Solution Manual Advanced Mechanics of Materials and Applied Elasticity Ugural Fenster 6th Edition
Complete Solution Manual With Answers
Sample Chapter is Posted Below
From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886)
Copyright © 2020 Pearson Education, Inc. All rights reserved.CONTENTS
Chapter 1 Analysis of Stress 1
Chapter 2 Strain and Material Properties 48
Chapter 3 Problem in Elasticity 83
Chapter 4 Failure Criteria 111
Chapter 5 Bending of Beams 133
Chapter 6 Torsion of Prismatic Bars 166
Chapter 7 Numerical Methods 186
Chapter 8 Thick-Walled Cylinders and Rotating Disks 227
Chapter 9 Beams on Elastic Foundations 248
Chapter 10 Applications of Energy Methods 259
Chapter 11 Stability of Columns 284
Chapter 12 Plastic Behavior of Materials 309
Chapter 13 Stresses in Plates and Shells 335
The Solutions Manual to accompany the text Advanced Mechanics of Materials and
Applied Elasticity 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 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 its 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
iii
From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886)
Copyright © 2020 Pearson Education, Inc. All rights reserved.CHAPTER 1
SOLUTION (1.1)
We have
A−
3 2 50 75 3.75(10 ) m
= × = ,
50=
o
θ , and AP
=
σ.
x
Equations (1.11), with o
θ :
50=
3
σ
= σ
‘ =
)10(700 2
o
=
σ
cos
413.050
x 18.110
x
x
or 6.35 kN
P=
P
and
3
τ σ σ = = = =
x y ‘ ‘ 560(10 ) sin 50 cos50 0.492 131.2o o
P
x x
Solving
4.27 kN
P= all P=
______________________________________________________________________________________
SOLUTION (1.2)
Normal stress is
P
3 125(10 )
σ ×
= = =
x A
0.05 0.05 50 MPa
( a ) Equations (1.11), with o
θ :
20=
2
‘ 50cos 20 44.15 MPao
x σ = =
x y τ = − ‘ ‘ 50sin 20 cos 20 16.08 MPao o
= −
2
σ = + =
y
‘ 50cos (20 90 ) 5.849 MPao o
5.849 MPa
y’
16.08 MPa
44.15 MPa
x’
20 o
x
( b ) Equations (1.11), with θ :
o 45=
2
‘ 50cos 45 25 MPao
x σ = =
x y τ = − ‘ ‘ 50sin 45 cos 45 25 MPao o
= −
2
σ = + =
y
‘ 50cos (45 90 ) 25 MPao o
25 MPa
25 MPa
y’
25 MPa
x’
45 o
x
______________________________________________________________________________________
1
From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886)
Copyright © 2020 Pearson Education, Inc. All rights reserved.______________________________________________________________________________________
SOLUTION (1.3)
From Eq. (1.11a),
σ
‘
75
σ−
= = = −
100 MPax
x
2 2
θ
cos cos 30
o
For o
50=
θ , Eqs. (1.11) give then
2
x σ = − ‘ 100cos 50 41.32 MPao
= −
τ
)100(‘ −=
sin
50
o
cos
yx 50
o
49.24 MPa=
Similarly, for o
θ :
140=
2
x σ = − ‘ 100cos 140 58.68 MPao
= −
x y τ = −
‘ ‘ 49.24 MPa
41.32 MPa
58.68 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 0=
σ and 0=
τ , become
y
xy
1
1
θσ 2
1
cos
=
sin
x += and θστ 2
‘ x
2
x
2
yx
‘ x
2
or
1(20 2 θ
+= A
P and θ 2
cos
)2
P
=
10 2 A
sin
The foregoing lead to
sin
2
2=− θ (a)
cos
12
By introducing trigonometric identities, Eq. (a) becomes
4 2
cosin
=− θ or 21
2
cos
0
tan=
θ . Hence
θ
56.26=
o
Thus,
gives
2(1300) 20 (1 0.6) P = +
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):
σ
1
P
3
−
= = = −
150(10 ) 76.4 MPa
A
π
