Shigley's Mechanical
Engineering Design 10th
Edition Solution Manual
Mechanical Systems Design
Oakland University (OU)
739 pag.
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Chapter 1
Problems 1-1 through 1-6 are for student research. No standard solutions are provided.
From Fig. 1-2, cost of grinding to ± 0.0005 in is 270%. Cost of turning to ± 0.003 in is
60%.
Relative cost of grinding vs. turning = 270/60 = 4.5 times
Ans.
______________________________________________________________________________
1-7
1-8
CA = CB,
10 + 0.8 P = 60 + 0.8 P − 0.005 P 2
⇒
P = 100 parts Ans.
P 2 = 50/0.005
______________________________________________________________________________
1-9
Max. load = 1.10 P
Min. area = (0.95)2A
Min. strength = 0.85 S
To offset the absolute uncertainties, the design factor, from Eq. (1-1) should be
nd =
1.10
= 1.43
Ans.
2
0.85 ( 0.95 )
______________________________________________________________________________
1-10 (a) X1 + X2:
x1 + x2 = X 1 + e1 + X 2 + e2
error = e = ( x1 + x2 ) − ( X 1 + X 2 )
= e1 + e2
(b) X1 − X2:
Ans.
x1 − x2 = X 1 + e1 − ( X 2 + e2 )
e = ( x1 − x2 ) − ( X1 − X 2 ) = e1 − e2
(c) X1 X2:
Ans.
x1 x2 = ( X 1 + e1 )( X 2 + e2 )
e = x1 x2 − X 1 X 2 = X 1e2 + X 2 e1 + e1e2
 e
e 
≈ X 1e2 + X 2 e1 = X 1 X 2  1 + 2 
 X1 X 2 
Shigley’s MED, 10th edition
Ans.
Chapter 1 Solutions, Page 1/12
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(d) X1/X2:
x1 X 1 + e1 X 1  1 + e1 X 1 
=
=


x2 X 2 + e2 X 2  1 + e2 X 2 
−1

e2 
e2
1 +
 ≈ 1−
X2
 X2 
then
 1 + e1 X 1  
e1 
e2 
e1 e2
−

 ≈ 1 +
1 −
 ≈ 1+
X 1  X 2 
X1 X 2
 1 + e2 X 2  
x1 X 1 X 1  e1 e2 
−
≈
−

 Ans.
x2 X 2 X 2  X 1 X 2 
______________________________________________________________________________
Thus,
e=
x1 = 7 = 2.645 751 311 1
(3 correct digits)
X1 = 2.64
x2 = 8 = 2.828 427 124 7
(3 correct digits)
X2 = 2.82
x1 + x2 = 5.474 178 435 8
e1 = x1 − X1 = 0.005 751 311 1
e2 = x2 − X2 = 0.008 427 124 7
e = e1 + e2 = 0.014 178 435 8
Sum = x1 + x2 = X1 + X2 + e
= 2.64 + 2.82 + 0.014 178 435 8 = 5.474 178 435 8
Checks
(b)
X1 = 2.65, X2 = 2.83 (3 digit significant numbers)
e1 = x1 − X1 = − 0.004 248 688 9
e2 = x2 − X2 = − 0.001 572 875 3
e = e1 + e2 = − 0.005 821 564 2
Sum = x1 + x2 = X1 + X2 + e
= 2.65 +2.83 − 0.001 572 875 3 = 5.474 178 435 8
Checks
______________________________________________________________________________
1-11 (a)
1-12 σ =
S
nd
⇒
3
32 (1000 ) 25 (10 )
=
πd3
2.5
Table A-17: d = 1 14 in
Factor of safety:
⇒ d = 1.006 in
Ans.
Ans.
n=
S
σ
=
25 (103 )
32 (1000 )
π (1.25 )
= 4.79
Ans.
3
______________________________________________________________________________
Shigley’s MED, 10th edition
Chapter 1 Solutions, Page 2/12
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1-13
(a)
f x2
7200
4900
19200
40500
80000
145200
86400
169000
156800
112500
51200
86700
64800
36100
0
44100
x
60
70
80
90
100
110
120
130
140
150
160
170
180
190
200
210
f
2
1
3
5
8
12
6
10
8
5
2
3
2
1
0
1
fx
120
70
240
450
800
1320
720
1300
1120
750
320
510
360
190
0
210
Σ
69
8480 1 104 600
x =
Eq. (1-6)
1
N
k
∑fx
i =1
i i
=
8480
= 122.9 kcycles
69
Eq. (1-7)
k
sx =
(b)
∑fx
i =1
2
i i
− N x2
N −1
Eq. (1-5)
1/ 2
1 104 600 − 69(122.9)2 
= 

69 − 1


z115 =
= 30.3 kcycles Ans.
x − µx
x − x 115 − 122.9
= 115
=
= −0.2607
sx
σˆ x
30.3
Interpolating from Table (A-10)
0.2600
0.2607
0.2700
0.3974
x
0.3936
⇒
NΦ(−0.2607) = 69 (0.3971) = 27.4 ≈ 27
x = 0.3971
Ans.
From the data, the number of instances less than 115 kcycles is
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2 + 1 + 3 + 5 + 8 + 12 = 31 (the data is not perfectly normal)
____________________________________________________________________________
1-14
x
174
182
190
198
206
214
222
Σ
Eq. (1-6)
x =
6
9
44
67
53
12
6
fx
1044
1638
8360
13266
10918
2568
1332
f x2
181656
298116
1588400
2626668
2249108
549552
295704
197
39126
7789204
f
1
N
k
∑fx
i =1
i i
k
∑fx
2
i i
=
39 126
= 198.61 kpsi
197
− N x2
12
 7 789 204 − 197(198.61) 2 
Eq. (1-7)
=
sx =
 = 9.68 kpsi Ans.
N −1
197 − 1


______________________________________________________________________________
i =1
1-15
L = 122.9 kcycles and sL = 30.3 kcycles
x − µ x x10 − L x10 − 122.9
=
=
sL
σˆ
30.3
Eq. (1-5)
z10 =
Thus,
x10 = 122.9 + 30.3 z10 = L10
From Table A-10, for 10 percent failure, z10 = −1.282. Thus,
L10 = 122.9 + 30.3(−1.282) = 84.1 kcycles Ans.
___________________________________________________________________________
Shigley’s MED, 10th edition
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1-16
Eq. (1-6)
x =
x
f
93
95
97
99
101
103
105
107
109
111
Σ
19
25
38
17
12
10
5
4
4
2
136
1
N
k
∑fx
i =1
i i
k
Eq. (1-7)
∑fx
sx =
i =1
2
i i
fx
1767
2375
3686
1683
1212
1030
525
428
436
222
13364
fx2
164331
225625
357542
166617
122412
106090
55125
45796
47524
24642
1315704
= 13 364 / 136 = 98.26471 = 98.26 kpsi
− N x2
N −1
12
 1 315 704 − 136(98.26471) 2 
=

136 − 1


= 4.30 kpsi
Note, for accuracy in the calculation given above, x needs to be of more significant
figures than the rounded value.
For a normal distribution, from Eq. (1-5), and a yield strength exceeded by 99 percent
(R = 0.99, pf = 0.01),
x − µx
x −x
x − 98.26
z0.01 =
= 0.01
= 0.01
4.30
sx
σˆ x
Solving for the yield strength gives
x0.01 = 98.26 + 4.30 z0.01
From Table A-10, z0.01 = − 2.326. Thus
x0.01 = 98.26 + 4.30(− 2.326) = 88.3 kpsi Ans.
______________________________________________________________________________
n
1-17
Eq. (1-9):
R = ∏ Ri = 0.98(0.96)0.94 = 0.88
i =1
Overall reliability = 88 percent
Ans.
______________________________________________________________________________
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1-18
Obtain the coefficients of variance for strength and stress
σˆ S
23.5
CS = sy =
= 0.07532
312
Ssy
Cσ =
σˆτ σˆT
145
=
=
= 0.09667
T
τ
1 500
For R = 0.99, from Table A-10, z = − 2.326.
Eq. (1-12):
n =
1 + 1 − (1 − z 2CS2 )(1 − z 2Cσ2 )
1 − z 2CS2
2
2
2
2
1 + 1 − 1 − ( −2.326 ) ( 0.07532 )  1 − ( −2.326 ) ( 0.09667 ) 



=
= 1.3229 =1.32
2
2
1 − ( −2.326 ) ( 0.07532 )
Ans.
From the given equation for stress,
τ max =
S sy
n
=
16T
π d3
Solving for d gives
1/ 3
1/ 3
 16 T n 
16(1500)1.3229 
=
d =
 = 0.0319 m = 31.9 mm Ans.
6
 π Ssy 
 π (312)10 


______________________________________________________________________________
1-19
Obtain the coefficients of variance for stress and strength
σˆ
σˆ
5
Cσ = σ = P =
= 0.09231
P
µσ
65
CS =
(a)
σˆ S σˆ S
6.59
=
=
= 0.06901
95.5
Sy
µS
y
n = 1.2
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n =
(b)
n C + Cσ
2
d
2
S
2
=−
1.2 − 1
1.2 ( 0.069012 ) + 0.092312
2
Interpolating Table A-10,
1.61
0.0537
1.6127
Φ
1.62
0.0526
⇒
R = 1 − 0.0534 = 0.9466
Ans.
Sy
σ
nd − 1
z = −
Eq. (1-11):
=
Sy
P / (π d / 4)
2
=
π d 2S y
4P
⇒ d =
= −1.6127
Φ = 0.0534
4Pn
=
π Sy
4 ( 65 )1.2
= 1.020 in Ans.
π ( 95.5)
n = 1.5
1.5 − 1
z = −
1.5 ( 0.069012 ) + 0.092312
2
3.6
3.605
3.7
0.000159
Φ
0.000108
R = 1 − 0.00015645 = 0.9998
⇒
Φ = 0.00015645
Ans.
4 ( 65 )1.5
= 1.140 in Ans.
π ( 95.5)
4 Pn
=
π Sy
d =
= −3.605
______________________________________________________________________________
1-20
µσ
max
= σ max = σ a + σ b = 90 + 383 = 473 MPa
From footnote 9, p. 25 of text,
σˆσ
max
(
= σˆσ2a + σˆσ2b
Cσ max =
CS y =
σˆσ
µσ
σˆ S
y
µS
y
max
=
max
=
σˆ S
)
1/ 2
= (8.42 + 22.32 )1/ 2 = 23.83 MPa
σˆσ
23.83
=
= 0.0504
σ max
473
max
y
Sy
=
42.7
= 0.0772
553
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n =
Eq. (1-11):
Sy
σ max
=
553
= 1.169 = 1.17 Ans.
473
z = −
From Table A-10,
nd − 1
nd2 CS2 + Cσ2
=−
1.169 − 1
1.1692 ( 0.07722 ) + 0.05042
= −1.635
Φ(− 1.635) = 0.05105
R = 1 − 0.05105 = 0.94895 = 94.9 percent Ans.
______________________________________________________________________________
1-21
(a)
a = 1.500 ± 0.001 in
b = 2.000 ± 0.003 in
c = 3.000 ± 0.004 in
d = 6.520 ± 0.010 in
w = d − a − b − c = 6.520 − 1.5 − 2 − 3 = 0.020 in
tw = ∑ tall = 0.001 + 0.003 + 0.004 +0.010 = 0.018
w = 0.020 ± 0.018 in
Ans.
(b) From part (a), wmin = 0.002 in. Thus, must add 0.008 in to d . Therefore,
d = 6.520 + 0.008 = 6.528 in
Ans.
______________________________________________________________________________
1-22 V = xyz, and x = a ± ∆ a, y = b ± ∆ b, z = c ± ∆ c,
V = abc
V = ( a ± ∆a )( b ± ∆b )( c ± ∆c )
= abc ± bc∆a ± ac∆b ± ab∆c ± a∆b∆c ± b∆c∆a ± c∆a∆b ± ∆a∆b∆c
The higher order terms in ∆ are negligible. Thus,
∆V ≈ bc∆a + ac∆b + ab∆c
and,
∆V bc∆a + ac∆b + ab∆c ∆a ∆b ∆c ∆a ∆b ∆c
≈
=
+
+
=
+
+
Ans.
V
abc
a
b
c
a
b
c
For the numerical values given, V = 1.500 (1.875 ) 3.000 = 8.4375 in 3
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∆V 0.002 0.003 0.004
≈
+
+
= 0.004267
V
1.500 1.875 3.000
V = 8.4375 ± 0.0360 in3
⇒
∆V ≈ 0.004267 ( 8.4375) = 0.0360 in 3
Ans.
8.4735
8.473551..
in, whereas, exact is V =
in
8.4015
8.401551..
______________________________________________________________________________
This answer yields V ≈
1-23
wmax = 0.05 in, wmin = 0.004 in
0.05 + 0.004
w=
= 0.027 in
2
Thus, ∆ w = 0.05 − 0.027 = 0.023 in, and then, w = 0.027 ± 0.023 in.
w= a −b −c
0.027 = a − 0.042 − 1.5
a = 1.569 in
tw =
Thus,
∑t
all
⇒
0.023 = ta + 0.002 + 0.005 ⇒ ta = 0.016 in
a = 1.569 ± 0.016 in
Ans.
______________________________________________________________________________
1-24
Do = Di + 2d = 3.734 + 2 ( 0.139 ) = 4.012 in
t Do = ∑ tall = 0.028 + 2 ( 0.004 ) = 0.036 in
Do = 4.012 ± 0.036 in
Ans.
______________________________________________________________________________
1-25 From O-Rings, Inc. (oringsusa.com), Di = 9.19 ± 0.13 mm, d = 2.62 ± 0.08 mm
Do = Di + 2d = 9.19 + 2 ( 2.62 ) = 14.43 mm
t Do = ∑ tall = 0.13 + 2 ( 0.08 ) = 0.29 mm
Do = 14.43 ± 0.29 mm
Ans.
______________________________________________________________________________
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1-26 From O-Rings, Inc. (oringsusa.com), Di = 34.52 ± 0.30 mm, d = 3.53 ± 0.10 mm
Do = Di + 2d = 34.52 + 2 ( 3.53) = 41.58 mm
t Do = ∑ tall = 0.30 + 2 ( 0.10 ) = 0.50 mm
Do = 41.58 ± 0.50 mm
Ans.
______________________________________________________________________________
1-27
From O-Rings, Inc. (oringsusa.com), Di = 5.237 ± 0.035 in, d = 0.103 ± 0.003 in
Do = Di + 2d = 5.237 + 2 ( 0.103) = 5.443 in
t Do = ∑ tall = 0.035 + 2 ( 0.003) = 0.041 in
Do = 5.443 ± 0.041 in
Ans.
______________________________________________________________________________
1-28 From O-Rings, Inc. (oringsusa.com), Di = 1.100 ± 0.012 in, d = 0.210 ± 0.005 in
Do = Di + 2d = 1.100 + 2 ( 0.210 ) = 1.520 in
t Do = ∑ tall = 0.012 + 2 ( 0.005 ) = 0.022 in
Do = 1.520 ± 0.022 in
Ans.
______________________________________________________________________________
1-29 From Table A-2,
(a) σ = 150/6.89 = 21.8 kpsi
Ans.
(b) F = 2 /4.45 = 0.449 kip = 449 lbf
Ans.
(c) M = 150/0.113 = 1330 lbf ⋅ in = 1.33 kip ⋅ in
(d) A = 1500/ 25.42 = 2.33 in2
(e) I = 750/2.544 = 18.0 in4
Ans.
Ans.
(f) E = 145/6.89 = 21.0 Mpsi
(g) v = 75/1.61 = 46.6 mi/h
Ans.
Ans.
Ans.
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(h) V = 1000/946 = 1.06 qt
Ans.
______________________________________________________________________________
1-30 From Table A-2,
(a) l = 5(0.305) = 1.53 m
Ans.
(b) σ = 90(6.89) = 620 MPa
Ans.
(c) p = 25(6.89) = 172 kPa
Ans.
(d) Z =12(16.4) = 197 cm3
Ans.
(e) w = 0.208(175) = 36.4 N/m
Ans.
(f) δ = 0.001 89(25.4) = 0.048 0 mm
(g) v = 1 200(0.0051) = 6.12 m/s
Ans.
Ans.
(h) ∫ = 0.002 15(1) = 0.002 15 mm/mm
(i) V = 1830(25.43) = 30.0 (106) mm3
Ans.
Ans.
______________________________________________________________________________
1-31
(a) σ = M /Z = 1770/0.934 = 1895 psi = 1.90 kpsi
(b) σ = F /A = 9440/23.8 = 397 psi
Ans.
Ans.
(c) y =Fl3/3EI = 270(31.5)3/[3(30)106(0.154)] = 0.609 in
Ans.
(d) θ = Tl /GJ = 9 740(9.85)/[11.3(106)(π /32)1.004] = 8.648(10−2) rad = 4.95°
Ans.
______________________________________________________________________________
1-32
(a) σ =F / wt = 1000/[25(5)] = 8 MPa
Ans.
(b) I = bh3 /12 = 10(25)3/12 = 13.0(103) mm4
Ans.
(c) I =π d4/64 = π (25.4)4/64 = 20.4(103) mm4
Ans.
(d) τ =16T /π d 3 = 16(25)103/[π (12.7)3] = 62.2 MPa Ans.
______________________________________________________________________________
1-33
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Chapter 1 Solutions, Page 11/12
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(a) τ =F /A = 2 700/[π (0.750)2/4] = 6110 psi = 6.11 kpsi
Ans.
(b) σ = 32Fa/π d 3 = 32(180)31.5/[π (1.25)3] = 29 570 psi = 29.6 kpsi
Ans.
(c) Z =π (do4 − di4)/(32 do) = π (1.504 − 1.004)/[32(1.50)] = 0.266 in3
Ans.
(d) k = (d 4G)/(8D 3 N) = 0.062 54(11.3)106/[8(0.760)3 32] = 1.53 lbf/in
Ans.
______________________________________________________________________________
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Chapter 2
From Tables A-20, A-21, A-22, and A-24c,
(a) UNS G10200 HR: Sut = 380 (55) MPa (kpsi), Syt = 210 (30) MPa (kpsi) Ans.
(b) SAE 1050 CD: Sut = 690 (100) MPa (kpsi), Syt = 580 (84) MPa (kpsi) Ans.
(c) AISI 1141 Q&T at 540°C (1000°F): Sut = 896 (130) MPa (kpsi), Syt = 765 (111)
MPa (kpsi) Ans.
(d) 2024-T4: Sut = 446 (64.8) MPa (kpsi), Syt = 296 (43.0) MPa (kpsi) Ans.
(e) Ti-6Al-4V annealed: Sut = 900 (130) MPa (kpsi), Syt = 830 (120) MPa (kpsi) Ans.
______________________________________________________________________________
2-1
2-2
(a) Maximize yield strength: Q&T at 425°C (800°F) Ans.
(b) Maximize elongation: Q&T at 650°C (1200°F) Ans.
______________________________________________________________________________
Conversion of kN/m3 to kg/ m3 multiply by 1(103) / 9.81 = 102
AISI 1018 CD steel: Tables A-20 and A-5
3
S y 370 (10 )
=
= 47.4 kN ⋅ m/kg
Ans.
ρ 76.5 (102 )
2011-T6 aluminum: Tables A-22 and A-5
3
S y 169 (10 )
=
= 62.3 kN ⋅ m/kg
Ans.
ρ 26.6 (102 )
Ti-6Al-4V titanium: Tables A-24c and A-5
3
S y 830 (10 )
=
= 187 kN ⋅ m/kg
Ans.
ρ 43.4 (102 )
ASTM No. 40 cast iron: Tables A-24a and A-5.Does not have a yield strength. Using the
ultimate strength in tension
3
Sut 42.5 ( 6.89 ) (10 )
=
= 40.7 kN ⋅ m/kg
Ans
ρ
70.6 (102 )
______________________________________________________________________________
2-3
2-4
AISI 1018 CD steel: Table A-5
6
E 30.0 (10 )
=
= 106 (106 ) in
γ
0.282
2011-T6 aluminum: Table A-5
6
E 10.4 (10 )
=
= 106 (106 ) in
γ
0.098
Ans.
Ans.
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Ti-6Al-6V titanium: Table A-5
6
E 16.5 (10 )
=
= 103 (106 ) in Ans.
γ
0.160
No. 40 cast iron: Table A-5
6
E 14.5 (10 )
=
= 55.8 (106 ) in Ans.
γ
0.260
______________________________________________________________________________
2-5
2G (1 + v) = E
⇒
v=
E − 2G
2G
Using values for E and G from Table A-5,
30.0 − 2 (11.5 )
Steel:
v=
= 0.304 Ans.
2 (11.5 )
The percent difference from the value in Table A-5 is
0.304 − 0.292
= 0.0411 = 4.11 percent
0.292
Ans.
10.4 − 2 ( 3.90)
= 0.333 Ans.
2 ( 3.90)
The percent difference from the value in Table A-5 is 0 percent Ans.
Aluminum:
v=
Beryllium copper:
v=
18.0 − 2 ( 7.0 )
2 ( 7.0 )
= 0.286
Ans.
The percent difference from the value in Table A-5 is
0.286 − 0.285
= 0.00351 = 0.351 percent
0.285
v=
14.5 − 2 ( 6.0 )
= 0.208
2 ( 6.0 )
The percent difference from the value in Table A-5 is
Gray cast iron:
Ans.
Ans.
0.208 − 0.211
= − 0.0142 = − 1.42 percent
Ans.
0.211
______________________________________________________________________________
2-6
(a) A0 = π (0.503)2/4 = 0.1987 in2, σ = Pi / A0
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For data in elastic range, ∫ = ∆ l / l0 = ∆ l / 2
A
∆l l − l0 l
For data in plastic range, ò =
=
= −1 = 0 −1
l0
l0
l0
A
On the next two pages, the data and plots are presented. Figure (a) shows the linear part of
the curve from data points 1-7. Figure (b) shows data points 1-12. Figure (c) shows the
complete range. Note: The exact value of A0 is used without rounding off.
(b) From Fig. (a) the slope of the line from a linear regression is E = 30.5 Mpsi
Ans.
From Fig. (b) the equation for the dotted offset line is found to be
σ = 30.5(106)∫ − 61 000
(1)
The equation for the line between data points 8 and 9 is
σ = 7.60(105)∫ + 42 900
(2)
Solving Eqs. (1) and (2) simultaneously yields σ = 45.6 kpsi which is the 0.2 percent
offset yield strength. Thus, Sy = 45.6 kpsi Ans.
The ultimate strength from Figure (c) is Su = 85.6 kpsi
Ans.
The reduction in area is given by Eq. (2-12) is
R=
A0 − Af
A0
(100 ) =
0.1987 − 0.1077
(100 ) = 45.8 %
0.1987
Data Point
Pi
∆l, Ai
∫
σ
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
0
0
0
1000
2000
3000
4000
7000
8400
8800
9200
8800
9200
9100
13200
15200
17000
16400
14800
0.0004
0.0006
0.001
0.0013
0.0023
0.0028
0.0036
0.0089
0.1984
0.1978
0.1963
0.1924
0.1875
0.1563
0.1307
0.1077
0.00020
0.00030
0.00050
0.00065
0.00115
0.00140
0.00180
0.00445
0.00158
0.00461
0.01229
0.03281
0.05980
0.27136
0.52037
0.84506
0
5032
10065
15097
20130
35227
42272
44285
46298
44285
46298
45795
66428
76492
85551
82531
74479
Shigley’s MED, 10th edition
Ans.
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50000
y = 3.05E+07x - 1.06E+01
Stress (psi)
40000
30000
Series1
20000
Linear (Series1)
10000
0
0.000
0.001
0.001
0.002
Strain
Stress (psi)
(a) Linear range
50000
Y
45000
40000
35000
30000
25000
20000
15000
10000
5000
0
0.000 0.002 0.004 0.006 0.008 0.010 0.012 0.014
Strain
(b) Offset yield
90000
80000
U
Stress (psi)
70000
60000
50000
40000
30000
20000
10000
0
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
Strain
(c) Complete range
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(c) The material is ductile since there is a large amount of deformation beyond yield.
(d) The closest material to the values of Sy, Sut, and R is SAE 1045 HR with Sy = 45 kpsi,
Sut = 82 kpsi, and R = 40 %. Ans.
______________________________________________________________________________
2-7
To plot σ true vs.ε, the following equations are applied to the data.
P
σ true =
A
Eq. (2-4)
l
ε = ln
for 0 ≤ ∆l ≤ 0.0028 in (0 ≤ P ≤ 8 400 lbf )
l0
ε = ln
where
A0 =
π (0.503)2
A0
A
for ∆l > 0.0028 in
(P > 8 400 lbf )
= 0.1987 in 2
4
The results are summarized in the table below and plotted on the next page. The last 5
points of data are used to plot log σ vs log ε
The curve fit gives
m = 0.2306
log σ0 = 5.1852 ⇒ σ0 = 153.2 kpsi
Ans.
For 20% cold work, Eq. (2-14) and Eq. (2-17) give,
A = A0 (1 – W) = 0.1987 (1 – 0.2) = 0.1590 in2
A0
0.1987
= ln
= 0.2231
0.1590
A
Eq. (2-18): S y′ = σ 0ε m = 153.2(0.2231)0.2306 = 108.4 kpsi
ε = ln
Ans.
Eq. (2-19), with Su = 85.6 from Prob. 2-6,
Su′ =
Su
85.6
=
= 107 kpsi
1 − W 1 − 0.2
Ans.
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P
0
1 000
2 000
3 000
4 000
7 000
8 400
8 800
9 200
9 100
13 200
15 200
17 000
16 400
14 800
∆l
0
0.000 4
0.000 6
0.001 0
0.001 3
0.002 3
0.002 8
A
0.198 7
0.198 7
0.198 7
0.198 7
0.198 7
0.198 7
0.198 7
0.198 4
0.197 8
0.196 3
0.192 4
0.187 5
0.156 3
0.130 7
0.107 7
ε
0
0.000 2
0.000 3
0.000 5
0.000 65
0.001 15
0.001 4
0.001 51
0.004 54
0.012 15
0.032 22
0.058 02
0.240 02
0.418 89
0.612 45
σtrue
0
5 032.71
10 065.4
15 098.1
20 130.9
35 229
42 274.8
44 354.8
46 511.6
46 357.6
68 607.1
81 066.7
108 765
125 478
137 419
Shigley’s MED, 10th edition
log ε
-3.699
-3.523
-3.301
-3.187
-2.940
-2.854
-2.821
-2.343
-1.915
-1.492
-1.236
-0.620
-0.378
-0.213
log σtrue
3.702
4.003
4.179
4.304
4.547
4.626
4.647
4.668
4.666
4.836
4.909
5.036
5.099
5.138
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______________________________________________________________________________
2-8
Tangent modulus at σ = 0 is
E=
∆σ
5000 − 0
≈
= 25 106 psi
−3
∆ò 0.2 10 − 0
(
)
( )
Ans.
At σ = 20 kpsi
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( 26 − 199 ) (103 )
E20 ≈
= 14.0 (106 ) psi
−3
(1.5 − 1) (10 )
Ans.
∫ (10-3) σ (kpsi)
0
0
0.20
5
0.44
10
0.80
16
1.0
19
1.5
26
2.0
32
2.8
40
3.4
46
4.0
49
5.0
54
_________________________
_______________________________________
____________________
2-9
W = 0.20,
(a) Before cold workin
ing: Annealed AISI 1018 steel. Table A-22, Sy = 32 kpsi, Su = 49.5
kpsi, σ0 = 90.0 kpsi, m = 0.25, ∫f = 1.05
After cold working: Eq.
q. (2-16), ∫u = m = 0.25
A0
1
1
Eq. (2-14),
=
=
= 1.25
Ai 1 − W 1 − 0.20
A
Eq. (2-17),
ε i = lnn 0 = ln1.25 = 0.223 < ε u
Ai
Eq. (2-18),
S y′ = σ 0ε im = 90 ( 0.223)
Eq. (2-19),
Su′ =
(b) Before:
Su 49.5
49
=
= 1.55
Sy
32
3
0.25
= 61.8 kpsi
Su
49.5
=
= 61.9 kpsi
1 − W 1 − 0.20
After:
Ans. 93%
3% increase
crease
Ans. 25% incr
Su′ 61.9
=
= 1.00
S y′ 61.8
Ans.
Ans.
Ans.
Lost most of its ductility
lity.
_________________________
______________________________________
____________________
2-10 W = 0.20,
(a) Before cold workin
ing: AISI 1212 HR steel. Table A-22, Sy = 28 kpsi, Su = 61.5 kpsi,
σ0 = 110 kpsi, m = 0.24,
24, ∫f = 0.85
After cold working: Eq.
q. (2-16), ∫u = m = 0.24
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Chapte
pter 2 Solutions, Page 8/22
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Eq. (2-14),
Eq. (2-17),
A0
1
1
=
=
= 1.25
Ai 1 − W 1 − 0.20
A
ε i = ln 0 = ln1.25 = 0.223 < ε u
Ai
Eq. (2-18),
S y′ = σ 0ε im = 110 ( 0.223)
Eq. (2-19),
Su′ =
(b) Before:
Su 61.5
=
= 2.20
Sy
28
0.24
= 76.7 kpsi
Su
61.5
=
= 76.9 kpsi
1 − W 1 − 0.20
After:
Ans. 174% increase
Ans. 25% increase
Su′ 76.9
=
= 1.00
S y′ 76.7
Ans.
Ans.
Ans.
Lost most of its ductility.
______________________________________________________________________________
2-11 W = 0.20,
(a) Before cold working: 2024-T4 aluminum alloy. Table A-22, Sy = 43.0 kpsi, Su =
64.8 kpsi, σ0 = 100 kpsi, m = 0.15, ∫f = 0.18
After cold working: Eq. (2-16), ∫u = m = 0.15
A0
1
1
Eq. (2-14),
=
=
= 1.25
Ai 1 − W 1 − 0.20
A
Eq. (2-17),
Ans.
ε i = ln 0 = ln1.25 = 0.223 > ε f Material fractures.
Ai
______________________________________________________________________________
Ans.
2-12 For HB = 275, Eq. (2-21), Su = 3.4(275) = 935 MPa
______________________________________________________________________________
2-13 Gray cast iron, HB = 200.
Eq. (2-22),
Su = 0.23(200) − 12.5 = 33.5 kpsi
Ans.
From Table A-24, this is probably ASTM No. 30 Gray cast iron
Ans.
______________________________________________________________________________
2-14 Eq. (2-21), 0.5HB = 100 ⇒ HB = 200
Ans.
______________________________________________________________________________
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2-15 For the data given, converting HB to Su using Eq. (2-21)
ΣSu =
Eq. (1-6)
Su =
∑S
u
N
=
Su2 (kpsi)
13225
13456
13456
13689
13806.25
13806.25
13806.25
13924
13924
14280.25
Su (kpsi)
115
116
116
117
117.5
117.5
117.5
118
118
119.5
HB
230
232
232
234
235
235
235
236
236
239
ΣSu2 = 137373
1172
1172
= 117.2 ≈ 117 kpsi Ans.
10
Eq. (1-7),
10
∑S
2
u
− NSu2
137373 − 10 (117.2 )
= 1.27 kpsi
Ans.
N −1
9
______________________________________________________________________________
s Su =
i =1
2
=
2-16 For the data given, converting HB to Su using Eq. (2-22)
Su (kpsi)
40.4
40.86
40.86
41.32
41.55
41.55
41.55
41.78
41.78
42.47
Su2 (kpsi)
1632.16
1669.54
1669.54
1707.342
1726.403
1726.403
1726.403
1745.568
1745.568
1803.701
ΣSu = 414.12
ΣSu2 = 17152.63
HB
230
232
232
234
235
235
235
236
236
239
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Eq. (1-6)
Su =
∑S
=
u
N
414.12
= 41.4 kpsi Ans.
10
Eq. (1-7),
10
∑S
2
u
− NSu2
17152.63 − 10 ( 41.4 )
=
= 1.20
s Su =
Ans.
N −1
9
______________________________________________________________________________
i =1
uR ≈
2-17 (a) Eq. (2-9)
2
45.62
= 34.7 in ⋅ lbf / in 3
2(30)
Ans.
(b) A0 = π(0.5032)/4 = 0.19871 in2
P
∆L
0
1 000
2 000
3 000
4 000
7 000
8 400
8 800
9 200
9 100
13 200
15 200
17 000
16 400
14 800
0
0.000 4
0.000 6
0.001 0
0.001 3
0.002 3
0.002 8
0.003 6
0.008 9
A
0.196 3
0.192 4
0.187 5
0.156 3
0.130 7
0.107 7
0.012 28
0.032 80
0.059 79
0.271 34
0.520 35
0.845 03
σ = P/A0
∫
(A0 / A) – 1
0
0.000 2
0.000 3
0.000 5
0.000 65
0.001 15
0.001 4
0.001 8
0.004 45
0.012 28
0.032 80
0.059 79
0.271 34
0.520 35
0.845 03
0
5 032.
10 070
15 100
20 130
35 230
42 270
44 290
46 300
45 800
66 430
76 500
85 550
82 530
74 480
From the figures on the next page,
5
1
uT ≈ ∑ Ai = (43 000)(0.001 5) + 45 000(0.004 45 − 0.001 5)
2
i =1
1
+ ( 45 000 + 76 500 ) (0.059 8 − 0.004 45)
2
+ 81 000 ( 0.4 − 0.059 8 ) + 80 000 ( 0.845 − 0.4 )
( )
≈ 66.7 103 in ⋅ lbf/in 3
Ans.
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ter 2 Solutions, Page 12/22
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2-18, 2-19
These problems are for student research. No standard solutions are provided.
______________________________________________________________________________
2-20 Appropriate tables: Young’s modulus and Density (Table A-5)1020 HR and CD (Table A20), 1040 and 4140 (Table A-21), Aluminum (Table A-24), Titanium (Table A-24c)
Appropriate equations:
F
F
For diameter, σ = =
= Sy
A (π / 4 ) d 2
⇒
d=
4F
π Sy
Weight/length = ρA, Cost/length = $/in = ($/lbf) Weight/length,
Deflection/length = δ /L = F/(AE)
With F = 100 kips = 100(103) lbf,
Young's
Material Modulus
units
Mpsi
1020 HR
1020 CD
1040
4140
Al
Ti
Density
lbf/in3
30
30
30
30
10.4
16.5
0.282
0.282
0.282
0.282
0.098
0.16
Yield
Weight/
Strength Cost/lbf Diameter length
kpsi
$/lbf
in
lbf/in
30
57
80
165
50
120
0.27
0.30
0.35
0.80
1.10
7.00
2.060
1.495
1.262
0.878
1.596
1.030
0.9400
0.4947
0.3525
0.1709
0.1960
0.1333
Cost/ Deflection/
length
length
$/in
in/in
0.25
0.15
0.12
0.14
0.22
$0.93
1.000E-03
1.900E-03
2.667E-03
5.500E-03
4.808E-03
7.273E-03
The selected materials with minimum values are shaded in the table above.
Ans.
______________________________________________________________________________
2-21
First, try to find the broad category of material (such as in Table A-5). Visual, magnetic,
and scratch tests are fast and inexpensive, so should all be done. Results from these three
would favor steel, cast iron, or maybe a less common ferrous material. The expectation
would likely be hot-rolled steel. If it is desired to confirm this, either a weight or bending
test could be done to check density or modulus of elasticity. The weight test is faster.
From the measured weight of 7.95 lbf, the unit weight is determined to be
w=
W
7.95 lbf
=
= 0.281 lbf/in 3
Al [π (1 in)2 / 4](36 in)
which agrees well with the unit weight of 0.282 lbf/in3 reported in Table A-5 for carbon
steel. Nickel steel and stainless steel have similar unit weights, but surface finish and
darker coloring do not favor their selection. To select a likely specification from Table
A-20, perform a Brinell hardness test, then use Eq. (2-21) to estimate an ultimate strength
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of Su = 0.5H B = 0.5(200) = 100 kpsi . Assuming the material is hot-rolled due to the
rough surface finish, appropriate choices from Table A-20 would be one of the higher
carbon steels, such as hot-rolled AISI 1050, 1060, or 1080.
Ans.
______________________________________________________________________________
2-22
First, try to find the broad category of material (such as in Table A-5). Visual, magnetic,
and scratch tests are fast and inexpensive, so should all be done. Results from these three
favor a softer, non-ferrous material like aluminum. If it is desired to confirm this, either a
weight or bending test could be done to check density or modulus of elasticity. The
weight test is faster. From the measured weight of 2.90 lbf, the unit weight is determined
to be
W
2.9 lbf
w=
=
= 0.103 lbf/in 3
Al [π (1 in)2 / 4](36 in)
which agrees reasonably well with the unit weight of 0.098 lbf/in3 reported in Table A-5
for aluminum. No other materials come close to this unit weight, so the material is likely
aluminum. Ans.
______________________________________________________________________________
2-23
First, try to find the broad category of material (such as in Table A-5). Visual, magnetic,
and scratch tests are fast and inexpensive, so should all be done. Results from these three
favor a softer, non-ferrous copper-based material such as copper, brass, or bronze. To
further distinguish the material, either a weight or bending test could be done to check
density or modulus of elasticity. The weight test is faster. From the measured weight of
9 lbf, the unit weight is determined to be
w=
9.0 lbf
W
=
= 0.318 lbf/in 3
Al [π (1 in)2 / 4](36 in)
which agrees reasonably well with the unit weight of 0.322 lbf/in3 reported in Table A-5
for copper. Brass is not far off (0.309 lbf/in3), so the deflection test could be used to gain
additional insight. From the measured deflection and utilizing the deflection equation for
an end-loaded cantilever beam from Table A-9, Young’s modulus is determined to be
100 ( 24 )
Fl 3
E=
=
= 17.7 Mpsi
3Iy 3 (π (1) 4 64 ) (17 / 32)
3
which agrees better with the modulus for copper (17.2 Mpsi) than with brass (15.4 Mpsi).
The conclusion is that the material is likely copper.
Ans.
______________________________________________________________________________
2-24 and 2-25 These problems are for student research. No standard solutions are provided.
______________________________________________________________________________
2-26 For strength, σ = F/A = S
⇒ A = F/S
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For mass, m = Alρ = (F/S) lρ
Thus,
f 3(M ) = ρ /S , and maximize S/ρ (β = 1)
In Fig. (2-19), draw lines parallel to S/ρ
The higher strength aluminum alloys have the greatest potential, as determined by
Ans.
comparing each material’s bubble to the S/ρ guidelines.
______________________________________________________________________________
2-27 For stiffness, k = AE/l ⇒ A = kl/E
For mass, m = Alρ = (kl/E) lρ =kl2 ρ /E
Thus,
f 3(M) = ρ /E , and maximize E/ρ (β = 1)
In Fig. (2-16), draw lines parallel to E/ρ
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From the list of materials given, tungsten carbide (WC) is best, closely followed by
aluminum alloys. They are close enough that other factors, like cost or availability, would
likely dictate the best choice. Polycarbonate polymer is clearly not a good choice
compared to the other candidate materials.
Ans.
______________________________________________________________________________
2-28 For strength,
σ = Fl/Z = S
(1)
where Fl is the bending moment and Z is the section modulus [see Eq. (3-26b), p. 104 ].
The section modulus is strictly a function of the dimensions of the cross section and has
the units in3 (ips) or m3 (SI). Thus, for a given cross section, Z =C (A)3/2, where C is a
(
number. For example, for a circular cross section, C = 4 π
)
−1
. Then, for strength, Eq.
(1) is
Fl
=S
CA3/2
⇒
 Fl 
A=

 CS 
2/3
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For mass,
Thus,
 Fl 
m = Al ρ = 

 CS 
2/3
F
lρ =  
C
2/3
 ρ 
l 5/3  2/3 
S 
f 3(M) = ρ /S 2/3, and maximize S 2/3/ρ (β = 2/3)
In Fig. (2-19), draw lines parallel to S 2/3/ρ
From the list of materials given, a higher strength aluminum alloy has the greatest
potential, followed closely by high carbon heat-treated steel. Tungsten carbide is clearly
not a good choice compared to the other candidate materials. .Ans.
______________________________________________________________________________
2-29 Eq. (2-26), p. 77, applies to a circular cross section. However, for any cross section shape
it can be shown that I = CA 2, where C is a constant. For example, consider a rectangular
section of height h and width b, where for a given scaled shape, h = cb, where c is a
constant. The moment of inertia is I = bh 3/12, and the area is A = bh. Then I = h(bh2)/12
= cb (bh2)/12 = (c/12)(bh)2 = CA 2, where C = c/12 (a constant).
Thus, Eq. (2-27) becomes
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1/2
 kl 3 
A=

 3CE 
and Eq. (2-29) becomes
 k  5/2  ρ 
m = Al ρ = 
 l  1/2 
 3C 
E 
1/2
Thus, minimize f 3 ( M ) =
ρ
E 1/2
, or maximize M =
E1/2
ρ
. From Fig. (2-16)
From the list of materials given, aluminum alloys are clearly the best followed by steels
and tungsten carbide. Polycarbonate polymer is not a good choice compared to the other
candidate materials. Ans.
______________________________________________________________________________
2-30 For stiffness, k = AE/l ⇒ A = kl/E
For mass, m = Alρ = (kl/E) lρ =kl2 ρ /E
So,
f 3(M) = ρ /E, and maximize E/ρ . Thus, β = 1. Ans.
______________________________________________________________________________
2-31 For strength, σ = F/A = S ⇒ A = F/S
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For mass, m = Alρ = (F/S) lρ
So, f 3(M ) = ρ /S, and maximize S/ρ . Thus, β = 1. Ans.
______________________________________________________________________________
2-32 Eq. (2-26), p. 77, applies to a circular cross section. However, for any cross section shape
it can be shown that I = CA 2, where C is a constant. For the circular cross section (see
p.77), C = (4π)−1. Another example, consider a rectangular section of height h and width
b, where for a given scaled shape, h = cb, where c is a constant. The moment of inertia is
I = bh 3/12, and the area is A = bh. Then I = h(bh2)/12 = cb (bh2)/12 = (c/12)(bh)2 = CA 2,
where C = c/12, a constant.
Thus, Eq. (2-27) becomes
1/2
 kl 3 
A=

 3CE 
and Eq. (2-29) becomes
 k  5/2  ρ 
m = Al ρ = 
 l  1/2 
 3C 
E 
1/2
So, minimize f 3 ( M ) =
ρ
E1/2
. Thus, β = 1/2. Ans.
ρ
E1/2
______________________________________________________________________________
2-33 For strength,
, or maximize M =
σ = Fl/Z = S
(1)
where Fl is the bending moment and Z is the section modulus [see Eq. (3-26b), p. 104 ].
The section modulus is strictly a function of the dimensions of the cross section and has
the units in3 (ips) or m3 (SI). The area of the cross section has the units in2 or m2. Thus, for
a given cross section, Z =C (A)3/2, where C is a number. For example, for a circular cross
(
section, Z = πd 3/(32)and the area is A = πd 2/4. This leads to C = 4 π
3/2
)
−1
. So, with
Z =C (A) , for strength, Eq. (1) is
Fl
=S
CA3/2
For mass,
 Fl 
A=

 CS 
⇒
 Fl 
m = Al ρ = 

 CS 
2/3
F
lρ =  
C
2/3
2/3
(2)
 ρ 
l 5/3  2/3 
S 
So, f 3(M) = ρ /S 2/3, and maximize S 2/3/ρ. Thus, β = 2/3. Ans.
______________________________________________________________________________
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2-34 For stiffness, k=AE/l, or, A = kl/E.
Thus, m = ρAl =ρ (kl/E )l = kl 2 ρ /E. Then, M = E /ρ and β = 1.
From Fig. 2-16, lines parallel to E /ρ for ductile materials include steel, titanium,
molybdenum, aluminum alloys, and composites.
For strength, S = F/A, or, A = F/S.
Thus, m = ρAl =ρ F/Sl = Fl ρ /S. Then, M = S/ρ and β = 1.
From Fig. 2-19, lines parallel to S/ρ give for ductile materials, steel, aluminum alloys,
nickel alloys, titanium, and composites.
Common to both stiffness and strength are steel, titanium, aluminum alloys, and
composites. Ans.
______________________________________________________________________________
2-35 See Prob. 1-13 solution for x = 122.9 kcycles and sx = 30.3 kcycles. Also, in that solution
it is observed that the number of instances less than 115 kcycles predicted by the normal
distribution is 27; whereas, the data indicates the number to be 31.
From Eq. (1-4), the probability density function (PDF), with µ = x and σˆ = sx , is
 1  x − x 2 
 1  x − 122.9  2 
1
1
f ( x) =
exp  − 
exp  − 
  =
 
sx 2π
 2  sx   30.3 2π
 2  30.3  
(1)
The discrete PDF is given by f /(Nw), where N = 69 and w = 10 kcycles. From the Eq. (1)
and the data of Prob. 1-13, the following plots are obtained.
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Range
midpoint
(kcycles)
Frequency
x
60
70
80
90
100
110
120
130
140
150
160
170
180
190
200
210
f
2
1
3
5
8
12
6
10
8
5
2
3
2
1
0
1
Observed
PDF
f /(Nw)
0.002898551
0.001449275
0.004347826
0.007246377
0.011594203
0.017391304
0.008695652
0.014492754
0.011594203
0.007246377
0.002898551
0.004347826
0.002898551
0.001449275
0
0.001449275
Normal
PDF
f (x)
0.001526493
0.002868043
0.004832507
0.007302224
0.009895407
0.012025636
0.013106245
0.012809861
0.011228104
0.008826008
0.006221829
0.003933396
0.002230043
0.001133847
0.000517001
0.00021141
Plots of the PDF’s are shown below.
0.02
0.018
0.016
Probability Density
0.014
0.012
0.01
Normal Distribution
Histogram
0.008
0.006
0.004
0.002
0
0
20 40 60 80 100 120 140 160 180 200 220
L (kcycles)
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It can be seen that the data is not perfectly normal and is skewed to the left indicating that
the number of instances below 115 kcycles for the data (31) would be higher than the
hypothetical normal distribution (27).
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Chapter 3
3-1
ΣM O = 0
18 RB − 6(100) = 0
RB = 33.3 lbf
Ans.
ΣFy = 0
RO + RB − 100 = 0
RO = 66.7 lbf
Ans.
RC = RB = 33.3 lbf
Ans.
______________________________________________________________________________
3-2
Body AB:
ΣFx = 0
RAx = RBx
ΣFy = 0
RAy = RBy
ΣM B = 0
RAy (10) − RAx (10) = 0
RAx = RAy
Body OAC:
ΣM O = 0
RAy (10) − 100(30) = 0
RAy = 300 lbf
Ans.
ΣFx = 0
ROx = − RAx = −300 lbf
ΣFy = 0
ROy + RAy − 100 = 0
ROy = −200 lbf
Ans.
Ans.
______________________________________________________________________________
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3-3
0.8
= 1.39 kN Ans.
tan 30o
0.8
RA =
= 1.6 kN Ans.
sin 30o
RO =
_________________________
______________________________________
___________________
3-4
Step 1: Find RA & RE
4.5
= 7.794 m
tan 30o
ΣM A = 0
h=
9 RE − 7.794(400 cos 300o )
−4.5(400sin 30o ) = 0
RE = 400 N Ans.
∑F
x
=0
RAx + 400 co
cos 30o = 0
RAx = −346.4
6.4 N
∑F
y
=0
RAy + 400 − 4400sin 30o = 0
RAy = −2000 N
RA = 346.42 + 2002 = 400
4 N
Ans.
Step 2: Find components
nts of RC and RD on link 4
∑M
C
=0
400(4.5) − ( 7.794 − 1.9
1 ) RD = 0
RD = 305.4 N
Ans
ns.
x
=0
⇒
( RCx )4 = 305.4 N
y
=0
⇒
( RCy ) 4 = −400 N
∑F
∑F
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Step 3: Find components of RC on link 2
∑ Fx = 0
46.4 = 0
( RCx )2 + 305.4 − 346
( RCx )2 = 41 N
∑F
y
=0
(R )
Cy 2
= 200 N
Ans.
______________________________________
________________________________________________________
____________________________
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3-5
ΣM C = 0
−1500 R1 + 300(5) + 1200(9) = 0
R1 = 8.2 kN
Ans.
ΣFy = 0
8.2 − 9 − 5 + R2 = 0
R2 = 5.8 kN
M 1 = 8.2(300) = 2460 N ⋅ m
Ans.
Ans.
M 2 = 2460 − 0.8(900) = 1740 N ⋅ m
Ans.
M 3 = 1740 − 5.8(300) = 0 checks!
_____________________________________________________________________________
3-6
ΣFy = 0
RO = 500 + 40(6) = 740 lbf
Ans.
ΣM O = 0
M O = 500(8) + 40(6)(17) = 8080 lbf ⋅ in
M 1 = −8080 + 740(8) = −2160 lbf ⋅ in
Ans.
Ans.
M 2 = −2160 + 240(6) = −720 lbf ⋅ in Ans.
1
M 3 = −720 + (240)(6) = 0 checks!
2
______________________________________________________________________________
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3-7
ΣM B = 0
−2.2 R1 + 1(2) − 1(4) = 0
R1 = −0.91 kN
ΣFy = 0
Ans.
−0.91 − 2 + R2 − 4 = 0
R2 = 6.91 kN
Ans.
M 1 = −0.91(1.2) = −1.09 kN ⋅ m
M 2 = −1.09 − 2.91(1) = −4 kN ⋅ m
Ans.
Ans.
M 3 = −4 + 4(1) = 0 checks!
______________________________________________________________________________
3-8
Break at the hinge at B
Beam OB:
From symmetry,
R1 = VB = 200 lbf
Ans.
Beam BD:
ΣM D = 0
200(12) − R2 (10) + 40(10)(5) = 0
R2 = 440 lbf
Ans.
ΣFy = 0
−200 + 440 − 40(10) + R3 = 0
R3 = 160 lbf
Ans.
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M 1 = 200(4) = 800 lbf ⋅ in
Ans.
M 2 = 800 − 200(4) = 0 checks at hinge
M 3 = 800 − 200(6) = −400 lbf ⋅ in Ans.
1
M 4 = −400 + (240)(6) = 320 lbf ⋅ in Ans.
2
1
M 5 = 320 − (160)(4) = 0 checks!
2
______________________________________________________________________________
3-9
q = R1 x
−1
− 9 x − 300
−1
− 5 x − 1200
0
−1
+ R2 x − 1500
0
V = R1 − 9 x − 300 − 5 x − 1200 + R2 x − 1500
1
1
−1
0
M = R1 x − 9 x − 300 − 5 x − 1200 + R2 x − 1500
(1)
1
(2)
At x = 1500+ V = M = 0. Applying Eqs. (1) and (2),
R1 − 9 − 5 + R2 = 0
⇒
R1 + R2 = 14
1500 R1 − 9(1500 − 300) − 5(1500 − 1200) = 0
R2 = 14 − 8.2 = 5.8 kN
⇒
R1 = 8.2 kN
Ans.
Ans.
V = 8.2 kN, M = 8.2 x N ⋅ m
0 ≤ x ≤ 300 :
300 ≤ x ≤ 1200 : V = 8.2 − 9 = −0.8 kN
M = 8.2 x − 9( x − 300) = −0.8 x + 2700 N ⋅ m
1200 ≤ x ≤ 1500 : V = 8.2 − 9 − 5 = −5.8 kN
M = 8.2 x − 9( x − 300) − 5( x − 1200) = −5.8 x + 8700 N ⋅ m
Plots of V and M are the same as in Prob. 3-5.
______________________________________________________________________________
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3-10
q = RO x
−1
− MO x
V = RO − M O x
−1
−2
− 500 x − 8
−1
0
− 40 x − 14 + 40 x − 20
0
1
− 500 x − 8 − 40 x − 14 + 40 x − 20
1
2
M = RO x − M O − 500 x − 8 − 20 x − 14 + 20 x − 20
0
1
(1)
2
(2)
at x = 20+ in, V = M = 0, Eqs. (1) and (2) give
RO − 500 − 40 ( 20 − 14 ) = 0
RO (20) − M O − 500(20 − 8) − 20(20 − 14) 2 = 0
0 ≤ x ≤ 8:
RO = 740 lbf
Ans.
M O = 8080 lbf ⋅ in
Ans.
⇒
⇒
V = 740 lbf, M = 740 x − 8080 lbf ⋅ in
8 ≤ x ≤ 14 : V = 740 − 500 = 240 lbf
M = 740 x − 8080 − 500( x − 8) = 240 x − 4080 lbf ⋅ in
14 ≤ x ≤ 20 : V = 740 − 500 − 40( x − 14) = −40 x + 800 lbf
M = 740 x − 8080 − 500( x − 8) − 20( x − 14)2 = −20 x 2 + 800 x − 8000 lbf ⋅ in
Plots of V and M are the same as in Prob. 3-6.
______________________________________________________________________________
3-11
q = R1 x
−1
− 2 x − 1.2
−1
+ R2 x − 2.2
0
−1
0
− 4 x − 3.2
V = R1 − 2 x − 1.2 + R2 x − 2.2 − 4 x − 3.2
1
1
−1
0
(1)
1
M = R1 x − 2 x − 1.2 + R2 x − 2.2 − 4 x − 3.2
at x = 3.2+, V = M = 0. Applying Eqs. (1) and (2),
(2)
R1 − 2 + R2 − 4 = 0
⇒
R1 + R2 = 6
(3)
3.2 R1 − 2(2) + R2 (1) = 0
⇒
3.2 R1 + R2 = 4
(4)
Solving Eqs. (3) and (4) simultaneously,
R1 = -0.91 kN, R2 = 6.91 kN Ans.
V = −0.91 kN, M = −0.91x kN ⋅ m
0 ≤ x ≤ 1.2 :
1.2 ≤ x ≤ 2.2 : V = −0.91 − 2 = −2.91 kN
M = −0.91x − 2( x − 1.2) = −2.91x + 2.4 kN ⋅ m
2.2 ≤ x ≤ 3.2 : V = −0.91 − 2 + 6.91 = 4 kN
M = −0.91x − 2( x −1.2) + 6.91( x − 2.2) = 4 x − 12.8 kN ⋅ m
Plots of V and M are the same as in Prob. 3-7.
______________________________________________________________________________
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3-12
q = R1 x
−1
− 400 x − 4
−1
0
+ R2 x − 10
−1
0
0
− 40 x − 10 + 40 x − 20 + R3 x − 20
0
1
1
V = R1 − 400 x − 4 + R2 x − 10 − 40 x − 10 + 40 x − 20 + R3 x − 20
1
1
2
0
2
M = R1 x − 400 x − 4 + R2 x − 10 − 20 x − 10 + 20 x − 20 + R3 x − 20
M = 0 at x = 8 in ∴ 8 R1 − 400(8 − 4) = 0
⇒
R1 = 200 lbf
−1
(1)
1
(2)
Ans.
at x = 20+, V =M = 0. Applying Eqs. (1) and (2),
R2 + R3 = 600
200 − 400 + R2 − 40(10) + R3 = 0
⇒
200(20) − 400(16) + R2 (10) − 20(10) 2 = 0 ⇒ R2 = 440 lbf
Ans.
R3 = 600 − 440 = 160 lbf
0 ≤ x ≤ 4:
Ans.
V = 200 lbf, M = 200 x lbf ⋅ in
4 ≤ x ≤ 10 : V = 200 − 400 = −200 lbf,
M = 200 x − 400( x − 4) = −200 x + 1600 lbf ⋅ in
10 ≤ x ≤ 20 : V = 200 − 400 + 440 − 40( x − 10) = 640 − 40 x lbf
M = 200 x − 400( x − 4) + 440( x − 10) − 20 ( x − 10 ) = −20 x 2 + 640 x − 4800 lbf ⋅ in
2
Plots of V and M are the same as in Prob. 3-8.
______________________________________________________________________________
3-13 Solution depends upon the beam selected.
______________________________________________________________________________
3-14
(a) Moment at center,
( l − 2a )
xc =
2
2
wl
 l   wl  l

