ) ( 0.5 in.)

Ans. Ans.

(b) Shear flow and shear stress at section 2–2: Calculate Q for the top flange, starting at section 2–2 and extending rightward to the free end of the flange.  7.25 in. − 0.5 in.   18.5 in. 0.8 in.  3 Q2−2 =  −  ( 0.8 in.)   = 23.90 in. 2 2 2    

q2− 2 =

( 47, 000 lb ) ( 23.90 in.3 ) 1,110.277 in.4

= 1, 012 lb/in.

( 47,000 lb ) ( 23.90 in.3 )  2− 2 = = 1, 264 psi (1,110.277 in.4 ) ( 0.8 in.)

Ans. Ans.

(c) Shear flow and shear stress at section 3–3: Calculate Q for the top flange, starting at section 3–3 and extending leftward to the free end of the flange.  18.5 in. 0.8 in.  3 Q3−3 = (1.75 in.)( 0.8 in.)  −  = 12.39 in. 2 2  

q3−3 =

( 47, 000 lb ) (12.39 in.3 ) 1,110.277 in.4

= 524 lb/in.

( 47,000 lb ) (12.39 in.3 )  3−3 = = 656 psi (1,110.277 in.4 ) ( 0.8 in.)

Ans. Ans.

P9.36 A plastic extrusion has the shape shown in Figure P9.36 where b1 = 80 mm, b2 = 50 mm, d = 65 mm, and t = 2 mm. A vertical shear force of V = 600 N acts on the cross section. Determine the shear flow magnitude and the shear stress magnitude that acts in the shape: (a) at section 1–1. (b) at section 2–2. (c) at section 3–3 for a = 15 mm. FIGURE P9.36

Solution

Centroid location in y direction: Shape

Width b

Depth d

Area Ai

(1) (2) (3)

(mm) 80 50 2

(mm) 2 2 61

(mm2) 160.0 100.0 122.0 382.0

yi Ai Ai

14,305.0 mm3 = 37.448 mm 382.0 mm 2 = 27.552 mm

Moment of inertia about the z axis: Shape Width b Depth d Area A (1) (2) (3)

(mm) 80 50 2

(mm) 2 2 61

(mm2) 160.000 100.000 122.000

yi (from bottom)

yi Ai (mm3) 10,240.0 100.0 3,965.0 14,305.0

(mm) 64 1 32.5

(from bottom of shape to centroid) (from top of shape to centroid)

d = y − yi

d²A

IC + d²A

(mm4) 53.333 33.333 37,830.167

(mm) –26.552 36.448 4.948

(mm4) 112,804.418 132,843.075 2,986.460

(mm4) 112,857.751 132,876.408 40,816.627 286,550.786

(a) Shear flow and shear stress at section 1–1: Calculate Q for the bottom flange, starting at section 1–1 and extending leftward to the free end of the flange. 2 mm   50 mm − 2 mm   3 Q1−1 =   ( 2 mm )  37.448 mm −  = 1, 749.487 mm 2 2    

q1−1 =

( 600 N ) (1, 749.487 mm3 )

= 3.66 N/mm

Ans.

( 600 N ) (1,749.487 mm3 ) 1−1 = = 1,832 kPa ( 286,550.786 mm4 ) ( 2 mm)

Ans.

286,550.786 mm 4

(b) Shear flow and shear stress at section 2–2: Calculate Q for the top flange, starting at section 2–2 and extending rightward to the free end of the flange. 2 mm   80 mm − 2 mm   3 Q2−2 =   ( 2 mm )  27.552 mm −  = 2, 071.084 mm 2 2    

q2− 2 =

( 600 N ) ( 2, 071.084 mm3 )

= 4.34 N/mm

Ans.

( 600 N ) ( 2,071.084 mm3 )  2− 2 = = 2,168 kPa ( 286,550.786 mm4 ) ( 2 mm)

Ans.

286,550.786 mm 4

(c) Shear flow and shear stress at section 3–3: Calculate Q for the top flange, starting at section 3–3 and extending rightward to the free end of the flange. 2 mm   3 Q3−3 = (15 mm )( 2 mm )  27.552 mm −  = 796.571 mm 2  

q3−3 =

( 600 N ) ( 796.571 mm3 )

= 1.668 N/mm

Ans.

( 600 N ) ( 796.571 mm3 )  3−3 = = 834 kPa ( 286,550.786 mm4 ) ( 2 mm)

Ans.

286,550.786 mm 4

P9.37 Three plates are welded together to form the section shown in Figure P9.37. Dimensions of the cross section are b = 4 in., d = 12 in., a = 3 in., and t = 0.5 in. For a vertical shear force of V = 35 kips, determine the shear flow through the welded surface at B.

FIGURE P9.37

Solution

Centroid location in y direction: Shape

Width b

Depth d

Area Ai

(1) (2) (3)

(in.) 0.5 4 4

(in.) 12 0.5 0.5

(in.2) 6.000 2.000 2.000 10.000

yi Ai Ai

49.0 in.3 = 4.900 in. 10.0 in.2

yi (from bottom) (in.) 6 3.25 3.25

(1) (2) (3)

(in.) 12 0.5 0.5

(in.3) 36.000 6.500 6.500 49.000

(from bottom of shape to centroid)

= 7.100 in. (from top of shape to centroid) Moment of inertia about the z axis: d = y − yi Shape Width b Depth d Area A IC

(in.) 0.5 4 4

yi Ai

(in.2) 6.000 2.000 2.000

(in.4) 72.000 0.042 0.042

(in.) –1.100 1.650 1.650

d²A

IC + d²A

(in.4) 7.260 5.445 5.445

(in.4) 79.260 5.487 5.487 90.233

Shear flow through welded surface B: Calculate Q for plate AB, starting at B and extending leftward. 0.5 in.   3 Q = ( 4 in.)( 0.5 in.)  4.900 in. − 3 in. −  = 3.30 in. 2  

qB =

( 35,000 lb ) (3.30 in.3 ) 90.233 in.4

= 1, 280 lb/in.

Ans.

P9.38 An aluminum extrusion has the shape shown in Figure P9.38 where b = 36 mm, d = 50 mm, a = 12 mm, and t = 3 mm. A vertical shear force of V = 1,700 N acts on the cross section. Determine the shear flow that acts in the shape at points A–D.

Solution

FIGURE P9.38

Centroid location in y direction: Shape

Width b

Depth d

Area Ai

(1) (2) (3) (4) (5)

(mm) 3 3 3 13.5 13.5

(mm) 50 12 12 3 3

(mm2) 150.0 36.0 36.0 40.5 40.5 303.0

yi Ai Ai

4,303.5 mm3 = 14.203 mm 303.0 mm 2

= 35.797 mm Moment of inertia about the z axis: Shape Width b Depth d Area A

(1) (2) (3) (4) (5)

(mm) 3 3 3 13.5 13.5

(mm) 50 12 12 3 3

(mm2) 150.0 36.0 36.0 40.5 40.5

yi (from bottom)

yi Ai (mm3) 3,750.0 216.0 216.0 60.8 60.8 4,303.5

(mm) 25 6 6 1.5 1.5

(from bottom of shape to centroid) (from top of shape to centroid)

d = y − yi

d²A

IC + d²A

(mm4) 31,250.000 432.000 432.000 30.375 30.375

(mm) –10.797 8.203 8.203 12.703 12.703

(mm4) 17,486.378 2,422.394 2,422.394 6,535.301 6,535.301

(mm4) 48,736.378 2,854.394 2,854.394 6,565.676 6,565.676 67,576.517

Shear flow at A: Calculate Q for a portion of area (2), starting at A and extending upward to the free end of the area. 9 mm   3 QA = ( 3 mm )(12 mm − 3 mm) 14.203 mm − 3 mm −  = 180.980 mm 2  

qA =

(1, 700 N ) (180.980 mm3 ) 67,576.517 mm4

= 4.55 N/mm

Ans.

Shear flow at B: Calculate Q for area (2). 12 mm   3 QB = ( 3 mm )(12 mm ) 14.203 mm −  = 295.307 mm 2  

qB =

(1, 700 N ) ( 295.307 mm3 ) 67,576.517 mm4

= 7.43 N/mm

Ans.

Shear flow at C: Calculate Q for areas (2) and (4). 12 mm   QC = ( 3 mm )(12 mm ) 14.203 mm −  2   3 mm   3 + (13.5 mm )( 3 mm ) 14.203 mm −  = 809.777 mm 2  

qC =

(1, 700 N ) (809.777 mm3 ) 67,576.517 mm 4

= 20.4 N/mm

Ans.

Shear flow at D: Calculate Q for areas (2) through (5) plus the bottom portion of area (1) that starts are D and extends downward. 12 mm   QD = 2 ( 3 mm )(12 mm ) 14.203 mm −  2   3 mm   +2 (13.5 mm )( 3 mm ) 14.203 mm −  2   3 mm   3 + ( 3 mm )( 3 mm ) 14.203 mm −  = 1, 733.881 mm 2  

qD =

(1, 700 N ) (1, 733.881 mm3 ) 67,576.517 mm4

= 43.6 N/mm

Ans.

P9.39 An aluminum extrusion has the shape shown in Figure P9.39 where b = 60 mm, d = 90 mm, a = 20 mm, and t = 3 mm. A vertical shear force of V = 13 kN acts on the cross section. Determine the shear stress in the shape: (a) at point A. (b) at section 1–1. (c) at section 2–2. (d) at section 3–3. (e) at section 4–4. (f) at the neutral axis.

FIGURE P9.39

Solution

Moment of inertia about the z axis: Shape Width b Depth d Area A (1) (2) (3) (4) (5) (6) (7)

(mm) 3 3 3 3 3 51 51

(mm) 90 20 20 20 20 3 3

(mm2) 270.0 60.0 60.0 60.0 60.0 153.0 153.0

d = y − yi

d²A

IC + d²A

(mm4) 182,250.0 2,000.0 2,000.0 2,000.0 2,000.0 114.8 114.8

(mm) 0.000 –35.000 –35.000 35.000 35.000 –43.500 43.500

(mm4) 0.0 73,500.0 73,500.0 73,500.0 73,500.0 289,514.3 289,514.3

(mm4) 182,250.0 75,500.0 75,500.0 75,500.0 75,500.0 289,629.0 289,629.0 1,063,508.0

(a) Shear stress at point A: Point A is a free surface; thus, the shear stress at A is zero.  A = 0 kPa

Ans.

(b) Shear stress at section 1–1: Calculate Q for a portion of area (4), starting at section 1–1 and extending upward to the free end of the area. 17 mm   3 Q1−1 = ( 3 mm )( 20 mm − 3 mm )  45 mm − 3 mm −  = 1, 708.500 mm 2   Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

(13,000 N ) (1.708.500 mm3 ) 1−1 = = 6.96 MPa (1,063,508 mm4 ) (3 mm )

Ans.

(c) Shear stress at section 2–2: Calculate Q for area (2). 20 mm   3 Q2−2 = ( 3 mm )( 20 mm )  45 mm −  = 2,100 mm 2  

(13,000 N ) ( 2,100 mm3 )  2−2 = = 8.56 MPa (1,063,508 mm4 ) (3 mm )

Ans.

(d) Shear stress at section 3–3: Calculate Q for area (4) and half of area (7). 20 mm   Q3−3 = ( 3 mm )( 20 mm )  45 mm −  2   3 mm   51 mm   3 +  ( 3 mm )  45 mm −  = 5, 427.75 mm 2   2  

(13,000 N ) (5, 427.75 mm3 )  3−3 = = 22.1 MPa (1,063,508 mm4 ) (3 mm )

Ans.

(e) Shear stress at section 4–4: Calculate Q for areas (2), (3), and (6) plus the uppermost portion of area (1). 20 mm   Q4− 4 = 2 ( 3 mm )( 20 mm )  45 mm −  2   3 mm   + ( 51 mm )( 3 mm )  45 mm −  2   3 mm   3 + ( 3 mm )( 3 mm )  45 mm −  = 11, 247.00 mm 2  

(13,000 N ) (11, 247.00 mm3 )  4− 4 = = 45.8 MPa (1,063,508 mm4 ) (3 mm)

Ans.

(f) Shear stress at the neutral axis: Calculate Q for areas (2), (3), and (6) plus the upper half of area (1). 20 mm   QNA = 2 ( 3 mm )( 20 mm )  45 mm −  2   3 mm   + ( 51 mm )( 3 mm )  45 mm −  2    45 mm  3 + ( 3 mm )( 45 mm )   = 13,893.00 mm 2  

(13,000 N ) (13,893.00 mm3 )  max = = 56.6 MPa (1,063,508 mm4 ) (3 mm)

Ans.

P9.40 The U shape shown in Figure P9.41 has dimensions of b = 250 mm, d = 120 mm, tf = 19 mm, and tw = 11 mm. A vertical shear force of V = 130 kN acts on the cross section. Determine the shear stress at points A–E. Assume a = 80 mm.

FIGURE P9.40

Solution

Centroid location in y direction: Width Shape Depth d Area Ai b (mm) 19 19 212

(1) (2) (3)

yi Ai Ai

(mm) 120 120 11

yi (from bottom)

(mm2) 2,280.0 2,280.0 2,332.0 6,892.0

(mm) 60 60 5.5

Moment of inertia about the z axis: Depth Shape Width b Area A d (1) (2) (3)

(mm3) 136,800.0 136,800.0 12,826.0 286,426.0

286, 426.0 mm 3 = 41.5592 mm (from bottom of shape to centroid) 6,892.0 mm 2 = 78.4408 mm

(mm) 19 19 212

yi Ai

(mm) 120 120 11

(mm2) 2,280.0 2,280.0 2,332.0

(from top of shape to centroid)

d = y − yi

d²A

IC + d²A

(mm4) 2,736,000.0 2,736,000.0 23,514.3

(mm) –18.441 –18.441 36.059

(mm4) 775,344.0 775,344.0 3,032,219.9

(mm4) 3,511,344.0 3,511,344.0 3,055,734.3 10,078,422.2

(a) Shear stress at point A: Point A is a free surface; thus, the shear stress at A is zero.  A = 0 MPa

Ans.

Shear stress at point B: Calculate Q for a portion of area (2), starting at point B and extending upward to the free end of the area. 109 mm   3 QB = (19 mm )(120 mm − 11 mm )  78.4408 mm −  = 49,581.4 mm 2  

Note that we could also calculate QB in the following manner:  78.4408 mm  QB = (19 mm )( 78.4408 mm )   2    41.5592 mm − 11 mm  3 − (19 mm )( 41.5592 mm − 11 mm )   = 49,581.4 mm 2  

This approach to calculating Q treats area above the neutral axis as positive areas and areas below the neutral axis as negative areas. While this approach is not necessary to calculate QB (or QC), it will be necessary for points D and E. (130,000 N ) ( 49,581.4 mm3 ) Ans. B = = 33.7 MPa (10,078, 422.2 mm4 ) (19 mm) Shear stress at point C: Calculate Q for area (2). 120 mm   3 QC = (19 mm )(120 mm )  78.4408 mm −  = 42, 045.0 mm 2  

Note that we could also calculate QC in the following manner:  78.4408 mm  QC = (19 mm )( 78.4408 mm )   2    41.5592 mm  3 − (19 mm )( 41.5592 mm )   = 42, 045.0 mm 2  

C =

(130,000 N ) ( 42,045.0 mm3 ) = 49.3 MPa (10,078, 422.2 mm4 ) (11 mm)

Ans.

Shear stress at point D: Calculate Q for area (2) and the portion of area (3) between C and D. Here, we will begin with the value that we previously obtained for point C and then add in the first moment of the area (which will be negative) between C and D. 11 mm   3 QD = 42, 045.0 mm3 − (80 mm − 19 mm )(11 mm )  41.5592 mm −  = 17,849.3 mm 2  

(130,000 N ) (17,849.3 mm3 ) D = = 20.9 MPa (10,078, 422.2 mm4 ) (11 mm)

Ans.

Shear stress at point E: Calculate Q for area (2) and the portion of area (3) between C and E. Here, we will begin with the value that we previously obtained for point C and then add in the first moment of the area (which will be negative) between C and E. 11 mm   250 mm   3 QE = 42, 045.0 mm3 −  − 19 mm  (11 mm )  41.5592 mm −  = 0 mm 2 2    

 E = 0 MPa

Ans.

The shear stress on the vertical axis of symmetry must be zero by virtue of that symmetry.

P9.41 The channel shape shown in Figure P9.41 has dimensions of b = 6.0 in., d = 3.0 in., a = 1.5 in., c = 2.0 in., and t = 0.25 in. A vertical shear force of V = 47,000 lb acts on the cross section. Determine: (a) the shear flow that acts in the shape at points A–G. (b) the resultant horizontal force that acts between points A and B.

FIGURE P9.41

Solution

Centroid location in y direction: Shape

Width b

Depth d

Area Ai

(in.) 0.25 0.25 1.25 1.25 5.5

(in.) 3 3 0.25 0.25 0.25

(in.2) 0.750 0.750 0.313 0.313 1.375 3.500

(1) (2) (3) (4) (5)

yi Ai Ai

yi Ai (in.3) 1.125 1.125 0.898 0.898 0.172 4.219

4.219 in.3 = 1.2054 in. 3.50 in.2

(from bottom of shape to centroid)

= 1.7946 in.

(from top of shape to centroid)

Moment of inertia about the z axis: Shape Width b Depth d Area A (1) (2) (3) (4) (5)

yi (from bottom) (in.) 1.5 1.5 2.875 2.875 0.125

(in.) 0.25 0.25 1.25 1.25 5.5

(in.) 3 3 0.25 0.25 0.25

(in.2) 0.750 0.750 0.313 0.313 1.375

d = y − yi

d²A

IC + d²A

(in.4) 0.563 0.563 0.002 0.002 0.007

(in.) –0.295 –0.295 –1.670 –1.670 1.080

(in.4) 0.065 0.065 0.871 0.871 1.605

(in.4) 0.628 0.628 0.873 0.873 1.612 4.613

(a) Shear flow at point A: Point A is a free surface; thus, the shear flow at A is zero. qA = 0 kips/in.

Ans.

Shear flow at point B: For this cross section, we must be careful in dealing with areas both above and below the neutral axis. Areas above the neutral axis will be positive areas and areas below the neutral axis are negative areas. Calculate Q for area (4): 0.25 in.   3 QB = (1.5 in. − 0.25 in.)( 0.25 in.) 1.7946 in. −  = 0.52176 in. 2  

qB =

( 47, 000 lb ) ( 0.52176 in.3 ) 4.613 in.4

= 5.32 kips/in.

Ans.

Shear flow at point C: We will begin with our value for QB and add the first moment of the area between B and C. 0.25 in.   3 QC = 0.52176 in.3 + ( 0.25 in.)( 0.25 in.) 1.7946 in. −  = 0.62612 in. 2  

qC =

( 47, 000 lb ) ( 0.62612 in.3 ) 4.613 in.4

= 6.38 kips/in.

Ans.

Shear flow at point D: We will begin with our value for QB and add the first moment of the area between B and D, assigning a positive value to the area above the neutral axis and a negative value to the area below the N.A.  1.7946 in.  QD = 0.52176 in.3 + ( 0.25 in.)(1.7946 in.)   2    1.2054 in. − 0.25  3 − ( 0.25 in.)(1.2054 in. − 0.25 in.)   = 0.81027 in. 2  

qD =

( 47, 000 lb ) ( 0.81027 in.3 ) 4.613 in.4

= 8.26 kips/in.

Ans.

Shear flow at point E: We will begin with our value for QB and add the first moment of the area between B and E, assigning a positive value to the area above the neutral axis and a negative value to the area below the N.A.  1.7946 in.  QE = 0.52176 in.3 + ( 0.25 in.)(1.7946 in.)   2    1.2054 in.  3 − ( 0.25 in.)(1.2054 in.)   = 0.74275 in. 2  

qE =

( 47, 000 lb ) ( 0.74275 in.3 ) 4.613 in.4

= 7.57 kips/in.

Ans.

Shear flow at point F: We will begin with our value for QE and add the first moment of the area between E and F, assigning a positive value to the area above the neutral axis and a negative value to the area below the N.A. 0.25 in.   3 QF = 0.74275 in.3 − ( 2.0 in. − 0.25 in.)( 0.25 in.) 1.2054 in. −  = 0.27009 in. 2  

qF =

( 47, 000 lb ) ( 0.27009 in.3 ) 4.613 in.4

= 2.75 kips/in.

Ans.

Shear flow at point G: We will begin with our value for QE and add the first moment of the area between E and G, assigning a positive value to the area above the neutral axis and a negative value to the area below the N.A. 0.25 in.   6.0 in.   3 QG = 0.74275 in.3 −  − 0.25 in. ( 0.25 in.) 1.2054 in. −  = 0 in. 2   2  

qG = 0 kips/in.

Ans.

The shear stress on the vertical axis of symmetry must be zero by virtue of symmetry. (b) Resultant horizontal force between points A and B: Since area (4) is parallel to the neutral axis, the shear flow will be linearly distributed. The horizontal force between A and B is thus equal to the average of the shear flows at A and B times the length of the element between the two end points: q A + qB 0 + 5.31625 kips/in. (a − t ) = (1.5 in. − 0.25 in.) = 3.32 kips 2 2 This force is directed from B towards A. F=

Ans.

P9.42 The hat shape extrusion shown in Figure P9.42 has dimensions of b = 50 mm, d = 40 mm, a = 30 mm, and t = 2 mm. A vertical shear force of V = 600 N acts on the cross section. Determine: (a) the shear stress at points A–D. (b) the resultant horizontal force that acts between points A and B. FIGURE P9.42

Solution

Centroid location in y direction: Shape

Width b

Depth d

(mm) 2 2 28 28 46

(mm) 40 40 2 2 2

(1) (2) (3) (4) (5)

yi Ai Ai

Area Ai (mm2) 80.0 80.0 56.0 56.0 92.0 364.0

7, 660.0 mm3 = 21.044 mm 364.0 mm 2 = 18.956 mm

Moment of inertia about the z axis: Shape Width b Depth d Area A (1) (2) (3) (4) (5)

(mm) 2 2 28 28 46

(mm) 40 40 2 2 2

(mm2) 80.0 80.0 56.0 56.0 92.0

yi (from bottom) (mm) 20 20 39 39 1

yi Ai (mm3) 1,600.0 1,600.0 2,184.0 2,184.0 92.0 7,660.0

(from bottom of shape to centroid) (from top of shape to centroid)

d = y − yi

d²A

IC + d²A

(mm4) 10,666.7 10,666.7 18.7 18.7 30.7

(mm) 1.044 1.044 –17.956 –17.956 20.044

(mm4) 87.2 87.2 18,055.5 18,055.5 36,961.9

(mm4) 10,753.9 10,753.9 18,074.2 18,074.2 36,992.6 94,648.6

(a) Shear stress at point A: Point A is a free surface; thus, the shear stress at A is zero.  A = 0 MPa

Ans.

Shear stress at point B: For this cross section, we must be careful in dealing with areas both above and below the neutral axis. Areas above the neutral axis will be positive areas and areas below the neutral axis are negative areas. Calculate Q for area (4): 2 mm   3 QB = ( 28 mm )( 2 mm ) 18.956 mm −  = 1, 005.536 mm 2  

( 600 N ) (1,000.536 mm3 ) B = = 3.19 MPa (94,648.6 mm4 ) ( 2 mm)

Ans.

Shear stress at point C: We will begin with our value for QB and add the first moment of area (2), assigning a positive value to the area above the neutral axis and a negative value to the area below the N.A.:  18.956 mm  QC = 1, 005.536 mm 3 + ( 2 mm )(18.956 mm )   2    21.044 mm  3 − ( 2 mm )( 21.044 mm )   = 922.016 mm 2  

( 600 N ) (922.016 mm3 ) C = = 2.92 MPa (94,648.6 mm4 ) ( 2 mm)

Ans.

Shear stress at point D: We will begin with our value for QC and add the first moment of half of area (5), assigning a negative value to the area below the N.A.: 2 mm   46 mm   3 QD = 922.016 mm3 −   ( 2 mm )  21.044 mm −  = 0 mm 2 2    

 D = 0 MPa

Ans.

( 600 N ) (1, 000.536 mm3 )

= 6.343 N/mm 94, 648.6 mm 4 q + qB 0 + 6.343 N/mm F= A (a − t ) = ( 30 mm − 2 mm ) = 88.8 N 2 2

Ans.

The resultant force will be directed from B towards A.

P9.43 Sheet metal is bent into the shape shown in Figure P9.43 and then used as a beam. The shape has dimensions of b = 4.5 in., d = 7.0 in., a = 1.5 in., and t = 0.14 in. A vertical shear force of V = 6,200 lb acts on the cross section. Determine the shear stress that acts in the shape at points A–G.

Solution

FIGURE P9.43

Centroid location in y direction: Shape

Width b

Depth d

Area Ai

(1) (2) (3) (4) (5) (6) (7)

(in.) 0.14 0.14 0.14 0.14 4.22 1.22 1.22

(in.) 7 7 6.86 6.86 0.14 0.14 0.14

(in.2) 0.980 0.980 0.960 0.960 0.591 0.171 0.171 4.813

yi Ai Ai

yi (from bottom) (in.) 3.5 3.5 3.43 3.43 6.93 0.07 0.07

yi Ai (in.3) 3.430 3.430 3.294 3.294 4.094 0.012 0.012 17.567

17.567 in.3 = 3.6497 in. 4.813 in.2

(from bottom of shape to centroid)

= 3.3503 in.

(from top of shape to centroid)

Moment of inertia about the z axis: Shape Width b Depth d Area A (1) (2) (3) (4) (5) (6) (7)

(in.) 0.14 0.14 0.14 0.14 4.22 1.22 1.22

(in.) 7 7 6.86 6.86 0.14 0.14 0.14

(in. ) 0.980 0.980 0.960 0.960 0.591 0.171 0.171

IC 4

(in. ) 4.002 4.002 3.766 3.766 0.001 0.000 0.000

d = y − yi

(in.) 0.150 0.150 0.220 0.220 –3.280 3.580 3.580

d²A

IC + d²A

(in.4) 4.024 4.024 3.813 3.813 6.358 2.189 2.189 26.409

(in. ) 0.022 0.022 0.046 0.046 6.357 2.189 2.189

Calculation strategy: At first, this shape seems very confusing to analyze. However, we will approach the calculations in a systematic manner that will make it very manageable. The shape is symmetrical about the y axis. Thus, we need only consider half of the shape in our calculations; for example, only the right half of the shape from point A to point G. However, we still use V = 6,200 lb and Iz = 26.409 in.4 in our calculations. We will begin at point A, a free surface, where we know with certainty that Q = 0 and  = 0. We will proceed in order from A to G, summing the first moment of area terms (i.e., the Q terms) as we go. Areas below the neutral axis will be treated as negative quantities. When calculating the shear stress, we will use the absolute values of Q in each instance since we are only interested in the magnitude of the shear stress, not the magnitude and direction. Shear stress at point A: Point A is a free surface; thus, the shear stress at A is zero.  A = 0 MPa

Ans.

Shear stress at point B: Add the Q term for the portion of area (4) from B to A:

( 7.0 in. − 0.14 in. − 3.6497 in.) = 0.7214 in.3 QB = 0 + ( 0.14 in.) 2

( 6, 200 lb ) ( 0.7214 in. ) = 1, 210 psi ( 26.409 in.4 ) ( 0.14 in.) 3

B =

Ans.

Shear stress at point C: Add the Q term for the portion of area (4) from C to B:

( 3.6497 in. − 0.14 in.) = −0.1409 in.3 QC = 0.7214 in. − ( 0.14 in.) 2

( 6, 200 lb ) ( 0.1409 in. ) = 236 psi ( 26.409 in.4 ) ( 0.14 in.) 3

C =

Ans.

Shear stress at point D: Add the Q term for area (7) and portions of areas (4) and (2): 0.14 in.   3 QD = −0.1409 in.3 − (1.5 in.)( 0.14 in.)  3.6497 in. −  = −0.8925 in. 2  

D =

( 6, 200 lb ) ( 0.8926 in.3 ) = 1, 497 psi ( 26.409 in.4 ) ( 0.14 in.)

Ans.

Shear stress at point E: Add the Q term for area (2) between points D and E: QE = −0.8925 in.3 − ( 0.14 in.)

( 3.6497 in. − 0.14 in.) = −1.7548 in.3 2

( 6, 200 lb ) (1.7548 in. ) = 2,940 psi ( 26.409 in.4 ) ( 0.14 in.) 3

E =

Ans.

Shear stress at point F: Add the Q term for area (2) between points E and F: QF = −1.7548 in.3 + ( 0.14 in.)

( 3.3503 in.) = −0.9690 in.3 2

( 6, 200 lb ) ( 0.9690 in. ) = 1,625 psi ( 26.409 in.4 ) ( 0.14 in.) 3

F =

Ans.

Shear stress at point G: Add the Q term for area (5) between points F and G. By virtue of symmetry, Q should be equal to zero at G. 0.14 in.   4.5 in.   3 QG = −0.9690 in.3 +  − 0.14 in.  ( 0.14 in.)  3.3503 in. −  = 0 in. 2   2  

 G = 0 psi

Ans.

P9.44 Sheet metal is bent into the shape shown in Figure P9.44 and then used as a beam. The shape has dimensions of b = 80 mm, d = 60 mm, a = 20 mm, and t = 3 mm. A vertical shear force of V = 5,400 N acts on the cross section. Determine the shear flow at points A–E.

Solution

FIGURE P9.44

Centroid location in y direction: Shape

Width b

Depth d

Area Ai

(1) (2) (3) (4) (5) (6) (7)

(mm) 3 3 3 3 14 14 74

(mm) 60 60 60 60 3 3 3

(mm2) 180.0 180.0 180.0 180.0 42.0 42.0 222.0 1,026.0

yi Ai Ai

26,847.0 mm 3 = 26.167 mm 1,026.0 mm 2 = 33.833 mm

yi (from bottom) (mm) 30 30 30 30 58.5 58.5 1.5

yi Ai (mm3) 5,400.0 5,400.0 5,400.0 5,400.0 2,457.0 2,457.0 333.0 26,847.0

(from bottom of shape to centroid) (from top of shape to centroid)

Moment of inertia about the z axis: Shape Width b Depth d Area A (1) (2) (3) (4) (5) (6) (7)

(mm) 3 3 3 3 14 14 74

(mm2) 180.0 180.0 180.0 180.0 42.0 42.0 222.0

(mm) 60 60 60 60 3 3 3

d = y − yi

d²A

IC + d²A

(mm4) 54,000.0 54,000.0 54,000.0 54,000.0 31.5 31.5 166.5

(mm) –3.833 –3.833 –3.833 –3.833 –32.333 –32.333 24.667

(mm4) 2,645.0 2,645.0 2,645.0 2,645.0 43,908.7 43,908.7 135,074.7

(mm4) 56,645.0 56,645.0 56,645.0 56,645.0 43,940.2 43,940.2 135,241.2 449,701.5

Calculation strategy: At first, this shape seems very confusing to analyze. However, we will approach the calculations in a systematic manner that will make it very manageable. The shape is symmetrical about the y axis. Thus, we need only consider half of the shape in our calculations; for example, only the left half of the shape from point A to point E. However, we still use V = 5,400 N and Iz = 449,701.5 mm4 in our calculations. We will begin at the lower end of area (1), a free surface, where we know with certainty that Q = 0 and q = 0. We will proceed in order from A to E, summing the first moment of area terms (i.e., the Q terms) as we go. Areas below the neutral axis will be treated as negative quantities. When calculating the shear flow, we will use the absolute values of Q in each instance since we are only interested in the magnitude of the shear flow. Shear flow at point A: Calculate Q for the portion of area (1) below the neutral axis:

( 26.167 mm ) = −1, 027.042 mm3 QA = − ( 3 mm ) 2

qA =

( 5, 400 N ) (1, 027.042 mm3 ) 449, 701.5 mm 4

= 12.33 N/mm

Ans.

Shear flow at point B: Add the Q term for the upper portion of area (1) and for area (5): QB = −1, 027.042 mm3 + ( 3 mm )

( 33.833 mm )

2 3 mm   3 +  20 mm − 2 ( 3 mm )  ( 3 mm )  33.833 mm −  = 2, 048.000 mm 2  

qB =

( 5, 400 N ) ( 2, 048.000 mm3 ) 449, 701.5 mm 4

= 24.6 N/mm

Ans.

Shear flow at point C: Add the Q term for the upper portion of area (2):

( 33.833 mm ) = 3, 765.042 mm3 QC = 2, 048.000 mm + ( 3 mm ) 2

( 5, 400 N ) ( 3, 765.042 mm )

qC =

449, 701.5 mm 4

= 45.2 N/mm

Ans.

Shear flow at point D: Add the Q term for the lower portion of area (2): QD = 3, 765.042 mm3 − ( 3 mm )

( 26.167 mm ) = 2, 738.000 mm3 2

( 5, 400 N ) ( 2, 738.000 mm )

qD =

449, 701.5 mm4

= 32.9 N/mm

Ans.

Shear flow at point E: Add the Q term for area (7) between points D and E. By virtue of symmetry, Q should be equal to zero at E. 3 mm   80 mm   3 QE = 2, 738.000 mm3 −  − 3 mm  ( 3 mm )  26.167 mm −  = 0 mm 2   2  

qE = 0 N/mm

Ans.

P9.45 Sheet metal is bent into the shape shown in Figure P9.45 and then used as a beam. The shape has centerline dimensions of a = 120 mm, b = 80 mm, and t = 3 mm. A vertical shear force of V = 6,400 N acts on the cross section. Determine the shear stress magnitude that acts in the shape at points A–D.

FIGURE P9.45

Solution

where b sin 30 = ( 80 mm ) sin 30 = 40 mm and b cos 30 = ( 80 mm ) cos 30 = 69.2820 mm

(1) (2) (3) (4)

yi Ai Ai

(mm) 120 40 40 120

(mm2) 360.0 240.0 240.0 360.0 1,200.0

81, 600.0 mm3 = 68.0 mm 1,200.0 mm 2 = 92.0 mm

(mm) 100 20 20 100

(mm3) 36,000.0 4,800.0 4,800.0 36,000.0 81,600.0

(from bottom of shape to centroid) (from top of shape to centroid)

Moment of inertia about the z axis: The moment of inertia of a parallelogram [i.e., areas (2) and (3)] about its centroid is bd3/12 where b will be 6 mm. d = y − yi Shape Width b Depth d Area A IC d²A IC + d²A (1) (2) (3) (4)

(mm) 3 6 6 3

(mm2) 360.0 240.0 240.0 360.0

(mm) 120 40 40 120

(mm4) 432,000.0 32,000.0 32,000.0 432,000.0

(mm) –32.000 48.000 48.000 –32.000

(mm4) 368,640.0 552,960.0 552,960.0 368,640.0

Shear stress at point A: Point A is a free surface; thus, the shear stress at A is zero.  A = 0 MPa

(mm4) 800,640.0 584,960.0 584,960.0 800,640.0 2,771,200.0

Ans.

Shear stress at point B: Calculate Q for the portion of area (1) above the neutral axis: QB = ( 3 mm )( 92 mm )

( 92 mm ) = 12, 696 mm3 2

( 6, 400 N ) (12,696 mm3 ) B = = 9.77 MPa ( 2,771, 200 mm4 ) (3 mm)

Ans.

Shear stress at point C: Add Q for the portion of area (1) below the neutral axis:

(120 mm − 92 mm ) = 11,520 mm3 QC = 12, 696 mm − ( 3 mm ) 2

( 6, 400 N ) (11,520 mm ) = 8.87 MPa ( 2,771, 200 mm4 ) (3 mm) 3

C =

Ans.

Shear stress at point D: Add Q for area (2), which is below the neutral axis. By virtue of symmetry, Q should be equal to zero at D.: 40 mm   3 QD = 11,520 mm3 − ( 6 mm )( 40 mm )  68 mm −  = 0 mm 2  

 D = 0 MPa

Ans.

P9.46 Sheet metal is bent into the shape shown in Figure P9.46 and then used as a beam. The shape has centerline dimensions of b = 3.0 in., d = 4.5 in., a = 1.8 in., and t = 0.12 in. A vertical shear force of V = 480 lb acts on the cross section. Determine: (a) the shear stress that acts in the shape at points A–D. (b) the resultant horizontal force acting on the cross-sectional element between A and B.

FIGURE P9.46

Solution

Centroid location in y direction: Since the shape has a very thin wall, we will approximate the sheet metal shape as shown above. We will treat areas (2) and (3) as parallelograms, each having a width of 0.1442 in. and a depth of d = 4.5 in. yi Shape Width b Depth d Area Ai yi Ai (from bottom) 2 (in.) (in.) (in. ) (in.) (in.3) (1) 1.8 0.12 0.2160 4.5 0.9720 (2) 0.1442 4.5 0.6489 2.25 1.4600 (3) 0.1442 4.5 0.6489 2.25 1.4600 (4) 1.8 0.12 0.2160 4.5 0.9720 1.7298 4.8641 y=

yi Ai Ai

4.8641 in.3 = 2.8119 in. 1.7298 in.2 = 1.6881 in.

(from bottom of shape to centroid) (from top of shape to centroid)

Moment of inertia about the z axis: The moment of inertia of a parallelogram [i.e., areas (2) and (3)] about its centroid is bd3/12 where b will be 0.1442 in. d = y − yi Shape Width b Depth d Area A IC d²A IC + d²A (1) (2) (3) (4)

(in.) 1.8000 0.1442 0.1442 1.8000

(in.) 0.1200 4.5000 4.5000 0.1200

(in.2) 0.2160 0.6489 0.6489 0.2160

(in.4) 0.0003 1.0950 1.0950 0.0003

(in.) –1.6881 0.5619 0.5619 –1.6881

(in.4) 0.6155 0.2049 0.2049 0.6155

(a) Shear stress at point A: Point A is a free surface; thus, the shear stress at A is zero.  A = 0 psi

(in.4) 0.6158 1.2999 1.2999 0.6158 3.8314

Ans.

Shear stress at point B: Calculate Q for area (1):

QB = (1.8 in.)( 0.12 in.)(1.6881 in.) = 0.3646 in.3

( 480 lb ) ( 0.3646 in.3 ) B = = 381 psi (3.8314 in.4 ) ( 0.12 in.)

Ans.

Shear stress at point C: Add Q for the portion of area (2) above the neutral axis:  1.6881 in.  3 QC = 0.3646 in.3 + ( 0.1442 in.)(1.6881 in.)   = 0.5701 in. 2  

( 480 lb ) ( 0.5701 in.3 ) C = = 595 psi (3.8314 in.4 ) ( 0.12 in.)

Ans.

Shear stress at point D: Add Q for that portion of area (2) which is below the neutral axis. By virtue of symmetry, Q should be equal to zero at D.:  2.8119 in.  3 QD = 0.5701 in.3 − ( 0.1442 in.)( 2.8119 in.)   = 0 in. 2  

 D = 0 psi

Ans.

(b) Resultant horizontal force acting between A and B: The shear flow at A is zero since A is a free surface. The shear flow at B is: ( 480 lb ) 0.3646 in.3 qB = = 45.6779 lb/in. Ans. 3.8314 in.4

(

)

Since area (1) is parallel to the neutral axis, the shear flow will be linearly distributed. The horizontal force between A and B is thus equal to the average of the shear flows at A and B times the length of the element between the two end points: q + qB 45.6779 lb/in. F= A a= Ans. (1.8 in.) = 41.1 lb 2 2 The force will be directed from B towards A.

P9.47 The thin-walled sheet pile cross section shown in Figure P9.47 is subjected to a vertical shear force of V = 175 kN. For the sheet pile cross section, the centerline dimensions are b = 160 mm, d = 350 mm, a = 120 mm, and t = 12 mm. Determine: (a) the shear stress in the shape at points A–E. (b) the resultant horizontal force acting on the cross-sectional element between A and B.

FIGURE P9.47

Solution

Centroid location in y direction: Since the shape has a very thin wall, we will approximate the sheet pile as shown above. We will treat areas (3) and (4) as parallelograms, each having a width of 12.686 mm and a depth of d = 350 mm. Our reference axis for the centroid calculation will be the horizontal centerline of areas (1) and (2). yi Width Shape Depth d Area Ai yi Ai b (from bottom) (1) (2) (3) (4) (5) (6)

(mm) 160 160 12.686 12.686 160 160

(mm) 12 12 350 350 12 12

(mm2) 1,920.0 1,920.0 4,440.1 4,440.1 1,920.0 1,920.0 16,560.2

yi Ai

2,898, 035 mm 3 y= = = 175.0 mm Ai 16,560.2 mm 2 = 175.0 mm

(mm) 0 0 175 175 350 350

(mm3) 0.0 0.0 777,017.5 777,017.5 672,000.0 672,000.0 2,898,035.0

(from bottom of shape to centroid) (from top of shape to centroid)

Moment of inertia about the z axis: The moment of inertia of a parallelogram [i.e., areas (3) and (4)] about its centroid is bd3/12 where b will be 0.1442 in. d = y − yi Shape Width b Depth d Area A IC d²A IC + d²A (1) (2) (3) (4) (5) (6)

(mm) 160 160 12.686 12.686 160 160

(mm) 12 12 350 350 12 12

(mm2) 1,920.0 1,920.0 4,440.1 4,440.1 1,920.0 1,920.0

(mm4) 23,040.0 23,040.0 45,326,020.8 45,326,020.8 23,040.0 23,040.0

(mm) 175.000 175.000 0.000 0.000 –175.000 –175.000

(mm4) 58,800,000.0 58,800,000.0 0.0 0.0 58,800,000.0 58,800,000.0

(a) Shear stress at point A: Point A is a free surface; thus, the shear stress at A is zero.  A = 0 MPa

(mm4) 58,823,040.0 58,823,040.0 45,326,020.8 45,326,020.8 58,823,040.0 58,823,040.0 325,944,201.7

Ans.

Shear stress at point B: Calculate Q for area (1). Since area (1) is below the neutral axis, we will treat it as a negative value. QB = − (160 mm)(12 mm)(175 mm) = −336,000 mm3

(175,000 N ) (336,000 mm3 ) B = = 15.03 MPa (325,944, 201.7 mm4 ) (12 mm )

Ans.

Shear stress at point C: Add Q for the portion of area (3) below the neutral axis:  175 mm  3 QC = −336, 000 mm3 − (12.686 mm )(175 mm )   = −530, 250 mm 2  

(175,000 N ) (530, 250 mm3 ) C = = 23.7 MPa (325,944, 201.7 mm4 ) (12 mm)

Ans.

Shear stress at point D: Add Q for that portion of area (3) which is above the neutral axis.  175 mm  3 QD = −530, 250 mm3 + (12.686 mm)(175 mm)   = −336, 000 mm 2  

(175,000 N ) (336,000 mm3 ) D = = 15.03 MPa (325,944, 201.7 mm4 ) (12 mm)

Ans.

Shear stress at point E: Add Q for area (5) which is above the neutral axis. By virtue of symmetry, Q should be equal to zero at E. QE = −336,000 mm3 + (160 mm)(12 mm)(175 mm) = 0 mm3

 E = 0 MPa

Ans.

(b) Resultant horizontal force acting between A and B: The shear flow at A is zero since A is a free surface. The shear flow at B is: (175, 000 N ) ( 336, 000 mm3 ) qB = = 180.400 N/mm 325,944, 201.7 mm 4 Since area (1) is parallel to the neutral axis, the shear flow will be linearly distributed. The horizontal force between A and B is thus equal to the average of the shear flows at A and B times the length of the element between the two end points: q + qB 180.400 N/mm Ans. F= A a= (160 mm ) = 14, 432 N = 14.43 kN 2 2 The force will be directed from B towards A.

P9.48 The beam cross section shown in Figure P9.44 is subjected to a shear force of V = 12 kips. Dimensions of the cross section are b = 18 in., d = 10 in., and t = 0.4 in. Using a = 3 in., calculate the shear stress magnitude at sections 1–1, 2–2, and 3–3.

FIGURE P9.48

Solution Moment of inertia about the z axis: Shape Width b Depth d (1) (2)

(in.) 18.0 17.2

(in.) 10.0 9.2

IC (in.4) 1500.0000 –1116.1195 383.8805

Shear stress at section 1–1: We will calculate Q for areas (1) and (2) of the box shape:  10 in. 0.4 in.  Q1−1 = 2 ( 3 in.)( 0.4 in.)  −  2   2

= 2 ( 5.760 in.3 ) = 11.52 in.3

(12,000 lb ) (11.52 in.3 ) 1−1 = = 450 psi (383.8805 in.4 ) ( 2  0.4 in.)

Ans.

Shear stress at section 2–2: We will calculate Q for areas (3) and (4) of the box shape:  10 in. 0.4 in.  Q2−2 = 2 ( 6 in.)( 0.4 in.)  −  2   2

= 2 (11.52 in.3 ) = 23.04 in.3

(12,000 lb ) ( 23.04 in.3 )  2− 2 = = 900 psi (383.8805 in.4 ) ( 2  0.4 in.)

Ans.

Shear stress at section 3–3: We will calculate Q for areas (5) and (6) of the box shape:  10 in. 0.4 in.  Q3−3 = 2 ( 8.6 in.)( 0.4 in.)  −  2   2

= 2 (16.512 in.3 ) = 33.024 in.3

(12,000 lb ) (33.024 in.3 )  3−3 = = 1, 290 psi (383.8805 in.4 ) ( 2  0.4 in.)

Ans.

P9.49 The beam cross section shown in Figure P9.49 is subjected to a shear force of V = 12 kips. Dimensions of the cross section are b = 4 in., d = 7 in., a = 2 in., and t = 0.25 in. Calculate the shear stress magnitude at sections 1–1, 2–2, and 3–3. Calculate the maximum shear stress.

FIGURE P9.49

Solution

Moment of inertia about the z axis: Shape Width b Depth d Area A

d = y − yi

d²A

IC + d²A

Outer rectangle Inner rectangle (1) (2)

(in.)

(in.2)

(in.4)

(in.)

(in.4)

4.00

7.00

28.0000

114.3333

3.50

6.50

–22.7500

–80.0990

0.25 0.25

1.75 1.75

0.4375 0.4375

0.1117 0.1117

–2.3750 2.3750

2.4678 2.4678

2.5794 2.5794 39.3932

Shear stress at section 1–1: Calculate Q for area (3) of the shape: 1.75 in.   7 in. Q1−1 = ( 0.25 in.)( 2.0 in. − 0.25 )  − 0.25 in. −  2   2

= 1.0391 in.3

1−1 =

(12 kips ) (1.0391 in.3 ) = 1.266 ksi (39.3932 in.4 ) ( 0.25 in.)

Ans.

Shear stress at section 2–2: Calculate Q for area (4) of the shape:  7 in. 2.0 in.  Q2−2 = ( 0.25 in.)( 2.0 in.)  −  2   2

= 1.25 in.3

(12 kips ) (1.25 in.3 )  2− 2 = = 0.762 ksi (39.3932 in.4 ) ( 2  0.25 in.)

Ans.

Shear stress at section 3–3: Calculate Q for areas (5) and (6) of the shape:  7 in. 0.25 in.  Q3−3 =  4.0 in. − 2 ( 0.25 in.)  ( 0.25 in.)  −  + Q1−1 2   2

= 2.9531 in.3 + 1.0391 in.3 = 3.9922 in.3

(12 kips ) ( 3.9922 in.3 )  3− 3 = = 2.43 ksi (39.3932 in.4 ) ( 2  0.25 in.)

Ans.

P9.50 An extruded plastic beam with the cross section shown in Figure P9.50 is subjected to a vertical shear force of V = 850 N. Centerline dimensions of the cross section are a = 20 mm, b = 40 mm, d = 50 mm, and t = 3 mm. Determine the shear stress magnitude that acts in the shape at points A through E.

FIGURE P9.50

Solution

(1) (2) (3) (4)

yi Ai Ai

(mm) 3 3 50 50

(mm2) 240.0 120.0 161.6 161.6 683.1

(mm) 50 0 25 25

(mm3) 12,000.0 0.0 4,038.8 4,038.8 20,077.5

20, 077.5 mm3 = 29.3917 mm 683.1 mm 2

[from centerline of shape (2) to centroid]

= 20.6083 mm

[from centroid to centerline of shape (1)]

Moment of inertia about the z axis: The moment of inertia of a parallelogram [i.e., areas (3) and (4)] about its centroid is bd3/12 where b will be 3.231 mm. d = y − yi Shape Width b Depth d Area A IC d²A IC + d²A (1) (2) (3) (4)

(mm) 80 40 3.231 3.231

(mm) 3 3 50 50

(mm2) 240.0 120.0 161.6 161.6

(mm4) 180.0 90.0 33,656.3 33,656.3

(mm) -20.608 29.392 4.392 4.392

(mm4) 101,928.1 103,665.0 3,115.9 3,115.9

(mm4) 102,108.1 103,755.0 36,772.1 36,772.1 279,407.3

Shear stress at point A: Point A lies on an axis of symmetry; thus, the shear stress at A is zero.  A = 0 MPa

Ans.

Shear stress at point B: Calculate Q for areas (5) and (6) of the trapezoidal box shape:

QB = 2 ( 40 mm )( 3 mm )( 20.6083 mm )

= 2 ( 2, 472.998 mm3 ) = 4,945.9968 mm3

(850 N ) ( 4,945.9968 mm3 ) B = = 2,510 kPa ( 279, 407.3 mm4 ) ( 2  3 mm )

Ans.

Shear stress at point C: Add Q for areas (7) and (8) to the value of Q obtained for point B:

QC = 4,945.9968 mm3  20.6083 mm  +2 ( 3.231 mm )( 20.6083 mm )   2  

= 4,945.9968 mm3 + 2 ( 686.1061 mm3 ) = 6,318.2538 mm3

(850 N ) ( 6,318.2538 mm3 ) C = = 3, 200 kPa ( 279, 407.3 mm4 ) ( 2  3 mm )

Ans.

Shear stress at point D: Add Q for areas (9) and (10) to the value of Q obtained for point C. Note that these areas are negative since they are located below the neutral axis. QD = 6,318.2538 mm3

 29.3917 mm  −2 ( 3.231 mm )( 29.3917 mm )   2  

= 6,318.2538 mm3 + 2 ( −1,395.5853 mm3 ) = 3,527.0016 mm3

(850 N ) (3,527.0016 mm3 ) D = = 1,788 kPa ( 279, 407.3 mm4 ) ( 2  3 mm)

Ans.

Shear stress at point E: Add Q for areas (11) and (12) to the value of Q obtained for point D. Note that these areas are negative since they are located below the neutral axis. Since point E lies on an axis of symmetry, Q should equal zero. QE = 3,527.0016 mm3

−2 ( 20 mm )( 3 mm )( 29.3917 mm )

= 3,527.0016 mm3 + 2 ( −1, 763.5020 mm3 ) = 0 mm3

 E = 0 kPa

Ans.

P9.51 The angle shown in Figure P9.51 is subjected to a vertical shear force of V = 3.5 kips. Sketch the distribution of shear flow along the leg AB. Indicate the numerical value at all peaks.

FIGURE P9.51

Solution Moment of inertia about the neutral axis: Recognizing that the wall thickness is thin, the moment of inertia for the channel shape can be calculated as: 0.25 in. b= = 0.3535534 in. sin 45 h = (5 in.) cos 45 = 3.5355339 in.  (0.3535534 in.)(3.5355339 in.)3  I NA = 2  = 2.6041667 in.4  12  

First moment of area Q:

 y  sin 45  y   Q = yA =  2.5 in. − + y  (0.25 in.)  2.5 in. −   sin 45  2 sin 45      y   (2.5 in.)sin 45 y   = − + y  (0.25 in.)  2.5 in. −  2 2 sin 45      y  = 0.8838835 in. +  0.625 in.2 − (0.3535534 in.)y  2  = 0.5524272 in.3 + (0.3125 in.2 ) y − (0.3125 in.2 ) y − (0.1767767 in.)y 2 = 0.5524272 in.3 − (0.1767767 in.)y 2

Shear flow: VQ q= I (3,500 lb) 0.5524272 in.3 − (0.1767767 in.)y 2 = 2.6041667 in.4

(

= 742.4621473 lb/in. − (237.5878818 lb/in.3 ) y 2

) Ans.

At y = 0, q = qmax = 742 lb/in.

Ans.

P9.52 The vertical shear force V acts on the thin-walled section shown in Figure P9.52. Sketch the shear flow diagram for the cross section. Assume that the wall thickness of the section is constant.

FIGURE P9.52

Solution Section Properties: y = r cos 

dA = t ds = t rd

dI = y dA = r cos  (trd ) = r 3t sin 2  d 2

2

I = r 3t  cos 2  d 0

2  cos 2 + 1 = r 3t    d 0  2

=  r 3t

Determine an expression for Q. dQ = ydA = r cos  (trd ) = r 2t cos  d 

Q = r 2t  cos  d = r 2t sin  0

Shear Flow: VQ Vr 2t sin  V q= = = sin  I  r 3t r V qmax = q  =  = r 2

P9.53 Determine the location of the shear center O for the cross section shown in Figure P9.53.

Solution Moment of inertia about the neutral axis: Recognizing that the wall thickness is thin, the moment of inertia for the channel shape can be calculated as: 2  (0.25 in.)(6 in.)3  6 in.   4 I NA = + 2 (3 in.)(0.25 in.)    = 18 in. 12  2   

FIGURE P9.53

To help the student visualize how the shear stress flows through the shape, the cross-sectional sketch below shows the shear stress resultants Ff and Fw acting in directions consistent with the direction of shear force V. However, the student is referred to Section 9.10 and in particular to Figures 9.26 and 9.27 to better understand the twisting effect created by the shear stress resultants. To determine the shear center location, it is sufficient to simply require that the magnitude of the external torsional moment created by Ve must equal the magnitude of the internal torsional moment created by the shear stress resultants. Shear flow in area (1): We will choose to sum moments about point A so that we need only determine one force: Ff. Begin by calculating the shear flow in the upper flange.  6 in.  3 Q = ( 3 in.)( 0.25 in.)   = 2.25 in.  2  Let the shear force V be equal to the moment of inertia INA, and express the shear flow q at the junction of the upper flange and the web as. 3 VQ (18 lb ) ( 2.25 in. ) q= = = 2.25 lb/in. I 18 in.4 Calculate the horizontal force Ff in the flange as: q 2.25 lb/in. Ff = ( 3 in.) = ( 3 in.) = 3.375 lb 2 2 Shear Center: Sum moments about point A to find the shear center: Ve = ( 6 in.) Ff e=

( 6 in.)( 3.375 lb ) = 1.125 in. 18 lb

Ans.

P9.54 Determine the location of the shear center O for the cross section shown in Figure P9.54. Assume a uniform thickness of t = 4 mm for all portions of the cross section. Use a = 70 mm, b = 40 mm, and c = 90 mm.

FIGURE P9.54

Solution (a) Moment of inertia about the neutral axis: Recognizing that the wall thickness is thin, the moment of inertia for the shape can be calculated as: 3 2  4 mm )(170 mm ) (  90 mm   I NA = + 2 ( 70 mm )( 4 mm )    12  2    = 1, 637, 666.67 mm 4 + 1,134, 000 mm 4 = 2, 771, 666.67 mm 4

Shear flow in area (1): We will choose to sum moments about point B so that we need only determine one force: F2. Begin by calculating the shear flow at point A.  90 mm  3 QA = ( 70 mm )( 4 mm )   = 12,600 mm  2  Let the shear force V be equal to the moment of inertia INA, and express the shear flow q at point A as. 3 VQA ( 2, 771, 666.67 N ) (12, 600 mm ) qA = = = 12, 600 N/mm I 2, 771, 666.67 mm 4 Calculate the horizontal force F2 as: q 12, 600 N/mm F2 = A ( 70 mm ) = ( 70 mm ) = 441, 000 N 2 2 Shear Center Ve = ( 90 mm) F2

( 90 mm )( 441,000 N ) = 14.32 mm 2,771,666.67 N

Ans.

P9.55 An extruded beam has the cross section shown in Figure P9.55. Determine (a) the location of the shear center O, and (b) the distribution of shear stress created by a shear force of V = 30 kN.

FIGURE P9.55

Solution (a) Moment of inertia about the neutral axis: Recognizing that the wall thickness is thin, the moment of inertia for the stiffened channel shape can be calculated as: 2  (3 mm)(80 mm)3  80 mm   I NA = + 2 (55 mm)(5 mm)    12  2     (3 mm)(30 mm)3  +2  + (3 mm)(30 mm)(25 mm) 2  12   = 128, 000 mm 4 + 880, 000 mm 4 + 126, 000 mm 4 = 1,134, 000 mm 4

To simplify the calculations, we will arbitrarily assign a value of INA to the shear forceV. The value that we use for V is immaterial because it will cancel out of the calculations in the end. The location of the shear center is not dependent on the magnitude of V. Thus, let V = INA. Shear Flow in Upper Stiffener: Derive an expression for Q for the upper stiffener as a function of a temporary variable v which originates at A and extends upward. v  Qs = 10 mm +  (3 mm)v = (30 mm2 )v + (1.5 mm)v 2 2  Express the shear flow q in the upper stiffener using Qs. VQs 1,134, 000 N  (30 mm 2 )v + (1.5 mm)v 2  qs = = 4  I 1,134, 000 mm = (1 N/mm 4 )  (30 mm 2 )v + (1.5 mm)v 2 

Integrate with respect to the temporary variable v to determine the resultant force in the upper stiffener.

Fs = 

30 mm

=

30 mm

qs dv

(1 N/mm ) (30 mm )v + (1.5 mm)v  dv 4

30 mm

 30 mm 2 2 1.5 mm 3  = (1 N/mm )  v + v  3  2 0 4

= 27, 000 N

Shear Flow in Upper Flange: Derive an expression for Q for the upper flange as a function of a temporary variable u which originates at B and extends toward C in the upper flange. Q f = (25 mm)(3 mm)(30 mm) + (40 mm)  (5 mm)u  = 2, 250 mm3 + (200 mm 2 )u

Express the shear flow q in the upper flange using Qf. VQ f 1,134, 000 N  2, 250 mm3 + (200 mm 2 )u  qf = = I 1,134, 000 mm4 

= (1 N/mm4 )  2, 250 mm3 + (200 mm2 )u 

Integrate with respect to the temporary variable u to determine the resultant force in the upper flange.

Ff = 

55 mm

q f du

(1 N/mm ) 2, 250 mm + (200 mm )u  du = (1 N/mm ) (2, 250 mm )u + (100 mm )u  = (1 N/mm ) 123, 750 mm + 302,500 mm  = 426, 250 N

=

55 mm

2 55 mm 0

Shear Flow in Web: Derive an expression for Q for the web as a function of a temporary variable v which originates at C and extends downward in the web. Qw = (25 mm)(3 mm)(30 mm) + (40 mm)(55 mm)(5 mm)

v  +  40 mm −  (3 mm)v  2  3 = 2, 250 mm + 11, 000 mm3 + (120 mm 2 )v − (1.5 mm)v 2

Express the shear flow q in the web using Qw. VQw 1,134, 000 N 13, 250 mm3 + (120 mm 2 )v − (1.5 mm)v 2  qw = = 4  I 1,134, 000 mm = (1 N/mm 4 ) 13, 250 mm3 + (120 mm 2 )v − (1.5 mm)v 2 

It is not required for this problem; however, for completeness, the resultant force in the web can be calculated as:

Fw = 2

40 mm

= 2

40 mm

qw dv

(1 N/mm ) 13, 250 mm + (120 mm )v − (1.5 mm)v  dv 4

40 mm

 120 mm 2 2 1.5 mm 3  = 2 (1 N/mm ) (13, 250 mm3 )v + v − v  2 3  0 4

= 2 (1 N/mm 4 ) 530, 000 mm 4 + 96, 000 mm 4 − 32, 000 mm 4  = 1,188, 000 N

Shear center: For the cross section to bend without twisting, the internal torsional moment created by the shear stress resultants in the various portions of the cross section must be counteracted by an external torsional moment produced by the force V acting at the shear center O. To help the student visualize how the shear stress flows through the shape, the cross-sectional sketch to the right shows the shear stress resultants Fs, Ff, and Fw acting in directions consistent with the direction of shear force V. However, the student is referred to Section 9.10 and in particular to Figures 9.26 and 9.27 to better understand the twisting effect created by the shear stress resultants. To determine the shear center location, it is sufficient to simply require that the magnitude of the external torsional moment created by Ve must equal the magnitude of the internal torsional moment created by the shear stress resultants. For this cross section, sum moments about the lower left corner of the channel shape. Ve = Ff (80 mm) + 2Fs (55 mm)

Ff (80 mm) + 2 Fs (55 mm)

V (426, 250 N)(80 mm) + 2(27, 000 N)(55 mm) = 1,134, 000 N = 32.7 mm

Ans.

(b) Shear Stress Distribution



Location

Element

q (N/mm)

t (mm)

(MPa)

Stiffener

59.5238

19.84

Flange

59.5238

11.90

Flange

350.5291

70.1

Web

350.5291

116.8

Web

414.0212

138.0

P9.56 An extruded beam has the cross section shown in Figure P9.56. Using dimensions of b = 30 mm, h = 36 mm, and t = 5 mm, calculate the location of the shear center O.

FIGURE P9.56

Solution Moment of inertia about the neutral axis: Recognizing that the wall thickness is thin, the moment of inertia for the shape can be calculated as: 2  (5 mm)(2  36 mm)3  36 mm   I NA = + 2 (30 mm)(5 mm)    12  2    = 155,520 mm 4 + 97, 200 mm 4 = 252, 720 mm 4 To simplify the calculations, we will arbitrarily assign a value of INA to the shear forceV. The value that we use for V is immaterial because it will cancel out of the calculations in the end. The location of the shear center is not dependent on the magnitude of V. Thus, let V = INA. Shear Flow in Upper Stiffener: Derive an expression for Q for the upper stiffener as a function of a temporary variable v which originates at the tip of the upper stiffener. v  Qs = yA =  36 mm −  (5 mm)v = (180 mm2 )v − (2.5 mm)v 2 2  Express the shear flow q in the upper stiffener using Qs. VQ 252, 720 N (180 mm 2 )v − (2.5 mm)v 2  q= = 4  I 252, 720 mm = (1 N/mm 4 ) (180 mm 2 )v − (2.5 mm)v 2 

Integrate with respect to the temporary variable v to determine the resultant force in the upper stiffener.

Fs = 

18 mm

=

18 mm

q dv

(1 N/mm ) (180 mm )v − (2.5 mm)v  dv 4

18 mm

2.5 mm 3   = (1 N/mm 4 ) (90 mm 2 )v 2 − v  3  0

= 24,300 N

Shear Flow in Upper Flange: Derive an expression for Q for the upper flange as a function of a temporary variable u which originates at the right edge of the upper flange. Q f = y A = (27 mm)(5 mm)(18 mm) + (18 mm)  (5 mm)u  = 2, 430 mm3 + (90 mm 2 )u

Express the shear flow q in the upper flange using Qf. VQ 252, 720 N  2, 430 mm3 + (90 mm 2 )u  q= = 4  I 252, 720 mm = (1 N/mm 4 )  2, 430 mm 3 + (90 mm 2 )u 

Integrate with respect to the temporary variable u to determine the resultant force in the upper flange.

Ff = 

30 mm

q du

(1 N/mm ) 2, 430 mm + (90 mm )u  du = (1 N/mm ) (2, 430 mm )u + (45 mm )u  = (1 N/mm ) 72,900 mm + 40,500 mm  = 113, 400 N

=

30 mm

2 30 mm

Shear center: For the cross section to bend without twisting, the internal torsional moment created by the shear stress resultants in the various portions of the cross section must be counteracted by an external torsional moment produced by the force V acting at the shear center O. To help the student visualize how the shear stress flows through the shape, the cross-sectional sketch to the right shows the shear stress resultants Fs, Ff, and Fw acting in directions consistent with the direction of shear force V. However, the student is referred to Section 9.10 and in particular to Figures 9.26 and 9.27 to better understand the twisting effect created by the shear stress resultants. To determine the shear center location, it is sufficient to simply require that the magnitude of the external torsional moment created by Ve must equal the magnitude of the internal torsional moment created by the shear stress resultants. To determine the shear center, sum moments about the lower left corner of the channel web. Ve = Ff (36 mm) − 2Fs (30 mm) Ff (36 mm) − 2 Fs (30 mm) e= V (113, 400 N)(36 mm) − 2(24,300 N)(30 mm) = 252, 720 N

= 16.1538444 mm − 5.7692308 mm = 10.38 mm

Ans.

P9.57 An extruded beam has the cross section shown in Figure P9.57. For this shape, use dimensions of b = 50 mm, h = 40 mm, and t = 3 mm. What is the distance e to the shear center O?

FIGURE P9.57

Solution Moment of inertia about the neutral axis: Recognizing that the wall thickness is thin, the moment of inertia for the stiffened channel shape can be calculated as: 2  (3 mm)(40 mm)3  40 mm   I NA = 2 + 2 (50 mm)(3 mm)    12  2    = 32, 000 mm 4 + 120, 000 mm 4 = 152, 000 mm 4 To simplify the calculations, we will arbitrarily assign a value of INA to the shear forceV. The value that we use for V is immaterial because it will cancel out of the calculations in the end. The location of the shear center is not dependent on the magnitude of V. Thus, let V = INA. Shear Flow in Upper Stiffener: Derive an expression for Q for the upper stiffener as a function of a temporary variable v which originates at the tip of the upper stiffener. v Qs = yA =   (3 mm)v  = (1.5 mm)v 2 2 Express the shear flow q in the upper stiffener using Qs. VQ 152, 000 N (1.5 mm)v 2  = (1 N/mm 4 ) (1.5 mm)v 2  q= = 4  I 152, 000 mm Integrate with respect to the temporary variable v to determine the resultant force in the upper stiffener. Fs = 

20 mm

=

20 mm

q dv

(1 N/mm ) (1.5 mm)v  dv 4

20 mm

1.5 mm 3  = (1 N/mm )  v   3 0 4

= 4, 000 N

Shear Flow in Upper Flange: Derive an expression for Q for the upper flange as a function of a temporary variable u which originates at the right edge of the upper flange. Q f = y A = (10 mm)(3 mm)(20 mm) + (20 mm)  (3 mm)u  = 600 mm3 + (60 mm 2 )u

Express the shear flow q in the upper flange using Qf. VQ 152, 000 N 600 mm3 + (60 mm 2 )u  q= = 4  I 152, 000 mm = (1 N/mm 4 ) 600 mm3 + (60 mm 2 )u 

Integrate with respect to the temporary variable u to determine the resultant force in the upper flange.

Ff = 

50 mm

q du

(1 N/mm ) 600 mm + (60 mm )u  du = (1 N/mm ) (600 mm )u + (30 mm )u  = (1 N/mm ) 30, 000 mm + 75, 000 mm  = 105, 000 N

=

50 mm

2 50 mm 0

Shear center: For the cross section to bend without twisting, the internal torsional moment created by the shear stress resultants in the various portions of the cross section must be counteracted by an external torsional moment produced by the force V acting at the shear center O. To help the student visualize how the shear stress flows through the shape, the cross-sectional sketch to the right shows the shear stress resultants Fs, Ff, and Fw acting in directions consistent with the direction of shear force V. However, the student is referred to Section 9.10 and in particular to Figures 9.26 and 9.27 to better understand the twisting effect created by the shear stress resultants. To determine the shear center location, it is sufficient to simply require that the magnitude of the external torsional moment created by Ve must equal the magnitude of the internal torsional moment created by the shear stress resultants. To determine the shear center, sum moments about the lower left corner of the channel web. Ve = Ff (40 mm) + 2Fs (50 mm) Ff (40 mm) + 2 Fs (50 mm) e= V (105, 000 N)(40 mm) + 2(4, 000 N)(50 mm) = 152, 000 N

= 27.6315800 mm + 2.6315800 mm = 30.3 mm

Ans.

P9.58 An extruded beam has the cross section shown in Figure P9.70. The dimensions of this shape are b = 45 mm, h = 75 mm, and t = 4 mm. Assume that the thickness t is constant for all portions of the cross section. What is the distance e from the left-most element to the shear center O?

FIGURE P9.58

Solution Moment of inertia about the neutral axis: Recognizing that the wall thickness is thin, the moment of inertia for the shape can be calculated as: 2 2   (4 mm)(150 mm)3  75 mm    150 mm   I NA = + 2 (45 mm)(4 mm)  + 2 (45 mm)(4 mm)       2   12 2     = 1,125,000 mm 4 + 506, 250 mm 4 + 2,025,000 mm 4 = 3,656, 250 mm 4 To help the student visualize how the shear stress flows through the shape, the cross-sectional sketch below shows the shear stress resultants Ff and Fw acting in directions consistent with the direction of shear force V. However, the student is referred to Section 9.10 and in particular to Figures 9.26 and 9.27 to better understand the twisting effect created by the shear stress resultants. To determine the shear center location, it is sufficient to simply require that the magnitude of the external torsional moment created by Ve must equal the magnitude of the internal torsional moment created by the shear stress resultants. We will choose to sum moments about point D so that we need only determine two forces: Fupper and Fmiddle. Begin by calculating the shear flow in the upper flange at A. QA = ( 45 mm)( 4 mm)( 75 mm) = 13,500 mm3 Let the shear force V be equal to the moment of inertia INA, and express the shear flow q at the junction of the upper flange and the web as. 3 VQA ( 3, 656, 250 N ) (13,500 mm ) qA = = = 13,500 N/mm I 3, 656, 250 mm 4 Calculate the horizontal force Fupper in the upper flange as: q 13,500 N/mm Fupper = A ( 45 mm ) = ( 45 mm ) = 303, 750 N 2 2 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Next, calculate the shear flow in the middle flange at B.  75 mm  3 QB = ( 45 mm )( 4 mm )   = 6,750 mm 2   The shear flow qB at the junction of the middle flange and the web is thus: 3 VQB ( 3, 656, 250 N ) ( 6, 750 mm ) qB = = = 6, 750 N/mm I 3, 656, 250 mm 4 Calculate the horizontal force Fmiddle in the middle flange as: q 6, 750 N/mm Fmiddle = B ( 45 mm ) = ( 45 mm ) = 151,875 N 2 2 Shear Center: Sum moments about point D to find the shear center: Ve = (150 mm) Fupper + (112.5 mm) Fmiddle − (37.5 mm) Fmiddle e=

(150 mm )( 303, 750 N ) + (112.5 mm )(151,875 N ) − ( 37.5 mm )(151,875 N )

= 15.58 mm

3, 656, 250 N

Ans.

P9.59 Determine the location of the shear center for the cross section shown in Figure P9.59. Use dimensions of a = 50 mm, b = 100 mm, h = 300 mm, and t = 5 mm. Assume that the thickness t is constant for all portions of the cross section.

FIGURE P9.59

Solution Moment of inertia about the neutral axis: Recognizing that the wall thickness is thin, the moment of inertia for the shape can be calculated as: 2  (5 mm)(300 mm)3  300 mm   I NA = + 2 (50 mm + 100 mm)(5 mm)     12 2    = 11, 250,000 mm 4 + 33,750,000 mm 4 = 45,000,000 mm 4 To simplify the calculations, we will arbitrarily assign a value of INA to the shear forceV. The value that we use for V is immaterial because it will cancel out of the calculations in the end. The location of the shear center is not dependent on the magnitude of V. Thus, let V = INA. Shear Flow in Right-side Flange: Derive an expression for Q for the right-side flange as a function of a temporary variable u which originates at the right end of the right-side flange. Qright = yA = (150 mm) (5 mm)u  = (750 mm2 )u Express the shear flow q in the right-side flange using Qright. VQright  (45, 000, 000 N)(750 mm 2 )  2 qright = =  = ( 750 N/mm ) u 4 I 45, 000, 000 mm   Integrate with respect to the temporary variable u to determine the resultant force in the right-side flange.

Fright = 

100 mm

qright du = 

100 mm

( 750 N/mm ) u du 2

100 mm

1  = ( 750 N/mm2 )  u 2   2 0

= 3, 750, 000 N

Shear Flow in Left-side Flange: Derive an expression for Q for the leftside flange as a function of a temporary variable v which originates at the left end of the left-side flange. Qleft = yA = (150 mm) (5 mm)v = (750 mm2 )v Express the shear flow q in the left-side flange using Qleft. VQleft  (45, 000, 000 N)(750 mm 2 )  2 qleft = =  v = ( 750 N/mm ) v 4 I 45, 000, 000 mm   Integrate with respect to the temporary variable v to determine the resultant force in the left-side flange.

Fleft = 

50 mm

=

50 mm

qleft dv

( 750 N/mm ) v dv 2

50 mm

1  = ( 750 N/mm )  v 2   2 0 2

= 937,500 N

Shear center: For the cross section to bend without twisting, the internal torsional moment created by the shear stress resultants in the various portions of the cross section must be counteracted by an external torsional moment produced by the force V acting at the shear center O. To help the student visualize how the shear stress flows through the shape, the cross-sectional sketch to the right shows the shear stress resultants Fleft, Fright, and Fw acting in directions consistent with the direction of shear force V. However, the student is referred to Section 9.10 and in particular to Figures 9.26 and 9.27 to better understand the twisting effect created by the shear stress resultants. To determine the shear center location, it is sufficient to simply require that the magnitude of the external torsional moment created by Ve must equal the magnitude of the internal torsional moment created by the shear stress resultants. To determine the shear center, sum moments about the point where the web connects to the lower flange. Ve = Fright (300 mm) − Fleft (300 mm) Fright (300 mm) − Fleft (300 mm) e= V (3, 750, 000 N)(300 mm) − (937,500 N)(300 mm) = 45, 000, 000 N

= 25 mm − 6.25 mm = 18.75 mm

Ans.

P9.60 Locate the shear center for the cross section shown in Figure P9.60. Assume that the web thickness is the same as the flange thickness.

FIGURE P9.60

Solution Section Properties:

Horizontal distance to centroid (measured from center of left-side flange). (5 in.)(10 in.)(0.375 in.) + (10 in.)(0.375 in.)(5 in.) z= (0.375 in.)(3 in.) + (10 in.)(0.375 in.) + (0.375 in.)(5 in.) =

18.75 in.3 + 18.75 in.3 37.5 in.3 = = 5.5555556 in. 1.125 in.2 + 3.75 in.2 + 1.875 in.2 6.75 in.2

Moment of inertia about z axis. 1 1 I z = (0.375 in.)(3 in.)3 + (0.375 in.)(5 in.)3 = 4.75 in.4 12 12 To simplify the calculations, we will arbitrarily assign a value of INA to the shear forceV. The value that we use for V is immaterial because it will cancel out of the calculations in the end. The location of the shear center is not dependent on the magnitude of V. Thus, let V = INA. Shear Flow Resultant: Consider the right-side flange. Q for the 5 in. deep flange can be expressed as:  2.5 in. + y  Q=  ( (0.375 in.)(2.5 in. − y ))  2

y  = 1.25 in. +  0.9375 in.2 − (0.375 in.)y   2 = 1.1718750 in.3 + (0.46875 in.2 ) y −(0.46875 in.2 ) y − (0.1875 in.)y 2 = 1.1718750 in.3 − (0.1875 in.)y 2

VQ 4.75 lb 1.1718750 in.3 − (0.1875 in.)y 2  = (1 lb/in.4 ) 1.1718750 in.3 − (0.1875 in.)y 2  = 4  I 4.75 in. Fright =  q dy

=

2.5 in.

−2.5 in.

(1 lb/in. ) 1.1718750 in. − (0.1875 in.)y  dy 4

2.5 in.

(0.1875 in.) 3   = (1 lb/in.4 ) (1.1718750 in.3 ) y − y  3   −2.5 in.

= (1 lb/in.4 )  2.9296875 in.4 − 0.9765625 in.4 − ( −2.9296875 in.4 ) − 0.9765625 in.4  = 3.9063 lb

Shear Center: For the cross section to bend without twisting, the internal torsional moment created by the shear stress resultants in the various portions of the cross section must be counteracted by an external torsional moment produced by the force V acting at the shear center O. To help the student visualize how the shear stress flows through the shape, the cross-sectional sketch to the right shows the shear stress resultants Fleft and Fright acting in directions consistent with the direction of shear force V. However, the student is referred to Section 9.10 and in particular to Figures 9.26 and 9.27 to better understand the twisting effect created by the shear stress resultants. To determine the shear center location, it is sufficient to simply require that the magnitude of the external torsional moment created by Ve must equal the magnitude of the internal torsional moment created by the shear stress resultants. Sum moments about the center of the left-side flange to find V ( z + e) = Fright (10 in.)

Fright (10 in.)

(3.9063 lb)(10 in.) = 8.223684 in. V 4.75 lb e = 8.223684 in. − 5.5555556 in. = 2.67 in.

z +e =

Ans.

P9.61 Show that the shear center for the zee-shaped section shown in Figure P9.61 is located at the centroid of the section.

FIGURE P9.61

Solution For the cross section to bend without twisting, the internal torsional moment created by the shear stress resultants in the various portions of the cross section must be counteracted by an external torsional moment produced by the force V acting at the shear center O. To help the student visualize how the shear stress flows through the shape, the cross-sectional sketch to the right shows the shear stress resultants F1, F2, and F3 acting in directions consistent with the direction of shear force V. However, the student is referred to Section 9.10 and in particular to Figures 9.26 and 9.27 to better understand the twisting effect created by the shear stress resultants. To determine the shear center location, it is sufficient to simply require that the magnitude of the external torsional moment created by Ve must equal the magnitude of the internal torsional moment created by the shear stress resultants. If the shear center O was located at some distance e away from the centroid of the zee shape, the shear flows through the zee shape would be directed as sketched to the right. The shape would not satisfy equilibrium in the horizontal direction because the sum of F1 and F3 would not equal zero. Therefore, to satisfy equilibrium, forces F1 and F3 must each equal zero. Hence, the sum of moments about the centroid gives: Ans. Ve = 0 e = 0

P9.62-P9.66 Determine the location of the shear center O of a thin-walled beam of uniform thickness having the cross section shown in Figures P9.62–P9.66.

FIGURE P9.62

Solution Section Properties: dA = t ds = t rd

y = r sin 

dI = y 2 dA = r 2 sin 2  (trd ) = r 3t sin 2  d 2

I = r 3t  sin 2  d 0

2  1 − cos 2  = r 3t   d 0 2   3 =r t

dQ = ydA = r sin  (trd ) = r 2t sin  d 

Q = r 2t  sin  d = r 2t (1 − cos  ) 0

Shear Flow Resultant: VQ Vr 2t (1 − cos  ) V q= = = (1 − cos  ) I  r 3t r F =  q ds =

2

V (1 − cos  ) rd r 2

 0

(1 − cos  ) d = 2V

Shear Center: Sum moments about the center of the split tube to find Ve = Fr

Fr 2Vr = = 2r V V

Ans.

P9.62-P9.66 Determine the location of the shear center O of a thin-walled beam of uniform thickness having the cross section shown in Figures P9.62–P9.66.

FIGURE P9.63

Solution Section Properties: For the arc y = r cos  dA = t ds = t rd dI arc = y 2 dA = r 2 cos 2  (trd ) = r 3t sin 2  d 

I arc = r 3t  cos 2  d 0

  cos 2 + 1 = r 3t    d 0  2

1 =  r 3t 2

The moment of inertia for the complete cross section is: 1 r 3t I = I arc + 2  rt (r ) 2  =  r 3t + 2r 3t = ( + 4) 2 2 Derive an expression for the value of Q for the upper horizontal portion in terms of a temporary variable u: Qupper = r (u  t ) = r 2 t

For locations in the arc, Q can be expressed as: Qarc = Qupper +  dQ where dQ = ydA = r cos  (trd ) = r 2t cos  d 

Qarc = r 2t + r 2t  cos  d = r 2t (1 + sin  ) 0

Shear Flow Resultant (for the upper portion): VQu Vrt 2V qupper = = 3 u= 2 u rt I r ( + 4) ( + 4) 2 Fupper =  qupper du 2V V u du = 0 r ( + 4) +4

=

Shear Flow Resultant (for the arc): VQarc Vr 2t (1 + sin  ) 2V qarc = = = (1 + sin  ) 3 rt I r ( + 4) ( + 4) 2

Farc =  qarc ds 

2V (1 + sin  ) r d 0 r ( + 4) 2V  2V ( + 2) = (1 + sin  ) d =  0 +4 +4

=

Shear Center: For the cross section to bend without twisting, the internal torsional moment created by the shear stress resultants in the various portions of the cross section must be counteracted by an external torsional moment produced by the force V acting at the shear center O. To help the student visualize how the shear stress flows through the shape, the cross-sectional sketch to the right shows the shear stress resultants dF and Fupper acting in directions consistent with the direction of shear force V. However, the student is referred to Section 9.10 and in particular to Figures 9.26 and 9.27 to better understand the twisting effect created by the shear stress resultants. To determine the shear center location, it is sufficient to simply require that the magnitude of the external torsional moment created by Ve must equal the magnitude of the internal torsional moment created by the shear stress resultants. Sum moments about the center of the arc to find Ve = 2 Fupper r + Farc r  V   2V ( + 2)  Ve = 2  r+ r    + 4   + 4  =

2Vr ( + 3) +4

  + 3 e =  2r   + 4 

Ans.

P9.62-P9.66 Determine the location of the shear center O of a thin-walled beam of uniform thickness having the cross section shown in Figures P9.62–P9.66.

FIGURE P9.64

Solution Moment of inertia about the neutral axis: Recognizing that the wall thickness is thin, the moment of inertia for the shape can be calculated as follows. For segment AB, I AB = bt (b sin60)2 = tb3 sin2 60 For segment BC, dA = t ds

y = s sin 60

dI = y 2 dA = ( s sin 60) 2 t ds = t sin 2 60 s 2 ds I BC = t sin 2 60 s 2 ds b

tb sin 60 3 Therefore, the moment of inertia about the horizontal centroidal axis is  tb3 sin 2 60  I = 2 tb3 sin 2 60 +  3   =

= 2tb3

Shear Flow in Flange: Derive an expression for Q for the flange as a function of a temporary variable s which originates at the tip of the flange. Q = yA = b sin60 st  = (tb sin60) s Express the shear flow q in the flange using Q. VQ V ( tb sin 60) s 3V q= = = s 3 I 2tb 4b2 Integrate with respect to the temporary variable s to determine the resultant force in the flange. b b 3V 3V 2 b 3  s  = Ff =  q ds =  s ds = V 2 2 0 0 0 4b 8b 8

Shear Center: For the cross section to bend without twisting, the internal torsional moment created by the shear stress resultants in the various portions of the cross section must be counteracted by an external torsional moment produced by the force V acting at the shear center O. To help the student visualize how the shear stress flows through the shape, the cross-sectional sketch to the right shows the shear stress resultants Fweb and Ff acting in directions consistent with the direction of shear force V. However, the student is referred to Section 9.10 and in particular to Figures 9.26 and 9.27 to better understand the twisting effect created by the shear stress resultants. To determine the shear center location, it is sufficient to simply require that the magnitude of the external torsional moment created by Ve must equal the magnitude of the internal torsional moment created by the shear stress resultants. Sum moments about C to find  3   3 Ve = 2 Ff (b sin 60) = 2  V    b  8  2 

3  e = b = 0.375b 8

Ans.

P9.62-P9.66 Determine the location of the shear center O of a thin-walled beam of uniform thickness having the cross section shown in Figures P9.62–P9.66.

FIGURE P9.65

Solution Moment of inertia about the neutral axis: Recognizing that the wall thickness is thin, the moment of inertia for the shape can be calculated in the following manner. For segment AB, 6 mm b= = 12 mm sin 30 h = 35 mm

35 mm = 70 mm sin 30 bh3 (12 mm)(35 mm)3 I AB = = = 171,500 mm 3 3 3 The moment of inertia for segment EF will be identical. AB =

The moment of inertia for segment BDE is simply (6 mm)(70 mm)3 I BDE = = 171,500 mm3 12 and so the total moment of inertia for the cross section is I = I BDE + 2I AB = 171,500 mm4 + 2(171,500 mm3 ) = 514,500 mm4 Shear Flow in Element AB: Derive an expression for Q for element AB as a function of a temporary variable s which originates at A and extends toward B. 1  QAB = y A =  s sin30  (6 mm)s  = (1.5 mm) s 2 2  Express the shear flow q in element AB using QAB. VQAB  V (1.5 mm)  2 qAB = = s  514,500 mm4  I Integrate with respect to the temporary variable s to determine the resultant force in element AB.

FAB = 

70 mm

=

70 mm

q AB ds  V (1.5 mm)  2  514,500 mm 4  s ds 70 mm

V (1.5 mm)  1 3  = s 514,500 mm 4  3  0

= 0.3333333V

Shear Center: For the cross section to bend without twisting, the internal torsional moment created by the shear stress resultants in the various portions of the cross section must be counteracted by an external torsional moment produced by the force V acting at the shear center O. To help the student visualize how the shear stress flows through the shape, the cross-sectional sketch to the right shows the shear stress resultants FAB and Fw acting in directions consistent with the direction of shear force V. However, the student is referred to Section 9.10 and in particular to Figures 9.26 and 9.27 to better understand the twisting effect created by the shear stress resultants. To determine the shear center location, it is sufficient to simply require that the magnitude of the external torsional moment created by Ve must equal the magnitude of the internal torsional moment created by the shear stress resultants. Sum moments about E to find Ve = (70 mm)FAB cos30

(70 mm)(0.3333333V )cos30 = 20.2 mm V

Ans.

P9.62-P9.66 Determine the location of the shear center O of a thin-walled beam of uniform thickness having the cross section shown in Figures P9.62–P9.66.

FIGURE P9.66

Solution Moment of inertia about the neutral axis: Recognizing that the wall thickness is thin, the moment of inertia for the shape can be calculated as: 3 in. tan  = = 0.75   = 36.8699 4 in. 0.25 in. b= = 0.41667 in. sin 36.8699 h = 3 in. 3 in. BD = = 5 in. sin 36.8699

I NA = 2 I AB + 2 I BD  (0.25 in.)(2 in.)3  = 2 + (0.25 in.)(2 in.)(4 in.) 2  12    (0.41667 in.)(3 in.)3  +2   3   4 4 = 16.3333 in. + 7.5 in. = 23.8333 in.4 Shear Flow in Element AB: Derive an expression for Q for element AB as a function of a temporary variable s which originates at A and extends toward B. s  QAB = yA =  5 in. −   (0.25 in.)s  = (1.25 in.2 )s − (0.125 in.)s 2  2 Express the shear flow q in element AB using QAB. VQAB  V  (1.25 in.2 )s − (0.125 in.)s 2  qAB = = 4   23.8333 in.  I

Integrate with respect to the temporary variable s to determine the resultant force in element AB.

FAB = 

2 in.

=

2 in.

q AB ds V   (1.25 in.2 ) s − (0.125 in.)s 2  ds  4  23.8333 in.  2 in.

1.25 in.2 2 0.125 in. 3  V = s − s   23.8333 in.4  2 3 0

= 0.0909092V

Shear Center: For the cross section to bend without twisting, the internal torsional moment created by the shear stress resultants in the various portions of the cross section must be counteracted by an external torsional moment produced by the force V acting at the shear center O. To help the student visualize how the shear stress flows through the shape, the cross-sectional sketch to the right shows the shear stress resultants FAB and FBD acting in directions consistent with the direction of shear force V. However, the student is referred to Section 9.10 and in particular to Figures 9.26 and 9.27 to better understand the twisting effect created by the shear stress resultants. To determine the shear center location, it is sufficient to simply require that the magnitude of the external torsional moment created by Ve must equal the magnitude of the internal torsional moment created by the shear stress resultants. Sum moments about D to find Ve = 2(4 in.)FAB

2(4 in.)(0.0909092V ) = 0.727 in. V

Ans.

P10.1 For the loading shown, use the double-integration method to determine (a) the equation of the elastic curve for the cantilever beam, (b) the deflection at the free end, and (c) the slope at the free end. Assume that EI is constant for each beam. FIGURE P10.1

Solution Integration of moment equation: d 2v EI 2 = M ( x ) = − M 0 dx dv EI = − M 0 x + C1 dx M x2 EI v = − 0 + C1 x + C2 2

(a) (b)

Boundary conditions: dv = 0 at x=0 dx v=0 at x=0 Evaluate constants: From Eq. (a), C1 = 0. From Eq. (b), C2 = 0 (a) Elastic curve equation:

M 0 x2 EI v = − 2

M 0 x2 v = − 2 EI

Ans.

(b) Deflection at the free end:

vB = −

M 0 ( L) 2 M L2 = − 0 2 EI 2 EI

Ans.

P10.2 For the loading shown, use the double-integration method to determine (a) the equation of the elastic curve for the cantilever beam, (b) the deflection at the free end, and (c) the slope at the free end. Assume that EI is constant for each beam. FIGURE P10.2

Solution Integration of moment equation: d 2v wx 2 EI 2 = M ( x ) = − dx 2 3 dv wx EI =− + C1 dx 6 wx 4 EI v = − + C1 x + C2 24

(a) (b)

Boundary conditions: dv = 0 at x=L dx v=0 at x=L Evaluate constants: Substitute x = L and dv/dx = 0 into Eq. (a) to determine C1: w( L)3 wL3 EI (0) = − + C1  C1 = 6 6 Substitute x = L and v = 0 into Eq. (b) to determine C2: w( L) 4 wL4 wL4 wL4 EI (0) = − + C1 ( L) + C2 = − + + C2  C2 = − 24 24 6 8 (a) Elastic curve equation: wx 4 wL3 x wL4 w  x 4 − 4L3 x + 3L4  EI v = − + − v = − 24 6 8 24 EI

Ans.

(b) Deflection at the free end: vA =

w 3wL4 wL4  −(0) 4 + 4 L3 (0) − 3L4  = − = − 24 EI 24 EI 8EI

Ans.

P10.3 For the loading shown, use the doubleintegration method to determine (a) the equation of the elastic curve for the cantilever beam, (b) the deflection at the free end, and (c) the slope at the free end. Assume that EI is constant for each beam. FIGURE P10.3

Solution Integration of moment equation: d 2v w x3 EI 2 = M ( x) = − 0 dx 6L 4 dv wx EI = − 0 + C1 dx 24 L w0 x5 EI v = − + C1 x + C2 120 L

(a) (b)

Boundary conditions: dv = 0 at x=L dx v=0 at x=L Evaluate constants: Substitute x = L and dv/dx = 0 into Eq. (a) to determine C1: w ( L) 4 w L3 EI (0) = − 0 + C1  C1 = 0 24 L 24 Substitute x = L and v = 0 into Eq. (b) to determine C2: w0 ( L)5 w0 L5 w0 L3 EI (0) = − + C1 ( L) + C2 = − + ( L) + C2 120 L 120 L 24 w L4 w L4 w L4  C2 = 0 − 0 = − 0 120 24 30 (a) Elastic curve equation: w0 x5 w0 L3 w0 L4 EI v = − + x− 120 L 24 30

v = −

w0  x5 − 5L4 x + 4 L5  120 L EI

Ans.

(b) Deflection at the free end: w0 w0 L4 5 4 5 (0) − 5L (0) + 4 L  = − vA = − 120 L EI  30 EI

Ans.

P10.4 For the beam and loading shown in Figure P10.4, use the double-integration method to determine (a) the equation of the elastic curve for segment AB of the beam, (b) the deflection at B, and (c) the slope at A. Assume that EI is constant for the beam. FIGURE P10.4

Solution Integration of moment equation: d 2v P EI 2 = M ( x) = x dx 2 2 dv P x EI = + C1 dx 4 EI v =

(a)

Px3 + C1 x + C2 12

(b)

Boundary conditions: v=0 at x=0 dv L = 0 at x= dx 2 Evaluate constants: Substitute x = L/2 and dv/dx = 0 into Eq. (a) to determine C1: P ( L / 2) 2 PL2 EI (0) = + C1  C1 = − 4 16 Substitute x = 0 and v = 0 into Eq. (b) to determine C2: P (0)3 PL2 (0) EI (0) = − + C2  C2 = 0 12 16 (a) Elastic curve equation: P x3 PL2 x EI v = − 12 16

v = −

Px 3L2 − 4 x 2  48EI

(b) Deflection at B: 2 P ( L / 2)  2 PL3  L  vB = − 3 L − 4 = −     48 EI  2  48 EI

(0  x 

L ) 2

Ans.

P10.5 For the beam and loading shown in Figure P10.5, use the doubleintegration method to determine (a) the equation of the elastic curve for the beam, (b) the slope at A, (c) the slope at B, and (d) the deflection at midspan. Assume that EI is constant for the beam. FIGURE P10.5

Solution Beam FBD: Fy = Ay + By = 0  Ay = − By M A = B y L − M 0 = 0  By =

M0 L

and

Ay = −

M0 L

Moment equation:

M a − a = M ( x) − Ay x − M 0 = M ( x) +  M ( x) = M 0 −

M0 x − M0 = 0 L

M0x L

Integration of moment equation: d 2v M x EI 2 = M ( x) = M 0 − 0 dx L 2 dv M x EI = M 0 x − 0 + C1 dx 2L 2 M x M x3 EI v = 0 − 0 + C1 x + C2 2 6L

(a) (b)

Boundary conditions: v=0 at x=0

v=0

x=L

Evaluate constants: Substitute x = 0 and v = 0 into Eq. (b) to determine C2: M (0) 2 M 0 (0)3 EI (0) = 0 − + C1 (0) + C2 2 6L Substitute x = L and v = 0 into Eq. (b) to determine C1:

 C2 = 0

M 0 ( L) 2 M 0 ( L)3 EI (0) = − + C1 ( L) 2 6L M L M L M L  C1 = 0 − 0 = − 0 6 2 3 (a) Elastic curve equation: M 0 x 2 M 0 x3 M 0 Lx EI v = − − 2 6L 3

v = −

M0x 2  x − 3Lx + 2 L2  6 L EI

(b) Slope at A: dv M (0)2 M 0 L M L =  A = M 0 (0) − 0 − = − 0 dx A 2 L EI 3EI 3EI (c) Slope at B: dv M ( L) M 0 ( L) 2 M 0 L M 0 M0L = B = 0 − − = 6 L − 3L − 2 L  =  dx B EI 2L EI 3EI 6 EI 6 EI (d) Deflection at midspan: 2  M 0 ( L / 2)   L  M 0 L2  L 2 vx = L / 2 = −    − 3L   + 2 L  = −  2 6 L EI   2  16 EI 

Ans.

P10.6 For the beam and loading shown in Figure P10.6, use the double-integration method to determine (a) the equation of the elastic curve for the beam, (b) the maximum deflection, and (c) the slope at A. Assume that EI is constant for the beam. FIGURE P10.6

Solution Moment equation:

wLx wx 2 + =0 2 2 wx 2 wLx  M ( x) = − + 2 2 Integration of moment equation: d 2v wx 2 wLx EI 2 = M ( x) = − + dx 2 2 3 2 dv wx wLx EI =− + + C1 dx 6 4 wx 4 wLx 3 EI v = − + + C1 x + C2 24 12 M a − a = M ( x) −

(a) (b)

Boundary conditions: v=0 at x=0

v=0

x=L

Evaluate constants: Substitute x = 0 and v = 0 into Eq. (b) to determine C2: w(0) 4 wL(0)3 EI (0) = − + + C1 (0) + C2  C2 = 0 24 12 Substitute x = L and v = 0 into Eq. (b) to determine C1: w( L) 4 wL( L)3 w( L) 4 w( L) 4 wL3 EI (0) = − + + C1 ( L)  C1 = − =− 24 12 24 L 12 L 24 (a) Elastic curve equation: wx 4 wLx3 wL3 x wx  x3 − 2 Lx 2 + L3  EI v = − + − v = − 24 12 24 24 EI  (b) Maximum deflection: At x = L/2: 3 2  w( L / 2)  L  wL  L3 L3 5wL4  L 2 2 vmax = −   − 2 L   + L  = −  − +L= −  2 24 EI  2  48 EI  8 2 384 EI  

Ans.

P10.7 For the simply supported steel beam [E = 200 GPa; I = 129 × 106 mm4] shown in Figure P10.7, use the double-integration method to determine the deflection at B. Assume L = 4 m, P = 60 kN, and w = 40 kN/m.

FIGURE P10.7

Solution Beam FBD:

L L M A = − wL   − P   + C y ( L) = 0 2 2 wL P Cy = + 2 2 Fy = Ay + C y − w( L) − P = 0  Ay =

wL P + 2 2

Moment equation: wx 2 wx 2 wLx Px M a − a = M ( x) + − Ay x = M ( x) + − − =0 2 2 2 2 wx 2 wLx Px  M ( x) = − + + 2 2 2 Integration of moment equation: d 2v wx 2 wLx Px EI 2 = M ( x) = − + + dx 2 2 2 3 2 2 dv wx wLx Px EI =− + + + C1 dx 6 4 4 wx 4 wLx 3 Px 3 EI v = − + + + C1 x + C2 24 12 12

(a) (b)

Boundary conditions: v=0 at x=0 dv L = 0 at x= dx 2 Evaluate constants: Substitute x = 0 and v = 0 into Eq. (b) to determine C2 = 0. Next, substitute x = L/2 and dv/dx = 0 into Eq. (b) to determine C1: w( L / 2)3 wL( L / 2) 2 P( L / 2) 2 EI (0) = − + + + C1 6 4 4 wL3 wL3 PL2 wL3 PL2  C1 = − − =− − 48 16 16 24 16 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Elastic curve equation: wx 4 wLx3 Px3 wL3 x PL2 x EI v = − + + − − 24 12 12 24 16 wx Px  x3 − 2 Lx 2 + L3  − 3L2 − 4 x 2  v = − 24 EI 48EI  Deflection at B: At x = L/2: 5wL4 PL3 vB = − − 384 EI 48 EI Let E = 200 GPa, I = 129 × 106 mm4, w = 40 kN/m, P = 60 kN, and L = 4 m. 5(40 N/mm)(4,000 mm) 4 (60,000 N)(4,000 mm)3 vB = − − 384(200,000 N/mm 2 )(129  106 mm 4 ) 48(200,000 N/mm 2 )(129  106 mm 4 ) = −5.1680 mm − 3.1008 mm

= −8.27 mm

Ans.

P10.8 For the cantilever steel beam [E = 200 GPa; I = 129 × 106 mm4] shown in Figure P10.8, use the double-integration method to determine the deflection at A. Assume L = 2.5 m, P = 50 kN-m, and w0 = 90 kN/m. FIGURE P10.8

Solution Moment equation:

M a − a = M ( x) +

w0 x  x  ( x)   + Px = 0 2L 3

w0 x3  M ( x) = − − Px 6L Integration of moment equation: d 2v w x3 EI 2 = M ( x) = − 0 − Px dx 6L 4 dv wx Px 2 EI =− 0 − + C1 dx 24 L 2 w x5 Px3 EI v = − 0 − + C1 x + C2 120 L 6

(a) (b)

Boundary conditions: v=0 at x=L dv = 0 at x=L dx Evaluate constants: Substitute x = L and dv/dx = 0 into Eq. (a) to determine C1: w ( L) 4 P ( L) 2 w L3 PL2 EI (0) = − 0 − + C1  C1 = 0 + 24 L 2 24 2 Substitute x = L and v = 0 into Eq. (b) to determine C2: w0 ( L)5 P( L)3 w0 L3 PL2 w0 L4 w0 L4 PL3 PL3 EI (0) = − − + ( L) + ( L) + C2 = − + − + + C2 120 L 6 24 2 120 24 6 2 w L4 PL3  C2 = − 0 − 30 3 Elastic curve equation: w x5 Px3 w0 L3 PL2 w L4 PL3 EI v = − 0 − + x+ x− 0 − 120 L 6 24 2 30 3 w0 P  x5 − 5L4 x + 4 L5  −  x3 − 3L2 x + 2 L3  v = − 120 L EI 6 EI

Deflection at A: Let E = 200 GPa, I = 129 × 106 mm4, w0 = 90 kN/m, P = 50 kN, and L = 2.5 m. w0 P (0)5 − 5 L4 (0) + 4 L5  − (0)3 − 3L2 (0) + 2 L3  vA = − 120 L EI 6 EI

w0 L4 PL3 − 30 EI 3EI (90 N/mm)(2,500 mm) 4 (50,000 N)(2,500 mm)3 =− − 30(200,000 N/mm 2 )(129  106 mm 4 ) 3(200,000 N/mm 2 )(129  106 mm 4 ) = −4.5422 mm − 10.0937 mm

=−

= −14.64 mm

Ans.

P10.9 For the beam and loading shown in Figure P10.9, use the double-integration method to determine (a) the equation of the elastic curve for the cantilever beam AB, (b) the deflection at the free end, and (c) the slope at the free end. Assume that EI is constant for each beam. FIGURE P10.9

Solution Beam FBD:

M A = − M A −

w0 L  L   =0 2 3

w0 L2 6 wL Fy = Ay − 0 = 0 2 w0 L  Ay = 2 MA = −

Moment equation: M a − a = − M ( x ) −

w0 ( L − x)3 =0 2L 3

w0 ( L − x )3 6L w = − 0 ( L3 − 3L2 x + 3Lx 2 − x 3 ) 6L w L2 w Lx w x 2 w x 3 =− 0 + 0 − 0 + 0 6 2 2 6L

M ( x) = −

Integration of moment equation: d 2v w x 3 w x 2 w Lx w L2 EI 2 = M ( x) = 0 − 0 + 0 − 0 dx 6L 2 2 6 4 3 2 2 dv w0 x wx w Lx wLx EI = − 0 + 0 − 0 + C1 dx 24 L 6 4 6 w0 x 5 w0 x 4 w0 Lx 3 w0 L2 x 2 EI v = − + − + C1 x + C2 120 L 24 12 12

(a) (b)

Boundary conditions: v=0 at x=0 dv = 0 at x=0 dx

Evaluate constants: Substitute x = 0 and v = 0 into Eq. (b) to determine C2 = 0. Next, substitute x = 0 and dv/dx = 0 into Eq. (b) to determine C1 = 0. (a) Elastic curve equation: w x5 w x 4 w Lx3 w0 L2 x 2 EI v = 0 − 0 + 0 − 120 L 24 12 12 v = −

w0 x 2  − x3 + 5Lx 2 − 10 L2 x + 10 L3  120 L EI

Ans.

(b) Deflection at the free end: vB =

w0 4w0 L5 w L4 ( L)5 − 5L( L) 4 + 10 L2 ( L)3 − 10 L3 ( L) 2  = − = − 0 120 L EI 120 L EI 30 EI

Ans.

(c) Slope at the free end: dv w ( L) 4 w0 ( L)3 w0 L( L) 2 w0 L2 ( L) w L3 = B = 0 − + − = − 0 dx B 24 L EI 6 EI 4 EI 6 EI 24 EI

Ans.

P10.10 For the beam and loading shown in Figure P10.10, use the double-integration method to determine (a) the equation of the elastic curve for the cantilever beam, (b) the deflection at B, (c) the deflection at the free end, and (d) the slope at the free end. Assume that EI is constant for the beam. FIGURE P10.10

Solution Beam FBD:

M A = − M A −

wL  L L   + =0 2 2 4

3wL2 8 wL Fy = Ay − =0 2 wL  Ay = 2 MA = −

Consider beam segment AB (0 ≤ x ≤ L/2) Moment equation:

3wL2 wL M a − a = M ( x) − M A − Ay x = M ( x) + − x=0 8 2 3wL2 wLx  M ( x) = − + 8 2 Integration of moment equation: d 2v 3wL2 wLx EI 2 = M ( x) = − + dx 8 2 2 2 dv 3wL x wLx EI =− + + C1 dx 8 4 3wL2 x 2 wLx 3 EI v = − + + C1 x + C2 16 12

(a) (b)

Boundary conditions: v=0 at x=0 dv = 0 at x=0 dx Evaluate constants: Substitute x = 0 and v = 0 into Eq. (b) to determine C2 = 0. Next, substitute x = 0 and dv/dx = 0 into Eq. (b) to determine C1 = 0.

Elastic curve equation for beam segment AB: 3wL2 x 2 wLx3 EI v = − + 16 12

v = −

wLx 2  9L − 4 x 48EI

(0  x  L / 2)

Slope at B: Let x = L/2 dv 3wL2 ( L / 2) wL( L / 2) 2 wL3 = B = − + =− dx B 8EI 4EI 8EI Deflection at B: Let x = L/2 wL( L / 2)2  7 wL4  L  vB = − 9 L − 4 = −   48EI  192 EI  2 

Consider beam segment BC (L/2 ≤ x ≤ L) Moment equation: w L M b − b = M ( x) − M A − Ay x +  x −  2 2

3wL2 wL w L = M ( x) + − x+ x−  =0 8 2 2 2 2

w L  wL 3wL2  M ( x) = −  x −  + x− 2 2 2 8 =−

wx 2 wL2 + wLx − 2 2

Integration of moment equation: d 2v wx 2 wL2 EI 2 = M ( x) = − + wLx − dx 2 2 3 2 2 dv wx wLx wL x EI =− + − + C3 dx 6 2 2 wx 4 wLx 3 wL2 x 2 EI v = − + − + C3 x + C4 24 6 4 Continuity conditions: 7 wL4 v=− at 192 EI dv wL3 =− at dx 8 EI

L 2 L x= 2 x=

Evaluate constants: Substitute the slope continuity condition into Eq. (c) for x = L/2 and solve for C3: dv w( L / 2)3 wL( L / 2) 2 wL2 ( L / 2) wL3 EI =− + − + C3 = − dx 6 2 2 8 3 wL  C3 = 48 Next, substitute the deflection continuity condition into Eq. (d) for x = L/2 and solve for C4 w( L / 2) 4 wL( L / 2)3 wL2 ( L / 2) 2 wL3 7 wL4 EI v = − + − + ( L / 2) + C4 = − 24 6 4 48 192 4 wL  C4 = − 384 Elastic curve equation for beam segment BC: wx 4 wLx 3 wL2 x 2 wL3 x wL4 EI v = − + − − − 24 6 4 48 384 w 16 x 4 − 64 Lx 3 + 96 L2 x 2 − 8 L3 x + L4  v = − 384 EI

( L / 2  x  L)

(a) Elastic curve equations for entire beam: v= −

wLx 2  9L − 4 x 48EI

v= −

w 16 x 4 − 64 Lx3 + 96 L2 x 2 − 8L3 x + L4  384 EI

(0  x  L / 2)

Ans. ( L / 2  x  L)

Ans.

(b) Deflection at B: vB = −

7 wL4 192 EI

Ans.

(c) Deflection at free end of cantilever: w 41wL4 4 3 2 2 3 4 16( L) − 64 L( L) + 96 L ( L) − 8L ( L) + L  = − vC = − 384 EI  384 EI

Ans.

(d) Slope at free end of cantilever: dv 8w( L)3 24wL( L) 2 24wL2 ( L) wL3 7 wL3 EI =− + − + =− dx 48 48 48 48 48



dv 7 wL3 = C = − dx C 48EI

Ans.

P10.11 For the beam and loading shown in Figure P10.11, use the double-integration method to determine (a) the equation of the elastic curve for the beam, and (b) the deflection at B. Assume that EI is constant for the beam. FIGURE P10.11

Solution Beam FBD:

wL  L   =0 2 4 wL Cy = 8 wL Fy = Ay + C y − =0 2 3wL  Ay = 8

M A = C y ( L) −

Consider beam segment AB (0 ≤ x ≤ L/2) Moment equation:

wx 2 wx 2 3wL − Ay x = M ( x) + − x=0 2 2 8 wx 2 3wLx  M ( x) = − + 2 8

M a − a = M ( x) +

Integration of moment equation: d 2v wx 2 3wLx EI 2 = M ( x) = − + dx 2 8 3 2 dv wx 3wLx EI =− + + C1 dx 6 16 wx 4 wLx 3 EI v = − + + C1 x + C2 24 16

(a) (b)

Boundary conditions: v=0 at x=0 Evaluate constants: Substitute x = 0 and v = 0 into Eq. (b) to determine C2 = 0. Slope at B: Let x = L/2 in Eq. (a).

dv w( L / 2)3 3wL( L / 2) 2 wL3 3wL3 5wL3 EI = EI B = − + + C1 = − + + C1 = + C1 dx B 6 16 48 64 192

(c)

Deflection at B: Let x = L/2 in Eq. (b).

w( L / 2) 4 wL( L / 2)3 wL4 wL4 C1L wL4 C1L EI vB = − + + C1 ( L / 2) = − + + = + 24 16 384 128 2 192 2

(d)

Consider beam segment BC (L/2 ≤ x ≤ L) Moment equation:

wL ( L − x) = 0 8 wL wL2 wLx  M ( x) = ( L − x) = − 8 8 8 Integration of moment equation: d 2v wLx wL2 EI 2 = M ( x) = − + dx 8 8 2 2 dv wLx wL x EI =− + + C3 dx 16 8 wLx 3 wL2 x 2 EI v = − + + C3 x + C4 48 16 M b − b = − M ( x) + C y ( L − x) = − M ( x) +

(e) (f)

Boundary conditions: v=0 at x=L Evaluate constants: Substitute x = L and v = 0 into Eq. (f) to find wL( L)3 wL2 ( L)2 EI (0) = − + + C3 ( L) + C4 48 16

wL4  C3 L + C4 = − 24

(g)

Slope at B: Let x = L/2 in Eq. (e).

dv wL( L / 2)2 wL2 ( L / 2) wL3 wL3 3wL3 = EI  B = − + + C3 = − + + C3 = + C3 dx B 16 8 64 16 64

(h)

Deflection at B: Let x = L/2 in Eq. (f).

EI vB = −

wL( L / 2)3 wL2 ( L / 2) 2 5wL4 C3 L + + C3 ( L / 2) + C4 = + + C4 48 16 384 2

Continuity conditions: Since the slope at B must be the same for both beam segments, equate Eqs. (c) and (h): 5wL3 3wL3 + C1 = + C3 192 64

(i)

(j)

Further, the deflection at B must be the same for both segments; therefore, equate Eqs. (d) and (i): wL4 C1L 5wL4 C3 L + = + + C4 192 2 384 2

(k)

Evaluate constants: Solve Eqs. (g), (j), and (k) simultaneously to determine the values of constants C1, C3, and C4: 9 wL3 17 wL3 wL4 C1 = − C3 = − C4 = 384 384 384 (a) Elastic curve equation for beam segment AB: wx 4 wLx 3 9wL3 x EI v = − + − 24 16 384 wx 16 x3 − 24 Lx 2 + 9 L3  v = − 384 EI (a) Elastic curve equation for beam segment BC: wLx 3 wL2 x 2 17 wL3 x wL4 EI v = − + − + 48 16 384 384 wL 8 x 3 − 24 Lx 2 + 17 L2 x − L3  v = − 384 EI

(0  x  L / 2)

Ans.

( L / 2  x  L)

Ans.

(b) Deflection at B: EI vB =

wL4 9wL4 5wL4 − =− 192 768 768

 vB = −

5wL4 768EI

Ans.

P10.12 For the beam and loading shown in Figure P10.12, use the double-integration method to determine (a) the equation of the elastic curve for the beam, (b) the location of the maximum deflection, and (c) the maximum beam deflection. Assume that EI is constant for the beam. FIGURE P10.12

Solution Beam FBD:

w0 L  2 L   =0 2  3  wL  By = 0 3 wL Fy = Ay + By − 0 = 0 2 wL  Ay = 0 6

M A = By L −

Moment equation:

M a − a = M ( x) +

w0 x 2  x    − Ay x 2L  3 

w0 x 2  x  w0 Lx =0  − 2L  3  6 w0 x 3 w0 Lx  M ( x) = − + 6L 6 = M ( x) +

Integration of moment equation: d 2v w x 3 w Lx EI 2 = M ( x) = − 0 + 0 dx 6L 6 4 2 dv wx w Lx EI =− 0 + 0 + C1 dx 24 L 12 w x 5 w Lx 3 EI v = − 0 + 0 + C1 x + C2 120 L 36

(a) (b)

Boundary conditions: v=0 at x=0

v=0

x=L

Evaluate constants: Substitute x = 0 and v = 0 into Eq. (b) to determine C2 = 0. Next, substitute x = L and v = 0 into Eq. (b) and solve for C1: w ( L)5 w0 L( L)3 7 w L3 EI (0) = − 0 + + C1 ( L)  C1 = − 0 120 L 36 360 (a) Elastic curve equation: w x5 w Lx3 7 w0 L3 x EI v = − 0 + 0 − 120 L 36 360

v = −

w0 x 3x 4 − 10 L2 x 2 + 7 L4  360 L EI

Ans.

(b) Location of maximum deflection: The maximum deflection occurs where the beam slope is zero. Therefore, set the beam slope equation [Eq. (a)] equal to zero: dv w x 4 w Lx 2 7 w0 L3 EI =− 0 + 0 − =0 dx 24 L 12 360 Multiply by −360L/w0 to obtain: 15 x 4 − 30 L2 x 2 + 7 L4 = 0 Solve this equation numerically to obtain: Ans. x = 0.51932962236L = 0.51933L (c) Maximum beam deflection: w (0.51933L) 3(0.51933L) 4 − 10 L2 (0.51933L) 2 + 7 L4  vmax = − 0 360 L EI

=−

w0 (0.51933) (0.0065222) w0 L4 w L4  4.52118L4  = − = −0.00652 0 360 EI EI EI

Ans.

P10.13 For the beam and loading shown in Figure P10.13, integrate the load distribution to determine (a) the equation of the elastic curve for the beam, and (b) the maximum deflection for the beam. Assume that EI is constant for the beam. FIGURE P10.13

Solution Integrate the load distribution: d 4v wx EI 4 = − 0 dx L 3 d v w x2 EI 3 = − 0 + C1 dx 2L 2 d v w0 x 3 EI 2 = − + C1 x + C2 dx 6L dv w x4 C x2 EI = − 0 + 1 + C2 x + C3 dx 24 L 2 5 wx C x3 C x 2 EI v = − 0 + 1 + 2 + C3 x + C4 120 L 6 2 Boundary conditions and evaluate constants: d 3v at x = 0, V = EI 3 = 0 dx d 2v at x = 0, M = EI 2 = 0 dx dv w0 ( L) 4 at x = L, =0 − + C3 = 0 dx 24 L w ( L)5 w0 L3 ( L) at x = L, v = 0 − 0 + + C4 = 0 120 L 24 (a) Elastic curve equation: w x5 w L3 x w0 L4 EI v = − 0 + 0 − 120 L 24 30

v = −

 C1 = 0  C2 = 0 w0 L3  C3 = 24 w L4  C4 = − 0 30

w0  x5 − 5L4 x + 4 L5  120 LEI

Ans.

(b) Maximum deflection: vmax = −

w0 w L4 (0)5 − 5L4 (0) + 4 L5  = − 0 120 LEI 30 EI

Ans.

P10.14 For the beam and loading shown in Figure P10.14, integrate the load distribution to determine (a) the equation of the elastic curve, (b) the deflection at the left end of the beam, and (c) the support reactions By and MB. Assume that EI is constant for the beam. FIGURE P10.14

Solution Integrate the load distribution: d 4v x EI 4 = − w0 cos dx 2L 3 d v 2w L x EI 3 = − 0 sin + C1 dx  2L d 2v 4 w L2 x EI 2 = 02 cos + C1 x + C2 dx  2L dv 8w0 L3  x C1 x 2 EI = sin + + C2 x + C3 dx 3 2L 2 16 w0 L4  x C1 x 3 C2 x 2 EI v = − cos + + + C3 x + C4 4 2L 6 2 Boundary conditions and evaluate constants: d 3v at x = 0, V = EI 3 = 0  C1 = 0 dx d 2v 4w0 L2  (0) 4 w0 L2 at x = 0, M = EI 2 = 0 cos + C2 = 0  C2 = − dx 2 2L 2 dv 8w0 L3  ( L) 4w0 L2 ( L) 4 w0 L3 at x = L, =0 sin − + C = 0  C = − (2 −  ) 3 3 dx 3 2L 2 3 16w0 L4  ( L) 4w0 L2 ( L) 2 4w0 L3 ( L) at x = L, v = 0 − cos − − (2 −  ) + C4 = 0 4 2L 2 2 3 2w0 L4  C4 = (4 −  ) 3 (a) Elastic curve equation: 16w0 L4  x 4w0 L2 x 2 4w0 L3 2w0 L4 EI v = − cos − − (2 −  ) + (4 −  ) 4 2L 2 2 3 3

v = −

w0  x  32 L4 cos + 4 2 L2 x 2 + 8 L3 x(2 −  ) − 4 L4 (4 −  )  4  2 EI  2L 

Ans.

(b) Deflection at left end of beam: w   (0)  v A = − 40 32 L4 cos + 4 2 L2 (0) 2 + 8 L3 (0)(2 −  ) − 4 L4 (4 −  )  2 EI  2L  =−

w0 w0 L4 4 4   32 L − 4  L (4 −  ) = −  32 − 4 (4 −  )  2 4 EI  2 4 EI

w0 L4 = −0.1089 EI

d 2v 4w0 L2  ( L) 4w0 L2 4 w0 L2 = cos − = − dx 2 x = L 2 2L 2 2

Ans.

 By =

2w0 L

MB =

4w0 L2



2



Ans. (cw)

Ans.

P10.15 For the beam and loading shown in Figure P10.15, integrate the load distribution to determine (a) the equation of the elastic curve, (b) the deflection midway between the supports, (c) the slope at the left end of the beam, and (d) the support reactions Ay and By. Assume that EI is constant for the beam. FIGURE P10.15

Solution Integrate the load distribution: d 4v x EI 4 = − w0 sin dx L 3 d v wL x EI 3 = 0 cos + C1 dx  L d 2v w L2 x EI 2 = 0 2 sin + C1 x + C2 dx  L dv w L3  x C1 x 2 EI = − 0 3 cos + + C2 x + C3 dx  L 2 w0 L4  x C1 x 3 C2 x 2 EI v = − 4 sin + + + C3 x + C4  L 6 2 Boundary conditions and evaluate constants: d 2v at x = 0, M = EI 2 = 0 dx d 2v w0 L2  ( L) at x = L, M = EI 2 = 0 sin + C1 ( L) = 0 2 dx  L at x = 0, v = 0 at x = L, v = 0

−

w0 L



sin

 ( L) L

 C2 = 0

 C1 = 0

 C4 = 0

+ C3 ( L) = 0

 C3 = 0

(a) Elastic curve equation: EI v = −

w0 L4

4

sin

x L

w0 L4 x  v = − 4 sin  EI L

Ans.

(b) Deflection midway between the supports: vx = L / 2 = −

w0 L4  ( L / 2) w0 L4 sin = −  4 EI L  4 EI

Ans.

dv w L3  (0) w L3 = EI  A = − 0 3 cos = − 03 dx A  L 

 A = −

w0 L3  3 EI

Ans.

(d) Support reactions Ay and By: d 3v wL  (0) w0 L VA = EI 3 = 0 cos = dx x = 0  L  VB = EI

d 3v wL  ( L) wL = 0 cos =− 0 3 dx x = L  L 

 Ay =

w0 L



Ans.

 By =

w0 L



Ans.

 

P10.16 For the beam and loading shown in Figure P10.16, integrate the load distribution to determine (a) the equation of the elastic curve, (b) the deflection midway between the supports, (c) the slope at the left end of the beam, and (d) the support reactions Ay and By. Assume that EI is constant for the beam. FIGURE P10.16

Solution Integrate the load distribution: d 4v x EI 4 = − w0 sin dx 2L 3 d v 2 w0 L x EI 3 = cos + C1 dx  2L d 2v 4 w L2 x EI 2 = 02 sin + C1 x + C2 dx  2L dv 8w L3  x C1 x 2 EI = − 03 cos + + C2 x + C3 dx  2L 2 16 w0 L4  x C1 x 3 C2 x 2 EI v = − sin + + + C3 x + C4 4 2L 6 2 Boundary conditions and evaluate constants: d 2v at x = 0, M = EI 2 = 0 dx d 2v 4w0 L2  ( L) at x = L, M = EI 2 = 0 sin + C1 ( L) = 0 2 dx  2L at x = 0, v = 0 at x = L, v = 0

−

16 w0 L

4

 ( L)

4 w L( L) sin − 0 2 + C3 ( L) = 0 2L 6

 C2 = 0

 C1 = −

4w0 L

2

 C4 = 0 2 w0 L3  C3 = (24 +  2 ) 4 3

(a) Elastic curve equation: 16w0 L4  x 4w0 Lx3 2w0 L3 x EI v = − sin − + (24 +  2 ) 4 2 4  2L 6 3

v = −

2w0  x  24 L4 sin +  2 Lx3 − (24 +  2 ) L3 x  4  3 EI  2L 

Ans.

(b) Deflection midway between the supports: 3 2w   ( L / 2)  L  L  vx = L / 2 = − 4 0  24 L4 sin +  2 L  − (24 +  2 ) L3     2  2   3 EI  2L 2 w L4    2 (24 +  2 )  = − 40  24sin + −  3 EI  4 8 2  2 w0 L4 = − 4 (1.2694611) 3 EI = −0.0086882

w0 L4 w L4 = −0.00869 0 EI EI

Ans.

(c) Slope at the left end of the beam: dv 8w0 L3  (0) 2w0 L(0) 2 2w0 L3 EI = EI  A = − 3 cos − + (24 +  2 ) 2 4 dx A  2L  3

=−

8w0 L3

3

2w0 L3 2   8 16 (24 +  2 ) = − w0 L3  3 − 4 − 2  = −0.026209w0 L3 4 3  3  

w0 L3   A = −0.0262 EI

Ans.

(d) Support reactions Ay and By: d 3v 2w L  (0) 4w0 L 2w0 L VA = EI 3 = 0 cos − 2 = ( − 2) dx x = 0  2L  2  Ay =

VB = EI

2w0 L

2

( − 2) 

Ans.

d 3v 2w L  ( L) 4w0 L 4w L = 0 cos − 2 = − 02 3 dx x = L  2L  

 By =

4w0 L

2



Ans.

P10.17 For the beam and loading shown, use discontinuity functions to compute the deflection of the beam at D. Assume a constant value of EI = 1,750 kip-ft2 for the beam.

FIGURE P10.17

Solution Support reactions: A FBD of the beam is shown to the right. M A = − (5 kips)(4 ft) − (3 kips)(13 ft) + C y (10 ft) = 0  C y = 5.90 kips Fy = Ay + C y − 5 kips − 3 kips = 0  Ay = 2.10 kips

Load function w(x): −1 −1 −1 −1 w( x) = 2.10 kips x − 0 ft − 5 kips x − 4 ft + 5.90 kips x − 10 ft − 3 kips x − 13 ft Shear-force function V(x) and bending-moment function M(x): 0 0 0 0 V ( x) = 2.10 kips x − 0 ft − 5 kips x − 4 ft + 5.90 kips x − 10 ft − 3 kips x − 13 ft

M ( x) = 2.10 kips x − 0 ft − 5 kips x − 4 ft + 5.90 kips x − 10 ft − 3 kips x − 13 ft 1

Equations for beam slope and beam deflection: From Eq. (10.1), we can write: d 2v 1 1 1 1 EI 2 = M ( x) = 2.10 kips x − 0 ft − 5 kips x − 4 ft + 5.90 kips x − 10 ft − 3 kips x − 13 ft dx Integrate the moment function to obtain an expression for the beam slope: dv 2.10 kips 5 kips 2 2 EI = x − 0 ft − x − 4 ft dx 2 2 5.90 kips 3 kips 2 2 (a) + x − 10 ft − x − 13 ft + C1 2 2 Integrate again to obtain the beam deflection function: 2.10 kips 5 kips 3 3 EI v = x − 0 ft − x − 4 ft 6 6 5.90 kips 3 kips 3 3 (b) + x − 10 ft − x − 13 ft + C1 x + C2 6 6 Evaluate constants using boundary conditions: Boundary conditions are specific values of deflection v or slope dv/dx that are known at particular locations along the beam span. For this beam, the deflection v is known at the pin support (x = 0 ft) and at the roller support (x = 10 ft). Substitute the boundary condition v = 0 at x = 0 ft into Eq. (b) to obtain: C2 = 0

Next, substitute the boundary condition v = 0 at x = 10 ft into Eq. (b) to obtain: 2.10 kips 5 kips EI v = (10 ft)3 − (6 ft)3 + C1 (10 ft) = 0 6 6  C1 = −17 kip-ft 2 The beam slope and elastic curve equations are now complete: dv 2.10 kips 5 kips 2 2 EI = x − 0 ft − x − 4 ft dx 2 2 5.90 kips 3 kips 2 2 + x − 10 ft − x − 13 ft − 17 kip-ft 2 2 2

2.10 kips 5 kips 3 3 x − 0 ft − x − 4 ft 6 6 5.90 kips 3 kips 3 3 + x − 10 ft − x − 13 ft − (17 kip-ft 2 ) x 6 6

(c)

EI v =

(d)

Beam deflection at D: At the tip of the overhang where x = 13 ft, the beam deflection is: 2.10 kips 5 kips 5.90 kips EI vD = (13 ft)3 − (9 ft)3 + (3 ft)3 − (17 kip-ft 2 )(13 ft) 6 6 6 = −33.000 kip-ft 3

 vD = −

33.000 kip-ft 3 = −0.018857 ft = 0.226 in.  1,750 kip-ft 2

Ans.

P10.18 The solid 30 mm diameter steel [E = 200 GPa] shaft shown in Figure P10.18 supports two pulleys. For the loading shown, use discontinuity functions to compute: (a) the shaft deflection at pulley B. (b) the shaft deflection at pulley C.

FIGURE P10.18

Solution Support reactions: A FBD of the beam is shown to the right. Fy = Ay − 800 N − 500 N = 0  Ay = 1,300 N M A = −(800 N)(250 mm) − (500 N)(600 mm) − M A = 0  M A = −500,000 N-mm Load function w(x): −2 −1 w( x) = −500,000 N-mm x − 0 mm + 1,300 N x − 0 mm

−800 N x − 250 mm

−1

− 500 N x − 600 mm

−1

Shear-force function V(x) and bending-moment function M(x): −1 0 V ( x) = −500,000 N-mm x − 0 mm + 1,300 N x − 0 mm

−800 N x − 250 mm − 500 N x − 600 mm 0

M ( x) = −500,000 N-mm x − 0 mm + 1,300 N x − 0 mm 0

−800 N x − 250 mm − 500 N x − 600 mm 1

Equations for beam slope and beam deflection: From Eq. (10.1), we can write: d 2v 0 1 EI 2 = M ( x) = −500,000 N-mm x − 0 mm + 1,300 N x − 0 mm dx −800 N x − 250 mm − 500 N x − 600 mm Integrate the moment function to obtain an expression for the beam slope: dv 1,300 N 1 2 EI = −500,000 N-mm x − 0 mm + x − 0 mm dx 2 800 N 500 N 2 2 − x − 250 mm − x − 600 mm + C1 2 2 Integrate again to obtain the beam deflection function: 500,000 N-mm 1,300 N 2 3 EI v = − x − 0 mm + x − 0 mm 2 6 800 N 500 N 3 3 − x − 250 mm − x − 600 mm + C1 x + C2 6 6 1

(a)

(b)

Evaluate constants using boundary conditions: Boundary conditions are specific values of deflection v or slope dv/dx that are known at particular locations along the beam span. For this beam, both the slope dv/dx and the deflection v are known at the fixed support (x = 0 mm). Substitute the boundary condition dv/dx = 0 at x = 0 mm into Eq. (a) to obtain: C1 = 0 Next, substitute the boundary condition v = 0 at x = 0 mm into Eq. (b) to obtain: C2 = 0 The beam slope and elastic curve equations are now complete: dv 1,300 N 1 2 EI = −500,000 N-mm x − 0 mm + x − 0 mm dx 2 800 N 500 N 2 2 − x − 250 mm − x − 600 mm 2 2

500,000 N-mm 1,300 N 2 3 x − 0 mm + x − 0 mm 2 6 800 N 500 N 3 3 − x − 250 mm − x − 600 mm 6 6

EI v = −

Section properties: I=



(30 mm) 4 = 39,750.782 mm 4 64  EI = 7.9522  109 N-mm 2

E = 200 GPa = 200,000 N/mm 2

(a) Beam deflection at B: The beam deflection at B where x = 250 mm is: 500,000 N-mm 1,300 N EI vB = − (250 mm)2 + (250 mm)3 2 6 9 3 12.2396  10 N-mm  vB = − = −1.5392 mm = 1.539 mm  7.9522  109 N-mm2 (b) Beam deflection at C: The beam deflection at C where x = 600 mm is: 500,000 N-mm 1,300 N 800 N EI vC = − (600 mm)2 + (600 mm)3 − (350 mm)3 2 6 6 9 3 48.9167  10 N-mm  vC = − = −6.1514 mm = 6.15 mm  7.9522  109 N-mm 2

Ans.

P10.19 For the beam and loading shown, use discontinuity functions to compute (a) the slope of the beam at C and (b) the deflection of the beam at C. Assume a constant value of EI = 560×106 N-mm2 for the beam. FIGURE P10.19

Solution Support reactions: A FBD of the beam is shown to the right. M A = (210 N-m)(1,000 mm/m) − (1, 400 N)(450 mm) + E y (700 mm) = 0  E y = 600 N Fy = Ay + E y − 1, 400 N = 0  Ay = 800 N

Load function w(x): −1 −2 w( x) = 800 N x − 0 mm − 210,000 N-mm x − 200 mm

−1,400 N x − 450 mm

−1

+ 600 N x − 700 mm

Shear-force function V(x) and bending-moment function M(x): 0 −1 V ( x) = 800 N x − 0 mm − 210,000 N-mm x − 200 mm

−1,400 N x − 450 mm + 600 N x − 700 mm 0

M ( x) = 800 N x − 0 mm − 210,000 N-mm x − 200 mm 1

−1,400 N x − 450 mm + 600 N x − 700 mm 1

Equations for beam slope and beam deflection: From Eq. (10.1), we can write: d 2v 1 0 EI 2 = M ( x) = 800 N x − 0 mm − 210,000 N-mm x − 200 mm dx −1, 400 N x − 450 mm + 600 N x − 700 mm Integrate the moment function to obtain an expression for the beam slope: dv 800 N 2 1 EI = x − 0 mm − 210,000 N-mm x − 200 mm dx 2 1,400 N 600 N 2 2 − x − 450 mm + x − 700 mm + C1 2 2 Integrate again to obtain the beam deflection function: 800 N 210,000 N-mm 3 2 EI v = x − 0 mm − x − 200 mm 6 2 1,400 N 600 N 3 3 − x − 450 mm + x − 700 mm + C1x + C2 6 6 1

(a)

(b)

Evaluate constants using boundary conditions: Boundary conditions are specific values of deflection v or slope dv/dx that are known at particular locations along the beam span. For this beam, the deflection v is known at the pin support (x = 0 mm) and at the roller support (x = 700 mm). Substitute the boundary condition v = 0 at x = 0 mm into Eq. (b) to obtain: C2 = 0 Next, substitute the boundary condition v = 0 at x = 700 mm into Eq. (b) to obtain: 800 N 210,000 N-mm 1,400 N 0= (700 mm)3 − (500 mm) 2 − (250 mm)3 + C1 (700 mm) 6 2 6 2  C1 = −22,625,000 N-mm The beam slope and elastic curve equations are now complete: dv 800 N 2 1 EI = x − 0 mm − 210,000 N-mm x − 200 mm dx 2 1,400 N 600 N 2 2 − x − 450 mm + x − 700 mm − 22,625,000 N-mm2 2 2

EI v =

800 N 210,000 N-mm 3 2 x − 0 mm − x − 200 mm 6 2 1,400 N 600 N 3 3 − x − 450 mm + x − 700 mm − (22,625,000 N-mm2 ) x 6 6

(a) Beam slope at C: The beam slope at C is: dv 800 N EI = (350 mm)2 − (210,000 N-mm)(150 mm) − 22,625,000 N-mm2 dx C 2

dv 5.125  106 N-mm2  =− = −0.009152 rad = −0.00915 rad dx C 560  106 N-mm2

Ans.

(b) Beam deflection at C: The beam deflection at C is: 800 N 210,000 N-mm EI vC = (350 mm)3 − (150 mm) 2 − (22,625,000 N-mm 2 )(350 mm) 6 2 9 3 = −4.564583  10 N-mm  vC = −

4.564583  109 N-mm3 = −8.1510 mm = 8.15 mm  560  106 N-mm 2

Ans.

P10.20 The solid 30 mm diameter steel [E = 200 GPa] shaft shown in Figure P10.20 supports two belt pulleys. Assume that the bearing at A can be idealized as a pin support and that the bearing at E can be idealized as a roller support. For the loading shown, use discontinuity functions to compute: (a) the shaft deflection at pulley B. (b) the shaft deflection at point C. FIGURE P10.20

Solution Support reactions: A FBD of the beam is shown to the right. M A = −(600 N)(300 mm) − (800 N)(800 mm) + E y (1,000 mm) = 0  E y = 820 N Fy = Ay + E y − 600 N − 800 N = 0  Ay = 580 N

Load function w(x): −1 −1 −1 −1 w( x) = 580 N x − 0 mm − 600 N x − 300 mm − 800 N x − 800 mm + 820 N x − 1,000 mm Shear-force function V(x) and bending-moment function M(x): 0 0 0 0 V ( x) = 580 N x − 0 mm − 600 N x − 300 mm − 800 N x − 800 mm + 820 N x − 1,000 mm

M ( x) = 580 N x − 0 mm − 600 N x − 300 mm − 800 N x − 800 mm + 820 N x − 1,000 mm 1

Equations for beam slope and beam deflection: From Eq. (10.1), we can write: d 2v 1 1 EI 2 = M ( x) = 580 N x − 0 mm − 600 N x − 300 mm dx

−800 N x − 800 mm + 820 N x − 1,000 mm Integrate the moment function to obtain an expression for the beam slope: dv 580 N 600 N 800 N 2 2 2 EI = x − 0 mm − x − 300 mm − x − 800 mm dx 2 2 2 820 N 2 + x − 1,000 mm + C1 2 Integrate again to obtain the beam deflection function: 580 N 600 N 800 N 3 3 3 EI v = x − 0 mm − x − 300 mm − x − 800 mm 6 6 6 820 N 3 + x − 1,000 mm + C1 x + C2 6 1

(a)

(b)

Evaluate constants using boundary conditions: Boundary conditions are specific values of deflection v or slope dv/dx that are known at particular locations along the beam span. For this beam, the deflection Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

v is known at the pin support (x = 0 mm) and at the roller support (x = 1,000 mm). Substitute the boundary condition v = 0 at x = 0 mm into Eq. (b) to obtain: C2 = 0 Next, substitute the boundary condition v = 0 at x = 1,000 mm into Eq. (b) to obtain: 580 N 600 N 800 N 0= (1,000 mm)3 − (700 mm)3 − (200 mm)3 + C1 (1,000 mm) 6 6 6 6 2  C1 = −61.3  10 N-mm The beam slope and elastic curve equations are now complete: dv 580 N 600 N 800 N 2 2 2 EI = x − 0 mm − x − 300 mm − x − 800 mm dx 2 2 2 820 N 2 + x − 1,000 mm − 61.3  106 N-mm 2 2

EI v =

580 N 600 N 800 N 3 3 3 x − 0 mm − x − 300 mm − x − 800 mm 6 6 6 820 N 3 + x − 1,000 mm − (61.3  106 N-mm 2 ) x 6

Section properties: I=



(30 mm) 4 = 39,750.782 mm 4

64  EI = 7.9522  109 N-mm 2

E = 200 GPa = 200,000 N/mm 2

(a) Beam deflection at B: The beam deflection at B where x = 300 mm is: 580 N EI vB = (300 mm)3 − (61.3  106 N-mm 2 )(300 mm) 6 15.7800  109 N-mm3  vB = − = −1.9844 mm = 1.984 mm  7.9522  109 N-mm2 (b) Beam deflection at C: The beam deflection at C where x = 500 mm is: 580 N 600 N EI vC = (500 mm)3 − (200 mm)3 − (61.3  106 N-mm2 )(500 mm) 6 6 9 3 19.3667  10 N-mm  vC = − = −2.4354 mm = 2.44 mm  7.9522  109 N-mm2

Ans.

P10.21 The cantilever beam shown in Figure P10.21 consists of a W530 × 74 structural steel wide-flange shape [E = 200 GPa; I = 410 × 106 mm4]. Use discontinuity functions to compute the deflection of the beam at C for the loading shown. FIGURE P10.21

Solution Support reactions: A FBD of the beam is shown to the right. Fy = Ay − (30 kN/m)(3 m) − 40 kN = 0  Ay = 130 kN M A = −(30 kN/m)(3 m)(1.5 m) − (40 kN)(5 m) − M A = 0  M A = −335 kN-m Load function w(x): −1 −2 w( x) = 130 kN x − 0 m − 335 kN-m x − 0 m

−30 kN/m x − 0 m + 30 kN/m x − 3 m − 40 kN x − 5 m 0

−1

Shear-force function V(x) and bending-moment function M(x): 0 −1 V ( x) = 130 kN x − 0 m − 335 kN-m x − 0 m

−30 kN/m x − 0 m + 30 kN/m x − 3 m − 40 kN x − 5 m 1

M ( x) = 130 kN x − 0 m − 335 kN-m x − 0 m 1

−

30 kN/m 30 kN/m 2 2 1 x−0 m + x − 3 m − 40 kN x − 5 m 2 2

Equations for beam slope and beam deflection: From Eq. (10.1), we can write: d 2v 1 0 EI 2 = M ( x) = 130 kN x − 0 m − 335 kN-m x − 0 m dx 30 kN/m 30 kN/m 2 2 1 − x−0 m + x − 3 m − 40 kN x − 5 m 2 2 Integrate the moment function to obtain an expression for the beam slope: dv 130 kN 2 1 EI = x − 0 m − 335 kN-m x − 0 m dx 2 30 kN/m 30 kN/m 40 kN 3 3 2 − x−0 m + x−3 m − x − 5 m + C1 6 6 2 Integrate again to obtain the beam deflection function: 130 kN 335 kN-m 3 2 EI v = x−0 m − x−0 m 6 2 30 kN/m 30 kN/m 40 kN 4 4 3 − x−0 m + x−3 m − x − 5 m + C1x + C2 24 24 6

(a)

(b)

Evaluate constants using boundary conditions: Boundary conditions are specific values of deflection v or slope dv/dx that are known at particular locations along the beam span. For this beam, both the slope dv/dx and the deflection v are known at the fixed support (x = 0 m). Substitute the boundary condition dv/dx = 0 at x = 0 m into Eq. (a) to obtain: C1 = 0 Next, substitute the boundary condition v = 0 at x = 0 m into Eq. (b) to obtain: C2 = 0 The beam slope and elastic curve equations are now complete: dv 130 kN 2 1 EI = x − 0 m − 335 kN-m x − 0 m dx 2 30 kN/m 30 kN/m 40 kN 3 3 2 − x−0 m + x−3 m − x−5 m 6 6 2

130 kN 335 kN-m 3 2 x−0 m − x−0 m 6 2 30 kN/m 30 kN/m 40 kN 4 4 3 − x−0 m + x−3 m − x−5 m 24 24 6

EI v =

Beam deflection at C: For the W530 × 74 structural steel wide-flange shape, EI = 82,000 kN-m2. At the tip of the overhang where x = 5 m, the beam deflection is: 130 kN 335 kN-m 30 kN/m 30 kN/m EI vC = (5 m)3 − (5 m) 2 − (5 m) 4 + (2 m) 4 6 2 24 24 3 = −2,240.416667 kN-m

2,240.416667 kN-m3  vC = − = −0.027322 m = 27.3 mm  82,000 kN-m 2

Ans.

P10.22 The cantilever beam shown in Figure P10.22 consists of a W21 × 50 structural steel wide-flange shape [E = 29,000 ksi; I = 984 in.4]. Use discontinuity functions to compute the deflection of the beam at D for the loading shown. FIGURE P10.22

Solution Support reactions: A FBD of the beam is shown to the right.

M A = −(9 kips)(4 ft)

− 12 ( 4 kips/ft )( 9 ft ) (13 ft) − M A = 0

 M A = −270.00 kip-ft

Fy = Ay − (9 kips) − 12 ( 4 kips/ft )( 9 ft ) = 0  Ay = 27.00 kips Load function w(x):

w( x) = −270 kip-ft x − 0 ft +

−2

+ 27 kips x − 0 ft

−1

− 9 kips x − 4 ft

−

4 kips/ft 1 x − 7 ft 9 ft

4 kips/ft 1 0 x − 16 ft + 4 kips/ft x − 16 ft 9 ft

Shear-force function V(x) and bending-moment function M(x): −

4 kips/ft 2 x − 7 ft 2(9 ft)

M ( x) = −270 kip-ft x − 0 ft + 27 kips x − 0 ft − 9 kips x − 4 ft −

4 kips/ft 3 x − 7 ft 6(9 ft)

V ( x) = −270 kip-ft x − 0 ft +

−1

4 kips/ft 2 1 x − 16 ft + 4 kips/ft x − 16 ft 2(9 ft) 0

+ 27 kips x − 0 ft − 9 kips x − 4 ft

4 kips/ft 4 kips/ft 3 2 x − 16 ft + x − 16 ft 6(9 ft) 2

Equations for beam slope and beam deflection: From Eq. (10.1), we can write: d 2v 0 1 1 EI 2 = M ( x) = −270 kip-ft x − 0 ft + 27 kips x − 0 ft − 9 kips x − 4 ft dx 4 kips/ft 4 kips/ft 4 kips/ft 3 3 2 − x − 7 ft + x − 16 ft + x − 16 ft 6(9 ft) 6(9 ft) 2 Integrate the moment function to obtain an expression for the beam slope:

dv 27 kips 9 kips 1 2 2 = −270 kip-ft x − 0 ft + x − 0 ft − x − 4 ft dx 2 2 4 kips/ft 4 kips/ft 4 kips/ft 4 4 3 − x − 7 ft + x − 16 ft + x − 16 ft + C1 24(9 ft) 24(9 ft) 6 Integrate again to obtain the beam deflection function: 270 kip-ft 27 kips 9 kips 2 3 3 EI v = − x − 0 ft + x − 0 ft − x − 4 ft 2 6 6 4 kips/ft 4 kips/ft 4 kips/ft 5 5 4 − x − 7 ft + x − 16 ft + x − 16 ft + C1 x + C2 120(9 ft) 120(9 ft) 24 EI

(a)

(b)

Evaluate constants using boundary conditions: Boundary conditions are specific values of deflection v or slope dv/dx that are known at particular locations along the beam span. For this beam, both the slope dv/dx and the deflection v are known at the fixed support (x = 0 ft). Substitute the boundary condition dv/dx = 0 at x = 0 ft into Eq. (a) to obtain: C1 = 0 Next, substitute the boundary condition v = 0 at x = 0 ft into Eq. (b) to obtain: C2 = 0 The beam slope and elastic curve equations are now complete: dv 27 kips 9 kips 4 kips/ft 1 2 2 4 EI = −270 kip-ft x − 0 ft + x − 0 ft − x − 4 ft − x − 7 ft dx 2 2 24(9 ft) 4 kips/ft 4 kips/ft 4 3 + x − 16 ft + x − 16 ft 24(9 ft) 6 270 kip-ft 27 kips 9 kips 4 kips/ft 2 3 3 5 x − 0 ft + x − 0 ft − x − 4 ft − x − 7 ft 2 6 6 120(9 ft) 4 kips/ft 4 kips/ft 5 4 + x − 16 ft + x − 16 ft 120(9 ft) 24

EI v = −

Beam deflection at D: For the W21 × 50 structural steel wide-flange shape, EI = 198,166.658 kip-ft2. At the tip of the overhang where x = 16 ft, the beam deflection is: 270 kip-ft 27 kips 9 kips 4 kips/ft EI vD = − (16 ft) 2 + (16 ft)3 − (12 ft)3 − (9 ft)5 2 6 6 120(9 ft) = −18,938.7 kip-ft 3  vD = −

18,938.7 kip-ft 3 = −0.095570 ft = 1.147 in.  198,166.658 kip-ft 2

Ans.

