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Chapter 5
5-1
σ1 − σ3 = S y /n
MSS:
⇒
Sy
σ1 − σ3
n=
Sy
σ
1/2 2
1/2
σ = σ A2 − σ A σ B + σ B2
= σx − σx σ y + σ y2 + 3τx2y
n=
DE:
(a) MSS:
σ1 = 12, σ2 = 6, σ3 = 0 kpsi
50
n=
= 4.17 Ans.
12
σ = (122 − 6(12) + 62 ) 1/2 = 10.39 kpsi,
12 2
12
+ (−8) 2 = 16, −4 kpsi
±
(b) σ A , σ B =
2
2
DE:
n=
50
= 4.81
10.39
Ans.
σ1 = 16, σ2 = 0, σ3 = −4 kpsi
MSS:
n=
50
= 2.5
16 − (−4)
Ans.
50
σ = (122 + 3(−82 )) 1/2 = 18.33 kpsi, n =
= 2.73 Ans.
18.33
−6 + 10 2
−6 − 10
+ (−5) 2 = −2.615, −13.385 kpsi
±
(c) σ A , σ B =
2
2
DE:
σ1 = 0, σ2 = −2.615, σ3 = −13.385 kpsi
MSS:
50
n=
= 3.74
0 − (−13.385)
␴B
Ans.
σ = [(−6) 2 − (−6)(−10) + (−10) 2 + 3(−5) 2 ]1/2
= 12.29 kpsi
50
= 4.07 Ans.
n=
12.29
12 − 4 2
12 + 4
+ 12 = 12.123, 3.877 kpsi
±
(d) σ A , σ B =
2
2
DE:
σ1 = 12.123, σ2 = 3.877, σ3 = 0 kpsi
MSS:
DE:
n=
50
= 4.12
12.123 − 0
Ans.
σ = [122 − 12(4) + 42 + 3(12 )]1/2 = 10.72 kpsi
n=
50
= 4.66
10.72
Ans.
␴A
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5-2 S y = 50 kpsi
σ1 − σ3 = S y /n
DE:
2
1/2
= S y /n
σ A − σ A σ B + σ B2
n=
DE:
[122
n=
[122
1/2
n = S y / σ A2 − σ A σ B + σ B2
⇒
50
= 4.17 Ans.
12 − 0
50
= 4.17
− (12)(12) + 122 ]1/2
Ans.
50
= 4.81
− (12)(6) + 62 ]1/2
Ans.
σ1 = 12 kpsi, σ3 = −12 kpsi, n =
(c) MSS:
n=
DE:
[122
n=
50
= 2.08 Ans.
12 − (−12)
50
= 2.41
− (12)(−12) + (−12) 2 ]1/3
σ1 = 0, σ3 = −12 kpsi, n =
DE:
Ans.
50
= 4.17
12
σ1 = 12 kpsi, σ3 = 0, n =
DE:
(d) MSS:
n=
σ1 = 12 kpsi, σ3 = 0, n =
(a) MSS:
(b) MSS:
⇒
Sy
σ1 − σ3
MSS:
[(−6) 2
Ans.
50
= 4.17 Ans.
−(−12)
50
= 4.81
− (−6)(−12) + (−12) 2 ]1/2
5-3 S y = 390 MPa
σ1 − σ3 = S y /n
DE:
2
1/2
= S y /n
σ A − σ A σ B + σ B2
(a) MSS:
⇒
n=
Sy
σ1 − σ3
MSS:
σ1 = 180 MPa, σ3 = 0, n =
⇒
1/2
n = S y / σ A2 − σ A σ B + σ B2
390
= 2.17 Ans.
180
390
= 2.50 Ans.
[1802 − 180(100) + 1002 ]1/2
180 2
180
+ 1002 = 224.5, −44.5 MPa = σ1 , σ3
±
(b) σ A , σ B =
2
2
DE:
n=
MSS:
n=
DE:
n=
390
= 1.45 Ans.
224.5 − (−44.5)
[1802
390
= 1.56
+ 3(1002 )]1/2
Ans.
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Chapter 5
160 2
160
−
+ 1002 = 48.06, −208.06 MPa = σ1 , σ3
±
(c) σ A , σ B = −
2
2
390
= 1.52 Ans.
48.06 − (−208.06)
390
= 1.65 Ans.
n=
2
[−160 + 3(1002 )]1/2
n=
MSS:
DE:
(d) σ A , σ B = 150, −150 MPa = σ1 , σ3
390
= 1.30 Ans.
n=
MSS:
150 − (−150)
390
= 1.50 Ans.
n=
DE:
[3(150) 2 ]1/2
5-4 S y = 220 MPa
(a) σ1 = 100, σ2 = 80, σ3 = 0 MPa
MSS:
DET:
220
= 2.20 Ans.
100 − 0
σ = [1002 − 100(80) + 802 ]1/2 = 91.65 MPa
220
n=
= 2.40 Ans.
91.65
n=
(b) σ1 = 100, σ2 = 10, σ3 = 0 MPa
MSS:
DET:
n=
220
= 2.20 Ans.
100
σ = [1002 − 100(10) + 102 ]1/2 = 95.39 MPa
n=
220
= 2.31 Ans.
95.39
(c) σ1 = 100, σ2 = 0, σ3 = −80 MPa
MSS:
DE:
n=
220
= 1.22 Ans.
100 − (−80)
σ = [1002 − 100(−80) + (−80) 2 ]1/2 = 156.2 MPa
220
n=
= 1.41 Ans.
156.2
(d) σ1 = 0, σ2 = −80, σ3 = −100 MPa
220
n=
= 2.20 Ans.
MSS:
0 − (−100)
σ = [(−80) 2 − (−80)(−100) + (−100) 2 ] = 91.65 MPa
DE:
n=
220
= 2.40 Ans.