M c =  ( l − 2a ) −    =
 − a
2  2
 2   2  4

At reaction, M r = wa2 2
a = 2.25, l = 10 in, w = 100 lbf/in
Mc =
100(10)  10

 − 2.25  = 125 lbf ⋅ in
2  4

Mr =
100 ( 2.252 )
= 253 lbf ⋅ in
2
(b) Optimal occurs when M c = M r
Ans.
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2
wl  l
 wa
−
a
=
⇒ a 2 + al − 0.25l 2 = 0


2 4
2

Taking the positive root
(
)
1
l
−l + l 2 + 4 ( 0.25l 2 )  =
2 − 1 = 0.207 l


 2
2
for l = 10 in, w = 100 lbf,
lbf a = 0.207(10) = 2.07 in
M min = (100 2 ) 2.07
072 = 214 lbf ⋅ in
a=
Ans.
_________________________
______________________________________
___________________
3-15
(a)
20 − 10
= 5 kpsi
2
20 + 10
CD =
= 15 kpsi
si
2
C=
R = 152 + 82 = 17 kpsi
si
σ 1 = 5 + 17 = 22 kpsi
σ 2 = 5 − 17 = −12 kpsii
1
8
φ p = tan −1   = 14.0
.04o cw
2
15
 
τ 1 = R = 17 kpsi
φs = 45o − 14.04o = 30.96
.96o ccw
(b)
9 + 16
= 12.5 kpsii
2
16 − 9
CD =
= 3.5 kpsi
si
2
C=
R = 52 + 3.52 = 6.10 kpsi
k
σ 1 = 12.5 + 6.1 = 18.6 kpsi
kp
σ 2 = 12.5 − 6.1 = 6.4 kpsi
kps
1
 5 
φ p = tan −1 
7.5o ccw
 = 27.
2
3.5


τ 1 = R = 6.10 kpsi
c
φs = 45o − 27.5o = 17.5o cw
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(c)
24 + 10
= 17 kpsi
2
24 − 10
CD =
= 7 kpsii
2
C=
R = 7 2 + 6 2 = 9.22 kps
psi
σ 1 = 17 + 9.22 = 26.22 kkpsi
σ 2 = 17 − 9.22 = 7.78 kpsi
kp
1
 7 
 
φ p = 90o + tan −1   = 69.7o ccw
2
6

τ 1 = R = 9.22 kpsi
φs = 69.7o − 45o = 24.7o ccw
(d)
−12 + 22
= 5 kpsii
2
12 + 22
CD =
si
= 17 kpsi
2
C=
R = 17 2 + 122 = 20.811 kpsi
σ 1 = 5 + 20.81 = 25.81 kkpsi
σ 2 = 5 − 20.81 = −15.81
81 kpsi
1
7 
 17
 

τ 1 = R = 20.81 kpsi
φ p = 90o + tan −1    = 72.39o cw
2
122
φs = 72.39 − 45 = 27.399o cw
_________________________
______________________________________
___________________
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3-16
(a)
−8 + 7
Pa
= −0.5 MPa
2
8+7
CD =
= 7.5 MPaa
2
C=
R = 7.52 + 62 = 9.60 M
MPa
σ 1 = 9.60 − 0.5 = 9.10 M
MPa
.1 Mpa
σ 2 = −0.5 − 9.6 = −10.1
1
 7.5  
φ p = 90o + tan −1 
70.67o cw
  = 70
2
6


τ 1 = R = 9.60 MPa
φs = 70.67o − 45o = 25.6
.67o cw
(b)
9−6
= 1.5 MPa
2
9+6
CD =
= 7.5 MPaa
2
C=
R = 7.52 + 32 = 8.0788 MPa
σ 1 = 1.5 + 8.078 = 9.588 M
MPa
.58 MPa
σ 2 = 1.5 − 8.078 = −6.58
1
 3 
φ p = tan −1 
0.9o cw
 = 10.9
2
7.5


τ 1 = R = 8.078 MPa
φs = 45o − 10.9o = 34.1o ccw
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(c)
12 − 4
= 4 MPa
2
12 + 4
CD =
= 8 MPa
2
C=
R = 82 + 7 2 = 10.63 M
MPa
σ 1 = 4 + 10.63 = 14.63 MPa
M
MPa
σ 2 = 4 − 10.63 = −6.633 M
1
 8 
 
φ p = 90o + tan −1   = 69.4o ccw
2
7

τ 1 = R = 10.63 MPa
φs = 69.4o − 45o = 24.44o ccw
(d)
6−5
= 0.5 MPa
2
6+5
CD =
= 5.5 MPaa
2
C=
R = 5.52 + 82 = 9.71 MPa
M
σ 1 = 0.5 + 9.71 = 10.211 M
MPa
σ 2 = 0.5 − 9.71 = −9.211 MPa
1
 8 
φ p = tan −1   = 27.7
7.75o ccw
2
5.5


τ 1 = R = 9.71 MPa
φs = 45o − 27.75o = 17.25
.25o cw
_________________________
______________________________________
___________________
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3-17
(a)
12 + 6
= 9 kpsi
2
12 − 6
= 3 kpsi
CD =
2
C=
R = 32 + 42 = 5 kpsi
σ 1 = 5 + 9 = 14 kpsi
σ 2 = 9 − 5 = 4 kpsi
1
 4
tan −1   = 26.6
.6o ccw
 3
2
τ 1 = R = 5 kpsi
φp =
φs = 45o − 26.6o = 18.44o ccw
(b)
30 − 10
= 10 kpsi
2
30 + 10
CD =
si
= 20 kpsi
2
C=
R = 202 + 102 = 22.36
36 kpsi
σ 1 = 10 + 22.36 = 32.366 kpsi
σ 2 = 10 − 22.36 = −12.3
.36 kpsi
1
 10 
3.28o ccw
 = 13.2
2
 20 
τ 1 = R = 22.36 kpsi
φ p = tan −1 
.72o cw
φs = 45o − 13.28o = 31.72
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(c)
−10 + 18
= 4 kpsii
2
10 + 18
CD =
= 14 kpsi
si
2
C=
R = 142 + 92 = 16.64 kpsi
k
σ 1 = 4 + 16.64 = 20.644 kpsi
k
64 kpsi
σ 2 = 4 − 16.64 = −12.64
1
4 
 14
φ p = 90o + tan −1   = 73.63o cw
2
 9 
τ 1 = R = 16.64 kpsi
φs = 73.63 − 45 = 28.633o cw
(d)
9 + 19
= 14 kpsi
2
19 − 9
CD =
= 5 kpsi
2
C=
R = 52 + 82 = 9.434 kpsi
kp
σ 1 = 14 + 9.43 = 23.43 kkpsi
kpsi
σ 2 = 14 − 9.43 = 4.57 kp
1
 5 
 
φ p = 90o + tan −1    = 61.0o cw
2
8

τ 1 = R = 9.34 kpsi
φs = 61o − 45o = 16o cw
w
_________________________
______________________________________
___________________
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3-18
(a)
−80 − 30
= −55 MP
Pa
2
80 − 30
Pa
= 25 MPa
CD =
2
C=
R = 252 + 202 = 32.02
02 MPa
σ 1 = 0 MPa
σ 2 = −55 + 32.02 = −22
22.98 = −23.0 MPa
.0 MPa
σ 3 = −55 − 32.0 = −87.0
τ1 2 =
23
= 11.5 MPa,
2
τ 2 3 = 32.0 MPa,
τ1 3 =
87
= 43.5 MPa
2
(b)
30 − 60
Pa
= −15 MPa
2
60 + 30
Pa
CD =
= 45 MPa
2
C=
R = 452 + 302 = 54.11 M
MPa
σ 1 = −15 + 54.1 = 39.1 M
MPa
σ 2 = 0 MPa
σ 3 = −15 − 54.1 = −69.1
.1 MPa
39.1 + 69.1
= 54.1
.1 MPa
2
39.1
= 19.6 MPaa
τ1 2 =
2
69.1
= 34.6 MPaa
τ2 3 =
2
τ1 3 =
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(c)
40 + 0
= 20 MPa
2
40 − 0
CD =
= 20 MPaa
2
C=
MPa
R = 202 + 202 = 28.33 M
σ 1 = 20 + 28.3 = 48.3 MPa
M
MPa
σ 2 = 20 − 28.3 = −8.3 M
σ 3 = σ z = −30 MPa
τ1 3 =
48.3 + 30
= 39.1 M
MPa,
2
τ 1 2 = 28.3 MPa,
τ2 3 =
30 − 8.3
= 10
10.9 MPa
2
(d)
50
= 25 MPa
2
50
CD =
= 25 MPa
2
C=
1 MPa
M
R = 252 + 302 = 39.1
σ 1 = 25 + 39.1 = 64.1 MPa
M
σ 2 = 25 − 39.1 = −14.11 M
MPa
σ 3 = σ z = −20 MPa
64.1 + 20
20 − 14.1
= 42.1 M
MPa, τ 1 2 = 39.1 MPa, τ 2 3 =
= 22.95 MPa
2
2
_________________________
______________________________________
___________________
τ1 3 =
3-19
(a)
Since there are no shear
ar stresses on the
stress element, the stres
ess element
already represents princ
ncipal stresses.
σ 1 = σ x = 10 kpsi
σ 2 = 0 kpsi
σ 3 = σ y = −4 kpsi
10 − (−4)
= 7 kpsi
si
2
10
τ 1 2 = = 5 kpsi
2
0 − (−4)
τ2 3 =
= 2 kpsii
2
τ1 3 =
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(b)
0 + 10
= 5 kpsi
2
10 − 0
CD =
= 5 kpsi
2
C=
R = 52 + 42 = 6.40 kps
psi
σ 1 = 5 + 6.40 = 11.40 kpsi
kp
σ 2 = 0 kpsi, σ 3 = 5 − 66.40 = −1.40 kpsi
τ 1 3 = R = 6.40 kpsi,
τ1 2 =
11.40
= 5.70 kpsi,
2
τ3 =
1.40
= 0.70 kpsi
kp
2
(c)
−2 − 8
= −5 kpsi
2
8−2
CD =
= 3 kpsi
2
C=
R = 32 + 42 = 5 kpsi
σ 1 = −5 + 5 = 0 kpsi, σ 2 = 0 kpsi
σ 3 = −5 − 5 = −10 kpsii
τ1 3 =
10
= 5 kpsi,
2
τ 1 2 = 0 kpsi,
τ 2 3 = 5 kpsi
(d)
10 − 30
= −10 kpsi
si
2
10 + 30
si
= 20 kpsi
CD =
2
C=
36 kpsi
R = 202 + 102 = 22.36
σ 1 = −10 + 22.36 = 12.3
.36 kpsi
σ 2 = 0 kpsi
σ 3 = −10 − 22.36 = −32.
32.36 kpsi
12.36
32.36
= 6.18 kpsi, τ 2 3 =
= 16.18
8 kpsi
2
2
_________________________
______________________________________
___________________
τ 1 3 = 22.36 kpsi,
τ1 2 =
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3-20
From Eq. (3-15),
σ 3 − (−6 + 18 − 12)σ 2 +  −6(18) + (−6)(−12) + 18(−12) − 92 − 62 − (−15)2  σ
−  −6(18)(−12) + 2(9)(6)(−15) − (−6)(6) 2 − 18(−15)2 − (−12)(9)2  = 0
σ 3 − 594σ + 3186 = 0
Roots are: 21.04, 5.67, –26.71 kpsi
Ans.
21.04 − 5.67
= 7.69 kpsi
2
5.67 + 26.71
τ2 3 =
= 16.19 kpsi
2
21.04 + 26.71
τ max = τ 1 3 =
= 23.88 kpsi
2
τ1 2 =
Ans.
_____________________________________________________________________________
3-21
From Eq. (3-15)
(
)
σ 3 − (20 + 0 + 20)σ 2 +  20(0) + 20(20) + 0(20) − 402 − −20 2 − 02  σ

(
)
(
−  20(0)(20) + 2(40) −20 2 (0) − 20 −20 2

σ 3 − 40σ 2 − 2 000σ + 48 000 = 0
Roots are: 60, 20, –40 kpsi
)
2
2

− 0(0) 2 − 20(40)2  = 0

Ans.
60 − 20
= 20 kpsi
2
20 + 40
= 30 kpsi
τ2 3 =
2
60 + 40
= 50 kpsi
τ max = τ 1 3 =
2
τ1 2 =
Ans.
_____________________________________________________________________________
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3-22 From Eq. (3-15)
2
2
σ 3 − (10 + 40 + 40)σ 2 + 10(40) + 10(40) + 40(40) − 202 − ( −40 ) − ( −20 )  σ


− 10(40)(40) + 2(20)( −40)(−20) − 10( −40) − 40(−20) − 40(20)  = 0
σ 3 − 90σ 2 = 0
2
Roots are: 90, 0, 0 MPa
2
2
Ans.
τ2 3 = 0
τ 1 2 = τ 1 3 = τ max =
90
= 45 MPa
2
Ans.
_____________________________________________________________________________
3-23
σ=
F
15000
=
= 33 950 psi = 34.0 kpsi
A (π 4 ) ( 0.752 )
δ=
60
FL
L
= σ = 33 950
= 0.0679 in
AE
E
30 (106 )
δ
0.0679
= 1130 (10−6 ) = 1130µ
60
L
From Table A-5, v = 0.292
ò1 =
=
ò2 = −vò1 = −0.292(1130) = −330 µ
∆d = ò2 d = −330 (10
−6
Ans.
Ans.
Ans.
Ans.
) (0.75) = −248 (10 ) in
−6
Ans.
_____________________________________________________________________________
3-24
σ=
3000
F
=
= 6790 psi = 6.79 kpsi
A (π 4 ) ( 0.752 )
δ=
60
FL
L
= σ = 6790
= 0.0392 in
AE
E
10.4 (106 )
δ
0.0392
= 653 (10−6 ) = 653µ
L
60
From Table A-5, v = 0.333
ò1 =
=
ò2 = −vò1 = −0.333(653) = −217 µ
Ans.
Ans.
Ans.
Ans.
∆d = ò2 d = −217 (10−6 ) (0.75) = −163 (10−6 ) in
Ans.
_____________________________________________________________________________
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3-25
∆d −0.0001d
=
= −0.0001
d
d
From Table A-5, v = 0.326, E = 119 GPa
−ò
−0.0001
ò1 = 2 =
= 306.7 (10−6 )
v
0.326
FL
F
and σ = , so
δ=
AE
A
δE
= ò1 E = 306.7 (10−6 ) (119) (109 ) = 36.5 MPa
σ=
L
ò2 =
F = σ A = 36.5 (10
)
π ( 0.03)
2
= 25 800 N = 25.8 kN
Ans.
4
Sy = 70 MPa > σ , so elastic deformation assumption is valid.
_____________________________________________________________________________
6
3-26
δ=
FL
L
8(12)
= σ = 20 000
= 0.185 in
AE
E
10.4 (106 )
Ans.
_____________________________________________________________________________
3-27
δ=
FL
L
3
= σ = 140 (106 )
= 0.00586 m = 5.86 mm
AE
E
71.7 (109 )
Ans.
_____________________________________________________________________________
3-28
δ=
10(12)
FL
L
= σ = 15 000
= 0.173 in
AE
E
10.4 (106 )
Ans.
_____________________________________________________________________________
3-29
With σ z = 0, solve the first two equations of Eq. (3-19) simulatenously. Place E on the
left-hand side of both equations, and using Cramer’s rule,
σx =
Eòx
Eòy
−v
1
1
−v
−v
1
=
Eòx + vEòy
1− v
2
=
E (òx + vòy )
1 − v2
Likewise,
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σy =
E (òy + ν òx )
1 − v2
From Table A-5, E = 207 GPa and ν = 0.292. Thus,
E (òx + vòy )
σx =
=
207 (109 )  0.0019 + 0.292 ( −0.000 72 ) 
1− v
1 − 0.292
9
207 (10 )  −0.000 72 + 0.292 ( 0.0019 ) 
2
2
(10 ) = 382 MPa
−6
Ans.
(10−6 ) = −37.4 MPa Ans.
1 − 0.2922
_____________________________________________________________________________
σy =
3-30
With σ z = 0, solve the first two equations of Eq. (3-19) simulatenously. Place E on the
left-hand side of both equations, and using Cramer’s rule,
Eòx
Eòy
σx =
−v
1
1 −v
−v 1
=
Eòx + vEòy
1− v
2
=
E (òx + vòy )
1 − v2
Likewise,
σy =
E (òy + ν òx )
1 − v2
From Table A-5, E = 71.7 GPa and ν = 0.333. Thus,
E (òx + vòy ) 71.7 (109 ) 0.0019 + 0.333 ( −0.000 72 ) 
σx =
10−6 ) = 134 MPa Ans.
=
(
2
2
1− v
1 − 0.333
9
71.7 (10 )  −0.000 72 + 0.333 ( 0.0019 ) 
σy =
(10−6 ) = −7.04 MPa Ans.
1 − 0.3332
_____________________________________________________________________________
3-31
c
ac
(a) R1 = F M max = R1a = F
l
l
6M
6 ac
σ bh 2l
F⇒F=
σ= 2 = 2
bh
bh l
6ac
Ans.
F
(σ σ )( bm b )( hm h ) ( lm l1 ) = 1( s)( s)2 ( s) = s 2
(b) m = m
F
( s )( s)
( am a )( cm c )
2
Ans.
For equal stress, the model load varies by the square of the scale factor.
_____________________________________________________________________________
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3-32
wl
w l  l  wl 2
, M max x =l /2 =
l −  =
2
2 2 2
8
2
2
6M
6 wl
3Wl
4 σ bh
σ= 2 = 2
=
⇒ W=
Ans.
bh
bh 8
4bh 2
3 l
(a)
(b)
R1 =
Wm (σ m / σ )(bm / b)(hm / h) 2 1( s )( s )2
=
=
= s2
W
lm / l
s
Ans.
wmlm
wm s 2
= s2 ⇒
= = s Ans.
wl
w
s
For equal stress, the model load w varies linearly with the scale factor.
_____________________________________________________________________________
3-33
(a)
Can solve by iteration or derive
equations for the general case. Find
maximum moment under wheel W3 .
WT = ΣW at centroid of W’s
l − x3 − d3
RA =
WT
l
Under wheel 3,
M 3 = RA x3 − W1a13 − W2 a23 =
For maximum,
( l − x3 − d3 ) W
T
l
dM 3
W
= 0 = ( l − d3 − 2 x3 ) T
dx3
l
Substitute into M ⇒ M 3 =
( l − d3 )
x3 − W1a13 − W2 a23
x3 =
⇒
l − d3
2
2
WT − W1a13 − W2 a23
4l
This means the midpoint of d 3 intersects the midpoint of the beam.
For wheel i,
l − di
xi =
,
2
Mi
( l − di )
=
4l
2
i −1
WT − ∑ W j a ji
j =1
Note for wheel 1: ΣW j a ji = 0
WT = 104.4,
W1 = W2 = W3 = W4 =
104.4
= 26.1 kips
4
476
(1200 − 238) 2
= 238 in, M 1 =
(104.4) = 20 128 kip ⋅ in
2
4(1200)
Wheel 2: d 2 = 238 − 84 = 154 in
Wheel 1: d1 =
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M2 =
(1200 − 154)) 2
(104.4) − 26.1(84) = 21 605 kip ⋅ in = M max
4(1200)
Ans.
Check if all of the whee
eels are on the rail.
(b)
xmax = 600 − 777 = 523 in Ans.
(c)
See above sketc
tch.
(d)
Inner axles
_________________________
______________________________________
__________________
3-34
(a) Let a = total area of entire envelope
Let b = area of sidee notch
A = a − 2b = 40(2)(3
)(37.5) − 25 ( 34) = 2150 mm2
1
1
3
3
( 40 )( 75) − ( 34 )( 25)
12
12
I = 1.36 (106 ) mm4 Ans.
I = I a − 2Ib =
Dimension
ions in mm.
(b)
Aa = 0.375(1.875) = 0.7
.703 125 in 2
Ab = 0.375(1.75) = 0.65
656 25 in 2
A = 2(0.703125) + 0.656
56 25 = 2.0625 in 2
2(0.703 125)(0.937
375) + 0.656 25(0.6875)
= 0.858 in Ans.
2
2.0625
0.375(1.875)3
Ia =
= 0.206
0
in 4
12
1.75(0.375)3
Ib =
= 0.0
.007 69 in 4
12
I1 = 2 0.206 + 0.703 12
125(0.0795) 2  + 0.00769 + 0.656 25(0.1705)) 2  = 0.448 in 4
y=
Ans.
(c)
Use two negative areas.
as.
2
Aa = 625 mm , Ab = 55625 mm 2 , Ac = 10 000 mm 2
A = 10 000 − 5625 − 62
625 = 3750 mm 2 ;
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ya = 6.25 mm, yb = 50 mm, yc = 50 mm
10 000(50) − 5625(50) − 625(6.25)
= 57.29 mm
3750
c1 = 100 − 57.29 = 42.71 mm Ans.
y=
Ans.
50(12.5)3
= 8138 mm 4
12
75(75)3
Ib =
= 2.637 (106 ) mm 4
12
100(100)3
Ic =
= 8.333 (106 ) in 4
12
2
2
I1 = 8.333 (106 ) + 10 000(7.29)2  −  2.637 (106 ) + 5625 ( 7.29 )  − 8138 + 625 ( 57.29 − 6.25 ) 

 

Ia =
I1 = 4.29 (106 ) in 4
Ans.
(d)
Aa = 4 ( 0.875 ) = 3.5 in 2
Ab = 2.5 ( 0.875 ) = 2.1875 in 2
A = Aa + Ab = 5.6875 in 2
2.9375 ( 3.5 ) + 1.25(2.1875)
= 2.288 in Ans.
5.6875
1
1
3
2
3
2
I = (4) ( 0.875 ) + 3.5 ( 2.9375 − 2.288 ) + ( 0.875 )( 2.5 ) + 2.1875 ( 2.288 − 1.25 )
12
12
I = 5.20 in 4 Ans.
_____________________________________________________________________________
y=
3-35
1
(20)(40)3 = 1.067 (105 ) mm4
12
A = 20(40) = 800 mm2
Mmax is at A. At the bottom of the section,
Mc 450 000(20)
σ max =
=
= 84.3 MPa Ans.
I
1.067 (105 )
I=
Due to V, τmax is between A and B at y = 0.
3 V 3  3000 
τ max =
= 
 = 5.63 MPa Ans.
2 A 2  800 
_____________________________________________________________________________
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3-36
1
(1)(2)3 = 0.6667 in 4
12
A = 1(2) = 2 in 2
I=
ΣM O = 0
8 RA − 100(8)(12) = 0
RA = 1200 lbf
RO = 1200 − 100(8) = 400 lbf
M max is at A. At the top of the beam,
σ max =
Mc 3200(0.5)
=
= 2400 psi
I
0.6667
Ans.
Due to V, τ max is at A, at y = 0.
3 V 3  800 
= 
 = 600 psi Ans.
2 A 2 2 
_____________________________________________________________________________
τ max =
3-37
1
(0.75)(2)3 = 0.5 in 4
12
A = (0.75)(2) = 1.5 in 2
I=
ΣM A = 0
15RB − 1000(20) = 0
RB = 1333.3 lbf
RA = 3000 − 1333.3 + 1000 = 2666.7 lbf
M max is at B. At the top of the beam,
σ max =
Mc 5000(1)
=
= 10 000 psi
I
0.5
Ans.
Due to V, τ max is between B and C at y = 0.
3 V 3  1000 
= 
 = 1000 psi Ans.
2 A 2  1.5 
_____________________________________________________________________________
τ max =
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3-38
I=
πd4
=
π (50) 4
( )
= 306.796 103 mm 4
64
64
π d 2 π (50) 2
=
= 1963 mm 2
A=
4
4
ΣM B = 0
6(300)(150) − 200 RA = 0
RA = 1350 kN
RB = 6(300) − 1350 = 450 kN
M max is at A. At the top,
Mc 30 000(25)
=
= 2.44 kN/mm 2 = 2.44 GPa
I
30 6796
Due to V, τ max is at A, at y = 0.
σ max =
Ans.
4 V 4  750 
2
= 
 = 0.509 kN/mm = 509 MPa Ans.
3 A 3  1963 
_____________________________________________________________________________
τ max =
3-39
8σ I
wl 2
wl 2 c
⇒ σ max =
⇒ w = max
8
8I
cl 2
(a) l = 48 in; Table A-8, I = 0.537 in 4
M max =
w=
8 (12 ) (103 ) ( 0.537 )
1( 482 )
= 22.38 lbf/in
Ans.
(b) l = 60 in, I ≈ (1 12 )( 2 ) ( 33 ) − (1 12 )(1.625 ) ( 2.6253 ) = 2.051 in 4
w=
8 (12 ) (103 ) ( 2.051)
(1.5) ( 602 )
= 36.5 lbf/in
Ans.
(c) l = 60 in; Table A-6, I = 2 ( 0.703 ) = 1.406 in 4
y = 0.717 in, cmax = 1.783 in
8 (12 ) (103 ) (1.406 )
w=
= 21.0 lbf/in Ans.
1.783 ( 602 )
(d) l = 60 in, Table A-7, I = 2.07 in 4
w=
8 (12 ) (103 ) ( 2.07 )
1.5 ( 602 )
= 36.8 lbf/in
Ans.
_____________________________________________________________________________
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3-40
I=
π
( 0.5 ) = 3.068 (10 ) in ,
64
4
−3
4
A=
π
( 0.5 ) = 0.1963 in
4
2
2
Model (c)
500(0.5) 500(0.75 / 2)
M =
+
= 218.75 lbf ⋅ in
2
2
Mc 218.75(0.25)
=
σ=
I
3.068 (10 −3 )
σ = 17 825 psi = 17.8 kpsi
τ max =
Ans.
4 V 4 500
=
= 3400 psi = 3.4 kpsi
3 A 3 0.1963
Ans.
Model (d)
M = 500(0.625) = 312.5 lbf ⋅ in
σ=
Mc 312.5(0.25)
=
I
3.068 (10−3 )
σ = 25 464 psi = 25.5 kpsi
τ max =
Ans.
4 V 4 500
=
= 3400 psi = 3.4 kpsi
3 A 3 0.1963
Ans.
Model (e)
M = 500(0.4375) = 218.75 lbf ⋅ in
σ=
Mc 218.75(0.25)
=
I
3.068 (10−3 )
σ = 17 825 psi = 17.8 kpsi
τ max =
Ans.
4 V 4 500
=
= 3400 psi = 3.4 kpsi
3 A 3 0.1963
Ans.
_____________________________________________________________________________
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3-41
I=
π
π
12 ) = 1018 mm , A = (12 ) = 113.1 mm
(
64
4
4
4
2
2
Model (c)
2000(6) 2000(9)
+
= 15 000 N ⋅ mm
M =
2
2
Mc 15 000(6)
=
σ=
I
1018
σ = 88.4 N/mm 2 = 88.4 MPa Ans.
τ max =
4 V 4  2000 
2
= 
 = 23.6 N/mm = 23.6 MPa
3 A 3  113.1 
Ans.
Model (d)
M = 2000(12) = 24 000 N ⋅ mm
Mc 24 000(6)
=
I
1018
σ = 141.5 N/mm 2 = 141.5 MPa Ans.
4 V 4  2000 
2
= 
τ max =
 = 23.6 N/mm = 23.6 MPa
3 A 3  113.1 
σ=
Ans.
Model (e)
M = 2000(7.5) = 15000 N ⋅ mm
Mc 15000(6)
=
I
1018
σ = 88.4 N/mm 2 = 88.4 MPa Ans.
4 V 4  2000 
2
τ max =
= 
 = 23.6 N/mm = 23.6 MPa
3 A 3  113.1 
σ=
Ans.
_____________________________________________________________________________
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3-42
(a) σ =
d=
(b) τ =
d=
(c) τ =
Mc M ( d / 2 ) 32 M
=
=
I
π d 4 / 64 π d 3
3
32M
=
πσ
3
32(218.75)
= 0.420 in Ans.
π (30 000)
V
V
=
A πd2 / 4
4V
πτ
=
4(500)
= 0.206 in
π (15000)
Ans.
4V 4
V
=
3 A 3 (π d 2 / 4 )
4 4V
4 4(500)
=
= 0.238 in Ans.
3 πτ
3 π (15000)
_____________________________________________________________________________
d=
3-43
p1 + p2
1
x − l + terms for x > l + a
a
p
p
+
1
2
2
V = − F + p1 x − l − 1
x − l + terms for x > l + a
2a
p
p + p2
2
3
M = − Fx + 1 x − l − 1
x − l + terms for x > l + a
2
6a
q = −F x
−1
0
+ p1 x − l −
At x = (l + a ) + , V = M = 0, terms for x > l + a = 0
p1 + p2 2
2F
a = 0 ⇒ p1 − p2 =
2a
a
2
6 F (l + a)
pa
p + p2 3
a =0
− F (l + a) + 1 − 1
⇒
2 p1 − p2 =
a2
2
6a
− F + p1a −
From (1) and (2)
p1 =
2F
(3l + 2a),
a2
p2 =
2F
(3l + a)
a2
Shigley’s MED, 10th edition
(1)
(2)
(3)
Chapter 3 Solutions, Page 29/100
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b
a
=
p2 p1 + p2
From similar triangless
⇒
b=
ap2
p1 + p2
(4)
Mmax occurs where V = 0
xmax = l + a − 2b
p1
p + p2
(a − 2b) 2 − 1
( a − 2b)3
2
6a
p
p + p2
= − Fl − F ( a − 2b) + 1 ( a − 2b) 2 − 1
(a − 2b)3
2
6a
Normally Mmax = − Fl
M max = − F (l + a − 2b) +
The fractional increasee in the magnitude is
∆=
F (a − 2b) − ( p1 2 ) (a − 2b)2 + ( p1 + p2 ) 6a  (a − 2b)3
Fl
(5)
For example, consider F = 1500 lbf, a = 1.2 in, l = 1.5 in
From (3)
From (4)
p1 =
2(1
(1500)
3 (1.5) + 2(1.2)  = 14 375 lbf/in
1. 2 
1.2
p2 =
2(1
(1500)
3(1.5) + 1.2 = 11 875 lbf/in
1 2 
1.2
b = 1.2(1
(11 875)/(14 375 + 11 875) = 0.5429 in
Substituting into (5) yie
ields
∆ = 0.036 89 or 3.7
3.7% higher than -Fl
_________________________
______________________________________
__________________
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Chapter 3 Solutions, Page 30/100
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3-44
300(30) 40
+ 1800 = 6900 lbf
2
30
300(30) 10
− 1800 = 3900 lbf
R2 =
2
30
3900
= 13 in
a=
300
R1 =
MB = −1800(10) = −18 000 lbf⋅in
Mx = 27 in = (1/2)3900(13) = 25 350 lbf⋅in
0.5(3) + 2.5(3)
= 1.5 in
6
1
I1 = (3)(13 ) = 0.25 in 4
12
1
I 2 = (1)(33 ) = 2.25 in 4
12
Applying the parallel-axis theorem,
y=
I z =  0.25 + 3(1.5 − 0.5) 2  +  2.25 + 3(2.5 − 1.5) 2  = 8.5 in 4
(a)
−18000(−1.5)
At x = 10 in, y = −1.5 in, σ x = −
= −3176 psi
8.5
−18000(2.5)
= 5294 psi
At x = 10 in, y = 2.5 in, σ x = −
8.5
25350( −1.5)
= 4474 psi
At x = 27 in, y = −1.5 in, σ x = −
8.5
25350(2.5)
At x = 27 in, y = 2.5 in, σ x = −
= −7456 psi
8.5
Max tension = 5294 psi
Ans.
Max compression = −7456 psi
Ans.
(b)
The maximum shear stress due to V is at B, at the neutral axis.
Vmax = 5100 lbf
Q = y′A′ = 1.25(2.5)(1) = 3.125 in 3
VQ 5100(3.125)
=
= 1875 psi
Ans.
Ib
8.5(1)
(c)
There are three potentially critical locations for the maximum shear stress, all at x
= 27 in: (i) at the top where the bending stress is maximum, (ii) at the neutral axis where
(τ max )V
=
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the transverse shear is maximum, or (iii) in the web just above the flange where bending
stress and shear stress are in their largest combination.
For (i):
The maximum bending stress was previously found to be − 7456 psi, and the shear
stress is zero. From Mohr’s circle,
σ
7456
τ max = max =
= 3728 psi
2
2
For (ii):
The bending stress is zero, and the transverse shear stress was found previously to be
1875 psi. Thus, τmax = 1875 psi.
For (iii):
The bending stress, at y = – 0.5 in, is
−18000( −0.5)
σx = −
= −1059 psi
8.5
The transverse shear stress is
Q = y′A′ = (1)(3)(1) = 3.0 in 3
τ=
VQ 5100(3.0)
=
= 1800 psi
Ib
8.5(1)
From Mohr’s circle,
2
 −1059 
2
 + 1800 = 1876 psi
 2 
Ans.
The critical location is at x = 27 in, at the top surface, where τmax = 3728 psi.
_____________________________________________________________________________
τ max = 
3-45
(a) L = 10 in. Element A:
My
−(1000)(10)(0.5)
=−
10−3 = 101.9 kpsi
I
(π / 64)(1) 4
VQ
τA =
, Q = 0 ⇒ τA = 0
Ib
(
σA = −
)
σ 
 101.9 
2
τ max =  A  + τ A2 = 
 + (0) = 50.9 kpsi
 2 
 2 
2
2
Ans.
Element B:
σB = −
My
, y=0
I
 4r
Q = y′A′ = 
 3π
⇒
σB = 0
2
3
4 ( 0.5 )
  π r  4r
=
=
= 1/ 12 in 3


2
6
6


3
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Chapter 3 Solutions, Page 32/100
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(
)
τB =
VQ (1000)(1/12)
=
10−3 = 1.698 kpsi
4
Ib (π / 64)(1) (1)
τ max
0
=   + 1.6982 = 1.698 kpsi
2
2
Ans.
Element C:
σC = −
My
−(1000)(10)(0.25)
=−
10−3 = 50.93 kpsi
(π / 64)(1)4
I
(
r
r
r
y1
y1
y1
)
)
(
Q = ∫ ydA = ∫ y (2 x ) dy = ∫ y 2 r 2 − y 2 dy
=−
(
2 2
r − y2
3
(
)
3/ 2
r
y1
(
2
= −  r2 − r2
3 
)
3/2
(
− r 2 − y12
)
3/2


)
3/ 2
2 2
r − y12
3
For C, y1 = r /2 =0.25 in
=
Q=
3/2
2
0.52 − 0.252 ) = 0.05413 in3
(
3
b = 2 x = 2 r 2 − y12 = 2 0.52 − 0.252 = 0.866 in
τC =
VQ
(1000)(0.05413)
=
10−3 ) = 1.273 kpsi
(
4
Ib (π / 64)(1) (0.866)
τ max
 50.93 
2
= 
 + (1.273) = 25.50 kpsi
 2 
2
Ans.
(b) Neglecting transverse shear stress:
Element A: Since the transverse shear stress at point A is zero, there is no change.
τ max = 50.9 kpsi Ans.
% error = 0% Ans.
Element B: Since the only stress at point B is transverse shear stress, neglecting
the transverse shear stress ignores the entire stress.
0
2
τ max =   = 0 psi
2
Ans.
 1.698 − 0 
% error = 
 * (100) = 100% Ans.
 1.698 
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Element C:
2
τ max
 50.93 
= 
 = 25.47 kpsi
 2 
Ans.
 25.50 − 25.47 
% error = 
 *(100) = 0.12% Ans.
25.50


(c) Repeating the process with different beam lengths produces the results in the table.
L = 10 in
A
B
C
L = 4 in
A
B
C
L = 1 in
A
B
C
L = 0.1in
A
B
C
Bending
stress,
σ (kpsi)
Transverse
shear stress,
τ (kpsi)
Max shear
stress,
τmax (kpsi)
Max shear
stress,
neglecting τ,
τmax (kpsi)
% error
102
0
50.9
0
1.70
1.27
50.9
1.70
25.50
50.9
0
25.47
0
100
0.12
40.7
0
20.4
0
1.70
1.27
20.4
1.70
10.26
20.4
0
10.19
0
100
0.77
10.2
0
5.09
0
1.70
1.27
5.09
1.70
2.85
5.09
0
2.55
0
100
10.6
1.02
0
0.509
0
1.70
1.27
0.509
1.70
1.30
0.509
0
0.255
0
100
80.4
Discussion:
The transverse shear stress is only significant in determining the critical stress element as
the length of the cantilever beam becomes smaller. As this length decreases, bending
stress reduces greatly and transverse shear stress stays the same. This causes the critical
element location to go from being at point A, on the surface, to point B, in the center. The
maximum shear stress is on the outer surface at point A for all cases except L = 0.1 in,
where it is at point B at the center. When the critical stress element is at point A, there is
no error from neglecting transverse shear stress, since it is zero at that location.
Neglecting the transverse shear stress has extreme significance at the stress element at the
center at point B, but that location is probably only of practical significance for very short
beam lengths.
_____________________________________________________________________________
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3-46
c
F
l
c
M = Fx 0 ≤ x ≤ a
l
6 M 6 ( c l ) Fx
σ= 2=
bh
bh 2
R1 =
h=
6 Fcx
lbσ max
0≤ x≤a
Ans.
_____________________________________________________________________________
3-47
c
From Problem 3-46, R1 = F = V , 0 ≤ x ≤ a
l
3 V 3 (c / l ) F
3 Fc
τ max =
=
⇒ h=
2 bh 2 bh
2 lbτ max
Ans.
6 Fcx
.
lbσ max
Sub in x = e and equate to h above.
3 Fc
6 Fce
=
2 lbτ max
lbσ max
From Problem 3-46, h( x) =
3 Fcσ max
Ans.
2
8 lbτ max
_____________________________________________________________________________
e=
3-48
(a)
x-z plane
ΣM O = 0 = 1.5(0.5) + 2(1.5) sin(30o )(2.25) − R2 z (3)
R2 z = 1.375 kN
Ans.
ΣFz = 0 = R1z − 1.5 − 2(1.5) sin(30o ) + 1.375
R1z = 1.625 kN Ans.
x-y plane
ΣM O = 0 = −2(1.5) cos(30o )(2.25) + R2 y (3)
R2 y = 1.949 kN
Ans.
ΣFy = 0 = R1 y − 2(1.5) cos(30o ) + 1.949
R1 y = 0.6491 kN
Ans.
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(b)
(c) The transverse shear and bending moments for most points of interest can readily be
taken straight from the diagrams. For 1.5 < x < 3, the bending moment equations are
parabolic, and are obtained by integrating the linear expressions for shear. For
convenience, use a coordinate shift of x′ = x – 1.5. Then, for 0 < x′ < 1.5,
Vz = x′ − 0.125
M y = ∫ Vz dx′ =
( x′ )
2
2
− 0.125x′ + C
At x′ = 0, M y = C = −0.9375 ⇒ M y = 0.5 ( x′ ) − 0.125 x′ + 0.9375
2
1.949
x′ + 0.6491 = −1.732 x′ + 0.6491
1.125
−1.732
2
Mz =
( x′ ) + 0.6491x′ + C
2
2
At x′ = 0, M z = C = 0.9737 ⇒ M z = −0.8662 ( x′ ) − 0.125 x′ − 0.9375
By programming these bending moment equations, we can find My, Mz, and their vector
combination at any point along the beam. The maximum combined bending moment is
found to be at x = 1.79 m, where M = 1.433 kN·m. The table below shows values at key
locations on the shear and bending moment diagrams.
Vy = −
x (m)
0
0.5−
1.5
1.625
1.875
3−
Vz (kN)
–1.625
–1.625
–0.1250
0
0.2500
1.375
Vy (kN)
0.6491
0.6491
0.6491
0.4327
0
–1.949
My
Mz
V (kN) (kN⋅m) (kN⋅m)
1.750
0
0
1.750 –0.8125 0.3246
0.6610 0.9375 0.9737
0.4327 –0.9453 1.041
0.2500 –0.9141 1.095
2.385
0
0
Shigley’s MED, 10th edition
M
(kN⋅m)
0
0.8749
1.352
1.406
1.427
0
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(d) The bending stress is obtained from Eq. (3-27),
−M z yA M y zA
+
σx =
Iz
Iy
The maximum tensile bending stress will be at point A in the cross section of Prob. 3-34
(a), where distances from the neutral axes for both bending moments will be maximum.
At A, for Mz, yA = –37.5 mm, and for My, zA = –20 mm.
40(75)3 34(25)3
Iz =
−
= 1.36(106 ) mm 4 = 1.36(10−6 ) m 4
12
12
 25(40)3  25(6)3
= 2.67(105 ) mm 4 = 2.67(10 −7 ) m 4
Iy = 2
+
12
12