P10.23 The simply supported beam shown in Figure P10.23 consists of a W410 × 85 structural steel wide-flange shape [E = 200 GPa; I = 316 × 106 mm4]. For the loading shown, use discontinuity functions to compute (a) the slope of the beam at A and (b) the deflection of the beam at midspan. FIGURE P10.23

Solution Support reactions: A FBD of the beam is shown to the right. M A = −(75 kN/m)(2.5 m)(1.25 m) −(75 kN/m)(2.5 m)(6.75 m) + Dy (8 m) = 0  Dy = 187.5 kN Fy = Ay + Dy − (75 kN/m)(2.5 m) − (75 kN/m)(2.5 m) = 0  Ay = 187.5 kN

Load function w(x): −1 0 0 w( x) = 187.5 kN x − 0 m − 75 kN/m x − 0 m + 75 kN/m x − 2.5 m

−75 kN/m x − 5.5 m + 75 kN/m x − 8 m + 187.5 kN x − 8 m 0

−1

Shear-force function V(x) and bending-moment function M(x): 0 1 1 V ( x) = 187.5 kN x − 0 m − 75 kN/m x − 0 m + 75 kN/m x − 2.5 m

−75 kN/m x − 5.5 m + 75 kN/m x − 8 m + 187.5 kN x − 8 m 75 kN/m 75 kN/m 1 2 2 M ( x) = 187.5 kN x − 0 m − x−0 m + x − 2.5 m 2 2 75 kN/m 75 kN/m 2 2 1 − x − 5.5 m + x − 8 m + 187.5 kN x − 8 m 2 2 1

Equations for beam slope and beam deflection: From Eq. (10.1), we can write: d 2v 75 kN/m 75 kN/m 1 2 2 EI 2 = M ( x) = 187.5 kN x − 0 m − x−0 m + x − 2.5 m dx 2 2 75 kN/m 75 kN/m 2 2 1 − x − 5.5 m + x − 8 m + 187.5 kN x − 8 m 2 2 Integrate the moment function to obtain an expression for the beam slope: dv 187.5 kN 75 kN/m 75 kN/m 2 3 3 EI = x−0 m − x−0 m + x − 2.5 m dx 2 6 6 75 kN/m 75 kN/m 187.5 kN 3 3 2 − x − 5.5 m + x −8 m + x − 8 m + C1 6 6 2 Integrate again to obtain the beam deflection function: 187.5 kN 75 kN/m 75 kN/m 3 4 4 EI v = x−0 m − x−0 m + x − 2.5 m 6 24 24 75 kN/m 75 kN/m 187.5 kN 4 4 3 − x − 5.5 m + x −8 m + x − 8 m + C1 x + C2 24 24 6

(a)

(b)

Evaluate constants using boundary conditions: Boundary conditions are specific values of deflection v or slope dv/dx that are known at particular locations along the beam span. For this beam, the deflection v is known at the pin support (x = 0 m) and at the roller support (x = 8 m). Substitute the boundary condition v = 0 at x = 0 m into Eq. (b) to obtain: C2 = 0 Next, substitute the boundary condition v = 0 at x = 8 m into Eq. (b) to obtain: 187.5 kN 75 kN/m 75 kN/m 75 kN/m 0= (8 m)3 − (8 m) 4 + (5.5 m) 4 − (2.5 m) 4 + C1 (8 m) 6 24 24 24 2  C1 = −742.1875 kN-m The beam slope and elastic curve equations are now complete: dv 187.5 kN 75 kN/m 75 kN/m 2 3 3 EI = x−0 m − x−0 m + x − 2.5 m dx 2 6 6 75 kN/m 75 kN/m 187.5 kN 3 3 2 − x − 5.5 m + x −8 m + x − 8 m − 742.1875 kN-m2 6 6 2

187.5 kN 75 kN/m 75 kN/m 3 4 4 x−0 m − x−0 m + x − 2.5 m 6 24 24 75 kN/m 75 kN/m 187.5 kN 4 4 3 − x − 5.5 m + x −8 m + x − 8 m − (742.1875 kN-m2 ) x 24 24 6

EI v =

(a) Beam slope at A: For the W410 × 85 structural steel wide-flange shape, EI = 63,200 kN-m2. The beam slope at A is: dv EI = −742.1875 kN-m 2 dx A

dv 742.1875 kN-m 2  =− = −0.011743 rad = −0.01174 rad dx A 63,200 kN-m2

Ans.

(b) Beam deflection at midspan: At midspan where x = 4 m, the beam deflection is: 187.5 kN 75 kN/m 75 kN/m EI vmidspan = (4 m)3 − (4 m) 4 + (1.5 m) 4 − (742.1875 kN-m 2 )(4 m) 6 24 24 3 = −1,752.929687 kN-m

 vmidspan = −

1,752.929687 kN-m3 = −0.027736 m = 27.7 mm  63,200 kN-m 2

Ans.

P10.24 The simply supported beam shown in Figure P10.24 consists of a W14 × 30 structural steel wideflange shape [E = 29,000 ksi; I = 291 in.4]. For the loading shown, use discontinuity functions to compute (a) the slope of the beam at A and (b) the deflection of the beam at midspan. FIGURE P10.24

Solution Support reactions: A FBD of the beam is shown to the right. M A = − (2.5 kips/ft)(12 ft)(12 ft) + Dy (24 ft) = 0  Dy = 15 kips Fy = Ay + Dy − (2.5 kips/ft)(12 ft) = 0  Ay = 15 kips

Load function w(x): −1 0 0 −1 w( x) = 15 kips x − 0 ft − 2.5 kips/ft x − 6 ft + 2.5 kips/ft x − 18 ft + 15 kips x − 24 ft Shear-force function V(x) and bending-moment function M(x): 0 1 1 0 V ( x) = 15 kips x − 0 ft − 2.5 kips/ft x − 6 ft + 2.5 kips/ft x − 18 ft + 15 kips x − 24 ft 2.5 kips/ft 2.5 kips/ft 1 2 2 1 M ( x) = 15 kips x − 0 ft − x − 6 ft + x − 18 ft + 15 kips x − 24 ft 2 2 Equations for beam slope and beam deflection: From Eq. (10.1), we can write: d 2v 2.5 kips/ft 1 2 EI 2 = M ( x) = 15 kips x − 0 ft − x − 6 ft dx 2 2.5 kips/ft 2 1 + x − 18 ft + 15 kips x − 24 ft 2 Integrate the moment function to obtain an expression for the beam slope: dv 15 kips 2.5 kips/ft 2 3 EI = x − 0 ft − x − 6 ft dx 2 6 2.5 kips/ft 15 kips 3 2 + x − 18 ft + x − 24 ft + C1 6 2 Integrate again to obtain the beam deflection function: 15 kips 2.5 kips/ft 3 4 EI v = x − 0 ft − x − 6 ft 6 24 2.5 kips/ft 15 kips 4 3 + x − 18 ft + x − 24 ft + C1x + C2 24 6

(a)

(b)

Evaluate constants using boundary conditions: Boundary conditions are specific values of deflection v or slope dv/dx that are known at particular locations along the beam span. For this beam, the deflection v is known at the pin support (x = 0 ft) and at the roller support (x = 24 ft). Substitute the boundary condition v = 0 at x = 0 ft into Eq. (b) to obtain: Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

C2 = 0 Next, substitute the boundary condition v = 0 at x = 24 ft into Eq. (b) to obtain: 15 kips 2.5 kips/ft 2.5 kips/ft 0= (24 ft)3 − (18 ft) 4 + (6 ft) 4 + C1 (24 ft) 6 24 24 2  C1 = −990 kip-ft

The beam slope and elastic curve equations are now complete: dv 15 kips 2.5 kips/ft 2 3 EI = x − 0 ft − x − 6 ft dx 2 6 2.5 kips/ft 15 kips 3 2 + x − 18 ft + x − 24 ft − 990 kip-ft 2 6 2

15 kips 2.5 kips/ft 3 4 x − 0 ft − x − 6 ft 6 24 2.5 kips/ft 15 kips 4 3 + x − 18 ft + x − 24 ft − (990 kip-ft 2 ) x 24 6

EI v =

(a) Beam slope at A: For the W14 × 30 structural steel wide-flange shape, EI = 58,604.164 kip-ft2. The beam slope at A is: dv EI = −990 kip-ft 2 dx A



dv 990 kip-ft 2 =− = −0.016893 rad = −0.01689 rad dx A 58,604.164 kip-ft 2

(b) Beam deflection at midspan: At midspan where x = 12 ft, the beam deflection is: 15 kips 2.5 kips/ft EI vmidspan = (12 ft)3 − (6 ft) 4 − (990 kip-ft 2 )(12 ft) = −7,695 kip-ft 3 6 24 7,695 kip-ft 3  vmidspan = − = −0.131305 ft = 1.576 in.  58,604.164 kip-ft 2

Ans.

P10.25 The simply supported beam shown in Figure P10.25 consists of a W21 × 50 structural steel wideflange shape [E = 29,000 ksi; I = 984 in.4]. For the loading shown, use discontinuity functions to compute (a) the slope of the beam at A and (b) the deflection of the beam at B. FIGURE P10.25

Solution Support reactions: A FBD of the beam is shown to the right. M A = −(7 kips/ft)(11 ft)(5.5 ft) − (4 kips/ft)(9 ft)(15.5 ft) +C y (20 ft) = 0  C y = 49.075 kips Fy = Ay + C y − (7 kips/ft)(11 ft) − (4 kips/ft)(9 ft) = 0  Ay = 63.925 kips

Load function w(x): −1 0 0 w( x) = 63.925 kips x − 0 ft − 7 kips/ft x − 0 ft + 7 kips/ft x − 11 ft

−4 kips/ft x − 11 ft + 4 kips/ft x − 20 ft + 49.075 kips x − 20 ft 0

−1

Shear-force function V(x) and bending-moment function M(x): 0 1 1 V ( x) = 63.925 kips x − 0 ft − 7 kips/ft x − 0 ft + 7 kips/ft x − 11 ft

−4 kips/ft x − 11 ft + 4 kips/ft x − 20 ft + 49.075 kips x − 20 ft 7 kips/ft 7 kips/ft 1 2 2 M ( x) = 63.925 kips x − 0 ft − x − 0 ft + x − 11 ft 2 2 4 kips/ft 4 kips/ft 2 2 1 − x − 11 ft + x − 20 ft + 49.075 kips x − 20 ft 2 2 1

Equations for beam slope and beam deflection: From Eq. (10.1), we can write: d 2v 7 kips/ft 7 kips/ft 1 2 2 EI 2 = M ( x) = 63.925 kips x − 0 ft − x − 0 ft + x − 11 ft dx 2 2 4 kips/ft 4 kips/ft 2 2 1 − x − 11 ft + x − 20 ft + 49.075 kips x − 20 ft 2 2 Integrate the moment function to obtain an expression for the beam slope: dv 63.925 kips 7 kips/ft 7 kips/ft 2 3 3 EI = x − 0 ft − x − 0 ft + x − 11 ft dx 2 6 6 4 kips/ft 4 kips/ft 49.075 kips 3 3 2 − x − 11 ft + x − 20 ft + x − 20 ft + C1 6 6 2 Integrate again to obtain the beam deflection function:

(a)

63.925 kips 7 kips/ft 7 kips/ft 3 4 4 x − 0 ft − x − 0 ft + x − 11 ft 6 24 24 4 kips/ft 4 kips/ft 49.075 kips 4 4 3 − x − 11 ft + x − 20 ft + x − 20 ft + C1x + C2 24 24 6

EI v =

(b)

Evaluate constants using boundary conditions: Boundary conditions are specific values of deflection v or slope dv/dx that are known at particular locations along the beam span. For this beam, the deflection v is known at the pin support (x = 0 ft) and at the roller support (x = 20 ft). Substitute the boundary condition v = 0 at x = 0 ft into Eq. (b) to obtain: C2 = 0 Next, substitute the boundary condition v = 0 at x = 20 ft into Eq. (b) to obtain: 63.925 kips 7 kips/ft 7 kips/ft 4 kips/ft 0= (20 ft)3 − (20 ft) 4 + (9 ft) 4 − (9 ft) 4 + C1 (20 ft) 6 24 24 24 2  C1 = −1,969.3396 kip-ft The beam slope and elastic curve equations are now complete: dv 63.925 kips 7 kips/ft 7 kips/ft 4 kips/ft 2 3 3 3 EI = x − 0 ft − x − 0 ft + x − 11 ft − x − 11 ft dx 2 6 6 6 4 kips/ft 49.075 kips 3 2 + x − 20 ft + x − 20 ft − 1,969.3396 kip-ft 2 6 2

63.925 kips 7 kips/ft 7 kips/ft 4 kips/ft 3 4 4 4 x − 0 ft − x − 0 ft + x − 11 ft − x − 11 ft 6 24 24 24 4 kips/ft 49.075 kips 4 3 + x − 20 ft + x − 20 ft − (1,969.3396 kip-ft 2 ) x 24 6

EI v =

(a) Beam slope at A: For the W21 × 50 structural steel wide-flange shape, EI = 198,166.658 kip-ft2. The beam slope at A is: dv EI = −1,969.3396 kip-ft 2 dx A

dv 1,969.3396 kip-ft 2  =− = −0.009938 rad = −0.00994 rad dx A 198,166.658 kip-ft 2

(b) Beam deflection at B: At midspan where x = 11 ft, the beam deflection is: 63.925 kips 7 kips/ft EI vB = (11 ft)3 − (11 ft)4 − (1,969.3396 kip-ft 2 )(11 ft) 6 24 11,752.33123 kip-ft 3  vB = − = −0.059305 ft = 0.712 in.  198,166.658 kip-ft 2

Ans.

P10.26 The simply supported beam shown in Figure P10.26 consists of a W200 × 59 structural steel wide-flange shape [E = 200 GPa; I = 60.8 × 106 mm4]. For the loading shown, use discontinuity functions to compute (a) the deflection of the beam at C and (b) the deflection of the beam at F. FIGURE P10.26

Solution Support reactions: A FBD of the beam is shown to the right. M A = −(20 kN)(2 m) − (8 kN/m)(6 m)(7 m) −(10 kN)(12 m) + Dy (8 m) = 0  Dy = 62 kN Fy = Ay + Dy − 20 kN − (8 kN/m)(6 m) − 10 kN = 0  Ay = 16 kN

Load function w(x): −1 −1 0 −1 w( x) = 16 kN x − 0 m − 20 kN x − 2 m − 8 kN/m x − 4 m + 62 kN x − 8 m −1

+8 kN/m x − 10 m − 10 kN x − 12 m 0

Shear-force function V(x) and bending-moment function M(x): 0 0 1 0 V ( x) = 16 kN x − 0 m − 20 kN x − 2 m − 8 kN/m x − 4 m + 62 kN x − 8 m

+8 kN/m x − 10 m − 10 kN x − 12 m 1

M ( x) = 16 kN x − 0 m − 20 kN x − 2 m − 1

8 kN/m 2 1 x − 4 m + 62 kN x − 8 m 2

8 kN/m 2 1 x − 10 m − 10 kN x − 12 m 2

Equations for beam slope and beam deflection: From Eq. (10.1), we can write: d 2v 8 kN/m 1 1 2 1 EI 2 = M ( x) = 16 kN x − 0 m − 20 kN x − 2 m − x − 4 m + 62 kN x − 8 m dx 2 8 kN/m 2 1 + x − 10 m − 10 kN x − 12 m 2 Integrate the moment function to obtain an expression for the beam slope: dv 16 kN 20 kN 8 kN/m 62 kN 2 2 3 2 EI = x−0 m − x−2 m − x−4 m + x −8 m dx 2 2 6 2 8 kN/m 10 kN 3 2 + x − 10 m − x − 12 m + C1 6 2 Integrate again to obtain the beam deflection function:

(a)

16 kN 20 kN 8 kN/m 62 kN 3 3 4 3 x−0 m − x−2 m − x−4 m + x −8 m 6 6 24 6 8 kN/m 10 kN 4 3 + x − 10 m − x − 12 m + C1x + C2 24 6

EI v =

(b)

Evaluate constants using boundary conditions: Boundary conditions are specific values of deflection v or slope dv/dx that are known at particular locations along the beam span. For this beam, the deflection v is known at the pin support (x = 0 m) and at the roller support (x = 8 m). Substitute the boundary condition v = 0 at x = 0 m into Eq. (b) to obtain: C2 = 0 Next, substitute the boundary condition v = 0 at x = 8 m into Eq. (b) to obtain: 16 kN 20 kN 8 kN/m 0= (8 m)3 − (6 m)3 − (4 m) 4 + C1 (8 m) 6 6 24 2  C1 = −70 kN-m The beam slope and elastic curve equations are now complete: dv 16 kN 20 kN 8 kN/m 62 kN 2 2 3 2 EI = x−0 m − x−2 m − x−4 m + x −8 m dx 2 2 6 2 8 kN/m 10 kN 3 2 + x − 10 m − x − 12 m − 70 kN-m2 6 2

16 kN 20 kN 8 kN/m 62 kN 3 3 4 3 x−0 m − x−2 m − x−4 m + x −8 m 6 6 24 6 8 kN/m 10 kN 4 3 + x − 10 m − x − 12 m − (70 kN-m2 ) x 24 6

EI v =

(a) Beam deflection at C: For the W200 × 59 structural steel wide-flange shape, EI = 12,160 kN-m2. At C where x = 4 m, the beam deflection is: 16 kN 20 kN EI vC = (4 m)3 − (2 m)3 − (70 kN-m 2 )(4 m) 6 6 3 = −136 kN-m

136 kN-m3  vC = − = −0.011184 m = 11.18 mm  12,160 kN-m 2

Ans.

(b) Beam deflection at F: At F where x = 12 m, the beam deflection is: 16 kN 20 kN 8 kN/m 62 kN EI vF = (12 m)3 − (10 m)3 − (8 m) 4 + (4 m)3 6 6 24 6 8 kN/m + (2 m) 4 − (70 kN-m 2 )(12 m) 24 = −264 kN-m3

 vF = −

264 kN-m3 = −0.021711 m = 21.7 mm  12,160 kN-m 2

Ans.

P10.27 The solid 0.50 in. diameter steel [E = 30,000 ksi] shaft shown in Figure P10.27 supports two belt pulleys. Assume that the bearing at B can be idealized as a pin support and that the bearing at D can be idealized as a roller support. For the loading shown, use discontinuity functions to compute: (a) the shaft deflection at pulley A. (b) the shaft deflection at pulley C. FIGURE P10.27

Solution Support reactions: A FBD of the beam is shown to the right. M B = (90 lb)(5 in.) − (120 lb)(10 in.) + Dy (20 in.) = 0  Dy = 37.5 lb Fy = By + Dy − 90 lb − 120 lb = 0  By = 172.5 lb

Load function w(x): −1 −1 −1 −1 w( x) = −90 lb x − 0 in. + 172.5 lb x − 5 in. − 120 lb x − 15 in. + 37.5 lb x − 25 in. Shear-force function V(x) and bending-moment function M(x): 0 0 0 0 V ( x) = −90 lb x − 0 in. + 172.5 lb x − 5 in. − 120 lb x − 15 in. + 37.5 lb x − 25 in.

M ( x) = −90 lb x − 0 in. + 172.5 lb x − 5 in. − 120 lb x − 15 in. + 37.5 lb x − 25 in. 1

Equations for beam slope and beam deflection: From Eq. (10.1), we can write: d 2v 1 1 1 1 EI 2 = M ( x) = −90 lb x − 0 in. + 172.5 lb x − 5 in. − 120 lb x − 15 in. + 37.5 lb x − 25 in. dx Integrate the moment function to obtain an expression for the beam slope: dv 90 lb 172.5 lb 120 lb 37.5 lb 2 2 2 2 EI =− x − 0 in. + x − 5 in. − x − 15 in. + x − 25 in. + C1 (a) dx 2 2 2 2 Integrate again to obtain the beam deflection function: 90 lb 172.5 lb 120 lb 3 3 3 EI v = − x − 0 in. + x − 5 in. − x − 15 in. 6 6 6 37.5 lb 3 (b) + x − 25 in. + C1 x + C2 6 Evaluate constants using boundary conditions: Boundary conditions are specific values of deflection v or slope dv/dx that are known at particular locations along the beam span. For this beam, the deflection v is known at the pin support (x = 5 in.) and at the roller support (x = 25 in.). Substitute the boundary condition v = 0 at x = 5 in. into Eq. (b) to obtain:

90 lb (5 in.)3 + C1 (5 in.) + C2 6  C1 (5 in.) + C2 = 1,875.0 lb-in.3

0=−

(c)

Next, substitute the boundary condition v = 0 at x = 25 in. into Eq. (b) to obtain: 90 lb 172.5 lb 120 lb 0=− (25 in.)3 + (20 in.)3 − (10 in.)3 + C1 (25 in.) + C2 6 6 6 3  C1 (25 in.) + C2 = 24,375.0 lb-in.

(d)

Solve Eqs. (c) and (d) simultaneously for the two constants of integration C1 and C2: C1 = 1,125 lb-in.2 and C2 = −3,750 lb-in.3

The beam slope and elastic curve equations are now complete: dv 90 lb 172.5 lb 120 lb 2 2 2 EI =− x − 0 in. + x − 5 in. − x − 15 in. dx 2 2 2 37.5 lb 2 + x − 25 in. + 1,125 lb-in.2 2

EI v = −

90 lb 172.5 lb 120 lb 3 3 3 x − 0 in. + x − 5 in. − x − 15 in. 6 6 6 37.5 lb 3 + x − 25 in. + (1,125 lb-in.2 ) x − 3,750 lb-in.3 6

Section properties: I=



(0.5 in.) 4 = 3.06796  10 −3 in.4

64  EI = 92.0388  103 lb-in.2

E = 30,000 ksi = 30  10 6 psi

(a) Beam deflection at A: The beam deflection at A where x = 0 in. is: EI vA = −3,750 lb-in.3  vA = −

3,750 lb-in.3 = −0.040744 in. = 0.0407 in.  92.0388  103 lb-in.2

(b) Beam deflection at C: The beam deflection at C where x = 15 in. is: 90 lb 172.5 lb EI vC = − (15 in.)3 + (10 in.)3 + (1,125 lb-in.2 )(15 in.) − 3,750 lb-in.3 6 6 3 8,750 lb-in.  vC = − = −0.095069 in. = 0.0951 in.  92.0388  103 lb-in.2

Ans.

P10.28 The cantilever beam shown in Figure P10.28 consists of a W8 × 31 structural steel wide-flange shape [E = 29,000 ksi; I = 110 in.4]. For the loading shown, use discontinuity functions to compute (a) the slope of the beam at A and (b) the deflection of the beam at A. FIGURE P10.28

Solution Support reactions: A FBD of the beam is shown to the right. Fy = C y − (3.5 kips/ft)(10 ft) = 0  C y = 35 kips M C = −75 kip-ft + (3.5 kips/ft)(10 ft)(5 ft) + M C = 0  M C = −100 kip-ft Load function w(x): −2 0 0 w( x) = 75 kip-ft x − 0 ft − 3.5 kips/ft x − 5 ft + 3.5 kips/ft x − 15 ft

+35 kips x − 15 ft

−1

+ 100 kip-ft x − 15 ft

−2

Shear-force function V(x) and bending-moment function M(x): −1 1 1 V ( x) = 75 kip-ft x − 0 ft − 3.5 kips/ft x − 5 ft + 3.5 kips/ft x − 15 ft −1

+35 kips x − 15 ft + 100 kip-ft x − 15 ft 3.5 kips/ft 3.5 kips/ft 0 2 2 M ( x) = 75 kip-ft x − 0 ft − x − 5 ft + x − 15 ft 2 2 1 0 +35 kips x − 15 ft + 100 kip-ft x − 15 ft 0

Equations for beam slope and beam deflection: From Eq. (10.1), we can write: d 2v 3.5 kips/ft 3.5 kips/ft 0 2 2 EI 2 = M ( x) = 75 kip-ft x − 0 ft − x − 5 ft + x − 15 ft dx 2 2 +35 kips x − 15 ft + 100 kip-ft x − 15 ft Integrate the moment function to obtain an expression for the beam slope: dv 3.5 kips/ft 3.5 kips/ft 1 3 3 EI = 75 kip-ft x − 0 ft − x − 5 ft + x − 15 ft dx 6 6 35 kips 2 1 + x − 15 ft + 100 kip-ft x − 15 ft + C1 2 Integrate again to obtain the beam deflection function: 75 kip-ft 3.5 kips/ft 3.5 kips/ft 2 4 4 EI v = x − 0 ft − x − 5 ft + x − 15 ft 2 24 24 35 kips 100 kip-ft 3 2 + x − 15 ft + x − 15 ft + C1 x + C2 6 2 1

(a)

(b)

Evaluate constants using boundary conditions: Boundary conditions are specific values of deflection v or slope dv/dx that are known at particular locations along the beam span. For this beam, both the slope dv/dx and the deflection v are known at the fixed support (x = 15 ft). Substitute the boundary condition dv/dx = 0 at x = 15 ft into Eq. (a) to obtain: 3.5 kips/ft 0 = (75 kip-ft)(15 ft)1 − (10 ft)3 + C1 6  C1 = −541.666667 kip-ft 2 Next, substitute the boundary condition v = 0 at x = 15 ft into Eq. (b) to obtain: 75 kip-ft 3.5 kips/ft 0= (15 ft)2 − (10 ft)4 + ( −541.666667 kip-ft 2 )(15 ft) + C2 2 24  C2 = 1,145.833334 kip-ft 3 The beam slope and elastic curve equations are now complete: dv 3.5 kips/ft 3.5 kips/ft 1 3 3 EI = 75 kip-ft x − 0 ft − x − 5 ft + x − 15 ft dx 6 6 35 kips 2 1 + x − 15 ft + 100 kip-ft x − 15 ft − 541.666667 kip-ft 2 2

75 kip-ft 3.5 kips/ft 3.5 kips/ft 2 4 4 x − 0 ft − x − 5 ft + x − 15 ft 2 24 24 35 kips 100 kip-ft 3 2 + x − 15 ft + x − 15 ft − (541.666667 kip-ft 2 ) x + 1,145.833334 kip-ft 3 6 2

EI v =

(a) Beam slope at A: For the W8 × 31 structural steel wide-flange shape, EI = 22,152.777 kip-ft2. At the tip of the overhang where x = 0 ft, the beam slope is: dv EI = −541.666667 kip-ft 2 dx A

dv 541.666667 kip-ft 2  =− = −0.024451 rad = −0.0245 rad dx A 22,152.777 kip-ft 2

Ans.

(b) Beam deflection at A: At the tip of the overhang where x = 0 ft, the beam deflection is: EI v A = 1,145.833334 kip-ft 3  vA =

1,145.833334 kip-ft 3 = 0.051724 ft = 0.621 in.  22,152.777 kip-ft 2

Ans.

P10.29 The simply supported beam shown in Figure P10.29 consists of a W14 × 34 structural steel wideflange shape [E = 29,000 ksi; I = 340 in.4]. For the loading shown, use discontinuity functions to compute (a) the slope of the beam at E and (b) the deflection of the beam at C. FIGURE P10.29

Solution Support reactions: A FBD of the beam is shown to the right. 8 ft M B = 12 (6 kips/ft)(8 ft) − (4 kips/ft)(10 ft)(13 ft) 3 + E y (22 ft) = 0  E y = 20.727 kips Fy = By + E y − 12 (6 kips/ft)(8 ft) − (4 kips/ft)(10 ft) = 0  By = 43.273 kips

Load function w(x): 6 kips/ft 6 kips/ft 1 1 0 −1 w( x) = − x − 0 ft + x − 8 ft + 6 kips/ft x − 8 ft + 43.273 kips x − 8 ft 8 ft 8 ft −4 kips/ft x − 16 ft

+ 4 kips/ft x − 26 ft

+ 20.727 kips x − 30 ft

−1

Shear-force function V(x) and bending-moment function M(x): 6 kips/ft 6 kips/ft 2 2 1 0 V ( x) = − x − 0 ft + x − 8 ft + 6 kips/ft x − 8 ft + 43.273 kips x − 8 ft 2(8 ft) 2(8 ft) −4 kips/ft x − 16 ft + 4 kips/ft x − 26 ft + 20.727 kips x − 30 ft 1

6 kips/ft 6 kips/ft 6 kips/ft 3 3 2 1 x − 0 ft + x − 8 ft + x − 8 ft + 43.273 kips x − 8 ft 6(8 ft) 6(8 ft) 2 4 kips/ft 4 kips/ft 2 2 1 − x − 16 ft + x − 26 ft + 20.727 kips x − 30 ft 2 2

M ( x) = −

Equations for beam slope and beam deflection: From Eq. (10.1), we can write: d 2v 6 kips/ft 6 kips/ft 6 kips/ft 3 3 2 EI 2 = M ( x) = − x − 0 ft + x − 8 ft + x − 8 ft dx 6(8 ft) 6(8 ft) 2 4 kips/ft 4 kips/ft 1 2 2 +43.273 kips x − 8 ft − x − 16 ft + x − 26 ft 2 2

+20.727 kips x − 30 ft

Integrate the moment function to obtain an expression for the beam slope:

dv 6 kips/ft 6 kips/ft 6 kips/ft 43.273 kips 4 4 3 2 =− x − 0 ft + x − 8 ft + x − 8 ft + x − 8 ft dx 24(8 ft) 24(8 ft) 6 2 4 kips/ft 4 kips/ft 20.727 kips 3 3 2 (a) − x − 16 ft + x − 26 ft + x − 30 ft + C1 6 6 2 Integrate again to obtain the beam deflection function: 6 kips/ft 6 kips/ft 6 kips/ft 43.273 kips 5 5 4 3 EI v = − x − 0 ft + x − 8 ft + x − 8 ft + x − 8 ft 120(8 ft) 120(8 ft) 24 6 4 kips/ft 4 kips/ft 20.727 kips 4 4 3 (b) − x − 16 ft + x − 26 ft + x − 30 ft + C1 x + C2 24 24 6 EI

Next, substitute the boundary condition v = 0 at x = 30 ft into Eq. (b) to obtain: 6 kips/ft 6 kips/ft 6 kips/ft 43.273 kips 0=− (30 ft)5 + (22 ft)5 + (22 ft) 4 + (22 ft)3 120(8 ft) 120(8 ft) 24 6 4 kips/ft 4 kips/ft − (14 ft) 4 + (4 ft) 4 + C1 (30 ft) + C2 24 24  C1 (30 ft) + C2 = −9,334.351 kip-ft 3

(c)

(d)

Solve Eqs. (c) and (d) simultaneously for the two constants of integration C1 and C2: C1 = −433.598 kip-ft 2 and C2 = 3,673.582 kip-ft3 The beam slope and elastic curve equations are now complete: dv 6 kips/ft 6 kips/ft 6 kips/ft 43.273 kips 4 4 3 2 EI =− x − 0 ft + x − 8 ft + x − 8 ft + x − 8 ft dx 24(8 ft) 24(8 ft) 6 2 4 kips/ft 4 kips/ft 20.727 kips 3 3 2 − x − 16 ft + x − 26 ft + x − 30 ft − 433.598 kip-ft 2 6 6 2 6 kips/ft 6 kips/ft 6 kips/ft 43.273 kips 5 5 4 3 x − 0 ft + x − 8 ft + x − 8 ft + x − 8 ft 120(8 ft) 120(8 ft) 24 6 4 kips/ft 4 kips/ft 20.727 kips 4 4 3 − x − 16 ft + x − 26 ft + x − 30 ft 24 24 6 2 3 − (433.598 kip-ft ) x + 3,673.582 kip-ft

EI v = −

(a) Beam slope at E: For the W14 × 34 structural steel wide-flange shape, EI = 68,472.219 kip-ft2. The beam slope at E is: dv 6 kips/ft 6 kips/ft 6 kips/ft 43.273 kips EI =− (30 ft) 4 + (22 ft) 4 + (22 ft)3 + (22 ft) 2 dx E 24(8 ft) 24(8 ft) 6 2

4 kips/ft 4 kips/ft (14 ft)3 + (4 ft)3 − 433.598 kip-ft 2 6 6 2 dv 907.801 kip-ft  = = 0.013258 rad = 0.01326 rad dx E 68, 472.219 kip-ft 2 −

Ans.

(b) Beam deflection at C: At C where x = 16 ft, the beam deflection is: 6 kips/ft 6 kips/ft 6 kips/ft 43.273 kips EI vC = − (16 ft)5 + (8 ft)5 + (8 ft) 4 + (8 ft)3 120(8 ft) 120(8 ft) 24 6 −(433.598 kip-ft 2 )(16 ft) + 3,673.582 kip-ft 3  vC = −

4,896.157 kip-ft 3 = −0.071506 ft = 0.858 in.  68, 472.219 kip-ft 2

Ans.

P10.30 For the beam and loading shown, use discontinuity functions to compute (a) the deflection of the beam at A and (b) the deflection of the beam at midspan (i.e., x = 2.5 m). Assume a constant value of EI = 1,500 kN-m2 for the beam. FIGURE P10.30

Solution Support reactions: A FBD of the beam is shown to the right.

18 kN/m 18 kN/m 1 1 −1 x −1 m − x − 4 m + 6 kN x − 4 m 3m 3m

Equations for beam slope and beam deflection: From Eq. (10.1), we can write: d 2v 18 kN/m 0 1 2 EI 2 = M ( x) = −9 kN-m x − 0 m + 21 kN x − 1 m − x −1 m dx 2 18 kN/m 18 kN/m 3 3 1 + x −1 m − x − 4 m + 6 kN x − 4 m 6(3 m) 6(3 m) Integrate the moment function to obtain an expression for the beam slope: dv 21 kN 18 kN/m 1 2 3 EI = −9 kN-m x − 0 m + x −1 m − x −1 m dx 2 6 18 kN/m 18 kN/m 6 kN 4 4 2 + x −1 m − x−4m + x − 4 m + C1 24(3 m) 24(3 m) 2

(a)

Integrate again to obtain the beam deflection function: 9 kN-m 21 kN 18 kN/m 2 3 4 EI v = − x−0 m + x −1 m − x −1 m 2 6 24 18 kN/m 18 kN/m 6 kN 5 5 3 + x −1 m − x−4m + x − 4 m + C1 x + C2 120(3 m) 120(3 m) 6

(b)

Evaluate constants using boundary conditions: Boundary conditions are specific values of deflection v or slope dv/dx that are known at particular locations along the beam span. For this beam, the deflection v is known at the pin support (x = 1 m) and at the roller support (x = 4 m). Substitute the boundary condition v = 0 at x = 1 m into Eq. (b) to obtain: 9 kN-m 0=− (1 m) 2 + C1 (1 m) + C2 2 (c)  C1 (1 m) + C2 = 4.5 kN-m3 Next, substitute the boundary condition v = 0 at x = 4 m into Eq. (b) to obtain: 9 kN-m 21 kN 18 kN/m 18 kN/m 0=− (4 m) 2 + (3 m)3 − (3 m) 4 + (3 m)5 + C1 (4 m) + C2 2 6 24 120(3 m)  C1 (4 m) + C2 = 26.10 kN-m3

(d)

Solve Eqs. (c) and (d) simultaneously for the two constants of integration C1 and C2: C1 = 7.2 kN-m2 and C2 = −2.7 kN-m3 The beam slope and elastic curve equations are now complete: dv 21 kN 18 kN/m 1 2 3 EI = −9 kN-m x − 0 m + x −1 m − x −1 m dx 2 6 18 kN/m 18 kN/m 6 kN 4 4 2 + x −1 m − x−4m + x − 4 m + 7.2 kN-m 2 24(3 m) 24(3 m) 2 9 kN-m 21 kN 18 kN/m 2 3 4 x−0 m + x −1 m − x −1 m 2 6 24 18 kN/m 18 kN/m 6 kN 5 5 3 + x −1 m − x−4m + x − 4 m + (7.2 kN-m 2 ) x − 2.7 kN-m3 120(3 m) 120(3 m) 6

EI v = −

(a) Beam deflection at A: The beam deflection at A is: EI v A = −2.7 kN-m3

 vA = −

2.7 kN-m3 = −0.001800 m = 1.800 mm  1,500 kN-m 2

Ans.

(b) Beam deflection at midspan: At x = 2.5 m, the beam deflection is:

9 kN-m 21 kN 18 kN/m (2.5 m) 2 + (1.5 m)3 − (1.5 m) 4 2 6 24 18 kN/m + (1.5 m)5 + (7.2 kN-m 2 )(2.5 m) − 2.7 kN-m3 120(3 m)

EI vmidspan = −

= −4.429688 kN-m3  vmidspan = −

4.429688 kN-m3 = −0.002953 m = 2.95 mm  1,500 kN-m 2

Ans.

P10.31 For the beam and loading shown, use discontinuity functions to compute (a) the slope of the beam at B and (b) the deflection of the beam at A. Assume a constant value of EI = 133,000 kip-ft2 for the beam. FIGURE P10.31

Solution Support reactions: A FBD of the beam is shown to the right.

Fy = C y − 12 (4 kips/ft)(9 ft) = 0  C y = 18 kips M C = 12 (4 kips/ft)(9 ft)(11 ft) + M C = 0  M C = −198 kip-ft Load function w(x): w( x) = −4 kips/ft x − 0 ft +18 kips x − 14 ft

−1

4 kips/ft 4 kips/ft 1 1 x − 0 ft − x − 9 ft 9 ft 9 ft −2

+ 198 kip-ft x − 14 ft

Shear-force function V(x) and bending-moment function M(x): 4 kips/ft 4 kips/ft 1 2 2 V ( x) = −4 kips/ft x − 0 ft + x − 0 ft − x − 9 ft 2(9 ft) 2(9 ft) +18 kips x − 14 ft M ( x) = −

+ 198 kip-ft x − 14 ft

−1

4 kips/ft 4 kips/ft 4 kips/ft 2 3 3 x − 0 ft + x − 0 ft − x − 9 ft 2 6(9 ft) 6(9 ft)

+18 kips x − 14 ft + 198 kip-ft x − 14 ft 1

Equations for beam slope and beam deflection: From Eq. (10.1), we can write: d 2v 4 kips/ft 4 kips/ft 4 kips/ft 2 3 3 EI 2 = M ( x) = − x − 0 ft + x − 0 ft − x − 9 ft dx 2 6(9 ft) 6(9 ft)

+18 kips x − 14 ft + 198 kip-ft x − 14 ft 1

Integrate the moment function to obtain an expression for the beam slope: dv 4 kips/ft 4 kips/ft 4 kips/ft 3 4 4 EI =− x − 0 ft + x − 0 ft − x − 9 ft dx 6 24(9 ft) 24(9 ft) 18 kips 2 1 + x − 14 ft + 198 kip-ft x − 14 ft + C1 2

(a)

Integrate again to obtain the beam deflection function: 4 kips/ft 4 kips/ft 4 kips/ft 4 5 5 EI v = − x − 0 ft + x − 0 ft − x − 9 ft 24 120(9 ft) 120(9 ft) 18 kips 198 kip-ft 3 2 + x − 14 ft + x − 14 ft + C1 x + C2 6 2

(b)

Evaluate constants using boundary conditions: Boundary conditions are specific values of deflection v or slope dv/dx that are known at particular locations along the beam span. For this beam, both the slope dv/dx and the deflection v are known at the fixed support (x = 14 ft). Substitute the boundary condition dv/dx = 0 at x = 14 ft into Eq. (a) to obtain: 4 kips/ft 4 kips/ft 4 kips/ft 0=− (14 ft)3 + (14 ft) 4 − (5 ft) 4 + C1 6 24(9 ft) 24(9 ft)  C1 = 1,129.5 kip-ft 2 Next, substitute the boundary condition v = 0 at x = 4 m into Eq. (b) to obtain: 4 kips/ft 4 kips/ft 4 kips/ft 0=− (14 ft) 4 + (14 ft)5 − (5 ft)5 + (1,129.5 kip-ft 2 )(14 ft) + C2 24 120(9 ft) 120(9 ft)

 C2 = −11,390.7 kip-ft 3

The beam slope and elastic curve equations are now complete: dv 4 kips/ft 4 kips/ft 4 kips/ft 3 4 4 EI =− x − 0 ft + x − 0 ft − x − 9 ft dx 6 24(9 ft) 24(9 ft) 18 kips 2 1 + x − 14 ft + 198 kip-ft x − 14 ft + 1,129.5 kip-ft 2 2 4 kips/ft 4 kips/ft 4 kips/ft 4 5 5 x − 0 ft + x − 0 ft − x − 9 ft 24 120(9 ft) 120(9 ft) 18 kips 198 kip-ft 3 2 + x − 14 ft + x − 14 ft + (1,129.5 kip-ft 2 ) x − 11,390.7 kip-ft 3 6 2

EI v = −

(a) Beam slope at B: The beam slope at B (i.e., x = 9 ft) is: dv 4 kips/ft 4 kips/ft EI =− (9 ft)3 + (9 ft)4 + 1,129.5 kip-ft 2 = 765 kip-ft 2 dx B 6 24(9 ft)



dv 765 kip-ft 2 = = 0.005752 rad = 0.00575 rad dx B 133,000 kip-ft 2

Ans.

(b) Beam deflection at A: The beam deflection at A is: EI v A = −11,390.7 kip-ft 3  vA = −

11,390.7 kip-ft 3 = −0.085644 ft = 1.028 in.  133,000 kip-ft 2

Ans.

P10.32 For the beam and loading shown, use discontinuity functions to compute (a) the slope of the beam at B and (b) the deflection of the beam at C. Assume a constant value of EI = 34×106 lb-ft2 for the beam. FIGURE P10.32

Solution Support reactions: A FBD of the beam is shown to the right. M B = − 12 (7,000 lb/ft)(9 ft)(3 ft) + Dy (14 ft) = 0

 Dy = 6,750 lbs Fy = By + Dy − 12 (7,000 lb/ft)(9 ft) = 0  By = 24,750 lbs Load function w(x):

w( x) = 24,750 lbs x − 4 ft −

−1

− 7,000 lb/ft x − 4 ft

7,000 lb/ft 1 x − 4 ft 9 ft

7,000 lb/ft 1 −1 x − 13 ft + 6,750 lbs x − 18 ft 9 ft

Shear-force function V(x) and bending-moment function M(x): 7,000 lb/ft 0 1 2 V ( x) = 24,750 lbs x − 4 ft − 7,000 lb/ft x − 4 ft + x − 4 ft 2(9 ft) 7,000 lb/ft 2 0 − x − 13 ft + 6,750 lbs x − 18 ft 2(9 ft) 7,000 lb/ft 7,000 lb/ft 1 2 3 M ( x) = 24,750 lbs x − 4 ft − x − 4 ft + x − 4 ft 2 6(9 ft) 7,000 lb/ft 3 1 − x − 13 ft + 6,750 lbs x − 18 ft 6(9 ft)

Equations for beam slope and beam deflection: From Eq. (10.1), we can write: d 2v 7,000 lb/ft 7,000 lb/ft 1 2 3 EI 2 = M ( x) = 24,750 lbs x − 4 ft − x − 4 ft + x − 4 ft dx 2 6(9 ft) 7,000 lb/ft 3 1 − x − 13 ft + 6,750 lbs x − 18 ft 6(9 ft) Integrate the moment function to obtain an expression for the beam slope: dv 24,750 lbs 7,000 lb/ft 7,000 lb/ft 2 3 4 EI = x − 4 ft − x − 4 ft + x − 4 ft dx 2 6 24(9 ft) 7,000 lb/ft 6,750 lbs 4 2 − x − 13 ft + x − 18 ft + C1 24(9 ft) 2

(a)

Integrate again to obtain the beam deflection function: 24,750 lbs 7,000 lb/ft 7,000 lb/ft 3 4 5 EI v = x − 4 ft − x − 4 ft + x − 4 ft 6 24 120(9 ft) 7,000 lb/ft 6,750 lbs 5 3 − x − 13 ft + x − 18 ft + C1 x + C2 120(9 ft) 6

(b)

 C1 (18 ft) + C2 = 3,579,975 lb-ft 3

(d)

Solve Eqs. (c) and (d) simultaneously for the two constants of integration C1 and C2: C1 = −255,712.5 lb-ft 2 and C2 = 1,022,850 lb-ft3 The beam slope and elastic curve equations are now complete: dv 24,750 lbs 7,000 lb/ft 7,000 lb/ft 2 3 4 EI = x − 4 ft − x − 4 ft + x − 4 ft dx 2 6 24(9 ft) 7,000 lb/ft 6,750 lbs 4 2 − x − 13 ft + x − 18 ft − 255,712.5 lb-ft 2 24(9 ft) 2 EI v =

24,750 lbs 7,000 lb/ft 7,000 lb/ft 3 4 5 x − 4 ft − x − 4 ft + x − 4 ft 6 24 120(9 ft) 7,000 lb/ft 6,750 lbs 5 3 − x − 13 ft + x − 18 ft − (255,712.5 lb-ft 2 ) x + 1,022,850 lb-ft 3 120(9 ft) 6

(a) Beam slope at B: The beam slope at B is: dv EI = −255,712.5 lb-ft 2 dx B



dv 255,712.5 lb-ft 2 =− = −0.0075210 rad = −0.00752 rad dx B 34  106 lb-ft 2

Ans.

(b) Beam deflection at C: At C where x = 13 ft, the beam deflection is:

EI vC =

24,750 lbs 7,000 lb/ft 7,000 lb/ft (9 ft)3 − (9 ft)4 + (9 ft)5 6 24 120(9 ft) −(255,712.5 lb-ft 2 )(13 ft) + 1,022,850 lb-ft 3

825,187.5 kip-ft 3 = −0.0242702 ft = 0.291 in.  34  106 lb-ft 2 P10.33 For the beam and loading shown, use discontinuity functions to compute (a) the slope of the beam at A and (b) the deflection of the beam at B. Assume a constant value of EI = 370,000 kip-ft2 for the beam.  vC = −

Ans.

FIGURE P10.33

Solution Support reactions: A FBD of the beam is shown to the right. M A = − 12 (8 kips/ft)(12 ft)(8 ft) − 12 (8 kips/ft)(12 ft)(16 ft) + C y (24 ft) = 0  C y = 48 kips Fy = Ay + C y − 12 (8 kips/ft)(24 ft) = 0  Ay = 48 kips

Load function w(x):

8 kips/ft 8 kips/ft 1 1 x − 0 ft + x − 12 ft 12 ft 12 ft 8 kips/ft 0 0 1 +8 kips/ft x − 12 ft − 8 kips/ft x − 12 ft + x − 12 ft 12 ft 8 kips/ft 1 −1 − x − 24 ft + 48 kips x − 24 ft 12 ft 8 kips/ft 2(8 kips/ft) −1 1 1 = 48 kips x − 0 ft − x − 0 ft + x − 12 ft 12 ft 12 ft 8 kips/ft 1 −1 − x − 24 ft + 48 kips x − 24 ft 12 ft

w( x) = 48 kips x − 0 ft

−1

−

Shear-force function V(x) and bending-moment function M(x): 8 kips/ft 2(8 kips/ft) 0 2 2 V ( x) = 48 kips x − 0 ft − x − 0 ft + x − 12 ft 2(12 ft) 2(12 ft) 8 kips/ft 2 0 − x − 24 ft + 48 kips x − 24 ft 2(12 ft) 8 kips/ft 2(8 kips/ft) 1 3 3 M ( x) = 48 kips x − 0 ft − x − 0 ft + x − 12 ft 6(12 ft) 6(12 ft) 8 kips/ft 3 1 − x − 24 ft + 48 kips x − 24 ft 6(12 ft) Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Equations for beam slope and beam deflection: From Eq. (10.1), we can write: d 2v 8 kips/ft 2(8 kips/ft) 1 3 3 EI 2 = M ( x) = 48 kips x − 0 ft − x − 0 ft + x − 12 ft dx 6(12 ft) 6(12 ft) 8 kips/ft 3 1 − x − 24 ft + 48 kips x − 24 ft 6(12 ft) Integrate the moment function to obtain an expression for the beam slope: dv 48 kips 8 kips/ft 2(8 kips/ft) 2 4 4 EI = x − 0 ft − x − 0 ft + x − 12 ft dx 2 24(12 ft) 24(12 ft) 8 kips/ft 48 kips 4 2 − x − 24 ft + x − 24 ft + C1 24(12 ft) 2 Integrate again to obtain the beam deflection function: 48 kips 8 kips/ft 2(8 kips/ft) 3 5 5 EI v = x − 0 ft − x − 0 ft + x − 12 ft 6 120(12 ft) 120(12 ft) 8 kips/ft 48 kips 5 3 − x − 24 ft + x − 24 ft + C1 x + C2 120(12 ft) 6

(a)

(b)

The beam slope and elastic curve equations are now complete: dv 48 kips 8 kips/ft 2(8 kips/ft) 2 4 4 EI = x − 0 ft − x − 0 ft + x − 12 ft dx 2 24(12 ft) 24(12 ft) 8 kips/ft 48 kips 4 2 − x − 24 ft + x − 24 ft − 2,880 kip-ft 2 24(12 ft) 2 EI v =

48 kips 8 kips/ft 2(8 kips/ft) 3 5 5 x − 0 ft − x − 0 ft + x − 12 ft 6 120(12 ft) 120(12 ft) 8 kips/ft 48 kips 5 3 − x − 24 ft + x − 24 ft − (2,880 kip-ft 2 ) x 120(12 ft) 6

(a) Beam slope at A: The beam slope at A is:

dv = −2,880 kip-ft 2 dx A 

dv 2,880 kip-ft 2 =− = −0.0077838 rad = −0.00778 rad dx A 370,000 kip-ft 2

Ans.

(b) Beam deflection at B: At B where x = 12 ft, the beam deflection is: 48 kips 8 kips/ft EI vB = (12 ft)3 − (12 ft)5 − (2,880 kip-ft 2 )(12 ft) = −22,118.4 kip-ft 3 6 120(12 ft)

 vB = −

22,118.4 kip-ft 3 kip-ft 3 = −0.0597795 ft = 0.717 in.  370,000 kip-ft 2

Ans.

P10.34 For the beam and loading shown, use discontinuity functions to compute (a) the slope of the beam at B and (b) the deflection of the beam at B. Assume a constant value of EI = 110,000 kN-m2 for the beam. FIGURE P10.34

Solution Support reactions: A FBD of the beam is shown to the right. Fy = Ay − (15 kN/m)(4 m) − 12 (25 kN/m)(4 m) = 0  Ay = 110 kN M A = − (15 kN/m)(4 m)(2 m) 2(4 m) − MA = 0 3  M A = −253.33 kN-m

− 12 (25 kN/m)(4 m)

Load function w(x):

w( x) = −253.33 kN-m x − 0 m

−2

+ 110 kN x − 0 m

−1

− 15 kN/m x − 0 m − 0

25 kN/m 1 x−0 m 4m

25 kN/m 1 0 x − 4 m + 25 kN/m x − 4 m 4m Shear-force function V(x) and bending-moment function M(x): +15 kN/m x − 4 m + 0

V ( x) = −253.33 kN-m x − 0 m

−1

+ 110 kN x − 0 m − 15 kN/m x − 0 m − 0

25 kN/m 2 x−0 m 2(4 m)

25 kN/m 2 1 x − 4 m + 25 kN/m x − 4 m 2(4 m) 15 kN/m 25 kN/m 0 1 2 3 M ( x) = −253.33 kN-m x − 0 m + 110 kN x − 0 m − x−0 m − x−0 m 2 6(4 m) 15 kN/m 25 kN/m 25 kN/m 2 3 2 + x−4m + x−4 m + x−4 m 2 6(4 m) 2 +15 kN/m x − 4 m + 1

Equations for beam slope and beam deflection: From Eq. (10.1), we can write: d 2v 15 kN/m 0 1 2 EI 2 = M ( x) = −253.33 kN-m x − 0 m + 110 kN x − 0 m − x−0 m dx 2 25 kN/m 15 kN/m 25 kN/m 25 kN/m 3 2 3 2 − x−0 m + x−4m + x−4m + x−4m 6(4 m) 2 6(4 m) 2 Integrate the moment function to obtain an expression for the beam slope:

dv 110 kN 15 kN/m 25 kN/m 1 2 3 4 = −253.33 kN-m x − 0 m + x−0 m − x−0 m − x−0 m dx 2 6 24(4 m) 15 kN/m 25 kN/m 25 kN/m 3 4 3 + x−4 m + x−4 m + x − 4 m + C1 6 24(4 m) 6

(a) Integrate again to obtain the beam deflection function: 253.33 kN-m 110 kN 15 kN/m 25 kN/m 2 3 4 5 EI v = − x−0 m + x−0 m − x−0 m − x−0 m 2 6 24 120(4 m) 15 kN/m 25 kN/m 25 kN/m 4 5 4 + x−4m + x−4m + x − 4 m + C1 x + C2 24 120(4 m) 24 (b) Evaluate constants using boundary conditions: Boundary conditions are specific values of deflection v or slope dv/dx that are known at particular locations along the beam span. For this beam, both the slope dv/dx and the deflection v are known at the fixed support (x = 0 m). Substitute the boundary condition dv/dx = 0 at x = 0 m into Eq. (a) to obtain: C1 = 0 Next, substitute the boundary condition v = 0 at x = 0 m into Eq. (b) to obtain: C2 = 0 The beam slope and elastic curve equations are now complete: dv 110 kN 15 kN/m 25 kN/m 1 2 3 4 EI = −253.33 kN-m x − 0 m + x−0 m − x−0 m − x−0 m dx 2 6 24(4 m) 15 kN/m 25 kN/m 25 kN/m 3 4 3 + x−4 m + x−4 m + x−4m 6 24(4 m) 6 253.33 kN-m 110 kN 15 kN/m 25 kN/m 2 3 4 5 x−0 m + x−0 m − x−0 m − x−0 m 2 6 24 120(4 m) 15 kN/m 25 kN/m 25 kN/m 4 5 4 + x−4m + x−4m + x−4m 24 120(4 m) 24

EI v = −

(a) Beam slope at B: The beam slope at B is: dv 110 kN 15 kN/m 25 kN/m EI = ( −253.33 kN-m)(4 m)1 + (4 m)2 − (4 m)3 − (4 m)4 dx B 2 6 24(4 m) = −360 kN-m 2 

dv 2,120 kN-m 2 =− = −0.003273 rad = −0.00327 rad dx B 110,000 kN-m 2

Ans.

(b) Beam deflection at B: The beam deflection at B is:

EI vB = −

253.33 kN-m 110 kN 15 kN/m 25 kN/m (4 m) 2 + (4 m)3 − (4 m) 4 − (4 m)5 2 6 24 120(4 m)

= −1,066.67 kN-m3  vB = −

1,066.67 kN-m3 = −0.009697 m = 9.70 mm  110,000 kN-m 2

Ans.

P10.35 For the beam and loading shown, use discontinuity functions to compute (a) the deflection of the beam at A and (b) the deflection of the beam at C. Assume a constant value of EI = 24,000 kNm2 for the beam. FIGURE P10.35

Solution Support reactions: A FBD of the beam is shown to the right. M B = (35 kN)(2.5 m) − (25 kN/m)(4.0 m)(2.0 m) − 12 (45 kN/m)(4.0 m)

2(4.0 m) + Dy (5.5 m) = 0 3

 Dy = 64.09 kN Fy = By + Dy − 35 kN − (25 kN/m)(4.0 m) − 12 (45 kN/m)(4.0 m) = 0  By = 160.91 kN

Load function w(x): −1 −1 0 w( x) = −35 kN x − 0 m + 160.91 kN x − 2.5 m − 25 kN/m x − 2.5 m +25 kN/m x − 6.5 m − 0

45 kN/m 45 kN/m 1 1 x − 2.5 m + x − 6.5 m 4.0 m 4.0 m

+45 kN/m x − 6.5 m + 64.09 kN x − 8 m 0

−1

Shear-force function V(x) and bending-moment function M(x): 0 0 1 V ( x) = −35 kN x − 0 m + 160.91 kN x − 2.5 m − 25 kN/m x − 2.5 m

+25 kN/m x − 6.5 m − 1

45 kN/m 45 kN/m 2 2 x − 2.5 m + x − 6.5 m 2(4.0 m) 2(4.0 m)

+45 kN/m x − 6.5 m + 64.09 kN x − 8 m 1

25 kN/m 2 x − 2.5 m 2 25 kN/m 45 kN/m 45 kN/m 2 3 3 + x − 6.5 m − x − 2.5 m + x − 6.5 m 2 6(4.0 m) 6(4.0 m) 45 kN/m 2 1 + x − 6.5 m + 64.09 kN x − 8 m 2 25 kN/m 45 kN/m 1 1 2 3 = −35 kN x − 0 m + 160.91 kN x − 2.5 m − x − 2.5 m − x − 2.5 m 2 6(4.0 m) 70 kN/m 45 kN/m 2 3 1 + x − 6.5 m + x − 6.5 m + 64.09 kN x − 8 m 2 6(4.0 m)

M ( x) = −35 kN x − 0 m + 160.91 kN x − 2.5 m − 1

Equations for beam slope and beam deflection: From Eq. (10.1), we can write: d 2v 25 kN/m 1 1 2 EI 2 = M ( x) = −35 kN x − 0 m + 160.91 kN x − 2.5 m − x − 2.5 m dx 2 45 kN/m 70 kN/m 45 kN/m 3 2 3 − x − 2.5 m + x − 6.5 m + x − 6.5 m 6(4.0 m) 2 6(4.0 m)

+64.09 kN x − 8 m Integrate the moment function to obtain an expression for the beam slope: dv 35 kN 160.91 kN 25 kN/m 2 2 3 EI =− x−0 m + x − 2.5 m − x − 2.5 m dx 2 2 6 45 kN/m 70 kN/m 45 kN/m 4 3 4 − x − 2.5 m + x − 6.5 m + x − 6.5 m 24(4.0 m) 6 24(4.0 m) 64.09 kN 2 + x − 8 m + C1 2 1

Integrate again to obtain the beam deflection function: 35 kN 160.91 kN 25 kN/m 3 3 4 EI v = − x−0 m + x − 2.5 m − x − 2.5 m 6 6 24 45 kN/m 70 kN/m 45 kN/m 5 4 5 − x − 2.5 m + x − 6.5 m + x − 6.5 m 120(4.0 m) 24 120(4.0 m) 64.09 kN 3 + x − 8 m + C1 x + C2 6

(a)

(b)

Evaluate constants using boundary conditions: Boundary conditions are specific values of deflection v or slope dv/dx that are known at particular locations along the beam span. For this beam, the deflection v is known at the pin support (x = 2.5 m) and at the roller support (x = 8 m). Substitute the boundary condition v = 0 at x = 2.5 m into Eq. (b) to obtain: 35 kN 0=− (2.5 m)3 + C1 (2.5 m) + C2 6 (c)  C1 (2.5 m) + C2 = 91.145833 kN-m3 Next, substitute the boundary condition v = 0 at x = 8 m into Eq. (b) to obtain: 35 kN 160.91 kN 25 kN/m 45 kN/m 0=− (8 m)3 + (5.5 m)3 − (5.5 m) 4 − (5.5 m)5 6 6 24 120(4.0 m) 70 kN/m 45 kN/m + (1.5 m)4 + (1.5 m)5 + C1 (8 m) + C2 24 120(4.0 m)  C1 (8 m) + C2 = −65.666667 kN-m3

(d)

Solve Eqs. (c) and (d) simultaneously for the two constants of integration C1 and C2: C1 = −28.511 kN-m2 and C2 = 162.424 kN-m3

The beam slope and elastic curve equations are now complete: dv 35 kN 160.91 kN 25 kN/m 2 2 3 EI =− x−0 m + x − 2.5 m − x − 2.5 m dx 2 2 6 45 kN/m 70 kN/m 45 kN/m 4 3 4 − x − 2.5 m + x − 6.5 m + x − 6.5 m 24(4.0 m) 6 24(4.0 m) 64.09 kN 2 + x − 8 m − 28.511 kN-m 2 2 35 kN 160.91 kN 25 kN/m 3 3 4 x−0 m + x − 2.5 m − x − 2.5 m 6 6 24 45 kN/m 70 kN/m 45 kN/m 5 4 5 − x − 2.5 m + x − 6.5 m + x − 6.5 m 120(4.0 m) 24 120(4.0 m) 64.09 kN 3 + x − 8 m − (28.511 kN-m 2 ) x + 162.424 kN-m3 6

EI v = −

(a) Beam deflection at A: The beam deflection at A is: EI v A = 162.424 kN-m3

 vA =

162.424 kN-m3 = 0.006768 m = 6.77 mm  24,000 kN-m 2

Ans.

(b) Beam deflection at C: At x = 6.5 m, the beam deflection is: 35 kN 160.91 kN 25 kN/m EI vC = − (6.5 m)3 + (4.0 m)3 − (4.0 m) 4 6 6 24 45 kN/m − (4.0 m)5 − (28.511 kN-m 2 )(6.5 m) + 162.424 kN-m3 120(4.0 m) = −271.1797 kN-m3 271.1797 kN-m3  vC = − = −0.0011299 m = 11.30 mm  24,000 kN-m 2

Ans.

P10.36 For the beam and loading shown, use discontinuity functions to compute (a) the slope of the beam at B and (b) the deflection of the beam at A. Assume a constant value of EI = 54,000 kN-m2 for the beam. FIGURE P10.36

Solution Support reactions: A FBD of the beam is shown to the right. Fy = C y − (20 kN/m)(3 m) − 12 (30 kN/m)(3 m) = 0  C y = 105 kN M C = (20 kN/m)(3 m)(2.5 m) + 12 (30 kN/m)(3 m)(2 m) + M C = 0  M C = −240 kN-m

Load function w(x): w( x) = −20 kN/m x − 0 m − 0

30 kN/m 30 kN/m 1 0 1 x − 0 m + 20 kN/m x − 3 m + x−3 m 3m 3m −1

−2

+30 kN/m x − 3 m + 105 kN x − 4 m + 240 kN-m x − 4 m 1

−1

20 kN/m 30 kN/m 20 kN/m 30 kN/m 2 3 2 3 x−0 m − x−0 m + x−3 m + x−3 m 2 6(3 m) 2 6(3 m) 30 kN/m 2 1 0 + x − 3 m + 105 kN x − 4 m + 240 kN-m x − 4 m 2

M ( x) = −

Equations for beam slope and beam deflection: From Eq. (10.1), we can write: d 2v 20 kN/m 30 kN/m 20 kN/m 2 3 2 EI 2 = M ( x) = − x−0 m − x−0 m + x−3 m dx 2 6(3 m) 2 30 kN/m 30 kN/m 3 2 1 0 + x−3 m + x − 3 m + 105 kN x − 4 m + 240 kN-m x − 4 m 6(3 m) 2 Integrate the moment function to obtain an expression for the beam slope: dv 20 kN/m 30 kN/m 20 kN/m 30 kN/m 3 4 3 4 EI =− x−0 m − x−0 m + x−3 m + x−3 m dx 6 24(3 m) 6 24(3 m) 30 kN/m 105 kN 3 2 1 (a) + x−3 m + x − 4 m + 240 kN-m x − 4 m + C1 6 2 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Integrate again to obtain the beam deflection function: 20 kN/m 30 kN/m 20 kN/m 30 kN/m 4 5 4 5 EI v = − x−0 m − x−0 m + x−3 m + x−3 m 24 120(3 m) 24 120(3 m) 30 kN/m 105 kN 240 kN-m 4 3 2 + x−3 m + x−4m + x − 4 m + C1 x + C2 24 6 2

(b)

Evaluate constants using boundary conditions: Boundary conditions are specific values of deflection v or slope dv/dx that are known at particular locations along the beam span. For this beam, both the slope dv/dx and the deflection v are known at the fixed support (x = 4 m). Substitute the boundary condition dv/dx = 0 at x = 4 m into Eq. (a) to obtain: 20 kN/m 30 kN/m 20 kN/m 30 kN/m 30 kN/m 0=− (4 m)3 − (4 m) 4 + (1 m)3 + (1 m) 4 + (1 m)3 + C1 6 24(3 m) 6 24(3 m) 6  C1 = 311.25 kN-m 2 Next, substitute the boundary condition v = 0 at x = 4 m into Eq. (b) to obtain: 20 kN/m 30 kN/m 20 kN/m 30 kN/m 0=− (4 m) 4 − (4 m)5 + (1 m) 4 + (1 m)5 24 120(3 m) 24 120(3 m)

30 kN/m (1 m)4 + (311.25 kN-m 2 )(4 m) + C2 24  C2 = −948.50 kN-m3 +

The beam slope and elastic curve equations are now complete: dv 20 kN/m 30 kN/m 20 kN/m 30 kN/m 3 4 3 4 EI =− x−0 m − x−0 m + x−3 m + x−3 m dx 6 24(3 m) 6 24(3 m) 30 kN/m 105 kN 3 2 1 + x−3 m + x − 4 m + 240 kN-m x − 4 m + 311.25 kN-m 2 6 2 20 kN/m 30 kN/m 20 kN/m 30 kN/m 4 5 4 5 x−0 m − x−0 m + x−3 m + x−3m 24 120(3 m) 24 120(3 m) 30 kN/m 105 kN 240 kN-m 4 3 2 + x−3 m + x−4m + x−4m 24 6 2 3 + (311.25 kN-m)x − 948.50 kN-m

EI v = −

(a) Beam slope at B: The beam slope at B is: dv 20 kN/m 30 kN/m EI =− (3 m)3 − (3 m) 4 + 311.25 kN-m 2 dx B 6 24(3 m) = 187.5 kN-m 2 

dv 187.5 kN-m 2 = = 0.003472 rad = 0.00347 rad dx B 54,000 kN-m 2

Ans.

(b) Beam deflection at A: The beam deflection at A is:

EI v A = −948.50 kN-m3 948.50 kN-m3  vA = − = −0.017565 m = 17.56 mm  54,000 kN-m 2

Ans.

P10.37a For the beams and loadings shown below, determine the beam deflection at point H. Assume that EI = 8 × 104 kN-m2 is constant for each beam.

FIGURE P10.37a

Solution Determine beam slope at A. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: ML A = (slope magnitude) 6 EI Values: M = 150 kN-m, L = 8 m, EI = 8 × 104 kN2 m Computation: ML (150 kN-m)(8 m) A = = = 0.00250 rad 6EI 6(8  104 kN-m2 ) Determine beam deflection at H. [Skill 1] vH = (3 m)(0.00250 rad) = 0.00750 m = 7.50 mm 

Ans.

P10.37b For the beams and loadings shown below, determine the beam deflection at point H. Assume that EI = 8 × 104 kN-m2 is constant for each beam.

FIGURE P10.37b

Solution Determine beam deflection at A. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL4 vA = − 8 EI Values: w = 6 kN/m, L = 4 m, EI = 8 × 104 kN-m2 Computation: wL4 (6 kN/m)(4 m)4 vA = − =− = −0.00240 m 8EI 8(8  104 kN-m2 ) Determine beam slope at A. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL3 A = (slope magnitude) 6 EI Values: w = 6 kN/m, L = 4 m, EI = 8 × 104 kN-m2 Computation: wL3 (6 kN/m)(4 m)3 A = = = 0.00080 rad 6 EI 6(8  104 kN-m2 ) Determine beam deflection at H. [Skill 2] vH = −0.00240 m − (2 m)(0.00080 rad) = −0.00400 m = 4.00 mm 

Ans.

P10.37c For the beams and loadings shown below, determine the beam deflection at point H. Assume that EI = 8 × 104 kN-m2 is constant for each beam.

FIGURE P10.37c

Solution Determine beam deflection at H. [Skill 3] [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pbx 2 vH = − ( L − b2 − x2 ) (elastic curve) 6LEI Values: P = 30 kN-m, L = 12 m, b = 4 m, x = 4 m, EI = 8 × 104 kN-m2 Computation: Pbx 2 vH = − ( L − b2 − x2 ) 6 LEI (30 kN)(4 m)(4 m) (12 m) 2 − (4 m) 2 − (4 m) 2  =− 4 2  6(12 m)(8  10 kN-m ) = 0.00933 m = 9.33 mm 

Ans.

P10.37d For the beams and loadings shown below, determine the beam deflection at point H. Assume that EI = 8 × 104 kN-m2 is constant for each beam.

FIGURE P10.37d

Solution Determine deflection of cantilever overhang. [Appendix C, Cantilever beam with concentrated load.] Relevant equation from Appendix C: PL3 vH ,cant = − (assuming fixed support at B) 3EI Values: P = 15 kN, L = 4 m, EI = 8 × 104 kN-m2 Computation:

PL3 (15 kN)(4 m)3 vH ,cant = − =− = −0.004000 m 3EI 3(8  104 kN-m2 ) Determine beam slope at B. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: ML B = (slope magnitude) 3EI Values: M = (15 kN)(4 m) = 60 kN-m, L = 8 m, EI = 8 × 104 kN-m2 Computation: ML (60 kN-m)(8 m) B = = = 0.002000 rad 3EI 3(8  104 kN-m2 ) Determine beam deflection at H. [Skill 4] vH = −0.00400 m − (4 m)(0.00200 rad) = −0.01200 m = 12.00 mm 

Ans.

P10.38a For the beams and loadings shown below, determine the beam deflection at point H. Assume that EI = 6 × 104 kN-m2 is constant for each beam.

FIGURE P10.38a

Solution Determine beam deflection at H. [Skill 3] [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: Mx vH = − (2L2 − 3Lx + x2 ) (elastic curve) 6LEI Values: M = −60 kN-m, L = 12 m, x = 6 m, EI = 6 × 104 kN-m2 Computation: Mx vH = − (2 L2 − 3Lx + x 2 ) 6 LEI ( −60 kN-m)(6 m)  2(12 m) 2 − 3(12 m)(6 m) + (6 m) 2  =− 6(12 m)(6  104 kN-m 2 )  = 0.009000 m = 9.00 mm 

Ans.

P10.38b For the beams and loadings shown below, determine the beam deflection at point H. Assume that EI = 6 × 104 kN-m2 is constant for each beam.

FIGURE P10.38b

Solution Determine deflection of cantilever overhang. [Appendix C, Cantilever beam with uniform load.] Relevant equation from Appendix C: wL4 vH ,cant = − (assuming fixed support at A) 8EI Values: w = 7.5 kN/m, L = 3 m, EI = 6 × 104 kN-m2 Computation:

vH ,cant = −

wL4 (7.5 kN/m)(3 m)4 =− = −0.00126563 m 8EI 8(6  104 kN-m2 )

Determine beam slope at A. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: ML A = (slope magnitude) 3EI Values: M = (7.5 kN/m)(3 m)(1.5 m) = 33.75 kN-m, L = 6 m, EI = 6 × 104 kN-m2 Computation: ML (33.75 kN-m)(6 m) A = = = 0.001125 rad 3EI 3(6  104 kN-m2 ) Determine beam deflection at H. [Skill 4] vH = −0.00126563 m − (3 m)(0.001125 rad) = −0.00464063 m = 4.64 mm 

Ans.

P10.38c For the beams and loadings shown below, determine the beam deflection at point H. Assume that EI = 6 × 104 kN-m2 is constant for each beam.

FIGURE P10.38c

Solution Determine beam deflection at B. [Appendix C, Cantilever beam with concentrated load.] Relevant equation from Appendix C: PL3 vB = − 3EI Values: P = 30 kN, L = 3 m, EI = 6 × 104 kN-m2 Computation: PL3 (30 kN)(3 m)3 vB = − =− = −0.004500 m 3EI 3(6  104 kN-m2 ) Determine beam slope at B. [Appendix C, Cantilever beam with concentrated load.] Relevant equation from Appendix C: PL2 B = (slope magnitude) 2 EI Values: P = 30 kN, L = 3 m, EI = 6 × 104 kN-m2 Computation: PL2 (30 kN)(3 m)2 B = = = 0.002250 rad 2 EI 2(6  104 kN-m2 ) Determine beam deflection at H. [Skill 2] vH = −0.004500 m − (3 m)(0.002250 rad) = −0.01125 m = 11.25 mm 

Ans.

Alternative solution for beam deflection at B. [Appendix C, Cantilever beam with concentrated load at midspan.] 5 PL3 Relevant equation from Appendix C: vH = − 48 EI Values: P = 30 kN, L = 6 m, EI = 6 × 104 kN-m2 5PL3 5(30 kN)(6 m)3 =− = −0.011250 m = 11.25 mm  Computation: vH = − 48EI 48(6  104 kN-m2 )

P10.38d For the beams and loadings shown below, determine the beam deflection at point H. Assume that EI = 6 × 104 kN-m2 is constant for each beam.

FIGURE P10.38d

Solution Determine beam slope at C. [Appendix C, SS beam with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C: wa 2 C = (2 L − a)2 (slope magnitude) 24 LEI Values: w = 5 kN/m, L = 9 m, a = 6 m, EI = 6 × 104 kN-m2 Computation: wa 2 (5 kN/m)(6 m)2 2 C = (2 L − a)2 = 2(9 m) − (6 m) = 0.00200 rad 4 2  24 LEI 24(9 m)(6  10 kN-m ) Determine beam deflection at H. [Skill 1] vH = (3 m)(0.00200 rad) = 0.00600 m = 6.00 mm 

Ans.

P10.39 The simply supported beam shown in Figure P10.39 consists of a W24 × 94 structural steel wide-flange shape [E = 29,000 ksi; I = 2,700 in.4]. For the loading shown, determine the beam deflection at point C. FIGURE P10.39

Solution Consider distributed load. [Appendix C, SS beam with uniformly distributed load over portion of span.] Relevant equation from Appendix C: wa3 vC = − (4 L2 − 7aL + 3a 2 ) 24 LEI Values: w = 3.2 kips/ft, L = 28 ft, a = 21 ft, EI = 7.830 × 107 kip-in.2 Computation: wa3 vC = − (4 L2 − 7aL + 3a 2 ) 24 LEI

=−

(3.2 kips/ft)(21 ft)3 (12 in./ft)3  4(28 ft)2 − 7(21 ft)(28 ft) + 3(21 ft) 2  = −0.333822 in. 7 2  24(28 ft)(7.830  10 kip-in. )

Consider concentrated load. [Appendix C, SS beam with concentrated load at midspan.] Relevant equation from Appendix C: Px vC = − (3L2 − 4 x 2 ) (elastic curve) 48EI Values: P = 36 kips, L = 28 ft, x = 7 ft, EI = 7.830 × 107 kip-in.2 Computation: Px vC = − (3L2 − 4 x 2 ) 48EI (36 kips)(7 ft)(12 in./ft)3 3(28 ft) 2 − 4(7 ft) 2  = −0.249799 in. =− 7 2 48(7.830  10 kip-in. ) Beam deflection at C vC = −0.333822 in. − 0.249799 in. = −0.583620 in. = 0.584 in. 

Ans.

P10.40 The simply supported beam shown in Figure P10.40 consists of a W410 × 60 structural steel wide-flange shape [E = 200 GPa; I = 216 × 106 mm4]. For the loading shown, determine the beam deflection at point B. FIGURE P10.40

Solution Consider concentrated load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pab 2 vB = − ( L − a 2 − b2 ) 6LEI Values: P = 60 kN, L = 9 m, a = 3 m, b = 6 m, EI = 4.32 × 104 kN-m2 Computation: Pab 2 vB = − ( L − a 2 − b2 ) 6 LEI (60 kN)(3 m)(6 m) (9 m) 2 − (3 m) 2 − (6 m) 2  = −0.016667 m =− 4 2  6(9 m)(4.32  10 kN-m ) Consider concentrated moment. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: Mx vB = − (2L2 − 3Lx + x 2 ) (elastic curve) 6LEI Values: M = −45 kN-m, L = 9 m, x = 6 m, EI = 4.32 × 104 kN-m2 Computation: Mx vB = − (2 L2 − 3Lx + x 2 ) 6 LEI (−45 kN-m)(6 m)  2(9 m) 2 − 3(9 m)(6 m) + (6 m) 2  = 0.004167 m =− 6(9 m)(4.32  104 kN-m 2 )  Beam deflection at B vB = −0.016667 m + 0.004167 m = −0.012500 m = 12.50 mm 

Ans.

P10.41 The cantilever beam shown in Figure P10.41 consists of a rectangular structural steel tube shape [E = 29,000 ksi; I = 476 in.4]. For the loading shown, determine: (a) the beam deflection at point B. (b) the beam deflection at point C. FIGURE P10.41

Solution (a) Beam deflection at point B Consider uniformly distributed load. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL4 vB = − 8 EI Values: w = 2 kips/ft, L = 6 ft, EI = 1.3804 × 107 kip-in.2 Computation: wL4 (2 kips/ft)(6 ft)4 (12 in./ft)3 vB = − =− = −0.040559 in. 8EI 8(1.3804  107 kip-in.2 ) Consider concentrated load. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: Px 2 vB = − (3L − x ) (elastic curve) 6 EI Values: P = 12 kips, L = 10 ft, x = 6 ft, EI = 1.3804 × 107 kip-in.2 Computation: Px 2 (12 kips)(6 ft)2 (12 in./ft)3 vB = − (3L − x) = − 3(10 ft) − (6 ft) = −0.216313 in. 6 EI 6(1.3804  107 kip-in.2 ) Beam deflection at B vB = −0.040559 in. − 0.216313 in. = −0.256872 in. = 0.257 in. 

Ans.

(b) Beam deflection at point C Consider uniformly distributed load. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL3 B = − 6 EI Values: w = 2 kips/ft, L = 6 ft, EI = 1.3804 × 107 kip-in.2 Computation: wL3 (2 kips/ft)(6 ft)3 (12 in./ft) 2 B = − =− = −751.0866  10−6 rad 6 EI 6(1.3804  107 kip-in.2 ) vC = −0.040559 in. − (4 ft)(12 in./ft)(751.0866  10−6 rad) = −0.076611 in.

Consider concentrated load. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: PL3 vC = − 3EI Values: P = 12 kips, L = 10 ft, EI = 1.3804 × 107 kip-in.2 Computation: PL3 (12 kips)(10 ft)3 (12 in./ft)3 vC = − =− = −0.500724 in. 3EI 3(1.3804  107 kip-in.2 ) Beam deflection at C vC = −0.076611 in. − 0.500724 in. = −0.577336 in. = 0.577 in. 

Ans.

P10.42 The solid 1.25 in. diameter steel [E = 29,000 ksi] shaft shown in Figure P10.42 supports two pulleys. For the loading shown, determine: (a) the shaft deflection at point B. (b) the shaft deflection at point C.

FIGURE P10.42

Solution Section properties: I=



(1.25 in.) 4 = 0.119842 in.4

(a) Shaft deflection at point B Consider concentrated load at pulley B. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: PL3 vB = − 3EI Values: P = 200 lb, L = 10 in., EI = 3.47543 × 106 lb-in.2 Computation: PL3 (200 lb)(10 in.)3 vB = − =− = −0.019182 in. 3EI 3(3.47543  106 lb-in.2 ) Consider concentrated load at pulley C. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: Px 2 vB = − (3L − x ) (elastic curve) 6 EI Values: P = 120 lb, L = 25 in., x = 10 in., EI = 3.47543 × 106 lb-in.2 Computation: Px 2 (120 lb)(10 in.)2 vB = − (3L − x) = − 3(25 in.) − (10 in.) = −0.037405 in. 6 EI 6(3.47543  106 lb-in.2 ) Shaft deflection at B vB = −0.019182 in. − 0.037405 in. = −0.056588 in. = 0.0566 in. 

Ans.

(b) Shaft deflection at point C Consider concentrated load at pulley B. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equations from Appendix C: PL3 PL2 vB = − and  B = (magnitude) 3EI 2 EI Values: P = 200 lb, L = 10 in., EI = 3.47543 × 106 lb-in.2 Computation: PL3 (200 lb)(10 in.)3 vB = − =− = −0.019182 in. 3EI 3(3.47543  106 lb-in.2 )

B =

PL2 (200 lb)(10 in.) 2 = = 0.0028773 rad 2 EI 2(3.47543  106 lb-in.2 )

vC = −0.019182 in. − (15 in.)(0.0028773 rad) = −0.062342 in.

Consider concentrated load at pulley C. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: PL3 vC = − 3EI Values: P = 120 lb, L = 25 in., EI = 3.47543 × 106 lb-in.2 Computation: PL3 (120 lb)(25 in.)3 vC = − =− = −0.179834 in. 3EI 3(3.47543  106 lb-in.2 ) Shaft deflection at C vC = −0.062342 in. − 0.179834 in. = −0.242176 in. = 0.242 in. 

Ans.

P10.43 The cantilever beam shown in Figure P10.43 consists of a rectangular structural steel tube shape [E = 29,000 ksi; I = 1,710 in.4]. For the loading shown, determine: (a) the beam deflection at point A. (b) the beam deflection at point B. FIGURE P10.43

Solution (a) Beam deflection at point A Consider concentrated moment. [Appendix C, Cantilever beam with concentrated moment.] Relevant equation from Appendix C: ML2 vA = − 2 EI Values: M = −200 kip-ft, L = 15 ft, EI = 4.959 × 107 kip-in.2 Computation: ML2 (−200 kip-ft)(15 ft)2 (12 in./ft)3 vA = − =− = 0.784029 in. 2 EI 2(4.959  107 kip-in.2 )

Consider concentrated load. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equations from Appendix C: PL3 PL2 vB = − and  B = (slope 3EI 2 EI magnitude) Values: P = −18 kips, L = 9 ft, EI = 4.959 × 107 kip-in.2 Computation: PL3 (−18 kips)(9 ft)3 (12 in./ft)3 vB = − =− = 0.152415 in. 3EI 3(4.959  107 kip-in.2 ) PL2 (18 kips)(9 ft) 2 (12 in./ft)2 B = = = 0.0021169 rad 2 EI 2(4.959  107 kip-in.2 ) v A = 0.152415 in. + (6 ft)(12 in./ft)(0.0021169 rad) = 0.304830 in.

Beam deflection at A vA = 0.784029 in. + 0.304830 in. = 1.088860 in. = 1.089 in. 

Ans.

(b) Beam deflection at point B Consider concentrated moment. [Appendix C, Cantilever beam with concentrated moment.] Relevant equation from Appendix C: Mx 2 vB = − (elastic curve) 2 EI Values: M = −200 kip-ft, L = 15 ft, x = 9 ft, EI = 4.959 × 107 kip-in.2 Computation: Mx 2 (−200 kip-ft)(9 ft)2 (12 in./ft)3 vB = − =− = 0.282250 in. 2 EI 2(4.959  107 kip-in.2 )

Consider concentrated load. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: PL3 vB = − 3EI Values: P = −18 kips, L = 9 ft, EI = 4.959 × 107 kip-in.2 Computation: PL3 (−18 kips)(9 ft)3 (12 in./ft)3 vB = − =− = 0.152415 in. 3EI 3(4.959  107 kip-in.2 ) Beam deflection at B vB = 0.282250 in. + 0.152415 in. = 0.434665 in. = 0.435 in. 

Ans.

P10.44 The simply supported beam shown in Figure P10.44 consists of a W21 × 44 structural steel wide-flange shape [E = 29,000 ksi; I = 843 in.4]. For the loading shown, determine: (a) the beam deflection at point A. (b) the beam deflection at point C. FIGURE P10.44

Solution (a) Beam deflection at point A Determine cantilever deflection due to uniformly distributed load on overhang. [Appendix C, Cantilever beam with uniform load.] Relevant equation from Appendix C: wL4 vA = − (assuming fixed support at B) 8 EI Values: w = 4 kips/ft, L = 8 ft, EI = 2.4447 × 107 kip2 in. Computation: wL4 (4 kips/ft)(8 ft) 4 (12 in./ft)3 vA = − =− = −0.144760 in. 8EI 8(2.4447  107 kip-in.2 ) Consider deflection at A resulting from rotation at B caused by distributed load on overhang. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: ML B = (slope magnitude) 3EI Values: M = (4 kips/ft)(8 ft)(4 ft) = 128 kip-ft, L = 22 ft, EI = 2.4447 × 107 kip-in.2 Computation: ML (128 kip-ft)(22 ft)(12 in./ft) 2 B = = = 0.0055290 rad 3EI 3(2.4447  10 7 kip-in.2 ) v A = (8 ft)(12 in./ft)(0.0055290 rad) = −0.530786 in.

Consider concentrated load. [Appendix C, SS beam with concentrated load at midspan.] Relevant equation from Appendix C: PL2 B = (slope magnitude) 16 EI Values: P = 45 kips, L = 22 ft, EI = 2.4447 × 107 kip2 in. Computation: PL2 (45 kips)(22 ft)2 (12 in./ft)2 B = = = 0.0080182 rad 16 EI 16(2.4447  107 kip-in.2 )

vA = (8 ft)(12 in./ft)(0.0080182 rad) = 0.769744 in. Beam deflection at A vA = −0.144760 in. − 0.530786 in. + 0.769744 in. = 0.094198 in. = 0.0942 in. 

Ans.

(b) Beam deflection at point C Consider distributed load on overhang. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: Mx vC = − (2L2 − 3Lx + x 2 ) (elastic curve) 6LEI Values: M = (4 kips/ft)(8 ft)(4 ft) = −128 kip-ft, L = 22 ft, x = 11 ft, EI = 2.4447 × 107 kip-in.2 Computation: Mx vC = − (2 L2 − 3Lx + x 2 ) 6 LEI =−

(−128 kip-ft)(11 ft)(12 in./ft) 3  2(22 ft) 2 − 3(22 ft)(11 ft) + (11 ft) 2  = 0.273687 in. 6(22 ft)(2.4447  107 kip-in.2 ) 

Consider concentrated load. [Appendix C, SS beam with concentrated load at midspan.] Relevant equation from Appendix C: PL3 vC = − 48 EI Values: P = 45 kips, L = 22 ft, EI = 2.4447 × 107 kipin.2 Computation: PL3 (45 kips)(22 ft)3 (12 in./ft)3 vC = − =− = −0.705598 in. 48EI 48(2.4447  107 kip-in.2 ) Beam deflection at C vC = 0.273687 in. − 0.705598 in. = −0.431912 in. = 0.432 in. 

Ans.

P10.45 The solid 30 mm diameter steel [E = 200 GPa] shaft shown in Figure P10.45 supports two belt pulleys. Assume that the bearing at B can be idealized as a roller support and that the bearing at D can be idealized as a pin support. For the loading shown, determine: (a) the shaft deflection at pulley A. (b) the shaft deflection at pulley C. FIGURE P10.45

Solution Section properties: I=



(30 mm) 4 = 39,760.78 mm 4

(a) Shaft deflection at pulley A Determine cantilever deflection due to pulley A load. [Appendix C, Cantilever beam with concentrated load.] Relevant equation from Appendix C: PL3 vA = − (assuming fixed support at B) 3EI Values: P = 700 N, L = 500 mm, EI = 7.95216 × 109 N-mm2 Computation: PL3 (700 N)(500 mm)3 vA = − =− = −3.6678 mm 3EI 3(7.95216  109 N-mm2 ) Consider deflection at A resulting from rotation at B caused by pulley A load. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: ML B = (slope magnitude) 3EI Values: M = (700 N)(500 mm) = 350,000 N-mm, L = 1,800 mm, EI = 7.95216 × 109 N-mm2 Computation: ML (350,000 N-mm)(1,800 mm) B = = = 0.0264079 rad 3EI 3(7.95216  109 N-mm 2 )

v A = −(500 mm)(0.0264079 rad) = −13.2040 mm Consider deflection at A resulting from rotation at B caused by pulley C load. [Appendix C, SS beam with concentrated load.]

Relevant equation from Appendix C: PL2 B = (slope magnitude) 16 EI Values: P = 1,000 N, L = 1,800 mm, EI = 7.95216 × 109 N-mm2 Computation: PL2 (1,000 N)(1,800 mm) 2 B = = = 0.0254648 rad 16 EI 16(7.95216  109 N-mm 2 ) v A = (500 mm)(0.0254648 rad) = 12.7324 mm

Shaft deflection at A vA = −3.6678 mm − 13.2040 mm + 12.7324 mm = −4.1393 mm = 4.14 mm 

Ans.

(b) Shaft deflection at pulley C Consider pulley A load. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: Mx vC = − (2L2 − 3Lx + x 2 ) (elastic curve) 6LEI Values: M = (700 N)(500 mm) = −350,000 N-mm, L = 1,800 mm, x = 900 mm, EI = 7.95216 × 109 N-mm2 Computation: Mx vC = − (2 L2 − 3Lx + x 2 ) 6 LEI (−350,000 N-mm)(900 mm)  2(1,800 mm) 2 − 3(1,800 mm)(900 mm) + (900 mm) 2  =− 9 2  6(1,800 mm)(7.95216  10 N-mm ) = 8.9127 mm

Consider pulley C load. [Appendix C, SS beam with concentrated load.] Relevant equation from Appendix C: PL3 vC = − 48 EI Values: P = 1,000 N, L = 1,800 mm, EI = 7.95216 × 109 N-mm2 Computation: PL3 (1,000 N)(1,800 mm)3 vC = − =− = −15.2789 mm 48EI 48(7.95216  109 N-mm 2 ) Shaft deflection at C vC = 8.9127 mm − 15.2789 mm = −6.3662 mm = 6.37 mm 

Ans.

P10.46 The cantilever beam shown in Figure P10.46 consists of a W530 × 92 structural steel wide-flange shape [E = 200 GPa; I = 552 × 106 mm4]. For the loading shown, determine: (a) the beam deflection at point A. (b) the beam deflection at point B. FIGURE P10.46

Solution (a) Beam deflection at point A Consider an upward 85 kN/m uniformly distributed load acting over entire 4-m span. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL4 vA = − 8 EI Values: w = −85 kN/m, L = 4 m, EI = 1.104 × 105 kN-m2 Computation: wL4 (−85 kN/m)(4 m)4 vA = − =− = 0.024638 m 8EI 8(1.104  105 kN-m2 ) Consider a downward 85 kN/m uniformly distributed load acting over span BC. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL4 wL3 vB = − and  B = (magnitude) 8 EI 6 EI Values: w = 85 kN/m, L = 2.5 m, EI = 1.104 × 105 kN-m2 Computation: wL4 (85 kN)(2.5 m) 4 vB = − =− = −0.003759 m 8 EI 8(1.104  105 kN-m 2 )

B =

wL3 (85 kN)(2.5 m)3 = = 0.0020050 rad 6 EI 6(1.104  105 kN-m 2 )

v A = −0.003759 m − (1.5 m)(0.0020050 rad) = −0.006767 m

Beam deflection at A vA = 0.024638 m − 0.006767 m = 0.017871 m = 17.87 mm 

Ans.

(b) Beam deflection at point B Consider an upward 85 kN/m uniformly distributed load acting over entire 4-m span. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wx 2 vB = − (6 L2 − 4 Lx + x 2 ) (elastic curve) 24 EI Values: w = −85 kN/m, L = 4 m, x = 2.5 m, EI = 1.104 × 105 kN-m2 Computation: wx 2 vB = − (6 L2 − 4 Lx + x 2 ) 24 EI (−85 kN/m)(2.5 m) 2 6(4 m) 2 − 4(4 m)(2.5 m) + (2.5 m) 2  = 0.012481 m =− 5 2  24(1.104  10 kN-m ) Consider a downward 85 kN/m uniformly distributed load acting over span BC. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL4 vB = − 8 EI Values: w = 85 kN/m, L = 2.5 m, EI = 1.104 × 105 kN-m2 Computation: wL4 (85 kN)(2.5 m)4 vB = − =− = −0.003759 m 8EI 8(1.104  105 kN-m2 ) Beam deflection at B vB = 0.012481 m − 0.003759 m = 0.008722 m = 8.72 mm 

Ans.

P10.47 The simply supported beam shown in Figure P10.47 consists of a W410 × 60 structural steel wide-flange shape [E = 200 GPa; I = 216 × 106 mm4]. For the loading shown, determine the beam deflection at point B.

FIGURE P10.47

Solution Beam deflection at point B Consider concentrated moment. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: Mx vB = − (2L2 − 3Lx + x 2 ) (elastic curve) 6LEI Values: M = −180 kN-m, L = 6 m, x = 1.5 m, EI = 4.32 × 104 kN-m2 Computation: Mx vB = − (2 L2 − 3Lx + x 2 ) 6 LEI

(−180 kN-m)(1.5 m)  2(6 m) 2 − 3(6 m)(1.5 m) + (1.5 m) 2  = 0.008203 m 4 2  6(6 m)(4.32  10 kN-m ) Consider concentrated load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pab 2 vB = − ( L − a 2 − b2 ) 6LEI Values: P = 70 kN, L = 6 m, a = 1.5 m, b = 4.5 m, EI = 4.32 × 104 kN-m2 Computation: Pab 2 vB = − ( L − a 2 − b2 ) 6 LEI (70 kN)(1.5 m)(4.5 m) (6 m) 2 − (1.5 m) 2 − (4.5 m) 2  = −0.004102 m =− 4 2  6(6 m)(4.32  10 kN-m ) =−

Consider uniformly distributed load. [Appendix C, SS beam with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C: wa 2 vB = − (2 x3 − 6 Lx 2 + a 2 x + 4 L2 x − a 2 L) 24 LEI Values: w = 80 kN/m, L = 6 m, a = 3 m, x = 4.5 m, EI = 4.32 × 104 kN-m2

Computation: Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

wa 2 vB = − (2 x3 − 6 Lx 2 + a 2 x + 4 L2 x − a 2 L) 24 LEI (80 kN/m)(3 m) 2  2(4.5)3 − 6(6)(4.5) 2 + (3) 2 (4.5) + 4(6) 2 (4.5) − (3) 2 (6)  =− 4 2  24(6.0 m)(4.32  10 kN-m ) = −0.010156 m Beam deflection at B

vB = 0.008203 m − 0.004102 m − 0.010156 m = −0.006055 m = 6.06 mm 

Ans.

P10.48 The simply supported beam shown in Figure P10.48 consists of a W410 × 60 structural steel wide-flange shape [E = 200 GPa; I = 216 × 106 mm4]. For the loading shown, determine the beam deflection at point C.

FIGURE P10.48

Solution Beam deflection at point C Consider concentrated moment. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: Mx vC = − (2L2 − 3Lx + x 2 ) (elastic curve) 6LEI Values: M = −180 kN-m, L = 6 m, x = 3.0 m, EI = 4.32 × 104 kN-m2 Computation: Mx vC = − (2 L2 − 3Lx + x 2 ) 6 LEI

=−

(−180 kN-m)(3.0 m)  2(6 m) 2 − 3(6 m)(3.0 m) + (3.0 m) 2  = 0.009375 m 4 2  6(6 m)(4.32  10 kN-m )

Consider concentrated load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pbx 2 vC = − ( L − b2 − x2 ) (elastic curve) 6LEI Values: P = 70 kN, L = 6 m, x = 3.0 m, b = 1.5 m, EI = 4.32 × 104 kN-m2 Computation: Pbx 2 vC = − ( L − b2 − x2 ) 6 LEI (70 kN)(1.5 m)(3.0 m) (6 m) 2 − (1.5 m) 2 − (3.0 m)2  = −0.005013 m =− 4 2  6(6 m)(4.32  10 kN-m )

Consider uniformly distributed load. [Appendix C, SS beam with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C: wa3 vC = − (4 L2 − 7aL + 3a 2 ) 24 LEI Values: w = 80 kN/m, L = 6 m, a = 3 m, EI = 4.32 × 104 kN-m2 Computation: wa3 vC = − (4 L2 − 7aL + 3a 2 ) 24 LEI =−

(80 kN/m)(3 m)3  4(6 m)2 − 7(3 m)(6 m) + 3(3 m)2  = −0.015625 m 4 2  24(6.0 m)(4.32  10 kN-m )

Beam deflection at C vC = 0.009375 m − 0.005013 m − 0.015625 m = −0.011263 m = 11.26 mm 

Ans.

P10.49 The simply supported beam shown in Figure P10.49 consists of a W530 × 66 structural steel wide-flange shape [E = 200 GPa; I = 351 × 106 mm4]. If w = 80 kN/m, determine (a) the beam deflection at point A. (b) the beam deflection at point C. FIGURE P10.49

Solution (a) Beam deflection at point A Determine cantilever deflection due to concentrated load on overhang AB. [Appendix C, Cantilever beam with concentrated load.] Relevant equation from Appendix C: PL3 vA = − (assuming fixed support at B) 3EI Values: P = 35 kN, L = 4 m, EI = 7.02 × 104 kN-m2 Computation: PL3 (35 kN)(4 m)3 vA = − =− = −0.0106363 m 3EI 3(7.02  104 kN-m2 ) Consider deflection at A resulting from rotation at B caused by concentrated load on overhang AB. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: ML B = (slope magnitude) 3EI Values: M = (35 kN)(4 m) = 140 kN-m, L = 8 m, EI = 7.02 × 104 kN-m2 Computation: ML (140 kN-m)(8 m) B = = = 0.0053181 rad 3EI 3(7.02  104 kN-m 2 )

vA = −(4 m)(0.0053181 rad) = −0.0212726 m Consider uniformly distributed loads between C and D. [Appendix C, SS beam with uniformly distributed load over portion of span.] Relevant equations from Appendix C: wa 2 B = (2 L2 − a 2 ) (slope magnitude) 24 LEI Values: w = 80 kN/m, L = 8 m, a = 4 m, EI = 7.02 × 104 kN-m2

Computation: wa 2 (80 kN/m)(4 m)2  2(8 m)2 − (4 m)2  = 0.0106363 rad B = (2 L2 − a 2 ) = 4 2  24 LEI 24(8 m)(7.02  10 kN-m )

vA = (4 m)(0.0106363 rad) = 0.0425451 m Consider deflection at A resulting from rotation at B caused by uniform load on overhang DE. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: ML B = (slope magnitude) 6 EI Values: M = (80 kN/m)(2 m)(1 m) = 160 kN-m, L = 8 m, EI = 7.02 × 104 kN-m2 Computation: ML (160 kN-m)(8 m) B = = = 0.0030389 rad 6 EI 6(7.02  104 kN-m 2 ) v A = −(4 m)(0.0030389 rad) = −0.0121557 m

Beam deflection at A vA = −0.0106363 m − 0.0212726 m + 0.0425451 m − 0.0121557 m = −0.0015195 m = 1.520 mm 

Ans.

(b) Beam deflection at point C Consider concentrated moment from overhang AB. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: Mx vC = − (2L2 − 3Lx + x 2 ) (elastic curve) 6LEI Values: M = (35 kN)(4 m) = −140 kN-m, L = 8 m, x = 4 m, EI = 7.02 × 104 kN-m2 Computation: Mx vC = − (2 L2 − 3Lx + x 2 ) 6 LEI (−140 kN-m)(4 m)  2(8 m) 2 − 3(8 m)(4 m) + (4 m) 2  = 0.0079772 m =− 6(8 m)(7.02  104 kN-m 2 ) 

Consider uniformly distributed loads between C and D. [Appendix C, SS beam with uniformly distributed load over portion of span.] Relevant equations from Appendix C: wa3 vC = − (4 L2 − 7aL + 3a 2 ) 24 LEI Values: w = 80 kN/m, L = 8 m, a = 4 m, EI = 7.02 × 104 kN-m2 Computation: wa3 vC = − (4 L2 − 7aL + 3a 2 ) 24 LEI

=−

(80 kN/m)(4 m)3  4(8 m)2 − 7(4 m)(8 m) + 3(4 m) 2  = −0.0303894 m 4 2  24(8 m)(7.02  10 kN-m )

Consider concentrated moment from overhang DE. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: Mx vC = − (2L2 − 3Lx + x 2 ) (elastic curve) 6LEI Values: M = (80 kN/m)(2 m)(1 m) = 160 kN-m, L = 8 m, x = 4 m, EI = 7.02 × 104 kN-m2 Computation: Mx vC = − (2 L2 − 3Lx + x 2 ) 6 LEI (−160 kN-m)(4 m)  2(8 m) 2 − 3(8 m)(4 m) + (4 m) 2  = 0.0091168 m =− 4 2  6(8 m)(7.02  10 kN-m ) Beam deflection at C

vC = 0.0079772 m − 0.0303894 m + 0.0091168 m = −0.0132954 m = 13.30 mm 

Ans.

P10.50 The simply supported beam shown in Figure P10.50 consists of a W530 × 66 structural steel wide-flange shape [E = 200 GPa; I = 351 × 106 mm4]. If w = 90 kN/m, determine: (a) the beam deflection at point C. (b) the beam deflection at point E. FIGURE P10.50

Solution (a) Beam deflection at point C Consider concentrated moment from overhang AB. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: Mx vC = − (2L2 − 3Lx + x 2 ) (elastic curve) 6LEI Values: M = (35 kN)(4 m) = −140 kN-m, L = 8 m, x = 4 m, EI = 7.02 × 104 kN-m2 Computation: Mx vC = − (2 L2 − 3Lx + x 2 ) 6 LEI (−140 kN-m)(4 m)  2(8 m) 2 − 3(8 m)(4 m) + (4 m) 2  = 0.0079772 m =− 4 2  6(8 m)(7.02  10 kN-m ) Consider uniformly distributed loads between C and D. [Appendix C, SS beam with uniformly distributed load over portion of span.] Relevant equations from Appendix C: wa3 vC = − (4 L2 − 7aL + 3a 2 ) 24 LEI Values: w = 90 kN/m, L = 8 m, a = 4 m, EI = 7.02 × 104 kN-m2 Computation: wa3 vC = − (4 L2 − 7aL + 3a 2 ) 24 LEI

=−

(90 kN/m)(4 m)3  4(8 m) 2 − 7(4 m)(8 m) + 3(4 m) 2  = −0.0341881 m 24(8 m)(7.02  104 kN-m 2 ) 

=−

( −180 kN-m)(4 m)  2(8 m) 2 − 3(8 m)(4 m) + (4 m) 2  = 0.0102564 m 4 2  6(8 m)(7.02  10 kN-m )

Beam deflection at C

vC = 0.0079772 m − 0.0341881 m + 0.0102564 m = −0.0159545 m = 15.95 mm 

Ans.

(b) Beam deflection at point E Consider deflection at E resulting from rotation at D caused by concentrated load on overhang AB. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: ML D = (slope magnitude) 6 EI Values: M = (35 kN)(4 m) = 140 kN-m, L = 8 m, EI = 7.02 × 104 kN-m2 Computation: ML (140 kN-m)(8 m) D = = = 0.0026591 rad 6 EI 6(7.02  104 kN-m 2 ) vE = −(2 m)(0.0026591 rad) = −0.0053181 m

Consider uniformly distributed loads between C and D. [Appendix C, SS beam with uniformly distributed load over portion of span.] Relevant equations from Appendix C: wa 2 D = (2 L − a) 2 (slope magnitude) 24 LEI Values: w = 90 kN/m, L = 8 m, a = 4 m, EI = 7.02 × 104 kN-m2 Computation: wa 2 (90 kN/m)(4 m)2 2 D = (2 L − a)2 = 2(8 m) − (4 m) = 0.0153846 rad 4 2  24 LEI 24(8 m)(7.02 10 kN-m )

vE = (2 m)(0.0153846 rad) = 0.0307692 m Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Consider deflection at E resulting from rotation at D caused by uniform load on overhang DE. [Appendix C, SS beam with concentrated moment.] Relevant equation from Appendix C: ML D = (slope magnitude) 3EI Values: M = (90 kN/m)(2 m)(1 m) = 180 kN-m, L = 8 m, EI = 7.02 × 104 kN-m2 Computation: ML (180 kN-m)(8 m) D = = = 0.0068376 rad 3EI 3(7.02  104 kN-m 2 ) vE = −(2 m)(0.0068376 rad) = −0.0136753 m

Determine cantilever deflection due to uniformly distributed load on overhang DE. [Appendix C, Cantilever beam with distributed load.] Relevant equation from Appendix C: wL4 vE = − (assuming fixed support at D) 8 EI Values: w = 90 kN/m, L = 2 m, EI = 7.02 × 104 kN-m2 wL4 (90 kN/m)(2 m)4 Computation: vE = − =− = −0.0025641 m 8EI 8(7.02  104 kN-m2 ) Beam deflection at E vE = −0.0053181 m + 0.0307692 m − 0.0136753 m − 0.0025641 m = 0.0092117 m = 9.21 mm 

Ans.

P10.51 The cantilever beam shown in Figure P10.51 consists of a rectangular structural steel tube shape [E = 200 GPa; I = 95 × 106 mm4]. For the loading shown, determine the beam deflection at point B. FIGURE P10.51

Consider an upward 25 kN/m uniformly distributed load acting over entire 5-m span. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wx 2 vB = − (6 L2 − 4 Lx + x 2 ) (elastic curve) 24 EI Values: w = −25 kN/m, L = 5 m, x = 2 m, EI = 1.9 × 104 kN-m2 Computation: wx 2 vB = − (6 L2 − 4 Lx + x 2 ) 24 EI (−25 kN/m)(2 m) 2 6(5 m) 2 − 4(5 m)(2 m) + (2 m) 2  = 0.0250000 m =− 4 2  24(1.9  10 kN-m ) Consider a downward 25 kN/m uniformly distributed load acting over span AB. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL4 vB = − 8 EI Values: w = 25 kN/m, L = 2 m, EI = 1.9 × 104 kN-m2 Computation:

vB = −

wL4 (25 kN/m)(2 m)4 =− = −0.0026316 m 8EI 8(1.9  104 kN-m2 )

Consider 20-kN concentrated load at B. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equations from Appendix C: PL3 vB = − 3EI Values: P = −20 kN, L = 2 m, EI = 1.9 × 104 kN-m2 Computation: PL3 (−20 kN)(2 m)3 vB = − =− = 0.0028070 m 3EI 3(1.9  104 kN-m2 ) Consider 50-kN concentrated load at C. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equations from Appendix C: Px 2 vB = − (3L − x ) (elastic curve) 6 EI Values: P = 50 kN, L = 5 m, x = 2 m, EI = 1.9 × 104 kN2 m Computation: Px 2 (50 kN)(2 m)2 vB = − (3L − x) = − 3(5 m) − (2 m) = −0.0228070 m 6 EI 6(1.9  104 kN-m2 ) Beam deflection at B vB = −0.0052632 m + 0.0250000 m − 0.0026316 m + 0.0028070 m − 0.0228070 m = −0.0028947 m = 2.89 mm 

Ans.

P10.52 The cantilever beam shown in Figure P10.52 consists of a rectangular structural steel tube shape [E = 200 GPa; I = 95 × 106 mm4]. For the loading shown, determine the beam deflection at point C. FIGURE P10.52

Solution Consider the downward 50 kN/m uniformly distributed load acting over span AB. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL3 wL4 vB = − and  B = (slope magnitude) 6 EI 8 EI Values: w = 50 kN/m, L = 2 m, EI = 1.9 × 104 kN-m2 Computation: wL4 (50 kN/m)(2 m)4 vB = − =− = −0.0052632 m 8EI 8(1.9  104 kN-m2 ) wL3 (50 kN/m)(2 m)3 B = = = 0.0035088 rad 6 EI 6(1.9  104 kN-m 2 ) vC = −0.0052632 m − (3 m)(0.0035088 rad) = −0.0157895 m

Consider an upward 25 kN/m uniformly distributed load acting over entire 5-m span. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL4 vC = − 8EI Values: w = −25 kN/m, L = 5 m, EI = 1.9 × 104 kN-m2 Computation: wL4 (−25 kN/m)(5 m)4 vC = − =− = 0.1027961 m 8EI 8(1.9  104 kN-m2 ) Consider a downward 25 kN/m uniformly distributed load acting over span AB. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL3 wL4 vB = − and  B = (slope magnitude) 6 EI 8 EI Values: w = 25 kN/m, L = 2 m, EI = 1.9 × 104 kN-m2

Computation: wL4 (25 kN/m)(2 m)4 vB = − =− = −0.0026316 m 8EI 8(1.9  104 kN-m2 ) wL3 (25 kN)(2 m)3 B = = = 0.0017544 rad 6 EI 6(1.9  104 kN-m 2 ) vC = −0.0026316 m − (3 m)(0.0017544 rad) = −0.0078948 m

Consider 20-kN concentrated load at B. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equations from Appendix C: PL2 PL3 vB = − and  B = (slope magnitude) 3EI 2 EI Values: P = −20 kN, L = 2 m, EI = 1.9 × 104 kN-m2 Computation: PL3 (−20 kN)(2 m)3 vB = − =− = 0.0028070 m 3EI 3(1.9  104 kN-m2 ) PL2 (20 kN)(2 m) 2 B = = = 0.0021053 rad 2 EI 2(1.9  10 4 kN-m 2 ) vC = 0.0028070 m + (3 m)(0.0021053 rad) = 0.0091228 m

Consider 50-kN concentrated load at C. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equations from Appendix C: PL3 vC = − 3EI Values: P = 50 kN, L = 5 m, EI = 1.9 × 104 kN-m2 Computation: PL3 (50 kN)(5 m)3 vC = − =− = −0.1096491 m 3EI 3(1.9  104 kN-m2 ) Beam deflection at C vC = −0.0157895 m + 0.1027961 m − 0.0078948 m + 0.0091228 m − 0.1096491 m = −0.0214145 m = 21.4 mm 

Ans.

P10.53 The simply supported beam shown in Figure P10.53 consists of a W10 × 30 structural steel wide-flange shape [E = 29,000 ksi; I = 170 in.4]. If w = 5 kips/ft, determine: (a) the beam deflection at point A. (b) the beam deflection at point C. FIGURE P10.53

Solution (a) Beam deflection at point A Consider cantilever beam deflection of 85 kip-ft concentrated moment. [Appendix C, Cantilever beam with concentrated moment at one end.] Relevant equation from Appendix C: ML2 vA = − 2 EI Values: M = 85 kip-ft, L = 3 ft, EI = 4.93 × 106 kip-in.2 Computation: ML2 (85 kip-ft)(3 ft)2 (12 in./ft)3 vA = − =− = −0.134069 in. 2 EI 2(4.93  106 kip-in.2 ) Consider rotation at B caused by 85 kip-ft concentrated moment. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: ML B = (slope magnitude) 3EI Values: M = 85 kip-ft, L = 15 ft, EI = 4.93 × 106 kip-in.2 Computation: ML (85 kip-ft)(15 ft)(12 in./ft) 2 B = = = 0.0124138 rad 3EI 3(4.93  106 kip-in.2 ) v A = −(3 ft)(12 in./ft)(0.0124138 rad) = −0.446897 in.

Consider cantilever beam deflection of 5 kips/ft uniformly distributed load. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL4 vA = − 8 EI Values: w = 5 kips/ft, L = 3 ft, EI = 4.93 × 106 kip-in.2

Computation: wL4 (5 kips/ft)(3 ft) 4 (12 in./ft)3 vA = − =− = −0.017744 in. 8EI 8(4.93  106 kip-in.2 ) Consider rotation at B caused by 5 kips/ft uniformly distributed load. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: ML B = (slope magnitude) 3EI Values: M = (5 kips/ft)(3 ft)(1.5 ft) = 22.5 kip-ft, L = 15 ft, EI = 4.93 × 106 kip-in.2 Computation: ML (22.5 kip-ft)(15 ft)(12 in./ft) 2 B = = = 0.0032860 rad 3EI 3(4.93  106 kip-in.2 ) v A = −(3 ft)(12 in./ft)(0.0032860 rad) = −0.118296 in.

Consider 5 kips/ft uniformly distributed load on segment BC. [Appendix C, SS beam with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C: wa 2 B = (2 L − a)2 (slope magnitude) 24 LEI Values: w = 5 kips/ft, L = 15 ft, a = 5 ft, EI = 4.93 × 106 kip-in.2 Computation: wa 2 (5 kips/ft)(5 ft) 2 (12 in./ft) 2 2 2 B = (2 L − a) = 2(15 ft) − (5 ft)  = 0.0063387 rad 6 2  24 LEI 24(15 ft)(4.93  10 kip-in. ) v A = (3 ft)(12 in./ft)(0.0063387 rad) = 0.228195 in.

Consider 25-kip concentrated load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pb( L2 − b2 ) (slope magnitude) B = 6 LEI Values: P = 25 kips, L = 15 ft, b = 5 ft, EI = 4.93 × 106 kip-in.2 Computation: Pb( L2 − b 2 ) (25 kips)(5 ft)(12 in./ft) 2 (15 ft) 2 − (5 ft) 2  = 0.0081136 rad B = = 6 LEI 6(15 ft)(4.93  106 kip-in.2 )  v A = (3 ft)(12 in./ft)(0.0081136 rad) = 0.292089 in.

Beam deflection at A vA = −0.134069 in. − 0.446897 in. − 0.017744 in. − 0.118296 in. + 0.228195 in. + 0.292089 in. = −0.196722 in. = 0.1967 in. 

Ans.

(b) Beam deflection at point C Consider 85 kip-ft concentrated moment. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: Mx vC = − (2L2 − 3Lx + x 2 ) (elastic curve) 6LEI Values: M = −85 kip-ft, L = 15 ft, x = 5 ft, EI = 4.93 × 106 kip-in.2 Computation: Mx vC = − (2 L2 − 3Lx + x 2 ) 6 LEI (−85 kip-ft)(5 ft)(12 in./ft)3  2(15 ft) 2 − 3(15 ft)(5 ft) + (5 ft) 2  = 0.413793 in. =− 6 2  6(15 ft)(4.93  10 kip-in. )

Consider moment at B caused by 5 kips/ft uniformly distributed load on overhang AB. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: Mx vC = − (2L2 − 3Lx + x 2 ) (elastic curve) 6LEI Values: M = −(5 kips/ft)(3 ft)(1.5 ft) = −22.5 kip-ft, L = 15 ft, x = 5 ft, EI = 4.93 × 106 kip-in.2 Computation: Mx vC = − (2 L2 − 3Lx + x 2 ) 6 LEI =−

(−22.5 kip-ft)(5 ft)(12 in./ft)3  2(15 ft) 2 − 3(15 ft)(5 ft) + (5 ft) 2  = 0.109533 in. 6 2 6(15 ft)(4.93  10 kip-in. )

Computation: wa3 vC = − (4 L2 − 7aL + 3a 2 ) 24 LEI =−

(5 kips/ft)(5 ft)3 (12 in./ft)3  4(15 ft) 2 − 7(5 ft)(15 ft) + 3(5 ft) 2  = −0.273834 in. 6 2  24(15 ft)(4.93  10 kip-in. )

Consider 25-kip concentrated load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pbx 2 vC = − ( L − b2 − x2 ) (elastic curve) 6LEI Values: P = 25 kips, L = 15 ft, b = 5 ft, x = 5 ft, EI = 4.93 × 106 kip-in.2 Computation: Pbx 2 vC = − ( L − b2 − x2 ) 6 LEI (25 kips)(5 ft)(5 ft)(12 in./ft)3 (15 ft) 2 − (5 ft) 2 − (5 ft) 2  = −0.425963 in. =− 6 2 6(15 ft)(4.93  10 kip-in. )

Beam deflection at C vC = 0.413793 in. + 0.109533 in. − 0.273834 in. − 0.425963 in. = −0.176471 in. = 0.1765 in. 

Ans.

P10.54 The simply supported beam shown in Figure P10.54 consists of a W10 × 30 structural steel wide-flange shape [E = 29,000 ksi; I = 170 in.4]. If w = 9 kips/ft, determine: (a) the beam deflection at point A. (b) the beam deflection at point D. FIGURE P10.54

Consider cantilever beam deflection of 9 kips/ft uniformly distributed load. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL4 vA = − 8 EI Values: w = 9 kips/ft, L = 3 ft, EI = 4.93 × 106 kip-in.2

Computation: wL4 (9 kips/ft)(3 ft)4 (12 in./ft)3 vA = − =− = −0.031939 in. 8EI 8(4.93  106 kip-in.2 ) Consider rotation at B caused by 9 kips/ft uniformly distributed load. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: ML B = (slope magnitude) 3EI Values: M = (9 kips/ft)(3 ft)(1.5 ft) = 40.5 kip-ft, L = 15 ft, EI = 4.93 × 106 kip-in.2 Computation: ML (40.5 kip-ft)(15 ft)(12 in./ft) 2 B = = = 0.0059148 rad 3EI 3(4.93  106 kip-in.2 ) v A = − (3 ft)(12 in./ft)(0.0059148 rad) = −0.212933 in.

Consider 9 kips/ft uniformly distributed load on segment BC. [Appendix C, SS beam with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C: wa 2 B = (2 L − a)2 (slope magnitude) 24 LEI Values: w = 9 kips/ft, L = 15 ft, a = 5 ft, EI = 4.93 × 106 kip-in.2 Computation:

B =

wa 2 (9 kips/ft)(5 ft) 2 (12 in./ft) 2 2 (2 L − a ) 2 = 2(15 ft) − (5 ft)  = 0.0114097 rad 6 2  24 LEI 24(15 ft)(4.93  10 kip-in. )

v A = (3 ft)(12 in./ft)(0.0114097 rad) = 0.410748 in.

Consider 25-kip concentrated load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pb( L2 − b2 ) (slope magnitude) B = 6 LEI Values: P = 25 kips, L = 15 ft, b = 5 ft, EI = 4.93 × 106 kip-in.2 Computation: Pb( L2 − b 2 ) (25 kips)(5 ft)(12 in./ft) 2 (15 ft) 2 − (5 ft) 2  = 0.0081136 rad B = = 6 2  6 LEI 6(15 ft)(4.93  10 kip-in. ) v A = (3 ft)(12 in./ft)(0.0081136 rad) = 0.292089 in.

Beam deflection at A vA = −0.134069 in. − 0.446897 in. − 0.031939 in. − 0.212933 in. + 0.410748 in. + 0.292089 in. = −0.123001 in. = 0.1230 in. 

Ans.

(b) Beam deflection at point D Consider 85 kip-ft concentrated moment. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: Mx vD = − (2L2 − 3Lx + x2 ) (elastic curve) 6LEI Values: M = −85 kip-ft, L = 15 ft, x = 10 ft, EI = 4.93 × 106 kip-in.2 Computation: Mx vD = − (2 L2 − 3Lx + x 2 ) 6 LEI (−85 kip-ft)(10 ft)(12 in./ft)3  2(15 ft) 2 − 3(15 ft)(10 ft) + (10 ft) 2  = 0.331034 in. =− 6 2 6(15 ft)(4.93  10 kip-in. )

Consider moment at B caused by 9 kips/ft uniformly distributed load on overhang AB. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: Mx vD = − (2L2 − 3Lx + x2 ) (elastic curve) 6LEI Values: M = −(9 kips/ft)(3 ft)(1.5 ft) = −40.5 kip-ft, L = 15 ft, x = 10 ft, EI = 4.93 × 106 kip-in.2 Computation: Mx vD = − (2 L2 − 3Lx + x 2 ) 6 LEI =−

( −40.5 kip-ft)(10 ft)(12 in./ft)3  2(15 ft) 2 − 3(15 ft)(10 ft) + (10 ft) 2  = 0.157729 in. 6(15 ft)(4.93  106 kip-in.2 ) 

Consider 9 kips/ft uniformly distributed load on segment BC. [Appendix C, SS beam with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C: wa 2 vD = − (2 x3 − 6 Lx 2 + a 2 x + 4L2 x − a 2 L) 24 LEI Values: w = 9 kips/ft, L = 15 ft, a = 5 ft, x = 10 ft, EI = 4.93 × 106 kip-in.2

Computation: wa 2 vD = − (2 x 3 − 6 Lx 2 + a 2 x + 4 L2 x − a 2 L) 24 LEI =−

(9 kips/ft)(5 ft) 2 (12 in./ft)3  2(10 ft)3 − 6(15 ft)(10 ft) 2 + (5 ft) 2 (10 ft) + 4(15 ft) 2 (10 ft) − (5 ft) 2 (15 ft)  24(15 ft)(4.93  106 kip-in.2 ) 

= −0.410751 in. Consider 25-kip concentrated load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pab 2 vD = − ( L − a 2 − b2 ) 6LEI Values: P = 25 kips, L = 15 ft, a = 10 ft, b = 5 ft, EI = 4.93 × 106 kip-in.2

Computation: Pab 2 vD = − ( L − a 2 − b2 ) 6 LEI =−

(25 kips)(10 ft)(5 ft)(12 in./ft)3 (15 ft) 2 − (10 ft) 2 − (5 ft) 2  = −0.486815 in. 6 2 6(15 ft)(4.93  10 kip-in. )

Beam deflection at D vD = 0.331034 in. + 0.157729 in. − 0.410751 in. − 0.486815 in. = −0.408803 in. = 0.409 in. 

Ans.

P10.55 The simply supported beam shown in Figure P10.55 consists of a W21 × 44 structural steel wide-flange shape [E = 29,000 ksi; I = 843 in.4]. For the loading shown, determine: (a) the beam deflection at point A. (b) the beam deflection at point C. FIGURE P10.55

Solution (a) Beam deflection at point A Consider cantilever beam deflection of downward 4 kips/ft uniform load over AB. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL4 vA = − 8 EI Values: w = 4 kips/ft, L = 12 ft, EI = 2.4447 × 107 kip-in.2 Computation: wL4 (4 kips/ft)(12 ft)4 (12 in./ft)3 vA = − =− = −0.732847 in. 8EI 8(2.4447  107 kip-in.2 ) Consider cantilever beam deflection of upward 4 kips/ft uniform load over 6-ft segment. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL4 wL3 v=− and = (slope magnitude) 8 EI 6 EI Values: w = −4 kips/ft, L = 6 ft, EI = 2.4447 × 107 kip-in.2 Computation: wL4 (−4 kips/ft)(6 ft) 4 (12 in./ft)3 v=− =− = 0.045803 in. 8 EI 8(2.4447  107 kip-in.2 )

=

wL3 (4 kips/ft)(6 ft)3 (12 in./ft) 2 = = 0.0008482 rad 6 EI 6(2.4447  107 kip-in.2 )

v A = 0.045803 in. + (6 ft)(12 in./ft)(0.0008482 rad) = 0.106873 in.

Consider rotation at B caused by downward 4 kips/ft uniform load. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: ML B = (slope magnitude) 3EI Values: M = (4 kips/ft)(6 ft)(9 ft) = 216 kip-ft, L = 24 ft, EI = 2.4447 × 107 kip-in.2 Computation: ML (216 kip-ft)(24 ft)(12 in./ft) 2 B = = = 0.0101784 rad 3EI 3(2.4447  107 kip-in.2 ) v A = −(12 ft)(12 in./ft)(0.0101784 rad) = −1.465693 in.

Consider 42-kip concentrated load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pb( L2 − b2 ) (slope magnitude) B = 6 LEI Values: P = 42 kips, L = 24 ft, b = 18 ft, EI = 2.4447 × 107 kip-in.2 Computation: Pb( L2 − b 2 ) (42 kips)(18 ft)(12 in./ft) 2 (24 ft) 2 − (18 ft) 2  = 0.0077929 rad B = = 7 2  6 LEI 6(24 ft)(2.4447  10 kip-in. ) v A = (12 ft)(12 in./ft)(0.0077929 rad) = 1.122172 in.

Consider 4 kips/ft uniformly distributed load on 6-ft segment near D. [Appendix C, SS beam with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C: wa 2 B = (2 L2 − a 2 ) (slope magnitude) 24 LEI Values: w = 4 kips/ft, L = 24 ft, a = 6 ft, EI = 2.4447 × 107 kip-in.2 Computation: wa 2 (4 kips/ft)(6 ft) 2 (12 in./ft) 2  2(24 ft) 2 − (6 ft) 2  = 0.0016434 rad B = (2 L2 − a 2 ) = 7 2  24 LEI 24(24 ft)(2.4447  10 kip-in. ) v A = (12 ft)(12 in./ft)(0.0016434 rad) = 0.236648 in.

Beam deflection at A vA = −0.732847 in. + 0.106873 in. − 1.465693 in. + 1.122172 in. + 0.236648 in. = −0.732847 in. = 0.733 in. 

Ans.

(b) Beam deflection at point C Consider moment at B caused by downward 4 kips/ft uniform load. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: Mx vC = − (2L2 − 3Lx + x 2 ) (elastic curve) 6LEI Values: M = −(4 kips/ft)(6 ft)(9 ft) = −216 kip-ft, L = 24 ft, x = 12 ft, EI = 2.4447 × 107 kip-in.2 Computation: Mx vC = − (2 L2 − 3Lx + x 2 ) 6 LEI =−

(−216 kip-ft)(12 ft)(12 in./ft) 3  2(24 ft) 2 − 3(24 ft)(12 ft) + (12 ft) 2  = 0.549635 in. 6(24 ft)(2.4447  107 kip-in.2 ) 

Consider 42-kip concentrated load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pbx 2 vC = − ( L − b2 − x2 ) (elastic curve) 6LEI Values: P = 42 kips, L = 24 ft, b = 6 ft, x = 12 ft, EI = 2.4447 × 107 kip-in.2 Computation: Pbx 2 vC = − ( L − b2 − x2 ) 6 LEI =−

(42 kips)(6 ft)(12 ft)(12 in./ft)3 (24 ft) 2 − (6 ft) 2 − (12 ft) 2  = −0.587804 in. 6(24 ft)(2.4447  107 kip-in.2 ) 

Computation: wa 2 vC = − (2 x3 − 6 Lx 2 + a 2 x + 4 L2 x − a 2 L) 24 LEI (4 kips/ft)(6 ft)2 (12 in./ft)3  2(12 ft)3 − 6(24 ft)(12 ft) 2 + (6 ft) 2 (12 ft)  =−   24(24 ft)(2.4447  107 kip-in.2 )  +4(24 ft) 2 (12 ft) − (6 ft) 2 (24 ft)  = −0.175578 in.

Beam deflection at C vC = 0.549635 in. − 0.587804 in. − 0.175578 in. = −0.213747 in. = 0.214 in. 

Ans.

P10.56 The simply supported beam shown in Figure P10.56 consists of a W530 × 66 structural steel wide-flange shape [E = 200 GPa; I = 351 × 106 mm4]. If w = 85 kN/m, determine the beam deflection at point B.

FIGURE P10.56

Solution Beam deflection at point B Consider 300 kN-m concentrated moment. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: Mx vB = − (2L2 − 3Lx + x 2 ) (elastic curve) 6LEI Values: M = −300 kN-m, L = 9 m, x = 4 m, EI = 7.02 × 104 kN-m2 Computation: Mx vB = − (2 L2 − 3Lx + x 2 ) 6 LEI (−300 kN-m)(4 m)  2(9 m) 2 − 3(9 m)(4 m) + (4 m) 2  = 0.022159 m =− 4 2  6(9 m)(7.02  10 kN-m ) Consider 85 kN/m uniformly distributed load on segment AB. [Appendix C, SS beam with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C: wa3 vB = − (4 L2 − 7aL + 3a 2 ) 24 LEI Values: w = 85 kN/m, L = 9 m, a = 4 m, EI = 7.02 × 104 kN-m2 Computation: wa3 vB = − (4 L2 − 7aL + 3a 2 ) 24 LEI =−

(85 kN/m)(4 m)3  4(9 m)2 − 7(4 m)(9 m) + 3(4 m) 2  = −0.043052 m 4 2  24(9 m)(7.02  10 kN-m )

Consider 140-kN concentrated load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pbx 2 vB = − ( L − b2 − x2 ) (elastic curve) 6LEI Values: P = 140 kN, L = 9 m, b = 3 m, x = 4 m, EI = 7.02 × 104 kN-m2 Computation: Pbx 2 vB = − ( L − b2 − x2 ) 6 LEI (140 kN)(3 m)(4 m) (9 m) 2 − (3 m) 2 − (4 m) 2  = −0.024818 m =− 4 2  6(9 m)(7.02  10 kN-m ) Consider 175 kN-m concentrated moment. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: Mx vB = − (2L2 − 3Lx + x 2 ) (elastic curve) 6LEI Values: M = −175 kN-m, L = 9 m, x = 5 m, EI = 7.02 × 104 kN-m2 Computation: Mx vB = − (2 L2 − 3Lx + x 2 ) 6 LEI (−175 kN-m)(5 m)  2(9 m) 2 − 3(9 m)(5 m) + (5 m) 2  = 0.012003 m =− 4 2  6(9 m)(7.02  10 kN-m ) Beam deflection at B vB = 0.022159 m − 0.043052 m − 0.024818 m + 0.012003 m = −0.033708 m = 33.7 mm 

Ans.

P10.57 The simply supported beam shown in Figure P10.57 consists of a W530 × 66 structural steel wide-flange shape [E = 200 GPa; I = 351 × 106 mm4]. If w = 115 kN/m, determine the beam deflection at point C.

FIGURE P10.57

Solution (b) Beam deflection at point C Consider 300 kN-m concentrated moment. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: Mx vC = − (2L2 − 3Lx + x 2 ) (elastic curve) 6LEI Values: M = −300 kN-m, L = 9 m, x = 6 m, EI = 7.02 × 104 kN-m2 Computation: Mx vC = − (2 L2 − 3Lx + x 2 ) 6 LEI (−300 kN-m)(6 m)  2(9 m) 2 − 3(9 m)(6 m) + (6 m) 2  = 0.017094 m =− 4 2  6(9 m)(7.02  10 kN-m ) Consider 115 kN/m uniformly distributed load on segment AB. [Appendix C, SS beam with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C: wa 2 vC = − (2 x3 − 6 Lx 2 + a 2 x + 4 L2 x − a 2 L) 24 LEI Values: w = 115 kN/m, L = 9 m, a = 4 m, x = 6 m, EI = 7.02 × 104 kN-m2 Computation: wa 2 vC = − (2 x 3 − 6 Lx 2 + a 2 x + 4 L2 x − a 2 L) 24 LEI (115 kN/m)(4 m) 2  2(6 m)3 − 6(9 m)(6 m) 2 + (4 m) 2 (6 m) + 4(9 m) 2 (6 m) − (4 m) 2 (9 m)  =− 4 2  24(9 m)(7.02  10 kN-m ) = −0.046597 m

Consider 140-kN concentrated load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pab 2 vC = − ( L − a 2 − b2 ) 6LEI Values: P = 140 kN, L = 9 m, a = 6 m, b = 3 m, EI = 7.02 × 104 kN-m2 Computation: Pab 2 vC = − ( L − a 2 − b2 ) 6 LEI (140 kN)(6 m)(3 m) (9 m) 2 − (6 m) 2 − (3 m) 2  = −0.023932 m =− 4 2  6(9 m)(7.02  10 kN-m ) Consider 175 kN-m concentrated moment. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: Mx vC = − (2L2 − 3Lx + x 2 ) (elastic curve) 6LEI Values: M = −175 kN-m, L = 9 m, x = 3 m, EI = 7.02 × 104 kN-m2 Computation: Mx vC = − (2 L2 − 3Lx + x 2 ) 6 LEI (−175 kN-m)(3 m)  2(9 m) 2 − 3(9 m)(3 m) + (3 m) 2  = 0.012464 m =− 4 2  6(9 m)(7.02  10 kN-m ) Beam deflection at C vC = 0.017094 m − 0.046597 m − 0.023932 m + 0.012464 m = −0.040971 m = 41.0 mm 

Ans.

P10.58 A 25-ft-long soldier beam is used as a key component of an earth retention system at an excavation site. The soldier beam is subjected to a soil loading that is linearly distributed from 520 lb/ft to 260 lb/ft, as shown in Figure P10.58. The soldier beam can be idealized as a cantilever with a fixed support at A. Added support is supplied by a tieback anchor at B, which exerts a force of 5,000 lb on the soldier beam. Determine the horizontal deflection of the soldier beam at point C. Assume EI = 5 × 108 lb-in.2. FIGURE P10.58

Solution Consider 260 lb/ft uniformly distributed load. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL4 vC = − 8EI Values: w = 260 lb/ft, L = 25 ft, EI = 5.0 × 108 lb-in.2 Computation: wL4 (260 lb/ft)(25 ft)4 (12 in./ft)3 vC = − =− = −43.875 in. 8EI 8(5.0  108 lb-in.2 )

Consider a linearly distributed load that varies from 260 lb/ft at A to 0 lb/ft at C. [Appendix C, Cantilever beam with linearly distributed load.] Relevant equation from Appendix C: w L4 vC = − 0 30 EI Values: w0 = 260 lb/ft, L = 25 ft, EI = 5.0 × 108 lb-in.2 Computation: w L4 (260 lb/ft)(25 ft)4 (12 in./ft)3 vC = − 0 = − = −11.700 in. 30 EI 30(5.0  108 lb-in.2 )

Consider 5,000-lb concentrated load. [Appendix C, Cantilever beam with concentrated load.] Relevant equations from Appendix C: PL3 PL2 vB = and B = (slope magnitude) 3EI 2 EI Values: P = 5,000 lb, L = 18 ft, EI = 5.0 × 108 lb-in.2 Computation: PL3 (5,000 lb)(18 ft)3 (12 in./ft)3 vB = = = 33.592320 in. 3EI 3(5.0  108 lb-in.2 ) PL2 (5,000 lb)(18 ft) 2 (12 in./ft) 2 B = = = 0.2332800 rad 2 EI 2(5.0  108 lb-in.2 ) vC = 33.592320 in. + (7 ft)(12 in./ft)(0.2332800 rad) = 53.187840 in.

Beam deflection at C

vC = −43.875 in. − 11.700 in. + 53.187840 in. = −2.387160 in. = 2.39 in. →

Ans.

P10.59 A 25-ft-long soldier beam is used as a key component of an earth retention system at an excavation site. The soldier beam is subjected to a uniformly distributed soil loading of 260 lb/ft, as shown in Figure P10.59. The soldier beam can be idealized as a cantilever with a fixed support at A. Added support is supplied by a tieback anchor at B, which exerts a force of 4,000 lb on the soldier beam. Determine the horizontal deflection of the soldier beam at point C. Assume EI = 5 × 108 lb-in.2. FIGURE P10.59

Consider 4,000-lb concentrated load. [Appendix C, Cantilever beam with concentrated load.] Relevant equations from Appendix C: PL3 PL2 vB = and B = (slope magnitude) 3EI 2 EI Values: P = 4,000 lb, L = 18 ft, EI = 5.0 × 108 lb-in.2 Computation: PL3 (4,000 lb)(18 ft)3 (12 in./ft)3 vB = = = 26.873856 in. 3EI 3(5.0  108 lb-in.2 )

B =

PL2 (4,000 lb)(18 ft) 2 (12 in./ft) 2 = = 0.1866240 rad 2 EI 2(5.0  108 lb-in.2 )

vC = 26.873856 in. + (7 ft)(12 in./ft)(0.1866240 rad) = 42.550272 in.

Beam deflection at C

vC = −43.875 in. + 42.550272 in. = −1.324728 in. = 1.325 in. →

Ans.

P11.1 A beam is loaded and supported as shown in Figure P11.1. Use the doubleintegration method to determine the magnitude of the moment M0 required to make the slope at the left end of the beam zero. FIGURE P11.1

Solution Moment equation:

 x M a − a = M ( x) + wx   − M 0 = 0 2 wx 2  M ( x) = M 0 − 2 Integration: d 2v wx 2 EI 2 = M ( x) = M 0 − dx 2 3 dv wx EI = M0x − + C1 dx 6 M x 2 wx 4 EI v = 0 − + C1 x + C2 2 24 Boundary conditions and evaluate constants: dv w( L)3 at x = L, = 0 M 0 ( L) − + C1 = 0 dx 6

 C1 =

wL3 − M 0L 6

Beam slope equation: dv wx 3 wL3 EI = M0x − + − M 0L dx 6 6 Constraint: At x = 0, the slope of the beam is to be zero; therefore, dv w(0)3 wL3 EI = M 0 (0) − + − M 0L = 0 dx A 6 6  M0 =

wL2 6

Ans.

P11.2 When the load P is applied to the right end of the cantilever beam shown in Figure P11.3, the deflection at the right end of the beam is zero. Use the double-integration method to determine the magnitude of the load P.

FIGURE P11.2

Solution Moment equation:

L−x M a − a = −w( L − x)   + P( L − x ) − M ( x) = 0  2  w  M ( x) = − ( L − x) 2 + P( L − x) 2

Integration: d 2v w EI 2 = M ( x) = − ( L − x) 2 + P ( L − x) dx 2 dv w P EI = ( L − x)3 − ( L − x) 2 + C1 dx 6 2 w P EI v = − ( L − x) 4 + ( L − x)3 + C1 x + C2 24 6 Boundary conditions and evaluate constants: dv w P at x = 0, = 0 ( L − 0)3 − ( L − 0) 2 + C1 = 0 dx 6 2 wL3 PL2  C1 = − + 6 2 w P at x = 0, v = 0 − ( L − 0) 4 + ( L − 0)3 + C1 (0) + C2 = 0 24 6

 C2 =

wL4 PL3 − 24 6

Beam elastic curve equation: w P wLx3 PL2 x wL4 PL3 4 3 EI v = − ( L − x) + ( L − x) − + + − 24 6 6 2 24 6 3 4 2 w wLx wL P PL x PL3 = − ( L − x) 4 − + + ( L − x )3 + − 24 6 24 6 2 6

Constraint: At x = L, the deflection of the beam is to be zero; therefore, w wL( L)3 wL4 P PL2 ( L) PL3 4 3 EI vB = − ( L − L) − + + ( L − L) + − =0 24 6 24 6 2 6 which simplifies to wL4 wL4 PL3 PL3 wL4 PL3 EI vB = − + + − =− + =0 6 24 2 6 8 3 Therefore, the magnitude of P is 3wL P= 8

Ans.

P11.3 A beam is loaded and supported as shown in Figure P11.3. (a) Use the double-integration method to determine the reactions at supports A and B. (b) Draw the shear-force and bendingmoment diagrams for the beam. FIGURE P11.3

Solution Beam FBD: Fy = Ay + By − wL = 0  L M A = M B + By L − wL   = 0  2 Moment equation: wx 2  x M a − a = M ( x) + wx   − Ay x = 0  M ( x) = − + Ay x  2 2 Integration: d 2v wx 2 EI 2 = M ( x) = − + Ay x dx 2 2 dv wx3 Ay x EI =− + + C1 dx 6 2 3 wx4 Ay x EI v = − + + C1x + C2 24 6 Boundary conditions and evaluate constants: 3 w(0)4 Ay (0) at x = 0, v = 0 − + + C1 (0) + C2 = 0 24 6 2 dv w( L)3 Ay ( L) at x = L, = 0 − + + C1 = 0 dx 6 2

 C2 = 0 2

wL3 Ay L  C1 = − 6 2

w( L) 4 Ay ( L) + + C1 ( L) = 0 24 6 Solve Eqs. (a) and (b) simultaneously to find: at x = L, v = 0

−

wL3 3wL 3wL C1 = − and Ay = =  48 8 8 Backsubstitute into equilibrium equations: 3wL 5wL Fy = Ay + By − wL = 0 By = wL − Ay = wL − = 8 8

 L M A = M B + By L − wL   = 0  2 MB = −

wL2 wL2 = (cw) 8 8

MB =

(a)

 C1 =

wL3 Ay L − 24 6

(b)

Ans.

 By =

5wL  8

Ans.

wL2 wL2 5wL2 wL2 − By L = − =− 2 2 8 8

Ans.

P11.4 A beam is loaded and supported as shown in Figure P11.4. Use the doubleintegration method to determine the reactions at supports A and B.

FIGURE P11.4

Solution Beam FBD: w0 L =0 2 w L  2L  M A = M B + By L − 0   = 0 2  3

Fy = Ay + By −

Moment equation: w0 x 2  x  M a − a = M ( x ) +   − Ay x = 0 2L  3  w0 x3  M ( x) = − + Ay x 6L

Integration: d 2v w x3 EI 2 = M ( x) = − 0 + Ay x dx 6L 4 Ay x2 dv w0 x EI =− + + C1 dx 24L 2 3 w x5 Ay x EI v = − 0 + + C1x + C2 120L 6 Boundary conditions and evaluate constants: 3 w0 (0)5 Ay (0) at x = 0, v = 0 − + + C1 (0) + C2 = 0 120L 6 2 dv w0 ( L) 4 Ay ( L) at x = L, = 0 − + + C1 = 0 dx 24 L 2 3

at x = L, v = 0

w0 ( L)5 Ay ( L) − + + C1 ( L) = 0 120 L 6

 C2 = 0 2

w L3 Ay L  C1 = 0 − 24 2

(a)

w0 L3 Ay L  C1 = − 120 6

(b)

Solve Eqs. (a) and (b) simultaneously to find: C1 = −

w0 L3 120

and

Ay =

w0 L w0 L =  10 10

Backsubstitute into equilibrium equations: wL wL w L w L 4w L Fy = Ay + By − 0 = 0 By = 0 − Ay = 0 − 0 = 0 2 2 2 10 10 M A = M B + By L −

w0 L  2 L    =0 2  3

w0 L2 w0 L2 MB = − = (cw) 15 15

MB =

Ans.

 By =

2w0 L  5

Ans.

w0 L2 w L2 2w L2 w L2 − By L = 0 − 0 = − 0 3 3 5 15

Ans.

P11.5 A beam is loaded and supported as shown in Figure P11.5. Use the fourth-order integration method to determine the reaction at roller support A.

FIGURE P11.5

Solution Integrate the load distribution: d 4v w0 x 2 EI 4 = − 2 dx L 3 d v w x3 EI 3 = − 0 2 + C1 dx 3L 2 d v w x4 EI 2 = − 0 2 + C1 x + C2 dx 12 L dv w x5 C x 2 EI = − 0 2 + 1 + C2 x + C3 dx 60 L 2 6 w0 x C1 x 3 C2 x 2 EI v = − + + + C3 x + C4 360 L2 6 2 Boundary conditions and evaluate constants: w (0)6 C (0)3 C2 (0) 2 at x = 0, v = 0 − 0 2 + 1 + + C3 (0) + C4 = 0 360 L 6 2 d 2v w0 (0) 4 at x = 0, M = EI 2 = 0 − + C1 (0) + C2 = 0 dx 12 L2

dv at x = L, =0 dx

w0 ( L)5 C1 ( L) 2 − + + C3 = 0 60 L2 2

at x = L, v = 0

−

 C4 = 0  C2 = 0

w0 L3  C1L + 2C3 = 30

(a)

w0 L3 60

(b)

w0 ( L)6 C1 ( L)3 + + C3 ( L) = 0 360 L2 6

 C1L2 + 6C3 =

Solve Eqs. (a) and (b) simultaneously to obtain:

w0 L3 w0 L3 w0 L3 −4C3 = − = 30 60 60 C1L2 =

w0 L3 w0 L3 5w0 L3 + = 30 120 120

w0 L3  C3 = − 240  C1 =

Roller reaction at A: d 3v w (0)3 w L w L VA = EI 3 =− 0 2 + 0 = 0 dx x = 0 3L 24 24

5w0 L w0 L = 120 24

 Ay =

w0 L  24

Ans.

P11.6 A beam is loaded and supported as shown in Figure P11.6. Use the fourthorder integration method to determine the reactions at supports A and B.

FIGURE P11.6

Solution Integrate the load distribution: d 4v x EI 4 = w0 sin dx L 3 d v wL x EI 3 = − 0 cos + C1 dx  L d 2v w0 L2 x EI 2 = − 2 sin + C1 x + C2 dx  L dv w0 L3  x C1 x 2 EI = 3 cos + + C2 x + C3 dx  L 2 w L4  x C1 x 3 C2 x 2 EI v = − 0 4 sin + + + C3 x + C4  L 6 2 Boundary conditions and evaluate constants: dv w0 L3  (0) C1 (0) 2 w0 L3 at x = 0, =0 cos + + C (0) + C = 0  C = − 2 3 3 dx 3 L 2 3 w0 L4  (0) C1 (0)3 C2 (0)2 at x = 0, v = 0 − 4 sin + + + C3 (0) + C4 = 0  C4 = 0  L 6 2

at x = L,

w0 L3

dv =0 dx

cos 3



at x = L, v = 0

−

w0 L4



 ( L)

sin 4

 ( L) L

C1 ( L) 2 w L3 + C2 ( L) − 0 3 = 0 2 

 C1L + 2C2 =

4w0 L2

(a)

C1 ( L)3 C2 ( L)2 w0 L3 + − 3 ( L) = 0 6 2 

 C1L + 3C2 =

6w0 L2

(b)

3

Solve Eqs. (a) and (b) simultaneously to obtain:

C2 =

6w0 L2

C1L =

3

−

4w0 L2

3

4w0 L2

 C2 =

3

 2w L2  − 2  03    

2w0 L2

3

 C1 = 0

Reactions at supports A and B d 3v wL  (0) wL VA = EI 3 = − 0 cos =− 0 dx x = 0  L   Ay =

w0 L





Ans.

VB = EI

d 3v wL  ( L) w0 L = − 0 cos = 3 dx x = L  L   By =

M A = EI

w0 L





Ans.

d 2v w0 L2  (0) 2 w0 L2 2w0 L2 = − sin + = dx 2 x = 0 2 L 3 3 MA =

2 w0 L2

3

(cw)

Ans.

d 2v w0 L2  ( L) 2w0 L2 2w0 L2 M B = EI 2 = − 2 sin + = dx x = L  L 3 3 MB =

2 w0 L2

3

(ccw)

Ans.