(50)
2
4
P
τ = =
max 38.2 MPa
2
A
______________________________________________________________________________________
2
From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886)
Copyright © 2020 Pearson Education, Inc. All rights reserved.______________________________________________________________________________________
SOLUTION (1.6)
Shaded transverse area:
3 2 2 2(10)(75) 1.5(10 ) mm A at= = =
Metal is capable of supporting the load
P A σ−
6 3 90(10 )(1.5 10 ) 135 kN
= = × =
Apply Eqs. (1.11):
P
6 2
o
σ
= =
P=
x
‘ 3 25(10 ) (cos 55 ) 1.5(10 )
, 114 kN
−
6
P
τ
= = −
x y
12(10 ) sin 55 cos55
o o
P=
‘ ‘ 3
, 38.3 kN
−
1.5(10 )
Thus,
all P=
38.3 kN
______________________________________________________________________________________
SOLUTION (1.7)
Use Eqs. (1.11):
P
6 2
o
σ
= =
P=
x
‘ 3 20(10 ) (cos 40 ) 1.5(10 )
, 51.1 kN
−
P
6
τ
= = −
x y
8(10 ) ‘ ‘ 3
−
sin 40 cos 40
o o
1.5(10 )
P=
, 24.4 kN
Thus,
all P=
24.4 kN
______________________________________________________________________________________
SOLUTION (1.8)
A = × =
15 30 450 mm
2
Apply Eqs. (1.11):
3
2
o
x σ
= =
‘ 6
120(10 ) (cos 40 ) 156 MPa
450 10
−
×
3
x y τ
= − = −
120(10 ) sin 40 cos 40 131 MPa
o o
‘ ‘ 6
450 10
−
×
______________________________________________________________________________________
SOLUTION (1.9)
A−
We have 6 2 450(10 ) m
= . Use Eqs. (1.11):
3
−
2
o
x σ
= = −
‘ 6
100(10 ) (cos 60 ) 55.6 MPa
450 10
−
×
3
−
x y τ
= − =
100(10 ) sin 60 cos 60 96.2 MPa
o o
‘ ‘ 6
450 10
−
×
______________________________________________________________________________________
3
From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886)
Copyright © 2020 Pearson Education, Inc. All rights reserved.______________________________________________________________________________________
SOLUTION (1.10)
θ
o 1309040 =+=
o
3
P
150(10 )
σ
= = − = −
31.83 MPa
x A π
2 2
−
(0.08 0.07 )
Equations (1.11):
2
x σ = − ‘ 31.83cos 130 13.15 MPao
= −
x y τ = ‘ ‘ 31.83sin130 cos130 15.67 MPao o
= −
x’
13.15 MPa
130 o
15.67 MPa
x
Plane of weld
y’
______________________________________________________________________________________
SOLUTION (1.11)
Use Eqs. (1.14),
0)()2()2( =+−+ x Fx
x
xy
0)0()2()( 2 =+−+− y
y
yz
Fx
0)2()0()4( =+−+− z Fz
z
xy
Solving, we have (in 3
MN m ):
xFx 23 +−= 2 2
xy
y F x y yz = − + + z
Fz += 4 (a)
xy
Substituting x=-0.01 m, y=0.03 m, and z=0.06 m, Eqs. (a) yield the following values
F F F= = =
3 3 3 29.4 kN m 14.5 kN m 58.8 kN m x y z
Resultant body force is thus
2 2 2 3
F F F F = + + =
67.32 kN m
x y z
______________________________________________________________________________________
SOLUTION (1.12)
Equations (1.14):
=+− yc
04,02 1
ycyc
1 ≠
1
zc
=+ zc
3 ≠
,0 3
0
0 =+
No. Eqs. (1.14) are not satisfied.
______________________________________________________________________________________
4
From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886)
Copyright © 2020 Pearson Education, Inc. All rights reserved.______________________________________________________________________________________
SOLUTION (1.13)
( a ) ( b ) No. Eqs. (1.14) are not satisfied.
Yes. Eqs. (1.14) are satisfied.
