CHAPTER 7
PR
ROBLEM 7.1
7
4 ksi
3 ksi
708
8 ksi
Foor the given sttate of stress, determine thee normal and shearing stressses exerted
onn the oblique face of the shhaded trianguular element shown.
s
Use a method of
anaalysis based on
o the equilibrrium of that ellement, as waas done in the derivations
of Sec. 7.1A.
SOLUTION
F  0:  A  8 A cos 20
2  cos 20  3 A cos 20 sin 20
2   3 A sin 200 cos 20  4 A sin 20 sin 20
2  0
  8cos 2 20  3cos 20 sin 20  3 sin 20 cos 20  4sin 2 20  0
  9.46 ksi 
F  0:  A  8 A cos 200 sin 20  3A
A cos 20 cos 20  3 A sin 200 sin 20  4A
A sin 20 cos 200  0
  8coos 20 sin 20  3(cos2 20  sin 2 20)  4sin
4 20 cos 200
  1.013 ksi 
PRO
OPRIETARY MAT
TERIAL. Copyrigght © 2015 McGrraw-Hill Educatioon. This is proprietary material soleely for authorizedd instructor use.
Not authorized
a
for salee or distribution inn any manner. Thiis document may not
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in whole or part.
1027
60 MPa
PRO
OBLEM 7.2
2
For th
he given statee of stress, dettermine the noormal and sheearing stressess exerted on
the oblique face off the shaded triangular
t
elem
ment shown. Use
U a methodd of analysis
based
d on the equilibbrium of that element,
e
as waas done in the derivations off Sec. 7.1A.
608
90 MPa
M
SO
OLUTION
F  0:  A  90
9 A sin 30 coss 30  90 A cos 30 sin 30  60 A cos 30 ccos 30  0
  180sin 30 cos
c 30  60 coos 2 30
  32.9
3
M Pa 
F  0:  A  900 A sin 30 sin 30
3   90 A cos 30  cos 30  60 A cos 30  sinn 30   0
  90(cos 2 30  sin 2 30)  60 cos 30 sin 30
  71.0
7
M Pa 
PRO
OPRIETARY MATERIAL. Copyriight © 2015 McG
Graw-Hill Educattion. This is proprrietary material soolely for authorizedd instructor use.
Not authorized for salle or distribution in
i any manner. Thhis document may not be copied, scaanned, duplicated,, forwarded, distributed, or posted
on a website, in wholee or part.
1028
PROBLEM
M 7.3
10 ksi
6 ksi
758
For the giveen state of sttress, determiine the normaal and sheariing stresses
exerted on thhe oblique faace of the shaaded triangulaar element shoown. Use a
method of annalysis based on the equilibbrium of that element, as was
w done in
the derivationns of Sec. 7.1A
A.
4 ksi
SOLUTION
F  0:  A  4 A cos15 sin15  100 A cos15 cos115  6 A sin15 sin15  4 A sin15 cos155  0
  4 coos15 sin15  10 cos 2 15  6sin 2 15  4sin15
4
 cos155
  10.93
1
ksi 
F  0:  A  4 A cos15 cos15  10 A cos15 sin 15  6 A sin15 cos15  4 A sin15 sin155  0
  4(ccos2 15  sin 2 15)  (10  6) cos15 sin15
  0.536
0
ksi 
PRO
OPRIETARY MAT
TERIAL. Copyrigght © 2015 McGrraw-Hill Educatioon. This is proprietary material soleely for authorizedd instructor use.
Not authorized
a
for salee or distribution inn any manner. Thiis document may not
n be copied, scannned, duplicated, forwarded, distribbuted, or posted
on a website,
w
in whole or part.
1029
PROBLEM
P
7.4
80 MPa
M
40 MPa
558
For
F the given state
s
of stress, determine thhe normal andd shearing streesses exerted
on
o the obliquee face of the shaded trianggular element shown. Use a method of
analysis
a
based on the equilibbrium of that element,
e
as was
w done in thee derivations
of
o Sec. 7.1A.
SO
OLUTION
Streesses
Areas
Forces
F  0:
0  A  80 A cos 55 cos555  40 A sin 55 sin 55  0
  80 cos 2 55  40sin 2 55
  0.5521 MPa 
F  0:
0  A  80 A cos 55 sin 55
5   40 A sin 55 cos 55 
  56.4
5 MPa 
  120
1 cos 55 sin 55
PRO
OPRIETARY MATERIAL. Copyriight © 2015 McG
Graw-Hill Educattion. This is proprrietary material soolely for authorizedd instructor use.
Not authorized for salle or distribution in
i any manner. Thhis document may not be copied, scaanned, duplicated,, forwarded, distributed, or posted
on a website, in wholee or part.
1030
PROBLEM 7.5
40 MPa
35 MPa
For the given state of stress, determine (a) the principal planes, (b) the principal
stresses.
60 MPa
SOLUTION
 x  60 MPa  y  40 MPa  xy  35 MPa
(a)
tan 2 p 
2 xy
x  y

(2)(35)
 3.50
60  40
2 p  74.05
(b)
 max, min 
x y
2
 p  37.0, 53.0 
2
x  y 
2
 
   xy
2


2
60  40
 60  40 
2

 
  (35)
2
2


 50  36.4 MPa
 max  13.60 MPa 
 min  86.4 MPa 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1031
PROBLEM 7.6
10 ksi
For the given state of stress, determine (a) the principal planes, (b) the principal
stresses.
2 ksi
3 ksi
SOLUTION
 x  2 ksi
(a)
tan 2 p 
2 xy
x  y
 y  10 ksi

 xy  3 ksi
(2)(3)
 0.750
2  10
2 p  36.87
(b)
 max,min 
x  y
2
 p  18.4, 108.4 ◄
2
x  y 
2
 
   xy
2


2
2  10
 2  10 
2

 
  (3)
2
 2 
 6  5 ksi
 max  11.00 ksi ◄
 min  1.000 ksi ◄
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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on a website, in whole or part.
1032
PROBLEM 7.7
30 MPa
For the given state of stress, determine (a) the principal planes, (b) the principal
stresses.
150 MPa
80 MPa
SOLUTION
 x  150 MPa,
(a)
tan 2 p 
2 xy
x   y

 y  30 MPa,
 xy  80 MPa
2(80 MPa)
 1.33333 MPa
(150 MPa  30 MPa)
2 p  53.130 and 126.870
 p  26.6 and  63.4 ◄
(b)
 max,min 
x  y
2
x  y 
2
 
   xy
2


2
150 MPa  30 MPa
 150 MPa  30 MPa 
2

 
  (80 MPa)
2
2


 90 MPa  100 MPa
 max  190.0 MPa ◄
 min  10.00 MPa ◄
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1033
PROBLEM 7.8
12 ksi
8 ksi
For the given state of stress, determine (a) the principal planes, (b) the principal
stresses.
18 ksi
SOLUTION
 y  12 ksi
 x  18 ksi
(a)
tan 2 p 
2 xy
x  y

 xy  8 ksi
(2)(8)
 0.5333
18  12
2 p  28.07
(b)
 max,min 
x   y
2
 p  14.0, 104.0 ◄
2
x  y 
2
 
   xy
2


2

18  12
 18  12 
2
 
  (8)
2
 2 
 3  17 ksi
 max  20.0 ksi ◄
 min  14.00 ksi ◄
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
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1034
PROBLEM 7.9
40 MPa
35 MPa
60 MPa
For the given state of stress, determine (a) the orientation of the planes of
maximum in-plane shearing stress, (b) the maximum in-plane shearing stress,
(c) the corresponding normal stress.
SOLUTION
 x  60 MPa  y  40 MPa  xy  35 MPa
(a)
tan 2 s  
x  y
60  40

 0.2857
2 xy
(2)(35)
2 s  15.95
 s  8.0, 98.0 
2
(b)
 max
 x  y 
2
 
   xy
2 

2
 60  40 
2
 
  (35)
2


(c)
    ave 
x y
2

 max  36.4 MPa 
60  40
2
   50.0 MPa 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1035
PROBLEM 7.10
10 ksi
For the given state of stress, determine (a) the orientation of the planes of
maximum in-plane shearing stress, (b) the maximum in-plane shearing stress,
(c) the corresponding normal stress.
2 ksi
3 ksi
SOLUTION
 y  10 ksi
 x  2 ksi
(a)
tan 2 s  
 xy  3 ksi
x  y
2  10

 1.33333
2 xy
(2)(3)
2 s  53.13
 s  26.6, 63.4 
2
(b)
x  y 
2
   xy
2


 max  
2

 2  10 
2

  (3)
2


 max  5.00 ksi 
(c)
    ave 
x  y
2

2  10
2
   6.00 ksi 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1036
PROBLEM 7.11
30 MPa
For the given state of stress, determine (a) the orientation of the planes of
maximum in-plane shearing stress, (b) the maximum in-plane shearing stress,
(c) the corresponding normal stress.
150 MPa
80 MPa
SOLUTION
 x  150 MPa,
(a)
tan 2 s  
 y  30 MPa,
 xy  80 MPa
x   y
150  30

 0.750
2 xy
2(80)
2 s  36.87 and 216.87
 s  18.4 and 108.4 
2
(b)
x  y 
2
   xy
2


 max  
2

 150  30 
2

  (80)
2


 max  100.0 MPa 
(c)
    ave 
x  y
2
 150  30 


2


   90.0 MPa 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1037
PROBLEM 7.12
12 ksi
8 ksi
For the given state of stress, determine (a) the orientation of the planes of
maximum in-plane shearing stress, (b) the maximum in-plane shearing stress,
(c) the corresponding normal stress.
18 ksi
SOLUTION
 y  12 ksi
 x  18 ksi
(a)
tan 2 s  
 xy  8 ksi
x  y
18  12

 1.875
2 xy
(2)(8)
2 s  61.93
 s  31.0, 59.0 
2
(b)
x  y 
2
   xy
2


 max  
2

 18  12 
2

  (8)
2


 max  17.00 ksi 
(c)
    ave 
x  y
2

18  12
2
   3.00 ksi 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
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1038
PROBLEM 7.13
8 ksi
5 ksi
For the given state of stress, determine the normal and shearing stresses after the
element shown has been rotated through (a) 25 clockwise, (b) 10 counterclockwise.
SOLUTION
 x  0  y  8 ksi  xy  5 ksi
x y
2
 x 
x y
2
 xy  
 y 
(a)
x  y
 4 ksi
2

x  y
2
x y
2

 4 ksi
x  y
2
sin 2 +  xy cos 2
x  y
2
cos 2   xy sin 2
  25 2  50
 x  4  4 cos (50°) + 5 sin (50°)
 xy  4 sin (50)  5 cos (50)
 x  2.40 ksi 
 xy  0.1498 ksi 
 y  4  4 cos (50)  5 sin (50)
(b)
cos 2 +  xy sin 2
 y  10.40 ksi 
  10 2  20
 x  4  4 cos (20°) + 5 sin (20°)
 x  1.951 ksi 
 xy  4 sin (20°) + 5 cos (20°)
 xy  6.07 ksi 
 y  4  4 cos (20°)  5 cos (20°)
 y  6.05 ksi 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
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1039
PROBLEM 7.14
90 MPa
30 MPa
For the given state of stress, determine the normal and shearing stresses after
the element shown has been rotated through (a) 25 clockwise, (b) 10
counterclockwise.
60 MPa
SOLUTION
 x  60 MPa  y  90 MPa  xy  30 MPa
x y
2
 x 
x y
2
 xy  
 y 
(a)
x  y
 15 MPa

x  y
2
x y
2

2
x  y
 75 MPa
cos 2 + xy sin 2
2
sin 2 +  xy cos 2
x  y
2
cos 2   xy sin 2
  25 2  50
 x  15  75 cos (50)  30 sin (50)
 xy  75 sin (50)  30 cos (50)
 y  15  75 cos (50)  30 sin (50)
(b)
 x  56.2 MPa 
 xy  38.2 MPa 
 y  86.2 MPa 
  10 2  20
 x  15  75 cos (20°) + 30 sin (20°)
 x  45.2 MPa 
 xy  75 sin (20°) + 30 cos (20°)
 xy  53.8 MPa 
 y  15  75 cos (20°)  30 sin (20°)
 y  75.2 MPa 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1040
PROBLEM 7.15
12 ksi
For the given state of stress, determine the normal and shearing stresses after
the element shown has been rotated through (a) 25 clockwise, (b) 10
counterclockwise.
8 ksi
6 ksi
SOLUTION
 x  8 ksi  y  12 ksi  xy  6 ksi
x y
2
 2 ksi
 x 
x y
2
 xy  
 y 
(a)
x  y
2

x  y
2
x y
2

 10 ksi
x  y
2
sin 2 +  xy cos 2
x  y
2

cos 2   xy sin 2
  25 2  50
 x  2  10 cos (50)  6 sin (50)
 x  9.02 ksi 
 xy  10 sin ( 50)  6 cos (50)
 xy  3.80 ksi 
 y  2  10 cos (50)  6 sin (50)
(b)
cos 2 +  xy sin 2
 y  13.02 ksi 
  10 2  20
 x  2  10 cos (20°)  6 sin (20°)
 x  5.34 ksi 
 xy  10 sin (20°)  6 cos (20°) 
 xy  9.06 ksi 
 y  2  10 cos (20°) + 6 sin (20°)
 y  9.34 ksi 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1041
PROBLEM 7.16
80 MPa
For the given state of stress, determine the normal and shearing stresses after the
element shown has been rotated through (a) 25 clockwise, (b) 10 counterclockwise.
50 MPa
SOLUTION
 x  0  y  80 MPa  xy  50 MPa
x y
2
 x 
x y
2
 xy  
 y 
(a)
x  y
 40 MPa

x  y
2
x y
2
2
x  y
2
cos 2   xy sin 2
sin 2 +  xy cos 2

x  y
2
cos 2   xy sin 2
  25 2  50
 x  40  40 cos (50)  50 sin (50°)
(b)
 40 MPa
 x  24.0 MPa 
 xy  40 sin (50°)  50 cos (50)
 xy  1.498 MPa 
 y  40  40 cos (50)  50 sin (50)
 y  104.0 MPa 
  10 2  20
 x  40  40 cos (20°)  50 sin (20°)
 x  19.51 MPa 
 xy  40 sin (20°)  50 cos (20°)
 xy  60.7 MPa 
 y  40  40 cos (20°) + 50 sin (20°)
 y  60.5 MPa 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1042
PROBLEM 7.17
250 psi
The grain of a wooden member forms an angle of 15° with the vertical. For the state
of stress shown, determine (a) the in-plane shearing stress parallel to the grain, (b) the
normal stress perpendicular to the grain.
158
SOLUTION
x  y  0
(a)
 xy  
 xy  250 psi
  15
x   y
sin 2   xy cos 2
2
 0  250cos(30)
 xy  217 psi 
(b)
 x 
x  y
x  y
cos 2   xy sin 2
2
2
 0  0  250sin(30)

 x  125.0 psi 
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1043
PROBLEM 7.18
1.8 MPa
3 MPa
The grain of a wooden member forms an angle of 15° with the vertical. For the
state of stress shown, determine (a) the in-plane shearing stress parallel to the
grain, (b) the normal stress perpendicular to the grain.
158
SOLUTION
 y  1.8 MPa
 x  3 MPa
  15
(a)
 xy  
 xy  0
2  30
x  y
sin 2   xy sin 2
2
3  1.8
sin(30)  0

2
 xy  0.300 MPa 
(b)
 x 
x  y
x  y
cos 2   xy sin 2
2
2
3  1.8 3  1.8
cos(30)  0


2
2

 x  2.92 MPa 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1044
P'
80 mm
m
P
PROBLEM
7
7.19
1200 mm
Tw
wo wooden members
m
of 800  120-mm uniform
u
rectanngular cross
seection are joined by the simpple glued scarrf splice shownn. Knowing
thhat   22 and
a that the maximum
m
allow
wable stresses in the joint
arre, respectivelly, 400 kPa inn tension (perrpendicular to the splice)
annd 600 kPa inn shear (parallel to the splicce), determinee the largest
ceentric load P thhat can be appplied.
b
P
SOLUTION
Forces
Areeas
A  (80) (120)  9.6  103 mm 2  9.6  103 m 2
N all   all
a A/sin  
(4400  103 )(9.6  103 )
 10.2251  103 N
sin 22
Fy  0: N  P sin   0
Sall   aall A/sin  
P
N
10.251  103

 27.4  10
1 3N
sinn 
sin 222
(6600  103 )(9.6  103 )
 15.3376  103 N
sin 22
Fx  0:: S  P cos   0
P
S
15.376  103

 16.58  103 N
coos 
cos 22
2 
P  16.58 kN 
Thee smaller valuee for P governns.
PRO
OPRIETARY MAT
TERIAL. Copyrigght © 2015 McGrraw-Hill Educatioon. This is proprietary material soleely for authorizedd instructor use.
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in whole or part.
1045
P'
PROBLEM 7.20
7
1220 mm
80 mm
b
P
Tw
wo wooden members
m
of 800  120-mm uniform
u
rectanngular cross
section are joineed by the simp
mple glued scarrf splice show
wn. Knowing
a that centriic loads of magnitude
m
P  10 kN are
that   25 and
t
in-plane
appplied to the members ass shown, dettermine (a) the
shhearing stresss parallel to the splice, (b) the noormal stress
peerpendicular too the splice.
SO
OLUTION
Forcess
A
Areas
A  (80)(1220)  9.6  103 mm
m 2  9.6  1003 m 2
(a)
Fx  0: S  P cos   0

(b)
N
(9.063  103 )sinn 25
 399  1003 Pa

A/sin 
9.6  103
Fy  0: N  P sin
n 0

S  P cos   (10  103 ) cos 25  9.063  103 N
  399 kPa 
N  P sin   (10  103 )sin 25  4.226  103 N
N
(4.226  103 )sin 25

 186.0  103 Pa
A/sin 
9.6  103
  186.0
1
kPa 

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OPRIETARY MATERIAL. Copyriight © 2015 McG
Graw-Hill Educattion. This is proprrietary material soolely for authorizedd instructor use.
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1046
PROBL
LEM 7.21
P
The centrric force P is applied to a short post as shown. Know
wing that the stresses on
plane a-a are   15 ksi and   5 ksi, determinne (a) the anggle  that planne a-a forms
h
(b) the maximum
m compressivee stress in the post.
p
with the horizontal,
a
a
SOLUTION
x  0
 xy  0
 y   P/ A
(a)
From the Mohr’s
M
circle,
tan  
 
(b)
5
 0.33333
15
  18.4 
P
P

cos 2 
2A 2A
P
2( )
(2)(115)


A 1  co
os 2 1  coss 2 
P
 16.67
1
ksi 
A
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in whole or part.
1047
PROBLEM 7.2
22
Two
o members of uniform crosss section 50  80 mm are glued
g
togetherr along plane
a-a that forms ann angle of 255 with the horizontal.
h
Knnowing that thhe allowable
a   600 kPa,
k
determinne the largest
stressses for the gluued joint are   800 kPa and
centtric load P thatt can be applieed.
a
a
25
50 mm
m
P
SO
OLUTION
Forr plane a-a,   65.
 x  0,
0  xy  0,  y 
P
A
   x cos 2    y sin 2   2 xy sin  cos   0 
P 2
sin 655  0
A
A
(50  103 )(80  103 )(800  103 )
 3.90  103 N

sin 2 65
sin
s 2 65
6 
P
  ( x   y )sin  cos    xy (cos
( 2   sin 2  )  sin 65 cos 65  0
A
3
3
(50  10 )((80  10 )(600  103 )
A
P
 6.277  103 N

sin
s 65 cos 65
sinn 65 cos 65
P
P  3.90 kN 
maller one.
Alllowable value of P is the sm
PRO
OPRIETARY MATERIAL. Copyriight © 2015 McG
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1048
PROBLEM 7.23
0.2 m
0.15 m
The axle of an automobile is acted upon by the forces and couple
shown. Knowing that the diameter of the solid axle is 32 mm,
determine (a) the principal planes and principal stresses at point H
located on top of the axle, (b) the maximum shearing stress at the
same point.
H
3 kN
350 N · m
3 kN
SOLUTION
c
Torsion:
 
Bending:
I 
1
1
d  (32)  16 mm  16  103 m
2
2
Tc
2T
2(350 N  m)


 54.399  106 Pa  54.399 MPa
3
J
c
 (16  10 3 m)3

4
c4 

4
(16  103 ) 4  51.472  109 m 4
M  (0.15m)(3  103 N)  450 N  m
 
(450)(16  103 )
My

 139.882  106 Pa  139.882 MPa
9
I
51.472  10
Top view:
Stresses:
 x  139.882 MPa
 ave 
y  0
 xy  54.399 MPa
1
1
( x   y )  (139.882  0)  69.941 MPa
2
2
2
R 
(a)
x  y 
2

   xy 
2


(69.941)2  (54.399)2  88.606 MPa
 max   ave  R  69.941  88.606
 max  18.67 MPa 
 min   ave  R  69.941  88.606
 min  158.5 MPa 
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1049
PROBLEM 7.23 (Continued)
tan 2 p 
2 xy
x  y

(2)(54.399)
 0.77778
139.882
2 p  37.88
 p  18.9 and 108.9° 
(b)
 max  R  88.6 MPa
 max  88.6 MPa 
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1050
6 in.
PROBLEM 7.24
C
H
A 400-lb vertical force is applied at D to a gear attached to the solid l-in.
diameter shaft AB. Determine the principal stresses and the maximum
shearing stress at point H located as shown on top of the shaft.
B
A
D
2 in.
400 lb
SOLUTION
Equivalent force-couple system at center of shaft in section at point H:
V  400 lb
M  (400)(6)  2400 lb  in.
T  (400)(2)  800 lb  in.
d  1 in. c 
Shaft cross section:
J 

2
1
d  0.5 in.
2
c 4  0.098175 in 4
I 
1
J  0.049087 in 4
2
Torsion:
 
Tc (800)(0.5)

 4.074  103 psi  4.074 ksi
J
0.098175
Bending:
 
Mc
(2400)(0.5)

 24.446  103 psi  24.446 ksi
I
0.049087
Transverse shear:
Stress at point H is zero.
 x  24.446 ksi,  y  0,  xy  4.074 ksi
 ave 
1
( x   y )  12.223 ksi
2
2
R 
x  y 
2

   xy 
2


(12.223) 2  (4.074) 2
 12.884 ksi

 a   ave  R
 a  25.1 ksi 
 b   ave  R
 b  0.661 ksi 
 max  R 
 max  12.88 ksi 
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1051
PROBLEM 7.25
H
E
A mechanic uses a crowfoot wrench to loosen a bolt at E. Knowing that
the mechanic applies a vertical 24-lb force at A, determine the principal
stresses and the maximum shearing stress at point H located as shown
as on top of the 34 -in. diameter shaft.
6 in.
B
24 lb
A
10 in.
SOLUTION
Equivalent force-couple system at center of shaft in section at point H:
V  24 lb M  (24)(6)  144 lb  in.
T  (24)(10)  240 lb  in.
1
d  0.375 in.
2
1

J  c 4  0.031063 in 4 I  J  0.015532 in 4
2
2
Shaft cross section:
d  0.75 in., c 
Torsion:
 
Tc (240)(0.375)

 2.897  103 psi  2.897 ksi
J
0.031063
Bending:
 
Mc (144)(0.375)

 3.477  103 psi  3.477 ksi
I
0.015532
Transverse shear:
At point H, the stress due to transverse shear is zero.
Resultant stresses:
 x  3.477 ksi,  y  0,  xy  2.897 ksi
 ave 
1
( x   y )  1.738 ksi
2
2
R 

x  y 
2
2
2

   xy  1.738  2.897  3.378 ksi
2


 a   ave  R
 a  5.12 ksi 
 b   ave  R
 b  1.640 ksi 
 max  R 
 max  3.38 ksi 
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1052
P
PROBLEM
7
7.26
y
m
6 mm
200 mm
Thhe steel pipe AB
A has a 1022-mm outer diameter and a 6-mm wall
thhickness. Knowing that arm
m CD is rigiddly attached to
t the pipe,
deetermine the principal
p
stressses and the maximum
m
shearing stress
att point K.
51 mm
A
A
T
D
10 kN
N
C
1 mm
150
H
K
B
x
z
SOLUTION
ro 
d o 102
1

 51 mm
2
2


ri  ro  t  45 mm

ro4  ri4  4.18555  106 mm 4
2
 4.18555  10 6 m 4
J
I
1
J  2.0927  10 6 m 4
2
Forcce-couple systtem at center of
o tube in the plane
p
containiing points H and
a K:
Fx  10 kN
 10  1003 N
M y  (10  103 )(200  103 )
 2000 N  m
M z  (10  103 )(150  103 )
 15000 N  m
Torsion:
At po
oint K, place local
l
x-axis in negative globbal z-directionn.
T  M y  2000 N  m
c  ro  511  103 m
Tc ((2000)(51  1003 )

J
4.1855  106
 24.37  106 Pa
 24.37 MPa
 xy 
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1053
PROBLEM 7.26 (Continued)
Transverse shear:
Stress due to transverse shear V  Fx is zero at point K.
Bending:
| y | 
|M z |c (1500)(51  103 )

 36.56  106 Pa  36.56 MPa
I
2.0927  106
Point K lies on compression side of neutral axis.
 y  36.56 MPa
Total stresses at point K:
 x  0,  y  36.56 MPa,  xy  24.37 MPa
1
2
 ave  ( x   y )  18.28 MPa
2
x  y 
2
R 
   xy  30.46 MPa
2



 max   ave  R  18.28  30.46
 max  12.18 MPa 
 min   ave  R  18.28  30.46
 min  48.7 MPa 
 max  R 
 max  30.5 MPa 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1054
y
PROBLEM 7.27
20 MPa
60 MPa
For the state of plane stress shown, determine the largest value of  y for which
the maximum in-plane shearing stress is equal to or less than 75 MPa.
SOLUTION
 x  60 MPa,  y  ?,  xy  20 MPa
Let
u
x  y
2
.
Then
 y   x  2u
2
R  u 2   xy
 75 MPa
2
u   R 2   xy
  752  202  72.284 MPa
 y   x  2u  60  (2)(72.284)  84.6 MPa or 205 MPa
Largest value of  y is required.
 y  205 MPa 
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1055
PROBLEM 7.28
8 ksi
xy
10 ksi
For the state of plane stress shown, determine (a) the largest value of  xy for which
the maximum in-plane shearing stress is equal to or less than 12 ksi, (b) the
corresponding principal stresses.
SOLUTION
 x  10 ksi,  y  8 ksi,  xy  ?
2
 max
2
x  y 
 10  (8) 
2
2
R 
   xy  
   xy
z
z




2
 92   xy
 12 ksi
(a)
 xy  122  9 2
(b)
 ave  ( x   y )  1 ksi
 xy  7.94 ksi 
1
2
 a   ave  R  1  12  13 ksi
 a  13.00 ksi 
 b   ave  R  1  12  11 ksi
 b  11.00 ksi 
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1056
P
PROBLEM
7.29
2 MPaa
xy
75
12 MPa
For the state of plane stress shown, determ
F
mine (a) the vaalue of  xy foor which the
inn-plane shearring stress paarallel to the weld is zeroo, (b) the corrresponding
p
principal
stressses.
SOLUTION
 x  12 MPa,  y  2 MPaa,  xy  ?
Sincce  xy  0, x-direction is a principal direection.
 p  15
tan 2 p 
(a)
1
2
2 xy
x y
1
2
 xy  ( x   y ) tan 2 p  (12  2)) tan(30)
 xy  2.89 MPa 
2
x  y 
2
2
2
R 
M
   xyy  5  2.899  5.7735 MPa
2


