Solution Manual for Advanced Mechanics of Materials and Applied Elasticity, 5th Edition by Ace Test Banks

Solutions Manual for

Advanced Mechanics of Materials and Applied Elasticity

Fifth Edition

Upper Saddle River, NJ • Boston • Indianapolis • San Francisco New York • Toronto • Montreal • London • Munich • Paris • Madrid Capetown • Sydney • Tokyo • Singapore • Mexico City

The author and publisher have taken care in the preparation of this book, but make no expressed or implied warranty of any kind and assume no responsibility for errors or omissions No liability is assumed for incidental or consequential damages in connection with or arising out of the use of the information or programs contained herein.

ISBN-10: 0-13-269049-7 ISBN-13: 978-0-13-269049-2

CONTENTS

Chapter 1

Analysis of Stress

1–1

Chapter 2

Strain and Material Properties

2–1

Chapter 3

Problem in Elasticity

3–1

Chapter 4

Failure Criteria

4–1

Chapter 5

Bending of Beams

5–1

Chapter 6

Torsion of Prismatic Bars

6–1

Chapter 7

Numerical Methods

7–1

Chapter 8

Axisymmetrically Loaded Members

8–1

Chapter 9

Beams on Elastic Foundations

9–1

Chapter 10

Applications of Energy Methods

10–1

Chapter 11

Stability of Columns

11–1

Chapter 12

Plastic Behavior of Materials

12–1

Chapter 13

Plates and Shells

13–1

NOTES TO THE INSTRUCTOR

The Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition supplements the study of stress and deformation analyses developed in the book. The main objective of the manual is to provide efficient solutions for problems dealing with variously loaded members. This manual can also serve to guide the instructor in the assignments of problems, in grading these problems, and in preparing lecture materials as well as examination questions. Every effort has been made to have a solutions manual that can cut through the clutter and is as self-explanatory as possible, thus reducing the work on the instructor. It is written and class-tested by the author, Ansel Ugural. As indicated in the book’s Preface, the text is designed for the senior and/or first year graduate level courses in stress analysis. In order to accommodate courses of varying emphasis, considerably more material has been presented in the book than can be covered effectively in a single three-credit course. The instructor has the choice of assigning a variety of problems in each chapter. Answers to selected problems are given at the end of the text. A description of the topics covered is given in the introduction of each chapter throughout the text. It is hoped that the foregoing materials will help instructor in organizing his or her course to best fit the needs of his or her students. Ansel C. Ugural Holmdel, NJ

CHAPTER 1 SOLUTION (1.1) We have

A = 50 " 75 = 3.75(10 !3 ) m 2 , ! = 90o " 40o = 50o , and ! x = P A . o

Equations (1.8), with ! = 50 :

! x ' = 700(10 3 ) = ! x cos 2 50o = 0.413! x = 110.18P

P = 6.35 kN

" x ' y ' = 560(10 3 )! x sin 50o cos 50o = 0.492! x = 131.2 P Solving

P = 4.27 kN = Pall

______________________________________________________________________________________ SOLUTION (1.2) Normal stress is 3

(10 ) " x = PA = 0125 .05!0.05 = 50 MPa

" 70o = 20o : ! x ' = 50 cos 2 20o = 44.15 MPa

( a ) Equations (1.11), with ! = 90

" x ' y ' = !50 sin 20o cos 20o = !16.08 MPa

! y ' = 50 cos 2 ( 20o + 90o ) = 5.849 MPa 5.849 MPa y’

44.15 MPa

16.08 MPa

x’ 20 o x o

( b ) Equations (1.11), with ! = 45 :

! x ' = 50 cos 2 45o = 25 MPa

" x ' y ' = !50 sin 45o cos 45o = !25 MPa ! y ' = 50 cos 2 ( 45o + 90o ) = 25 MPa 25 MPa y’ 25 MPa

25 MPa x’ 45 o x

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (1.3) From Eq. (1.11a),

# x = cos# x2'" = cos!27530o = !100 MPa o

For ! = 50 , Eqs. (1.11) give then

" x ' = !100 cos 2 50o = !41.32 MPa

" x ' y ' = !( !100) sin 50o cos 50o = 49.24 MPa o Similarly, for ! = 140 : " x ' = !100 cos 2 140o = !58.68 MPa " x ' y ' = !49.24 MPa

58.68 MPa

41.32 MPa 50 o

49.24 MPa

______________________________________________________________________________________ SOLUTION (1.4) Refer to Fig. 1.6c. Equations (1.11) by substituting the double angle-trigonometric relations, or Eqs. (1.18) with ! y = 0 and ! xy = 0 , become or

" x ' = 12 " x + 12 " x cos 2!

and

# x ' y ' = 12 " x sin 2!

20 = 2PA (1 + cos 2! )

and

10 = 2PA sin 2!

The foregoing lead to

2 sin 2" ! cos 2" = 1

(a)

By introducing trigonometric identities, Eq. (a) becomes

Thus, gives

4 sin " cos " ! 2 cos 2 " = 0 or tan ! = 1 2 . Hence ! = 26.56o P 20 = 2 (1300 ) = (1 + 0.6)

P = 32.5 kN

It can be shown that use of Mohr’s circle yields readily the same result. ______________________________________________________________________________________ SOLUTION (1.5) Equations (1.12):

P #150(103 ) = = #76.4 MPa " A 2 (50) 4 P ! max = = 38.2 MPa 2A

!1 =

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (1.6) Shaded transverse area:

A = 2at = 2(10)(75) = 1.5(103 ) mm 2 Metal is capable of supporting the load

P = ! A = 90(106 )(1.5 #10"3 ) = 135 kN

Apply Eqs. (1.11):

P (cos 2 55o ) , P = 114 kN "3 1.5(10 ) P ! x ' y ' = 12(106 ) = " sin 55o cos 55o , P = 38.3 kN "3 1.5(10 )

! x ' = 25(106 ) =

Thus,

Pall = 38.3 kN

______________________________________________________________________________________ SOLUTION (1.7) Use Eqs. (1.11):

P (cos 2 40o ) , P = 51.1 kN 3 1.5(10 ) P ! x ' y ' = 8(106 ) = " sin 40o cos 40o , P = 24.4 kN 1.5(103 )

! x ' = 20(106 ) =

Thus,

Pall = 24.4 kN

______________________________________________________________________________________ SOLUTION (1.8)

A = 15 ! 30 = 450 mm 2 Apply Eqs. (1.11):

120(103 ) (cos 2 40o ) = 156 MPa 450 #10"6 120(103 ) sin 40o cos 40o = "131 MPa ! x' y' = " 450 #10"6

! x' =

______________________________________________________________________________________ SOLUTION (1.9) We have A = 450(10

) m2 .

Use Eqs. (1.11):

"100(10 ) (cos 2 60o ) = "55.6 MPa "6 450 #10 "100(103 ) sin 60o cos 60o = 96.2 MPa ! x' y' = " "6 450 #10

! x' =

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (1.10)

! = 40o + 90o = 130o (103 ) # x = PA = ! " ( 0150 = !31.83 MPa .082 !0.07 2 ) Equations (1.11):

" x ' = !31.83 cos 2 130o = !13.15 MPa

" x ' y ' = 31.83 sin 130o cos 130o = !15.67 MPa x’

13.15 MPa

130 o

15.67 MPa

x Plane of weld y’

______________________________________________________________________________________ SOLUTION (1.11) Use Eqs. (1.14),

( 2 x ) + ( !2 xy ) + ( x ) + Fx = 0 ( ! y 2 ) + ( !2 yz + x ) + (0) + Fy = 0

( z ! 4 xy ) + (0) + ( !2 z ) + Fz = 0 3

Solving, we have (in MN m ):

Fx = !3x + 2 xy

Fy = ! x + y 2 + 2 xz

Fz = 4 xy + z

(a)

Substituting x=-0.01 m, y=0.03 m, and z=0.06 m, Eqs. (a) yield the following values

Fx = 29.4 kN m 3

Fy = 14.5 kN m 3

Fz = 58.8 kN m 3

Resultant body force is thus

F = Fx2 + Fy2 + Fz2 = 67.32 kN m 3 ______________________________________________________________________________________ SOLUTION (1.12) Equations (1.14):

" 2c1 y " 2c1 y + 0 + 0 = 0, 0 + c3 z + 0 + 0 = 0, 0+0+0+0=0

4c1 y ! 0 c3 z ! 0

No. Eqs. (1.14) are not satisfied. ______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (1.13) ( a ) No. Eqs. (1.14) are not satisfied. ( b ) Yes. Eqs. (1.14) are satisfied. ______________________________________________________________________________________ SOLUTION (1.14) Eqs. (1.14) for the given stress field yield:

Fx = Fy = Fz = 0

______________________________________________________________________________________ SOLUTION (1.15) y

! x ' y ' "A ! x ' "A

y’ 40 !A cos20o

50 !A cos20o

20o

x’

x 50 !A sin20o 60 !A sin20o

$F = 0:

! x ' "A + 40 cos 2 20o # 60"A sin 2 20o

$F = 0: !

x' y'

!2(50"A sin 20o cos 20o ) = 0 ! x ' = "35.32 + 7.02 + 32.14 = 3.84 MPa

"A # 40"A sin 20o cos 20o

!60"A sin 20o cos 20o ! 50"A cos 2 20o +50!A sin 2 20o = 0 ! x ' y ' = 12.86 + 19.28 + 44.15 " 5.85 = 70.4 MPa

______________________________________________________________________________________ SOLUTION (1.16) 50 !A cos25o y’

15 !A cos25o 15 !A sin25o

25o

90 !A sin25o

! x ' "A 2

# F = 0 : ! "A + 50"A cos 25 x'

! x ' y ' "A x’

!90"A sin 2 25o ! 2(15"A sin 25o cos 25o ) = 0 ! x ' = "41.7 + 16.07 + 11.49 = "12.9 MPa

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 1.16 (CONT.)

$F = 0: ! y'

x' y'

"A # 50"A sin 25o cos 25o

!90"A sin 25o cos 25o ! 15"A cos 2 25o +15!A sin 2 25o = 0 ! x ' y ' = 19.15 + 34.47 + 12.32 " 2.68 = 63.3 MPa

______________________________________________________________________________________ SOLUTION (1.17) y y’ 40 MPa

! x' y'

! x' 20

x’

! =20o

! x

50 MPa 60 MPa

1 1 ! x ' = ("40 + 60) + ("40 " 60) cos 40o + 50sin 40o 2 2 = 10 ! 38.3 + 32.1 = !3.8 MPa 1 ! x ' y ' = " ("40 " 60) sin 40o + 50 cos 40o 2 = 32.14 + 38.3 = 70.4 MPa ______________________________________________________________________________________ SOLUTION (1.18)

! x' x’

! y’

! x' y' 90 MPa 25o 15 MPa

! =115o

50 Mpa

1 1 ! x ' = (90 " 50) + (90 + 50) cos 230o " 15sin 230o 2 2 = 20 ! 45 + 11.5 = !13.5 MPa 1 ! x ' y ' = " (90 + 50) sin 230o " 15cos 230o 2 = 53.62 + 9.64 = 63.3 MPa ______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (1.19) o

Transform from ! = 40 to ! = 0 . For convenience in computations, Let

! x = #160 MPa,

! y = #80 MPa,

" xy = 40 MPa and ! = "40o

Then

1 1 ! x ' = (! x + ! y ) + (! x $ ! y ) cos 2" + # xy sin 2" 2 2 1 1 = (!160 ! 80) + (!160 + 80) cos(!80o ) + 40sin(!80o ) 2 2 = !138.6 MPa

1 ! x ' y ' = $ (" x $ " y ) sin 2# + ! xy cos 2# 2 1 = ! (!160 + 80) sin(!80o ) + 40 cos(!80o ) 2 = !32.4 MPa

! y ' = ! x + ! y " ! x ' = "160 " 80 + 138.6 = "101.4 MPa

For ! = 0 :

101.4 MPa 32.4 MPa 138.6 MPa x

______________________________________________________________________________________ SOLUTION (1.20)

4 = 53.1o 3 45 + 90 45 " 90 + cos106.2o ! x' = 2 2 = 67.5 + 6.28 = 73.8 MPa 45 " 90 sin106.2o = 43.2 MPa ! x' y' = " 2

! = tan "1

43.2 MPa y’

73.8 MPa x’

53.1o x ______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (1.21)

" = 70o

! xy = 0 (a)

" x ' y ' = #30 = #

(b)

! x ' = 80 =

! # 60 sin140o 2

! = 153.3 MPa

! + 60 ! " 60 + cos140o 2 2

! = 231 MPa

______________________________________________________________________________________ SOLUTION (1.22) ! (MPa)

60 MPa

20o

(60, 50)

50 MPa

O #

40 MPa (-40, -50)

" (MPa)

40o x’

50 = 39.8o 60 1 2 r = (60 + 502 ) 2 = 78.1

! = tan "1

! x ' y ' = sin 79.8o (78.1) = 76.9 MPa

! x ' = cos 79.8o (78.1) = "13.83 MPa Sketch of results is as shown in solution of Prob. 1.15. ______________________________________________________________________________________ SOLUTION (1.23) ! (MPa) "’=20 50 MPa r

15 MPa 25

(-50, -15)

90 MPa x

O 62.1o

y’ #

(90, 15)

" (MPa)

("x’, ! x’y’)

15 ! = tan = 12.1o 70 1 2 r = (15 + 702 ) 2 = 73.14 ! x ' y ' = 73.14sin 62.1o = 64.6 MPa

x’

! x ' = "73.14 cos 62.1o + 20 = !14.22 MPa

y’

14.22 MPa

25o x’

64.6 MPa 90 MPa 15 MPa

50 MPa ______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (1.24) ! (MPa)

! ' =67.5

x’

r =22.5 r

45 MPa

53.1o

90 MPa

" (MPa) 73.8o x’ ("x’, $! x’ y’ )

106.2o

! x ' y ' = 22.5sin 73.8o = 21.6 MPa

! x ' = 67.5 + 22.5cos 73.8o = 73.8 MPa

Sketch of results is as shown in solution of Prob. 1.20. ______________________________________________________________________________________ SOLUTION (1.25) ! (MPa) (a) x’ 1 60 MPa 140o

x’ O

70o

(! " 60)

" (MPa)

! # 60 sin 40o ; ! = 153.3 MPa 2 ! " 60 ! x ' = 80 = 60 + (1 " cos 40o ) 2 ! = 231 MPa

" x ' y ' = #30 = (b)

______________________________________________________________________________________ SOLUTION (1.26) ( a ) From Mohr’s circle, Fig. (a):

# 1 = 121 MPa o

" p ' = !19.3

" s ' = 25.7

! (MPa) !max

# 2 = !71 MPa

A(100,60) 2! p '

" max = 96 MPa

" (MPa)

B (CONT.) Figure (a) ______________________________________________________________________________________

______________________________________________________________________________________ 1.26 (CONT.) By applying Eq. (1.20):

! 1, 2 = 502 ± [22,4500 + 36000]2 = 25 ± 96

" 1 = 121 MPa

" 2 = !71 MPa

Using Eq. (1.19):

tan 2" p = ! 12 15 = !0.8

" p ' = !19.3o

" s ' = 25.7 o

( b ) From Mohr’s circle, Fig. (b):

# 1 = 200 MPa

! p ' = 26.55o ! (MPa)

# 2 = !50 MPa

" max = 125 MPa

! s ' = 71.55o !max

2%p’

" (MPa)

A(150,-100) Figure (b) Through the use of Eq. (1.20),

! 1, 2 = 75 ± [22,4500 + 10,000]2 = 75 ± 125 1

" 1 = 200 MPa " 2 = !50 MPa Using Eq. (1.19), tan 2! p = 4 3 : ! p ' = 26.55o

! s ' = 71.55o

______________________________________________________________________________________ SOLUTION (1.27) Referring to Mohr’s circle, Fig. 1.15:

" x ' = " 1 +2" 2 + " 1 #2" 2 cos 2!

(a)

" y ' = " 1 +2" 2 # " 1 #2" 2 cos 2!

" x ' y ' = # 1 $2# 2 sin 2!

(b)

From Eqs. (a),

! x' + ! y' = ! 1 + ! 2 2! + sin 2 2! = 1 , and Eqs. (a) and (b), we have # x ' ! # y ' " $ x2' y ' = # 1 ! # 2 = const .

By using cos

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (1.28) We have 2#

70 ) tan 2$ p = " x !xy" y = 502 (!!(190 ) = !0.583

2" p = !30.26o

and " p = !15.13

Equations (1.18):

" x ' = 50!2190 + 50+2190 cos( !30.26o ) ! 70 sin( !30.26o ) = "70 + 103.65 + 35.275 = 68.92 MPa = ! 1 ! y ' = ! x + ! y " ! x ' = "208.9 MPa = ! 2 208.9 MPa 15.13o 68.92 MPa

______________________________________________________________________________________ SOLUTION (1.29) " #"

! max = ( x 2 y ) 2 + ! xy2 Substituting the given values 2

140 2 = (60+2100 ) + ! xy2

! xy ,max = 114.19 MPa

______________________________________________________________________________________ SOLUTION (1.30) o

Transform from ! = 60 to ! = 0 with " x ' = !20 MPa , " y ' = 60 MPa ,

" x ' y ' = !22 MPa , and " = !60o . Use Eqs. (1.18): " x = !202+60 + !202!60 cos 2( !60o ) ! 22 sin 2( !60o ) = 59 MPa " y = " x ' + " y ' ! " x = !19 MPa

" xy = !23.6 MPa y

19 MPa 23.6 MPa 59 MPa x

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (1.31) "y

"ycos60o

14 MPa !xy

!xycos60o "xsin60o

30 MPa Area =1

30 MPa

!xysin60o Figure (b)

Figure (a) ( a ) Figure (a):

! y = 14 sin 60o = 12.12 MPa

! xy = 14 cos 60o = 7 MPa Figure (b):

! xy = 7 MPa (as before) o

! F = "# sin 60 + 30 + 7 cos 60 = 0 x

" F = 12.12 cos 60 ! # sin 60 = 0 x

! x = 38.68 MPa

( b ) Equation (1.20) is therefore:

" 1, 2 = 38.68+212.12 ± [( 38.682!12.12 ) 2 + 7 2 ]2 1

or Also,

! 1 = 40.41 MPa ,

! 2 = 10.39 MPa

" p = 12 tan !1 38.682 (!712) .12 = 13.9 o

Note: Eq. (1.18a) gives, ! x ' = 40.41 MPa . Thus,

! p ' = 13.9 o 10.39 MPa 40.41 MPa x’ %p’

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (1.32) "y !xy Figure (a)

Figure (a):

! x = 100 cos 45o = 70.7 MPa ! y = 100 sin 45o = 70.7 MPa

! xy = 100 cos 45o = 70.7 MPa Now, Eqs. (1.18) give (Fig. b):

! x ' = 70.7 + 0 + 70.7 sin 240o = 9.47 MPa

" x ' y ' = !0 + 70.7 cos 240o = !35.35 MPa " y ' = 70.7 ! 0 ! 70.7 sin 240o = 131.9 MPa "y

"x’ !xy

x’ "x y’

!x’y’

Figure (b)

n 30o

"y’

______________________________________________________________________________________ SOLUTION (1.33)

" y = !70 sin 30o = !35 MPa

! xy = 70 cos 30o = 60.6 MPa ( a ) Figure (a):

" F = !150 + 0.5# + 60.6(0.866) = 0 x

! x = 195 MPa

35cos30o

150 MPa

60.6cos30o

30o Area=1

60.6sin30o "xsin30o

Figure (a)

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 1.33 (CONT.) ( b ) Equation (1.20):

" 1, 2 = 1952!30 ± [( 1952+35 ) 2 + 60.6 2 ]2 1

or Also,

" 1 = 210 MPa

" 2 = !50 MPa

2 ( 60.6 ) o " p = 12 tan !1 195 + 35 = 13.89

Equation (1.18a):

! x ' = 80 + 115 cos 2(13.89 o ) + 60.6 sin 2(13.89 o ) = 210 MPa

Thus,

! p ' = 13.89 o

50 MPa 210 MPa x’ %p’

______________________________________________________________________________________ SOLUTION (1.34) For pure shear, ! 1 = "! 2 : pr t

from which

! 1 = pr t

= " 2prt + 2!Prt

" 2 = 2prt # 2!Prt

P = 3!pr 2

______________________________________________________________________________________ SOLUTION (1.35) "! y Table C.1:

A = 2!rt J = 2!r 3t

! +!a ! xy

Figure (a)

Stresses are (Fig. a): 3

# = !AP = 2" !( 030.12(10)( 0)."005) ! 25 MPa 6

! a = 2prt = 4 (102 ( 5))120 = 48 MPa ! " = 2! a = 96 MPa 3

) # xy = !JTr = 2" (!010.12"2()(100.005 = !69.4 MPa )

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 1.35 (CONT.) Hence,

" x = 48 ! 25 = 23 MPa

! y = 96 MPa

Therefore, we have

" max = ±[( 23!296 ) 2 + 69.4 2 ]2 = ±78.4 MPa 1

Also and

y’

! ' = 12 ( 23 + 96) = 59.5 MPa o !96 " s = 12 tan !1 2 (23!69 .4 ) = !13.87

!s ' '

59.4 MPa

Equation (1.18b) with " s = !13.87 :

78.4 MPa

" x ' y ' = !16.99 ! 61.42 = !78.4 MPa Thus,

x x’

Figure (b)

! s ' ' = 13.87 o

______________________________________________________________________________________ SOLUTION (1.36)

A = 2!rt = 2! (60)( 4) = 1508 mm 2 J = 2"r 3t = 2" (60) 3 ( 4) = 5.429 ! 10 6 mm 4

! y = prt = 5( 460 ) = 75 MPa

( 0.05 ) " xy = ! TrJ = ! 5600 .429 (10! 6 )

Thus

= !5.526 MPa P ! x = 2prt + PA = 37.5 + 1508 ( P in newtons )

"1 =

" x +" y 2

!x ! xy

" #"

+ ( x 2 y ) 2 + ! xy2

Substituting the numerical values gives

[

] 1

80 = 56.3 + 331.6 " 10 !6 P + ( !18.75 + 331.6 " 10 !6 P ) 2 + ( !5.526) 2 2 Solving,

P = 64.01 kN

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (1.37) At point A, T = 8 kN and P = 400 kN

!x = $

P 4(400 #103 ) =$ = $50.9 MPa A " (0.1) 2

! xy A

16T 16(8 #10 ) = = 40.7 MPa ! xy = "d3 " (0.1)3

Hence

" max = (

!x 2 2 ) + " xy 2

= (!25.45) 2 + (40.7) 2 = 48 MPa = R

! y

! ! avg = x = "25.45 MPa 2 tan 2! p " =

40.7 , 25.45

! 2 2%p"

! p " = 29o

O "avg

______________________________________________________________________________________ SOLUTION (1.38)

! = " + 90 = 50 + 90 = 140o x'

!x =

! xy =

y' Equations(1.18):

! w = ! x' = and

16T "d3

120 #103 = 95.5 MPa " 2 (0.04) 4 16(1.5 $10#3 ) = 119.4 MPa != " (0.04)3

!x =

P A

95.5 95.5 cos 280o + 119.4sin 280o = 61.5 MPa + 2 2

! w = ! x' y' = " x’ 61.5 MPa 26.3 MPa

95.5 sin 280o " 119.4 cos 280o = 47.02 " 20.73 = 26.3 MPa 2 140o x

y’ ______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (1.39)

A = #4 (180 2 " 120 2 ) = 11.31 ! 10 3 mm 2 # J = 32 (180 4 " 120 4 ) = 82.70 ! 10 6 mm 4

" x = ! PA = ! 11700 .31 = !61.89 MPa ,

"y =0

( 90 ) # xy = TrJ = 8220 = 21.76 MPa .70"10! 6

Equation (1.20) is therefore

" max = " 1 = ! 612.89 + ( 612.89 ) 2 + ( 21.76) 2 = 6.885 MPa ______________________________________________________________________________________ SOLUTION (1.40)

P = ! 0 Lt

! 0 Lt

M = ! 0 Lth A = 2ht I = 121 t ( 2h ) 3 = 23 th 3

P M

! L

Axial stress: " a = PA = 20h

Bending stress: " b = Mc I = Point A

3! 0 L 2h

!0 A

" a + " b = !20hL

From Eqs. (1.20) and (1.22), we obtain

" 1, 2 = ! 0hL ± ! 0 ( Lh ) 2 + 1 ! max = ! 0 ( Lh ) 2 + 1 Point B B

" b # " a = ! 0hL

Hence

! 1 = " 0hL

!2 = 0

! max = 2! 0h

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (1.41) y

P M

A = " (30 2 ! 152 ) = 2.121(10 !3 ) m 2 I = "4 (30 4 ! 154 ) = 0.596(10 !6 ) m 4

J = 2I

!a +!b

! xy

We have 3

" a = PA = 2.5012((1010!)3 ) = 23.58 MPa ( 0.03) " b = MrI = 0200 = 10.07 MPa .596 (10! 6 ) 500 ( 0.03) " xy = !JTr = 1!.192 = !12.58 MPa (10! 6 )

Thus,

! x = 23.58 + 10.07 = 33.65 MPa

! ' = 16.83 MPa

! (MPa)

(33.65, 12.58) 2%p’ " (MPa)

Figure (a)

16.83

From Mohr’s circle (Fig. a):

r = 12.582 + 16.832 = 21.01 MPa

o .58 " p ' = 12 tan !1 12 16.83 = 18.39

! 1 = 16.83 + 21.01 = 37.84 MPa " 2 = !4.18 MPa Results are shown in Fig. (b). %p’

x x’

Figure (b)

37.84 MPa 4.18 MPa ______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (1.42) A

y 30

20o

! 0 = ! xy ( a ) At " = !60

0= or

C !0

"x’

-60o

y’ Figure (a)

x’

(Fig. a): # x +# y 2

# x !# y 2

cos 2( !60o ) + " 0 sin 2( !60o )

0 = 0.5" x + 1.5" y # 1.732! 0

(a)

We also have " !"

# 0 = ! x 2 y sin 2( !60o ) + cos 2( !60o ) or

! x = 3.464" 0 + ! y

(b)

Substituting Eq. (b) into (a), we obtain ! y = 0 . Results are shown in Fig. b. y

! xy = ! 0

" x = 3.464! 0

Figure (b)

x Alternatively, using an element ABC (Fig. c): A

"x !0

!0 30o

Figure (c)

Area=1

! F = 0.5$ " 0.866# " 0.866# = 0 x

or " x = 3.464! 0 , as before. o

Stresses on planes at 20 , taking " = !70 (Fig. b): o o 3.464 " 20o = [ 3.464 2 + 2 cos( #140 ) + sin( #140 )]! 0 = #0.237! 0 o o ! 20o = [ " 3.464 2 sin( "140 ) + cos( "140 )]! 0 = 0.347! 0

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 1.42 (CONT.) ( b ) Principal stresses: 2 2 2 " 1, 2 = 3.464 ( 3.464 2 ±[ 2 ) +!0 ] 1

" 1 = 3.732! 0

" 2 = #0.268! 0

The maximum principal stress is on plane inclined at "0 o # p ' = 12 tan !1 1.732 " 0 = 15

______________________________________________________________________________________ SOLUTION (1.43) At a critical point on the shaft surface, the state of stress of stress is as shown in Fig. (a). y

! max

!p''

! xy

Figure (a)

Figure (b)

We have

" x = ! PA ! MrI 3

) )( 0.075 ) = ! "81( 0(.10 ! 13"((100.075 = !43.818 MPa 075 ) 2 )4 4 3

) 0.075 $ xy = ! TrJ = ! (15# .(60".10 = !23.54 MPa 075 ) 4 2

Therefore,

" 1, 2 = !432.818 ± [( 43.2818 ) 2 + ( !23.54) 2 ]2 1

and

" 1 = 10.248 MPa, " 2 = !54.066 MPa 1 # max = 2 (" 1 ! " 2 ) = 32.157 MPa ( 23.54 ) o " p ' ' = 12 tan !1 243 .818 = 23.53

Results are shown in Fig. (b). ______________________________________________________________________________________ SOLUTION (1.44) o

Apply Eqs. (1.20) to Fig. P1.44b, for " = !30 :

" xb = !40 sin 2( !30o ) = 20 3 MPa

" yb = !20 3 MPa

(b) o

" xyb = !40 cos 2( !30 ) = !20 MPa (CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 1.44 (CONT.) o

Now apply Eqs. (1.18) to Fig. P1.44c, for " = !60 :

" xc = 10 sin 2( !60o ) = !5 3 MPa

! yc = 5 3 MPa

" xyc = 10 cos 2( !60 ) = !5 MPa Superposing stresses in Eqs. (b) and (c) and those in Fig. P1.44a, we obtain Fig. (a). y

15 3 MPa 15 3 MPa x

45 MPa Figure (a) Referring to Fig. (a):

[

] 1

" 1, 2 = 0 ± (15 3 ) 2 + ( !45) 2 2 or When

" 1 = 51.96 MPa

" 2 = !51.96 MPa

" p ' = 12 tan !1 22((15!453)) = !30o

is substituted into Eq. (1.18a), we have 51.96 MPa (Fig. b). x

!p' x’

51.96 MPa 51.96 MPa Figure (b) ______________________________________________________________________________________ SOLUTION (1.45) o

Apply Eqs. (1.18) to Fig. P1.45a, for " = !15 , to obtain stresses in Fig. (a):

" xa = ! 302 ! 302 cos 2( !15o ) = !27.99 MPa

" ya = !15 + 15 cos 2( !15o ) = !2.01 MPa

" xya = 15 sin 2( !15o ) = !7.5 MPa Superposition of stresses in Figs. (a) and P1.45b gives Fig. (b).

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 1.45 (CONT.) 2.01 MPa

47.99 MPa 27.99 MPa

27.99 MPa

30o

30o 7.5 MPa

7.5 MPa

Figure (b)

Figure (a) Apply Eq. (1.20) to Fig. (b):

" 1, 2 = !27.992+ 47.99 ± [14 ( !27.99 ! 47.99) 2 + ( !7.5) 2 ]2 1

or When

" 1 = 48.72 MPa,

" 2 = !28.72 MPa

5) o " p = 12 tan !1 !( 272.(99!7+.47 .99 ) = 5.58

is substituted into Eq. (1.18a), we obtain –28.72 MPa (Fig. c). 48.72 MPa

28.72 MPa 35.58o x

Figure (c) ______________________________________________________________________________________ SOLUTION (1.46) o

Equations (1.18) are applied to Fig. P1.46a, for " = !30 :

" xa = 20+2 30 + 20!2 30 cos 2( !30o ) = 22.5 MPa " ya = 25 ! ( !5) cos 2( !30o ) = 27.5 MPa " xya = !( !5) sin 2( !30o ) = !4.33 MPa These stresses and that of Fig. P1.46b are superimposed to yield Fig. (a). y

14.33 MPa 37.5 MPa x 22.5 MPa Figure (a)

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 1.46 (CONT.) Principal stresses are thus

" 1, 2 = 37.5+2 22.5 ± [( 37.5!2 22.5 ) 2 + 14.332 ]2 1

or Hence We have

! 1 = 46.17 MPa

! 2 = 13.83 MPa

# max = 12 (" 1 ! " 2 ) = 16.17 MPa " p = 12 tan !1 237( !.514!22.33.5) = !31.2 o

Equation (1.18a) results in

" x ' = 37.5+2 22.5 + 37.5!2 22.5 cos( !62.4 o ) ! 14.33 sin( !62.4 o ) = 46.17 MPa

Therefore

! p ' = 31.2 o

Results are shown in a properly oriented element in Fig. (b). 13.83 MPa

!p' x’

Figure (b)

46.17 MPa ______________________________________________________________________________________ SOLUTION (1.47) State of stress is represented by Mohr’s circle in Fig. (a). From this circle, we determine

" x = !40 MPa ! y = 20 MPa " p ' = tan 1 2

!1 4 3

= 26.57

-60 (-"x,-40)

! (MPa) ("y,40) O

" (MPa)

2%!p’ 40

Figure (a)

Results are shown in Fig. (b). 50 MPa

!p''

Figure (b)

60 MPa

40 MPa ______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (1.48) 27 MPa

y’

x’

21 MPa

! x

! x + ! y = ! x' + ! y' 45 " 30 = "27 + ! ; ! = 42 MPa 45 # 30 45 + 30 + cos 2" + 15sin 2" ! x ' = 42 = 2 2

49.5 = 37.5cos 2! + 15sin 2! ! x ' y ' = #21 = #37.5sin 2" + 15cos 2"

(1)

Multiply this by ! 2.5 :

52.5 = 93.75sin 2! " 37.5cos 2!

(2)

Add Eqs. (1) and (2),

102 = 108.75sin 2! , 2! = 69.71o

! = 34.9o

______________________________________________________________________________________ SOLUTION (1.49) State of stress is represented by Mohr’s circle in Fig. (a). ! (MPa)

(100,!xy) 2%p’ 60

110

" (MPa)

Figure (a)

(20,-!xy) Referring to this circle, we obtain the results (Fig. b).

" p ' = 12 tan !1 43 = 18.43o 20 MPa

100 MPa 30 MPa

50 MPa

10 MPa

x x’

!p'

110 MPa

(b) (a) Figure (b) ______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (1.50)

" x = 60 MPa

" y = !18 MPa

# xy = !15 MPa

" x ' = 30 MPa

From Equation (1.18a):

30 = 602"18 + 602+18 cos 2!1 " 15 sin 2!1

13 cos 2"1 ! 5 sin 2"1 ! 3 = 0

Solving

2!1 = 56.52 o

!1 = 28.26o

We have

" y ' = " x + " y ! " x ' = 12 MPa

Equation (1.14b) gives

" x ' y ' = ! 602+18 sin 56.52 o ! 15 cos 56.52 o = !40.8 MPa

______________________________________________________________________________________ SOLUTION (1.51) We have 3

" = 4!Mr 3 = 4 (!21( 0!.1))103 = 84 MPa State of stress is represented by Mohr’s circle in Fig. (a). ! (MPa)

56 (84,!)

! O

84 MPa

" (MPa)

Figure (a) 1

" = (56 2 ! 42 2 ) 2 = 37.04 MPa Thus 3

Hence

T = #rJ = ( 37.04 )2" ( 0.1 ) = 58.18 kN ! m P = 2!fT = 2! ( 20)58.18 = 7311 kW

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (1.52) ! "

!xy

60o "2

"1=2"2

r = 2! 22"! 2 = !22

From Mohr’s circle,

" x ' y ' = ( !22 ) sin 60o = 0.433! 2

Therefore We have Solving

30 = 0.433! 2

! 2 = 69.28 MPa

! 2 = 2prt = 69.28

p( 250 2 ) = 69.28

p = 2.771 MPa

______________________________________________________________________________________ SOLUTION (1.53) pr t

A pr 2t

+ 2!Prt

Mohr’s circle representing stress at point A is shown in Fig. (a). ! (MPa)

(84,!xy) 60o

" (MPa)

(56,-!xy) Figure (a) From this circle: Then gives

.45 ) 42 = p0(.0005 = 90 p

p = 467 kPa

98(103 ) = 902P + 2! ( 0.45P)( 0.005)

P = 1069 kN

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (1.54) "’=70 ! (MPa)

y’

(40,-!xy)

35o

" (MPa)

70o

x’ 30

(a)

(100,!xy)

r = cos3035o = 36.62 MPa

" xy = !36.62 sin 35o = !21 MPa

( b ) Because of symmetry:

" x ' y ' = !" xy = 21 MPa

and

! x + ! y = ! x ' + ! y ' = 140 MPa

gives

! y ' = 40 MPa

______________________________________________________________________________________ SOLUTION (1.55) State of stress is represented by Mohr’s circle in Fig. (a). ! (MPa) (-12,20) -14

" (MPa) ("x,-20)

Figure (a) ( a ) Using this circle, we write

[

] 1

" max = ( ! x 2+12 ) 2 + 20 2 2

and

# max = 14 + OC = 14 + 12 (" x ! 12)

Solving,

! x = 186 MPa

Note that, alternately,

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 1.55 (CONT.)

[

] 1

! 14 = " x 2!12 ! ( " x 2+12 ) 2 + 20 2 2 yields ! x = 186 MPa , as before. ( b ) We have

" 1, 2 = 1862!12 ± [( 1862+12 ) 2 + 20 2 ]2 1

or and Also

" 1 = 188 MPa

" 2 = !14 MPa

# max = 12 (" 1 ! " 2 ) = 101 MPa 2 ( 20 ) o " p ' = 12 tan !1 186 +12 = 5.71

Results are shown in Fig. (b). 14 MPa 101 MPa

188 MPa

!p'

Figure (b) ______________________________________________________________________________________ SOLUTION (1.56) (a)

! 1 = 96.05 MPa

! 2 = 23.05 MPa

(b)

(# 12 ) max = 12 (" 1 ! " 2 ) = 36.05 MPa

!3 = 0

(# 13 ) max = 12 (" 1 ! " 3 ) = 48.03 MPa (# 23 ) max = 12 (" 2 ! " 3 ) = 11.98 MPa Plane of (! 12 ) max is shown in Fig. (a). Other maximum shear planes are sketched similarly. "2 (!12)max

"1 "3

28.15o x

Figure (a)

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (1.57)

A = 2!rt = 2! (60)( 4) = 1508 mm 2 ! y = prt = 2 (460 ) = 30 MPa 3

50 (10 ) " x = ! PA + 2prt = ! 1508 + 15 = 11,684 MPa (10! 6 )

! (MPa) "’ y

"x’

30 MPa

64o 11.684 MPa

11.684

" (MPa) r

! ' = (30 + 11.684) = 20.84 MPa r = 12 (30 ! 11.684) = 9.158 MPa 1 2

(a)

" x ' = " '! r cos 64 o = 16.82 MPa

(b)

! x ' y ' = r sin 64 o = 8.231 MPa

______________________________________________________________________________________ SOLUTION (1.58) 0.009& MN

! 1 = prt ! " 2 = 2prt # 0.2009 !rt

x’

60o

y’

50p Figure (a)

25p-2.5

Equilibrium of x ' and y ' directed forces results in (Fig. a): or

21 ! 50 p( 23 ) 2 ! ( 25 p ! 2.5)( 12 ) 2 = 0

pall = 494 kPa

and

7 + ( 25 p ! 2.5)( 43 ) ! 50 p( 43 ) = 0

from which

p = 547 kPa

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (1.59) Direction cosines are:

l1 = 3 2 m1 = 1 2 n1 = 0 l 2 = ! 1 2 m2 = 3 2 n 2 = 0 l3 = 0 m3 = 0 n3 = 1 Equation (1.28a) is thus

! x ' = 20( 43 ) + 0 + 0 + 2(12)( 23 )( 12 ) + 0 + 0 = 25.392 MPa Similarly, applying Eqs. (1.28b) through (1.28e), we obtain [! i ' j ' ] :

& 25.342 ' 2.66 ' 7.99# $ ' 2.66 ' 5.392 16.16! MPa $ ! 16.16 6 !" $% ' 7.99 Then, Eqs. (1.34) result in

I 1 = I 1 ' = 26 MPa I 2 = I 2 ' = !349 ( MPa ) 2 I 3 = I 3 ' = !6464 ( MPa ) 3 ______________________________________________________________________________________ SOLUTION (1.60) Direction cosines are:

l1 = 3 2 m1 = 1 2 l2 = ! 1 2 l3 = 0

n1 = 0

m2 = 3 2 n 2 = 0 m3 = 0 n3 = 1

Equation (1.28a) is therefore

! x ' = 60( 43 ) + 0 + 0 + 2( 40)( 23 )( 12 ) + 0 + 0 = 20[( 94 ) + 3 ] = 79.64 MPa Similarly, applying Eqs. (1.28b) through (1.28e), we obtain [! i ' j ' ] :

& 79.64 ' 5.98 ' 44.64# $ ' 5.98 ' 19.64 ' 2.68! MPa $ ! 20 !" $% ' 44.64 ' 2.68 Then, Eqs. (1.34) lead to

I 1 = I 1 ' = 80 MPa I 2 = I 2 ' = !2400 ( MPa ) 2 I 3 = I 3 ' = 8000 ( MPa ) 3

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (1.61) Referring to Appendix B:

" 1 = 13.212 MPa

and Thus,

l1 = 0.9556

" 2 = 5.684 MPa

m1 = 0.1688

" 3 = !8.896 MPa

n1 = 0.2416

# max = 12 (" 1 ! " 3 ) = 11.054 MPa

______________________________________________________________________________________ SOLUTION (1.62) Referring to Appendix B:

" 1 = 66.016 MPa

and

l1 = 0.9556

" 2 = 28.418 MPa

m1 = 0.1688

" 3 = !44.479 MPa

n1 = 0.2416

______________________________________________________________________________________ SOLUTION (1.63) Referring to Appendix B:

" 1 = 30.493 MPa

Thus,

" 2 = 12.485 MPa

" 3 = !16.979 MPa

# max = 12 (" 1 ! " 3 ) = 23.736 MPa

______________________________________________________________________________________ SOLUTION (1.64) Referring to Appendix B:

! 1 = 24.747 MPa

and

l1 = 0.6467

! 2 = 8.480 MPa

m1 = 0.3958

! 3 = 2.773 MPa

n1 = 0.6421

______________________________________________________________________________________ SOLUTION (1.65) ( a ) Equation (1.32) becomes

(30 ! " p ) 0 0 !" p 20

20 0 =0 !" p

Expanding,

! " p [" p (" p ! 30) ! 400] = 0

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 1.65 (CONT.) or

" p = 0,

Thus

" p = !10,

" 1 = 40 MPa ,

" p = 40

" 2 = 0,

" 3 = !10 MPa

( b ) For ! 1 = 40 MPa :

(30 ! 40)l1 + (0)m1 + 20n1 = 0, (0)m1 = 0, 2

l1 = 2n1 m1 = 0

The condition l1 + 0 + n1 = 1 gives

( 2n1 ) 2 + n12 = 1, Thus

l1 = 25 ,

n1 = 15

m1 = 0,

n1 = 15

______________________________________________________________________________________ SOLUTION (1.66) y 60o

y’

x,x’

30o z

60o

z’

Figure (a)

( a ) At point (3,1,5) with respect to xyz axis, we have [! ij ] :

&10 $ 0 $ $% 0

0 '4 0

0# 0 ! MPa ! 8 !"

Then, Eqs. (1.34) result in

I 1 = 14 MPa

(a)

I 2 = 8 ( MPa ) 2

I 3 = !320 ( MPa ) 3

Direction cosines of x’ y’ z’, referring to Fig. (a) are

l1 = 1 m1 = 0 l 2 = 0 m2 = 1 2

n1 = 0 n2 = 3 2

l 3 = 0 m3 = ! 3 2 n 3 = 1 2 Now Eqs. (1.28) and (a) give [! i ' j ' ] :

0 0 # &10 $0 5 3 3 ! MPa $ ! $% 0 3 3 ' 1 !"

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 1.66 (CONT.) Thus, Eqs. (1.34) yield

I 2 ' = 8 ( MPa ) 2

I 1 ' = 14 MPa

I 3 ' = !320 ( MPa ) 3

as before. y

y” x

26.56o

Figure (b)

x”

z,z” ( b ) Direction cosines are (Fig. b):

l1 = 2

5 m1 = ! 1

l2 = 1 5 l3 = 0

m2 = 2 m3 = 0

5 n1 = 0 5

n2 = 0 n3 = 1

With these and Eq. (a), Eqs. (1.28) yield [! i ' j ' ] :

5.6 0 # &7.2 $5.6 ' 1.2 0 ! MPa $ ! 0 8 !" $%0 Thus, Eqs. (1.34) result in

I 1 ' ' = 14 MPa

I 2 ' ' = 8 ( MPa ) 2

I 3 ' ' = !320 ( MPa ) 3

The I’s are thus invariants. ______________________________________________________________________________________ SOLUTION (1.67) Introducing the given data into Eq. (1.28a), we obtain

! x ' = 12( 12 ) 2 + 10( 23 ) 2 + 14(0) + 2[6( 12 )( 23 )] + 0 + 0 = 15.696 MPa Remaining stress components are determined in a like manner. The result, [! i ' j ' ] , is

7.089# & 15.696 ' 3.866 $' 3.866 6.304 ' 6.294! MPa $ ! $% 7.089 ' 6.294 14. !" ______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (1.68) Equations (1.34) become

I 2 = " x $ " y # ! xy2

I1 = ! x + ! y

I3 = 0

Equation (1.33) is then

" 3p ! (" x + " y )" p2 + (" x" y ! # xy2 )" p = 0

# p2 ! (# x + # y )# p + (# x# y ! " xy2 ) = 0

Solution of this quadratic equation is 1

" 1, 2 =

" x +" y 2

± 12 [" x2 + 2" x" y + " y2 # 4(" x + " y # ! xy2 )] 2

" x +" y 2

± [( x 2 y ) 2 + ! xy2 ]

" #"

1 2

______________________________________________________________________________________ SOLUTION (1.69) Referring to Appendix B, we obtain the following values.

! 1 = 12.049 MPa

(a) and

l1 = 0.6184

" 2 = !1.521 MPa

and

n1 = 0.5772

m1 = 0.5333

! 1 = 19.238 MPa

(b)

l1 = 0.3339

" 3 = !4.528 MPa

! 2 = 13.704 MPa

m1 = 0.3862

! 3 = 4.648 MPa

n1 = 0.8599

______________________________________________________________________________________ SOLUTION (1.70) ( a ) Direction cosines are: y C

l = 4 5 = 0.8 m = 3 5 = 0.6 n=0

" x’ B

Equation (1.40) is thus

! = 100(0.8) 2 + 60(0.6) 2 + 2( 40)(0.8)(0.6) = 124 MPa

Equations (1.26) yield

p x = 100(0.8) + 40(0.6) = 104 MPa

p y = 40(0.8) + 60(0.6) = 68 MPa

p z = 80(0.6) = 48 MPa

Equation (1.41) is then

" = [104 2 + 682 + 482 ! 124 2 ] 2 = 48.66 MPa (CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 1.70 (CONT.) ( b ) Direction cosines are:

y "

y’ z

l=0 m=2

20 = 0.447

n=4

20 = 0.894

B 2

Equation (1.40) results in

! = 60(0.447) 2 + 20(0.894) 2 + 2(80)(0.447)(0.894) = 91.913 MPa

Equations (1.26) yield

p x = 40(0.447) = 17.88 MPa

p y = 60(0.447) + 80(0.894) = 98.34 MPa

p z = 80(0.447) + 20(0.894) = 53.64 MPa Equation (1.41) leads to 1

" = [17.882 + 98.34 2 + 53.66 2 ! 91.9132 ] 2 = 66.481 MPa ( c ) Direction cosines are:

l = 0.512

m = 0.384

n = 0.768

Equation (1.40) is therefore

! = 100(0.512) 2 + 60(0.384) 2 + 20(0.768) 2 + 2[40(0.512)(0.384) + 80(0.384)(0.768)] = 109.77 MPa

Equations (1.26) yield

p x = 100(0.512) + 40(0.384) = 66.56 MPa p y = 40(0.512) + 60(0.384) + 80(0.768) = 104.96 MPa

p z = 80(0.384) + 20(0.768) = 46.08 MPa Equation (1.41) gives 1

" = [66.56 2 + 104.96 2 + 46.082 ! 109.77 2 ] 2 = 74.3 MPa ______________________________________________________________________________________ SOLUTION (1.71) (a)

Direction cosines are: y C 2 A

x’

l=2 !

13 = 0.555

m = 3 13 = 0.882 n=0

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 1.71 (CONT.) Equation (1.40) is then

! = 100(0.555) 2 + 60(0.832) 2 + 2( 40)(0.555)(0.832) = 109.277 MPa

Equations (1.26) lead to

p x = 100(0.555) + 40(0.832) = 88.78 MPa

p y = 40(0.555) + 60(0.832) = 72.12 MPa

p z = 80(0.832) = 66.56 MPa Equation (1.41) gives then 1

" = [88.782 + 72.12 2 + 66.56 2 ! 109.277 2 ] 2 = 74.67 MPa ( b ) Direction cosines are:

y "

y’ z

B 1

G z’

l=0 m =1

5 = 0.447

n=2

5 = 0.894

Thus, the results are the same as those obtained in Solution of Prob. 1.70b.

r r

( c ) We have rg = 3i , re = 2 j , ra = k . Equation (P1.70) is therefore

& x ' 3 y z# $ ' 3 2 0! = 0 $ ! $% ' 3 0 1!" or

2 x + 3 y + 6z = 6

Direction cosines are:

2 2 2 + 32 + 62

= 72

m = 73

n = 76

With these and given stresses, Eqs. (1.40) and (1.26) yield

! = 102.449 MPa

and

p x = 45.714 MPa

p y = 105.714 MPa

p z = 51.429 MPa

Substituting the above values into Eq. (1.41), we obtain

! = 73.582 MPa

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (1.72) See: Hint, Prob. 1.70:

2 32 +12 + 2 2

= 214

1 14

n = ! 314

Equation (1.40) gives

" = 20( 214 ) 2 + 30( 114 ) 2 + 50( ! 314 ) 2 + 2[10( 214 )( 114 ) ! 10( 214 )( ! 314 )] = 51.43 MPa

Equation (1.41):

" = {[20( 214 ) + 10( 114 ) ! 10( ! 314 )]2 1

+ [10( 214 ) + 30( 114 ) + 0]2 + [ !10( 214 ) + 0 + 50( ! 314 )]2 ! 51.34 2 } 2

= 7.413 MPa

______________________________________________________________________________________ SOLUTION (1.73) Direction cosines are

l = cos 35o = 0.8192

m = cos 60o = 0.5

n = cos 73.6o = 0.2823

Equation (1.40) results in

" = 60(0.8192) 2 ! 40(0.5) 2 + 30(0.2823) 2 + 2[20(0.8192)(0.5) ! 5(0.5)(0.2823) + 10(0.8192)(0.2823)] = 52.25 MPa

Equations (1.26):

p x = 60(0.8192) + 20(0.5) + 10(0.2823) = 61.9725 MPa

p y = 20(0.8192) ! 40(0.5) ! 5(0.2823) = !5.0287 MPa

p z = 10(0.8192) ! 5(0.5) + 30(0.2823) = 14.1618 MPa Equation (1.41) is thus 1

" = [(61.9725) 2 + ( !5.0287) 2 + (14.1618) 2 ! (52.25) 2 ] 2 = 36.56 MPa ______________________________________________________________________________________ SOLUTION (1.74) Direction cosines are

l = cos 40o = 0.766

m = cos 75o = 0.259

n = cos 54 o = 0.588

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 1.74 (CONT.) Equation (1.40):

! = 40(0.766) 2 + 20(0.259) 2 + 20(0.588) 2 + 2[40(0.766)(0.259) + 0 + 30(0.766)(0.588)] = 23.47 + 1.34 + 6.91 + 42.9 = 74.62 MPa

Equation (1.41) gives

" = {[40(0.766) + 40(0.259) ! 30(0.588)]2 1

+ [40(0.766) + 20(0.259) + 0]2 + [30(0.766) + 0 + 20(0.588)]2 ! 74.62 2 } 2 1

= [3436.3 + 1282.9 + 1206.7 ! 5568.1] 2 = 18.93 MPa ______________________________________________________________________________________ SOLUTION (1.75) Note: Planes of maximum shear stresses can be determined upon following a procedure similar to that used in Solution of Prob. 1.56. ( a ) From Problem 1.69a: Thus,

" 1 = 12.049 MPa " 2 = !1.521 MPa 1 (# 13 ) max = 2 (" 1 ! " 3 ) = 8.288 MPa (# 12 ) max = 12 (" 1 ! " 2 ) = 6.785 MPa (# 23 ) max = 12 (" 2 ! " 3 ) = 1.503 MPa

( b ) From Problem 1.69b: Thus,

! 1 = 19.237 MPa ! 2 = 13.704 MPa (# 13 ) max = 12 (" 1 ! " 3 ) = 7.294 MPa (# 12 ) max = 12 (" 1 ! " 2 ) = 2.766 MPa (# 23 ) max = 12 (" 2 ! " 3 ) = 4.528 MPa

" 3 = !4.528 MPa

! 3 = 4.648 MPa

______________________________________________________________________________________ SOLUTION (1.76) We have

" 1 = 48 MPa

" 2 = 36 MPa

" 3 = !72 MPa

( a ) From Eqs. (1.43) and (1.44): 1

" oct = 13 [( 48 ! 36) 2 + (36 + 72) 2 + ( !72 ! 48) 2 ] 2 = 53.96 MPa " oct = 13 ( 48 + 36 ! 72) = 4 MPa ( b ) Using Eq. (1.45),

" max = 12 ( 48 ! 36) = 6 MPa

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (1.77)

( !100 ! " p ) 0 ! 80 0 ( 20 ! " p ) 0 =0 (a) ! 80 0 ( 20 ! " p ) Expanding,

( 20 ! " )[(" + 100)(" ! 20) ! 6400] = 0

and

" 1 = 60 MPa ,

" 2 = 20 MPa ,

" 3 = !140 MPa

( b ) Apply Eqs. (1.43), (1.44), and (1.45): 1

" oct = 13 [(60 ! 20) 2 + ( 20 + 140) 2 + ( !140 ! 60) 2 ] 2 = 86.41 MPa " oct = 13 (60 + 20 ! 140) = !20 MPa ! max = 12 (60 + 140) = 100 MPa

______________________________________________________________________________________ SOLUTION (1.78) Octahedral and shearing stresses are given by

2 " oct = 19 [(! 1 # ! 2 ) 2 + (! 2 # ! 3 ) 2 + (! 3 # ! 1 ) 2 ]

2 " max = 14 (! 1 # ! 3 ) 2 2

Let us say, ! max > ! oct . Then

( ! 1 "2! 3 ) 2 > 19 [(! 1 " ! 2 ) 2 + (! 2 " ! 3 ) 2 + (! 3 " ! 1 ) 2 ] or 9 4

(! 1 " ! 3 ) > [(! 1 " ! 2 ) 2 + (! 2 " ! 3 ) 2 + (! 3 " ! 1 ) 2 ] 2

Subtracting (! 1 " ! 3 ) from both sides and noting that (! 1 " ! 3 ) we have 5 4

But

= (! 3 " ! 1 ) 2 ,

(! 1 " ! 3 ) 2 > (! 1 " ! 3 ) 2 + (! 2 " ! 3 ) 2

(! 1 " ! 3 ) 2 > (! 1 " ! 2 ) 2 + (! 2 " ! 3 ) 2

Thus, 5 4

(! 1 " ! 3 ) 2 > (! 1 " ! 2 ) 2 + (! 2 " ! 3 ) 2

(a)

The squares of the difference between ! 1 and ! 3 will always be greater then the sum of the squares of the difference between ! 1 and ! 2 , ! 2 and ! 3 , since ! 1 > ! 2 > ! 3 . Hence, Eq. (a) is true and our assumption is correct. That is

! max > ! oct ______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (1.79) From Solution of Problem 1.64:

! 1 = 24.747 MPa

! 2 = 8.48 MPa

! 3 = 2.773 MPa

Applying Eqs. (1.43) and (1.44): 1

and

" oct = 13 [( 24.747 ! 8.48) 2 + (8.48 ! 2.773) 2 + ( 2.773 ! 24.747) 2 ] 2 = 9.31 MPa ! oct = 13 ( 24.747 + 8.48 + 2.773) = 12 MPa

Therefore

poct = (9.312 + 12 2 ) 2 = 15.19 MPa

______________________________________________________________________________________ SOLUTION (1.80) Shearing stress, in terms of principal stresses, is given by

" 2 = ! 12 l 2 + ! 2 m 2 + ! 32 n 2 # (! 1l 2 + ! 2 m 2 + ! 3n 2 ) 2 2

(a)

We substitute n = 1 ! m ! l into Eq. (a), calculate its derivatives with respect to l and m , and equate these derivatives to zero: "$ "l

= l[(# 1 ! # 3 )l 2 + (# 2 ! # 3 )m 2 ! 12 (# 3 ! # 1 )] = 0

(b)

"$ "m

= m[(# 1 ! # 3 )l 2 + (# 2 ! # 3 )m 2 ! 12 (# 2 ! # 3 )] = 0

(c)

One solution is l = m = 0 . Solutions for the direction cosines of planes for which ! is a maximum of minimum can also be found as follows. Take l = 0 :

Eq. (c) gives m = ±

Take m = 0 :

Eq. (c) gives l = ±

There are, in general, no solutions of Eqs. (b) and (c) in which l and m are both different from zero, for this case the expressions in brackets cannot both vanish. By the above procedure we can form the following table. Direction cosines for planes of ! max and ! min

l= 0 0 ±1 0 m = 0 ±1 0 ± 1 2 n = ±1 0 0 ± 12

± 12 0 ± 12

± 12 ± 12 0

The first three columns define the planes for ! min , where ! = 0 . The last three columns give planes through each principal axes bisecting the angles between the two other principal axes. Substituting the latter direction cosines into Eq. (a), we have

(" 23 ) max = ± ! 2 #2! 3

(" 13 ) max = ± ! 1 #2! 3

(" 12 ) max = ± ! 1 #2! 2 Similarly, introducing the direction cosines given in the above table into Eq. (1.37), we obtain the normal stresses associated with the maximum shearing stresses:

! 12 ' = ! 1 +2! 2

! 13 ' = ! 1 +2! 3

! 23 ' = ! 2 +2! 3

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (1.81)

1 ! avg = (! x + ! y ) 2 1 = (100 + 20) = 60 MPa 2 ! x #! y 2 2 r= ( ) + " xy 2 100 ! 20 2 = ( ) + 602 2 = 72.1 MPa

! (MPa) "avg =60

" (MPa)

! 1 = ! avg + R = 132.1 MPa

! 3 = "12.1 MPa (a)

! 2 = 30 MPa ! 1 = 132.1 MPa ! 3 = "12.1 MPa 1 (! max ) a = (" 1 # " 3 ) 2 1 = [132.1 ! (!12.1)] = 72.1 MPa 2

(b)

! 3 = "30 MPa ! 1 = 132.1 MPa 1 (! max ) a = (" 1 # " 3 ) 2 1 = [132.1 ! (!30)] = 81 MPa 2

! 2 = "12.1 MPa

______________________________________________________________________________________ SOLUTION (1.82) ! (MPa)

-50

O 100+ 60 2

= 80

o r 45

" (MPa)

(100,-20)

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 1.82 (CONT.) From Mohr’s circle, we have

r = 202 + 202 = 28.3 MPa ! 1 = 108 MPa ! 2 = 51.7 MPa # 3 = !50 MPa " p ' = 22.5o ______________________________________________________________________________________ SOLUTION (1.83) ! (MPa)

-60

C1 80 2

= 40

" (MPa)

45o r

(80,-40)

Referring to Mohr’s circle:

r = 402 + 402 = 56.57 MPa ! p ' = 22.5o

! 1 = 96.57 MPa

! 2 = 16.57 MPa

______________________________________________________________________________________ SOLUTION (1.84) ( a ) In the yz plane: 50 MPa

"=25

15 MPa

!(MPa)

C r

O "1

" (MPa)

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 1.84 (CONT.) We have

r = 252 + 152 = 29.16 MPa Thus

" 1 = r ! " = 29.16 ! 25 = 4.16 MPa " 3 = ! r ! " = !29.16 ! 25 = !54.16 MPa " 2 = !20 MPa

( b ) Using Eqs. (1.43) and (1.44): 1

" oct = 13 [( 4.16 + 20) 2 + ( !20 + 54.16) 2 + ( !54.16 ! 4.16) 2 ] 2 = 23.92 MPa " oct = 13 ( 4.16 ! 20 ! 54.16) = !23.33 MPa From Eq. (1.45),

! max = 12 ( 4.16 + 54.16) = 29.16 MPa

______________________________________________________________________________________ SOLUTION (1.85) It is noted that l

+ m2 + n2 = 1 .

Applying Eq. (1.37), we have

" = 35( 133 ) ! 14( 131 ) ! 28( 139 ) = !12.39 MPa

Equation (1.39), substituting the given data and the direction cosines determined above, gives

! = 26.2 MPa

Surface tractions.

Equations (1.48) give

p x = ! 1l = 35(

3 13

) = 16.81 MPa

p y = " 2 m = !14(

1 13

p z = " 3n = !28(

9 13

Check: p

) = !3.88 MPa

) = !23.30 MPa

= p x2 + p 2y + p z2 = " 2 + ! 2 = 840 ( MPa ) 2

Observe that Approach I is more conveniently leads to results. ______________________________________________________________________________________ SOLUTION (1.86)

! oct = 13 ( 40 + 25 + 15) = 26.667 MPa 1

" oct = 13 [( 40 ! 25) 2 + ( 25 ! 15) 2 + (15 ! 40) 2 ] 2 = 10.274 MPa

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 1.86 (CONT.) We have l = m = n = 1

3 . Note that l 2 + m 2 + n 2 = 1 .

Surface tractions. Equations (1.48) give

p x = ! 1l = 40( 13 ) = 23.09 MPa

p y = ! 2 m = 25( 13 ) = 14.43 MPa p z = ! 3n = 15( 13 ) = 8.66 MPa Check:

p 2 = p x2 + p 2y + p z2 = " 2 + ! 2 = 816.7 ( MPa ) 2 End of Chapter 1

CHAPTER 2 SOLUTION (2.1) ( a ) Yes. Eqs. (2.12) are satisfied. ( b ) No. Eqs. (2.12) are not satisfied. ______________________________________________________________________________________ SOLUTION (2.2) Apply Eqs. (2.4):

# x = 2c

(a)

# y = !6cy

u AB = ! " x dx = 4c = 0.4 mm 1

v AD = "1 # y dy = ! 6c y2 2

Thus,

(b)

" xy = 2c( x + y )

2 1

= !11.25c = !1.125 mm

LA' B ' = 2000.4 mm

LA' D ' = 1498.875 mm

! xy = 2c(1 + 12 ) = 300 µ

( c ) We have and

u A = c( 2 ! 1 + 14 ) = 2.25c

v A = c(12 " 3 ! 14 ) = 0.25c

x A' = 1 + 2.25c = 1000.225 mm y A' = 0.5 + 0.25c = 500.025 mm

______________________________________________________________________________________ SOLUTION (2.3) Equations (2.4), for the given displacement field, yield [! ij ] :

0 'y 2 # & 2x $ 0 2z ( 2 y ' x ) 2! c $ ! 2z $% ' y 2 ( 2 y ' x ) 2 !" At point (0,2,1), we have [! ij ] :

0 ' 100# & 0 $ 0 200 200 ! µ $ ! $% ' 100 200 200 !"

______________________________________________________________________________________ Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 2–1

______________________________________________________________________________________ SOLUTION (2.4) First two of Eqs. (2.4) give

! x = 2a0 xy 2 + a1 y 2 + 2a 2 xy ! y = b0 x 2 + b1 x

Equation (2.11):

( 4a0 + 2a1 ) + ( 2b0 ) = 2c0 x + c1

2( 2a0 ! c0 ) x + 2( a1 + b0 ) ! c1 = 0

This is satisfied if x ! 0 :

2a0 ! c0 = 0,

c0 = 2 a 0

2( a1 + b0 ) ! c1 = 0,

c1 = 2( a1 + b0 )

______________________________________________________________________________________ SOLUTION (2.5) Equation (2.11) yields

2a1 + 12 y 2 + 2b1 + 12 x 2 = 3c1 ( x 2 + y 2 ) + c1c2

Solving,

c1 = 4

c2 = 12 ( a1 + b1 )

______________________________________________________________________________________ SOLUTION (2.6) The change in length of bar AB is

! AB = " AB LAB = (#500 $10#6 )(400) = #0.2 mm From triangles B’BF and EE’F: 250 mm B’ 0.2 mm B

F D’

500 mm E

E’

x = 117.2 mm

132.8 632.8 , = 3 !D

! D = 0.63 mm

From triangles DD’F and EE’F:

!D 3 mm

0.2 3 = , x 750 ! x

Thus

" CD =

! D 0.63 = = 1260 µ LCD 500

______________________________________________________________________________________ SOLUTION (2.7) C

"LAB 2.5 =# 1800 LAB = !1389 µ

! AB = 3.0 m

2.4 m

!LBC

! A

2.5 mm

" BC =

1..8 m

#$LBC #0.0025cos ! = 3.0 LBC

!0.0025(1.8 3.0) = !500 µ 3.0

______________________________________________________________________________________ SOLUTION (2.8) (a)

(b)

2.8 = 1400 µ 2000 "1.3 = "1300 µ !y = 1000 ! xy = 0

!x =

ACB = 90o

1.0014 = 90.1547 o 0.9987 ! = 90o " 90.1547 o = "0.1547o = "2.7 # 10"3 rad = !2700 µ A ' C ' B ' = 2 tan !1

______________________________________________________________________________________ SOLUTION (2.9) ( a ) Equations (2.4) give

"0.075 ! x = ##ux = 0.175150 = 667 µ

"( "0.05 ) ! y = ##yv = 0.025100 = 750 µ

and

"( "0.05 )] 0.075 ! xy = 0"100 + [ "0.125150 = "1250 µ

( b ) Equation (2.16) is therefore 1

" 1, 2 = 667+2 750 ± [( 667!2 750 ) 2 + 6252 ] 2 or When

! 1 = 1335 µ ! 2 = 82 µ !1 !1250 1 " p = 2 tan 667!750 = 43.1o

and stresses are substituted into Eq. (2.14a), we obtain ! x ' = 82 µ . Thus,

! p ' ' = 43.1o ______________________________________________________________________________________ Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 2–3

______________________________________________________________________________________ SOLUTION (2.10) Use Eq. (2.16) with the strains obtained in Example 2.1: 1

" 1, 2 = 1250!22000 ± [( 1250+22000 ) 2 + 750 2 ] 2 or ! 1 = 1415 µ Apply Eq. (2.15):

! 2 = "2165 µ

o " p = 12 tan !1 12501500 ! 2000 = 12.39

Substituting this angle and the stresses, Eq. (2.14a), gives 1415 µ . Thus,

! p ' = 12.39 o

______________________________________________________________________________________ SOLUTION (2.11) We have, using Eqs. (2.3):

004 ! x = ""ux = 0.20 = 200 µ #v "0.003 ! y = #y = 12 = "250 µ

and

! xy = ##vx + ##uy = "1000 " 500 = "1500 µ Then, Eq. (2.16) yields 1

" 1, 2 = 200!2 250 ± [( 200+2 250 ) 2 + ( !750) 2 ] 2 or

! 1 = 758 µ

! 2 = "808 µ

Apply Eq. (2.15):

o !1500 " p = 12 tan !1 200 + 250 = 36.65

For this angle, Eq. (2.14a) gives ! x ' = 758 µ . Therefore,

" p ' = !36.65o ______________________________________________________________________________________ SOLUTION (2.12) y

20 mm

A x”

x’ B 12 mm

We have, from geometry:

AC = QB = 23.32 mm !1 = 30.96o ! 2 = 149.04 o

Equation (2.14a) leads to

! x ' = 300+2 500 + 300"2500 cos 61.92 o + 100 sin 61.92 o = 441 µ

(CONT.) ______________________________________________________________________________________ Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 2–4

______________________________________________________________________________________ 2.12 (CONT.) Thus,

! QB = 441 µ ( 23.32) = 0.01 mm

Similarly,

! x " = 400 " 100 cos 298.08o + 100 sin 298.08o = 265 µ

and

! AC = 265 µ ( 23.32) = 0.006 mm

______________________________________________________________________________________ SOLUTION (2.13) y x’ 30 mm A B !

x”

15 mm !