P11.7 A beam is loaded and supported as shown in Figure P11.7. (a) Use the double-integration method to determine the reactions at supports A and C. (b) Draw the shear-force and bendingmoment diagrams for the beam. (c) Determine the deflection in the middle of the span. FIGURE P11.7

Solution Beam FBD: from symmetry, P Ay = C y = 2 and M A = MC Moment equation: M a − a = M ( x) − M A −

P x=0 2

 M ( x) =

Px + MA 2

Integration: d 2v Px EI 2 = M ( x) = + MA dx 2 dv Px 2 EI = + M A x + C1 dx 4 Px 3 M A x 2 EI v = + + C1 x + C2 12 2 Boundary conditions and evaluate constants: dv P (0) 2 at x = 0, = 0 + M A (0) + C1 = 0 dx 4 P (0)3 M A (0) 2 at x = 0, v = 0 + + C2 = 0 12 2 (a) Beam reaction forces: P Ay = C y = 2

 C1 = 0  C2 = 0

Ans.

(a) Beam reaction moments: L dv at x = , = 0 2 dx

MA = −

P( L / 2) 2  L + MA   = 0  2 4

PL PL = (ccw) 8 8

MC =

PL (cw) 8

Ans.

Elastic curve equation: Px3 M A x 2 Px3 PLx 2 Px 2 EI v = + = − =−  3L − 4 x  12 2 12 16 48 Px 2 v = −  3L − 4 x  48EI (c) Midspan deflection: vB = −

P( L / 2) 2 PL3 3L − 4( L / 2) = −  48EI 192 EI

Ans.

P11.8 A beam is loaded and supported as shown in Figure P11.8. (a) Use the double-integration method to determine the reactions at supports A and B. (b) Draw the shear-force and bendingmoment diagrams for the beam. (c) Determine the deflection in the middle of the span. FIGURE P11.8

Solution Beam FBD: from symmetry, wL Ay = B y = 2 and MA = MB Moment equation:

 x   wL  M a − a = M ( x) − M A + wx   −  x = 0 2  2  wx 2 wLx  M ( x) = − + + MA 2 2

Integration: d 2v wx 2 wLx EI 2 = M ( x) = − + + MA dx 2 2 dv wx 3 wLx 2 EI =− + + M A x + C1 dx 6 4 wx 4 wLx 3 M A x 2 EI v = − + + + C1 x + C2 24 12 2 Boundary conditions and evaluate constants: dv w(0)3 wL(0) 2 at x = 0, = 0 − + + M A (0) + C1 = 0 dx 6 4 w(0) 4 wL(0)3 M A (0) 2 at x = 0, v = 0 − + + + C2 = 0 24 12 2 (a) Beam reaction forces: wL Ay = By = 2

 C1 = 0  C2 = 0

Ans.

(a) Beam reaction moments: L dv at x = , = 0 2 dx

MA = −

−

w( L / 2)3 wL( L / 2) 2  L + + MA  = 0  2 6 4

wL2 wL2 = (ccw) 12 12

MB = −

wL2 wL2 = (cw) 12 12

Ans.

Elastic curve equation: wx 4 wLx3 M A x 2 wx 4 wLx 3 wL2 x 2 wx 2 2 wx 2 2   EI v = − + + =− + − =− x − 2 Lx + L  = − ( x − L) 2  24 12 2 24 12 24 24 24 2 wx v = − ( x − L) 2 24 EI (c) Midspan deflection: 2

w( L / 2) 2  L   wL4 vx = L / 2 = −   − L)  = − 24 EI  2  384 EI 

Ans.

P11.9 A beam is loaded and supported as shown in Figure P11.9. (a) Use the double-integration method to determine the reactions at supports A and C. (b) Determine the deflection in the middle of the span. FIGURE P11.9

Solution Beam FBD: from symmetry, wL Ay = C y = 0 2 and M A = MC Moment equation: M a − a = M ( x ) − M A +  M ( x) = −

w0 x 2  x   w0 L   − x = 0 2L  3   2 

w0 x 3 w0 Lx + + MA 6L 2

Integration: d 2v w x 3 w Lx EI 2 = M ( x) = − 0 + 0 + M A dx 6L 2 4 2 dv wx w Lx EI =− 0 + 0 + M A x + C1 dx 24 L 4 w x 5 w Lx 3 M A x 2 EI v = − 0 + 0 + + C1 x + C2 120 L 12 2 Boundary conditions and evaluate constants: dv w (0) 4 w0 L(0) 2 at x = 0, = 0 − 0 + + M A (0) + C1 = 0 dx 24 L 4 w0 (0)5 w0 L(0)3 M A (0) 2 at x = 0, v = 0 − + + + C2 = 0 120 L 12 2 (a) Beam reaction forces: wL Ay = C y = 0 2

 C1 = 0  C2 = 0

Ans.

(a) Beam reaction moments: dv at x = L, = 0 dx

−

w0 ( L)4 w0 L( L) 2 + + M A ( L) = 0 24 L 4

5w0 L2 5w0 L2 MA = − = (ccw) 24 24

5w0 L2 5w0 L2 MC = − = (cw) 24 24

Ans.

Elastic curve equation: w x5 w Lx 3 M A x 2 w x 5 w Lx 3 5w0 L2 x 2 EI v = − 0 + 0 + =− 0 + 0 − 120 L 12 2 120 L 12 48 5 2 3 3 2 2 2w x 20 w0 L x 25w0 L x wx =− 0 + − = − 0  2 x3 − 20 L2 x + 25 L3  240 L 240 L 240 L 240 L 2 w0 x  2 x3 − 20 L2 x + 25L3  v = − 240 L EI (c) Midspan deflection: vB = −

w0 ( L) 2 7 w L4  2( L)3 − 20 L2 ( L) + 25 L3  = − 0 240 L EI 240 EI

Ans.

P11.10 A beam is loaded and supported as shown in Figure P11.10. (a) Use the double-integration method to determine the reactions at supports A and C. (b) Draw the shear-force and bendingmoment diagrams for the beam. FIGURE P11.10

Solution Beam FBD: Fy = Ay + C y = 0

 C y = − Ay

M A = − M A + C y L − M 0 = 0 Moment equation: M a − a = M ( x) − Ay x − M A = 0  M ( x) = Ay x + M A

L   0  x   2

M b − b = M ( x) − Ay x − M A − M 0 = 0  M ( x) = Ay x + M A + M 0

Integration: For beam segment AB: d 2v EI 2 = M ( x) = Ay x + M A dx 2 dv Ay x EI = + M A x + C1 dx 2 Ay x3 M A x 2 EI v = + + C1x + C2 6 2

L    x  L 2

For beam segment BC: d 2v EI 2 = M ( x) = Ay x + M A + M 0 dx 2 dv Ay x EI = + M A x + M 0 x + C3 dx 2 Ay x3 M A x2 M 0 x 2 EI v = + + + C3 x + C4 6 2 2

Boundary conditions and evaluate constants for segment AB: Ay (0)3 M A (0)2 at x = 0, v = 0 + + C1 (0) + C2 = 0 6 2 Ay (0)2 dv at x = 0, = 0 + M A (0) + C1 = 0 dx 2

 C2 = 0  C1 = 0

Slope continuity condition at x = L/2: L dv dv at x = , = 2 dx AB dx BC Ay x 2 2

+ M Ax =

Ay x 2 2

+ M A x + M 0 x + C3

 C3 = −

M 0L 2

Deflection continuity condition at x = L/2: L at x = , vB AB = vB BC 2 3 3 Ay x M A x 2 Ay x M x 2 M x 2 M Lx + = + A + 0 − 0 + C4 6 2 6 2 2 2  C4 =

M 0 L2 8

Boundary condition for segment BC: Ay ( L)3 M A ( L)2 M 0 ( L)2 M 0 L M L2 at x = L, v = 0 + + − ( L) + 0 = 0 6 2 2 2 8 Also, the beam moment equilibrium equation can be written as: Ay L + M A = −M 0 (a) Beam Reactions: Solve these two equations simultaneously to obtain: M M0 9M 0 9M 0 MA = 0 = (cw) Ay = − =  8 8 8L 8L

 Ay L + 3M A = −

Cy =

9M 0  8L

3M 0 4

Ans.

P11.11 A beam is loaded and supported as shown in Figure P11.11. (a) Use the double-integration method to determine the reactions at supports A and C. (b) Draw the shear-force and bending-moment diagrams for the beam. FIGURE P11.11

Solution Beam FBD: wL =0 2 wL  L  M A = M C + C y L −   =0 2  4 Fy = Ay + C y −

Moment equation:  x M a − a = M ( x) + wx   − Ay x = 0  2 wx 2  M ( x) = − + Ay x 2 M b − b = M ( x ) +  M ( x) = −

L   0  x   2

wL  L  x −  − Ay x = 0 2 4

wL  L  x −  + Ay x 2 4

Integration: For beam segment AB: d 2v wx 2 EI 2 = M ( x) = − + Ay x dx 2 2 dv wx3 Ay x EI =− + + C1 dx 6 2 3 wx4 Ay x EI v = − + + C1x + C2 24 6

L    x  L 2

For beam segment BC: d 2v wL  L EI 2 = M ( x) = −  x −  + Ay x dx 2 4 2

Ay x dv wL  L EI =− + C3  x −  + dx 4 4 2 3

Ay x wL  L EI v = − + C3 x + C4  x −  + 12 4 6

Boundary conditions and evaluate constants for segment AB: 3 w(0)4 Ay (0) at x = 0, v = 0 − + + C1 (0) + C2 = 0 24 6

 C2 = 0

Slope continuity condition at x = L/2: Equate the slope expressions for the two beam segments: 2 2 Ay x 2 wx3 Ay x wL  L − + + C1 = − + C3  x −  + 6 2 4  4 2 Set x = L/2 and solve for the constant C1: 2

wx 3 wL  L w( L / 2)3 wL  L L  wL3 wL3 C1 = C3 + − x − = C + − − = C + −     3 3 6 4  4 6 4  2 4 48 64 wL3  C1 = C3 + 192

Deflection continuity condition at x = L/2: Equate the deflection expressions for the two beam segments: 3 3 Ay x3 wx 4 Ay x wL  L − + + C1 x = − x − + + C3 x + C4   24 6 12  4 6 Set x = L/2 and solve for the constant C4: 3 w( L / 2) 4  wL3   L  wL  L L   L − + C3 +  −  + C3   + C4    = − 24 192  2 12 2 4 2 

−

4 wL4 wL4  L  wL  L + C3   + =− + C3   + C4  2  384  2 384 768

 C4 =

wL4 768

Boundary conditions and evaluate constants for segment BC: 2

Ay ( L) wL  L − + C3 = 0  L −  + 4 4 2

dv at x = L, = 0 dx

9wL3 Ay L  C3 = − 64 2

at x = L, v = 0 3 Ay ( L)3 wL  L wL4 − L − + + C ( L ) + =0   3 12  4 6 768 3 2 27 wL4 Ay L  9wL3 Ay L  wL4 − + + − =0  ( L) + 768 6 2  768  64 3

27 wL4 Ay L 9 wL4 Ay L wL4 − + + − + =0 768 6 64 2 768 Ay L3 3 Ay L3 26 wL4 108wL4 − = − 6 6 768 768 3 4 Ay L 82 wL 41wL =  Ay = 3 768 128

Solve for C3:

9wL3 41wL3 5wL3 C3 = − = − 64 256 256 and for C1: 5wL3 wL3 11wL3 C1 = − + = − 256 192 768 (a) Beam force reactions: 41wL Ay = 128 wL wL 41wL 23wL − Ay = − = 2 2 128 128 Beam moment reaction: Cy =

MC =

wL2 wL2 23wL2 7 wL2 − Cy L = − =− 8 8 128 128

Ans.  Cy =

23wL  128

 MC = −

Ans.

7 wL2 7 wL2 = (cw) 128 128

Ans.

P11.12 A propped cantilever beam is loaded as shown in Figure P11.12. Assume EI = 200,000 kN·m2. Use discontinuity functions to determine: (a) the reactions at A and C. (b) the beam deflection at B. FIGURE P11.12

Solution Moment equation: (a) Support reactions: Fy = Ay + C y − 150 kN = 0

 Ay + C y = 150 kN

(a)

M A = −(150 kN)(7 m) + C y (12 m) − M A = 0  M A − C y (12 m) = −1,050 kN-m

(b)

Discontinuity expressions:

w( x) = M A x − 0 m

−2

+ Ay x − 0 m

V ( x) =  w ( x ) dx = M A x − 0 m

−1

+ Ay x − 0 m − 150 kN x − 7 m + C y x − 12 m

−1

− 150 kN x − 7 m

+ C y x − 12 m

M ( x ) =  V ( x)dx = M A x − 0 m + Ay x − 0 m − 150 kN x − 7 m + C y x − 12 m 0

−1

d 2v 0 1 1 1 = M ( x) = M A x − 0 m + Ay x − 0 m − 150 kN x − 7 m + C y x − 12 m 2 dx Ay Cy dv 150 kN 1 2 2 2 EI = MA x − 0 m + x−0 m − x−7 m + x − 12 m + C1 dx 2 2 2 A C M 150 kN 2 3 3 3 y y EI v = A x − 0 m + x−0 m − x−7 m + x − 12 m + C1 x + C2 2 6 6 6 EI

Boundary conditions and evaluate constants: dv at x = 0 m, = 0  C1 = 0 dx at x = 0 m, v = 0  C2 = 0 Ay MA 150 kN (12 m) 2 + (12 m)3 − (5 m)3 2 6 6 2 3 3  M A (72 m ) + Ay (288 m ) = 3,125 kN-m

at x = 12 m, v = 0

(e)

(a) Solve for Ay, Cy, and MA: Solve equations (a), (b), and (c) simultaneously to obtain the results: Ay = 88.3247 kN = 88.3 kN  M A = −309.8958 kN-m = 310 kN-m (ccw)

C y = 61.6753 kN = 61.7 kN 

Ans.

(b) Beam deflection at B: From Eq. (d), the beam deflection at B (x = 7 m) is computed as follows: 309.8958 kN-m 88.3247 kN EI vB = − (7 m) 2 + (7 m)3 2 6 = −2,543.2219 kN-m3

 vB = −

2,543.2219 kN-m3 = −0.0127161 m = 12.72 mm  200, 000 kN-m 2

Ans.

P11.13 A propped cantilever beam is loaded as shown in Figure P11.13. Assume EI = 200,000 kN·m2. Use discontinuity functions to determine: (a) the reactions at A and B. (b) the beam deflection at C. FIGURE P11.13

Solution Moment equation: (a) Support reactions: Fy = Ay + By = 0

 Ay = − By

(a)

M A = −750 kN-m + By (5 m) − M A = 0  M A − By (5 m) = −750 kN-m

(b)

Discontinuity expressions:

w( x) = M A x − 0 m

−2

+ Ay x − 0 m

V ( x) =  w ( x ) dx = M A x − 0 m

−1

+ Ay x − 0 m + By x − 5 m + 750 kN-m x − 7.5 m

−1

+ By x − 5 m

+ 750 kN-m x − 7.5 m

M ( x ) =  V ( x)dx = M A x − 0 m + Ay x − 0 m + By x − 5 m + 750 kN-m x − 7.5 m 0

−2

−1

d 2v 0 1 1 0 = M ( x) = M A x − 0 m + Ay x − 0 m + By x − 5 m + 750 kN-m x − 7.5 m 2 dx Ay By dv 1 2 2 1 EI = MA x − 0 m + x−0 m + x − 5 m + 750 kN-m x − 7.5 m + C1 dx 2 2 A B M 750 kN-m 2 3 3 2 y y EI v = A x − 0 m + x−0 m + x−5 m + x − 7.5 m + C1 x + C2 2 6 6 2 EI

Boundary conditions and evaluate constants: dv at x = 0 m, = 0  C1 = 0 dx at x = 0 m, v = 0  C2 = 0 Ay MA (5 m) 2 + (5 m)3 2 6 2 3  M A (12.5 m ) + Ay (20.83333 m ) = 0

at x = 5 m, v = 0

(e)

(a) Solve for Ay, By, and MA: Solve equations (a), (b), and (c) simultaneously to obtain the results: Ay = −225.000 kN = 225 kN  M A = 375.000 kN-m = 375 kN-m (cw)

By = 225.000 kN = 225 kN 

Ans.

(b) Beam deflection at C: From Eq. (d), the beam deflection at C (x = 7.5 m) is computed as follows: Ay By M EI vC = A (7.5 m) 2 + (7.5 m)3 + (2.5 m)3 2 6 6 3 = −4,687.500 kN-m

4,687.500 kN-m3  vC = − = −0.023438 m = 23.4 mm  200,000 kN-m 2

Ans.

P11.14 A propped cantilever beam is loaded as shown in Figure P11.14. Assume EI = 100,000 kip·ft2. Use discontinuity functions to determine: (a) the reactions at A and E. (b) the beam deflection at C.

FIGURE P11.14

Solution Moment equation: (a) Support reactions: Fy = Ay + E y − 20 kips − 30 kips − 20 kips = 0

 Ay + E y = 70 kips

(a)

M E = − Ay (28 ft) + 20 kips(21 ft) +30 kips(14 ft) + 20 kips(7 ft) + M E = 0  − Ay (28 ft) + M E = −980 kip-ft

(b)

Discontinuity expressions: −1 −1 −1 −1 w ( x ) = Ay x − 0 ft − 20 kips x − 7 ft − 30 kips x − 14 ft − 20 kips x − 21 ft − M E x − 28 ft V ( x) =  w ( x ) dx = Ay x − 0 ft

−2

− 20 kips x − 7 ft

− M E x − 28 ft

−1

+ E y x − 28 ft

− 30 kips x − 14 ft

− 20 kips x − 21 ft

M ( x ) =  V ( x)dx = Ay x − 0 ft − 20 kips x − 7 ft − 30 kips x − 14 ft − 20 kips x − 21 ft 1

− M E x − 28 ft EI

+ E y x − 28 ft

d 2v 1 1 1 1 = M ( x) = Ay x − 0 ft − 20 kips x − 7 ft − 30 kips x − 14 ft − 20 kips x − 21 ft 2 dx − M E x − 28 ft

+ E y x − 28 ft

dv Ay 20 kips 30 kips 20 kips 2 2 2 2 = x − 0 ft − x − 7 ft − x − 14 ft − x − 21 ft dx 2 2 2 2 Ey 1 2 − M E x − 28 ft + x − 28 ft + C1 2 Ay 20 kips 30 kips 20 kips 3 3 3 3 EI v = x − 0 ft − x − 7 ft − x − 14 ft − x − 21 ft 6 6 6 6 Ey M 2 3 − E x − 28 ft + x − 28 ft + C1 x + C2 2 6 EI

(c)

(d)

Boundary conditions and evaluate constants: at x = 0 ft, v = 0  C2 = 0

at x = 28 ft, v = 0 

Ay 6

(28 ft)3 −

20 kips 30 kips 20 kips (21 ft)3 − (14 ft)3 − (7 ft)3 + C1 (28 ft) = 0 6 6 6

(e)

dv =0 dx Ay 20 kips 30 kips 20 kips  (28 ft)2 − (21 ft)2 − (14 ft) 2 − (7 ft) 2 + C1 = 0 2 2 2 2 at x = 28 ft,

(f)

(a) Solve for Ay, Ey, and ME: Solve equations (e) and (f) simultaneously to obtain: C1 = −1,470.000 kip-ft 2 Ay = 23.7500 kips = 23.8 kips 

Ans.

With the value of Ay, calculate Ey and ME from equations (a) and (b), respectively.

E y = 46.2500 kips = 46.3 kips  M E = −315.000 kip-ft = 315 kip-ft (cw)

Ans.

(b) Beam deflection at C: From Eq. (d), the beam deflection at C (x = 14 ft) is computed as follows: 23.75 kips 20 kips EI vC = (14 ft)3 − (7 ft)3 − (1,470 kip-ft 2 )(14 ft) 6 6 3 = −10,861.6667 kip-ft

 vC = −

10,861.6667 kip-ft 3 = −0.108617 ft = −1.3034 in. = 1.303 in.  100,000 kip-ft 2

Ans.

P11.15 A propped cantilever beam is loaded as shown in Figure P11.15. Assume EI = 100,000 kip·ft2. Use discontinuity functions to determine: (a) the reactions at A and B. (b) the beam deflection at x = 7 ft. FIGURE P11.15

Solution Moment equation: (a) Support reactions: 1 Fy = Ay + By − (12 kips/ft)(16 ft) = 0 2  Ay + By = 96 kips

(a)

1 2(16 ft) M B = − Ay (16 ft) + (12 kips/ft)(16 ft) 2 3 +M B = 0  − Ay (16 ft) + M B = −1,024 kip-ft

(b)

Discontinuity expressions: w ( x ) = Ay x − 0 ft

−1

− 12 kips/ft x − 0 ft

− M B x − 16 ft V ( x) =  w ( x ) dx = Ay x − 0 ft

−2

−1

+ By x − 16 ft

− 12 kips/ft x − 0 ft +

− M B x − 16 ft M ( x ) =  V ( x)dx = Ay x − 0 ft − 1

− M B x − 16 ft

−1

12 kips/ft 1 x − 0 ft 16 ft

+ By x − 16 ft

12 kips/ft 2 x − 0 ft 2(16 ft)

12 kips/ft 12 kips/ft 2 3 x − 0 ft + x − 0 ft 2 6(16 ft) 0

+ By x − 16 ft

d 2v 12 kips/ft 12 kips/ft 1 2 3 EI 2 = M ( x) = Ay x − 0 ft − x − 0 ft + x − 0 ft dx 2 6(16 ft) − M B x − 16 ft

+ By x − 16 ft

dv Ay 12 kips/ft 12 kips/ft 2 3 4 = x − 0 ft − x − 0 ft + x − 0 ft dx 2 6 24(16 ft) By 1 2 − M B x − 16 ft + x − 16 ft + C1 2 Ay 12 kips/ft 12 kips/ft 3 4 5 EI v = x − 0 ft − x − 0 ft + x − 0 ft 6 24 120(16 ft) By M 2 3 − B x − 16 ft + x − 16 ft + C1 x + C2 2 6 EI

(c)

(d)

Boundary conditions and evaluate constants: at x = 0 ft, v = 0  C2 = 0 at x = 16 ft, v = 0 Ay 12 kips/ft 12 kips/ft  (16 ft)3 − (16 ft) 4 + (16 ft)5 + C1 (16 ft) = 0 6 24 120(16 ft) dv at x = 16 ft, = 0 dx Ay 12 kips/ft 12 kips/ft  (16 ft) 2 − (16 ft)3 + (16 ft) 4 + C1 = 0 2 6 24(16 ft)

(e)

(f)

(a) Solve for Ay, By, and MB: Solve equations (e) and (f) simultaneously to obtain: C1 = −614.4000 kip-ft 2 Ay = 52.8000 kips = 52.8 kips 

Ans.

With the value of Ay, calculate By and MB from equations (a) and (b), respectively.

By = 43.200 kips = 43.2 kips  M B = −179.200 kip-ft = 179.2 kip-ft (cw)

Ans.

(b) Beam deflection at x = 7 ft: From Eq. (d), the beam deflection at x = 7 ft is computed as follows: 52.8 kips 12 kips/ft 12 kips/ft EI v = (7 ft)3 − (7 ft) 4 + (7 ft)5 − (614.400 kip-ft 2 )(7 ft) 6 24 120(16 ft)

= −2,377.85625 kip-ft 3 v = −

2,377.85625 kip-ft 3 = −0.023779 ft = −0.2853 in. = 0.285 in.  100,000 kip-ft 2

Ans.

P11.16 A propped cantilever beam is loaded as shown in Figure P11.16. Assume EI = 200,000 kN·m2. Use discontinuity functions to determine: (a) the reactions at A and B. (b) the beam deflection at C. FIGURE P11.16

Solution Moment equation: (a) Support reactions: 1 Fy = Ay + By − (120 kN/m)(8 m) = 0 2  Ay + By = 480 kN

(a)

1 2(8 m) M A = − (120 kN/m)(8 m) + By (6 m) 2 3 −M A = 0  By (6 m) − M A = 2,560 kN-m

(b)

Discontinuity expressions: 120 kN/m 1 −1 x − 0 m + By x − 6 m 8m 120 kN/m −1 0 2 0 V ( x) =  w ( x ) dx = M A x − 0 m + Ay x − 0 m − x − 0 m + By x − 6 m 2(8 m) 120 kN/m 0 1 3 1 M ( x ) =  V ( x)dx = M A x − 0 m + Ay x − 0 m − x − 0 m + By x − 6 m 6(8 m) w( x) = M A x − 0 m

−2

+ Ay x − 0 m

−1

−

d 2v 120 kN/m 0 1 3 1 = M ( x) = M A x − 0 m + Ay x − 0 m − x − 0 m + By x − 6 m 2 dx 6(8 m) Ay By dv 120 kN/m 1 2 4 2 EI = MA x − 0 m + x−0 m − x−0 m + x − 6 m + C1 dx 2 24(8 m) 2 Ay By M 120 kN/m 2 3 5 3 EI v = A x − 0 m + x−0 m − x−0 m + x − 6 m + C1x + C2 2 6 120(8 m) 6 EI

Boundary conditions and evaluate constants: dv at x = 0 m, = 0  C1 = 0 dx at x = 0 m, v = 0  C2 = 0

at x = 6 m, v = 0



Ay MA 120 kN/m (6 m)2 + (6 m)3 − (6 m)5 = 0 2 6 120(8 m)

(e)

(a) Solve for Ay, By, and MA: Solve equations (a), (b), and (e) simultaneously to obtain: Ay = 66.5000 kN = 66.5 kN 

By = 413.5000 kN = 414 kN 

M A = −79.0000 kN-m = 79.0 kN-m (ccw)

Ans.

(b) Beam deflection at C: From Eq. (d), the beam deflection at C (x = 8 m) is computed as follows: 79.0 kN-m 66.5 kN 120 kN/m 413.5 kN EI v = − (8 m)2 + (8 m)3 − (8 m)5 + (2 m)3 2 6 120(8 m) 6

= −398.0000 kN-m3 v = −

398.0000 kN-m3 = −0.001990 m = −1.990 mm = 1.990 mm  200, 000 kN-m 2

Ans.

P11.17 For the beam shown in Figure P11.17, assume EI = 200,000 kN·m2 and use discontinuity functions to determine: (a) the reactions at A, C, and D. (b) the beam deflection at B. FIGURE P11.17

Solution Moment equation: (a) Support reactions: Fy = Ay + C y + Dy − (120 kN/m)(6 m) = 0

 Ay + C y + Dy = 720 kN

(a)

M A = −(120 kN/m)(6 m)(3 m) + C y (6 m) + Dy (10 m) = 0  C y (6 m) + Dy (10 m) = 2,160 kN-m

(b)

Discontinuity expressions: −1 0 0 w ( x ) = Ay x − 0 m − 120 kN/m x − 0 m + 120 kN/m x − 6 m −1

+C y x − 6 m

−1

+ Dy x − 10 m

V ( x) =  w ( x ) dx = Ay x − 0 m − 120 kN/m x − 0 m + 120 kN/m x − 6 m 0

+ C y x − 6 m + Dy x − 10 m 0

M ( x ) =  V ( x)dx = Ay x − 0 m − 1

120 kN/m 120 kN/m 2 2 x−0 m + x−6 m 2 2

+ C y x − 6 m + Dy x − 10 m 1

d 2v 120 kN/m 120 kN/m 1 2 2 = M ( x) = Ay x − 0 m − x−0 m + x−6 m 2 dx 2 2 + C y x − 6 m + Dy x − 10 m 1

dv Ay 120 kN/m 120 kN/m 2 3 3 = x−0 m − x−0 m + x−6 m dx 2 6 6 Cy D 2 2 y + x−6 m + x − 10 m + C1 2 2 Ay 120 kN/m 120 kN/m 3 4 4 EI v = x−0 m − x−0 m + x−6 m 6 24 24 Cy Dy 3 3 + x−6 m + x − 10 m + C1 x + C2 6 6

(c)

(d)

Boundary conditions and evaluate constants: at x = 0 m, v = 0  C2 = 0 at x = 6 m, v = 0



Ay 6

(6 m)3 −

120 kN/m (6 m) 4 + C1 (6 m) = 0 24

(e)

at x = 10 m, v = 0 

Ay 6

(10 m)3 −

Cy 120 kN/m 120 kN/m (10 m)4 + (4 m) 4 + (4 m)3 + C1 (10 m) = 0 24 24 6

(f)

(a) Solve for Ay, Cy, and Dy: Solve equations (a), (b), (e), and (f) simultaneously to obtain: C1 = −756.000 kN-m 2 Ay = 306.0000 kN = 306 kN  C y = 495.0000 kN = 495 kN  Dy = −81.0000 kN = 81.0 kN 

Ans.

(b) Beam deflection at B: From Eq. (d), the beam deflection at B (x = 3 m) is computed as follows: 306.00 kN 120 kN/m EI vB = (3 m)3 − (3 m)4 − (756.000 kN-m 2 )(3 m) 6 24 3 = −1,296.0000 kN-m

1,296.0000 kN-m3  vB = − = −0.006480 m = −6.48 mm = 6.48 mm  200,000 kN-m2

Ans.

P11.18 For the beam shown in Figure P11.18, assume EI = 100,000 kip·ft2 and use discontinuity functions to determine: (a) the reactions at A, C, and D. (b) the beam deflection at B. FIGURE P11.18

Solution Moment equation: (a) Support reactions: Fy = Ay + C y + Dy − 75 kips − (7 kips/ft)(16 ft) = 0

 Ay + C y + Dy = 187 kips

(a)

M A = −(75 kips)(8 ft) − (7 kips/ft)(16 ft)(24 ft) +C y (16 ft) + Dy (32 ft) = 0  C y (16 ft) + Dy (32 ft) = 3,288 kip-ft

(b)

Discontinuity expressions: −1 −1 −1 w ( x ) = Ay x − 0 ft − 75 kips x − 8 ft + C y x − 16 ft −7 kips/ft x − 16 ft V ( x) =  w ( x ) dx = Ay x − 0 ft

+ 7 kips/ft x − 32 ft

− 75 kips x − 8 ft

+ Dy x − 32 ft

+ C y x − 16 ft

−7 kips/ft x − 16 ft + 7 kips/ft x − 32 ft + Dy x − 32 ft 1

M ( x ) =  V ( x)dx = Ay x − 0 ft − 75 kips x − 8 ft + C y x − 16 ft 1

−

−1

7 kips/ft 7 kips/ft 2 2 1 x − 16 ft + x − 32 ft + Dy x − 32 ft 2 2

d 2v 1 1 1 EI 2 = M ( x) = Ay x − 0 ft − 75 kips x − 8 ft + C y x − 16 ft dx 7 kips/ft 7 kips/ft 2 2 1 − x − 16 ft + x − 32 ft + Dy x − 32 ft 2 2 Cy dv Ay 75 kips 2 2 2 EI = x − 0 ft − x − 8 ft + x − 16 ft dx 2 2 2 Dy 7 kips/ft 7 kips/ft 3 3 2 − x − 16 ft + x − 32 ft + x − 32 ft + C1 6 6 2 Ay Cy 75 kips 3 3 3 EI v = x − 0 ft − x − 8 ft + x − 16 ft 6 6 6 Dy 7 kips/ft 7 kips/ft 4 4 3 − x − 16 ft + x − 32 ft + x − 32 ft + C1 x + C2 24 24 6

(c)

(d)

Boundary conditions and evaluate constants: at x = 0 ft, v = 0  C2 = 0 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

at x = 16 ft, v = 0



at x = 32 ft, v = 0 

Ay 6

(32 ft)3 −

(16 ft)3 −

75 kips (8 ft)3 + C1 (16 ft) = 0 6

Cy 75 kips 7 kips/ft (24 ft)3 + (16 ft)3 − (16 ft)4 + C1 (32 ft) = 0 6 6 24

(e)

(f)

(a) Solve for Ay, Cy, and Dy: Solve equations (a), (b), (e), and (f) simultaneously to obtain: C1 = −601.3333 kip-ft 2 Ay = 23.4688 kips = 23.5 kips  C y = 121.5625 kips = 121.6 kips  Dy = 41.9688 kips = 42.0 kips 

Ans.

(b) Beam deflection at B: From Eq. (d), the beam deflection at B (x = 8 ft) is computed as follows: 23.4688 kips EI vB = (8 ft)3 − (601.3333 kip-ft 2 )(8 ft) 6 = −2,808.0000 kip-ft 3

2,808.0000 kip-ft 3  vB = − = −0.028080 ft = −0.3370 in. = 0.337 in.  100,000 kip-ft 2

Ans.

P11.19 For the propped cantilever beam shown in Figure P11.19, assume EI = 100,000 kip·ft2 and use discontinuity functions to determine: (a) the reactions at B and D. (b) the beam deflection at C. FIGURE P11.19

Solution Moment equation: (a) Support reactions: 1 Fy = By + Dy − (5 kips/ft)(10 ft) 2 −(5 kips/ft)(12 ft) = 0  By + Dy = 85 kips

M D =

(a)

1 10 ft   (5 kips/ft)(10 ft)  20 ft +   2 3  + (5 kips/ft)(12 ft)(14 ft) − By (20 ft) + M D = 0

 By (20 ft) − M D = 1, 423.3333 kip-ft

(b)

Discontinuity expressions: 5 kips/ft 5 kips/ft 1 1 0 w( x) = − x − 0 ft + x − 10 ft + 5 kips/ft x − 10 ft 10 ft 10 ft + By x − 10 ft

−1

− M D x − 30 ft =−

−2

+ Dy x − 30 ft

+ 5 kips/ft x − 22 ft

−1

5 kips/ft 5 kips/ft 1 1 0 x − 0 ft + x − 10 ft + 5 kips/ft x − 22 ft 10 ft 10 ft

− M D x − 30 ft V ( x) =  w ( x ) dx = −

− 5 kips/ft x − 10 ft

−2

+ Dy x − 30 ft

−1

5 kips/ft 5 kips/ft 2 2 0 x − 0 ft + x − 10 ft + By x − 10 ft 2(10 ft) 2(10 ft)

+5 kips/ft x − 22 ft − M D x − 30 ft 1

−1

+ Dy x − 30 ft

5 kips/ft 5 kips/ft 3 3 1 x − 0 ft + x − 10 ft + By x − 10 ft 6(10 ft) 6(10 ft) 5 kips/ft 2 0 1 + x − 22 ft − M D x − 30 ft + Dy x − 30 ft 2 2 d v 5 kips/ft 5 kips/ft 3 3 1 EI 2 = M ( x) = − x − 0 ft + x − 10 ft + By x − 10 ft dx 6(10 ft) 6(10 ft) 5 kips/ft 2 0 1 + x − 22 ft − M D x − 30 ft + Dy x − 30 ft 2

M ( x ) =  V ( x)dx = −

By dv 5 kips/ft 5 kips/ft 5 kips/ft 4 4 2 3 =− x − 0 ft + x − 10 ft + x − 10 ft + x − 22 ft dx 24(10 ft) 24(10 ft) 2 6 Dy 1 2 − M D x − 30 ft + x − 30 ft + C1 2 By 5 kips/ft 5 kips/ft 5 kips/ft 5 5 3 4 EI v = − x − 0 ft + x − 10 ft + x − 10 ft + x − 22 ft 120(10 ft) 120(10 ft) 6 24 Dy M 2 3 − D x − 30 ft + x − 30 ft + C1 x + C2 2 6

(c)

(d)

Boundary conditions and evaluate constants: at x = 10 ft, v = 0

5 kips/ft (10 ft)5 + C1 (10 ft) + C2 = 0 120(10 ft) at x = 30 ft, v = 0 −

(e)

By 5 kips/ft 5 kips/ft (30 ft)5 + (20 ft)5 + (20 ft)3 120(10 ft) 120(10 ft) 6 5 kips/ft + (8 ft) 4 + C1 (30 ft) + C2 = 0 24 dv at x = 30 ft, = 0 dx By 5 kips/ft 5 kips/ft 5 kips/ft − (30 ft) 4 + (20 ft) 4 + (20 ft) 2 + (8 ft)3 + C1 = 0 24(10 ft) 24(10 ft) 2 6 −

(f)

(g)

(a) Solve for By, Dy, and MD: Solve equations (a), (b), (e), (f), and (g) simultaneously to obtain: C1 = −59.0000 kip-ft 2

C2 = 1,006.6667 kip-ft 3 By = 65.8700 kips = 65.9 kips  Dy = 19.1300 kips = 19.13 kips  M D = −105.9333 kip-ft = 105.9 kip-ft (cw)

Ans.

(b) Beam deflection at C: From Eq. (d), the beam deflection at C (x = 22 ft) is computed as follows: 5 kips/ft 5 kips/ft 65.8700 kips EI vC = − (22 ft)5 + (12 ft)5 + (12 ft)3 120(10 ft) 120(10 ft) 6

−(59.0000 kip-ft 2 )(22 ft) + 1,006.6667 kip-ft 3 = −1,757.4400 kip-ft 3  vC = −

1,757.4400 kip-ft 3 = −0.017574 ft = −0.2109 in. = 0.211 in.  100,000 kip-ft 2

Ans.

P11.20 For the beam shown in Figure P11.20, assume EI = 200,000 kN·m2 and use discontinuity functions to determine: (a) the reactions at B, C, and D. (b) the beam deflection at A. FIGURE P11.20

Solution Moment equation: (a) Support reactions: Fy = By + C y + Dy − 120 kN − (60 kN/m)(12 m) = 0

 By + C y + Dy = 840 kN

(a)

M B = (120 kN)(3 m) − (60 kN/m)(12 m)(6 m) +C y (6 m) + Dy (12 m) = 0  C y (6 m) + Dy (12 m) = 3,960 kN-m

(b)

Discontinuity expressions: −1 −1 0 w ( x ) = −120 kN x − 0 m + By x − 3 m − 60 kN/m x − 3 m +C y x − 9 m

−1

+ Dy x − 15 m

V ( x) =  w ( x ) dx = −120 kN x − 0 m + By x − 3 m − 60 kN/m x − 3 m 0

+ C y x − 9 m + Dy x − 15 m 0

M ( x ) =  V ( x)dx = −120 kN x − 0 m + By x − 3 m − 1

+ C y x − 9 m + Dy x − 15 m 1

60 kN/m 2 x−3 m 2

d 2v 60 kN/m 1 1 2 = M ( x) = −120 kN x − 0 m + By x − 3 m − x−3 m 2 dx 2 + C y x − 9 m + Dy x − 15 m 1

By dv 120 kN 60 kN/m 2 2 3 =− x−0 m + x−3 m − x−3 m dx 2 2 6 Cy Dy 2 2 + x−9 m + x − 15 m + C1 2 2 By 120 kN 60 kN/m 3 3 4 EI v = − x−0 m + x−3 m − x−3 m 6 6 24 Cy Dy 3 3 + x−9 m + x − 15 m + C1 x + C2 6 6 EI

(c)

(d)

Boundary conditions and evaluate constants: at x = 3 m, v = 0

−

120 kN (3 m)3 + C1 (3 m) + C2 = 0 6

(e)

at x = 9 m, v = 0 By 120 kN 60 kN/m (9 m)3 + (6 m)3 − (6 m)4 + C1 (9 m) + C2 = 0 6 6 24 at x = 15 m, v = 0 −

−

(f)

By Cy 120 kN 60 kN/m (15 m)3 + (12 m)3 − (12 m)4 + (6 m)3 + C1 (15 m) + C2 = 0 6 6 24 6

(g)

(a) Solve for By, Cy, and Dy: Solve equations (a), (b), (e), (f), and (g) simultaneously to obtain: C1 = 900.0000 kN-m 2

C2 = −2,160.0000 kN-m3 By = 330.0000 kN = 330 kN  C y = 360.0000 kN = 360 kN  Dy = 150.0000 kN = 150.0 kN 

Ans.

(b) Beam deflection at A: From Eq. (d), the beam deflection at A (x = 0 m) is computed as follows: EI v A = −2,160.0000 kN-m3  vA = −

2,160.0000 kN-m3 = −0.010800 m = −10.80 mm = 10.80 mm  200,000 kN-m 2

Ans.

P11.21 For the beam shown in Figure P11.21, assume EI = 200,000 kN·m2 and use discontinuity functions to determine: (a) the reactions at B, C, and D. (b) the beam deflection at A. FIGURE P11.21

Solution Moment equation: (a) Support reactions: Fy = By + C y + Dy − (60 kN/m)(6 m) = 0

 By + C y + Dy = 360 kN

(a)

M B = 420 kN-m − (60 kN/m)(6 m)(3 m) +C y (6 m) + Dy (12 m) = 0  C y (6 m) + Dy (12 m) = 660 kN-m

(b)

Discontinuity expressions: −2 −1 0 w ( x ) = −420 kN-m x − 0 m + By x − 3 m − 60 kN/m x − 3 m

+C y x − 9 m

−1

+ 60 kN/m x − 9 m + Dy x − 15 m

V ( x) =  w ( x ) dx = −420 kN-m x − 0 m

−1

+ By x − 3 m − 60 kN/m x − 3 m 0

+C y x − 9 m + 60 kN/m x − 9 m + Dy x − 15 m 0

M ( x ) =  V ( x)dx = −420 kN-m x − 0 m + By x − 3 m − 0

60 kN/m 2 x−3 m 2

60 kN/m 2 1 x − 9 m + Dy x − 15 m 2 2 d v 60 kN/m 0 1 2 EI 2 = M ( x) = −420 kN-m x − 0 m + By x − 3 m − x−3 m dx 2 60 kN/m 1 2 1 +C y x − 9 m + x − 9 m + Dy x − 15 m 2 By dv 60 kN/m 1 2 3 EI = −420 kN-m x − 0 m + x−3 m − x−3 m dx 2 6 Cy Dy 60 kN/m 2 3 2 + x−9 m + x−9 m + x − 15 m + C1 2 6 2 By 420 kN-m 60 kN/m 2 3 4 EI v = − x−0 m + x−3 m − x−3 m 2 6 24 Cy Dy 60 kN/m 3 4 3 + x−9 m + x−9 m + x − 15 m + C1 x + C2 6 24 6 +C y x − 9 m + 1

(c)

(d)

Boundary conditions and evaluate constants: at x = 3 m, v = 0

−

420 kN-m (3 m)2 + C1 (3 m) + C2 = 0 2

(e)

at x = 9 m, v = 0 By 420 kN-m 60 kN/m (9 m)2 + (6 m)3 − (6 m)4 + C1 (9 m) + C2 = 0 2 6 24 at x = 15 m, v = 0 By 420 kN-m 60 kN/m − (15 m) 2 + (12 m)3 − (12 m) 4 2 6 24 Cy 60 kN/m + (6 m)3 + (6 m) 4 + C1 (15 m) + C2 = 0 6 24 −

(f)

(g)

(a) Solve for By, Cy, and Dy: Solve equations (a), (b), (e), (f), and (g) simultaneously to obtain: C1 = 1,590.0000 kN-m 2

C2 = −2,880.0000 kN-m3 By = 245.0000 kN = 245 kN  C y = 120.0000 kN = 120 kN  Dy = −5.0000 kN = 5.00 kN 

Ans.

(b) Beam deflection at A: From Eq. (d), the beam deflection at A (x = 0 m) is computed as follows: EI v A = −2,880.0000 kN-m3  vA = −

2,880.0000 kN-m3 = −0.014400 m = −14.40 mm = 14.40 mm  200,000 kN-m 2

Ans.

P11.22a For the beams and loadings shown below, assume that EI = 5.0 × 104 kN·m2 is constant for each beam. (a) For the beam in Figure P11.22a, determine the concentrated downward force P required to make the total beam deflection at B equal to zero (i.e., vB = 0). FIGURE P11.22a

Solution Upward deflection at B due to 105 kN·m concentrated moment. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: Mx vB = − (2L2 − 3Lx + x 2 ) (elastic curve) 6LEI Values: M = −105 kN·m, L = 8 m, x = 4 m, EI = 5.0 × 104 kN·m2 Computation: Mx vB = − (2 L2 − 3Lx + x 2 ) 6 LEI

=−

(−105 kN-m)(4 m) 420 kN-m3  2(8 m) 2 − 3(8 m)(4 m) + (4 m) 2  = 6(8 m)EI EI

Downward deflection at B due to concentrated load P. [Appendix C, SS beam with concentrated load at midspan.] Relevant equation from Appendix C: PL3 vB = − 48 EI Values: L = 8 m, EI = 5.0 × 104 kN·m2 Computation:

PL3 P(8 m)3 P(10.666667 m3 ) vB = − =− = − 48EI 48EI EI Compatibility equation at B: 420 kN-m3 P(10.666667 m3 ) − =0 EI EI 420 kN-m3 P = = 39.375 kN = 39.4 kN 10.666667 m3

Ans.

P11.22b For the beams and loadings shown below, assume that EI = 5.0 × 104 kN·m2 is constant for each beam. (b) For the beam in Figure P11.22b, determine the concentrated moment M required to make the total beam slope at A equal to zero (i.e., A = 0). FIGURE P11.22b

Solution Slope at A due to 6 kN/m uniformly distributed load. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL3 A = (slope magnitude) 6 EI Values: w = 6 kN/m, L = 5 m, EI = 5.0 × 104 kN·m2

Computation: wL3 (6 kN/m)(5 m)3 125 kN-m 2 A = = = 6 EI 6 EI EI

(positive slope by inspection)

Slope at A due to concentrated moment M. [Appendix C, Cantilever beam with concentrated moment at tip.] Relevant equation from Appendix C: ML (slope magnitude) A = EI Values: L = 5 m, EI = 5.0 × 104 kN·m2 Computation: ML M (5 m) A = = EI EI

(negative slope by inspection)

Compatibility equation at A: 125 kN-m 2 M (5 m) − =0 EI EI 125 kN-m 2 M = = 25.0 kN-m 5m

Ans.

P11.23a For the beams and loadings shown below, assume that EI = 8.0 × 106 kip·in.2 is constant for each beam. (a) For the beam in Figure P11.23a, determine the concentrated downward force P required to make the total beam deflection at B equal to zero (i.e., vB = 0). FIGURE P11.23a

Solution Upward deflection at B due to 125 kip·ft concentrated moment. [Appendix C, Cantilever beam with concentrated moment at tip.] Relevant equation from Appendix C: ML2 vB = − 2 EI Values: M = −125 kip·ft, L = 15 ft, EI = 8.0 × 106 kip·in.2 Computation:

vB = −

ML2 (−125 kip-ft)(15 ft)2 14,062.5 kip-ft 3 =− = 2 EI 2 EI EI

Downward deflection at B due to concentrated load P. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: PL3 vB = − 3EI Values: L = 15 ft, EI = 8.0 × 106 kip·in.2 Computation:

vB = −

PL3 P(15 ft)3 P(1,125 ft 3 ) =− = − 3EI 3EI EI

Compatibility equation at B: 14,062.5 kip-ft 3 P(1,125 ft 3 ) − =0 EI EI 14,062.5 kip-ft 3 P = = 12.50 kips 1,125 ft 3

Ans.

P11.23b For the beams and loadings shown below, assume that EI = 8.0 × 106 kip·in.2 is constant for each beam. (b) For the beam in Figure P11.23b, determine the concentrated moment M required to make the total beam slope at A equal to zero (i.e., A = 0). FIGURE P11.23b

Solution Slope at A due to 7 kips/ft uniformly distributed load. [Appendix C, SS beam with uniformly distributed load over portion of span.] Relevant equation from Appendix C: wa 2 A = (2 L2 − a 2 ) (slope magnitude) 24 LEI Values: w = 7 kips/ft, L = 23 ft, a = 15 ft, EI = 8.0 × 106 kip·in.2 Computation: wa 2 A = (2 L2 − a 2 ) 24 LEI =

(7 kips/ft)(15 ft) 2  2(23 ft) 2 − (15 ft) 2  24(23 ft)EI

= −

2,376.766304 kip-ft 2 EI

(negative slope by inspection)

Slope at A due to concentrated moment M. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: ML (slope magnitude) A = 3EI Values: L = 23 ft, EI = 8.0 × 106 kip·in.2 Computation: ML M (23 ft) M (7.666667 ft) A = = = 3EI 3EI EI Compatibility equation at A: 2,376.766304 kip-ft 2 M (7.666667 ft) − + =0 EI EI 2,376.766304 kip-ft 2 M = = 310 kip-ft 7.666667 ft

(positive slope by inspection)

Ans.

P11.24 For the beam and loading shown below, derive an expression for the reactions at supports A and B. Assume that EI is constant for the beam.

FIGURE P11.24

Solution Choose the reaction force at B as the redundant; therefore, the released beam is a cantilever. Consider downward deflection of cantilever beam at B due to linearly distributed load. [Appendix C, Cantilever beam with linearly distributed load.] Relevant equation from Appendix C: w L4 vB = − 0 30 EI

Consider upward deflection of cantilever beam at B due to concentrated load By. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: 3 PL3 By L vB = = 3EI 3EI

Compatibility equation for deflection at B: 3 w L4 By L wL − 0 + =0  By = 0  30 EI 3EI 10

Ans.

Equilibrium equations for entire beam: wL w L w L 4w L 2w0 L Fy = Ay + By − 0 = 0  Ay = 0 − 0 = 0 =  2 2 10 10 5 w L  L M A = − M A − 0   + By L = 0 2  3 w0 L2 w0 L w0 L2 w0 L2 w0 L2  M A = By L − = ( L) − =− = 6 10 6 15 15

(ccw)

Ans.

P11.25 For the beam and loading shown below, derive an expression for the reactions at supports A and B. Assume that EI is constant for the beam.

FIGURE P11.25

Solution Choose the reaction force at A as the redundant; therefore, the released beam is a cantilever. Consider downward deflection of cantilever beam at A due to uniformly distributed load. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wx 2 vA = − (6 L2 − 4 Lx + x 2 ) (elastic curve) 24 EI

Let x = L, L =

3L 2

2  w( L) 2   3L  17 wL4  3L  2  vA = −  6   − 4   ( L) + ( L)  = − 24 EI   2  48EI  2  

Consider upward deflection of cantilever beam at A due to concentrated load Ay. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: 3 PL3 Ay L vA = = 3EI 3EI

Compatibility equation for deflection at A: 3 17 wL4 Ay L 17 wL − + =0  Ay =  48EI 3EI 16

Ans.

Equilibrium

equations

for

entire

beam:

3wL 3wL 17 wL 7 wL 7 wL =0  By = − = =  2 2 16 16 16  3wL   3L  M B = M B − Ay L +  =0  2   4  Fy = Ay + By −

 M B = Ay L −

9wL2 17 wL 9wL2 wL2 wL2 = ( L) − =− = 8 16 8 16 16

(cw)

Ans.

P11.26 For the beam and loading shown below, derive an expression for the reactions at supports A and B. Assume that EI is constant for the beam.

FIGURE P11.26

Solution Choose the reaction force at A as the redundant; therefore, the released beam is a cantilever. Consider downward deflection of cantilever beam at A due to concentrated load P. [Appendix C, Cantilever beam with concentrated load at midspan.] Relevant equation from Appendix C: 5 PL3 vA = − 48 EI Let L = 2 L

 vA = −

5P(2 L)3 5PL3 =− 48EI 6 EI

Consider upward deflection of cantilever beam at A due to concentrated load Ay. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: PL3 vA = − 3EI Let L = 2 L, P = − Ay  vA = −

( − Ay )(2 L)3 3EI

8 Ay L3 3EI

Compatibility equation for deflection at A: 3 5PL3 8 Ay L 5P − + =0  Ay =  6 EI 3EI 16

Ans.

Equilibrium equations for entire beam: Fy = Ay + By − P = 0

5P 11P 11P = =  16 16 16 M B = M B − Ay (2L) + P(L) = 0  By = P −

 M B = Ay (2 L) − P( L) =

Ans.

5PL 3PL 3PL − PL = − = (cw) 8 8 8

Ans.

P11.27 For the beam and loading shown below, derive an expression for the reactions at supports A and C. Assume that EI is constant for the beam.

FIGURE P11.27

Solution Choose the reaction force at C as the redundant; therefore, the released beam is a cantilever. Consider downward deflection of cantilever beam at C due to uniformly distributed load. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equations from Appendix C: wL4 wL3 vB = − and  B = − 8 EI 6 EI 4 3 wL wL 7 wL4  vC = − − ( L) = − 8 EI 6 EI 24 EI Consider upward deflection of cantilever beam at C due to concentrated load Cy. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: PL3 vC = − 3EI Let L = 2 L, P = −C y  vC = −

( −C y )(2 L)3 3EI

8C y L3 3EI

Compatibility equation for deflection at C: 3 7 wL4 8C y L 7 wL − + =0  Cy =  24 EI 3EI 64

Ans.

Equilibrium equations for entire beam: Fy = Ay + C y − wL = 0

7 wL 57 wL =  64 64  L M A = − M A − wL   + Cy (2L) = 0  2  Ay = wL −

Ans.

wL2 7 wL wL2 18wL2 9wL2 9wL2  M A = C y (2 L) − = (2 L) − =− =− = 2 64 2 64 32 32

(ccw)

Ans.

P11.28 For the beam and loading shown below, derive an expression for the reaction force at B. Assume that EI is constant for the beam. (Reminder: The roller symbol implies that both upward and downward displacement is restrained.) FIGURE P11.28

Solution Choose the reaction force at B as the redundant; therefore, the released beam is simply supported. Consider upward deflection of simply supported beam at B due to M0. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: Mx vB = − (2L2 − 3Lx + x2 ) (elastic curve) 6LEI Let L = 2 L, x = L, M = − M 0  vB = −

(− M 0 )( L) M0 M L2  2(2 L) 2 − 3(2 L)( L) + ( L) 2  = 3L2  = 0 6(2 L) EI 12 EI 4 EI

Consider upward deflection of simply supported beam at B due to concentrated load By. [Appendix C, SS beam with concentrated load at midspan.] Relevant equation from Appendix C: PL3 vB = − 48 EI Let L = 2 L, P = − By  vB = −

( − By )(2 L)3 48 EI

By L3 6 EI

Compatibility equation for deflection at B: 3 M 0 L2 By L 3M 0 3M 0 + =0  By = − =  4 EI 6 EI 2L 2L

Ans.

P11.29 For the beam and loading shown below, derive an expression for the reaction force at B. Assume that EI is constant for the beam.

FIGURE P11.29

Solution Choose the reaction force at B as the redundant; therefore, the released beam is simply supported. Consider downward deflection of simply supported beam at B due to uniformly distributed load. [Appendix C, SS beam with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C: wa3 vB = − (4 L2 − 7aL + 3a 2 ) 24 LEI

Let L = 3L, a = 2 L  vB = −

w(2 L)3 wL2 2wL4  4(3L) 2 − 7(2 L)(3L) + 3(2 L) 2  = − 6 L2  = − 24(3L) EI 9 EI 3EI

Let L = 3L, a = L, b = 2 L, P = − By  vB = −

( − By )( L)(2 L) 6(3L) EI

By L

4 By L3

(3L) − ( L) − (2 L)  = 4 L  = 9 EI   9 EI 2

Compatibility equation for deflection at B: 3 2wL4 4 By L 3wL 3wL − + =0  By = =  3EI 9EI 2 2

Ans.

P11.30 For the beam and loading shown below, derive an expression for the reaction force at B. Assume that EI is constant for the beam.

FIGURE P11.30

Solution Choose the reaction force at B as the redundant; therefore, the released beam is simply supported. Consider downward deflection of simply supported beam at B due to one concentrated load P. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pbx 2 vB = − ( L − b2 − x 2 ) (elastic curve) 6LEI

Let L = 4 L, b = L, x = 2 L  vB = −

P( L)(2 L) PL 11PL3 (4 L) 2 − ( L) 2 − (2 L) 2  = − 11L2  = − 6(4 L) EI 12 EI 12 EI

The second concentrated load will cause an additional deflection at B of the same magnitude. Consider upward deflection of simply supported beam at B due to concentrated load By. [Appendix C, SS beam with concentrated load at midspan.] Relevant equation from Appendix C: PL3 vB = − 48 EI Let L = 4 L, P = − By  vB = −

( − By )(4 L)3 48EI

64 By L3 48EI

16 By L3 12 EI

Compatibility equation for deflection at B: 3 11PL3 11PL3 16 By L − − + =0 12 EI 12 EI 12 EI

 By =

11P  8

Ans.

P11.31 The beam shown in Figure P11.31 consists of a W360 × 79 structural steel wide-flange shape [E = 200 GPa; I = 225 × 106 mm4]. For the loading shown, determine: (a) the reactions at A, B, and C. (b) the magnitude of the maximum bending stress in the beam. FIGURE P11.31

Solution (a) Reactions at A, B, and C. Choose the reaction force at B as the redundant; therefore, the released beam is simply supported between A and C. Consider downward deflection of simply supported beam at B due to uniformly distributed load. [Appendix C, SS beam with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C: wa3 vB = − (4 L2 − 7aL + 3a 2 ) 24 LEI Values: w = 90 kN/m, L = 9 m, a = 6 m Calculation: wa3 vB = − (4 L2 − 7aL + 3a 2 ) 24 LEI

=−

(90 kN/m)(6 m)3 4,860 kN-m3  4(9 m)2 − 7(6 m)(9 m) + 3(6 m)2  = − 24(9 m) EI EI

Consider downward deflection of simply supported beam at B due to concentrated moment. [Appendix C, SS beam with concentrated moment at one end of span.] Relevant equation from Appendix C: Mx vB = − (2L2 − 3Lx + x2 ) (elastic curve) 6LEI Values: M = 180 kN·m, L = 9 m, x = 3 m Calculation: Mx vB = − (2 L2 − 3Lx + x 2 ) 6 LEI

=−

(180 kN-m)(3 m) 900 kN-m3  2(9 m)2 − 3(9 m)(3 m) + (3 m) 2  = − 6(9 m) EI EI

Consider upward deflection of simply supported beam at B due to concentrated load By. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pab 2 vB = − ( L − a 2 − b2 ) 6LEI Values: P = −By, L = 9 m, a = 3 m, b = 6 m

Calculation: Pab 2 vB = − ( L − a 2 − b2 ) 6 LEI

=−

( − By )(3 m)(6 m) 6(9 m) EI

(9 m) 2 − (3 m) 2 − (6 m) 2  =

(12 m3 ) By EI

Compatibility equation for deflection at B: 3 4,860 kN-m3 900 kN-m3 (12 m ) By − − + =0 EI EI EI 5,760 kN-m3  By = = 480 kN = 480 kN  12 m3

Ans.

Equilibrium equations for entire beam: M A = By (3 m) + Cy (9 m) − 180 kN-m − (90 kN/m)(6 m)(6 m) = 0  Cy =

180 kN-m + (90 kN/m)(6 m)(6 m) − (480 kN)(3 m) 9m

= 220.0 kN = 220 kN 

Ans.

Fy = Ay + By + Cy − (90 kN/m)(6 m) = 0

 Ay = (90 kN/m)(6 m) − 480 kN − 220 kN = −160.0 kN = 160.0 kN 

Ans.

Shear-force and bending-moment diagrams (b) Magnitude of maximum bending stress: Section properties (from Appendix B): I = 225  106 mm 4 d = 353 mm

S = 1,270  103 mm3 Maximum bending moment magnitude Mmax = 300 kN·m Bending stresses at maximum moment (300 kN-m)(353 mm/2)(1,000) 2 x = 225  106 mm 4

= 235 MPa

Ans.

or using the tabulated section modulus value: (300 kN-m)(1,000) 2 x = 1, 270  103 mm3 = 236 MPa

Ans.

P11.32 The beam shown in Figure P11.32 consists of a W610 × 140 structural steel wide-flange shape [E = 200 GPa; I = 1,120 × 106 mm4]. For the loading shown, determine: (a) the reactions at A, B, and D. (b) the magnitude of the maximum bending stress in the beam. FIGURE P11.32

Solution (a) Reactions at A, B, and D. Choose the reaction force at B as the redundant; therefore, the released beam is simply supported between A and D. Consider downward deflection of simply supported beam at B due to uniformly distributed load. [Appendix C, SS beam with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C:

wx 3 ( L − 2 Lx 2 + x3 ) (elastic curve) 24 EI Values: w = 60 kN/m, L = 7.5 m, x = 1.5 m vB = −

Calculation: wx vB = − ( L3 − 2 Lx 2 + x 3 ) 24 EI =−

(60 kN/m)(1.5 m) 1, 468.125 kN-m3 (7.5 m)3 − 2(7.5 m)(1.5 m) 2 + (1.5 m)3  = − 24 EI EI

Consider downward deflection of simply supported beam at B due to concentrated load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pbx 2 vB = − ( L − b2 − x 2 ) (elastic curve) 6LEI Values: P = 125 kN, L = 7.5 m, b = 2.5 m, x = 1.5 m Calculation: Pbx 2 vB = − ( L − b2 − x 2 ) 6 LEI

(125 kN)(2.5 m)(1.5 m) 497.396 kN-m3 2 2 2   =− (7.5 m) − (2.5 m) − (1.5 m)  = − 6(7.5 m) EI EI

Calculation: Pab 2 vB = − ( L − a 2 − b2 ) 6 LEI

=−

( − By )(1.5 m)(6 m) 6(7.5 m) EI

(7.5 m) 2 − (1.5 m) 2 − (6 m) 2  =

(3.6 m3 ) By EI

Compatibility equation for deflection at B: 3 1, 468.125 kN-m3 497.396 kN-m 3 (3.6 m ) By − − + =0 EI EI EI 1,965.521 kN-m3  By = = 545.978 kN = 546 kN  3.6 m3

Ans.

Equilibrium equations for entire beam: M A = By (1.5 m) + Dy (7.5 m) − (60 kN/m)(7.5 m)(3.75 m) − (125 kN)(5 m) = 0  Dy =

(60 kN/m)(7.5 m)(3.75 m) + (125 kN)(5 m) − (545.978 kN)(1.5 m) 7.5 m

= 199.138 kN = 199.1 kN 

Ans.

Fy = Ay + By + Dy − (60 kN/m)(7.5 m) − 125 kN = 0  Ay = (60 kN/m)(7.5 m) + 125 kN − 545.978 kN − 199.138 kN = −170.116 kN = 170.1 kN 

Ans.

Shear-force and bending-moment diagrams (b) Magnitude of maximum bending stress: Section properties (from Appendix B): I = 1,120  106 mm 4 d = 617 mm

S = 3,640  103 mm3 Maximum bending moment magnitude Mmax = 322.67 kN·m Bending stresses at maximum moment (322.67 kN-m)(617 mm/2)(1,000) 2 x = 1,120  106 mm 4 = 88.9 MPa

Ans.

or using the tabulated section modulus value: (322.67 kN-m)(1,000) 2 x = 3,640  103 mm3 = 88.6 MPa

Ans.

P11.33 A propped cantilever beam is loaded as shown in Figure P11.33. Assume EI = 24 × 106 kip·in.2. Determine: (a) the reactions at B and C for the beam. (b) the beam deflection at A. FIGURE P11.33

Solution Choose the reaction force at B as the redundant; therefore, the released beam is a cantilever. Consider downward deflection of cantilever beam at B due to uniformly distributed load. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL4 vB = − 8 EI Values: w = 8 kips/ft, L = 24 ft Calculation:

vB = −

wL4 (8 kips/ft)(24 ft)4 331,776 kip-ft 3 =− = − 8EI 8EI EI

(a)

Consider downward deflection of cantilever beam at B due to the 40-kip concentrated load. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: Px 2 vB = − (3L − x) (elastic curve) 6 EI Values: P = 40 kips, L = 36 ft, x = 24 ft Calculation:

Px 2 (40 kips)(24 ft) 2 322,560 kip-ft 3 vB = − (3L − x) = −  3(36 ft) − (24 ft) = − 6 EI 6 EI EI

Consider upward deflection of cantilever beam at B due to concentrated load By. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: PL3 vB = − 3EI Values: P = −By, L = 24 ft Calculation: ( − By )(24 ft) (4,608 ft ) By PL3 vB = − =− = 3EI 3EI EI 3

Compatibility equation for deflection at B: 3 331,776 kip-ft 3 322,560 kip-ft 3 (4,608 ft ) By − − + =0 EI EI EI 654,336 kip-ft 3  By = = 142.0 kips = 142.0 kips  4,608 ft 3

(b)

Ans.

Equilibrium equations for entire beam: Fy = By + C y − 40 kips − (8 kips/ft)(24 ft) = 0  C y = 40 kips + (8 kips/ft)(24 ft) − 142.0 kips = 90.0 kips 

Ans.

MC = MC + (40 kips)(36 ft) + (8 kips/ft)(24 ft)(12 ft) − By (24 ft) = 0  M C = (142.0 kips)(24 ft) − (40 kips)(36 ft) − (8 kips/ft)(24 ft)(12 ft) = −336.0 kip-ft = 336 kip-ft (cw)

Ans.

(b) Beam deflection at A: Consider downward deflection of cantilever beam at A due to uniformly distributed load. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equations from Appendix C: wL4 wL3 and  B = (slope magnitude) 8 EI 6 EI Values: w = 8 kips/ft, L = 24 ft, EI = 24 × 106 kip·in.2 vB = −

Calculation: 331,776 kip-ft 3 vB = − calculated previously in Eq. (a) EI wL3 (8 kips/ft)(24 ft)3 18, 432 kip-ft 2 B = = = 6 EI 6 EI EI vA = −

 18, 432 kip-ft 2  331,776 kip-ft 3 552,960 kip-ft 3 − (12 ft)  = −  EI EI EI 

Consider downward deflection of cantilever beam at A due to the 40-kip concentrated load. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: PL3 vA = − 3EI Values: P = 40 kips, L = 36 ft, EI = 24 × 106 kip·in.2 Calculation:

PL3 (40 kips)(36 ft)3 622,080 kip-ft 3 vA = − =− = − 3EI 3EI EI Consider upward deflection of cantilever beam at B due to concentrated load By. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equations from Appendix C: PL3 PL2 vB = − and  B = (slope magnitude) 3EI 2 EI Values: P = −By = 142 kips, L = 24 ft, EI = 24 × 106 kip·in.2

Calculation: (4,608 ft 3 )(142.0 kips) 654,336 kip-ft 3 vB = = EI EI 2 2 PL (142.0 kips)(24 ft) 40,896 kip-ft 2 B = = = 2 EI 2 EI EI vA =

using the results from Eq. (b)

 40,896 kip-ft 2  1,145,088 kip-ft 3 654,336 kip-ft 3 + (12 ft)   = EI EI EI 

Beam deflection at A. 552,960 kip-ft 3 622,080 kip-ft 3 1,145,088 kip-ft 3 vA = − − + EI EI EI 3 3 29,952 kip-ft (29,952 kip-ft )(12 in./ft) 3 =− =− = −2.156544 in. = 2.16 in.  EI 24  106 kip-in.2

Ans.

P11.34 The beam shown in Figure P11.34 consists of a W610 × 82 structural steel wide-flange shape [E = 200 GPa; I = 562 × 106 mm4]. For the loading shown, determine: (a) the reaction force at C. (b) the beam deflection at A. FIGURE P11.34

Solution (a) Reaction force at C. Choose the reaction force at C as the redundant; therefore, the released beam is simply supported between B and D. Consider upward deflection of simply supported beam at C due to uniformly distributed load on overhang AB. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C:

Mx (2L2 − 3Lx + x2 ) (elastic curve) 6LEI Values: M = −(105 kN/m)(3 m)(1.5 m) = −472.5 kN·m, L = 14 m, x = 7 m vC = −

Calculation: Mx vC = − (2 L2 − 3Lx + x 2 ) 6 LEI

=−

( −472.5 kN-m)(7 m) 5,788.125 kN-m3  2(14 m) 2 − 3(14 m)(7 m) + (7 m) 2  = 6(14 m)EI EI

Consider downward deflection of simply supported beam at C due to uniformly distributed load. [Appendix C, SS beam with uniformly distributed load.] Relevant equation from Appendix C: 5wL4 vC = − 384 EI Values: w = 105 kN/m, L = 14 m Calculation:

vC = −

5wL4 5(105 kN/m)(14 m) 4 52,521.875 kN-m3 =− = − 384 EI 384 EI EI

Consider upward deflection of simply supported beam at C due to concentrated load Cy. [Appendix C, SS beam with concentrated load at midspan.] Relevant equation from Appendix C: PL3 vC = − 48 EI Values: P = −Cy, L = 14 m Calculation: ( −C y )(14 m) (57.1667 m )C y PL3 vC = − =− = 48EI 48EI EI 3

Compatibility equation for deflection at C: 3 5,788.125 kN-m3 52,521.875 kN-m3 (57.1667 m )C y − + =0 EI EI EI 46,733.750 kN-m3  Cy = = 817.5 kN = 818 kN  57.1667 m3

Ans.

(b) Beam deflection at A. Consider downward cantilever beam deflection caused by uniformly distributed load on overhang AB. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wL4 vA = − 8 EI Values: w = 105 kN/m, L = 3 m Calculation:

vA = −

wL4 (105 kN/m)(3 m) 4 1,063.125 kN-m3 =− = − 8EI 8EI EI

Consider downward deflection at A resulting from rotation at B caused by concentrated load on overhang AB. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: ML B = (slope magnitude) 3EI Values: M = (105 kN/m)(3 m)(1.5 m) = 472.5 kN·m, L = 14 m Computation: ML (472.5 kN-m)(14 m) 2,205 kN-m 2 B = = = 3EI 3EI EI

 2, 205 kN-m 2  6, 615 kN-m3 vA = −(3 m)  = −  EI EI   Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Consider upward deflection at A due to uniformly distributed load between B and D. [Appendix C, SS beam with uniformly distributed load.] Relevant equation from Appendix C: wL3 B = (slope magnitude) 24 EI Values: w = 105 kN/m, L = 14 m Calculation: wL3 (105 kN/m)(14 m)3 12, 005 kN-m 2 B = = = 24 EI 24 EI EI

 12, 005 kN-m 2  36, 015 kN-m3 vA = (3 m)  = EI EI   Consider downward deflection at A due to concentrated load Cy. [Appendix C, SS beam with concentrated load at midspan.] Relevant equation from Appendix C: PL2 B = (slope magnitude) 16 EI Values: P = −Cy = −817.5 kN, L = 14 m Calculation: PL2 (817.5 kN)(14 m)2 10, 014.375 kN-m2 B = = = 16 EI 16 EI EI

 10, 014.375 kN-m2  30, 043.125 kN-m3 vA = −(3 m)  = − EI EI   Beam deflection at A. 1,063.125 kN-m3 6,615 kN-m3 36,015 kN-m 3 30,043.125 kN-m3 vA = − − + − EI EI EI EI 3 3 −1,706.25 kN-m −1,706.25 kN-m = = = −0.015180 m = 15.18 mm  EI 112,400 kN-m 2

Ans.

P11.35 The beam shown in Figure P11.35 consists of a W8 × 15 structural steel wideflange shape [E = 29,000 ksi; I = 48 in.4]. For the loading shown, determine: (a) the reactions at A and B. (b) the magnitude of the maximum bending stress in the beam. (Reminder: The roller symbol implies that both upward and downward displacement is restrained.) FIGURE P11.35

Solution Choose the reaction force at B as the redundant; therefore, the released beam is a cantilever. Consider deflection of cantilever beam at B due to uniformly distributed load over entire beam span. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wx 2 vB = − (6 L2 − 4 Lx + x 2 ) (elastic curve) 24 EI Values: w = −80 lb/in., L = 150 in., x = 100 in. Calculation: wx 2 vB = − (6 L2 − 4 Lx + x 2 ) 24 EI (−80 lb/in.)(100 in.) 2 6(150 in.) 2 − 4(150 in.)(100 in.) + (100 in.) 2  =− 24 EI =

2.8333  109 lb-in.3 EI

Consider deflection of cantilever beam at B due to the force caused by the linear portion of the distributed load. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: PL3 vB = − 3EI Values: P = −½(50 in.)(60 lb/in.) = −1,500 lb, L = 100 in. Calculation: PL3 (−1,500 lb)(100 in.)3 500  106 lb-in.3 vB = − =− = 3EI 3EI EI

Consider deflection of cantilever beam at B due to the moment caused by the linear portion of the distributed load. [Appendix C, Cantilever beam with concentrated moment at tip.] Relevant equation from Appendix C: ML2 vB = − 2 EI Values: M = −½(50 in.)(60 lb/in.)[⅔(50 in.)] = −50,000 lb·in., L = 100 in. Calculation:

vB = −

ML2 (−50,000 lb-in.)(100 in.) 2 250  106 lb-in.3 =− = 2 EI 2 EI EI

Consider deflection of cantilever beam at B due to concentrated load By. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: PL3 vB = − 3EI Values: P = −By, L = 100 in.

Calculation: ( − By )(100 in.) (333.3333 in. ) By PL3 vB = − =− = 3EI 3EI EI 3

Compatibility equation for deflection at B: 3 3 2.8333  109 lb-in.3 500  106 lb-in.3 250  106 lb-in.3 (333.3333  10 in. ) By + + + =0 EI EI EI EI 3.5833  109 lb-in.3  By = − = −10,750 lb = 10,750 lb  333.3333  103 in.3

Ans.

Equilibrium equations for entire beam: 1 Fy = Ay + By + (80 lb/in.)(150 in.) + (140 lb/in. − 80 lb/in.)(50 in.) = 0 2 (60 lb/in.)(50 in.)  Ay = −(80 lb/in.)(150 in.) − − ( −10,750 lb) 2

= −2,750 lb = 2, 750 lb 

Ans.

M A = − M A + (80 lb/in.)(150 in.)(75 in.) 2  (140 lb/in. − 80 lb/in.)(50 in.)    + 100 in. + (50 in.) + By (100 in.) = 0   2 3    (60 lb/in.)(50 in.)  M A = (80 lb/in.)(150 in.)(75 in.) + (133.3333 in.) + ( −10,750 in.)(100 in.) 2 = 25,000 lb-in. = 25,000 lb-in. (cw)

Ans.

(b) Magnitude of maximum bending stress: Section properties (from Appendix B): I = 48 in.4 d = 8.11 in. S = 11.8 in.3 Maximum bending moment magnitude Mmax = 150,000 lb·in. (at B) Bending stresses at maximum moment (150,000 lb-in.)(8.11 in./2) x = = 12,671.875 psi = 12,670 psi 48 in.4 or, using the tabulated value for the section modulus: 150,000 lb-in. x = = 12,711.864 psi = 12,710 psi 11.8 in.3

Ans.

P11.36 The solid 1.00-in.-diameter steel [E = 29,000 ksi] shaft shown in Figure P11.36 supports three belt pulleys. Assume that the bearing at A can be idealized as a pin support and that the bearings at C and E can be idealized as roller supports. For the loading shown, determine: (a) the reaction forces at bearings A, C, and E. (b) the magnitude of the maximum bending stress in the shaft.

FIGURE P11.36

Solution (a) Reaction forces at A, C, and E. Choose the reaction force at C as the redundant; therefore, the released beam is simply supported between A and E. Consider downward deflection of simply supported beam at C due to pulley B load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pbx 2 vC = − ( L − b2 − x 2 ) (elastic curve) 6LEI Values: P = 200 lb, L = 60 in., b = 15 in., x = 30 in. Calculation: Pbx 2 vC = − ( L − b2 − x2 ) 6 LEI

=−

(200 lb)(15 in.)(30 in.) (60 in.)2 − (15 in.)2 − (30 in.)2  6(60 in.)EI

618,750 lb-in.3 = − EI Consider downward deflection of simply supported beam at C due to pulley D load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pbx 2 vC = − ( L − b2 − x 2 ) (elastic curve) 6LEI Values: P = 280 lb, L = 60 in., b = 15 in., x = 30 in. Calculation: Pbx 2 vC = − ( L − b2 − x2 ) 6 LEI (280 lb)(15 in.)(30 in.)  (60 in.) 2 − (15 in.) 2 − (30 in.) 2  =− 6(60 in.)EI = −

866, 250 lb-in.3 EI

Consider upward deflection of simply supported beam at C due to pulley F load. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C:

Mx (2L2 − 3Lx + x2 ) (elastic curve) 6LEI Values: M = −(120 lb)(10 in.) = −1,200 lb·in., L = 60 in., x = 30 in. vC = −

Calculation: Mx vC = − (2 L2 − 3Lx + x 2 ) 6 LEI

=−

(−1,200 lb-in.)(30 in.)  2(60 in.)2 − 3(60 in.)(30 in.) + (30 in.)2  6(60 in.)EI

270,000 lb-in.3 = EI Consider upward deflection of simply supported beam at C due to concentrated load Cy. [Appendix C, SS beam with concentrated load at midspan.] Relevant equation from Appendix C: PL3 vC = − 48 EI Values: P = −Cy, L = 60 in. Calculation: ( −C y )(60 in.) (4,500 in. )C y PL3 =− = 48EI 48EI EI 3

vC = −

Compatibility equation for deflection at C: 3 618,750 lb-in.3 866, 250 lb-in.3 270,000 lb-in.3 (4,500 in. )C y − − + + =0 EI EI EI EI 1, 215,000 lb-in.3  Cy = = 270 lb = 270 lb  4,500 in.3

Ans.

Equilibrium equations for entire beam: M A = −(200 lb)(15 in.) − (280 lb)(45 in.) − (120 lb)(70 in.) + Cy (30 in.) + Ey (60 in.) = 0 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

 Ey =

(200 lb)(15 in.) + (280 lb)(45 in.) + (120 lb)(70 in.) − (215 lb)(30 in.) 60 in.

= 265.0 lb = 265 lb 

Ans.

Fy = Ay + C y + E y − 200 lb − 280 lb − 120 lb = 0  Ay = 200 lb + 280 lb + 120 lb − 270 lb − 265 lb = 65.0 lb 

Ans.

Shear-force and bending-moment diagrams (b) Magnitude of maximum bending stress: Section properties:



(1.00 in.)4 = 0.0490874 in.4

Maximum bending moment magnitude Mmax = 1,200 lb·in. Bending stresses at maximum moment (1,200 lb-in.)(1.00 in./2) x = 0.0490874 in.4 = 12,223.1 psi

= 12,220 psi

Ans.

P11.37 The solid 1.00-in.-diameter steel [E = 29,000 ksi] shaft shown in Figure P11.37 supports two belt pulleys. Assume that the bearing at E can be idealized as a pin support and that the bearings at B and C can be idealized as roller supports. For the loading shown, determine: (a) the reaction forces at bearings B, C, and E. (b) the magnitude of the maximum bending stress in the shaft.

FIGURE P11.37

Solution (a) Reaction forces at B, C, and E. Choose the reaction force at C as the redundant; therefore, the released beam is simply supported between B and E. Consider upward deflection of simply supported beam at C due to pulley A load. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: Mx vC = − (2L2 − 3Lx + x2 ) (elastic curve) 6LEI Values: M = −(90 lb)(7 in.) = −630 lb·in., L = 45 in., x = 15 in. Calculation: Mx vC = − (2 L2 − 3Lx + x 2 ) 6 LEI ( −630 lb-in.)(15 in.)  2(45 in.) 2 − 3(45 in.)(15 in.) + (15 in.) 2  =− 6(45 in.)EI 78,750 lb-in.3 = EI

Consider downward deflection of simply supported beam at C due to pulley D load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pbx 2 vC = − ( L − b2 − x 2 ) (elastic curve) 6LEI Values: P = 240 lb, L = 45 in., b = 15 in., x = 15 in.

Calculation: Pbx 2 vC = − (L − b2 − x2 ) 6 LEI (240 lb)(15 in.)(15 in.) (45 in.) 2 − (15 in.) 2 − (15 in.) 2  =− 6(45 in.)EI = −

315,000 lb-in.3 EI

Consider upward deflection of simply supported beam at C due to concentrated load Cy. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pab 2 vC = − ( L − a 2 − b2 ) 6LEI Values: P = −Cy, L = 45 in., a = 15 in., b = 30 in. Calculation: Pab 2 vC = − ( L − a 2 − b2 ) 6 LEI

=−

( −C y )(15 in.)(30 in.) 6(45 in.)EI

(45 in.) − (15 in.) − (30 in.)  = 2

(1,500 in.3 )C y

Compatibility equation for deflection at C: 3 78,750 lb-in.3 315,000 lb-in.3 (1,500 in. )C y − + =0 EI EI EI 236, 250 lb-in.3  Cy = = 157.50 lb = 157.5 lb  1,500 in.3

Ans.

Equilibrium equations for entire beam: M E = (240 lb)(15 in.) + (90 lb)(52 in.) − By (45 in.) − Cy (30 in.) = 0  By =

(240 lb)(15 in.) + (90 lb)(52 in.) − (157.5 lb)(30 in.) 45 in.

= 79.0 lb = 79.0 lb 

Ans.

Fy = By + C y + E y − 90 lb − 240 lb = 0  E y = 90 lb + 240 lb − 157.5 lb − 79.0 lb = 93.5 lb = 93.5 lb 

Ans.

Shear-force and bending-moment diagrams (b) Magnitude of maximum bending stress: Section properties:



(1.00 in.)4 = 0.0490874 in.4

Maximum bending moment magnitude Mmax = 1,402.5 lb·in. Bending stresses at maximum moment (1, 402.5 lb-in.)(1.00 in./2) x = 0.0490874 in.4 = 14, 285.74 psi

= 14, 290 psi

Ans.

P11.38 The beam shown in Figure P11.38 consists of a W360 × 101 structural steel wideflange shape [E = 200 GPa; I = 301 × 106 mm4]. For the loading shown, determine: (a) the reactions at A and B. (b) the magnitude of the maximum bending stress in the beam. FIGURE P11.38

Solution Choose the reaction force at B as the redundant; therefore, the released beam is a cantilever. Consider deflection of cantilever beam at B due to uniformly distributed load over entire beam span. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equations from Appendix C: wx 2 vB = − (6 L2 − 4 Lx + x 2 ) (elastic curve) 24 EI Values: w = 30 kN/m, L = 8 m, x = 5.5 m Calculation: wx 2 vB = − (6 L2 − 4 Lx + x 2 ) 24 EI (30 kN/m)(5.5 m)2 6(8 m) 2 − 4(8 m)(5.5 m) + (5.5 m) 2  =− 24 EI = −

9,008.828125 kN-m3 EI

Consider deflection of cantilever beam at B due a linearly distributed load. [Appendix C, Cantilever beam with linearly distributed load.] Relevant equation from Appendix C: w x2 vB = − 0 (10 L3 − 10 L2 x + 5Lx 2 − x3 ) (elastic 120 LEI curve) Values: w0 = 60 kN/m, L = 8 m, x = 5.5 m

Calculation: w x2 vB = − 0 (10 L3 − 10 L2 x + 5Lx 2 − x3 ) 120 LEI =−

(60 kN/m)(5.5 m)2 10(8 m)3 − 10(8 m) 2 (5.5 m) + 5(8 m)(5.5 m)2 − (5.5 m)3  120(8 m)EI

= −

4,998.103516 kN-m3 EI

Compatibility equation for deflection at B: 3 9,008.828125 kN-m3 4,998.103516 kN-m3 (55.458333 m ) By − − + =0 EI EI EI 14,006.93164 kN-m3  By = = 252.56676 kN = 253 kN  55.458333 m3

Ans.

Equilibrium equations for entire beam: 1 Fy = Ay + By − (30 kN/m)(8 m) − (90 kN/m − 30 kN/m)(8 m) = 0 2 1  Ay = (30 kN/m)(8 m) + (90 kN/m − 30 kN/m)(8 m) − 252.56676 kN 2 = 227.43324 kN = 227 kN 

Ans.

 (90 kN/m − 30 kN/m)(8 m)   8 m  M A = − M A − (30 kN/m)(8 m)(4 m) −     + By (5.5 m) = 0  2 3 

 M A = −960 kN-m − 640 kN-m + (252.56676 kN)(5.5 m) = −210.88281 kN-m = 211 kN-m (ccw)

Ans.

Shear-force and bending-moment diagrams (b) Magnitude of maximum bending stress: Section properties (from Appendix B): I = 301  106 mm 4 d = 356 mm S = 1,690  103 mm3

Maximum bending moment magnitude Mmax = 210.88281 kN·m Bending stresses at maximum moment (210.88281 kN-m)(356 mm/2)(1,000)2 x = 301  106 mm 4

= 124.7 MPa or, using the tabulated value for the section modulus: (210.88281 kN-m)(1,000) 2 x = 1,690  103 mm 3 = 124.8 MPa

Ans.

P11.39 A W530 × 92 structural steel wideflange shape [E = 200 GPa; I = 554 × 106 mm4] is loaded and supported as shown in Figure P11.39. Determine: (a) the force and moment reactions at supports A and C. (b) the maximum bending stress in the beam. (c) the deflection of the beam at B. FIGURE P11.39

Solution (a) Reactions at A and C. Choose the moment reactions at A and C as the redundants. This will leave a simply supported beam between A and C as the released beam. Determine the slopes at A and C caused by the 150-kN concentrated load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equations from Appendix C: Pb( L2 − b2 ) Pa( L2 − a 2 ) A = − and C = 6 LEI 6 LEI Values: P = 150 kN, L = 10 m, a = 6 m, b = 4 m Calculation:

A = − C =

Pb( L2 − b 2 ) (150 kN)(4 m) 840 kN-m 2 (10 m) 2 − (4 m) 2  = − =− 6 LEI 6(10 m)EI EI

Pa ( L2 − a 2 ) (150 kN)(6 m) 960 kN-m 2 (10 m) 2 − (6 m) 2  = = 6 LEI 6(10 m)EI EI

Determine the slopes at A and C caused by moment reaction MA. [Appendix C, SS beam with concentrated moment at one end.] Relevant equations from Appendix C: ML ML A = − and C = 3EI 6 EI Values: M = MA, L = 10 m Calculation: ML M (10 m) (3.333333 m)M A A = − =− A = − 3EI 3EI EI

C =

ML M A (10 m) (1.666667 m)M A = = 6 EI 6 EI EI

Determine the slopes at A and C caused by moment reaction MC. [Appendix C, SS beam with concentrated moment at one end.] Relevant equations from Appendix C: ML ML A = − and C = 6 EI 3EI Values: M = MC, L = 10 m Calculation: ML M (10 m) (1.666667 m)M C A = − =− C = − 6 EI 6 EI EI

C =

ML M C (10 m) (3.333333 m)M C = = 3EI 3EI EI

Compatibility equation for slope at A: 840 kN-m 2 (3.333333 m)M A (1.666667 m)M C − − − =0 EI EI EI Compatibility equation for slope at C: 960 kN-m 2 (1.666667 m)M A (3.333333 m)M C + + =0 EI EI EI

(a)

(b)

Solve Equations (a) and (b). Equations (a) and (b) can be rewritten as: (3.333333 m)M A + (1.666667 m)M C = −840 kN-m 2

(1.666667 m)M A + (3.333333 m)M C = −960 kN-m 2 and solved simultaneously for MA and MC: M A = −144 kN-m = 144 kN-m (ccw) M C = −216 kN-m = 216 kN-m (cw)

Ans. Ans.

Equilibrium equations for entire beam: M A = −M A + MC − (150 kN)(6 m) + Cy (10 m) = 0

 Cy =

(150 kN)(6 m) + ( −144 kN-m) − ( −216 kN-m) 10 m

= 97.2 kN = 97.2 kN  Fy = Ay + Cy − 150 kN = 0

 Ay = 150 kN − 97.2 kN = 52.8 kN = 52.8 kN 

Ans. Ans.

Shear-force and bending-moment diagrams (b) Magnitude of maximum bending stress: Section properties (from Appendix B): I = 554  106 mm 4 d = 533 mm S = 2,080  103 mm 3

Maximum bending moment magnitude Mmax = 216 kN·m (at C) Bending stresses at maximum moment (216 kN-m)(533 mm/2)(1,000) 2 x = 554  106 mm 4

= 103.9 MPa Ans. or using the tabulated value for the section modulus: (216 kN-m)(1,000) 2 x = 2,080  103 mm3 = 103.8 MPa

Ans.

(c) Beam deflection at B: Determine the deflection at B caused by the 150-kN concentrated load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pab 2 vB = − ( L − a 2 − b2 ) 6LEI Values: P = 150 kN, L = 10 m, a = 6 m, b=4m Calculation:

vB = −

Pab 2 (150 kN)(6 m)(4 m) 2,880 kN-m3 (10 m)2 − (6 m)2 − (4 m)2  = − ( L − a 2 − b2 ) = − 6 LEI 6(10 m)EI EI

Determine the deflection at B caused by MA. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: Mx vB = − (2L2 − 3Lx + x 2 ) 6LEI Values: M = −144 kN·m, L = 10 m, x = 6 m

Calculation: Mx vB = − (2 L2 − 3Lx + x 2 ) 6 LEI

=−

(−144 kN-m)(6 m) 806.4 kN-m3  2(10 m)2 − 3(10 m)(6 m) + (6 m)2  = 6(10 m)EI EI

Determine the deflection at B caused by MC. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: Mx vB = − (2L2 − 3Lx + x 2 ) 6LEI Values: M = −216 kN·m, L = 10 m, x = 4 m Calculation: Mx vB = − (2 L2 − 3Lx + x 2 ) 6 LEI

=−

(−216 kN-m)(4 m) 1,382.4 kN-m3  2(10 m) 2 − 3(10 m)(4 m) + (4 m) 2  = 6(10 m)EI EI

Beam deflection vB: 2,880 kN-m3 806.4 kN-m3 1,382.4 kN-m 3 vB = − + + EI EI EI 3 3 691.2 kN-m 691.2 kN-m =− =− = −0.006238 m = 6.24 mm  EI 110,800 kN-m 2

Ans.

P11.40 A W530 × 92 structural steel wideflange shape [E = 200 GPa; I = 554 × 106 mm4] is loaded and supported as shown in Figure P11.40. Determine: (a) the force and moment reactions at supports A and C. (b) the maximum bending stress in the beam. (c) the deflection of the beam at B. FIGURE P11.40

Solution (a) Reactions at A and C. Choose the moment reactions at A and C as the redundants. This will leave a simply supported beam between A and C as the released beam. Determine the slopes at A and C caused by the 80 kN/m uniformly distributed load. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equations from Appendix C: wa 2 A = − (2 L − a) 2 and 24 LEI

C =

wa 2 (2 L2 − a 2 ) 24 LEI

Values: w = 80 kN/m, L = 9 m, a = 4.5 m Calculation:

A = − C =

wa 2 (80 kN/m)(4.5 m) 2 1,366.875 kN-m 2 2 (2 L − a) 2 = − 2(9 m) − (4.5 m) = −   24 LEI 24(9 m)EI EI

wa 2 (80 kN/m)(4.5 m) 2 1,063.125 kN-m 2  2(9 m) 2 − (4.5 m) 2  = (2 L2 − a 2 ) = 24 LEI 24(9 m)EI EI

Determine the slopes at A and C caused by moment reaction MA. [Appendix C, SS beam with concentrated moment at one end.] Relevant equations from Appendix C: ML ML A = − and C = 3EI 6 EI Values: M = MA, L = 9 m Calculation: ML M (9 m) (3 m)M A A = − =− A = − 3EI 3EI EI

C =

ML M A (9 m) (1.5 m)M A = = 6 EI 6 EI EI

Determine the slopes at A and C caused by moment reaction MC. [Appendix C, SS beam with concentrated moment at one end.] Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Relevant equations from Appendix C: ML ML A = − and C = 6 EI 3EI Values: M = MC, L = 9 m Calculation: ML M (9 m) (1.5 m)M C A = − =− C = − 6 EI 6 EI EI

C =

ML M C (9 m) (3 m)M C = = 3EI 3EI EI

Compatibility equation for slope at A: 1,366.875 kN-m 2 (3 m)M A (1.5 m)M C − − − =0 EI EI EI Compatibility equation for slope at C: 1,063.125 kN-m 2 (1.5 m)M A (3 m)M C + + =0 EI EI EI

(a)

(b)

Solve Equations (a) and (b). Equations (a) and (b) can be rewritten as: (3 m)M A + (1.5 m)M C = −1,366.875 kN-m 2

(1.5 m)M A + (3 m)M C = −1,063.125 kN-m 2 and solved simultaneously for MA and MC: M A = −371.25 kN-m = 371.25 kN-m (ccw)

Ans.

M C = −168.75 kN-m = 168.8 kN-m (cw)

Ans.

Equilibrium equations for entire beam: M A = −M A + MC − (80 kN/m)(4.5 m)(2.25 m) + Cy (9 m) = 0

 Cy =

(80 kN/m)(4.5 m)(2.25 m) + ( −371.25 kN-m) − ( −168.75 kN-m) 9m

= 67.5 kN = 67.5 kN 

Ans.

Fy = Ay + Cy − (80 kN/m)(4.5 m) = 0

 Ay = (80 kN/m)(4.5 m) − 67.5 kN = 292.5 kN = 293 kN 

Ans.

Shear-force and bending-moment diagrams (b) Magnitude of maximum bending stress: Section properties (from Appendix B): I = 554  106 mm 4 d = 533 mm S = 2,080  103 mm 3

Maximum bending moment magnitude Mmax = 371.25 kN·m (at A) Bending stresses at maximum moment (371.25 kN-m)(533 mm/2)(1,000)2 x = 554  106 mm 4

= 178.6 MPa Ans. or using the tabulated value for the section modulus: (371.25 kN-m)(1,000) 2 x = 2,080  103 mm3 = 178.5 MPa

Ans.

(c) Beam deflection at B: Determine the deflection at B caused by the 80 kN/m uniformly distributed load. [Appendix C, SS beam with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C: wa3 vB = − (4 L2 − 7aL + 3a 2 ) 24 LEI Values: w = 80 kN/m, L = 9 m, a = 4.5 m Calculation: wa 3 vB = − (4 L2 − 7aL + 3a 2 ) 24 LEI

=−

(80 kN/m)(4.5 m)3 3,417.1875 kN-m3  4(9 m) 2 − 7(4.5 m)(9 m) + 3(4.5 m) 2  = − 24(9 m)EI EI

Determine the deflection at B caused by MA. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: Mx vB = − (2L2 − 3Lx + x 2 ) 6LEI Values: M = −371.25 kN·m, L = 9 m, x = 4.5 m Calculation: Mx vB = − (2 L2 − 3Lx + x 2 ) 6 LEI

=−

(−371.25 kN-m)(4.5 m) 1,879.4531 kN-m3  2(9 m) 2 − 3(9 m)(4.5 m) + (4.5 m) 2  = 6(9 m)EI EI

Determine the deflection at B caused by MC. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: Mx vB = − (2L2 − 3Lx + x 2 ) 6LEI Values: M = −168.75 kN·m, L = 9 m, x = 4.5 m Calculation: Mx vB = − (2 L2 − 3Lx + x 2 ) 6 LEI

=−

(−168.75 kN-m)(4.5 m) 854.2969 kN-m3  2(9 m) 2 − 3(9 m)(4.5 m) + (4.5 m) 2  = 6(9 m)EI EI

Beam deflection vB: 3, 417.1875 kN-m3 1,879.4531 kN-m3 854.2969 kN-m3 vB = − + + EI EI EI 3 3 683.4375 kN-m 683.4375 kN-m =− =− = −0.006168 m = 6.17 mm  EI 110,800 kN-m 2

Ans.

P11.41 A W360 × 72 structural steel [E = 200 GPa] wide-flange shape is loaded and supported as shown in Figure P11.41. The beam is supported at B by 20-mm-diameter solid aluminum [E = 70 GPa] rod. After a concentrated load of 40 kN is applied to the tip of the cantilever, determine: (a) the force produced in the aluminum rod. (b) the maximum bending stress in the beam. (c) the deflection of the beam at B. FIGURE P11.41

Solution Section properties: W360 × 72: I beam = 201  106 mm 4

d = 351 mm

Ebeam = 200,000 MPa

S beam = 1,150  103 mm3 Rod (1):

A1 =

 4

(20 mm) 2 = 314.159266 mm 2

E1 = 70,000 MPa

Downward deflection of W360 × 72 beam at B due to 40-kN concentrated load. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: Px 2 vB = − (3L − x) (elastic curve) 6 EI Values: P = 40 kN, L = 5 m, x = 3.6 m, EI = 40,200 kN·m2 Calculation: Px 2 (40 kN)(3.6 m)2 vB = − (3L − x) = − 3(5 m) − (3.6 m) = −24.50149  10−3 m 2  6 EI 6(40,200 kN-m )

Upward deflection of W360 × 72 beam at B due to force F1 in rod (1). [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: PL3 vB = − 3EI Values: P = −F1, L = 3.6 m, EI = 40,200 kN·m2 Calculation: PL3 (− F1 )(3.6 m)3 vB = − =− = (386.8657  10−6 m/kN)F1 3EI 3(40,200 kN-m2 ) Elongation of aluminum rod (1) due to force F1. FL (3 m) 1 = 1 1 = F1 = (136.4185  10−6 m/kN)F1 2 2 A1E1 (314.159266 mm )(70,000 N/mm )(1 kN/1,000 N) The deflection of the W360 × 72 beam at B will not equal zero in this instance because the rod that supports the beam at B will elongate, thus permitting the W360 × 72 beam to deflect downward. Therefore,

vB = −1 = −(136.4185  10−6 m/kN)F1 Compatibility equation at B: The sum of the downward deflection caused by the 40-kN concentrated load and the upward deflection caused by the force in rod (1) must equal the elongation of the aluminum rod. Elongation of the aluminum rod will produce a downward (i.e., negative) deflection of the W360 × 72 beam at B.

−24.50149  10−3 m + (386.8657  10−6 m/kN)F1 = −(136.4185  10−6 m/kN)F1  F1 =

24.50149  10−3 m = 46.82254 kN = 46.8 kN (T) 523.2842  10−6 m/kN

Ans.

(b) Determine maximum bending stress in the W360 × 72 beam: Maximum bending moment magnitude Mmax = 56 kN·m (at B) Bending stress at maximum moment (56 kN-m)(351 mm/2)(1,000) 2 x = 201  106 mm 4

= 48.9 MPa Ans. or using the tabulated value for the section modulus of the W360 × 72 beam: (56 kN-m)(1,000)2 Ans. x = = 48.7 MPa 1,150  103 mm3

(c) Beam deflection at B. The beam deflection at B is equal to the elongation of the aluminum rod: FL (46,822.54 N)(3,000 mm) vB = −1 = − 1 1 = − = −6.38746 mm = 6.39 mm  A1E1 (314.159266 mm2 )(70,000 N/mm2 )

Ans.

P11.42 A W18 × 55 structural steel [E = 29,000 ksi] wide-flange shape is loaded and supported as shown in Figure P11.42. The beam is supported at C by a ¾-in.-diameter aluminum [E = 10,000 ksi] rod, which has no load before the distributed load is applied to the beam. After a distributed load of 4 kips/ft is applied to the beam, determine: (a) the force carried by the aluminum rod. (b) the maximum bending stress in the steel beam. (c) the deflection of the beam at C. FIGURE P11.42

Solution Section properties: Beam: I beam = 890 in.4

d = 18.1 in.

Ebeam = 29,000 ksi

S beam = 98.3 in.

Rod (1):

A1 =

 4

(0.75 in.) 2 = 0.4417865 in.2

E1 = 10,000 ksi

(a) Force in the aluminum rod. The reaction force from rod (1) will be taken as the redundant, leaving a cantilever beam as the released beam. For this analysis, a tension force is assumed to exist in axial member (1). Downward deflection of W18 × 55 beam at C due to 4 kips/ft uniformly distributed load. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equations from Appendix C: wL4 wL3 vB = − and B = − 8 EI 6 EI Values: w = 4 kips/ft, L = 11 ft, EI = 25.81×106 kip-in.2 Calculation: wL4 (4 kips/ft)(11 ft) 4 (12 in./ft)3 vB = − =− = −0.490113 in. 8 EI 8(25.81  106 kip-in.2 )

B = −

wL3 (4 kips/ft)(11 ft)3 (12 in./ft) 2 =− = −4.950639  10−3 rad 6 2 6 EI 6(25.81  10 kip-in. )

vC = −0.490113 in. + (5 ft)(12 in./ft)( − 4.950639  10−3 rad) = −0.787152 in.

Upward deflection of W18 × 55 beam at C due to force F1 in rod (1). [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: PL3 vC = − 3EI Values: P = −F1, L = 16 ft, EI = 25.81×106 kip-in.2 Calculation: PL3 (− F1 )(16 ft)3 (12 in./ft)3 vC = − =− = (91.41015  10−3 in./kip)F1 3EI 3(25.81  106 kip-in.2 ) Elongation of aluminum rod (1) due to force F1. FL (14 ft)(12 in./ft) 1 = 1 1 = F1 = (38.02742  10−3 in./kip)F1 2 A1E1 (0.4417865 in. )(10,000 ksi) The deflection of the W18 × 55 beam at C will not equal zero in this instance because the rod that supports the beam at C elongates, thus permitting the W18 × 55 beam to deflect downward. Therefore,

vC = −1 = −(38.02742  10−3 in./kip)F1 Compatibility equation at C: The sum of the downward deflection caused by the uniformly distributed load and the upward deflection caused by the force in rod (1) must equal the elongation of the aluminum rod. Elongation of the aluminum rod will produce a downward (i.e., negative) deflection of the W18 × 55 beam at C.

−0.787152 in. + (91.41015  10−3 in./kip)F1 = −(38.02742  10−3 in./kip)F1  F1 =

0.787152 in. = 6.081323 kips = 6.08 kips (T) 129.4376  10−3 in./kip

Ans.

(b) Determine maximum bending stress in W18 × 55 beam: Maximum bending moment magnitude Mmax = 144.70 kip·ft (at A) Bending stress at maximum moment (144.70 kip-ft)(18.1 in./2)(12 in./ft) x = 890 in.4 Ans. = 17.66 ksi or using the tabulated value for the section modulus of the W18 × 55 beam: (144.70 kip-ft)(12 in./ft) x = 98.3 in.3 Ans. = 17.66 ksi

(c) Beam deflection at C. The beam deflection at C is equal to the elongation of the aluminum rod: FL (6.081323 kips)(14 ft)(12 in./ft) vC = −1 = − 1 1 = − = −0.231257 in. = 0.231 in.  A1E1 (0.4417865 in.2 )(10,000 ksi)

Ans.

P11.43 A W250 × 32.7 structural steel [E = 200 GPa] wide-flange shape is loaded and supported as shown in Figure P11.43. A uniformly distributed load of 16 kN/m is applied to the beam, causing the roller support at B to settle downward (i.e., displace downward) by 15 mm. Determine: (a) the reactions at supports A, B, and C. (b) the maximum bending stress in the beam. FIGURE P11.43

Solution Section properties: W 250  32.7 :

I = 49.1 106 mm4

d = 259 mm

S = 380  103 mm3

(a) Reactions at A, B, and C. Choose the reaction force at B as the redundant; therefore, the released beam is simply supported between A and C. Consider downward deflection of simply supported beam at B due to uniformly distributed load. [Appendix C, SS beam with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C: wx 3 vB = − ( L − 2 Lx 2 + x3 ) (elastic curve) 24 EI Values: w = 16 kN/m, L = 10 m, x = 4 m Calculation: wx vB = − ( L3 − 2 Lx 2 + x 3 ) 24 EI =−

(16 kN/m)(4 m) 1,984 kN-m 3 (10 m)3 − 2(10 m)(4 m) 2 + (4 m)3  = − 24 EI EI

=−

( − By )(4 m)(6 m) 6(10 m) EI

(10 m) 2 − (4 m) 2 − (6 m) 2  =

(19.2 m3 ) By EI

Compatibility equation for deflection at B: EI = (200,000 N/mm 2 )(49.1  106 mm 4 ) = 9.82  1012 N-mm 2 = 9.82  103 kN-m2 3

1,984 kN-m3 (19.2 m ) By − + = −0.015 m EI EI ( −0.015 m)(9.82  103 kN-m 2 ) + 1,984 kN-m3  By = = 95.6615 kN = 95.7 kN  19.2 m3

Ans.

Equilibrium equations for entire beam: M A = By (4 m) + Cy (10 m) − (16 kN/m)(10 m)(5 m) = 0

 Cy =

(16 kN/m)(10 m)(5 m) − (95.6615 kN)(4 m) 10 m

= 41.7354 kN = 41.7 kN 

Ans.

Fy = Ay + By + Cy − (16 kN/m)(10 m) = 0

 Ay = (16 kN/m)(10 m) − 95.6615 kN − 41.7354 kN = 22.6031 kN = 22.6 kN 

Ans.

Shear-force and bending-moment diagrams (b) Determine maximum bending stress in W250 × 32.7 beam: Maximum bending moment magnitude Mmax = 54.4327 kN·m Bending stress at maximum moment (54.4327 kN-m)(259 mm/2)(1,000) 2 x = 49.1  106 mm 4

= 143.6 MPa Ans. or using the tabulated value for the section modulus of the W250 × 32.7 beam: (54.4327 kN-m)(1,000)2 x = 380  103 mm3 = 143.2 MPa Ans.

P11.44 A timber [E = 12 GPa] beam is loaded and supported as shown in Figure P11.44. The cross section of the timber beam is 100-mm. wide and 300-mm deep. The beam is supported at B by a 12-mm-diameter steel [E = 200 GPa] rod, which has no load before the distributed load is applied to the beam. After a distributed load of 7 kN/m is applied to the beam, determine: (a) the force carried by the steel rod. (b) the maximum bending stress in the timber beam. (c) the deflection of the beam at B. FIGURE P11.44

Solution Section properties:

Beam:

I beam =

Rod (1):

A1 =

 4

(100 mm)(300 mm)3 = 225  106 mm 4 12

Ebeam = 12,000 MPa

(12 mm)2 = 113.097336 mm2

E1 = 200,000 MPa

(a) Force carried by the steel rod. The reaction force from rod (1) will be taken as the redundant, leaving a simply supported beam between A and C as the released beam. For this analysis, a tension force is assumed to exist in axial member (1).

Downward deflection of wood beam at B due to 7 kN/m uniformly distributed load. [Appendix C, SS beam with uniformly distributed load.] Relevant equation from Appendix C: wx 3 vB = − ( L − 2 Lx 2 + x3 ) (elastic curve) 24 EI Values: w = 7 kN/m, L = 6 m, x = 4 m, EI = 2,700 kN·m2 Calculation: wx vB = − ( L3 − 2 Lx 2 + x3 ) 24 EI (7 kN/m)(4 m) (6 m)3 − 2(6 m)(4 m) 2 + (4 m)3  = −38.02469  10−3 m =− 2  24(2,700 kN-m )

Upward deflection of wood beam at B due to force F1 in rod (1). [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pab 2 vB = − ( L − a 2 − b2 ) 6LEI Values: P = −F1, L = 6 m, a = 4 m, b = 2 m, EI = 2,700 kN·m2 Calculation: Pab 2 vB = − (L − a 2 − b2 ) 6 LEI (− F1 )(4 m)(2 m) (6 m) 2 − (4 m) 2 − (2 m) 2  = (1.316872  10−3 m/kN)F1 =− 6(6 m)(2,700 kN-m 2 )  Elongation of steel rod (1) due to force F1. FL (5 m) 1 = 1 1 = F1 = (221.0485  10−6 m/kN)F1 2 2 A1E1 (113.097336 mm )(200,000 N/mm )(1 kN/1,000 N) The deflection of the wood beam at B will not equal zero in this instance because the rod that supports the beam at B will elongate, thus permitting the wood beam to deflect downward. Therefore,

vB = −1 = −(221.0485  10−6 m/kN)F1 Compatibility equation at B: The sum of the downward deflection caused by the uniformly distributed load and the upward deflection caused by the force in rod (1) must equal the elongation of the steel rod. Elongation of the steel rod will produce a downward (i.e., negative) deflection of the wood beam at B.

−38.02469  10−3 m + (1.316872  10−3 m/kN)F1 = −(221.0485  10−6 m/kN)F1 38.02469  10−3 m  F1 = = 24.72474 kN = 24.7 kN (T) 1.537921  10−3 m/kN

Ans.

(b) Determine maximum bending stress in wood beam: Maximum bending moment magnitude Mmax = 11.6270 kN·m Bending stress at maximum moment (11.6270 kN-m)(300 mm/2)(1,000) 2 x = 225  106 mm 4

= 7.75 MPa

Ans.

(c) Beam deflection at B. The beam deflection at B is equal to the elongation of the steel rod: FL vB = −1 = − 1 1 A1E1 =−

(24.72474 kN)(5 m)(1,000 N/kN)(1,000 mm/m) (113.097336 mm 2 )(200,000 N/mm 2 )

= −5.465367 mm = 5.47 mm 

Ans.

P11.45 A W360 × 72 structural steel [E = 200 GPa] wide-flange shape is loaded and supported as shown in Figure P11.45. The beam is supported at B by a timber [E = 12 GPa] post having a cross-sectional area of 20,000 mm2. After a uniformly distributed load of 50 kN/m is applied to the beam, determine: (a) the reactions at supports A, B, and C. (b) the maximum bending stress in the beam. (c) the deflection of the beam at B. FIGURE P11.45

Solution Section properties: W360 × 72: I beam = 201  106 mm 4

Sbeam = 1,150  10 mm 3

Post (1):

d = 351 mm

Ebeam = 200,000 MPa

A1 = 20,000 mm2

E1 = 12,000 MPa

(a) Reactions at supports A, B, and C. The reaction force from post (1) will be taken as the redundant, leaving a simply supported beam as the released beam. To be consistent with earlier sign conventions (e.g., Chapter 5), we will assume that the force in the axial member is tension (even though intuitively we recognize that the post must be in compression).

Downward deflection of W360 × 72 beam at B due to 50 kN/m uniformly distributed load. [Appendix C, SS beam with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C: wa3 vB = − (4 L2 − 7aL + 3a 2 ) 24 LEI Values: P = 50 kN/m, L = 13 m, a = 6 m, EI = 40,200 kN·m2 Calculation: wa3 vB = − (4 L2 − 7aL + 3a 2 ) 24 LEI

=−

(50 kN/m)(6 m)3  4(13 m)2 − 7(6 m)(13 m) + 3(6 m) 2  = −204.937  10−3 m 2  24(13 m)(40,200 kN-m )

Since we are assuming that a tension force exists in post (1), the tension force from the post will cause a downward deflection of W360 × 72 beam at B. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pab 2 vB = − ( L − a 2 − b2 ) 6LEI Values: P = F1, L = 13 m, a = 6 m, b = 7 m, EI = 40,200 kN·m2 Calculation: Pab 2 vB = − ( L − a 2 − b2 ) 6 LEI ( F1 )(6 m)(7 m) (13 m) 2 − (6 m) 2 − (7 m) 2  = −(1.12514  10−3 m/kN)F1 =− 6(13 m)(40, 200 kN-m 2 )  Deformation of wood post (1) due to force F1. FL (5 m) 1 = 1 1 = F1 = (20.8333  10−6 m/kN)F1 2 2 A1E1 (20,000 mm )(12,000 N/mm )(1 kN/1,000 N) The deflection of the W360 × 72 beam at B will not equal zero in this instance because the post deforms. If we are consistent and assume that there is tension in the post, then the steel beam must deflect upward at B. Therefore,

vB = 1 = (20.8333  10−6 m/kN)F1 Compatibility equation at B: The sum of the downward deflection caused by the 50 kN/m uniformly distributed load and the downward deflection caused by the force in post (1) must equal the deformation of the wood post.

−204.937  10−3 m − (1.12514  10 −3 m/kN)F1 = (20.8333  10 −6 m/kN)F1  F1 =

204.937  10−3 m = −178.832 kN = 178.8 kN (C) = By −1.14598  10−3 m/kN

Ans.

Equilibrium equations for entire beam: M A = − F1 (6 m) + Cy (13 m) − (50 kN/m)(6 m)(3 m) = 0

 Cy =

(50 kN/m)(6 m)(3 m) + ( −178.832 kN)(6 m) = −13.3071 kN = 13.31 kN  13 m

Ans.

Fy = Ay − F1 + Cy − (50 kN/m)(6 m) = 0 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

 Ay = (50 kN/m)(6 m) + ( −178.832 kN) − ( −13.3071 kN) = 134.475 kN = 134.5 kN 

Ans. (b) Determine maximum bending stress in the W360 × 72 beam: Maximum bending moment magnitude Mmax = 180.836 kN·m Bending stress at maximum moment (180.836 kN-m)(351 mm/2)(1,000) 2 x = 201  106 mm 4

= 157.9 MPa Ans. or using the tabulated value for the section modulus of the W360 × 72 beam: (180.836 kN-m)(1,000) 2 x = 1,150  103 mm 3 = 157.2 MPa

Ans.

(c) Beam deflection at B. The beam deflection at B is equal to the deformation of the wood post: FL ( − 178.832 kN)(5 m)(1,000 N/kN)(1,000 mm/m) vB =  1 = 1 1 = A1E1 (20,000 mm 2 )(12,000 N/mm 2 ) = −3.7257 mm = 3.73 mm 

Ans.

P11.46 A W530 × 66 structural steel [E = 200 GPa] wide-flange shape is loaded and supported as shown in Figure P11.46. A uniformly distributed load of 70 kN/m is applied to the beam, causing the roller support at B to settle downward (i.e., displace downward) by 10 mm. Determine: (a) the reactions at supports A and B. (b) the maximum bending stress in the beam.

FIGURE P11.46

Solution Section properties: W530  66: I = 351  106 mm 4

d = 526 mm

S = 1,340  103 mm3

(a) Reactions at supports A and B. The reaction force at B will be taken as the redundant, leaving a cantilever beam between A and C as the released beam. Downward deflection of beam at B due to 70 kN/m uniformly distributed load. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wx 2 vB = − (6 L2 − 4 Lx + x 2 ) (elastic curve) 24 EI Values: w = 70 kN/m, L = 6 m, x = 4.5 m, EI = 70,200 kN·m2 Calculation: wx 2 vB = − (6 L2 − 4 Lx + x 2 ) 24 EI (70 kN/m)(4.5 m) 2 6(6 m) 2 − 4(6 m)(4.5 m) + (4.5 m) 2  = −0.107903 m =− 24(70, 200 kN-m 2 )  Upward deflection of beam at B due to reaction force By. [Appendix C, SS beam with concentrated load at tip.] Relevant equation from Appendix C: PL3 vB = − 3EI Values: P = −By, L = 4.5 m, EI = 70,200 kN·m2 Calculation: PL3 vB = − 3EI ( − By )(4.5 m)3 =− = (432.6923  10 −6 m/kN)By 3(70, 200 kN-m 2 ) Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Compatibility equation at B: The sum of the downward deflection caused by the uniformly distributed load and the upward deflection caused by the reaction force By must equal the support settlement:

−0.107903 m + (432.6923  10 −6 m/kN)By = −0.010 m  By =

97.90264  10 −3 m = 226.2639 kN = 226 kN  432.6923  10 −6 m/kN

Ans.

Equilibrium equations for entire beam: M A = −M A + By (4.5 m) − (70 kN/m)(6 m)(3 m) = 0  M A = (226.2639 kN)(4.5 m) − (70 kN/m)(6 m)(3 m) = −241.8125 kN-m = 242 kN-m (ccw)

Ans.

Fy = Ay + By − (70 kN/m)(6 m) = 0

 Ay = (70 kN/m)(6 m) − 226.2639 kN = 193.7361 kN = 193.7 kN 

Ans.

(b) Determine maximum bending stress in beam: Maximum bending moment magnitude Mmax = 241.8125 kN·m (at A) Bending stress at maximum moment (241.8125 kN-m)(526 mm/2)(1,000) 2 x = 351  106 mm 4

= 181.187 MPa = 181.2 MPa Ans. or using the tabulated value for the section modulus: (241.8125 kN-m)(1,000) 2 x = 1,340  103 mm3 = 180.457 MPa = 180.5 MPa

Ans.

P11.47 Steel beam (1) carries a concentrated load of P = 13 kips that is applied at midspan, as shown in Figure P11.47/48. The steel beam is supported at ends A and B by nondeflecting supports and at its middle by simply supported timber beam (2). In the unloaded condition, steel beam (1) touches but exerts no force on timber beam (2). The length of the steel beam is L1 = 30 ft, and its flexural rigidity is EI1 = 7.2 × 106 kip-in.2. The length and the flexural rigidity of the timber beam are L2 = 20 ft and EI2 = 1.0 × 106 kip-in.2, respectively. Determine the vertical reaction force that acts (a) on the steel beam at A, and (b) on the timber beam at C.

FIGURE P11.47/48

Solution Case 1—Simply Supported Steel Beam with Downward Concentrated Load at Midspan Remove wood beam (2) and consider simply supported steel beam (1) subjected to a concentrated load of P = 13 kips at midspan. The deflection of this beam must be determined at point D. From Appendix C, the deflection of beam (1) at midspan is given by: PL13 (a) vD = − 48E1I1 Case 2—Simply Supported Steel Beam with Upward Concentrated Load at Midspan Timber beam (2) exerts an upward reaction force on the steel beam at D. Consider steel beam (1) subjected to this upward reaction force Dy. Using the appropriate beam deflection formula from Appendix C, the midspan deflection of simply supported beam (1) due to the upward reaction force Dy is given by: ( − Dy ) L13 Dy L13 (b) vD = − = 48E1I1 48E1I1 Case 3—Simply Supported Timber Beam with Downward Concentrated Load at Midspan Timber beam (2) supplies an upward force to the steel beam at D. Conversely, steel beam (1) exerts an equal magnitude force on the timber beam, causing it to deflect downward. The downward deflection of beam (2) that is produced by reaction force Dy is given by: Dy L32 (c) vD = − 48E2 I 2 Compatibility Equation The sum of the downward deflection of the steel beam due to the concentrated load [Equation (a)] and the upward deflection produced by the reaction force supplied by the timber beam [Equation (b)] must equal the downward deflection of the timber beam [Equation (c)]. These three equations for the deflection at D are combined in a compatibility equation: Dy L13 Dy L32 PL13 (d) − + =− 48E1I1 48E1I1 48E2 I 2 The only unknown term in this equation is the reaction force Dy. Rearrange this equation to obtain:  L3 L3  PL3 Dy  1 + 2  = 1 (e)  E1I1 E2 I 2  E1I1

(b) Reaction at C. The reaction at C on timber beam (2) can now be calculated from: L  M E = C y L2 − Dy  2  = 0  2 Dy 4.149 kips  Cy = = = 2.074 kips = 2.07 kips  2 2

Ans.

P11.48 In Figure P11.47/48, a W10 × 45 steel beam (1) carries a concentrated load of P = 9 kips that is applied at midspan. The steel beam is supported at ends A and B by nondeflecting supports and at its middle by simply supported timber beam (2) that is 8-in.-wide and 12-in.-deep. In the unloaded condition, steel beam (1) touches but exerts no force on timber beam (2). The length of the steel beam is L1 = 24 ft, and its modulus of elasticity is E1 = 29 × 103 ksi. The length and the modulus of elasticity of the timber beam are L2 = 15 ft and E2 = 1.8 × 103 ksi, respectively. Determine the maximum flexural stress (a) in the steel beam. (b) in the timber beam. (c) in the steel beam if the timber beam is removed. FIGURE P11.47/48

Solution Case 1—Simply Supported Steel Beam with Downward Concentrated Load at Midspan Remove timber beam (2) and consider simply supported steel beam (1) subjected to a concentrated load of P = 9 kips at midspan. The deflection of this beam must be determined at point D. From Appendix C, the deflection of beam (1) at midspan is given by: PL13 (a) vD = − 48E1I1 Case 2—Simply Supported Steel Beam with Upward Concentrated Load at Midspan Timber beam (2) exerts an upward reaction force on the steel beam at D. Consider steel beam (1) subjected to this upward reaction force Dy. Using the appropriate beam deflection formula from Appendix C, the midspan deflection of simply supported beam (1) due to the upward reaction force Dy is given by: ( − Dy ) L13 Dy L13 (b) vD = − = 48E1I1 48E1I1 Case 3—Simply Supported Timber Beam with Downward Concentrated Load Timber beam (2) supplies an upward force to the steel beam at D. Conversely, steel beam (1) exerts an equal magnitude force on the timber beam, causing it to deflect downward. The downward deflection of beam (2) that is produced by reaction force Dy is given by: Dy L32 (c) vD = − 48E2 I 2 Compatibility Equation The sum of the downward deflection of the steel beam due to the concentrated load [Equation (a)] and the upward deflection produced by the reaction force supplied by the timber beam [Equation (b)] must equal the downward deflection of the timber beam [Equation (c)]. These three equations for the deflection at D are combined in a compatibility equation: Dy L13 Dy L32 PL13 (d) − + =− 48E1I1 48E1I1 48E2 I 2 The only unknown term in this equation is the reaction force Dy. Rearrange this equation to obtain:  L3 L3  PL3 Dy  1 + 2  = 1 (e)  E1I1 E2 I 2  E1I1 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Before beginning the calculation, pay special attention to the distinction between those properties that apply to the steel beam (i.e., L1, I1, and E1) and those that apply to the timber beam (i.e., L2, I2, and E2). For instance, the flexural stiffness term EI appears in each term, but EI for the timber beam is much different than EI for the steel beam. Calculate the reaction force exerted on steel beam (1) using the following values: Steel Beam (1) Properties Timber Beam (2) Properties P = 9 kips L2 = 15 ft = 180 in. L1 = 24 ft = 288 in. (8 in.)(12 in.)3 4 = 1,152 in.4 I = 2 I1 = 248 in. (from Appendix B for W10 × 45) 12 E1 = 29×103 ksi E2 = 1.8×103 ksi Substitute these values into Equation (e) and compute Dy = 4.873 kips. (a) Maximum flexural stress in the steel beam. The reaction at A on steel beam (1) can now be calculated from: L L M B = Ay L1 − P  1  + Dy  1  = 0  2  2  24 ft   24 ft  (9 kips)  − (4.873 kips)    2   2   Ay = 24 ft (9 kips − 4.873 kips) = = 2.063 kips  2 The maximum bending moment occurs at midspan for the steel beam. Using the reaction force at A, calculate the maximum bending moment as: L  24 ft  M max = Ay  1  = (2.063 kips)  = 24.760 kip-ft = 297.116 kip-in.  2  2  The depth of the W10 × 45 shape is 10.1 in. (from Appendix B). The bending stress in steel beam (1) at midspan is calculated as:  10.1 in. (297.116 kip-in.)   2  M c  max = max = = 6.05 ksi Ans. I 248 in.4

(b) Maximum flexural stress in the timber beam. The reaction at C on timber beam (2) can now be calculated from: L  M E = C y L2 − Dy  2  = 0  2 Dy

4.873 kips = 2.437 kips  2 2 The maximum bending moment in the timber beam occurs at D. Using the reaction force at C, calculate the maximum bending moment as: L   15 ft  M max = Cy  2  = (2.437 kips)  = 18.275 kip-ft = 219.302 kip-in.  2  2  The depth of the timber beam is 12 in. The bending stress in the timber beam at D is calculated as:  Cy =

 12 in. (219.302 kip-in.)   2  M max c  max = = = 1.142 ksi I 1,152 in.4

Ans.

(c) Maximum flexural stress in the steel beam if the timber beam is removed. The reaction at A on steel beam (1) if the timber beam is removed is simply P/2 = 4.5 kips. The maximum bending moment is thus: L  24 ft  M max = Ay  1  = (4.5 kips)  = 54 kip-ft = 648 kip-in.  2  2  and the bending stress in the steel beam at midspan is:  10.1 in. (648 kip-in.)   2  M max c  max = = = 13.20 ksi Ans. I 248 in.4

P11.49 Two steel beams support a concentrated load of P = 45 kN, as shown in Figure P11.49/50. Beam (1) is supported by a fixed support at A and by a simply supported beam (2) at D. In the unloaded condition, beam (1) touches but exerts no force on beam (2). The beam lengths are a = 4.0 m, b = 1.5 m, and L2 = 6 m. The flexural rigidities of the beams are EI1 = 40,000 kN-m2 and EI2 = 14,000 kN-m2. Determine the deflection of beam (1) (a) at D and (b) at B.

FIGURE P11.49/50

Solution Case 1—Cantilever Beam with Downward Concentrated Load at Tip Remove beam (2) and consider cantilever beam (1) subjected to a concentrated load of P = 45 kN at its tip. The deflection of this beam must be determined at point D. From Appendix C, the elastic curve equation can be adapted to give the deflection of beam (1) at D: Pa 2 Pa 2 (a) vD = − (3(a + b) − a) = − (2a + 3b) 6 E1I1 6 E1I1 Case 2—Cantilever Beam with Upward Concentrated Load at D Beam (2) exerts an upward reaction force on the beam (1) at D. Consider beam (1) subjected to this upward reaction force Dy. Using the appropriate beam deflection formula from Appendix C, the deflection of the cantilever beam at the point of application of the upward reaction force Dy is given by: ( − Dy ) a 3 D y a 3 (b) vD = − = 3E1I1 3E1I1 Case 3—Simply Supported Beam with Downward Concentrated Load at Midspan Beam (2) supplies an upward force to beam (1) at D. Conversely, beam (1) exerts an equal magnitude force on beam (2), causing it to deflect downward. The downward deflection of beam (2) that is produced by reaction force Dy is given by: Dy L32 (c) vD = − 48E2 I 2 Compatibility Equation The sum of the downward deflection of beam (1) due to the concentrated load [Equation (a)] and the upward deflection produced by the reaction force supplied by beam (2) [Equation (b)] must equal the downward deflection of beam (2) [Equation (c)]. These three equations for the deflection at D are combined in a compatibility equation: Dy a 3 Dy L32 Pa 2 (d) − (2a + 3b) + =− 6 E1I1 3E1I1 48E2 I 2 The only unknown term in this equation is the reaction force Dy. Rearrange this equation to obtain:  a3 L32  Pa 2 Dy  + (2a + 3b) (e) =  3E1I1 48E2 I 2  6 E1I1 Before beginning the calculation, pay special attention to the distinction between those properties that apply to beam (1) (i.e., L1, I1, and E1) and those that apply to beam (2) (i.e., L2, I2, and E2). For instance, Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

the flexural stiffness term EI appears in each term, but EI for beam (2) is much different than EI for beam (1). Calculate the reaction force exerted on beam (1) using the following values: Beam (1) Properties Beam (2) Properties P = 45 kN L2 = 6 m a = 4.0 m EI2 = 14,000 kN-m2 b = 1.5 m EI1 = 40,000 kN-m2 Substitute these values into Equation (e) and compute Dy = 43.872 kN. (a) Determine the deflection of beam (1) at D. The deflection of beam (1) at D is equal to the deflection of beam (2) at D; therefore, calculate the deflection from: Dy L32 (43.872 kN)(6 m)3 Ans. vD = − =− = −0.01410 m = 14.10 mm  48E2 I 2 48(14,000 kN-m2 ) (b) Determine the deflection of beam (1) at B. Consider downward deflection of cantilever beam at B due to P. [Appendix C, Cantilever beam with concentrated load at tip.] P ( a + b) 3 (45 kN)(4.0 m + 1.5 m)3 vB = − =− = −0.062391 m 3E1I1 3(40,000 kN-m2 ) Consider upward deflection of cantilever beam at B due to concentrated load Dy. [Appendix C, Cantilever beam with concentrated load at tip.] Dy a 3 (43.872 kN)(4.0 m)3 vD = = = 0.023398 m 3E1I1 3(40,000 kN-m 2 )

D =

Dy a 2 2 E1I1

(43.872 kN)(4.0 m) 2 = 0.00877440 rad 2(40,000 kN-m 2 )

vB = vD +  D b = 0.023398 m + (0.00877440 rad)(1.5 m) = 0.036560 m Deflection at B. vB = −0.062391 m + 0.036560 m = −0.025831 m = 25.8 mm  P11.50 Two steel beams support a concentrated load of P = 60 kN, as shown in Figure P11.49/50. Beam (1) is supported by a fixed support at A and by a simply supported beam (2) at D. In the unloaded condition, beam (1) touches but exerts no force on beam (2). The beam lengths are a = 5.0 m, b = 2.0 m, and L2 = 8 m. The flexural rigidities of the beams are EI1 = 40,000 kN-m2 and EI2 = 25,000 kN-m2. Determine (a) the reactions that act on beam (1) at A, and (b) the reaction on beam (2) at C.

Ans.

FIGURE P11.49/50

Solution Case 1—Cantilever Beam with Downward Concentrated Load at Tip Remove beam (2) and consider cantilever beam (1) subjected to a concentrated load of P = 45 kN at its tip. The deflection of this beam must be determined at point D. From Appendix C, the elastic curve equation can be adapted to give the deflection of beam (1) at D: Pa 2 Pa 2 (a) vD = − (3(a + b) − a) = − (2a + 3b) 6 E1I1 6 E1I1 Case 2—Cantilever Beam with Upward Concentrated Load at D Beam (2) exerts an upward reaction force on the beam (1) at D. Consider beam (1) subjected to this upward reaction force Dy. Using the appropriate beam deflection formula from Appendix C, the deflection of the cantilever beam at the point of application of the upward reaction force Dy is given by: ( − Dy ) a 3 D y a 3 (b) vD = − = 3E1I1 3E1I1 Case 3—Simply Supported Beam with Downward Concentrated Load at Midspan Beam (2) supplies an upward force to beam (1) at D. Conversely, beam (1) exerts an equal magnitude force on beam (2), causing it to deflect downward. The downward deflection of beam (2) that is produced by reaction force Dy is given by: Dy L32 (c) vD = − 48E2 I 2 Compatibility Equation The sum of the downward deflection of beam (1) due to the concentrated load [Equation (a)] and the upward deflection produced by the reaction force supplied by beam (2) [Equation (b)] must equal the downward deflection of beam (2) [Equation (c)]. These three equations for the deflection at D are combined in a compatibility equation: Dy a 3 Dy L32 Pa 2 (d) − (2a + 3b) + =− 6 E1I1 3E1I1 48E2 I 2 The only unknown term in this equation is the reaction force Dy. Rearrange this equation to obtain:  a3 L32  Pa 2 (e) Dy  + (2a + 3b) = 3 E I 48 E I 6 E I  11 2 2 1 1

Before beginning the calculation, pay special attention to the distinction between those properties that apply to beam (1) (i.e., L1, I1, and E1) and those that apply to beam (2) (i.e., L2, I2, and E2). For instance, the flexural stiffness term EI appears in each term, but EI for beam (2) is much different than EI for beam (1). Calculate the reaction force exerted on beam (1) using the following values: Beam (1) Properties Beam (2) Properties P = 60 kN L2 = 8 m a = 5.0 m EI2 = 25,000 kN-m2 b = 2.0 m EI1 = 40,000 kN-m2 Substitute these values into Equation (e) and compute Dy = 68.1044 kN. Equilibrium Equations (a) Reactions at A. The reactions at A on beam (1) can now be calculated from: M A = M A − P ( a + b ) + Dy ( a ) = 0  M A = (60 kN)(5.0 m + 2.0 m) − (68.1044 kN)(5.0 m) = 79.478 kN-m = 79.5 kN-m (ccw)

Ans.

Fy = Ay − P + Dy = 0  Ay = 60 kN − 68.1044 kN = −8.1044 kN = 8.10 kN 

Ans.

(b) Reaction at C. The reaction at C on beam (2) can now be calculated from: L  M E = C y L2 − Dy  2  = 0  2  Cy =

Dy 2

68.104 kN = 34.052 kips = 34.1 kN  2

Ans.

P12.1 A compound shaft consists of segment (1), which has a diameter of 1.25 in., and segment (2), which has a diameter of 1.75 in. The shaft is subjected to a tensile axial load of P = 9 kips and torques TA = 7 kip·in. and TB = 28 kip·in., which act in the directions shown in Figure P12.1. Determine the normal and shear stresses at (a) point H and (b) point K. For each point, show the stresses on a stress element.

FIGURE P12.1

Solution Equilibrium: M x = TA − T1 = 7 kip  in. − T1 = 0

T1 = 7.0 kip  in.

M x = TA − TB − T2 = 7 kip  in. − 28 kip  in. − T2 = 0

Section properties:

A1 = A2 =



 4



(1.25 in.) = 1.2272 in.2

J1 =

(1.75 in.) = 2.4053 in.2

J2 =

Normal and shear stress magnitudes: F 9 kips 1 = 1 = = 7.3339 ksi (T) A1 1.2272 in.2

 T2 = −21 kip  in.



Tc J1

(1.25 in.) = 0.2397 in.4 4

(1.75 in.) = 0.9208 in.4

1 = 1 1 =

( 7.0 kip  in.)(1.25 in. / 2 ) = 18.2532 ksi

F2 9 kips Tc = = 3.7418 ksi (T) 2 = 2 2 = 2 A2 2.4053 in. J2 (sense of shear stresses to be determined by inspection)

2 =

0.2397 in.4 ( 21.0 kip  in.)(1.75 in. / 2) 0.9208 in.4

= 19.9561 ksi

Stress elements for points H and K are shown below.

Summary of stresses at H:  x = 7.33 ksi

 y = 0 ksi  xy = 18.25 ksi

Ans.

Summary of stresses at K:  x = 3.74 ksi

 y = 0 ksi  xy = −19.96 ksi

Ans.

P12.2 The flanged member shown in Figure P12.2 is subjected to an internal axial force of P = 6,300 lb, an internal shear force of V = 5,500 lb, and an internal bending moment of M = 77,000 lb-ft, acting in the directions shown. The dimensions of the cross section are bf = 10.00 in., tf = 0.68 in., d = 12.00 in., and tw = 0.32 in. Determine the normal and shear stresses at point H, where a = 2.50 in.

FIGURE P12.2a

FIGURE P12.2b

Moment of inertia about the z axis: Shape Width b Depth d

Area A

d = y − yi

d²A

IC + d²A

(in.)

(in.2)

(in.4)

(in.)

(in.4)

0.68

6.800

0.262

–5.660

217.842

218.104

0.32

10.64

3.405

32.121

0.000

32.121

0.68

6.800

0.262

5.660

217.842

218.104

Area (in.2) =

17.005

Moment of inertia about the z axis (in.4) =

468.330

Left flange Web Right flange

Axial stress at point H: F −6,300 lb x = = = −370.484 psi A 17.005 in.2 Bending stress at point H: (y = a = 2.50 in.) ( −77,000 lb  ft )( 2.50 in.)(12 in./ft ) = 4,932.424 psi My x = − =− Iz 468.330 in.4 Transverse and horizontal shear stress at point H:  12.00 in. 0.68 in.  Q = (10 in.)( 0.68 in.)  −  2 2   + ( 0.32 in.)( 2.82 in.)( 3.91 in.) = 42.016 in.3 3 VQ ( 5,500 lb ) ( 42.016 in. ) = = = 1,541.984 psi It ( 468.330 in.4 ) ( 0.32 in.)

(Note: Sense to be determined by inspection)

Summary of stresses at H:  x = 4,560 psi

 y = 0 psi  xy = −1,542 psi

Ans.

P12.3 The doubly-symmetric beam cross section shown in Figure P12.3b has been proposed for a short foot bridge. The cross section will consist of two identical steel pipes that are securely welded to a steel web plate that has a thickness of t = 9.5 mm. Each pipe has an outside diameter of d = 70 mm and a wall thickness is 5 mm. The distance between the centers of the two pipes is h = 370 mm. Internal forces of P = 13 kN, V = 25 kN, and M = 9 kN·m act in the directions shown in Figure P12.3a. Determine the stresses acting on horizontal and vertical planes: (a) at point H, which is located at a distance of yH = 120 mm above the z centroidal axis. (b) at point K, which is located at a distance of yK = 80 mm below the z centroidal axis. Show the stresses on a stress element for each point.

FIGURE P12.3b

FIGURE P12.3a

Solution Moment of inertia about the z axis: (i.e., horizontal axis) Width Depth d = y − yi Shape Area A IC b d Upper pipe Web Lower pipe

(mm)

9.5

300 Area =

(mm2) 1,021.0 2,850.0 1,021.0 4,892.0

d²A

(mm4) (mm) (mm4) 542,415.6 –185.000 34,944,327.4 21,375,000.0 0.000 0.0 542,415.6 185.000 34,944,327.4 Moment of inertia about the z axis =

IC + d²A (mm4) 35,486,743.0 21,375,000.0 35,486,743.0 92,348,486.0

Axial stress at point H: F 13,000 N x = = = 2.657 MPa A 4,892.0 mm 2 Bending stress at point H: (y = 120 mm) 9  106 N  mm ) (120 mm ) ( My x = − =− = −11.695 MPa Iz 92,348,486 mm 4 Transverse and horizontal shear stress at point H: Q = 1,021.0 mm2 (185 mm ) + ( 9.5 mm )( 30 mm )(135 mm ) = 227,360 mm3

(

=

)

3 VQ ( 25,000 N ) ( 227,360 mm ) = = 6.479 MPa It ( 92,348,486 mm4 ) ( 9.5 mm )

(Sense to be determined by inspection)

Summary of stresses at H:  x = −9.04 MPa

 y = 0 MPa  xy = 6.48 MPa

Ans.

Axial stress at point K: F 13,000 N x = = = 2.657 MPa A 4,892.0 mm 2 Bending stress at point K: (y = −80 mm) 9  106 N  mm ) ( −80 mm ) ( My x = − =− = 7.797 MPa Iz 92,348,486 mm 4 Transverse and horizontal shear stress at point K: Q = 1,021.0 mm2 (185 mm ) + ( 9.5 mm )( 70 mm )(115 mm ) = 265,360 mm3

(

)

3 VQ ( 25,000 N ) ( 265,360 mm ) = = = 7.562 MPa It ( 92,348,486 mm4 ) ( 9.5 mm )

(Sense to be determined by inspection)

Summary of stresses at K:  x = 10.45 MPa

 y = 0 MPa  xy = 7.56 MPa

Ans.

P12.4 A hat-shaped flexural member is subjected to an internal axial force of P = 7,200 N, an internal shear force of V = 6,000 N, and an internal bending moment of M = 1,300 N·m, acting as shown in Figure P12.4a. Dimensions of the cross section (Figure P12.4b) are a = 20 mm, b = 100 mm, d = 55 mm, and t = 4 mm. Determine the stresses acting on horizontal and vertical planes: (a) at point H, which is located at a distance of yH = 20 mm above the z centroidal axis. (b) at point K, which is located at a distance of yK = 12 mm below the z centroidal axis. Show the stresses on a stress element for each point.

FIGURE P12.4a

FIGURE P12.4b

Solution

Centroid location in y direction: (reference axis at bottom of shape) yi Shape Width b Depth d Area Ai yi Ai (from bottom) (mm) 16 4 92 4 16

(1) (2) (3) (4) (5)

yi A i A i

(mm) 4 55 4 55 4

(mm2) 64.0 220.0 368.0 220.0 64.0 936.0

(mm) 53 27.5 2 27.5 53

(mm3) 3,392.0 6,050.0 736.0 6,050.0 3,392.0 19,620.0

19,620 mm3 = 20.9615 mm (from bottom to centroid) 936.0 mm 2 = 34.0385 mm (from centroid to top)

Moment of inertia about the z axis: (i.e., horizontal axis) Shape Width b Depth d Area A IC (1) (2) (3) (4) (5)

(mm) 16 4 92 4 16

(mm) 4 55 4 55 4 Area =

(mm2) 64.0 220.0 368.0 220.0 64.0 936.0

d = y − yi

d²A

(mm4) (mm) (mm4) 85.3 -32.038 65,693.6 55,458.3 -6.538 9,405.3 490.7 18.962 132,310.7 55,458.3 -6.538 9,405.3 85.3 -32.038 65,693.6 Moment of inertia about the z axis =

IC + d²A (mm4) 65,779.0 64,863.7 132,801.4 64,863.7 65,779.0 394,086.6

Axial stress at point H: F 7,200 N x = = = 7.692 MPa A 936.0 mm2 Bending stress at point H: (y = 20 mm) 1,300  103 N  mm ) ( 20 mm ) ( My x = − =− = −65.975 MPa Iz 394,086.6 mm 4 Transverse and horizontal shear stress at point H: Q = 2 (16 mm )( 4 mm )( 53 mm − 20.9615 mm )

 34.0385 mm + 20 mm  3 +2 ( 4 mm )( 34.0385 mm − 20 mm )   = 7,135.391 mm 2   3 VQ ( 6,000 N ) ( 7,135.391 mm ) = = = 13.580 MPa (Sense to be determined by inspection) It ( 394,086.6 mm4 ) ( 2  4 mm )

Summary of stresses at H:  x = −58.3 MPa

 y = 0 MPa  xy = −13.58 MPa

Ans.

Axial stress at point K: F 7,200 N x = = = 7.692 MPa A 936.0 mm2 Bending stress at point K: (y = −12 mm) 1,300  103 N  mm ) ( −12 mm ) ( My x = − =− = 39.585 MPa Iz 394,086.6 mm 4

Transverse and horizontal shear stress at point K: Q = ( 92 mm )( 4 mm )( 20.9615 mm − 2 mm )

 20.9615 mm + 12 mm  3 +2 ( 4 mm )( 20.9615 mm − 12 mm )   = 8,159.370 mm 2   3 VQ ( 6,000 N ) (8,159.370 mm ) = = = 15.528 MPa (Sense to be determined by inspection) It ( 394,086.6 mm4 ) ( 2  4 mm )

Summary of stresses at K:  x = 47.3 MPa

 y = 0 MPa  xy = 15.53 MPa

Ans.

P12.5 An extruded polymer flexural member is subjected to an internal axial force of P = 580 lb, an internal shear force of V = 420 lb, and an internal bending moment of M = 6,400 lb·in., acting in the directions shown in Figure P12.5a. The cross-sectional dimensions (Figure P12.5b) of the extrusion are b1 = 2.0 in., t1 = 0.6 in., b2 = 4.0 in., t2 = 0.4 in., d = 4.5 in., and tw = 0.4 in. Determine the normal and shear stresses acting on horizontal and vertical planes: (a) at point H, which is located at a distance of yH = 0.8 in. above the z centroidal axis. (b) at point K, which is located at a distance of yK = 1.1 in. below the z centroidal axis. Show the stresses on a stress element for each point.

FIGURE P12.5a

FIGURE P12.5b

Solution

Centroid location in y direction: (reference axis at bottom of shape) Shape

Width b

Depth d

Area Ai

(1) (2) (3)

(in.) 1.6 3.6 0.4

(in.) 0.6 0.4 4.5

(in.2) 0.960 1.440 1.800 4.200

yi Ai A i

yi (from bottom) (in.) 4.2 0.2 2.25

yi Ai (in.3) 4.0320 0.2880 4.0500 8.3700

8.3700 in.3 = 1.9929 in. (from bottom to centroid) 4.200 in.2 = 2.5071 in. (from centroid to top)

Moment of inertia about the z axis: (i.e., horizontal axis) d = y − yi Shape Width b Depth d Area A IC (1) (2) (3)

(in.) 1.6 3.6 0.4

(in.) 0.6 0.4 4.5 Area =

(in.2) 0.960 1.440 1.800 4.200

d²A

(in.4) (in.) (in.4) 0.0288 -2.2071 4.6766 0.0192 1.7929 4.6286 3.0375 -0.2571 0.1190 Moment of inertia about the z axis =

IC + d²A (in.4) 4.7054 4.6478 3.1565 12.5098

Axial stress at point H: F −580 lb x = = = −138.095 psi A 4.20 in.2 Bending stress at point H: (y = 0.8 in.) ( 6,400 lb  in.)( 0.8 in.) = −409.280 psi My x = − =− Iz 12.5098 in.4 Transverse and horizontal shear stress at point H: Q = (1.6 in.)( 0.6 in.)( 2.5071 in. − 0.3 in.)

2.5071 in. − 0.8 in.   3 + ( 0.4 in.)( 2.5071 in. − 0.8 in.)  0.8 in. +  = 3.248 in. 2   3 ( 420 lb ) ( 3.248 in. ) VQ = = = 272.619 psi (Sense to be determined by inspection) It (12.5098 in.4 ) ( 0.4 in.)

Summary of stresses at H:  x = −547 psi

 y = 0 psi  xy = −273 psi

Ans.

Axial stress at point K: F −580 lb x = = = −138.095 psi A 4.20 in.2 Bending stress at point K: (y = −1.1 in.) ( 6,400 lb  in.)( −1.1 in.) = 562.760 psi My x = − =− Iz 12.5098 in.4

Transverse and horizontal shear stress at point K: Q = ( 3.6 in.)( 0.4 in.)(1.9929 in. − 0.2 in.)

