solucionario shigley
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Chapter 6 6-1
− σ = S /n ⇒ n = σ S− σ S n= DE: σ = σ − σ σ + σ + 3τ σ = σ − σ σ + σ σ = 12, σ = 6, σ = 0 kpsi (a) MSS: 50 = 4.17 Ans. n= 12 MSS:
y
σ 1
y
3
1
3
y
2
A
1
σ
DE: (b) σ A , σ B
12
=
2
2 1/2
A B
B
2
2 1/2
2
± +− 12
2
2
y
x y
n
= 1050.39 = 4.81
( 8) 2
3
Ans.
50 = (12 + 3(−8 )) = 18.33 kp = 2.73 Ans. kpsi si,, n = 18.33 = −6 − 10 ± −6 + 10 + (−5) = −2.615, −13.38 385 5 kp kpsi si 2
2
2
1/2
2
2
2
2
3
2
2
− ± + 2
1
B
2
2
3
2
2
2
1/2
B
2 1/2
+4 12 4 1 = 12.123 23,, 3.87 877 7 kp kpsi si (d) σ , σ = 2 2 σ = 12.123, σ = 3.877, σ = 0 kpsi 50 = 4.12 Ans. n= MSS: 12.123 − 0 σ = [12 − 12(4) + 4 + 3(1 )] = 10.72 kp kpsi si DE: 50 n= = 4.66 Ans. 10.72 12
A
Ans.
= 16, −4 kps psii
= 0, σ = −2.615, σ = −13.38 385 5 kp kpsi si 50 = 3.74 Ans. n= MSS: 0 − ( −13.385) DE: σ = [( −6) − ( −6)( −10) + ( −10) + 3( −5) ] = 12.29 kp kpsi si 50 n= = 4.07 Ans. 12.29
σ 1
1/2
2
2
σ
(c) σ A , σ B
x
= (12 − 6(12) + 6 ) = 10.39 kp kpsi si,,
2
DE:
x y
3
= 16, σ = 0, σ = −4 kpsi 50 n= = 2.5 MSS: 16 − ( −4)
σ 1
2
A
150
Solutions Manual • Instructor’ Instructor’s s Solution Manual to Accompany Mechanical Mechanical Engineering Design Design
6-2 S y
= 50 kp kpsi si
MSS:
σ 1
DE:
− σ = S /n ⇒
σ A2
(a) MSS:
A B
n
DE:
(b) MSS:
2 1/2
− σ σ + σ σ 1
σ 1
n
y
3
B
= σ S− σ y
1
3
= S /n ⇒
=S /
n
y
y
50 = 12 kp = 4.17 kpsi si,, σ = 0, n = 12 − 0
50 = 12 kp = 4.17 kpsi si,, σ = 0 , n = 12 3
A B
(c) MSS:
2 1/2
2
1
3
50 = [12 − (12)(−12) + (−12) ] = 2.41 50 σ = 0, σ = −12 kp kpsi si,, n = −(−12) = 4.17 n
DE: (d) MSS:
DE:
n
Ans.
Ans.
2 1/3
2
1
B
Ans .
50 = [12 − (12)(6) + 6 ] = 4.81 Ans. 50 σ = 12 kpsi, σ = −12 kp = 2.08 kpsi si,, n = 12 − ( −12) n
DE:
2 1/2
− σ σ + σ
Ans.
2 1/2
2
σ A2
Ans.
3
50 = [12 − (12)(12) + 12 ] = 4.17
Ans.
3
= [(−6) − (−6)( −5012) + (−12) ] = 4.81 2 1/2
2
= 39 390 0 MP MPaa σ − σ = S / n ⇒ MSS:
6-3 S y
1
DE: (a) MSS:
σ A2
DE:
(b) σ A , σ B
A B
n
=
2 1/2
− σ σ + σ σ 1
B
= σ S− σ y
1
3
= S /n ⇒
n
y
=S / y
390 = 18 = 2.17 180 0 MP MPa, a, σ = 0, n = 180 3
390 = [180 − 180(100) + 100 ] = 2.50
180 2
n
y
3
2 1/2
2
± + 180
1002
A B
Ans. Ans.
= 224.5, −44.5 MPa = σ , σ 1
MSS:
n
= 224.5 −390(−44.5) = 1.45
DE:
n
= [180 +390 = 1.56 3(100 )] 2
1/2
Ans.
Ans.
2 1/2
− σ σ + σ
2
2
2
σ A2
3
B
150
Solutions Manual • Instructor’ Instructor’s s Solution Manual to Accompany Mechanical Mechanical Engineering Design Design
6-2 S y
= 50 kp kpsi si
MSS:
σ 1
DE:
− σ = S /n ⇒
σ A2
(a) MSS:
A B
n
DE:
(b) MSS:
2 1/2
− σ σ + σ σ 1
σ 1
n
y
3
B
= σ S− σ y
1
3
= S /n ⇒
=S /
n
y
y
50 = 12 kp = 4.17 kpsi si,, σ = 0, n = 12 − 0
50 = 12 kp = 4.17 kpsi si,, σ = 0 , n = 12 3
A B
(c) MSS:
2 1/2
2
1
3
50 = [12 − (12)(−12) + (−12) ] = 2.41 50 σ = 0, σ = −12 kp kpsi si,, n = −(−12) = 4.17 n
DE: (d) MSS:
DE:
n
Ans.
Ans.
2 1/3
2
1
B
Ans .
50 = [12 − (12)(6) + 6 ] = 4.81 Ans. 50 σ = 12 kpsi, σ = −12 kp = 2.08 kpsi si,, n = 12 − ( −12) n
DE:
2 1/2
− σ σ + σ
Ans.
2 1/2
2
σ A2
Ans.
3
50 = [12 − (12)(12) + 12 ] = 4.17
Ans.
3
= [(−6) − (−6)( −5012) + (−12) ] = 4.81 2 1/2
2
= 39 390 0 MP MPaa σ − σ = S / n ⇒ MSS:
6-3 S y
1
DE: (a) MSS:
σ A2
DE:
(b) σ A , σ B
A B
n
=
2 1/2
− σ σ + σ σ 1
B
= σ S− σ y
1
3
= S /n ⇒
n
y
=S / y
390 = 18 = 2.17 180 0 MP MPa, a, σ = 0, n = 180 3
390 = [180 − 180(100) + 100 ] = 2.50
180 2
n
y
3
2 1/2
2
± + 180
1002
A B
Ans. Ans.