1
 ave  ( x   y )  7 MPa
M
2
(b)
 a   ave  R  7  5.77735
 a  12..77 MPa 
 b   ave  R  7  5.77735
 b  1.2226 MPa 
PRO
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1057
PROBLEM 7.30
15 ksi
8 ksi
Determine the range of values of  x for which the maximum in-plane shearing stress
is equal to or less than 10 ksi.
x
SOLUTION
 x  ?,  y  15 ksi,  xy  8 ksi
Let u 
x  y
2
 x   y  2u
2
R  u 2   xy
  max  10 ksi
2
u   R 2   xy
  102  82 z  6 ksi
 x   y  2u  15  (2)(6)  27 ksi or 3 ksi
3 ksi   x  27 ksi 
Allowable range:
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1058
PR
ROBLEM 7.31
7
40 MPaa
Soolve Probs. 7.55 and 7.9, usinng Mohr’s circcle.
355 MPa
PR
ROBLEM 7.55 through 7..8 For the givven state of stress,
s
determ
mine (a) the
priincipal planes, (b) the principal stresses.
60 MPa
PR
ROBLEM 7.99 through 7.12 For the giiven state of stress, determ
mine (a) the
oriientation of thhe planes of maximum
m
in-pplane shearing stress, (b) thee maximum
in--plane shearinng stress, (c) thhe correspondiing normal strress.
SOLUTION
 x  60
6 MPa,
 y  40
4 MPa,
 xy  355 MPa
 ave 
x y
2
 50 MPa
Plottted points forr Mohr’s circlee:
X : ( x ,  xy )  (60 MPa,  35 MPa)
Y : ( y ,  xy )  ( 40 MPa, 35 MPa)
C : ( ave , 0)  (50 MPa, 0)
(a)
X 35
GX

 3.5000
CG
G 10
  744.05
tan  
1
2
  180    105.995
1
 a    52.97
2
b  37.0 
b     37.03
2
 a  53.0 
2
R  CG
C  GX  10 2  352  36.4 MPa
(b)
(a)
(b)
(c)
 min   ave
a  R  50  36.4
 min  866.4 MPa 
 max   ave
a  R  50  36.4
 max  133.60 MPa 
 d   B  45  7.97
 d  8.0 
e   A  45  97.977
e  98.0 
 max  R  36.4 MPa
 max  366.4 MPa 
    ave
a
a  50 MPa
   500.0 MPa 
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OPRIETARY MAT
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1059
PROBLEM 7.32
30 MPa
Solve Probs. 7.7 and 7.11, using Mohr’s circle.
150 MPa
80 MPa
PROBLEM 7.5 through 7.8 For the given state of stress, determine (a) the
principal planes, (b) the principal stresses.
PROBLEM 7.9 through 7.12 For the given state of stress, determine (a) the
orientation of the planes of maximum in-plane shearing stress, (b) the maximum
in-plane shearing stress, (c) the corresponding normal stress.
SOLUTION
 x  150 MPa
 y  30 MPa
 xy  80 MPa
 ave 
x  y
2
 90 MPa
Plotted points for Mohr’s circle:
X : ( x ,  xy )  (150 MPa, 80 MPa)
Y : ( y ,  xy )  (30 MPa,  80 MPa)
C : ( ave , 0)  (90 MPa, 0)
x  y
2

(150  30)
 60
2
R  (60)2  (80)2  100
(a)
tan 2 p 
80
60
2 p  53.130
(b)
 p  26.6 and 63.4 
 max   ave  R  90  100
 max  190.0 MPa 
 min   ave  R  90  100
 min  10.00 MPa 
(a′)
 s   p  45
(b′)
 max  R
(c′)
    ave
 s  18.4 and 108.4 
 max  100.0 MPa 
   90.0 MPa 
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1060
PROBLEM 7.33
10 ksi
Solve Prob. 7.10, using Mohr’s circle.
2 ksi
PROBLEM 7.9 through 7.12 For the given state of stress, determine (a) the
orientation of the planes of maximum in-plane shearing stress, (b) the maximum
in-plane shearing stress, (c) the corresponding normal stress.
3 ksi
SOLUTION
 x  2 ksi
x  y
 ave 
2
 y  10 ksi

 xy  3 ksi
2  10
 6 ksi
2
Plotted points for Mohr’s circle:
X : ( x ,  xy )  (2 ksi, 3 ksi)
Y : ( y ,  xy )  (10 ksi, 3 ksi)
C : ( ave , 0)  (6 ksi, 0)
tan  
FX
3
  0.75
4
FC
  36.87 
1
 B    18.43
2
(a)
 D   B  45  26.6
 D  26.6 
 E   B  45  63.4
R
2
CF  FX
2
(b)
 max  R  5.00 ksi
(c)
    ave  6.00 ksi

 E  63.4 
42  32  5 ksi
 max  5.00 ksi 
   6.00 ksi 
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1061
PROBLEM 7.34
12 ksi
8 ksi
Solve Prob. 7.12, using Mohr’s circle.
18 ksi
PROBLEM 7.9 through 7.12 For the given state of stress, determine (a) the
orientation of the planes of maximum in-plane shearing stress, (b) the
maximum in-plane shearing stress, (c) the corresponding normal stress.
SOLUTION
 x  18 ksi
 ave 
x  y
2
 y  12 ksi
 xy  8 ksi
 3 ksi
Plotted points for Mohr’s circle:
X : ( x ,  xy )  (18 ksi, 8 ksi)
Y : ( y ,  xy )  (12 ksi, 8 ksi)
C : ( ave , 0)  (3 ksi, 0)
tan  
FX
8

 0.5333
15
CF
  28.07 
1
 A    14.04
2
(a)
 D   A  45  59.0
 D  59.0 
 E   A  45  30.1
2
R  CF  FX
2
 E  30.1 
 152  82  17 ksi
(b)
 max  R  17.00 ksi
(c)
    ave  3.00 ksi
 max  17.00 ksi 
   3.00 ksi 
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1062
PROBLEM 7.35
8 ksi
5 ksi
Solve Prrob. 7.13, usinng Mohr’s circcle.
PROBL
LEM 7.13 through 7.16 For the given staate of stress, determine
d
the normal and
shearingg stresses afterr the element shown has beeen rotated thrrough (a) 25 clockwise,
(b) 10 counterclockw
c
wise.
SOLUTION
 x  0,
0
 y  8 ksi,
 xy  5 ksi
 ave 
x y
2
 4 kssi
Plottted points forr Mohr’s circlee:
X : (0,  5 ksi)
k 5 ksi)
Y : (8 ksi,
k 0)
C : (4 ksi,
FX 5
  1.25
1
FC 4
2 p  51.34
tan 2 p 
2
2
R  FC  FX
X  42  52  6.4031 ksi
(a)
  25
.
2  50
  51.34  50  1.34
 x   ave  R cos 
 xy  R sin 
 x  2.40 ksi 
 xy  0.1497 ksi 
 y   ave  R cos 
(b)
  10
.
 y  10.40 ksi 
2  20
  51.34  20  71.34
 x   ave  R cos 
 x  1.951
1
ksi 
 xy  R sin 
 xy  6.07 ksi 
 y   ave  R cos  
 y  6.05
6
ksi 
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OPRIETARY MAT
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1063
PROBLEM
M 7.36
90 MP
Pa
3 MPa
30
Solve Prob. 7.14, using Mohr’s
M
circle.
60 MPa
PROBLEM
M 7.13 througgh 7.16 For thhe given statee of stress, deetermine the
normal and shearing
s
stresses after the element
e
shownn has been rotaated through
(a) 25 clockkwise, (b) 10 counterclockkwise.
SO
OLUTION
 x  60 MP
Pa,
 y  90 MPa,,
 xy  30 MPa
 ave 
x y
2
 15 MPa
Plootted points for Mohr’s circlle:
X : (60 MPa,  30 MPa)
Y : (90 MPa, 300 MPa)
C : (15 MPa, 0)
tan 2 p 
FX 30

 0.4
0
FC 75
2 p  21.80  P  10.90
2
2
R  FC  FX
X  752  300 2  80.78 MP
Pa
(a)
  25
2  50
5
.
  2  2 P  50  21.80  288.20
 x   ave  R cos 
 x  56.2
5
MPa 
 xy  R
 sin 
 xy  38.2
3
MPa 
 y   ave  R cos 
 y  86.2
8 MPa 
PRO
OPRIETARY MATERIAL. Copyriight © 2015 McG
Graw-Hill Educattion. This is proprrietary material soolely for authorizedd instructor use.
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1064
PROB
BLEM 7.36 (Continued)
(
d)
(b)
  10
2  200
.
  2 p  2  21.880  20  41.80
 x   ave  R cos 
 x  455.2 MPa 
 xy  R sin 
 xy  533.8 MPa 
 y   ave  R cos 
 y  755.2 MPa 

PRO
OPRIETARY MAT
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1065
PR
ROBLEM 7..37
12 ksi
Solv
ve Prob. 7.15, using Mohr’ss circle.
8 ksi
6 ksi
PRO
OBLEM 7.133 through 7.16 For the giveen state of stress, determinee the normal
and shearing streesses after thee element shoown has beenn rotated through (a) 25
clocckwise, (b) 10 counterclockkwise.
SO
OLUTION
 x  8 ksi,
 y  12 ksi,
 xy  6 ksi
 ave 
x y
2
 2 ksi
k
Plootted points for Mohr’s circlle:
X : (8 ksi, 6 ksi)
Y : (12 ksi,  6 ksi))
C : (2 ksi, 0)
FX
6
 0.6

CF 100
2 p  30.96
tan 2 p 
2
2
R  CF  FX  102  62  11.66 ksi
k
(a)
  25
.
2  50
5
  50
5   30.96  19.04
 x   ave  R cos 
 x  9.02 ksi 
 xy  R sin 
 y   ave  R cos 
(b)
  10
.
 xy  3.80 ksi 
 y  13.02 ksi 
2  20
2
  30.96
3
  20  50.96
 x   ave  R cos 
 x  5.34 ksi 
 xy   R sin 
 xy  9.06 ksi 
 y   ave  R cos 
 y  9.34 ksi 
PRO
OPRIETARY MATERIAL. Copyriight © 2015 McG
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1066
80 MPa
PROB
BLEM 7.38
Solve Prob.
P
7.16, usiing Mohr’s cirrcle.
50 MPa
PROBL
LEM 7.13 thrrough 7.16 Foor the given sttate of stress, determine
d
the normal and
shearinng stresses afteer the elementt shown has been rotated thhrough (a) 25 clockwise,
(b) 10 counterclockkwise.
SOLUTION
 x  0,
M
 y  80 MPa,
M
 xy  50 MPa
 ave 
x y
2
 40 MPa
Plotted points for Moohr’s circle:
X : (0, 50 MPa)
M
MPa, 50 MPa))
Y : (80 M
M 0)
C : ( 40 MPa,
FX 50

 1.25
CF 40
2 p  51.34
tann 2 p 
2
2
R  CF  FX  402  502
 64.031 MPa
(a)
  25
.
2  50
  51.34  50  1.34
 x   ave  R cos 
(b)   10
 x  244.0 MPa 
 xy   R sinn 
 xy  1.4497 MPa 
 y   ave  R cos 
 y  1044.0 MPa 
2  20
.
  51.34  20  71.34
 x   ave  R cos 
 xy   R sinn 
 y   ave  R cos 
 x  199.51 MPa 
 xy  600.7 MPa 
 y  600.5 MPa 
PRO
OPRIETARY MAT
TERIAL. Copyrigght © 2015 McGrraw-Hill Educatioon. This is proprietary material soleely for authorizedd instructor use.
Not authorized
a
for salee or distribution inn any manner. Thiis document may not
n be copied, scannned, duplicated, forwarded, distribbuted, or posted
on a website,
w
in whole or part.
1067
250 psi
PROBLEM 7.39
Solve Prob. 7.17, using Mohr’s circle.
158
PROBLEM 7.17 The grain of a wooden member forms an angle of 15° with the
vertical. For the state of stress shown, determine (a) the in-plane shearing stress
parallel to the grain, (b) the normal stress perpendicular to the grain.
SOLUTION
x  y  0
 xy  250 psi
Plotted points for Mohr’s circle:
X  (0, 250 psi)
Y  (0, 250 psi)
C  (0, 0)
(a)
 xy  R cos 2
 (250 psi)cos30
 217 psi
 xy  217 psi 
(b)
 x   R sin 2
 (250 psi) sin 30
 125.0 psi
 x  125.0 psi 
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1068
1.8 MPa
PROBLEM 7.40
Solve Prob. 7.18, using Mohr’s circle.
3 MPa
158
PROBLEM 7.18 The grain of a wooden member forms an angle of 15° with the
vertical. For the state of stress shown, determine (a) the in-plane shearing stress
parallel to the grain, (b) the normal stress perpendicular to the grain.
SOLUTION
 y  1.8 MPa
 x  3 MPa
 ave 
x   y
2
 xy  0
 2.4 MPa
Points.
X : ( x ,  xy )  (3 MPa, 0)
Y : ( y ,  xy )  (1.8 MPa, 0)
C : ( ave , 0)  (2.4 MPa, 0)
  15
CX  0.6 MPa
2  30
R  0.6 MPa
(a)
 xy  CX  sin 30   R sin 30  0.6sin 30  0.300 MPa
(b)
 x   ave  CX  cos 30  2.4  0.6 cos 30  2.92 MPa
 xy  0.300 MPa 
 x  2.92 MPa 
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1069
PROBLEM 7.41
P'
80 mm
Solve Prob. 7.19, using Mohr’s circle.
120 mm
b
P
PROBLEM 7.19 Two wooden members of 80  120-mm uniform
rectangular cross section are joined by the simple glued scarf
splice shown. Knowing that   22 and that the maximum
allowable stresses in the joint are, respectively, 400 kPa in tension
(perpendicular to the splice) and 600 kPa in shear (parallel to the
splice), determine the largest centric load P that can be applied.
SOLUTION
x 
P
,
A
y  0
 xy  0
Plotted points for Mohr’s circle:
P 
X :  , 0,
A 
Y : (0, 0)
P 
C:  , 0
2 
R  CX 
P
2A
Coordinates of point Y′:
P
(1  cos 2 )
2A
P
 
sin 2
2A
 
Data:
A  (80)(120)  9.6  103 mm 2  9.6  103 m 2
If
  400 kPa  400  103 Pa,
P 
2 A
(2)(9.6  103 )(400  103 )

1  cos 2
(1  cos 44)
 27.4  103 N  27.4 kN
If
  600 kPa  600  103 Pa,
P
2 A
(2)(9.6  103 )(600  103 )

sin 2
(sin 44)
 16.58  103 N  16.58 kN
The smaller value of P governs.
P  16.58 kN 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1070
PROBLEM 7.42
P'
80 mm
Solve Prob. 7.20, using Mohr’s circle.
120 mm
b
P
PROBLEM 7.20 Two wooden members of 80  120-mm uniform
rectangular cross section are joined by the simple glued scarf
splice shown. Knowing that   25 and that centric loads of
magnitude P  10 kN are applied to the members as shown,
determine (a) the in-plane shearing stress parallel to the splice,
(b) the normal stress perpendicular to the splice.
SOLUTION
x 
P
A
y  0
 xy  0
Plotted points for Mohr’s circle:
P 
X :  , 0
A 
Y : (0, 0)
 P 
C:  , 0
 2A 
R  CX 
P
2A
Coordinates of point Y:
P
(1  cos 2 )
2A
P
 
sin 2
2A
 
Data:
A  (80)(120)  9.6  103 mm 2  9.6  103 m 2
(a)
 
(10  103 )sin 50
 399  103 Pa  399 kPa
(2)(9.6  103 )

(b)
 
(10  103 )(1  cos 50)
 186.0  103 Pa  186.0 kPa
(2)(9.6  103 )

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1071
PROBLEM 7.43
P
Solve Prob. 7.21, using Mohr’s circle.
a
␤
a
PROBLEM 7.21 The centric force P is applied to a short post as shown. Knowing that
the stresses on plane a-a are   15 ksi and   5 ksi, determine (a) the angle  that
plane a-a forms with the horizontal, (b) the maximum compressive stress in the post.
SOLUTION
x  0
 xy  0
y  
P
A
(a)
From the Mohr’s circle,
5
 0.3333
15
P
P
 

cos 2
2A 2A
tan  
(b)
  18.4 
P
2( )
(2)(15)


A 1  cos 2 1  cos 2
 16.67 ksi
16.67 ksi 
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1072
PROBLEM 7.44
Solve Prob. 7.22, using Mohr’s circle.
a
a
25
50 mm
PROBLEM 7.22 Two members of uniform cross section 50  80 mm are glued
together along plane a-a that forms an angle of 25 with the horizontal. Knowing
that the allowable stresses for the glued joint are   800 kPa and   600 kPa,
determine the largest centric load P that can be applied.
P
SOLUTION
x  0
 xy  0
 y  P/A
A  (50  103 )(80  103 )
 4  103 m 2
P
(1  cos50)
2A
2 A
P
1  cos 50

(2)(4  103 )(800  103 )
1  cos 50
P  3.90  103 N
P

P
2 A
(2)(4  103 )(600  103 )
sin 50 P 

 6.27  103 N
2A
sin 50
sin 50
P  3.90 kN 
Choosing the smaller value,
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1073
PROBLEM 7.45
0.2 m
0.15 m
Solve Prob. 7.23, using Mohr’s circle.
H
3 kN
PROBLEM 7.23 The axle of an automobile is acted upon by
the forces and couple shown. Knowing that the diameter of the
solid axle is 32 mm, determine (a) the principal planes and
principal stresses at point H located on top of the axle, (b) the
maximum shearing stress at the same point.
350 N · m
3 kN
SOLUTION
Torsion:
c
1
1
d  (32)  16 mm  16  103 m
2
2
 
Tc
2T

J
 c3
 
2(350 N  m)
 54.399  106 Pa  54.399 MPa
3
3
 (16  10 m)
I 
Bending:

4
c4 

4
(16  103 )4  51.472  109 m 4
M  (0.15m)(3  103 N)  450 N  m
 
My
(450)(16  103 )

 139.882  106 Pa  139.882 MPa
9
I
51.472  10
Top view
Stresses
 x  139.882 MPa,
Plotted points:
X : (139.882, 54.399);
 ave 
 y  0,
 xy  54.399 MPa
Y: (0, 54.399); C: (69.941, 0)
1
( x   y )  69.941 MPa
2
2
R
x  y 
2

   xy
2


2

 139.882 
2

  (54.399)  88.606 MPa
2


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1074
PROBLEM 7.45 (Continued)
tan 2 p 
2 xy
x  y

(2)(54.399)
139.882
 0.77778
 a  18.9 , b  108.9
(a)
(b)

 a   ave  R  69.941  88.606
 a  158.5 MPa 
 b   ave  R  69.941  88.606
 b  18.67 MPa 
 max  R 
 max  88.6 MPa 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1075
PROBLEM 7.46
6 in.
C
H
Solve Prob. 7.24, using Mohr’s circle.
B
PROBLEM 7.24 A 400-lb vertical force is applied at D to a gear
attached to the solid 1-in.-diameter shaft AB. Determine the principal
stresses and the maximum shearing stress at point H located as shown
on top of the shaft.
A
D
2 in.
400 lb
SOLUTION
Equivalent force-couple system at center of shaft in section at point H:
V  400 lb
M  (400)(6)  2400 lb  in.
T  (400)(2)  800 lb  in.
Shaft cross section:
d  1 in.
J 

2
c
1
d  0.5 in.
2
c 4  0.098175 in 4
1
J  0.049087 in 4
2
Tc
(800)(0.5)

 4.074  103 psi  4.074 ksi
J
0.098175
Torsion:
 
Bending:
 
Transverse shear:
Stress at point H is zero.
Resultant stresses:
 x  24.446 ksi,
 ave 
I 
Mc (2400)(0.5)

 24.446  103 psi  24.446 ksi
I
0.049087
 y  0,
 xy  4.074 ksi
1
( x   y )  12.223 ksi
2
2
R


x  y 
2

   xy
2


(12.223) 2  (4.074) 2  12.884 ksi
 a   ave  R
 a  25.1 ksi 
 b   ave  R
 b  0.661 ksi 
 max  R 
 max  12.88 ksi 

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1076
PROBLEM 7.47
H
Solve Prob. 7.25, using Mohr’s circle.
E
PROBLEM 7.25 A mechanic uses a crowfoot wrench to loosen a
bolt at E. Knowing that the mechanic applies a vertical 24-lb force at A,
determine the principal stresses and the maximum shearing stress at
point H located as shown as on top of the 34 -in.-diameter shaft.
6 in.
B
24 lb
10 in.
A
SOLUTION
Equivalent force-couple system at center of shaft in section at point H:
V  24 lb
M  (24)(6)  144 lb  in.
T  (24)(10)  240 lb  in.
Shaft cross section:
d  0.75 in.
J 

2
c
1
d  0.375 in.
2
c 4  0.031063 in 4
I 
1
J  0.015532 in 4
2
Tc
(240)(0.375)

 2.897  103 psi  2.897 ksi
J
0.031063
Torsion:
 
Bending:
 
Transverse shear:
At point H, stress due to transverse shear is zero.
Resultant stresses:
 x  3.477 ksi,
 ave 
Mc (144)(0.375)

 3.477  103 psi  3.477 ksi
I
0.015532
 y  0,
 xy  2.897 ksi
1
( x   y )  1.738 ksi
2
2
R
x   y 
2

   xy
2


 1.7382  2.8972  3.378 ksi

 a   ave  R
 a  5.12 ksi 
 b   ave  R
 b  1.640 ksi 
 max  R 
 max  3.38 ksi 
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1077
PROBLEM 7.48
y
6 mm
200 mm
Solve Prob. 7.26, using Mohr’s circle.
51 mm
A
A
PROBLEM 7.26 The steel pipe AB has a 102-mm outer
diameter and a 6-mm wall thickness. Knowing that arm CD is
rigidly attached to the pipe, determine the principal stresses and
the maximum shearing stress at point K.
T
D
10 kN
C
150 mm
H
K
B
z
x
SOLUTION
ro 
J 
I 
d o 102

 51 mm
2
2

r
2
4
o
ri  ro  t  45 mm

 ri4  4.1855  106 mm 4  4.1855  106 m 4
1
J  2.0927  10 6 m 4
2
Force-couple system at center of tube in the plane containing points H and K:
Fx  10  103 N
M y  (10  103 )(200  103 )  2000 N  m
M z  (10  103 )(150  103 )  1500 N  m
Torsion:
T  M y  2000 N  m
c  ro  51  103 m
 xy 
Tc
(2000)(51  103 )

 24.37 MPa
J
4.1855  106
Note that the local x-axis is taken along a negative global z direction.
Transverse shear:
Stress due to V  Fx is zero at point K.
Bending:
y 
Mz c
I

(1500)(51  103 )
 36.56 MPa
2.0927  106
Point K lies on compression side of neutral axis.
 y  36.56 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1078
PROBLEM 7.48 (Continued)
Total stresses at point K:
 x  0,
 ave 
 y  36.56 MPa,
 xy  24.37 MPa
1
( x   y )  18.28 MPa
2
2
R 
x  y 
2

   xy  30.46 MPa
2


 max   ave  R  18.28  30.46
 max  12.18 MPa 

 min   ave  R  18.28  30.46
 min  48.7 MPa 
 max  R
 max  30.5 MPa 

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1079
y
PROBLEM 7.49
20 MPa
60 MPa
Solve Prob. 7.27, using Mohr’s circle.
PROBLEM 7.27 For the state of plane stress shown, determine the largest
value of  y for which the maximum in-plane shearing stress is equal to or
less than 75 MPa.
SOLUTION
 x  60 MPa,  y  ?,  xy  20 MPa
Given:
 max  R  75 MPa
XY  2 R  150 MPa
DY  (2)( xy )  40 MPa
2
2
XD  XY  DY  1502  402  144.6 MPa
 y   x  XD  60  144.6
 y  205 MPa 
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1080
PROBLEM 7.50
8 ksi
xy
10 ksi
Solve Prob. 7.28, using Mohr’s circle.
PROBLEM 7.28 For the state of plane stress shown, determine (a) the largest
value of  xy for which the maximum in-plane shearing stress is equal to or less
than 12 ksi, (b) the corresponding principal stresses.
SOLUTION
The center of the Mohr’s circle lies at point C with coordinates


  x   y  10  8
, 0 
, 0  (1 ksi, 0).

