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Chapter 6
6-1
MSS:
σ
1
− σ
3
= S
y
/n ⇒ n =
S
y
σ
1
− σ
3
DE:
n =
S
y
σ
σ
=
σ
2
A
− σ
A
σ
B
+ σ
2
B
1/2
=
σ
2
x
− σ
x
σ
y
+ σ
2
y
+ 3τ
2
xy
1/2
(a) MSS:
σ
1
= 12, σ
2
= 6, σ
3
= 0
kpsi
n =
50
12
= 4.17
Ans.
DE:
σ
= (12
2
− 6(12) + 6
2
)
1/2
= 10.39 kpsi, n =
50
10.39
= 4.81
Ans.
(b)
σ
A
, σ
B
=
12
2
±
12
2
2
+ (−8)
2
= 16, −4 kpsi
σ
1
= 16, σ
2
= 0, σ
3
=−4 kpsi
MSS:
n =
50
16 − (−4)
= 2.5
Ans.
DE:
σ
= (12
2
+ 3(−8
2
))
1/2
= 18.33 kpsi, n =
50
18.33
= 2.73
Ans.
(c)
σ
A
, σ
B
=
−6 − 10
2
±
−6 + 10
2
2
+ (−5)
2
=−2.615, −13.385 kpsi
σ
1
= 0, σ
2
=−2.615, σ
3
=−13.385 kpsi
MSS:
n =
50
0 − (−13.385)
= 3.74
Ans.
DE:
σ
= [(−6)
2
− (−6)(−10) + (−10)
2
+ 3(−5)
2
]
1/2
= 12.29 kpsi
n =
50
12.29
= 4.07
Ans.
(d)
σ
A
, σ
B
=
12 + 4
2
±
12 − 4
2
2
+ 1
2
= 12.123, 3.877 kpsi
σ
1
= 12.123, σ
2
= 3.877, σ
3
= 0 kpsi
MSS:
n =
50
12.123 −0
= 4.12
Ans.
DE:
σ
= [12
2
− 12(4) + 4
2
+ 3(1
2
)]
1/2
= 10.72 kpsi
n =
50
10.72
= 4.66
Ans.
B
A
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150 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
6-2
S
y
= 50 kpsi
MSS:
σ
1
− σ
3
= S
y
/n ⇒ n =
S
y
σ
1
− σ
3
DE:
σ
2
A
− σ
A
σ
B
+ σ
2
B
1/2
= S
y
/n ⇒ n = S
y
/
σ
2
A
− σ
A
σ
B
+ σ
2
B
1/2
(a) MSS:
σ
1
= 12 kpsi, σ
3
= 0, n =
50
12 − 0
= 4.17 Ans.
DE:
n =
50
[12
2
− (12)(12) + 12
2
]
1/2
= 4.17
Ans.
(b) MSS:
σ
1
= 12 kpsi, σ
3
= 0
,
n =
50
12
= 4.17 Ans.
DE:
n =
50
[12
2
− (12)(6) + 6
2
]
1/2
= 4.81
Ans.
(c) MSS:
σ
1
= 12 kpsi, σ
3
=−12 kpsi
,
n =
50
12 − (−12)
= 2.08 Ans.
DE:
n =
50
[12
2
− (12)(−12) + (−12)
2
]
1/3
= 2.41
Ans.
(d) MSS:
σ
1
= 0, σ
3
=−12 kpsi,
n =
50
−(−12)
= 4.17 Ans.
DE:
n =
50
[(−6)
2
− (−6)(−12) + (−12)
2
]
1/2
= 4.81
6-3
S
y
= 390
MPa
MSS:
σ
1
− σ
3
= S
y
/n ⇒ n =
S
y
σ
1
− σ
3
DE:
σ
2
A
− σ
A
σ
B
+ σ
2
B
1/2
= S
y
/n ⇒ n = S
y
/
σ
2
A
− σ
A
σ
B
+ σ
2
B
1/2
(a) MSS:
σ
1
= 180 MPa, σ
3
= 0, n =
390
180
= 2.17 Ans.
DE:
n =
390
[180
2
− 180(100) + 100
2
]
1/2
= 2.50
Ans.
(b)
σ
A
, σ
B
=
180
2
±
180
2
2
+ 100
2
= 224.5, −44.5MPa= σ
1
, σ
3
MSS:
n =
390
224.5 − (−44.5)
= 1.45 Ans.
DE:
n =
390
[180
2
+ 3(100
2
)]
1/2
= 1.56
Ans.