1.9929 in. − 1.1 in.   3 + ( 0.4 in.)(1.9929 in. − 1.1 in.) 1.1 in. +  = 3.134 in. 2   3 ( 420 lb ) ( 3.134 in. ) VQ = = = 263.051 psi (Sense to be determined by inspection) It (12.5098 in.4 ) ( 0.4 in.)

Summary of stresses at K:  x = 425 psi

 y = 0 psi  xy = −263 psi

Ans.

P12.6 The machine component shown in Figure P12.6 is subjected to a load of P = 450 lb that acts at an angle of  = 25°. Dimensions of the component are a = 4.0 in., c = 2.0 in., b = 0.625 in., and d = 2.5 in. Determine the normal and shear stresses acting on horizontal and vertical planes: (a) at point H, which is located at a distance of yH = 0.5 in. above the z centroidal axis. (b) at point K, which is located at a distance of yK = 0.75 in. below the z centroidal axis. Show the stresses on a stress element for each point

FIGURE P12.6

Solution Section properties: A = ( 0.625 in.)( 2.5 in.) = 1.5625 in.2

( 0.625 in.)( 2.5 in.) = 0.8138 in.4 I = 3

0.75 in.   3 QH = ( 0.625 in.)( 0.75 in.)  0.5 in. +  = 0.4102 in. 2   0.50 in.   3 QK = ( 0.625 in.)( 0.50 in.)  0.75 in. +  = 0.3125 in. 2  

Internal forces at section containing H and K:

Fx = F − P sin  = F − ( 450 lb ) sin 25 = 0  F = 190.1782 lb Fy = −V − P cos  = −V − ( 450 lb ) cos 25 = 0 V = −407.8385 lb M = M + aP cos  − cP sin  = M + P ( a cos  − c sin  ) = M + ( 450 lb ) ( 4.0 in.) cos 25 − ( 2.0 in.) sin 25  = 0  M = −1, 250.9976 lb  in.

Axial stress at H: 190.1782 lb x = = 121.7141 psi 1.5625 in.2 Bending stress at H: ( −1, 250.9976 lb  in.)( 0.5 in.) = 768.6129 psi x = − 0.8138 in.4 Shear stress at H: ( 407.8385 lb ) ( 0.4102 in.3 ) H = = 328.917 psi ( 0.8138 in.4 ) ( 0.625 in.)

Summary of stresses at H:  x = 890 psi

 y = 0 psi  xy = 329 psi

Ans.

Axial stress at K: 190.1782 lb x = = 121.7141 psi 1.5625 in.2 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Bending stress at K: ( −1, 250.9976 lb  in.)( −0.75 in.) = −1,152.9194 psi x = − 0.8138 in.4 Shear stress at K: ( 407.8385 lb ) ( 0.3125 in.3 ) K = = 250.5760 psi ( 0.8138 in.4 ) ( 0.625 in.)

Summary of stresses at K:  x = −1,031 psi

 y = 0 psi  xy = 251 psi

Ans.

P12.7 A load of P = 900 N at an angle of  = 30° acts on the machine part shown in Figure P12.7a. Dimensions of the machine part are a = 180 mm, b = 60 mm, c = 100 mm, d = 40 mm, and e = 10 mm. The machine part has a uniform thickness of 8 mm (i.e., 8 mm thickness in the z direction). Determine the normal and shear stresses acting on horizontal and vertical planes at points H and K, which are shown in detail in Figure P12.7b. For each point, show these stresses on a stress element.

FIGURE P12.7a

FIGURE P12.7b Detail at pin B.

Solution Section properties:

A = (8 mm )( 40 mm ) = 320 mm

(8 mm )( 40 mm ) = 42,666.667 mm4 I = 3

Equilibrium: Fy = By − P cos  = 0

By = ( 900 N ) cos30 = 779.423 N M B = Ax ( a + b ) + ( P cos  )( b + c ) = 0 ( 900 N )( cos30 )  ( 60 mm + 100 mm ) Ax = −  = −519.615 N 180 mm + 60 mm Fx = Ax + Bx − P sin  = 0 Bx = ( 900 N ) sin 0 − ( −519.6 N ) = 969.615 N

Internal forces at H:

Fx = FH − P sin  = 0

 FH = ( 900 N ) sin 30 = 450 N

Fy = −VH − P cos  = 0 VH = − ( 900 N ) cos30 = −779.423 N M H = M H + cP cos30 = 0

 M H = − (100 mm )( 900 N ) cos30 = −77,942.286 N  mm

Axial stress at H: 450 N  axial = = 1.406 MPa (T) 320 mm2 Shear stress at H: QH = ( 8 mm )(10 mm )(15 mm ) = 1,200 mm3

( 779.423 N ) (1,200 mm3 ) H = = 2.740 MPa ( 42,666.667 mm4 ) (8 mm )

Bending stress at H: ( 77,942.286 N  mm )(10 mm ) = 18.268 MPa (T)  bend = 42,666.667 mm4

(by inspection)

Summary of stresses at H:  x = 1.406 MPa + 18.268 MPa = 19.67 MPa

 y = 0 MPa  xy = 2.74 MPa

Internal forces at K: Fx = VK + Ax = 0

VK = − Ax = 519.615 N Fy = FK = 0 M K = M K + Ax a = 0

 M K = − Ax a = − ( −519.615 N )(180 mm ) = 93,530.744 N  mm

Axial stress at K:  axial = 0 MPa Shear stress at K: QK = ( 8 mm )(10 mm )(15 mm ) = 1,200 mm3

( 519.615 N ) (1,200 mm3 ) K = = 1.827 MPa ( 42,666.667 mm4 ) (8 mm )

Bending stress at K: ( 93,530.744 N  mm )(10 mm ) = 21.921 MPa (C)  bend = 42,666.667 mm4

(by inspection)

Summary of stresses at K:  x = 0 MPa

 y = 0 MPa − 21.921 MPa = −21.9 MPa  xy = 1.827 MPa

P12.8 A hollow aluminum pipe is subjected to a vertical force of Px = 1,500 N, a horizontal force of Py = 2,400 N, and a concentrated torque of T = 80 N·m, acting as shown in Figure P12.8. The outside and inside diameters of the pipe are 64 mm and 56 mm, respectively. Assume a = 90 mm. Determine the normal and shear stresses on horizontal and vertical planes at (a) point H and (b) point K. For each point, show these stresses on a stress element.

FIGURE P12.8

Solution Section properties:  2 2 A = ( 64 mm ) − ( 56 mm )  = 753.982 mm 2   4   4 4 J= ( 64 mm ) − ( 56 mm )  = 681,599.942 mm 4  32   4 4 I y = Iz = ( 64 mm ) − ( 56 mm )  = 340, 799.971 mm 4  64 1 3 3 Q = ( 64 mm ) − ( 56 mm )  = 7, 210.667 mm 3   12 Equivalent forces at H and K: Fx = 1,500 N Fy = −2,400 N

M x = −80,000 N  mm

M y = 0 N  mm

Fz = 0 N

M z = − ( 2,400 N )( 90 mm ) = −216,000 N  mm

Axial stress magnitude at H due to Px: 1,500 N x = = 1.989 MPa 753.982 mm 2 Shear stress magnitude at H due to Py: ( 2,400 N ) ( 7,210.667 mm3 )  xy = = 6.347 MPa (340,799.971 mm4 ) ( 64 mm − 56 mm ) Torsion shear stress magnitude at H due to Mx: M c (80,000 N  mm )( 64 mm / 2 )  xy = x = = 3.756 MPa J 681,599.942 mm4

(a) Summary of stresses at H:  x = 1.989 MPa

 y = 0 MPa  xy = 3.756 MPa − 6.347 MPa = −2.592 MPa

Ans.

Axial stress magnitude at K due to Fy: 1,500 N x = = 1.989 MPa 753.982 mm 2 Bending stress magnitude at K due to Mz: M y ( 216,000 N  mm )( 64 mm / 2 ) x = z = = 20.282 MPa Iz 340,799.971 mm4 Torsion shear stress magnitude at K due to Mx: M c (80,000 N  mm )( 64 mm / 2 )  xz = x = = 3.756 MPa J 681,599.942 mm4

(b) Summary of stresses at K:  x = 1.989 MPa + 20.282 MPa = 22.3 MPa

 z = 0 MPa  xz = −3.76 MPa

Ans.

P12.9 A steel pipe with an outside diameter of 10.75 in. and an inside diameter of 9.50 in. is subjected to Px = 16 kips, Pz = 5 kips, and T= 23 kip·ft, acting as shown in Figure P12.9. Assume a = 4 ft. Determine the normal and shear stresses on horizontal and vertical planes at (a) point H and (b) point K. For each point, show these stresses on a stress element.

FIGURE P12.9

Solution Section properties: A= J=



(10.75 in.) 2 − (9.50 in.) 2  = 19.880 in.2 4



(10.75 in.) 4 − (9.50 in.) 4  = 511.454 in.4 32 

I y = Iz = Q=



(10.75 in.) 4 − (9.50 in.) 4  = 255.727 in.4 64 

1 (10.75 in.)3 − (9.50 in.)3  = 32.077 in.3 12 

Equivalent forces at H and K: Fx = − Px = −16 kips = −16,000 lb Fy = 0 lb Fz = Pz = 5 kips = 5,000 lb

Equivalent moments at H and K: M x = −T = −23 kip  ft = −276  103 lb  in. M y = − Pz a = − ( 5 kips )( 4 ft )

= −20 kip  ft = −240  103 lb  in. M z = 0 lb  in.

Each of the non-zero forces and moments will be evaluated to determine whether stresses are created at the point of interest. (a) Consider point H. Force Fx creates an axial stress at H. The magnitude of this normal stress is: 16,000 lb x = = 804.813 psi 19.880 in.2

Force Fz creates a transverse shear stress in the xz plane at H. The magnitude of this shear stress is: ( 5,000 lb ) (32.077 in.3 )  xz = = 501.735 psi ( 255.727 in.4 ) (10.75 in. − 9.50 in.) Moment Mx, which is a torque, creates a torsion shear stress in the xz plane at H. The magnitude of this shear stress is: 3 M xc ( 276  10 lb  in.) (10.75 in. / 2 )  xz = = = 2,900.554 psi J 511.454 in.4 Moment My creates bending stress in the pipe, but since point H lies on the neutral axis for My, the bending stress at H is zero.

Summary of stresses at H:  x = −805 psi

 z = 0 psi  xz = 501.735 psi − 2,900.554 psi = −2, 400 psi

Ans.

(b) Consider point K. Force Fx creates an axial stress at K. The magnitude of this normal stress is: 16,000 lb x = = 804.813 psi 19.880 in.2 Force Fz creates no stress at K. Moment Mx, which is a torque, creates a torsion shear stress in the xy plane at K. The magnitude of this shear stress is: 3 M xc ( 276  10 lb  in.) (10.75 in. / 2 )  xy = = = 2,900.554 psi J 511.454 in.4 Moment My creates a tensile bending stress at K. The magnitude of this normal stress is: ( 240 103 lb  in.) (10.75 in. / 2) = 5,044.441 psi x = 255.727 in.4

Summary of stresses at K:  x = −804.813 psi + 5,044.441 psi = 4, 240 psi

 y = 0 psi  xy = −2,900 psi

Ans.

P12.10 Concentrated loads of Px = 3,300 N, Py = 2,100 N, and Pz = 2,800 N are applied to the cantilever beam in the locations and directions shown in Figure P12.10a. The beam cross section shown in Figure P12.10b has dimensions of b = 100 mm and d = 40 mm. Using a value of a = 75 mm, determine the normal and shear stresses at (a) point H and (b) point K. Show these stresses on a stress element.

FIGURE P12.10a

FIGURE P12.10b

Solution Section properties: A = (100 mm )( 40 mm ) = 4,000 mm 2

(100 mm )( 40 mm ) = 533,333.333 mm 4 I = 3

( 40 mm )(100 mm ) = 3,333,333.333 mm 4 I = 3

Equivalent forces and moments at the section that contains points H and K: F = Pz = 2,800 N Vx = Px = 3,300 N Vy = Py = 2,100 N

M x = −2 Py a = −2 ( 2,100 N )( 75 mm ) = −315  103 N  mm M y = 4 Px a = 4 ( 3,300 N )( 75 mm ) = 990  103 N  mm (a) Consider point H. Axial stress at H due to F: 2,800 N  axial = = 0.700 MPa (T) 4,000 mm 2 Shear stress at H due to Vx: QH = ( 40 mm )( 30 mm )( 35 mm ) = 42,000 mm3

( 3,300 N ) ( 42,000 mm3 ) H = = 1.040 MPa (3,333,333.333 mm4 ) ( 40 mm )

Shear stress at H due to Vy: QH = 0 mm3  H = 0 MPa Bending stress at H due to Mx: 3 M x y ( 315  10 N  mm ) ( 40 mm / 2 )  bend x = = = 11.813 MPa (C) Ix 533,333.333 mm 4 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Bending stress at H due to My:

(990  10 N  mm )  1005mm  3

 bend y =

M yx Iy

3,333,333.333 mm 4

= 5.940 MPa (T)

Summary of stresses at H:  x = 0 MPa

 z = 0.700 MPa − 11.813 MPa + 5.940 MPa = −5.17 MPa

 xz = 1.040 MPa (b) Consider point K. Axial stress at K due to F: 2,800 N  axial = = 0.700 MPa (T) 4,000 mm 2 Shear stress at K due to Vx: QK = ( 40 mm )(16.667 mm )( 41.667 mm ) = 27,778.556 mm3

( 3,300 N ) ( 27,778.556 mm3 ) K = = 0.687 MPa (3,333,333.333 mm4 ) ( 40 mm )

Shear stress at K due to Vy: QK = 0 mm3  K = 0 MPa Bending stress at K due to Mx: 3 M y ( 315  10 N  mm ) ( 40 mm / 2 )  bend x = x = = 11.813 MPa (C) Ix 533,333.333 mm 4 Bending stress at K due to My:  100 mm  990  103 N  mm )  (  M x 3   = 9.900 MPa (C)  bend y = y = 4 Iy 3,333,333.333 mm Summary of stresses at K:  x = 0 MPa

 z = 0.700 MPa − 11.813 MPa − 9.900 MPa = −21.0 MPa

 xz = 0.687 MPa

P12.11 The stresses shown in the figure act at a point in a stressed body. Using the equilibrium equation approach, determine the normal and shear stresses at this point on the inclined plane shown.

FIGURE P12.11

Solution Fn =  n dA − (75 MPa) cos 40(dA cos 40) + (210 MPa)sin 40(dA sin 40) = 0

 n = −42.755 MPa = 42.8 MPa (C)

Ans.

Ft =  nt dA − (75 MPa)sin 40(dA cos 40) − (210 MPa)cos 40( dA sin 40) = 0

 nt = 140.335 MPa = 140.3 MPa

Ans.

P12.12 The stresses shown in the figure act at a point in a stressed body. Using the equilibrium equation approach, determine the normal and shear stresses at this point on the inclined plane shown. FIGURE P12.12

Solution Fn =  n dA − (185 MPa)cos30(dA cos30) −(110 MPa)sin 30(dA cos30) − (110 MPa)cos30(dA sin 30) = 0

 n = 234.013 MPa = 234 MPa (T)

Ans.

Ft =  nt dA + (185 MPa)sin 30(dA cos30) −(110 MPa)cos30(dA cos30) + (110 MPa)sin 30( dAsin 30) = 0

 nt = −25.107 MPa = −25.1 MPa

Ans.

P12.13 The stresses shown in the figure act at a point in a stressed body. Determine the normal and shear stresses at this point on the inclined plane shown.

FIGURE P12.13

Solution Fn =  n dA + (2,885 psi)cos 40(dA cos 40) − (1,305 psi)sin 40(dA sin 40) −(2,115 psi)sin 40(dA cos 40) − (2,115 psi)cos 40(dA sin 40) = 0

 n = 929.075 psi = 929 psi (T)

Ans.

Ft =  nt dA + (2,885 psi)sin 40(dA cos 40) + (1,305 psi)cos 40(dA sin 40) +(2,115 psi)cos 40(dA cos 40) − (2,115 psi)sin 40(dA sin 40) = 0

 nt = −2,430.438 psi = −2,430 psi

Ans.

P12.14 The stresses shown in the figure act at a point in a stressed body. Determine the normal and shear stresses at this point on the inclined plane shown.

FIGURE P12.14

Solution Fn =  n dA − (45 MPa)cos68.1986(dA cos68.1986) + (86 MPa)sin 68.1986(dA sin 68.1986) −(58 MPa)sin 68.1986(dA cos68.1986) − (58 MPa)cos68.1986(dA sin 68.1986) = 0

 n = 29.931 MPa = 29.9 MPa (C)

Ans.

Ft =  nt dA + (45 MPa)sin 68.1986(dA cos68.1986) + (86 MPa)cos68.1986(dA sin 68.1986) −(58 MPa)cos68.1986(dA cos68.1986) + (58 MPa)sin 68.1986(dA sin 68.1986) = 0

 nt = −87.172 MPa = −87.2 MPa

Ans.

P12.15 The stresses shown in the figure act at a point in a stressed body. Determine the normal and shear stresses at this point on the inclined plane shown.

FIGURE P12.15

Solution The given stress values are:  x = −7.2 ksi,  y = 13.6 ksi,  xy = 0 ksi,  = −20 The normal stress transformation equation [Eq. (12.3)] gives n:  n =  x cos2  +  y sin 2  + 2 xy sin  cos

= ( −7.2 ksi ) cos2 ( −20 ) + (13.6 ksi ) sin 2 ( −20 ) + 2 ( 0 ksi ) sin ( −20 ) cos ( −20 ) = −4.767 ksi = 4.77 ksi (C)

Ans.

The shear stress transformation equation [Eq. (12.4)] gives nt:  nt = − ( x −  y ) sin  cos +  xy ( cos 2  − sin 2  ) = − ( −7.2 ksi ) − (13.6 ksi )  sin ( −20 ) cos ( −20 ) + ( 0 ksi ) cos 2 ( −20 ) − sin 2 ( −20 )  = −6.685 ksi = −6.69 ksi

Ans.

P12.16 The stresses shown in the figure act at a point in a stressed body. Determine the normal and shear stresses at this point on the inclined plane shown.

FIGURE P12.16

Solution The given stress values are:  x = 30 MPa,  y = 90 MPa,  xy = 105 MPa,  = −50 The normal stress transformation equation [Eq. (12.3)] gives n:  n =  x cos2  +  y sin 2  + 2 xy sin  cos

= ( 30 MPa ) cos 2 ( −50 ) + ( 90 MPa ) sin 2 ( −50 ) + 2 (105 MPa ) sin ( −50 ) cos ( −50 ) = −38.1954 MPa = 38.2 MPa (C)

Ans.

The shear stress transformation equation [Eq. (12.4)] gives nt:  nt = − ( x −  y ) sin  cos +  xy ( cos 2  − sin 2  ) = − ( 30 MPa ) − ( 90 MPa )  sin ( −50 ) cos ( −50 ) + (105 MPa ) cos 2 ( −50 ) − sin 2 ( −50 )  = −47.7773 MPa = −47.8 MPa

Ans.

P12.17 The stresses shown in the figure act at a point in a stressed body. Determine the normal and shear stresses at this point on the inclined plane shown.

FIGURE P12.17

Solution The given stress values are:  x = 85 MPa,  y = 0 MPa,  xy = −147 MPa,  = 21.8 The normal stress transformation equation [Eq. (12.3)] gives n:  n =  x cos 2  +  y sin 2  + 2 xy sin  cos

= ( 85 MPa ) cos 2 ( 21.8 ) + ( 0 MPa ) sin 2 ( 21.8 ) + 2 ( −147 MPa ) sin ( 21.8 ) cos ( 21.8 ) = −28.10968 MPa = 28.1 MPa (C)

Ans.

The shear stress transformation equation [Eq. (12.4)] gives nt:  nt = − ( x −  y ) sin  cos +  xy ( cos 2  − sin 2  ) = − ( 85 MPa ) − ( 0 MPa )  sin ( 21.8 ) cos ( 21.8 ) + ( −147 MPa ) cos 2 ( 21.8 ) − sin 2 ( 21.8 )  = −135.7621 MPa = −135.8 MPa

Ans.

P12.18 The stresses shown in the figure act at a point in a stressed body. Determine the normal and shear stresses at this point on the inclined plane shown.

FIGURE P12.18

Solution The given stress values are:  x = 136.5 MPa,  y = 94.5 MPa,  xy = −63 MPa,  = 40 The normal stress transformation equation [Eq. (12.3)] gives n:  n =  x cos2  +  y sin 2  + 2 xy sin  cos

= (136.5 MPa ) cos2 ( 40 ) + ( 94.5 MPa ) sin 2 ( 40 ) + 2 ( −63 MPa ) sin ( 40 ) cos ( 40 ) = 57.1037 MPa = 57.1 MPa (T)

Ans.

The shear stress transformation equation [Eq. (12.4)] gives nt:  nt = − ( x −  y ) sin  cos +  xy ( cos 2  − sin 2  ) = − (136.5 MPa ) − ( 94.5 MPa )  sin ( 40 ) cos ( 40 ) + ( −63 MPa ) cos 2 ( 40 ) − sin 2 ( 40 )  = −31.6208 MPa = −31.6 MPa

Ans.

P12.19 Two steel plates of uniform cross section are welded together as shown in Figure P12.19. The plate dimensions are b = 5.5 in. and t = 0.375 in. An axial force of P = 28.0 kips acts in the member. If a = 2.25 in., determine the magnitude of (a) the normal stress that acts perpendicular to the weld seam. (b) the shear stress that acts parallel to the weld seam.

FIGURE P12.19

Solution Begin by calculating the value of : a 2.25 in. tan  = = = 0.40909  = 22.2490 b 5.5 in. Since this angle turns clockwise from the x axis to the n axis (see below), we use a negative value for .

The normal stress in the x direction is: P 28.0 kips x = = = 13.5758 ksi A ( 0.375 in.)( 5.5 in.) Therefore, stress values to be considered here are:  x = 13.5758 ksi,  y = 0 ksi,  xy = 0 ksi,  = −22.2490 (a) The normal stress perpendicular to the weld seam is found using Eq. (12.3):  n =  x cos 2  +  y sin 2  + 2 xy sin  cos = (13.5758 ksi ) cos 2 ( −22.2490 ) + ( 0 ksi ) sin 2 ( −22.2490 ) + 2 ( 0 ksi ) sin ( −22.2490 ) cos ( −22.2490 )

= (13.5758 ksi ) cos2 ( −22.2490 ) = 11.6295 ksi = 11.63 ksi (T)

Ans.

(b) The shear stress parallel to the weld seams is found using Eq. (12.4):  nt = − ( x −  y ) sin  cos +  xy ( cos 2  − sin 2  )

= − (13.5758 ksi ) − ( 0 ksi )]sin ( −22.2490 ) cos ( −22.2490 ) + ( 0 ksi )[cos 2 ( −22.2490 ) − sin 2 ( −22.2490 )  = ( −13.5758 ksi ) sin ( −22.2490 ) cos ( −22.2490 ) = 4.7575 ksi = 4.76 ksi

Ans.

P12.20 The cylinder in Figure P12.20 consists of spirally wrapped steel plates that are welded at the seams. The cylinder has an outside diameter of 280 mm and a wall thickness of 7 mm. The cylinder is subjected to axial loads of P = 90 kN. For a seam angle of  = 35°, determine: (a) the normal stress perpendicular to the weld seams. (b) the shear stress parallel to the weld seams.

FIGURE P12.20

Solution The value of  to be used in Eqs. (12.3) and (12.4) is equal in magnitude to ; however,  will be negative.

The cross-sectional area of the cylinder is:



2 2 ( 280 mm ) − ( 266 mm )  = 6,003.584 mm2  4 Thus, the normal stress in the x direction is: P 90,000 N x = = = 14.991 MPa A 6,003.584 mm 2 Therefore, stress values to be considered here are:  x = 14.991 MPa,  y = 0 MPa,  xy = 0 MPa,  = −35

(a) The normal stress perpendicular to the weld seam is found using Eq. (12.3):  n =  x cos2  +  y sin 2  + 2 xy sin  cos

= (14.991 MPa ) cos 2 ( −35 ) + ( 0 MPa ) sin 2 ( −35 ) + 2 ( 0 MPa ) sin ( −35 ) cos ( −35 ) = 10.059 MPa = 10.06 MPa (T)

Ans.

(b) The shear stress parallel to the weld seams is found using Eq. (12.4):  nt = − ( x −  y ) sin  cos +  xy ( cos 2  − sin 2  ) = − (14.991 MPa ) − ( 0 MPa )  sin ( −35 ) cos ( −35 ) + ( 0 MPa ) cos 2 ( −35 ) − sin 2 ( −35 )  = 7.043 MPa = 7.04 MPa

Ans.

P12.21 A rectangular polymer plate of dimensions a = 150 mm and b = 360 mm is formed by gluing two triangular plates as shown in Figure P12.21. The plate is subjected to a tensile normal stress of 10 MPa in the long direction and a compressive normal stress of 6 MPa in the short direction. Determine: (a) the normal stress perpendicular to the glue seam. (b) the shear stress parallel to the glue seam.

FIGURE P12.21

Solution The value of  to be used in Eqs. (12.3) and (12.4) is calculated from the plate dimensions: b 360 mm tan  = = = 2.4  = 67.3801 a 150 mm Since this angle turns clockwise from the x axis to the n axis (see below), we use a negative value for . Therefore, stress values needed for this calculation are:  x = 10 MPa,  y = −6 MPa,

 xy = 0 MPa,  = −67.3801

(a) The normal stress perpendicular to the weld seam is found using Eq. (12.3):  n =  x cos 2  +  y sin 2  + 2 xy sin  cos = (10 MPa ) cos 2 ( −67.3801 ) + ( −6 MPa ) sin 2 ( −67.3801 ) +2 ( 0 MPa ) sin ( −67.3801 ) cos ( −67.3801 ) = −3.6331 MPa = 3.63 MPa (C)

Ans.

(b) The shear stress parallel to the weld seams is found using Eq. (12.4):  nt = − ( x −  y ) sin  cos +  xy ( cos 2  − sin 2  ) = − (10 MPa ) − ( −6 MPa )  sin ( −67.3801 ) cos ( −67.3801 ) + ( 0 MPa ) cos 2 ( −67.3801 ) − sin 2 ( −67.3801 )  = 5.6806 MPa = 5.68 MPa

Ans.

P12.22 The state of stress at a point in a solid component is shown in Figure P12.22. The stress magnitudes are n = 80 MPa, t = 40 MPa, and nt = 32 MPa, acting in the directions indicated in the figure. Determine the normal and shear stress components acting on plane AB for a value of  = 35°.

FIGURE P12.22

Solution Let’s rename the axes given for this problem. We’ll rename the n and t axes as the x and y axes. Further, we’ll call the horizontal axis the n axis and the vertical axis will be renamed as the t axis. Our revised stress element is shown to the right. Our task now is to determine the normal stress in the n direction for a counterclockwise rotation of  =  = 35° from the x axis. Therefore, stress values needed for this calculation are:  x = −80 MPa,  y = −40 MPa,

 xy = 32 MPa,  = 35

The normal stress transformation equation [Eq. (12.3)] gives the normal stress acting on plane AB:  n =  x cos 2  +  y sin 2  + 2 xy sin  cos = ( −80 MPa ) cos 2 ( 35 ) + ( −40 MPa ) sin 2 ( 35 ) +2 ( 32 MPa ) sin ( 35 ) cos ( 35 ) = −36.7702 MPa = 36.8 MPa (C)

Ans.

The shear stress acting on plane AB is found using Eq. (12.4):  nt = − ( x −  y ) sin  cos +  xy ( cos 2  − sin 2  ) = − ( −80 MPa ) − ( −40 MPa )  sin ( 35 ) cos ( 35 ) + ( 32 MPa ) cos 2 ( 35 ) − sin 2 ( 35 )  = 29.7385 MPa = 29.7 MPa

Ans.

P12.23 The stresses shown in Figure P12.23a act at a point on the free surface of a stressed body. Determine the normal stresses n and t and the shear stress nt at this point if they act on the rotated stress element shown in Figure P12.23b. (a)

(b) FIGURE P12.23

Solution The given stress values are:  x = 86 MPa,  y = −32 MPa,  xy = −71 MPa,  = −36 The normal stress transformation equation [Eq. (12.3)] gives n:  n =  x cos 2  +  y sin 2  + 2 xy sin  cos  = (86 MPa) cos 2 ( −36) + ( −32 MPa)sin 2 ( −36) + 2( −71 MPa)sin( −36) cos( −36) = 112.757 MPa = 112.8 MPa (T)

Ans.

To find t, add 90° to the value of  used in Eq. (12.3):  t =  x cos 2  +  y sin 2  + 2 xy sin  cos  = (86 MPa) cos 2 ( −36 + 90) + ( −32 MPa)sin 2 ( −36 + 90) +2( −71 MPa)sin( −36 + 90) cos( −36 + 90) = (86 MPa) cos 2 (54) + ( −32 MPa)sin 2 (54) + 2( −71 MPa)sin(54) cos(54) = −58.757 MPa = 58.8 MPa (C)

Ans.

The shear stress transformation equation [Eq. (12.4)] gives nt:  nt = −( x −  y )sin  cos  +  xy (cos 2  − sin 2  )

= −[(86 MPa) − ( −32 MPa)]sin( −36)cos( −36) + ( −71 MPa)[cos 2 (−36) − sin 2 (−36)] = 34.172 MPa = 34.2 MPa

Ans.

P12.24 The stresses shown in Figure P12.24a act at a point on the free surface of a stressed body. Determine the normal stresses n and t and the shear stress nt at this point if they act on the rotated stress element shown in Figure P12.24b.

(a)

(b) FIGURE P12.24

Solution The given stress values are:  x = 2,900 psi,  y = 1,100 psi,  xy = 1,750 psi,  = 20 The normal stress transformation equation [Eq. (12.3)] gives n:  n =  x cos 2  +  y sin 2  + 2 xy sin  cos  = (2,900 psi)cos 2 (20) + (1,100 psi)sin 2 (20) + 2(1, 750 psi)sin(20)cos(20) = 3,814.318 psi = 3,810 psi (T)

Ans.

To find t, add 90° to the value of  used in Eq. (12.3):  n =  x cos 2  +  y sin 2  + 2 xy sin  cos  = (2,900 psi)cos 2 (20 + 90) + (1,100 psi)sin 2 (20 + 90) +2(1,750 psi)sin(20 + 90)cos(20 + 90) = (2,900 psi)cos 2 (110) + (1,100 psi)sin 2 (110) + 2(1,750 psi)sin(110)cos(110) = 185.682 psi = 185.7 psi (T)

Ans.

The shear stress transformation equation [Eq. (12.4)] gives nt:  nt = − ( x −  y )sin  cos  +  xy (cos 2  − sin 2  ) = −[(2,900 psi) − (1,100 psi)]sin(20) cos(20) + (1,750 psi)[cos 2 (20) − sin 2 (20)] = 762.069 psi = 762 psi

Ans.

P12.25 The stresses shown in Figure P12.25 act at a point on the free surface of a machine component. Determine the normal stresses x and y and the shear stress xy at the point.

FIGURE P12.25

Solution Redefine the axes, calling the rotated axes x and y. The angle from the rotated element to the unrotated element is now a positive value (since it is counterclockwise). Thus, the given stress values can be expressed as:  x = 59 MPa,  y = −48 MPa,  xy = −82 MPa,  = 30 The normal stress transformation equation [Eq. (12.3)] gives n, which is actually the normal stress in the horizontal direction (i.e., the original x direction) on the unrotated element:  n =  x cos 2  +  y sin 2  + 2 xy sin  cos  = (59 MPa)cos 2 (30) + ( −48 MPa)sin 2 (30) + 2( −82 MPa)sin(30)cos(30) = −38.764 MPa = 38.8 MPa (C)

Ans.

To find t, which is actually the normal stress in the vertical direction (i.e., the original y direction) on the unrotated element, add 90° to the value of  used in Eq. (12.3):  t =  x cos 2  +  y sin 2  + 2 xy sin  cos  = (59 MPa)cos 2 (30 + 90) + ( −48 MPa)sin 2 (30 + 90) + 2( −82 MPa)sin(30 + 90)cos(30 + 90)

= (59 MPa)cos 2 (120) + ( −48 MPa)sin 2 (120) + 2( −82 MPa)sin(120)cos(120) = 49.764 MPa = 49.8 MPa (T)

Ans.

The shear stress transformation equation [Eq. (12.4)] gives nt, which is actually the shear stress on the horizontal and vertical faces of the unrotated element:  nt = −( x −  y )sin  cos  +  xy (cos 2  − sin 2  )

= −[(59 MPa) − ( −48 MPa)]sin(30)cos(30) + ( −82 MPa)[cos 2 (30) − sin 2 (30)] = −87.332 MPa = −87.3 MPa

Ans.

P12.26 The stresses shown in Figure P12.26 act at a point on the free surface of a machine component. Determine the normal stresses x and y and the shear stress xy at the point.

FIGURE P12.26

Solution Redefine the axes, calling the rotated axes x and y. The angle from the rotated element to the unrotated element is now a negative value (since it is clockwise) . Thus, the given stress values can be expressed as:  x = 19.1 ksi,  y = 4.7 ksi,  xy = 13.8 ksi,  = −24 The normal stress transformation equation [Eq. (12.3)] gives n, which is actually the normal stress in the horizontal direction (i.e., the original x direction) on the unrotated element:  n =  x cos 2  +  y sin 2  + 2 xy sin  cos  = (19.1 ksi)cos 2 ( −24) + (4.7 ksi)sin 2 ( −24) + 2(13.8 ksi)sin( −24)cos( −24) = 6.462 ksi = 6.46 ksi (T)

Ans.

To find t, which is actually the normal stress in the vertical direction (i.e., the original y direction) on the unrotated element, add 90° to the value of  used in Eq. (12.3):  t =  x cos 2  +  y sin 2  + 2 xy sin  cos  = (19.1 ksi) cos 2 ( −24 + 90) + (4.7 ksi)sin 2 ( −24 + 90) +2(13.8 ksi)sin( −24 + 90) cos( −24 + 90) = (19.1 ksi) cos 2 (66) + (4.7 ksi)sin 2 (66) + 2(13.8 ksi)sin(66) cos(66) = 17.338 ksi = 17.34 ksi (T)

Ans.

= −[(19.1 ksi) − (4.7 ksi)]sin( −24)cos( −24) + (13.8 ksi)[cos 2 ( −24) − sin 2 ( −24)] = 14.585 ksi = 14.59 ksi

Ans.

P12.27 Consider a point in a structural member that is subjected to plane stress. Normal and shear stresses acting on horizontal and vertical planes at the point are shown. (a) Determine the principal stresses and the maximum in-plane shear stress acting at the point. (b) Show these stresses on an appropriate sketch (e.g., see Figure 12.15 or Figure 12.16) (c) Compute the absolute shear stress at the point. Instructors: Problems P12.27-P12.30 should be assigned as a set.

FIGURE P12.27

Solution The given stress values are:  x = −62 MPa,  y = −94 MPa,  xy = −42 MPa The principal stress magnitudes can be computed from Eq. (12.12):

 p1, p 2 =

x + y 2

x −y    +  xy2  2   2

( −62 MPa) + ( −94 MPa)  ( −62 MPa) − ( −94 MPa)  2 =    + ( −42 MPa)  2 2 2

= −78.0000 MPa  44.9444 MPa

 p1 = −33.1 MPa and  p 2 = −122.9 MPa

Ans.

 max = 44.9 MPa

Ans.

(maximum in-plane shear stress)

 avg = 78.0 MPa (C) tan 2 p =

(normal stress on planes of maximum in-plane shear stress)

Ans.

 xy −42 MPa = = −2.6250 ( x −  y ) / 2 [( −62 MPa) − ( −94 MPa)] / 2

 p = −34.6

(clockwise from the x axis to the direction of  p1 )

Ans.

P12.28 Consider a point in a structural member that is subjected to plane stress. Normal and shear stresses acting on horizontal and vertical planes at the point are shown. (a) Determine the principal stresses and the maximum in-plane shear stress acting at the point. (b) Show these stresses on an appropriate sketch (e.g., see Figure 12.15 or Figure 12.16) (c) Compute the absolute shear stress at the point. Instructors: Problems P12.27-P12.30 should be assigned as a set.

FIGURE P12.28

Solution The given stress values are:  x = 31 MPa,  y = 67 MPa,  xy = −17 MPa The principal stress magnitudes can be computed from Eq. (12.12):

 p1, p 2 =

x + y 2

x −y    +  xy2  2   2

(31 MPa) + (67 MPa)  (31 MPa) − (67 MPa)  2 =    + ( −17 MPa)  2 2 2

= 49.0000 MPa  24.7588 MPa

 p1 = 73.8 MPa and  p 2 = 24.2 MPa

Ans.

 max = 24.8 MPa

Ans.

(maximum in-plane shear stress)

 avg = 49.0 MPa (T) tan 2 p =

(normal stress on planes of maximum in-plane shear stress)

Ans.

 xy −17 MPa = = 0.9444 ( x −  y ) / 2 [(31 MPa) − (67 MPa)] / 2

 p = 21.7

(counterclockwise from the x axis to the direction of  p 2 )

Ans.

P12.29 Consider a point in a structural member that is subjected to plane stress. Normal and shear stresses acting on horizontal and vertical planes at the point are shown. (a) Determine the principal stresses and the maximum in-plane shear stress acting at the point. (b) Show these stresses on an appropriate sketch (e.g., see Figure 12.15 or Figure 12.16) (c) Compute the absolute shear stress at the point. Instructors: Problems P12.27-P12.30 should be assigned as a set.

FIGURE P12.29

Solution The given stress values are:  x = −106 MPa,  y = 172 MPa,  xy = 144 MPa The principal stress magnitudes can be computed from Eq. (12.12):

 p1, p 2 =

x + y 2

x −y    +  xy2  2   2

( −106 MPa) + (172 MPa)  ( −106 MPa) − (172 MPa)  2 =    + (144 MPa)  2 2 2

= 33.0000 MPa  200.1424 MPa

 p1 = 233 MPa and  p 2 = −167.1 MPa

Ans.

 max = 200 MPa

Ans.

(maximum in-plane shear stress)

 avg = 33.0 MPa (T) tan 2 p =

(normal stress on planes of maximum in-plane shear stress)

Ans.

 xy 144 MPa = = −1.0360 ( x −  y ) / 2 [( −106 MPa) − (172 MPa)] / 2

 p = −23.0

(clockwise from the x axis to the direction of  p 2 )

Ans.

P12.30 Consider a point in a structural member that is subjected to plane stress. Normal and shear stresses acting on horizontal and vertical planes at the point are shown. (a) Determine the principal stresses and the maximum in-plane shear stress acting at the point. (b) Show these stresses on an appropriate sketch (e.g., see Figure 12.15 or Figure 12.16) (c) Compute the absolute shear stress at the point. Instructors: Problems P12.27-P12.30 should be assigned as a set.

FIGURE P12.30

Solution The given stress values are:  x = 29 ksi,  y = 17 ksi,  xy = 7 ksi The principal stress magnitudes can be computed from Eq. (12.12):

 p1, p 2 =

x + y 2

x −y    +  xy2  2   2

(29 ksi) + (17 ksi)  (29 ksi) − (17 ksi)  2 =    + (7 ksi)  2 2 2

= 23.0000 ksi  9.2195 ksi

 p1 = 32.2 ksi and  p 2 = 13.78 ksi

Ans.

 max = 9.22 ksi

Ans.

(maximum in-plane shear stress)

 avg = 23.0 ksi (T) tan 2 p =

(normal stress on planes of maximum in-plane shear stress)

Ans.

 xy 7 ksi = = 1.1667 ( x −  y ) / 2 [(29 ksi) − (17 ksi)] / 2

 p = 24.7

(counterclockwise from the x axis to the direction of  p1 )

Ans.

P12.31 Consider a point in a structural member that is subjected to plane stress. Normal and shear stresses acting on horizontal and vertical planes at the point are shown. (a) Determine the principal stresses and the maximum in-plane shear stress acting at the point. (b) Show these stresses on an appropriate sketch (e.g., see Figure 12.15 or Figure 12.16) (c) Compute the absolute maximum shear stress at the point. Instructors: Problems P12.31-P12.34 should be assigned as a set.

FIGURE P12.31

Solution The given stress values are:  x = −3.5 ksi,  y = −12.5 ksi,  xy = −10 ksi The principal stress magnitudes can be computed from Eq. (12.12):

 p1, p 2 =

x + y 2

x −y    +  xy2  2   2

( −3.5 ksi) + ( −12.5 ksi)  ( −3.5 ksi) − ( −12.5 ksi)  2 =    + ( −10 ksi)  2 2 2

= −8.0000 ksi  10.9659 ksi

 p1 = 2.97 ksi and  p 2 = −18.97 ksi

Ans.

 max = 10.97 ksi

Ans.

 avg = 8.00 ksi (C) tan 2 p =

(maximum in-plane shear stress) (normal stress on planes of maximum in-plane shear stress)

Ans.

 xy −10 ksi = = −2.2222 ( x −  y ) / 2 [( −3.5 ksi) − ( −12.5 ksi)] / 2

 p = −32.9

(clockwise from the x axis to the direction of  p1 )

Ans.

P12.32 Consider a point in a structural member that is subjected to plane stress. Normal and shear stresses acting on horizontal and vertical planes at the point are shown. (a) Determine the principal stresses and the maximum in-plane shear stress acting at the point. (b) Show these stresses on an appropriate sketch (e.g., see Figure 12.15 or Figure 12.16) (c) Compute the absolute maximum shear stress at the point. Instructors: Problems P12.31-P12.34 should be assigned as a set.

FIGURE P12.32

Solution The given stress values are:  x = 9 ksi,  y = 13.4 ksi,  xy = −5.6 ksi The principal stress magnitudes can be computed from Eq. (12.12):

 p1, p 2 =

x + y 2

x −y    +  xy2  2   2

(9 ksi) + (13.4 ksi)  (9 ksi) − (13.4 ksi)  2 =    + ( −5.6 ksi)  2 2 2

= 11.2000 ksi  6.0166 ksi

 p1 = 17.22 ksi and  p 2 = 5.18 ksi

Ans.

 max = 6.02 ksi

Ans.

(maximum in-plane shear stress)

 avg = 11.20 ksi (T) tan 2 p =

(normal stress on planes of maximum in-plane shear stress)

Ans.

 xy −5.6 ksi = = 2.5455 ( x −  y ) / 2 [(9 ksi) − (13.4 ksi)] / 2

 p = 34.3

(counterclockwise from the x axis to the direction of  p 2 )

Ans.

P12.33 Consider a point in a structural member that is subjected to plane stress. Normal and shear stresses acting on horizontal and vertical planes at the point are shown. (a) Determine the principal stresses and the maximum in-plane shear stress acting at the point. (b) Show these stresses on an appropriate sketch (e.g., see Figure 12.15 or Figure 12.16) (c) Compute the absolute maximum shear stress at the point. Instructors: Problems P12.31-P12.34 should be assigned as a set.

FIGURE P12.33

Solution The given stress values are:  x = 17.6 ksi,  y = 20.4 ksi,  xy = 13.8 ksi The principal stress magnitudes can be computed from Eq. (12.12):

 p1, p 2 =

x + y 2

x −y    +  xy2  2   2

(17.6 ksi) + (20.4 ksi)  (17.6 ksi) − (20.4 ksi)  2 =    + (13.8 ksi)  2 2 2

= 19.0000 ksi  13.8708 ksi

 p1 = 32.9 ksi and  p 2 = 5.13 ksi

Ans.

 max = 13.87 ksi

Ans.

(maximum in-plane shear stress)

 avg = 19.00 ksi (T) tan 2 p =

(normal stress on planes of maximum in-plane shear stress)

Ans.

 xy 13.8 ksi = = −9.8571 ( x −  y ) / 2 [(17.6 ksi) − (20.4 ksi)] / 2

 p = −42.1

(clockwise from the x axis to the direction of  p 2 )

Ans.

P12.34 Consider a point in a structural member that is subjected to plane stress. Normal and shear stresses acting on horizontal and vertical planes at the point are shown. (a) Determine the principal stresses and the maximum in-plane shear stress acting at the point. (b) Show these stresses on an appropriate sketch (e.g., see Figure 12.15 or Figure 12.16) (c) Compute the absolute maximum shear stress at the point. Instructors: Problems P12.31-P12.34 should be assigned as a set.

FIGURE P12.34

Solution The given stress values are:  x = −114.8 MPa,  y = −154.8 MPa,  xy = 87 MPa The principal stress magnitudes can be computed from Eq. (12.12):

 p1, p 2 =

x + y 2

x −y    +  xy2  2   2

( −114.8 MPa) + ( −154.8 MPa)  ( −114.8 MPa) − ( −154.8 MPa)  2 =    + (87 MPa)  2 2 2

= −134.8000 MPa  89.2693 MPa

 p1 = −45.5 MPa and  p 2 = −224 MPa

Ans.

 max = 89.3 MPa

Ans.

(maximum in-plane shear stress)

 avg = 134.8 MPa (C) tan 2 p =

(normal stress on planes of maximum in-plane shear stress)

Ans.

 xy 87 MPa = = 4.3500 ( x −  y ) / 2 [( −114.8 MPa) − ( −154.8 MPa)] / 2

 p = 38.5

(counterclockwise from the x axis to the direction of  p1 )

Ans.

P12.35 A shear wall in a reinforced concrete building is subjected to a vertical uniform load of intensity w and a horizontal force H, as shown in Figure P12.35a. As a consequence of these loads, the stresses at point A on the surface of the wall have the magnitudes y = 115.0 MPa and xy = 60.0 MPa, acting in the directions shown on the stress element in Figure P12.35b. (a) Determine the largest tension normal stress that acts at point A. (b) What is the orientation of this stress with respect to the x axis?

FIGURE P12.35a

FIGURE P12.35b

Solution The given stress values are:  x = 0 MPa,  y = −115.0 MPa,  xy = −60.0 MPa The principal stress magnitudes can be computed from Eq. (12.12):

 p1, p 2 =

x +y 2

x − y  2    +  xy  2  2

(0 MPa) + (−115 MPa)  (−0 MPa) − (−115 MPa)  2 =    + (−60 MPa) 2 2   = −57.5 MPa  83.1039 MPa 2

 p1 = 25.6039 MPa and  p 2 = −140.6039 MPa (a) Largest tension normal stress: The largest tension normal stress that acts at point A is:  p1 = 25.6 MPa

Ans.

(b) What is the orientation of this stress with respect to the x axis? 2 xy 2 ( −60 MPa ) tan 2 p = = = −1.043478  x −  y 0 MPa − ( −115 MPa )

 p = 23.1

(clockwise from the x axis to the direction of  p1 )

Ans.

P12.36 A 2 in. diameter shaft is supported by bearings at A and D, as shown in Figure P12.36. The bearings provide vertical reactions only. The shaft is subjected to an axial load of P = 3,600 lb and a transverse load of Q = 250 lb. The shaft length is L = 48 in. For point B located on top of the shaft, determine: (a) the principal stresses. (b) the maximum in-plane shear stress. Show these results on a stress element.

FIGURE P12.36

Solution Shear-force and bending-moment diagram:

Bending moment at B = 1,500 lb·in.

Section properties:



( 2.0 in.) = 3.1416 in.2 2

 64

( 2.0 in.) = 0.7854 in.4 4

Axial stress at point B: P 3,600 lb x = = = 1,145.9156 psi (T) A 3.1416 in.2 Bending stress at point B: (1,500 lb  in.)(1 in.) = 1,909.8594 psi (C) Mc x = =− I 0.7854 in.4 Shear stress at point B: There is zero shear stress on top of the shaft at B.

Summary of stresses at H:  x = 1,145.9156 psi − 1,909.8594 psi = −763.9438 psi

 z = 0 psi  xz = 0 psi (a) Principal stresses: Since the shear stresses are zero on the x and z planes, we know that the normal shown above are also principal stresses. Ans.  p1 = 0 psi and  p 2 = −764 psi (b) Maximum in-plane shear stress:  −  p 2 0 psi − ( −763.9438 psi )  max = p1 = = 382 psi 2 2

(maximum in-plane shear stress)

Ans.

P12.37 The principal compressive stress on a vertical plane through a point in a wooden block is equal to three times the principal compression stress on a horizontal plane. The plane of the grain is 25° clockwise from the vertical plane. If the normal and shear stresses must not exceed 400 psi (C) and 90 psi shear, determine the maximum allowable compressive stress on the horizontal plane.

Solution The principal compressive stress on a vertical plane (that is, the x face of a stress element) is equal to three times the principal compression stress on a horizontal plane (that is, the y face of a stress element). Thus, from the problem statement, we know that x = 3y. Since we are told that the stresses on the x and y faces are principal stress, we also know that xy = 0. The plane of the wood grain is oriented 25° clockwise from the vertical plane; therefore,  = −25°. We are told that the normal stress on the plane of the wood grain must not exceed −400 psi, or in other words, n ≤ −400 psi. The normal stress transformation equation [Eq. (12-3)], which gives n, can be rearranged to solve for y:  n =  x cos 2  +  y sin 2  + 2 xy sin  cos 

−400 psi  3 y cos 2 ( −25) +  y sin 2 ( −25) + 2(0 psi)sin( −25) cos(−25) −400 psi   y [3cos 2 ( −25) + sin 2 ( −25)]  y 

−400 psi −400 psi = = −151.3546 psi 2 [3cos (−25) + sin ( −25)] 2.6428 2

(a)

A second condition of the stresses acting on the plane of the wood grain is that the shear stress must not exceed 90 psi, or in other words, nt ≤ 90 psi. The shear stress transformation equation [Eq. (12-4)], which gives nt, can be rearranged to solve for nt:  nt = −( x −  y )sin  cos  +  xy (cos 2  − sin 2  )

90 psi  −[3 y −  y ]sin(−25) cos(−25) + (0 psi)[cos 2 ( −25) − sin 2 (−25)] 90 psi  −2 y sin(−25) cos(−25) 90 psi 90 psi = = 117.4935 psi (b) −2sin(−25) cos(−25) 0.7660 Since we are told that y is a compressive normal stress, it is clear that we must choose the negative value for y.  y 

Compare the two limits found in Eqs. (a) and (b) to find that the maximum compression stress that may be applied to the horizontal plane is Ans.  y  −117.5 psi

P12.38 At a point on the free surface of a stressed body, a normal stress of 64 MPa (C) and an unknown positive shear stress exist on a horizontal plane. One principal stress at the point is 8 MPa (C). The absolute maximum shear stress at the point has a magnitude of 95 MPa. Determine the unknown stresses on the horizontal and vertical planes and the unknown principal stress at the point.

Solution The absolute maximum shear stress can be found from Eq. (12-18)  −  min  abs max = max 2 The absolute maximum shear stress at the point has a magnitude of 95 MPa. Suppose we assume that the given principal stress of −8 MPa is min. If this assumption is true, then  max =  min + 2 abs max = −8 MPa + 2(95 MPa) = 182 MPa However, this assumption cannot be true because the normal stress on the horizontal plane is y = −64 MPa, which is more negative than the given principal stress of −8 MPa. Therefore, we now know that the second principal stress must be negative and its magnitude must be greater than 64 MPa. The point in question occurs on the free surface of a stressed body. From this information, we can know that a state of plane stress exists at the point. Therefore,  z =  p3 = 0 (since it is a free surface) Since both of the in-plane principal stresses must be negative, max = p3 = 0. The minimum principal stress can now be determined from the absolute maximum shear stress:  min =  max − 2 abs max = 0 MPa − 2(95 MPa) = −190 MPa Thus, the two in-plane principal stresses are: Ans.  p1 = −8 MPa and  p 2 = −190 MPa Since y is given, x can easily be determined from the principal of stress invariance:  x +  y =  p1 +  p 2   x =  p1 +  p 2 −  y = ( −8 MPa) + ( −190 MPa) − ( −64 MPa) = −134 MPa

Ans.

The maximum in-plane shear stress can be found from  −  p 2 (−8 MPa) − (−190 MPa)  max = p1 = = 91 MPa 2 2 Since x, y, and max are known, the magnitude of xy can be found from the expression

  − y  2  max =  x  +  xy  2  2

 (−134 MPa) − (−64 MPa)  2 91 MPa =   +  xy 2    xy = 84 MPa 2

The problem states that a positive shear stress exists on a horizontal plane; therefore  xy = 84 MPa

Ans.

P12.39 At a point on the free surface of a stressed body, the normal stresses are 20 ksi (T) on a vertical plane and 30 ksi (C) on a horizontal plane. An unknown negative shear stress exists on the vertical plane. The absolute maximum shear stress at the point has a magnitude of 32 ksi. Determine the principal stresses and the shear stress on the vertical plane at the point.

Solution Since x and y have opposite signs, the absolute maximum shear stress is equal to the maximum in-plane shear stress:  max =  abs max = 32 ksi Since x, y, and max are known, the magnitude of xy can be found from the expression

  − y  2  max =  x  +  xy  2  2

 (20 ksi) − ( −30 ksi)  2 32 ksi =   +  xy 2    xy = 19.9750 ksi 2

The problem states that a negative shear stress exists on the vertical plane; therefore  xy = −19.98 ksi

Ans.

From the principal of stress invariance:  x +  y =  p1 +  p 2

 p1 +  p 2 = (20 ksi) + (−30 ksi) = −10 ksi

(a)

The maximum in-plane shear stress is equal to one-half of the difference between the two in-plane principal stresses  −  p2  max = p1 2  p1 −  p 2 = 2 max = 2(32 ksi) = 64 ksi (b) Add Eqs. (a) and (b) to find p1: 2 p1 = 54 ksi  p1 = 27 ksi = 27 ksi (T)

Ans.

and subtract Eq. (b) from Eq. (a) to find p2: 2 p 2 = −74 ksi  p2 = −37 ksi = 37 ksi (C)

Ans.

The point in question occurs on the free surface of a stressed body. From this information, we can know that a state of plane stress exists at the point. Therefore,  z =  p3 = 0 (since the point is on a free surface) Ans.

P12.40 At a point on the free surface of a stressed body, a normal stress of 75 MPa (T) and an unknown negative shear stress exist on a horizontal plane. One principal stress at the point is 200 MPa (T). The maximum in-plane shear stress at the point has a magnitude of 85 MPa. Determine the unknown stresses on the vertical plane, the unknown principal stress, and the absolute maximum shear stress at the point.

Solution Since y = 75 MPa is less than the given principal stress, we will assume that p1 = 200 MPa. If this assumption is true, then p2 can be found from p1 and max:  −  p2  max = p1 2   p 2 =  p1 − 2 max = 200 MPa − 2(85 MPa) = 30 MPa

Ans.

The maximum in-plane shear stress at the point has a magnitude of 85 MPa. However, this assumption cannot be true because the normal stress on the horizontal plane is y = −64 MPa, which is more negative than the given principal stress of −8 MPa. Therefore, we now know that the second principal stress must be negative and its magnitude must be greater than 64 MPa. From the principal of stress invariance:  x +  y =  p1 +  p 2   x = (200 MPa) + (30 MPa) − (75 MPa) = 155 MPa

Ans.

Since x, y, and max are known, the magnitude of xy can be found from the expression

  − y  2  max =  x  +  xy  2  2

 (155 MPa) − (75 MPa)  2 85 MPa =   +  xy 2    xy = 75 MPa 2

The problem states that a negative shear stress exists on the vertical plane; therefore  xy = −75 MPa

Ans.

The point in question occurs on the free surface of a stressed body. From this information, we can know that a state of plane stress exists at the point. Therefore,  z =  p3 = 0 (since it is a free surface) Since both in-plane principal stresses are positive, the absolute maximum shear stress is found from  −  p 3  p1 − 0 200 MPa  abs max = p1 = = = 100 MPa Ans. 2 2 2

P12.41 For the state of plane stress shown, determine (a) the largest value of xy for which the maximum in-plane shear stress is equal to or less than 150 MPa and (b) the corresponding principal stresses.

FIGURE P12.41

Solution Since x, y, and max are known, the magnitude of xy can be found from the expression   − y  2  max =  x  +  xy 2   2

 (120 MPa) − (−70 MPa)  2 150 MPa    +  xy 2   2

 xy2  (150 MPa)2 − (95 MPa)2  xy  116.1 MPa

Ans.

The corresponding principal stresses are:

 p1, p 2 =

x + y 2

  x − y  2    +  xy 2   2

(120 MPa) + (−70 MPa)  (120 MPa) − (−70 MPa)  2 =    + (116.1 MPa) 2 2   = 25 MPa  150 MPa 2

 p1 = 175 MPa (T) and  p 2 = 125 MPa (C)

Ans.

P12.42 Mohr’s circle is shown for a point in a physical object that is subjected to plane stress. (a) Determine the stresses x, y, and xy and show them on a stress element. (b) Determine the principal stresses and the maximum in-plane shear stress acting at the point and show these stresses on an appropriate sketch (e.g., see Figure 12.15 or Figure 12.16).

FIGURE P12.42

Solution x = ( 2,000 psi, 3,000 psi )

 p1 = C + R = 500 psi + 3,354.102 psi = 3,854.102 psi = 3,850 psi

y = ( −1,000 psi, 3,000 psi )

 p 2 = C − R = 500 psi − 3,354.102 psi = −2,854.102 psi = −2,850 psi

C = 500 psi

 max = R = 3,350 psi

(1,500 psi ) + ( 3,000 psi ) = 3,354.102 psi 2

 avg = C = 500 psi

The magnitude of the angle 2p between point x (i.e., the x face of the stress element) and p1 is found from: 3,000 psi 3,000 psi tan 2 p = = = 2.0  2 p = 63.43 thus,  p = 31.72 2,000 psi − 500 psi 1,500 psi By inspection, the angle p from point x to p1 is turned clockwise.

P12.43 Mohr’s circle is shown for a point in a physical object that is subjected to plane stress. (a) Determine the stresses x, y, and xy and show them on a stress element. (b) Determine the principal stresses and the maximum in-plane shear stress acting at the point and show these stresses on an appropriate sketch (e.g., see Figure 12.15 or Figure 12.16).

FIGURE P12.43

Solution x = (10 ksi, 15 ksi ) y = ( −30 ksi, 15 ksi ) C = −10 ksi R=

( 20 ksi ) + (15 ksi ) = 25 ksi 2

 p1 = C + R = −10 ksi + 25 ksi = 15 ksi  p 2 = C − R = −10 ksi − 25 ksi = −35 ksi  max = R = 25 ksi  avg = C = −10 ksi The magnitude of the angle 2p between point x (i.e., the x face of the stress element) and p1 is found from: 15 ksi 15 ksi tan 2 p = = = 0.75  2 p = 36.87 thus,  p = 18.43 10 ksi − ( −10 ksi ) 20 ksi By inspection, the angle p from point x to p1 is turned counterclockwise.

P12.44 Mohr’s circle is shown for a point in a physical object that is subjected to plane stress. (a) Determine the stresses x, y, and xy and show them on a stress element. (b) Determine the principal stresses and the maximum in-plane shear stress acting at the point and show these stresses on an appropriate sketch (e.g., see Figure 12.15 or Figure 12.16).

FIGURE P12.44

Solution x = ( 80 MPa, 70 MPa )

 p1 = C + R = 30 MPa + 86.023 MPa = 116.0 MPa

y = ( −20 MPa, 70 MPa )

 p 2 = C − R = 30 MPa − 86.023 MPa = −56.0 MPa

C = 30 MPa

 max = R = 86.0 MPa

( 50 MPa ) + ( 70 MPa ) = 86.023 MPa 2

 avg = C = 30 MPa

The magnitude of the angle 2p between point x (i.e., the x face of the stress element) and p1 is found from: 70 MPa 70 MPa tan 2 p = = = 1.4  2 p = 54.46 thus,  p = 27.23 80 MPa − 30 MPa 50 MPa By inspection, the angle p from point x to p1 is turned counterclockwise.

P12.45 Mohr’s circle is shown for a point in a physical object that is subjected to plane stress. (a) Determine the stresses x, y, and xy and show them on a stress element. (b) Determine the principal stresses and the maximum in-plane shear stress acting at the point and show these stresses on an appropriate sketch (e.g., see Figure 12.15 or Figure 12.16).

FIGURE P12.45

Solution x = ( 0 MPa, 20 MPa )

 p1 = C + R = 15 MPa + 25 MPa = 40 MPa

y = ( 30 MPa, 20 MPa )

 p 2 = C − R = 15 MPa − 25 MPa = −10 MPa

C = 15 MPa

 max = R = 25 MPa

(15 MPa ) + ( 20 MPa ) = 25 MPa 2

 avg = C = 15 MPa

The magnitude of the angle 2p between point x (i.e., the x face of the stress element) and p2 is found from: 20 MPa 20 MPa tan 2 p = = = 1.3333  2 p = 53.13 thus,  p = 26.57 0 MPa − 15 MPa 15 MPa By inspection, the angle p from point x to p2 is turned clockwise.

P12.46 Mohr’s circle is shown for a point in a physical object that is subjected to plane stress. (a) Determine the stresses x, y, and xy and show them on a stress element. (b) Determine the stresses n, t, and nt and show them on a stress element that is properly rotated with respect to the x-y element. The sketch must include the magnitude of the angle between the x and n axes and an indication of the rotation direction (i.e., either clockwise or counterclockwise).

FIGURE P12.46

Solution x = ( 20 MPa, 60 MPa )

y = ( −60 MPa, 60 MPa )

n = ( 50 MPa, 20 MPa) )

t = ( −90 MPa, 20 MPa )

C = −20 MPa

( 40 MPa ) + ( 60 MPa ) = 72.1 MPa 2

The magnitude of the angle 2p between point x (i.e., the x face of the stress element) and p1 is found from: 60 MPa 60 MPa tan 2 p = = = 1.50  2 p = 56.3 20 MPa − ( −20 MPa ) 40 MPa The magnitude of the angle  between p1 and point n is found from: 20 MPa 20 MPa tan  = = = 0.2857  = 15.9 50 MPa − ( −20 MPa ) 70 MPa Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

The angle  between point x and point n is thus 72.2°. Since angles in Mohr’s circle are doubled, the actual angle between the x face and the n face is half of this magnitude: 36.1°. By inspection, the 72.2° angle from point x to point n is turned in a clockwise direction. The correct stresses on the n and t faces are shown in the sketch below.

P12.47 Mohr’s circle is shown for a point in a physical object that is subjected to plane stress. (a) Determine the stresses x, y, and xy and show them on a stress element. (b) Determine the stresses n, t, and nt and show them on a stress element that is properly rotated with respect to the x-y element. The sketch must include the magnitude of the angle between the x and n axes and an indication of the rotation direction (i.e., either clockwise or counterclockwise).

FIGURE P12.47

Solution x = (12 MPa, 9 MPa )

y = ( −36 MPa, 9 MPa )

n = ( −30 MPa, 18 MPa )

t = ( 6 MPa, 18 MPa )

C = −12 MPa

( 24 MPa ) + ( 9 MPa ) = 25.6 MPa 2

The magnitude of the angle 2p between point x (i.e., the x face of the stress element) and p1 is found from: 9 MPa 9 MPa tan 2 p = = = 0.375  2 p = 20.6 12 MPa − ( −12 MPa ) 24 MPa The magnitude of the angle  between point n and p2 is found from: 18 MPa 18 MPa tan  = = = 1.0  = 45 ( −30 MPa ) − ( −12 MPa ) 18 MPa Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

The angle  between point x and point n is thus  = 180 − 2 p −  = 180 − 20.6 − 45 = 114.4 Since angles in Mohr’s circle are doubled, the actual angle between the x face and the n face is half of this magnitude: 57.2°. By inspection, the 57.2° angle from point x to point n is turned in a clockwise direction. The correct stresses on the n and t faces are shown in the sketch below.

P12.48 Consider a point in a structural member that is subjected to plane stress. Normal and shear stresses acting on horizontal and vertical planes at the point are shown. (a) Draw Mohr’s circle for this state of stress. (b) Determine the principal stresses and the maximum in-plane shear stress acting at the point. (c) Show these stresses on an appropriate sketch (e.g., see Fig. 12.15 or Fig. 12.16). (d) Determine the absolute maximum shear stress at the point. Instructors: Problems P12.48-P12.51 should be assigned as a set.

FIGURE P12.48

Solution (b) The basic Mohr’s circle is shown.

(−36 ksi) + (−18 ksi) = −27 ksi 2

R = (9 ksi) 2 + (12 ksi) 2 = 15 ksi

 p1 = C + R = −27 ksi + 15 ksi = −12 ksi  p 2 = C − R = −27 ksi − 15 ksi = −42 ksi  max = R = 15 ksi  avg = C = 27 ksi (C) The magnitude of the angle 2p between point x (i.e., the x face of the stress element) and point 2 (i.e., the principal plane subjected to p2) is found from:

tan 2 p =

12 ksi 12 ksi = = 1.3333 (−36 ksi) − (−27 ksi) 9 ksi

 2 p = 53.130

thus,  p = 26.57

By inspection, the angle p from point x to point 2 is turned counterclockwise. (c) The orientation of the principal stresses and the maximum in-plane shear stress is shown on the sketch below.

(d) Since the point in a structural member is subjected to plane stress  z =  p3 = 0 Three Mohr’s circles can be constructed to show stress combinations in the p1–p2 plane, the p1–p3 plane, and the p2–p3 plane. These three circles are shown below.

In this case, the absolute maximum shear stress occurs in the p2–p3 plane; therefore,  p 2 −  p3 −42 ksi − 0 ksi  abs max = = = 21 ksi 2 2

Ans.

P12.49 Consider a point in a structural member that is subjected to plane stress. Normal and shear stresses acting on horizontal and vertical planes at the point are shown. (a) Draw Mohr’s circle for this state of stress. (b) Determine the principal stresses and the maximum in-plane shear stress acting at the point. (c) Show these stresses on an appropriate sketch (e.g., see Fig. 12.15 or Fig. 12.16). (d) Determine the absolute maximum shear stress at the point. Instructors: Problems P12.48-P12.51 should be assigned as a set.

FIGURE P12.49

Solution (b) The basic Mohr’s circle construction is shown.

(6 ksi) + (18 ksi) = 12 ksi 2

R = (6 ksi) 2 + (30 ksi) 2 = 30.59 ksi

 p1 = C + R = 12 ksi + 30.59 ksi = 42.59 ksi  p 2 = C − R = 12 ksi − 30.59 ksi = −18.59 ksi  max = R = 30.59 ksi  avg = C = 12 ksi The magnitude of the angle 2p between point x (i.e., the x face of the stress element) and point 2 (i.e., the principal plane subjected to p2) is found from: Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

tan 2 p =

30 ksi 30 ksi = =5 (6 ksi) − (12 ksi) 6 ksi

 2 p = 78.690

thus,  p = 39.35

By inspection, the angle p from point x to point 2 is turned clockwise. (c) The orientation of the principal stresses and the maximum in-plane shear stress is shown on the sketch below.

In this case, the absolute maximum shear stress occurs in the p1–p2 plane (which is also the x-y plane). Therefore Ans.  abs max =  max = 30.59 ksi

P12.50 Consider a point in a structural member that is subjected to plane stress. Normal and shear stresses acting on horizontal and vertical planes at the point are shown. (a) Draw Mohr’s circle for this state of stress. (b) Determine the principal stresses and the maximum in-plane shear stress acting at the point. (c) Show these stresses on an appropriate sketch (e.g., see Fig. 12.15 or Fig. 12.16). (d) Determine the absolute maximum shear stress at the point. Instructors: Problems P12.48-P12.51 should be assigned as a set.

FIGURE P12.50

Solution (b) The basic Mohr’s circle is shown.

(−35 MPa) + (−65 MPa) = −50 MPa 2

R = (15 MPa) 2 + (24 MPa) 2 = 28.30 MPa

 p1 = C + R = −50 MPa + 28.30 MPa = −21.70 MPa  p 2 = C − R = −50 MPa − 28.30 MPa = −78.30 MPa  max = R = 28.30 MPa  avg = C = 50 MPa (C) The magnitude of the angle 2p between point x (i.e., the x face of the stress element) and point 1 (i.e., the principal plane subjected to p1) is found from:

tan 2 p =

24 MPa 24 MPa = = 1.6 (−35 MPa) − (−50 MPa) 15 MPa

 2 p = 57.995

thus,  p = 29.00

By inspection, the angle p from point x to point 1 is turned counterclockwise. (c) The orientation of the principal stresses and the maximum in-plane shear stress is shown on the sketch below.

In this case, the absolute maximum shear stress occurs in the p2–p3 plane. Therefore  p 2 −  p3 −78.30 MPa − 0 MPa  abs max = = = 39.15 MPa 2 2

Ans.

P12.51 Consider a point in a structural member that is subjected to plane stress. Normal and shear stresses acting on horizontal and vertical planes at the point are shown. (a) Draw Mohr’s circle for this state of stress. (b) Determine the principal stresses and the maximum in-plane shear stress acting at the point. (c) Show these stresses on an appropriate sketch (e.g., see Fig. 12.15 or Fig. 12.16). (d) Determine the absolute maximum shear stress at the point. Instructors: Problems P12.48-P12.51 should be assigned as a set.

FIGURE P12.51

Solution (b) The basic Mohr’s circle is shown.

(0 MPa) + (−45 MPa) = −22.5 MPa 2

R = (22.5 MPa) 2 + (25 MPa) 2 = 33.63 MPa

 p1 = C + R = −22.5 MPa + 33.63 MPa = 11.13 MPa  p 2 = C − R = −22.5 MPa − 33.63 MPa = −56.13 MPa  max = R = 33.63 MPa  avg = C = 22.5 MPa (C) The magnitude of the angle 2p between point x (i.e., the x face of the stress element) and point 1 (i.e., the principal plane subjected to p1) is found from:

tan 2 p =

25 MPa 25 MPa = = 1.1111 (0 MPa) − (−22.5 MPa) 22.5 MPa

 2 p = 48.013

thus,  p = 24.01

By inspection, the angle p from point x to point 1 is turned clockwise. (c) The orientation of the principal stresses and the maximum in-plane shear stress is shown on the sketch below.

In this case, the absolute maximum shear stress occurs in the p1–p2 plane (which is also the x-y plane). Therefore Ans.  abs max =  max = 33.63 MPa Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

P12.52 Consider a point in a structural member that is subjected to plane stress. Normal and shear stresses acting on horizontal and vertical planes at the point are shown. (a) Draw Mohr’s circle for this state of stress. (b) Determine the principal stresses and the maximum in-plane shear stress acting at the point. (c) Show these stresses on an appropriate sketch (e.g., see Figure 12.15 or Figure 12.16). (d) Determine the absolute maximum shear stress at the point. Instructors: Problems P12.52-P12.55 should be assigned as a set.

FIGURE P12.52

Solution (b) The basic Mohr’s circle is shown.

(60.5 MPa) + (8.4 MPa) = 34.45 MPa 2

R = (26.05 MPa) 2 + (44.8 MPa) 2 = 51.8232 MPa

 p1 = C + R = 34.45 MPa + 51.8232 MPa = 86.3 MPa  p 2 = C − R = 34.45 MPa − 51.8232 MPa = −17.37 MPa  max = R = 51.8 MPa  avg = C = 34.5 MPa (T) The magnitude of the angle 2p between point x (i.e., the x face of the stress element) and point 2 (i.e., the principal plane subjected to p2) is found from: 44.8 MPa 44.8 MPa tan 2 p = = = 1.7198  2 p = 59.8231 thus,  p = 29.9 (60.5 MPa) − (34.45 MPa) 26.05 MPa By inspection, the angle p from point x to point 1 is turned counterclockwise.

In this case, the absolute maximum shear stress occurs in the p1–p2 plane (which is also the x-y plane). Therefore Ans.  abs max =  max = 51.8 MPa

P12.53 Consider a point in a structural member that is subjected to plane stress. Normal and shear stresses acting on horizontal and vertical planes at the point are shown. (a) Draw Mohr’s circle for this state of stress. (b) Determine the principal stresses and the maximum in-plane shear stress acting at the point. (c) Show these stresses on an appropriate sketch (e.g., see Figure 12.15 or Figure 12.16). (d) Determine the absolute maximum shear stress at the point. Instructors: Problems P12.52-P12.55 should be assigned as a set.

FIGURE P12.53

Solution (b) The basic Mohr’s circle is shown.

(96 MPa) + (126 MPa) = 111 MPa 2

R = (15 MPa) 2 + (66 MPa) 2 = 67.6831 MPa

 p1 = C + R = 111 MPa + 67.6831 MPa = 178.7 MPa  p 2 = C − R = 111 MPa − 67.6831 MPa = 43.3 MPa  max = R = 67.7 MPa  avg = C = 111 MPa (T) The magnitude of the angle 2p between point x (i.e., the x face of the stress element) and point 2 (i.e., the principal plane subjected to p2) is found from: 66 MPa 66 MPa tan 2 p = = = 4.400 2 p = 77.1957 thus,  p = 38.6 (96 MPa) − (111 MPa) 15 MPa By inspection, the angle p from point x to point 2 is turned clockwise. (c) The orientation of the principal stresses and the maximum in-plane shear stress is shown in the sketch below.

In this case, the absolute maximum shear stress occurs in the p1–p3 plane; therefore,  p1 −  p 3 178.68 MPa − 0 MPa  abs max = = = 89.3 MPa 2 2

Ans.

P12.54 Consider a point in a structural member that is subjected to plane stress. Normal and shear stresses acting on horizontal and vertical planes at the point are shown. (a) Draw Mohr’s circle for this state of stress. (b) Determine the principal stresses and the maximum in-plane shear stress acting at the point. (c) Show these stresses on an appropriate sketch (e.g., see Figure 12.15 or Figure 12.16). (d) Determine the absolute maximum shear stress at the point. Instructors: Problems P12.52-P12.55 should be assigned as a set.

FIGURE P12.54

Solution (b) The basic Mohr’s circle is shown.

(17.5 ksi) + (30.0 ksi) = 23.75ksi 2

R = (6.25 ksi) 2 + (5.5 ksi) 2 = 8.3254 ksi

 p1 = C + R = 23.75 ksi + 8.3254 ksi = 32.1 ksi  p 2 = C − R = 23.75 ksi − 8.3254 ksi = 15.42 ksi  max = R = 8.33 ksi  avg = C = 23.75 ksi (T) The magnitude of the angle 2p between point x (i.e., the x face of the stress element) and point 2 (i.e., the principal plane subjected to p2) is found from:

tan 2 p =

5.5 ksi 5.5 ksi = = 0.8800  2 p = 41.3478 (17.5 ksi) − (23.75 ksi) 6.25 ksi

thus,  p = 20.7

By inspection, the angle p from point x to point 2 is turned counterclockwise. (c) The orientation of the principal stresses and the maximum in-plane shear stress is shown in the sketch below.

In this case, the absolute maximum shear stress occurs in the p1–p3 plane; therefore,  p1 −  p 3 32.0754 ksi − 0 ksi  abs max = = = 16.04 ksi 2 2

Ans.

P12.55 Consider a point in a structural member that is subjected to plane stress. Normal and shear stresses acting on horizontal and vertical planes at the point are shown. (a) Draw Mohr’s circle for this state of stress. (b) Determine the principal stresses and the maximum in-plane shear stress acting at the point. (c) Show these stresses on an appropriate sketch (e.g., see Figure 12.15 or Figure 12.16). (d) Determine the absolute maximum shear stress at the point. Instructors: Problems P12.52-P12.55 should be assigned as a set.

FIGURE P12.55

Solution (b) The basic Mohr’s circle is shown.

( −950 psi) + ( −2, 250 psi) = −1,600 psi 2

R = (950 psi) 2 + (680 psi) 2 = 940.6912 psi

 p1 = C + R = −1,600 psi + 940.6912 psi = −659 psi  p 2 = C − R = −1,600 psi − 940.6912 psi = −2,540 psi  max = R = 941 psi  avg = C = 1,600 psi (C) The magnitude of the angle 2p between point x (i.e., the x face of the stress element) and point 2 (i.e., the principal plane subjected to p2) is found from: 680 psi 680 psi tan 2 p = = = 1.0462  2 p = 46.2922 thus,  p = 23.1 ( −950 psi) − (−1,600 psi) 650 psi By inspection, the angle p from point x to point 1 is turned clockwise. (c) The orientation of the principal stresses and the maximum in-plane shear stress is shown in the sketch below.

In this case, the absolute maximum shear stress occurs in the p2–p3 plane; therefore,  p 2 −  p3 −2,540.69128 psi − 0 psi  abs max = = = 1,270 psi 2 2

Ans.

P12.56 Consider a point in a structural member that is subjected to plane stress. Normal and shear stresses acting on horizontal and vertical planes at the point are shown. (a) Draw Mohr’s circle for this state of stress. (b) Determine the principal stresses and the maximum in-plane shear stress acting at the point. Show these stresses on an appropriate sketch (e.g., see Figure 12.15 or Figure 12.16). (c) Determine the normal and shear stresses on the indicated plane and show these stresses on a sketch. (d) Determine the absolute maximum shear stress at the point. Instructors: Problems P12.56-P12.59 should be assigned as a set.

FIGURE P12.56

Solution (b) The basic Mohr’s circle is shown.

( −3.8 ksi ) + ( 9.4 ksi ) = 2.80 ksi

( 6.6 ksi ) + ( 5.7 ksi ) = 8.72 ksi

2 2

 p1 = C + R = 2.80 ksi + 8.72 ksi = 11.52 ksi  p 2 = C − R = 2.80 ksi − 8.72 ksi = −5.92 ksi  max = R = 8.72 ksi  avg = C = 2.80 ksi (T)

The magnitude of the angle 2p between point x (i.e., the x face of the stress element) and point p2 (i.e., the principal plane subjected to p2) is found from:

tan 2 p =

5.70 ksi 5.70 ksi = = 0.86364 ( −3.80 ksi ) − ( 2.80 ksi ) 6.60 ksi

 2 p = 40.82

thus,  p = 20.41

By inspection, the angle p from point x to point p2 is turned clockwise. The orientation of the principal stresses and the maximum in-plane shear stress is shown in the sketch below.

(c) To determine the normal and shear stresses on the indicated plane, we must first determine the orientation of the n plane relative to the x face of the stress element. Looking at the stress element, we observe that the normal to the indicated plane is oriented 38.66° counterclockwise from the x axis. In Mohr’s circle, all angle measures are doubled; therefore, point n (which represents the state of stress on the n plane) on Mohr’s circle is rotated 2(38.66°) = 77.32° counterclockwise from point x. The angle between point n and point p1 is  = 180 − 40.82 − 77.32 = 61.86 The  coordinate of point n is found from:  n = C + R cos  = 2.80 ksi + ( 8.72 ksi ) cos 61.86 = 6.91 ksi = 6.91 ksi (T)

Ans.

The  coordinate of point n is found from:  nt = R sin  = ( 8.72 ksi ) sin 61.86 = 7.69 ksi

Ans.

Since point n is below the  axis, the shear stress acting on the plane surface tends to rotate the stress element counterclockwise. (d) Since the point in a structural member is subjected to plane stress  z =  p3 = 0 Three Mohr’s circles can be constructed to show stress combinations in the p1–p2 plane, the p1–p3 plane, and the p2–p3 plane. These three circles are shown below.

In this case, the absolute maximum shear stress occurs in the p1–p2 plane (which is also the x-y plane). Therefore Ans.  abs max =  max = 8.72 ksi

P12.57 Consider a point in a structural member that is subjected to plane stress. Normal and shear stresses acting on horizontal and vertical planes at the point are shown. (a) Draw Mohr’s circle for this state of stress. (b) Determine the principal stresses and the maximum in-plane shear stress acting at the point. Show these stresses on an appropriate sketch (e.g., see Figure 12.15 or Figure 12.16). (c) Determine the normal and shear stresses on the indicated plane and show these stresses on a sketch. (d) Determine the absolute maximum shear stress at the point. Instructors: Problems P12.56-P12.59 should be assigned as a set.

FIGURE P12.57

Solution (b) The basic Mohr’s circle is shown.

(108 MPa ) + ( −14 MPa ) = 47 MPa

(108 MPa − 47 MPa ) + ( 72 MPa ) = 94.37 MPa

 p1 = C + R = 47 MPa + 94.37 MPa = 141.4 MPa  p 2 = C − R = 47 MPa − 94.37 MPa = −47.4 MPa  max = R = 94.4 MPa  avg = C = 47 MPa (T)

The magnitude of the angle 2p between point x (i.e., the x face of the stress element) and point p1 (i.e., the principal plane subjected to p1) is found from:

tan 2 p =

72 MPa 72 MPa = = 1.18032 (108 MPa ) − ( 47 MPa ) 61 MPa

 2 p = 49.73

thus,  p = 24.86

By inspection, the angle p from point x to point p1 is turned in a clockwise direction. The orientation of the principal stresses and the maximum in-plane shear stress is shown in the sketch below.

(c) To determine the normal and shear stresses on the indicated plane, we must first determine the orientation of the n plane relative to the x face of the stress element. Looking at the stress element, we observe that the normal to the indicated plane is oriented 50° counterclockwise from the x axis. In Mohr’s circle, all angle measures are doubled; therefore, point n (which represents the state of stress on the n plane) on Mohr’s circle is rotated 2(50°) = 100° counterclockwise from point x. The angle between point n and point p2 is  = 180 − 49.73 − 100 = 30.27 The  coordinate of point n is found from:  n = C − R cos  = 47 MPa − ( 94.37 MPa ) cos30.27 = −34.50 MPa = 34.5 MPa (C)

Ans.

The  coordinate of point n is found from:  nt = R sin  = ( 94.37 MPa ) sin 30.27 = 47.6 MPa

Ans.

Since point n is above the  axis, the shear stress acting on the plane surface tends to rotate the stress element clockwise. (d) Since the point in a structural member is subjected to plane stress  z =  p3 = 0 Three Mohr’s circles can be constructed to show stress combinations in the p1–p2 plane, the p1–p3 plane, and the p2–p3 plane. These three circles are shown below.

In this case, the absolute maximum shear stress occurs in the p1–p2 plane (which is also the x-y plane). Therefore Ans.  abs max =  max = 94.4 MPa

P12.58 Consider a point in a structural member that is subjected to plane stress. Normal and shear stresses acting on horizontal and vertical planes at the point are shown. (a) Draw Mohr’s circle for this state of stress. (b) Determine the principal stresses and the maximum in-plane shear stress acting at the point. Show these stresses on an appropriate sketch (e.g., see Figure 12.15 or Figure 12.16). (c) Determine the normal and shear stresses on the indicated plane and show these stresses on a sketch. (d) Determine the absolute maximum shear stress at the point. Instructors: Problems P12.56-P12.59 should be assigned as a set.

FIGURE P12.58

Solution (b) The basic Mohr’s circle is shown. C=

( 28 MPa ) + ( −42 MPa ) = −7 MPa

( 28 MPa − ( −7 MPa ) ) + ( 20 MPa ) = 40.31 MPa

 p1 = C + R = −7 MPa + 40.31 MPa = 33.3 MPa  p 2 = C − R = −7 MPa − 40.31 MPa = −47.3 MPa  max = R = 40.3 MPa  avg = C = 7 MPa (C)

The magnitude of the angle 2p between point x (i.e., the x face of the stress element) and point p1 (i.e., the principal plane subjected to p1) is found from:

tan 2 p =

20 MPa 20 MPa = = 0.57143 ( 28 MPa ) − ( −7 MPa ) 35 MPa

 2 p = 29.74

thus,  p = 14.87

By inspection, the angle p from point x to point p1 is turned in a counterclockwise direction. The orientation of the principal stresses and the maximum in-plane shear stress is shown in the sketch below.

(c) To determine the normal and shear stresses on the indicated plane, we must first determine the orientation of the n plane relative to the x face of the stress element. Looking at the stress element, we observe that the normal to the indicated plane is oriented 68.20° counterclockwise from the x axis. In Mohr’s circle, all angle measures are doubled; therefore, point n (which represents the state of stress on the n plane) on Mohr’s circle is rotated 2(68.20°) = 136.40° counterclockwise from point x. The angle between point n and point p1 is 136.40 − 29.74 = 106.66 Consequently, the angle between point n and point p2 is  = 180 − 106.66 = 73.34 The  coordinate of point n is found from:  n = C − R cos  = −7 MPa − ( 40.31 MPa ) cos 73.34 = −18.55 MPa = 18.55 MPa (C)

Ans.

The  coordinate of point n is found from:  nt = R sin  = ( 40.31 MPa ) sin 73.34 = 38.6 MPa

Ans.

Since point n is above the  axis, the shear stress acting on the plane surface tends to rotate the stress element clockwise.

In this case, the absolute maximum shear stress occurs in the p1–p2 plane (which is also the x-y plane). Therefore Ans.  abs max =  max = 40.3 MPa

P12.59 Consider a point in a structural member that is subjected to plane stress. Normal and shear stresses acting on horizontal and vertical planes at the point are shown. (a) Draw Mohr’s circle for this state of stress. (b) Determine the principal stresses and the maximum in-plane shear stress acting at the point. Show these stresses on an appropriate sketch (e.g., see Figure 12.15 or Figure 12.16). (c) Determine the normal and shear stresses on the indicated plane and show these stresses on a sketch. (d) Determine the absolute maximum shear stress at the point. Instructors: Problems P12.56-P12.59 should be assigned as a set.

FIGURE P12.59

Solution (b) The basic Mohr’s circle is shown.

( −2,150 psi ) + (860 psi ) = −645 psi

( −2,150 psi − ( −645 psi ) ) + (1, 460 psi ) = 2,096.8 psi

 p1 = C + R = −645 psi + 2,096.8 psi = 1, 451.8 psi = 1, 452 psi  p 2 = C − R = −645 psi − 2,096.8 psi = −2,741.8 psi = −2,740 psi  max = R = 2,100 psi  avg = C = −645 psi

The magnitude of the angle 2p between point x (i.e., the x face of the stress element) and point p2 (i.e., the principal plane subjected to p2) is found from:

tan 2 p =

1,460 psi 1,460 psi = = 0.97010 ( −2,150 psi ) − ( −645.0 psi ) 1,505 psi

 2 p = 44.13

thus,  p = 22.1

By inspection, the angle p from point x to point p2 is turned in a counterclockwise direction. The orientation of the principal stresses and the maximum in-plane shear stress is shown in the sketch below.

(c) To determine the normal and shear stresses on the indicated plane, we must first determine the orientation of the n plane relative to the x face of the stress element. Looking at the stress element, we observe that the normal to the indicated plane is oriented 40° clockwise from the x axis. In Mohr’s circle, all angle measures are doubled; therefore, point n (which represents the state of stress on the n plane) on Mohr’s circle is rotated 2(40°) = 80° clockwise from point x. The angle between point n and point p1 is  = 180 − 44.13 − 80 = 55.87 The  coordinate of point n is found from:  n = C + R cos 

= −645 psi + ( 2,096.8 psi ) cos55.87 = 531.5 psi (T)

Ans.

The  coordinate of point n is found from:  nt = R sin  = ( 2,096.8 psi ) sin 55.87 = 1,735.7 psi = 1,736 psi

Ans.

Since point n is above the  axis, the shear stress acting on the plane surface tends to rotate the stress element clockwise.

In this case, the absolute maximum shear stress occurs in the p1–p2 plane (which is also the x-y plane). Therefore Ans.  abs max =  max = 2,100 psi

P12.60 At a point in a stressed body, the principal stresses are oriented as shown in Figure P12.60. Use Mohr’s circle to determine: (a) the stresses on plane a-a. (b) the stresses on the horizontal and vertical planes at the point. (c) the absolute maximum shear stress at the point. FIGURE P12.60

Solution The center of Mohr’s circle can be found from the two principal stresses:  p1 +  p 2 (150 MPa) + (30 MPa) C= = = 90 MPa 2 2 The radius of the circle is  p1 −  p 2 (150 MPa) − (30 MPa) R= = = 60 MPa 2 2 (a) The stresses on plane a-a are found by rotating 270° counterclockwise from the p1 point on Mohr’s circle. Therefore, the point at the bottom of the circle directly underneath the center corresponds to the state of stress on plane a-a. Ans.  a − a = C = 90 MPa = 90 MPa (T)

 a − a = R = 60 MPa

(shear stress rotates the wedge element counterclockwise)

Ans.

(b) The angle p shown on the problem statement sketch is 1  3  p = tan −1   = 18.4349  4 2 The p1 principal plane is rotated 18.4349° counterclockwise from the x face of the stress element. We need to find the point on Mohr’s circle that corresponds to the x face of the stress element. Since we know the location of p1 on Mohr’s circle, we can begin there and rotate 2p in the opposite direction to find point x. Therefore, beginning at point p1, rotate 2(18.4349°) = 36.8698° clockwise to locate point x. The  coordinate of point x is found from:  x = C + R cos(2 p )

= 90 MPa + (60 MPa)cos(36.8698) = 138.0 MPa (T)

Ans.

The  coordinate of point x is found from:  nt = R sin(2 p )

= (60 MPa)sin(36.8698) = 36.0 MPa Similarly, the  coordinate of point y is found from:  y = C − R cos(2 p )

(rotates element counterclockwise)

= 90 MPa − (60 MPa)cos(36.8698) = 42.0 MPa (T)

Ans.

The  coordinate of point y is also 42.0 MPa, and the shear stress on the y face rotates the stress element clockwise. The stresses on the vertical and horizontal faces of the stress element are shown below.

(c) Since both p1 and p2 are positive, the absolute maximum shear stress will be larger than the maximum in-plane shear stress. The radius of the largest Mohr’s circle gives the absolute maximum shear stress. In this case, the absolute maximum shear stress occurs in the p1–p3 plane; therefore,  −  p 3 150 MPa − 0 MPa  abs max = p1 = = 75.0 MPa Ans. 2 2

P12.61 At a point in a stressed body, the principal stresses are oriented as shown in Figure P12.61. Use Mohr’s circle to determine: (a) the stresses on plane a-a. (b) the stresses on the horizontal and vertical planes at the point. (c) the absolute maximum shear stress at the point. FIGURE P12.61

Solution The center of Mohr’s circle can be found from the two principal stresses:  p1 +  p 2 ( −7.82 ksi) + ( −30.18 ksi) C= = = −19.00 ksi 2 2 The radius of the circle is  p1 −  p 2 ( −7.82 ksi) − ( −30.18 ksi) R= = = 11.18 ksi 2 2 (a) The stresses on plane a-a are found by rotating 270° counterclockwise from the p2 point on Mohr’s circle. Therefore, the point at the top of the circle directly above the center corresponds to the state of stress on plane a-a. Ans.  a − a = C = −19.00 ksi = 19.00 ksi (C)

 a − a = R = 11.18 ksi

(shear stress rotates the wedge element clockwise)

Ans.

(b) The angle p shown in the problem statement sketch is 1  p = tan −1 (2) = 31.7175 2 The p2 principal plane is rotated 31.7175° clockwise from the x face of the stress element. We need to find the point on Mohr’s circle that corresponds to the x face of the stress element. Since we know the location of p2 on Mohr’s circle, we can begin there and rotate 2p in the opposite direction to find point x. Therefore, beginning at point p2, rotate 2(31.7175°) = 63.4349° counterclockwise to locate point x. The  coordinate of point x is found from:  x = C − R cos(2 p )

= −19 ksi − (11.18 ksi)cos(63.4349) = 24.0 ksi (C)

Ans.

The  coordinate of point x is found from:  nt = R sin(2 p )

= (11.18 ksi)sin(63.4349) = 10.00 ksi Similarly, the  coordinate of point y is found from:  y = C + R cos(2 p )

(rotates element counterclockwise)

= −19 ksi + (11.18 ksi)cos(63.4349) = 14.00 ksi (C)

Ans.

The  coordinate of point y is also 10.00 ksi, and the shear stress on the y face rotates the stress element clockwise. The stresses on the vertical and horizontal faces of the stress element are shown below.

(c) Since both p1 and p2 are negative, the absolute maximum shear stress will be larger than the maximum in-plane shear stress. The radius of the largest Mohr’s circle gives the absolute maximum shear stress. In this case, the absolute maximum shear stress occurs in the p2–p3 plane; therefore,  p 2 −  p3 −30.1803 ksi − 0 ksi Ans.  abs max = = = 15.09 ksi 2 2

P12.62 A solid 1.50 in. diameter shaft is subjected to a torque of T = 330 lb·ft and an axial load P acting as shown in Figure P12.62/63. If the largest tensile normal stress in the shaft must be limited to 12,000 psi, what is the largest load P that can be applied to the shaft?

FIGURE P12.62/63

Solution Section properties:



(1.50 in.) = 1.76715 in.2

 32

(1.50 in.) = 0.49701 in.4 4

Shear stress magnitudes: Tc ( 330 lb  ft )(1.50 in. / 2 )(12 in./ft ) = = = 5,975.744 psi J 0.49701 in.4 (sense of shear stress determined by inspection) Mohr circle

P P P = = 2 2 A 2 (1.76715 in.) 3.53430 in.2 2

P lb   P   Tc     R=  +  5,975.744 2   +  =  2  in.   2A   J   3.53430 in.  

 p1 = C + R =

2 P P   +  + ( 5,975.744 lb/in.2 )  12,000 lb/in.2 2 2  3.53430 in.  3.53430 in.  2

2 P P   12,000 lb/in. −   + ( 5,975.744 lb/in.2 ) 2 2  3.53430 in.  3.53430 in.  2

P + (12,000 lb/in. ) − 2 (12,000 lb/in. ) 3.53430 in. 2 2

 2

( 3.53430 in. ) ( 3.53430 in. ) 2

2 2

+ ( 5,975.744 lb/in.2 )

2 2 3.53430 in. P  (12,000 lb/in.2 ) − ( 5,975.744 lb/in.2 )  = 15,947 lb   24,000 lb/in.2

P  15,950 lb

Ans.

P12.63 A solid 20 mm diameter shaft is subjected to an axial load of P = 30 kN and a torque of T acting as shown in Figure P12.62/63. If the largest shear stress in the shaft must be limited to 90 MPa, what is the largest torque T that can be applied to the shaft?

FIGURE P12.62/63

Solution Section properties:



( 20 mm ) = 314.159 mm2

 32

( 20 mm ) = 15,707.963 mm 4 4

Normal stress magnitude: P ( 30 kN )(1,000 N/kN ) = = = 95.493 MPa A 314.159 mm 2 Mohr circle

P 30,000 N = = 47.746 MPa 2 A 2 ( 314.159 mm )2 2

 P   Tc  R=   +  =  2A   J 

( 47.746 MPa ) +  2

Tc    J 

 Tc   max = R = ( 47.746 MPa ) +    90 MPa  J  2

2 2  Tc     ( 90 MPa ) − ( 47.746 MPa )  J 

Tc  J

( 90 MPa ) − ( 47.746 MPa ) = 76.291 MPa 2

( 76.291 N/mm )(15,707.963 mm ) = 119,837.649 N  mm = 119.838 N  m T 2

20 mm / 2

T  119.8 N  m

Ans.

P12.64 At a point in a solid body subjected to plane stress, x = 130 MPa and y = 48 MPa, acting as shown in Figure P12.64a. For plane n shown in Figure P12.64b, determine: (a) the resultant stress S. (b) the normal stress n and the shear stress nt.

FIGURE P12.64a

FIGURE P12.64b

Solution The known stresses are  x = 130 MPa  y = 48 MPa

 xy = 0 MPa

 z = 0 MPa

 yz = 0 MPa

 zx = 0 MPa

(a) The plane of interest is defined by its direction cosines: l = cos90 = 0 m = cos45 = 0.70711

n = cos45 = 0.70711

The three orthogonal components of the resultant stress are: S x =  x  l +  xy  m +  zx  n = (130 )( 0 ) + ( 0 )( 0.70711) + ( 0 )( 0.70711) = 0 MPa S y =  xy  l +  y  m +  yz  n = ( 0 )( 0 ) + ( 48 )( 0.70711) + ( 0 )( 0.70711) = 33.941 MPa S z =  zx  l +  yz  m +  z  n = ( 0 )( 0 ) + ( 0 )( 0.70711) + ( 0 )( 0.70711) = 0 MPa

The magnitude of the resultant stress is S = S x2 + S y2 + S z2 =

( 0 ) + ( 33.941) + ( 0 ) = 33.941 MPa = 33.9 MPa 2

Ans.

(b) The normal component n of the resultant stress is  n = Sx  l + S y  m + Sz  n

= ( 0 )( 0 ) + ( 33.941)( 0.70711) + ( 0 )( 0.70711) = 24.0 MPa = 24.0 MPa (T)

Ans.

The shear stress nt on the oblique plane can be obtained from the relation S 2 =  n2 + nt2 .

 nt =

( S −  ) = (33.941) − ( 24.0) = 24.0 MPa 2

2 n

Ans.

P12.65 At a point in a solid body subjected to plane stress, x = 9.2 ksi, y = 35.8 ksi, and xy = 6.4 ksi, acting as shown in Figure P12.65a. For plane n shown in Figure P12.65b, determine: (a) the resultant stress S. (b) the normal stress n and the shear stress nt.

FIGURE P12.65a

FIGURE P12.65b

Solution The known stresses are  x = 9.2 ksi  y = 35.8 ksi

 xy = 6.4 ksi

 yz = 0 ksi

 z = 0 ksi  zx = 0 ksi

(a) The plane of interest is defined by its direction cosines: l = cos45 = 0.70711 m = cos90 = 0

n = cos45 = 0.70711

The three orthogonal components of the resultant stress are: S x =  x  l +  xy  m +  zx  n = ( 9.2 )( 0.70711) + ( 6.4 )( 0 ) + ( 0 )( 0.70711) = 6.505 ksi S y =  xy  l +  y  m +  yz  n = ( 6.4 )( 0.70711) + ( 35.8)( 0 ) + ( 0 )( 0.70711) = 4.525 ksi S z =  zx  l +  yz  m +  z  n = ( 0 )( 0.70711) + ( 0 )( 0 ) + ( 0 )( 0.70711) = 0 ksi

The magnitude of the resultant stress is S = S x2 + S y2 + S z2 =

( 6.505 ) + ( 4.525) + ( 0 ) = 7.9246 ksi = 7.92 ksi 2

Ans.

(b) The normal component n of the resultant stress is  n = Sx  l + S y  m + Sz  n

= ( 6.505)( 0.70711) + ( 4.525)( 0 ) + ( 0 )( 0.70711) = 4.60 ksi (T)

Ans.

The shear stress nt on the oblique plane can be obtained from the relation S 2 =  n2 + nt2 .

 nt =

( S −  ) = ( 7.9246) − ( 4.60) = 6.45 ksi 2

2 n

Ans.

P12.66 The stresses at point O in a solid body that is subjected to plane stress are x = 16.5 ksi, y = −5.8 ksi, and xy = 8.7 ksi. Stresses are to be determined on the inclined plane shown in Figure P12.66/67. The normal to the inclined surface is defined by vector OA, where point A has coordinates of a = 1.0 in., b = 1.4 in., and c = 2.5 in. Determine: (a) the direction cosines for vector OA. (b) the resultant stress S on the inclined plane. (c) the normal and shear stresses on the inclined plane.

FIGURE P12.66/P12.67

Solution (a) The normal to the plane of interest is defined by the vector that begins at the origin O and passes through point A. OA = ai + bj + ck = 1.0i + 1.4 j + 2.5k The length of this vector is calculated as: OA =

(1.0 ) + (1.4 ) + ( 2.5) = 3.0348 2

Create a unit vector in the direction of OA: OA 1.0 1.4 2.5 = i+ j+ k = 0.32951i + 0.46132 j + 0.82378k OA 3.0348 3.0348 3.0348 The direction cosines for the plane of interest are thus: l = 0.32951 m = 0.46132 n = 0.82378 (b) The known stresses are  x = 16.5 ksi  y = −5.8 ksi

 xy = 8.7 ksi

 yz = 0 ksi

Ans.

 z = 0 ksi  zx = 0 ksi

The three orthogonal components of the resultant stress are: S x =  x  l +  xy  m +  zx  n = (16.5 )( 0.32951) + ( 8.7 )( 0.46132 ) + ( 0 )( 0.82378 ) = 9.4504 ksi S y =  xy  l +  y  m +  yz  n = ( 8.7 )( 0.32951) + ( −5.8 )( 0.46132 ) + ( 0 )( 0.82378 ) = 0.1911 ksi S z =  zx  l +  yz  m +  z  n = ( 0 )( 0.32951) + ( 0 )( 0.46132 ) + ( 0 )( 0.82378 ) = 0 ksi

The magnitude of the resultant stress is S = S x2 + S y2 + S z2 =

( 9.4504 ) + ( 0.1911) + ( 0 ) = 9.4523 ksi = 9.45 ksi 2

Ans.

= ( 9.4504 )( 0.32951) + ( 0.1911)( 0.46132 ) + ( 0 )( 0.82378 ) = 3.20217 = 3.20 ksi (T)

Ans.

The shear stress nt on the oblique plane can be obtained from the relation S 2 =  n2 + nt2 .

 nt =

( S −  ) = (9.4523) − (3.20217) = 8.89 ksi 2

2 n

Ans.

P12.67 The stresses at point O in a solid body are x = −80 MPa, y = −40 MPa, z = −40 MPa, xy = 23 MPa, xz = −65 MPa, and yz = 49 MPa. Stresses are to be determined on the inclined plane shown in Figure P12.66/67. The normal to the inclined surface is defined by vector OA, where point A has coordinates of a = 90 mm, b = 160 mm, and c = 220 mm. Determine: (a) the direction cosines for vector OA. (b) the resultant stress S on the inclined plane. (c) the normal and shear stresses on the inclined plane.

FIGURE P12.66/P12.67

Solution (a) The normal to the plane of interest is defined by the vector that begins at the origin O and passes through point A. OA = ai + bj + ck = 90i + 160 j + 220k The length of this vector is calculated as: OA =

( 90 ) + (160 ) + ( 220 ) = 286.531 2

Create a unit vector in the direction of OA: OA 90 160 220 = i+ j+ k = 0.31410i + 0.55840 j + 0.76781k OA 286.531 286.531 286.531 The direction cosines for the plane of interest are thus: l = 0.31410 m = 0.55840 n = 0.76781 (b) The known stresses are  x = −80 MPa  y = −40 MPa

 xy = 23 MPa

Ans.

 z = −40 MPa

 yz = 49 MPa

 zx = −65 MPa

The three orthogonal components of the resultant stress are: S x =  x  l +  xy  m +  zx  n = ( −80 )( 0.31410 ) + ( 23)( 0.55840 ) + ( −65 )( 0.76781) = −62.1922 MPa S y =  xy  l +  y  m +  yz  n = ( 23)( 0.31410 ) + ( −40 )( 0.55840 ) + ( 49 )( 0.76781) = 22.5107 MPa S z =  zx  l +  yz  m +  z  n = ( −65 )( 0.31410 ) + ( 49 )( 0.55840 ) + ( −40 )( 0.76781) = −23.7671 MPa

The magnitude of the resultant stress is S = S x2 + S y2 + S z2 =

( −62.1922 ) + ( 22.5107 ) + ( −23.7671) = 70.2814 MPa = 70.3 MPa 2

Ans.

= ( −62.1922 )( 0.31410 ) + ( 22.5107 )( 0.55840 ) + ( −23.7671)( 0.76781) = −25.21315 = 25.2 MPa (C)

Ans.

The shear stress nt on the oblique plane can be obtained from the relation S 2 =  n2 + nt2 .

 nt =

( S −  ) = ( 70.2814) − ( −25.21315) = 65.6 MPa 2

2 n

Ans.

P12.68 At a point in a stressed body, the known stresses are x = 10 MPa, y = 0, z = 0, xy = 24 MPa, xz = 38 MPa, and yz = 15 MPa. Determine the normal and shear stresses on a plane whose outward normal makes equal angles with the x, y, and z axes.

Solution We are looking for a plane whose outward normal makes equal angles with x, y, and z axes. We can define this normal using the origin and point A, where point A has the coordinates A(1, 1, 1). The normal to the plane of interest is defined by the vector that begins at the origin O and passes through point A. OA = ai + bj + ck = 1i + 1 j + 1k The length of this vector is calculated as: OA =

(1) + (1) + (1) = 3 2

Create a unit vector in the direction of OA: OA 1 1 1 = i+ j+ k = 0.57735i + 0.57735 j + 0.57735k OA 3 3 3 The direction cosines for the plane of interest are thus: l = 0.57735, m = 0.57735, n = 0.57735 The known stresses are  x = 10 MPa  y = 0 MPa

 xy = 24 MPa

 z = 0 MPa

 yz = 15 MPa

 zx = 38 MPa

The three orthogonal components of the resultant stress are: S x =  x  l +  xy  m +  zx  n = (10 )( 0.57735 ) + ( 24 )( 0.57735 ) + ( 38 )( 0.57735 ) = 41.5692 MPa S y =  xy  l +  y  m +  yz  n = ( 24 )( 0.57735 ) + ( 0 )( 0.57735 ) + (15 )( 0.57735 ) = 22.5167 MPa S z =  zx  l +  yz  m +  z  n = ( 38 )( 0.57735 ) + (15 )( 0.57735 ) + ( 0 )( 0.57735 ) = 30.5996 MPa

The normal component n of the resultant stress is  n = Sx  l + S y  m + Sz  n

= ( 41.5692 )( 0.57735) + ( 22.5167 )( 0.57735) + ( 30.5996 )( 0.57735 ) = 54.6667 = 54.7 MPa (T)

Ans.

The shear stress nt on the oblique plane can be obtained from the relation S 2 =  n2 + nt2 .

 nt =

( S −  ) = (56.3146) − (54.6667) = 13.52 MPa 2

2 n

Ans.

P12.69 At a point in a stressed body, the known stresses are x = 9 ksi (C), y = 13 ksi (T), z = 22 ksi (T), xy = 4 ksi, xz = −19 ksi, and yz = 8 ksi. Determine: (a) the stress invariants. (b) the principal stresses and the absolute maximum shear stress at the point.

Solution The known stresses are  x = −9 ksi  y = 13 ksi

 xy = 4 ksi

 z = 22 ksi

 yz = 8 ksi

 zx = −19 ksi

(a) The three invariants have values of I1 =  x +  y +  z = ( −9 ) + (13) + ( 22 ) = 26

Ans.

I 2 =  x y +  y z +  z x −  xy2 −  yz2 −  zx2 = ( −9 )(13) + (13)( 22 ) + ( 22 )( −9 ) − ( 4 ) − (8 ) − ( −19 ) = −470 2

Ans.

I 3 =  x y z + 2 xy yz zx − (  +   +   ) 2 x yz

2 y zx

2 z xy

2 2 2 = ( −9 )(13)( 22 ) + 2 ( 4 )( 8 )( −19 ) − ( −9 )( 8 ) + (13)( −19 ) + ( 22 )( 4 )  = −8, 259  

(b) Eq. (12.27) can now be written as  3p − ( 26) p2 + ( −470) p − ( −8,259) = 0 To finding the three roots of Equation (12.27), begin by calculating the following constants: Q = 13 I1I 2 − I 3 − 272 I13 = 13 ( 26 )( −470 ) − ( −8, 259 ) − 272 ( 26 ) = 2,883.74074 3

R = 13 I12 − 3I 2 = 13

( 26 ) − 3 ( −470 ) = 15.22425 2

 2,883.74074   Q  −1 −  = 1.99173 rad = cos  3  2 (15.22425 )3   2R   

 = cos −1  −

The roots of Equation (12.27) are calculated as     I   1.99173 rad   26  a = 2 R cos    + 1 = 2 (15.22425) cos  = 32.6475  + 3  3   3  3  

   2   I1   1.99173 rad 2   26  b = 2 R cos  + + = −19.5725   + = 2 (15.22425)  cos   + 

3

3 





3 

   4   I1   1.99173 rad 4   26  c = 2 R cos  + + = 12.9250   + = 2 (15.22425) cos   + 

3

3 





3 

Ans.

Sort these three values in descending order to obtain the principal stresses:  p1 = 32.6475 ksi = 32.6 ksi (T)

 p 2 = 19.9250 ksi = 12.93 ksi (T)

Ans.

 p 3 = −19.5725 ksi = 19.57 ksi (C) The absolute maximum shear stress at the point is found from  −  min 32.6475 − (−19.5725)  abs max = max = = 26.1 ksi 2 2

Ans.

P12.70 At a point in a stressed body, the known stresses are x = 200 MPa (C), y = 90 MPa (C), z = 140 MPa (C), xy = −24 MPa, xz = 120 MPa, and yz = 56 MPa. Determine: (a) the stress invariants. (b) the principal stresses and the absolute maximum shear stress at the point.

Solution The known stresses are  x = −200 MPa  y = −90 MPa

 xy = −24 MPa

 z = −140 MPa

 yz = 56 MPa

 zx = 120 MPa

(a) The three invariants have values of I1 =  x +  y +  z = ( −200 ) + ( −90 ) + ( −140 ) = −430

Ans.

I 2 =  x y +  y z +  z x −  xy2 −  yz2 −  zx2 = ( −200 )( −90 ) + ( −90 )( −140 ) + ( −140 )( −200 ) − ( −24 ) − ( 56 ) − (120 ) = 40,488 2

Ans.

I 3 =  x y z + 2 xy yz zx − (  +   +   ) 2 x yz

2 y zx

2 z xy

2 2 2 = ( −200 )( −90 )( −140 ) + 2 ( −24 )( 56 )(120 ) − ( −200 )( 56 ) + ( −90 )(120 ) + ( −140 )( −24 )   

= −838,720

Ans.

(b) Eq. (12.27) can now be written as  3p − ( −430) p2 + ( 40,488) p − ( −838,720) = 0 To finding the three roots of Equation (12.27), begin by calculating the following constants: Q = 13 I1I 2 − I 3 − 272 I13 = 13 ( −430 )( 40, 488 ) − ( −838,720 ) − 272 ( −430 ) = 924,847.407 3

R = 13 I12 − 3I 2 = 13

( −430 ) − 3 ( 40, 488) = 83.955 2

 924,847.407   Q  −1 −  = 2.46778 rad = cos 3   2 ( 83.955 )3   2R   

 = cos −1  −

The roots of Equation (12.27) are calculated as     I   2.46778 rad   −430  a = 2 R cos    + 1 = 2 (83.955 ) cos  = −29.1003  + 3 3    3  3  

   2   I1   2.46778 rad 2   −430  b = 2 R cos  + + = −307.0258   + = 2 ( 83.955 ) cos   + 

3

3 





3 

   4   I1   2.46778 rad 4   −430  c = 2 R cos  + + = −93.8739   + = 2 ( 83.955 ) cos   + 

3

3 





3 

Sort these three values in descending order to obtain the principal stresses:  p1 = −29.100 MPa = 29.1 MPa (C)

 p 2 = −93.874 MPa = 93.9 MPa (C)

Ans.

 p 3 = −307.026 MPa = 307 MPa (C) The absolute maximum shear stress at the point is found from  −  min −29.100 − (−307.026)  abs max = max = = 139.0 MPa 2 2

Ans.

P12.71 At a point in a stressed body, the known stresses are x = 75 MPa (T), y = 30 MPa (C), z = 60 MPa (T), xy = 35 MPa, xz = 15 MPa, and yz = −20 MPa. Determine: (a) the stress invariants. (b) the principal stresses and the absolute maximum shear stress at the point. (c) the orientation of the plane on which the maximum tensile normal stress acts.

Solution The known stresses are  x = 75 MPa  y = −30 MPa

 xy = 35 MPa

 yz = −20 MPa

 z = 60 MPa  zx = 15 MPa

(a) The three invariants have values of I1 =  x +  y +  z = ( 75 ) + ( −30 ) + ( 60 ) = 105

Ans.

I 2 =  x y +  y z +  z x −  xy2 −  yz2 −  zx2 = ( 75)( −30 ) + ( −30 )( 60 ) + ( 60 )( 75) − ( 35) − ( −20 ) − (15) = −1,400 2

Ans.

I 3 =  x y z + 2 xy yz zx − ( x yz2 +  y zx2 +  z xy2 ) 2 2 2 = ( 75 )( −30 )( 60 ) + 2 ( 35 )( −20 )(15 ) − ( 75 )( −20 ) + ( −30 )(15 ) + ( 60 )( 35 )   

= −252,750

Ans.

(b) Eq. (12.27) can now be written as  3p − (105) p2 + ( −1,400) p − ( −252,750) = 0 To finding the three roots of Equation (12.27), begin by calculating the following constants: Q = 13 I1I 2 − I 3 − 272 I13 = 13 (105 )( −1, 400 ) − ( −252,750 ) − 272 (105 ) = 118,000 3

R = 13 I12 − 3I 2 = 13

(105) − 3 ( −1, 400 ) = 41.1299 2

 118,000   Q  −1   = 2.58294 rad = cos −  3  2 ( 41.1299 )3   2R   

 = cos −1  −

The roots of Equation (12.27) are calculated as     I   2.58294 rad   105  a = 2 R cos    + 1 = 2 ( 41.1299 ) cos  = 88.6083  + 3 3    3  3  

   2   I1   2.58294 rad 2   105  b = 2 R cos  + + = −45.8376   + = 2 ( 41.1299 ) cos   + 

3

3 





3 

   4   I1   2.58294 rad 4   105  c = 2 R cos  + + = 62.2293   + = 2 ( 41.1299 ) cos   + 

3

3 





3 

Sort these three values in descending order to obtain the principal stresses:  p1 = 88.6083 MPa = 88.6 MPa (T)

 p 2 = 62.2293 MPa = 62.2 MPa (T)

Ans.

 p 3 = −45.8376 MPa = 45.8 MPa (C) The absolute maximum shear stress at the point is found from  −  min 88.608 − (−45.838)  abs max = max = = 67.2 MPa 2 2

Ans.

(c) Orientation of the plane on which the maximum tensile normal stress acts. To determine the direction cosines that define each principal plane, begin by rewriting Equation (b) in Section 12.10 in matrix form.  x −  pi  xy  xz   li      y −  pi  yz  mi  = 0   xy   xz  yz  z −  pi   ni   We are interested in the maximum tensile normal stress; therefore, i = 1 and pi = p1. The direction cosines {li, mi, ni} are {l1, m1, n1}, which are the directions cosines corresponding to p1. 35 15  ( 75) − ( 88.608)   l1     35 −20 ( −30 ) − (88.608)   m1  = 0  15 −20 ( 60 ) − (88.608)   n1  or, simplifying: 35 15   l1   −13.608    35 −118.608 −20  m1  = 0   15 −20 −28.608   n1  Calculate the cofactors of the elements on the first row of this matrix.  y −  p1  yz −118.608 −20 a1 = = = 2,993.1785  yz  z −  p1 −20 −28.608 b1 = −

 xy  xz

 yz =−  z −  p1

c1 =

 xy  xz

 y −  p1 =  yz

Calculate the term k1 as: 1 k1 = = 2 a1 + b12 + c12

−20

−28.608

−118.608

−20

= 701.2897 = 1,079.1242

( 2,993.1785) + ( 701.2897 ) + (1,079.1242 ) 2

1 3, 258.11343

The direction cosines giving the orientation of the maximum tensile normal stress are thus 2,993.1785 l1 = a1k1 = = 0.91868 3, 258.13343 701.2897 m1 = b1k1 = = 0.21524 3, 258.13343 1, 079.1242 n1 = c1k1 = = 0.33121 3, 258.13343

Ans.

P12.72 At a point in a stressed body, the known stresses are x = 80 MPa (C), y = 80 MPa (C), z = 120 MPa (C), xy = 40 MPa, xz = −60 MPa, and yz = 50 MPa. Determine: (a) the stress invariants. (b) the principal stresses and the absolute maximum shear stress at the point. (c) the orientation of the plane on which the maximum compressive normal stress acts.

Solution The known stresses are  x = −80 MPa  y = −80 MPa

 xy = 40 MPa

 yz = 50 MPa

 z = −120 MPa  zx = −60 MPa

(a) The three invariants have values of I1 =  x +  y +  z = ( −80 ) + ( −80 ) + ( −120 ) = −280

Ans.

I 2 =  x y +  y z +  z x −  xy2 −  yz2 −  zx2 = ( −80 )( −80 ) + ( −80 )( −120 ) + ( −120 )( −80 ) − ( 40 ) − ( 50 ) − ( −60 ) = 17,900 2

Ans.

I 3 =  x y z + 2 xy yz zx − ( x yz2 +  y zx2 +  z xy2 ) 2 2 2 = ( −80 )( −80 )( −120 ) + 2 ( 40 )( 50 )( −60 ) − ( −80 )( 50 ) + ( −80 )( −60 ) + ( −120 )( 40 )   

= −328, 000

Ans.

(b) Eq. (12.27) can now be written as  3p − ( −280) p2 + (17,900) p − ( −328,000) = 0 To finding the three roots of Equation (12.27), begin by calculating the following constants:

Q = 13 I1I 2 − I 3 − 272 I13 = 13 ( −280 )(17,900 ) − ( −328, 000 ) − 272 ( −280 ) = 283, 407.4074 3

R = 13 I12 − 3I 2 = 13

( −280 ) − 3 (17,900 ) = 52.3874 2

 283, 407.4074   Q  −1 −  = 2.97167 rad = cos 3   2 ( 52.3874 )3   2R   

 = cos −1  −

The roots of Equation (12.27) are calculated as     I   2.97167 rad   −280  a = 2 R cos    + 1 = 2 ( 52.3874 ) cos  = −35.8931  + 3 3    3  3      2   I1   2.97167 rad 2   −280  b = 2 R cos  + + = −197.9402   + = 2 ( 52.3874 ) cos   + 3 3  3   3 3  3      4   I1   2.97167 rad 4   −280  c = 2 R cos  + + = −46.1667   + = 2 ( 52.3874 ) cos   + 3 3   3   3 3  3  

Sort these three values in descending order to obtain the principal stresses:  p1 = −35.893 MPa = 35.9 MPa (C)

 p 2 = −46.167 MPa = 46.2 MPa (C)

Ans.

 p 3 = −197.940 MPa = 197.9 MPa (C) The absolute maximum shear stress at the point is found from  −  min −35.893 − (−197.940)  abs max = max = = 81.0 MPa 2 2

Ans.

(c) Orientation of the plane on which the maximum compressive normal stress acts. To determine the direction cosines that define each principal plane, begin by rewriting Equation (b) in Section 12.10 in matrix form.  x −  pi  xy  xz   li      y −  pi  yz  mi  = 0   xy   xz  yz  z −  pi   ni   We are interested in the maximum compressive normal stress; therefore, i = 3 and pi = p3. The direction cosines {li, mi, ni} are {l3, m3, n3}, which are the directions cosines corresponding to p3. 40 −60  ( −80 ) − ( −197.940 )   l3     40 50 ( −80 ) − ( −197.940)   m3  = 0  −60 50 ( −120) − ( −197.940)  n3  or, simplifying: 40 −60   l3  117.940    40 117.940 50  m3  = 0   −60 50 77.940   n3  Calculate the cofactors of the elements on the first row of this matrix.  y −  p3  yz 117.940 50 a3 = = = 6, 692.2817  yz  z −  p3 50 77.940 b3 = −

 xy  xz

 yz =−  z −  p3

c3 =

 xy  xz

 y −  p3 =  yz

Calculate the term k3 as: 1 k3 = = 2 a3 + b32 + c32

−60

77.940

117.940

−60

= −6,117.6078 = 9, 076.4117

( 6, 692.2817 ) + ( −6,117.6078) + ( 9, 076.4117 ) 2

1 12,829.3807

The direction cosines giving the orientation of the maximum compressive normal stress are thus 6, 692.2817 l3 = a3k3 = = 0.52164 12,829.3807 −6,117.6078 m3 = b3k3 = = −0.47684 12,829.3807 9, 076.4417 n3 = c3k3 = = 0.70747 12,829.3807

Ans.

P13.1 The thin rectangular plate shown in Figure P13.1/2 is uniformly deformed such that x = 230 , y = −480 , and xy = –760 rad. Using dimensions of a = 20 mm and b = 25 mm, determine the normal strain in the plate in the direction defined by (a) points O and A. (b) points O and C. FIGURE P13.1/2

Solution (a) From the geometry of the plate: 2a 40 mm tan OA = = = 0.5333 3b 75 mm The given strain values are:  x = 230 με  y = −480 με

OA = 28.0725

 xy = −760 μrad

Substitute these values into the normal strain transformation equation [Eq. (13.3)] to obtain the strain in the direction of line OA:  OA =  x cos 2  +  y sin 2  +  xy sin  cos 

= (230 με) cos 2 (28.0725) + ( −480 με)sin 2 (28.0725) + ( −760 μrad) sin(28.0725) cos(28.0725) = −242.8028 με = −243 με (b) From the geometry of the plate: − a −20 tan OC = = = −0.2000 4b 100

Ans.

OC = −11.3099

Substitute this angle and the given strains into the normal strain transformation equation [Eq. (13.3)] to obtain in the direction of line OC:  OC =  x cos 2  +  y sin 2  +  xy sin  cos 

= (230 με) cos 2 ( −11.3099) + ( −480 με)sin 2 ( −11.3099) + ( −760 μrad) sin( −11.3099) cos( −11.3099) = 348.8462 με = 349 με

Ans.

P13.2 The thin rectangular plate shown in Figure P13.1/2 is uniformly deformed such that x = –360 , y = 770 , and xy = 940 rad. Using dimensions of a = 25 mm and b = 40 mm, determine the normal strain in the plate in the direction defined by (a) points O and B. (b) points O and D.

FIGURE P13.1/2

Solution (a) From the geometry of the plate: 2a 50 mm tan OB = = = 1.2500 b 40 mm The given strain values are:  x = −360 με  y = 770 με

OB = 51.3402

 xy = 940 μrad

Substitute these values into the normal strain transformation equation [Eq. (13.3)] to obtain the strain in the direction of line OB:  OB =  x cos 2  +  y sin 2  +  xy sin  cos 

= ( −360 με) cos 2 (51.3402) + (770 με)sin 2 (51.3402) + (940 μrad) sin(51.3402) cos(51.3402) = 787.5610 με = 788 με (b) From the geometry of the plate: −2a −50 tan OD = = = −0.6250 2b 80

Ans.

OD = −32.0054

Substitute this angle and the given strains into the normal strain transformation equation [Eq. (13.3)] to obtain in the direction of line OD:  OD =  x cos 2  +  y sin 2  +  xy sin  cos 

= ( −360 με) cos 2 ( −32.0054) + (770 με)sin 2 ( −32.0054) + (940 μrad) sin( −32.0054) cos( −32.0054) = −465.0562 με = −465 με

Ans.

P13.3 The thin rectangular plate shown in Figure P13.3/4 is uniformly deformed such that x = 120 , y = –860 , and xy = 1,100 rad. If a = 25 mm, determine (a) the normal strain n in the plate. (b) the normal strain t in the plate. (c) the shear strain nt in the plate.

FIGURE P13.3/4

Solution (a) From the geometry of the plate, the n axis is oriented at an angle of a 1 tan  = =  = 18.4349 3a 3 The given strain values are:  x = 120 με  y = −860 με  xy = 1,100 μrad Substitute these values into the normal strain transformation equation [Eq. (13.3)] to obtain the strain in the n direction:  n =  x cos 2  +  y sin 2  +  xy sin  cos 

= (120 με) cos 2 (18.4349) + ( −860 με)sin 2 (18.4349) + (1,100 μrad)sin(18.4349) cos(18.4349) = 352.0000 με = 352 με

Ans.

(b) To determine the normal strain in the t direction, use  + 90° in the normal strain transformation equation [Eq. (13.3)]:  t = (120 με) cos 2 (18.4349 + 90) + ( −860 με) sin 2 (18.4349 + 90) + (1,100 μrad) sin(18.4349 + 90) cos(18.4349 + 90) = −1, 092.0000 με = −1, 092 με

Ans.

(c) The shear strain nt is found from the shear strain transformation equation [Eq. (13.5)]:  nt = −2( x −  y )sin  cos  +  xy (cos 2  − sin 2  )

= −2[(120 με) − ( −860 με)]sin(18.4349) cos(18.4349) + (1,100 μrad)[ cos 2 (18.4349) − sin 2 (18.4349)] = 292.0000 μrad = 292 μrad

Ans.

P13.4 The thin rectangular plate shown in Figure P13.3/4 is uniformly deformed such that x = –890 , y = 440 , and xy = –310 rad. If a = 50 mm, determine (a) the normal strain n' in the plate. (b) the normal strain t' in the plate. (c) the shear strain n't' in the plate.

FIGURE P13.3/4

Solution (a) From the geometry of the plate, the n' axis is oriented at an angle of −2a 2 tan  = =−  = −33.6901 3a 3 The given strain values are:  x = −890 με  y = 440 με  xy = −310 μrad Substitute these values into the normal strain transformation equation [Eq. (13.3)] to obtain the strain in the n' direction:  n ' =  x cos 2  +  y sin 2  +  xy sin  cos 

= ( −890 με) cos 2 ( −33.6901) + (440 με)sin 2 ( −33.6901) + ( −310 μrad)sin( −33.6901) cos( −33.6901) = −337.6923 με = −338 με

Ans.

(b) To determine the normal strain in the t' direction, use  + 90° in the normal strain transformation equation [Eq. (13.3)]:  t ' = ( −890 με) cos 2 ( −33.6901 + 90) + (440 με) sin 2 (−33.6901 + 90) + ( −310 μrad) sin( −33.6901 + 90) cos(−33.6901 + 90) = −112.3077 με = −112.3 με

Ans.

(c) The shear strain n't' is found from the shear strain transformation equation [Eq. (13.5)]:  n 't ' = −2[( −890 με) − (440 με)]sin( −33.6901) cos(−33.6901) + ( −310 μrad)[ cos 2 ( −33.6901) − sin 2 ( −33.6901)] = −1,346.9231 μrad = −1,347 μrad

Ans.

P13.5 The thin square plate shown in Figure P13.5/6 is uniformly deformed such that n = 660 , t = 910 , and nt = 830 rad. Determine (a) the normal strain x in the plate. (b) the normal strain y in the plate. (c) the shear strain xy in the plate.

FIGURE P13.5/6

Solution The given strain values are:  n = 660 με  t = 910 με

 nt = 830 μrad (a) Normal strain x: The x axis is rotated 45° counterclockwise from the n axis; therefore, we can use  = +45° (from the n to the x axis) with the strains associated with the n and t axes to determine the strains in the x and y directions.  x =  n cos 2  +  t sin 2  +  nt sin  cos  = (660 με) cos 2 (45) + (910 με) sin 2 (45) + (830 μrad) sin(45) cos(45) = 1, 200 με

Ans.

(b) Normal strain y: The normal strain in the y direction is found by setting  = 45° + 90° = 135° in the normal strain transformation equation:  y = (660 με) cos 2 (135) + (910 με)sin 2 (135) + (830 μrad)sin(135) cos(135) = 370 με

Ans.

(c) Shear strain xy: The shear strain xy is found from the shear strain transformation equation [Eq. (13.5)]:  xy = −2( n −  t ) sin  cos  +  nt (cos 2  − sin 2  ) = −2[(660 με) − (910 με)]sin(45) cos(45) + (830 μrad)[ cos 2 (45) − sin 2 (45)] = 250 μrad

Ans.

P13.6 The thin square plate shown in Figure P13.5/6 is uniformly deformed such that x = 0 , y = 0 , and xy = –1,850 rad. Using a = 650 mm, determine the deformed length of (a) diagonal AC and (b) diagonal BD.

FIGURE P13.5/6

Solution The given strain values are:  x = 0 με  y = 0 με

 xy = −1,850 μrad

(a) For diagonal AC,  = 45°. Substitute these values into the normal strain transformation equation [Eq. (13.3)] to obtain the strain in the direction defined by AC:  AC =  x cos2  +  y sin 2  +  xy sin  cos 

= (0 με) cos 2 (45) + (0 με)sin 2 (45) + ( −1,850 μrad)sin(45) cos(45) = −925 με The original length of diagonal AC is

LAC = (650 mm)2 + (650 mm)2 = 919.2388 mm The deformation of this diagonal is  AC =  AC LAC = (−925 10−6 )(919.2388 mm) = −0.8503 mm Thus, the deformed length of diagonal AC is LAC  = 919.2388 mm − 0.8503 mm = 918.3885 mm = 918 mm

Ans.

(b) For diagonal BD,  = –45°. Substitute this value into the normal strain transformation equation [Eq. (13.3)] to obtain the strain in the direction defined by BD:  BD =  x cos 2  +  y sin 2  +  xy sin  cos 

= (0 με) cos 2 ( −45) + (0 με)sin 2 ( −45) + ( −1,850 μrad)sin(−45) cos( −45) = 925 με The deformation of this diagonal is  AC =  AC LAC = (925 10−6 )(919.2388 mm) = 0.8503 mm Thus, the deformed length of diagonal BD is LBD  = 919.2388 mm + 0.8503 mm = 920.0891 mm = 920 mm

Ans.

P13.7 The strain components x, y, and xy are given for a point in a body subjected to plane strain. Determine the strain components n, t, and nt at the point if the n-t axes are rotated with respect to the x-y axes by the amount and in the direction indicated by the angle  shown in either Figure P13.7 or Figure P13.8. Sketch the deformed shape of the element. x = –1,050  y = 400  xy = 1,360 rad  = 36°

FIGURE P13.7

Solution The n axis is rotated counterclockwise from the x axis; therefore,  = +36°. Use the normal strain transformation equation [Eq. (13.3)] to obtain the strain in the n direction:  n =  x cos 2  +  y sin 2  +  xy sin  cos  = ( −1, 050 με) cos 2 (36) + (400 με) sin 2 (36) + (1,360 μrad) sin(36) cos(36) = 97.6811 με = 97.7 με

Ans.

The normal strain in the t direction is found by setting  = 36° + 90° = +126° in the normal strain transformation equation: t = (−1,050 με) cos2 (126) + (400 με)sin 2 (126) + (1,360 μrad)sin(126) cos(126)

= −747.6811 με = −748 με

Ans.

The shear strain nt is found from the shear strain transformation equation [Eq. (13.5)]:  nt = −2( x −  y ) sin  cos  +  xy (cos 2  − sin 2  ) = −2[( −1, 050 με) − (400 με)]sin(36) cos(36) + (1,360 μrad)[ cos 2 (36) − sin 2 (36)] = 1, 799.2951 μrad = 1, 799 μrad

Ans.

P13.8 The strain components x, y, and xy are given for a point in a body subjected to plane strain. Determine the strain components n, t, and nt at the point if the n-t axes are rotated with respect to the x-y axes by the amount and in the direction indicated by the angle  shown in either Figure P13.7 or Figure P13.8. Sketch the deformed shape of the element. x = –350  y = 1,650  xy = 720 rad  = 14° FIGURE P13.8

Solution The n axis is rotated clockwise from the x axis; therefore,  = –14°. Use the normal strain transformation equation [Eq. (13.3)] to obtain the strain in the n direction:  n =  x cos 2  +  y sin 2  +  xy sin  cos  = ( −350 με) cos 2 ( −14) + (1, 650 με) sin 2 ( −14) + (720 μrad) sin( −14) cos( −14) = −401.9574 με = −402 με

Ans.

The normal strain in the t direction is found by setting  = –14° + 90° = +76° in the normal strain transformation equation: t = (−350 με) cos2 (76) + (1,650 με)sin 2 (76) + (720 μrad)sin(76) cos(76)

= 1,701.9574 με = 1,702 με

Ans.

The shear strain nt is found from the shear strain transformation equation [Eq. (13.5)]:  nt = −2( x −  y )sin  cos  +  xy (cos 2  − sin 2  ) = −2[( −350 με) − (1,650 με)]sin( −14)cos( −14) + (720 μrad)[cos 2 ( −14) − sin 2 ( −14)] = −303.2209 μrad = −303 μrad

Ans.

P13.9 The strain components x, y, and xy are given for a point in a body subjected to plane strain. Determine the strain components n, t, and nt at the point if the n-t axes are rotated with respect to the x-y axes by the amount and in the direction indicated by the angle  shown in either Figure P13.7 or Figure P13.8. Sketch the deformed shape of the element. x = –1,375  y = –1,825  xy = 650 rad  = 15°

FIGURE P13.7

Solution The n axis is rotated counterclockwise from the x axis; therefore,  = +15°. Use the normal strain transformation equation [Eq. (13.3)] to obtain the strain in the n direction:  n =  x cos 2  +  y sin 2  +  xy sin  cos  = ( −1,375 με) cos 2 (15) + ( −1,825 με)sin 2 (15) + (650 μrad)sin(15) cos(15) = −1, 242.6443 με = −1, 243 με

Ans.

The normal strain in the t direction is found by setting  = 15° + 90° = +105° in the normal strain transformation equation: t = (−1,375 με)cos2 (105) + (−1,825 με)sin 2 (105) + (650 μrad)sin(105)cos(105)

= −1,957.3557 με = −1,957 με

Ans.

The shear strain nt is found from the shear strain transformation equation [Eq. (13.5)]:  nt = −2( x −  y )sin  cos  +  xy (cos 2  − sin 2  ) = −2[( −1,375 με) − ( −1,825 με)]sin(15)cos(15) + (650 μrad)[cos 2 (15) − sin 2 (15)] = 337.9165 μrad = 338 μrad

Ans.

P13.10 The strain components x, y, and xy are given for a point in a body subjected to plane strain. Determine the strain components n, t, and nt at the point if the n-t axes are rotated with respect to the x-y axes by the amount and in the direction indicated by the angle  shown in either Figure P13.7 or Figure P13.8. Sketch the deformed shape of the element. x = 590  y = –1,670  xy = –1,185 rad  = 23° FIGURE P13.8

Solution The n axis is rotated clockwise from the x axis; therefore,  = –23°. Use the normal strain transformation equation [Eq. (13.3)] to obtain the strain in the n direction:  n =  x cos 2  +  y sin 2  +  xy sin  cos  = (590 με)cos 2 ( −23) + ( −1,670 με)sin 2 ( −23) + ( −1,185 μrad)sin( −23)cos( −23) = 671.1728 με = 671 με

Ans.

The normal strain in the t direction is found by setting  = –23° + 90° = +67° in the normal strain transformation equation: t = (590 με)cos2 (67) + (−1,670 με)sin 2 (67) + (−1,185 μrad)sin(67)cos(67)

= −1,751.1728 με = −1,751 με

Ans.

The shear strain nt is found from the shear strain transformation equation [Eq. (13.5)]:  nt = −2( x −  y )sin  cos  +  xy (cos 2  − sin 2  ) = −2[(590 με) − ( −1,670 με)]sin( −23)cos( −23) + ( −1,185 μrad)[cos 2 (−23) − sin 2 ( −23)] = 802.5378 μrad = 803 μrad

Ans.

P13.11 The strain components x, y, and xy are given for a point in a body subjected to plane strain. Determine the principal strains, the maximum in-plane shear strain, and the absolute maximum shear strain at the point. Show the angle p, the principal strain deformations, and the maximum in-plane shear strain distortion on a sketch. x = –550  y = −285  xy = 940 rad

Solution The principal strain magnitudes can be computed from Eq. (13.10):

 p1, p 2 =

x +  y 2

  x −  y    xy    +  2   2  2

( −550 μ) + ( −285 μ)  ( −550 μ) − ( −285 μ)   940 μ  =    +    2 2 2  2

= −417.5 μ  488.3198 μ

 p1 = 70.8 με and  p 2 = −906 με

Ans.

 max = 977 μrad

Ans.

tan 2 p =

(maximum in-plane shear strain)

 xy 940 μ = = −3.54717 ( x −  y ) [( −550 μ) − ( −285 μ)]

 p = −37.1

(clockwise from the x axis to the direction of  p 2 )

Ans.

For plane strain, z = p3 = 0. Since p1 is positive and p2 is negative, the absolute maximum shear strain is the maximum in-plane shear strain:  abs max =  p1 −  p 2 = 70.8198 μ − (−905.8198 μ) = 977 μrad Ans.

P13.12 The strain components x, y, and xy are given for a point in a body subjected to plane strain. Determine the principal strains, the maximum in-plane shear strain, and the absolute maximum shear strain at the point. Show the angle p, the principal strain deformations, and the maximum in-plane shear strain distortion on a sketch. x = 940  y = –360  xy = 830 rad

Solution The principal strain magnitudes can be computed from Eq. (13.10):

 p1, p 2 =

x +  y 2

  x −  y    xy    +  2   2  2

(940 μ) + ( −360 μ)  (940 μ) − ( −360 μ)   830 μ  =    +    2 2 2  2

= 290 μ  771.1842 μ

 p1 = 1,061 με and  p 2 = −481 με

Ans.

 max = 1,542 μrad

Ans.

tan 2 p =

(maximum in-plane shear strain)

 xy 830 μ = = 0.63846 ( x −  y ) [(940 μ) − ( −360 μ)]

 p = 16.28

(counterclockwise from the x axis to the direction of  p1 )

Ans.

For plane strain, z = p3 = 0. Since p1 is positive and p2 is negative, the absolute maximum shear strain is the maximum in-plane shear strain:  abs max =  p1 −  p 2 = 1,061.1842 μ − (−481.1842 μ) = 1,542 μrad Ans.

P13.13 The strain components x, y, and xy are given for a point in a body subjected to plane strain. Determine the principal strains, the maximum in-plane shear strain, and the absolute maximum shear strain at the point. Show the angle p, the principal strain deformations, and the maximum in-plane shear strain distortion on a sketch. x = −270  y = 510  xy = 1,150 rad

Solution The principal strain magnitudes can be computed from Eq. (13.10):

 p1, p 2 =

x +  y 2

  x −  y    xy    +  2   2  2

( −270 μ) + (510 μ)  ( −270 μ) − (510 μ)   1,150 μ  =    +    2 2 2  2

= 120 μ  694.7841 μ

 p1 = 815 με and  p 2 = −575 με

Ans.

 max = 1,390 μrad

Ans.

tan 2 p =

(maximum in-plane shear strain)

 xy 1,150 μ = = −1.47436 ( x −  y ) [( −270 μ) − (510 μ)]

 p = −27.9

(clockwise from the x axis to the direction of  p 2 )

Ans.

For plane strain, z = p3 = 0. Since p1 is positive and p2 is negative, the absolute maximum shear strain is the maximum in-plane shear strain:  abs max =  p1 −  p 2 = 814.7841 μ − (−574.7841 μ) = 1,390 μrad Ans.

P13.14 The strain components x, y, and xy are given for a point in a body subjected to plane strain. Determine the principal strains, the maximum in-plane shear strain, and the absolute maximum shear strain at the point. Show the angle p, the principal strain deformations, and the maximum in-plane shear strain distortion on a sketch. x = 670  y = –280  xy = −800 rad

Solution The principal strain magnitudes can be computed from Eq. (13.10):

 p1, p 2 =

x +  y 2

  x −  y    xy    +  2   2  2

(670 μ) + ( −280 μ)  (670 μ) − ( −280 μ)   −800 μ  =    +    2 2 2  2

= 195 μ  620.9871 μ

 p1 = 816 με and  p 2 = −426 με

Ans.

 max = 1,242 μrad

Ans.

tan 2 p =

(maximum in-plane shear strain)

 xy −800 μ = = −0.84211 ( x −  y ) [(670 μ) − ( −280 μ)]

  p = −20.1

(clockwise from the x axis to the direction of  p1 )

Ans.

For plane strain, z = p3 = 0. Since p1 is positive and p2 is negative, the absolute maximum shear strain is the maximum in-plane shear strain:  abs max =  p1 −  p 2 = 815.9871 μ − (−425.9871 μ) = 1,242 μrad Ans.

The principal strains are given for a point in a body subjected to plane strain. Construct Mohr’s circle and use it to (a) determine the strains x, y, and xy. (Assume x > y) (b) determine the maximum in-plane shear strain and the absolute maximum shear strain. (c) draw a sketch showing the angle p, the principal strain deformations, and the maximum in-plane shear strain distortions. P13.16 p1 = 780  p2 = 590  p = 35.66°

Solution The center of Mohr’s circle is at  p1 +  p 2 (780 με) + (590 με) C= = = 685 με 2 2 and the radius of Mohr’s circle is equal to  p1 −  p 2 (780 με) − (590 με) R= = = 95.0 μ 2 2 The angle p is given as 35.66°, and we are told to assume that x > y. From this information, we know that the p1 principal plane is rotated 35.66° in a counterclockwise direction from the x face. Thus, to locate point x on Mohr’s circle, begin at p1 and rotate 2(35.66°) = 71.32° in a clockwise direction. The Mohr’s circle is shown. (a) The normal strain in the x direction is computed as:  x = C + R cos 2 p

= 685 μ + (95 μ)cos(71.32) = 715.4268 με = 715 με

Ans.

and the shear strain xy is computed from:

 xy 2

= R sin 2 p = (95 μ)sin(71.32) = 89.9956 μrad   xy = 179.9912 μrad = 180.0 μrad

(positive since point x plots below the  axis)

Ans.

The normal strain in the y direction is computed from:  y = C − R cos 2 p

= 685 μ − (95 μ)cos(71.32) = 654.5732 με = 655 με

Ans.

(b) The maximum shear strain is simply two times the radius of Mohr’s circle:  max = 2R = 2(95 μ) = 190.0 μrad

Ans.

Since p1 and p2 are both positive, the absolute maximum shear strain is numerically equal to p1:  abs max = 780 μrad

Ans.

The principal strains are given for a point in a body subjected to plane strain. Construct Mohr’s circle and use it to (a) determine the strains x, y, and xy. (Assume x > y) (b) determine the maximum in-plane shear strain and the absolute maximum shear strain. (c) draw a sketch showing the angle p, the principal strain deformations, and the maximum in-plane shear strain distortions. P13.17 p1 = −350  p2 = −890  p = −19.50°

Solution The center of Mohr’s circle is at  p1 +  p 2 ( −350 με) + ( −890 με) C= = = −620 με 2 2 and the radius of Mohr’s circle is equal to  p1 −  p 2 ( −350 με) − ( −890 με) R= = = 270 μ 2 2 The angle p is given as −19.50°, and we are told to assume that x > y. From this information, we know that the p1 principal plane is rotated 19.50° in a clockwise direction from the x face. Thus, to locate point x on Mohr’s circle, begin at p1 and rotate 2(19.50°) = 39° in a counterclockwise direction. The Mohr’s circle is shown. (a) The normal strain in the x direction is computed as:  x = C + R cos 2 p

= −620 μ + (270 μ)cos(39) = −410.1706 με = −410 με

Ans.

and the shear strain xy is computed from:

 xy 2

= R sin 2 p = (270 μ)sin(39) = 169.9165 μrad   xy = 339.8330 μrad = −340 μrad

(negative since point x plots above the  axis) Ans.

The normal strain in the y direction is computed from:  y = C − R cos 2 p

= −620 μ − (270 μ)cos(39) = −829.8294 με = −830 με

Ans.

(b) The maximum shear strain is simply two times the radius of Mohr’s circle:  max = 2R = 2(270 μ) = 540 μrad

Ans.

Since p1 and p2 are both negative, the absolute maximum shear strain is numerically equal to the absolute value of p2: Ans.  abs max = 890 μrad (c) A sketch of the principal strain deformations and the maximum in-plane shear strain distortions is shown below.

The strain components x, y, and xy are given for a point in a body subjected to plane strain. Using Mohr’s circle, determine the principal strains, the maximum in-plane shear strain, and the absolute maximum shear strain at the point. Show the angle p, the principal strain deformations, and the maximum in-plane shear strain distortion in a sketch. P13.18 x = 380  y = −770  xy = −650 rad

Solution The basic Mohr’s circle is shown. (380 μ) + ( −770 μ) C= = −195 με 2 R = (575 μ) 2 + (325 μ) 2 = 660.4922 με

 p1 = C + R = −195 με + 660.4922 με = 465 με  p 2 = C − R = −195 με − 660.4922 με = −855 με  max = 2 R = 1,321 μrad The magnitude of the angle 2p between point x and point 1 (i.e., the principal plane associated with p1) is found from:

tan 2 p =

650 μ 650 μ = = 0.56522 (380 μ) − ( −770 μ) 1,150 μ

 2 p = 29.4759

thus,  p = 14.74

By inspection, the angle p from point x to point 1 is turned clockwise. Since p1 is positive and p2 is negative, the absolute maximum shear strain is the maximum in-plane shear strain:  abs max =  max = 1,321 μrad Ans. A sketch of the principal strain deformations and the maximum in-plane shear strain distortions is shown below.

