345021434-Mechanics-of-Materials-7th-Edition-Beer-Johnson-Chapter-7.pdf
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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
8cos 2 20
3cos 20 sin 20
3 A cos 20 sin 20 2 3 sin 20 cos 20
3 A sin 200 cos 20 4sin 2 20
4 A sin 20 sin 20 2
0
0
9.46 ksi F
0:
A
8 A cos 200 sin 20
8coos 20 sin 20
3(cos2 20
3A A cos 20 cos 20
sin 2 20 )
3 A sin 200 sin 20
4A A sin 20 cos 200
0
4 20 cos 200 4sin
1.013 ksi
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PRO OBLEM 7.2 2
60 MPa
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:
90 9 A sin 30 coss 30
A
180sin 30 cos c 30
90 A cos 30 sin 30
60 A cos 30 ccos 30
0
60 coos 2 30 3 32.9 M Pa
F
0:
A
90 0 A sin 30 sin 30 3
90(cos 2 30
sin 2 30 )
90 A cos 30 cos 30
60 A cos 30 sinn 30
0
60 cos 30 sin 30 7 71.0 M Pa
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PROBLEM M 7.3
10 ksi
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.
6 ksi
758
4 ksi
SOLUTION
F
0:
A
4 A cos15 sin15
4 co os15 sin15
10 cos 2 15
100 A cos15 cos115
6sin 2 15
6 A sin15 sin15
4 A sin15 cos155
0
4 4sin15 cos155 1 10.93 ksi
F
0:
A
4(ccos2 15
4 A cos15 cos15
10 A cos15 sin 15
sin 2 15 )
6) cos15 sin15
(10
6 A sin15 cos15
4 A sin15 sin155
0
0 0.536 ksi
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80 MPa M
PROBLEM P 7.4 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.
40 MPa 558
SO OLUTION
Streesses
F
0 0:
A
80 A cos 55 cos555
80 cos 2 55
F
0 0:
A
Areas
Forces
40 A sin 55 sin 55
40sin 2 55
80 A cos 55 sin 55 5
0 0.5521 MPa
40 A sin 55 cos 55 5 MPa 56.4
1 cos 55 sin 55 120
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PROBLEM 7.5
40 MPa 35 MPa 60 MPa
For the given state of stress, determine (a) the principal planes, (b) the principal stresses.
SOLUTION 60 MPa
x
(a)
tan 2
2
xy
p x
2
p
y
(2)(35) 60 40
y
40 MPa
xy
35 MPa
3.50
74.05
37.0 , 53.0
p 2
(b)
x max, min
y
x
2 60 40 2
y
2 xy
2 60 40 2
2
(35)2
50 36.4 MPa max min
13.60 MPa 86.4 MPa
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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
(a)
tan 2
2 ksi
2 p
p
y
3 ksi
xy
(2)( 3) 2 10
xy
x
2
10 ksi
y
0.750
36.87
p
18.4 , 108.4 ◄
2
(b)
x
max,min
x
y
2 2
10 2
6
y
2 2
10 2
2 xy
2
( 3)2
5 ksi
max
11.00 ksi ◄
min
1.000 ksi ◄
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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
(a)
tan 2
150 MPa,
2 p
2
p
y
53.130
80 MPa
xy
2( 80 MPa) ( 150 MPa 30 MPa)
xy
x
30 MPa,
y
1.33333 MPa
and 126.870 p
(b)
max,min
x
y
x
2
y
2
150 MPa
30 MPa 2
90 MPa
26.6 and 63.4 ◄
2 xy
150 MPa
30 MPa 2
2
( 80 MPa)2
100 MPa
max min
190.0 MPa ◄ 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 x
(a)
tan 2
18 ksi
2 p
2
p
(2)(8) 18 12
xy
x
12 ksi
y
y
xy
8 ksi
0.5333
28.07
14.0 , 104.0 ◄
p
2
(b)
max,min
x
y
x
2 18
12 2
y
2 18
12 2
2 xy
2
(8)2
3 17 ksi max min
20.0 ksi ◄ 14.00 ksi ◄
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PROBLEM 7.9
40 MPa 35 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.
60 MPa
SOLUTION x
(a)
tan 2 2
x s
s
60 40 (2)(35)
y
2
xy
60 MPa
y
40 MPa
xy
35 MPa
0.2857
15.95
s
8.0 , 98.0
2
(b)
(c)
x max
y
2 xy
2 60 40 2
2
x
y
ave
2
(35) 2
max
60 40 2
36.4 MPa
50.0 MPa
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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 2 ksi
x
(a)
tan 2 2
y
x
s
2 10 (2)( 3)
y
2
10 ksi
xy
xy
3 ksi
1.33333
53.13
s
s
26.6 , 63.4
2
(b)
x
max
y
2 xy
2 2
10
2
( 3)2
2
max
(c)
ave
x
y
2
2
5.00 ksi
10 2
6.00 ksi
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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 150 MPa,
x
(a)
tan 2 2
x
s
30 MPa,
150 30 2( 80)
y
2
y
xy
xy
80 MPa
0.750
36.87 and 216.87
s
s
18.4 and 108.4
2
(b)
x
max
y
2 xy
2 150
30
2
( 80)2
2
max
(c)
ave
x
100.0 MPa
y
2 150 30 2
90.0 MPa
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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 18 ksi
x
(a)
tan 2 2
x
s
18 12 (2)(8)
y
2
12 ksi
y
xy
xy
8 ksi
1.875
61.93
s
31.0 , 59.0
s
2
(b)
x
max
y
2 xy
2 18
12
2
(8)2
2
max
(c)
ave
x
y
2
17.00 ksi
18 12 2
3.00 ksi
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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 x
y
2
0
8 ksi
y
x
4 ksi y
25
2
x
xy y
y
2
x
y
2
4 sin ( 50 ) 5 cos ( 50 )
xy
xy
sin 2
cos 2
cos 2
xy
sin 2
2.40 ksi
x
0.1498 ksi
xy
4 4 cos ( 50 ) 5 sin ( 50)
y
2
cos 2 +
sin 2 +
4 4 cos ( 50°) + 5 sin ( 50°)
xy
10
y
50
x
(b)
y
2 x
4 ksi
2
x
(a)
x
2
xy
y
y
2
x x
5 ksi
xy
10.40 ksi
y
20 4 4 cos (20°) + 5 sin (20°)
4 sin (20°) + 5 cos (20°) 4 4 cos (20°)
5 cos (20°)
x
1.951 ksi
xy
6.07 ksi
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 on a website, in whole or part. 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 x
y
2
60 MPa
x
y
(a)
25
2
x
2
sin 2 +
y
2
x
y
2
xy
xy
sin 2
cos 2
cos 2
xy
sin 2
x
56.2 MPa
xy
38.2 MPa
15 75 cos ( 50 ) 30 sin ( 50 )
y
86.2 MPa
20 15 75 cos (20°) + 30 sin (20°)
75 sin (20°) + 30 cos (20°)
xy y
y
75 sin ( 50 ) 30 cos ( 50 )
y
2
cos 2 +
15 75 cos ( 50 ) 30 sin ( 50 )
xy
10
y
50
x
(b)
x
30 MPa
xy
75 MPa
2
x
x
y
2
2
xy
y
x
15 MPa
x
90 MPa
y
15 75 cos (20°) 30 sin (20°)
x
45.2 MPa
xy
53.8 MPa
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 x
y
2
8 ksi
x
2 ksi x
x
y
2
x
2
y
2
xy
xy
sin 2
cos 2
cos 2
xy
sin 2
2 10 cos ( 50 ) 6 sin ( 50 )
9.02 ksi
x
10 sin ( 50 ) 6 cos ( 50 )
y
2
y
sin 2 +
cos 2 +
2 10 cos ( 50 ) 6 sin ( 50 )
xy
10
y
6 ksi
50
x
(b)
y
2 2
x
25
x
xy
10 ksi
2
x
(a)
y
2
xy
y
12 ksi
y
xy
3.80 ksi
13.02 ksi
y
20
x xy
y
2 10 cos (20°) 6 sin (20°)
x
10 sin (20°) 6 cos (20°)
2 10 cos (20°) + 6 sin (20°)
5.34 ksi
xy
9.06 ksi
y
9.34 ksi
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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 x
y
2
0
80 MPa
y
x x
y
(a)
25
2
2
sin 2 +
y
2
x
y
2
xy
xy
sin 2
xy
sin 2
cos 2
cos 2
40 sin ( 50°) 50 cos ( 50 ) 40 40 cos ( 50 ) 50 sin ( 50 )
y
2
y
cos 2
40 40 cos ( 50 ) 50 sin ( 50°)
xy
10
y
40 MPa
50 x
(b)
x
2
x
x
y
2
2
xy
y
x
40 MPa
50 MPa
xy
x
24.0 MPa 1.498 MPa
xy
y
104.0 MPa
x
19.51 MPa
xy
60.7 MPa
y
60.5 MPa
20 x
xy y
40 40 cos (20°) 50 sin (20°)
40 sin (20°) 50 cos (20°) 40 40 cos (20°) + 50 sin (20°)
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. 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
(a)
0
y
x
xy
0
250 psi
xy
y
sin 2 2 250cos( 30 )
15
xy cos 2
xy
(b)
x
x
0
y
2 0
x
y
cos 2 2 250sin( 30 )
217 psi
xy sin 2
x
125.0 psi
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. 1043
PROBLEM 7.18
1.8 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.
3 MPa
158
SOLUTION x
3 MPa
15 (a)
2
x
xy
1.8 MPa
y
y
2 3
1.8 2
xy
0
30
sin 2
xy sin 2
sin( 30 )
0
0.300 MPa
xy
(b)
x
x
y
x
2 3
y
2 1.8
2
3
1.8 2
cos 2 cos( 30 )
xy sin 2
0 x
2.92 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. 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 22 and thhat 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 N all
a all
Fy
0: N
Sall
Fx
aall
A/sin
P sin
A/sin
0:: S
P cos
9.6 10 3 m 2
(4400 103 )(9.6 10 3 ) 10.2251 103 N sin 22 0
P
N sinn
10.251 103 sin 222
27.4 1103 N
(6600 103 )(9.6 10 3 ) 15.3376 103 N sin 22
0
P
S coos
Thee smaller valuee for P governns.
15.376 103 cos 22 2
16.58 103 N P 16.58 kN
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. 1045
P'
PROBLEM 7.20 7
1220 mm
80 mm
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 25 and a that centriic loads of magnitude m P 10 kN are that 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.
b P
SO OLUTION
Forcess
A Areas
A (80)(1220) 9.6 103 mm m 2 (a)
Fx
0: S N A/sin
(b)
Fy
0: N N A/sin
P cos
0
S
P cos
(9.063 103 )sinn 25 9.6 10 3
P sin n
0
N
P sin
(4.226 103 )sin 25 9.6 103
9.6 100 3 m 2
(10 103 ) cos 25
9.063 103 N
399 1003 Pa
(10 103 )sin 25 186.0 103 Pa
399 kPa
4.226 103 N 1 186.0 kPa
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. 1046
P
PROBL LEM 7.21 The centrric force P is applied to a short post as shown. Know wing that the stresses on 5 ksi, determinne (a) the anggle that planne a-a forms 15 ksi and plane a-a are with the horizontal, h (b) the maximum m compressivee stress in the post. p
a
!
a
SOLUTION
x
0
xy
0
y
(a)
P/ A
From the Mohr’s M circle,
tan
5 15
P 2A (b)
P A
2( ) 1 co os 2
0.33333
18.4
P cos 2 2A
(2)(115) 1 coss 2
P 16.67 1 ksi A
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. 1047
PROBLEM 7.2 22 a a
25"
50 mm m
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 800 kPa and a 600 kPa, k determinne the largest stressses for the gluued joint are centtric load P thatt can be applieed.
P
SO OLUTION Forr plane a-a,
65 .
x
0, 0 x
P
0,
cos 2
x
y
y
P A
sin 2
2
xy
sin
cos
0
(50 10 3 )(80 10 3 )(800 103 ) sin 2 65 6
A sin s 2 65 (
P
xy
y )sin
A sin s 65 cos 65
P 2 sin 655 A
0
3.90 103 N
P sin 65 cos 65 0 A (50 10 3 )((80 10 3 )(600 103 ) 6.277 103 N sinn 65 cos 65
cos
( xy (cos
Alllowable value of P is the sm maller one.
2
sin 2 )
P
3.90 kN
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. 1048
PROBLEM 7.23
0.2 m 0.15 m 3 kN
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
350 N · m 3 kN
SOLUTION 1 d 2
c Tc J
Torsion:
I
Bending:
4
2T c3
c4
4
1 (32) 2
2(350 N m) (16 10 3 m)3
(16 10 3 )4
(0.15m)(3 103 N)
M
16 mm 16 10 3 m 54.399 MPa
51.472 10 9 m 4 450 N m
(450)(16 10 3 ) 51.472 10 9
My I
54.399 106 Pa
Top view:
139.882 106 Pa
139.882 MPa
Stresses:
x
139.882 MPa 1 ( 2
x
y)
R
x
y
ave
y
0
1 ( 139.882 2
xy
0)
54.399 MPa 69.941 MPa
2
(a)
2
max
ave
R
69.941
88.606
min
ave
R
69.941
88.606
2 xy
( 69.941)2
( 54.399) 2
88.606 MPa
max min
18.67 MPa 158.5 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. 1049
PROBLEM 7.23 (Continued)
tan 2
2 p x
xy y
(2)( 54.399) 139.882
0.77778
2
p
37.88
p
(b)
max
R
88.6 MPa
18.9
max
and 108.9°
88.6 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. 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:
Shaft cross section:
V
400 lb
T
(400)(2)
d
1 in. c
J
c4
2
1 d 2
(400)(6)
2400 lb in.
800 lb in.
0.5 in.
0.098175 in 4
Torsion:
Tc J
(800)(0.5) 0.098175
Bending:
Mc I
(2400)(0.5) 0.049087
Transverse shear:
M
1 J 2
I
0.049087 in 4
4.074 103 psi
4.074 ksi
24.446 103 psi
24.446 ksi
Stress at point H is zero. x ave
24.446 ksi, 1 ( 2
x
y)
x
y
y
0,
xy
4.074 ksi
12.223 ksi 2
R
2
2 xy
(12.223) 2
(4.074) 2
12.884 ksi a
ave
R
b
ave
R
max
R
a
25.1 ksi
b
0.661 ksi
max
12.88 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. 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:
Shaft cross section:
d
V
24 lb M
T
(24)(10)
1 d 2
0.75 in., c
J
2
c4
Tc J
(240)(0.375) 0.031063
Bending:
Mc I
(144)(0.375) 0.015532
Transverse shear: Resultant stresses:
144 lb in.
240 lb in.
0.375 in. 1 J 2
0.031063 in 4 I
Torsion:
(24)(6)
0.015532 in 4
2.897 103 psi
2.897 ksi
3.477 103 psi 3.477 ksi
At point H, the stress due to transverse shear is zero. x ave
3.477 ksi, 1 ( 2
y
x
y)
x
y
0,
2.897 ksi
xy
1.738 ksi 2
R
2
a
ave
R
b
ave
R
max
2 xy
1.7382
2.897 2
3.378 ksi
a
1.640 ksi
b
R
5.12 ksi
max
3.38 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. 1052
P PROBLEM 7 7.26
y
m 6 mm A 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 T
D
10 kN N C
1 mm 150 H
K
B x
z
SOLUTION ro J
I
do 2
1102 2
51 mm
ri
ro
t
45 mm
ro4 ri4 4.18555 106 mm 4 2 4.18555 10 6 m 4 1 J 2
2.0927 10 6 m 4
Forcce-couple systtem at center of o tube in the plane p containiing points H and a K:
Fx
10 kN 10 1003 N
My
(10 103 )(200 10 3 ) 2000 N m
Mz
(10 103 )(150 10 3 ) 15000 N m
Torsion:
At po oint K, place local l x-axis in negative globbal z-directionn. T
My
c
ro
xy
2000 N m 511 10 3 m
((2000)(51 100 3 ) 4.1855 106 24.37 106 Pa 24.37 MPa Tc J
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. 1053
PROBLEM 7.26 (Continued)
Transverse shear:
Stress due to transverse shear V
Fx is zero at point K.
Bending: |
y|
(1500)(51 10 3 ) 2.0927 10 6
|M z |c I
36.56 106 Pa
36.56 MPa
Point K lies on compression side of neutral axis.
36.56 MPa
y
Total stresses at point K: x ave
0,
36.56 MPa,
y
1 ( 2
x
y)
x
y
xy
24.37 MPa
18.28 MPa 2
R
2
2 xy
30.46 MPa
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. 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. 1054
PROBLEM 7.27
#y 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
y
x
R
u2
2
.