______________________________________________________________________________________
SOLUTION (1.14)
Eqs. (1.14) for the given stress field yield:
zyx F
0=
______________________________________________________________________________________
SOLUTION (1.15)
y
y’
τ ∆
A
‘ ‘ x y
σ ∆
‘
A
x
40 A ∆ cos20o
50 A ∆ cos20o 20o
x’
20o
x
50 A ∆ sin20o
60 A ∆ sin20o
F A A σ = ∆ + − ∆
∑
2 2
‘ ‘ 0 : 40cos 20 60 sin 20o o
x x
A − ∆ =
2(50 sin 20 cos 20 ) 0o o
‘ 35.32 7.02 32.14 3.8 MPa
x σ = − + + =
F A A τ = ∆ − ∆
∑
‘ ‘ ‘ 0 : 40 sin 20 cos 20o o
y x y
2 60 sin 20 cos 20 50 cos 20o o o
A A − ∆ − ∆
2 50 sin 20 0o
A + ∆ =
‘ ‘ 12.86 19.28 44.15 5.85 70.4 MPa
x y τ = + + − =
______________________________________________________________________________________
SOLUTION (1.16)
50 A ∆ cos25o
15 A ∆ cos25o
y’
15 A ∆ sin25o
25o
τ ∆
A
‘ ‘ x y
90 A ∆ sin25o
σ ∆
‘
A
x
x’
F A A σ = ∆ + ∆
∑
2
‘ ‘ 0 : 50 cos 25o
x x
A A − ∆ − ∆ =
2 90 sin 25 2(15 sin 25 cos 25 ) 0o o o
‘ 41.07 16.07 11.49 13.5 MPa
x σ = − + + = −
(CONT.)
______________________________________________________________________________________
5
From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886)
Copyright © 2020 Pearson Education, Inc. All rights reserved.______________________________________________________________________________________
1.16 (CONT.)
F A A τ = ∆ − ∆
∑
‘ ‘ ‘ 0 : 50 sin 25 cos 25o o
y x y
2 90 sin 25 cos 25 15 cos 25o o o
A A − ∆ − ∆ 2 15 sin 25 0o
A + ∆ =
‘ ‘ 19.15 34.47 12.32 2.68 63.3 MPa
x y τ = + + − =
______________________________________________________________________________________
SOLUTION (1.17)
y
y’
‘ ‘ x y τ
x σ
‘
50 MPa 20o
60 MPa θ
x’
x
θ=20o
80 MPa
o o
1 1 ( 40 60) ( 40 60)cos 40 50sin 40 2 2
x σ = − + + − − +
‘
10 38.3 32.1 3.8 MPa = − + =
x y τ = − − − +
‘ ‘
1 ( 40 60)sin 40 50cos 40
o o
2
32.14 38.3 70.4 MPa = + =
x’
y’
90 MPa
θ=115o
______________________________________________________________________________________
SOLUTION (1.18)
x σ
‘
‘ ‘ x y τ
θ
25o
x
15 MPa
50 MPa
o o
1 1 (90 50) (90 50)cos 230 15sin 230 2 2
x σ = − + + −
‘
20 45 11.5 13.5 MPa = − + = −
x y τ = − + −
‘ ‘
1 (90 50)sin 230 15cos 230
o o
2
53.62 9.64 63.3 MPa = + =
______________________________________________________________________________________
6
From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886)
Copyright © 2020 Pearson Education, Inc. All rights reserved.______________________________________________________________________________________
SOLUTION (1.19)
Transform from 40o
θ= to 0
θ= . For convenience in computations, Let
160 MPa, 80 MPa, 40 MPa
θ= −
x y xy σ σ τ = − = − = and 40o
Then
1 1 ( ) ( )cos 2 sin 2 2 2x x y x y xy σ σ σ σ σ θ τ θ = + + − +
‘
o o = − − + − + − + −
1 1 ( 160 80) ( 160 80)cos( 80 ) 40sin( 80 ) 2 2
166.3 MPa= −
x y x y xy τ σ σ θ τ θ = − − +
‘ ‘
1 ( )sin 2 cos 2
2
o o = − − + − + −
1 ( 160 80)sin( 80 ) 40cos( 80 )
2
32.4 MPa= −
So ‘ ‘ 160 80 166.3 73.7 MPa
y x y x σ σ σ σ = + − = − − + = −
For 0o
θ= :
y
73.7 MPa
32.4 MPa
166.3 MPa
x
______________________________________________________________________________________
SOLUTION (1.20)
1 4 tan 53.1
θ−
o
= =
3
+ −
x σ
‘
= +
45 90 45 90 cos106.2
o
2 2
67.5 6.28 73.8 MPa = + =
−
‘ ‘
x y τ
= − =
45 90 sin106.2 21.6 MPa
o
2
21.6 MPa
73.8 MPa
x’
53.1o
y’
x
______________________________________________________________________________________
7
