seed_sol

Strength of Materials: An Undergraduate Text Graham M. Seed Saxe-Coburg Publications, 2001 ISBN: 1-874672-12-1 Chapter Solutions

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 G.M. Seed, 2001. All rights reserved. No part of this document may be transmitted, in any form or by any means whatsoever, without the written permission of the copyright owner.

Chapter 1 Solutions 1.1 The cross-sectional area, A, of the bar is

(

A = π 50x10 −3

)

2

= 7.85x10 −3 m2

The axial stress, σ, due to the axial load P=30kN is P 30x103 σ= = = 3.82MPa A 7.85x10−3 1.2 From (1.6) the axial strain, e, is σ 3.82x106 e= = = 18.2 x10−6 9 E 210x10 From (1.3) the axial extension, δ, is δ = eL = 18.2 x10−6 x0.5 = 9.1x10 −6 m 1.3 From (1.16) the in-plane strains are 1 1 ε xx = σ xx − νσ yy , ε yy = σ − νσ xx E E yy Re-arranging the first equation for σxx σ xx = Eε xx + νσ yy

[

]

[

]

and substituting into the second equation then σyy is found to be E 70x109 σ yy = ε + νε = [60 + 0.28x40]x10−6 = 5.4 MPa yy xx 1− ν2 1 − 0.282 Substituting σyy into εxx and re-arranging for σxx E 70x109 σ xx = ε + νε = [40 + 0.28x60]x10−6 = 4.3MPa xx yy 1− ν2 1 − 0.282

[

]

[

]

1.4 From (1.23) the bulk modulus is E 210x109 K= = = 175GPa 3(1 − 2ν ) 3(1 − 2 x0.3) 1.5 With E=σ/ε, σ=P/A, ε=δ/L and δ=α(∆T) then P is given by EAα ( ∆T ) 120x109 x7.1x10−4 x11x10−6 (100) P= = = 125kN L 0.75 illustrating that a compressive axial load of 125kN is required to cancel out the extension due to thermal expansion.

1.6 The compressive stresses of each bar are P 150x103 σ copper = = Acopper π 25x10 −3

(

P 150x103 = Asteel π 37.5x10−3 The contractions of each bar are σ steel =

(

)

) 2

2

= 76MPa = 34 MPa

1

δ copper = δ steel =

PLcopper E copper Acopper

=

150x103 x0.5

(

120x109 xπ 25x10

PLsteel 150x103 x0.6 = E steel Asteel 200x109 xπ 37.5x10−3

(

)

−3 2

)

2

= 318x10−6 m

= 102x10−6 m

The total contraction of the composite bar is δ=δcopper+δsteel=420x10-6m.

1.7 From (1.31,1.32,1.36) the in-plane strains εxx, εyy and γxy are given by ∂u 1 ∂v 1 = = ε xx = σ xx − νσ yy , ε yy = σ yy − νσ xx ∂x E ∂y E

(

γ xy =

)

(

)

1  ∂u ∂v   + =0 2  ∂y ∂x 

2

Chapter 2 Solutions 2.1 Figure Sol2.1 illustrates a circle of radius R with an elemental strip of thickness dy at distance y from the x-axis. The area, dA, of the elemental strip is therefore dA = 2 xdy = 2 R 2 − y 2 dy The area of the strip is therefore given by, (2.1) +R

A = ∫∫ dA = 2 ∫ R 2 − y 2 dy A

−R

The integration is assisted by making the substitution y=Rsinθ, dy=Rcosθdθ π /2

π /2

π /2

1  ϑ sin 2ϑ  A = 4 R ∫ cos dϑ = 4 R ∫  (1 + cos 2ϑ )dϑ = 4 R 2  + πR 2  2 2 2   0 0 0  The first moment of area Qx is, (2.2) +R +R 3/ 2 + R 2 2 2 =0 Qx = ∫∫ ydA = ∫ y 2 R − y dy = 2 ∫ y R 2 − y 2 dy = − R2 − y 2 A 3 −R −R −R Similarly, it is found that Qy=0. Therefore, as expected the coordinates of the centroid are xc=Qy/A=0 and yc=Qx/A=0.

2

2

2

(

[(

)

)

]

y

dx

2x dy R θ

x

2y

Figure Sol2.1. A circle of radius R and elemental strip dy.

2.2 The derivation of the area, first moments of area and centroid are analogous to those outlined in Exercise 2.1 except that the range of integration is now [0:R]. Thus, the area of the semicircle is π /2 R π /2 πR 2 ϑ sin 2ϑ  A = ∫∫ dA = 2∫ R 2 − y 2 dy = 2 R 2 ∫ cos2 ϑdϑ = 2 R 2  + = A 2  0 2 2 0 0 The first moment of area Qx is, using the elemental strip dy shown in Figure Sol2.1

3

R

Qx = ∫∫ dA = 2 ∫ y R 2 − y 2 dy = − A

0

[(

2 R2 − y 2 3

)

]

3/ 2 R 0

=

2R3 3

The y-coordinate of the centroid is therefore Q 2R3 / 3 4R yc = x = = A πR 2 / 2 3π The first moment of area Qy is, using the elemental strip dx shown in Figure Sol2.1 in which dA=ydx +R 3/ 2 + R 1 Qy = ∫∫ xdA = ∫ x R 2 − x 2 dx = − R2 − x 2 =0 A 3 −R −R and hence xc=Qy/A=0.

[(

)

]

2.3 Figure Sol2.3 illustrates an ellipse with elemental strip of length 2x and width dy.

The area A of the ellipse is therefore, noting that the equation of an ellipse is x2/a2+y2/b2=1 A=

b

b

−b

−b

∫∫ dA = ∫ 2 xdy = 2a ∫ A

1−

y2 2a b dy = b 2 − y 2 dy 2 b ∫−b b

Using the standard indefinite integral

∫

a 2 − x 2 dx =

a2  x 1 sin −1   + x a 2 − x 2  a 2 2

then the area is found to be b

 2a  b 2  y 1 A= sin −1   + y b 2 − y 2  =   b 2 b 2  −b  b 2   2a  πb 2 πb 2  2 a  b 2 −1 −1 + = πab  sin (1) −  sin ( −1)  =  b  2 4   2  b  4 The second moment of area Ix is, with dA=2xdy =

b

b

−b

−b

I x = ∫∫ y 2 dA = ∫ y 2 2 xdy = ∫ y 2 2a 1 − A

b y2 y2 2 2 1 dy = a y − dy ∫−b b2 b2

Using the standard indefinite integral x ∫ x ax + cdx = 4a then Ix is 2

2

(ax

2

+c

)

3

 −a  cx c2 2 sin −1  x − ax + c −  ; a<0 8a c  8a − a 

3   y y2  y  I x = 2a  1 − 2  − 2  b  8 −1 / b 2  − 4 / b Similarly, it can be shown that Iy=πa3b/4.

(

)

(

b

 1 / b 2   1 πab 3 y −1   1− 2 − sin  y  = 1   4 b −8 / b 2 1 / b 2   −b 2

)

(

)

4

y +b

dy

2x

x +a

-a -b

Figure Sol2.3. An ellipse with half major and minor axes a and b.

2.4 The centroidal cordinates for rectangles 1 and 2 are (xc1,yc1)=(27.5,2.5) and (xc2,yc2)=(2.5,25). The areas of rectangles 1 and 2 are A1=225mm2 and A2=250mm2 with the total cross-sectional area equal to A=A1+A2=475mm2. From (2.2) and (2.3) Qy and xc are Qy Q y = ∑ x ci Ai = x c1 A1 + x c2 A2 = 6,812 mm 3 , x c = = 14.34 mm A i and similarly with Qx and yc given by Q Qx = ∑ y ci Ai = y c1 A1 + y c 2 A2 = 6,812 mm 3 , y c = x = 14.34 mm A i with xc=yc and Qx=Qy due to the symmetry of the bracket about the (x,y) axes.

From (2.5) and (2.6) the second moments of area of rectangles 1 and 2 are given by, with respect to the centroidal axes 45(5) 3 5(45) 3 = 468mm 4 , I yc1 = = 37,968mm 4 I xc1 = 12 12 3 50(5) 3 5(50) = 52,083mm 4 , I yc2 = = 520mm 4 I xc2 = 12 12 Use of the parallel-axis theorem (2.8) and (2.9) for rectangle 1 gives Ix and Iy with respect to axes (x,y) I x1 = I xc1 + A1 y c21 = 1,875mm 4 , I y1 = I yc1 + A1 x c21 = 208,125mm 4 and similarly for rectangle 2 I x 2 = I xc 2 + A2 y c22 = 208,333mm 4 , I y 2 = I yc2 + A2 x c22 = 2,083mm 4 Finally, Ix and Iy are given by I x = I x1 + I x 2 = 210,208mm 4 , I y = I y1 + I y 2 = 210,208mm 4 with Ix=Iy as expected.

2.5 From (2.10) with dA=bdy then Ixy is given by h/2

 y2  I xy = ∫∫ xydA = bx ∫ ydy = bx   =0 A −h/ 2  2  −h/2 and is seen to be equal to zero due to the symmetry about the coordinate axes. h/ 2

2.6 From (2.13) the polar second moment of area for the ellipse of Exercise 2.3 is

5

I p = Ix + I y =

πa 3b πa 3b πab 2 + = a + b2 4 4 4

(

)

2.7 From (2.15) the radii of gyration rx and ry for the ellipse of Exercise 2.3 are Iy Ix πab 3 / 4 b πa 3b / 4 a , ry = rx = = = = = 2 2 A πab A πab

6

Chapter 3 Solutions 3.1 From (3.12) the maximum shear stress is given by 16T 16(10x103 ) τ max = = = 407 MPa πD 3 π 50x10−3 3

(

)

3.2 From (3.1) the polar moment of area, J, is

(

)

4

−3 πD 4 π 50x10 J= = = 6136 . x10−6 m4 32 32 and from (3.11) the angle of twist is TL 10x103 x125 . θ= = = 2.55x10−3 radians = 015 . o −6 9 GJ 80x10 6136 . x10

(

)

3.3 From Example 3.1 then the maximum shear stress occurs at the smallest diameter of d1=50mm 16 12x103 16T τ max = 3 = = 489MPa πd1 π 50x10−3 3

(

(

) )

and the angle of twist is 1 32TL 1 θ= − 3  3 3πG(d 2 − d1 )  d1 d 2  =

 ( ) 1 1  − 3πx80x10 (75x10 − 50x10 )  (50x10 ) (75x10 )  32 12 x103 x1 −3

9

−3

−3 3

−3

  = 01147 . radians = 6.57 o 3  

3.4 From (3.16) the mean shear stress is 100 T τm = = = 163MPa 2 −3 2 −3 2πRm t 2π 3125 . x10 . x10 01

(

)(

)

3.5 Re-arranging (3.16) for wall-thickness t T 15 . x103 t= = = 2.3mm 2πRm2τ m 2π 40x10−3 2 65x106

(

)(

)

3.6 From (3.1) and (3.2) the polar moments of area of the inner solid bar, JA, and outer tube, JB, are

(

)

4

−3 πRA4 π 12.5x10 JA = = = 383 . x10−8 m4 2 2 4 π 4 π JB = RB − RA4 = 25x10−3 − 12.5x10−3 2 2 From (3.21) the angle of twist is

(

)

[(

) (

) ] = 5.75x10 4

−7

m4

7

θ=

5x103 (0.75) TL . radians = 11324 . o = = 01976 −8 9 −7 G A J A + GB J B 45x109 383 . x10 + 30x10 5.75x10

(

)

(

)

3.7 From (3.20) the torques in the inner solid bar, TA, and outer tube, TB, are   . x10 −8 46x109 383   GA J A   5x103 = 454 Nm TA =  T =  9 9 −8 −7   G A J A + GB J B  . x10 + 30x10 5.75x10   46x10 383  GB J B TB =   G A J A + GB J B

( ) ( ) ( )   30x10 (5.75x10 )   5x10  T =   . x10 ) + 30x10 (5.75x10 )   46x10 ( 383 9

9

−7

−8

9

−7

3

= 4.55kNm

and from (3.22) the corresponding maximum shear stresses are −3 TA R A 454 12.5x10 τ A,max = = = 148MPa JA . x10 −8 383

(

τ B ,max

)

(

)

3 −3 TB RB 4.55x10 25x10 = = = 198MPa JB 5.75x10 −7

8

Chapter 4 Solutions 4.1 From (4.2) and (4.4) the principal stresses in the pipe are σ pr 50x6.895x10 3 x0.8 σ1 = = = 18.4 MPa , σ 2 = 1 = 9.2 MPa −3 t 15x10 2 4.2 From the Hookian equations (1.16) we have 1 1 ε zz = [σ zz − νσ θθ ] , ε θθ = [σ θθ − νσ zz ] E E Solving these for the in-plane stresses then we have E 70x10 9 σ zz = ε νε 429 + 0.3x1,821]x10 −6 = 75MPa + = θθ ] 2 [ zz 2 [ 1−ν 1 − 0.3 E 70x10 9 σ θθ = ε + νε = [1821 + 0.3x429]x10−6 = 150MPa [ ] θθ zz 1−ν 2 1 − 0.32 The maximum in-plane shear stress is, (4.7) σ 1 − σ 2 σ θθ − σ zz (150 − 75) x106 τ max = = = = 37.5MPa 2 2 2 4.3 With the pressure p equal to ρgh and σθθ=σ2 not exceeding the maximum allowable stress σallow=300MPa/S where S(=10) is the safety factor then from (4.11) we find the maximum permissible depth of water to be −3 6 2tσ allow 2 25x10 300x10 = = 153m h= ρgrS 1000x9.81x1x10

(

)

4.4 From (4.5) and (4.12) the circumferential strains for the cylinder, εc, and hemispherical end caps, εs, are pr  ν  pr εc = (1 − ν ) 1 −  , ε s =   2 2 Et s Et c where E and ν denote Young’s modulus and Poisson’s ratio respectively, p is the internal pressure, r is the radius of the vessel and tc and ts are the wall thicknesses of the cylinder and hemi-spherical ends respectively. Equivalence of the circumferential strains εc and εs yields  1− ν  ts =  t 2−ν c and with tc=1mm then ts=0.4mm.