Then
Largest value of
y
2 xy
R2
u y
2u
x
2u
75 MPa 2 xy
752
202
60 (2)(72.284)
is required.
72.284 MPa 84.6 MPa or 205 MPa y
205 MPa
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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,
8 ksi,
y
xy
2 max
x
R
z 92
122
(a)
xy
(b)
ave
1 ( 2
y
2 xy
2 xy
? 10 ( 8) z
2 2 xy
12 ksi
92
xy y)
x
7.94 ksi
1 ksi
a
ave
R 1 12 13 ksi
b
ave
R 1 12
a
11 ksi
b
13.00 ksi 11.00 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. 1056
P PROBLEM 7.29
2 MPaa
$xy
12 MPa
75"
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
Sincce
xy
12 MPa,
y
tan 2
?
15 2
1 ( 2
x
y)
tan 2
x
y
2 xyy
y)
7 MPa M
p
xy
p x
xy
xy
0, x -direction is a principal direection. p
(a)
2 MPaa,
y
1 (12 2)) tan( 30 ) 2
xy
2.89 MPa
2
R
ave
(b)
2 1 ( 2
x
52
2.8992
5.7735 MPa M
a
ave
R
7 5.77735
a
12..77 MPa
b
ave
R
7 5.77735
b
1.2 226 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. 1057
PROBLEM 7.30
15 ksi 8 ksi
Determine the range of values of is equal to or less than 10 ksi.
x
for which the maximum in-plane shearing stress
#x
SOLUTION x
Let u
R
x
x
2 u2
2 xy
R2
u x
y
y
max 2 xy
y
?,
y
15 ksi,
xy
8 ksi
2u
10 ksi 102
2u 15 (2)(6)
82 z
6 ksi
27 ksi or 3 ksi 3 ksi
Allowable range:
x
27 ksi
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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
6 MPa, 60
y
4 MPa, 40 355 MPa
xy
x ave
y
2
50 MPa
Plottted points forr Mohr’s circlee:
(a)
a
R
(a )
(b ) (c )
x,
xy )
Y:(
y,
xy )
( 40 MPa, 35 MPa)
C:(
ave ,
0)
( 50 MPa, 0)
X 35 GX CG G 10 74 4.05
tan
b
(b)
X :(
1 2 180 1 2
( 60 MPa, 35 MPa)
3.5000
37.03 105.995 52.97
CG C
2
GX
2
10 2
ave a
R
50 36.4
max
a ave
R
50 36.4
d
B
45
7.97
e
A
45
97.977
352
36.4 MPa min max
86 6.4 MPa 13 3.60 MPa d e
R 36.4 MPa ave a
53.0
a
min
max
37.0
b
max
50 MPaa
8.0 98.0
36 6.4 MPa 50 0.0 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. 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 80 MPa
xy
x
ave
y
2
90 MPa
Plotted points for Mohr’s circle:
x
tan 2
p
2
p
y , xy )
C:(
ave ,
30)
(60)2
xy )
0)
(80)2
100
p
90
100
min
ave
R
90
100
max
(90 MPa, 0)
53.130 R
(b′)
(30 MPa, 80 MPa)
80 60
ave
s
(150 MPa, 80 MPa)
60
max
(a′)
(c )
Y:(
2
R
(b)
x,
(150
y
2
(a)
X :(
p
26.6 max min
45
s
R
18.4 max
and 63.4 190.0 MPa 10.00 MPa and 108.4 100.0 MPa
90.0 MPa
ave
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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 y
x
ave
10 ksi
y
2
2
10 2
xy
3 ksi
6 ksi
Plotted points for Mohr’s circle:
FX FC
tan
X: (
x,
Y: (
y , xy )
(10 ksi, 3 ksi)
C: (
ave ,
(6 ksi, 0)
3 4
xy )
0)
(2 ksi, 3 ksi)
0.75
36.87 B
(a)
(b) (c)
1 2
18.43
D
B
45
E
B
45
R
CF
max
R
5.00 ksi
ave
6.00 ksi
2
26.6 63.4
FX
2
26.6
D E
42
32
63.4
5 ksi max
5.00 ksi
6.00 ksi
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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 18 ksi
x
x
ave
12 ksi
y
y
xy
8 ksi
3 ksi
2
Plotted points for Mohr’s circle:
FX CF
tan
X: (
x,
Y: (
y,
xy )
C: (
ave ,
0)
8 15
xy )
(18 ksi, 8 ksi) ( 12 ksi, 8 ksi) (3 ksi, 0)
0.5333
28.07 A
(a)
(b) (c)
1 2
14.04
D
A
45
E
A
45
R
CF
max
R
17.00 ksi
ave
3.00 ksi
2
59.0
D
30.1
FX
2
E
152
82
59.0 30.1
17 ksi max
17.00 ksi
3.00 ksi
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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 x
ave
y
4 kssi
2
Plottted points forr Mohr’s circlee:
X : (0, 5 ksi) k 5 ksi) Y : (8 ksi, k 0) C : (4 ksi, tan 2
p
2
p
FX 5 1 1.25 FC 4 51.34
R (a)
25
FC .
2
X FX
2
xy
y
10
.
42
52
6.4031 ksi
50
x
(b)
2
2
51.34
50
ave
R cos
1.34
R sin ave
2.40 ksi
x
0.1497 ksi
xy
R cos
y
10.40 ksi
20
51.34 x xy y
ave
20 R cos
R sin ave
R cos
71.34 x
1.951 1 ksi
xy
6.07 ksi
y
6 6.05 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. 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 60 MP Pa,
x y
90 MPa,,
xy
30 MPa x
ave
y
15 MPa
2
Plootted points for Mohr’s circlle:
X : ( 60 MPa, 30 MPa) Y : (90 MPa, 300 MPa) C : (15 MPa, 0) tan 2
p
2
p
FX FC
21.80
R
(a)
25
.
30 75
FC
2
0 0.4
10.90
P
FX X
2
752
300 2
80.78 MP Pa
50 5
2
2 x
xy y
2 ave
P
50
21.80
288.20
R cos
R sin ave
R cos
x
5 56.2 MPa
xy
3 38.2 MPa y
8 MPa 86.2
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PROB BLEM 7.36 (Continued) ( d)
(b)
10
.
2
200
2 x
p ave
xy
R sin
y
ave
2
21.880
20
41.80
R cos
R cos
x
45 5.2 MPa
xy
53 3.8 MPa
y
755.2 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. 1065
PR ROBLEM 7..37
12 ksi
8 ksi
6 ksi
Solv ve Prob. 7.15, using Mohr’ss circle.
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 8 ksi,
x
12 ksi,
y
6 ksi
xy
x
y
ave
2 ksi k
2
Plootted points for Mohr’s circlle:
X : (8 ksi, 6 ksi) Y : ( 12 ksi, 6 ksi)) C : ( 2 ksi, 0)
(a)
tan 2
p
2
p
25
FX 6 CF 100 30.96
R
CF
.
2
2
0.6
FX
2
x
xy y
10
.
62
11.66 ksi k
5 50
5 50
(b)
102
2
ave
30.96
19.04
R cos
R sin ave
R cos
x
9.02 ksi
xy
3.80 ksi 13.02 ksi
y
2 20
3 30.96 x
xy y
20 R cos
ave
R sin ave
50.96
R cos
x
5.34 ksi
xy
9.06 ksi
y
9.34 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. 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,
y
M 80 MPa,
xy
50 MPa M x
ave
y
40 MPa
2
Plotted points for Moohr’s circle:
X : (0, 50 MPa) M MPa, 50 MPa)) Y : ( 80 M M 0) C : ( 40 MPa,
tann 2
p
2
p
R (a)
25
.
2
FX 50 1.25 CF 40 51.34 2
FX CF 64.031 MPa
x
xy y
10
.
402
502
50
51.34
(b)
2
2
x
xy y
50
1.34
R cos
ave
x
R sinn
1.4497 MPa
xy
R cos
ave
244.0 MPa
y
1044.0 MPa
x
199.51 MPa
xy
600.7 MPa
y
0.5 MPa 60
20
51.34
20
ave
R cos
R sinn ave
R cos
71.34
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 xy
y
0
250 psi
Plotted points for Mohr’s circle:
(a)
xy
X
(0, 250 psi)
Y
(0, 250 psi)
C
(0, 0)
R cos 2 (250 psi)cos30 217 psi xy
(b)
x
217 psi
R sin 2 (250 psi) sin 30 125.0 psi x
125.0 psi
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. 1068
PROBLEM 7.40
1.8 MPa
Solve Prob. 7.18, using Mohr’s circle. 3 MPa
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.
158
SOLUTION x
ave
3 MPa x
1.8 MPa
y
y
xy
0
2.4 MPa
2
Points. X: (
x,
xy )
( 3 MPa, 0)
Y: (
y,
xy )
( 1.8 MPa, 0)
C: (
ave ,
0)
( 2.4 MPa, 0)
15 CX
(a)
xy
CX sin 30
(b)
x
ave
CX cos 30
2
0.6 MPa
R sin 30 2.4
30
R
0.6 MPa
0.6sin 30 0.6 cos 30
0.300 MPa 2.92 MPa
0.300 MPa
xy x
2.92 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. 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 22 and that the maximum splice shown. Knowing that 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 P , A
x
0
y
xy
0
Plotted points for Mohr’s circle:
X:
P ,0 , A
C:
P ,0 2
R
CX
Y : (0, 0)
P 2A
Coordinates of point Y : P (1 cos 2 ) 2A P sin 2 2A
Data:
A
If
(80)(120) 400 kPa
P
9.6 103 mm 2 400 103 Pa,
(2)(9.6 10 3 )(400 103 ) (1 cos 44 )
2A 1 cos 2 27.4 103 N
If
600 kPa
P
2A sin 2
9.6 10 3 m 2
27.4 kN
600 103 Pa,
(2)(9.6 10 3 )(600 103 ) (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. 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. 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 25 and that centric loads of splice shown. Knowing that 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
0
y
xy
0
Plotted points for Mohr’s circle:
X:
P ,0 A
Y : (0, 0)
C:
P ,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
(a)
(10 103 )sin 50 (2)(9.6 10 3 )
(b)
(10 103 )(1 cos 50 ) (2)(9.6 10 3 )
9.6 10 3 m 2
399 103 Pa
399 kPa
186.0 103 Pa
186.0 kPa
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. 1071
PROBLEM 7.43
P
Solve Prob. 7.21, using Mohr’s circle. PROBLEM 7.21 The centric force P is applied to a short post as shown. Knowing that 5 ksi, determine (a) the angle that 15 ksi and the stresses on plane a-a are plane a-a forms with the horizontal, (b) the maximum compressive stress in the post.
a
!
a
SOLUTION
x
0
xy
0
y
P A
(a)
From the Mohr’s circle, tan
(b)
P A
5 0.3333 15 P P cos 2 2A 2A
1
2( ) cos 2
1
18.4
(2)(15) cos 2
16.67 ksi 16.67 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. 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
P (1 cos50 ) 2A 2A 1 cos 50
(2)(4 10 3 )(800 103 ) 1 cos 50 P 3.90 103 N P
P sin 50 2A
P
2A sin 50
Choosing the smaller value,
(2)(4 10 3 )(600 103 ) sin 50
6.27 103 N P
3.90 kN
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. 1073
PROBLEM 7.45
0.2 m 0.15 m 3 kN
Solve Prob. 7.23, using Mohr’s circle.
H
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 c
Torsion:
1 d 2
1 (32) 2
Tc J
2T c3
16 mm 16 10 3 m
2(350 N m) (16 10 3 m)3 Bending:
I
M
4
c4
4
54.399 106 Pa
(16 10 3 )4
(0.15m)(3 103 N) My I
54.399 MPa
51.472 10 9 m 4
450 N m
(450)(16 10 3 ) 51.472 10 9
139.882 106 Pa
Top view
Stresses
x
Plotted points:
139.882 MPa
139.882 MPa,
y
X : ( 139.882, 54.399); ave
1 ( 2
0,
xy
54.399 MPa
Y: (0, 54.399); C: ( 69.941, 0)
x
y)
x
y
69.941 MPa 2
R
2 xy
2 139.882 2
2
(54.399)2
88.606 MPa
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PROBLEM 7.45 (Continued)
tan 2
2 p
xy
x
y
(2)( 54.399) 139.882
0.77778 (a)
(b)
a a
ave
R
69.941
88.606
b
ave
R
69.941
88.606
max
R
18.9
,
b
108.9 158.5 MPa
a b
max
18.67 MPa 88.6 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. 1075
6 in.
PROBLEM 7.46
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
T
(400)(2)
800 lb in.
Shaft cross section:
(400)(6)
2400 lb in.
d
1 in.
J
c4
Resultant stresses:
(800)(0.5) 0.098175
Mc I
Bending:
0.5 in.
0.098175 in 4
Tc J
Torsion:
Transverse shear:
2
1 d 2
c
I
1 J 2
0.049087 in 4
4.074 103 psi
(2400)(0.5) 0.049087
4.074 ksi
24.446 103 psi
24.446 ksi
Stress at point H is zero. x ave
24.446 ksi, 1 ( 2
x
y)
x
y
y
0,
xy
4.074 ksi
12.223 ksi 2
R
2 (12.223) 2
a
ave
R
b
ave
R
max
R
2 xy
(4.074) 2
12.884 ksi a
25.1 ksi
b
0.661 ksi
max
12.88 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. 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
T
(24)(10)
240 lb in.
Shaft cross section:
(24)(6)
144 lb in.
d
0.75 in.
J
c4
Resultant stresses:
1 J 2
I
(240)(0.375) 0.031063
Mc I
Bending:
0.375 in.
0.031063 in 4
Tc J
Torsion:
Transverse shear:
2
1 d 2
c
0.015532 in 4
2.897 103 psi
(144)(0.375) 0.015532
3.477 103 psi
2.897 ksi
3.477 ksi
At point H, stress due to transverse shear is zero. x ave
3.477 ksi, 1 ( 2
0,
y
x
y)
x
y
xy
2.897 ksi
1.738 ksi 2
R
2 1.7382
a
ave
R
b
ave
R
max
R
2.8972
xy
2
3.378 ksi a
5.12 ksi
1.640 ksi
b max
3.38 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. 1077
PROBLEM 7.48
y
6 mm A 200 mm
Solve Prob. 7.26, using Mohr’s circle.
51 mm
A T
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.
D
10 kN C
150 mm H
K
B z
x
SOLUTION ro
J I
do 2
102 2
ro4
2
51 mm
ri4
1 J 2
ri
ro
t
4.1855 106 mm 4
45 mm
4.1855 10
6
m4
2.0927 10 6 m 4
Force-couple system at center of tube in the plane containing points H and K: Fx My Mz Torsion:
10 103 N (10 103 )(200 10 3 ) (10 103 )(150 10 3 )
T
My
c
ro
xy
2000 N m 1500 N m
2000 N m 51 10 3 m (2000)(51 10 3 ) 4.1855 10 6
Tc J
24.37 MPa
Note that the local x-axis is taken along a negative global z direction. Transverse shear: Bending:
Stress due to V y
Fx is zero at point K. (1500)(51 10 3 ) 2.0927 10 6
Mz c I
Point K lies on compression side of neutral axis.
y
36.56 MPa
36.56 MPa
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PROBLEM 7.48 (Continued)
Total stresses at point K:
x
ave
0, 1 ( 2
36.56 MPa,
y
x
y)
x
y
xy
24.37 MPa
18.28 MPa 2
R max
min
max
2 ave
ave
R
R
R
2 xy
18.28
18.28
30.46 MPa
30.46 max
12.18 MPa
min
48.7 MPa
30.46
max
30.5 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. 1079
PROBLEM 7.49
#y 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
XY
2 R 150 MPa
DY
(2)(
XD y
75 MPa
XY x
xy ) 2
40 MPa DY
2
1502
XD 60 144.6
402
144.6 MPa y
205 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. 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
2 The radius of the circle is
max (in-plane)
10 8 , 0 2
,0
(1 ksi, 0).
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 2
(b)
The principal stresses are
is obtained from triangle CDX. 2
DX 1
DX 2
xy
122
a
1 12
b
1 12
2
CX 1
CD
2
92
xy a b
7.94 ksi 13.00 ksi 11.00 ksi
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2 MPa
PROBLEM 7.51 $xy
Solve Prob. 7.29, using Mohr’s circle. 12 MPa
75"
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 that y 2 MPa. The coordinates of C are
so that
2 12 , 0 2
x
12 MPa. Likewise, point Y lies on line Y Y
(7 MPa, 0).
