Advanced Mechanics Of Materials And Applied Elasticity 5th Ed

Advanced Mechanics Of Materials And Applied Elasticity 5th Edition Ugural Solutions Manual Full Download: https://testbanklive.com/download/advanced-mechanics-of-materials-and-applied-elasticity-5th-edition-ugural-solu

Solutions Manual for

Advanced Mechanics of Materials and Applied Elasticity Fifth Edition

Ansel C. Ugural Saul K. Fenster

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

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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.

Visit us on the Web: InformIT.com/ph Copyright Š 2012 Pearson Education, Inc. This work is protected by United States copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning. Dissemination or sale of any part of this work (including on the World Wide Web) will destroy the integrity of the work and is not permitted. The work and materials from it should never be made available to students except by instructors using the accompanying text in their classes. All recipients of this work are expected to abide by these restrictions and to honor the intended pedagogical purposes and the needs of other instructors who rely on these materials.

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 .

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

Equations (1.8), with or

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

"x =

P A

=

125(103 ) 0.05!0.05

= 50 MPa

! = 90o " 70o = 20o : ! x ' = 50 cos 2 20o = 44.15 MPa

( a ) Equations (1.11), with

" 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

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

( b ) Equations (1.11), with

" 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

______________________________________________________________________________________

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

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

#x = For

# x'

cos2 "

=

!75 cos2 30o

= !100 MPa

! = 50o , 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 =

and

10 =

P 2A

(1 + cos 2! )

P 2A

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 20 =

P 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):

!1 = ! max

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

______________________________________________________________________________________

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

______________________________________________________________________________________ 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 = 12(106 ) = " sin 55o cos 55o , P = 38.3 kN "3 1.5(10 )

! x ' = 25(106 ) =

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

! x ' = 20(106 ) = ! x' y' 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 =" 450 #10"6

! x' =

! x' y'

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

!6

) m2 .

Use Eqs. (1.11):

3

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

! x' = ! x' y'

______________________________________________________________________________________

Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 1–3

______________________________________________________________________________________ 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 MN m 3 ): Fy = ! x + y 2 + 2 xz Fx = !3x + 2 xy

Solving, we have (in

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. ______________________________________________________________________________________

Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 1–4

______________________________________________________________________________________ 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

20o

x’

x 50 !A sin20o 60 !A sin20o

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

$F

= 0:

$F

= 0 : ! x ' y ' "A # 40"A sin 20o cos 20o

x'

y'

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

! x' y'

!60"A sin 20o cos 20o ! 50"A cos 2 20o +50!A sin 2 20o = 0 = 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

#F

x'

! x ' y ' "A

! x ' "A 2

o

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

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.) ______________________________________________________________________________________

Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 1–5

______________________________________________________________________________________ 1.16 (CONT.)

$F

y'

= 0 : ! 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

o

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

x

! =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 ______________________________________________________________________________________

Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 1–6

______________________________________________________________________________________ SOLUTION (1.19)

! = 40o to ! = 0 . For convenience in computations, Let ! x = #160 MPa, ! y = #80 MPa, " xy = 40 MPa and ! = "40o

Transform from 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

So For

! = 0o :

y

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 ______________________________________________________________________________________

Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 1–7

______________________________________________________________________________________ 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)

10

50 MPa

r

O #

40 MPa (-40, -50)

C

" (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

o

(-50, -15)

90 MPa x

O 62.1o

y’ #

(90, 15)

C

" (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’

"1

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

y’

14.22 MPa

25o x’

64.6 MPa 90 MPa 15 MPa

x

50 MPa ______________________________________________________________________________________

Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 1–8

______________________________________________________________________________________ SOLUTION (1.24) ! (MPa)

! ' =67.5

x’

r =22.5 r

45 MPa

90

45

53.1o

O

x

C

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

C

r

r=

"

2

(! " 60)

" (MPa)

x

! # 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 '

O

2

C

" max = 96 MPa

o

"

1

" (MPa)

B (CONT.) Figure (a) ______________________________________________________________________________________

Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 1–9

______________________________________________________________________________________ 1.26 (CONT.) By applying Eq. (1.20): 50 2

! 1, 2 = or

± [22,4500 + 36000]2 = 25 ± 96 1

" 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

O

# 2 = !50 MPa

" max = 125 MPa

! s ' = 71.55o !max

C

2%p’

"

" (MPa)

1

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

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

or

" 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' = " y' =

" x' y' =

" 1 +" 2 2 " 1 +" 2 2 # 1 $# 2 2

+ " 1 #2" 2 cos 2!

(a)

# " 1 #2" 2 cos 2! sin 2!

(b)

From Eqs. (a),

! x' + ! y' = ! 1 + ! 2 By using

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

______________________________________________________________________________________

Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 1–10

Advanced Mechanics Of Materials And Applied Elasticity 5th Edition Ugural Solutions Manual Full Download: https://testbanklive.com/download/advanced-mechanics-of-materials-and-applied-elasticity-5th-edition-ugural-solu

______________________________________________________________________________________ SOLUTION (1.28) We have

tan 2$ p =

2# xy " x !" y

2 ( !70 ) 50!(190 )

=

2" p = !30.26o

and

= !0.583

" p = !15.13o

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 #" y 2 2

) + ! xy2

Substituting the given values 2

140 2 = (60+2100 ) + ! xy2

or

! xy ,max = 114.19 MPa

______________________________________________________________________________________ SOLUTION (1.30) Transform from

! = 60o

to

! = 0o

with

" x ' = !20 MPa , " y ' = 60 MPa ,

o

" x ' y ' = !22 MPa , and " = !60 . 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

______________________________________________________________________________________

Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 1–11

Full download all chapters instantly please go to Solutions Manual, Test Bank site: TestBankLive.com