We find, from geometry:

AC = QB = 33.54 mm !1 = 26.56o ! 2 = 153.44 o

Equation (2.14a) gives

! x ' = 400+2 200 + 400"2 200 cos 53.12 o + 150 sin 53.12 o = 240 µ

Hence,

! QB = 240 µ (33.54) = 0.008 mm

In a like manner,

! x " = 300 + 100 cos 306.88o " 150 sin 306.88o = 480 µ

and

! AC = 480 µ (33.54) = 0.016 mm

______________________________________________________________________________________ SOLUTION (2.14) x’ x” C 60 mm B ! x’’’ 40 mm 30o ! x D A 3

We obtain, from geometry:

AC = 96.73 mm BD = 32.29 mm o ! 2 = 11.93 ! 3 = 141.73o

From Solution of Problem 1.42:

" x = 3.464! 0

"y =0

! xy = ! 0

Thus, (CONT.) ______________________________________________________________________________________ Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 2–5

______________________________________________________________________________________ 2.14 (CONT.) 6

( 70"10 ) ! x = 3.464 = 1212 µ 200 (109 )

! y = "0.3(1212) = "364 µ 6

0.3)( 70"10 ) ! xy = 2 (1+200 = 910 µ (109 )

(a)

! x ' = 12122"364 + 12122+364 cos 60o + 455 sin 60o = 1212 µ ! AB = 1212 µ ( 40) = 0.05 mm

(b)

! x " = 424 + 788 cos 23.86o + 455 sin 23.86o = 1328.7 µ ! AC = 1328.7 µ (96.73) = 0.13 mm Similarly,

! x ''' = 424 + 788 cos 283.46o + 455 sin 283.46o = 164.92 µ ! BD = 164.92(32.29) = 0.005 mm 1

" 1, 2 = 12122!364 ± [( 12122+364 ) 2 + 4552 ] 2

! 1 = 1334 µ

! 2 = 486 µ

Substitution of

" p ' = 12 tan !1 1212910+364 = 15o into Eq. (2.14a) yields 1334 µ .

______________________________________________________________________________________ SOLUTION (2.15) ( a ) We have

! max = 400 " 200 = 200 µ

Maximum shearing strain occurs on a plane oriented at 45 o from the plane of principal strains. ! 2

(106 )

100

(" x , 12 ! xy ) 60o

200

400

! (10 6 )

(" y ,# 12 ! xy ) Figure (a) ( b ) From Mohr’s circle Fig. (a):

! x = 300 + 100 cos 60o = 350 µ ! x = 300 " 100 cos 60o = 250 µ

! xy = "( 400 " 200) sin 60o = "173 µ ______________________________________________________________________________________ Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 2–6

______________________________________________________________________________________ SOLUTION (2.16) We have u B = 3 mm and v B = 1.5 mm . ( a ) Take: Hence,

u = c1 xy

v = c2 xy c1 = 500(10 !6 )

3(10 "3 ) = c1 (3 ! 2) ,

c2 = 250(10 !6 )

1.5(10 "3 ) = c2 (3 ! 2) , Thus,

u = 500(10 !6 ) xy

( b ) Using Eqs. (2.3),

" x = 500 µy

v = 250(10 !6 ) xy

" y = 250 µx

! xy = 250 µ ( 2 x + y )

which satisfy Eq. (2.11): the strain field is possible. At point B, we thus have

(" x ) B = 1000 µ

( c ) We obtain " = tan

(" y ) B = 750 µ

( ! xy ) B = 2000 µ

= 33.69 o .

!1 2 3

Equation (2.14a) is therefore

! x ' = 875 + 125 cos 67.38o + 1000 sin 67.38o = 1846 µ

______________________________________________________________________________________ SOLUTION (2.17) (-300,450) C

2! p '

(-900,-450)

! 2

(106 )

! (10 6 )

-600

Figure (a) Refer to Mohr’s circle (Fig. a): 1

" 1, 2 = !600 ± [(300) 2 + ( 450) 2 ] 2 or and or

! 1 = "59 µ

! 2 = "1141 µ

o 450 ( 2 ) 2" p ' ' = tan !1 !( 900 !300 ) = !56.31

! p ' = 61.85o

______________________________________________________________________________________ SOLUTION (2.18) ! 2

(106 )

(300,450)

2! p '

!2 C

! (10 6 )

!1 (900,-450)

600

Figure (a) Referring to Mohr’s circle shown in Fig. (a), we obtain 1

! 1, 2 = 600 ± [(300) 2 + ( 450) 2 ] 2 from which and or

! 1 = 1141 µ

! 2 = 59 µ

o 450 ( 2 ) 2" p ' ' = tan !1 900 !300 = 56.31

" p ' = !61.85o

______________________________________________________________________________________ SOLUTION (2.19) Referring to Fig. P2.19,

c1 = !83.3(10 !6 )

u A = !0.0005 = c1 (3 " 1 " 2),

c2 = 50(10 !6 )

v A = 0.0003 = c2 (6),

c3 = !100(10 !6 )

w A = !0.0006 = c3 (6), Hence,

u = !83.3(10 !6 ) xyz

( a ) Using Eqs. (2.4), we thus have

! x = "83.3 µyz

v = 50(10 !6 ) xyz

! y = 50 µxz

! xy = ( "83.3xz + 50 yz ) µ

w = !100(10 !6 ) xyz

! z = "100 µxy

! xz = ( "100 yz " 83.4 xy ) µ

! yz = (50 xy " 100 xz ) µ This forgoing expressions satisfy Eqs. (2.12): the strain field is possible. Substitute x=-3 m, y=1 m, and z=2 m into the above equations to obtain strains at A, [! ij ] :

& ' 167 ' 200 ' 225# $ ' 200 300 ' 225! µ $ ! %$ ' 225 ' 225 ' 300"!

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 2.19 (CONT.) ( b ) Let, for example, x’-axis lie along the line from A to B (Fig. P2.19). The direction of cosines of AB are

l1 = ! 313

m1 = 0

n1 = ! 213

Thus, Eq. (2.18a):

" x ' = " x l12 + " z n12 + ! xz l1n1 = !167( 139 ) ! 300( 134 ) ! 450( !133 )( !132 ) = !416 µ

( c ) Let y’-axis be placed along AC (Fig. P2.19). The direction cosines of AC are

m 2 = !1

l2 = 0

n2 = 0

Equation (2.18b) is therefore

! x ' y ' = ! xy l1m2 + ! yz n1m2 = !400( !133 )( !1) ! 450( !132 )( !1)

= !582 µ Negative sign shows that the angle BAC has increased. ______________________________________________________________________________________ SOLUTION (2.20) We now have (Fig. P2.19):

c1 = 100(10 !6 )

u A = 0.0006 = c1 (3 " 1 " 2),

Hence,

v A = !0.0003 = c2 (6),

c2 = !50(10 !6 )

w A = !0.0004 = c3 (6),

c3 = !66.7(10 !6 )

u = 100(10 !6 ) xyz

v = !50(10 !6 ) xyz

( a ) Applying Eqs. (2.4), we obtain

! x = 100 µyz

! y = "50 µxz

w = !66.7(10 !6 ) xyz

! z = "66.7 µxy

! xz = ( "66.7 yz + 100 xy ) µ

! xy = (100 xz " 50 yz ) µ

! yz = (100 xy " 66.7 xz ) µ These expressions satisfy Eqs. (2.12): the strain field is possible. Introducing x=3 m, y=1 m, and z=2 m into the above equations we find strains at A, [! ij ] :

250 &200 $250 ' 300 $ %$ 83.5 ' 275

83.5# ' 275 ! µ ! ' 200 "!

( b ) Let, for instance, x’-axis lie along the line from A to B (Fig. P2.19). The direction of cosines of AB are

l1 = ! 313

m1 = 0

n1 = ! 213

______________________________________________________________________________________ 2.20 (CONT.) Therefore, Eq. (2.18a):

" x ' = " x l12 + " z n12 + ! xz l1n1 = 200( 139 ) ! 200( 134 ) ! 167( !133 )( !132 ) = !154 µ

( c ) Let y’-axis be placed along (Fig. P2.19). The direction cosines of AC are

m 2 = !1

l2 = 0

Equation (2.18b) is thus

n2 = 0

! x ' y ' = ! xy l1m2 + ! yz n1m2 = 500( !133 )( !1) ! 550( !132 )( !1)

= 111 µ Positive sign means that the angle BAC has decreased. ______________________________________________________________________________________ SOLUTION (2.21) ( a ) Applying Eqs. (2.21),

J 1 = 200 ! 100 ! 400 = !300 µ

and

J 2 = ( !2 ! 8 + 4 ! 9 ! 4 ! 25)(10 4 ) = !44(10 4 ) ( µ ) 2 200 J 3 = 300 200

300 ! 100 500

200 500 = 58(106 ) ( µ ) 3 ! 400

( b ) Table of direction cosines:

x' y'

x 3 2 !1 2

y z 12 0 3 2 0 0

Thus, using Eqs. (2.18a),

" x ' = " x l12 + " y m12 + ! xy l1m1 = 200( 23 ) 2 ! 100( 12 ) 2 + 600( 23 )( 12 ) = 385 µ

( c ) Use Table B.1 (with " # ! and " # !

! 1 = 598 µ (d)

! 2 = "126 µ

2 ):

! 3 = "772 µ

! max = 598 + 772 = 1370 µ

______________________________________________________________________________________ SOLUTION (2.22) ( a ) Applying Eqs. (2.21),

J 1 = 400 + 0 + 600 = 1(103 ) µ

and

J 2 = (0 + 24 + 0 ! 1 ! 4 ! 0)(10 4 ) = 19(10 4 ) ( µ ) 2 400 J 3 = 100 0

100 0 ! 200

0 ! 200 = !22(106 ) ( µ ) 3 600

( b ) Using Eq. (2.18a),

! x ' = 400( 23 ) 2 + 200( 23 )( 12 ) = 387 µ ( c ) Use Table B.1 (with " # ! and " # !

! 1 = 664 µ

2 ):

! 2 = 416 µ

! 3 = "80 µ

! max = 664 + 80 = 744 µ

(d)

______________________________________________________________________________________ SOLUTION (2.23)

! 1 = 1807 µ

2 ): ! 2 = "228 µ

! 3 = "679 µ

l1 = 0.6184

m1 = 0.5333

n1 = 0.5772

Use Table B.1 (with " # ! and " # ! and

______________________________________________________________________________________ SOLUTION (2.24)

! = "L = 050.10 = 2000 µ 3

) " = PA = (! 164 )((10 = 141.5 MPa 0.012 ) 2 !6

(a)

.5(10 ) E = #" = 141 = 70.8 GPa 2000 (10! 6 )

(b)

#& = $%d = 0.33( 2000 " 10 !6 )(12) = 7.92(10 !3 ) mm e = $ (1 ! 2# ) = 2000(10 !6 )(1 ! 2 " 0.33) = 6.80(10 !4 )

______________________________________________________________________________________ SOLUTION (2.25) Nominal strain

025 ! 0 = 0.75 = 333 µ

______________________________________________________________________________________ 2.25 (CONT.) Nominal stress 3

10 ) " 0 = ! ( 09.(012 = 79.577 MPa )2 4

Modulus of elasticity 6

.577 (10 ) E = 79333 = 238.97 GPa (10! 6 )

True strain

! = ln(1 + 0.000333) = 333 µ

True stress

! = 79.577(1 + 0.000333) = 79.603 MPa

______________________________________________________________________________________ SOLUTION (2.26) d 5 kN

5 kN L

!x =

2 mm

5(103 ) 20(103 ) = = 160 #106 "d2 4 "d2 d min = 0.0063 m = 6.3 mm

Also,

E= or

160 !106 = 210 !109 0.002 L

Lmin = 2.625 m

______________________________________________________________________________________ SOLUTION (2.27) (a) Axial stress in the bar ! s = ! a = ! is

! = " s Es = 600(210) = 126 MPa Hence

! P = " A = 126[ (402 )] = 158.3 kN 4 (b)

Axial strain in aluminum equals

"a = Therefore

! a 126(106 ) = = 1,800 µ Ea 70(109 )

! = " a La + " s Ls = [1,800(0.5) + 600(1.5)]10!6 = 1.8 mm

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (2.28)

A = 2(! " 402 4) = 2.513! mm 2 (a)

P 600! (103 ) "= = = 239 MPa A 2513! (10#6 ) Since ! < ! yp , the result is valid. Thus,

(b)

! 239(106 ) = = 1195 µ E 200(109 ) #L = L! = 5(103 )(1195 $10"6 ) = 5.98 mm

! t = #"! = #0.3(1195)(10#6 ) = #358 µ "d = !358(10!6 )(40) = !0.014 mm

______________________________________________________________________________________ SOLUTION (2.29)

! P $150(103 ) = !758 µ = = E AE 70(109 )(# 4)(0.12 $ 0.082 )

(a)

#L = ! L = "758(10"6 )(0.4 $ 103 ) = !0.303 mm

(b)

$D = ! (" D) = 0.3(758)(10#6 )100 = 0.023 mm #6

(c) $t = ! (" t ) = 0.3(758)(10 )10 = 0.0023 mm ______________________________________________________________________________________ SOLUTION (2.30) (a) According to assumption 1, the rubber is in triaxial stress:

! x = ! z = # p,

!y = #

" 2 d 4

4Q "d2

Strains are: ! x = ! z = 0. The first of Eqs. (2.34) gives

!x = 0 = or

1 [" x $# (" y + " z )] E

0 = p #! ( # Solving,

4Q # p) "d2

4! Q " d 2 (1 #! )

!x !z

______________________________________________________________________________________ 2.30 (CONT.) (b) Substituting the data,

4(0.3)(5 "103 ) = 1.091 MPa (C ) ! (0.05) 2 (1 # 0.3)

______________________________________________________________________________________ SOLUTION (2.31) p=160 MPa= -!

! x = ! y = ! z = "!

d=250 mm

4 4! (1253 ) = 8.18(106 ) mm3 Vo = ! r 3 = 3 3 (a)

" x = $ E1 [! $# (! + ! )] = $ !E (1 $ 2# ) !160(106 ) = (1 ! 0.6) = !914 µ 70 "109 #d = ! x d = "914(10"6 )250 = !0.229 mm Decrease in circumference:

! ("d ) = #0.229! = #0.72 mm

(b)

#V = (1 $ 2! )" xVo = (0.4)(!914)(10!6 )(8.18 "106 ) = !2291 mm3

______________________________________________________________________________________ SOLUTION (2.32)

500 " 100 = 200 µ 2 1 75 ! p ' = tan "1 = 7o 2 300 ! s " = 45 + 7 = 52o

!'=

(a) ! 2 (µ)

!' y

r O

!1 C

2! p '

" (µ)

(500, -75)

r = 752 + 3002 = 309 µ ! max = 2 R = 618 µ !1 = 200 + 309 = 509 µ ! 2 = "390 + 200 = "109 µ

______________________________________________________________________________________ 2.32 (CONT.) 509 µ

y’

x’

109 µ

220 µ

x’ 7o (b)

!3 = ! z = " Thus,

200 µ

618 µ

52o

0.3 (500 " 100) = "172 µ 0.7

(! max )t = 509 + 172 = 681 µ

______________________________________________________________________________________ SOLUTION (2.33) ! 2 (µ)

1 ! avg = (! x + ! y ) 2 1 = (480 + 800) = 640 µ 2

480 r

" (µ)

640

(a)

r = (640 ! 800) 2 + 5602

= 582 µ

!1,2 = ! avg ± r

!1 = 640 + 582 = 1222 µ ! 2 = 640 " 582 = 58 µ (b)

! max = 2r = 2(582) = 1164 µ

(c)

(! max )t = "1 # " 3 = "1 # 0 = 1222 µ

______________________________________________________________________________________ SOLUTION (2.34) (a)

P 25(103 ) !x = = = 69.4 MPa A (0.06 " 0.006) ! 69.4(106 ) = 347 µ "x = x = 200(109 ) E ! y = #"! x = #(0.3)(347 $10#6 ) = #104 µ

______________________________________________________________________________________ 2.34 (CONT.) (b)

(c)

1 1 ! x ' = (! x + ! y ) + (! x # ! y ) cos 2" 2 2 o For ! = "30 : 1 1 ! x ' = (347 " 104) + (347 + 104) cos("60o ) = 182 µ 2 2 1 1 ! y ' = (347 " 104) " (347 + 104) cos("60o ) = 8.8 µ 2 2 o

For ! = "30 :

! x ' y ' = $(" x $ " y ) sin 2# = !(347 + 104) sin(!60o ) = 391 µ

______________________________________________________________________________________ SOLUTION (2.35) o

Substituting ! a = 0 , ! b = 45 , and ! c = 90 , Eqs (2.44) give

1 ! a = ! x , ! c = ! y , ! b = (! x + ! x + " xy ) 2 ! x = ! a , ! y = ! c , " xy = 2! b # (! a + ! c ) Thus, we have

! x = #800 µ ,

! y = 400 µ ,

(P2.35)

" xy = 2(#1000) # (#800 + 400) = #1600 µ

Thus,

"800 + 400 1200 2 1600 2 ± (" ) + (" ) = !200 ± 1000 2 2 2 !1 = 800 µ , ! 2 = "1200 µ "1600 ] = 26.6o ! p = 12 tan "1[ "800 " 400 ! x ' = "200 " 600 cos 53.2o " 800sin 53.2o = "1200 µ = ! 2

!1,2 =

Thus,

! p " = 26.6o y’

1200 µ

800 µ

x’ 26.6o x ______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (2.36)

DB = 0.005 2 m

10 MPa D y’

5 MPa

50 mm 50

x’ 45o

20 MPa x

( a ) Hooke’s law gives

# y = (10"0E.3!20 ) = E4

# x = ( 20"0E.3!10 ) = 17E

! xy = 5( 2E.6 ) = 13E Then, using Eq. (2.14a) with ! = ! + " 2 ,

" y ' = 21E (17 + 4) ! 0 ! 213E sin 90o = E4 Thus,

! BD = (" y ' )( BD ) = 0.283 m E

( b ) Applying Eqs. (1.18a,c):

! x ' = 302 + 102 cos 90o + 5 sin 90o = 20 MPa

" y ' = 15 ! 5 cos 90o ! 5 sin 90o = 10 MPa As before, Hooke’s law yields

# y ' = (10"0E.3!20 ) = E4

and

! BD = (" y ' )( BD ) = 0.283 m E

______________________________________________________________________________________ SOLUTION (2.37) The generalized Hooke’s law gives

# x = 0 = E1 [! x $ " (! y + ! z )]

# z = 0 = E1 [! z $ " (! x + ! y )] or

" x # !" z = !" y

# !" x + " z = !" y

Solving,

! x = ! z = 1#1" ! y Thus

" x = " z = # 1!#p!

______________________________________________________________________________________ SOLUTION (2.38) Using Eqs. (2.46): 1

" 1, 2 = 12 {( !100 + 100) ± [( !200) 2 + 100 2 ] 2 } = 12 (0 ± 224) (CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 2.38 (CONT.)

! 1 = 112 µ

! 2 = "112 µ

! 1, 2 = 200(210 ) [0 ± ( 224 1.3 )] or

" 1 = 17.2 MPa

and

" 2 = !17.2 MPa

o " p = 12 tan !1 !100 200 = !13.28

Equations (2.44),

! x = !a

! y = !c

" xy = 2! b # ! a # ! c

Equation (2.14a) gives

! x ' = 0 " 100 cos( "26.56o ) + 50 sin( "26.56o ) = "112 µ

Thus,

" p ' ' = !13.28o

______________________________________________________________________________________ SOLUTION (2.39) We have

! x + ! y = ! a + ! c = 1200 µ

and the first two of Eqs. (2.44):

1000 = # x cos 2 ( !15o ) + # y sin 2 ( !15o ) + " xy sin( !15o ) cos( !15o ) # 250 = " x cos 2 30o + " y sin 2 30o + ! xy sin 30o cos 30o

These may be written

! y = 1200 " ! x

1000 = 0.933" x + 0.067" y # 0.25! xy # 250 = 0.75" x + 0.25" y + 0.433! xy Solving,

" x = 522 µ

" y = 678 µ

! xy = #1873 µ

______________________________________________________________________________________ SOLUTION (2.40) ( a ) We have

! c = ! y = "50 µ

First two of Eqs. (2.44):

400 = " x ( 43 ) # 50( 14 ) + ! xy ( 12 )( 23 ) 300 = # x ( 43 ) ! 50( 14 ) + " xy ( ! 12 )( 23 ) Solving,

" x = 482 µ

! xy = 116 µ

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 2.40 (CONT.) Applying Eq. (2.16), 1

" 1, 2 = 4832!50 ± [( 4832+50 ) 2 + 582 ] 2 or Thus,

! 1 = 489 µ

! 2 = "56 µ

" max = ! 1 # ! 2 = 545 µ

( b ) Using Eq. (3.11b) of Chap.3, Hence,

! z = # 1"#" ( 483.50) = #217 µ = ! 3 (" max ) t = ! 1 # ! 3 = 706 µ

______________________________________________________________________________________ SOLUTION (2.41) From Eqs. (2.35), (2.37), and (2.38), we have 9

(10 ) # = 2200 ! 1 = 0.25 ( 80"109 ) 9

and

) " = 0.251!.25200!0(.10 = 80(10 9 ) 5

e = 200 + 300 = 500 µ

Then, Eqs. (2.36) lead to the following stress components, [! ij ] :

&72 16 0# $16 88 64! MPa $ ! $% 0 64 40!" ______________________________________________________________________________________ SOLUTION (2.42) Equation (2.35) yields 9

(10 ) # = 2200 ! 0.25 ( 80"109 )

Then, introducing the given data into the generalized Hooke’s law, Eqs. (2.34), we calculate the following strain components, [! ij ] :

31.25# & 81.25 ' 25 $ ' 25 ' 43.75 62.5 ! µ $ ! 62.5 50 !" $% 31.25 ______________________________________________________________________________________ Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 2–19

______________________________________________________________________________________ SOLUTION (2.43) Using Eq. (2.35), 9

G = 279(1(+100.3)) = 26.92 GPa For x=1 m, y=2 m, and z=4 m, we obtain [! ij ] :

&34 48 8# $ 48 5 1! MPa $ ! $% 8 1 5!" Then, Eqs. (2.34) yield [! ij ] : & 443 892 149# $892 ' 96 19! µ $ ! 19 ' 96!" $%149

______________________________________________________________________________________ SOLUTION (2.44) Substituting x=3/4 m, y=1/4 m, and z=1/2 m into Eqs. (d) of Example 1.2, we have [! ij ] :

& ' 0.359 2.625 0.234# $ 0.234 0.875 0 ! MPa $ ! 0.125"! %$ 0.234 0

Equation (2.35) gives, 9

(10 ) G = 2200 (1+ 0.25 ) = 80 GPa

Applying Hooke’s law, we compute the strain components [! ij ] :

& ' 3 33 3# $ 33 5 0! µ $ ! $% 3 0 0!"

______________________________________________________________________________________ SOLUTION (2.45) We have ! z = 0 . Using Hooke’s law:

# x = E1 (" x $ !" y ) = 70%1103 (60 $ 903 ) = 1286 µ # y = E1 (" y $ !" z ) = 70%1103 (90 $ 603 ) = 1571 µ

" z = $ #E (! x + ! y ) = $ 2101%103 (60 + 90) = $714 µ (a)

" AB = $ y b = (1571 ! 10 #6 )( 400) = 0.628 mm

(b)

e = ! x + ! y + ! z = 1286 + 1571 " 714 = 2143 µ V0 = 300 ! 400 ! 10 = 1.2 ! 10 6 mm 3 "V = eV0 = ( 2143 ! 10 #6 )(1.2 ! 10 6 ) = 2,571.6 mm 3

______________________________________________________________________________________ SOLUTION (2.46) ( a ) Using generalized Hooke’s law, 6

! x = 20010(109 ) [ "60 " 0.3( "50 " 40)] = "165 µ

! y = 200(1103 ) [ "50 " 0.3( "60 " 40)] = "100 µ ! z = 200(1103 ) [ "40 " 0.3( "60 " 50)] = "35 µ Thus,

"a = a# x = !0.04125 mm "b = b# y = !0.02 mm

"c = c# z = !0.00525 mm e = ! x + ! y + ! z = "300 µ

(b) and

"V = e( abc ) = !2250 mm 3

______________________________________________________________________________________ SOLUTION (2.47) Applying generalized Hooke’s law, 6

! x = 7010(109 ) [70 " 13 ( "30 " 15)] = 1214 µ

! y = 70(1103 ) [ "30 " 13 (70 " 15)] = "691 µ

! z = 70(1103 ) [ "15 " 13 (70 " 30)] = "405 µ Thus,

!a = a" x = 0.1821 mm "b = b# y = !0.0691 mm

"c = c# z = !0.0304 mm e = ! x + ! y + ! z = 118 µ

(b) and

!V = e( abc ) = 132.75 mm 3

______________________________________________________________________________________ SOLUTION (2.48) We have 9

(10 ) G = 200 2 (1+ 0.3) = 76.96 GPa 9

(10 ) " = 0.31!.200 = 111.11(10 9 ) 3( 0.4 )

Let

! 1 = 5c

! 2 = 4c

! 3 = 3c

We then obtain e=12c.

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 2.48 (CONT.) Using the first of Eqs. (2.36),

# 1 = 2G" 1 + !e

140(10 6 ) = 2(76.96 ! 10 9 )(5c ) + 111.11(10 9 )(12c )

This yields

c = 66.587(10 !6 )

Hence,

! 1 = 332.935 µ

! 2 = 266.348 µ

! 3 = 199.761 µ

Now applying Eqs. (2.36), we calculate the principal stresses as follows:

! 1 = 51.165 + 88.782 = 139.947 MPa ! 2 = 40.975 + 88.782 = 129.757 MPa ! 3 = 30.731 + 88.782 = 119.513 MPa

Therefore

! 3 : ! 2 : ! 1 = 119.513 : 129.757 : 139.947

! 3 : ! 2 : ! 1 = 1 : 1.086 : 1.171

______________________________________________________________________________________ SOLUTION (2.49)

% x = $E = !!ux

(a)

% y = " #$E = !!yv

Integrating; Then

u = #E x + f ( y ) !u !y

gives

v = ! "#E y + g ( x )

(a)

+ !!vx = 0

!f ( y ) !y

+ !g!(xx ) = 0 or

Integration leads to

g ( x ) = cx + d

"g ( x ) "x

= ! "f"(yy ) = c

f ( y ) = !cy + e

Equations (a) are thus

u = #E x ! cy + e

v = ! "#E y + cx + d

Boundary conditions u(0,0)=0 and v(0,0)=0 result in c=d=e=0. Thus

u = "E x

(b)

v = # !"E y

$ x = #E = const. Hence ! x = u x and " y = #!

$ y = ! "#E = const. y . Therefore

u = !E x and v = # !"E y ______________________________________________________________________________________ Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 2–22

______________________________________________________________________________________ SOLUTION (2.50) Equations (2.34) become 2

" x = cy +E!cx

" y = # cx E+!cy

Integrating,

Given:

u( x, y ) = ! # x dx = 3cE (3 y 2 x + "x 3 ) + g1 ( y )

(1)

v ( x, y ) = ! # y dy = 3cE (3 y 2 x + "y 2 ) + g 2 ( x )

(2)

" xy = 0

" xy = !!uy + !!vx = 0

(3)

From Eq. (1), !u !y

= dgdy1 + 2 Ec xy

(a)

Equation (2) leads to "v "x

= dgdx2 ! 2 Ec xy

dg1 dy

= 2 Ec xy + a1

(b)

Substituting Eqs. (a) and (b) into Eq. (3): dg 2 dx

= !2 Ec xy + a1

from which, after integration,

g1 ( y ) = Ec xy 2 + a1 y + a 2

(4)

g 2 ( x ) = ! Ec yx 2 ! a1 x ! a 2

Constants a1 and a 2 are obtained upon satisfying the prescribed boundary conditions. Substitution of Eqs. (4) into Eqs. (1) and (2) yields the solution for displacement field. ______________________________________________________________________________________ SOLUTION (2.51) Equations (2.39) and (2.35) give

K = 3(1"E2! ) = 23G(1("12+!! ))

Also,

K = 3(1"E2! ) = 3(12+E! ()(1+1!"2)! ) = 3E(1(+3!! +)(11""22!!)) = (1+! !)(E1"2! ) + 3(1E+! )

= ! + ( 2G3 )

Equations (2.39) and (2.35) yield 2

G = 3 K2 ((11+"!2!) ) = ( 3"1E+ 2(!1") 2(!1") 2! ) = E {[E3 (1(1!!22"")])!1}

= 9E

3E 2 3 E (1! 2" )

1 3 1! 2"

= 93KKE! E

Equations (2.35) and (2.39) give,

E = 2G (1 + " )

and G ( 3+ 2 G ) " +G

E 2 (1+! )

E = 3K (1 ! 2 E )

[3" + 1+E! ]

" + 2 (1E+! )

The above expression, after substituting ! from Eq. (2.38) and simplifying, reduce to E. (CONT.) ______________________________________________________________________________________ Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 2–23

______________________________________________________________________________________ 2.51 (CONT.) From formula (2.39),

(1 ! 2" ) = 3EK

" = 12 ! 6EK = 3 K6 K! E

We also have or

G = 2EG ! 1

(1 + " ) = 2EG

" = 2EG ! 1

This expression is written as

" = 3 K (21G!2" ) ! 1 = 32KG ! 62KG" ! 1

from which or

" + 62KG" = " (1 + 26"G ) = 32KG ! 1

" = 23( K3 K!+2GG ) Finally, we write

" = 2 (1+!2!)(E1#2! ) = 12#!2G! or

(1 ! 2" )# ! 2"G = 0

This yields,

" = 2 ( !!+G ) ______________________________________________________________________________________ SOLUTION (2.52) Equations (2.4) yield,

$ y = ! "# ( aE! x )

# x = " ( aE! x )

# xy = ! "#Ey + "#Ey = 0

The stresses are therefore,

% x = 1!E# 2 ($ x + #$ y ) = " ( a ! x )

$ y = 1!E" 2 (# y + "# x ) = 0

" xy = G! xy = 0 At x=0 and x=a, we have

! x = "a

!x =0

Applying Eqs. (1.48) at x=a, we obtain

' p x $ -0 0 0* '1 $ '0$ ! ! + (! ! ! ! & p y # = +0 0 0( &0# = &0# ! p ! +0 0 0( !0! !0! % z" , )% " % " This is, boundary conditions at x=a are satisfied.

______________________________________________________________________________________ 2.52 (CONT.) At y = ± b , stresses are

# x = $ (a ! x )

# y = " xy = 0

' p x $ -( x . a )/ ! ! + & py # = + 0 !p ! + 0 % z" ,

0 0* ' 0 $ '0$ ! ! ! ! 0 0( &± 1# = &0# ( 0 0() !% 0 !" !%0!"

and

We see that boundary conditions at y = ± b are also satisfied. ______________________________________________________________________________________ SOLUTION (2.53) Equations (2.4) give,

# x = $ !"Ez , " z = !Ez ,

# y = $ !"Ez

(a)

! xy = ! yz = ! xz = 0

Equations (2.12) are satisfied by the above strains. Equations (2.36) and (a) yield,

# x = 2G! x + " (! x + ! y + ! z ) = 0

and

# z = "z

#y =0

Note that

Fx = Fy = 0

" xy = ! yx = ! xz = 0

and

Fz = "!

Thus, the first of Eqs. (1.14) are satisfied, and the third leads to

0 + 0 + !!"zz + Fz = 0 or " ! " = 0 At ends, since x’ and z are parallel, n=cos(x’,z)= ± 1. Thus, boundary conditions (1.48) at z=L:

px = 0 + 0 + 0 = 0

py = 0 + 0 + 0 = 0

p z = 0 + 0 + !L(1) = !L

as required. Similarly, at z=0:

px = 0

py = 0

p z = " (0)( !1) = 0

Therefore, all equations of elasticity are satisfied. ______________________________________________________________________________________ SOLUTION (2.54)

Using Eq. (2.59), with P1 = P1 + P2 : 2

P2 ) L U = ( P12+ AE

______________________________________________________________________________________ 2.54 (CONT.) From Example 2.8, with P1 = P2 = P :

U v = 12! E ( AL) = 3PAEL 2

Hence,

5! 5P L U d = 12 E ( AL ) = 3 AE 2

U = U v + U d = 2 PAEL 2

______________________________________________________________________________________ SOLUTION (2.55) We have 2

L 4) U 2 = P 2(AE + P2 E( 3( 2LA4)) = 85 U 1

U 1 = 2PAEL And

L 8) U 3 = P 2(AE + P2 E( 7( 3LA8)) = 125 U 1

Comparison of these results show that strain energy decreases as the volume of the bar is increased, although all three bars have the same maximum stress. ______________________________________________________________________________________ SOLUTION (2.56) Stress field is described by

# x = # y = # z = ! p,

( a ) Equation (2.37) reduces to

e = " 3(1"E2! ) p ;

" xy = " xz = " yz = 0

1!( 2 3)] p ! 0.005 = ! 3[110 (109 )

p = 550 MPa

( b ) Equation (2.52) becomes 2

U 0 = 21E ( p 2 + p 2 + p 2 ) ! "E ( p 2 + P 2 + p 2 ) = 3 p 2(1E"2! ) 6 2

"10 ) = 32((550 (1 ! 23 ) = 1.375 MPa 110"109 )

Thus,

U = U 0V0 = 1.375(10 6 )( 43" ! 0.153 ) = 19.439 kN ! m

______________________________________________________________________________________ SOLUTION (2.57) Substituting the given data into Eq. (2.52), we have 3

(10 ) U 0 = 2 (10200 ) (60 2 + 50 2 + 40 2 ) ! 0.3200 (3000 + 2400 + 2000) = 8.15 kPa 3

Thus,

U = U 0 ( abc ) = 8.15(10 3 )(0.25 ! 0.2 ! 0.15) = 61.125 N ! m

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (2.58) Equation (2.59) leads to 2

U n = P 2( 3AEL 4 ) + 2P( n(2LA)4E) = 8PAEL (3 + n12 ) 2

Hence,

= 1+43nn2 2PAEL

We have, Thus,

for n = 1 :

U 1 = 2PAEL

for n = 12 :

U 1 2 = 7U4 1

for n = 2 :

U 2 = 1316U1

U1 2 > U1

and

U 2 < U1

______________________________________________________________________________________ SOLUTION (2.59) (a)

U = 21E ! dx ! " 2 dA = 21E ! dx ! ( PA + MyI ) 2 dA Since

! ydA = 0 , this becomes: U =! +! =U +U P 2 dx 2 AE

Here

M 2 dx 2 EI

a b 2 2 U b = 2 1EI & ' M AD dx + ' M DB dx # $% 0 !" 0 a b = 2 1EI & ( ( ' ML0 x ) 2 dx + ( ( ML0 x ' ) 2 dx '# $% 0 !" 0 2

= 6MEIL0 2 ( a 3 + b 3 ) 2

U = 2PAEL + 6MEIL0 2 ( a 3 + b 3 ) 2

Thus,

( b ) Substituting the given data: 3 2

3 2

) ( 2!10 ) ( 0.027 + 0.729 ) U = 2 ( 7.(58!!1010"3))((701.2!10 9 + ) 6 ( 70!109 )( 0.075!0.13 12 )(1.2 ) 2

= 0.0731 + 0.8 = 0.8731 N ! m ______________________________________________________________________________________ SOLUTION (2.60) 2

U = ! 2TJGL = T "( L 42 ) + T "( L 42 ) 2

d1 ) G

d2 )G

= 8!TGL ( d14 + d14 ) 1

(a)

TL # = ! JG = T"( L 42 ) + T"( L 42 ) 32

d1 G

= 16!GTL ( d14 + d14 ) 1

d2 G

(b)

Solving T from Eq. (b) and substituting into Eq. (a) result in Eq. (P2.60). ______________________________________________________________________________________ Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 2–27

______________________________________________________________________________________ SOLUTION (2.61) We have for segment AB, T AB = 3T and for segment BC , TBC = T (a)

U AB = ( 2TJGL ) AB = ! (91T.4 d(1)4.2Ga )16 = 44.977 !Td 4aG 2

U BC = ( 2TJGL ) BC = 16!dT4Ga 2

and

The total energy is thus

U = 60.981 !Td 4aG 2

( b ) Substituting the given data into the above equation, 3 2

U = 60.981 # ((10..402"10)4 ()42("010.5)9 ) = 2.831 kN ! m ______________________________________________________________________________________ SOLUTION (2.62)

M = 2p ( Lx ! x 2 ) b h x

pL 2

The maximum bending moment occurs at the midspan: 2

c ( pL 8 )( h 2 ) pL ! max = M max = 34 bh 2 I bh 3 12

Maximum strain energy density, 2

2 4

9p L U 0,max = !2max E = 32 Eb 2 h 4

Using Eq. (2.63), we obtain L

U = 2 1EI ! ( 2p ) 2 ( Lx " x 2 ) 2 dx 0

p 2 L5 240 EI

2 5

L = 20pEbh 3

It is required to find c: U 0,max = cU or V Thus, and

c = U 0,max UV

2 4

p L bhl c = 329 Eb = 458 2 4 h p 2 L5 20 Ebh 3

U 0,max = 458VU

______________________________________________________________________________________ SOLUTION (2.63) P

x a

x’ L

Pa L

M AB = Px M BC = Pax L

______________________________________________________________________________________ 2.63 (CONT.) We have a

U AB = 2 1EI ! ( Px )dx = P6 EIa

2 3

0 a

2 P a L U BC = 2 1EI ! ( Pax L ) dx ' = 6 EI 2 2

Therefore

U = U AB + U BC = P6 EIa ( L + a ) 2 2

______________________________________________________________________________________ SOLUTION (2.64)

A P 2

x L/2

! MLo

M AC = ( P2 ! MLo ) x

B x’ L/2

P 2

+ MLo

M BC = ( P2 + MLo ) x '

We have L

2 U AB = 2 1EI " M AB dx = ( P2 ! ML0 ) 2 ( 48LEI ) + M 0 ( P2 ! ML0 )( 8LEI ) + M4 EI0 L 2

Thus

U BC = 2 1EI ! M BC dx ' = ( P2 + ML0 ) 2 ( 48LEI ) 2

U = U AB + U BC 2

PM 0 L M0 L P L = 96 EI + 16 EI + 6 EI 2 3

______________________________________________________________________________________ SOLUTION (2.65) Equation (2.66) yields

" oct = 13 [( !19 ! 4.6) 2 + ( 4.6 + 8.3) 2 + ( !8.3 + 19) 2 1

+ 6.452 + 11.82 ] 2 = 15.11 MPa Applying Eqs. (2.65) and (2.64), respectively,

U 0 d = 4 ( 76.93!109 ) (15.11) 2 = 2226.71 Pa

and 2

.6"8.3) (10 ) U 0 v = ( "1918+(4166 = 171.76 Pa .67!109 )

It follows that U0 d U0 v

= 12.96 ! 13

______________________________________________________________________________________ SOLUTION (2.66) Dilatational stress is

+ 40 ) " m = ( 200!50 = 63.33 MPa 3

Distortional stresses are then

" x ! " m = 200 ! 63.33 = 136.6 MPa " y ! " m = !50 ! 63.33 = !113.3 MPa " z ! " m = 40 ! 63.33 = !23.33 MPa

Dilatational and deviator stress matrices are, respectively,

0 # &63.33 0 $ 0 63.33 0 ! MPa $ ! 0 63.33!" $% 0

and

20 10 # &136.6 $ 20 ' 113.3 0 ! MPa $ ! 0 ' 23.33!" $% 10 Next, substitute the above deviator stresses into Eq. (1.33). Then, solve the resulting equation for principal deviator stresses, using Table B.1:

! 1d = 138.8 MPa

" 2 d = !23.9 MPa

" 3d = !114.9 MPa

______________________________________________________________________________________ SOLUTION (2.67) Only existing stresses are: " x = "

! xy = !

Here, 3

!10 ) # = "2rT3 = 2"( 20 = 58.95 MPa ( 0.06 )3 3

!10 ) # = 4"Mr 3 = 4"(15 = 88.42 MPa ( 0.06 )3

We have

G = 25E

K = 23E

Equations (2.64) and (2.65): 2

) (10 ) U 0 v = 18! K = 12! E = (8812.42( 200 ) 2

= 3.258 kPa 2

U 0 d = 125E (" 2 + 3! 2 ) = 125((10200)) (7818.1 + 10,425.3)

= 38.007 kPa Thus,

U 0 = U 0 v + U od = 41.265 kPa

______________________________________________________________________________________ SOLUTION (2.68) Principal stresses are 1

! 1, 2 = !2 ± [ !4 + " 2 ] 2 , 2

!3 = 0

and 1

! oct = 13 ( 2" 2 + 6! 2 ) 2 Equations (2.64) and (2.65) are therefore

U 0 v = 181K (! 12 + ! 22 ) = 18! K = 1#6 2E" ! 2 2

2 U 0 d = 43G ! oct = 13+E# (" 2 + 3! 2 )

______________________________________________________________________________________ SOLUTION (2.69) Only existing stress components are: " x = "

! xy = !

where,

# = !Pr 2

" = !2rT3

The area properties are:

A = !r 2

J = !2r

Thus, Eqs. (2.64) and (2.65) become

U 0 v = 18! K = 1#6 2E" ! 2 = 12! E 2

2 5" 5! U 0 d = 43G ! oct = 12 E + 4E 2

The components of strain energy are

U v = ! U 0 v dV = 121E " PA dx = 12P!rL2 E 2

U d = ! U 0 d dV = 125E ! PE dx + 45E ! TJ dx 2

= 125!Pr 2LE + 25!Tr 4LE Total strain energy is therefore

U = 2!Lr 2 E ( P 2 + 5 Tr 2 ) 2

End of Chapter 2

CHAPTER 3 SOLUTION (3.1) ( a ) We obtain " 4# "x 4

= !12 pxy

" 4# "y 4

= 6 pxy

" 4# "x 2"y 2

Thus, # " = !12 pxy + 2(6 pxy ) = 0 and the given stress field represents a possible solution. (b)

= pxy 3 ! 2 px 3 y

" 2# "x 2

Integrating twice 3 3

" = px6 y ! px10 y + f1 ( y ) x + f 2 ( y ) 4

The above is substituted into " ! = 0 to obtain d 4 f1 ( y ) dy 4

x + d dyf 2 4( y ) = 0

This is possible only if d 4 f1 ( y ) dy 4

d 4 f2 ( y )

dy 4

We find then

f 1 = c 4 y 3 + c5 y 2 + c 6 y + c 7 f 2 = c8 y 3 + c9 y 2 + c10 y + c11

Therefore, 3 3

" = px6 y ! px10 y + ( c4 y 3 + c5 y 2 + c6 y + c7 ) x + c8 y 3 + c9 y 2 + c10 y + c11

( c ) Edge y=0:

"a a

Vx = # ! xy tdx = # ( px2 + c3 )tdx = pa5 t + 2c3 at Py = ! # y tdx = ! (0)tdx = 0

Edge y=b:

Vx = " (! 32 px 2b 2 + c1b 2 + px2 + c3 )tdx !a

= ! pa 3 (b 2 ! a5 )t + 2a ( c1b 2 + c3 )t 2

Py = " ( pxb3 ! 2 px 3b)tdx = 0 !a

______________________________________________________________________________________ SOLUTION (3.2) Edge x = ± a :

" xy = 0 :

! 23 pa 2 y 2 + c1 y 2 + 12 pa 4 + c3 = 0

" xy = 0 :

! 23 pa 2 y 2 + c1 y 2 + 12 pa 4 + c3 = 0

Adding, ( !3 pa

+ 2c1 ) y 2 + pa 4 + 2c3 = 0

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 3.2 (CONT.)

c1 = 23 pa 2 Edge x = a : "x = 0:

c3 = ! 12 pa 4

pa 3 y ! 2c1ay + c2 y = 0

c2 = 2 pa 3

______________________________________________________________________________________ SOLUTION (3.3) ( a ) Equations (3.6) become !# x !x

!" xy !x

+ !yxy = 0

Substituting the given stresses, we have

c 2 y ! 2 c3 y = 0

Thus

(b)

c 2 = 2 c3

c1 = arbitrary

# x = c1 y + c2 xy

" xy = c22 (b 2 ! y 2 )

Assume c1 > 0 and c2 > 0 . y

" xy =

c2 2

(b ! y )

! x = c1 y

" xy = c22 (b 2 ! y 2 ) x

! x = ( c1 + c2 a ) y

a ______________________________________________________________________________________ SOLUTION (3.4) Boundary conditions, Eq. (3.6): !" x !x

+ !yxy = 0

!# xy !x

+ !yy = 0

( 2ab ! 2ab) x = 0 ( !2ab + 2ab) y = 0 or are fulfilled. However, equation of compatibility: 2

( !!x 2 + !!y 2 )(" x + " y ) = 0 or 4ab ! 0 is not satisfied. Thus, the stress field given does not meet requirements for solution. ______________________________________________________________________________________ SOLUTION (3.5) It is readily shown that

" 4!1 = 0 4

" !2 = 0

is satisfied is satisfied

______________________________________________________________________________________ 3.5 (CONT.) We have 2

% x = ""y#21 = 2c,

% y = ""x#21 = 2a,

Thus, stresses are uniform over the body. Similarly, for ! 2 :

# x = 2cx + 6dy

# y = 6ax + 2by

$ xy = ! ""x#"y1 = !b

" xy = !2bx ! 2cy

Thus, stresses vary linearly with respect to x and y over the body. ______________________________________________________________________________________ SOLUTION (3.6) Note: Since ! z = 0 and ! y = 0 , we have plane stress in xy plane and plane strain in xz plane, respectively. Equations of compatibility and equilibrium are satisfied by

" y = !c

" x = !" 0

"z = 0

! xy = ! yz = ! xz = 0 We have

!y = 0

(a)

(b)

Stress-strain relations become

#x =

(" x $!" y ) E

#z =

!% ($ x +$ y ) E

#y = ,

(" y $!" x ) E

(c)

" xy = " yz = " xz = 0

Substituting Eqs. (a,b) into Eqs. (c), and solving

! y = $"! 0 2

" x = ! (1#!0$E )

# z = " (1+E" )! 0 "y = 0

Then, Eqs. (2.3) yield, after integrating: 2

u = ! (1!# E)" 0 x

v=0

w = " (1+"E )! 0 z ______________________________________________________________________________________ SOLUTION (3.7) Equations of equilibrium, !"

2axy + 2axy = 0

ay 2 ! ay 2 = 0

!# x !x

+ !yxy = 0,

"$ xy "y

+ "xy = 0,

are satisfied. Equation (3.12) gives

( ##x 2 + ##y 2 )($ x + $ y ) = "4ay ! 0 2

Compatibility is violated; solution is not valid. ______________________________________________________________________________________ Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 3–3

______________________________________________________________________________________ SOLUTION (3.8) We have " 2$ x

" 2$ y

"y 2

"x 2

" 2# xy "x"y

= !2ay

= 2ay

Equation of compatibility, Eq. (3.8) is satisfied. Stresses are

# x = 1$E! 2 (" x + !" y ) = 1$aE! 2 ( x 3 + !x 2 y )

# y = 1$E! 2 (" y + !" x ) = 1$aE! 2 ( x 2 y + !x 3 ) 2 # xy = G" xy = 2 (aE 1+! ) xy

Equations (3.6) become aE 1!" 2

(3x 2 + 2"xy ) + 1aE +" xy = 0

aE 1!" 2

y 2 + 1!aE" 2 x 2 = 0

These cannot be true for all values of x and y. Thus, Solution is not valid. ______________________________________________________________________________________ SOLUTION (3.9)

# x = ""ux = !2$cx

# y = ""yv = 2ax

# xy = ""uy + ""vx = !2cy + 2cy = 0 Thus

% x = 1&E# 2 ($ x + #$ y ) = 0

" xy = G! xy

$ y = 1!E" 2 (# y + "# x ) = 2 Ecx

2Eac

2b 2Eac

Note that this is a state of pure bending. ______________________________________________________________________________________ SOLUTION (3.10) (a)

% y = 6 pxy

% x = ""y#2 = 0,

$ xy = !3 px 2

Note that " ! = 0 is satisfied. (b)

! y = 6 pbx y

!x = 0 b ! xy = 0

! xy = 3px 2

!x = 0 a

!y =0 ( c ) Edge x = 0 : Edge x = a :

! xy = 3pa 2 x

! xy = 3px 2

V y = Px = 0

Px = 0

______________________________________________________________________________________ 3.10 (CONT.) b

V y = " # xy tdy = 3 pa 2 bt !

Edge y = 0 :

Py = 0

V x = " # xy tdx = pa 3t ! 0

V x = pa 3t !

Edge y = b : a

Py = " # y tdx = 3 pa 2 bt ! 0

______________________________________________________________________________________ SOLUTION (3.11) 4

( a ) We have # " ! 0 is not satisfied. 2

% y = !!x"2 = pya 2 , 2

(b)

! xy = 2p (1 + 4 ax )

!x = 0

! x = p(1 + ay )

! xy = 2pya 2 ( 4a + y )

" y = 0, ! xy = 0 p a

Vy = ! 2pya2 dy = 16 pat 0

Edge x = a :

$ xy = # p ( 42xya+2 y )

!y = p py 2 a " xy = ! 2a 2

( c ) Edge x = 0 :

% x = p ( xa 2+ xy ) ,

Px = 0

V y = " # xy tdy = 76 pat ! 0 a

Px = " # x tdy = 23 pat ! 0

Edge y = 0 :

Vx = 0

Edge y = a :

Py = 0

V x = " # xy tdx = 23 pat ! 0 a

Py = " # y ptdx = pat ! 0

______________________________________________________________________________________ SOLUTION (3.12) 4

( a ) We have " ! = 0 is satisfied. The stresses are

$ x = ""y#2 = ! bpx3 (6b ! 12 y ) 2

$ xy = ! ""x"#y = 6bpy3 (b ! y )

$ y = ""x#2 = 0

(CONT.)

______________________________________________________________________________________ 3.12 (CONT.) (b)

" x = ! bpa3 (6b ! 12 y )

! xy

a ______________________________________________________________________________________ SOLUTION (3.13) We have "# "y

= ! $P [tan !1 xy + x 2xy+ y 2 ],

" 2# "y 2

= ! $P [ x 2 +x y 2 + ( x (+xy2 +) yx2!)22 y x ]

"# "x

!y = ! Py $ x2 + y2

The stresses are thus, 2

% x = !!y"2 = # 2$P ( x 2 +x y 2 )2 2

% y = !!x"2 = # 2$P ( x 2xy+ y 2 )2 2

% xy = # !!x!"y = # 2$P ( x 2x+ yy2 )2 2

!x 2 P !L

! xy

______________________________________________________________________________________ SOLUTION (3.14) Various derivatives of ! are: "# "x

" 4# "x 2"y 2

= 0,

" 2# "y 2

$0 4h

! " !x 4

= $40 ( y ! yh ! hy2 ), =

= 0,

" 2# "x"y

( !2 x ! 4

! " !y 4

6 xy h

" 2# "x 2

=0 2

= $40 (1 ! 2hy ! 3hy2 )

+ 2L +

6 Ly h2

)

(a)

It is clear that Eqs. (a) satisfy Eq. (3.17). On the basis of Eq. (a) and (3.16), we obtain (CONT.) ______________________________________________________________________________________ Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 3–6

______________________________________________________________________________________ 3.14 (CONT.)

" x = 4# 0h ( !2 x ! 6hxy + 2 L + 6 hLy ),

"y =0

(b)

" xy = ! "40 (1 ! 2hy ! 3hy2 )

From Eqs. (b), we determine "y =0 Edge y = h :

! xy = ! 0

Edge y = ! h :

#y =0

" xy = 0

x = L:

# x = 0,

" xy = ! "40 (1 ! 2hy ! 3hy2 )

Edge

It is observed from the above that boundary conditions are satisfied at y = ± h ,

but not at x = L . ______________________________________________________________________________________ SOLUTION (3.15) 4

For # " = 0, e = !5d and a, b, c are arbitrary.

(a)

" = ax 2 + bx 2 y + cy 3 + d ( y 5 ! 5 x 2 y 3 )

Thus

(1)

( b ) The stresses:

$ x = ""y#2 = 6cy + 10d ( 2 y 3 ! 3x 2 y )

(2)

$ y = ""x#2 = 2a + 2by ! 10dy 3

(3)

$ xy = ! ""x"#y = !2bx ! 30dxy 2

(4)

Boundary conditions:

# y = !p

" xy = 0

(at y=h)

(5)

Equations (3), (4), and (5) give

b = !15dh 2

2a ! 40dh 3 = ! p

! $ dy = 0 "h

! y$ dy = 0

Equations (2), (4), and (7) yield Similarly

! # dy = 0 "h

(at x=0)

c = !2dh 2

"y =0

give

(6)

(8)

! xy = 0

(at y=-h)

a = 20dh 3

(9)

Solution of Eqs. (6), (8), and (9) results in

a = ! 4p

(7)

b = ! 163ph

The stresses are therefore

c = ! 40ph

d = 80ph3

e = ! 16ph3

(10)

" x = ! 320pyh + 8hp3 ( 2 y 3 ! 3x 2 y ) 3

" y = ! 2p ! 38pyh ! 8pyh3 2

" xy = 38pxh (1 ! hy2 )

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (3.16) We obtain 2

$ xy = ! ""x"#y = !

2 pxy a2

$ x = ""y#2 = p ( x a!22 y ) 2

# y = !!x"2 = pya 2 2

(a)

Taking higher derivatives of ! , it is seen that Eq. (3.17) is not satisfied. Stress field along the edges of the plate, as determined from Eqs. (a), is sketched bellow. y

!y = p

! xy = 2 p ax

! xy = 2 p ay

! xy = 0

! x = 2 p ay2

" x = p(1 ! 2 ay2 )

a 2

" y = 0, ! xy = 0

______________________________________________________________________________________ SOLUTION (3.17) The first of Eqs. (3.6) with Fx = 0 !" xy !y

= pxy I

Integrating, 2

! xy = pxy 2 I + f1 ( x )

(a)

The boundary condition, 2

(! xy ) y =h = 0 = pxh 2 I + f1 ( x ) gives f 1 ( x ) = ! pxh

" xy = !

px 2I

2 I . Equation (a) becomes (h ! y 2 ) 2

(b)

Clearly, (" xy ) y = ! h = 0 is satisfied by Eq. (b). Then, the second of Eqs. (3.6) with Fy = 0 results in "# y "y

= p ( h2 !I y )

Integrating, 2

" y = 2pI y ( h 2 ! y3 ) + f 2 ( x )

(c)

Boundary condition, with t = 3I 2h , (CONT.) ______________________________________________________________________________________ Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 3–8

______________________________________________________________________________________ 3.17 (CONT.)

(" y ) y = ! h = ! pt = ! 2phI ( h 2 ! h3 ) + f 2 ( x ) 3

gives f 2 ( x ) = ! ph

3I . Equation (c) is thus " y = 6pI (3h 2 y ! y 3 ! 2h 3 )

(d)

This satisfies the condition that (! y ) y =h = 0 . Note that Eq. (3.12) is not satisfied: the solution obtained does not provide a compatible displacement field. ______________________________________________________________________________________ SOLUTION (3.18) Substituting the stresses from Eqs. (3.10) into Eqs. (3.6) and taking Fx = Fy = 0 :

E 1"% 2

( !!$xx + % !xy ) + G !yxy = 0

E 1"% 2

( !yy + % !!$yx ) + G !xxy = 0

E 2 (1+# )

[ 1"2# !!x#2 + 12"## !!x!#y + !!yu2 + !!x!#y ] = 0

E 2 (1+# )

[ 1!2# ""y#2 + 12!## ""x"uy ! ""x"uy + ""x#2 ] = 0

The foregoing become " 2u "x 2

+ 11+!## ""xu2 ! 11+!## ""x"vy + ""yu2 = 0

! 2v !y 2

+ 11+"## !!yv2 + 11+"## !!x!vy + !!yu2 = 0

! 2u !x 2

+ !!yu2 + 11+"## !!x ( !!ux + !!yv ) = 0

! 2v !y 2

+ !!xv2 + 11+"## !!y ( !!yv + !!ux ) = 0

or 2

______________________________________________________________________________________ SOLUTION (3.19) It is readily found that

And

"# xy "y

"$ x "x

= ! pxy I

= pxy I

"$ y "y

= ! pI ( !h 2 + y 2 )

"# xy "x

= ! 2pI ( h 2 ! y 2 )

Thus, Eqs. (3.6) are satisfied: stress field is possible. We have xQ = 1.5 m, yQ = 0.05 m , and 3

I = 2 th3 = 2 ( 0.043)( 0.1) = 2.67(10 !5 ) m 4 3

Substituting the given data, Eqs. (P3.19) yield at Q: "10 "10 # x = 10(!210 (5 " 105 + 2 " 0.12 )0.05 + 3( 2!.10 (1.25 " 10 !4 ) .67"10! 5 ) 67"10! 5 ) 3

= !21.09 MPa (CONT.) ______________________________________________________________________________________ Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 3–9

______________________________________________________________________________________ 3.19 (CONT.)

# xy = 2 ( 2!.1067""1010!5 ) (0.12 ! 0.052 ) = !2.106 MPa 3

# y = 6( 2!.1067""1010!5 ) ( 2 " 0.13 ! 3 " 0.0005 + 0.053 ) = !0.039 MPa 3

Applying Eq. (2.29), we have 9

(10 ) G = 200 2 (1+ 0.3) = 76.9 MPa

Hooke’s law is therefore

! x = 200(1109 ) ( "21.09 + 0.3 # 0.039)10 6 = "105 µ ! y = 200(1109 ) ( #0.039 + 0.3 " 21.09)106 = 31.4 µ 6

2.106 (10 ) ! xy = "76 = "27.4 µ .9 (109 )

Principal strains are 1

" 1, 2 = !1052+31.4 ± [( 1362 .4 ) 2 + ( !272 .4 ) 2 ] 2

! 1 = 32.8 µ ! 2 = "106 µ o !1 !27.4 1 We have " p = 2 tan !105!31.4 = 5.65

For this angle, Eq. (2.14a) yield ! x ' = 32.8 µ . Thus,

! p ' = 5.65o ______________________________________________________________________________________ SOLUTION (3.20) Assume " x = " y = 0,

! z = 0,

! x = ! y = constant

which satisfy Eqs. (3.6) and (3.27). Hooke’s law becomes

% x = E1 ($ x ! #$ y ) + "T1 = 0

(a)

% y = E1 ($ y ! #$ x ) + "T1 = 0

(b)

$ z = ( !# x ! # y ) + "T1 = 0

(c)

% E

From Eqs. (a) and (b), we obtain ! x = ! y . Therefore,

$ x = %Ex (1 ! # ) + "T1 = 0

This yields

$ x = $ y = E"#!T11

Then, Eq. (c) becomes

" z = 21#!$#T1 + !T1 We also have: ! xy = ! yz = ! xz = 0 and ! xy = ! yz = ! xz . ______________________________________________________________________________________ SOLUTION (3.21) The nonzero strain components are

" x = " y = " z = !T

______________________________________________________________________________________ 3.21 (CONT.) Compatibility equations reduce to ! 2# x !y 2

! 2# y

! 2#

! 2"

+ !x 2y = !y!xyx

!z 2

+ !!y#2z = !y!yzz

! 2# z !x 2

+ !!z#2x = !!x"!xzz

Adding the above equations and substituting the given data: 2

2" ( !!xT2 + !!yT2 + !!zT2 ) = 0 or ! 2T !x 2

+ !!yT2 + !!zT2 = 0

This equation, for a time independent temperature field, has the solution

T = c1 ' x + c2 ' y + c3 ' z + c4 '

which may be written as

!T = c1 x + c2 y + c3 z + c4

______________________________________________________________________________________ SOLUTION (3.22) Stress is ! x = "! 0 regardless of T1 and still we have ! y = 0 . The second of Eqs. (3.26a) is thus

$ y = E1 (# y + "# 0 ) + !T1 = 0 # y = $"# 0 $ E!T1

(a)

The first of Eqs. (3.26a) and (a) result in

$ x = % 1%E" # 02 + (1 + " )!T1 2

(b)

Now, Hooke’s law

# z = % $E (" x % " y ) + !T1

leads to

$ z = " (1E+" ) # 0 + (1 + " )!T1

(c)

Then Equations (2.4) yield, after integration,

u = !xx

v=0

w = !zz

Here, ! x and ! z are given by Eqs. (b) and (c). ______________________________________________________________________________________ SOLUTION (3.23)

Equation (3.27) reduces to d2 dy 2

(" x + !ET ) = 0

______________________________________________________________________________________ 3.23 (CONT.) from which

# x = !"ET + c1 y + c2 = !"ET ( a1 y + a 2 ) + c1 y + c2 Referring to Part (b) of Example 3.2: c1 = c2 = 0 and # x = !"E ( a1 y + a 2 )

We have h

Px = $ " x tdy = #2 E!hta 2 #h

M z = " $ x tydy = ! E#t a13y + a22y !h

h !h

= " E!th a1 2 3

The Px and M z are opposite to that shown above. ______________________________________________________________________________________ SOLUTION (3.24) Assume a stress distribution:

" xy = " yz = " xz = ! y = ! z = 0

! x = constant

(a)

which satisfy Eqs. (3.6) and (3.27). Then, Hooke’s law becomes

" x = $Ex + !T ,

" y = " z = % #$E x + !T

(b,c)

! xy = ! yz = ! xz = 0

(d)

Due to constraint imposed by the walls and because of the uniformity of the temperature distribution: we take ! x = 0 . Equations (a) and (b) give

" x = #!ET Equation (c) is then

# y = # z = "T (1 + ! ) = constant

The compressive force P on the tube is

Px = " x A = # E!At

Substituting the data given

Px = !120(10 9 )(16.8 " 10 !6 ) " (800 " 10 !6 )(100) = !161.3 kN

______________________________________________________________________________________ SOLUTION (3.25) Derivatives of the given stress function are $ 2% $r 2

= 0,

1 $% r $r

= # P"!r sin ! ,

1 $ 2% r 2 $! 2

= P"!r sin ! # 2"Pr cos !

Equation (3.40) becomes

% 4 $ = ( &&r 2 + 1r &&r + r12 &&! 2 )( # 2"Pr cos ! ) 2

______________________________________________________________________________________ 3.25 (CONT.) 4

After performing the derivatives, we obtain " ! = 0 . Substituting the derivatives obtained above, into Eqs. (3.32):

" r = $ P#!r sin ! + P#!r sin ! $ 2#Pr cos ! = # 2"Pr cos ! Similarly, we find " ! = 0 and " r! = 0 .

______________________________________________________________________________________ SOLUTION (3.26) Refer to Fig. P3.26a. Let ABC = 1 and hence AAB = cos ! , AAC = sin ! .

!F = 0: x

# x = # r cos ! cos ! + # ! sin ! sin ! $ 2" r! sin ! cos !

!F = 0: y

" xy = # r cos ! sin ! $ # ! sin ! cos ! + " r! cos ! cos ! # " r! sin ! sin !