= 224.5, −44.5 MPa = σ , σ 1
MSS:
n
= 224.5 −390(−44.5) = 1.45
DE:
n
= [180 +390 = 1.56 3(100 )] 2
1/2
Ans.
Ans.
2 1/2
− σ σ + σ
2
2
2
σ A2
3
B
151
Chapter 6
(c) σ A , σ B
=−
MSS: DE: (d) σ A , σ B
MSS:
n
160 2
± − + 160
2
2
1002
= 48.06, −208.06 MP MPaa = σ , σ
= 48.06 −390 = 1.52 (−208.06)
1
Ans.
= [−160 +390 = 1.65 3(100 )] = 150, −15 150 0 MP MPaa = σ , σ 380 = 1.27 Ans. n= 150 − ( −150) n
2
2
1
DE:
n
1/2
3
390 = [3(150) = 1.50 ] 2 1/2
Ans.
Ans.
= 22 220 0 MP MPaa (a) σ = 100, σ = 80, σ = 0 MPa 220 = 2.20 Ans. n= MSS: 100 − 0 σ = [100 − 100(80) + 80 ] = 91.65 MP DET: MPaa 220 n= = 2.40 Ans. 91.65 (b) σ = 100, σ = 10, σ = 0 MPa 220 = 2.20 Ans. n= MSS: 100 = 95.39 MP σ = [100 − 100(10) + 10 ] MPaa DET: 220 = 2.31 Ans. n= 95.39 MPaa (c) σ = 100, σ = 0, σ = −80 MP 220 = 1.22 Ans. n= MSS: 100 − ( −80) = 156.2 MPa σ = [100 − 100( −80) + ( −80) ] DE: 220 = 1.41 Ans. n= 156.2 100 0 MP MPaa (d) σ = 0, σ = −80, σ = −10 220 = 2.20 Ans. n= MSS: 0 − (−100) σ = [( −80) − ( −80)( −100) + ( −100) ] = 91.65 MPa DE: 220 n= = 2.40 Ans. 91.65
6-4 S y
1
2
3
2 1/2
2
1
2
3
2 1/2
2
1
2
3
2 1/2
2
1
2
3
2
2
3
152
Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
6-5 (a) MSS:
DE: (b) MSS:
DE:
n
= OO BA = 21..23 = 2.1 08
n
2.56 = OC = = 2.4 OA 1.08
n
= OO DE = 11..65 = 1.5 10
n
= OO DF = 11..81 = 1.6 B
(a) C B
A
Scale 1" 200 MPa
O
A
D E F (b)
J K
G
L
(d )
H I
(c)
(c) MSS:
DE: (d) MSS:
DE:
n
= OO GH = 11..68 = 1.6 05
n
= OOGI = 11..85 = 1.8 05
n
= OOK J = 11..38 = 1.3 05
n
= OO LJ = 11..62 = 1.5 05
153
Chapter 6
6-6 S y
= 220 MPa
(a) MSS:
DE: (b) MSS:
DE:
n
= OO BA = 21..823 = 2.2
n
3.1 = OC = = 2.4 OA 1.3
n
= OO DE = 21.2 = 2.2
n
= OO DF = 2.133 = 2.3 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 = OO H = = 1.2 G 1.3
n
= OOGI = 11..83 = 1.4
n
= OOK J = 21..823 = 2.2
n
= OO LJ = 31..13 = 2.4
(b)
154
Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
6-7 Sut
= 30 kpsi, S = 100 kpsi; σ = 20 kpsi, σ = 6 kpsi uc
A
B
(a) MNS: Eq. (6-30 a)
n
= Sσ = 30 = 1.5 20 ut
Ans.
x
BCM: Eq. (6-31a )
n
= 30 = 1.5 20
Ans.
M1M: Eq. (6-32a )
n
= 30 = 1.5 20
Ans.
M2M: Eq. (6-33a )
n
= 30 = 1.5 20
Ans.
(b) σ x
= 12 kpsi, τ = −8 kpsi x y
σ A , σ B
=
12 2
12
( 8) 2
2
= 16, −4 kpsi
= 30 = 1.88 Ans. 16 1 16 (−4) = − ⇒ n 30 100
BCM: Eq. (6-31 b)
n
M1M: Eq. (6-32a)
= 30 = 1.88 16
n
= 1.74
Ans.
Ans.
= 30 = 1.88 Ans. 16 = −6 kpsi, σ = −10 kpsi, τ = −5 kpsi −6 − 10 ± −6 + 10 + (−5) = −2.61, −13.39 kpsi σ , σ = 2 2
M2M: Eq. (6-33a)
n
y
A
x y
B
MNS: Eq. (6-30b)
n
BCM: Eq. (6-31 c)
n
M1M: Eq. (6-32c)
n
M2M: Eq. (6-33c)
n
(d) σ x
2
n
MNS: Eq. (6-30a)
(c) σ x
± +−
= −12 kpsi, τ = 8 kpsi x y
σ A , σ B
MNS: Eq. (6-30b)
12
=−2 n
2
2
= −−100 = 7.47 13.39 = − −100 = 7.47 13.39
= −−100 = 7.47 13.39
= − −100 = 7.47 13.39
± − + 12 2
= −−100 = 6.25 16
Ans. Ans. Ans. Ans.
2
82
Ans.
= 4, −16 kpsi
155
Chapter 6
BCM: Eq. (6-31b) M1M: Eq. (6-32b)
4 ( −16) = − ⇒ n = 3.41 Ans. n 30 100 1 = (100 − 30)4 − −16 ⇒ n = 3.95 1
100(30)
n
+
100
2
− + 30 = 1 n M2M: Eq. (6-33b) 30 30 − 100 n − 1.1979n − 15.625 = 0 Reduces to 1.1979 + 1.1979 + 4(15.625) = 4.60 n= 4
n ( 16)
2
1"
2
2
Ans.
B
B
20 kpsi
(a)
A O
A
C
E D
K F
G
H
I
L
(c)
J
(d )
(b)
Ans.
