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Chapter 5
5-1
Sy
MSS: Ã1 - Ã3 = Sy/n Ò! n =
Ã1 - Ã3
Sy
DE: n =

Ã
1/2 1/2
2 2 2 2 2
à = ÃA - ÃAÃB + ÃB = Ãx - ÃxÃy + Ãy + 3Äxy
(a) MSS: Ã1 = 12, Ã2 = 6, Ã3 = 0 kpsi
50
n = = 4.17 Ans.
12
50

DE: Ã = (122 - 6(12) + 62)1/2 = 10.39 kpsi, n = = 4.81 Ans.
10.39
2
12 12
(b) ÃA, ÃB = Ä… + (-8)2 = 16, -4 kpsi
2 2
Ã1 = 16, Ã2 = 0, Ã3 =-4 kpsi
50
MSS: n = = 2.5 Ans.
16 - (-4)
50

DE: Ã = (122 + 3(-82))1/2 = 18.33 kpsi, n = = 2.73 Ans.
18.33

2
-6 - 10 -6 + 10
(c) ÃA, ÃB = Ä… + (-5)2 =-2.615, -13.385 kpsi
2 2
Ã1 = 0, Ã2 =-2.615, Ã3 =-13.385 kpsi
B
50
MSS: n = = 3.74 Ans.
0 - (-13.385)

DE: Ã = [(-6)2 - (-6)(-10) + (-10)2 + 3(-5)2]1/2
A
= 12.29 kpsi
50
n = = 4.07 Ans.
12.29
2
12 + 4 12 - 4
(d) ÃA, ÃB = Ä… + 12 = 12.123, 3.877 kpsi
2 2
Ã1 = 12.123, Ã2 = 3.877, Ã3 = 0 kpsi
50
MSS: n = = 4.12 Ans.
12.123 - 0

DE: Ã = [122 - 12(4) + 42 + 3(12)]1/2 = 10.72 kpsi
50
n = = 4.66 Ans.
10.72
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5-2 Sy = 50 kpsi
Sy
MSS: Ã1 - Ã3 = Sy/n Ò! n =
Ã1 - Ã3
1/2 1/2
2 2 2 2
DE: ÃA - ÃAÃB + ÃB = Sy/n Ò! n = Sy/ ÃA - ÃAÃB + ÃB
50
(a) MSS: Ã1 = 12 kpsi, Ã3 = 0, n = = 4.17 Ans.
12 - 0
50
DE: n = = 4.17 Ans.
[122 - (12)(12) + 122]1/2
50
(b) MSS: Ã1 = 12 kpsi, Ã3 = 0, n = = 4.17 Ans.
12
50
DE: n = = 4.81 Ans.
[122 - (12)(6) + 62]1/2
50
(c) MSS: Ã1 = 12 kpsi, Ã3 =-12 kpsi, n = = 2.08 Ans.
12 - (-12)
50
DE: n = = 2.41 Ans.
[122 - (12)(-12) + (-12)2]1/3
50
(d) MSS: Ã1 = 0, Ã3 =-12 kpsi, n = = 4.17 Ans.
-(-12)
50
DE: n = = 4.81
[(-6)2 - (-6)(-12) + (-12)2]1/2
5-3 Sy = 390 MPa
Sy
MSS: Ã1 - Ã3 = Sy/n Ò! n =
Ã1 - Ã3
1/2 1/2
2 2 2 2
DE: ÃA - ÃAÃB + ÃB = Sy/n Ò! n = Sy/ ÃA - ÃAÃB + ÃB
390
(a) MSS: Ã1 = 180 MPa, Ã3 = 0, n = = 2.17 Ans.
180
390
DE: n = = 2.50 Ans.
[1802 - 180(100) + 1002]1/2
2
180 180
(b) ÃA, ÃB = Ä… + 1002 = 224.5, -44.5MPa = Ã1, Ã3
2 2
390
MSS: n = = 1.45 Ans.
224.5 - (-44.5)
390
DE: n = = 1.56 Ans.
[1802 + 3(1002)]1/2
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Chapter 5 117
2
160 160
(c) ÃA, ÃB =- Ä… - + 1002 = 48.06, -208.06 MPa = Ã1, Ã3
2 2
390
MSS: n = = 1.52 Ans.
48.06 - (-208.06)
390
DE: n = = 1.65 Ans.
[-1602 + 3(1002)]1/2
(d) ÃA, ÃB = 150, -150 MPa = Ã1, Ã3
390
MSS: n = = 1.30 Ans.
150 - (-150)
390
DE: n = = 1.50 Ans.
[3(150)2]1/2
5-4 Sy = 220 MPa
(a) Ã1 = 100, Ã2 = 80, Ã3 = 0MPa
220
MSS: n = = 2.20 Ans.
100 - 0

DET: Ã = [1002 - 100(80) + 802]1/2 = 91.65 MPa
220
n = = 2.40 Ans.
91.65
(b) Ã1 = 100, Ã2 = 10, Ã3 = 0MPa
220
MSS: n = = 2.20 Ans.
100

DET: Ã = [1002 - 100(10) + 102]1/2 = 95.39 MPa
220
n = = 2.31 Ans.
95.39
(c) Ã1 = 100, Ã2 = 0, Ã3 =-80 MPa
220
MSS: n = = 1.22 Ans.
100 - (-80)

DE: Ã = [1002 - 100(-80) + (-80)2]1/2 = 156.2MPa
220
n = = 1.41 Ans.
156.2
(d) Ã1 = 0, Ã2 =-80, Ã3 =-100 MPa
220
MSS: n = = 2.20 Ans.
0 - (-100)