Similarly, from Fig. P3.26b:

!F = 0: y

# y = # r sin 2 ! + # ! cos 2 ! + 2" r! sin ! cos !

Check:

!F = 0: x

" xy = # r sin ! cos ! $ # ! sin ! cos ! + " r! (cos 2 ! $ sin 2 ! )

Thus, quoted equations are derived. ______________________________________________________________________________________ SOLUTION (3.27) Apply the chain rule (Sec. 3.9): and

"# "x

= ""#r cos ! $ 1r ""#! sin !

" 2# "x 2

! = ""r#2 cos 2 ! $ 2 ""!"#r sin! rcos! + ""#r sinr ! $ 2 ""#r sin!rcos + ""!#2 sinr 2 ! 2 2

(a)

" 2# "y 2

! = ""r#2 sin 2 ! + 2 ""!"#r sin! rcos! + ""#r cosr ! + 2 ""#! sin!rcos + ""!#2 cosr 2 ! 2

(b)

Similarly,

Adding Eqs. (a) and (b), we have " 2# "x 2

+ ""y#2 = ""r#2 + 1r ""#r + r12 ""!#2

(c)

By referring to the identity ! 4" !x 4

+ 2 !x!2!"y 2 + !!y"4 = ( !!x 2 + !!y 2 )( !!x"2 + !!y"2 )

and Eq. (c), we can readily write the equation quoted, Eq. (3.40). ______________________________________________________________________________________ SOLUTION (3.28) Equation (3.28) is written as

( drd 2 + 1r drd )( ddr!2 + 1r ddr! ) + "ET ( drd 2 + 1r drd ) = 0 2

______________________________________________________________________________________ 3.28 (CONT.) or or

d 2! dr 2

+ 1r ddr! + "ET = 0

1 d r dr

( r ddr! ) + "ET = 0

______________________________________________________________________________________ SOLUTION (3.29) ( a ) Let C = 2 (sin 2! ""2M! cos 2! ) , and

$ = C (sin 2" # 2" cos 2! ) Various derivatives of ! are:

and

! 2" !r 2

!" !r

$% $"

= 2C cos 2" # 2C cos 2!

# 2$ #! 2

= "4C sin 2!

2! $ 2 # = r12 %%!#2 = " 4C sin r2 2

We thus obtain

2$ 2$ # 4 " = C[ 8 sinr 42$ ! 24 sin + 16 sin ]=0 r4 r4

(b)

"! = 0

2! " r = r12 $$!%2 = # 4C sin r2

# r" = $ %%r ( 1r %%&" ) = 2rC2 (cos 2" $ cos 2! ) (c)

Letting " = ! 2 : C = # M 2! . It follows that

#! = 0

2! # r = 2 M"sin r2

" # r" = 2rC2 (cos 2" + 1) = $ 2 M!cos r2 2

where cos ! = (1 + cos 2! ) 2 ______________________________________________________________________________________ SOLUTION (3.30) Using Eq. (3.48) and Fig. P3.30: "

Fx = ! ($ r rd# ) sin # = ! ( 2"P cos # sin # )d# 2

= 2!P 12 sin 2 "

= !P

Similarly, #

Fy = ! # (% r rd$ ) cos $ = ! # ( 2#P cos 2 $ )d$ 2

" 2

= 2"P #2 + 14 sin 2# !2" = P 2

______________________________________________________________________________________ SOLUTION (3.31) NOTE: (In Probs. 3.31, 3.32, and 3.33), the P, L, and ! are constants. It can readily be verified that, the maximum values of the functions in parentheses occur:

(sin ! cos 3 ! ) = 0 , or tan 2 ! = 13

when ! = ±30

d d!

(sin 2 ! cos 2 ! ) = 0 , or tan 2 ! = 1

when ! = ±45

d d!

Maximum stresses, using Eqs. (3.37) and (3.43):

(" x ) elast . = L (! + 1Psin 2! )

(a)

(# xy ) elast . =

P sin " cos3 " L (! + 12 sin 2! )

Elementary solution of maximum stresses are P (" x ) elem. = 2 L tan # ,

(! xy ) elem. = 0

(b)

( a ) For ! = 15 , (! + 12 sin 2! = 0.512) :

Thus,

(! x ) elast . = P 0.512 L

at ! = 0

(! xy ) elast . = P 2.195L

at ! = 15

(! x ) elem. = P 0.536 L

at any !

(! x ) elast . = 1.047(! x ) elem. o

( b ) For ! = 60 , (! + 12 sin 2! = 1.48) :

Thus,

(! x ) elast . = P 1.48L

at ! = 0

(! xy ) elast . = P 4.557 L

at ! = 30

(! x ) elem. = P 3.464 L

at any !

(! x ) elast . = 2.341(! x ) elem.

______________________________________________________________________________________ SOLUTION (3.32) See: NOTE, solution of Prob. 3.31. o

Substitute ! = 30 into Eqs. (a) and (b) of Solution of Prob. 3.31.

Thus,

(! x ) elast . = P 0.957 L

at ! = 0

(! xy ) elast . = P 2.946 L

at ! = 30

(! x ) elem. = P 1.155L

at any !

(! x ) elast . = 1.207(! x ) elem.

______________________________________________________________________________________ SOLUTION (3.33) See: NOTE, solution of Prob. 3.31. We have r = L cos " , hmn

2 = c = L # tan ! , and I = 2c 3 3 .

Equations (3.43) give 3

cos " (# x ) elast . = LF(!sin$"1 sin 2! ) 2

(# xy ) elast . =

F sin 2 " cos2 " L (! $ 12 sin 2! )

Elementary solution:

(" x ) elem. = 3FL 2c 2 ,

(! xy ) elem. = 3F 4c

( a ) For ! = 15 , (! + 12 sin 2! = 0.512) :

Thus,

(! x ) elast . = 19.43F L

at ! = 15

(! xy ) elast . = 5.21F L

at ! = 15

(! x ) elem. = 20.89 F L

at ! = 15

(! xy ) elem. = 2.8F L

at ! = 0

o o

(! x ) elast . = 0.93(! x ) elem. (! xy ) elast . = 1.86(! xy ) elem. o

( b ) For " = 60 , (" ! 12 sin 2" = 0.614) :

Thus,

(! x ) elast . = 0.529 F L

at ! = 30

(! xy ) elast . = 0.407 F L

at ! = 45

(! x ) elem. = 0.5F L

at ! = 60

(! xy ) elem. = 0.433F L

at ! = 0

o o

(! x ) elast . = 1.058(! x ) elem. (! xy ) elast . = 0.94(! xy ) elem.

______________________________________________________________________________________ SOLUTION (3.34) With x = r " cos ! , Eqs. (3.50) become

" x = $( 2#Pr ) cos 3 !

" y = $( 2#Pr ) sin 2 ! cos ! " xy = $( 2#Pr ) sin ! cos 2 ! Substituting,

prd! 2p 3 2 d" x = $ #2r ( cos ! ) cos ! = $ # cos !d!

d# y = % 2$p sin 2 ! ,

d" xy = % $p sin 2!d!

______________________________________________________________________________________ 3.34 (CONT.) Integrating, !2

" x = $ 2#p % cos 2 !d! = $ 2p# [2(! 2 $ !1 ) + (sin 2! 2 $ sin 2!1 )] !1

"y =$

p 2#

[2(! 2 $ !1 ) $ (sin 2! 2 $ sin 2!1 )]

" xy = 2p# [cos 2! 2 $ cos 2!1 ] ______________________________________________________________________________________ SOLUTION (3.35) APPROACH (a): 2

( drd 2 + 1r drd )( ddr 2f1 + 1r dfdr1 ) = 0 2

We have

(d) 4

d dr

( ddr 2f1 ) = ddrf31 ,

d dr

( 1r dfdr1 ) = ! r !2 dfdr1 + r !1 ddr 2f1

d2 dr 2

( 1r dfdr1 ) = r23 dfdr1 ! 2r ddr 2f1 + 1r ddr f31

d2 dr 2

( ddr 2f1 ) = ddr 4f1 2

1 d r dr

( ddr 2f1 ) = 1r ddrf31

1 d r dr

( 1r dfdr1 ) = 1r ( ! r12 dfdr1 + 1r ddr 2f1 )

Then, Eq. (d) becomes d 4 f1

dr 4

+ 2r ddrf31 ! r12 ddr 2f1 + r13 dfdr1 = 0

1 d r dr

{r drd [ r !1 drd ( r dfdr1 )]} = 0

1 d r dr

{r drd [ r !1 dfdr1 + ddr 2f1 ]} = 0

1 d r dr

{r[ ! r !2 dfdr1 + r !1 ddr 2f1 + ddrf31 ]} = 0

1 d r dr

{! r !1 dfdr1 + ddr 2f1 + r ddr f31 } = 0

(d’)

The first equation of Problem 3.35 may be written as:

1 r

{r !2 dfdr1 ! r !1 ddr 2f1 + 2 ddrf31 + r ddr 4f1 } = 0

1 df1 r 3 dr

! r12 ddr 2f1 + 2r ddrf31 + ddr 4f1 } = 0

which is the same as Eq. (d’). Now let us integrate the expression: 1 d r dr

{r drd [ 1r drd ( r dfdr1 )]} = 0

r drd [ 1r drd ( r dfdr1 )] = c1 1 d r dr

( r dfdr1 ) = c1 ln r + c2

r dfdr1 = c1 ! r ln rdr + c2 ! rdr

r dfdr1 = c1 [ r2 ln r ! r4 ] + c2 r2 + c3 2

df1 dr

= c1r ln r + c2 r 2 + c3 ln r + c4

______________________________________________________________________________________ 3.35 (CONT.) or

f1 = c1r 2 ln r + c2 r 2 + c3 ln r + c4 Expression (e) may be treated in a like manner.

APPROACH (b): Letting t = ln r , we have df1 dr

= dfdt1 drdt = 1r dfdr1

d 2 f1 dr

= r12 ( ddt 2f1 ! dfdt1 )

d 3 f1 dr 3

= r13 ( ddt 3f1 ! 3 ddt 2f1 + 2 dfdt1 )

d 4 f1

= 1r ( ddt 4f1 ! 6 ddt 3f1 + 11 ddt 2f1 ! 6 dfdt1 )

Substituting these derivatives into Eq. (d’), we obtain: d 4 f1 dt

! 4 ddt 3f1 + 4 ddt 2f1 = 0

This is an ordinary differential equation with constant coefficients. It has a solution

f1 = c1r 2 ln r + c2 r 2 + c3 ln r + c4

In a like manner, it can be shown that, 2

( drd 2 + 1r drd ! r42 )( ddrf22 + 1r drdf22 ! 4rf22 ) = 0 2

is solved to yield Eq. (g) of Section 3.12. ______________________________________________________________________________________ SOLUTION (3.36) (a)

!0 =

2! 0 2

" r1 = "20 [(1 # ar 2 ) + (1 + 3ra4 # 4ra2 ) cos 2! ]

" ! 1 = "20 [(1 + ar 2 ) # (1 + 3ra4 ) cos 2! ] 4

and

" r! 1 = $ #20 (1 $ 3ra4 + 2ra2 ) sin 2! # r 2 = #20 [(1 ! ar 2 ) + (1 + 3ra4 ! 4ra2 ) cos 2(" + 90o )] 2

# " 2 = #20 [(1 + ar 2 ) ! (1 + 3ra4 ) cos 2(" + 90o )] 2

# r" 2 = ! $20 (1 ! 3ra4 + 2ra2 ) cos 2(" + 90o ) 4

______________________________________________________________________________________ 3.36 (CONT.) We have, by superposition:

# r = # r1 + # r 2 # ! = # !1 + # ! 2 Hence, at r=a and " = ! 2 , ! r1 = 0 ! r2 = 0 ! " 1 = 3! 0 ! " 2 = #! 0 " r! 1 = 0 " r! 2 = 0 lead to the solution:

#r = 0

# ! = 2# 0

" r! = " r! 1 + " r! 2

" r! = 0

( b ) Referring to the results of part ( a ), we write

! r1 = 0 ! r2 = 0 ! " 1 = 3! 0 !"2 = ! 0 " r! 1 = 0 " r! 2 = 0

Thus,

#r = 0

# ! = 4# 0

" r! = 0

______________________________________________________________________________________ SOLUTION (3.37) We have

d D

= 13

Then, from Fig. D.8: K ! 2.3 . Hence 3

180(10 ) ! max = K PA = 2.3 (150 "50)20 = 207 MPa

______________________________________________________________________________________ SOLUTION (3.38)

! max 130 = = 1.625 ! nom 80 From Fig. D.1: r d = 0.25 . Then D = 2r + d K=

gives

40 = 2(0.25d ) + d = 1.5d d = 26.7 mm r = 6.67 mm

______________________________________________________________________________________ SOLUTION (3.39) For

r = 0.15 : d D = 2r + d ; 40 = 2(0.15d ) + d = 1.3d

______________________________________________________________________________________ 3.39 (CONT.) or d = 30.76 We thus have

D = 1.3 d Figure D.1 gives K ! 1.7. Hence,

! max = 250(106 ) = 1.7

Pall (20)(30.76)

Pall = 90.5 kN

______________________________________________________________________________________ SOLUTION (3.40)

TBC = 1 kN ! m

TAB = 2 kN ! m

! max = 2T " c3 yields

Then,

2(1#103 ) = 79.6 MPa ! BC = " (0.02)3

! AB =

2(2 #103 ) = 81.5 MPa " (0.025)3

Thus

! max = K! AB = (1.6)(81.5) = 130.4 MPa

______________________________________________________________________________________ SOLUTION (3.41) (a)

We have

D 40 r 2 = = 1.14 = = 0.057 d 35 d 35 Hence, by Fig. D.4, K ! 1.6 .

We have

! ! (d ) 4 = (35) 4 = 147.3(103 ) mm 4 32 32 Tc 100(0.0175) ] = 19 MPa ! yp = K = 1.6[ J 147.3(10"9 ) J=

______________________________________________________________________________________ SOLUTION (3.42) We have

D 40 r 8 = = 2.5 = = 0.2 d 16 d 16 From Fig. D.6 , K ! 1.31 Tc 16T 250(106 ) = 1.31[ ] ! max = ! all = K ; J " (0.016)3 or

T = 153.5 N ! m

______________________________________________________________________________________ SOLUTION (3.43) 5 r = 37.5 25 = 1.5, d = 25 = 0.2 Figure D.1, K = 1.72 . ! max 1.4 = 210 Hence ! nom = 1.72 1.72 = 87.2 MPa D d

and

Pall = A! nom = (25 "10)(87.2) = 21.8 kN

______________________________________________________________________________________ SOLUTION (3.44) At a section through B

M B = 400(0.3) = 120 N ! m M c 120(0.02) ! nom = B = = 37.5 MPa 1 I 3 (0.012)(0.04) 12

(a)

r 5 = = 0.125 d 40

D 60 = = 1.5 : d 40

K ! 1.65

(Fig. D.2)

! max = 1.65(37.5) = 61.9 MPa

(b)

r 10 D 60 = = 0.25 = = 1.5 : d 40 d 40 ! max = 1.41(37.5) = 52.8 MPa

K ! 1.41 (Fig. D.2)

______________________________________________________________________________________ SOLUTION (3.45)

"! !

!a Without hole:

! " = pd 2t

! a = pd 4t

With hole:

2! a

!a (1)

(2)

______________________________________________________________________________________ 3.45 (CONT.) We use Eq. (3.55b), with r=a. 2

" ! 1 = "2a [(1 + aa 2 ) # (1 + 3aa4 ) cos 2! ] = " a [1 # 2 cos 2! ]

# " 2 = 2# a [1 ! 2 cos 2(" + 90o )] o

For ! = 0 :

! " 2 = 6! a

! " 1 = #! a For " = ! 2 : ! " 1 = 3! a

! " 2 = #2! a o

Therefore, superposing the results at ! = 0 :

! " = 5! a = 5 pd 4t at " = ! 2 : ! " = ! a = pd 4t

______________________________________________________________________________________ SOLUTION (3.46) ( a ) We have D/d=1.1 and r/d=0.05. Then, we find from Figs. D.6, D.7, and D.5 that

K t = 1.64

K b = 2.2

K a = 2.3

Then, Eqs. (b) of Example 3.5 yield 3

) #10 ) " x = 2.3 !50(10 + 2.2 4(10 = 4.42 MPa (0.2)2 ! (0.2)3 3

#10 ) " xy = 1.64 2(20 = 2.61 MPa ! (0.2)3

Equation (a) of Example 3.4 is therefore 1

2 2 4.42 2 ! 1,2 = 4.42 2 ± [( 2 ) + (2.61) ]

! 1 = 5.63 MPa

! 2 = "1.21 MPa

(b)

! max = 12 (5.63 + 1.21) = 3.42 MPa

(c)

! oct = 13 (5.63 " 1.21) = 1.47 MPa 1

! oct = 13 [(5.63 + 1.21) 2 + ("1.21) 2 + ("5.63) 2 ] 2 = 2.98 MPa

______________________________________________________________________________________ SOLUTION (3.47) ( a ) We have D/d=2 and r/d=0.04. Then, we find from Figs. D.7 and D.6 that

K b = 2.6

K t = 1.9

Equation (b) of Example 3.5 is therefore 3

#10 ) " x = 2.6 !4(20 = 33.9 MPa (0.125)3 3

#10 ) " xy = 1.9 !2(5 = 9.73 MPa (0.125)3

______________________________________________________________________________________ 3.47 (CONT.) Equation (a) of Example 3.5: 1

2 2 33.9 2 ! 1,2 = 33.9 2 ± [( 2 ) + (9.73) ]

! 1 = 36.5 MPa

! 2 = "2.59 MPa

(b)

! max = 12 (36.5 + 2.59) = 16.96 MPa

(c)

! oct = 13 (36.5 " 2.59) = 11.3 MPa 1

! oct = 13 [(36.5 + 2.59) 2 + ("2.59) 2 + ("36.5) 2 ] 2 = 17.85 MPa

______________________________________________________________________________________ SOLUTION (3.48) We apply Eqs. (3.63). 1

(a)

)( 0.025!0.0375 ) 3 a = 0.88[ 2 ((500 ] = 0.635 mm 200!109 )( 0.0125 )

(b)

2 3 0.0125 " c = 0.62[500( 200 ! 109 ) 2 ( 2!0.025 !0.0375 ) ] = 596.1 MPa

(c)

# = 1.54[(500) 2 ( 2 ( 200"109 0)2.0125 )] 3 = 5.339(10 !3 ) mm ( 0.025"0.0375 )

______________________________________________________________________________________ SOLUTION (3.49) ( a ) Use Eq. (3.62): 1

9 2

!10 ) 3 " c = 0.62[ 5004((200 ] = 1233 MPa 0.025 ) 2

( b ) Apply Eqs. (3.60) and (3.59) for r1 = r2 = r and E1 = E 2 = E to obtain the formula 1

! c = 0.617[ PEr 2 ] 3 Thus,

9 2

!10 ) 3 " c = 0.617[ 500((0200 ] = 1959 MPa .025 ) 2

______________________________________________________________________________________ SOLUTION (3.50) Using Eqs. (3.73), (3.74), and the second of (3.70), we have

m = (1 0.4 )+4(1 0.25) = 0.6154 9

"10 ) n = 43((200 = 2.9304(1011 ) 1!0.32 ) ) cos " = ± ((11 00..44 ))+!((11 00..25 25 ) = 0.2308

Interpolating Table 3.3:

ca = 1.1774

! = 76.66o

cb = 0.8616

Apply Eqs. (3.69): 1

)( 0.6154 ) 3 a = 1.774[ 4 (210.9304 ] = 2.393 mm !1011 1

Thus,

)( 0.6154 ) 3 b = 0.8616[ 4 (210.9304 ] = 1.752 mm !1011 3

) $ c = 1.5 # ( 2.3934"(10 = 455.5 MPa 1.752 )10! 6

______________________________________________________________________________________ SOLUTION (3.51) Use Eqs. (3.67): 3

)( 200!10 ) 2 " c = 0.418[ 2.5(10 ] = 418 MPa 0.1( 0.005 ) 1

5(10 )( 0.005 ) 2 2b = 2{1.52[ 2.200 ] } = 2(0.038) = 0.076 mm !109 ( 0.1)

______________________________________________________________________________________ SOLUTION (3.52) Equations (3.65), for ! 1 = ! 2 = 0.25, E1 = E 2 = E , r1 = r2 = r : 6

!10 )( 2 ) 2 " c = 0.412[ 2 (10 )( 200 ] = 824 MPa 0.2 1

(10 )( 0.2 ) 2 2b = 2{1.545[ 2200 ] } = 2(1.545) = 3.09 mm !109 ( 2 )

______________________________________________________________________________________ SOLUTION (3.53) Refer to Example 3.6.

1 r1 ' = 0 1 r2 ' = 0 " =! 2 4 m = (1 0.5)+(1 0.2 ) = 0.5714 9

"10 ) n = 43((210 = 2.9867(1011 ) 1!0.252 )

cos " = ± ((11 00..55))+!((11 00..22 )) = 0.4286 Table 3.3: ca = 1.3862

! = 64.62 o

cb = 0.7758 3

( 0.5714 ) 3 a = 1.3862[ 5!210 ] = 2.994 mm .9867!1011 ( 0.5714 ) 3 b = 0.7558[ 5!210 ] = 1.605 mm .9867!1011

Thus,

) $ c = 1.5 # ( 2.9945("101.605 = 505.2 MPa )10! 6

______________________________________________________________________________________ SOLUTION (3.54) Refer to Example 3.6. We now have r1 = r2 = r. Thus, Equations (3.75) and (3.76) become

m = (1 r )+4 (1 r ) = 2r = 2(0.2) = 0.4

cos " = ((11 rr ))+!((11 rr )) = 0,

" = 90o

From Table 3.3 it can be concluded that surface of contact has a circular boundary: ca = cb = 1. 9

"10 ) n = 43((210 = 2.98667(1011 ) 1!0.252 ) 1

Thus,

(10 )( 0.4 ) 3 a = b = 1[ 25.98667 ] = 1.885 mm !1011 3

(10 ) # c = 1.5 " (1.5885 = 671.9 MPa ) 2 10! 6

______________________________________________________________________________________ SOLUTION (3.55) Case B ( 1st column ), Table 3.2 with r1 = r2 , (a)

1 E

m= + =

2 E

1 r

E1 = E2 .

2 r

0.17 a = 0.88 3 F Er = 0.88 3 400 210(10 = 0.604 mm 9 )

(b)

400 p0 = 1.5 !Pa2 = 1.5 ! (0.604) = 349 MPa 2 (10"6 )

( c ) Equations (3.68) at z=0:

" x = " y = # p0 [(1 +! ) # 12 ] = # 1+22! p0 = #0.8(349) = #279 MPa

! z = " p0 = "349 MPa

! yz = ! xz = 12 (" x # " z ) = # p20 (0.8 # 1) = 34.9 MPa ______________________________________________________________________________________ SOLUTION (3.56) Refer 2nd column of C, Table 3.2.

E1 = E2 = E = 210 GPa

Hence

r1 = 8 mm

F1 = 240 kN m

1 1 n = 0.008 " 0.05 = 105

2 1 ! = E2 = 210(10 , 9 = ) 105(109 ) 3

r2 = 50 mm

(a)

240(10 ) a = 1.076[ 105(10 ] 2 = 0.159 mm 9 )(105)

(b)

240(10 ) F p0 = !2 aL = !2 0.159(10 "3 = 961 MPa )

______________________________________________________________________________________ SOLUTION (3.57) o

We have r1 = 0.48 m, r2 = 0.34 m, r1 ' = !, r2 ' = !, and ! = 90 . Thus, using Eqs.(3.72) through (3.74):

4 = 0.796 1 + 0.34 A= , B = ± 12 ( r11 ! r12 )

1 0.48 2 m

0.48 "1 0.34 cos ! = BA = ± 11 0.48 +1 0.34 = 0.143,

Interpolating from Table 3.3: ca = 1.104, 3

1 3

!10 ) 11 n = 4(210 3(1" 0.09) = 3.0769(10 )

! = 81.78o cb = 0.911 . Hence

)(0.796) a = 1.104[ 4(10 ] = 2.4058 mm 3.0769(1011 ) )(0.796) 3 b = 0.911[ 4(10 ] = 1.9852 mm 3.0769(1011 )

Thus

) p0 = 1.5 !Fab = 1.5 ! (2.40584(10 = 400 MPa #1.9852#10"6 )

______________________________________________________________________________________ SOLUTION (3.58) Given quantities are:

Thus,

r1 = r1 ' = 0.025 m r2 = !0.03 m E = 200 GPa r2 ' = !0.125 m " = 0.3

4 1 1 1 1 + ! ! 0.025 0.025 0.3 0.125

= 0.10345

"10 ) 9 n = 43((200 1!0.09 ) = 293.04029(10 )

Also

2 A = m2 = 0.10345 = 19.33301 1

B = 12 [(0) 2 + ( r12 ! r1' ) + 2(0)] 2 = 12.66667 2

" = cos

= 49.06645o

!1 12.66667 19.33301

From Table 3.3, we find

ca = 1.78611

cb = 0.63409

The semiaxes are then 1

1800 ( 0.10345 ) 3 a = 1.78611[ 293 ] = 0.00154 m = 1.54 mm .04029 (109 ) 1

1800 ( 0.10345 ) 3 b = 0.67409[ 293 ] = 0.00055 m = 0.55 mm .04029 (109 )

Maximum contact pressure is therefore

$ c = 1.5 # (1.54"1800 = 1014.7 MPa 0.55 )(10! 6 )

______________________________________________________________________________________ SOLUTION (3.59) Now given data is as follows:

r1 = r1 ' = 0.02 m r2 ' = !0.125 m

r2 = !0.022 m E = 200 GPa " = 0.3

Therefore,

4 1 1 1 1 + ! ! 0.02 0.02 0.022 0.125

= 0.08594

9 "10 ) n = 43((200 1!0.09 ) = 293.0403(10 )

We have

A = m2 = 23.2721 1 1 B = ± 12 [ ! 0.022 + 0.125 ] = 18.7273 o .7273 " = cos !1 18 23.2721 = 36.42

Using Table 3.3

ca = 2.323

cb = 0.541

______________________________________________________________________________________ 3.59 (CONT.) Then, the semiaxes are: 1

( 0.08594 ) 3 a = 2.323[ 1800 ] = 0.00188 m = 1.88 mm 293.0403!109 1

( 0.08594 ) 3 b = 0.541[ 1800 ] = 0.00044 m = 0.44 mm 293.0403!109

Maximum contact stress is now obtained as

$ c = 1.5 # (1.881800 = 1039 MPa "0.44 )10! 6 End of Chapter 3

CHAPTER 4 SOLUTION (4.1)

!1 = 0

60 mm

M=1.5 kN ! m 3

P Mc 4P 32(1.5 #10 ) $ =$ $ 2 A I " (0.06) " (0.06)3 = !353.7 ! 70.74 "106

!2 = $

We have 2 ! 12 " ! 1! 2 + ! 22 = ! yp

! 2 = ! yp

So,

353.7 + 70.74 !106 = 250 !106 ,

Pall = 707 kN

______________________________________________________________________________________ SOLUTION (4.2)

p = 7.86(9.81)(! d 2 4) = 60.6d 2 kN m p=60.6d2 T=325 N ! m

2.5 m

!x =

32 M "d3

! = 16T " d 3

32( pL2 8) 32(60.6d 2 )(25)103 1.93 #106 = = 8" d 3 d "d3 3 16(325) 1.66 #10 =$ ! =$ 3 d3 "d

!x =

Equation (4.9a), ! x + 3"

2 = ! all

1.93 !106 2 "1.66 !103 2 ( ) + 3( ) = (100 !106 ) 2 d d3

3.725 !106 8.267 + 6 = 100 !108 2 d d d = 32.8 mm. Solving by trial & error: Use a 33 ! mm diameter shaft. ______________________________________________________________________________________ Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 4–1

______________________________________________________________________________________ SOLUTION (4.3)

! y = 100 MPa

" xy = #80 MPa

!x = 0

! yp = 250 MPa (Table D.1) Hence

! 1,2 =

!y 2

± (

#! y 2 ) + " xy2 2

= 50 ± (!50) 2 + (!80) 2 = 50 ± 94.3 or (a)

! 1 = 144.3 MPa

! 2 = "43.3 MPa

Maximum shear stress memory:

!1 " ! 2 #

! yp n

144.3 ! (!43.3) " 187.6 > 166.7

250 1.5

Failure will occur. (b)

Maximum energy of distortion theory:

! 12 " ! 1! 2 + ! 22 # (

! yp n

)

(144.3) 2 ! (144.3)(!43.3) ! (!43.3) 2

" 166.7

160.1 < 166.7

Failure will not occur. ______________________________________________________________________________________ SOLUTION (4.4) State of stress is given by

! 1 = ! = "32( 0.M1)3 + " (40P.1)2 ,

Refer to Table 4.1 and Eq. (2.66): 0.47! yp = 0.47! or Thus, Solving,

221(10 ) =

32 (17 ) ! ( 0.1)3

!2 = !3 = 0 ! yp = !

+ ! (40P.1)2

P = 375.7 kN

______________________________________________________________________________________ SOLUTION (4.5) From Table D.1: ! yp = 250 MPa We have 1

" 1, 2 = !302+90 ± [( !302!90 ) 2 + ( !40) 2 ] 2 (CONT.) ______________________________________________________________________________________ Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 4–2

______________________________________________________________________________________ 4.5 (CONT.)

" 1 = 102.1 MPa

" 2 = !42.1 MPa

( a ) Equation (4.2a) gives

102.1 + 42.1 = 250 n n = 1.73

( b ) Equation (4.5a) leads to

2 (102.1) 2 ! (102.1)( !42.1) + ( !42.1) 2 = ( 250 n )

from which

n = 1.95

______________________________________________________________________________________ SOLUTION (4.6) Referring to Appendix B, we obtain

" 1 = 101.3 MPa

" 3 = !51.32 MPa

! 1 " ! 3 = ! yp

(a)

! yp = 101.3 + 51.32 = 152.6 MPa

or (b)

"2 = 0

! oct = 13 [(101.3) 2 + (101.3 + 51.32) 2 + (51.32) 2 ] 2 = 63.41 MPa

Thus,

! yp = 63.41 0.47 = 134.9 MPa

______________________________________________________________________________________ SOLUTION (4.7) Using Eq. (4.6), we have 1

Here

0.47 nyp = 13 [(! 1 " ! 2 ) 2 + ! 12 + ! 22 ] 2

(a)

" 1, 2 = "2 ± 12 [" 2 + 4! 2 ] 2 # = 32!dM3 + !4dP2 ,

and

T " = 16 !d 3

Substituting the given data: 3

" 1, 2 = 12 [ 32! ((04.#1210)3) + 4!((450.#1210)2) ] 3

± 12 {[ 32! ((04."1210)3) + 4!((450."1210)2) ]2 + 4[ 16!(11( 0..212"10)3 ) ]2 } 2 Or " 1 = 49.55 MPa Equation (a) is then

" 2 = !21.99 MPa 1

Solving,

2 2 2 2 1.41 280 n = [( 71.54) + ( 49.55) + ( !21.99) ]

n = 4.4

______________________________________________________________________________________ SOLUTION (4.8) Maximum stresses, occurring at the fixed end are:

! = McI = 4502(t04.253 t ) = 168t 3.75

! = 23 450 = 337t 2 .5 2t 2 From Eq. (4.9a), we have

2 # yp = # 2 + 3" 2 = ( 280 ! 10 6 ) 2 Therefore, at neutral axis ! = 0 : " yp = 3! gives t = 1.45 mm At the extreme fibers, ! = 0 : ! yp = ! gives t = 8.45 mm

Allowable width is thus

t all = 8.45 mm

______________________________________________________________________________________ SOLUTION (4.9) We have "

! yp = 2yp = 175 MPa , " 1 = #" 2 = ! 175 1.5

(a) or

) = 16!(d500 3

d = 27.95 mm

" 1, 2 = "2 ± 12 " 2 + 4! 2

(b) Here

" = 32!dM3 = 32 (!pLd 3 8) = 8"32d 3 [ 14 "d 2 (77 ! 10 3 )10 2 ]

= 77(105 ) d ) T " = 16 = 16!(d500 = 2547.77 d 3 3 !d 3 Using Eq. (4.8a): " yp n

= " 2 + 4! 2

350 (106 ) 1.5

.77 2 2 = [( 77!d10 ) 2 + 4( 2547 ) ] d3 5

5.444(1016 ) = 5.929d(210 ) + 2.594d (610 ) Solving, by trial and error:

d = 0.0368 m = 36.8 mm

______________________________________________________________________________________ SOLUTION (4.10) We have ! all = 90 1.2 = 75 MPa (a)

" all = " 1 ! " 3 = 63.4 + 12.2 = 75.6 > 75 (b)

Failure occurs

2 2" all = (63.4 ! 0.53) 2 + (0.53 + 12.2) 2 + ( !12.2 ! 63.4) 2

! all = 70 < 75

No failure

______________________________________________________________________________________ SOLUTION (4.11) ( a ) Using the torsion formula,

" = ! T( 0(.005.05)4) 2 = 5093T

and " 1 = !" 2 = 5093T Equation (4.5a) yields then

[280(10 6 )]2 = (5093T ) 2 + (5093T )(5093T ) + ( !5093T ) 2

Solving,

T = 31.74 kN ! m

( b ) We now have 3

(10 )! " = 400 = 160 MPa ! ( 0.05 ) 2

! = 5093T

(as before)

Principal stresses are:

" 1, 2 = 160(210 ) ± 12 [(160 ! 10 6 ) 2 + 4(5093T ) 2 ] 2 123 1444442444443 a

=a±b With this notation, Eq. (4.5a) becomes

2 " yp = ( a + b) 2 ! ( a 2 ! b 2 ) + ( a ! b) 2 = a 2 + 3b 2

Thus,

( 280 ! 10 6 ) 2 = ( 160!210 ) 2 + 43 [(160 ! 10 6 ) 2 + 4(5093T ) 2 ] 6

Solving,

T = 26.05 kN ! m

______________________________________________________________________________________ SOLUTION (4.12) Maximum moment is 2

M = PL8 = 6(18.5) = 1.688 kN ! m 2

and hence

( 0.125 )10 ! = ! 1 = MyI = 1.0688 = 1.62 MPa .1( 0.25 )3 12

______________________________________________________________________________________ 4.12 (CONT.) ! yp n

= ! 12 " 0 = ! 1 28 n = 1.62

(a)

from which

n = 17.3 !

! 1 " 0 = nyp n = 17.3

(b)

or ______________________________________________________________________________________ SOLUTION (4.13) Referring to Appendix B, we compute

" 1 = 12.05 MPa

" 2 = !1.521 MPa

" 3 = !4.528 MPa

Using Eq. (4.5a),

2 2" yp = (12.05 + 1.521) 2 + ( !1.521 + 4.528) 2 + ( !4.528 ! 12.05) 2

Solving,

= 468

! yp = 15.3 MPa

Hence,

! yp = 15.3(0.577) = 8.828 MPa

Therefore

9 ! y = 2( 8.828 ) = 2.039 MPa

9 ! x = 3( 8.828 ) = 3.058 MPa

______________________________________________________________________________________ SOLUTION (4.14) y A

P T

1.2 m d

P T

! We have P = 50 R, T = 0.8 R, and M = 1.2 R . Stresses are (1.2 R ) " b = 32!dM3 = 32 = 97,784.8R ! ( 0.05 )3

T # = ! 16 = !"16( 0(.005.8)R3 ) = !32,595R "d 3

" a = ! ( 050.05R)2 4 = 25,464.8R ! x = ! a + ! b = 123,249.6 R (CONT.) ______________________________________________________________________________________ Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 4–6

______________________________________________________________________________________ 4.14 (CONT.) ( a ) Equation (4.8a): 1

260 (106 ) 2

= R[(123,249.6) 2 + 4( !32,595) 2 ] 2 R = 932 N

( b ) Equation (4.9a): 1

130 " 10 6 = R[(123,249.6) 2 + 3( !32,595) 2 ] 2 R = 959 N

or ______________________________________________________________________________________ SOLUTION (4.15) Referring to Appendix B,

" 1 = 197.4 MPa

" 2 = !14.44 MPa

" 3 = !72.96 MPa

Applying Eq. (4.4b):

2 2" yp = (197.4 + 14.44) 2 + ( !14.44 + 72.96) 2 + ( !72.96 ! 197.4) 2

! yp = 246.4 MPa

Hence,

! yp = 346.4(0.577) = 142.2 MPa

Thus,

140 ! y = 40 142 .2 = 39.38 MPa

140 ! x = 50 142 .2 = 49.23 MPa

______________________________________________________________________________________ SOLUTION (4.16) Stresses are

.25 ) ! 1 = prt = p0(.0005 = 50 p

! 2 = 2prt = 25 p ( a ) Applying Eq. (4.5a),

(50 p ) 2 ! 1250 p 2 + ( 25 p ) 2 = ( 280) 2 p = 6.466 MPa

or ( b ) Using Eq. (4.2a),

50 p ! 0 = 280 or

p = 5.6 MPa

______________________________________________________________________________________ SOLUTION (4.17) We have (! yp ) all = 182.2 = 68.33 MPa . Referring to Appendix B:

" 1 = 63.44 MPa

" 2 = 0.533 MPa

" 3 = !12.17 MPa

______________________________________________________________________________________ 4.17 (CONT.) ( a ) Using Eq. (4.1),

! yp = 63.44 + 12.17 = 75.6 > 68.33 :

Failure occurs

( b ) Applying Eq. (4.4b)

2 2" yp = (63.44 ! 0.533) 2 + (0.533 + 12.17) 2 + ( !12.17 ! 63.44) 2

! yp = 70.13 > 68.33 :

Failure occurs

______________________________________________________________________________________ SOLUTION (4.18) Referring to Appendix B:

! 1 = 162.4 MPa

! 2 = 46.15 MPa

! 3 = 1.468 MPa

( a ) Equation (4.1) yields

n = 162.4300 !1.468 = 1.86

( b ) Equation (4.4b) gives

2" 2

) n 2 = (" !" )2 +(" !"yp )2 +(" !" )2 = (116.25)2 +( 442 (.300 682 ) 2 + (160.932 ) 2 1

or n = 2.09 ______________________________________________________________________________________ SOLUTION (4.19) Referring to Appendix B:

! 1 = 156.2 MPa

! 2 = 42.13 MPa

! 3 = 11.7 MPa

( a ) Equation (4.1) gives

n = 156.220 2 !11.7 = 1.52

( b ) Equation (4.4b) yields

2" 2

) n 2 = (" !" )2 +(" !"yp )2 +(" !" )2 = (114.07 )2 +(230( 220 .43) 2 + ( !144.5 ) 2 1

n = 1.67

______________________________________________________________________________________ SOLUTION (4.20)

J " 2! r 3t pr 5(106 )(105) "! = = t 10 = 52.5 MPa pr !a = = 26.25 MPa 2t

! =$

Tc 50 #103 (0.21) =$ = $144.4 MPa J 2" (0.105)3 (0.01)

! 1,2 =

26.25 + 52.5 26.25 " 52.5 2 ± ( ) + ("144.4) 2 = 39.4 ± 145 2 2

______________________________________________________________________________________ 4.20 (CONT.)

! 1 = 184.4 MPa ! Thus, ! 1 = u ; n !u ' ; and !2 = n

! 2 = "105.6 MPa 250 184.4 = , n = 1.4 n "520 , " 105.6 = n = 4.9 n

______________________________________________________________________________________ SOLUTION (4.21)

RB = 250 kN

" M = 0 : 150(2)(1) ! R (1.2) = 0 , ! F = 0 : R = 50 kN A

150 kN/m z

1.2 m

50 kN

0.8 m

C b

250 kN 120

V, kN 50

x -130

3V 3 130 "103 97.5 "103 = = 2 A 2 2b 2 b2 ! max = " 1 = #" 2

! max = And Thus,

97.5 #103 = 120 #106 , ! max = " all ; 2 b

b = 28.5 mm

Use a 30 mm by 60 mm rectangular beam. ______________________________________________________________________________________ SOLUTION (4.22)

-420

! 2 = "4! 1

260

______________________________________________________________________________________ 4.22 (CONT.) ( a ) Upon following the procedure described in Sec. 4.11, Mohr’s circle is constructed as shown in the sketch above. The circle representing the given loading is then drawn by a trial and error procedure, as is indicated by the dashed lines. From the diagram, we measure the following values:

" 1 = 77 MPa

" 2 = !308 MPa

( b ) Applying Eq. (4.12a), or Solving,

"1 "u

! ""u2' = 1

"1 260

4" 1 ! !420 =1

" 1 = 75 MPa

" 2 = !300 MPa

______________________________________________________________________________________ SOLUTION (4.23) Principal stresses are 1

2 2 180 2 " 1, 2 = !180 2 ± [( 2 ) + 200 ]

" 1 = 129.3 MPa,

" 2 = !309.3 MPa

( a ) Equation (4.11a):

! 1 < 290 MPa But since

! 2 > 290 MPa :

failure occurs

( b ) Equation (4.12a):

.3 ! !309 650 = 1 gives 0.446 + 0.476 = 0.922 < 1 129.3 290

Thus, no fracture Note that Coulomb-Mohr theory is the most reliable when ! u ' >> ! u , as in this example. ______________________________________________________________________________________ SOLUTION (4.24)

" x = 2prt + 2!Prt 6

(10 ) = 2.8(210( 5))125 + 2! ( 045 .125 )( 0.005 ) = 46.46 MPa 6

! y = prt = 2.8(105 )125 = 70 MPa 3

.36 (10 ) " = 2!Trr 3t = 2! ( 31 = 63.89 MPa 0.1252 )( 0.005 )

Thus, 1

" 1, 2 = 12 ( 46.46 + 70) ± [ 14 ( 46.46 ! 70) 2 + (63.89) 2 ] 2 (CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 4.24 (CONT.)

" 1 = 123.2 MPa

" 2 = !6.74 MPa

( a ) Equation (4.12a),

6.74 ! !500 =1 gives 0.587 + 0.013 = 0.6 < 1 123.2 210

Thus,

no fracture

( b ) Equations (4.11a) shows

123.2 < 210, 6.14 < 210,

no fracture

no fracture ______________________________________________________________________________________ SOLUTION (4.25) State of stress is represented by Mohr’s circle shown below.

! !

3 8

!u ( 43 " u , ! )

O 3 4

2! p '

5 8

!u From the circle, we obtain

" = ! u ( 85 ) 2 # ( 83 ) 2 = 12 ! u " p ' = 12 tan !1 13 28 = 26.57 o

and

Orientation of the fracture plane is shown below. Fracture surface T

T !p' P P ______________________________________________________________________________________ SOLUTION (4.26) Principal stresses are 1

! 1, 2 = 12 ( 200 + 20) ± [(90) 2 + (150 2 )] 2 or

(a)

" 1 = 284.9 MPa

" 2 = !64.9 MPa

284.9 = 420 n ,

n = 1.47

64.9 =

n = 6.47

420 n

______________________________________________________________________________________ 4.26 (CONT.) !64.9 .9 1 ( b ) Equation (4.12a): 284 420 ! 900 = n

Solving, n = 1.33 ______________________________________________________________________________________ SOLUTION (4.27) Uniform shear stress ! acts on a typical element as shown.

" = P !dt " 1 = #" 3 = ! !1 = ! u

(a) (b)

P "td

= !u,

!1 !u

# !!u3' = 1;

!3 = !u

P = "td! u

" (1 + !!uu ' ) = ! u

P = (1+!tdu! u! u ')

______________________________________________________________________________________ SOLUTION (4.28) Table 4.3: K c = 23 1000 MPa

! yp = 444 MPa

Note that the values of crack length a and plate thickness t satisfy Table 4.3. 25 = 0.2 ! = 1.37 Table 4.2: wa = 125 Equation (4.18), with n=1:

1000 # all = " K!c a = (1.23 = 59.91 MPa 37 ) ! ( 25 )

Therefore

Pall = " all ( wt ) = 59.91(125 ! 25) = 187.2 kN

Note, the nominal stress at fracture: 3

.2 (10 ) " = t ( wP!a ) = 20187(125 ! 25 ) = 93.6 MPa < 444 MPa

______________________________________________________________________________________ SOLUTION (4.29) We have a w = 0.005 5 = 0.001 and ! = 1 by Table 4.2. From Table 4.3:

K c = 59 MPa m (a)

! yp = 1503 ksi

K = "# !a = (1)(100) ! (0.03) = 30.7 MPa m n = KKc = 3059.7 = 1.92

( b ) Using Eq. (4.18) with n = 1 :

# = " K!c a = (1) !59( 0.03) = 192.2 MPa This is well below the yield strength. ______________________________________________________________________________________ Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 4–12

______________________________________________________________________________________ SOLUTION (4.30) By Table 4.3: K c = 66 We have

a w

10 65

1000 MPa mm and ! yp = 1149 MPa. Table 4.2:

= 0.15

! = 1.02

1000 # = "nKc!a = (1.0266 = 165.9 MPa )( 2.2 ) ! (10 )

Thus 3

(10 ) t req = 2 wP! = 2 (200 65 )(165.9 ) = 9.27 mm

A thickness of 9.3 mm should be used. Note that both values of a and t satisfy Table 4.3. ______________________________________________________________________________________ SOLUTION (4.31) 8 = 0.2 . Refer to Example 4.5. We now have wa = 40

!a = 1.37

Table 4.2:

!b = 1.06

K c = 59 1000 MPa mm ! yp = 1503 MPa Equation (a) of Example 4.5, with M = 0.17 P : P) !" = (1.37) ( 0.04 )(P 0.01) + 1.06 ( 06.01( 0)(.170.04 = 71,000 P )2

Table 4.3:

By Eq. (4.18):

"# = n K!c a ;

1000 (10 ) 71,000 P = 59 1.8 ! ( 0.008 )

from which

P = 92.09 kN

The nominal stress at fracture: 3

" = t ( wP!a ) = ( 0.0192)(.009.04(10!0.)008) = 287.8 MPa < 1503 MPa ______________________________________________________________________________________ SOLUTION (4.32) From Table 4.3: K c = 23 1000 MPa

. , Case B of Table 4.2: a w = 016

mm and ! yp = 444 MPa

! = 112 .

By Eq.(4.18), with n = 1 : 23 1000 # = ! K"c a = 1.12 = 74.79 MPa " (24)

It follows that

P = ! ( wt ) = 74.79(150 " 30) = 337 kN

Then 3

) ! = ( w"Pa )t = (0.15337(10 " 0.024)(0.03) = 89.15 MPa

= 89.15 MPa < ! yp ______________________________________________________________________________________ Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 4–13

______________________________________________________________________________________ SOLUTION (4.33) Case A of Table 4.2 and Table 4.3:

! yp = 1503 MPa

K c = 59 1000 MPa mm ! = 1.01 (assumed) By Eq.(4.18):

59 1000 # = n!Kc" a = (2)(1.01) = 213 MPa < # yp " (6)

p r

p f = !rt = 213(10) = 71 MPa 30

p r

p f = 12 (71) = 35.5 MPa

(a)

! = tf ,

(b)

! = 2ft ,

______________________________________________________________________________________ SOLUTION (4.34) Table 4.3: K c = 31 MPa

! yp = 392 MPa

Case D of Table 4.2: wa = 0.4

" ! = 1.32 2

Using Eq.(4.18) with n = 1 and ! = 6 M tw : 6M 31(106 ) = 1.32 0.03(0.12) " (0.048) 2

K c = ! 6twM2 " a ; Solving

M = 4.35 kN ! m

______________________________________________________________________________________ SOLUTION (4.35) 3

(10 ) " m = 120 = 48 MPa 25(10! 4 )

and

!a ! cr

+ !! mf = 1;

1200 FA

48 (10 ) + 700 =1 240 (10 ) (106 ) 6

or FA = 186.3 kN ______________________________________________________________________________________ SOLUTION (4.36)

! 1a " ! 3a = 15 p " 0 = ! ea ! 1m " ! 3m = 9 p " 0 = ! em Then, or

15 p 250 (106 )

p + 3009(10 =1 6 )

p = 11.11 MPa

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (4.37) We have

! cr = 1"(!!ma ! u )

(a)

where,

" m = Fmax2+AFmin ,

" a = Fmax2!AFmin

(b)

Substituting Eqs. (b) into (a): Solving,

max " Fmin ) 2 A ! cr = 1"[ ((FFmax + Fmin ) 2 A! u ]

A = !1cr [ Fmax " 12 ( Fmax + Fmin )(1 " !!cru )] ______________________________________________________________________________________ SOLUTION (4.38) We have ! m = ! a . Using Table 4.4: !m 510 1.5

!m + 1050 1.5 = 1

from which ! m = 228.8 MPa. At the fixed end:

M max = PL = 10(0.05) = 0.5 N ! m

Hence,

M a ,m = ( M max 2± M min ) = 0.25 N ! m and Solving,

0.25 ) ! a = ! m = 6btM2m = 06.(005 = 300 = 228.8(10 6 ) t2 t2

t = 1.145 mm

______________________________________________________________________________________ SOLUTION (4.39) We have ! a = ! m . From Table 4.4: !m 740 2.5

!m + 1500 2.5 = 1

! m = 198.2 MPa.

At the center of the beam:

M max = PL 4 = (0.25)( 20)(0.125) = 0.625 N ! m

Hence,

M a ,m = ( M max 2± M min ) = 0.3125 N ! m

and Solving,

) ! a = ! m = 6btM2m = 6(00..013125 = 187t 2.5 = 198.2(10 6 ) t2

t = 0.973 mm

______________________________________________________________________________________ SOLUTION (4.40)

! max = ! min = McI , and

! m = 4"Mr 3 ,

" max = TrJ ,

" min = 0,

"x =",

! xy = ! ,

!a = 0

" m = " a = !Trr 3 " y = " z = ! xz = ! yz = 0

Equations (4.21) yield

0 + 3" a2 = ! ea ,

! m2 + 3" m2 = ! em

Then, Soderberg relation becomes

! cr = or

3" a

! m2 + 3" m2 1# ! yp

" cr = "" cryp " m2 + 3! m2 + 3 ! a

or 1

" cr = "" cryp [( 4!Mr 3 ) 2 + 3( !Tr 3 ) 2 ] 2 + !3rT3 Solving this expression for r, we obtain Eq. (P4.40). ______________________________________________________________________________________ SOLUTION (4.41) " xa +" ya 2

" 1a , 2 a =

2 ± [( xa 2 ya ) 2 + ! xya ]2 2

2 = "2xa ± [ "4xa + ! xya ] 4

" 1a = 942.44 MPa,

Similarly,

" 1m = 161.8 MPa ,

Thus,

81(10 ) = 900 + 4(10 4 )] 2 2 ±[ 4

" 2 a = !42.44 MPa " 2 m = !61.8 MPa

" ea = " 1a ! " 2 a = 984.88 MPa ! em = ! 1m + ! 2 m = 233.6 MPa

( a ) Modified Goodman relation:

984.88 " cr = 1!( 223 .6 2400 ) = 1086 MPa

"2400 800 ) b = ln( 0ln(.910 = !0.0863 3 108 ) !11.587 N cr = 103 ( 0.91086 = 2.89(106 ) cycles "2400 )

( b ) Soderberg criterion:

948.88 " cr = 1!( 223 .6 1600 ) = 1145 MPa !11.587 N cr = 103 ( 0.91145 = 1.57(106 ) cycles "2400 )

______________________________________________________________________________________ 4.41 (CONT.) ( c ) The SAE criterion:

" cr = 1086 MPa , N cr = 1(

1086 !16.778 2400

)

b = !0.0596 = 0.59(106 ) cycles

( d ) Gerber criterion:

" cr = 1!( 223948.6.882400 )2 = 993.51 MPa .51 !11.587 N cr = 103 ( 0.993 = 8.12(106 ) cycles 9"2400 )

______________________________________________________________________________________ SOLUTION (4.42)

! xa = (800 + 600) 2 = 700 MPa " xm = (800 ! 600) 2 = 100 MPa ! ya = (500 + 300) 2 = 400 MPa " ym = (500 ! 300) 2 = 100 MPa ! xya = ( 200 + 150) 2 = 175 MPa

" xym = ( 200 ! 150) 2 = 25 MPa Equations (4.21) give then

2" ea2 = (700 ! 400) 2 + 400 2 + 700 2 + 6(175) 2 2 2" em = ( 200 ! 100) 2 + 100 2 + 100 2 + 6( 25) 2

! ea = 679.61 MPa,

! em = 108.97 MPa

( a ) Modified Goodman criterion:

" cr = 1!(108679.97.611600 ) = 729.27 MPa "1600 0.5"1600 ) b = ln( 0.9ln( = !0.08509 103 108 ) .27 !11.752 N cr = 103 ( 0729 = 2.97(106 ) cycles .9"1600 )

( b ) Soderberg criterion:

" cr = 1!(108679.97.611000 ) = 762.72 MPa .72 !11.752 N cr = 103 ( 0762 = 1.75(106 ) cycles .9"1600 )

______________________________________________________________________________________ SOLUTION (4.43)

" a = ( M max 2!IM min ) c ! m = ( M max 2+IM min ) c ,

M = PL

Substituting these into Soderberg relation, we obtain

M max = (21"+ !cr)Ic + 11+#!! M min (CONT.) ______________________________________________________________________________________ Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 4–17

______________________________________________________________________________________ 4.43 (CONT.) where

" = !! cryp = 23

Thus,

13 200!10 )( 0.05!0.1 ) Pmax = c (21"+cr! I) L + 11+#!! Pmin = 2 (12 ( 5 3)( 0.05 )(1.2 ) + 5 3 (10,000)

= 16,667 + 2000 = 18.7 kN ______________________________________________________________________________________ SOLUTION (4.44) We have W = mg = 80 ! 9.81 = 784.8 N From Eq. (4.29);

" max = (1 + 1 + !2sth ) WA Solving, with ! st = WL AE , we obtain max h = L2"EW ( A" max ! 2W )

(a)

Substituting the given data: 6

( 2 )( 350"10 ) h = 2 (105 ( 250 " 350 ! 1569.6) = 0.365 m "109 )( 784.8 )

______________________________________________________________________________________ SOLUTION (4.45) Through the use of Eq. (4.29): A [ " max # 1]2 = 1 + 2h! st W

" max = WA [1 + 1 + !2sth ], from which 2 A2 " max W2

hAE ! 2"Wmax A + 1 = 1 + 2WL

Solving for W, we obtain Eq. (P4.45). ______________________________________________________________________________________ SOLUTION (4.46) The kinetic energy is 2 2

E k = 12 T!

where T =

E k = W$2 gr = 1090( 240"22( 9#.8160)) ( 0.35) = 4300 N ! m But Thus,

GJ! L

! = 2 LEk JG

(a)

Substituting the data given: 1

)2 # = [ 80.52(!101.95)("4300 ] 2 = 0.08309 rad = 4.76o ( 0.0625 ) 4

Introducing !L TL " = GJ = !rJ GJL = rG

into Eq. (a), we obtain

! = 4GE k AL

where A = !c

Substituting the numerical values, 9

5!10 ) 4300 2 # = [ 4"((800..0625 ] = 274.3 MPa ) 4 (1.5 )

______________________________________________________________________________________ SOLUTION (4.47) W 0.75 m 1.2 m We have ! yp = Mc I = PLc I from which ! I

.05 12 ) P = Lcyp = 280(110.2 ()(0.0025 = 4.86 kN )

End deflection is 3

(1.2 ) (12 ) " max = 3(4860 = 26.86 mm 200!109 )( 0.05 ) 4

But Solving,

W (0.75 + 0.02686) = 12 ( 4860)(0.02686)

W = 84.02 N End of Chapter 4

CHAPTER 5 SOLUTION (5.1) Y 15 mm

A1 z

N.A. 150 mm

y A2

15 mm Z A

135 mm .5+ (135!15 )[15+ (135 2 ) z = A1Az11++ AA22z2 = (150!15) 7150 !15+135!15

or Then,

z = y = 43 mm

I y = 121 (150)(15) 3 + (150 ! 15)(35.5) 2 + 121 (15)(135) 3

+ (135 ! 15)(39.5) 2 I y = I z = 9.11(10 6 ) mm 4

I yz = (150 ! 15)( "32)( "35.5) + (135 ! 15)(35.5)(39.5)

= 5.4(10 6 ) mm 4 We have the moment components:

M y = 0,

Thus,

M z = "11.25(0.9) = "10.125 kN ! m

) +10125( 9.11)( 0.043) (" x ) A = !10125( 5[(.49)(.110.)107 = !35 MPa 2 !( 5.4 ) 2 ](10! 6 )

Equation (5.15) gives or

tan ! = 95..114 = 0.593

! = 30.66o

______________________________________________________________________________________ SOLUTION (5.2) 80 mm

N.A 150 mm

A z

! = 63.8o 1 kN 30o y

I y = 121 hb 3 = 121 (150)(80) 3 = 6.4 ! 10 6 mm 4 (CONT.) ______________________________________________________________________________________ Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 5–1

______________________________________________________________________________________ 5.2 (CONT.)

I z = 121 bh 3 = 121 (80)(150) 3 = 22.5 ! 10 6 mm 4 M y = ( P sin " ) L = 600 N ! m

M z = ( P cos " ) L = 1,039.2 N ! m y d = !75 mm z d = 40 mm ( a ) Equation (5.15) with I yz = 0 : M

tan " = II zy M yz = II zy tan #

! " = 63.8o

( b ) Thus, maximum tensile stress is at point A. Equation (5.16) gives

( 0.04 ) # max = 600 ! 1,0396..42"(10!0!.6075) = 13.25 MPa 22.5"10! 6

______________________________________________________________________________________ SOLUTION (5.3)

I y = 121 hb 3 = 121 ( 40)(100) 3 = 3.33 ! 10 6 mm 4

I z = 121 bh 3 = 121 (100)( 40) 3 = 0.533 ! 10 6 mm 4 M y = M o cos " = 800(cos 25o ) = 725 N ! m

M z = M o sin " = 800(sin 25o ) = 338.1 N ! m ( a ) Equation (5.15) with I yz = 0 : M

tan " = II zy M yz = II zy cot #

! " = 18.9 o

( b ) There, maximum compressive stress is at A. 100 mm 40 mm

N.A A

Equation (5.16) results in

25o y

! = 18.9 o 800 N ! m

( !0.05 ) .1( 0.02 ) " max = 725 ! 0338 = !23.57 MPa 3.33(10! 6 ) .533(10! 6 )

______________________________________________________________________________________ SOLUTION (5.4)

I y = 2[ 3012!8 + 240 ! 34 2 ] + 8!1276 = 85(10 4 ) mm 4 3

I z = 2[ 3012!8 + 240 ! 19 2 ] + 7612!8 = 21.2(10 4 ) mm 4 3

I yz = [0 + 240(19)( !34)] + [0 + 240( !19)(34) + 0 = !31(10 4 ) mm 4 Using Eq. (5.14):

M o [21.2 + 1.5( "31)]z = M o [ "31 + 1.5 ! 85] y

______________________________________________________________________________________ 5.4 (CONT.)

z = !3.81 y or Thus, point A is the farthest from the N.A., as shown. A

N.A.

y Hence, Eq. (5.13) gives

31)]0.03![ !31+1.5( 85 )]( !0.034 ) (# x ) A = 80(10 6 ) = [ 21.2+1.5( ![85 Mo "21.2 !( !31) 2 ]10!8

from which

M o = 266.8 N ! m

______________________________________________________________________________________ SOLUTION (5.5) Bending moment at the midspan is

M z = "24( 4) " 24( 2) = "48 kN ! m

From Fig. 5.4 and Example 5.1:

z E = 0.105 m zD = 0 y E = !0.045 m y D = !0.045 m I y = I z = 11.596(10 !6 ) m 4

I yz = !6.79(10 !6 ) m 4 Then, Eq. (5.13) with M = 0 :

)(11.596 )( !0.045 ) (" x ) D = 0![(( !1148000 = !283.4 MPa .596 ) 2 !( 6.79 ) 2 ]10! 6

Similarly,

( 0.105 ) !11.596 ( !0.045 )] (" x ) E = !48000[([ !116..79 = 103.8 MPa 596 ) 2 !( 6.79 ) 2 ]10! 6

______________________________________________________________________________________ SOLUTION (5.6)

Y 60 mm 80 mm

z 20 mm

N . A. for ! = 15o N . A. for ! = 30o

20 mm

Y (CONT.) ______________________________________________________________________________________ Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 5–3

______________________________________________________________________________________ 5.6 (CONT.) !20 ( 50 ) z = 80!2080(!1020)++60 = 27.14 mm 60!20 !20 !60 I y = 8012 + 20 ! 80(17.14) 2 + 2012 + 20 ! 60( 22.86) 2 3

= 15.1048(105 ) mm 4 !20 !30 I z = 8012 + 2[ 2012 + 20 ! 30( 25) 2 ] = 8.933(105 ) mm 4 3

Due to the symmetry I yz = 0. ( a ) We have

! = 0o

My = 0

M z = 1.5P

Equation (5.13) is thus, 290 (106 ) 1.2

5 P ( 0.04 ) = 1.5I zPy = 81..933 (10! 7 )

P = 3.6 kN

( b ) Now we have ! = 15 and

M z = 1.5P cos 15o = 1.4489 P M y = 1.5P sin 15o = 0.3882 P

Equation (5.14):

0.3882 P(8.933) z = 1.4489 P(15.1048) y

from which

z = 6.3105 y

The farthest point from the N.A. is A. Equation (5.13): 6

) 0.02714 ) !1.4889 P (15.1048 ) 0.04 (" x ) A = 2901(.10 = 0.3882 P (8.933)(8.!933 2 (15.1048 )10! 7

Solving, P = 3.36 kN ______________________________________________________________________________________ SOLUTION (5.7) o

Given ! = 30 and

M z = 1.5P cos 30o = 1.299 P M y = 1.5P sin 30o = 0.75P

The area properties are already found in Solution of Prob. 5.6. Equation (5.14) gives

0.75P(8.933) z = 1.299 P(15.1048) y z = 2.9287 y

As before, the maximum stress occurs at A. Equation (5.13): 6

) 0.02714 ) !1.299 P (15.1048 ) 0.04 (" x ) A = 2901(.10 = 0.75 P (8.933)(8!.933 2 (15.1048 )10! 7

Solving,

P = 3.37 kN

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (5.8) N.A A z

y We have M y = 0

and

M z = PL

Equation (5.14) becomes

I yz z = I y y or

! th 3 z = 23 th 3 y

y = ! 3z2

Point A is the farthest from the N.A. Thus, with y A = !h ! 2t and z A = ! 2t Equation (5.13) yields 3

(" x ) A = PL[ ! (th2 th( !3 t32)()8!th( 23th3)!3()!hth(3!)h2 !t 2 ) or

(! x ) A = 3 PL (72th.53t + 2 h ) = ! max

______________________________________________________________________________________ SOLUTION (5.9)

A 80 mm

C y

N.A

I y = bh36 = 80(3690 ) = 1.62(10 6 ) mm 4

30 mm 90 mm

I z = hb48 = 90(4880 ) = 0.96(10 6 ) mm 4 M y = "3(10 3 ) sin 20o = "1.026 kN ! m

M z = 3(10 3 ) cos 20o = 2.819 kN ! m

( a ) Equation (5.15):

96 " = tan !1 [ 10..62 tan( !20o )] = !12.17 o

( b ) Point A is the farthest from the N.A. Equation (5.16):

( !0.03) ( !0.04 ) " A = !1026 ! 2819 = 136.5 MPa 1.62 (10! 6 ) 0.96 (10! 6 )

______________________________________________________________________________________ SOLUTION (5.10) (a)

$ x = ""y#2 = !c4 xy ! 23 c5 x 3 y ! c6 xy 3 2

$ y = ""x#2 = c2 x ! c3 xy ! 23 c5 xy 3 2

$ xy = ! ""x"#y = c1 + c23 x 2 + c24 y 2 + c5 x 2 y 2 + c46 y 4 2

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 5.10 (CONT.) " 4# "x 4

Thus,

" 4 ! = 6c6 xy + 8c5 xy = 0

" 4# "y 4

= !6c6 xy ,

and

= 0,

" 4# "x 2"y 2

= !4c5 xy

6c6 + 8c5 = 0

At y = h 2 :

(a)

!y = 0: 3

5h c2 ! c23h ! c12 =0 At y = ! h 2 : " y = ! px Lt :

(b)

( c2 + c3 y ! 2 c35 y ) x = ! px Lt 3

5h c2 + c23h + c12 = ! Ltp

(c)

Adding Eqs. (b) and (c),

c2 = ! 2pLt

Substituting this into Eq. (b): 3

c3h 2

5h + c12 = ! 2pLt On y = ± h2 , ! xy = 0 :

(d)

c6 ! c1 ! c23 x 2 ! c84 h 2 ! c45 h 2 x 2 ! 64 h4 = 0 2

6h ( c23 + c54h ) x 2 + ( c1 + c48h + c64 )=0 This is of form A ! B + C = 0, where A, B, C are independent. Thus,

c3 2

+ c54h = 0

(f) 2

(e)

6h c1 + c48h + c64 =0

(g)

Multiply Eq. (f) by h and subtract it from Eq. (d) to find

c5 = ! 3Ltp

Then, Eqs. (g) and (a) give p c3 = 23htL ,

On x = 0 : h

c6 = th43pL

V = 0: h

c y c y " h # xy tdy = " h ( !c1 ! 42 ! 64 )tdy = 0 2

! 2

h c1 + c4 ( ) + c6 ( 320 )=0 h2 24

(h)

Subtracting Eq. (h) from (g), together with the value of c6 already obtained, we have p c4 = ! 53htL

Finally, substitute c4 and c6 into Eq. (g) to determine

c1 = 80phtL (CONT.) ______________________________________________________________________________________ Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 5–6

______________________________________________________________________________________ 5.10 (CONT.) The stress function is thus, 3

3 3

x " = Ltp [ 80h xy ! 12 ! x4 hy + 10xyh + xhy3 ! 5xyh3 ] 3

and the stresses are

" x = Ltp [ 53h xy + h23 x 3 y ! h43 xy 3 ]

" y = Ltp [ ! 2x ! 23h xy + h23 xy 3 ] 2

2 3

3y 3x y y " xy = Ltp [ 80h ! 34xh + 10 h + h3 ! h3 ] 2

px ! x = McI = ( px th63L12)( h 2 ) = Lth 2

(b)

( c ) The maximum stresses are 3

(" x ) elast . = 10pth [6 y + 2000 y ! 40h 2y ] 3

(! x ) elem. = p10(10hh3t) = 100( pt ) Thus,

(! x ) elast . = 99.8( pt )

at y = ± h2

(! x ) elast . = 0.998(! x ) elem.