2
2


The radius of the circle is  max (in-plane)  12 ksi.
The stress point ( x ,  xy ) lies along the line X1 X 2 of the Mohr circle diagram. The extreme points with
R  12 ksi are X 1 and X 2 .
(a)
The largest allowable value of  xy is obtained from triangle CDX.
2
2
2
DX 1  DX 2  CX 1  CD
(b)
The principal stresses are
2
 xy  122  9 2
 xy  7.94 ksi 
 a  1  12
 a  13.00 ksi 
 b  1  12
 b  11.00 ksi 
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1081
PROBLEM 7.51
2 MPa
xy
75
Solve Prob. 7.29, using Mohr’s circle.
12 MPa
PROBLEM 7.29 For the state of plane stress shown, determine (a) the value
of  xy for which the in-plane shearing stress parallel to the weld is zero,
(b) the corresponding principal stresses.
SOLUTION
Point X of Mohr’s circle must lie on X X  so that  x  12 MPa. Likewise, point Y lies on line Y Y  so
that  y  2 MPa. The coordinates of C are
2  12 , 0  (7 MPa, 0).
2
Counterclockwise rotation through 150° brings line CX to CB, where   0.
R
(a)
(b)
 xy  
x  y
2
sec 30 
12  2
sec 30  5.7735 MPa
2
x  y
tan 30
2
12  2

tan 30
2
 xy  2.89 MPa 
 a   ave  R  7  5.7735
 a  12.77 MPa 
 b   ave  R  7  5.7735
 b  1.226 MPa 
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1082
PROBLEM 7.52
15 ksi
8 ksi
Solve Prob. 7.30, using Mohr’s circle.
x
PROBLEM 7.30 Determine the range of values of  x for which the maximum
in-plane shearing stress is equal to or less than 10 ksi.
SOLUTION
For the Mohr’s circle, point Y lies at (15 ksi, 8 ksi). The radius of limiting circles is R  10 ksi.
Let C1 be the location of the leftmost limiting circle and C2 be that of the rightmost one.
C1Y  10 ksi
C2Y  10 ksi
Noting right triangles C1 DY and C2 DY ,
2
2
C1D  DY  C1Y
2
2
C1D  82  102
C1D  6 ksi
Coordinates of point C1 are (0, 15  6)  (0, 9 ksi).
Likewise, coordinates of point C2 are (0, 15  6)  (0, 21 ksi).
Coordinates of point X1: (9  6, 8)  (3 ksi, 8 ksi)
Coordinates of point X2: (21  6, 8)  (27 ksi, 8 ksi)
The point ( x ,  xy ) must lie on the line X1 X2.
3 ksi   x  27 ksi 
Thus,
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1083
PROBLEM 7.53
2 MPa
xy
75
Solve Problem 7.29, using Mohr’s circle and assuming that the weld forms
an angle of 60 with the horizontal.
12 MPa
PROBLEM 7.29 For the state of plane stress shown, determine (a) the value
of  xy for which the in-plane shearing stress parallel to the weld is zero,
(b) the corresponding principal stresses.
SOLUTION
Locate point C at  
12  2
 7 MPa with   0 .
2
Angle XCB  120
x  y
2
12  2
2
 5 MPa

R  5sec 60
 10 MPa
 xy  5 tan 60
 xy  8.66 MPa 
 a   ave  R
 7  10
 a  17.00 MPa 
 b   ave  R
 7  10
 b  3.00 MPa 
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1084
3 ksi
6 ksi
5 ksi
+
458
PROBLEM 7.54
Determine the principal planes and the principal
stresses for the state of plane stress resulting
from the superposition of the two states of stress
shown.
2 ksi
4 ksi
SOLUTION
Consider state of stress on the right. We shall express it in terms of horizontal and vertical components.
We now can add the two stress elements by superposition.
Principal planes and principal stresses:
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1085
PROBLEM 7.54 (Continued)
 ave 
x  y
2

1
(6  2)  2
2
1
(6  2)  4
2
(4)2  (3)2  5
R
tan 2 p 
3
4
2 p  36.87
 p  18.4, 108.4 
 max   ave  R  2  5
 max  7.00 ksi 
 min   ave  R  2  5
 min  3.00 ksi 
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1086
PROBLEM 7.55
100 MPa
50 MPa
+
50 MPa
308
75 MPa
Determine the principal planes and the principal
stresses for the state of plane stress resulting
from the superposition of the two states of
stress shown.
SOLUTION
Consider the state of stress on the left. We shall express it in terms of horizontal and vertical components.
 x  50 cos 30
 43.30
 y  43.30
 xy  50sin 30
 25.0
Principal axes and principal stress:
 ave 
x  y
2

R
tan 2 p 
1
(118.3  56.7)  87.5
2
1
(118.3  56.7)  30.8
2
(30.8)2  (75)2  81.08
75
30.8
 p  33.8 , and 123.8
2 p  67.67
 max   ave  R  87.5  81.08

 max  168.6 MPa 
 min   ave  R  87.5  81.08
 min  6.42 MPa 
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1087
0
0
0
PROBLEM
M 7.56
0
Determine thhe principal planes
p
and thhe principal
stresses for the
t state of pllane stress ressulting from
the superposiition of the twoo states of streess shown.
30
30
SO
OLUTION
Exppress each state of stress in terms of horizzontal and verrtical componeents.
s
of stresss,
Addding the two states
 p  0 and
a 90°

 max   0 
 min
m   0 

PRO
OPRIETARY MATERIAL. Copyriight © 2015 McG
Graw-Hill Educattion. This is proprrietary material soolely for authorizedd instructor use.
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i any manner. Thhis document may not be copied, scaanned, duplicated,, forwarded, distributed, or posted
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1088
PROBLEM 7.57
0
0
30
+
Determine the principal planes and the principal
stresses for the state of plane stress resulting from the
superposition of the two states of stress shown.
SOLUTION
Mohr’s circle for 2nd state of stress:
 x  0
 y  0
 xy    0
 x   0 sin 60  
 xy   0 cos 60 
3
0
2
 y   0 sin 60 
3
0
2
1
0
2
Resultant stresses:
3
3
0   0
2
2
1
3
 0  0  0
2
2
x  0 
 xy
 ave 
tan 2 p 
3
3
0 
0
2
2
1
( x   y )  0
2
2
R
y  0 
x  y 
2

   xy 
2


2 xy
x  y
2
2
 3 
3 
 0     0  

2 
 2

3 0
3
(2)  
2   3

 3
2 p  60
b  30
 a  60 
 a   ave  R
 a  3 0 
 b   ave  R
 b   3  0 
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1089
PROBLEM 7.58
120 MPa
xy
For the element shown, determine the range of values of  xy for which the
maximum tensile stress is equal to or less than 60 MPa.
20 MPa
SOLUTION
 x  20 MPa  y  120 MPa
1
2
 ave  ( x   y )  70 MPa
Set
 max  60 MPa   ave  R
R   max   ave  130 MPa
But
2
 x 
2
R  x
   xy
2


 xy
 x 
 R  x

2


2
2
 1302  502
 120.0 MPa
Range of  xy :
120.0 MPa   xy  120.0 MPa 
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1090
PROBLEM 7.59
120 MPa
xy
For the element shown, determine the range of values of  xy for which the
maximum in-plane shearing stress is equal to or less than 150 MPa.
20 MPa
SOLUTION
 x  20 MPa  y  120 MPa
1 (   )  50 MPa
y
2 x
Set
 max (in-plane)  R  150 MPa
2
But
x  y 
2
R 
   xy
2


 xy
x  y 
 R 

2


2
2
 1502  502
 141.4 MPa
Range of  xy :
141.4 MPa   xy  141.4 MPa 
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1091
␴y'
6 ksi
␶x'y'
PROBLEM 7.60
␴x'
␪
16 ksi
For the state of stress shown, determine the
range of values of  for which the magnitude
of the shearing stress  xy is equal to or less
than 8 ksi.
SOLUTION
 x  16 ksi,  y  0
 xy  6 ksi
 ave 
1
( x   y )  8 ksi
2
2
R

tan 2 p 
x  y 
2

   xy
2


(8)2  (6) 2  10 ksi
2 xy
x  y

(2)(6)
 0.75
16
2 p  36.870
b  18.435
 xy  8 ksi for states of stress corresponding to arcs HBK and UAV of Mohr’s circle. The angle  is
calculated from
R sin 2  8
2  53.130
sin 2 
8
 0.8
10
  26.565
 k  b    18.435  26.565  45
 k  b    18.435  26.565  8.13
u   h  90  45
v   k  90  98.13
Permissible range of  :
Also,
h    k
45    8.13 
u     v
45    98.13 
135    188.13 and 225    278.13
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1092
PROBLEM 7.61
y'
x'
90 MPa
x'y'
For the state of stress shown, determine the
range of values of  for which the normal
stress  x is equal to or less than 50 MPa.
60 MPa
SOLUTION
 x  90 MPa,  y  0
 xy  60 MPa
1
( x   y )  45 MPa
2
 ave 
2
x  y 
2

   xy
2


R
452  602  75 MPa

tan 2 p 
2 xy
x  y

(2)(60)
4

90
3
2 p  53.13
 a  26.565
 x  50 MPa for states of stress corresponding to the arc HBK of Mohr’s circle. From the circle,
R cos 2  50  45  5 MPa
cos 2 
5
 0.066667
75
2  86.177
  43.089
 h   a    26.565  43.089  16.524
2 k  2 h  360  4  32.524  360  172.355  220.169
 k  110.085
Permissible range of  :
h    k
16.5     110.1  
Also,
196.5    290.1
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1093
PROBLEM 7.62
y'
x'
90 MPa
x'y'
For the state of stress shown, determine the
range of values of  for which the normal
stress  x is equal to or less than 100 MPa.
60 MPa
SOLUTION
 x  90 MPa,  y  0
 xy  60 MPa
1
( x   y )  45 MPa
2
 ave 
2
x  y 
2

   xy
2


R
452  602  75 MPa

tan 2 p 
2 xy
x  y

(2)(60)
4

90
3
2 p  53.13
 a  26.565
 x  100 MPa for states of stress corresponding to arc HBK of Mohr’s circle. From the circle,
R cos 2  100  45  55 MPa
cos 2 
55
 0.73333
75
2  42.833
  21.417
 h   a    26.565  21.417  5.15
2 k  2 h  360  4  10.297  360  85.666  264.037
 k  132.02
Permissible range of  is
h    k
5.1    132.0
Also,

174.8    312.0 
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1094
y
PROBLEM 7.63
xy
x
For the state of stress shown, it is known that the normal and shearing stresses are
directed as shown and that  x  14 ksi,  y  9 ksi, and  min  5 ksi. Determine
(a) the orientation of the principal planes, (b) the principal stress  max, (c) the
maximum in-plane shearing stress.
SOLUTION
1
2
 x  14 ksi,  y  9 ksi,  ave  ( x   y )  11.5 ksi
 min   ave  R

R   ave   min
 11.5  5  6.5 ksi
2
x  y 
2
R 
   xy
2


2
x  y 
2
2
   6.5  2.5   6 ksi
2


 xy   R 2  
But it is given that  xy is positive, thus  xy   6 ksi.
(a)
tan 2 p 
2 xy
x  y
(2)(6)
 2.4
5
2 p  67.38

 a  33.7 
b  123.7 
(b)
 max   ave  R
 max  18.00 ksi 
(c)
 max (in-plane)  R
 max (in-plane)  6.50 ksi 
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1095
␶
PROBLEM 7.64
␴y
␴y'
Y
Y'
C
O
2␪p
2␪
␶x'y'
X'
␴
␶xy
The Mohr’s circle shown corresponds to the state of stress given
in Fig. 7.5a and b. Noting that  x  OC  (CX )cos (2 p  2) and
that  xy  (CX )sin (2 p  2 ), derive the expressions for  x and
 xy given in Eqs. (7.5) and (7.6), respectively. [Hint: Use
sin( A  B)  sin A cos B  cos A sin B and cos ( A  B)  cos A cos B 
sin A sin B.]
X
␴x
␴x'
SOLUTION
OC 
1
( x   y )
2
CX   CX
CX  cos 2 p  CX cos 2 p 
x y
2
CX  sin 2 p  CX sin 2 p   xy
 x  OC  CX  cos (2 p  2 )
 OC  CX  (cos 2 p cos 2  sin 2 p sin 2 )
 OC  CX  cos 2 p cos 2  CX  sin 2 p sin 2

x y
2

x  y
2
cos 2   xy sin 2

 xy   CX  sin (2 p  2 )  CX  (sin 2 p cos 2  cos 2 p sin 2 )
 CX  sin 2 p cos 2  CX  cos 2 p sin 2
  xy cos 2 
x  y
2
sin 2

PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1096
PROBLEM 7.65
(a) Prove that the expression  x y   x2y , where  x ,  y , and  xy are components of the stress along the
rectangular axes x and y , is independent of the orientation of these axes. Also, show that the given
expression represents the square of the tangent drawn from the origin of the coordinates to Mohr’s circle.
(b) Using the invariance property established in part a, express the shearing stress  xy in terms of  x ,  y , and
the principal stresses  max and  min .
SOLUTION
(a)
From Mohr’s circle,
 xy  R sin 2 p
 x   ave  R cos 2 p
 y   ave  R cos 2 p
 x y   x2y
2
  ave
 R 2 cos2 2 p  R 2 sin 2 2 p
2
  ave
 R 2 ; independent of  p .
Draw line OK from origin tangent to the circle at K. Triangle OCK is a right triangle.
2
2
2
2
2
2
OC  OK  CK
OK  OC  CK
2
  ave
 R2
  x y   x2y
(b)
Applying above to  x ,  y , and  xy , and to  a ,  b ,
2
2
2
 x y   xy
  a b   ab
  ave
 R2
But
 ab  0,  a   max ,  b   min
2
 x y   xy
  max min
2
 xy
  x y   max min
 xy    x y   max min

The sign cannot be determined from above equation.
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1097
PROBLEM 7.66
y
σy
For the state of plane stress shown, determine the maximum shearing
stress when (a) x  14 ksi and y  4 ksi, (b) x  21 ksi and y  14 ksi.
(Hint: Consider both in-plane and out-of-plane shearing stresses.)
12 ksi
σx
z
x
SOLUTION
(a)
1
(14  4)  9
2
1
1
( x   y )  (14  4)  5
2
2
 ave 
R
(5)2  (12)2  13
 max   ave  R  9  13  22
 min   ave  R  9  13  4
Since  max and  min have opposite signs, the maximum shearing stress is equal to the maximum inplane shearing stress.
 max  R  13.00 ksi
 max  13.00 ksi 
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1098
PROBLEM 7.66 (Continued)
(b)
1
(21  14)  17.5
2
1
1
( x   y )  (21  14)  3.5
2
2
 ave 
(3.5)2  (12)2  12.5
R
 max   ave  R  17.5  12.5  30
 min   ave  R  17.5  12.5  5
Since  max and  min have the same sign,  max is out of the plane of stress. Using Mohr’s circle through
O and A, we have
 max 
1
1
 max  (30 ksi)
2
2
 max  15.00 ksi 
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1099
PROBLEM 7.67
y
σy
For the state of plane stress shown, determine the maximum shearing
stress when (a) x  20 ksi and y  10 ksi, (b) x  12 ksi and y  5 ksi.
(Hint: Consider both in-plane and out-of-plane shearing stresses.)
12 ksi
σx
z
x
SOLUTION
(a)
 ave 
1
(20  10)  15
2
1
1
( x   y )  (20  10)  5
2
2
R
(5)2  (12)2  13
 max   ave  R  15  13  28
 min   ave  R  15  13  2
Since  max and  min have the same sign,  max is out of the plane of stress. Using Mohr’s circle through
O and A, we have
 max 
1
1
 max  (28 ksi)
2
2
 max  14.00 ksi 
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1100
PROBLEM 7.67 (Continued)
(b)
1
(12  5)  8.5
2
1
1
( x   y )  (12  5)  3.5
2
2
 ave 
R
(3.5)2  (12)2  12.5
 max   ave  R  8.5  12.5  21
 min   ave  R  8.5  12.5  4
Since  max and  min have opposite signs, the maximum shearing stress is equal to the maximum
in-plane shearing stress.
 max  R  12.50 ksi 
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1101
PROBLEM 7.68
y
σy
For the state of stress shown, determine the maximum shearing stress
when (a)  y  40 MPa, (b)  y  120 MPa. (Hint: Consider both in-plane
and out-of-plane shearing stresses.)
80 MPa
140 MPa
z
x
SOLUTION
(a)
 x  140 MPa,  y  40 MPa,  xy  80 MPa
1
2
 ave  ( x   y )  90 MPa
2
x  y 
2
2
2
R 
   xy  50  80  94.34 MPa
2


 a   ave  R  184.34 MPa (max)
 b   ave  R  4.34 MPa (min)
c  0
1
2
1
1
 max  ( max   min )  ( a   b )  94.3 MPa
2
2
 x  140 MPa,  y  120 MPa,  xy  80 MPa
 max (in-plane)  ( a   b )  R  94.34 MPa
(b)
 max  94.3 MPa 
1
2
 ave  ( x   y )  130 MPa
2
 x  y 
2
2
2
R 
   xy  10  80  80.62 MPa
2 

 a   ave  R  210.62 MPa (max)
 b   ave  R  49.38 MPa
 c  0 (min)
 max   a  210.62 MPa  min   c  0
 max (in-plane)  R  86.62 MPa
1
2
 max  105.3 MPa 
 max  ( max   min )  105.3 MPa
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1102
PROBLEM 7.69
y
σy
For the state of stress shown, determine the maximum shearing stress
when (a)  y  20 MPa, (b)  y  140 MPa. (Hint: Consider both in-plane
and out-of-plane shearing stresses.)
80 MPa
140 MPa
z
x
SOLUTION
(a)
 x  140 MPa,  y  20 MPa,  xy  80 MPa
1
2
 ave  ( x   y )  80 MPa
2
 x  y 
2
2
2
R 
   xy  60  80  100 MPa
2


 a   ave  R  80  100  180 MPa (max)
 b   ave  R  80  100  20 MPa (min)
c  0
1
2
1
 max  ( max   min )  100 MPa
2
 x  140 MPa,  y  140 MPa,  xy  80 MPa
 max (in-plane)  ( a   b )  100 MPa
(b)
 max  100.0 MPa 
1
2
 ave  ( x   y )  140 MPa
2
 x  y 
2
2
R 
   xy  0  80  80 MPa
2


 a   ave  R  220 MPa (max)
 b   ave  R  60 MPa
 c  0 (min)
1
2
1
 ( max   min )  110 MPa
2
 max (in-plane)  ( a   b )  80 MPa
 max
 max  110.0 MPa 
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1103
PROBLEM 7.70
y
For the state of stress shown, determine the maximum shearing stress when
(a) z  0, (b) z  60 MPa, (c) z  60 MPa.
100 MPa
84 MPa
σz
30 MPa
x
z
SOLUTION
The z axis is a principal axis. We determine the other two principal axes by drawing Mohr’s circle for a
rotation in the x y plane.
1
(30  100)  65
2
1
1
( x   y )  (30  100)  35
2
2
 ave 
R  (35)2  (84)2  91
 A   ave  R  65  91  156 MPa
 B   ave  R  65  91  26 MPa
(a)
 z  0. Point Z corresponding to the z axis is located at O between A and B. Therefore, the largest of
the 3 Mohr’s circles is the circle we drew through A and B. We have
 max  R  91.0 MPa 
(b)
 z   60 MPa. Point Z is located between A and B. The largest of the 3 circles is still the circle
through A and B, and we still have
 max  R  91.0 MPa 
(c)
 z   60 MPa. Point Z is now outside the circle through A and B. The largest circle is the circle
through Z and A.
 max 
1
1
( ZH )  (60  156)
2
2
 max  108.0 MPa 
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1104
PROBLEM 7.71
y
For the state of stress shown, determine the maximum shearing stress
when (a) z  0, (b) z  60 MPa, (c) z  60 MPa.
100 MPa
84 MPa
z
170 MPa
x
z
SOLUTION
1
(170  100)  135
2
1
1
( x   y )  (170  100)  35
2
2
 ave 
R
(35)2  (84)2  91
 A  135  91  226 MPa
 B  135  91  44 MPa
(a)
 z  0. Point Z corresponding to the z axis is located at O, outside the circle drawn through A and B.
The largest of the 3 Mohr’s circles is the circle through O and A. We have
 max 
(b)
1
1
1
(OA)   A  (226)
2
2
2
 max  113.0 MPa 
 z   60 MPa. Point Z is located between B and A. The largest of the 3 circles is the one drawn
through A and B.
 max  R  91.0 MPa 
(c)
 z   60 MPa. Point Z is located outside the circle drawn through A and B. The largest of the 8
Mohr’s circles is the circle through Z and A. We have
 max 
1
1
( ZA)  (60  226)
2
2
 max  143.0 MPa 
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1105
PRO
OBLEM 7.72
y
For thhe state of strress shown, determine the maximum sheearing stress
when (a)  yz  17.5 ksi, (b)  yz  8 ksi, (c)  yz  0.
τyz
12 ksi
3 ksi
x
z
SO
OLUTION

(a)
 yz  17.55 ksi  x  3 ksi
R  (6) 2  (17.5) 2  18.5
 A  6  18.5  24.5
 B  6  18.5  12.5
 max   A  24.5 ksi
 min   B  12.5 ksi
1
2
 max  ( max   min )



(b)
 max  18.50 ksi 
 yz  8 kssi  x  3 kssi
R  (6))2  (8) 2  10
 A  6  10
1  16
 B  6  10
1  4
 max   A  16 ksi
 min   B  4 ksi
1
2
 max  ( max   min )



(c)
 max  10.00 ksi 
 yz  0  x  3 ksi
 max   z  12 ksi
 min   x  3 ksi
1
2
 max  ( max   min )
 max  7.50 ksi 

PRO
OPRIETARY MATERIAL. Copyriight © 2015 McG
Graw-Hill Educattion. This is proprrietary material soolely for authorizedd instructor use.
Not authorized for salle or distribution in
i any manner. Thhis document may not be copied, scaanned, duplicated,, forwarded, distributed, or posted
on a website, in wholee or part.
1106
PROBLEM 7.73
y
For the state of stress shown, determine the maximum shearing stress
when (a)  yz  17.5 ksi, (b)  yz  8 ksi, (c)  yz  0.
τyz
12 ksi
10 ksi
x
z
SOLUTION
(a)
 yz  17.5 ksi
R
(6)2  (17.5)2  18.5
 A  6  18.5  24.5
 B  6  18.5  12.5
 max   A  24.5 ksi
 min   B  12.5 ksi
 max 
1
( max   min )
2
 max  18.50 ksi 
(b)
 yz  8 ksi
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1107
PROBLEM 7.73 (Continued)
R
(6)2  (8)2  10
 A  6  10  16
 B  6  10  4
 max   A  16 ksi
 min   x  10 ksi
 max 
1
( max   min )
2
 max  13.00 ksi 
(c)
 yz  0
 max   z  12 ksi
 min   x  10 ksi
 max 
1
( max   min )
2
 max  11.00 ksi 
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1108
PROBLEM 7.74
y
For the state of stress shown, determine the value of xy for which the
maximum shearing stress is (a) 9 ksi, (b) 12 ksi.
6 ksi
τ xy
15 ksi
z
x
SOLUTION
 y  6 ksi
 x  15 ksi
1
 ave  ( x   y )  10.5 ksi
2
x  y
u
 4.5 ksi
2
 (ksi)
(a)
For  max  9 ksi,
center of Mohr’s circle lies at point C.
Lines marked (a) show the limits on  max .
Limit on  max is  max  2 max  18 ksi .
The Mohr’s circle  a   max corresponds
to point Aa.
R   a   ave  18  10.5  7.5 ksi
R
2
u 2   xy
 xy   R 2  u 2
  7.52  4.52
 6.00 ksi
(b)
 xy  6.00 ksi 
For  max  12 ksi,
center of Mohr’s circle lies at point C.
R  12 ksi
 xy   R 2  u 2
 11.24 ksi
Checking,  a  10.5  12  22.5 ksi
 b  10.5  12  1.5 ksi
c  0
1
 max  ( max   min )  12 ksi
2
 xy  11.24 ksi 
(o.k.)
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1109
PROBLEM 7.75
y
For the state of stress shown, determine the value of xy for which the
maximum shearing stress is 80 MPa.
70 MPa
τ xy
120 MPa
z
x
SOLUTION
 x  120 MPa
 ave 
x   y
2
Assume

 y  70 MPa
1
( x   y )  95 MPa
2
120  70
 25 MPa
2
 min  0
 max  2 max  160 MPa
 a   max   ave  R
R   max   ave  160  95  65 MPa
2
x   y 
2
R 
   xy
2


2
2
2
 xy
x  y 
2
2
2
 R 
  65  25  60
2


2
 xy  60.0 MPa 
 b   a  2 R  160  130  30 MPa  0 (o.k.)
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1110
PROBLEM 7.76
y
σy
For the state of stress shown, determine two values of  y for which the
maximum shearing stress is 73 MPa.
48 MPa
50 MPa
z
x
SOLUTION
 x  50 MPa,  xy  48 MPa
Let
u 
 ave 
R 
 max  R  73 MPa,
Case (1)
(1a)
y  x
2
 y  2u   x
1
( x   y )   x  u
2
2
u 2   xy
2
u   R 2   xy
u   732  482  55 MPa
u  55 MPa  y  2u   x  60 MPa
1
( x   y )  5 MPa
2
 a   ave  R  78 MPa,  b   ave  R  68 MPa
 ave 
 a  0  max  78 MPa,  min  68 MPa,  max  73 MPa
(1b)
u  55 MPa  y  2u   x  160 MPa (reject)
1
( x   y )  105 MPa,  a   ave  R  32 MPa
2
 b   ave  R  178 MPa,  c  0,  max  0
 ave 
 min  178 MPa,  max 
1
( max   min )  89 MPa  73 MPa
2
 y  60.0 MPa 



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1111
PROBLEM 7.76 (Continued)
Assume  max  0.
Case (2)
 max 
1
( max   min )  73 MPa
2
 min  146 MPa   b
2
 b   ave  R   x  u  u 2   xy
2
u 2   xy
  x  u   b
2
u 2   xy
 ( x   b )2  2( x   b )u  u 2
2u 
2
 xy
 ( x   b )2
(48)2  (50  146)2

 72 MPa
50  146
 x  b
u  36 MPa
R 
 y  2u   x  122 MPa
2
u 2   xy
 60 MPa
 a   b  2R  146  120  26 MPa (o.k.)
 y  122.0 MPa 

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1112
PROBLEM 7.77
y
σy
For the state of stress shown, determine two values of  y for which the
maximum shearing stress is 10 ksi.
8 ksi
14 ksi
z
x
SOLUTION
 x  14 ksi,  xy  8 ksi,  max  10 ksi
Let
u 
 ave 
y  x
2
1
( x   y )   x  u
2
R 
(1a)
2
u   R 2   xy
u  6 ksi
u  6 ksi  y  2u   x  26 ksi (reject)
 ave 
1
( x   y )  20 ksi,  a   ave  R  30 ksi,  b   ave  R  10 ksi
2
 max  30 ksi,  min  0,  max 
(1b)
2
u 2   xy
 max  R  10 ksi,
Case (1)
 y  2u   x
1
( max   min )  15 ksi  7.5 ksi
2
u  6 ksi  y  2u   x  2 ksi
 ave 
1
( x   y )  8 ksi,  a   ave  R  18 ksi,  b   ave  R  2 ksi
2
 max  18 ksi,  min  2 ksi,  max 
1
( max   min )  10 ksi (o.k.)
2
 y  2.00 ksi 





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1113
PROBLEM 7.77 (Continued)
Assume  min  0.
Case (2)
 max  2 max  20 ksi =  a
2
 a   ave  R   x  u  u 2   xy
a  x  u 
2
u 2   xy
2
( a   x  u)2  u 2   xy
2
( a   x )2  2( a   x )u  u 2  u 2   xy
2u 
2
( a   x )2   xy
a   x

(20  14)2  82
 4.6667 ksi
20  14
u  2.3333 ksi  y  2u   x  9.3333 ksi
 ave 
1
( x   y )  11.6667 ksi R 
2
 a   ave  R  20 ksi

2
u 2   xy
 8.3333 ksi
 b   ave  R  3.3334 ksi
 max  20 ksi,  min  0,  max  10 ksi 
 y  9.33 ksi 