The strain components x, y, and xy are given for a point in a body subjected to plane strain. Using Mohr’s circle, determine the principal strains, the maximum in-plane shear strain, and the absolute maximum shear strain at the point. Show the angle p, the principal strain deformations, and the maximum in-plane shear strain distortion in a sketch. P13.19 x = 760  y = 590  xy = −360 rad

Solution The basic Mohr’s circle is shown. (760 μ) + (590 μ) C= = 675 με 2 R = (85 μ) 2 + ( −180 μ) 2 = 199.0603 με

 p1 = C + R = 675 με + 199.0603 με = 874 με  p 2 = C − R = 675 με − 199.0603 με = 476 με  max = 2 R = 398 μrad The magnitude of the angle 2p between point x and point 1 (i.e., the principal plane associated with p1) is found from: 360 μ 360 μ tan 2 p = = = 2.11765  2 p = 64.7223 thus,  p = 32.4 (760 μ) − (590 μ) 170 μ By inspection, the angle p from point x to point 1 is turned clockwise. Since both p1 and p2 are positive, the absolute maximum shear strain is greater than the maximum in-plane shear strain:  abs max =  p1 −  p3 = 874.0603 μ − (0 μ) = 874 μrad Ans. A sketch of the principal strain deformations and the maximum in-plane shear strain distortions is shown below.

The strain components x, y, and xy are given for a point in a body subjected to plane strain. Using Mohr’s circle, determine the principal strains, the maximum in-plane shear strain, and the absolute maximum shear strain at the point. Show the angle p, the principal strain deformations, and the maximum in-plane shear strain distortion in a sketch. P13.20 x = −1,570  y = −430  xy = −950 rad

Solution The basic Mohr’s circle is shown. ( −1,570 μ) + ( −430 μ) C= = −1,000 με 2 R = ( −570 μ) 2 + (475 μ) 2 = 741.9737 με

 p1 = C + R = −1,000 με + 741.9737 με = −258 με  p 2 = C − R = −1,000 με − 741.9737 με = −1,742 με  max = 2 R = 1, 484 μrad The magnitude of the angle 2p between point x and point 2 (i.e., the principal plane associated with p2) is found from: 950 μ 950 μ tan 2 p = = = 0.83333 2 p = 39.8056 ( −1,570 μ) − ( −430 μ) 1,140 μ

thus,  p = 19.90

By inspection, the angle p from point x to point 2 is turned counterclockwise. Since both p1 and p2 are negative, the absolute maximum shear strain is greater than the maximum in-plane shear strain:  abs max =  p3 −  p 2 = 0 μ − (−1,741.9737 μ) = 1,742 μrad Ans. A sketch of the principal strain deformations and the maximum in-plane shear strain distortions is shown below.

The strain components x, y, and xy are given for a point in a body subjected to plane strain. Using Mohr’s circle, determine the principal strains, the maximum in-plane shear strain, and the absolute maximum shear strain at the point. Show the angle p, the principal strain deformations, and the maximum in-plane shear strain distortion in a sketch. P13.21 x = 475  y = 685  xy = −150 rad

Solution The basic Mohr’s circle is shown. (475 μ) + (685 μ) C= = 580 με 2 R = ( −105 μ) 2 + (75 μ) 2 = 129.0349 με

 p1 = C + R = 580 με + 129.0349 με = 709 με  p 2 = C − R = 580 με − 129.0349 με = 451 με  max = 2 R = 709 μrad The magnitude of the angle 2p between point x and point 2 (i.e., the principal plane associated with p2) is found from: 150 μ 150 μ tan 2 p = = = 0.71429  2 p = 35.5377 (475 μ) − (685 μ) 210 μ

thus,  p = 17.77

By inspection, the angle p from point x to point 2 is turned counterclockwise. Since both p1 and p2 are positive, the absolute maximum shear strain is greater than the maximum in-plane shear strain:  abs max =  p1 −  p3 = 709.0349 μ − (0 μ) = 709 μrad Ans. A sketch of the principal strain deformations and the maximum in-plane shear strain distortions is shown below.

The strain components x, y, and xy are given for a point in a body subjected to plane strain. Using Mohr’s circle, determine the principal strains, the maximum in-plane shear strain, and the absolute maximum shear strain at the point. Show the angle p, the principal strain deformations, and the maximum in-plane shear strain distortion in a sketch. P13.22 x = 670  y = 455  xy = −900 rad

Solution The basic Mohr’s circle is shown. (670 μ) + (455 μ) C= = 562.5 με 2 R = (107.5 μ) 2 + (450 μ) 2 = 462.6621 με

 p1 = C + R = 562.5 με + 462.6621 με = 1,025 με  p 2 = C − R = 562.5 με − 462.6621 με = 99.8 με  max = 2 R = 925 μrad The magnitude of the angle 2p between point x and point 1 (i.e., the principal plane associated with p1) is found from:

tan 2 p =

900 μ 900 μ = = 4.18605 (670 μ) − (455 μ) 215 μ

 2 p = 76.5645 thus,  p = 38.3

By inspection, the angle p from point x to point 1 is turned clockwise. Since both p1 and p2 are positive, the absolute maximum shear strain is greater than the maximum in-plane shear strain:  abs max =  p1 −  p3 = 1,025.1621 μ − (0 μ) = 1,025 μrad Ans. A sketch of the principal strain deformations and the maximum in-plane shear strain distortions is shown below.

The strain components x, y, and xy are given for a point in a body subjected to plane strain. Using Mohr’s circle, determine the principal strains, the maximum in-plane shear strain, and the absolute maximum shear strain at the point. Show the angle p, the principal strain deformations, and the maximum in-plane shear strain distortion in a sketch. P13.23 x = 0  y = 320  xy = 260 rad

Solution The basic Mohr’s circle is shown. (0 μ) + (320 μ) C= = 160 με 2 R = ( −160 μ) 2 + (130 μ) 2 = 206.1553 με

 p1 = C + R = 160 με + 206.1553 με = 366 με  p 2 = C − R = 160 με − 206.1553 με = −46.2 με  max = 2 R = 412 μrad The magnitude of the angle 2p between point x and point 2 (i.e., the principal plane associated with p2) is found from: 260 μ 260 μ tan 2 p = = = 0.81250  2 p = 39.0939 (0 μ) − ( −320 μ) 320 μ

thus,  p = 19.55

By inspection, the angle p from point x to point 2 is turned clockwise. Since p1 is positive and p2 is negative, the absolute maximum shear strain is the maximum in-plane shear strain:  abs max =  max = 412 μrad Ans. A sketch of the principal strain deformations and the maximum in-plane shear strain distortions is shown below.

The strain components x, y, and xy are given for a point in a body subjected to plane strain. Using Mohr’s circle, determine the principal strains, the maximum in-plane shear strain, and the absolute maximum shear strain at the point. Show the angle p, the principal strain deformations, and the maximum in-plane shear strain distortion in a sketch. P13.24 x = −180  y = −1,480  xy = 425 rad

Solution The basic Mohr’s circle is shown. ( −180 μ) + ( −1, 480 μ) C= = −830 με 2 R = (630 μ) 2 + (212.5 μ) 2 = 683.8540 με

 p1 = C + R = −830 με + 683.8540 με = −146.1 με  p 2 = C − R = −830 με − 683.8540 με = −1,514 με  max = 2 R = 1,368 μrad The magnitude of the angle 2p between point x and point 1 (i.e., the principal plane associated with p1) is found from: 425 μ 425 μ tan 2 p = = = 0.32692 2 p = 18.1038 ( −180 μ) − ( −1,480 μ) 1300 μ

thus,  p = 9.05

By inspection, the angle p from point x to point 1 is turned counterclockwise. Since both p1 and p2 are negative, the absolute maximum shear strain is greater than the maximum in-plane shear strain:  abs max =  p3 −  p 2 = 0 μ − (−1,513.8540 μ) = 1,514 μrad Ans. A sketch of the principal strain deformations and the maximum in-plane shear strain distortions is shown below.

P13.25 The strain rosette shown in the figure was used to obtain normal strain data at a point on the free surface of a machine part. (a) Determine the strain components x, y, and xy at the point. (b) Determine the principal strains and the maximum in-plane shear strain at the point. (c) Draw a sketch showing the angle p, the principal strain deformations, and the maximum in-plane shear strain distortions. (d) Determine the magnitude of the absolute maximum shear strain. FIGURE P13.25

a = 410 , b = −540 , c = −330 ,  = 0.30

Solution (a) Write three normal strain transformation equations [Eq. (13.3)], one for each strain gage, where n is the measured normal strain. In each equation, the angle  associated with each strain gage will be referenced from the positive x axis. (a) 410 με =  x cos2 (0) +  y sin 2 (0) +  xy sin(0)cos(0)

−540 με =  x cos2 (45) +  y sin 2 (45) +  xy sin(45)cos(45)

(b)

−330 με =  x cos2 (90) +  y sin 2 (90) +  xy sin(90)cos(90) From Eq. (a):  x = 410 με

and from Eq. (c):  y = −330 με

Ans.

Using these two results, solve Eq. (b) to find xy: −540 με = (410 με)cos 2 (45) + ( −330 με)sin 2 (45) +  xy sin(45)cos(45)   xy = −1,160 μrad

Ans.

(b) Using these results, the principal strain magnitudes can be computed from Eq. (13.10):

 p1, p 2 =

x +  y 2

  x −  y    xy    +  2   2  2

(410 μ) + ( −330 μ)  (410 μ) − ( −330 μ)   −1,160 μ  =    +    2 2 2 2

= 40 μ  687.9680 μ

 p1 = 728 με and  p 2 = −648 με

Ans.

 max = 1,376 μrad

Ans.

tan 2 p =

(maximum in-plane shear strain)

 xy −1,160 μ −1,160 μ = = = −1.5676 ( x −  y ) [(410 μ) − ( −330 μ)] 740 μ

 p = −28.7

(clockwise from the x axis to the direction of  p1 )

Ans.

(d) The problem states that the strain readings were obtained from the free surface of a machine part. From this statement, we can conclude that a state of plane stress exists. For plane stress, the third principal strain z = p3 is not equal to zero. The normal strain in the z direction can be computed from Eq. 13.15:  0.30 z = − ( x +  y ) = − [(410 μ) + ( −330 μ)] = −34.2857 με 1− 1 − 0.30 Since p1 is positive and p2 is negative, the absolute maximum shear strain is the maximum in-plane shear strain: Ans.  abs max =  max = 1,376 μrad

P13.26 The strain rosette shown in the figure was used to obtain normal strain data at a point on the free surface of a machine part. (a) Determine the strain components x, y, and xy at the point. (b) Determine the principal strains and the maximum in-plane shear strain at the point. (c) Draw a sketch showing the angle p, the principal strain deformations, and the maximum in-plane shear strain distortions. (d) Determine the magnitude of the absolute maximum shear strain.

a = 215 , b = −710 , c = −760 ,  = 0.12 FIGURE P13.26

Solution (a) Write three normal strain transformation equations [Eq. (13.3)], one for each strain gage, where n is the measured normal strain. In each equation, the angle  associated with each strain gage will be referenced from the positive x axis. (a) 215 με =  x cos2 (270) +  y sin 2 (270) +  xy sin(270)cos(270)

−710 με =  x cos2 (0) +  y sin 2 (0) +  xy sin(0)cos(0)

(b)

−760 με =  x cos2 (135) +  y sin 2 (135) +  xy sin(135)cos(135) From Eq. (a):  y = 215 με

Ans.

and from Eq. (b):  x = −710 με

Ans.

(c)

Using these two results, solve Eq. (c) to find xy: −760 με = ( −710 με)cos 2 (135) + (215 με)sin 2 (135) +  xy sin(135)cos(135)

  xy = 1,025 μrad

Ans.

(b) Using these results, the principal strain magnitudes can be computed from Eq. (13.10):

 p1, p 2 =

x +  y 2

  x −  y    xy    +  2   2  2

( −710 μ) + (215 μ)  ( −710 μ) − (215 μ)   1,025 μ  =    +    2 2 2  2

= −247.5 μ  690.3351 μ

 p1 = 443 με and  p 2 = −938 με

Ans.

 max = 1,381 μrad

Ans.

tan 2 p =

(maximum in-plane shear strain)

 xy 1,025 μ 1,025 μ = = = −1.1081 ( x −  y ) [( −710 μ) − (215 μ)] −925 μ

 p = −24.0

(clockwise from the x axis to the direction of  p 2 )

Ans.

(c) The principal strain deformations and the maximum in-plane shear strain distortions are shown on the sketch below. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

(d) The problem states that the strain readings were obtained from the free surface of a machine part. From this statement, we can conclude that a state of plane stress exists. For plane stress, the third principal strain z = p3 is not equal to zero. From Eq. 13.15, the normal strain in the z direction can be computed as:  0.12 z = − ( x +  y ) = − [( −710 μ) + (215 μ)] = 67.5 με 1− 1 − 0.12 Since p1 is positive and p2 is negative, the absolute maximum shear strain is the maximum in-plane shear strain: Ans.  abs max =  max = 1,381 μrad

P13.27 The strain rosette shown in the figure was used to obtain normal strain data at a point on the free surface of a machine part. (a) Determine the strain components x, y, and xy at the point. (b) Determine the principal strains and the maximum in-plane shear strain at the point. (c) Draw a sketch showing the angle p, the principal strain deformations, and the maximum in-plane shear strain distortions. (d) Determine the magnitude of the absolute maximum shear strain.

a = 510 , b = 415 , c = 430 ,  = 0.33 FIGURE P13.27

Solution (a) Write three normal strain transformation equations [Eq. (13.3)], one for each strain gage, where n is the measured normal strain. In each equation, the angle  associated with each strain gage will be referenced from the positive x axis. (a) 510 με =  x cos2 (45) +  y sin 2 (45) +  xy sin(45)cos(45)

415 με =  x cos2 (90) +  y sin 2 (90) +  xy sin(90)cos(90)

(b)

430 με =  x cos (135) +  y sin (135) +  xy sin(135)cos(135) From Eq. (b):  y = 415 με

Using this result, solve Eqs. (a) and (c) simultaneously to obtain:  x = 525 με and  xy = 80 μrad

Ans.

(b) The principal strain magnitudes can be computed from Eq. (13.10):

 p1, p 2 =

x +  y 2

  x −  y    xy    +  2   2  2

(525 μ) + (415 μ)  (525 μ) − (415 μ)   80 μ  =    +    2 2 2  2

= 470 μ  68.0074 μ

 p1 = 538 με and  p 2 = 402 με

Ans.

 max = 136.0 μrad

Ans.

tan 2 p =

(maximum in-plane shear strain)

 xy 80 μ 80 μ = = = 0.7273 ( x −  y ) [(525 μ) − (415 μ)] 110 μ

 p = 18.01

(counterclockwise from the x axis to the direction of  p1 )

Ans.

(d) The problem states that the strain readings were obtained from the free surface of a machine part. From this statement, we can conclude that a state of plane stress exists. For plane stress, the third principal strain z = p3 is not equal to zero. From Eq. 13.15, the normal strain in the z direction can be computed as:  0.33 z = − ( x +  y ) = − [(525 μ) + (415 μ)] = −462.9851 με 1− 1 − 0.33 Since both p1 and p2 are positive, the absolute maximum shear strain will be greater than the maximum in-plane shear strain. Since this is a plane stress situation, we must remember to take into account the non-zero value of p3: Ans.  abs max =  p1 −  p3 = 538.0074 μ − (−462.9851 μ) = 1,001 μrad

P13.28 The strain rosette shown in the figure was used to obtain normal strain data at a point on the free surface of a machine part. (a) Determine the strain components x, y, and xy at the point. (b) Determine the principal strains and the maximum in-plane shear strain at the point. (c) Draw a sketch showing the angle p, the principal strain deformations, and the maximum in-plane shear strain distortions. (d) Determine the magnitude of the absolute maximum shear strain.

a = −960 , b = −815 , c = −505 ,  = 0.33

FIGURE P13.28

Solution (a) Write three normal strain transformation equations [Eq. (13.3)], one for each strain gage, where n is the measured normal strain. In each equation, the angle  associated with each strain gage will be referenced from the positive x axis. (a) −960 με =  x cos2 (0) +  y sin 2 (0) +  xy sin(0)cos(0)

−815 με =  x cos2 (60) +  y sin 2 (60) +  xy sin(60)cos(60)

(b)

−505 με =  x cos2 (120) +  y sin 2 (120) +  xy sin(120)cos(120) From Eq. (a):  x = −960 με

Using this result, solve Eqs. (b) and (c) simultaneously to obtain:  y = −560.0000 με = −560 με and  xy = −357.9572 μrad = −358 μrad

Ans.

(b) The principal strain magnitudes can be computed from Eq. (13.10):

 p1, p 2 =

x +  y 2

  x −  y    xy    +  2   2  2

( −960 μ) + ( −560 μ)  ( −960 μ) − ( −560 μ)   −357.9572 μ  =    +    2 2 2 2

= −760 μ  268.3903 μ

 p1 = −491 με and  p 2 = −1,028 με

Ans.

 max = 537 μrad

Ans.

tan 2 p =

(maximum in-plane shear strain)

 xy −357.9572 μ −357.9572 μ = = = 0.8949 ( x −  y ) [( −960 μ) − ( −560 μ)] 400 μ

 p = 20.9

(counterclockwise from the x axis to the direction of  p 2 )

Ans.

(d) The problem states that the strain readings were obtained from the free surface of a machine part. From this statement, we can conclude that a state of plane stress exists. For plane stress, the third principal strain z = p3 is not equal to zero. From Eq. 13.15, the normal strain in the z direction can be computed as:  0.33 z = − ( x +  y ) = − [( −960 μ) + ( −560 μ)] = 748.6567 με 1− 1 − 0.33 Since both p1 and p2 are negative, the absolute maximum shear strain will be greater than the maximum in-plane shear strain. Since this is a plane stress situation, we must remember to take into account the non-zero value of p3: Ans.  abs max =  p3 −  p 2 = 748.6567 μ − (−1,028.3903 μ) = 1,777 μrad

P13.29 The strain rosette shown in the figure was used to obtain normal strain data at a point on the free surface of a machine part. (a) Determine the strain components x, y, and xy at the point. (b) Determine the principal strains and the maximum in-plane shear strain at the point. (c) Draw a sketch showing the angle p, the principal strain deformations, and the maximum in-plane shear strain distortions. (d) Determine the magnitude of the absolute maximum shear strain.

a = −360 , b = −230 , c = 815 ,  = 0.15 FIGURE P13.29

Solution (a) Write three normal strain transformation equations [Eq. (13.3)], one for each strain gage, where n is the measured normal strain. In each equation, the angle  associated with each strain gage will be referenced from the positive x axis. (a) −360 με =  x cos2 (0) +  y sin 2 (0) +  xy sin(0)cos(0)

−230 με =  x cos2 (120) +  y sin 2 (120) +  xy sin(120)cos(120)

(b)

815 με =  x cos (240) +  y sin (240) +  xy sin(240)cos(240) From Eq. (a):  x = −360 με

Using this result, solve Eqs. (b) and (c) simultaneously to obtain:  y = 510 με and  xy = 1,206.6621 μrad = 1,207 μrad

Ans.

(b) Using these results, the principal strain magnitudes can be computed from Eq. (13.10):

 p1, p 2 =

x +  y 2

  x −  y    xy    +  2   2  2

( −360 μ) + (510 μ)  ( −360 μ) − (510 μ)   1, 206.6621 μ     +    2 2 2 2

= 75 μ  743.7966 μ

 p1 = 819 με and  p 2 = −669 με

Ans.

 max = 1,488 μrad

Ans.

tan 2 p =

(maximum in-plane shear strain)

 xy 1, 206.6621 μ 1, 206.6621 μ = = = −1.3870 ( x −  y ) [( −360 μ) − (510 μ)] −870 μ

 p = −27.1

(clockwise from the x axis to the direction of  p 2 )

Ans.

(d) The problem states that the strain readings were obtained from the free surface of a machine part. From this statement, we can conclude that a state of plane stress exists. For plane stress, the third principal strain z = p3 is not equal to zero. From Eq. 13.15, the normal strain in the z direction can be computed as:  0.15 z = − ( x +  y ) = − [( −360 μ) + (510 μ)] = −26.4706 με 1− 1 − 0.15 Since p1 is positive and p2 is negative, the absolute maximum shear strain is the maximum in-plane shear strain: Ans.  abs max =  max = 1,488 μrad

P13.30 The strain rosette shown in the figure was used to obtain normal strain data at a point on the free surface of a machine part. (a) Determine the strain components x, y, and xy at the point. (b) Determine the principal strains and the maximum in-plane shear strain at the point. (c) Draw a sketch showing the angle p, the principal strain deformations, and the maximum in-plane shear strain distortions. (d) Determine the magnitude of the absolute maximum shear strain. FIGURE P13.30

a = 775 , b = −515 , c = 415 ,  = 0.30

Solution (a) Write three normal strain transformation equations [Eq. (13.3)], one for each strain gage, where n is the measured normal strain. In each equation, the angle  associated with each strain gage will be referenced from the positive x axis. (a) 775 με =  x cos2 (0) +  y sin 2 (0) +  xy sin(0)cos(0)

−515 με =  x cos2 (120) +  y sin 2 (120) +  xy sin(120)cos(120)

(b)

415 με =  x cos2 (60) +  y sin 2 (60) +  xy sin(60)cos(60) From Eq. (a):  x = 775 με

Ans.

Using this result, solve Eqs. (b) and (c) simultaneously to obtain:  y = −325 με and  xy = 1,073.8715 μrad = 1,074 μrad

Ans.

(c)

(b) Using these results, the principal strain magnitudes can be computed from Eq. (13.10):

 p1, p 2 =

x +  y 2

  x −  y    xy    +  2   2  2

(775 μ) + ( −325 μ)  (775 μ) − ( −325 μ)   1,073.8715 μ  =    +    2 2 2 2

= 225 μ  768.6352 μ

 p1 = 994 με and  p 2 = −544 με

Ans.

 max = 1,537 μrad

Ans.

tan 2 p =

(maximum in-plane shear strain)

 xy 1,073.8715 μ 1,073.8715 μ = = = 0.9762 ( x −  y ) [(775 μ) − ( −325 μ)] 1,100 μ

 p = 22.2

(counterclockwise from the x axis to the direction of  p1 )

Ans.

(d) The problem states that the strain readings were obtained from the free surface of a machine part. From this statement, we can conclude that a state of plane stress exists. For plane stress, the third principal strain z = p3 is not equal to zero. From Eq. 13.15, the normal strain in the z direction can be computed as:  0.30 z = − ( x +  y ) = − [(775 μ) + ( −325 μ)] = −192.8571 με 1− 1 − 0.30 Since p1 is positive and p2 is negative, the absolute maximum shear strain is the maximum in-plane shear strain: Ans.  abs max =  max = 1,537 μrad

P13.31 An 8-mm-thick brass [E =83 GPa;  = 0.33] plate is subjected to biaxial stress with x = 180 MPa and y = 65 MPa. The plate dimensions are b = 350 mm and h = 175 mm (see Figure P13.31). Determine (a) the change in length of edges AB and AD. (b) the change in length of diagonal AC. (c) the change in thickness of the plate. FIGURE P13.31

Solution (a) From the generalized Hooke’s law equations for plane stress, the normal strains produced in the plate can be computed from Eqs. (13.21): 1 1  x = ( x −  y ) = [180 MPa − (0.33)(65 MPa)] = 1,910.24  10−6 mm/mm E 83,000 MPa 1 1  y = ( y −  x ) = [65 MPa − (0.33)(180 MPa)] = 67.47  10−6 mm/mm E 83,000 MPa  0.33  z = − ( x +  y ) = − [180 MPa + 65 MPa] = −974.10  10−6 mm/mm E 83,000 MPa Plate edge AB is aligned with the x direction; therefore, the change in length of edge AB can be computed from the product of x and plate dimension b: Ans.  AB =  xb = (1,910.24  10−6 mm/mm)(375 mm) = 0.66858 mm = 0.669 mm Plate edge AD is aligned with the y direction; therefore, the change in length of edge AD can be computed from the product of y and plate dimension h: Ans.  AD =  y h = (67.47  10−6 mm/mm)(175 mm) = 0.01181 mm = 0.01181 mm (b) A strain transformation equation [Eq. (13.3)]  n =  x cos2  +  y sin 2  +  xy sin  cos can be written to determine the normal strain in the direction of diagonal AC. Since there is no shear stress acting on the plate, we know that xy = 0 (since Hooke’s law relating shear stress and shear strain is xy = Gxy). The angle  between edge AB and diagonal AC is: 175 mm tan  = = 0.5   = 26.565 350 mm Therefore, the normal strain in the direction of diagonal AC is:  AC = (1,910.24  10−6 mm/mm)cos 2 (26.565) + (67.47  10 −6 mm/mm)sin 2 (26.565)

= 1,541.69  10−6 mm/mm The initial length of diagonal AC is: LAC = (350 mm)2 + (175 mm)2 = 391.312 mm

The change in length of diagonal AC is computed from the product of AC and the initial diagonal length: Ans.  AC =  AC LAC = (1,541.69  10−6 mm/mm)(391.312 mm) = 0.6034 mm = 0.603 mm Alternate Method: The change in length of diagonal AC can also be computed using the Pythagorean theorem. After deformation, the final length of side AB of the plate is 350.66868 mm. The final length of side AD is 175.01181 mm. From the Pythagorean theorem, the deformed length of diagonal AC is: LAC  = (350.66868 mm)2 + (175.01181 mm) 2 = 391.9153 mm and therefore, the elongation of diagonal AC is

 AC = 391.9153 mm − 391.3119 mm = 0.6034mm = 0.603 mm (c) The change in plate thickness is computed from the product of z and the plate thickness:  thick =  z (thickness) = (−974.0964  10−6 mm/mm)(8 mm) = −0.00779 mm

Ans.

P13.32 A 0.75-in.-thick polymer [E = 470,000 psi;  = 0.37] casting is subjected to biaxial stresses of x = 2,500 psi and y = 8,300 psi, acting in the directions shown in Figure P13.32. The dimensions of the casting are b = 12.0 in. and h = 8.0 in. Determine (a) the change in length of edges AB and AD. (b) the change in length of diagonal AC. (c) the change in thickness of the plate. FIGURE P13.32

Solution (a) From the generalized Hooke’s law equations for plane stress, the normal strains produced in the plate can be computed from Eqs. (13.21): 1 1  x = ( x −  y ) = [2,500 psi − (0.37)( −8,300 psi)] = 11,853.19  10−6 in./in. E 470,000 psi 1 1  y = ( y −  x ) = [ −8,300 psi − (0.37)(2,500 psi)] = −19,627.66  10−6 in./in. E 470,000 psi  0.37  z = − ( x +  y ) = − [2,500 psi + ( −8,300 psi)] = 4,565.96  10 −6 in./in. E 470,000 psi Plate edge AB is aligned with the x direction; therefore, the change in length of edge AB can be computed from the product of x and plate dimension b: Ans.  AB =  xb = (11,853.19  10−6 in./in.)(12 in.) = 0.1422 in. Plate edge AD is aligned with the y direction; therefore, the change in length of edge AD can be computed from the product of y and plate dimension h: Ans.  AD =  y h = (−19,627.66  10−6 in./in.)(8.00 in.) = −0.1570 in. (b) The change in length of diagonal AC can also be computed using the Pythagorean Theorem. After deformation, the final length of side AB of the plate is 12.14224 in. The final length of side AD is 7.84298 in. From the Pythagorean Theorem, the deformed length of diagonal AC is: LAC  = (12.14224 in.)2 + (7.84298 in.)2 = 14.45497 in. and therefore, the elongation of diagonal AC is

 AC = 14.45497 in. − 14.42221 in. = 0.00328 in. (c) The change in plate thickness is computed from the product of z and the plate thickness:  thick =  z (thickness) = (4,565.96  10−6 in./in.)(0.750 in.) = 0.00342 in.

Ans.

P13.33 A 100 mm by 100 mm square plate with a circle inscribed is subjected to normal stresses x = 200 MPa and y = 90 MPa, acting as shown in Figure P13.33. The plate material has an elastic modulus E = 70 GPa and a Poisson’s ratio  = 0.33. Assuming plane stress, determine the major and minor axes of the ellipse formed after deformation of the plate.

FIGURE P13.33

Solution From the generalized Hooke’s law equations for plane stress, the normal strains produced in the plate can be computed from Eqs. (13.21): 1 1  x = ( x − y ) =  −200 MPa − ( 0.33)( 90 MPa )  = −3, 281.426 10 −6 mm/mm E 70, 000 MPa  1 1  y = ( y − x ) = 90 MPa − ( 0.33)( −200 MPa )  = 2, 228.57110 −6 mm/mm E 70, 000 MPa Consider the 100 mm circle diameter that is aligned in the x direction. The change in length of this diameter is computed from the product of x and circle diameter:  x =  x (100 mm ) = ( −3, 281.426 10−6 mm/mm ) (100 mm ) = −0.32814 mm Next, consider the 100 mm circle diameter that is aligned in the y direction. The change in length of this diameter is computed from the product of y and circle diameter:  y =  y (100 mm ) = ( 2, 228.57110−6 mm/mm ) (100 mm ) = 0.22286 mm The major and minor axes of the ellipse are thus: Major axis = 100.223 mm

Minor axis = 99.672 mm

Ans. Ans.

P13.34 A thin steel [E = 30,000 ksi;  = 0.3] in a state of plane stress has dimensions of 8 in. in the x direction and 4 in. in the y direction. The plate increases in length in the x direction by 0.0015 in. and decreases in the y direction by 0.00028 in. Compute the normal stresses x and y that produced these deformations. Assume that xy = 0.

Solution Compute the normal strains in the x and y directions from the specified deformations and the specified plate dimensions: 0.0015 in. x = = 187.5 10−6 in./in. 8 in. −0.00028 in. y = = −70.0 10−6 in./in. 4 in. The normal stresses can now be computed from Eq. (13.23): E x = ( x + y ) 1 − 2 30, 000 ksi (187.5  10−6 in./in.) + ( 0.3) ( −70.0  10−6 in./in.)  =  1 − (0.3) 2  = 5.489 ksi = 5.49 ksi

Ans.

and E ( y + x ) 1 − 2 30, 000 ksi ( −70.0 10−6 in./in.) + ( 0.3) (187.5 10−6 in./in.)  =  1 − (0.3) 2 

y =

= −0.4533 ksi = −0.453 ksi

Ans.

P13.35 A thin aluminum [E = 10,000 ksi; G = 3,800 ksi] plate is subjected to biaxial stress (Figure P13.35/54). The strains measured in the plate are x = 810  and z = 1,350 . Determine x and z.

FIGURE P13.35/36

Solution Derive an expression for  from Eq. (13.18): E G= 2(1 +  ) E 1 + = 2G E = −1 2G Determine Poisson’s ratio from this expression: E 10, 000 ksi = −1 = − 1 = 0.3158 2G 2(3,800 ksi) The normal stresses can now be computed from Eq. (13.23): E x = ( x +  z ) 1 − 2 10,000 ksi = [(810  10−6 in./in.) + (0.3158)(1,350  10 −6 in./in.)] 1 − (0.3158) 2

= 13.733 ksi = 13.73 ksi

Ans.

and

E ( z +  x ) 1 − 2 10,000 ksi = [(1,350  10−6 in./in.) + (0.3158)(810  10 −6 in./in.)] 2 1 − (0.3158)

z =

= 17.837 ksi = 17.84 ksi

Ans.

P13.36 A thin stainless steel plate [E = 190 GPa; G = 86 GPa] plate is subjected to biaxial stress (Figure P13.35/54). The strains measured in the plate are x = 275  and z = 1,150 . Determine x and z.

FIGURE P13.35/36

Solution Derive an expression for  from Eq. (13.18): E G= 2(1 +  ) E 1 + = 2G E = −1 2G Determine Poisson’s ratio from this expression: E 190 GPa = −1 = − 1 = 0.10465 2G 2(86 GPa) The normal stresses can now be computed from Eq. (13.23): E x = ( x +  z ) 1 − 2 190,000 MPa = [(275  10−6 mm/mm) + (0.10465)(1,150  10−6 mm/mm)] 2 1 − (0.10465)

= 75.948 MPa = 75.9 MPa

Ans.

and

E ( z +  x ) 1 − 2 190,000 MPa = [(1,150  10−6 mm/mm) + (0.10465)(275  10−6 mm/mm)] 2 1 − (0.10465)

z =

= 226.448 MPa = 226 MPa

Ans.

P13.37 The thin brass [E = 16,700 ksi;  = 0.307] bar shown in Figure P13.37/38 is subjected to a normal stress of x = 19 ksi. A strain gage is mounted on the bar at an orientation of  = 25° as shown in the figure. What normal strain reading would be expected from the strain gage at the specified stress? FIGURE P13.37/38

Solution The stresses in the bar are x = 19 ksi, y = 0, and xy = 0. The normal strains in the x and y directions can be computed from Eqs. (13.21): 1 1  x = ( x −  y ) = [19 ksi − (0.307)(0 ksi)] E 16,700 ksi = 1,137.725  10 −6 in./in. 1 1  y = ( y −  x ) = [0 ksi − (0.307)(19 ksi)] E 16,700 ksi = −349.281  10 −6 in./in.

Since xy = 0, the shear strain xy = 0. Using the strain transformation equation [Eq. (13.3)]  n =  x cos2  +  y sin 2  +  xy sin  cos the expected normal strain in the direction of strain gage can be calculated:  n = (1,137.725  10−6 in./in.)cos 2 (25) + ( −349.281  10−6 in./in.)sin 2 (25)

= 872.136  10−6 in./in. = 872 με

Ans.

P13.38 A strain gage is mounted on a thin brass [E = 12,000 ksi;  = 0.33] bar at an angle of  = 35° as shown in Figure P13.37/38. If the strain gage records a normal strain of n = 470 , what is the magnitude of the normal stress x? FIGURE P13.37/38

Solution We observe that y = 0 and xy = 0 for this bar. The normal strains in the x and y directions for the bar can be expressed as: 1 1  x = ( x −  y ) =  x E E 1   y = ( y −  x ) = −  x E E Since xy = 0, the shear strain xy = 0. Write a strain transformation equation [Eq. (13.3)] for the normal strain in the n direction using these results:  n =  x cos 2  +  y sin 2  +  xy sin  cos  =

1   x cos 2 (35) −  x sin 2 (35) E E

x

cos 2 (35) −  sin 2 (35)  E  Solve this expression for x: (470  10−6 in./in.)(12,000 ksi) x = = 10.03 ksi cos2 (35) − (0.33)sin 2 (35) =

Ans.

P13.39 A thin brass [E = 100 GPa; G = 39 GPa] plate is subjected to biaxial stress as shown in Figure P13.39/40. The normal stress in the y direction is known to be y = 160 MPa. The strain gage measures a normal strain of 920  at an orientation of  = 35° in the indicated direction. What is the magnitude of x that acts on the plate?

FIGURE P13.39/40

Solution Derive an expression for  from Eq. (13.18): E E G=  = −1 2(1 +  ) 2G Determine Poisson’s ratio from this expression: E 100 GPa = −1 = − 1 = 0.28205 2G 2(39 GPa) A strain transformation equation [Eq. (13.3)]  n =  x cos2  +  y sin 2  +  xy sin  cos can be written for the normal strain in the direction of strain gage: 920 με =  x cos2 (35) +  y sin 2 (35) +  xy sin(35)cos(35) Note that the shear strain xy is related to the shear stress xy by Eq. (13.22): 1  xy =  xy G Since xy = 0, the shear strain xy must also equal zero, and the strain transformation equation reduces to: 920 με = 920  10−6 mm/mm =  x cos2 (35) +  y sin 2 (35) Substitute Eqs. (13.21) for x and y to obtain an expression in terms of x and y: 920  10−6 =  x cos 2 (35) +  y sin 2 (35)

1 1 ( x −  y )cos 2 (35) + ( y −  x )sin 2 (35) E E 1 1 = [ x cos 2 (35) −  x sin 2 (35)] + [ y sin 2 (35) −  y cos 2 (35)] E E

x

[cos 2 (35) −  sin 2 (35)] +

y

[sin 2 (35) −  cos 2 (35)] E E Substitute the known value of y = 160 MPa and solve for x: 100,000 MPa 160 MPa   x = 920  10 −6 − [sin 2 (35) − (0.28205)cos 2 (35)] 2 2  [cos (35) − (0.28205)sin (35)]  100,000 MPa  =

= 120.444 MPa = 120.4 MPa Ans.

P13.40 A thin brass [E = 14,500 ksi; G = 5,500 ksi] plate is subjected to biaxial stress (Figure P13.39/40). The normal stress in the x direction is known to be twice as large as the normal stress in the y direction. The strain gage measures a normal strain of 775  at an orientation of  = 50° in the indicated direction. Determine the magnitudes of the normal stresses x and y acting on the plate.

FIGURE P13.39/40

Solution Derive an expression for  from Eq. (13.18): E E G=  = −1 2(1 +  ) 2G Determine Poisson’s ratio from this expression: E 14,500 ksi = −1 = − 1 = 0.3182 2G 2(5,500 ksi) (a) A strain transformation equation [Eq. (13.3)]  n =  x cos2  +  y sin 2  +  xy sin  cos can be written for the normal strain in the direction of strain gage: 775 με =  x cos2 (50) +  y sin 2 (50) +  xy sin(50)cos(50) Note that the shear strain xy is related to the shear stress xy by Eq. (13.22): 1  xy =  xy G Since xy = 0, the shear strain xy must also equal zero, and the strain transformation equation reduces to: 775 με = 775  10−6 in./in. =  x cos2 (50) +  y sin 2 (50) Substitute Eqs. (13.21) for x and y to obtain an expression in terms of x and y: 775  10−6 =  x cos 2 (50) +  y sin 2 (50) 1 1 ( x −  y )cos 2 (50) + ( y −  x )sin 2 (50) E E It is known that x = 2y. Make this substitution to obtain the following expression: (775  10 −6 ) E = (2 y −  y ) cos 2 (50) + ( y − 2 y )sin 2 (50) =

=  y [(2 −  ) cos 2 (50) + (1 − 2 )sin 2 (50)]

 y =

(775  10−6 ) E (2 −  )cos2 (50) + (1 − 2 )sin 2 (50)

Compute y:

y =

(775  10−6 )(14,500 ksi) = 12.3723 ksi = 12.37 ksi (2 − 0.3182)cos2 (50) + [1 − 2(0.3182)]sin 2 (50)

Ans.

and x:

 x = 2 y = 2(12.3723 ksi) = 24.7446 ksi = 24.7 ksi

Ans.

P13.41 On the free surface of an aluminum [E = 10,000 ksi;  = 0.33] component, the strain rosette shown in Figure P13.41 was used to obtain the following normal strain data: a = 440  b = 550 , and c = 870 . Determine (a) the normal stress x. (b) the normal stress y. (c) the shear stress xy. FIGURE P13.41

Solution Write three normal strain transformation equations [Eq. (13.3)], one for each strain gage, where n is the measured normal strain. In each equation, the angle  associated with each strain gage will be referenced from the positive x axis. (a) 440 με =  x cos2 (315) +  y sin 2 (315) +  xy sin(315)cos(315)

550 με =  x cos2 (0) +  y sin 2 (0) +  xy sin(0)cos(0)

(b)

870 με =  x cos (45) +  y sin (45) +  xy sin(45)cos(45) From Eq. (b):  x = 550 με Using this result, solve Eqs. (a) and (c) simultaneously to obtain:  y = 760 με and  xy = 430 μrad

(c)

(a) From Eqs. (13.23), compute x: E 10,000 ksi x = ( x +  y ) = [(550  10−6 in./in.) + (0.33)(760  10−6 in./in.)] 2 2 1− 1 − (0.33)

= 8.9866 ksi = 8.99 ksi (T)

Ans.

(b) From Eqs. (13.23), compute y: E 10,000 ksi y = ( y +  x ) = [(760  10−6 in./in.) + (0.33)(550  10 −6 in./in.)] 2 2 1− 1 − (0.33)

= 10.5656 ksi = 10.57 ksi (T) (c) From Eq. (13.18), determine the shear modulus G: E 10, 000 ksi G= = = 3, 759.4 ksi 2(1 +  ) 2(1 + 0.33) and compute the shear stress xy from Eq. (13.20):  xy = G xy = (3,759.4 ksi)(430  10−6 rad) = 1.6165 ksi = 1.617 ksi

Ans.

P13.42 On the free surface of an aluminum [E = 70 GPa;  = 0.35] component, the strain rosette shown in Figure P13.42 was used to obtain the following normal strain data: a = –300   b = 735 , and c = 410 . Determine (a) the normal stress x. (b) the normal stress y. (c) the shear stress xy. FIGURE P13.42

Solution Write three normal strain transformation equations [Eq. (13.3)], one for each strain gage, where n is the measured normal strain. In each equation, the angle  associated with each strain gage will be referenced from the positive x axis. (a) −300 με =  x cos2 (30) +  y sin 2 (30) +  xy sin(30)cos(30)

735 με =  x cos2 (90) +  y sin 2 (90) +  xy sin(90)cos(90)

(b)

410 με =  x cos2 (150) +  y sin 2 (150) +  xy sin(150)cos(150) From Eq. (b):  y = 735 με Using this result, solve Eqs. (a) and (c) simultaneously to obtain:  x = −171.667 με and  xy = −819.837 μrad

(c)

(a) From Eqs. (13.23), compute x: E 70, 000 MPa x = ( x + y ) = [(−171.667 10−6 ) + (0.35)(735 10−6 )] 2 2 1 − 1 − (0.35)

= 6.8272 MPa = 6.83 MPa (T)

Ans.

(b) From Eqs. (13.23), compute y: E 70, 000 MPa y = ( y + x ) = [(735 10−6 ) + (0.35)(−171.667 10−6 )] 2 2 1 − 1 − (0.35)

= 53.8395 MPa = 53.8 MPa (T) (c) From Eq. (13.18), determine the shear modulus G: E 70,000 MPa G= = = 25,925.926 MPa 2(1 +  ) 2(1 + 0.35) and compute the shear stress xy from Eq. (13.20):  xy = G xy = (25,925.926 MPa)(−819.8374 10−6 rad) = −21.2550 MPa = −21.3 MPa

Ans.

P13.43 On the free surface of a steel [E = 207 GPa;  = 0.29] component, a strain rosette located at point A in Figure P13.43 was used to obtain the following normal strain data: a = 133   b = –92 , and c = –319 . If  = 50°, determine the stresses n, t, and nt that act at point A.

FIGURE P13.43

Solution Write three normal strain transformation equations [Eq. (13.3)], one for each strain gage, using the measured normal strain. In each equation, the angle  associated with each strain gage will be referenced from the positive x axis. (a) 133 με =  x cos2 (0) +  y sin 2 (0) +  xy sin(0)cos(0)

−92 με =  x cos2 (120) +  y sin 2 (120) +  xy sin(120)cos(120) −319 με =  x cos2 (60) +  y sin 2 (60) +  xy sin(60)cos(60) From Eq. (a):  x = 133 με Using this result, solve Eqs. (b) and (c) simultaneously to obtain:  y = −318.333 με and  xy = −262.117 μrad

(b) (c)

From Eqs. (13.23), compute x: E 207,000 MPa x = ( x +  y ) = [(133  10−6 ) + (0.29)( −318.333  10 −6 )] 2 1− 1 − (0.29)2 and y:

= 9.195 MPa E 207,000 MPa ( y +  x ) = [( −318.333  10−6 ) + (0.29)(133  10−6 )] 2 2 1− 1 − (0.29) = −63.229 MPa

y =

From Eq. (13.18), determine the shear modulus G: E 207,000 MPa G= = = 80,232.558 MPa 2(1 +  ) 2(1 + 0.29) and compute the shear stress xy from Eq. (13.20):  xy = G xy = (80,232.558 MPa)(-262.117  10−6 rad) = −21.030 MPa To summarize, normal and shear stresses in the x-y plane are:  x = 9.195 MPa,  y = −63.229 MPa,  xy = −21.030 MPa

A normal stress transformation equation [Eq. (12-3)] can now be written to determine the normal stress at an orientation of  = 50°:  n =  x cos 2  +  y sin 2  + 2 xy sin  cos  = (9.195 MPa) cos 2 (50) + ( −63.229 MPa)sin 2 (50) + 2( −21.030 MPa)sin(50) cos(50) = −54.016 MPa = 54.0 MPa (C)

Ans.

For t, use  = 140°:  t =  x cos 2  +  y sin 2  + 2 xy sin  cos  = (9.195 MPa)cos 2 (140) + ( −63.229 MPa)sin 2 (140) + 2( −21.030 MPa)sin(140)cos(140) = −0.01800 MPa = 0.01800 MPa (C)

Ans.

A shear stress transformation equation [Eq. (12-4)] can now be written to determine the shear stress at an orientation of  = 50°:  nt = −( x −  y )sin  cos  + 2 xy (cos 2  − sin 2  ) = [(9.195 MPa) − ( −63.229 MPa)]sin(50)cos(50) + 2( −21.030 MPa)[cos 2 (50) − sin 2 (50)] = −32.010 MPa = −32.0 MPa

Ans.

P13.44–P13.46 The strain components x, y, and xy are given for a point on the free surface of a machine component. Determine the stresses x, y, and xy at the point. Problem E x y xy  680  −320  −840 rad P13.44 16,500 ksi 0.33

Solution The normal stresses can now be computed from Eq. (13.23): E x = ( x + y ) 1 − 2 16,500 ksi  680 10−6 in./in.) + ( 0.33) ( −320 10−6 in./in.)  = 2 (  1 − 0.33

(

)

= 10.6358 ksi = 10.64 ksi

Ans.

and

E ( y + x ) 1 − 2 16,500 ksi  −320 10−6 in./in.) + ( 0.33) ( 680 10−6 in./in.)  = 2 (  1 − 0.33

y =

(

)

= −1.7702 ksi = −1.770 ksi

Ans.

From Eq. (13.18), determine the shear modulus G: E 16,500 ksi G= = = 6, 203.0075 ksi 2 (1 + ) 2 (1 + 0.33) and compute the shear stress xy from Eq. (13.20):  xy = G xy = ( 6, 203.0075 ksi ) ( −840 10−6 rad ) = −5.2105 ksi = −5.21 ksi

Ans.

P13.44–P13.46 The strain components x, y, and xy are given for a point on the free surface of a machine component. Determine the stresses x, y, and xy at the point. Problem E x y xy  −1,120  −890  1,300 rad P13.45 207 GPa 0.30

Solution The normal stresses can now be computed from Eq. (13.23): E x = ( x + y ) 1 − 2 207, 000 MPa ( −1,120 10−6 mm/mm ) + ( 0.3) ( −890 10−6 mm/mm )  = 2   1 − 0.3

(

)

= −315.5044 MPa = −316 MPa

Ans.

and

E ( y + x ) 1 − 2 207, 000 MPa ( −890 10−6 mm/mm ) + ( 0.3) ( −1,120 10−6 mm/mm )  = 2   1 − 0.3

y =

(

)

= −278.8813 MPa = −279 MPa

Ans.

From Eq. (13.18), determine the shear modulus G: E 207,000 MPa G= = = 79,615.3846 MPa 2 (1 + ) 2 (1 + 0.3) and compute the shear stress xy from Eq. (13.20):  xy = G xy = ( 79,615.3846 MPa ) (1,300 10−6 rad ) = 103.5000 MPa = 103.5 MPa

Ans.

P13.44–P13.46 The strain components x, y, and xy are given for a point on the free surface of a machine component. Determine the stresses x, y, and xy at the point. Problem E x y xy  1,500  2,100  950 rad P13.46 28,000 ksi 0.12

Solution The normal stresses can now be computed from Eq. (13.23): E x = ( x + y ) 1 − 2 28, 000 ksi  1,500 10−6 in./in.) + ( 0.12 ) ( 2,100 10−6 in./in.)  = 2 (  1 − 0.12

(

)

= 49.7727 ksi = 49.8 ksi

Ans.

and

E ( y + x ) 1 − 2 28, 000 ksi  2,100 10−6 in./in.) + ( 0.12 ) (1,500 10−6 in./in.)  = 2 (  1 − 0.12

y =

(

)

= 64.7727 ksi = 64.8 ksi

Ans.

From Eq. (13.18), determine the shear modulus G: E 28,000 ksi G= = = 12,500 ksi 2 (1 + ) 2 (1 + 0.12 ) and compute the shear stress xy from Eq. (13.20):  xy = G xy = (12,500 ksi ) ( 950 10−6 rad ) = 11.8750 ksi = 11.88 ksi

Ans.

The strain rosette shown in the Figures P13.47–P13.49 was used to obtain normal strain data at a point on the free surface of a machine component. Determine: (a) the stress components x, y, and xy at the point. (b) the principal stresses and the maximum in-plane shear stress at the point; show these stresses on an appropriate sketch that indicates the orientation of the principal planes and the planes of maximum in-plane shear stress. (c) the magnitude of the absolute maximum shear stress at the point. Problem E a b c  −870  340  P13.47 −1,320  100 GPa 0.28

FIGURE P13.47

Solution (a) Write three normal strain transformation equations [Eq. (13.3)], one for each strain gage, where n is the measured normal strain. In each equation, the angle  associated with each strain gage will be referenced from the positive x axis. (a) −1320 με =  x cos2 (0) +  y sin 2 (0) +  xy sin(0)cos(0)

−870 με =  x cos2 (90) +  y sin 2 (90) +  xy sin(90)cos(90)

(b)

340 με =  x cos2 (225) +  y sin 2 (225) +  xy sin(225)cos(225)

(c)

From Eqs. (a) and (b):  x = −1,320 με and  y = −870 με Solve Eq. (c) to find:  xy = 2,870 μrad From Eqs. (13.23), compute x: E 100, 000 MPa x = ( x + y ) = [(−1,320 10−6 ) + (0.28)(−870 10−6 )] 2 2 1 − 1 − (0.28)

= −169.6615 MPa = 169.7 MPa (C)

Ans.

and y:

y =

E 100, 000 MPa ( y + x ) = [(−870 10−6 ) + (0.28)(−1,320 10−6 )] 2 2 1 − 1 − (0.28)

= −134.5052 MPa = 134.5 MPa (C) From Eq. (13.18), determine the shear modulus G: E 100, 000 MPa G= = = 39, 062.5 MPa 2(1 + ) 2(1 + 0.28) and compute the shear stress xy from Eq. (13.20):  xy = G xy = (39,062.5 MPa)(2,870 10−6 rad) = 112.1094 MPa = 112.1 MPa

Ans.

(b) The principal stress magnitudes can be computed from Eq. (12.12):

 p1, p 2 =

x + y 2

  x − y  2    +  xy  2  2

(−169.6615) + (−134.5052) 2  ( −169.6615) − ( −134.5052)  =    + (112.1094 ) 2 2   = −152.0833 MPa  113.4791 MPa  p1 = −38.6 MPa and  p 2 = −266 MPa

Ans.

 max = 113.5 MPa

Ans.

(maximum in-plane shear stress)

 avg = 152.1 MPa (C) tan 2 p =

2 xy

 x − y

(normal stress on planes of maximum in-plane shear stress)

Ans.

2 (112.1094 ) 224.2188 = = −6.3778 ( −169.6615) − ( −134.5052 ) −35.1563

 p = 40.5

(clockwise from the x axis to the direction of  p 2 )

(c) For plane stress, z = p3 = 0. Since both p1 and p2 are negative, 0 − ( −265.5624 )  −  min  abs max = max = = 132.8 MPa 2 2

Ans.

The strain rosette shown in the Figures P13.47–P13.49 was used to obtain normal strain data at a point on the free surface of a machine component. Determine: (a) the stress components x, y, and xy at the point. (b) the principal stresses and the maximum in-plane shear stress at the point; show these stresses on an appropriate sketch that indicates the orientation of the principal planes and the planes of maximum in-plane shear stress. (c) the magnitude of the absolute maximum shear stress at the point. Problem E a b c  1,400  560  −1,270  210 GPa P13.48 0.31

FIGURE P13.48

Solution (a) Write three normal strain transformation equations [Eq. (13.3)], one for each strain gage, where n is the measured normal strain. In each equation, the angle  associated with each strain gage will be referenced from the positive x axis. (a) 1, 400 με =  x cos2 (180) +  y sin 2 (180) +  xy sin(180)cos(180)

560 με =  x cos2 (300) +  y sin 2 (300) +  xy sin(300)cos(300)

(b)

−1, 270 με =  x cos2 (60) +  y sin 2 (60) +  xy sin(60)cos(60)

(c)

From Eq. (a):  x = 1, 400 με Solve Eqs. (b) and (c) simultaneously to find:  y = −940 με and  xy = −2,113.102 μrad From Eqs. (13.23), compute x: E 210, 000 MPa x = ( x + y ) = [(1, 400 10−6 ) + (0.31)(−940 10−6 )] 2 2 1 − 1 − (0.31)

= 257.5573 MPa = 258 MPa (T)

Ans.

and y:

y =

E 210, 000 MPa ( y + x ) = [(−940 10−6 ) + (0.31)(1, 400 10−6 )] 2 2 1 − 1 − (0.31)

= −117.5573 MPa = 117.6 MPa (C) From Eq. (13.18), determine the shear modulus G: E 210,000 MPa G= = = 80,152.7 MPa 2(1 +  ) 2(1 + 0.31) and compute the shear stress xy from Eq. (13.20):  xy = G xy = (80,152.7 MPa)( − 2,113.102 10−6 rad) = −169.3708 MPa = −169.4 MPa

Ans.

(b) The principal stress magnitudes can be computed from Eq. (12.12):

 p1, p 2 =

x + y 2

  x − y  2    +  xy  2  2

(257.5573) + (−117.5573)  (257.5573) − ( −117.5573)  2 =    + (−169.3708) 2 2   = 70 MPa  252.7136 MPa  p1 = 323 MPa and  p 2 = −182.7 MPa

Ans.

 max = 253 MPa

Ans.

(maximum in-plane shear stress)

 avg = 70 MPa (T) tan 2 p =

2 xy

 x − y

(normal stress on planes of maximum in-plane shear stress)

Ans.

2 ( −169.3708 ) −338.7415 = = −0.9030 ( 257.5573) − ( −117.5573) 375.1145

 p = 21.04

(clockwise from the x axis to the direction of  p1 )

Ans.

The strain rosette shown in the Figures P13.47–P13.49 was used to obtain normal strain data at a point on the free surface of a machine component. Determine: (a) the stress components x, y, and xy at the point. (b) the principal stresses and the maximum in-plane shear stress at the point; show these stresses on an appropriate sketch that indicates the orientation of the principal planes and the planes of maximum in-plane shear stress. (c) the magnitude of the absolute maximum shear stress at the point. Problem E a b c  910  720  1,200  15,000 ksi P13.49 0.15

FIGURE P13.49

Solution (a) Write three normal strain transformation equations [Eq. (13.3)], one for each strain gage, where n is the measured normal strain. In each equation, the angle  associated with each strain gage will be referenced from the positive x axis. (a) 910 με =  x cos2 (180) +  y sin 2 (180) +  xy sin(180)cos(180)

720 με =  x cos2 (135) +  y sin 2 (135) +  xy sin(135)cos(135)

(b)

1, 200 με =  x cos2 (90) +  y sin 2 (90) +  xy sin(90)cos(90)

(c)

From Eqs. (a) and (c):  x = 910 με and  y = 1, 200 με Solve Eq. (b) to find:  xy = 670 μrad From Eqs. (13.23), compute x: E 15, 000 ksi x = ( x + y ) = [(910 10−6 ) + (0.15)(1, 200 10−6 )] 2 2 1 − 1 − (0.15)

= 16.7263 ksi = 16.73 ksi (T)

Ans.

and y:

y =

E 15, 000 ksi ( y + x ) = [(1, 200 10−6 ) + (0.15)(910 10−6 )] 2 2 1 − 1 − (0.15)

= 20.5090 ksi = 20.5 ksi (T) From Eq. (13.18), determine the shear modulus G: E 15,000 ksi G= = = 6,521.7 ksi 2(1 +  ) 2(1 + 0.15) and compute the shear stress xy from Eq. (13.20):  xy = G xy = (6,521.7 ksi)(670 10−6 rad) = 4.3696 ksi = 4.37 ksi

Ans.

(b) The principal stress magnitudes can be computed from Eq. (12.12):

 p1, p 2 =

x + y 2

  x − y  2    +  xy  2  2

(16.7263) + (20.5090)  (16.7263) − (20.5090)  2 =    + (4.3696) 2 2   = 18.6176 ksi  4.7613 ksi  p1 = 23.4 ksi and  p 2 = 13.86 ksi

Ans.

 max = 4.76 ksi

Ans.

(maximum in-plane shear stress)

 avg = 18.62 ksi (T) tan 2 p =

2 xy

 x − y

(normal stress on planes of maximum in-plane shear stress)

Ans.

2 ( 4.3696 ) 8.7391 = = −2.3103 (16.7263) − ( 20.5090 ) −3.7826

 p = 33.30

(clockwise from the x axis to the direction of  p 2 )

(c) For plane stress, z = p3 = 0. Since both p1 and p2 are positive,  −  min 23.3790 − 0  abs max = max = = 11.69 ksi 2 2

Ans.

The strain rosette shown in the Figures P13.50–P13.52 was used to obtain normal strain data at a point on the free surface of a machine component. (a) Determine the strain components x, y, and xy at the point. (b) Determine the principal strains and the maximum in-plane shear strain at the point. (c) Using the results from part (b), determine the principal stresses and the maximum in-plane shear stress. Show these stresses on an appropriate sketch that indicates the orientation of the principal planes and the planes of maximum in-plane shear stress. (d) Determine the magnitude of the absolute maximum shear stress at the point. Problem E a b c  −840  −1,775  665  9,000 ksi P13.50 0.24

FIGURE P13.50

Solution (a) Write three normal strain transformation equations [Eq. (13.3)], one for each strain gage, where n is the measured normal strain. In each equation, the angle  associated with each strain gage will be referenced from the positive x axis. (a) −840 με =  x cos2 (30) +  y sin 2 (30) +  xy sin(30)cos(30)

−1,775 με =  x cos2 (150) +  y sin 2 (150) +  xy sin(150)cos(150)

(b)

665 με =  x cos (90) +  y sin (90) +  xy sin(90)cos(90) From Eq. (c):  y = 665 με

Ans.

Solve Eqs. (a) and (b) simultaneously to obtain:  x = −1,965 με and  xy = 1, 079.6450 μrad = 1, 080 μrad

Ans.

(c)

(b) Using these results, the principal strain magnitudes can be computed from Eq. (13.10):

 p1, p 2 =

x +y 2

  −  y    xy    x  +   2   2  2

(−1,965 μ) + (665 μ)  (−1,965 μ) − (665 μ)   1, 079.6450 μ  =    +  2 2 2     = −650 μ  1, 421.4898 μ 2

 p1 = 771.4898 με = 771 με and  p 2 = −2, 071.4898 με = −2, 070 με

Ans.

 max = 2,842.9797 μrad = 2,840 μrad

Ans.

tan 2 p =

(maximum in-plane shear strain)

 xy 1, 079.6450 μ 1, 079.6450 μ = = = −0.4105  x −  y (−1,965 μ) − (665 μ) −2, 630 μ

 p = 11.16

(clockwise from the x axis to the direction of  p 2 )

Ans.

(c) Use p1 and p2 in Eqs. (13.23) to compute p1: E 9, 000 ksi  p1 = ( p1 + p 2 ) = [(771.4898 10−6 ) + (0.24)(−2, 071.4898 10−6 )] 2 1 − 1 − (0.24)2

= 2.6199 ksi = 2.62 ksi (T)

Ans.

and p2:

 p2 =

E 9, 000 ksi ( p 2 + p1 ) = [(−2, 071.4898 10−6 ) + (0.24)(771.4898 10−6 )] 2 1 − 1 − (0.24)2

= −18.0146 ksi = 18.01 ksi (C) The maximum in-plane shear stress can be computed from the two principal stresses:  p1 −  p 2 (2.6199 ksi) − (−18.0146 ksi)  max = = = 10.3173 ksi = 10.32 ksi 2 2 and the normal stress on the plane of maximum shear stress is  +  p 2 (2.6199 ksi) + (−18.0146 ksi)  avg = p1 = = −7.6974 ksi = 7.70 ksi (C) 2 2

Ans.

Note: For completeness, the stresses x, y, and xy would be calculated as: E 9, 000 ksi  −1,965 10−6 ) + ( 0.24 ) ( 665  10−6 )  = −17.2417 ksi x = ( x + y ) = 2 ( 2  1 − 1 − ( 0.24 )

y =

E 9, 000 ksi  665 10−6 ) + ( 0.24 ) ( −1,965  10−6 )  = 1.8470 ksi ( y + x ) = 2 ( 2  1 − 1 − ( 0.24 )

E 9, 000 ksi = = 3, 629.0323 ksi 2 (1 + ) 2 (1 + 0.24 )

 xy = G xy = ( 3, 629.0323 ksi ) (1, 079.6450  10−6 rad ) = 3.9181 ksi

(d) For plane stress, z = p3 = 0. Since p1 and p2 are of opposite signs,  abs max =  max = 10.32 ksi

Ans.

The strain rosette shown in the Figures P13.50–P13.52 was used to obtain normal strain data at a point on the free surface of a machine component. (a) Determine the strain components x, y, and xy at the point. (b) Determine the principal strains and the maximum in-plane shear strain at the point. (c) Using the results from part (b), determine the principal stresses and the maximum in-plane shear stress. Show these stresses on an appropriate sketch that indicates the orientation of the principal planes and the planes of maximum in-plane shear stress. (d) Determine the magnitude of the absolute maximum shear stress at the point. Problem E a b c  –680  220  –80  17,000 ksi P13.51 0.18

FIGURE P13.51

Solution (a) Write three normal strain transformation equations [Eq. (13.3)], one for each strain gage, where n is the measured normal strain. In each equation, the angle  associated with each strain gage will be referenced from the positive x axis. (a) −680 με =  x cos2 (300) +  y sin 2 (300) +  xy sin(300)cos(300)

220 με =  x cos2 (0) +  y sin 2 (0) +  xy sin(0)cos(0)

(b)

−80 με =  x cos (60) +  y sin (60) +  xy sin(60)cos(60) From Eq. (b):  x = 220 με

Solve Eqs. (a) and (c) simultaneously to obtain:  y = −580 με and  xy = 692.8203 μrad = 693 μrad

Ans.

(b) Using these results, the principal strain magnitudes can be computed from Eq. (13.10):

 p1, p 2 =

x +  y 2

  x −  y    xy    +  2   2  2

(220 μ) + ( −580 μ)  (220 μ) − ( −580 μ)   692.8203 μ  =    +    2 2 2 2

= −180 μ  529.1503 μ

 p1 = 349.1503 με = 349 με and  p 2 = −709.1503 με = −709 με

Ans.

 max = 1,058.3005 μrad = 1,058 μrad

Ans.

tan 2 p =

(maximum in-plane shear strain)

 xy 692.8203 μ 692.8203 μ = = = 0.8660 ( x −  y ) [(220 μ) − ( −580 μ)] 800 μ

 p = 20.45

(counterclockwise from the x axis to the direction of  p1 )

Ans.

(c) Use p1 and p2 in Eqs. (13.23) to compute p1: E 17,000 ksi  p1 = ( p1 +  p 2 ) = [(349.1503  10−6 ) + (0.18)( −709.1503  10−6 )] 2 1− 1 − (0.18)2

= 3.8916 ksi = 3.89 ksi (T)

Ans.

and p2:

 p2 =

E 17,000 ksi ( p 2 +  p1 ) = [( −709.1503  10−6 ) + (0.18)(349.1503  10−6 )] 2 1− 1 − (0.18)2

= −11.3551 ksi = 11.36 ksi (C) The maximum in-plane shear stress can be computed from the two principal stresses:  p1 −  p 2 (3.8916 ksi) − (−11.3551 ksi)  max = = = 7.6234 ksi = 7.62 ksi 2 2 and the normal stress on the plane of maximum shear stress is  +  p 2 (3.8916 ksi) + ( −11.3551 ksi)  avg = p1 = = −3.7317 ksi = 3.73 ksi (C) 2 2

Ans.

Note: For completeness, the stresses x, y, and xy would be calculated as: E 17, 000 ksi  220 10−6 ) + ( 0.18 ) ( −580 10 −6 )  = 2.0310 ksi x = ( x + y ) = 2 ( 2  1 − 1 − ( 0.18 )

y =

E 17, 000 ksi  −580 10−6 ) + ( 0.18 ) ( 220 10 −6 )  = −9.4944 ksi ( y + x ) = 2 ( 2  1 − 1 − ( 0.18 )

E 17, 000 ksi = = 7, 203.3898 ksi 2 (1 + ) 2 (1 + 0.18 )

 xy = G xy = ( 7, 203.3898 ksi ) ( 692.9203 10−6 rad ) = 4.9907 ksi

(d) For plane stress, z = p3 = 0. Since p1 and p2 are of opposite signs,  abs max =  max = 7.62 ksi

Ans.

a

b

c

55 

–110 

–35 

E 212 GPa

FIGURE P13.52

 0.30

Solution (a) Write three normal strain transformation equations [Eq. (13.3)], one for each strain gage, where n is the measured normal strain. In each equation, the angle  associated with each strain gage will be referenced from the positive x axis. (a) 55 με =  x cos2 (315) +  y sin 2 (315) +  xy sin(315)cos(315)

−110 με =  x cos2 (0) +  y sin 2 (0) +  xy sin(0)cos(0)

(b)

−35 με =  x cos (45) +  y sin (45) +  xy sin(45)cos(45) From Eq. (b):  x = −110 με

(c)

Using this result, solve Eqs. (a) and (c) simultaneously to obtain:  y = 130 με and  xy = −90 μrad (b) Using these results, the principal strain magnitudes can be computed from Eq. (13.10):

 p1, p 2 =

x +  y 2

  x −  y    xy    +  2   2  2

( −110 μ) + (130 μ)  ( −110 μ) − (130 μ)   −90 μ     +    2 2 2  2

= 10 μ  128.1601 μ

 p1 = 138.1601 με = 138.2 με and  p 2 = −118.1601 με = −118.2 με

Ans.

 max = 256.3201 μrad = 256 μrad

Ans.

tan 2 p =

(maximum in-plane shear strain)

 xy −90 μ −90 μ = = = 0.3750 ( x −  y ) [( −110 μ) − (130 μ)] −240 μ

 p = 10.28

(counterclockwise from the x axis to the direction of  p 2 )

Ans.

(c) Use p1 and p2 in Eqs. (13.23) to compute p1: E 212,000 MPa  p1 = ( p1 +  p 2 ) = [(138.1601  10−6 ) + (0.3)( −118.1601  10−6 )] 2 1− 1 − (0.3)2

= 23.9285 MPa = 23.9 MPa (T)

Ans.

and p2:

 p2 =

E 212,000 MPa ( p 2 +  p1 ) = [( −118.1601  10−6 ) + (0.3)(138.1601  10−6 )] 2 1− 1 − (0.3)2

= −17.8714 MPa = 17.87 MPa (C) The maximum in-plane shear stress can be computed from the two principal stresses:  p1 −  p 2 (23.9285 MPa) − ( −17.8714 MPa)  max = = = 20.8999 MPa = 20.9 MPa 2 2 and the normal stress on the plane of maximum shear stress is  +  p 2 (23.9285 MPa) + ( −17.8714 MPa)  avg = p1 = = 3.0286 MPa = 3.03 MPa (T) 2 2

(d) For plane stress, z = p3 = 0. Since p1 and p2 are of opposite signs,  abs max =  max = 20.9 MPa

Ans.

P13.53 A block of 2014-T4 aluminum [E = 73 GPa;  = 0.33] has a width a = 640 mm, a height b = 200 mm, and a thickness t = 160 mm. The block is constrained between two rigid, perfectly smooth surfaces as shown in Figure P13.53/54/55. The block is compressed by a normal stress x = 210 MPa. Assuming plane stress, determine (a) the average normal stress in the y direction. (b) the change in the width a of the block. (c) the change in the thickness t of the block. FIGURE P13.53/54/55

Solution (a) Average normal stress in the y direction: The block is subjected to stress only in the x–y plane. Since the block is constrained by rigid surfaces in the y direction, we know that y = 0. From the problem statement, we know that the normal stress in x direction compresses the block. From this, we can infer that x = –210 MPa. Use Eqs. (13.24) to solve for y: 1  y = ( y − x ) E 1 0 =  y − ( 0.33)( −210 MPa )  E Ans.  y = ( 0.33)( −210 MPa ) = −69.3 MPa = 69.3 MPa (C) (b) Change in the width a of the block: Now that x and y are known, calculate x: 1  x = ( x − y ) E 1 ( −210 MPa ) − ( 0.33)( −69.3 MPa )  = −2,563.4384 10−6 mm/mm = 73, 000 MPa  The change in width a is thus  x =  x a = ( −2,563.4384 10−6 mm/mm ) ( 640 mm ) = −1.641 mm

Ans.

z = −



( +  ) x

0.33 ( −210 MPa ) + ( −69.3 MPa )  = 1, 262.5890 10−6 mm/mm 73, 000 MPa The change in thickness t is thus  z =  z t = (1, 262.5890 10−6 mm/mm ) (160 mm ) = 0.202 mm =−

Ans.

P13.54 A plate of stainless steel [E = 28,000 ksi;  = 0.12] has a width a = 18 in. and a height b = 9 in. The plate is constrained between two rigid, perfectly smooth surfaces as shown in Figure P13.53/54/55. After a normal stress x was applied, the width a of the plate was found to have decreased by 0.10 in. Assuming plane stress, determine the average normal stresses acting on the plate in the x and y directions. FIGURE P13.53/54/55

Solution The block is subjected to stress only in the x–y plane. Since the block is constrained by rigid surfaces in the y direction, we know that y = 0. From the deformation specified for the width a, we can calculate x. Then, x and y can be calculated from Eqs. (13.24):  −0.10 in. x = x = = −5,555.5556 10−6 in./in. a 18 in. E ( x + y ) 1 − 2 28, 000 ksi  −5,555.5556 10−6 in./in.) + ( 0.12 )( 0 )  = 2 (  1 − 0.12

x =

(

)

= −157.8283 ksi = 157.8 ksi (C)

Ans.

E ( y + x ) 1 − 2 28, 000 ksi  0 + ( 0.12 ) ( −5,555.5556  10−6 in./in.)  = 2 ( )  1 − 0.12

y =

(

)

= −18.9394 ksi = 18.94 ksi (C)

Ans.

P13.55 A plate of ductile cast iron [E = 168 GPa;  = 0.32;  = 10.8×10–6/°C] has a width a = 420 mm, a height b = 250 mm, and a thickness t = 30 mm. As shown in Figure P13.53/54/55, the plate is constrained between two rigid, perfectly smooth surfaces at an ambient temperature of 20°C. The plate is compressed by a constant normal stress x = 300 MPa. Assume that plane stress conditions exist. (a) Determine the normal strains x and z and the average normal stress y at the ambient temperature. (b) Determine the normal strains x and z and the average normal stress y at a temperature of 150°C. (c) At what temperature will the average normal stress in the y direction be reduced to zero? FIGURE P13.53/54/55

Solution (a) Normal strains x and z and average normal stress y at the ambient temperature: The block is subjected to stress only in the x–y plane. Since the block is constrained by rigid surfaces in the y direction, we know that y = 0. From the problem statement, we know that the normal stress in x direction compresses the block. From this, we can infer that x = –300 MPa. Use Eqs. (13.24) to solve for y: 1  y = ( y − x ) E 1 0 =  y − ( 0.32 )( −300 MPa )  E Ans.  y = ( 0.32 )( −300 MPa ) = −96 MPa = 96 MPa (C) Now that x and y are known, calculate x: 1  x = ( x − y ) E 1 = ( −300 MPa ) − ( 0.32 )( −96 MPa )  168, 000 MPa 

= −1, 602.857110−6 mm/mm = −1, 603 με

Ans.

and z:

z = − =−

 E

( +  ) x

0.32 ( −300 MPa ) + ( −96 MPa )  168, 000 MPa 

= 754.2857 10−6 mm/mm = 754 με

Ans.

(b) Normal strains x and z and average normal stress y at a temperature of 150°C: The change in temperature is T = 150°C – 20°C =130°C. Using Eqs. (13.30), solve for y, knowing that x = –300 MPa and y = 0: Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

1 ( y − x ) + T E 1 0 =  y − ( 0.32 )( −300 MPa )  + (10.8 10−6 /°C ) (130C ) E  y = − (168, 000 MPa ) (10.8 10−6 /°C ) (130C ) + ( 0.32 )( −300 MPa )

y =

 y = −331.8720 MPa = 332 MPa (C)

Ans.

Now that x and y are known, calculate x: 1  x = ( x − y ) + T E 1 = ( −300 MPa ) − ( 0.32 )( −331.8720 MPa )  + (10.8 10−6 /°C ) (130C ) 168, 000 MPa

= 250.4429 10−6 mm/mm = 250 με

Ans.

and z:

z = − =−

 E

( +  ) + T x

0.32 ( −300 MPa ) + ( −331.8720 MPa )  + (10.8 10−6 /°C ) (130C ) 168, 000 MPa 

= 2, 607.5657 10−6 mm/mm = 2, 610 με

Ans.

(c) At what temperature will the average normal stress in the y direction be reduced to zero? Use Eqs. (13.30) with x = –300 MPa, y = 0 MPa, and y = 0. Solve for T: 1  y = ( y − x ) + T E 1 0 = ( 0 ) − ( 0.32 )( −300 MPa )  + (10.8 10 −6 /°C ) T E ( 0.32 )( −300 MPa ) T = = −52.91C (10.8 10−6 /°C ) (168, 000 MPa ) Tfinal = Tinitial + T = 20C + ( −52.9101°C ) = −32.9101°C = −32.9C

Ans.

P13.56 A thin aluminum alloy [E = 69 GPa;  = 0.33;  = 23.6×10–6/°C] plate with dimensions a = 1,700 mm and b = 1,000 mm is set in a rigid frictionless cavity as shown in Figure P13.56. At room temperature, there is a gap of c = 2 mm between the rigid cavity and three sides of the plate as shown in the figure. The plate is not constrained in the z direction. After the temperature of the plate has been raised by 140°C, what are the normal stresses x and y in the plate? FIGURE P13.56

Solution If the plate were totally free to expand, the normal strain in the plate would be equal to the thermal strain: T = T = ( 23.6 10−6 /°C) (140°C) = 3,304.0 10−6 mm/mm The plate can expand, but only by an amount equal to the gaps. From the size of the plate and the width of the gaps, we can determine the normal strains in the x and y directions that are possible:  2c 2 ( 2 mm ) (a) x = x = = = 2,352.9412 10−6 mm/mm a a 1,700 mm

y

c 2 mm (b) = = 2,000 10−6 mm/mm b b 1,000 mm Since both of these strains are less than the thermal strain, we know that the plate will indeed contact the rigid frictionless cavity walls. When the plate contacts the cavity walls, stresses will be created. These stresses will act in the x–y plane only.

y =

Substitute the normal strains computed in Eqs. (a) and (b) into Eqs. (13.31) to calculate the normal stresses x and y that act in the plate: E E x =  + y ) −  T 2 ( x 1 − 1 − 69, 000 MPa  2,352.9412 10−6 ) + ( 0.33) ( 2, 000 10−6 )  = 2 (  1 − 0.33

(

)

−

69, 000 MPa 23.6 10−6 / C ) (140C ) ( 1 − 0.33

= −106.9635 MPa = 107.0 MPa (C)

Ans.

E E  + x ) −  T 2 ( y 1 − 1 − 69, 000 MPa  2, 000 10−6 ) + ( 0.33) ( 2,352.9412 10−6 )  = 2 (  1 − 0.33

y =

(

)

−

69, 000 MPa 23.6 10−6 /C ) (140C ) ( 1 − 0.33

= −125.2739 MPa = 125.3 MPa (C)

Ans.

P13.57 At a point on the free surface of a ductile cast iron [E = 168 GPa;  = 0.32;  = 10.8×10–6/°C] machine part, the measured strains resulting from both a temperature decrease of 65°C and externally applied loads are x = −370 , y = −715 , and xy = 0. Determine the stresses x and y at the point.

Solution Since we are told that the strains act on a free surface, we know that the point is subjected to plane stress. Substitute the specified normal strains along with the temperature change into Eqs. (13.31) to calculate the normal stresses x and y that act at the point: E E x =  + y ) −  T 2 ( x 1 − 1 − 168, 000 MPa ( −370 10−6 ) + ( 0.32 ) ( −715  10−6 )  = 2   1 − 0.32

(

)

−

168, 000 MPa 10.8 10−6 / C ) ( −65C ) ( 1 − 0.32

= 61.3604 MPa = 61.4 MPa (T)

Ans.

E E  + x ) −  T 2 ( y 1 − 1 − 168, 000 MPa ( −715 10−6 ) + ( 0.32 ) ( −370  10−6 )  = 2   1 − 0.32

y =

(

)

−

168, 000 MPa (10.8 10−6 /°C ) ( −65C ) 1 − 0.32

= 17.4513 MPa = 17.45 MPa (T)

Ans.

P13.58 A thin plate (z = 0) of an aluminum alloy [E = 10,000 ksi;  = 0.33;  = 13.1×10–6/°F] is stretched until the strains in the x and y directions are 0.0010 in./in. and 0.0015 in./in., respectively. Then, the plate is rigidly held in the deformed position and heated until its temperature has increased by 80°F. Determine the final stresses in the plate.

Solution Use the specified strains and the specified temperature increase in Eqs. (13.31) to compute the final stresses: E E x =  + y ) − T 2 ( x 1 − 1 − 10, 000 ksi 10, 000 ksi = 13.110−6 /°C ) ( 80F )  0.0010 in./in.) + ( 0.33)( 0.0015 in./in.)  − ( 2 ( 1 − 0.33 1 − ( 0.33) = 1.1352 ksi = 1.135 ksi (T)

Ans.

E E  + x ) − T 2 ( y 1 − 1 − 10, 000 ksi 10, 000 ksi = 13.110−6 /°C ) ( 80F )  0.0015 in./in.) + ( 0.33)( 0.0010 in./in.)  − ( 2 ( 1 − 0.33 1 − ( 0.33)

y =

= 4.8946 ksi = 4.89 ksi (T)

Ans.

P13.59 A thin plate (z = 0) of an aluminum alloy [E = 10,000 ksi;  = 0.33;  = 13.1×10–6/°F] laying in the x–y plane is heated from an ambient temperature of 65°F to a final temperature of 350°F. Then, the plate is rigidly clamped in position so that it is restrained in both the x and y directions. Determine the absolute maximum shear stress in the plate when its temperature cools down to the ambient temperature.

Solution Initially, the plate is totally free to expand; thus, the normal strain in the plate is equal to the thermal strain: T = T = (13.110−6 /°F) ( 350°F − 65°F) = 3,733.5 10−6 in./in. After the free expansion, the plate is clamped in position. Thus, we know that the normal strains in the x and y directions are equal to the thermal strain. When the plate cools down to the ambient temperature, the normal strains in the x and y directions remain equal to the thermal strain. Use the specified strains and the specified temperature decrease in Eqs. (13.26) to compute the final stresses: E x = ( x + y ) 1 − 2 10, 000 ksi  3, 733.5 10−6 in./in.) + ( 0.33) ( 3, 733.5 10 −6 in./in.)  = 2 (  1 − 0.33

(

)

= 55.7239 ksi The normal stress in the y direction is exactly the same value as the normal stress in the x direction:  y = 55.7239 ksi There are no shear stresses in the plane of the plate, meaning that xy = 0. From this observation, we can conclude that x and y are principal stresses. Furthermore, the plate is in a state of plane stress—there are no stresses acting in the z direction. Consequently, z = p3 = 0. The absolute maximum shear stress in the plate is thus:  −  min 55.7239 ksi − 0  abs max = max = = 27.9 ksi 2 2

Ans.

P13.60 The principal stresses at a point are x = 92 MPa, y = 78 MPa, and z = 66 MPa, acting as shown in Figure P13.60. The material is red brass, for which E = 115 GPa and  = 0.307. Determine (a) the principal strains. (b) the dilatation.

FIGURE P13.60

Solution (a) Principal strains: Using the specified stresses, calculate the normal strains in the x, y, and z directions from Eqs. (13.16): 1  x = [ x − ( y +  z )] E 1 = ( 92 MPa ) − ( 0.307 ) ( −78 MPa ) + ( −66 MPa )  115, 000 MPa





= 1,184.4174 10 −6 mm/mm 1  y = [ y − ( x +  z )] E 1 = ( −78 MPa ) − ( 0.307 ) ( 92 MPa ) + ( −66 MPa ) 115, 000 MPa





= −747.6696 10−6 mm/mm 1  z = [ z − ( x +  y )] E 1 = ( −66 MPa ) − ( 0.307 ) ( 92 MPa ) + ( −78 MPa ) 115, 000 MPa





= −611.2870 10−6 mm/mm

We are told in the problem statement that the x, y, and z normal stresses are principal stresses. Therefore, the strains in the x, y, and z directions must also be principal strains. Therefore:  p1 = 1,184 με  p 2 = −611 με  p3 = −748 με

Ans.

(b) The dilatation: The dilatation is computed from Eq. (13.21): e = x +y +z = 1,184.4174 10−6 − 747.6696 10 −6 − 611.2870 10 −6 = −174.539110−6 = −174.5 10−6

Ans.

P13.61 The titanium [E = 16,500 ksi;  = 0.33] block shown in Figure P13.61/62/63/64 has dimensions a = 6 in., b = 4 in., and c = 2 in. The block is subjected to triaxial stresses x = −45 ksi, y = −25 ksi, and z = 15 ksi, acting on the x, y, and z faces, respectively. Determine (a) the changes a, b, and c in the dimensions of the block and (b) the change V in the volume of the block. FIGURE P13.61/62/63/64

Solution (a) Determine the change in length of the block dimensions a, b, and c: From Eqs. (13.16), calculate the normal strains: 1  x =  x − ( y +  z )  E 1 = ( −45 ksi ) − ( 0.33) ( −25 ksi ) + (15 ksi ) = −2,527.273 10−6 in./in. 16,500 ksi 1  y =  y − ( x +  z )  E 1 = ( −25 ksi ) − ( 0.33) ( −45 ksi ) + (15 ksi ) = −915.152 10−6 in./in. 16,500 ksi 1  z =  z − ( x +  y )  E 1 = (15 ksi ) − ( 0.33) ( −45 ksi ) + ( −25 ksi ) = 2,309.09110−6 in./in. 16,500 ksi













The changes in the dimensions of the block are thus: a =  x a = ( −2,527.273 10−6 in./in.) ( 6 in.) = −0.01516 in. b =  y b = ( −915.152 10−6 in./in.) ( 4 in.) = −0.00366 in.

c =  z c = ( 2,309.09110−6 in./in.) ( 2 in.) = 0.00462 in.

Ans. Ans. Ans.

(b) Change V in the volume of the block: Calculate the dilatation of the block: e = x +y +z = ( −2,527.273 10−6 ) + ( −915.152 10−6 ) + ( 2,309.09110−6 ) = −1.133333 10−3 The change in volume is computed from the dilatation as: V =e V

V = Ve = ( abc ) e = ( 6 in.)( 4 in.)( 2 in.)  ( −1.133333  10−6 ) = −0.0544 in.3

Ans.

P13.62 The malleable cast iron [E = 26,000 ksi;  = 0.27] block shown in Figure P13.61/62/63/64 has dimensions a = 9 in., b = 6 in., and c = 3 in. The block is subjected to normal stresses y = −19 ksi and z = −12 ksi; however, the block is constrained at its ends against displacement in the x direction. What normal stress develops in the x direction, and what are the strains in the y and z directions? FIGURE P13.61/62/63/64

Solution Normal stress in the x direction: The normal strain in the x direction is x = 0. Thus, Eqs. (13.16), can be expressed as: 1  x =  x − ( y +  z )  = 0 E From this equation, calculate the normal stress in the x direction:  x =  ( y +  z )

= ( 0.27 ) ( −19 ksi ) + ( −12 ksi )  = −8.37 ksi

Ans.

Normal strains in the y and z directions: Now that all normal stresses are known, calculate the strains in the y and z directions from Eq. (13.16): 1  y = [ y − ( x +  z )] E 1 = ( −19 ksi ) − ( 0.27 ) ( −8.37 ksi ) + ( −12 ksi ) 26, 000 ksi





= −519.2345 10−6 in./in. = −519 με

Ans.

1 [ z − ( x +  y )] E 1 = ( −12 ksi ) − ( 0.27 ) ( −8.37 ksi ) + ( −19 ksi ) 26, 000 ksi

z =



= −177.3115 10−6 in./in. = −177.3 με

 Ans.

P13.63 The polymer [E = 3.7 GPa;  = 0.33;  = 85×10–6/°C] block shown in Figure P13.61/62/63/64 has dimensions a = 400 mm, b = 250 mm, and c = 50 mm. The block is subjected to a uniform temperature increase of 45°C. The block is completely free to expand in the y and z directions but is constrained at its ends against displacement in the x direction. Determine (a) the changes b and c in the dimensions of the block and (b) the normal stress in the x direction. FIGURE P13.61/62/63/64

Solution Since the block is completely free to expand in the y and z directions, we can conclude that the normal stresses in the y and z directions are zero: y = z = 0. We are told that the block is constrained in the x direction; thus, x = 0. Use these facts in Eqs. (13.28) to compute x: 1  x =  x − ( y +  z ) +  T E

1  x − ( 0 )  +  T E  x = − E T = − ( 3, 700 MPa ) ( 85 10−6 /°C ) ( 45°C ) = −14.1525 MPa = 14.15 MPa (C) 0=

Ans.

Changes in the block dimensions: Calculate the strains in the y and z directions from Eqs. (13.28): 1  y =  y − ( x +  z )  +  T E 1 = ( 0 ) − ( 0.33) ( −14.1525 MPa ) + 0 + (85 10−6 /°C ) ( 45°C ) 3, 700 MPa





= 5, 087.25 10−6 mm/mm 1  z =  z − ( x +  y )  +  T E 1 = ( 0 ) − ( 0.33) ( −14.1525 MPa ) + 0 + (85 10−6 /°C ) ( 45°C ) 3, 700 MPa





= 5, 087.25 10−6 mm/mm

The changes in dimensions of the block are: b =  y b = ( 5,087.25 10−6 mm/mm ) ( 250 mm ) = 1.272 mm

c =  z c = ( 5,087.25 10−6 mm/mm ) ( 50 mm ) = 0.254 mm

Ans. Ans.

P13.64 The polymer [E = 3.7 GPa;  = 0.33;  = 85×10–6/°C] block shown in Figure P13.61/62/63/64 has dimensions a = 400 mm, b = 250 mm, and c = 50 mm. The block is subjected to a uniform temperature increase of 45°C. The block is completely free to expand in the z direction but is constrained at its ends against displacement in the x and y directions. Determine (a) the normal stresses in the x and y directions and (b) the change in the length of the block in the z direction. FIGURE P13.61/62/63/64

Solution (a) Normal stresses in the x and y directions: Since the block is completely free to expand in the z direction, we can conclude that the normal stresses in the z direction is zero: z = 0. We are told that the block is constrained in the x and y directions; thus, x = y =0. Since x =0 and y =0, Eqs. (13.28) simplify to: 1  x =  x − ( y +  z )  +  T E 1  0 =  x − y  +  T (a) E and: 1  y =  y − ( x +  z )  +  T E 1  0 =  y − x  +  T (b) E Solve Eqs. (a) and (b) simultaneously to find: 1 1  x − y  +  T =  y − x  +  T E E  x − y =  y − x

 x (1 + ) =  y (1 + )  x =  y

(c)

Backsubstitute this result into Eq. (a) to calculate x: 1 0 =  x − x  +  T E  x (1 − ) = − E T

( 3, 700 MPa ) (85 10−6 /°C ) ( 45°C ) E T x = − =− = −21.1231 MPa = 21.1 MPa (C) 1 − 1 − 0.33

and from Eq. (c):  y = −21.1231 MPa = 21.1 MPa (C)

Ans.

(b) Change in length of the block in the z direction: Calculate the normal strain in the z direction from Eqs. (13.28): 1  z =  z − ( x +  y )  +  T E 1 = ( 0 ) − ( 0.33) ( −21.1231 MPa ) + ( −21.1231 MPa )  + ( 85 10−6 /°C ) ( 45°C ) 3, 700 MPa





= 7,592.9104 10−6 mm/mm

The change in length of the block in the z direction is: c =  z c = ( 7,592.9104 10−6 mm/mm ) ( 50 mm ) = 0.380 mm

Ans.

P13.65 A solid plastic [E = 45 MPa;  = 0.33] rod with a diameter d = 100 mm is placed in a D = 101 mm diameter hole with rigid walls as shown in Figure P13.65. The rod has a length L = 400 mm. Determine the change in length of the rod after a load P = 32 kN is applied.

FIGURE P13.65

Solution Denote the longitudinal axis of the rod as the y axis. The radial directions of the rod will be the x and z axes. By symmetry, the normal stresses and strains in the x and z directions will be equal. The cross-sectional area of the 100 mm diameter rod is:



(100 mm ) = 7,853.9816 mm2 2

4 and so, the normal stress in the longitudinal direction (i.e., the y direction) is: P −32, 000 N y = = = −4.0744 MPa A 7,853.9816 mm 2 We will assume that the compressive force acting on the rod will cause the rod to expand enough so that it contacts the rigid sides of the hole. If this assumption is true, then the deformation in the x and z directions will be equal to the gap between the rod and the hole:  x =  z = D − d = 101 mm − 100 mm = 1 mm and thus, the normal strains in the x and z directions are:  1 mm x = x = = 0.01 mm/mm d 100 mm  z =  x = 0.01 mm/mm From Eqs. (13.16) and knowing that x = z by virtue of symmetry, calculate the stress in the x direction: 1 1  x =  x − ( y +  z )  =  x − ( y +  x )  E E  (1 − )  y 1  x =  x − y − x  = x − = 0.01 mm/mm E E E  y  E  x = 0.01 mm/mm +   1 −  E 

x =

( 0.33)( −4.0744 MPa )  = −1.3351 MPa 45 MPa  0.01 mm/mm +  1 − 0.33  45 MPa 

Since x is negative, our original assumption that the loaded rod contacts the rigid walls is confirmed.

All three normal stresses are now known, and thus, we can compute the normal strain in the y direction (i.e., the longitudinal axis of the rod) from Eqs. (13.16): 1  y =  y − ( x +  z )  E 1 = ( −4.0744 MPa ) − ( 0.33) ( −1.3351 MPa ) + ( −1.3351 MPa ) 45 MPa = −70,959.4876 10−6 mm/mm





The change in length of the rod after a load P = 32 kN is applied is thus: y = yL

= ( −70,959.4876 10−6 mm/mm ) ( 400 mm ) = −28.3838 mm = −28.4 mm

Ans.

P13.66 A plate of epoxy reinforced with unidirectional Kevlar 49 fibers is loaded with stresses x = 350 MPa, y = 17 MPa, and xy = 12 MPa. The fibers are aligned in the x direction as shown in Figure P13.66/67/68. Determine the inplane normal and shear strains for this loading.

FIGURE P13.66/67/68

Solution From Table 13.3, the material properties for unidirectional Kevlar 49 fibers are: Ex = 76 GPa Ey = 5.5 GPa Gxy = 2.1 GPa  xy = 0.34 The minor Poisson’s ratio yx is calculated from Eq. (13.34) as E 5.5 GPa  yx =  xy y = ( 0.34 ) = 0.0246 Ex 76 GPa From Eqs. (13.32), the in-plane normal strains can be calculated as:   350 MPa 17 MPa  x = x − yx y = − ( 0.0246 ) = 4,529.2 10−6 mm/mm = 4,530 με Ex E y 76, 000 MPa 5,500 MPa

y =

y Ey

− xy

x Ex

17 MPa 350 MPa − ( 0.34 ) = 1,525.110−6 mm/mm = 1,525 με 5,500 MPa 76, 000 MPa

From Eqs. (13.32), the in-plane shear strain is:  12 MPa  xy = xy = = 5, 714.3 10−6 rad = 5, 710 μrad Gxy 2,100 MPa

Ans. Ans.

Ans.

P13.67 A plate of epoxy reinforced with unidirectional T-300 fibers is loaded with stresses x = 40 ksi, y = −10 ksi, and xy = 3 ksi. The fibers are aligned in the x direction as shown in Figure P13.66/67/68. Determine the in-plane normal and shear strains for this loading.

FIGURE P13.66/67/68

Solution From Table 13.3, the material properties for unidirectional T-300 fibers are: Ex = 19, 200 ksi Ey = 1,500 ksi Gxy = 950 ksi  xy = 0.25 The minor Poisson’s ratio yx is calculated from Eq. (13.34) as E 1,500 ksi  yx =  xy y = ( 0.25) = 0.0195 Ex 19, 200 ksi From Eqs. (13.32), the in-plane normal strains can be calculated as:   40 ksi −10 ksi  x = x − yx y = − ( 0.0195) = 2, 213.5 10−6 in./in. = 2, 210 με Ex E y 19, 200 ksi 1, 500 ksi

y =

y Ey

− xy

x Ex

−10 ksi 40 ksi − ( 0.25) = −7,187.5 10−6 in./in. = −7,190 με 1,500 ksi 19, 200 ksi

From Eqs. (13.32), the in-plane shear strain is:  3 ksi  xy = xy = = 3,157.9 10−6 rad = 3,160 μrad Gxy 950 ksi

Ans. Ans.

Ans.

P13.68 A plate of epoxy reinforced with unidirectional T-300 fibers is subjected to strains x = 1,380 , y = 5,250 , and xy = 1,750 rad. The fibers are aligned in the x direction as shown in Figure P13.66/67/68. Determine the in-plane normal and shear stresses (in ksi) in the composite plate.

FIGURE P13.66/67/68

Solution From Table 13.3, the material properties for unidirectional T-300 fibers are: Ex = 19, 200 ksi Ey = 1,500 ksi Gxy = 950 ksi  xy = 0.25 The minor Poisson’s ratio yx is calculated from Eq. (13.34) as E 1,500 ksi  yx =  xy y = ( 0.25) = 0.0195 Ex 19, 200 ksi From Eqs. (13.33), the in-plane normal stresses can be calculated as: Ex x = ( x + yx y ) 1 − xy yx

19, 200 ksi (1,380 10−6 ) + ( 0.0195) ( 5, 250 10−6 )   1 − ( 0.25 )( 0.0195) 

= 28.5199 ksi = 28.5 ksi (T)

y = =

Ey 1 − xy yx

Ans.

(  +  ) y

xy x

19, 200 ksi ( 5, 250 10−6 ) + ( 0.25 ) (1,380  10−6 )   1 − ( 0.25 )( 0.0195 ) 

= 8.3993 ksi = 8.40 ksi (T)

From Eqs. (13.33), the in-plane shear strain is:  xy = Gxy xy = ( 950 ksi ) (1,750 10−6 rad ) = 1.6625 ksi = 1.663 ksi

Ans.

P15.1 A solid 1.50 in. diameter shaft is subjected to a torque of T = 225 lb·ft and an axial load of P = 5,500 lb, as shown in Figure P15.1/2. (a) Determine the principal stresses and the maximum shear stress at point H on the surface of the shaft. (b) Show the stresses of part (a) and their directions on an appropriate sketch.

FIGURE P15.1/2

Solution Section properties:

 (1.5 in.) 2 = 1.76715 in.2 4  J = (1.5 in.)4 = 0.49701 in.4 32 A=

Normal and shear stress magnitudes: P 5,500 lb = = = 3.112.363 ksi (C) A 1.76715 in.2 Tc (225 lb  ft)(1.5 in. / 2)(12 in./ft) = = = 4,074.367 psi (sense determined by inspection) J 0.49701 in.4

Principal stress calculations: The principal stress magnitudes can be computed from Eq. (12.12):

 p1, p 2 =

x + y 2

x −y    +  xy2  2   2

( −3,112.363) + (0)  ( −3,112.363) − (0)  =   + ( −4,074.367) 2    2 2 2

= −1,556.182  4,361.441

 p1 = 2,810 psi and  p 2 = −5,920 psi

Ans.

 max = 4,360 psi

Ans.

(maximum in-plane shear stress)

 avg = 1,556 psi (C) tan 2 p =

2 xy

x − y

(normal stress on planes of maximum in-plane shear stress)

2 ( −4,074.367 )

( −3,112.363) − ( 0 )

 p = 34.55

Ans.

−8,148.734 = 2.6182 −3,112.363

(counterclockwise from the x axis to the direction of  p 2 )

Ans.

Principal Stresses and Maximum In-plane Shear Stress

P15.2 A solid 19 mm diameter aluminum alloy [E = 70 GPa;  = 0.33] shaft is subjected to a torque of T = 60 N·m and an axial load of P = 15 kN, as shown in Figure P15.1/2. At point H on the outer surface of the shaft, determine: (a) the strains x, y , and xy. (b) the principal strains p1 and p2. (c) the absolute maximum shear strain. FIGURE P15.1/2

Solution Section properties:

 (19 mm) 2 = 283.52874 mm 2 4  J = (19 mm)4 = 12,794.23426 mm 4 32 A=

Normal and shear stress magnitudes: P −15,000 N = = = 52.905 MPa (C) A 283.52874 mm2 Tc (60 N  m)(19 mm / 2)(1,000 mm/m) = = = 44.551 MPa (sense determined by inspection) J 12,794.23426 mm 4

(a) Normal and shear strain magnitudes: From Eqs. (13.21), the normal strains in the x and y directions are 1 1  x = ( x −  y ) = [ −52.905 MPa − (0.33)(0 MPa)] E 70,000 MPa

= −755.781  10−6 mm/mm = −756 με

y =

Ans.

1 1 ( y −  x ) = [0 MPa − (0.33)(−52.905 MPa)] E 70,000 MPa

= 249.408  10−6 mm/mm = 249 με From Eq. (13.18), determine the shear modulus G: E 70, 000 MPa G= = = 26,315.8 MPa 2(1 + ) 2(1 + 0.33) and compute the shear strain xy from Eq. (13.22):  −44.551 MPa  xy = xy = = −1,692.950  10−6 rad = −1,693 μrad G 26,315.8 MPa

Ans.

(b) Principal Strains p1 and p2: Using these results, the principal strain magnitudes can be computed from Eq. (13.10):

 p1, p 2 =

x +  y 2

  x −  y    xy    +  2   2  2

( −755.781 μ) + (249.408 μ)  ( −755.781 μ) − (249.408 μ)   −1,692.950 μ  =    +    2 2 2 2

= −253.187 μ  984.440 μ

 p1 = 731.253 με = 731 με and  p 2 = −1,237.626 με = −1,238 με

Ans.

z = −

 E

( x +  y ) = −

0.33 [−52.905 MPa + 0 MPa] = 249.408  10−6 mm/mm 70,000 MPa

The third principal strain is p3 = z. Since p2 ≤ p3 ≤ p1, the absolute maximum shear strain is the same as the maximum in-plane shear strain.

 abs max = 731.253 μrad − (−1,237.626 μrad) = 1,969 μrad

(absolute maximum shear strain)

Ans.

P15.3 A hollow bronze [E = 15,200 ksi;  = 0.34] shaft with an outside diameter of 2.50 in. and a wall thickness of 0.125 in. is subjected to a torque of T = 720 lb·ft and an axial load of P = 1,900 lb, as shown in Figure P15.3/4. At point H on the outer surface of the shaft, determine: (a) the strains x, y , and xy. (b) the principal strains p1 and p2. (c) the absolute maximum shear strain.

FIGURE P15.3/4

Solution Section properties:

 (2.50 in.) 2 − (2.25 in.) 2  = 0.93266 in.2 4  J = (2.50 in.) 4 − (2.25 in.) 4  = 1.31884 in.4 32 A=

Normal and shear stress magnitudes: P 1,900 lb = = = 2,037.183 psi (T) A 0.93266 in.2 Tc (720 lb  ft)(2.50 in. / 2)(12 in./ft) = = = 8,189.015 psi (sense determined by inspection) J 1.31884 in.4

(a) Normal and shear strain magnitudes: From Eqs. (13.21), the normal strains in the x and y directions are 1 1  x = ( x −  y ) = [2,037.183 psi − (0.34)(0 psi)] E 15,200,000 psi

= 134.025  10−6 in./in. = 134.0 με

y =

Ans.

1 1 ( y −  x ) = [0 psi − (0.34)(2,037.183 psi)] E 15,200,000 psi

= −45.569  10−6 in./in. = −45.6 με From Eq. (13.18), determine the shear modulus G: E 15,200,000 psi G= = = 5,671,642 psi 2(1 +  ) 2(1 + 0.34) and compute the shear strain xy from Eq. (13.22):  8,189.015 psi  xy = xy = = 1,443.853  10−6 rad = 1,444 μrad G 5,671,642 psi

Ans.

(b) Principal Strains p1 and p2: Using these results, the principal strain magnitudes can be computed from Eq. (13.10):

 p1, p 2 =

x +  y 2

  x −  y    xy    +  2   2  2

(134.025 μ) + ( −45.569 μ)  (134.025 μ) − ( −45.569 μ)   1, 443.853 μ  =    +    2 2 2 2

= 44.228 μ  727.490 μ

 p1 = 771.718 με = 772 με and  p 2 = −683.261 με = −683 με

Ans.

z = −

 E

( x +  y ) = −

0.34 [2,037.183 psi + 0 psi] = −45.459  10−6 in./in. 15,200,000 psi

The third principal strain is p3 = z. Since p2 ≤ p3 ≤ p1, the absolute maximum shear strain is the same as the maximum in-plane shear strain.

 abs max = 771.718 μrad − (−583.261 μrad) = 1,455 μrad

(absolute maximum shear strain)

Ans.

P15.4 A hollow bronze shaft with an outside diameter of 80 mm and a wall thickness of 5 mm is subjected to a torque of T = 620 N·m and an axial load of P = 9,500 N, as shown in Figure P15.3/4. (a) Determine the principal stresses and the maximum shear stress at point H on the surface of the shaft. (b) Show the stresses of part (a) and their directions on an appropriate sketch.

FIGURE P15.3/4

Solution Section properties:

 (80 mm) 2 − (70 mm) 2  = 1,178.097 mm 2 4  J = (80 mm)4 − (70 mm) 4  = 1,664,062.359 mm4 32 A=

Normal and shear stress magnitudes: P 9,500 N = = = 8.064 MPa (T) A 1,178.097 mm2 Tc (620 N  m)(80 mm / 2)(1,000 mm/m) = = = 14.903 MPa J 1,664,062.359 mm 4 (sense of shear stress determined by inspection)

Principal stress calculations: The principal stress magnitudes can be computed from Eq. (12.12):

 p1, p 2 =

x + y 2

x −y    +  xy2  2   2

(8.064) + (0)  (8.064) − (0)  =   + (14.903)2    2 2 2

= 4.032  15.439

 p1 = 19.47 MPa and  p 2 = −11.41 MPa

Ans.

 max = 15.44 MPa

Ans.

 avg = 4.03 MPa (T)

(maximum in-plane shear stress) (normal stress on planes of maximum in-plane shear stress)

Ans.

tan 2 p =

2 xy

x − y

2 (14.903)

(8.064 ) − ( 0 )

 p = 37.43

29.806 = 3.6953 4.032

(counterclockwise from the x axis to the direction of  p1 )

Ans.

Principal Stresses and Maximum In-plane Shear Stress

P15.5 A solid 1.50 in. diameter shaft is used in an aircraft engine to transmit 160 hp at 2,800 rpm to a propeller that develops a thrust of 1,800 lb. Determine the magnitudes of the principal stresses and the maximum shear stress at any point on the outside surface of the shaft.

Solution Section properties:



(1.5 in.)2 = 1.767146 in.2

 32

(1.5 in.)4 = 0.497010 in.4

Normal and shear stress magnitudes: P 1,800 lb = = = 1,018.592 psi (T) A 1.767146 in.2 The torque in the propeller shaft is:  550 lb  ft/s  (160 hp )   P  1 hp  T= = = 300.121 lb  ft   2,800 rev  2 rad  1 min       min  1 rev  60 s 

Tc (300.121 lb  ft)(1.5 in. / 2)(12 in./ft) = = 5,434.676 psi J 0.497010 in.4 (sense of shear stress cannot be established definitively from the information given)

=

Principal stress calculations: The principal stress magnitudes can be computed from Eq. (12.12):

 p1, p 2 =

x + y 2

x −y    +  xy2  2   2

(1,018.592) + (0)  (1,018.592) − (0)  2    + (5, 434.676)  2 2 2

= 509.296  5.458.487

 p1 = 5,970 psi and  p 2 = −4,950 psi

Ans.

 max = 5,460 psi

Ans.

 avg = 509 psi (T)

(maximum in-plane shear stress) (normal stress on planes of maximum in-plane shear stress)

Ans.

P15.6 A short cast-iron shaft 2 in. in diameter is subjected to a tensile force of 22,000 lb combined with a torque T. Find the torque T which the shaft can resist (a) if the allowable shear stress is 6,000 psi and (b) if the allowable tensile stress is 10,000 psi.

Solution Section properties:

 (2 in.) 2 = 3.141593 in.2 4

 (2 in.) 4 = 1.570796 in.4 32

Normal and shear stress magnitudes: P 22, 000 lb = = = 7, 002.817 psi (T) A 3.141593 in.2 Tc T (2 in. / 2) = = J 1.570796 in.4 (sense of shear stress cannot be established definitively from the information given) (a) Maximum shear stress calculations: The maximum in-plane shear stress must not exceed 6,000 psi; therefore:

  − y  2  max =   x  +  xy  2  2

 7, 002.817 psi − 0  2 6, 000 psi     +  xy 2   Square both sides of the equation to obtain: 2 2  7, 002.817 psi  2 ( 6, 000 psi )    +  xy 2   and further reduce this equation to compute xy: 2

7, 002.817 psi  2 ( 6, 000 psi ) −   = 4,872.385 psi 2   Note that the axial and torsional loads create plane stress conditions and that the absolute maximum shear stress is equal to the in-plane maximum shear stress.  xy 

 p1, p 2 =

z + y 2

  − y  2   z  +  yz 2   2

7, 002.817 psi + 0 2  7, 002.817 psi − 0  =    + ( 4,872.385 psi ) 2 2   = 3,501.409  6000 psi  p1 = 9,501.408 psi and  p 2 = −2, 498.591 psi  −  min 9,501.408 − (−2, 498.591)  abs max = max = = 6, 000 psi 2 2  abs max =  max = 6, 000 psi The torque applied to the shaft must be limited to: T   J / c = (4,872.385 psi)(1.570796 in.4 ) / (2 in. / 2) = 7,650 lb  in. = 638 lb  ft 2

Ans.

(b) Principal stress calculations: The tensile principal stress must not exceed 10,000 psi; therefore:

 p1 =

x + y 2

  − y  2 +  x  +  xy  2  2

7, 002.817 psi + 0  7, 002.817 psi − 0  2 +   +  xy 2 2   Rearrange and then square both sides of the equation to obtain: 2 2 7, 002.817 psi   7, 002.817 psi   2 10, 000 psi −      +  xy 2 2     and further reduce this equation to compute xy: 2

10, 000 psi 

7, 002.817 psi   7, 002.817 psi    xy  10, 000 psi −  −  = 5, 474.653 psi 2 2     The torque applied to the shaft must be limited to: T   J / c = (5, 474.653 psi)(1.570796 in.4 ) / (2 in. / 2) = 8,600 lb  in. = 717 lb  ft

Ans.

P15.7 A compound shaft consists of two tube segments. Segment (1) has an outside diameter of 40 mm and a wall thickness of 3.2 mm. Segment (2) has an outside diameter of 70 mm and a wall thickness of 4.0 mm. The shaft is subjected to a tensile load of P = 15 kN and torques TA = 350 N·m and TB = 900 N·m, which act in the directions shown in Figure P15.7/8. (a) Determine the principal stresses and the maximum shear stress at point K on the surface of the shaft. (b) Show these stresses on an appropriate sketch.

FIGURE P15.7/8

Solution Equilibrium: M z = −TA + TB − T2 = 0

T2 = −350 N  m + 900 N  m = 550 N  m

Section properties:

A2 =



J2 =

 

70 mm ) − ( 62 mm )  = 829.3805 mm 2 ( 

70 mm ) − ( 62 mm )  = 906,512.8434 mm 4 (  4

Normal and shear stress magnitudes: F 15, 000 N 2 = 2 = = 18.0858 MPa (T) A2 829.3805 mm2

( 550 N  m )( 70 mm / 2 )(1,000 mm/m ) = 21.2352 MPa Tc J2 906,512.8434 mm4 (sense of shear stresses to be determined by inspection) 2 = 2 2 =

The stress element for point K is shown below.

Principal stress calculations for point K:

 p1, p 2 =

 z + y 2

  − y  2   z  +  yz  2  2

(18.0858) + ( 0 )   (18.0858 ) − ( 0 )  + 21.2352 2 = )   ( 2





= 9.0429  23.0805

 p1 = 32.1 MPa

and

 p 2 = −14.04 MPa

Ans.

 max = 23.1 MPa

(maximum in-plane shear stress)

Ans.

 avg = 9.04 MPa (T) tan 2 p =

2 yz

 z − y

(normal stress on planes of maximum in-plane shear stress)

Ans.

2 ( 21.2352 ) 42.4704 = = 2.3483 (18.0858) − ( 0 ) 18.0858

 p = 33.5

( clockwise from the z axis to the direction of  ) p1

Ans.

Principal Stresses and Maximum In-plane Shear Stress

P15.8 A compound shaft consists of two stainless steel tube segments. Segment (1) has an outside diameter of 1.5 in. and a wall thickness of 0.120 in. Segment (2) has an outside diameter of 3.0 in. and a wall thickness of 0.120 in. The shaft is subjected to a tensile load of P = 6,200 lb and torques TA = 420 lb·ft and TB = 1,060 lb·ft, which act in the directions shown in Figure P15.7/8. Determine the maximum compressive normal stress in the tube wall at (a) point H and (b) point K.

FIGURE P15.7/8

Solution Equilibrium: M z = −TA − T1 = 0

T1 = −420 lb  ft

M z = −TA + TB − T2 = 0

T2 = −420 lb  ft + 1,060 lb  ft = 640 lb  ft

Section properties:

A1 =



J1 =

 

1.50 in.) − (1.260 in.)  = 0.5202 in.2 ( 

1.50 in.) − (1.260 in.)  = 0.2496 in.4 (  4

A2 =



J2 =

 

3.0 in.) − ( 2.760 in.)  = 1.0857 in.2 ( 

3.0 in.) − ( 2.760 in.)  = 2.2553 in.4 (  4

Normal and shear stress magnitudes: F 6, 200 lb 1 = 1 = = 11,917.3996 psi (T) A1 0.5202 in.2 Tc J1

1 = 1 1 =

2 =

( 420 lb  ft )(1.50 in. / 2 )(12 in./ft ) = 15,146.4862 psi 0.2496 in.4

F2 6, 200 lb = = 5, 710.4207 psi (T) A2 1.0857 in.2

T2 c2 ( 640 lb  ft )( 3.0 in. / 2 )(12 in./ft ) = = 5,107.9963 psi J2 2.2553 in.4 (sense of shear stresses to be determined by inspection)

2 =

The stress elements for points H and K are shown below.

(a) Principal stress calculations for point H:

 p1, p 2 =

z + y 2

  z − y  2    +  yz  2  2

(11,917.3996 ) + ( 0 )   (11,917.3996 ) − ( 0 )  + −15,146.4862 2 = )   ( 2

 = 5,948.6998  16, 276.4292 2



Maximum compressive stress = p 2 = 10,320 psi (C)

Ans.

(b) Principal stress calculations for point K:

 p1, p 2 =

z + y 2

  z − y  2    +  yz  2  2

( 5, 710.4207 ) + ( 0 )   ( 5, 710.4207 ) − ( 0 )  + 5,107.9963 2 = )   ( 2

 = 2,855.2103  5,851.8247 2



Maximum compressive stress = p 2 = 3,000 psi (C)

Ans.

P15.9 A hollow shaft is subjected to an axial load P and a torque T, acting in the directions shown in Figure P15.9. The shaft is made of bronze [E = 105 GPa;  = 0.34], and it has an outside diameter of 55 mm and an inside diameter of 45 mm. A strain gage is mounted at an angle of  = 40° with respect to the longitudinal axis of the shaft, as shown in Figure P15.9. (a) If P = 13,000 N and T = 260 N·m, what is the strain reading that would be expected from the gage? (b) If the strain gage gives a reading of –195  when the axial load has a magnitude of P = 6,200 N, what is the magnitude of the torque T applied to the shaft?

FIGURE P15.9

Solution Section properties:



(55 mm) 2 − (45 mm) 2  = 785.398 mm2 4



(55 mm)4 − (45 mm)4  = 495,782.591 mm4 32 

(a) Normal and shear stress magnitudes: P 13,000 N = = = 16.552 MPa (T) A 785.398 mm2 Tc (260 N  m)(55 mm / 2)(1,000 mm/m) = = = 14.422 MPa J 495,782.591 mm 4 (sense of shear stress to be determined by inspection) A stress element showing the normal and shear stresses is shown below.

Generalized Hooke’s Law: Since the stresses are known, the normal strains in the x and y directions can be computed from Eqs. (13.21): 1 1  x = ( x −  y ) = [16.552 MPa − (0.34)(0)] E 105,000 MPa

= 157.639  10−6 mm/mm = 157.639 με 1 1  y = ( y −  x ) = [0 − (0.34)(16.552 MPa)] E 105,000 MPa = −53.597  10−6 mm/mm = −53.597 με

From Eq. (13.18), determine the shear modulus G: E 105,000 MPa G= = = 39,179.104 MPa 2(1 +  ) 2(1 + 0.34) and compute the shear strain xy from Eq. (13.20):  −14.422 MPa  xy = xy = = −368.095  10−6 rad G 39,179.104 MPa Strain transformation equation: Determine the strain that should be measured by the strain gage oriented at  = 40°:  n =  x cos 2  +  y sin 2  +  xy sin  cos  = (157.639 με) cos 2 (40) + ( −53.597 με)sin 2 (40) + ( −368.095 μrad)sin(40) cos(40) = −110.890 με = −110.9 με

Ans.

(b) Normal stress magnitude: For an applied load of P = 6,200 N: P 6,200 N = = = 7.894 MPa A 785.398 mm 2 Generalized Hooke’s Law: The normal stresses are known; therefore, the normal strains in the x and y directions can be computed from Eqs. (13.21): 1 1  x = ( x −  y ) = [7.894 MPa − (0.34)(0)] E 105,000 MPa

= 75.182  10−6 mm/mm = 75.182 με 1 1  y = ( y −  x ) = [0 − (0.34)(7.894 MPa)] E 105,000 MPa = −25.562  10−6 mm/mm = −25.562 με Strain transformation equation: The strain gage measures a normal strain of n = –195  at an orientation of  = +40°. From the normal strain transformation equation, determine the shear strain:  n =  x cos 2  +  y sin 2  +  xy sin  cos  −195 με = (75.182 με) cos 2 (40) + ( −25.562 με)sin 2 (40) +  xy sin(40) cos(40)

therefore:

(75.182 με)cos2 (40) + (−25.562 με)sin 2 (40) + 195 με = −464.166 μrad sin(40)cos(40) Shear stress: The shear stress xy can be computed from Eq. (13.20):  xy = G xy = (39,179.104 MPa)(−464.166  10−6 rad) = −18.186 MPa

 xy = −

Torque: The torque required to produce this shear stress magnitude is found from:  J (18.186 N/mm 2 )(495,782.591 mm 4 ) T= = = 327,858 N  mm = 328 N  m c 55 mm/2

Ans.

P15.10 A hollow shaft is subjected to an axial load P and a torque T, acting in the directions shown in Figure P15.10. The shaft is made of bronze [E = 15,200 ksi;  = 0.34], and it has an outside diameter of 2.50 in. and an inside diameter of 2.00 in. Strain gages a and b are mounted on the shaft at the orientations shown in Figure P15.18 where  has a magnitude of 25°. (a) If P = 6 kips and T = 17 kip-in., determine the strain readings that would be expected from the gages. (b) If the strain gage readings are a = –1,100  and b = 720 , determine the axial force P and the torque T applied to the shaft.

FIGURE P15.10

Solution Section properties:



(2.5 in.) 2 − (2.0 in.) 2  = 1.767146 in.2 4



(2.5 in.)4 − (2.0 in.)4  = 2.264156 in.4 32

(a) Normal and shear stress magnitudes: P 6 kips = = = 3.395 ksi (C) A 1.767146 in.2 Tc (17 kip  in.)(2.5 in. / 2) = = = 9.385 ksi (sense of shear stress determined by inspection) J 2.264156 in.4 A stress element showing the normal and shear stresses is shown below.

Generalized Hooke’s Law: Since the stresses are known, the normal strains in the x and y directions can be computed from Eqs. (13.21): 1 1  x = ( x −  y ) = [ −3.395 ksi − (0.34)(0)] E 15,200 ksi

= −223.375  10−6 in./in. = −223.375 με 1 1  y = ( y −  x ) = [0 − (0.34)( −3.395 ksi)] E 15,200 ksi = 75.948  10−6 in./in. = 75.948 με From Eq. (13.18), determine the shear modulus G: E 15,200 ksi G= = = 5,671.624 ksi 2(1 +  ) 2(1 + 0.34) Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

and compute the shear stress xy from Eq. (13.20):  9.385 ksi  xy = xy = = 1,654.794  10−6 rad G 5,671.624 ksi Strain transformation equation: Determine the strain that should be measured by strain gage a, which is oriented at a = –25°:  a =  x cos 2  a +  y sin 2  a +  xy sin  a cos  a = ( −223.375 με) cos 2 ( −25) + (75.948 με)sin 2 ( −25) + (1, 654.794 μrad)sin( −25) cos( −25) = −803.737 με = −804 με

Ans.

Determine the strain that should be measured by strain gage b, which is oriented at b = +65°:  b =  x cos 2 b +  y sin 2 b +  xy sin b cos b = ( −223.375 με) cos 2 (65) + (75.948 με)sin 2 (65) + (1, 654.794 μrad)sin(65)cos(65) = 656.309 με = 656 με

Ans.

(b) Generalized Hooke’s Law: Using the generalized Hooke’s Law and recognizing that the normal stress in the y direction is y = 0, the normal strains in the x and y directions can be expressed as:   1 1  y = ( y − x ) = − x  x = ( x − y ) = x and E E E E Write a strain transformation equation for each strain gage. For gage a:  a =  x cos 2  a +  y sin 2  a +  xy sin  a cos  a =

 x cos 2 ( −25)

−

 x sin 2 ( −25)

 xy sin( −25)cos( −25)

E E G which can be further simplified with Eq. (13.18): E G= 2(1 + ) to  cos 2 ( −25)  x sin 2 ( −25) 2(1 +  ) xy sin( −25) cos( −25) a = x − + E E E A similar expression can be written for gage b:  x cos 2 (65)  x sin 2 (65) 2(1 +  ) xy sin(65) cos(65) b = − + E E E

We now have two equations for the two unknowns x and xy: (15,200 ksi)( − 1,100  10−6 ) =  x cos 2 ( −25) − (0.34)sin 2 ( −25) +  xy  2(1 + 0.34)sin(−25)cos(−25)  (15,200 ksi)(720  10−6 ) =  x cos 2 (65) − (0.34)sin 2 (65)  +  xy  2(1 + 0.34)sin(65)cos(65) 

Solve simultaneously for x and xy:  x = −8.752 ksi  xy = 9.803 ksi

Axial load: P =  A = (−8.752 ksi)(1.767146 in.2 ) = 15.465 kips (C) = 15.47 kips (C)

Torque:  J (9.803 ksi)(2.264156 in.4 ) T= = = 17.757 kip  in. = 17.76 kip  in. c 2.5 in. / 2

Ans.

P15.11 The cylinder in Figure P15.11 is fabricated from spirallywrapped steel plates that are welded at the seams in the orientation shown. The cylinder has an outside diameter of 1,060 mm and a wall thickness of 10 mm. The cylinder is subjected to a compressive load of P = 2,200 kN and a torque of T = 850 kN·m, which acts in the direction shown. For a seam angle of  = 30°, determine: (a) the normal stress perpendicular to the weld seams. (b) the shear stress parallel to the weld seams. (c) the principal stresses and the maximum shear stress on the outside surface of the cylinder. FIGURE P15.11

Solution Section properties:



 

2 2 (1, 060 mm ) − (1, 040 mm )  = 32.9867 103 mm2  4

4 4 (1, 060 mm ) − (1, 040 mm )  = 9.0928 109 mm4  32

Normal and shear stress magnitudes: P ( 2, 200 kN )(1, 000 N/kN ) = = = 66.6935 MPa (C) A 32.9867 103 mm 2 Tc ( 850 kN  m )(1,060 mm / 2 )(1, 000 ) = = = 49.5447 MPa J 9.0928 109 mm 4 (sense of shear stresses to be determined by inspection) 2

A stress element for a point on the weld seam is shown below.

(a) Normal stress perpendicular to the weld seams: The normal stress perpendicular to the weld seam can be determined by Mohr’s circle or the stress transformation equations directly for a value of  = 60°. From the normal stress transformation equation Eq. (12.3):

 n =  x cos 2  +  y sin 2  + 2 xy sin  cos  = ( 0 ) cos 2 60 + ( −66.6935 MPa ) sin 2 60 + 2 ( 49.5447 MPa ) sin 60 cos 60 = −7.1131 MPa = 7.11 MPa (C)

Ans.

(b) Shear stress parallel to the weld seams: The shear stress parallel to the weld seam can be determined from the shear stress transformation equation [Eq. (12.4)]:  nt = −( x −  y )sin  cos +  xy (cos2  − sin 2  )

= − ( 0 ) − ( −66.6935 MPa )  sin 60 cos 60 + ( 49.5447 MPa ) cos 2 ( 60 ) − sin 2 ( 60 )  = −53.6515 MPa = −53.7 MPa

Ans.

 p1, p 2 =

x + y

  − y  2   x  +  xy  2  2

( 0 ) + ( −66.6935)   ( 0 ) − ( −66.6935)  + 49.5447 2 = )   ( 2





= −33.3467  59.7218

 p1 = 26.4 MPa

and

 p 2 = −93.1 MPa

Ans.

 max = 59.7 MPa

( maximum in-plane shear stress )

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P15.12 The extruded flexural member shown in Figure P15.12a is subjected to a shear force V = 3,600 N and a bending moment M = 550 N·m. The cross-sectional dimensions of the shape are shown in Figure P15.12b where a = 15 mm, b = 8 mm, c = 30 mm, and d = 42 mm. Determine the principal stresses and the maximum shear stress acting at point H, which is located at h = 20 mm below the upper surface of the member.

FIGURE P15.12a

FIGURE P15.12b

Solution

Centroid location in y direction: (reference axis at bottom of shape) Shape

Width b

Depth d

Area Ai

(1) (2) (3) (4)

(mm) 8 8 8 114

(mm) 42 42 42 8

(mm2) 336 336 336 912 1,920

yi Ai Ai

yi (from bottom)

yi Ai (mm3) 9,744 9,744 9,744 3,648 32,880

(mm) 29 29 29 4

32,880 mm3 = 17.125 mm (measured upward from bottom edge of bottom flange) 1,920 mm2

32.875 mm (measured downward from top edge) Moment of inertia about the z axis: (i.e., horizontal axis) d = y − yi Shape Width b Depth d Area A IC (1) (2) (3) (4)

(mm) 8 8 8 114

(mm) 42 42 42 8

(mm2) 336 336 336 912

(mm4) 49,392 49,392 49,392 4,864

(mm) –11.875 –11.875 –11.875 13.125

d²A

IC + d²A

(mm4) 47,381.25 47,381.25 47,381.25 157,106.25

(mm4) 96,773.25 96,773.25 96,773.25 161,970.25

Mechanics of Materials: An Integrated Learning System, 4th Ed. 1,920

Area =

Timothy A. Philpot

Moment of inertia about the z axis =

452,290.00

Bending stress at point H: (y = 32.875 mm – 17.125 mm = 12.875 mm) ( 550 N  m )(12.875 mm )(1,000 mm/m ) = −15.6564 MPa My x = − =− Iz 452,290 mm4 Transverse and horizontal shear stress at point H: 20 mm   3 Q = 3 (8 mm )( 20 mm ) 12.875 mm +  = 10,980 mm 2   3 ( 3,600 N ) 10,980 mm VQ = = = 3.6415 MPa (Sense to be determined by inspection) It 452, 290 mm4 ( 3  8 mm )

(

)

Stress element at H:

Principal stress calculations: The principal stress magnitudes can be computed from Eq. (12.12):

 p1, p 2 =

x + y 2

  − y  2   x  +  xy  2  2

( −15.6564 ) + ( 0 )   ( −15.6564 ) − ( 0 )  + −3.6415 2 = )   ( 2





= −7.8282  8.6337

 p1 = 0.806 MPa

and

 p 2 = −16.46 MPa

Ans.

 max = 8.63 MPa

(maximum in-plane shear stress)

Ans.

( normal stress on planes of maximum in-plane shear stress ) 2 xy 2 ( −3.6415 ) −7.2829 tan 2 p = = = = 0.4652  x −  y ( −15.6564 ) − ( 0 ) −15.6564

 avg = 7.83 MPa (C)

 p = 12.47

(counterclockwise from the x axis to the direction of  p 2 )

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Principal stresses and maximum in-plane shear stress.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P15.13 The beam cross section shown in Figure P15.13a/14a will consist of two square tubes that are welded to a rectangular web plate. Dimensions of the cross section as shown in Figure P15.13b/14b are d = 240 mm, tw = 12 mm, b = 80 mm, and t = 8.0 mm. The beam is subjected to axial force P = 30 kN, shear force V = 180 kN, and bending moment M = 75 kN·m. Determine the principal stresses and the maximum shear stress acting at point H, located at yH = 90 mm above the z centroidal axis. Show these stresses on an appropriate sketch.

FIGURE P15.13a/14a

FIGURE P15.13b/14b

Solution Moment of inertia about the z axis: (i.e., horizontal axis) 2 2 Atube = ( 80 mm ) − ( 64 mm ) = 2,304 mm 2

(80 mm ) − ( 64 mm ) = 2, 015, 232 mm4 = 4

I tube

Shape Top Tube Web Plate Bottom Tube

Width b

Depth d

Area A

d = y − yi

d²A

IC + d²A

(mm)

(mm2)

(mm4)

(mm)

(mm4)

2,304

2,015,232

–160

58,982,400

60,997,632

240

2,880

13,824,000

2,304

2,015,232

160

58,982,400

60,997,632

Area =

7,488

Moment of inertia about the z axis =

135,819,264

Axial stress at point H: F −30, 000 N x = = = −4.0064 MPa A 7,488 mm 2 Bending stress at point H: (yH = 90 mm) ( −75 106 N  mm ) ( 90 mm ) = 49.6984 MPa My x = − =− Iz 135,819,264 mm4

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Transverse and horizontal shear stress at point H: 80 mm   3 Q = ( 2,304 mm2 ) 120 mm +  + (12 mm)( 30 mm )(105 mm ) = 406, 440 mm 2   3 VQ (180,000 N ) 406, 440 mm = = = 44.8876 MPa (Sense to be determined by inspection) It 135,819, 264 mm4 (12 mm )

(

)

Stress element at H:

Principal stress calculations: The principal stress magnitudes can be computed from Eq. (12.12):

 p1, p 2 =

x + y 2

  − y  2   x  +  xy 2   2

( 45.6920 ) + ( 0 )   ( 45.6920 ) − ( 0 )  + 44.8876 2 = )   ( 2





= 22.8460  50.3670

 p1 = 73.2 MPa

and

 p 2 = −27.5 MPa

Ans.

( maximum in-plane shear stress )  avg = 22.8 MPa (T) ( normal stress on planes of maximum in-plane shear stress ) 2 xy 2 ( 44.8876 ) 89.7752 tan 2 p = = = = 1.9648  x −  y ( 45.6920 ) − ( 0 ) 45.6920

 max = 50.4 MPa

 p = 31.5

( counterclockwise from the x axis to the direction of  ) p1

Ans. Ans.

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Principal stresses and maximum in-plane shear stress.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P15.14 The beam cross section shown in Figure P15.13a/14a will consist of two square tubes that are welded to a rectangular web plate. Dimensions of the cross section as shown in Figure P15.13b/14b are d = 240 mm, tw = 12 mm, b = 80 mm, and t = 8.0 mm. The beam is subjected to axial force P = 30 kN, shear force V = 180 kN, and bending moment M = 75 kN·m. Determine the principal stresses and the maximum shear stress acting at point K, located at yK = 60 mm below the z centroidal axis. Show these stresses on an appropriate sketch.

FIGURE P15.13a/14a

FIGURE P15.13b/14b

Solution Atube = ( 80 mm ) − ( 64 mm ) = 2,304 mm 2 2

(80 mm ) − ( 64 mm ) = 2, 015, 232 mm4 = 4

I tube

Moment of inertia about the z axis: (i.e., horizontal axis) d = y − yi Shape Width b Depth d Area A IC Top Tube Web Plate Bottom Tube

d²A

IC + d²A

(mm)

(mm2)

(mm4)

(mm)

(mm4)

2,304

2,015,232

–160

58,982,400

60,997,632

240

2,880

13,824,000

2,304

2,015,232

160

58,982,400

60,997,632

Area =

7,488

Moment of inertia about the z axis =

135,819,264

Axial stress at point K: F −30, 000 N x = = = −4.0064 MPa A 7,488 mm 2 Bending stress at point K: (yK = –60 mm) ( −75 106 N  mm ) ( −60 mm ) = −33.1323 MPa My x = − =− Iz 135,819,264 mm 4

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Transverse and horizontal shear stress at point K: 80 mm   3 Q = ( 2,304 mm2 ) 120 mm +  + (12 mm )( 60 mm )( 90 mm ) = 433, 440 mm 2   3 VQ (180,000 N ) 433, 440 mm = = = 47.8695 MPa (Sense to be determined by inspection) It 135,819, 264 mm4 (12 mm )

(

)

Stress element at K:

Principal stress calculations: The principal stress magnitudes can be computed from Eq. (12.12):

 p1, p 2 =

x + y

  − y  2   x  +  xy 2   2

( −37.1387 ) + ( 0 )   ( −37.1387 ) − ( 0 )  + 47.8695 2 = )   ( 2





= −18.5693  51.3450

 p1 = 32.8 MPa

and

 p 2 = −69.9 MPa

Ans.

( maximum in-plane shear stress )  avg = 18.57 MPa (C) ( normal stress on planes of maximum in-plane shear stress ) 2 xy 2 ( 47.8695 ) 95.7390 tan 2 p = = = = −2.5779  x −  y ( −37.1387 ) − ( 0 ) −37.1387

 max = 51.3 MPa

 p = 34.4

( clockwise from the x axis to the direction of  ) p2

Ans. Ans.

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Principal stresses and maximum in-plane shear stress.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P15.15 The cantilever beam shown in Figure P15.15a is subjected to concentrated loads Px = 18 kips and Py = 32 kips. The cross-sectional dimensions of the rectangular tube shape shown in Figure P15.15b are b = 8 in., d = 12 in., and t = 0.25 in. Point H is located at a distance of a = 24 in. to the left of the concentrated loads. Calculate the principal stresses and maximum in-plane shear stress at point H for the following values of h: (a) h = 4.0 in. (b) h = 6.0 in. (c) h = 8.0 in.

FIGURE P15.15a

FIGURE P15.15b

Solution Section Properties: A = ( 8.0 in.)(12 in.) − ( 7.5 in.)(11.5 in.) = 9.75 in.2

( 8.0 in.)(12 in.) − ( 7.5 in.)(11.5 in.) = 201.4531 in.4 I = 3

Equilibrium: Fx = Px − F = 0

 F = Px = 18 kips Fy = V − Py = 0 V = Py = 32 kips M = − M − Py a = 0  M = − Py a = − ( 32 kips )( 24 in.) = −768 kip  in. (a) Axial stress at point H: F 18 kips x = = = 1.8462 ksi A 9.75 in.2 Bending stress at point H for h = 4.0 in.: (yH = –6.0 in. + 4.0 in. = –2.0 in.) ( −768 kip  in.)( −2.0 in.) = −7.6246 ksi x = − 201.4531 in.4

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Transverse and horizontal shear stress at point H: Q = ( 7.5 in.)( 0.25 in.)( 5.875 in.) +2 ( 0.25 in.)( 4 in.)( 4 in.) = 19.0156 in.3

( 32 kips ) (19.0156 in.3 ) VQ = = = 6.0411 ksi I t ( 201.4531 in.4 ) ( 2  0.250 in.) (Note: Sense to be determined by inspection) Stress element at H:

Principal stress calculations: The principal stress magnitudes can be computed from Eq. (12.12):

 p1, p 2 =

x + y 2

  − y  2   x  +  xy  2  2

( −5.7784 ) + ( 0 )   ( −5.7784 ) − ( 0 )  + −6.0411 2 = )   ( 2





= −2.8892  6.6965

 p1 = 3.81 ksi

and

 p 2 = −9.59 ksi

Ans.

( maximum in-plane shear stress )  avg = 2.89 ksi (C) ( normal stress on planes of maximum in-plane shear stress ) 2 xy 2 ( −6.0411) −12.0822 tan 2 p = = = = 2.0909  x −  y ( −5.7784 ) − ( 0 ) −5.7784

 max = 6.70 ksi

 p = 32.2

( counterclockwise from the x axis to the direction of  ) p2

Ans. Ans.

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Principal stresses and maximum in-plane shear stress.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

(b) Axial stress at point H: F 18 kips x = = = 1.8462 ksi A 9.75 in.2 Bending stress at point H for h = 6.0 in.: (yH = –6.0 in. + 6.0 in. = 0.0 in.) ( −768 kip  in.)( 0 in.) = 0 ksi x = − 201.4531 in.4 Transverse and horizontal shear stress at point H: Q = ( 7.5 in.)( 0.25 in.)( 5.875 in.) +2 ( 0.25 in.)( 6 in.)( 3 in.) = 20.0156 in.3

( 32 kips ) ( 20.0156 in.3 ) VQ = = = 6.3588 ksi I t ( 201.4531 in.4 ) ( 2  0.250 in.) (Note: Sense to be determined by inspection) Stress element at H:

Principal stress calculations: The principal stress magnitudes can be computed from Eq. (12.12):

 p1, p 2 =

x + y 2

  − y  2   x  +  xy  2  2

(1.8462 ) + ( 0 )   (1.8462 ) − ( 0 )  + −6.3588 2 = )   ( 2





= 0.9231  6.4254

 p1 = 7.35 ksi

and

 p 2 = −5.50 ksi

Ans.

( maximum in-plane shear stress )  avg = 0.923 ksi (T) ( normal stress on planes of maximum in-plane shear stress ) 2 xy 2 ( −6.3588 ) −12.7176 tan 2 p = = = = −6.8887  x −  y (1.8462 ) − ( 0 ) 1.8462

 max = 6.43 ksi

 p = 40.9

( clockwise from the x axis to the direction of  ) p1

Ans. Ans.

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Principal stresses and maximum in-plane shear stress.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

(c) Axial stress at point H: F 18 kips x = = = 1.8462 ksi A 9.75 in.2 Bending stress at point H for h = 8.0 in.: (yH = –6.0 in. + 8.0 in. = 2.0 in.) ( −768 kip  in.)( 2.0 in.) = 7.6246 ksi x = − 201.4531 in.4 Transverse and horizontal shear stress at point H: Q = ( 7.5 in.)( 0.25 in.)( 5.875 in.) +2 ( 0.25 in.)( 4 in.)( 4 in.) = 19.0156 in.3

( 32 kips ) (19.0156 in.3 ) VQ = = = 6.0411 ksi I t ( 201.4531 in.4 ) ( 2  0.250 in.) (Note: Sense to be determined by inspection) Stress element at H:

Principal stress calculations: The principal stress magnitudes can be computed from Eq. (12.12):

 p1, p 2 =

 x + y 2

  − y  2   x  +  xy  2  2

( 9.4708) + ( 0 )   ( 9.4708) − ( 0 )  + −6.0411 2 = )   ( 2





= 4.7354  7.6759

 p1 = 12.41 ksi

and

 p 2 = −2.94 ksi

Ans.

( maximum in-plane shear stress )  avg = 4.74 ksi (T) ( normal stress on planes of maximum in-plane shear stress ) 2 xy 2 ( −6.0411) −12.0822 tan 2 p = = = = −1.2757  x −  y ( 9.4708 ) − ( 0 ) 9.4708

 max = 7.68 ksi

 p = 26.0

( clockwise from the x axis to the direction of  ) p1

Ans. Ans.

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Principal stresses and maximum in-plane shear stress.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P15.16 Loads of Px = 13 kips and Py = 20 kips act on the titanium alloy [E = 16,500 ksi;  = 0.33] bar shown in Figure P15.16/17. The bar dimensions are d = 4.0 in., t = 0.75 in., and L = 18 in. Gages a and b are oriented at  = 40° as shown in the figure. Point H is located a distance of h = 1.5 in. above the bottom surface of the bar. Determine the normal strains expected in gages a and b.

FIGURE P15.16/17

Solution Section Properties: A = ( 0.75 in.)( 4.0 in.) = 3.0 in.2

( 0.75 in.)( 4.0 in.) = 4.0 in.4 I = 3

Equilibrium: Fx = Px − F = 0

 F = Px = 13 kips Fy = V − Py = 0 V = Py = 20 kips L M = − M − Py   = 0 2 L  18 in.   M = − Py   = − ( 20 kips )   = −180 kip  in. 2  2  Axial stress at point H: F 13 kips x = = = 4.3333 ksi A 3 in.2 Bending stress at point H: (yH = –2.0 in. + 1.5 in. = –0.5 in.) ( −180 kip  in.)( −0.5 in.) = −22.5 ksi x = − 4.0 in.4

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Transverse and horizontal shear stress at point H: Q = ( 0.75 in.)(1.5 in.)(1.25 in.) = 1.40625 in.3

VQ ( 20 kips ) (1.40625 in. ) = = = 9.375 ksi It ( 4.0 in.4 ) ( 0.75 in.) 3

(Note: Sense to be determined by inspection) Stress element at H:

Generalized Hooke’s law: From the generalized Hooke’s law for plane stress, calculate x, y, and xy. 1 1  x = ( x − y ) = ( −18.1667 ksi ) − ( 0.33 )( 0 )  E 16,500 ksi  = −1,101.010110−6 in./in. = −1,101.0101 με

y =

1 1  y − x ) = ( 0 ) − ( 0.33)( −18.1667 ksi )  ( E 16,500 ksi 

= 363.3333 10−6 in./in. = 363.3333 με E 16,500 ksi G= = = 6, 203.0075 ksi 2 (1 + ) 2 (1 + 0.33) 1 −9.3750 ksi  xy =  xy = = −1,511.3637  10−6 rad = −1,511.3637 μrad G 6, 203.0075 ksi

Strain transformation equations: Use the strain transformations equations to determine the expected strains in gages a and b.  +  −   a = x y + x y cos ( 2  40 ) + xy sin ( 2  40 ) 2 2 2 ( −1,101.0101 με ) + ( 363.3333 με ) + ( −1,101.0101 με ) − ( 363.3333 με ) cos 80 = ( ) 2 2 1,511.3637 μrad − sin ( 80 ) 2 = -1240 με Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

x +y

x −y

 xy

sin ( 2  140 ) 2 2 2 ( −1,101.0101 με ) + ( 363.3333 με ) + ( −1,101.0101 με ) − ( 363.3333 με ) cos 280 = ( ) 2 2 −1,511.3637 μrad + sin ( 280 ) 2 = 248 με

b =

cos ( 2  140 ) +

Timothy A. Philpot

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P15.17 Normal strain values of a = 740  and b = −180  were recorded with the two strain gages mounted at point H on the aluminum alloy [E = 10,000 ksi;  = 0.33] bar shown in Figure P15.16/17. The bar dimensions are d = 2.50 in., t = 0.375 in., and L = 12 in. Gages a and b are oriented at  = 35° as shown in the figure. Point H is located a distance of h = 1.0 in. above the bottom surface of the bar. Determine the magnitudes of loads Px and Py.

FIGURE P15.16/17

Solution Section Properties: A = ( 0.375 in.)( 2.5 in.) = 0.9375 in.2

( 0.375 in.)( 2.5 in.) = 0.4883 in.4 I = 3

Equilibrium: Fx = Px − F = 0

 F = Px Fy = V − Py = 0 V = Py L M = − M − Py   = 0 2 L  12 in.   M = − Py   = − Py   = − ( 6 in.) Py 2  2  Axial stress at point H: Px F x = = = 1.0667 in.−2 Px A 0.9375 in.2

(

)

Bending stress at point H: (yH = –1.25 in. + 1.0 in. = –0.25 in.) ( −6 in.) Py ( −0.25 in.) x = − = − 3.0719 in.−2 Py 4 0.4883 in.

(

)

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Normal stresses at point H:  x = (1.0667 in.−2 ) Px − ( 3.0719 in.−2 ) Py

y = 0 Transverse and horizontal shear stress at point H: Q = ( 0.375 in.)(1.0 in.)( 0.75 in.) = 0.2813 in.3

0.2813 in.3 ) Py ( VQ = = = (1.5359 in.−2 ) Py I t ( 0.4883 in.4 ) ( 0.375 in.) (Note: Sense to be determined by inspection)

Generalized Hooke’s law: From the generalized Hooke’s law for plane stress, express x, y, and xy in terms of Px and Py. 1  x = ( x − y ) E 1 (1.0667 in.−2 ) Px − ( 3.0719 in.−2 ) Py  − ( 0.33 )( 0 ) =  10, 000 ksi 





(1.0667 in. ) P − ( 3.0719 in. ) P = −2

−2

10, 000 ksi

Py Px − 9,374.7070 kips 3, 255.3143 kips 1  y = ( y − x ) E 1 = ( 0 ) − ( 0.33) (1.0667 in.−2 ) Px − ( 3.0719 in.−2 ) Py  10, 000 ksi −0.33 (1.0667 in.−2 ) Px − ( 3.0719 in.−2 ) Py  =  10, 000 ksi  Py Px =− + 28, 408.2032 kips 9,864.5888 kips E 10,000 ksi G= = = 3,759.3985 ksi 2 (1 + ) 2 (1 + 0.33) =





1.5359 in.−2 ) Py Py ( 1  xy =  xy = = G 3, 759.3985 ksi 2, 447.6844 kips

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Strain transformation equations: Write strain transformations equations for gages a and b.  a =  x cos 2 35 +  y sin 2 35 +  xy sin 35 cos 35 Py   Px = −  ( 0.67101)  9,374.7070 kips 3, 255.3143 kips  Py   Px + − +  ( 0.32899 )  28, 408.2032 kips 9,864.5888 kips  Py   +  ( 0.46985 )  2, 447.6844 kips  Py Px = + 16, 667.8259 kips 52,137.6991 kips

(a)

and

 b =  x cos 2 145 +  y sin 2 145 +  xy sin145 cos145 Py   Px = −  ( 0.67101)  9,374.7070 kips 3, 255.3143 kips  Py   Px + − +  ( 0.32899 )  28, 408.2032 kips 9,864.5888 kips  Py   +  ( −0.46985 )  2, 447.6844 kips  Py Px = − 16, 667.8259 kips 2, 741.7249 kips

(b)

Subtract Eq. (b) from Eq. (a) to determine Py. Py Py Px Px a − b = + − + 16, 667.8259 kips 52,137.6991 kips 16, 667.8259 kips 2, 741.7249 kips Py Py 740 10−6 − ( −180 10−6 ) = + 52,137.6991 kips 2, 741.7249 kips Py 920 10−6 = 2, 604.7509 kips  Py = 2.3964 kips = 2.40 kips

Ans.

Backsubstitute this result into Eq. (a) to compute Px. Py Px a = + 16, 667.8259 kips 52,137.6991 kips Px 2.3964 kips 740  10−6 = + 16, 667.8259 kips 52,137.6991 kips Px 740  10−6 − 45.9629  10−6 = 16, 667.8259 kips  Px = 11.5681 kips = 11.57 kips

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P15.18 The beam shown in Figure P15.18/19 spans a distance of L = 30 in., and its cross-sectional dimensions are b = 1.0 in. and d = 4.0 in. A load of P = 12,000 lb is applied at midspan. Point K is located at a distance of a = 6 in. from the roller support at C. Calculate the maximum compressive normal stress at point K for the following values of k: (a) k = 1.5 in. (b) k = 2.0 in. (c) k = 2.5 in.