From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886)
Copyright © 2020 Pearson Education, Inc. All rights reserved.______________________________________________________________________________________
SOLUTION (1.21)
xy τ θ = =
0 70o
σ
−
(a) ‘ ‘
60 30 sin140
τ
= − = − o
x y
2
σ =
153.3 MPa
σ
(b) ‘
x
σ σ
+ −
60 60
80 cos140
o
= = + 231 MPa
σ =
2 2
______________________________________________________________________________________
SOLUTION (1.22)
θ=
Equations(1.18) with 60o
, 110 MPa
x σ =
, 0
σ =
, 50 MPa
y
xy τ = give
1 1
x σ = + + =
‘ 2 2 (110) (110)cos120 50sin120 70.8 MPao o
1
x y τ = − + = −
‘ ‘ 2 (110)sin120 50cos120 72.6 MPao o
1 1
σ = − − =
‘ 2 2 (110) (110)cos120 50sin120 39.2 MPao o
y
______________________________________________________________________________________
SOLUTION (1.23)
θ=
Equations(1.18) with 30o
, 110 MPa
x σ =
, 0
σ =
, 50 MPa
y
xy τ = result in
1
x σ = + + =
‘ 2 (110) 55cos 60 50sin 60 125.8 MPao o
1
x y τ = − + = −
‘ ‘ 2 (110)sin 60 50cos 60 22.6 MPao o
1
σ = − − = −
y
‘ 2 (110) 55cos 60 50sin 60 15.8 MPao o
10 MPa
x
______________________________________________________________________________________
SOLUTION (1.24)
We have
x σ
‘
x’
θ= + =
25 90 115o
x σ = −
10 MPa
‘ ‘ x y τ
θ
25o
σ =
y
30 MPa
y’
xy τ =
0
30 MPa
1 1 ( ) ( )cos 2 2 2 x x y x y σ σ σ σ σ θ = + + −
(a) ‘
o = − + + − − =
1 1 ( 10 30) ( 10 30)cos 230 22.86 MPa 2 2
Thus,
w x σ σ = =
‘ 22.86 MPa
(CONT.)
______________________________________________________________________________________
8
From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886)
Copyright © 2020 Pearson Education, Inc. All rights reserved.______________________________________________________________________________________
1.24 (CONT.)
(b) ‘ ‘
1 ( )sin 2
x y x y τ σ σ θ = − −
2
o
= − − − = −
1 ( 10 30)sin 230 15.32 MPa
2
So
w τ
w x y τ τ = = −
‘ ‘ 15.32 MPa
______________________________________________________________________________________
SOLUTION (1.25)
0 50 0 50
+ −
(a) 2 2
= = + +
σ τ
1
80 ( )
2 2
τ =
49 MPa
−
(b) 2 2
50 ( ) 49 55 MPa 2
τ
= + =
max
50 ‘ 25 MPa
σ = =
2
−
s θ−
o
= − =
1 0 50 2 tan [ ] 27 2(49)
x y τ = + =
‘ ‘
50 sin 27 49cos 27 55 MPa
o o
2
Thus,
s θ =
‘ 13.5o
25 MPa
y’
25 MPa
x’
13.5o
x
55 MPa
______________________________________________________________________________________
SOLUTION (1.26)
80 MPa
40 MPa
θ=-30o
x
(CONT.)
______________________________________________________________________________________
9
From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886)
Copyright © 2020 Pearson Education, Inc. All rights reserved.______________________________________________________________________________________
1.26 (CONT.)
+ −
o
x σ
= + − = − =
40 80 40 80 cos( 60 ) 60 10 50 MPa
2 2
σ = + =
60 10 70 MPa
y
−
o
xy τ
= − − = −
40 80 sin( 60 ) 17.32 MPa
2
70 MPa
70 MPa
50 MPa
50 MPa
+ =
17.32 MPa 20 MPa
2.68 MPa
+ −
2 2
50 70 50 70 ( ) 2.68 2 2
σ
= ± + 60 10.35 = ±
1,2
1 2 70.35 MPa 49.65 MPa σ σ = =
1 2(2.68) 2 tan [ ] 15 50 70
p θ−
o
= = −
−
50 70 60 −
o o
x σ
‘
= + − + −
cos( 15 ) 2.68sin( 15 ) 2
60 9.66 0.694 49.65 MPa= − − =
Thus,
p θ = −
” 7.5o
x
7.5o
49.65 MPa
70.35 MPa
______________________________________________________________________________________
SOLUTION (1.27)
60 MPa
τ
(MPa)
10
(60, 50)
20o
50 MPa
40 MPa
O
R
C
σ (MPa)
(CONT.)