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1114
PROBLEM 7.78
y
For the state of stress shown, determine the range of values of  xz for
which the maximum shearing stress is equal to or less than 60 MPa.
σ y 100 MPa
60 MPa
x
τ xz
z
SOLUTION
 x  60 MPa,  z  0,
 y  100 MPa
For Mohr’s circle of stresses in zx plane,
1
( x   z )  30 MPa
2
  z
u  x
 30 MPa
2
 ave 
Assume
 max   y  100 MPa
 min   b   max  2 max
 100  (2)(60)  20 MPa
R   ave   b
 30  (20)  50 MPa
 a   ave  R
 30  50  80 MPa <  y
R
2
u 2   xz
 xz   R 2  u 2
  502  302   40 MPa
40.0 MPa   xz  40.0 MPa 
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1115
PROBLEM 7.79
y
For the state of stress shown, determine two values of  y for which the
maximum shearing stress is 80 MPa.
σy
90 MPa
x
z
60 MPa
SOLUTION
 x  90 MPa,  z  0,
 xz  60 MPa
Mohr’s circle of stresses in zx plane:
 ave 
1
( x   z )  45 MPa
2
R
x  y 
2

   zx 
2


2
452  602  75 MPa
 a   ave  R  120 MPa,  b   ave  R  30 MPa
Assume
 max   a  120 MPa.
 y   min   max  2 max
 y  40.0 MPa 
 120  (2)(80)
Assume
 min   b  30 MPa.
 y   max   min  2 max
 y  130.0 MPa 
 30  (2)(80)
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1116
PROBLEM 7.80*
y
σy
For the state of stress of Prob. 7.69, determine (a) the value of  y
for which the maximum shearing stress is as small as possible,
(b) the corresponding value of the shearing stress.
80 MPa
140 MPa
z
x
SOLUTION
u
Let
x  y
2
 y   x  2u
1
2
 ave  ( x   y )   x  u
2
R  u 2   xy
2
 a   ave  R   x  u  u 2   xy
2
 b   ave  R   x  u  u 2   xy
Assume  max is the in-plane shearing stress.  max  R
Then  max (in-plane) is minimum if u  0.
 y   x  2u   x  140 MPa,
 ave   x  u  140 MPa
R   xy  80 MPa
 a   ave  R  140  80  220 MPa
 b   ave  R  140  80  60 MPa
 max  220 MPa,  min  0,
1
2
 max  ( max   min )  110 MPa
Assumption is incorrect.
Assume
2
 max   a   ave  R   x  u  u 2   xy
 min  0
1
2
1
2
 max  ( max   min )   a
d a
u
 1 
0
2
2
du
u   xy
(no minimum)
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1117
PROBLEM 7.80* (Continued)
Optimum value for u occurs when  max (out-of-plane)   max (in-plane)
1
2
( a  R)  R or  a  R or  x  u  u 2   xy
2
2
2
2
( x  u )2   x2  2u x  u  u   xy
2u 
2
 x2   xy
140 2  80 2

 94.286 MPa
140
x
(a)
 y   x  2u  140  94.286
(b)
2
R  u 2   xy
  max  92.857 MPa
u  47.143 MPa
 y  45.7 MPa 
 max  92.857 MPa 
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1118
PROBLEM 7.81
σ0
100 MPa
σ0
The state of plane stress shown occurs in a machine component made of a steel with
 Y  325 MPa. Using the maximum-distortion-energy criterion, determine whether
yield will occur when (a)  0  200 MPa, (b)  0  240 MPa, (c)  0  280 MPa. If
yield does not occur, determine the corresponding factor of safety.
SOLUTION
2
 ave    0
(a)
 0  200 MPa
 x  y 
2
R 
   xy  100 MPa
2


 ave   200 MPa
 a   ave  R  100 MPa,
 b   ave  R  300 MPa
 a2   b2   a b  264.56 MPa < 325 MPa
F . S. 
(b)
 0  240 MPa
325
264.56
F . S .  1.228 
 ave  240 MPa
 a   ave  R  140 MPa,
 b   ave  R  340 MPa
 a2   b2   a b  295.97 MPa < 325 MPa
F . S. 
(c)
 0  280 MPa
(No yielding)
325
295.97
(No yielding)
F . S .  1.098 
 ave  280 MPa
 a   ave  R  180 MPa,
 a2   b2   a b  329.24 MPa > 325 MPa
 b   ave  R  380 MPa
(Yielding occurs) 
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1119
PROBLEM 7.82
σ0
100 MPa
σ0
Solve Prob. 7.81, using the maximum-shearing-stress criterion.
PROBLEM 7.81 The state of plane stress shown occurs in a machine component
made of a steel with  Y  325 MPa. Using the maximum-distortion-energy criterion,
determine whether yield will occur when (a)  0  200 MPa, (b)  0  240 MPa,
(c)  0  280 MPa. If yield does not occur, determine the corresponding factor of
safety.
SOLUTION
2
 ave   0
(a)
 0  200 MPa:
 x  y 
2
R 
   xy  100 MPa
2


 ave  200 MPa
 a   ave  R  100 MPa
 max  0,
 b   ave  R  300 MPa
 min  300 MPa
2 max   max   min  300 MPa  325 MPa
F . S. 
(b)
 0  240 MPa:
325
300
F . S .  1.083 
 ave  240 MPa
 a   ave  R  140 MPa,
 max  0,
 b   ave  R  340 MPa
 min  340 MPa
2 max   max   min  340 MPa > 325 MPa
(c)
 0  280 MPa:
(No yielding)
(Yielding occurs) 
 ave  280 MPa
 a   ave  R  180 MPa,
 max  0,
 b   ave  R  380 MPa
 min  380 MPa
2 max   max   min  380 MPa  325 MPa
(Yielding occurs) 
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1120
PROBLEM 7.83
21 ksi
τ xy
The state of plane stress shown occurs in a machine component made of a steel with
 Y  45 ksi. Using the maximum-distortion-energy criterion, determine whether
yield will occur when (a)  xy  9 ksi, (b)  xy  18 ksi, (c)  xy  20 ksi. If yield does
not occur, determine the corresponding factor of safety.
36 ksi
SOLUTION
 x  36 ksi,  y  21 ksi,  z  0
 ave 
For stresses in xy plane,
x  y
2
1
( x   y )  28.5 ksi
2
 7.5 ksi
2
(a)
 xy  9 ksi
x  y 
2

   xy 
2


R 
(7.5)2  (9) 2  11.715 ksi
 a   ave  R  40.215 ksi,  b   ave  R  16.875 ksi
 a2   b2   a  b  34.977 ksi  45 ksi
F .S . 
(No yielding)
45
39.977
F .S .  1.287 
2
(b)
 xy  18 ksi R 
x  y 
2

   xy 
2


(7.5)2  (18)2  19.5 ksi
 a   ave  R  48 ksi,  b   ave  R  9 ksi
 a2   b2   a  b  44.193 ksi  45 ksi
F .S . 
(No yielding)
45
44.193
F .S .  1.018 
2
(c)
 xy  20 ksi
R
x  y 
2

   xy 
2


(7.5) 2  (20) 2  21.36 ksi
 a   ave  R  49.86 ksi,  b   ave  R  7.14 ksi
 a2   b2   a  b  46.732 ksi  45 ksi
(Yielding occurs) 
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1121
PROBLEM 7.84
21 ksi
Solve Prob. 7.83, using the maximum-shearing-stress criterion.
τ xy
36 ksi
PROBLEM 7.83 The state of plane stress shown occurs in a machine component
made of a steel with  Y  45 ksi. Using the maximum-distortion-energy criterion,
determine whether yield will occur when (a)  xy  9 ksi, (b)  xy  18 ksi,
(c)  xy  20 ksi. If yield does not occur, determine the corresponding factor of
safety.
SOLUTION
 x  36 ksi,  y  21 ksi,  z  0
 ave 
For stress in xy plane,
1
( x   y )  28.5 ksi
2
x   y
2
 7.5 ksi
2
(a)
 xy  9 ksi
x  y 
2

   xy  11.715 ksi
2


R
 a   ave  R  40.215 ksi,  b   ave  R  16.875 ksi
 max  34.977 ksi,  min  0
2 max   max   min  40.215 ksi  45 ksi
F .S. 
(No yielding)
45
40.215
F .S .  1.119 
2
(b)
x  y 
2
   xy  19.5 ksi
2


 xy  18 ksi R  
 a   ave  R  48 ksi,  b   ave  R  9 ksi
 max  48 ksi  min  0
2 max   max   min  48 ksi  45 ksi
(Yielding occurs) 
2
(c)
 xy  20 ksi R 
x  y 
2

   xy  21.36 ksi
2


 a   ave  R  49.86 ksi  b   ave  R  7.14 ksi
 max  49.86 ksi  min  0
2 max   max   min  49.86 ksi  45 ksi
(Yielding occurs) 
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1122
PROBLEM 7.85
The 38-mm-diameter shaft AB is made of a grade of steel for which the yield
strength is  Y  250 MPa. Using the maximum-shearing-stress criterion,
determine the magnitude of the torque T for which yield occurs when
P  240 kN.
B
T
P
A
d = 38 mm
SOLUTION
P  240  103 N
A

d2 
4

4
(38) 2  1.13411  103 mm 2  1.13411  10 3 m 2
P
240  103

 211.62  106 Pa  211.62 MPa
A 1.13411  103
y  0
x 
 ave 
1
1
( x   y )   x
2
2
2
x  y 
2

   xy 
2


R
1 2
2
 x   xy
4
2
2 max  2R   x2  4 xy
 y
2
4 xy
  Y2   x2
 xy 
1
1
250 2  211.62 2
 Y2   x2 
2
2
 66.553 MPa  66.553  106 Pa
From torsion:
 xy 
J 
c 
Tc
J

2
c4 
T 
  38 
J xy
c
4
9 4
3
4
   204.71  10 mm  204.71  10 m
2 2 
1
d  19  10 3 m
2
(204.71  109 )(66.553  106 )
19  103
 717 N  m
T 

PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1123
PROBLEM 7.86
B
T
P
A
d = 38 mm
Solve Prob. 7.85, using the maximum-distortion-energy criterion.
PROBLEM 7.85 The 38-mm-diameter shaft AB is made of a grade of steel
for which the yield strength is  Y  250 MPa. Using the maximum-shearingstress criterion, determine the magnitude of the torque T for which yield
occurs when P  240 kN.
SOLUTION
P  240  103 N
A

4
d2 

4
(38) 2  1.13411  103 mm 2  1.13411  103 m 2
P
240  103

 211.62  106 Pa  211.62 MPa
A 1.13411  103
y  0
x 
 ave 
1
1
( x   y )   x
2
2
2
R
x  y 
2

   xy 
2


1 2
2
 x   xy
4
 a   ave  R 
1
x 
2
1 2
2
 x   xy
4
 b   ave  R 
1
x 
2
1 2
2
 x   xy
4
 a2   b2   a b 
1 2
1
1
2
2
 x   x  x2   xy
  x2   xy
4
4
4
1
1 2
1
1
1
2
2
2
  x2   x
 x   xy
  x2   xy
  x2   x2   xy
4
4
4
4
4
2
  x2  3 xy
  Y2
 xy


1 2
 Y   x2
3
1
2502  211.622  76.848 MPa  76.848  106 Pa

3
2
 xy

PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1124
PROBLEM 7.86 (Continued)
From torsion,
 xy 
J 
c 
Tc
J

2
c4 
T 
  38 
J xy
c
4
9 4
3
4
   204.71  10 mm  204.71  10 m
2 2 
1
d  19  10 3 m
2
(204.71  109 )(76.848  106 )
19  103
 828 N  m
T 

PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1125
PROBLEM 7.87
P
T
A
1.5 in.
The 1.5-in.-diameter shaft AB is made of a grade of steel with a 42-ksi tensile yield
stress. Using the maximum-shearing-stress criterion, determine the magnitude of the
torque T for which yield occurs when P  60 kips.
B
SOLUTION
P  60 kips
A

4
d2 

(1.5) 2  1.76715 in 2
4
P
60
x    
 33.953 ksi
1.76715
A
y  0
 ave 
1
1
( x   y )   x
2
2
2
x  y 
2

   xy 
2


R
1 2
2
 x   xy
4
2
2 max  2R   x2  4 xy
 Y
2
4 xy
  Y2   x2
 xy 
1
1
422  33.9532
 Y2   x2 
2
2
 12.3612 ksi
From torsion,
J xy
Tc
T 
c
J
1
c  d  0.75 in.
2
 xy 
J 

2
c 4  0.49701 in 4
(0.49701)(12.3612)
0.75
 8.19 kip  in.
T 

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Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1126
PROBLEM 7.88
P
T
A
Solve Prob. 7.87, using the maximum-distortion-energy criterion.
PROBLEM 7.87 The 1.5-in.-diameter shaft AB is made of a grade of steel with a
42-ksi tensile yield stress. Using the maximum-shearing-stress criterion, determine
the magnitude of the torque T for which yield occurs when P  60 kips.
1.5 in.
B
SOLUTION
P  60 kips
 2 
A
(1.5) 2  1.76715 in 2
4
p
60
x    
 33.953 ksi
1.76715
A
y  0
 ave 
4
d 
1
1
( x   y )   x
2
2
2
x  y 
2

   xy 
2


 a   ave  R
 b   ave  R
R
1 2
2
 x   xy
4
 a2   b2   a b  ( ave  R) 2  ( ave  R)2  ( ave  R)( ave  R)
2
2
2
  ave
 2 ave R  R 2   ave
 2 ave R  R 2   ave
 R2
2
  ave
 3R 2
1 2
1
2 
2
2
2
 x  3   x2   xy
   x  3 xy   Y
4
4

  Y2   x2

2
3 xy
 xy 


1
1
422  33.9532
 Y2   x2 
3
3
 14.2734 ksi
From torsion,
J xy
Tc
T 
c
J
1
c  d  0.75 in.
2
 xy 

c4 

(0.75) 4  0.49701 in 4
2
2
(0.49701)(14.2734)
T 
 9.46 kip  in.
0.75
J 

PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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on a website, in whole or part.
1127
PRO
OBLEM 7.89
100 MPa
The state
s
of plane stress shownn is expected to occur in an
a aluminum
8 MPa and
castinng. Knowing that for the aluuminum alloy used  UT  80
d
wheether rupture
 UC  200 MPa annd using Mohrr’s criterion, determine
of thee casting will occur.
o
60 MPa
M
10 MPa
SO
OLUTION
 x  10 MPa,
M
 y  1000 MPa,
 xy  60 MPa
 ave 
x y
2

10  1000
 45 MPaa
2
2
x y 
2
2
2
R 
   xy  (55)  (60)  81.399 MPa
2


 a   avee  R  45  81.39  36.39 MPa
 b   avee  R  45  81.39
8
 126.39 MPa
Equuation of 4th quadrant
q
of bo
oundary:
a

 b 1
 UT  UC
36.39 (1226.39)
 1.087  1

80
2000
Rupture will
w occur. 
PRO
OPRIETARY MATERIAL. Copyriight © 2015 McG
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i any manner. Thhis document may not be copied, scaanned, duplicated,, forwarded, distributed, or posted
on a website, in wholee or part.
1128
75 MPa
32 MPa
PROBLEM
M 7.90
The state of plane stress shown
s
is expeected to occur in an aluminuum casting.
Knowing thatt for the alumiinum alloy useed  UT  80 MPa
M and  UC  200 MPa
and using Mohr’s
M
criterioon, determinee whether ruppture of the casting
c
will
occur.
SOLUTION
 x  32 MPa,
M
 y  0,
M
 xy  75 MPa
1
2
 ave  ( x   y )  16 MPa
M
2
x  y 
2
2
2
R 
M
   xy  (16)  (775)  76.69 MPa
2


 a   ave  R  16  766.69  60.69 MPa
M
 b   ave  R  16  766.69  92.69 MPa
Equuation of 4th quadrant
q
of bouundary:
a

 b 1
 UT  UC
60.69 (92..69)

 1.222  1
80
200
Rupture will
w occur. 
PRO
OPRIETARY MAT
TERIAL. Copyrigght © 2015 McGrraw-Hill Educatioon. This is proprietary material soleely for authorizedd instructor use.
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a
for salee or distribution inn any manner. Thiis document may not
n be copied, scannned, duplicated, forwarded, distribbuted, or posted
on a website,
w
in whole or part.
1129
PROBLEM
M 7.91
7 ksi
The state off plane stress shown
s
is expeected to occurr in an aluminnum casting.
1 ksi and  UC  30 ksi
Knowing thaat for the alum
minum alloy used  UT  10
and using Mohr’s
M
criterionn, determine whether
w
rupturee of the castingg will occur.
8 ksi
SO
OLUTION
 x  8 ksi,
 y  0,
 xy  7 ksi
1
2
 ave  ( x   y )  4 ksi
2
x  y 
2
2
2
R 
   xy  4  7  8.062 ksi
2


 a   ave
2 ksi
a  R  4  8.062  4.062
 b   ave
062 ksi
a  R  4  8.062  12.0
Equuation of 4th quadrant
q
of bo
oundary:
a

 b 1
 UT  UC
4.062 (122.062)

 0.8088  1
10
330
No rupture. 
PRO
OPRIETARY MATERIAL. Copyriight © 2015 McG
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i any manner. Thhis document may not be copied, scaanned, duplicated,, forwarded, distributed, or posted
on a website, in wholee or part.
1130
PROBLEM
M 7.92
15 ksi
k
The state of plane
p
stress shhown is expeccted to occur in an aluminuum casting.
Knowing thatt for the alum
minum alloy used
u
 UT  10 ksi and  UC
U  30 ksi
and using Mohr’s criterion, determine whhether rupture of the casting will occur.
9 ksi
2 ksi
SOLUTION
 x  2 ksi,
 y  15 ksi,
 xy  9 ksi
1
2
 ave
( x   y )  6.5 ksii
a 
2
x  y 
2
2
2
R 
1
ksi
   xy  8.5  9  12.379
2


 a   ave  R  5.879 ksi
 b   ave  R  18.879 ksii
q
of bouundary:
Equuation of 4th quadrant
a

 b 1
 UT  UC
5.879 (18.879)

 1.217  1
10
30
Rupture will occur. 
PRO
OPRIETARY MAT
TERIAL. Copyrigght © 2015 McGrraw-Hill Educatioon. This is proprietary material soleely for authorizedd instructor use.
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a
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n be copied, scannned, duplicated, forwarded, distribbuted, or posted
on a website,
w
in whole or part.
1131
PROBLEM
M 7.93
8 ksi
t0
The state of plane stress shown
s
will occcur at a critical point in an
a aluminum
casting that is made of ann alloy for whhich  UT  10 ksi and  UC
U  25 ksi.
Using Mohrr’s criterion, determine
d
thee shearing stress  0 for which
w
failure
should be exppected.
SO
OLUTION
 x  8 ksi,
 y  0,
 xyy   0
1
2
 avee  ( x   y )  4 ksi
2
x  y 
2
2
2
R 
   xy  4   0
2


 0   R 2  42
 a   ave  R  (4  R) ksi
 b   ave  R  (4  R) ksi
Sinnce  ave < R, stress point lies in 4th quaddrant. Equatioon of 4th quaddrant boundaryy is
a

 b 1
 UT  UC
4 R 4R

1
10
25
1 
4
4
 1
 10  25  R  1  10  25


R  5.429 ksi
 0   5.42992  42
 0  3.67 ksi 
PRO
OPRIETARY MATERIAL. Copyriight © 2015 McG
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on a website, in wholee or part.
1132
PROBLEM
M 7.94
80 MPa
␶0
The state off plane stress shown
s
will occcur at a criticaal point in a piipe made of
an aluminum
m alloy for which
w
Using
 UT  75 MPa and  UC  150 MPa.
M
Mohr’s criteerion, determinne the shearinng stress  0 foor which failurre should be
expected.
SOLUTION
 x  80 MPa,
 y  0,
 xy   0
1
2
 ave  ( x   y )  40 MPa
2
x  y 
2
2
2
R 
   xy  40   0 MPa
2


 a   ave  R
 b   ave  R
 0   R 2  402
Sincce  ave < R, stress point lies in 4th quaddrant. Equationn of 4th quadrrant boundary is
a

 b 1
 UT  UC
40  R 40  R

1
75
150
R
R
40 40


1
 1.2667
75 150
75 150
R  63.33 MP
Pa,
 0   63.332  402
 0  8.49 MPa 
PRO
OPRIETARY MAT
TERIAL. Copyrigght © 2015 McGrraw-Hill Educatioon. This is proprietary material soleely for authorizedd instructor use.
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a
for salee or distribution inn any manner. Thiis document may not
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w
in whole or part.
1133
PR
ROBLEM 7.95
7
T'
t0
T
Thee cast-aluminnum rod shoown is made of an alloyy for which
a  UC  1775 MPa. Know
wing that the magnitude
m
T
 UTT  70 MPa and
of the
t applied torrques is slowlly increased annd using Mohr’s criterion,
dettermine the shearing stress  0 that shouldd be expected at
a rupture.
SO
OLUTION
x  0
y  0
 xy   0
1
2
 ave  ( x   y )  0
2
x  y 
2
2
R 
   xy  0   xyy   xy
2


 a   ave
a RR
 b   ave
a  R  R
Sinnce  ave < R, stress point lies in 4th quaddrant. Equatioon of boundaryy of 4th quadrrant is
a

 b 1
 UT  UC
R R

1
700 175
1 
 1
R 1
 70  175
1


R  50 MPa
M
0  R
 0  50.0
5
MPa 
PRO
OPRIETARY MATERIAL. Copyriight © 2015 McG
Graw-Hill Educattion. This is proprrietary material soolely for authorizedd instructor use.
Not authorized for salle or distribution in
i any manner. Thhis document may not be copied, scaanned, duplicated,, forwarded, distributed, or posted
on a website, in wholee or part.
1134
P
PROBLEM
7.96
T cast-alum
The
minum rod shhown is madee of an alloyy for which
U
Mohr’s criterion,
 UT  60 MPa and  UC  120 MPa. Using
d
determine
the magnitude off the torque T for which faiilure should
b expected.
be
32 mm
B
T
A
26 kN
SOLUTION
P  26  103 N
x 
A

4
(32) 2  804.25 mm 2  804.25  106 m 2
P
26  1003

 32.3288  106 Pa  322.328 MPa
6
A 804.25  10
1
1
2
2
x  y 1
 (32.328  0))  16.164 MP
Pa
2
2
 ave  ( x   y )  (32.328  0)
0  16.164 MP
Pa
 a   ave  R  16..164  R MPa
 b   ave  R  16.164  R MPa
Sincce  ave < R, stress point lies in the 4th quadrant.
q
Equaation of the 4thh quadrant is
a

16.1644  R 16.1644  R
 b 1

1
 UT  UCC
1200
600
1 
16.1664 16.164
 1
 60  120  R  1  60  120


x  y  2
R 
   xy
2


R  34
4.612 MPa
2
 xy
x  y 
2
2
 R 
1
 30.6606 MPa
  34.612  16.164
2


2
6  106 Pa
 30.606
For torsion,
 xy 
T
Tc 2T

J  c3

2
c3 xy
x 
wherre c 

2
1
d  166 mm  16  1003 m
2
(16  1003 )3 (30.606  106 )
T  196
6.9 N  m 
PRO
OPRIETARY MAT
TERIAL. Copyrigght © 2015 McGrraw-Hill Educatioon. This is proprietary material soleely for authorizedd instructor use.
Not authorized
a
for salee or distribution inn any manner. Thiis document may not
n be copied, scannned, duplicated, forwarded, distribbuted, or posted
on a website,
w
in whole or part.
1135
1␴
2 0
1␴
2 0
␴0
(a)
PROBLEM
M 7.97
1␴
2 0
␴0
(b)
␴0
A machine component
c
is made of a grade
g
of cast
k and  UC  20 ksi. For
iron for whicch  UT  8 ksi
each of the states
s
of stress shown, and using
u
Mohr’s
criterion, dettermine the normal stress  0 at which
rupture of thee component should
s
be expeected.
(c)
SO
OLUTION
(a)
a  0
1
2
b  0
Stress poinnt lies in 1st quadrant.
q
 a   0   UT
(b)
 0  8.00 ksi 
a  0
1
2
b   0
Stress poinnt lies in 4th quadrant.
q
Equaation of 4th quuadrant bounddary is
a

 b 1
 UT  UCC
0
8
(c)

 12  0
20
1
 0  6.67 ksi 
1
2
 a   0 ,  b   0 , 4th quadrannt
1

2 0
8

 0
1
20
 0  8.89 ksi 
PRO
OPRIETARY MATERIAL. Copyriight © 2015 McG
Graw-Hill Educattion. This is proprrietary material soolely for authorizedd instructor use.
Not authorized for salle or distribution in
i any manner. Thhis document may not be copied, scaanned, duplicated,, forwarded, distributed, or posted
on a website, in wholee or part.
1136
PROBLEM 7.98
A spherical pressure vessel has an outer diameter of 3 m and a wall thickness of 12 mm. Knowing that for
the steel used all  80 MPa, E  200 GPa, and   0.29, determine (a) the allowable gage pressure, (b) the
corresponding increase in the diameter of the vessel.
SOLUTION
r 
1
1
d  t  (3)  12  103  1.488 m
2
2
1   2   all  80  106 Pa
(a)
1   2 
p
pr
2t
2t1
(2)(12  103 )(80  106 )

r
1.488
p  1.290 MPa 
p  1.290  106 Pa
(b)
1
( 1   2 )
E
1
1  0.29
(80  106 )  284  106

1 
9
E
200  10
1 
 d  d 1  (3)(284  106 )  852  106 m
 d  0.852 mm 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1137
PROBLEM 7.99
A spherical gas container having an inner diameter of 5 m and a wall thickness of 24 mm is made of steel
for which E  200 GPa and   0.29. Knowing that the gage pressure in the container is increased from zero
to 1.8 MPa, determine (a) the maximum normal stress in the container, (b) the corresponding increase in the
diameter of the container.
SOLUTION
p  1.8 MPa
r 
(a)
1
1
d  t  (5)  24  10 3  2.476 m
2
2
1   2 
pr
(1.8)(2.476)

 92.850 MPa
2t
(2)(24  103 )
  92.9 MPa 
1 
(b)
1
1
1  0.29
( 1   2 ) 
(92.85  106 )  329.6 
1 
E
E
200  109
 d  d 1  (5)(329.6  106 )  1.648  103 m
 d  1.648 mm 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1138
PROBLEM 7.100
The maximum gage pressure is known to be 1150 psi in a spherical steel pressure vessel having a 10-in. outer
diameter and a 0.25-in. wall thickness. Knowing that the ultimate stress in the steel used is U  60 ksi,
determine the factor of safety with respect to tensile failure.
SOLUTION
d
t
2
10 in.

 0.25 in.
2
 4.75 in.
r 
pr
2t
(1150 psi)(4.75 in.)