91.65
117
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5-5
(a) MSS:
DE:
(b) MSS:
DE:
n=
2.23
OB
=
= 2.1
OA
1.08
n=
2.56
OC
=
= 2.4
OA
1.08
n=
OE
1.65
=
= 1.5
OD
1.10
n=
1.8
OF
=
= 1.6
OD
1.1
␴B
(a)
C
B
A
Scale
1" ⫽ 200 MPa
␴A
O
D
E F
J
K
G
L
(d)
H
I
(c)
(c) MSS:
DE:
(d) MSS:
DE:
n=
OH
1.68
=
= 1.6
OG
1.05
n=
1.85
OI
=
= 1.8
OG
1.05
n=
1.38
OK
=
= 1.3
OJ
1.05
n=
OL
1.62
=
= 1.5
OJ
1.05
(b)
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Chapter 5
5-6 S y = 220 MPa
OB
2.82
=
= 2.2
OA
1.3
3.1
OC
=
= 2.4
n=
OA
1.3
2.2
OE
=
= 2.2
n=
OD
1
n=
(a) MSS:
DE:
(b) MSS:
n=
DE:
2.33
OF
=
= 2.3
OD
1
␴B
(a)
C
B
A
1" ⫽ 100 MPa
E
D
F
O
␴A
G
J
H
I
(c)
K
L
(d)
(c) MSS:
DE:
(d) MSS:
DE:
n=
1.55
OH
=
= 1.2
OG
1.3
n=
OI
1.8
=
= 1.4
OG
1.3
OK
2.82
=
= 2.2
OJ
1.3
3.1
OL
=
= 2.4
n=
OJ
1.3
n=
(b)
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5-7 Sut = 30 kpsi, Suc = 100 kpsi; σ A = 20 kpsi, σ B = 6 kpsi
(a) MNS: Eq. (5-30a)
n=
BCM: Eq. (5-31a)
n=
Sut
30
=
= 1.5 Ans.
σx
20
30
= 1.5 Ans.
20
30
n=
= 1.5 Ans.
20
MM: Eq. (5-32a)
(b) σx = 12 kpsi,τx y = −8 kpsi
12
±
σ A, σB =
2
12
2
2
+ (−8) 2 = 16, −4 kpsi
30
= 1.88 Ans.
16
16 (−4)
1
=
−
⇒
n
30
100
30
= 1.88 Ans.
n=
16
n=
MNS: Eq. (5-30a)
BCM: Eq. (5-31b)
MM: Eq. (5-32a)
n = 1.74 Ans.
(c) σx = −6 kpsi, σ y = −10 kpsi,τx y = −5 kpsi
−6 + 10 2
−6 − 10
+ (−5) 2 = −2.61, −13.39 kpsi
±
σ A, σB =
2
2
MNS: Eq. (5-30b)
n=−
100
= 7.47 Ans.
−13.39
BCM: Eq. (5-31c)
n=−
100
= 7.47 Ans.
−13.39
MM: Eq. (5-32c)
n=−
100
= 7.47 Ans.
−13.39
(d) σx = −12 kpsi,τx y = 8 kpsi
12 2
12
−
σ A, σB = − ±
+ 82 = 4, −16 kpsi
2
2
MNS: Eq. (5-30b)
n=
−100
= 6.25 Ans.
−16
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Chapter 5
BCM: Eq. (5-31b)
4
(−16)
1
=
−
n
30
100
MM: Eq. (5-32b)
(100 − 30)4 −16
1
=
−
n
100(30)
100
⇒
n = 3.41 Ans.
⇒
n = 3.95 Ans.
␴B
B
1" ⫽ 20 kpsi
(a)
A
O
␴A
C
E
D
K
F
G
H
L
(c)
J
(d)
(b)
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5-8 See Prob. 5-7 for plot.
(a) For all methods:
n=
1.55
OB
=
= 1.5
OA
1.03
(b) BCM:
n=
1.4
OD
=
= 1.75
OC
0.8
n=
OE
1.55
=
= 1.9
OC
0.8
(c) For all methods:
n=
5.2
OL
=
= 7.6
OK
0.68
(d) MNS:
n=
5.12
OJ
=
= 6.2
OF
0.82
BCM:
n=
2.85
OG
=
= 3.5
OF
0.82
MM:
n=
3.3
OH
=
= 4.0
OF
0.82
All other methods:
5-9 Given: S y = 42 kpsi, Sut = 66.2 kpsi, ε f = 0.90. Since ε f > 0.05, the material is ductile and
thus we may follow convention by setting S yc = S yt .
Use DE theory for analytical solution. For σ , use Eq. (5-13) or (5-15) for plane stress and
Eq. (5-12) or (5-14) for general 3-D.
(a) σ = [92 − 9(−5) + (−5) 2 ]1/2 = 12.29 kpsi
n=
42
= 3.42 Ans.
12.29
(b) σ = [122 + 3(32 )]1/2 = 13.08 kpsi
n=
42
= 3.21 Ans.
13.08
(c) σ = [(−4) 2 − (−4)(−9) + (−9) 2 + 3(52 )]1/2 = 11.66 kpsi
n=
42
= 3.60 Ans.
11.66
(d) σ = [112 − (11)(4) + 42 + 3(12 )]1/2 = 9.798
n=
42
= 4.29 Ans.
9.798
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123
Chapter 5
␴B
(d)
H
1 cm ⫽ 10 kpsi
G
O
C
D
(b)
␴A
A
E
B (a)
F
(c)
For graphical solution, plot load lines on DE envelope as shown.
σ A = 9, σ B = −5 kpsi
OB
3.5
n=
=
= 3.5 Ans.
OA
1
12 2
12
+ 32 = 12.7, −0.708 kpsi
±
(b) σ A , σ B =
2
2
(a)
4.2
OD
=
= 3.23
OC
1.3
4−9 2
−4 − 9
+ 52 = −0.910, −12.09 kpsi
±
(c) σ A , σ B =
2
2
n=
4.5
OF
=
= 3.6 Ans.