______________________________________________________________________________________ SOLUTION (5.11) 4

( a ) We can show that given ! satisfies " ! = 0. From Eqs. (3.13):

" x = 0.p43 {2[0.78 ! tan !1 xy ] ! x 22+xyy 2 }

(a)

" y = 0.p43 {2[0.78 ! tan !1 xy ] ! 2 + x 22+xyy 2 }

(b)

" xy = ! 0.p43 x22 +y y 2

(c)

To test which stress is maximum we rewrite ! y in the form: 2

" y = 0.p43 {2[0.78 ! tan !1 xy ] ! 2 (xx2 ++yy2 ) + x 22+xyy 2 } 2

= 0.p43 {2[0.78 ! tan !1 xy ] ! x 22+xyy 2 ! 2x( 2x+! yy2) } Comparing this with Eq. (a), noting 2( x ! y )

( x 2 + y 2 ) > 0, we conclude that ! x > ! y .

When y = 0, Eqs. (a) to (c) yield # x = 3.63 p, # y = ! p 0.43 , " xy = ! p 0.43 Maximum stress occurs at y = 0 :

! x ,max = 3.63( 2) = 7.26 MPa (b)

px 2 ( x )

(! x ) elem. = McI = 2 x 3 32 = 3 p = 6 MPa Thus,

(! x ) elast . = 1.21(! x ) elem.

______________________________________________________________________________________ SOLUTION (5.12) ! bottom ! top

and Thus

c1 c2

c1 = 22.5 mm c2 = 67.5 mm

15 mm

y Ay y = c1 = ! Ai i !i (15 )( 90 )( 45 ) + ( b !15 )(15 )( 7.5 ) = (15)( 90 )+( b!15)(15) = 22.5 mm

y = c1 c2

90 mm

15 mm

+ ( b !15 )(112.5 ) = 601,750 = 22.5 mm , 350+ ( b !15 )(15 )

Solving,

b = 150 mm

______________________________________________________________________________________ SOLUTION (5.13) pL ( th 2 )( h 4 ) 3 pL ! max = VQ Ib = 2 ( th 3 12 ) t = 4 ht 2

! max = McI = pL8 thh3 212 = 43 pL th 2 p

t h

L Thus, " max ! max

= Lh

from which h L = "!max = 8.4 (00.7.15) = 1.8 m max

Then,

p = 43 " maxL( th ) = 43 700( 0.105.8!0.15) = 3.88 kN ! m ______________________________________________________________________________________ SOLUTION (5.14) P 4500

R A = P2 ! 2250

RB = P2 + 6750

V(N)

( N ! m)

4500

0.75P-3375

P 2

+ 2250

6750

______________________________________________________________________________________ 5.14 (CONT.) We have 3

0.25 ) I = 0.2 (12 ! 0.1512( 0.2 ) = 160.417(10 !6 ) m 4

Then,

6 ! = VQ Ib = 0.7(10 )

P 2 ) + 2250 ) = 160(.417 (0.2 ! 0.025 ! 0.1125 + 0.1 ! 0.05 ! 0.05)( 2) !10" 6 ( 0.05 )

Solving,

P = 9.32 kN

Similarly, or Thus,

3375 )( 0.125 ) # = McI = 7(10 6 ) = ( 0.75160P !.417 "10! 6 P = 16.478 kN Pall = 9.32 kN

______________________________________________________________________________________ SOLUTION (5.15) y

! all = 150 MPa

and

bh3 1 ! (b ! 2t )(h ! 2t )3 12 12 1 1 = (120)(170)3 ! (100)(150)3 12 12 6 4 = 21.005(10 ) mm

Iz =

(a)

M z

C t

Therefore

Mc ! all = , Iz

t b

I 21.005(10"6 ) M = ! all = 150(106 ) = 37.1 kN # m c 0.085 1 M 37,100 = = = 0.0252 9 rx EI (70 "10 )(21.005)(10!6 ) rx = 39.68 m

or (b)

______________________________________________________________________________________ SOLUTION (5.16)

I= (a)

1 ! ( D 4 " d 4 ) = (604 " 404 ) = 510.509(103 ) mm 4 64 64

y A = 30 mm = 0.03 m My A (600)(0.03) !A = = = 35.3 MPa 510.509(10"9 ) I

______________________________________________________________________________________ 5.16 (CONT.) (b)

(c)

yB = 20 mm = 0.02 m MyB (600)(0.02) !B = = = 23.5 MPa 510.509(10"9 ) I

1 M = rx EI

600 70(10 )(510.509 "10!9 ) = 0.0168 =

rx = 59.56 m

M C B A d D

Hence,

r 59.56 = "205.48 m rz = " x = " 0.29 !

______________________________________________________________________________________ SOLUTION (5.17)

! A y = 2A y + A y 2A + A !A i

2(150 ! 25)(75) + (350 ! 25)(137.5) = c2 2(150 ! 25) + (350 ! 25) z C c1 25 = 108.65 mm 25 So c1 = 108.65 mm and c2 = 41.35 mm 350 mm 1 3 2 I z = 2[ (25)(150) + (25 !150)(33.65) ] + 12 1 (350)(25)3 + (350 ! 25)(28.85) 2 = 30.29(106 ) mm 4 12 1 1 M z = M = PL = P(2.4) = 1.2 P 2 2

150 mm

Therefore

Mc1 1.2 P(0.10865) ; 60(106 ) = , P = 14.4 MPa 30.29(10"6 ) Iz Mc2 1.2 P(0.04153) 60(106 ) = , P = 36.6 kN !c = 30.29(10"6 ) Iz

!t =

Hence Pall = 14.4 kN ______________________________________________________________________________________ Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 5–10

______________________________________________________________________________________ SOLUTION (5.18)

M max =

1 2 1 pL = (12)(3) 2 = 13.5 kN ! m 8 8

1 (80)(120)3 = 11.52 !106 mm 4 12 Mc 13.5 #103 (45 #10"3 ) = = 52.7 MPa ! nom = I 11.52 #10"6 ! 95 K = all = = 1.8 ! nom 52.7 D 120 = = 1.33 : Use Fig. D.3. For K = 1.8, d 90 r = 0.14 d I=

Thus,

rmin = 0.14(90) = 12.6 mm

______________________________________________________________________________________ SOLUTION (5.19) p (a) h L pL/2 pL/2 b V, kN

pL/2 x

M kN ! m

Mmax= pL2/8

-pL/2 x

3 V 3 pL 2 3 pL = = 2 A 2 bh 4 bh 2 Mc pL 8(h 2) 3 pL2 ! max = = = I bh3 12 4 bh 2

! max =

Thus,

" max ! max = h L

(1) (2)

(3)

For example, if L = 10h , the above ratio is 1 10 . (b)

From Eq. (3), we have:

L=h

! all 9 = 0.16( ) = 1.029 m 1.4 " all

______________________________________________________________________________________ 5.19 (CONT.) Equation (1) gives then

pall =

4 bh 4 0.05 " 0.16 (1.4 "106 ) = 10.34 kN m ! all = 3 L 3 1.029

______________________________________________________________________________________ SOLUTION (5.20) P

300 mm 60

M = 3P n = E s Et = 20

I t = ( 520 )(12300 ) = 1170(10 6 ) mm 4 400

y =150

z y

The allowable stress in the transformed section is 12 20

= 6 MPa < 7 MPa

Thus, the stress in the steel is the controlling stress. Hence,

M max = 3Pmax = ! maxc I t 6

(10 ) = 6(10 )(1170 0.15 )

Pmax = 15.6 kN

______________________________________________________________________________________ SOLUTION (5.21) 180 mm 25 kN/m 10

4m 2

z 300 mm

M max = pL8 = 25"108 ( 4 ) = 50 kN ! m 3

I t = 121 (180)(300) 3 + 2[ 126012(10 ) + (1260)(10)(155) 2 ] = 1010.64(10 6 ) mm 4 180 mm

C y

180x7=1260 mm 3

50"10 ( 0.15 ) # w,max = Mc I t = 1010.64 (10! 6 ) = 7.4 MPa 3

7"50"10 ( 0.16 ) # a ,max = 7 Mc I t = 1010.64 (10! 6 ) = 55.4 MPa

______________________________________________________________________________________ SOLUTION (5.22) Equation (5.54) becomes

20 20 ( kd ) 2 + ( kd )( 300 )(1200) ! ( 300 )(500)(1200) = 0

or ( kd ) Solving,

+ 180kd ! 40(10 3 ) = 0

kd = 164 mm

Hence,

500 ! kd = 336 mm

From Eqs. (e) of Example 5.5:

M c = 12 " c (bkd )( d ! kd3 ) = 12 (12 " 10 6 )(0.3 " 0.164)(0.5 ! 0.164 3 ) = 89.5 kN ! m M s = " s As ( d ! kd3 )

and

= 150(10 6 )(1200 " 10 !6 )(0.445) = 81.9 kN ! m Thus,

M all = 81.9 MPa

______________________________________________________________________________________ SOLUTION (5.23) 5 C

d A

E c1=kd D B

80/8 The stresses in concrete and the equivalent of the steel (Fig. 514b) have the values shown in the preceding figure. From the similarity of !ECD and !EAB we find c1 d

= (80 58)+5 = 155

c1 = 515d = 155

Equation (d) of Example 5.5: 1 2 1 2

! c ( c1b) = ! s As (5)( 500 3 )300 = 80 As

As = 1563 mm 2

Then, we have from Eq. (e) of Example 5.5,

M = " s As ( d ! c31 )

= 80(10 6 )(1563 " 10 !6 )(0.5 ! 09.5 ) = 55.6 kN ! m ______________________________________________________________________________________ Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 5–13

______________________________________________________________________________________ SOLUTION (5.24) The shearing stress at a distance s is given by V Q

$ = Iyz bz = I zy ! R cos # (tRd# ) 0

2V y "Rt

sin !

This shows that ! = 0 at the free ends and ! max at the neutral axis, same as for a rectangular section. The shearing stress produces the following twisting moment about O:

T = ! $RdA = !

" 2V sin # y

"Rt

R(Rtd# )

4 RV y !

By applying the principle of moment at O: V y e = M .

Thus,

4R !

______________________________________________________________________________________ SOLUTION (5.25) Shearing stress in the web is neglected. Moment of the forces about S: e1 e2

V1e1 = V2 e2 ;

= VV12

(a)

Let M 1 and M 2 be bending moments on flanges 1 and 2, respectively. Curvature-moment are related by 1 r1

M1 = EI , 1

M2 = EI 2

1 r2

(b)

By assuming r1 = r2 , we have M1 EI1

M2 = EI ; 2

dM 1 I1

= dMI 2 2

Introducing dM dx = !V , Eq. (c) gives V1 V2

= II12

e1 e2

= II12

Equation (a) now becomes

Since Thus, where

e1 + e2 = h;

[ ( h !e1e1 ) ] = II12

e1 = ( I1I+2hI 2 ) 3

I 1 = b121 t1

I 2 = b122 t2

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (5.26) A 1

y Y

Fig. (a) Location of centroid C (Fig. a):

+ 9.5!125( 62.5 ) z = 15.515!75.5!( "7537+.95.)5+!0250 = 6.481 mm + 9.5!125 ) + 0+ 9.5"125( !125 ) y = 15.515"75.5"( !75250 = !124.339 mm + 9.5"250+ 9.5"125

Moment of inertia:

I y = 121 (15.5)(75) 3 + (15.5 ! 75)( 43.98) 2

+ 121 ( 250)(9.5) 3 + 250(9.5)(6.48) 2 + 121 (9.5)(125) 3 + 9.5(125)(56.01) 2

= 8.18(10 6 ) mm 4 I z = 121 (75)(15.5) 3 + (75 ! 15.5)(125.6) 2 + 121 (9.5)( 250) 3 + (9.5 ! 250)(0.65) 2 + 121 (125)(9.5) 3 + (125 ! 9.5)(124.3) 2

= 49.10(106 ) mm 4 I yz = (15.5 " 75)( !125)( !43.98)

+ (9.5 " 250)( !0.65)( !6.48) + (9.5 ! 125)(124.3)(56) = 14.70(106 ) mm 4 .7 ) o " p = 12 tan !1 [ ! 8.218(14!49 .1 ] = 17.85

Then,

I y ' = 10 6 [ 8.18+2 49.1 + 8.18!2 49.1 cos 35.7 o ! 14.7 sin 35.7 o ] = 3.45(10 6 ) mm 4

Similarly, we compute

I 1 = I z ' = 53.85(10 6 ) mm 4 I 2 = I y ' = 3.45(10 6 ) mm 4

From geometry of section (Fig. b):

HB = 144.57 mm

BC = 39.05 mm

Thus, V

" xz = I zy't' = [ st (0.1445 ! 2s sin 17.85o ]

______________________________________________________________________________________ 5.26 (CONT.) Shear force due to V y ' (Fig. b): s

We write

V ( 0.0155 )

F1 = ! $ xz tds = 54y ' .1#10"6 ! 0

0.075

[0.1445s " s2 sin 17.85o ]ds = 0.1108V y ' 2

V y ' ! ez ' = 0.25F1 = 0.25(0.1108V y ' )

e z ' = 0.0277 m = 27.7 mm

Shear force due to V z ' . Now assume that the

direction of F1 shown in the figure is reversed. Then,

F1 = VIzy''t !

0.075

[0.03905s " s2 cos 17.85o ]ds 2

= 0.1928Vz '

We write

s F1

144.57

Vy’ Vz’ ey’

Vz ' ! e y ' = 0.25(0.1928Vz ' )

s 2

C B

17.85

z’ z o

y y’ ez’

Figure (b)

e y ' = 0.0482 m = 48.2 mm

______________________________________________________________________________________ SOLUTION (5.27) 1 2

x 3

( p0 Lx ) x

p0 Lx

M A = p03L

R A = p20 L '

We have EIv = M = 16 p0 x

Integrating

! 12 p0 Lx + 13 p0 L2 2

EIv ' = 24p0L x 4 ! p40 L x 2 + po3L x + c1 v ' (0) = 0; c1 = 0. 2

EIv ' = 24p0L x 4 ! p40 L x 2 + po3L x

(a)

Integrating

0 EIv = 120pEIL x 5 ! p120 L x 3 + po6L x 2 + c2 v (0) = 0; c2 = 0.

(a)

0 v = 120pEIL ( x 5 ! 10 L2 x 3 + 20 L3 x 2 )

11 p L4

( b ) Let x=L in this equation: v B = 1200EI p L3

( c ) Let x=L in Eq. ( a ): ! B = 80EI

______________________________________________________________________________________ SOLUTION (5.28)

EIv ' ' ' ' = p EIv ' ' ' = px + c1 2 EIv ' ' = 12 px + c1 x + c2 EIv ' = 16 px 3 + 12 c1 x 2 + c2 x + c3 EIv = 241 px 4 + 16 c1 x 3 + 12 c2 x 2 + c3 x + c4 Boundary conditions:

EIv (0) = 0; EIv ' ' (0) = 0;

c4 = 0 c2 = 0

EIv ( L) = 0;

pL3 24 3

+ c16L + c3 = 0 3

EIv ' ( L) = ! EI 96pLEI = pL6 +

c1L2 2

(a)

+ c3

c3 = ! 1796pL ! c12L

(b)

Substituting Eq. (b) into (a), we obtain reaction at right end:

c1 = ! 1332pL = R

______________________________________________________________________________________ SOLUTION (5.29) c

MA RA

b=L-c L

MB B

y Segment AD:

EIv1 ' ' = M A ! R A x

EIv1 ' = M A x ! 12 R A x 2 + c1

EIv1 = 12 M A x 2 ! 16 R A x 3 + c1 x + c2

(a)

Segment BD:

EIv 2 ' ' = M A ! R A x + P( x ! c ) EIv 2 ' = M A x ! 12 R A x 2 + 12 P ( x ! c ) 2 + c3 EIv 2 = 12 M A x 2 ! 16 R A x 3 + 16 P( x ! c ) 3 + c3 x + c4

Boundary conditions:

v1 (0) = 0; v1 ' (0) = 0;

(b)

c2 = 0 c1 = 0

______________________________________________________________________________________ 5.29 (CONT.)

and

v1 ( c ) = v2 ( c );

c3 c + c 4 = 0

v1 ' ( c ) = v2 ' ( c );

c3 = 0,

c4 = 0

v 2 ' ( L) = 0 = M A L ! 12 R A L2 + 12 Pb 2

(c)

v 2 ( L) = 0 = 12 M A L2 ! 16 R A L3 + 16 Pb 3

(d)

Solving Eqs.(c) and (d), 2

R A = PbL3 (3c + b) = P ( LL!3 c ) ( 2c + L) 3

Substituting of this into Eq. (c) gives

M A = Pcb = Pc ( LL2!c ) L2 2

Thus, introducing the values of M A , R A , and c1 = c2 = c3 = c4 = 0 into Eqs. (a) and (b) we obtain the deflections. For 0 ! x ! c : 2

c) x v1 = P ( L6 !EIL (3cL ! 2cx ! Lx ) 3

For c ! x ! L : c) x v2 = P ( L6 !EIL (3cL ! 2cx ! Lx ) + 6PEI ( x ! c ) 3 3

______________________________________________________________________________________ SOLUTION (5.30)

L-x

The reactions are statically indeterminate. We have

EIv ' ' = M = RB x ! RB L + M 0

EIv ' = 12 RB x 2 ! RB Lx + M 0 x + c1

v ' (0) = 0;

c1 = 0.

EIv = 16 RB x 3 ! 12 RB Lx 2 + 12 M 0 x 2 + c2

v (0) = 0;

c2 = 0

v ( L) = 0;

RB = 32ML0

The preceding equation gives 2

( x!L) v = M 0 x4 EIL

______________________________________________________________________________________ SOLUTION (5.31) r

M1 M2

1 2

( a ) Static equilibrium gives

P1 = P2 = P

M 1 + M 2 = Ph

Equation (5.9):

M 1 = E1rI1

M 2 = E2rI 2

Interface strains must be the same:

#1"T + 123 due to temp . increase

Fig. (a) Bent beam

= # 2 "T ! EP22h ! hr2

E1h {

r {

due to bending ( from Eq . 5.9 )

due to axial force

This yields an expression for the interface curvature: 1 r

= 12h ((14# 2+"n#+11) !nT)

where n = E1 E 2 .

( b ) At the interface

! 1 = Ph + E21rh = (! yp )1

! 2 = " Ph " E22rh = "(! yp ) 2 ( c ) Summing these equations.

(" yp )1 ! (" yp ) 2 = ! h ( E21 !r E2 )

It follows that ($ yp )1 "($ yp ) 2 E1 " E2

= h2 24h ((14#2+"n#+11) !nT)

from which

+ n +1 n "T = 14 6 ($ 2 !$1 )

(# yp )1 !(# yp ) 2 E1 ! E2

______________________________________________________________________________________ SOLUTION (5.32) Case B h

b t

h y

Q = Qoutside ! Qinside = b2 ( h 2 ! y12 ) ! b!22 t [( h ! t ) 2 ! y12 ] = bht ! 12 bt 2 + th 2 + t 3 ! ty12

______________________________________________________________________________________ 5.32 (CONT.) Q " = IA2 ! ( width dA )2 2

A I2

= or where

bt '! h )t ( bht +th 2 ) bt )2 ht 2 +t 3 )ty12 )2 $! + th 2 ) 2 ht 2 + t 3 )ty12 ) 2 h ( bht ) 2 2 2 [ 2 tdy bdy ] + & (0 1 1 # (h)t b2 b2 !% !" A " = Aweb = 1 + 2 ( hb!t )

A = 2bh ! (b ! 2t )( 2h ! 2t ) = 2bt + 4th ! 4t 2 Aweb = 2( 2h ! 2t )t = 4ht ! 4t 2 I = 23 bh 3 ! 23 (b ! 2t )( h ! t ) 3 b

Case B

Q = ( h ! t ! y1 )t[ y1 + 12 ( h ! t ! y1 )] = th2 ! ht 2 + t2 + bht ! bt2 ! 2t y12 2

A I2

or where

bh bt ty1 t 2 '! h )t ( bht + ht 2 ) bh ) bt ) by1 + t )2 $! ) ) + ) h ( bht + ht 2 + 2 2 2 2 2 2 2 2 2 [ tdy bdy ] + 2 2 & (0 # 1 1 ( b t h )t !% !" A " = Aweb = 1 + hb!t

A = 2bt + ( 2h ! 2t )t = 2bt + 2th ! 2t 2 Aweb = ( 2h ! 2t )t = 2ht ! 2t 2 I = 121 t ( 2h ! 2t ) 3 + 2[ 121 bt 3 + bt ( h ! 2t ) 2 ]

Note: For thin-walled sections flange and web thicknesses are small with compared with to unity and their products are neglected. Case C O !/2 r

r sin !2

y = r cos !2 A*

y (CONT.) ______________________________________________________________________________________ Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 5–20

______________________________________________________________________________________ 5.32 (CONT.) We write A = r2 (! " sin ! ) =area of segment 2

dA = r2 (1 " cos ! )d! = r 2 (1 " cos 2 !2 )d! 2

Q = ! * ydA = ! * ( r cos #2 ) r 2 (1 " cos 2 #2 )d# A

= r 3 ! (cos #2 " cos 3 #2 )d# = r 3 [2 sin "2 ! 23 sin "2 (cos 2 "2 + 2)] 0

= 23 r 3 [sin !2 " sin !2 cos 2 !2 ] 2

Q " = IA2 ! ( width dA )2

= Since,

A I2

2# [

A = !r 2

2 #r 3 $ $ $ (sin "sin cos2 )] 3 2 2 2 $ ( 2 r sin ) 2 2 2 ! 2 r8 16

r 2 (1 " cos 2 $2 )d$

I =

We have, after simplification: 2#

% = #16r 6 " ( r9 )(1 ! cos 2 $2 ) 3 d$ = 109 6

Case D

!/2

"/2 y

Q = Q2 ! Q1 = 23 r22 [sin "2 # sin "2 cos 2 "2 ] # 23 r12` [sin !2 # sin !2 cos 2 !2 ]

& = IA2 [ !

% ( 2 r2 sin ) 2 2

Letting

Q22

r22 (1 " cos 2 %2 )d% " !

A = " ( r22 ! r12 ),

after integration, we obtain

Q12

2 1 $ ( 2 r1 sin ) 2 2

r (1 " cos 2 $2 )d$ ]

I 2 = "16 ( r24 ! r14 ) 2

! =2

______________________________________________________________________________________ SOLUTION (5.33) Since stress is symmetrical, Eq. (3.40) reduces to 2

( ""r 2 + 1r ""r )( ""r#2 + 1r ""#r ) = ""r#4 + 2r ""r#3 ! r12 ""r#2 + r13 ""#r = 0 The given ! satisfies this equation. Equations (3.32) lead to $ r = 1r ""#r = rA2 + B(1 + 2 ! ln r ) + 2C

& % = ##r$2 = ( " rA2 ) + B (3 + 2 ! ln r ) + 2C 2

The constants A, B, and C are determined from the boundary conditions (b) and (d) of Sec. 5.13. In so doing, we arrive at the solution given by Eq. (5.67). ______________________________________________________________________________________ Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 5–21

______________________________________________________________________________________ SOLUTION (5.34) 46 mm 6 mm

z z 30 mm

6 mm

y A2

We have

z = ( A(1Az11++ AA22z)2 ) = 16.14 mm

r = 136.14 mm ri = 120 mm ro = 166 mm A 420 R= = 126 = 134.7045 mm, e = r ! R = 1.4355 mm 166 " # dAr # 30rdr + # 6rdr 120

126

We have M = !(30 + 120 + 16.14) P = !166.14 P Outer Edge. Applying Eq. (5.73),

P (134.7045 !166) P !80 mmN 2 = 420 + 166.14 420(1.4355)(166)

P = 1.614 kN = Pall

Inner Edge. Using Eq. (5.73),

P (134.7045 !126) P 80 mmN 2 = 420 + 166.14 420(1.4355)(126)

Solving,

P = 3.73 kN

______________________________________________________________________________________ SOLUTION (5.35) P M B A

Figure (a) ( a ) Equation (5.66) yields

N = (1 ! 00..212 ) 2 ! 4 00..212 ln 2 00..21 = 0.082 2

M = PR = 70(10 3 ) " 0.15 = 10.5 kN ! m Thus, from Eq. (5.67):

, 500 ) (# " ) A = 0.054( 0(10 [(1 ! 00..212 )(1 + ln 1) ! (1 + 1) ln 00..21 ] .2 ) 2 ( 0.082 ) 2

= 163 MPa 2 2 , 500 ) (# " ) B = 0.054( 0(10 [(1 ! 00..212 )(1 + ln 2) ! (1 + 00..212 ) ln 2] .2 ) 2 ( 0.082 ) = 103 MPa (CONT.) ______________________________________________________________________________________ Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 5–22

______________________________________________________________________________________ 5.35 (CONT.) Referring to Fig. (a),

( b ) We have

# max = (# " ) A ! PA = !163 ! 705 = !177 MPa # min = (# " ) B ! PA = 103 ! 14 = 89 MPa

ri = 100 mm, ro = 200 mm, R = 100 ln 2 = 144.2695 A = 5000 mm 2 , e = 5.7305

Thus,

(" ! ) A = # PA [1 + r ( Rer#i ri ) ] = #176.2 MPa (" ! ) B = # PA [1 + r ( Rer#o ro ) ] = 88.1 MPa

______________________________________________________________________________________ SOLUTION (5.36)

A = 20b + 2(60 !10) = 20b + 2400 We have rA = 60 mm and rB = 140 mm . Applying Eq. (5.70): (" M )( R " rA ) (" M )( R " rB ) "! A = ! B = =" AerA AerB from which or Then

rB ( R ! rA ) = !rA ( R ! rB ) , 140( R ! 60) = !60( R ! 140) R = 134 mm A R= dA !" r 8 14 " ! 80 bdr 140 2(10)dr " ! A = 84 # ' +' = 84 b ln + 20 ln #% 80 r $& 6 8 $& % 60 r = 24.1653b + 1880.3091 20b + 2400 = 24.1653b + 1880.3091 b = 124.8 mm

Hence or ______________________________________________________________________________________ SOLUTION (5.37)

1 r = b + c = 25 mm We have c = d = 10 mm 2 1 R = [r + r 2 ! c 2 ] (by Table 5.2) 2 1 = [25 + 252 ! 102 ] = 23.9564 mm 2

e = r ! R = 1.0436 mm A = ! c 2 = ! (10) 2 = 314.16 mm 2

r O

P M=-P(a+ r ) B

______________________________________________________________________________________ 5.37 (CONT.) (a)

From Eq.(5.74):

(a + r )( R " rA ) P M ( R " rA ) P " = [1 + ] A AerA A erA 800 (50)(23.9564 ! 15) = [1 + ] = 75.4 MPa 314.16 (1.0436)(15)

!A =

(b)

Using Eq.(5.74):

(a + r )( R " rB ) P M ( R " rB ) P " = [1 + A AerB A erB 800 50(29.1421 ! 35) = [1 + ] = !34.2 MPa 314.16 1.0436(35)

!B =

______________________________________________________________________________________ SOLUTION (5.38) Locate centroid : 150 mm 50 mm

C A1

75 mm

100 mm

A1r1 + A2 r2 A1 + A2 (5625)(150) + (3750)(200) = = 170 mm 5625 + 3750 1 A = (50 + 75)(150) = 9375 mm 2 2

1 2 h (b1 + b2 ) 2 R= r (b1ro ! b2 ri ) ln o ! h(b1 ! b2 ) ri (0.5)(150) 2 (75 + 50) = 158.9165 mm 25 [(75)(250) ! (50)(100)]ln ! (150)(75 ! 50) 10 e = r ! R = 11.0825 mm =

(a)

Use Eq.(5.75a),

(" ! ) A = #

r ( R # rA ) P M ( R # rA ) P # = # [1 + ] A AerA A erA

50(103 ) ! (170)(158.9175 # 100) " =# 1+ % 9375 &$ (11.0825)(100) ' = !53.5 MPa

B M=P r

______________________________________________________________________________________ 5.38 (CONT.) (b)

From Eq.(5.75b):

(" ! ) B = #

r ( R # rB ) P [1 + ] A erB

50(103 ) ! (170)(158.9175 # 250) " =# 1+ % = #24.5 ksi 9375 $& 11.0825(250) '

______________________________________________________________________________________ SOLUTION (5.39)

1 A = bh 2

The section width w varies linearly with r. Thus w = c0 + c1r (1) Since

w = b (at r = ri ) w = 0 (at r = ro )

Substituting Eq.(1) into Eq.(2);

Then

br c0 = o h

b c1 = ! h

(2) w

ro w c +c r dA !A r = !ri r dr = 0 r 1 dr Inserting c1 and c0 into this, after integrating and rearranging,

we have

r O

b h

ri r

" r = b( h ln r ! 1) Therefore

1 h A 2 = dA ro ro " r h ln ri ! 1

______________________________________________________________________________________ r SOLUTION (5.40) r

A = ! c2

Through use of the polar coordinates we write:

w = 2c sin ! r = r " c cos ! dr = c sin ! d! dA = wdr = 2c 2 sin 2 ! d!

dA ! 2c 2 sin 2 " $ r =$0 r # c cos " d"

c ccos!

______________________________________________________________________________________ 5.40 (CONT.) 2 2 2 2 2 ! r $ c cos " $ ( r $ c ) c 2 (1 $ cos 2 " ) = 2% 0 0 r $ c cos " r $ c cos # ! ! d" = 2 $ (r + c cos " )d" # 2(r 2 # c 2 ) $ 0 0 r # c cos "

= 2%

This gives

= 2r " 0 + 2c sin " 0 # 2(r 2 # c 2 )

r #c

r 2 # c 2 tan

tan #1

r +c

" 2

# r = 2! (r " r " c )

Hence, it can be shown that

A 1 = (r + r 2 + c 2 ) dA 2 !r

______________________________________________________________________________________ SOLUTION (5.41)

1 A = (b1 + b2 )h 2

The section width w varies linearly with r as

w = c0 + c1r

We have

(1)

w = b1 (at r = ri )

(2)

w = b2 (at r = ro )

Introduce Eq.(1) into Eq.(2), then solve for c0 and c1

r b ! rb c0 = o 1 i 2 h

Then, we write

b !b c1 = ! 1 2 h

c0 + c1r r dr = c0 ln o + c1 (ro ! ri ) ri r ri

" r = " r dr = " ri

This gives, substituting Eqs.(3):

"r = Hence

(3)

rob1 ! rb r i 2 ln o ! (b1 ! b2 ) h ri

1 2 h (b1 + b2 ) A 2 R= = dA r " r (rob1 ! rbi 2 ) ln roi ! (b1 ! b2 )

______________________________________________________________________________________ SOLUTION (5.42) P z M A2 A B A1

z A2

P We have

z = ( A(1Az11++ AA22z)2 ) = 12.5 mm

A = 0.0008 m 2

r = 52.5 mm

dA r

800 50

50 dr 40 r

2(5) dr r 50

ri = 40 mm

ro = 80 mm

= 50.4502 mm, e = r ! R = 2.0498 mm

M = (0.06 + 0.04 + 0.0125) P = 337.5 N ! m Inner Edge. Using Eq. (5.73), #3

337.5(50.4502 # 40)10 3000 (" ! ) A = # 0.0008 # 0.0008(2.0498)(40)10 #6

= !57.5 MPa

Outer Edge. Applying Eq. (5.73), #3

337.5(50.4502 #80)10 3000 (" ! ) B = # 0.0008 # 0.0008(2.0498)(80)10 #6

= 74.6 MPa ______________________________________________________________________________________ SOLUTION (5.43) Using Eq. (P5.44), at section A-B of Fig. P5.43:

M = !0.182 Pr = !0.182(0.75b) P = !0.136 Pb

( a ) Equation (5.66) yields,

N = (1 ! 14 ) 2 ! 4( 14 )(ln 2) 2 = 0.082

Point A. Applying Eq. (5.67),

($ # ) A = tb4 2MN [(1 " 14 )(1 + ln 1) " 2 ! ln 2] + btP Pb ) 3 = 4tb( !20(.0136 [ 4 ! 1.3863] + btP .082 )

= 5.22 P bt Point B. Using Eq. (5.67),

Pb ) 3 (# " ) B = 4tb( !20(.0136 [ 4 (1 + ln 2) ! 45 ln 2] + btP .082 )

= ! 1.68P bt (CONT.) ______________________________________________________________________________________ Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 5–27

______________________________________________________________________________________ 5.43 (CONT.) ( b ) We have r = 0.75b, ri = 0.5b, and ro = b . Thus, R = ( b 2 ) ln 2 = 0.7213b Points A and B. Apply Eq. (5.74) with P = P 2 and M = !0.136 Pb :

(" ! ) A = btP + 4.2 btP = 5.2 P bt (" ! ) B = btP # 2.64 btP = #1.64 P bt (" ! ) A,B = ( 2PA ) ± ( McI )

c = b4

I = tb96

Thus,

(b 4) (" ! ) A = btP + 0.136tbPb = btP + 3.26 btP = 4.26 P bt 2 96

and

(# " ) B = btP ! 3.26 btP = ! 2.26 P bt

Comparing the result, we observe the following differences: At the point A: Elasticity vs. hyperbolic 0.4 % Elasticity vs. linear 18 % At the point B: Elasticity vs. hyperbolic 2.4 % Elasticity vs. Linear 34.5 % ______________________________________________________________________________________ SOLUTION (5.44) From the condition of symmetry, the distribution of stress in any quadrant is known to be the same as the other.

r O

P 2

cos !

P/2 At any angle ! , the bending moment referring to this figure is expressed as follows

M ! = M a " 12 Pr (1 " cos ! )

(a)

The problem is statically indeterminate and value of M a is found first. Note that shear component of the load will be omitted in our solution. Applying Castigliano’s theorem, we have: "U "M a

= 4!

0 = EI4 $

# 2

0 ! 2

1 EI

M $ ""MM$a ds

[ M a # Pr2 (1 # cos " )]rd"

0 = EI4 [ M a " Pr2 ( !2 " 1)]

______________________________________________________________________________________ 5.44 (CONT.) solving,

M a = Pr ( 12 " !1 ) = 0.182 Pr

Then, Eq. (a) becomes

M ! = 0.182 Pr " 12 Pr (1 " cos ! )

(b)

Therefore stress at any point of a section, using Eq. (5.74), is expressed in the form

" ! = # P2 cosA! + MA! ( Rer# r ) Here, the moment M ! is given by Eq. (b). ______________________________________________________________________________________ SOLUTION (5.45) ( a ) Using Eq. (P5.44) at " = ! 4 (Fig. P5.44): We have

M $ = 0.182(1.5) P " 12 P(1.5)(1 " cos #4 ) = 0.00533P N ! m

A = 0.05(0.1) = 0.005 m 2

and

h 100 = 2 = 144.2695 ro ln ri ln 1

e = r ! R = 150 ! 144.2695 = 5.7355 mm At inner fiber, ri = !0.1 m : ! 4) P 144.2695 $100 (# " )! 4 = $ ( P 2)cos( $ 0.00533 0.005 0.005 [ 5.7355(0.1) ] = $182.3P Pa

(b) M V

! O

At ! = 0

(M ) =0 P/2 !

: M " = 0.182 PR

N" = ! P 2

At any angle:

M ! = 0.182 Pr " 0.5 Pr (1 " cos ! ) = "0.318 Pr + 0.5Pr cos ! "M ! " ( P 2)

= #0.636r + r cos !

N ! = " P2 cos ! , V! = " P2 sin ! ,

#N! #( P 2) #V! #( P 2)

= " cos !

= " sin !

______________________________________________________________________________________ 5.46 (CONT.) Based on the symmetry of the ring, we have

$ = 2&

" 2 N

%N! ! AE % ( P 2)

+ EI! % ( P !2) + AG! % ( P!2) ]rd!

Substituting the values of M ! , N ! , and V! , this expression becomes

# P = 2&

! 2

2 " P P [ AE cos 2 $ + 2PrEI (%0.636 + cos $ ) 2 + AG 2 sin $ ]rd$ 2

! 2

= 2[ 2PrAE "2 + 14 sin 2" 0 + 2PrEI 0.904" # 1.27 sin " + sin42" 0 ! 2

from which

Pr # 1 + 2"AG 2 $ 4 sin 2# 0 ] 3

" Pr ! # P = 2[ 2PrAE ( !4 ) + 2PrEI (0.15) + AG 2 ( 2 )]

For the given rectangular cross section:

r = 0.15 m ! = 65 = 1.2

A = 0.005 m 2 G = 25 E = 0.4 E

I = 121 (0.05)(0.1) 3 = 4.17(10 !6 ) m 4 The deflection is therefore 3

( 0.15 )! P ( 0.15 ) ( 0.15 ) P ( 0.15 )! 1.2 " P = 2[ 0P.005 ] 8 E ( 8 ) + 2 E ( 4.17$10# 6 ) + 0.4 E ( 0.005 )

This results in

! P = 215.65P E m End of Chapter 5

CHAPTER 6 SOLUTION (6.1)

! all = c

b. or 6

or 100(10 )[1.5(10 Solving,

" 4 4 (c # b ) 2 100(106 ) =

2(3 "103 )(0.035) ! [(0.035) 4 # b 4 ]

) ! b 4 ] = 66.845

b = 0.0302 m = 30.2 mm

______________________________________________________________________________________ SOLUTION (6.2) D t = 25

d=40 mm

! max = or

T (d 2) T ( D 2) = 4 " d 32 " ( D 4 # Di4 ) 32

23 Di = 25 D

23 4 ) ] 25 (40)3 = D 3 (0.2836) D = 60.9 mm d 3 = D 3 [1 ! (

______________________________________________________________________________________ SOLUTION (6.3)

Di = D2 d

! max =

or Thus

16T T ( D 2) 16T = = 3 1 " "d ( D 4 # Di4 ) " D 3 (1 # ) 32 16 15 d 3 = D 3 ( ), D = 1.0217 d 16 d=

D 60 = = 58.7 mm 1.0217 1.0217

______________________________________________________________________________________ SOLUTION (6.4) Apply the method of sections between the change of load points:

TCD = 0.5 kN ! m

TBC = 1.5 kN ! m

TAB = 2 kN ! m 3

Therefore, ! max = 2T " c :

2(2 #103 ) = 81.5 MPa " (0.025)3 2(1.5 #103 ) = 119.4 MPa ! BC = " (0.02)3

! AB =

! CD =

2(0.5 #103 ) = 94.3 MPa " (0.015)3

______________________________________________________________________________________ SOLUTION (6.5) We have, applying the method of sections:

TCD = 0.5 kN ! m

TBC = 1.5 kN ! m

TAB = 2 kN ! m Hence,

! max =

2Tc " (c 4 # b 4 )

gives,

2(2 #103 )(0.025) = 82.4 MPa ! AB = " [(0.025) 4 $ (0.008) 4 ]

! BC =

2(1.5 #103 )(0.02) = 122.5 MPa " [(0.02) 4 $ (0.008) 4 ]

! CD =

2(0.5 #103 )(0.015) = 102.6 MPa " [(0.015) 4 $ (0.008) 4 ]

______________________________________________________________________________________ SOLUTION (6.6)

TBC = 1 kN ! m Then,

TAB = 2 kN ! m

! max = 2T " c3 yields

2(1#103 ) = 79.58 MPa " (0.02)3 2(2 #103 ) = 81.49 MPa ! AB = " (0.025)3

! BC =

Therefore,

(! max ) B = K! AB = 1.6(81.49) = 130.4 MPa

______________________________________________________________________________________ SOLUTION (6.7) We have

TBC = 4 kN ! m

TAB = 4 kN ! m

Then

!C = " (TL GJ ) yields 4(103 )

0.35 0.7 1 0.35 0.7 ]= ( ) # # 4 3 " 3.5" 6.25 150 (28 $109 ) (0.05) (0.1) 4 2 2 !3 o = 4.67 "10 rad = 0.27

!C =

[

______________________________________________________________________________________ SOLUTION (6.8)

! = 50 + 90 = 140o ! J = [(0.05) 4 " (0.045) 4 ] = 3.376 #10"6 m 4 2 o From Eq. (6.5a) at ! = 140 : ! x ' = " sin 280o = #0.985" Tc T (0.05) 120 "106 = 0.985 = 0.985 J 3.376 "10!6 T = 8.23 kN ! m or Similarly, Eq. (6.5b):

! x ' y ' = ! cos 280o = 0.174!

gives

50 "106 = 0.174 Thus,

T (0.05) , 3.376 " 10!6

T = 19.4 kN # m

Tall = 8.23 kN ! m

______________________________________________________________________________________ SOLUTION (6.9) From solution of Prob. 6.8:

J = 3.376 "10!6 m 4 o

From Equations (6.5) at ! = 35 + 90 = 125 ,

or and

! x ' = " sin 250o = #0.94" Tc T (0.05) 120 " (106 ) = 0.94 = 0.94 J 3.376 " 10!6 T = 8.62 kN ! m

! x ' y ' = ! cos 250o = "0.342!

______________________________________________________________________________________ 6.9 (CONT.)

T (0.05) , 3.376 "10!6

50 "106 = 0.342 Thus,

T = 9.87 kN # m

Tall = 8.62 kN ! m

______________________________________________________________________________________ SOLUTION (6.10) 3 kN ! m

2 kN ! m

30 mm

1 kN ! m

Apply the method of sections:

TAB = 2 kN ! m

TBC = 1 kN ! m

Then,

2 #103 J ! d13 T = = = 16 " all 50 #106 c d1 = 0.05884 m = 58.84 mm Also,

! = " (TL GJ ) :

103

1(1) 2(2) " 4 ] 4 ! (0.04) d1 (39 #109 ) 32 4 76576.32 = 390625 ! 4 d1 0.02 =

Solving

[

d1 = 59.74 "10!3 m = 59.74 mm We therefore use

d1 = 60 mm

______________________________________________________________________________________ SOLUTION (6.11) 40 mm

2 kN m

1.5 kN ! m

A 3m

Apply the method of sections:

TAB = 1.5 kN ! m

0.5 kN ! m

TBC = 0.5 kN ! m

______________________________________________________________________________________ 6.11 (CONT.) We have

J ! d13 T 1.5 #103 = = = 16 " all 50 #106 c d1 = 53.46 "10!3 m = 53.5 mm Also,

! = " (TL GJ ) : or

Solving

103

0.5(1) 1.5(3) ] " 4 4 (0.04) d 9 ! 1 (40 #10 ) 32 4.5 78539.82 = 195,312.5 ! 4 d1

0.02 =

[

d1 = 78.79 "10!3 m = 78.8 mm Use

d1 = 78.8 mm

______________________________________________________________________________________ SOLUTION (6.12) (a)

D 60 = = 1.2 d 50

r 1 = = 0.02 d 50

Figure D.4: K = 2.1 .

2T 2(2 #103 ) = = 81.5 MPa " c3 " (0.025)3 ! max = 2.1(81.5) = 171.2 MPa

! nom =

(b)

r 5 D = = 0.1, = 1.2; K = 1.33 d 50 d ! max = 1.33(81.5) = 108.4 MPa

(Fig. D.4)

______________________________________________________________________________________ SOLUTION (6.13)

r 10 = = 0.2 d 50

D 100 = = 2.0; K = 1.25 (from Fig. D.4) d 50 For smaller part of the shaft, c = d 2 = 25 mm : 16T 16T ] ! max = ! all = K ( 3 ) ; 80(106 ) = 1.25[ ! (0.05)3 "d Solving

T = 1.571 kN ! m

______________________________________________________________________________________ SOLUTION (6.14) b

b a

( a ) For circular bar:

" c = #2bT3 = G! c b

! c = G2#Tb4 For elliptical bar: 2

! e = T"(aa3b+3bG ) ,

T = "aa2 +bbG2 ! e 3 3

# e = "2abT2 = 20a!2e+bab2G

We have "e "c

= ( a 22+!beba2 )!GbG c

Setting ! e = ! c : !e !c

= ( a22 +a b2 )

Since a > b, ( a

+ b 2 ) < 2a 2 , and ! e ! c > 1, or

!e >!c 4

Tc = " c#2b G ;

(b)

G" c b = ! c

" c !b G " c!b Tc = Gb 2 = 2 4

3 3

Te = ! ea"2a+bb2G ,

# e = 2a!2eba+b2G

Rearranging,

" e = ! e 2( aba+2Gb ) Thus,

Te = " e!2ab

We obtain

Tc Te

= (("" c!!abb2 22)) e

Setting ! e = ! c :

Tc Te = b a , or

Te > Tc ______________________________________________________________________________________ SOLUTION (6.15) We have " ( c (a)

! b 2 ) = "a 2 ; 4

a 2 = c 2 ! b2 .

Th = J#c a = "# a ( 2c c!b ) Ts = J"aa = !"2a a

______________________________________________________________________________________ 6.15 (CONT.) T

Hence, Th = c ca!b3

c 4 !b4 c ( c 2 !b2 )

For c = 1.4a : Th Ts

(b)

8416 = 12..3168 = 2.16 4

Th = JGL#a = ( GL#a )[ " ( c 2!b ) ] Ts = ( GL"a )( !a2 ) 4

Hence, Th = c !2b

= ( cc2 !!bb2 )2 = cc2 +!bb2

For c = 1.4b : Th Ts

96 = 02..96 = 3.08

______________________________________________________________________________________ SOLUTION (6.16) Use Eqs. (f) of Example 6.2:

TA = 1+( aJTb bJ a ) =

T 0.4 (15 ) 4 (! 32 )

= (1+0.T6328) = 0.6124T

0.2 ( 20 ) 4 (! 32 )

TB = T ! TA = 0.3876T Based on shear in segment AC:

T) " a = 16!dT3A = 16!( 0( .06124 = 150(10 6 ) .02 )3 a

T = 384.7 N ! m

Based on shear in segment CB: 16TB !d b3

T) = 16!((00.3876 = 150(106 ) .015 )3

T = 256.5 N ! m = Tall

______________________________________________________________________________________ SOLUTION (6.17) Use Eqs. (f) of Example 6.2:

TA =

T 0.8 (15 ) 4 (! 32 )

= 1+0.T2074 = 0.8283T

0.5( 25 ) 4 (! 32 )

TB = T ! TA = 0.1717T Based on shear in segment AC: 16TA !d a3

T) = 16!((00.8283 = 70(106 ) .025 )3

T = 259.3 N ! m = Tall

Based on shear in segment CB: 16TB

!d b3

T) = 16!((00.1717 = 70(10 6 ) .015 )3

T = 270.2 N ! m

______________________________________________________________________________________ SOLUTION (6.18) " 2# "x 2

= k [2( a 2 b ! a 2 ) + 2 y 2 (b 2 + 1) ! 12bx 2 ]

" 2# "y 2

= k [2( a 2 b ! a 2 ) + 2 x 2 (b 2 + 1) ! 12by 2 ]

Substituting these in Eq. (6.9):

k = 2 a 2 ( b!1)+( b2G!"6b+1)( x 2 + y 2 )

In order k be a constant: b

G" 2 a 2 ( b !1)

! 6b + 1 = 0 . Thus,

______________________________________________________________________________________ SOLUTION (6.19)

u = !"z ( y ! b), # x = # y = # z = 0,

v = "z ( x ! z ), w = w( x , y ) " xy = !!vx + !!uy = 0

$ xz = ""wx ! # ( y ! b), Thus

$ yz = ""wy + # ( x ! a )

% xz = G[ !!wx # $ ( y # b)] = !!"y % yz = G[ !!wy # $ ( x # a )] = # !!"x

Substituting these in Eqs. (6.6), (6.7), and (6.11), we obtain $" xz $y

# $xyz = #2G! ,

$ 2% $x 2

+ $$y%2 = #2G! 2

T = !! [( x " a )# yz " ( y " b)# xz ]dxdy

and

= [ $ !! ( x $ a ) ""#x $ !! ( y $ b) ""#y ]dxdy = 2 !! " dxdy

We observe that characteristic equations remain unchanged. ______________________________________________________________________________________ SOLUTION (6.20)

Here

T = !! ( x% yz $ y% xz )dxdy = $ !! ( x ""#x $ y ""#y )dxdy x2

# ! dy ! x $$"x dx = # ! xd" dy # [ # !! " dxdy ] x1

= #c ! ( x2 # x1 )dy + !! " dxdy = # c !! dxdy + !! " dxdy Note that ( x 2 ! x1 )dy is area and equals

Thus,

!! dxdy . Similarly,

# ! dx ! y $$"y dy = #c !! dxdy + !! " dxdy T = 2 !! (# " c )dxdy

______________________________________________________________________________________ SOLUTION (6.21)

T! = T cos 2 ! Hence, " dz " = ! TGJ = GJ1 ! T cos 2 "ad"

This gives, at sections A and B: Ta # A = GJ !

" 2

0 "

( 12 + cos22# )d# = 2Ta r 4G

2 Ta # B = GJ ! cos #d# = rTa4G 0

______________________________________________________________________________________ SOLUTION (6.22) b=a c 2a

From Table 6.2: " = 0.246, ! = 0.229 T T " r = !ab 2 = 0.492 a 3

" c = TcJ = !2cT3 Thus,

T 0.492 a 3

= !2cT3 ;

c = 0.684a

Similarly,

! r = 0.229 (T2 a ) a 3G = 0.458T a 4G Then,

T " c = JG = !2cT4G T 0.458 a 4G

= !2c 4TG ;

c = 0.736a = call

______________________________________________________________________________________ SOLUTION (6.23) " 2# "y 2

= k [( x + h3 ) + ( x + 3 y ! 23 h ) + ( x + h3 ) + ( x ! 3 y ! 23 h ) + 2( x ! 23 h )]

" 2# "y 2

= k [!3( x + h3 ) ! 3( x + h3 )]

Substituting these in Eq. (6.9),

" 4kh = "2G! ;

Thus,

k = G2 !h

" = !G# [ 12 ( x 2 + y 2 ) ! 21h ( x 3 ! 3xy 2 ) ! 272 h 2 ]

Along the x axis ! xz = 0, due the symmetry. Equation (6.8) is therefore,

for y = 0 :

% yz = ! ""#x = 32Gh$ ( 23hx ! x 2 ) (CONT.) ______________________________________________________________________________________ Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 6–9

______________________________________________________________________________________ 6.23 (CONT.) When

x = 0:

! yz = 0

x = 23h :

! yz = 0

x = # h3 :

" yz = " max = G2!h

Next, substitute the preceding value of ! into Eq. (6.11) to obtain

T = 2 !! " dxdy

= "4G# !

" 3 y + 23 h

[ 12 ( x 2 + y 2 ) " 21h ( x 3 " 3xy 2 ) " 272 h 2 ]dxdy

G!h = 15 3

! = 15h4G3T

This gives The shear stress is thus

! max = ( 2 h20T3 )3 = 152 h33T ______________________________________________________________________________________ SOLUTION (6.24) For a circular bar:

T = G!J = C c!

where

C c = !r2G 4

(a)

For an elliptical bar (from Example 6.3): 2

H = # 2T "( aa 3b+3b ) = #2G! T = a"2a+bb2 G! = C e! 3 3

C e = a!2a+bb2 G 3 3

(b)

For an equilateral bar (from Prob. 6.23): G!h T = 15 = !C t 3 4

Ct = 15h G3 4

(c)

Bars have equal areas:

Ac = !r 2 ,

Ae = !ab,

At = h 3 2

Setting Ac = Ae = At :

r 4 = a 2 b 2 = 3h! 2 4

Then, Eqs. (a), (b), and (c) give Ce 2 a 3b3 Cc r 4 ( a 2 + b 2 )

= a22 +abb2

and Ct Cc

= 15h G3 !r24G = 2!15 3

______________________________________________________________________________________ SOLUTION (6.25) do

di t

For the seamless tube, from ! 1 = T GJ

#1 = "d 4 (1!32d T4 d 4 )G o

For the split tube, referring to Eq. (6.17):

# 2 = 3GT

We have "1 "2

1 do + di do !di 3 )( ) 2 2

= 32((dd2o+!ddi2)) o

(a)

For very thin tubes d o + d i !1 !2

! 2d o2 , and Eq. (a) becomes

= 43 ( dto )

______________________________________________________________________________________ SOLUTION (6.26) Apply Eqs. (6.17) and (6.19):

) ! max = bt3T2 = 0.1253((80 = 76.8 MPa 0.005 ) 2 3( 80 ) " = bt33TG = 0.125( 0.005 = 0.192 rad m )3 ( 80!109 )

______________________________________________________________________________________ SOLUTION (6.27) Referring to Table 6.1, we have

% = $L = "abTL2G

T # max = !ab 2

Thus $ #

= bGL "!

(1)

For a b = 24 16 = 1.5 : ! = 0.231 and ! = 0.196 Introducing given data into Eq. (1): 1.5(# 180 ) " max

Solving,

4 ( 0.231) = ( 0.024 )(0.80 !109 )( 0.196 )

! max = 106.6 MPa

______________________________________________________________________________________ SOLUTION (6.28) From Eq. (1) of solution of Prob. 6.27: L " b = $ max G # !

(a)

By Table 6.1: ! = 0.208 and ! = 0.141. Substituting the numerical values into Eq. (a), we obtain 6

(10 ) 0.204 2 b = 120 = 10.1 mm 79 (109 ) 25(! 180 ) 0.141

______________________________________________________________________________________ SOLUTION (6.29) Equation (6.20) yields

J e = ! 13 bt 3 = 13 (100)(10) 3 + 13 (115)( 4) 3

= 3.5787(10 4 ) mm 4

Maximum shear stress occurs on the lower leg:

500 ( 0.1) " max = TtJ e1 = 3.5787 = 139.7 MPa (10!8 )

Angle of twist per unit length is

# = JTeG = 3.5787 (10500 "8 )( 200!109 ) = 69.86(10 !3 ) rad m = 4.00o per meter ______________________________________________________________________________________ SOLUTION (6.30) Refer to Table 6.2. (a)

T = a2#20max = ( 0.045) 20( 50"10 ) = 228 N ! m

Also

T = a246G".2all = ( 0.045) (80#4610.2 )(1.5! 180 ) = 186 N ! m = Tall (b)

.5 " = 46.2GTmax [ aL14 + aL24 ] = 4680.2!(101869 ) [ ( 0.206.5)4 + ( 0.1045 ] )4 1

= 1.0742(10 )[192,901.235 + 364,797.897] = 0.06 rad = 3.44 o ______________________________________________________________________________________ SOLUTION (6.31) Referring to Fig. P6.31, an expression for t is written as

t = t0 (1 ! by )

Substitute this into the given stress function to obtain, 2

We have

" = G# [ t40 (1 ! by ) 2 ! x 2 ] T = 2 !! # dxdy = 4 !

t0 2

= 4G# !

t0 2

Let

(1! 2 x ) t0

b (1" 2t x ) t 2 0 0 4 0

[ (1 " by ) 2 " x 2 ]dxdy

{ t40 [b(1 " 2t0x ) " b(1 " 2tox ) 2 + b3 (1 " 2t0x ) 3 ] " bx 3 (1 " 2t0x )}dx

x = (1!2u ) to

dx = ! t02du

du = ! 2tdx0 ,

Then, the preceding expression for the torque becomes 3

T = 4G t08b ! (u 2 " 23 u 3 )du

Integrating,

T = G!t03 12

______________________________________________________________________________________ SOLUTION (6.32) 3

( a ) From Table 6.2: T = ( 2"r tG )! = C! or

C = 2!r 3tG = 2! (0.02375) 3 (0.0025)G = 2.1(10 !7 )G " max = 2!Tr 2t = 2! ( 0.02375T )2 ( 0.0025) = 112.860T

( b ) Equation (6.16): 3

C = bt3G = ( 0.02375)(30.0025) G = 1.24(10 !10 )G 3

Equation (6.18):

! max = bt3T2 = ( 0.023753)(T0.0025)2 = 20,210,526T

( c ) From Table 6.2: for a = b, t = t1 :

T = (0.02375) 3 tG! = 1.34(10 "5 )0.0025G! = 3.35(10 "8 )G! = C! ! max = 2 Ta 2t = 2 ( 0.02375T)2 ( 0.0025) = 354,571T ______________________________________________________________________________________ SOLUTION (6.33) Referring to Table 6.2:

! A = 20a 3T ;

420 ( 2 ) 3

= ( 020.05T)3

Solving, T = 1.75 kN ! m Also 3

46.2 (1.75!10 ) # = 46a 4.2GT = ( 6.25 = 0.1617 rad m !10" 6 )( 80!109 )

______________________________________________________________________________________ SOLUTION (6.34) For a thin-walled tube, letting

ro ! ri ! ravg . = r 4

J = ! ( ro2" ri ) = !2 ( ro2 + ri 2 )( ro + ri )( ro " ri ) = 2!rt

Equation (6.2):

! # = TJ! = 2T"!r 3t = 2TArt

Since r = ! , we obtain Eq. (6.22): ! = T 2 At . ______________________________________________________________________________________ SOLUTION (6.35) For a regular hexagon, we can write

A = 3 23 a 3

ds = 6a

Equation (6.22), substituting the value of A, results in the shear stress

! = 2TAt = T9 a 23t

Angle of twist per unit length, using Eq. (6.23):

______________________________________________________________________________________ 6.35 (CONT.)

6a ) " = !2(GA = 23ATa2Gt

2T ! = 9Ga 3 t

______________________________________________________________________________________ SOLUTION (6.36) 0.25

0.25 t3

t2 A1 t1

0.25

t2 Given:

t1 = 0.012 m,

t 2 = t3 = 0.006 m,

t 4 = t5 = 0.01 m,

G = 28 GPa

T = 56.5 kN ! m, A1 = A2 = A = 0.0625 m 2 s1 = s2 = s3 = s4 = s5 = 0.25 m. We write or

T = 2 A1h1 + 2 A2 h2 = 2 A1t1! 1 + 2 A2 t3! 3 2! 1 + ! 3 = 75.3(10 6 )

(1)

The shear flow yields

! 1t1 = ! 2 t 2 ! 2 t 2 = ! 3t 3 + ! 5 t5 ! 3t 3 = ! 4 t 4

Also,

(2) (3) (4)

" 1 s1 + 2" 2 s2 + " 5 s5 = 2G!A1 # " 5 s5 + 2" 3 s3 + " 4 s4 = 2G!A2

(5) (6)

Simultaneous solution of Eqs. (1) through (6), after substitution of the given numerical values yields:

! 1 = 19.1 MPa ! 3 = 37.2 MPa

and

! 2 = 38.2 MPa ! 4 = 22.3 MPa

! 5 = 0.62 MPa

" = 6.86(10 !3 ) rad m

______________________________________________________________________________________ SOLUTION (6.37) a

t=3.5 mm

a=50 mm We have o

Then,

sin 60 ) A = ( 0.05)( 0.05 = 1.0825(10 !3 ) m 2 . 2 40 h = 2TA = 2 ( 0.0010825 ) = 18,475.209 N m , 209 ! = ht = 180, 475 = 5.279 MPa .0035 6

5.279 (10 ) # $ = 2GA " ds = 2( 28!109 )(0.0010825) [0.05 + 0.05 + 0.05]

= 13.063(10 !3 ) rad m

______________________________________________________________________________________ SOLUTION (6.38) t

c 1

For circular tube: Area enclosed by c, Am1 = !c

Area of the section, A1 = 2!ct 3

Polar moment of inertia, J 1 = 2!c t For square tube: Enclosed area by a, Am 2 = a

Area of the section, A2 = 4at

Polar moment of inertia (from Table 6.2 with t = t1 and a = b ): 2 2

J 2 = (2attt1+abtb1 ) = a 3t Then, A1 = A2 gives: a = !c 2 and hence, from Eq. (6.22) we obtain "1 "2

= AAmm12 = !ac2 = 0.785 2

Similarly, from ! = T GJ : "1 "2

= JJ12 = 2a!c3 = 0.617 3

______________________________________________________________________________________ SOLUTION (6.39) t1

t2 A2

t3 t2

t1 We have 1

s1 = [0.252 + 0.03752 ] 2 = 0.2528 m 1

s2 = [0.52 + 0.052 ] 2 = 0.5025 m s3 = 0.15 m, s4 = 0.05 m, and

s5 = 0.075 m

! 1t1 = ! 3t3 + ! 2 t 2 ! 5t5 = ! 1t1 , ! 2 t2 = ! 4 t4 T = 2 A1t1! 1 + 2 A2 t 2! 2

(1) (2,3) (4)

Using Eq. (6.23),

" 5 s5 + 2" 1 s1 + " 3 s3 = 2G!A1 2" 2 s2 + " 4 s4 # " 3 s3 = 2G!A2

(5) (6)

Given: t1 = t 2 = t 4 = t5 = 0.0005 m, t3 = 0.00075 m, G = 28 GPa , and

T = 4 kN ! m.

The cell areas are calculated as

A1 = 0.028125 m 2

A2 = 0.05 m 2

Substituting the numerical values and solving Eqs. (1) through (6):

! 1 = ! 5 = 50.6 MPa, " 2 = " 4 = 50.88 MPa,

! 3 = 0.593 MPa ! = 0.0194 rad m

______________________________________________________________________________________ SOLUTION (6.40) ( a ) We have ! s = ! c and Eq. (6.37) becomes Ps Rs3 Gs

Solving,

= PGc Rcc ;

Ps ( 0.062 )3 9

79 (10 )

c ( 0.05 ) = P41 (109 )

Ps = 1.011Pc

Assume that copper controls

" c = 16!Pdc3Rc ;

0.05 ) 300(10 6 ) = 16! P(c0(.01 )3

from which

Pc = 1178.1 N

Then,

Ps = 1188.4 N

______________________________________________________________________________________ 6.40 (CONT.) Check the assumption:

.4 )( 0.062 ) " s = 16(1188 = 375.25 MPa ! ( 0.01)3

Since ! s < 500 MPa , the assumption is correct. Thus, the total force is

Pt = Pc + Ps = 2366.5 N

( b ) For the condition specified, Ps ks

= kPcc

from which ks kc

= PPcs 1.011 End of Chapter 6

CHAPTER 7 SOLUTION (7.1) Refer to Fig. 7.2 and Eq. (7.10): "4w "x 4

= h14 ( w9 ! 4 w1 + 6w0 + w11 )

"4w "y 4

= h14 ( w10 ! 4 w2 + 6w0 ! 4 w4 + w12 )

"4w "x 2"y 2

= h14 [ w5 ! w6 + w7 + w8 ! 2( w1 ! w2 ! w3 ! w4 + 2 w0 )] 4

Substituting these into ! w :

h 4 " 4 w = 20w0 ! 8( w1 + w2 + w3 + w4 ) ! 2( w5 + w6 + w7 + w8 )

+ w9 ! w10 ! w11 ! w12 ______________________________________________________________________________________ SOLUTION (7.2) y

x h=a/2

4a Only a quarter of the section need be considered. Using the symmetry the nodal points are labeled as shown in the preceding figure, and ! = 0 at the boundary. The finite difference equations for the nodes 1 through 8 are, respectively:

" 4# 1 + 2# 2 + 2# 3 = "2G!h 2

(1)

(2)

(3)

# 2 + # 3 " 4# 4 + # 6 = "2G!h 2

(4)

# 3 " 4# 5 + 2# 6 + # 7 = "2G!h

(5)

# 4 + # 5 " 4# 6 + # 8 = "2G!h

(6)

# 5 " 4# 7 + 2# 8 = "2G!h 2

(7)

# 6 + # 7 " 4# 8 = "2G!h 2

(8)

# 1 " 4# 2 + 2# 4 = "2G!h

# 1 " 4# 3 + 2# 4 + # 5 = "2G!h

Solving,

" 1 = 3.587G!h 2

" 2 = 2.708G!h 2

" 3 = 3.467G!h 2

" 4 = 2.622G!h 2

" 5 = 3.037G!h 2

" 6 = 2.313G!h 2

" 7 = 2.055G!h 2

" 8 = 1.592G!h 2

______________________________________________________________________________________ 7.2 (CONT.) Tables of differences are in P7.2a and P7.2b. Table P7.2a (when y=b, ! = ! 1 ; y=h,

y 0

# G! h 3.587

2.708

! = ! 2 , etc.)