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
= OO BA = 11..55 = 1.5 03
(b) BCM:
n
D = OOC = 10..48 = 1.75
n
E = OOC = 10..558 = 1.9
(c) For all methods:
n
= OOK L = 05..682 = 7.6
(d) MNS:
n
= OO F J = 50..12 = 6.2 82
BCM:
n
= OO GF = 20..85 = 3.5 82
M1M:
n
3.3 = OO H = = 4.0 F 0.82
M2M:
n
= OOF I = 30..82 = 4.7 82
All other methods:
=
=
=
42 kpsi, Sut 66.2 kpsi, ε f 0.90. Since ε f > 0.05, the material is ductile and 6-9 Given: S y S yt . thus we may follow convention by setting S yc 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
=
2 1/2
2
= [9 − 9(−5) + (−5) ] = 12.29 kpsi 42 n= = 3.42 Ans. 12.29 = 13.08 kpsi (b) σ = [12 + 3(3 )] 42 = 3.21 Ans. n= 13.08 = 11.66 kpsi (c) σ = [( −4) − ( −4)( −9) + ( −9) + 3(5 )] 42 n= = 3.60 Ans. 11.66 = 9.798 (d) σ = [11 − (11)(4) + 4 + 3(1 )] 42 = 4.29 Ans. n= 9.798 2
2
1/2
2
2
2
2
2
2
1/2
1/2
157
Chapter 6 B
1 cm
(d ) H
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. (a)
= 9, σ = −5 kpsi OB 3.5 = = 3.5 n= OA 1
σ A
(b) σ A , σ B n
(c) σ A , σ B n
(d) σ A , σ B
n
6-10
B
=
12 2
± + 12
Ans.
2
2
32
= 12.7, −0.708 kpsi
D = OOC = 41..23 = 3.23
= −4 − 9 2
− ± + 4
2
4.5 = OO F = = 3.6 E 1.25
= 2+ 4 11
2
9
52
Ans.
− ± + 11
= −0.910, −12.09 kpsi
4
2
5.0 = OO H = = 4.35 G 1.15
=
2
12
= 11.14, 3.86 kpsi
Ans.
=
=
275 kpsi and ε f 0.06. The steel is This heat-treated steel exhibits S yt 235 kpsi, S yc 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.
158
Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
=
=−
90 kpsi, σ y 50 kpsi, σ z (a) σ x fourth quadrant, from Eq. (6-13)
=0
σ A
= 90 kpsi and σ = −50 kpsi. For the B
= (σ /S ) −1 (σ /S ) = (90/235) −1(−50/275) = 1.77 Ans. = 120 kpsi, τ = −30 kpsi ccw. σ , σ = 127.1, −7.08 kpsi. For the fourth n
A
(b) σ x quadrant
yt
B
uc
x y
A
B
=−
= (127.1/235) −1 (−7.08/275) = 1.76 Ans. = −90 kpsi, τ = 50 kpsi. σ , σ = −9.10, −120.9 kpsi.
=
= − Sσ = − −275 = 2.27 Ans. 120.9 = 40 kpsi, τ = 10 kpsi cw. σ , σ = 111.4, 38.6 kpsi. For the
n
40 kpsi, σ y (c) σ x x y A Although no solution exists for the third quadrant, use
B
yc
n
y
110 kpsi, σ y (d) σ x first quadrant
x y
n
A
235 = Sσ = 111 = 2.11 .4 yt
B
Ans.
A
Graphical Solution: OB
B
1.82
(a) n
= O A = 1.02 = 1.78
(b) n
D = OOC = 21..24 = 1.75 28
(c) n
2.75 = OO F = = 2.22 E 1.24
(d) n
= OO GH = 21..46 = 2.08 18
(d )
H
1 in
100 kpsi
G
A
O C
D
A
B (a) E
F (c)
(b)
159
Chapter 6
6-11
The material is brittle and exhibits unequal tensile and compressive strengths. Decision: Use the Modified II-Mohr theory as shown in Fig. 6-28 which is limited to first and fourth quadrants. Sut 22 kpsi, Suc 83 kpsi
=
Parabolic failure segment: S A
=
= 22
S B
−22 −30 −40 −50 = 9 kpsi, σ = −5
(a) σ x Eq. (6-33a)
− 1
2
S A
S B
S A
22.0 21.6 20.1 17.4
−60 −70 −80 −83
13.5 8.4 2.3 0
kpsi. σ A , σ B
y
+ 22 22 − 83
S B
= 9, −5
kpsi. For the fourth quadrant, use
= Sσ = 229 = 2.44 Ans. = −3 kpsi ccw. σ , σ = 12.7, 0.708 kpsi. For the first quadrant, S 22 = = 1.73 Ans. n= σ 12.7 = −9 kpsi, τ = 5 kpsi . σ , σ = −0.910, −12.09 kpsi. For the ut
n
A
(b) σ x
= 12 kpsi, τ
x y
A
B
ut
A
=−
4 kpsi, σ y (c) σ x x y A B third quadrant, no solution exists; however, use Eq. (6-33 c)
= −−1283.09 = 6.87 Ans. = 4 kpsi,τ = 1 kpsi. σ , σ = 11.14, 3.86 kpsi. For the firstquadrant S S 22 = = = 1.97 Ans. n= σ σ 11.14 n
(d) σ x
= 11 kpsi,σ
y
x y
A
A
yt
A
A
B
B
30 Sut
30
–50
Sut
83
–90
22
A
160
6-12
Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
Since ε f < 0.05, the material is brittle. Thus, Sut basically the same as MNS.
= 9, −5 kpsi 35 = 3.89 Ans. n= 9 (b) σ , σ = 12.7, −0.708 kpsi 35 = 2.76 Ans. n= 12.7 (c) σ , σ = −0.910, −12.09 kpsi 36 = 2.98 Ans. n= 12.09 (d) σ , σ = 11.14, 3.86 kpsi 35 = 3.14 Ans. n= 11.14
=. S
uc
and we may use M2M which is
(a) σ A , σ B
A
B
A
B
A
(3rd quadrant) B
B
1 cm
10 kpsi
Graphical Solution: (a) n
= O A = 1 = 4.0
Ans.
= OC = 1.28 = 2.70
(c) n
3.7 = OO F = = 2.85 E 1.3
Sut
O H
O
Ans.
B
Ans. (3rd quadrant) Ans.
(c)
= 30 kpsi, S = 109 kpsi uc
Use M2M:
= 20, 20 kpsi
Eq. (6-33a): (b) σ A , σ B
Eq. (6-33a) (c) σ A , σ B
n
=±
(15) 2
= 30 = 1.5 20
= 15, −15 kpsi 30 =2 n= 15
Ans.