It is apparent the maximum bending moment, and thus the maximum stress, will be in the
parabolic section of the bending moment diagrams. Programming Eq. (3-27) with the
bending moment equations previously derived, the maximum tensile bending stress is
found at x = 1.77 m, where My = – 0.9408 kN·m, Mz = 1.075 kN·m, and σx = 100.1 MPa.
Ans.
_____________________________________________________________________________
3-49
(a) x-z plane
3
600
(10) + M Oy
ΣM O = 0 = (1000)(4) −
5
2
M Oy = 1842.6 lbf ⋅ in Ans.
3
600
ΣFz = 0 = ROz − (1000) +
5
2
ROz = 175.7 lbf Ans.
x-y plane
4
600
ΣM O = 0 = − (1000)(4) −
(10) + M Oz
5
2
M Oz = 7442.5 lbf ⋅ in Ans.
4
600
ΣFy = 0 = ROy − (1000) −
5
2
ROy = 1224.3 lbf Ans.
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(b)
(
(c)
1/2
V ( x ) = Vy ( x ) 2 + Vz ( x) 2 
M ( x ) =  M y ( x ) 2 + M z ( x ) 2 
x (m)
0
4−
10−
Vz (kN)
–175.7
–175.7
424.3
1/2
Vy (kN)
1224.3
1224.3
424.3
V (kN)
1237
1237
600
My (kN⋅m) Mz (kN⋅m) M (kN⋅m)
–1842.6
–7442.6
7667
–2545.4
–2545.4
3600
0
0
0
(d) The maximum tensile bending stress will be at the outer corner of the cross section in
the positive y, negative z quadrant, where y = 1.5 in and z = –1 in.
2(3)3 (1.625)(2.625)3
Iz =
−
= 2.051 in 4
12
12
3(2)3 (2.625)(1.625)3
Iy =
−
= 1.601 in 4
12
12
At x = 0, using Eq. (3-27),
M y M z
σx = − z + y
Iz
Iy
( −7442.6)(1.5) ( −1842.6)( −1)
σx = −
+
= 6594 psi
2.051
1.601
Check at x = 4 in,
( −2545.4)(1.5) ( −2545.4)( −1)
σx = −
+
= 2706 psi
2.051
1.601
The critical location is at x = 0, where σx = 6594 psi. Ans.
_____________________________________________________________________________
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3-50
(a) The area within thee wall median line, Am, is
Square: Am = (b − t ) 2 . From
F
Eq. (3-45)
Tsq = 2 Amtτ all = 2((b − t ) 2 tτ all
Round: Am = π (b − t ) 2 / 4
Trd = 2π (b − t ) 2 tτ allll / 4
Ratio of Torques
Tsq
2(b − t ) 2 tτ all
4
=
= = 1.27 Ans.
2
Trd π (b − t ) tτ alll / 2 π
gth from Eq. (3-46) is
(b) Twist per unit length
θ1 =
TLm
2 A tτ L
τ L
L
= m all2 m = all m = C m
2
4GAmt
4GAmt
2G Am
Am
Square:
θsq = C
Round:
θrd = C
4(b − t )
(b − t )2
π (b − t )
π (b − t ) / 4
2
=C
4(b − t )
(b − t )2
θ sq
= 1 . Twists are thee same. Ans.
θ rd
_________________________
______________________________________
__________________
3-51
(a) The area enclosedd bby the section median line is Am = (1 − 0.0625
25)2 = 0.8789 in2 and
the length of the section
on median line is Lm = 4(1 − 0.0625) = 3.75 in. From Eq. (3-45),
T = 2 Am tτ = 2(0
(0.8789)(0.0625)(12 000) = 1318 lbf ⋅ in
From Eq. (3-46),
θ = θ1l =
TLml
4GAm2 t
=
(1318)(3.75) ( 36 )
( )
4 (11.5 ) 10 (0.8789) ( 0.0625 )
6
2
Ans
An .
= 0.0801
1 rad = 4.59o Ans.
(b) The radius at the median
me
line is rm = 0.125 + (0.5)(0.0625) = 0.15
.15625 in. The area enclosed
by the section median line
li is Am = (1 − 0.0625)2 – 4(0.15625)2 + 4(π
π /4)(0.15625)2 = 0.8579
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in2. The length of the section median line is Lm = 4[1 – 0.0625 – 2(0.15625)] + 2π(0.15625) =
3.482 in.
From Eq. (3-45),
T = 2 Amtτ = 2(0.8579)(0.0625)(12 000) = 1287 lbf ⋅ in
Ans.
From Eq. (3-46),
(1287)(3.482) ( 36 )
TLm l
θ = θ1l =
=
= 0.0762 rad = 4.37o Ans.
2
6
2
4GAm t 4 (11.5 ) 10 (0.8579) ( 0.0625 )
( )
_____________________________________________________________________________
3-52
θ1 =
3Ti
Ti =
⇒
GLi ci3
T = T1 + T2 + T3 =
θ1G
3
θ1GLi ci3
3
3
∑ Li ci3
i =1
Ans.
From Eq. (3-47), τ = Gθ1c
G and θ1 are constant, therefore the largest shear stress occurs when c is a maximum.
τ max = Gθ1cmax
Ans.
_____________________________________________________________________________
3-53
(b) Solve part (b) first since the angle of twist per unit length is needed for part (a).
τ max = τ allow = 12 ( 6.89 ) = 82.7 MPa
θ1 =
τ max
Gcmax
=
( ) = 0.348 rad/m
79.3 (10 ) (0.003)
82.7 106
9
(a)
T1 =
T2 =
T3 =
θ1GL1c13
3
θ 2GL2 c23
3
=
( )
0.348(79.3) 109 (0.020)(0.0023 )
=
θ3GL3c33
Ans.
3
0.348(79.3) (109 ) (0.030)(0.0033 )
3
9
0.348(79.3) (10 ) (0)(03 )
= 1.47 N ⋅ m
Ans.
= 7.45 N ⋅ m Ans.
=
= 0 Ans.
3
3
T = T1 + T2 + T3 = 1.47 + 7.45 + 0 = 8.92 N ⋅ m Ans.
_____________________________________________________________________________
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3-54
(b) Solve part (b) first since the angle of twist per unit length is needed for part (a).
τ
12 000
θ1 = max =
= 8.35 10−3 rad/in Ans.
Gcmax 11.5 106 (0.125)
(
( )
(a)
T1 =
T2 =
θ1GL1c13
3
=
θ 2GL2 c23
)
( 8.35) (10−3 ) (11.5) (106 ) ( 0.75) ( 0.06253 )
3
−3
( 8.35) (10 ) (11.5) (106 ) (1) ( 0.1253 )
=
= 5.86 lbf ⋅ in Ans.
= 62.52 lbf ⋅ in Ans.
3
6
3
−3
θ GL c3 ( 8.35) (10 ) (11.5) (10 ) ( 0.625 ) ( 0.0625 )
T3 = 3 3 3 =
= 4.88 lbf ⋅ in Ans.
3
3
T = T1 + T2 + T3 = 5.86 + 62.52 + 4.88 = 73.3 lbf ⋅ in Ans.
_____________________________________________________________________________
3
3-55
(b) Solve part (b) first since the angle of twist per unit length is needed for part (a).
τ max = τ allow = 12 ( 6.89 ) = 82.7 MPa
θ1 =
τ max
Gcmax
=
( ) = 0.348 rad/m
79.3 (10 ) (0.003)
82.7 106
9
(a)
T1 =
T2 =
T3 =
θ1GL1c13
3
θ 2GL2 c23
3
=
( )
0.348(79.3) 109 (0.020)(0.0023 )
=
θ3GL3c33
Ans.
3
0.348(79.3) (109 ) (0.030)(0.0033 )
3
9
0.348(79.3) (10 ) (0.025)(0.0023 )
= 1.47 N ⋅ m
= 7.45 N ⋅ m
Ans.
Ans.
=
= 1.84 N ⋅ m Ans.
3
3
T = T1 + T2 + T3 = 1.47 + 7.45 + 1.84 = 10.8 N ⋅ m Ans.
_____________________________________________________________________________
3-56
(a) From Eq. (3-40), with two 2-mm strips,
( 80 ) (106 ) ( 0.030 ) ( 0.0022 )
T=
=
= 3.08 N ⋅ m
3 + 1.8 / (b / c)
3 + 1.8 / ( 0.030 / 0.002 )
τ max bc 2
Tmax = 2(3.08) = 6.16 N ⋅ m
Ans.
Shigley’s MED, 10th edition
Chapter 3 Solutions, Page 41/100
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From the table on p. 116, with b/c = 30/2 = 15, α = β and has a value between 0.313 and 0.333.
From Eq. (3-40),
1
α≈
= 0.321
3 + 1.8 / (30 / 2)
From Eq. (3-41),
Tl
3.08(0.3)
θ=
=
= 0.151 rad
3
β bc G 0.321( 0.030 ) 0.0023 ( 79.3) 109
(
kt =
T
θ
=
6.16
= 40.8 N ⋅ m
0.151
)
( )
Ans.
Ans.
From Eq. (3-40), with a single 4-mm strip,
(80 ) (106 ) ( 0.030) ( 0.0042 )
Tmax =
=
= 11.9 N ⋅ m
3 + 1.8 / (b / c)
3 + 1.8 / ( 0.030 / 0.004 )
τ max bc 2
Ans.
Interpolating from the table on p. 116, with b/c = 30/4 = 7.5,
7.5 − 6
β=
(0.307 − 0.299) + 0.299 = 0.305
8−6
From Eq. (3-41)
Tl
11.9(0.3)
θ=
=
= 0.0769 rad
3
β bc G 0.305 ( 0.030 ) 0.0043 ( 79.3) 109
(
kt =
T
θ
=
)
11.9
= 155 N ⋅ m
0.0769
( )
Ans.
Ans.
(b) From Eq. (3-47), with two 2-mm strips,
(
)
( )
2
6
Lc 2τ ( 0.030 ) 0.002 ( 80 ) 10
=
= 3.20 N ⋅ m
T=
3
3
Tmax = 2(3.20) = 6.40 N ⋅ m Ans.
3Tl
3(3.20)(0.3)
θ= 3 =
= 0.151 rad
Lc G ( 0.030 ) 0.0023 ( 79.3) 109
(
)
( )
kt = T θ = 6.40 0.151 = 42.4 N ⋅ m
Ans.
Ans.
From Eq. (3-47), with a single 4-mm strip,
Tmax
(
)
( )
2
6
Lc 2τ ( 0.030 ) 0.004 ( 80 ) 10
=
=
= 12.8 N ⋅ m
3
3
Shigley’s MED, 10th edition
Ans.
Chapter 3 Solutions, Page 42/100
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θ=
3Tl
3(12.8)(0.3)
=
= 0.0757 rad
3
Lc G ( 0.030 ) 0.0043 ( 79.3) 109
(
)
( )
kt = T θ = 12.8 0.0757 = 169 N ⋅ m
Ans.
Ans.
The results for the spring constants when using Eq. (3-47) are slightly larger than when using
Eq. (3-40) and Eq. (3-41) because the strips are not infinitesimally thin (i.e. b/c does not equal
infinity). The spring constants when considering one solid strip are significantly larger (almost
four times larger) than when considering two thin strips because two thin strips would be able
to slip along the center plane.
_____________________________________________________________________________
3-57
(a) Obtain the torque from the given power and speed using Eq. (3-44).
H
(40000)
T = 9.55 = 9.55
= 152.8 N ⋅ m
n
2500
Tr 16T
τ max = = 3
J πd
13
 16 152.8 
(
)  = 0.0223 m = 22.3 mm
=
6
 π ( 70 ) 10 


H
(40000)
= 1528 N ⋅ m
(b) T = 9.55 = 9.55
n
250
13
 16T 
d =

 πτ max 
( )
Ans.
13


16(1528) 

= 0.0481 m = 48.1 mm Ans.
d=
 π ( 70 ) 106 


_____________________________________________________________________________
( )
3-58
(a) Obtain the torque from the given power and speed using Eq. (3-42).
63025H 63025(50)
=
= 1261 lbf ⋅ in
T=
n
2500
Tr 16T
τ max = = 3
J πd
 16T 
 16 (1261) 
d =
 =
 = 0.685 in
 πτ max 
 π (20 000) 
63025H 63025(50)
=
= 12610 lbf ⋅ in
(b) T =
n
250
13
13
Ans.
13
16(12 610) 
d=
 = 1.48 in Ans.
 π (20 000) 
_____________________________________________________________________________
Shigley’s MED, 10th edition
Chapter 3 Solutions, Page 43/100
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3-59
τ max
16T
=
π d3
( 50 ) (106 ) π ( 0.033 )
=
= 265 N ⋅ m
16
16
Tn 265(2000)
=
= 55.5 103 W = 55.5 kW Ans.
Eq. (3-44), H =
9.55
9.55
_____________________________________________________________________________
3-60
16T
π
π
τ = 3 ⇒ T = τ d 3 = (110 ) 106 0.0203 = 173 N ⋅ m
16
16
πd
π 

π 0.0204 ( 79.3) 109 15
4

π d Gθ
Tl
 180 
θ=
⇒ l=
=
32T
32(173)
JG
l = 1.89 m Ans.
_____________________________________________________________________________
⇒
T=
τ maxπ d 3
( )
( )(
(
3-61
)
)
( )
16T
π
π
⇒ T = τ d 3 = ( 30 000 ) 0.753 = 2485 lbf ⋅ in
3
16
16
πd
Tl
32Tl
32(2485)(24)
θ=
=
=
= 0.167 rad = 9.57o Ans.
JG π d 4G π 0.754 (11.5 ) 106
(
τ=
(
)
)
( )
_____________________________________________________________________________
3-62
(a) Tsolid =
Jτ max π do4τ max
=
r
16d o
Thollow =
Jτ max π (d o4 − di4 )τ max
=
r
16d o
( )
( )
364
Tsolid − Thollow
di4
% ∆T =
(100%) = 4 (100%) =
(100%) = 65.6%
Tsolid
do
404
(b) Wsolid = kd o2 ,
(
Whollow = k d o 2 − di 2
)
Ans.
( )
( )
362
Wsolid − Whollow
di2
%∆W =
(100%) = 2 (100%) =
(100%) = 81.0%
Wsolid
do
402
Ans.
_____________________________________________________________________________
3-63
4
 4
Jτ max π d 4τ max
Jτ max π  d − ( xd )  τ max
Thollow =
=
=
(a) Tsolid =
r
16d
r
16d
4
T
−T
( xd )
%∆T = solid hollow (100%) =
(100%) = x 4 (100%) Ans.
4
Tsolid
d
Shigley’s MED, 10th edition
Chapter 3 Solutions, Page 44/100
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(
Whollow = k d 2 − ( xd )
(b) Wsolid = kd 2
2
)
Wsolid − Whollow
( xd ) (100%) = x 2 (100%)
(100%) =
Wsolid
d2
2
%∆W =
Ans.
Plot %∆T and %∆W versus x.
Percent Reduction in
Torque and Weight
Percent Reduction
100
80
60
Weight
40
Torque
20
difference
0
0
0.2
0.4
0.6
0.8
1
x
The value of greatest difference in percent reduction of weight and torque is 25% and
occurs at x = 2 2 .
_____________________________________________________________________________
3-64
Tc
(a) τ =
J
( )=
⇒
120 10
( )
4200 ( d 2 )
6
(π
32 )  d 4 − ( 0.70d )


4
=
( )
2.8149 104
d
3
13
 2.8149 104 
 = 6.17 10−2 m = 61.7 mm
d =
 120(106 ) 


From Table A-17, the next preferred size is d = 80 mm.
Ans.
di = 0.7d = 56 mm. The next preferred size smaller is di = 50 mm
(
)
Ans.
(b)
4200 ( d i 2 )
4200 ( 0.050 2 )
Tc
=
=
= 30.8 MPa
Ans.
J (π 32 )  d 4 − ( d )4  (π 32 ) ( 0.080 )4 − ( 0.050 )4 
i




_____________________________________________________________________________
τ=
Shigley’s MED, 10th edition
Chapter 3 Solutions, Page 45/100
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3-65
T = 9.55
H
(1500)
= 9.55
= 1433 N ⋅ m
n
10
13
 16 1433 
(
)  = 0.045 m = 45 mm
=
 π ( 80 ) 106 


From Table A-17, select 50 mm. Ans.
16 ( 2 )(1433)
= 117 (106 ) Pa = 117 MPa Ans.
(a) τ start =
π ( 0.0503 )
16T
τ= 3
π dC
13
 16T 
⇒ dC = 

 πτ 
( )
(b) Design activity
_____________________________________________________________________________
3-66
T=
63 025 H 63 025(1)
=
= 7880 lbf ⋅ in
n
8
16T
τ= 3
π dC
13
⇒
 16T 
dC = 

 πτ 
 16 ( 7880 ) 
=

 π (15 000 ) 
13
= 1.39 in
From Table A-17, select 1.40 in. Ans.
_____________________________________________________________________________
3-67
For a square cross section with side length b, and a circular section with diameter d,
π
π
Asquare = Acircular ⇒ b 2 = d 2
⇒ b=
d
4
2
From Eq. (3-40) with b = c,
(τ max )square =
3
T 
1.8  T  1.8  T  2 
T
3+
(4.8) = 6.896 3
 = 3 3+
= 3

2 
bc 
b/c b 
1  d  π 
d
For the circular cross section,
16T
T
(τ max )circular = 3 = 5.093 3
d
πd
T
(τ max )square 6.896 d 3
=
= 1.354
(τ max )circular 5.093 T
d3
The shear stress in the square cross section is 35.4% greater.
Ans.
(b) For the square cross section, from the table on p. 116, β = 0.141. From Eq. (3-41),
Tl
Tl
Tl
Tl
θsquare =
=
=
= 11.50 4
4
3
4
β bc G β b G
d G
 π 
d G
0.141
 2 
Shigley’s MED, 10th edition
Chapter 3 Solutions, Page 46/100
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For the circular cross section,
Tl
Tl
Tl
θ rd =
=
= 10.19 4
4
GJ G (π d 32 )
d G
Tl
θ sq 11.50 d 4G
=
= 1.129
θ rd 10.19 Tl
4
d G
The angle of twist in the square cross section is 12.9% greater. Ans.
_____________________________________________________________________________
3-68 (a)
T1 = 0.15T2
∑ T = 0 = (500 − 75)(4) − (T
2
1700 − 4.25T2 = 0
⇒
T1 = 0.15 ( 400 ) = 60 lbf
(b)
∑M
O
− T1 )( 5 ) = 1700 − (T2 − 0.15T2 )( 5 )
T2 = 400 lbf
Ans.
Ans.
= 0 = −575(10) + 460(28) − RC (40)
RC = 178.25 lbf
∑F =0=R
O
Ans.
+ 575 − 460 + 178.25
RO = −293.25 lbf
Ans.
(c)
(d) The maximum bending moment is at x = 10 in, and is M = 2932.5 lbf·in. Since the
shaft rotates, each stress element will experience both positive and negative bending
stress as it moves from tension to compression. The torque transmitted through the shaft
Shigley’s MED, 10th edition
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from A to B is T = (500 − 75)(4) = 1700 lbf·in. For a stress element on the outer surface
where the bending stress and the torsional stress are both maximum,
Mc 32 M 32 ( 2932.5 )
=
=
= 15 294 psi = 15.3 kpsi Ans.
I
πd3
π (1.25)3
Tr 16T 16(1700)
τ= = 3=
= 4433 psi = 4.43 kpsi
Ans.
J πd
π (1.25)3
σ=
(e)
σx
2
15.3
2
σ 
 15.3 
±  x  + (τ xy ) =
± 
σ1, σ 2 =
 + ( 4.43)
2
2
 2 
 2 
Ans.
σ 1 = 16.5 kpsi
2
σ 2 = −1.19 kpsi
2
Ans.
2
2
σ 
 15.3 
τ max =  x  + (τ xy ) = 
Ans.
 + ( 4.43) = 8.84 kpsi
 2 
 2 
_____________________________________________________________________________
3-69 (a)
T2 = 0.15T1
2
2
∑ T = 0 = (1800 − 270 ) (200) + (T
2
306 (103 ) − 106.25T1 = 0
⇒
− T1 ) (125) = 306 (103 ) + 125 ( 0.15T1 − T1 )
T1 = 2880 N Ans.
T2 = 0.15 ( 2880 ) = 432 N Ans.
(b)
∑M
O
= 0 = 3312(230) + RC (510) − 2070(810)
RC = 1794 N Ans.
∑F
y
= 0 = RO + 3312 + 1794 − 2070
RO = −3036 N Ans.
(c)
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Chapter 3 Solutions, Page 48/100
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(d) The maximum bending moment is at x = 230 mm, and is M = –698.3 N·m. Since the
shaft rotates, each stress element will experience both positive and negative bending
stress as it moves from tension to compression. The torque transmitted through the shaft
from A to B is T = (1800 − 270)(0.200) = 306 N·m. For a stress element on the outer
surface where the bending stress and the torsional stress are both maximum,
Mc 32 M 32 ( 698.3 )
=
=
= 263 (103 ) Pa = 263 MPa Ans.
I
π d 3 π (0.030)3
Tr 16T
16(306)
τ= = 3=
= 57.7 106 Pa = 57.7MPa
Ans.
J πd
π (0.030)3
σ=
( )
(e)
2
263
2
σ 
 263 
±  x  + (τ xy ) =
± 
 + ( 57.7 )
2
2
2
2


 
σ 1 = 275 MPa
Ans.
σ1 , σ 2 =
σx
2
σ 2 = −12.1 MPa
2
Ans.
2
2
σ 
 263 
τ max =  x  + (τ xy ) = 
 + ( 57.7 ) = 144 MPa
 2 
 2 
2
2
Ans.
_____________________________________________________________________________
3-70
(a)
T2 = 0.15T1
∑ T = 0 = ( 300 − 50 ) (4) + (T
− T1 ) (3) = 1000 + ( 0.15T1 − T1 ) (3)
1000 − 2.55T1 = 0
T1 = 392.16 lbf Ans.
2
⇒
T2 = 0.15 ( 392.16 ) = 58.82 lbf Ans.
(b)
∑M
Oy
= 0 = −450.98(16) − RC z (22)
RC z = −327.99 lbf Ans.
∑F
z
= 0 = RO z + 450.98 − 327.99
RO z = −122.99 lbf Ans.
∑M
Oz
= 0 =350(8) + RC y (22)
RC y = −127.27 lbf Ans.
∑F
y
= 0 = RO y + 350 − 127.27
RO y = −222.73 lbf Ans.
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(c)
(d) Combine the bending moments from both planes at A and B to find the critical
location.
M A = (983.92) 2 + (−1781.84) 2 = 2035 lbf ⋅ in
M B = (1967.84) 2 + (−763.65) 2 = 2111 lbf ⋅ in
The critical location is at B. The torque transmitted through the shaft from A to B is T =
(300 − 50)(4) = 1000 lbf·in. For a stress element on the outer surface where the bending
stress and the torsional stress are both maximum,
Mc 32 M 32 ( 2111)
=
=
= 21 502 psi = 21.5 kpsi Ans.
I
πd3
π (1)3
Tr 16T 16(1000)
τ= = 3=
= 5093 psi = 5.09 kpsi
Ans.
J πd
π (1)3
σ=
(e)
2
21.5
2
σ 
 21.5 
±  x  + (τ xy ) =
± 
 + ( 5.09 )
2
2
 2 
 2 
σ 1 = 22.6 kpsi
Ans.
σ1 , σ 2 =
σx
2
σ 2 = −1.14 kpsi
2
Ans.
2
2
σ 
 21.5 
τ max =  x  + (τ xy ) = 
 + ( 5.09 ) = 11.9 kpsi
2
2


 
2
2
Ans.
_____________________________________________________________________________
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3-71
(a)
T2 = 0.15T1
∑ T = 0 = ( 300 − 45 ) (125) + (T
2
31 875 − 127.5T1 = 0
⇒
− T1 ) (150) = 31 875 + ( 0.15T1 − T1 ) (150)
T1 = 250 N ⋅ mm Ans.
T2 = 0.15 ( 250 ) = 37.5 N ⋅ mm Ans.
(b)
∑M
Oy
= 0 = 345sin 45o (300) − 287.5(700) − RC z (850)
RC z = −150.7 N
∑F
z
= 0 = RO z − 345 cos 45o + 287.5 − 150.7
RO z = 107.2 N
∑M
Oz
y
Ans.
= 0 = 345sin 45o (300) + RC y (850)
RC y = −86.10 N
∑F
Ans.
Ans.
= 0 = RO y + 345 cos 45o − 86.10
RO y = −157.9 N
Ans.
(c)
(d) From the bending moment diagrams, it is clear that the critical location is at A where
both planes have the maximum bending moment. Combining the bending moments from
the two planes,
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( −47.37 ) + ( −32.16 )
2
M=
2
= 57.26 N ⋅ m
The torque transmitted through the shaft from A to B is T = (300 − 45)(0.125) = 31.88
N·m. For a stress element on the outer surface where the bending stress and the torsional
stress are both maximum,
Mc 32M 32 ( 57.26 )
σ=
=
=
= 72.9 (106 ) Pa = 72.9 MPa
Ans.
π d 3 π (0.020)3
I
Tr 16T 16(31.88)
τ= = 3=
= 20.3 (106 ) Pa = 20.3 MPa
Ans.
3
π (0.020)
J πd
(e)
σx
2
72.9
2
σ 
 72.9 
σ 1 , σ 2 = ±  x  + (τ xy ) =
± 
 + ( 20.3)
2
2
 2 
 2 
σ 1 = 78.2 MPa
Ans.
2
σ 2 = −5.27 MPa
2
Ans.
2
2
σ 
 72.9 
τ max =  x  + (τ xy ) = 
 + ( 20.3) = 41.7 MPa
 2 
 2 
2
2
Ans.
_____________________________________________________________________________
3-72
(a)
∑ T = 0 = − 300(cos 20º )(10) + FB (cos 20º )(4)
FB = 750 lbf Ans.
(b)
∑M
Oz
= 0 = 300(cos 20º )(16) − 750(sin 20º )(39) +RC y (30)
RC y = 183.1 lbf Ans.
∑F
y
= 0 =RO y + 300(cos 20º ) + 183.1 − 750(sin 20º )
RO y = −208.5 lbf Ans.
∑M
Oy
= 0 = 300(sin 20º )(16) − RC z (30) − 750(cos 20º )(39)
RC z = −861.5 lbf Ans.
∑F
z
= 0 =RO z − 300(sin 20º ) − 861.5 + 750(cos 20º )
RO z = 259.3 lbf Ans.
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(c)
(d) Combine the bending moments from both planes at A and C to find the critical
location.
M A = (−3336) 2 + (−4149) 2 = 5324 lbf ⋅ in
M C = (−2308) 2 + (−6343)2 = 6750 lbf ⋅ in
The critical location is at C. The torque transmitted through the shaft from A to B is
T = 300 cos ( 20º )(10 ) = 2819 lbf ⋅ in . For a stress element on the outer surface where the
bending stress and the torsional stress are both maximum,
Mc 32 M 32 ( 6750 )
=
=
= 35 203 psi = 35.2 kpsi
Ans.
I
πd3
π (1.25)3
Tr 16T 16(2819)
τ= = 3=
= 7351 psi = 7.35 kpsi
Ans.
J πd
π (1.25)3
σ=
(e)
σx
2
35.2
2
σ 
 35.2 
±  x  + (τ xy ) =
± 
σ1, σ 2 =
 + ( 7.35 )
2
2
 2 
 2 
σ 1 = 36.7 kpsi
Ans.
2
σ 2 = −1.47 kpsi
2
Ans.
2
2
σ 
 35.2 
τ max =  x  + (τ xy ) = 
 + ( 7.35 ) = 19.1 kpsi
 2 
 2 
2
2
Ans.
_____________________________________________________________________________
Shigley’s MED, 10th edition
Chapter 3 Solutions, Page 53/100
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3-73
(a)
∑ T = 0 = −11 000(cos 20º )(300) + F (cos 25º )(150)
B
FB = 22 810 N
(b)
∑M
Oz
= 0 = −11 000(sin 20º )(400) − 22 810(sin 25º )(750) +RC y (1050)
RC y = 8319 N
∑F
y
Ans.
= 0 =RO y − 11000(sin 20º ) − 22 810sin(25º ) + 8319
RO y = 5083 N
∑M
Ans.
Oy
Ans.
= 0 =11 000(cos 20º )(400) − 22 810(cos 25º )(750) − RC z (1050)
RC z = −10 830 N Ans.
∑F
z
= 0 =RO z − 11 000(cos 20º ) + 22 810(cos 25º ) − 10 830
RO z = 494 N
Ans.
(c)
(d) From the bending moment diagrams, it is clear that the critical location is at B where
both planes have the maximum bending moment. Combining the bending moments from
the two planes,
M=
( 2496 ) + ( 3249 )
2
2
= 4097 N ⋅ m
The torque transmitted through the shaft from A to B is
T = 11 000 cos ( 20º )( 0.3) = 3101 N ⋅ m .
For a stress element on the outer surface where the bending stress and the torsional stress
are both maximum,
Shigley’s MED, 10th edition
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Mc 32 M 32 ( 4097 )
=
=
= 333.9 (10 6 ) Pa = 333.9 MPa
Ans.
π d 3 π (0.050)3
I
Tr 16T 16(3101)
τ= = 3=
= 126.3 106 Pa = 126.3 MPa
Ans.
3
J πd
π (0.050)
σ=
( )
(e)
σx
2
333.9
2
σ 
 333.9 
±  x  + (τ xy ) =
± 
σ1 , σ 2 =
 + (126.3)
2
2
 2 
 2 
σ 1 = 376 MPa
Ans.
2
σ 2 = −42.4 MPa
2
Ans.
2
2
σ 
 333.9 
τ max =  x  + (τ xy ) = 
Ans.
 + (126.3) = 209 MPa
 2 
 2 
_____________________________________________________________________________
3-74
(a)
( ΣM D ) z = 6.13Cx − 3.8(92.8) − 3.88(362.8) = 0
2
C x = 287.2 lbf
2
Ans.
( ΣM C ) z = 6.13Dx + 2.33(92.8) − 3.88(362.8) = 0
Dx = 194.4 lbf
3.8
(808) = 500.9 lbf
6.13
2.33
= 0 ⇒ Dz =
(808) = 307.1 lbf
6.13
( ΣM D ) x = 0
( ΣM C ) x
Ans.
⇒ Cz =
Ans.
Ans.
(b) For DQC, let x′, y′, z′ correspond to the original − y, x, z axes.
Shigley’s MED, 10th edition
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(c) The critical stress element is just to the right of Q, where the bending moment in both
planes is a maximum, and where the torsional and axial loads exist.
T = 808(3.88) = 3135 lbf ⋅ in
M = 669.22 + 11672 = 1345 lbf ⋅ in
16T 16(3135)
τ= 3=
= 11 070 psi Ans.
πd
π (1.133 )
σb = ±
32 M
32(1345)
=±
= ± 9495 psi
3
πd
π 1.133
Ans.
σa = −
F
362.8
=−
= −362 psi
A
(π / 4) 1.132
Ans.
(
)
(
)
(d) The critical stress element will be where the bending stress and axial stress are both in
compression.
σ max = −9495 − 362 = −9857 psi
2
τ max
 −9857 
2
= 
 + 11 070 = 12 118 psi = 12.1 kpsi
 2 
Ans.
2
−9857
 −9857 
2
± 
 + 11 070
2
2


σ 1 = 7189 psi = 7.19 kpsi Ans.
σ1,σ 2 =
σ 2 = −17 046 psi = −17.0 kpsi
Ans.
_____________________________________________________________________________
3-75
(a)
( ΣM D ) z = 0
6.13Cx − 3.8(46.6) − 3.88(140) = 0
Cx = 117.5 lbf
Ans.
( ΣM C ) z = 0
−6.13Dx − 2.33(46.6) + 3.88(140) = 0
Dx = 70.9 lbf
3.8
(406) = 251.7 lbf
6.13
2.33
= 0 ⇒ Dz =
(406) = 154.3 lbf
6.13
( ΣM D ) x = 0
( ΣM C ) x
Ans.
⇒ Cz =
Ans.
Ans.
Shigley’s MED, 10th edition
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(b) For DQC, let x′, y′, z′ correspond to the original − y, x, z axes.
(c) The critical stress element is just to the right of Q, where the bending moment in both
planes is a maximum, and where the torsional and axial loads exist.
T = 406(3.88) = 1575 lbf ⋅ in
M = 273.82 + 586.32 = 647.1 lbf ⋅ in
16T 16(1575)
τ= 3=
= 8021 psi Ans.
πd
π (13 )
σb = ±
32 M
32(647.1)
=±
= ± 6591 psi
3
πd
π 13
σa = −
F
140
=−
= −178.3 psi
A
(π / 4) 12
( )
( )
Ans.
Ans.
(d) The critical stress element will be where the bending stress and axial stress are both in
compression.
σ max = −6591 − 178.3 = −6769 psi
2
 −6769 
2
 + 8021 = 8706 psi = 8.71 kpsi
2


τ max = 
Ans.
2
−6769
 −6769 
2
± 
σ1 ,σ 2 =
 + 8021
2
 2 
σ 1 = 5321 psi = 5.32 kpsi Ans.
Shigley’s MED, 10th edition
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σ 2 = −12 090 psi = −12.1 kpsi
Ans.
_____________________________________________________________________________
3-76
( ΣM B ) z = −5.62(362.8) + 1.3(92.8) + 3 Ay = 0
Ay = 639.4 lbf
Ans.
( ΣM A ) z = −2.62(362.8) + 1.3(92.8) + 3By = 0
By = 276.6 lbf
5.62
(808) = 1513.7 lbf
3
2.62
= 0 ⇒ Bz =
(808) = 705.7 lbf
3
( ΣM B ) y = 0
( ΣM A ) y
Ans.
⇒ Az =
Ans.
Ans.
(b)
(c) The critical stress element is just to the left of A, where the bending moment in both
planes is a maximum, and where the torsional and axial loads exist.
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T = 808(1.3) = 1050 lbf ⋅ in
16(1050)
τ=
= 7847 psi Ans.
π 0.883
(
)
M = (829.8) 2 + (2117) 2 = 2274 lbf ⋅ in
32 M
32(2274)
σb = ±
=±
= ±33 990 psi
3
πd
π 0.883
(
σa = −
)
F
92.8
=−
= −153 psi
A
(π / 4) 0.882
(
)
Ans.
Ans.
(d) The critical stress will occur when the bending stress and axial stress are both in
compression.
σ max = −33 990 − 153 = −34 143 psi
2
 −34 143 
2
 + 7847 = 18 789 psi = 18.8 kpsi
2


τ max = 
Ans.
2
−34143
 −34143 
2
± 
 + 7847
2
2 

σ 1 = 1717 psi = 1.72 kpsi Ans.
σ1,σ 2 =
σ 2 = −35 860 psi = −35.9 kpsi
Ans.
_____________________________________________________________________________
3-77
Ft =
T
100
=
= 1600 N
c / 2 0.125 / 2
Fn = 1600 tan 20 = 582.4 N
TC = Ft ( b 2 ) = 1600 ( 0.250 2 ) = 200 N ⋅ m
P=
TC
200
=
= 2667 N
2
0.150
2)
a
( ) (
∑ ( M A )z = 0
450 RDy − 582.4(325) − 2667(75) = 0
RDy = 865.1 N
∑ ( M A ) y = 0 = −450 RDz + 1600(325) ⇒ RDz = 1156 N
∑ Fy = 0 = RAy + 865.1 − 582.4 − 2667 ⇒ RAy = 2384 N
∑ Fz = 0 = RAz + 1156 − 1600 ⇒ RAz = 444 N
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Chapter 3 Solutions, Page 59/100
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AB The maximum bending moment will either be at B or C. If this is not obvious, sketch
the shear and bending moment diagrams. We will directly obtain the combined moments
from each plane.
M B = AB RA2 y + RA2z = 0.075 23842 + 4442 = 181.9 N ⋅ m
M C = CD RD2 y + RD2 z = 0.125 865.12 + 11562 = 180.5 N ⋅ m
The stresses at B and C are almost identical, but the maximum stresses occur at B.
32 M B 32(181.9)
σB =
=
= 68.6 106 Pa = 68.6 MPa
3
3
πd
π 0.030
(
τB =
)
Ans.
( )
( )
16TB
16(200)
=
= 37.7 106 Pa = 37.7 MPa
3
3
πd
π 0.030
σ max =
(
)
2
σB
2
68.6
σ 
 68.6 
2
+  B  + τ B2 =
+ 
 + 37.7 = 85.3 MPa
2
2
2
2




2
Ans.
2
σ 
 68.6 
2
τ max =  B  + τ B2 = 
 + 37.7 = 51.0 MPa Ans.
 2 
 2 
_____________________________________________________________________________
3-78
Ft =
T
100
=
= 1600 N
c / 2 0.125 / 2
Fn = 1600 tan 20 = 582.4 N
TC = Ft ( b 2 ) = 1600 ( 0.250 2 ) = 200 N ⋅ m
P=
TC
200
=
= 2667 N
( a 2 ) ( 0.150 2 )
⇒ RDy = 420.6 N
∑ ( M A ) z = 0 = 450 RDy − 582.4(325)
∑ ( M A ) y = 0 = −450 RDz + 1600(325) − 2667(75) ⇒ RDz = 711.1 N
∑ Fy = 0 = RAy + 420.6 − 582.4
∑ Fz = 0 = RAz + 711.1 − 1600 + 2667
⇒ RAy = 161.8 N
⇒ RAz = −1778 N
Shigley’s MED, 10th edition
Chapter 3 Solutions, Page 60/100
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The maximum bending moment will either be at B or C. If this is not obvious, sketch
shear and bending moment diagrams. We will directly obtain the combined moments
from each plane.
M B = AB RA2 y + RA2z = 0.075 161.82 + ( −1778 ) = 133.9 N ⋅ m
2
M C = CD RD2 y + RD2 z = 0.125 420.62 + 711.12 = 103.3 N ⋅ m
The maximum stresses occur at B. Ans.
32M B 32(133.9)
σB =
=
= 50.5 106 Pa = 50.5 MPa
3
3
πd
π 0.030
(
τB =
)
( )
( )
16TB
16(200)
=
= 37.7 106 Pa = 37.7 MPa
3
3
πd
π 0.030
σ max =
(
)
2
σB
2
50.5
σ 
 50.5 
2
+  B  + τ B2 =
+ 
 + 37.7 = 70.6 MPa
2
2
2
2




2
Ans.
2
σ 
 50.5 
2
τ max =  B  + τ B2 = 
 + 37.7 = 45.4 MPa Ans.
 2 
 2 
_____________________________________________________________________________
3-79
T
900
=
= 180 lbf
c / 2 10 / 2
Fn = 180 tan 20 = 65.5 lbf
Ft =
TC = Ft ( b 2 ) = 180 ( 5 2 ) = 450 lbf ⋅ in
P=
TC
450
=
= 150 lbf
( a 2) ( 6 2)
∑ ( M A ) z = 0 = 20 RDy − 65.5(14) − 150(4) ⇒ RDy = 75.9 lbf
⇒ RDz = 126 lbf
∑ ( M A ) y = 0 = −20 RDz + 180(14)
∑ Fy = 0 = RAy + 75.9 − 65.5 − 150
∑ Fz = 0 = RAz + 126 − 180
⇒ RAy = 140 lbf
⇒ RAz = 54.0 lbf
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Chapter 3 Solutions, Page 61/100
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The maximum bending moment will either be at B or C. If this is not obvious, sketch
shear and bending moment diagrams. We will directly obtain the combined moments
from each plane.
M B = AB RA2 y + RA2z = 4 1402 + 542 = 600 lbf ⋅ in
M C = CD RD2 y + RD2 z = 6 75.92 + 1262 = 883 lbf ⋅ in
The maximum stresses occur at C.
σC =
32 M C
τC =
16TC
πd
πd
=
3
3
σ max =
=
32(883)
(
π 1.3753
16(450)
(
π 1.3753
)
)
Ans.
= 3460 psi
= 882 psi
2
σC
2
3460
σ 
 3460 
2
+  C  + τ C2 =
+ 
 + 882 = 3670 psi
2
2
 2 
 2 
2
Ans.
2
σ 
 3460 
2
τ max =  C  + τ C2 = 
 + 882 = 1940 psi
2
2




Ans.
_____________________________________________________________________________
3-80
(a) Rod AB experiences constant torsion throughout its length, and maximum bending
moment at the wall. Both torsional shear stress and bending stress will be a maximum on
the outer surface. The transverse shear will be very small compared to bending and
torsion, due to the reasonably high length to diameter ratio, so it will not dominate the
determination of the critical location. The critical stress element will be at the wall, at
either the top (compression) or the bottom (tension) on the y axis. We will select the
bottom element for this analysis.
(b) Transverse shear is zero at the critical stress elements on the top and bottom surfaces.
Mc M ( d / 2 ) 32 M 32 ( 8 )( 200 )
σx =
=
=
=
= 16 297 psi = 16.3 kpsi
3
I
π d 4 / 64 π d 3
π (1)
τ xz =
Tr T ( d / 2 ) 16T 16 ( 5 )( 200 )
=
=
=
= 5093 psi = 5.09 kpsi
3
J π d 4 / 32 π d 3
π (1)
Shigley’s MED, 10th edition
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(c)
σx
16.3
2
2
σ 
 16.3 
σ1, σ 2 =
±  x  + (τ xz ) =
± 
 + ( 5.09 )
2
2
 2 
 2 
σ 1 = 17.8 kpsi
Ans.
2
σ 2 = −1.46 kpsi
2
Ans.
2
2
σ 
 16.3 
τ max =  x  + (τ xz ) = 
 + ( 5.09 ) = 9.61 kpsi
 2 
 2 
2
2
Ans.
_____________________________________________________________________________
3-81
(a) Rod AB experiences constant torsion throughout its length, and maximum bending
moments at the wall in both planes of bending. Both torsional shear stress and bending
stress will be a maximum on the outer surface. The transverse shear will be very small
compared to bending and torsion, due to the reasonably high length to diameter ratio, so
it will not dominate the determination of the critical location. The critical stress element
will be on the outer surface at the wall, with its critical location determined by the plane
of the combined bending moments.
My = – (100)(8) = – 800 lbf·in
Mz = (175)(8) = 1400 lbf·in
M tot = M y2 + M z2
=
( −800 )
2
+ 14002 = 1612 lbf ⋅ in
 My 
−1  800 
 = tan 
 = 29.7º
M
1400


z


The combined bending moment vector is at an angle of 29.7º CCW from the z axis. The
critical bending stress location, and thus the critical stress element, will be ±90º from this
vector, as shown. There are two equally critical stress elements, one in tension (119.7º
CCW from the z axis) and the other in compression (60.3º CW from the z axis). We’ll
continue the analysis with the element in tension.
(b) Transverse shear is zero at the critical stress elements on the outer surfaces.
M c M ( d / 2 ) 32M tot 32 (1612 )
σ x = tot = tot 4
=
=
= 16 420 psi = 16.4 kpsi
3
π d / 64
πd3
I
π (1)
θ = tan −1 
τ=
Tr T ( d / 2 ) 16T 16 ( 5 )(175 )
=
=
=
= 4456 psi = 4.46 kpsi
3
J π d 4 / 32 π d 3
π (1)
Shigley’s MED, 10th edition
Chapter 3 Solutions, Page 63/100
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(c)
σx
16.4
2
σ 
 16.4 
σ1, σ 2 =
±  x  +τ 2 =
± 
 + ( 4.46 )
2
2
 2 
 2 
σ 1 = 17.5 kpsi
Ans.
2
σ 2 = −1.13 kpsi
2
Ans.
2
σ 
 16.4 
τ max =  x  + τ 2 = 
 + ( 4.46 ) = 9.33 kpsi
 2 
 2 
2
2
Ans.
_____________________________________________________________________________
3-82
(a) Rod AB experiences constant torsion and constant axial tension throughout its length,
and maximum bending moments at the wall from both planes of bending. Both torsional
shear stress and bending stress will be maximum on the outer surface. The transverse
shear will be very small compared to bending and torsion, due to the reasonably high
length to diameter ratio, so it will not dominate the determination of the critical location.
The critical stress element will be on the outer surface at the wall, with its critical
location determined by the plane of the combined bending moments.
My = – (100)(8) – (75)(5) = – 1175 lbf·in
Mz = (–200)(8) = –1600 lbf·in
M tot = M y2 + M z2
=
( −1175 ) + ( −1600 )
2
2
= 1985 lbf ⋅ in
 My 
−1  1175 
 = tan 
 = 36.3º
 1600 
 Mz 
The combined bending moment vector is at an angle of 36.3º CW from the negative z
axis. The critical bending stress location will be ±90º from this vector, as shown. Since
there is an axial stress in tension, the critical stress element will be where the bending is
also in tension. The critical stress element is therefore on the outer surface at the wall, at
an angle of 36.3º CW from the y axis.
(b) Transverse shear is zero at the critical stress element on the outer surface.
M c M ( d / 2 ) 32M tot 32 (1985 )
σ x ,bend = tot = tot 4
=
=
= 20 220 psi = 20.2 kpsi
3
π d / 64
πd3
I
π (1)
θ = tan −1 
σ x ,axial =
Fx
Fx
75
=
=
= 95.5 psi = 0.1 kpsi , which is essentially negligible
2
A π d / 4 π (1) 2 / 4
σ x = σ x ,axial + σ x ,bend = 20 220 + 95.5 = 20 316 psi = 20.3 kpsi
τ=
Tr 16T 16 ( 5 )( 200 )
=
=
= 5093 psi = 5.09 kpsi
3
J πd3
π (1)
Shigley’s MED, 10th edition
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(c)
20.3
2
σ 
 20.3 
±  x  +τ 2 =
± 
 + ( 5.09 )
2
2
2
2


 
σ 1 = 21.5 kpsi
Ans.
σ1, σ 2 =
σx
2
σ 2 = −1.20 kpsi
2
Ans.
2
σ 
 20.3 
τ max =  x  + τ 2 = 
Ans.
 + ( 5.09 ) = 11.4 kpsi
 2 
 2 
_____________________________________________________________________________
3-83
T = (2)(200) = 400 lbf·in
The maximum shear stress due to torsion occurs in the middle of the longest side of the
rectangular cross section. From the table on p. 116, with b/c = 1.5/0.25 = 6, α = 0.299.
From Eq. (3-40),
400
T
τ max =
=
= 14 270 psi = 14.3 kpsi
Ans.
2
α bc ( 0.299 )(1.5 )( 0.25 )2
2
2
____________________________________________________________________________
3-84
(a) The cross section at A will experience bending, torsion, and transverse shear. Both
torsional shear stress and bending stress will be a maximum on the outer surface. The
transverse shear will be very small compared to bending and torsion, due to the
reasonably high length to diameter ratio, so it will not dominate the determination of the
critical location. The critical stress element will be at either the top (compression) or the
bottom (tension) on the y axis. We’ll select the bottom element for this analysis.
(b) Transverse shear is zero at the critical stress elements on the top and bottom surfaces.
Mc M ( d / 2 ) 32 M 32 (11)( 250 )
σx =
=
=
=
= 28 011 psi = 28.0 kpsi
3
π d 4 / 64 π d 3
I
π (1)
τ xz =
Tr T ( d / 2 ) 16T 16 (12 )( 250 )
=
=
=
= 15 279 psi = 15.3 kpsi
3
J π d 4 / 32 π d 3
π (1)
Shigley’s MED, 10th edition
Chapter 3 Solutions, Page 65/100
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(c)
σx
28.0
2
2
σ 
 28.0 
σ1 , σ 2 =
±  x  + (τ xz ) =
± 
 + (15.3)
2
2
 2 
 2 
σ 1 = 34.7 kpsi
Ans.
2
σ 2 = −6.7 kpsi
2
Ans.
2
2
σ 
 28.0 
τ max =  x  + (τ xz ) = 
 + (15.3) = 20.7 kpsi
 2 
 2 
2
2
Ans.
____________________________________________________________________________
3-85
(a) The cross section at A will experience bending, torsion, axial, and transverse shear.
Both torsional shear stress and bending stress will be a maximum on the outer surface.
The transverse shear will be very small compared to bending and torsion, due to the
reasonably high length to diameter ratio, so it will not dominate the determination of the
critical location. The critical stress element will be on the outer surface, with its critical
location determined by the plane of the combined bending moments.
My = (300)(12) = 3600 lbf·in
Mz = (250)(11) = 2750 lbf·in
M tot = M y2 + M z2
=
( 3600 ) + ( 2750 )
2
 M
z
 My

θ = tan −1 
2
= 4530 lbf ⋅ in

 2750 
 = tan −1 
 = 37.4º

 3600 

The combined bending moment vector is at an angle of 37.4º CCW from the y axis. The
critical bending stress location will be 90º CCW from this vector, where the tensile
bending stress is additive with the tensile axial stress. The critical stress element is
therefore on the outer surface, at an angle of 37.4º CCW from the z axis.
(b)
M c M ( d / 2 ) 32 M tot 32 ( 4530 )
σ x ,bend = tot = tot 4
=
=
= 46 142 psi = 46.1 kpsi
3
π d / 64
πd3
I
π (1)
σ x ,axial =
Fx
Fx
300
=
=
= 382 psi = 0.382 kpsi
2
A π d / 4 π (1)2 / 4
σ x = σ x ,axial + σ x ,bend = 46 142 + 382 = 46 524 psi = 46.5 kpsi
τ=
Tr 16T 16 (12 )( 250 )
=
=
= 15 279 psi = 15.3 kpsi
3
J πd3
π (1)
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(c)
46.5
2
σ 
 46.5 
±  x  +τ 2 =
± 
 + (15.3)
2
2
2
2


 
σ 1 = 51.1 kpsi
Ans.
σ1, σ 2 =
σx
2
σ 2 = −4.58 kpsi
2
Ans.
2
σ 
 46.5 
=  x  +τ 2 = 
 + (15.3) = 27.8 kpsi
 2 
 2 
2
τ max
2
Ans.
____________________________________________________________________________
3-86
(a) The cross section at A will experience bending, torsion, axial, and transverse shear.
Both torsional shear stress and bending stress will be a maximum on the outer surface.
The transverse shear will be very small compared to bending and torsion, due to the
reasonably high length to diameter ratio, so it will not dominate the determination of the
critical location. The critical stress element will be on the outer surface, with its critical
location determined by the plane of the combined bending moments.
My = (300)(12) – (–100)(11) = 4700 lbf·in
Mz = (250)(11) = 2750 lbf·in
M tot = M y2 + M z2
=
( 4700 ) + ( 2750 )
2
 M
z
 My

θ = tan −1 
2
= 5445 lbf ⋅ in

 2750 
 = tan −1 
 = 30.3º

 4700 

The combined bending moment vector is at an angle of 30.3º CCW from the y axis. The
critical bending stress location will be 90º CCW from this vector, where the tensile
bending stress is additive with the tensile axial stress. The critical stress element is
therefore on the outer surface, at an angle of 30.3º CCW from the z axis.
(b)
M c M ( d / 2 ) 32 M tot 32 ( 5445 )
σ x ,bend = tot = tot 4
=
=
= 55 462 psi = 55.5 kpsi
3
I
π d / 64
πd3
π (1)
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σ x ,axial =
Fx
Fx
300
=
=
= 382 psi = 0.382 kpsi
A π d 2 / 4 π (1)2 / 4
σ x = σ x ,axial + σ x ,bend = 55 462 + 382 = 55 844 psi = 55.8 kpsi
τ=
Tr 16T 16 (12 )( 250 )
=
=
= 15 279 psi = 15.3 kpsi
3
J πd3
π (1)
(c)
55.8
2
σ 
 55.8 
±  x  +τ 2 =
± 
 + (15.3)
2
2
 2 
 2 
σ 1 = 59.7 kpsi
Ans.
σ1 , σ 2 =
σx
2
σ 2 = −3.92 kpsi
2
Ans.
2
σ 
 55.8 
=  x  +τ 2 = 
 + (15.3) = 31.8 kpsi
 2 
 2 
2
τ max
2
Ans.
____________________________________________________________________________
3-87
(a) The cross section at A will experience bending, torsion, and transverse shear. Both
torsional shear stress and bending stress will be a maximum on the outer surface, where
the stress concentration will also be applicable. The transverse shear will be very small
compared to bending and torsion, due to the reasonably high length to diameter ratio, so
it will not dominate the determination of the critical location. The critical stress element
will be at either the top (compression) or the bottom (tension) on the y axis. We’ll select
the bottom element for this analysis.
(b) Transverse shear is zero at the critical stress elements on the top and bottom surfaces.
r / d = 0.125 /1 = 0.125
D / d = 1.5 /1 = 1.5
Fig. A-15-8
K t ,torsion = 1.39
K t ,bend = 1.59
σ x = K t ,bend
Fig. A-15-9
32 (11)( 250 )
32 M
Mc
= K t ,bend
= (1.59)
= 44 538 psi = 44.5 kpsi
3
3
I
πd
π (1)
τ xz = K t ,torsion
16 (12 )( 250 )
Tr
16T
= K t ,torsion
= (1.39)
= 21 238 psi = 21.2 kpsi
3
3
J
πd
π (1)
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(c)
44.5
2
2
σ 
 44.5 
±  x  + (τ xz ) =
± 
 + ( 21.2 )
2
2
 2 
 2 
Ans.
σ 1 = 53.0 kpsi
σ1, σ 2 =
σx
2
σ 2 = −8.48 kpsi
2
Ans.
2
2
σ 
 44.5 
τ max =  x  + (τ xz ) = 
 + ( 21.2 ) = 30.7 kpsi
2
2


 
2
2
Ans.
____________________________________________________________________________
3-88
(a) The cross section at A will experience bending, torsion, axial, and transverse shear.
Both torsional shear stress and bending stress will be a maximum on the outer surface,
where the stress concentration will also be applicable. The transverse shear will be very
small compared to bending and torsion, due to the reasonably high length to diameter
ratio, so it will not dominate the determination of the critical location. The critical stress
element will be on the outer surface, with its critical location determined by the plane of
the combined bending moments.
My = (300)(12) = 3600 lbf·in
Mz = (250)(11) = 2750 lbf·in
M tot = M y2 + M z2
=
( 3600 ) + ( 2750 )
2
 M
z
 My