DE: Ã = [(-80)2 - (-80)(-100) + (-100)2] = 91.65 MPa
220
n = = 2.40 Ans.
91.65
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5-5
OB 2.23
(a) MSS: n = = = 2.1
OA 1.08
OC 2.56
DE: n = = = 2.4
OA 1.08
OE 1.65
(b) MSS: n = = = 1.5
OD 1.10
OF 1.8
DE: n = = = 1.6
OD 1.1
B
(a)
C
B
A
Scale
1" 200 MPa
O A
D
(b)
E
F
J
K
G
L
(d)
H
I
(c)
OH 1.68
(c) MSS: n = = = 1.6
OG 1.05
OI 1.85
DE: n = = = 1.8
OG 1.05
OK 1.38
(d) MSS: n = = = 1.3
OJ 1.05
OL 1.62
DE: n = = = 1.5
OJ 1.05
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Chapter 5 119
5-6 Sy = 220 MPa
OB 2.82
(a) MSS: n = = = 2.2
OA 1.3
OC 3.1
DE: n = = = 2.4
OA 1.3
OE 2.2
(b) MSS: n = = = 2.2
OD 1
OF 2.33
DE: n = = = 2.3
OD 1
B
(a)
C
B
1" 100 MPa
E
(b)
D
F
O
A
G
H
J
I
(c)
K
L
(d)
OH 1.55
(c) MSS: n = = = 1.2
OG 1.3
OI 1.8
DE: n = = = 1.4
OG 1.3
OK 2.82
(d) MSS: n = = = 2.2
OJ 1.3
OL 3.1
DE: n = = = 2.4
OJ 1.3
A
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5-7 Sut = 30 kpsi, Suc = 100 kpsi; ÃA = 20 kpsi, ÃB = 6 kpsi
Sut 30
(a) MNS: Eq. (5-30a) n = = = 1.5 Ans.
Ãx 20
30
BCM: Eq. (5-31a) n = = 1.5 Ans.
20
30
MM: Eq. (5-32a) n = = 1.5 Ans.
20
(b) Ãx = 12 kpsi,Äxy =-8 kpsi
2
12 12
ÃA, ÃB = Ä… + (-8)2 = 16, -4 kpsi
2 2
30
MNS: Eq. (5-30a) n = = 1.88 Ans.
16
1 16 (-4)
BCM: Eq. (5-31b) = - Ò! n = 1.74 Ans.
n 30 100
30
MM: Eq. (5-32a) n = = 1.88 Ans.
16
(c) Ãx =-6 kpsi, Ãy =-10 kpsi,Äxy =-5 kpsi

2
-6 - 10 -6 + 10
ÃA, ÃB = Ä… + (-5)2 =-2.61, -13.39 kpsi
2 2
100
MNS: Eq. (5-30b) n =- = 7.47 Ans.
-13.39
100
BCM: Eq. (5-31c) n =- = 7.47 Ans.
-13.39
100
MM: Eq. (5-32c) n =- = 7.47 Ans.
-13.39
(d) Ãx =-12 kpsi,Äxy = 8 kpsi
2
12 12
ÃA, ÃB =- Ä… - + 82 = 4, -16 kpsi
2 2
-100
MNS: Eq. (5-30b) n = = 6.25 Ans.
-16
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Chapter 5 121
1 4 (-16)
BCM: Eq. (5-31b) = - Ò! n = 3.41 Ans.
n 30 100
1 (100 - 30)4 -16
MM: Eq. (5-32b) = - Ò! n = 3.95 Ans.
n 100(30) 100
B
B
(a)
1" 20 kpsi
A
O
A
C
E
D (b)
K
F
G
H
J
L
(d)
(c)
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5-8 See Prob. 5-7 for plot.
OB 1.55
(a) For all methods: n = = = 1.5
OA 1.03
OD 1.4
(b) BCM: n = = = 1.75
OC 0.8
OE 1.55
All other methods: n = = = 1.9
OC 0.8
OL 5.2
(c) For all methods: n = = = 7.6
OK 0.68
OJ 5.12
(d) MNS: n = = = 6.2
OF 0.82
OG 2.85
BCM: n = = = 3.5
OF 0.82
OH 3.3
MM: n = = = 4.0
OF 0.82
5-9 Given: Sy = 42 kpsi, Sut = 66.2 kpsi, µf = 0.90. Since µf > 0.05, the material is ductile and
thus we may follow convention by setting Syc = Syt.

Use DE theory for analytical solution. For à , use Eq. (5-13) or (5-15) for plane stress and
Eq. (5-12) or (5-14) for general 3-D.

(a) Ã = [92 - 9(-5) + (-5)2]1/2 = 12.29 kpsi
42
n = = 3.42 Ans.
12.29

(b) Ã = [122 + 3(32)]1/2 = 13.08 kpsi
42
n = = 3.21 Ans.
13.08

(c) Ã = [(-4)2 - (-4)(-9) + (-9)2 + 3(52)]1/2 = 11.66 kpsi
42
n = = 3.60 Ans.
11.66