4.5 From (4.9) the absolute maximum shear is equal to σ pr τ max = 1 = σ 2 = 2 2t The cylinder pressure, p, is P P p= = 2 A πr where P is the force acting on the piston and r is the piston radius. Substituting p into τmax then

9

τY P = S 2πrt where τY is the yield stress in pure shear and S is the safety factor. Re-arranging for the cylinder wall thickness t SP 2(50x103 ) t= = = 2.65mm 2πrτ Y 2π (40x10−3 )150x106 τ max =

4.6 Re-arranging (4.11) for internal pressure p with σUTS denoting the ultimate tensile stress then 2σ UTS t 2 x737 x106 (10−3 ) p= = = 227 MPa 65x10−3 r 4.7 From (4.19) Ix and Iy are given by I x = I y = πr 3 t = πx253 x1 = 49,087 mm 4

10

Chapter 5 Solutions 5.1 From the equations of equilibrium the two unknown reactions RA and RB are found ∑ F v = 0 : R A + RB − 30(2) − 40 = 0 ⇒ R A + RB = 100kN

∑M = 0:

RB (7) − 60(4) − 40(8) = 0 ⇒ R A = 80kN and ∴ R A = 20kN The shear force, Vx, and bending moment, Mx, for the four intervals i) 0<x<3, ii) 3<x<5, iii) 5<x<7 and iv) 7<x<8 are: i)

0<x<3

Vx = RA = 20kN M x = R A x = 20 xkNm

ii)

3<x<5

Vx = RA − 30( x − 3) = 20 − 30( x − 3)kN M x = R A x − 30

iii)

5<x<7

( x − 3)2 2

= 20 x − 30

( x − 3)2 kNm 2

Vx = R A − 60 = −20kN M x = R A x − 60( x − 4) = 20 x − 60( x − 4)kNm

iv)

7<x<8 Vx = R A + RB − 60 = 40kN

M x = R A x + RB ( x − 7) − 6( x − 4) = 20 x + 80( x − 7) − 6( x − 4)kNm with Vx and Mx illustrated in Figure Sol5.1.

11

30kN/m A

40kN B

RA

RB

Vx 40 20 0 -20 -40

60 40

Mx

20 0 -20 -40 Figure Sol5.1. Shear force and bending moment diagrams for the simply supported beam of Exercise 5.1.

5.2 From the equations of equilibrium the two unknown reactions are WL 5x10 3 x5 R A = RB = = = 12.5kN 2 2 Taking moments at an arbitrary cut at a distance x from the left hand end x Wx 2 WLx Wx 2 M xx + Wx − R A x = 0 ⇒ M xx = R A x − = − 2 2 2 2 Differentiating Mxx with respect to x and setting dMxx/dx=0 we find that the maximum bending moment occurs at x=L/2. Substituting x=L/2 into Mxx then the maximum bending moment is given by WL2 5x10 3 x52 M max = = = 15,625Nm 8 8 For the rectangular section the second moment of area is given by, Example 2.3

12

(50x10 )(75x10 ) I=

−3 3

−3

From (5.30) the y=±h/2=±37.5mm

maximum

σ max = ±

12 tensile

M max ( h / 2)

=±

. x10 −6 m 4 = 17578

and

compressive

(

15,625 37.5x10 −3 −6

stresses

occur

at

) = ±333MPa

I 17578 . x10 5.3 From Example 5.2 the shear force, Vx, at a distance x from the left hand end of the beam is L  V x = W  − x 2  and attains maximum and minimum values at x=0 and x=L respectively WL 5x10 3 x5 Vmax = = = 12.5kN , Vmin = −Vmax = −12.5kN 2 2 From the beam shear formula (5.40) and the rectangular cross-section examined in Example 5.4 then the maximum shear stress at a given section is Vx h 2 τ max = 8I and occurs at the neutral axis of the beam. Substituting Vmax and Vmin we have τ max,min = ±

(

12.5x10 3 75x10 −3

(

8 17578 . x10 −6

)

)

2

= ±5MPa

and are seen to be considerably less than the maximum tensile and compressive bending stresses of Exercise 5.2.

5.4 From the equations of equilibrium the two unknown reactions RA and RB are found WL ∑ F v = 0 : R A + RB = 2 WL  2 L  WL WL ∑ M = 0 : RB L − 2  3  = 0 ⇒ RB = 3 ⇒ RA = 6 where the centre of gravity of the distributed load acts at x=2L/3. From the equations of equilibrium for the free body diagram of Figure Sol5.4 the bending moment is given by Wx 2  x  WLx Wx 3 = + − = ⇒ = − M 0 : M R x 0 M   ∑ x A x 2L  3  6 6L

13

Wx/L Mx A x

Vx

RA 2

Wx /2L Figure Sol5.4. Free body diagram for the beam of Figure 5.27 cut at a distance x from support A. Substituting the bending moment into (5.41) we have WLx Wx 3 − EIv' ' = 6 6L Integrating with respect to x then the slope of the beam is WLx 2 Wx 4 EIv' = − + C1 12 24 L and integrating once more then the deflection is WLx 3 Wx 5 EIv = − + C1 x + C2 36 120 L The boundary condition v=0 at x=0 reveals that C2=0 and the boundary condition v=0 at x=L gives C1=-21WL3/1080. Substituting C1 and C2 into the above expressions for v and v′ and re-arranging we have Wx 3 x 4 − 10 L2 x 2 + 7 L4 v=− 360 LEI W 7 L4 − 30 L2 x 2 + 15 x 4 v' = − 360 LEI The maximum deflection, δmax, can be determined from the condition that the slope of the beam will be horizontal at the point of maximum deflection, xmax. Substituting v′=0 in the above and re-arranging we have the following quadratic equation for unknown x2 15 x 4 − 30 L2 x 2 + 7 L4 = 0 2 Solving for x  480  2 x 2 = 1 ± L 30   which gives the two solutions x=1.3154L and x=0.5193L. Since x≤L then the maximum deflection occurs at xmax=0.5193L. Finally, substituting xmax into v we arrive at δmax WL4 δ max = −v(0.5193L ) = 0.00652 EI

(

(

)

)

5.5 From the equations of equilibrium for the entire beam WL WL 3 L R A + RB − =0 , − RA L = 0 2 2 4 we find that the reactions RA and RB are given by

14

3WL WL , RB = 8 8 As noted in Example 5.8, when using Macaulay's method, if a distributed load does not extend to the right hand end of the beam then we need to extend and counterbalance the distributed load to the right hand end of the beam. The bending moment at a cut x-x (L/2<x<L) is, Figure 5.27b) x {x − L / 2} M xx + Wx − R A x − W {x − L / 2} =0 2 2 3WLx W {x − L / 2} 2 Wx 2 ⇒ M xx = + − 8 2 2 Substituting Mxx into the flexure formula (5.41) 3WLx W {x − L / 2}2 Wx 2 EIv' ' = + − 8 2 2 and integrating twice 3WLx 2 W {x − L / 2} 3 Wx 3 EIv' = + − + C1 16 6 6 3WLx 3 W {x − L / 2}4 Wx 4 EIv = + − + C1 x + C2 48 24 24 From the two boundary conditions v=0 at x=0 and x=L we find that the constants of integration C1 and C2 are 9WL3 C1 = − , C2 = 0 384 and upon substitution into v we arrive at the following expression for the deflection of the entire beam WLx 3 W {x − L / 2} 4 Wx 4 9WL3 x EIv = + − − 16 24 24 384 Macaulay's method informs us that a term in curly brackets is ignored if it is either zero or negative. Therefore, when x≤L/2 we have Wx L v=− 16 x 3 − 24 Lx 2 + 9 L3 ; 0 ≤ x ≤ 384 EI 2 RA =

(

)

5.6 The beam of Figure 5.28 has three unknown reactions (RA, RB and MA) but only two independent equations equilibrium can be written and as a result the beam is statically indeterminate to the first degree. Let reaction RB be the redundant reaction.

From the equations of equilibrium R A + RB − WL = 0 ⇒ R A = WL − RB L WL2 = 0 ⇒ MA = − RB L 2 2 The bending moment at an arbitrary cut x-x is x Wx 2 M xx + M A + Wx − R A x = 0 ⇒ M xx = R A x − M A − 2 2 Substituting RA and MA into Mxx and using the flexure formula (5.41) we have WL2 Wx 2 EIv' ' = WLx − RB x − + RB L − 2 2 Performing two integrations we have M A + R B L − WL

15

WLx 2 RB x 2 WL2 x Wx 3 EIv' = − − + RB Lx − + C1 2 2 2 6 WLx 3 RB x 3 WL2 x 2 RB Lx 2 Wx 4 EIv = − − + − + C1 x + C2 6 6 4 2 24 Applying the boundary conditions v'=0 at x=0 and v=0 at x=0 we find that C1=0 and C2=0. Applying the boundary condition v=0 at x=L gives the redundant reaction 3WL RB = 8 The unknown reactions RA and MA are now given by 5WL WL2 RA = , MA = 8 8 Substituting RB, C1 and C2 into v and collecting terms then the deflection for the beam is given by Wx 2 v=− 3 L2 + 2 x 2 − 5 Lx 48 EI

(

)

5.7 The beam has three unknown reactions (RA, RB and MA) but only two independent equations of equilibrium and is therefore statically indeterminate to the first degree.

Letting RB be the redundant reaction then from the equations of equilibrium WL2 R A = WL − RB , M A = − RB L 2 Using the method of superposition we now proceed to apply the distributed load W to the released beam, Figure Sol5.7b), and the redundant reaction RA to the released beam, Figure Sol5.7c). Since the deflection of the original beam is zero at end B then we have the following compatibility equation δ B = δ B1 − δ B 2 = 0 The force-displacement relations give the deflections δB1 and δB2 R L3 WL4 δ B1 = , δ B2 = B 8 EI 3EI which upon substituting into the compatibility equation gives WL4 RB L3 δB = − =0 8 EI 3EI and solving for RB 3WL RB = 8 The remaining reactions RA and MA are found from the above equilibrium equations 5WL WL2 RA = , MA = 8 8

16

W B

a) MA

A RA

RB

L

W b)

c)

δB

δB

1

2

RB

Figure Sol5.7. Propped cantilever beam of Exercise 5.7. a) The propped cantilever beam subject to a uniformly distributed load. b) The released beam with redundant reaction RB and load W applied. c) The released beam with RB applied.

17

Chapter 6 Solutions 6.1 With I=πd4/64 for a solid circular cross-section of diameter d and substituting I into (6.10) and re-arranging for d we have  64 Pcr L2  d =  3  π E 

1/ 4

 64 x200x10 3 x4 2  =   π 3 x210x10 9 

1/ 4

= 75mm

6.2 The second moments of area with respect to centroidal coordinates (x,y) for the square, circle and equilateral triangle shown in Figure Sol6.2 are Ab 2 I A square: I = I xx = I yy = , A = b2 , r 2 = = ≈ 0.083& A 12 A 12 Ab 2 I A circle: I = I xx = I yy = , A = πb 2 , r 2 = = ≈ 0.0797 A 4 A 4π 3 2 Ab 2 I A equilateral triangle: I = I xx = I yy = , A= b , r2 = = ≈ 0.0722 A 36 4 A 8 3 From (6.14) σcr is given by π 2 Er 2 L2 with σcr seen to be proportional to r2. Therefore, the struts from largest to smallest σcr are the square, circular and equilateral triangle cross-sections. σ cr =

y

y y x

b

x

b

b

x

b b Figure Sol6.2. Square, circle and equilateral triangle.

b

6.3 With reference to Figure Sol6.3 the second moments of area Ix and Iy are, (2.5) and (2.6) I x = ∫∫ y 2 dA = 4 ∫ (a cosθ ) (tds) = 4 ∫ 2

A

π /2

0

I y = ∫∫ x 2 dA = 4 ∫ (a sin θ ) ( tds) = 4 ∫ 2

A

π /2

0

(a cosθ ) 2 (tadθ ) = πa 3 t

(a sin θ ) 2 (tadθ ) = πa 3 t

with Ix=Iy due to symmetry. The area of the section is A=

∫∫ dA = 4∫ tds = 4∫ A

π /2

0

t ( adθ ) = 2πat

Finally, the radius of gyration is, (6.12) r=

I = A

πa 3 t a = 2πat 2

18

y

t

ds a dθ θ

x

Figure Sol6.3. One quadrant of a circular tubular cross-section.

6.4 From the equations of equilibrium the compressive load acting on the strut is equal to √3W. The cross-sectional area and second moment of area of the strut are

(

A = 50x10 −3

(

)

2

= 2.5x10 −3 m 2

)(

1 50x10 −3 50x10 −3 12 From (6.12) the radius of gyration is I=

)

3

= 5.2083& x10 −7 m 4

I 5.2083x10 −7 = = 0.0144 m A 2.5x10 -3 The strut is built-in at one end and pin-jointed at the other end where the load W acts so that from (6.42) the effective length of the strut is Le=0.7L=1.75m and the critical buckling load is, (6.31) π 2 EA π 2 x210x10 9 x2.5x10 −3 = = 351kN Pcr = 2 . / 0.0144) (175 ( Le / r ) 2 Thus, the strut will fail due to buckling when W exceeds 351kN. r=

6.5 The area and second moment of area of the rectangular tube are . ) − (018 . x0.08) = 5.6x10 −3 m 2 A = (0.2 x01

[

]

1 . 3 − 018 . x0.08 3 = 8.986& x10 −6 m 4 0.2 x01 12 From (6.32) the critical buckling load is 4π 2 EI 4π 2 x210x10 9 x8.986x10 −6 Pcr = = = 993.3kN L2 52 and the critical stress is Pcr 993.3x10 3 σ cr = = = 177.4 MPa A 5.6x10 −3 I=

6.6 The stress and eccentricity ratio are

19

P 350x10 3 = = 107 MPa A 3270x10 −6 ey eA 0.025x3270x10 −6 = = = 0.5677 S r2 144 x10 −6 The critical buckling load is, (6.10) π 2 EI π 2 x210x10 9 x11x10 −6 Pcr = = = 912 kN L2 52 From (6.52) and (6.57) the maximum deflection and stress are  π P   δ = e sec  − 1 = 19.5mm   2 Pcr   π P  P  ey   = 215MPa σ max = 1 + 2 sec A  r  2 Pcr   6.7 From Table 6.2 the constant a is equal to 1/7500 and the slenderness ratio is 2 L l= = = 50.505 r 39.6x10 −3 Therefore, from (6.61) the critical stress according to the Rankine-Gordon formula is σY 300x10 6 = = 224 MPa σR = 1 + al 2 50.5052 1+ 7500

20

Chapter 7 Solutions 7.1 The σxx stress is P 75x103 =− = −50MPa A 1500x10−6 From (7.8) and (7.12) the local direct and shear stresses on the cut plane ab are 1 σ x 'x ' = σ xx (1 + cos 2ϑ ) = σ xx cos2 ϑ = −37.5MPa 2 1 σ y'y' = σ xx (1 − cos 2ϑ ) = −12.5MPa 2 1 . MPa τ x ' y ' = −σ xx sin 2ϑ = −σ xx sin ϑ cos ϑ = 2165 2 σ xx =

7.2 From the stress transformation equations we have σ x 'x ' = 195MPa , σ y ' y ' = 105MPa , τ x ' y ' = 50MPa observing that the sum of the global and local stresses are equal, (7.13) σ xx + σ yy = σ x 'x ' + σ y ' y ' 7.3 The centre C, point A, point B and radius R of Mohr’s circle are, see §7.5.1  σ + σ yy  C =  xx ,0 = [ −25,0] 2  

( ) ( B(ϑ = 90 ) = (σ

)

A ϑ = 0o = σ xx , τ xy = (50,−50) o

yy

)

,−τ xy = ( −100,50)

 σ − σ yy  2 R =  xx  + τ xy = 90.14 2   Mohr’s circle can now be plotted and is shown in Figure Sol7.3.