Ans.
= −80, −80 kpsi
For the 3rd quadrant, there is no solution but use Eq. (6-33 c). Eq. (6-33c):
D
(b)
E
F
3.6
= O G = 1.15 = 3.13
(a) σ A , σ B
C A
3.45
OD
(d )
G
4
OB
(b) n
(d) n
6-13
H
n
= − −109 = 1.36 80
Ans.
(a)
A
161
Chapter 6
(d) σ A , σ B
= 15, −25 kpsi n (15)
Eq. (6-33b):
30 n
+ −
= 1.90
(a) n
= OO BA = 42..25 = 1.50 83
(b) n
D = OOC = 42..24 = 2.00 12
(c) n
15.5 = OO F = = 1.37 (3rd quadrant) E 11.3
(d) n
= OO GH = 52..39 = 1.83
2
+ 30 = 1 30 − 109 25n
Ans.
B
B (a)
A
O
A
C
1 cm
10 kpsi G D
(b)
H
(d )
E
F
(c)
6-14
=
= 8.0 kN, and T = 30 N · m, applying the = 280 MPa )(0.1) 4(8)(10 ) = 32π d Fl + π4d P = 32(0π.55)(10 + π (0 .020 ) (0.020 ) = 95.49(10 ) Pa = 95.49 MPa
Given: AISI 1006 CD steel, F 0.55 N, P DE theory to stress elements A and B with S y
3
A:
σ x
3
2
6
3
3
2
162
Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
τ x y σ
n
B:
σ x
T 16(30) = 16 = = 19.10(10 ) Pa = 19.10 MPa π d π (0 .020 ) 6
3
=
σ x2
3
2
+ 3τ
x y
1/2
280 = σ S = 101 = 2.77 .1 y
2 1/2
2
= [95.49 + 3(19.1) ] = 101.1 MPa Ans.
4(8)(103 )
4P
6
= π d = π (0.020 ) = 25.47(10 ) Pa = 25.47 MPa 3
2
16T
4 V
16(30)
4
0.55(103 )
= π d + 3 A = π (0.020 ) + 3 (π/4)(0.020 ) = 21.43(10 ) Pa = 21.43 MPa = 45.02 MPa σ = [25.47 + 3(21.43 )] 280 n= = 6.22 Ans. 45.02
τ x y
3
3
6
2
6-15
2
2
1/2
Design decisions required: • Material and condition • Design factor • Failure model • Diameter of pin Using F
= 416 lbf from Ex. 6-3 σ max
d
M = 32 π d 3
1/3
= 32 M
π σ max
Decision 1: Select the same material and condition of Ex. 6-3 (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
σ max
= nS = S1 = 81000 psi y
y
d
d
=
32(416)(15) π (81 000)
1/3
= 0.922 in
163
Chapter 6
Choose preferred size of d
= 1.000 in π (1) (81 000) = 530 lbf F = 32(15) 3
= 530 = 1.274 416
n
Set design factor to n d Adequacy Assessment:
= 1.274 000 = nS = 81 = 63580 psi 1.274 y
σ max
d
d
6-16
=
32(416)(15) π (63 580)
π (1) 3 (81 000)
F
=
n
= 530 = 1.274 416
32(15)
1/3
= 1.000 in
= 530 lbf (OK)
For a thin walled cylinder made of AISI 1018 steel, S y The state of stress is σ t
(OK )
p (8) = pd = = 40 p, 4t 4(0.05)
σ l
= 54 kpsi, S = 64 kpsi. ut
= pd = 20 p, 8t
σ r
= − p
These three are all principal stresses. Therefore, σ
= √ 1 [(σ − σ ) + (σ − σ ) + (σ − σ ) ] 2
1
2
2
2
2
3
3
1
2 1/2
= √ 1 [(40 p − 20 p) + (20 p + p) + (− p − 40 p) ] 2 = 35.51 p = 54 ⇒ p = 1.52 kpsi (for yield) Ans. . . For rupture, 35.51 p = 64 ⇒ p = 1.80 kpsi Ans. 2
6-17
2
2
For hot-forged AISI steel w 0.282 lbf/in3 , S y 30 kpsi and ν 2 2 0.282/386 lbf s /in r i 3 in r o 5 in r i 9 r o2 25 3 ν
·
Eq. (4-56) for r
= ; = ; =
= r becomes σ = ρω
= = 0.292. Then ρ = w/g = ; = ; = ; + = 3.292; 1 + 3ν = 1.876.
i
t
2
+ 3
ν
8
2
2r o
+ r
Rearranging and substituting the above values: S y ω2
0.282
2
i
− 1
+ 3ν 3+ν
1
+ −
= 386 3.292 8 = 0.016 19
50
9 1
1.876 3.292
164
Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
Setting the tangential stress equal to the yield stress, ω
=
1/2
30 000 0.016 19
= 1361 rad/s
= 60ω/2π = 60(1361)/(2π ) = 13 000rev/min Now check the stresses at r = (r r ) , or r = [5(3)] = 3.873 in 3+ν σ = ρω (r − r ) or
n
1/2
o i
2
r
1/2
o
8
=
0.282ω2
3.292
386
8
= 0.001203ω
i
(5
2
− 3)
2
2
Applying Eq. (4-56) for σ t σ t
=ω
2
+ 0.282
3.292
386
8
= 0.012 16ω
9
+
25
9(25) 15
−
1.876(15) 3.292
2
Using the Distortion-Energy theory σ
Solving
ω
σ t 2
= =
6-18
r t
30 000 0.011 61
So the inner radius governs and n
2 1/2
− σ σ + σ
r
1/2
= 0.011 61ω
2
= 1607 rad/s
= 13 000rev/min
Ans.