(d) Ã = [112 - (11)(4) + 42 + 3(12)]1/2 = 9.798
42
n = = 4.29 Ans.
9.798
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Chapter 5 123
B
(d)
H
1 cm 10 kpsi
G
O
A
(b)
C
D
A
E
B
(a)
F
(c)
For graphical solution, plot load lines on DE envelope as shown.
(a) ÃA = 9, ÃB =-5 kpsi
OB 3.5
n = = = 3.5 Ans.
OA 1
2
12 12
(b) ÃA, ÃB = Ä… + 32 = 12.7, -0.708 kpsi
2 2
OD 4.2
n = = = 3.23
OC 1.3
2
-4 - 9 4 - 9
(c) ÃA, ÃB = Ä… + 52 =-0.910, -12.09 kpsi
2 2
OF 4.5
n = = = 3.6 Ans.
OE 1.25
2
11 + 4 11 - 4
(d) ÃA, ÃB = Ä… + 12 = 11.14, 3.86 kpsi
2 2
OH 5.0
n = = = 4.35 Ans.
OG 1.15
5-10 This heat-treated steel exhibits Syt = 235 kpsi, Syc = 275 kpsi and µf = 0.06. The steel is
ductile (µf > 0.05) but of unequal yield strengths. The Ductile Coulomb-Mohr hypothesis
(DCM) of Fig. 5-19 applies  confine its use to first and fourth quadrants.
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(a) Ãx = 90 kpsi, Ãy =-50 kpsi, Ãz = 0 ÃA = 90 kpsi and ÃB =-50 kpsi. For the
fourth quadrant, from Eq. (5-31b)
1 1
n = = = 1.77 Ans.
(ÃA/Syt) - (ÃB/Suc) (90/235) - (-50/275)
(b) Ãx = 120 kpsi, Äxy =-30 kpsi ccw. ÃA, ÃB = 127.1, -7.08 kpsi. For the fourth
quadrant
1
n = = 1.76 Ans.
(127.1/235) - (-7.08/275)
(c) Ãx =-40 kpsi, Ãy =-90 kpsi, Äxy = 50 kpsi. ÃA, ÃB =-9.10, -120.9 kpsi.
Although no solution exists for the third quadrant, use
Syc 275
n =- =- = 2.27 Ans.
Ãy -120.9
(d) Ãx = 110 kpsi, Ãy = 40 kpsi, Äxy = 10 kpsi cw. ÃA, ÃB = 111.4, 38.6 kpsi. For the
first quadrant
Syt 235
n = = = 2.11 Ans.
ÃA 111.4
Graphical Solution:
B
OB 1.82
(a) n = = = 1.78
OA 1.02
OD 2.24
(b) n = = = 1.75
OC 1.28
OF 2.75
(c) n = = = 2.22
OE 1.24
OH 2.46
(d) n = = = 2.08
H (d)
OG 1.18
1 in 100 kpsi
G
A
O
C (b)
D
A
B (a)
E
F
(c)
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Chapter 5 125
5-11 The material is brittle and exhibits unequal tensile and compressive strengths. Decision:
Use the Modified Mohr theory.
Sut = 22 kpsi, Suc = 83 kpsi
(a) Ãx = 9 kpsi, Ãy =-5 kpsi. ÃA, ÃB = 9, -5 kpsi. For the fourth quadrant,
5
|ÃB | = < 1, use Eq. (5-32a)
ÃA 9
Sut 22
n = = = 2.44 Ans.
ÃA 9
(b) Ãx = 12 kpsi, Äxy =-3 kpsi ccw. ÃA, ÃB = 12.7, -0.708 kpsi. For the fourth quad-
0.708
rant, |ÃB | = < 1,
ÃA 12.7
Sut 22
n = = = 1.73 Ans.
ÃA 12.7
(c) Ãx =-4 kpsi, Ãy =-9 kpsi, Äxy = 5 kpsi. ÃA, ÃB =-0.910, -12.09 kpsi. For the
third quadrant, no solution exists; however, use Eq. (6-32c)
-83
n = = 6.87 Ans.
-12.09
(d) Ãx = 11 kpsi, Ãy = 4 kpsi,Äxy = 1 kpsi. ÃA, ÃB = 11.14, 3.86 kpsi. Forthefirstquadrant
SA Syt 22
n = = = = 1.97 Ans.
ÃA ÃA 11.14
B
30
Sut 22
(d )
A
30
(b)
(a)
 50
Sut 83
(c)
 90
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.
5-12 Since µf < 0.05, the material is brittle. Thus, Sut = Suc and we may use MM which is
basically the same as MNS.
(a) ÃA, ÃB = 9, -5 kpsi
35
n = = 3.89 Ans.
9
(b) ÃA, ÃB = 12.7, -0.708 kpsi
35
n = = 2.76 Ans.
12.7
(c) ÃA, ÃB =-0.910, -12.09 kpsi (3rd quadrant)
36
n = = 2.98 Ans.
B
12.09
(d) ÃA, ÃB = 11.14, 3.86 kpsi
35
n = = 3.14 Ans.
11.14
1 cm 10 kpsi
H
(d)
Graphical Solution:
G
OB 4
(a) n = = = 4.0 Ans. A
O
(b)
OA 1 C
D
A
OD 3.45
E
(b) n = = = 2.70 Ans.
OC 1.28
B
OF 3.7
(a)
(c) n = = = 2.85 Ans. (3rd quadrant)
OE 1.3
F
OH 3.6
(d) n = = = 3.13 Ans.
(c)
OG 1.15
5-13 Sut = 30 kpsi, Suc = 109 kpsi
Use MM:
(a) ÃA, ÃB = 20, 20 kpsi
30
Eq. (5-32a): n = = 1.5 Ans.
20

(b) ÃA, ÃB =Ä… (15)2 = 15, -15 kpsi
30
Eq. (5-32a) n = = 2 Ans.
15
(c) ÃA, ÃB =-80, -80 kpsi
For the 3rd quadrant, there is no solution but use Eq. (5-32c).
109
Eq. (5-32c): n =- = 1.36 Ans.
-80
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Chapter 5 127
(d) ÃA, ÃB = 15, -25 kpsi, |ÃB|ÃA| =25/15 > 1,
(109 - 30)15 -25 1
Eq. (5-32b): - =
109(30) 109 n
n = 1.69 Ans.
OB 4.25
(a) n = = = 1.50
OA 2.83
OD 4.24 B
(a)
(b) n = = = 2.00
B
OC 2.12
OF 15.5
A
(c) n = = = 1.37 (3rd quadrant)
OE 11.3
OH 4.9
(d) n = = = 1.69
OG 2.9
O
A
C
1 cm 10 kpsi
G
D
(b)
H
(d)
E
F
(c)
5-14 Given: AISI 1006 CD steel, F = 0.55 N, P = 8.0 kN, and T = 30 N · m, applying the
DE theory to stress elements A and B with Sy = 280 MPa
32Fl 4P 32(0.55)(103)(0.1) 4(8)(103)
A: Ãx = + = +
Ä„d3 Ä„d2 Ä„(0.0203) Ä„(0.0202)
= 95.49(106) Pa = 95.49 MPa
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16T 16(30)
Äxy = = = 19.10(106) Pa = 19.10 MPa
Ä„d3 Ä„(0.0203)
1/2
2 2
à = Ãx + 3Äxy = [95.492 + 3(19.1)2]1/2 = 101.1MPa
Sy 280
n = = = 2.77 Ans.

à 101.1
4P 4(8)(103)
B: Ãx = = = 25.47(106) Pa = 25.47 MPa
Ä„d3 Ä„(0.0202)

16T 4 V 16(30) 4 0.55(103)
Äxy = + = +
Ä„d3 3 A Ä„(0.0203) 3 (Ä„/4)(0.0202)
= 21.43(106) Pa = 21.43 MPa

à = [25.472 + 3(21.432)]1/2 = 45.02 MPa
280
n = = 6.22 Ans.
45.02
5-15 Sy = 32 kpsi
At A, M = 6(190) = 1 140 lbf·in, T = 4(190) = 760 lbf · in.
32M 32(1140)
Ãx = = = 27 520 psi
Ä„d3 Ä„(3/4)3
16T 16(760)
Äzx = = = 9175 psi
Ä„d3 Ä„(3/4)3
2
27 520
Ämax = + 91752 = 16 540 psi
2
Sy 32
n = = = 0.967 Ans.
2Ämax 2(16.54)
MSS predicts yielding
5-16 From Prob. 4-15, Ãx = 27.52 kpsi, Äzx = 9.175 kpsi. For Eq. (5-15), adjusted for coordinates,
1/2

à = 27.522 + 3(9.175)2 = 31.78 kpsi
Sy 32
n = = = 1.01 Ans.