FIGURE P15.18/19

Solution Section Properties:

(1.0 in.)( 4.0 in.) = 5.3333 in.4 I = 3

Shear-force and bending-moment diagrams. At point K, V = −6,000 lb

M = 36,000 lb  in.

(a) k = 1.5 in. Bending stress at point H: (yH = –2.0 in. + 1.5 in. = –0.5 in.) ( 36, 000 lb  in.)( −0.5 in.) = 3,375.0 psi x = − 5.3333 in.4

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Transverse and horizontal shear stress at point H: Q = (1.0 in.)(1.5 in.)(1.25 in.) = 1.875 in.3

3 VQ ( 6,000 lb ) (1.875 in. ) = = = 2,109.3750 psi It (5.3333 in.4 ) (1.0 in.)

Stress element at H:

Mohr’s circle

The maximum compressive normal stress at point K for k = 1.5 in. is:  p 2 = −1, 014 psi

Ans.

(b) k = 2.0 in. Bending stress at point H: (yH = –2.0 in. + 2.0 in. = 0 in.) x = 0

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Transverse and horizontal shear stress at point H: Q = (1.0 in.)( 2.0 in.)(1.0 in.) = 2.0 in.3 3 VQ ( 6,000 lb ) ( 2.0 in. ) = = = 2, 250 psi It ( 5.3333 in.4 ) (1.0 in.)

Stress element at H:

Mohr’s circle

The maximum compressive normal stress at point K for k = 2.0 in. is:  p 2 = −2, 250 psi

Ans.

(c) k = 2.5 in. Bending stress at point H: (yH = –2.0 in. + 2.5 in. = 0.5 in.) ( 36, 000 lb  in.)( 0.5 in.) = −3,375.0 psi x = − 5.3333 in.4

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Transverse and horizontal shear stress at point H: Q = (1.0 in.)(1.5 in.)(1.25 in.) = 1.875 in.3

3 VQ ( 6,000 lb ) (1.875 in. ) = = = 2,109.3750 psi It (5.3333 in.4 ) (1.0 in.)

Stress element at H:

Mohr’s circle

The maximum compressive normal stress at point K for k = 2.5 in. is:  p 2 = −4,390 psi

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P15.19 A single strain gage is mounted on a simply supported beam at point H as shown in Figure P15.18/19. The beam is made of an aluminum alloy [E = 70 GPa;  = 0.33], and it spans a distance of L = 1.0 m. The cross-sectional dimensions are b = 20 mm and d = 60 mm. Point H is located at a distance of a = 0.3 m from the pin support at A, and the gage is aligned at an angle of  = 45° as shown. If a load of P = 9,000 N is applied at midspan, what strains should be expected in the gage for the following values of h: (a) h = 20 mm. (b) h = 30 mm. (c) h = 40 mm.

FIGURE P15.18/19

Solution Section Properties:

( 20 mm )( 60 mm ) = 360, 000 mm4 I = 3

Shear-force and bending-moment diagrams. At point H, V = 4,500 N

M = 1,350,000 N  mm

(a) h = 20 mm Bending stress at point H: (yH = –30 mm + 20 mm = –10 mm) (1,350,000 N  mm )( −10 mm ) = 37.5 MPa x = − 360,000 mm4 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Transverse and horizontal shear stress at point H: Q = ( 20 mm)( 20 mm)( 20 mm) = 8,000 mm3 3 VQ ( 4,500 N ) (8,000 mm ) = = = 5.0 MPa It ( 360,000 mm4 ) ( 20 mm )

Stress element at H:

Generalized Hooke’s law: 1 1  x = ( x − y ) = ( 37.5 MPa ) − ( 0.33)( 0 )  E 70, 000 MPa  = 535.7143 10−6 mm/mm = 535.7143 με

y =

1 1  y − x ) = ( 0 ) − ( 0.33)( 37.5 MPa )  ( E 70, 000 MPa 

= −176.7857  10−6 mm/mm = −176.7857 με E 70,000 MPa G= = = 26,315.7895 MPa 2 (1 + ) 2 (1 + 0.33) 1 −5.0 MPa  xy =  xy = = −190.0  10−6 rad = −190.0 μrad G 26,315.7895 MPa

Strain transformation equations: Use the strain transformations equations to determine the expected strain in the gage.  +  −   gage at H = x y + x y cos ( 2  45 ) + xy sin ( 2  45 ) 2 2 2 ( 535.7143 με ) + ( −176.7857 με ) + ( 535.7143 με ) − ( −176.7857 με ) cos 90 = ( ) 2 2 −190.0 μrad + sin ( 90 ) 2 = 84.5 με Ans. (b) h = 30 mm Bending stress at point H: (yH = –30 mm + 30 mm = 0 mm)  x = 0 MPa

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Transverse and horizontal shear stress at point H: Q = ( 20 mm)( 30 mm)(15 mm) = 9,000 mm3 3 VQ ( 4,500 N ) ( 9,000 mm ) = = = 5.625 MPa It ( 360,000 mm4 ) ( 20 mm )

Stress element at H:

Generalized Hooke’s law: x = 0

y = 0  xy =

1 −5.625 MPa  xy = = −213.75  10−6 rad = −213.75 μrad G 26,315.7895 MPa

Strain transformation equations: Use the strain transformations equations to determine the expected strain in the gage.  +  −   gage at H = x y + x y cos ( 2  45 ) + xy sin ( 2  45 ) 2 2 2 −213.75 μrad = sin ( 90 ) 2 Ans. = −106.9 με (c) h = 40 mm Bending stress at point H: (yH = –30 mm + 40 mm = 10 mm) (1,350,000 N  mm )(10 mm ) = −37.5 MPa x = − 360,000 mm4 Transverse and horizontal shear stress at point H: Q = ( 20 mm)( 20 mm)( 20 mm) = 8,000 mm3

=

3 VQ ( 4,500 N ) (8,000 mm ) = = 5.0 MPa It ( 360,000 mm4 ) ( 20 mm )

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Stress element at H:

Generalized Hooke’s law: 1 1  x = ( x − y ) = ( −37.5 MPa ) − ( 0.33)( 0 )  E 70, 000 MPa  = −535.7143 10−6 mm/mm = −535.7143 με

y =

1 1  y − x ) = ( 0 ) − ( 0.33)( −37.5 MPa )  ( E 70, 000 MPa 

= 176.7857 10−6 mm/mm = 176.7857 με 1 −5.0 MPa  xy =  xy = = −190.0  10−6 rad = −190.0 μrad G 26,315.7895 MPa

Strain transformation equations: Use the strain transformations equations to determine the expected strain in the gage.  +  −   gage at H = x y + x y cos ( 2  45 ) + xy sin ( 2  45 ) 2 2 2 ( −535.7143 με ) + (176.7857 με ) + ( −535.7143 με ) − (176.7857 με ) cos 90 = ( ) 2 2 −190.0 μrad + sin ( 90 ) 2 = −274 με Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P15.20 The simply supported beam shown in Figure P15.20a/21a supports three concentrated loads. The loads at B and C each have a magnitude of P = 25 kips, and the load at D is Q = 60 kips. The beam span is L = 32 ft. The cross-sectional dimensions of the beam as shown in Figure P15.20b/21b are bf = 12.0 in., tf = 0.85 in., d = 20.0 in., tw = 0.50 in., and yH = 3.5 in. Determine the principal stresses and the maximum shear stress acting at point H, which is located at xH = 4 ft. Show these stresses on an appropriate sketch.

FIGURE P15.20a/21a

FIGURE P15.20b/21b

Solution Moment of inertia about the z axis:

Shape

Width b

Depth d

Area A

d = y − yi

d²A

IC + d²A

(1) (2) (3)

(in.) 11.5 0.5 11.5

(in.) 0.85 20 0.85

(in.2) 9.775 10.000 9.775

(in.4) 0.5885 333.3333 0.5885

(in.) –9.575 0.000 9.575

(in.4) 896.1781 0.0000 896.1781

(in.4) 896.7666 333.3333 896.7666

Area =

29.550

Moment of inertia about the z axis =

2,126.8666

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Shear-force and bending-moment diagram: Shear force at H: V = 46.25 kips Bending moment at H: M = 185 kip·ft

Bending stress at point H: (y = 3.50 in.) (185 kip  ft )( 3.5 in.)(12 in./ft ) = −3.6533 ksi My x = − =− Iz 2,126.8666 in.4 Transverse and horizontal shear stress at point H: Q = (11.5 in.)( 0.85 in.)(9.575 in.) + ( 0.5 in.)( 6.50 in.)( 6.75 in.) = 115.5331 in.3 3 VQ ( 46.25 kips ) (115.5331 in. ) = = = 5.0247 ksi It ( 2,126.8666 in.4 ) ( 0.50 in.)

(Sense to be determined by inspection)

Stress element at H:

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Mohr’s circle

Principal stress calculations: The principal stress magnitudes can be computed from Eq. (12.12):

 p1, p 2 =

x + y 2

  − y  2   x  +  xy 2   2

( −3.6533) + (0)   ( −3.6533) − (0)  + −5.0247 2 = )   ( 2





= −1.8266  5.3464

 p1 = 3.52 ksi

and

 p 2 = −7.17 ksi

Ans.

( maximum in-plane shear stress )  avg = 1.827 ksi (C) ( normal stress on planes of maximum in-plane shear stress ) 2 xy 2 ( −5.0247 ) −10.0494 tan 2 p = = = = 2.7508  x −  y ( −3.6533) − ( 0 ) −3.6533

 max = 5.35 ksi

 p = 35.0

( counterclockwise from the x axis to the direction of  ) p2

Ans. Ans.

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Principal stresses and maximum in-plane shear stress.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P15.21 The simply supported beam shown in Figure P15.20a/21a supports three concentrated loads. The loads at B and C each have a magnitude of P = 70 kN, and the load at D is Q = 260 kN. The beam span is L = 10 m. The cross-sectional dimensions of the beam as shown in Figure P15.20b/21b are bf = 300 mm, tf = 20 mm, d = 500 mm, tw = 12 mm, and yK = 120 mm. Determine the principal stresses and the maximum shear stress acting at point K, which is located at xK = 1.75 m. Show these stresses on an appropriate sketch.

FIGURE P15.20a/21a

FIGURE P15.20b/21b

Solution Moment of inertia about the z axis:

Shape

Width b

Depth d

Area A

(1) (2) (3)

(mm) 288 12 288

(mm) 20 500 20 Area =

(mm2) 5,760 6,000 5,760 17,520

d = y − yi

d²A

(mm4) (mm) (mm4) 192,000 –240 331,776,000 125,000,000 0 0 192,000 240 331,776,000 Moment of inertia about the z axis =

IC + d²A (mm4) 331,968,000 125,000,000 331,968,000 788,936,000

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Shear-force and bending-moment diagram: Shear force at K: V = –247.5 kN Bending moment at K: M = 433.125 kN·m

Bending stress at point K: (y = –120 mm)

( 433.125 kN  m )( −120 mm )(1000 ) = 65.8799 MPa My x = − =− Iz 788,936,000 mm4 2

Transverse and horizontal shear stress at point K: Q = ( 288 mm)( 20 mm)( 240 mm) + (12 mm)(130 mm)(185 mm) = 1,671,000 mm3 3 VQ ( 247,500 N ) (1,671,000 mm ) = = = 43.6846 MPa It ( 788,936,000 mm4 ) (12 mm )

(Sense to be determined by inspection.)

Stress element at K:

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Mohr’s circle

Principal stress calculations: The principal stress magnitudes can be computed from Eq. (12.12):

 p1, p 2 =

 x + y 2

  − y  2   x  +  xy  2  2

( 65.8799 ) + (0)   ( 65.8799 ) − (0)  + 43.6846 2 = )   ( 2





= 32.9399  54.7118

 p1 = 87.7 MPa

and

 p 2 = −21.8 MPa

Ans.

( maximum in-plane shear stress )  avg = 32.9 MPa (T) ( normal stress on planes of maximum in-plane shear stress ) 2 xy 2 ( 43.6846 ) 87.3693 tan 2 p = = = = 1.3262  x −  y ( 65.8799 ) − ( 0 ) 65.8799

 max = 54.7 MPa

 p = 26.5

( counterclockwise from the x axis to the direction of  ) p1

Ans. Ans.

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Principal stresses and maximum in-plane shear stress.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P15.22 The beam shown in Figure P15.22a is supported by a tie bar at B and by a pin connection at C. The beam span is L = 7 m and the uniformly distributed load is w = 22 kN/m. The tie bar at B has an orientation of  = 25°. The cross-sectional dimensions of the beam shown in Figure P15.22b are bf = 130 mm, tf = 12 mm, d = 360 mm, tw = 6 mm, and yH = 50 mm. Determine the principal stresses and the maximum shear stress acting at point H. Show these stresses on an appropriate sketch.

FIGURE P15.22a

FIGURE P15.22b

Solution Moment of inertia about the z axis: d = y − yi Shape Width b Height h IC d²A 4 (mm) (mm) (mm ) (mm) (mm4) top flange 130 12 18,720.0 –174.0 47,230,560.0 web 6 336 18,966,528.0 0.0 0.0 bottom flange 130 12 18,720.0 174.0 47,230,560.0 Moment of inertia about the z axis (mm4) =

IC + d²A (mm4) 47,249,280.0 18,966,528.0 47,249,280.0 113,465,088.0

Shear-force and bending-moment diagram: Shear force at H: V = –46.20 kN Bending moment at H: M = 86.24 kN·m

Axial force at H: By 77.00 kN P= = = 165.127 kN tan(25) tan(25)

Axial stress at H: 165,127 N x = = 32.151 MPa 5,136 mm2

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Bending stress at point H: (y = −50 mm) My (86.240 kN-m)( − 50 mm)(1,000 N/kN)(1,000 mm/m) x = − =− = 38.003 MPa Iz 113,465,088.0 mm4 Transverse and horizontal shear stress at point H: Q = (130 mm)(12 mm)(174 mm) + (6 mm)(118 mm)(109 mm) = 348,612 mm3

=

VQ (46,200 N)(348,612 mm3 ) = = 23.658 MPa (Sense to be determined by inspection) It (113,465,088.0 mm4 )(6 mm)

Stress element at H:

Principal stress calculations: The principal stress magnitudes can be computed from Eq. (12.12):

 p1, p 2 =

x + y 2

x −y    +  xy2 2   2

(70.154) + (0)  (70.154) − (0)  =   + (23.658) 2    2 2 2

= 35.077  42.309  p1 = 77.4 MPa and

 max = 42.3 MPa

 p 2 = −7.23 MPa

Ans.

(maximum in-plane shear stress)

Ans.

 avg = 35.1 MPa (T) tan 2 p =

2 xy

 x − y

(normal stress on planes of maximum in-plane shear stress)

Ans.

2 ( 23.658 ) 47.315 = = 0.6744 ( 70.154 ) − ( 0 ) 70.154

 p = 17.00

( counterclockwise from the x axis to the direction of  ) p1

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Mohr’s circle:

Principal stresses and maximum in-plane shear stress:

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P15.23 The bracket shown in Figure P15.23a/24a supports a pulley that has a diameter of D = 120 mm. The tension in the pulley belt is P = 200 N, and the angle of the pulley belt is  = 30°as shown. The bracket has dimensions a = 90 mm and c = 150 mm. The cross-sectional dimensions of the bracket are b = 10 mm and d = 30 mm as shown in Figure P15.23b/24b. Determine the principal stresses and the maximum shear stress acting at point H. Show these stresses on an appropriate sketch.

FIGURE P15.23a/24a

FIGURE P15.23b/24b

Solution Equilibrium: Begin with a free-body diagram of the pulley and determine the reaction forces Rx and Ry. Fx = P + P sin  − Rx = 0

 Rx = 200 N + ( 200 N ) sin 30 = 300 N

Fy = P cos  − Ry = 0  Ry = ( 200 N ) cos 30 = 173.2051 N Next, determine the axial force F, the shear force V, and the bending moment M at the section that contains points H and K. From the FBD, we find: Fx = Rx − V = 0

V = 300 N Fy = Ry − F = 0  F = 173.2051 N d  M = − Rx c + Ry  a +  + M = 0 2  30 mm    M = ( 300 N )(150 mm ) − (173.2051 N )  90 mm +  2   = 26,813.4665 N  mm Section Properties: A = (10 mm )( 30 mm ) = 300 mm 2

(10 mm )( 30 mm ) = 22,500 mm 4 I = 3

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Axial stress at point H: F 173.2051 N y = = = 0.5774 MPa (T) A 300 mm 2 Bending stress at point H: (xH = –5 mm) ( 26,813.4665 N  mm )( −5 mm ) = 5.9585 MPa (T) y = − 22,500 mm4 Transverse and horizontal shear stress at point H: Q = (10 mm)(10 mm)(10 mm) = 1,000 mm3

( 300 N ) (1, 000 mm3 ) VQ = = = 1.3333 MPa I t ( 22,500 mm 4 ) (10 mm )

( Note: Sense to be determined by inspection.) Stress element at H:

Principal stress calculations: The principal stress magnitudes can be computed from Eq. (12.12):

 p1, p 2 =

x + y 2

  − y  2   x  +  xy  2  2

( 0 ) + ( 6.5359 )   ( 0 ) − ( 6.5359 )  + 1.3333 2 = )   ( 2





= 3.2679  3.5295

 p1 = 6.80 MPa

and

 p 2 = −0.262 MPa

Ans.

( maximum in-plane shear stress )  avg = 3.27 MPa (T) ( normal stress on planes of maximum in-plane shear stress ) 2 xy 2 (1.3333) 2.6667 tan 2 p = = = = −0.4080  x −  y ( 0 ) − ( 6.5359 ) −6.5359

 max = 3.53 MPa

 p = 11.10

( clockwise from the x axis to the direction of  ) p2

Ans. Ans.

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Principal stresses and maximum in-plane shear stress.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P15.24 The bracket shown in Figure P15.23a/24a supports a pulley that has a diameter of D = 6 in. The tension in the pulley belt is P = 75 lb, and the angle of the pulley belt is  = 30°as shown. The bracket has dimensions a = 4.5 in. and c = 7.5 in. The cross-sectional dimensions of the bracket are b = 0.5 in. and d = 1.5 in. as shown in Figure P15.23b/24b. Determine the principal stresses and the maximum shear stress acting at point K. Show these stresses on an appropriate sketch.

FIGURE P15.23a/24a

FIGURE P15.23b/24b

Solution Equilibrium: Begin with a free-body diagram of the pulley and determine the reaction forces Rx and Ry. Fx = P + P sin  − Rx = 0

 Rx = 75 lb + ( 75 lb ) sin 30 = 112.5 lb

Fy = P cos  − Ry = 0  Ry = ( 75 lb ) cos 30 = 64.9519 lb Next, determine the axial force F, the shear force V, and the bending moment M at the section that contains points H and K. From the FBD, we find: Fx = Rx − V = 0

V = 112.5 lb Fy = Ry − F = 0  F = 64.9519 lb d  M = − Rx c + Ry  a +  + M = 0 2  1.5 in.    M = (112.5 lb )( 7.5 in.) − ( 64.9519 lb )  4.5 in. +  2   = 502.7525 lb  in. Section Properties: A = ( 0.5 in.)(1.5 in.) = 0.75 in.2

( 0.5 in.)(1.5 in.) = 0.140625 in.4 I = 3

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Axial stress at point K: F 64.9519 lb y = = = 86.6025 psi (T) A 0.75 in.2 Bending stress at point K: (xK = 0.375 in.) ( 502.7525 lb  in.)( 0.375 in.) = −1,340.6734 psi (C) y = − 0.140625 in.4 Transverse and horizontal shear stress at point K: Q = ( 0.5 in.)( 0.375 in.)( 0.5625 in.) = 0.105469 in.3 3 VQ (112.5 lb ) ( 0.105469 in. ) = = = 168.75 psi It ( 0.140625 in.4 ) ( 0.5 in.)

( Note: Sense to be determined by inspection.) Stress element at H:

Principal stress calculations: The principal stress magnitudes can be computed from Eq. (12.12):

 p1, p 2 =

x + y 2

  − y  2   x  +  xy 2   2

( 0 ) + ( −1, 254.0708 )   ( 0 ) − ( −1, 254.0708 )  + 168.75 2 = )   ( 2





= −627.0354  649.3458

 p1 = 22.3 psi

and

 p 2 = −1, 276 psi

Ans.

( maximum in-plane shear stress )  avg = 627 psi (C) ( normal stress on planes of maximum in-plane shear stress ) 2 xy 2 (168.75 ) 337.5 tan 2 p = = = = 0.2691  x −  y ( 0 ) − ( −1, 254.0708 ) 1, 254.0708

 max = 649 psi

 p = 7.53

( counterclockwise from the x axis to the direction of  ) p1

Ans. Ans.

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Mohr’s circle for element K:

Principal stresses and maximum in-plane shear stress.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P15.25 A load of P = 75 kN acting at an angle of  = 35° is supported by the structure shown in Figure P15.25a. The overall dimensions of the structure are a = 2.4 m, b = 0.6 m, c = 1.5 m, e = 0.32 m, and x1 = 2.2 m. The cross-sectional dimensions of member BC (shown in Figure P15.25b) are bf = 160 mm, tf = 15 mm, d = 300 mm, tw = 10 mm, and h = 90 mm. Determine the principal stresses and the maximum shear stress acting at point H in member BC. Show these stresses on an appropriate sketch.

FIGURE P15.25a

FIGURE P15.25b

Solution Moment of inertia about the z axis:

Shape

(1) (2) (3)

Width b

Depth d

Area A

(mm) 150 10 150

(mm) 15 300 15 Area =

(mm2) 2,250 3,000 2,250 7,500

d = y − yi

d²A

(mm4) (mm) (mm4) 42,187.5 –142.5 45,689,062.5 22,500,000.0 0.0 0.0 42,187.5 142.5 45,689,062.5 Moment of inertia about the z axis =

IC + d²A (mm4) 45,731,250 22,500,000 45,731,250 113,962,500

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Equilibrium: Fx = Bx − FDE cos 45 + P sin  = 0

Fy = By − FDE sin 45 − P cos  = 0 M B = x1 FDE cos 45 + eFDE sin 45

−aP cos  − ( b + c ) P sin  = 0

 FDE = 133.4443 kN Bx = 51.3412 kN By = 155.7958 kN

Internal forces at H: Fx = Bx + VH = 0

VH = −51.3412 kN

Fy = FH + By = 0

 FH = −155.7958 kN

M H = M H + Bx ( 600 mm ) = 0

 M H = −30.8047 106 N  mm

Axial stress at H: −155, 795.8 N y = = 20.7728 MPa (C) 7,500 mm 2 Bending stress at H: (30.8047 106 N  mm ) ( 60 mm ) = 16.2183 MPa (C) y = 113,962,500 mm4

( by inspection )

Shear stress at H: QH = (150 mm )(15 mm )(142.5 mm ) + (10 mm )( 90 mm )(105 mm ) = 415,125 mm3

( 51,341.2 N ) ( 415,125 mm3 ) H = = 18.7018 MPa (113,962,500 mm4 ) (10 mm )

Summary of stresses at H:  x = 0 MPa

 y = −20.7728 MPa − 16.2183 MPa = −36.9911 MPa  xy = −18.7018 MPa

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Principal stress magnitudes: The principal stress magnitudes can be computed from Eq. (12.12):

 p1, p 2 =

x + y 2

  − y  2   x  +  xy 2   2

( 0 ) + ( −36.9911)   ( 0 ) − ( −36.9911)  + −18.7018 2 = )   ( 2



= −18.4956  26.3029  p1 = 7.81 MPa and

 max = 26.3 MPa



 p 2 = −44.8 MPa

Ans.

(maximum in-plane shear stress)

Ans.

( normal stress on planes of maximum in-plane shear stress ) 2 xy 2 ( −18.7018 ) −37.4035 tan 2 p = = = = −1.0111  x −  y ( 0 ) − ( −36.9911) 36.9911

 avg = 18.50 MPa (C)

 p = 22.7

( clockwise from the x axis to the direction of  ) p1

Ans.

Mohr’s circle for element H:

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Principal stresses and maximum in-plane shear stress.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P15.26 A short rectangular post supports a compressive load of P = 35 kN as shown in Figure P15.26a. A top view of the post showing the location where load P is applied to the top of the post is shown in Figure P15.26b. The cross-sectional dimensions of the post are b = 240 mm and d = 160 mm. The load P is applied at offset distances of yP = 60 mm and zP = 50 mm from the centroid of the post. Determine the normal stresses at corners A, B, C, and D of the post.

FIGURE P15.26a

FIGURE P15.26b Top view of post.

Solution Section properties: A = (160 mm)(240 mm) = 38, 400 mm 2 (240 mm)(160 mm)3 = 81.92  106 mm 4 12 (160 mm)(240 mm)3 Iz = = 184.32  106 mm 4 12

Iy =

Equivalent forces at base of post: F = 35 kN = 35,000 N

M y = (35 kN)(50 mm) = 1,750 kN-mm = 1.75  106 N-mm M z = (35 kN)( −60 mm) = −2,100 kN-mm = −2.10  106 N-mm Axial stress due to F: 35,000 N  axial = = 0.911 MPa (C) 38,400 mm2 Bending stress due to My: M y z (1.75  106 N-mm)(  80 mm)  bend y = = = 1.709 MPa Iy 81.92  106 mm4 Bending stress due to Mz: M y (2.10  106 N-mm)(  120 mm)  bend z = z = = 1.367 MPa Iz 184.32  106 mm4 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Normal stresses at corners A, B, C, and D: The normal stresses acting at each of the four corners of the post can be determined by superimposing the results obtained above. In all instances, the normal stresses act in the vertical direction; that is, the x direction. The sense of the stress, either tension or compression, can be determined by inspection. Corner A:  A = 0.911 MPa (C) + 1.709 MPa (T) + 1.367 MPa (C) = −0.911 MPa + 1.709 MPa − 1.367 MPa = −0.570 MPa = 0.570 MPa (C)

Ans.

Corner B:  B = 0.911 MPa (C) + 1.709 MPa (C) + 1.367 MPa (C) = −0.911 MPa − 1.709 MPa − 1.367 MPa = −3.988 MPa = 3.99 MPa (C)

Ans.

Corner C:  C = 0.911 MPa (C) + 1.709 MPa (C) + 1.367 MPa (T) = −0.911 MPa − 1.709 MPa + 1.367 MPa = −1.253 MPa = 1.253 MPa (C)

Ans.

Corner D:  D = 0.911 MPa (C) + 1.709 MPa (T) + 1.367 MPa (T) = −0.911 MPa + 1.709 MPa + 1.367 MPa = 2.165 MPa = 2.17 MPa (T)

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P15.27 Three loads are applied to the rectangular tube shown in Figure P15.27a/28a. The load magnitudes are P = 175 kN, Q = 60 kN, and R = 85 kN. The dimensions of the tube, as shown in Figure P15.27b/28b, are b = 150 mm, d = 200 mm, t = 10 mm, and zK = 50 mm. The location of load P is also shown in this figure. Use a = 125 mm and determine the normal and shear stresses that act at point H. Show these stresses on a stress element.

FIGURE P15.27a/28a

FIGURE P15.27b/28b

Solution Section properties: A = (150 mm )( 200 mm ) − (130 mm )(180 mm ) = 6, 600 mm 2

(150 mm )( 200 mm ) − (130 mm )(180 mm ) = 36.820 106 mm 4 I = 3

( 200 mm )(150 mm ) − (180 mm )(130 mm ) = 23.295 106 mm4 I = 3

Equivalent forces at H: F = −175 kN = −175, 000 N

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Shear stress at H due to Vx: QH = 2 (10 mm )( 75 mm )( 37.5 mm ) + (180 mm )(10 mm )( 70 mm ) = 182, 250 mm3

(85, 000 N ) (182, 250 mm3 ) H = = 33.2502 MPa ( 23.295 106 mm4 ) ( 2 10 mm )

Shear stress at H due to Vz: QH = 0 mm3  H = 0 MPa Bending stress at H due to Mx: 6 M x z (1.625 10 N  mm ) (100 mm )  bend x = = = 4.4134 MPa (T) Ix 36.820 106 mm 4 Bending stress at H due to Mz: 6 M z x ( −22.875 10 N  mm ) ( 0 mm )  bend z = = = 0 MPa Iz 23.295 106 mm 4

Summary of stresses at H:  x = 0 MPa

 y = −26.5152 MPa + 4.4134 MPa = −22.1018 MPa  xy = 33.2502 MPa

Principal stress calculations: The principal stress magnitudes can be computed from Eq. (12.12):

 p1, p 2 =

x + y

  − y  2   x  +  xy  2  2

( 0 ) + ( −22.1018)   ( 0 ) − ( −22.1018)  + 33.2502 2 = )   ( 2





= −11.0509  35.0385

 p1 = 24.0 MPa

and

 p 2 = −46.1 MPa

Ans.

( maximum in-plane shear stress )  avg = 11.05 MPa (C) ( normal stress on planes of maximum in-plane shear stress ) 2 xy 2 ( 33.2502 ) 66.5003 tan 2 p = = = = 3.0088  x −  y ( 0 ) − ( −22.1018 ) 22.1018

 max = 35.0 MPa

 p = 35.8

( counterclockwise from the x axis to the direction of  ) p1

Ans. Ans.

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Mohr’s circle for element H.

Principal stresses and maximum in-plane shear stress.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P15.28 Three loads are applied to the rectangular tube shown in Figure P15.27a/28a. The load magnitudes are P = 175 kN, Q = 60 kN, and R = 85 kN. The dimensions of the tube, as shown in Figure P15.27b/28b, are b = 150 mm, d = 200 mm, t = 10 mm, and zK = 50 mm. The location of load P is also shown in this figure. Use a = 125 mm and determine the normal and shear stresses that act at point K. Show these stresses on a stress element.

FIGURE P15.27a/28a

FIGURE P15.27b/28b

Solution Section properties: A = (150 mm )( 200 mm ) − (130 mm )(180 mm ) = 6, 600 mm 2

(150 mm )( 200 mm ) − (130 mm )(180 mm ) = 36.820 106 mm 4 I = 3

( 200 mm )(150 mm ) − (180 mm )(130 mm ) = 23.295 106 mm4 I = 3

Equivalent forces at K: F = −175 kN = −175, 000 N

Vx = 85 kN = 85, 000 N Vz = 60 kN = 60, 000 N M x = − (175 kN )( 95 mm ) + ( 60 kN )( 250 mm ) = −1, 625 kN  mm = −1.625  106 N  mm M z = − (175 kN )( 70 mm ) − ( 85 kN )(125 mm ) = −22,875 kN  mm = −22.875 106 N  mm Axial stress at K due to F: 175, 000 N  axial = = 26.5152 MPa (C) 6, 600 mm 2 Shear stress at K due to Vx: QK = 0 mm3  K = 0 MPa Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Shear stress at K due to Vz: QK = 2 (10 mm )( 50 mm )( 75 mm ) + (130 mm )(10 mm )( 95 mm ) = 198,500 mm3

( 60, 000 N ) (198,500 mm3 ) K = = 16.1733 MPa (36.820 106 mm4 ) ( 2 10 mm )

Bending stress at K due to Mx: 6 M x z (1.625 10 N  mm ) ( 50 mm )  bend x = = = 2.2067 MPa (T) Ix 36.820 106 mm 4 Bending stress at K due to Mz: 6 M z x ( 22.875 10 N  mm ) ( 75 mm )  bend z = = = 73.6478 MPa (C) Iz 23.295 106 mm 4

Summary of stresses at K:  x = 0 MPa

 y = −26.5152 MPa + 2.2067 MPa − 73.6478 MPa = −97.9562 MPa

 xy = −16.1733 MPa Principal stress calculations: The principal stress magnitudes can be computed from Eq. (12.12):

 p1, p 2 =

x + y

  − y  2   x  +  xy  2  2

( 0 ) + ( −97.9562 )   ( 0 ) − ( −97.9562 )  + −16.1733 2 = )   ( 2





= −48.9781  51.5794

 p1 = 2.60 MPa

and

 p 2 = −100.6 MPa

Ans.

( maximum in-plane shear stress )  avg = 49.0 MPa (C) ( normal stress on planes of maximum in-plane shear stress ) 2 xy 2 ( −16.1733) −33.3466 tan 2 p = = = = −0.3302  x −  y ( 0 ) − ( −97.9562 ) 97.9562

 max = 51.6 MPa

 p = 9.14

( clockwise from the x axis to the direction of  ) p1

Ans. Ans.

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Mohr’s circle for element K.

Principal stresses and maximum in-plane shear stress.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P15.29 Concentrated loads of Px = 33 kips, Py = 29 kips, and Pz = 46 kips are applied to the cantilever beam in the locations and directions shown in Figure P15.29a/30a. The beam cross section shown in Figure P15.29b/30b has dimensions of b = 9 in. and d = 4 in. Using a value of a = 6.4 in., determine the normal and shear stresses at point H. Show these stresses on a stress element.

FIGURE P15.29a/30a

FIGURE P15.29b /30b

Solution Section properties: A = ( 9 in.)( 4 in.) = 36 in.2

( 9 in.)( 4 in.) = 48 in.4 I = 3

( 4 in.)( 9 in.) = 243 in.4 I = 3

Equivalent forces at H: F = −46, 000 lb Vx = 33, 000 lb

Vy = 0 lb

M y = ( 33, 000 lb )( 6.4 in.) = 211, 200 lb  in.

Axial stress at H due to F: −46, 000 lb  axial = = 1, 277.7778 psi (C) 36 in.2 Shear stress at H due to Vx: QH = ( 4 in.)(1.5 in.)( 3.75 in.) = 22.5 in.3

( 33, 000 lb ) ( 22.5 in.3 ) H = = 763.8889 psi ( 243 in.4 ) ( 4 in.)

Shear stress at H due to Vy: QH = 0 in.3  H = 0 psi Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Bending stress at H due to Mx: M y ( 0 lb  in.)( 2 in.)  bend x = x = =0 Ix 48 in.4 Bending stress at H due to My: M x ( 211, 200 lb  in.)( 3.0 in.)  bend y = y = = 2, 607.4075 psi (C) Iy 243 in.4 Summary of stresses at H:  x = 0 psi

 z = −1, 277.7778 psi − 2, 607.4075 psi = −3,885.1854 psi = 3,890 psi (C)

 xz = 763.8889 psi = 764 psi

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P15.30 Concentrated loads of Px = 33 kips, Py = 29 kips, and Pz = 46 kips are applied to the cantilever beam in the locations and directions shown in Figure P15.29a/30a. The beam cross section shown in Figure P15.29b/30b has dimensions of b = 9 in. and d = 4 in. Using a value of a = 6.4 in., determine the normal and shear stresses at point K. Show these stresses on a stress element.

FIGURE P15.29a/30a

FIGURE P15.29b /30b

Solution Section properties: A = ( 9 in.)( 4 in.) = 36 in.2

( 9 in.)( 4 in.) = 48 in.4 I = 3

( 4 in.)( 9 in.) = 243 in.4 I = 3

Equivalent forces at K: F = −46, 000 lb Vx = 33, 000 lb

Vy = −29, 000 lb

M x = ( 29, 000 lb )( 6.4 in.) = 185, 600 lb  in. M y = ( 33, 000 lb )(19.2 in.) = 633, 600 lb  in. Axial stress at K due to F: −46, 000 lb  axial = = 1, 277.7778 psi (C) 36 in.2 Shear stress at K due to Vx: QK = ( 4 in.)( 2.7 in.)( 3.15 in.) = 34.02 in.3

( 33, 000 lb ) ( 34.02 in.3 ) K = = 1,155 psi ( 243 in.4 ) ( 4 in.)

Shear stress at K due to Vy: QK = 0 in.3  K = 0 psi Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Bending stress at K due to Mx: M y (185, 600 lb  in.)( 2 in.)  bend x = x = = 7, 733.3333 psi (T) Ix 48 in.4 Bending stress at K due to My: M x ( 633, 600 lb  in.)(1.8 in.)  bend y = y = = 4, 693.3333 psi (T) Iy 243 in.4 Summary of stresses at K:  x = 0 psi

 z = −1, 277.7778 psi + 12, 426.6666 psi = 11,148.8889 psi = 11,150 psi (T)

 xz = 1,155 psi

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P15.31 For the cantilever beam shown in Figure P15.31a/32a, determine the normal and shear stresses acting at point K. The cross-sectional dimensions of the beam cross section and the location of point K are shown in Figure P15.31b/32b. Use the following values: a = 2.15 m, b = 0.85 m, Py = 13 kN, Pz = 6 kN.

FIGURE P15.31a/32a

FIGURE P15.31b/32b

Solution Section properties: A = 5,422 mm2 QH = 2(12 mm)(36 mm)(62 mm) = 53,568 mm3 QK = 2(12 mm)(54 mm)(53 mm) = 68,688 mm3

Moment of inertia about the z axis: Shape top flange web bottom flange

Width b (mm) 160 7 160

Height h (mm) 12 226 12

top flange web bottom flange

(mm) 160 7 160

d²A

(mm4) 27,212,160.0 6,733,519.3 27,212,160.0 61,157,839.3

d = z − zi

IC + d²A

(mm ) (mm) (mm ) 23,040.0 –119.0 27,189,120.0 6,733,519.3 0.0 0.0 23,040.0 119.0 27,189,120.0 Moment of inertia about the z axis (mm4) =

Moment of inertia about the y axis: Shape Width b Height h (mm) 12 226 12

d = y − yi

d²A

(mm4) (mm) (mm4) 4,096,000.0 0 0 6,459.8 0 0 4,096,000.0 0 0 Moment of inertia about the y axis (mm4) =

Equivalent forces at H and K: Fx = 0 kN Fy = 13 kN

(mm4) 4,096,000.0 6,459.8 4,096,000.0 8,198,459.8

Fz = 6 kN

M x = 0 kN-m M y = −(6 kN)(2.15 m) = −12.90 kN-m M z = (13 kN)(3 m) = 39.00 kN-m Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Shear stress magnitude at K due to Fz: (6,000 N)(68,688 mm3 )  xz = = 2.095 MPa (8,198,459.8 mm4 )(2  12 mm) Bending stress at K due to My: M y z (12.9  106 N-mm)(26 mm)  bend y = = = 40.910 MPa (C) Iy 8,198,459.8 mm4 Bending stress at K due to Mz: M z y (39.0  106 N-mm)(113 mm)  bend z = = = 72.059 MPa (T) Iz 61,157,839.3 mm4 Summary of stresses at K:  x = −40.910 MPa + 72.059 MPa = 31.149 MPa = 31.1 MPa (T)

 z = 0 MPa  xz = 2.095 MPa = 2.10 MPa

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P15.32 For the cantilever beam shown in Figure P15.31a/32a, determine the normal and shear stresses acting at point H. The cross-sectional dimensions of the beam cross section and the location of point H are shown in Figure P15.31b/32b. Use the following values: a = 2.15 m, b = 0.85 m, Py = 13 kN, Pz = 6 kN.

FIGURE P15.31a/32a

FIGURE P15.31b/32b

Solution Section properties: A = 5,422 mm2 QH = 2(12 mm)(36 mm)(62 mm) = 53,568 mm3 QK = 2(12 mm)(54 mm)(53 mm) = 68,688 mm3

Moment of inertia about the z axis: Shape top flange web bottom flange

Width b (mm) 160 7 160

Height h (mm) 12 226 12

top flange web bottom flange

(mm) 160 7 160

d²A

(mm4) 27,212,160.0 6,733,519.3 27,212,160.0 61,157,839.3

d = z − zi

IC + d²A

(mm ) (mm) (mm ) 23,040.0 –119.0 27,189,120.0 6,733,519.3 0.0 0.0 23,040.0 119.0 27,189,120.0 Moment of inertia about the z axis (mm4) =

Moment of inertia about the y axis: Shape Width b Height h (mm) 12 226 12

d = y − yi

d²A

(mm4) (mm) (mm4) 4,096,000.0 0 0 6,459.8 0 0 4,096,000.0 0 0 Moment of inertia about the y axis (mm4) =

Equivalent forces at H and K: Fx = 0 kN Fy = 13 kN

(mm4) 4,096,000.0 6,459.8 4,096,000.0 8,198,459.8

Fz = 6 kN

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Shear stress magnitude at H due to Fx: (6,000 N)(53,568 mm3 )  xz = = 1.633 MPa (8,198,459.8 mm4 )(2  12 mm) Bending stress at H due to My: M y z (12.9  106 N-mm)(44 mm)  bend y = = = 69.233 MPa (T) Iy 8,198,459.8 mm4 Bending stress at H due to Mz: M z y (39.0  106 N-mm)(125 mm)  bend z = = = 79.712 MPa (C) Iz 61,157,839.3 mm4 Summary of stresses at H:  x = 69.233 MPa − 79.712 MPa = −10.479 MPa = 10.48 MPa (C)

 z = 0 MPa  xz = 1.633 MPa

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P15.33 A circular tube with an outside diameter of 4.0 in. and a wall thickness of 0.083 in. is subjected to loads Px = 1,360 lb, Pz = 950 lb, and T = 8,420 lb·in. as shown in Figure P15.33. Using a = 6 in., determine the normal and shear stresses that exist at (a) point H and (b) point K.

FIGURE P15.33

Solution Section properties: A=

 4

4.00 in.) − ( 3.834 in.)  = 1.0214 in.2 ( 

Ix = Iz = J=

 

  64

4.00 in.) − ( 3.834 in.)  = 1.9597 in.4 (  4

4.00 in.) − ( 3.834 in.)  = 3.9194 in.4 (  4

32 1 3 3 Q = ( 4.00 in.) − ( 3.834 in.)  = 0.6368 in.3  12 

Equivalent forces at H and K: F = 0 lb Vx = 1,360 lb

M x = ( 950 lb )( 6 in.) = 5, 700 lb  in.

Vz = 950 lb

M y = 8, 420 lb  in. M z = − (1,360 lb )( 6 in.) = −8,160 lb  in.

(a) Determine the normal and shear stresses that exist at point H. Shear stress at H due to Vx: (1,360 lb ) ( 0.6368 in.3 ) H = = 2,662.3027 psi (1.9597 in.4 ) ( 4 in. − 3.834 in.) Shear stress at H due to Vz: QH = 0 in.3  H = 0 psi Shear stress at H due to My (i.e, torque T): ( 8, 420 lb  in.)( 4.0 in. / 2 ) = 4, 296.5361 psi H = 3.9194 in.4

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Bending stress at H due to Mx: M z ( 5,700 lb  in.)( 4 in. / 2 )  bend x = x = = 5,817.1628 psi (C) Ix 1.9597 in.4 Bending stress at H due to Mz: M x ( 8,160 lb  in.)( 0 in.)  bend z = z = = 0 psi Iz 1.9597 in.4 Summary of stresses at H:  x = 0 psi

 y = −5,817.1628 psi = −5,820 psi

 xy = 2, 662.3027 psi + 4, 296.5361 psi = 6,958.8387 psi = 6,960 psi

Ans.

(b) Determine the normal and shear stresses that exist at point K. Shear stress at K due to Vx: QK = 0 in.3  K = 0 psi Shear stress at K due to Vz: ( 950 lb ) ( 0.6368 in.3 ) K = = 1,859.6967 psi (1.9597 in.4 ) ( 4 in. − 3.834 in.)

Shear stress at K due to My (i.e, torque T): ( 8, 420 lb  in.)( 4.0 in. / 2 ) = 4, 296.5361 psi K = 3.9194 in.4 Bending stress at K due to Mx: M z ( 5,700 lb  in.)( 0 )  bend x = x = =0 Ix 1.9597 in.4 Bending stress at K due to Mz: M x ( 8,160 lb  in.)( 4 in. / 2 )  bend z = z = = 8,327.7279 psi (C) Iz 1.9597 in.4

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Summary of stresses at K:  z = 0 psi

 y = −8,327.7279 psi = −8,330 psi

 yz = 1,859.6967 psi − 4, 296.5361 psi = −2, 436.8394 psi = −2, 440 psi

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P15.34 Forces of Px = 580 lb and Py = 220 lb act on the teeth of the gear shown in Figure P15.34. The gear has a radius of R = 6 in., and the solid gear shaft has a diameter of 2.0 in. Using a = 5 in., determine the normal and shear stresses at (a) point H and (b) point K. Show the orientation of these stresses on an appropriate sketch.

FIGURE P15.34

Solution Section properties:  2 A = ( 2.0 in.) = 3.1416 in.2 4 J=

 32

( 2.0 in.) = 1.5708 in. 4

Ix = I y =

 64

( 2.0 in.) = 0.7854 in.4 4

( 2.0 in.) = 0.6667 in.3 Q= 3

Equivalent forces at H and K: Fx = −580 lb Fy = −220 lb Fz = 0 lb

Equivalent moments at H and K: M x = ( 220 lb )( 5 in.) = 1,100 lb  in. M y = − ( 580 lb )( 5 in.) = −2,900 lb  in. M z = ( 580 lb )( 6 in.) = 3, 480 lb  in.

Each of the non-zero forces and moments will be evaluated to determine whether stresses are created at the point of interest. (a) Determine the normal and shear stresses that exist at point H. Force Fx creates a shear stress at H. The magnitude of this shear stress is: ( 580 lb ) ( 0.6667 in.3 )  xz = = 246.1603 psi ( 0.7854 in.4 ) ( 2.0 in.) Force Fy does not cause either a normal stress or a shear stress at H. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Moment Mx, which is simply a bending moment, creates bending stress at H. The magnitude of this stress is: M y (1,100 lb  in.)( 2.0 in. / 2 ) z = x = = 1, 400.5636 psi Ix 0.7854 in.4 Moment My does not cause either a normal stress or a shear stress at H. Moment Mz, which is a torque, creates a torsion shear stress in the xz plane at H. The magnitude of this shear stress is: M c ( 3, 480 lb  in.)( 2.0 in. / 2 )  xz = z = = 2, 215.4316 psi J 1.5708 in.4 Summary of stresses at H:  x = 0 psi

 z = 1, 400.5636 psi = 1, 401 psi (T)  xz = −246.1603 psi − 2, 215.4369 psi = −2, 461.5966 psi = −2, 460 psi

Ans.

(b) Determine the normal and shear stresses that exist at point K. Force Fx does not cause either a normal stress or a shear stress at K. Force Fy creates a shear stress at K. The magnitude of this shear stress is: ( 220 lb ) ( 0.6667 in.3 )  yz = = 93.3709 psi ( 0.7854 in.4 ) ( 2.0 in.) Moment Mx does not cause either a normal stress or a shear stress at K. Moment My, which is simply a bending moment, creates bending stress at K. The magnitude of this stress is: M x ( 2,900 lb  in.)( 2.0 in. / 2 ) z = y = = 3, 692.3948 psi Iy 0.7854 in.4 Moment Mz, which is a torque, creates a torsion shear stress in the yz plane at K. The magnitude of this shear stress is: M c ( 3, 480 lb  in.)( 2.0 in. / 2 )  yz = z = = 2, 215.4316 psi J 1.5708 in.4 Summary of stresses at K:  y = 0 psi

 z = 3, 692.3948 psi = 3, 690 psi (T)  yz = −93.3709 psi + 2, 215.4369 psi = 2,122.0660 psi = 2,120 psi

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P15.35 A 7 in. diameter pulley is attached to a solid 1.0 in. diameter steel shaft. The tension forces in the pulley belt are T2 = 160 lb and T1 = 30 lb, each acting in the x′-y′ plane, as shown in Figure P15.35. Tension T2 is oriented at  = 30° with respect to the y′ axis. Using a = 7 in., determine the normal and shear stresses at (a) point H and (b) point K. Show the orientation of these stresses on an appropriate sketch.

FIGURE P15.35

Solution Section properties:  2 A = (1.0 in.) = 0.785398 in.2 4 J=

 32

(1.0 in.) = 0.098175 in.4 4

Ix = I y =

 64

(1.0 in.) = 0.049087 in.4 4

(1.0 in.) = 0.083333 in.3 3

Equivalent forces at H and K: Fx = T1 + T2 sin 

= 30 lb + (160 lb ) sin 30 = 110 lb

Fy = T2 cos  = (160 lb ) cos 30 = 138.5641 lb Fz = 0 lb

Equivalent moments at H and K: M x = − (138.5641 lb )( 7 in.) = −969.9485 lb  in. M y = (110 lb )( 7 in.) = 770 lb  in.  7 in.  M z = ( 30 lb − 160 lb )   = −455 lb  in.  2 

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Force Fy does not cause either a normal stress or a shear stress at H. Moment Mx, which is simply a bending moment, creates bending stress at H. The magnitude of this stress is: M y ( 969.9485 lb  in.)(1.0 in. / 2 ) z = x = = 9,879.8142 psi Ix 0.049087 in.4 Moment My does not cause either a normal stress or a shear stress at H. Moment Mz, which is a torque, creates a torsion shear stress in the xz plane at H. The magnitude of this shear stress is: M c ( 455 lb  in.)(1.0 in. / 2 )  xz = z = = 2,317.2961 psi J 0.098175 in.4 Summary of stresses at H:  x = 0 psi

 z = −9,879.8142 psi = −9,880 psi  xz = 186.7418 psi + 2,317.2961 psi = 2, 504.0379 psi = 2, 500 psi

Ans.

(b) Determine the normal and shear stresses that exist at point K. Force Fx does not cause either a normal stress or a shear stress at K. Force Fy creates a shear stress at K. The magnitude of this shear stress is: (138.5641 lb ) ( 0.083333 in.3 )  yz = = 235.2337 psi ( 0.049087 in.4 ) (1.0 in.) Moment Mx does not cause either a normal stress or a shear stress at K. Moment My, which is simply a bending moment, creates bending stress at K. The magnitude of this stress is: M x ( 770 lb  in.)(1.0 in. / 2 ) z = y = = 7,843.1559 psi Iy 0.049087 in.4 Moment Mz, which is a torque, creates a torsion shear stress in the yz plane at K. The magnitude of this shear stress is: M c ( 455 lb  in.)(1.0 in. / 2 )  yz = z = = 2,317.2961 psi J 0.098175 in.4 Summary of stresses at K:  y = 0 psi

 z = −7,843.1559 psi = −7,840 psi  yz = 235.2337 psi − 2,317.2961 psi = −2, 082.0624 psi = −2, 080 psi

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P15.36 A force of Pz = 780 N is exerted on control arm BC, which is securely attached to the solid 30 mm diameter shaft shown in Figure P15.36. Dimensions of the assembly are a = 150 mm and b = 120 mm. Determine the absolute maximum shear stress in the shaft at (a) point H and (b) point K.

FIGURE P15.36

Solution Section properties:



( 30 mm ) = 706.8583 mm2 2

( 30 mm ) = 2, 250 mm3

 32

( 30 mm ) = 79,521.5640 mm 4

I y = Iz =

 64

( 30 mm ) = 39, 760.7820 mm 4 4

Equivalent forces at H and K: Fx = 0 Fy = 0 Fz = 780 N

Equivalent moments at H and K: M x = − ( 780 N )(120 mm ) = −93, 600 N  mm M y = − ( 780 N )(150 mm ) = −117, 000 N  mm Mz = 0

Each of the non-zero forces and moments will be evaluated to determine whether stresses are created at the point of interest. (a) Consider point H. Force Fz creates a shear stress at H. The magnitude of this shear stress is: ( 780 N ) ( 2,250 mm3 )  xz = = 1.4713 MPa (39,760.7820 mm4 ) (30 mm ) Moment Mx, which is a torque, creates a torsion shear stress in the xz plane at H. The magnitude of this shear stress is: M c ( 93,600 N  mm )( 30 mm / 2 )  xz = x = = 17.6556 MPa J 79,521.5640 mm4 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Moment My does not create bending stress at H because H is located on the neutral axis for bending about the y axis. Summary of stresses at H:  x = 0 MPa

 z = 0 MPa  xz = 1.4713 MPa − 17.6556 MPa = −16.1843 MPa

Absolute maximum shear stress calculation: The principal stress magnitudes can be computed from Eq. (12.12):  p1 ,  p 2 = 0  16.1843 = 16.1843 MPa, − 16.1843 MPa

 p3 = 0  abs max =

16.1843 MPa − ( −16.1843 MPa ) = 16.18 MPa 2

Ans.

(b) Consider point K. Force Fz does not create normal or shear stress at K. Moment Mx, which is a torque, creates a torsion shear stress in the xy plane at K. The magnitude of this shear stress is: M c ( 93,600 N  mm )( 30 mm / 2 )  xy = x = = 17.6556 MPa J 79,521.5640 mm4 Moment My, which is simply a bending moment, creates bending stress at K. The magnitude of this stress is: M z (117, 000 N  mm )( 30 mm / 2 ) x = y = = 44.1390 MPa Iy 39, 760.7820 mm 4 Summary of stresses at K:  x = −44.1390 MPa

 y = 0 MPa  xy = 17.6556 MPa Absolute maximum shear stress calculations: The principal stress magnitudes can be computed from Eq. (12.12):

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

( −44.1390 ) + ( 0 )   ( −44.1390 ) − ( 0 )  + 17.6556 2  p1 ,  p 2 = )   ( 2





= −22.0695  28.2627 = 6.1932 MPa, − 50.3322 MPa

 p3 = 0  abs max =

6.1932 MPa − ( −50.3322 MPa ) = 28.3 MPa 2

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P15.37 Solid shaft AB of the pulley assembly shown in Figure P15.37/38 has a diameter of 35 mm. The length dimensions of the assembly are a = 80 mm, b = 150 mm, and c = 115 mm, and the diameter of the pulley is 130 mm. The pulley belt tension is P = 950 N. For point H on top of the shaft, determine the principal stresses and the maximum in-plane shear stress.

FIGURE P15.37/38 Solution Section properties:



( 35 mm ) = 962.1128 mm 2 2

 32

( 35 mm ) = 3,572.9167 mm3

( 35 mm ) = 147,323.5149 mm 4 4

I y = Iz =

 64

( 35 mm ) = 73, 661.7574 mm 4 4

Equivalent forces at H and K: Fx = 0 Fy = 950 N Fz = 950 N

Equivalent moments at H and K: M x = − ( 950 N )(115 mm ) = −109, 250 N  mm M y = − ( 950 N )( 230 mm ) = −218,500 N  mm M z = ( 950 N )( 230 mm ) = 218,500 N  mm

Each of the non-zero forces and moments will be evaluated to determine whether stresses are created at the point of interest. Stresses at point H. Force Fy does not cause either a normal stress or a shear stress at H.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Force Fz creates a shear stress at H. The magnitude of this shear stress is: ( 950 N ) (3,572.9167 mm3 )  xz = = 1.3165 MPa ( 73,661.7574 mm4 ) (35 mm ) Moment Mx, which is a torque, creates a torsion shear stress in the xz plane at H. The magnitude of this shear stress is: M c (109, 250 N  mm )( 35 mm / 2 )  xz = x = = 12.9774 MPa J 147,323.5149 mm4 Moment My does not create bending stress at H because H is located on the neutral axis for bending about the y axis. Moment Mz, which is simply a bending moment, creates bending stress at H. The magnitude of this stress is: M y ( 218,500 N  mm )( 35 mm / 2 ) x = z = = 51.9096 MPa Iz 73, 661.7574 mm4

Summary of stresses at H:  x = −51.9096 MPa

 z = 0 MPa  xz = 1.3165 MPa − 12.9974 MPa = −11.6608 MPa Principal stress calculations: The principal stress magnitudes can be computed from Eq. (12.12):

( −51.9096 ) + ( 0 )   ( −51.9096 ) − ( 0 )  + −11.6608 2  p1 ,  p 2 = )   ( 2





= −25.9548  28.4539

 p1 = 2.50 MPa

and

 p 2 = −54.4 MPa

Ans.

 max = 28.5 MPa

( maximum in-plane shear stress )

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P15.38 Solid shaft AB of the pulley assembly shown in Figure P15.37/38 has a diameter of 1.75 in. The length dimensions of the assembly are a = 3.0 in., b = 5.5 in., and c = 4.5 in., and the diameter of the pulley is 5 in. The pulley belt tension is P = 250 lb. Determine the absolute maximum shear stress for point K, which is located on the side of the shaft.

FIGURE P15.37/38 Solution Section properties:



(1.75 in.) = 2.40528 in.2 2

(1.75 in.) = 0.44661 in.3

 32

(1.75 in.) = 0.92077 in.4

I y = Iz =

 64

(1.75 in.) = 0.46039 in.4 4

Equivalent forces at H and K: Fx = 0 Fy = 250 lb Fz = 250 lb

Equivalent moments at H and K: M x = − ( 250 lb )( 4.5 in.) = −1,125 lb  in. M y = − ( 250 lb )( 8.5 in.) = −2,125 lb  in. M z = ( 250 lb )( 8.5 in.) = 2,125 lb  in.

Each of the non-zero forces and moments will be evaluated to determine whether stresses are created at the point of interest. Stresses at K. Force Fy creates a shear stress at K. The magnitude of this shear stress is: ( 250 lb ) ( 0.44661 in.3 )  xy = = 138.5839 psi ( 0.46039 in.4 ) (1.75 in.) Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Force Fz does not cause either a normal stress or a shear stress at K. Moment Mx, which is a torque, creates a torsion shear stress in the xy plane at K. The magnitude of this shear stress is: M c (1,125 lb  in.)(1.75 in. / 2 )  xy = x = = 1, 069.0758 psi J 0.92077 in.4 Moment My, which is simply a bending moment, creates bending stress at K. The magnitude of this stress is: M z ( 2,125 lb  in.)(1.75 in. / 2 ) x = y = = 4, 038.7309 psi Iy 0.46039 in.4 Moment Mz does not create bending stress at K because K is located on the neutral axis for bending about the z axis.

Summary of stresses at K:  x = −4, 038.7309 psi

 y = 0 psi  xy = 138.5839 psi + 1, 069.0758 psi = 1, 207.6597 psi Principal stress calculations: The principal stress magnitudes can be computed from Eq. (12.12):

( −4, 038.7309 ) + ( 0 )   ( −4, 038.7309 ) − ( 0)  + 1, 207.6597 2  p1 ,  p 2 = )   ( 2

 = −2, 019.3654  2,352.9298 2



= 333.5644 psi, − 4,372.2952 psi

 p3 = 0  abs max =

333.5644 psi − ( −4,372.2952 psi ) = 2,350 psi 2

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P15.39 The tube shown in Figure P15.39/40 has an outside diameter of 3.50 in., a wall thickness of 0.12 in., and length dimensions of a = 15 in. and b = 7 in. The load magnitudes are Px = 850 lb, Py = 375 lb, and Pz = 550 lb. Determine the principal stresses on the side of the tube at point K, and show these results on a properly oriented stress element.

FIGURE P15.39/40

Solution Section properties:



( 4

 

3.50 in.) − ( 3.26 in.)  = 1.274230 in.2 

( 32 

 

1  3 3 ( 3.50 in.) − ( 3.26 in.)  = 0.685752 in.3  12

Ix = I y =

( 64 

3.50 in.) − ( 3.26 in.)  = 1.821958 in.4  4

3.50 in.) − ( 3.26 in.)  = 3.643915 in.4  4

Equivalent forces at H and K: Fx = 850 lb Fy = −375 lb Fz = 550 lb

Equivalent moments at H and K: M x = ( 375 lb )(15 in.) − ( 550 lb )( 7 in.) = 1, 775 lb  in. M y = ( 850 lb )(15 in.) = 12, 750 lb  in. M z = ( 850 lb )( 7 in ) = 5,950 lb  in.

Each of the non-zero forces and moments will be evaluated to determine whether stresses are created at the point of interest.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Consider point K. Force Fx does not cause either a normal stress or a shear stress at K. Force Fy creates a transverse shear stress in the yz plane at K. The magnitude of this shear stress is: ( 375 lb ) ( 0.685752 in.3 )  yz = = 588.0968 psi (1.821958 in.4 ) (3.50 in.) − ( 3.26 in.) Force Fz creates an axial stress at K. The magnitude of this normal stress is: 550 lb z = = 431.6332 psi 1.274230 in.2 Moment Mx does not create bending stress at K because K is located on the neutral axis for bending about the x axis. Moment My creates bending stress at K. The magnitude of this stress is: M x (12, 750 lb  in.)( 3.50 in. / 2 ) z = y = = 12, 246.4427 psi Iy 1.821958 in.4 Moment Mz, which is a torque, creates a torsion shear stress in the yz plane at K. The magnitude of this shear stress is: M c ( 5,950 lb  in.)( 3.50 in. / 2 )  yz = z = = 2,857.5003 psi J 3.643915 in.4 Summary of stresses at K: y = 0

 z = 431.6332 psi − 12, 246.4427 psi = −11,814.8094 psi

 yz = −588.0968 psi + 2,857.5033 psi = 2, 269.4065 psi

Principal stress calculations:

( −11,814.8094 ) + ( 0 )   ( −11,814.8094 ) − ( 0 )  + 2, 269.4065 2  p1 ,  p 2 = )   ( 2

 = −5,907.4047  6,328.3202 2

 p1 = 421 psi

and

 p 2 = −12, 240 psi



Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P15.40 The tube shown in Figure P15.39/40 has an outside diameter of 3.50 in., a wall thickness of 0.12 in., and length dimensions of a = 15 in. and b = 7 in. The load magnitudes are Px = 850 lb, Py = 375 lb, and Pz = 550 lb. Determine the principal stresses on the underside of the tube at point H, and show these results on a properly oriented stress element.

FIGURE P15.39/40

Solution Section properties:



( 4

 

3.50 in.) − ( 3.26 in.)  = 1.274230 in.2 

( 32 

 

1  3 3 ( 3.50 in.) − ( 3.26 in.)  = 0.685752 in.3  12

Ix = I y =

( 64 

3.50 in.) − ( 3.26 in.)  = 1.821958 in.4  4

3.50 in.) − ( 3.26 in.)  = 3.643915 in.4  4

Equivalent forces at H and K: Fx = 850 lb Fy = −375 lb Fz = 550 lb

Equivalent moments at H and K: M x = ( 375 lb )(15 in.) − ( 550 lb )( 7 in.) = 1, 775 lb  in. M y = ( 850 lb )(15 in.) = 12, 750 lb  in. M z = ( 850 lb )( 7 in ) = 5,950 lb  in.

Consider point H. Each of the non-zero forces and moments will be evaluated to determine whether stresses are created at the point of interest.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Force Fx creates a transverse shear stress in the xz plane at H. The magnitude of this shear stress is: (850 lb ) ( 0.685752 in.3 )  xz = = 1,333.0195 psi (1.821958 in.4 ) (3.50 in.) − (3.26 in.) Force Fy does not cause either a normal stress or a shear stress at H. Force Fz creates an axial stress at H. The magnitude of this normal stress is: 550 lb z = = 431.6332 psi 1.274230 in.2 Moment Mx creates bending stress at H. The magnitude of this stress is: M y (1, 775 lb  in.)( 3.50 in. / 2 ) z = x = = 1, 704.8969 psi Ix 1.821958 in.4 Moment My does not create bending stress at H because H is located on the neutral axis for bending about the y axis. Moment Mz, which is a torque, creates a torsion shear stress in the xz plane at H. The magnitude of this shear stress is: M c ( 5,950 lb  in.)( 3.50 in. / 2 )  xz = z = = 2,857.5033 psi J 3.643915 in.4 Summary of stresses at H:  x = 0 psi

 z = 431.6332 psi − 1, 704.8969 psi = −1, 273.2637 psi

 xz = 1,333.0195 psi + 2,857.5033 psi = 4,190.5227 psi Principal stress calculations:

( −1, 273.2637 ) + ( 0 )   ( −1, 273.2637 ) − ( 0 )  + 4,190.5227 2  p1 ,  p 2 = )   ( 2

 = −636.6318  4, 238.6060 2

 p1 = 3,600 psi

and

 p 2 = −4,880 psi



Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P15.41 A solid steel shaft with an outside diameter of 25 mm is supported in flexible bearings at A and D as shown in Figure P15.41/42. Two pulleys are keyed to the shaft at B and C. Pulley B has a diameter of DB = 350 mm and belt tensions of T1 = 842 N and T2 = 234 N. Pulley C has a diameter of DC = 160 mm and belt tensions of T3 = 1,660 N and T4 = 330 N. Length dimensions of the shaft are x1 = 250 mm, x2 = 120 mm, x3 = 80 mm, and x4 = 160 mm. Determine the principal stresses and the absolute maximum shear stress at point H on the shaft. FIGURE P15.41/42

Solution Section properties:



( 25 mm ) = 490.874 mm2 2

( 25 mm ) = 1,302.083 mm3

 32

( 25 mm ) = 38,349.520 mm 4

I y = Iz =

 64

( 25 mm ) = 19,174.760 mm 4 4

Equilibrium of entire shaft: (Since no forces act in the z direction, the z reaction forces have been omitted from the FBD.) M A, z axis = − (T1 + T2 ) x1 + (T3 + T4 )( x1 + x2 + x3 ) + Dy ( x1 + x2 + x3 + x4 ) = 0 Dy = Dy =

(T1 + T2 ) x1 − (T3 + T4 )( x1 + x2 + x3 ) x1 + x2 + x3 + x4

( 842 N + 234 N )( 250 mm ) − (1, 660 N + 330 N )( 250 mm + 120 mm + 80 mm ) 250 mm + 120 mm + 80 mm + 160 mm Dy = −1, 027.0492 N

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Fy = Ay + Dy − T1 − T2 + T3 + T4 = 0 Ay = T1 + T2 − T3 − T4 − Dy Ay = 842 N + 234 N − 1, 660 N − 330 N − ( −1, 027.0492 N ) Ay = 113.0492 N

Detail of FBD at H and K: Equivalent Forces

Fx = 0 Fy = T3 + T4 + Dy = 1, 660 N + 330 N + ( −1, 027.0492 N ) = 962.9508 N Fz = 0

Detail of FBD at H and K: Equivalent Moments

Mx =

(T4 − T3 ) DC

2 ( 330 N ) − (1, 660 N )  (160 mm )

2 = −106, 400 N  mm

My = 0 M z = Dy ( x3 + x4 ) + (T3 + T4 ) x3 = ( −1, 027.0492 N )( 80 mm + 160 mm ) + (1, 660 N + 330 N )( 80 mm )

= −87, 291.8033 N  mm Each of the non-zero forces and moments will be evaluated to determine whether stresses are created at the point of interest. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Consider point H. Force Fy creates a transverse shear stress in the xy plane at H. The magnitude of this shear stress is: ( 962.9508 N ) 1,302.083 mm3  xy = = 2.6156 MPa 19,174.760 mm3 ( 25 mm )

(

)

Moment Mx, which is a torque, creates a torsion shear stress in the xy plane at H. The magnitude of this shear stress is: M c (106, 400 N  mm )( 25 mm / 2 )  xy = x = = 34.6810 MPa J 38,349.520 mm4 Moment Mz does not create bending stress at H because H is located on the neutral axis for bending about the z axis. Summary of stresses at H:  x = 0 MPa

 y = 0 MPa  xy = 2.6156 MPa + 34.6810 MPa = 37.2966 MPa Principal stress calculations: The principal stress magnitudes can be computed from Eq. (12.12).

( 0 ) + ( 0 )   ( 0 ) − ( 0 )  + 37.2966 2  p1 ,  p 2 = )   ( 2



= 0  37.2966  p1 = 37.3 MPa and

 abs max =



 p 2 = −37.3 MPa

37.3 MPa − ( −37.3 MPa ) = 37.3 MPa 2

Ans. Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P15.42 A solid steel shaft with an outside diameter of 25 mm is supported in flexible bearings at A and D as shown in Figure P15.41/42. Two pulleys are keyed to the shaft at B and C. Pulley B has a diameter of DB = 350 mm and belt tensions of T1 = 842 N and T2 = 234 N. Pulley C has a diameter of DC = 160 mm and belt tensions of T3 = 1,660 N and T4 = 330 N. Length dimensions of the shaft are x1 = 250 mm, x2 = 120 mm, x3 = 80 mm, and x4 = 160 mm. Determine the principal stresses and the absolute maximum shear stress at point K on the shaft. FIGURE P15.41/42

Solution Section properties:



( 25 mm ) = 490.874 mm2 2

( 25 mm ) = 1,302.083 mm3

 32

( 25 mm ) = 38,349.520 mm 4

I y = Iz =

 64

( 25 mm ) = 19,174.760 mm 4 4

(T1 + T2 ) x1 − (T3 + T4 )( x1 + x2 + x3 ) x1 + x2 + x3 + x4

( 842 N + 234 N )( 250 mm ) − (1, 660 N + 330 N )( 250 mm + 120 mm + 80 mm ) 250 mm + 120 mm + 80 mm + 160 mm Dy = −1, 027.0492 N

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Fy = Ay + Dy − T1 − T2 + T3 + T4 = 0 Ay = T1 + T2 − T3 − T4 − Dy Ay = 842 N + 234 N − 1, 660 N − 330 N − ( −1, 027.0492 N ) Ay = 113.0492 N

Detail of FBD at H and K: Equivalent Forces

Fx = 0 Fy = T3 + T4 + Dy = 1, 660 N + 330 N + ( −1, 027.0492 N ) = 962.9508 N Fz = 0

Detail of FBD at H and K: Equivalent Moments

Mx =

(T4 − T3 ) DC

2 ( 330 N ) − (1, 660 N )  (160 mm )

2 = −106, 400 N  mm

My = 0 M z = Dy ( x3 + x4 ) + (T3 + T4 ) x3 = ( −1, 027.0492 N )( 80 mm + 160 mm ) + (1, 660 N + 330 N )( 80 mm )

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Consider point K. Force Fy does not create a normal or a shear stress at point K. Moment Mx, which is a torque, creates a torsion shear stress in the xz plane at K. The magnitude of this shear stress is: M c (106, 400 N  mm )( 25 mm / 2 )  xz = x = = 34.6810 MPa J 38,349.520 mm4 Moment Mz creates bending stress at K. The magnitude of this stress is: M y ( 87, 291.8033 N  mm )( 25 mm / 2 ) x = z = = 56.9054 MPa Iz 19,174.760 mm 4

Summary of stresses at K:  x = −56.9054 MPa

 z = 0 MPa  xz = 34.6810 MPa Principal stress calculations: The principal stress magnitudes can be computed from Eq. (12.12).

( −56.9054 ) + ( 0 )   ( −56.9054 ) − ( 0 )  + 34.6810 2  p1 ,  p 2 = )   ( 2





= −28.4527  44.8590  p1 = 16.41 MPa and  p 2 = −73.3 MPa

Ans.

 abs max =

Ans.

16.41 MPa − ( −73.3 MPa ) = 44.9 MPa 2

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P15.43 A solid steel shaft with an outside diameter of 1.5 in. is supported in flexible bearings at A and C as shown in Figure P15.43/44. Two pulleys are keyed to the shaft at B and D. Pulley B has a diameter of DB = 14 in. and belt tensions of T1 = 100 lb and T2 = 220 lb. Pulley D has a diameter of DD = 7 in. and belt tensions of T3 = 60 lb and T4 = 300 lb. Length dimensions of the shaft are x1 = 8 in., x2 = 4 in., x3 = 3 in., and x4 = 6 in. Determine the normal and shear stresses at point K on the shaft. Show the orientation of these stresses on an appropriate sketch. FIGURE P15.43/44

Solution Section properties:



(1.5 in.) = 1.767146 in.2 2

(1.5 in.) = 0.281250 in.3

 32

(1.5 in.) = 0.497010 in.4 4

Equilibrium of entire shaft: M A, z axis = − (T1 + T2 ) x1 + C y ( x1 + x2 + x3 ) = 0

(T + T ) x Cy = 1 2 1 x1 + x2 + x3

(100 lb + 220 lb )(8 in.) C = y

8 in. + 4 in. + 3 in.

I y = Iz =

 64

(1.5 in.) = 0.248505 in.4 4

Fy = Ay + C y − T1 − T2 = 0 Ay = T1 + T2 − C y Ay = 100 lb + 220 lb − 170.6667 lb Ay = 149.3333 lb

C y = 170.6667 lb

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

M A, y axis = − (T3 + T4 )( x1 + x2 + x3 + x4 ) − C z ( x1 + x2 + x3 ) = 0 Cz = − Cz = −

Fz = Az + Cz + T3 + T4 = 0

(T3 + T4 )( x1 + x2 + x3 + x4 )

Az = −T3 − T4 − Cz

x1 + x2 + x3

Az = −60 lb − 300 lb − ( −504 lb )

( 60 lb + 300 lb )(8 in. + 4 in. + 3 in. + 6 in.) 8 in. + 4 in. + 3 in.

Az = 144 lb Cz = −504 lb

Detail of FBD at H and K: Equivalent Forces

Fx = 0 Fy = C y = 170.6667 lb Fz = Cz + T3 + T4 = −504 lb + 60 lb + 300 lb = −144 lb

Detail of FBD at H and K: Equivalent Moments

Mx =

(T4 − T3 ) DD = ( 300 lb ) − ( 60 lb ) ( 7 in.)

2 = 840 lb  in.

M y = −Cz x3 − (T3 + T4 )( x3 + x4 ) = − ( −504 lb )( 3 in.) − ( 60 lb + 300 lb )( 3 in. + 6 in.) = −1, 728 lb  in. M z = C y ( x3 ) = (170.6667 lb )( 3 in.) = 512 lb  in.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Each of the non-zero forces and moments will be evaluated to determine whether stresses are created at the point of interest. Consider point K. Force Fy does not create a normal or a shear stress at point K. Force Fz creates a transverse shear stress in the xz plane at K. The magnitude of this shear stress is: (144 lb ) 0.281250 in.3  xz = = 108.6498 psi 0.248505 in.4 (1.5 in.)

(

)

Moment Mx, which is a torque, creates a torsion shear stress in the xz plane at K. The magnitude of this shear stress is: M c ( 840 lb  in.)(1.5 in. / 2 )  xz = x = = 1, 267.5808 psi J 0.497010 in.4 Moment My does not create bending stress at K because K is located on the neutral axis for bending about the y axis. Moment Mz creates bending stress at K. The magnitude of this stress is: M y ( 512 lb  in.)(1.5 in. / 2 ) x = z = = 1,545.2413 psi Iz 0.248505 in.4 Summary of stresses at K:  x = 1,545.2413 psi = 1,545 psi

 z = 0 psi  xz = −108.6498 psi − 1, 267.5808 psi = −1,376.2305 psi = −1,376 psi

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P15.44 A solid steel shaft with an outside diameter of 1.5 in. is supported in flexible bearings at A and C as shown in Figure P15.43/44. Two pulleys are keyed to the shaft at B and D. Pulley B has a diameter of DB = 14 in. and belt tensions of T1 = 100 lb and T2 = 220 lb. Pulley D has a diameter of DD = 7 in. and belt tensions of T3 = 60 lb and T4 = 300 lb. Length dimensions of the shaft are x1 = 8 in., x2 = 4 in., x3 = 3 in., and x4 = 6 in. Determine the normal and shear stresses at point H on the shaft. Show the orientation of these stresses on an appropriate sketch. FIGURE P15.43/44

Solution Section properties:



(1.5 in.) = 1.767146 in.2 2

(1.5 in.) = 0.281250 in.3

 32

(1.5 in.) = 0.497010 in.4 4

Equilibrium of entire shaft: M A, z axis = − (T1 + T2 ) x1 + C y ( x1 + x2 + x3 ) = 0

(T + T ) x Cy = 1 2 1 x1 + x2 + x3

(100 lb + 220 lb )(8 in.) C = y

8 in. + 4 in. + 3 in.

I y = Iz =

 64

(1.5 in.) = 0.248505 in.4 4

Fy = Ay + C y − T1 − T2 = 0 Ay = T1 + T2 − C y Ay = 100 lb + 220 lb − 170.6667 lb Ay = 149.3333 lb

C y = 170.6667 lb

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

M A, y axis = − (T3 + T4 )( x1 + x2 + x3 + x4 ) − C z ( x1 + x2 + x3 ) = 0 Cz = − Cz = −

Fz = Az + Cz + T3 + T4 = 0

(T3 + T4 )( x1 + x2 + x3 + x4 )

Az = −T3 − T4 − Cz

x1 + x2 + x3

Az = −60 lb − 300 lb − ( −504 lb )

( 60 lb + 300 lb )(8 in. + 4 in. + 3 in. + 6 in.) 8 in. + 4 in. + 3 in.

Az = 144 lb Cz = −504 lb

Detail of FBD at H and K: Equivalent Forces

Fx = 0 Fy = C y = 170.6667 lb Fz = Cz + T3 + T4 = −504 lb + 60 lb + 300 lb = −144 lb

Detail of FBD at H and K: Equivalent Moments

Mx =

(T4 − T3 ) DD = ( 300 lb ) − ( 60 lb ) ( 7 in.)

2 = 840 lb  in.

M y = −Cz x3 − (T3 + T4 )( x3 + x4 ) = − ( −504 lb )( 3 in.) − ( 60 lb + 300 lb )( 3 in. + 6 in.) = −1, 728 lb  in. M z = C y ( x3 ) = (170.6667 lb )( 3 in.) = 512 lb  in.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

Each of the non-zero forces and moments will be evaluated to determine whether stresses are created at the point of interest. Consider point H. Force Fy creates a transverse shear stress in the xy plane at H. The magnitude of this shear stress is: (170.6667 lb ) 0.281250 in.3  xy = = 128.7701 psi 0.248505 in.4 (1.5 in.)

(

)

Force Fz does not create a normal or a shear stress at point H. Moment Mx, which is a torque, creates a torsion shear stress in the xy plane at H. The magnitude of this shear stress is: M c ( 840 lb  in.)(1.5 in. / 2 )  xy = x = = 1, 267.5808 psi J 0.497010 in.4 Moment My creates bending stress at H. The magnitude of this stress is: M z (1, 728 lb  in.)(1.5 in. / 2 ) x = y = = 5, 215.1894 psi Iy 0.248505 in.4 Moment Mz does not create bending stress at H because H is located on the neutral axis for bending about the z axis. Summary of stresses at H:  x = −5, 215.1894 psi = −5, 220 psi

 y = 0 psi  xy = 128.7701 psi − 1, 267.5808 psi = −1,138.8106 psi = −1,139 psi

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed. P15.45 A stainless steel pipe (Figure P15.45) with an outside diameter of 2.375 in. and a wall thickness of 0.109 in. is subjected to a bending moment of M = 50 lb·ft and an internal pressure of 180 psi. Determine the absolute maximum shear stress on the outer surface of the pipe.

Timothy A. Philpot

FIGURE P15.45

Solution Section properties: d = 2.375 in. − 2 ( 0.109 in.) = 2.157 in.



 

2.375 in.) − ( 2.157 in.)  = 0.775955 in.2 ( 

2.375 in.) − ( 2.157 in.)  = 0.499195 in.4 ( 

Bending moment M creates a normal stress of: Mc ( 50 lb  ft )( 2.375 in. / 2 )(12 in./ft ) = = = 1, 427.2993 psi I 0.499195 in.4 Stresses due to internal pressure: The 180 psi internal fluid pressure creates tension normal stresses in the wall of the pipe. The longitudinal stress in the pipe wall is: pd (180 psi )( 2.157 in.)  long = = = 890.5046 psi (T) 4t 4 ( 0.109 in.) and the circumferential stress is: pd (180 psi )( 2.157 in.)  hoop = = = 1, 781.0092 psi (T) 2t 2 ( 0.109 in.) Summary of stresses: Let the longitudinal axis of the pipe be denoted by x and the circumferential direction be denoted by y. The normal and shear stresses at a point on the bottom of the pipe (on the outside surface) are thus:  x = 1, 427.2993 psi + 890.5046 psi = 2,317.8039 psi

 y = 1, 781.0092 psi  xy = 0 psi Since xy = 0, the stresses x and y are principal stresses. The outer surface of the pipe is in plane stress, and thus, the third principal stress is zero:  p1 = 2,317.8039 psi

 p 2 = 1, 781.0092 psi  p3 = 0 The absolute maximum shear stress is accordingly:  −  min 2,317.8039 psi − 0  abs max = max = = 1,159 psi 2

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P15.46 The piping assembly shown in Figure P15.46/47/48/49 consists of stainless steel pipe that has an outside diameter of 6.675 in. and a wall thickness of 0.28 in. The assembly is subjected to concentrated loads Px = 320 lb, Py = 410 lb, and Pz = 180 lb as well as an internal fluid pressure of 100 psi that acts in all pipes. Dimensions of the assembly are y1 = 2.0 ft, y2 = 4.5 ft, z1 = 3.5 ft, and z2 = 3.0 ft. Determine the normal and shear stresses on the outer surface of the pipe at (a) point A and (b) point D. Show these stresses on an appropriate sketch.

FIGURE P15.46/47/48/49

Solution Section properties: A=



 

6.675 in.) − ( 6.115 in.)  = 5.6253 in.2 ( 

6.675 in.) − ( 6.115 in.)  = 57.6237 in.4 ( 

 

6.675 in.) − ( 6.115 in.)  = 28.8119 in.4 (  4

64 1 3 3 Q = ( 6.675 in.) − ( 6.115 in.)  = 5.7291 in.3   12

Detail of FBD at A and B: Equivalent Forces

Detail of FBD at A and B: Equivalent Moments

Mechanics of Materials: An Integrated Learning System, 4th Ed. Fx = Px = 320 lb Fy = − Py = −410 lb Fz = Pz = 180 lb

Timothy A. Philpot

M x = Py z2 + Pz y2 = ( 410 lb )( 3.0 ft ) + (180 lb )( 4.5 ft ) = 2, 040 lb  ft = 24, 480 lb  in. M y = Px z2 = ( 320 lb )( 3.0 ft ) = 960 lb  ft = 11,520 lb  in. M z = − Px y2

= − ( 320 lb )( 4.5 ft ) = −1, 440 lb  ft = −17, 280 lb  in.

Detail of FBD at C and D: Equivalent Forces

Detail of FBD at C and D: Equivalent Moments

Fx = Px = 320 lb

M x = Py ( z1 + z2 ) + Pz ( y1 + y2 )

Fy = − Py = −410 lb

= ( 410 lb )( 3.5 ft + 3.0 ft )

Fz = Pz = 180 lb

+ (180 lb )( 2.0 ft + 4.5 ft ) = 3,835 lb  ft = 46, 020 lb  in. M y = Px ( z1 + z2 ) = ( 320 lb )( 3.5 ft + 3.0 ft ) = 2, 080 lb  ft = 24,960 lb  in. M z = − Px ( y1 + y2 ) = − ( 320 lb )( 2.0 ft + 4.5 ft ) = −2, 080 lb  ft = −24,960 lb  in.

Stresses due to internal pressure: The 100 psi internal fluid pressure creates tensile normal stresses in the 0.28 in. thick wall of the pipe. The longitudinal stress in the pipe wall is: pd (100 psi )( 6.115 in.)  long = = = 545.9821 psi (T) 4t 4 ( 0.28 in.) and the circumferential stress is: pd (100 psi )( 6.115 in.)  hoop = = = 1, 091.9643 psi (T) 2t 2 ( 0.28 in.) Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

(a) Consider point A. Force Fx creates a transverse shear stress in the xy plane at A. The magnitude of this shear stress is: ( 320 lb ) 5.7291 in.3  xy = = 113.6259 psi 28.8119 in.4 ( 6.675 in. − 6.115 in.)

(

)

(

)

This shear stress acts in the positive x direction on the positive y face of the stress element. Force Fy creates an axial stress in the y direction at A. The magnitude of this compressive normal stress is: 410 lb y = = 72.8845 psi 5.6253 in.2 Force Fz does not create a normal or a shear stress at point A. Moment Mx creates bending stress that acts in the y direction at A. The magnitude of this compressive normal stress is: M z ( 24, 480 lb  in.)( 6.675 in. / 2 ) y = x = = 2,835.7071 psi Ix 28.8119 in.4 Moment My is a torque that creates shear stress in the xy plane at A. The magnitude of this shear stress is: M c (11,520 lb  in.)( 6.675 in. / 2 )  xy = y = = 667.2252 psi J 57.6237 in.4 This shear stress acts in the positive x direction on the positive y face of the stress element. Moment Mz does not create bending stress at A because A is located on the neutral axis for bending about the z axis. Stresses due to internal pressure: At A, the longitudinal stress acts in the y direction, and the hoop stress acts in the x direction. Summary of stresses at A:  x = 1, 091.9643 psi = 1, 092 psi

 y = −72.8845 psi − 2,835.7071 psi + 545.9821 psi = −2, 362.6095 psi = −2,360 psi

 xy = 113.6259 psi + 667.2252 psi = 780.8511 psi = 781 psi

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

(b) Consider point D. Force Fx does not create a normal or a shear stress at point D. Force Fy creates a transverse shear stress in the yz plane at D. The magnitude of this shear stress is: ( 410 lb ) 5.7291 in.3  yz = = 145.5831 psi 28.8119 in.4 ( 6.675 in. − 6.115 in.)

(

)

(

)

This shear stress acts in the negative y direction on the positive z face of the stress element. Force Fz creates an axial stress in the z direction at D. The magnitude of this tensile normal stress is: 180 lb z = = 31.9981 psi 5.6253 in.2 Moment Mx does not create bending stress at D because D is located on the neutral axis for bending about the x axis. Moment My creates bending stress in the z direction at D. The magnitude of this compressive normal stress is: M x ( 24,960 lb  in.)( 6.675 in. / 2 ) z = y = = 2,891.3092 psi Iy 28.8119 in.4 Moment Mz is a torque that creates shear stress in the yz plane at D. The magnitude of this shear stress is: M c ( 24,960 lb  in.)( 6.675 in. / 2 )  yz = z = = 1, 445.6546 psi J 57.6237 in.4 This shear stress acts in the negative y direction on the positive z face of the stress element. Stresses due to internal pressure: At D, the longitudinal stress acts in the z direction, and the hoop stress acts in the y direction. Summary of stresses at D:  y = 1, 091.9643 psi = 1, 092 psi

 z = 31.9981 psi − 2,891.3092 psi + 545.9821 psi = −2,313.3290 psi = −2,313 psi

 yz = −145.5831 psi − 1, 445.6546 psi = −1,591.2377 psi = −1,591 psi

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P15.47 The piping assembly shown in Figure P15.46/47/48/49 consists of stainless steel pipe that has an outside diameter of 6.675 in. and a wall thickness of 0.28 in. The assembly is subjected to concentrated loads Px = 320 lb, Py = 410 lb, and Pz = 180 lb as well as an internal fluid pressure of 100 psi that acts in all pipes. Dimensions of the assembly are y1 = 2.0 ft, y2 = 4.5 ft, z1 = 3.5 ft, and z2 = 3.0 ft. Determine the normal and shear stresses on the outer surface of the pipe at (a) point B and (b) point C. Show these stresses on an appropriate sketch.

FIGURE P15.46/47/48/49

Solution Section properties: A=



 

6.675 in.) − ( 6.115 in.)  = 5.6253 in.2 ( 

6.675 in.) − ( 6.115 in.)  = 57.6237 in.4 ( 

 

6.675 in.) − ( 6.115 in.)  = 28.8119 in.4 (  4

64 1 3 3 Q = ( 6.675 in.) − ( 6.115 in.)  = 5.7291 in.3   12

Detail of FBD at A and B: Equivalent Forces

Detail of FBD at A and B: Equivalent Moments

Mechanics of Materials: An Integrated Learning System, 4th Ed. Fx = Px = 320 lb Fy = − Py = −410 lb Fz = Pz = 180 lb

Timothy A. Philpot

M x = Py z2 + Pz y2 = ( 410 lb )( 3.0 ft ) + (180 lb )( 4.5 ft ) = 2, 040 lb  ft = 24, 480 lb  in. M y = Px z2 = ( 320 lb )( 3.0 ft ) = 960 lb  ft = 11,520 lb  in. M z = − Px y2

= − ( 320 lb )( 4.5 ft ) = −1, 440 lb  ft = −17, 280 lb  in.

Detail of FBD at C and D: Equivalent Forces

Detail of FBD at C and D: Equivalent Moments

Fx = Px = 320 lb

M x = Py ( z1 + z2 ) + Pz ( y1 + y2 )

Fy = − Py = −410 lb

= ( 410 lb )( 3.5 ft + 3.0 ft )

Fz = Pz = 180 lb

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

(a) Consider point B. Force Fx does not create a normal or a shear stress at point B. Force Fy creates an axial stress in the y direction at B. The magnitude of this compressive normal stress is: 410 lb y = = 72.8845 psi 5.6253 in.2 Force Fz creates a transverse shear stress in the yz plane at B. The magnitude of this shear stress is: (180 lb ) 5.7291 in.3  yz = = 63.9145 psi 28.8119 in.4 ( 6.675 in. − 6.115 in.)

(

)

(

)

This shear stress acts in the positive z direction on the positive y face of the stress element. Moment Mx does not create bending stress at B because B is located on the neutral axis for bending about the x axis. Moment My is a torque that creates shear stress in the yz plane at B. The magnitude of this shear stress is: M c (11,520 lb  in.)( 6.675 in. / 2 )  yz = y = = 667.2252 psi J 57.6237 in.4 This shear stress acts in the negative z direction on the positive y face of the stress element. Moment Mz creates bending stress that acts in the y direction at B. The magnitude of this compressive normal stress is: M x (17, 280 lb  in.)( 6.675 in. / 2 ) y = z = = 2, 001.6756 psi Iz 28.8119 in.4 Stresses due to internal pressure: At B, the longitudinal stress acts in the y direction, and the hoop stress acts in the z direction. Summary of stresses at B:  y = −72.8845 psi − 2, 001.6756 psi + 545.9821 psi = −1,528.5780 psi = −1,529 psi

 z = 1, 091.9643 psi = 1, 092 psi  yz = 63.9145 psi − 667.2252 psi = −603.3107 psi = −603 psi

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

(b) Consider point C. Force Fx creates a transverse shear stress in the xz plane at C. The magnitude of this shear stress is: ( 320 lb ) 5.7291 in.3  xz = = 113.6259 psi 28.8119 in.4 ( 6.675 in. − 6.115 in.)

(

)

(

)

This shear stress acts in the positive x direction on the positive z face of the stress element. Force Fy does not create a normal or a shear stress at point C. Force Fz creates an axial stress in the z direction at C. The magnitude of this tensile normal stress is: 180 lb z = = 31.9981 psi 5.6253 in.2 Moment Mx creates bending stress in the z direction at C. The magnitude of this tensile normal stress is: M y ( 46, 020 lb  in.)( 6.675 in. / 2 ) z = x = = 5,330.8514 psi Ix 28.8119 in.4 Moment My does not create bending stress at C because C is located on the neutral axis for bending about the y axis. Moment Mz is a torque that creates shear stress in the xz plane at C. The magnitude of this shear stress is: M c ( 24,960 lb  in.)( 6.675 in. / 2 )  xz = z = = 1, 445.6546 psi J 57.6237 in.4 This shear stress acts in the positive x direction on the positive z face of the stress element. Stresses due to internal pressure: At C, the longitudinal stress acts in the z direction, and the hoop stress acts in the x direction. Summary of stresses at C:  x = 1, 091.9643 psi = 1, 092 psi

 z = 31.9981 psi + 5,330.8514 psi + 545.9821 psi = 5,908.8316 psi = 5,910 psi

 xz = 113.6259 psi + 1, 445.6546 psi = 1,559.2805 psi = 1,559 psi

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P15.48 The piping assembly shown in Figure P15.46/47/48/49 consists of stainless steel pipe that has an outside diameter of 200 mm and a wall thickness of 8 mm. The assembly is subjected to concentrated loads Px = 2,400 N, Py = 0, and Pz = 1,100 N as well as an internal fluid pressure of 900 kPa that acts in all pipes. Dimensions of the assembly are y1 = 0.7 m, y2 = 1.8 m, z1 = 1.3 m, and z2 = 1.1 m. Determine the normal and shear stresses on the outer surface of the pipe at (a) point B and (b) point C. Show these stresses on an appropriate sketch.

FIGURE P15.46/47/48/49

Solution Section properties: A=



 

200 mm ) − (184 mm )  = 4,825.4863 mm 2 ( 

200 mm ) − (184 mm )  = 44,548,890 mm 4 ( 

 

200 mm ) − (184 mm )  = 22, 274, 445 mm 4 (  4

64 1 3 3 Q = ( 200 mm ) − (184 mm )  = 147,541.33 mm3   12

Detail of FBD at A and B: Equivalent Forces

Detail of FBD at A and B: Equivalent Moments

Mechanics of Materials: An Integrated Learning System, 4th Ed. Fx = Px = 2, 400 N Fy = − Py = 0 Fz = Pz = 1,100 N

Timothy A. Philpot

M x = Py z2 + Pz y2 = ( 0 )(1.1 m ) + (1,100 N )(1.8 m ) = 1,980 N  m = 1,980, 000 N  mm M y = Px z2 = ( 2, 400 N )(1.1 m ) = 2, 640 N  m = 2, 640, 000 N  mm M z = − Px y2

= − ( 2, 400 N )(1.8 m ) = −4,320 N  m = −4,320, 000 N  mm

Detail of FBD at C and D: Equivalent Forces

Fx = Px = 2, 400 N Fy = − Py = 0 Fz = Pz = 1,100 N

Detail of FBD at C and D: Equivalent Moments

M x = Py ( z1 + z2 ) + Pz ( y1 + y2 ) = ( 0 )(1.3 m + 1.1 m ) + (1,100 N )( 0.7 m + 1.8 m ) = 2, 750 N  m = 2, 750, 000 N  mm M y = Px ( z1 + z2 ) = ( 2, 400 N )(1.3 m + 1.1 m ) = 5, 760 N  m = 5, 760, 000 N  mm M z = − Px ( y1 + y2 ) = − ( 2, 400 N )( 0.7 m + 1.8 m ) = −6, 000 N  m = −6, 000, 000 N  mm

Stresses due to internal pressure: The 900 kPa internal fluid pressure creates tensile normal stresses in the 8 mm thick wall of the pipe. The longitudinal stress in the pipe wall is: pd ( 0.9 MPa )(184 mm )  long = = = 5.175 MPa (T) 4t 4 ( 8 mm ) and the circumferential stress is: pd ( 0.9 MPa )(184 mm )  hoop = = = 10.35 MPa (T) 2t 2 (8 mm ) Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

(a) Consider point B. Force Fx does not create a normal or a shear stress at point B. Force Fz creates a transverse shear stress in the yz plane at B. The magnitude of this shear stress is: (1,100 N ) 147,541.33 mm3  yz = = 0.4554 MPa 22, 274, 445 mm4 ( 200 mm − 184 mm )

(

)

This shear stress acts in the positive z direction on the positive y face of the stress element. Moment Mx does not create bending stress at B because B is located on the neutral axis for bending about the x axis. Moment My is a torque that creates shear stress in the yz plane at B. The magnitude of this shear stress is: M c ( 2,640,000 N  mm )( 200 mm / 2 )  yz = y = = 5.9261 MPa J 44,548,890 mm4 This shear stress acts in the negative z direction on the positive y face of the stress element. Moment Mz creates bending stress that acts in the y direction at B. The magnitude of this compressive normal stress is: M x ( 4,320, 000 N  mm )( 200 mm / 2 ) y = z = = 19.3944 MPa Iz 22, 274, 445 mm4 Stresses due to internal pressure: At B, the longitudinal stress acts in the y direction, and the hoop stress acts in the z direction. Summary of stresses at B:  y = −19.3944 MPa + 5.175 MPa = −14.2194 MPa = −14.22 MPa

 z = 10.35 MPa  yz = 0.4554 MPa − 5.9261 MPa = −5.4707 MPa = −5.47 MPa

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

(b) Consider point C. Force Fx creates a transverse shear stress in the xz plane at C. The magnitude of this shear stress is: ( 2, 400 N ) 147,541.33 mm3  xz = = 0.9936 MPa 22, 274, 445 mm4 ( 200 mm − 184 mm )

(

)

This shear stress acts in the positive x direction on the positive z face of the stress element. Force Fz creates an axial stress in the z direction at C. The magnitude of this tensile normal stress is: 1,100 N z = = 0.2280 MPa 4,825.4863 mm 2 Moment Mx creates bending stress in the z direction at C. The magnitude of this tensile normal stress is: M y ( 2, 750, 000 N  mm )( 200 mm / 2 ) z = x = = 12.3460 MPa Ix 22, 274, 445 mm4 Moment My does not create bending stress at C because C is located on the neutral axis for bending about the y axis. Moment Mz is a torque that creates shear stress in the xz plane at C. The magnitude of this shear stress is: M c ( 6,000,000 N  mm )( 200 mm / 2 )  xz = z = = 13.4683 MPa J 44,548,890 mm4 This shear stress acts in the positive x direction on the positive z face of the stress element. Stresses due to internal pressure: At C, the longitudinal stress acts in the z direction, and the hoop stress acts in the x direction. Summary of stresses at C:  x = 10.35 MPa

 z = 0.2280 MPa + 12.3460 MPa + 5.175 MPa = 17.7489 MPa = 17.75 MPa

 xz = 0.9936 MPa + 13.4683 MPa = 14.4619 MPa = 14.46 MPa

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P15.49 The piping assembly shown in Figure P15.46/47/48/49 consists of stainless steel pipe that has an outside diameter of 200 mm and a wall thickness of 8 mm. The assembly is subjected to concentrated loads Px = 2,400 N, Py = 0, and Pz = 1,100 N as well as an internal fluid pressure of 900 kPa that acts in all pipes. Dimensions of the assembly are y1 = 0.7 m, y2 = 1.8 m, z1 = 1.3 m, and z2 = 1.1 m. Determine the normal and shear stresses on the outer surface of the pipe at (a) point A and (b) point D. Show these stresses on an appropriate sketch.

FIGURE P15.46/47/48/49

Solution Section properties: A=



 

200 mm ) − (184 mm )  = 4,825.4863 mm 2 ( 

200 mm ) − (184 mm )  = 44,548,890 mm 4 ( 

 

200 mm ) − (184 mm )  = 22, 274, 445 mm 4 (  4

64 1 3 3 Q = ( 200 mm ) − (184 mm )  = 147,541.33 mm3   12

Detail of FBD at A and B: Equivalent Forces

Detail of FBD at A and B: Equivalent Moments

Mechanics of Materials: An Integrated Learning System, 4th Ed. Fx = Px = 2, 400 N Fy = − Py = 0 Fz = Pz = 1,100 N

Timothy A. Philpot

M x = Py z2 + Pz y2 = ( 0 )(1.1 m ) + (1,100 N )(1.8 m ) = 1,980 N  m = 1,980, 000 N  mm M y = Px z2 = ( 2, 400 N )(1.1 m ) = 2, 640 N  m = 2, 640, 000 N  mm M z = − Px y2

= − ( 2, 400 N )(1.8 m ) = −4,320 N  m = −4,320, 000 N  mm

Detail of FBD at C and D: Equivalent Forces

Fx = Px = 2, 400 N Fy = − Py = 0 Fz = Pz = 1,100 N

Detail of FBD at C and D: Equivalent Moments

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

(a) Consider point A. Force Fx creates a transverse shear stress in the xy plane at A. The magnitude of this shear stress is: ( 2, 400 N ) 147,541.33 mm3  xy = = 0.9936 MPa 22, 274, 445 mm4 ( 200 mm − 184 mm )

(

)

This shear stress acts in the positive x direction on the positive y face of the stress element. Force Fz does not create a normal or a shear stress at point A. Moment Mx creates bending stress that acts in the y direction at A. The magnitude of this compressive normal stress is: M z (1,980, 000 N  mm )( 200 mm / 2 ) y = x = = 8.8891 MPa Ix 22, 274, 445 mm4 Moment My is a torque that creates shear stress in the xy plane at A. The magnitude of this shear stress is: M c ( 2,640,000 N  mm )( 200 mm / 2 )  xy = y = = 5.9261 MPa J 44,548,890 mm4 This shear stress acts in the positive x direction on the positive y face of the stress element. Moment Mz does not create bending stress at A because A is located on the neutral axis for bending about the z axis. Stresses due to internal pressure: At A, the longitudinal stress acts in the y direction, and the hoop stress acts in the x direction. Summary of stresses at A:  x = 10.35 MPa

 y = −8.8891 MPa + 5.175 MPa = −3.7141 MPa = −3.71 MPa

 xy = 0.9936 MPa + 5.9261 MPa = 6.9196 MPa = 6.92 MPa

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

(b) Consider point D. Force Fx does not create a normal or a shear stress at point D. Force Fz creates an axial stress in the z direction at D. The magnitude of this tensile normal stress is: 1,100 N z = = 0.2280 MPa 4,825.4863 mm 2 Moment Mx does not create bending stress at D because D is located on the neutral axis for bending about the x axis. Moment My creates bending stress in the z direction at D. The magnitude of this compressive normal stress is: M x ( 5, 760, 000 N  mm )( 200 mm / 2 ) z = y = = 25.8592 MPa Iy 22, 274, 445 mm4 Moment Mz is a torque that creates shear stress in the yz plane at D. The magnitude of this shear stress is: M c ( 6,000,000 N  mm )( 200 mm / 2 )  yz = z = = 13.4683 MPa J 44,548,890 mm4 This shear stress acts in the negative y direction on the positive z face of the stress element. Stresses due to internal pressure: At D, the longitudinal stress acts in the z direction, and the hoop stress acts in the y direction.

Summary of stresses at D:  y = 10.35 MPa

 z = 0.2280 MPa − 25.8592 MPa + 5.175 MPa = −20.4562 MPa = −20.5 MPa

 yz = −13.4683 MPa = −13.47 MPa

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P15.50 The piping assembly shown in Figure P15.50/51 consists of stainless steel pipe that has an outside diameter of 275 mm and a wall thickness of 9 mm. The assembly is subjected to concentrated loads Px = 3.2 kN, Py = 5.4 kN, and Pz = 1.3 kN as well as an internal fluid pressure of 1,400 kPa that acts in all pipes. Dimensions of the assembly are x1 = 1.35 m, y1 = 1.0 m, y2 = 2.3 m, and z1 = 1.6 m. Determine the normal and shear stresses on the outer surface of the pipe at (a) point A and (b) point C. Show these stresses on an appropriate sketch.

FIGURE P15.50/51

Solution Section properties: A=



 

275 mm ) − ( 257 mm )  = 7,520.973 mm 2 ( 

275 mm ) − ( 257 mm )  = 133,190, 790 mm 4 ( 

 

275 mm ) − ( 257 mm )  = 66,595,394 mm 4 (  4

64 1 3 3 Q = ( 275 mm ) − ( 257 mm )  = 318,523.5 mm 3  12 

Detail of FBD at A and B: Equivalent Forces

Detail of FBD at A and B: Equivalent Moments

Mechanics of Materials: An Integrated Learning System, 4th Ed. Fx = Px = 3, 200 N Fy = − Py = −5, 400 N Fz = Pz = 1,300 N

Timothy A. Philpot

M x = Py z1 + Pz y2 = ( 5, 400 N )(1.6 m ) + (1,300 N )( 2.3 m ) = 11, 630 N  m = 11, 630, 000 N  mm M y = Px z1 = ( 3, 200 N )(1.6 m ) = 5,120 N  m = 5,120, 000 N  mm M z = − Px y2

= − ( 3, 200 N )( 2.3 m ) = −7,360 N  m = −7,360, 000 N  mm

Detail of FBD at C and D: Equivalent Forces

Detail of FBD at C and D: Equivalent Moments

Fx = Px = 3, 200 N

M x = Py ( z1 ) + Pz ( y1 + y2 )

Fy = − Py = −5, 400 N

= ( 5, 400 N )(1.6 m )

Fz = Pz = 1,300 N

+ (1,300 N )(1.0 m + 2.3 m ) = 12,930 N  m = 12,930, 000 N  mm M y = Px ( z1 ) − Pz ( x1 ) = ( 3, 200 N )(1.6 m ) − (1,300 N )(1.35 m ) = 3,365 N  m = 3,365, 000 N  mm M z = − Px ( y1 + y2 ) − Py ( x1 ) = − ( 3, 200 N )(1.0 m + 2.3 m ) − ( 5, 400 N )(1.35 m ) = −17,850 N  m = −17,850, 000 N  mm

Stresses due to internal pressure: The 1.4 MPa internal fluid pressure creates tensile normal stresses in the 9 mm thick wall of the pipe. The longitudinal stress in the pipe wall is: pd (1.4 MPa )( 257 mm )  long = = = 9.9944 MPa (T) 4t 4 ( 9 mm ) and the circumferential stress is: pd (1.4 MPa )( 257 mm )  hoop = = = 19.9889 MPa (T) 2t 2 ( 9 mm ) Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

(a) Consider point A. Force Fx creates a transverse shear stress in the xy plane at A. The magnitude of this shear stress is: ( 3, 200 N ) 318,523.5 mm3  xy = = 0.8503 MPa 66,595,394 mm4 ( 275 mm − 257 mm )

(

)

This shear stress acts in the positive x direction on the positive y face of the stress element. Force Fy creates an axial stress in the y direction at A. The magnitude of this compressive normal stress is: 5, 400 N y = = 0.7180 MPa 7,520.973 mm 2 Force Fz does not create a normal or a shear stress at point A. Moment Mx creates bending stress that acts in the y direction at A. The magnitude of this compressive normal stress is: M z (11, 630, 000 N  mm )( 275 mm / 2 ) y = x = = 24.0125 MPa Ix 66,595,394 mm4 Moment My is a torque that creates shear stress in the xy plane at A. The magnitude of this shear stress is: M c ( 5,120,000 N  mm )( 275 mm / 2 )  xy = y = = 5.2857 MPa J 133,190,790 mm4 This shear stress acts in the positive x direction on the positive y face of the stress element. Moment Mz does not create bending stress at A because A is located on the neutral axis for bending about the z axis. Stresses due to internal pressure: At A, the longitudinal stress acts in the y direction, and the hoop stress acts in the x direction. Summary of stresses at A:  x = 19.99 MPa

 y = −0.7180 MPa − 24.0125 MPa + 9.9944 MPa = −14.7361 MPa = −14.74 MPa

 xy = 0.8503 MPa + 5.2857 MPa = 6.1360 MPa = 6.14 MPa

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

(b) Consider point C. Force Fx creates an axial stress in the x direction at C. The magnitude of this tensile normal stress is: 3, 200 N x = = 0.4255 MPa 7,520.973 mm 2 Force Fy does not create a normal or a shear stress at point C. Force Fz creates a transverse shear stress in the xz plane at C. The magnitude of this shear stress is: (1,300 N ) 318,523.5 mm3  xz = = 0.3454 MPa 66,595,394 mm4 ( 275 mm − 257 mm )

(

)

This shear stress acts in the positive z direction on the positive x face of the stress element. Moment Mx is a torque that creates shear stress in the xz plane at C. The magnitude of this shear stress is: M c (12,930,000 N  mm )( 275 mm / 2 )  xz = z = = 13.3483 MPa J 133,190,790 mm4 This shear stress acts in the positive z direction on the positive x face of the stress element. Moment My does not create bending stress at C because C is located on the neutral axis for bending about the y axis. Moment Mz creates bending stress in the x direction at C. The magnitude of this tensile normal stress is: M y (17,850, 000 N  mm )( 275 mm / 2 ) x = z = = 36.8550 MPa Iz 66,595,394 mm4 Stresses due to internal pressure: At C, the longitudinal stress acts in the x direction, and the hoop stress acts in the z direction. Summary of stresses at C:  x = 0.4255 MPa + 36.8550 MPa + 9.9944 MPa = 47.2749 MPa = 47.3 MPa

 z = 19.9899 MPa = 20.0 MPa  xz = 0.3454 MPa + 13.3483 MPa = 13.6938 MPa = 13.7 MPa

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P15.51 The piping assembly shown in Figure P15.50/51 consists of stainless steel pipe that has an outside diameter of 275 mm and a wall thickness of 9 mm. The assembly is subjected to concentrated loads Px = 3.2 kN, Py = 5.4 kN, and Pz = 1.3 kN as well as an internal fluid pressure of 1,400 kPa that acts in all pipes. Dimensions of the assembly are x1 = 1.35 m, y1 = 1.0 m, y2 = 2.3 m, and z1 = 1.6 m. Determine the normal and shear stresses on the outer surface of the pipe at (a) point B and (b) point D. Show these stresses on an appropriate sketch.

FIGURE P15.50/51

Solution Section properties: A=



 

275 mm ) − ( 257 mm )  = 7,520.973 mm 2 ( 

275 mm ) − ( 257 mm )  = 133,190, 790 mm 4 ( 

 

275 mm ) − ( 257 mm )  = 66,595,394 mm 4 (  4

64 1 3 3 Q = ( 275 mm ) − ( 257 mm )  = 318,523.5 mm 3  12 

Detail of FBD at A and B: Equivalent Forces

Detail of FBD at A and B: Equivalent Moments

Mechanics of Materials: An Integrated Learning System, 4th Ed. Fx = Px = 3, 200 N Fy = − Py = −5, 400 N Fz = Pz = 1,300 N

Timothy A. Philpot

M x = Py z1 + Pz y2 = ( 5, 400 N )(1.6 m ) + (1,300 N )( 2.3 m ) = 11, 630 N  m = 11, 630, 000 N  mm M y = Px z1 = ( 3, 200 N )(1.6 m ) = 5,120 N  m = 5,120, 000 N  mm M z = − Px y2

= − ( 3, 200 N )( 2.3 m ) = −7,360 N  m = −7,360, 000 N  mm

Detail of FBD at C and D: Equivalent Forces

Detail of FBD at C and D: Equivalent Moments

Fx = Px = 3, 200 N

M x = Py ( z1 ) + Pz ( y1 + y2 )

Fy = − Py = −5, 400 N

= ( 5, 400 N )(1.6 m )

Fz = Pz = 1,300 N

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

(a) Consider point B. Force Fx does not create a normal or a shear stress at point B. Force Fy creates an axial stress in the y direction at B. The magnitude of this compressive normal stress is: 5, 400 N y = = 0.7180 MPa 7,520.973 mm 2 Force Fz creates a transverse shear stress in the yz plane at B. The magnitude of this shear stress is: (1,300 N ) 318,523.5 mm3  yz = = 0.3454 MPa 66,595,394 mm4 ( 275 mm − 257 mm )

(

)

This shear stress acts in the positive z direction on the positive y face of the stress element. Moment Mx does not create bending stress at B because B is located on the neutral axis for bending about the x axis. Moment My is a torque that creates shear stress in the yz plane at B. The magnitude of this shear stress is: M c ( 5,120,000 N  mm )( 275 mm / 2 )  yz = y = = 5.2857 MPa J 133,190,790 mm4 This shear stress acts in the negative z direction on the positive y face of the stress element. Moment Mz creates bending stress that acts in the y direction at B. The magnitude of this compressive normal stress is: M x ( 7,360,000 N  mm )( 275 mm / 2 ) y = z = = 15.1962 MPa Iz 66,595,394 mm4 Stresses due to internal pressure: At B, the longitudinal stress acts in the y direction, and the hoop stress acts in the z direction. Summary of stresses at B:  y = −0.7180 MPa − 15.1962 MPa + 9.9944 MPa = −5.9198 MPa = −5.92 MPa

 z = 19.99 MPa  yz = 0.3454 MPa − 5.2857 MPa = −4.9403 MPa = −4.94 MPa

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

(b) Consider point D. Force Fx creates an axial stress in the x direction at D. The magnitude of this tensile normal stress is: 3, 200 N x = = 0.4255 MPa 7,520.973 mm 2 Force Fy creates a transverse shear stress in the xy plane at D. The magnitude of this shear stress is: ( 5,400 N ) 318,523.5 mm3  xy = = 1.4349 MPa 66,595,394 mm4 ( 275 mm − 257 mm )

(

)

This shear stress acts in the negative y direction on the positive x face of the stress element. Force Fz does not create a normal or a shear stress at point D. Moment Mx is a torque that creates shear stress in the xy plane at D. The magnitude of this shear stress is: M c (12,930,000 N  mm )( 275 mm / 2 )  xy = x = = 13.3483 MPa J 133,190,790 mm4 This shear stress acts in the negative y direction on the positive x face of the stress element. Moment My creates bending stress in the x direction at D. The magnitude of this tensile normal stress is: M z ( 3,365,000 N  mm )( 275 mm / 2 ) x = y = = 6.9477 MPa Iy 66,595,394 mm4 Moment Mz does not create bending stress at D because D is located on the neutral axis for bending about the z axis. Stresses due to internal pressure: At D, the longitudinal stress acts in the x direction, and the hoop stress acts in the y direction. Summary of stresses at D:  x = 0.4255 MPa + 6.9477 MPa + 9.9944 MPa = 17.3677 MPa = 17.37 MPa

 y = 19.99 MPa  xy = −1.4349 MPa − 13.3483 MPa = −14.7832 MPa = −14.78 MPa

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P15.52 The stresses on the surface of a structural-steel beam are x = 38 ksi, y = 0, and xy = 16.5 ksi. The steel has a yield strength of Y = 50 ksi. (a) What is the factor of safety predicted by the maximum-shear-stress theory of failure for the stress state shown? Does the beam fail according to this theory? (b) What is the value of the Mises equivalent stress for the given state of plane stress? (c) What is the factor of safety predicted by the failure criterion of the maximum-distortion-energy theory of failure? Does the beam fail according to this theory?

Solution Principal stresses:

( 38) + ( 0 )   ( 38) − ( 0 )  + 16.5 2 = 19  25.1645  p1 ,  p 2 = )   ( 2





therefore,  p1 = 44.1645 ksi

 p 2 = −6.1645 ksi (a) Maximum-shear-stress theory: Since p1 is positive and p2 is negative, failure will occur if  p1 −  p 2   Y . For the principal stresses existing in the beam:

 p1 −  p 2 = 44.1645 ksi − ( −6.1645 ksi ) = 50.3289 ksi  50 ksi

N.G.

Therefore, the beam fails according to the maximum-shear-stress theory. The factor of safety associated with this state of stress can be calculated as: 50 ksi FS = = 0.994 Ans. 50.3289 ksi (b) Mises equivalent stress: The Mises equivalent stress M associated with the maximum-distortionenergy theory can be calculated from Eq. (15.8) for the plane stress state considered here. 1/2

 M =  p21 −  p1 p 2 +  p2 2 

2 1/2

= ( 44.1645 ksi ) − ( 44.1645 ksi )( −6.1645 ksi ) + ( −6.1645 ksi )    2

= 47.5473 ksi = 47.5 ksi

Ans.

(c) Maximum-distortion-energy theory factor of safety: The factor of safety for the maximumdistortion-energy theory can be calculated from the Mises equivalent stress: 50 ksi FS = = 1.052 Ans. 47.5473 ksi According to the maximum-distortion-energy theory, the beam does not fail.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P15.53 The stresses on the surface of a cast iron machine component are x = 85 MPa, y = −230 MPa, and xy = 125 MPa. The cast iron has a yield strength of Y = 350 MPa. (a) What is the factor of safety predicted by the maximum-shear-stress theory of failure for the stress state shown? Does the component fail according to this theory? (b) What is the value of the Mises equivalent stress for the given state of plane stress? (c) What is the factor of safety predicted by the failure criterion of the maximum-distortion-energy theory of failure? Does the component fail according to this theory?

Solution Principal stresses:

(85) + ( −230 )   (85) − ( −230 )  + 125 2 = −72.5  201.0752  p1 ,  p 2 = )   ( 2





therefore,  p1 = 128.5752 MPa

 p 2 = −273.5752 MPa (a) Maximum-shear-stress theory: Since p1 is positive and p2 is negative, failure will occur if  p1 −  p 2   Y . For the principal stresses existing in the component:

 p1 −  p 2 = 128.5752 MPa − ( −273.5752 MPa ) = 402.1505 MPa  350 MPa

N.G.

Therefore, the component fails according to the maximum-shear-stress theory. The factor of safety associated with this state of stress can be calculated as: 350 MPa FS = = 0.870 Ans. 402.1505 (b) Mises equivalent stress: The Mises equivalent stress M associated with the maximum-distortionenergy theory can be calculated from Eq. (15.8) for the plane stress state considered here. 1/2

 M =  p21 −  p1 p 2 +  p2 2 

2 1/2

= (128.5752 ) − (128.5752 )( −273.5752 ) + ( −273.5752)    2

= 355.7387 MPa = 356 MPa

Ans.

(c) Maximum-distortion-energy theory factor of safety: The factor of safety for the maximumdistortion-energy theory can be calculated from the Mises equivalent stress: 350 MPa FS = = 0.984 Ans. 355.7387 MPa According to the maximum-distortion-energy theory, the component fails.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P15.54 A thin-walled cylindrical pressure vessel of inside diameter d = 36 in. is fabricated from a material with a tensile yield strength of 40 ksi. The internal pressure in the cylinder is 1,600 psi. Assuming that the material obeys the von Mises criterion of yielding, and that there is to be a safety factor against yielding of 3.0, determine the necessary wall thickness remote from the ends of the vessel.

Solution Hoop and longitudinal normal shear stresses on the outside of the pressure vessel: The longitudinal and circumferential normal stresses in the cylinder wall due to an internal pressure p are expressed as: pd (1.6 ksi )( 36 in.) 14.4 kip/in.  long = = = 4t 4t t pd (1.6 ksi )( 36 in.) 28.8 kip/in.  hoop = = = 2t 2t t These stresses are the principal stresses for the pressure vessel.  p1 =  hoop

 p 2 =  long Allowable tensile normal stress:  40 ksi  allow = Y = = 13.3333 ksi FS 3.0 Maximum-distortion-energy theory: The Mises equivalent stress M associated with the maximumdistortion-energy theory can be calculated from Eq. (15.8) for the plane stress state considered here. 1/2

 M =  p21 −  p1 p 2 +  p2 2  2

 28.8 kip/in.   28.8 kip/in.   14.4 kip/in.   14.4 kip/in.  =  +    − t t t    t    

829.44 kip 2 /in.2 414.72 kip 2 /in.2 207.36 kip 2 /in.2 = − + t2 t2 t2 =

622.08 kip 2 /in.2 t2

Equate M and allow and solve for t:  M   allow 622.08 kip 2 /in.2  13.3333 ksi t2 622.08 kip 2 /in.2  177.7769 kip 2 /in.4 2 t t

622.08 kip 2 /in.2 = 1.871 in. 177.7769 kip 2 /in.4

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P15.55 A thin-walled cylindrical pressure vessel of inside diameter d = 500 mm and wall thickness t = 5 mm is fabricated from a material with a tensile yield strength of 280 MPa. Determine the maximum internal pressure p that may be used in the cylinder according to (a) the maximum-shear-stress theory. (b) the maximum-distortion-energy theory.

 hoop =

pd p ( 500 mm ) = = 50 p 2t 2 ( 5 mm )

These stresses are the principal stresses for the pressure vessel.  p1 = 50 p

 p 2 = 25 p (a) Maximum-shear-stress theory: Since both p1 and p2 are positive values, failure will occur if p1 > Y for the pressure vessel. Equate p1 and Y and solve for p:  p1   Y

50 p  280 MPa  p  5.60 MPa

Ans.

(b) Maximum-distortion-energy theory: The Mises equivalent stress M associated with the maximumdistortion-energy theory can be calculated from Eq. (15.8) for the plane stress state considered here. 1/2

 M =  p21 −  p1 p 2 +  p2 2 

2 1/2

= ( 50 p ) − ( 50 p )( 25 p ) + ( 25 p )    = 43.30127 p 2

Equate M and Y and solve for p:  M  Y 43.30127 p  280 MPa p  6.46632 MPa = 6.47 MPa

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P15.56 A solid 80 mm diameter circular shaft made of cold-rolled steel is subjected to the simultaneous action of a torque T = 19 kN·m, a bending moment M = 8.5 kN·m, and a compressive axial force P = 250 kN. The cold-rolled steel has a yield strength of Y = 420 MPa, both in tension and compression. On the basis of (a) the maximum-shear-stress theory and (b) the maximum-distortion-energy theory, is the shaft overstressed?

Solution Section properties:

A= J= I=



d2 =

 32



d4 = d4 =

(80 mm ) = 5, 026.55 mm 2 2



(80 mm ) = 4, 021, 238.60 mm 4 4



(80 mm ) = 2, 010, 619.30 mm 4 4

64 64 Normal and shear stresses on the outside of the shaft: Assume that the axial direction of the shaft is the x direction. The normal stress due to the tensile axial force P is: ( 250 kN )(1,000 N/kN ) = −49.7359 MPa P x = − = − A 5,026.55 mm2 The normal stress due to the bending moment M is: (8.5 kN  m )(80 mm / 2)(1,000 N/kN )(1,000 mm/m ) = 169.1021 MPa Mc x = = I 2,010,619.30 mm4 The normal stress in the y direction (i.e., the circumferential direction for the shaft) is y = 0. The torsional shear stress is calculated as: Tc (19 kN  m )(80 mm / 2 )(1,000 N/kN )(1,000 mm/m )  xy = = = 188.9965 MPa J 4,021, 238.60 mm4 Principal stresses:

( −49.7359 − 169.1021) + ( 0)   ( −49.7359 − 169.1021) − ( 0)  + 188.9965 2  p1 ,  p 2 = )   ( 2

 = −109.4190 MPa  218.3854 MPa 2



= 108.9664 MPa; − 327.8045 MPa (a) Maximum-shear-stress theory: From the preceding calculation, p1 is a positive value and p2 is a negative value. Accordingly, failure will occur if  p1 −  p 2   Y .

 p1 −  p 2  420 MPa 108.9664 MPa − ( −327.8045 MPa )  420 MPa 436.7709 MPa  420 MPa

 Fails!

Ans.

Therefore, the shaft is overstressed according to the maximum-shear-stress theory. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

 M =  p21 −  p1 p 2 +  p2 2 

2 1/2

= (108.9664 MPa ) − (108.9664 MPa )( −327.8045 MPa ) + ( −327.8045 MPa )    = 393.7628 MPa 2

Since the M < Y, the component is not overstressed according to the maximum-distortion-energy theory.  M  Y

393.7628 MPa  420 MPa

 Acceptable!

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P15.57 A 75 mm diameter solid circular shaft rotating at 800 rpm transmits 400 kW and carries a tensile axial force P = 140 kN. The material that comprises the shaft has a tensile yield strength of Y = 110 MPa. On the basis of (a) the maximum-shear-stress theory and (b) the maximum-distortion-energy theory, is the shaft overstressed? State the factors of safety for each theory.

Solution Section properties:

A= J=



d2 =



d4 =

( 75 mm ) = 4, 417.86 mm2



( 75 mm ) = 3,106,311.10 mm4 4

32 32 Normal and shear stresses on the outside of the shaft: Assume that the axial direction of the shaft is the x direction. The normal stress due to the tensile axial force P is: P (140 kN )(1,000 N/kN ) x = = = 31.6895 MPa A 4, 417.86 mm2 The normal stress in the y direction (i.e., the circumferential direction for the shaft) is y = 0. The torque in the shaft is P 400, 000 N  m/s T= = = 4, 774.6483 N  m   800 rev  2 rad  1 min       min  1 rev  60 s  The torsional shear stress is thus: Tc ( 4,774.6483 N  m )(1000 mm / m )( 75 mm / 2 )  xy = = = 57.6405 MPa J 3,106,311.10 mm4 Principal stresses:

( 31.6895) + ( 0 )   ( 31.6895) − ( 0 )  + 57.6405 2  p1 ,  p 2 = )   ( 2

2  = 15.8448 MPa  59.7786 MPa 2



= 75.6234 MPa; − 43.9339 MPa (a) Maximum-shear-stress theory: From the preceding calculation, p1 is a positive value and p2 is a negative value. Accordingly, failure will occur if  p1 −  p 2   Y .

 p1 −  p 2  110 MPa 75.6234 MPa − ( −43.9339 MPa )  110 MPa Ans. 119.5572 MPa  110 MPa  Fails! Therefore, the shaft is overstressed according to the maximum-shear-stress theory. The factor of safety is: 110 MPa FS = = 0.920 Ans. 119.5572 MPa Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

 M =  p21 −  p1 p 2 +  p2 2 

2 1/2

= ( 75.6234 MPa ) − ( 75.6234 MPa )( −43.9339 MPa ) + ( −43.9339 MPa )    = 104.7450 MPa 2

Since the M < Y, the component is not overstressed according to the maximum-distortion-energy theory.  M  Y

104.7450 MPa  110 MPa The factor of safety is: 110 MPa FS = = 1.050 104.7450 MPa

 Acceptable!

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P15.58 A hollow circular shaft made of an aluminum alloy is subjected to a bending moment M = 1.2 kN·m and a torque T. The shaft has an outside diameter of 60 mm and an inside diameter of 40 mm. Assume that the aluminum alloy has a yield strength of Y = 275 MPa. Use (a) the maximum-shear-stress theory and (b) the maximum-distortion-energy theory to determine the value of the torque T so that the shaft does not fail by yielding.

Solution Section properties:

J= I=



 D − d ) = ( 60 mm ) − ( 40 mm )  = 1, 021, 017.60 mm (  32 32  4

 D − d ) = ( 60 mm ) − ( 40 mm )  = 510,508.81 mm (  64 64  4

Normal and shear stresses on the outside of the shaft: Assume that the axial direction of the shaft is the x direction. The normal stress due to the bending moment M is: (1.2 kN  m )( 60 mm / 2)(1,000 N/kN )(1,000 mm/m ) = 70.5179 MPa Mc x = = I 510,508.81 mm4 The normal stress in the y direction (i.e., the circumferential direction for the shaft) is y = 0. The torsional shear stress can be expressed as: T ( 60 mm / 2 ) Tc T  xy = = = 4 J 1,021,017.60 mm 34,033.92 mm3 Principal stresses:

( 70.5179 MPa ) + ( 0 )   ( 70.5179 MPa ) − ( 0 )  +   p1 ,  p 2 =    2



= 35.2589 MPa 



T  3   34, 033.92 mm 

T   2 ( 35.2589 MPa ) +  3   34, 033.92 mm 

(a) Maximum-shear-stress theory: From this result, it is apparent that p1 will be a positive value and p2 will be a negative value. Accordingly, failure will occur if  p1 −  p 2   Y . For the principal stresses existing in the component: T    p1 −  p 2 = 35.2589 MPa + ( 35.2589 MPa ) +  3   34, 033.92 mm 

T   −35.2589 MPa + ( 35.2589 MPa ) +  3   34, 033.92 mm 

T   = 2 ( 35.2589 MPa ) +  3   34, 033.92 mm 

Set this expression equal to Y and solve for the largest value of T that will not cause failure:

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

T   2 ( 35.2589 MPa ) +   275 MPa 3   34, 033.92 mm  2

T   2  (137.5 MPa ) ( 35.2589 MPa ) +  3   34, 033.92 mm  T 2 2  (137.5 MPa ) − ( 35.2589 MPa ) 3 34, 033.92 mm 2

T  (132.9024 MPa ) ( 34, 033.092 mm3 ) = 4,523, 081 N  mm

To keep the shaft from failing according to the maximum-shear-stress theory, the torque should be no larger than: Ans. T  4.52 kN  m (b) Maximum-distortion-energy theory: The Mises equivalent stress M associated with the maximumdistortion-energy theory can be calculated from Eq. (15.9) for the plane stress state considered here.

 M =  x2 −  x y +  y2 + 3 xy2 =  x2 + 3 xy2 T   = ( 70.5179 MPa ) + 3  3   34, 033.92 mm 

Set the Mises equivalent stress equal to the yield stress and solve for the largest torque T that can be applied without causing failure in the material. 2

T    275 MPa ( 70.5179 MPa ) + 3  3   34, 033.92 mm  2

T  34, 033.92 mm3

( 275 MPa ) − ( 70.5179 MPa ) 2

T  (153.4625 MPa ) ( 34, 033.92 mm 3 ) = 5, 222,930.8 N  mm

To keep the shaft from failing according to the maximum-distortion-energy theory, the torque should be no larger than: Ans. T  5.22 kN  m

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P15.59 A 3 in. diameter solid circular shaft made of stainless steel is subjected to the simultaneous action of a torque T = 6.0 kip·ft, a tensile axial force P = 55 kips, and a bending moment M. The stainless steel has a yield strength of Y = 36 ksi, both in tension and compression. Use (a) the maximum-shear-stress theory and (b) the maximum-distortion-energy theory to determine the value of the moment M so that the shaft does not fail by yielding.

Solution Section properties:

A= J= I=



d2 =

 32



d4 = d4 =

( 3 in.) = 7.0686 in.2 2



( 3 in.) = 7.9522 in.4 4



( 3 in.) = 3.9761 in.4 4

64 64 Normal and shear stresses on the outside of the shaft: Assume that the axial direction of the shaft is the x direction. The normal stress due to the tensile axial force is: P 55 kips x = = = 7.7809 ksi A 7.0686 in.2 The normal stress due to the bending moment M can be expressed as:  3 in.  M  Mc M 2   x = = = 4 I 3.9761 in. 2.6507 in.3 The normal stress in the y direction (i.e., the circumferential direction for the shaft) is y = 0. The torsional shear stress can be expressed as: Tc ( 6.0 kip  ft )(12 in./ft )( 3 in. / 2 )  xy = = = 13.5812 ksi J 7.9522 in 4 Principal stresses:

M   + 0  7.7809 ksi + 3  ( ) 2.6507 in.    p1 ,  p 2 =  2

 M     7.7809 ksi + 2.6507 in.3  − ( 0 )  2    + (13.5812 ksi )  2  2

M M 2     =  3.8905 ksi +  3.8905 ksi + + (13.5812 ksi ) 3  3 5.3014 in.  5.3014 in.    (a) Maximum-shear-stress theory: From this result, it is apparent that p1 will be a positive value and p2 will be a negative value. Accordingly, failure will occur if  p1 −  p 2   Y . For the principal stresses existing in the component:

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot 2

M M 2      p1 −  p 2 =  3.8905 ksi + + 3.8905 ksi + + (13.5812 ksi ) 3  3 5.3014 in.  5.3014 in.    2

M M 2     −  3.8905 ksi + + 3.8905 ksi + + (13.5812 ksi ) 3  3 5.3014 in.  5.3014 in.    2

M 2   = 2 3.8905 ksi + + (13.5812 ksi ) 3 5.3014 in.   kips  M2  = 2 15.1360 ksi + 1.46773 M+ + 184.44899 ksi 2 5  6 in.  28.10484 in.  2

kips  M2  = 2 199.58499 ksi +  1.46773 M + in.5  28.10484 in.6  2

Set this expression equal to Y and reduce the equation:

kips  M2  2 199.58499 ksi 2 + 1.46773 M + = 36 ksi  in.5  28.10484 in.6  kips  M2 2  199.58499 ksi 2 + 1.46773 M + = (18 ksi ) 5  6 in.  28.10484 in.  2 M kips  kips 2  + 1.46773 M − 124.41501 =0   28.10484 in.6  in.5  in.4 M 2 + ( 41.25031 kip  in.) M − 3, 496.66395 kips 2  in.2 = 0

(a)

Use the quadratic formula to calculate M from Eq. (a): M=

− ( 41.25031) 

( 41.25031) − 4 (1)( −3, 496.66395 ) 2 (1) 2

−41.25031  125.25272 2 = 42.00121 kip  in.; − 83.25152 kip  in.

Thus, to keep the shaft from failing according to the maximum-shear-stress theory, the moment should be no larger than: Ans. M  42.0 kip  in.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

(b) Maximum-distortion-energy theory: The Mises equivalent stress M associated with the maximumdistortion-energy theory can be calculated from Eq. (15.9) for the plane stress state considered here.

 M =  x2 −  x y +  y2 + 3 xy2 =  x2 + 3 xy2 2

M 2   = 7.7809 ksi + + 3 (13.5812 ksi ) 3 2.6507 in.  

Set the Mises equivalent stress equal to the yield stress and solve for the largest moment M that can be applied without causing failure in the material. 2

M 2   7.7809 ksi + 2.6507 in.3  + 3 (13.5812 ksi ) = 36 ksi 60.54240

kips 2  kips  M2 kips 2 kips 2 + 5.87083 M + + 553.34698 = 1, 296   in.4  in.5  7.02621 in.6 in.4 in.4 M2 kips  kips 2  +  5.87083 =0  M − 682.11062 7.02621 in.6  in.5  in.4 M 2 + ( 41.24969 kip  in.) M − 4, 792.65246 kip 2  in.2 = 0

(b)

Use the quadratic formula to calculate M from Eq. (b): M=

− ( 41.24969 ) 

( 41.24969 ) − 4 (1)( −4, 792.65246 ) 2 (1) 2

−41.24969  144.47196 2 = 51.61113 kip  in.; − 92.86082 kip  in.

To keep the shaft from failing according to the maximum-distortion-energy theory, the moment M should be no larger than: Ans. M  51.6 kip  in.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P15.60 A solid circular shaft made of steel is subjected to a torque T = 780 lb·ft and a tensile axial force P = 13,500 lb. Assume that the steel has a yield strength of Y = 30,000 psi. Use (a) the maximum-shear-stress theory and (b) the maximum-distortion-energy theory to determine the required diameter of the shaft so that the shaft does not fail by yielding.

Solution Normal and shear stresses on the outside of the shaft: Assume that the axial force acts in the x direction. The normal stress due to the tensile axial force can be expressed as: P P 4P x = = = 2  A d2 d 4 The normal stress in the y direction (i.e., the circumferential direction for the shaft) is y = 0. The torsional shear stress can be expressed as: d  T  Tc 16T 2  xy = =  =  4 d3 J d 32 Principal stresses: 2

  4P      d 2  − ( 0 )   16T 2    + 3  2   d     

 4P   2  + ( 0) d   p1 ,  p 2 =   2 2

2P  2 P   16T    2  + 3  2 d d  d 

2P 4  2 64T 2  =  P + 2  d2  2 d 4  d  =

2P 2  2 d d2

P2 +

64T 2 d2

 p1 −  p 2 = =

2P 2 + 2 d d2

P2 +

4 d2

64T 2 d2

P2 +

64T 2 2 P 2 − + 2 2 d d d2

P2 +

64T 2 d2

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

To avoid failure, this expression must be less than Y. Solving for d gives: 4 d2 4 d2

(13,500 lb ) + 2

P2 +

64T 2  Y d2

64 ( 780 lb  ft )(12 in./ft )  d2

 30, 000 psi

 ( 30, 000 psi )  d 2  64 ( 9,360 lb  in.) 13,500 lb +  ( )   d2 4   2

182.25  106 lb 2 +

2 5, 607.0144 106 lb 2in.2  6 lb  4  555.16524  10  d d2 in.4  

 lb 2  4 5, 607.0144  106 lb 2in.2 182.25  106 lb 2   555.16524  106 d − in.4  d2  10.099721 in.6 0  d4 − − 0.32828063 in.4 2 d Using trial-and-error, we can quickly determine that d must be in the range of 1.4 in. < d < 1.6 in. d Eq. (a) 1 –9.4280 1.2 –5.2684 1.4 –1.6396 1.6 2.2801 1.8 7.0521 2 13.1468

(a)

Further refinement in this interval using the bisection algorithm gives a result of d = 1.4873 in. d Eq. (a) 1.4 –1.6396 1.5 0.2455 1.45 –0.7114 1.475 –0.2372 1.4875 0.0030 1.4813 –0.1164 1.4844 –0.0567 1.486 –0.0259 1.4867 –0.0124 1.4871 –0.0047 1.4873 –0.0008

d  1.4873 in. = 1.487 in.

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

 M =  x2 −  x y +  y2 + 3 xy2 =  x2 + 3 xy2 16  2 48T 2   4P   16T  =  2  + 3 3  = P + 2   2 d 4  d  d  d  2

4 48T 2 2 = P + 2 d2 d Set the Mises equivalent stress equal to the yield stress 4 d2 4 d2

P2 +

48T 2  Y d2

48 ( 9,360 lb  in.)  30, 000 psi (13,500 lb ) + d2 2

 ( 30, 000 psi )  d 2  48 ( 9,360 lb  in.)  (13,500 lb ) +   d2 4   2 4, 205.2608  106 lb 2in.2  6 lb  4 182.25  106 lb 2 +  555.16524  10  d d2 in.4   2

2  4, 205.2608  106 lb 2in.2 6 lb  4 182.25  10 lb   555.16524 10 d − in.4  d2  7.5747911 in.6 (b) 0  d4 − − 0.32828063 in.4 d2 Using trial-and-error, we can quickly determine that d must be in the range of 1.4 in. < d < 1.6 in. d Eq. (b) 1 –6.9031 1.2 –3.5150 1.4 –0.3514 1.6 3.2664 Further refinement in this interval using the bisection algorithm gives a result of d = 1.4211 in. d Eq. (b) 1.4 -0.3514 1.5 1.3676 1.45 0.4895 1.425 0.0649 1.4125 -0.1442 1.4188 -0.0391 1.4219 0.0128 1.4203 -0.0140 1.4211 -0.0006 6

d  1.4211 in. = 1.421 in.

Ans.

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P15.61 The stresses on the surface of a machine component will be in a state of plane stress with x = y = −40 ksi and xy = 85 ksi. The ultimate failure strengths for this material are 100 ksi in tension and 190 ksi in compression. Use the Mohr failure criterion to determine whether this component is safe for the state of stress given. Support your answer with appropriate documentation.

Solution Principal stresses:

( −40 ksi ) + ( −40 ksi )   ( −40 ksi ) − ( −40 ksi )  + 85 ksi 2  p1 ,  p 2 = )   ( 2





= −40 ksi  85 ksi therefore,

 p1 = 45 ksi and

 p 2 = −125 ksi

Mohr failure criterion: If p1 is positive and p2 is negative, then failure will occur if the following interaction equation is greater than or equal to 1:

 p1  p 2 − 1.  UT  UC

For the principal stresses existing in the component:  p1  p 2 45 ksi −125 ksi − = −  UT  UC 100 ksi 190 ksi

= 0.45 − ( −0.6579 )

= 1.108  1 N.G. Therefore, the component fails according to the Mohr failure criterion.

Ans.

The factor of safety corresponding to the Mohr failure criterion is: 1 FS = = 0.903 1.108

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P15.62 The stresses on the surface of a machine component will be in a state of plane stress with x = 510 MPa, y = −140 MPa, and xy = 375 MPa. The ultimate failure strengths for this material are 700 MPa in tension and 420 MPa in compression. Use the Mohr failure criterion to determine whether this component is safe for the state of stress given. Support your answer with appropriate documentation.

Solution Principal stresses:

( 510 MPa ) + ( −140 MPa )   ( 510 MPa ) − ( −140 MPa )  + 375 MPa 2  p1 ,  p 2 = )   ( 2





= 185 MPa  496.236 MPa therefore,

 p1 = 681.236 MPa and

 p 2 = −311.236 MPa

Mohr failure criterion: If p1 is positive and p2 is negative, then failure will occur if the following interaction equation is greater than or equal to 1:

 p1  p 2 − 1.  UT  UC

For the principal stresses existing in the component:  p1  p 2 681.236 MPa −311.236 MPa − = −  UT  UC 700 MPa 420 MPa

= 0.9732 − ( −0.7410 )

= 1.714  1 N.G. Therefore, the component fails according to the Mohr failure criterion.

Ans.

The factor of safety corresponding to the Mohr failure criterion is: 1 FS = = 0.583 1.714

Mechanics of Materials: An Integrated Learning System, 4th Ed.

Timothy A. Philpot

P15.63 The state of stress at a point in a cast iron [UT = 290 MPa; UC = 650 MPa] component is x = 0, y = –180 MPa, and xy = 200 MPa. Determine whether failure occurs at the point according to (a) the maximum-normal-stress theory and (b) the Mohr failure criterion.

Solution Principal stresses:

( 0 ) + ( −180 MPa )   ( 0 ) − ( −180 MPa )  + 200 MPa 2  p1 ,  p 2 = )   ( 2

 = −90 MPa  219.317 MPa 2

therefore,

 p1 = 129.317 MPa

and



 p 2 = −309.317 MPa

(a) Maximum-normal-stress theory: Compare the maximum tensile normal stress with UT to find:  max tension   UT 129.317 MPa  290 MPa

 OK

Next, compare the maximum compressive normal stress with UC to find:  max compression  UC

−309.317 MPa  −650 MPa

 OK

Since both checks are satisfied, the component does not fail.

Ans.

(b) Mohr failure criterion: If p1 is positive and p2 is negative, then failure will occur if the following interaction equation is greater than or equal to 1:

 p1  p 2 − 1.  UT  UC

For the principal stresses existing in the component:  p1  p 2 129.317 MPa −309.317 MPa − = −  UT  UC 290 MPa 650 MPa

= 0.4459 − ( −0.4759 )

= 0.922  1 OK Therefore, the component does not fail according to the Mohr failure criterion.

Ans.

The factor of safety corresponding to the Mohr failure criterion is: 1 FS = = 1.085 0.922

P16.1 Determine the slenderness ratio and the Euler buckling load for round wooden dowels that are 1 m long and have a diameter of (a) 16 mm and (b) 25 mm. Assume E = 10 GPa.

Solution (a) 16 mm dowels:

I= A=



 4

(16 mm) 4 = 3, 216.991 mm 4 (16 mm) 2 = 201.062 mm 2

3, 216.991 mm 4 r= = 4.000 mm 201.062 mm 2 Slenderness ratio: L 1, 000 mm = = 250 r 4.000 mm Euler buckling load:  2 EI  2 (10,000 N/mm2 )(3, 216.991 mm4 ) Pcr = 2 = = 317.504 N = 318 N L (1,000 mm)2

Ans.

(b) 25 mm dowels:

I= A= r=



 4

(25 mm)4 = 19,174.760 mm 4 (25 mm)2 = 490.874 mm 2

19,174.760 mm4 = 6.250 mm 490.874 mm 2

Slenderness ratio: L 1, 000 mm = = 160 r 6.250 mm Euler buckling load:  2 EI  2 (10,000 N/mm2 )(19,174.760 mm2 ) Pcr = 2 = = 1,892.473 N = 1,892 N L (1,000 mm)2

Ans.

P16.2 An aluminum alloy tube with an outside diameter of 3.50 in. and a wall thickness of 0.30 in. is used as a 14 ft long column. Assume that E = 10,000 ksi and that pinned connections are used at each end of the column. Determine the slenderness ratio and the Euler buckling load for the column.

Solution  I=

(3.500 in.)4 − (2.900 in.)4  = 3.894318 in.4 64



(3.500 in.)2 − (2.900 in.)2  = 3.015929 in.2 4

3.894318 in.4 r= = 1.136 in. 3.015929 in.2 Slenderness ratio: L (14 ft)(12 in./ft) = = 147.8 r 1.136 in.

Ans.

Euler buckling load:  2 EI  2 (10, 000 ksi)(3.894318 in.4 ) Pcr = 2 = = 13.618 kips = 13.62 kips 2 L (14 ft)(12 in./ft)

Ans.

P16.3 A WT205 × 30 structural steel section (see Appendix B for cross-sectional properties) is used for a 6.5 m column. Assume pinned connections at each end of the column. Determine: (a) the slenderness ratio. (b) the Euler buckling load. Use E = 200 GPa for the steel. (c) the axial stress in the column when the Euler load is applied.

Solution The following section properties for a standard steel WT205 × 30 shape are given in Appendix B: A = 3,800 mm2, Ix = 13.8×106 mm4, rx = 87.7 mm, Iy = 5.99×106 mm4, ry = 39.6 mm (a) Slenderness ratio: L 6,500 mm = = 164.1 r 39.6 mm (b) Euler buckling load:  2 EI  2 (200,000 N/mm2 )(5.99 106 mm4 ) Pcr = 2 = = 279,853 N = 280 kN L (6,500 mm)2 (c) Axial stress at Pcr: 279,853 N = = 73.6 MPa 3,800 mm 2

Ans.

P16.4 Determine the maximum compressive load that a HSS6 × 4 × 1/4 structural steel column (see Appendix B for cross-sectional properties) can support if it is 24 ft long and a factor of safety of 1.92 is specified. Use E = 29,000 ksi for the steel.

Solution The following section properties for a standard steel HSS6 × 4 × 1/4 shape are given in Appendix B: A = 4.30 in.2, Ix = 20.9 in.4, rx = 2.20 in., Iy = 11.1 in.4, ry = 1.61 in. Euler buckling load:  2 EI  2 (29, 000 ksi)(11.1 in.4 ) Pcr = 2 = = 38.303 kips = 38.3 kips 2 L (24 ft)(12 in./ft) Allowable column load: P 38.303 kips Pallow = cr = = 19.95 kips FS 1.92

Ans.