2

S2 D’

τm

θ(+ve.) in o

A(θ=0 )

2θs 2 2θp 2

P2

C

R B(θ=90

o

)

-2θ O p1

P1

σx ’ x ’

β 2θs 1 o

S1

τm

D(θ=45 ) a x

τx ’ y ’

Figure Sol7.3. Mohr’s circle for the stress element of Exercise 7.3.

21

The principal stresses and associated planes are . = 6514 . σ 1 = OC + R = −25 + 9014 . σ 2 = OC − R = −25 − 90.14 = −11514  2 . o 2ϑ p 2 = 180o − ∠ACP1 = 180o − tan −1   = 146.31o ; ϑ p 2 = 7316  3 . o or − 16.84o 2ϑ p1 = 180o + 2ϑ p 2 ; ϑ p1 = 16316 The maximum and minimum shear stresses and associated planes are τ max = R = 90.14 τ min = − R = −9014 .  3 2ϑ s2 = ∠ACS2 = tan −1   = 56.31o ; ϑ s2 = 2815 . o  2 2ϑ s1 = 180o + 2ϑ s 2 ; ϑ s1 = 11815 . o The stresses on the plane θ=-45° are represented by points D and D′ on Mohr’s circle shown in Figure Sol7.3. With angle β given by  2 β = 2ϑ − ∠ACP1 = 90o − tan −1   = 90o − 33.69o = 56.31o  3 . cos 56.31o = 25 σ x 'x ' ( D) = OC + R cos β = −25 + 9014 τ x ' y ' ( D) = R sin β = 90.14 sin 56.31o = 75 σ y ' y ' = σ x 'x ' ( D' ) = OC − R cos β = −25 − 90.14 cos 56.31o = −75 7.4 Mohr’s circle can be schematically constructed by positioning the centre, C, of the circle and determining its radius, R, as follows for the three requested cases: a) Uniaxial (σ σxx=σ σ, σyy=ττxy=0)  σ + σ yy   σ  C =  xx ,0 =  ,0 0   2   σ − σ yy  σ 2 R =  xx  + τ xy = 2 2   2

b) Equi-biaxial (σ σxx=σ σyy=σ σ, τ xy=0) C = [σ ,0] R=0 noting that Mohr’s circle reduces to a point. c) Pure Shear (σ σxx=σ σyy=0, τ xy=ττ) C = [0,0] R=τ

22

Each of the above cases is correspondingly illustrated in Figure Sol7.4.

σx ’ x ’

C[σ/2,0]

O

C[σ,0]

O

σx ’ x ’

σ/2

σx ’ x ’

C[0,0]

τ

τx ’ y ’

τx ’ y ’

τx ’ y ’

a)

b)

c)

Figure Sol7.4. Mohr’s circle for the cases of a) uniaxial, b) equi-biaxial and c) pure shear loadings.

7.5 (i) With I1, I2 and θp given by I1,2 =

Ix + Iy 2

2 I xy  I − Iy  2 ±  x  + I xy , tan2ϑ p = −  2  Ix − Iy 2

and since Ix=Iy then 4 1 4  4  9π 2 + 18π − 128  4 π I1 = I x + I xy =  − + − r =  r  16 9π 8 9π  144π   4 1 4  4  π − 2 4 π + + I 2 = I x − I xy =  − r =  r  16 9π 8 9π   16  tan 2ϑ p = ∞ ; 2ϑ p = 90o ; ϑ p = 45o (ii) With r=10mm then from the Ix, Iy and Ixy expressions given I x = I y = 549 mm4 , I xy = −165mm4 and from the transformation equations  I + Iy   Ix − I  4 4 I x ', y ' =  x  cos 2ϑ m I xy sin 2ϑ = I x m I xy sin 2ϑ = 692 mm and 406mm  ± 2 2     I x ' y ' = I xy cos 2ϑ = −82.5mm4 7.6 With θ=30° then cos2θ=1/2 and sin2θ=√3/2 and substituting Ix, Iy and Ixy into (7.39) and (7.41) we find that Ix′=Iy, Iy′=Ix and Ix′y′=Ixy as required.

7.7 From (7.54) the shear modulus, G, for the aluminium alloy and steel are 70 Eal Gal = = = 26GPa 2(1 + ν al ) 2(1 + 0.33) Gsteel =

210 Esteel = = 81GPa 2(1 + ν steel ) 2(1 + 0.3)

23

Chapter 8 Solutions 8.1 Since θ is taken as positive for an anticlockwise rotation then θ=-30° in the present case. From (8.10) the local strains are 200x10 −6  500 + (-300)   500 − ( −300)  −6 −6 o x 10 ± x 10 − 60 ± ε x ' x ', y ' y ' =  cos( ) sin( −60 o )        2 2 2 = 213x10 −6 ,−13.4 x10 −6 γ

−6

200x10  500 − ( −300)  −6 o = − cos( −60 o ) = 396.4 x10 −6  x10 sin( −60 ) +   2 2 2 -6 so that γx′y′ is equal to 793x10 . x' y '

8.2 From (8.11) the principal strains ε1 and ε2 are ε xx + ε yy

 ε − ε yy   γ xy  −6 −6 ±  xx ε 1,2 =  +  = −190x10 ,−360x10 2  2   2  From (8.11) we obtain the planes of the principal strains 80   −1  8  2ϑ p = tan −1   = tan    −350 − ( −200)  15 2

2

which has the two roots of θp=-14° and 76° for θp in the range 0≤θp≤180°. To establish which angle is associated with either ε1 and ε2 then let us examine, say, θp=14° for εx′x′, (8.10) −350 + ( −200) −350 − ( −200) 80 ε x ' x ' ( −14 o ) = x10 −6 + x10 −6 cos( −28 o ) + x10 −6 sin( −28 o ) 2 2 2 −6 = −360x10 observing that εx′x′ (-14°)=ε2 so we conclude that θp2=-14° and θp1=76°. The principal strains are illustrated graphically in Figure Sol8.2a). y y’

y’

y

o

θp 2 =76

εa v g dy’

ε2 dy’

x’ o

θs 1 =31 εa v g dx’

x

x

o

θp 1 =14 ε1 dx’

x’

b) a) Figure Sol8.2. Schematic illustration of strains. a) Principal stress element. b) Maximum shear strain element.

The maximum shear strain is, (8.12)

24

2 2  γ x 'y'   ε xx − ε yy   γ xy   80   −350 − ( −200)  −6 =   +   = 85x10    +  =   2    2  max  2   2  2 2

2

and therefore γmax=170x10-6 and is illustrated graphicaly in Figure Sol8.2b).

8.3 With θa=0°, θb=60° and θc=120° we find that the system of equations (8.18) reduces to 60x10−6 = ε xx 135x10−6 = 0.25ε xx + 0.75ε yy + 0.433γ 264x10

−6

xy

= 0.25ε xx + 0.75ε yy − 0.433γ xy

which result in the global strains εxx=60x10-6, εyy=246x10-6 and γxy=-149x10-6.

We will use the transformations equations to determine the principal strains and their associated planes. From (8.11) the principal strains are  60 + 246   6 − 246   −149  −6 −6 ε 1,2 =  ±   +  = 272 x10 , 34 x10       2 2 2 and with principal planes  γ xy  1 1 −149  1 o −1  = tan −1  θ p = tan −1   = tan (0.8) = 19.33   2 60 − 246 2  ε xx − ε yy  2 Inserting 19.33° into (8.10) we find that θp2=19.33° and therefore θp1=90°+θp2=109.33°. These results can be compared with Example 8.2 which alternatively determined the principal strains and planes using Mohr's circle.

2

2

8.4 Points A and B, centre C and radius R of Mohr’s circle for the in-plane strains εxx=250x10-6, εyy=-150x10-6 and γxy=120x10-6 are as follows A ϑ = 0o = ε xx , γ xy / 2 = 250x10−6 ,60x10−6

( ) ( ) ( B(ϑ = 90 ) = (ε ,−γ / 2) = ( −150x10 C = [ε ,0] = [50x10 ,0] o

yy

xy

)

−6

,−60x10 −6

)

−6

avg

 γ xy   ε − ε yy  −6 R =  xx  = 209 x10  + 2  2    with Mohr's strain circle illustrated in Figure Sol8.4. From Mohr’s circle we find the principal strains and planes ε 1 = OC + R = (50 + 209) x10−6 = 259 x10−6 2

2

ε 2 = OC − R = (50 − 209) x10−6 = −159 x10−6  60  o o 2ϑ p1 = tan −1   = 16.7 ; ϑ p1 = 8.35  250 − 50  2ϑ p 2 = 180o + 2ϑ p1 ; ϑ p 2 = 98.35o The maximum shear strain (γ/2)max=R so that γmax=418x10-6. From Mohr’s circle the planes of the maximum and minimum shear strains are 2ϑ s1 = − 90o − 2ϑ p1 = −73.3o ; ϑ s1 = −36.65o

(

)

2ϑ s2 = 180o − 2ϑ s1 = 106.7 o ; ϑ s 2 = 53.35o

25

o

B(θ=90 ) (-150,-60)

- 6

ε2

2θ p 2

O

εx ’ x ’ x10

ε1

C[50,0] 2θ p 1 -2θs 1

R=209

o

A(θ =0 ) (250,60)

(γx ’ y ’ /2)x10

-6

Figure Sol8.4. Mohr’s circle for the in-plane strains εxx=250x10-6, εyy=-150x10-6 and γxy=120x10-6.

8.5 The average normal strain εavg, centre C, points A and B and radius R of Mohr’s strain circle are ε + ε yy ε avg = xx = −200x10−6 2 C = ε avg ,0 = −200 x10−6 ,0

] [ ] A(ϑ = 0 ) = (ε , γ / 2) = ( −300x10 ,50x10 ) B(ϑ = 90 ) = (ε ,−γ / 2) = ( −100x10 ,−50x10 ) [

−6

o

xx

−6

xy

−6

o

yy

−6

xy

 γ xy   ε − ε yy  . x10 −6 R =  xx  = 1118  + 2  2    2

2

Mohr’s circle can now be constructed and is shown in Figure Sol8.5. Since we are required to determine the strain components on an element that is rotated by 20° in a clockwise direction then θ=-20°. From Figure Sol8.5 the principal plane θp2 is given by 1 50 1   −1 1 = −13.28o ϑ p2 = − tan −1   = − tan   2 300 − 200 2 2 so that β=2θ-2θp2=40°-26.57°=13.43°. From triangle DEC we find the local strains ε x 'x ' = −(OC + R cos β ) = − 200 + 1118 . cos 13.43o x10−6 = −309 x10−6

(

γ x 'y'

)

= − R sin β = −1118 . x10−6 sin 13.43o = −25.97x10−6 ; γ x ' y ' = 5193 . x10 −6

2 and from triangle D′FC we find the remaining local strain εy′y′

26

ε y ' y ' = OC + R cos β = (−200 + 1118 . cos 13.43o ) x10−6 = −9126 . x10−6

o

B(θ=90 ) (-100,-50)

D(εx ’ x ’ ,-γx ’ y ’ /2) -2θ β E

C[-200,0]

-6

O

F

εx ’ x ’ x10

-2θp 2

D’(εy ’ y ’ ,γx ’ y ’ /2)

o

A(θ=0 ) (-300,50)

(γ x ’ y ’ /2)m

a x

(γ x ’ y ’ /2)x10

=R=111.8

-6

Figure Sol8.5. Mohr’s circle for the in-plane strains εxx=300x10-6, εyy=-100x10-6 and γxy=100x10-6.

8.6 From (8.10) the local strains are 100x10 −6  200 − 400   200 + 400  −6 −6 o ε x ' x ', y ' y ' =  sin 80 o  x10 cos 80 ±  x10 ±      2 2 2 = 332 x10 −6 , 268x10 −6 γ x' y ' 100x10 −6  200 − 400  −6 o = − cos 80 o = 107 x10 −6  x10 sin 80 +   2 2 2 -6 so that γx′y′ is equal to 214x10 .

From the Hookian equations (1.16) we have τ x'y' 1 1 ε x'x' = σ x ' x ' − νσ y ' y ' , ε y ' y ' = σ y ' y ' − νσ x ' x ' , γ x ' y ' = E E G Solving these for the in-plane stresses then we have E 210x10 9 σ x ' x' = ε + νε = [332 + 0.3x268]x10 −6 = 95MPa x' x ' y' y ' 1−ν 2 1 − 0.32 E 210x10 9 + = 268 + 0.3x332]x10 −6 = 85MPa σ y'y' = ε νε y'y' x ' x' 2 2 [ 1−ν 1 − 0.3 210x10 9 E 214 x10 −6 = 17 MPa τ x ' y ' = Gγ x ' y ' = γ x' y ' = 2(1 + ν ) 1(1 + 0.3)

[

]

[

[

]

[

]

]

8.7 With θa=0°, θb=45° and θc=90° then the system of equations (8.18) reduce to 65x10 −6 = ε xx 95x10 −6 = 0.5ε xx + 0.5ε yy + 0.5γ 25x10

−6

xy

= ε yy

27

which result in the global strains εxx=65x10-6, εyy=25x10-6 and γxy=100x10-6. From (8.11) the principal strains are ε xx + ε yy

 ε xx − ε yy   γ xy  −6 −6 ε 1,2 = ±   +  = 98.85x10 , − 8.85x10 2 2    2  and from (8.11) we obtain the planes of the principal strains  100  −1 o 2ϑ p = tan −1   = tan (2.5) = 68.2  65 − 25 which has the two roots of θp=34.1° and 124.1° for θp in the range 0≤θp≤180°.