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Chapter 6 151
(c)
σ
A
, σ
B
=−
160
2
±
−
160
2
2
+ 100
2
= 48.06, −208.06 MPa = σ
1
, σ
3
MSS:
n =
390
48.06 − (−208.06)
= 1.52 Ans.
DE:
n =
390
[−160
2
+ 3(100
2
)]
1/2
= 1.65
Ans.
(d)
σ
A
, σ
B
= 150, −150 MPa = σ
1
, σ
3
MSS:
n =
380
150 − (−150)
= 1.27 Ans.
DE:
n =
390
[3(150)
2
]
1/2
= 1.50
Ans.
6-4
S
y
= 220 MPa
(a)
σ
1
= 100, σ
2
= 80, σ
3
= 0MPa
MSS:
n =
220
100 − 0
= 2.20 Ans.
DET:
σ
= [100
2
− 100(80) + 80
2
]
1/2
= 91.65 MPa
n =
220
91.65
= 2.40 Ans.
(b)
σ
1
= 100, σ
2
= 10, σ
3
= 0MPa
MSS:
n =
220
100
= 2.20 Ans.
DET:
σ
= [100
2
− 100(10) + 10
2
]
1/2
= 95.39 MPa
n =
220
95.39
= 2.31 Ans.
(c)
σ
1
= 100, σ
2
= 0, σ
3
=−80 MPa
MSS:
n =
220
100 − (−80)
= 1.22 Ans.
DE:
σ
= [100
2
− 100(−80) + (−80)
2
]
1/2
= 156.2MPa
n =
220
156.2
= 1.41 Ans.
(d)
σ
1
= 0, σ
2
=−80, σ
3
=−100 MPa
MSS:
n =
220
0 − (−100)
= 2.20 Ans.
DE:
σ
= [(−80)
2
− (−80)(−100) + (−100)
2
] = 91.65
MPa
n =
220
91.65
= 2.40 Ans.
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152 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
6-5
(a) MSS:
n =
OB
OA
=
2.23
1.08
= 2.1
DE:
n =
OC
OA
=
2.56
1.08
= 2.4
(b) MSS:
n =
OE
OD
=
1.65
1.10
= 1.5
DE:
n =
OF
OD
=
1.8
1.1
= 1.6
(c) MSS:
n =
OH
OG
=
1.68
1.05
= 1.6
DE:
n =
OI
OG
=
1.85
1.05
= 1.8
(d) MSS:
n =
OK
OJ
=
1.38
1.05
= 1.3
DE:
n =
OL
OJ
=
1.62
1.05
= 1.5
O
(a)
(b)
(d)
(c)
H
I
G
J
K
L
F
E
D
A
B
C
Scale
1" ϭ 200 MPa
B
A
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Chapter 6 153
6-6
S
y
= 220
MPa
(a) MSS:
n =
OB
OA
=
2.82
1.3
= 2.2
DE:
n =
OC
OA
=
3.1
1.3
= 2.4
(b) MSS:
n =
OE
OD
=
2.2
1
= 2.2
DE:
n =
OF
OD
=
2.33
1
= 2.3
(c) MSS:
n =
OH
OG
=
1.55
1.3
= 1.2
DE:
n =
OI
OG
=
1.8
1.3
= 1.4
(d) MSS:
n =
OK
OJ
=
2.82
1.3
= 2.2
DE:
n =
OL
OJ
=
3.1
1.3
= 2.4
B
A
O
(a)
(b)
(c)
(d)
H
G
J
K
L
I
F
E
D
A
B
C
1" ϭ 100 MPa
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154 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
6-7
S
ut
= 30
kpsi,
S
uc
= 100
kpsi;
σ
A
= 20 kpsi, σ
B
= 6 kpsi
(a) MNS: Eq. (6-30a)
n =
S
ut
σ
x
=
30
20
= 1.5 Ans.
BCM: Eq. (6-31a)
n =
30
20
= 1.5 Ans.
M1M: Eq. (6-32a)
n =
30
20
= 1.5 Ans.
M2M: Eq. (6-33a)
n =
30
20
= 1.5 Ans.
(b)
σ
x
= 12 kpsi,τ
xy
=−8 kpsi
σ
A
, σ
B
=
12
2
±
12
2
2
+ (−8)
2
= 16, −4 kpsi
MNS: Eq. (6-30a)
n =
30
16
= 1.88 Ans.
BCM: Eq. (6-31b)
1
n
=
16
30
−
(−4)
100
⇒ n = 1.74 Ans.