" G! h 2 ! 0.380

"2 G!h 2 ! 1.828

! 2.708

Table P7.2b (when x=0, ! = ! 1 ; x=h,

! = ! 3 , etc.)

x 0

# G!h 2 3.587

" G!h 2 ! 0.120

"2 G!h 2 ! 0.310

"3 G!h 2 ! 0.242

h 2h 3h 4h

3.467 3.037 2.055 0

! 0.430 ! 0.982 ! 2.055

! 0.552 ! 1.074

! 0.522

"4 G!h 2 ! 0.28

We have $" $x

= #"h 0 + #2 h"20 [2 x ! h ] + #6 h"30 [3x ! 6 xh + 2h 2 ] 4

+ #24"h 40 [4 x 3 ! 18 x 2 h + 22 xh 2 ! 6h 3 ] 2

( ##!x ) x =4 h = "!h 0 + "2 h!20 (7h ) + "6 h!30 ( 26h 2 ) + "24!h 40 (50h 3 ) Using Table P7.2b, we obtain ( ##$x ) x =4 h = "1.418G!a. That is

(" ) x = ±2 a = #1.418G!a Similarly, using Table P7.2a: 2

( %%#y ) y =2 h = $#h 0 + $2 h#20 ( 2 y " h ) = "1.811G!a

(" ) y = ±2 h = #1.811G!a

Hence, the error using this method is [(1.86 ! 1.811) 1.86]100 = 2.63 % ______________________________________________________________________________________ SOLUTION (7.3) Equation (7.19) is applied at points b, c, d, respectively:

2# c + 1.178# g " 4.857# b = "2G!h 2

# b + 1.178# f + # d " 4.857# c = "2G!h 2

1.178# c + 1.178# e " 5.714# d = "2G!h 2 (CONT.) ______________________________________________________________________________________ Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 7–2

______________________________________________________________________________________ 7.3 (CONT.) Similarly, Eq. (7.16) is applied at points e, f, and g, respectively:

2# d + # f " 4# e = "2G!h 2 2# c + # g + # e " 4# f = "2G!h 2 2# b + 2# f " 4# g = "2G!h 2 In matrix form these equations are written as:

0 0 0 2 / 4 +'( b $ . 2 ! ! , 0 2 0 1 1 )!( c ! /4 ) , 0 2 1 0 ) !!( d !! /4 , 0 )& # , 1.178 / 5.714 1.178 0 0 )! ( e ! , 0 , 1 1 0 1.178 0 ) !( f ! / 4.857 )! ! , 2 0 0 0 1.178* !%( g !" - / 4.857 = "2G!h 2 {1, 1, 1, 1, 1, 1} Solution is

" b = 1.612G!h 2

" c = 1.487G!h 2

" d = 0.975G!h 2

" e = 1.547G!h 2

" f = 2.236G!h 2

" g = 2.424G!h 2

The finite differences are then o

$ = " e # " B = 1.547G!h 2 o

$2 = # f " 2# e + # B = "0.858G!h 2 o

$3 = " g # 3" f + 3" e # " B = 0.357G!h 2 o

$4 = # f " 4# g + 6# f " 4# B + # B = "0.232G!h 2 o

$5 = # e " 5# f + 10# g " 10# f + 5# e " # B = "0.018G!h 2 o

$6 = " B # 6" e + 15" f # 20" g + 15" f # 6" e + " B = 0.036G!h 2 Hence,

" B = ( $$%x ) B = h1 ( # & #2 + #3 & #4 + #5 & #6 )G!h 2 0.357 0.232 0.018 0.036 = (1.547 + 0.858 2 + 3 + 4 " 5 " 6 )G!h 2

" B = 2.144G!h = 0.0107G!

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (7.4) Applying Eq. (7.16) at points b, c, e, f, and g, we obtain

# g + 2# c " 4# b = "2G!h 2

# b + # d + # f " 4# c = "2G!h 2

2# d + # f " 4# e = "2G!h 2

# g + # e + 2# c " 4# f = "2G!h 2

2# f + 2# b " 4# g = "2G!h 2 At point d, we apply Eq. (7.19):

1.308# c + # e " 5.77# d = "2G!h 2

Solving these equations, we have

" b = 2.238G!h 2

" c = 2.000G!h 2

" d = 1.096G!h 2

" e = 1.715G!h 2

" f = 2.667G!h 2

" g = 2.953G!h 2

The finite differences are the computed as shown in Table P7.4a. Thus,

" A = ( $$%x ) A = h1 ( # & #2 + #3 & #4 )G!h 2 0.093 0.0186 = ( 2.238 + 1.523 2 + 3 + 4 )G!h = 3.035G!h = 0.0129G! 2

Table P7.4a

y 0

# G!h 2 0

" G!h 2 2.238

"2 G!h 2 ! 1.523

"3 G!h 2 0.093

h 2h 3h 4h

2.238 2.953 2.238 0

0.715 ! 0.715 ! 2.238

! 1.430 ! 1.523

! 0.003

"4 G!h 2 ! 0.019

Alternatively, we construct the table as shown in Fig. P7.4b: Table P7.4b

y 0

# G! h 2 2.953

" G! h 2 ! 0.715

h 2h

2.238 0

! 2.238

"2 G!h 2 ! 1.523

Then we obtain, 2

" A = ( &&$y ) y =2 h = %$h 0 + 23 % h$ 0 = [ #0.715 # 23 (1.523)]G!h = 3.0G!h This result is approximately the same as calculated before. Clearly now computation involved are reduced considerably. ______________________________________________________________________________________ Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 7–4

______________________________________________________________________________________ SOLUTION (7.5) y

1.5a A

b=a 1

h=a/4 x

Applying Eq. (7.16) at nodes in to 6, we obtain

0 1 0 0 , ')1 $ '1$ /( 4 1 ! ! !1! * - 1 (4 1 0 1 0 )2 !! *! ! 2 (4 0 0 1 * !) 3 ! - 0 2 !1! * & # = (2G0h & # 0 0 (4 1 0 * !) 4 ! !1! - 2 !1! - 0 2 0 1 ( 4 1 * !) 5 ! !! *! ! 0 2 0 2 ( 4 + %) 6 " %1" . 0 Solving,

" 1 = 1.551G!h 2

" 2 = 2.206G!h 2

" 3 = 2.387G!h 2

" 4 = 1.997G!h 2

" 5 = 2.887G!h 2

" 6 = 3.137G!h 2

Then, the differences are as shown in Table P7.5. Table P7.5

x 0

# G!h 2 0

" G!h 2 2.387

"2 G!h 2 ! 1.637

"3 G!h 2 0.137

h 2h 3h 4h

2.387 3.137 2.387 0

0.75 ! 0.75 ! 2.387

! 1.50 ! 1.637

! 0.137

"4 G!h 2 ! 0.274

At point A, we have 2

( $$"x ) x =0 = h1 ( #" 0 % # 2" 0 + # 3" 0 % # 4" 0 )G!h 2

Substituting the first row of Table P7.5: or

0.137 0.274 ( ""#x ) A = ( 2.387 + 1.637 2 + 3 + 4 )G!h

( ##$x ) A = 0.83G"a = ! max

This differs 2.12 % from the exact solution 0.848G!a (Table 6.2) ______________________________________________________________________________________ Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 7–5

______________________________________________________________________________________ SOLUTION (7.6) L/2

P v-1

O y

L/2 v2

Boundary conditions yield,

v (0) = 0; v0 = 0

v ' (0) = 0; v1 = v !1

and

Apply Eq. (7.22) at nodes 0, 1, and 2, respectively: 2

v !1 ! 2v0 + v0 = h 2(EIPL ) ;

v1 = 4hEI PL 2

v 2 ! 2v1 + v0 = h (22EIPL 3) ;

v2 = 65hEI PL 2

3) v3 ! 2v2 + v1 = h (2PL EI

(a)

Substituting v1 and v 2 into Eq. (a) we obtain

v3 = 327 PL EI

______________________________________________________________________________________ SOLUTION (7.7) L/2

Mo A

3EI

O y

h=L/4 2C

L/2 3

B 4

RA M

Mo 3Mo/4 M

Let C = ! EI0 h

Mo/2

Mo/4

v0 = v 4 = 0

Applying Eq. (7.22) at 1, 2, 3: o

Solving,

v2 ! 2v1 + v0 = C4 v1 = ! 83 C

v3 ! 2v2 + v1 = C4

v2 = ! 12 C

(a)

" C = " 3 = 21h ( v3 ! v1 ) = 0

(b)

vC = v 2 = 361 MEI0 L

v 4 ! 2v3 + v2 = C4

v3 = ! 83 C

Results are the same as the exact solutions. ______________________________________________________________________________________ Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 7–6

______________________________________________________________________________________ SOLUTION (7.8) 4

Let C = ( p EI )h . We have h = a 4 = L 8 and v0 = v8 = 0. p -1

EI 1

O y

2EI

C L=2a

8 5

-7

Apply Eq. (7.23) at 1 through 7:

v3 ! 4v2 + 5v1 = C

(1)

v 4 ! 4v3 + 6v 2 ! 4v1 = C

(2)

v5 ! 4v4 + 6v3 ! 4v2 + v1 = C

(3)

v 6 ! 4 v 5 + 6v 4 ! 4 v 3 + v 2 =

C 1.5

(4)

v7 ! 4v6 + 6v5 ! 4v4 + v3 = C2

(5)

! 4 v 7 + 6v 6 ! 4 v 5 + v 4 =

C 2

(6)

5v7 ! 4v6 + v5 = C2

(7)

Solving, 4

Note:

pa vC = v 4 = 0.00966 pL EI = 0.1546 EI

( vC ) exact = 5 pa 4 32 EI

______________________________________________________________________________________ SOLUTION (7.9) po po/4 4

Let N = p0 L

-2

-1

A 0

po/2

h=L/4

3po/4

Boundary conditions at A:

v ' (0) = 0; v (0) = 0;

v1 = v !1 v0 = 0

v ' ' ' ' (0) = 0;

v2 ! 8v1 + v !2 = 0

Apply Eq. (7.23) at 1 through 4:

7v1 ! 4v 2 + v3 = (0.25) 5 N

! 4v1 + 6v 2 ! 4v3 + v 4 = (0.5) 9 N v1 ! 4v2 + 6v3 ! 4v4 + v5 = 3(0.5)10 N v 2 ! 4v3 + 6v 4 ! 4v5 + v6 = (0.25) 4 N (CONT.) ______________________________________________________________________________________ Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 7–7

______________________________________________________________________________________ 7.9 (CONT.) Boundary conditions at B:

v ' ' ' ( L) = 0;

! v 2 + 2v 3 ! 2v5 + v 6 = 0

v ' ' ( L) = 0;

v 3 ! 2v 4 + v5 = 0

The foregoing six equations are solved to yield

v1 = 0.010742 N v3 = 0.066406 N v5 = 0.132812 N

The error is therefore 0.099609 "0.091667 0.091667

v2 = 0.035156 N v4 = 0.099609 N v2 = 0.167969 N

! 100 = 8.7 % .

______________________________________________________________________________________ SOLUTION (7.10) 0

L/3

h=L/6

2P/3

B P/3

Figure (a)

Application of Eq. (7.22) at points 1 through 5 gives, respectively:

v 2 ! 2v1 + v0 = ! 9PLEI h 2

2 v3 ! 2v2 + v1 = ! 29 PL EI h 2 v 4 ! 2v3 + v 2 = ! PL EI h

v5 ! 2v4 + v3 = ! PL EI h

(a)

2 v6 ! 2v5 + v 4 = ! PL EI h

The boundary conditions are v0 = v6 = 0. Then Eqs. (a) may be represented in matrix form as

0 0 0 * ' v1 $ '2 $ -. 2 1 + 1 .2 1 0 0 ( !v 2 ! !4 ! ( !! !! !! !! + 1 .2 1 0 ( & v3 # = & 3# C + 0 ( + 0 1 . 2 1 ( !v 4 ! !2 ! + 0 ! ! ! ! +, 0 0 0 1 . 2() !%v5 !" !%1 !" 3

648EI . Solving the foregoing; v1 = !6.67C , v 2 = !11.33, v 4 = !9.67C , v5 = !5.33C , and v3 = !12C , or

where C = ! PL

v3 = 0.01852 PL EI

Note: The exact value of the deflection at the center (see Table D.4) is 0.01775PL EI . ______________________________________________________________________________________ Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 7–8

______________________________________________________________________________________ SOLUTION (7.11) p 1

-1

n=0

n=4

h=L/4

Figure (a)

We observe that because of symmetry only half of the beam span need be considered "

as shown in Fig. (a). From the boundary conditions v (0) = v (0) = 0 (Fig. 7.6) and symmetry:

v0 = v 4 = 0

v1 = ! v !1

v1 = v3

(a)

By Eq. (7.23) at points 1 and 2,

v3 ! 4v2 + 6v1 ! 4v0 + v !1 = 0

(b)

v 4 ! 4v3 + 6v 2 ! 4v1 + v0 = phEI

Substituting Eqs. (a) into Eqs. (b), we obtain

6v1 ! 4v 2 = 0

! 4v1 + 3v2 = phEI

Solving the above and setting h = L 4 yields v1 = 0.0039 pL

EI and

v2 = v max = 0.0059 pL EI

______________________________________________________________________________________ SOLUTION (7.12)

A y

P 2

P 4 3

h=a/2

P P

Pa x

Equation (7.22) is applied at nodes 1, 2, 3, respectively:

0 ! 2v1 + v 2 = 0; v1 ! 2v2 + v3 = 0;

v1 = v2 2 v 2 = 2v 3 3

v 2 ! 2v3 + 0 = ( h 2 EI )( Pa 2) or

v3 = ! 3Pah 2 8EI

______________________________________________________________________________________ 7.12 (CONT.) Thus,

v B = v 2 = ! 161 PaEI = !0.0625 PaEI 3

and

v1 = ! Pa 3 32 EI

Slop at A is then

" A = v12!hv!1 = vh1 = ! 16PaEI 2

We have v B = 0.08333 Pa 0.08333"0.0625 0.08333

EI as exact solution. Error is thus ! 100 = 25 %

______________________________________________________________________________________ SOLUTION (7.13) Boundary conditions yield

and or

v (0) = 0;

v0 = 0

v ' (0) = 0;

v1 = v !1

v ' ' ( L ) = 0 = v 4 ! 2v 3 + v 2 v 4 = 2v 3 ! v 2

(1)

v ' ' ' ( L) = 0 = v5 ! 2v4 + 2v2 ! v1 or

v5 = 4v3 ! 4v2 + v1

(2)

Apply Eq. (7.23) at points 1, 2, and 3, with v0 = 0 and v1 = v !1 : 4

v3 ! 4v 2 + 7v1 = 162pLEI

(3)

v 4 ! 4v3 ! 6v 2 ! 4v1 = 81pLEI

(4)

v5 ! 4v4 + 6v3 ! 4v2 + v1 = 81pLEI

(5)

Solving Eqs. (1) to (6),

v3 = 7 pL4 54 EI

and

v1 = 4v3 21

v 2 = 4v 3 7

______________________________________________________________________________________ SOLUTION (7.14)

C = ph 4 EI

h=L 4

v1 = v !1 p

-1

A y

-3

______________________________________________________________________________________ 7.14 (CONT.) Apply Eq. (7.22) at 1, 2, 3:

v3 ! 4v 2 + 7v1 = C

! 4v3 + v 2 ! 4v1 = C

5v3 ! 4v2 + v1 = C ph 4

v3 = 1.36364 phEI

Solving, v1 = 0.90909 EI

v2 = v max = 1.68182 phEI

= 0.00657 pL EI !

" B = " max = 21h ( v5 ! v3 ) = ! vh3 3

= !0.02131 pL EI

______________________________________________________________________________________ SOLUTION (7.15) P A B MA=M M 0 3 4 1 2 RA=P/2 RB=P/2 L/2 L/2 y PL/2 P PL/4 Mp x MR

-PL/8

-PL/4

-3PL/8

-PL/2

v1 = v !1 , v1 = v3 (due to symmetry), v3 = v5 v0 = v4 = 0, C = L3 16 EI . Apply Eq.(7.22) at 1through 4: o

v 2 ! 2v1 + v0 = ( ! PL 8 + M )C

(1)

v3 ! 2v2 + v1 = ( ! PL4 + M )C

(2)

v 4 ! 2v3 + v2 = ( PL4 ! 3 PL 8 + M )C o

v5 ! 2v4 + v3 = ( PL4 ! PL2 + M )C

(3) (4)

Equations (4), (3), (2) lead to

2v3 = MC

v2 = ( 2 M ! PL 8 )C v1 = ( 92M ! PL2 )C

Then, Eq. (1) gives

M = 81 PL

Note:

M exact = 81 PL

______________________________________________________________________________________ SOLUTION (7.16) p L h=L/4 0

y Boundary conditions are v0 = v 4 = 0 and from symmetry v1 = v3 . Then, using Eq. (7.23) at nodes 1 and 2: 4

v1 + 6v1 ! 4v2 + v3 = phEI 4

Solving,

! 4v1 + 6v 2 ! 4v3 = phEI 4

v 2 = phEI = EIP ( L4 ) 4 = 0.00391 pL EI

The exact solution is

v2 = 0.002604 pL EI

The slope is zero at nodes 0, 2, and 4. Thus, 4

) 4 pL "1 = " 3 = v22!hv0 = ( 0.0039 2L EI 3

= 0.078 pL EI

______________________________________________________________________________________ SOLUTION (7.17) F41

AE 13.5(10!4 )(69 "109 ) = L 1.7 = 54.79 !106 N m

1 150o L=1.7 m 4

c = cos150o = ! 3 2 (a)

F41

s = sin150o = 1 2

Equation (7.38):

! 34 3 4" # 3 4 #3 4 $ % 14 3 4 #1 4 % 6 $# 3 4 [k ]e = 54.79 &10 $ % 34 # 3 4% $ #3 4 # 3 4 $ % #1 4 # 3 4 14 ( ' 3 4

______________________________________________________________________________________ 7.17 (CONT.) or

(b)

Equation (7.39) with i=4 and j=1:

or (c)

" 0.75 !0.433 !0.75 0.433 # $ !0.433 0.25 0.433 !0.25 %% $ [k ]e = 54.79 MN m $ !0.75 !0.433 0.75 !0.433% $ % & 0.433 !0.25 !0.433 0.75 '

" 3 F41 = 54.79(106 ) ( ! , 2 F1 = F41 = !73.14 kN

1 # $ 2 + 1.1 % !3 ' (10 ) )& 2 - *!1.5 + 1.2 +

Then {! }e = [T ]{! }e , give

0.5 0 0 % " !1.1# " 0.35 # "u4 # $ !0.866 & v & ' !0.5 !0.866 0 0 (( &&!1.2 && && 1.59 && & 4& ' = ) * ) *=) * mm ' ( u 0 0 0.866 0.5 2 0.98 ! ! 1 & & & & & & ( &- v1 .& +' 0 0 !0.5 !0.866 , &- 1.5 &. &- !0.3 &. ______________________________________________________________________________________ SOLUTION (7.18) We have

(

AE AE )1 = L L

(

AE 2 AE )2 = L L

Equation (7.25a):

[k ]1 =

(a)

AE " 1 !1# L &$ !1 1 '%

[k ]2 =

(

AE 3 AE )3 = L L

AE " 2 !2 # L &$ !2 2 '%

[k ]3 =

AE " 3 !3# L &$ !3 3 '%

System stiffness matrix, [ k ] = [ k ]1 + [ k ]2 + [ k ]3 is order of 4 ! 4 . Thus

" 1 !1 0 0 # $ % AE $ !1 3 !2 0 % [K ] = L $ 0 !2 5 !3% $ % & 0 0 !3 3 ' (CONT.) ______________________________________________________________________________________ Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 7–13

______________________________________________________________________________________ 7.18 (CONT.) (b)

" F1x # $ 1 !1 0 0 % " u1 # &F & ' !1 3 !2 0 ( &u & & 2 x & AE ' ( &) 2 &* ) *= & F3 x & L ' 0 !2 5 !3( &u3 & ' ( + 0 0 !3 3 , &-u4 &. -& F4 x .&

(1)

Boundary conditions are u1 = u4 = 0 . We have F3x = ! P . Thus

Solving

" 0 # AE $ 3 !2 % "u2 # & '= ( )& ' *! P + L , !2 5 - *u3 +

u2 = ! (c)

2 PL 11AE

u3 = !

3PL 11AE

Equations (1) :

" F1x # $ 1 !1 0 0 % " 0 # "2 P 11# &F & ' !1 3 !2 0 ( &! 2 P 11& & & & 2 x & AE ' & & L & 0 & ( =) ) *= ) * * & F3 x & L ' 0 !2 5 !3( & ! 3P 11& AE & ! P & ' ( &-9 P 11&. + 0 0 !3 3 , &- 0 &. -& F4 x .& The reactions are

R1 =

2 P! 11

R4 =

9 P! 11

______________________________________________________________________________________ SOLUTION (7.19) We have

AE A(2 E ) 6 AE )1,2 = = L L3 L AE 2 A( E ) 6 AE ( )3 = = L L3 L (

There are four displacement components ( u1 , u2 , u3 , u4 ) and so the order of the system matrix is 4x4. Using Eq. (7.25a):

[k ]1 = [k ]2 = [k ]3 =

6 AE " 1 !1# L $& !1 1 %'

______________________________________________________________________________________ 7.19 (CONT.)

(a)

!1 0 0# "1 " 1 !1 0 0 # $ !1 (1 + 1) % !1 0 % 6 AE $$ !1 2 !1 0 %% 6 AE $ [K ] = = !1 (1 + 1) !1% L $0 L $ 0 !1 2 !1% $ % $ % 0 1' !1 &0 & 0 0 !1 1 '

(b)

" F1x # $ 1 !1 0 0 % " u1 # &F & ' (& & & 2 x & AE ' !1 2 !1 0 ( &u2 & ) *= ) * & F3 x & L ' 0 !1 2 !1( &u3 & ' ( &- F4 x &. + 0 0 !1 1 , &-u4 &.

(1)

Boundary conditions are u1 = u4 = 0 and Fx 3 = P . Equation (1) is then

Solving

" 0 # AE $ 2 !1% "u2 # & '= ( )& ' * P + L , !1 2 - *u3 +

u2 = (c)

PL 9 AE

u3 =

PL 18 AE

Equations (1) result in

0 " F1x # $ 1 !1 0 0 % " # " !2 P 3# &F & ' !1 2 !1 0 ( & PL 9 AE & & & & 2 x & 6 AE ' & & L & P & ( = = ) * ) * ) * L ' 0 !1 2 !1( & PL 18 AE & AE & 0 & & F3 x & ' ( 0 + 0 0 !1 1 , -& .& -& ! P 3 .& -& F4 x .& The reactions are

R1 =

2 P! 3

1 R4 = P ! 3

______________________________________________________________________________________ SOLUTION (7.20)

(

AE AE )1 = L L

Equation (7.25a):

[k ]1 = [k ]2 =

(

AE AE )2 = L L

AE " 1 !1# L $& !1 1 %'

(

AE 4 AE AE )3 = = 0.8 L 5L L

[k ]3 =

AE " 0.8 !0.8# L $& !0.8 0.8 %'

______________________________________________________________________________________ 7.20 (CONT.) (a) System matrix, [ k ] = [ k ]1 + [ k ]2 + [ k ]3 , is then

!1 0 0 # "1 $ !1 (1 + 1) !1 0 %% AE $ [K ] = !1 (1 + 0.8) !0.8% L $0 $ % 0 0.8 ' !0.8 &0 u1

(b)

0 % " u1 # " F1x # $ 1 !1 0 &F & ' 0 (( &&u2 && !1 & 2 x & AE ' !1 2 = ) * ) * & F3 x & L ' 0 !1 1.8 !0.8( &u3 & ' ( &- F4 x &. + 0 0 !0.8 0.8 , &-u4 &.

(1)

Boundary conditions are u1 = u4 = 0 , We have Fx 3 = P . Thus

Solving

" 0 # AE $ 2 !1 % "u2 # & '= ( )& ' * P + L , !1 1.8- *u3 +

u2 = ! (c)

PL 2.6 AE

u3 =

PL 1.3 AE

Equations (1) yield

0 %" 0 # " F1x # $ 1 !1 0 " ! P 2.6 # &F & ' !1 2 ( & & & 0 ( & P 2.6 & L && 0 !1 & 2 x & AE ' & =) ) *= ) * * P & F3 x & L ' 0 !1 1.8 !0.8( & P 1.3 & AE & & ' ( &- F4 x &. + 0 0 !0.8 0.8 , -& 0 &. -&!1.6 P 2.6 .& The reactions are

R1 =

1 P! 2.6

R4 =

0.8 P! 1.3

______________________________________________________________________________________ SOLUTION (7.21) Table P7.21 Data for the truss of Fig P7.21 Element Length(m) ! c s

1 2 3 4

7.5 6 4.5 4.5

36.9o 0o 90o 0o

4.5 2

135o

0.8 1 0 1

0.6 0 1 0

"0.707 0.707

c2 cs s2 0.639 0.48 0.36 1 0 0 0 0 1 1 0 0 0.5

"0.5

0.5

______________________________________________________________________________________ 7.21 (CONT.)

" 0.639 0.48 !0.639 !0.48 # $ 0.361 !0.48 !0.361%% AE $ 0.48 [k ]1 = 0.639 0.48 % 7.5 $ !0.639 !0.48 $ % 0.361' & !0.48 !0.361 0.48 u1 v1 u3 v3 & 1 0 '1 0 # $ 0 0 !! AE $ 0 0 [k ]2 = 1 0! 6.0 $' 1 0 $ ! 0 0" % 0 0 u2 v2 u3 v3 0 &0 $0 1 AE $ [ k ]3 = 0 4.5 $0 $ %0 ' 1 u3 v3 & 1 0 $ AE $ 0 0 [k ]4 = 4.5 $' 1 0 $ % 0 0

0 0# 0 ' 1 !! 0 0! ! 0 1" u4 v 4 '1 0 # 0 0 !! 1 0! ! 0 0"

u2 v2 u4 v4 " 0.5 !0.5 !0.5 0.5# $ % AE $ !0.5 0.5 0.5 !0.5% [k ]5 = 4.5 2 $ !0.5 0.5 0.5 !0.5% $ % & 0.5 !0.5 !0.5 0.5'

______________________________________________________________________________________ SOLUTION (7.22) Table P7.22 Data for the truss of Fig.P7.22

Element 1

Length (m) 2.4

2 3 4

2.6 1.0 2.4

2.6

! 0o 22.62o "90o 0o 22.62o

c 1

s 0

c2 1

cs 0

s2 0

0.923 0.385 0.852 0.355 0.148 0 1 0 0 1 1 0 1 0 0 0.923 0.385 0.852 0.355 0.148

______________________________________________________________________________________ 7.22 (CONT.) (a)

Use Eq.(7.38);

u A v A uC v C " 1 0 !1 0 # $ 0 0 %% AE $ 0 0 [k ]1 = 1 0% 2.4 $ !1 0 $ % 0 0' & 0 0 uA vA uB vB " 0.852 0.355 !0.852 !0.355# $ % AE $ 0.355 0.148 !0.355 !0.148% [k ]2 = 2.6 $ !0.852 !0.355 0.852 0.355% $ % & !0.355 !0.148 0.355 0.148' uB v B 0 !0 #0 1 AE # [k ]3 = 0 1.0 #0 # &0 %1

uC v C 0 0" 0 %1 $$ 0 0$ $ 0 1'

uB v B " 1 0 $ AE $ 0 0 [k ]4 = 2.4 $ !1 0 $ & 0 0

uD v D !1 0 # 0 0 %% 1 0% % 0 0'

uC vC uD vD " 0.852 0.355 !0.852 !0.355# $ % AE $ 0.355 0.148 !0.355 !0.148% [k ]5 = 2.6 $ !0.852 !0.355 0.852 0.355% $ % & !0.355 !0.148 0.355 0.148'

______________________________________________________________________________________ 7.22 (CONT.) (b)

Global Stiffness Matrix [ K ] =

uA vA uB vB uC vC uD vD 0 0 " 0.745 0.137 !0.328 !0.137 !0.417 0 # $ 0.137 !0.057 !0.137 !0.057 0 % 0 0 0 $ % $ !0.328 !0.137 0.745 0.137 0 % 0 !0.833 0 $ % !0.137 !0.057 0.137 1.057 0 !2 0 0 $ % AE $ !0.417 0 0 0 0.745 0.137 !0.328 !0.137 % $ % 0 0 !2 0.137 1.057 !0.137 !0.057 % $ 0 $ 0 0 !0.833 0 !0.655 !0.273 0.745 0.137 % $ % !0.137 !0.057 0.137 0.057 '% 0 0 0 &$ 0 ! RAx % !0% #R # #0# # Ay # # # # 0 # # uB # # # # # #v B # # 0 # & = [K ] " & " #uC # # 0 # #vC # # 0 # # # # # #uD # # P # #$ 0 #' # RDy # ' $

{F } = [ K ]{! };

______________________________________________________________________________________ SOLUTION (7.23) (a)

Element 1

! = 135o

Equation (7.38):

c 2 = 0.5 v1

cs = "0.5 u2 v2

s 2 = 0.5

" 0.5 !0.5 !0.5 0.5# $ % AE $ !0.5 0.5 0.5 !0.5% [k ]1 = L $ !0.5 0.5 0.5 !0.5% $ % & 0.5 !0.5 !0.5 0.5'

! = 180o c 2 = 1.0 cs = 0 u1 v1 u 3 v3 " 1 0 !1 0 # $ 0 0 %% AE $ 0 0 [k ]2 = 1 0% L $ !1 0 $ % 0 0' & 0 0

Element 2

s2 = 0

______________________________________________________________________________________ 7.23 (CONT.)

! = 270o c2 = 0 cs = 0 u1 v1 u4 v4 0 0 0 " !0 #0 1.0 0 %1.0 $$ # [k ]3 = k ' #0 0 0 0 $ # $ 1.0 ' &0 %1.0 0 where k ' = kL AE

s 2 = 1.0

Element 3

(b)

System matrix is 8x8. Superposition. [ K ] =

! [k ] results in u 3 v3 u 4

" 1.5 !0.5 !0.5 0.5 !1 $ !0.5 0.5 0.5 !0.5 0 $ $ !0.5 0.5 0.5 !0.5 0 $ AE $ 0.5 !0.5 !0.5 0.5 0 [K ] = 0 0 0 1 L $ !1 $ 0 0 0 0 $ 0 $ 0 0 0 0 0 $ 0 0 0 &$ 0 !k '

0 0 0 0 0 0 0 0

0 0# 0 !k '%% 0 0% % 0 0% 0 0% % 0 0% 0 0% % k '%' 0

where k ' = kL AE . (c)

Boundary conditions are

u2 = v2 = u3 = v3 = u4 = v4 = 0

Therefore

! 0 " ! u1 " #$P # #u # # # # 2# # F2 x # #0# # # # # # F2 y # #0# % & = [K ]% & # F3 x # #0# # F3 y # #0# # # # # # F4 x # #0# # F4 y # #0# ' ( ' (

The total 7 ! 7 stiffness matrix can be reduced to a 7 ! 4 matrix by using the following antisymmetrical conditions: u1 = !u5 , u2 = !u4 , and v2 = v1 ______________________________________________________________________________________ Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 7–20

______________________________________________________________________________________ SOLUTION (7.24) Table P7.24 Data for the truss of Fig.P7.24

Element 1

Length(m) ! 5 53.13o

2 4 ( a ) Use Eq.(7.38):

180o

c s c2 cs s2 0.6 0.8 0.36 0.48 0.64 "1 0 1 0 0

u1 v1 u2 v2 0.48 ' 0.36 ' 0.48# & 0.36 $ 0.48 0.64 ' 0.48 ' 0.64!! AE $ [k ]1 = 0.36 0.48! 5 $' 0.36 ' 0.48 $ ! 0.48 0.64" %' 0.48 ' 0.64 u2 v 2 u3 v3 & 1 0 '1 0 # $ 0 0 !! AE $ 0 0 [k ]2 = 1 0! 4 $' 1 0 $ ! 0 0" % 0 0

(b)

Global Stiffness Matrix

u1 v1 u2 v2 u3 v3 0.096 ' 0.072 ' 0.096 0 0# & 0.072 $ 0.096 0.128 ' 0.096 ' 0.128 0 0 !! $ $' 0.072 ' 0.096 0.322 0.096 ' 0.25 0 ! [ K ] = AE $ ! 0.096 0.128 0 0! $' 0.096 ' 0.128 $ 0 0 ' 0.25 0 0.25 0 ! ! $ 0 0 0 0 0 !" $% 0

(c)

! 0 " #0.322 0.096 $ !u2 " % & = AE ' (% & *)6000 + , 0.093 0.128- * v2 + "u2 # 1 $ 4 & '= 6 ( , v2 - 10(10 ) * !3

!3 % " 0 # " 0.9 # !3 & '=& ' (10 ) m ) 10.063+ ,!6000 - ,!3.02 -

(d)

" R1x # $ !0.072 !0.096 % " 4.5024 # "u2 # & & & & ' ( ) R1 y * = AE ' !0.096 !0.128 ( ) * = ) 6.0032 * kN + v2 , &!4.5 &R & & 0 + 3x , -' !0.25 .( + ,

______________________________________________________________________________________ 7.24 (CONT.) ( e ) Use Eq.(7.39)

!u " ! 0.0018 " F12 = AE [0.6 0.8]$ 2 % = 20(106) [0.6 0.8]$ % = #75 kN (C ) &#0.00604 ' & v2 ' "!u # " !0.0018 # F23 = AE [!1 0]$ 2 % = 20(106) [!1 0]$ % = 36 kN (T ) & 0.00604 ' & !v2 ' ______________________________________________________________________________________ SOLUTION (7.25) We have E = 105 GPa

A = 10 "10!4 m 2

Table P7.25 Data for the truss of Fig.P7.25 Element Length(m) ! c s

1 2

53.13 o 90 o

5 4

c2 cs s2 0.6 0.8 0.36 0.48 0.64 0 1 0 0 1

( a ) Apply Eq.(7.38):

u1 v1 u2 v2 0.48 ' 0.36 ' 0.48# & 0.36 $ 0.48 0.64 ' 0.48 ' 0.64!! AE $ [k ]1 = 0.36 0.48! 5 $' 0.36 ' 0.48 $ ! 0.48 0.64" %' 0.48 ' 0.64 u1 &0 $ AE $0 [k ]2 = 4 $0 $ %0

v1 0 1 0 '1

( b ) Global Stiffness Matrix

u3 v3 0 0# 0 ' 1 !! 0 0! ! 0 1"

u1 v1 u2 v2 u3 1.008 ' 0.756 ' 1.008 0 & 0.756 $ 1.008 3.969 ' 1.008 ' 1.344 0 $ $' 0.756 ' 1.008 0.756 1.008 0 [ K ] = 10 7 $ 1.008 1.344 0 $ ' 1.008 ' 1.344 $ 0 0 0 0 0 $ ' 2.625 0 0 0 $% 0

v3 0 # ' 2.625!! ! 0 ! 0 ! ! 0 ! 2.625!"

______________________________________________________________________________________ 7.25 (CONT.) (c)

1.008 % "!0.015# " F1x # 7 $ 0.756 & ' = 10 ( ' )& *10, 000 + ,1.008 3.969 - * v1 + 10 ! 103 = (1.008 ! 107 )("0.015) + 3.696 !107 v1 7

F1x = (0.756 !10 )("0.015) + 1.008 !10 v1 (d)

or or

v1 = 0.0044 m

F1x = !69.4 kN

Support reactions:

" F2 x # $ !0.756 !1.008 % " 69 # &F & ' !1.008 !1.344 ( !0.015 & & " # & 92.1& & 2y & 7 ' ( ) * = 10 ) *=) * kN ' 0 ( + 0.0044 , & 0 0 & & F3 x & ' ( &+ F3 y &, &+!115.5&, !2.625 . - 0

______________________________________________________________________________________ SOLUTION (7.26) Equation (7.38):

" 12 6 L !12 6 L # $ 2 2% EI 6 L 4 L !6 L 2 L % [k ] = [ K ] = 3 $ 12 !6 L % L $ !12 !6 L $ 2 2% & 6 L 2 L !6 L 4 L '

(a)

(b)

Boundary conditions are v2 = ! 2 = 0 . Equation (7.45a) becomes

#" P $ EI % 12 6 L & # v1 $ ' (= 2 ) 2*' ( + 0 , L -6 L 4 L . +!1 , PL2 PL3 v1 = " Solving, !1 = 2 EI 3EI

Equation (7.45a):

Solving

where

" F1 y # $ 12 6 L !12 6 L % "! PL3 3EI # & &M & ' 2 2(& 2 & 1 & EI ' 6 L 4 L !6 L 2 L ( & PL 2 EI & ) *= 3 ) * 12 !6 L ( & 0 & F2 y & L ' !12 !6 L & ' 2 2( 0 + 6 L 2 L !6 L 4 L , -& -& M 2 .& .& " F1 y # " ! P # $M $ $ 0 $ $ 1$ $ $ % &=% & $ F2 y $ $ P $ '$ M 2 ($ $'! PL $( F2 y = R2 = P .

______________________________________________________________________________________ SOLUTION (7.27) Due to the symmetry only one-half of the beam need be considered. Referring to Case 3 of Table D.5, equivalent nodal forces obtained (see Fig. a). pL/4

Boundary conditions are v1 = ! 2 = 0

pL/4

We have M 1 = ! pL

pL /48 1

L 2 Figure (a) Equivalent nodal forces

pL2/48

L1 =

Inverting

$ pL2 48% EI " 4 L12 ' (= 3 ) & pL 4 + , L1 - &6 L1 $!1 % L3 "12 L2 & '= ( +v2 , 24 EI - 3 L

48 and

F2 y = ! pL 4 Equation (7.45a), with element of length L1 = L 2 .

&6 L1 # $!1 % 8 EI " L12 *' ( = 3 ) L - $3L1 12 . +v2 ,

$3L # %!1 & *' ( 12 . +v2 ,

3 L # $ pL2 48% ' )& 1 . + * pL 4 ,

"!1 # " $ pL3 24 EI # % &=% & 4 'v2 ( '$5 pL 384 EI (

This is the exact solution (See Table D.4) ______________________________________________________________________________________ SOLUTION (7.28) Due to symmetry, only one-half of the beam need be considered.

" 12 6 L !12 6 L # $ 2 2% EI 6 L 4 L !6 L 2 L % [k ]1 = 3 $ 12 !6 L % L $ !12 !6 L $ 2 2% & 6 L 2 L !6 L 4 L '

P/2

1 1

Therefore

k/2

! 0 # 3 EI kL EI [k ]2 = 3 # L # 0 # % 0

0 0 0 0

0" 0 $$ 0$ $ 0&

!12 6 L # " 12 6 L $ 6 L 4 L2 !6 L 2 L2 %% EI $ [K ] = 3 $ % kL3 L $ !12 !6 L 12 + !6 L % EI $ % 2 !6 L 4 L2 %' $& 6 L 2 L

______________________________________________________________________________________ 7.28 (CONT.) (a)

Boundary conditions are v1 = 0 and ! 2 = 0. Equation (7.45a) with F2 y = ! P 2 and M 1 = 0 :

" 4 L2 $6 L # %!1 & % 0 & EI ' 3( ) *= 3 ' kL ( )+v2 *, +$ P 2 , L $6 L 12 + 'EI (.

Introduce the data and solve:

v2 = "7.9338 mm

(b)

!1 = "2.9752 rad

!12 6 L # " 12 6 L $ F1 y % $ 0 % & 6 L 4 L2 2' ! 6 L 2 L ( 0 ( ( ' (!2.9752 (( !3 ( ( EI & 3 ') ) *= 3 & * (10 ) kL !6 L ' ( !7.9338( ( F2 y ( L & !12 !6 L 12 + EI & '( 0 ( (+ M 2 (, 2 , !6 L 4 L2 .' + -& 6 L 2 L ! F1 y " ! 5.20 kN " # # # # % F2 y & = % $3.80 kN & # # # # ( M 2 ) ($20.8 kN ' m )

Fspring = 200(7.9338) = 1.587 kN (C ) From symmetry: F1 y = F3 y = 5.20 kN ! ______________________________________________________________________________________ SOLUTION (7.29) P/2 PL/8

P/4

Figure (a). Equivalent Nodal forces.

PL/8

Boundary conditions are

v1 = 0

!1 = 0

We have

PL 8 P F1 y = F2 y = ! 2

M 2 = !M1 =

Equation (7.45a) reduces to

#" P 2 $ EI % 12 "6 L & # v1 $ ' (= 3 ) ( 2*' + PL 8 , L - "6 L 4 L . +! 2 , Inverting

% v2 & L # L2 3L $ %" P 2 & !#5 PL3 48 EI " ' (= (=$ % ) *' 2 +! 2 , 6 EI -3L 6 . + PL 8 , & # PL 8 EI ' ______________________________________________________________________________________ Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 7–25

______________________________________________________________________________________ SOLUTION (7.30)

EI EI 4 EI )2 = 3 = 3 L L 4L So, the stiffness materices [ k ]1 + [ k ]2 and [ K ] are the same as obtained in

We have (

EI EI )1 = 3 L L

(

and

Example 7.10. Boundary conditions are v1 = 0, !1 = 0, and v3 = 0. Equation (f) of Example 7.10 becomes

18 "30 & # " P $ # v2 $ % 28 L L2 ( ' ' ' ' 18 51 L " 39 L )) *3PL + *! 2 + = ( '! ' 276 EI ( "30 " 39 L 111 L ) ' 0 ' , 3. /, " 8 # " !2 # 3 PL3 $ $ PL $ $ = % 84 L & = % 21 L & 276 EI $ $ 69 EI $!12 L $ '! 48 L ( ' ( Substituting the given data:

# v2 $ # "8 $ # "1.936 mm $ (30 %103 )(1.2)3 & & & & & & '! 2 ( = '168 ( = ' 0.041 rad ( 3 &! & 69(207 % 10 )(15) &"96 & &"0.023 rad & ) 3* ) * ) *

______________________________________________________________________________________ SOLUTION (7.31) (a)

The three element stiffness are identical. Equation (7.46):

" 12 6 L !12 6 L # $ 2 2% EI $ 6 L 4 L !6 L 2 L % [k ]1 = [k ]2 = [k ]3 = 3 12 !6 L % L $ !12 !6 L $ 2 2% & 6 L 2 L !6 L 4 L '

(b) The beam stiffness matrix, [ K ] = [ k ]1 + [ k ]2 + [ k ]3 , is assembled as

!1 6L " 12 $ 6L 4 L2 $ $ !12 !6 L $ 6L 2 L2 EI [K ] = 3 $ 0 L $ 0 $ 0 $ 0 $ 0 0 $ 0 $& 0

!2 !12 6L !6 L 2 L2 24 0 0 8L2 !12 L !6 L 6L 2 L2 0 0 0 0

!3 0 0 0 0 !12 L 6L !6 L 2 L2 24 0 0 8L2 !12 !6 L 6L 2 L2

!4 0 0 # 0 0 %% 0 0 % % 0 0 % 6L % !12 % 2 L2 % !6 L 12 !6 L % % 4 L2 %' !6 L

______________________________________________________________________________________ 7.31 (CONT.) Boundary conditions are :

v2 = v3 = v4 = ! 4 = 0

System governing relations, from . (7.40):

# "P$ # v1 $ % 0% % % % % %!1 % % Fy 2 % %0% % % % % %M 2 % %! 2 % & ' = [K ] & ' % Fy 3 % %0% %M3 % %!3 % % % % % % Fy 4 % %0% %M % %0% ( ) ( 4)

______________________________________________________________________________________ SOLUTION (7.32) (a)

Use Eq.(7.45a):

v1 !1 v2 !2 " 12 6 L !12 6 L # $ 2 2% EI 6 L 4 L !6 L 2 L % [k ]1 = 3 $ 12 !6 L % L $ !12 !6 L $ 2 2% & 6 L 2 L !6 L 4 L ' v2

" 12 $ EI $ 6 L [k ]2 = 3 L $ !12 $ & 6L (b)

!2 v3 !3 6 L !12 6 L # 4 L2 !6 L 2 L2 %% 12 !6 L % !6 L % 2 2 L !6 L 4 L2 '

Assemble the global stiffness matrix of the beam: [ K ] = [ k ]1 + [ k ]2 . Then,

6 L "12 6 L 0 0 & # v1 $ # R1 $ % 12 'M ' ( 2 2 0 0 )) ''!1 '' ' 1' ( 6 L 4 L "6 L 2 L 0 "12 6 L ) '' v2 '' '' R2 '' EI ( "12 "6 L 24 * += 3 ( )* + 2 0 8L2 "6 L 2 L2 ) '! 2 ' 'M 2 ' L ( 6L 2L ' F3 y ' ( 0 0 "12 "6 L 12 "6 L ) ' v2 ' ' ' ( )' ' 0 6 L 2 L2 "6 L 4 L2 - .'!3 /' , 0 .' M 3 /'

(1)

______________________________________________________________________________________ 7.32 (CONT.) (c)

The boundary conditions are v1 = 0, !1 = 0, and v2 = 0 , Hence

#" P $ % 12 6 L ' ' EI ( 2 * 0 + = 3 (6 L 4 L ' 0 ' L (6 L 2 L2 , .

6 L & # v3 $ ' ' 2 L2 )) *!3 + 8 L2 )/ ',! 2 '-

Solving

v3 = " (d)

7 PL3 , 12 EI

!3 =

3PL2 , 4 EI

!2 =

PL2 4 EI

Introducing these equations into Eq. (1), after multiplying :

F3 y = ! P,

M 3 = 0,

R2 =

M 2 = ! PL,

3 R1 = ! P, 2

5 P 2

M1 =

(e)

1 PL 2 P ki

3 P

-3P/2 M PL/2

x PL

______________________________________________________________________________________ SOLUTION (7.33) Table P7.33 Data for the truss of Fig P7.33 Element Length( m) ! c s

1 2 3 4 5

7.5 6 4.5 4.5 4.5 2

36.9 o 0o 90 o 0o 135o

c2 cs s2 0.8 0.6 0.639 0.48 0.36 1 0 1 0 0 0 1 0 0 1 1 0 1 0 0 " 0.707 0.707 0.5 " 0.5 0.5

______________________________________________________________________________________ 7.33 (CONT.) Apply Eq.(7.38):

u1 v1 u2 v2 0.48 ' 0.639 ' 0.48 # u1 & 0.639 $ 0.48 0.361 ' 0.48 ' 0.361! v1 AE $ ! [k ]1 = 0.639 0.48 ! u2 7.5 $' 0.639 ' 0.48 $ ! 0.48 0.361" v 2 % ' 0.48 ' 0.361

u1 " 1 $ AE $ 0 [k ]2 = 6.0 $! 1 $ # 0

v1 u3 0 !1 0 0 0 1 0 0

u2 v 2 0 "0 $0 1 AE $ [k ]3 = 0 4.5 $0 $ #0 ! 1 u3 " 1 $ AE $ 0 [k ]4 = 4.5 $! 1 $ # 0

v3 0 % u1 0 ' v1 ' 0 ' u3 ' 0 & v3

u3 v3 0 0 % u2 0 ! 1 ' v2 ' 0 0 ' u3 ' 0 1 & v3

v3 u4 v 4 0 ! 1 0 % u3 0 0 0 ' v3 ' 0 1 0 ' u4 ' 0 0 0 & v4

u2 v2 u4 v4 0.5# u2 & 0.5 ' 0.5 ' 0.5 $ ' 0.5 0.5 0.5 ' 0.5! v2 AE $ ! [k ]5 = 0.5 0.5 ' 0.5! u4 4.5 2 $ ' 0.5 $ ! 0.5" v4 % 0.5 ' 0.5 ' 0.5

______________________________________________________________________________________ SOLUTION (7.34) Table P7.34 Data for the truss of Fig.P7.34

Element 1 2 3 4 5

! 0o 60 o 120 o 0o 60 o

c 1 0.5

0 1 0.866 0.25

0 0.433

0 0.75

" 0.5 0.866 0.25 " 0.433 0.75 1 0 1 0 0 0.5 0.866 0.25 0.433 0.75

( a ) Use Eq.(7.38):

u1 " 1 $ AE $ 0 [k ]1 = L $! 1 $ # 0

v1 u4 0 !1 0 0 0 1 0 0

v4 0 % u1 0 ' v1 ' 0 ' u4 ' 0 & v4

u1 v1 u2 v2 0.433 ! 0.25 ! 0.433% u1 " 0.25 $ 0.433 0.75 ! 0.433 ! 0.75 ' v1 AE $ ' [k ]2 = 0.25 0.433' u2 L $! 0.25 ! 0.433 $ ' 0.433 0.75 & v 2 #! 0.433 ! 0.75 u2 v2 u4 v4 0.433% u2 " 0.25 ! 0.433 ! 0.25 $! 0.433 0.75 0.433 ! 0.75 ' v 2 AE $ ' [k ]3 = 0.433 0.25 ! 0.433' u4 L $! 0.25 $ ' 0.75 & v 4 # 0.433 ! 0.75 ! 0.433 u2 " 1 $ AE $ 0 [k ]4 = L $! 1 $ # 0

v 2 u3 0 !1 0 0 0 1 0 0

v3 0 % u2 0 ' v2 ' 0 ' u3 ' 0 & v3

______________________________________________________________________________________ 7.34 (CONT.)

u3 v3 u4 v4 0.433 ! 0.25 ! 0.433% u3 " 0.25 $ 0.433 0.75 ! 0.433 ! 0.75 ' v3 AE $ ' [k ]5 = 0.25 0.433' u4 L $! 0.25 ! 0.433 $ ' 0.433 0.75 & v 4 #! 0.433 ! 0.75 ( b ) Global Stiffness Matrix [ K ] =

u1 v1 . 0.433 " 125 $ 0.433 0.75 $ $! 0.25 ! 0.433 $ AE $! 0.433 ! 0.75 0 L $ 0 $ 0 $ 0 $! 1 0 $ 0 $# 0

{F } = [ K ]{! };

u2 v2 u3 v3 ! 0.25 ! 0.433 0 0 ! 0.433 ! 0.75 0 0 15 . 0 !1 0 0 15 . 0 0 !1 0 125 . 0.433 0 0 0.433 0.75 ! 0.25 0.433 ! 0.25 ! 0.433 0.433 ! 0.75 ! 0.433 ! 0.75

u4 v4 !1 0 % u1 ' v 0 0 ' 1 ! 0.25 0.433' u2 ' 0.433 ! 0.75 ' v 2 ! 0.25 ! 0.433' u3 ' ! 0.433 ! 0.75 ' v3 'u 15 . 0 ' 4 . '& v 4 0 15

! R1x % !0% #R # #0# # 1y # # # # 0 # #u2 # # # # # #v 2 # # P # " & = [K ]" & # u3 # #Q# # v3 # # 0 # # # # # #0# # R4 x # #$ 0 #' # R4 y # $ '

______________________________________________________________________________________ SOLUTION (7.35) We have AE=30 MN. Table P7.35 Data for the truss of Fig.P7.35

Element 1 2

Length ! 5 5313 . o 4 0o

c s c2 cs s2 0.6 0.8 0.36 0.48 0.64 1 0 1 0 0

______________________________________________________________________________________ 7.35 (CONT.) ( a ) Use Eq.(7.38):

u1 v1 u2 v2 0.48 ! 0.36 ! 0.48% u1 " 0.36 $ 0.48 0.64 ! 0.48 ! 0.64' v1 AE $ ' [k ]1 = 0.36 0.48' u2 5 $! 0.36 ! 0.48 ' $ 0.48 0.64& v 2 #! 0.48 ! 0.64 u2 v 2 u3 v3 " 1 0 ! 1 0 % u2 $ 0 0 ' v2 AE $ 0 0 ' [k ]2 = 1 0 ' u3 4 $! 1 0 $ ' 0 0 & v3 # 0 0

( b ) Global Stiffness Matrix

u1 v1 u2 v2 u3 v3 0.096 ! 0.072 ! 0.096 0 0 % u1 " 0.072 $ 0.096 0128 . . 0 0 ' v1 ! 0.096 ! 0128 $ ' $! 0.072 ! 0.096 0.322 0.096 ! 0.25 0 ' u2 [ K ] = AE $ ' . 0.096 0128 . 0 0 ' v2 $! 0.096 ! 0128 $ 0 0 ! 0.250 0 0.25 0 ' u3 $ ' 0 0 0 0 0 & v3 # 0

( c ) Boundary Conditions: u1 = v1 = u3 = v3 = 0.

! F2 x $ '0.322 0.096* !u2 $ " % = AE ) " % . ,+ #v 2 & (0.096 0128 # F2 y &

' 3 + ! 0 $ ! 0.0010 $ !u2 $ 1 (4 " %= %=" %m * -" #v 2 & AE )' 3 10.063, #' 10,000& #' 0.0039& (d)

' F1 x $ ' 7632 $ . ( 0.072 ( 0.096+ !F ! ! ! , ( 0.096 ( 0.128) 0.0010 ' $ ! 10176 ! ! 1y ! , ) & #=& & # = AE #N 0 ) %( 0.0039 " !( 7500 ! , ( 0.250 ! F3 x ! , ) !% F3 y !" !% 0 !" 0 * - 0

______________________________________________________________________________________ 7.35 (CONT.) ( e ) Use Eq.(7.39)

F12 =

! 0.001 " AE [0.6 0.8]$#0.0039% = #15.12 kN (C ) 5 & '

F23 =

" !0.001# AE [1 0]$0.0039% = !7.5 kN (C ) 4 & '

______________________________________________________________________________________ SOLUTIONS (7.36 through 7.39) It is important to take into account any condition of symmetry which may exist. Use a twodimensional finite element program or ANSYS. ______________________________________________________________________________________ SOLUTION (7.40) ( a ) Refer to Fig. 7.25a: p !p

( 2 p j + pm ) 6

p !p

( 2 pm + p j ) 6

Q j = 12 p j h1t + 13 m 2 j h1t = h1t Qm = 12 p j h1t + 23 m 2 j h1t = h1t ( b ) Refer to Fig. 7.25b: 2 2 3P " xy = PQ It = 4 th 3 ( h ! y )

(a)

Substituting Eq. (a) into the given expression for Qm :

Qm = ym 1" y j !

3P 3 y j 4h

( h 2 " y 2 )( y " y j )dy 3

= ym 1! y j 34Ph [ ! y j ( y ! 3yh 2 )

ym yj

+ ( y2 ! 4yh 2 )

ym yj

]

This leads to the first of Eqs. (7.79). Similarly, ym

Q j = " # xy tdy ! Qm yj

(b)

Substituting Eq. (a) and Qm (from the first of Eqs. 7.79), Eq. (b) leads to the second of Eqs. (7.79).

______________________________________________________________________________________ SOLUTION (7.41) Qy3 3 1 cm

Qy4 4

(b)

1 cm

(a)

4 cm

x P=5 kN 2

Body force effects:

{Q}ba = 13 At{Q x1 , Q x 2 , Q x 3 , Q y1 , Q y 2 , Q y 3 }

= 13 ( 4 " 0.3){0, 0, 0, ! 0.077, ! 0.077, ! 0.077} N {Q}ba = {0, 0, 0, 0, ! 0.0308, ! 0.0308, ! 0.0308, ! 0.0308, 0} N

Similarly,

{Q}bb = {0, 0, 0, 0, 0, ! 0.0308, ! 0.0308, ! 0.0308} N

Hence, or

{Q}b = {Q}ba + {Q}bb

{Q}b = {0, 0, 0, 0, ! 0.0308, ! 0.0616, ! 0.0616, ! 0.0308} N

Effect of shear force, P: From Case Study 7.1,

{Q}P = {0, 0, 0, 0, ! 2500, ! 2500} N

Surface traction effects, p: Total load ( 4 ! 0.3)700 = 840 N is equally divided between nodes 3 and 4. Thus,

{Q}P = {0, 0, 0, 0, 0, 0, ! 420, ! 420} N Thermal strain effects: We have ! 0 = 12(10

Hence,

)50 = 600 µ and

& b1 1 $ [ B ]a = 0 2A $ $%a1

b2 0

b3 0

0 a1

0 a2

0# a3 ! ! b3 !"

&' 2 1$ = 0 8$ %$ ' 4

2 0 0

0 0 4

0 '4 '2

0 0 2

0# 4! ! 0 "!

{Q}ta = [ B ]Ta [ D ]{! 0 }( At )

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 7.41 (CONT.)

-. 2 + 2 + 1+ 0 = + 8+ 0 + 0 + , 0

0 0 0 .4 0 4

.4* 0( ( 0.3 0 * '600 µ $ -1 4 ( 2(10 7 ) + ! ! 0.3 1 0 ( &600 µ #(1.2) ( + ( . 2 ( 0.91 0 0.35() !% 0 !" +,0 2( ( 0)

After multiplication, this gives

{Q}ta = {!5142.85, 5142.85, 0, ! 10285.7, 0, 10285.7} N

{Q}ta = {!5142.85, 5142.85, 0, 0, ! 10285.7, 0, 10285.7, 0} N Similarly,

Thus,

or Hence, or

& b2 1 $ [ B ]b = 0 2A $ $%a 2

b4 0 a4

b3 0 a3

0 a2 b2

0 a4 b4

& 0 1$ = 0 8$ $% ' 4

2 0 4

'2 0 0

0 '4 0

0 4 2

0# a3 ! ! b3 !" 0# 0! ! ' 2 !"

{Q}t = [ B ]Ta [ D ]{! 0 }( At ) = {0, 5142.85, ! 5142.85, ! 10285.7, 10285.7, 0} N {Q}tb = {0, 0, ! 5142.85, 5142.85, 0, ! 10285.7, 0, 10285.7} N {Q}t = {Q}ta + {Q}tb

{Q}t = {!5142.85, 5142.85, ! 5142.85, 5142.85, ! 10285.7, ! 10285.7, 10285.7, 10285.7} N

The system nodal force matrix:

{Q} = {Q}b + {Q}P + {Q} p + {Q}t = {!5142.85, 5142.85, ! 5142.85, 5142.85, ! 10285.7, ! 12785.76, 986.64, 7365.67} N

System equation: where

{Q} = [ K ]{0, u2 , 0, u4 , 0, v 2 , 0, v4 }

[K ] is given by Eq. (n) of Case Study 7.1.

______________________________________________________________________________________ 7.41 (CONT.) Since we have only 4 unknown quantities u2 , u4 , v 2 , v 4 (and 8 equations are available), there are redundant equations. Examination of these system of equations shows that: [K ] is reduced by crossing out; row 1 and column 1 for u1 = 0, row 3 and column 3 for u3 = 0, row 5 and column 5 for v1 = 0, row 7 and column 7 for v3 = 0.