Counterclockwise rotation through 150° brings line CX to CB, where
R (a)
x xy
y
2
x
y
2
sec 30
12 2 sec 30 2
0.
5.7735 MPa
tan 30
12 2 tan 30 2
(b)
so
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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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 in-plane shearing stress is equal to or less than 10 ksi.
x
for which the maximum
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 ,
C1D
2
DY
2
C1Y
2
C1D
2
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
Thus,
x
27 ksi
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2 MPa
PROBLEM 7.53 $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 12
Locate point C at Angle XCB x
y
2
2
7 MPa with
2
0.
120
12
2
2 5 MPa
R
5sec 60 10 MPa
5 tan 60
xy
8.66 MPa
xy ave
a
7
10 ave
b
7
R
10
a
17.00 MPa
b
3.00 MPa
R
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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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PROBLEM 7.54 (Continued)
ave
x
y
2
1 (6 2
2)
1 (6 2
2)
(4)2
R tan 2
p
2
p
2
4
(3)2
5
3 4
36.87 p
max
ave
2
R
18.4 , 108.4
5 max
min
ave
R
2
7.00 ksi
5 min
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 on a website, in whole or part. 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
y
x
2 R tan 2
p
1 (118.3 2
56.7)
1 (118.3 2
56.7)
(30.8)2
(75)2
75 30.8
2
p
87.5
30.8 81.08
67.67
max
ave
R
87.5
81.08
min
ave
R
87.5
81.08
p
33.8
, and 123.8 max min
168.6 MPa 6.42 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. 1087
#0 #0
PROBLEM M 7.56
#0 #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 m min
0 0
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. 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 0
xy
xy
3 2
0 sin 60
x
1 2
0 cos 60
0
y
0
y
3 2
0 sin 60
0
0
Resultant stresses: x
3 2 1 2
0
xy
0
ave
1 ( 2
3 2
0
3 2
0
x
y)
x
y
tan 2
2 x
2
2 xy
(2)
xy y
p
60
a
ave
R
b
ave
R
0
0
2
p
3 2
0
0
2
R
3 2
0
3 2 3
3 2
2 0
3 2
2 0
3
0
3 b
30
a
a b
60
3
0
3
0
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PROBLEM 7.58
120 MPa
$xy
For the element shown, determine the range of values of maximum tensile stress is equal to or less than 60 MPa.
xy
for which the
20 MPa
SOLUTION x ave
Set
max
R
20 MPa 1 ( 2
y)
x
60 MPa max
120 MPa
y
70 MPa
R
ave
130 MPa
ave
But 2 x
R
2 xy
x
2
2 xy
R
2
x
x
2
1302 502 120.0 MPa Range of
xy :
120.0 MPa
xy
120.0 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. 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 20 MPa
x
1( 2
y)
x
Set
max (in-plane)
But
R
120 MPa
y
50 MPa
R 150 MPa 2 x
y
2 xy
2
2 xy
R
2
1502
x
y
2 502
141.4 MPa
Range of
xy :
141.4 MPa
xy
141.4 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. 1091
!y' 6 ksi
PROBLEM 7.60
"x'y'
!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 xy ave
16 ksi,
0
y
6 ksi 1 ( 2
x
y)
x
y
8 ksi 2
R
( 8)2 tan 2
2 p x
2
2 xy
2 (6) 2
xy y
p
36.870
b
18.435
10 ksi
(2)(6) 16
0.75
8 ksi for states of stress corresponding to arcs HBK and UAV of Mohr’s circle. The angle
xy
is
calculated from R sin 2
2
8
8 10
sin 2
53.130
0.8
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
u
v
135
45 45
188.13
and 225
8.13 98.13
278.13
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PROBLEM 7.61
#y' #x'
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.
%
90 MPa
$x'y' 60 MPa
SOLUTION x
90 MPa,
0
60 MPa
xy ave
y
1 ( 2
x
y)
x
y
45 MPa 2
R
452 2
tan 2
p
602
xy
x
2
x
2 xy
2
y
p
53.13
a
26.565
75 MPa (2)( 60) 90
4 3
50 MPa for states of stress corresponding to the arc HBK of Mohr’s circle. From the circle,
R cos 2
50
cos 2
5 75
2
5 MPa
0.066667
86.177 h
2
45
43.089 26.565
a
k
2
k
110.085
h
360
Permissible range of
:
4
43.089
16.524
32.524
360
h
172.355
220.169
k
16.5
Also,
196.5
110.1
290.1
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. 1093
PROBLEM 7.62
#y' #x'
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.
%
90 MPa
$x'y' 60 MPa
SOLUTION x xy ave
90 MPa,
y
0
60 MPa 1 ( 2
x
y)
x
y
45 MPa 2
R
2 452 2
tan 2
p
xy
x
2
x
602
y
p
53.13
a
26.565
2 xy
75 MPa (2)( 60) 90
4 3
100 MPa for states of stress corresponding to arc HBK of Mohr’s circle. From the circle, R cos 2
100
45
cos 2
55 75
0.73333
2
42.833 h
2
55 MPa
21.417 26.565
a
k
2
k
132.02
h
Permissible range of
Also,
360
4
21.417 10.297
is
h
5.15 360
85.666
264.037
k
5.1
132.0
174.8
312.0
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. 1094
PROBLEM 7.63
#y $xy
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.
#x
SOLUTION 14 ksi,
x min
9 ksi,
y
R
ave
1 ( 2
ave
R
ave
x
y)
11.5 ksi
min
11.5 5 6.5 ksi 2 x
R
y
2 xy
2
2 xy
But it is given that (a)
tan 2
x
R2
2 xy
2
is positive, thus
6.52
xy
2.52
6 ksi
6 ksi.
xy
p x
2
y
p
y
(2)(6) 5 67.38
2.4
a b
(b)
max
ave
max (in-plane)
123.7
R max
(c)
33.7
18.00 ksi
R max (in-plane)
6.50 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. 1095
!
PROBLEM 7.64 "y "y'
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 given in Eqs. (7.5) and (7.6), respectively. [Hint: Use xy sin( A B) sin A cos B cos A sin B and cos ( A B) cos A cos B sin A sin B.]
Y Y' C
O
2#p 2#
!x'y' X'
" !xy
X
"x "x'
SOLUTION OC
1 ( 2
y)
x
CX
CX
x
y
CX cos 2
p
CX cos 2
p
CX sin 2
p
CX sin 2
p
x
OC CX cos (2
p
2 )
OC CX (cos 2
p
cos 2
2 xy
OC CX cos 2 x
y
x
2 xy
p
CX sin 2
p
cos 2
cos 2
y
2
CX sin (2
xy
p
cos 2
sin 2
p
sin 2 )
CX sin 2 xy
p
sin 2
2 )
CX (sin 2
p
cos 2
cos 2
CX cos 2
p
sin 2
x
y
2
sin 2
cos 2
p
sin 2 )
sin 2
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. 1096
PROBLEM 7.65 2 (a) Prove that the expression x y x 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, R sin 2
xy x y
x
p
ave
R cos 2
p
ave
R cos 2
p
2 xy
y 2 ave
R 2 cos2 2
R 2 sin 2 2
2 ave
R 2 ; independent of
p
p
p.
Draw line OK from origin tangent to the circle at K. Triangle OCK is a right triangle. OC OK
2 2
OK OC
2 2
CK
2 ave x
(b)
2
CK
2
R2 2 xy
y
x,
Applying above to x
y
But x
2 xy ab
y
a
max
2 xy
x
xy
xy
2 ave
R2
max ,
a
2 xy
and
2 ab
b
0,
y,
b
, and to
a,
b,
min
min max
y x
y
min max
min
The sign cannot be determined from above equation.
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. 1097
PROBLEM 7.66
y
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.)
σy
12 ksi
σx
z
x
SOLUTION (a)
ave
1 ( 2
x
y)
1 (14 2 1 (14 2
4)
9
4)
5
(5)2
R
(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
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. 1098
PROBLEM 7.66 (Continued)
(b)
ave
1 ( 2
x
y)
1 (21 14) 2 1 (21 14) 2
(3.5)2
R
17.5 3.5
(12)2
12.5
max
ave
R
17.5
12.5
30
min
ave
R
17.5
12.5
5
Since max and min have the same sign, O and A, we have max
1 2
max
max
is out of the plane of stress. Using Mohr’s circle through
1 (30 ksi) 2
max
15.00 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. 1099
PROBLEM 7.67
y
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.)
σy
12 ksi
σx
z
x
SOLUTION
(a)
ave
1 (20 2 1 ( x 2
10)
15
y)
1 (20 2
R
(5)2
10)
5
(12)2
13
max
ave
R
15
13
28
min
ave
R
15
13
2
Since max and min have the same sign, O and A, we have
max
max
1 2
max
is out of the plane of stress. Using Mohr’s circle through 1 (28 ksi) 2
max
14.00 ksi
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PROBLEM 7.67 (Continued)
(b)
ave
1 (12 2 1 ( 2
x
5)
8.5
y)
1 (12 2
R
5)
(3.5)2
3.5
(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
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. 1101
y
PROBLEM 7.68 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 ave
140 MPa, 1 ( 2
y
x
y)
x
y
40 MPa,
80 MPa
xy
90 MPa 2
R
502 802
a
ave
R 184.34 MPa (max)
b
ave
R
c
4.34 MPa
x ave
94.34 MPa
(min)
0 1 ( 2 1 ( max 2 140 MPa,
max (in-plane)
(b)
2 xy
2
1 ( 2
b)
a
y y)
x
y
1 ( 2 120 MPa, min )
max
x
94.34 MPa
R
b)
a xy
94.3 MPa
max
94.3 MPa
80 MPa
130 MPa 2
R
2 xy
2
102 802
a
ave
R
210.62 MPa (max)
b
ave
R
49.38 MPa
c max
80.62 MPa
0 (min) a
max (in-plane)
max
210.62 MPa
min
c
0
R 86.62 MPa 1 ( 2
max
min )
105.3 MPa
max
105.3 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. 1102
y
PROBLEM 7.69 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 ave
140 MPa, 1 ( 2
y
x
y)
x
y
20 MPa,
xy
80 MPa
80 MPa 2
R
602
802
100 MPa
a
ave
R 80 100 180 MPa (max)
b
ave
R 80 100
c
x ave
20 MPa (min)
0 1 ( 2 1 ( max 2 140 MPa,
max (in-plane)
(b)
2 xy
2
1 ( 2
b)
a
min )
max y
x
y)
x
y
100 MPa 100 MPa
140 MPa,
xy
max
100.0 MPa
max
110.0 MPa
80 MPa
140 MPa 2
R
2 xy
2
0 802
a
ave
R
220 MPa (max)
b
ave
R
60 MPa
c
80 MPa
0 (min)
max (in-plane)
max
1 ( 2 1 ( 2
a
max
b)
80 MPa min )
110 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. 1103
PROBLEM 7.70
y
For the state of stress shown, determine the maximum shearing stress when 60 MPa, (c) z 60 MPa. (a) z 0, (b) z
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.
ave
1 ( 2
x
y)
1 (30 2 1 (30 2
R
(35)2
A
ave
B
ave
100) 100)
(84)2 R R
65 65
65 35
91 91 91
156 MPa 26 MPa
(a)
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 R 91.0 MPa max
(b)
60 MPa. Point Z is located between A and B. The largest of the 3 circles is still the circle z through A and B, and we still have R 91.0 MPa max
(c)
60 MPa. Point Z is now outside the circle through A and B. The largest circle is the circle through Z and A. 1 1 108.0 MPa ( ZH ) (60 156) max max 2 2
z
z
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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 170 MPa
z
x
z
SOLUTION
ave
1 ( 2
x
y)
1 (170 2 1 (170 2
(35)2
R A B
(a)
135
100)
35
(84)2
91 91
91
226 MPa 44 MPa
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 z
max
(b)
135 135
100)
1 (OA) 2
1 2
A
1 (226) 2
max
60 MPa. Point Z is located between B and A. The largest of the 3 circles is the one drawn z through A and B. max
(c)
113.0 MPa
R
91.0 MPa
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 z
max
1 ( ZA) 2
1 (60 2
226)
max
143.0 MPa
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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)
(b)
17.55 ksi
yz
R
(6) 2
A
6 18.5
B
6 18.5 A
min
B
max
1 ( 2
24.5 12.5
min )
max
(6))2
(8) 2
6 10 1 16
B
6 10 1
18.50 ksi
max
10.00 ksi
10
4
max
A
16 ksi
min
B
4 ksi
max
1 ( 2
0
max
3 kssi
x
A
yz
18.5
12.5 ksi
8 kssi
R
(c)
(17.5) 2
24.5 ksi
max
yz
3 ksi
x
min )
max
3 ksi
x
max
z
12 ksi
min
x
3 ksi
max
1 ( 2
max
min )
max
7.50 ksi
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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)
17.5 ksi
yz
(6)2
R A B
max
6 6 A
min
B
max
1 ( 2
(17.5)2
18.5
18.5 24.5 18.5 12.5 24.5 ksi 12.5 ksi max
min ) max
(b)
yz
18.50 ksi
8 ksi
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PROBLEM 7.73 (Continued)
(6)2
R
(c)
yz
(8)2
A
6
10
16
B
6
10
4
max
A
min
x
max
1 ( 2
10
16 ksi
10 ksi min )
max
max
13.00 ksi
max
11.00 ksi
0
max
z
min
x
max
1 ( 2
12 ksi 10 ksi max
min )
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PROBLEM 7.74
y
For the state of stress shown, determine the value of maximum shearing stress is (a) 9 ksi, (b) 12 ksi.
6 ksi
xy
for which the
τ xy 15 ksi z x
SOLUTION 15 ksi 1 ( x 2
x
(ksi)
ave
y)
x
u (a)
y
y
R xy
a
u
18
ave
2
2 xy
R2
u2
R2
c max
xy
6.00 ksi
u2
11.24 ksi
b
7.5 ksi
For max 12 ksi, center of Mohr’s circle lies at point C. R 12 ksi xy
10.5 12 10.5 12 0 1 ( max 2
10.5
4.52
6.00 ksi
a
4.5 ksi
2
7.52
Checking,
10.5 ksi
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
(b)
6 ksi
xy
11.24 ksi
22.5 ksi 1.5 ksi min )
12 ksi
(o.k.)
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PROBLEM 7.75
y
For the state of stress shown, determine the value of maximum shearing stress is 80 MPa.
70 MPa
xy
for which the
τ xy 120 MPa z x
SOLUTION 120 MPa
x
1 ( 2
ave
x
2 Assume
min
0
max
2
y)
x
120
y
y
70 MPa
95 MPa
70
25 MPa
2
160 MPa
max
a
max
ave
R
R
max
ave
160
95
65 MPa
652
252
2 x
R2
y
2 xy
2
2 2 xy
R
x
2
y
2
602 xy
b
a
2R
160
130
30 MPa
0
60.0 MPa
(o.k.)
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y
PROBLEM 7.76 σy
For the state of stress shown, determine two values of maximum shearing stress is 73 MPa.
y
for which the
48 MPa 50 MPa z x
SOLUTION 50 MPa,
x
y
u
Let
1 ( 2
(1a)
u ave
55 MPa 1 ( 2
a
a
(1b)
u ave b
min
R
ave
0
max
ave
78 MPa,
min
2u
y
R
178 MPa,
2 xy
u R2
u
732
u
x
482
2 xy
55 MPa
5 MPa b
y)
x
y)
x
60 MPa
x
78 MPa,
55 MPa 1 ( 2
2u
y y)
x
73 MPa,
R
max
2u
y
x
u2
R
Case (1)
x
2
ave
48 MPa
xy
ave
1 ( 2
max
73 MPa
160 MPa (reject)
105 MPa,
max
68 MPa
68 MPa,
x
178 MPa,
R
a c
max
R
ave
0,
0
max min )
32 MPa
89 MPa
73 MPa y
60.0 MPa
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PROBLEM 7.76 (Continued)
Assume
Case (2)
0.
x
u
u2
2 xy
2(
x
b )u
u2
146 MPa
min b
ave
u2
2 xy
x
u2
2 xy
2u
max
(
2 xy
min )
73 MPa
b
b)
x
max
b
R u
1 ( 2
max
(
2
b)
x
x
2
(48)2
b
u
36 MPa
R
u2
a
b
2 xy
2R
y
2u
( 50 146)2 50 146 x
72 MPa
122 MPa
60 MPa
146
120
26 MPa
(o.k.) y
122.0 MPa
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y
PROBLEM 7.77 σy
For the state of stress shown, determine two values of maximum shearing stress is 10 ksi.
y
for which the
y
2.00 ksi
8 ksi 14 ksi z x
SOLUTION 14 ksi,
x
y
u
Let
1 ( 2
u ave
(1b)
6 ksi 1 ( 2
x
max
30 ksi,
u
6 ksi
ave
max
1 ( 2
x
18 ksi,
y y)
min
y y)
min
2u
20 ksi,
2u
a
max
2 ksi,
u
x
R2
u
2 xy
6 ksi
u
1 ( 2
30 ksi,
R
ave
min )
max
b
15 ksi
R
ave
10 ksi
7.5 ksi
2 ksi
x
8 ksi,
x
26 ksi (reject)
x
0,
2 xy
10 ksi
max
2u
y)
10 ksi,
R
max
8 ksi, y
x
u2
R
(1a)
x
2
ave
Case (1)
xy
a
ave
max
R
1 ( 2
18 ksi,
max
min )
b
ave
R
2 ksi
10 ksi (o.k.)