α
(-40, -50)
40o
x’
______________________________________________________________________________________
10
From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886)
Copyright © 2020 Pearson Education, Inc. All rights reserved.______________________________________________________________________________________
1.27 (CONT.)
1 50 tan 45
α−
o
= =
50
1
2
R = + =
2 2 (50 50 ) 70.7
‘ ‘ sin85 (70.7) 70.4 MPao
x y τ = =
x σ = − =
‘ 10 cos85 (70.7) 3.84 MPao
______________________________________________________________________________________
SOLUTION (1.28)
τ (MPa)
50 MPa
15 MPa
25o
O
σ’=20
y’
α
R
(90, 15)
90 MPa
(-50, -15)
62.1o
(σx’, τ x’y’)
x’
σ (MPa)
63.3 MPa
90 MPa
25o
C
x
1 15 tan 12.1
α−
o
= =
70
y’ 13.5 MPa
1
2
R = + =
2 2 (15 70 ) 71.6
‘ ‘ 71.6sin 62.1 63.3 MPao
x y τ = =
‘ 71.6cos 62.1 20o
x σ = − +
13.5 MPa= −
x
x’ 15 MPa
50 MPa
______________________________________________________________________________________
SOLUTION (1.29)
τ (MPa)
R =22.5
x’
R
90
‘
σ =67.5
45 MPa
45 53.1o
O
C
σ (MPa)
x
90 MPa 106.2o
73.8o
x’
(σx’, −τ x’ y’ )
(CONT.)
______________________________________________________________________________________
11
From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886)
Copyright © 2020 Pearson Education, Inc. All rights reserved.x’
140o
O
σ
60
C
R
σ (MPa)
______________________________________________________________________________________
1.29 (CONT.)
‘ ‘ 22.5sin 73.8 21.6 MPao
x y τ = =
‘ 67.5 22.5cos 73.8 73.8 MPao
x σ = + =
Sketch of results is as shown in solution of Prob. 1.20.
______________________________________________________________________________________
SOLUTION (1.30)
τ (MPa)
(a)
60 MPa
R σ = −
1 ( 60)
2
x’
σ
70o
x
σ
−
o
60 30 sin( 40 ); 153.3 MPa 2
τ σ
= − = − =
‘ ‘
x y
σ
−
o
60 80 60 [1 cos( 40 )] 2
σ
(b) ‘
= = + − −
x
σ =
231 MPa
______________________________________________________________________________________
SOLUTION (1.31)
( a ) From Mohr’s circle, Fig. (a):
1 2 max 121 MPa 71 MPa 96 MPa σ σ τ = = − =
θ
o
−= θ
p 7.25’3.19′=
s
o
τ
(MPa)
τmax
A(100,60)
‘2 p θ
σ2 O
C
B
Figure (a)
By applying Eq. (1.20):
1
50
22,500
2
σ = ± + = ±
1,2 2 4 3600 25 96
or 1 2 121 MPa 71 MPa σ σ = = −
Using Eq. (1.19):
2
tan 15
p θ
12 −=−=
8.0
θ
o
−= θ
p 7.25’3.19′=
s
o
(CONT.)
______________________________________________________________________________________
12
From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886)
Copyright © 2020 Pearson Education, Inc. All rights reserved.
σ (MPa)
σ1______________________________________________________________________________________
1.31 (CONT.)
( b ) From Mohr’s circle, Fig. (b):
1 2 max 200 MPa 50 MPa 125 MPa σ σ τ = = − =
θ
o
= θ
p 55.71’55.26′=
o
s
τ
(MPa)
τmax
σ2 O
σ (MPa)
C
σ1
A(150,-100)
2θp
’
Figure (b)
Through the use of Eq. (1.20),
1
500,22
σ
75 2
2,1 ±=+±=
[ ] 12575000,10
4
or
1 2 200 MPa 50 MPa σ σ = = −
tan=
Using Eq. (1.19), 342
p θ :
‘ 26.57 ‘ 71.57o o
p s θ θ = =
______________________________________________________________________________________
SOLUTION (1.32)
Referring to Mohr’s circle, Fig. 1.15:
σ
x (a)
‘
+ +=
2121−
σ 2
cos
θ
2
2
σ
‘
y
+
2121−
−=
σ 2
cos
θ
2
2
τ
21−
=
yx (b)
‘
σ 2
sin
θ
2
From Eqs. (a),
21′ σ +=+ yx
cos 2
2
2 =+ sin
By using 12
2
=⋅=−⋅ 21
const
‘ στσ.
yxyx
θ , and Eqs. (a) and (b), we have
______________________________________________________________________________________
SOLUTION (1.33)
We have
τ
2 2( 70)
θ−
tan 2 0.583xy
= = = −
p
σ σ
− − −
x y
50 ( 190)
2 30.24o
p θ = − and 15.12o
p θ = −
(CONT.)