2(0.25 in.)
1   2 
 10.925 ksi
F.S. 
60 ksi
U

 max 10.925 ksi
F.S.  5.49 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
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1139
PROBLEM 7.101
A spherical pressure vessel of 750-mm outer diameter is to be fabricated from a steel having an ultimate stress
U  400 MPa. Knowing that a factor of safety of 4.0 is desired and that the gage pressure can reach 4.2 MPa,
determine the smallest wall thickness that should be used.
SOLUTION
1
d t
2
1
 (0.750 m)  t
2
 0.375t (m)
r 
We have
and
 max  1   2 
F.S. 
pr
2t
U
 max
Combining these two equations gives
F.S. 
or
2t U
pr
2 U t  (F.S.) pr
Substituting for r gives
2(400  106 Pa)t  (4)(4.2  106 Pa)(0.375  t )
816.80  106 t  6.30  106
t  7.71  103 m
t  7.71 mm 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
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1140
PROBLEM 7.102
A spherical gas container made of steel has a 20-ft outer diameter and a wall thickness of 167 in. Knowing that
the internal pressure is 75 psi, determine the maximum normal stress and the maximum shearing stress in the
container.
SOLUTION
d  20 ft  240 in.
7
in.  0.4375 in.
16
1
r  d  t  119.56 in.
2
(75)(119.56)
pr
 

 10.25  103 psi
2t
(2)(0.4375)
t 
  10.25 ksi 
 max  10.25 ksi
 min  0
 max 
(Neglecting small radial stress)
1
( max   min )
2
 max  5.12 ksi 
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1141
PROBLEM 7.103
A basketball has a 300-mm outer diameter and a 3-mm wall thickness. Determine the normal stress in the wall
when the basketball is inflated to a 120-kPa gage pressure.
SOLUTION
1
d t
2
1
 (300 mm)  3
2
r 
 147 mm
1   2 

or
147  103 m
pr
2t
(120  103 Pa)(147  103 m)
2(3  103 m)
 2.9400  106 Pa
  2.94 MPa 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1142
PROBLEM 7.104
8m
14.5 m
h
The unpressurized cylindrical storage tank shown has a 5-mm wall thickness
and is made of steel having a 400-MPa ultimate strength in tension.
Determine the maximum height h to which it can be filled with water if a
factor of safety of 4.0 is desired. (Density of water  1000 kg/m3.)
SOLUTION
d0  8 m
t  5 mm  0.005 m
1
d  t  4  0.005  3.995 m
2

400 MPa
 U 
 100 MPa  100  106 Pa
F.S.
4.0
pr

t
r 
 all
 all
p 
but
t all
(0.005 m)(100  106 Pa)

 125.156  103 Pa
r
3.995 m
p   gh,
h
125.156  103 Pa
p

 12.7580 m
 g (1000 kg/m3 )(9.81 m/s2 )
h  12.76 m 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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on a website, in whole or part.
1143
PROBLEM 7.105
8m
14.5 m
For the storage tank of Prob. 7.104, determine the maximum normal stress
and the maximum shearing stress in the cylindrical wall when the tank is
filled to capacity (h  14.5 m).
h
PROBLEM 7.104 The unpressurized cylindrical storage tank shown has a
5-mm wall thickness and is made of steel having a 400-MPa ultimate strength
in tension. Determine the maximum height h to which it can be filled with
water if a factor of safety of 4.0 is desired. (Density of water  1000 kg/m3.)
SOLUTION
d0  8 m
t  5 mm  0.005 m
1
r  d  t  4  0.005  3.995 m
2
p   gh  (1000 kg/m3 )(9.81 m/s 2 )(14.5 m)
 142.245  103 Pa
1 
(142.245  103 Pa)(3.995 m)
pr

0.005 m
t
 113.654  106 Pa
 max  1
 max  113.7 MPa 
 min  0
 max 
1
( max   min )
2
 max  56.8 MPa 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1144
PROBLEM 7.106
The bulk storage tank shown in Photo 7.3 has an outer diameter of 3.3 m and a wall thickness of 18 mm. At a
time when the internal pressure of the tank is 1.5 MPa, determine the maximum normal stress and the
maximum shearing stress in the tank.
SOLUTION
r 
1 
d
3.3
t 
 18  103  1.632 m,
2
2
t  18  103 m
pr
(1.5  106 Pa)(1.632 m)

 136  106 Pa
t
18  103 m
 max  1  136  106 Pa
 max  136.0 MPa 
 min   p  0
 max 
1
( max   min )  68  106 Pa
2
 max  68.0 MPa 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1145
PROBLEM 7.107
A standard-weight steel pipe of 12-in. nominal diameter carries water under a pressure of 400 psi.
(a) Knowing that the outside diameter is 12.75 in. and the wall thickness is 0.375 in., determine the maximum
tensile stress in the pipe. (b) Solve part a, assuming an extra-strong pipe is used of 12.75-in. outside diameter
and 0.5-in. wall thickness.
SOLUTION
(a)
d 0  12.75 in. t  0.375 in. r 

(b)
pr (400)(6.00)

 6400 psi
t
0.375
d 0  12.75 in. t  0.500 in. r 

1
d 0  t  6.00 in.
2
  6.40 ksi 
1
d 0  t  5.875 in.
2
pr (400)(5.875)

 4700 psi
t
0.500
  4.70 ksi 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
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1146
PROBLEM 7.108
A cylindrical storage tank contains liquefied propane under a pressure of 1.5 MPa at a temperature of 38C.
Knowing that the tank has an outer diameter of 320 mm and a wall thickness of 3 mm, determine the
maximum normal stress and the maximum shearing stress in the tank.
SOLUTION
r 
d
320
t 
 3  157 mm  157  103 m
2
2
t  3  103 m
1 
pr
(1.5  106 Pa)(157  103 m)

 78.5  106 Pa
3
t
3  10 m
 max  1  78.5  106 Pa
 max  78.5 MPa 
 min   p  0
 max 
1
( max   min )  39.25  106 Pa
2
 max  39.3 MPa 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1147
PROBLEM 7.109
Determine the largest internal pressure that can be applied to a cylindrical tank of 5.5-ft outer diameter and
5
-in. wall thickness if the ultimate normal stress of the steel used is 65 ksi and a factor of safety of 5.0
8
is desired.
SOLUTION
U
65 ksi
 13 ksi  13  103 psi
F .S.
5.0
d
(5.5)(12)
r  t 
 0.625  32.375 in.
2
2
1 
1 
pr
t

p
t1 (0.625)(13  103 )

r
32.375
p  251 psi 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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on a website, in whole or part.
1148
PROBLEM 7.110
A
A steel penstock has a 36-in. outer diameter, a 0.5-in. wall
thickness, and connects a reservoir at A with a generating station
at B. Knowing that the specific weight of water is 62.4 lb/ft3,
determine the maximum normal stress and the maximum shearing
stress in the penstock under static conditions.
500 ft
B
36 in.
SOLUTION
r 
1
1
d  t  (36)  0.5  17.5 in.
2
2
p  rh  (62.4 lb/ft 3 )(500 ft)  31.2  103 lb/ft 2
 216.67 psi
1 
pr
(216.67)(17.5)

 7583 psi
t
0.5
 max  1  7583 psi
 max  7.58 ksi 
 min   p  217 psi
 max 
1
( max   min )  3900 psi
2
 max  3.90 ksi 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1149
PROBLEM 7.111
A
A steel penstock has a 36-in. outer diameter and connects a reservoir
at A with a generating station at B. Knowing that the specific weight
of water is 62.4 lb/ft3 and that the allowable normal stress in the
steel is 12.5 ksi, determine the smallest thickness that can be used
for the penstock.
500 ft
B
36 in.
SOLUTION
p   h  (62.4 lb/ft 3 )(500 ft)  31.2  103 lb/ft 2
 216.67 psi
1  12.5 ksi  12.5  103 psi
1
d  t  18  t
2
pr
r 1
1 
,

t
t
p
r 
18  t 12.5  103

 57.692
216.67
t
18
 58.692
t
t  0.307 in. 
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1150
PROBLEM 7.112
600 mm
b
The cylindrical portion of the compressed-air tank shown is fabricated of 8-mm-thick
plate welded along a helix forming an angle   30° with the horizontal. Knowing
that the allowable stress normal to the weld is 75 MPa, determine the largest gage
pressure that can be used in the tank.
1.8 m
SOLUTION
1
1
d  t  (600)  6  292 mm
2
2
pr
1 
t
1 pr
2 
2 t
1
3 pr
 ave  (1   2 ) 
2
4 t
   2 1 pr
R 1

2
4 t
 w   ave  R cos 60
r 
5 pr
8 t
8  wt
p
5 r

p
8 (75)(8)
 3.29 MPa 
5 292
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1151
PROBLEM 7.113
600 mm
For the compressed-air tank of Prob. 7.112, determine the gage pressure that will
cause a shearing stress parallel to the weld of 30 MPa.
b
1.8 m
PROBLEM 7.112 The cylindrical portion of the compressed-air tank shown is
fabricated of 8-mm-thick plate welded along a helix forming an angle   30° with
the horizontal. Knowing that the allowable stress normal to the weld is 75 MPa,
determine the largest gage pressure that can be used in the tank.
SOLUTION
r 
1 
2 
R
w 
1
1
d  t  (600)  8  292 mm
2
2
pr
t
1 pr
2 t
1   2 1 pr

2
4 t
R sin 60
3 pr
8 t
8 w t
p 
3 R

p
8 (30)(8)
 3.80 MPa 
3 292
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1152
PROBLEM 7.114
The steel pressure tank shown has a 750-mm inner diameter and a 9-mm
wall thickness. Knowing that the butt-welded seams form an angle   50
with the longitudinal axis of the tank and that the gage pressure in the tank is
1.5 MPa, determine (a) the normal stress perpendicular to the weld, (b) the
shearing stress parallel to the weld.
SOLUTION
r 
d
 375 mm  0.375 m
2
1 
pr
(1.5  106 Pa  0.375 m)

 62.5  106 Pa  62.5 MPa
t
0.009 m
2 
1
1  31.25 MPa
2
2  100
1
( 1   2 )  46.875 MPa
2
  2
R 1
 15.625 MPa
2
 ave 
(a)
 w   ave  R cos100
 w  44.2 MPa 
(b)
 w  R sin100
 w  15.39 MPa 
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1153
PROBLEM 7.115
The pressurized tank shown was fabricated by welding strips of plate along
a helix forming an angle  with a transverse plane. Determine the largest
value of  that can be used if the normal stress perpendicular to the weld is
not to be larger than 85 percent of the maximum stress in the tank.
SOLUTION
1 
pr
t
2 
pr
2t
1
2
3 pr
4 t
 1   2 1 pr
R

2
4 t
 w   ave  R cos 2 
 ave  ( 1   2 ) 
0.85
pr  3 1
 pr
   cos 2  
t
4 4
 t
3

cos 2  4  0.85    0.4
4

2  113.6
  56.8 
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1154
12 ft
PROBLEM 7.116
12 ft
45
20 ft
(a)
Square plates, each of 0.5-in. thickness, can be bent and
welded together in either of the two ways shown to form the
cylindrical portion of a compressed-air tank. Knowing that the
allowable normal stress perpendicular to the weld is 12 ksi,
determine the largest allowable gage pressure in each case.
(b)
SOLUTION
1
d  t  71.5 in.
2
pr
2 
2t
d  12ft  144 in. r 
1 
(a)
pr
t
1  12 ksi
p
1t (12)(0.5)

 0.0839 ksi
71.5
r
p  83.9 psi 
(b)
1
3 pr
( 1   2 ) 
2
4 t
1   2 1 pr

R
2
4 t
   45
 ave 
 w   ave  R cos 

p
3 pr
4 t
4  wt  4  (12)(0.5)
 
 0.1119 ksi
3 r
 3  71.5
p  111.9 psi 
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1155
PROBLEM 7.117
3m
1.6 m
The pressure tank shown has a 0.375-in. wall thickness and butt-welded
seams forming an angle   20° with a transverse plane. For a gage
pressure of 85 psi, determine (a) the normal stress perpendicular to the
weld, (b) the shearing stress parallel to the weld.
SOLUTION
d  5 ft  60 in.
1
d  t  30  0.375  29.625 in.
2
(85)(29.625)
pr
1 

 6715 psi
0.375
t
1
 2  1  3357.5 psi
2
1
 ave  ( 1   2 )  5036.2 psi
2
1   2
R
 1678.75 psi
2
r 
(a)
 w   ave  R cos 40  3750 psi

(b)
 w  R sin 40  1079 psi

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1156
PROBLEM 7.118
3m
1.6 m
For the tank of Prob. 7.117, determine the largest allowable gage pressure,
knowing that the allowable normal stress perpendicular to the weld is
18 ksi and the allowable shearing stress parallel to the weld is 10 ksi.
PROBLEM 7.117 The pressure tank shown has a 0.375-in. wall thickness
and butt-welded seams forming an angle   20° with a transverse plane.
For a gage pressure of 85 psi, determine (a) the normal stress
perpendicular to the weld, (b) the shearing stress parallel to the weld.
SOLUTION
d  5 ft  60 in.
r 
1 
2 
 ave 
R
1
d  t  30  0.375  29.625 in.
2
pr
t
pr
2t
1
3 pr
( 1   2 ) 
2
4 t
1   2 1 pr

2
4 t
 w   ave  R cos 50
3 1
 pr
   cos 50 
4 4
 t
 0.58930
p
w t
0.5893r

pr
t
(18)(0.375)
 0.38664 ksi  387 psi
(0.58930)(29.625)
pr
t
w t
(10)(0.375)

 0.66097 ksi  661 psi
p
0.191511r
(0.191511)(29.625)
 w  R sin 50  0.191511
p  387 psi 
Allowable gage pressure is the smaller value.
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1157
PROBLEM 7.119
3m
1.6 m
For the tank of Prob. 7.117, determine the range of values of  that can be
used if the shearing stress parallel to the weld is not to exceed 1350 psi
when the gage pressure is 85 psi.
PROBLEM 7.117 The pressure tank shown has a 0.375-in. wall thickness
and butt-welded seams forming an angle   20° with a transverse plane.
For a gage pressure of 85 psi, determine (a) the normal stress perpendicular
to the weld, (b) the shearing stress parallel to the weld.
SOLUTION
d  5 ft  60 in.
1
3
d  t  30   29.625 in.
2
8
(85)(29.625)
pr
1 

 6715 psi
0.375
t
1
 2  1  3357.5 psi
2
  2
R 1
 1678.75
2
r 
 w  R sin 2   all
sin 2 a 
2 a  53.53
 a  26.8
2b  53.53
b  26.8 
2 c  53.53  180  126.47
 c  63.2 

 d  116.8
2 d  53.53  180  233.53
w
R

1350
 0.80417
1678.75
  26.8    
63.2    


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1158
PRO
OBLEM 7.120
4 ft
A preessure vessel of
o 10-in. inner diameter annd 0.25-in. waall thickness
is fabbricated from a 4-ft sectioon of spirally--welded pipe AB and is
equippped with two rigid end plattes. The gage pressure
p
insidde the vessel
is 3000 psi and 10-kkip centric axxial forces P and
a P are appplied to the
end plates.
p
Determ
mine (a) the noormal stress peerpendicular to
t the weld,
(b) the shearing streess parallel to the weld.
P'
A
P
35
B
SOLUTION
1
1
d  (10)  5 in.
t  0.25 in.
2
2
pr (3000)(5)
 6000 psi

1 
p  6 ksi
t
0.225
pr (3000)(5)
 3000 psi

2 
p  3 ksi
2t (2)(00.25)
r0  r  t  5  0.25  5.25 inn.
r


A   r02  r 2   (5.252  5.002 )  8.05003 in 2
 
Totaal stresses.
Longitudinal:
100  103
P
 12442 psi  1.242 ksi

8.0803
A
 x  3  1.242  1.7588 ksi
Circumferential:  y  6 ksi
k
Shear:
 xy  0
Plottted points forr Mohr’s circlee:
X : (1.758, 0)
Y : (6, 0)
C : (3.879)
1
2
 ave  ( x   y )  3.8879 ksi
2
x  y 
2
R 
   xy
2


2
 ((1.758  6) 
 
  0  2.121 kssi
2


(a)
(b)
 x   avee  R cos 70  3.879  2.1221 cos 70
| xy |  R siin 70  2.1211 sin 70
 x  3.15 ksi 
| xy  |  1.993
1
ksi 
PRO
OPRIETARY MAT
TERIAL. Copyrigght © 2015 McGrraw-Hill Educatioon. This is proprietary material soleely for authorizedd instructor use.
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a
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n be copied, scannned, duplicated, forwarded, distribbuted, or posted
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w
in whole or part.
1159
PROB
BLEM 7.12
21
Solve Prob. 7.120, assuming thatt the magnituude P of the tw
wo forces is
increassed to 30 kips.
4 ft
P'
A
PROB
BLEM 7.120 A pressure vesssel of 10-in. inner diameterr and 0.25-in.
wall thhickness is faabricated from
m a 4-ft sectioon of spirally--welded pipe
AB andd is equipped with two rigiid end plates. The gage preessure inside
the vesssel is 300 psii and 10-kip centric
c
axial foorces P and P are applied
to the end plates. Determine
D
(a) the normal stress perpenddicular to the
weld, (b)
( the shearinng stress parallel to the weldd.
P
35
B
SO
OLUTION
1
1
d  (10)
(  5 in.
t  0.25 in.
2
2
pr (3000)(5)
1 

 6000 psi  6 ksi
t
0
0.25
pr (300)(5)
2 

 30000 psi  3 ksi
2t (2))(0.25)
r0  r  t  5  0.25  5.25 in.
r


A   r02  r 2   (5.252  52 )  8.05033 in 2
 
Tottal stresses.
30  103
P
 37727 psi  3.7727 ksi

A
8.0503
 x  3  3.727  0.7727 ksi
Longitudinall:
Circumferen
ntial:  y  6 ksi
 xy  0
Shear:
Plootted points for Mohr’s circlle:
X : (0.727, 0)
0
Y : (6, 0)
C : (2.66365, 0)
1
2
 avve  ( x   y )  2.6365 kssi
2
x  y 
2
R 
   xy
2


2
 0.7277  6 
3
ksi
 
  0  3.3635
2


(a)
 x   ave  R cos 70  2.6365  3.3635 cos 70
 x  1.486 ksi 
(b)
| xy  |  R sin 70  3.36
635 sin 70
| xy |  3.16 ksi 
PRO
OPRIETARY MATERIAL. Copyriight © 2015 McG
Graw-Hill Educattion. This is proprrietary material soolely for authorizedd instructor use.
Not authorized for salle or distribution in
i any manner. Thhis document may not be copied, scaanned, duplicated,, forwarded, distributed, or posted
on a website, in wholee or part.
1160
PROBLEM 7.122
2
T
A torquue of magnituude T  12 kN
nd of a tank containing
N  m is applied to the en
compresssed air under a pressure off 8 MPa. Know
wing that the tank
t
has a 1800-mm inner
diameterr and a 12-mm
m wall thicknness, determinne the maximuum normal strress and the
maximuum shearing strress in the tank.
SOLUTION

d  180 mm
m
r 

1
d  90 mm t  12 mm
2
Torssion:


c1  90 mm
m c2  90  12  102 mm
m

J 

 



2
c
4
2

 c14  66.9668  106 mm 4  66.968  106 m 4
Tc (12  103 )(1022  103 )

 188.277 MPa
J
66.968  106
Presssure:

1 
pr
pr
(8)(90)

 600 MPa  2 
 30 MP
Pa
12
2t
t
Sum
mmary of stresses:
 x  60 MPa,
M
 y  30 MPa,  xy  18.277 MPa



1
2
 ave  ( x   y )  45 MPa
2
x y 
2
R 
Pa
   xy
x  23.64 MP
2


 a   avee  R  68.64 MPa
M
 b   avee  R  21.36 MPa
M
c  0
 max  688.6 MPa 
 min  0
1
2
 max  ( max   min )
 max  344.3 MPa 
PRO
OPRIETARY MAT
TERIAL. Copyrigght © 2015 McGrraw-Hill Educatioon. This is proprietary material soleely for authorizedd instructor use.
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a
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in whole or part.
1161
PROBLEM 7.123
T
The tank shown has a 180-mm inner diameter and a 12-mm wall thickness. Knowing
that the tank contains compressed air under a pressure of 8 MPa, determine the
magnitude T of the applied torque for which the maximum normal stress is 75 MPa.
SOLUTION
1
1
d    (180)  90 mm
2
2
pr (8)(90)
1 

 60 MPa
t
12
pr
2 
 30 MPa
2t
1
 ave  ( 1   y )  45 MPa
2
 max  75 MPa
r
t  12 mm
R   max   ave  30 MPa
2
  2 
2
2
2
R  1
   xy  15   xy
2


 xy  R 2  152  302  152  25.98 MPa
 25.98  106 Pa
Torsion:
c1  90 mm
c2  90  12  102 mm
 xy


T 4
c2  c14  66.968  106 mm 4  66.968  106 m 4
2
J xy (66.968  106 )(25.98  106 )
Tc


 17.06  103 N  m
T
3
J
c
102  10
J
T  17.06 kN  m 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1162
PROBLEM 7.124
y
150 mm
The compressed-air tank AB has a 250-mm outside diameter and an 8-mm
wall thickness. It is fitted with a collar by which a 40-kN force P is
applied at B in the horizontal direction. Knowing that the gage pressure
inside the tank is 5 MPa, determine the maximum normal stress and the
maximum shearing stress at point K.
B
P
600 mm
K
L
A
z
150 mm
x
SOLUTION
Consider element at point K.
Stresses due to internal pressure:
p  5 MPa  5  106 Pa
1
250
r  d t 
 8  117 mm
2
2
pr (5  106 )(117  103 )
x 

 73.125 MPa
t
(8  103 )
y 
Stress due to bending moment:
pr (5  106 )(117  103 )

 36.563 MPa
2t
(2)(8  103 )
Point K is on the neutral axis.
y  0
Stress due to transverse shear:
V  P  40  103 N
1
c2  d  125 mm
2
c1  c2  t  117 mm
2
2
Q  c23  c13  (1253  1173 )
3
3
3
 234.34  10 mm3  234.34  106 m3



c
4
 xy 
4
2

 c14 

(1254  117 4 )
4
 44.573  106 mm 4  44.573  106 m 4
I
(40  103 )(234.34  106 )
VQ
PQ


It
I (2t ) (44.573  106 )(16  103 )
 13.1436  106 Pa  13.1436 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1163
PROBLEM 7.124 (Continued)
Total stresses:
 x  73.125 MPa,  y  36.563 MPa,  xy  13.1436 MPa
Mohr’s circle:
 ave  ( x   y )  54.844 MPa
1
2
2
x  y 
2
R 
   xy
2


 (18.281) 2  (13.1436) 2  22.516 MPa
 a   ave  R  77.360 MPa
 b   ave  R  32.328 MPa
 a  77.4 MPa,  b  32.3 MPa 
Principal stresses:
z  0 
The 3rd principal stress is the radial stress.
 max  77.4 MPa,  min  0
Maximum shearing stress:
1
2
 max  ( max   min )
 max  77.4 MPa 
 max  38.7 MPa 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1164
PROB
BLEM 7.12
25
y
150
0 mm
In Probb. 7.124, deterrmine the maxximum normall stress and thee maximum
shearinng stress at poiint L.
B
P
PROB
BLEM 7.124 The
T compresssed-air tank AB
B has a 250-m
mm outside
diametter and an 8-m
mm wall thicknness. It is fitteed with a collaar by which
a 40-kN
N force P is appplied at B in the horizontal direction. Knnowing that
the gagge pressure innside the tankk is 5 MPa, determine
d
thee maximum
normall stress and thee maximum shhearing stress at point K.
600 mm
K
L
A
z
m
150 mm
x
SOLUTION
Connsider elementt at point L.
Streesses due to in
nternal pressurre:
p  5 MPa  5  106 Pa
1
250
r  d t 
 8  1177 mm
2
2
pr (5  106 )(117  1003 )
x 

 73.125 MPa
t
8  103
pr (5  103 )(117  1003 )
y 

 36.563 MPa
2t
(2)(8  103 )
ding moment:
Streess due to bend
M  (40 kN)(600
k
mm)  24,000 N  m
1
d  125 mm
2
c1  c2  t  125  8  117 mm
c2 

c
4
y  
4
2

 c14 

(1254  117 4 )
4
 44.573  106 mm 4  44.573  106 m 4
I
Mc
(24, 000)(125  103 )
 67.305 MP
Pa

I
44.573  106
PRO
OPRIETARY MAT
TERIAL. Copyrigght © 2015 McGrraw-Hill Educatioon. This is proprietary material soleely for authorizedd instructor use.
Not authorized
a
for salee or distribution inn any manner. Thiis document may not
n be copied, scannned, duplicated, forwarded, distribbuted, or posted
on a website,
w
in whole or part.
1165
PROBLEM 7.125 (Continued)
Stress due to transverse shear:
Point L lies in a plane of symmetry.
 xy  0
 x  73.125 MPa,  y  30.742 MPa,  xy  0
Total stresses:
Principal stresses:
Since  xy  0,  x and  y are principal stresses. The 3rd principal stress is in the
radial direction,  z  0.
 max  73.125 MPa,  min  0,  a  73.1 MPa,  b  30.7 MPa,  z  0
 max  73.1 MPa 
Maximum stress:
Maximum shearing stress:
1
2
 max  ( max   min )
 max  51.9 MPa 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1166
PROBLEM 7.126
1.5 in.
STEEL
ts 81 in.
Es 29 106 psi
ss 6.5 10–6/F
5 in.
BRASS
tb 14 in.
Eb 15 106 psi
bs 11.6 10–6/F
A brass ring of 5-in. outer diameter and 0.25-in. thickness fits exactly
inside a steel ring of 5-in. inner diameter and 0.125-in. thickness
when the temperature of both rings is 50F. Knowing that the
temperature of both rings is then raised to 125F, determine (a) the
tensile stress in the steel ring, (b) the corresponding pressure exerted
by the brass ring on the steel ring.
SOLUTION
Let p be the contact pressure between the rings. Subscript s refers to the steel ring. Subscript b refers to the
brass ring.
Steel ring.
Internal pressure p:
s 
pr
ts
s
(1)
Corresponding strain:
 sp 
Strain due to temperature change:
 sT   s T
s 
Total strain:
Es

pr
Es t s
pr
  s T
Es t s
Change in length of circumference:

 pr
Ls  2 r s  2 r 
  s T 
 Es ts

Brass ring.
External pressure p:
Corresponding strains:
b  
pr
tb
 bp  
pr
,  bT   b T
Eb tb
Change in length of circumference:
 pr

Lb  2 r b  2 r  
  b T 
 Eb tb

Equating Ls to Lb ,
pr
pr
  s T  
  b T
Es t s
Eb tb
 r
r 


 p  (b   s ) T
 Es ts Eb tb 
(2)
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1167
PROBLEM 7.126 (Continued)
T  125F  50F  75F
Data:
r
From Equation (2),
1
1
d  (5)  2.5 in.
2
2


2.5
2.5
6


 p  (11.6  6.5)(10 )(75)
6
6
 (29  10 )(0.125) (15  10 )(0.25) 
1.35632  106 p  382.5  106
p  282.0 psi
From Equation (1),
s 
pr (282.0)(2.5)

 5.64  103 psi
ts
0.125
(a)  s  5.64 ksi 
(b) p  282 psi 
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1168
PROBLEM 7.127
1.5 in.
Solve Prob. 7.126, assuming that the brass ring is 0.125 in. thick and
the steel ring is 0.25 in. thick.
STEEL
ts 81 in.
Es 29 106 psi
ss 6.5 10–6/F
5 in.
BRASS
tb 14 in.
Eb 15 106 psi
bs 11.6 10–6/F
PROBLEM 7.126 A brass ring of 5-in. outer diameter and 0.25-in.
thickness fits exactly inside a steel ring of 5-in. inner diameter and
0.125-in. thickness when the temperature of both rings is 50F.
Knowing that the temperature of both rings is then raised to 125F,
determine (a) the tensile stress in the steel ring, (b) the corresponding
pressure exerted by the brass ring on the steel ring.
SOLUTION
Let p be the contact pressure between the rings. Subscript s refers to the steel ring. Subscript b refers to the
brass ring.
Steel ring.
Internal pressure p:
s 
pr
ts
s
(1)
Corresponding strain:
 sp 
Strain due to temperature change:
 sT   s T
s 
Total strain:
Es

pr
Es t s
pr
  s T
Es t s
Change in length of circumference:
 pr

Ls  2 r s  2 r 
  s T 
 Es t s

Brass ring.
External pressure p:
Corresponding strains:
b  
pr
tb
 bp  
pr
,  bT   b T
Eb tb
Change in length of circumference:
 pr

Lb  2 r b  2 r  
  b T 
 Eb tb

Equating Ls to Lb ,
pr
pr
  s T  
  b T
Es t s
Eb tb
 r
r 


 p  (b   s )T
 Es ts Eb tb 
(2)
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1169
PROBLEM 7.127 (Continued)
Data:
From Equation (2),
T  125F  50F  75F
1
1
r  d  (5)  2.5 in.
2
2