2

2

28

Chapter 9 Solutions 9.1 From (9.10) the in-plane strains are 1 1 ε xx = σ xx − νσ yy , ε yy = σ − νσ xx E E yy and substituting into the strain energy density, (9.9), we have 1 1 U 0 = σ xx ε xx + σ yy ε yy = σ 2xx − 2νσ xxσ yy + σ 2yy 2 2E Integrating U0 throughout the entire volume of the plate then the strain energy is given by V U = ∫ U 0 dV = σ 2 − 2νσ xxσ yy + σ 2yy V 2 E xx

[

[

]

[

]

]

[

]

(

)

9.2 From (9.4) the strain energy density, U0, is equal to σε/2 so that the strain energy is given by 2 L σ ( x) U = ∫ U 0 dV = ∫ A( x )dx V 0 2E where A is the cross-sectional area of the bar and is a function of x. With σ(x)=W/A(x) then U is W 2 L dx U= 2 E ∫0 A( x) It remains to find A(x). Linearly interpolating across the length of the bar from d1 to d2 then πd12   d 2  x   d1  d 2 d1  x  A( x ) = πy = π  +  −   = 1 +  − 1  2  L 4   d1  L  2  2 and substituting A(x) into U dx 2W 2 L U= 2 2 ∫0 Eπd1   d2  x  1 +  − 1    d1  L  The integral is seen to be of the following general form dx 1 −2 −1 ∫ (ax + b)2 = ∫ (ax + b) dx = − a (ax + b) 2

2

2

Performing the integration then U is found to be L    2W 2 L 2W 2  L 1 U= −   = Eπd12  d 2 / d1 − 1  1 + ( d 2 / d1 − 1)( x / L)  0  Eπd1d 2   as required.

From Castigiliano's second theorem then the displacement of the bar is, (9.59) ∂U 4WL δ= = ∂W Eπd1d 2 When d1=d2=d then the bar is of constant cross-section and δ is given by 4WL δ= Eπd 2 and is seen to agree with δ=WL/AE where A=πd2/4.

29

9.3 Resolving forces vertically at joint B then the force, F, in each member is P F= 2 cosϑ and the length, L, of each member is d/cosθ. Considering member BC then the strain energy is, (9.12) L σ 2  σ 2 AL U BC = ∫ U odV = ∫  xx  Adx = xx 2E  2E  V 0 with x taken along the member axis. Substituting for σxx and L then P2d U BC = 8EA cos3 ϑ Due to symmetry UAB=UBC so that the total strain energy, U, of the frame is P 2d U = U AB + U BC = 2U BC = 4 EA cos3 ϑ From Castigilano’s second theorem (9.59) the displacement, δB, at joint B is  ∂U ∂  P2d Pd δB = =  = 3 ∂P ∂P  4 EA cos ϑ  2 EA cos3 ϑ 9.4 Re-arranging (9.49) for applied load W δd 4 cos α 10 x54 cos 25o W= = = 50.4 N 2  cos 2 25o 2 sin 2 25o  2 sin 2 α  3  cos α 3 + + 8D n 8x50 x10  3 E  2261 . x10 3   G  85x10 The shear stress is given by (9.51) M (d / 2) (WD / 2) cos α (d / 2) 50.4(50 / 2) cos 25o (5 / 2) τ = x4 = = = 46.53MPa πd / 32 πd 4 / 32 π 54 / 32 and the bending stress is given by (9.52) M y (d / 2) (WD / 2) sin α (d / 2) 50.4(50 / 2) sin 25o (5 / 2) σ= = = = 43.39 MPa πd 4 / 64 πd 4 / 64 π 54 / 64 9.5 Letting ls(=nd) denote the solid length then from (9.40) we have W Gd 4 45x10 9 d 4 k= = ⇒ 2 .

5 = ⇒ D 3 = 45x10 6 d 5 3 3 δ 8D n 8 D (l s / d ) From (9.42) with τ≤120MPa and re-arranging for D τπd 3 120x10 6 πd 3 D= = = 1346 . x10 6 d 3 8W 8x35 Eliminating D from these two equations we have 45x10 6 d= = 2.07 mm 2.439 x1018 It follows that the mean coil diameter is 4

(

. x10 6 2.07 x10 −3 D = 1346

)

3

. mm = 1194

Finally, the number of coils is equal to n=

ls 50 = ≈ 24 d 2.07

30

9.6 Considering the right hand side of the ring shown in Figure 9.19 then the bending moment, M, at angle θ is seen to be M = ( R − R cos ϑ ) P = RP(1 − cos ϑ ) For pure bending the strain energy, U, of a beam is given by (9.35) which in the present case is L π M 2 dx M 2 Rdϑ U=∫ =∫ 2 EI 2 EI 0 0 substituting M we have π π 2 1 R3 P 2 U= RP(1 − cos ϑ ) Rdϑ = 1 − 2 cos ϑ + cos2 ϑ dϑ 2 EI ∫0 2 EI ∫0 Performing the integration π R3 P2  ϑ sin 2ϑ  3πR 3 P 2 U= ϑ − 2 sin ϑ + + = 2 EI  2 4  0 4 EI An application of Castigliano’s second theorem, (9.59), gives the displacement, δu, at the point at which the point load P acts ∂U 3πR 3 P δu = = ∂P 2 EI Since the total gap, δ, between the two opposing point forces is equal to twice δu then 3πR 3 P δ = 2δ u = EI as required.

To determine the value of P required to produce a total gap of δ=10mm we can re-arrange the above expression for P with I=w4/12, where w=2.5mm is the width of the square section EIδ 210x109 x39.0625x10 −12 x10x10−3 P= = = 955 .N 3πR 3 3x(45x10 −3 ) 3 Inspection of the expression for bending moment M we observe that M obtains a maximum at θ=180° (position perpendicular to the applied forces P, as expected) and is equal to Mmax=2RP=2x45x10-3x95.5=8.595N. Thus, the maximum bending stress is, (5.30)  2.5x10 −3  8.595x  2   M max y σ max = − =− = −275MPa I 39.0625x10 −12 which is less in magnitude than the tensile yield stress of σY=300MPa.

[

]

(

)

9.7 From the given beam deflection equation the maximum static deflection, δst, occurring at x=L is WL3 1xgx13 δ st = = = 0.2307 mm 3EI 3x210x109 x6.75x10−8 The impact factor F is, (9.89) 2h 2x0.5 = 1+ 1+ = 66.85 F = 1+ 1+ δ st 2.307 x10 −4 Therefore, the maximum displacement, δmax, due to the impact of the falling mass is

31

δ max = Fδ st = 15.42mm

32

Chapter 10 Solutions 10.1 Refer to §10.2.

10.2 Refer to §10.3.

10.3 Refer to §10.4.

10.4 To determine the slope at the free end of the cantilever beam then we can apply a virtual unit couple moment 1Nm at the free end of the beam, as shown in Figure Sol10.4. The virtual bending moment, Mv, is Mv =1 0≤ x≤ L with the bending moment due to the applied load P M =0 0≤ x ≤b M = P( x − b )

b≤ x≤ L

An analogous equation to (10.36) can be written for the angle of rotation, θ 1 θ= M v Mdx ∫ L EI In the present example we have 1 θ= EI

 P  x2 P  L2 − 2bL + b 2  P ( L − b) Pa 2 P( x − b)dx = = − bx  =  = EI  2 2 2 2 EI  EI  b EI  L

∫

L

b

2

P 1Nm θ

b

a

x

Figure Sol10.4. A cantilever beam with a virtual unit couple moment applied at the free end.

10.5 With a virtual unit load applied at the free end, Figure Sol10.5, then the associated bending moment is M v = −( R − R sinθ ) whereas the bending moment due to the real applied point force P is M = − PR cosθ Thus, from (10.36) the horizontal deflection, δ, is 1 π /2 v 1 π /2 δ= M Mds = − R(1 − sin θ )( − PR cosθ ) Rdθ = EI ∫0 EI ∫0 PR 3 = EI

π /2

1   sin θ + 4 cos 2θ  0

PR 3 = 2 EI

33

P 1

ds

R-Rsinθ

δ

R dθ

Rsinθ

θ

R-Rcosθ Rcosθ Figure Sol10.5. A quarter circle beam subject to a concentrated force P and virtual unit force at the free end.

10.6 To determine the vertical deflection at the free end of the beam we add a virtual unit load at the free end, see Figure Sol10.6. For the virtual unit load the bending moment is Mv = x 0≤ x≤7 For the real applied loading M =0 0 ≤ x ≤ x3 M = 10( x − 3)

3≤ x ≤ 4

M = 10( x − 3) + 20( x − 4)

4≤x≤7

From (10.36) we have

[

]

EIδ = ∫ M v Mdx = ∫ 10( x − 3) xdx + ∫ 10( x − 3) + 20( x − 4) xdx = L

4

4

7

3

4

7  x 3 3x 2  = 10 − + 10 x 3 − 55x 2 4 = 9,768.3&  2 3 3 Re-arranging for δ we have 9,768.3& δ= = 9.53mm 205x10 9 5x10 −3

[

]

(

)

34

20kN

10kN 3m

1m

1kN δ

3m x

Figure Sol10.6. A cantilever beam subject to real concentrated loads of 10kN and 20kN and a virtual concentrated unit load at the free end of the beam.

10.7 To determine the deflection at the free end of the beam we apply a unit virtual point force at this point. With x measured from the free end of the beam then for the virtual unit load the virtual bending moment, Mv, is M v = x 0 ≤ x ≤ 3a For the real applied loading system we have the following bending moments, M M = 0 0≤ x ≤a M = W ( x − a ) a ≤ x ≤ 2a M = W ( x − a ) + W ( x − 2a) 2a ≤ x ≤ 3a From (10.36) the displacement, δ, at the free end is 2a

3a

a

2a

EIδ = ∫ W ( x − a) xdx +

∫ [W ( x − a) + W ( x − 2a)]xdx

with no integral in the interval 0≤x≤a because M=0. Performing the integrations we find the desired solution 6Wa 3 δ= EI

35

Chapter 11 Solutions 11.1 From (11.27) we have 2 2 2 σ θθ − σ rr p  (b / r ) + 1 (b / r ) − 1 2 p  (b / r )  =  + =    r r  (b / a) 2 − 1 (b / a ) 2 − 1 r  (b / a) 2 − 1 Differentiating the radial stress with respect to r we have 2 ∂σ rr p 2 p  (b / r )  2 3 b r =− − = / 2   ∂r r  (b / a) 2 − 1 (b / a) 2 − 1 which is equivalent to the above equation and therefore satisfies the equilibrium equation (11.3).

(

)

11.2 With a=75mm, b=250mm and p=75MPa then from (11.27) the radial and circumferential stresses on the inner surface, r=75mm, are  (b / r ) 2 − 1   (250 / 75) 2 − 1 σ rr = − p   = −75  = −75MPa 2 2  (b / a ) − 1  (250 / 75) − 1  (250 / 75) 2 + 1  (b / r ) 2 + 1  σ θθ = p   = 90MPa  = 75 2 2  (250 / 75) − 1  (b / a ) − 1 Assuming closed ends then from (11.33) the axial stress is 75 p σ zz = = = 7.4MPa 2 (b / a) − 1 (250 / 75) 2 − 1 11.3 With a=0.5m, b=1m, pi=5MPa, po=100kPa=0.1MPa then the constants A and B in (11.20) are 2 2 b / a ) po − pi (1 / 0.5) x01 . x106 − 5x106 ( A= = = 153 . & x106 2 2 1 − (b / a ) 1 − (1 / 0.5) B=

b 2 ( po − pi )

=

(

12 01 . x106 − 5x106

) = 163 . & x10

6

1 − (b / a ) 1 − (1 / 0.5) From (11.21) the radial and circumferential stresses at the radius r=0.75m are . & x106 B 163 6 & . x10 − . MPa σ rr = A − 2 = 153 == −137 r 0.752 B 163 . & x10 6 . & x106 + σ θθ = A + 2 = 153 = 4.44 MPa r 0.752 From (11.23) the axial stress is 2 2 po (b / a) − pi 0.1(1 / 0.5) − 5 . & σ zz = = = 153MPa 2 2 1 − (b / a) 1 − (1 / 0.5) 2

2

11.4 Consider the first vessel with the boundary conditions σ rr = −45MPa at r = 75mm σ rr = 0 at r = 100mm From Lame’s equations, (11.16), we find the constants A and B to be given by A = 586 . x106 , B = 578.6x103 36

For the inner surface, in which σθθ will be maximum, then the circumferential stress is 578.6x106 B σ θθ = A + 2 = 57.86x106 + = 160.72 MPa −3 2 r 75x10

(

)

For a safety factor of 2 then the maximum allowable cylindrical stress for the second cylindrical pressure vessel will be 160.72/2=80.36MPa. For the second pressure vessel then our boundary condition (σrr=0 at the inner surface) combined with the maximum design circumferential stress give the two simultaneous equations, (11.16) B' B' σ θθ = A'+ 2 ⇒ 80.36x10 6 = A'+ 2 ri 75x10 −3

(

σ rr = A'−

B' r2

⇒

0 = A '−

(75x10

)

B' −3

+ 50x10 −3

)

2

noting the two new constants A´ and B´. Solving for A´ and B´ we have A' = 2127 . x106 , B' = 332.34x103 Re-arranging the radial stress component of Lame’s equations (11.16) for applied internal pressure p then we have at the inner surface, r=75mm   332.34 x103  B'   . x106 − p = −  A'− 2  = − 2127 = 37.8MPa = 378bar −3 2  r    75 10 x   Thus, the maximum safe working pressure for the second pressure vessel is 378bar.

(

)

11.5 We first need to determine the interference pressure, p, so that the maximum stress (σθθ) at the sleeve-collar interface does not exceed 300x106. Therefore, from Lame's equations B B σ θθ ( Rci ) = 300x106 = A + , σ rr ( Rco ) = 0 = A − 2 −3 2 49.5x10 100x10−3

(

)

(

)

Solving for A and B we find A = 59 x10 6 , B = 59 x10 4 The radial stress at r=Rci is σ rr ( Rci ) = A −

59x104 B 6 = 59 x 10 − Rci2 49.5x10−3

(

)

2

= −182MPa

Therefore, the interference pressure is 182MPa. From (11.40) the radial compression on the shaft is pR 182 x106 (50 x10 −3 ) us = − s (1 − ν ) = − (1 − 0.3) = −30.3& x10−6 m 9 E 210x10 From (11.44) the radial expansion of the collar is 2  Rco   Rci  pRci2   uc = 1 − ν + (1 + ν )   E  Rco2 − Rci2    Rci     2 49.5x10−3  182 x106 (49.5x10 −3 ) 2    100   −6 =   = 85x10 m 1 − 0.2 + (1 + 0.3) 210x109  (100x10 −3 ) 2 − (49.5x10 −3 ) 2   49.5  Therefore, the total radial interference is, (11.45)

37

δ = uc + us = 115.3& x10 −6 m 11.6 Let the inner and outer cylinders be denoted by vessels 1 and 2 as in §11.10.