OE
1.25
11 − 4 2
11 + 4
+ 12 = 11.14, 3.86 kpsi
±
(d) σ A , σ B =
2
2
n=
n=
5-10
OH
5.0
=
= 4.35 Ans.
OG
1.15
This heat-treated steel exhibits S yt = 235 kpsi, S yc = 275 kpsi and ε f = 0.06. The steel is
ductile (ε f > 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) σx = 90 kpsi, σ y = −50 kpsi, σz = 0 σ A = 90 kpsi and σ B = −50 kpsi. For the
fourth quadrant, from Eq. (5-31b)
1
1
n=
=
= 1.77 Ans.
(σ A /S yt ) − (σ B /Suc )
(90/235) − (−50/275)
(b) σx = 120 kpsi, τx y = −30 kpsi ccw. σ A , σ B = 127.1, −7.08 kpsi. For the fourth
quadrant
1
n=
= 1.76 Ans.
(127.1/235) − (−7.08/275)
(c) σx = −40 kpsi, σ y = −90 kpsi, τx y = 50 kpsi . σ A , σ B = −9.10, −120.9 kpsi.
Although no solution exists for the third quadrant, use
S yc
275
n=−
=−
= 2.27 Ans.
σy
−120.9
(d) σx = 110 kpsi, σ y = 40 kpsi, τx y = 10 kpsi cw. σ A , σ B = 111.4, 38.6 kpsi. For the
first quadrant
S yt
235
n=
=
= 2.11 Ans.
σA
111.4
Graphical Solution:
␴B
OB
1.82
=
= 1.78
(a) n =
OA
1.02
(b) n =
OD
2.24
=
= 1.75
OC
1.28
(c) n =
2.75
OF
=
= 2.22
OE
1.24
(d) n =
2.46
OH
=
= 2.08
OG
1.18
(d)
H
1 in ⫽ 100 kpsi
G
␴A
O
C
D
A
B (a)
E
F
(c)
(b)
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125
Chapter 5
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) σx = 9 kpsi, σ y = −5 kpsi. σ A , σ B = 9, −5 kpsi. For the fourth quadrant,
| σσ BA | = 59 < 1, use Eq. (5-32a)
22
Sut
=
= 2.44 Ans.
n=
σA
9
(b) σx = 12 kpsi, τx y = −3 kpsi ccw. σ A , σ B = 12.7, −0.708 kpsi. For the fourth quad.708
rant, | σσ BA | = 012
.7 < 1,
n=
Sut
22
=
= 1.73 Ans.
σA
12.7
(c) σx = −4 kpsi, σ y = −9 kpsi, τx y = 5 kpsi . σ A , σ B = −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) σx = 11 kpsi, σ y = 4 kpsi,τx y = 1 kpsi. σ A , σ B = 11.14, 3.86 kpsi. For the first quadrant
n=
S yt
22
SA
=
=
= 1.97 Ans.
σA
σA
11.14
␴B
30
Sut ⫽ 22
(d )
30
␴A
(b)
(a)
–50
Sut ⫽ 83
(c) –90
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.
Since ε f < 0.05, the material is brittle. Thus, Sut = Suc and we may use MM which is
basically the same as MNS.
(a) σ A , σ B = 9, −5 kpsi
35
= 3.89 Ans.
9
(b) σ A , σ B = 12.7, −0.708 kpsi
n=
35
= 2.76 Ans.
12.7
(c) σ A , σ B = −0.910, −12.09 kpsi (3rd quadrant)
n=
n=
36
= 2.98 Ans.
12.09
␴B
(d) σ A , σ B = 11.14, 3.86 kpsi
n=
35
= 3.14 Ans.
11.14
1 cm ⫽ 10 kpsi
Graphical Solution:
O
3.45
OD
=
= 2.70 Ans.
OC
1.28
(c) n =
3.7
OF
=
= 2.85 Ans. (3rd quadrant)
OE
1.3
D
(b)
B
(a)
E
F
OH
3.6
=
= 3.13 Ans.
(d) n =
OG
1.15
(c)
Sut = 30 kpsi, Suc = 109 kpsi
Use MM:
(a) σ A , σ B = 20, 20 kpsi
30
n=
= 1.5 Ans.
Eq. (5-32a):
20
(b) σ A , σ B = ± (15) 2 = 15, −15 kpsi
n=
30
= 2 Ans.
15
(c) σ A , σ B = −80, −80 kpsi
For the 3rd quadrant, there is no solution but use Eq. (5-32c).
Eq. (5-32c):
C
A
(b) n =
Eq. (5-32a)
(d)
G
4
OB
= = 4.0 Ans.
(a) n =
OA
1
5-13
H
n=−
109
= 1.36 Ans.
−80
␴A
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127
Chapter 5
(d) σ A , σ B = 15, −25 kpsi, |σ B |σ A | = 25/15 > 1,
1
(109 − 30)15 −25
−
=
109(30)
109
n
Eq. (5-32b):
n = 1.69 Ans.
OB
4.25
=
= 1.50
OA
2.83
4.24
OD
=
= 2.00
(b) n =
OC
2.12
(a) n =
(c) n =
15.5
OF
=
= 1.37 (3rd quadrant)
OE
11.3
(d) n =
4.9
OH
=
= 1.69
OG
2.9
␴B
B (a)
A
␴A
O
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
A:
σx =
32Fl
4P
32(0.55)(103 )(0.1)
4(8)(103 )
+
=
+
πd 3
πd 2
π(0.0203 )
π(0.0202 )
= 95.49(106 ) Pa = 95.49 MPa
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16T
16(30)
=
= 19.10(106 ) Pa = 19.10 MPa
3
πd
π(0.0203 )
1/2
= [95.492 + 3(19.1) 2 ]1/2 = 101.1 MPa
σ = σx2 + 3τx2y
τx y =
n=
B:
Sy
280
=
= 2.77
σ
101.1
Ans.