θ = tan −1 
2
= 4530 lbf ⋅ in

 2750 
 = tan −1 
 = 37.4º

3600



The combined bending moment vector is at an angle of 37.4º CCW from the y axis. The
critical bending stress location will be 90º CCW from this vector, where the tensile
bending stress is additive with the tensile axial stress. The critical stress element is
therefore on the outer surface, at an angle of 37.4º CCW from the z axis.
(b)
r / d = 0.125 /1 = 0.125
D / d = 1.5 /1 = 1.5
Fig. A-15-7
K t , axial = 1.75
K t ,torsion = 1.39
Fig. A-15-8
K t ,bend = 1.59
Fig. A-15-9
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σ x ,bend = K t ,bend
32 ( 4530 )
Mc
32 M
= K t ,bend
= (1.59)
= 73 366 psi = 73.4 kpsi
3
3
I
πd
π (1)
σ x ,axial = K t ,axial
Fx
300
= (1.75 )
= 668 psi = 0.668 kpsi
2
A
π (1) / 4
σ x = σ x ,axial + σ x ,bend = 73 366 + 668 = 74 034 psi = 74.0 kpsi
τ = K t ,torsion
16 (12 )( 250 )
Tr
16T
= K t ,torsion
= (1.39)
= 21 238 psi = 21.2 kpsi
3
3
J
πd
π (1)
(c)
σx
74.0
2
σ 
 74.0 
± 
σ1 , σ 2 = ±  x  + τ 2 =
 + ( 21.2 )
2
2
 2 
 2 
Ans.
σ 1 = 79.6 kpsi
2
σ 2 = −5.64 kpsi
2
Ans.
2
σ 
 74.0 
=  x  +τ 2 = 
 + ( 21.2 ) = 42.6 kpsi
 2 
 2 
2
τ max
2
Ans.
____________________________________________________________________________
3-89
(a) The cross section at A will experience bending, torsion, axial, and transverse shear.
Both torsional shear stress and bending stress will be a maximum on the outer surface,
where the stress concentration is also applicable. The transverse shear will be very small
compared to bending and torsion, due to the reasonably high length to diameter ratio, so
it will not dominate the determination of the critical location. The critical stress element
will be on the outer surface, with its critical location determined by the plane of the
combined bending moments.
My = (300)(12) – (–100)(11) = 4700 lbf·in
Mz = (250)(11) = 2750 lbf·in
M tot = M y2 + M z2
=
( 4700 ) + ( 2750 )
2
 M
z
 My

θ = tan −1 
2
= 5445 lbf ⋅ in

 2750 
 = tan −1 
 = 30.3º

 4700 

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The combined bending moment vector is at an angle of 30.3º CCW from the y axis. The
critical bending stress location will be 90º CCW from this vector, where the tensile
bending stress is additive with the tensile axial stress. The critical stress element is
therefore on the outer surface, at an angle of 30.3º CCW from the z axis.
(b)
r / d = 0.125 /1 = 0.125
D / d = 1.5 /1 = 1.5
Fig. A-15-7
K t , axial = 1.75
K t ,torsion = 1.39
Fig. A-15-8
K t ,bend = 1.59
Fig. A-15-9
σ x ,bend = K t ,bend
32 ( 5445 )
Mc
32 M
= K t ,bend
= (1.59)
= 88185 psi = 88.2 kpsi
3
3
I
πd
π (1)
σ x ,axial = K t ,axial
Fx
300
= (1.75 )
= 668 psi = 0.668 kpsi
2
A
π (1) / 4
σ x = σ x ,axial + σ x ,bend = 88 185 + 668 = 88 853 psi = 88.9 kpsi
τ = K t ,torsion
16 (12 )( 250 )
16T
Tr
= K t ,torsion
= (1.39)
= 21 238 psi = 21.2 kpsi
3
3
J
πd
π (1)
(c)
88.9
2
σ 
 88.9 
±  x  +τ 2 =
± 
 + ( 21.2 )
2
2
 2 
 2 
σ 1 = 93.7 kpsi
Ans.
σ1 , σ 2 =
σx
2
σ 2 = −4.80 kpsi
2
Ans.
2
σ 
 88.9 
τ max =  x  + τ 2 = 
Ans.
 + ( 21.2 ) = 49.2 kpsi
 2 
 2 
____________________________________________________________________________
2
3-90
2
(a) M = F(p / 4), c = p / 4, I = bh3 / 12, b = π dr nt, h = p / 2
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σb = ±
 F ( p / 4 )  ( p / 4 )
Mc
Fp 2
=±
=
±
3
I
bh3 /12
16 (π d r nt )( p / 2 ) / 12
6F
Ans.
π d r nt p
F
F
4F
(b) σ a = − = − 2
=− 2
π dr / 4
π dr
A
σb = ±
Ans.
Tr T ( d r / 2 ) 16T
=
=
Ans.
J
π d r4 / 32 π d r3
(c) The bending stress causes compression in the x direction. The axial stress causes
compression in the y direction. The torsional stress shears across the y face in the negative z
direction.
τt =
(d) Analyze the stress element from part (c) using the equations developed in parts (a) and (b).
d r = d − p = 1.5 − 0.25 = 1.25 in
σ x = σb = −
σ y = σa = −
6 (1500 )
6F
=−
= − 4584 psi = − 4.584 kpsi
π d r nt p
π (1.25 )( 2 )( 0.25 )
4 (1500 )
4F
=−
= − 1222 psi = − 1.222 kpsi
2
π dr
π (1.252 )
τ yz = −τ t = −
16 ( 235 )
16T
=−
= − 612.8 psi = − 0.6128 kpsi
3
π dr
π (1.253 )
Use Eq. (3-15) for the three-dimensional stress element.
2
2
σ 3 − ( −4.584 − 1.222 ) σ 2 + ( −4.584 )( −1.222 ) − ( −0.6128 )  σ −  − ( −4.584 )( −0.6128 )  = 0

σ + 5.806σ + 5.226σ − 1.721 = 0
3



2
The roots are at 0.2543, – 4.584, and –1.476. Thus, the ordered principal stresses are
σ1 = 0.2543 kpsi, σ2 = –1.476 kpsi, and σ3 = – 4.584 kpsi.
Ans.
From Eq. (3-16), the principal shear stresses are
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0.2543 − ( −1.476 )
= 0.8652 kpsi Ans.
2
2
σ − σ 3 ( −1.476 ) − ( −4.584 )
τ 2/3 = 2
Ans.
=
= 1.554 kpsi
2
2
0.2543 − ( −4.584 )
σ −σ
Ans.
τ 1/3 = 1 3 =
= 2.419 kpsi
2
2
____________________________________________________________________________
τ 1/2 =
3-91
σ1 − σ 2
=
As shown in Fig. 3-32, the maximum stresses occur at the inside fiber where r = ri.
Therefore, from Eq. (3-50)
ri2 pi  ro2 
1 + 
ro2 − ri2  ri2 
 r2 + r2 
= pi  o2 i2  Ans.
 r −r 
 o i 
r2 p  r2 
σ r ,max = 2i i 2 1 − o2  = − pi Ans.
ro − ri  ri 
______________________________________________________________________________
σ t ,max =
3-92
If pi = 0, Eq. (3-49) becomes
σt =
− po ro2 − ri 2 ro2 po / r 2
ro2 − ri 2
po ro2  ri 2 
1 + 
ro2 − ri 2  r 2 
The maximum tangential stress occurs at r = ri. So
=−
σ t ,max = −
2 po ro2
Ans.
ro2 − ri 2
For σr, we have
σr =
− po ro2 − ri 2 ro2 po / r 2
ro2 − ri 2
po ro2  ri 2 
 − 1
ro2 − ri 2  r 2 
So σr = 0 at r = ri. Thus at r = ro
p r2  r2 − r2 
σ r ,max = 2 o o 2  i 2 o  = − po Ans.
ro − ri  ro 
______________________________________________________________________________
=
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3-93
The force due to the pressure on half of the sphere is resisted by the stress that is
distributed around the center plane of the sphere. All planes are the same, so
(σ t )av = σ 1 = σ 2 =
p (π / 4 ) di2
π di t
=
pdi
4t
Ans.
The radial stress on the inner surface of the shell is, σ3 = − p Ans.
______________________________________________________________________________
3-94
σt > σl > σr
τmax = (σt − σr)/2 at r = ri
ro2 pi
1  ri 2 pi  ro2  ri 2 pi  ro2  
+
−
−
=
1
1





2  ro2 − ri 2  ri 2  ro2 − ri 2  ri 2   ro2 − ri 2
τ max = 
ro2 − ri 2
32 − 2.752
τ
=
(10 000) = 1597 psi Ans.
max
ro 2
32
______________________________________________________________________________
⇒
3-95
pi =
σt > σl > σr
τmax = (σt − σr)/2 at r = ri
r2 p  r2 
r2 p
1  r 2 p  r 2  r 2 p  r 2 
τ max =  2i i 2 1 + o2  − 2i i 2 1 − o2   = 2i i 2  o2  = 2o i 2
2  ro − ri  ri  ro − ri  ri   ro − ri  ri  ro − ri
⇒
ri = ro
(τ max − pi )
= 100
τ max
(25 − 4)106
= 91.7 mm
25 (106 )
t = ro − ri = 100 − 91.7 = 8.3 mm Ans.
______________________________________________________________________________
3-96
σt > σl > σr
τmax = (σt − σr)/2 at r = ri
1  ri 2 pi  ro2
1 +
2  ro2 − ri 2  ri 2
τ max = 
 ri 2 pi  ro2
 − 2 2 1 − 2
 ro − ri  ri

ri 2 pi  ro2
=

2
2  2
  ro − ri  ri

ro2 pi
=

2
2
 ro − ri
42 (500)
= 4129 psi Ans.
42 − 3.752
______________________________________________________________________________
=
3-97
From Eq. (3-49) with pi = 0,
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σt = −
ro2 po  ri 2 
1 + 
ro2 − ri 2  r 2 
ro2 po  ri 2 
1 − 
ro2 − ri 2  r 2 
σt > σl > σr, and since σt and σr are negative,
τmax = (σr − σt)/2 at r = ro
σr = −
r2 p  r2  r2 p
1  ro2 po  ri 2  ro2 po  ri 2  
1 − 2  + 2 2 1 + 2   = 2o o 2  i 2  = 2i o 2
2
2 
2  ro − ri  ro  ro − ri  ro   ro − ri  ro  ro − ri
τ max =  −
ro2 − ri 2
32 − 2.752
τ max =
(10 000) = 1900 psi Ans.
2
ri
2.752
______________________________________________________________________________
⇒
3-98
po =
From Eq. (3-49) with pi = 0,
r2 p  r2 
σ t = − 2o o 2 1 + i 2 
ro − ri  r 
σr = −
ro2 po  ri 2 
1 − 
ro2 − ri 2  r 2 
σt > σl > σr, and since σt and σr are negative,
τmax = (σr − σt)/2 at r = ro
1  ro2 po  ri 2  ro2 po  ri 2   ro2 po  ri 2  ri 2 po
1 −  +
1 +   =
 =
2  ro2 − ri 2  ro 2  ro2 − ri 2  ro 2   ro2 − ri 2  ro 2  ro2 − ri 2
τ max =  −
⇒
τ max
ri = ro
(τ max + po )
= 100
25 (106 )
( 25 + 4 )106
= 92.8 mm
t = ro − ri = 100 − 92.8 = 7.2 mm Ans.
______________________________________________________________________________
3-99
From Eq. (3-49) with pi = 0,
σt = −
ro2 po  ri 2 
1 + 
ro2 − ri 2  r 2 
σr = −
ro2 po  ri 2 
1 − 
ro2 − ri 2  r 2 
σt > σl > σr, and since σt and σr are negative,
τmax = (σr − σt)/2 at r = ro
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ro2 po  ri 2  ri 2 po
1  ro2 po  ri 2  ro2 po  ri 2  
1
1
−
+
+
=




 =
2  ro2 − ri 2  ro 2  ro2 − ri 2  ro 2   ro2 − ri 2  ro 2  ro2 − ri 2
τ max =  −
3.752 (500)
= 3629 psi Ans.
42 − 3.752
______________________________________________________________________________
=
3-100 From Table A-20, Sy=490 MPa
From Eq. (3-49) with pi = 0,
σt = −
ro2 po  ri 2 
1 + 
ro2 − ri 2  r 2 
Maximum will occur at r = ri
[0.8(−490)] ( 25 − 19 )
σ (r − r )
2ro2 po
⇒ po = − t ,max o2 i = −
= 82.8 MPa Ans.
2
2
ro − ri
2ro
2(252 )
______________________________________________________________________________
2
σ t ,max = −
2
2
2
3-101 From Table A-20, Sy = 71 kpsi
From Eq. (3-49) with pi = 0,
σt = −
ro2 po  ri 2 
1 + 
ro2 − ri 2  r 2 
Maximum will occur at r = ri
σ t ,max ( ro − ri )
[ 0.8(−71)] (1 − 0.75 )
2r 2 p
σ t ,max = − 2 o o2 ⇒ po = −
=−
= 12.4 kpsi Ans.
2
ro − ri
2ro
2(12 )
______________________________________________________________________________
2
2
2
2
3-102 From Table A-20, Sy=490 MPa
From Eq. (3-50)
σt =
ri 2 pi  ro2 
1 + 
ro2 − ri 2  r 2 
Maximum will occur at r = ri
σ t ,max =
⇒
2
2
ri 2 pi  ro2  pi ( ro + ri )
1
+
=


ro2 − ri 2  ri 2 
ro2 − ri 2
pi =
σ t ,max (ro2 − ri 2 )
=
[ 0.8(490)] (252 − 192 ) = 105 MPa
Ans.
ro2 + ri 2
(252 + 192 )
______________________________________________________________________________
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3-103 From Table A-20, Sy =71 MPa
From Eq. (3-50)
σt =
ri 2 pi  ro2 
1 + 
ro2 − ri 2  r 2 
Maximum will occur at r = ri
r 2 p  r 2  p (r 2 + r 2 )
σ t ,max = 2i i 2 1 + o2  = i 2o 2i
ro − ri  ri 
ro − ri
⇒
pi =
σ t ,max (ro2 − ri 2 )
=
[ 0.8(71)] (12 − 0.752 ) = 15.9 ksi
Ans.
ro2 + ri 2
(12 + 0.752 )
______________________________________________________________________________
3-104 The longitudinal stress will be due to the weight of the vessel above the maximum stress
point. From Table A-5, the unit weight of steel is γs = 0.282 lbf/in3. The area of the wall
is
Awall = (π /4)(3602 − 358.52) = 846. 5 in2
The volume of the wall and dome are
Vwall = Awall h = 846.5 (720) = 609.5 (103) in3
Vdome = (2π /3)(1803 − 179.253) = 152.0 (103) in3
The weight of the structure on the wall area at the tank bottom is
W = γs Vtotal = 0.282(609.5 +152.0) (103) = 214.7(103) lbf
214.7 (103 )
W
σl = −
=−
= −254 psi
846.5
Awall
The maximum pressure will occur at the bottom of the tank, pi = γwater h. From Eq. (3-50)
with r = ri
 ro2 + ri 2 
ri 2 pi  ro2 
p
+
=
1



i
2
2 
ro2 − ri 2  ri 2 
 ro − ri 

 1 ft 2    1802 + 179.252 
= 62.4(55) 
= 5708 5710 psi Ans.
2  
2
2 
 144 in    180 − 179.25 

 1 ft 2 
r2 p  r2 
= −23.8 psi Ans.
σ r = 2i i 2 1 − o2  = − pi = −62.4(55) 
2 
ro − ri  ri 
 144 in 
Note: These stresses are very idealized as the floor of the tank will restrict the values
calculated.
σt =
Since σ1 ≥ σ2 ≥ σ3, σ1 = σt = 5708 psi, σ2 = σr = − 24 psi andσ3 = σl = − 254 psi. From
Eq. (3-16),
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5708 + 254
= 2981 ≈ 2980 psi
2
5708 + 24
τ1 2 =
= 2866 ≈ 2870 psi
Ans.
2
−24 + 254
τ2 3 =
= 115 psi
2
______________________________________________________________________________
τ1 3 =
3-105 Stresses from additional pressure are,
Eq. (3-51),
50 (179.252 )
= 5963 psi
(σ l )50psi = 2
180 − 179.252
(σr)50 psi = − 50 psi
Eq. (3-50)
1802 + 179.252
= 11 975 psi
(σ t )50psi = 50 2
180 − 179.252
Adding these to the stresses found in Prob. 3-104 gives
σt = 5708 + 11 975 = 17683 psi = 17.7 kpsi Ans.
σr = − 23.8 − 50 = − 73.8 psi Ans.
σl = − 254 + 5963 = 5709 psi Ans.
Note: These stresses are very idealized as the floor of the tank will restrict the values
calculated.
From Eq. (3-16)
17 683 + 73.8
τ1 3 =
= 8879 psi
2
17 683 − 5709
Ans.
= 5987 psi
τ1 2 =
2
5709 + 23.8
= 2866 psi
τ2 3 =
2
______________________________________________________________________________
3-106 Since σt and σr are both positive and σt > σr
τ max = (σ t ) max 2
From Eq. (3-55), σt is maximum at r = ri = 0.3125 in. The term
 3 +ν  0.282  2π ( 5000 )   3 + 0.292 
ρω 

 
=
 = 82.42 lbf/in
60
8
 8  386 

 
2
2
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
(σ t )max = 82.42 0.31252 + 2.752 +


= 1260 psi
τ max =
1260
= 630 psi
2
Ans.

r 2r 2
( 0.3125 )( 2.75 ) − 1 + 3(0.292)
2
2
3 + 0.292
0.31252
(

0.31252 


)
Radial stress:

σ r = k  ri2 + ro2 − i 2o − r 2 
r


Maxima:
 r 2r 2

dσ r
= k  2 i 3o − 2r  = 0 ⇒ r = ri ro = 0.3125(2.75) = 0.927 in
dr
 r



0.31252 2.752
2
2
2

−
0.927
(σ r )max = 82.42 0.3125 + 2.75 −


0.927 2


= 490 psi Ans.
______________________________________________________________________________
(
)
3-107 ω = 2π (2000)/60 = 209.4 rad/s, ρ = 3320 20 kg/m3, ν = 0.24, ri = 0.01 m, ro = 0.125 m
Using Eq. (3-55)
2
 3 + 0.24  
2
2 1 + 3(0.24)
( 0.01)2  (10)−6
 ( 0.01) + (0.125) + (0.125) −
3 + 0.24
 8 

= 1.85 MPa Ans.
______________________________________________________________________________
σ t = 3320(209.4)2 
3-108 ω = 2π (12 000)/60 = 1256.6 rad/s,
( 5 / 16 )
ρ=
= 6.749 10−4 lbf ⋅ s 2 / in 4
2
2
386 (1 16 )(π 4 ) 5 − 0.75
(
)
(
)
The maximum shear stress occurs at bore where τmax = σt /2. From Eq. (3-55)
1 + 3(0.20)
2  3 + 0.20  

2
2
2
(σ t )max = 6.749(10−4 ) (1256.6 ) 
(0.375)2 
  0.375 + 2.5 + 2.5 −
3 + 0.20
 8 

= 5360 psi
τmax = 5360 / 2 = 2680 psi
Ans.
______________________________________________________________________________
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3-109 ω = 2π (3500)/60 = 366.5 rad/s,
mass of blade = m = ρV = (0.282 / 386) [1.25(30)(0.125)] = 3.425(10−3) lbf⋅s2/in
F = (m/2) ω 2r
= [3.425(10−3)/2]( 366.52)(7.5)
= 1725 lbf
Anom = (1.25 − 0.5)(1/8) = 0.093 75 in2
σnom = F/ Anom = 1725/0.093 75 = 18 400 psi Ans.
Note: Stress concentration Fig. A-15-1 gives Kt = 2.25 which increases σmax and fatigue.
______________________________________________________________________________
3-110 ν = 0.292, E = 207 GPa, ri = 0, R = 25 mm, ro = 50 mm
Eq. (3-57),
207(109 )δ  (0.052 − 0.0252 )(0.0252 − 0)  −9
3

 10 = 3.105(10 )δ
2(0.025)3 
(0.052 − 0)

where p is in MPa and δ is in mm.
(
p=
)
(1)
Maximum interference,
1
δ max = [50.042 − 50.000] = 0.021 mm Ans.
2
Minimum interference,
1
δ min = [50.026 − 50.025] = 0.0005 mm Ans.
2
From Eq. (1)
pmax = 3.105(103)(0.021) = 65.2 MPa Ans.
pmin = 3.105(103)(0.0005) = 1.55 MPa Ans.
______________________________________________________________________________
3-111 ν = 0.292, E = 30 Mpsi, ri = 0, R = 1 in, ro = 2 in
Eq. (3-57),
30(106 )δ  (22 − 12 )(12 − 0) 
7
p=

 = 1.125(10 )δ
3
2
2(1 ) 
(2 − 0)

(1)
where p is in psi and δ is in inches.
Maximum interference,
1
δ max = [2.0016 − 2.0000] = 0.0008 in
2
Minimum interference,
Ans.
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1
2
δ min = [2.0010 − 2.0010] = 0
Ans.
From Eq. (1),
pmax = 1.125(107)(0.0008) = 9 000 psi
Ans.
pmin = 1.125(107)(0) = 0 Ans.
______________________________________________________________________________
3-112 ν = 0.292, E = 207 GPa, ri = 0, R = 25 mm, ro = 50 mm
Eq. (3-57),
207(109 )δ  (0.052 − 0.0252 )(0.0252 − 0)  −9
3
 10 = 3.105(10 )δ
3 
2
2(0.025) 
(0.05 − 0)

where p is in MPa and δ is in mm.
(
p=
)
(1)
Maximum interference,
1
δ max = [50.059 − 50.000] = 0.0295 mm Ans.
2
Minimum interference,
1
δ min = [50.043 − 50.025] = 0.009 mm Ans.
2
From Eq. (1)
pmax = 3.105(103)(0.0295) = 91.6 MPa Ans.
pmin = 3.105(103)(0.009) = 27.9 MPa Ans.
______________________________________________________________________________
3-113 ν = 0.292, E = 30 Mpsi, ri = 0, R = 1 in, ro = 2 in
Eq. (3-57),
30(106 )δ  (22 − 12 )(12 − 0) 
7
p=

 = 1.125(10 )δ
2(13 ) 
(22 − 0)

(1)
where p is in psi and δ is in inches.
Maximum interference,
1
δ max = [2.0023 − 2.0000] = 0.00115 in
2
Minimum interference,
1
2
δ min = [2.0017 − 2.0010] = 0.00035
Ans.
Ans.
From Eq. (1),
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pmax = 1.125(107)(0.00115) = 12 940 psi
pmin = 1.125(107)(0.00035) = 3 938
Ans.
Ans.
______________________________________________________________________________
3-114 ν = 0.292, E = 207 GPa, ri = 0, R = 25 mm, ro = 50 mm
Eq. (3-57),
207(109 )δ  (0.052 − 0.0252 )(0.0252 − 0)  −9
3

 10 = 3.105(10 )δ
2(0.025)3 
(0.052 − 0)

where p is in MPa and δ is in mm.
(
p=
)
(1)
Maximum interference,
1
δ max = [50.086 − 50.000] = 0.043 mm Ans.
2
Minimum interference,
1
δ min = [50.070 − 50.025] = 0.0225 mm Ans.
2
From Eq. (1)
pmax = 3.105(103)(0.043) = 134 MPa Ans.
pmin = 3.105(103)(0.0225) = 69.9 MPa Ans.
______________________________________________________________________________
3-115 ν = 0.292, E = 30 Mpsi, ri = 0, R = 1 in, ro = 2 in
Eq. (3-57),
30(106 )δ  (22 − 12 )(12 − 0) 
7
p=

 = 1.125(10 )δ
3
2
2(1 ) 
(2 − 0)

(1)
where p is in psi and δ is in inches.
Maximum interference,
1
δ max = [2.0034 − 2.0000] = 0.0017 in
2
Minimum interference,
1
2
δ min = [2.0028 − 2.0010] = 0.0009
Ans.
Ans.
From Eq. (1),
pmax = 1.125(107)(0.0017) = 19 130 psi
Ans.
pmin = 1.125(107)(0.0009) = 10 130 Ans.
______________________________________________________________________________
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3-116 From Table A-5, Ei = Eo = 30 Mpsi, νi = νo = 0.292. ri = 0, R = 1 in, ro = 1.5 in
1
The radial interference is δ = ( 2.002 − 2.000 ) = 0.001 in Ans.
2
Eq. (3-57),
(
)(
2
2
2
2
Eδ  ro − R R − ri
p= 3
2R 
ro2 − ri 2

= 8333 psi = 8.33 kpsi
)  = 30 (10 ) 0.001  (1.5 − 1 )(1 − 0 ) 
2 (1 )


(1.5 − 0) 
6
2
3
2
2
2
Ans.
The tangential stresses at the interface for the inner and outer members are given by Eqs.
(3-58) and (3-59), respectively.
R2 + r 2
12 + 02
(σ t )i r = R = − p 2 i 2 = −(8333) 2 2 = −8333 psi = −8.33 kpsi Ans.
R − ri
1 −0
ro2 + R 2
1.52 + 12
(8333)
=
= 21 670 psi = 21.7 kpsi Ans.
r=R
ro2 − R 2
1.52 − 12
______________________________________________________________________________
(σ t )o
=p
3-117 From Table A-5, Ei = 30 Mpsi, Eo =14.5 Mpsi, νi = 0.292, νo = 0.211.
ri = 0, R = 1 in, ro = 1.5 in
1
The radial interference is δ = ( 2.002 − 2.000 ) = 0.001 in Ans.
2
Eq. (3-56),
p=
p=
δ
 1 r +R
 1
+ν o  +
R 
2
 Ei
 Eo  r − R
2
o
2
o
2
 R 2 + ri 2

 2 2 −ν i  
 R − ri

0.001

 1.52 + 12

 12 + 02

1
1
+
+
−
1
0.211
0.292


2
2
2
6 
6  2
14.5 10  1.5 − 1
 30 10  1 − 0
 
( )
= 4599 psi
Ans.
( )
The tangential stresses at the interface for the inner and outer members are given by Eqs.
(3-58) and (3-59), respectively.
R2 + r 2
12 + 02
(σ t )i r = R = − p 2 i2 = −(4599) 2 2 = −4599 psi Ans.
R − ri
1 −0
ro2 + R 2
1.52 + 12
(4599)
=
= 11 960 psi Ans.
r=R
ro2 − R 2
1.52 − 12
______________________________________________________________________________
(σ t )o
=p
3-118 From Table A-5, Ei = Eo = 30 Mpsi, νi = νo = 0.292. ri = 0, R = 0.5 in, ro = 1 in
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The minimum and maximum radial interferences are
1
δ min = (1.002 − 1.002 ) = 0.000 in Ans.
2
1
δ max = (1.003 − 1.001) = 0.001 in Ans.
2
Since the minimum interference is zero, the minimum pressure and tangential stresses are
zero.
Ans.
The maximum pressure is obtained from Eq. (3-57).
(
)(
2
2
2
2
Eδ  ro − R R − ri
p= 3 
ro2 − ri 2
2R 

( )
( )
(
) 

)(
30 106 0.001  12 − 0.52 0.52 − 0

p=
2 0.53
12 − 0

(
)
)  = 22 500 psi

Ans
The maximum tangential stresses at the interface for the inner and outer members are
given by Eqs. (3-58) and (3-59), respectively.
R2 + r 2
0.52 + 02
(σ t )i r = R = − p 2 i 2 = −(22 500) 2 2 = −22 500 psi Ans.
R − ri
0.5 − 0
ro2 + R 2
12 + 0.52
=
= 37 500 psi Ans.
(22
500)
r =R
ro2 − R 2
12 − 0.52
______________________________________________________________________________
(σ t )o
=p
3-119 From Table A-5, Ei = 10.4 Mpsi, Eo =30 Mpsi, νi = 0.333, νo = 0.292.
ri = 0, R = 1 in, ro = 1.5 in
The minimum and maximum radial interferences are
1
δ min = [2.003 − 2.002] = 0.0005 in Ans.
2
1
δ max = [2.006 − 2.000] = 0.003 in Ans.
2
Eq. (3-56),
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p=
p=
δ
1
R
 Eo
r +R
 1
+ν o  +

2
r −R
 Ei
2
o
2
o
2
 R 2 + ri 2

 2 2 −ν i  
 R − ri

δ
 1
 1.52 + 12

 12 + 02

1
1
+
0.292
+
−
0.333




2
2
2
2
6
6
 30 10  1.5 − 1
 10.4 10  1 − 0
 
( )
p = 6.229 (10 ) δ psi
p = 6.229 (10 ) δ
p = 6.229 (10 ) δ
( )
6
6
min
min
6
max
max
Ans.
= 6.229 (10 6 ) ( 0.0005 ) = 3114.6 psi = 3.11 kpsi
= 6.229 (106 ) ( 0.003) = 18 687 psi = 18.7 kpsi
Ans.
Ans.
The tangential stresses at the interface for the inner and outer members are given by Eqs.
(3-58) and (3-59), respectively.
Minimum interference:
R2 + r 2
12 + 02
(σ t )i min = − pmin 2 i 2 = −(3.11) 2 2 = −3.11 kpsi Ans.
R − ri
1 −0
(σ t )o
min
= pmin
ro2 + R 2
1.52 + 12
=
= 8.09 kpsi
(3.11)
ro2 − R 2
1.52 − 12
Ans.
Maximum interference:
R2 + r 2
12 + 02
(σ t )i max = − pmax 2 i 2 = −(18.7) 2 2 = −18.7 kpsi Ans.
R − ri
1 −0
ro2 + R 2
1.52 + 12
(18.7)
=
= 48.6 kpsi Ans.
max
ro2 − R 2
1.52 − 12
______________________________________________________________________________
(σ t )o
= pmax
3-120 d = 20 mm, ri = 37.5 mm, ro = 57.5 mm
From Table 3-4, for R = 10 mm,
rc = 37.5 + 10 = 47.5 mm
rn =
(
102
2 47.5 − 47.52 − 102
)
= 46.96772 mm
e = rc − rn = 47.5 − 46.96772 = 0.53228 mm
ci = rn − ri = 46.9677 − 37.5 = 9.4677 mm
co = ro − rn = 57.5 − 46.9677 = 10.5323 mm
A = π d 2 / 4 = π (20) 2 / 4 = 314.16 mm 2
M = Frc = 4000(47.5) = 190 000 N ⋅ mm
Using Eq. (3-65) for the bending stress, and combining with the axial stress,
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4000
190 000(9.4677)
F Mci
+
=
+
= 300 MPa
A Aeri 314.16 314.16(0.53228)(37.5)
σi =
Ans.
4000
190 000(10.5323)
F Mco
−
=
−
= −195 MPa Ans.
A Aero 314.16 314.16(0.53228)(57.5)
______________________________________________________________________________
σo =
3-121 d = 0.75 in, ri = 1.25 in, ro = 2.0 in
From Table 3-4, for R = 0.375 in,
rc = 1.25 + 0.375 = 1.625 in
rn =
0.3752
(
2 1.625 − 1.6252 − 0.3752
)
= 1.60307 in
e = rc − rn = 1.625 − 1.60307 = 0.02193 in
ci = rn − ri = 1.60307 − 1.25 = 0.35307 in
co = ro − rn = 2.0 − 1.60307 = 0.39693 in
A = π d 2 / 4 = π (0.75) 2 / 4 = 0.44179 in 2
M = Frc = 750(1.625) = 1218.8 lbf ⋅ in
Using Eq. (3-65) for the bending stress, and combining with the axial stress,
σi =
750
1218.8(0.35307)
F Mci
+
=
+
= 37 230 psi = 37.2 kpsi
A Aeri 0.44179 0.44179(0.02193)(1.25)
Ans.
750
1218.8(0.39693)
F Mco
−
=
−
= −23 269 psi = −23.3 kpsi Ans.
A Aero 0.44179 0.44179(0.02193)(2.0)
______________________________________________________________________________
σo =
3-122 d = 6 mm, ri = 10 mm, ro = 16 mm
From Table 3-4, for R = 3 mm,
rc = 10 + 3 = 13 mm
rn =
(
32
2 13 − 132 − 32
)
= 12.82456 mm
e = rc − rn = 13 − 12.82456 = 0.17544 mm
ci = rn − ri = 12.82456 − 10 = 2.82456 mm
co = ro − rn = 16 − 12.82456 = 3.17544 mm
A = π d 2 / 4 = π (6) 2 / 4 = 28.2743 mm 2
M = Frc = 300(13) = 3900 N ⋅ mm
Using Eq. (3-65) for the bending stress, and combining with the axial stress,
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300
3900(2.82456)
F Mci
+
=
+
= 233 MPa
A Aeri 28.2743 28.2743(0.17544)(10)
σi =
Ans.
300
3900(3.17544)
F Mco
−
=
−
= −145 MPa Ans.
A Aero 28.2743 28.2743(0.17544)(16)
______________________________________________________________________________
σo =
3-123
d = 6 mm, ri = 10 mm, ro = 16 mm
From Table 3-4, for R = 3 mm,
rc = 10 + 3 = 13 mm
rn =
(
32
2 13 − 132 − 32
)
= 12.82456 mm
e = rc − rn = 13 − 12.82456 = 0.17544 mm
ci = rn − ri = 12.82456 − 10 = 2.82456 mm
co = ro − rn = 16 − 12.82456 = 3.17544 mm
A = π d 2 / 4 = π (6) 2 / 4 = 28.2743 mm 2
The angle θ of the line of radius centers is
 R+d /2 
−1  10 + 6 / 2 
o
θ = sin −1 
 = sin 
 = 30
 R+d +R
 10 + 6 + 10 
M = F ( R + d / 2 ) sin θ = 300 (10 + 6 / 2 ) sin 30o = 1950 N ⋅ mm
Using Eq. (3-65) for the bending stress, and combining with the axial stress,
F sin θ Mci 300 sin 30o
1950(2.82456)
+
=
+
= 116 MPa
A
Aeri
28.2743
28.2743(0.17544)(10)
σi =
Ans.
1950(3.17544)
F sin θ Mco 300 sin 30o
−
=
−
= −72.7 MPa Ans.
A
Aero
28.2743
28.2743(0.17544)(16)
Note that the shear stress due to the shear force is zero at the surface.
______________________________________________________________________________
σo =
3-124 d = 0.25 in, ri = 0.5 in, ro = 0.75 in
From Table 3-4, for R = 0.125 in,
rc = 0.5 + 0.125 = 0.625 in
rn =
(
0.1252
2 0.625 − 0.6252 − 0.1252
)
= 0.618686 in
e = rc − rn = 0.625 − 0.618686 = 0.006314 in
ci = rn − ri = 0.618686 − 0.5 = 0.118686 in
co = ro − rn = 0.75 − 0.618686 = 0.131314 in
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A = π d 2 / 4 = π (0.25)2 / 4 = 0.049087 in 2
M = Frc = 75(0.625) = 46.875 lbf ⋅ in
Using Eq. (3-65) for the bending stress, and combining with the axial stress,
σi =
75
46.875(0.118686)
F Mci
+
=
+
= 37 428 psi = 37.4 kpsi
A Aeri 0.049087 0.049087(0.006314)(0.5)
Ans.
75
46.875(0.131314)
F Mco
−
=
−
= −24 952 psi = −25.0 kpsi Ans.
A Aero 0.049087 0.049087(0.006314)(0.75)
______________________________________________________________________________
σo =
3-125 d = 0.25 in, ri = 0.5 in, ro = 0.75 in
From Table 3-4, for R = 0.125 in,
rc = 0.5 + 0.125 = 0.625 in
rn =
(
0.1252
2 0.625 − 0.6252 − 0.1252
)
= 0.618686 in
e = rc − rn = 0.625 − 0.618686 = 0.006314 in
ci = rn − ri = 0.618686 − 0.5 = 0.118686 in
co = ro − rn = 0.75 − 0.618686 = 0.131314 in
A = π d 2 / 4 = π (0.25)2 / 4 = 0.049087 in 2
The angle θ of the line of radius centers is
 R+d /2 
−1  0.5 + 0.25 / 2 
o
θ = sin −1 
 = sin 
 = 30
 R+d +R
 0.5 + 0.25 + 0.5 
M = F ( R + d / 2 ) sin θ = 75 ( 0.5 + 0.25 / 2 ) sin 30o = 23.44 lbf ⋅ in
Using Eq. (3-65) for the bending stress, and combining with the axial stress,
σi =
23.44(0.118686)
F sin θ Mci 75sin 30o
+
=
+
= 18 716 psi = 18.7 kpsi
A
Aeri 0.049087 0.049087(0.006314)(0.5)
Ans.
23.44(0.131314)
F sin θ Mco 75sin 30o
−
=
−
= −12 478 psi = −12.5 kpsi Ans.
A
Aero 0.049087 0.049087(0.006314)(0.75)
Note that the shear stress due to the shear force is zero at the surface.
______________________________________________________________________________
σo =
3-126
(a) σ = ±
[3(4)][0.5(0.1094)] = ±8021 psi = ±8.02 kpsi
Mc
=±
I
(0.75) ( 0.10943 ) /12
Ans.
(b) ri = 0.125 in, ro = ri + h = 0.125 + 0.1094 = 0.2344 in
From Table 3-4,
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rc = 0.125 + (0.5)(0.1094) = 0.1797 in
0.1094
= 0.174006 in
ln(0.2344 / 0.125)
e = rc − rn = 0.1797 − 0.174006 = 0.005694 in
rn =
ci = rn − ri = 0.174006 − 0.125 = 0.049006 in
co = ro − rn = 0.2344 − 0.174006 = 0.060394 in
A = bh = 0.75(0.1094) = 0.08205 in 2
M = −3(4) = −12 lbf ⋅ in
The negative sign on the bending moment is due to the sign convention shown in Fig. 3-34. Using
Eq. (3-65),
σi =
Mci
−12(0.049006)
=
= −10 070 psi = −10.1 kpsi
Aeri 0.08205(0.005694)(0.125)
σo = −
Mco
−12(0.060394)
=−
= 6618 psi = 6.62 kpsi
Aero
0.08205(0.005694)(0.2344)
σ i −10.1
=
= 1.26
σ −8.02
σ
6.62
Ko = o =
= 0.825
σ 8.02
(c) K i =
Ans.
Ans.
Ans.
Ans.
______________________________________________________________________________
3-127
(a) σ = ±
[3(4)][0.5(0.1406)] = ±4856 psi = ±4.86 kpsi
Mc
=±
I
(0.75) ( 0.14063 ) / 12
Ans.
(b) ri = 0.125 in, ro = ri + h = 0.125 + 0.1406 = 0.2656 in
From Table 3-4,
rc = 0.125 + (0.5)(0.1406) = 0.1953 in
0.1406
= 0.186552 in
ln(0.2656 / 0.125)
e = rc − rn = 0.1953 − 0.186552 = 0.008748 in
rn =
ci = rn − ri = 0.186552 − 0.125 = 0.061552 in
co = ro − rn = 0.2656 − 0.186552 = 0.079048 in
A = bh = 0.75(0.1406) = 0.10545 in 2
M = −3(4) = −12 lbf ⋅ in
The negative sign on the bending moment is due to the sign convention shown in Fig. 3-34. Using
Eq. (3-65),
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σi =
Mci
−12(0.061552)
=
= −6406 psi = −6.41 kpsi
Aeri 0.10545(0.008748)(0.125)
σo = −
Ans.
Mco
−12(0.079048)
=−
= 3872 psi = 3.87 kpsi
Aero
0.10545(0.008748)(0.2656)
Ans.
σ i −6.41
=
= 1.32
Ans.
σ −4.86
σ
3.87
Ko = o =
= 0.80
Ans.
σ 4.86
______________________________________________________________________________
(c) K i =
3-128
(a) σ = ±
[3(4)][0.5(0.1094)] = ±8021 psi = ±8.02 kpsi
Mc
=±
I
(0.75) ( 0.10943 ) /12
Ans.
(b) ri = 0.25 in, ro = ri + h = 0.25 + 0.1094 = 0.3594 in
From Table 3-4,
rc = 0.25 + (0.5)(0.1094) = 0.3047 in
0.1094
= 0.301398 in
ln(0.3594 / 0.25)
e = rc − rn = 0.3047 − 0.301398 = 0.003302 in
rn =
ci = rn − ri = 0.301398 − 0.25 = 0.051398 in
co = ro − rn = 0.3594 − 0.301398 = 0.058002 in
A = bh = 0.75(0.1094) = 0.08205 in 2
M = −3(4) = −12 lbf ⋅ in
The negative sign on the bending moment is due to the sign convention shown in Fig. 3-34. Using
Eq. (3-65),
σi =
Mci
−12(0.051398)
=
= −9106 psi = −9.11 kpsi
Aeri 0.08205(0.003302)(0.25)
σo = −
Ans.
Mco
−12(0.058002)
=−
= 7148 psi = 7.15 kpsi
Aero
0.08205(0.003302)(0.3594)
Ans.
σ i −9.11
=
= 1.14
Ans.
σ −8.02
σ
7.15
Ko = o =
= 0.89
Ans.
σ 8.02
______________________________________________________________________________
(c) K i =
3-129 ri = 25 mm, ro = ri + h = 25 + 87 = 112 mm, rc = 25 + 87/2 = 68.5 mm
The radius of the neutral axis is found from Eq. (3-63), given below.
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rn =
A
∫ ( dA / r )
For a rectangular area with constant width b, the denominator is
ro  bdr 
ro
∫ri  r  = b ln ri
Applying this equation over each of the four rectangular areas,
dA
 45 
 54.5 
 92   112 
= 9  ln  + 31 ln
 + 31 ln
 + 9  ln
 = 16.3769
r
45 
 25 

 82.5   92 
A = 2 [ 20(9) + 31(9.5)] = 949 mm2
∫
rn =
A
∫ ( dA / r )
=
949
= 57.9475 mm
16.3769
e = rc − rn = 68.5 − 57.9475 = 10.5525 mm
ci = rn − ri = 57.9475 − 25 = 32.9475 mm
co = ro − rn = 112 − 57.9475 = 54.0525 mm
M = 150F2 = 150(3.2) = 480 kN∙mm
We need to find the forces transmitted through the section in order to determine the axial
stress. It is not immediately obvious which plane should be used for resolving the axial
versus shear directions. It is convenient to use the plane containing the reaction force at
the bushing, which assumes its contribution resolves entirely into shear force. To find the
angle of this plane, find the resultant of F1 and F2.
Fx = F1x + F2 x = 2.4 cos 60o + 3.2 cos 0o = 4.40 kN
Fy = F1 y + F2 y = 2.4sin 60o + 3.2sin 0o = 2.08 kN
F = ( 4.402 + 2.082 )
12
= 4.87 kN
This is the pin force on the lever which acts in a direction
θ = tan −1
Fy
Fx
= tan −1
2.08
= 25.3o
4.40
On the surface 25.3° from the horizontal, find the internal forces in the tangential and
normal directions. Resolving F1 into components,
Ft = 2.4 cos ( 60o − 25.3o ) = 1.97 kN
Fn = 2.4sin ( 60o − 25.3o ) = 1.37 kN
The transverse shear stress is zero at the inner and outer surfaces. Using Eq. (3-65) for
the bending stress, and combining with the axial stress due to Fn,
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σi =
Fn Mci 1370 ( 3200 )(150 )  (32.9475)
+
=
+
= 64.6 MPa
A Aeri 949
949(10.5525)(25)
Ans.
Fn Mco 1370 ( 3200 )(150 )  (54.0525)
−
=
−
= −21.7 MPa Ans.
A Aero 949
949(10.5525)(112)
______________________________________________________________________________
σo =
3-130 ri = 2 in, ro = ri + h = 2 + 4 = 6 in, rc = 2 + 0.5(4) = 4 in
A = (6 − 2 − 0.75)(0.75) = 2.4375 in 2
Similar to Prob. 3-129,
dA
3.625
6
= 0.75 ln
+ 0.75 ln
= 0.682 920 in
r
2
4.375
A
2.4375
rn =
=
= 3.56923 in
∫ (dA / r ) 0.682 920
∫
e = rc − rn = 4 − 3.56923 = 0.43077 in
ci = rn − ri = 3.56923 − 2 = 1.56923 in
co = ro − rn = 6 − 3.56923 = 2.43077 in
M = Frc = 6000(4) = 24 000 lbf ⋅ in
Using Eq. (3-65) for the bending stress, and combining with the axial stress,
σi =
6000
24 000(1.56923)
F Mci
+
=
+
= 20 396 psi = 20.4 kpsi
A Aeri 2.4375 2.4375(0.43077)(2)
Ans.
6000
24 000(2.43077)
F Mco
−
=
−
= −6 799 psi = −6.80 kpsi Ans.
A Aero 2.4375 2.4375(0.43077)(6)
______________________________________________________________________________
σo =
3-131
ri = 12 in, ro = ri + h = 12 + 3 = 15 in, rc = 12 + 3/2 = 13.5 in
π
π
I = a 3b = (1.53 )(0.75) = 1.988 in 4
4
4
A = π ab = π (1.5)(0.75) = 3.534
M = 20(3 + 1.5) = 90 kip ⋅ in
Since the radius is large compared to the cross section, assume Eq. 3-67 is applicable for
the bending stress. Combining the bending stress and the axial stress,
20
90(1.5)(13.5)
F Mci rc
+
=
+
= 82.1 kpsi Ans.
A
Iri
3.534 (1.988)(12)
20
90(1.5)(13.5)
F Mc r
σo = − o c =
−
= −55.5 kpsi Ans.
A
Iro
3.534
1.988(15)
σi =
______________________________________________________________________________
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3-132
ri = 1.25 in, ro = ri + h = 1.25 + 0.5 + 1 + 0.5 = 3.25 in
Ans.
rc = (ri + ro) / 2 = (1.25 + 3.25)/2 = 2.25 in
ro
 dA 
For outer rectangle,  ∫
 = b ln
ri
 r 



A 
r2

 , A = π r2
 =
For circle, 
O

2
2 
 ∫ ( dA r ) 
 2 rc − rc − r  O
O
 dA 
∴ ∫  = 2π ( rc − rc2 − r 2 )
 r O
Combine the integrals subtracting the circle from the rectangle
)
(
(
)
dA
3.25
= 1.25ln
− 2π 2.25 − 2.252 − 0.52 = 0.840 904 in
r
1.25
A = 1.25(2) − π (0.52 ) = 1.714 60 in 2
Ans.
A
1.714 60
rn =
=
= 2.0390 in Ans.
∑ ∫ (dA / r ) 0.840 904
Σ∫
e = rc − rn = 2.25 − 2.0390 = 0.2110 in Ans.
ci = rn − ri = 2.0390 − 1.25 = 0.7890 in
co = ro − rn = 3.25 − 2.0390 = 1.2110 in
M = 2000(4.5 + 1.25 + 0.5 + 0.5) = 13 500 lbf ⋅ in
2000
13 500(0.7890)
F Mci
σi = +
=
+
= 20 720 psi = 20.7 kpsi Ans.
A Aeri 1.7146 1.7146(0.2110)(1.25)
2000
13 500(1.2110)
F Mco
σo = −
=
−
= −12 738 psi = −12.7 kpsi
Ans.
A Aero 1.7146 1.7146(0.2110)(3.25)
______________________________________________________________________________
3-133 From Eq. (3-68),

2 (1 −ν 2 ) E  
1 3  3  
= F  

2 (1 d )
 8 

13
a = KF 1 3
Use ν = 0.292, F in newtons, E in N/mm2 and d in mm, then
1/3
 3  [(1 − 0.2922 ) / 207 000] 
K =  

1 / 30
 8 

= 0.03685
From Eq. (3-69),
pmax =
3F
3F
3F 1/3
3F 1/3
=
=
=
= 352 F 1/3 MPa
2
1/3 2
2
2
2π a
2π ( KF )
2π K
2π (0.03685)
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From Eq. (3-71), the maximum principal stress occurs on the surface where z = 0, and is
equal to – pmax.
σ max = σ z = − pmax = −352 F 1/3 MPa
Ans.
From Fig. 3-37,
τ max = 0.3 pmax = 106 F 1/3 MPa
Ans.
______________________________________________________________________________
3-134 From Eq. (3-68),
2
2
 3F  (1 −ν 1 ) E1 + (1 −ν 2 ) E2
a=3

1 d1 + 1 d 2
 8 
 3 (10 )  (1 − 0.292
a=3

 8 
2
) ( 207 000 ) + (1 − 0.333 ) ( 71 700 ) = 0.0990 mm
2
1 25 + 1 40
From Eq. (3-69),
3 (10 )
3F
pmax =
=
= 487.2 MPa
2
2π a
2π ( 0.09902 )
From Fig. 3-37, the maximum shear stress occurs at a depth of z = 0.48 a.
z = 0.48a = 0.48 ( 0.0990 ) = 0.0475 mm
Ans.
The principal stresses are obtained from Eqs. (3-70) and (3-71) at a depth of z/a = 0.48.

σ 1 = σ 2 = −487.2  1 − 0.48 tan −1 (1/ 0.48)  (1 + 0.333) −

σ3 =

1
 = −101.3 MPa
2 (1 + 0.482 ) 
−487.2
= −396.0 MPa
1 + 0.482
From Eq. (3-72),
σ −σ
( −101.3) − ( −396.0 ) = 147.4 MPa Ans.
τ max = 1 3 =
2
2
Note that if a closer examination of the applicability of the depth assumption from Fig. 337 is desired, implementing Eqs. (3-70), (3-71), and (3-72) on a spreadsheet will allow
for calculating and plotting the stresses versus the depth for specific values of ν. For ν =
0.333 for aluminum, the maximum shear stress occurs at a depth of z = 0.492a with τmax
= 0.3025 pmax.
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This gives τmax = 0.3025 pmax = (0.3025)(487.2) = 147.38 MPa. Even though the depth
assumption was a little off, it did not have significant effect on the the maximum shear
stress.
______________________________________________________________________________
3-135 From the solution to Prob. 3-134, a = 0.0990 mm and pmax = 487.2 MPa. Assuming
applicability of Fig. 3-37, the maximum shear stress occurs at a depth of z = 0.48 a =
0.0475 mm. Ans.
The principal stresses are obtained from Eqs. (3-70) and (3-71) at a depth of z/a = 0.48.