From (11.52), (11.51), (11.48) and (11.27) the total radial interference is given by b 1 2 δ = u2 − u1 = σ θθ ,2 − νσ rr ,2 ) − (σ θθ ,1 − νσ rr ,1 ) = b (1 − ν )( A2 − A1 ) + (1 + ν )( B2 − B1 ) ( E Eb where pb 4 a 2 − c 2 pb 2 a 2 − c 2 , B2 − B1 = 2 = b 2 ( A2 − A1 ) A2 − A1 = 2 2 2 2 2 2 2 a −b c −b a −b c −b

[

]

(

(

)(

)

)

[

(

(

]

)

)(

Substituting (A2-A1) and (B2-B1) into δ we finally arrive at a 2 − c2 b( A2 − A1 ) pb 3 δ= = E E a 2 − b2 c2 − b2

(

(

)

)(

)

)

Re-arranging for p then the interference pressure is 2 2 2 2 210x109 100x10−6  502 − 752 1002 − 752 Eδ a − b c − b  = p= 3 3 b 502 − 1002 a2 − c2  75x10 −3

(

(

)(

)

)

(

(

)

) (

(

)(

)

) x10 

−6

= 91MPa

11.7 From (11.72) with p=100MPa, a=100mm, b=175mm and r=100mm then the circumferential stress is 3 3 p  2 + (b / r )  100  2 + (175 / 100)  σ θθ =   = 68.66MPa  = 2  (b / a) 3 − 1 2  (175 / 100) 3 − 1 

38

Chapter 12 Solutions 12.1 Letting the total strain, ε, be the sum of the elastic, εe, and plastic, εp, strains and with εp=εe/5 then ε is ε 6ε ε = εe + ε p = εe + e = e 5 5 From Hooke's law εe=σe/E and at the point of yielding then σe=σY and the total strain is 6σ Y ε= 5E Substituting this total strain into the constitutive equation E  6σ Y  σY =   200  5E 

1/ 5

Solving for σY then we arrive at σ Y = 139 . x10 −3 E =

E 719

12.2 From (12.9) the mean yield stress, σm, is σ  1 1 ε  ε σ  σ m = ∫ σ Y dε = ∫ σ Y + B ε − Y  dε = σ Y + B − Y   2 E  ε ε 0 E  12.3 To determine the empirical constants C and n of Ludwik's power law from the given engineering stress and strain data then we require relations (12.11) and (12.14) which relate the true and engineering components, that is σ = σ 0 (1 + e0 ) = 340(1 + 0.3) = 442 MPa

ε = ln(1 + e0 ) = ln(1 + 0.3) = 0.2624 Inserting these into the Ludwik power law, (12.1), we have n 442 = C (0.2624) At the point of plastic instability we know from (12.15) that σ=dσ/dε, or from Example 12.2 that σ=Cnn and ε=n for Ludwik's power law. Therefore, n=ε=0.2624 and upon substitution into the previous equation and solving for C we have 442 = C( 0.2624)

0.2624

⇒

C = 627.9

12.4 The maximum bending stress, σ, and shear stress, τmax, for the circular crosssection bar are, (5.30) and (3.11) My M ( d / 2) 32 M TR Td 16T σ= = = = = , τ max = 3 I I πd J 2 J πd 3 and substituting into (12.61) 2 2  32 M   16T  + 3 3  = σ Y2  3   πd   πd  Substituting M=cT and dividing by τmax we arrive at the required result σY = 3 + 4c 2 τ max

39

12.5 Tresca's yield criterion is given by (12.29) with the difference in principal stresses obtained from (12.31). Therefore, the yield stress is σY =

(σ

xx

− σ yy

)

2

(500 − 100) 2 + 4(100) 2

+ 4τ 2xy =

= 447MPa

To determine the value of the yield stress according to the Huber-von Mises yield criterion we will first evaluate the principal stresses, (12.30)  σ xx + σ yy   σ xx − σ yy  2 = ±   + τ xy 2 2     2

σ 1,2

 500 + 100   500 − 100  2 = ±   + 100 = 523MPa , 76MPa     2 2 From (12.46) the yield stress is 2

σ Y = σ 12 − σ 1σ 2 + σ 22 = 5232 − (523)(76) + 76 2 = 489MPa 12.6 From (12.66) the total torque, T, consisting of the elastic torque, TE, and plastic torque, TP, is  1  Rp  3  2  1  16  3  2 −3 3 3 6 T = TE + TP = πkR 1 −    = πx175x10 25x10 1 −    = 5.352kNm 3  4  25   4  R   3 with R=50/2=25mm and Rp=25-9=16mm. At first yield Rp=R and the torque is equal to π TY = kR 3 = 4.295kNm 2 and when the entire section is fully plastic then Rp=0 and the torque is equal to 2 TFP = πkR 3 = 5.727kNm 3

(

)

12.7 From (12.90) the applied bending moment, MY, at first yield is

(

)

−2 −2 bh 2 σ Y 2.5x10 4 x10 MY = = 6 6 From (12.96) the value of applied moment, M, depth of 1cm is  1. 6. x10 3  3   2   3 MY  1  y0   M= 1 −   = 2  3  h / 2   2

2

x250x10 6

. = 166 . kNm

to cause the plasticity to extend to a

 1  1x10 −2  2  1 −   = 2.29 kNm −2   3  2 x10   For the plasticity to spread throughout the entire cross-section then y0=0 at which point the bending moment is .

3 1. 6 x10 3    3MY M FP = = = 2.5kNm 2 2

40

Chapter 13 Solutions 13.1 At the nodes i and j we have, (13.23) φ i = α 1 + α 2 xi , φ j = α 1 + α 2 x j Solving for α1 and α2 we find α1 =

φ i x j − φ j xi

, α2 =

φ j − φi

L L and substituting into the interpolation function (13.23) then φ is φ j − φi x  xj − x φ x − φ j xi  x − xi  φ= i j + =  φ = Niφ i + N jφ j φ i +   L  j L L  L  where Ni and Nj are the shape functions of the element and have the following properties:

(

• • • •

)

Ni=1 at x=xi and Ni=0 at x=xj.

Nj=1 at x=xj and Nj=0 at x=xi.

The sum of Ni and Nj is always equal to unity for x within the range xi≤x≤xj.

The shape functions are of the same order as the interpolation function.

With xi=2 and xj=6 then the L=xj-xi=4. With φi=10 at xi=2 and φj=20 at xj=6 then from the above interpolation function at x=3 we have Ni=3/4 and Nj=1/4 with φ=12.5. The value of φ at x=3 is seen to be a linear interpolation of the nodal values.

13.2 From (13.33) the D matrix for plane stress is 0 1 ν   1 0.28 0  E   9 0 1 0  [ D] = 1 − ν 2 ν 1  = 75.95x10 0.28   0 0 (1 − ν ) / 2   0 0 0.36 and from (13.36) the D matrix for plane strain is ν 0 0   1 − ν 0.72 0.28 E   9 0 0  [ D] = 1 + ν 1 − 2ν  ν 1 − ν  = 124.29 x10 0.28 0.72 ( )( ) 0  0 0 0 0.22  (1 − 2ν ) / 2  From (13.32) the stress vector for plane stress is σ xx  ε xx   1 0.28 0  60 6.26          −6 9 0  80x10 = 7.35x10 6 σ yy  = [ D]ε yy  = 75.95x10 0.28 1 τ  γ   15   0 0 0.36 55  .   xy   xy  and similarly for plane strain ε xx  σ xx  0  60 .  815 0.72 0.28          9 −6 6 σ yy  = [ D]ε yy  = 124.29 x10 0.28 0.721 0  80x10 = 9.25x10 τ  γ   15   0 0 0.22 55  .   xy   xy  13.3 The cross-sectional areas of elements 1 and 2 are πD12 π 80 2 πD22 π 50 2 A1 = = = 5,027 mm 2 , A2 = = = 1,964 mm 2 4 4 4 4 The element stiffness matrices are, (13.44)

41

(

)

(

)

5,027 200x10 3  1 −1  1 −1 = = 2010.8x10 3     N / mm 500  −1 1   −1 1 

[K ] 1

1964 120x10 3  1 −1 1 −1 3 336 7 10 = .

K = x  −1 1  −1 1  N / mm 700     The structure stiffness matrix and force vector are, using node ordering (1,2,3) −2010.8 0   0   2010.8   [ K ] s = −2010.8 2018. + 336.6 −336.7 N / mm , {F } s =  0 x103 N 185  0 −336.7 336.7    Incorporating the boundary condition u1=0 then the structure system of equations to be solved for U is 0   0  0 + R1   2010.8 −2010.8     3 10 −2010.8 2347.5 −336.7 u2  =  0 x10 3  0 −336.7 336.7  u3   −185  where R1 is the reaction at node 1. Performing row multiplications we have 10 3 ( −2010.8u2 ) = R1

[ ] 2

10 3 ( 2347.5u2 − 336.7u3 ) = 0

10 3 ( −336.7u2 + 336.7u3 ) = −185x10 3 Solving these equations we find u1=0, u2=-0.092mm and u3=-0.6415mm. Both u2 and u3 are negative and u3<u2 as expected. An additional check illustrates that R1 is equal and opposite to the applied force of -185kN. From (13.30) the element strains are 1 1 ε 1xx = ( − u1 + u2 ) = (0 − 0.092) = −184x10−6 L 500 1 1 ε 2xx = ( − u2 + u3 ) = (0.092 − 0.6415) = −785x10 −6 L 700 From (13.32) the element stresses are σ 1xx = E1ε 1xx = 200x10 3 −184 x10 −6 = −36.8N / mm 2 σ 2xx = E 2 ε 2xx

( ) = 120x10 ( −785x10 ) = −94.28N / mm 3

−6

2

and are found to agree exactly with the theoretical estimates P 185x10 3 σ 1xx = =− = −36.8N / mm 2 A1 5,027 σ

2 xx

P 185x10 3 = =− = −94.28N / mm 2 A2 1,964

13.4 The stiffness matrices of elements 1 and 2 are, (13.81)  0.25   0.25   −0.433 0.75    0.75 sym. sym.  , K 2 =  0.433  K1 =   −0.25 0.433 0.25   −0.25 −0.433 0.25       0.433 −0.75 −0.433 0.75 −0.433 −0.75 0.433 0.75 Inserting into the global stiffness matrix we find the structure system of equations

[ ]

[ ]

42

 0   0   0.25  0   0  −0.433 0.75        u     − 0 .

25 0 .

433 0 .

25 sym .

0 2 346.41x106    =   0 15 .  v2  −1000  0.433 −0.75  0   0   0 0 −0.25 −0.433 0.25      0 −0.433 −0.75 0.433 0.75  0   0   0 with node ordering 1, 2 and 3. Multiplying rows 3 and 4 we find 0.25u2 = 0 346.41x106 (15 . v2 ) = −1000 -6 leading to u2=0 and v2=-1.9245x10 m; with u2=0 due to symmetry. Similarly, the axial strain, stress and force components in elements 1 and 2 are equivalent and so we will consider only element 1. The axial strain is, from (13.83) u1  v  1  1 1 ε = [ − C − S C S ]  L1 u2  v 2 

{ }

 0   0  1   −6 -0.5 0.866 0.5 -0.866] . x10 −6 = [ x10 = 14433 0 .

11547    −19245 .

The axial stress and force are σ 1 = Eε 1 = 0.2887MPa , F 1 = σ 1 A = 577.32N From the free-body diagram in Figure 13.28 we observe that 2Tcos30°=1kN and therefore T=577.35N which agrees exactly with the finite element prediction. The displacement v2 is found by an application of Castigliano’s second theorem as in Exercise 9.4 Pd v2 = − = −19245 . x10 −6 m 3 2 EA cos ϑ which, again, is exact agreement with the finite element estimate.

13.5 The cross-sectional area, A, and perimeter, P, for both elements are 25x10-6m2 and 20x10-3m respectively. Letting elements 1 and 2 have nodes (1,2) and (2,3) respectively then the stiffness matrix for element 1 has contributions due to conduction and perimeter convection, (13.118) . −4.9167 k A  1 −1 hPL1 2 1  516 −3 K 1 = xx  + =  x10    .

516 L1 −1 1  6 1 2 −4.9167  with the force vector due to {Fh} only, (13.119) hT PL 1 0 F1 = ∞ 1   =   2 1 0 Element 2 is identical to element 1 except that it also experiences end-convection through node 2 so that the stiffness matrix of element 3 is, (13.118)

[ ]

{ }

43

−4.9167  . −4.9167 0 0  516 −3 = x10 −3 + hA    x10 516 .

0 1 − 4 .

9167 5185 .      Similarly, adding the end-convection term to the force vector of element 2 we have 0 0 0 F 3 =   + hT∞ A  =   0 1 0 Assembling the element contributions into the structure stiffness matrix and force vector we find −4.9167 0  T1  0 .  516     −3  10 −4.9167 516 −4.9167 T2  = 0 . + 516 .     −4.9167 5185 .  0  T3  0 with node ordering 1 to 3. We have a prescribed temperature of 100oC at node 1 which results in a non-homogeneous boundary condition. The stiffness matrix and force vector are modified by first setting all non-diagonal terms in the first row and column of the stiffness matrix to zero. Also, the term (-4.9167)x100oC=-491.67 on the left hand side of the second equation is transposed to the right hand side as +491.67.

The resulting systems of equations is now given by 0 0  T1   100  1     −3  .  −4.9167 T2  = 49167 10 0 10.32 0 −4.9167  T3   0  .

5185 The second through to third equations are now solved in the usual manner, with the solution vector given by {T}={100,86.87,82.34}.

[ K ] = −4.9167 3

516 .

{ }

13.6 All three elements experience no perimeter convection (P=0) with element 3 experiencing convection at node 4. Assuming a unit cross-sectional area for all three elements then the stiffness matrix and force vector for element 1 are, (13.118) and (13.119)  1 −1 k A  1 −1 0 . , F1 =   = 01 K 1 = xx    L1  −1 1   −1 1  0 and similarly for element 2 k A  1 −1  1 −1 0 . , F2 =   = 01 K 2 = xx    L2 −1 1   −1 1  0 Element 3 consists of an additional end-convection term . −01 . 0  0  k A  1 −1 0 0  01 K 3 = xx  + hA = , F 3 = hT∞ A  =      . 501 . L3  − 1 1  0 1 −01 1 −250 Assembling the elements into the structure stiffness matrix and force vector gives, with node ordering 1 to 4 . −01 .

0 0   T1   0   01  −01 .

0  T2   0   . 0.2 −01  =   0 −01 . 0.2 −01 .  T3   0    0 −01 . 501 .  T4  −250  0

[ ]

[ ]

{ }

[ ]

{ }

{ }

Incorporating the prescribed boundary condition T1=25oC then the system of equations is modified as follows

44

0 0  T1   21  1 0 0 0.2 −01 .