For a thin-walled pressure vessel,
= 3.5 − 2(0.065) = 3.37in p( d + t ) σ = 2t 500(3.37 + 0.065) = 13 212 psi σ = d i
i
t
t
2(0.065)
500(3.37) = pd = = 6481 psi 4t 4(0.065) σ = − p = −500 psi σ l
r
i
i
165
Chapter 6
These are all principal stresses, thus,
= √ 1 {(13 212 − 6481) + [6481 − (−500)] + (−500 − 13 212) } 2 σ = 11 876 psi S 46 000 46 000 = n= = σ σ 11 876 = 3.87 Ans. σ
2
2 1/2
2
y
6-19
Table A-20 gives S y as 320 MPa. The maximum significant stress condition occurs at r i σ r 0, σ 2 0, and σ 3 σ t . From Eq. (4-50) for r r i where σ 1
= =
=
=
=
2r o2 po
2(1502 ) po
= −r − r = − 150 − 100 = −3.6 p σ = 3.6 p = S = 320 320 = 88.9 MPa Ans. p = 3.6 σ t
2
2
o
2
i
o
o
2
y
o
6-20
Sut 30 kpsi, w 0.260 lbf/in3 , ν radius, from Prob. 6-17
=
=
σ t ω2
Here r o2
=ρ
= 0.211, 3 + ν = 3.211, 1 + 3ν = 1.633. At the inner
+ 3
ν
8
+ 3ν r 2r + r − 3+ν 2 o
2 i
1
2 i
2 i
= 25, r = 9, and so σ t ω2
=
0.260 386
3.211
50
8
+9−
1.633(9) 3.211
=
0.0147
Since σ r is of the same sign, we use M2M failure criteria in the first quadrant. From Table A-24, Sut 31 kpsi , thus,
=
ω
= 31 000
1/2
0.0147
= 1452 rad/s
= 60ω/(2π ) = 60(1452)/(2π ) = 13 866 rev/min Using the grade number of 30 for S = 30 000 kpsi gives a bursting speed of 13640 rev/min. rpm
ut
6-21
T C
= (360 − 27)(3) = 1000 lbf · in ,
T B
= (300 − 50)(4) = 1000 lbf · in
y
223 lbf A
127 lbf B
8"
C
8" 350 lbf xy plane
6"
D
166
Solutions Manual
• Instructor’s Solution Manual to Accompany Mechanical Engineering Design
In x y plane, M B = 223(8) = 1784 lbf · in and M C = 127(6) = 762 lbf · in. 387 lbf 8"
A
8"
6"
B
D
C
106 lbf
281 lbf xz plane
In the x z plane, M B = 848lbf · in and M C = 1686 lbf · in. The resultants are M B = [(1784) 2 + (848) 2 ]1/2 = 1975 lbf · in M C = [(1686) 2 + (762) 2 ]1/2 = 1850 lbf · in
So point B governs and the stresses are τ x y = σ x =
Then σ A , σ B =
σ A , σ B =
=
16T
=
π d 3
32 M B
σ x
2 1 d 3
π d 3
16(1000) π d 3
=
32(1975) π d 3
σ A =
d 3
d 3
±
2
+ τ x y
2
20.12 2
psi
1/2
2
σ x
psi
20 120
=
20.12
±
d 3
+ (5 .09)
kpsi · in3
10.06 + 11.27 d 3
=
1/2
2
2
(10.06 ± 11.27)
Then
5093
=
21.33 d 3
2
kpsi
and σ B =
10.06 − 11.27 d 3
=−
1.21 d 3
kpsi
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. (6-31b) to arrive at 21.33 25d 3
−
−1.21
97d 3
Solving gives d = 1.34 in. So use d = 1 3/8 in
=
1 2.8
Ans.
Note that this has been solved as a statics problem. Fatigue will be considered in the next chapter. 6-22
As in Prob. 6-21, we will assume this to be statics problem. Since the proportions are unchanged, the bearing reactions will be the same as in Prob. 6-21. Thus x y plane:
M B = 223(4) = 892lbf · in
x z plane:
M B = 106(4) = 424lbf · in
167
Chapter 6
So 2 1/2
2
= [(892) + (424) ] = 988lbf · in 32(988) 10 060 = 32π M = = psi π d d d
M max
B
σ x
3
3
3
Since the torsional stress is unchanged, 3
= 5.09/d kpsi
τ x z
+ ±
1
= d
σ A , σ B
3
σ A
10.06
10.06
2
2
3
= 12.19/d
and
1/2
2
(5.09)
2
3
= −2.13/d
σ B
Using the Brittle-Coulomb-Mohr, as was used in Prob. 6-21, gives
− −972.d 13 = 21.8
12.19 25d 3 Solving gives d
3
= 1 1/8 in. Now compare to Modified II-Mohr theory
Ans.
= 300cos20 = 281.9 lbf , ( F ) = 300 sin 20 = 102.6 lbf 3383 = 676.6lbf T = 281.9(12) = 3383 lbf · in, ( F ) = 5 ( F ) = 676.6tan20 = 246.3 lbf
6-23
( F A ) t
A r
C t
C r
y ROy = 193.7 lbf O
ROz = 233.5 lbf
246.3 lbf
A
C
B
20"
16"
x
10"
O
R By = 158.1 lbf
281.9 lbf
20"
M B σ x τ x y σ
102.6 lbf
2
2
2
2
3
(maximum)
3
17 230 = 16(3383) = π d d 3
=
σ x2
3
2
+ 3τ
x y
1/2
d 3
2
= Sn
y
3
17 230 d 3
2
1/2
= 79d 180 = 603000 .5
so use a standard diameter size of 1.75 in
10"
xz plane
= 20 193.7 + 233.5 = 6068 lbf · in = 10 246.3 + 676.6 = 7200 lbf · in 73 340 = 32(7200) = π d d
73 340
= 1.665 in
C
R Bz = 807.5 lbf
z
+ d
B
16"
xy plane
M A
676.6 lbf A
3
Ans.
x
168
6-24
Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
From Prob. 6-23, 1/2
= + + = τ max
73 340
2
1/2
2
17 230
= 2Sn y
τ x2y
2
2d 3
6-25
2
σ x
40 516
d 3
d 3
d
= 1.678 in
T
= (270 − 50)(0.150) = 33 N · m, S = 370 MPa ( T − 0.15T )(0 .125) = 33 ⇒ T = 310.6 N, T = 0.15(310.6) = 46.6 N ( T + T ) cos 45 = 252.6 N
Ans.
so use 1.75in
y
1
1
1
2
1
2
107.0 N
y
163.4 N
252.6 N
300
O
89.2 N
400
A
300
150 B
252.6 N 400
z
C
M A M B σ x τ x y
320 N
d
2
2
= 0.3 163.4 + 107 = 58.59 N · m = 0.15 89.2 + 174.4 = 29.38 N · m .59) 596.8 = 32(58 = π d d 2
174.4 N
(maximum)
2
3
3
168.1 = 16(33) = π d d 3
=
σ
150
xz plane
xy plane
6-26
000 = 60 2(3.5)
2
σ x
3
2
+ 3τ
= +
x y
2
168.1
3
d 3
= 17.5(10− ) m = 17.5 mm, 3
2
596.8
1/2
1/2
=
d 3
so use 18 mm
664.0 d 3
Ans.