M1M: Eq. (6-32a)
n =
30
16
= 1.88 Ans.
M2M: Eq. (6-33a)
n =
30
16
= 1.88 Ans.
(c)
σ
x
=−6 kpsi, σ
y
=−10 kpsi,τ
xy
=−5 kpsi
σ
A
, σ
B
=
−6 − 10
2
±
−6 + 10
2
2
+ (−5)
2
=−2.61, −13.39 kpsi
MNS: Eq. (6-30b)
n =−
100
−13.39
= 7.47 Ans.
BCM: Eq. (6-31c)
n =−
100
−13.39
= 7.47 Ans.
M1M: Eq. (6-32c)
n =−
100
−13.39
= 7.47 Ans.
M2M: Eq. (6-33c)
n =−
100
−13.39
= 7.47 Ans.
(d)
σ
x
=−12 kpsi,τ
xy
= 8 kpsi
σ
A
, σ
B
=−
12
2
±
−
12
2
2
+ 8
2
= 4, −16 kpsi
MNS: Eq. (6-30b)
n =
−100
−16
= 6.25 Ans.
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Chapter 6 155
BCM: Eq. (6-31b)
1
n
=
4
30
−
(−16)
100
⇒ n = 3.41 Ans.
M1M: Eq. (6-32b)
1
n
=
(100 −30)4
100(30)
−
−16
100
⇒ n = 3.95 Ans.
M2M: Eq. (6-33b)
n
4
30
+
n(−16) + 30
30 − 100
2
= 1
Reduces to
n
2
− 1.1979n − 15.625 = 0
n =
1.1979 +
1.1979
2
+ 4(15.625)
2
= 4.60
Ans.
(c)
L
(d)
J
(b)
(a)
I
H
G
K
F
O
C
D
E
A
B
1" ϭ 20 kpsi
B
A
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156 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
6-8 See Prob. 6-7 for plot.
(a) For all methods:
n =
OB
OA
=
1.55
1.03
= 1.5
(b) BCM:
n =
OD
OC
=
1.4
0.8
= 1.75
All other methods:
n =
OE
OC
=
1.55
0.8
= 1.9
(c) For all methods:
n =
OL
OK
=
5.2
0.68
= 7.6
(d) MNS:
n =
OJ
OF
=
5.12
0.82
= 6.2
BCM:
n =
OG
OF
=
2.85
0.82
= 3.5
M1M:
n =
OH
OF
=
3.3
0.82
= 4.0
M2M:
n =
OI
OF
=
3.82
0.82
= 4.7
6-9 Given:
S
y
= 42 kpsi, S
ut
= 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. (6-13) or (6-15) for plane stress and
Eq. (6-12) or (6-14) for general 3-D.
(a)
σ
= [9
2
− 9(−5) + (−5)
2
]
1/2
= 12.29 kpsi
n =
42
12.29
= 3.42 Ans.
(b)
σ
= [12
2
+ 3(3
2
)]
1/2
= 13.08
kpsi
n =
42
13.08
= 3.21 Ans.
(c)
σ
= [(−4)
2
− (−4)(−9) + (−9)
2
+ 3(5
2
)]
1/2
= 11.66
kpsi
n =
42
11.66
= 3.60 Ans.
(d)
σ
= [11
2
− (11)(4) + 4
2
+ 3(1
2
)]
1/2
= 9.798
n =
42
9.798
= 4.29 Ans.
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Chapter 6 157
For graphical solution, plot load lines on DE envelope as shown.
(a)
σ
A
= 9, σ
B
=−5
kpsi
n =
OB
OA
=
3.5
1
= 3.5 Ans.
(b)
σ
A
, σ
B
=
12
2
±
12
2
2
+ 3
2
= 12.7, −0.708
kpsi
n =
OD
OC
=
4.2
1.3
= 3.23
(c)
σ
A
, σ
B
=
−4 − 9
2
±
4 − 9
2
2
+ 5
2
=−0.910, −12.09
kpsi
n =
OF
OE
=
4.5
1.25
= 3.6 Ans.
(d)
σ
A
, σ
B
=
11 + 4
2
±
11 − 4
2
2
+ 1
2
= 11.14, 3.86
kpsi
n =
OH
OG
=
5.0
1.15
= 4.35 Ans.
6-10 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. 6-27 applies — confine its use to first and fourth quadrants.