à 31.78
DE predicts no yielding, but it is extremely close. Shaft size should be increased.
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Chapter 5 129
5-17 Design decisions required:
" Material and condition
" Design factor
" Failure model
" Diameter of pin
Using F = 416 lbf from Ex. 5-3
32M
Ãmax =
Ä„d3
1/3
32M
d =
Ä„Ãmax
Decision 1: Select the same material and condition of Ex. 5-3 (AISI 1035 steel, Sy =
81 000).
Decision 2: Since we prefer the pin to yield, set nd a little larger than 1. Further explana-
tion will follow.
Decision 3: Use the Distortion Energy static failure theory.
Decision 4: Initially set nd = 1
Sy Sy
Ãmax = = = 81 000 psi
nd 1
1/3
32(416)(15)
d = = 0.922 in
Ä„(81 000)
Choose preferred size of d = 1.000 in
Ä„(1)3(81 000)
F = = 530 lbf
32(15)
530
n = = 1.274
416
Set design factor to nd = 1.274
Adequacy Assessment:
Sy 81 000
Ãmax = = = 63 580 psi
nd 1.274
1/3
32(416)(15)
d = = 1.000 in (OK )
Ä„(63 580)
Ä„(1)3(81 000)
F = = 530 lbf
32(15)
530
n = = 1.274 (OK)
416
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5-18 For a thin walled cylinder made of AISI 1018 steel, Sy = 54 kpsi, Sut = 64 kpsi.
The state of stress is
pd p(8) pd
Ãt = = = 40p, Ãl = = 20p, Ãr =-p
4t 4(0.05) 8t
These three are all principal stresses. Therefore,
1

à = " [(Ã1 - Ã2)2 + (Ã2 - Ã3)2 + (Ã3 - Ã1)2]1/2
2
1
= " [(40p - 20p)2 + (20p + p)2 + (-p - 40p)2]
2
= 35.51p = 54 Ò! p = 1.52 kpsi (for yield) Ans.
. .
For rupture, 35.51p = 64 Ò! p = 1.80 kpsi Ans.
5-19 For hot-forged AISI steel w = 0.282 lbf/in3, Sy = 30 kpsi and ½ = 0.292. Then Á = w/g =
2
0.282/386 lbf · s2/in; ri = 3in; ro = 5in; ri2 = 9; ro = 25; 3 + ½ = 3.292; 1 + 3½ = 1.876.
Eq. (3-55) for r = ri becomes

3 + ½ 1 + 3½
2
Ãt = ÁÉ2 2ro + ri2 1 -
8 3 + ½
Rearranging and substituting the above values:

Sy 0.282 3.292 1.876
= 50 + 9 1 -
É2 386 8 3.292
= 0.016 19
Setting the tangential stress equal to the yield stress,
1/2
30 000
É = = 1361 rad/s
0.016 19
or n = 60É/2Ä„ = 60(1361)/(2Ä„)
= 13 000 rev/min
Now check the stresses at r = (rori)1/2 , or r = [5(3)]1/2 = 3.873 in

3 + ½
Ãr = ÁÉ2 (ro - ri)2
8

0.282É2 3.292
= (5 - 3)2
386 8
= 0.001 203É2
Applying Eq. (3-55) for Ãt

0.282 3.292 9(25) 1.876(15)
Ãt = É2 9 + 25 + -
386 8 15 3.292
= 0.012 16É2
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Using the Distortion-Energy theory
1/2

à = Ãt2 - ÃrÃt + Ãr2 = 0.011 61É2
1/2
30 000
Solving É = = 1607 rad/s
0.011 61
So the inner radius governs and n = 13 000 rev/min Ans.
5-20 For a thin-walled pressure vessel,
di = 3.5 - 2(0.065) = 3.37 in
p(di + t)
Ãt =
2t
500(3.37 + 0.065)
Ãt = = 13 212 psi
2(0.065)
pdi 500(3.37)
Ãl = = = 6481 psi
4t 4(0.065)
Ãr =-pi =-500 psi
These are all principal stresses, thus,
1

à = " {(13 212 - 6481)2 + [6481 - (-500)]2 + (-500 - 13 212)2}1/2
2

à = 11 876 psi
Sy 46 000 46 000
n = = =

à à 11 876
= 3.87 Ans.
5-21 Table A-20 gives Sy as 320 MPa. The maximum significant stress condition occurs at ri
where Ã1 = Ãr = 0, Ã2 = 0, and Ã3 = Ãt. From Eq. (3-49) for r = ri , pi = 0,
2
2ro po 2(1502) po
Ãt =- =- =-3.6po
2
ro - ri2 1502 - 1002

à = 3.6po = Sy = 320
320
po = = 88.9MPa Ans.
3.6
5-22 Sut = 30 kpsi, w = 0.260 lbf/in3 , ½ = 0.211, 3 + ½ = 3.211, 1 + 3½ = 1.633. At the inner
radius, from Prob. 5-19

Ãt 3 + ½ 1 + 3½
2
= Á 2ro + ri2 - ri2
É2 8 3 + ½
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2
Here ro = 25, ri2 = 9, and so