2.5
2.5
6


 p  (11.6  6.5)(10 )(75)
6
6
(29
10
)(0.25)
(15
10
)(0.125)




1.67816  106 p  382.5  106
p  227.93 psi
From Equation (1),
s 
pr (227.93)(2.5)

 2279 psi
ts
0.25
(a)  s  2.28 ksi 
(b) p  228 psi 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1170
PROBLEM 7.128
y
y'
x'
x
For the given state of plane strain, use the method of Sec. 7.7A to
determine the state of plane strain associated with axes x and y 
rotated through the given angle  .
 x  800 ,  y  450 ,  xy  200 ,   25
SOLUTION
  25
x  y
2
x  y
 175
 x 
x  y
2
x   y
 625
 xy
2
 100
sin 2
2
2
2
 175  (625 ) cos (50)  (100  )sin (50)
 y 
x  y

x  y
cos 2 
 xy
cos 2 
 xy
sin 2
2
2
2
 175  (625 ) cos (50)  (100 )sin (50)

 x  653 
 y  303 
 xy  ( x   y )sin 2   xy cos 2
 (800  450 )sin (50)  (200 ) cos (50)
 xy  829 
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1171
PROBLEM 7.129
y
y'
x'
x
For the given state of plane strain, use the method of Sec. 7.7A to
determine the state of plane strain associated with axes x and y 
rotated through the given angle  .
 x  240 ,  y  160 ,  xy  150 ,   60
SOLUTION
  60
x  y
2
x  y
 200
 x 
x  y
2
x   y
 xy
 40
 xy
 75
sin 2
2
2
2
 200  40 cos (120)  75 sin (120)
 y 
x  y

x  y
cos 2 
2
cos 2 
 xy
sin 2
2
2
2
 200  40cos (120)  75sin (120)

 x  115.0  
 y  285 
 xy  ( x   y )sin 2   xy cos 2
 (240  160)sin (120)  150 cos (120)
 xy  5.72 
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1172
PROBLEM 7.130
y
y'
x'
␪
x
For the given state of plane strain, use the method of Sec. 7.7A to
determine the state of plane strain associated with axes x and y 
rotated through the given angle  .
 x  500  ,  y  250  ,  xy  0,   15
SOLUTION
  15
x   y
2
x   y
 125
 x 
x  y
2

x  y
 375
cos 2 
 xy
2
2
2
 125  (375 ) cos 30  0
 y 
x  y

x   y
cos 2 
 xy
2
2
2
 125  (375 ) cos 30  0
 xy
2
0
sin 2
 x  450  
sin 2
 y  199.8 
 xy  ( x   y )sin 2   xy cos 2
 (500   250  )sin 30  0
 xy  375 
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1173
PROBLEM 7.131
y
y'
For the given state of plane strain, use the method of Sec 7.7A to
determine the state of plane strain associated with axes x and y
rotated through the given angle  .
x'
x
␪
 x  0,  y  320  ,  xy  100  ,   30
SOLUTION
  30
x  y
2
x  y
 160 
 x 
x   y
2

2
x  y
2
 160
cos 2 
 160  160 cos 60 
 y 
x  y
2

x  y
2
2
sin 2
100
sin 60
2
cos 2 
 160  160cos 60 
 xy
 xy
2
 x  36.7  
sin 2
100
sin 60
2
 y   283 
 xy  ( x   y )sin 2   xy cos 2
 (0  320)sin 60  100 cos 60
 xy  227  
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1174
PROBLEM 7.132
y
y'
x'
x
␪
For the given state of plane strain, use Mohr’s circle to determine the
state of plane strain associated with axes x and y  rotated through
the given angle  .
 x  800  ,  y   450  ,  xy  200  ,   25
SOLUTION
Plotted points:
X : (800 , 100 )
Y : (450 ,  100 )
C : (175 , 0)
tan  
100
625
  9.09
R  (625 ) 2  (100  ) 2  632.95
  2    50  9.09  40.91
 x   ave  R cos   175  632.95 cos 40.91
 x  653 
 y   ave  R cos   175  632.95 cos 40.91
 y  303 
1
 xy   R sin   632.95 sin 40.91
2
 xy  829  
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1175
PR
ROBLEM 7.133
7
y
y'
x'
␪
x
For the given staate of plane sttrain, use Mohhr’s circle to determine
d
the
staate of plane sttrain associateed with axes x and y rotaated through
thee given angle  .
 x  240  ,  y  160  ,  xy  150  ,   60
SO
OLUTION
Plootted points for Mohr’s circlle:
X : (240
2  ,  75  )
Y : (160
1  , 75  )
C : (200
2  , 0)
tan
n 
75
 1.875   61.933
40
R  (40  )2  (75 )2  85
8 


  2    120  61.93  181.93
 x   ave  R cos
c   200   (85 ) cos (181.93)
 x  115.0  
 y   ave  R cos
c   200   (85 ) cos ( 181.93)
 yy  285  
1
 xy   R sin   85 sin (181.93)  2.86
2 
2
 xy  5.72  
PRO
OPRIETARY MATERIAL. Copyriight © 2015 McG
Graw-Hill Educattion. This is proprrietary material soolely for authorizedd instructor use.
Not authorized for salle or distribution in
i any manner. Thhis document may not be copied, scaanned, duplicated,, forwarded, distributed, or posted
on a website, in wholee or part.
1176
PROBLEM 7.134
y
y'
x'
x
␪
For the given state of plane strain, use Mohr’s circle to determine the
state of plane strain associated with axes x and y rotated through
the given angle  .
 x  500  ,  y  250  ,  xy  0,   15
SOLUTION
Plotted points:
X : (500 ,0)
Y : (250 , 0)
C : (125 , 0)
R  375



 x   ave  R cos 2  125  375cos 30 
 x  450  
 y   ave  R cos 2  125  375cos 30 
 y  199.8  
1
 xy  R sin 2  375sin 30 
2
 xy  375  
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
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1177
PR
ROBLEM 7.135
7
y
y'
x'
␪
x
For the given staate of plane sttrain, use Mohhr’s circle to determine
d
the
staate of plane strrain associated with axes x and y  rotaated through
thee given angle  .
 x  0,  y  3200  ,  xy  100  ,   30
SO
OLUTION
Plootted points for Mohr’s circlle:
X : (0, 50 )
Y : (320 ,  50 )
C : (160 , 0)
tan
n 
50
  17.35
160
R  (160  ) 2  (50  ) 2  167.63
  2    60  17.35  42.65
 x   ave  R cos
c   160   (167.63 ) coos 42.65
 x  36.7  
 y   ave  R cos
c   160   (167.63 ) coos 42.65
 y  283 
1
 xy   R sin   (167.63 )ssin 42.65
2
 xy   227  
PRO
OPRIETARY MATERIAL. Copyriight © 2015 McG
Graw-Hill Educattion. This is proprrietary material soolely for authorizedd instructor use.
Not authorized for salle or distribution in
i any manner. Thhis document may not be copied, scaanned, duplicated,, forwarded, distributed, or posted
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1178
PR
ROBLEM 7.136
Thee following staate of strain haas been measuured on the surrface of a thinn plate. Knowiing that the suurface of the
platte is unstresseed, determine (a) the direction and magnnitude of the principal
p
strains, (b) the maaximum inplanne shearing strrain, (c) the maximum
m
shearring strain. (U
Use v  13 . )
 x  2600  ,  y  600  ,  xy  4880 
SOLUTION
For Mohr’s circlee of strain, plot points:
X : (2660 ,  240 )
Y : (600 , 240 )
C : (1660 , 0)
tan 2 p 
 xy
480

 2.4
 x   y 260  60
2 p  67.38
b  33.7 
 a  56.3 
R  (100  )  (240 )
2
2
R  260
(a)
(b)
 a   ave  R  160   260 
 a  100  
 b   ave  R  160   260 
 b  420  
1
 max (in-plaane)  R  max ((in-plane)  2R
2
v
v
1/3
c  
( a   b )  
( x   y )  
(260  60)
1 v
1 v
2/3
 160
 c  160 
 max
 160   min  420 
m
(c)
 max (in-plane))  520  
 max   maxx   min  160   420 
 maxx  580  
PRO
OPRIETARY MAT
TERIAL. Copyrigght © 2015 McGrraw-Hill Educatioon. This is proprietary material soleely for authorizedd instructor use.
Not authorized
a
for salee or distribution inn any manner. Thiis document may not
n be copied, scannned, duplicated, forwarded, distribbuted, or posted
on a website,
w
in whole or part.
1179
PR
ROBLEM 7.137
7
Thee following sttate of strain has
h been measuured on the suurface of a thinn plate. Know
wing that the suurface of the
plaate is unstresseed, determine (a) the direction and magnnitude of the principal straains, (b) the maximum
m
inplaane shearing sttrain, (c) the maximum
m
sheaaring strain. (U
Use v  13 . )
 x  6000  ,  y  4000 ,  xy  350
3 
SO
OLUTION
Plootted points for Mohr’s circlle:
X : (600  , 175 )
Y : (400  , 175 )
C : (500  , 0)
0
tan 2 p  
175
100
2 p  60.26
b  30.1 
 a  59.9 
R  (100  ) 2  (175 ) 2
 201.6 
(a)
(b)
 a   avee  R  500   201.6 
 a  298 
 b   avee  R  500   201.6 
 b  702  
 max (in-pllane)  2R
c  
 max (in-planee)  403 
v
v
1/3
(600   400  )
( a   b )  
( x   y )  
2/3
1 v
1 v
 c  500  
 max  500   min  702 
(c)
 max   max   min  500
5   702 
 max  1202  
PRO
OPRIETARY MATERIAL. Copyriight © 2015 McG
Graw-Hill Educattion. This is proprrietary material soolely for authorizedd instructor use.
Not authorized for salle or distribution in
i any manner. Thhis document may not be copied, scaanned, duplicated,, forwarded, distributed, or posted
on a website, in wholee or part.
1180
PR
ROBLEM 7.138
Thee following staate of strain haas been measuured on the surrface of a thinn plate. Knowiing that the suurface of the
platte is unstresseed, determine (a) the direction and magnnitude of the principal
p
strains, (b) the maaximum inplanne shearing strrain, (c) the maximum
m
shearring strain. (U
Use v  13 . )
 x  160  ,  y  4800 ,  xy  600
6 
SOLUTION
(a)
For Mohr’ss circle of straain, plot pointss:
X : (160  , 300  )
Y : (480  ,  300  )
C : (160  , 0)
0
(a)
tan 2 p 
 xy
3000
 0.9375

x  y
3200
2 p  43.15  p  21.58 andd 21.58  900  68.42
 a  21.6 
b  68.4 

R  (320  ) 2  (3000  ) 2  438.66 
 a   ave
0   438.6 
a  R  160
 a  279 
 b   ave
0   438.6 
a  R  160
 b  599 
(b)
1
 (max, in-plaane)  R  (maxx, in-plane)  2R
2
(c)
c  
 (max, in-plane))  877  
v
v
1/3
( a   b )  
( x   y )  
(160   480  )
1 v
1 v
2/3
 c  160.0  
 max  2778.6   min  598.6 
 max   max
78.6   598.6 
m   min  27
 max  877  
PRO
OPRIETARY MAT
TERIAL. Copyrigght © 2015 McGrraw-Hill Educatioon. This is proprietary material soleely for authorizedd instructor use.
Not authorized
a
for salee or distribution inn any manner. Thiis document may not
n be copied, scannned, duplicated, forwarded, distribbuted, or posted
on a website,
w
in whole or part.
1181
PR
ROBLEM 7.139
7
Thee following sttate of strain has
h been measuured on the suurface of a thinn plate. Know
wing that the suurface of the
plaate is unstresseed, determine (a) the direction and magnnitude of the principal straains, (b) the maximum
m
inplaane shearing sttrain, (c) the maximum
m
sheaaring strain. (U
Use v  13 .)
 x  30  ,  y  5700  ,  xy  720
7 
SO
OLUTION
Plootted points for Mohr’s circlle:
X : (30 , 360 )
Y : (570 , 360 )
C : (300 , 0)
360
 1.3333
270
2 p  53.13
tan 2 p 
b  26.6 
(a)
 a  64.4 
R  (2770  )2  (360 ) 2  450 
(b)
 a   ave  R  300   450 
 a  750  
 b   ave  R  300   450 
 b  150.0  
 max (in-planee)  2R
c  
 max (in-planee)  900 
v
1/3
( a   b )  
(750   150  )
2/33
1 v
 c  300  
 maax   a  750 ,  min   c  300 
(c)
 max   maax   min  7500   ( 300  )
 max  1050  
PRO
OPRIETARY MATERIAL. Copyriight © 2015 McG
Graw-Hill Educattion. This is proprrietary material soolely for authorizedd instructor use.
Not authorized for salle or distribution in
i any manner. Thhis document may not be copied, scaanned, duplicated,, forwarded, distributed, or posted
on a website, in wholee or part.
1182
PR
ROBLEM 7.140
For the given staate of plane sttrain, use Mohhr’s circle to determine (a)) the orientatioon and magniitude of the
prinncipal strains, (b) the maxim
mum in-plane strain,
s
(c) the maximum
m
sheearing strain.
 x  60  ,  y  2400 ,  xy  50
5 
SOLUTION
Plottted points:
X : (60 , 25
2 )
Y : (240 , 25 )
C : (150  , 0)
tan 2 p 
 xy
50

 0.277778
0
 x   y 60  240
2 p  15.52
 a  97.8 
b  7.8 
R  (90  ) 2  (25 ) 2  933.4 
(a)
 a   ave  R  150   933.4 
 a  243 
 b   ave  R  150   933.4 
 b  56.6  
(b)
 max (in-plane))  2R
(c)
 c  0,  max
m  243.4  ,  min  0
 max (in-plane)  186.8 
c  0 
 maxx  243  
 max
  max   m
m
min
PRO
OPRIETARY MAT
TERIAL. Copyrigght © 2015 McGrraw-Hill Educatioon. This is proprietary material soleely for authorizedd instructor use.
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a
for salee or distribution inn any manner. Thiis document may not
n be copied, scannned, duplicated, forwarded, distribbuted, or posted
on a website,
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in whole or part.
1183
PR
ROBLEM 7.141
7
Forr the given staate of plane strain,
s
use Moohr’s circle to determine (aa) the orientation and magnnitude of the
prinncipal strains,, (b) the maxim
mum in-plane strain, (c) the maximum shearing strain.
 x  4000 ,  y  200  ,  xy  375
3 
SO
OLUTION
Plootted points for Mohr’s circlle:
X : (400  , 187.5 )
Y : (200  , 187.5 )
C : (300  , 0)
0
tan 2 p 
 xy
375

 1.875
 x   y 400  200
2 p  61.93
 a  31.0 
b  121.0 
R  (100  )2  (187.5 ) 2  212.5
(a)
 a   ave  R  300   212.5
2

 a  513 
 b   ave  R  300   212.5
2

 b  87.5 
(b)
 max (in-planee)  2R
(c)
 c  0  max  512.5  min  0
 max (in-planee)  425 
c  0 
 max
m  513 
 max   max   min
PRO
OPRIETARY MATERIAL. Copyriight © 2015 McG
Graw-Hill Educattion. This is proprrietary material soolely for authorizedd instructor use.
Not authorized for salle or distribution in
i any manner. Thhis document may not be copied, scaanned, duplicated,, forwarded, distributed, or posted
on a website, in wholee or part.
1184
PR
ROBLEM 7.142
For the given staate of plane sttrain, use Mohhr’s circle to determine (a)) the orientatioon and magniitude of the
prinncipal strains, (b) the maxim
mum in-plane strain,
s
(c) the maximum
m
sheearing strain.
 x  3000  ,  y  600  ,  xy  1000 
SOLUTION
X : (300 , 500 )
Y : (60 , 50 )
C : (180 , 0)
tan 2 p 
 xy
x  y

100
300  60
2 p  22.62
 a  11.3 

b  101.3 

R  (120 )2  (50  ) 2  130
1 
(a)
 a   ave  R  180   1330 
 a  310  
 b   ave  R  180   1330 
 b  50.0 
(b)
 max (in-plane))  2R
(c)
 c  0,  max
 310  ,  min  0
m
 max (in-plane))  260  
c  0 
 max
  max   m
m
min
 maxx  310  
PRO
OPRIETARY MAT
TERIAL. Copyrigght © 2015 McGrraw-Hill Educatioon. This is proprietary material soleely for authorizedd instructor use.
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a
for salee or distribution inn any manner. Thiis document may not
n be copied, scannned, duplicated, forwarded, distribbuted, or posted
on a website,
w
in whole or part.
1185
PR
ROBLEM 7.143
7
Forr the given staate of plane strain,
s
use Moohr’s circle to determine (aa) the orientation and magnnitude of the
prinncipal strains,, (b) the maxim
mum in-plane strain, (c) the maximum shearing strain.
 x  1800 ,  y  2660  ,  xy  315
3 
SO
OLUTION
Plootted points for Mohr’s circlle:
X : (180  , 157.5 )
Y : (260 , 157.5 )
C : (220 , 0)
0
(a)
tan 2 p 
 xy
315
5

 3.9375
 x   y 80
2 p  75.75
7
 a  37.9 
b  127.9 
R  (40  )2  (15
57.5 )2  162..5
 a   ave  R   2220   162.5
 a  57.5 
 b  383 
 b   ave  R  2220   162.5
(b)
 max (in-planee)  2 R  325
(c)
 c  0,  max  0,  minn  382.5

c  0 
 max
m  383 
 max   max   min  0  382.5
PRO
OPRIETARY MATERIAL. Copyriight © 2015 McG
Graw-Hill Educattion. This is proprrietary material soolely for authorizedd instructor use.
Not authorized for salle or distribution in
i any manner. Thhis document may not be copied, scaanned, duplicated,, forwarded, distributed, or posted
on a website, in wholee or part.
1186
PRO
OBLEM 7.1
144
45
3
2
Deterrmine the strain  x , knowingg that the folloowing strains have been dettermined by
use of
o the rosette shown:
s
30
x
1  4800  2  1220  3  800
15
1
SOLUTION
1  15
 2  30
3  75
 x cos
c 2 1   y sinn 2 1   xy sinn 1 cos 1  1
0.9330 x  0.06699 y  0.25 xy  4880
(1)
 x coos2  2   y sinn 2  2   xy sin  2 cos  2   2
0.75 x  0.25 y  0..4330 xy  120
1 
(2)
 x coos2 3   y sinn 2 3   xy sin 3 cos 3   3
0.06699 x  0.9330 y  0.25 xy  800
(3)
Solvving (1), (2), and
a (3) simultaaneously,
 x  253 ,
 y  307  ,
 xy  8893
 x  253 
PRO
OPRIETARY MAT
TERIAL. Copyrigght © 2015 McGrraw-Hill Educatioon. This is proprietary material soleely for authorizedd instructor use.
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a
for salee or distribution inn any manner. Thiis document may not
n be copied, scannned, duplicated, forwarded, distribbuted, or posted
on a website,
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in whole or part.
1187
PROB
BLEM 7.145
y
The strrains determinned by the usee of the rosettte shown durring the test of
o a machine
elemen
nt are
30
3
2
1
30
1   600  2  4500  3   755
x
Determ
mine (a) the in--plane principal strains, (b) the in-plane maximum
m
sheaaring strain.
SO
OLUTION
1  30
 2  1500
3  90
 x coos 2 1   y sinn 2 1   xy sin 1 cos1  1
0.75 x  0.25 y  0.433301 xy  6000
(1)
 x coos2  2   y sin 2  2   xy sin  2 cos  2   2
0.75 x  0.25 y  0.433301 xy  4500
(2)
 x cos2 3   y sinn 2 3   xy sin 3 cos 3   3
0   y  0  755
(3)
Sollving (1), (2), and (3) simulttaneously,
 x  725
7  ,  y  75 ,  xy  173.21 
1
2
 ave  ( x   y )  325
2
2
2
2
  x   y    xy 
 725  75   173.21 

R 
4

  
 
  2   409.3
2
 


 2   2 
(a)
 a   ave  R  734 
 a  734  
 b   ave  R  84.3
(b)
 b  84.3 
 max (in-plane))  2 R  819
 max (in-planee)  819 
PRO
OPRIETARY MATERIAL. Copyriight © 2015 McG
Graw-Hill Educattion. This is proprrietary material soolely for authorizedd instructor use.
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i any manner. Thhis document may not be copied, scaanned, duplicated,, forwarded, distributed, or posted
on a website, in wholee or part.
1188
PROBLEM 7.146
4
45
3
45
The rosette shown has been used to determine the following strains at a point
on the surface of a crane hook:
2
1  420  106 in./in.  2  45  106 in./in.  4  165  106 in./in.
45
1
x
(a) What should be the reading of gage 3? (b) Determine the principal strains
and the maximum in-plane shearing strain.
SOLUTION
(a) Gages 2 and 4 are 90 apart.
1
2
1
 (45  106  165  106 )  60  106 in./in.
2
 ave  ( 2   4 )
 ave
Gages 1 and 3 are also 90 apart.
1
2
 3  2 ave  1  (2)(60  106 )  420  106
 ave  (1   3 )
 3  300  106 in./in. 
(b)  x  1  420  10 6 in./in.  y   3  300  10 6 in./in.
 xy  2 2  1   3  (2)(45  106 )  420  106  (300  106 )
 210  106 in./in.
2
2
2
 420  106  (300  106 )   210  106
  x   y    xy 
R 
  
 
  
2
2
 2   2 

 



2
 375  106 in./in.
 a   ave  R  60  106  375  106
 a  435  106 in./in. 
 b   ave  R  60  106  375  106
 b  315  106 in./in. 
 max (in-plane)  750  106 in./in. 
 max (in-plane)  2 R
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1189
␥
2
PROBLEM 7.147
⑀2
3
45⬚
Using a 45 rosette, the strains 1,  2 , and  3 have
been determined at a given point. Using Mohr’s
circle, show that the principal strains are:
⑀3
2
B
O
45⬚
A
⑀
C
1
1
1
(1   3 ) 
[(1   2 )2  ( 2   3 ) 2 ] 2
2
2
(Hint: The shaded triangles are congruent.)
 max,min 
⑀ min
1
⑀1
⑀ max
SOLUTION
Since gage directions 1 and 3 are 90 apart,
1
2
 ave  (1   3 )
Let
1
(1   3 )
2
1
  2  (1   3 )
2
u  1   ave 
v   2   ave
R2  u 2  v2

1
1
(1   3 )2   22   2 (1   3 )  (1   3 ) 2
4
4

1 2 1
1
1
1
1
1  1 3   32   22   21   2 3  12  1 3   32
4
2
4
4
2
4
1 2
1
1   21   22   2 3   32
2
2
1
1
 (1   2 ) 2  ( 2   3 )2
2
2
1
R
[(1   2 ) 2  ( 2   3 ) 2 ]1/2
2

 max, min   ave  R gives the required formula.
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1190
PROBLEM 7.148
2
3
Show that the sum of the three strain measurements made with a 60
rosette is independent of the orientation of the rosette and equal to
60
60
1   2   3  3 avg
1
where  avg is the abscissa of the center of the corresponding Mohr’s
circle.
x
SOLUTION
1   ave 
 2   ave 
  ave 

 xy
2
x   y
2
x  y
2
x  y
2
cos 2 
 xy
2
sin 2
cos (2  120) 
 xy
2
(1)
sin (2  120)
(cos 120 cos 2  sin 120 sin 2 )
(cos 120 sin 2  sin 120 cos 2 )
x  y  1

3
sin 2 
  cos 2 

2  2
2

 xy  1

3

cos 2 
  sin 2 

2  2
2

  ave 
 3   ave 
  ave 

 xy
2
x  y
2
x  y
2
cos (2  240) 
 xy
2
sin (2  240)
(cos 240 cos 2  sin 240 sin 2 )
(cos 240 sin 2  sin 240 cos 2 )
x  y  1

3
sin 2 
  cos 2 

2  2
2

 xy  1

3
cos 2 

  sin 2 

2  2
2

  ave 
(2)
(3)
Adding (1), (2), and (3),
1   2   3  3 ave  0  0
3 ave  1   2   3

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1191
PROBLEM 7.149
The strains determined by the use of the rosette attached as shown during the test of a
machine element are
3
75
1  93.1  106 in./in.
2
x
 2  385  106 in./in.
75
 3  210  106 in./in.
1
Determine (a) the orientation and magnitude of the principal strains in the plane of the
rosette, (b) the maximum in-plane shearing strain.
SOLUTION
Use  x 
 xy
1
1
( x   y )  ( x   y ) cos 2 
sin 2
2
2
2
where
and
  75
for gage 1,
 0
for gage 2,
  75
for gage 3.
1
2
1
2
1
2
1
2
1
2
1
2
1  ( x   y )  ( x   y ) cos (150) 
 2  ( x   y )  ( x   y ) cos 0 
 xy
2
 xy
2
sin (150)
sin 0
 3  ( x   y )  ( x   y ) cos (150) 
 xy
2
sin (150)
(1)
(2)
(3)
 x   z  385  106 in./in.
From Eq. (2),
Adding Eqs. (1) and (3),
1   3  ( x   y )  ( x   y ) cos 150
  x (1  cos 150)   y (1  cos 150)
     (1  cos 150)
y  1 3 x
(1  cos 150)

93.1  106  210  106  385  106 (1  cos 150)
1  cos 150
 35.0  106 in./in.
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1192
PROBLEM 7.149 (Continued)
Subtracting Eq. (1) from Eq. (3),
 3  1   xy sin 150
 xy 
 3  1
sin 150

210  106  (93.1  106 )
sin 150
 606.2  106 in./in.
tan 2 p 
 xy
606.2  106

 1.732
 x   y 385  106  35.0  106
(a)  a  30.0, b  120.0 
1
1
2
2
6
 210  10 in./in.
 ave  ( x   y )  (385  106  35.0  106 )
2
  x   y    xy 
R 
 

 2   2 
2
 385  106  35.0  106
 
2

(b)
 max (in-plane)
2
2
  606.2 2
6
  
  350.0  10
2



 a   ave  R  210  106  350.0  106
 a  560  106 in./in. 
 b   ave  R  210  106  350.0  106
 b  140.0  106 in./in. 
 R  350.0  106 in./in.
 max (in-plane)  700  10 6 in./in. 
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1193
PROBLE
EM 7.150
y
1 in
n.
A centric axial
a
force P and
a a horizonttal force Qx arre both applieed at point C
of the rectaangular bar shown. A 45 sttrain rosette on the surface of the bar at
point A indicates the folloowing strains::
P
Qx
C
1  60  106 in./in.
x
 2  240  106 in./in.
12 in.
i
 3  200  106 in./in.
3
A
3 in.
mine the magnitudes of P
Knowing thhat E  29  106 psi and v  0.30, determ
and Qx.
2
45⬚
1
3 in..
SO
OLUTION
 x  1  60  106
 y   3  200  106
 xy  2 2  1   3  3400  106
29
E
[60
( x  v y ) 
6  (0.3)(200)]  0
2
1 v
1  (0.3)2
29
E
[2000  (0.3)(60)]  5.8  103 psi
y 
p
( y  v x ) 
2
2
1 v
1  (0.3)
P
  y P  A y  (22)(6)(5.8  103 )
A
 69.6  103 lb
x 
G
P  69.6 kips 
E
29  1006