σ 1 = σ 2 = −487.2  1 − 0.48 tan −1 (1/ 0.48)  (1 + 0.292 ) −

σ3 =

1
 = −92.09 MPa
2 (1 + 0.482 ) 
−487.2
= −396.0 MPa
1 + 0.482
From Eq. (3-72),
σ −σ
( −92.09 ) − ( −396.0 ) = 152.0 MPa Ans.
τ max = 1 3 =
2
2
Note that if a closer examination of the applicability of the depth assumption from Fig. 337 is desired, implementing Eqs. (3-70), (3-71), and (3-72) on a spreadsheet will allow
for calculating and plotting the stresses versus the depth for specific values of ν. For ν =
0.292 for steel, the maximum shear stress occurs at a depth of z = 0.478a with τmax =
0.3119 pmax.
______________________________________________________________________________
3-136 From Eq. (3-68),
2
 3F  2 (1 −ν ) E
a= 

 8  1 d1 + 1 d 2
3
2
 3 ( 20 )  2 (1 − 0.292 ) ( 207 000 )
a=3
= 0.1258 mm

1 30 + 1 ∞
 8 
From Eq. (3-69),
3 ( 20 )
3F
pmax =
=
= 603.4 MPa
2
2π a
2π ( 0.12582 )
From Fig. 3-37, the maximum shear stress occurs at a depth of
z = 0.48a = 0.48 ( 0.1258) = 0.0604 mm
Ans.
Also from Fig. 3-37, the maximum shear stress is
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τ max = 0.3 pmax = 0.3(603.4) = 181 MPa
Ans.
______________________________________________________________________________
3-137 Aluminum Plate-Ball interface: From Eq. (3-68),
2
2
 3F  (1 −ν 1 ) E1 + (1 −ν 2 ) E2
a= 

1 d1 + 1 d 2
 8 
3
 3F  (1 − 0.292
a= 

 8 
3
2
) ( 30 ) (10 ) + (1 − 0.333 ) (10.4 ) (10 )
6
2
6
1 1+1 ∞
= 3.517 (10−3 ) F 1/3 in
From Eq. (3-69),
3F
3F
pmax =
=
= 3.860 104 F 1/3 psi
2
2
−3
1/3
2π a
2π 3.517 10 F 
By examination of Eqs. (3-70), (3-71), and (3-72), it can be seen that the only difference
in the maximum shear stress for the plate and the ball will be due to poisson’s ratio in Eq.
(3-70). The larger poisson’s ratio will create the greater maximum shear stress, so the
aluminum plate will be the critical element in this interface. Applying the equations for
the aluminum plate,
(
( )
)

σ 1 = −3.86 (104 ) F 1/3  1 − 0.48 tan −1 (1/ 0.48)  (1 + 0.333) −

−3.86 (10 ) F 1/3
4
σ3 =
1 + 0.482
From Eq. (3-72),
τ max =
σ1 − σ 3

1
= −8025 F 1/3 psi
2 
2 (1 + 0.48 ) 
= −3.137 (104 ) F 1/3 psi
( −8025F ) − ( −3.137 (10 ) F )
=
= 1.167 (10 ) F
2
1/3
4
1/3
4
2
Comparing this stress to the allowable stress, and solving for F,
1/3
psi
3
 20 000 
 = 5.03 lbf
F =
4
1.167 (10 ) 
Table-Ball interface: From Eq. (3-68),
2
6
2
6




 3F  (1 − 0.292 ) ( 30 ) (10 )  + (1 − 0.211 ) (14.5 ) (10 ) 
a= 
= 3.306 (10−3 ) F 1/3 in

1 1+1 ∞
 8 
3
From Eq. (3-69),
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( )
3F
3F
=
= 4.369 104 F 1/3 psi
2
2
−
3
1/3
2π a
2π 3.306 10 F 
The steel ball has a higher poisson’s ratio than the cast iron table, so it will dominate.
pmax =
(
)

σ 1 = −4.369 (104 ) F 1/3  1 − 0.48 tan −1 (1/ 0.48)  (1 + 0.292 ) −

σ3 =
−4.369 (104 ) F 1/3
1 + 0.48
From Eq. (3-72),
τ max =
σ1 − σ 3
2

1
1/3
 = −8258 F psi
2 (1 + 0.482 ) 
= −3.551(104 ) F 1/3 psi
( −8258F ) − ( −3.551(10 ) F )
=
= 1.363 (10 ) F
2
1/3
4
1/3
4
1/3
2
Comparing this stress to the allowable stress, and solving for F,
psi
3
 20 000 
 = 3.16 lbf
F =
4
1.363 (10 ) 
The steel ball is critical, with F = 3.16 lbf.
Ans.
______________________________________________________________________________
3-138 v1 = 0.333, E1 = 10.4 Mpsi, l = 2 in, d1 = 1.25 in, v2 = 0.211, E2 = 14.5 Mpsi, d2 = –12 in.
From Eq. 3-73, with b = KcF1/2
 2 (1 − 0.3332 ) 10.4 (106 )  + (1 − 0.2112 ) 14.5 (106 )  




Kc = 
 π (2)

1/1.25 + 1/ ( −12 )


12
= 2.593 (10−4 )
By examination of Eqs. (3-75), (3-76), and (3-77), it can be seen that the only difference
in the maximum shear stress for the two materials will be due to poisson’s ratio in Eq. (375). The larger poisson’s ratio will create the greater maximum shear stress, so the
aluminum roller will be the critical element in this interface. Instead of applying these
equations, we will assume the poisson’s ratio for aluminum of 0.333 is close enough to
0.3 to make Fig. 3-39 applicable.
τ max = 0.3 pmax
pmax =
4000
= 13 300 psi
0.3
From Eq. (3-74), pmax = 2F / (πbl ), so we have
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Chapter 3 Solutions, Page 97/100
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pmax =
2F
2F 1 2
=
π lK c F 1 2 π lK c
So,
 π lK c pmax 
F =

2


2
 π (2)(2.593) (10−4 ) (13 300) 

=


2


= 117.4 lbf Ans.
______________________________________________________________________________
2
3-139
v = 0.292, E = 30 Mpsi, l = 0.75 in, d1 = 2(0.47) = 0.94 in, d2 = 2(0.62) = 1.24 in.
Eq. (3-73):
 2(40) 2 (1 − 0.2922 ) 30 (106 )  


b=
 π (0.75)

1/ 0.94 + 1/ 1.24


12
Eq. (3-74):
pmax =
= 1.052 (10−3 ) in
2 ( 40 )
2F
==
= 32 275 psi = 32.3 kpsi
π bl
π (1.052 ) (10−3 ) ( 0.75 )
Ans.
From Fig. 3-39,
τ max = 0.3 pmax = 0.3 ( 32 275) = 9682.5 psi = 9.68 kpsi
Ans.
______________________________________________________________________________
3-140 Use Eqs. (3-73) through (3-77).
1/2
 2 F (1 −ν 12 ) / E1 + (1 − v22 ) / E2 
b=

(1 / d1 ) + (1 / d 2 )
 πl

1/2
 2(600) (1 − 0.2922 ) / (30(106 )) + (1 − 0.2922 ) / (30(106 )) 
=

1/ 5 + 1 / ∞
 π (2)

b = 0.007 631 in
2F
2(600)
=
= 25 028 psi
π bl π (0.007 631)(2)

z2 z 
σ x = −2ν pmax  1 + 2 −  = −2 ( 0.292 )( 25 028 )

b
b 

Ans.
= −7102 psi = −7.10 kpsi
pmax =
(
Shigley’s MED, 10th edition
1 + 0.7862 − 0.786
)
Chapter 3 Solutions, Page 98/100
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

z2
 1 + 2 0.7862

 1+ 2 2

(
)
z
b


−2
= −25 028
− 2 ( 0.786 ) 
σ y = − pmax
2


2


b
z
 1 + ( 0.786 )


 1+ 2

b


= −4 646 psi = −4.65 kpsi Ans.
− pmax
−25 028
σz =
=
= −19 677 psi = −19.7 kpsi
Ans.
2
z
1 + 0.7862
1+ 2
b
σ − σ z −4 646 − ( −19 677 )
τ max = y
=
= 7 516 psi = 7.52 kpsi
Ans.
2
2
______________________________________________________________________________
3-141 Use Eqs. (3-73) through (3-77).
 2 F (1 −ν 12 ) E1 + (1 −ν 2 2 ) E2 

b=
 πl

1/ d1 + 1/ d 2


12
 2(2000) (1 − 0.2922 )  207 (103 )  + (1 − 0.2112 ) 100 (103 )  




=
 π (40)

1/ 150 + 1/ ∞


b = 0.2583 mm
12
pmax =
2F
2(2000)
=
= 123.2 MPa
π bl π (0.2583)(40)
(

z2 z 
−  = −2 ( 0.292 )(123.2 ) 1 + 0.7862 − 0.786
2

b
b 

Ans.
= −35.0 MPa
2


z
 1 + 2 0.7862

 1+ 2 2

(
)
z
b


σ y = − pmax
−2
= −123.2
− 2 ( 0.786 ) 
 1 + 0.7862


b
z2
(
)


 1+ 2

b


= −22.9 MPa
Ans.
− pmax
−123.2
σz =
=
= −96.9 MPa
Ans.
2
z
1 + 0.7862
1+ 2
b
σ − σ z −22.9 − ( −96.9 )
=
= 37.0 MPa
Ans.
τ max = y
2
2
σ x = −2ν pmax  1 +
)
______________________________________________________________________________
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3-142
Use Eqs. (3-73) through (3-77).
 2 F (1 −ν 12 ) E1 + (1 −ν 2 2 ) E2 

b=
 πl

1/ d1 + 1/ d 2


12
 2(250) (1 − 0.2112 ) 14.5 (106 )  + (1 − 0.2112 ) 14.5 (106 )  




=
 π (1.25)

1/ 3 + 1/ ∞


b = 0.007 095 in
12
2F
2(250)
=
= 17 946 psi
π bl π (0.007 095)(1.25)

z2 z 
σ x = −2ν pmax  1 + 2 −  = −2 ( 0.211)(17 946 )

b
b 

Ans.
= −3 680 psi = −3.68 kpsi
pmax =
(
1 + 0.7862 − 0.786
)


z2
 1 + 2 0.7862

 1+ 2 2

(
)
z
b


−2
= −17 946
− 2 ( 0.786 ) 
σ y = − pmax
2
2




b
z
 1 + ( 0.786 )


 1+ 2

b


= −3 332 psi = −3.33 kpsi Ans.
− pmax
−17 946
σz =
=
= −14 109 psi = −14.1 kpsi
Ans.
2
z
1 + 0.7862
1+ 2
b
σ − σ z −3 332 − ( −14 109 )
=
= 5 389 psi = 5.39 kpsi
Ans.
τ max = y
2
2
______________________________________________________________________________
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Chapter 4
For a torsion bar, kT = T/θ = Fl/θ, and so θ = Fl/kT. For a cantilever, kl = F/δ ,δ = F/kl. For
the assembly, k = F/y, or, y = F/k = lθ + δ
Thus
F Fl 2 F
y= =
+
k
kT kl
Solving for k
kk
1
= 2l T
k= 2
Ans.
1 kl l + kT
l
+
kT kl
______________________________________________________________________________
4-1
For a torsion bar, kT = T/θ = Fl/θ, and so θ = Fl/kT. For each cantilever, kl = F/δl, δl = F/kl,
and,δL = F/kL. For the assembly, k = F/y, or, y = F/k = lθ + δl +δL.
Thus
F Fl 2 F F
y= =
+ +
k
kT kl k L
Solving for k
k L kl kT
1
k= 2
Ans.
=
2
l
1 1 kl k Ll + kT k L + kT kl
+ +
kT kl k L
______________________________________________________________________________
4-2
4-3
(a) For a torsion bar, k =T/θ =GJ/l.
Two springs in parallel, with J =πdi 4/32,
and d1 = d2 = d,
J G J G π  d4 d4 
k = 1 + 2 = G 1 + 2 
x
l − x 32  x l − x 
π
1 
1
Gd 4  +

32
 x l−x
Deflection equation,
=
θ=
Ans. (1)
T1 x T2 ( l − x )
=
JG
JG
T2 ( l − x )
(2)
x
From statics, T1 + T2 = T = 1500. Substitute Eq. (2)
results in
T1 =
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Chapter 4 Solutions, Page 1/80
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l−x
T2 
 + T2 = 1500
 x 
T2 = 1500
⇒
x
l
Ans.
(3)
l−x
Ans. (4)
l
π
1 
1
3
(b) From Eq. (1), k = ( 0.54 )11.5 (106 )  +
 = 28.2 (10 ) lbf ⋅ in/rad
32
 5 10 − 5 
10 − 5
From Eq. (4), T1 = 1500
= 750 lbf ⋅ in Ans.
10
5
From Eq. (3), T2 = 1500 = 750 lbf ⋅ in
Ans.
10
16Ti 16 (1500 )
=
= 30.6 (103 ) psi = 30.6 kpsi
Ans.
From either section, τ =
π di3 π ( 0.53 )
T1 = 1500
Substitute into Eq. (2) resulting in
Ans.
______________________________________________________________________________
4-4
Deflection to be the same as Prob. 4-3 where T1 = 750 lbf⋅in, l1 = l / 2 = 5 in, and d1 = 0.5
in
θ1=θ2=θ
T1 ( 4 )
π
32
Or,
4
1
d G
=
T2 ( 6 )
π
32
=
π
4
2
d G
32
T1 = 15 (103 ) d14
( 0.5 ) G
4T1 6T2
= 4 = 60 (103 )
d14
d2
⇒
4
(1)
(2)
T2 = 10 (103 ) d 24
Equal stress, τ 1 = τ 2 ⇒
750 ( 5 )
(3)
16T1 16T2
T
T
=
⇒ 13 = 23
3
3
d1 d 2
π d1 π d 2
(4)
Divide Eq. (4) by the first two equations of Eq.(1) results in
T1
T2
3
d1
d3
= 2 ⇒ d 2 = 1.5d1 (5)
4T1 6T2
d14
d 24
Statics, T1 + T2 = 1500
(6)
Substitute in Eqs. (2) and (3), with Eq. (5) gives
15 (103 ) d14 + 10 (103 ) (1.5d1 ) = 1500
4
Solving for d1 and substituting it back into Eq. (5) gives
d1 = 0.388 8 in, d2 = 0.583 2 in
Ans.
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Chapter 4 Solutions, Page 2/80
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From Eqs. (2) and (3),
T1 = 15(103)(0.388 8)4 = 343 lbf⋅in
T2 = 10(103)(0.583 2)4 = 1 157 lbf⋅in
Deflection of T is
θ1 =
Spring constant is
k=
The stress in d1 is
τ1 =
Ans.
Ans.
343 ( 4 )
T1l1
=
= 0.053 18 rad
J1G (π / 32 ) ( 0.388 84 )11.5 (106 )
T
θ1
=
1500
= 28.2 (103 ) lbf ⋅ in
0.053 18
Ans.
16 ( 343)
16T1
=
= 29.7 (103 ) psi = 29.7 kpsi
3
3
π d1 π ( 0.388 8 )
Ans.
16 (1 157 )
16T2
=
= 29.7 (103 ) psi = 29.7 kpsi
Ans.
3
π d 2 π ( 0.583 2 )3
______________________________________________________________________________
The stress in d1 is
4-5
τ2 =
(a) Let the radii of the straight sections be r1 = d1 /2 and r2 = d2 /2. Let the angle of the
taper be α where tan α = (r2 − r1)/2. Thus, the radius in the taper as a function of x is
r = r1 + x tan α, and the area is A = π (r1 + x tan α)2. The deflection of the tapered portion
is
l
l
F
F
dx
F
1
δ =∫
=−
dx =
2
∫
π E 0 ( r1 + x tan α )
π E ( r1 + x tan α ) tan α
AE
0
=

1 1
1
F  1
F
−

=
 − 
π E  r1 tan α tan α ( r1 + l tan α )  π E tan α  r1 r2 
=
F
r2 − r1
F
l tan α
Fl
=
=
π E tan α r1r2
π E tan α r1r2
π r1r2 E
=
4 Fl
π d1d 2 E
l
0
Ans.
(b) For section 1,
δ1 =
4 Fl
4(1000)(2)
Fl
=
=
= 3.40(10−4 ) in
AE π d12 E π (0.52 )(30)(106 )
For the tapered section,
4 Fl
4
1000(2)
δ=
=
= 2.26(10−4 ) in
π d1d 2 E π (0.5)(0.75)(30)(106 )
For section 2,
Shigley’s MED, 10th edition
Ans.
Ans.
Chapter 4 Solutions, Page 3/80
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Fl
4 Fl
4(1000)(2)
=
=
= 1.51(10−4 ) in Ans.
AE π d12 E π (0.752 )(30)(106 )
______________________________________________________________________________
δ2 =
4-6
(a) Let the radii of the straight sections be r1 = d1 /2 and r2 = d2 /2. Let the angle of the
taper be α where tan α = (r2 − r1)/2. Thus, the radius in the taper as a function of x is
r = r1 + x tan α, and the polar second area moment is J = (π /2) (r1 + x tan α)4. The angular
deflection of the tapered portion is
l
l
l
T
2T
dx
1 2T
1
θ = ∫ dx =
=−
4
∫
GJ
π G 0 ( r1 + x tan α )
3 π G ( r1 + x tan α )3 tan α
0
=
0
 2
T 1 1
2 T  1
1
−
=
 3
 − 
3
3π G  r1 tan α tan α ( r1 + l tan α )  3π G tan α  r13 r23 
2
2
2
2 T  l  r23 − r13
2 Tl ( r1 + r1r2 + r2 )
T
r23 − r13
=
=
=


r13r23
3π G tan α r13 r23
3π G  r2 − r1  r13r23
3π G
=
2
2
32 Tl ( d1 + d1d 2 + d 2 )
d13d 23
3π G
Ans.
(b) The deflections, in degrees, are:
Section 1,
Tl  180  32Tl  180 
32(1500)(2)  180 

=

=

 = 2.44 deg
4
GJ  π  π d1 G  π  π (0.54 )11.5(106 )  π 
θ1 =
Ans.
Tapered section,
θ=
32 Tl (d12 + d1d 2 + d 2 2 )  180 


3π
Gd13 d 23
 π 
2
2
32 (1500)(2) 0.5 + (0.5)(0.75) + 0.75   180 
=

 = 1.14 deg
3π
11.5(106 )(0.53 )(.753 )
 π 
Ans.
Section 2,
Tl  180  32Tl  180 
32(1500)(2)  180 
Ans.

=

=

 = 0.481 deg
4
GJ  π  π d 2 G  π  π (0.754 )11.5(106 )  π 
______________________________________________________________________________
θ2 =
4-7
The area and the elastic modulus remain constant. However, the force changes with
respect to x. From Table A-5, the unit weight of steel is γ = 0.282 lbf/in3 and the elastic
modulus is E = 30 Mpsi. Starting from the top of the cable (i.e. x = 0, at the top).
F = γ(A)(l−x)
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1 
Fdx w l
γ 
γ l 2 0.282 [500(12) ]
= ∫ (l − x)dx =  lx − x 2  =
=
= 0.169 in
2  0 2E
2(30)106
AE E 0
E
2
l
δc = ∫
l
o
From the weight at the bottom of the cable,
4(5000) [500(12)]
4Wl
Wl
=
=
= 5.093 in
2
π (0.52 )30(106 )
AE π d E
δ = δ c + δW = 0.169 + 5.093 = 5.262 in
Ans.
δW =
The percentage of total elongation due to the cable’s own weight
0.169
(100) = 3.21%
Ans.
5.262
______________________________________________________________________________
4-8
ΣFy = 0 = R 1 − F ⇒ R 1 = F
ΣMA = 0 = M1 − Fa ⇒ M1 = Fa
VAB = F, MAB =F (x − a ), VBC = MBC = 0
Section AB:

1
F  x2
θ AB =
F
x
−
a
dx
=
(
)
 − ax  + C1
∫
EI
EI  2

θAB = 0 at x = 0 ⇒ C1 = 0
y AB =

F  x2
F  x3
x2 
−
ax
dx
=
−
a



 + C2
EI ∫  2
EI  6
2

(1)
(2)
yAB = 0 at x = 0 ⇒ C2 = 0, and
Fx 2
y AB =
Ans.
( x − 3a )
6 EI
Section BC:
θ BC =
1
EI
∫ ( 0 ) dx = 0 + C
3
From Eq. (1), at x = a (with C1 = 0), θ =
θ BC = −

F  a2
Fa 2
−
a
(
a
)
=
−
= C3. Thus,


EI  2
2 EI

Fa 2
2 EI
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yBC = −
Fa 2
Fa 2
dx
=
−
x + C4
2 EI ∫
2 EI
(3)
From Eq. (2), at x = a (with C2 = 0), y =
−
Fa 2
Fa 3
a + C4 = −
2 EI
3EI
yBC = −
C4 =
⇒
F  a3
a2 
Fa3
a
−
=
−
. Thus, from Eq. (3)


EI  6
2
3EI
Fa 3
6 EI
Fa 2
Fa 3 Fa 2
x+
=
( a − 3x )
2 EI
6 EI 6 EI
Substitute into Eq. (3), obtaining
Ans.
The maximum deflection occurs at x= l,
Fa 2
Ans.
( a − 3l )
6 EI
______________________________________________________________________________
ymax =
4-9
ΣMC = 0 = F (l /2) − R1 l ⇒ R1 = F /2
ΣFy = 0 = F /2 + R 2 − F ⇒
R 2 = F /2
Break at 0 ≤ x ≤ l /2:
VAB = R 1 = F /2,
MAB = R 1 x = Fx /2
Break at l /2 ≤ x ≤ l :
VBC = R 1 − F = − R 2 = − F /2,
MBC = R 1 x − F ( x − l / 2) = F (l − x) /2
Section AB:
θ AB =
1
EI
Fx
F x2
dx
=
+ C1
∫2
EI 4
2
l
F 
 2 +C = 0
From symmetry, θAB = 0 at x = l /2 ⇒
1
4 EI
θ AB =
F x 2 Fl 2
F
−
=
4 x2 − l 2 )
(
EI 4 16 EI 16 EI
⇒
Fl 2
. Thus,
C1 = −
16 EI
(1)
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y AB =
F
16 EI
yAB = 0 at x = 0
y AB =
∫ ( 4x
⇒
Fx
( 4 x 2 − 3l 2 )
48 EI
2
− l 2 ) dx =
F  4 x3 2 
− l x  + C2

16 EI  3

C2 = 0, and,
(2)
yBC is not given, because with symmetry, Eq. (2) can be used in this region. The
maximum deflection occurs at x =l /2,
ymax
l
F 
2

Fl 3
2  l
=    4   − 3l 2  = −
48 EI   2 
48 EI

Ans.
______________________________________________________________________________
4-10 From Table A-6, for each angle, I1-1 = 207 cm4. Thus, I = 2(207) (104) = 4.14(106) mm4
From Table A-9, use beam 2 with F = 2500 N, a = 2000 mm, and l = 3000 mm; and beam
3 with w = 1 N/mm and l = 3000 mm.
Fa 2
w l4
(a − 3l ) −
6 EI
8 EI
2500(2000) 2
(1)(3000) 4
2000
3(3000)
=
−
−
[
]
6(207)103 (4.14)106
8(207)(103 )(4.14)(106 )
= −25.4 mm
Ans.
ymax =
M O = − Fa − ( wl 2 / 2)
= − 2500(2000) − [1(30002)/2] = − 9.5(106) N⋅mm
From Table A-6, from centroid to upper surface is y = 29 mm. From centroid to bottom
surface is y = 29.0 − 100= − 71 mm. The maximum stress is compressive at the bottom of
the beam at the wall. This stress is
My
−9.5(106 )(−71)
=−
= −163 MPa Ans.
I
4.14(106 )
______________________________________________________________________________
σ max = −
Shigley’s MED, 10th edition
Chapter 4 Solutions, Page 7/80
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4-11
14
10
(450) + (300) = 465 lbf
20
20
6
10
(450) + (300) = 285 lbf
RC =
20
20
RO =
M1 = 465(6)12 = 33.48(103) lbf⋅in
M2 = 33.48(103) +15(4)12
= 34.20(103) lbf⋅in
M max
34.2
Z = 2.28 in 3
⇒ 15 =
Z
Z
For deflections, use beams 5 and 6 of Table A-9
2
F1a[l − (l / 2)]  l 
l  F2l 3
2
y x =10ft =
  + a − 2l  −
6 EIl
2  48 EI
 2 
450(72)(120)
300(2403 )
2
2
2
120
72
240
−0.5 =
+
−
−
(
)
6(30)(106 ) I (240)
48(30)(106 ) I
σ max =
I = 12.60 in 4 ⇒ I / 2 = 6.30 in 4
Select two 5 in-6.7 lbf/ft channels from Table A-7, I = 2(7.49) = 14.98 in4, Z =2(3.00) =
6.00 in3
12.60  1 
ymidspan =
 −  = −0.421 in
14.98  2 
34.2
σ max =
= 5.70 kpsi
6.00
______________________________________________________________________________
4-12
I=
π
(1.54 ) = 0.2485 in 4
64
From Table A-9 by superposition of beams 6 and 7, at x = a = 15 in, with b = 24 in and
l = 39 in
Fba 2
wa
y=
[a + b 2 − l 2 ] +
(2la 2 − a 3 − l 3 )
6 EIl
24 EI
yA =
340(24)15
152 + 242 − 392 
6
6(30)10 (0.2485)39
(150 / 12)(15)
 2(39)(152 ) − 153 − 393  = −0.0978 in
+
24(30)106 (0.2485) 
Ans.
At x = l /2 = 19.5 in
2
2
3

Fa[l − (l / 2)]  l 
l  w(l / 2)   l   l 
2
3
y=
  + a − 2l  +
 2l   −   − l 
6 EIl
2  24 EI   2   2 
 2 

Shigley’s MED, 10th edition
Chapter 4 Solutions, Page 8/80
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y=
340(15)(19.5)
19.52 + 152 − 392 
6
6(30)(10 )(0.2485)(39)
(150 /12)(19.5)
 2(39)(19.52 ) − 19.53 − 393  = −0.1027 in
+
6
24(30)(10 )(0.2485)
Ans.
−0.1027 + 0.0978
(100) = 5.01% Ans.
−0.0978
______________________________________________________________________________
% difference =
4-13 I =
( )
1
(6)(323 ) = 16.384 103 mm 4
12
From Table A-9-10, beam 10
Fa 2
yC = −
(l + a )
3EI
Fax 2
y AB =
(l − x2 )
6 EIl
dy AB
Fa 2
=
(l − 3x 2 )
dx
6 EIl
dy AB
= θA
dx
Fal 2 Fal
θA =
=
6 EIl 6 EI
Fa 2l
yO = −θ A a = −
6 EI
At x = 0,
With both loads,
Fa 2l Fa 2
−
(l + a )
6 EI 3EI
400(3002 )
Fa 2
=−
(3l + 2a) = −
[3(500) + 2(300)] = −3.72 mm Ans.
6 EI
6(207)103 (16.384)103
At midspan,
2
2 Fa (l / 2)  2  l   3 Fal 2
3
400(300)(5002 )
yE =
=
= 1.11 mm Ans.
l −    =
6 EIl 
24 207 (103 )16.384 (103 )
 2   24 EI
_____________________________________________________________________________
π
4-14 I = (24 − 1.54 ) = 0.5369 in 4
64
yO = −
From Table A-5, E = 10.4 Mpsi
From Table A-9, beams 1 and 2, by superposition
Shigley’s MED, 10th edition
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−200 [ 4(12)]
300 [ 2(12) ]
F l 3 F a2
yB = − B + A (a − 3l ) =
+
[ 2(12) − 3(4)(12)]
6
3EI 6 EI
3(10.4)10 (0.5369) 6(10.4)106 (0.5369)
yB = −1.94 in Ans.
______________________________________________________________________________
3
2
4-15 From Table A-7, I = 2(1.85) = 3.70 in4
From Table A-5, E = 30.0 Mpsi
From Table A-9, beams 1 and 3, by superposition
[5 + 2(5 / 12)] (60 ) = −0.182 in Ans.
Fl 3 ( w + wc )l 4
150(603 )
yA = −
−
=−
−
6
3EI
8 EI
3(30)10 (3.70)
8(30)106 (3.70)
______________________________________________________________________________
π 4
4-16 I =
d
64
From Table A-5, E = 207(103 ) MPa
From Table A-9, beams 5 and 9, with FC = FA = F, by superposition
F l3
Fa
1
 − FB l 3 + 2 Fa (4a 2 − 3l 2 ) 
yB = − B +
(4a 2 − 3l 2 ) ⇒ I =
48 EI 24 EI
48 EyB
4
I=
{
}
1
−550(10003 ) + 2 ( 375) (250)  4(2502 ) − 3(10002 ) 
3
48(207)10 ( −2 )
( )
= 53.624 103 mm 4
d=
4
64
π
I =
4
64
π
(53.624)103 = 32.3 mm
Ans.
______________________________________________________________________________
4-17 From Table A-9, beams 8 (region BC for this beam with a = 0) and 10 (with a = a), by
superposition
(
)
(
)
MA 3
Fax 2
x − 3lx 2 + 2l 2 x +
l − x2
6 EIl
6 EIl
1 
=
M A x 3 − 3lx 2 + 2l 2 x + Fax l 2 − x 2 
6 EIl 
y AB =
(
)
(
)
Ans.
d M
F (x − l)

yBC =   A x3 − 3lx 2 + 2l 2 x   ( x − l ) +
[( x − l )2 − a(3x − l )]
dx
6
EIl
6
EI
  x =l
 
M l
F (x − l)
= − A (x − l) +
[( x − l )2 − a (3x − l )]
6 EI
6 EI
(x − l)
=
− M Al + F ( x − l ) 2 − a (3 x − l ) 
Ans.
6 EI
______________________________________________________________________________
(
{
)
}
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4-18 Note to the instructor: Beams with discontinuous loading are better solved using
singularity functions. This eliminates matching the slopes and displacements at the
discontinuity as is done in this solution.
a
wa

∑ M C = 0 = R1l − wa  l − a + 2  ⇒ R1 = 2l ( 2l − a ) Ans.
wa
wa 2
w
F
=
0
=
2
l
−
a
+
R
−
a
⇒
R
=
Ans.
(
) 2
∑ y
2
2l
2l
wa
w
VAB = R1 − wx =
( 2l − a ) − wx =  2l ( a − x ) − a 2  Ans.
2l
2l
2
wa
VBC = − R2 = −
Ans.
2l

w 
x2 
M AB = ∫ VAB dx =  2l  ax −  − a 2 x  + C1
2l  
2 

M AB = 0 at x = 0 ∴ C1 = 0
⇒
M AB =
wx
 2al − a 2 − lx  Ans.
2l 
wa 2
wa 2
dx = −
x + C2
2l
2l
wa 2
wa 2
M BC = 0 at x = l ∴ C2 =
⇒ M BC =
(l − x)
Ans.
2
2l
M
1 wx
1 w  2 1 2 2 1 3

2al − a 2 − lx ) dx =
θ AB = ∫ AB dx =
(
 alx − a x − lx  + C3 

∫
EI
EI 2l
EI  2l 
2
3 

M BC = ∫ VBC dx = ∫ −

1 w 2 1 2 2 1 3
 alx − a x − lx  + C3  dx

∫
EI  2l 
2
3 


1 w 1 3 1 2 3 1 4
=
 alx − a x − lx  + C3 x + C4 

EI  2l  3
6
12 

y AB = ∫ θ AB dx =
y AB = 0 at x = 0 ∴ C4 = 0
θ BC = ∫
M BC
1
dx =
EI
EI

wa 2
1  wa 2 
1 2
−
=
(
)
l
x
dx
 lx − x  + C5 

∫ 2l
2 
EI  2l 

θ AB = θ BC at x = a ∴

 1  wa 2 
1 w 2 1 4 1 3
1 2
wa 3
+ C5
 ala − a − la  + C3  =
 la − a  + C5  ⇒ C3 =


2
3 
2 
6
EI  2l 
 EI  2l 

Shigley’s MED, 10th edition
(1)
Chapter 4 Solutions, Page 11/80
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

1  wa 2 
1 2
1  wa 2  1 2 1 3 
lx
−
x
+
C
dx
=
5


 lx − x  + C5 x + C6 


∫
EI  2l 
EI  2l  2
2 
6 


2 2
wa l
= 0 at x = l ∴ C6 = −
− C5l
6

1  wa 2  1 2 1 3 1 3 
=
 lx − x − l  + C5 ( x − l ) 

EI  2l  2
6
3 

yBC = ∫ θ BC dx =
yBC
yBC
y AB = yBC at x = a ∴
w 1
wa 2  1 2 1 3 1 3 
1 5 1 4
3
ala
−
a
−
la
+
C
a
=
3


 la − a − l  + C5 (a − l )
2l  3
6
12
2l  2
6
3 

2
wa
C3 a =
3la 2 − 4l 3 ) + C5 (a − l )
(2)
(
24l
wa 2
Substituting (1) into (2) yields C5 =
( −a 2 − 4l 2 ) . Substituting this back into (2) gives
24l
wa 2
C3 =
( 4al − a 2 − 4l 2 ) . Thus,
24l
w
y AB =
( 4alx3 − 2a 2 x3 − lx 4 + 4a3lx − a 4 x − 4a 2l 2 x )
24 EIl
wx 
2
⇒ y AB =
2ax 2 (2l − a ) − lx 3 − a 2 ( 2l − a ) 
Ans.

24 EIl 
w
yBC =
6a 2lx 2 − 2a 2 x 3 − a 4 x − 4a 2l 2 x + a 4l )
Ans.
(
24 EIl
This result is sufficient for yBC. However, this can be shown to be equivalent to
w
w
yBC =
4alx 3 − 2a 2 x 3 − lx 4 − 4a 2l 2 x + 4a 3lx − a 4 x ) +
( x − a)4
(
24 EIl
24 EI
w
yBC = y AB +
( x − a)4
Ans.
24 EI
by expanding this or by solving the problem using singularity functions.
______________________________________________________________________________
4-19
The beam can be broken up into a uniform load w downward from points A to C and a
uniform load w upward from points A to B. Using the results of Prob. 4-18, with b = a for
A to C and a = a for A to B, results in
wx 
wx 
2
2
y AB =
2bx 2 (2l − b) − lx 3 − b 2 ( 2l − b )  −
2ax 2 (2l − a) − lx 3 − a 2 ( 2l − a ) 




24 EIl
24 EIl
wx 
2
2
=
2bx 2 (2l − b) − b 2 ( 2l − b ) − 2ax 2 (2l − a) + a 2 ( 2l − a ) 
Ans.

24 EIl 
w 
2
yBC =
2bx 3 (2l − b) − lx 4 − b 2 x ( 2l − b )
24 EIl 
− ( 4alx 3 − 2a 2 x 3 − lx 4 − 4a 2l 2 x + 4a 3lx − a 4 x ) − l ( x − a) 4 
Shigley’s MED, 10th edition
Ans.
Chapter 4 Solutions, Page 12/80
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w
 4blx3 − 2b 2 x3 − lx 4 − 4b 2l 2 x + 4b3lx − b 4 x + l ( x − b)4 
24 EIl
w
 4alx3 − 2a 2 x3 − lx 4 − 4a 2l 2 x + 4a 3lx − a 4 x + l ( x − a )4 
−
24 EIl 
w
( x − b) 4 − ( x − a ) 4  + y AB
=
Ans.
24 EI
______________________________________________________________________________
yCD =
4-20
Note to the instructor: See the note in the solution for Problem 4-18.
wa 2
wa
F
=
0
=
R
−
− wa ⇒ RB =
( 2l + a ) Ans.
∑ y
B
2l
2l
For region BC, isolate right-hand element of length (l + a − x)
wa 2
VAB = − RA = −
,
VBC = w ( l + a − x )
Ans.
2l
wa 2
w
2
M AB = − RA x = −
x,
M BC = − ( l + a − x )
Ans.
2l
2
wa 2 2
EIθ AB = ∫ M AB dx = −
x + C1
4l
wa 2 3
EIy AB = −
x + C1 x + C2
12l
wa 2 3
yAB = 0 at x = 0 ⇒ C2 = 0 ∴ EIy AB = −
x + C1 x
12l
w a 2l
yAB = 0 at x = l ⇒ C1 =
∴
12
w a 2 3 w a 2l
wa2 x 2
wa 2 x 2
2
EIy AB = −
x +
x=
l
−
x
⇒
y
=
(
)
( l − x2 ) Ans.
AB
12l
12
12l
12 EIl
w
3
EIθ BC = ∫ M BC dx = − ( l + a − x ) + C3
6
w
4
EIy BC = − ( l + a − x ) + C3 x + C4
24
wa 4
wa 4
(1)
yBC = 0 at x = l ⇒ −
+ C3l + C4 = 0 ⇒ C4 =
− C3l
24
24
wa 2l wa 2l wa 3
wa 2
θAB = θBC at x = l ⇒ −
+
=
+ C3 ⇒ C3 = −
(l + a )
4
12
6
6
wa 2 2
 a + 4l ( l + a )  . Substitute back into yBC
Substitute C3 into Eq. (1) gives C4 =
24 

wa 2
wa 4 wa 2l
1  w
4
yBC =
−
l
+
a
−
x
−
x
l
+
a
+
+
(
)
(
)
( l + a )

EI  24
6
24
6

=−
w 
4
l + a − x ) − 4a 2 ( l − x )( l + a ) − a 4 
(

24 EI 
Shigley’s MED, 10th edition
Ans.
Chapter 4 Solutions, Page 13/80
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Table A-9, beam 7,
w l 100(10)
R1 = R2 =
=
= 500 lbf ↑
2
2
wx
100 x
 2(10) x 2 − x 3 − 103 
2lx 2 − x 3 − l 3 ) =
y AB =
(
6
24 EI
24 ( 30 )10 ( 0.05 )
4-21
= 2.7778 (10−6 ) x ( 20 x 2 − x 3 − 1000 )
Slope:
θ AB =
w
wl 3
2
3
3
6
l
l
−
4
l
−
l
=
(
)
x =l
24 EI
24 EI
3
100 (103 )
wl
(x −l) =
(x −l) =
( x − 10 ) = 2.7778 (10−3 ) ( x − 10 )
x =l
24 EI
24(30)106 (0.05)
At x = l, θ AB
yBC = θ AB
d y AB
w
=
( 6lx 2 − 4 x3 − l 3 )
24 EI
dx
=
From Prob. 4-20,
2
wa 2 100 ( 4 )
RA =
=
= 80 lbf ↓
2l
2(10)
RB =
100 ( 4 )
wa
( 2l + a ) =
[ 2(10) + 4] = 480 lbf ↑
2l
2(10)
100 ( 42 ) x
wa 2 x 2
2
l
−
x
=
(
) 12 ( 30 )106 ( 0.05 ) (102 − x 2 ) = 8.8889 (10−6 ) x (100 − x 2 )
12 EIl
w 
4
=−
( l + a − x ) − 4a 2 ( l − x )( l + a ) − a 4 

24 EI
100
(10 + 4 − x )4 − 4 ( 42 ) (10 − x )(10 + 4 ) − 44 
=−
6

24 ( 30 )10 ( 0.05 ) 
y AB =
yBC
4
= −2.7778 (10−6 ) (14 − x ) + 896 x − 9216 


Superposition,
RA = 500 − 80 = 420 lbf ↑
RB = 500 + 480 = 980 lbf ↑
y AB = 2.7778 (10 −6 ) x ( 20 x 2 − x 3 − 1000 ) + 8.8889 (10 −6 ) x (100 − x 2 )
Ans.
Ans.
yBC = 2.7778 (10−3 ) ( x − 10 ) − 2.7778 (10−6 ) (14 − x ) + 896 x − 9216
Ans.


The deflection equations can be simplified further. However, they are sufficient for
plotting.
Using a spreadsheet,
4
x
y
0
0.5
1
1.5
2
2.5
3
3.5
0.000000 -0.000939 -0.001845 -0.002690 -0.003449 -0.004102 -0.004632 -0.005027
x
y
4
4.5
5
5.5
6
6.5
7
7.5
-0.005280 -0.005387 -0.005347 -0.005167 -0.004853 -0.004421 -0.003885 -0.003268
Shigley’s MED, 10th edition
Chapter 4 Solutions, Page 14/80
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x
y
x
y
8
8.5
9
9.5
10
10.5
11
11.5
-0.002596 -0.001897 -0.001205 -0.000559 0.000000 0.000439 0.000775 0.001036
12
0.001244
12.5
0.001419
13
13.5
14
0.001575 0.001722 0.001867
0.002
Beam Deflection, Prob. 4-21
0.001
0.000
-0.001
y (in) -0.002
-0.003
-0.004
-0.005
-0.006
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
x (in)
______________________________________________________________________________
4-22
(a) Useful relations
k=
F 48 EI
= 3
y
l
I=
3
kl 3 1800 ( 36 )
=
= 0.05832 in 4
48 E 48(30)106
From I = bh 3/12, and b = 10 h, then I = 5 h 4/6, or,
h=
4
6 I 4 6(0.05832)
=
= 0.514 in
5
5
h is close to 1/2 in and 9/16 in, while b is close to 5.14 in. Changing the height drastically
changes the spring rate, so changing the base will make finding a close solution easier.
Trial and error was applied to find the combination of values from Table A-17 that
yielded the closet desired spring rate.
h (in)
1/2
1/2
1/2
9/16
9/16
b (in)
5
5½
5¾
5
4
b/h
10
11
11.5
8.89
7.11
k (lbf/in)
1608
1768
1849
2289
1831
Shigley’s MED, 10th edition
Chapter 4 Solutions, Page 15/80
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h = ½ in, b = 5 ½ in should be selected because it results in a close spring rate and b/h is
still reasonably close to 10.
(b) I = 5.5(0.5)3 / 12 = 0.05729 in 4
Mc ( Fl / 4)c
4σ I 4(60)103 (0.05729)
σ=
=
⇒ F=
=
= 1528 lbf
I
I
lc
( 36 ) (0.25)
(1528) ( 36 )
Fl 3
=
= 0.864 in Ans.
48EI 48(30)106 (0.05729)
______________________________________________________________________________
3
y=
4-23 From the solutions to Prob. 3-68, T1 = 60 lbf and T2 = 400 lbf
I=
πd4
64
=
π (1.25)4
64
= 0.1198 in 4
From Table A-9, beam 6,
Fb x
Fb x

z A =  1 1 ( x 2 + b12 − l 2 ) + 2 2 ( x 2 + b2 2 − l 2 ) 
6 EIl
 6 EIl
 x =10in
(−575)(30)(10)
=
(102 + 302 − 402 )
6(30)106 (0.1198)(40)
460(12)(10)
+
(102 + 122 − 402 ) = 0.0332 in
6(30)106 (0.1198)(40)
Ans.
dz
 d Fb x
Fb x

= −   1 1 ( x 2 + b12 − l 2 ) + 2 2 ( x 2 + b2 2 − l 2 )  

6 EIl
 dx  x =10in
  x =10in
 dx  6 EIl
Fb
 Fb

= −  1 1 (3 x 2 + b12 − l 2 ) + 2 2 (3 x 2 + b2 2 − l 2 ) 
6
6
EIl
EIl

 x =10in
(θ A ) y = − 
=−
(575)(30)
3 (102 ) + 302 − 402 

6(30)106 (0.1198)(40) 
−460(12)
3 (102 ) + 122 − 402 
−
6

6(30)10 (0.1198)(40) 
Ans.
= 6.02(10−4 ) rad
______________________________________________________________________________
4-24 From the solutions to Prob. 3-69, T1 = 2880 N and T2 = 432 N
I=
πd4
64
=
π (30)4
64
( )
= 39.76 103 mm 4
Shigley’s MED, 10th edition
Chapter 4 Solutions, Page 16/80
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The load in between the supports supplies an angle to the overhanging end of the beam.
That angle is found by taking the derivative of the deflection from that load. From Table
A-9, beams 6 (subscript 1) and 10 (subscript 2),
y A = θ BC
θ BC
C
C
( a2 ) beam 6 + ( y A )beam10
(1)
 d  F a ( l − x ) 2
 
 Fa

x + a12 − 2lx )   =  1 1 ( 6lx − 3 x 2 − a12 − 2l 2 ) 
=  1 1
(
6
6
dx
EIl
EIl
 x =l
  x =l 
 
F1a1 2
( l − a12 )
6 EIl
Equation (1) is thus
=
F1a1 2
Fa2
l − a12 ) a2 − 2 2 (l + a2 )
(
6 EIl
3EI
2070(3002 )
−3312(230)
2
2
510
230
300
=
−
−
(
)
( 510 + 300 )
(
)
6(207)103 (39.76)103 (510)
3(207)103 (39.76)103
Ans.
= −7.99 mm
yA =
The slope at A, relative to the z axis is
(θ A ) z =
F1a1 2
 d  F (x − l)

( x − l )2 − a2 (3x − l )   
(l − a12 ) +   2

6 EIl
  x =l + a2
 dx  6 EI
F1a1 2
F
l − a12 ) + 2 3( x − l ) 2 − 3a2 ( x − l ) − a2 (3 x − l ) 
(
x = l + a2
6 EIl
6 EI
Fa
F
= 1 1 (l 2 − a12 ) − 2 ( 3a2 2 + 2la2 )
6 EIl
6 EI
−3312(230)
=
5102 − 2302 )
(
3
3
6(207)10 (39.76)10 (510)
2070
3(3002 ) + 2(510)(300) 
−
3
3 
6(207)10 (39.76)10
= −0.0304 rad
Ans.
______________________________________________________________________________
=
4-25 From the solutions to Prob. 3-70, T1 = 392.16 lbf and T2 = 58.82 lbf
I=
πd4
=
64
From Table A-9, beam 6,
π (1) 4
64
= 0.049 09 in 4
Shigley’s MED, 10th edition
Chapter 4 Solutions, Page 17/80
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( −350)(14)(8)
Fb x

y A =  1 1 ( x 2 + b12 − l 2 ) 
82 + 14 2 − 22 2 ) = 0.0452 in Ans.
=
(
6
 6 EIl
 x =8in 6(30)10 (0.049 09)(22)
( −450.98)(6)(8)
F b x

z A =  2 2 ( x 2 + b2 2 − l 2 ) 
82 + 6 2 − 22 2 ) = 0.0428 in Ans.
=
(
6
6
EIl
6(30)10
(0.049
09)(22)

 x =8in
The displacement magnitude is δ = y A2 + z A2 = 0.04522 + 0.04282 = 0.0622 in
Ans.
d y
 d  Fb x
Fb

=   1 1 ( x 2 + b12 − l 2 )  
= 1 1 (3a12 + b12 − l 2 )

  x = a1 6 EIl
 d x  x = a1  dx  6 EIl
(θ A ) z = 
=
(−350)(14)
3 ( 82 ) + 142 − 222  = 0.00242 rad
6

6(30)10 (0.04909)(22) 
Ans.
 dz
 d  −F b x
Fb

= −   2 2 ( x 2 + b22 − l 2 )  
= 2 2 ( 3a12 + b22 − l 2 )

  x = a1 6 EIl
 dx  6 EIl
 d x  x = a1
(θ A ) y =  −
=
(450.98)(6)
3 ( 82 ) + 62 − 222  = −0.00356 rad

6(30)106 (0.04909)(22) 
Ans.
The slope magnitude is Θ A = 0.002422 + ( −0.00356 ) = 0.00430 rad Ans.
2
______________________________________________________________________________
4-26 From the solutions to Prob. 3-71, T1 = 250 N and T2 = 37.5 N
I=
πd4
64
=
π (20) 4
64
= 7 854 mm 4
−345sin 45o ) (550)(300)
(
 F1 y b1 x 2
2
2 
( x + b1 − l ) 
=
yA = 
( 3002 + 5502 − 8502 )
3
6
6(207)10
(7
854)(850)
EIl

 x =300mm
= 1.60 mm Ans.
Fb x
F b x

z A =  1z 1 ( x 2 + b12 − l 2 ) + 2 2 ( x 2 + b2 2 − l 2 ) 
6 EIl
 6 EIl
 x =300mm
( 345cos 45 ) (550)(300)
=
o
( 3002 + 5502 − 8502 )
6(207)103 (7 854)(850)
−287.5(150)(300)
+
( 3002 + 1502 − 8502 ) = −0.650 mm
6(207)103 (7 854)(850)
Ans.
The displacement magnitude is δ = y A2 + z A2 = 1.602 + ( −0.650 ) = 1.73 mm
2
Shigley’s MED, 10th edition
Ans.
Chapter 4 Solutions, Page 18/80
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 d  F b x
F b
 
d y
=   1 y 1 ( x 2 + b12 − l 2 )   = 1 y 1 (3a12 + b12 − l 2 )

 d x  x = a1  dx  6 EIl
  x = a1 6 EIl
(θ A ) z = 
=
− ( 345sin 45o ) (550)
3 ( 3002 ) + 5502 − 8502  = 0.00243 rad

6(207)10 (7 854)(850) 
Ans.
3
dz
 d F b x
Fb x

= −   1z 1 ( x 2 + b12 − l 2 ) + 2 2 ( x 2 + b2 2 − l 2 )  

6 EIl
 x = a1
 dx  6 EIl
 d x  x = a1
(θ A ) y = − 
F1z b1
Fb
3a12 + b12 − l 2 ) − 2 2 ( 3a12 + b22 − l 2 )
(
6 EIl
6 EIl
o
( 345cos 45 ) (550) 3 3002 + 5502 − 8502 
=−
( )

6(207)103 (7 854)(850) 
=−
−
−287.5(150)
3 ( 3002 ) + 1502 − 8502  = 1.91 ⋅10−4 rad