P16.5 Two C12 × 25 structural steel channels (see Appendix B for cross-sectional properties) are used for a column that is 35 ft long. Assume pinned connections at each end of the column and use E = 29,000 ksi for the steel. Determine the total compressive load required to buckle the two members if: (a) they act independently of each other. (b) they are latticed back-to-back as shown in Figure P16.5.

FIGURE P16.5

Solution The following section properties for a standard steel C12 × 25 shape are given in Appendix B: A = 7.34 in.2, Ix = 144 in.4, Iy = 4.45 in.4, x = 0.674 in. (a) Independent channels: Consider buckling about horizontal cross-sectional axis: I = 2(144 in.4 ) = 288 in.4 Pcr =

 2 EI L2

 2 (29,000 ksi)(288 in.4 )

(35 ft)(12 in./ft)

= 467.296 kips

(a)

Consider buckling about vertical cross-sectional axis: I = 2(4.45 in.4 ) = 8.90 in.4 Pcr =

 2 EI L2

 2 (29,000 ksi)(8.90 in.4 )

(35 ft)(12 in./ft)

= 14.44 kips

Euler buckling load for independent channels: Pcr = 14.44 kips

Ans.

(b) Latticed channels: Consider buckling about horizontal cross-sectional axis: Pcr = 467.296 kips calculated previously in Eq. (a) Consider buckling about vertical cross-sectional axis: I = 2  4.45 in.4 + (3 in. + 0.674 in.) 2 (7.34 in.2 )  = 207.055 in.4

Pcr =

 2 EI L2

 2 (29,000 ksi)(207.055 in.4 )

(35 ft)(12 in./ft)

Euler buckling load for latticed channels: Pcr = 336 kips

= 335.957 kips

Ans.

P16.6 Two L102 × 76 × 9.5 structural steel angles (see Appendix B for cross-sectional properties) are used as a compression member that is 4.5 m long. The angles are separated at intervals by spacer blocks and connected by bolts (as shown in Figure P16.6), which ensure that the double-angle shape acts as a unified structural member. Assume pinned connections at each end of the column and use E = 200 GPa for the steel. Determine the Euler buckling load for the double-angle column if the spacer block thickness is (a) 5 mm or (b) 20 mm. FIGURE P16.6

Solution The following section properties for a standard steel L102 × 76 × 9.5 shape are given in Appendix B: A = 1,600 mm2, Ix = 1.64×106 mm4, Iy = 0.787×106 mm4, x = 19.7 mm (a) 5 mm spacer block thickness: Consider buckling about horizontal cross-sectional axis: I = 2(1.64  106 mm 4 ) = 3.28  106 mm 4 Pcr =

 2 EI

 2 (200,000 N/mm 2 )(3.28  106 mm 4 )

= 319,726 N L2 (4,500 mm) 2 Consider buckling about vertical cross-sectional axis: I = 2 0.787  106 mm 4 + (5 mm/2 + 19.7 mm)2 (1,600 mm 2 )  = 3.151  106 mm4

Pcr =

 2 EI

 2 (200,000 N/mm 2 )(3.151  106 mm 4 )

L2 (4,500 mm)2 Euler buckling load for 5 mm spacer blocks: Pcr = 307 kN

(a)

= 307,160 N

Ans.

(b) 20 mm spacer block thickness: Consider buckling about horizontal cross-sectional axis: Pcr = 319,726 N calculated previously in Eq. (a) Consider buckling about vertical cross-sectional axis: I = 2 0.787  106 mm 4 + (20 mm/2 + 19.7 mm) 2 (1,600 mm 2 )  = 4.3967  106 mm 4 Pcr =

 2 EI

 2 (200,000 N/mm 2 )(4.3967  106 mm 4 )

= L2 (4,500 mm) 2 Euler buckling load for 20 mm spacer blocks: Pcr = 320 kN

= 428,578 N

Ans.

P16.7 A solid 0.5 in. diameter cold-rolled steel rod is pinned to fixed supports at A and B. The length of the rod is L = 24 in., its elastic modulus is E = 30,000 ksi, and its coefficient of thermal expansion is  = 6.6 × 10−6 /°F. Determine the temperature increase T that will cause the rod to buckle. FIGURE P16.7

Solution Section properties:

A= I=



(0.5 in.)2 = 0.1963495 in.2 (0.5 in.)4 = 0.00306796 in.4

Euler buckling load:  2 EI  2 (30,000 ksi)(0.00306796 in.4 ) Pcr = 2 = = 1.57706 kips (C) L (24 in.)2 Force-Temperature-Deformation Relationship The relationship between internal force, temperature change, and deformation of an axial member is: FL e= +  T L AE Since the rod is attached to rigid supports, e = 0. FL +  T L = 0 AE Set F = Pcr Pcr L +  T L = 0 AE and solve for T: P T = − cr  AE =−

(6.6  10

−6

−1.57706 kips / F)(0.1963495 in.2 )(30, 000 ksi)

= 40.565F = 40.6F

Ans.

P16.8 Rigid beam ABC is supported by a pinned connection at A and by a timber post that is pinconnected at B and D, as shown in Figure P16.8. A distributed load of w = 2 kips/ft acts on the 14 ft long beam, which has length dimensions of x1 = 8 ft and x2 = 6 ft. The timber post has a length of L = 10 ft, an elastic modulus of E = 1,800 ksi, and a square cross section. If a factor of safety of 2.0 with respect to buckling is specified, determine the minimum width required for the square post.

FIGURE P16.8

Solution Equilibrium of rigid beam ABC:

M A = −(2 kips/ft)(14 ft)(7ft) − F1 (8 ft) = 0  F1 = −24.5 kips = 24.5 kips (C)

Minimum required moment of inertia for square post: Since a factor of safety of 2.0 is required with respect to buckling, the post must have a critical buckling load of Pcr = 2(24.5 kips) = 49 kips The minimum required moment of inertia is thus:  2 EI Pcr = 2  49 kips L (49 kips)  (10 ft)(12 in./ft) 

I 

 (1,800 ksi) 2

Required post dimensions: For a square cross section of width b: b4 I =  39.7179 in.4 12  b  4.67 in.

= 39.7179 in.4

Ans.

P16.9 Rigid beam ABC is supported by a pinned connection at A and by a 180 mm by 180 mm square timber post that is pin-connected at B and D, as shown in Figure P16.9. The length dimensions of the beam are x1 = 3.6 m and x2 = 2.8 m. The timber post has a length of L = 4 m and an elastic modulus of E = 12 GPa. If a factor of safety of 2.0 with respect to buckling is specified, determine the magnitude of the maximum distributed load w that may be supported by the beam.

FIGURE P16.9

Solution Euler buckling load: (180 mm) 4 I= = 87.48  106 mm 4 12  2 EI  2 (12,000 N/mm 2 )(87.48  106 mm 4 ) Pcr = 2 = = 647,544.745 N L (4,000 mm) 2 Allowable column load: P 647,544.745 N Pallow = cr = = 323,772.373 N = 323.772 kN (C) FS 2.0

Equilibrium of rigid beam ABC: M A = − w(6.4 m)(3.2 m) − (−323.772 kN)(3.6 m) = 0 1,165.57920 kN-m w = = 56.9 kN/m 20.480 m 2

Ans.

P16.10 A rigid beam is supported by a pinned connection at B and by an inclined strut that is pinconnected at A and C as shown in Figure P16.10a. Dimensions of the structure are a = 3.5 m, b = 2.5 m, c = 3.8 m, and e = 200 mm. Loads on the structure are P = 13 kN and w = 150 kN/m. The strut is fabricated from two steel [E = 200 GPa] L127 × 76 × 12.7 angles, which are oriented with the long legs back-to-back, as shown in Figure P16.10b. The angles are separated and connected by spacer blocks, which are s = 40 mm thick. Determine (a) the compressive force in the strut created by the loads acting on the beam. (b) the slenderness ratios for the strut about the strong and weak axes of the double-angle shape. (c) the minimum factor of safety in the strut with respect to buckling.

FIGURE P16.10a

FIGURE P16.10b

Solution (a) Compression force in strut: Member AC is a two-force member that is oriented at  with respect to the horizontal axis: c−e tan  = a+b 3.8 m − 0.2 m = = 0.6 3.5 m + 2.5 m  = 30.964 From a FBD of rigid bar ABC, write the following equilibrium equation: M B = FAC ( 6.0 m ) sin 30.964 + FAC ( 0.2 m ) cos 30.964

 2.5 m  + (13 kN )( 6.0 m ) + (150 kN/m )( 2.5 m )  =0  2  and compute FAC = –101.016 kN = 101.016 kN (C). FAC = −167.793 kN = 167.8 kN (C)

Ans.

(b) Slenderness ratios: The following section properties for a standard steel L127 × 76 × 12.7 shape are given in Appendix B: A = 2,420 mm2, Ix = 3.930×106 mm4, Iy = 1.060×106 mm4, x = 18.9 mm

Consider strong axis (i.e., buckling about horizontal cross-sectional axis): L=

( 6.0 m ) + ( 3.6 m ) = 6.997142 m = 6,997.142 mm 2

A = 2 ( 2, 420 mm 2 ) = 4,840 mm 2 I = 2 ( 3.930 106 mm 4 ) = 7.860 106 mm 4 7.860 106 mm 4 = 40.298 mm 4,840 mm 2 L 6,997.142 mm = = 173.6 r 40.298 mm Consider weak axis (i.e., buckling about vertical cross-sectional axis): 2    40 mm  6 4 I = 2 1.060 10 mm +  + 18.9 mm  ( 2, 420 mm 2 )  = 9.44394  106 mm 4  2    r=

Ans.

9.44394 106 mm 4 r= = 44.173 mm 4,840 mm 2 L 6,997.142 mm = = 158.4 r 44.173 mm

Ans.

(c) Minimum factor of safety with respect to buckling: 2 2 6 4  2 EI  ( 200, 000 N/mm )( 7.860 10 mm ) Pcr = 2 = = 316,892 N = 316.892 kN 2 L ( 6,997.142 mm )

FS =

316.892 kN = 1.889 167.793 kN

Ans.

P16.11 An assembly consisting of tie rod (1) and pipe strut (2) is used to support an 80 kip load, which is applied to joint B. Strut (2) is a pinconnected steel [E = 29,000 ksi] pipe with an outside diameter of 8.625 in. and a wall thickness of 0.322 in. For the loading shown in Figure P16.11, determine the factor of safety with respect to buckling for member (2).

FIGURE P16.11

Solution Equilibrium of joint B: 12 ft tan  AB = = 0.5 24 ft 30 ft tan  BC = = 1.25 24 ft

 AB = 26.565  BC = 51.340

Fx = − F1 cos(26.565) − F2 cos(51.340) = 0 Fy = F1 sin(26.565) − F2 sin(51.340) − P = 0 Note: Tension assumed in each truss member. Solve these equations simultaneously to obtain: F1 = 51.110 kips F2 = −73.179 kips Euler buckling load for member (2): d = 8.625 in. − 2(0.322 in.) = 7.981 in.



(8.625 in.) 4 − (7.981 in.) 4  = 72.489241 in.4 64

L2 = (24 ft) 2 + (30 ft) 2 = 38.419 ft = 461.025 in.

Pcr =

 2 EI L2

 2 (29,000 ksi)(72.489241 in.4 ) (461.025 in.)2

Factor of safety for member (2): 97.616 kips FS = = 1.334 73.179 kips

= 97.616 kips

Ans.

P16.12 A tie rod (1) and a structural steel WT shape (2) are used to support a load P as shown in Figure P16.12. Tie rod (1) is a solid 1.125 in. diameter steel rod and member (2) is a WT8 × 20 structural shape oriented so that the tee stem points upward. Both the tie rod and the WT shape have an elastic modulus of 29,000 ksi and a yield strength of 36 ksi. Determine the maximum load P that can be applied to the structure if a factor of safety of 2.0 with respect to failure by yielding and a factor of safety of 3.0 with respect to failure by buckling are specified. FIGURE P16.12

Solution Equilibrium of joint B: 12 ft tan  BC = = 0.75  BC = 36.870 16 ft Fy = F1 sin(36.870) − P = 0  F1 = 1.666667 P

Fx = − F1 cos(36.870) − F2 = 0

 F2 = −1.333333P

Consider yielding of tie rod (1): The solid 1.125 in. diameter steel tie rod has an area of A1 = 0.9940196 in.2. The allowable force in the tie rod is:  36 ksi  2 F1,allow =  allow A1 =   (0.9940196 in. ) = 17.8924 kips  2.0  Therefore, the maximum load P that may be applied to the structure at B is: 17.8924 kips (a) Pallow  = 10.7354 kips 1.666667 Consider buckling of member (2): The following section properties for a standard steel WT8 × 20 shape are given in Appendix B: A = 5.89 in.2, Ix = 33.1 in.4, Iy = 14.4 in.4 The critical buckling load for member (2) is:  2 EI  2 (29,000 ksi)(14.4 in.4 ) Pcr = 2 = = 111.8041 kips 2 L (16 ft)(12 in./ft) The allowable load for member (2) based on the minimum required factor of safety is: 111.8041 kips F2,allow = = 37.2680 kips 3.0 Therefore, the maximum load P that may be applied to the structure at B is: 37.2680 kips Pallow  = 27.9510 kips 1.333333

(b)

Consider yielding of member (2): The allowable force in member (2) is:  36 ksi  2 F2,allow =  allow A2 =   (5.89 in. ) = 106.0200 kips  2.0  Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Therefore, the maximum load P that may be applied to the structure at B is: 106.0200 kips Pallow  = 79.5150 kips 1.333333 Allowable load P: Compare the values in Eqs. (a), (b), and (c) to obtain: Pallow = 10.74 kips

(c)

Ans.

P16.13 A simple pin-connected truss is loaded and supported as shown in Figure P16.13. All members of the truss are aluminum [E = 10,000 ksi] pipes with an outside diameter of 4.00 in. and a wall thickness of 0.226 in. Consider all compression members and determine the minimum factor of safety for the truss with respect to failure by buckling.

FIGURE P16.13

Solution Section properties: d = 4.00 in. − 2(0.226 in.) = 3.548 in.



(4.00 in.) 4 − (3.548 in.)4  = 4.787719 in.4 64

Euler buckling load:  2 EI  2 (10,000 ksi)(4.787719 in.4 ) 472,528.925 kip-in.2 Pcr = 2 = = L L2 L2

(a)

Truss analysis results: In the table below, Eq. (a) will be used to compute the Euler buckling load for each compression truss member.

(kips)

Member length L (in.)

(kips)

17.25 (C)

91.151

5.28

17.25 (C)

91.151

5.28

2.25 (T)

N.A.

12.00 (C)

91.151

12.00 (T)

N.A.

15.00 (C)

120

32.815

0 (T)

N.A.

8.75 (T)

120

N.A.

7.00 (C)

51.273

Member

Axial force

Minimum factor of safety: FSmin = 2.19

Pcr

7.60

2.19

7.33

Ans.

P16.14 A simple pin-connected wood truss is loaded and supported as shown in Figure P16.14. The members of the truss are 150 mm by 150 mm square Douglas fir timbers that have an elastic modulus of E = 11 GPa. Consider all compression members and determine the minimum factor of safety for the truss with respect to failure by buckling. FIGURE P16.14

Solution Section properties: (150 mm) 4 I= = 42,187,500 mm 4 12 Euler buckling load:  2 EI  2 (11,000 N/mm 2 )(42,187,500 mm 4 ) Pcr = 2 = L L2 4.5801132  1012 N-mm 2 4,580.1133 kN-m 2 (a) = = L2 L2 Truss analysis results: In the table below, Eq. (a) will be used to compute the Euler buckling load for each compression truss member.

(kN)

Member length L (m)

(kN)

17.25 (T)

6.0

N.A.

17.25 (T)

6.0

N.A.

15.75 (T)

6.0

N.A.

15.75 (T)

6.0

N.A.

28.75 (C)

10.0

49.965

1.738

24.75 (C)

6.0

138.791

5.61

24.75 (C)

6.0

138.791

5.61

26.25 (C)

10.0

49.965

1.903

13.00 (T)

8.0

N.A.

12.50 (T)

10.0

N.A.

8.0

N.A.

15.00 (T)

10.0

N.A.

9.0 (T)

8.0

N.A.

Member

Axial force

Minimum factor of safety: FSmin = 1.738

Pcr

Ans.

P16.15 A HSS152.4 × 101.6 × 6.4 structural steel [E = 200 GPa] section (see Appendix B for crosssectional properties) is used as a column with an actual length of 6 m. The column is supported only at its ends and it may buckle in any direction. If a factor of safety of 2 with respect to failure by buckling is specified, determine the maximum safe load for the column for the following end conditions: (a) pinned-pinned (b) fixed-free (c) fixed-pinned (d) fixed-fixed

Solution The following section properties for a standard steel HSS152.4 × 101.6 × 6.4 shape are given in Appendix B: A = 2,770 mm2, Ix = 8.70×106 mm4, Iy = 4.62×106 mm4 (a) Pinned-pinned column:  2 EI  2 (200,000 N/mm 2 )(4.62  106 mm 4 ) Pcr = = = 253,320 N = 253.320 kN 2 ( KL) 2  (1.0)(6,000 mm) Pallow =

253.320 kN = 126.660 kN = 126.7 kN 2

Ans.

(b) Fixed-free column:  2 EI  2 (200,000 N/mm 2 )(4.62  106 mm 4 ) Pcr = = = 63,330 N = 63.330 kN 2 ( KL) 2  (2.0)(6,000 mm) Pallow =

63.330 kN = 31.665 kN = 31.7 kN 2

Ans.

(c) Fixed-pinned column:  2 EI  2 (200,000 N/mm 2 )(4.62  106 mm 4 ) Pcr = = = 516,979 N = 516.979 kN 2 ( KL) 2 (0.7)(6,000 mm) Pallow =

516.979 kN = 258.490 kN = 258 kN 2

Ans.

(d) Fixed-fixed column:  2 EI  2 (200,000 N/mm 2 )(4.62  106 mm 4 ) Pcr = = = 1,013, 279 N = 1,013.279 kN 2 ( KL) 2 (0.5)(6,000 mm) Pallow =

1,013.279 kN = 506.640 kN = 507 kN 2

Ans.

P16.16 A W250 × 80 structural steel [E = 200 GPa] section (see Appendix B for crosssectional properties) is used as a column with an actual length of L = 12 m. The column is supported only at its ends and it may buckle in any direction. The column is fixed at its base and pinned at its upper end. Determine the maximum load P that may be supported by the column if a factor of safety of 2.5 with respect to buckling is specified.

FIGURE P16.16

Solution The following section properties for a standard steel W250 × 80 shape are given in Appendix B: A = 10,200 mm2, Ix = 126×106 mm4, Iy = 42.9×106 mm4 Maximum allowable column load:  2 EI  2 (200,000 N/mm 2 )(42.9  106 mm 4 ) Pcr = = = 1, 200,130 N = 1, 200.130 kN 2 ( KL) 2 (0.7)(12,000 mm) Pallow =

1, 200.130 kN = 480.052 kN = 480 kN 2.5

Ans.

P16.17 A W14 × 53 structural steel [E = 29,000 ksi] section (see Appendix B for cross-sectional properties) is used as a column with an actual length of L = 16 ft. The column is fixed at its base and unrestrained at its upper end. Determine the maximum load P that may be supported by the column if a factor of safety of 2.5 with respect to buckling is specified.

FIGURE P16.17

Solution The following section properties for a standard steel W14 × 53 shape are given in Appendix B: A = 15.6 in.2, Ix = 541 in.4, Iy = 57.7 in.4 Maximum allowable column load:  2 EI  2 (29,000 ksi)(60.9 in.4 ) Pcr = = = 111.998 kips 2 ( KL) 2  (2.0)(16 ft)(12 in./ft) Pallow =

111.998 kips = 44.799 kips = 44.8 kips 2.5

Ans.

P16.18 A long, slender structural steel [E = 29,000 ksi] HSS8 × 4 × ¼ shape (see Appendix B for crosssectional properties) is used as a 32 ft long column. The column is supported in the x direction at base A and pinned at ends A and C against translation in the y and z directions. Lateral support is provided to the column so that deflection in the x-z plane is restrained at midheight B; however, the column is free to deflect in the x-y plane at B (Figure P16.18). Determine the maximum compressive load the column can support if a factor of safety of 1.92 is required. In your analysis, consider the possibility that buckling could occur about either the strong axis (i.e., the z axis) or the weak axis (i.e., the y axis) of the steel column.

FIGURE P16.18

Solution The following section properties for a standard steel HSS8 × 4 × ¼ shape are given in Appendix B: A = 5.24 in.2, Ix = 42.5 in.4, Iy = 14.4 in.4 Consider buckling about z axis:  2 EI z  2 (29,000 ksi)(42.5 in.4 ) Pcr = = = 82.494 kips 2 ( KL) 2z  (1.0)(32 ft)(12 in./ft)  Pallow =

82.494 kips = 42.966 kips = 43.0 kips 1.92

Consider buckling about y axis:  2 EI y  2 (29,000 ksi)(14.4 in.4 ) Pcr = = = 111.804 kips 2 ( KL)2y (1.0)(16 ft)(12 in./ft)

Pallow =

111.804 kips = 58.231 kips = 58.2 kips 1.92

Maximum compressive load P: Pallow = 43.0 kips

Ans.

P16.19 The aluminum column shown in Figure P16.19 has a rectangular cross section and supports an axial load of P. The base of the column is fixed. The support at the top allows rotation of the column in the x-y plane (i.e., bending about the strong axis) but prevents rotation in the x-z plane (i.e., bending about the weak axis). (a) Determine the critical buckling load of the column for the following parameters: L = 50 in., b = 0.50 in., h = 0.875 in., and E = 10,000 ksi. (b) Determine the ratio b/h for which the critical buckling load about both the strong and weak axes is the same.

FIGURE P16.19

Solution Section properties: (0.50 in.)(0.875 in.)3 I strong = = 0.027913 in.4 12

I weak =

(0.875 in.)(0.50 in.)3 = 0.009115 in.4 12

(a) Critical buckling load: Consider buckling about strong axis: K = 0.7 (fixed-pinned column) Pcr =

 2 EI

 2 (10,000,000 psi)(0.027913 in.4 )

( KL)

(0.7)(50 in.)

= 2

= 2, 248.933 lb

Consider buckling about weak axis: K = 0.5 (fixed-fixed column) Pcr =

 2 EI

 2 (10,000,000 psi)(0.009115 in.4 )

( KL)

(0.5)(50 in.)

= 2

Critical load: Pcr = 1,439 lb

= 1, 439.317 lb

Ans.

(b) Ratio of b/h to give same buckling load: Equate the two buckling equations:  bh3   hb3  2  2 (10,000,000 psi)   (10,000,000 psi)     12  =  12  2 2 (0.7)(50 in.) (0.5)(50 in.) bh3 hb3 = (0.7) 2 (0.5) 2 (0.5) 2 = b2 / h2 (0.7) 2  b / h = 0.714

Ans.

P16.20 The steel compression link shown in Figure P16.20 has a rectangular cross section and supports an axial load of P. The supports allow rotation about the strong axis of the link cross section but prevents rotation about the weak axis. Determine the allowable compression load P if a factor of safety of 2.0 is specified. Use the following parameters: L = 1,200 mm, b = 15 mm, h = 40 mm, and E = 200 GPa.

FIGURE P16.20

Solution Section properties: (15 mm)(40 mm)3 I strong = = 80,000 mm 4 12

I weak =

(40 mm)(15 mm)3 = 11, 250 mm 4 12

Critical buckling load: Consider buckling about strong axis: K = 1.0 (pinned-pinned column) Pcr =

 2 EI

 2 (200,000 N/mm 2 )(80,000 mm 4 )

( KL)

 (1.0)(1, 200 mm)

= 2

= 109,662.3 N = 109.662 kN

Consider buckling about weak axis: K = 0.5 (fixed-fixed column) Pcr =

 2 EI

 2 (200,000 N/mm 2 )(11, 250 mm 4 )

( KL)

(0.5)(1, 200 mm)

= 2

Allowable compression load P: 61.685 kN Pcr = = 30.843 kN = 30.8 kN 2.0

= 61,685.0 N = 61.685 kN

Ans.

P16.21 A stainless steel pipe with an outside diameter of 100 mm and a wall thickness of 8 mm is rigidly attached to fixed supports at A and B. The length of the pipe is L = 8 m, its elastic modulus is E = 190 GPa, and its coefficient of thermal expansion is  = 17.3 × 10−6 /°C. Determine the temperature increase T that will cause the pipe to buckle. FIGURE P16.21

Solution Section properties: d = 100 mm − 2(8 mm) = 84 mm A= I=



(100 mm) 2 − (84 mm) 2  = 2,312.212 mm 2 4



(100 mm) 4 − (84 mm) 4  = 2, 464,818 mm 4 64

Critical buckling load: K = 0.5 Pcr =

(fixed-fixed column)

 2 EI

 2 (190,000 N/mm 2 )(2, 464,818 mm 4 )

( KL)

(0.5)(8,000 mm)

= 2

= 288,880.5 N

Force-Temperature-Deformation Relationship The relationship between internal force, temperature change, and deformation of an axial member is: FL e= +  T L AE Since the rod is attached to rigid supports, e = 0. FL +  T L = 0 AE Set F = Pcr Pcr L +  T L = 0 AE and solve for T: P T = − cr  AE

=−

(17.3  10

−6

−288,880.5 N / C)(2,312.212 mm 2 )(190,000 N/mm 2 )

= 38.009C = 38.0C

Ans.

P16.22 An axial load P is applied to a solid 30 mm diameter steel rod AB as shown in Figure P16.22. For L = 1.5 m, P = 18 kN, and e = 3.0 mm, determine (a) the lateral deflection midway between A and B and (b) the maximum stress in the rod. Use E = 200 GPa.

FIGURE P16.22

Solution Section properties: A= r=



(30 mm) 2 = 706.858 mm 2

 64

(30 mm) 4 = 39,760.782 mm 4

39,760.782 mm = 7.500 mm 706.858 mm 2

(a) Lateral deflection midway between A and B:  L P   vmax = e sec   − 1   2 EI     1,500 mm   18,000 N = (3.0 mm) sec   − 1 2 (200,000 N/mm 2 )(39,760.782 mm 4 )     = 4.0073 mm = 4.01 mm

Ans.

(b) Maximum stress in the rod:  L P  P  ec  max = 1 + 2 sec    A  r 2 r EA     18,000 N   (3.0 mm)(30 mm/2)   1,500 mm 18,000 N = 1 + sec      2 2   706.858 mm 2   (7.500 mm)2   2(7.500 mm) (200,000 N/mm )(706.858 mm )   =

 18,000 N   (3.0 mm)(30 mm/2)  1+   (2.3358)  2  2 706.858 mm   (7.500 mm)  

18,000 N  2.8686 706.858 mm 2 = 73.0487 MPa = 73.0 MPa

Ans.

P16.23 A 4 ft long steel [E = 29,000 ksi; Y = 36 ksi] tube supports an eccentrically applied axial load P, as shown in Figure P16.23. The tube has an outside diameter of 2.00 in. and a wall thickness of 0.15 in. For an eccentricity of e = 0.25 in., determine (a) the maximum load P which can be applied without causing either buckling or yielding of the tube and (b) the corresponding maximum deflection midway between A and B.

FIGURE P16.23

Solution Section properties:  A = (2.00 in.) 2 − (1.70 in.) 2  = 0.871792 in.2 4 I=



(2.00 in.) 4 − (1.70 in.) 4  = 0.375415 in.4 64

d = 2.00 in. − 2(0.15 in.) = 1.70 in. r=

0.375415 in.4 = 0.656220 in. 0.871792 in.2

Euler buckling load:  2 EI  2 (29,000 ksi)(0.375415 in.4 ) Pcr = 2 = = 46.63665 kips 2 L (4 ft)(12 in./ft)

(a)

Allowable eccentric load P based on yield stress: The secant formula is:  KL P   P  ec  max = 1 + 2 sec    A  r 2 r EA    It is convenient to calculate the eccentricity ratio and the slenderness ratio: ec (0.25 in.)(2 in./2) KL (1.0)(48 in.) = = 0.58055 = = 73.146 2 2 r (0.656220 in.) r 0.656220 in. For this column, the secant formula can be written as:    P (48 in.) P  max = 1 + 0.58055 sec ( )    2(0.656220 in.) (29,000 ksi)(0.871792 in.2 )   0.871792 in.2     which can be further simplified to:    P P  max = 1 + ( 0.58055) sec  (36.5731)  2  0.871792 in.  25,281.968 kips    By trial-and-error, determine the value of P that gives max = 36 ksi. From Fig. 16.10 for a slenderness ratio of 73 and an eccentricity ratio of 0.6, we can approximate the average compression stress as P/A = 18 ksi. Thus, we will begin with a trial value of P that corresponds to 18 ksi, i.e., P = (18 ksi)(0.871792 in.2) = 15.7 kips.

Trial value of P (kips) 15.7 15.8 15.9 16.0 16.1 16.05 P16.019955

Corresponding

max

(ksi) 35.074 35.362 35.651 35.942 36.234 36.088 36.000

Thus, the allowable eccentric load is: P = 16.01995 kips = 16.02 kips

(b)

(a) Maximum load P: The maximum load P which can be applied without causing either buckling [see Eq. (a)] or yielding [see Eq. (b)] of the tube is thus: Ans. P = 16.01995 kips = 16.02 kips (b) Corresponding maximum deflection midway between A and B:   P   vmax = e sec   − 1    2 Pcr      16.01995 kips   = (0.25 in.) sec   − 1   2 46.63665 kips   = 0.16301 in. = 0.1630 in.

Ans.

P16.24 A square tube shape made of an aluminum alloy supports an eccentric compression load P that is applied at an eccentricity of e = 4.0 in. from the centerline of the shape (Figure P16.24). The width of the square tube is 3 in. and its wall thickness is 0.12 in. The column is fixed at its base, free at its upper end, and its length is L = 8 ft. For an applied load of P = 900 lb, determine (a) the lateral deflection at the upper end of the column and (b) the maximum stress in the square tube. Use E = 10 × 106 psi. FIGURE P16.24

Solution Section properties: A = (3 in.) 2 − (2.76 in.) 2 = 1.3824 in.2 r=

(3 in.) 4 − (2.76 in.) 4 = 1.914348 in.4 12

1.914348 in.4 = 1.176775 in. 1.3824 in.2

(a) Lateral deflection at the upper end of the column:   KL P   vmax = e sec   − 1   2 EI     (2)(8 ft)(12 in./ft)   900 lb = (4.0 in.) sec   − 1 2 (10  106 psi)(1.914348 in.4 )     = 1.056429 in. = 1.056 in.

(b) Maximum stress in the square tube:  KL P   P  ec  max = 1 + 2 sec    A  r  2r EA    900 lb   (4.0 in.)(3.0 in./2)   (2)(96 in.) 900 lb = 1 + sec      1.3824 in.2   (1.176775 in.)2   2(1.176775 in.) (10  106 psi)(1.3824 in.2 )    900 lb   (4.0 in.)(3.0 in./2)  = 1+  (1.264107)  2  2  1.3824 in.   (1.176775 in.)   900 lb = 6.477071 1.3824 in.2 = 4,216.84 psi = 4,220 psi

Ans.

P16.25 A steel pipe (outside diameter = 130 mm; wall thickness = 12.5 mm) supports an axial load of P = 25 kN, which is applied at an eccentricity of e = 175 mm from the pipe centerline (Figure P16.25). The column is fixed at its base, free at its upper end, and its length is L = 4.0 m. Determine (a) the lateral deflection at the upper end of the column and (b) the maximum stress in the pipe. Use E = 200 GPa.

FIGURE P16.25

Solution Section properties: A=



(130 mm) 2 − (105 mm) 2  = 4,614.2 mm 2 4

d = 130 mm − 2(12.5 mm) = 105 mm



8,053, 246 mm 4 r= = 41.777 mm 4,614.2 mm 2

(130 mm) 4 − (105 mm) 4  = 8,053, 246 mm 4 I= 64

(a) Lateral deflection at the upper end of the column:   KL P   vmax = e sec   − 1   2 EI     (2)(4,000 mm)   25,000 N = (175 mm) sec   − 1 2 (200,000 N/mm 2 )(8,053, 246 mm 4 )     = 24.232 mm = 24.2 mm

Ans.

(b) Maximum stress in the pipe:  KL P   P  ec  max = 1 + 2 sec    A  r  2r EA    25,000 N   (175 mm)(130 mm/2)   (2)(4,000 mm) 25,000 N = 1 + sec      2 2   4,614.2 mm 2   (41.777 mm) 2   2(41.777 mm) (200,000 N/mm )(4,614.2 mm )   =

 25,000 N   (175 mm)(130 mm/2)  1+   (1.138466)  2  2 4,614.2 mm   (41.777 mm)  

25,000 N 8.4199 4,614.2 mm 2

= 45.6194 MPa = 45.6 MPa

Ans.

P16.26 A steel [E = 200 GPa] pipe with an outside diameter of 170 mm and a wall thickness of 7 mm supports an axial load of P, which is applied at an eccentricity of e = 150 mm from the pipe centerline (Figure P16.26). The column is fixed at its base, free at its upper end, and its length is L = 4.0 m. The maximum compression stress in the column must be limited to max = 80 MPa. (a) Use a trial-and-error approach or an iterative numerical solution to determine the allowable eccentric load P that can be applied. (b) Determine the lateral deflection at the upper end of the column for the allowable load P. FIGURE P16.26

Solution Section properties: A= I=



(170 mm) 2 − (156 mm) 2  = 3,584.6 mm 2 4



(170 mm) 4 − (156 mm) 4  = 11,926,718 mm 4 64

d = 170 mm − 2(7 mm) = 156 mm r=

11,926,718 mm 4 = 57.682 mm 3,584.6 mm 2

(a) Allowable eccentric load P: The secant formula is:  KL P   P  ec  max = 1 + 2 sec    A  r  2r EA   It is convenient to calculate the eccentricity ratio for use in the secant formula. ec (150 mm)(170 mm/2) = = 3.8320 r2 (57.682 mm)2 For this column, the secant formula can be written as:   (2)(4,000 mm)  P P  max = 1 + ( 3.8320 ) sec  2  2 2   3,584.6 mm   2(57.682 mm) (200,000 N/mm )(3,584.6 mm )   which can be further simplified to:    P P  max = 1 + 3.8320 sec (69.3457) ( )     3,584.6 mm2  716,911,444 N  

By trial-and-error, determine the value of P that gives max = 80 MPa. We will begin with a trial value of P that corresponds to 25% of max, i.e., P = 0.25(80 MPa)(3,584.557 mm2) = 71,691 N. Trial value of P (N) 71,691 60,000 50,000 55,000 52,500 51,250 51,200 51,230 51,211.265

Corresponding

max

(MPa) 119.656 96.375 77.814 86.947 82.344 80.070 79.980 80.034 80.000

Thus, the allowable eccentric load is: P = 51,211.265 N = 51.2 kN

Ans.

(b) Lateral deflection at the upper end of the column:   KL P   vmax = e sec   − 1 2 EI       (2)(4,000 mm)   51, 211.265 N = (150 mm) sec  − 1 2 4  2 (200,000 N/mm )(11,926,718 mm )     = 30.049 mm = 30.0 mm

Ans.

P16.27 Use the AISC equations to determine the allowable axial load Pallow that may be supported by a W8 × 48 wide-flange column for the following effective lengths: (a) KL = 13 ft and (b) KL = 26 ft. Assume E = 29,000 ksi and Y = 50 ksi.

Solution The following section properties for a standard steel W8 × 48 shape are given in Appendix B: A = 14.1 in.2, rx = 3.61 in., ry = 2.08 in. Limiting slenderness ratio E 29, 000 ksi 4.71 = 4.71 = 113.432 Y 50 ksi (a) Slenderness ratios for KL = 13 ft: KL / rx = (13 ft)(12 in./ft)/(3.61 in.) = 43.213 KL / ry = (13 ft)(12 in./ft)/(2.08 in.) = 75.000

Elastic critical buckling stress  2E  2 (29, 000 ksi) e = = = 50.883 ksi 2 ( 75.000)2  KL    r  Critical and allowable buckling stresses KL E Since  4.71 r y  Y   50 ksi          50.883 ksi  e  cr = 0.658   Y = 0.658  (50 ksi) = 33.140 ksi      33.140 ksi  allow = cr = = 19.844 ksi 1.67 1.67

Allowable axial load: Pallow =  allow A = (19.844 ksi)(14.1 in.2 ) = 280 kips

Ans.

(b) Slenderness ratios for KL = 26 ft: KL / rx = (26 ft)(12 in./ft)/(3.61 in.) = 86.427 KL / ry = (26 ft)(12 in./ft)/(2.08 in.) = 150.000

Elastic critical buckling stress  2E  2 (29, 000 ksi) e = = = 12.721 ksi 2 (150.000)2  KL    r  Critical and allowable buckling stresses KL E Since  4.71 r y

 cr = 0.877 e = (0.877)(12.721 ksi) = 11.156 ksi  11.156 ksi  allow = cr = = 6.680 ksi 1.67

1.67

Allowable axial load: Pallow =  allow A = (6.680 ksi)(14.1 in.2 ) = 94.2 kips

Ans.

P16.28 Use the AISC equations to determine the allowable axial load Pallow that may be supported by a HSS152.4 × 101.6 × 6.4 column for the following effective lengths: (a) KL = 3.75 m and (b) KL = 7.5 m. Assume E = 200 GPa and Y = 320 MPa.

Solution The following section properties for a standard steel HSS152.4 × 101.6 × 6.4 shape are given in Appendix B: A = 2,770 mm2, rx = 55.9 mm, ry = 40.9 mm Limiting slenderness ratio E 200, 000 MPa 4.71 = 4.71 = 117.750 Y 320 MPa (a) Slenderness ratios for KL = 3.75 m: KL / rx = (3,750 mm)/(55.9 mm) = 67.084 KL / ry = (3,750 mm)/(40.9 mm) = 91.687

Elastic critical buckling stress  2E  2 (200, 000 MPa) e = = = 234.809 MPa 2 ( 91.687)2  KL    r  Critical and allowable buckling stresses KL E Since  4.71 r y  Y   320 MPa          234.809 MPa  e    cr = 0.658  Y = 0.658  (320 MPa) = 180.895 MPa      180.895 MPa  allow = cr = = 108.320 MPa 1.67 1.67

Allowable axial load: Pallow =  allow A = (108.320 N/mm2 )(2,770 mm2 ) = 300,047 N = 300 kN

Ans.

(b) Slenderness ratios for KL = 7.5 m: KL / rx = (7,500 mm)/(55.9 mm) = 134.168 KL / ry = (7,500 mm)/(40.9 mm) = 183.374

Elastic critical buckling stress  2E  2 (200, 000 MPa) e = = = 58.702 MPa 2 (183.374)2  KL    r  Critical and allowable buckling stresses KL E Since  4.71 r y

 cr = 0.877 e = (0.877)(58.702 MPa) = 51.482 MPa  51.482 MPa  allow = cr = = 30.827 MPa 1.67

1.67

Allowable axial load: Pallow =  allow A = (30.827 N/mm2 )(2,770 mm2 ) = 85,392 N = 85.4 kN

Ans.

P16.29 The 10 m long HSS304.8 × 203.2 × 9.5 (see Appendix B for cross-sectional properties) column shown in Figure P16.29 is fixed at base A with respect to bending about both the strong and weak axes of the HSS cross section. At upper end B, the column is restrained against rotation and translation in the x-z plane (i.e., bending about the weak axis), and it is restrained against translation in the x-y plane (i.e., free to rotate about the strong axis). Use the AISC equations to determine the allowable axial load Pallow that may be supported by the column based on (a) buckling in the x-y plane and (b) buckling in the x-z plane. Assume E = 200 GPa and Y = 320 MPa. FIGURE P16.29

Solution The following section properties for a standard steel HSS304.8 × 203.2 × 9.5 shape are given in Appendix B: A = 8,520 mm2, rx = 114 mm, ry = 83.1 mm Limiting slenderness ratio E 200, 000 MPa 4.71 = 4.71 = 117.750 Y 320 MPa (a) Consider buckling about strong axis: K x L (0.7)(10,000 mm) = = 61.404 rx 114 mm Elastic critical buckling stress  2E  2 (200, 000 MPa) e = = = 523.532 MPa 2 ( 61.404)2  KL    r  Critical and allowable buckling stresses KL E Since  4.71 r y  Y   320 MPa          cr = 0.658  e    Y = 0.658 523.532 MPa   (320 MPa) = 247.767 MPa      247.767 MPa  allow = cr = = 148.364 MPa 1.67 1.67

Allowable axial load based on buckling in the x-y plane: Pallow =  allow A = (148.364 N/mm2 )(8,520 mm2 ) = 1, 264,057 N = 1, 264 kN

Ans.

(b) Consider buckling about weak axis: K y L (0.5)(10,000 mm) = = 60.168 ry 83.1 mm Elastic critical buckling stress  2E  2 (200, 000 MPa) e = = = 545.245 MPa 2 ( 60.168)2  KL    r  Critical and allowable buckling stresses KL E Since  4.71 r y  Y   320 MPa          545.245 MPa  e  cr = 0.658   Y = 0.658  (320 MPa) = 250.304 MPa      250.304 MPa  allow = cr = = 149.883 MPa 1.67 1.67

Allowable axial load based on buckling in the x-z plane: Pallow =  allow A = (149.883 N/mm2 )(8,520 mm2 ) = 1, 277,001 N = 1, 277 kN

Ans.

P16.30 The 25 ft long HSS6 × 4 × 1/8 (see Appendix B for cross-sectional properties) column shown in Figure P16.30 is fixed at base A with respect to bending about both the strong and weak axes of the HSS cross section. At upper end B, the column is restrained against rotation and translation in the x-z plane (i.e., bending about the weak axis), and it is restrained against translation in the x-y plane (i.e., free to rotate about the strong axis). Use the AISC equations to determine the allowable axial load Pallow that may be supported by the column based on (a) buckling in the x-y plane and (b) buckling in the x-z plane. Assume E = 29,000 ksi and Y = 46 ksi. FIGURE P16.30

Solution The following section properties for a standard steel HSS6 × 4 × 1/8 shape are given in Appendix B: A = 2.23 in.2, rx = 2.26 in., ry = 1.66 in. Limiting slenderness ratio E 29, 000 ksi 4.71 = 4.71 = 118.261 Y 46 ksi (a) Consider buckling about strong axis: K x L (0.7)(25 ft)(12 in./ft) = = 92.920 rx 2.26 in. Elastic critical buckling stress  2E  2 (29, 000 ksi) e = = = 33.149 ksi 2 ( 92.920) 2  KL    r  Critical and allowable buckling stresses KL E Since  4.71 r y  Y   46 ksi          cr = 0.658  e    Y = 0.658 33.149 ksi   (46 ksi) = 25.735 ksi      25.735 ksi  allow = cr = = 15.410 ksi 1.67 1.67

Allowable axial load based on buckling in the x-y plane: Pallow =  allow A = (15.410 ksi)(2.230 in.2 ) = 34.4 kips

Ans.

(b) Consider buckling about weak axis: K y L (0.5)(25 ft)(12 in./ft) = = 90.361 ry 1.66 in. Elastic critical buckling stress  2E  2 (29, 000 ksi) e = = = 35.054 ksi 2 ( 90.361) 2  KL    r  Critical and allowable buckling stresses KL E Since  4.71 r y  Y   46 ksi          35.054 ksi  e  cr = 0.658   Y = 0.658  (46 ksi) = 26.559 ksi      26.559 ksi  allow = cr = = 15.904 ksi 1.67 1.67

Allowable axial load based on buckling in the x-z plane: Pallow =  allow A = (15.904 ksi)(2.23 in.2 ) = 35.5 kips

Ans.

P16.31 A column with an effective length of 28 ft is fabricated by connecting two C15 × 40 steel channels (see Appendix B for cross-sectional properties) with lacing bars as shown in Figure P16.31. Use the AISC equations to determine the allowable axial load Pallow that may be supported by the column if d = 10 in. Assume E = 29,000 ksi and Y = 36 ksi.

Solution The following section properties for a standard steel C15 × 40 shape are given in Appendix B: A = 11.8 in.2, rx = 5.45 in. Iy = 9.17 in.4, x = 0.778 in. FIGURE P16.31 Limiting slenderness ratio E 29, 000 ksi 4.71 = 4.71 = 133.681 Y 36 ksi Slenderness ratio about horizontal cross-sectional axis: K x L (1.0)(28 ft)(12 in./ft) = = 61.651 rx 5.45 in. Slenderness ratio about vertical cross-sectional axis: I = 2 9.17 in.4 + (5 in. − 0.778 in.) 2 (11.8 in.2 )  = 439.017 in.4 KyL ry

ry =

439.017 in.4 = 4.313 in. 2(11.8 in.2 )

(1.0)(28 ft)(12 in./ft) = 77.903 4.313 in.

Controlling slenderness ratio: K y L / ry = 77.903 Elastic critical buckling stress  2E  2 (29, 000 ksi) e = = = 47.162 ksi 2 ( 77.903) 2  KL    r  Critical and allowable buckling stresses KL E Since  4.71 r y  Y   36 ksi          47.162 ksi  e    cr = 0.658  Y = 0.658  (36 ksi) = 26.155 ksi      26.155 ksi  allow = cr = = 15.661 ksi 1.67 1.67

Allowable axial load: Pallow =  allow A = (15.661 ksi)(2  11.8 in.2 ) = 370 kips

Ans.

P16.32 A column is fabricated by connecting two C310 × 45 steel channels (see Appendix B for cross-sectional properties) with lacing bars as shown in Figure P16.32. (a) Determine the distance d required so that the moments of inertia for the section about the two principal axes are equal. (b) For a column with an effective length of KL = 9.5 m, determine the allowable axial load Pallow that may be supported by the column using the value of d determined in part (a). Use the AISC equations and assume E = 200 GPa and Y = 340 MPa. FIGURE P16.32

Solution The following section properties for a standard steel C310 × 45 shape are given in Appendix B: A = 5,680 mm2, Ix = 67.4×106 mm4, rx = 109 mm, Iy = 2.13×106 mm4, x = 17.1 mm (a) Determine distance d: I x = 2(67.4 106 mm4 ) = 134.8 106 mm4 I y = 2 2.13  106 mm4 + (d / 2 − 17.1 mm)2 (5,680 mm2 ) 

Equate these two moment of inertia expressions and solve for d: d = 248.594 mm = 249 mm (b) Determine the allowable axial load: Limiting slenderness ratio E 200, 000 MPa 4.71 = 4.71 = 114.234 Y 340 MPa Slenderness ratio: 134.8  106 mm 4 r= = 108.932 mm 2(5,680 mm 2 )



KL 9,500 mm = = 87.210 r 108.932 mm

Elastic critical buckling stress  2E  2 (200, 000 MPa) e = = = 259.534 MPa 2 (87.210)2  KL    r 

Ans.

Critical and allowable buckling stresses KL E Since  4.71 r y  Y   340 MPa          259.534 MPa  e  cr = 0.658   Y = 0.658  (340 MPa) = 196.493 MPa      196.493 MPa  allow = cr = = 117.661 MPa 1.67 1.67

Allowable axial load: Pallow =  allow A = (117.661 N/mm2 )(2  5,680 mm2 ) = 1,337 kN

Ans.

P16.33 A compression chord of a small truss consists of two L127 × 76 × 12.7 steel angles arranged with long legs back-to-back as shown in Figure P16.33. The angles are separated at intervals by spacer blocks. (a) Determine the spacer thickness required so that the moments of inertia for the section about the two principal axes are equal. (b) For a compression chord with an effective length of KL = 7 m, determine the allowable axial load Pallow that may be supported by the column using the spacer thickness determined in part (a). Use the AISC equations and assume E = 200 GPa and Y = 340 MPa. FIGURE P16.33

Solution The following section properties for a standard steel L127 × 76 × 12.7 shape are given in Appendix B: A = 2,420 mm2, Ix = 3.93×106 mm4, rx = 40.1 mm, Iy = 1.06×106 mm4, x = 18.9 mm (a) Determine spacer thickness b: I x = 2(3.93 106 mm4 ) = 7.86 106 mm4 I y = 2 1.06  106 mm4 + (b / 2 + 18.9 mm)2 (2,420 mm2 ) 

Equate these two moment of inertia expressions and solve for the block thickness b: b = 31.075 mm = 31.1 mm

Ans.

(b) Determine the allowable axial load: Limiting slenderness ratio E 200, 000 MPa 4.71 = 4.71 = 114.234 Y 340 MPa Slenderness ratio: 7.86  106 mm 4 KL 7,000 mm r= = 40.298 mm  = = 173.704 2 2(2, 420 mm ) r 40.298 mm Elastic critical buckling stress  2E  2 (200, 000 MPa) e = = = 65.420 MPa 2 (173.704)2  KL    r  Critical and allowable buckling stresses KL E Since  4.71 r y

 cr = 0.877 e = (0.877)(65.420 MPa) = 57.373 MPa  57.373 MPa  allow = cr = = 34.355 MPa 1.67

1.67

Allowable axial load: Pallow =  allow A = (34.355 N/mm2 )(2  2, 420 mm2 ) = 166.3 kN

Ans.

P16.34 Develop a list of three acceptable structural steel WT shapes (from those listed in Appendix B) that can be used as an 18 ft long pin-ended column to carry an axial compression load of 30 kips. Include the most economical WT8, WT9, and WT10.5 shapes on the list of possibilities and select the most economical shape from the available alternatives. Use the AISC equation for long columns and assume E = 29,000 ksi and Y = 50 ksi.

Solution Limiting slenderness ratio E 29, 000 ksi 4.71 = 4.71 = 113.432 Y 50 ksi Allowable buckling stress KL E If  4.71 r y

 allow =

 cr 1.67

0.877 e 0.877 2 E 5.183020 E 150,307.5741 ksi = = = 2 2 2 1.67  KL   KL   KL  1.67       r  r  r 

Investigate various shapes:

allow

Pallow

(ksi)

(kips)

184.615

4.410

20.110

N.G.

1.56

138.462

7.840

46.178

O.K.

5.15

1.22

177.049

4.795

24.694

N.G.

WT9×20

5.88

1.27

170.079

5.196

30.553

O.K.

WT10.5×22

6.49

1.26

171.429

5.115

33.194

O.K.

Designation

Area A

(in.2)

(in.)

WT8×15.5

4.56

1.17

WT8×20

5.89

WT9×17.5

KL/ry

Lightest shape is WT8×20. Other acceptable shapes are noted above.

Ans.

P16.35 A 6061-T6 aluminum-alloy pipe column with pinned ends has an outside diameter of 4.50 in. and a wall thickness of 0.237 in. Determine the allowable axial load Pallow that may be supported by the aluminum pipe column for the following effective lengths: (a) KL = 7.5 ft and (b) KL = 15 ft. Use the Aluminum Association column design formulas.

Solution Section properties: d = 4.50 in. − 2(0.237 in.) = 4.026 in. A= I=



(4.50 in.) 2 − (4.026 in.) 2  = 3.174048 in.2 4



(4.50 in.) 4 − (4.026 in.) 4  = 7.232600 in.4 64

7.232600 in.4 r= = 1.509526 in. 3.174048 in.2

(a) KL = 7.5 ft: Effective-slenderness ratio: KL (7.5 ft)(12 in./ft) = = 59.621 r 1.509526 in. Aluminum Association column design formula:

 allow =  20.2 − 0.125( KL / r ) ksi

where 9.5 

=  20.2 − 0.125(59.621) ksi

KL  66 r

= 12.747 ksi Allowable axial load Pallow: Pallow =  allow A = (12.747 ksi)(3.174048 in.2 ) = 40.5 kips

Ans.

(b) KL = 15 ft: Effective-slenderness ratio: KL (15 ft)(12 in./ft) = = 119.243 r 1.509526 in. Aluminum Association column design formula: 51,000 KL  allow = ksi where  66 2 ( KL / r ) r 51,000 = ksi (119.243) 2 = 3.587 ksi Allowable axial load Pallow: Pallow =  allow A = (3.587 ksi)(3.174048 in.2 ) = 11.39 kips

Ans.

P16.36 A 6061-T6 aluminum-alloy rectangular tube shape has cross-sectional dimensions of b = 100 mm, d = 150 mm, and t = 5 mm as shown in Figure P16.36. The rectangular tube is used as a compression member that is 7.5 m long. For buckling about the z axis, assume that both ends of the column are pinned. For buckling about the y axis, however, assume that both ends of the column are fixed. Determine the allowable axial load Pallow that may be supported by the rectangular tube. Use the Aluminum Association column design formulas.

FIGURE P16.36

Solution Section properties: A = (100 mm )(150 mm ) − ( 90 mm )(140 mm ) = 2, 400 mm 2

(100 mm )(150 mm ) − ( 90 mm )(140 mm ) = 7.545  106 mm 4 I = 3

rz =

7.545  10 mm = 56.069 mm 2, 400 mm 2 6

(150 mm )(100 mm ) − (140 mm )( 90 mm ) = 3.995  106 mm 4 I = 3

ry =

3.995  106 mm 4 = 40.799 mm 2, 400 mm 2

Effective-slenderness ratios: KL (1.00 )( 7,500 mm ) = = 133.763 rz 56.059 mm

KL ( 0.5 )( 7,500 mm ) = = 91.913 ry 40.799 mm

Aluminum Association column design formula: 351,000 KL  allow = MPa where  66 2 ( KL / r ) r 351,000 = MPa 2 (133.763)

= 19.617 MPa Allowable axial load Pallow: Pallow =  allow A = (19.617 N/mm2 )( 2,400 mm2 ) = 47,080.8 N = 47.1 kN

Ans.

P16.37 The aluminum column shown in Figure P16.37 has a rectangular cross section and supports a compressive axial load P. The base of the column is fixed. The support at the top allows rotation of the column in the x-y plane (i.e., bending about the strong axis) but prevents rotation in the x-z plane (i.e., bending about the weak axis). Determine the allowable axial load Pallow that may be applied to the column for the following parameters: L = 60 in., b = 1.25 in., and h = 2.00 in. Use the Aluminum Association column design formulas.

FIGURE P16.37

Solution Section properties: A = (1.25 in.)(2.00 in.) = 2.50 in.2

Iz =

(1.25 in.)(2.00 in.)3 = 0.833333 in.4 12

rz =

(2.00 in.)(1.25 in.)3 Iy = = 0.325521 in.4 12 Effective-slenderness ratio: K z L (0.7)(60 in.) = = 72.746 rz 0.577350 in.

0.833333 in.4 = 0.577350 in. 2.50 in.2

0.325521 in.4 ry = = 0.360844 in. 2.50 in.2

KyL ry

(0.5)(60 in.) = 83.138 0.360844 in.

Aluminum Association column design formula: 51,000 KL  allow = ksi where  66 2 ( KL / r ) r 51,000 = ksi (83.138) 2 = 7.378 ksi Allowable axial load Pallow: Pallow =  allow A = (7.378 ksi)(2.50 in.2 ) = 18.45 kips

Ans.

P16.38 A 6061-T6 aluminum-alloy wide-flange shape is used as a column of length L = 5.5 m. The column is fixed at base A. Pin-connected lateral bracing is present at B so that deflection in the x-z plane is restrained at the upper end of the column; however, the column is free to deflect in the x-y plane at B (see Figure P16.38a). The cross-sectional dimensions of the shape as shown in Figure P16.38b are bf = 130 mm, tf = 9 mm, d = 200 mm, and tw = 6 mm. Use the Aluminum Association column design formulas to determine the allowable compressive load Pallow that the column can support. In your analysis, consider the possibility that buckling could occur about either the strong axis (i.e., the z axis) or the weak axis (i.e., the y axis) of the aluminum column.

FIGURE P16.38b

FIGURE P16.38a

Solution Section properties: A = 2 (130 mm )( 9 mm ) + ( 6 mm )(182 mm ) = 3, 432 mm 2

(130 mm )( 200 mm ) − (124 mm )(182 mm ) = 24,371, 464 mm 4 I = 3

rz =

24,371, 464 mm 4 = 84.269 mm 3, 432 mm 2

 ( 9 mm )(130 mm )3  (182 mm )( 6 mm )3 Iy = 2 = 3, 298,776 mm 4 + 12 12   3, 298,776 mm 4 ry = = 31.003 mm 3, 432 mm 2 Effective-slenderness ratios: K z L ( 2.0 )( 5,500 mm ) = = 130.535 rz 84.269 mm

KL ( 0.7 )( 5,500 mm ) = = 124.182 ry 31.003 mm

Aluminum Association column design formula: 351,000 KL  allow = MPa where  66 2 ( KL / r ) r 351,000 = MPa = 20.599 MPa 2 (130.535) Allowable axial load Pallow: Pallow =  allow A = ( 20.599 N/mm2 )( 3,432 mm2 ) = 70,697.4 N = 70.7 kN

Ans.

P16.39 A wood post of rectangular cross section (Figure P16.39) consists of Select Structural grade Douglas fir lumber  = 580,000 psi). The finished dimensions (Fc = 1,150 psi; Emin of the post are b = 5.5 in. and d = 7.5 in. Assume pinned connections at each end of the post. Determine the allowable axial load Pallow that may be supported by the post for the following column lengths: (a) L = 10 ft, (b) L = 16 ft, and (c) L = 24 ft. Use the NFPA NDS column design formula.

FIGURE P16.39

Solution (a) L = 10 ft KL (1.0 )(10 ft )(12 in./ft ) = = 21.8182 d 5.5 in. 0.822 ( 580,000 psi )  0.822 Emin FcE = = = 1,001.527 psi 2 2 ( KL / d ) ( 21.8182 ) 1 + ( F / F ) cE c − 2 c 

 allow = Fc 

FcE 1,001.527 psi = = 0.8709 Fc 1,150 psi

2 FcE / Fc  1 + ( FcE / Fc )  −    2c c   

2 1 + 0.8709 1 + ( 0.8709 )  0.8709  )  ( = (1,150 psi )  −   = 737.634 psi  − 0.8   2(0.8)   2(0.8)  Pallow = ( 737.634 psi )( 5.5 in.)( 7.5 in.) = 30,427 lb = 30,400 lb

(b) L = 16 ft KL (1.0 )(16 ft )(12 in./ft ) = = 34.9091 d 5.5 in. 0.822 ( 580,000 psi )  0.822 Emin FcE = = = 391.222 psi 2 2 ( KL / d ) ( 34.9091) 1 + ( F / F ) cE c − 2 c 

 allow = Fc 

Ans.

FcE 391.222 psi = = 0.3402 Fc 1,150 psi

2 FcE / Fc  1 + ( FcE / Fc )  −    2c c   

2 1 + 0.3402 1 + ( 0.3402 )  0.3402  )  ( = (1,150 psi )  −   = 358.701 psi  − 0.8   2(0.8)   2(0.8)  Pallow = ( 358.701 psi )( 5.5 in.)( 7.5 in.) = 14,796 lb = 14,800 lb

Ans.

(c) L = 24 ft KL (1.0 )( 24 ft )(12 in./ft ) = = 52.3636 d 5.5 in. 0.822 ( 580,000 psi )  0.822 Emin FcE = = = 173.876 psi 2 2 ( KL / d ) ( 52.3636 ) 1 + ( F / F ) cE c  allow = Fc  − 2 c 

FcE 173.876 psi = = 0.1512 Fc 1,150 psi

2 1 + ( FcE / Fc )  FcE / Fc     − 2c c   

2 1 + 0.1512 1 + ( 0.1512 )  0.1512  )  ( = (1,150 psi )  −   = 168.119 psi  − 2(0.8) 2(0.8) 0.8     Pallow = (168.119 psi )( 5.5 in.)( 7.5 in.) = 6,934.9 lb = 6,930 lb

Ans.

P16.40 A Select Structural grade Hem-Fir (Fc  = 4.0 GPa) wood column of = 10.3 MPa; Emin rectangular cross section has finished dimensions of b = 100 mm and d = 235 mm. The length of the column is L = 4.75 m. The column is fixed at base A. Pin-connected lateral bracing is present at B so that deflection in the x-z plane is restrained at the upper end of the column; however, the column is free to deflect in the x-y plane at B (see Figure P16.40). Use the NFPA NDS column design formula to determine the allowable compressive load Pallow that the column can support. In your analysis, consider the possibility that buckling could occur about either the strong axis (i.e., the z axis) or the weak axis (i.e., the y axis) of the wood column.

Solution KL ( 0.7 )( 4,750 mm ) = = 33.250 d1 100 mm FcE =

0.822 ( 4,000 MPa )  0.822 Emin = = 2.0120 MPa 2 2 ( KL / d ) ( 40.426 ) 1 + ( F / F ) cE c − 2 c 

 allow = Fc 

KL ( 2.0 )( 4,750 mm ) = = 40.426 d2 235 mm FcE 2.0120 MPa = = 0.1953 Fc 10.3 MPa

2 1 + ( FcE / Fc )  FcE / Fc     − 2c c   

2 1 + 0.1953 1 + ( 0.1953)  0.1953  )  ( = (10.3 MPa )  −   = 1.9236 MPa  − 0.8   2(0.8)   2(0.8)  2 Pallow = (1.9236 N/mm ) (100 mm )( 235 mm ) = 45,205 N = 45.2 kN

Ans.

P16.41 A simple pin-connected wood truss is loaded and supported as shown in Figure P16.41. The members of the truss are square Douglas fir timbers (finished dimensions = 3.5 in. by 3.5 in.)  = 580,000 psi. with Fc = 1,350 psi and Emin (a) For the loads shown, determine the axial forces produced in chord members AF, FG, GH, and EH and in web members BG and DG. (b) Use the NFPA NDS column design formula to determine the allowable compressive load Pallow for each of these members. (c) Report the ratio Pallow /Pactual for each of these members.

FIGURE P16.41

Solution Truss analysis results Member Member Length (ft)

(a) Axial Force Pactual (lb)

2,175 (T)

3,625 (C)

3,450 (T)

2,900 (T)

2,125 (C)

3,450 (T)

2,400 (T)

2,925 (T)

875 (C)

3,900 (T)

FG GH

(b) Allowable Force Pallow (lb)

4,611.7

1.272

4,611.7

2.17

4,611.7

5.27

4,875 (C)

4,611.7

0.946

2,175 (C)

10,344.8

4.76

2,925 (C)

10,344.8

3.54

P16.42 The structural steel column shown in Figure P16.42 is fixed at its base and free at its upper end. At the top of the column, a load P is applied to the stiffened seat support at an eccentricity of e = 9 in. from the centroidal axis of the wide-flange shape. Use the AISC equations given in Section 16.5 and assume that E = 29,000 ksi and Y = 36 ksi. Employ the allowable stress method to determine: (a) whether the column is safe for a load of P = 15 kips. Report the results in the form of the stress ratio x /allow. (b) the magnitude of the largest eccentric load P that may be applied to the column.

Solution Section properties: A = 2(0.50 in.)(8 in.) + (7 in.)(0.35 in.) = 10.450 in.2 Iz =

FIGURE P16.42

(8 in.)(8 in.)3 (7.65 in.)(7 in.)3 − = 122.671 in.4 12 12

 (0.50 in.)(8 in.)3  (7 in.)(0.35 in.)3 Iy = 2 = 42.692 in.4 + 12 12   Limiting slenderness ratio E 29, 000 ksi 4.71 = 4.71 = 133.681 Y 36 ksi

K z L (2)(12 ft)(12 in./ft) = = 84.058 rz 3.426 in.

KyL ry

rz =

122.671 in.4 = 3.426 in. 10.450 in.2

ry =

42.692 in.4 = 2.021 in. 10.450 in.2

(2)(12 ft)(12 in./ft) = 142.488 2.021 in.

Elastic critical buckling stress  2E  2 (29, 000 ksi) e = = = 14.097 ksi 2 (142.488) 2  KL    r  Critical and allowable buckling stresses KL E Since  4.71 r y

 cr = 0.877 e = (0.877)(14.097 ksi) = 12.363 ksi  12.363 ksi  allow = cr = = 7.403 ksi 1.67

1.67

(a) Allowable stress method with P = 15 kips: P M c 15 kips (15 kips)(9 in.)(8 in./2) x = + z = + = 1.435 ksi + 4.402 ksi = 5.837 ksi 2 A Iz 10.450 in. 122.671 in.4 x 5.837 ksi = = 0.788  Safe  allow 7.403 ksi

Ans.

(b) Largest eccentric load P: 1 (9 in.)(8 in./2)   P +  7.403 ksi 2 122.671 in.4  10.450 in.

 P  19.02 kips

Ans.

P16.43 The structural steel column shown in Figure P16.43 is fixed at its base and free at its upper end. At the top of the column, a load P is applied to the stiffened seat support at an eccentricity of e from the centroidal axis of the wide-flange shape. Using the allowable stress method, determine the maximum allowable eccentricity e if (a) P = 15 kips and (b) P = 35 kips. Apply the AISC equations given in Section 16.5 and assume that E = 29,000 ksi and Y = 50 ksi.

Solution Section properties: A = 2(0.50 in.)(8 in.) + (7 in.)(0.35 in.) = 10.450 in.2 Iz =

FIGURE P16.43

(8 in.)(8 in.)3 (7.65 in.)(7 in.)3 − = 122.671 in.4 12 12

rz =

 (0.50 in.)(8 in.)3  (7 in.)(0.35 in.)3 Iy = 2 = 42.692 in.4 + 12 12  

122.671 in.4 = 3.426 in. 10.450 in.2

42.692 in.4 ry = = 2.021 in. 10.450 in.2

Limiting slenderness ratio E 29, 000 ksi 4.71 = 4.71 = 113.432 Y 50 ksi

K z L (2)(12 ft)(12 in./ft) = = 84.058 rz 3.426 in.

KyL ry

(2)(12 ft)(12 in./ft) = 142.488 2.021 in.

Elastic critical buckling stress  2E  2 (29, 000 ksi) e = = = 14.097 ksi 2 (142.488) 2  KL    r  Critical and allowable buckling stresses KL E Since  4.71 r y

 cr = 0.877 e = (0.877)(14.097 ksi) = 12.363 ksi  12.363 ksi  allow = cr = = 7.403 ksi 1.67

1.67

(a) Allowable eccentricity for P = 15 kips: P Pec +   allow A Iz P I  e   allow −  z A  Pc  15 kips  122.671 in.4   (7.403 ksi) − = 12.20 in. 10.450 in.2  (15 kips)(8 in./2) 

Ans.

(b) Allowable eccentricity for P = 35 kips: P I  e   allow −  z A  Pc  35 kips  122.671 in.4   (7.403 ksi) − = 3.55 in. 10.450 in.2  (35 kips)(8 in./2) 

Ans.

P16.44 A W200 × 46.1 structural steel shape (see Appendix B for cross-sectional properties) is used as a column to support an eccentric axial load P. The column is 3.6 m long and it is fixed at its base and free at its upper end. At the upper end of the column (see Figure P16.44), the load P is applied to a bracket at a distance of e = 170 mm from the x axis, creating a bending moment about the strong axis of the W200 × 46.1 shape (i.e., the z axis). Apply the AISC equations given in Section 16.5 and assume E = 200 GPa and Y = 250 MPa. Based on the allowable stress method, determine: (a) whether the column is safe for a load of P = 125 kN. Report the results in the form of the stress ratio x /allow. (b) the magnitude of the largest eccentric load P that may be applied to the column. FIGURE P16.44

Solution The following section properties for a standard steel W10 × 54 shape are given in Appendix B: A = 5,880 mm2, d = 203 mm, Ix = 45.8×106 mm4, rx = 88.1 mm, ry = 51.3 mm Limiting slenderness ratio E 200, 000 MPa 4.71 = 4.71 = 133.219 Y 250 MPa

K z L (2)(3,600 mm) = = 81.725 rz 88.1 mm

KyL ry

(2)(3,600 mm) = 140.351 51.3 mm

Elastic critical buckling stress  2E  2 (200, 000 MPa) e = = = 100.207 MPa 2 (140.351) 2  KL    r  Critical and allowable buckling stresses KL E Since  4.71 r y

 cr = 0.877 e = (0.877)(100.207 MPa) = 87.882 MPa  87.882 MPa  allow = cr = = 52.624 MPa 1.67

1.67

(a) Allowable stress method with P = 125 kN: P M c 125,000 N (125,000 N)(170 mm)(203 mm/2) x = + z = + A Iz 5,880 mm 2 45.8  106 mm 4 = 21.259 MPa + 47.093 MPa = 68.352 MPa x 68.352 MPa = = 1.299  Not Safe  allow 52.624 MPa

Ans.

(b) Largest eccentric load P: 1 (170 mm)(203 mm/2)   P +  52.624 MPa 2 45.8  10 6 mm 4   5,880 mm  P  96, 237 N = 96.2 kN

Ans.

P16.45 An eccentric compression load of P = 32 kN is applied at an eccentricity of e = 12 mm from the centerline of a solid 45-mm-diameter 6061-T6 aluminumalloy rod (see Figure P16.45). Using the interaction method and an allowable bending stress of 150 MPa, determine the longest effective length L that can be used.

Solution Section properties:  A = (45 mm) 2 = 1,590.431 mm 2 4 r=

 64

(45 mm) 4 = 201, 288.959 mm 4

201, 288.959 mm = 11.25 mm 1,590.431 mm 2

FIGURE P16.45 Interaction equation: The interaction equation has the form P/ A Mc/ I + =1 ( allow )a ( allow )b The unknown in this equation is (allow)a. Move the bending stress terms to the right-hand side of the equation: P/ A Mc/ I =1− ( allow )a ( allow )b and solve for (allow)a P/ A P/ A ( allow )a = = Mc/ I Pec / I 1− 1− ( allow )b ( allow )b For this column: 32,000 N 20.120 MPa 1,590.431 mm 2 ( allow ) a = = = 28.186 MPa (32,000 N)(12 mm)(45 mm/2) 1 − 0.285156 201,288.959 mm 4 1− 150 MPa Aluminum Association column design formula: Assume KL/r > 66; therefore, 351,000 KL  allow = MPa where  66 2 ( KL / r ) r Solve for maximum L: 351,000 MPa 28.186 MPa  ( KL / r )2

 KL / r 

351,000 MPa = 111.593 28.186 MPa

L  (111.593)(11.25 mm)/(1.0) = 1,255.424 mm = 1,255 mm

Ans.

P16.46 An eccentric compression load of P = 13 kips is applied at an eccentricity of e = 0.75 in. from the centerline of a solid 6061-T6 aluminum-alloy rod (see Figure P16.46). The rod has an effective length of 45 in. Using the interaction method and an allowable bending stress of 21 ksi, determine the smallest diameter that can be used.

Solution Section properties:

 4

 64

 d

d2 d = =  2 16 4 d 4

FIGURE P16.46

Aluminum Association column design formula: If KL/r > 66, then the allowable axial stress for this column can be expressed as: 51,000 ksi 51,000 ksi  allow = = = 1.5741d 2 2 2  KL   45 in.    r  d /4 Interaction equation: The interaction equation has the form P/ A Mc/ I P/ A Pec / I + = + =1 ( allow )a ( allow )b ( allow )a ( allow )b For a solid rod, the interaction equation can be expressed as: ec / I  1 ec   1/ A  + + P = P    ( allow ) a ( allow )b   A( allow ) a I ( allow )b  4 32ed 8e    4P  d + + = P 2 = 4   =1 3    d ( allow ) a  d ( allow )b   d  ( allow ) a ( allow )b  For this column, the interaction equation becomes: 4(13 kips)  d 8(0.75 in.)  52 kips  1  = + + 0.285714 in.−1  = 1 3 3 2    d  d 1.5741d 21 ksi  1.5741d 

Solve this equation by trial-and-error: d (in.)

52 kips  1  + 0.285714 in.−1  3   d 1.5741d 

1.0 2.0 2.5 2.25 2.125 2.131

15.624 1.250 0.572 0.826 1.009 1.000

Therefore, the minimum diameter for the column is dmin = 2.13 in.

Ans.

P16.47 A sawn wood post of rectangular cross section (Figure P16.47) consists of Select Structural Spruce-PineFir lumber (Fc = 700 psi; E′min = 440,000 psi). The finished dimensions of the post are b = 5.5 in. and h = 7.25 in. The post is 12 ft. long and the ends of the post are pinned. Using the interaction method and an allowable bending stress of 1,000 psi, determine the maximum allowable load that can be supported by the post if the load P acts at an eccentricity of e = 6 in. from the centerline of the post. Use the NFPA NDS column design formula. FIGURE P16.47

Solution (7.25 in.)(5.5 in.)3 Iy = = 100.5182 in.4 12 KL (1.0)(12 ft)(12 in./ft) = = 26.1818 d 5.5 in.  0.822 Emin 0.822(440,000 psi) FcE = = = 527.6244 psi 2 ( KL / d ) (26.1818)2 1 + ( F / F ) cE c − 2 c 

 allow = Fc 

FcE 527.6244 psi = = 0.7537 Fc 700 psi

2 1 + ( FcE / Fc )  FcE / Fc     − 2c c   

1 + 0.7537 )−  ( = (700 psi)   2(0.8)

2 1 + ( 0.7537 )  0.7537   = 410.8601 psi   − 0.8   2(0.8)  

Interaction equation: 1  (6.0 in.)(5.5 in. / 2)   (5.5 in.)(7.25 in.)  4 + = P + 100.5182 in.  1 ( allow ) a ( allow )b 1,000 psi  410.8601 psi  P A

Mc I

1 1   + P  1 16,383.0465 lb 6,092.0121 lb   P  4, 440.73 lb = 4, 440 lb

Ans.

P17.1 Determine the modulus of resilience for each of the following aluminum alloys: (a) 7075-T651 E = 71.7 GPa, Y = 503 MPa (b) 5082-H112 E = 70.3 GPa, Y = 190 MPa (c) 6262-T651 E = 69.0 GPa, Y = 241 MPa

Solution E = 71.7 GPa, Y = 503 MPa  (503  106 N/m2 )2 ur = = = 1,764.36  103 N-m/m3 = 1,764 kJ/m3 9 2 2E 2(71.7  10 N/m )

(a) 7075-T651

2 Y

E = 70.3 GPa, Y = 190 MPa  (190  106 N/m2 )2 ur = = = 256.76  103 N-m/m3 = 257 kJ/m3 9 2 2E 2(70.3  10 N/m )

Ans.

(b) 5082-H112

2 Y

E = 69.0 GPa, Y = 241 MPa  (241  106 N/m2 )2 ur = = = 420.88  103 N-m/m3 = 421 kJ/m3 9 2 2E 2(69.0  10 N/m )

Ans.

2 Y

Ans.

P17.2 For each of the following metals, calculate the modulus of resilience: (a) Red Brass UNS C23000 E = 115 GPa, Y = 125 MPa (b) Titanium Ti-6Al-4V (Grade 5) Annealed E = 114 GPa, Y = 830 MPa (c) 304 Stainless Steel E = 193 GPa, Y = 215 MPa

Solution (a) Red Brass UNS C23000 E = 115 GPa, Y = 125 MPa 2 6 2 2  (125  10 N/m ) ur = Y = = 67.935  103 N-m/m3 = 67.9 kJ/m3 2E 2(115  109 N/m2 ) (b) Titanium Ti-6Al-4V (Grade 5) Annealed E = 114 GPa, Y = 830 MPa 2 6 2 2  (830  10 N/m ) ur = Y = = 3,021.49  103 N-m/m3 = 3,020 kJ/m3 9 2 2E 2(114  10 N/m ) (c) 304 Stainless Steel E = 193 GPa, Y = 215 MPa 2 6 2 2  (215  10 N/m ) ur = Y = = 119.754  103 N-m/m3 = 119.8 kJ/m3 9 2 2E 2(193  10 N/m )

Ans.

P17.3 The compound solid steel rod shown in Figure P17.3/4 is subjected to a tensile force P. Assume that E = 29,000 ksi, d1 = 0.50 in., L1 =18 in., d2 = 0.875 in., L2 = 27 in., and P = 5.5 kips. Determine (a) the elastic strain energy in rod ABC. (b) the corresponding strain-energy density in segments (1) and (2) of the rod.

FIGURE P17.3/4

Solution Compute the cross-sectional areas of segments (1) and (2). A1 = A2 =



 4

(0.5 in.) 2 = 0.19635 in.2 (0.875 in.) 2 = 0.60132 in.2

(a) Elastic Strain Energy: Use Equation (17.14) to calculate the strain energy of each segment as well as the total strain energy in the brass rod. n F 2L F 2L F 2L U = i i = 1 1 + 2 2 2 A1 E1 2 A2 E2 i =1 2 Ai Ei

(5,500 lb) 2 (18 in.) (5,500 lb) 2 (27 in.) + 2(0.19635 in.2 )(29  106 psi) 2(0.60132 in.2 )(29  106 psi) = 47.81234 lb-in. + 23.41829 lb-in.

= 71.23063 lb-in. U = 71.2 lb-in. (b) Strain-Energy Density: From Equation (17.4), 5,500 lb 1 = = 28, 011.27 psi 0.19635 in.2  2 (28, 011.27 psi)2 u1 = 1 = = 13.52813 lb-in./in.3 = 13.53 lb-in./in.3 2 E1 2(29  106 psi) 5,500 lb 2 = = 9,146.54 psi 0.60132 in.2  2 (9,146.54 psi) 2 u2 = 2 = = 1.44240 lb-in./in.3 = 1.442 lb-in./in.3 2 E2 2(29  106 psi)

Ans.

P17.4 In Figure P17.3/4, the compound solid aluminum rod is subjected to a tensile force P. Make the assumption that E = 69 GPa, d1 = 16 mm, L1 = 600 mm, d2 = 25 mm, L2 = 900 mm, and Y = 276 MPa. Calculate the largest amount of strain energy that can be stored in the rod without causing any yielding.

FIGURE P17.3/4

Solution Compute the cross-sectional areas of segments (1) and (2). A1 = A2 =



 4

(16 mm) 2 = 201.062 mm 2 (25 mm) 2 = 490.874 mm 2

Maximum load without causing any yielding: PY = Y Amin = (276 N/mm2 )(201.062 mm2 ) = 55, 493.093 N Elastic Strain Energy: Use Equation (17.14) to calculate the strain energy of each segment as well as the total strain energy in the brass rod. n Fi 2 Li F12 L1 F22 L2 U = = + 2 A1 E1 2 A2 E2 i =1 2 Ai Ei

(55, 493.093 N) 2 (600 mm) (55, 493.093 N) 2 (900 mm) + 2(201.062 mm 2 )(69, 000 N/mm 2 ) 2(490.874 mm 2 )(69, 000 N/mm 2 ) = 66,591.711 N-mm + 40,913.947 N-mm

= 107,505.658 N-mm = 107.506 N-m U = 107.5 J

Ans.

P17.5 A solid 2.5 m long stainless steel rod has a yield strength of 276 MPa and an elastic modulus of 193 GPa. A strain energy of U = 13 N·m must be stored in the rod when a tensile load P is applied to rod. What is (a) the maximum strain-energy density that can be stored in the solid rod if a factor of safety of 4.0 with respect to yielding is specified? (b) the minimum diameter d required for the solid rod?

Solution (a) Maximum Strain-Energy Density: Allowable normal stress:  276 MPa  allow = Y = = 69 MPa FS 4.0 Strain-energy density: 2  allow (69  106 N/m2 )2 u= = = 12,334.197 J/m3 = 12.33 kJ/m3 9 2 2E 2(193  10 N/m )

Ans.

(b) Minimum diameter: Strain energy: U = u AL  13 N-m

A

 4

(13 N-m)(1,000 mm/m) 2 = 421.592 mm 2 (12,334.197 N-m/m3 )(2.5 m)

d 2  421.592 mm 2 d 



(421.592 mm 2 ) = 23.169 mm = 23.2 mm

Ans.

P17.6 A solid stepped shaft made of AISI 1020 cold-rolled steel [G = 11,600 ksi] is shown in Figure P17.6/7/8. The diameters of segments (1) and (2) are d1 = 2.25 in. and d2 = 1.00 in., respectively. The segment lengths are L1 =36 in. and L2 = 27 in. Determine the elastic strain energy U stored in the shaft if the torque TC produces a rotation angle of 4° at C. FIGURE P17.6/7/8

Solution Polar moment of inertia for segments (1) and (2):

J1 =



(2.25 in.)4 = 2.516112 in.4

J2 =

 32

(1.00 in.)4 = 0.0981748 in.4

Shaft torque.

C = TC =

T1L1 T2 L2 TC  L1 L2  + =  +  J1G1 J 2G2 G  J1 J 2 

G L1 L2 + J1 J 2

(4 deg)( rad/180 deg)(11,600,000 psi) = 2,799.017 lb-in. 36 in. 27 in. + 2.516112 in.4 0.0981748 in.4

Total Strain Energy U n T 2L U = i i i =1 2 J i Gi

(2,799.017 lb-in.) 2 (36 in.) (2,799.017 lb-in.)2 (27 in.) = + 2(2.516112 in.4 )(11,600,000 lb/in.2 ) 2(0.0981748 in.4 )(11,600,000 lb/in.2 ) = 4.832 lb-in. + 92.872 lb-in. = 97.7 lb-in. Ans.

P17.7 In Figure P17.6/7/8, a solid stepped shaft made of AISI 1020 cold-rolled steel [G = 80 GPa] has diameters for segments (1) and (2) of d1 = 30 mm and d2 = 15 mm, respectively, and segment lengths of L1 = 320 mm and L2 = 250 mm. What is the maximum torque TC that can be applied to the shaft if the elastic strain energy must be limited to U = 5.0 J? FIGURE P17.6/7/8

Solution Polar moment of inertia for segments (1) and (2):

J1 = J2 =



(30 mm)4 = 79,521.564 mm4 (15 mm)4 = 4,970.098 mm4

32 Maximum torque TC: n Ti 2 Li TC2 n Li U = =  2G i =1 J i i =1 2 J i Gi 5 N-m = 5, 000 N-mm =  TC =

TC2 320 mm 250 mm   + 2  4 4 2(80, 000 N/mm )  79,521.564 mm 4,970.098 mm 

2(5, 000 N-mm)(80,000 N/mm 2 ) = 121,351.620 N-mm = 121.4 N-m 320 mm 250 mm + 79,521.564 mm 4 4,970.098 mm 4

Ans.

P17.8 Figure P17.6/7/8 shows a solid stepped shaft made of 2014-T4 aluminum [G = 28 GPa] that has diameters for segments (1) and (2) of d1 = 20 mm and d2 = 12 mm. The segment lengths are L1 = 240 mm and L2 = 180 mm. Determine the elastic strain energy stored in the shaft when the maximum shear stress is 130 MPa. FIGURE P17.6/7/8

Solution Polar moment of inertia for segments (1) and (2):

J1 = J2 =



(20 mm)4 = 15, 707.963 mm4 (12 mm)4 = 2, 035.752 mm4

32 Maximum shear stress: Tc = J  J (130 N/mm 2 )(2, 035.752 mm 4 )  TC = max 2 = = 44,107.961 N-mm c2 12 mm / 2 Maximum strain energy n T 2L T2 n L U = i i = C  i 2G i =1 J i i =1 2 J i Gi (44,107.961 N-mm) 2  240 mm 180 mm  = + 2 4 4  2(28, 000 N/mm ) 15, 707.963 mm 2, 035.752 mm  = 3, 602.612 N-mm = 3.60 J

Ans.

P17.11 Determine the elastic strain energy of the prismatic beam AB shown in Figure P17.11 if w = 6 kN/m, L = 5 m, and EI = 3×107 N·m2.

FIGURE P17.11

Solution Moment equation: Here, it is convenient to redefine the coordinate axis so that x originates at the free end of the cantilever (i.e., end B) and increases to the left. w M a − a = − x 2 − M = 0 2 wx 2 M = − 0 x L 2 Elastic strain energy: 2

L 1  M2 wx 2  U = dx =  − dx 0 2 EI 0 2 EI   2  L

w2 L 4 x dx 8 EI 0 L w2  x5  = 0 40 EI w2 L5 = 40 EI Therefore: w2 L5 (6,000 N/m)2 (5 m)5 U= = = 93.750 N-m = 93.8 J 40 EI 40(3  107 N-m2 ) =

Ans.

P17.10 In Figure P17.10, what is the elastic strain energy of the prismatic beam if w = 4,000 lb/ft, L = 18 ft, and EI = 1.33×108 lb·ft2? FIGURE P17.10

Solution Moment equation: Draw a free-body diagram that cuts through the beam at section a–a between A and B and derive a moment equation M from this freebody diagram. wL w M a − a = − x + x2 + M = 0 2 2 w wL wx  M = − x2 + x= 0 x L ( L − x) 2 2 2 Elastic strain energy: 2 L M U = dx 0 2 EI 2

1  wx w2 L 2  = x ( L − x ) 2 dx ( L − x ) dx =   0 2 EI  2 0 8 EI L

w2 L 2 2 = ( L x − 2 Lx 3 + x 4 )dx  0 8 EI L

w2  L2 x 3 2 Lx 4 1 5  w2  L5 2 L5 L5  = − + x = − +  8 EI  3 4 5  0 8 EI  3 4 5 w2  20 L5 30 L5 12 L5  w2 L5 = − + = 8 EI  60 60 60  240 EI Therefore: w2 L5 U= 240 EI (4, 000 lb/ft) 2 (18 ft)5 = 240(1.33  108 lb-ft 2 )

= 947.152 lb-ft = 947 lb-ft

Ans.

P17.11 Determine the elastic strain energy of the prismatic beam shown in Figure P17.11 if P = 75 kN, L = 8 m, and EI = 5.10×107 N·m2.

FIGURE P17.11

Solution Moment equation: Draw a free-body diagram that cuts through the beam at section a–a between A and B and derive a moment equation M from this freebody diagram. P M a − a = − x + M = 0 2 P M = x 0 x L/2 2 This equation can also be used for segment BC, where x originates at C. Elastic strain energy: 2 L M U = dx 0 2 EI 2

1 P  P 2 L2 2 = 2 x dx  x dx = 0 2 EI  2  4 EI 0 L 2

L P2 P 2 L3 3 2 x  = = 12 EI   0 96 EI Therefore: P 2 L3 U= 96 EI (75, 000 N) 2 (8 m)3 = 96(5.1  107 N-m 2 )

= 588.235 N-m = 588 J

Ans.

P17.12 A 19 mm diameter steel [E = 200 GPa] rod is required to absorb the energy of a 25 kg collar that falls h = 75 mm, as shown in Figure P17.12/13. Determine the minimum required rod length L so that the maximum stress in the rod does not exceed 210 MPa.

FIGURE P17.12/13

Solution Compute static normal stress:



(19 mm)2 = 283.5287 mm 2 4 F (25 kg)(9.8067 m/s 2 )  st = st = = 0.8647 MPa A 283.5287 mm 2 Compute impact factor n:  210 MPa n = max = = 242.8588  st 0.8647 MPa Derive expression for static deformation in terms of unknown length L: FL (25 kg)(9.8067 m/s 2 ) L  st = st = = (4.3235  10−6 ) L 2 AE (283.5287 mm )(200,000 MPa) The formula for the impact factor is: 2h n = 1+ 1+

 st

Substitute the expression for n and for the static deformation st and solve for L: 2(75 mm) 242.8588 = 1 + 1 + (4.3235  10 −6 ) L 2(75 mm) (4.3235  10 −6 ) L 2(75 mm) L= = 593.1132 mm = 593 mm (4.3235  10 −6 )(58, 494.6791)

(242.8588 − 1) 2 − 1 =

Ans.

P17.13 In Figure P17.12/13, a 500 mm long steel [E = 200 GPa] rod is required to absorb the energy of a 16 kg mass that falls a distance of h. If the rod diameter is 10 mm, what is the maximum drop height h so that the maximum stress in the rod does not exceed 210 MPa?

FIGURE P17.12/13

Solution Calculate static normal stress:  A = (10 mm) 2 = 78.5398 mm 2 4 F (16 kg)(9.8067 m/s 2 )  st = st = = 1.9978 MPa A 78.5398 mm2 Calculate impact factor n:  210 MPa n = max = = 105.1159  st 1.9978 MPa Calculate static deformation: F L (16 kg)(9.8067 m/s 2 )(500 mm)  st = st = = 4.99449  10−3 mm 2 AE (78.5398 mm )(200,000 MPa) The formula for the impact factor is: 2h n = 1+ 1+

 st Knowing n and st, we can now solve for h:

2h 4.99449  10−3 mm 2h (105.1159 − 1) 2 − 1 = 4.99449  10−3 mm (105.1159 − 1) 2 − 1 (4.99449  10 −3 mm) h=  = 27.0679 mm = 27.1 mm 2 105.1159 = 1 + 1 +

Ans.

P17.14 A weight W = 4,000 lbs falls from a height of h = 18 in. onto the top of a 10 in. diameter wood pole, as shown in Figure P17.14. The pole has a length of L = 24 ft and a modulus of elasticity of E = 1.5×106 psi. For this problem, disregard any potential buckling effects. Calculate (a) the impact factor n. (b) the maximum shortening of the pole. (c) the maximum compression stress in the pole.

FIGURE P17.14

Solution (a)

Impact factor n  A = (10 in.) 2 = 78.5398 in.2 4 F L (4, 000 lb)(24 ft)(12 in./ft)  st = st = = 9.7785  10 −3 in. 2 6 AE (78.5398 in. )(1.5  10 psi)

n = 1+ 1+ = 1+ 1+

 st 2(18 in.) 9.7785  10 −3 in.

= 61.6840 = 61.7 (b)

(c)

Ans.

Maximum shortening of the pole  max = nst = 61.6840(9.7785  10−3 in.) = 0.6032 in. = 0.603 in.

Ans.

Maximum compression stress F 4, 000 lb  st = st = = 50.9296 psi A 78.5398 in.2  max = n st = 61.6840(50.9296 psi) = 3,141.5429 psi = 3,140 psi

Ans.

P17.15 As seen in Figure P17.15/16, collar D is released from rest and slides without friction downward a distance of h = 300 mm where it strikes a head fixed to the end of compound rod ABC. Rod segment (1) is made of aluminum [E1 = 70 GPa], and it has a length of L1 = 800 mm and a diameter of d1 = 12 mm. Rod segment (2) is made of bronze [E2 = 105 GPa], and it has a length of L2 = 1,300 mm and a diameter of d2 = 16 mm. What is the allowable mass for collar D if the maximum normal stress in the aluminum rod segment must be limited to 200 MPa?

Solution  A1 =

(12 mm) 2 = 113.0973 mm2

A2 =

 4

(16 mm)2 = 201.0619 mm 2 FIGURE P17.15/16

The total static deformation is  L FL F L L   st = 1 1 + 2 2 = Fst  1 + 2  A1 E1 A2 E2  A1 E1 A2 E2 

  800 mm 1,300 mm = Fst  + 2 2 2 2   (113.0973 mm )(70, 000 N/mm ) (201.0619 mm )(105, 000 N/mm )  Fst = 6,148.9798 N/mm and the impact factor can be expressed as  max 200 MPa (200 N/mm2 )(113.0973 mm2 ) 22,619.4600 N n= = = =  st Fst / A1 Fst Fst Substitute these two expressions into 2h n = 1+ 1+

 st

and solve for Fst.

22,619.4600 N 2(300 mm) 3,689,387.880 N = 1+ 1+ = 1+ 1+ Fst Fst / (6,148.9798 N/mm) Fst 2

 22,619.4600 N  3,689,387.880 N − 1 = 1 +  F F  st

1 F + 3,689,387.880 N 2 22,619.4600 N − Fst ) = st 2 ( Fst Fst

( 22,619.4600 N − Fst ) = Fst ( Fst + 3,689,387.880 N ) 2

2(22,619.4600 N) Fst = (22,619.4600 N) 2 − (3,689,387.880 N)Fst Fst =

(22,619.4600 N) 2 = 136.9990 N 2(22,619.4600 N) + 3,689,387.880 N

Allowable mass: F 136.9990 N m = st = = 13.9700 kg = 13.97 kg g 9.806650 m/s2

Ans.

P17.16 Collar D shown in Figure P17.15/16 has a mass of 11 kg. When released from rest, the collar slides without friction downward a distance of h where it strikes a head fixed to the end of compound rod ABC. Rod segment (1) is made of aluminum [E1 = 70 GPa] and has a length of L1 = 600 mm and a diameter of d1 = 12 mm. Rod segment (2) is made of bronze [E2 = 105 GPa] and has a length of L2 = 1,000 mm and a diameter of d2 = 16 mm. If the maximum normal stress in the aluminum rod segment must be limited to 250 MPa, determine the largest acceptable drop height h.

FIGURE P17.15/16

Solution A1 =



(12 mm)2 = 113.0973 mm2

4 In the aluminum rod segment: Fst = (11 kg)(9.806650 m/s 2 ) = 107.8732 N

 st,1 = n=

A2 =

 4

(16 mm)2 = 201.0619 mm2

Fst 107.8732 N = = 0.9538 MPa A1 113.0973 mm 2

 max 250 MPa = = 262.1072  st 0.9538 MPa

The total static deformation is  L FL F L L   st = 1 1 + 2 2 = Fst  1 + 2  A1 E1 A2 E2  A1 E1 A2 E2 

  600 mm 1, 000 mm = (107.8732 N)  + 2 2 2 2   (113.0973 mm )(70, 000 N/mm ) (201.0619 mm )(105, 000 N/mm )  107.8732 N = = 0.013285 mm 8,119.8067 N/mm Solve for the drop height h: 2h n = 1+ 1+

 st

2h 0.013285 mm 2h (262.1072 − 1)2 − 1 = 0.013285 mm (262.1072 − 1)2 − 1 (0.013285 mm) h=  = 452.8653 mm = 453 mm 2 262.1072 = 1 + 1 +

Ans.

P17.17 In Figure P17.17, the 12 kg mass is falling at a velocity of v = 1.5 m/s at the instant it is h = 300 mm above the spring and post assembly. The solid bronze post has a length of L = 450 mm, a diameter of 60 mm, and a modulus of elasticity of E = 105 GPa. Compute the maximum stress in the bronze post and the impact factor (a) if the spring has a stiffness of k = 5,000 N/mm. (b) if the spring has a stiffness of k = 500 N/mm.

FIGURE P17.17

Solution (a)

Spring stiffness k = 5,000 N/mm



(60 mm)2 = 2,827.4334 mm2

4 W = (12 kg)(9.806650 m/s 2 ) = 117.6798 N Conservation of energy: 1 2 1 1 2 2 mv + W (h +  max ) = kspring  spring + kpost  post 2 2 2 where max = spring +  post

(1) (2)

From equilibrium, Pspring = Ppost

kspring  spring = kpost  post Note that post is the dynamic deformation of the post. The spring constant of the post is: AE (2,827.4334 mm 2 )(105, 000 N/mm 2 ) kpost = = = 659, 734.4573 N/mm L 450 mm therefore, the dynamic deflection spring can be expressed as: (5, 000 N/mm) spring = (659, 734.4573 N/mm) post

659, 734.4573 N/mm  post = 131.9469 post 5, 000 N/mm Substitute Equation (3) into Equation (2): max = spring +  post = 131.9469 post +  post = 132.9469 post Next, substitute Equations (3) and (4) into Equation (1): 1 2 1 1 2 mv + W (h + 132.9469 post ) = kspring (131.9469 post ) 2 + kpost post 2 2 2 Substitute values for m, v, W, kspring, and kpost: 1 (12 kg)(1.5 m/s) 2 + (117.6798 N)(300 mm + 132.9469 post ) 2 1 1 2 = (5, 000 N/mm)(131.9469 post ) 2 + (659, 734.4573 N/mm) post 2 2   spring =

(3) (4)

Use quadratic formula to solve for post:  post = 0.033538 mm The dynamic force in the post can be computed as: Fpost = kpost post = (659,734.4573 N/mm)(0.033538 mm) = 22,126.3670 N and thus, the maximum dynamic stress is F 22,126.3670 N  max = post = = 7.8256 MPa = 7.83 MPa A 2,827.4334 mm2

Ans.

(b) Spring stiffness k = 500 N/mm The dynamic deflection spring can be expressed as: (500 N/mm) spring = (659, 734.4573 N/mm) post 659, 734.4573 N/mm  post = 1,319.4689 post 500 N/mm Substitute Equation (3) into Equation (2): max = spring +  post = 1,319.469 post +  post = 1,320.469 post   spring =

Next, substitute Equations (3) and (4) into Equation (1): 1 2 1 1 2 mv + W (h + 1,320.469 post ) = kspring (1,320.469 post ) 2 + kpost post 2 2 2 Substitute values for m, v, W, kspring, and kpost: 1 (12 kg)(1.5 m/s) 2 + (117.6798 N)(300 mm + 1,320.469 post ) 2 1 1 2 = (5, 000 N/mm)(1,320.469 post ) 2 + (659, 734.4573 N/mm) post 2 2 Use quadratic formula to solve for post:  post = 0.010765 mm The dynamic force in the post can be computed as: Fpost = kpost post = (659,734.4573 N/mm)(0.010765 mm) = 7,102.0072 N and thus, the maximum dynamic stress is F 7,102.0072 N  max = post = = 2.5118 MPa = 2.51 MPa A 2,827.4334 mm2

(3) (4)

Ans.

P17.18 The 32 mm diameter rod AB shown in Figure P17.18 has a length of L = 1.5 m. The rod is made of bronze [E = 105 GPa] that has a yield stress of Y = 330 MPa. Collar C moves along the rod at a speed of v0 = 3.5 m/s until it strikes the rod end at B. If a factor of safety of 4 with respect to yield is required for the maximum normal stress in the rod, determine the maximum allowable mass for collar C.

FIGURE P17.18

Solution A=



(32 mm)2 = 804.2477 mm2 4 Allowable dynamic stress:  330 MPa  max = Y = = 82.5 MPa FS 4 Allowable dynamic force: Fmax =  max A = (82.5 N/mm2 )(804.2477 mm2 ) = 66,350.4368 N Strain energy in rod at impact from Equation (17.12): F2 L (66,350.4368 N)2 (1,500 mm) U = max = = 39,099.3655 N-mm 2 AE 2(804.2477 mm2 )(105,000 N/mm2 ) Equate kinetic energy and strain energy: 2 L 1 2 Fmax mv = 2 2 AE and compute mass m. 2(39, 099.3655 N-mm) m= = 6.3836 kg = 6.38 kg (3.5 m/s) 2 (1, 000 mm/m)

Ans.

P17.19 The block E has a horizontal velocity of v0 = 9 ft/s when it squarely strikes the yoke BD that is connected to the 3/4 in. diameter rods AB and CD. (See Figure P17.19/20.) The rods are made of 6061-T6 aluminum that has a yield strength of Y = 40 ksi and an elastic modulus of E = 10,000 ksi. Both rods have a length of L = 5 ft. Yoke BD may be assumed to be rigid. What is the maximum allowable weight of block E if a factor of safety of 3 with respect to yield is required for the maximum normal stress in the rods? FIGURE P17.19/20

Solution A=



(0.75 in.) 2 = 0.441786 in.2

4 Allowable dynamic stress:  40 ksi  max = Y = = 13.3333 ksi FS 3 Total allowable dynamic force in one rod: Fmax =  max A = (13.3333 ksi)(0.441786 in.2 ) = 5.8905 kips Strain energy in two rods at impact from Equation (17.12):  F 2 L  (5.8905 kips)2 (5 ft)(1,000 lbs/kip)  U = 2  max  = 2   = 39.2699 lb-ft 2  2 AE   2(0.441786 in. )(10,000 ksi)  Equate kinetic energy and strain energy: 1 2 mv = U 2 and compute the maximum weight of block E. 2(39.2699 lb-ft) W = mg =  32.1740 ft/s2 = 31.1968 lb = 31.2 lb 2 (9 ft/s)

Ans.

P17.20 In Figure P17.19/20, the 20 lb block E possesses a horizontal velocity v0 when it hits squarely the yoke BD that is connected to the 1/4 in. diameter rods AB and CD. Both rods are made of 6061-T6 aluminum that has a yield strength of Y = 40 ksi and an elastic modulus of E = 10,000 ksi, and both have a length of L = 30 in. Yoke BD may be assumed to be rigid. Calculate the maximum allowable velocity v0 of block E if a factor of safety of 3 with respect to yield is required for the maximum normal stress in the rods. FIGURE P17.19/20

Solution A=



(0.25 in.) 2 = 0.049087 in.2 4 Allowable dynamic stress:  40 ksi  max = Y = = 13.3333 ksi FS 3 Allowable dynamic force: Fmax =  max A = (13.3333 ksi)(0.049087 in.2 ) = 0.654498 kips = 654.498 lb Strain energy in two rods at impact from Equation (17.12): 2  Fmax   L (654.498 lb) 2 (30 in.)(1 ft/12 in.) U = 2 = 2  = 2.1817 lb-ft 2   2 AE   2(0.049087 in. )(10,000 ksi)(1,000 psi/ksi)  Equate kinetic energy and strain energy: 1 2 mv = U 2 and compute the maximum velocity of block E. 2(2.1817 lb-ft) v= = 2.6494 ft/s = 2.65 ft/s 20 lb     32.1740 ft/s 2 

Ans.

P17.21 The 120 kg block C shown in Figure P17.21 is dropped from a height of h onto a wide-flange steel beam that spans L = 6 m. The steel beam has a moment of inertia of I = 125×106 mm4, a depth of d = 300 mm, a yield stress of Y = 340 MPa, and an elastic modulus of E = 200 GPa. A factor of safety of 2.5 with respect to the yield stress is required for the maximum dynamic bending stress. If the falling block produces the maximum allowable dynamic bending stress, determine (a) the equivalent static load. (b) the maximum dynamic beam deflection at A. (c) the maximum height h from which the 120 kg block C can be dropped.

FIGURE P17.21

Solution (a) Equivalent static load: Maximum allowable dynamic moment:  340 MPa  allow = Y = = 136 MPa FS 3.5 Mc =  136 MPa I (136 N/mm 2 )(125  106 mm 4 ) M  = 113.333  106 N-mm = 113.333 kN-m (300 mm / 2) Equivalent static load at A: M = PL = 113.333 kN-m 113.333 kN-m  Pequiv = Pmax = = 18.8889 kN = 18.89 kN 6m (b) Maximum beam deflection: PL3 (18.8889 kN)(6,000 mm)3 (1,000 N/kN) vmax = = = 54.40 mm = 54.4 mm 3EI 3(200,000 N/mm2 )(125  106 mm4 ) (c) Maximum drop height h: Static deflection: PL3 (120 kg)(9.806650 m/s2 )(6,000 mm)3 vst = = = 3.389178 mm 3EI 3(200,000 N/mm2 )(125  106 mm4 ) Compute the impact factor n and then solve for h: v 54.40 mm n = max = = 16.0511 vst 3.389178 mm

n = 1+ 1+

Ans.

2h vst

(n − 1)2 − 1 (16.0511 − 1) 2 − 1 h = vst = (3.389178 mm) = 382.1896 mm = 382 mm 2 2

Ans.

P17.22 The overhanging beam ABC shown in Figure P17.22/23 is made from an aluminum I-shape, which has a moment of inertia of I = 25×106 mm4, a depth of d = 200 mm, and an elastic modulus of E = 70 GPa. The beam spans are a = 2.5 m and b = 1.5 m. A block D with a mass of 90 kg is dropped from a height h = 1.5 m onto the free end of the overhang at C. Calculate (a) the maximum bending stress in the beam. (b) the maximum beam deflection at C due to the falling block.

FIGURE P17.22/23

Solution Static deflection at C: P = (90 kg)(9.806650 m/s 2 ) = 882.5985 N

Pb 2 (a + b) (882.5985 N)(1,500 mm) 2 (2,500 mm + 1,500 mm) = = 1.513026 mm 3EI 3(70, 000 N/mm 2 )(25  106 mm 4 ) Compute the impact factor n: 2h n = 1+ 1+ vst vst =

= 1+ 1+

2(1,500 mm) 1.513026 mm

= 45.5397 Dynamic load: Pmax = nPst = 45.5397(882.5985 N) = 40,193.2377 N = 40.193 kN Bending moment due to dynamic load: M = Pb = (40.193 kN)(1.5 m) = 60.2899 kN-m (a) Maximum bending stress: Mc (60.2899 kN-m)(200 mm / 2)(1,000 N/kN)(1, 000 mm/m) = = I 25  106 mm 4

= 241.1594 MPa = 241 MPa (b) Maximum beam deflection: vmax = nvst = 45.5397(1.513026 mm) = 68.8027 mm = 68.8 mm

Ans. Ans.

P17.23 In Figure P17.22/23, the overhanging beam ABC, made from an aluminum I-shape, has a moment of inertia of I = 25×106 mm4, a depth of d = 200 mm, and an elastic modulus of E = 70 GPa. The beam spans are a = 3.5 m and b = 1.75 m. A block D with a mass of 110 kg is dropped from a height h onto the free end of the overhang at C. If the maximum bending stress due to impact must not exceed 125 MPa, compute (a) the maximum dynamic load allowed at C. (b) the impact factor n. (c) the maximum height h from which the 110 kg block D can be dropped.

FIGURE P17.22/23

Solution (a) Maximum dynamic load allowed at C: Maximum allowable dynamic moment: Mc =  125 MPa I (125 N/mm 2 )(25  106 mm 4 ) M  = 31.25  106 N-mm = 31.25 kN-m (200 mm / 2) Maximum allowable dynamic load at C: M = Pb = 31.25 kN-m

 Pequiv = Pmax =

31.25 kN-m = 17.8571 kN = 17.86 kN 1.75 m

(b) Impact factor n: Compute the impact factor n: P 17,857.1 N n = max = = 16.5538 = 16.55 Pst (110 kg)(9.806650 m/s2 )

Ans.

vst =

Pb 2 (a + b) (1, 078.7315 N)(1,750 mm) 2 (3,500 mm + 1, 750 mm) = = 3.303615 mm 3EI 3(70, 000 N/mm 2 )(25  106 mm 4 )

n = 1+ 1+

2h vst

(n − 1) 2 − 1 (16.5538 − 1) 2 − 1 h = vst = (3.303615 mm) = 397.9564 mm = 398 mm 2 2

Ans.

P17.24 Figure P17.24 shows block D, weighing 200 lb, dropped from a height of h = 6 ft onto a wide-flange steel beam that spans L = 24 ft with a = 8 ft and b = 16 ft. The steel beam has a moment of inertia of I = 300 in.4, a depth of d = 12 in., and an elastic modulus of E = 29,000 ksi. Determine (a) the dynamic load applied to the beam. (b) the maximum bending stress in the beam. (c) the beam deflection at B due to the falling block.

FIGURE P17.24

Solution (a) Dynamic load applied to the beam by block D: Static deflection at B: Pa 2b2 (200 lb)(8 ft)2 (16 ft)2 (12 in./ft)3 vst = = = 9.03945  10−3 in. 6 4 3LEI 3(24 ft)(29  10 psi)(300 in. ) Compute the impact factor n: 2h n = 1+ 1+ vst 2(6 ft)(12 in./ft) 9.03945  10−3 in. = 127.2188 Dynamic load: Pmax = nPst = 127.2188(200 lb) = 25, 433.7518 lb = 25.4338 kips = 25.4 kips = 1+ 1+

Ans.

(b) Maximum bending stress: Bending moment due to dynamic load: Pab (25.4338 kips)(8 ft)(16 ft) M= = = 135.7000 kip-ft L 24 ft The bending stress produced by this moment is: Mc (135.7000 kip-ft)(12 in. / 2)(12 in./ft) = = I 300 in.4

= 32.5680 ksi = 32.6 ksi (c) Beam deflection at B due to falling block: vmax = nvst = 127.2188(9.03945  10−3 in.) = 1.1500 in. = 1.150 in.

Ans.

P17.25 A 75 lb block D at rest is dropped from a height of h = 2 ft onto the top of the simply supported wood beam. (See Figure P17.25.) The cross section of the beam is square—8 in. wide by 8 in. deep—and the modulus of elasticity of the wood is E = 1,600 ksi. The beam spans L = 14 ft, and it is supported at A and C by springs that each have a stiffness of k = 1,000 lb/in. Assume that the springs at A and C do not restrain beam rotation. Compute (a) the maximum beam deflection at B due to the falling block. (b) the equivalent static load required to produce the same deflection. (c) the maximum bending stress in the timber beam.

FIGURE P17.25

Solution (a) Beam deflection at B due to falling block: The moment of inertia of the timber beam is (8 in.)(8 in.)3 I= = 341.3333 in.4 12 From Appendix C, for a simply-support beam with a concentrated load acting at midspan: PL3 vB = 48 EI The spring constant of the beam can be considered as: P 48EI 48(1,600 ksi)(341.3333 in.4 )(1,000 lb/kip) kbeam = = 3 = = 5,528.5604 lb/in. vB L (14 ft)3 (12 in./ft)3 From equilibrium, the forces at the spring supports at A and C are related to the load on the beam by: 2Fspring = Pbeam and thus 2kspring  spring = kbeam vbeam 2(1, 000 lb/in.) spring = (5,528.5604 lb/in.)vbeam

(1)  spring = 2.764280vbeam Equate the work done by the falling block D to the energy stored in the spring supports and in the flexing beam: 1  1 2 2 (2) WD (h + spring + vbeam ) = 2  kspring  spring  + kbeamvbeam 2 2 Substitute Equation (1) into Equation (2) and note that h = 2 ft = 24 in.: 1 2 1 2 (75 lb) ( 24 in. + 2.764280vbeam + vbeam ) = 2  kspring ( 2.764280vbeam )  + kbeamvbeam 2  2 This quadratic equation can be solved for the beam deflection: 2 2 1,800 lb-in. + (282.321 lb)vbeam = (7, 641.243918 lb/in.)vbeam + (2, 764.280200 lb/in.)vbeam 2 (10, 405.52412 lb/in.)vbeam − (282.321 lb)vbeam − 1,800 lb-in. = 0

282.321  ( −282.321)2 − 4(10, 405.52412)( − 1,800) vbeam = = 0.429702 in., − 0.402570 in. 2(10, 405.52412)

Use the positive value for the beam deflection; thus, the maximum dynamic beam deflection is Ans. vmax = 0.429702 in. = 0.430 in. Note that vmax pertains to the beam deflection only. There is also contraction in the spring supports at A and C, and these contractions cause the entire beam to displace downward. (b) Equivalent static load: The equivalent static load required to produce this deflection is: Pmax = kbeamvbeam = (5,528.5604 lb/in.)(0.429702 in.) = 2,375.6322 lb = 2,380 lb

Ans.

(c) Maximum bending stress: Bending moment due to dynamic load: PL (2,375.6322 lb)(14 ft) M= = = 8,314.7129 lb-ft 4 4 The bending stress from this moment is: Mc (8,314.7129 lb-ft)(8 in. / 2)(12 in./ft) = = I 341.3333 in.4

= 1,169.2565 psi = 1,169 psi

Ans.

P17.26 The 120 kg block (Figure P17.26) is falling at 1.25 m/s when it is h = 1,400 mm above the spring that is located at midspan of the simply supported steel beam. The steel beam’s moment of inertia is I = 70×106 mm4, its depth is d = 250 mm, and its elastic modulus is E = 200 GPa. L = 5.5 m is the beam span. The spring constant is k = 100 kN/m. Calculate (a) the maximum beam deflection at B due to the falling block. (b) the equivalent static load required to produce the same deflection. (c) the maximum bending stress in the steel beam.

FIGURE P17.26

Solution (a) Beam deflection at B due to falling block: From Appendix C, for a simply-support beam with a concentrated load acting at midspan: PL3 vB = 48 EI The spring constant of the beam can be considered as: P 48EI 48(200  109 N/m 2 )(70  106 mm 4 )(1 m/1, 000 mm) 4 kbeam = = 3 = vB L (5.5 m)3

= 4, 039, 068.3696 N/m = 4, 039.0684 kN/m From equilibrium, the force in the spring and the load on the beam are equal: Fspring = Pbeam and thus kspring  spring = kbeam vbeam (100 kN/m) spring = (4, 039.0684 kN/m)vbeam

(1)  spring = 40.390684vbeam From conservation of energy, the kinetic energy of the block plus the work done by the falling block must equal the energy stored in the spring and in the flexing beam: 1 2 1 1 2 2 mv + W (h +  spring + vbeam ) = kspring  spring + kbeam vbeam (2) 2 2 2 Substitute Equation (1) into Equation (2) and note that h = 1,400 mm = 1.4 m: 1 (120 kg)(1.25 m/s) 2 + (120 kg)(9.806650 m/s 2 ) (1.4 m + 40.390684vbeam + vbeam ) 2 1 1 2 2 = kspring ( 40.390684vbeam ) + kbeam vbeam 2 2

This quadratic equation 93.75 N-m + 1, 647.5172 N-m + (48, 708.47415 N)vbeam 2 2 = (81.570352 106 N/m)vbeam + (2.019534 106 N/m)vbeam 2 (83.589886 106 N/m)vbeam − (48, 708.47415 N)vbeam − 1, 741.267200 N-m = 0 can be solved for the beam deflection:

vbeam =

48, 708.47415  (−48, 708.47415) 2 − 4(83.589886 106 )( − 1, 741.267200) 2(83.589886 106 )

= 4.86475110−3 m, − 4.282043 10−3 m Use the positive value for the beam deflection; thus, the maximum dynamic beam deflection is vmax = 4.864751 mm = 4.86 mm

Ans.

(b) Equivalent static load: The equivalent static load required to produce this deflection is: Pmax = kbeamvbeam = (4,039.0684 kN/m)(4.86475110−3 m) = 19.6491 kN = 19.65 kN

Ans.

(c) Maximum bending stress: Bending moment due to dynamic load: PL (19.6491 kN)(5.5 m) M= = = 27.0175 kN-m 4 4 The bending stress from this moment is: Mc (27.0175 kN-m)(250 mm / 2)(1,000 N/kN)(1,000 mm/m) = = I 70 106 mm4

= 48.2455 MPa = 48.2 MPa

Ans.

P17.27 The post AB shown in Figure P17.27/28 has a length of L = 2.25 m. The post is made from a steel HSS that has a moment of inertia of I = 8.7×106 mm4, a depth of d = 150 mm, a yield strength of Y = 315 MPa, and an elastic modulus of E = 200 GPa. A block with a mass of m = 25 kg moves horizontally with a velocity of v0 and strikes the HSS post squarely at B. If a factor of safety of 1.5 is specified for the maximum bending stress, what is the largest acceptable velocity v0 for the block?

FIGURE P17.27/28

Solution Maximum allowable dynamic moment:  315 MPa  allow = Y = = 210 MPa FS 1.5 Mc =  210 MPa I (210 N/mm 2 )(8.7 106 mm 4 ) M  = 24.3600  106 N-mm = 24.3600 kN-m (150 mm / 2) Equivalent static load at B: M = PL = 24.3600 kN-m 24.360 kN-m  Pequiv = Pmax = = 10.8267 kN 2.25 m Maximum beam deflection: PL3 (10.8267 kN)(2,250 mm)3 (1,000 N/kN) vmax = = = 23.625 mm 3EI 3(200,000 N/mm2 )(8.7 106 mm4 ) By the conservation of energy, the work that is performed on the post must equal the kinetic energy of the block 1 1 Pmax vmax = mv02 2 2 Therefore, the largest acceptable velocity for the block is

v0 =

Pmax vmax = m

(10,826.7 N)(0.023625 m) = 3.1986 m/s = 3.20 m/s 25 kg

Ans.

P17.28 In Figure P17.27/28, the post AB, length of L = 4.2 m, is made from a steel HSS with a moment of inertia of I = 24.4×106 mm4, a depth of d = 200 mm, a yield strength of Y = 315 MPa, and an elastic modulus of E = 200 GPa. A block with a mass of m moves horizontally with a velocity of v0 = 4.5 m/s and strikes the HSS post squarely at B. A factor of safety of 1.75 is specified for the maximum bending stress; determine the largest acceptable mass m for the block.

FIGURE P17.27/28

Solution Maximum allowable dynamic moment:  315 MPa  allow = Y = = 180 MPa FS 1.75 Mc =  180 MPa I (180 N/mm 2 )(24.4  106 mm 4 ) M  = 43.9200 106 N-mm = 43.9200 kN-m (200 mm / 2) Equivalent static load at B: M = PL = 43.9200 kN-m 43.9200 kN-m  Pequiv = Pmax = = 10.4571 kN 4.2 m Maximum beam deflection: PL3 (10.4571 kN)(4,200 mm)3 (1,000 N/kN) vmax = = = 52.92 mm 3EI 3(200,000 N/mm2 )(24.4 106 mm4 ) By the conservation of energy, the work that is performed on the post must equal the kinetic energy of the block 1 1 Pmax vmax = mv02 2 2 Therefore, the largest acceptable mass for the block is P v (10, 457.1 N)(0.05292 m) Ans. m = max 2 max = = 27.3280 kg = 27.3 kg v0 (4.5 m/s)2

P17.29 The simply supported steel beam shown in Figure P17.29 is struck squarely at midspan by a 180 kg block moving horizontally with a velocity of v0 = 2.5 m/s. The beam’s span is L = 4 m, its moment of inertia is I = 15×106 mm4, its depth is d = 155 mm, and its elastic modulus is E = 200 GPa. Compute (a) the maximum dynamic load applied to the beam. (b) the maximum bending stress in the steel beam. (c) the maximum beam deflection at B due to the moving block.

FIGURE P17.29

Solution (a) Maximum dynamic load By the conservation of energy, the work that is performed on the beam must equal the kinetic energy of the block 1 1 Pmax vmax = mv02 2 2 From Appendix C, the midspan deflection of the beam is expressed by PL3 vmax = 48 EI Combine these two expressions  Pmax L3  1 2 1 Pmax   = mv0 2  48 EI  2 and calculate the maximum dynamic load 48EI mv02 Pmax = L3 =

48(200  109 N/m 2 )(15  10−6 m 4 )(180 kg)(2.5 m/s) 2 (4 m)3

= 50.3115  103 N = 50.3115 kN = 50.3 kN

Ans.

(b) Maximum bending stress: The maximum bending moment is P L (50.3115 kN)(4 m) M max = max = = 50.3115 kN-m 4 4 and thus the maximum bending stress is Mc = I (50.3115 kN-m)(155 mm / 2)(1, 000 N/kN)(1,000 mm/m) = 15 106 mm 4 = 259.9429 MPa = 260 MPa

Ans.

(c) Maximum beam deflection: P L3 (50.3115 kN)(4 m)3 (1,000 N/kN)(1,000 mm/m)3 vmax = max = = 22.4 mm 48EI 48(200,000 N/mm2 )(15 106 mm4 )

P17.30 In Figure P17.30, compute the horizontal displacement of joint B of the two-bar assembly if P = 80 kN and if x1 = 3.0 m, y1 = 3.5 m, and x2 = 2.0 m. Assume that A1E1 = 9.0×104 kN and A2E2 = 38.0×104 kN.

FIGURE P17.30

Solution The length of bar (1) is:

L1 = (3.0 m)2 + (3.5 m)2 = 4.6098 m The length of bar (2) is:

L2 = (2.0 m)2 + (3.5 m)2 = 4.0311 m Write equilibrium equations for the sum of forces in the horizontal and vertical directions. Note: Bars (1) and (2) are two-force members. 3.0 2.0 Fx = − F1 + F2 + 80 kN = 0 (a) 4.6098 4.0311 3.5 3.5 Fy = − F1 − F2 = 0 (b) 4.6098 4.0311 Solving (a) and (b) simultaneously gives F1 = 73.7564 kN F2 = −64.4981 kN The strain energy in bar (1) is F 2L (73.7564 kN)2 (4.6098 m)(1,000 N/kN) U1 = 1 1 = = 139.3176 N-m 2 A1E1 2(9.0  104 kN) The strain energy in bar (2) is F22 L2 (−64.4981 kN)2 (4.0311 m)(1,000 N/kN) U2 = = = 22.0651 N-m 2 A2 E2 2(38.0  104 kN) The total strain energy of the two-bar assembly is therefore U = U1 + U 2 = 139.3176 N-m + 22.0651 N-m = 161.3827 N-m External work of the 80 kN load: W = 12 (80, 000 N)  From conservation of energy 1 (80,000 N)  = (161.3827 N-m)(1,000 mm/m) 2

  = 4.03 mm →

Ans.

P17.31 If P = 215 kN. a = 3.5 m, and b = 2.75 m, determine the vertical displacement of joint C of the truss in Figure P17.31. Assume that AE = 8.50×105 kN for all members.

FIGURE P17.31

Solution Truss Reaction Forces Ax = 0 kN Ay = 215 kN  Member AB AD BC BD BE CE DE

L (mm) 5,500 4,451.1235 5,500 4,451.1235 4,451.1235 4,451.1235 5,500

By = 430 kN 

F (kN) –168.9286 273.4262 –168.9286 –273.4262 –273.4262 273.4262 337.8571

F2L (kN2·mm) 156,952,795 332,774,391 156,952,795 332,774,391 332,774,391 332,774,391 627,810,810

F L =

2,272,813,967

Total strain energy  Fj2 L j  1 2, 272,813,967 kN 2 -mm 2 U =  = 1,336.9494 kN-mm  = 2 AE  F j L j = 2(8.5  105 kN) j  2 Aj E j  j From conservation of energy 1 2 (215 kN)  C = 1,336.9494 kN-m m   C = 12.44 mm 

Ans.

P17.32 Rigid bar BCD in Figure P17.32 is supported by a pin at C and by steel rod (1). A concentrated load P = 2.5 kips is applied to the lower end of aluminum rod (2), which is attached to the rigid bar at D. For this structure, a = 20 in. and b = 30 in. For steel rod (1), L1 = 50 in., A1 = 0.4 in.2, and E1 = 30,000 ksi. For aluminum rod (2), L2 = 100 in., A2 = 0.2 in.2, and E2 = 10,000 ksi. What is the vertical displacement of point E?

FIGURE P17.32

Solution Consider the equilibrium of the rigid bar, and write a moment equilibrium equation about C to determine the force in steel rod (1): M C = (20 in.)F1 − (30 in.)(2.5 kips) = 0

 F1 = 3.75 kips The strain energy in steel rod (1) is F12 L1 (3.75 kips)2 (50 in.)(1,000 lb/kip) U1 = = = 29.2969 lb-in. 2 A1E1 2(0.4 in.2 )(30, 000 ksi) The strain energy in bar (2) is F 2L (2.5 kips)2 (100 in.)(1,000 lb/kip) U2 = 2 2 = = 156.2500 lb-in. 2 A2 E2 2(0.2 in.2 )(10, 000 ksi) The total strain energy of the assembly is therefore U = U1 + U 2 = 29.2969 lb-in. + 156.2500 lb-in. = 185.5469 lb-in. External work of the 2.5 kip load: W = 12 (2,500 lb)  E From conservation of energy 1 2 (2,500 lb)  E = 185.5469 lb-in.   E = 0.1484 in. 

Ans.

P17.33 In Figure P17.33, bronze rod (1) and aluminum rod (2) support rigid bar ABC. A concentrated load P = 90 kN is applied to the free end of aluminum rod (3). For this structure, a = 800 mm and b = 500 mm. For bronze rod (1), L1 = 1.8 m, d1 = 15 mm, and E1 = 100 GPa. For aluminum rod (2), L2 = 2.5 m, d2 = 25 mm, and E2 = 70 GPa. For aluminum rod (3), L3 = 1.0 m, d3 = 25 mm, and E3 = 70 GPa. Calculate the vertical displacement of point D.

FIGURE P17.33

Solution From a FBD cut through rod (3), equilibrium requires that the internal force in rod (3) is F3 = P = 90 kN. From a FBD of the rigid bar, write two equilibrium equations: M A = (1,300 mm)F2 − (800 mm)(90 kN) = 0

 F2 = 55.3846 kN Fy = F1 + F2 − F3 = 0  F1 = F3 − F2 = 90 kN − 55.3846 kN = 34.6154 kN

Areas:

A1 =



(15 mm)2 = 176.7146 mm 2

A2 =



(25 mm) 2 = 490.8739 mm 2 = A3

4 4 The strain energy in bronze rod (1) is F12 L1 (34, 615.4 N)2 (1.8 m) U1 = = = 61.0251 N-m 2 A1E1 2(176.7146 mm2 )(100, 000 N/mm2 ) The strain energy in aluminum rod (2) is F 2L (55,384.6 N)2 (2.5 m) U2 = 2 2 = = 111.5887 N-m 2 A2 E2 2(490.8739 mm2 )(70, 000 N/mm2 ) The strain energy in aluminum rod (3) is F 2L (90, 000 N)2 (1.0 m) U3 = 3 3 = = 117.8656 N-m 2 A3 E3 2(490.8739 mm2 )(70, 000 N/mm2 ) The total strain energy of the assembly is therefore U = U1 + U 2 + U 3 = 61.0251 N-m + 111.5887 N-m + 117.8656 N-m = 290.4794 N-m External work of the 90 kN load: W = 12 (90, 000 N)  D From conservation of energy 1 2 (90, 000 N)  D = (290.4794 N-m)(1,000 mm/m)   D = 6.46 mm 

Ans.

P17.34 Use the virtual-work method to determine the vertical displacement of joint B for the truss shown in Figure P17.34/35. Assume that each member has a cross-sectional area of A = 1.25 in.2 and an elastic modulus of E = 10,000 ksi. The loads acting on the truss are P = 21 kips and Q = 7 kips. FIGURE P17.34/35

Solution Real Truss Reaction Forces Ay = 11.870 kips  Cx = 7.000 kips →

Cy = 9.130 kips 

Virtual Load Assume virtual load applied in –y direction at B. The reaction forces for the virtual load are: Ay = 0.43478 kips  Cx = 0.0 kips Cy = 0.56522 kips  Member AB AC BC

L (in.) 94.8683 138.0000 80.7217

F (kips) –20.8527 17.1449 –13.6486

f (kips) –0.7638 0.6280 –0.8449

f (FL) (kips2-in.) 1,511.0747 1,485.8955 930.8701

 f (FL) =

3,927.8404

Equation (17.30) can now be applied.  Fj L j  1 1  =  f j  =  f j Fj L j   Aj E j  AE j j

(

)

From the tabulated results: (1 kip)   B =

(3,927.8404 kips 2 -in.) (1.25 in.2 )(10, 000 ksi)

 B = 0.314 in. 

Ans.

Since the virtual load was applied in a downward direction at B, the positive value of the result confirms that joint B does displace downward.

P17.35 Calculate the horizontal displacement of joint B for the truss in Figure P17.34/35 by applying the virtual-work method. Make the assumptions that each member has a crosssectional area of A = 1.25 in.2 and an elastic modulus of E = 10,000 ksi, and that the loads acting on the truss are P = 21 kips and Q = 7 kips. FIGURE P17.34/35

Solution Real Truss Reaction Forces Ay = 11.870 kips  Cx = 7.000 kips →

Cy = 9.130 kips 

Virtual Load Assume 1 kip virtual load applied in –x direction at B. The reaction forces for the virtual load are: Ay = 0.3913 kips  Cx = 1.0 kips → Cy = 0.3913 kips  Member AB AC BC

L (in.) 94.8683 138.0000 80.7217

F (kips) –20.8527 17.1449 –13.6486

f (kips) –0.6875 0.5652 0.5849

f (FL) (kips2-in.) 1,359.9554 1,337.3107 –644.4511

 f (FL) =

2,052.8150

Apply Equation (17.30)  Fj L j  1 1  =  f j  =  f j Fj L j   Aj E j  AE j j to get the deflection from the tabulated results: (2, 052.8150 kips 2 -in.) (1 kip)   B = (1.25 in.2 )(10, 000 ksi)

(

 B = 0.1642 in. 

)

Ans.

Since the virtual load was applied in the –x direction at B, the positive value of the result confirms that joint B does displace to the left.

P17.36 Employing the virtual-work method, determine the vertical displacement of joint D for the truss shown in Figure P17.36/37. Each member is assumed to have a cross-sectional area of A = 1.60 in.2 and an elastic modulus of E = 29,000 ksi. The loads acting on the truss are P = 20 kips and Q = 30 kips. FIGURE P17.36/37

Solution Real Truss Reaction Forces Ax = 30.000 kips  Ay = 36.667 kips 

Cy = 76.667 kips 

Virtual Load Assume that a 1 kip virtual load is applied in the –y direction at D. The reaction forces for the virtual load are: Ax = 0.00 kips Ay = 0.667 kips  Cy = 1.667 kips  Member AB AC BC BD

L (in.) 259.5997 216.0000 227.6840 216.0000

F (kips) 44.0679 5.5556 –59.7319 43.3333

f (kips) 0.8012 –0.4444 –0.7027 0.6667

f (FL) (kips2-in.) 9,166.0714 –533.3284 9,557.1278 6,240.0307

259.5997

–24.0370

–1.20185

7,499.5445

 f (FL) =

31,929.4460

Apply Equation (17.30)  Fj L j  1 1  =  f j  =  f j Fj L j   Aj E j  AE j j to get the deflection at B from the tabulated results: (31,929.4460 kips 2 -in.) (1 kip)   D = (1.60 in.2 )(29, 000 ksi)

(

 D = 0.688 in. 

)

Ans.

Since the virtual load was applied in a downward direction at D, the positive value of the result confirms that joint D does displace downward.

P17.37 Calculate the horizontal displacement of joint D for the truss in Figure P17.36/37 by using the virtual-work method. Assume that the loads acting on the truss are P = 20 kips and Q = 30 kips and that each member has a cross-sectional area of A = 1.60 in.2 and an elastic modulus of E = 29,000 ksi. FIGURE P17.36/37

Solution Real Truss Reaction Forces Ax = 30.000 kips  Ay = 36.667 kips 

Cy = 76.667 kips 

Virtual Load Assume that a 1 kip virtual load is applied in the +x direction at D. The reaction forces for the virtual load are: Ax = 1.00 kips  Ay = 1.00 kips  Cy = 1.00 kips  Member AB AC BC BD

L (in.) 259.5997 216.0000 227.6840 216.0000

F (kips) 44.0679 5.5556 –59.7319 43.3333

f (kips) 1.2019 0.3333 –1.0541 1.0000

f (FL) (kips2-in.) 13,749.1643 399.9963 14,335.6237 9,359.9993

259.5997

–24.0370

0.0000

 f (FL) =

37,844.7836

Equation (17.30) can now be applied.  Fj L j  1 1  =  f j  =  f j Fj L j   Aj E j  AE j j

(

)

From the tabulated results: (1 kip)   D =

(37,844.7836 kips 2 -in.) (1.60 in.2 )(29, 000 ksi)

 D = 0.816 in. →

Ans.

Since the virtual load was applied in the +x direction at D, the positive value of the result confirms that joint D does displace to the right.

P17.38 Determine the horizontal displacement of joint A for the truss in Figure P17.38/39/40 by applying the virtual-work method. Make the assumption that each member has a cross-sectional area of A = 750 mm2 and an elastic modulus of E = 70 GPa.

FIGURE P17.38/39/40

Solution Real Truss Reaction Forces Dx = 102.800 kN  Ex = 102.800 kN →

Ey = 54.000 kN 

Virtual Load Assume that a 1 kN virtual load is applied in the +x direction at A. The reaction forces for the virtual load are: Dx = 1.000 kN  Ex = 0.000 kN Ey = 0.000 kN Member AB AC

L (mm) 4,800.00 7,683.75

F (kN) –27.2000 43.5413

f (kN) –1.0000 0.0000

f (FL) (kN2-mm) 130,560.00 0.00

BC BD CD CE DE

6,000.00 8,400.00 10,322.79 8,400.00 6,000.00

20.0000 –27.2000 –92.9051 102.8000 54.0000

0.0000 –1.0000 0.0000 0.0000 0.0000

0.00 228,480.00 0.00 0.00 0.00

 f (FL) =

359,040.00

Apply Equation (17.30)  Fj L j  1 1  =  f j  =  f j Fj L j   Aj E j  AE j j to get the deflection at B from the tabulated results: (359, 040.00 kN 2 -mm)(1,000 N/kN) (1 kN)   A = (750 mm 2 )(70, 000 N/mm 2 )

(

 A = 6.84 mm →

)

Ans.

Since the virtual load was applied in a +x direction at A, the positive value of the result confirms that joint A does displace horizontally to the right.

P17.39 In Figure P17.38/39/40, calculate the vertical displacement of joint B for the truss, employing the virtual-work method. Each member is assume to have a cross-sectional area of A = 750 mm2 and an elastic modulus of E = 70 GPa.

FIGURE P17.38/39/40

Solution Real Truss Reaction Forces Dx = 102.800 kN  Ex = 102.800 kN →

Ey = 54.000 kN 

Virtual Load Assume that a 1 kN virtual load is applied in the –y direction at B. The reaction forces for the virtual load are: Dx = 1.400 kN  Ex = 1.400 kN → Ey = 1.000 kN  Member AB AC

L (mm) 4,800.00 7,683.75

F (kN) –27.2000 43.5413

f (kN) 0.0000 0.0000

f (FL) (kN2-mm) 0.00 0.00

BC BD CD CE DE

6,000.00 8,400.00 10,322.79 8,400.00 6,000.00

20.0000 –27.2000 –92.9051 102.8000 54.0000

1.0000 0.0000 –1.7205 1.4000 1.0000

120,000.00 0.00 1,649,999.50 1,208,928.00 324,000.00

 f (FL) =

3,302,927.50

Apply Equation (17.30)  Fj L j  1 1  =  f j  =  f j Fj L j   Aj E j  AE j j to get the deflection at B from the tabulated results: (3,302,927.50 kN 2 -mm)(1,000 N/kN) (1 kN)   B = (750 mm2 )(70, 000 N/mm2 )

(

 B = 62.9 mm 

)

Ans.

Since the virtual load was applied in a –y direction at B, the positive value of the result confirms that joint B does displace downward.

P17.40 The truss shown in Figure P17.38/39/40 is constructed from aluminum [E = 70 GPa;  = 23.6×10–6/°C] members that each have a cross-sectional area of A = 750 mm2. Use the virtual-work method to determine the vertical displacement of joint A for the following two conditions: (a) T = 0°C. (b) T = +40°C.

FIGURE P17.38/39/40

Solution Real Truss Reaction Forces Dx = 102.800 kN  Ex = 102.800 kN →

Ey = 54.000 kN 

Virtual Load Assume that a 1 kN virtual load is applied in the –y direction at A. The reaction forces for the virtual load are: Dx = 2.200 kN  Ex = 2.200 kN → Ey = 1.000 kN  (a) T = 0°C. Member

AB AC BC BD CD CE DE

(mm) 4,800.00 7,683.75 6,000.00 8,400.00 10,322.79 8,400.00 6,000.00

(kN) –27.2000 43.5413 20.0000 –27.2000 –92.9051 102.8000 54.0000

FL AE (mm) –2.4869 6.3726 2.2857 –4.3520 –18.2674 16.4480 6.1714

 T L

(mm) 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000

(kN) –0.8000 1.2806 0.0000 –0.8000 –1.7205 2.2000 1.0000  FL   f  AE +  T L =

 FL  f +  T L  AE  (kN-mm) 1.9895 8.1608 0.0000 3.4816 31.4286 36.1856 6.1714 87.4175

Apply Equation (17.33)  Fj L j  1  =  f j  +  j T j L j + L j   Aj E j  j to get the deflection at A from the tabulated results: (1 kN)   A = 87.4175 kN-mm

 A = 87.4 mm 

Ans.

Since the virtual load was applied in a –y direction at A, the positive value of the result confirms that joint A does displace downward. (b) T = 40°C. Member

AB AC BC BD CD

(mm) 4,800.00 7,683.75 6,000.00 8,400.00 10,322.79

(kN) –27.2000 43.5413 20.0000 –27.2000 –92.9051

CE DE

8,400.00 6,000.00

102.8000 54.0000

FL AE (mm) –2.4869 6.3726 2.2857 –4.3520 –18.2674 16.4480 6.1714

 T L

(mm) 4.5312 7.2535 5.6640 7.9296 9.7447

(kN) –0.8000 1.2806 0.0000 –0.8000 –1.7205

7.9296 5.6640

2.2000 1.0000  FL   f  AE +  T L =

 FL  f +  T L  AE  (kN-mm) –1.6355 17.4498 0.0000 –2.8621 14.6631 53.6307 11.8354 93.0814

Apply Equation (17.33)  Fj L j  1  =  f j  +  j T j L j + L j   Aj E j  j to get the deflection at A from the tabulated results: (1 kN)   A = 93.0814 kN-mm

 A = 93.1 mm 

Ans.

Since the virtual load was applied in a –y direction at A, the positive value of the result confirms that joint A does displace downward.

P17.41 Figure P17.41/42 shows a truss subjected to concentrated loads P = 320 kN and Q = 60 kN. Its members AB, BC, DE, and EF each have a cross-sectional area of A = 2,700 mm2, with all other members each having a cross-sectional area of A = 1,060 mm2. All members are made of steel [E = 200 GPa]. For the given loads, utilize the virtual-work method to calculate the horizontal displacement of (a) joint F and (b) joint B. FIGURE P17.41/42

Solution Real Truss Reaction Forces Ay = 440.000 kN  Dx = 120.000 kN →

Dy = 200.000 kN 

(a) Horizontal Displacement of Joint F Virtual Load Assume that a 1 kN virtual load is applied in the –x direction at F. The reaction forces for the virtual load are: Ay = 1.33333 kN  Dx = 1.000 kN → Dy = 1.33333 kN  Member

(mm) 6,000.00

(mm2) 2,700

(kN) –360.0000

AD AE BC BE BF CF DE EF

9,000.00 10,816.65 6,000.00 9,000.00 10,816.65 9,000.00 6000.00 6000.00

1,060 1,060 2,700 1,060 1,060 1,060 2,700 2,700

120.0000 –144.2221 –320.0000 60.0000 –72.1110 0.0000 –200.0000 –280.0000

FL AE (mm) –4.0000 5.0943 –7.3585 –3.5556 2.5472 –3.6792 0.0000 –2.2222 –3.1111

f (kN) –0.6667 1.0000 –1.2019 0.0000 1.0000 –1.2019 0.0000 1.3333 0.6667  FL   f  AE  =

 FL  f  AE  (kN-mm) 2.6667 5.0943 8.8438 0.0000 2.5472 4.4219 0.0000 –2.9630 –2.0741 18.5369

The deflection at F is thus  Fj L j  1  =  f j    Aj E j  j (1 kN)   F = 18.5369 kN-mm  F = 18.54 mm 

Ans.

Since the virtual load was applied in the –x direction at F, the positive value of the result confirms that joint F does displace to the left. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

(b) Horizontal Displacement of Joint B Virtual Load Assume that a 1 kN virtual load is applied in the –x direction at B. The reaction forces for the virtual load are: Ay = 0.66667 kN  Dx = 1.000 kN → Dy = 0.66667 kN  Member

AB AD AE BC BE

(mm) 6,000.00 9,000.00 10,816.65 6,000.00 9,000.00

(mm2) 2,700 1,060 1,060 2,700 1,060

(kN) –360.0000 120.0000 –144.2221 –320.0000 60.0000

10,816.65

1,060

–72.1110

CF DE EF

9,000.00 6,000.00 6,000.00

1,060 2,700 2,700

0.0000 –200.0000 –280.0000

FL AE (mm) –4.0000 5.0943 –7.3585 –3.5556 2.5472 –3.6792 0.0000 –2.2222 –3.1111

f (kN) 0.0000 1.0000 –1.2019 0.0000 1.0000 0.0000 0.0000 0.6667 0.0000  FL   f  AE  =

 FL  f  AE  (kN-mm) 0.0000 5.0943 8.8438 0.0000 2.5472 0.0000 0.0000 –1.4815 0.0000 15.0038

The deflection at B is thus  Fj L j  1  =  f j    Aj E j  j (1 kN)   B = 15.0038 kN-mm  B = 15.00 mm 

Ans.

Since the virtual load was applied in the –x direction at B, the positive value of the result confirms that joint B does displace to the left.

P17.42 In Figure P17.41/42, the truss is subjected to concentrated loads P = 200 kN and Q = 35 kN. Members AB, BC, DE, and EF each have a cross-sectional area of A = 2,700 mm2, and all other members each have a cross-sectional area of A = 1,060 mm2. All members are made of steel [E = 200 GPa]. During construction, it was discovered that members AE and BF were fabricated 15 mm shorter than their intended length. For the given loads and the two member misfits, employ the virtual-work method to determine the horizontal displacement of (a) joint F and (b) joint B. FIGURE P17.41/42

Solution Real Truss Reaction Forces Ay = 270.000 kN  Dx = 70.000 kN →

Dy = 130.000 kN 

L

(kN)

FL AE (mm)

(mm)

(kN)

 FL  f + L  AE  (kN-mm)

2,700

–223.3333

–2.4815

0.00

–0.6667

1.6543

9,000.00

1,060

70.0000

2.9717

0.00

1.0000

2.9717

10,816.65

1,060

–84.1295

–4.2925

–15.00

–1.2019

23.1866

6,000.00

2,700

–200.0000

–2.2222

0.00

0.0000

9,000.00

1,060

35.0000

1.4858

0.00

1.0000

1.4858

10,816.65

1,060

–42.0648

–2.1462

–15.00

–1.2019

20.6072

9,000.00

1,060

0.0000

0.00

0.0000

6000.00

2,700

–130.0000

–1.4444

0.00

1.3333

–1.9259

6000.00

2,700

–176.6667

–1.9630

0.00

0.6667

–1.3086

Member

(mm)

(mm2)

6,000.00

 FL



 f  AE + L =

46.6711

The deflection at F is thus  Fj L j  1  =  f j  + L j   Aj E j  j (1 kN)   F = 46.6711 kN-mm  F = 46.7 mm 

Ans.

L

(kN)

FL AE (mm)

(mm)

(kN)

 FL  f + L  AE  (kN-mm)

2,700

–223.3333

–2.4815

0.00

0.0000

9,000.00

1,060

70.0000

2.9717

0.00

1.0000

2.9717

10,816.65

1,060

–84.1295

–4.2925

–15.00

–1.2019

23.1866

6,000.00

2,700

–200.0000

–2.2222

0.00

0.0000

9,000.00

1,060

35.0000

1.4858

0.00

1.0000

1.4858

10,816.65

1,060

–42.0648

–2.1462

–15.00

0.0000

9,000.00

1,060

0.0000

0.00

0.0000

6000.00

2,700

–130.0000

–1.4444

0.00

0.6667

–0.9630

6000.00

2,700

–176.6667

–1.9630

0.00

0.0000

Member

(mm)

(mm2)

6,000.00

 FL



 f  AE + L =

26.6812

The deflection at B is thus  Fj L j  1  =  f j  + L j   Aj E j  j (1 kN)   B = 26.6812 kN-mm  B = 26.7 mm 

Ans.

Since the virtual load was applied in the –x direction at B, the positive value of the result confirms that joint B does displace to the left.

P17.43 The truss seen in Figure P17.43/44/45 is subjected to concentrated loads of P = 160 kN and 2P = 320 kN. All members are made of steel [E = 200 GPa], and each has a cross-sectional area of A = 3,500 mm2. Use the virtualwork method to determine (a) the horizontal displacement of joint A. (b) the vertical displacement of joint A. FIGURE P17.43/44/45

Solution Real Truss Reaction Forces Bx = 0.000 kN By = 640.000 kN 

Fy = 640.000 kN 

(a) Horizontal Displacement of Joint A Virtual Load Assume that a 1 kN virtual load is applied in the –x direction at A. The reaction forces for the virtual load are: Bx = 1.000 kN → By = 0.375 kN  Fy = 0.375 kN  Member AB AC BC

L (mm) 10,000 8,000 6,000

F (kN) –266.67 213.33 –480.00

f (kN) 0.0000 1.0000 –0.3750

f (FL) (kN2-mm) 0.00 1,706,666.64 1,080,000.00

BD CD CE DE DF DG EG

8,000 10,000 8,000 6,000 8,000 10,000 8,000

–213.33 266.67 0.00 –320.00 –213.33 266.67 0.00

–1.0000 0.6250 0.5000 0.0000 0.0000 –0.6250 0.5000

1,706,666.64 1,666,666.69 0.00 0.00 0.00 –1,666,666.69 0.00

FG FH

6,000 10,000

–480.00 –266.67

0.3750 0.0000

–1,080,000.00 0.00

8,000

213.33

0.0000

0.00

 f (FL) =

3,413,333.28

Equation (17.30) can now be applied.  Fj L j  1 1  =  f j  =  f j Fj L j   Aj E j  AE j j

(

)

From the tabulated results: (1 kN)   A =

(3, 413,333.28 kN 2 -mm)(1, 000 N/kN) (3,500 mm 2 )(200, 000 N/mm 2 )

 A = 4.88 mm 

Ans.