{Q} is reduced by crossing out Q x1 , Q x 3 , Q y1 , Q y 3 . for

u1 = u3 = v1 = v3 = 0 Thus, from the reduced equations, we obtain

0.180 0.252 0.247+ ' 5142.85 $ 'u2 $ .0.429 !u ! ,0.180 0.483 ( 0.256 ( 0.351) !! 5142.85 !! ! 4! (6 , )& & # = 10 # v ( 0 . 252 0 . 256 1 . 366 1 . 373 , ) !( 12785.76 ! ! 2! , ) !%v 4 !" 1.373 1.546* !% 7365.67 !" -0.247 ( 0.351 (u2 % ( 1728 % "u " " 4098 " " 4" " " !6 ' $=' $ 10 cm v ! 7302 2 " " " " "&v 4 "# "&! 6702 "#

______________________________________________________________________________________ SOLUTION (7.42) 3 1000 N/cm 1 1 cm (a) (b)

1 cm

4 cm

x P=4 kN 4 500 N/cm

We have 7

(10 ) n = 27.1(10 =3 6 ) 7

(10 ) m = 27.8(10 = 0.4 6 )

Then

0.3 0 # &3 7(10 6 ) $ [ D] = 0.3 1 0 ! 2 $ ! 1 ' 3(0.1) 0 0.39!" $%0 [ D * ] = t 4[ DA] = ( 2 716)[ D ]

0.3 0 # &3 = 0.13(10 6 ) $0.3 1 0 ! $ ! 0 0.39!" $%0

(a)

______________________________________________________________________________________ 7.42 (CONT.) Element a: Equations (7.68), with reference to the preceding figure yield

a i = a 2 = !4

bi = b2 = 0

a j = a3 = 0

b j = b3 = 2

a m = a1 = 4

bm = b1 = !2

(b)

Substituting Eqs. (a) and (b) into Eqs. (7.54), we obtain (in 10 ):

k uu ,11 = 0.13[3( 4) + 0.39(16) + 0] = 2.37 k uu ,12 = 0.13[0 + 0.39( !16) + 0] = !0.81 k uu ,13 = 0.13[3( !4) + 0 + 0] = !1.56 k uu , 21 = 0.13[0 + 0.39( !16) + 0] = !0.81 k uu , 22 = 0.13[0 + 0.39( !16) + 0] = 0.81 k uu , 23 = 0.13[0 + 0 + 0] = 0 k uu ,31 = 0.13[3( !4) + 0 + 0] = !1.56 k uu ,32 = 0.13[0 + 0 + 0] = 0 k uu ,33 = 0.13[3( 4) + 0 + 0] = 1.56 6

Similarly, remaining matrices are determined. Stiffness matrix for the element a (in 10 ) is:

0.31 0.41# & 2.37 ' 0.81 ' 1.56 ' 0.72 $ ' 0.81 0.81 0 0.41 0 ' 0.41! ! $ 0 1.56 0.31 ' 0.31 0 ! $ ' 1.56 [ k ]a = $ ! 0.41 0.31 2.28 ' 2.08 ' 0.20! $ ' 0.72 $ 0.31 0 ' 0.31 ' 2.08 2.08 0 ! ! $ 0 ' 0.20 0 0.20" % 0.41 ' 0.41

& 2.37 ' 0.81 ' 1.56 $ ' 0.81 0.81 0 $ 0 1.56 $ ' 1.56 $ 0 0 0 [k ]a = $ $ ' 0.72 0.41 0.31 $ 0 ' 0.31 $ 0.31 $ 0.41 ' 0.41 0 $ 0 0 % 0

0 0

0.72 0.41

0 0

0.31 0

0 0

2.28 ' 2.08

0 0

' 0.20 0

0.31 0.41 0 # 0 ' 0.41 0 ! ! ' 0.31 0 0! ! 0 0 0! ' 2.08 ' 0.20 0 ! ! 2.08 0 0! 0 0.20 0 ! ! 0 0 0"

______________________________________________________________________________________ 7.42 (CONT.) Element b: Equations (7.68), referring to the preceding figure, now give

a j = a2 = 0

bi = b2 = !2

a j = a 4 = !4

b j = b4 = 2

am = a3 = 4

bm = b3 = 0

(c)

Introducing Eqs. (a) and (c) into Eqs. (7.54), we obtain (in 10 ):

k uu , 22 = 0.13[3( 4) + 0 + 0] = 1.56 k uu , 23 = 0.13[0 + 0 + 0] = 0 k uu , 24 = 0.13[3( !4) + 0 + 0] = !1.56 The remaining terms are obtained in a like manner. The stiffness matrix for the 6

element b (in 10 ):

' 1.56 ' 0.31 0 0 0.31# & 1.56 $ 0 0.81 ' 0.81 ' 0.41 0 0.41! ! $ 2.37 0.41 0.31 ' 0.72! $ ' 1.56 ' 0.81 [k ]b = $ ! ' 0.41 0.41 0.2 0 ' 0.2 ! $ 0 $ ' 0.31 0 0.31 0 2.08 ' 2.08! ! $ 0.41 ' 0.72 ' 0.2 ' 2.08 2.28" % 0.31 or

&0 $0 $ $0 $ 0 [k ]b = $ $0 $ $0 $0 $ %0

0 1.56 0

0 0 0 0 ' 1.56 0 0.81 ' 0.81 0

' 1.56 ' 0.81 0 0 0 ' 0.41 ' 0.31 0.31

2.37 0 0 0 0.41 0

0 0.31 0 0.41 ' 0.72 0

0 0 0 ' 0.31 ' 0.41 0

0 # 0.31! ! 0.41! ! 0.41 0.31 ' 0.72! 0 0 0 ! ! 0.20 0 ' 0.20! 0 2.08 ' 2.08! ! ' 0.20 ' 2.08 2.28"

The system matrix is found by addition of the matrices of the elements a and b:

[ K ] = [ k ] a + [ k ]b 6

That is (in 10 ), (CONT.) ______________________________________________________________________________________ Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 7–38

______________________________________________________________________________________ 7.42 (CONT.)

0 ' 0.72 0.31 0.41 0 # & 2.37 ' 0.81 ' 1.56 $' 0.81 2.37 0 ' 1.56 0.41 0 0.72 0.31! ! $ 0 2.37 ' 0.81 0.31 ' 0.72 0 0.41! $ ' 1.56 ! $ 0 ' 1.56 ' 0.81 2.37 0 0.41 0.31 ' 0.72! $ [K ] = $' 0.72 0.41 0.31 0 2.28 ' 2.08 ' 0.20 0 ! ! $ 0 ' 0.72 0.41 ' 2.08 2.28 0 ' 0.20! $ 0.31 $ 0.41 ' 0.72 0 0.31 ' 0.20 0 2.28 ' 2.08! ! $ 0.31 0.41 ' 0.72 0 ' 0.20 ' 2.08 2.28" % 0 Nodal force matrix: Equations (7.78) lead to

Q j = Q x 4 = ! 2[ 2 ( 5006)+1000 ] = !667 N

Qm = Q x 3 = ! 2[ 2 (10006 )+500 ] = !833 N Due to the shear, we also have

Q y 3 = !2000 N ,

Thus,

Q y 4 = !2000 N

{Q} = {0, 0, ! 833, ! 667, 0, 0, ! 2000, ! 2000} N

Nodal displacement equation:

{! } = {0, 0, u3 , u4 , 0, 0, v3 , v 4 }

The system equation: The redundant equations are eliminated by crossing out the 1 st, 2nd , 5th , and 6th rows and columns. This leaves

0 0.41* 'u3 $ ' . 833 $ - 2.37 . 0.81 ! . 667 ! + . 0.81 2.37 0.31 . 0.72( !!u4 !! ! ! 6+ (& # 10 = & # . 2000 0 0 . 31 2 . 28 2 . 08 . + ( ! v3 ! ! ! + ( !%. 2000 !" 2.28) !% v4 !" , 0.41 . 0.72 . 2.08 Solving,

u3 = 0.001226 cm

u4 = !0.002324 cm

v3 = !0.013092 cm

v4 = !0.013742 cm

Stress in element a: The strain matrix is (CONT.) ______________________________________________________________________________________ Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 7–39

______________________________________________________________________________________ 7.42 (CONT.)

'/x $ - bi ! ! 1 + &/ y # = +0 !. ! 2 A +a % xy " , i

bj 0 aj

. 0 10 (6 , 0 = 8 , ,- ( 4

Then,

bm 0 am

2 0 0

0 ai bi

(2 0 4

' ui $ !u ! 0 *! j ! ( !u ! am ( & m # v bm () ! i ! !v j ! ! ! %vm "

0 aj bj

0 (4 0

0 0 2

' 0 $ ! 1226 ! ! 0 +! ! 0 ! ) 4 & # ) 0 ! ! ( 2 )* !( 13,092 ! ! ! % 0 "

' 307 $ ! ! = & 0 #µ !( 3273! % "

)" x & # # ( " y % = [ D ]{! }a #" # ' xy $ a

0.3 0 + ( 307 % .3 70(10 9 ) , " " = 0.3 1 0 ) ' 0 $10 !6 , ) 0.97 0 0.39)* "&! 3273"# ,-0 = {66.46, 6.65, ! 92.12} MPa Stress in element b: The strain matrix is

'0x $ .( 2 10 (6 , ! ! 0 &0 y # = 8 , !/ ! ,- 0 % xy " b

2 0

0 0

0 (4

= {!581, 325, ! 1661} µ

' 0 $ ! 2324 ! ! 0 +! 1226 ( ! ! 4 )& # ) 0 ! ! 0 *) !( 13,742 ! ! ! % 13,092 "

______________________________________________________________________________________ 7.42 (CONT.)

Thus,

'0 x $ 0.3 ( 581 + ' ( 581 $ .3 70(10 3 ) , ! ! ! ! 0.3 1 325 ) & 325 # &0 y # = ) 0.97 , !/ ! ! ! 0 0 1661 ( , xy *) %( 1661 " % " b

= {!118.75, 10.88, ! 46.75} MPa

______________________________________________________________________________________ SOLUTION (7.43) y 3

4 (a)

1 cm

(b) 4 cm

1 cm

Stiffness matrix of element a: Let i = 1, j = 4, m = 3. Then,

x1 = 0

x4 = 4

x3 = 0

y1 = !1

y4 = 4

y3 = 1

Equations (7.68) give

a1 = 0 ! 4 = !4 a4 = 0 ! 0 = 0 a3 = 4 ! 0 = 4

b1 = 1 ! 1 = 0 b4 = 1 + 1 = 2 b3 = !1 ! 1 = !2

We have

&3.33 0.99 0 # 10 6 $ [D ] = 0.99 3.3 0 ! $ ! 8 0 1.16!" $%0 *

Equations (7.76) in 10 are thus,

k uu ,11 = [0 + 1.16(16)] 8 = 2.32

k uu ,14 = [0 + 0 + 0] 8 = 0

k uu ,13 = [0 + 1.16( !16) + 0] 8 = !2.32 k uu , 43 = [3.3( !4) + 0 + 0] 8 = !1.65

k uu , 44 = [3.3( 4) + 0 + 0] 8 = 1.65 k uu ,33 = [3.3( 4) + 0 + 0] 8 = 3.97

These can be written as follows:

0 # & 2.32 ' 2.32 $ k uu = ' 2.32 3.97 ' 1.65!(106 ) $ ! 1.65!" ' 1.65 $% 0

______________________________________________________________________________________ 7.43 (CONT.) Similarly, we find submatrices k vv and k uv . In so doing and after assembling these matrices, 6

we obtain the stiffness matrix for the element a (in 10 ):

0 0 1.16 ' 1.16# & 2.32 ' 2.32 $' 2.32 3.97 ' 1.65 0.99 ' 2.15 1.16! $ ! ' 1.65 1.65 ' 0.99 0.99 0 ! $ 0 [k ]a = $ ! 0.99 ' 0.99 6.6 ' 6.6 0 ! $ 0 $ 1.16 ' 2.15 0.99 ' 6.6 7.18 ' 0.58! $ ! 1.16 0 0 ' 0.58 0.58" % ' 1.16

Stiffness matrix for element b: Let i = 1, j = 2, m = 4. Then,

a1 = 4 ! 4 = 0 a 2 = 0 ! 4 = !4 a4 = 4 ! 0 = 4

b1 = !1 ! 1 = !2 b2 = 1 + 1 = 2 b4 = !1 + 1 = 0 6

Then, Eqs. (7.54) yield (in 10 ):

k uu ,11 = [3.3( 4) + 0 + 0] 8 = 1.65 k uu ,12 = [3.3( !4) + 0 + 0] 8 = !1.65 k uu ,14 = [0 + 0 + 0] 8 = 0 k uu , 22 = [3.3( 4) + 1.16(16) + 0] 8 = 3.97 k uu , 24 = [0 + 1.16( !16) + 0] 8 = !2.32 k uu , 44 = [0 + 1.16(16) + 0] 8 = 2.32

0 # & 1.65 ' 1.65 $ k uu = ' 1.65 3.97 ' 2.32!(10 6 ) $ ! 2.32!" ' 2.32 $% 0

Similarly, we obtain submatrices k vv and k uv . In so doing and after assembling these matrices, 6

we determine the stiffness matrix for the element b (in 10 ):

0 0 0.99 ' 0.99# & 1.65 ' 1.65 $ ' 1.65 3.97 ' 2.32 1.16 ' 2.15 0.99! $ ! ' 2.32 2.32 ' 1.16 1.16 0 ! $ 0 [k ]b = $ ! 1.16 ' 1.16 0.58 ' 0.58 0 ! $ 0 $ 0.99 ' 2.15 1.16 ' 0.58 7.18 ' 6.6 ! $ ! 0.99 0 0 ' 6.6 6.6 " % ' 0.99

______________________________________________________________________________________ 7.43 (CONT.) Prior to addition: the 2nd and 6th rows and columns of zeros are added to the matrix [k ]a ; the 3rd and 7th rows and columns of zeros are added to the matrix [k ]b . The system matrix:

[ K ] = [ k ] a + [ k ]b 6

is then (in 10 ):

0 0 0.99 1.16 ' 2.15# & 3.97 ' 1.65 ' 2.32 $ ' 1.65 3.97 0 ' 2.32 1.16 ' 2.15 0 0.99! ! $ 0 3.97 ' 1.65 0.99 0 ' 2.15 1.16! $' 2.32 ! $ 0 ' 2.32 ' 1.65 3.97 ' 2.15 1.16 0.99 0 ! $ [K ] = $ 0 1.16 0.99 ' 2.15 7.18 ' 0.58 ' 6.6 0 ! ! $ 0 1.16 ' 0.58 7.18 0 ' 6.6 ! $ 0.99 ' 2.15 $ 1.66 0 ' 2.15 0.99 ' 6.6 0 7.18 ' 0.58! ! $ 0.99 1.16 0 0 ' 6.6 ' 0.58 7.18" % ' 2.15 The force-displacement relation, Eq. (q) of Case Study 7.1 becomes

0.99* 'u2 $ ' 0 $ - 3.97 . 2.32 . 2.15 ! 0 ! + . 2.32 3.97 1.16 0 ( !!u4 !! ! ! + (& # & #= . 1 . 2 2 . 15 1 . 16 7 . 18 6 . 6 . . + ( !v2 ! ! ! + ( !%. 2.5 !" , 0.99 0 7.18) !% v4 !" . 6.6 This yields

179.9 357.9 225.8+ ' 0 $ 'u2 $ .483.2 !u ! ,178.6 ( 246.8 429 251.7) !! 0 !! ! 4! (6 , )& & # = 10 # v ( ( 357 . 9 247 1546 1373 , ) !( 2.5! ! 2! , ) !%v 4 !" ( 1366 * !%( 2.5!" -255.8 ( 251.7 ( 1373

(! 1534.35 % " 1246.325" " " !6 =' $(10 ) cm ! 433 " " "& 17.125"# (CONT.) ______________________________________________________________________________________ Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 7–43

______________________________________________________________________________________ 7.43 (CONT.) Stresses in element b:

' u1 $ !u ! 2 '/x $ - b1 b2 b4 0 0 0 * ! ! u ! ! 1 + ! ! 0 0 0 a1 a 2 a 4 ( & 4 # &/ y # = ( v1 2A + !. ! ! ! a a a b b b + 2 4 1 2 4( , 1 ) !v ! % xy " b 2 ! ! %v4 " 0 $ ' !( 1534.35! ! 2 0 0 0 0 +! .( 2 ! 1246.325 ! 10 (6 , ) 0 0 0 0 (4 4 & = # ) 0 8 , ! ! 2 0 )* ,- 0 ( 4 4 ( 2 ! ( 433 ! ! ! % 17.125 " = {!383.6, 225, 1282} µ Thus,

'0 x $ 0.3 0 + ' ( 581 $ .1 200(103 ) , ! ! ! ! 0.3 1 0 ) & 325 # &0 y # = , ) 0.91 !/ ! 0 0.35)* !%( 1661 !" ,-0 % xy " b = {!118.75, 10.88, ! 46.75} MPa Stresses in element a:

'0x $ - 0 10 .6 + ! ! 0 &0 y # = + 8 !/ ! +, . 4 % xy " a

.2 0

2 0

0 .4

0 4

= {311.5, 0, 4.28} µ

0 ' $ ! ! 0 ! ! 0* 1246 . 325 ! ! 0 (& # ( 0 ! ! 2 )( ! ! 0 ! ! % 17.125 "

'/ x $ 0.3 0 * '311.5 $ -1 200(10 3 ) + ! ! ! ! 0.3 1 0 (& 0 # &/ y # = + ( 0.91 !. ! 0 0.35() !% 4.28!" +,0 % xy " a = {68.46, 20.54, 0.33} MPa End of Chapter 7

CHAPTER 8 SOLUTION (8.1) ( a ) From Eq. (8.13), we have 2

# " ,min = ba2 !pai2 (1 + bb2 ) = pi b22 !aa 2 Hence,

# " ,max = ba2 !pai2 (1 + ab2 ) = pi ba2 !+ab2 " ! , max " ! , min

= a2+a b2 = a +2(1a.21a ) = 1.105 2

( b ) Using Eq. (8.16),

# " ,max = !2 p0 b2b!a 2 2

# " ,min = ! p0 ba2 !+ab2 2

Thus,

# " , max # " , min

= b22!ba 2 = 2 (21..2121aa2) = 1.1 2

______________________________________________________________________________________ SOLUTION (8.2) Equation (8.20): 2

" z = bp2 i!aa 2 = 102.!60.p6i2 = 0.5625 pi = 140 pi = 248.9 MPa

Equation (8.13):

# " ,max = bb2 +!aa 2 pi = 112 +!00..662 pi = 2.12 pi = 140 2

pi = 65.9 MPa

Equation (8.10): 2

" max = bp2 i!ba 2 = 12 !10.62 pi = 1.5625 pi = 80 2

pi = 51.2 MPa = pall

______________________________________________________________________________________ SOLUTION (8.3) ( a ) Initial maximum tangential stress, from Eq. (8.13),

# " = pi bb2 +!aa 2 = nn 2 +!11 pi 2

pi = # " nn 2 +!11 2

After boring, denoting the inner radius by rx , we have 2 2

2 2

"$ # + $ # = pi nn 2aa 2 +! rrx2 = nn 2 +!11 $ # nn 2aa 2 +! rrx2 x

( "$ # + $ # )( n 2 + 1)( n 2 a 2 ! rx2 ) = ( n 2 ! 1)$ # ( n 2 a 2 + rx2 )

______________________________________________________________________________________ 8.3 (CONT.) or

( $" ! + " ! )( n 2 + 1)n 2 a 2 # ( n 2 # 1)" ! n 2 a 2 = rx2 [( $" ! + " ! )( n 2 + 1) + ( n 2 # 1)" ! ]

This gives the new inner radius in the form 2 2

2 2

rx = [ 2 n a#"" !(+n#2 +"1!)(+n2"+1n) n2 a ] 2 !

(b) 2

rx = [ 2 ( 4 )( 0.0250.1) ""!!( 5+)0+.12""!!((54))4 ( 0.025) ] 2

= 0.02712 m = 27.12 mm ______________________________________________________________________________________ SOLUTION (8.4) Using Eq. (8.13)

# " ,max = pi ba2 !+ab2 2

280 2

= 7 (b02.!6()0.+6b)2

Solving, b = 0.6308 m. Therefore,

t = b ! a = 630.8 ! 600 = 30.8 mm ______________________________________________________________________________________ SOLUTION (8.5)

( a ) Equation (8.13) and (8.16) give at r=a:

# " 1 = pi bb2 +!aa 2 , 2

Then,

# " 2 = !2 po b2b!a 2

" !1 = " ! 2 ;

pi bb2 +!aa 2 = 2 po b2b!a 2

pi po

= a 22 +bb2 = a22(+44aa )2 = 1.6 2

( b ) By neglecting the strain ! L in the longitudinal direction, Then,

# $ 1 = E1 (" $ 1 + !pi ),

# $ 2 = E1 (" $ 2 + !po )

"! 1 = "! 2 gives

" # 1 + !p i = " # 2 $ !p o

(a)

But

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 8.5(CONT.) 2

Hence,

= ba2 !+ab2 = a4 a+2 !4 aa 2 = 1.66

#" 1 pi

" ! 1 = 1.66 pi

(b)

We also have #" 2 po

= b22!ba 2 = 42a( 42 !aa )2 = 2.66 2

" ! 2 = 2.66 po

(c)

Substituting Eqs. (b) and (c) into (a) and letting ! = 1 3 : pi po

= 1.16

______________________________________________________________________________________ SOLUTION (8.6) Equation (8.14), substituting the given data yields

u = apEi ( ba2 "+ab2 + ! ) 2

6 ( 7"10 ) 0.6 + 0.6308 = 0.200 ( 0.63082 !0.62 + 0.3) "109 2

= 0.426 mm

______________________________________________________________________________________ SOLUTION (8.7) ( a ) We have " ! = u r , where u is defined by Eq. (8.14). Thus, at r=a:

" # ,max = pEi [ bb2 +$aa 2 + ! ] 2

(a)

( b ) Introducing ! " , ! r , and ! z from Eqs. (8.12), (8.13) and (8.20) into Hooke’s law we have

# $ = E1 [! $ % " (! r + ! z )] 2

" # ,max = pEi [ b +b(21$$a!2) a + ! ]

(b)

Substituting the data, Eq. (a): 6

60 (10 ) 4 + a 0.001 = 200 [ + 13 ] (109 ) 4 ! a 2 2

Solving, a = 1.41 m. Then,

t = 2 ! 1.41 = 0.59 m = t req. Similarly, Eq. (b) yields 2

60 (10 ) 4 + 2 a 3 0.001 = 200 [ 4!a 2 + 13 ] (109 )

a = 1.48 m; t = 0.52 m or ______________________________________________________________________________________ SOLUTION (8.8) Equation (8.13) at r=a: + 0.05 # " ,max = 00..25 (60) = 65 MPa 252 !0.052 2

______________________________________________________________________________________ 8.8 (CONT.) Equation (8.12) at r=a:

" r ,max = ! pi = !60 MPa

Equation (8.10): 2

( 0.25 ) " max = 060 = 62.5 MPa .252 !0.052

Equation (8.20) with po = 0 :

" z = 00.25.05!(060.05)2 = 2.5 MPa

Stress-strain relationship is given by

$ % = E1 [" % ! # (" r + " z )] = ur

(a)

Substituting Eqs. (8.12), (8.13), and (8.20), this expression results in at r=a:

u = E ( bap2 "i a 2 ) [(1 " 2! )a 2 + (1 + ! )b 2 ]

(P8.8)

Introducing the given numerical values, we obtain 6

( 60"10 ) u = 72"100.05 [0.4(0.05) 2 + 1.3(0.25) 2 ] 6 ( 0.252 !0.052 )

= 0.0571 mm The change in the internal diameter is therefore,

!d = 2u = 0.1142 mm

______________________________________________________________________________________ SOLUTION (8.9) Equation (8.19): 2

60 ) # " ,max = ! 20( .025.252 !) 0(.05 = !125 MPa

Equation (8.15) at r=b:

" r ,max = ! po = !60 MPa

Equation (8.20) for pi = 0 : 2

" z = ! 00.25.252 !(060.05)2 = !62.5 MPa

Equation (8.9) at r=a and pi = 0 : 2

( 0.25 ) " max = ! bp2 o!ba 2 = ! 060 = !62.5 MPa .252 !0.052

Substitution of Eqs. (8.15), (8.16), and (8.20), into Eq. (a) of Solution of Prob. 8.8 leads to for r=a: 2

b u = " E (apb2o"a 2 (2 " ! ) )

(P8.9)

)( 0.25 ) = ! 720."0510(960( 0".10 ( 2 ! 0.3) = !0.0738 mm 252 !0.052 )

"d = 2u = !0.1476 mm Thus, ______________________________________________________________________________________ SOLUTION (8.10) Given:

b = 0.1 m c = 0.3 m ! = 0.001(0.1) = 0.0001 m

a=0

Using Eq. (8.23):

______________________________________________________________________________________ 8.10 (CONT.) 9

p = Eb# c 2!cb2 = 200"100.(10.0001) 02.09( 0!.090.01 ) 2

= 88.89 MPa Then, letting p = pi = 88.89 MPa , Eq. (8.8) gives at r=b: 2

# " = 0 = b cp2!!cb2po + (cp2!!pbo2 )

After substituting the numerical values, this equation results in

po = 49.38 MPa

______________________________________________________________________________________ SOLUTION (8.11) ( a ) Using Eq. (8.18), # " , max pi

or Then, Hence,

= ba2 !+ab2 = 43

b = 2.65a b a +t a = a 2.65; t 2a

t = 1.65a

= 12.6aa = 0.825

( b ) Neglect longitudinal strain and consider

" r = ! pi = 6.3 MPa

" ! = 4 ( 63.3) = 8.4 MPa Therefore, for a = 0.075 m and b = 2.65a : 6

"10 )( 0.075 ) 4 !d = % d ( 2a ) = 2 Epi a [ ba2 #+ab2 + $ ] = 2 ( 6.3210 [ 3 + 13 ] (109 ) 2

= 7.4(10 !3 ) mm Alternatively, use Eq. (8.14) and let !d = 2u. ______________________________________________________________________________________ SOLUTION (8.12) Introducing various values of P, as given in Fig. 8.4, into Eqs. (a) and (8.21) of Sec. 8.3, it is seen that " ! and S values as shown in the figure are found. For example, let P = 1 or pi = po . Then,

# " = pi 1R!2R!1 + pi b 2 ( R12!!11) r ; 2

And

# " = ! pi

S = ""!!oi = 11++00 = 1

" !i = " !o

S =1

______________________________________________________________________________________ SOLUTION (8.13) ( a ) For % z = 0; [# z ! $ (# r + # " ) E = 0, using Eqs. (8.8), (CONT.) ______________________________________________________________________________________ Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 8–5

______________________________________________________________________________________ 8.13 (CONT.) 2

$ z = " ($ r + $ # ) = 2" ( ab2p!i !a b2 po ) ( b ) Similarly, for ! z = 0 : 2

% z = ! "E ($ r + $ # ) = ! 2" (Ea( bp2i!!ab2 )po ) ______________________________________________________________________________________ SOLUTION (8.14) At r=a: from Eq. (8.18),

# " ,max = 44 aa 2 +!aa 2 pi = 53 pi 2

and from Eq. (8.12),

" r ,max = ! pi .

Energy of distortion theory: 1

pi [( 53 ) 2 " ( 53 )( "1) + ( "1) 2 ] 2 = ! yp or

pi = 0.429! yp

Maximum shearing stress theory: 5 3

pi " ( " pi ) = ! yp

pi = 0.375! yp

______________________________________________________________________________________ SOLUTION (8.15) We have, at r=a:

# " ,max = bb2 +!aa 2 pi = 99 aa 2 +!aa 2 pi = 45 pi 2

" r ,max = ! pi ! u = 45 pi

(a) or And

pi = 0.8(350) = 280 MPa

! u = pi ,

pi = 350 MPa

( b ) Using Eq. (4.12a), 5 pi 4 ( 350 )

Solving,

! !630pi = 1

pi = 193.8 MPa

______________________________________________________________________________________ SOLUTION (8.16) From Eq. (8.18) for r=a: 2

25 ) !( 0.05 ) pi = 35 (( 00..25 = 32.31 MPa ) 2 + ( 0.05 ) 2

Total radial force on the contact surface is 2!aLp and total frictional force equals 2!apLµ . (CONT.) ______________________________________________________________________________________ Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 8–6

______________________________________________________________________________________ 8.16 (CONT.) Hence, Torque = 2!apLµ ( a )

= 2" (0.053 )(32.31 ! 10 6 )(0.2) = 5.073 kN ! m ______________________________________________________________________________________ SOLUTION (8.17) Let ! s1 , ! c1 , and ! s 2 , ! c 2 be initial and final compressive and tensile strains in the shaft and in the cylinder, respectively. Also, let " ! 1 , p1 and " ! 2 , p2 denote the initial and final maximum stresses and contact pressures, respectively. Then,

! s 1 + ! c1 = ! s 2 + ! c 2 Here,

(a)

# s1 = E1 ( p1 ! "p1 ) = 32E p1 # c1 = E1 (" $ 1 + !p1 ) = pE1 ( 2 + 13 ) = 37E p1 # s 2 = E1 [( p2 $ "p2 ) + "!paL2 ]

= E1 [ p2 (1 " 13 ) + 13 ! ( 04..055 )2 ] = E1 [ 23 p2 + 19093 .86 ]

From the condition of linearity, "! 2 "! 1

and

= pp12 ;

" ! 2 = p2 "p!11 = p2 2pp11 = 2 p2

# c 2 = E1 (" $ 2 + !p2 ) = E1 ( 2 p2 + !p2 ) = 73pE2

Equation (a) is thus, 2 p1 3E

from which Hence,

+ 73pE1 = E1 ( 23 p2 + 19093 .86 ) + 73pE2

p1 = p2 + 636.62

"p = p2 ! p1 = !636.62 kPa

______________________________________________________________________________________ SOLUTION (8.18) From Eq. (8.22),

a +b " = Ebpb ( bc2 #+bc2 + ! b ) + bp Es ( b2 #a 2 # ! s )

b[ ( 1 Eb

c 2 +b2 c 2 #b2

" 2 2 + ! b ) + E1s ( ba2 #+ab2 # ! s )]

Substituting the given data, this gives p = 9.01 MPa Stresses in the steel cylinder, using Eq. (8.16), 6

"10 )( 0.08 ) ($ # ) r =a = ! 2 ((90..01 = !24.03 MPa 08 ) 2 !( 0.04 ) 2

______________________________________________________________________________________ 8.18 (CONT.) 2

) + ( 0.04 ) (# " ) r =0.08 = !9.01(10 6 ) (( 00..08 08 ) 2 !( 0.04 ) 2

= !15.02 MPa Stresses in the brass cylinder, from Eq. (8.13): 2

) + ( 0.08 ) (# " ) r =0.08 = 9.01(106 ) (( 00..14 14 ) 2 !( 0.08 ) 2

= 17.75 MPa 6

"10 )( 0.08 ) ($ # ) r =0.14 = 2 ((90..01 = 8.74 MPa 14 ) 2 !( 0.08 ) 2

______________________________________________________________________________________ SOLUTION (8.19) ( a ) Using Eq. (P8.18),

( 0.5 2 )(10'3 ) & * 1'0.29 # 1 - 0.152 + 0.12 + 0.1$ + 0.33 (( + 9 + 2 2 9 ! ) 200 (10 ) "! %$ 72 (10 ) , 0.15 '0.1

= 56.49 MPa ( b ) From Eq. (8.14), for r=b: 2

"10 ) u = E 2( ba2 !bpai2 ) = 2 ( 072.1") 10( 09.(150.)(15562 !.49 0.12 )

= 0.1883 mm ! = 2u = 0.3766 mm ______________________________________________________________________________________ SOLUTION (8.20) ( a ) Using Eq. (8.14), we have at r=a: "0 a

= E ( bpa [(1 # ! )a 2 + (1 + ! )b 2 ] 2 #a 2 )

from which 2

p = a 2 [(1"#0!E)(ab2 +#(a1+!) ) b2 ]

(P8.20)

( b ) Substitution of Eq. (P8.20) into Eqs. (8.12) result in

$ r = (1!" ) a#2 0+E(1+" ) b2 (1 ! br 2 ) 2

% $ = (1!" ) a#2 0+E(1+" ) b2 (1 + br 2 ) 2

______________________________________________________________________________________ SOLUTION (8.21) Equation (8.14) at r=b:

u = #20 = bpE ( bc2 !+bc2 + " ) = bpE ( b2.25+ 2b.225!bb2 + 13 ) 2

Solving, p = 5.860b

Then, Eq. (8.17) yields at r=b:

u = bpE (1 # " ) = 3(25!.860 ) = 0.114! 0 Hence,

"d = 2u = 0.23! 0

______________________________________________________________________________________ SOLUTION (8.22) ( a ) Intial difference in diameter: where,

Thus,

" = 2b(! 1 + ! 2 )

! 1 =tangential (comp.) strain in the shaft ! 2 =tangential (tens.) strain in the cylinder % = 2Eb [( p $ !p ) + (" # ,max + !p )] = 2Eb [(" ! ,max + p ) = 2Eb ( 2 p + p ) = 6Epb

( b ) Compressive (uniform) strain ! L , due to axial load P, is

" L = !bP2 E

We now have

# 1 = E1 ( p1 $ "p1 $ !"bP2 ) # 2 = E1 (" $ 1,max + !p1 )

(comp.) (tens.)

where " ! 1,max is the increased tangential stress. Thus,

'1 = 2b(% 1 + % 2 ) = 2Eb ($ # 1,max + p1 & !"bP2 ) Based on the linearity condition: or

" ! 1, max p1

" ! , max P

" ! 1,max = 2 p1

Setting ! = ! 1 6 pb E

= 2Eb ( 2 p1 + p1 " 3!Pb2 )

P = 9"b 2 ( p1 ! p )

______________________________________________________________________________________ SOLUTION (8.23) Radial strain is

! = ! b,comp + ! s ,tens. = E1b ( p $ !p ) + E1s (" # ,max + !p ) We also have

" = (T2 # T1 )(! b # ! s ) = !T (19.5 " 11.7)10 "6 = 7.8(10 "6 ) !T

Note that from Eq. (8.18) at r=a: # " , max p

= 44bb2 +!bb2 = 53

______________________________________________________________________________________ 8.23 (CONT.) Thus,

7.8(10 !6 ) "T = E1b ( p ! 3p ) + E1s ( 53p + 3p )

p = 11( E.7 (+!3TE) E)10b E6s s

Hence, Eb E s ( T2 !T1 ) # " ,max = 1.95 ( E + 3 E )105 s

______________________________________________________________________________________ SOLUTION (8.24) Axial Stress, Eq. (8.20):

!z =

pi a 2 0.82 pi = = 0.8 pi = 80 , b 2 " a 2 1.22 " 0.82

pi = 100 MPa

Tangential stress, Eq. (8.13):

(" ! ) max =

b2 + a 2 1.22 + 0.82 p = pi = 2.6 pi = 80 , i b2 # a 2 1.22 # 0.82

pi = 30.8 MPa

Shear Stress, Eq. (8.10):

Solving

pi b 2 1.22 ! max = 2 = pi = 1.8 pi = 50 b " a 2 1.22 " 0.82 pi = 27.8 MPa = pall

______________________________________________________________________________________ SOLUTION (8.25) The maximum radial displacement umax occurs at the inner edge of the tank.

So, Eq. (8.14) for r=a, results in

umax =

api a 2 + b2 =( 2 +! ) E b " a2

Introducing the given data

umax =

0.5(60 !106 ) 0.52 + 0.82 ( 2 + 0.3) = 1.107(10!3 ) m = 1.12 mm 9 2 70(10 ) 0.8 " 0.5

______________________________________________________________________________________ SOLUTION (8.26) Equation (8.18) and Eq.(8.12) at r=a:

" 1 = " # ,max = pi ba2 !+ab2 = 45 pi 2

pi " (" pi ) = 350,

(a)

! 1 " ! 2 = ! yp ;

(b)

2 ! 12 " ! 1! 2 + ! 22 = ! yp

5 4

" 2 = " r ,max = ! pi

pi [( 54 ) 2 ! ( 54 )(!1) + (!1) 2 ] 2 = 350,

pi = 155.6 MPa

pi = 179.3 MPa

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (8.27) 2

!" ,max = pi bb2 +#aa 2 = 53 pi = !1 , ! r ,max = # pi = ! 2 (a)

! u = 53 pi , and

(b)

!1 !u

pi = 53 (320) = 192 MPa

! u = pi ,

pi = 320 MPa

5 pi 3(320)

" !!u2' = 1;

(governs)

" "620pi = 1

pi = 146.8 MPa

______________________________________________________________________________________ SOLUTION (8.28) We have a=20 mm, b=30 mm, c=60 mm. Equation (8.22): p 60 + 30 + 20 0.05 = 21030!10 + 0.3] + 10530!10p 9 [ 30 " 0.3], 9 [ 602 "302 302 " 202 2

p = 53.3 MPa

Steel, Eq.(8.18): 2

+ 30 " ! ,max = p cc2 +#bb2 = 53.3 60 = 88.8 MPa 602 #302

Bronze, Eq.(8.19): 2

" ! ,max = #2 p b2b# a2 = #2(53.3) 30230# 202 = #191.9 MPa ______________________________________________________________________________________ SOLUTION (8.29) ( a ) Use Eq.(8.18) with pi = p, 2

a = b, b = c:

" ! ,max = p cc2 +#bb2 = 80 MPa, By Eq.(8.22), with a = 0,

c +b c 2 #b2

b = 50,

(Fig. 8.6) 80(106 ) p

(a)

! = 0.06 mm :

p p 80!10 0.06 = 10050!10 + 0.3] + 20050!10 9 [ 9 [1 " 0.3] p 6

60 !106 = 40 !106 + 0.15 p + 0.175 p,

p = 61.5 MPa

( b ) Equation (a) becomes 80 61.5

0.005 = cc2 +!0.005 2 ,

c = 138 mm

2c = 276 mm

______________________________________________________________________________________ SOLUTION (8.30) Equation (8.22) with a=0, b=50 mm, c=140 mm. bp

! = Ec [ bc2 #+bc2 + " c ] + E s [1 # " s ]

______________________________________________________________________________________ 8.30 (CONT.) p p 140 + 50 0.04 = 12050!10 + 0.25] + 21050!10 9 [ 9 (0.7) 1402 "502 2

Solving p = 49.4 MPa

Shaft: " ! = " r = # p = #49.4 MPa Cylinder: 2

+ 50 " ! ,max = p cc2 +#bb2 = 49.4[ 140 ] = 63.8 MPa 1402 #502

! r ,max = " p = "49.4 MPa ______________________________________________________________________________________ SOLUTION (8.31)

! o from Equation (8.22): ! 2

bp = (ud ) r =b = bpE [ bc2 #+bc2 +" ] = bpE ( 17 15 + 0.3) = 1.433 E 2

E! p = 2.866 b

Then, ! i from Eq.(8.22) with a=0, 0.7 ! us = bpE (1 #" ) = 2.866 = 0.244!

Therefore,

"d s = 0.489!

______________________________________________________________________________________ SOLUTION (8.32) We have (Fig. 8.6):

a = 0.05 m

b = 0.1 m

c = 0.15 m

Maximum tangential stress occurs at r=0.1 m in gear wheel:

" max = p cc2 +!bb2 2

where, p is the internal pressure exerted by the contact surfaces. Substituting the given data into this equation, we have + 0.1 0.21 = p 00..15 152 !0.12 2

Solving,

p = 0.081 MPa = 81 kPa

This interface pressure produces the torque at the contact surface. Area of contact is

2!bL = 2! (0.1)(0.1) = 0.02!

The torque transmitted is thus,

T = [(0.2)(81 " 10 3 )(0.02! )](0.1) = 101.79 N ! m

______________________________________________________________________________________ SOLUTION (8.33) From the torsion, formula, ! = Tb (" b 3

2) :

b T = ! "2b = ! (120) = 188.5b3 2

Also, Eq.(8.35b):

T = 2!b 2 fpl = 2!b 2 fp(3b) = 6!fb 3 p

And

6! fb3 p = 188.5b3 , We have E h = E s = E , 2 bpc

p = 6188.5 ! (0.2) = 50 MPa

! h = ! s = ! , and a=0. Equation (8.22) becomes 2

2 bp ( 4 b )

8 pb

! = E ( c 2 "b 2 ) = E ( 4 b 2 "b 2 ) = 3 E Hence

8(50) b ! = 3(200 = 0.6667(10"3 )b #103 )

______________________________________________________________________________________ SOLUTION (8.34) ( a ) a=10 mm, b=40 mm

120 F! = Ta = 0.01 = 12 kN

Thus or

12(103 ) = 2! afpt = 2! (0.01)(0.16)(0.04) p p = 29.84 MPa

(Eq.8.35a)

Equation (8.22) gives then

! = 10(20.84) [ 50 +10 + 0.3] + 10(29.84) (1 " 0.3) = (4.128 + 1.044)10"3 = 0.005 mm 100(103 ) 502 "102 200#103 2

(b)

+10 " ! ,max = p bb2 +# aa2 = 29.84[ 50 ] = 32.3 MPa 502 #102

______________________________________________________________________________________ SOLUTION (8.35) From Eq. (8.28a), "& r "r

= 3+8% #$ 2 [b 2 + a 2 ! ""r ( r 2 ) ! a 2 b 2 ""r ( r12 )] = 0

2 a 2b2 r3

! 2r = 0;

r = ab

This value of r is substituted into Eq. (8.28a) to yield

$ r ,max = "# 2 (b ! a ) 2

We also find from Eq. (8.28b), for r=a:

% " ,max = 2 3+8! #$ 2 (b 2 + 13&+!! )

Thus, # $ , max # r , max

1!" 2 a ) 3+" 2 ( b!a )

2 ( b2 +

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (8.36) At r=0: $ % = $ r = (3 + # )b 2

!" 2 8 . Thus,

2 ! " # ! r! " + ! r2 = ! yp

$ all = b1

8# yp ( 3+" ) !

(P8.36) 1 2

!10 ) 1 = 0.125 [ 108((2260 ] = 3845.9 rad sec .7!103 ) 3

= 36,726 rpm ______________________________________________________________________________________ SOLUTION (8.37) We have at inner edge:

$ max = # "2!0

" ! = 2" ! = 2(90) = 180 MPa Equation (8.28b) with r = 0.03 m is thus 180(106 ) = 108 3 [(0.03) 2 + (0.1) " 101+13 (0.03) 2 + (0.1) 2 ](7800)! 2 or

from which

! = 525.8 rad s = 5021 rpm

______________________________________________________________________________________ SOLUTION (8.38) ( a ) When p = 0, at interface,

ud ! u s = 0.05(10 !3 )

(a)

Use Eq. (8.28b) with r=a: 2

(" % ) d = #$4 [(3 + ! )b 2 + (1 & ! )a 2 ] 3

= 7.8(104 )! [3.3(0.1252 ) + 0.7(0.0252 )] = 101.473! 2 Radial displacement of disk (with ! r = 0 ): ! a (ud ) r =a = (" #E)d a = 101200.473 (109 ) 2

For shaft, using Eq. (8.30b) with r=a: 2

(" # ) s = 3.3( 08.025) (1 $ 13..93 )7.8(103 )! 2 = 0.8531! 2 Radial displacement of shaft (with ! r = 0 ): 2

! ( 0.025 ) (u s ) r =a = (" #E) s a = 0.8531 200 (106 )

Thus, Eq. (a) gives 101.473" 2 ( 0.025 ) 200 (106 )

" ( 0.025 ) ! 0.8531 = 0.05 200 (106 )

! = 1,933.884 rad sec = 19,040 rpm

( b ) Maximum stress occurs in disk

(" ! ) max = 101.473(1,993.884) 2 = 403 MPa

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (8.39) We have

a = 0.5c

b = 2c

r = ab = c

( a ) Equation (8.28a):

50(10 6 ) = 3.38c (0.25 + 4 " 1 " 12 )7.8(10 3 )(5000 260! ) 2 2

or Thus,

c = 0.1587 m t r = 1.5c = 238.1 mm

( b ) Equation (8.28b) at r=a:

# " ,max = 3.38c [0.25 + 4 $ 13..93 (0.25) + 4]7.8(10 3 )(5000 260! ) 2 = 180.1 MPa 2

______________________________________________________________________________________ SOLUTION (8.40) ( a ) Equation (8.23), with a=0: 9

0.000075 ) 0.16!0.005625 p = 210(102 ( 0)(.075 = 101.31 MPa ) 0.16

Then, Eq. (8.18) gives

+ 0.005625 # " ,max = 101.31(10 6 ) 00..16 16!0.005625 = 108.68 MPa

( b ) Applying Eq. (8.28c),

.3( 0.7 ) 0.000075 = 8(3210 [0.005625 + 0.16 ! 13..33 (0.005625) "109 )

Solving,

+ 10..37 (0.16)](7.8 " 10 3 )(0.075)! 2

! = 450 rad sec = 4300 rpm

______________________________________________________________________________________ SOLUTION (8.41) From Eqs. (8.30a) and (8.30b), for r=0 and ! = 1 3 :

# max = 3+8$ ! ("b) 2 = 125 !v 2 ______________________________________________________________________________________ SOLUTION (8.42) Apply Eq. (8.28c) with u = 0.04 mm at r = 25 mm . Substituting the given data:

.3( 0.7 ) 0.04(10 !3 ) = 8(3210 [(0.025) 2 + (0.25) 2 ! 13..33 (0.025) 2 "109 )

Solving,

1.3 0.7

(0.25) 2 ](7800)! 2 (0.025)

! = 913.1 rad s = 8719 rpm

______________________________________________________________________________________ SOLUTION (8.43) We have ti to

= ( ab ) s ;

0.125 0.0625

s = ( 00..625 125 )

Solving, s=0.431. Then, 1

0.431 2 2 m1, 2 = " 0.431 2 ± [( 2 ) + (1 + 0.3 ! 0.431)]

m1 = 0.869

m2 = !1.3

Equation (8.39) gives

# r = ct11 r 0.3 + ct12 r $1.869 $ 0.50169 " (!r ) 2

Given conditions are:

(! r ) r =0.625 = 0,

(a)

(! r ) r =0.125 = 0

Thus, substituting the given data into Eq. (a) and solving: c1 t1

= 0.2322 !" 2 ,

c2 t1

= #0.0024 !" 2

The second of Eqs. (8.39), at the bore r=0.125 m, leads to

140(10 6 ) = 0.2322(0.1250.3 )(0.869) !" 2

) 2 + ( #0.0024)( #1.3)(0.125#1.869 ) !" 2 # (81#+(03.9.3)()(00.125 .431) !"

!" 2 = 549(10 6 )

It follows, from the second of Eqs. (8.39), that

(" ! ) r =0.625 = [0.2322(0.6250.3 )(0.869) 2

) 6 " 0.0024( "1.3)(0.625"1.869 ) " 81".93(.30(.625 0.431) ](549 ! 10 )

= 37.5 MPa Circumferential force is thus

37,500(0.0625) = 2344 kN m

______________________________________________________________________________________ SOLUTION (8.44) ( a ) Uniform thickness

Centrifugal force due to the blades, m! r , is 540 9.81

000! 2 ( 10,60 ) (0.575) = 8.677(10 6 ) N

The pressure at b is then 6

(10 ) po = 2!8.(667 0.5 )( 0.05 ) = 55.24 MPa

We have

000" 2 #$ 2 = 7.8(10 !3 )( 10,60 ) = 2.1384(10 3 )

The condition of zero pressure at the bore is satisfied by, using Eq. (8.27b): 2

2 2

(" r ) r =a = 0 = 1%E! 2 [ % ( 3+! )(1%8!E ) #$ a + (1 + ! )c1 % (1 % ! ) ac22 ]

______________________________________________________________________________________ 8.44 (CONT.) 2

0 = $ ( 3+! )8a "# + [ E1($1!+!2 ) c1 ] + [ $ E1($1!$!2 ) c2 ] a12 1 424 3 14243

0 = A1 + A2 a12 $

( 3+# ) a 2 !" 2 8

Substituting the data, we have

0 = A1 + 256 A2 ! 3.4456

(a )

The condition at the outer circumference is satisfied by: 2

(" r ) r =b = 55.24 = A1 + ( 0A.52)2 ! 3.3( 0.5) 8( 2138.4 ) or

55.24 = A1 + 4 A2 ! 220.52

(b)

A1 = 280.08

(c)

Solution of Eqs. (a) and (b) gives

A2 = !1.08

Equation (8.27c) is also written as follows 2 2

% $ = A1 & A2 r12 & (1+3# )8!" r

(d)

where A1 and A2 are given by (c). At r=a, we have from Eq. (d):

(# " ) r =a = 280.08 + 276.48 ! 1.98 = 554.58 MPa = ! max

Similarly, Eq. (d) gives r=b:

(# " ) r =b = 280.08 + 4.32 ! 126.97 = 157.43 MPa

Note that ! r ,max occurs at r =

ab = 0.1768 m. Thus Eq. (8.27):

(" r ) r =0.1768 = !27.67 + 280 ! 34.55 = 217.88 MPa ( b ) Hyperbolic section We have Then,

0.4 0.05

.5 = ( 0.00625 )s ;

s =1 1

m1, 2 = ! 12 ± [0.25 + 1.3] 2 or Letting

m1 = 0.745 c1 t1

= B1

m2 = 1.745 c2 t1

= B2

Eq. (8.39) becomes then 2

2 $ r = B1r m1 + s %1 + B2 r m2 + s %1 % (83%+(!3)+"# ! )s r

______________________________________________________________________________________ 8.44 (CONT.)

(! r ) r =a = 0

and

= B1 (0.0625) 0.745 + B2 (0.0625) #1.745 # 0.7021(0.0625) 2 !" 2 or Also

0 = B1 + 996.35B2 # 0.0216 !" 2

(e)

(# r ) r =b = 55.24 $ B1 (0.5) 0.745 + B2 (0.5) $1.745 $ 0.702(0.5) 2 !" 2

55.24 = 0.55967 B1 + 3.352 B2 # 0.1755!" 2

(f)

Substituting the given data and solving Eqs. (e) and (f):

B1 = 725.4

B2 = !0.6814

(g)

The second of Eqs. (8.33), 2

% $ = B1m1r m1 + s &1 + B2 m2 r m2 + s &1 & (18+&3(!3+)!"#) s r 2 gives at r=a:

(" ! ) r =a = 725.14(0.745)(0.0625) 0.745

# 0.6814( #1.745)(0.0625) #1.745 # 0.4043(0.0625) 2 !" 2 or

(! " ) r =a = 215.2 MPa = ! max

Also similarly we obtain that

(" ! ) r =b = 110.2 MPa

( c ) Uniform stress Using Eq. (8.41), where,

to t1

= e $ ( "# 2! ) b

Thus,

.59 ) ( 0.5 ) " ( $% 2 2# )b 2 = "7.8!10 (2523 ( 84 ) = !3.182

to = 0.02425e !3.182 = 0.001 m = 1 mm

and

t = 0.02425e !12.728 r

______________________________________________________________________________________ SOLUTION (8.45) Because u=0 at r=b, set c2 = 0 in Eq. (8.43). Thus, Eq. (8.43) with c2 = 0, T =constant, and u=0 at r=b: b

0 = c1b + (1+b# ) "T ! rdr 0

c1 = # (1+"2)!T

______________________________________________________________________________________ 8.45 (CONT.) Substituting this into Eq. (d) and (e) of Sec. 8.10, we obtain 2 #ET " r = $ #ET 2 $ 2 (1$! 2 ) (1 + ! )

2 "ET # $ = "ET 2 % "ET % 2 (1%! 2 ) (1 + ! )

Letting 1 " !

= (1 + ! )(1 " ! ) and simplifying we obtain $ r = $ # = % "1ET %!

______________________________________________________________________________________ SOLUTION (8.46) We substitute the given T into Eqs. (8.44) to obtain 2

E (Ta #Tb ) a ( r #b ) b b " r = !2ln( b a ) [ # ln a # r 2 ( b 2 # a 2 ) ln a ]

Then, $" r $r

! (Ta #Tb ) d b a d b b = 0 = E2ln( b a ) { dr (ln a ) # b 2 # a 2 [1 # dr ( r 2 )]ln a } 2

yields, after differentiation: 1

r = ab( b2 !2a 2 ln ab ) 2

It is noted that, Eq. (8.53) gives the same result. ______________________________________________________________________________________ SOLUTION (8.47) Using Eq. P8.46, we have 1

1.5 ) r = 0.01(0.015)[ ( 0.0152 ln( ] 2 = 12.1 mm ) 2 !( 0.01) 2

Equation (8.53) are therefore !6

(12.1 !15 ) (# r ) r =12.1 = 10.4 (102 ( 0.7)()90ln("110.5))( !8) [ ! ln 1215.1 ! 100 ln(1.5)] 12.12 (152 !102 )

= 1.319(10 7 )[0.041] = 0.541 MPa 2

(10 +15 ) (" r ) r =10 = 1.319(10 7 )[1 ! ln(1.5) ! 100 ln(1.5)] 100 (152 !102 )

= 6.045 MPa 2

100 (15 +15 ) (# " ) r =15 = 1.319(10 7 )[1 ! ln(1) ! 100 ln(1.5)] (152 !102 )

= !4.64 MPa 2

100 (12.1 +15 ) (# " ) r =12.1 = 1.319(10 7 )[1 ! ln( 1215.1 ) ! 12 ln(1.5)] .52 (152 !102 )

= 0.488 MPa Similarly, 2

(" z ) r =10 = !1.319(10 7 )[1 ! 2 ln(1.5) ! 152 (210!10)2 ln(1.5)] = 6.055 MPa = ! max

(" z ) r =15 = !1.319(107 )[1 ! 2 ln(1) ! 152 (210!10)2 ln(1.5)]

= !4.64 MPa

______________________________________________________________________________________ SOLUTION (8.48) Introducing

T ( r ) = T0 (b ! r ) b

into Eqs. (8.45), after integration, we obtain the following expressions for stresses

" r = 13 T0 ( br # 1)!E " # = 13 T0 ( 2br $ 1)!E

From these we observe that, at r=b: the radial stress vanishes while tangential stress assumes its maximum value. ______________________________________________________________________________________ SOLUTION (8.49) z 3

p=1.4 kN/cm

1 cm 0 1 cm

20 cm

4 cm

Equivalent nodal force matrix We have

r1 = 0 r2 = 4 r3 = 0

z1 = !1 z 2 = !1 z3 = 1

r =4 3 z = !1 3

Weight of the element is

2!r A(0.077) = 2! ( 4 3)( 4)(0.077) = 2.58 N

Body forces:

{Q}e = {Qr1 , Qr 2 , Qr 3 , Q z1 , Q z 2 , Q z 3 }

Therefore,

{Q}be = {0, 0, 0, ! 0.86, ! 0.86, ! 0.86} N

Surface forces:

q2 = q3 = 14002 20 = 3130.4 N cm qr 2 = qr 3 = !3130.4( 220 ) = !1400 N cm q z 2 = q z 3 = !3130.4( 420 ) = !2800 N cm Since Then,

Qr = 2!r qr

Q z = 2!r q z

Qr 2 = Qr 3 = !11.729 kN Q z 2 = Q z 3 = !23.457 kN

______________________________________________________________________________________ 8.49 (CONT.) Therefore,

{Q}ep = {0, ! 11.729, ! 11.729, 0, ! 23.457, ! 23.457} kN

Thermal forces:

{Q}te = 2"r A[ B ]T [ D ][ B ]{! 0 }

Here,

&0.3 $ 6 $0.7 [ D ] = 38.46(10 ) $0.3 $ %0

We also have

ai = a1 = 4

b1 = !2

a j = a2 = 0

b2 = 2

am = a3 = 4

b3 = 0

0.7 0.3 0 # 0.3 0.3 0 ! ! 0.3 0.7 0 ! ! 0 0 0.2"

c1 = !4

d 1 = ar1 + b1 + c1rz = 2

c2 = 0 c3 = 4

d2 = 2 d3 = 2

Therefore,

&' 2 $ 1$ 0 [B ] = 8$ 2 $ %' 4

2 0 2 0

0 0 2 4

0 '4 0 '2

0 0 0 2

0# 4! ! 0! ! 0"

It follows that

'0.0006$ !0.0006! ! ! {Q}te = 2( ( 43 )4[ B ]T [ D ][ B ]& # !0.0006! !% 0 !"

= {0, 502.3, 251.2, ! 502.3, 0, 502.3} kN

Equivalent nodal force matrix is thus,

{Q}e = {Q}be + {Q}ep + {Q}te b

where {Q}e , {Q}e , {Q}e are already obtained. Element stiffness matrix:

[k ]e = 2!r A[ B ]T [ D ][ B ] = 2! ( 43 )4[ B ]T [ D ][ B ]

______________________________________________________________________________________ 8.49 (CONT.) Substituting the values of [D ] and [B ], after carrying out the matrix matrix multiplications, we obtain:

1.6 ( 1.6 ' 4.8 % 1.6 6.4 4 % 4 6 % ( 1.6 [ k ]e = % (4 % 4.8 ( 8 % ( 1.6 0 1.6 % 8 2.4 & ( 3.2

4.8

( 1.6 ( 3.2$ 0 8 " (8 " 1.6 2.4" (4 6 " ( 20.14 ! 10 ) 5.6 ( 0.8 ( 4.8" 0.8 0 " ( 0.8 " 0 4.8# ( 4.8

Element governing equation

{Q}e = [k ]e {! }e Here,

(a)

(! ) e = {u1 , u2 , u3 , w1 , w2 , w3 }

When boundary conditions are given, (! ) e is computed from Eq. (a), as illustrated in Chap. 7. ______________________________________________________________________________________ SOLUTION (8.50 through 8.53) Owing to the symmetry, only any one-quarter of the cylinder need be analyzed. Use a finite element computer program, such as ANSYS.

End of Chapter 8

CHAPTER 9 SOLUTION (9.1) Using Eq. (9.3), 1

# = ( 4 kEI ) 4 = [ 4 ( 200"101.49 ()(105.04) "10!6 ) ] 4 = 0.7676 m !1 Equation (9.8) yields, for x = 0 : M max = "4 !P f 3 ( !x ) = " 4P! f 3 (0) = " 4P! Thus, 6

10 ) 4 ( 0.7676 ) P = $ max Ic( 4 # ) = 210"10 ( 5.040."0635 = 51.18 kN

______________________________________________________________________________________ SOLUTION (9.2) b

I = b ( 2.5b) 3 12 = 1.302b 4 6

2.5b

I M max = " max = 250!101.25(1.b302 b ) = 260.4(10 6 )b 3 c

(a)

1 4

" = [ 4 ( 200!2010(910)(1.)302 b4 ) ] = 0.0662 b Equation (9.8):

M max = 4P!

(b)

From Eqs. (a) and (b), 3

260.4(10 6 )b 3 = 440( 0(.100662) b) or

b = 0.0241 m = 24.1 mm

______________________________________________________________________________________ SOLUTION (9.3) Select a particular solution of the form v p = a sin 2L!x a = const. Introduce this into Eq. (9.1),

( 2L! ) 4 a + EIk a = EIp1

Thus,

= k [1+ 4 (p"1 !L )4 ]

16" 4 EI L4

p1 16! 4 EI

sin 2L!x

(a)

General solution is

v = e "x [ A cos "x + B sin "x ] + e ! "x [c cos "x + D sin "x ] + v p

______________________________________________________________________________________ 9.3 (CONT.) Boundary conditions are:

v ( !) = 0;

A= B=0 Loading repeats itself periodically, p1 sin( 2!x L). But v = e " !x [C cos !x + D sin !x ]

cannot repeat periodically and represents a damped wave. Thus, in order deflection to repeat itself with the same wave length as loading it is required that C = D = 0. Accordingly, solution is

v = vc + v p = 0 + v p = v p

given by Eq. (a). ______________________________________________________________________________________ SOLUTION (9.4) p Q

a b

From Example 9.1: b

" !x " !x vQ = # pdx [cos !x + sin !x ] " # pdx [cos !x + sin !x ] 2 k !e 2 k !e

0 p 2k

" !x

cos !a " e

" !b

cos !x ]

Note that, if a=0 and b=L (large):

vQ ! 2pk

When a and b increase (large):

vQ ! 0

(a)

While according to the result of Example 9.1, when a=0 and b=L (large):

vQ ! 2pk

When a and b increase (large);

vQ ! 2pk

(b)

the answers differ, as observed by comparing Eqs. (a) and (b). ______________________________________________________________________________________ SOLUTION (9.5) Applying Eqs. (9.3) and (9.8) we obtain 1

!1 .8 4 " = [ 4 kEI ] 4 = [ 4 (16 8.437 ) ] = 0.84 m

v (0) = v max = P2!k ( 0.84 ) = 0.135 = 3.375(10 !3 ) m 2 (16.8 )

and

M (0) = M max = 4P" = 4 (135 0.84 ) = 40.179 kN ! m

Maximum stress is therefore 3

(10 ) " max = MSmax = 403..9179 = 103.02 MPa (10! 4 )

______________________________________________________________________________________ SOLUTION (9.6) Refer to Solution of Prob. 9.1: 1 k 14 12(106 ) ) =[ ]4 9 "6 4 EI 4(200 #10 )(5.04 #10 ) !1 = 1.3135 m Equation (9.8) gives, for x = 0 : P M max = " 4!

! =(

Therefore

! max I (4 " ) (210 2.5)106 (5.04 $10#6 )4(1.3135) = c 0.0635 = 35 kN

______________________________________________________________________________________ SOLUTION (9.7)

I = b 4 12

M max =

! max I (260 1.8)106 (b 4 12) = c b2

(a)

= 24.074(106 )b3

! =[

1 7(106 ) ] 4 = 0.1316 b 9 4 4(70 "10 )(b 12)

Equation (9.8):

M max =

P 4!

(b)

From Eqs. (a) and (b),

24.074(106 )b3 = Solving,

50(10)3 b 4(0.1316)

b = 0.0628 m = 62.8 mm

______________________________________________________________________________________ SOLUTION (9.8) We have

1 k 14 15 ] =[ ] 4 = 0.8165 m "1 4 EI 4(8.437) P! v(0) = vmax = 2k 0.25(0.8165) = = 6.8(10!3 ) m = 6.8 mm 2(15)

! =[

______________________________________________________________________________________ 9.8 (CONT.) And

M (0) = M max =

P 250 = = 76.55 kN " m 4 ! 4(0.8165)

Maximum stress is thus

! max =

M max 76.55(103 ) = = 196.3 MPa S 3.9

______________________________________________________________________________________ SOLUTION (9.9) Equations (9.3) and (9.8): 1

! = [ 4 kEI ] 4 , (a)

v max = P2!k ,

M max = 4P!

k is 1.25 times the actual value.

! changes by 4 1.25 = 1.057. v max changes by 1.057 1.25 = 0.846. ! max (or M max ) changes by 1 1.057 = 0.946. (b)

k is 1.4 times the actual value.

! changes by 4 1.4 = 1.088. v max changes by 1.088 1.4 = 0.777. ! max (or M max ) changes by 1 1.088 = 0.919. Note that stress calculation is not affected appreciably in both cases. ______________________________________________________________________________________ SOLUTION (9.10) P1 P2 O 0.83 m

x 0.83 m

y By using the principle of superposition, deflection of any point, say O, of rail is expressed as the algebraic sum of v1 and v 2 caused by P1 and P2 , respectively. Thus, from Eq. (9.8), we have

v0 = P21k! f1 ( !x1 ) + P22k! f1 ( !x 2 )

(a)

. Then, f1 ( !x1 ) and f1 ( !x 2 ) are found from Table 9.1 for x1 = 0.83 m and x 2 = !0.83 m. We have !x1 = 0.697 and "x2 = !0.697. Thus f1 ( !x1 ) = 0.702. where, P1 = P2 = P, " = 0.84 m

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 9.10 (CONT.) Since, from symmetry f 1 ( !x1 ) has the same value for a (+) or (-) value of !x , Eq. (a) may be written as

v0 = P2!k (0.702 + 0.702) = 1.404 P2!k

Resultant bending moment at O, from Eq. (9.8), is

M 0 = 4P! f 3 ( !x1 ) + 4P! f 3 ( !x 2 )

= 4P! [2 f 3 (0.697)] = 0.125 4P! It may be verified by comparing the results of Problems 9.5 and 9.7 that addition of one or more load (reduces appreciably value of maximum moment) causes a large increase in the maximum deflection of the rail. ______________________________________________________________________________________ SOLUTION (9.11) x px

po x

Q y

Expression for loading are:

p x = pL0 ( a ! x ) Px =

p0 L

(a + x )

(segment BQ) (segment QC)

Deflection at Q is obtained by substituting ( p x dx ) for P in Eq. (9.6). That is a

vQ = 2p0kL# {! ( a " x )e " #x [cos #x + sin #x ]dx 0

Integrating, we have

+ ! ( a + x )e "x [cos "x + sin "x ]dx} 0

vQ = 4p!0k [ f 3 ( !a ) " f 3 ( !b) " 2 !Lf 4 ( !b) + 4 !a ] ______________________________________________________________________________________ SOLUTION (9.12) p A

x y p

Figure (a)

A R

p/k x

Figure (b)

Figure (c)

______________________________________________________________________________________ 9.12 (CONT.) Referring to Fig. (b):

v A1 = ! kp

" A1 = 0

Referring to Fig. (c) and using Eq. (9.12),

v A2 = 2k! [ " Rf 4 (0) + !M 0 f 3 (0)] 2

" A2 = # 2 !k [ # Rf1 (0) + 2 !M 0 f 4 (0)] The problem may be separated into two different cases both semi-infinite beams (as is seen in the figure above) provided that deflection at A is zero. Thus,

solving,

v A1 " v A2 = 0;

" kp = 2k! [ " R + !M 0 ]

" A1 # " A2 = 0;

2! 2 k

M 0 = 2 ! 2 EI kp ,

[ # R + 2 !M 0 ]

R = 4 ! 2 EI kp = !p

______________________________________________________________________________________ SOLUTION (9.13) ( a ) Using Eq. (9.12), v = 2

v max = 2 ! kM A

2! 2 k

M A f 3 ( !x )

at x = 0

Scan Table 9.1,

(down)

f 3 ( "x ) = !0.2079 at "x = !2 , and 2

v min = 2 "k M A ( !0.2079)

(up)

Therefore

vmax vmin

1 = !0.2079 = !4.81

( b ) Applying Eq. (9.12), M = M A f 1 ( !x )

v max = M A

at x = 0

f1 ( "x ) = !0.0432 at "x = ! , and M min = !0.0432 M A

Scan Table 9.1, Thus

M max M min

1 = !0.0432 = !23.15

______________________________________________________________________________________ SOLUTION (9.14) ML A

x y

Equation (9.11) with v (0) = 0 gives

0 = 2 ! ( " R A + !M L ) k

______________________________________________________________________________________ 9.14 (CONT.) from which R A = M L . When R A is substituted for ! P into Eq. (9.12):

v = " 2 !M2LEI e " ! sin !x = " 2 !M2LEI f 2 ( !x )

Successive differentiations of this expression give:

v ' = " 2M!EIL f 3 ( !x )

M = 2MEIL f 4 ( !x ) V = " 2MEIL f1 ( !x )

______________________________________________________________________________________ SOLUTION (9.15) K

We have k = as = 0180 .625 = 288 kPa 1

(10 ) = [ 4 ( 200!288 ] 4 = 0.51697 m !1 109 )( 5.04!106 )

Using Eq. (9.8),

.5697 ) v (0) = v max = P2!k = 6.752((0288 = 5.61 mm )

.75 M (0) = M max = 4P" = 4 ( 06.51697 ) = 3.26421 kN ! m

Thus,

0.0625 ) # max = McI = 32645..0421"(10 = 41.13 MPa !6

______________________________________________________________________________________ SOLUTION (9.16) From Eq. (9.3), 1

18 (10 ) # = [ 4 ( 0.375"200 ] 4 = 0.824 m !1 "109 )( 5.2"10! 7 )

Since

"L = 0.824(0.74) = 0.618 < ! 4 the beam can be considered rigid.

( a ) Uniform deflection is

v = 3FK = 3540 (18 ) = 10 mm P K

( b ) We may replace the given beam by the beams shown in Figs. (a) and (b). 0.25

0.125

0.375 Figure (a)

P F+PB1

F+PB2

F+PB3

Figure (b)

______________________________________________________________________________________ 9.16 (CONT.) Note that F is the direct load on each spring and PBi is the bending loads with the values:

F = P 3 = 540 3 = 180 N and Here

PBi = M Brr2i !i

(i = 1, 2, 3)

)( 0.375 ) PB1 = 5402((015.125 = 90 N , PB 2 = 0, !0.125 ) 2

PB 3 = ! PB1

End deflections are thus,

v1 = (180 + 90) 18 = 15 mm v 2 = (180 ! 90) 18 = 5 mm

______________________________________________________________________________________ SOLUTION (9.17) Using Eq. (9.3), 1

" = [ 4 (148.4 ) ] 4 = 0.8034 m !1 !L = 0.8034(0.6) = 0.482 rad = 27.62 o Hence, from Eq. (9.13):

( 0.8034 ) 2 + cos 27.62 + cosh 27.62 vc = 4500 2 (14!106 ) sin 27.62o +sinh 27.62o o

= 186(10 !6 ) m = 0.186 mm Note: Slope may be found as in Prob. 9.15, with p=0. ______________________________________________________________________________________ SOLUTION (9.18) P

!L < 1

E C L/2

L/2

Referring to Sec. 9.8 and Table D.4 4

5[( P + pL ) L ] L 5 pL vc = 48PLEI + 384 EI ! 384 EI 3

(0.15) = 48(4.8 [9000 + 5(75008"0.15) ! 5(90008+1125) ] "10!6 )

= 2.81(10 !8 ) m Similarly, 3

0.15](0.15) ! E = 16PLEI + 24pLEI " [(9000+1125) 24 EI 2

Substituting the given data,

" E = 5(10 !7 ) rad

______________________________________________________________________________________ SOLUTION (9.19) Equation (9.3),

k 14 8 14 ] =[ ] = 0.6687 m "1 4 EI 4(10) ! L = 0.6687(0.8) = 0.535 rad = 30.65o

! =[

From Eq. (9.13):

15(103 )(0.6687) 2 + cos 30.65o + cosh 0.535 2(8 !106 ) sin 30.65o + sinh 0.535 = 629.9[18.4208](10!6 ) = 5, 279(10!6 ) m = 5.28 mm

vc =

______________________________________________________________________________________ SOLUTION (9.20) Refer to Sec. 9.8 and Table D.4, we have I = bh

12 = 46.8(10!6 ) m 4 ,

EI = 3.4 MN ! m 2 , and ! L < 1 .