6(207)103 (7 854)(850) 
Ans.
The slope magnitude is Θ A = 0.002432 + 0.0001912 = 0.00244 rad Ans.
______________________________________________________________________________
4-27 From the solutions to Prob. 3-72, FB = 750 lbf
I=
πd4
64
=
π (1.25)4
64
= 0.1198 in 4
From Table A-9, beams 6 (subscript 1) and 10 (subscript 2)
F2 y a2 x 2
 F1 y b1 x 2

yA = 
x + b12 − l 2 ) +
l − x2 )
(
(
6 EIl
 6 EIl
 x =16in
( −300 cos 20 ) (14)(16)
o
=
6(30)106 (0.119 8)(30)
Ans.
= 0.0805 in
(162 + 142 − 302 ) +
( 750sin 20 ) (9)(16)
o
6(30)106 (0.119 8)(30)
( 30
2
− 162 )
F ax
F b x

z A =  1z 1 ( x 2 + b12 − l 2 ) + 2 z 2 ( l 2 − x 2 ) 
6 EIl
 6 EIl
 x =16in
( 300sin 20 ) (14)(16)
o
=
6(30)106 (0.119 8)(30)
= −0.1169 in
(16
2
+ 14 − 30
2
2
)
( −750 cos 20 ) (9)(16)
+
o
6(30)106 (0.119 8)(30)
( 30
2
− 162 )
Ans.
The displacement magnitude is δ = y A2 + z A2 = 0.08052 + ( −0.1169 ) = 0.142 in
2
Shigley’s MED, 10th edition
Ans.
Chapter 4 Solutions, Page 19/80
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 d  F b x
F ax
 
d y
=   1 y 1 ( x 2 + b12 − l 2 ) + 2 y 2 ( l 2 − x 2 )  

6 EIl
 d x  x = a1  dx  6 EIl
  x = a1
(θ A ) z = 
( 3a + b − l ) + 6EIl ( l − 3a )
6 EIl
( −300 cos 20 ) (14) 3 16 + 14 − 30 
=
( )

6(30)10 (0.119 8)(30) 
=
F1 y b1
2
1
2
1
2
F2 y a2
2
2
1
o
2
2
2
6
( 750sin 20 ) (9)
o
+
302 − 3 (162 )  = 8.06 (10−5 ) rad

6(30)106 (0.119 8)(30) 
Ans.
dz
 d F b x
F ax

= −   1z 1 ( x 2 + b12 − l 2 ) + 2 z 2 ( l 2 − x 2 )  

6 EIl
  x = a1
 dx  6 EIl
 d x  x = a1
(θ A ) y = − 
F1z b1
F a
3a12 + b12 − l 2 ) − 2 z 2 ( l 2 − 3a12 )
(
6 EIl
6 EIl
o
−750 cos 20o ) (9)
300sin 20 ) (14)
(
(
2
2
2
2
 2

3 (16 ) + 14 − 30  −
=−
 6(30)106 (0.119 8)(30) 30 − 3 (16 ) 
6(30)106 (0.119 8)(30) 
=−
= 0.00115 rad
Ans.
The slope magnitude is Θ A = 8.06 (10−5 )  + 0.001152 = 0.00115 rad Ans.
______________________________________________________________________________
2
4-28 From the solutions to Prob. 3-73, FB = 22.8 (103) N
4
π d 4 π ( 50 )
=
= 306.8 (103 ) mm 4
I=
64
64
From Table A-9, beam 6,
F bx
F b x

y A =  1 y 1 ( x 2 + b12 − l 2 ) + 2 y 2 ( x 2 + b2 2 − l 2 ) 
6 EIl
 6 EIl
 x = 400mm
11(103 ) sin 20o  (650)(400)

=
( 4002 + 6502 − 10502 )
6(207)103 (306.8)103 (1050)
 22.8 (103 ) sin 25o  (300)(400)

+
( 4002 + 3002 − 10502 )
6(207)103 (306.8)103 (1050)
= −3.735 mm
Ans.
Shigley’s MED, 10th edition
Chapter 4 Solutions, Page 20/80
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F bx
F b x

z A =  1z 1 ( x 2 + b12 − l 2 ) + 2 z 2 ( x 2 + b2 2 − l 2 ) 
6 EIl
 6 EIl
 x = 400mm
11(103 ) cos 20o  (650)(400)

4002 + 6502 − 10502 )
=
(
3
3
6(207)10 (306.8)10 (1050)
 −22.8 (103 ) cos 25o  (300)(400)

+
( 4002 + 3002 − 10502 ) = 1.791 mm
6(207)103 (306.8)103 (1050)
The displacement magnitude is δ = y A2 + z A2 =
( −3.735)
2
+ 1.7912 = 4.14 mm
Ans.
Ans.
d y
 d F b x
F bx

=   1z 1 ( x 2 + b12 − l 2 ) + 2 z 2 ( x 2 + b2 2 − l 2 )  

6 EIl
  x = a1
 d x  x = a1  dx  6 EIl
(θ A ) z = 
( 3a + b − l ) + 6 EIl ( 3a + b − l )
6 EIl
11(10 ) sin 20  (650)

3 ( 400 ) + 650
= 
6(207)10 (306.8)10 (1050) 
=
F1 y b1
2
1
3
2
1
F2 y b2
2
2
1
2
2
o
2
3
2
3
2
− 10502 
 22.8 (103 ) sin 25o  (300)

3 ( 4002 ) + 3002 − 10502 
+ 

6(207)103 (306.8)103 (1050) 
Ans.
= −0.00507 rad
dz
 d F b x
F bx

= −   1z 1 ( x 2 + b12 − l 2 ) + 2 z 2 ( x 2 + b2 2 − l 2 )  

6 EIl
  x = a1
 dx  6 EIl
 d x  x = a1
(θ A ) y = − 
F1z b1
F b
3a12 + b12 − l 2 ) − 2 z 2 ( 3a12 + b22 − l 2 )
(
6 EIl
6 EIl
3
o
11(10 ) cos 20  (650)

3 ( 4002 ) + 6502 − 10502 
=− 
3
3

6(207)10 (306.8)10 (1050) 
=−
 −22.8 (103 ) cos 25o  (300)

3 ( 4002 ) + 3002 − 10502 
− 
3
3

6(207)10 (306.8)10 (1050) 
= −0.00489 rad
Ans.
The slope magnitude is Θ A =
( −0.00507 ) + ( −0.00489 )
2
2
= 0.00704 rad Ans.
______________________________________________________________________________
4-29 From the solutions to Prob. 3-68, T1 = 60 lbf and T2 = 400 lbf , and Prob. 4-23, I = 0.119 8
in4. From Table A-9, beam 6,
Shigley’s MED, 10th edition
Chapter 4 Solutions, Page 21/80
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dz
 d  F1z b1 x 2
F bx

x + b12 − l 2 ) + 2 z 2 ( x 2 + b22 − l 2 )  
(
 = − 
6 EIl
  x =0
 dx  6 EIl
 d x  x =0
−575(30)
F b
F b
302 − 402 )
= − 1z 1 ( b12 − l 2 ) − 2 z 2 ( b2 2 − l 2 ) = −
(
6
6 EIl
6 EIl
6(30)10 (0.119 8)(40)
(θO ) y = − 
−
460(12)
122 − 402 ) = −0.00468 rad
(
6
6(30)10 (0.119 8)(40)
Ans.
 d  F1z a1 ( l − x ) 2
F a (l − x ) 2
 
dz
x + a12 − 2lx ) + 2 z 2
x + a22 − 2lx )  
(
(
 = − 
6 EIl
 dx  6 EIl
 d x  x =l
  x =l
F a
F a

= −  1z 1 ( 6lx − 2l 2 − 3 x 2 − a12 ) + 2 z 2 ( 6lx − 2l 2 − 3 x 2 − a22 ) 
6 EIl
 6 EIl
 x =l
(θC ) y = − 
F1z a1 2
F a
l − a12 ) − 2 z 2 ( l 2 − a22 )
(
6 EIl
6 EIl
2
−575(10) ( 40 − 102 )
460(28) ( 402 − 282 )
=−
−
= −0.00219 rad
Ans.
6(30)106 (0.119 8)(40) 6(30)106 (0.119 8)(40)
______________________________________________________________________________
=−
4-30 From the solutions to Prob. 3-69, T1 = 2 880 N and T2 = 432 N, and Prob. 4-24, I = 39.76
(103) mm4. From Table A-9, beams 6 and 10
d y
 d Fb x

Fa x
(θO ) z =   =   1 1 ( x 2 + b12 − l 2 ) + 2 2 (l 2 − x 2 )  
6 EIl
  x =0
 d x  x = 0  dx  6 EIl
Fa
Fb
Fal
 Fb

=  1 1 (3 x 2 + b12 − l 2 ) + 2 2 (l 2 − 3x 2 )  = 1 1 (b12 − l 2 ) + 2 2
6 EIl
6 EI
 6 EIl
 x = 0 6 EIl
2 070(300)(510)
−3 312(280)
2802 − 5102 ) +
=
(
3
3
6(207)10 (39.76)10 (510)
6(207)103 (39.76)103
= 0.0131 rad
Ans.
d y
Fa x
 d  F1a1 (l − x) 2

( x + a12 − 2lx) + 2 2 (l 2 − x 2 )  
 = 
6 EIl
  x =l
 d x  x =l  dx  6 EIl
(θC ) z = 
Fa
Fa
Fal
 Fa

=  1 1 (6lx − 2l 2 − 3x 2 − a12 ) + 2 2 (l 2 − 3x 2 )  = 1 1 (l 2 − a12 ) − 2 2
6
6
6
3EI
EIl
EIl
EIl

 x =l
2 070(300)(510)
−3 312(230)
(5102 − 2302 ) −
=
3
3
6(207)10 (39.76)10 (510)
3(207)103 (39.76)103
= −0.0191 rad
Ans.
______________________________________________________________________________
4-31 From the solutions to Prob. 3-70, T1 = 392.19 lbf and T2 = 58.82 lbf , and Prob. 4-25, I =
0.0490 9 in4. From Table A-9, beam 6
Shigley’s MED, 10th edition
Chapter 4 Solutions, Page 22/80
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F1 y b1 2 2
 
d y
 d  F1 y b1 x 2
x + b12 − l 2 )   =
(b1 − l )
(
 = 
 d x  x = 0  dx  6 EIl
  x = 0 6 EIl
−350(14)
=
(142 − 222 ) = 0.00726 rad Ans.
6(30)106 (0.04909)(22)
(θO ) z = 
dz
 d  F2 z b2 x 2
F b

x + b22 − l 2 )   = − 2 z 2 ( b22 − l 2 )
(
 = − 
6 EIl
 x =0
 dx  6 EIl
 d x  x =0
−450.98(6)
=−
( 62 − 222 )
6(30)106 (0.04909)(22)
(θO ) y = − 
= −0.00624 rad
Ans.
The slope magnitude is ΘO = 0.007262 + ( −0.00624 ) = 0.00957 rad Ans.
2
 
d y
 d  F1 y a1 (l − x) 2
x + a12 − 2lx )  
(
 = 
 d x  x =l  dx  6 EIl
  x =l
F a
F a

=  1 y 1 ( 6lx − 2l 2 − 3 x 2 − a12 )  = 1 y 1 (l 2 − a12 )
 6 EIl
 x =l 6 EIl
(θC ) z = 
=
−350(8)
( 222 − 82 ) = −0.00605 rad
6(30)106 (0.0491)(22)
Ans.
dz
 d  F2 z a2 (l − x) 2

x + a22 − 2lx )  
(
 = − 
  x =l
 dx  6 EIl
 d x  x =l
(θC ) y = − 
F a
F a

= −  2 z 2 ( 6lx − 2l 2 − 3 x 2 − a22 )  = − 2 z 2 ( l 2 − a22 )
6 EIl
 6 EIl
 x =l
−450.98(16)
=−
222 − 162 ) = 0.00846 rad
(
6
6(30)10 (0.04909)(22)
The slope magnitude is ΘC =
( −0.00605)
2
Ans.
+ 0.008462 = 0.0104 rad Ans.
______________________________________________________________________________
4-32 From the solutions to Prob. 3-71, T1 =250 N and T1 =37.5 N, and Prob. 4-26, I = 7 854
mm4. From Table A-9, beam 6
 d  F1 y b1 x 2
F b
 
d y
x + b12 − l 2 )   = 1 y 1 (b12 − l 2 )
(
 = 
 d x  x = 0  dx  6 EIl
  x = 0 6 EIl
(θO ) z = 
 −345sin 45o  (550)
=
( 5502 − 8502 ) = 0.00680 rad
6(207)103 (7 854)(850)
Shigley’s MED, 10th edition
Ans.
Chapter 4 Solutions, Page 23/80
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dz
F bx
 d  F1z b1 x 2

x + b12 − l 2 ) + 2 z 2 ( x 2 + b22 − l 2 )  
(
 = − 
6 EIl
 x =0
 dx  6 EIl
 d x  x=0
(θO ) y = − 
345cos 45o  (550)
F1z b1 2 2 F2 z b2 2 2
b1 − l ) −
b2 − l ) = −
5502 − 8502 )
=−
(
(
(
3
6 EIl
6 EIl
6(207)10 (7 854)(850)
−287.5(150)
−
(1502 − 8502 ) = 0.00316 rad Ans.
6(207)103 (7 854)(850)
The slope magnitude is ΘO = 0.006802 + 0.003162 = 0.00750 rad Ans.
 
 F1 y a1

d y
 d  F1 y a1 (l − x) 2
x + a12 − 2lx )   = 
6lx − 2l 2 − 3 x 2 − a12 ) 
(
(
 = 
 d x  x =l  dx  6 EIl
  x =l  6 EIl
 x =l
(θC ) z = 
 −345sin 45o  (300)
=
(l − a ) =
8502 − 3002 ) = −0.00558 rad
(
3
6 EIl
6(207)10 (7 854)(850)
F1 y a1
2
2
1
Ans.
dz
 d  F1z a1 (l − x) 2
F a (l − x) 2

x + a12 − 2lx ) + 2 z 2
x + a22 − 2lx )  
(
(
 = − 
6 EIl
  x =l
 dx  6 EIl
 d x  x =l
(θC ) y = − 
345cos 45o  (300)
F1z a1 2
F2 z a2 2
2
2
=−
( l − a1 ) − 6EIl ( l − a2 ) = − 6(207)103 (7 854)(850) (8502 − 3002 )
6 EIl
−287.5(700)
−
(8502 − 7002 ) = 6.04 (10−5 ) rad Ans.
6(207)103 (7 854)(850)
The slope magnitude is ΘC =
( −0.00558)
2
+ 6.04 (10−5 )  = 0.00558 rad Ans.
2
________________________________________________________________________
4-33 From the solutions to Prob. 3-72, FB = 750 lbf, and Prob. 4-27, I = 0.119 8 in4. From Table
A-9, beams 6 and 10
 d F b x
F ax
d y

(θO ) z =   =   1 y 1 ( x 2 + b12 − l 2 ) + 2 y 2 ( l 2 − x 2 ) 
6 EIl
 d x  x = 0  dx  6 EIl
 x = 0
F2 y a2 2
F1 y b1 2 2 F2 y a2l
 F1 y b1

3 x 2 + b12 − l 2 ) +
l − 3x 2 ) =
=
(
(
( b1 − l ) + 6EI
6 EIl
 6 EIl
 x =0 6 EIl
750sin 20o  (9)(30)
 −300 cos 20o  (14)
2
2
=
(14 − 30 ) + 6(30)106 (0.119 8) = 0.00751 rad
6(30)106 (0.119 8)(30)
Shigley’s MED, 10th edition
Ans.
Chapter 4 Solutions, Page 24/80
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dz
 d  F1z b1 x 2
F ax

x + b12 − l 2 ) + 2 z 2 ( l 2 − x 2 )  
(
 = − 
6 EIl
  x =0
 dx  6 EIl
 d x  x =0
(θO ) y = − 
F a
F b
F al
F b

= −  1z 1 ( 3 x 2 + b12 − l 2 ) + 2 z 2 ( l 2 − 3x 2 )  = − 1z 1 ( b12 − l 2 ) − 2 z 2
6 EIl
6 EIl
6 EI
 6 EIl
 x=0
300sin 20o  (14)
 −750 cos 20o  (9)(30)
2
2
=−
(14 − 30 ) − 6(30)106 (0.119 8) = 0.0104 rad
6(30)106 (0.119 8)(30)
Ans.
The slope magnitude is ΘO = 0.007512 + 0.0104 2 = 0.0128 rad Ans.
F2 y a2 x 2
 
 dy 
 d  F1 y a1 (l − x) 2
x + a12 − 2lx ) +
l − x 2 ) 
(
(
 = 
6 EIl
 dx  x =l  dx  6 EIl
 x =l
F2 y a2 2
F1 y a1 2
F2 y a2l
 F1 y a1

6lx − 2l 2 − 3x 2 − a12 ) +
(l − a12 ) −
l − 3x 2 ) =
=
(
(
6 EIl
3EI
 6 EIl
 x =l 6 EIl
(θC ) z = 
 −300 cos 20o  (16)
750sin 20o  (9)(30)
2
2
=
( 30 − 16 ) − 3(30)106 (0.119 8) = −0.0109 rad
6(30)106 (0.119 8)(30)
Ans.
dz
 d  F1z a1 (l − x) 2
F ax

x + a12 − 2lx ) + 2 z 2 ( l 2 − x 2 )  
(
 = − 
6 EIl
  x =l
 dx  6 EIl
 d x  x =l
(θC ) y = − 
F a
F a
F al
F a

= −  1z 1 ( 6lx − 2l 2 − 3 x 2 − a12 ) + 2 z 2 ( l 2 − 3 x 2 )  = − 1z 1 ( l 2 − a12 ) + 2 z 2
6 EIl
6 EIl
3EI
 6 EIl
 x =l
300sin 20o  (16)
 −750 cos 20o  (9)(30)
2
2
=−
( 30 − 16 ) + 3(30)106 (0.119 8) = −0.0193 rad
6(30)106 (0.119 8)(30)
( −0.0109 ) + ( −0.0193)
2
The slope magnitude is ΘC =
2
Ans.
= 0.0222 rad Ans.
______________________________________________________________________________
4-34 From the solutions to Prob. 3-73, FB = 22.8 kN, and Prob. 4-28, I = 306.8 (103) mm4.
From Table A-9, beam 6
 d  F1 y b1 x 2
F bx
 
d y
x + b12 − l 2 ) + 2 y 2 ( x 2 + b22 − l 2 )  
(
 = 
6 EIl
 d x  x = 0  dx  6 EIl
 x = 0
(θO ) z = 
11(103 ) sin 20o  (650)


=
( b − l ) + 6EIl ( b − l ) = 6(207)10
( 6502 − 10502 )
3
6 EIl
(306.8)103 (1050)
F1 y b1
2
1
2
F2 y b2
2
2
2
 22.8 (103 ) sin 25o  (300)

+ 
( 3002 − 10502 ) = −0.0115 rad
6(207)103 (306.8)103 (1050)
Shigley’s MED, 10th edition
Ans.
Chapter 4 Solutions, Page 25/80
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dz
 d  F1z b1 x 2
F bx

x + b12 − l 2 ) + 2 z 2 ( x 2 + b22 − l 2 )  
(
 = − 
6 EIl
 x =0
 dx  6 EIl
 d x  x =0
F b
F b
= − 1z 1 ( b12 − l 2 ) − 2 z 2 ( b22 − l 2 )
6 EIl
6 EIl
3
11(10 ) cos 20o  (650)

=− 
6502 − 10502 )
(
3
3
6(207)10 (306.8)10 (1050)
(θO ) y = − 
 −22.8 (103 ) cos 25o  (300)

−
3002 − 10502 ) = −0.00427 rad
(
3
3
6(207)10 (306.8)10 (1050)
( −0.0115) + ( −0.00427 )
2
The slope magnitude is ΘO =
2
Ans.
= 0.0123 rad Ans.
 d  F1 y a1 (l − x) 2
F2 y a2 (l − x) 2
d y
 
x + a12 − 2lx ) +
x + a22 − 2lx )  
(
(
 = 
6 EIl
 d x  x =l  dx  6 EIl
  x =l
F a
F a

=  1 y 1 (6lx − 2l 2 − 3 x 2 − a12 ) + 2 y 2 ( 6lx − 2l 2 − 3x 2 − a22 ) 
6 EIl
 6 EIl
 x =l
(θC ) z = 
11(103 ) sin 20o  (400)

10502 − 4002 )
l −a )+
l −a ) = 
=
(
(
(
3
3
6(207)10 (306.8)10 (1050)
6 EIl
6 EIl
F1 y a1
2
2
1
F2 y a2
2
2
2
 22.8 (103 ) sin 25o  (750)

10502 − 7502 ) = 0.0133 rad
+ 
(
3
3
6(207)10 (306.8)10 (1050)
Ans.
dz
 d  F1z a1 (l − x) 2
F a (l − x) 2

x + a12 − 2lx ) + 2 z 2
x + a22 − 2lx )  
(
(
 = − 
6 EIl
  x =l
 dx  6 EIl
 d x  x =l
(θC ) y = − 
F a
F a

= −  1z 1 ( 6lx − 2l 2 − 3x 2 − a12 ) + 2 z 2 ( 6lx − 2l 2 − 3 x 2 − a22 ) 
6 EIl
 6 EIl
 x =l
11(103 ) cos 20o  (400)
F1z a1 2
F2 z a2 2

2
2
l − a1 ) −
l − a2 ) = − 
10502 − 4002 )
=−
(
(
(
3
3
6 EIl
6 EIl
6(207)10 (306.8)10 (1050)
 −22.8 (103 ) cos 25o  (750)

−
(10502 − 7502 ) = 0.0112 rad
6(207)103 (306.8)103 (1050)
Ans.
The slope magnitude is ΘC = 0.01332 + 0.0112 2 = 0.0174 rad Ans.
______________________________________________________________________________
4-35 The required new slope in radians is θ new = 0.06(π /180) = 0.00105 rad.
In Prob. 4-29, I = 0.119 8 in4, and it was found that the greater angle occurs at the bearing
at O where (θO)y = − 0.00468 rad.
Sinceθ is inversely proportional to I,
Shigley’s MED, 10th edition
Chapter 4 Solutions, Page 26/80
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θ new Inew = θ old Iold
⇒
4
Inew = π d new
/64 = θ old Iold /θ new
1/4
or,
d new
 64 θ old

= 
I old 
 π θ new

The absolute sign is used as the old slope may be negative.
1/4
 64 −0.00468

d new = 
0.119 8  = 1.82 in
Ans.
 π 0.00105

______________________________________________________________________________
4-36 The required new slope in radians is θ new = 0.06(π /180) = 0.00105 rad.
In Prob. 4-30, I = 39.76 (103) mm4, and it was found that the greater angle occurs at the
bearing at C where (θC)y = − 0.0191 rad.
See the solution to Prob. 4-35 for the development of the equation
1/4
d new
 64 θ old

I old 
= 
 π θ new

1/4
 64 −0.0191

d new = 
39.76 (103 )  = 62.0 mm
Ans.
 π 0.00105

______________________________________________________________________________
4-37 The required new slope in radians is θ new = 0.06(π /180) = 0.00105 rad.
In Prob. 4-31, I = 0.0491 in4, and the maximum slope is ΘC = 0.0104 rad.
See the solution to Prob. 4-35 for the development of the equation
1/4
d new
 64 θ old

I old 
=
π θ

new


1/4
 64 0.0104

d new = 
0.0491 = 1.77 in
Ans.
 π 0.00105

______________________________________________________________________________
4-38 The required new slope in radians is θ new = 0.06(π /180) = 0.00105 rad.
In Prob. 4-32, I = 7 854 mm4, and the maximum slope is ΘO = 0.00750 rad.
See the solution to Prob. 4-35 for the development of the equation
Shigley’s MED, 10th edition
Chapter 4 Solutions, Page 27/80
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1/4
d new
 64 θ old

=
I old 
π θ

new


1/4
 64 0.00750

d new = 
7 854  = 32.7 mm
Ans.
 π 0.00105

______________________________________________________________________________
4-39 The required new slope in radians is θ new = 0.06(π /180) = 0.00105 rad.
In Prob. 4-33, I = 0.119 8 in4, and the maximum slope Θ = 0.0222 rad.
See the solution to Prob. 4-35 for the development of the equation
1/4
d new
 64 θ old

= 
I old 
 π θ new

1/4
 64 0.0222

d new = 
0.119 8  = 2.68 in
Ans.
 π 0.00105

______________________________________________________________________________
4-40 The required new slope in radians is θ new = 0.06(π /180) = 0.00105 rad.
In Prob. 4-34, I = 306.8 (103) mm4, and the maximum slope is ΘC = 0.0174 rad.
See the solution to Prob. 4-35 for the development of the equation
1/4
d new
 64 θ old

I old 
= 
 π θ new

1/4
 64 0.0174

d new = 
306.8 (103 )  = 100.9 mm
Ans.
π
0.00105


______________________________________________________________________________
4-41 IAB = π 14/64 = 0.04909 in4, JAB = 2 IAB = 0.09818 in4, IBC = (0.25)(1.5)3/12 = 0.07031 in4,
ICD = π (3/4)4/64 = 0.01553 in4. For Eq. (3-41), p. 116, b/c = 1.5/0.25 = 6 ⇒ β = 0.299.
The deflection can be broken down into several parts
1. The vertical deflection of B due to force and moment acting on B (y1).
2. The vertical deflection due to the slope at B, θB1, due to the force and moment acting on
B (y2 = CD θB1 = 2θB1).
Shigley’s MED, 10th edition
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3. The vertical deflection due to the rotation at B, θB2, due to the torsion acting at B (y3 =
BC θB1 = 5θB1).
4. The vertical deflection of C due to the force acting on C (y4).
5. The rotation at C, θC, due to the torsion acting at C (y3 = CD θC = 2θC).
6. The vertical deflection of D due to the force acting on D (y5).
1. From Table A-9, beams 1 and 4 with F = − 200 lbf and MB = 2(200) = 400 lbf⋅in
400 ( 62 )
−200 ( 63 )
y1 = −
+
= 0.01467 in
3 ( 30 )106 ( 0.04909 ) 2 ( 30 )106 ( 0.04909 )
2. From Table A-9, beams 1 and 4
 d  Fx 2
M x2  
M x
Fx
θ B1 =  
( x − 3l ) + B   =  ( 3x − 6l ) + B 
2 EI   x =l  6 EI
EI  x =l
 dx  6 EI
6
 l
=
 − ( −200 )( 6 ) + 2 ( 400 )  = 0.004074 rad
[ − Fl + 2M B ] =
6
 2 EI
 2 ( 30 )10 ( 0.04909 )
y 2 = 2(0.004072) = 0.00815 in
3. The torsion at B is TB = 5(200) = 1000 lbf⋅in. From Eq. (4-5)
1000 ( 6 )
 TL 
= 0.005314 rad
θB2 = 
 =
6
 JG  AB 0.09818 (11.5 )10
y 3 = 5(0.005314) = 0.02657 in
4. For bending of BC, from Table A-9, beam 1
y4 = −
−200 ( 53 )
3 ( 30 )106 ( 0.07031)
= 0.00395 in
5. For twist of BC, from Eq. (3-41), p. 116, with T = 2(200) = 400 lbf⋅in
θC =
400 ( 5 )
0.299 (1.5 ) 0.253 (11.5 )106
= 0.02482 rad
y 5 = 2(0.02482) = 0.04964 in
6. For bending of CD, from Table A-9, beam 1
y6 = −
−200 ( 23 )
3 ( 30 )106 ( 0.01553)
= 0.00114 in
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Summing the deflections results in
6
yD = ∑ yi = 0.01467 + 0.00815 + 0.02657 + 0.00395 + 0.04964 + 0.00114 = 0.1041 in Ans.
i =1
This problem is solved more easily using Castigliano’s theorem. See Prob. 4-71.
______________________________________________________________________________
4-42 The deflection of D in the x direction due to Fz is from:
1. The deflection due to the slope at B, θB1, due to the force and moment acting on B (x1 =
BC θB1 = 5θB1).
2. The deflection due to the moment acting on C (x2).
1. For AB, IAB = π 14/64 = 0.04909 in4. From Table A-9, beams 1 and 4
 d  Fx 2
M B x 2  
M x
 Fx
x
l
−
+
3
(
)
( 3x − 6l ) + B 

 = 
EI  x =l
2 EI   x =l  6 EI
 dx  6 EI
θ B1 = 
6
 l
=
 − (100 )( 6 ) + 2 ( −200 )  = −0.002037 rad
[ − Fl + 2M B ] =
6
 2 EI
 2 ( 30 )10 ( 0.04909 )
x 1 = 5(− 0.002037) = − 0.01019 in
2. For BC, IBC = (1.5)(0.25)3/12 = 0.001953 in4. From Table A-9, beam 4
x2 =
2 ( −100 ) 5
MCl2
=
= −0.04267 in
2 EI
2 ( 30 )106 ( 0.001953)
The deflection of D in the x direction due to Fx is from:
3. The elongation of AB due to the tension. For AB, the area is A = π 12/4 = 0.7854 in2
−150 ( 6 )
 Fl 
x3 = 
= −3.82 (10−5 ) in
 =
6
 AE  AB 0.7854 ( 30 )10
4. The deflection due to the slope at B, θB2, due to the moment acting on B (x1 = BC θB2 =
5θB2). With IAB = 0.04907 in4,
θB2 =
5 ( −150 ) 6
M Bl
=
= −0.003056 rad
EI
30 (106 ) 0.04909
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x4 = 5(− 0.003056) = − 0.01528 in
5. The deflection at C due to the bending force acting on C. With IBC = 0.001953 in4
150 ( 53 )
 Fl 3 
= −0.10667 in
x5 =  −
 =−
3 ( 30 )106 ( 0.001953)
 3EI  BC
6. The elongation of CD due to the tension. For CD, the area is A = π (0.752)/4 = 0.4418
in2
−150 ( 2 )
 Fl 
x6 = 
= −2.26 (10−5 ) in
 =
6
AE
0.4418
30
10
( )

CD
Summing the deflections results in
xD = ∑ xi = −0.01019 − 0.04267 − 3.82 (10−5 )
6
i =1
− 0.01528 − 0.10667 − 2.26 (10−5 ) = −0.1749 in Ans.
______________________________________________________________________________
4-43
JOA = JBC = π (1.54)/32 = 0.4970 in4, JAB = π (14)/32 = 0.09817 in4, IAB = π (14)/64 =
0.04909 in4, and ICD = π (0.754)/64 = 0.01553 in4.
T  lOA l AB lBC 
 Tl 
 Tl 
 Tl 
+
+
θ =

 +
 +
 = 
 GJ OA  GJ  AB  GJ  BC G  J OA J AB J BC 
250(12)  2
9
2 
+
+
 = 0.0260 rad
6 
11.5(10 )  0.4970 0.09817 0.4970 
Simplified
Tl
250(12)(13)
θs =
=
GJ 11.5 (106 ) ( 0.09817 )
=
Ans.
θ s = 0.0345 rad
Ans.
Simplified is 0.0345/0.0260 = 1.33 times greater Ans.
yD =
Fy lOC 3
3EI AB
+ θ s ( lCD ) +
Fy lCD 3
3EI CD
=
250 (133 )
3(30)106 ( 0.04909 )
+ 0.0345(12) +
250 (123 )
3(30)106 ( 0.01553)
yD = 0.847 in
Ans.
______________________________________________________________________________
4-44 Reverse the deflection equation of beam 7 of Table A-9. Using units in lbf, inches
Shigley’s MED, 10th edition
Chapter 4 Solutions, Page 31/80
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y=−
( 3000 / 12 ) x 2 25 x 2 − x3 − 25 12 3
wx
2lx 2 − x3 − l 3 ) = −
( )
(
 ( )
24 EI
24 ( 30 )106 ( 485 )
{
= 7.159 (10−10 ) x  27 (106 ) − 600 x 2 + x3 
}
Ans.
The maximum height occurs at x = 25(12)/2 = 150 in
ymax = 7.159 (10−10 )150  27 (106 ) − 600 (1502 ) + 1503  = 1.812 in
Ans.
______________________________________________________________________________
4-45 From Table A-9, beam 6,
Fbx 2
yL =
( x + b2 − l 2 )
6 EIl
Fb 3
yL =
x + b2 x − l 2 x )
(
6 EIl
dyL
Fb
=
( 3x 2 + b2 − l 2 )
dx 6 EIl
dyL
dx
Let ξ =
dyL
dx
=
Fb ( b 2 − l 2 )
6 EIl
x =0
and set I =
x =0
dL =
π d L4
64
. Thus,
32 Fb ( b 2 − l 2 )
1/4
Ans.
3π Elξ
For the other end view, observe beam 6 of Table A-9 from the back of the page, noting
that a and b interchange as do x and –x
dR =
32 Fa ( l 2 − a 2 )
3π Elξ
1/4
Ans.
For a uniform diameter shaft the necessary diameter is the larger of d L and d R .
______________________________________________________________________________
4-46 The maximum slope will occur at the left bearing. Incorporating a design factor into the
solution for d L of Prob. 4-45,
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 32nFb ( l 2 − b 2 ) 

d=
3π Elξ


1/4
d=
4
32(1.28)(3000)(200) ( 3002 − 2002 )
3π (207)103 (300)(0.001)
d = 38.1 mm
I=
π ( 38.14 )
Ans.
= 103.4 (103 ) mm 4
64
From Table A-9, beam 6, the maximum deflection will occur in BC where dyBC /dx = 0

d  Fa ( l − x ) 2
x + a 2 − 2lx )  = 0 ⇒ 3 x 2 − 6lx + ( a 2 + 2l 2 ) = 0
(

dx  6 EIl

3x 2 − 6 ( 300 ) x + 1002 + 2 ( 3002 )  = 0 ⇒ x 2 − 600 x + 63333 = 0
x=
1
600 ± 6002 − 4(1)63 333  = 463.3, 136.7 mm

2
x = 136.7 mm is acceptable.
 Fa ( l − x ) 2

ymax = 
x + a 2 − 2lx ) 
(
 6 EIl
 x =136.7 mm
3 (103 )100 ( 300 − 136.7 )
136.7 2 + 1002 − 2 ( 300 )136.7  = −0.0678 mm Ans.
6 ( 207 )10 (103.4 )10 ( 300 )
______________________________________________________________________________
=
3
3
4-47 I = π (1.254)/64 = 0.1198 in4. From Table A-9, beam 6
2
 F1a1 (l − x) 2
 F b x

( x + a12 − 2lx)  +  2 2 ( x 2 + b2 2 − l 2 ) 
EIl
6 EIl
6
 

2
δ= 

2
 

150(5)(20 − 8)
2
2
= 
(8 + 5 − 2(20)(8) )
6

  6(30)10 ( 0.1198 ) (20)


250(10)(8)
+
82 + 102 − 202 ) 
(
6
 6(30)10 ( 0.1198 ) (20)

2
1/2



= 0.0120 in
Ans.
______________________________________________________________________________
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Chapter 4 Solutions, Page 33/80
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4-48 I = π (1.254)/64 = 0.1198 in4. For both forces use beam 6 of Table A-9.
For F1 = 150 lbf:
0≤x≤5
150 (15 ) x
Fb x
y = 1 1 ( x 2 + b12 − l 2 ) =
( x 2 + 152 − 202 )
6 EIl
6 ( 30 )106 ( 0.1198 )( 20 )
= 5.217 (10−6 ) x ( x 2 − 175 )
5 ≤ x ≤ 20
y=
(1)
150 ( 5 )( 20 − x )
F1a1 ( l − x ) 2
 x 2 + 52 − 2 ( 20 ) x 
x + a12 − 2lx ) =
(
6
6 EIl
6 ( 30 )10 ( 0.1198 )( 20 )
= 1.739 (10−6 ) ( 20 − x ) ( x 2 − 40 x + 25 )
(2)
For F2 = 250 lbf:
0 ≤ x ≤ 10
250 (10 ) x
Fb x
z = 2 2 ( x 2 + b22 − l 2 ) =
( x 2 + 102 − 202 )
6 EIl
6 ( 30 )106 ( 0.1198 )( 20 )
= 5.797 (10−6 ) x ( x 2 − 300 )
(3)
10 ≤ x ≤ 20
250 (10 )( 20 − x )
F a (l − x ) 2
 x 2 + 102 − 2 ( 20 ) x 
z= 2 2
x + a22 − 2lx ) =
(
6
6 EIl
6 ( 30 )10 ( 0.1198 )( 20 )
= 5.797 (10−6 ) ( 20 − x ) ( x 2 − 40 x + 100 )
(4)
Plot Eqs. (1) to (4) for each 0.1 in using a spreadsheet. There are 201 data points, too
numerous to tabulate here but the plot is shown below, where the maximum deflection of
δ = 0.01255 in occurs at x = 9.9 in. Ans.
0.015
0.01
0.005
y (in)
0
Displacement (in)
z (in)
-0.005
Total (in)
-0.01
-0.015
0
5
10
15
20
x (in)
______________________________________________________________________________
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4-49 The larger slope will occur at the left end.
From Table A-9, beam 8
MBx 2
( x + 3a 2 − 6al + 2l 2 )
6 EIl
dy AB M B
=
(3x 2 + 3a 2 − 6al + 2l 2 )
dx
6 EIl
y AB =
With I = π d 4/64, the slope at the left bearing is
dy AB
dx
= θA =
x=0
MB
(3a 2 − 6al + 2l 2 )
4
6 E (π d / 64 ) l
Solving for d
32 M B
32(1000)
3(42 ) − 6(4)(10) + 2 (102 ) 
3a 2 − 6al + 2l 2 ) = 4
d=4
(
6

3π Eθ Al
3π (30)10 (0.002)(10) 
= 0.461 in
Ans.
______________________________________________________________________________
4-50 From Table A-5, E = 10.4 Mpsi
ΣMO = 0 = 18 FBC − 6(100) ⇒ FBC = 33.33 lbf
The cross sectional area of rod BC is A = π (0.52)/4 = 0.1963 in2.
The deflection at point B will be equal to the elongation of the rod BC.
33.33(12)
 FL 
yB = 
= 6.79 (10−5 ) in Ans.
 =
6
 AE  BC ( 0.1963) 30 (10 )
______________________________________________________________________________
4-51 ΣMO = 0 = 6 FAC − 11(100) ⇒
FAC = 183.3 lbf
The deflection at point A in the negative y direction is equal to the elongation of the rod
AC. From Table A-5, Es = 30 Mpsi.
183.3 (12 )
 FL 
yA = − 
= −3.735 (10−4 ) in
 =−
2
6
AE



 AC
π ( 0.5 ) / 4  30 (10 )
By similar triangles the deflection at B due to the elongation of the rod AC is
y A yB1
=
6
18
⇒
yB1 = 3 y A = 3( −3.735)10−4 = −0.00112 in
From Table A-5, Ea = 10.4 Mpsi
The bar can then be treated as a simply supported beam with an overhang AB. From Table
A-9, beam 10,
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 dy
 Fa 2
Fa 2
 d  F (x − l)

2


yB 2 = BD  BC
−
(
l
+
a
)
=
7
(
x
−
l
)
−
a
(3
x
−
l
)
−
(l + a )
 


 
  x =l + a 3EI
 dx  6 EI
 dx x =l + a  3EI
F
Fa 2
7 Fa
Fa 2
3( x − l ) 2 − 3a( x − l ) − a (3 x − l )  |x =l + a −
(l + a ) = −
(2l + 3a ) −
(l + a )
=7
6 EI
3EI
6 EI
3EI
100 ( 52 )
7 (100 ) 5
=−
(6 + 5)
[ 2(6) + 3(5)] −
6(10.4)106 ( 0.25(23 ) / 12 )
3(10.4)106 ( 0.25(23 ) /12 )
( )
= −0.01438 in
yB = yB1 + yB2 = − 0.00112 − 0.01438 = − 0.0155 in
Ans.
______________________________________________________________________________
4-52 From Table A-5, E = 207 GPa, and G = 79.3 GPa.
Fl AB 3
FlOC l AB 2
Fl AC l AB 2
Fl AB 3
 Tl 
 Tl 
yB = 
l
l
+
+
=
+
+
 AB 
 AB
3EI AB G (π dOC 4 / 32 ) G (π d AC 4 / 32 ) 3E (π d34 / 64 )
 GJ OC
 GJ  AC
l
32 Fl AB 2  lOC
2l AB 
=
+ AC 4 +

4
4
π
 GdOC Gd AC 3Ed AB 
The spring rate is k = F/ |yB|. Thus
2
 32l
k =  AB
 π
 lOC
l
2l AB  
+ AC 4 +

4
4 
 GdOC Gd AC 3Ed AB  
−1
−1
 32 ( 2002 ) 
 
2 ( 200 )
200
200


=
+
+
3
4
3
4
3
4
π
 79.3 (10 )18 79.3 (10 )12 3 ( 207 )10 ( 8 )  

= 8.10 N/mm
Ans.
_____________________________________________________________________________
4-53 For the beam deflection, use beam 5 of Table A-9.
F
R1 = R2 =
2
F
F
δ1 =
, and δ 2 =
2k1
2k 2
y AB = −δ1 +
δ1 − δ 2
l
x+
Fx
(4 x 2 − 3l 3 )
48EI
 1 k2 − k1

x
y AB = F  −
x+
(4 x 2 − 3l 3 ) 
+
48 EI
 2k1 2k1k2l

Ans.
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For BC, since Table A-9 does not have an equation (because of symmetry) an equation
will need to be developed as the problem is no longer symmetric. This can be done easily
using beam 6 of Table A-9 with a = l /2
F ( l / 2 )( l − x )  2 l 2

− F Fk2 − Fk1
yBC =
x+
+
 x + − 2lx 
EIl
2k1
2k1k 2l
4


 1 k 2 − k1
( l − x ) 4 x 2 + l 2 − 8lx  Ans.
= F −
+
x+
(
)
48 EI
 2k1 2k1k2l

______________________________________________________________________________
4-54
Fa
F
, and R2 = (l + a )
l
l
Fa
F
, and δ 2 =
(l + a)
δ1 =
lk1
lk2
R1 =
y AB = −δ1 +
δ1 − δ 2
l
x+
Fax 2
(l − x 2 )
6 EIl
 a

x
ax 2
 k a − k1 ( l + a )  +
y AB = F  − +
(l − x 2 ) 
Ans.
2  2
6 EIl
 k1l k1k2l

δ −δ
F (x − l)
( x − l )2 − a(3 x − l ) 
yBC = −δ1 + 1 2 x +
l
6 EI 
 a

x
(x − l)
( x − l ) 2 − a (3x − l )  
yBC = F − +
Ans.
 k a − k1 ( l + a )  +
2  2

6 EI
 k1l k1k2l

______________________________________________________________________________
4-55 Let the load be at x ≥ l/2. The maximum deflection will be in Section AB
(Table A-9, beam 6)
y AB =
Fbx 2
x + b2 − l 2 )
(
6 EIl
dy AB
Fb
=
( 3x 2 + b 2 − l 2 ) = 0
dx
6 EIl
x=
⇒
l 2 − b2
l2
, xmax =
= 0.577l
3
3
3x 2 + b2 − l 2 = 0
Ans.
For x ≤ l/2, xmin = l − 0.577l = 0.423l Ans.
______________________________________________________________________________
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4-56
M O = 1(3000)(1500) + 2500(2000)
= 9.5 (106 ) N·mm
RO = 1(3000) + 2500 = 5 500 N
From Prob. 4-10,
I = 4.14(106 ) mm 4
x2
1
− 2500 x - 2 000
2
dy
x3
2
EI
= −9.5 (106 ) x + 2 750 x 2 − − 1250 x − 2000 + C1
dx
6
M = −9.5 (106 ) + 5500 x −
dy
= 0 at x = 0 ∴ C1 = 0
dx
dy
x3
2
EI
= −9.5 (106 ) x + 2 750 x 2 − − 1250 x − 2000
dx
6
x4
3
EIy = −4.75 (106 ) x 2 + 916.67 x 3 − − 416.67 x − 2000 + C2
24
y = 0 at x = 0 ∴ C2 = 0 , and therefore
y=−
1 
3
114 (106 ) x 2 − 22 (103 ) x 3 + x 4 + 10 (103 ) x − 2000 


24 EI
yB = −
1
114 (106 ) 30002 − 22 (103 ) 30003
3
6 
24 ( 207 )10 ( 4.14 )10
3
+30004 + 10 (103 ) ( 3000 − 2000 ) 

= −25.4 mm
Ans.
MO = 9.5 (106) N⋅m. The maximum stress is compressive at the bottom of the beam where
y = 29.0 − 100 = − 71 mm
−9.5 (106 ) (−71)
My
=−
= −163 (106 ) Pa = −163MPa Ans.
σ max = −
6
I
4.14(10 )
The solutions are the same as Prob. 4-10.
______________________________________________________________________________
4-57 See Prob. 4-11 for reactions: RO = 465 lbf and RC = 285 lbf. Using lbf and inch units
M = 465 x − 450 ⟨ x − 72⟩1 − 300 ⟨x − 120⟩1
Shigley’s MED, 10th edition
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dy
2
2
= 232.5 x 2 − 225 x − 72 − 150 x − 120 + C1
dx
EIy = 77.5 x3 − 75 ⟨x − 72⟩ 3 − 50 ⟨x − 120⟩3 − C1x
EI
y = 0 at x = 0
⇒ C2 = 0
y = 0 at x = 240 in
0 = 77.5(2403) − 75(240 −72)3 − 50(240 − 120)3 + C1 x ⇒
and,
EIy = 77.5 x3 − 75 ⟨x − 72⟩ 3 − 50 ⟨x − 120⟩3 −2.622(106) x
C1 = − 2.622(106) lbf⋅in2
Substituting y = − 0.5 in at x = 120 in gives
30(106) I (− 0.5) = 77.5 (1203) − 75(120 − 72)3 − 50(120 − 120)3 −2.622(106)(120)
I = 12.60 in4
Select two 5 in × 6.7 lbf/ft channels; from Table A-7, I = 2(7.49) = 14.98 in4
12.60  1 
Ans.
 −  = −0.421 in
14.98  2 
The maximum moment occurs at x = 120 in where Mmax = 34.2(103) lbf⋅in
ymidspan =
Mc 34.2(103 )(2.5)
=
= 5 710 psi O.K.
I
14.98
The solutions are the same as Prob. 4-17.
______________________________________________________________________________
σ max =
4-58 I = π (1.54)/64 = 0.2485 in4, and w = 150/12 = 12.5 lbf/in.
1
24
RO = (12.5 ) 39 + (340) = 453.0 lbf
2
39
12.5 2
1
M = 453.0 x −
x − 340 x − 15
2
dy
12.5 3
2
EI
= 226.5 x 2 −
x − 170 x − 15 + C1
dx
6
3
3
EIy = 75.5 x − 0.5208 x 4 − 56.67 x − 15 + C1 x + C2
y = 0 at x = 0 ⇒ C2 = 0
y = 0 at x = 39 in
y=
⇒
C1 = −6.385(10 4 ) lbf ⋅ in 2 Thus,
1 
3
75.5 x 3 − 0.5208 x 4 − 56.67 x − 15 − 6.385 (104 ) x 


EI
Evaluating at x = 15 in,
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1
75.5 (153 ) − 0.5208 (154 ) − 56.67 (15 − 15 )3 − 6.385 (104 ) (15) 

30(10 )(0.2485) 
= −0.0978 in Ans.
yA =
6
1
75.5 (19.53 ) − 0.5208 (19.54 ) − 56.67 (19.5 − 15 )3 − 6.385 (104 ) (19.5) 

30(10 )(0.2485) 
= − 0.1027 in Ans.
ymidspan =
6
5 % difference
Ans.
The solutions are the same as Prob. 4-12.
______________________________________________________________________________
3 (14 )100
7 (14 )100
4-59 I = 0.05 in4, RA =
= 420 lbf ↑ and RB =
= 980 lbf ↑
10
10
M = 420 x − 50 x2 + 980 ⟨ x − 10 ⟩1
EI
dy
2
= 210 x 2 − 16.667 x 3 + 490 x − 10 + C1
dx
3
EIy = 70 x 3 − 4.167 x 4 + 163.3 x − 10 + C1 x + C2
y = 0 at x = 0 ⇒ C2 = 0
y = 0 at x = 10 in ⇒ C1 = − 2 833 lbf⋅in2. Thus,
y=
1
70 x 3 − 4.167 x 4 + 163.3 x − 10 3 − 2833 x 
6

30 (10 ) 0.05 
3
= 6.667 (10−7 ) 70 x 3 − 4.167 x 4 + 163.3 x − 10 − 2833 x 
Ans.