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PROBLEM 7.77 (Continued)
Assume
Case (2) a
ave
a
x
x
(
a
x)
2u
u2
u)2
a
2
x)
u ave
a max
y)
x
ave
R
20 ksi,
2 xy
u2
u2
(20
x
2.3333 ksi 1 ( 2
x )u
2 xy
a
u2
y
20 ksi =
a
2 xy
14)2 82 20 14
2u
x
0,
b max
4.6667 ksi
9.3333 ksi u2
11.6667 ksi R
20 ksi min
max
2 xy
a
2
2
max
2 xy
u2
2(
a
u
x
u
(
(
R
0.
min
ave
2 xy
R
8.3333 ksi
3.3334 ksi
10 ksi y
9.33 ksi
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PROBLEM 7.78
y
σ y & 100 MPa
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.
60 MPa z
x
τ xz
SOLUTION
60 MPa,
x
0,
z
y
100 MPa
For Mohr’s circle of stresses in zx plane,
Assume
max
y
min
b
30
(2)(60) ( 20)
30
xz
u
x
z)
x z
2
30 MPa
30 MPa
max
20 MPa
b
50 MPa
R
ave
R
2
max
ave
a
1 ( 2
100 MPa
100 R
ave
50
u2 R2 502
80 MPa <
y
2 xz
u2 302
40 MPa 40.0 MPa
xz
40.0 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. 1115
PROBLEM 7.79
y
For the state of stress shown, determine two values of maximum shearing stress is 80 MPa.
σy
y
for which the
90 MPa z
x 60 MPa
SOLUTION x
90 MPa,
0,
z
xz
60 MPa
Mohr’s circle of stresses in zx plane: ave
1 ( 2
x
z)
x
y
45 MPa 2
R
a
Assume
ave
a
max y
min
120 Assume
min y
b max
30
R
120 MPa,
b
ave
2
R
2 zx
452
602
75 MPa
30 MPa
120 MPa. max
2
max
(2)(80)
40.0 MPa
y
30 MPa. min
2
max
(2)(80)
y
130.0 MPa
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y
PROBLEM 7.80* 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 x
u
Let
1 ( 2
ave
Assume Then
max
y y
2 y)
x
u
x
u2
a
ave
R
x
u
u2
2 xy
b
ave
R
x
u
u2
2 xy
is minimum if u
2 xy
R
max
0.
2u
140 MPa,
y
x
R
xy
80 MPa
a
ave
R 140 80
b
ave
R 140 80 60 MPa
max
2u
R
is the in-plane shearing stress.
max (in-plane)
x
x
220 MPa,
x
u 140 MPa
220 MPa
0,
min
ave
max
1 ( 2
max
min )
110 MPa
Assumption is incorrect. Assume
max min
d a du
a
R
ave
0
max
1
u u
2
2 xy
x
1 ( 2
u max
0
u2 min )
2 xy
1 2
a
(no minimum)
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PROBLEM 7.80* (Continued)
Optimum value for u occurs when
1 ( 2 (
a
x
R) u )2 2u
max (out-of-plane)
R or 2 x
2u
2 x
2 xy x
(a) (b)
y
R
x
u2
a
x
max (in-plane)
R or u
2
u
u
x 2
140 2 80 2 140
u2
2 xy
2 xy
94.286 MPa
u
2u 140 94.286 2 xy
max
47.143 MPa y
92.857 MPa
max
45.7 MPa 92.857 MPa
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σ0
PROBLEM 7.81 100 MPa
σ0
The state of plane stress shown occurs in a machine component made of a steel with 325 MPa. Using the maximum-distortion-energy criterion, determine whether Y 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
(a)
0
200 MPa
ave a
2 a
F . S.
(b)
0
240 MPa
ave a
2 a
F . S.
(c)
0
280 MPa
ave a
2 a
2 b
a
b
R
0
x
y
2 xy
2
100 MPa
200 MPa ave
2 b
R a
100 MPa, b
b
ave
R
300 MPa
264.56 MPa < 325 MPa
(No yielding)
325 264.56
F . S . 1.228
240 MPa ave
2 b
R a
140 MPa, b
b
ave
R
340 MPa
295.97 MPa < 325 MPa
(No yielding)
325 295.97
F . S . 1.098
280 MPa ave
R
180 MPa,
329.24 MPa > 325 MPa
b
ave
R
380 MPa (Yielding occurs)
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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
(a)
0
200 MPa:
ave a max
2
max
F . S.
(b)
0
240 MPa:
ave a max
2 (c)
0
280 MPa:
max ave a max
2
max
R
0
x
y
2 xy
2
100 MPa
200 MPa ave
0,
R
100 MPa
ave
R
300 MPa
300 MPa
min max
b
min
300 MPa
325 MPa
(No yielding)
325 300
F . S . 1.083
240 MPa ave
0,
R
140 MPa,
min
max
min
b
ave
R
340 MPa
340 MPa 340 MPa > 325 MPa
(Yielding occurs)
280 MPa ave
0, max
R
180 MPa,
min min
b
ave
R
380 MPa
380 MPa 380 MPa
325 MPa
(Yielding occurs)
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PROBLEM 7.83
21 ksi
τ xy
The state of plane stress shown occurs in a machine component made of a steel with 45 ksi. Using the maximum-distortion-energy criterion, determine whether Y 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
36 ksi,
x
For stresses in xy plane,
1 ( 2
ave
x
y
21 ksi,
y y)
x
z
0
28.5 ksi
7.5 ksi
2 2
(a)
xy
9 ksi
a
ave
2 a
F .S .
2 b
x
R R a
y
2 xy
2 40.215 ksi,
ave
b
34.977 ksi
b
(7.5) 2 R
(9) 2
11.715 ksi
16.875 ksi
45 ksi
(No yielding)
45 39.977
F .S .
1.287
2
(b)
x
18 ksi R
xy
ave
a
2 a
F .S .
2 b
R a
y
2 xy
2 48 ksi,
ave
b
44.193 ksi
b
(7.5)2 R
(18)2
19.5 ksi
9 ksi
45 ksi
(No yielding)
45 44.193
F .S .
1.018
2
(c)
xy
20 ksi
a
ave
2 a
2 b
R R a
x
2 xy
2
49.86 ksi, b
y
b
46.732 ksi
ave
(7.5) 2 R
(20) 2
21.36 ksi
7.14 ksi
45 ksi
(Yielding occurs)
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PROBLEM 7.84
21 ksi
Solve Prob. 7.83, using the maximum-shearing-stress criterion.
τ xy
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.
36 ksi
SOLUTION
36 ksi,
x
21 ksi,
y
0
z
For stress in xy plane,
1 ( 2
ave
y)
x
28.5 ksi
x
y
2
7.5 ksi
2
(a)
xy
9 ksi
a
ave
max
2
max
F .S.
x
R R
2 40.215 ksi,
34.977 ksi, max
y
b
11.715 ksi
ave
R
16.875 ksi
0
min
40.215 ksi
min
2 xy
(No yielding)
45 ksi
45 40.215
F .S .
1.119
2
(b)
xy
a
2
x
18 ksi R ave
max
48 ksi
max
max
y
2 xy
2
R
48 ksi,
9 ksi
0
min min
R
ave
b
19.5 ksi
48 ksi
(Yielding occurs)
45 ksi 2
(c)
xy
a max
2
max
x
20 ksi R ave
R
min min
2 xy
2 49.86 ksi
49.86 ksi max
y
b
ave
21.36 ksi R
7.14 ksi
0
49.86 ksi
(Yielding occurs)
45 ksi
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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 A
x y ave
240 103 N d2
4
4
1.13411 103 mm 2
240 103 1.13411 10
P A 0 1 ( 2
(38) 2
x
y)
x
y
3
1 2
211.62 106 Pa
2
2R
2 xy
2 Y
4
xy
From torsion:
xy
J c
T
1 2
1 4
2 xy
2
max
2 x
2 xy
4
2 x
2 x
1 250 2 2
2 x
2 Y
211.62 2
66.553 106 Pa
Tc J
J
c4
1 d 2
2 xy
y
66.553 MPa
2
211.62 MPa
x
2
R
1.13411 10 3 m 2
T
2
38 2
xy
c 4
204.71 103 mm 4
204.71 10 9 m 4
19 10 3 m
(204.71 10 9 )(66.553 106 ) 19 10 3 717 N m
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PROBLEM 7.86 Solve Prob. 7.85, using the maximum-distortion-energy criterion.
B
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.
T P
A
d = 38 mm
SOLUTION P A
x y ave
240 103 N 4
d2
(38) 2
4
1.13411 103 mm 2
240 103 1.13411 10
P A 0 1 ( 2
x
y)
x
y
1 2
211.62 106 Pa
3
2 a
2 b
1 4
2 xy
2
2 xy
2 x
a
ave
R
1 2
x
1 4
2 x
2 xy
b
ave
R
1 2
x
1 4
2 x
2 xy
a b
1 4
2 x
1 4 2 x 2 xy
xy
1 4
x
2 x
3
x
2 xy
1 4
2 xy
2 x
1 4
211.62 MPa
x
2
R
1.13411 10 3 m 2
2 x
2 xy
1 4
2 x
2 xy 2 x
2 xy
1 4
2 x
1 4
2 x
2 xy
2 Y
1 2 2 Y x 3 1 2502 211.622 3
76.848 MPa
76.848 106 Pa
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PROBLEM 7.86 (Continued)
From torsion,
xy
J c
T
Tc J
2
c4
1 d 2
T 38 2 2
J
xy
c
4
204.71 103 mm 4
204.71 10 9 m 4
19 10 3 m
(204.71 10 9 )(76.848 106 ) 19 10 3 828 N m
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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
d2
4
y ave
4
1.76715 in 2
60 1.76715
P A
x
(1.5) 2
33.953 ksi
0 1 ( 2
x
y)
x
y
1 2
x
2
R 2
2
max
2R
2 xy
2 Y
4
xy
1 2
2 x
1 4
2 xy
4
2 xy
2 x
2 xy
Y
2 x 2 Y
2 x
1 422 2
33.9532
12.3612 ksi From torsion,
xy
c
J
T
Tc J 1 d 2
2
c4
T
J
xy
c
0.75 in.
0.49701 in 4
(0.49701)(12.3612) 0.75 8.19 kip in.
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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
A
d2
4
ave
1.76715 in 2
60 1.76715
p A
x
y
4
(1.5) 2
33.953 ksi
0 1 ( 2
x
y)
x
y
1 2
x
2
R
2
a
ave
b
2 a
2 b
a b
ave
(
ave 2 ave 2 ave
1 4
3
2 xy
2 Y
xy
1 3
1 4
2 xy
2 x
2 xy
R R
R) 2 2 3R
2 x
3
(
ave R 2
1 4
R) 2
ave 2
R
2 x
(
2 ave
2 xy
2 2 x
R)(
ave
ave 2
ave R
R
2 xy
2 Y
3
R) 2 ave
R2
2 x 2 Y
2 x
1 422 3
33.9532
14.2734 ksi From torsion,
xy
c J
T
Tc J 1 d 2
T
J
xy
c
0.75 in.
c4 (0.75) 4 0.49701 in 4 2 2 (0.49701)(14.2734) 9.46 kip in. 0.75
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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 nd using Mohrr’s criterion, determine d wheether rupture 200 MPa an UC of thee casting will occur. o
60 MPa
M 10 MPa
SO OLUTION x y xy
10 MPa, M 10 00 MPa, 60 MPa x
ave
10 1000 2
y
2
45 MPaa
2 x
R
y
2 xy
2
(55) 2
(60)2
a
avee
R
45 81.39 36.39 MPa
b
avee
R
45 81.39 8
81.399 MPa
126.39 MPa
Equuation of 4th quadrant q of bo oundary:
36.39 80
a
b
UT
UC
1
( 1226.39) 1.087 1 2000 Rupture will w occur.
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PROBLEM M 7.90
75 MPa
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.
32 MPa
SOLUTION x y xy ave
32 MPa, M 0, M 75 MPa 1 ( 2
x
y)
x
y
M 16 MPa 2
R
(16) 2
2 xy
2
(775) 2
a
ave
R
16 766.69 60.69 MPa M
b
ave
R
16 766.69
76.69 MPa M
92.69 MPa
Equuation of 4th quadrant q of bouundary:
60.69 80
a
b
UT
UC
1
( 92..69) 1.222 1 200 Rupture will w occur.
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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 y xy ave
8 ksi, 0, 7 ksi 1 ( 2
x
y)
x
y
4 ksi 2
R
2 xy
2
42
a
ave a
R
4 8.062
b
ave a
R
4 8.062
72
8.062 ksi
4.0622 ksi 12.0062 ksi
Equuation of 4th quadrant q of bo oundary:
4.062 10
a
b
UT
UC
( 122.062) 330
1 0.8088 1 No rupture.
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PROBLEM M 7.92
15 ksi k
The state of plane p stress shhown is expeccted to occur in an aluminuum casting. 30 ksi Knowing thatt for the alum minum alloy used u 10 ksi and UC U UT and using Mohr’s criterion, determine whhether rupture of the casting will occur.
9 ksi
2 ksi
SOLUTION x
2 ksi, 15 ksi,
y xy ave a
9 ksi 1 ( 2
x
y)
x
y
6.5 ksii 2
R
2 xy
2
a
ave
R 5.879 ksi
b
ave
R
8.52
92
1 12.379 ksi
18.879 ksii
q of bouundary: Equuation of 4th quadrant
5.879 10
a
b
UT
UC
1
( 18.879) 1.217 1 30 Rupture will occur.
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PROBLEM M 7.93 The state of plane stress shown s will occcur at a critical point in an a aluminum 25 ksi. casting that is made of ann alloy for whhich UT 10 ksi and UC U Using Mohrr’s criterion, determine d thee shearing stress 0 for which w failure should be exppected.
8 ksi
t0
SO OLUTION x
8 ksi,
y
0,
xyy
0
avee
1 ( 2
x
y)
x
y
4 ksi 2
R
R2
0
Sinnce
ave
42
2 xy
2
2 0
42
a
ave
R
(4 R) ksi
b
ave
R
(4 R) ksi
< R, stress point lies in 4th quaddrant. Equatioon of 4th quaddrant boundaryy is a
b
UT
UC
4 R 10 1 10
4 R 25
1 1
1 4 R 1 25 10 R
4 25
5.429 ksi
0
5.42992
42
0
3.67 ksi
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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 75 MPa and UC 150 MPa. M UT Mohr’s criteerion, determinne the shearinng stress 0 foor which failurre should be expected.