 11.1538  106 psi
2(1  v) (2)(1.300)
 xy  G xy  (11.1538)((340)  3.79233  103 psi
1 3 1
bbh  (2)(6))3  36 in 4
12
12
ˆ
Q  A y  (2)(3)(1.5)  9 in 3
t  2 in.
ˆ
VQ
 xy 
It
It xy (36)(2)(3..7923  103 )
 30.338  103 lb

l
V
9
Qˆ
I
Q  30.3 kips 
Q V
PRO
OPRIETARY MATERIAL. Copyriight © 2015 McG
Graw-Hill Educattion. This is proprrietary material soolely for authorizedd instructor use.
Not authorized for salle or distribution in
i any manner. Thhis document may not be copied, scaanned, duplicated,, forwarded, distributed, or posted
on a website, in wholee or part.
1194
PROBLEM 7.151
y
1 in.
Solve Prob. 7.150, assuming that the rosette at point A indicates the
following strains:
P
1  30  106 in./in.
Qx
C
 2  250  106 in./in.
x
 3  100  106 in./in.
12 in.
PROBLEM 7.150 A centric axial force P and a horizontal force Qx are both
applied at point C of the rectangular bar shown. A 45 strain rosette on the
surface of the bar at point A indicates the following strains:
3
A
3 in.
1  60  106 in./in.
2
45
 2  240  106 in./in.
1
3 in.
 3  200  106 in./in.
Knowing that E  29  106 psi and v  0.30, determine the magnitudes of
P and Qx.
SOLUTION
 x  1  30  106
 y   3  100  106
 xy  2 2  1   3  430  106
E
29
( x  v y ) 
[30  (0.3)(100)]
2
1 v
1  (0.3)2
0
E
29
y 
( y  v x ) 
[100  (0.3)(30)]
2
1 v
1  (0.3)2
x 
 2.9  103 psi
P
y
A
P  A y  (2)(6)(2.9  103 )
 34.8  103 lb
P  34.8 kips 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1195
PROB
BLEM 7.151 (Continue
ed)
G
E
29  106
 111.1538  106 pssi

2(1  v) (2)(1.30)
 xyy  G xy  (11.1538)(430)  4.7962  103 psi
1 3 1
bh  (2)(6)3  36 in 4
12
12
ˆ
Q  A y  (2)(33)(1.5)  9 in 3
t  2 in.
I
VQˆ
It
It xy (366)(2)(4.7962  103 )
V
 38.37  103 lb

ˆ
9
Q
 xyy 
Q V
Q  38.4
3
kips 
PRO
OPRIETARY MATERIAL. Copyriight © 2015 McG
Graw-Hill Educattion. This is proprrietary material soolely for authorizedd instructor use.
Not authorized for salle or distribution in
i any manner. Thhis document may not be copied, scaanned, duplicated,, forwarded, distributed, or posted
on a website, in wholee or part.
1196
PROBL
LEM 7.152
T'
T
A single strain gage iss cemented to a solid 4-in.-diameter steell shaft at an
angle   25 with a line parallel to the axis off the shaft. Knnowing that
G  11.5  106 psi, deetermine the toorque T indicaated by a gagee reading of
300  1006 in./in.
2 in.
SOLUTION
For torsion,
 x   y  0,    0
1
( x  v y )  0
E
1
 y  ( y  v x )  0
E
0 1

 xy  0
 xy 
G 2
2G
x 
Draaw the Mohr’s circle for straain.
R
0
2G
 x  R sin 2 
But
0
2
2G
sin 2 
0 
2G  x
Tc 2T


J  c3 ssin 2 
T
 c3G x
sin 2

 (2)3(11.5  106 )(300  106 )
sinn 50
 113.2  103 lbb  in.
T  113.2
2 kip  in. 
PRO
OPRIETARY MAT
TERIAL. Copyrigght © 2015 McGrraw-Hill Educatioon. This is proprietary material soleely for authorizedd instructor use.
Not authorized
a
for salee or distribution inn any manner. Thiis document may not
n be copied, scannned, duplicated, forwarded, distribbuted, or posted
on a website,
w
in whole or part.
1197
PROB
BLEM 7.153
3
T'
Solve Prob.
P
7.152, asssuming that thhe gage formss an angle   35 with a
line parrallel to the axxis of the shaftt.
T
PROBL
LEM 7.152 A single gagee is cementedd to a solid 4--in.-diameter
steel shhaft at an anglee   25 with a line paralllel to the axis of the shaft.
Knowinng that G  11.5  106 psi, determine thee torque T inddicated by a
gage reaading of 300  106 in./in.
2 in.
SO
OLUTION
Forr torsion,
 x  0,
0  y  0,  xy   0
1
( x  v y )  0
E
1
 y  ( y  v x )  0
E
0 1

 xy 
 xy  0
G 2
2G
x 
Draaw Mohr’s cirrcle for strain.
R
0
2G
 x  R sin 2 
0
2
2G
sin 2 
But
0 
2G x
Tc 2T


J  c3 ssin 2 
T
 c3G x  (2)3(11.5  106 )(300  106 )

sin 2 
7 
sin 70
 92.3  103 lb  in.
T  92.3 kip  in. 
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1198
PROBLEM 7.15
54
A singgle strain gage forming an angle   18 with a hoorizontal planee is used to
determ
mine the gage pressure in thhe cylindrical steel tank shoown. The cylinndrical wall
of the tank is 6 mm thick, has a 600-mm
6
insidee diameter, annd is made of a steel with
E  200
2 GPa and v  0.30. Dettermine the prressure in the tank
t
indicatedd by a strain
gage reading of 280.
SOLUTION
 x  1 
pr
t
1
2
1
 v 
 x  ( x  v y  v z )  1   x
E
 2 E
 y  x, z  0
 0.85
y 
E
1
1

(v x   y  v z )    v  x
E
2
 E
 0.20
 xy 
x
 xy
G
x
E
0
Draaw Mohr’s circcle for strain.
x
1
E
2

1
R  ( x   y )  0.325 x
E
2
 ave
( x   y )  0.525
a 
 x   ave  R cos 2  (0.5225  0.325cos 2 )
Data:
p
tE x
t x

r
r (00.525  0.325ccos 2  )
r
1
1
m  0.300 m
d  (6600)  300 mm
2
2
x
E
E  200  109 Pa,  x  280  106
t  6  103 mm
m
p
3
9
6
(6  10 )(200
)
 10 )(2880  10 )
 1..421  106 Pa
(0.300)((0.525  0.325 cos 36)
  18
p  1.4
421 MPa 
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1199
PRO
OBLEM 7.155
Solvee Prob. 7.154, assuming thaat the gage form
ms an angle   35 with a horizontal
planee.
PROBLEM 7.1544 A single straain gage formiing an angle   18 with a horizontal
planee is used to deetermine the gaage pressure in
i the cylindriical steel tank shown. The
cylind
drical wall off the tank is 6 mm thick, has a 600-mm
m inside diam
meter, and is
madee of a steel witth E  200 GP
Pa and v  0.30.
Determinne the pressurre in the tank
0
indicaated by a straiin gage reading of 280.
SO
OLUTION
 x  1 
pr
t
1
2

1
 v 
 x  ( x  v y  v z )  1   x  0.85 x
E
E
 2 E
 y  x, z  0
y 
 xy 

1
1

(v x   y  v z )    v  x  0.20 x
E
E
2
 E
 xy
x
G
0
Draaw Mohr’s cirrcle for strain.
x
1
E
2
x
1
R  ( x   y )  0.325
E
2

cos
2
 x   ave
R

a
 ave  ( x   y )  0.525
 (0
0.525  0.325 cos
c 2 )
x
E
tE x
t x

p
r
r (0.525  0.325 cos 2 )
Data:
r
1
1
d  (600)  300 mm  0.300 m
2
2
t  6  103 m E  200  109 Pa,  x  280  10
1 6
p
3
6
9
(6
6  10 )(200  10 )(280  100 )
 1.761  106 Pa
 0.325 cos 70)
(0.300)(0.525
(
  35
p  1.761 MPa 
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1200
PROBL
LEM 7.156
150 MPa
The given state of planne stress is knoown to exist on the surface of
o a machine component.
c
G , determ
mine the direection and
Knowingg that E  200 GPa andd G  77.2 GPa
magnitudde of the threee principal strrains (a) by determining
d
thhe correspondiing state of
strain [usse Eq. (2.43) and Eq. (2.38)] and then using Mohr’ss circle for strrain, (b) by
using Moohr’s circle foor stress to deetermine the principal
p
plannes and princippal stresses
and then determining the
t correspondding strains.
75 MP
Pa
SOLUTION
(a)
 x  0,  y  150  1006 Pa,  xy  75  106 Pa
E  200  109 Pa G  77  109 Pa
E
E
v
 1  0.2987
2(1  v)
2G
1
1
[0  (00.2987)(150  10
 x  ( x  v y ) 
1 6 )]
E
2000  109
 224
1
1
[(1500  106 )  0]
 y  ( y  v x ) 
E
2000  109
 7500
 xy 75  106
 xy 

 974 
G
77  109
G
 xy
2
 4877.0 
1
2
 x   y  974
 ave  ( x   y )  2633
taan 2 a 
 xyy
x   y

974
 1.000
974
2 a  45.0
 a  22.5 
2
2
  x   y    xy 
R 
  689 
 
 2   2 
b  67.5 
 a   ave  R
 a  426  
 b   ave  R
 b  952  
c  
v
(0.2987)(0  150
1  106 )
( x   y )  
E
200  10
1 9
 c  224  
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1201
PROB
BLEM 7.156
6 (Continued
d)
(b)
1
2
 ave  ( x   y )  75 MPa
2
2
x  y 
 0  1150 
2
 752
R 




xy


2
 2 


 1006.07 MPa
 a   ave  R  31.07 MPa
 b   ave  R  1811.07 MPa
1
( a  v b )
E
1

[31.07  106  (0.29987)(181.07  106 )]
200  109
a 
 4226  10 6


tan 2 a 
2 xy
x  y
 a  426  
2 a  45 
 1.0
000 
 a  22.5 

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1202
PR
ROBLEM 7.157
Thee following staate of strain haas been determ
mined on the suurface of a casst-iron machinne part:
 x  720  y  400  xy  660
6 
Knoowing that E  69 GPa annd G  28 GP
Pa, determinee the principaal planes and principal streesses (a) by
deteermining the corresponding
c
g state of planee stress [use Eq.
E (2.36), Eqq. (2.43), and the
t first two equations
e
of
Probb. 2.73] and then
t
using Mohr’s
M
circle for
fo stress, (b) by using Moohr’s circle forr strain to dettermine the
orieentation and magnitude
m
of thhe principal strrains and thenn determining the corresponding stresses.
SOLUTION
Thee 3rd principall stress is  z  0.
E
69
E
1 
 1  0.2321
v
2(1  v)
2G
56
E
6
69
 72.933 GPa

1  v 2 1  (0..232) 2
G
(a)
E
( x  v y )
1  v2
 (72..93  109 )[720  106  (0.2232)(400  1006 )]
x 
 59
9.28 MPa
E
y 
( y  v x )
1  v2
 (72..93  109 )[4000  106  (0.22321)(720  10
1 6 )]
 41.36 MPa
9
6
 xy  G xy
x  (28  10 )(660  10 )
 18.4
48 MPa
1
2
 ave  ( x   y )  500.32 MPa
tan 2b 
2 xy
x  y
 2.06225
2b  64
4.1,
b  32.1,  a  57.9 
2
x  y 
2
R 
Pa
   xy  20.54 MP
2


 a   avee  R
 a  29.8 MPa 
 b   avee  R
 b  700.9 MPa 
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1203
PROB
BLEM 7.157 (Continue
ed)
(b)
1
2
 ave  ( x   y )  560
5 
tan 2b 
 xy
 2.0625
x y
2b  64.1
6 , b  32.1,  a  57.9
5

2
2
  x   y    xy 
R 
 
  366..74 
 2   2 
 a   avve  R  193..26 
 b   avve  R  926..74 
a 
E
( a  v b )
1  v2
 a  29.8
2
MPa 
b 
E
( b  v a )
1  v2
 b  70.9
7
MPa 
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1204
PROBLEM
M 7.158
P
T
1
4
A steel pipe of 12-in. outter diameter iss fabricated frrom 14 -in. -thiick plate by
w a plane peerpendicular
welding alonng a helix thatt forms an anggle of 22.5 with
to the axis off the pipe. Knoowing that a 40-kip
4
axial foorce P and an 80-kip  in.
torque T, eaach directed as shown, arre applied to the pipe, dettermine the
normal and in-plane
i
shearring stresses in
i directions, respectively, normal and
tangential to the weld.
in.
Weld
22.5
SOLUTION
1
d 2  6 in., t  0.25 in.
2
c1  c2  t  5.75 in.
d 2  12 in., c2 


J  c
2


 c   (66
2
A   c22  c12   (62  5.752 )  9.2284
9
in 2
4
2
4
1
4
3
in 4
 5.754 )  318.67
Streesses:
 
P
A
40
 4.33444 ksi
9.22284
Tc
 2
J
(80))(6)

 1.5063 ksi
k
318.67
 x  0,  y  4.33444 ksi,  xy  1..5063 ksi

Chooose the x an
nd y  axes, resspectively, tanngential and noormal to the weld.
w
Theen
 w   y and  w   xy   22.5
 y 
x y
x y
cos 2   xy
x sin 2
2
2
(4.3344) [(4.3344)]

cos 45  1.5063 sin

s 45°
2
2
4
ksi
 4.76
 xy  

 w  4.76 ksi 
x  y
sin 2   xy cos 2
2
[(4.3344)]

s 45  1.50663 cos 45
sin
2
0
ksi
 0.467
 w  0.467
0
ksi 
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1205
100 kN
PROBLEM 7.159
Two steel plates of uniform cross section 10  80 mm are welded
together as shown. Knowing that centric 100-kN forces are applied to
the welded plates and that   25 , determine (a) the in-plane shearing
stress parallel to the weld, (b) the normal stress perpendicular to the
weld.
80 mm
100 kN
SOLUTION
Area of weld:
Aw 
(10  103 )(80  103 )
cos 25
 882.7  106 m 2
(a)
 Fs  0: Fs  100sin 25  0
w 
(b)
Fs
42.26  103

 47.9  106 Pa
6
Aw
882.7  10
 Fn  0: Fn  100 cos 25  0
w 
Fs  42.26 kN
 w  47.9 MPa 
Fn  90.63 kN
Fn
90.63  103

 102.7  106 Pa
Aw
882.7  106
 w  102.7 MPa 
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1206
100 kN
PROBLEM 7.160
Two steel plates of uniform cross section 10  80 mm are welded
together as shown. Knowing that centric 100-kN forces are applied to
the welded plates and that the in-plane shearing stress parallel to the
weld is 30 MPa, determine (a) the angle , (b) the corresponding normal
stress perpendicular to the weld.
80 mm
100 kN
SOLUTION
Area of weld:
Aw 

(a)
 Fs  0: Fs  100sin   0
w 
Fs
Aw
30  106 
sin  cos  
(b)

(10  103 )(80  103 )
cos 
800  106 2
m
cos 
Fs  100sin  kN  100  103 sin  N
100  103 sin 
 125  106 sin  cos 
800  106 / cos 
1
30  106
sin 2 
 0.240
2
125  106
 Fn  0: Fn  100 cos   0
  14.34 
Fn  100cos14.34  96.88 kN 
Aw 
800  106
 825.74  106 m 2
cos14.34
 
Fn
96.88  103

 117.3  106 Pa 
6
Aw
825.74  10
  117.3 MPa 
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1207
0
+
0
PROBLEM 7.161
Determine the principal planes and the principal
stresses for the state of plane stress resulting from the
superposition of the two states of stress shown.
SOLUTION
Mohr’s circle for 2nd stress state:
1
1
2
2
1
1
 y   0   0 cos 2
2
2
1
 xy   0 sin 2
2
 x   0   0 cos 2
Resultant stresses:
1
2
1
2
3
2
1
2
 x   0   0   0 cos 2   0   0 cos 2
1
1
1
2
2
2
1
1
 xy  0   0 sin 2   0 sin 2
2
2
1
 ave  ( x   y )   0
2
2 xy
 0 sin 2
tan 2 p 

 x   y  0   0 cos 2
1
2
 y  0   0   0 cos 2   0   0 cos 2

sin 2
 tan 
1  cos 2
1
2
p   
2
2
2
x  y 
1
1
 1

2
R 
   xy    0   0 cos 2     0 sin 2 
2
2
2
 2



p 
(   )

2
1
2
  0 1  2 cos 2 + cos 2 2  sin 2 2 
 0 1  cos 2   0 |cos  |
2
2
 a   ave  R
 a   0   0 cos  
 b   ave  R
 b   0   0 cos  
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1208
PROBLEM 7.162
y
For the state of stress shown, determine the maximum shearing stress
when (a)  z  4 ksi, (b)  z  4 ksi, (c)  z  0.
2 ksi
6 ksi
σz
7 ksi
z
x
SOLUTION
 x  7 ksi,
 y  2 ksi,
 xy  6 ksi
1
2
 ave  ( x   y )  4.5 ksi
2
x  y 
2
R 
   xy
2


 2.52  (6) 2  6.5 ksi
 a   ave  R  11 ksi
 b   ave  R  2 ksi
(a)
 z  4 ksi,  a  11 ksi,  b  2 ksi
1
2
 max  11 ksi,  min  2 ksi,  max  ( max   min )
(b)
 z  4 ksi,  a  11 ksi,  b  2 ksi
 max  11 ksi,  min  4 ksi,
(c)
 max  6.50 ksi 
1
2
 max  7.50 ksi 
1
2
 max  6.50 ksi 
 max  ( max   min )
 z  0,  a  11 ksi,  b  2 ksi
 max  11 ksi,  min  2 ksi,
 max  ( max   min )
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1209
PROBLEM 7.163
y
For the state of stress shown, determine the value of  xy for which the
40 MPa
maximum shearing stress is (a) 60 MPa, (b) 78 MPa.
τ xy
100 MPa
z
x
SOLUTION
 x  100 MPa,  y  40 MPa,  z  0
1
2
 ave  ( x   y )  70 MPa
(a)
 max  60 MPa.
If  z is  min , then  max   min  2 max .
 max  0  (2)(60)  120 MPa
 max   ave  R
R   max   ave  120  70  50 MPa
 b   max  2 R  20 MPa > 0
2
x  y 
2
2
2
R 
   xy  30   xy  50 MPa
2


 xy  40.0 MPa 
 xy  502  302
(b)
 max  78 MPa.
If  z is  min , then  max   min  2 max  0  (2)(78)  156 MPa.
 max   ave  R
R   max   ave  156  70  86 MPa >  max  78 MPa
Set
R   max  78 MPa.
 min   ave  R  8 MPa < 0
2
x  y 
2
2
2
R 
   xy  30   xy
2


 xy  72.0 MPa 
 xy  782  302
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1210
PROBLEM 7.164
14 ksi
xy
24 ksi
The state of plane stress shown occurs in a machine component made of a
steel with  Y  30 ksi. Using the maximum-distortion-energy criterion,
determine whether yield will occur when (a)  xy  6 ksi, (b)  xy  12 ksi,
(c)  xy  14 ksi. If yield does not occur, determine the corresponding factor
of safety.
SOLUTION
 x  24 ksi
1
2
 ave  ( x   y )  19 ksi
For stresses in xy-plane,
(a)
 y  14 ksi
z  0
x  y
2
 5 ksi
 xy  6 ksi
2
x  y 
2
2
2
R 
   xy  (5)  (6)  7.810 ksi
2


 a   ave  R  26.810 ksi,  b   ave  R  11.190 ksi
 a2   b2   a b  23.324 ksi < 30 ksi
F .S . 
(b)
30
23.324
(No yielding)
F .S .  1.286 
 xy  12 ksi
2
x  y 
2
2
2
R 
   xy  (5)  (12)  13 ksi
2


 a   ave  R  32 ksi,  b   ave  R  6 ksi
 a2   b2   a b  29.462 ksi < 30 ksi
F .S . 
(c)
30
29.462
(No yielding)
F .S .  1.018 
 xy  14 ksi
2
x  y 
2
2
2
R 
   xy  (5)  (14)  14.866 ksi
2


 a   ave  R  33.866,  b   ave  R  4.134 ksi
 a2   b2   a b  32.00 ksi > 30 ksi
(Yielding occurs) 
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1211
PROBLEM 7.165
750 mm
750 mm
The compressed-air tank AB has an inner diameter of 450 mm and a
uniform wall thickness of 6 mm. Knowing that the gage pressure
inside the tank is 1.2 MPa, determine the maximum normal stress
and the maximum in-plane shearing stress at point a on the top of the
tank.
b
a
B
D
A
5 kN
500 mm
SOLUTION
Internal pressure:
Torsion:
1
d  225 mm t  6 mm
2
pr
(1.2)(225)
1 

 45 MPa
t
6
pr
2 
 22.5 MPa
2t
r 
c1  225 mm, c2  225  6  231 mm
J 

2
c
4
2

 c14  446.9  106 mm 4  446.9  106 m 4
T  (5  103 )(500  103 )  2500 N  m
 
Tc (2500)(231  103 )

J
446.9  106
 1.29224  106 Pa  1.29224 MPa
Transverse shear:
  0 at point a.
Bending:
I 
At point a,
1
J  223.45  106 m 4 , c  231  103 m
2
M  (5  103 )(750  10 3 )  3750 N  m
 
Mc (3750)(231  103 )

 3.8767 MPa
I
223.45  106
Total stresses (MPa).
Longitudinal:
 x  22.5  3.8767  26.377 MPa
Circumferential:
 y  45 MPa
Shear:
 xy  1.29224 MPa
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1212
PROBLEM 7.165 (Continued)
 ave 
1
( x   y )  35.688 MPa
2
2
R
x   y 
2

   xy  9.4007 MPa
2


 max   ave  R  45.1 MPa
 max(in-plane)  R  9.40 MPa
 max  45.1 MPa 
 max (in-plane)  9.40 MPa 
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1213
PROBLEM 7.166
750 mm
750 mm
b
a
D
A
5 kN
500 mm
B
For the compressed-air tank and loading of Prob. 7.165, determine
the maximum normal stress and the maximum in-plane shearing
stress at point b on the top of the tank.
PROBLEM 7.165 The compressed-air tank AB has an inner
diameter of 450 mm and a uniform wall thickness of 6 mm. Knowing
that the gage pressure inside the tank is 1.2 MPa, determine the
maximum normal stress and the maximum in-plane shearing stress at
point a on the top of the tank.
SOLUTION
Internal pressure:
Torsion:
1
d  225 mm t  6 mm
2
pr
(1.2)(225)

 45 MPa
1 
t
6
pr
 22.5 MPa
2 
2t
r 
c1  225 mm, c2  225  6  231 mm
J 

2
c
4
2

 c14  446.9  106 mm 4  446.9  106 m 4
T  (5  103 )(500  103 )  2500 N  m
 
Tc (2500)(231  103 )

 1.29224  106 Pa  1.29224 MPa
J
446.9  106
Transverse shear:
  0 at point b.
Bending:
I 
At point b,
1
J  223.45  106 m 4 , c  231  103 m
2
M  (5  103 )(2  750  103 )  7500 N  m
 
Mc (7500)(231  103 )

 7.7534 MPa
I
223.45  106
Total stresses (MPa).
Longitudinal:
 x  22.5  7.7534  30.253 MPa
Circumferential:
 y  45 MPa
Shear:
 xy  1.29224 MPa
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
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1214
PROBLEM 7.166 (Continued)
 ave 
1
( x   y )  37.626 MPa
2
2
R
x   y 
2

   xy  7.4859 MPa
2


 max   ave  R  45.1 MPa
 max (in-plane)  R  7.49 MPa
 max  45.1 MPa 
 max (in-plane)  7.49 MPa 
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1215
0.12 in.
A
PROBLEM 7.167
The brass pipe AD is fitted with a jacket used to apply a hydrostatic pressure of 500 psi
to portion BC of the pipe. Knowing that the pressure inside the pipe is 100 psi,
determine the maximum normal stress in the pipe.
B
0.15 in.
C
D
2 in.
4 in.
SOLUTION
The only stress to be considered is the hoop stress. This stress can be obtained by applying
1 
pr
t
Using successively the inside and outside pressures (the latter of which causes a compressive stress),
pi  100 psi, ri  1  0.12  0.88 in., t  0.12 in.
( max )i 
pi ri
(100)(0.88)

 733.33 psi
t
0.12
po  500 psi, ro  1 in.,
t  0.12 in.
po ro
(500)(1)

 4166.7 psi
t
0.12
 733.33  4166.7  3433.4 psi
( max )o  
 max
 max  3.43 ksi (compression) 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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1216
0.12 in.
A
PROB
BLEM 7.168
8
For the assembly of Prob. 7.167, determine the normal streess in the jackket (a) in a
o the jacket, (b) in a directtion parallel
directionn perpendicular to the longitudinal axis of
to that axis.
a
B
PROBL
LEM 7.167 Thhe brass pipe AD
A is fitted with
w a jacket ussed to apply a hydrostatic
pressuree of 500 psi too portion BC of the pipe. Knnowing that thee pressure inside the pipe
is 100 psi, determine the
t maximum
m normal stresss in the pipe.
0.15 in.
C
D
2 in.
4 in.
SOLUTION
(a)
Hoop stress.
0 1.85 in.
p  500 psii, t  0.15 inn., r  2  0.15
(1) 
((500)(1.85)
pr
 6166.7 psi

0.15
t
 1  6.17 ksi 
(b)
Longitudin
nal stress.
Free body of portion of jacket
j
above a horizontal seection, consideering vertical forces
f
only:
 Fy  0:
A f  A  2 dA j  0
A f p dA
j
pA f   2 A j  0
2  p

  r

   [(2)
Af
Aj
(1)
A f   r22  r12   [(1.85) 2  (11)2 ]  7.6105 in
i 2
Areas :
Aj
2
3
 r22
2
 (1.855)2 ]  1.814277 in 2
Recalling Eq.
E (1),
2  p
Af
Aj
 (500)
7.6105
 20097.4 psi
1
1.81427
 2  2.10
2
ksi 
PRO
OPRIETARY MAT
TERIAL. Copyrigght © 2015 McGrraw-Hill Educatioon. This is proprietary material soleely for authorizedd instructor use.
Not authorized
a
for salee or distribution inn any manner. Thiis document may not
n be copied, scannned, duplicated, forwarded, distribbuted, or posted
on a website,
w
in whole or part.
1217
PROBLEM
P
7.169
2
1
3
45⬚
Determine
D
the largest in-planne normal straain, knowing that the follow
wing strains
haave been obtaiined by the usse of the rosettte shown:
1  50  1006 in./in.
45⬚
x
 2  360  106 in./in.
 3  315  10
1 6 in./in.
SO
OLUTION
1  455,  2  45, 3  0
 x cos 2 1   y sin 2 1   xy sin 1 cos 1  1
0  xy  50  106
0.55 x  0.5 y  0.5
(1)
 x cos 2  2   y sin 2  2   xy sin  2 cos  2   2
0.55 x  0.5 y  0.5 xy  360  106
(2)
 x cos 2 3   y sin 2 3   xy sin 3 cos 3   3
 x  0  0  315  10 6
 x  315
3  106 in.//in.
Froom (3),
 xy  50 106  360  106  410
4 106 in./iin.
Eq. (1)  Eq. (2):
Eq. (1)  Eq. (2):
(3)
 x   y  1   2
 y  1   2   x  50  106  360
3  106  3115  10 6  5  106 in./in.
1
2
 ave  ( x   y )  155
1  106 in./inn.
2
R
  xy 
 x   y 