The tabular results and plot are exactly the same as Prob. 4-21.
______________________________________________________________________________
4-60 RA = RB = 400 N, and I = 6(323) /12 = 16 384 mm4.
First half of beam,
M = − 400 x + 400 ⟨ x − 300 ⟩1
dy
2
EI
= −200 x 2 + 200 x − 300 + C1
dx
From symmetry, dy/dx = 0 at x = 550 mm
⇒
⇒
0 = − 200(5502) + 200(550 – 300) 2 + C1
C1 = 48(106) N·mm2
EIy = − 66.67 x3 + 66.67 ⟨ x − 300 ⟩3 + 48(106) x + C2
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y = 0 at x = 300 mm
C2 = − 12.60(109) N·mm3.
⇒
The term (EI)−1 = [207(103)16 384] −1 = 2.949 (10−10 ) Thus
y = 2.949 (10−10) [− 66.67 x3 + 66.67 ⟨ x − 300 ⟩3 + 48(106) x − 12.60(109)]
yO = − 3.72 mm Ans.
y|x = 550 mm =2.949 (10−10) [− 66.67 (5503) + 66.67 (550 − 300)3
+ 48(106) 550 − 12.60(109)] = 1.11 mm Ans.
The solutions are the same as Prob. 4-13.
______________________________________________________________________________
4-61
1
( M A − Fa )
l
1
∑ M A = 0 = M A + R2l − F (l + a) ⇒ R2 = l ( Fl + Fa − M A )
∑M
B
= 0 = R1l + Fa − M A
M = R1 x − M A + R2 x − l
⇒ R1 =
1
dy 1
1
2
= R1 x 2 − M A x + R2 x − l + C1
2
dx 2
1
1
1
3
EIy = R1 x 3 − M A x 2 + R2 x − l + C1 x + C2
6
2
6
EI
y = 0 at x = 0
⇒
C2 = 0
y = 0 at x = l
⇒
1
1
C1 = − R1l 2 + M Al . Thus,
6
2
EIy =
y=
1
1
1
1
3
 1

R1 x 3 − M A x 2 + R2 x − l +  − R1l 2 + M Al  x
6
2
6
2
 6

1 
3
( M A − Fa ) x3 − 3M A x 2l + ( Fl + Fa − M A ) x − l + ( Fal 2 + 2M Al 2 ) x 

6 EIl
Ans.
In regions,
1
( M A − Fa ) x 3 − 3M A x 2l + ( Fal 2 + 2M Al 2 ) x 

6 EIl 
x
 M A ( x 2 − 3lx + 2l 2 ) + Fa ( l 2 − x 2 ) 
Ans.
=

6 EIl 
y AB =
Shigley’s MED, 10th edition
Chapter 4 Solutions, Page 41/80
Document shared on www.docsity.com
1 
3
( M A − Fa ) x3 − 3M A x 2l + ( Fl + Fa − M A )( x − l ) + ( Fal 2 + 2M Al 2 ) x 

6 EIl
1
3
3
M A  x3 − 3 x 2l − ( x − l ) + 2 xl 2  + F  −ax3 + ( l + a )( x − l ) + axl 2 
=




6 EIl
yBC =
{
1
=
{−M ( x − l ) l + Fl ( x − l ) ( x − l ) − a (3x − l )}
6 EIl
( x − l ) −M l + F  x − l − a 3x − l 
Ans.
=
) (
) }
{
(
6 EI
}
2
2
A
2
A
The solutions reduce to the same as Prob. 4-17.
______________________________________________________________________________
w (b − a )
1


4-62 ∑ M D = 0 = R1l − w ( b − a ) l − b + ( b − a ) 
⇒ R1 =
( 2l − b − a )
2
2l


w
w
2
2
x−a +
x−b
2
2
dy 1
w
w
3
3
2
EI
= R1 x −
x−a +
x − b + C1
dx 2
6
6
1
w
w
4
4
EIy = R1 x 3 −
x−a +
x − b + C1 x + C2
6
24
24
M = R1 x −
y = 0 at x = 0
y = 0 at x = l
⇒
C2 = 0
1 1
w
w
4
4
C1 = −  R1l 3 − ( l − a ) + ( l − b ) 
24
24
l 6

y=
1
EI
 1 w (b − a )
w
w
4
( 2l − b − a ) x3 − x − a + x − b

2l
24
24
6
4
1  1 w (b − a )
w
w
4
4 

−x 
( 2l − b − a ) l 3 − ( l − a ) + ( l − b )  
l 6
2l
24
24
 
=
{
w
4
2 ( b − a )( 2l − b − a ) x3 − l x − a + l x − b
24 EIl
4
}
4
4
− x  2 ( b − a )( 2l − b − a ) l 2 − ( l − a ) + ( l − b ) 


Ans.
The above answer is sufficient. In regions,
Shigley’s MED, 10th edition
Chapter 4 Solutions, Page 42/80
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}
w
4
4
2 ( b − a )( 2l − b − a ) x 3 − x  2 ( b − a )( 2l − b − a ) l 2 − ( l − a ) + ( l − b ) 
{


24 EIl
wx 
4
4
2 ( b − a )( 2l − b − a ) x 2 − 2 ( b − a )( 2l − b − a ) l 2 + ( l − a ) − ( l − b ) 
=


24 EIl
y AB =
yBC =
{
w
4
2 ( b − a )( 2l − b − a ) x 3 − l ( x − a )
24 EIl
}
4
4
− x  2 ( b − a )( 2l − b − a ) l 2 − ( l − a ) + ( l − b ) 


yCD =
{
w
4
4
2 ( b − a )( 2l − b − a ) x 3 − l ( x − a ) + l ( x − b )
24 EIl
}
4
4
− x  2 ( b − a )( 2l − b − a ) l 2 − ( l − a ) + ( l − b ) 


These equations can be shown to be equivalent to the results found in Prob. 4-19.
______________________________________________________________________________
4-63 I1 = π (1.3754)/64 = 0.1755 in4, I2 = π (1.754)/64 = 0.4604 in4,
R1 = 0.5(180)(10) = 900 lbf
Since the loading and geometry are symmetric, we will only write the equations for the
first half of the beam
2
For 0 ≤ x ≤ 8 in M = 900 x − 90 x − 3
At x = 3, M = 2700 lbf⋅in
Writing an equation for M / I, as seen in the figure,
the magnitude and slope reduce since I 2 > I 1.
To reduce the magnitude at x = 3 in, we add the
term, − 2700(1/I 1 − 1/ I 2)⟨ x − 3 ⟩0. The slope of 900 at x = 3 in is also reduced. We
account for this with a ramp function, ⟨ x − 3⟩1 . Thus,
1 1
1 1
M 900 x
90
0
1
x −3
=
− 2700  −  x − 3 − 900  −  x − 3 −
I
I1
I
I
I
I
I2
2 
2 
 1
 1
0
1
= 5128 x − 9520 x − 3 − 3173 x − 3 − 195.5 x − 3
E
2
2
dy
1
2
3
= 2564 x 2 − 9520 x − 3 − 1587 x − 3 − 65.17 x − 3 + C1
dx
Boundary Condition:
dy
= 0 at x = 8 in
dx
Shigley’s MED, 10th edition
Chapter 4 Solutions, Page 43/80
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0 = 2564 ( 8 ) − 9520 ( 8 − 3 ) − 1587 ( 8 − 3 ) − 65.17 ( 8 − 3) + C1 ⇒
2
2
3
3
4
C1 = − 68.67 (103) lbf/in2
2
Ey = 854.7 x 3 − 4760 x − 3 − 529 x − 3 − 16.29 x − 3 − 68.67(103 ) x + C2
⇒
y = 0 at x = 0
C2 = 0
Thus, for 0 ≤ x ≤ 8 in
1 
2
3
4
y=
854.7 x 3 − 4760 x − 3 − 529 x − 3 − 16.29 x − 3 − 68.7(103 ) x 
6 

30(10 )
Ans.
Using a spreadsheet, the following graph represents the deflection equation found above
Beam Deflection
0
0
1
2
3
4
5
6
7
8
-0.002
-0.004
y (in)
-0.006
-0.008
-0.01
-0.012
x (in)
The maximum is ymax = −0.0102 in at x = 8 in Ans.
______________________________________________________________________________
4-64 The force and moment reactions at the left support
are F and Fl respectively. The bending moment
equation is
M = Fx − Fl
Plots for M and M /I are shown.
M /I can be expressed using singularity functions
Shigley’s MED, 10th edition
Chapter 4 Solutions, Page 44/80
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M
F
Fl Fl
l
=
x−
−
x−
2 I1
2 I1 4 I1
2
I
0
+
F
l
x−
2 I1
2
1
where the step down and increase in slope at x = l /2 are given by the last two terms.
Integrate
dy F 2 Fl
Fl
l
E
=
x −
x−
x−
dx 4 I1
2 I1
4 I1
2
dy/dx = 0 at x = 0 ⇒ C1 = 0
1
F
l
+
x−
4 I1
2
2
+ C1
2
3
F 3 Fl 2 Fl
l
F
l
x −
x −
x−
+
x−
+ C2
12 I1
4 I1
8I1
2
12 I1
2
y = 0 at x = 0 ⇒ C2 = 0
2
3
F  3
l
l 
2
y=
+2 x−
 2 x − 6lx − 3l x −

24 EI1 
2
2 
3
2

F  l
5Fl 3
l
y x =l /2 =
Ans.
 2   − 6l   − 3l (0) + 2(0)  = −
24 EI1   2 
96 EI1
2

2
3
F 
l 
3Fl 3
3
 l

2
y x =l =
2
l
6
l
l
3
l
l
2
x
Ans.
−
−
−
+
−
=
−
(
)
( )  2   2   16 EI

24 EI1 
1

Ey =
The answers are identical to Ex. 4-10.
______________________________________________________________________________
4-65 Place a dummy force, Q, at the center. The reaction, R1 = wl / 2 + Q / 2
∂M x
wx2
 wl Q 
M = + x−
=
2
∂Q 2
 2 2
Integrating for half the beam and doubling the results
 1
ymax =  2
 EI
l /2
∫
0
 ∂M
M
 ∂Q
 
2
 dx  =
EI
 Q = 0
l /2
∫
0
 wl 
wx 2   x 
x
−
 
  dx
2  2 
 2 
Note, after differentiating with respect to Q, it can be set to zero
ymax
w
=
2 EI
l /2
∫
0
l /2
w  x 3l x 4 
5w
x ( l − x )dx =
−  =

2 EI  3
4 0
384 EI
2
Ans.
______________________________________________________________________________
4-66 Place a fictitious force Q pointing downwards at the end. Use the variable x originating at
the free end and positive to the left
Shigley’s MED, 10th edition
Chapter 4 Solutions, Page 45/80
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∂M
= −x
∂Q
l
l
 1 l  ∂M  
1  wx 2 
w
dx
x
dx
x3dx
=  ∫M 
=
−
−
=

( )
 
∫
∫
2 EI 0
 EI 0  δ Q   Q =0 EI 0  2 
M = −Qx −
ymax
wx 2
2
wl 4
Ans.
8 EI
______________________________________________________________________________
=
4-67 From Table A-7, I1-1 = 1.85 in4. Thus, I = 2(1.85) = 3.70 in4
First treat the end force as a variable, F.
Adding weight of channels of 2(5)/12 =
0.833 lbf/in. Using the variable x as
shown in the figure
M = −F x −
∂M
= −x
∂F
δA =
1
EI
∫
60
0
5.833 2
x = − F x − 2.917 x 2
2
M
∂M
1
dx=
∂F
EI
∫
60
0
( F x + 2.917 x 2 )( x ) d x =
60
1
( Fx 3 / 3 + 2.917 x 4 / 4)
0
EI
(150)(603 ) / 3 + (2.917)(604 ) / 4
= 0.182 in in the direction of the 150 lbf force
30(106 )(3.70)
∴ y A = − 0.182 in Ans.
______________________________________________________________________________
=
4-68 The energy includes torsion in AC, torsion in CO, and bending in AB.
Neglecting transverse shear in AB
M = Fx,
∂M
=x
∂F
In AC and CO,
∂T
T = Fl AB ,
= l AB
∂F
The total energy is
 T 2l 
 T 2l 
U =
 +
 +
 2GJ  AC  2GJ CO
l AB
∫
0
M2
dx
2 EI AB
Shigley’s MED, 10th edition
Chapter 4 Solutions, Page 46/80
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The deflection at the tip is
∂U Tl AC ∂T TlCO ∂T
=
+
+
δ=
∂F GJ AC ∂F GJ CO ∂F
δ=
=
k=
l AB
∫
0
Tl l
Tl l
M ∂M
1
dx = AC AB + CO AB +
EI 3 ∂F
GJ AC
GJ CO
EI AB
l AB
∫ Fx dx
2
0
2
2
3
3
Tl AC l AB TlCO l AB Fl AB
Fl AC l AB
FlCO l AB
Fl AB
+
+
=
+
+
4
4
4
GJ AC
GJ CO 3EI AB G (π d AC
/ 32 ) G (π d CO
/ 32 ) 3E (π d AB
/ 64 )
2
 l AC
32 Fl AB
2l AB 
l
+ CO4 +

4
4 
π  Gd AC GdCO 3Ed AB

F
δ
=
π
2
32l AB
 l AC
l
2l AB 
+ CO4 +

4
4 
 Gd AC GdCO 3Ed AB 
−1
−1


2 ( 200 )
200
200

 = 8.10 N/mm Ans.
=
+
+
32 ( 2002 )  79.3 (103 )184 79.3 (103 )124 3 ( 207 )103 ( 84 ) 
______________________________________________________________________________
π
4-69 I1 = π (1.3754)/64 = 0.1755 in4, I2 = π (1.754)/64 = 0.4604 in4
Place a fictitious force Q pointing downwards at the midspan of the beam, x = 8 in
R1 =
1
1
(10)180 + Q = 900 + 0.5Q
2
2
For 0 ≤ x ≤ 3 in M = ( 900 + 0.5Q ) x
∂M
= 0.5 x
∂Q
∂M
= 0.5 x
∂Q
For 3 ≤ x ≤ 13 in M = ( 900 + 0.5Q ) x − 90( x − 3)2
By symmetry it is equivalent to use twice the integral from 0 to 8

8

0
δ =  2∫
3
8
M ∂M 
1
1 
2
2
dx  =
900
x
dx
900 x − 90 ( x − 3)  x dx
+
∫
∫


EI ∂Q Q =0 EI1 0
EI 2 3
3
8
300 x 3
1 
1
9

300 x3 − 90( x 4 − 2 x3 + x 2 ) 
=
+

4
2
EI1 0 EI 2 
3
120.2 (103 )
8100 1
8100
3
3
145.5 (10 ) − 25.31(10 )  =
=
+
 30 106 0.1755 + 30 106 0.4604
EI1 EI 2 
( )
( )
= 0.0102 in
Ans.
______________________________________________________________________________
Shigley’s MED, 10th edition
Chapter 4 Solutions, Page 47/80
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4-70 I = π (0.54)/64 = 3.068 (10−3) in4, J = 2 I = 6.136 (10−3) in4, A =π (0.52)/4 = 0.1963 in2.
Consider x to be in the direction of OA, y vertically upward, and z in the direction of AB.
Resolve the force F into components in the x and y directions obtaining 0.6 F in the
horizontal direction and 0.8 F in the negative vertical direction. The 0.6 F force creates
strain energy in the form of bending in AB and OA, and tension in OA. The 0.8 F force
creates strain energy in the form of bending in AB and OA, and torsion in OA. Use the
dummy variable x to originate at the end where the loads are applied on each segment,
0.6 F: AB
OA
∂M
= 0.6 x
∂F
M = 0.6 F x
∂M
= 4.2
∂F
M = 4.2 F
∂Fa
= 0.6
∂F
Fa = 0.6 F
0.8 F: AB
M = 0.8 F x
∂M
= 0.8 x
∂F
OA
M = 0.8 F x
∂M
= 0.8 x
∂F
∂T
= 5.6
∂F
Once the derivatives are taken the value of F = 15 lbf can be substituted in. The deflection
of B in the direction of F is*
T = 5.6 F
(δ B ) F
∂U  Fa L  ∂Fa  TL  ∂T 1
∂M
=
+
+
M
dx
∑


∫
∂F  AE OA ∂F  JG OA ∂F EI
∂F
0.6 (15 )15
5.6 (15 )15
=
0.6 ) +
( 5.6 )
6 (
0.1963 ( 30 )10
6.136 (10−3 )11.5 (106 )
=
7
15
15 ( 4.22 )
15
2
x
d
x
dx+
+
+
0.6
(
)
30 (106 ) 3.068 (10−3 ) ∫0
30 (106 ) 3.068 (10−3 ) ∫0
+
7
15
15
15
2
2
( 0.8 x ) d x +
( 0.8 x ) d x
−3 ∫
6
6
−3 ∫
30 (10 ) 3.068 (10 ) 0
30 (10 ) 3.068 (10 ) 0
= 1.38 (10−5 ) + 0.1000 + 6.71(10−3 ) + 0.0431 + 0.0119 + 0.1173
= 0.279 in
Ans.
Shigley’s MED, 10th edition
Chapter 4 Solutions, Page 48/80
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*Note. Be careful, this is not the actual deflection of point B. For this, dummy forces must
be placed on B in the x, y, and z directions. Determine the energy due to each, take
derivatives, and then substitute the values of Fx = 9 lbf, Fy = − 12 lbf, and Fz = 0. This can
be done separately and then added by vector addition. The actual deflections of B are
found to be
δB = 0.0831 i − 0.2862 j − 0.00770 k in
From this, the deflection of B in the direction of F is
(δ B ) F = 0.6 ( 0.0831) + 0.8 ( 0.2862 ) = 0.279 in
which agrees with our result.
______________________________________________________________________________
4-71 Strain energy. AB: Bending and torsion, BC: Bending and torsion, CD: Bending.
IAB = π (14)/64 = 0.04909 in4, JAB = 2 IAB = 0.09818 in4, IBC = 0.25(1.53)/12 = 0.07031 in4,
ICD = π (0.754)/64 = 0.01553 in4.
For the torsion of bar BC, Eq. (3-41) is in the form of θ =TL/(JG), where the equivalent of
J is Jeq = βbc 3. With b/c = 1.5/0.25 = 6, JBC = βbc 3 = 0.299(1.5)0.253 = 7.008 (10−3) in4.
Use the dummy variable x to originate at the end where the loads are applied on each
segment,
∂M
AB: Bending M = F x + 2 F
= x +2
∂F
∂T
Torsion T = 5 F
=5
∂F
∂M
BC: Bending M = F x
=x
∂F
∂T
Torsion T = 2 F
=2
∂F
∂M
CD: Bending M = F x
=x
∂F
∂U
Tl ∂T
1
∂M
=∑
+∑ ∫M
dx
∂F
JG ∂F
EI
∂F
6
5F ( 6)
2F (5)
1
2
=
+
+
5
2
F ( x + 2) d x
(
)
∫
6
6
−3
0.09818 (11.5 )106
7.008 (10 )11.5 (10 )
30 (10 ) 0.04909 0
δD =
+
5
2
1
1
F x 2d x +
F x 2d x
∫
6
6
30 (10 ) 0.07031 0
30 (10 ) 0.01553 ∫0
= 1.329 (10−4 ) F + 2.482 (10−4 ) F + 1.141(10−4 ) F + 1.98 (10−5 ) F + 5.72 (10−6 ) F
= 5.207 (10−4 ) F = 5.207 (10−4 ) 200 = 0.104 in
Ans.
______________________________________________________________________________
Shigley’s MED, 10th edition
Chapter 4 Solutions, Page 49/80
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4-72 AAB = π (12)/4 = 0.7854 in2, IAB = π (14)/64 = 0.04909 in4, IBC = 1.5 (0.253)/12 = 1.953
(10−3) in4, ACD = π (0.752)/4 = 0.4418 in2, IAB = π (0.754)/64 = 0.01553 in4. For (δD )x let
F = Fx = − 150 lbf and Fz = − 100 lbf . Use the dummy variable x to originate at the end
where the loads are applied on each segment,
∂M y
CD:
M y = Fz x
=0
∂F
∂Fa
=1
Fa = F
∂F
∂M y
BC:
M y = F x + 2 Fz
=x
∂F
∂Fa
=0
Fa = Fz
∂F
∂M y
AB:
M y = 5F + 2 Fz + Fz x
=5
∂F
∂Fa
Fa = F
=1
∂F
5
∂Fa
1
∂U
FL
+
(δ D ) x = =  
( F x + 2 Fz ) x d x
∂F  AE CD ∂F EI BC ∫0
+
=
1
EI AB
6
∫ ( 5F + 2 F
z
0
 FL  ∂Fa
+ Fz x )( 5 ) d x + 

 AE  AB ∂F
F ( 2)
1
3
F
1 +
( 5) + Fz ( 52 ) 
6 ( )
6
−3 
0.4418 ( 30 )10
30 (10 )1.953 (10 )  3

+
F ( 6)
1
F


25 F ( 6 ) + 10 Fz ( 6 ) + z ( 62 ) 5 +
1
6 ( )

2
30 (10 ) 0.04909 
 0.7854 ( 30 )10
6
= 1.509 (10−7 ) F + 7.112 (10−4 ) F + 4.267 (10−4 ) Fz + 1.019 (10−4 ) F
+ 1.019 (10−4 ) Fz + 2.546 (10−7 ) F = 8.135 (10−4 ) F + 5.286 (10−4 ) Fz
Substituting F = Fx = − 150 lbf and Fz = − 100 lbf gives
(δ D ) x = 8.135 (10−4 ) ( −150 ) + 5.286 (10−4 ) ( −100 ) = −0.1749 in
Ans.
______________________________________________________________________________
4-73 IOA = IBC = π (1.54)/64 = 0.2485 in4, JOA = JBC = 2 IOA = 0.4970 in4, IAB = π (14)/64 =
0.04909 in4, JAB = 2 IAB = 0.09818 in4, ICD = π (0.754)/64 = 0.01553 in4
Let Fy = F, and use the dummy variable x to originate at the end where the loads are
applied on each segment,
Shigley’s MED, 10th edition
Chapter 4 Solutions, Page 50/80
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OC:
M =Fx
∂M
= x,
∂F
DC:
M =Fx
∂M
=x
∂F
T = 12 F
∂T
= 12
∂F
∂U
1
∂M
 TL  ∂T
dx
= ∑
+∑ ∫M

∂F
∂F
EI
 JG OC ∂F
The terms involving the torsion and bending moments in OC must be split up because of
the changing second-area moments.
(δ D ) y =
(δ D ) y =
2
12 F ( 4 )
12 F ( 9 )
1
12
12
F x 2d x
+
+
(
)
(
)
0.4970 (11.5 )106
0.09818 (11.5 )106
30 (106 ) 0.2485 ∫0
+
11
13
12
1
1
1
F x 2d x +
F x 2d x +
F x 2d x
∫
∫
∫
6
6
6
30 (10 ) 0.04909 2
30 (10 ) 0.2485 11
30 (10 ) 0.01553 0
= 1.008 (10−4 ) F + 1.148 (10−3 ) F + 3.58 (10−7 ) F
+ 2.994 (10−4 ) F + 3.872 (10−5 ) F + 1.2363 (10−3 ) F
= 2.824 (10−3 ) F = 2.824 (10−3 ) 250 = 0.706 in
Ans.
For the simplified shaft OC,
(δ B ) y =
13
12
12 F (13)
1
1
2
+
F
x
d
x
+
F x 2d x
12
(
)
0.09818 (11.5 )106
30 (106 ) 0.04909 ∫0
30 (106 ) 0.01553 ∫0
= 1.6580 (10−3 ) F + 4.973 (10−4 ) F + 1.2363 (10−3 ) F = 3.392 (10−3 ) F = 3.392 (10−3 ) 250
= 0.848 in
Ans.
Simplified is 0.848/0.706 = 1.20 times greater Ans.
______________________________________________________________________________
4-74 Place a dummy force Q pointing downwards at point B. The reaction at C is
RC = Q + (6/18)100 = Q + 33.33
This is the axial force in member BC. Isolating the beam, we find that the moment is not a
function of Q, and thus does not contribute to the strain energy. Thus, only energy in the
member BC needs to be considered. Let the axial force in BC be F, where
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∂F
=1
∂Q
F = Q + 33.33
 FL  ∂F 
( 0 + 33.33)12
= 
(1) = 6.79 (10−5 ) in Ans.

 =
2
6
AE
Q
∂


 BC

 Q =0 π ( 0.5 ) / 4  30 (10 )
Q =0
______________________________________________________________________________
δB =
4-75
∂U
∂Q
IOB = 0.25(23)/12 = 0.1667 in4
AAC = π (0.52)/4 = 0.1963 in2
ΣMO = 0 = 6 RC − 11(100) − 18 Q
RC = 3Q + 183.3
ΣMA = 0 = 6 RO − 5(100) − 12 Q
⇒ RO = 2Q + 83.33
Bending in OB.
BD: Bending in BD is only due to Q which when set to zero after differentiation
gives no contribution.
AD: Using the variable x as shown in the figure above
∂M
= − (7 + x )
∂Q
OA: Using the variable x as shown in the figure above
M = −100 x − Q ( 7 + x )
M = − ( 2Q + 83.33) x
Axial in AC:
F = 3Q + 183.3
∂M
= −2 x
∂Q
∂F
=3
∂Q
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Chapter 4 Solutions, Page 52/80
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 ∂U 
 FL  ∂F 
∂M
 1

δB = 
 = 

 +  ∑ M ∂Q dx 
 ∂Q Q = 0  AE  ∂Q  Q = 0  EI
Q = 0
=
183.3 (12 )
1
( 3) +
EI
0.1963 ( 30 )106
5
∫ (100 x )( 7 + x ) d x + ∫ 2 (83.33) x dx
6
2
0
0
5
6


2
= 1.121(10 ) +
+
+
100
7
166.7
x
x
d
x
x
dx
(
)

∫0 
10.4 (106 ) 0.1667  ∫0
1
−3
= 1.121(10−3 ) + 5.768 (10−7 ) 100 (129.2 ) + 166.7 ( 72 )  = 0.0155 in
Ans.
______________________________________________________________________________
4-76
There is no bending in AB. Using the variableθ, rotating counterclockwise from B
M = PR sin θ
Fr = P cos θ
Fθ = P sin θ
∂MFθ
= 2 PR sin 2 θ
∂P
∂M
= R sin θ
∂P
∂Fr
= cos θ
∂P
∂Fθ
= sin θ
∂P
A = 6(4) = 24 mm 2 , ro = 40 + 12 (6) = 43 mm, ri = 40 − 12 (6) = 37 mm,
From Table 3-4, p. 135, for a rectangular cross section
6
rn =
= 39.92489 mm
ln(43 / 37)
From Eq. (4-33), the eccentricity is e = R − rn =40 − 39.92489 = 0.07511 mm
From Table A-5, E = 207(103) MPa, G = 79.3(103) MPa
From Table 4-1, C = 1.2
From Eq. (4-38)
π
π
π
π
1 ∂ ( MFθ )
M  ∂M 
2
2 F R  ∂F 
2
2 CFr R  ∂Fr
δ =∫
dθ + ∫ θ  θ  dθ − ∫
dθ + ∫



0 AeE
0
0
0
AE  ∂P 
AE ∂P
AG  ∂P
 ∂P 

 dθ

2
π
π
π
P ( R sin θ )
2 PR ( sin θ )
2 2 PR sin θ
2 CPR ( cos θ )
dθ + ∫
dθ − ∫
dθ + ∫
dθ
0
0
0
0
AeE
AE
AE
AG
EC 
π PR  R
π (10)(40)  40
(207 ⋅103 )(1.2) 
1
2
1
2
=
+
−
+
=
+
−
+




4 AE  e
G  4(24)(207 ⋅103 )  0.07511
79.3 ⋅103 
δ = 0.0338 mm
Ans.
______________________________________________________________________________
=∫
π
2
2
2
2
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4-77
Place a dummy force Q pointing downwards at point A. Bending in AB is only due to Q
which when set to zero after differentiation gives no contribution. For section BC use the
variableθ, rotating counterclockwise from B
∂M
M = PR sin θ + Q ( R + R sin θ )
= R (1 + sin θ )
∂Q
∂Fr
= cos θ
Fr = ( P + Q ) cos θ
∂Q
∂Fθ
Fθ = ( P + Q ) sin θ
= sin θ
∂Q
MFθ =  PR sin θ + QR (1 + sin θ )  ( P + Q ) sin θ
∂MFθ
= PR sin 2 θ + + PR sin θ (1 + sin θ ) + 2QR sin θ (1 + sin θ )
∂Q
But after differentiation, we can set Q = 0. Thus,
∂MFθ
= PR sin θ (1 + 2sin θ )
∂Q
A = 6(4) = 24 mm 2 , ro = 40 + 12 (6) = 43 mm, ri = 40 − 12 (6) = 37 mm,
From Table 3-4, p.135, for a rectangular cross section
6
rn =
= 39.92489 mm
ln(43 / 37)
From Eq. (4-33), the eccentricity is e = R − rn =40 − 39.92489 = 0.07511 mm
From Table A-5, E = 207(103) MPa, G = 79.3(103) MPa
From Table 4-1, C = 1.2
From Eq. (4-38),
π
π
π

1 ∂ ( MFθ )
2 Fθ R  ∂Fθ 
2
2 CFr R  ∂Fr 
d
θ
d
θ
d
θ
+
−
+




 dθ
∫
∫
∫
0
0 AE
0 AE
0
∂Q
AG  ∂Q 

 ∂Q 
PR 2 π2
PR π2 2
PR π2
d
d
=
θ
+
θ
θ
+
θ
θ
−
sin
1
sin
sin
sin θ (1 + 2sin θ ) dθ
(
)
AeE ∫0
AE ∫0
AE ∫0
CPR π2
+
cos 2 θ dθ
AG ∫0
2
π CE 
π
 PR π PR  π
 PR π CPR PR  π
R
=  + 1
+
−  + 2
+
=
 + 1 − 2 +

4 G 
 4  AeE 4 AE  4
 AE 4 AG AE  4  e
δ =∫
π
2
M  ∂M

AeE  ∂Q
3
 π
π 1.2 ( 207 )10 
 40
 + 1

−2+
4 79.3 (103 ) 
 4  0.07511

= 0.0766 mm
Ans.
______________________________________________________________________________
=
10 ( 40 )
24 ( 207 )103
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4-78
A = 3(2.25) −2.25(1.5) = 3.375 in2
(1 + 1.5)(3)(2.25) − (1 + 0.75 + 1.125)(1.5)(2.25)
R=
= 2.125 in
3.375
Section is equivalent to the “T” section of Table 3-4, p. 135,
2.25(0.75) + 0.75(2.25)
= 1.7960 in
2.25ln[(1 + 0.75) /1] + 0.75ln[(1 + 3) / (1 + 0.75)]
e = R − rn = 2.125 − 1.7960 = 0.329 in
rn =
For the straight section
1
I z = (2.25) ( 33 ) + 2.25(3)(1.5 − 1.125)2
12
2
1
2.25

 
−  (1.5) ( 2.253 ) + 1.5(2.25)  0.75 +
− 1.125  
2

 
12
= 2.689 in 4
For 0 ≤ x ≤ 4 in
M = − Fx
∂M
= − x,
∂F
V =F
∂V
=1
∂F
For θ ≤ π /2
Fr = F cos θ
∂Fr
= cos θ ,
∂F
Fθ = F sin θ
∂Fθ
= sin θ
∂F
∂M
= (4 + 2.125sin θ )
∂F
∂MFθ
= 2 F (4 + 2.365sin θ ) sin θ
MFθ = F (4 + 2.125sin θ ) F sin θ
∂F
M = F (4 + 2.125sin θ )
Use Eqs. (4-31) and (4-24) (with C = 1) for the straight part, and Eq. (4-38) for the
curved part, integrating from 0 to π/2, and double the results
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δ=
π /2
2 1 4 2
F (4)(1)
(4 + 2.125sin θ ) 2
+∫ F
dθ
 ∫0 Fx dx +
E I
3.375(G / E ) 0
3.375(0.329)
π /2 2 F (4 + 2.125sin θ ) sin θ
F sin 2 θ (2.125)
dθ − ∫
dθ
0
0
3.375
3.375
2
π /2 (1) F cos θ (2.125)
+∫
dθ
0
3.375(G / E )
+∫
π /2
Substitute
I = 2.689 in4, F = 6700 lbf, E = 30 (106) psi, G = 11.5 (106) psi
2 ( 6700 )  43
 π 
4
1
 π 
δ=
+
+
16   + 17(1) + 4.516   

6 
30 (10 )  3 ( 2.689 ) 3.375(11.5 / 30) 3.375(0.329)   2 
 4 
+
2.125  π 
2 
2.125
 π 
 π  
4 (1) + 2.125    +
 −
 

3.375  4  3.375 
 4   3.375 (11.5 / 30 )  4  
= 0.0226 in
Ans.
______________________________________________________________________________
4-79 Since R/h = 35/4.5 = 7.78 use Eq. (4-38), integrate from 0 to π , and double the results
∂M
M = FR (1 − cosθ )
= R (1 − cosθ )
∂F
∂Fr
Fr = F sin θ
= sin θ
∂F
∂Fθ
Fθ = F cos θ
= cos θ
∂F
MFθ = F 2 Rcosθ (1 − cosθ )
∂ ( MFθ )
= 2 FRcosθ (1 − cosθ )
∂F
From Eq. (4-38),
 FR 2 π
FR π
(1 − cos θ ) 2 dθ +
cos 2 θ dθ
∫
∫
0
0
AE
 AeE
δ = 2
−
=
2 FR π
1.2 FR π 2

cos θ (1 − cos θ ) dθ +
sin θ dθ 
∫
AE 0
AG ∫0

E
2 FR  3π R 3π
+
+ 0.6π 

AE  2 e 2
G
A = 4.5(3) = 13.5 mm2, E = 207 (103) N/mm2, G = 79.3 (103) N/mm2, and from Table 3-4,
p. 135,
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h
4.5
=
= 34.95173 mm
ro
37.25
ln
ln
32.75
ri
and e = R − rn = 35 − 34.95173 = 0.04827 mm. Thus,
2 F ( 35 )  3π
35
3π
207 
+
+ 0.6π
δ=
 = 0.08583F
3 
13.5 ( 207 )10  2 0.04827 2
79.3 
1
where F is in N. For δ = 1 mm, F =
= 11.65 N
Ans.
0.08583
Note: The first term in the equation for δ dominates and this is from the bending moment.
Try Eq. (4-41), and compare the results.
______________________________________________________________________________
rn =
4-80 R/h = 20 > 10 so Eq. (4-41) can be used to determine deflections. Consider the horizontal
reaction to be applied at B and subject to the constraint (δ B ) H = 0.
M=
∂M
= − R sin θ
∂H
FR
(1 − cos θ ) − HR sin θ
2
0 <θ <
π
2
By symmetry, we may consider only half of the wire form and use twice the strain energy
Eq. (4-41) then becomes,
(δ B ) H =
∫
π /2
0
∂U
2
≈
∂H EI
π /2

∂M 
∫  M ∂H  Rdθ = 0
0
 FR

 2 (1 − cos θ ) − HR sin θ  (− R sin θ ) R dθ = 0
F F
π
F 30
− + +H =0 ⇒ H = =
= 9.55 N
2 4
4
π π
Ans.
The reaction at A is the same where H goes to the left. Substituting H into the moment
equation we get,
M=
FR
[π (1 − cos θ ) − 2 sin θ ]
2π
∂M
R
=
[π (1 − cos θ ) − 2sin θ ]
∂F 2π
Shigley’s MED, 10th edition
0 <θ <
π
2
Chapter 4 Solutions, Page 57/80
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δP =
∂U
2  ∂M
≈∫
M
∂P
EI  ∂F
FR 3
2π 2 EI
FR 3
=
2π 2 EI
=
∫
π /2
0
2

Rdθ =
EI

∫
π /2
0
FR 2
[π (1 − cos θ ) − 2sin θ ]2 R dθ
4π 2
(π 2 + π 2 cos 2 θ + 4sin 2 θ − 2π 2 cos θ − 4π sin θ + 4π sin θ cos θ ) dθ
 2 π 

π 
2π 
2
π  2  + π  4  + 4  4  − 2π − 4π + 2π 
 
 
  

2
3
2
(3π − 8π − 4)
(30)(403 )
(3π − 8π − 4) FR
= 0.224 mm
=
=
EI
8π
8π
207 (103 ) π ( 24 ) / 64 
Ans.
______________________________________________________________________________
4-81 The radius is sufficiently large compared to the wire diameter to use Eq. (4-41) for the
curved beam portion. The shear and axial components will be negligible compared to
bending.
Place a fictitious force Q pointing to the left at point A.
∂M
M = PR sin θ + Q( R sin θ + l )
= R sin θ + l
∂Q
Note that the strain energy in the straight portion is zero since there is no real force in that
section.
From Eq. (4-41),


1  ∂M
M
EI  ∂Q
PR 2
EI
∫ ( R sin
π /2
δ = ∫
0
=
π /2
2
0


1
Rdθ  =

 Q = 0 EI
θ + l sin θ ) dθ =
∫
π /2
0
PR sin θ ( R sin θ + l )Rdθ
PR 2  π
1(52 )

π

R
l
+
=


 (5) + 4 
6
4
EI  4
 30 (10 ) π ( 0.125 ) / 64   4

= 0.551 in
Ans.
______________________________________________________________________________
4-82 Both the radius and the length are sufficiently large to use Eq. (4-41) for the curved beam
portion and to neglect transverse shear stress for the straight portion.
∂M AB
=x
∂P
Straight portion:
M AB = Px
Curved portion:
M BC = P [ R (1 − cos θ ) + l ]
∂M BC
= [ R (1 − cos θ ) + l ]
∂P
From Eq. (4-41) with the addition of the bending strain energy in the straight portion of
the wire,
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π /2 1 
∂M BC 
∂M AB 
1 
M AB
dx + ∫
M BC
Rdθ



0 EI
0
EI 
∂P 
∂P 

P l 2
PR π /2
2
x dx +
=
[ R(1 − cos θ ) + l ] dθ
∫
∫
0
0
EI
EI
3
/2
π
Pl
PR
 R 2 (1 − 2 cos θ + cos 2 θ ) + 2 Rl (1 − cos θ ) + l 2 dθ
=
+
∫

0
3EI EI
Pl 3 PR π /2 2
 R cos 2 θ − ( 2 R 2 + 2 Rl ) cos θ + ( R + l )2 dθ
=
+
∫


0
3EI EI
3
Pl
PR  π 2
π

=
+
R − ( 2 R 2 + 2 Rl ) + ( R + l ) 2 
2
3EI EI  4

δ =∫
l
l3 π 3
π
2
2
 3 + 4 R − R ( 2 R + 2 Rl ) + 2 R ( R + l ) 


=
P
EI
=
 43 π 3
π
1
2
2
 + (5 ) − 5  2(5 ) + 2(5)(4)  + ( 5 )( 5 + 4 ) 
6
4
2
30 (10 ) π ( 0.125 ) / 64  3 4

= 0.850 in Ans.
______________________________________________________________________________
4-83 Both the radius and the length are sufficiently large to use Eq. (4-41) for the curved beam
portion and to neglect transverse shear stress for the straight portion.
Place a dummy force, Q, at A vertically downward. The only load in the straight section is
the axial force, Q. Since this will be zero, there is no contribution.
In the curved section
∂M
= R (1 − cos θ )
∂Q
M = PR sin θ + QR (1 − cos θ )
From Eq. (4-41)
 π /2 1  ∂M 

1
δ = ∫
M
Rdθ  =


0
EI  ∂Q 

 Q =0 EI
PR 3
=
EI
=
∫
π /2
0
∫
π /2
0
PR sin θ  R (1 − cos θ ) Rdθ
PR3  1  PR3
( sin θ − sin θ cos θ ) dθ =
1 −  =
EI  2  2 EI
1( 53 )
2 ( 30 )106 π ( 0.1254 ) / 64 
= 0.174 in
Ans.
______________________________________________________________________________
4-84 Both the radius and the length are sufficiently large to use Eq. (4-41) for the curved beam
portion and to neglect transverse shear stress for the straight portion.
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Chapter 4 Solutions, Page 59/80
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Place a dummy force, Q, at A vertically downward. The load in the straight section is the
axial force, Q, whereas the bending moment is only a function of P and is not a function
of Q. When setting Q = 0, there is no axial or bending contribution.
In the curved section
M = P  R (1 − cos θ ) + l  − QR sin θ
∂M
= − R sin θ
∂Q
From Eq. (4-41)

π /2

0
δ = ∫
=−
=−

1  ∂M 
1
M
Rdθ  =
EI  ∂Q 
 Q =0 EI
PR 2
EI
π /2
∫
π /2
0
P  R (1 − cos θ ) + l  ( − R sin θ ) Rdθ
∫ ( R sin θ − R sin θ cos θ + l sin θ ) dθ = −
0
1( 52 )
2 ( 30 )106 π ( 0.1254 ) / 64 
PR 2 
PR 2
1 
R
l
R
+
−
=
−
( R + 2l )


EI 
2 
2 EI
5 + 2 ( 4 )  = −0.452 in
Since the deflection is negative, δ is in the opposite direction of Q. Thus the deflection is
δ = 0.452 in ↑
Ans.
______________________________________________________________________________
4-85 Consider the force of the mass to be F, where F = 9.81(1) = 9.81 N. The load in AB is
tension
∂FAB
FAB = F
=1
∂F
For the curved section, the radius is sufficiently large to use Eq. (4-41). There is no
bending in section DE. For section BCD, let θ be counterclockwise originating at D
∂M
M = FR sin θ
= R sin θ
0 ≤θ ≤π
∂F
Using Eqs. (4-29) and (4-41)
3
π 1 
π FR
∂M 
Fl
 Fl  ∂FAB
+∫
M
Rdθ =
1) + ∫
sin 2 θ dθ
(



0
0
EI  ∂F 
AE
EI
 AE  AB ∂F
δ =
π ( 403 ) 
Fl π FR 3 F  l π R 3 
9.81 
80

=
+
=  +
+
=
AE 2 EI
E  A 2 I  207 (103 )  π ( 22 ) / 4  2 π ( 24 ) / 64  




= 6.067 mm
Ans.
______________________________________________________________________________
Shigley’s MED, 10th edition
Chapter 4 Solutions, Page 60/80
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4-86 AOA = 2(0.25) = 0.5 in2,
IOAB = 0.25(23)/12 = 0.1667 in4,
IAC = π (0.54)/64 = 3.068 (10-3) in4
Applying a force F at point B, using
statics (AC is a two-force member),
the reaction forces at O and C are as
shown.
∂FOA
OA: Axial FOA = 3F
=3
∂F
∂M OA
= −2 x
∂F
Bending M OA = −2 Fx
AB: Bending
∂M AB
= −x
∂F
M AB = − F x
AC: Isolating the upper curved section
∂M AC
= 3R ( sin θ + cos θ − 1)
∂F
10
20
1
1
 Fl  ∂FOA
2
δ =
+
F x 2d x
4 Fx dx +

∫
∫
( EI )OAB 0
 AE OA ∂F ( EI )OAB 0
M AC = 3FR ( sin θ + cos θ − 1)
+
9 FR 3
( EI ) AC
π /2
∫ ( sin θ + cosθ − 1)
2
dθ
0
F ( 203 )
4 F (103 )
3F (10 )
=
+
( 3) +
0.5 (10.4 )106
3 (10.4 )106 ( 0.1667 ) 3 (10.4 )106 ( 0.1667 )
+
9 F (103 )
30 (10 ) 3.068 (10
6
π /2
−3
( sin
)∫
2
θ + 2sin θ cos θ − 2sin θ + cos 2 θ − 2 cos θ + 1) dθ
0
π
π
π
= 1.731(10−5 ) F + 7.691(10−4 ) F + 1.538 (10−3 ) F + 0.09778 F  + 1 − 2 + − 2 + 
4
2
4
= 0.0162 F = 0.0162 (100 ) = 1.62 in
Ans.
_____________________________________________________________________________
4-87 AOA = 2(0.25) = 0.5 in2,
IOAB = 0.25(23)/12 = 0.1667 in4,
IAC = π (0.54)/64 = 3.068 (10-3) in4
Applying a vertical dummy force, Q, at A,
from statics the reactions are as shown. The
dummy force is transmitted through section
OA and member AC.
Shigley’s MED, 10th edition
Chapter 4 Solutions, Page 61/80
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∂FOA
=1
∂Q
OA: FOA = 3F + Q
M AC = ( 3F + Q ) R sin θ − ( 3F + Q ) R (1 − cos θ )
AC:
∂M AC
= R ( sin θ + cos θ − 1)
∂Q
 Fl   ∂FOA   1  π / 2

∂M AC
δ = 
Rdθ 
 +   ∫ M AC
 
∂Q
 AE OA  ∂Q   EI  AC 0
 Q=0
=
=
3FlOA
3FR 3
+
( AE )OA ( EI ) AC
3 (100 )10
10.4 (10 ) 0.5
6
+
π /2
∫ ( sin θ + cos θ − 1)
2
dθ
0
3 (100 )103
π
π
π
 + 1 − 2 + − 2 +  = 0.462 in
4
2
30 (10 ) 3.068 (10 )  4
6
−3
Ans.
______________________________________________________________________________
4-88
I = π (64)/64 = 63.62 mm4
0≤θ ≤π/2
∂M
= R sin θ
∂F
∂T
T = FR (1 − cos θ )
= R(1 − cos θ )
∂F
According to Castigliano’s theorem, a positive
∂ U/∂ F will yield a deflection of A in the negative y direction. Thus the deflection in the
positive y direction is
M = FR sin θ
(δ A ) y = −
∂U
1
= −
∂F
 EI
∫
π /2
0
F ( R sin θ ) 2 R dθ +
1
GJ
∫
π /2
0

F [ R (1 − cos θ )]2 R dθ 

Integrating and substituting J = 2 I and G = E /  2 (1 + ν ) 
π
FR 3
 3π

+
(1
+
)
−
2
=
−
4
−
8
+
(3
−
8)
ν
π
π
ν
[
]


4
4 EI
 4


(250)(80)3
= −[4π − 8 + (3π − 8)(0.29)]
= − 12.5 mm Ans.
4(200)103 ( 63.62 )
______________________________________________________________________________
(δ A ) y = −
FR 3
EI
4-89 The force applied to the copper and steel wire assembly is
Fc + Fs = 400 lbf
(1)
Since the deflections are equal, δ c = δ s
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 Fl   Fl 

 =

 AE c  AE  s
Fc l
Fs l
=
2
6
3(π / 4)(0.1019) (17.2)10
(π / 4)(0.1055) 2 (30)106
Yields, Fc = 1.6046 Fs . Substituting this into Eq. (1) gives
1.604 Fs + Fs = 2.6046Fs = 400
⇒
Fs = 153.6 lbf
Fc = 1.6046 Fs = 246.5 lbf
F
246.5
σc = c =
= 10 075 psi = 10.1 kpsi Ans.
Ac 3(π / 4)(0.1019) 2
F
153.6
σs = s =
= 17 571 psi = 17.6 kpsi Ans.
As (π / 4)(0.10552 )
 Fl 
153.6(100)(12)
= 0.703 in
δ =
 =
2
6
 AE  s (π / 4)(0.1055) (30)10
Ans.
______________________________________________________________________________
4-90 (a) Bolt stress
σ b = 0.75(65) = 48.8 kpsi
Ans.
π 
Fb = 6σ b Ab = 6(48.8)   (0.52 ) = 57.5 kips
4
F
57.43
Cylinder stress
σc = − b =
= −13.9 kpsi Ans.
Ac (π / 4)(5.52 − 52 )
(b) Force from pressure
π D2
π (52 )
(500) = 9817 lbf = 9.82 kip
P=
p=
4
4
ΣFx = 0
Total bolt force
Pb + Pc = 9.82
(1)
Since δ c = δ b ,
Pc l
Pbl
=
2
2
(π / 4)(5.5 − 5 ) E 6(π / 4)(0.52 ) E
Pc = 3.5 Pb
(2)
Substituting this into Eq. (1)
Pb + 3.5 Pb = 4.5 Pb = 9.82 ⇒ Pb = 2.182 kip. From Eq. (2), Pc = 7.638 kip
Using the results of (a) above, the total bolt and cylinder stresses are
2.182
σ b = 48.8 +
= 50.7 kpsi Ans.
6(π / 4)(0.52 )
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7.638
= −12.0 kpsi Ans.
(π / 4)(5.52 − 52 )
______________________________________________________________________________
σ c = −13.9 +
4-91
(1)
Tc + Ts = T
θc = θs
Tc l
Tl
= s
( JG )c ( JG )s
⇒
Substitute this into Eq. (1)
( JG )c
T +T = T
( JG )s s s
⇒
Tc =
⇒
Ts =
( JG )c
T
( JG )s s
(2)
( JG )s
T
( JG )s + ( JG )c
The percentage of the total torque carried by the shell is
% Torque =
100 ( JG ) s
( JG )s + ( JG )c
Ans.
______________________________________________________________________________
4-92 RO + RB = W
(1)
δOA = δAB
 Fl 
 Fl 

 =

 AE OA  AE  AB
400 RO 600 RB
3
=
⇒ RO = RB
AE
AE
2
Substitute this unto Eq. (1)
3
RB + RB = 4
2
⇒
RB = 1.6 kN
(2)
Ans.
3
RO = 1.6 = 2.4 kN Ans.
2
2 400(400)
 Fl 
= 0.0223 mm Ans.
δA = 
 =
3
 AE OA 10(60)(71.7)(10 )
______________________________________________________________________________
From Eq. (2)
4-93 See figure in Prob. 4-92 solution.
Procedure 1:
1. Let RB be the redundant reaction.
Shigley’s MED, 10th edition
Chapter 4 Solutions, Page 64/80
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⇒ RO = 4 000 − RB
2. Statics. RO + RB = 4 000 N
3. Deflection of point B. δ B =
(1)
RB ( 600 ) ( RB − 4000 )( 400 )
+
=0
AE
AE
(2)
4. From Eq. (2), AE cancels and RB = 1 600 N Ans.
Ans.
and from Eq. (1), RO = 4 000 − 1 600 = 2 400 N
2 400(400)
 Fl 
= 0.0223 mm Ans.
 =
3
 AE OA 10(60)(71.7)(10 )
______________________________________________________________________________
δA = 
4-94 (a) Without the right-hand wall, the deflection of point C would be
δC = ∑
5 (103 ) 8
2 (103 ) 5
Fl
=
+
AE (π / 4 ) 0.752 (10.4 )106 ( π / 4 ) 0.52 (10.4 )106
= 0.01360 in > 0.005 in ∴ Hits wall
Ans.
(b) Let RC be the reaction of the wall at C acting to the left (←). Thus, the deflection of
point C is now
5 (103 ) − RC  8
 2 (103 ) − RC  5



+ 
δC =
2
6
2
(π / 4 ) 0.75 (10.4 )10 (π / 4 ) 0.5 (10.4 )106
= 0.01360 −
4 RC
5 
 8
+ 2  = 0.005
6 
2
π (10.4 )10  0.75 0.5 
or,
0.01360 − 4.190 (10 −6 ) RC = 0.005
⇒
RC = 2 053 lbf = 2.05 kip ← Ans.
Σ Fx = 5 000 + RA − 2 053 = 0 ⇒ RA = −2 947 lbf, ⇒ RA = 2.95 kip ← Ans.
Deflection. AB is 2 947 lbf in tension. Thus
δ B = δ AB =
RA ( 8 )
=
2947 ( 8 )
= 5.13 (10−3 ) in → Ans.
AAB E (π / 4 ) 0.75 (10.4 )10
______________________________________________________________________________
2
6
4-95 Since θOA = θAB,
TOA (4) TAB (6)
=
JG
JG
⇒
3
TOA = TAB
2
Shigley’s MED, 10th edition
(1)
Chapter 4 Solutions, Page 65/80
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Statics. TOA + TAB = 200
(2)
Substitute Eq. (1) into Eq. (2),
3
5
TAB + TAB = TAB = 200
⇒
2
2
From Eq. (1)
TAB = 80 lbf ⋅ in
Ans.
3
3
TOA = TAB = 80 = 120 lbf ⋅ in
Ans.
2
2
80 ( 6 )
180
 TL 
θA = 
= 0.3900
 =
4
6
π
π
JG
/
32
0.5
11.5
10
(
)
(
)