4P
4(8)(103 )
=
= 25.47(106 ) Pa = 25.47 MPa
σx =
3
2
πd
π(0.020 )
0.55(103 )
16T
4V
16(30)
4
+
=
+
τx y =
πd 3 3 A
π(0.0203 ) 3 (π/4)(0.0202 )
= 21.43(106 ) Pa = 21.43 MPa
σ = [25.472 + 3(21.432 )]1/2 = 45.02 MPa
n=
5-15
280
= 6.22
45.02
Ans.
S y = 32 kpsi
At A, M = 6(190) = 1 140 lbf·in, T = 4(190) = 760 lbf · in.
σx =
32M
32(1140)
=
= 27 520 psi
3
πd
π(3/4) 3
τzx =
16T
16(760)
=
= 9175 psi
πd 3
π(3/4) 3
τmax =
n=
27 520
2
2
+ 91752 = 16 540 psi
Sy
32
=
= 0.967
2τmax
2(16.54)
Ans.
MSS predicts yielding
5-16
From Prob. 4-15, σx = 27.52 kpsi, τzx = 9.175 kpsi. For Eq. (5-15), adjusted for coordinates,
σ = 27.522 + 3(9.175) 2
n=
1/2
Sy
32
=
= 1.01
σ
31.78
= 31.78 kpsi
Ans.
DE predicts no yielding, but it is extremely close. Shaft size should be increased.
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Chapter 5
5-17
129
Design decisions required:
•
•
•
•
Material and condition
Design factor
Failure model
Diameter of pin
Using F = 416 lbf from Ex. 5-3
32M
πd 3
32M 1/3
d=
πσmax
σmax =
Decision 1: Select the same material and condition of Ex. 5-3 (AISI 1035 steel, S y =
81 000).
Decision 2: Since we prefer the pin to yield, set n d a little larger than 1. Further explanation will follow.
Decision 3: Use the Distortion Energy static failure theory.
Decision 4: Initially set n d = 1
Sy
Sy
=
= 81 000 psi
nd
1
32(416)(15) 1/3
= 0.922 in
d=
π(81 000)
σmax =
Choose preferred size of d = 1.000 in
F=
π(1) 3 (81 000)
= 530 lbf
32(15)
n=
530
= 1.274
416
Set design factor to n d = 1.274
Adequacy Assessment:
Sy
81 000
=
= 63 580 psi
nd
1.274
32(416)(15) 1/3
= 1.000 in (OK )
d=
π(63 580)
σmax =
F=
π(1) 3 (81 000)
= 530 lbf
32(15)
n=
530
= 1.274 (OK)
416
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For a thin walled cylinder made of AISI 1018 steel, S y = 54 kpsi, Sut = 64 kpsi.
The state of stress is
pd
pd
p(8)
=
= 40 p, σl =
= 20 p,
4t
4(0.05)
8t
These three are all principal stresses. Therefore,
σt =
σr = − p
1
σ = √ [(σ1 − σ2 ) 2 + (σ2 − σ3 ) 2 + (σ3 − σ1 ) 2 ]1/2
2
1
= √ [(40 p − 20 p) 2 + (20 p + p) 2 + (− p − 40 p) 2 ]
2
= 35.51 p = 54 ⇒ p = 1.52 kpsi (for yield) Ans.
.
.
For rupture, 35.51 p = 64 ⇒ p = 1.80 kpsi Ans.
5-19
For hot-forged AISI steel w = 0.282 lbf/in3 , S y = 30 kpsi and ν = 0.292. Then ρ = w/g =
0.282/386 lbf · s2 /in; ri = 3 in; ro = 5 in; ri2 = 9; ro2 = 25; 3 + ν = 3.292; 1 + 3ν = 1.876.
Eq. (3-55) for r = ri becomes
σt = ρω
2
3+ν
8
1 + 3ν
2
2
2ro + ri 1 −
3+ν
Rearranging and substituting the above values:
Sy
1.876
0.282 3.292
50 + 9 1 −
=
ω2
386
8
3.292
= 0.016 19
Setting the tangential stress equal to the yield stress,
30 000 1/2
ω=
= 1361 rad/s
0.016 19
or
n = 60ω/2π = 60(1361)/(2π)
= 13 000 rev/min
Now check the stresses at r = (rori ) 1/2 , or r = [5(3)]1/2 = 3.873 in
2 3+ν
(ro − ri ) 2
σr = ρω
8
0.282ω2 3.292
(5 − 3) 2
=
386
8
= 0.001 203ω2
Applying Eq. (3-55) for σt
3.292
9(25) 1.876(15)
2 0.282
9 + 25 +
σt = ω
−
386
8
15
3.292
= 0.012 16ω2
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Chapter 5
Using the Distortion-Energy theory
1/2
= 0.011 61ω2
σ = σt2 − σr σt + σr2
ω=
Solving
30 000
0.011 61
1/2
= 1607 rad/s
So the inner radius governs and n = 13 000 rev/min
5-20
Ans.
For a thin-walled pressure vessel,
di = 3.5 − 2(0.065) = 3.37 in
σt =
p(di + t)
2t
σt =
500(3.37 + 0.065)
= 13 212 psi
2(0.065)
σl =
pdi
500(3.37)
=
= 6481 psi
4t
4(0.065)
σr = − pi = −500 psi
These are all principal stresses, thus,
1
σ = √ {(13 212 − 6481) 2 + [6481 − (−500)]2 + (−500 − 13 212) 2 }1/2
2
σ = 11 876 psi
n=
Sy
46 000
46 000
=
=
σ
σ
11 876
= 3.87 Ans.
5-21
Table A-20 gives S y as 320 MPa. The maximum significant stress condition occurs at ri
where σ1 = σr = 0, σ2 = 0, and σ3 = σt . From Eq. (3-49) for r = ri , pi = 0,
σt = −
2ro2 po
2(1502 ) po
=
−
= −3.6 po
1502 − 1002
ro2 − ri2
σ = 3.6 po = S y = 320
po =
5-22
320
= 88.9 MPa Ans.