0  T2   2.5    =  0 −01 . 0.2 −01 .  T3   0    . 501 .  T4  −250 0 0 −01 where the right hand side of the first equation is set to 25oC. The term (-0.1)x25oC=2.5 on the left hand side of the second equation is transposed to the right hand side as +2.5. Solution of the system of equations yields {T}={25,15,5,-5}. From Fourier's law the heat flux for an element of length L and nodes i and j is dT  1 1   Ti  q x = − k xx = k xx [ B]{T} = − k xx −   dx  L L  Tj  Evaluation of qx for all three elements reveals that the heat flux is constant and equal to Qx=qxA=qx=1 for all three elements.

13.7 The stiffness matrices for all three elements are equivalent and given by (13.136) Ak xx  1 −1 K1 = K2 = K 3 = L  −1 1 

[ ] [ ] [ ] . x10 )1x10 (31416 =

−2

 1 −1  1 −1 = 9.425x10 −4     0.33& −1 1   −1 1  Since there are no sources or sinks and no applied surface flow rates then both Q and q are equal to zero in (13.138) so that the element force vectors are 0 F1 = F 2 = F 3 =   0 Assembling the element components we have the following structure system of equations for unknown fluid heads p1, …, p4  1 −1 0 0   p1  0 −1 2 −1 0   p  0   2  =   9.425x10 −4   0 −1 2 −1  p3  0    0 0 −1 1   p4  0 02

{ } { } { }

Incorporating the non-homogeneous boundary conditions of p1=0.2m and p4=0.1m in a similar manner to that discussed in Example 13.4 we arrive at . x10 −4  1 0 0 0  p1  1885  0 2 −1 0  p   . x10 −4   2  1885 −4   9.425x10  =  0 −1 2 0  p3  9.425x10 −5      −5 0 0 0   p4  9.425x10  Solving the second and third equations for p2 and p3 then the solution vector is {P}={0.2,0.16,0.13,0.2}. From (13.130) the element velocity for an element of length L with nodes i and j is  1 1   pi  v x = − k xx { g} = − k xx [ B]{ P} = − k xx  −    L L   p j  For example, for element 1 we have  1 v x = −1x10 −2  −  0.33&

1   0.2  −3  &  = 1x10 m / s  & .  0.33  016 45

with equivalent velocities for elements 2 and 3.

46

Chapter 14 Solutions 14.1 From §1.11 the in-plane strains are ∂u ∂v ε xx = = 2axy 2 , ε yy = = bx 3 , γ ∂x ∂y Differentiating the strains ∂ 2 ε xx ∂ = (4axy ) = 4ax , 2 ∂y ∂y ∂ 2γ

xy

=

1  ∂u ∂v  1 2 2  +  = 2ax y + 3bx y 2  ∂y ∂x  2

∂ 2 ε yy ∂x

2

(

=

)

∂ 3bx 2 = 6bx ∂x

(

)

1 ∂ 1 (4axy + 6bxy) = (4ax + 6bx)  ∂x∂y ∂y  2  2 Upon substituting these into the compatibility equation, (14.21), we observe that u and v are compatible. xy

=

14.2 Differentiating φ we find ∂φ = 2 Ax + By , ∂x ∂φ = 2Cy + Bx , ∂y

∂ 2φ ∂ 3φ A , = 2 =0 , L ∂x 2 ∂x 3 ∂ 2φ ∂ 3φ = 2C , =0 , L ∂y 2 ∂y 3

∂ 2φ =0 , L ∂x 2 ∂y Thus, we observe that φ satisfies ∇4φ=0 since all terms of φ are less than power 4.

The stresses follow immediately from the Airy stresses, (14.46) ∂ 2φ ∂ σ xx = 2 = ( Bx + 2Cy ) = 2C ∂y ∂y ∂ 2φ ∂ = (2 Ax + Bx ) = 2 A ∂x 2 ∂x ∂ 2φ ∂ = − (2 Ax + By ) = − B τ xy = − ∂x∂y ∂x It follows that φ provides a solution for a plate subject to uniform stresses along its sides of σxx=2C, σyy=2A and τxy=-B. From the Hookian equations for a state of plane stress, (14.42), the in-plane strains are 1 2 ε xx = σ xx − νσ yy = (C − νA) E E 1 2 ε yy = σ yy − νσ xx = ( A − νC ) E E 2(1 + ν ) 2(1 + ν ) B γ xy = τ xy = − E E Integrating the strains we find the displacements 2 2 u = ∫ ε xx dx = (C − νA)dx = ( C − νA) x + f ( y ) E E 2 2 v = ∫ ε yy dy = ( A − νC )dy = ( A − νC ) y + g ( x ) E E σ yy =

[

]

[

]

47

where the functions f(y) and g(x) are determined from the boundary conditions.

14.3 Let the resultant stress be S(Sx,Sy,Sz) with S 2 = S x2 + S y2 + S z2 This resultant stress consists of both normal, SN, and shear, SS, components S 2 = S N2 + S S2 If the direction cosines of ABC are l=cosα, m=cosβ and n=cosγ then S N = lS x + mS y + nS z where (Sx,Sy,Sz) in terms of the coordinate components are S x = lσ xx + mτ xy + nτ xz S y = lτ yx + mσ yy + nτ yz S z = lτ zx + mτ zy + nσ zz Substituting Sx, Sy and Sz into SN we have

(

S N = l 2σ xx + m 2 σ yy + n 2 σ zz + 2 lmτ xy + mnτ yz + lnτ zx

)

, S S = S 2 − S N2

Substituting σij and (l,m,n) into Sx, Sy and Sz we have 1 30 Sx = (10 + 5 + 15) = = 17.32 3 3 1 50 Sy = (5 + 20 + 25) = = 28.87 3 3 1 70 Sz = (15 + 25 + 30) = = 40.41 3 3 with S equal to 2

2

2

 30   50   70  S =   +   +   = 52.6  3  3  3 The normal and shear stresses are 1  30  1  50  1  70  2 2 SN =   = 50 , S S = 52.6 − 50 = 16.33  +  + 3  3 3  3 3  3 The direction of SN acts normal to plane ABC and is therefore defined by (l,m,n)=(1/√3,1/√3,1/√3) with α=β=γ=cos-1(1/√3)=54.74°. A useful check is to ensure that l2+m2+n2=1 which is the case. The direction of SS acts parallel to the plane ABC and let it be defined by (ls,ms,ns) where ls=cosαs, ms=cosβs and ns=cosγs. The components (Sx,Sy,Sz) can now alternatively be defined as, resolving SN and SS S x = S N cos α + S S cos α s = lS N + l s S S S y = S N cos β + S S cos β s = mS N + ms S S S z = S N cos γ + S S cos γ s = nS N + n s S S from which it follows ls = ( S x − lS N ) / S S = −

(

)

1 2

ms = S y − mS N / S S = 0 ns = ( S z − nS N ) / S S =

1 2

48

confirming that ls2 + ms2 + ns2 = 1 . The angles αs, βs and γs are αs=cos-1(-1/√2)=135°, βs=cos-1(0)=90° and γs=cos-1(1/√2)=45°. Finally, we can check that the two direction vectors of SN and SS are orthogonal by ensuring that the dot product of (l,m,n) and (ls,ms,ns) is equal to zero, that is (1/√3,1/√3,1/√3).( -1/√2,0, 1/√2)=0.

14.4 From (14.100) the C matrix is 0.3 1 − 0.3 210x10 9  [C] = 1 + 0.3 1 − 2x0.3  0.3 1 − 0.3 ( )( ) 0 0 

0  0.7 0.3 0    9 0  = 404 x10  0.3 0.7 0   0 0 0.4  (1 − 2x0.3) / 2 T -6 With [ε] =[-19,64,3]x10 then the stress vector is σ xx  0.7 0.3 0  −19  2.38        9 −6 . MPa σ yy  = 404 x10  0.3 0.7 0   64 x10 = 158 τ  0.48  0 0 0.4  3     xy 

14.5 From (14.190) the shear stresses τxz and τyz are ∂φ ∂φ 3Gθ  2ax  τ xz = = − Gθy[1 − x ] , τ yz = − = − x2 + y2   2a  3 ∂y ∂x  The centroid of the triangle is at (x,y)=(0,0) and the three corners are at (2a/3,0), (a/3, a / 3 ) and (-a/3,- a / 3 ). Substituting these coordinates into τyz above we observe that τyz is equal to zero at the centroid and three corners.

14.6 Since the hole in the plate is circular, a/b=1, then from (14.212) the stress concentration factor is Kt=3. The maximum of σP that can be applied to the plate is therefore σ 300MPa = 100MPa σ P (max) = Y = Kt 3 From (14.210) the σyy stress at a distance x=a/2=6.25mm from the notch root (notch root radius of ρ=b2/a=a=12.5mm) is ρ 12.5 σ yy = σ P K t = 100x10 6 x3 = 173MPa ρ + 4x 12.5 + 4(6.25) 14.7 From (7.19) the principal stresses are given by  σ xx + σ yy   σ xx − σ yy  2 = ±   + τ xy 2 2     2

σ 1,2

With the stresses given by (14.223) for the semi-infinite Boussinesq wedge then the three main terms in the above expression are σ xx + σ yy f = − 4 x 3 + xy 2 2 πr

(

)

2  σ xx − σ yy   f  2 3  =  4  xy − x   πr  2   2

(

)

2

2

 f  τ 2xy =  4  4 x 4 y 2  πr 

49

Noting that the square root term reduces to 2 2  σ xx − σ yy   f  2 2 2 2   + τ xy =  2  x x + y  πr  2  

(

)

and substituting into the principal stresses we finally arrive at the required result 2 fx σ1 = 0 , σ2 = − 2 πr

50

Chapter 15 Solutions 15.1 Refer to §15.2 for a discussion on equivalent stress and strain.

15.2 Refer to §15.2 for a discussion on the constancy of volume condition and its application to illustrate that Poission’s ratio is equal to ½ for incompressible materials.

15.3 For a thin-walled pressure vessel with no shear stresses then the circumferential, axial and radial coordinate stresses σθθ(=pd/2t), σzz(=σθθ/2) and σrr are equivalent to the principal stresses σ1, σ2 and σ3; where p is the internal pressure, d is the mean diameter and t is the wall thickness. From (15.1) the equivalent stress is 1 3 σ= (σ θθ − σ zz )2 + σ 2zz + σ θθ2 = 2 σ θθ 2 The equivalent plastic strain is given by (15.2) with the plastic strain increments dε 1p , dε 2p and dε 3p determined from the Levy-Mises flow rule. The deviatoric component of σθθ is 1   σ + σ zz + σ rr  2  ' σ θθ = σ θθ −  θθ  = σ θθ − (σ zz + σ rr )   3 3 2  The plastic strain increments are, from (15.22) 2λ  1  λ p dε 1p = dε θθ σ σ σ = − + ( ) θθ zz rr  = 2 σ θθ 3  2 2λ  1  dε 2p = dε zzp = σ zz − (σ θθ + σ rr ) = 0  3  2  λ 2λ  1  σ rr − (σ θθ + σ zz ) = − σ θθ  3  2 2  with the equivalent plastic strain given by, (15.2) 2 2 2 p p dε p = dε θθ − dε zzp + dε zzp − dε rrp + dε θθ − dε rrp 3 dε 3p = dε rrp =

(

) (

) (

)

2

15.4 In the absence of shearing stresses, τrz, then the principal stresses are equal to the coordinate stresses σθθ, σrr and σzz. With the external pressure po=0 and the internal presure pi=p then we have from Lame's equations, (15.34) σ rr = − p σ θθ

 k 2 + 1 p =  2  ; k =b/a  k − 1

σ zz = 0 (open ends) Let us examine both the Tresca and Huber-von Mises yield criteria starting with Tresca's criterion.

Tresca's Yield Criterion Since the coordinate stresses are equivalent to the principal stresses, (15.33) σ θθ − σ rr = σ Y

51

Substituting the radial and circumferential stresses we find that first yield occurs when σ  1 p = Y 1 − 2  2  k  Re-arranging for k  (σ Y / p)  k=   (σ Y / p) − 2  With p=100MPa and σY=250MPa then k=√5=2.24.

1/ 2

Huber-von Mises Yield Criterion From the Huber-von Mises yield criterion

2 2 (σ rr − σ θθ ) + (σ θθ − σ zz ) + (σ zz − σ rr )2 = 2σ Y2

With σzz=0 for open ends then this equation reduces to 2 σ rr2 − σ rr σ θθ + σ θθ = σ Y2 Substituting the radial and circumferential stresses and re-arranging for p we arrive at k2 − 1 p =σY 3k 4 + 1 Re-arranging for k we arrive at the following quadratic equation with unknown k2 3 p 2 − σ Y2 k 4 + 2σ Y2 k 2 + p 2 − σ Y2 = 0

[

]

(

)

Solving, then k is given by

(

)(

)

1/ 2

 −σ 2 ± σ 4 − 3 p 2 − σ 2 p 2 − σ 2  Y Y Y Y  k= 2 2   3p − σ Y   With p=100MPa and σY=250MPa then the two solutions of k are k=0.69 and k=1.83.

Since k>1 then k=1.83 and is approximately 22% less than the Tresca prediction and results in approximately 70% difference in cross-sectional area.