Ãt 0.260 3.211 1.633(9)
= 50 + 9 - = 0.0147
É2 386 8 3.211
Since Ãr is of the same sign, we use M2M failure criteria in the first quadrant. From Table
A-24, Sut = 31 kpsi, thus,
1/2
31 000
É = = 1452 rad/s
0.0147
rpm = 60É/(2Ä„) = 60(1452)/(2Ä„)
= 13 866 rev/min
Using the grade number of 30 for Sut = 30 000 kpsi gives a bursting speed of 13640 rev/min.
5-23 TC = (360 - 27)(3) = 1000 lbf · in, TB = (300 - 50)(4) = 1000 lbf · in
y
223 lbf 127 lbf
B C
AD
8" 8" 6"
350 lbf
xy plane
In xy plane, MB = 223(8) = 1784 lbf · in and MC = 127(6) = 762 lbf · in.
387 lbf
8" 8" 6"
AD
B C
106 lbf 281 lbf
xz plane
In the xz plane, MB = 848 lbf · in and MC = 1686 lbf · in. The resultants are
MB = [(1784)2 + (848)2]1/2 = 1975 lbf · in
MC = [(1686)2 + (762)2]1/2 = 1850 lbf · in
So point B governs and the stresses are
16T 16(1000) 5093
Äxy = = = psi
Ä„d3 Ä„d3 d3
32MB 32(1975) 20 120
Ãx = = = psi
Ä„d3 Ä„d3 d3
2 1/2
Then
Ãx Ãx 2
ÃA, ÃB = Ä… + Äxy
2 2
Å„Å‚
1/2üÅ‚
òÅ‚20.12 20.12 2 żł
1
ÃA, ÃB = Ä… + (5.09)2
ół þÅ‚
d3 2 2
(10.06 Ä… 11.27)
= kpsi · in3
d3
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Chapter 5 133
Then
10.06 + 11.27 21.33
ÃA = = kpsi
d3 d3
and
10.06 - 11.27 1.21
ÃB = =- kpsi
d3 d3
For this state of stress, use the Brittle-Coulomb-Mohr theory for illustration. Here we use
Sut(min) = 25 kpsi, Suc(min) = 97 kpsi, and Eq. (5-31b) to arrive at
21.33 -1.21 1
- =
25d3 97d3 2.8
Solving gives d = 1.34 in. So use d = 13/8 in Ans.
Note that this has been solved as a statics problem. Fatigue will be considered in the next
chapter.
5-24 As in Prob. 5-23, we will assume this to be statics problem. Since the proportions are un-
changed, the bearing reactions will be the same as in Prob. 5-23. Thus
xy plane: MB = 223(4) = 892 lbf · in
xz plane: MB = 106(4) = 424 lbf · in
So
Mmax = [(892)2 + (424)2]1/2 = 988 lbf · in
32MB 32(988) 10 060
Ãx = = = psi
Ä„d3 Ä„d3 d3
Since the torsional stress is unchanged,
Äxz = 5.09/d3 kpsi
Å„Å‚
1/2üÅ‚
òÅ‚ 10.06 10.06 2 żł
1
ÃA, ÃB = Ä… + (5.09)2
ół þÅ‚
d3 2 2
ÃA = 12.19/d3 and ÃB =-2.13/d3
Using the Brittle-Coulomb-Mohr, as was used in Prob. 5-23, gives
12.19 -2.13 1
- =
25d3 97d3 2.8
Solving gives d = 11/8 in. Ans.
5-25 (FA)t = 300 cos 20 = 281.9 lbf, (FA)r = 300 sin 20 = 102.6 lbf
3383
T = 281.9(12) = 3383 lbf · in, (FC)t = = 676.6 lbf
5
(FC)r = 676.6 tan 20 = 246.3 lbf
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y
ROy = 193.7 lbf 246.3 lbf ROz = 233.5 lbf 676.6 lbf
A B C A B C
x x
O O
20" 16" 10" 20" 16" 10"
RBy = 158.1 lbf RBz = 807.5 lbf
z
281.9 lbf 102.6 lbf
xy plane xz plane

MA = 20 193.72 + 233.52 = 6068 lbf · in

MB = 10 246.32 + 676.62 = 7200 lbf · in (maximum)
32(7200) 73 340
Ãx = =
Ä„d3 d3
16(3383) 17 230
Äxy = =
Ä„d3 d3
1/2
Sy
2 2
à = Ãx + 3Äxy =
n
2 2 1/2
73 340 17 230 79 180 60 000
+ 3 = =
d3 d3 d3 3.5
d = 1.665 in so use a standard diameter size of 1.75 in Ans.
5-26 From Prob. 5-25,
2 1/2
Ãx Sy
2
Ämax = + Äxy =
2 2n
2 2 1/2
73 340 17 230 40 516 60 000
+ = =
2d3 d3 d3 2(3.5)
d = 1.678 in so use 1.75 in Ans.
5-27 T = (270 - 50)(0.150) = 33 N · m, Sy = 370 MPa
(T1 - 0.15T1)(0.125) = 33 Ò! T1 = 310.6N, T2 = 0.15(310.6) = 46.6N
(T1 + T2) cos 45 = 252.6N
y 107.0 N 252.6 N
163.4 N 252.6 N 89.2 N
300 400 150
300 400 150
A
O z
C
320 N 174.4 N
xy plane
xz plane
B
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Chapter 5 135

MA = 0.3 163.42 + 1072 = 58.59 N · m (maximum)

MB = 0.15 89.22 + 174.42 = 29.38 N · m
32(58.59) 596.8
Ãx = =
Ä„d3 d3
16(33) 168.1
Äxy = =
Ä„d3 d3
2 2 1/2
1/2
596.8 168.1 664.0 370(106)
2 2
à = Ãx + 3Äxy = + 3 = =
d3 d3 d3 3.0
d = 17.5(10-3) m= 17.5mm, so use 18 mm Ans.
5-28 From Prob. 5-27,
2 1/2
Ãx Sy
2
Ämax = + Äxy =
2 2n
2 2 1/2
596.8 168.1 342.5 370(106)
+ = =
2d3 d3 d3 2(3.0)
d = 17.7(10-3) m= 17.7mm, so use 18 mm Ans.
5-29 For the loading scheme shown in Figure (c),
V

F a b 4.4
Mmax = + = (6 + 4.5)
2 2 4 2
M
= 23.1N· m
y
For a stress element at A: B
C x
32M 32(23.1)(103)
A
Ãx = = = 136.2MPa
Ä„d3 Ä„(12)3
The shear at C is
4(F/2) 4(4.4/2)(103)
Äxy = = = 25.94 MPa
3Ä„d2/4 3Ä„(12)2/4
2 1/2
136.2
Ämax = = 68.1MPa
2
Since Sy = 220 MPa, Ssy = 220/2 = 110 MPa, and
Ssy 110
n = = = 1.62 Ans.
Ämax 68.1
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For the loading scheme depicted in Figure (d)
2
F a + b F 1 b F a b
Mmax = - = +
2 2 2 2 2 2 2 4
This result is the same as that obtained for Figure (c). At point B, we also have a surface
compression of
-F -F -4.4(103)
Ãy = = - =-20.4MPa
A bd 18(12)
With Ãx =-136.2MPa. From a Mohrs circle diagram, Ämax = 136.2/2 = 68.1MPa.
110
n = = 1.62 MPa Ans.
68.1
5-30 Based on Figure (c) and using Eq. (5-15)
1/2
2
à = Ãx
= (136.22)1/2 = 136.2MPa
Sy 220
n = = = 1.62 Ans.