SOLUTION x y
80 MPa, 0,
xy
0
ave
1 ( 2
x
y)
x
y
40 MPa 2
R
a
ave
R
b
ave
R
0
Sincce
ave
2 xy
2
R2
402
2 0
MPa
40 2
< R, stress point lies in 4th quaddrant. Equationn of 4th quadrrant boundary is a
b
UT
UC
40 R 75 R 75
R 150
1
R 63.33 MP Pa,
40 R 150 40 75
40 150 0
1 1 1.2667
63.332
402
0
8.49 MPa
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PR ROBLEM 7.95 7 T'
Thee cast-aluminnum rod shoown is made of an alloyy for which a wing that the magnitude m T 70 MPa and 1775 MPa. Know UT T UC 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.
t0 T
SO OLUTION x
0
y
0
xy
0
ave
1 ( 2
x
y)
x
y
0 2
R
Sinnce
ave
2 xy
2
a
ave a
R
b
ave a
R
2 xyy
0
xy
R R
< R, stress point lies in 4th quaddrant. Equatioon of boundaryy of 4th quadrrant is a
b
UT
UC
R 700 1 70
1
R 1 175
1 R 1 1175 R 50 M MPa 0
R
0
5 50.0 MPa
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P PROBLEM 7.96 T cast-alum The minum rod shhown is madee of an alloyy for which U Mohr’s criterion, 60 MPa and UC 120 MPa. Using UT 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
y
2
Sincce
(32) 2
804.25 mm 2
32.3288 106 Pa
6
804.25 10 6 m 2
322.328 MPa
1 1 ( x (32.328 0) 0 16.164 MP Pa y) 2 2 1 (32.328 0)) 16.164 MP Pa 2
ave x
4
26 1003 804.25 10
P A
x
A
a
ave
R 16..164 R MPa
b
ave
R 16.164 R MPa
< R, stress point lies in the 4th quadrant. q Equaation of the 4thh quadrant is
ave a
b
UT
UC C
1 60
1
16.1644 600
R
1 16.1664 R 1 60 120
16.1644 R 1200
1
16.164 120
R
34 4.612 MPa
2
y
x
R
2 xy
2
R
xy
2
x
y
34.6122 116.1642
2
30.6606 MPa
30.606 6 106 Pa For torsion,
xy
T
Tc J 2
c3
2T c3 x xy
wherre c
2
1 d 2
166 mm 16 100 3 m
(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
!0
(a)
PROBLEM M 7.97
1! 2 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.
0
UT
(c)
SO OLUTION (a)
a b
0
1 2
0
Stress poinnt lies in 1st quadrant. q a
(b)
a
0
b
1 2
0
8.00 ksi
0
Stress poinnt lies in 4th quadrant. q Equaation of 4th quuadrant bounddary is a
b
UT
C UC 1 2
0
8
(c)
a
1 2
0,
b
0,
1
0
1
0
6.67 ksi
0
1
0
8.89 ksi
20
4th quadrannt 1 2
0
8
20
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
(a)
(b)
r
1 d 2
1
2
all
1
2
pr 2t
p
2t r
p
1.290 106 Pa
1
d
1 ( E 1 E
d
1
1 (3) 2
t
12 10
3
1.488 m
80 106 Pa
(2)(12 10 3 )(80 106 ) 1.488
1
p
1.290 MPa
d
0.852 mm
2)
1
1
1 0.29 (80 106 ) 9 200 10
(3)(284 10 6 )
284 10
6
852 10 6 m
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. 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 0.29. Knowing that the gage pressure in the container is increased from zero for which E 200 GPa and to 1.8 MPa, determine (a) the maximum normal stress in the container, (b) the corresponding increase in the diameter of the container.
SOLUTION
(a)
p
1.8 MPa
r
1 d 2
1
2
1 (5) 2
t
pr 2t
24 10
3
(1.8)(2.476) (2)(24 10 3 )
2.476 m
92.850 MPa 92.9 MPa
1
(b)
d
1 ( E
d
2)
1
1
1 E
1
(5)(329.6 10 6 )
1 0.29 (92.85 106 ) 200 109
1.648 10 3 m
329.6 d
1.648 mm
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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 r
1
d t 2 10 in. 0.25 in. 2 4.75 in. pr 2t (1150 psi)(4.75 in.) 2(0.25 in.) 2
10.925 ksi F.S.
U max
60 ksi 10.925 ksi F.S.
5.49
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PROBLEM 7.101 A spherical pressure vessel of 750-mm outer diameter is to be fabricated from a steel having an ultimate stress 400 MPa. Knowing that a factor of safety of 4.0 is desired and that the gage pressure can reach 4.2 MPa, U determine the smallest wall thickness that should be used.
SOLUTION r
We have and
max
F.S.
1 d t 2 1 (0.750 m) 2 0.375t (m) 1
t
pr 2t
2 U
max
Combining these two equations gives
F.S. or
2
Ut
2t U pr (F.S.) pr
Substituting for r gives 2(400 106 Pa)t 6
816.80 10 t t
(4)(4.2 106 Pa)(0.375 6.30 10
t)
6
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 on a website, in whole or part. 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 t r
20 ft
240 in.
7 in. 0.4375 in. 16 1 d t 119.56 in. 2 (75)(119.56) pr 10.25 103 psi 2t (2)(0.4375)
max
10.25 ksi
min
0
(Neglecting small radial stress)
1 ( 2
max
max
min )
10.25 ksi
max
5.12 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. 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 r
1 d t 2 1 (300 mm) 2 147 mm
1
2
3 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
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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
t
5 mm
r
1 d 2
all
all
p
but
8m
U
F.S. pr t t
all
r
p
gh,
h
p g
0.005 m t
4
0.005
400 MPa 4.0
3.995 m 100 MPa
(0.005 m)(100 106 Pa) 3.995 m
125.156 103 Pa (1000 kg/m3 )(9.81 m/s 2 )
100 106 Pa
125.156 103 Pa
12.7580 m h
12.76 m
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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 t r
p
8m 5 mm 0.005 m 1 d t 4 0.005 2
gh
3.995 m
(1000 kg/m3 )(9.81 m/s2 )(14.5 m) 142.245 103 Pa
1
pr t
(142.245 103 Pa)(3.995 m) 0.005 m
113.654 106 Pa max min max
1
max
113.7 MPa
0 1 ( 2
max
min )
max
56.8 MPa
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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
d 2
1
pr t
max
1
min
p
0
1 ( 2
max
max
3.3 2
t
18 10
3
1.632 m,
(1.5 106 Pa)(1.632 m) 18 10 3 m 136 106 Pa
min )
t
18 10 3 m
136 106 Pa
max
68 106 Pa
max
136.0 MPa
68.0 MPa
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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)
d0
12.75 in. t pr t
(b)
d0
(400)(6.00) 0.375
12.75 in. t pr t
0.375 in. r
t
6.00 in.
6400 psi
0.500 in. r
(400)(5.875) 0.500
1 d0 2
6.40 ksi
1 d0 2
t
5.875 in.
4700 psi
4.70 ksi
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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 2
t
3 10 3 m
320 2
t
157 mm
157 10 3 m
(1.5 106 Pa)(157 10 3 m) 3 10 3 m
1
pr t
max
1
min
p
0
1 ( 2
max
max
3
78.5 106 Pa
min )
78.5 106 Pa
max
78.5 MPa
max
39.3 MPa
39.25 106 Pa
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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 1
r
1
65 ksi 13 ksi 13 103 psi F .S. 5.0 d (5.5)(12) t 0.625 32.375 in. 2 2 U
pr t
p
t
1
r
(0.625)(13 103 ) 32.375
p
251 psi
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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 d 2
t
1 (36) 2
p
rh
(62.4 lb/ft 3 )(500 ft)
0.5
17.5 in.
31.2 103 lb/ft 2
216.67 psi 1
pr t
max
1
min
p
max
1 ( 2
(216.67)(17.5) 0.5
7583 psi
7583 psi
max
7.58 ksi
max
3.90 ksi
217 psi max
min )
3900 psi
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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
r 1
18
t t 18 t
12.5 ksi
12.5 103 psi
1 d t 18 t 2 pr r 1 , t t p 12.5 103 216.67
57.692
t
58.692
0.307 in.
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. 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 r 1
2
ave
R w
p
1 d t 2 pr t 1 pr 2 t 1 ( 1 2
1 (600) 2
6
292 mm
3 pr 4 t 1 pr 1 2 2 4 t R cos 60 ave 2)
5 pr 8 t 8 wt 5 r p
8 (75)(8) 5 292
3.29 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. 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
p
1 d t 2 pr t 1 pr 2 t 1
2
2 R sin 60
1 (600) 2
8
292 mm
1 pr 4 t
3 pr 8 t 8 wt 3 R p
8 (30)(8) 3 292
3.80 MPa
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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 2
1
pr t
2
1 2
375 mm
0.375 m
(1.5 106 Pa 0.375 m) 0.009 m 1
31.25 MPa
62.5 106 Pa
2
100
ave
1 ( 2
R
1
(a)
2)
1 2
2 w
62.5 MPa
46.875 MPa
15.625 MPa ave
R cos100 w
(b)
w
44.2 MPa
R sin100 w
15.39 MPa
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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
ave
1 ( 2
R w
0.85
pr t
cos 2 2
2
pr 2t
3 pr 4 t 1 pr 1 2 2 4 t R cos 2 ave 2)
1
3 4
1 cos 2 4
4 0.85
3 4
pr t
0.4
113.6
56.8
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12 ft
PROBLEM 7.116
12 ft
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.
45" 20 ft
(a)
(b)
SOLUTION d
12ft
1
pr t
(a)
1
144 in. r 2
1 d 2 pr 2t
t
71.5 in.
12 ksi 1t
p
(12)(0.5) 71.5
r
0.0839 ksi p
(b)
ave
1 ( 2
R
1
w
ave
2)
1
2 45
2
83.9 psi
3 pr 4 t 1 pr 4 t
R cos
3 pr 4 t p
4 wt 3 r
4 (12)(0.5) 3 71.5
0.1119 ksi
p
111.9 psi
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. 1155
PROBLEM 7.117
3m
The pressure tank shown has a 0.375-in. wall thickness and butt-welded 20° with a transverse plane. For a gage seams forming an angle pressure of 85 psi, determine (a) the normal stress perpendicular to the weld, (b) the shearing stress parallel to the weld.
1.6 m
!
SOLUTION d r 1
2
ave
R
(a)
w
(b)
w
ave
R cos 40
R sin 40
5 ft
60 in.
1 d t 30 0.375 29.625 in. 2 (85)(29.625) pr 6715 psi 0.375 t 1 3357.5 psi 1 2 1 ( 1 5036.2 psi 2) 2 1
2
2
1678.75 psi
3750 psi
1079 psi
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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 20° with a transverse plane. and butt-welded seams forming an angle 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 2 pr t pr 2t 1 ( 2
r 1
2
ave
t
30
1
w
ave
2
R cos 50 1 cos 50 4
0.58930
p w
p
wt
0.5893r
(18)(0.375) (0.58930)(29.625)
R sin 50
0.191511
wt
0.191511r
0.38664 ksi
pr t (10)(0.375) (0.191511)(29.625)
3 pr 4 t 1 pr 4 t
2 3 4
29.625 in.
2)
1
R
0.375
pr t
pr t
387 psi
0.66097 ksi
661 psi
p
Allowable gage pressure is the smaller value.
387 psi
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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 20° with a transverse plane. and butt-welded seams forming an angle 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 r 1
2
R w
sin 2 2
a
53.53
a
2
b
53.53
b
26.8
2
c
53.53
c
63.2
2
d
d
116.8
53.53
180 180
126.47 233.53
26.8
a
5 ft
60 in.
1 3 29.625 in. d t 30 2 8 (85)(29.625) pr 6715 psi 0.375 t 1 3357.5 psi 1 2 1
2
2 R sin 2 w
R
1678.75 all
1350 1678.75
0.80417
26.8 63.2
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PRO OBLEM 7.120 4 ft
P'
A
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 35"
B
SOLUTION
r0
1 d 2 pr t pr 2t r t
A
r02
r 1
2
P A Totaal stresses.
1 t 0.25 in. (10) 5 in. 2 (3000)(5) 6000 psi p 6 ksi 0.225 (3000)(5) 3000 psi p 3 ksi (2)(00.25) 5 0.25 5.25 inn. (5.252
r2
100 103 8.0803
5.002 ) 8.05003 in 2
12442 psi
Longitudinal:
x
3 1.242 1.7588 ksi
Circumferential:
y
6 ksi k
Shear:
xy
1.242 ksi
0
Plottted points forr Mohr’s circlee:
X : (1.758, 0) Y : (6, 0) C : (3.879) ave
1 ( 2
x
y)
x
y
3.8879 ksi 2
R
2 xy
2 ((1.758 6) 2
(a) (b)
x
|
xy |
avee
R cos 70
R siin 70
2
0
2.121 kssi
3.879 2.1221 cos 70
2.1211 sin 70
x
|
xy
3.15 ksi
| 1.993 1 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. 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
r0
1 d 2 pr t pr 2t r t
A
r02
r 1
2
Tottal stresses.
1 ( (10) 5 in. t 0.25 in. 2 (3000)(5) 6000 psi 6 ksi 0 0.25 (300)(5) 30000 psi 3 ksi (2))(0.25) 5 0.25 5.25 in.
r2
(5.252
52 ) 8.05033 in 2
P A
30 103 8.0503
37727 psi
Longitudinall:
x
3 3.727
0.7727 ksi
Circumferen ntial:
y
6 ksi
Shear:
3.7727 ksi
0
xy
Plootted points for Mohr’s circlle:
X : ( 0.727, 0) 0 Y : (6, 0) C : (2.66365, 0) av ve
1 ( 2
x
y)
x
y
2.6365 kssi 2
R
0.7277 6 2 (a) (b)
x
|
xy
ave
|
R cos 70
R sin 70
2 xy
2 2
0
3 3.3635 ksi
2.6365 3.3635 cos 70
3.36 635 sin 70
x
|
xy
1.486 ksi
| 3.16 ksi
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T
PROBLEM 7.122 2 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
1 d 2
180 mm m r
90 mm t
12 mm
Torssion: c1
90 mm m c2
J
c24
2
90
c14
12
102 mm m
66.9668 106 mm 4
(12 103 )(1022 10 3 ) 66.968 10 6
Tc J
66.968 10 6 m 4
188.277 MPa
Presssure: pr t
1
(8)(90) 12
600 MPa
2
pr 2t
30 MP Pa
Sum mmary of stresses: x
ave
60 MPa, M 1 ( 2
30 MPa,
y
x
y)
x
y
xy
18.277 MPa
45 MPa 2
R
2 xxy
2
a
avee
R
M 68.64 MPa
b
avee
R
21.36 MPa M
c
min
max
Pa 23.64 MP
0 max
688.6 MPa
max
344.3 MPa
0
1 ( 2
max
min )
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. 1161
T
PROBLEM 7.123 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 r 1
2
ave max
R
1 1 d (180) 90 mm 2 2 pr (8)(90) 60 MPa t 12 pr 30 MPa 2t 1 ( 1 45 MPa y) 2 75 MPa max
t 12 mm
30 MPa
ave 2
R xy
1
2
2 R 2 152
2 xy
152
302 152
2 xy
25.98 MPa
6
25.98 10 Pa
Torsion:
c1
90 mm
c2
90 12 102 mm
J 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 T 17.06 103 N m 3 J c 102 10 T
17.06 kN m
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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 r x
y
Stress due to bending moment:
5 MPa 5 106 Pa 1 250 d t 8 117 mm 2 2 pr (5 106 )(117 10 3 ) 73.125 MPa t (8 10 3 ) (5 106 )(117 10 3 ) (2)(8 10 3 )
pr 2t
Point K is on the neutral axis. 0
y
Stress due to transverse shear:
36.563 MPa
V c2 c1 Q
I
xy
P 40 103 N 1 d 125 mm 2 c2 t 117 mm 2 3 3 2 c2 c1 (1253 1173 ) 3 3 234.34 103 mm3 234.34 10 6 m3 (1254 117 4 ) c24 c14 4 4 44.573 106 mm 4 44.573 10 6 m 4 VQ It
PQ I (2t )
(40 103 )(234.34 10 6 ) (44.573 10 6 )(16 10 3 )
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PROBLEM 7.124 (Continued)
Total stresses:
x
Mohr’s circle:
ave
73.125 MPa, 1 ( 2
x
y)
36.563 MPa,
y
xy
13.1436 MPa
54.844 MPa 2 x
R
y
2 xy
2 (18.281) 2
(13.1436) 2
a
ave
R
77.360 MPa
b
ave
R
32.328 MPa
22.516 MPa
Principal stresses:
a
77.4 MPa,
b
The 3rd principal stress is the radial stress.
z max
Maximum shearing stress:
32.3 MPa
max
77.4 MPa,
1 ( 2
max
min
min )
0
max
max
0
77.4 MPa 38.7 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. 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.
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.
B
P
600 mm K
L
A z
m 150 mm x
SOLUTION Connsider elementt at point L. Streesses due to in nternal pressurre:
p r x
y
ding moment: Streess due to bend
5 MPa 5 106 Pa 1 250 d t 8 1177 mm 2 2 pr (5 106 )(117 100 3 ) 73.125 MPa t 8 10 3 pr (5 103 )(117 100 3 ) 36.563 MPa 2t (2)(8 10 3 )
M c2 c1 I
y
(40 kN)(600 k mm) 24,000 N m 1 d 125 mm 2 c2 t 125 8 117 mm c24 c14 (1254 117 4 ) 4 4 44.573 106 mm 4 44.573 10 6 m 4 Mc I
(24, 000)(125 10 3 ) 44.573 10 6
Pa 67.305 MP
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PROBLEM 7.125 (Continued)
Stress due to transverse shear:
Point L lies in a plane of symmetry. xy
Total stresses:
x
Principal stresses:
max
73.125 MPa,
0 30.742 MPa,
y
xy
0
Since xy 0, x and y are principal stresses. The 3rd principal stress is in the radial direction, z 0. 73.125 MPa,
min
0,
a
73.1 MPa,
b
Maximum stress: Maximum shearing stress:
max
1 ( 2
max
min )
30.7 MPa,
z
0 max
73.1 MPa
max
51.9 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. 1166
PROBLEM 7.126
1.5 in.