______________________________________________________________________________________
13
From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886)
Copyright © 2020 Pearson Education, Inc. All rights reserved.______________________________________________________________________________________
1.33 (CONT.)
Equations (1.18):
19050
−
σ
‘
+= +
19050
2
2
cos(
o
70)26.30
sin(
x−
o
)26.30
= − + + = =
70 103.65 35.275 68.93 MPa σ
1
= + − = − =
y x y x σ σ σ σ σ
‘ ‘ 2 208.9 MPa
208.9 MPa
15.13o
68.92 MPa
______________________________________________________________________________________
SOLUTION (1.34)
τ
=−
σ
2
yx τ
+
max )( xy
2
Substituting the given values
2
2
10060
+= +
( )2
τ
140 xy
2
or
xy τ =
,max 114.89 MPa
______________________________________________________________________________________
SOLUTION (1.35)
Transform from o
θ to o
60=
0=
θ with ‘ ‘ 20 MPa, 60 MPa x y σ σ = − =
,
‘ ‘ 22 MPa
60−=
x y τ = − , and o
θ . Use Eqs. (1.18):
20 60 20 60
x σ − + − − = + − − − =
2 2 cos 2( 60 ) 22sin 2( 60 ) 59 MPao o
‘ ‘ 19 MPa
y x y x σ σ σ σ = + − = −
xy τ = −
23.6 MPa
y
19 MPa
23.6 MPa
59 MPa
x
______________________________________________________________________________________
14
From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886)
Copyright © 2020 Pearson Education, Inc. All rights reserved.______________________________________________________________________________________
SOLUTION (1.36)
σycos60o
σy
14 MPa
τxy
σxsin60o
30 MPa
τxycos60o
30 MPa
Area =1
60o
Figure (a)
τxysin60o
Figure (b)
( a ) Figure (a):
σ = =
14sin 60 12.12 MPao
y
14cos 60 7 MPao
xy τ = =
Figure (b):
=
60
o
F τ
−
sin
∑o
12.12=
cos
060
y
xy
or
7 MPa
xy τ = (as before)
o
sino
−= 060
73060
+
∑=
x F σ
cos
x
or
x σ =
38.68 MPa
( b ) Equation (1.20) is therefore:
1
12.1268.38
+
σ
±=−
12.1268.38
2,1 7)( +
2
2
[ ]2
2
or 1 2 40.41 MPa, 10.39 MPa σ σ = =
Also,
θ
−
1
)7(2
=
1 =
tan 12.1268.38
p 9.13
o
2
−
Note: Eq. (1.18a) gives, ‘ 40.41 MPa
x σ =
.
Thus,
θ
p 9.13′=
o
10.39 MPa
40.41 MPa
x’
x
θp
’
______________________________________________________________________________________
15
From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886)
Copyright © 2020 Pearson Education, Inc. All rights reserved.______________________________________________________________________________________
SOLUTION (1.37)
σy
τxy
σx Figure (a)
Figure (a):
x σ = =
100cos 45 70.7 MPao
σ = =
100sin 45 70.7 MPao
y
xy τ = =
100cos 45 70.7 MPao
Now, Eqs. (1.18) give (Fig. b):
‘ 70.7 0 70.7sin 240 9.47 MPao
x σ = + + =
x y τ = − + ‘ ‘ 0 70.7 cos 240 35.35 MPao
= −
σ = − − =
‘ 70.7 0 70.7sin 240 131.9 MPao
y
y σy
σx’ τx’y’
x’
σx
y’
σy’
τxy
Figure (b)
n
30o
x
m
______________________________________________________________________________________
SOLUTION (1.38)
σ = − = −
70sin 30 35 MPao
y
xy τ = =
70cos30 60.6 MPao
( a ) Figure (a):
∑x
x F σ
+−=
0)866.0(6.605.0150=
or 195 MPa
x σ =
35cos30o
60.6cos30o
30o
60.6sin30o
150 MPa Figure (a)
σxsin30o
Area=1
(CONT.)
______________________________________________________________________________________
16
From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886)
Copyright © 2020 Pearson Education, Inc. All rights reserved.
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