à 136.2
Based on Figure (d) and using Eq. (5-15) and the solution of Prob. 5-29,
1/2
2 2
à = Ãx - ÃxÃy + Ãy
= [(-136.2)2 - (-136.2)(-20.4) + (-20.4)2]1/2
= 127.2MPa
Sy 220
n = = = 1.73 Ans.

à 127.2
5-31
When the ring is set, the hoop tension in the ring is
w
equal to the screw tension.
dF

2
ri2 pi ro
Ãt = 1 +
2
r2
ro - ri2
r
We have the hoop tension at any radius. The differential hoop tension dF is
dF = wÃt dr

ro 2
wri2 pi ro ro
F = wÃt dr = 1 + dr = wri pi (1)
2
r2
ro - ri2
ri ri
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Chapter 5 137
The screw equation is
T
Fi = (2)
0.2d
From Eqs. (1) and (2)
piri d
F T
pi = =
wri 0.2dwri
dFx
dFx = f piri d¸

2Ä„ 2Ä„
f T w
Fx = f piwri d¸ = ri d¸
0.2dwri o
o
2Ä„ f T
= Ans.
0.2d
5-32
(a) From Prob. 5-31, T = 0.2Fid
T 190
Fi = = = 3800 lbf Ans.
0.2d 0.2(0.25)
(b) From Prob. 5-31, F = wri pi
F Fi 3800
pi = = = = 15 200 psi Ans.
wri wri 0.5(0.5)


2
2
pi ri2 + ro
ri2 pi ro
(c) Ãt = 1 + =
2 2
r
ro - ri2 ro - ri2
r=ri
15 200(0.52 + 12)
= = 25 333 psi Ans.
12 - 0.52
Ãr =-pi =-15 200 psi
Ã1 - Ã3 Ãt - Ãr
(d) Ämax = =
2 2
25 333 - (-15 200)
= = 20 267 psi Ans.
2
1/2
2 2
à = ÃA + ÃB - ÃAÃB
= [25 3332 + (-15 200)2 - 25 333(-15 200)]1/2
= 35 466 psi Ans.
(e) Maximum Shear hypothesis
Ssy 0.5Sy 0.5(63)
n = = = = 1.55 Ans.
Ämax Ämax 20.267
Distortion Energy theory
Sy 63
n = = = 1.78 Ans.

à 35 466
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5-33
re The moment about the center caused by force F
1"R
is Fre where re is the effective radius. This is
1"
R
2
balanced by the moment about the center
caused by the tangential (hoop) stress.

ro
Fre = rÃtw dr
ri
t

2
wpiri2 ro ro
r
= r + dr
2
r
ro - ri2
ri

2
wpiri2 ro - ri2 ro
2

re = + ro ln
2
2 ri
F ro - ri2
From Prob. 5-31, F = wri pi. Therefore,

ro
ri 2 - ri2 ro
2
re = + ro ln
2
2 ri
ro - ri2
For the conditions of Prob. 5-31, ri = 0.5 and ro = 1in

0.5 12 - 0.52 1
re = + 12 ln = 0.712 in
12 - 0.52 2 0.5
5-34 ´nom = 0.0005 in
(a) From Eq. (3-57)

30(106)(0.0005) (1.52 - 12)(12 - 0.52)
p = = 3516 psi Ans.
(13) 2(1.52 - 0.52)
Inner member:

R2 + ri2 12 + 0.52
Eq. (3-58) (Ãt)i =-p =-3516 =-5860 psi
R2 - ri2 12 - 0.52
(Ãr)i =-p =-3516 psi
1/2
2 2
Eq. (5-13) Ãi = ÃA - ÃAÃB + ÃB
= [(-5860)2 - (-5860)(-3516) + (-3516)2]1/2
= 5110 psi Ans.
Outer member:

1.52 + 12
Eq. (3-59) (Ãt)o = 3516 = 9142 psi
1.52 - 12
(Ãr)o =-p =-3516 psi

Ão = [91422 - 9142(-3516) + (-3516)2]1/2
Eq. (5-13)
= 11 320 psi Ans.
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Chapter 5 139
(b) For a solid inner tube,

30(106)(0.0005) (1.52 - 12)(12)
p = = 4167 psi Ans.
1 2(12)(1.52)
(Ãt)i =-p =-4167 psi, (Ãr)i =-4167 psi
Ãi = [(-4167)2 - (-4167)(-4167) + (-4167)2]1/2 = 4167 psi Ans.

1.52 + 12
(Ãt)o = 4167 = 10 830 psi, (Ãr)o =-4167 psi
1.52 - 12

Ão = [10 8302 - 10 830(-4167) + (-4167)2]1/2 = 13 410 psi Ans.
5-35 Using Eq. (3-57) with diametral values,

207(103)(0.02) (752 - 502)(502 - 252)
p = = 19.41 MPa Ans.
(503) 2(752 - 252)

502 + 252
Eq. (3-58) (Ãt)i =-19.41 =-32.35 MPa
502 - 252
(Ãr)i =-19.41 MPa
Ãi = [(-32.35)2 - (-32.35)(-19.41) + (-19.41)2]1/2
Eq. (5-13)
= 28.20 MPa Ans.