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.
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
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 pr Es t s
s
Corresponding strain:
sp
Es
Strain due to temperature change:
sT
s
Total strain:
(1)
T
s
pr Es t s
Ls
2 r
b
pr tb
s
T
Change in length of circumference:
Brass ring.
External pressure p:
Corresponding strains:
s
2 r
pr Es ts
bT
b
pr , Eb tb
bp
s
T
T
Change in length of circumference:
Lb Equating
Ls to Lb ,
pr Es t s
r Es t s
s
2 r
(
b
pr Eb tb
2 r
pr Eb tb
T
r p Eb tb
b
b
s)
T
b
T
T
(2)
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PROBLEM 7.126 (Continued)
T 125 F 50 F 75 F
Data:
r From Equation (2),
2.5 (29 106 )(0.125)
1 d 2
1 (5) 2
2.5 in.
2.5 p (15 106 )(0.25)
(11.6 6.5)(10 6 )(75)
1.35632 10 6 p 382.5 10 p From Equation (1),
s
pr ts
(282.0)(2.5) 0.125
6
282.0 psi
5.64 103 psi
(a)
s
(b) p
5.64 ksi
282 psi
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PROBLEM 7.127
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
Solve Prob. 7.126, assuming that the brass ring is 0.125 in. thick and the steel ring is 0.25 in. thick. 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 pr Es t s
s
Corresponding strain:
sp
Es
Strain due to temperature change:
sT
s
Total strain:
(1)
T
s
pr Es t s
Ls
2 r
s
T
Change in length of circumference:
Brass ring.
External pressure p:
2 r
pr Es t s
bT
b
s
T
pr tb
b
Corresponding strains:
s
pr , Eb tb
bp
T
Change in length of circumference:
Lb Equating
Ls to Lb ,
pr Es t s
r Es ts
s
2 r
b
pr Eb tb
T
r p ( Eb tb
b
pr Eb tb
2 r
b
s)
T
b
T
T (2)
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PROBLEM 7.127 (Continued)
Data:
T
125 F 50 F 75 F 1 1 (5) 2.5 in. r d 2 2
From Equation (2),
2.5 (29 106 )(0.25)
2.5 p (15 106 )(0.125)
(11.6 6.5)(10 6 )(75)
1.67816 10 6 p 382.5 10 p From Equation (1),
s
pr ts
(227.93)(2.5) 0.25
6
227.93 psi
2279 psi
(a)
s
(b) p
2.28 ksi
228 psi
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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 .
800 ,
x
450 ,
y
xy
200 ,
25
SOLUTION 25 x
y
2
x x
xy
y
y
2 175 (
y
x
( 800
x
xy
625
2
2 175 x
y
x
175
y
2
100
xy
cos 2 sin 2 2 2 ( 625 ) cos ( 50 ) (100 )sin ( 50 ) x
y
xy
cos 2 sin 2 2 2 ( 625 ) cos ( 50 ) (100 )sin ( 50 ) y )sin
2
653
x
xy
y
303
cos 2
450 )sin ( 50 ) ( 200 ) cos ( 50 )
xy
829
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PROBLEM 7.129 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
240 ,
x
160 ,
y
xy
150 ,
60
SOLUTION 60 x
y
2
x
x
xy
y
2 y
x
y
xy
40
y
x
y
cos 2
2
75
xy
sin 2 2 2 2 200 40 cos ( 120 ) 75 sin ( 120 ) x
y
x
200
xy
sin 2 2 2 2 200 40cos ( 120 ) 75sin ( 120 ) (
x
y )sin
cos 2
2
xy
115.0
x
y
285
cos 2
(240 160)sin ( 120 ) 150 cos ( 120 )
xy
5.72
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PROBLEM 7.130 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
500 ,
x
250 ,
y
xy
0,
15
SOLUTION 15 x
y
2
x x
xy
2 125
x
y
y
x
( 500
x
y
cos 2
cos 2
2 ( 375 ) cos 30 y )sin
2
xy
375
2 ( 375 ) cos 30
2 125 (
y
2 y
x y
x
125
xy
2 xy
2 0 xy
2 0
0
sin 2 x
450
y
199.8
sin 2
cos 2
250 )sin 30
0
xy
375
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PROBLEM 7.131 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 . 0,
x
y
320 ,
xy
100 ,
30
SOLUTION 30 x
y
2
x
160
2
x x
y
y
x
2
y
2
x
y
2
x
y
2
160 160cos 60 xy
(
x
y )sin
cos 2
xy
2
sin 2
100 sin 60 2
160 160 cos 60
y
160
2
(0 320)sin 60
cos 2
xy
2
100 sin 60 2 xy
x
36.7
sin 2 y
283
xy
227
cos 2
100 cos 60
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PROBLEM 7.132 y
y'
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' x
!
800 ,
x
450 ,
y
xy
200 ,
25
SOLUTION Plotted points:
X : ( 800 , 100 ) Y : ( 450 , 100 ) C : ( 175 , 0) 100 625
tan
(625 ) 2
R
x
ave
9.09 (100 ) 2
2
50
R cos
175
9.09
632.95
40.91
632.95 cos 40.91 653
x y
ave
R cos
175
632.95 cos 40.91 y
1 2
xy
R sin
632.95 sin 40.91
xy
303 829
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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 . 240 ,
x
y
160 ,
xy
150 ,
60
SO OLUTION Plootted points for Mohr’s circlle:
X : ( 240 2 , 75 ) Y : ( 1160 , 75 ) C : ( 200 2 , 0) 75 1.875 40
tan n
(40 )2
R 2
1 2
61.933 (75 )2
8 85
x
ave
181.93 120 61.93 R cos c 200 (85 ) cos ( 181.93 )
y
ave
R cos c
xy
R sin
200
(85 ) cos ( 181.93 )
85 sin ( 181.93 )
2 2.86
115.0
x y xy
285 5.72
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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 . 500 ,
x
y
250 ,
xy
0,
15
SOLUTION
Plotted points:
X : ( 500 ,0) Y : ( 250 , 0) C : ( 125 , 0)
1 2
R
375
x
ave
R cos 2
125 375cos 30
y
ave
R cos 2
125 375cos 30
xy
R sin 2
375sin 30
x
y
xy
450 199.8 375
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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 . 0,
x
y
3200 ,
xy
100 ,
30
SO OLUTION Plootted points for Mohr’s circlle:
X : (0, 50 ) Y : (320 , 50 ) C : (160 , 0) 50 160
tan n
17.35
(160 ) 2
R
2
1 2
(50 ) 2
60
167.63
17.35
42.65
x
ave
R cos c
160
(167.63 ) coos 42.65
y
ave
R cos c
160
(167.63 ) coos 42.65
xy
R sin
(167.63 )ssin 42.65
x
36.7
y
283
xy
227
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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 . ) 2600 ,
x
600 ,
y
xy
4880
SOLUTION For Mohr’s circlee of strain, plot points:
X : ( 2660 , 240 ) Y : ( 600 , 240 ) C : ( 1660 , 0) tan 2
x
2
ave
R
160
260
b
ave
R
160
260
max (in-plaane)
R
max m
(c)
max
max x
1 v
(
160 min
a
min
160
2
(240 )
a
56.3
a
100
2
420
b
2R
max ((in-plane)
v c
33.7
260
a
1 2
2.4 b
(100 )
R
(b)
y
67.38
p
R
(a)
480 260 60
xy p
max (in-plane))
v
b)
1 v
(
x
y)
1/3 ( 260 60) 2/3 160
420
c
420
520
maxx
160 580
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 . ) 6000 ,
x
y
4000 ,
xy
3 350
SO OLUTION Plootted points for Mohr’s circlle:
X : ( 600 , 175 ) Y : ( 400 , 175 ) C : ( 500 , 0) 0 tan 2
p
2
p
175 100
60.26 30.1
b a
(100 ) 2
R
59.9
(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 v ( 1 v
c
(c)
max
max
500
max
min
a
min
500 5
v ( 1 v
b)
x
y)
1/3 ( 600 2/3
max (in-planee)
403
c
500
400 )
702 702
max
1202
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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 . ) 160 ,
x
4800 ,
y
xy
6 600
SOLUTION (a)
For Mohr’ss circle of straain, plot pointss:
X : (160 , 300 ) Y : ( 480 , 300 ) C : ( 160 , 0) 0 (a)
tan 2
x
2
3000 3200
xy p
p
y
43.15
0.9375
21.58
p
andd
21.58 900
68.42
21.6
a b
R
(b) (c)
1 2 c
(320 ) 2
(3000 ) 2
68.4
438.66
a
ave a
R
1600
438.6
a
279
b
ave a
R
1600
438.6
b
599
R
(max, in-plaane)
v ( 1 v
a
2R
(maxx, in-plane)
b)
v ( 1 v
max
2778.6
max
m max
1/3 (160 2/3
y)
x
min min
(max, in-plane))
480 )
c
877 160.0
598.6
2778.6
598.6
max
877
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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
7 720
SO OLUTION Plootted points for Mohr’s circlle:
X : (30 , 360 ) Y : (570 , 360 ) C : (300 , 0) tan 2
p
2
p
360 270 53.13
1.3333
(a)
26.6
b a
(b)
R
(2770 )2
a
ave
R
300
450
b
ave
R
300
450
max (in-planee)
(360 ) 2
(c)
max
450 a
max (in-planee)
1 v
maax
a
maax
min
(
a
750 , 750 0
b)
min
1/3 (750 2/33 c
150 )
750
150.0
b
2R v
c
64.4
900
c
300
max
1050
300
( 300 )
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. 60 ,
x
2400 ,
y
xy
5 50
SOLUTION Plottted points:
X : (60 , 25 2 ) Y : (240 , 25 ) C : (150 , 0) tan 2
xy p x
2
y
50 60 240
0 0.277778
15.52
p
97.8
a
7.8
b
(90 ) 2
R (a)
(b)
(c)
a
ave
R 150
933.4
b
ave
R 150
933.4
max (in-plane))
c
0,
max m
(25 ) 2
933.4 a b
2R
max (in-plane)
243.4 ,
max m
max
min
0
243 56.6
186.8
c max x
m min
0
243
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. 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. 4000 ,
x
200 ,
y
xy
3 375
SO OLUTION Plootted points for Mohr’s circlle:
X : ( 400 , 187.5 ) Y : ( 200 , 187.5 ) C : ( 300 , 0) 0 tan 2
xy p x
2
y
375 400 200
1.875
61.93
p
a
121.0
b
(100 )2
R (a)
a
ave
R
300
212.5 2
b
ave
R
300
212.5 2
(b)
max (in-planee)
(c)
c
0
(187.5 ) 2
212.5 a b
2R
max (in-planee)
max
512.5
max
max
min
31.0
0
513 87.5
425
c m max
min
0
513
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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. 3000 ,
x
600 ,
y
xy
1000
SOLUTION X : (300 , 500 ) Y : (60 , 50 ) C : (180 , 0) tan 2
xy p x
2
p
y
100 300 60
22.62 a b
R (a)
(120 )2
(50 ) 2
11.3 101.3
1 130
a
ave
R 180
1330
a
310
b
ave
R 180
1330
b
50.0
max (in-plane))
260
(b)
max (in-plane))
(c)
c
0,
max m
2R
310 ,
max m
max
min
0
c
m min
maxx
0
310
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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. 1800 ,
x
y
2660 ,
xy
3 315
SO OLUTION
Plootted points for Mohr’s circlle:
X : ( 180 , 157.5 ) Y : ( 260 , 157.5 ) C : ( 220 , 0) 0 (a)
tan 2
x
2
315 5 80
xy p
p
y
3.9375
7 75.75
a b
(40 )2
R
(15 57.5 )2
ave
R
22 20
162.5
b
ave
R
22 20
162.5
max (in-planee)
(c)
c
0,
max
2R
57.5
a
b
383
325
0,
max
127.9
162..5
a
(b)
37.9
max
min n min
382.5
c
0 382.5
m max
0
383
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3
PRO OBLEM 7.1 144
45"
2
Deterrmine the strain x , knowingg that the folloowing strains have been dettermined by use of o the rosette shown: s
30" 15"
x
1
1
4800
1
15
2
1220
3
800
SOLUTION
c x cos
2
n y sin
1
0.9330
os x co
2
0.75
os x co
2
3
2
x
n y sin
0.06699
30
3
75
1
n 1 cos 1 xy sin
0.06699
x
n y sin
2
2
2
x
2
0.25
y
xy
2
xy sin 2 cos 2
0.25
y
3
0..4330
xy
xy sin 3 cos 3
0.9330
y
0.25
xy
1
4880
(1)
2
(2)
1 120 3
(3)
800
Solvving (1), (2), and a (3) simultaaneously, x
253 ,
y
307 ,
xy
8893 x
253
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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"
600
1
x
4500
2
755
3
Determ mine (a) the in--plane principal strains, (b) the in-plane maximum m sheaaring strain.
SO OLUTION
os x co
2
n y sin
1
0.75 x
co os2
2
y
0.75 x
cos2
y
3
2
sin 2 x
2
2
1500
3
90 xy sin 1 cos 1
1
0.433301
xy
6000
cos
2
2
0.433301
xy
4500
y
xy
0.25
sinn 2
30
1
0.25
x
1
3
sin
y
xy
2
sin
0
3 cos 3 y
(1)
(2)
3
0
755
(3)
Sollving (1), (2), and (3) simulttaneously, x ave
725 7 , 1 ( 2
75 ,
y
x
y)
x
y
(a)
(b)
a
ave
R
b
ave
R
max (in-plane))
2
173.21
325 2
R
xy
2
725 75 2
xy
2
2
173.21 2
2
4 409.3
734
a
84.3
b
2 R 819
max (in-planee)
734 84.3
819
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PROBLEM 7.146 4
3
45"
45"
The rosette shown has been used to determine the following strains at a point on the surface of a crane hook:
2 45" 1
420 10 6 in./in.
1
x
45 10 6 in./in.
2
165 10 6 in./in.
4
(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 4) 2 1 ( 45 10 2
ave
ave
6
165 10 6 )
60 10 6 in./in.
Gages 1 and 3 are also 90 apart. 1 ( 1 2 2 ave
ave 3
3) 1
(2)(60 10 6 ) 420 10
6
300 10 6 in./in.
3
(b)
x
xy
420 10 6 in./in.
1
2
2
1
3
y
300 10 6 in./in.
3
(2)( 45 10 6 ) 420 10
6
( 300 10 6 )
210 10 6 in./in. 2 x
R
y
2 xy
2
420 10
2
6
( 300 10 6 ) 2
2
210 10 2
6
2
375 10 6 in./in. a
ave
R
60 10
6
375 10
6
b
ave
R
60 10
6
375 10
6
max (in-plane)
a b
2R
max (in-plane)
435 10 6 in./in. 315 10 6 in./in. 750 10 6 in./in.
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. 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 2 ( 1 [( 1 ( 3) 2) 2 2 (Hint: The shaded triangles are congruent.) max,min
! min
1
!1 ! max
2
3)
SOLUTION Since gage directions 1 and 3 are 90 apart,
1 ( 2
ave
Let
u
1
ave
1 ( 2
v
2
ave
2
R2
1 ( 2 u2
1 4
R max, min
ave
3)
1
1 ( 4
3)
1
3)
1
v2 3)
1
1 2
2 1
1 2 1 2 1 ( 1 2 1 [( 2
2
2 2
1
1 4
1 3
2 3
2 2
2 1 2)
2( 1
2)
2
2 2
(
2
2
2 1
1 2
2 3
1 ( 2
2
1 ( 4
3)
3)
1
2 3
3)
2
1 4
2 1
1 2
1 3
1 4
2 3
2 3
2
3)
2 1/2
]
R gives the required formula.