PL3 5 pL4 5[( P + pL) L]L4 + ! 48EI 384 EI 384 EI 3 (0.4) 5(7000 ! 0.4) 5(8000 + 2800) = [8000 + " ] 6 48(3.4 !10 ) 8 8 = 0.00276(10!3 ) m = 2.8 mm

vc =

Likewise,

PL2 pL3 [(8000 + 2800) 0.4](0.4)3 + " !E = 16 EI 24 EI 24 EI 2 (0.4) 7000(0.4) 10,800 [8000 + ] = ! !6 16(3.4 "10 ) 32 32 = 0.29(10!3 ) rad ______________________________________________________________________________________ SOLUTION (9.21) L p x Figure (a) y -v1 -v1 z 0 v1 Figure (b) -v2 -v1 v1=v2 z Figure (c) 0 v1 v2 -v1 -v3 v1=v3 z Figure (d) 0 v1 v2 v3 Boundary conditions are:

v (0) = v ( L) = 0,

v ' ' ( 0) = v ' ' ( L ) = 0

(a) (CONT.) ______________________________________________________________________________________ Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 9–9

______________________________________________________________________________________ 9.21 (CONT.) These are transformed into the following central difference conditions by using Eq. (7.4) and (7.7):

v0 = 0,

! v !1 = v1 ,

v n = 0,

v n !1 = !vn +1

(b)

For m=2 (Fig. b):

v n !2 ! 4v n !1 + 6( mm4+1 )vn ! 4v n +1 + v n + 2 = 1m.64 4

! v1 + 6 224+1 v1 ! v1 = 12.46 ;

v1 = 21.4 mm

For m=3 (Fig. c):

! 4v1 + 6 334+1 v2 ! v 2 = 13.46 4

! v1 + 6 334+1 v1 ! 4v 2 = 13.46 4

Solving,

v1 = v2 = 17.2 mm

For m=4 (Fig. d):

! v1 + 6 444+1 v1 ! 4v1 + v1 = 14.46 4

! 4v1 + 6 444+1 v 2 ! 4v1 = 14.46 4

Solving,

v 3 = v1 = 10.2 mm v2 = 13.9 mm

______________________________________________________________________________________ SOLUTION (9.22) From Example 9.1:

v p = 2pk ( 2 " e " !a cos !a " e " !a cos !b)

where

k = 48aLEI3

Thus,

" !a t v p = 2 (paL cos !a " e " !b cos !b) 48 ) EI [ 2 " e

= 96paLEIt [2 " f 4 ( !a ) " f 4 ( !b)] We have " = 3.936 6 = 0.656 m At midspan, we have !a = !b

and

f 4 ( "a ) ! 0

paL3t

Thus,

v p = 48 EI

Since,

v m = 48paLEIt = R48ccEILt

Solving,

Rcc = ap = 0.3 p End of Chapter 9

CHAPTER 10 SOLUTION (10.1) P

B x

D x

a A Defections at point D: We write

M DB = Px

M BA = Pb + Rx

Using Eq. (10.5), b

U DB = ! [ ( 2PxEI) + 2RAE ]dx = P6 EIb + 2RAEb 0 a

2 3

U BA = ! [ ( Pb2+EIRx ) + 2PAE ]dx 2

= 2 1EI [ p 2 ab 2 + PRa 2 b + 13 R 2 a 3 ] + 2PAEa 2

Total strain energy U = U BD + U BA . Vertical deflection at D is thus Pa " v = !!UP = EI1 [ Ra2 b + P( ab 2 + b3 )] + AE 2

Horizontal deflection at D:

Rb " h = !!UR = EI1 [ Ra2 b + Ra3 ] + AE

Angular rotation of D: Introduce a couple moment C at D as shown in the figure. Then, and

M DB = Px + C b

M BA = Pb + Rx + C

U DB = ! [ ( Px2+EIC ) + 2RAE ]dx 2

= 2 1EI [ P 3b + C 2 b + PCb 2 ] + 2RAEb 2 3

U BA = 2 1EI ! [ Pb + Rx + C ]2 dx + 2RAEb 2

= 2 1EI [ P 2 ab 2 + R 3a + Ca 2 + PRa 2 b + 2 PCab + CRa 2 ] + 2PAEa 2 3

Hence,

" D = !!UC C =0 = 2 1EI [ Pb 2 + 2 Pab + Ra 2 ]

Note that displacements of a simple (straight) cantilever beam may be readily found by setting b=0 in the foregoing results. ______________________________________________________________________________________ Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 10–1

______________________________________________________________________________________ SOLUTION (10.2) Equations of statics are applied to obtain the reactions:

! F =0 : ! M =0 : ! M =0 :

R Ax = 0

RC = 14 P1 + 12 P2 + 43 P3

R Ay = 43 P1 + 12 P2 + 14 P3

The axial force in each member is obtained by applying the method sections, as required. The results are as follows:

N DC =

P1 + 22 P2 + 3 4 2 P3

2 4

N AE = 3 4 2 P1 + 22 P2 + 42 P3

N DE =

1 2

P1 +

P2 +

1 2

N BE = ! 42 P1 + 22 P2 + 42 P3 N BD =

2 4

P1 + 22 P2 ! 42 P3

N AB =

3 4

P1 + 12 P2 +

1 4

N BC =

1 4

P1 +

3 4

1 2

P2 +

Numerical results are determined for P1 = P2 = P3 = 45 kN and L = 3 m. These are tabulated bellow.

Bar Axial force ( kN )

!N !P2

Length ( m )

DC AE DE BE BD AB BC

2 2 2 2 1 2 2 2 2 12 12

2.125 2.125 3.000 2.125 2.125 3.000 3.000

N DC = 99.5 N AE = 99.5 N DE = 90 N BE = 31.8 N BD = 31.8 N AB = 67.5 N BC = 67.5

Thus, the vertical deflection at B: 7

1 # B = AE ! N j "P2j L j j =1

103 AE

[2(67.5)(0.5)(3) + 2(31.8)(0.707)( 2.125) + 90(1)(3) + 2(99.5)(0.707)2.125] = 686,773.675 AE m

______________________________________________________________________________________ SOLUTION (10.3) The moment is expressed by

M = P( 2a ! x );

"M "P = 2a ! x

Castigliano’s theorem gives then

2 M #M P v P = ! EI #P dx = ! E ( 2a " x ) ( c1 x + c 2 ) dx + ! 4

P EI 2

( 2a " x ) 2 dx

v P = 1112PcE1a + 7 Pc3 E2a + 3PaEI 2 3

______________________________________________________________________________________ SOLUTION (10.4) P C

y Deflection at A

(with M = Px ): L

v A = EI1 ! M ""MP dx = EI1 ! Px ( x )dx 0

vA =

PL3 3 EI

(with M = Px + C ):

Slope at A

# A = EI1 ! M ""MC dx = EI1 ! ( Px + C )(1)dx

Setting C = 0 and integrating

! A = 2PLEI

______________________________________________________________________________________ SOLUTION (10.5) p

P C

y Deflection at A

(with M = Px + 12 px ): L

# A = EI1 ! M ""MP dx = EI1 ! ( Px + 12 px 2 )( x )dx 0

= Slope at A

PL3 3 EI

pL4 8 EI

(with M = Px + 12 px

+ C ):

# A = EI1 ! M ""MC dx = EI1 ! ( Px + 12 px 2 + C )dx

Setting C = 0 and integrating

! A = 2PLEI + 6pLEI 2

______________________________________________________________________________________ SOLUTION (10.6) P C

A y

Deflection at A

I B L/2

L/2

(with M = Px ):

# A = EI1 ! M ""MP dx = EI1 !

( Px )( x )dx + 2 1EI ! ( Px )( x )dx L2

Integrating, 3

7 PL 3 PL ! A = 24PLEI + 48 EI = 16 EI Slope at A (with M = Px + C ):

# A = EI1 ! M ""MC dx = EI1 !

( Px + C )(1)dx + 2 1EI ! ( Px + C )dx L2

Setting C = 0 and integrating 2

3 PL ! A = 8PLEI + 16 EI

5 PL ! A = 16 EI

______________________________________________________________________________________ SOLUTION (10.7)

R ds=Rd!

Q Mo

( a ) Horizontal deflection

M " = PR(1 ! cos " ) + QR sin " + M 0

and

$ h = EI1 ! M # ""MQ# ds = EI1 !

# 2

R2 2 EI

[ PR 3 (1 " cos $ ) sin $ + QR 3 sin 2 $ + M 0 R 2 sin $ ]d$

( 12 PR + M 0 )

Vertical deflection (Q=0)

$ v = EI1 ! M # ""MP# ds = EI1 !

# 2

[ PR 3 (1 " cos $ ) 2 + MR 2 (1 " cos $ )]d$

= 4REI [ PR(3" ! 8) + 2 M 0 (" ! 2)] 2

( b ) Rotation of the free end (Q=0)

% = EI1 ! M % ##MM%0 ds = EI1 !

$ 2

[ PR(1 " cos % ) + M 0 ]d%

= 2REI [ PR(! " 2) + !M 0 ]

______________________________________________________________________________________ SOLUTION (10.8) We now have

M = " FR sin ! , N = F sin ! , Applying Eq. (10.6), with ! = 6 5 and T = 0 :

V = " F cos !

2 2 R $ h = REI ! F sin 2 #d# + AE ! F sin #d# + 56AGR ! F cos #d# 3

Integrating,

3!FR " h = !2FREI + !2FR EA + 5 AG

Substituting the given data: 3

)( 0.05 ) 4000 )( 0.05 ) )( 0.05 ) # h = " ( 4000 + " (400 + 3"5((4000 400 ( 6.67 ) (105 ) 2 80!105 ) 2

= (0.59 + 0.008 + 0.02)10 !3 = 0.62 mm The error, if N and V are omitted, is 4.5 %. ______________________________________________________________________________________ SOLUTION (10.9) Introducing a rightward horizontal force Q at point D, we write

M 1 = !Qx M 2 = ! 12 Fx ! Qa M 3 = ! 12 Fx ! Qa + F ( x ! b2 ) M 4 = ! 12 Fb + 12 Fb ! Q ( a ! x )

0! x!a 0 ! x ! b2 b 2

!x!b

0! x!a

Applying Castigliano’s theorem, after setting Q=0, we have

# h = EI1 !

1 2

!h =

Integrating,

Faxdx + EI1 ! ( " 12 Fax + 12 Fab)dx b2

Fab 2 8 EI

______________________________________________________________________________________ SOLUTION (10.10) a x

! R

From symmetry,

M 1 = ! Px M 2 = " Pa " PR sin !

Hence, or

$ 2

0! x!a 0 # " # !2

# 2

= EI1 ! M 1 ""MP1 dx + EI1 ! a

" 2

0 2

$ = EI2 [ ! Px 2 dx + !

M 2 ""MP2 Rd%

P( a + R sin # ) 2 Rd# ]

= 6PEI ( 4a 3 + 6!Ra + 24 R 2 a + 3!R 3 )

______________________________________________________________________________________ SOLUTION (10.11) Referring to Fig. P10.11, we write

M = PR sin ! m = R sin !

T = PR(1 " cos ! ) t = R(1 " cos ! )

Here t denotes torque caused by a unit load. Applying Eq. (10.13), deflection at the free end (perpendicular to the plane of the ring):

% = EI1 !

# 2

( PR 2 sin 2 $ ) Rd$ + JG1 !

# 2

! 2

[ PR 2 (1 " cos $ ) 2 ]Rd$

! 2

" sin 2" " sin 2" = PR + PR EI 2 # 4 0 JG " # 2 sin " + 2 + 4 0 3

= PR4 ( EI" + 3"JG!8 ) Letting J = 2 I = !r

! =

PR 3 r4

2 , we have ( + ) 0.226 G

1 E

______________________________________________________________________________________ SOLUTION (10.12) Referring to Fig. P10.12:

M = " PR sin ! T = PR(1 " cos ! )

m = " R sin ! t = R(1 " cos ! )

where t denotes torque caused by a unit load. We have sin(2" # ! ) = # sin ! cos(2" # ! ) = cos ! and Applying Eq. (10.13), deflection at the free end (perpendicular to the plane of the ring): 2#

% = PR sin 2 $d$ + PR (1 " cos $ ) 2 d$ EI ! JG ! 3

Integrating,

" = PR 3! [ EI1 + JG3 ]

______________________________________________________________________________________ SOLUTION (10.13) Q A

p a

Segment AC

M 1 = !Qx

Segment BC

M 2 = "Qx " Thus,

"M 1 "Q = ! x

p ( x " a)2 2

!M 2 = "x !Q

3 a M !M M 1 !M 1 2 2 dx + " dx 0 EI !Q a EI !Q

vA = "

Let Q=0. Hence

______________________________________________________________________________________ 10.13 (CONT.)

vA =

p 3a p 3a 3 ( x ! a ) 2 xdx = ( x ! 2ax 2 + a 2 x)dx " 2 EI a 2 EI "a 3a

p x 4 2ax 3 a 2 x 2 = ! + 2 EI 4 3 2 a

10 pa 4 ! 3 EI

______________________________________________________________________________________ SOLUTION (10.14) B

LAB = LCD = 2 L

L A

LBC = LAC = LBD = L

The method of joints(or sections) is applied:

N AB = 2W Thus,

N BC = !W

N CD = ! 2W

N BD = W

N AC = !W

"N j 1 Nj Lj # AE "W WL WL = [ 2( 2) 2 + 1 + 1 + 1 + 2( 2) 2] = 8.657 ! AE AE

!V =

______________________________________________________________________________________ SOLUTION (10.15) P x x’ A

Pa L

C L

Segment AB

M1 = ! P Segment BC

a C x! x L L

M 2 = ! Px '! C

(a)

For this case C=0:

vC =

1 L Pax ax 1 a (! )(! )dx + ! Px '(! x ')dx ' " EI 0 L L EI "0 L

Pa 2 Pa 2 x 3 P x '3 = ( L + a) ! = + EIL2 3 0 EI 3 0 3EI

______________________________________________________________________________________ 10.15 (CONT.) (b)

!M 1 x =" , !C L

For C=0, we have :

!C =

!M 2 = "1. !C

1 L Pax x 1 a ( )( ) dx " " + " Px '(" x ')dx ' EI #0 L L EI #0 L

Pa x 3 P x '2 = + EIL2 3 0 EI 2 0 =

Pa (2 L + 3a ) 6 EI

______________________________________________________________________________________ SOLUTION (10.16) x” Q p A RA=Q/2 Segment AC

x’

1 M 1 = Qx 2

Let Q=0:

C a

RB=2pa+Q/2

!M 1 x = !Q 2

!M 1 1 a M1 dx = 0 " 0 EI !Q

Segment CB

1 1 M 2 = Q(a + x ') ! Qx '! px '2 2 2 Q 1 = (a ! x ') ! px '2 2 2 !M 2 1 = (a " x ') !Q 2

Let Q=0:

!M 2 1 1 a px '2 1 pa 4 ' ( ) ( ') ' M dx = " a " x dx = ! 2 2 2 EI # !Q EI #0 48 EI Segment BD

M3 = "

1 px ''2 2

!M 3 =0 !Q

Therefore

!M i pa 4 vC = # M i dx = " !Q 48 EI ______________________________________________________________________________________ Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 10–8

______________________________________________________________________________________ SOLUTION (10.17) Q p L/2

pL 8

+ Q2

L/2 x’

Segment AC

3 pL 8

+ Q2

!M 1 x = !Q 2

pL Q + )x 8 2

M1 = (

Segment CB

3 pL Q px '2 M2 = ( + ) x '" 8 2 2

!M 2 x ' = !Q 2

Let Q=0: L 2 3 pLx ' pLx x px '2 x ' dx + " ( ! ) dx ' 0 0 8 2 8 2 2 2 2 L 2 pLx L 2 3 pLx ' px '3 pL4 = p" dx + p " ( ! )dx ' = 0 0 16 16 4 96 4 pL vC = ! 96 EI

EIvC = "

______________________________________________________________________________________ SOLUTION (10.18) Consider RD as redundant. C 4

A 2.25 m

"F = 0: N x

where

1.5 m

NAB

AE! D = # N j L j

!N BD =1 !RD

!F = 0: y

N BC = 1.25 P ! 1.25 RD

= !0.75 P + 0.75 RD

!N BC !RD = "1.25

Thus,

NBD =RD

NBC

!N AB !RD = 0.75

"N j

=0 "RD = (1.25 P ! 1.25 RD )(7.5)(!1.25) +(!0.75 P + 0.75 RD )(4.5)(0.75) + RD (3)1 = 0

= !14.25 P + 17.25 RD = 0 or

RD = 0.826 P

Then

N BD = 0.826 P

N AB = !0.131P

N BC = 0.22 P

______________________________________________________________________________________ SOLUTION (10.19) Let RA and M A be redundant. P

C L RA

1 lb

B C

1 lb ! in.

Mm dx EI

!A = "

m = !1(a + x)

M = ! Px

vA = "

m ' = !1

Mm ' dx EI

1 b Pb 2 a b ( ! Px )[ ! ( a + x )] dx = ( + ) EI "0 EI 2 3 1 b Pb 2 ( Px )( 1) dx ! AP = " " = EI #0 2 EI Let now RA and M A are applied. v AP =

x MA

A RA

M = RA x ! M A

1 lb ! in.

1 lb

m=x

m' =1

( RA x ! M A ) x 1 RA L3 M A L2 v AR = " dx = ( ! ) 0 EI EI 3 2 L ( R x " M )1 1 RA L2 A A ! AR = # dx = ( " M A L) 0 EI EI 2 v AP = v AR ! AP = ! AR Since L

Solving,

1 RA L3 M A L2 Pb 2 a b ( + )= ( ) ! 2 EI 2 3 EI 3 1 RA L2 Pb 2 ( = ! M A L) 2 EI EI 3 Pb 2 RA = 3 (3a + b) ! L

Pab 2 MA = 2 L

______________________________________________________________________________________ SOLUTION (10.20) x

R = k! A

Refer to the above figures, we write:

M = Rx !

1 2 px 2

m = !1x

Hence, deflection spring at A:

Mm R dx = EI k 1 L 1 R = ( Rx ! px 2 )(! x)dx = " EI 0 2 k

!A = "

Solving

3 pL 8 1 + (3EI kL3 )

Reactions at B may then be found from statics. ______________________________________________________________________________________ SOLUTION (10.21) Consider RB as redundant. P

x A

Segment BC:

M 1 = ! Px

!M 1 !RB = 0

Segment AB:

M 2 = ! Px + RB ( x ! a )

"M 2 "RB = x ! a

Thus, a

EIvB = 0 = " (! Px)(0)dx + " [! Px + RB ( x ! a )]( x ! a )dx 3a

= " [! Px 2 + RB x 2 ! 2 RB ax + Pax + RB a 2 ]dx a

Px 3 RB x3 Pax 2 =! + ! RB ax 2 + + RB xa 2 3 3 2 a =!

14 3 P + RB 3 8

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 10.21 (CONT.) from which

7 P! 4

RB = Statics:

MA =

1 Pa 2

RA =

3 P! 4

______________________________________________________________________________________ SOLUTION (10.22) Consider RA and M A as redundants. y MA

C L

!M 1 =x !RA

!M 1 = "1 !M A

RA Segment AC:

M 1 = RA x " M A Segment BC:

M 2 = RA x ! M A ! P ( x ! a )

!M 2 !RA = x

!M 2 !M A = "1

Thus,

EIv A = 0 = " ( RA x ! M A ) xdx + " [ RA x ! M A ! P ( x ! a )]xdx a

= RA

Simplifying,

Similarly

a a L ! a3 L2 ! a 2 L3 ! a 3 ! M A + RA !MA !P 3 2 3 2 3 L2 ! a 2 + Pa =0 2

1 1 P RA L3 ! M A L2 ! (3a + 2b)b 2 = 0 3 2 6 a

(1)

EI! A = 0 = # ( RA x " M A )("1)dx + # [( RA x " M A " P ( x " a )]("1)dx a

= ! RA

a L !a L2 ! a 2 + M A a ! RA + M A ( L ! a) + P 2 2 2 ! Pa ( L ! a ) = 0

This reduces to

1 1 RA L2 ! M A L ! Pb 2 = 0 2 2

(2)

______________________________________________________________________________________ 10.22 (CONT.) Solving Eqs.(1) and (2):

MA = Statics:

RB =

Pab 2 L2

RA =

Pb 2 (3a + b) ! L3

Pa 2 (a + 3b) ! L3

MB =

Pa 2b L2

______________________________________________________________________________________ SOLUTION (10.23) Consider RB as redundant. M0 MA

M = ! RB x + M 0 !M !RB = " x

B RB

Therefore

Solving

L 1 1 EIvB = 0 = " ( RB x 2 ! M 0 x)dx = RB L3 ! M 0 L2 0 3 2

RB =

3 M0 ! 2 L

RA =

3 M0 ! 2 L

Statics:

MA =

1 M0 2

______________________________________________________________________________________ SOLUTION (10.24) We introduce a couple moment C at point B. The expression for the moments are then,

M 1 = ! RD x M 2 = ! RD x + P( x ! L2 ) + C

0 ! x ! L2 L 2

!x!L

Applying Eq. (10.6), " B = !U !P : L

# B = EI1 {! [" RD x + P( x " L2 ) + C ]( x " L2 )dx} L2

Setting C=0 and integrating,

" B = ! 48PLEI [ 1!( 325EI16kL3 ) ! 2] 3

Similarly, applying Eq. (10.7), " B = !U !C : L

# B = EI1 {! [" RD x + P( x " L2 ) + C ]dx} L2

Setting C=0 and integrating, 2

15 PL 1 " B = ! 128 EI 1!( 3 EI kL3 )

______________________________________________________________________________________ SOLUTION (10.25) Apply Eq. (10.7) to write a

$ A = JG1 [ " TA ##TTAA dx + " (TA ! T ) # (T#AT!A T ) ] = 0 Integrating and simplifying:

TA a + (TA ! T )b = 0

TA = Lb T

Condition of equilibrium gives

TB = T ! TA = aL T

______________________________________________________________________________________ SOLUTION (10.26) We write

M 1 = Qx M 2 = Q ( a + R sin ! ) + PR(1 " cos ! ) Applying Eq. (10.6), with ! Q = 0 : "U "Q

Integrating,

0! x!a 0 #" #!

= 0 = EI1 ! Qx 2 dx + ! [Q ( a + R sin $ ) + PR(1 " cos $ )]( a + R sin $ ) Rd$ 3

QR 2 2 ! PR 0 = QL 3 EI + EI (!a + 4 Ra + 2 R ) + EI (!a + 2 R ) 2

This may be written in the following form: 3

(!a + 2 R ) Q = 3 1 "!PR R R 2 ! R 3

a [ + +4( ) + ( ) ] a 3 a 2 a

______________________________________________________________________________________ SOLUTION (10.27) po

RA MA

B L

We have Then,

M = ! RB x + pL0 x 2x 3x = ! RB x + p06x L

$ B = 0 = EI1 ! M ##RMB dx = EI1 ! ( RB x " p60Lx ) xdx 0

This yields, after integrating,

RB = p100 L Then, from equations of statics: p0 L R A = 410

M A = p150 L

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (10.28) Moments are expressed by

M 1 = ! Rx M 2 = ! Rx + M 0

0 ! x ! L2 L 2

!x!L

Applying Eq. (10.6), with ! R = 0 :

= 0 = EI1 !

#U #R

Rx 2 dx + EI1 ! ( " Rx + M 0 )( " x )dx L2

from which

R = 98ML0

Using Eq. (10.7), slope at C: L

$ C = ##MU = 0 + EI1 ! ( " Rx + M 0 )dx L2

5M 0L 64 EI

______________________________________________________________________________________ SOLUTION (10.29) p E1I1

E2I2

x A

L1 VB

M1 = H Ax M 2 = !V A x + H A L2 + 12 px 2

HC VC

0 ! x ! L2 0 ! x ! L1

Applying Eq. (10.6), with ! Av = 0 and ! Ah = 0, respectively:

Letting

#U #V A

= E11I1 " ( !V A x + H A L2 + px2 )( ! x )dx = 0

#U #H A

= E21I 2 " H A x ( x )dx + E11I1 " ( !V A x + H A L2 + 12 px 2 ) L2 dx = 0

(a) (b)

! = EE12II12LL21

Equations (a) and (b) become

8V A ! 12 H A L2 = 3 pL12

3!V A L1 " 2(3! + 1) H A L2 = !pL12 V A = 23((3!!++14)) pL1 = R Av 2

1 H A = 4 ( 3!! +4 ) pL L2 = R Ah

The remaining reactions may then be found by using the equations of equilibrium. ______________________________________________________________________________________ Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 10–15

______________________________________________________________________________________ SOLUTION (10.30) MC

B x

! R

P/2

a/2

Clearly, the problem is statically in determined. We have

M x = !M C

0 ! x ! a2

M x = " M C + 12 PR(1 " cos ! )

0 # " # !2

Slope at C is zero: #U #M C

Integrating,

= EI1 !

$ 2

( " M C R + PR2 " PR2 cos % )d% + EI1 ! 2

a 2

" M C dx = 0

(! " 2 ) M C = PR 2 ( a +!R )

Hence, for 0 # " # !2 : 2

(" # 2 ) PR PR M = # PR 2 ( a +"R ) + 2 # 2 cos !

For 0 ! x ! a : 2

(! " 2 ) M = PR 2 ( a +!R )

Therefore, M is maximum along a, for ! = 0. ______________________________________________________________________________________ SOLUTION (10.31) ( a ) Introduce Q at point E, as shown. (disregard C/L’s). A

B L

L D

2L C L

Q C L

Joint E

N EB N ED

N ED = ! 2 P (C )

E 1

N EB = 2 P + Q (T )

Similarly we obtain the remaining member forces. Joint B

Joint D

N BD = ! P (C )

N DA = 2P 2

N BA = 2 P + Q (T )

N DC = !3P (C )

#E = !

N j L j "N j AE "Q

1 = AE [( 2 P )(5)(1) + ( 2 P )(5)(1) + 0 + 0 + 0 + 0] = 10AE2P

______________________________________________________________________________________ 10.31 (CONT.) ( b ) Introduce couple C/L applied perpendicular to line DE at points D and E as shown in the figure ( now disregard Q ). Joint E

N ED = !( P + C2L ) 2 (C )

P E

N EB

N ED

N EB = P + 22CL (T )

1 C L

Similarly, Joint D

N DA = 2 2 P

N DC = !3P ! 22CL (C )

Thus

1 # DE = AE ! N j "Cj = AEP ( 22 + 22 + 2 + 62 ) = AEP ( 102 + 2 )

______________________________________________________________________________________ SOLUTION (10.32) The complementary energy for the ith member of length Li , from Eq. (2.49) is "i " 3

U 0*i = !

d" = 4"Ki 3 = 4 K1 3 ( NAii ) 4

The complementary energy of the truss is thus 6

U * = ! 4 Kj 3j ( A jj ) 4 j =1

Equation (10.16) is then 6

# E = ! K j3 ( A jj ) 3 "Pj

(P10.32)

j =1

Applying the method of joints, as needed, we obtain

N 1 = 32P

N 2 = ! 54P

N 4 = 54P

N 5 = N 6 = ! 34P

N3 = 0

We have L1 = L5 = L6 = 3 m, L3 = 4 m, L2 = L4 = 5 m and Ai = A. Equation (P10.32) becomes 1 3 # E = ( AK ) {L1 N 13 ( !!NP1 ) + L2 N 23 ( !!NP2 ) + " " " " " + L6 N 63 ( !!NP6 )}

P 3 " E = ( AK ) {3( 23 ) 3 ( 23 ) + 5( ! 45 ) 3 ( 45 ) + 0

+ ( 45 ) 3 ( 45 ) + 3( 43 ) 3 ( 43 ) + 3( ! 43 ) 3 ( ! 43 )} or

P 3 48!81+5!625!2 + 3!81!2 = ( AK ) 256 P 3 ! E = 41.5( AK )

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (10.33) C

A P

L AC = 6.403 m, LAB = 8 m

P 5

LCB = 5 m

N AB

P/2

B P/2

P/2

N AC

Joint A

!N = 0: !N = 0: y

N AC =

N AB = 83 P

41 8

N BC = ! 85 P

Joint B

A horizontal unit load is applied a C: Thus

n AB = 83

n BC = ! 85

n AC =

41 8

1 " C = AE ! n j N jLj P = AE [( 841 )( 841 )( 6.403) + ( ! 85 )( 85 )(5) + ( 83 )( 83 )(8)]

P = 718 . AE !

______________________________________________________________________________________ SOLUTION (10.34) ( a ) Refer to Fig. (a) with no unit load:

M AB = 0 M BE = Rx M EC = Rx ! P ( x ! L) M CD = R ( 2 L) ! PL

C L x

The vertical deflection at A is zero. Thus

# v = EI1 ! M i ""MRi dx = 0

E L

B x

A R

1 Figure (a)

= ! 0dx + ! Rx ( x )dx + ! [ Rx " P ( x " L)] xdx + ! ( 2 LR " PL)( 2 L)dx = 0

Integrating, we have 29 R = 64 P

( b ) Introducing a horizontal unit load at A, Fig. (a), we write

m AB = x

m BE = L

m EC = L

mCD = L ! x

Hence

" h = EI1 ! M i mi dx

______________________________________________________________________________________ 10.34 (CONT.) L

L 2L

= EI1 {! 0dx + ! Rx ( L)dx + ! [ Rx " P ( x " L)]Ldx + ! ( 2 LR " PL)( L " x )dx} 0

Integrating, after substituting R = 29P 64 , results in 3

PL ! h = 13 32 EI

______________________________________________________________________________________ SOLUTION (10.35) x

C B

Introduce a vertical upward unit force 1 N at point C. Then Segment CB:

M1 = Rx

m1 = x

Segment BA:

M 2 = Rb = M 0

Thus

m2 = b

EI!C = 0 = " M1 m1 dx + " M 2 m2 dx 0 b

0 b

= " Rx dx + " ( Rb ! M 0 )( b)dx 0

= Rb + Rab 2 ! M 0 ba 1 3

3M a

R = b( 3a +0 b ) ! ______________________________________________________________________________________ SOLUTION (10.36) We have ! = 6 5 (Table 5.1), dx = ds = Rd! Let a downward unit load of 1 N in addition to load P is applied at the free end in Fig. P17.25. We write

N = P cos ! V = P sin ! M = PR(1 ! cos " )

n = cos ! v = sin ! m = R(1 ! cos " )

(a)

Thus, 3"

1 % v = EI1 ! MmRd$ + AG ! #Vvdx + AE1 ! 2

Nnds

Substitution of Eqs.(a) gives 3

PR !v = PR (1 $ cos " ) 2 d" + 56AG EI % % sin "d" + AEPR % cos 2 "d"

Integrating

3"PR 3.6 PR !v = ( 94" + 2) PR EI + 4 AG + 4 AE #

______________________________________________________________________________________ SOLUTION (10.37) P/2 1N A A A NA MA ! 1N m R C P/2 Actual loading

Dummy loading

Because of symmetry about the vertical axis, only one half of the circle need be analyzed. Referring to the figure above, we have the following moments: For 0 # " # !2 :

M 1 = M A + N A R(1 " cos ! ) " 12 PR sin !

(a)

m1' = 1

m1 = R(1 ! cos " ), For !2 # " # ! :

M 2 = M A + N A R(1 ! cos " ) ! 12 PR

(b)

' 2

m2 = R (1 ! cos " ),

m =1

Horizontal deflection and slope are zero at A. Thus, from Eqs. (10.13) and (10.14): #

% Ah = EI1 " MR(1 ! cos $ ) Rd$ = 0 0

" M (1 ! cos $ )d$ = 0 # = ! Mm dx = ! M (1)dx = 0

(c)

1 EI

! M = 0 , where M represents M and M . 1

(d)

Substituting Eq. (d) into Eq. (c), the latter reduces to "

! M cos #d# = 0

(e)

Introducing Eqs. (a) and (b) into Eqs. (d) and (e), we have #

0 # 2

M A ! d$ + N A R ! (1 " cos $ )d$

! 12 PR "

and

sin $d$ ! 12 PR " d$ = 0 # 2

M A ! cos $d$ + N A R ! (1 " cos $ ) cos $d$

" 12 PR !

# 2

Integrating and solving,

N A = 2P!

sin $ cos $d$ " 12 PR ! cos $d$ = 0 # 2

M A = PR 4

______________________________________________________________________________________ SOLUTION (10.38) From Eq. (P10.38), we have

U = GJ2 [( !aC ) 2 a + ( !bC ) 2 b]

Using Eq. (10.22): !U !" C

= T;

GJ" C ( a1 + 1b ) = T

Solving, Then and

Tab ! C = GJL

TA = GJa ! C = Lb T TB = GJb ! C = aL T

______________________________________________________________________________________ SOLUTION (10.39)

LAD = LDC = 4.24 m U = 12 ALE# 2 = 12 ALE (" v cosL! ) 2 Vertical load of the joint, using Eq. (10.22), "U "$ v

= P = ! ELi Ai i $ v cos 2 # i =1

P = EA! v [ cosLAD45 + cosLDC45 + L1BD ] = 0.5689 AE! V 2

Substituting the numerical values,

P = 373.34 kN

______________________________________________________________________________________ SOLUTION (10.40) p x L y The deflection curve may be expressed by Eq. (10.23) and the bending strain energy is given by Eq. (10.25). The strain energy of deformation of the foundation is (Chap. 9): #

U 2 = 12 k " v 2 dx = 14 kL ! a n2 0

n =1

Work done by the uniform load: #

W = 12 " pvdx = p( $L )! n1 a n cos n$Lx 0 0

n =1

W = 2$pL ! ann n =1, 3,###

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 10.40 (CONT.) Then, principle of virtual work yields ! 4 EI 2 L3

n 4 a n + kL2 a n = 2npL ! 4

a n = n! ( n 44! pL 4 EI + kL4 ) Substituting this into Eq. (10.23) we obtain the required deflection curve. ______________________________________________________________________________________ SOLUTION (10.41) P L x A y We obtain

v ' ' = 3 aL13 ( L ! x )

and

1 U = EI2 ! ( v ' ' ) 2 dx = 92EIa ( L " x ) 2 dx L6 !

= Also

3 EI 2 L3

2 1

(a)

!W = P " !v A

(b)

Thus, !W = !U gives,

P " !a1 = 32EIL3 ( 2a1!a1 )

v A = a1 = 3PLEI

______________________________________________________________________________________ SOLUTION (10.42) Strain energy, by Eq. (c) of Sec. 10.10:

U = !64EI a2 L3 4

Work done by the load P is L

W = ! pvdx = pa ! dx " pa ! cos 2#xL dx 0

= pa ( L "

2L !

)

Applying the Ritz method, we obtain d" da

Solving

= dad (U ! W ) = #64EI ! pL(1 ! 2#L ) = 0 L3 4

a = 0.1194 pL EI

Substitution of this into Eq. (P10.31) and letting x=L result in the deflection v A at the free end of the beam. ______________________________________________________________________________________ Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 10–22

______________________________________________________________________________________ SOLUTION (10.43)

W = P ! v A = P ( a1 L2 + a 2 L3 ) = PL2 ( a1 + a 2 L) L

Thus,

Solving,

U = EI2 ! ( v ' ' ) 2 dx = 2 EIL( a12 + 3a1a 2 L + 3a 23 L2 ) 0

!" !a1

= 0:

PL2 = 2 EIL( 2a1 + 3a 2 L)

!" !a2

= 0:

PL3 = 6 EIL2 ( a1 + 2a 2 L)

a1 = 2PLEI

a 2 = ! 6PEI

Substituting back into Eq. (P10.32):

v = 6PxEI (3L ! x ) 2

______________________________________________________________________________________ SOLUTION (10.44) c P B L

y We have L

Here

# = EI2 $ ( v ' ' ) 2 dx " P ! v B

(a)

v = ax ( L ! x ) = axL ! ax 2 v ' = aL ! 2ax, v ' ' = !2 a

Substituting these into Eq. (a), after integration, we obtain and or

" = 2a 2 EIL ! PacL + Pac 2 d" da

= 4aEIL ! PcL + Pc 2 = 0

L!c ) a = Pc4(EIL

The deflection of the beam at point B is therefore 2

L!c ) v B = Pc 4(EIL

______________________________________________________________________________________ SOLUTION (10.45) The assumed deflection is of the general cubic form:

v = a1 x 3 + a 2 x 2 + a3 x + a 4

For the left hand portion of the beam: (at x = 0 ); v=0

v' = 0

(at x = L 2 );

(a)

v' = 0 v=!

(at x = 0 )

(at x = L 2 );

(1, 2) (3, 4)

Here ! is, deflection at midspan, to be determined. Eqs. (1) and (2) give a 3 = a 4 = 0 (CONT.) ______________________________________________________________________________________ Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 10–23

______________________________________________________________________________________ 10.45 (CONT.) Eq. (3) yields

a 2 = ! 3a41L

Eq. (4) gives

a1 = " 16L3!

Introducing these into Eq. (a):

v = 4 #L3x (3L " 4 x )

0 ! x ! L2

(b)

Strain energy is

U = 2( EI2 2 ) !

( v ' ' ) 2 dx + 2( EI2 ) ! ( v " ) 2 dx

(c)

! Inserting Eq. (b) into Eq. (c) and integrating, we obtain: U = 72 EI L3

We have

! # = 72 EI " P! L3 2

The midspan deflection, from "# "! = 0, is thus PL ! = 144 EI = v max 3

______________________________________________________________________________________ SOLUTION (10.46) We obtain "

v ' ' = ! a n ( nL# ) 2 cos n#Lx n =1

U = EI2 " ( v ' ' ) 2 dx = EI2 ! a n2 ( nL$ ) 4 L2 0

n =1

W = P ( v ) x = L = P " 2a n 2

( n = 2, 4, 6, ! ! ! )

From the minimizing condition, !" !a n = 0, we obtain or

EIa n ( nL" ) 4 L2 ! 2 P = 0 #

a n = " n44$PL4 EI Therefore,

( n = 2, 4, 6, ! ! ! )

1 v max = v ( L2 ) = $8 PL 4 EI " n 4 3

( n = 2, 4, 6, ! ! ! )

End of Chapter 10

CHAPTER 11 SOLUTION (11.1) Pcr 1

F 2

Pcr

F 2 = Pcr 1

! 4 4 ! (c " b ) = (254 " 204 ) = 18.11!104 mm 4 4 4

Pcr ! 2 EI ! 2 (200 #109 )(0.1811#10"6 ) = = = 143 kN n nL2 2.5(1) 2 F = 2.5(143) = 357.5 kN Justification of the formula used:

A = ! (c 2 " b 2 ) = ! (252 " 202 ) = 706.86 mm 2

r= and

I 181,100 = = 5.06 mm A 706.86

L 1000 = = 197.6 O.K. r 5.06

______________________________________________________________________________________ SOLUTION (11.2) (a)

Same area:

! 2 (d o " di2 ) = bo2 " bi2 4 ! ! bi2 = bo2 " (d o2 " di2 ) = 502 " (502 " 402 ) 4 4 bi = 42.35 mm

(b)

Circular bar

1 t = (bo ! bi ) = 3.83 mm 2

! 4 ! (d o " di4 ) = (504 " 404 ) = 181#10"9 in.4 64 64 2 ! EI ! 2 (72 #109 )(181#10"9 ) Pcr = 2 = = 14.29 kN Le (3) 2

______________________________________________________________________________________ 11.2 (CONT.) Square bar

1 4 4 1 (bo ! bi ) = (504 ! 42.354 ) = 252.8 "10!9 m 4 12 12 2 ! EI ! 2 (72 #106 )(252.8 #10"9 ) Pcr = 2 = = 19.96 kN Le (3) 2

______________________________________________________________________________________ SOLUTION (11.3)

LBC = 22 + 12 = 2.236 m I=

d 50 = = 12.5 mm 4 4

! 4 ! d = (40) 4 = 125.7(10"9 ) m 4 , 64 64

LAB = 12 + 12 = 1.414 m

Bar BC ( L r = 2, 236 12.5) = 178 . Thus

( FBC ) all =

Pcr ! 2 (70)(125.7) = = 9.65 kN 1.8(2.236) 2 n

( FAB ) all =

Pcr ! 2 (70)(125.7) = = 24.1 kN 1.8(1.414) 2 n

Bar AB

Joint B F

FAB

" F = 0 : 2 F ! 5 F = 0, F = 0.791F x

B 1 2

" F = 0 : 2 F + 5 F ! F = 0, F = 1.061F y

FBC

Solving

F = 1.061FAB

F = 1.341FBC

The allowable value for F:

F < 1.341(9.65) = 12.93 kN F < 1.061(24.1) = 25.6 kN

Thus

Fall = 12.93 kN

______________________________________________________________________________________ SOLUTION (11.4) Differential equation for a free-body of a segment of x is given by Eq.(11.1):

d 2v EI 2 + Pv = 0 dx

______________________________________________________________________________________ 11.4 (CONT.) or

d 2v + p 2v = 0 2 dx

P EI

General solution is

v = A sin px + B cos px

(1)

Boundary conditions :

v(0) = 0 : 0 + 0 + B = 0, B=0 v '( L) = 0 : AP cos pL = 0, Ap = 0

Since A ! 0 :

cos For n = 1, Therefore

Pcr =

P L = 0 or EI !2 P = 2 EI 4 L

! P L = n( ) EI 2

! 2 EI 4 L2

______________________________________________________________________________________ SOLUTION (11.5)

1 (240 !1203 " 190 ! 703 ) = 29.13 !106 mm 4 12 A = 240 !120 " 190 ! 70 = 15.5 ! 103 mm 2

I min =

rmin = I min A = 43.35 mm

Le = 0.5 L = 4.5 m

Le r = 4500 43.35 = 103.8 Hence,

" cr =

! 2E ! 2 (200 #109 ) = = 183.2 MPa ( Le r ) 2 (103.8) 2

______________________________________________________________________________________ SOLUTION (11.6)

Le = 0.7 L = 6.3 m . From solution of Prob.11.5: rmin = 43.35 mm . We now have

Le r = 6300 43.35 = 145.3

Hence,

! 2E ! 2 (200 #109 ) " cr = = = 93.5 MPa ( Le r ) (145.3) 2

______________________________________________________________________________________ SOLUTION (11.7) 30 kN

B 1.8 m

FBD

1.3 m 1

" M = 0 : ! 30(3.1) + 2 F (1.8) = 0 , A

I = b 4 12 = 504 12 = 520.8 ! 103 mm 4 ,

r= So,

I = 14.4 mm A

" cr =

FBD = 73.1 kN

A = 50 ! 50 = 2.5 ! 103 mm 2

L 1800 = = 125 r 14.4

! 2E ! 2 (210 #109 ) = = 132.6 MPa ( L r )2 (125) 2

We have

! cr < ! yp

(solution is valid)

( FBD )cr = 132.6 "106 (2.5 "10!3 ) = 331.5 kN and

( FBD )cr 331.5 = = 4.5 73.1 FBD

______________________________________________________________________________________ SOLUTION (11.8) A solid circular section of diameter d:

! 2 d 4

Using Eq. (11.13a), we have

Solving,

! 4 d 64

I d = A 4

4 Pcr 1 ! yp 4 Le 2 ( ) = # ! yp "d2 E 2" d 2 2 ! yp Le Pcr + 2 )1 2 d = 2( "! yp " E

Q.E.D.

A rectangular section of height h and width b:

A = bh

By Eq. (11.13a), we obtain

1 3 bh 12

r2 =

h2 12

Pcr 12 Le 2 1 ! ) = ! yp # ( yp bh E 2" h

______________________________________________________________________________________ 11.8 (CONT.) from which

Pcr 3L2! h! yp (1 # 2e yp2 ) " Eh It is assumed that h ! b . b=

Q.E.D.

______________________________________________________________________________________ SOLUTION (11.9)

P 90 #103 = 30 MPa ! all = = A 3 #10"3 Cc =

(1)

2! 2 (200 "109 ) = 125.7 250 " 106

Assuming L r ! Cc , use Eq. (11.8) and (1):

" all =

! 2 (200 #109 ) = 30 #106 , 2 1.92( L r )

L = 185 r

Our assumption was correct. Hence,

L = (185)r = (185)(25) = 4, 625 mm

L = 4, 625 m

______________________________________________________________________________________ SOLUTION (11.10) P

2! 2 (210 "109 ) = 121.7 280 "106 Le 4.2 = = 42 r 0.1

Cc = 6m

Le=0.7L

Since L ry < Cc , use Eq. (11.9) and (11.8):

5 3 42 1 42 3 n= + ( )! ( ) = 1.75 3 8 121.7 8 121.7 1 " (42) 2 [2(121.7) 2 ] (280 #106 ) = 150.5 MPa ! all = 1.75

So,

Pall = 150.5(27 !103 ) = 4063 kN ______________________________________________________________________________________ Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 11–5

______________________________________________________________________________________ SOLUTION (11.11) P

!"1

k 1

!"2

L L

!"3

L k 3

The moment expression about joints 1 and 2:

" M = 0; " M = 0; or, in general,

PL#$1 ! k (#$1 ! #$ 2 ) = 0

PL(#$1 ! #$ 2 ) ! k (#$ 2 ! #$ 3 ) = 0

PL( " #$ i ) ! k#$ n = 0

(P11.11)

i =1

In matrix form, we have

. PL / k , PL , , ( , , ( ,- PL

k PL / k ( ( PL

0 k

0 0 ( ( ( ( PL PL

( ( ( ( (

( 0 + '011 $ '0$ ( 0 ) !01 2 ! !0! !! !! !! ) !! ( ( )& ( # = & ( # ) ( ( )! ( ! ! ( ! ! ! ! ! ( PL / k )* !%01 n !" !%0!"

That is

[ A]{!" } = {0} Determinant A = 0, yields Pcr . In our case, n = 3. ______________________________________________________________________________________ SOLUTION (11.12) (a)

Pcr is proportional to I . We have I s = !4r

15 "r 15 I h = "4 [ r 4 ! ( 2r ) 4 ] = 16 4 = 16 I s 4

Thus, reduction in Pcr is 6.25 % ( b ) Substituting the given data: 4

I h = 15" ( 064.015) = 37.276(10 !9 ) m 4 2

Pcr = # LEI2 h = # (110"10(1)(.537)2.276"10 ) = 17.99 kN e

______________________________________________________________________________________ SOLUTION (11.13) (a)

Pcr = 2(100) = 200 kN 2

(10 )( 2 ) I = P"cr2LE = 200 = 7.369(10 6 ) mm 4 " 2 (11!109 ) a4 12

= 7.369(10 6 );

a = 96.97 mm

We have 3

(10 ) ! = PA = (100 = 10.63 MPa < 15 MPa 0.09697 ) 2

Thus, a cross section of 97 ! 97 mm is acceptable. (b)

Pcr = 2( 200) = 400 kN 3

(10 )( 2 ) I = 400 = 14.738(10 6 ) mm 4 ; " 2 (11!109 )

a = 115.32 mm

We obtain 3

(10 ) ! = PA = ( 0200 = 15.04 MPa > 15 MPa .11532 ) 2

Dimension is not acceptable. Therefore, 3

(10 ) a 2 = A = 200 ; 15(106 )

a = 115.5 mm

Use a cross section of

116 ! 116 mm

______________________________________________________________________________________ SOLUTION (11.14) We have 4

Hence

) 6 4 I = 1( 50 12 = 0.521 ! 10 mm . 2

Pcr = " LEI = " ( 200!(103)2 )( 0.521) = 114.3 kN 2 2

Thus

.3 Pall = Pncr = 114 2.2 = 52 kN

F = P2all.5 = 252.5 = 20.8 kN

______________________________________________________________________________________ SOLUTION (11.15) ( a ) Using Eq. (11.5), 2

!0.05 12 ) Pcr = " ( 210!010.49)(( 30.6.075 = 127.5 kN )2 ( 2 ) 3

.5(10 ) " cr = 127 = 34 MPa 3.79 (10! 3 )

( b ) We have and

I min = 0.07512( 0.05)

!4 r 2 = I min A = 2.08(10 )

______________________________________________________________________________________ 11.15 (CONT.) Thus, Le r

( 3.6 ) = 0.7r L = 00.7.0144 = 175 1

C c = [ 2!" ypE ] 2 = 121.673 2

Also,

As 121.673 < 175, use Eq. (11.13): 2

( 210!10 ) # all = "1.92 = 35.25 MPa (175 ) 2

Note that 3

(10 ) ! = PA = ( 0.450 05 )( 0.075 ) = 120 MPa

and yielding does not occur. But member fails as a column, since Pcr < P. ______________________________________________________________________________________ SOLUTION (11.16) y P A

B P

P C

Symmetrical buckling shown in the figure, creates relative bending moments M e which resist free rotation of the ends of the member AB and CD. Thus, for member AB:

EI ddx 2v = ! Pv + M e 2

with the general solution

v = C1 cos !x + C 2 sin !x + MP2

where,

(a)

!2 = EIP

Boundary conditions are

v ( 0) = 0

v ' ( L2 ) = 0

v ' (0) = ! = M2 EIe L Introducing v from Eq. (a) into these: C1 + MPe = 0, ! C1" sin "2L + C 2 " cos "2L = 0 C 2 ! = M2 EIe L

The foregoing lead to the following trancendental equation or

tan !2L + 2PL !EI = 0 tan !2L + !2L = 0

from which !L 2 = 2.029. Thus, EI Pcr = 16.47 = ( 0.!774EIL )2 L2 2

The effective length in the situation described is therefore equal to 0.774L. ______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (11.17) P

Let P EI

!1 = M !2 = EI

The governing differential equation is

EIv ' ' = M ! Pv

v ' '+ !1v = !22

y L

P=W

Solution is

v = A cos !1 x + B sin !2 x + ( !!12 ) 2

Boundary conditions give:

v ' ( 0) = 0 = B v (0) = A + ( !!12 ) 2 = 0;

A1 = "( !!12 ) 2

v ' ( L) = ( !!12 ) 2 sin !1 L = 0

sin #1 L = 0; Choose "1 L = ! . Then,

#1 L = 0, " , 2" , ! ! !

"1 = !L = ( EIP ) 2 ; Thus,

Le = L

P = ! LEI 2 2

______________________________________________________________________________________ SOLUTION (11.18) For the vertical bar, from Eq. (11.5):

Pcr = !4 EI L 2

The midspan deflection of the beam is thus 3

Pcr L 5 PL " = 384 EI ! 48 EI 4

2 5 PL 5 PL ! L L " = 384 EI # 192 = 192 ( 2 EI # ! ) 4

______________________________________________________________________________________ SOLUTION (11.19) ( a ) We have " ( #T ) L = ! 2 or (b) where

#T = 2"!L PL " ( $T ) L # AE = !2

P = !4 LEI2 2

Thus,

# " I L 1 $T = 2#!L + "4 LEI2 AE !L = 2!L + 4 L2 A! 2

______________________________________________________________________________________ SOLUTION (11.20) L/4 L/2 L/4

For a fixed ended column Le may be determined as follows. Inflection points and midpoint divide the bar into 4 equal portions. Each portion is the same as the fundamental case (Fig. 11.2a). Thus Le = L 4 . For the dimension given, we obtain L r

932 = 01..0294 = 65.99

Equation (11.6) gives then

" cr = (4L! r E)2 = 0.0091E 2

Substituting E = 174(10 ),

" cr = 0.0091(175 ! 109 ) = 1592.5 MPa Since, from Fig. P11.20, E is valid only up to 175 MPa, the 1592,5 MPa cannot be

critical stress.

Similarly, for inelastic range, from Eq. (11.7):

! cr = 0.0091Et1

= 0.0091( 46.7 ! 10 9 ) = 425 MPa Applying the same reasoning as before, this value is also not a critical stress. For E t 2 :

" cr = 0.0091( 28 ! 10 9 ) = 254.8 MPa We now observe from the sketch that 254.8 falls in the stress range for which E t 2 is valid. Thus, the buckling load:

Pt = ! cr A = 254.8(0.0323) = 823 kN

______________________________________________________________________________________ SOLUTION (11.21) A P C

For bar AB:

1.41P = 4200(5.4 ! 10 "5 ) = 2268

______________________________________________________________________________________ 11.21 (CONT.)

P = 1604 N or For bar BC: 2

= # ( 210"106.25)( 3.9"10 ) P = Pcr = ! LEI 2 2

And

= 1297 N " cr = 5.41297 = 24 MPa (10! 5 )

Conclusion:

Bar BC fails as a column, Pcr = 1297 N .

______________________________________________________________________________________ SOLUTION (11.22) Referring to Fig. P11.22, we write

I y = 2 Arc2 + 2 Ad 2

= 2(0.0225) 2 + 2(1.719 ! 10 "3 )(0.0125 + 0.2325) 2 = 6.13(10 !6 ) m 4 and 1

ry = ( 2 Ay ) 2 = ( rc2 + d 2 ) 2 1

= (0.02252 + 0.035752 ) 2 = 42.24 mm We see that ry for two channels is the same as for one channel. But rz = rc = 0.0225 m. Thus, rz < ry . Columns tends to buckle with respect to z axis. The slenderness ratios: Le rz

Le rz

.10 = 02.0225 ! 94,

From Sec. 11.7: 2

.20 = 04.0225 ! 187

( 210!10 ) C c2 = 2#" ypE = 2" 203 = 20,420; (106 ) 2

( a ) In this case 0 < ( Le

C c = 143

r ) < C c . Then, substituting the given data,

the first of Eqs. (11.8) yields

! all = 97.685 MPa

( b ) Now we have: C c ! ( Le

r ) ! 200 and the second of Eqs. (11.13) gives

! all = 30.87 MPa ______________________________________________________________________________________ SOLUTION (11.23) From Fig. P11.22, we observe that

I y > Iz .

Therefore the column buckles with respect to the z axis. Since, L r

We obtain

.2 = 0.40225 = 186.7 2

210!10 ) " cr = ( L! Er )2 = " ((186 = 59.46 MPa .7 ) 2 2

______________________________________________________________________________________ SOLUTION (11.24) From Eq. (11.6), 2

We have

(140!10 ) 2 ( Lr ) lim it = [ " 280 ] = 70.2 !106

I = bh12 = Ar 2 3

I = ( 0.05)(120.025) = 1.25 " 10 !3 r 2 Solving, r = 7.216 mm 1.2 Hence, ( Lr ) actual = = 166.3 7.216 (10! 3 ) or

Thus, elastic buckling occurs, since ( L r ) lim it < ( L r ) actual . We have ! = " (!T ) L and # = !L = " ( !T ) .

The condition that

% = $ ( "T ) ! ( L# r )2 = 0 2

results in

"T = $ (#L r )2 = 10(10!6#)(166.3)2 = 35.7 o C 2

______________________________________________________________________________________ SOLUTION (11.25) 3

(10 ) " all = PA = 3125 = 40.85 MPa .06 (10! 3 ) 2

( 200!10 ) 2 C c = [ 2" 250 ] = 125.7 (106 )

Also

Assuming ( L r ) ! C c : # all = Solving, Lr = 158.6

" 2 ( 200!109 ) 1.92 ( L r ) 2

) = 1028( L.08r ()10 2

O.K.

Choosing the smallest radii of gyration: L ry

L = 0.0246 = 158.6

from which

L = 3.9 m

______________________________________________________________________________________ SOLUTION (11.26) P A 40 mm

vmax 0.7 m B e

Figure (a)

A = ! (20) 2 = 1256.64 mm 2 ! I = (20) 4 = 125.66 "103 mm 4 4 ! 2 EI Pcr = 2 L 2 ! (200 #109 )(125.66 #10"9 ) = (0.7) 2 = 506.2 kN

______________________________________________________________________________________ 11.26 (CONT.) (a)

Using Eq. (11.18):

# %! 60 & $ o 1'10"3 = e (sec ** ++ " 1) = e #,sec(31 ) " 1$- , 2 506.2 (, . / )(b)

e = 6 mm

Referring to Fig. (a):

M = P (vmax + e) = 80(1 + 6) = 560 N ! m

Hence,

80 "103 560(20)10!3 P Mc = + + 1256.64 "10!6 125.66 "10!9 A I = 63.66 + 89.13 = 152.8 MPa

! max =

______________________________________________________________________________________ SOLUTION (11.27)

Le = 2 L = 1.4 m . Refer to solution of Prob. 11.26

Pcr = (a)

! 2 EI ! 2 (200 #109 )(125.66 #10"9 ) = 126.6 kN = L2e (1.4) 2

Equation (11.18):

# %! 60 & $ o 1'10"3 = e (sec ** ++ " 1) = e[sec(62 ) " 1] , 2 126.6 - /) .( , (b)

e = 0.88 mm

M = P (vmax + e) = 80(1 + 0.88) = 150.4 N ! m . Therefore,

! max =

150.4(20 "10!3 ) P Mc = 63.66 + = 63.66 + 23.9 = 87.6 MPa + 125.66 "10!9 A I

______________________________________________________________________________________ SOLUTION (11.28) A

100 mmC 20 mm

Le = 2 L = 3.6 m A = 200 !100 " 160 ! 60 = 10.4 !103 mm 2

200 mm D

r = I A = 36.4 mm e = c = 50 mm

1 (200 !1003 " 160 ! 603 ) 12 = 13.79 !106 mm 4

Thus,

ec 50 ! 50 = = 1.89 r 2 (36.4) 2

Le = 49.45 2r

______________________________________________________________________________________ 11.28 (CONT.) Use Eq.(11.19) with L = Le :

! max =

% &$ 250 '103 # 250 '103 ( + 1 1.89sec 49.45 * + ) = 98.7 MPa * 10.4 '10"3 ( 70 '109 (10.4 '10"3 ) -+ ) , . /

______________________________________________________________________________________ SOLUTION (11.29)

200 mm

A P

20 mm

Le = 2 L = 3.6 m A = 200 !100 " 160 ! 60 = 10.4 !103 mm 2

D 100 mm B

1 (100 ! 2003 " 60 !1603 ) 12 = 46.19 !106 mm 4

e = c = 100 mm r = I A = 66.6 mm Therefore,

ec 100 !100 = = 2.25 r2 (66.6) 2

Le = 27.03 2r

Apply Eq.(11.19) with L = Le :

! max =

% &$ 250 '103 # 250 '103 ( + 1 2.25sec 27.03 * + ) = 85.7 MPa "3 + 9 * ' ' 10.4 '10"3 ( 70 10 (10.4 10 ) , - )/ .

______________________________________________________________________________________ SOLUTION (11.30)

I = 121 (50)( 25) 3 = 65.1 ! 10 3 mm 4 2

)( 65.1) Pcr = ! LEI = ! ( 210 = 59.97 kN 2 (1.5 ) 2 2

e = h2 = 252 = 12.5 mm,

P = 10 kN

Equation (11.13) gives

v max = 0.0125[sec( "2 Thus

10 59.97

) ! 1] = 3.10 mm

M max = P( e + v max ) = 10(12.5 + 3.10) = 156 N ! m

______________________________________________________________________________________ SOLUTION (11.31)

p = 77(10 3 )(0.05 ! 0.05) = 192.5 N m 4

I = ( 0.1205) = 5.2(10 !7 ) m 4 (a)

" max = McI = 192.58( 9 ) 5.20(.025 = 93.705 MPa 10! 7 ) 4

5 pL v max = 384 EI 4

5(192.5 )( 9 ) = 384 ( 210 = 0.1506 m = 150.6 mm "109 )( 5.2"10! 7 )

( b ) Taking 4

5 pL a0 = 384 EI = 150.6 mm

We obtain, using Eq. (11.15):

v max =

= 227.5 mm

0.1506 4500 ( 9 ) 2 9

( 210"10 )( 5.2"10 )

Applying Eq. (11.16):

0.025 " max = 04500 .025 [1 + 2275 5.2(10! 7 ) 0.025 ]

= 51.019 MPa ______________________________________________________________________________________ SOLUTION (11.32) Given e=0.05 m, c=0.1016 m

rz2 = IAz

and

!3 .66 = 45 5880 = 7.765(10 ) mm

Then,

Pcr =

" 2 EI y L2

= " ( 210( 4!.510)2 )15.4 = 1575 kN

Also, 2

Pcr = " ( 210( 4!.105)2) 45.66 = 4669 kN Thus, Eq. (11.19) becomes 1016 ) # 210(10 6 ) = 588(P10"6 ) [1 + 07..05765( 0(.10 " 3 sec( 2 )

P 4669!103

)]

from which, by trial and error,

P ! 700 kN = Pmax

______________________________________________________________________________________ SOLUTION (11.33) Governing equation is

EIv1 ' ' = !( v0 + v1 )

where,

v0 = a1 sin !Lx + 5a1 sin 2L!x

This may be written

v1 ' '+ "2 v1 = #"2 a1 sin !Lx # 5"2 a1 sin 2L!x

(a)

______________________________________________________________________________________ 11.33 (CONT.) We have

v1 = c1 cos !x + c2 sin !x + v p

(b)

Particular solution is

v p = A sin !Lx + B cos !Lx + D sin 2L!x + E cos 2L!x

(c)

Substituting Eq. (c) into Eq. (a): 2

A = #"2 a1 (

B=0

! "2 )

D = 45#"2 a1 (

E =0

! "2 )

v p = (! "L )a21#"2 sin !Lx + 4 (!5"L )a21#"2 sin 2L!x

Thus,

Boundary conditions

v1 (0) = 0 yield c1 = c2 = 0.

v1 ( L) = 0

General solution is

v = v1 + v p = v0 + v p

Letting

b = "!2 L2 = !PL2 EI 2

we obtain

v = 1a"1b sin !Lx + 204"ab1 sin 2L!x

Solution of the b is found from a1 1"b

Thus,

v( ) = 0 = b = 0.89 2 b = !PL2 EI ,

and

P = 0.89 ! LEI 2

3L 4

sin

3L ) 4 L

+ 20

a1 1"b

sin

3L ) 4 L

2! (

P = b!L2EI 2

______________________________________________________________________________________ SOLUTION (11.34) x ds dv dx v p(x) V

P M

V+dV P

M+dM dv dx

+ !!!

An element isolated from the beam is shown in a deformed state in the figure above. Assume sin " ! " , cos " ! 1, and ds ! dx. Then

"F = 0: y

p = ! dV dx

! V + pbx + (V + dV ) = 0 (a)

______________________________________________________________________________________ 11.34 (CONT.)

"M = 0:

M ! Pdv ! Vdx + pdx dx2 ! ( M + dM ) = 0

Neglecting terms of second order, this becomes dv V = ! dM dx ! p dx

(b)

M = EI ddx 2v

(c)

If the shear and axial deformations are neglected, the moment at any point is 2

Substituting of Eqs. (c) and (b) into Eq. (a) gives

( EI ddx 2v ) + P ddx 2v = p For EI = constant, 2 d 2v + EIP ddx 2v = EIp dx 2 d2 dx 2

(d) (e)

Homogeneous solution of this equation is

v = c1 sin

P EI

x + c2 cos

x + c3 + c 4

P EI

where, c1 through c4 will require for evaluation, four boundary conditions. ______________________________________________________________________________________ SOLUTION (11.35) Given

v = a0 [1 ! ( 4Lx2 )] 2

and hence,

v ' = ! 8 Lxa2 0

v ' ' = ! 8La20

Potential energy function is

# = 2!

1 2

EI ( v ' ' ) 2 dx " 2 !

64 EIa0 64 Pa0 12 L L3

Thus,

"! "a 0

Pcr = 12LEI 2

1 2

P ( v ' ) 2 dx =

32 EIa02 3

32 Pa0 ! 12 L

______________________________________________________________________________________ SOLUTION (11.36) Assume v = v0 sin( !Lx ). Then,

U = 2!

EI 1 (1 + Lx2 )( v ' ' ) 2 dx

= 2 EI 1v0 ' ' "L4 [ ! 4

sin 2 "Lx dx + L2 !

= 2 EI 1v02 !L4 [ L4 + !2 L2 ( !16 + 14 )] 4

x sin 2 "Lx dx ] (a)

We also have L

Thus,

yields

" (v' ) dx = v 0

2 !2 0 L2

" cos 0

2 !x L

2 2

dx = v20!L

2 2 ! EI (v' ' ) dx = P ! (v' ) dx 2

Pcr = ! LEI2 1 ( 23 + !22 ) = 1.7 ! LEI 2 2

______________________________________________________________________________________ SOLUTION (11.37)

v ' ' = 2 Lv12

We have v ' = 2v1 Lx2

Potential energy function is L

# = EI21 ! (1 " 2xL ) 4Lv41 dx " P2 ! 4Lv41 x 2 dx = 23 EIL13v1 ! 23 v1LP Hence, gives

#" #v1

0 3 EI1v1 L3

4 v1P 3L

Pcr = 94EIL21

______________________________________________________________________________________ SOLUTION (11.38) Assume v =

! a sin( ). Then, U= !n a W = ! P ( v ) dx + ! pvdx = !n a + !a n#x L

" 4 EI 4 L3

L 1 2 0

and

2 n

' 2

" 2P 4L

2 PL "

1 n n

Hence, from !U = !W , by letting

b = PPcr = !PL2 EI 2

we obtain "

1 a n = %4 PL 5 EI ! n 3 ( n 2 $b ) 4

1, 3, # # #

Thus, #

1 v = !4 PL sin n!Lx 5 EI " n 3 ( n 2 %b ) 4

1, 3, $ $ $

______________________________________________________________________________________ SOLUTION (11.39) x

P L/2

L/4

Let c1 = L4 and c2 = L2 . Deflection curve is expressed by "

v = ! a n sin n#Lx

(a)

and

U = EI2 ! ( v ' ' ) 2 dx 0

= F ! a n sin n#Lc1 + 2 F ! a n sin n#Lc2 + P2 " ( v ' ) 2 dx 0

______________________________________________________________________________________ 11.39 (CONT.) Therefore,

! (U "W ) !an

gives ! 4 EI 2 L3

n 4 a n"a n = F"a n sin n!Lc1 + 2 F"a n sin n!Lc2 + P2!L a n n 2 2

a n = F sin("n2"EIn4 )2+2 F sin(P n" 2 ) 2 L3

( n2 !