 
 2 
 2 
2
2

 410  106 
 315  106  5  106 
  


2
2




2
 260  106 in../in.
 max   ave  R  1555  106  260  106
 max  415  1106 in./in. 
PRO
OPRIETARY MATERIAL. Copyriight © 2015 McG
Graw-Hill Educattion. This is proprrietary material soolely for authorizedd instructor use.
Not authorized for salle or distribution in
i any manner. Thhis document may not be copied, scaanned, duplicated,, forwarded, distributed, or posted
on a website, in wholee or part.
1218
y'
y
␴y
PROBLEM 7.C1
1
y
␪
␴y'
␶x'y'
␶xy
x
Q
␴x
z
␪
␴x'
Q
x
x'
x
z
(a)
(
(b)
A statee of plane streess is defined by
b the stress
compoonents  x ,  y , and  xy assoociated with
the eleement shown inn Fig. P7.C1a. (a) Write a
compuuter program that can be
b used to
calculaate the stress components  x y , and
 x y  asssociated withh the element after it has
rotatedd through an angle
a
 abouut the z axis
(Fig. P.7C1
P
b). (b) Use
U this prograam to solve
Probs. 7.13 through 7.16.
7
SOLUTION
Proggram followin
ng equations:
x y
Equuation (7.5), Paage 427:
 x 
Equuation (7.7), Paage 427:
 y 
Equuation (7.6), Paage. 427:
 xy  
2
x y
2

2
x  y

x  y
2
x  y
2
c 2   xy sinn 2
cos
c 2   xy sinn 2
cos
sin 2   xy
x cos 2
Enteer  x ,  y ,  xyy and 
Prinnt values obtaiined for  x ,  y and  xy
Prooblem Outputts
Probblem 7.13
 x  0 ksi
 x  8 ksi
 xy  5 ksi
Rootation of elem
ment
(+ counterclockw
c
wise)
  25
Rotation off element
(+ countercllockwise)
  10
 x  2.40 ksi
 y  10.40 ksi
 x  1.995 ksi
 y  6.05 ksi
 xy  0.15 ksi
 xy  6..07 ksi
PRO
OPRIETARY MAT
TERIAL. Copyrigght © 2015 McGrraw-Hill Educatioon. This is proprietary material soleely for authorizedd instructor use.
Not authorized
a
for salee or distribution inn any manner. Thiis document may not
n be copied, scannned, duplicated, forwarded, distribbuted, or posted
on a website,
w
in whole or part.
1219
PROB
BLEM 7.C1 (Continued
d)
Prooblem 7.14
 x  60 MPa
M
 y  90 MPa
Pa
 xy  30 MP
Ro
otation of Elem
ment
( counterclockw
wise)
  25
Rotation off Element
( countercclockwise)
  10
1
 x  56.19 MP
Pa
 y  86.19 MP
Pa
 x  455.22 MPa
 y  755.22 MPa
 xy  38.17 MP
Pa
 xy  53.84 MPa
Prooblem 7.15
 x  8 ksii
 y  12 ksi
 xy  6 ksi
Rotation off Element
( countercclockwise)
  10
1
Ro
otation of Elem
ment
( counterclockw
wise)
  25
 x  9.02 kssi
 y  13.02 kssi
 x  5.344 ksi MPa
 y  9.344 ksi MPa
 xy 
 xy  9.066 ksi MPa
3.80 kssi
Prooblem 7.16
 x  0 MPa
M
 y  80 MPa
M
 xy  50 MPa
M
Ro
otation of Elem
ment
( counterclockw
wise)
  25
Rotation off Element
( countercllockwise)
  10
 x  24.01 MPa
M
 y  104.01 MPa
M
 x  19.51 MPa
 y  60..49 MPa
 xy   1.50 MPa
M
 xy  60.67 MPa
PRO
OPRIETARY MATERIAL. Copyriight © 2015 McG
Graw-Hill Educattion. This is proprrietary material soolely for authorizedd instructor use.
Not authorized for salle or distribution in
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on a website, in wholee or part.
1220
PROBLE
EM 7.C2
y
␴y
␶xy
x
Q
x
␴x
A state of plane
p
stress iss defined by thhe stress compponents  x ,  y , and  xy
associated with the elem
ment shown inn Fig. P7.C1aa. (a) Write a computer
program thhat can be used to calcullate the princcipal axes, thhe principal
stresses, thee maximum inn-plane sheariing stress, andd the maximuum shearing
stress. (b) Use
U this prograam to solve Prrobs. 7.5, 7.9, 7.68, and 7.699.
z
SOLUTION
Proggram followin
ng equations:
 ave 
Equuation (7.10)
x y
2
Equuation (7.14)
 max   ave  R
J min   ave  R
Equuation (7.12)
 p  taan 1
Equuation (7.15)
 s  taan 1 
Sheearing stress:
Theen
2
 x  y 
2
: R 
   xy
2


2 xy
x  y
x  y
2 xy
If  max
 0 and  min
m
m  0:
 maax(in-plane)  R;  max(out-of-planne)  R
If  max
 0 and  min
m
m  0:
Theen
If  max
m  0 and  min
m
Theen
1
2
 0:
 max(in-plane)
 R;  max(out-of-planne)   max
m
1
2
 max(in-plane)
 R;  max(out-of-plaane)  | min |
m
Proogram Outputts
Probblems 7.5 and
d 7.9
 x  60.00 MPa
 y  40.00 MPa
 xy  35.00 MPa
PRO
OPRIETARY MAT
TERIAL. Copyrigght © 2015 McGrraw-Hill Educatioon. This is proprietary material soleely for authorizedd instructor use.
Not authorized
a
for salee or distribution inn any manner. Thiis document may not
n be copied, scannned, duplicated, forwarded, distribbuted, or posted
on a website,
w
in whole or part.
1221
PROB
BLEM 7.C2 (Continued
d)
Proogram Outpu
uts (Continued
d)
Anngle between xy
x axes and priincipal axes (+
+ counterclockkwise):
 p  37.003 and 522.97°
 max  13.660 MPa
 min  86.440 MPa
x axis and plaanes of maxim
mum in-plane shearing
s
stresss (+ countercloockwise):
Anngle between xy
 s  7.97 and 97.977°
 max (in-plane)  36.400 MPa
 max  43.200 MPa
Prooblem 7.68
 x  140.000 MPa
 y  40.000 MPa
 xy  80.000 MPa
Anngle between xy
x axes and priincipal axes (
 counterclockkwise):
 p  29.000 and 1199°
 max  1844.34 MPa
 min  4.344 MPa
Anngle between xy
x axis and plaanes of maxim
mum in-plane in-plane sheariing stress ( counterclockw
c
wise):
 s  74.000 and 1644.00°
 max (in-plane)
 94.334 MPa
(
 max (out--of-plane)  94.334 MPa
 x  140.000 MPa
 y  120.000 MPa
 xy  80.000 MPa
x axes and priincipal axes (+
+ counterclockkwise):
Anngle between xy
 p  41.444 and 1311.44°
 max  210.62 MPa
 min  49.338 MPa
x axis and plaanes of maxim
mum in-plane in-plane sheariing stress (+ counterclockw
c
wise):
Anngle between xy
 s  86.444 and 1766.44°
 80.662 MPa
 max (in-plane)
(
 max (out--of-plane)  105.331 MPa
PRO
OPRIETARY MATERIAL. Copyriight © 2015 McG
Graw-Hill Educattion. This is proprrietary material soolely for authorizedd instructor use.
Not authorized for salle or distribution in
i any manner. Thhis document may not be copied, scaanned, duplicated,, forwarded, distributed, or posted
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1222
PROBLEM 7.C2 (Continued)
Program Outputs (Continued)
Problem 7.69
 x  140.00 MPa
 y  20.00 MPa
 xy  80.00 MPa
Angle between xy axes and principal axes (+ counterclockwise):
 p  26.57 and 116.57
 max  180.00 MPa
 min  20.00 MPa
Angle between xy axis and planes of maximum in-plane in-plane shearing stress (+ counterclockwise):
 s  71.57 and 161.57
 max (in-plane)  100.00 MPa
 max (out-of-plane)  100.00 MPa
 x  140.00 MPa
 y  140.00 MPa
 xy  80.00 MPa
Angle between xy axes and principal axes (+ counterclockwise):
 p  45 and 135.00
 max  220.00 MPa
 min  60.00 MPa
Angle between xy axis and planes of maximum in-plane in-plane shearing stress ( counterclockwise):
 s  90.00 and 180.00°
 max (in-plane)  80.00 MPa
 max (out-of-plane)  110.00 MPa 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1223
PR
ROBLEM 7.C3
7
(a) Write a com
mputer program
m that, for a given
g
state off plane stress and a given yield
y
strength of a ductile
mine whether the
t material will
w yield. The program shouuld use both thhe maximum
maaterial, can be used to determ
sheearing-strengthh criterion and
d the maximum
m-distortion-ennergy criterionn. It should allso print the values
v
of the
prinncipal stressess and, if the material
m
does noot yield, calculate the factor of safety. (b) Use this proggram to solve
Proobs. 7.81, 7.82, and 7.164.
SO
OLUTION
Principal stressess.
 avve 
x y
2
2
x y 
2
; R 
   xy
2


 a   ave  R
 b   ave  R
Maaximum-shearring-stress criterion.
1
2
y  y
1
2
 maax   a
If  a and  b havve same sign,
If  max   y , yieelding occurs.
If  max   y , no yielding occu
urs, and factor of safety 
y
 max
m
Maaximum-distorrtion-energy criterion.
Compute radicall   a2   a b   b2
r
  y , yielding occu
urs.
If radical
If radical
r
  y , no yielding occcurs, and facttor of safety 
y
Radical
Proogram Outpu
uts
Prooblems 7.81a and
a 7.82a
Yield strengthh  325 MPa
 x  200.00 MPa
M
 y  200.00 MPa
M
 xy  100.00 MPa
M
 maxx  100.00 MPa
M
 minn  300.00 MPa
M
PRO
OPRIETARY MATERIAL. Copyriight © 2015 McG
Graw-Hill Educattion. This is proprrietary material soolely for authorizedd instructor use.
Not authorized for salle or distribution in
i any manner. Thhis document may not be copied, scaanned, duplicated,, forwarded, distributed, or posted
on a website, in wholee or part.
1224
PROB
BLEM 7.C3 (Continued
d)
Proogram Outputts (Continuedd)
Usinng the maximu
um-shearing-sstress criterionn,
matterial will not yield.
y
F .S .  1.0083
Usinng the maximu
um-distortion-energy criteriion,
matterial will not yield.
y
F .S .  1.2228
a 7.82b
Probblems 7.81b and
Yield strenngth  325 MP
Pa
 x  240.000 MPa
 y  240.000 MPa
 xy  100.00 MPa
 max  140.000 MPa
 min  340.000 MPa
um-shearing-sstress criterionn,
Usinng the maximu
matterial will yield
d.
Usinng the maximu
um-distortion-energy criteriion,
matterial will not yield.
y
F .S .  1.098
Probblems 7.81c and 7.82c
Yield strenngth  325 MP
Pa
 x  280.000 MPa
 y  280.000 MPa
 xy  100.000 MPa
 max  180.000 MPa
 min  380.000 MPa
um-shearing-sstress criterionn,
Usinng the maximu
matterial will yield
d.
Usinng the maximu
um-distortion-energy criteriion,
matterial will yield
d.
Probblem 7.164a
Yield strenngth  30 ksi
 x  24.00 ksi
k
k
 y  14.00 ksi
 xy  6.00 kksi
 max  26.81 ksi
k
 min  11.19 ksi
k
PRO
OPRIETARY MAT
TERIAL. Copyrigght © 2015 McGrraw-Hill Educatioon. This is proprietary material soleely for authorizedd instructor use.
Not authorized
a
for salee or distribution inn any manner. Thiis document may not
n be copied, scannned, duplicated, forwarded, distribbuted, or posted
on a website,
w
in whole or part.
1225
PROB
BLEM 7.C3 (Continued
d)
Proogram Outpu
uts (Continued
d)
(a)
Using the maximum-sheearing-stress criterion,
c
material will
w not yield.
F .S .  1.119
(b)
Using the maximum-disstortion-energgy criterion,
material will
w not yield.
F .S .  1.286
Prooblem 7.164b
Yield strenngth  30 ksi
 x  24.00 ksi
k
 y  14.00 kssi
k
 xy  12.00 ksi
 max  32.00 ksi
k
 min  6.00 kssi
(a)
Using the maximum-sheearing-stress criterion,
c
material will
w yield.
(b)
Using the maximum-disstortion-energyy criterion,
material will
w not yield.
F .S .  1.018
Prooblem 7.164c
Yield strength  30 ksi
 x  24.00 kssi
 y  14.00 kssi
 xy  14.00 kssi
 max  33.87 kssi
 min  4.13 kssi
(a)
Using the maximum-sheearing-stress criterion,
c
material will
w yield.
(b)
Using the maximum-disstortion-energyy criterion,
material will
w yield.

PRO
OPRIETARY MATERIAL. Copyriight © 2015 McG
Graw-Hill Educattion. This is proprrietary material soolely for authorizedd instructor use.
Not authorized for salle or distribution in
i any manner. Thhis document may not be copied, scaanned, duplicated,, forwarded, distributed, or posted
on a website, in wholee or part.
1226
PR
ROBLEM 7.C4
(a) Write
W
a computer program based on Mohhr’s fracture criterion
c
for brrittle materialss that, for a givven state of
planne stress and given
g
values of
o the ultimatee strength of thhe material inn tension and compression,
c
c be used
can
to determine
d
wheether rupture will
w occur. Thhe program shhould also print the values of the princippal stresses.
(b) Use
U this progrram to solve Probs.
P
7.89 andd 7.90 and to check
c
the answ
wers to Probs. 7.93 and 7.944.
SOLUTION
Prinncipal stresses.
 avee 
x y
2
2
x  y 
2
R 
   xy
2


 a   ave  R
 b   ave  R
c
Mohhr’s fracture criterion.
If  a and  b hav
ve same sign, and
a
 a   UT and  b   UC , no faailure;
 a   UT or  b   UC , failuree.
If  a  0 and  b  0 :
Connsider fourth quadrant
q
of Figgure 7.47.
For no rupture to occur, point ( a ,  b ) mustt lie within Moohr’s envelope (Figure 7.477).
n,
If  b  Criterion
thenn rupture occu
urs.
If  b  Criterion
n,
thenn no rupture occcurs.
PRO
OPRIETARY MAT
TERIAL. Copyrigght © 2015 McGrraw-Hill Educatioon. This is proprietary material soleely for authorizedd instructor use.
Not authorized
a
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n be copied, scannned, duplicated, forwarded, distribbuted, or posted
on a website,
w
in whole or part.
1227
PROB
BLEM 7.C4 (Continued
d)
Proogram Outpu
uts
Prooblem 7.89
 x  10.00 MPa
 y  100.00 MPa
 xy  60 MPa
 80 MPaa
Ulttimate strengthh in tension
Ulttimate strengthh in compressiion  200 MP
Pa
 max   a 336.39 MPa
 min   b  1226.39 MPa
Rupture will occcur
 x  32.000 MPa
 y  0.00 MPa
M
Prooblem 7.90
 xy  75.00 MPa
M
 80 MP
P
Ulttimate strengthh in tension
Ulttimate strengthh in compressiion  200 M
MP
 max   a 60.69 MPa
 min   b  92.69 MPa
Rupture will not occur.
To check answerrs to the follow
wing problemss, we check foor rupture usinng given answeers and an adjacent value.
Annswer:
Ruppture occurs for
f  0  3.67 ksi.
k
 x  8.00 ksii
 y  0.00 ksii
Prooblem 7.93
  xy  3.67 ksii 
Ulttimate strengthh in tension
 10 ksi
Ulttimate strengthh in compressiion  25 ksi
 max   a 9.443 ksi
 min   b  1.443 ksi
PRO
OPRIETARY MATERIAL. Copyriight © 2015 McG
Graw-Hill Educattion. This is proprrietary material soolely for authorizedd instructor use.
Not authorized for salle or distribution in
i any manner. Thhis document may not be copied, scaanned, duplicated,, forwarded, distributed, or posted
on a website, in wholee or part.
1228
PROB
BLEM 7.C4 (Continued
d)
Proogram Outputts (Continuedd)
Ruppture will not occur.
o
 x  8.00 ksi
 y  0.00 ksi
  xy  3.68 ksi 
Ultiimate strength
h in tension
 10 ksi
Ultiimate strength
h in compressioon  25 ksi
 max
4 ksi
m   a 9.44
 min
4 ksi
m   b  1.44
Ruppture will occu
ur.
Ansswer:
Ruppture occurs fo
or  0  49.1 MPa.
M
Probblem 7.94
 x  80.00 MPa
 y  0.00 MP
Pa
  xy  49.10 MPa
M 
Ultiimate strength
h in tension
 75 MPaa
Ultiimate strength
h in compressioon  150 MP
Pa
 max   a  23.33 MPa
 min   b  1103.33 MPa
Ruppture will not occur.
o
 x  80.00 MPa
 y  0.00 MP
Pa
  xy  49.20 MPa
M 
Ultiimate strength
h in tension
 75 MPaa
Ultiimate strength
h in compressioon  150 MP
Pa
 max   a  233.41 MPa
 min   b  103.41
1
MPa
Ruppture will occu
ur.
PRO
OPRIETARY MAT
TERIAL. Copyrigght © 2015 McGrraw-Hill Educatioon. This is proprietary material soleely for authorizedd instructor use.
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a
for salee or distribution inn any manner. Thiis document may not
n be copied, scannned, duplicated, forwarded, distribbuted, or posted
on a website,
w
in whole or part.
1229
PROBLEM 7.C5
y
y'
x'
x
A state of plane strain is defined by the strain components  x ,  y ,
and  xy associated with the x and y axes. (a) Write a computer
program that can be used to calculate the strain components  x ,  y  ,
and  xy associated with the frame of reference xy obtained by
rotating the x and y axes through an angle θ. (b) Use this program to
solve Probs 7.129 and 7.131.
SOLUTION
Program following equations:
x  y
x   y
1
cos 2   xy sin 2
2
Equation (7.44):
 x 
Equation (7.45):
 y 
Equation (7.46):
 xy  ( x   y )sin 2   xy cos 2
Enter
2
x  y
2


2
x  y
2
1
sin 2   xy cos 2
2
 x ,  y ,  xy , and  .
Print values obtained for  x ,  y , and  xy .
Program Outputs
Problem 7.129
 x  240 micro meters
 y  160 micro meters
 xy  150 micro radians
Rotation of element, in degrees (+ counterclockwise):
  60
 x  115.05 micro meters
 y  284.95 micro meters
 xy   5.72 micro radians
Problem 7.131
x 
0 micro meters
 y  320 micro meters
 xy  100 micro radians
Rotation of element, in degrees (+ counterclockwise):
  30
 x  36.70 micro meters
 y  283.30 micro meters 
 xy  227.13 micro radians
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1230
PROBLEM 7.C6
A state of strain is defined by the strain components  x ,  y , and  xy associated with the x and y axes.
(a) Write a computer program that can be used to determine the orientation and magnitude of the principal
strains, the maximum in-plane shearing strain, and the maximum shearing strain. (b) Use this program to
solve Probs 7.136 through 7.139.
SOLUTION
Program following equations:
 ave 
Equation (7.50):
x  y
2
2
  x   y    xy 
R 
 

 2   2 
2
 max   ave  R  min   ave  R
Equation (7.51):
 p  tan 1
Equation (7.52):
Shearing strains:
 xy
x   y
Maximum in-plane shearing strain
 max (in-plane)  2R
Calculate out-of-plane shearing strain and check whether it is the maximum shearing strain.
Let  a   max
 b   min
Calculate
c  
v
( a   b )
1 v
If  a   b   c ,  out-of-plane   a   c
If  a   c   b ,  out-of-plane   a   b  2 R
If  c   a   b ,  out-of-plane   c   b
Program Printout
Problem 7.136
 x  260 micro meters
 y  60 micro meters
 xy  480 micro radians
  0.333
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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on a website, in whole or part.
1231
PROBLEM 7.C6 (Continued)
Program Printout (Continued)
Angle between xy axes and principal axes (counterclockwise):
 p  33.69
 a  100.00 micro meters
 b  420.00 micro meters
 c  159.98 micro meters
 max (in-plane)  520.00 microradians
 max  579.98 microradians
 x   600 micrometers
 y  400 micrometers
Problem 7.137
 xy  350 microradians
  0.333
Angle between xy axes and principal axes (+ = counterclockwise):
 p  30.13
 a  298.44 micrometers
 b  701.56 micrometers
 c  500.00 micrometers
 max(in-plane)  403.11 microradians
 max  1201.56 microradians
 x  160 micrometers
 y  480 micrometers
Problem 7.138
 xy  600.00 microradians
  0.333
Angle between xy axes and principal axes ( counterclockwise):
 p  21.58
 a  278.63 micrometers
 b  598.63 micrometers
 c  159.98 micrometers
 max(in-plane)  877.27 microradians
 max  877.27 microradians
Problem 7.139
 x  30 micrometers
 y  570 micrometers
 xy  720 microradians
  0.333
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1232
PROBLEM 7.C6 (Continued)
Angle between xy axes and principal axes (  counterclockwise):
 p   26.57
 a  750.00 micrometers
 b  150.00 micrometers
 c  300.00 micrometers
 max(in-plane)  900.00 microradians

 max  1050.00 microradians
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1233
PROBLEM 7.C7
A state of plane strain is defined by the strain components  x ,  y , and  xy measured at a point. (a) Write a
computer program that can be used to determine the orientation and magnitude of the principal strains, the
maximum in-plane shearing strain, and the magnitude of the shearing strain. (b) Use this program to solve
Probs 7.140 through 7.143.
SOLUTION
Program following equations:
x   y
2
  x   y    xy 
R 
 

2

  2 
Equation (7.50)
 ave 
Equation (7.51)
 max   ave  R  min   ave  R
2
 p  tan 1
Equation (7.52)
Shearing strains:
2
 xy
x   y
Maximum in-plane shearing strain
 xy (in-plane)  2 R
Calculate out-of-plane-shearing strain and check whether it is the maximum shearing strain.
Let  a   max
 b   min
 c  0 (Plain strain)
If  a   b   c ,  out-of-plane   a   c
If  a   c   b ,  out-of-plane   a   b  2 R
If  c   a   b ,  out-of-plane   c   b
Program Printout
 x  60 micrometers
 y  240 micrometers
Problem 7.140
 xy  50 microradians
  0.000
Angle between xy axes and principal axes (+ = counterclockwise):
 p  7.76 and  82.24
 a  243.41 micrometers
 b  56.59 micrometers
 c  0.00 micrometers
 max(in-plane)  186.82 microradians
 max  243.41 microradians
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1234
PROBLEM 7.C7 (Continued)
Program Printout (Continued)
x
y
 xy

Problem 7.141
 400 micrometers
 200 micrometers
 375 microradians
 0.000
Angle between xy axes and principal axes (  counterclockwise):
p
a
b
c
 30.96 and 59.04
 512.50 micrometers
 87.50 micrometers
 0.00 micrometers
 425.00 microradians
 max(in-plane)
 max  512.50 microradians
Problem 7.142
 x  300 micrometers
 y  60 micrometers
 xy  100 microradians
  0.000
Angle between xy axes and principal axes (+ = counterclockwise):
 p  11.31 and  78.69
 a  310.00 micrometers
 b  50.00 micrometers
 c  0.00 micrometers
 max(in-plane)  260.00 microradians
 max  310.00 microradians
 x  180 micrometers
 y  260 micrometers
Problem 7.143
 xy  315 microradians
  0.000
Angle between xy axes and principal axes (+ = counterclockwise):
 p  37.87 and  52.13
 a  57.50 micrometers
 b  382.50 micrometers

 c  0.00 micrometers
 max(in-plane)  325.00 microradians
 max  382.50 microradians
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1235
PROBLEM 7.C8
A rosette consisting of three gages forming, respectively, angles of 1 ,  2 , and 3 with the x axis is attached
to the free surface of a machine component made of a material with a given Poisson’s ratio v. (a) Write a
computer program that, for given readings 1 ,  2 , and  3 of the gages, can be used to calculate the strain
components associated with the x and y axes and to determine the orientation and magnitude of the three
principal strains, the maximum in-plane shearing strain, and the maximum shearing strain. (b) Use this
program to solve Probs 7.144, 7.145, 7.146, and 7.169.
SOLUTION
For n  1 to 3, enter  n and  n .
Enter: NU  V
Solve Equation (7.60) for  x ,  y , and  xy using method of determinates or any other method.
2
x  y 
2
; R 
   xy
2
2


 a   max   ave  R
 ave 
Enter
x   y
 b   max   avg  R
c  
V
( a   b )
1V
1
2
 p  tan 1
Shearing strains:
 xy
x   y
Maximum in-plane shearing strain
 max (in  plane)  2R
Calculate out-of-plane shearing strain, and check whether it is the maximum shearing strain.
If  c   b ,
 out-of-plane   a   c
If  c   a ,
 out-of-plane   c   b
Otherwise,
 out-of-plane  2 R
Problem Outputs
Problem 7.144
Gage
Theta Degrees
Epsilon Micro Meters
1
–15
480
2
30
–120
3
75
80
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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on a website, in whole or part.
1236
PROBLEM 7.C8 (Continued)
Program Outputs (Continued)
 x  253.21 micrometers
 y  306.79 micrometers
 xy  892.82 microradians
 a  727.21 micrometers
 b  167.21 micrometers
 max (in-plane)  894.43 microradians
Problem 7.145
Gage
Theta Degrees
Epsilon Micro Meters
1
2
3
30
–30
90
600
450
–75
 x  725.00 micrometers
 y  75.000 micrometers
 xy  173.205 microradians
 a  734.268 micrometers
 b  84.268 micrometers
 max (in-plane)  818.535 microradians
Problem 7.146
Observe that Gage 3 is orientated along the y axis. Therefore,
enter  4 and  4 as  3 and  3 , the value of  y that is obtained is also the expected reading of Gage 3.
Gage
Theta Degrees
Epsilon in./in.
1
0
420
2
45
–45
4
135
165
 x  420.00 in./in.
 y  300.00 in./in.
 xy  210.00 microradians
 a  435.00 in./in.
 b  315.00 in./in.
 max (in-plane)  750.00 microradians
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1237
PROBLEM 7.C8 (Continued)
Program Outputs (Continued)
Problem 7.169
Gage
Theta Degrees
Epsilon in./in.
1
2
3
45
–45
0
–50
360
315
 x  315.000 in./in.
 y  5.000 in./in.
 xy  410.000 microradians
 a  415.048 in./in.
 b  105.048 in./in.
 max (in-plane)  520.096 microradians
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
1238