(c)
(a)
(b)
(d)
E
C
G
H
D
B
A
O
F
1 cm ϭ 10 kpsi
B
A
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158 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
(a)
σ
x
= 90
kpsi,
σ
y
=−50
kpsi,
σ
z
= 0
І
σ
A
= 90
kpsi and
σ
B
=−50
kpsi. For the
fourth quadrant, from Eq. (6-13)
n =
1
(σ
A
/S
yt
) − (σ
B
/S
uc
)
=
1
(90/235) − (−50/275)
= 1.77 Ans.
(b)
σ
x
= 120
kpsi,
τ
xy
=−30
kpsi ccw.
σ
A
, σ
B
= 127.1, −7.08
kpsi. For the fourth
quadrant
n =
1
(127.1/235) − (−7.08/275)
= 1.76 Ans.
(c)
σ
x
=−40 kpsi, σ
y
=−90 kpsi, τ
xy
= 50 kpsi
.
σ
A
, σ
B
=−9.10, −120.9 kpsi.
Although no solution exists for the third quadrant, use
n =−
S
yc
σ
y
=−
275
−120.9
= 2.27 Ans.
(d)
σ
x
= 110
kpsi,
σ
y
= 40
kpsi,
τ
xy
= 10
kpsi cw.
σ
A
, σ
B
= 111.4, 38.6
kpsi. For the
first quadrant
n =
S
yt
σ
A
=
235
111.4
= 2.11 Ans.
Graphical Solution:
(a)
n =
OB
OA
=
1.82
1.02
= 1.78
(b)
n =
OD
OC
=
2.24
1.28
= 1.75
(c)
n =
OF
OE
=
2.75
1.24
= 2.22
(d)
n =
OH
OG
=
2.46
1.18
= 2.08
O
(d)
(b)
(a)
(c)
E
F
B
D
G
C
A
H
1 in ϭ 100 kpsi
B
A
shi20396_ch06.qxd 8/18/03 12:22 PM Page 158
[...]... = 67 6 .6 tan 20 = 2 46. 3 lbf y ROy = 193.7 lbf O ROz = 233.5 lbf 2 46. 3 lbf A C B 20" 16" 10" x RBy = 158.1 lbf 281.9 lbf O 67 6 .6 lbf A B 20" 16" z 102 .6 lbf C 10" RBz = 807.5 lbf xz plane xy plane M A = 20 193.72 + 233.52 = 60 68 lbf · in M B = 10 2 46. 32 + 67 6 .62 = 7200 lbf · in (maximum) σx = 32(7200) 73 340 = 3 πd d3 16( 3383) 17 230 = 3 πd d3 Sy 2 2 1/2 σ = σx + 3τx y = n τx y = 73 340 d3 d = 1 .66 5 in... 0.011 61 ω2 = 160 7 rad/s So the inner radius governs and n = 13 000 rev/min 6- 18 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 = 500(3.37) pdi = = 64 81 psi 4t 4(0. 065 ) σr = − pi = −500 psi shi203 96_ ch 06. qxd 8/18/03 12:22 PM Page 165 165 Chapter 6 These are all principal stresses, thus, 1 σ = √ {(13 212 − 64 81) 2 + [64 81... 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 252 .6 N 300 O A 89.2 N 400 300 150 400 z C B 252 .6 N 150 320 N 174.4 N xz plane xy plane M A = 0.3 163 .42 + 1072 = 58.59 N · m (maximum) M B = 0.15 89.22 + 174.42 = 29.38 N · m 32(58.59) 5 96. 8 = 3 πd d3 16( 33) 168 .1 = = 3 πd d3 σx = τx y σ = σx + 3τx y 2 2 1/2 5 96. 8 d3 = d = 17.5(10−3 ) m = 17.5 mm, 6- 26 2 168 .1 +3... ]1/2 Eq (6- 13) = 11 320 psi Ans (b) For a solid inner tube, p= 30(1 06 )(0.0005) (1.52 − 12 )(12 ) = 4 167 psi Ans 1 2(12 )(1.52 ) (σt ) i = − p = −4 167 psi, (σr ) i = −4 167 psi σi = [(−4 167 ) 2 − (−4 167 )(−4 167 ) + (−4 167 ) 2 ]1/2 = 4 167 psi Ans (σt ) o = 4 167 1.52 + 12 1.52 − 12 = 10 830 psi, (σr ) o = −4 167 psi σo = [10 8302 − 10 830(−4 167 ) + (−4 167 ) 2 ]1/2 = 13 410 psi Ans 6- 33 Using Eq (4 -60 ) with diametral... 