3.6
Sut = 30 kpsi, w = 0.260 lbf/in3 , ν = 0.211, 3 + ν = 3.211, 1 + 3ν = 1.633. At the inner
radius, from Prob. 5-19
σt
3+ν
1 + 3ν 2
2
2
2ro + ri −
=ρ
r
ω2
8
3+ν i
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Here ro2 = 25, ri2 = 9, and so
σt
0.260
=
2
ω
386
3.211
8
1.633(9)
50 + 9 −
= 0.0147
3.211
Since σr is of the same sign, we use M2M failure criteria in the first quadrant. From Table
A-24, Sut = 31 kpsi, thus,
31 000 1/2
ω=
= 1452 rad/s
0.0147
rpm = 60ω/(2π) = 60(1452)/(2π)
= 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
127 lbf
223 lbf
A
B
8"
C
8"
6"
D
350 lbf
xy plane
In x y plane, M B = 223(8) = 1784 lbf · in and MC = 127(6) = 762 lbf · in.
387 lbf
A
8"
8"
B
6"
C
106 lbf
D
281 lbf
xz plane
In the x z plane, M B = 848 lbf · in and MC = 1686 lbf · in. The resultants are
M B = [(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
=
= 3 psi
3
3
πd
πd
d
32M B
32(1975)
20 120
=
=
psi
σx =
πd 3
πd 3
d3
1/2
2
σx
σx
2
+ τx y
±
σ A, σB =
2
2

1  20.12
20.12 2
σ A, σB = 3
+ (5.09) 2
±

d
2
2
τx y =
Then
=
(10.06 ± 11.27)
kpsi · in3
d3

1/2 

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Chapter 5
Then
σA =
10.06 + 11.27
21.33
=
kpsi
3
d
d3
σB =
10.06 − 11.27
1.21
= − 3 kpsi
3
d
d
and
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
−
=
3
3
25d
97d
2.8
Solving gives d = 1.34 in. So use d = 1 3/8 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 unchanged, the bearing reactions will be the same as in Prob. 5-23. Thus
M B = 223(4) = 892 lbf · in
M B = 106(4) = 424 lbf · in
x y plane:
x z plane:
So
Mmax = [(892) 2 + (424) 2 ]1/2 = 988 lbf · in
σx =
32M B
32(988)
10 060
=
=
psi
3
3
πd
πd
d3
Since the torsional stress is unchanged,
τx z = 5.09/d 3 kpsi

10.06 2
1  10.06
±
σ A, σB = 3
+ (5.09) 2
d 
2
2
σ A = 12.19/d 3

1/2 

and σ B = −2.13/d 3
Using the Brittle-Coulomb-Mohr, as was used in Prob. 5-23, gives
12.19 −2.13
1
−
=
3
3
25d
97d
2.8
Solving gives d = 1 1/8 in.
5-25
Ans.
( FA ) t = 300 cos 20 = 281.9 lbf, ( FA )r = 300 sin 20 = 102.6 lbf
T = 281.9(12) = 3383 lbf · in, ( FC ) t =
( FC )r = 676.6 tan 20 = 246.3 lbf
3383
= 676.6 lbf
5
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y
ROy = 193.7 lbf
O
ROz = 233.5 lbf
246.3 lbf
A
C
B
20"
16"
x
10"
O
RBy = 158.1 lbf
281.9 lbf
676.6 lbf
A
B
20"
C
16"
10"
x
RBz = 807.5 lbf
z
102.6 lbf
xz plane
xy plane
M A = 20 193.72 + 233.52 = 6068 lbf · in
M B = 10 246.32 + 676.62 = 7200 lbf · in (maximum)
σx =
32(7200)
73 340
=
3
πd
d3
16(3383)
17 230
=
πd 3
d3
1/2
Sy
σ = σx2 + 3τx2y
=
n
2
73 340
17 230 2
+3
d3
d3
τx y =
d = 1.665 in
5-26
1/2
=
79 180
60 000
=
3
d
3.5
so use a standard diameter size of 1.75 in
Ans.
From Prob. 5-25,
τmax =
73 340
2d 3
2
σx
2
2
17 230
+
d3
1/2
+ τx y
=
2
2
1/2
=
Sy
2n
40 516
60 000
=
d3
2(3.5)
d = 1.678 in so use 1.75 in Ans.
5-27
T = (270 − 50)(0.150) = 33 N · m , S y = 370 MPa
(T1 − 0.15T1 )(0.125) = 33
⇒
T1 = 310.6 N,
T2 = 0.15(310.6) = 46.6 N
(T1 + T2 ) cos 45 = 252.6 N
107.0 N
y
163.4 N
O
300
252.6 N
A
89.2 N
400
150
B
xy plane
C
300
252.6 N
400
z
320 N
xz plane
150
174.4 N
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Chapter 5
M A = 0.3 163.42 + 1072 = 58.59 N · m
M B = 0.15 89.22 + 174.42 = 29.38 N · m
(maximum)
32(58.59)
596.8
=
3
πd
d3
16(33)
168.1
τx y =
=
3
πd
d3
2
596.8 2
168.1 2
2 1/2
σ = σx + 3τx y
=
+3
d3
d3
σx =
d = 17.5(10−3 ) m = 17.5 mm,
5-28
664.0
370(106 )
=
=
d3
3.0
so use 18 mm Ans.
From Prob. 5-27,
τmax =
596.8
2d 3
2
σx
2
168.1
+
d3
d = 17.7(10−3 ) m = 17.7 mm,
5-29
1/2
2
1/2
+ τx y
=
2
2
1/2
=
Sy
2n
342.5
370(106 )
=
d3
2(3.0)
so use 18 mm Ans.