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1
2 2 ]
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
1
&
where circle.
x
2
3
3
avg
is the abscissa of the center of the corresponding Mohr’s
avg
SOLUTION x 1
ave
2
ave
y
2 x
y
2 x
ave
y
2 xy
2
x
cos (2
2
y
2 x
ave
y
2 xy
cos (2
2 xy
2
120 )
sin 120 sin 2 )
3 sin 2 2 (2) xy
240 )
2
(cos 240 cos 2
y
sin (2
3 cos 2 2
(cos 240 sin 2 x
ave
xy
2
sin 120 cos 2 )
1 cos 2 2
y
x ave
(1)
120 )
1 sin 2 2
xy
sin 2
(cos 120 cos 2
2
2
2
(cos 120 sin 2
ave
3
xy
cos 2
240 )
sin 240 sin 2 )
sin 240 cos 2 )
1 cos 2 2
1 sin 2 2
sin (2
3 sin 2 2
3 cos 2 2
(3)
Adding (1), (2), and (3), 1
2
3
3 ave
3
0 0
ave
1
2
3
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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"
2
x
75" 1
1
93.1 10 6 in./in.
2
385 10 6 in./in.
3
210 10 6 in./in.
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
1 ( 2
x
y)
1 ( 2
where
sin 2
for gage 2,
0
and
2
for gage 1,
75
for gage 3.
75
From Eq. (2),
xy
y ) cos 2
x
1
1 ( 2
x
y)
1 ( 2
x
y ) cos (
2
1 ( 2
x
y)
1 ( 2
x
y ) cos
0
3
1 ( 2
x
y)
1 ( 2
x
y ) cos
(150 )
x
z
xy
150 ) xy
2
2
sin ( 150 )
sin 0 xy
2
sin (150 )
(1) (2) (3)
385 10 6 in./in.
Adding Eqs. (1) and (3), 1
3
(
x (1 y
y)
x
1
(
y ) cos 150
x
cos 150 )
y (1
cos 150 )
x (1 cos 150 ) (1 cos 150 )
3
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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PROBLEM 7.149 (Continued)
Subtracting Eq. (1) from Eq. (3), 3
1
sin 150
xy 3
xy
210 10
1
sin 150
6
( 93.1 10 6 ) sin 150
606.2 10 6 in./in. tan 2
606.2 10 6 385 10 6 35.0 10
xy p x
y
1 1 ( x (385 10 y) 2 2 210 10 6 in./in.
ave
2 x
R
6
2
30.0 ,
b
120.0
xy
2
2 6
35.0 10
6
2
606.2 2
a
ave
R
210 10
6
350.0 10
6
b
ave
R
210 10
6
350.0 10
6
R
a
35.0 10 6 )
2
max (in-plane)
(a)
2
y
385 10
(b)
1.732
6
350.0 10 6 in./in.
2
350.0 10
6
a b
max (in-plane)
560 10 6 in./in. 140.0 10 6 in./in. 700 10 6 in./in.
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. 1193
PROBLE EM 7.150
y 1 in n.
P
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::
Qx
C
x 12 in. i
3 A 3 in.
45!
60 10 6 in./in.
2
240 10 6 in./in.
3
200 10 6 in./in.
29 106 psi and v
Knowing thhat E and Qx.
2
1
0.30, determ mine the magnitudes of P
1 3 in..
SO OLUTION x
1
60 10
6
y
3
200 10
6
xy x
y
P A
2
2
1
E ( 1 v2 E ( 1 v2 P
y
3400 10
3
x
v y)
y
v x) A
y
6
29 [ 60 6 (0.3)(200)] 0 1 (0.3)2 29 [2000 (0.3)( 60)] 5.8 103 psi p 2 1 (0.3) (22)(6)(5.8 103 ) 69.6 103 lb
G xy
I Qˆ xy
V
E 2(1 v) G
xy
69.6 kips
Q
30.3 kips
29 1006 11.1538 106 psi (2)(1.300)
(11.1538)((340) 3.79233 103 psi
1 3 1 (2)(6))3 36 in 4 bbh 12 12 A y (2)(3)(1.5) 9 in 3 t ˆ VQ It It xy Qˆ
P
(36)(2)(3..7923 103 ) 9
Q V
2 in.
30.338 103 lb l
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PROBLEM 7.151 y 1 in.
Solve Prob. 7.150, assuming that the rosette at point A indicates the following strains:
P Qx
C
x 12 in.
1
30 10 6 in./in.
2
250 10 6 in./in.
3
100 10 6 in./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.
2 45" 1 3 in.
Knowing that E P and Qx.
1
60 10 6 in./in.
2
240 10 6 in./in.
3
200 10 6 in./in.
29 106 psi and v
0.30, determine the magnitudes of
SOLUTION 6
x
1
30 10
y
3
100 10
xy x
y
2
2
1
E ( 1 v2 0 E ( 1 v2
6
430 10
3
6
x
v y)
29 [ 30 (0.3)(100)] 1 (0.3)2
y
v x)
29 [100 (0.3)( 30)] 1 (0.3)2
2.9 103 psi P A
y
P
A
y
(2)(6)(2.9 103 )
34.8 103 lb
P
34.8 kips
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PROB BLEM 7.151 (Continue ed)
G xyy
I Qˆ t xyy
V
E 2(1 v) G
xy
29 106 (2)(1.30)
111.1538 106 pssi
(11.1538)(430)
4.7962 103 psi
1 3 1 bh (2)(6)3 36 in 4 12 12 A y (2)(33)(1.5) 9 in 3 2 in.
VQˆ It It xy Qˆ
(366)(2)(4.7962 103 ) 9
Q V
38.37 103 lb Q
3 38.4 kips
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PROBL LEM 7.152 T'
A single strain gage iss cemented to a solid 4-in.-diameter steell shaft at an 25 with a line parallel to the axis off the shaft. Knnowing that angle G 11.5 106 psi, deetermine the toorque T indicaated by a gagee reading of 300 100 6 in./in.
! T
2 in.
SOLUTION For torsion,
x
0,
y
1 ( E 1 ( E
x
y
x
v
y)
0
y
v
x)
0
1 2
0
xy
0
G
xy
0
2G
Draaw the Mohr’s circle for straain. R x
But
0
T
0
2G R sin 2 Tc J
c 3G sin 2
2T c3
0
2 2G
sin 2
2G x ssin 2
x
(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 line parrallel to the axxis of the shaftt.
!
35 with a
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.
T
2 in.
SO OLUTION Forr torsion,
0 0,
x
1 ( E 1 ( E
x
y
G
xy
0
x
v
y)
0
y
v
x)
0
1 2
0
xy
0,
y
xy
0
2G
Draaw Mohr’s cirrcle for strain. R x
0
2G R sin 2
0
2 2G
sin 2
But 0
T
Tc J c 3G sin 2
2T c3 x
2G x ssin 2 (2)3(11.5 106 )(300 10 6 ) 7 sin 70
92.3 103 lb in.
T
92.3 kip in.
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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
pr t
1
1 x, 2 1 ( x E
y
x
0.85
v
0.20 xy xy
z)
v
y
v 2
1
x
E
x
E
1 ( v E
y
0
z
x
v
y
z)
1 2
v
x
E
x
E 0
G
Draaw Mohr’s circcle for strain. ave a
R
1 ( 2 1 ( 2
x
p Data:
ave
t
x
r
x
y)
0.525
x
y)
0.325
R cos 2
x
E
(0.5225 0.325cos 2 )
x
E
tE x r (00.525 0.325ccos 2 )
r
1 d 2
t
6 10 3 m mm E
1 (6600) 2
3
p
x
E
300 mm m
0.300 m
200 109 Pa, 9
x
280 10
6
18
6
(6 10 )(200 ) 10 )(2880 10 ) 1..421 106 Pa (0.300)((0.525 0.325 cos 36 )
p 1.4 421 MPa
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PRO OBLEM 7.155 Solvee Prob. 7.154, assuming thaat the gage form ms an angle planee. !
35 with a horizontal
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
y
x
y
pr t
1
1 x, 2 1 ( x E
v
1 ( v E xy x
xy
0
z
x
v
y
y
z)
v
v 2
1 1 2
z)
x
0.85
E x
v
E
x
E
0.20
x
E
0
G
Draaw Mohr’s cirrcle for strain. ave
R
1 ( 2 1 ( 2
x
ave a
x
y)
0.525
x
y)
0.325
x
E x
E
R cos 2
0.525 0.325 cos c 2 ) (0 p Data:
t
x
r
tE x r (0.525 0.325 cos 2 )
r
1 d 2
t
6 10 3 m E
1 (600) 2
3
p
x
E
300 mm
0.300 m
200 109 Pa,
x
280 10 1
6
35
6
9
(6 6 10 )(200 10 )(280 100 ) 1.761 106 Pa (0.300)(0.525 ( 0.325 cos 70 )
p 1.761 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. 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
E G x
y
xy xy
2 ave x
y
taan 2
0,
4877.0 1 ( x 2 974
y)
2633 974 974
xyy y
1.000
45.0 x
R
y
a
ave
R
b
ave
R x
2 xy
2
v ( E
22.5
a
2
c
75 106 Pa
E E v 1 0.2987 2(1 v) 2G 1 1 ( x v y) [0 (00.2987)(150 1106 )] E 2000 109 224 1 1 ( y v x) [( 1500 106 ) 0] E 2000 109 7500 75 106 xy 974 G 77 109
a
a
xy
77 109 Pa
200 109 Pa G
x
2
150 1006 Pa,
y
2
y)
689
(0.2987)(0 150 1 106 ) 200 10 1 9
b
67.5
a
426
b
952
c
224
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PROB BLEM 7.156 6 (Continued d)
(b)
ave
1 ( 2
x
y)
x
y
75 MPa 2
R
0 1150 2
2 xy
2
2
752
1006.07 MPa a
ave
R
b
ave
R
a
1 ( E
31.07 MPa 181 1.07 MPa v
a
1 200 109 4226 10
tan 2
2
xy
[31.07 106
(0.29987)( 181.07 106 )]
6
000 1.0
a x
b)
2
a
426
a
45
y a
22.5
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PR ROBLEM 7.157 Thee following staate of strain haas been determ mined on the suurface of a casst-iron machinne part: 720
x
y
400
xy
6 660
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 6 69 72.933 GPa 1 (0..232) 2
G E 1 v2
(a)
x
y
E ( x v y) 1 v2 (72..93 109 )[ 720 10 59 9.28 MPa E ( y v x) 1 v2 (72..93 109 )[ 4000 10
6
(0.2232)( 400 100 6 )]
6
(0.22321)( 720 10 1 6 )]
41.36 MPa xy
G
(28 109 )(660 10 6 )
xy x
18.4 48 MPa ave
tan 2
1 ( 2 2
xy
2.06225
b y
x
2
b
500.32 MPa
y)
x
64 4.1 ,
b
32.1 ,
a
57.9
2 x
R
y
2
2 xy
20.54 MP Pa
a
avee
R
a
29.8 MPa
b
avee
R
b
700.9 MPa
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PROB BLEM 7.157 (Continue ed)
(b)
ave
tan 2
1 ( 2
xy
b
560 5 2.0625
b x
2
y)
x
y
6 64.1 ,
32.1 ,
b
2 x
R
y
a
5 57.9
2 xy
2
2
a
avve
R
193..26
b
avve
R
926..74
366..74
a
E ( 1 v2
a
v b)
a
2 29.8 MPa
b
E ( 1 v2
b
v a)
b
7 70.9 MPa
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P
PROBLEM M 7.158
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 d2 2 5.75 in.
d2
12 in., c2
c1
c2
t
A
c22
c12
J
c24
c14
2
(62 2
6 in., t
0.25 in.
5.752 )
9 9.2284 in 2
(664
5.754 ) 318.67 3 in 4
Streesses: P A
40 4.33444 ksi 9.22284 Tc2 J (80))(6) 1.5063 ksi k 318.67 0, 4.33444 ksi, y
x
xy
1..5063 ksi
Chooose the x an nd y axes, resspectively, tanngential and noormal to the weld. w Theen
w
y
y
and
w
xy
x
y
x
y
2 ( 4.3344) 2 4 4.76 ksi x
xy
22.5
cos 2 x sin 2 xy 2 [ ( 4.3344)] cos 45 1.5063 sin s 45° 2 w
4.76 ksi
y
sin 2 xy cos 2 2 [ ( 4.3344)] s 45 1.50663 cos 45 sin 2 0 0.467 ksi
w
0.467 0 ksi
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100 kN
!
80 mm
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.
SOLUTION Area of weld: Aw
(10 10 3 )(80 10 3 ) cos 25 882.7 10 6 m 2
(a)
Fs w
(b)
Fn w
0: Fs
Fs Aw
100sin 25
42.26 103 882.7 10 6
0: Fn
Fn Aw
100 cos 25
90.63 103 882.7 10 6
0
Fs
42.26 kN
47.9 106 Pa 0
Fn
w
47.9 MPa
90.63 kN
102.7 106 Pa
w
102.7 MPa
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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
(10 10 3 )(80 10 3 ) cos 800 10 cos
(a)
Fs w
0: Fs Fs Aw
sin cos (b)
100sin
Aw
800 10 6 cos14.34 Fn Aw
100sin
100 10 sin 800 10 6 / cos 30 106 125 106
1 sin 2 2
0: Fn
Fs 3
30 106
Fn
0
100 cos
0
kN
100 103 sin
m2
N
125 106 sin cos
0.240
Fn
6
14.34
100cos14.34
96.88 kN
825.74 10 6 m 2
96.88 103 825.74 10 6
117.3 106 Pa
117.3 MPa
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PROBLEM 7.161
'0 &
+
'0
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: x
y
xy
1 2 1 2 1 2
1 2 1 2
0
0
0
cos 2
0
cos 2
sin 2
0
Resultant stresses: x
y
0
xy
0
1 2 1 2
1 ( 2 2
ave
tan 2
1 2
0
1 2
0
1 2
0
y)
x xy
1 2
0 y
1 2 0
1 2
0
1 2
0
0
0
cos 2
cos 2
sin 2
0
p 0
sin 2 1 cos 2
3 2
cos 2
cos 2
0
sin 2
0
x
0
sin 2 0 cos 2
tan
p
2 x
R
y
2 1 2
0
2 xy
1 2
1 2 cos 2 + cos 2 2
0
1 2
2 0
sin 2 2
cos 2 2 2
0
1 2
2
(
0 sin 2
1 cos 2
1 2
p
0
) 2
|cos |
a
ave
R
a
0
0
cos
b
ave
R
b
0
0
cos
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y
PROBLEM 7.162 2 ksi
For the state of stress shown, determine the maximum shearing stress when (a) z 4 ksi, (b) z 4 ksi, (c) z 0. 6 ksi
σz
7 ksi
z x
SOLUTION 7 ksi,
x
ave
1 ( 2
x
y)
x
y
y
2 ksi,
xy
6 ksi
4.5 ksi 2
R
2.52
(a)
(b)
( 6) 2
6.5 ksi
a
ave
R 11 ksi
b
ave
R
z
4 ksi,
max
11 ksi,
z
max
2 ksi
a
a
11 ksi,
min
11 ksi,
11 ksi,
min
2 ksi
b
2 ksi,
min
11 ksi, 0,
11 ksi,
a
4 ksi,
z
max
(c)
2 xy
2
4 ksi, b
2 ksi,
1 ( 2
max
min )
max
max
6.50 ksi
2 ksi
b
max
1 ( 2
max
min )
max
7.50 ksi
1 ( 2
max
min )
max
6.50 ksi
2 ksi max
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y
PROBLEM 7.163 40 MPa
For the state of stress shown, determine the value of
xy
for which the
maximum shearing stress is (a) 60 MPa, (b) 78 MPa.
τ xy 100 MPa z x
SOLUTION x
100 MPa, 1 ( 2
ave
(a)
y)
x
40 MPa,
0
z
70 MPa
60 MPa.
max
If
y
z
is
min ,
then
max max max
2
min
max .
0 (2)(60) 120 MPa R
ave
R
max
b
max
ave
2R
120 70 50 MPa 20 MPa > 0 2
x
R xy
(b)
2 xy
2 502
302
2 xy
50 MPa
302
xy
40.0 MPa
xy
72.0 MPa
78 MPa.
max
If
y
z
is
min ,
then
max
min
2
max
ave
R
R
Set
max
0 (2)(78) 156 MPa.
max
156 70 86 MPa >
ave
R
max
78 MPa.
min
ave
R
max
78 MPa
8 MPa < 0 2
R xy
x
y
2 782
2 xy
302
302
2 xy
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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 24 ksi
x
For stresses in xy-plane, (a)
xy
ave
1 ( 2
x
y
14 ksi
y y)
x
0
z x
19 ksi
y
2
5 ksi
6 ksi 2
R a 2 a
2 b
a
b
F .S .