752 + 502
(Ãt)o = 19.41 = 50.47 MPa,
Eq. (3-59)
752 - 502
(Ãr)o =-19.41 MPa

Ão = [50.472 - 50.47(-19.41) + (-19.41)2]1/2 = 62.48 MPa Ans.
5-36 Max. shrink-fit conditions: Diametral interference ´d = 50.01 - 49.97 = 0.04 mm. Equa-
tion (3-57) using diametral values:

207(103)0.04 (752 - 502)(502 - 252)
p = = 38.81 MPa Ans.
503 2(752 - 252)

502 + 252
Eq. (3-58): (Ãt)i =-38.81 =-64.68 MPa
502 - 252
(Ãr)i =-38.81 MPa
Eq. (5-13):
1/2
Ãi = (-64.68)2 - (-64.68)(-38.81) + (-38.81)2
= 56.39 MPa Ans.
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5-37
1.9998 1.999
´ = - = 0.0004 in
2 2
Eq. (3-56)

p(1) 22 + 12 p(1) 12 + 0
0.0004 = + 0.211 + - 0.292
14.5(106) 22 - 12 30(106) 12 - 0
p = 2613 psi
Applying Eq. (4-58) at R,

22 + 12
(Ãt)o = 2613 = 4355 psi
22 - 12
(Ãr)o =-2613 psi, Sut = 20 kpsi, Suc = 83 kpsi


Ão 2613

= < 1, 4" use Eq. (5-32a)
ÃA
4355
h = Sut/ÃA = 20/4.355 = 4.59 Ans.
5-38 E = 30(106) psi, ½ = 0.292, I = (Ä„/64)(24 - 1.54) = 0.5369 in4
Eq. (3-57) can be written in terms of diameters,



do
E´d 2 - D2 D2 - di2
30(106) (22 - 1.752)(1.752 - 1.52)

p = = (0.002 46)
2
D 1.75 2(1.752)(22 - 1.52)
2D2 do - di2
= 2997 psi = 2.997 kpsi
Outer member:
1.752(2.997)
Outer radius: (Ãt)o = (2) = 19.58 kpsi, (Ãr)o = 0
22 - 1.752

1.752(2.997) 22
Inner radius: (Ãt)i = 1 + = 22.58 kpsi, (Ãr)i =-2.997 kpsi
22 - 1.752 1.752
Bending:
6.000(2/2)
ro: (Ãx)o = = 11.18 kpsi
0.5369
6.000(1.75/2)
ri: (Ãx)i = = 9.78 kpsi
0.5369
Torsion: J = 2I = 1.0738 in4
8.000(2/2)
ro: (Äxy)o = = 7.45 kpsi
1.0738
8.000(1.75/2)
ri: (Äxy)i = = 6.52 kpsi
1.0738
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Chapter 5 141
Outer radius is plane stress
Ãx = 11.18 kpsi, Ãy = 19.58 kpsi, Äxy = 7.45 kpsi
Sy 60

Eq. (5-15) Ã = [11.182 - (11.18)(19.58) + 19.582 + 3(7.452)]1/2 = =
no no
60
21.35 = Ò! no = 2.81 Ans.
no
z
Inner radius, 3D state of stress
 2.997 kpsi
9.78 kpsi
22.58 kpsi
x y
6.52 kpsi
From Eq. (5-14) with Äyz = Äzx = 0
1 60

à = " [(9.78 - 22.58)2 + (22.58 + 2.997)2 + (-2.997 - 9.78)2 + 6(6.52)2]1/2 =
ni
2
60
24.86 = Ò! ni = 2.41 Ans.
ni
5-39 From Prob. 5-38: p = 2.997 kpsi, I = 0.5369 in4, J = 1.0738 in4
Inner member:

(0.8752 + 0.752)
Outer radius: (Ãt)o =-2.997 =-19.60 kpsi
(0.8752 - 0.752)
(Ãr)o =-2.997 kpsi
2(2.997)(0.8752)
Inner radius: (Ãt)i =- =-22.59 kpsi
0.8752 - 0.752
(Ãr)i = 0
Bending:
6(0.875)
ro: (Ãx)o = = 9.78 kpsi
0.5369
6(0.75)
ri: (Ãx)i = = 8.38 kpsi
0.5369
Torsion:
8(0.875)
ro: (Äxy)o = = 6.52 kpsi
1.0738
8(0.75)
ri: (Äxy)i = = 5.59 kpsi
1.0738
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The inner radius is in plane stress: Ãx = 8.38 kpsi, Ãy =-22.59 kpsi, Äxy = 5.59 kpsi
Ãi = [8.382 - (8.38)(-22.59) + (-22.59)2 + 3(5.592)]1/2 = 29.4 kpsi
Sy 60
ni = = = 2.04 Ans.
Ãi 29.4
Outer radius experiences a radial stress, Ãr
1/2
1

Ão = " (-19.60 + 2.997)2 + (-2.997 - 9.78)2 + (9.78 + 19.60)2 + 6(6.52)2
2
= 27.9 kpsi
60
no = = 2.15 Ans.
27.9
5-40

2
1 KI ¸ KI ¸ ¸ 3¸
Ãp = 2" cos Ä… " sin cos sin
2 2 2 2 2
2Ä„r 2Ä„r
1/2
2
KI ¸ ¸ 3¸
+ " sin cos cos
2 2 2
2Ä„r

1/2
KI ¸ ¸ ¸ 3¸ ¸ ¸ 3¸
= " cos Ä… sin2 cos2 sin2 + sin2 cos2 cos2
2 2 2 2 2 2 2
2Ä„r

KI ¸ ¸ ¸ KI ¸ ¸
= " cos Ä… cos sin = " cos 1 Ä… sin
2 2 2 2 2
2Ä„r 2Ä„r
Plane stress: The third principal stress is zero and

KI ¸ ¸ KI ¸ ¸
Ã1 = " cos 1 + sin , Ã2 = " cos 1 - sin , Ã3 = 0 Ans.
2 2 2 2
2Ä„r 2Ä„r
Plane strain: Ã1 and Ã2 equations still valid however,
KI ¸
Ã3 = ½(Ãx + Ãy) = 2½ " cos Ans.
2
2Ä„r
5-41 For ¸ = 0 and plane strain, the principal stress equations of Prob. 5-40 give
KI KI
Ã1 = Ã2 = " , Ã3 = 2½ " = 2½Ã1
2Ä„r 2Ä„r
1
(a) DE: " [(Ã1 - Ã1)2 + (Ã1 - 2½Ã1)2 + (2½Ã1 - Ã1)2]1/2 = Sy
2
Ã1 - 2½Ã1 = Sy