15.5 The required applied internal pressure, p, to produce an elastic-plastic boundary to a depth of c=70mm can be found by setting r=a and p=-σrr in (15.40)   70  1   70  2     c  1   c  2  p = −σ rr = σ Y ln  + 1 −     = 300ln  + 1 −     = 177.45MPa   50  2   100      a  2   b    The fully plastic condition is reached when c=b and the required pressure, pY, is  b  100  pY = σ Y ln  = 300 ln  = 207.94MPa  a  50  15.6 From the material power law

n −1 dσ = nC ε P P dε and upon substitution into the plastic instability condition (15.48) we have n −1 n n nC ε P = 3σ = 3C ε P ⇒ εP = = 0.26 3 From (15.49) the mean radius and wall thickness at the point of plastic instability are P P r = r0 e 3ε / 2 = 0.45e 3 (0.26)/ 2 = 0.56m , t = t 0 e − 3ε / 2 = 1e − 3 ( 0.26)/ 2 = 0.8mm

( )

( )

( )

52

15.7 From Exercise 15.6 the equivalent plastic strain remains the same. From (15.57) the mean radius and wall thickness at the point of plastic instability are P P r = r0 e ε / 2 = 0.45e 0.26 / 2 = 0.51m , t = t 0 e − ε / 2 = 1e −0.26/ 2 = 0.88mm

53

Chapter 16 Solutions 16.1 From (16.5) and (16.6) the constants a, b and c and area A are ai = 8 , bi = −2 , ci = −1 a j = −2 , b j = 3 , c j = −1 a k = −1 , bk = −1 , ck = 2 A = 2.5 From (16.8) the shape functions at point p are 1 3 1 Ni = , N j = , Nk = 5 5 5 From (16.11) the displacement vector at point p is  ui  v   i 0 Nk 0  u j  u  N i 0 N j  =  = 0 N k  v j  v   0 N i 0 N j u k    v k   1  1   1 / 5 0 1 / 5 0 3 / 5 0    3  2  =   =    0 1 / 5 0 1 / 5 0 3 / 5 4 2.2 2   2 16.2 Denoting nodes (1,2,3) by (i,j,k) to assist in the use of the required formulae then from (16.5) and (16.6) the constants a, b and c and area A are ai = 9 , bi = −2 , ci = −1 a j = −3 , b j = 2 , c j = −1 a k = −2 , bk = 0 , c k = 2 A=2 From (16.8) the shape functions are 1 1 1 N i = [9 − 2 x − y ] , N j = [ −3 + 2 x − y ] , N k = [ −2 + 2 y ] 4 4 4 From (16.11) the [N] matrix is 0 N 0 N 0  N [ N ] =  0i N 0 j N 0k N  i j k  From (16.17) the [B] matrix is −2 0 2 0 0 0 1 [ B] = 4  0 −1 0 −1 0 2  −1 −2 −1 2 2 0 From (13.33) the [D] matrix is

54

 1 0.3 0  [ D] = 219 x10 0.3 1 0  N / mm2  0 0 0.35 From (16.17) the strain vector is 1  1  ε xx  −2 0 2 0 0 0    0.5  2  1   {ε } = ε yy  = [ B]{U } =  0 −1 0 −1 0 2  =  1.5 x10−3 4 3 2.25 γ    − − − 1 2 1 2 2 0 xy      4     5  From (16.18) the stress vector is σ xx   1 0.3 0   0.5   209       3 −3 {σ } = σ yy  = [ D]{ε } = 220x10 0.3 1 0   1.5 x10 =  363 N / mm 2 τ  173.25  0 0 0.35 2.25    xy  3

16.3 With xi=3, yi=3, xj=4, yj=1, xk=5 and yk=3 then the constants a, b and c and the area, A, of the element are ai = 7 x10 −6 m 2 , bi = −2 x10 −3 m , ci = 1x10 −3 m a j = 6x10 −6 m 2 , b j = 0m , c j = −2 x10 −3 m a k = −9 x10 −6 m 2 , bk = 2 x10 −3 m , ck = 1x10 −3 m A = 2x10 -6 m 2 From (16.17) the [B] matrix is bi 0 b j 0 bk 0   −2 0 0 0 2 0  1   [ B] = 2 A  0 ci 0 c j 0 ck  = 250 0 1 0 −2 0 1 ci bi c j b j c k bk   1 2 −2 0 1 2    and from (13.33) the [D] matrix is 0 1 ν   1 0.3 0  E   11  . x 0.3 1 0  0 [ D] = 1 − ν 2 ν 1  = 2 1978 10    0 0 (1 − ν ) / 2  0 0 0.35 The stiffness matrix is, (16.34) [ K ] = [ B]T [ D][ B]tA = 1  −2 0 0 1 2   0 2 0  1 0.3 0   −2 0 0 0 0 −2     11 = 250 0 250 0 1 0 −2 0 1(2 x10 −3 )(2 x10 −6 ) (2.1978x10 ) 0.3 1  0 −2 0   0  1 2 −2 0 1 2  0 0.35 2 0 1   1 2 0 Performing the multiplications we find

55

 4.35  01 .

2.4   0.7 14 .

14 . sym. [ K ] = 5.4945x107  12 0 4 −2  . −4.35 −01 . −0.7 −12 .  . −0.4 −0.4 −2  −13 noting that the matrix is symmetric (Kij=Kji), the principal non-zero (Kii>0) and the sum of all terms in either a

       4.35  13 .

2.4 terms are all positive and row or column are zero

( ∑ j =1 Kij = 0 ).

6

To determine the force vector we require just the contribution due to the edge pressure. From (16.42) the normal pressure term for edge (i,k) is  px   0  p     y  −200 L t0  0  { F } = {F } pressure = ik  0  =   2    0   px   0       p y  −200 16.4 Measuring the length coordinate ξ from node 1 then ξ=3/4 for point p. From (16.60) the shape functions are 2 1  3  3 2 N 1 = 2ξ − 2ξ + 1 = 2  − 2  + 1 = −  4  4 8 2

3 3  3 N 2 = 2ξ − ξ = 2  − =  4 4 8 2

2

3  3  3 N 3 = 4ξ − 4ξ 2 = 4  − 4  =  4  4 4 The displacement u at point p is, (16.59)  1  3  3 u = N 1 u1 + N 2 u2 + N 3 u3 =  −  2 +   2.25 +   2.55 = 2.51  8  8  4 16.5 To determine the element shape functions (16.84) we first require the area coordinates of point p. From (16.6) we find 5 1 1 3 A= , A1 = , A2 = , A3 = 2 2 2 2 From (16.43) the area coordinates of p are A A A 1 3 1 ξ1 = 1 = , ξ2 = 2 = , ξ3 = 3 = A 5 A 5 A 5 confirming that ξ3=1-ξ1-ξ2. Substituting ξ1, ξ2 and ξ3 into (16.84)

56

3  1 1 N 1 = (2ξ 1 − 1)ξ 1 =  2 − 1 = −  5 5 25 3  1 1 N 2 = (2ξ 2 − 1)ξ 2 =  2 − 1 = −  5 5 25  3 1 3 N 3 = (2ξ 3 − 1)ξ 3 =  2 − 1 =  5  5 25 1 1 4 N 4 = 4ξ 1ξ 2 = 4 ⋅ = 5 5 25 1 3 12 N 5 = 4ξ 2 ξ 3 = 4 ⋅ = 5 5 25 1 3 12 N 4 = 4ξ 1ξ 3 = 4 ⋅ = 5 5 25 confirming that ΣNi=1. The displacement (u,v) at point p is u1  v  1 .  u  N 1 0 L N 6 0    315 M =      =  .  v   0 N 1 L 0 N 6  u   38  6 v 6  16.6 From (16.60) the element shape functions are N 1 = 2ξ 2 − 3ξ + 1 , N 2 = 2ξ 2 − ξ , N 3 = 4ξ − 4ξ 2 so that the derivatives with respect to ξ are ∂N 1 ∂N 2 ∂N 3 = 4ξ − 3 , = 4ξ − 1 , = 4 − 8ξ ∂ξ ∂ξ ∂ξ From (16.70) the Jacobian matrix is ∂N ∂N ∂x ∂N [ J ] = ∂ξ = ∂ξ1 x1 + ∂ξ2 x2 + ∂ξ3 x3 = (4ξ − 3)1 + (4ξ − 1)5 + (4 − 8ξ )3 = 4 16.7 From (16.53) 1 1

n

0 0

i =1

I y = ∫∫ x dA = ∫ ∫ f (ξ 1 , ξ 2 ) J dξ 1dξ 2 = 2 A∑ wi f (ξ 1 , ξ 2 ) 2

A

noting that |J|=2A. With reference to Table 16.2 the x coordinate at the three integration points a, b and c are 3  2 1  1  1 x a = ∑ ξ ai xi = 0  + 3  + 0  =  3 2  6   6 i =1 3  1  1 1  2 xb = ∑ ξ bi xi = 0  + 3  + 0  =  3  6  6 2 i =1 3  1  1  2 x c = ∑ ξ ic x i = 0  + 3  + 0  = 2  6  3  6 i =1 With A=6 and wi=1/6 for all three integration points then Iy is  1  1  2 1  1  2 1 2  I y = 2 x 6   +   + 2  = 9 6  2 6   6  2 

57

An exact evaluation of Iy can be performed by referring to Example 2.1. Considering an elemental strip dy at a distance y from the x-axis then hx y =h− b With dA=ydx then Iy is b b hx  h b hb 3  I y = ∫∫ x 2 dA = ∫ x 2 ydx = ∫ x 2  h −  dx = ∫ bx 2 − x 3 dx = 0 0 A  b b 0 12 3 With b=3 and h=4 then Iy=4(3) /12=9 which agrees exactly with the evaluation of Iy using Gaussian integration.

(

)

58

Chapter 17 Solutions 17.1 Refer to §17.2 for a discussion of the stress intensity factor, §17.3 for a discussion of the T-stress and §17.5 for a discussion of several well-known stress intensity and T-stress expressions.

17.2 For a symmetrically cracked circular hole in a plate with far-field uniform loading consider the two limits of a→0 and a→∞ in the given expression. For a→0 we have 2 .4  R    → 1 and K I → 3.365σ πa  R + a and for the limit a→∞ we have 2 .4  R    → 0 and K I → σ πa  R + a Thus, as the cracks grow beyond the influence of the circular hole then KI tends to the case of a centrally cracked plate.

The case of a→0 requires further analysis. When a crack is short (a<<R) then the crack is approximately equivalent to an edge crack in a semi-infinite plate but with the applied stress modified by the circular hole stress concentration factor of 3σ. In this case KI is given by K I = 11215 .

(3σ ) πa = 3.3645σ πa which is essentially equivalent to the given equation by letting a→0. Similarly, for the T-stress as a→0 we have T = −0.5258( 3σ ) = −15774 . σ 17.3 Refer to §17.8 for a discussion of plane strain fracture toughness and its experimental determination.

17.4 The total crack length is 2a=25mm so that a=12.5mm. Neglecting finite plate effects with Y=1 then for case i) we have

(

)

K IC = σ f πa = 220 π 12.5x10−3 = 43.6MNm−3/ 2 For case ii) with α=1/π for Irwin's model then from (17.91) KIC is 2  σ f   K IC = σ f πa 1 + απ    = 49.1MNm−3/ 2  σ Y    The discrepancy between the elastic and plasticity correction estimates increases as σf/σY increases although the small-scale yielding assumption becomes increasingly invalid.

17.5 The rotor rotational speed in radians per second is  2π  ω = 12 x10 3   = 1,257 radians / second  60  At the critical crack length then KI=KIC and a=acritical

59

K IC

ω 2 R 2  3 − 2ν  = Yρ   πa critical 8  1− ν 

1,257 2 (0.35)  3 − 2(0.33)  ⇒ 85x10 = 0.55 8x10   πa critical 8  1 − 0.33  Solving for acritical we have a critical = 16.7 mm 6

(

3

2

)

Comparing KI with ∆K = Y∆σ πa then we observe that the cyclic stress is 2 − 2( 0.33)  ω 2 R 2  3 − 2ν  3 1,257 (0.35)  3 8 x 10 . x10 6 MPa =   = 6755     8 1− ν 8  1 − 0.33  Since when operating the rotational speed is constant then the only way in which the stress cycles is from the starting-stopping of the rotor. Using this assumption then ∆N will provide us with the number of times that the rotor can be run up to speed. From (17.159) the number of cycles is  ai1− m/ 2 − a 1f− m/ 2  1 ∆N = =  C∆σ mπ m/ 2 Y m  m / 2 − 1   0.012 1−3/ 2 − 0.017 1− 3/ 2  1 =  = 250 cycles 3 3  0.5 4.11x10 −11 ( 6755 . ) π 3/ 2 (0.55)   2

∆σ = ρ

17.6 Refer to §17.11 for a discussion on long and short fatigue cracks. From Hobson’s growth law (17.164) with α=0 and re-arranging for the number of cycles we have af 1 a f da 1 ∆N = ∫ = − ln(d − a ) a i C ai d − a C With ∆σ=638MPa then C is found to be 1 11.14 C = 164 . x10 −34 (638) = 2.887 x10 −3 ; = 346 C With the long crack condition ∆Kth=6MPa√m and Y=2/π then from ∆K = Y∆σ πa the total threshold crack length is

[

]

2(6) ∆K 2 a th = 2 2 th 2 = = 138x10 −6 m = 138µm ≡ d 2 πY ∆σ 2  2 π   (638) π 2

With ai=10µm and af=84% of d which is equal to 116µm then the number of cycles is ∆N = −346[ln(138 − 116) − ln(138 − 10)] = 609 cycles 17.7 For plane strain conditions then (κ+1)/8µ=(1-ν2)/E=1/E′. Therefore, from (17.183) the mode I stress intensity factor is KI =

(

)

E ' J = 231x10 9 2.4 x10 3 = 2355 . MPa m

60

Chapter 18 Solutions 18.1 Refer to sections 18.1 to 18.8.

18.2 From (18.7) and taking logarithms we have ln ε& s = ln A + n ln σ From a ln ε& s versus lnσ plot with two 1 (σ=100MPa) and 2 (σ=200MPa) then the slope is −4 −6 ln ε& s 2 − ln ε& s1 ln 7.6x10 − ln 4.4 x10 = 7.4328 n= = ln σ 2 − ln σ 1 ln 200 − ln 100 Assuming n to be an integer and equal to 7 then the constant A is, (18.7) ε& s 4.4 x10 −6 A= n = = 4.4 x10 −20 7 σ 100 in which the first test point has been used.

(

)

(

)

18.3 From (18.10) and taking logarithms we have Q RT From a ln ε& s versus 1/T plot with two points 1 (T=160K) and 2 (T=200K) then the slope is . x10 −3 − ln 7.24 x10 −6 Q ln ε& s 2 − ln ε& s1 ln 819 = = −5,625 − = 1 / 200 − 1 / 160 R 1 / T2 − 1 / T1 from which Q is found to be Q = 198 . x5,625 = 11kcal / mol Finally, solving for D at T=160K we have Q 11x10 3 ln D = ln ε& s + = ln 7.24 x10 −6 + ⇒ D = 8.7 x10 9 RT 198 . (160) ln ε& s = ln D −

(

)

(

(

)

)

18.4 For the equivalents strains in the two tests then t1 e − Q / RT1 = t 2 e − Q / RT2 Taking logarithms we have ln t1 + ( − Q / RT1 ) = ln t 2 + ( − Q / RT2 ) and re-arranging for Q we arrive at the required result RT1T2 Q= [ln t1 − ln t 2 ] T2 − T1 18.5 Re-arranging (18.29) for t/tf we have n

 A  3 = 1 −   = 1 −   = 0.5781  4 tf  A0  t

3

18.6 From (18.43) the skeletal radius is  J rsk =    Jc 

n / ( n −1)

 (1 + 3n) R 4  = (1+ 3n )/ n   4nR 

n / ( n −1)

 (1 + 3x4)50 4  = (1+ 3 x 4 )/ 4   4 x4 x50 

4 /( 4 −1)

= 38mm

61

18.7 The elastic second moment of area for the rectangular beam is b( 2h)

3

50x753 = 1,757,812 mm 4 12 12 The creep second moment of area is, (18.47) ( 2 x 5+1)/ 5 2bn ( 2n +1)/ n 2 x50x5  75 = 131,961 Ic = h =   2n + 1 2 x5 + 1  2  Thus, the skeletal depth is given by, (18.52) I=

y sk

 I =   Ic 

n /( n −1)

=

 1,757,812  =   131,961 

5/ ( 5−1)

= 25.45mm

62

Chapter 19 Solutions 19.1 From (19.6) the strain after 100s using the Maxwell model is  t 1 1   100 3 −6 ε =  + σ 0 =  +  5x10 = 517 x10 6  1000x10 0.3x10 9  µ E and according to the Voigt model, (9.16) σ 5x10 3 ε = 1 − e − Et / µ 0 = 1 − exp −100x10 3 x100 / 10x10 6 = 32 x10 −6 3 E 100x10

(

(

)

])

[

19.2 The total strain, ε, is the sum of the strains due to the Maxwell, εm, and Voigt, εv, models from (19.6) and (19.16)  t σ 1  − E t /µ ε = εm + εv =  + σ 0 + 1 − e v v 0 Ev  µ m Em 

(

)

At t=0 the instantaneous elastic strain is ε=σ0/Em. Differentiating ε with respect to time then the steady creep rate is  σ  µ ε& = 0 1 + m e − Ev t / µ v  µm  µv  As t→∞ then ε& → σ 0 / µ m . The variation of ε against t is shown in Figure Sol19.2.