For the loading scheme shown in Figure (c),
4.4
F a b
=
Mmax =
+
(6 + 4.5)
2 2 4
2
V
M
= 23.1 N · m
y
For a stress element at A:
B
32M
32(23.1)(103 )
σx =
=
= 136.2 MPa
πd 3
π(12) 3
The shear at C is
τx y =
4( F/2)
4(4.4/2)(103 )
=
= 25.94 MPa
3πd 2 /4
3π(12) 2 /4
τmax =
136.2
2
2
1/2
= 68.1 MPa
Since S y = 220 MPa, Ssy = 220/2 = 110 MPa, and
n=
Ssy
110
=
= 1.62 Ans.
τmax
68.1
C
A
x
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For the loading scheme depicted in Figure (d)
2
F 1
b
F a+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 )
σy =
=
−
= −20.4 MPa
A
bd
18(12)
With σx = −136.2 MPa. From a Mohrs circle diagram, τmax = 136.2/2 = 68.1 MPa.
n=
5-30
110
= 1.62 MPa
68.1
Ans.
Based on Figure (c) and using Eq. (5-15)
1/2
σ = σx2
= (136.22 ) 1/2 = 136.2 MPa
n=
Sy
220
=
= 1.62 Ans.
σ
136.2
Based on Figure (d) and using Eq. (5-15) and the solution of Prob. 5-29,
1/2
σ = σx2 − σx σ y + σ y2
= [(−136.2) 2 − (−136.2)(−20.4) + (−20.4) 2 ]1/2
= 127.2 MPa
n=
Sy
220
=
= 1.73 Ans.
σ
127.2
5-31
w
dF
r
When the ring is set, the hoop tension in the ring is
equal to the screw tension.
ri2 pi
ro2
1+ 2
σt = 2
r
ro − ri2
We have the hoop tension at any radius. The differential hoop tension d F is
d F = wσt dr
ro ro
wri2 pi
ro2
1 + 2 dr = wri pi
wσt dr = 2
F=
r
ro − ri2 ri
ri
(1)
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Chapter 5
The screw equation is
Fi =
T
0.2d
(2)
From Eqs. (1) and (2)
F
T
=
wri
0.2dwri
d Fx = f pi ri dθ
2π
Fx =
f pi wri dθ =
pi ri d␪
pi =
dFx
o
=
2π f T
0.2d
f Tw
ri
0.2dwri
2π
dθ
o
Ans.
5-32
T = 0.2Fi d
T
190
Fi =
=
= 3800 lbf
0.2d
0.2(0.25)
(a) From Prob. 5-31,
F = wri pi
(b) From Prob. 5-31,
pi =
(c)
Ans.
F
Fi
3800
=
=
= 15 200 psi
wri
wri
0.5(0.5)
Ans.
pi ri2 + ro2
ri2 pi
ro2
1+
σt = 2
=
r r=ri
ro − ri2
ro2 − ri2
15 200(0.52 + 12 )
= 25 333 psi Ans.
12 − 0.52
σr = − pi = −15 200 psi
σ1 − σ3
σt − σr
τmax =
=
2
2
25 333 − (−15 200)
= 20 267 psi Ans.
=
2
1/2
σ = σ A2 + σ B2 − σ A σ B
=
(d)
= [25 3332 + (−15 200) 2 − 25 333(−15 200)]1/2
= 35 466 psi Ans.
(e) Maximum Shear hypothesis
n=
0.5S y
Ssy
0.5(63)
=
=
= 1.55 Ans.
τmax
τmax
20.267
Distortion Energy theory
n=
Sy
63
=
= 1.78 Ans.
σ
35 466
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5-33
1"R
The moment about the center caused by force F
is Fre where re is the effective radius. This is
balanced by the moment about the center
caused by the tangential (hoop) stress.
ro
rσt w dr
Fre =
re
1"
R
2
ri
␴t
ro2
dr
r+
r
ri
wpi ri2
ro2 − ri2
r
o
re = 2
+ ro2 ln
2
ri
F ro − ri2
wpi r 2
= 2 i2
r o − ri
r
ro
From Prob. 5-31, F = wri pi . Therefore,
ro2 − ri2
ri
r
o
re = 2
+ ro2 ln
2
2
ri
r o − ri
For the conditions of Prob. 5-31, ri = 0.5 and ro = 1 in
2
1 − 0.52
0.5
1
2
= 0.712 in
re = 2
+ 1 ln
1 − 0.52
2
0.5
5-34
δnom = 0.0005 in
(a) From Eq. (3-57)
30(106 )(0.0005) (1.52 − 12 )(12 − 0.52 )
= 3516 psi Ans.
p=
(13 )
2(1.52 − 0.52 )
Inner member:
Eq. (3-58)
Eq. (5-13)
2
R 2 + ri2
1 + 0.52
(σt ) i = − p 2
= −5860 psi
= −3516 2
1 − 0.52
R − ri2
(σr ) i = − p = −3516 psi
1/2
σi = σ A2 − σ A σ B + σ B2
= [(−5860) 2 − (−5860)(−3516) + (−3516) 2 ]1/2
= 5110 psi Ans.
Outer member:
Eq. (3-59)
1.52 + 12
(σt ) o = 3516
1.52 − 12
= 9142 psi
(σr ) o = − p = −3516 psi
Eq. (5-13)
σo = [91422 − 9142(−3516) + (−3516) 2 ]1/2
= 11 320 psi Ans.
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Chapter 5
(b) For a solid inner tube,
30(106 )(0.0005) (1.52 − 12 )(12 )
= 4167 psi Ans.
p=
1
2(12 )(1.52 )
(σt ) i = − p = −4167 psi,
(σr ) i = −4167 psi
σi = [(−4167) 2 − (−4167)(−4167) + (−4167) 2 ]1/2 = 4167 psi Ans.
2
1.5 + 12
= 10 830 psi, (σr ) o = −4167 psi
(σt ) o = 4167
1.52 − 12
σo = [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 )
= 19.41 MPa Ans.
p=
(503 )
2(752 − 252 )
2
50 + 252
(σt ) i = −19.41
= −32.35 MPa
Eq. (3-58)
502 − 252
(σr ) i = −19.41 MPa
Eq. (5-13)
Eq. (3-59)
σi = [(−32.35) 2 − (−32.35)(−19.41) + (−19.41) 2 ]1/2
= 28.20 MPa Ans.