1 1
For ½ = , 1 - 2 Ã1 = Sy Ò! Ã1 = 3Sy Ans.
3 3
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Chapter 5 143
(b) MSS: Ã1 - Ã3 = Sy Ò! Ã1 - 2½Ã1 = Sy
1
½ = Ò! Ã1 = 3Sy Ans.
3
2
Ã3 = Ã1
3

Radius of largest circle


2
1 1, 2
1 2 Ã1
3
R = Ã1 - Ã1 =
2 3 6
5-42 (a) Ignoring stress concentration
F = Sy A = 160(4)(0.5) = 320 kips Ans.
(b) From Fig. 6-36: h/b = 1, a/b = 0.625/4 = 0.1563, ² = 1.3

F
Eq. (6-51) 70 = 1.3 Ä„(0.625)
4(0.5)
F = 76.9 kips Ans.
"
5-43 Given: a = 12.5mm, KIc = 80 MPa · m, Sy = 1200 MPa, Sut = 1350 MPa
350 350 - 50
ro = = 175 mm, ri = = 150 mm
2 2
12.5
a/(ro - ri) = = 0.5
175 - 150
150
ri/ro = = 0.857
175
.
Fig. 5-30: ² = 2.5
"
Eq. (5-37): KIc = ²Ã Ä„a

80 = 2.5Ã Ä„(0.0125)
à = 161.5MPa
Eq. (3-50) at r = ro:
ri2 pi
Ãt = (2)
2
ro - ri2
1502 pi(2)
161.5 =
1752 - 1502
pi = 29.2MPa Ans.
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5-44
(a) First convert the data to radial dimensions to agree with the formulations of Fig. 3-33.
Thus
ro = 0.5625 Ä… 0.001in
ri = 0.1875 Ä… 0.001 in
Ro = 0.375 Ä… 0.0002 in
Ri = 0.376 Ä… 0.0002 in
The stochastic nature of the dimensions affects the ´ =|Ri| -|Ro| relation in
Eq. (3-57) but not the others. Set R = (1/2)(Ri + Ro) = 0.3755. From Eq. (3-57)
2

ro
E´ - R2 R2 - ri2

p =
2
R
2R2 ro - ri2
Substituting and solving with E = 30 Mpsi gives
p = 18.70(106) ´
Since ´ = Ri - Ro
Å» Å» Å»
´ = Ri - Ro = 0.376 - 0.375 = 0.001 in
and
2 2 1/2
0.0002 0.0002
ô = +
Ć
4 4
= 0.000 070 7 in
Then
ô 0.000 070 7
Ć
C´ = = = 0.0707
Å»
0.001
´
The tangential inner-cylinder stress at the shrink-fit surface is given by
Å»
R2 +Ż2
ri
Ãit =-p
Å»
R2 -Ż2
ri

0.37552 + 0.18752
=-18.70(106) ´
0.37552 - 0.18752
=-31.1(106) ´
Å»
Ãit =-31.1(106) ´ =-31.1(106)(0.001)
Å»
=-31.1(103) psi
Also
ÃÃit =|C´Ãit| =0.0707(-31.1)103
Ć Ż
= 2899 psi
Ãit = N(-31 100, 2899) psi Ans.
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Chapter 5 145
(b) The tangential stress for the outer cylinder at the shrink-fit surface is given by

ro + R2
Ż2 Ż
Ãot = p
ro - R2
Ż2 Ż

0.56252 + 0.37552
= 18.70(106) ´
0.56252 - 0.37552
= 48.76(106) ´ psi
Ãot = 48.76(106)(0.001) = 48.76(103) psi
Å»
ÃÃot = C´Ãot = 0.0707(48.76)(103) = 34.45 psi
Ć Ż
Ãot = N(48 760, 3445) psi Ans.
5-45 From Prob. 5-44, at the fit surface Ãot = N(48.8, 3.45) kpsi. The radial stress is the fit
pressure which was found to be
p = 18.70(106) ´
p = 18.70(106)(0.001) = 18.7(103) psi
Å»
Ãp = C´ p = 0.0707(18.70)(103)
Ć Ż
= 1322 psi
and so
p = N(18.7, 1.32) kpsi
and
Ãor =-N(18.7, 1.32) kpsi
These represent the principal stresses. The von Mises stress is next assessed.
ÃA = 48.8 kpsi, ÃB =-18.7 kpsi
Å» Å»
k = ÃB/ÃA =-18.7/48.8 =-0.383
Å» Å»

à = ÃA(1 - k + k2)1/2
Å» Å»
= 48.8[1 - (-0.383) + (-0.383)2]1/2
= 60.4 kpsi


Ãà = Cpà = 0.0707(60.4) = 4.27 kpsi
Ć Ż
Using the interference equation
Å»
S -Å»
Ã
z =-
1/2
2 2
ÃS +Ć
Ć ÃÃ

95.5 - 60.4
=- =-4.5
[(6.59)2 + (4.27)2]1/2
pf = Ä… = 0.000 003 40,
or about 3 chances in a million. Ans.
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146 Solutions Manual " Instructor s Solution Manual to Accompany Mechanical Engineering Design
5-46
pd 6000N(1, 0.083 33)(0.75)
Ãt = =
2t 2(0.125)
= 18N(1, 0.083 33) kpsi
pd 6000N(1, 0.083 33)(0.75)
Ãl = =
4t 4(0.125)
= 9N(1, 0.083 33) kpsi
Ãr =-p =-6000N(1, 0.083 33) kpsi
These three stresses are principal stresses whose variability is due to the loading. From
Eq. (5-12), we find the von Mises stress to be
1/2
(18 - 9)2 + [9 - (-6)]2 + (-6 - 18)2

à =
2
= 21.0 kpsi


Ãà = Cpà = 0.083 33(21.0) = 1.75 kpsi
Ć Ż
Å»
S -Å»
Ã
z =-
1/2
2 2
ÃS +Ć
Ć ÃÃ

50 - 21.0
= =-6.5
(4.12 + 1.752)1/2
The reliability is very high
.
R = 1 - (6.5) = 1 - 4.02(10-11) = 1 Ans.