Since the virtual load was applied in the –x direction at A, the positive value of the result confirms that joint A does displace to the left. (b) Vertical Displacement of Joint A Virtual Load Assume that a 1 kN virtual load is applied in the –y direction at A. The reaction forces for the virtual load are: Bx = 0.00 kN By = 1.50 kN  Fy = 0.50 kN  Member AB AC BC BD CD CE DE DF

L (mm) 10,000 8,000 6,000 8,000 10,000 8,000 6,000 8,000

F (kN) –266.67 213.33 –480.00 –213.33 266.67 0.00 –320.00 –213.33

f (kN) –1.6667 1.3333 –0.5000 –1.3333 0.8333 0.6667 0.0000 0.0000

f (FL) (kN2-mm) 4,444,453.39 2,275,549.83 1,440,000.00 2,275,549.83 2,222,213.36 0.00 0.00 0.00

DG EG FG FH GH

10,000 8,000 6,000 10,000 8,000

266.67 0.00 –480.00 –266.67 213.33

–0.8333 0.6667 0.5000 0.0000 0.0000

–2,222,213.36 0.00 –1,440,000.00 0.00 0.00

 f (FL) =

8,995,553.05

Equation (17.30) can now be applied.  Fj L j  1 1  =  f j  =  f j Fj L j   Aj E j  AE j j

(

)

From the tabulated results: (1 kN)   A =

(8,995,553.05 kN 2 -mm)(1, 000 N/kN) (3,500 mm 2 )(200, 000 N/mm 2 )

 A = 12.85 mm 

Ans.

Since the virtual load was applied in the –y direction at A, the positive value of the result confirms that joint A does displace downward.

P17.44 Figure P17.43/44/45 shows a truss subjected to concentrated loads of P = 160 kN and 2P = 320 kN. Each member has a cross-sectional area of A = 3,500 mm2, with all members being made of steel [E = 200 GPa;  = 11.7×10–6/°C]. If the temperature of the truss increases by 30°C, use the virtual-work method to compute (a) the horizontal displacement of joint A. (b) the vertical displacement of joint A.

FIGURE P17.43/44/45

Solution Real Truss Reaction Forces Bx = 0.000 kN By = 640.000 kN 

Fy = 640.000 kN 

AB AC

(mm) 10,000 8,000

(kN) –266.67 213.33

BC BD CD CE DE DF DG EG FG

6,000 8,000 10,000 8,000 6,000 8,000 10,000 8,000 6,000

–480.00 –213.33 266.67 0.00 –320.00 –213.33 266.67 0.00 –480.00

FH GH

10,000 8,000

–266.67 213.33

FL AE (mm) –3.8095 2.4381 –4.1143 –2.4381 3.8095 0.0000 –2.7429 –2.4381 3.8095 0.0000 –4.1143 –3.8095 2.4381

 T L

(mm) 3.5100 2.8080

(kN) 0.0000 1.0000

2.1060 2.8080 3.5100 2.8080 2.1060 2.8080 3.5100 2.8080 2.1060

–0.3750 –1.0000 0.6250 0.5000 0.0000 0.0000 –0.6250 0.5000 0.3750

3.5100 2.8080

0.0000 0.0000  FL   f  AE +  T L =

 FL  f +  T L  AE  (kN-mm) 0.0000 5.2461 0.7531 –0.3699 4.5747 1.4040 0.0000 0.0000 –4.5747 1.4040 –0.7531 0.0000 0.0000 7.6842

Apply Equation (17.33)  Fj L j  1  =  f j  +  j T j L j + L j   Aj E j  j to get the deflection at A from the tabulated results: Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

(1 kN)   A = 7.6842 kN-mm  A = 7.68 mm 

Ans.

Since the virtual load was applied in the –x direction at A, the positive value of the result confirms that joint A does displace to the left.

(b) Vertical Displacement of Joint A Virtual Load Assume that a 1 kN virtual load is applied in the –y direction at A. The reaction forces for the virtual load are: Bx = 0.00 kN By = 1.50 kN  Fy = 0.50 kN  Member

AB AC BC BD CD CE DE DF

(mm) 10,000 8,000 6,000 8,000 10,000 8,000 6,000 8,000

(kN) –266.67 213.33 –480.00 –213.33 266.67 0.00 –320.00 –213.33

DG EG FG FH GH

10,000 8,000 6,000 10,000 8,000

266.67 0.00 –480.00 –266.67 213.33

FL AE (mm) –3.8095 2.4381 –4.1143 –2.4381 3.8095 0.0000 –2.7429 –2.4381 3.8095 0.0000 –4.1143 –3.8095 2.4381

 T L

(mm) 3.5100 2.8080 2.106 2.8080 3.5100 2.8080 2.1060 2.8080

(kN) –1.6667 1.3333 –0.5000 –1.3333 0.8333 0.6667 0.0000 0.0000

3.5100 2.8080 2.1060 3.5100 2.8080

–0.8333 0.6667 0.5000 0.0000 0.0000  FL   f  AE +  T L =

 FL  f +  T L  AE  (kN-mm) 0.4992 6.9948 1.0041 –0.4932 6.0996 1.8720 0.0000 0.0000 –6.0996 1.8720 –1.0041 0.0000 0.0000 10.7448

Apply Equation (17.33)  Fj L j  1  =  f j  +  j T j L j + L j   Aj E j  j to get the deflection at A from the tabulated results: (1 kN)   A = 10.7448 kN-mm

 A = 10.74 mm 

Ans.

Since the virtual load was applied in the –y direction at A, the positive value of the result confirms that joint A does displace downward.

P17.45 The truss in Figure P17.43/44/45 is subjected to concentrated loads of P = 160 kN and 2P = 320 kN. All members are made of steel [E = 200 GPa;  = 11.7×10–6/°C], and each has a cross-sectional area of A = 3,500 mm2. Utilizing the virtual-work method, calculate (a) the vertical displacement of joint D. (b) the vertical displacement of joint D if the temperature of the truss decreases by 40°C. FIGURE P17.43/44/45

Solution Real Truss Reaction Forces Bx = 0.000 kN By = 640.000 kN 

Fy = 640.000 kN 

(a) Vertical Displacement of Joint D Virtual Load Assume that a 1 kN virtual load is applied in the –y direction at D. The reaction forces for the virtual load are: Bx = 0.00 kN By = 0.500 kN  Fy = 0.500 kN 

AB AC BC

L (mm) 10,000 8,000 6,000

F (kN) –266.67 213.33 –480.00

f (kN) 0.0000 0.0000 -0.5000

BD CD CE DE DF DG EG FG FH

8,000 10,000 8,000 6,000 8,000 10,000 8,000 6,000 10,000

–213.33 266.67 0.00 –320.00 –213.33 266.67 0.00 –480.00 –266.67

0.0000 0.8333 -0.6667 0.0000 0.0000 0.8333 -0.6667 -0.5000 0.0000

8,000

213.33

0.0000

f ( FL) (kN2-mm) 0.000 0.000 1,440,000.000 0.000 2,222,213.361 0.000 0.000 0.000 2,222,213.361 0.000 1,440,000.000 0.000 0.000

 f ( FL) =

7,324,426.722

Member

Equation (17.30) can now be applied.  Fj L j  1 1  =  f j  =  f j Fj L j   Aj E j  AE j j

(

)

From the tabulated results:

(7,324, 426.722 kN 2 -mm)(1, 000 N/kN) (1 kN)   D = (3,500 mm 2 )(200, 000 N/mm 2 )  D = 10.46 mm 

Ans.

Since the virtual load was applied in the –y direction at D, the positive value of the result confirms that joint D does displace downward. (b) Vertical Displacement of Joint D for T = –40°C Member

(mm) 10,000

(kN) –266.67

AC BC

8,000 6,000

213.33 –480.00

BD CD CE DE DF DG EG

8,000 10,000 8,000 6,000 8,000 10,000 8,000

–213.33 266.67 0.00 –320.00 –213.33 266.67 0.00

FG FH GH

6,000 10,000 8,000

–480.00 –266.67 213.33

FL AE (mm) –3.8095 2.4381 –4.1143 –2.4381 3.8095 0.0000 –2.7429 –2.4381 3.8095 0.0000 –4.1143 –3.8095 2.4381

 T L

(mm) –4.6800

(kN) 0.0000

–3.7440 –2.8080

0.0000 –0.5000

–3.7440 –4.6800 –3.7440 –2.8080 –3.7440 –4.6800 –3.7440

0.0000 0.8333 –0.6667 0.0000 0.0000 0.8333 –0.6667

–2.8080 –4.6800 –3.7440

–0.5000 0.0000 0.0000  FL   f  AE +  T L =

 FL  f +  T L  AE  (kN-mm) 0.0000 0.0000 3.4611 0.0000 –0.7254 2.4960 0.0000 0.0000 –0.7254 2.4960 3.4611 0.0000 0.0000 10.4635

Apply Equation (17.33)  Fj L j  1  =  f j  +  j T j L j + L j   Aj E j  j to get the deflection at A from the tabulated results: (1 kN)   D = 10.4635 kN-mm

 D = 10.46 mm 

Ans.

Since the virtual load was applied in the –y direction at D, the positive value of the result confirms that joint D does displace downward.

P17.46 In Figure P17.46, compute the deflection of the beam at B for the loading, employing the virtualwork method. Assume that EI is constant for the beam.

FIGURE P17.46

Solution Virtual Moment m: To determine the deflection of the simply-supported beam, first remove the real load P from the beam and apply a virtual unit load downward at B. For this beam, two sets of equations must be derived.

Cut a free-body diagram through the beam at section a–a. Derive a moment equation m from this free-body diagram. b M a − a = − x1 + m = 0 L b  m = x1 0  x1  a L Cut a free-body diagram through the beam at section b–b. Derive a moment equation m from this free-body diagram. a M b −b = x2 − m = 0 L a  m = x2 0  x2  b L

Real Moment M: Remove the virtual load and reapply the real load P. The free-body diagram for the beam with the real load is shown.

Draw a free-body diagram that cuts through the beam at section a–a and derive a moment equation M from this free-body diagram. Pb M a − a = − x1 + M = 0 L Pb M = x1 0  x1  a L

Cut a free-body diagram through the beam at section b–b. Derive a moment equation M from this free-body diagram. Pa M b −b = x2 − M = 0 L Pa M = x2 0  x2  b L Virtual-Work Equation for Beam Deflection: From Equation (17.35), the beam deflection at B can now be determined. L M 1  B =  m   dx 0  EI  a b b a   Pb    Pa  =   x1   x1  dx1 +   x2   x dx 0  L 0  L   LEI    LEI 2  2

Pb 2 a 2 Pa 2 b 2 Pb 2 1 3 a Pa 2 1 3 b  x1  +  x2  x dx + x dx = 1 1 2 2 L2 EI 0 L2 EI 0 L2 EI 3   0 L2 EI 3   0 Pa 3b 2 Pa 2b3 Pa 2b 2 Pa 2b 2 = 2 + 2 = 2 ( a + b) = 3L EI 3L EI 3L EI 3LEI

 B =

Pa 2b 2  3LEI

Ans.

P17.47 Employ the virtual-work method to calculate the slope of the beam at A for the loading seen in Figure P17.47. Assume that EI is constant for the beam.

FIGURE P17.47

Solution Virtual Moment m: To determine the slope of the simply-supported beam, first remove the real load from the beam and apply a virtual unit moment clockwise at A. For this beam, two equations will be derived for the virtual moment. Cut a free-body diagram through the beam at section a–a. Derive a moment equation m from this free-body diagram. 1 M a − a = x1 − 1 + m = 0 L 1 L  m = 1 − x1 0  x1  L 2 Cut a second free-body diagram through the beam at section b–b that includes roller C. 1 M b −b = x2 − m = 0 L 1 L  m = x2 0  x2  L 2 Real Moment M: Remove the virtual moment and reapply the real load w0. The beam free-body diagram for the real load w0 is shown.

Draw a free-body diagram that cuts through the beam at section a–a and derive a moment equation M from this free-body diagram. w 3w L M a − a = 0 x12 − 0 x1 + M = 0 2 8 w 3w L L  M = − 0 x12 + 0 x1 0  x1  2 8 2 Draw a second free-body diagram that cuts through the beam at section b–b and derive a moment equation M from this free-body diagram. wL M b −b = 0 x2 − M = 0 8 wL L  M = 0 x2 0  x2  8 2 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Virtual-Work Equation for Beam Slope: From Equation (17.36), the beam slope at A can now be determined. L L L 1   w0 2 3w0 L  M 1 w L  2 1  A =  m   dx =   1 − x1   − x1 + x1  dx1 +  2  x2   0 x2  dx2 0 0 0  EI   L   2 EI  L   8 EI  8 EI  w0 3 3w0 2  1 L2  w0 2 3w0 L 1 L2  w0 2  = x1 + x1 + x1 − x1  dx1 + −  x2  dx2  EI 0  2 8 2L 8 EI 0  8  =

w0 L2  w0 L2 2 4 3 2 − 7 x + 3 Lx + x dx + x2 dx2  1 1 1 1 8 EI 0  L  8 EI 0 L

L w0  7 3 3L 2 1 4  2 w0 3 2   = − x1 + x1 + x1  + x2 8 EI  3 2 L  0 24 EI   0

w0 7 L3 w0 L3 7 w0 L3 2w0 L3 9w0 L3 = + = + = 8 EI 48 24 EI 8 384 EI 384 EI 384 EI  A =

3w0 L3 (CW) 128 EI

Ans.

P17.48 What are the slope and deflection of the beam at C for the loading shown in Figure P17.48? Utilize the virtual-work method and assume that EI is constant for the beam.

FIGURE P17.48

Solution Determine Slope at C Virtual Moment m: To determine the slope of the simply-supported beam, first remove the real load P from the beam and apply a virtual unit moment clockwise at C. The free-body diagram of the beam for this loading is shown. Cut a free-body diagram through the beam at section a–a. Derive a moment equation m from this free-body diagram. 1 1 M a − a = x1 + m = 0  m = − x1 0  x1  L L L Cut a second free-body diagram through the beam at section b–b. Derive a moment equation m from this free-body diagram. L M b −b = −1 − m = 0  m = −1 0  x2  2 Real Moment M: Remove the virtual moment and reapply the real load P. The free-body diagram for the beam with the real load is shown.

Draw a free-body diagram that cuts through the beam at section a–a and derive a moment equation M from this free-body diagram. P P M a − a = x1 + M = 0  M = − x1 0  x1  L 2 2 Draw a second free-body diagram that cuts through the beam at section b–b and derive a moment equation M from this free-body diagram. L M b −b = − Px2 − M = 0  M = − Px2 0  x2  2

Virtual-Work Equation for Beam Slope: From Equation (17.36), the beam slope at C can now be determined. L L L 1  P  M  P  1 C =  m   dx =   − x1   − x1  dx1 +  2 ( −1)  − x dx 0 0  0  EI   EI 2  2 L   2 EI  L

L P P 2 P 1 3 L P 1 2 L2 2  x1  +  x2  x dx + x dx = 1 1 2 2 2 LEI 0 EI 0 2 LEI 3   0 EI 2   0

PL2 PL2 7 PL2 + = 6 EI 8 EI 24 EI

C =

7 PL2 24 EI

(CW)

Ans.

Determine Deflection at C Virtual Moment m: To determine the deflection of the simply-supported beam, first remove the real load P from the beam and apply a virtual unit load downward at C. The free-body diagram of the beam for this loading is shown.

Cut a free-body diagram through the beam at section a–a. Derive a moment equation m from this free-body diagram. 1 1 M a − a = x1 + m = 0  m = − x1 0  x1  L 2 2 Cut a second free-body diagram through the beam at section b–b. Derive a moment equation m from this free-body diagram. L M b −b = −1x2 − m = 0  m = −1x2 0  x2  2 Virtual-Work Equation for Beam Deflection: From Equation (17.35), the beam deflection at C can now be determined. L L L 1  P  M  P  1  C =  m   dx =   − x1   − x1  dx1 +  2 ( −1x2 )  − x dx 0 0  0  EI   EI 2  2 2   2 EI  L

P L 2 P 2 2 P 1 3 L P 1 3 L2  x1  +  x2  x dx + x dx = 1 1 2 2 4 EI 0 EI 0 4 EI 3   0 EI 3   0 PL3 PL3 3PL3 = + = 12 EI 24 EI 24 EI

PL3  C =  8EI

Ans.

P17.49 Use the virtual-work method to determine the deflection of the compound rod at C for the loading shown in Figure P17.49. Between A and B, the rod has a diameter of 30 mm. Between B and C, the rod diameter is 15 mm. Assume that E = 200 GPa for both segments of the compound rod. FIGURE P17.49

Solution Virtual Moment m: To determine the deflection at C, first remove the real load and then apply a virtual unit load downward at C.

Cut a free-body diagram through the compound rod at section a–a. Derive a moment equation m from this free-body diagram. M a − a = −(1 N)x1 − m = 0

 m = −(1 N)x1

0  x1  320 mm

A second free-body diagram through the compound rod at section b–b is not necessary in this instance; however, it is included here for completeness. M b −b = −(1 N)x2 − m = 0

 m = −(1 N)x2

320 mm  x2  960 mm

Real Moment M: Remove the virtual load and reapply the real load. Draw a free-body diagram that cuts through the compound rod at section a–a and derive a moment equation M from this free-body diagram. M a − a = −(500 N)x1 − M = 0

 M = −(500 N)x1

0  x1  320 mm

Again, a second free-body diagram through the compound rod at section b–b is not necessary in this instance; however, it is included here for completeness. M b −b = −(500 N)x2 − M = 0

 M = −(500 N)x2

320 mm  x2  960 mm

Moments of Inertia: The moment of inertia for segment AB is:  I AB = (30 mm)4 = 39, 760.782 mm4 64 and the moment of inertia for segment BC is:  I BC = (15 mm)4 = 2, 485.049 mm 4 64 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Virtual-Work Equation for Beam Deflection: From Equation (17.35), the compound rod deflection at C can now be determined. L M (1 N) C =  m   dx 0  EI  =

320 mm





( −(1 N)x1 )  − (200, 000 N/mm 2 )(2, 485.049 mm 4 ) x1  dx1 500 N





( −(1 N)x2 )  − (200, 000 N/mm 2 )(39, 760.782 mm 4 ) x2  dx2 320 mm

+

960 mm

500 N

320 mm 500 N x12 dx1 2 4 0 (200, 000 N/mm )(2, 485.049 mm ) 960 mm 500 N + (1 N) x 2 dx2 2 4 320 mm 2 (200, 000 N/mm )(39, 760.782 mm ) 500 N 1 3 320 mm  x1  = (1 N) 2 4 (200, 000 N/mm )(2, 485.049 mm ) 3   0 500 N 1 3 960 mm  x2  + (1 N) 2 4 (200, 000 N/mm )(39, 760.782 mm ) 3   320 mm

= (1 N)

= (1 N)(335.3388  10 −9 mm −2 )(32.7680  106 mm3 ) + (1 N)(20.9587  10 −9 mm −2 )(851.9680  106 mm 3 ) = (1 N) 10.9884 mm + 17.8561 mm    C = 28.8 mm 

Ans.

P17.50 Figure P17.50/51 shows a compound steel [E = 200 GPa] rod that has a diameter of 15 mm in segments AB and DE and a diameter of 30 mm in segments BC and CD. For the given loading, apply the virtual-work method to find the slope of the compound rod at A. FIGURE P17.50/51

Solution Virtual Moment m: To determine the slope at A, first remove the real load and then apply a virtual unit moment clockwise at A. The beam free-body diagram for this loading is shown. Cut a free-body diagram through the compound rod at section a–a. Derive a moment equation m from this free-body diagram.  1  M a − a =  N x1 − (1 N-mm) + m = 0  330   1   m = (1 N-mm) −  N x1 0  x1  230 mm  330  A second free-body diagram through the compound rod at section b–b is not necessary in this instance; however, it is included here for completeness.  1  M b − b =  N x − (1 N-mm) + m = 0  330  2  1   m = (1 N-mm) −  N x2 0  x2  230 mm  330  Cut a third free-body diagram through the compound rod at section c–c. Derive a moment equation m from this free-body diagram.  1  M c − c =  N x3 − m = 0  330 

 1  m =  N x  330  3

0  x3  100 mm

Real Moment M: Remove the virtual load and reapply the real load. The beam free-body diagram is shown.

Draw a free-body diagram that cuts through the compound rod at section a–a and derive a moment equation M from this free-body diagram. M a − a = −(7,800 N) x1 + M = 0

 M = (7,800 N) x1

0  x1  100 mm

Cut a second free-body diagram through the compound rod at section b–b, and derive a second moment equation M. (120 N/mm) M b −b = −(7,800 N) x2 + (x2 − 100 mm) 2 + M = 0 2 (120 N/mm) M = − (x2 − 100 mm) 2 + (7,800 N) x2 2 100 mm  x2  230 mm Cut a third free-body diagram through the compound rod at section c–c, and derive a third moment equation M. M c −c = (7,800 N) x3 − M = 0

 M = (7,800 N) x3

0 mm  x3  100 mm

Moments of Inertia: The moment of inertia for 15 mm diameter segments AB and DE is:  I AB = I DE = (15 mm) 4 = 2, 485.049 mm 4 64 and the moment of inertia for 30 mm diameter segments BC and CD is:  I BC = I CD = (30 mm)4 = 39, 760.782 mm 4 64 Virtual-Work Equation for Beam Slope: From Equation (17.36), the beam slope at A can now be determined. M (1 N-mm)   A =  m   dx  EI  =

100

x1   7,800   x1 dx1 1 −  330   EI AB  x   60 7,800   1− 2   − (x2 − 100) 2 + x2 dx2  100  330   EI BC EI BC 

+

230

+

100

 x3   7,800  x3 dx3   330   EI DE 

7,800 100 7, 800 100 2 x1dx1 − x1 dx1  EI AB 0 330 EI AB 0 − + +

(

)

1 230 60(x2 − 100) 2 − 7,800 x2 dx2  100 EI BC

(

)

230 1 x2 60(x2 − 100) 2 − 7,800 x2 dx2  100 330 EI BC

7,800 100 2 x3 dx3 330 EI DE 0

7,800 100 7,800 100 2 x1dx1 − x1 dx1  EI AB 0 330 EI AB 0 − + +

60 230 7,800 230 (x2 − 100) 2 dx2 + x2 dx2  EI BC 100 EI BC 100 230 60 7,800 230 2 x2 (x2 − 100) 2 dx2 − x2 dx2  330 EI BC 100 330 EI BC 100

7,800 100 2 x3 dx3 330 EI DE 0

7,800 (100)2 7,800 (100)3 = − EI AB 2 330 EI AB 3 −

60 (130)3 7,800 (2302 − 1002 ) + EI BC 3 EI BC 2

60  (130)4 (100)(130)3  7,800 (230)3 − (100)3 + −  330 EI 330 EI BC  4 3 3  BC

7,800 (100)3 + 330 EI DE 3 39, 000, 000 7,878, 787.88 43,940, 000 167,310, 000 = − − + EI AB EI AB EI BC EI BC +

26, 297, 424.24 87,982, 424.24 7,878, 787.88 − + EI BC EI BC EI DE

39, 000, 000 61, 685, 000 + EI AB EI BC

39, 000, 000 61, 685, 000 + (200, 000)(2, 485.049) (200, 000)(39, 760.782)

= 78.469277  10−3 + 7.757015  10−3 = 86.226293  10−3  A = 0.0862 rad

(CW)

Ans.

P17.51 The compound steel [E = 200 GPa] rod in Figure P17.50/51 has a diameter of 15 mm in segments AB and DE and a diameter of 30 mm in segments BC and CD. For the given loading, compute the deflection of the compound rod at C, employing the virtual-work method. FIGURE P17.50/51 Solution Virtual Moment m: To determine the deflection at C, remove the real loading and apply a virtual unit load downward at C. The beam free-body diagram for this loading is shown.

Cut a free-body diagram through the compound rod at section a–a. Derive a moment equation m from this free-body diagram. M a − a = −(0.5 N) x1 + m = 0

 m = (0.5 N) x1 0  x1  165 mm A second free-body diagram through the compound rod at section b–b is not necessary in this instance; however, it is included here for completeness. M b −b = −(0.5 N) x2 + m = 0

 m = (0.5 N) x2

0  x2  165 mm

Real Moment M: Remove the virtual load and reapply the real load. The beam free-body diagram is shown.

Draw a free-body diagram that cuts through the compound rod at section a–a and derive a moment equation M from this free-body diagram. M a − a = −(7,800 N) x1 + M = 0

 M = (7,800 N) x1

0  x1  100 mm

Cut a third free-body diagram through the compound rod at section c–c, and derive a third moment equation M. M c −c = (7,800 N) x3 − M = 0

 M = (7,800 N) x3

0 mm  x3  100 mm

Moments of Inertia: The moment of inertia for 15 mm diameter segments AB and DE is:  I AB = I DE = (15 mm) 4 = 2, 485.049 mm 4 64 and the moment of inertia for 30 mm diameter segments BC and CD is:  I BC = I CD = (30 mm)4 = 39, 760.782 mm 4 64 Virtual-Work Equation for Beam Deflection: From Equation (17.35), the beam deflection at C can now be determined. Since both the virtual and real loadings are symmetrical, the integrals will be written for the interval [0,165] and then doubled. 165 M (1 N)   C = 2 m   dx 0  EI 

= 2

100

165  60  7,800  7,800  (0.5 x1 )  x1  dx1 + 2 (0.5 x2 )  − (x2 − 100) 2 + x2 dx2 1 00 EI BC   EI AB   EI BC

(

)

(

)

7, 800 100 2 1 165 x1 dx1 − x2 60(x2 − 100) 2 − 7, 800 x2 dx2  EI AB 0 EI BC 100 7,800 100 2 1 165 = x1 dx1 − x2 60(x2 − 100) 2 − 7,800 x2 dx2   0 100 EI AB EI BC

(

)

7,800 100 2 60 165 7,800 165 2 x1 dx1 − x2 (x2 − 100) 2 dx2 + x2 dx2   EI AB 0 EI BC 100 EI BC 100

7,800 (100)3 60  (65) 4 (100)(65)3  7,800 (165)3 − (100)3 = − + + EI AB 3 EI BC  4 3 3  EI BC =

2, 600  106 817.009  106 9, 079.525  106 − + EI AB EI BC EI BC

2, 600  106 8, 262.516  106 + (200, 000)(2, 485.049) (200, 000)(39, 760.782) = 5.231285 + 1.039028 = 6.270313

  C = 6.27 mm 

Ans.

P17.52 Figure P17.52 shows a simply supported beam. Assume that EI = 15×106 kip·in.2 for the beam. Apply the virtual-work method and calculate (a) the deflection at A. (b) the slope at C. FIGURE P17.52

Solution (a) Deflection at A Virtual Moment m: To determine the deflection of the beam, first remove the real load from the beam and apply a virtual unit load downward at A. The free-body diagram of the beam with this loading is shown. Cut a free-body diagram through the beam at section a–a. Derive a moment equation m from this free-body diagram. M a − a = (1 kip) x1 + m = 0

 m = −(1 kip) x1

0  x1  8 ft

Cut a second free-body diagram through the beam at section b–b. Derive a moment equation m from this free-body diagram. M b −b = −(0.4 kips) x2 − m = 0

 m = −(0.4 kips) x2

0  x2  20 ft

Real Moment M: Remove the virtual load and reapply the real load. A free-body of the beam is shown.

Draw a free-body diagram that cuts through the beam at section a–a and derive a moment equation M from this free-body diagram. 3.5 kips/ft 2 M a − a = x1 + M = 0 2 3.5 kips/ft 2 M = − x1 0  x1  8 ft 2 Draw a second free-body diagram that cuts through the beam at section b–b and derive a moment equation M from this free-body diagram. 3.5 kips/ft 2 M b − b = − x2 + (29.4 kips) x2 − M = 0 2 3.5 kips/ft 2 M = − x2 + (29.4 kips) x2 0 ft  x2  20 ft 2

Virtual-Work Equation for Beam Deflection: From Equation (17.35), the beam deflection at A can now be determined. L M (1 kip)   A =  m   dx 0  EI  1 8 1 20  3.5 2   3.5 2  ( −1x1 )  − x1  dx1 + ( −0.4 x2 )  − x2 + 29.4 x2  dx2   0 0     EI 2 EI 2 1.75 4 8 0.7 4 20 11.76 3 20  x1  +  x2  −  x2  = 4 EI   0 4 EI   0 3EI   0 1, 792 28, 000 31,360 = + − EI EI EI 1,568 (1,568 kip 2 -ft 3 )(12 in./ft)3 =− =− = −0.180634 kip-in. EI 15  106 kip-in.2

  A = 0.1806 in. 

Ans.

(b) Slope at C Virtual Moment m: To determine the slope of the beam, first remove the real load from the beam and apply a virtual unit moment counterclockwise at C. The free-body diagram of the beam with this loading is shown. Cut a free-body diagram through the beam at section a–a. Derive a moment equation m from this free-body diagram. M a − a = m = 0 0  x1  8 ft Cut a second free-body diagram through the beam at section b–b. Derive a moment equation m from this free-body diagram. M b −b = −(0.05 kips) x2 + (1 kip-ft) − m = 0

 m = −(0.05 kips) x2 + (1 kip-ft)

0  x2  20 ft

Virtual-Work Equation for Beam Slope: From Equation (17.36), the beam slope at C can now be determined. L M (1 kip-ft)  C =  m   dx 0  EI  =

1 8  3.5 2  1 20  3.5 2  (0) − x dx + ( −0.05 x2 + 1)  − x2 + 29.4 x2  dx2   1 1   0 0  2   2  EI EI

1 20 1 20  3.5 2  3.5 2   ( −0.05 x2 )  − x2 + 29.4 x2  dx2 + (1)  − x2 + 29.4 x2  dx2   0 0     EI 2 EI 2 0.0875 20 3 1.47 20 2 1.75 20 2 29.4 20 = x dx − x dx − x dx + x2 dx2 2 2 2 2 2 2 EI 0 EI 0 EI 0 EI 0 0.0875 4 20 1.47 3 20 1.75 3 20 29.4 2 20  x2  −  x2  −  x2  +  x2  = 4 EI   0 3EI   0 3EI   0 2 EI   0 3,500 3,920 4, 666.667 5,880 = − − + EI EI EI EI 2 3 793.333 (793.333 kip -ft )(12 in./ft) 2 = = = 0.007616 kip-ft EI 15  106 kip-in.2

 C = 0.00762 rad CCW

Ans.

P17.53 A cantilever beam is loaded as shown in Figure P17.53. Assume that EI = 74×103 kN·m 2 for the beam, and use the virtual-work method to find (a) the slope at C. (b) the deflection at C. FIGURE P17.53

Solution (a) Slope at C Virtual Moment m: To determine the slope of the beam, first remove the real load from the beam and apply a virtual unit moment clockwise at C. The beam with this loading is shown. Cut a free-body diagram through the beam at section a–a. Derive a moment equation m from this free-body diagram. M a − a = −(1 kN-m) − m = 0

 m = −(1 kN-m)

0  x1  5 m

Real Moment M: Remove the virtual moment and reapply the real loading. Draw a free-body diagram that cuts through the beam at section a–a and derive a moment equation M from this free-body diagram. 40 kN/m 2 M a − a = − x1 − M = 0 2 40 kN/m 2 M = − x1 0  x1  2 m 2 Draw a second free-body diagram that cuts through the beam at section b–b and derive a moment equation M from this free-body diagram. M b −b = −(40 kN/m)(2 m)(x2 − 1 m) − M = 0

 M = −(40 kN/m)(2 m)(x2 − 1 m)

2 m  x2  5 m

Virtual-Work Equation for Beam Slope: From Equation (17.36), the beam slope at C can now be determined. L M (1 kN-m)  C =  m   dx 0  EI  1 2 1 5  40 2  ( − 1) − x dx + ( −1) ( −80(x2 − 1) ) dx2   1 1  2  EI 0 EI 2 20 2 2 80 5 = x1 dx1 + (x2 − 1)dx2  EI 0 EI 2 5 20 3 2 80  x1  + ( x2 − 1) 2  = 0 2 3EI 2 EI

53.3333 600 + EI EI 653.3333 653.3333 kN 2 -m3 = = = 0.0088288 kN-m EI 74  103 kN-m2

C = 0.00883 rad CW

Ans.

(b) Deflection at C Virtual Moment m: To determine the deflection of the beam at C, remove the real load and apply a virtual unit load downward at C. The beam with this loading is shown. Cut a free-body diagram through the beam at section a–a. Derive a moment equation m from this free-body diagram. M a − a = −(1 kN)x1 − m = 0

 m = −(1 kN)x1

0  x1  5 m

Virtual-Work Equation for Beam Deflection: From Equation (17.35), the beam deflection at C can now be determined. L M (1 kN)   C =  m   dx 0  EI  1 2 1 5  40 2  ( − 1 x ) − x dx + ( −1x2 ) ( −80(x2 − 1) ) dx2   1  2 1  1 EI 2 EI 0 20 2 3 80 5 = x1 dx1 + x2 (x2 − 1)dx2  EI 0 EI 2

20 4 2 80  1 1  3  x1  + = ( x − 1) + ( x2 − 1) 2  2  0 4 EI EI  3 2 2 80 2, 280 + EI EI 2,360 2,360 kN 2 -m3 = = = 0.0318919 kN-m EI 74  103 kN-m 2

  C = 31.9 mm 

Ans.

P17.54 Utilize the virtual-work method to determine the minimum moment of inertia I required for the beam in Figure P17.54 if the maximum beam deflection must not exceed 35 mm. Assume that E = 200 GPa.

FIGURE P17.54

Solution Virtual Moment m: The maximum beam deflection will occur at midspan. To determine the deflection of the beam at midspan, first remove the real loads from the beam and then apply a virtual unit load downward at midspan. The free-body diagram of the beam with this loading is shown. Cut a free-body diagram through the beam at section a–a between A and B. Derive a moment equation m from this free-body diagram. M a − a = −(0.5 kN) x1 + m = 0

 m = (0.5 kN) x1

0  x1  4 m

Cut a second free-body diagram through the beam at section b–b between B and midspan. Derive a moment equation m from this free-body diagram. M b −b = −(0.5 kN) x2 + m = 0

 m = (0.5 kN) x2

4 m  x2  6 m

Real Moment M: Remove the virtual load and reapply the real loads. A free-body of the beam is shown.

Draw a free-body diagram that cuts through the beam at section a–a between A and B and derive a moment equation M from this free-body diagram. M a − a = −(125 kN) x1 + M = 0

 M = (125 kN) x1

0  x1  4 m

Draw a second free-body diagram that cuts through the beam at section b– b between B and midspan and derive a moment equation M from this freebody diagram. M b −b = −(125 kN) x2 + (125 kN)( x2 − 4 m) + M = 0  M = (125 kN) x2 − (125 kN)( x2 − 4 m) 4 m  x2  6 m

Virtual-Work Equation for Beam Deflection: From Equation (17.35), an expression for the beam deflection at midspan can now be developed. Owing to the symmetry of the beam, expressions will be written for one-half of the span and then doubled. L M (1 kN)   midspan =  m   dx 0  EI  2 4 2 6 = (0.5 x1 ) (125 x1 ) dx1 + (0.5 x2 ) 125 x2 − 125( x2 − 4)  dx2  EI 0 EI 4 125 4 2 500 6 = x1 dx1 + x2 dx2  EI 0 EI 4 125 3 4 500 2 6  x1  +  x2  = 3EI   0 2 EI   4 2, 666.6667 5, 000 = + EI EI 2 7, 666.6667 kN -m3 = EI The maximum beam deflection must be limited to 35 mm. 7, 666.6667 kN-m 3  midspan =  35 mm = 0.035 m EI Solving for the minimum moment of inertia gives (7,666.6667 kN-m3 )(1,000 N/kN) I = 1.095238  10−3 m4 9 2 (200  10 N/m )(0.035 m) or in terms of mm4, the minimum moment of inertia is 4 −3 4  1, 000 mm  I  (1.095238  10 m )  = 1.095  109 mm 4  1 m  Ans.

P17.55 In Figure P17.55, calculate the minimum moment of inertia I required for the beam if the maximum beam deflection must not exceed 0.5 in. Assume that E = 29,000 ksi, and employ the virtualwork method. FIGURE P17.55

Solution Virtual Moment m: The maximum beam deflection will occur at B. To determine the deflection of the beam at B, first remove the real loads from the beam and then apply a virtual unit load downward at B. The beam with this loading is shown. Cut a free-body diagram through the beam at section a–a between A and B. Derive a moment equation m from this free-body diagram. M a − a = −(1 kip) x − m = 0

 m = −(1 kip) x

0  x  15 ft

Real Moment M: Remove the virtual load and reapply the real loads. Draw a free-body diagram that cuts through the beam at section a–a between A and B and derive a moment equation M from this free-body diagram.  1.5 kips/ft  2 M a − a = −   x − (75 kip-ft) − M = 0  2  1.5 kips/ft  2 M = − 0  x  15 ft  x − (75 kip-ft)  2 Virtual-Work Equation for Beam Deflection: From Equation (17.35), an expression for the beam deflection at B can now be developed. L M (1 kip)   B =  m   dx 0  EI  =

1 15   1.5  2  0.75 15 3 75 15 ( − 1 x ) − x − 75 dx = x dx + xdx     2  EI 0 EI 0 EI 0  

15 0.75 4 15 75 9, 492.1875 8, 437.5  x  +  x 2  = + 0 0 4 EI 2 EI EI EI 2 3 3 (17,929.6875 kip -ft )(12 in./ft) 30,982,500 kip 2 -in.3 = = EI EI The maximum beam deflection must be limited to 0.5 in. 30,982,500 kip 2 -in.3 B =  0.5 in. EI Solving for the minimum moment of inertia gives 30,982,500 kip-in.3 I = 2,136.7241 in.4 = 2,140 in.4 (29,000 ksi)(0.5 in.)

Ans.

P17.56 Use Castigliano’s second theorem to determine the vertical displacement of joint B for the truss shown in Figure P17.56/57. Assume that each member has a cross-sectional area of A = 0.85 in.2 and an elastic modulus of E = 10,000 ksi. The loads acting on the truss are P = 17 kips and Q = 9 kips. FIGURE P17.56/57

Solution Truss Reaction Forces Ay = 0.4348P + 3.5217 kips 

Member

AB AC BC

(kips) –0.7638P – 6.1871 0.6280P + 5.0870 –0.8449P + 5.2645

Cx = 9.0000 kips → Cy = 0.5652P − 3.5217 kips  F P –0.7638 0.6280 –0.8449

F (for P = 17 kips) (kips) –19.1722 15.7633 –9.0991

L (in.) 94.8683 138.0000 80.7217

 F 

  P  FL =

 F    FL P (kip-in.) 1,389.2259 1,366.1106 620.5750 3,375.9115

Equation (17.39) can now be applied.

=

1  F     FL AE  P 

From the tabulated results: B =

(3,375.9115 kip-in.) = 0.397 in.  (0.85 in.2 )(10, 000 ksi)

Ans.

Since the external load was applied in a downward direction at B, the positive value of the result confirms that joint B does displace downward.

P17.57 Applying Castigliano’s second theorem, find the horizontal displacement of joint B for the truss in Figure P17.56/57. Assume that each member has a cross-sectional area of A = 0.85 in.2 and an elastic modulus of E = 10,000 ksi, and that the loads acting on the truss are P = 17 kips and Q = 9 kips. FIGURE P17.56/57

Solution Truss Reaction Forces Ay = 0.3913Q + 7.3913 kips 

Member

AB AC BC

(kips) –0.6875Q – 12.9852 0.5652Q + 10.6763 0.5849Q – 14.3635

Cx = Q →

Cy = −0.3913Q + 9.6087 kips 

F Q

F (for Q = 9 kips)

–0.6875 0.5652 0.5849

(kips) –19.1722 15.7633 –9.0991

(in.) 94.8683 138.0000 80.7217

 F 

  Q  FL =

 F   Q  FL (kip-in.) 1,250.3578

1,229.5431 –429.6356 2,050.2653

Apply Equation (17.39)

 F  1 FL  AE  Q  to calculate the deflection from the tabulated results: (2, 050.2653 kip-in.) B = = 0.241 in.  (0.85 in.2 )(10, 000 ksi) =

Ans.

Since the external load was applied in the –x direction at B, the positive value of the result confirms that joint B does displace to the left.

P17.58 Compute the vertical displacement of joint D for the truss in Figure P17.58/59. Assume each member has a cross-sectional area of A = 2.25 in.2 and an elastic modulus of E = 29,000 ksi. The loads acting on the truss are P = 13 kips and Q = 25 kips. Employ Castigliano’s second theorem. FIGURE P17.58/59

Solution Designate the vertical load at D as P' for these calculations. Truss Reaction Forces Ax = 25 kips  Ay = 0.6667 P  + 20.6667 kips  C y = 1.6667 P  + 33.6667 kips 

F P 

Member

AB AC

(kips) 0.8012P' + 24.8382 –0.4444P' + 11.2222

BC BD CD

–0.7027P' – 35.4878 0.6667P' + 25.0000 –1.2019P'

0.8012 –0.4444

F (for P' = 13 kips) (kips) 35.2543 5.4444

(in.) 259.5997 216.0000

–0.7027 0.6667 –1.2019

–44.6233 33.6667 –15.6241

227.6840 216.0000 259.5997

 F 

  P  FL =

 F    FL P  (kip-in.) 7,332.5867 –522.6101 7,139.4398 4,848.2472 4,874.9203 23,672.5839

Equation (17.39) can now be applied.

=

1  F     FL AE  P 

From the tabulated results: D =

(23,672.5839 kip-in.) = 0.363 in.  (2.25 in.2 )(29,000 ksi)

Ans.

Since the external load was applied in a downward direction at D, the positive value of the result confirms that joint D does displace downward.

P17.59 In Figure P17.58/59, use Castigliano’s second theorem to find the horizontal displacement of joint D for the truss. The assumptions are that each member has a crosssectional area of A = 2.25 in.2 and an elastic modulus of E = 29,000 ksi and that the loads acting on the truss are P = 13 kips and Q = 25 kips. FIGURE P17.58/59

Solution Truss Reaction Forces Ax = Q  Ay = Q + 4.3333 kips 

Member

AB AC BC BD CD

(kips) 1.2019Q + 5.2080 0.3333Q – 2.8889 –1.0541Q – 18.2709 Q + 8.6667 –15.6241

Cy = Q + 30.3333 kips 

F Q

F (for Q = 25 kips)

1.2019 0.3333 –1.0541 1.0000 0.0000

(kips) 35.2543 5.4444 –44.6233 33.6667 –15.6241

(in.) 259.5997 216.0000 227.6840 216.0000 259.5997

 F 

  Q  FL =

 F   Q  FL (kip-in.) 10,999.7953

391.9576 10,709.6677 7,272.0072 0.0000 29,373.4278

Apply Equation (17.39)

 F  1 FL  AE  Q  to calculate the deflection from the tabulated results: (29,373.4278 kip-in.) D = = 0.450 in. → (2.25 in.2 )(29, 000 ksi) =

Ans.

Since the external load was applied in the +x direction at D, the positive value of the result confirms that joint D does displace to the right.

P17.60 Calculate the horizontal displacement of joint A for the truss in Figure P17.60/61 employing Castigliano’s second theorem. Make the assumption that each member has a cross-sectional area of A = 1,600 mm2 and an elastic modulus of E = 200 GPa.

FIGURE P17.60/61

Solution Add a dummy load in the horizontal direction at joint A and designate it as P. Truss Reaction Forces Dx = P + 233.4000 kN  Ex = 233.4000 kN → E y = 137.0000 kN 

Member

AB AC BC BD CD CE DE

(kN) –P – 41.6000 66.5925 85.0000 –P – 41.6000 –235.7037 233.4000 137.0000

F P –1.0000 0.0000 0.0000 –1.0000 0.0000 0.0000 0.0000

F (for P = 0 kN) (kN) –41.6000 66.5925 85.0000 –41.6000 –235.7037 233.4000 137.0000

L (mm) 6,000.0000 9,604.6864 7,500.0000 10,500.0000 12,903.4879 10,500.0000 7,500.0000

 F 

  P  FL =

 F    FL P (kN-mm) 249,600.0000 0.0000 0.0000 436,800.0000 0.0000 0.0000 0.0000 686,400.0000

Apply Equation (17.39)

1  F     FL AE  P  to calculate the deflection from the tabulated results: (686, 400.0000 kN-mm)(1,000 N/kN) A = = 2.15 mm → (1, 600 mm 2 )(200, 000 N/mm 2 ) =

Ans.

P17.61 In Figure P17.60/61, use Castigliano’s second theorem to compute the vertical displacement of joint B for the truss. Each member is assumed to have a crosssectional area of A = 1,600 mm2 and an elastic modulus of E = 200 GPa.

FIGURE P17.60/61

Solution Designate the 85 kN load as P. Truss Reaction Forces Dx = 1.4P + 114.4000 kN 

Member

AB AC

(kN) –41.6000 66.5925

BC BD CD CE DE

P –41.6000 –1.7205P – 89.4642 1.4P + 114.4000 P + 52.0000

Ex = 1.4P + 114.4000 kN → F P

Ey = P + 52.0000 kN 

0.0000 0.0000

F (for P = 85 kN) (kN) –41.6000 66.5925

(mm) 6,000.0000 9,604.6864

1.0000 0.0000 –1.7205 1.4000 1.0000

85.0000 –41.6000 –235.7037 233.4000 137.0000

7,500.0000 10,500.0000 12,903.4879 10,500.0000 7,500.0000

 F 

  P  FL =

 F    FL P (kN-mm) 0.0000 0.0000 637,500.0000 0.0000 5,232,728.4266 3,430,980.0000 1,027,500.0000 10,328,708.4266

Apply Equation (17.39)

1  F     FL AE  P  to calculate the deflection from the tabulated results: (10,328, 708.4266 kN-mm)(1,000 N/kN) B = = 32.3 mm  (1, 600 mm 2 )(200, 000 N/mm 2 ) =

Ans.

P17.62 In Figure P17.62/63, the truss is subjected to concentrated loads P = 200 kN and Q = 40 kN. Members AB, BC, DE, and EF each have a crosssectional area of A = 2,700 mm2, with all other members each having a cross-sectional area of A = 1,060 mm2. All members are made of steel [E = 200 GPa]. For the given loads, calculate the horizontal displacement of joint F by applying Castigliano’s second theorem. FIGURE P17.62/63

Solution Designate the horizontal force at F as Q'. Truss Reaction Forces Ay = 1.3333Q  + 226.6667 kN  Dx = 1.0Q  + 40.0000 kN → Dy = −1.3333Q  + 173.3333 kN 

F Q 

F (for Q' = 40 kN)

AB AD AE BC BE BF

(kN) –0.6667Q' – 200.0000 1.0000Q' + 40.0000 –1.2019Q' – 48.0740 –200.0000 1.0000Q' –1.2019 Q'

–0.6667 1.0000 –1.2019 0.0000 1.0000 –1.2019

(kN) –226.6667 80.0000 –96.1480 –200.0000 40.0000 –48.0740

(mm) 6,000.0000 9,000.0000 10,816.6538 6,000.0000 9,000.0000 10,816.6538

(mm2) 2,700 1,060 1,060 2,700 1,060 1,060

CF DE EF

0.0000 1.3333Q' – 173.3333 0.6667Q' – 200.0000

0.0000 1.3333 0.6667

0.0000 –120.0000 –173.3333

9,000.0000 6,000.0000 6,000.0000

1,060 2,700 2,700

Member

 F   FL 

  Q   AE  =

 F   FL   Q    AE  (mm) 1.6790

3.3962 5.8959 0.0000 1.6981 2.9479 0.0000 –1.7778 –1.2840 12.5554

Apply Equation (17.39)

 F   FL   =     Q    AE  to calculate the deflection from the tabulated results:  F = 12.56 mm 

Ans.

P17.63 The truss in Figure P17.62/63 is subjected to concentrated loads P = 200 kN and Q = 40 kN. Members AB, BC, DE, and EF each have a cross-sectional area of A = 2,700 mm2. All other members each have a cross-sectional area of A = 1,060 mm2. All members are made of steel [E = 200 GPa]. For the given loads, utilize Castigliano’s second theorem to determine the horizontal displacement of joint B.

FIGURE P17.62/63

Solution Add a dummy horizontal force at B and designate it as Q'. Truss Reaction Forces Ay = 0.6667Q  + 280.0000 kN  Dx = 1.0Q  + 80.0000 kN → Dy = −0.6667Q  + 120.0000 kN 

F Q 

F (for Q' = 0 kN)

AB AD AE BC BE BF

(kN) –226.6667 1.0000Q' + 80.0000 –1.2019Q' – 96.1480 –200.0000 1.0000Q' + 40.0000 –48.0740

0.0000 1.0000 –1.2019 0.0000 1.0000 0.0000

(kN) –226.6667 80.0000 –96.1480 –200.0000 40.0000 –48.0740

(mm) 6,000.0000 9,000.0000 10,816.6538 6,000.0000 9,000.0000 10,816.6538

(mm2) 2,700 1,060 1,060 2,700 1,060 1,060

CF DE EF

0.0000 0.6667Q' – 120.0000 –173.3333

0.0000 0.6667 0.0000

0.0000 –120.0000 –173.3333

9,000.0000 6,000.0000 6,000.0000

1,060 2,700 2,700

Member

 F   FL 

  Q   AE  =

 F   FL   Q    AE  (mm) 0.0000

3.3962 5.8959 0.0000 1.6981 0.0000 0.0000 –0.8889 0.0000 10.1013

Apply Equation (17.39)

 F   FL   =     Q    AE  to calculate the deflection from the tabulated results:  B = 10.10 mm 

Ans.

P17.64 The truss in Figure P17.64 is subjected to concentrated loads of P = 130 kN and 2P = 260 kN. All members are made of steel [E = 200 GPa], and each has a cross-sectional area of A = 4,200 mm2. Use Castigliano’s second theorem to determine (a) the horizontal displacement of joint A. (b) the vertical displacement of joint D. FIGURE P17.64

Solution (a) Horizontal Displacement of Joint A Add a horizontal dummy load at A and designate it as Q. Truss Reaction Forces Bx = 1.0000Q → By = 0.375Q + 520.0000 kN  Fy = −0.375Q + 520.0000 kN 

F Q

F (Q = 0 kN)

AB AC BC BD CD CE DE DF

(kN) –216.6667 1.0000Q + 173.3333 –0.3750Q – 390.0000 –1.0000Q – 173.3333 0.6250Q + 216.6667 0.5000Q –260.0000 –173.3333

0.0000 1.0000 –0.3750 –1.0000 0.6250 0.5000 0.0000 0.0000

(kN) –216.6667 173.3333 –390.0000 –173.3333 216.6667 0.0000 –260.0000 –173.3333

(mm) 10,000.0000 8,000.0000 6,000.0000 8,000.0000 10,000.0000 8,000.0000 6,000.0000 8,000.0000

1,386,666.4000 877,500.0000 1,386,666.4000 1,354,166.8750 0.0000 0.0000 0.0000

DG EG FG FH GH

–0.6250Q + 216.6667 0.5000Q 0.3750Q – 390.0000 –216.6667 173.3333

–0.6250 0.5000 0.3750 0.0000 0.0000

216.6667 0.0000 –390.0000 –216.6667 173.3333

10,000.0000 8,000.0000 6,000.0000 10,000.0000 8,000.0000

–1,354,166.8750 0.0000 –877,500.0000 0.0000 0.0000

Member

 F 

  Q  FL =

 F   Q  FL (kN-mm) 0.0000

2,773,332.8000

Apply Equation (17.39)

 F  1 FL  AE  Q  to calculate the deflection from the tabulated results: (2, 773,332.8 kN-mm)(1,000 N/kN) A = = 3.30 mm  (4, 200 mm 2 )(200, 000 N/mm 2 ) =

Ans.

(b) Vertical Displacement of Joint D Add a vertical dummy load at D and designate it as PD. Truss Reaction Forces Bx = 0.0 By = 0.5000 PD + 520.0000 kN  Fy = 0.5000 PD + 520.0000 kN 

F PD

F (PD = 0 kN)

AB AC BC BD

(kN) –216.6667 173.3333 –0.5000PD – 390.0000 –173.3333

0.0000 0.0000 –0.5000 0.0000

(kN) –216.6667 173.3333 –390.0000 –173.3333

(mm) 10,000.0000 8,000.0000 6,000.0000 8,000.0000

0.0000 1,170,000.0000 0.0000

CD CE DE DF DG EG FG FH GH

0.8333PD + 216.6667 –0.6667PD –260.0000 –173.3333 0.8333PD + 216.6667 –0.6667PD –0.5000PD – 390.0000 –216.6667 173.3333

0.8333 –0.6667 0.0000 0.0000 0.8333 –0.6667 –0.5000 0.0000 0.0000

216.6667 0.0000 –260.0000 –173.3333 216.6667 0.0000 –390.0000 –216.6667 173.3333

10,000.0000 8,000.0000 6,000.0000 8,000.0000 10,000.0000 8,000.0000 6,000.0000 10,000.0000 8,000.0000

1,805,483.6111 0.0000 0.0000 0.0000 1,805,483.6111 0.0000 1,170,000.0000 0.0000 0.0000

Member

 F 

  P  FL = 



 F    FL  PD  (kN-mm) 0.0000

5,950,967.2222

Apply Equation (17.39)

 F  1 FL  AE  PD  to calculate the deflection from the tabulated results: (5,950,967.2222 kN-mm)(1,000 N/kN) D = = 7.08 mm  (4, 200 mm 2 )(200, 000 N/mm 2 ) =

Ans.

P17.65 Employing Castigliano’s second theorem, calculate the slope of the beam at A for the loading shown in Figure P17.65. Assume that EI is constant for the beam.

FIGURE P17.65

Solution Designate the concentrated moment at A as M'. The free-body diagram for the beam with this load is shown.

Draw a free-body diagram that cuts through the beam at section a–a and derive a moment equation M from this free-body diagram. M M M a − a = x−M =0 M = x L L Differentiate this expression to obtain ∂M/∂M'. M x = M  L Substitute M' = M0 into the bending-moment equation to obtain M M= 0x L Castigliano’s second theorem applied to beam slopes is expressed by Equation (17.41). When the expressions derived for ∂M/∂M' and M are substituted, Equation (17.41) becomes: L  M  M L  x   M x M0 L 2 0 =  dx = dx = x dx  0  L   EI L  0  M   EI EI L2 0 Integrate this expression over the beam length L to determine the beam slope at A. M 0 L3 M L Ans. A = = 0 (CW) 2 3EI L 3EI

P17.66 Determine the deflection of the beam at A for the loading in Figure P17.66, utilizing Castigliano’s second theorem. Assume that EI is constant for the beam.

FIGURE P17.66

Solution To determine the deflection of the cantilever beam at A, a dummy load Q will be applied downward at A.

Draw a free-body diagram that cuts through the beam at section a–a and derive a moment equation M from this free-body diagram. M a − a = Qx + M = 0  M = −Qx 0 xa Differentiate this expression to obtain ∂M/∂Q. M = −x Q Substitute Q = 0 into the bending moment equation to obtain M =0 0 xa Draw a second free-body diagram that cuts through the beam at section b–b and derive a moment equation M from this free-body diagram. M b −b = Qx + P( x − a) + M = 0

 M = −Qx − P( x − a) axL Differentiate this expression to obtain ∂M/∂Q. M = −x Q Substitute Q = 0 into the bending moment equation to obtain M = − P( x − a ) axL Castigliano’s second theorem applied to beam deflections can be expressed by Equation (17.40). L  M  M =   EI dx 0  Q 

Thus: a L P L 2  P   A =  ( − x )( 0) dx +  ( − x )  − ( x − a )  dx = x − xa dx 0 a EI a  EI  P L 2 Pa L = x dx − x dx  EI a EI a P 1 3 L Pa 1 2 L x  − x  = EI 3   a EI 2   a P 3 Pa 2 = L − a3 − L − a2 3EI 2 EI 2 PL3 2 Pa 3 3PaL2 3Pa 3 = − − + 6 EI 6 EI 6 EI 6 EI

(

A =

)

(

)

P 2 L3 − 3aL2 + a 3  6 EI

Ans.

This expression can be further simplified by substituting a + b for L. After substantial manipulation, this expression can be rewritten as

A =

Pb 2 ( 3L − b ) 6 EI



P17.67 Calculate the slope and the deflection of the beam at B for the loading shown in Figure P17.67. Use Castigliano’s second theorem, and assume that EI is constant for the beam.

FIGURE P17.67

Solution To determine the slope of the cantilever beam, first apply a dummy concentrated moment M' clockwise at B. The free-body diagram for the beam with this load is shown. For this beam, one equation will be derived for the bending moment. Draw a free-body diagram that cuts through the beam at section a–a and derive a moment equation M from this free-body diagram. w M a − a = − M  − 0 x 2 − M = 0 2 w  M = −M  − 0 x2 0 x L 2 Differentiate this expression to obtain ∂M/∂M'. M = −1 M  Substitute M' = 0 into the bending-moment equation to obtain w M = − 0 x2 2 Castigliano’s second theorem applied to beam slopes is expressed by Equation (17.41). L  M  M =  dx  0  M   EI Thus L  w   B =  ( −1)  − 0 x 2  dx 0  2 EI  =

w0 L 2 w0 L3 x dx = 2 EI 0 6 EI

w0 L3  B = 6 EI

(CW)

Ans.

To determine the deflection of the cantilever beam, remove the dummy concentrated moment and apply a dummy concentrated load downward at B. The free-body diagram for the beam with this load is shown. For this beam, one equation will be derived for the bending moment.

Cut a free-body diagram through the beam at section a–a. Derive a moment equation M from this free-body diagram. w M a − a = − Px − 0 x 2 − M = 0 2 w  M = − Px − 0 x 2 0 x L 2 Differentiate this expression to obtain ∂M/∂P. M = −x P Substitute P = 0 into the bending-moment equation to obtain w M = − 0 x2 2 Castigliano’s second theorem applied to beam deflections can be expressed by Equation (17.40). L  M  M =  dx  0  P  EI Thus L w L  w   B =  ( − x )  − 0 x 2  dx = 0  x 3dx 0  2 EI  2 EI 0

w0 1 4 L w0 L4 x  = 2 EI 4   0 8 EI

 B =

w0 L4 8 EI



Ans.

P17.68 Apply Castigliano’s second theorem to compute the slope and deflection of the beam at

C for the loading shown in Figure P17.68. Assume that EI is constant for the beam.

FIGURE P17.68

Solution Determine Slope at C To determine the slope of the simply-supported beam, first apply a dummy concentrated moment clockwise at C. The free-body diagram of the beam for this loading is shown.

Cut a free-body diagram through the beam at section a–a. Derive a moment equation M from this free-body diagram.  P M  M a − a =  +  x1 + M = 0 2 L   P M  M = − + 0  x1  L  x1 2 L  Differentiate this expression to obtain ∂M/∂M'. M 1 = − x1 M  L Substitute M' = 0 into the bending-moment equation to obtain P M = − x1 2 Cut a second free-body diagram through the beam at section b–b. Derive a moment equation m from this free-body diagram. M b −b = − M  − Px2 − M = 0

 M = − M  − Px2

0  x2 

L 2

Differentiate this expression to obtain ∂M/∂M'. M = −1 M  Substitute M' = 0 into the bending-moment equation to obtain M = − Px2 Castigliano’s second theorem applied to beam slopes is expressed by Equation (17.41). L  M  M =  dx  0  M   EI Using this formula, the beam slope at C can now be determined.

 1  P   P  C =   − x1   − x1  dx1 +  2 ( −1)  − x dx 0  0  EI 2  2 L   2 EI  L

L P P 2 P 1 3 L P 1 2 L2 2  x1  +  x2  x dx + x dx = 1 1 2 2 2 LEI 0 EI 0 2 LEI 3   0 EI 2   0

PL2 PL2 7 PL2 + = 6 EI 8 EI 24 EI

7 PL2 (CW) 24 EI Determine Deflection at C C =

Ans.

To determine the deflection of the simply-supported beam at C, consider the beam free-body diagram shown.

Cut a free-body diagram through the beam at section a–a. Derive a moment equation M from this free-body diagram. P P M a − a = x1 + M = 0  M = − x1 0  x1  L 2 2 Differentiate this expression to obtain ∂M/∂P. M 1 = − x1 P 2 Cut a second free-body diagram through the beam at section b–b. Derive a moment equation M from this free-body diagram. L M b −b = − Px2 − M = 0  M = − Px2 0  x2  2 Differentiate this expression to obtain ∂M/∂P. M = − x2 P Castigliano’s second theorem applied to beam deflections can be expressed by Equation (17.40). L  M  M =  dx  0  P  EI The beam deflection at C can now be determined. L L 1  P   P   C =   − x1   − x1  dx1 +  2 ( −1x2 )  − x dx 0  0  EI 2  2 2   2 EI  L

P L 2 P 2 2 P 1 3 L P 1 3 L2  x1  +  x2  = x1 dx1 + x2 dx2 = 4 EI 0 EI 0 4 EI 3   0 EI 3   0 PL3 PL3 3PL3 = + = 12 EI 24 EI 24 EI  C =

PL3 8 EI

 Ans.

P17.69 What is the deflection of the compound rod at C for the loading shown in Figure P17.69? Between A and B, the rod’s diameter is 35 mm, and between B and C, its diameter is 20 mm. Assume that E = 200 GPa for both segments of the compound rod, and use Castigliano’s second theorem. FIGURE P17.69

Solution Designate the concentrated load at C as P. Draw a free-body diagram that cuts through the compound rod at section a–a and derive a moment equation M from this free-body diagram. M a − a = − Px1 − M = 0

 M = − Px1 0  x1  320 mm Differentiate this expression to obtain ∂M/∂P. M = − x1 P Substitute P = 500 N into the bending-moment equation to obtain M = − (500 N) x1 Strictly speaking, a second free-body diagram through the compound rod at section b–b is not necessary in this instance; however, it is included here to emphasize that the integration will need to be performed in two steps to account for the change in moment of inertia of the compound rod. M b −b = − Px2 − M = 0  M = − Px2 320 mm  x2  960 mm Differentiate this expression to obtain ∂M/∂P. M = − x2 P Substitute P = 500 N into the bending-moment equation to obtain M = −(500 N) x2 Moments of Inertia: The moment of inertia for segment AB is:

I AB =



(35 mm) 4 = 73, 661.76 mm4

and the moment of inertia for segment BC is:

I BC =



(20 mm) 4 = 7,853.98 mm 4 64 Castigliano’s second theorem applied to beam deflections can be expressed by Equation (17.40). L  M  M =  dx  0  P  EI The beam deflection at C can now be determined.

C = 

320 mm





( − x1 )  − (200, 000 N/mm 2 )(7,853.98 mm 4 ) x1  dx1 500 N





( − x2 )  − (200, 000 N/mm 2 )(73, 661.76 mm 4 ) x2  dx2 320 mm

+

960 mm

500 N

320 mm 500 N x12 dx1 2 4 0 (200, 000 N/mm )(7,853.98 mm ) 960 mm 500 N + x 2 dx2 2 4 320 mm 2 (200, 000 N/mm )(73, 661.76 mm ) 500 N 1 3 320 mm  x1  = 2 4 (200, 000 N/mm )(7,853.98 mm ) 3   0 500 N 1 3 960 mm  x2  + 2 4 (200, 000 N/mm )(73, 661.76 mm ) 3   320 mm

= (106.103  10 −9 mm −2 )(32.7680  106 mm 3 ) + (11.3130  10 −9 mm −2 )(851.9680  106 mm3 ) =  3.4768 mm + 9.6383 mm    C = 13.12 mm 

Ans.

P17.70 The compound steel [E = 200 GPa] rod shown in Figure P17.70/71 has a diameter of 20 mm in segments AB and DE, and a diameter of 35 mm in segments BC and CD. For the given loading, employ Castigliano’s second theorem to find the slope of the compound rod at A. FIGURE P17.70/71

Solution To determine the slope of the simply-supported beam, first apply a dummy concentrated moment M' clockwise at A.

Draw a free-body diagram that cuts through the compound rod at section a–a and derive a moment equation M from this free-body diagram. M   M a − a = − M  −  7,800 N −  x1 + M = 0  330 mm  M    M = M  +  7,800 N − 0  x1  100 mm  x1  330 mm  Differentiate this expression to obtain ∂M/∂M'. x M = 1− 1 M  330 Substitute M' = 0 into the bending-moment equation to obtain M = (7,800 N) x1

Cut a second free-body diagram through the compound rod at section b–b, and derive a second moment equation M.

M  (120 N/mm)  M b −b = − M  −  7,800 N − x2 + (x2 − 100 mm) 2 + M = 0    330 mm 2 (120 N/mm) M   (x2 − 100 mm) 2 +  7,800 N −  x2 + M   2 330 mm  100 mm  x2  230 mm Differentiate this expression to obtain ∂M/∂M'. x M = 1− 2 M  330 Substitute M' = 0 into the bending-moment equation to obtain (120 N/mm) M =− (x2 − 100 mm)2 + (7,800 N) x2 2 M = −

Cut a third free-body diagram through the compound rod at section c–c, and derive a third moment equation M. M   M c − c =  7,800 N+  x3 − M = 0  330 mm  M    M =  7,800 N+ 0 mm  x3  100 mm  x3  330 mm  Differentiate this expression to obtain ∂M/∂M'. x M = 3 M  330 Substitute M' = 0 into the bending-moment equation to obtain M = (7,800 N) x3 Moments of Inertia: The moment of inertia for 20 mm diameter segments AB and DE is:

I AB = I DE =



(20 mm) 4 = 7,853.98 mm 4

and the moment of inertia for 35 mm diameter segments BC and CD is:

I BC = I CD =



(35 mm) 4 = 73, 661.76 mm 4

64 Castigliano’s second theorem applied to beam slopes is expressed by Equation (17.41). L  M  M =  dx  0  M   EI From this formula, the beam slope at A can now be determined. 100  x   7,800   A =  1 − 1   x1 dx1 0  330   EI AB  x   60 7,800   1− 2   − (x2 − 100) 2 + x2 dx2  100  330   EI BC EI BC 

+

230

+

100

 x3   7,800  x3 dx3   330   EI DE 

7,800 100 7,800 100 2 x1dx1 − x1 dx1  EI AB 0 330 EI AB 0 − + +

(

)

1 230 60(x2 − 100) 2 − 7, 800 x2 dx2 EI BC 100

(

)

230 1 x2 60(x2 − 100) 2 − 7, 800 x2 dx2  1 00 330 EI BC

7,800 100 2 x3 dx3 330 EI DE 0

7,800 100 7,800 100 2 x1dx1 − x1 dx1  EI AB 0 330 EI AB 0 − + +

60 230 7,800 230 (x2 − 100) 2 dx2 + x2 dx2  EI BC 100 EI BC 100 230 60 7,800 230 2 x2 (x2 − 100) 2 dx2 − x2 dx2  330 EI BC 100 330 EI BC 100

7,800 100 2 x3 dx3 330 EI DE 0

7,800 (100)2 7,800 (100)3 = − EI AB 2 330 EI AB 3 −

60 (130)3 7,800 (2302 − 1002 ) + EI BC 3 EI BC 2

60  (130)4 (100)(130)3  7,800 (230)3 − (100)3 + −  330 EI 330 EI BC  4 3 3  BC

7,800 (100)3 + 330 EI DE 3 39, 000, 000 7,878, 787.88 43,940, 000 167,310, 000 = − − + EI AB EI AB EI BC EI BC +

26, 297, 424.24 87,982, 424.24 7,878, 787.88 − + EI BC EI BC EI DE

39, 000, 000 61, 685, 000 + EI AB EI BC

39, 000, 000 61, 685, 000 + (200, 000)(7,853.98) (200, 000)(73, 661.76)

= 24.828176  10 −3 + 4.1870436  10 −3 = 29.015220  10−3  A = 0.0290 rad

(CW)

Ans.

P17.71 Figure P17.70/71 shows a compound steel [E = 200 GPa] rod with a diameter of 20 mm in segments AB and DE, and a diameter of 35 mm in segments BC and CD. For the given loading, calculate the deflection of the compound rod at C, using Castigliano’s second theorem. FIGURE P17.70/71

Solution To determine the deflection of the simply-supported beam, first apply a dummy concentrated load P downward at C. The beam free-body diagram for this loading is shown.

Draw a free-body diagram that cuts through the compound rod at section a–a and derive a moment equation M from this free-body diagram. P  M a − a = −  7,800 N +  x1 + M = 0  2 P   M =  7,800 N +  x1 0  x1  100 mm  2 Differentiate this expression to obtain ∂M/∂P. M = 0.5 x1 P Substitute P = 0 into the bending-moment equation to obtain M = (7,800 N) x1

Cut a second free-body diagram through the compound rod at section b–b, and derive a second moment equation M.

P (120 N/mm)  M b −b = −  7,800 N +  x2 + (x2 − 100 mm) 2 + M = 0  2 2 (120 N/mm) P  (x2 − 100 mm) 2 +  7,800 N +  x2  2 2 Differentiate this expression to obtain ∂M/∂P. M = 0.5 x2 P Substitute P = 0 into the bending-moment equation to obtain (120 N/mm) M =− (x2 − 100 mm)2 + (7,800 N) x2 2 M = −

100 mm  x2  230 mm

Cut a third free-body diagram through the compound rod at section c–c, and derive a third moment equation M. P  M c − c =  7,800 N +  x3 − M = 0  2 P   M =  7,800 N +  x3 0 mm  x3  100 mm  2 Differentiate this expression to obtain ∂M/∂P. M = 0.5 x3 P Substitute P = 0 into the bending-moment equation to obtain M = (7,800 N) x3 Moments of Inertia: The moment of inertia for 20 mm diameter segments AB and DE is:

I AB = I DE =



(20 mm) 4 = 7,853.98 mm 4

and the moment of inertia for 35 mm diameter segments BC and CD is:

I BC = I CD =



(35 mm) 4 = 73, 661.76 mm 4

64 Castigliano’s second theorem applied to beam deflections can be expressed by Equation (17.40). L  M  M =  dx  0  P  EI Since the loading is symmetrical on the beam, the integrals will be written for the interval [0,165] and then doubled. The beam deflection at C can now be determined.

 C = 2

100

165  60  7,800  7,800  (0.5 x1 )  x1  dx1 + 2 (0.5 x2 )  − (x2 − 100) 2 + x2 dx2 100 EI BC   EI AB   EI BC

(

)

(

)

7,800 100 2 1 165 x d x − x2 60(x2 − 100) 2 − 7,800 x2 dx2 1 1   0 1 00 EI AB EI BC 7,800 100 2 1 165 = x1 dx1 − x2 60(x2 − 100) 2 − 7,800 x2 dx2   0 100 EI AB EI BC

(

)

7,800 100 2 60 165 7,800 165 2 x1 dx1 − x2 (x2 − 100) 2 dx2 + x2 dx2   EI AB 0 EI BC 100 EI BC 100

7,800 (100)3 60  (65) 4 (100)(65)3  7,800 (165)3 − (100)3 − + + EI AB 3 EI BC  4 3 3  EI BC

2, 600  106 817.009  106 9, 079.525  106 − + EI AB EI BC EI BC

2, 600  106 8, 262.516  106 + (200, 000)(7,853.98) (200, 000)(73, 661.76) = 1.655212 + 0.560842 = 2.216053

  C = 2.22 mm 

Ans.

P17.72 Figure P17.72 shows a simply supported beam. Assume that EI = 15×106 kip·in. 2 for the beam. Use Castigliano’s second theorem to determine (a) the deflection at A. (b) the slope at C. FIGURE P17.72

Solution (a) Deflection at A To determine the deflection of the simply-supported beam, first apply a dummy concentrated load P downward at A. A free-body diagram of the beam is shown. Draw a free-body diagram that cuts through the beam at section a–a and derive a moment equation M from this free-body diagram. 3.5 kips/ft 2 M a − a = Px1 + x1 + M = 0 2 3.5 kips/ft 2 M = − x1 − Px1 0  x1  8 ft 2 Differentiate this expression to obtain ∂M/∂P. M = − x1 P Substitute P = 0 into the bending-moment equation to obtain 3.5 kips/ft 2 M =− x1 2 Draw a second free-body diagram that cuts through the beam at section b–b and derive a moment equation M from this free-body diagram. 3.5 kips/ft 2 M b − b = − x2 + ( 29.4 kips − 0.4 P ) x2 − M = 0 2 3.5 kips/ft 2 M = − x2 + ( 29.4 kips − 0.4 P ) x2 0 ft  x2  20 ft 2 Differentiate this expression to obtain ∂M/∂P. M = −0.4 x2 P Substitute P = 0 into the bending-moment equation to obtain 3.5 kips/ft 2 M =− x2 + (29.4 kips) x2 2 Castigliano’s second theorem applied to beam deflections can be expressed by Equation (17.40). L  M  M =  dx  0  P  EI The beam deflection at A can now be determined.

1 8 1 20  3.5 2   3.5 2  ( − x ) − x dx + ( −0.4 x2 )  − x2 + 29.4 x2  dx2   1 1 1   0 0  2   2  EI EI 1.75 4 8 0.7 4 20 11.76 3 20  x1  +  x2  −  x2  = 4 EI   0 4 EI   0 3EI   0 1, 792 28, 000 31,360 = + − EI EI EI 1,568 (1,568 kip-ft 3 )(12 in./ft)3 =− =− = −0.180634 in. EI 15  106 kip-in.2

A =

  A = 0.1806 in. 

Ans.

(b) Slope at C To determine the slope of the simply-supported beam, first apply a dummy concentrated moment M' counterclockwise at C. The free-body diagram for this beam is shown. Draw a free-body diagram that cuts through the beam at section a–a and derive a moment equation M from this free-body diagram. 3.5 kips/ft 2 M a − a = x1 + M = 0 2 3.5 kips/ft 2 M = − x1 0  x1  8 ft 2 Differentiate this expression to obtain ∂M/∂M'. M =0 M  Substitute M' = 0 into the bending-moment equation to obtain 3.5 kips/ft 2 M =− x1 2 Draw a second free-body diagram that cuts through the beam at section b–b and derive a moment equation M from this free-body diagram. 3.5 kips/ft 2  M  M b −b = − x2 +  29.4 kips −  x2 + M  − M = 0  2 20 ft  3.5 kips/ft 2  M  x2 +  29.4 kips −  x2 + M   2 20 ft  0 ft  x2  20 ft Differentiate this expression to obtain ∂M/∂M'. M = −0.05 x2 + 1 M  Substitute M' = 0 into the bending-moment equation to obtain 3.5 kips/ft 2 M =− x2 + (29.4 kips) x2 2 Castigliano’s second theorem applied to beam slopes is expressed by Equation (17.41). L  M  M =  dx  0  M   EI M = −

Thus, the beam slope at C is: 1 8  3.5 2  1 20  3.5 2  C = (0)  − x1  dx1 + ( −0.05 x2 + 1)  − x2 + 29.4 x2  dx2    2   2  EI 0 EI 0 1 20 1 20  3.5 2  3.5 2   ( −0.05 x2 )  − x2 + 29.4 x2  dx2 + (1)  − x2 + 29.4 x2  dx2    2   2  EI 0 EI 0 0.0875 20 3 1.47 20 2 1.75 20 2 29.4 20 = x2 dx2 − x2 dx2 − x2 dx2 + x2 dx2    0 EI EI 0 EI 0 EI 0 0.0875 4 20 1.47 3 20 1.75 3 20 29.4 2 20  x2  −  x2  −  x2  +  x2  = 4 EI   0 3EI   0 3EI   0 2 EI   0 3,500 3,920 4, 666.667 5,880 = − − + EI EI EI EI 2 793.333 (793.333 kip-ft )(12 in./ft) 2 = = = 0.007616 rad EI 15  106 kip-in.2

 C = 0.00762 rad CCW

Ans.

P17.73 A cantilever beam is loaded as shown in Figure P17.73. Assume that EI = 74×103 kN·m 2 for the beam, and employ Castigliano’s second theorem to find (a) the slope at C. (b) the deflection at C. FIGURE P17.73

Solution (a) Slope at C To determine the slope of the cantilever beam at C, first apply a dummy concentrated moment M' clockwise at C. The free-body diagram for this beam is shown. Draw a free-body diagram that cuts through the beam at section a–a and derive a moment equation M from this free-body diagram. 40 kN/m 2 M a − a = − M  − x1 − M = 0 2 40 kN/m 2 M = − x1 − M  0  x1  2 m 2 Differentiate this expression to obtain ∂M/∂M'. M = −1 M  Substitute M' = 0 into the bending-moment equation to obtain 40 kN/m 2 M =− x1 2 Draw a second free-body diagram that cuts through the beam at section b–b and derive a moment equation M from this free-body diagram. M b −b = − M  − (40 kN/m)(2 m)(x2 − 1 m) − M = 0  M = − (40 kN/m)(2 m)(x2 − 1 m) − M  2 m  x2  5 m Differentiate this expression to obtain ∂M/∂M'. M = −1 M  Substitute M' = 0 into the bending-moment equation to obtain M = −(40 kN/m)(2 m)(x2 − 1 m)

Castigliano’s second theorem applied to beam slopes is expressed by Equation (17.41). L  M  M =  dx  0  M   EI Thus, the beam slope at C is: 1 2 1 5  40 2  C = ( −1)  − x1  dx1 + ( −1) ( −80(x2 − 1) ) dx2  0  2  EI EI 2 20 2 2 80 5 = x1 dx1 + (x2 − 1)dx2  EI 0 EI 2 5 20 3 2 80  x1  + ( x2 − 1) 2  = 0 2 3EI 2 EI 53.3333 600 = + EI EI 653.3333 653.3333 kN-m 2 = = = 0.0088288 rad EI 74  103 kN-m 2  C = 0.00883 rad CW

Ans.

(b) Deflection at C To determine the deflection of the cantilever beam, first apply a dummy concentrated load P downward at C. A free-body diagram of the beam is shown.

Draw a free-body diagram that cuts through the beam at section a–a and derive a moment equation M from this free-body diagram. 40 kN/m 2 M a − a = − x1 − Px1 − M = 0 2 40 kN/m 2 M = − x1 − Px1 0  x1  2 m 2 Differentiate this expression to obtain ∂M/∂P. M = − x1 P Substitute P = 0 into the bending-moment equation to obtain 40 kN/m 2 M =− x1 2 Draw a second free-body diagram that cuts through the beam at section b–b and derive a moment equation M from this free-body diagram. M b −b = −(40 kN/m)(2 m)(x2 − 1 m) − Px2 − M = 0  M = −(40 kN/m)(2 m)(x2 − 1 m) − Px2 2 m  x2  5 m Differentiate this expression to obtain ∂M/∂P. M = − x2 P Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Substitute P = 0 into the bending-moment equation to obtain M = −(40 kN/m)(2 m)(x2 − 1 m) Castigliano’s second theorem applied to beam deflections can be expressed by Equation (17.40). L  M  M =  dx  0  P  EI The beam deflection at C can now be determined. 1 2 1 5  40  C = ( − x1 )  − x12  dx1 + ( − x2 ) ( −80(x2 − 1) ) dx2   2  EI 0 EI 2 20 2 3 80 5 = x dx + x2 (x2 − 1)dx2 1 1 EI 0 EI 2 5

20 4 2 80  1 1   x1  + = ( x2 − 1)3 + ( x2 − 1) 2   0 4 EI EI  3 2 2 80 2, 280 + EI EI 2,360 2,360 kN-m3 = = = 0.0318919 m EI 74  103 kN-m 2

  C = 31.9 mm 

Ans.

P17.74 Compute the minimum moment of inertia I required for the beam in Figure P17.74 if the maximum beam deflection must not exceed 35 mm. Assuming that E = 200 GPa, employ Castigliano’s second theorem.

FIGURE P17.74

Solution To determine the maximum deflection of the simplysupported beam, first apply a dummy concentrated load P downward at midspan. A free-body diagram of the beam is shown.

Draw a free-body diagram that cuts through the beam at section a–a between A and B and derive a moment equation M from this free-body diagram. M a − a = − (125 kN + 0.5P ) x1 + M = 0  M = (125 kN + 0.5P ) x1

0  x1  4 m

Differentiate this expression to obtain ∂M/∂P. M = 0.5 x1 P Substitute P = 0 into the bending-moment equation to obtain M = (125 kN) x1 Draw a second free-body diagram that cuts through the beam at section b– b between B and midspan and derive a moment equation M from this freebody diagram. M b−b = − (125 kN + 0.5P ) x2 + (125 kN)( x2 − 4 m) + M = 0

 M = (125 kN + 0.5P ) x2 − (125 kN)( x2 − 4 m) 4 m  x2  6 m Differentiate this expression to obtain ∂M/∂P. M = 0.5 x2 P Substitute P = 0 into the bending-moment equation to obtain M = (125 kN) x2 − (125 kN)( x2 − 4 m) Castigliano’s second theorem applied to beam deflections can be expressed by Equation (17.40). L  M  M =  dx  0  P  EI

An expression for the beam deflection at midspan can now be developed. Owing to the symmetry of the beam, expressions will be written for one-half of the span and then doubled.

2 4 2 6 (0.5 x1 ) (125 x1 ) dx1 + (0.5 x2 ) 125 x2 − 125( x2 − 4)  dx2  EI 0 EI 4 125 4 2 500 6 = x dx + x2 dx2 1 1 EI 0 EI 4 125 3 4 500 2 6  x1  +  x2  = 3EI   0 2 EI   4 2,666.6667 5,000 = + EI EI 7,666.6667 kN-m 3 = EI The maximum beam deflection must be limited to 35 mm. 7, 666.6667 kN-m 3  midspan =  35 mm = 0.035 m EI Solving for the minimum moment of inertia gives (7,666.6667 kN-m3 )(1,000 N/kN) I = 1.095238  10−3 m4 9 2 (200  10 N/m )(0.035 m) or in terms of mm4, the minimum moment of inertia is 4 −3 4  1, 000 mm  I  (1.095238  10 m )  = 1.095  109 mm 4   1m   midspan =

Ans.

P17.75 In Figure P17.75, if the maximum beam deflection must not exceed 0.5 in., what is the minimum moment of inertia I required for the beam? Utilize Castigliano’s second theorem, and assume that E = 29,000 ksi. FIGURE P17.75

Solution To determine the maximum deflection of the cantilever beam, first apply a dummy concentrated load P downward at B. A free-body diagram of the beam is shown.

Draw a free-body diagram that cuts through the beam at section a–a between A and B and derive a moment equation M from this free-body diagram.  1.5 kips/ft  2 M a − a = −   x − (75 kip-ft) − Px − M = 0 2    1.5 kips/ft  2 M = − 0  x  15 ft  x − (75 kip-ft) − Px 2   Differentiate this expression to obtain ∂M/∂P. M = −x P Substitute P = 0 into the bending-moment equation to obtain  1.5 kips/ft  2 M = −  x − (75 kip-ft) 2   Castigliano’s second theorem applied to beam deflections can be expressed by Equation (17.40). L  M  M =  dx  0  P  EI

The beam deflection at A can now be determined. 1 15   1.5  2  0.75 15 3 75 15 B = ( − x ) − x − 75 dx = x dx + xdx     2  EI 0 EI 0 EI 0   0.75 4 15 75 2 15 9, 492.1875 8, 437.5 x  + x  = + 4 EI   0 2 EI   0 EI EI 3 3 (17,929.6875 kip-ft )(12 in./ft) 30,982,500 kip-in.3 = = EI EI The maximum beam deflection must be limited to 0.5 in. 30,982,500 kip-in.3 B =  0.5 in. EI Solving for the minimum moment of inertia gives 30,982,500 kip-in.3 I = 2,136.7241 in.4 = 2,140 in.4 (29,000 ksi)(0.5 in.) =

Ans.

P14.1 Determine the normal stress in a ball (Figure P14.1), which has an outside diameter of 185 mm and a wall thickness of 3 mm, when the ball is inflated to a gage pressure of 80 kPa.

FIGURE P14.1

Solution D = 185 mm t = 3 mm d = 185 mm − 2(3 mm) = 179 mm

t =

pd (0.110 MPa)(179 mm) = = 1.193 MPa 4t 4(3 mm)

Ans.

P14.2 A spherical gas-storage tank with an inside diameter of 21 ft is being constructed to store gas under an internal pressure of 160 psi. The tank will be constructed from steel that has a yield strength of 50 ksi. If a factor of safety of 3.0 with respect to the yield strength is required, determine the minimum wall thickness required for the spherical tank.

Solution  allow =

Y

FS pd t  4t

50 ksi = 16.667 ksi 3.0 pd (160 psi)(21 ft)(12 in./ft) t  = = 0.605 in. 4 allow 4(16,667 psi)

Ans.

P14.3 A spherical pressure vessel has an inside diameter of 3.25 m and a wall thickness of 8 mm. The tank will be constructed from structural steel that has a yield strength of 370 MPa. If a factor of safety of 3.0 with respect to the yield strength is required, determine the maximum allowable internal pressure.

Solution  allow =

Y

FS pd t  4t

370 MPa = 123.333 MPa 3.0 4 t 4 (123.333 MPa )(8 mm )  pmax  allow = = 1.214 MPa d ( 3.25 m )(1, 000 mm/m )

Ans.

P14.4 A spherical pressure vessel has an inside diameter of 6 m and a wall thickness of 15 mm. The vessel will be constructed from steel [E = 200 GPa;  = 0.29] that has a yield strength of 340 MPa. If the internal pressure in the vessel is 1,750 kPa, determine (a) the normal stress in the vessel wall, (b) the factor of safety with respect to the yield strength, (c) the normal strain in the sphere and (d) the increase in the outside diameter of the vessel.

Solution (a) Normal stress in the vessel wall pd (1.750 MPa)(6,000 mm) t = = = 175.0 MPa 4t 4(15 mm)

Ans.

(b) Factor of safety with respect to the yield strength  340 MPa FS = Y = = 1.943  t 175 MPa

Ans.

(c) Normal strain in the sphere 1  x = ( x −  y ) E 1 = [175 MPa − (0.29)(175 MPa)] = 621.25  10−6 mm/mm = 621 με 200,000 MPa (d) Increase in outside diameter D =  D = (621.25  10−6 mm/mm) 6,000 mm + 2(15 mm) = 3.75 mm

Ans.

P14.5 The normal strain measured on the outside surface of a spherical pressure vessel is 670 . The sphere has an outside diameter of 1.20 m and a wall thickness of 10 mm, and it will be fabricated from an aluminum alloy [E = 73 GPa;  = 0.33]. Determine (a) the normal stress in the vessel wall and (b) the internal pressure in the vessel.

Solution (a) Normal stress in the vessel wall E x = ( x + y ) 1 − 2 73, 000 MPa  670 10−6 mm/mm ) + ( 0.33) ( 670 10−6 mm/mm )  = 2 (  1 − (0.33) = 73.000 MPa

 t = 73.0 MPa

Ans.

(b) Internal pressure D = 1, 200 mm t = 10 mm d = 1, 200 mm − 2(10 mm) = 1,180 mm

t =

pd 4t

p=

4 t t 4 ( 73.000 MPa )(10 mm ) = = 2.4746 MPa = 2.47 MPa d 1,180 mm

Ans.

P14.6 A typical aluminum-alloy scuba diving tank is shown in Figure P14.6. The outside diameter of the tank is 7.20 in. and the wall thickness is 0.55 in. If the air in the tank is pressurized to 3,500 psi, determine: (a) the longitudinal and hoop stresses in the wall of the tank. (b) the maximum shear stress in the plane of the cylinder wall. (c) the absolute maximum shear stress on the outer surface of the cylinder wall. FIGURE P14.6

Solution (a) Longitudinal and hoop stresses D = 7.20 in. t = 0.55 in. d = 7.20 in. − 2(0.55 in.) = 6.10 in. pd (3,500 psi)(6.10 in.)  long = = = 9,704.545 psi = 9,700 psi 4t 4(0.55 in.) pd (3,500 psi)(6.10 in.)  hoop = = = 19, 409.091 psi = 19, 410 psi 2t 2(0.55 in.)

Ans. Ans.

(b) Maximum shear stress in the plane of the cylinder wall If the longitudinal axis of the cylinder is defined as the x axis and the circumferential direction is defined as the y axis, the normal and shear stresses on longitudinal and circumferential faces of a stress element are:  x = 9.704.545 psi,  y = 19, 409.091 psi,  xy = 0 psi Since the shear stresses on the x and y faces (i.e., in the plane of the cylinder wall) are zero, these normal stresses are, by definition, principal stresses. Therefore,  p1 = 19, 410 psi and  p 2 = 9,700 psi The maximum in-plane shear stress can be computed from Eq. (12.16):  −  p 2 19, 409.091 psi − 9, 704.545 psi  max = p1 = = 4,852.273 psi = 4,850 psi 2 2

Ans.

(c) Absolute maximum shear stress on the outer surface of the cylinder wall The outer surface of the cylinder wall is in a state of plane stress since the pressure acting on the outer surface of the cylinder is simply atmospheric pressure (i.e., gage pressure = 0). Therefore, z = p3 = 0. Since p1 and p2 are both positive,  19, 409.091 psi  abs max = p1 = = 9,700 psi Ans. 2 2

P14.7 A cylindrical boiler with an outside diameter of 2.75 m and a wall thickness of 32 mm is made of a steel alloy that has a yield stress of 340 MPa. Determine: (a) the maximum normal stress produced by an internal pressure of 2.3 MPa. (b) the maximum allowable pressure if a factor of safety of 2.5 with respect to yield is required.

Solution (a) Maximum normal stress D = 2,750 mm t = 32 mm pd (2.30 MPa)(2,686 mm)  hoop = = = 96.5 MPa 2t 2(32 mm)

d = 2,750 mm − 2(32 mm) = 2,686 mm

(b) Maximum allowable pressure if FS = 2.5  340 MPa  allow = Y = = 136.0 MPa FS 2.5 pd 2 t 2(136.0 MPa)(32 mm)  allow   hoop =  p  allow = = 3.24 MPa 2t d 2,686 mm

Ans.

P14.8 When filled to capacity, the unpressurized storage tank shown in Figure P14.8 contains water to a height of h = 30 ft. The outside diameter of the tank is 12 ft and the wall thickness is 0.375 in. Determine the maximum normal stress and the absolute maximum shear stress on the outer surface of the tank at its base. (Weight density of water = 62.4 lb/ft3.)

FIGURE P14.8

Solution Water pressure p =  h = (62.4 lb/ft 3 )(30 ft) = 1,872.0 lb/ft 2 = 13.00 psi Hoop stress D = (12 ft)(12 in./ft) = 144 in. t = 0.375 in. d = 144 in. − 2(0.375 in.) = 143.25 in. pd (13.00 psi)(143.25 in.) Ans.  hoop = = = 2,482.992 psi = 2,480 psi 2t 2(0.375 in.) Principal stresses  p1 =  hoop = 2,482.992 psi

 p2 =  long = 0 psi

(since the tank is unpressurized)

Maximum shear stress The outer surface of the tank wall is in a state of plane stress since the pressure acting on the outer surface is simply atmospheric pressure (i.e., gage pressure = 0). Therefore, z = p3 = 0. Therefore,  2,482.992 psi  abs max = p1 = = 1, 241 psi Ans. 2 2

P14.9 A tall open-topped standpipe (Figure P14.9) has an inside diameter of 2,750 mm and a wall thickness of 6 mm. The standpipe contains water, which has a mass density of 1,000 kg/m3. (a) What height h of water will produce a circumferential stress of 16 MPa in the wall of the standpipe? (b) What is the axial stress in the wall of the standpipe due to the water pressure?

FIGURE P14.9

Solution Longitudinal and hoop stresses pd p(2,750 mm)  hoop = = = 16 MPa 2t 2(6 mm)

 p = 69.818  10−3 MPa (a) Height h of water p =  gh = 69.818  10 −3 MPa h =

69.818  103 N/m 2 = 7.122684 m = 7.12 m (1,000 kg/m3 )(9.81 m/s 2 )

Ans.

(b) Axial stress in the wall of the standpipe due to water pressure Since the standpipe is open to the atmosphere at its upper end, the fluid pressure will not create stress in the longitudinal direction of the standpipe; therefore, Ans.  long = 0

P14.10 The pressure tank in Figure P14.10/11 is fabricated from spirally wrapped metal plates that are welded at the seams in the orientation shown where  = 50°. The tank has an inside diameter of 28.0 in. and a wall thickness of 0.375 in. Determine the largest gage pressure that can be used inside the tank if the allowable normal stress perpendicular to the weld is 18 ksi and the allowable shear stress parallel to the weld is 10 ksi. FIGURE P14.10/11

Solution The longitudinal axis of the cylinder is defined as the x axis and the circumferential direction is defined as the y axis; therefore, the normal and shear stresses on longitudinal and circumferential faces of a stress element are: pd pd x = , y = ,  xy = 0 4t 2t The weld is oriented at 50° as shown; however, the angle  required for the stress transformation equations is the angle normal to the weld. Thus,  = 50° + 90° = 140° (or  = 50° − 90° = −40°). Using this value of , the normal stress transformation equation [Eq. (12.3)] can be used to compute the normal stress perpendicular to the weld:  n =  x cos 2  +  y sin 2  + 2 xy sin  cos  =

pd pd cos 2 (140) + sin 2 (140) + 2(0 ksi) sin(140) cos(140) 4t 2t

pd pd cos 2 (140) + sin 2 (140) 4t 2t The normal stress magnitude perpendicular to the weld n must not exceed 18.0 ksi; thus, pd pd 18, 000 psi  cos 2 (140) + sin 2 (140) 4t 2t =



pd  1  cos 2 (140) + sin 2 (140)   2t  2 



p(28 in.)  0.58682  + 0.41318  2(0.375 in.)  2 

 (26.37928) p Based on the allowable normal stress, p  682.35 psi

(a)

Similarly, the shear stress transformation equation [Eq. (12.4)] can be used to compute the shear stress parallel to the weld:  nt = −( x −  y ) sin  cos  +  xy (cos 2  − sin 2  )

 pd pd  = − − sin(140) cos(140) + (0 ksi)[cos 2 (140) − sin 2 (140)]  2t   4t  pd pd  = − − sin(140) cos(140) 2t   4t Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

The shear stress parallel to the weld nt must not exceed a magnitude of 10.0 ksi; thus,  pd pd  10, 000 psi  −  − sin(140) cos(140) 2t   4t −

pd  1   − 1 sin(140) cos(140) 2t  2 

−

p(28.0 in.)  1  − (−0.49240)   2(0.375 in.)  2 

 9.19154 p Based on the allowable shear stress, p  1, 097.96 psi

Compare the results in Eqs. (a) and (b) to find that the maximum allowable gage pressure is pallow = 682 psi

(b)

Ans.

P14.11 The pressure tank in Figure P14.10/11 is fabricated from spirally wrapped metal plates that are welded at the seams in the orientation shown where  = 40°. The tank has an inside diameter of 720 mm and a wall thickness of 8 mm. For a gage pressure of 2.15 MPa, determine (a) the normal stress perpendicular to the weld and (b) the shear stress parallel to the weld. FIGURE P14.10/11

Solution (a) Normal stress perpendicular to the weld pd (2.15 MPa)(720 mm)  long = = = 48.375 MPa 4t 4(8 mm) pd (2.15 MPa)(720 mm)  hoop = = = 96.750 MPa 2t 2(8 mm) The longitudinal axis of the cylinder is defined as the x axis and the circumferential direction is defined as the y axis; therefore, the normal and shear stresses on longitudinal and circumferential faces of a stress element are:  x = 48.375 MPa,  y = 96.750 MPa,  xy = 0 MPa The weld is oriented at 40° as shown; however, the angle  required for the stress transformation equations is the angle normal to the weld. Thus,  = 40° + 90° = 130° (or  = 40° − 90° = −50°). Using this value of , the normal stress transformation equation [Eq. (12.3)] can be used to compute the normal stress perpendicular to the weld:  n =  x cos 2  +  y sin 2  + 2 xy sin  cos  = (48.375 MPa) cos 2 (130) + (96.750 MPa)sin 2 (130) + 2(0 MPa)sin(130) cos(130) = 76.763 MPa = 76.8 MPa (T)

Ans.

(b) Shear stress parallel to the weld Similarly, the shear stress transformation equation [Eq. (12.4)] gives nt:  nt = −( x −  y )sin  cos  +  xy (cos 2  − sin 2  )

= −[(48.375 MPa) − (96.750 MPa)]sin(130)cos(130) + (0 MPa)[cos 2 (130) − sin 2 (130)] = −23.820 MPa = −23.8 MPa

Ans.

P14.12 The pressure tank in Figure P14.12/13 is fabricated from spirally wrapped metal plates that are welded at the seams in the orientation shown where  = 35°. The tank has an outside diameter of 30.0 in. and a wall thickness of 7/16 in. For a gage pressure of 500 psi, determine (a) the normal stress perpendicular to the weld and (b) the shear stress parallel to the weld.

FIGURE P14.12/13

Solution (a) Normal stress perpendicular to the weld pd (500 psi)(29.125 in.)  long = = = 8,321.429 psi 4t 4(0.4375 in.) pd (500 psi)(29.125 in.)  hoop = = = 16,642.857 psi 2t 2(0.4375 in.) The longitudinal axis of the cylinder is defined as the y axis and the circumferential direction is defined as the x axis; therefore, the normal and shear stresses on longitudinal and circumferential faces of a stress element are:  x = 16,642.857 psi,  y = 8,321.429 psi,  xy = 0 psi The weld is oriented at 35° as shown. Relative to the positive x axis, this orientation is defined by an angle of  = 180° − 35° = 145°. Using this value of , the normal stress transformation equation [Eq. (12.3)] can be used to compute the normal stress perpendicular to the weld:  n =  x cos 2  +  y sin 2  + 2 xy sin  cos  = (16, 642.857 psi) cos 2 (145) + (8,321.429 psi) sin 2 (145) + 2(0 psi) sin(145) cos(145) = 13, 905.191 psi = 13,910 psi (T)

Ans.

(b) Shear stress parallel to the weld Similarly, the shear stress transformation equation [Eq. (12.4)] gives nt:  nt = −( x −  y ) sin  cos  +  xy (cos 2  − sin 2  ) = −[(16, 642.857 psi) − (8,321.429 psi)]sin(145) cos(145) + (0 psi)[cos 2 (145) − sin 2 (145)] = 3,909.793 psi = 3,910 psi

Ans.

P14.13 The pressure tank in Figure P14.12/13 is fabricated from spirally wrapped metal plates that are welded at the seams in the orientation shown where  = 55°. The tank has an inside diameter of 60 in. and a wall thickness of 0.25 in. Determine the largest allowable gage pressure if the allowable normal stresses perpendicular to the weld is 12 ksi and the allowable shear stress parallel to the weld is 7 ksi. FIGURE P14.12/13

Solution The longitudinal axis of the cylinder is defined as the y axis and the circumferential direction is defined as the x axis; therefore, the normal and shear stresses on longitudinal and circumferential faces of a stress element are: pd pd x = , y = ,  xy = 0 2t 4t The weld is oriented at 55° as shown. Relative to the positive x axis, this orientation is defined by an angle of  = 180° − 55° = 125°. Using this value of , the normal stress transformation equation [Eq. (12.3)] can be used to compute the normal stress perpendicular to the weld:  n =  x cos 2  +  y sin 2  + 2 xy sin  cos  pd pd cos 2 (125) + sin 2 (125) + 2(0 ksi)sin(125)cos(125) 2t 4t pd pd = cos 2 (125) + sin 2 (125) 2t 4t

The normal stress magnitude perpendicular to the weld n must not exceed 12 ksi; thus, pd pd 2 12 ksi  cos 2 (125) + sin (125) 2t 4t



pd  2 1 2  cos (125  ) + sin (125)   2t  2 



p(60 in.)  0.671010  0.328990 +   2(0.25 in.)  2

 (79.739396) p Based on the allowable normal stress, p  0.150490 ksi

(a)

Similarly, the shear stress transformation equation [Eq. (12.4)] can be used to compute the shear stress parallel to the weld:  nt = −( x −  y )sin  cos  +  xy (cos 2  − sin 2  )  pd pd  2 2 = − −  sin(125)cos(125) + (0 ksi)[cos (125) − sin (125)] 2 t 4 t    pd pd  = − −  sin(125)cos(125) 4t   2t Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

The shear stress parallel to the weld nt must not exceed a magnitude of 7 ksi; thus,  pd pd  7 ksi  −  − sin(125)cos(125) 4t   2t  −

pd  1  1 −  sin(125)cos(125) 2t  2 

 −

p(60 in.)  1  ( −0.469846)   2(0.25 in.)  2 

 28.190779p Based on the allowable shear stress, p  0.248308 ksi

Compare the results in Eqs. (a) and (b) to find that the maximum allowable gage pressure is pallow = 150.5 psi

(b)

Ans.

P14.14 A strain gage is mounted to the outer surface of a thin-walled boiler as shown in Figure P14.14. The boiler has an inside diameter of 1,800 mm and a wall thickness of 20 mm, and it is made of stainless steel [E = 193 GPa;  = 0.27]. Determine: (a) the internal pressure in the boiler when the strain gage reads 190 . (b) the maximum shear strain in the plane of the boiler wall. (c) the absolute maximum shear strain on the outer surface of the boiler.

FIGURE P14.14

Solution (a) Internal pressure in the boiler The longitudinal axis of the cylinder is defined as the x axis and the circumferential direction is defined as the y axis; therefore, the normal and shear stresses on longitudinal and circumferential faces of a stress element are: pd pd x = , y = ,  xy = 0 4t 2t From the generalized Hooke’s Law equations for plane stress, the normal strains on the outer surface of the boiler can be computed from Eqs. (13.21):  pd   1 1  pd  x = ( x −  y ) =  −   E E  4t  2t   Thus, d  d  E x = p  −      2t    4t E x (193,000 MPa)(190  10 −6 mm/mm) p= = = 3.543 MPa = 3.54 MPa d 1 1,800 mm  1   −  − 0.27 2t  2 2(20 mm)  2  

Ans.

(b) Maximum shear strain in the plane of the boiler wall The strain in the longitudinal direction is given as x = 190×10−6 mm/mm. The strain in the circumferential direction (i.e., the y direction) can be expressed with the generalized Hooke’s Law equations as:  pd   pd    1 1  pd  y = ( y −  x ) =  −   = 1 −  E E  2t  4t   2tE  2  From the pressure computed in part (a), the strain in the y direction is: (3.543 MPa)(1,800 mm)  0.27  y = 1− = 714.565  10−6 mm/mm   2(20 mm)(193,000 MPa)  2  Since the longitudinal and hoop stresses are principal stresses, the corresponding strains are also principal strains.  p1 =  y = 714.565  10−6 mm/mm

 p 2 =  x = 190  10−6 mm/mm Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

The maximum shear strain in the plane of the boiler wall can be calculated from Eq. (13.12):  max =  p1 −  p 2 = 714.565  10−6 mm/mm − 190  10−6 mm/mm

= 524.565  10−6 rad = 525 μrad

Ans.

(c) Absolute maximum shear strain on the outer surface of the boiler The strain in the radial direction (i.e., the out-of-plane direction) can be expressed with the generalized Hooke’s Law equations as:    pd pd   pd  1   z = − ( x +  y ) = −  + +1 =− E E  4t 2t  2tE  2  From the pressure computed in part (a), the strain in the z direction is: (0.27)(3.543 MPa)(1,800 mm)  1  z = − + 1 = −334.565  10−6 mm/mm  2(20 mm)(193,000 MPa)  2  The strain in the z direction is also a principal strain; therefore,  p3 =  z = −334.565  10−6 mm/mm Since p1 is positive and p3 is negative, the absolute maximum shear strain is  abs max = 714.565  10−6 mm/mm − ( −334.565  10−6 mm/mm)

= 1,049.130  10−6 rad = 1,049 μrad

Ans.

P14.15 A closed cylindrical tank containing a pressurized fluid has an inside diameter of 830 mm and a wall thickness of 10 mm. The stresses in the wall of the tank acting on a rotated element have the values shown in Figure P14.15. What is the fluid pressure in the tank? FIGURE P14.15

Solution Let the given stresses be designated as:  x = 51 MPa,  y = 66 MPa,  xy = 18 MPa The principal stress magnitudes can be computed from Eq. (12.12):

 p1, p 2 =

x + y 2

x −y    +  xy2  2   2

(51 MPa) + (66 MPa)  (51 MPa) − (66 MPa)  2 =    + (18 MPa)  2 2 2

= 58.50 MPa  19.50 MPa  p1 = 78.0 MPa and  p 2 = 39.0 MPa

Since this is a cylindrical pressure vessel subjected to internal pressure only, we know that the principal stresses occur in the hoop and longitudinal directions. Thus, we can assert that: pd pd  p1 =  hoop = and  p 2 =  long = 2t 4t The internal pressure can be calculated from either expression: pd = 78 MPa 2t 2(10 mm)(78 MPa) Ans. p= = 1.880 MPa 830 mm

P14.16 A closed cylindrical vessel (Figure P14.16) contains a fluid at a pressure of 5.0 MPa. The cylinder, which has an outside diameter of 2,500 mm and a wall thickness of 20 mm, is fabricated from stainless steel [E = 193 GPa;  = 0.27]. Determine the increase in both the diameter and the length of the cylinder.

FIGURE P14.16

Solution D = 2,500 mm t = 20 mm d = 2,500 mm − 2(20 mm) = 2,460 mm pd (5 MPa)(2,460 mm)  long = = = 153.750 MPa 4t 4(20 mm) pd (5 MPa)(2,460 mm)  hoop = = = 307.500 MPa 2t 2(20 mm) From the generalized Hooke’s law, the strain in the longitudinal direction is: 1  x = ( x −  y ) E 1 = ( long −  hoop ) E 1 = 153.750 MPa − (0.27)(307.500 MPa)  193,000 MPa = 366.451  10 −6 mm/mm

Therefore, the change in length of the cylinder is: L =  x L = (366.451  10−6 mm/mm)(6,000 mm) = 2.20 mm

Ans.

Similarly, the strain in the circumferential direction is: 1  y = ( y −  x ) E 1 = ( hoop −  long ) E 1 =  307.500 MPa − (0.27)(153.750 MPa) 193,000 MPa

= 1,378.174  10 −6 mm/mm The change in diameter of the cylinder is: D =  y D = (1,378.174  10−6 mm/mm)(2,500 mm) = 3.45 mm

Ans.

P14.17 A strain gage is mounted at an angle of  = 25° with respect to the longitudinal axis of the cylindrical pressure vessel shown in Figure P14.17/18. The pressure vessel is fabricated from aluminum [E = 73 GPa;  = 0.33], and it has an inside diameter of 1.6 m and a wall thickness of 8 mm. If the strain gage measures a normal strain of 885 , determine: (a) the internal pressure in the cylinder. (b) the absolute maximum shear stress on the outer surface of the cylinder. (c) the absolute maximum shear stress on the inner surface of the cylinder.

FIGURE P14.17/18

Solution (a) Internal pressure in the cylinder A strain transformation equation [Eq. (13.3)]  n =  x cos2  +  y sin 2  +  xy sin  cos can be written for the normal strain in the direction of the strain gage: 885 με =  x cos2 (25) +  y sin 2 (25) +  xy sin(25)cos(25) Since this is a cylindrical pressure vessel, the shear stress xy must equal zero, and hence, the shear strain xy must also equal zero. The strain transformation equation reduces to: 885 με = 885 10−6 mm/mm =  x cos2 (25) +  y sin 2 (25) Substitute Eqs. (13.21) for x and y to obtain an expression in terms of x and y: 885 10−6 =  x cos 2 (25) +  y sin 2 (25) 1 1 ( x − y ) cos 2 (25) + ( y − x ) sin 2 (25) E E 1 1 = [ x cos 2 (25) − x sin 2 (25)] + [ y sin 2 (25) − y cos 2 (25)] E E

x

[cos 2 (25) − sin 2 (25)] +

y

[sin 2 (25) − cos 2 (25)]

E E The normal stress x is the longitudinal stress caused by the internal pressure, and y is the hoop stress. Substitute expressions for long and hoop to obtain:

885 10−6 =

 long

[cos 2 (25) − sin 2 (25)] +

 hoop

[sin 2 (25) − cos 2 (25)]

E E pd pd = [cos 2 (25) − sin 2 (25)] + [sin 2 (25) − cos 2 (25)] 4tE 2tE

pd cos 2 (25) − sin 2 (25) + 2sin 2 (25) − 2 cos 2 (25)  4tE Thus, the pressure p can be expressed as: 4tE (885 10−6 ) p= d cos2 (25) − sin 2 (25) + 2sin 2 (25) − 2 cos 2 (25)  =

Compute the internal pressure: 4(8 mm)(73,000 MPa)(885  10−6 mm/mm) p= (1,600 mm) cos 2 (25) − (0.33)sin 2 (25) + 2sin 2 (25) − 2(0.33)cos 2 (25)  =

4(8 mm)(73,000 MPa)(885  10−6 mm/mm) (1,600 mm)(0.577546)

= 2.237 MPa = 2.24 MPa

Ans.

(b) Absolute maximum shear stress on the outer surface of the cylinder The principal stresses are: pd (2.2372 MPa)(1,600 mm)  long = = = 111.861 MPa =  p 2 4t 4(8 mm) pd (2.2372 MPa)(1,600 mm)  hoop = = = 223.722 MPa =  p1 2t 2(8 mm) The outer surface of the cylinder is in plane stress; therefore, the absolute maximum shear stress is:  −  p 3 223.722 MPa − 0 MPa  abs max = p1 = = 111.9 MPa Ans. 2 2 (c) Absolute maximum shear stress on the inner surface of the cylinder Inside the cylinder, the pressure creates a stress in the radial direction; therefore,  p3 =  radial = − p = −2.237 MPa The absolute maximum shear stress inside the cylinder is  −  p 3 223.722 MPa − (−2.237 MPa)  abs max = p1 = = 113.0 MPa 2 2

Ans.

P14.18 A strain gage is mounted at an angle of  = 30° with respect to the longitudinal axis of the cylindrical pressure shown in Figure P14.17/18. The pressure vessel is fabricated from steel [E = 30,000 ksi;  = 0.30], and it has an outside diameter of 37 in. and a wall thickness of 3/16 in. If the internal pressure in the cylinder is 200 psi, determine: (a) the expected strain gage reading (in ). (b) the principal strains, the maximum shear strain, and the absolute maximum shear strain on the outer surface of the cylinder. FIGURE P14.17/18

Solution (a) Expected strain gage reading A strain transformation equation [Eq. (13.3)]  n =  x cos2  +  y sin 2  +  xy sin  cos can be written for the normal strain in the direction of the strain gage:  n =  x cos2 (30) +  y sin 2 (30) +  xy sin(30)cos(30) Since this is a cylindrical pressure vessel, the shear stress xy must equal zero, and hence, the shear strain xy must also equal zero. The strain transformation equation reduces to:  n =  x cos2 (30) +  y sin 2 (30) Substitute Eqs. (13.21) for x and y to obtain an expression in terms of x and y:  n =  x cos 2 (30) +  y sin 2 (30) 1 1 ( x −  y )cos 2 (30) + ( y −  x )sin 2 (30) E E 1 1 = [ x cos 2 (30) −  x sin 2 (30)] + [ y sin 2 (30) −  y cos 2 (30)] E E

x

[cos 2 (30) −  sin 2 (30)] +

y

[sin 2 (30) −  cos 2 (30)]

E E The normal stress x is the longitudinal stress caused by the internal pressure, and y is the hoop stress. Substitute expressions for long and hoop to obtain:

n =

 long

[cos 2 (30) −  sin 2 (30)] +

 hoop

[sin 2 (30) −  cos 2 (30)]

E E pd pd = [cos 2 (30) −  sin 2 (30)] + [sin 2 (30) −  cos 2 (30)] 4tE 2tE =

pd cos 2 (30) −  sin 2 (30) + 2sin 2 (30) − 2 cos 2 (30)  4tE

The expected strain gage reading is thus: (200 psi)(36.625 in.) cos2 (30) − (0.30)sin 2 (30) + 2sin 2 (30) − 2(0.30)cos 2 (30)  n = 4(0.188 in.)(30,000,000 psi) 

= 236.028  10−6 in./in. = 236 με

Ans.

(b) Principal strains on outer surface of cylinder The principal stresses are: pd (200 psi)(36.625 in.)  long = = = 9,766.667 psi =  p 2 4t 4(0.188 in.) pd (200 psi)(36.625 in.)  hoop = = = 19,533.333 psi =  p1 2t 2(0.188 in.) From the generalized Hooke’s Law equations for plane stress, the normal strains produced in the plate can be computed from Eqs. (13.21): 1 1  x = ( long −  hoop ) = [9,766.667 psi − (0.30)(19,533.333 psi)] = 130.222  10−6 in./in. E 30  106 psi 1 1  y = ( hoop −  long ) = [19,533.333 psi − (0.30)(9,766.667 psi)] = 553.444  10−6 in./in. E 30  106 psi  0.30  z = − ( long +  hoop ) = − [9,766.667 psi + 19,533.333 psi] = −293.000  10−6 in./in. 6 E 30  10 psi Therefore: Ans.  p1 = 553 με  p 2 = 130.2 με  p3 = −293 με Maximum shear strain  max =  p1 −  p 2 = 553.444  10−6 − 130.222.600  10−6 = 423 μrad

Ans.

Absolute maximum shear strain  abs max =  p1 −  p3 = 553.444  10−6 − (−293.000  10−6 ) = 846 μrad

Ans.

P14.19 The pressure vessel in Figure P14.19 consists of spirally wrapped steel plates that are welded at the seams in the orientation shown where  = 35°. The cylinder has an inside diameter of 16 in. and a wall thickness of 0.375 in. The ends of the cylinder are capped by two rigid end plates. The gage pressure inside the cylinder is 260 psi, and tensile axial loads of P = 15 kips are applied to the rigid end caps. Determine: (a) the normal stress perpendicular to the weld seams. (b) the shear stress parallel to the weld seams. (c) the absolute maximum shear stress in the cylinder. FIGURE P14.19

Solution (a) Normal stress perpendicular to the weld pd (260 psi)(16 in.)  long = = = 2,773.333 psi 4t 4(0.375 in.) pd (260 psi)(16 in.)  hoop = = = 5,546.667 psi 2t 2(0.375 in.) The compressive axial load also creates a normal stress in the x direction.



 D 2 − d 2  = (16.75 in.)2 − (16 in.)2  = 19.291 in.2 4 4

 axial =

P −15,000 lb = = −777.550 psi A 19.291 in.2

The longitudinal axis of the cylinder is defined as the x axis and the circumferential direction is defined as the y axis; therefore, the normal and shear stresses on longitudinal and circumferential faces of a stress element are:  x = 3,550.884 psi,  y = 5,546.667 psi,  xy = 0 psi The weld is oriented at 35° as shown. The angle  required for the stress transformation equations is the angle normal to the weld, which is also 35°. Using this value of , the normal stress transformation equation [Eq. (12.3)] can be used to compute the normal stress perpendicular to the weld:  n =  x cos 2  +  y sin 2  + 2 xy sin  cos = (3,550.884 psi)cos 2 (35) + (5,546.667 psi)sin 2 (35) + 2(0 psi)sin(35)cos(35) = 4, 207.476 psi = 4, 210 psi (T)

Ans.

(b) Shear stress parallel to the weld Similarly, the shear stress transformation equation [Eq. (12.4)] gives nt:  nt = −( x −  y )sin  cos +  xy (cos 2  − sin 2  ) = −[(3,550.884 psi) − (5,546.667 psi)]sin(35)cos(35) + (0 psi)[cos 2 (35) − sin 2 (35)] = 937.711 psi = 938 psi

Ans.

(c) Absolute maximum shear stress in the cylinder The principal stresses on the outside of the cylinder are:  p1 =  hoop = 5,546.667 psi  p 2 =  long +  axial = 3,550.884 psi

 p3 = 0

On the outside surface of the cylinder:  − 0 5,546.667 psi  abs max = p1 = = 2,773.333 psi 2 2 Inside the cylinder, the third principal stress is equal in magnitude to the internal pressure:  p3 =  radial = − p = −260 psi On the inside surface of the cylinder:  −  p 3 5,546.667 psi − (−260 psi)  abs max = p1 = = 2,903.333 psi 2 2 Thus, the absolute maximum shear stress in the cylinder is  abs max = 2,900 psi

Ans.

P14.20 The cylindrical pressure vessel shown in Figure P14.20/21 has an inside diameter of 610 mm and a wall thickness of 3 mm. The cylinder is made of an aluminum alloy that has an elastic modulus of E = 70 GPa and a shear modulus of G = 26.3 GPa. Two strain gages are mounted on the exterior surface of the cylinder at right angles to each other; however, the angle  is not known. If the strains measured by the two gages are a = 360  and b = 975 , what is the pressure in the vessel? Notice that when two orthogonal strains are measured, the angle  is not needed to determine the normal stresses. FIGURE P14.20/21

Solution Strain invariance: From strain invariance [Eq. (13.8)], we can state  a + b =  x +  y Let’s first focus on the right-hand side of this equation. From Eqs. (13.21), the sum of the strains in the x and y directions can be expressed as 1 1  x +  y = ( x −  y ) + ( y −  x ) E E 1 1 =  x −  y +  y −  x  =  x (1 −  ) +  y (1 −  )  E E 1−  x +  y  = E  The longitudinal axis of the cylinder is defined as the x axis and the circumferential direction is defined as the y axis; therefore: pd pd  x =  long = and  y =  hoop = 4t 2t Substitution of these expressions gives 1 −   pd pd  1 −  pd 3(1 −  ) pd x +  y = + 1 + 2 =   = E  4t 2t  E 4t E 4t Now, based on strain invariance, we can equate  x +  y and  a +  b

3(1 −  ) pd =  a + b E 4t Before calculating p, we need to derive an expression for  from Eq. (13.18): E E G=  = −1 2(1 +  ) 2G Determine Poisson’s ratio from this expression: E 70 GPa = −1 = − 1 = 0.331 2G 2(26.3 GPa) Having calculated the value of Poisson’s ratio, we can now calculate the internal pressure p: 4( a + b ) Et 4(360  10−6 + 975  10−6 )(70,000 MPa)(3 mm) p= = = 0.916 MPa 3(1 −  )d 3(1 − 0.331)(610 mm)

Ans.

P14.21 The cylindrical pressure vessel shown in Figure P14.20/21 has an inside diameter of 900 mm and a wall thickness of 12 mm. The cylinder is made of an aluminum alloy that has an elastic modulus of E = 70 GPa and a shear modulus of G = 26.3 GPa. Two strain gages are mounted on the exterior surface of the cylinder at right angles to each other. The angle  is 25°. If the pressure in the vessel is 1.75 MPa, determine (a) the strains that act in the x and y directions. (b) the strains expected in gages a and b. (c) the normal stresses n and t. (d) the shear stress nt. FIGURE P14.20/21

Solution (a) Strains in the x and y directions pd (1.75 MPa)(900 mm)  x =  long = = = 32.813 MPa 4t 4(12 mm) pd (1.75 MPa)(900 mm)  y =  hoop = = = 65.625 MPa 2t 2(12 mm) Before proceeding, we need to derive an expression for  from Eq. (13.18): E E G=  = −1 2(1 +  ) 2G Determine Poisson’s ratio from this expression: E 70 GPa = −1 = − 1 = 0.331 2G 2(26.3 GPa) From the generalized Hooke’s Law, the strain in the x direction is: 1  x = ( x −  y ) E 1 =  32.813 MPa − (0.331)(65.625 MPa) 70,000 MPa

= 158.626  10−6 mm/mm = 158.6 με

Ans.

Similarly, the strain in the y direction is: 1  y = ( y −  x ) E 1 =  65.625 MPa − (0.331)(32.813 MPa) 70,000 MPa

= 782.438  10−6 mm/mm = 782 με

Ans.

(b) Expected strain gage readings in gages a and b A strain transformation equation [Eq. (13.3)]  n =  x cos2  +  y sin 2  +  xy sin  cos can be written for the normal strain in the direction of strain gage a: a =  x cos2 (25) +  y sin 2 (25) +  xy sin(25)cos(25) Since this is a cylindrical pressure vessel, the shear stress xy must equal zero, and hence, the shear strain xy must also equal zero. The strain transformation equation reduces to:  a =  x cos2 (25) +  y sin 2 (25) The expected strain reading in gage a is thus:  a =  x cos2 (25) +  y sin 2 (25)

= (158.626 με)cos 2 (25) + (782.438 με)sin 2 (25) = 270 με

Ans.

Use a value of  = 25° + 90° = 115° to obtain the strain reading expected in gage b:  b =  x cos 2 (115) +  y sin 2 (115)

= (158.626 με)cos 2 (115) + (782.438 με)sin 2 (115) = 671 με

Ans.

(c) Normal stresses n and t. Use the normal stress transformation equation [Eq. (12.3)] to calculate n:  n =  x cos 2  +  y sin 2  + 2 xy sin  cos  = (32.813 MPa)cos 2 (25) + (65.625 MPa)sin 2 (25) + 2(0 MPa)sin(25)cos(25) = 38.673 MPa = 38.7 MPa (T)

Ans.

and t:  t =  x cos 2 ( + 90) +  y sin 2 ( + 90) + 2 xy sin( + 90)cos( + 90) = (32.813 MPa)cos 2 (115) + (65.625 MPa)sin 2 (115) + 2(0 MPa)sin(115)cos(115) = 59.764 MPa = 59.8 MPa (T)

Ans.

(d) Shear stress nt. The shear stress transformation equation [Eq. (12.4)] gives nt:  nt = −( x −  y )sin  cos  +  xy (cos 2  − sin 2  )

= −[(32.813 MPa) − (65.625 MPa)]sin(25)cos(25) + (0 MPa)[cos 2 (25) − sin 2 (25)] = 12.568 MPa = 12.57 MPa

Ans.

P14.22 A thick-walled cylindrical tank with closed ends has an inside radius of a = 250 mm and an outside radius of b = 290 mm is submerged to a depth of 40 m. Assume that the density of seawater is 1,100 kg/m3. The tank also has an internal pressure of pi = 2 MPa. Determine the maximum values of r, , and long as well as the maximum shear stress in the walls of the cylinder remote from its ends.

Solution The external pressure of the seawater at a depth of 40 m is: po =  gh = (1,100 kg/m3 )( 9.806650 m/s2 ) ( 40 m ) = 431,492.60 N/m2 = 0.4315 MPa The internal pressure in the tank is pi = 2 MPa Since the internal pressure is greater than the external pressure: The maximum radial stress occurs on the inner surface of the tank at r = a. From Eq. (14.17): ( r )max = − pi = −2 MPa

Ans.

The maximum circumferential stress occurs on the inner surface of the tank at r = a. From Eq. (14.18): b 2 + a 2 ) pi − 2b 2 po ( (  )max = b2 − a 2 ( 290 mm )2 + ( 250 mm )2  ( 2 MPa ) − 2 ( 290 mm )2 ( 0.4315 MPa )  = 2 2 ( 290 mm ) − ( 250 mm )

= 10.2140 MPa = 10.21 MPa

Ans.

The longitudinal stress away from the ends of the cylinder is given by Eq. (14.27): a 2 pi − b 2 po  long = b2 − a 2

( 250 mm ) ( 2 MPa ) − ( 290 mm ) ( 0.4315 MPa ) = 2 2 ( 290 mm ) − ( 250 mm ) 2

= 4.1070 MPa = 4.11 MPa The maximum shear stress can be found from Eq. (14.20): 1 1  max = (  −  r ) = 10.2140 MPa − ( −2 MPa )  = 6.1070 MPa = 6.11 MPa 2 2

Ans.

P14.23 An open-ended thick-walled cylindrical pressure vessel with an inside diameter of 250 mm and an outside diameter of 400 mm is subjected to an internal pressure of 75 MPa. Determine (a) The circumferential stress at a point on the inside surface of the cylinder. (b) The circumferential stress at a point on the outside surface of the cylinder. (c) The maximum shear stress in the cylinder.

Solution For this cylinder 250 mm a= = 125 mm 2

400 mm = 200 mm 2

(a) Circumferential stress at a point on the inside surface of the cylinder. From Eq. (14.24) and using r = a = 125 mm: a 2 p  b2    = 2 i 2 1 + 2  b −a  r 

(125 mm ) ( 75 MPa ) 1 + ( 200 mm )  = 171.1538 MPa = 171.2 MPa = 2 2 2 ( 200 mm ) − (125 mm )  (125 mm )  2

Ans.

(b) Circumferential stress at a point on the outside surface of the cylinder. From Eq. (14.24) and using r = b = 200 mm: a 2 pi  b 2    = 2 2 1 + 2  b −a  r 

(125 mm ) ( 75 MPa ) 1 + ( 200 mm )  = 96.1538 MPa = 96.2 MPa = 2 2 2 ( 200 mm ) − (125 mm )  ( 200 mm )  2

Ans.

(c) The maximum shear stress in the cylinder. The maximum shear stress occurs at r = a = 125 mm; thus, from Eq. (14.20) we find: a 2b 2 ( pi − po )  max = ( b2 − a 2 ) r 2

(125 mm ) ( 200 mm ) ( 75 MPa − 0 ) = ( 200 mm )2 − (125 mm )2  (125 mm )2   2

( 200 mm ) ( 75 MPa ) = 123.0769 MPa = 123.1 MPa = 2 2 ( 200 mm ) − (125 mm ) 2

Ans.

P14.24

A steel cylinder with an inside diameter of 8 in. and an outside diameter of 14 in. is subjected to an internal pressure of 25,000 psi. Determine (a) The maximum shear stress in the cylinder. (b) The radial and circumferential stresses at a point in the cylinder midway between the inside and outside surfaces.

Solution For this cylinder 8 in. a= = 4 in. 2

14 in. = 7 in. 2

(a) Maximum shear stress in the cylinder. The maximum shear stress is found at the inner surface where r = a = 4 in. From Eq. (14.20): a 2b 2 ( pi − po )  max = ( b2 − a 2 ) r 2

( 4 in.) ( 7 in.) ( 25,000 psi − 0 ) = 37,121.21 psi = 37,100 psi = ( 7 in.)2 − ( 4 in.)2  ( 4 in.)2   2

Ans.

(b) The radial and circumferential stresses at a point in the cylinder midway between the inside and outside surfaces. Midway between the inside and outside surfaces corresponds to a radius of 4 in. + 7 in. r= = 5.5 in. 2 From Eq. (14.23), the radial normal stress is: a 2 p  b2   r = 2 i 2 1 − 2  b −a  r 

( 4 in.) ( 25,000 psi ) 1 − ( 7 in.)  = −7,513.15 psi = −7,510 psi = 2 2 2 ( 7 in.) − ( 4 in.)  ( 5.5 in.)  2

Ans.

From Eq. (14.24), the circumferential normal stress is: a 2 p  b2    = 2 i 2 1 + 2  b −a  r 

( 4 in.) ( 25,000 psi ) 1 + ( 7 in.)  = 31,755.57 psi = 31,800 psi = 2 2 2 ( 7 in.) − ( 4 in.)  ( 5.5 in.)  2

Ans.

P14.25

A hydraulic cylinder with an inside diameter of 200 mm and an outside diameter of 450 mm is made of steel [E = 210 GPa;  = 0.30]. For an internal pressure of 125 MPa, determine (a) The maximum tensile stress in the cylinder. (b) The change in internal diameter of the cylinder.

Solution For the hydraulic cylinder 200 mm a= = 100 mm 2

450 mm = 225 mm 2

(a) Maximum tensile stress in the cylinder. For internal pressure only, the maximum tensile stress is the circumferential stress at r = a (See Figure 14.11). a 2 p  b2    = 2 i 2 1 + 2  b −a  r 

(100 mm ) (125 MPa ) 1 + ( 225 mm )  = 186.5385 MPa = 186.5 MPa = 2 2 2 ( 225 mm ) − (100 mm )  (100 mm )  2

Ans.

(b) The change in internal diameter of the cylinder. For internal pressure only, the radial displacement is calculated from Eq. (14.29) using r = a = 100 mm: a 2 pi (1 −  ) r 2 + (1 +  ) b 2  r = 2 2 (b − a ) rE

(100 mm ) (125 MPa ) (1 − 0.3)(100mm )2 + (1 + 0.3)( 225 mm )2  = 2 2  ( 225 mm ) − (100 mm )  (100 mm )( 210,000 MPa )    2

= 0.1067 mm

The change is diameter is d = 2 r = 2 ( 0.1067 mm) = 0.213 mm

Ans.

P14.26 A closed-ended cylindrical pressure vessel is to be designed to have an internal diameter of 125 mm and to be subjected to an internal gage pressure of 10 MPa. The maximum circumferential normal stress is not to exceed 100 MPa and the maximum shear stress is not to exceed 60 MPa. Determine the minimum external diameter that may be used for the vessel.

Solution For this cylinder 125 mm a= = 62.5 mm 2 For a cylinder subjected only to internal pressure, the maximum circumferential stress occurs at r = a (See Figure 14.11). Solve for the minimum value of b required to keep the circumferential stress less than 100 MPa. b 2 + a 2 pi   b2 − a 2

(

)

  ( b 2 − a 2 )  ( b 2 + a 2 ) pi

b 2  − a 2   b 2 pi + a 2 pi b 2  − b 2 pi  a 2  + a 2 pi b2  a 2

  + pi   − pi

ba

  + pi 100 MPa + 10 MPa 11 = ( 62.5 mm ) = ( 62.5 mm ) = 69.0963 mm   − pi 100 MPa − 10 MPa 9

The maximum shear stress also occurs at the inner surface where r = a. From Eq. (14.20), solve for the minimum value of b required to keep the maximum shear stress less than 60 MPa. b 2 pi  max  2 2 b −a

 max ( b 2 − a 2 )  b 2 pi

b 2 max − a 2 max  b 2 pi b 2 max − b 2 pi  a 2 max b 2 ( max − pi )  a 2 max b2  a 2 ba

 max

 max − pi  max

 max − pi

= ( 62.5 mm )

60 MPa 6 = ( 62.5 mm ) = 68.4653 mm 60 MPa − 10 MPa 5

Therefore, the minimum external radius that may be used for the vessel is

b  69.0963 mm

and accordingly, the minimum external diameter is D  138.1927 mm = 138.2 mm

Ans.

P14.27 A cylindrical tank having an external diameter of 16 in. is to resist an internal pressure of 2,000 psi without the circumferential stress exceeding 20,000 psi. Determine the minimum wall thickness required for the tank.

Solution For this tank 16 in. b= = 8 in. 2 For a cylinder subjected only to internal pressure, the maximum circumferential stress occurs at r = a (See Figure 14.11). Solve for the maximum value of a required to keep the circumferential stress less than 20,000 psi. b 2 + a 2 pi   b2 − a 2

(

)

  ( b 2 − a 2 )  ( b 2 + a 2 ) pi

b 2  − a 2   b 2 pi + a 2 pi b 2  − b 2 pi  a 2  + a 2 pi a 2  b2

  − pi   + pi

ab

  − pi 20,000 psi − 2,000 psi 18 = ( 8 in.) = ( 8 in.) = 7.2363 in.   + pi 20,000 psi + 2,000 psi 22

The minimum wall thickness is thus tmin  b − a = 8 in. − 7.2363 in. = 0.7637 in. = 0.764 in.

Ans.

PP14.28

A cylindrical steel tank having an internal diameter of 4 ft is submerged in seawater (weight density = 64 lb/ft3) to a depth of 10,000 ft. The internal gage pressure in the tank is 2,000 psi. (a) If the allowable stress of the steel is 36 ksi, what is the minimum allowable outside diameter of the tank? (b) What is the maximum tensile stress before the tank is immersed?

Solution For this tank 48 in. a= = 24 in. 2 (a) Minimum allowable outside diameter of the tank. The external pressure of the seawater at a depth of 10,000 ft is: 2

 1 ft  po =  wh = ( 64 lb/ft ) (10,000 ft )   = 4,444.44 psi  12 in.  The internal pressure in the tank is pi = 2,000 psi 3

Since the external pressure is greater than the internal pressure: The maximum radial stress occurs on the outer surface of the tank at r = b. From Eq. (14.19):

( r )max = − po = −4,444.44 psi

Since the largest radial stress is only 4,444 psi in magnitude, it is clear that radial stress will not control the design. The maximum circumferential stress occurs on the inner surface of the tank at r = a, but the circumferential stress will be compressive. Begin with Eq. (14.18) and set the circumferential stress equal to the allowable stress of the tank material and solve for the minimum outer radius b: b 2 + a 2 ) pi − 2b 2 po ( (  )max =  allow  b2 − a 2  allow ( b 2 − a 2 )  ( b 2 + a 2 ) pi − 2b 2 po

b 2 allow − a 2 allow  b 2 pi + a 2 pi − 2b 2 po b 2 allow − b 2 pi + 2b 2 po  a 2 allow + a 2 pi b2  a 2 ba

 allow + pi

 allow − pi + 2 po  allow + pi

 allow − pi + 2 po

 ( 24 in.)

−36,000 psi + 2,000 psi −36,000 psi − 2,000 psi + 2 ( 4, 444.44 psi )

 ( 24 in.)

−34,000 −29,111.11

= 25.9371 in.

The minimum allowable outside diameter of the tank is thus D  2b = 2 ( 25.9371 in.) = 51.8742 in. = 51.9 in.

Ans.

(b) What is the maximum tensile stress before the tank is immersed? For internal pressure only, the maximum tensile stress is the circumferential stress at r = a (See Figure 14.11). a 2 p  b2    = 2 i 2 1 + 2  b −a  r 

( 24 in.) ( 2,000 psi ) 1 + ( 25.9371 in.)  = 25,818.18 psi = 25,800 psi = 2 2 2 ( 25.9371 in.) − ( 24 in.)  ( 24 in.)  2

Ans.

P14.29

A thick-walled cylinder with closed ends is subjected only to internal pressure pi. Let a = 36 in. and b = 40 in. and determine the maximum permissible internal pressure pi that may be contained in the cylinder if the allowable tensile stress is 22 ksi and the allowable shear stress is 12 ksi.

Solution Consider allowable tensile stress of 22 ksi: The maximum tensile stress is found at the inner surface where r = a = 36 in. See Figure 14.11a. b2 + a 2 (  )max = 2 2 pi b −a b2 + a 2  allow  2 2 pi b −a

( 40 in.) − ( 36 in.) 22,000 psi = 2,309.39 psi b2 − a 2 pi  2  allow = ) 2 2 ( 2 b +a ( 40 in.) + ( 36 in.) 2

Consider allowable shear stress of 12 ksi: The maximum shear stress is found at the inner surface where r = a = 36 in. From Eq. (14.20): a 2b 2 ( pi − po ) b2  max = = 2 pi b − a2 b2 − a 2 r 2

(

 allow 

)

b2 pi b2 − a 2

( 40 in.) − ( 36 in.) 12,000 psi = 2,280 psi b2 − a 2 pi   allow == ( ) 2 2 b ( 40 in.) 2

Therefore, the maximum permissible internal pressure is pmax = 2,280 psi

Ans.

P14.30

A thick-walled cylindrical pressure vessel with an inside diameter of 200 mm and an outside diameter of 300 mm is made of a steel that has a yield strength of 430 MPa. Determine the maximum internal pressure that may be applied to the vessel if a factor of safety of 3.0 with respect to failure by yielding is required.

Solution Calculate allowable tensile stress: The allowable stress is  430 MPa  allow = Y = = 143.333 MPa FS 3.0 Radii: For this pressure vessel 200 mm a= = 100 mm 2

300 mm = 150 mm 2

Determine maximum allowable internal pressure: The maximum tensile stress is found at the inner surface where r = a = 100 mm. See Figure 14.11a. b2 + a 2  = (  )max 2 2 pi b −a b2 + a 2  allow  2 2 pi b −a

(150mm ) − (100mm ) 143.333 MPa b2 − a 2 pi  2  allow = ) 2 2 ( 2 b +a (150mm ) + (100mm ) 2

= 55.1282 MPa = 55.1 MPa

Ans.

P14.31

An open-ended compound thick-walled cylinder is made by shrink-fitting an exterior cylindrical steel jacket onto an interior steel tube. The following data are provided: E = 30,000 ksi, a = 20 in., b = 30 in., and c = 40 in. The radial interference is 0.05 in. Determine the maximum values of r and  produced in the compound cylinder by the shrink-fitting operation.

Solution Determine contact pressure: From Eq. (14.33), the contact pressure between the jacket and the tube is: E ( c 2 − b 2 )( b 2 − a 2 ) pc = 2b3 ( c 2 − a 2 )

( 30,000,000 psi )( 0.05 in.) ( 40 in.) − (30 in.)  (30 in.) − ( 20 in.)  2

2 ( 30 in.) ( 40 in.) − ( 20 in.)    3

= 8,101.852 psi

Determine stresses in the tube: The tube is subjected to an external pressure (See Figure 14.11b). The radial stress is equal in magnitude to the contact pressure. The largest radial stress occurs on the outer surface of the tube at r = b, and the radial stress is compressive. Thus,  r = − pc = −8,101.852 psi = −8,100 psi Ans. The circumferential stress in the tube is determined from Eq. (14.26). The largest circumferential stress occurs at r = a. This stress will be compressive.

2 ( 30 in.) (8,101.852 psi ) 2b2 pc  = − 2 2 = − = −29,166.67 psi = −29,200 psi 2 2 b −a ( 30 in.) − ( 20 in.) 2

Ans.

Determine stresses in the jacket: The jacket is subjected to an internal pressure (See Figure 14.11a). The radial stress is equal in magnitude to the contact pressure. The largest radial stress occurs on the inner surface of the jacket at r = b, and the radial stress is compressive. Thus,  r = − pc = −8,101.852 psi = −8,100 psi Ans. The circumferential stress in the jacket is determined from Eq. (14.24). The largest circumferential stress occurs at r = b. This stress will be tensile. 2 2 b 2 pc  c 2  c + b pc   = 2 2 1 + 2  = c −b  b  c2 − b2

(

)

( 40 in.)2 + ( 30 in.)2  ( 8,101.852 psi )  = = 28,935.19 psi = 28,900 psi 2 2 ( 40 in.) − ( 30 in.)

Ans.

A closed-ended compound cylinder is formed by shrinking a steel [E = 200 GPa;  = 0.3] tube that has an external diameter of 320 mm and an internal diameter of 250 mm onto a brass [E = 100 GPa;  = 0.35] tube that has an internal diameter of 200 mm. The difference between the diameters of the contacting surfaces before assembly is 0.10 mm. Determine the largest circumferential stress in the finished assembly after an internal pressure of pi = 50 MPa is applied.

P14.32

Solution Radii: a=

200 mm = 100 mm 2

250 mm = 125 mm 2

320 mm = 160 mm 2

Determine contact pressure: From Eq. (14.35), the contact pressure between the steel tube and the brass tube is: pc =



 1 b +a  1  c2 + b2  b  2 −  + J   2 T + 2 2  EJ  c − b   ET  b − a 0.10 mm 2 = 2 2   125 + 100   160 2 + 1252  1 1 − 0.35 + + 0.30 (125 mm )      2 2 2 2  200,000 MPa  160 − 125  100,000 MPa  125 − 100 2

= 6.2286 MPa

Determine circumferential stress in the brass tube: The brass tube is subjected to both an internal pressure of 50 MPa and an external pressure, which equals the contact pressure of 6.2286 MPa. Since the internal pressure is greater than the external pressure, the maximum circumferential stress in the brass tube is determined from Eq. (14.18). The largest circumferential stress occurs at r = a, and this stress will be tensile. b 2 + a 2 ) pi − 2b 2 po (  = b2 − a 2 (125 mm )2 + (100 mm )2  ( 50 MPa ) − 2 (125 mm )2 ( 6.2286 MPa )  = 2 2 (125 mm ) − (100 mm )

= 193.1744 MPa

Determine circumferential stress in the steel tube: The steel tube is only subjected to an internal pressure (See Figure 14.11a). The circumferential stress in the steel tube is determined from Eq. (14.24). The largest circumferential stress occurs at r = b. This stress will be tensile. 2 2 b 2 pc  c 2  c + b pc   = 2 2 1 + 2  = c −b  b  c2 − b2

(

)

(160 mm )2 + (125 mm )2  ( 6.2286 MPa )  = = 25.7418 MPa 2 2 (160 mm ) − (125 mm ) The largest circumferential stress in the finished assembly is thus  = 193.2 MPa

Ans.

A compound thick-walled steel [E = 200 GPa;  = 0.30] cylinder consists of a jacket with an inside diameter of 250 mm and an outside diameter of 300 mm shrunk onto a steel tube with an inside diameter of 150 mm. The initial shrinkage contact pressure is 28 MPa. Determine (a) The initial radial interference required to produce the initial shrinkage pressure. (b) The maximum tensile stress in the finished assembly after an internal pressure of pi = 200 MPa is applied.

P14.33

Solution Radii: a=

150 mm = 75 mm 2

250 mm = 125 mm 2

300 mm = 150 mm 2

(a) Determine the required initial radial interference: The required radial interference  between the steel jacket and the steel tube is determined from Eq. (14.32). 2b3 pc ( c 2 − a 2 ) = E ( c 2 − b 2 )( b 2 − a 2 ) 3 2 2 2 (125 mm ) ( 28 MPa ) (150 mm ) − ( 75 mm )    = 2 2 2 2 ( 200,000 MPa ) (150 mm ) − (125 mm )  (125 mm ) − ( 75 mm ) 

= 0.1342 mm

Ans.

(b) Determine maximum tensile stress in the finished assembly. Circumferential stress in the steel tube: The steel tube is subjected to both an internal pressure of 200 MPa and an external pressure, which equals the contact pressure of 28 MPa. Since the internal pressure is greater than the external pressure, the maximum circumferential stress in the steel tube is determined from Eq. (14.18). The largest circumferential stress occurs at r = a, and this stress will be tensile. b 2 + a 2 ) pi − 2b 2 po (  = b2 − a 2 (125 mm )2 + ( 75 mm )2  ( 200 MPa ) − 2 (125 mm )2 ( 28 MPa )  = 2 2 (125 mm ) − ( 75 mm )

= 337.500 MPa

Circumferential stress in the steel jacket: The steel jacket is subjected to an internal pressure only (See Figure 14.11a). The circumferential stress in the steel jacket is determined from Eq. (14.24). The largest circumferential stress occurs at r = b. This stress will be tensile. 2 2 b 2 pc  c 2  c + b pc   = 2 2 1 + 2  = c −b  b  c2 − b2

(

)

(150 mm )2 + (125 mm )2  ( 28 MPa )  = = 155.273 MPa 2 2 (150 mm ) − (125 mm ) The largest circumferential stress in the finished assembly is thus  = 338 MPa

Ans.

P14.34

A solid 40 mm diameter steel [E = 200 GPa] shaft is pressed into a steel gear hub that has an outside diameter of 120 mm. (a) If a contact pressure of 180 MPa is required so that the gear will not slip on the shaft, what is the minimum radial interference needed to produce this shrinkage pressure? (b) What is the maximum radial stress in the shaft? (c) What is the largest circumferential stress in the gear hub?

Solution Radii: a=0

40 mm = 20 mm 2

120 mm = 60 mm 2

(a) Determine the required initial interference: The contact pressure between a thick cylinder and a solid shaft is given in Eq. (14.34). Solve this equation for the required radial interference . E ( c 2 − b 2 ) pc = 2bc 2

2 ( 20 mm )( 60 mm ) (180 MPa ) 2bc 2 pc  = = = 0.0405 mm 2 2 E ( c − b ) ( 200, 000 MPa ) ( 60 mm )2 − ( 20 mm )2    2

Ans.

(b) Determine maximum radial stress in the shaft: The maximum radial stress in the shaft is simply equal to the contact pressure; therefore, Ans.  r = − pc = −180 MPa

(c) Circumferential stress in the gear hub: The steel hub is subjected to internal pressure only. The maximum circumferential stress in the steel hub is determined from Eq. (14.24). The largest circumferential stress occurs at r = b, and this stress will be tensile. 2 2 b2 pc  c 2  ( 20 mm ) (180 MPa )  ( 60 mm )    = 2 2 1 + 2  = 1 +  = 225 MPa Ans. c − b  b  ( 60 mm )2 − ( 20 mm )2  ( 20 mm )2 