Type:
Solution Manual
Resource:
Advanced Mechanics of Materials and Applied Elasticity
Edition:
5th Edition
Author(s):
Ansel Ugural Saul Fenster
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
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 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
o
" 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
______________________________________________________________________________________
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 = 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 =
______________________________________________________________________________________
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 ! 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
!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 ! x' y' = " "6 450 #10
! x' =
______________________________________________________________________________________
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 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. ______________________________________________________________________________________
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
$F = 0:
! x ' "A + 40 cos 2 20o # 60"A sin 2 20o
$F = 0: !
x' y'
x'
!2(50"A sin 20o cos 20o ) = 0 ! x ' = "35.32 + 7.02 + 32.14 = 3.84 MPa
y'
"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
o
# F = 0 : ! "A + 50"A cos 25 x'
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.) ______________________________________________________________________________________
Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 1–5
______________________________________________________________________________________ 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
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) 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
So
o
For ! = 0 :
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):
! 1, 2 = 502 ± [22,4500 + 36000]2 = 25 ± 96
or
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 ' = " 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
2
______________________________________________________________________________________
Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 1–10
______________________________________________________________________________________ SOLUTION (1.28) We have 2#
70 ) tan 2$ p = " x !xy" y = 502 (!!(190 ) = !0.583
2" p = !30.26o
o
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
or
! xy ,max = 114.19 MPa
______________________________________________________________________________________ SOLUTION (1.30) o
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
______________________________________________________________________________________
Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 1–11
______________________________________________________________________________________ SOLUTION (1.31) "y
"ycos60o
14 MPa !xy
!xycos60o "xsin60o
60
30 MPa Area =1
30 MPa
o
!xysin60o Figure (b)
Figure (a) ( a ) Figure (a):
! y = 14 sin 60o = 12.12 MPa
! xy = 14 cos 60o = 7 MPa Figure (b):
o
y
or
xy
! xy = 7 MPa (as before) o
o
! F = "# sin 60 + 30 + 7 cos 60 = 0 x
or
o
" 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’
x
______________________________________________________________________________________
Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 1–12
______________________________________________________________________________________ SOLUTION (1.32) "y !xy Figure (a)
"x
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
y
"x’ !xy
x’ "x y’
!x’y’
Figure (b)
n 30o
m
"y’
x
______________________________________________________________________________________ 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
or
x
! x = 195 MPa
35cos30o
150 MPa
60.6cos30o
30o Area=1
60.6sin30o "xsin30o
Figure (a)
(CONT.) ______________________________________________________________________________________
Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 1–13
______________________________________________________________________________________ 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’
x
______________________________________________________________________________________ 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)
x
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.) ______________________________________________________________________________________
Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 1–14
______________________________________________________________________________________ 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
o
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
!y
( 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
______________________________________________________________________________________
Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 1–15
______________________________________________________________________________________ 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
3
16T 16(8 #10 ) = = 40.7 MPa ! xy = "d3 " (0.1)3
!x
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
C
! 2 2%p"
! p " = 29o
O "avg
!1
"
x
______________________________________________________________________________________ SOLUTION (1.38)
! = " + 90 = 50 + 90 = 140o x'
!x =
!
! xy =
y' Equations(1.18):
! w = ! x' = and
16T "d3
x
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’ ______________________________________________________________________________________
Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 1–16
______________________________________________________________________________________ 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
A
2h
B
L
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
______________________________________________________________________________________
Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 1–17
______________________________________________________________________________________ SOLUTION (1.41) y
T
P
P M
M
A
T
x
A = " (30 2 ! 152 ) = 2.121(10 !3 ) m 2 I = "4 (30 4 ! 154 ) = 0.596(10 !6 ) m 4
J = 2I
y
!a +!b
A
! xy
x
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)
O
C
(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’
A
x x’
Figure (b)
37.84 MPa 4.18 MPa ______________________________________________________________________________________
Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 1–18
______________________________________________________________________________________ SOLUTION (1.42) A
y 30
20o
o
x
! 0 = ! xy ( a ) At " = !60
o
0= or
B
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
!0
C
Figure (c)
Area=1
B
! F = 0.5$ " 0.866# " 0.866# = 0 x
x
0
0
or " x = 3.464! 0 , as before. o
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.) ______________________________________________________________________________________
Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, Fifth Edition, © 2012 Pearson Education, Inc. 1–19