Clearly, when µm=0 then the model is equivalent to the standard linear solid model. ε σ 0 / µm 1 ((Em +Ev )/Em Ev )σ0

σ0 /Em t

Figure Sol19.2. Creep response for the model of Exercise 19.2.

19.3 From the solution of Exercise 19.2 the strain is given by  t σ 1  − E t /µ ε = εm + εv =  + σ 0 + 1 − e v v 0 Ev  µ m Em 

(

)

After the removal of stress σ0 at time t=τ then the elastic strain σ0/E is instantaneously recovered followed by the recovery strain of, from (19.6) and (19.22) σ t' σ ε = 0 + 0 1 − e − Evτ / µ v e − Ev t ' /µ v µm Ev As t→∞ then ε→σ0t′/µm. The strain-time curve is illustrated in Figure Sol19.3.

(

)

63

ε

σ0 /Em

((Em +Ev )/Em Ev )σ0

σ0 /Em

σ0 t’/µm t

τ Figure Sol19.3. Strain-time curve for the model of Exercise 19.2 with removal of stress σ0 at time t=τ.

19.4 From (9.33) the relaxation stress at time t=10s is 0.8x10 9 (10x10 −6 ) σ (10s) = 0.2 x10 9 + 0.8x10 9 exp[ − (0.8 + 0.2) x10 9 x10 / 5x10 9 = 2.466kPa (0.8 + 0.2) x10 9

[

]

19.5 From (19.6) and (19.10) the creep compliance, C(t), and relaxation modulus, G(t), for the Maxwell model are t 1 C(t ) = + , G (t ) = Ee − Et / µ µ E and from (19.16) the creep compliance function for the Voigt model is 1 − e − Et / µ C(t ) = E with G(t)=0.

19.6 Following a similar procedure as used in the derivation of (19.6) then the total strain rate is 1 ε& = ε& s + ε& d = σ& + Aσ n E For constant stress σ=σ0 then ε& = Aσ n0 or ε = Aσ n0 + B where B is a constant of integration. If at time t=0 the instantaneous elastic strain is σ0/E then B=σ0/E and ε is given by σ ε = 0 + Aσ no t E 19.7 From (19.42) the strain at time t=3500s is ε (3500) = 0.5x10 6 (152 . x10 −9 )( 3500) 0.15 + 0.25x10 6 (152 . x10 −9 )( 3500 − 1000) 0.15 + 0.25x10 6 (152 . x 10 −9 )( 3500 − 2000) 0.15 − . x10 −9 )(3500 − 3000) = 115 . x10 −3 1x10 6 x(152

64

Chapter 20 Solutions 20.1 Refer to §20.2 for a summary of different types of damage.

20.2 From (20.1) the damage parameter is A 826 ω= = = 0.81 A0 1,018 20.3 From (20.2) the continuity parameter is ψ = 1 − ω = 1 − 0.81 = 019 .

20.4 From (20.2) and (20.9) E' 67 ω = 1− ψ = 1− = 1− = 0.65 E 190 20.5 From (20.15)

σ 0− n = A(n + 1)t f = Bt f Letting m=-n and taking natural logarithms we have ln t f = m ln σ 0 − ln B

From a (lnσ0, lntf) plot then m is given by, using the first and last data points ln 2.17 − ln 97.11 m= = −13.32 ; n = − m = 13.32 ln 14 − ln 10 and with B given by ln B = − n ln σ 0 + ln t f = −(13.32 ln 14 + ln 2.17) = −35.9269

(

)

from which B is found

B = exp( −35.9269) = 2.5x10 −16

The constant A now follows A=

B 2.5x10 −16 = = 175 . x10 −17 n + 1 13.32 + 1

20.6 From (20.17) ψ is given by

ψ = (1 − 0.8) and ω is equal to ω=1-ψ=1-0.8937=0.1063.

1/(13.32 +1)

20.7 From (20.30) λ is λ=

= 0.8937

εf Aσ n t f

At the start of tertiary creep t=0 and ω=0 then from (20.23) ε& s = Aσ n Substituting ε& s then λ is given by εf λ= ε& s t f as required.

65

Chapter 21 Solutions 21.1 From (21.1) and (21.3) the maximum volume fractions (r=R) for square and hexagonal fibre configurations are 2 πr π V f ,max ( sq ) =   = ≈ 0.785 4  R 4 π r π ≈ 0.907 V f ,max (hex ) =   = 2 3  R 2 3 From (21.13) and (21.18) E|| and E⊥ for the square and hexagonal configurations at Vf,max are found to be E || (hex ,V f ,max ) 0.907 + 0.093n E ⊥ (hex ,V f ,max ) 0.215 + 0.785n , = = 0.785 + 0.215n 0.093 + 0.907n E || ( sq ,V f ,max ) E ⊥ ( sq ,V f ,max ) where n=Em/Ef. For example, when n=0.1 then E||(hex,Vf,max)/E||(sq,Vf,max)=1.14 and E⊥(hex,Vf,max)/E⊥(sq,Vf,max)=1.6.

2

21.2 From the rule of mixtures (21.13) E|| = E f V f + E m 1 − V f = 76(0.45) + 4(1 − 0.45) = 36.4GPa

(

)

21.3 From (21.19) E⊥ =

(

E f Em

)

E f 1 − V f + Em E f

=

76x4 = 6.97GPa 76(1 − 0.45) + 4( 0.45)

21.4 With E1=E||=36.4GPa from Exercise 21.2 and E2=E⊥=6.97GPa from Exercise 21.3 then E , G and ν from (21.34) are 3 5 3 5 E = E || + E ⊥ = (36.4) + (6.97) = 18GPa 8 8 8 8 1 1 1 1 G = E || + E ⊥ = (36.4) + (6.97) = 6.29GPa 8 4 8 4 E 18 ν= −1= − 1 = 0.43 2G 2(6.29) 21.5 From (21.38) the volume fraction at cross-over between low and high Vf is σ *m 75 V f' = * = = 013 . ' * σ f − σ f + σ m 2,000 − 1,500 + 75 If V f > V f' then fibre strength dominates.

21.6 From (21.43) the transverse failure strength, σ *⊥ , is  Vf     = 65 1 − 2 0.3  = 24.83MPa σ *⊥ = σ *m  1 − 2 π  π    21.7 From (21.45) the critical embedded fibre length is

66

lce =

σ *f r 2τ

=

(

750x10 6 0.2 x10 −3

(

2 45x10 6

)

) = 17. mm

67

Chapter 22 Solutions 22.1 Resolving Pr we find that P=Prsinθ and Q=Prcosθ and upon substitution into (22.7) and (22.18) and superimposing the stresses gives the stresses in the half-plane.

22.2 The stresses for a point force at x=b can be found by replacing x by (x-b) in (22.7) 2 2 P( x − b ) y σ xx = − 2 2 π ( x − b) + y 2

[

σ yy = −

τ xy = −

]

2 Py

[

3

π ( x − b) + y 2

[

2

2 P( x − b ) y 2

π ( x − b) + y 2 2

]

]

2

2

To show that τxy is zero along the line x=b/2 write expressions for τxy when P is at x=0 and x=b, from (22.7) and the above equation Pby 2 Pby 2 , P at x = b = τ xy ( P at x = 0) = − τ ( ) xy 2 2 2 π (b / 2) + y 2 π (b / 2) + y 2

[

[

]

]

which cancel if superimposed.

22.3 Referring to Figure 22.26 then the applied pressure distribution is given by p0 ( a − x ) p( x ) = ; x ≤a a Performing the integrations in (22.20) then the stresses are found to be p   r r  σ xx = 0 ( x − a )θ 1 + ( x + a )θ 2 − 2 xθ + 2 y ln 1 22    r  πa  p0 {( x − a)θ 1 + ( x + a )θ 2 − 2 xθ } πa p y τ xy = − 0 (θ 1 + θ 2 − 2θ ) πa where r, r1, r2, θ, θ1 and θ2 are given by 2 2 r = x 2 + y 2 , r12 = ( x − a ) + y 2 , r22 = ( x + a ) + y 2 σ yy =

tan θ =

y x

,

tan θ 1 =

y x−a

,

tan θ 2 =

y x+a

22.4 For a uniform pressure (N=P and T=0) then Φ(z) and Ψ(z) are t =a p a dt p p  z − a = Φ( z ) = − ln( z − t ) t =− a = ln  ∫ − a 2πi z − t 2πi 2πi  z + a 

[

Ψ( z) = −

]

zp a dt paz =− 2 ∫ 2πi − a ( t − z ) πi z 2 − a 2

(

)

68

With reference to Figure Sol22.4 then let z-t=re-iθ where r=|z-t|. From (14.148) and (14.152) the stress components are 2p σ xx + σ yy = 4 Re Φ( z ) = − (θ 1 − θ 2 ) π 2 pa  z − z  4 pay =− σ yy − σ xx + 2iτ xy = 2[ z Φ' ( z ) + Ψ ( z )] =  2 a πi  z − a  π z2 − a2

(

Solving for σxx, σyy and τxy we arrive at

(

)

σ xx

2 pay x 2 − y 2 − a 2 p = − (θ 1 − θ 2 ) + 2 π π x 2 + y 2 − a 2 + 4a 2 y 2

σ yy

2 pay x 2 − y 2 − a 2 p = − (θ 1 − θ 2 ) − 2 π π x 2 + y 2 − a 2 + 4a 2 y 2

[(

(

[(

τ xy =

[(

4 paxy 2

π x2 + y2 − a2

)

2

)

+ 4a 2 y 2

)

)

)

] ]

]

y p -a

t

O θ

a θ

θ

x 1

r

2

r

r 2

1

z

Figure Sol22.4. A half-plane subject to a uniform normal pressure.

22.5 From (22.55) the pressure beneath the centre of the punch is P 5x10 3 p(0) = = = 0127 . MPa πa π 12.5x10 −3

(

)

22.6 From (22.70) the radius of the contact radius, a, and total displacement, δ, are  3 P R1 R2  1 − ν 12 1 − ν 22   a= +   E2    4 R1 + R2  E1 9 P2 R + R 1 2 δ=  16 R1 R2

1/ 3

2  1 − ν 12 1 − ν 22   +    E2    E1 

1/ 3

69

Letting body 2 be the half-plane, R1=∞, and R=R2 then R1R2/(R1+R2) is found to reduce to the following R1 R2 R1 R2 = = R2 = R as R1 → ∞ R1 + R2 R1 (1 + R2 / R1 ) In addition, letting E=E1=E2 and ν=ν1=ν2 for both bodies having the same elastic properties then the above expressions for a and δ are found to reduce to  3 PR  1 − ν 2   a=    2  E 

1/ 3

9P2  1− ν 2  2  , δ =     16 R  E  

1/ 3

as required.

22.7 We can obtain expressions for a, δ and p0 for the circular ball in a circular seat from (22.70) by simply making R1 negative.

70

Chapter 23 Solutions 23.1 Table Sol23.1 summarises the failure strengths of the tested composite material. failure strength 650 680 700 710 740 750

ab. frequency 1 2 2 3 1 1

rel. frequency 0.1 0.2 0.2 0.3 0.1 0.1

cum. Ab. frequency 1 3 5 8 9 10

cum. rel. frequency 0.1 0.3 0.5 0.8 0.9 1

Table Sol23.1. Sample of 10 values of the tensile failure strength (MPa) of glasspolyester unidirectional laminae composite specimens.

23.2 From (23.4) the mean is x=

1 N

N

∑x

i

= 703MPa

i =1

From (23.5) the variance is 1 N . MPa ( xi − x ) 2 = 84555 ∑ N − 1 i =1 and the standard deviation is the square root of the variance s2 =

s = s 2 = 29.08MPa 23.3 Since a thrown die can result in a number which is both even and a multiple of 3 then both events can occur simultaneously and therefore the events are arbitrary.

Letting A represent an even number and B represent a number which is a multiple of 3 then A={2,4,6} and B={3,6}. The required probabilities for arbitrary events are P(A)=3/6=1/2, P(B)=2/6=1/3 and P(A∩B)=1/6 so that from (23.13) we have 1 1 1 2 P ( A ∪ B) = P( A) + P( B) − P( A ∩ B) = + − = 2 3 6 3 23.4 From (23.28) the mean is 3

b

3

a

0

x = ∫ xf ( x )dx = ∫

 x3  9  x x   dx =   =  2  6 0 2

From (23.30) the variance is σ = 2

∫ (x − x) b

a

2

f ( x )dx = ∫

3

 x 4 3x 3 81x 2  9  x 243  x dx = − + = −        2  2 2 16  0 16 8 2

23.5 For case i) then x =1 and σ2=1 so that we can obtain the probability directly from Table 23.2, and is found to 0.9772. For case ii) with x =0.4 and σ2=4 then the standardised variable is x − x 2 − 0.4 z= = = 0.8 σ 2 From Table 23.2 we have Φ(0.8)=0.7881.

23.6 From (23.53) and Table 23.3 the mean strength is

71

σ=

σ0  1 650 1  Γ1 +  = Γ  1 +  = 718MPa 1/ m 1/10  10   m  0.225 V

23.7 Letting the specimen of Exercise 23.6 be denoted by specimen 1 and the new specimen by 2 then from (23.65) the expected mean strength of the new specimen is, with σ 1 =718MPa σ1 718 σ2 = = = 670MPa 1/ m (0.45 / 0.225)1/10 (V2 / V1 )

72