Pcr

)

Solution is found by substituting this into Eq. (a). ______________________________________________________________________________________ SOLUTION (11.40) P x L

Boundary conditions are: v (0) = v ( L) = 0 We have Thus,

Hence,

v = aLx 2 + bLx 3 ! ax 3 ! bx 4

v ' = 2aLx + 3bLx 2 ! 3ax 2 ! 4bx 3 v ' ' = 2aL + 6 BLx ! 6ax ! 12bx 2 L

# = EI2 ! ( v ' ' ) 2 dx " P2 ! ( v ' ) 2 dx

= EI2 [4a 2 L3 + 8abL4 + 245 b 2 L5 ] ! P2 [ 152 a 2 L5 + 15 abL6 + 353 b 2 L7 ]

It is required that "# "a

= EI2 (8aL3 + 8abL4 ) ! P2 ( 154 aL5 + bL5 ) = 0

"# "b

= EI2 (8aL4 + 485 bL5 ) ! P2 ( 15 aL6 + bL35 ) = 0

Letting ! = PL EI , these become

( 4 ! 235" )a + ( 4 ! 10" )bL = 0 ( 4 ! 10" )a + ( 245 ! 335" )bL = 0 Since a ! 0 and b ! 0 : ! ( 4 ! 10" ) 2 + ( 4 ! 215" )( 245 ! 335" ) = 0 or

Solving, Hence,

"2 ! 128" + 2240 = 0 !1 = 20.9

!2 = 107.1

Pcr = 20.L9 EI = 2.12 ! LEI 2 2

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (11.41)

L/2

EI1

EI2

0 L/3

L/3

From Eq. (11.29):

vm +1 + ( "1h 2 ! 2)vm + vm !1 = 0

Here

!12 = EIP1

(a)

!22 = EIP2

Applying Eq. (a) at 1 and 2:

,.1h 2 - 2 1 ) & v1 # &0# * '% " = % " .2 h 2 - 2( $v2 ! $0! + 1 Thus, or

( "1h 2 ! 2)( "2 h 2 ! 2) ! 1 = 0

"1"2 h 4 ! 2"1h 2 ! 2"2 h 2 + 3 = 0 This is written as

P 2 [ 81EL2 I I ] ! P[ 92EIL1 + 92EIL 2 ] + 3 = 0 4

1 2

Solution is 1

( P1, 2 ) cr = 9LE2 ( I 1 + I 2 ) ± 9LE2 [ I 12 ! I 1 I 2 + I 22 ] 2 When I 1 = I 2 :

Pcr = 9LEI2

______________________________________________________________________________________ SOLUTION (11.42) L/2 0

L/2 1

y Boundary conditions yield

v ' (0) = 0; v (0) = 0; v ' ' ( L) = 0; v ' ' ' ( L) = 0;

v0 = 0 v1 = v !1 v3 ! 2v 2 + 2v1 = 0

(1)

v 4 ! 2v3 + 2v1 = 0

(2)

______________________________________________________________________________________ 11.42 (CONT.) Then, applying Eq. (l) of Example 11.8 at points 1, 2, 0, respectively:

(7 ! 2"2 h 2 )v1 + ( "2 h 2 ! 4)v2 + v3 = 0

(3)

( "2 h 2 ! 4)v1 + (6 ! 2"2 h 2 )v 2 + ( "2 h 2 ! 4)v3 + v 4 = 0

(4)

( " h ! 4)v1 + v 2 = 0

(5)

From Eq. (5):

v2 = ! v1 ( "2 h 2 ! 4)

Equation (1) becomes

v3 = 2v1 ! v1 = !2v1 ( "2 h 2 ! 4)v1 ! 2v1 ( "2 h 2 ! 4) ! v1

Substituting this into Eq. (3),

(7 ! 2"2 h 2 )v1 ! ( "2 h 2 ! 4)( "2 h 2 ! 4)v1 ! 2v1 ( "2 h 2 ! 4) ! v1 = 0

from which

"4 h 4 ! 4"2 h 2 + 2 = 0

!2 h 2 = 0.59 = PhEI

Thus,

Pcr = 2.36L2EI

______________________________________________________________________________________ SOLUTION (11.43) L 1 2 3 P P x 4 0

y From symmetry: v1 = v 3 . We have

I ( x ) = (1 + 2Lx ) EI 1

0 ! x ! L2

I ( x ) = (3 ! 2Lx ) EI 1

L 2

Equation (11.29), gives at 0, 1, 2, 3:

!x!L

v1 + v !1 = 0; v1 = !v !1 Ph 2 v2 + [ ( 3 2 ) EI1 ! 2]v1 = 0 v1 + [ Ph EI1 ! 2]v 2 + v1 = 0 2

[ ( 3 Ph 2 ) EI1 ! 2]v1 + v 2 = 0 2

The foregoing equations lead to

, 23 .2 h 2 - 2 ) & v1 # &0# 1 '% " = % " * 1 2 2 v 2 $0! 2 . h - 2( $ 2 ! + from which, since v1 ! 0 and v 2 ! 0 :

( "2 h 2 ! 6)( "2 h 2 ! 1) = 0 Thus,

Pcr = 16 EIL21

______________________________________________________________________________________ SOLUTION (11.44) P

L/4

3L/4

EI1

EI2

P 3

y We have

!1 = LL 44 = 1

! 2 = LL 24 = 2.

Applying Eq. (11.30) at 1 and 2:

[ 16PLEI1 ! 2]v1 + v2 = 0

(a)

v1 [ 16PLEI1 ! 1]v 2 = 0

(b)

2 3

Let

k1 = 16PLEI1 ! 1 2

k 2 = 16PLEI 2 2

Equations (a) and (b) yield then

k1 ! 2

2 3

k2 ! 1

from which or

( k1 ! 2)( k 2 ! 1) ! 23 = 0 L P 2 [ 256 EL2 I I ] ! P[ 162EI + 16LEI1 ] + 43 = 0 2 4

1 2

Solution of this quadratic equation gives the critical load as follows: 1

Pcr = 8 E ( 2LI21 + I 2 ) ! 8LE2 [4 I 12 + I 22 ! 1.24 I 1 I 2 ] 2 In a special case, for I 1 = I 2 = I , the preceding reduces to

Pcr = 24LEI ! 15.L32EI = 8.7 EI 2 L2

End of Chapter 11

CHAPTER 12 SOLUTION (12.1) Components of stress are

! x = PA + MrI ( 0.05 ) = ! ( 090.05)2 + !3375 = 45.84 MPa ( 0.05 ) 4 4 ( 0.05 ) " xy = TrJ = !4500 = 22.92 MPa ( 0.05 ) 4 2

and

" y = " z = ! xz = ! yz = 0

Thus, from Sec. 2.13 (in MPa):

& 2(3 x $ $) xy $0 %

) xy '

(x 3

0 # &30.56 22.92 0 # ! $ 0 ! = 22.92 ' 15.28 0 ! $ ! ' (3x !" $% 0 0 ' 15.28!"

______________________________________________________________________________________ SOLUTION (12.2) A

a B !

D P At instability, member AD (or DC) and BD become in length:

L' AD = L' DC = LAD + 1!n1n1 LAD = 1L!ADn1

L' BD = LBD + 1!nn2 2 LBD = 1L!BDn2 Therefore, Initially:

( 1L!ADn1 ) 2 = a 2 + ( 1L!BDn2 ) 2

(a)

L2AD = a 2 + L2BD

(b)

Eliminating a from Eqs. (a) and (b), we obtain

cos " = LLBD = ( 11!!nn12 ) AD For We have

n1 = 0.2

and

n1 ( 2 ! n1 ) n2 ( 2 ! n2 )

n2 = 0.3

cos ! = 00..78 [ 00..23((11..78)) ] 2 = 0.7351 or

! = 42.68o

______________________________________________________________________________________ SOLUTION (12.3) p PL We write x 2 1 1 2

PL2

M ( x ) = 2 p( L ! x )

Equation (12.7) is then 1

v ' ' = ( KIMn ) n = ( 2 KIP n ) n ( L ! x ) n

v ' = ! "2( L! x2)

2 +1 n

+ c1

( +1)( + 2 ) n n

2 +2

" ( L! x ) n 2 2 ( +1)( + 2 ) n n

+ c1 x + c2

(a)

Boundary conditions yield 2 +1 n

v ' (0) = 0;

c1 = !2L n

v (0) = 0;

2 +2 n

c2 = ! 2 "L 2

Substituting these and Eq. (g) of Sec. 12.5 into Eq. (a), we obtain

v=(

p 2 K !n

1 n

) [

2 +2

( L! x ) n 2 2 ( +1)( + 2 ) n n

2 +2

! 2 Ln 2

( +1)( + 2 ) n n

For n = 1, K = E , and x = L : 4

( +1)( + 2 ) n n

2 +1

+ L2n x ] n

v = 6pLEI ! 24pLEI = 8pLEI

______________________________________________________________________________________ SOLUTION (12.4) P x L

Hence, 1

We write

M = P( L ! x )

v ' ' = ( KIPn ) n ( L ! x ) n = " ( L ! x ) n

v ' = ! " ( L1! x ) n

v= (a)

1 +1 n

1 +2

" ( L! x ) n 1 1 ( +1)( + 2 ) n n

+ c1

+ c1 x + c2

Boundary conditions yield

v ' (0) = 0;

1 +1 n

c1 = !1L n

v (0) = 0;

1 +2 n

c2 = ! 1 "L 1

Substituting these onto Eq. (a), we find an expression for the deflection.

( +1)( + 2 ) n n

For a special case of n = 1 and K = E , the deflection at free end:

v ( L) = 2PLEI ! 6PLEI = 3PLEI 3

______________________________________________________________________________________ SOLUTION (12.5) 1

We have " = K! 4 and I = 23 bh . Here

! = hy ! max = ay ,

a = ! max h

Moment is thus, h

M = ! #ydA = ! yK ( ay ) 4 bdy = 89 Ka 4 "h

1 4

But

" max = K! max = Ka 4 h 4

Hence,

M = ( 89 )bh 2! max

! max = 89bhM2 = 34MhI

______________________________________________________________________________________ SOLUTION (12.6)

M = ! Px Hence,

0! x!a 1 n

1 n

v1 ' ' = ( " KIPxn ) = "!x ,

v1 = ! 1

1 +2 "x n

1 ( +1)( + 2 ) n n

v1 ' = ! "1x n

1 +1 n

+ c1

+ c1 x + c2

(a)

Similarly, M = ! Pa , at a " x " ( L ! a ) : 1

v 2 ' ' = ( " KIPan ) n = "!a n , 1

v 2 ' = ! "a n x + c 3

v2 = !"a n x2 + c3 x + c4 2

Boundary conditions yield,

(b)

v1 (0) = 0;

c2 = 0

v 2 ( L2 ) = 0;

c3 = !a n L2

v1 ' ( a ) = v2 ' ( a );

c1 = "a n L2 ! "a1

1 +1 n

n ( +1) n

1 +2 n

v1 ( a ) = v2 ( a );

1 +2

c 4 = "a 2 ! 1 "a n 1

( +1)( + 2 ) n n

1 +2

! "a1n

n ( +1) n

Substitution of these of these constants into Eqs. (a) and (b) gives the required solution. For a special case of n = 1 and K = E , the midspan deflection is Pa v ( L2 ) = v max = PaL 8 EI ! 6 EI 2

v max = 24PaEI (3L2 ! 4a 2 )

We compute

y = 23.19 mm (measured from top surface) and I = 4.4(10 !7 ) m 4 .

Therefore

) v max = 24 ( 4.04."4510(!8000 [3(1.2) 2 ! 4(0.45) 2 ] 7 ) 200"109

= 0.00598 m = 6 mm ______________________________________________________________________________________ Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 12–3

______________________________________________________________________________________ SOLUTION (12.7)

! (0.06) 2 = 2827.44 #10"6 m 2 4 ! ABC = (0.05) 2 = 1965.5 #10"6 m 2 4

AAB =

(a) Since ABC < AAB :

Pmax = ! yp ABC = 240 #106 (1963.5 #10"6 )

= 471.2 kN (b) Loading Segment AB deforms elastically:

! AB =

PL 471.2 #103 (1.5) = = 1.19 mm AE 2827.44 #10"6 (210 #109 )

Segment BC deforms plastically:

(! BC ) max = ! AC " ! AB = 10 " 1.19 = 8.81 mm

After unloading Segment BC remains elastic. Thus

(! AB ) p = 0

Segment BC remains plastic. We have:

" yp = # yp L =

! yp E

240 $106 (1) = 1.15 mm 210 $109

P Pmax

!yp

!BC

(!BC)p (!BC)max Referring to the above figure:

(! BC ) p = (! BC ) max " ! yp = 8.81 ! 1.15 = 7.66 mm

______________________________________________________________________________________ SOLUTION (12.8)

! (0.03) 2 = 706.85 #10"6 m 2 4 ! ABC = (0.02) 2 = 314.16 #10"6 m 2 4 AAB =

______________________________________________________________________________________ 12.8 (CONT.) (a) Since ABC < AAB :

Pmax = ! yp ABC = 280 #106 (314.16 #10"6 )

= 87.96 kN (b) Loading Segment AB deforms elastically:

! AB =

PL 87.96 #103 (1.5) = = 0.89 mm AE 706.85 #10"6 (210 #109 )

Segment BC deforms plastically:

(! BC ) max = ! AC " ! AB = 10 " 0.89 = 9.11 mm

After loading Segment AB remains elastic. Thus

(! AB ) p = 0

Segment BC remains plastic. We have:

280 #106 (1) = 1.33 mm L= " yp = 210 #109 E

! yp

Referring to the solution of prob. 12.1:

(! BC ) p = (! BC ) max " ! y

= 9.11 ! 1.33 = 7.78 mm ______________________________________________________________________________________ SOLUTION (12.9) B

A L

40o 40

C o

Using Eqs. (P12.9):

P cos 2 40o N AD = N CD = 1 + 2 cos3 40o = 0.309 Pu

N BD =

P = 0.527 P 1 + 2 cos3 40o

Since N BD > N AD , we have

0.527 Pu = ! yp A = 250 #106 (400 #10"6 ) or

Pu = 189.8 kN

______________________________________________________________________________________ SOLUTION (12.10) See Eqs.(i) & (j) of Example 12.3:

Ps' = 4.32 Pa'

2 Pa' + Ps' = 400

(1, 2)

Solving,

Pa' = 63.3 kN

Ps' = 189.9 kN

Then

189.9 #103 = 253 MPa !s = 750(10"6 )

63.3 #103 = 126.6 MPa !a = 500(10"6 )

Since 253 MPa > 240 MPa and 126.6 MPa < 320 MPa, steel bar yields while aluminum bars remain elastic. Thus

(! s ) res = 253 " 240 = 13 ksi

(! a ) res = 0

______________________________________________________________________________________ SOLUTION (12.11) Rod begins to yield at:

( Pr ) yp = (! r ) yp Ar = (250)(45) = 11.25 kN

(" r ) = (# r ) yp L =

(! r ) yp Er

250 $106 L= (1.2) = 1.5 mm 200 $ 109

The result is shown in Fig. (a). Here Yr corresponds to the onset of yield in the rod. Tube begins to yield at:

( Pt ) yp = (! t ) yp At = (310)(60) = 18.6 kN

(" t ) yp =

(! t ) yp Et

310 #106 (1.2) = 3.72 mm 100 # 109

The result is shown in Fig. (b), where Yt represents the onset of yield in the tube. Total P-" of the rod-tube combination:

P = Pr + Pt

! = ! r = !t

The result is in Fig. (c).

P (kN)

Pt (kN) Pr (kN) 11.25

29.85

18.6

3.72

!t (mm) 0

Yt Yr

Yt 1.5

(a)

!r (mm) 0

1.5

(b)

3.72 ! (mm)

1.5

(c)

______________________________________________________________________________________ SOLUTION (12.12) We have e = 6 " 2 = 4 mm, I = (1 12)(12)

= 1.728 ! 10 3 mm 4 and c = 6 mm.

( a ) Applying Eq. (12.9):

M yp = or

" yp I c

= 350!61.728

M yp = 100.8 N ! m

( b ) Equation (12.10) results in

M u = a" yp ( a 2 ! e3 ) = (0.012)(350)(12 2 ! 43 ) 2

= 582.4 N ! m ______________________________________________________________________________________ SOLUTION (12.13)

! yp

! yp c

h-c

Referring to this figure,

(! yp ) ca2 = ( h " c )a! yp

or Hence,

c = 2h3 M = 12 2 3ha ! yp 23 23h + h3 a! yp h6 11 = ( 54 )ah 2! yp

______________________________________________________________________________________ SOLUTION (12.14) L/2

p +

C L

L/2

pL2 2

Deflection at C (Case 7 of Tables D.4), in elastic range: 4

v max = 384pLEI Start of yielding:

p = p yp = Thus, M L

12 M yp L2 ( 2 bh 2!

3) L2

! L2

v max = 32ypEI = 32 E ( 2ypbh3 3) = 32ypEh

______________________________________________________________________________________ SOLUTION (12.15) Equation (12.9): (a)

(b)

2 M yp = bh 2! yp 3 2 ! 0.06(0.04) 2 = (240 !106 ) = 15.36 kN " m 3 Equation (12.10b):

3 1 20 M = (15.36 !103 )[1 " ( ) 2 ] = 21.12 kN ! m 2 3 40

______________________________________________________________________________________ SOLUTION (12.16) (a)

Equations (12.9) and (12.10b):

M 3 1 e = 1.3 = [1 ! ( ) 2 ] 2 3 h M yp or

1 e2 0.867 = 1 ! ( 2 ), 3 h

e = 0.632h

(b) The residual stress pattern will be as in Fig. 12.16c. ______________________________________________________________________________________ SOLUTION (12.17)

A = 2(10)(40) + (10)(30) = 1100 mm 2 Neutral axis divides section into two equal areas:

(30)(10) + 2(10)(h1 ) = Solving

h1 = 12.5 mm

Therefore,

Z= where

y1 =

A = 550 mm 2 2

h2 = 27.5 mm

A( y1 + y2 ) 2

y 10

h1 z

h2 10

" A y = 1 [2(h )(10)( h ) + 10(30)(h ! 5) ] = 6.93 mm 2 "A A 2 i

y2 =

h 1 [2(10)(h2 )( 2 )] = 13.75 mm A2 2

______________________________________________________________________________________ 12.17 (CONT.) So

1100 (6.93 + 13.75) = 11,374 mm3 2 M u = ! yp Z = (260 #106 )(11,374 #10"9 ) Z=

= 2.96 kN ! m ______________________________________________________________________________________ SOLUTION (12.18)

! c3 4 2 ! c 4c Z = 2 Ay = 2 2 3! 3 4c = 3

y= C z

4c 3!

Thus,

f =

! c2 2

Z 16 = " 1.7 S 3!

______________________________________________________________________________________ SOLUTION (12.19) y A1 h

h/2

b/2

bh h bh h 7bh 2 ! ]= 2 4 2(4) 8 32 I 2 1 b h S= = [bh3 ! ( )3 ] h 2 h 12 2 2

Z = 2 A1 y1 ! 2 A2 y2 = 2[

bh 2 1 15 (1 ! ) = bh 2 6 16 96

Thus,

f =

Z 7bh 2 96 = = 1.4 S 32 15bh 2

______________________________________________________________________________________ SOLUTION (12.20)

1 (70)(120)3 12 = 10 !106 mm 4 I M yp = ! yp c 10 "10!6 = (250 "106 ) 0.06 = 41.7 kN ! m

120 mm

z 70 mm

M u = fM yp =

3 (41.7) = 62.6 kN ! m 2

Elastic rebound stress

Mc 62.6 #103 (0.06) = I 10 #10"6 = 376 kPa

! 'max =

The results are sketched (in MPa) below.

-250

Mu =0

116

376

250

-250

60 mm 250 Loading

-376

-116

Unloading

Residual stresses

______________________________________________________________________________________ SOLUTION (12.21) Initial Yielding: where

" yp = !Nr12 + 4!Mr 31 N yp = !r 2" yp ,

(a)

M yp = " yp!r 3 4

______________________________________________________________________________________ 12.21 (CONT.) We express Eq. (a) in the from N1 N yp

+ MMyp1 = 1

(1)

Fully Plastic Deformation:

A2 e

dy y

From a Mathematics Handbook table: 1

A2 = "2r ! [e( r 2 ! e 2 ) 2 + r 2 sin !1 ( er )] 2

Thus,

A1 = e( r 2 ! e 2 ) 2 + r 2 sin !1 ( er ) 1

N 2 = 2" yp [e( r 2 ! e 2 ) 2 + r 2 sin !1 ( er )]

(2)

We also write 1

dA = 2( r 2 ! y 2 ) 2 dy r

Q = " 2 y ( r 2 ! y 2 ) 2 dy = 23 ( r 2 ! e 2 ) 2 e

Here Q is the first moment of the area A2 . Hence, 3

M 2 = 2Q! yp = 43 ( r 2 " e 2 ) 2 ! yp

(3)

M u = 43 r 3" yp = 316! M yp Solving Eq.(3), 2

e = [ r 2 " ( 43 !Myp2 ) 3 ] 2 Substituting this into Eq. (2): 2

N 2 = 2! yp [ r 2 " ( 43 !Myp2 ) 3 ] 2 ( 43 !Myp2 ) 3 + 2r 2! yp sin "1 [1 " ( 43r 2 !Myp2 ) 3 ] 2 The foregoing results in !N 2 2 N yp

= ( 316! ) 3 ( MMyp2 ) 3 [1 " ( 316! ) 3 ( MMyp2 ) 3 ] 2 + sin "1 [1 " ( 316! ) 3 ( MMyp2 ) 3 ] 2

(4)

The governing equations for yielding to impend and for fully plastic deformation are given by Eqs. (1) and (4). A sketch of these, interaction curves, are shown below. (CONT.) ______________________________________________________________________________________ Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 12–11

______________________________________________________________________________________ 12.21 (CONT.) N N yp

1.0

0.8

r M

Initial yielding

0.6 0.4

Fully plastic N M yp N yp N

0.2 0

= 201

0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6

M M yp

Figure P12.21 ______________________________________________________________________________________ SOLUTION (12.22) Let P=N. Then referring to Fig. P12.22, we have Thus, Also, and

M = Nd = N (0.05 + 0.05 + 0.025) = 0.125N N M = 1 0.125 M yp N yp

("r 3 4 )! yp

"r 3! yp

= 0.025 4

M M yp = 20( N N yp )

Now referring to Fig. P12.21, we find that B (1.68, 0.084) . Therefore,

N 2 = 0.084 N yp = 0.084("r 3! yp )

= 0.084" (0.025) 2 ( 280 ! 10 6 ) = 46.18 kN ______________________________________________________________________________________ SOLUTION (12.23) The plastic hinges for at 1 and 2, Fig. (a). x L" x

! Pu 1

2 x

L-x

Figure (a)

Apply the principle of virtual work:

Pu ( x"# ) = M u [2("# + Lx"# ! x )]

from which

Pu = x ( 2LL! x ) M u

The condition dPu

dx = 0 gives x = L 2 . Minimum magnitude of the ultimate load is thus

Pu = 8 ML u

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (12.24) b h

h1 e b1 2

h1 ! e ! h ! yp

0 ! e ! h1 ! yp

b1 2

Figure (b)

Figure (a)

Figure(c)

Initial Yielding:

! yp = NA1 + MI1c = 2 ( bhN!1b1h1 ) + ( 2 3)(Mbh13h!b h3 )

(a)

1 1

Here,

N yp = 2(bh " b1h1 )! yp M yp = 32h (bh 3 " b1h13 )! yp

Equation (a) may now be written N1 N yp

+ MMyp1 = 1

(1)

Fully Plastic Deformation: For 0 ! e ! h1 (Fig. b):

N 2 = 2 A1! yp = 2(b " b1 )e! yp

from which

N2 2 ( b "b1 )! yp

Figure (d)

We have (Fig. d):

A2 = b( h ! h1 ) + ( h1 ! e)(b ! b1 ) = bh ! eb + eb1 ! b1h1 h ! h1

b ( h ! h1 )( +1) 2 2 2 2 Ay 2 y=" = = 1 bh !b1h1 !e b+ e b1

Thus, For

bh ! eb + eb1 !b1h1

M 2 = 2! yp A2 y = (bh 2 " b1h12 " e 2 b 2 + e 2 b1 )! yp 2

e = 0:

(b)

2 1 1

M 2 = M u = (bh " b h )! yp .

Substituting the given data, the preceding expressions become

N yp = 1.68h 2! yp ,

M yp = 0.922h 3! yp

M u = 1.113h 3! yp and

Mu M yp

= 1.21,

M yp = 0.55hN yp

Hence, Eq. (b) leads to M2 1.21M yp

= 1 ! 3.116 ( NNyp2 ) 2

This is valid for

0 ! 0 ! h1

(2a)

0 ! NNyp2 ! 1.67

______________________________________________________________________________________ 12.24 (CONT.) For h1 ! e ! h2 (Figs. c and e):

N 2 = 2[0.7h + 0.2h + ( e " 0.7)2h ]! yp = ( "2.52h 2 + 4eh )! yp

e = 4 hN!2yp + 0.63h

M 2 = 2h( h " e)2( e + h 2"e )! yp 3

= 2h ! yp " 2he ! yp

Figure (e)

Hence, M2

1 M2 2.17 M yp

N2 = 1 ! he 2 = 1 ! 16 hN4"2 2 ! 0.h315 ! 0.397 2 " 2

2 h 3" yp

= 0.603 ! 0.176(

which is valid for

h1 ! e ! h

N2 2 N yp

) ! 0.529( NNyp2 )

(2b)

1.67 ! NNyp2 ! 1

Equations (1) and (2) are the governing expressions of the plastic bending. A sketch of these, interaction curves, are shown in the figure given below. N N yp

1.0

0.8

0.6 Fully plastic

Initial yielding

0.4 0.2 0

0.2 0.4 0.6 0.8 1.0 1.2

M M yp

Figure P12.24 ______________________________________________________________________________________ SOLUTION (12.25) P 1

2 #1

We have

v 3 L/3

2L/3

v = 43 L!1 = 23 L! 2

!v = 43 L!" 1 = 23 L!" 2 from which 2!" 1 = !" 2 .

and

Applying the principle of virtual work:

Pu!v = M u!" 1 + M u (!" 1 + !" 2 ) + M u!" 2

or Solving,

4 3

Pu L!" 1 = 6 M!" 1 Pu = 92MLu

______________________________________________________________________________________ SOLUTION (12.26) x p p px (a) L L/2 L/2 e 1 #1 1

v 2 #2

3 3 #3

(b)

Figure P12.26

v 4

(c)

We have two different modes to be checked for collapse loading. Mode A, Fig. P12.26b:

!1 = ! 2 ( Le "e ) ;

!1 + ! 2 = L!e 2

Applying the principle of virtual work: e

! $ ("1 x ) Lx pdx + ! $ ["1e # " 2 ( x # e)] Lx pdx = M u!" 1 + M u!" 2 + M u!" 2 e

0 e

# " $1 x pdx + # " [$1e ! $ 2 ( x ! e)] Lx pdx = M u#$ 2 ( Le + 1)

x L

(a)

Integrating the left hand side of this equation, carrying out the algebra and simplifying:

"# 2 p[ L !6 e ] = M u"# 2 ( Le + 1) 2

Solving,

p = e 6( LM!ue )

(b)

Then, dp de = 0 gives,

0 = ! e26(ML!ue ) + e 6( LM!ue ) ;

e = L2

Introducing this value of e into Eq. (b), the collapse load is

Pu = 24LM2 u Mode B, Fig. P12.26c: From symmetry ! 3 = ! 4 = ! . Principle of virtual work gives: L

! $ [" L2 # " ( x # L2 )] pdx = 3M u!" L2

! $ ["L # "x ] pdx = 3M u!" L2

Integrating, 2

!" L8p = 3M u!" Pu = 24LM u

Note that the collapse load is the same for modes A and B. ______________________________________________________________________________________ Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 12–15

______________________________________________________________________________________ SOLUTION (12.27) Assume that plastic hinges force at 1 and 2, as shown in Fig. (a). On the average, plastic limit load a# b

pu 1

# a

3 #

Figure (a)

Note that pu goes through a virtual displacement of a!" 2 . Thus, applying the principle of virtual work: Solving,

( pu L) 12 a!" = M u!" + M u (1 + ab )!" 2 L!a pu = 2 ML u ( La ) !a 2

The unknown distance a is determined from dpu

da = 0. In so doing and

simplifying the result, we obtain

a = (2 ! 2 ) L and

b = L ! a = ( 2 ! 1) L. ______________________________________________________________________________________ SOLUTION (12.28) ! yp y x dy y h h x 3

! yp

b ( a ) We have M yp = ! yp I From geometry, x b2

Then,

= ( h h2 )2! y ;

I = 2!

h x = b ( h2!h2 y )

( 2 x )dy ( y 2 ) = bh48

Hence, total yielding moment

M u = 2 bh48 h1 ! yp = bh24 ! yp 2

Also,

M u = 12 (area of rhombus) ! (distance between centroids) (! yp ) 2

= bh4 ( h3 )! yp = bh12 ! yp Thus,

Mu M yp

______________________________________________________________________________________ 12.28 (CONT.) y (b) r +

I = " ( b 4!a ) !

M yp = "4 (b 4 # a 4 ) byp = 4"b (b 4 # a 4 )! yp Also, referring to the figure: Hence,

M x = 2 " " r 2 sin $drd$ = 43 (b 3 ! a 3 )

y c = MAx = 43 (" 2b)(!ba2 !a 2 ) = 34" bb2 !!aa 2 3

We therefore have, Mu M yp

(" 2 )( b 2 # a 2 )!

= (" 4 b )( b4 #a 4 )!yp c = 163"b bb4 !!aa 4 yp

______________________________________________________________________________________ SOLUTION (12.29) (a)

! max L 6000 $10#6 (0.5) = = 0.10 rad c 0.03 ! 180 #106 = 2600 µ < 6000 µ " y = yp = 70 # 109 G

and shaft is yielded.

6000 µ 2600 µ

60 mm 2$0

(b)

From similar triangles:

!0 30 = 2600 6000

!0 = 13 mm

Elastic core. Use Eq.(b) of Sec. 12.9:

T1 =

!"03 ! # yp = (0.013)3 (180 $106 ) = 621 N % m 2 2

Outer part. Use Eq.(c) of Sec. 12.9:

2! 3 2! (c $ "03 )# yp = (0.033 $ 0.0133 )(180 %106 ) 3 3 = 7.01 kN ! m

T2 = Thus,

T = T1 + T2 = 7.63 kN ! m

______________________________________________________________________________________ SOLUTION (12.30)

G = 26 "106 GPa, (a)

! yp = 140 MPa (Table D.1)

For partially plastic shaft, using Eq.(12.19):

(

!0 3 3T 6T ) = 4$ = 4$ 3 c Typ " c # yp

Substituting the given values

!0 3 6(4.5 #103 ) ( ) = 4$ = 0.0711 0.025 " (0.025)3 (140 #106 ) !0 = 12.4 mm (b)

$y =

"0# ! yp = , L G

! yp L G "0

140 "106 (1.2) = 0.5211 rad = 29.9o 26 " 109 (0.0124)

______________________________________________________________________________________ SOLUTION (12.31) 80 mm

a=2 m

50 mm

b=1.5 m

240 #106 = = 3000 µ " yp = 80 #109 G c! 0.04(0.25) (" max ) AC = = = 5000 µ a 2 0.025(0.25) (! max )CB = = 4167 µ 1.5

! yp

Both segments are yielded and partially plastic. Segment AC

"0 =

%max

c! y

! max

0.04(3000 "10!6 ) = = 24 mm 5000 " 10!6 ! c3 ! (0.04)3 Typ = " yp = (240 #106 ) = 24.127 kN $ m 2 2

Use Eq. (12.19):

TAC =

4 1 24 (24,127)[1 ! ( )3 ] = 30.43 kN " m 3 4 40

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 12.31 (CONT.) Segment BC

"0 = Typ =

c! yp

! max

0.025(3000 $10#6 ) = 1.8 mm 4167 $10#6

!c ! (0.025)3 " yp = (240 #106 ) = 5.89 kN $ m 2 2

Use Eq. (12.19):

4 1 18 TBC = (5.89)[1 ! ( )3 ] = 7.12 kN " m 3 4 25

Total applied torque is therefore

T = TAC + TBC = 37.55 kN ! m

______________________________________________________________________________________ SOLUTION (12.32) For r=0, Eq. (f) of Sec. 12.11 leads to c1 = 0. Then, in plastic zone 2 2

! r = ! yp % #$3 r ,

! " = ! yp

If the plastic zone extends to radius c: 2 2

# c = # yp $ !"3 c

which may be found directly from Eq. (12.30) by setting a=0. The outer elastic zone is represented by an annular disk yielding at the inner radius c, wherein radial stress is ! c . We follow a procedure similar to that described in Sec. 12.11 for an annular disk. Boundary conditions:

( u ) r =0 = 0 are substituted into Eqs. (8.37) to obtain c1 and c2 . We then determine the stresses in the (! r ) r =c = ! c ,

elastic region as follows: #

# r = 24 Nyp 2 [3(1 + " ) ! (1 + 3" ) rc2b2 ](1 ! br2 ) "

where

" # = 24 Nyp 2 [ bc4 (1 + br 4 )(1 + 3! ) + 3! (3 + ! ) $ 3(1 + 3! ) br2 ] 2

c b ) !1] N 2 = 8+(1+3" )[( 24

______________________________________________________________________________________ SOLUTION (12.33) h a b The sand volume is

V = 12 (b ! a )ah + 2( 13 a 2 h2 )

______________________________________________________________________________________ 12.33 (CONT.) Slope

! yp = (ah2 )

The ultimate torque is thus 2

Tu = ( 3b"6a ) a ! yp The yield torque given in Table 6.2, by letting ! = ! yp . ______________________________________________________________________________________ SOLUTION (12.34)

a 3

a 3 3

3 ) = 33 ha 2

Volume = 13 h ( 12 ! 2a ! a

h = a 3 3 ! yp

Slope = a h3 3 = a3h3 = ! yp ,

Tu = 2V = 23 a 3! yp

(a)

( b ) From Table 6.2: ! A = 20 T3 = 20T3 a1

Thus,

T yp = 820a ! yp 3

( c ) Referring to the preceding results in items (a) and (b): Tu T yp

= 53

______________________________________________________________________________________ SOLUTION (12.35) Equilibrium condition, from Eq. (8.2): d dr

(t" r ) ! t (" # r!" r ) = 0

Profile is, using Eq. (8.36) with s = 1, rt = at a . For full plasticity

! " # ! r = ! yp Thus,

t# r = " atra

# yp r

# at

dr = ! ypr a + c1 ! at

c1 = ypb a Then, noting that " r = ! pi at t = t a : Since, (! r ) r =b = 0; pi " yp

= a ( 1r ! 1b )

pi = a ( brb" r ) ! yp

______________________________________________________________________________________ SOLUTION (12.36) We have ! = 1 2 and substituting the given data into Eq. (12.42): 6

pr0 (14!10 )( 0.5 ) t0 = ( 2 K0.606 = ( 2!9000.606 3 )( n 3 ) !106 1.73)( 0.2 1.73)0.2

= 0.0063 m = 6.3 mm ______________________________________________________________________________________ SOLUTION (12.37) In this case, we have " z > " ! . The total force is where

P = 2"rt! 1 "1 = " z " 2 = "!

(a)

The values of r and t are given by Eqs. (g) and (f) of Sec. 12.12. Substituting these into Eq. (a):

P = 2"r0 e $#1! 1 At instant of stability,

dP = ( !!#P1 )d# 1 + ( !!"P1 )d" 1 = 0 d! 1 d"1

= !1

Equations (12.28) has the form

# 1 = f (" )! 1n

from which stability condition is Then,

!1 = n

Solving,

n = ( "K1 ) n (! 2 # ! + 1)( 2#2! ) ( 1# n )

" 1 = K ( 2n ) n ( ! 2 #1! +1 ) 2 ( 2#1! ) n

(b)

Since ! 1 + ! 2 + ! 3 = 0, the maximum principal strain is

! 1 = ln LL0 = ln tt0 + ln rr0 Minimum strain is

t = t0 ln "1 ! 3

! 3 = ln tt0 ; Thus,

# 1 = 2"Prt = 2"r ln $1 !P t ln $1 ! o

2 0

Substituting Eqs. (12.38b) and (12.38c) into this equation:

#1 = or

t0 =

2$r0t0 ln !1 [( 2$r0# 1 ln !1 [(

P #1 1 2 2 1! n " ( !1)] ) (" !" +1) K 2 2

where, ! 1 is given by Eq. (b).

Results of the preceding expressions are simplified by setting ! = 1 2 . ______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (12.38) Since the mean radius does not change, we have d! 0 = 0. We also take ! 3 = ! r = 0. Material is incompressible d! L = d! 3 , or ! L = "! 3 . The Levy-Mises equation is thus d# L " L $" !

Hence, And

= d"#!L

" L = 2" !

! e = 23 ! L ! e = ( 23 )d! L = "( 23 )d! 3

Radius r0 = constant. Therefore,

P = 2"r0 # t! L

At instant of instability, dP = 0 : Hence,

d" L "L

= # dtt = #d! 3 = d! L

! e = 23 n = 23 ! L

" e = K ( 23 ) n = 23 " L ,

t = t0 e ! n

At instant of instability, the load is then given by Eq. (P12.38). ______________________________________________________________________________________ SOLUTION (12.39) ( a ) Use Eq. (8.10), with " max = ! yp " yp 2n

= bp2 i!ba 2 =

pi a 1!( ) 2 b

2n :

;

Substituting the given data:

50 =

60 50 2 ) b

1!(

5 6

= 1 ! ( 50b ) 2

Solving, b = 122.5 mm ( b ) Apply Eq. (12.47) with k = 1 and n = 3 : !

pu = nyp ln( ab ); or Solving,

50 b

! ( 150 ) 250

50 ! 60 = 250 2.5 ln( b )

= 0.549

b = 91.11 mm

______________________________________________________________________________________ SOLUTION (12.40) Apply Eq. (12.61a): with c = 1.4a and k = 2

2 3

. We have pi = ! r at r = a. Thus

pi = 23 ( 260)[ln( 1.a4 a ) ! ( 2 a 2) (!2(a1).24 a ) ]

= 300.2[ !0.336 ! 0.255] = !177.4 MPa ______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (12.41) ( a ) Using Eq. (8.11):

36! 25 p yp = 420 3 2 ( 36 ) = 21.39 MPa We have k = 1, and thus,

6 pu = nyp ln ab = 420 3 ln 5 = 25.53 MPa

(b)

( nyp ) 2 = ! "2 # ! " ! r + ! r2 or

+ 25 2 36+ 25 (140) 2 = p 2yp [( 36 36! 25 ) + 36! 25 + 1]

Solving, p yp = 22.92 MPa Now k = 2

3 , and hence, pu = 23 ( 25.53) = 29.48 MPa

______________________________________________________________________________________ SOLUTION (12.42) Using Eqs. (12.57) and (12.58) with k = 1 and r = 0.25 m,

pu = 400 ln( 00..23 ) = 162.2 MPa

and

.3 " r = !400 ln( 00.25 ) = !72.93 MPa 0.3 # " = 400[1 ! ln 0.25 ] = 327.1 MPa " z = 12 (" r + " ! ) = 127.1 MPa

Unloading from pu . At r = 0.25 m, Eqs. (8.12) and (8.20): 2

.2 ) .3 " r = 00.2.32(!162 (1 ! 00.25 2 ) = !57.09 MPa 0.2 2 .2 ) .3 # " = 00.2.32(!162 (1 + 00.25 2 ) = 316.6 MPa 0.2 2

" z = 162.2 0.302 .!20.22 = 129.8 MPa 2

Residual stresses at r=0.25 m:

(# " ) res. = 327.1 ! 316.6 = 10.5 MPa (" r ) res. = !72.93 ! 57.09 = !15.84 MPa (" z ) res. = 127.1 ! 129.8 = !2.7 MPa

______________________________________________________________________________________ SOLUTION (12.43) We have k = 1. Refer to Eq. (12.57). Inner cylinder, at r = b :

" pb = " pu + (! yp ) i ln ab Outer cylinder, at r = c : 0 = " pb + (! yp ) o ln bc

(a) (b)

______________________________________________________________________________________ 12.43 (CONT.) From Eqs. (a) and (b), after eliminating pb , we obtain

pu = (! yp ) i ln ab + (! yp ) o ln bc 50 = 280 ln 30 20 + 400 ln 30 = 317.9 MPa

______________________________________________________________________________________ SOLUTION (12.44) ( a ) Equation (12.60):

pc = k! yp 32 ("32)2 = 0.2778k! yp 2

( b ) Equation (12.61a):

(! r ) r =a = k! yp [ln 12 " 0.2778] = "0.9709k! yp

( c ) Equation (12.59b):

(! " ) r =b = Equation (12.59b):

(! " ) r =c = Equation (12.61b):

0.2778 k! yp ( 32 ) 32 # 2 2 0.2778 k! yp ( 32 ) 32 # 2 2

(1 + 332 ) = 0.444k! yp 2

(1 + 232 ) = 0.7222k! yp 2

(! " ) r =a = k! yp (1 + ln 12 # 0.2778) = 0.0291k! yp

We see from these results that the maximum stress occurs at the elastic-plastic boundary.

End of Chapter 12

CHAPTER 13 SOLUTION (13.1)

a x b y

( a ) Boundary conditions at y=0 and y=b:

w=0

dw dy

(a)

We have 2

w' ' ' ' = pD0

w' ' = p20Dy + c1 y + c2

w' = p60Dy + 12 c1 y 2 + c2 y + c3 4

p0 y c1 y c2 y w = 24 D + 6 + 2 + c3 y + c 4 Conditions (a) yield c4 = 0, c3 = 0, 2

c1 = ! 2p0Db Thus,

p0b c2 = 12 D

p0b y 4 y 3 y 2 w = 24 D [( b ) ! 2( b ) + ( b ) ]

( b ) Differentiating twice the foregoing expression, we have 4

p0b 2 12 12 2 w' ' = 24 D [ b 4 y ! b3 y + b 2 ]

For y = b2 :

d 2w dy 2

p0b = ! 24 D 4

Hence,

0b M y = p24

Thus,

! y ,max = p40 ( bt ) 2 ,

! x ,max = 12p0 ( bt ) 2

Similarly, for y=0: d 2w dy 2

and

p0b = ! 12 D ,

M y = p120b

! y ,max = 12 p0 ( bt ) 2 = ! max ! x ,max = 16 p0 ( bt ) 2

______________________________________________________________________________________ SOLUTION (13.2) We have rx = r ,

ry = !

and r = 0.12 m, Equation (13.3b):

t = 0.3(10 !3 ) m

.3 ! max = 2tr = 2 (0120 ) = 1250 µ

Equation (13.5): 9

"10 ( 0.3) $ max = ! 2 (1!Et# 2 ) r = ! 200 2 ( 0.91)120 = 274.7 MPa

______________________________________________________________________________________ SOLUTION (13.3) Using Eq. (13.8), 9

)( 0.012 ) D = 9 ( 200!10 = 32,400 12 ( 8 )

From Example 13.1:

w = ( !b ) 4 pD0 sin( !by ) !3 "10 wmax = ( 0#.6 ) 4 20 32 , 400 = 0.82(10 ) m = 0.82 mm 3

M y = " D ddyw2 = "( !b ) 2 p0 sin !by 2

Thus,

! y ,max =

6 M y , max t2

= 0.61 p0 ( bt ) 2

.6 2 = 0.61 p0 ( 20 ! 10 3 )( 0.0012 ) = 30.4 MPa

" x ,max = ! (30.4) = 10.13 MPa Then,

and

# y ,max = E1 (" y ,max $ !" x ,max ) = 2001"103 (30.4 ! 103.13 ) = 135 µ !10 ry = 2" yt,max = 0.2012 (135 ) = 44.41 m 6

______________________________________________________________________________________ SOLUTION (13.4)

M y = Ma

M x = Mb y

a ( a ) Using Eq. (13.7), #2w #x 2 2

" w "y 2

= " MD (b1""!!M2 a) M a !#M b

= ! D (1!# 2 ) ,

(a) 2

" w "x"y

Integrating these equations,

w = ! 2MDb(!1"!M" 2a) x 2 ! 2MDa(!1"!M" 2b) y 2 + c1 x + c2 y + c3 If the origin of xyz is located at the center and midplane of the plate, the c’s will vanish, and

w = " 2MDb("1!"M! 2a) x 2 " 2MDa("1!"M! 2b) y 2

(b)

( b ) By setting M a = ! M b in Eq. (a): "2w "x 2

= ! ""yw2 = D (M1!a# ) = r1x = ! r1y 2

Integrating and locating the origin of xyz, as in item (a):

w = 2 DM(1a!" ) ( x 2 ! y 2 ) ______________________________________________________________________________________ Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 13–2

______________________________________________________________________________________ SOLUTION (13.5) Equation (13.19) becomes, a

P pmn = 144 [ (b " y ) sin n#by dy ]( a " x ) sin ma#x dx a 4b 4 ! !

Integrating by parts,

P b a P pmn = 144 = mn144 ! 2 a 2b2 a 4 b 4 n! m! 2

(a)

Substituting Eqs. (13.18) into ! w = p D :

a mn = ! 4 D ( m2 pamn2 + n 2 b2 )2

(b)

Inserting Eqs. (a) and (b) into Eq. (13.18b), we find the required expression for the deflection. ______________________________________________________________________________________ SOLUTION (13.6) ( a ) From Eq. (13.19), we obtain

pmn = p0

Then, Eq. (13.20) becomes for a square plate (a=b): sin m"x sin

n"y

w = "p04aD !! ( ma2 + n 2 )b

(a)

At x = y = a 2 , p0 a 4

wmax = # 4 D !! (b)

m + n "1

( "1) 2 ( m2 + n 2 )2

) D = 21012"10(1!(00.3.025 = 300,480.77 2 )

Equation (a) is then, 4

p0 ( 3) 1 8(10 !3 ) = " 4 ( 300 [ 1 ! 100 ] , 480.77 ) 4

p0 = 12.05 kPa

______________________________________________________________________________________ SOLUTION (13.7) The flexural rigidity of the plate is

D = 12 (Et1"# 2 ) = 12200(1!"100.09t ) = 18.315 ! 10 9 t 3 3

9 3

The maximum deflection occurs at the center of the plate. Equation (13.27) is thus 4

p0 a wmax = 64 D ;

Solving,

(10 )( 0.05 ) 1.5 ! 10 "3 = 6410(18 .315!109 t 3 )

t = 3.29(10 !3 ) m = 3.29 mm

______________________________________________________________________________________ SOLUTION (13.8) a

M0 t

______________________________________________________________________________________ 12.8 (CONT.) Since Q = 0, Eq. (13.24c) becomes d dr

[ 1r drd ( r dw dr )] = 0

from which, after integration, 2

w = ! c14r ! c2 ln ar + c3

(a)

M r = D[ c21 ! cr 22 + " ( c21 + cr 22 )]

(b)

Substituting this into Eq. (13.24a): Boundary conditions

( M r ) r =a = M 0

w( 0 ) = 0

( dw dr ) r =0 = 0

yield c2 = 0 and 2

c1 = D2(1M+!0 )

c3 = 2 DM(01a+! )

Equation (a) becomes then,

w = 2 DM(10+" ) ( a 2 ! r 2 ) Introducing this into Eqs. (13.24a) and (13.24b) yields M r = M ! = M 0 . Hence,

" r ,max = " ! ,max = 6 tM2 0

______________________________________________________________________________________ SOLUTION (13.9) We have Qr = 0 and Eq. (13.24c) after integration gives 2

w = ! c14r ! c2 ln ar + c3

(a)

M r = D[ c21 ! cr 22 + " ( c21 + cr 22 )

(b)

Introducing this into Eq. (13.24a): Boundary conditions

w( a ) = 0

M r ( b) = M 0

and Eqs. (a) and (b) result in 2

c1 = (1+"2)(b bM2 !0 a 2 ) ,

2 2

c2 = (1!"a )(bbM2 !0a 2 )

2 2

c3 = 2 (1+"a ) bD (Ma 20 !b2 )

Carrying these into Eq. (a), we have the equation for deflection. ______________________________________________________________________________________ SOLUTION (13.10) From Example 13.3:

" 1 = " r ,max = ! 43 p0 ( at ) 2 ,

" 2 = ! 14 p0 ( at ) 2 ,

"3 = 0

Maximum shearing stress is then

# max = 12 (" 1 ! 0) = ! 83 p0 ( at ) 2

According to maximum shear stress theory: ! yp n

= 43 p0 ( at ) 2

Introducing the given data, 100 (106 ) n

= (0.4 ! 10 6 )( 00..022 ) 2 ( 43 ) n = 3.33

or ______________________________________________________________________________________ Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 13–4

______________________________________________________________________________________ SOLUTION (13.11) p 2a Solution proceeds as in Example 13.3. Boundary conditions are

( w) r = a = 0

( M r ) r =a = 0

(a)

We have D p

r w = 64 ! c24r + c4 4

(b)

Carrying Eqs. (a) and (b) into Eq. (13.24a), we obtain two equations. From these c2 and c4 are evaluated. In so doing, and substituting the values obtained into Eq. (b): 2

w = p ( a64 !Dr ) ( 15++"" a 2 ! r 2 )

______________________________________________________________________________________ SOLUTION (13.12) 2

then rx = rxy = ! ,

Let r = ry = ! w !y ,

M = My,

M x = M xy = 0

Hence, Eq.(13.5), for z= ! t z : 9

)(0.006) 120(106 ) = 70(10 , 2(0.91) r

! max = 2(1#Et" 2 )r ;

r=1.923 m, d= 3.85 m

Equation (13.9),

6 M max 120(106 ) = 6 Mt 2max = (0.006) 2 ;

M max = 320 N

______________________________________________________________________________________ SOLUTION (13.13)

Dw IV = p0,

(a) and

Dw' ' = 12 p0 y 2 + c1 y + c2

Dw' ' ' = p0 y + c1 ,

Dw' = 16 p0 y 3 + 12 c1 y 2 + c2 y + c3 Dw = 241 p0 y 4 + 16 c1 y 3 + 12 c2 y 2 + c3 y + c4

Boundary conditions:

w( 0) = 0: c4 = 0,

w' ' ( 0) = 0: c2 = 0

w( b) = 0:

p0 b4 24

+ b6 c1 + c3b = 0

w' ( b) = 0:

p0 b3 6

+ b2 c1 + c3 = 0

Solving, c1 = ! 83 b Equation (a) is thus

(a)

b c3 = 48

p b4

4 y3

9 y2

w = 480 D [( b ) ! 3( b )3 + 2( b ) 4 ]

(b)

( b ) We have dw dy

= 240 D [ b4 ! 2 b3 + 21b ],

d 2w dy 2

At y = b: 2

M max = ! D ddyw2 = ! p0 b8 ,

3 yb

= 2 D0 [ y 2 ! 4 ]

! y ,max =

6 M max t2

= "0.75 p0 ( bt )2

______________________________________________________________________________________ SOLUTION (13.14) y

Let W = 1 ! ax ! a .

a a

w = cx 2 y 2W .

Then !w !x

y=a!x

= 2cxy 2W 2 + 2cx 2 y 2W ( " a1 )

y ! 2w !x!y

( a ) At x=0: At y=0:

= 4cxyW + 4cxy 2W ( " a1 ) + 4cx 2 y ( " a1 ) + 2cx 2 y 2W "1 ( " a1 )

!w !y

= 2cx 2 yW + 2cx 2 y 2W ( " a1 )

! 2w !x 2

= 2cy 2W 2 + 4cxy 2W ( " a1 ) + 4cxy 2 ( " a1 ) + 2cx 2 y 2W "1 ( " a1 ) 2

! 2w !y 2

= 2cx 2W 2 + 4cx 2 yW ( " a1 ) + 4cx 2 y ( " a1 ) + 4cx 2 y 2W "1 ( " a1 ) 2

w=0, w=0,

At y=a-x: w=0,

!w !x

!w !y !w !x

= 0, !!wy = 0

( b ) At x=0, y=a: 2

! y = $ 2(1Et$" 2 ) [ ##yw2 + " ##xw2 ] = % % % % % % % = 0 , x = a2:

At y=0,

! 2w !y 2

! 2w !x 2

= ca8 , 2

! y ,max = # 2(1Et#" 2 ) [ ca8 ] = # 16Ecta , (1#" 2 )

and

! 2w !x!y

= 0,

! xy = 2(1E+" ) ##x#wy = $ $ $ $ $ $ $ = 0 =0

! xy = 0

______________________________________________________________________________________ SOLUTION (13.15) We have 2

M xy = ! D(1 ! " ) ##x#wy = M 0 Let,

! 2w !x!y

= D (1#"0 ) = k

Integrating with respect to x : !w !y

= kx + f ( y ) + c1

Then, integrating the above with respect to y gives

w = kxy + ! f1 ( y )dy + c2 Due to the symmetry in deflection :

where

f1 ( y ) = f ( y ) + c1

! f ( y )dy = 0. 1

Also, owing to the symmetry, center (a/2, a/2) should be free of displacement,

! c2 = " 14 ka 2

w = 0 = 14 ka 2 + c2

It follows that

w = ! D (1!0" ) ( xy ! a4 ) We observe that this solution satisfies boundary conditions, M x = 0 and M y = 0 at plate edges. ______________________________________________________________________________________ Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 13–6

______________________________________________________________________________________ SOLUTION (13.16) Cylinder end can be approximated as a clamped edge plate subjected to uniform loading. ( a ) Equation (13.29), 3p

)!4 ( at ) 2 = 135(10 = 120 3!1.5(106 )

! r ,max = 4 0 ( at ) 2 ; or a/t=10.954. Hence,

200 t = 10.954 = 18.26 mm

( b ) Then, Eq.(13.27) for r=0, gives 4

p0 a 1.5!10 (0.2) !12(1" 0.3 ) wmax = 64 D = 64!200!109 (0.01826)3 = 0.336 mm

______________________________________________________________________________________ SOLUTION (13.17) ( a ) From Eq. (13.29):

2 ! max = 34 p0 ( at ) 2 = 34 p yp ( 125 10 ) = 117.19! yp

Setting ! max = p yp !

yp 345 p yp = 117.19 = 117.19 = 2.944 MPa

We have D =

Et 3 12(1"! 2 )

)(0.01) = 200(10 = 18.315 kPa 12(1" 0.32 )

Eq. (13.27) for r=0 is then p a4

)(0.125) wmax = 64ypD = 2.944(10 = 0.613 mm 64(18.315!103 )

(b)

pallow = nyp = 2.944 1.2 = 2.45 MPa

______________________________________________________________________________________ SOLUTION (13.18) Referring to Example 13.2:

! max = 6 Mt 2max = 6(0.0534 p0 )( 502 ) 2 Thus,

240(10 6 ) = 200.25 p0

p0 = 1.2 MPa

Similarly, 4

wmax = 0.0454 p0 Eta 3 = 0.0454(1.2 ! 10 6 ) 70!10( 09.5( 0) .02 )3 4

= 0.0608 m = 6.1 mm ______________________________________________________________________________________ SOLUTION (13.19) We observe that a

n"y n"y ! sin ma"x sin ma"x dx = ! sin b sin b dy = 0

if m ! m' and n ! n ' Therefore, integrating, we consider only the squares of the terms in the parenthesis in Eq. (b) of Sec. 13.9. Using the formula:

______________________________________________________________________________________ 12.19 (CONT.) a

! ! sin

2 m"x a

sin 2 n"by dxdy = ab4

calculation of the first term of the integral in Eq. (b) gives # 4 abD 8

2 mn

!! a (

+ nb2 ) 2

m2 a2

Also, the second term of the integral in Eq. (b): a

p0 ! ! a mn sin ma"x sin n"by dxdy ab = !p20mn a mn (1 " cos m! )(1 " cos n! )

= 4! p20mnab a mn

( m, n = 1, 3, . . . )

______________________________________________________________________________________ SOLUTION (13.20) Deflection is given by Eq. (13.18b). Loading is expressed as follows:

p = 2 p0 ax

0 ! x ! a2

p = 2 p0 ! 2 p0 ax

a 2

Potential energy, Eq. (13.33):

W = 2!! "

a 2

a 2p x 0

!x!a

(a)

sin ma#x sin na#y dxdy

= !! m8 p2n0a" 3 a mn sin m2"

(b)

Strain energy, Eq. (13.32): a

0 4

U = D2 ! ! [a mn ( ma"2 + nb"2 ) sin ma"x sin na"y ]2 dxdy 2 2

2 2

2 = 81 D" a 2 !! a mn ( ma 2 + an 2 ) 2 2

We thus have or

"# "amn

= D$4 a a mn ( ma 2 + an 2 ) 2 ! m8 p2n0a$ 3 sin m2$ = 0 4 2

m! 2 ) a mn = m322np!0a7 Dsin( ( m2 + n 2 )2

( m, n = 1, 3, . )

(c)

Substitute this into Eq. (13.18b) to obtain deflection. ______________________________________________________________________________________ SOLUTION (13.21) Let p represent the pressure differential. Cylinder:

" ! = apt

" ! = 1506 p = 25 p 6

From the above, p = " ! / 25 = 15 # 10 / 25 = 600 kPa Sphere: Thus,

! = pa 2 t = 12.5 p p = ! 12.5 = 15 "106 12.5 = 1.2 MPa

______________________________________________________________________________________ SOLUTION (13.22)

2t! all 2(0.05)(30 "106 ) 2 = = 1.5 MPa 2.5 "12 r

We have

9.81(103 )h = 1.5 !106 ,

h = 152.9 m

______________________________________________________________________________________ SOLUTION (13.23) y p r

Total pressure at any depth:

p = 500(103 ) + ! (h " y ) (a) At y = h :

h x

p = 500(103 ) Pa

Liquid pressure Therefore,

pr 500(103 )4000 t= = = 11.11 mm 180(106 ) ! all (b)

At y = h 4 :

p = 500(103 ) + 15(103 )(13.5) = 702.5(103 ) Pa

t= (c)

702.5(103 )4000 = 15.6 mm 180(106 )

At y = 0 :

p = 500(103 ) + 15(103 )(18) = 770(103 ) Pa

770(103 )4000 = 17.1 mm 180(106 )

______________________________________________________________________________________ SOLUTION (13.24) Total pressure at any depth:

p = 200(103 ) + ! (h " y )

(a)

At y = h :

p = 200(103 ) Pa

pr 200(103 )4000 = = 4.44 mm 180 "106 ! all

______________________________________________________________________________________ 12.24 (CONT.) (b)

At y = h 4 :

p = 200(103 ) + 15(103 )(13.5) = 402.5(103 ) Pa

t= (c)

At y = 0 : :

402.5(103 )4000 = 8.94 mm 180 !106

p = 200(103 ) + 15(103 )18 = 470(103 ) Pa

470(103 )4000 = 10.4 mm 180 !106

______________________________________________________________________________________ SOLUTION (13.25) The thickness for circumferential stress:

pr 1.2(103 )(0.6) t= = = 12 mm 24.0 2.4 ! all

The thickness for axial stress:

pr 12 = = 6 mm 2! all 2 treq = 12 mm

t= Thus,

______________________________________________________________________________________ SOLUTION (13.26)

pr p =#h t pr ! hr 9.81(103 )(150)(400) = 8.83 mm treq = = = " all " all 120(106 ) 1.8

"! =

______________________________________________________________________________________ SOLUTION (13.27)

pr ! hr 9.81(103 )(150)(400) = 5.89 mm treq = = = " all " all 100(106 )

______________________________________________________________________________________ SOLUTION (13.28)

! 1 # ! 2 1 pr pr pr pd = ( # )= = 2 2 t 2t 4t 8t 6 8t! 8 "12 " 35 "10 d = max = = 336 mm p 10 "106

(a)

" max =

(b)

pr 10 "106 "168 = = 140 MPa !1 = t 12

______________________________________________________________________________________ SOLUTION (13.29)

F N! The given numerical values are:

r" = 180 ! 1 = 179 mm

r" = 80 ! 1 = 79 mm

ri = 80 ! 2 = 78 F = !ri 2 p

p = !0.08 MPa Equation (13.46b) is therefore 2

) ( 0.08$10 ) N " = 2!rF# (1) = ! ( 0.078 2! ( 0.079 )

= 3.081 kN m and

" ! = 03081 .002 = 1.541 MPa

Using Eq. (13.46a), or

"! 0.079

0.08 + 10..541 179 = 0.02

! " = 2.48 MPa = ! max

______________________________________________________________________________________ SOLUTION (13.30) ! b

p "

r! r0

Consider the portion of shell defined by ! . Vertical equilibrium of force yields,

2"r0 N # sin # = "p( r02 ! b 2 ) from which 2

pa ( r0 + b ) !b ) N " = p2(rr00sin " = 2 r0

N ! = b+ apasin ! ( a2 sin ! + b) Substituting N ! into Eq. (13.46a), setting p z = ! p, and r! = a :

N " = pr" 2( rr00!b ) = pa2 Since sin # = r0

r" = ( r0 ! b) a , from symmetry.

Note that N ! is constant throughout the shell from the condition of symmetry. ______________________________________________________________________________________ Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 13–11

______________________________________________________________________________________ SOLUTION (13.31) Referring to Solution of Prob. 13.30, we have

2a = (1 ! 0.7) 2 , 2b = (1 + 0.7) 2,

a = 0.075 m b = 0.425 m

At point A (crotch):

" A = " ! ,max = pa ( r0 + b) ( 2r0 t )

t = pa ( r0 + b) ( 2r0" ! ,max ) 6

.075 )( 0.35+ 0.425 ) = 2 (102)(( 00.35 )( 210!106 )

= 0.791 mm = t req. Similarly,

" ! = pa 2t

or 6

(10 ) 75 t = 22( 210 = 0.357 mm !106 )

______________________________________________________________________________________ SOLUTION (13.32) N

! r2 s

# #

t y

O Pressure at any level st is p r = " ( a ! y ) . We have

" = #2 + !

r0 = y tan !

Thus, the first of Eqs. (13.49) becomes and

y ) y tan ! N # = " ( a $cos !

tan ! $ # = " ( a%t y ) y cos !

The load F is equal to the weight of liquid in cylindrical portion stuv:

F = $"#y 2 ( a $ y + 3y ) tan 2 !

Then second of Eqs. (13.49) gives

N # = "y ( a $ 23 y ) tan ! 2 cos !

and

tan ! $ # = "y ( a %22t y 3) cos !

______________________________________________________________________________________ SOLUTION (13.33) Expressions for the components of pressure are:

p" = ! p cos "

pr = p sin "

px = 0

Thus,

N x# = " ! [ " p cos # + a1 ( " pa cos # )]dx + f1 (# )

= 2 px cos ! + f1 (! ) N ! = " pa sin !

N x = " ! a2 ( " px sin # )dx +

(a) (b) df1 d#

+ f 2 (# )

= a1 px 2 sin ! + dfd!1 + f 2 (! )

Boundary conditions: N x = 0 2

pL a

(c)

(at x=0 and x=L)

give

f 2 = 0 and

sin ! + L dfd!1 = 0

f1 (" ) = ! pL2 sin " + c Note that no torque is applied to the shell; c = 0. Hence,

N ! = " pa sin !

N x = " La" x px sin !

N x! = "( L " x ) p cos ! ______________________________________________________________________________________ SOLUTION (13.34) Now the cylinder length does not change:

!L 2

( N x ! $N # )dx = 0

Substituting Eqs. (b) of Example 13.7 into this, taking the resulting expression, we have

f1 = 0 and integrating

f 2 (! ) = "#a 2 (1 $ cos ! ) $ #24L cos ! Referring to Example 13.7, the solution is thus,

N ! = "a 2 (1 # cos ! ) 2

N x! = "ax sin ! 2

N x = #x2 cos ! + "#a 2 (1 $ cos ! ) $ #24L cos ! ______________________________________________________________________________________ SOLUTION (13.35) Referring to Fig. 13.13b:

p x = p sin !

p z = p cos !

r0 = x cos !

Stress resultants due to weight:

F = 2"r # p sin ! # rd!

______________________________________________________________________________________ 13.35 (CONT.) Since

rd! = dx and r = x cot ! . Then x

F = 2# ! x cot "p sin "dx = 2#p cos " ( x2 ) + c 2

For a cone supported at its edge c = 0 , since F = 0 at x = 0. Therefore, Eq. (13.46b) gives px N ! = " 2 sin !

(1)

Equation (13.46a): 2

p z r0 cos ! N " = sin ! = # px sin !

(2)

Stress resultants due to pressure: Equation (13.46a) yields,

cos ! N " = # p r xsin ! = # p r x cot !

(3)

We now have

F = ( 2"p r sin !rd! ) cos !

Following a procedure similar to that the preceding, we obtain

F = 2"p r cos 2 ! ( x2 ) 2

Equation (13.46a) leads to

N ! = " p2r xsincos! ! = " 12 p r x cot !

(4)

Solution is determined by the superposition of the preceding results: adding Eqs. (4) and (1), and (3) and (2). In so doing, we have

N x = " 2 sinx ! ( p + p r cos ! ) and

N # = ! x cot " ( p cos " + 12 p r ) End of Chapter 13