From Prob. 6-25,
= + + = τ max
596.8
2
2d 3
d
= 17.7(10− ) m = 17.7 mm, 3
σ x
2
2
1/2
d 3
so use 18 mm
= 2Sn y
τ x2y
2
168.1
1/2
342.5 d 3
Ans.
=
370(106 ) 2(3.0)
=
370(106 ) 3.0
169
Chapter 6
6-27
For the loading scheme shown in Figure ( c), M max
+ =
= F 2 a2 b4 = 23.1 N · m
4.4 2
(6
V
+ 4.5) M
y
For a stress element at A: σ x
=
32 M π d 3
=
B
32(23.1)(103 ) π (12) 3
C
x
A
= 136.2MPa
The shear at C is 3
τ x y
τ max
Since S y
/2)(10 ) = 34(πF d //2)4 = 4(43π.4(12) = 25.94 MPa /4 2
2
= 2
136.2
1/2
= 68.1 MPa
2
= 220 MPa, S = 220/2 = 110MPa, and sy
n
110 = τ S = 68 = 1.62 .1 sy
Ans.
max
For the loading scheme depicted in Figure ( d ) M max
=2
2
+ − = +
F a
b
2
F 1
b
2
2
2
F a
b
2
4
2
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 ) σ y 20.4 MPa A bd 18(12)
=− =− −−
With σ x
6-28
=−
= −136.2 MPa. From a Mohrs circle diagram, τ = 136.2/2 = 68.1 MPa. 110 = 1.62 MPa Ans . n= 68.1 max
Based on Figure ( c) and using Eq. (6-15) σ
2 1/2 x
= σ = (136.2 ) = 136.2 MPa S 220 = 1.62 Ans. n= = σ 136.2 y
2 1/2
170
Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
Based on Figure ( d ) and using Eq. (6-15) and the solution of Prob. 6-27, σ
2
2 1/2
= σ − σ σ + σ = [(−136.2) − (−136.2)( −20.4) + (−20.4) ] = 127.2 MPa S 220 n= = = 1.73 Ans. σ 127.2 x y
x
y
2
2 1/2
y
6-29
When the ring is set, the hoop tension in the ring is equal to the screw tension.
w
dF
σ t
r i2 pi
= r − r
1
2
2
o
r
r o2
+
i
r 2
We have the hoop tension at any radius. The differential hoop tension d F is
= wσ dr wr p wσ dr = F = r − r
d F
t
2 i i 2 2 o i
r o
t
r i
r o
r o2
+ 1
r 2
r i
dr
= wr p i
i
(1)
The screw equation is F i
= 0.T 2d
(2)
From Eqs. (1) and (2)
= wF r = 0.2d T wr d F = f p r d θ f T w F = f p wr d θ = r 0.2d wr
pi r i d
pi
i
dF x
x
i
i i
2π
x
i
i
i
i
o
= 20π. f2d T
2π
d θ
o
Ans.
6-30 (a) From Prob. 6-29,
= 0.2 F d T 190 = = 3800 lbf F = 0.2d 0.2(0.25) F = wr p T
i
i
(b) From Prob. 6-29, pi
i
Ans.
i
= wF r = wF r = 0.3800 = 15 200 psi 5(0.5) i
i
i
Ans .
171
Chapter 6
(c)
σ t
r i2 pi
= r − r
1
2
2
o
r o2
+
i
r
15 200(0.52
=
=
r r i
2
pi r i2
r o2
+
r o2
2 i
− r
+ 1 ) = 25 333 psi
= 1 − 0.5 Ans . σ = − p = −15200psi = σ −2 σ = σ −2 σ τ = 25 333 − (−15 200) = 20 267 psi Ans. 2
r
i
1
(d)
2
t
3
r
max
2
σ
2
1/2
2
= σ + σ − σ σ = [25333 + (−15 200) − 25 333(−15 200)] = 35 466 psi Ans. A
A B
B
2
2
1/2
(e) Maximum Shear hypothesis n
= τ S = 0τ .5S = 020.5(63) = 1.55 .267 sy
y
max
Ans.
max
Distortion Energy theory n
= Sσ = 3563466 = 1.78 y
Ans.
6-31 1" R
The moment about the center caused by force F is Fr e where r e is the effective radius. This is balanced by the moment about the center caused by the tangential (hoop) stress.
r e
1" R 2
r o
=
Fr e
r σ t w dr
r i
t
w pi r i2
r
r o
r
r i2
r o2
From Prob. 6-29, F
2
w pi r i
dr
2
r o
r i2
F r o2
r
r i
2
r e
r o2
= − + − = − + − r i
2
r o2 ln
= wr p . Therefore, i
i
r e
r i
= r − r
r e
0.5
= 1 − 0.5 2
2
2 o
For the conditions of Prob. 6-29, r i
2
r i2
r o2
i
2
+ r ln r r
o
2 o
i
= 0.5 and r = 1 in 1 − 0.5 1 + = 0.712 in 1 ln 2 0.5
o
2
2
2
r o r i
172
6-32
Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
δnom
= 0.0005 in
(a) From Eq. (4-60) p
=
30(106 )(0.0005)
1
(1 .52
2
2
2
− 1 )(1 − 0.5 ) = 3516 psi 2(1 )(1.5 − 0.5 ) 2
Inner member:
R 2 p R 2
2
2
2
2
2
2
2
Ans.
+ r = −3516 1 + 0.5 = −5860 psi ( σ ) = − 1 − 0.5 − r (σ ) = − p = −3516 psi σ = σ − σ σ + σ = [(−5860) − (−5860)(−3516) + (−3516) ] = 5110 psi Ans.