(b)
xy
2 xy
2 R
ave
(5)2
26.810 ksi,
(6)2 ave
b
7.810 ksi R 11.190 ksi
23.324 ksi < 30 ksi
(No yielding)
30 23.324
F .S . 1.286
12 ksi 2 x
R a 2 a
2 b
a
b
F .S .
(c)
xy
y
2 xy
2 R
ave
32 ksi,
(5)2 b
ave
(12)2 R
13 ksi 6 ksi
29.462 ksi < 30 ksi
(No yielding)
30 29.462
F .S . 1.018
14 ksi 2 x
R a 2 a
2 b
a
b
y
2 xy
2 ave
R
33.866,
32.00 ksi > 30 ksi
(5)2 b
ave
(14)2 R
14.866 ksi 4.134 ksi (Yielding occurs)
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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:
r 1
2
Torsion:
c1
J T
1 d 2 pr t pr 2t
225 mm t
6 mm
(1.2)(225) 6
45 MPa
22.5 MPa
225 mm, c2
2
c24
c14
225
6
446.9 106 mm 4
(5 103 )(500 10 3 ) Tc J
231 mm
2500 N m 3
(2500)(231 10 ) 446.9 10 6 1.29224 106 Pa
At point a,
1.29224 MPa
0 at point a.
Transverse shear: Bending:
446.9 10 6 m 4
I M
1 J 2
223.45 10 6 m 4 , c
(5 103 )(750 10 3 )
231 10 3 m
3750 N m
Mc I
(3750)(231 10 3 ) 223.45 10 6
3.8767
3.8767 MPa
Total stresses (MPa). Longitudinal:
x
22.5
Circumferential:
y
45 MPa
Shear:
xy
26.377 MPa
1.29224 MPa
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PROBLEM 7.165 (Continued)
ave
1 ( 2
x
y)
x
y
35.688 MPa 2
R max
max(in-plane)
2 xy
2 ave
R
R
45.1 MPa
9.40 MPa
9.4007 MPa max
45.1 MPa
max (in-plane)
9.40 MPa
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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:
r 1
2
Torsion:
c1
J T
1 d 2 pr t pr 2t
Bending: At point b,
(1.2)(225) 6
2
6 mm 45 MPa
22.5 MPa
225 mm, c2
c24
225
c14
6
231 mm
446.9 106 mm 4
(5 103 )(500 10 3 ) Tc J
Transverse shear:
225 mm t
446.9 10 6 m 4
2500 N m 3
(2500)(231 10 ) 446.9 10 6
1.29224 106 Pa
1.29224 MPa
0 at point b.
I
M
1 J 2
223.45 10
(5 103 )(2
6
m4 , c
750 10 3 )
Mc I
(7500)(231 10 3 ) 223.45 10 6 7.7534
231 10 3 m
7500 N m 7.7534 MPa
Total stresses (MPa). Longitudinal:
x
22.5
Circumferential:
y
45 MPa
Shear:
xy
30.253 MPa
1.29224 MPa
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PROBLEM 7.166 (Continued)
ave
1 ( x 2
y)
37.626 MPa 2
x
R max
max (in-plane)
y
2 xy
2 ave
R
R
45.1 MPa
7.49 MPa
7.4859 MPa max
45.1 MPa
max (in-plane)
7.49 MPa
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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
( po
max )i
1
pi ri t
500 psi, ro
(
max )o max
0.12
0.88 in., t
(100)(0.88) 0.12 1 in.,
po ro t 733.33
t
0.12 in.
733.33 psi
0.12 in.
(500)(1) 0.12 4166.7
4166.7 psi 3433.4 psi max
3.43 ksi (compression)
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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.
p ( 1)
500 psii, t pr t
0.15 inn., r
((500)(1.85) 0.15
2
0 0.15 1.85 in.
6166.7 psi 1
(b)
6.17 ksi
Longitudin nal stress. Free body of portion of jacket j above a horizontal seection, consideering vertical forces f only:
Fy Af
0: Af p dA
Aj
pA f
Areas :
Af
r22
r12
[(1.85) 2
Aj
r32
r22
[(2) 2
(11)2 ] (1.855)2 ]
2 dA j
0
2 Aj
0
2
p
Af
(1)
Aj
7.6105 in i 2 1.814277 in 2
Recalling Eq. E (1), 2
p
Af Aj
(500)
7.6105 1 1.81427
20097.4 psi 2
2 2.10 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 1
2
Determine D the largest in-planne normal straain, knowing that the follow wing strains haave been obtaiined by the usse of the rosettte shown:
3
1
50 100 6 in./in.
2
360 10 6 in./in.
3
1 6 in./in. 315 10
45! x
45!
SO OLUTION 455 ,
1
x
cos 2
1
sin 2
y
1
0.55 cos 2
x
y
2
sin 2
x
cos 2
3
y
sin 2
sin
xy
0.5
x
2
0.55
45 ,
2
xy
y
sin
0.5
x
3
xy
2 y
sin
cos
1
3
Eq. (1)
Eq. (2):
Eq. (1)
Eq. (2):
x
x
315 3 10
6
xy
50 10
6
y
1
2
y
1
2
ave
1 ( 2
xy
cos
2
0.5
xy
cos
3
360 10
y)
50 10
6
(1)
6
(2)
6
(3)
2
360 10 3
0 0
315 10
4 410 10 6 in./iin.
6
3 360 10
6
3115 10
6
5 10 6 in./in.
1 155 10 6 in./inn. 2
y
xy
2 315 10
6
50 10
2 x
R
1
in.//in.
x
x
1
0 0.5
x
Froom (3),
0
3
2 6
5 10
6
2
410 10 2
2
6
2
260 10 6 in../in. max
ave
R 1555 10
6
260 10
6
max
415 110 6 in./in.
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y'
y
"y
PROBLEM 7.C1 1
y
#
"y'
!x'y'
!xy x Q
z
#
"x'
Q
x
"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 ssociated withh the element after it has x y as 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
Equuation (7.5), Paage 427:
x
Equuation (7.7), Paage 427:
y
x,
y,
xyy
x
2 y
x
2
y
2
x xy
y
2
x
Equuation (7.6), Paage. 427: Enteer
y
y
2
sin 2
c 2 cos
xy
sinn 2
c 2 cos
xy
sinn 2
x xy
cos 2
and
Prinnt values obtaiined for
x
,
y
and
xy
Prooblem Outputts Probblem 7.13 x x xy
0 ksi 8 ksi 5 ksi
Rootation of elem ment (+ counterclockw c wise) 25
Rotation off element (+ countercllockwise) 10
x
2.40 ksi
x
1.995 ksi
y
10.40 ksi
y
6.05 ksi
xy
6..07 ksi
xy
0.15 ksi
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PROB BLEM 7.C1 (Continued d) Prooblem 7.14 y
60 MPa M 90 MPa
xy
Pa 30 MP
x
Ro otation of Elem ment ( counterclockw wise) 25
Rotation off Element ( countercclockwise) 1 10
x
56.19 MP Pa
x
455.22 MPa
y
86.19 MP Pa
y
755.22 MPa
xy
38.17 MP Pa
xy
53.84 MPa
Prooblem 7.15 x
8 ksii
y
12 ksi
xy
6 ksi Rotation off Element ( countercclockwise) 1 10
Ro otation of Elem ment ( counterclockw wise) 25 x
9.02 kssi
x
5.344 ksi MPa
y
13.02 kssi
y
9.344 ksi MPa
xy
9.066 ksi MPa
xy
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
x
19.51 MPa
y
104.01 MPa M
y
60..49 MPa
xy
60.67 MPa
xy
1.50 MPa M
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PROBLE EM 7.C2
y
"y
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.
!xy x Q
x
"x z
SOLUTION Proggram followin ng equations: 2 x
y
Equuation (7.10)
ave
Equuation (7.14)
max
ave
R
J min
ave
R
Equuation (7.12)
p
2
taan
: R
2
1
s
Sheearing stress: Theen
If
m max
2 xy
y x
2
y xy
0 and
m min
0 and
m min
max(out-of-planne)
m max
R;
max(in-plane) m
1
2
max(out-of-planne)
R;
max(in-plane) m
If
Theen
m max
R;
maax(in-plane)
If
Theen
taan
y
xy
x
Equuation (7.15)
x
0 and
m min
max(out-of-plaane)
0: R
0:
1 2 0: 1 | 2
max
min |
Proogram Outputts Probblems 7.5 and d 7.9 x
60.00 MPa
y
40.00 MPa
xy
35.00 MPa
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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°
max (in-plane) (
80.662 MPa
max (out--of-plane)
105.331 MPa
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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
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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. 2 x avve
Maaximum-shearring-stress criterion. If
a
and
b
havve same sign,
y
2
a
ave
R
b
ave
R
y
1 2
y
maax
1 2
a
If
max
y,
yieelding occurs.
If
max
y,
no yielding occu urs, and factor of safety
x
; R
y
2
2 xy
y m max
Maaximum-distorrtion-energy criterion. 2 a
Compute radicall
a
b
2 b
r If radical
y,
urs. yielding occu
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
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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
y
240.000 MPa 240.000 MPa
xy
100.00 MPa
x
max min
140.000 MPa 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
y
280.000 MPa 280.000 MPa
xy
100.000 MPa
x
max min
180.000 MPa 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
y
24.00 ksi k 14.00 ksi k
xy
6.00 ksi k
x
max min
26.81 ksi k 11.19 ksi k
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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
(a)
Yield strenngth
30 ksi
x
24.00 ksi k
y
14.00 kssi
xy
12.00 ksi k
max
32.00 ksi k
min
6.00 kssi
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
(a)
Yield strength
30 ksi
x
24.00 kssi
y
14.00 kssi
xy
14.00 kssi
max
33.87 kssi
min
4.13 kssi
Using the maximum-sheearing-stress criterion, c material will w yield.
(b)
Using the maximum-disstortion-energyy criterion, material will w yield.
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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. 2 x avee
y
2
a
ave
R
b
ave
R
R
x
y
2
2 xy
c Mohhr’s fracture criterion. If
If
and
b
a
UT
and
a
UT
or
a
a
0 and
hav ve same sign, and a UC ,
b
UC ,
b b
no faailure;
failuree.
0:
Connsider fourth quadrant q of Figgure 7.47. For no rupture to occur, point ( If
b
a,
b)
mustt lie within Moohr’s envelope (Figure 7.477).
Criterion n,
thenn rupture occu urs. If
b
Criterion n,
thenn no rupture occcurs.
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PROB BLEM 7.C4 (Continued d)
Proogram Outpu uts Prooblem 7.89
10.00 MPa
x
100.00 MPa
y
60 MPa
xy
Ulttimate strengthh in tension
80 MPaa
Ulttimate strengthh in compressiion
200 MP Pa
max
a
336.39 MPa
min
b
1226.39 MPa
Rupture will occcur Prooblem 7.90
32.000 MPa
x y xy
0.00 MPa M 75.00 MPa M
Ulttimate strengthh in tension
80 MP P
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
Prooblem 7.93
x
8.00 ksii
y
0.00 ksii
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
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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
m max
a
9.444 ksi
m min
b
1.444 ksi
Ruppture will occu ur. Ansswer: Ruppture occurs fo or
0
49.1 MPa. M
Probblem 7.94
x y xy
80.00 MPa Pa 0.00 MP 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 y xy
80.00 MPa Pa 0.00 MP 49.20 MPa M
Ultiimate strength h in tension
75 MPaa
Ultiimate strength h in compressioon
150 MP Pa
max
a
min
b
233.41 MPa 1 103.41 MPa
Ruppture will occu ur.
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PROBLEM 7.C5 y
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 . (b) Use this program to solve Probs 7.129 and 7.131.
y' x'
&
x
SOLUTION Program following equations: x
Equation (7.44):
x
Equation (7.45):
y
Equation (7.46):
xy
Enter
x,
y,
y
2 y
x
2 (
y
2
x
xy ,
x
y
2
cos 2
1 2
xy
sin 2
sin 2
1 2
xy
cos 2
y )sin 2
x
xy
cos 2
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):
Problem 7.131
x
60 115.05 micro meters
y
284.95 micro meters
xy
5.72 micro radians x
0 micro meters
y
320 micro meters
xy
100 micro radians
Rotation of element, in degrees (+ counterclockwise): x
30 36.70 micro meters
y
283.30 micro meters
xy
227.13 micro radians
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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: 2 x
y
Equation (7.50):
ave
Equation (7.51):
max
ave
Equation (7.52):
p
tan
R
2 R
min
y
2 ave
2 xy
2 R
xy
1
x
Shearing strains:
x
y
Maximum in-plane shearing strain 2R
max (in-plane)
Calculate out-of-plane shearing strain and check whether it is the maximum shearing strain.
Let
a
max
b
min
v
Calculate
c
1 v
If
a
b
c,
out-of-plane
a
c
If
a
c
b,
out-of-plane
a
b
If
c
a
c
b
b,
out-of-plane
(
a
b)
2R
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. 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. 1231
PROBLEM 7.C6 (Continued) Program Printout (Continued)
Angle between xy axes and principal axes ( p a b c
33.69 100.00 micro meters 420.00 micro meters 159.98 micro meters
max (in-plane)
520.00 microradians
max
579.98 microradians
Problem 7.137
counterclockwise):
x
600 micrometers
y
400 micrometers
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
Problem 7.138
x y xy
1201.56 microradians 160 micrometers 480 micrometers 600.00 microradians 0.333
Angle between xy axes and principal axes ( p a b c
counterclockwise):
21.58 278.63 micrometers 598.63 micrometers 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
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PROBLEM 7.C6 (Continued) Angle between xy axes and principal axes ( p a
counterclockwise):
26.57 750.00 micrometers
b
150.00 micrometers
c
300.00 micrometers
max(in-plane) max
900.00 microradians 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: 2 x
y
Equation (7.50)
ave
Equation (7.51)
max
ave
Equation (7.52)
p
tan
2 R
y
2
min
ave
2 xy
2 R
xy
1 x
Shearing strains:
x
R
y
Maximum in-plane shearing strain 2R
xy (in-plane)
Calculate out-of-plane-shearing strain and check whether it is the maximum shearing strain. Let
a b c
If
a
max min
(Plain strain)
0
b
c,
out-of-plane
a
c
out-of-plane
a
b
out-of-plane
c
b
If
a
c
b,
If
c
a
b,
2R
Program Printout
Problem 7.140
x
60 micrometers
y
240 micrometers
xy
50 microradians 0.000
Angle between xy axes and principal axes (+ = counterclockwise): 7.76 and 82.24 p a
243.41 micrometers
b
56.59 micrometers
max(in-plane)
0.00 micrometers 186.82 microradians
max
243.41 microradians
c
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PROBLEM 7.C7 (Continued) Program Printout (Continued) Problem 7.141
y
400 micrometers 200 micrometers
xy
375 microradians
x
0.000 Angle between xy axes and principal axes (
counterclockwise):
30.96 and 59.04
p
max(in-plane)
512.50 micrometers 87.50 micrometers 0.00 micrometers 425.00 microradians
max
512.50 microradians
a b c
Problem 7.142
x
300 micrometers
y
60 micrometers 100 microradians
xy
0.000
Angle between xy axes and principal axes (+ = counterclockwise): p
11.31 and
78.69
a
310.00 micrometers
b
50.00 micrometers
max(in-plane)
0.00 micrometers 260.00 microradians
max
310.00 microradians
c
Problem 7.143
x
180 micrometers
y
260 micrometers
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 Enter: NU
n
and
n.
V
Solve Equation (7.60) for
x,
y,
and
xy
using method of determinates or any other method. 2
x
Enter
y
ave
2
a
max
ave
R
b
max
avg
R
V c
p
Shearing strains:
x
; R
1 V 1 tan 2
(
y
2 xy
2
b)
a xy
1 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
2R
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. 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. 1236
PROBLEM 7.C8 (Continued) Program Outputs (Continued) x
253.21 micrometers
y
306.79 micrometers 892.82 microradians
xy
727.21 micrometers
a
167.21 micrometers
b
894.43 microradians
max (in-plane)
Problem 7.145
Gage
Theta Degrees
Epsilon Micro Meters
1 2 3
30 –30 90
600 450 –75
725.00 micrometers
x 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 Gage
y
that is obtained is also the expected reading of Gage 3.
Theta Degrees
Epsilon in./in.
1
0
420
2
45
–45
4
135
165
x y xy a b max (in-plane)
420.00 in./in. 300.00 in./in. 210.00 microradians 435.00 in./in. 315.00 in./in. 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 a b max (in-plane)
410.000 microradians 415.048 in./in. 105.048 in./in. 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
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