2
75 + 502
(σt ) o = 19.41
= 50.47 MPa,
752 − 502
(σr ) o = −19.41 MPa
σo = [50.472 − 50.47(−19.41) + (−19.41) 2 ]1/2 = 62.48 MPa Ans.
5-36
Max. shrink-fit conditions: Diametral interference δd = 50.01 − 49.97 = 0.04 mm. Equation (3-57) using diametral values:
207(103 )0.04 (752 − 502 )(502 − 252 )
= 38.81 MPa
p=
Ans.
503
2(752 − 252 )
2
50 + 252
(σt ) i = −38.81
= −64.68 MPa
Eq. (3-58):
502 − 252
(σr ) i = −38.81 MPa
Eq. (5-13):
σi = (−64.68) 2 − (−64.68)(−38.81) + (−38.81) 2
= 56.39 MPa
Ans.
1/2
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5-37
δ=
1.9998 1.999
−
= 0.0004 in
2
2
Eq. (3-56)
2
2
2 + 12
1 +0
p(1)
p(1)
+ 0.211 +
− 0.292
0.0004 =
14.5(106 ) 22 − 12
30(106 ) 12 − 0
p = 2613 psi
Applying Eq. (4-58) at R,
22 + 12
= 4355 psi
(σt ) o = 2613 2
2 − 12
(σr ) o = −2613 psi, Sut = 20 kpsi, Suc = 83 kpsi
σo 2613
=
σ 4355 < 1, ∴ use Eq. (5-32a)
A
h = Sut /σ A = 20/4.355 = 4.59 Ans.
5-38
E = 30(106 ) psi, ν = 0.292, I = (π/64)(24 − 1.54 ) = 0.5369 in4
Eq. (3-57) can be written in terms of diameters,
Eδd
p=
D
2
2
do − D 2 D 2 − di2
30(106 )
(2 − 1.752 )(1.752 − 1.52 )
=
(0.002 46)
1.75
2(1.752 )(22 − 1.52 )
2D 2 do2 − di2
= 2997 psi = 2.997 kpsi
Outer member:
Outer radius:
Inner radius:
1.752 (2.997)
(2) = 19.58 kpsi, (σr ) o = 0
22 − 1.752
22
1.752 (2.997)
1+
= 22.58 kpsi, (σr ) i = −2.997 kpsi
(σt ) i = 2
2 − 1.752
1.752
(σt ) o =
Bending:
ro :
(σx ) o =
6.000(2/2)
= 11.18 kpsi
0.5369
ri :
(σx ) i =
6.000(1.75/2)
= 9.78 kpsi
0.5369
Torsion:
J = 2I = 1.0738 in4
ro :
(τx y ) o =
8.000(2/2)
= 7.45 kpsi
1.0738
ri :
(τx y ) i =
8.000(1.75/2)
= 6.52 kpsi
1.0738
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Chapter 5
Outer radius is plane stress
σx = 11.18 kpsi,
σ y = 19.58 kpsi,
τx y = 7.45 kpsi
σ = [11.182 − (11.18)(19.58) + 19.582 + 3(7.452 )]1/2 =
Eq. (5-15)
21.35 =
60
no
⇒
Inner radius, 3D state of stress
Sy
60
=
no
no
n o = 2.81 Ans.
z
—2.997 kpsi
9.78 kpsi
22.58 kpsi
x
6.52 kpsi
y
From Eq. (5-14) with τ yz = τzx = 0
1
60
σ = √ [(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 =
⇒ n i = 2.41 Ans.
ni
5-39
From Prob. 5-38: p = 2.997 kpsi, I = 0.5369 in4 , J = 1.0738 in4
Inner member:
Outer radius:
(0.8752 + 0.752 )
= −19.60 kpsi
(σt ) o = −2.997
(0.8752 − 0.752 )
(σr ) o = −2.997 kpsi
Inner radius:
(σt ) i = −
2(2.997)(0.8752 )
= −22.59 kpsi
0.8752 − 0.752
(σr ) i = 0
Bending:
ro :
(σx ) o =
6(0.875)
= 9.78 kpsi
0.5369
ri :
(σx ) i =
6(0.75)
= 8.38 kpsi
0.5369
ro :
(τx y ) o =
8(0.875)
= 6.52 kpsi
1.0738
ri :
(τx y ) i =
8(0.75)
= 5.59 kpsi
1.0738
Torsion:
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The inner radius is in plane stress: σx = 8.38 kpsi, σ y = −22.59 kpsi, τx y = 5.59 kpsi
σi = [8.382 − (8.38)(−22.59) + (−22.59) 2 + 3(5.592 )]1/2 = 29.4 kpsi
ni =
Sy
60
=
= 2.04 Ans.
σi
29.4
Outer radius experiences a radial stress, σr
1
σo = √ (−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
KI
KI
θ
θ
1
θ
3θ 2
2√
±
cos
sin cos sin
σp =
√
2
2
2
2
2
2πr
2πr
1/2
θ
KI
θ
3θ 2
sin cos cos
+ √
2
2
2
2πr
1/2
θ
KI
2 θ
2 θ
2 3θ
2 θ
2 θ
2 3θ
cos ± sin cos sin
=√
+ sin cos cos
2
2
2
2
2
2
2
2πr
KI
θ
θ
θ
KI
θ
θ
=√
1 ± sin
cos ± cos sin
cos
=√
2
2
2
2
2
2πr
2πr
Plane stress: The third principal stress is zero and
θ
θ
θ
θ
KI
KI
σ1 = √
1 + sin
, σ2 = √
1 − sin
,
cos
cos
2
2
2
2
2πr
2πr
σ3 = 0 Ans.