Eq. (4-57)
t i
i 2 i
r i
Eq. (6-13)
i
2
2 1/2
A B
A
B
2
2 1/2
Outer member:
1.52
2
+ 1 = 9142 psi −1
= 3516 1.5 ( σ ) = − p = −3516 psi σ = [9142 − 9142( −3516) + ( −3516) ] = 11320 psi Ans. ( σ t ) o
Eq. (4-58)
2
2
r o
Eq. (6-13)
2 1/2
2
o
(b) For a solid inner tube, p
=
30(106 )(0.0005)
1
(1 .52
2
2
− 1 )(1 ) = 4167 psi 2(1 )(1.5 ) 2
2
Ans.
= − p = −4167 psi, ( σ ) = −4167 psi = 4167 psi Ans. σ = [( −4167) − ( −4167)( −4167) + ( −4167) ] 1.5 + 1 = 10 830 psi, ( σ ) = −4167 psi (σ ) = 4167 1.5 − 1 = 13410 psi Ans. σ = [10830 − 10 830(−4167) + ( −4167) ] ( σ t )i
r i
t o
2
2
2
2
r o
2 1/2
2
o
6-33
2 1/2
2
i
Using Eq. (4-60) with diametral values, p
Eq. (4-57)
=
207(103 )(0.02) 50
(752
2
2
2
− 50 )(50 − 25 ) = 19.41 MPa 2(50 )(75 − 25 ) 50 + 25 = −32.35 MPa 50 − 25 2
2
2
2
2
= −19.41 (σ ) = −19.41 MPa σ = [( −32.35) − ( −32.35)( −19.41) + (−19.41) ] = 28.20 MPa Ans. (σ t )i
2
2
r i
Eq. (6-13)
2
i
2 1/2
Ans.
173
Chapter 6
752
2
+ 50 = 50.47 MPa, − 50
= 19.41 75 ( σ ) = −19.41 MPa σ = [50.47 − 50.47(−19.41) + ( −19.41) ] = 62.48 MPa ( σ t ) o
Eq. (4-58)
2
2
r o
2 1/2
2
o
Ans.
6-34
1.999 = 1.9998 − = 0.0004 in 2 2
δ
Eq. (4-59) p (1)
= 14.5(10 ) p = 2613 psi
0.0004
6
22 22
2
+ 1 + 0.211 + p(1) −1 30(10 ) 2
6
12 12
+ 0 − 0.292 −0
Applying Eq. (4-58) at R, ( σ t ) o (σ r ) o
E
22
2
2
2
= 2613 2 +− 11 = 4355 psi = −2613 psi , S = 20 kpsi, S = 83 kpsi −2613n + 20 000 = 1 n (4355) + 20 000 20 000 − 83 000 n = 4.52 Ans. ut
uc
2
Eq. (6-33b)
6-35
6
4
4
4
= 30(10 ) psi, ν = 0.292, I = (π/64)(2 − 1.5 ) = 0.5369 in
Eq. (4-60) can be written in terms of diameters, p
=
E δd D
2
d o
−D
2 D 2
2
2
2
6
− d = 30(10 ) (0.002 46) 1.75 d − d D
i
2
2
o
i
= 2997 psi = 2.997 kpsi
(2 2
2
2
2
− 1.75 )(1.75 − 1.5 ) 2(1.75 )(2 − 1.5 ) 2
2
2
Outer member:
Outer radius:
( σ t ) o
=
1.752 (2 .997) 22
2
− 1.75
1.752 (2 .997)
(2)
= 19.58 kpsi, ( σ ) = 0 r o
22
+ =
( σ t ) i
=
r o :
( σ x ) o
/2) = 6.000(2 = 11.18 kpsi 0.5369
r i :
( σ x ) i
.75/2) = 6.000(1 = 9.78 kpsi 0.5369
Inner radius:
22
2
− 1.75
1
1.752
Bending:
22.58 kpsi, ( σ r ) i
= −2.997 kpsi
174
Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
Torsion:
J
r o :
( τ x y ) o
r i :
( τ x y )i
4
= 2 I = 1.0738 in /2) = 8.000(2 = 7.45 kpsi 1.0738 .75/2) = 8.000(1 = 6.52 kpsi 1.0738
Outer radius is plane stress
= 11.18 kpsi, σ = 19.58 kpsi, τ = 7.45 kpsi = nS = 60 σ = [11.18 − (11.18)(19.58) + 19.58 + 3(7.45 )] n
σ x
y
x y
2
Eq. (6-15)
2
1/2
2
y
o
21.35
= 60 ⇒ n
no
o
Inner radius, 3D state of stress
= 2.81
o
Ans.
z
–2.997 kpsi 9.78 kpsi
22.58 kpsi x
From Eq. (6-14) with τ yz σ
y
6.52 kpsi
= τ = 0 zx
= √ 1 [(9.78 − 22.58) + (22.58 + 2.997) + (−2.997 − 9.78) + 6(6.52) ] = 60 n 2
2
2
2
i
24.86
= 60 ⇒ n
ni
i
6-36
2 1/2
From Prob. 6-35: p
= 2.41
Ans.
4
4
= 2.997 kpsi, I = 0.5369 in , J = 1.0738 in
Inner member:
Outer radius:
( σ t )o ( σ r )o
(0 .8752
2
= −2.997 (0.875 = −2.997 kpsi .997)(0.875 ) = − 2(2 = −22.59 kpsi 0.875 − 0.75 =0 2
2
2
Inner radius:
( σ t )i ( σ r )i
+ 0.75 ) = −19.60 kpsi − 0.75 )
2
2
Bending: r o :
( σ x ) o
.875) = 6(0 = 9.78 kpsi 0.5369
r i :
( σ x ) i
.75) = 6(0 = 8.38 kpsi 0.5369
175
Chapter 6
Torsion: r o :
( τ x y ) o
.875) = 8(0 = 6.52 kpsi 1.0738
r i :
( τ x y )i
.75) = 8(0 = 5.59 kpsi 1.0738
= 8.38 kpsi, σ = −22.59 kpsi, τ = 5.59 kpsi − (8.38)( −22.59) + (−22.59) + 3(5.59 )] = 29.4 kpsi S 60 = 2.04 Ans. n = = σ 29.4
The inner radius is in plane stress: σ x σ i
2
= [8.38
y
x y
2
1/2
2
y
i
i
Outer radius experiences a radial stress, σ r σ o
= √ 1 (−19.60 + 2.997) + (−2.997 − 9.78) + (9.78 + 19.60) + 6(6.52) 2 = 27.9 kpsi 60 n = = 2.15 Ans. 27.9 2
2
2
2 1/2
o
6-37 σ p
1
=2
± √ + √ ± ± = √
√ 2
K I
cos
2π r
θ
K I
2
2π r
sin
K I
= √ 2π r K I
= √ 2π r
θ
cos
2
cos
θ
2
θ
2
sin
cos
2
θ
2
2
cos θ
2
2
3θ
θ
2
θ
cos
2
θ
K I
θ
2
2π r
sin
cos
2
2
cos
2
+ sin
2
3θ
sin
2
cos
sin
θ
2
sin
2π r
K I
θ
θ
2
2
3θ 2 2
cos
2
θ
1/2
2
cos
2
3θ 2
1/2
± 1
sin
θ
2
Plane stress: The third principal stress is zero and σ 1
K I
θ
= √ 2π r cos 2
+ 1
sin
θ
,
2
σ 2
K I
θ
= √ 2π r cos 2
− 1
sin
θ
2
,
σ 3
Plane strain: σ 1 and σ 2 equations still valid however, σ 3
6-38
For θ
= ν (σ + σ ) = 2ν √ K 2π r cos θ2 I
x
y
Ans.