17.5 mm, 6- 26 2 168 .1 +3 d3 2 1/2 66 4.0 370(1 06 ) = = d3 3.0 so use 18 mm Ans From Prob 6- 25, τmax = 5 96. 8 2d 3 2 d = 17.7(10−3 ) m = 17.7 mm, σx 2 168 .1 + d3 1/2 2 + τx y = 2 2 1/2 = Sy 2n 342.5 370(1 06 ) = d3 2(3.0) so use 18 mm Ans shi203 96_ ch 06. qxd 8/18/03 12:22 PM Page 169 169 Chapter 6 6-27 For the loading scheme shown in Figure (c), Mmax F = 2 a b + 2 4 V 4.4 = (6 + 4.5) 2 M = 23.1 N · m y For... With σx = −1 36. 2 MPa From a Mohrs circle diagram, τmax = 1 36. 2/2 = 68 .1 MPa n= 6- 28 110 = 1 .62 MPa 68 .1 Ans Based on Figure (c) and using Eq (6- 15) 2 σ = σx 1/2 = (1 36. 22 ) 1/2 = 1 36. 2 MPa n= Sy 220 = 1 .62 Ans = σ 1 36. 2 shi203 96_ ch 06. qxd 170 8/18/03 12:22 PM Page 170 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Based on Figure (d) and using Eq (6- 15) and the... speed of 1 364 0 rev/min 6- 21 TC = ( 360 − 27)(3) = 1000 lbf · in , TB = (300 − 50)(4) = 1000 lbf · in y 223 lbf A 127 lbf B 8" C 8" 350 lbf xy plane 6" D shi203 96_ ch 06. qxd 166 8/27/03 4:38 PM Page 166 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design In x y plane, M B = 223(8) = 1784 lbf · in and MC = 127 (6) = 762 lbf · in 387 lbf 8" A 8" B 6" D C 1 06 lbf 281 lbf... 2 ¯2 ¯ ro − R 2 ¯2 = 18.70(1 06 ) δ 0. 562 52 + 0.37552 0. 562 52 − 0.37552 = 48. 76( 1 06 ) δ psi σot = 48. 76( 1 06 )(0.001) = 48. 76( 103 ) psi ¯ σσot = Cδ σot = 0.0707(48. 76) (103 ) = 34.45 psi ˆ ¯ І σot = N(48 760 , 3445) psi Ans 6- 42 From Prob 6- 41, 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(1 06 ) δ p = 18.70(1 06 )(0.001) = 18.7(103 ) psi ¯... = 35 16 psi Ans 1 2(12 )(1.52 − 0.52 ) Inner member: (σt ) i = − p Eq (4-57) R 2 + ri2 12 + 0.52 = −35 16 2 1 − 0.52 R 2 − ri2 = −5 860 psi (σr ) i = − p = −35 16 psi 2 2 σi = σ A − σ A σ B + σ B Eq (6- 13) 1/2 = [(−5 860 ) 2 − (−5 860 )(−35 16) + (−35 16) 2 ]1/2 = 5110 psi Ans Outer member: (σt ) o = 35 16 Eq (4-58) 1.52 + 12 1.52 − 12 = 9142 psi (σr ) o = − p = −35 16 psi σo = [91422 − 9142(−35 16) + (−35 16) 2... kpsi For the 3rd quadrant, there is no solution but use Eq (6- 33c) Eq (6- 33c): C A OD 3.45 = = 2.70 Ans OC 1.28 Eq (6- 33a) (b) B O (b) n = Eq (6- 33a): D G OB 4 = = 4.0 Ans (a) n = OA 1 6- 13 (d) n=− 109 = 1. 36 Ans −80 A shi203 96_ ch 06. qxd 8/18/03 12:22 PM Page 161 161 Chapter 6 (d) σ A , σ B = 15, −25 kpsi n(15) −25n + 30 + 30 30 − 109 Eq (6- 33b): 2 =1 n = 1.90 Ans OB 4.25 = = 1.50 OA 2.83 OD 4.24 = . 4, − 16 kpsi
MNS: Eq. (6- 30b)
n =
−100
− 16
= 6. 25 Ans.
shi203 96_ ch 06. qxd 8/18/03 12:22 PM Page 154
Chapter 6 155
BCM: Eq. (6- 31b)
1
n
=
4
30
−
(− 16)
100
⇒. lbf
20" 16& quot; 10"
2 46. 3 lbf
O
xz plane
x
z
ABC
R
Oz
= 233.5 lbf
R
Bz
= 807.5 lbf
O
102 .6 lbf
20" 16& quot; 10"
67 6 .6 lbf
shi203 96_ ch 06. qxd
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