Plane strain: σ1 and σ2 equations still valid however,
θ
KI
cos
σ3 = ν(σx + σ y ) = 2ν √
2
2πr
5-41
1/2
Ans.
For θ = 0 and plane strain, the principal stress equations of Prob. 5-40 give
KI
,
σ1 = σ2 = √
2πr
KI
σ3 = 2ν √
= 2νσ1
2πr
1
√ [(σ1 − σ1 ) 2 + (σ1 − 2νσ1 ) 2 + (2νσ1 − σ1 ) 2 ]1/2 = S y
2
σ1 − 2νσ1 = S y
1
1
σ1 = S y ⇒ σ1 = 3S y Ans.
1−2
For ν = ,
3
3
(a) DE:
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Chapter 5
σ1 − σ3 = S y
(b) MSS:
ν=
σ1 − 2νσ1 = S y
σ1 = 3S y
⇒
Ans.
2
σ3 = σ1
3
␶
2␴
3 1
5-42
1
3
⇒
␴1, ␴2
␴
Radius of largest circle
σ1
1
2
σ1 − σ1 =
R=
2
3
6
(a) Ignoring stress concentration
F = S y A = 160(4)(0.5) = 320 kips Ans.
(b) From Fig. 6-36: h/b = 1, a/b = 0.625/4 = 0.1563, β = 1.3
70 = 1.3
Eq. (6-51)
F π(0.625)
4(0.5)
F = 76.9 kips Ans.
5-43
Given: a = 12.5 mm, K I c = 80 MPa ·
ro =
a/(ro − ri ) =
ri /ro =
Fig. 5-30:
Eq. (5-37):
√
m, S y = 1200 MPa, Sut = 1350 MPa
350
= 175 mm,
2
ri =
350 − 50
= 150 mm
2
12.5
= 0.5
175 − 150
150
= 0.857
175
.
β = 2.5
√
K I c = βσ πa
80 = 2.5σ π(0.0125)
σ = 161.5 MPa
Eq. (3-50) at r = ro :
σt =
161.5 =
ri2 pi
(2)
ro2 − ri2
1502 pi (2)
1752 − 1502
pi = 29.2 MPa 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 ± 0.001in
ri = 0.1875 ± 0.001 in
Ro = 0.375 ± 0.0002 in
Ri = 0.376 ± 0.0002 in
The stochastic nature of the dimensions affects the δ = |Ri | − |Ro | relation in
Eq. (3-57) but not the others. Set R = (1/2)( Ri + Ro ) = 0.3755. From Eq. (3-57)
Eδ ro2 − R 2 R 2 − ri2
p=
R
2R 2 ro2 − ri2
Substituting and solving with E = 30 Mpsi gives
p = 18.70(106 ) δ
Since δ = Ri − Ro
δ̄ = R̄i − R̄o = 0.376 − 0.375 = 0.001 in
and
σ̂δ =
0.0002
4
2
0.0002
+
4
2
1/2
= 0.000 070 7 in
Then
Cδ =
0.000 070 7
σ̂δ
=
= 0.0707
0.001
δ̄
The tangential inner-cylinder stress at the shrink-fit surface is given by
σit = −p
R̄ 2 + r̄i2
R̄ 2 − r̄i2
0.37552 + 0.18752
= −18.70(10 ) δ
0.37552 − 0.18752
6
= −31.1(106 ) δ
σ̄it = −31.1(106 ) δ̄ = −31.1(106 )(0.001)
= −31.1(103 ) psi
Also
σ̂σit = |Cδ σ̄it | = 0.0707(−31.1)103
= 2899 psi
σit = N(−31 100, 2899) psi Ans.
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145
(b) The tangential stress for the outer cylinder at the shrink-fit surface is given by
2
r̄o + R̄ 2
σot = p 2
r̄o − R̄ 2
0.56252 + 0.37552
6
= 18.70(10 ) δ
0.56252 − 0.37552
= 48.76(106 ) δ psi
σ̄ot = 48.76(106 )(0.001) = 48.76(103 ) psi
σ̂σot = Cδ σ̄ot = 0.0707(48.76)(103 ) = 34.45 psi
σot = N(48 760, 3445) psi Ans.
5-45
From Prob. 5-44, at the fit surface σot = N(48.8, 3.45) kpsi. The radial stress is the fit
pressure which was found to be
p = 18.70(106 ) δ
p̄ = 18.70(106 )(0.001) = 18.7(103 ) psi
σ̂ p = Cδ p̄ = 0.0707(18.70)(103 )
= 1322 psi
and so
p = N(18.7, 1.32) kpsi
and
σor = −N(18.7, 1.32) kpsi
These represent the principal stresses. The von Mises stress is next assessed.
σ̄ A = 48.8 kpsi,
σ̄ B = −18.7 kpsi
k = σ̄ B /σ̄ A = −18.7/48.8 = −0.383
σ̄ = σ̄ A (1 − k + k 2 ) 1/2
= 48.8[1 − (−0.383) + (−0.383) 2 ]1/2
= 60.4 kpsi
σ̂σ = C p σ̄ = 0.0707(60.4) = 4.27 kpsi
Using the interference equation
S̄ − σ̄ z = −
1/2
σ̂ S2 + σ̂σ2
=−
95.5 − 60.4
= −4.5
[(6.59) 2 + (4.27) 2 ]1/2
p f = α = 0.000 003 40,
or about 3 chances in a million.
Ans.
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5-46
σt =
pd
6000N(1, 0.083 33)(0.75)
=
2t
2(0.125)
= 18N(1, 0.083 33) kpsi
σl =
pd
6000N(1, 0.083 33)(0.75)
=
4t
4(0.125)
= 9N(1, 0.083 33) kpsi
σr = −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
σ =
2
= 21.0 kpsi
σ̂σ = C p σ̄ = 0.083 33(21.0) = 1.75 kpsi
S̄ − σ̄ z = −
1/2
σ̂ S2 + σ̂σ2
=
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.