= 0 and plane strain, the principal stress equations of Prob. 6-37 give K K σ = σ = √ = , σ = 2ν √ 2νσ 2π r 2π r I
1
2
I
3
1
=0
Ans.
176
Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
√ 1 [(σ − σ ) + (σ − 2νσ ) + (2νσ − σ ) ] = S 2 σ − 2νσ = S 1 1 ⇒ σ = 3S Ans. σ = S 1−2 For ν = , 3 3 σ − σ = S ⇒ σ − 2νσ = S MSS: 1 ⇒ σ = 3S Ans. ν= 3
(a) DE:
1
2
1
1
2
1
1
1
2 1/2
1
(b)
1
y
1
1
y
y
1
y
3
1
y
1
y
y
1
= 23 σ
σ 3
1
Radius of largest circle 2 3 1
6-39
1, 2
R
1
=2
σ 1
2
− 3 σ
1
=
6
(a) Ignoring stress concentration
= S A = 160(4)(0.5) = 320 kips Ans. From Fig. 6-36: h /b = 1, a /b = 0.625/4 = 0.1563, β = 1.3 F
(b)
y
70
Eq. (6-51)
F
= 1.3 4(0.5)
Given: a
= 12.5 mm, K
I c
Eq. (6-51):
ut
i
2
− r ) = 17512−.5150 = 0.5 i
r i /r o
Fig. 6-40:
π (0 .625)
y
o
a /(r o
= 76.9 kips Ans. = 80 MPa · √ m, S = 1200 MPa, S = 1350 MPa 350 = 175 mm, r = 350 − 50 = 150 mm r = F
6-40
σ 1
= 150 = 0.857 175
=. 2.5 √ K = βσ π a 80 = 2.5σ π (0 .0125) σ = 161.5 MPa β
I c
2
177
Chapter 6
Eq. (4-51) at r
= r : o
σ
161.5 pi
r i2 pi
= r − r (2) 2 i
2 o
1502 pi (2)
= 175 − 150 = 29.2 MPa Ans. 2
2
6-41 (a) First convert the data to radial dimensions to agree with the formulations of Fig. 4-25. Thus r o 0.5625 0.001in
= ± r = 0.1875 ± 0.001in R = 0.375 ± 0.0002in R = 0.376 ± 0.0002 in i
o i
= | |−| |
Ri Ro relation in The stochastic nature of the dimensions affects the δ (1/2)( Ri Ro ) 0.3755. From Eq. (4-60) Eq. (4-60) but not the others. Set R
=
p
δ = E R
Substituting and solving with E
+
2
r o
−R
2
=
2
− − R
2 R 2 r o2
2
r i
r i2
= 30 Mpsi gives p = 18.70(10 ) δ 6
Since δ and
=R −R i
o
¯ = R¯ − R¯ = 0.376 − 0.375 = 0.001 in
δ
i
o
ˆ = + 2
0.0002
σδ
2
0.0002
4
1/2
4
= 0.000 070 7 in
Then C δ
= σˆδ¯ = 0.0000707 = 0.0707 0.001 δ
The tangential inner-cylinder stress at the shrink-fit surface is given by
¯ + r¯ σ = − ¯ − r¯ + 0.1875 = −18.70(10 ) δ 00..3755 3755 − 0.1875 = −31.1(10 ) δ σ¯ = −31.1(10 ) δ¯ = −31.1(10 )(0.001) = −31.1(10 ) psi it
R 2 p R 2
2 i 2 i
6
6
i t
6
3
2
2
2
2
6
178
Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
Also 3
ˆ = |C σ¯ | = 0.0707( −31.1)10 = 2899 psi σ = N( −31100, 2899) psi Ans.
σσ i t
δ it
it
(b) The tangential stress for the outer cylinder at the shrink-fit surface is given by σ ot
=p
¯ + ¯ ¯−¯ r o2
R 2
r o2
R 2 6
= 18.70(10 ) δ
0.56252 0.56252
6
+ 0.3755 − 0.3755
2 2
= 48.76(10 ) δ psi σ¯ = 48.76(10 )(0 .001) = 48.76(10 ) psi σˆ = C σ¯ = 0.0707(48.76)(10 ) = 34.45 psi σ = N(48 760, 3445) psi Ans. 6
ot
σ ot
6-42
3
3
δ ot
ot
From Prob. 6-41, at the fit surface pressure which was found to be
σ ot
= N(48.8, 3.45) kpsi.
The radial stress is the fit
6
= 18.70(10 ) δ p¯ = 18.70(10 )(0.001) = 18.7(10 ) psi σˆ = C p¯ = 0.0707(18.70)(10 ) = 1322 psi p
6
p
3
3
δ
and so p
= N(18.7, 1.32) kpsi
and σ or
= −N(18.7, 1.32) kpsi
These represent the principal stresses. The von Mises stress is next assessed.
¯ = 48.8 kpsi, σ¯ = −18.7 kpsi k = σ¯ /σ¯ = −18.7/48.8 = −0.383 σ¯ = σ¯ (1 − k + k ) = 48.8[1 − (−0.383) + (−0.383) ] = 60.4 kpsi σˆ = C σ¯ = 0.0707(60.4) = 4.27 kpsi σ A
B
B
A
A
2 1/2
2 1/2
σ
p
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