 AB
τ max =
16T
πd3
⇒
τ AB =
16 ( 80 )
π ( 0.53 )
16 (120 )
τ OA =
π ( 0.53 )
Ans.
= 4890 psi = 4.89 kpsi
= 3260 psi = 3.26 kpsi
Ans.
Ans.
______________________________________________________________________________
4-96 Since θOA = θAB,
TOA (4)
TAB (6)
=
4
(π / 32 ) 0.5 G (π / 32 ) 0.754 G
Statics. TOA + TAB = 200
⇒
TOA = 0.2963 TAB
(1)
(2)
Substitute Eq. (1) into Eq. (2),
0.2963TAB + TAB = 1.2963TAB = 200
From Eq. (1)
TAB = 154.3 lbf ⋅ in
⇒
TOA = 0.2963TAB = 0.2963 (154.3) = 45.7 lbf ⋅ in
θA =
154.3 ( 6 )
180
(π / 32 ) 0.75 (11.5)10 π
τ max =
4
16T
πd3
⇒
τ AB =
6
τ OA =
= 0.1480
16 ( 45.7 )
π ( 0.53 )
Ans.
Ans.
Ans.
= 1862 psi =1.86 kpsi
Ans.
16 (154.3)
= 1862 psi =1.86 kpsi
Ans.
π ( 0.753 )
______________________________________________________________________________
Shigley’s MED, 10th edition
Chapter 4 Solutions, Page 66/80
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4-97 Procedure 1
1. Arbitrarily, choose RC as a redundant reaction.
2. Statics. ΣFx = 0,
12(103) − 6(103) − RO − RC = 0
RO = 6(103) − RC
(1)
3. The deflection of point C.
12(103 ) − 6(103 ) − RC  (20) 6(103 ) + RC  (10) R (15)
δC =
−
− C
=0
AE
AE
AE
4. The deflection equation simplifies to
− 45 RC + 60(103) = 0
⇒ RC = 1 333 lbf = 1.33 kip Ans.
RO = 6(103) − 1 333 = 4 667 lbf = 4.67 kip Ans.
From Eq. (1),
FAB = FB + RC = 6 +1.333 = 7.333 kips compression
FAB −7.333
=
= −14.7 kpsi Ans.
A
(0.5)(1)
Deflection of A. Since OA is in tension,
R l
4 667(20)
δ A = δ OA = O OA =
= 0.00622 in Ans.
AE
(0.5)(1)(30)106
______________________________________________________________________________
σ AB =
4-98 Procedure 1
1. Choose RB as redundant reaction.
2. Statics. RC = wl − RB
(1)
1
w l 2 − RB ( l − a )
(2)
2
3. Deflection equation for point B. Superposition of beams 2 and 3 of Table A-9,
MC =
RB ( l − a ) w ( l − a )
 4l ( l − a ) − ( l − a )2 − 6l 2  = 0
+

3EI
24 EI 
3
yB =
2
4. Solving for RB.
w  2
2
RB =
6l − 4l ( l − a ) + ( l − a ) 

8(l − a ) 
=
w
3l 2 + 2al + a 2 )
(
8 (l − a )
Ans.
Substituting this into Eqs. (1) and (2) gives
Shigley’s MED, 10th edition
Chapter 4 Solutions, Page 67/80
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RC = wl − RB =
w
5l 2 − 10al − a 2 )
(
8 (l − a )
Ans.
1 2
w
wl − RB ( l − a ) = ( l 2 − 2al − a 2 )
Ans.
2
8
______________________________________________________________________________
MC =
4-99 See figure in Prob. 4-98 solution.
Procedure 1
1. Choose RB as redundant reaction.
2. Statics. RC = wl − RB
(1)
1
M C = wl 2 − RB ( l − a )
(2)
2
3. Deflection equation for point B. Let the variable x start at point A and to the right. Using
singularity functions, the bending moment as a function of x is
1
M = − w x 2 + RB x − a
2
1
∂M
= x−a
∂RB
1
1
∂U
∂M
M
dx
=
∫
∂RB EI 0 ∂RB
l
yB =
l
=
l
1
1
1  1 2

− w x 2 ( 0 ) dx +
− w x + RB ( x − a )  ( x − a ) dx = 0
∫
∫

EI 0 2
EI a  2

or,
1 1
a
3
3
 R
− w  ( l 4 − a 4 ) − ( l 3 − a 3 )  + B ( l − a ) − ( a − a )  = 0


2 4
3
 3
Solving for RB gives
RB =
w
8 (l − a )
3
w
2
2
3 ( l 4 − a 4 ) − 4a ( l 3 − a 3 )  =

 8 ( l − a ) ( 3l + 2al + a )
Ans.
From Eqs. (1) and (2)
RC = wl − RB =
w
( 5l 2 − 10al − a 2 )
8 (l − a )
Ans.
w
1 2
wl − RB ( l − a ) = ( l 2 − 2al − a 2 )
Ans.
2
8
______________________________________________________________________________
MC =
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4-100 Note: When setting up the equations for this problem, no rounding of numbers was made
in the calculations. It turns out that the deflection equation is very sensitive to rounding.
Procedure 2
1. Statics.
R1 + R2 = wl
1 2
wl
2
2. Bending moment equation.
R2l + M 1 =
(1)
(2)
1
M = R1 x − w x 2 − M 1
2
dy 1
1
EI
= R1 x 2 − w x 3 − M 1 x + C1
dx 2
6
1
1
1
EIy = R1 x3 − w x 4 − M 1 x 2 + C1 x + C2
6
24
2
(3)
(4)
EI = 30(106)(0.85) = 25.5(106) lbf⋅in2.
3. Boundary condition 1. At x = 0, y = − R1/k1 = − R1/[1.5(106)]. Substitute into Eq. (4)
with value of EI yields C2 = − 17 R1.
Boundary condition 2. At x = 0, dy /dx = − M1/k2 = − M1/[2.5(106)]. Substitute into Eq.
(3) with value of EI yields C1 = − 10.2 M1.
Boundary condition 3. At x = l, y = − R2/k3 = − R1/[2.0(106)]. Substitute into Eq. (4)
with value of EI yields
1 3 1
1
R1l − wl 4 − M 1l 2 − 10.2 M 1l − 17 R1
(5)
6
24
2
Equations (1), (2), and (5), written in matrix form with w = 500/12 lbf/in and l = 24 in,
are
−12.75R2 =
1
0 
 1


24
1 
 0
 2287 12.75 −532.8 


 R1   1 
  
 3
 R2  =  12  (10 )
 M  576 
 1 

Solving, the simultaneous equations yields
R1 = 554.59 lbf, R2 = 445.41.59 lbf, M1 = 1310.1 lbf⋅in
Ans.
For the deflection at x = l /2 = 12 in, Eq. (4) gives
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y x =12in =
1
1 500 4 1
1
12 − (1310.1)122
( 554.59 )123 −
6 
24 12
2
25.5 (10 )  6
−10.2 (1310.1)12 − 17 ( 554.59 ) 
= −5.51(10−3 ) in
Ans.
______________________________________________________________________________
4-101 Cable area, A =
π
4
(0.52 ) = 0.1963 in 2
Procedure 2
1. Statics. RA + FBE + FDF = 5(103)
3 FDF + FBE = 10(103)
(1)
(2)
2. Bending moment equation.
1
M = RA x + FBE x − 16 − 5000 x − 32
1
dy 1
1
2
2
= RA x 2 + FBE x − 16 − 2500 x − 32 + C1
2
dx 2
1
1
2500
3
3
EIy = RA x 3 + FBE x − 16 −
x − 32 + C1 x + C2
6
6
3
EI
3. B.C. 1: At x = 0, y = 0 ⇒
(3)
(4)
C2 = 0
B.C. 2: At x = 16 in,
FBE (38)
 Fl 
yB = − 
= −6.453(10−6 ) FBE
 =−
0.1963(30)106
 AE  BE
Substituting into Eq. (4) and evaluating at x = 16 in
1
EIyB = 30(106 )(1.2)(−6.453)(10−6 ) FBE = RA (163 ) + C1 (16)
6
(5)
Simplifying gives 682.7 RA + 232.3 FBE + 16 C1 = 0
B.C. 2: At x = 48 in,
FDF (38)
 Fl 
= −6.453(10−6 ) FDF
yD = − 
 =−
6
0.1963(30)10
 AE  DF
Substituting into Eq. (4) and evaluating at x = 48 in,
1
1
2500
RA ( 483 ) + FBE (48 − 16)3 −
(48 − 32)3 + 48C1
6
6
3
18 432 RA + 5 461 FBE + 232.3 FDF + 48 C1 = 3.413(106)
EIyD = −232.3FDF =
Simplifying gives
Shigley’s MED, 10th edition
(6)
Chapter 4 Solutions, Page 70/80
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Equations (1), (2), (5) and (6) in matrix form are
1
1
0   RA   5000 
 1

  F   10 000 
0
1
3
0


  BE  = 

0
 682.7 232.3

0
16  FDF  



6
18 432 5 461 232.3 48   C1  3.413 (10 ) 
Solve simultaneously or use software. The results are
RA = − 970.5 lbf, FBE = 3956 lbf, FDF = 2015 lbf, and C1 = − 16 020 lbf⋅in2.
3956
2015
σ BE =
= 20.2 kpsi, σ DF =
= 10.3 kpsi Ans.
0.1963
0.1963
EI = 30(106)(1.2) = 36(106) lbf⋅in2
y=
=
1
2500
3
3
 970.5 3 3956

−
x +
x − 16 −
x − 32 − 16 020 x 
6 
6
6
3
36 10 

( )
−161.8 x + 659.3 x − 16
( )(
1
36 106
B: x = 16 in,
3
3
3
− 833.3 x − 32 − 16 020 x
)
yB =
1
 −161.8 (163 ) − 16 020 (16 )  = −0.0255 in
6 

36 (10 )
yC =
1
 −161.8 ( 323 ) + 659.3 ( 32 − 16 )3 − 16 020 ( 32 ) 
6 

36 (10 )
Ans.
C: x = 32 in,
= −0.0865 in
Ans.
D: x = 48 in,
yD =
( )
1
 −161.8 483 + 659.3 ( 48 − 16 )3 − 833.3 ( 48 − 32 )3 − 16 020 ( 48 ) 
6 

36 10
( )
= −0.0131 in Ans.
______________________________________________________________________________
4-102 Beam: EI = 207(103)21(103)
= 4.347(109) N⋅mm2.
Rods: A = (π /4)82 = 50.27 mm2.
Procedure 2
1. Statics.
Shigley’s MED, 10th edition
Chapter 4 Solutions, Page 71/80
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RC + FBE − FDF = 2 000
(1)
RC + 2FBE = 6 000
(2)
2. Bending moment equation.
M = − 2 000 x + FBE ⟨x − 75 ⟩1 + RC ⟨x − 150 ⟩1
dy
1
1
2
2
= −1000 x 2 + FBE x − 75 + RC x − 150 + C1
2
2
dx
1000 3 1
1
3
3
EIy = −
x + FBE x − 75 + RC x − 150 + C1 x + C2
3
6
6
EI
(3)
(4)
3. B.C 1. At x = 75 mm,
FBE ( 50 )
 Fl 
= −4.805 (10−6 ) FBE
yB = − 
 =−
50.27 ( 207 )103
 AE  BE
Substituting into Eq. (4) at x = 75 mm,
4.347 (109 )  −4.805 (10−6 ) FBE  = −
1000
753 ) + C1 ( 75 ) + C2
(
3
Simplifying gives
20.89 (103 ) FBE + 75C1 + C2 = 140.6 (10 6 )
(5)
B.C 2. At x = 150 mm, y = 0. From Eq. (4),
−
or,
1000
1
3
1503 ) + FBE (150 − 75 ) + C1 (150 ) + C2 = 0
(
3
6
70.31(103 ) FBE + 150C1 + C2 = 1.125 (109 )
(6)
B.C 3. At x = 225 mm,
FDF ( 65 )
 Fl 
= 6.246 (10−6 ) FDF
yD = 
 =
3
 AE  DF 50.27 ( 207 )10
Substituting into Eq. (4) at x = 225 mm,
Shigley’s MED, 10th edition
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4.347 (109 ) 6.246 (10−6 ) FDF  = −
1000
1
3
2253 ) + FBE ( 225 − 75 )
(
3
6
1
3
+ RC ( 225 − 150 ) + C1 ( 225 ) + C2
6
Simplifying gives
70.31(103 ) RC + 562.5 (103 ) FBE − 27.15 (103 ) FDF + 225C1 + C2 = 3.797 (109 )
(7)
Equations (1), (2), (5), (6), and (7) in matrix form are
 2 (103 ) 
−1
1
1
0 0



  RC  
3
1
2
0
0
0




6
10
(
)

 FBE


 
3


6 
0
20.89 (10 )
0
75 1 F
=
140.6
10
( )

  DF  
3






9
0
70.31(10 )
0
150 1 C1
 1.125 (10 ) 



 70.31(103 ) 562.5 (103 ) −27.15 (103 ) 225 1   C2  
9


3.797 (10 ) 
Solve simultaneously or use software. The results are
RC = − 2378 N, FBE = 4189 N, FDF = − 189.2 N
Ans.
and C1 = 1.036 (107) N⋅mm2, C2 = − 7.243 (108) N⋅mm3.
The bolt stresses are σBE = 4189/50.27 = 83.3 MPa, σDF = − 189/50.27= − 3.8 MPa Ans.
The deflections are
From Eq. (4)
yA =
1
 −7.243 (108 )  = −0.167 mm
9 

4.347 (10 )
Ans.
For points B and D use the axial deflection equations*.
4189 ( 50 )
 Fl 
= −0.0201 mm
yB = − 
 =−
50.27 ( 207 )103
 AE  BE
Ans.
−189 ( 65 )
 Fl 
= −1.18 (10−3 ) mm
yD = 
Ans.
 =
3
 AE  DF 50.27 ( 207 )10
*Note. The terms in Eq. (4) are quite large, and due to rounding are not very accurate for
calculating the very small deflections, especially for point D.
______________________________________________________________________________
4-103 (a) The cross section at A does not rotate. Thus, for a single quadrant we have
Shigley’s MED, 10th edition
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∂U
=0
∂M A
The bending moment at an angle θ to the x axis is
FR
∂M
=1
M = MA −
(1 − cos θ )
2
∂M A
The rotation at A is
π /2
∂U
1
∂M
M
Rdθ = 0
θA =
=
∫
∂M A EI 0
∂M A
Thus,
1
EI
π /2

∫  M
A
−
0
FR
(1 − cos θ )  (1) Rdθ = 0
2

⇒
FR  π FR

=0
MA −
 +
2 2
2

or,
FR  2 
1 − 
2  π
Substituting this into the equation for M gives
2
FR 
M=
 cos θ −  (1)
2 
π
The maximum occurs at B where θ = π /2
MA =
M max = M B = −
FR
Ans.
π
(b) Assume B is supported on a knife edge. The deflection of point D is ∂ U/∂ F. We will
deal with the quarter-ring segment and multiply the results by 4. From Eq. (1)
∂M R 
2
=  cos θ − 
∂F 2 
π
Thus,
∂U
4
=
δD =
∂F EI
π /2
∫
0
∂M
FR3
M
Rdθ =
∂F
EI
π /2
∫
0
2
2
FR 3  π 2 

cos
−
=
θ
θ
d

 − 
π 
EI  4 π 

3
FR
(π 2 − 8) Ans.
4π EI
______________________________________________________________________________
=
4-104
Pcr =
I=
Cπ 2 EI
l2
π
( D4 − d 4 ) =
π D4
(1 − K )
64
64
2
4

Cπ E  π D
Pcr = 2 
1 − K 4 )
(
l
 64

4
where K =
Shigley’s MED, 10th edition
d
D
Chapter 4 Solutions, Page 74/80
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1/4


64 Pcr l 2

D= 3
Ans.
4
 π CE (1 − K ) 
______________________________________________________________________________
π
D 2 (1 − K 2 ) ,
π
I=
D 4 (1 − K 4 ) =
π
D 4 (1 − K 2 )(1 + K 2 ) , where K = d / D.
4
64
64
The radius of gyration, k, is given by
I D2
k2 = =
1+ K 2 )
(
A 16
From Eq. (4-46)
S y2l 2
S y2l 2
Pcr
=
−
=
−
S
S
y
y
4π 2 k 2CE
4π 2 ( D 2 / 16 )(1 + K 2 ) CE
(π / 4 ) D 2 (1 − K 2 )
4-105 A =
4 Pcr = π D (1 − K
2
2
)S
y
−
π D 2 (1 − K 2 ) S y = 4 Pcr +
4 S y2l 2π D 2 (1 − K 2 )
π 2 D 2 (1 + K 2 ) CE
4S y2l 2 (1 − K 2 )
π (1 + K 2 ) CE


4 S y2l 2 (1 − K 2 )
4 Pcr

D=
+
2
2
2
 π S y (1 − K ) π (1 + K ) CEπ (1 − K ) S y 
1/2
1/2


S yl 2
Pcr

= 2
+
Ans.
2
2
2
 π S y (1 − K ) π CE (1 + K ) 
______________________________________________________________________________
0.9
4-106 (a) ΣM A = 0, (0.75)(800) −
0.92 + 0.52
FBO (0.5) = 0 ⇒ FBO = 1373 N
Using nd = 4, design for Fcr = nd FBO = 4(1373) = 5492 N
l = 0.92 + 0.52 = 1.03 m, S y = 165 MPa
In-plane:
1/2
1/2
 bh3 / 12 
I 
k =  =

 A
 bh 
l
1.03
=
= 142.7
k 0.007218
= 0.2887 h = 0.2887(0.025) = 0.007 218 m,
C = 1.0
1/2
2
9
 l   2π (207)(10 ) 

  =
6
 k 1  165(10 ) 
= 157.4
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Since (l / k )1 > (l / k ) use Johnson formula.
Try 25 mm x 12 mm,
2


165 (106 )

1


6
Pcr = 0.025(0.012) 165 (10 ) − 
(142.7) 
= 29.1 kN
9 
 2π
 1(207)10 



This is significantly greater than the design load of 5492 N found earlier. Check out-ofplane.
Out-of-plane: k = 0.2887(0.012) = 0.003 464 in,
l
1.03
=
= 297.3
k 0.003 464
Since (l / k )1 < (l / k ) use Euler equation.
Pcr = 0.025(0.012)
C = 1.2
1.2π 2 ( 207 )109
= 8321 N
297.32
This is greater than the design load of 5492 N found earlier. It is also significantly less
than the in-plane Pcr found earlier, so the out-of-plane condition will dominate. Iterate
the process to find the minimum h that gives Pcr greater than the design load.
With h = 0.010, Pcr = 4815 N (too small)
h = 0.011, Pcr = 6409 N (acceptable)
Use 25 mm x 11 mm. If standard size is preferred, use 25 mm x 12 mm. Ans.
( )
P
1373
=−
= −10.4 106 Pa = −10.4 MPa
0.012(0.011)
dh
No, bearing stress is not significant. Ans.
______________________________________________________________________________
(b) σ b = −
4-107 This is an open-ended design problem with no one distinct solution.
______________________________________________________________________________
F = 1500(π /4)22 = 4712 lbf. From Table A-20, Sy = 37.5 kpsi
Pcr = nd F = 2.5(4712) = 11 780 lbf
4-108
(a) Assume Euler with C = 1
1/4
1/4
 64 (11790 ) 502 
 64 Pcr l 2 

⇒
d = 3
 = 3
6
 π (1) 30 (10 ) 
 π CE 
Use d = 1.25 in. The radius of gyration, k = ( I / A)1/2 = d /4 = 0.3125 in
π
P l2
I = d 4 = cr 2
Cπ E
64
Shigley’s MED, 10th edition
= 1.193 in
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l
50
=
= 160
k 0.3125
 2π 2 (1)30 (106 ) 
 = 126
=
 37.5 (103 ) 


2
6
4
π ( 30 )10 (π / 64 )1.25
= 14194 lbf
Pcr =
502
1/2
1/2
2
 l   2π CE 
=


  
 k 1  S y 
Since 14 194 lbf > 11 780 lbf, d = 1.25 in is satisfactory.
∴ use Euler
Ans.
1/4
 64 (11780 )162 
 = 0.675 in, so use d = 0.750 in
d= 3
6
 π (1) 30 (10 ) 
k = 0.750/4 = 0.1875 in
l
16
=
= 85.33
use Johnson
k 0.1875
(b)
2


 37.5 (103 )

1


3
Pcr = ( 0.750 ) 37.5 (10 ) − 
85.33
= 12 748 lbf
6
4
2
1
30
10
π
(
)






Use d = 0.75 in.
π
2
(c)
14194
= 3.01
Ans.
4 712
12 748
n(b ) =
= 2.71
Ans.
4 712
______________________________________________________________________________
n( a ) =
4-109 From Table A-20, Sy = 180 MPa
4F sinθ = 2 943
735.8
sin θ
In range of operation, F is maximum when θ = 15°
735.8
Fmax =
= 2843 N per bar
sin15o
F=
Pcr = ndFmax = 3.50 (2 843) = 9 951 N
l = 350 mm, h = 30 mm
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Try b = 5 mm. Out of plane, k = b / 12 = 5/ 12 = 1.443 mm
350
l
=
= 242.6
k 1.443
2
9
 l   2π (1.4 ) 207 (10 ) 


=
 
180 (106 )
 k 1 

1/2
Pcr = A
Cπ 2 E
(l / k )
2
= 5(30)
= 178.3
1.4π 2 ( 207 )103
( 242.6 )
2
∴ use Euler
= 7 290 N
Too low. Try b = 6 mm. k = 6/ 12 = 1.732 mm
350
l
=
= 202.1
k 1.732
1.4π 2 ( 207 )103
Cπ 2 E
6(30)
=
= 12 605 N
Pcr = A
2
2
( 202.1)
(l / k )
O.K. Use 25 × 6 mm bars Ans. The factor of safety is
12 605
= 4.43 Ans.
2843
______________________________________________________________________________
n=
4-110 P = 1 500 + 9 000 = 10 500 lbf Ans.
ΣMA = 10 500 (4.5/2) − 9 000 (4.5) +M = 0
M = 16 874 lbf⋅in
e = M / P = 16 874/10 500 = 1.607 in Ans.
From Table A-8, A = 2.160 in2, and I = 2.059 in4. The stresses are determined using Eq.
(4-55)
k2 =
I 2.059
=
= 0.953 in 2
A 2.160
P  ec 
10500  1.607 ( 3 / 2 ) 
1 +
 = −17157 psi = −17.16 kpsi Ans.
1 + 2  = −
A k 
2.160 
0.953 
______________________________________________________________________________
σc = −
4-111 This is a design problem which has no single distinct solution.
______________________________________________________________________________
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4-112 Loss of potential energy of weight = W (h +δ )
1
Increase in potential energy of spring = kδ 2
2
1 2
W (h +δ ) = kδ
2
2
W
2
W
or, δ 2 −
h = 0 . W = 30 lbf, k = 100 lbf/in, h = 2 in yields
δ−
k
k
δ 2 − 0.6 δ − 1.2 = 0
Taking the positive root (see discussion on p. 206 of text)
1
δ max = 0.6 + (−0.6) 2 + 4(1.2)  = 1.436 in

2
Ans.
Fmax = k δ max = 100 (1.436) = 143.6 lbf
Ans.
______________________________________________________________________________
4-113 The drop of weight W1 converts potential energy, W1 h, to kinetic energy
1 W1 2
v1 .
2 g
Equating these provides the velocity of W1 at impact with W2.
1 W1 2
(1)
v1
⇒
v1 = 2 gh
2 g
Since the collision is inelastic, momentum is conserved. That is, (m1 + m2) v2 = m1 v1,
where v2 is the velocity of W1 + W2 after impact. Thus
W1h =
W1 + W2
W
v2 = 1 v1
g
g
⇒
v2 =
W1
W1
v1 =
2 gh
W1 + W2
W1 + W2
(2)
The kinetic and potential energies of W1 + W2 are then converted to potential energy of
the spring. Thus,
1 W1 + W2 2
1
v2 + (W1 + W2 ) δ = kδ 2
2
g
2
Substituting in Eq. (1) and rearranging results in
W1 + W2
W12 h
(3)
δ −2
=0
k
W1 + W2 k
Solving for the positive root (see discussion on p. 206)
δ2 −2
2
1  W1 + W2
W12 h 
 W + W2 

8
+ 4 1
+

2
k
k
W1 + W2 k 




δ = 2
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W1 = 40 N, W2 = 400 N, h = 200 mm, k = 32 kN/m = 32 N/mm.
2
1   40 + 400 
402 200 
 40 + 400 

 = 29.06 mm
δ = 2
 + 4
 +8
2   32 
32 
40 + 400 32 



Ans.
Fmax = kδ = 32(29.06) = 930 N Ans.
______________________________________________________________________________
1
4-114 The initial potential energy of the k1 spring is Vi = k1a 2 . The movement of the weight
2
1
1
2
W the distance y gives a final potential of Vf = k1 ( a − y ) + k 2 y 2 . Equating the two
2
2
energies give
1 2 1
1
2
k1a = k1 ( a − y ) + k2 y 2
2
2
2
Simplifying gives
( k1 + k2 ) y 2 − 2ak1 y = 0
2k1a
. Without damping the weight will vibrate between
k1 + k2
2k1a
these two limits. The maximum displacement is thus y max =
Ans.
k1 + k2
With W = 5 lbf, k1 = 10 lbf/in, k2 = 20 lbf/in, and a = 0.25 in
This has two roots, y = 0,
2 ( 0.25 )10
= 0.1667 in
Ans.
10 + 20
______________________________________________________________________________
ymax =
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Chapter 5
5-1
Sy = 350 MPa.
MSS: σ1 − σ3 = Sy /n
DE:
σ ′ = (σ A2 − σ Aσ B + σ
n=
n=
⇒
)
2 1/2
B
Sy
(σ 1 − σ 3 )
= (σ x2 − σ xσ y + σ y2 + 3τ xy2 )
1/2
Sy
σ′
(a) MSS:
σ1 = 100 MPa,σ2 = 100 MPa,σ3 = 0
n=
DE:
(b) MSS:
Ans.
σ ′ = (1002 − 100(100) + 1002 )1/2 = 100 MPa,
350
= 3.5
100 − 0
350
= 3.5
100
Ans.
350
= 4.04
86.6
Ans.
n=
σ1 = 100 MPa,σ2 = 50 MPa,σ3 = 0
n=
DE:
350
= 3.5
100 − 0
Ans.
σ ′ = (1002 − 100(50) + 502 )1/2 = 86.6 MPa,
n=
2
100
 100 
2
(c) σ A , σ B =
± 
 + (−75) = 140, − 40 MPa
2
 2 
σ 1 = 140, σ 2 = 0, σ 3 = −40 MPa
350
MSS: n =
= 1.94 Ans.
140 − (−40)
DE:
σ ′ = 1002 + 3 ( −752 )
1/2
= 164 MPa,
n=
350
= 2.13
164
Ans.
2
−50 − 75
 −50 + 75 
2
± 
 + (−50) = −11.0, −114.0 MPa
2
2


σ 1 = 0, σ 2 = −11.0, σ 3 = −114.0 MPa
350
MSS:
n=
= 3.07 Ans.
0 − (−114.0)
σ ′ = [(−50) 2 − (−50)(−75) + (−75) 2 + 3(−50) 2 ]1/ 2 = 109.0 MPa
DE:
350
n=
= 3.21 Ans.
109.0
(d) σ A , σ B =
2
100 + 20
2
 100 − 20 
(e) σ A , σ B =
± 
 + ( −20 ) = 104.7, 15.3 MPa
2
2


Shigley’s MED, 10th edition
Chapter 5 Solutions, Page 1/52
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σ 1 = 104.7, σ 2 = 15.3, σ 3 = 0 MPa
MSS:
n=
350
= 3.34
104.7 − 0
Ans.
1/2
2
σ ′ = 1002 − 100(20) + 202 + 3 ( −20 ) 
= 98.0 MPa


350
n=
= 3.57 Ans.
98.0
______________________________________________________________________________
DE:
5-2
Sy = 350 MPa.
MSS: σ1 − σ3 = Sy /n
DE:
n=
⇒
σ ′ = (σ A2 − σ Aσ B + σ B2 )
(a) MSS:
DE:
(b) MSS:
1/2
=
Sy
(σ 1 − σ 3 )
Sy
n
⇒
n=
σ 1 = 100 MPa, σ 3 = 0 ⇒ n =
n=
Sy
σ′
350
= 3.5
100 − 0
350
= 3.5
[100 − (100)(100) + 1002 ]1/2
Ans.
2
σ 1 = 100 , σ 3 = −100 MPa ⇒
n=
Ans.
350
= 1.75
100 − (−100)
Ans.
350
= 2.02 Ans.
1/2
100 − (100)(−100) + ( −100 )2 


350
(c) MSS:
σ 1 = 100 MPa, σ 3 = 0 ⇒ n =
= 3.5 Ans.
100 − 0
350
DE:
n=
= 4.04 Ans.
1/2
2
100 − (100)(50) + 502 
350
(d) MSS:
σ 1 = 100, σ 3 = −50 MPa ⇒ n =
= 2.33 Ans.
100 − (−50)
350
DE:
n=
= 2.65
Ans.
1/2
1002 − (100)(−50) + ( −50 )2 


350
(e) MSS: σ 1 = 0, σ 3 = −100 MPa ⇒ n =
= 3.5 Ans.
0 − (−100)
350
DE:
n=
= 4.04
Ans.
1/2
( −50 ) 2 − (−50)(−100) + ( −100 ) 2 


______________________________________________________________________________
DE:
n=
2
5-3
From Table A-20, Sy = 37.5 kpsi
Shigley’s MED, 10th edition
Chapter 5 Solutions, Page 2/52
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MSS: σ1 − σ3 = Sy /n
σ ′ = (σ A2 − σ Aσ B + σ
DE:
n=
n=
⇒
)
2 1/2
B
Sy
(σ 1 − σ 3 )
= (σ x2 − σ xσ y + σ y2 + 3τ xy2 )
1/2
Sy
σ′
σ 1 = 25 kpsi, σ 3 = 0
(a) MSS:
n=
DE:
⇒
37.5
1/2
 25 − (25)(15) + 152 
2
n=
= 1.72
σ 1 = 15 kpsi, σ 3 = −15 ⇒
(b) MSS:
n=
DE:
n=
37.5
1/2
15 − (15)(−15) + ( −15 )2 


2
37.5
= 1.5
25 − 0
Ans.
Ans.
37.5
= 1.25
15 − (−15)
= 1.44
Ans.
Ans.
2
20
 20 
±   + (−10) 2 = 24.1, − 4.1 kpsi
2
 2 
σ 1 = 24.1, σ 2 = 0, σ 3 = −4.1 kpsi
37.5
n=
= 1.33 Ans.
MSS:
24.1 − (−4.1)
(c) σ A , σ B =
σ ′ =  202 + 3 ( −102 )
DE:
1/2
= 26.5 kpsi
⇒
n=
37.5
= 1.42
26.5
Ans.
2
−12 + 15
 −12 − 15 
2
± 
 + (−9) = 17.7, −14.7 kpsi
2
2


σ 1 = 17.7, σ 2 = 0, σ 3 = −14.7 kpsi
37.5
n=
= 1.16 Ans.
MSS:
17.7 − (−14.7)
(d) σ A , σ B =
1/ 2
σ ′ = (−12) 2 − (−12)(15) + 152 + 3(−9)2 
DE:
n=
37.5
= 1.33
28.1
= 28.1 kpsi
Ans.
2
−24 − 24
2
 −24 + 24 
(e) σ A , σ B =
± 
 + ( −15 ) = −9, − 39 kpsi
2
2


σ 1 = 0, σ 2 = −9, σ 3 = −39 kpsi
37.5
MSS:
n=
= 0.96 Ans.
0 − ( −39 )
Shigley’s MED, 10th edition
Chapter 5 Solutions, Page 3/52
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2
2
2
σ ′ = ( −24 ) − ( −24 )( −24 ) + ( −24 ) + 3 ( −15 ) 
1/ 2
= 35.4 kpsi


37.5
n=
= 1.06 Ans.
35.4
______________________________________________________________________________
DE:
5-4
From Table A-20, Sy = 47 kpsi.
MSS: σ1 − σ3 = Sy /n
n=
⇒
σ ′ = (σ A2 − σ Aσ B + σ B2 )
Sy
(σ 1 − σ 3 )
Sy
n
σ′
47
(a) MSS:
σ 1 = 30 kpsi, σ 3 = 0 ⇒ n =
= 1.57 Ans.
30 − 0
47
DE:
n=
= 1.57 Ans.
2
[30 − (30)(30) + 302 ]1/2
47
(b) MSS:
σ 1 = 30 , σ 3 = −30 kpsi ⇒ n =
= 0.78 Ans.
30 − (−30)
47
DE:
n=
= 0.90 Ans.
1/2
302 − (30)(−30) + ( −30 )2 


47
(c) MSS:
σ 1 = 30 kpsi, σ 3 = 0 ⇒ n =
= 1.57 Ans.
30 − 0
47
DE:
n=
= 1.81 Ans.
1/2
302 − (30)(15) + 152 
47
(d) MSS:
σ 1 = 0, σ 3 = −30 kpsi ⇒ n =
= 1.57 Ans.
0 − (−30)
47
DE:
n=
= 1.81
Ans.
1/2
( −30 )2 − (−30)(−15) + ( −15 )2 


47
(e) MSS: σ 1 = 10, σ 3 = −50 kpsi ⇒ n =
= 0.78 Ans.
10 − (−50)
47
DE:
n=
= 0.84
Ans.
1/2
2
( −50 ) − (−50)(10) + 102 


______________________________________________________________________________
DE:
1/2
=
Sy
⇒
n=
Shigley’s MED, 10th edition
Chapter 5 Solutions, Page 4/52
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5-5
Note: The drawing in this manual may not be to the scale of original drawing. The
measurements were taken from the original drawing.
(a) MSS and DE:
n=
OB 4.95"
=
= 3.51 Ans.
OA 1.41"
(b) MSS:
n=
OD 3.91"
=
= 3.49 Ans.
OC 1.12"
DE:
n=
OE 4.51"
=
= 4.03 Ans.
OC 1.12"
(c) MSS:
n=
OG 2.50"
=
= 2.00 Ans.
OF 1.25"
DE: n =
OH 2.86"
=
= 2.29 Ans.
OF 1.25"
(d) MSS: n =
OJ 3.51"
OK 3.65"
=
= 3.05 Ans. , DE: n =
=
= 3.17 Ans.
OI 1.15"
OI 1.15"
OM 3.54"
ON 3.77"
=
= 3.34 Ans. , DE: n =
=
= 3.56 Ans.
OL 1.06"
OL 1.06"
______________________________________________________________________________
(e) MSS: n =
Shigley’s MED, 10th edition
Chapter 5 Solutions, Page 5/52
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5-6
Note: The drawing in this manual may not be to the scale of original drawing. The
measurements were taken from the original drawing.
(a) σA = 25 kpsi, σB = 15 kpsi
MSS:
n=
OB 4.37"
=
= 1.50
OA 2.92"
Ans.
DE:
OC 5.02"
=
= 1.72 Ans.
OA 2.92"
(b) σA = 15 kpsi, σB = − 15 kpsi
n=
MSS:
n=
OE 2.66"
=
= 1.25
OD 2.12"
Ans.
n=
OF 3.05"
=
= 1.44
OD 2.12"
Ans.
DE:
2
20
 20 
(c) σ A , σ B =
±   + (−10) 2 = 24.1, − 4.1 kpsi
2
 2 
MSS: n =
OH 3.25"
=
= 1.34
OG 2.43"
Ans. DE: n =
OI 3.46"
=
= 1.42
OG 2.43"
Ans.
2
(d) σ A , σ B =
MSS: n =
−12 + 15
 −12 − 15 
2
± 
 + (−9) = 17.7, −14.7 MPa
2
2


OK 2.67"
=
= 1.16
OJ 2.30"
Ans. DE: n =
OL 3.06"
=
= 1.33
OJ 2.30"
Ans.
2
(e) σ A , σ B =
−24 − 24
2
 −24 + 24 
± 
 + ( −15 ) = −9, − 39 kpsi
2
2


ON 3.85"
OP 4.23"
=
= 0.96 Ans. DE: n =
=
= 1.06 Ans.
OM 4.00"
OM 4.00"
______________________________________________________________________________
MSS: n =
Shigley’s MED, 10th edition
Chapter 5 Solutions, Page 6/52
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5-7
Sy = 295 MPa, σA = 75 MPa, σB = − 35 MPa,
(a) n =
Sy
(σ − σ Aσ B + σ
2
A
)
2 1/2
B
=
295
1/2
752 − 75 ( −35 ) + ( −35 )2 


= 3.03
Ans.
(b) Note: The drawing in this manual may not be to the scale of original drawing. The
measurements were taken from the original drawing.
n=
OB 2.50"
=
= 3.01 Ans.
OA 0.83"
______________________________________________________________________________
Shigley’s MED, 10th edition
Chapter 5 Solutions, Page 7/52
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5-8
Sy = 295 MPa, σA = 30 MPa, σB = − 100 MPa,
(a) n =
Sy
(σ − σ Aσ B + σ
2
A
)
2 1/2
B
=
295
1/2
302 − 30 ( −100 ) + ( −100 )2 


= 2.50
Ans.
(b) Note: The drawing in this manual may not be to the scale of original drawing. The
measurements were taken from the original drawing.
n=
OB 2.50"
=
= 3.01 Ans.
OA 0.83"
______________________________________________________________________________
Shigley’s MED, 10th edition
Chapter 5 Solutions, Page 8/52
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2
5-9
100
 100 
2
Sy = 295 MPa, σ A , σ B =
± 
 + (−25) = 105.9, − 5.9 MPa
2
 2 
Sy
295
(a) n =
=
= 2.71
1/2
1/2
(σ A2 − σ Aσ B + σ B2 ) 105.92 − 105.9 ( −5.9) + ( −5.9 )2 
Ans.
(b) Note: The drawing in this manual may not be to the scale of original drawing. The
measurements were taken from the original drawing.
n=
OB 2.87"
=
= 2.71 Ans.
OA 1.06"
______________________________________________________________________________
Shigley’s MED, 10th edition
Chapter 5 Solutions, Page 9/52
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2
5-10
−30 − 65
 −30 + 65 
2
Sy = 295 MPa, σ A , σ B =
± 
 + 40 = −3.8, −91.2 MPa
2
2


Sy
295
(a) n =
=
= 3.30
1/2
1/2
(σ A2 − σ Aσ B + σ B2 ) ( −3.8)2 − ( −3.8)( −91.2 ) + ( −91.2 )2 
Ans.
(b) Note: The drawing in this manual may not be to the scale of original drawing. The
measurements were taken from the original drawing.
n=
OB 3.00"
=
= 3.33 Ans.
OA 0.90"
______________________________________________________________________________
Shigley’s MED, 10th edition
Chapter 5 Solutions, Page 10/52
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2
5-11
−80 + 30
2
 −80 − 30 
Sy = 295 MPa, σ A , σ B =
± 
 + ( −10 ) = 30.9, −80.9 MPa
2
2


Sy
295
(a) n =
=
= 2.95 Ans.
1/2
1/2
(σ A2 − σ Aσ B + σ B2 ) 30.92 − 30.9 ( −80.9 ) + ( −80.9 )2 
(b) Note: The drawing in this manual may not be to the scale of original drawing. The
measurements were taken from the original drawing.
n=
OB 2.55"
=
= 2.93 Ans.
OA 0.87"
______________________________________________________________________________
5-12
Syt = 60 kpsi, Syc = 75 kpsi. Eq. (5-26) for yield is
σ
σ 
n= 1 − 3 
S

 yt S yc 
−1
−1
(a) σ 1 = 25 kpsi, σ 3 = 0
 25 0 
n =  −  = 2.40
 60 75 
⇒
(b) σ 1 = 15, σ 3 = −15 kpsi
Ans.
−1
⇒
 15 −15 
n= −
 = 2.22
 60 75 
Shigley’s MED, 10th edition
Ans.
Chapter 5 Solutions, Page 11/52
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2
20
 20 
(c) σ A , σ B =
±   + (−10) 2 = 24.1, − 4.1 kpsi ,
2
 2 
σ 1 = 24.1, σ 2 = 0, σ 3 = −4.1 kpsi ⇒
−1
 24.1 −4.1 
n=
−
 = 2.19
75 
 60
Ans.
2
−12 + 15
 −12 − 15 
2
(d) σ A , σ B =
± 
 + (−9) = 17.7, −14.7 kpsi
2
2


σ 1 = 17.7, σ 2 = 0, σ 3 = −14.7 kpsi
−1
⇒
 17.7 −14.7 
−
n=
 = 2.04
75 
 60
Ans.
2
(e) σ A , σ B =
−24 − 24
2
 −24 + 24 
± 
 + ( −15 ) = −9, − 39 kpsi
2
2


−1
 0 −39 
n= −
Ans.
 = 1.92
 60 75 
______________________________________________________________________________
σ 1 = 0, σ 2 = −9, σ 3 = −39 kpsi ⇒
5-13
Note: The drawing in this manual may not be to the scale of original drawing. The
measurements were taken from the original drawing.
(a) σ A = 25, σ B = 15 kpsi
n=
OB 3.49"
=
= 2.39 Ans.
OA 1.46"
(b) σ A = 15, σ B = −15 kpsi
n=
OD 2.36"
=
= 2.23 Ans.
OC 1.06"
(c)
2
σ A, σ B =
n=
20
 20 
±   + ( −10) 2 = 24.1, − 4.1 kpsi
2
 2 
OF 2.67"
=
= 2.19 Ans.
OE 1.22"
(d)
2
σ A ,σ B =
−12 + 15
 −12 − 15 
2
± 
 + (−9) = 17.7, −14.7 kpsi
2
2


Shigley’s MED, 10th edition
Chapter 5 Solutions, Page 12/52
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n=
OH 2.34"
=
= 2.03 Ans.
OG 1.15"
(e)
2
−24 − 24
2
 −24 + 24 
± 
 + ( −15 ) = −9, − 39 kpsi
2
2


OJ 3.85"
n=
=
= 1.93 Ans.
OI 2.00"
______________________________________________________________________________
σ A ,σ B =
5-14
Since εf > 0.05, and Syt ≠ Syc, the Coulomb-Mohr theory for ductile materials will be
used.
(a) From Eq. (5-26),
−1
−1
σ
σ 
 150 −50 
n= 1 − 3  =
−
 = 1.23 Ans.
 S yt S yc 
 235 285 


(b) Plots for Problems 5-14 to 5-18 are found here. Note: The drawing in this manual
may not be to the scale of original drawing. The measurements were taken from the
original drawing.
n=
OB 1.94"
=
= 1.23 Ans.
OA 1.58"
______________________________________________________________________________
5-15
(a) From Eq. (5-26),
−1
−1
σ
σ 
 50 −150 
−
n= 1 − 3  =
 = 1.35 Ans.
S

S
235
285


yt
yc


Shigley’s MED, 10th edition
Chapter 5 Solutions, Page 13/52
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(b) The plot for this problem is found on the page for Prob. 5-14. Note: The drawing in
this manual may not be to the scale of original drawing. The measurements were taken
from the original drawing.
OD 2.14"
=
= 1.35 Ans.
OC 1.58"
______________________________________________________________________________
n=
2
5-16
125
 125 
2
± 
 + (−75) = 160, − 35 kpsi
2
2


(a) From Eq. (5-26),
σ A, σ B =
−1
−1
σ
σ 
 160 −35 
−
n= 1 − 3  =
 = 1.24 Ans.
 S yt S yc 
 235 285 


(b) The plot for this problem is found on the page for Prob. 5-14. Note: The drawing in
this manual may not be to the scale of original drawing. The measurements were taken
from the original drawing.
OF 2.04"
=
= 1.24 Ans.
OE 1.64"
______________________________________________________________________________
n=
2
5-17
−80 − 125
 −80 + 125 
2
± 
 + 50 = −47.7, − 157.3 kpsi
2
2


(a) From Eq. (5-26),
σ A, σ B =
−1
−1
σ
σ 
−157.3 
 0
−
n= 1 − 3  =
 = 1.81 Ans.
S

285 
 235
 yt S yc 
(b) The plot for this problem is found on the page for Prob. 5-14. Note: The drawing in
this manual may not be to the scale of original drawing. The measurements were taken
from the original drawing.
OH 2.99"
=
= 1.82 Ans.
OG 1.64"
______________________________________________________________________________
n=
2
5-18
125 + 80
 125 − 80 
2
σ A, σ B =
± 
 + (−75) = 180.8, 24.2 kpsi
2
2


(a) From Eq. (5-26),
Shigley’s MED, 10th edition
Chapter 5 Solutions, Page 14/52
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−1
−1
σ
σ 
0 
 180.8
−
n= 1 − 3  =
 = 1.30 Ans.
S

 235 285 
 yt S yc 
(b) The plot for this problem is found on the page for Prob. 5-14. Note: The drawing in
this manual may not be to the scale of original drawing. The measurements were taken
from the original drawing.
OJ 2.37"
=
= 1.30 Ans.
OI 1.83"
______________________________________________________________________________
n=
5-19
Sut = 30 kpsi, Suc = 90 kpsi
BCM: Eqs. (5-31), p. 250
MM: see Eqs. (5-32), p. 250
(a) σA = 25 kpsi, σB = 15 kpsi
BCM : Eq. (5-31a), n =
Sut
σA
Sut
MM: Eq. (5-32a), n =
=
=
σA
(b) σA = 15 kpsi, σB = −15 kpsi,
30
= 1.2 Ans.
25
30
= 1.2 Ans.
25
−1
 15 −15 
BCM: Eq. (5-31a), n =  −
 = 1.5 Ans.
 30 90 
MM: σA ≥ 0 ≥ σB, and |σB /σA | ≤ 1, Eq. (5-32a), n =
Sut
σA
=
30
= 2.0 Ans.
15
=
30
= 1.24 Ans.
24.14
2
20
 20 
(c) σ A , σ B =
±   + (−10) 2 = 24.14, − 4.14 kpsi
2
 2 
−1
 24.14 −4.14 
BCM: Eq. (5-31b), n = 
−
 = 1.18 Ans.
90 
 30
MM: σA ≥ 0 ≥ σB, and |σB /σA | ≤ 1, Eq. (5-32a), n =
Sut
σA
2
(d) σ A , σ B =
−15 + 10
 −15 − 10 
2
± 
 + (−15) = 17.03, − 22.03 kpsi
2
2


Shigley’s MED, 10th edition
Chapter 5 Solutions, Page 15/52
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−1
 17.03 −22.03 
BCM: Eq. (5-31b), n = 
−
 = 1.23 Ans.
90 
 30
MM: σA ≥ 0 ≥ σB, and |σB /σA | ≥ 1, Eq. (5-32b),
−1
−1
 ( 90 − 30 )17.03 −22.03 
 ( S − Sut ) σ A σ B 
n =  uc
−
−
 = 1.60 Ans.
 =
Suc Sut
Suc 
90 ( 30 )
90 


2
(e) σ A , σ B =
−20 − 20
 −20 + 20 
2
± 
 + (−15) = −5, − 35 kpsi
2
2


BCM: Eq. (5-31c), n = −
Suc
σB
=−
90
= 2.57 Ans.
−35
90
= 2.57 Ans.
σB
−35
______________________________________________________________________________
MM: Eq. (5-32c), n = −
5-20
Suc
=−
Note: The drawing in this manual may not be to the scale of original drawing. The
measurements were taken from the original drawing.
(a) σA = 25, σB = 15 kpsi
BCM & MM:
OB 1.74"
n=
=
= 1.19 Ans.
OA 1.46"
(b) σA = 15, σB = − 15 kpsi
OC 1.59"
=
= 1.5 Ans.
OD 1.06"
OE 2.12"
MM: n =
=
= 2.0 Ans.
OC 1.06"
BCM: n =
2
20
 20 
(c) σ A , σ B =
±   + (−10) 2
2
 2 
= 24.14, − 4.14 kpsi
OG 1.44"
BCM: n =
=
= 1.18 Ans.
OF 1.22"
OH 1.52"
MM: n =
=
= 1.25 Ans.
OF 1.22"
Shigley’s MED, 10th edition
Chapter 5 Solutions, Page 16/52
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2
−15 + 10
 −15 − 10 
2
(d) σ A , σ B =
± 
 + (−15) = 17.03, − 22.03 kpsi
2
2


OJ 1.72"
BCM: n =
=
= 1.24 Ans.
OI 1.39"
OK 2.24"
MM: n =
=
= 1.61 Ans.
OI 1.39"
2
−20 − 20
 −20 + 20 
2
± 
 + ( −15) = −5, − 35 kpsi
2
2


OM 4.55"
BCM and MM: n =
=
= 2.57 Ans.
OL 1.77"
______________________________________________________________________________
(e) σ A , σ B =
5-21
From Table A-24, Sut = 31 kpsi, Suc = 109 kpsi
BCM: Eqs. (5-31), MM: Eqs. (5-32)
(a) σA = 15, σB = 10 kpsi.
31
= 2.07 Ans.
σ A 15
S
31
MM: Eq. (5-32a), n = ut =
= 2.07 Ans.
σ A 15
(b), (c) The plot is shown below is for Probs. 5-21 to 5-25. Note: The drawing in this
manual may not be to the scale of original drawing. The measurements were taken from
the original drawing.
BCM: Eq. (5-31a), n =
Sut
=
BCM and MM:
n=
OB 1.86"
=
= 2.07 Ans.
OA 0.90"
______________________________________________________________________________
5-22 Sut = 31 kpsi, Suc = 109 kpsi
Shigley’s MED, 10th edition
Chapter 5 Solutions, Page 17/52
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BCM: Eq. (5-31), MM: Eqs. (5-32)
(a) σA = 15, σB = − 50 kpsi, |σB /σA | > 1
−1
 15 −50 
BCM: Eq. (5-31b), n =  −
 = 1.06 Ans.
 31 109 
−1
−1
 (109 − 31)15 −50 
 ( S − Sut ) σ A σ B 
MM: Eq. (3-32b), n =  uc
−
−
 = 1.24 Ans.
 =
Suc Sut
Suc 
109
31
109
(
)



(b), (c) The plot is shown in the solution to Prob. 5-21.
OD 2.78"
=
= 1.07 Ans.
OC 2.61"
OE 3.25"
MM: n =
=
= 1.25 Ans.
OC 2.61"
______________________________________________________________________________
BCM: n =
5-23
From Table A-24, Sut = 31 kpsi, Suc = 109 kpsi
BCM: Eq. (5-31), MM: Eqs. (5-32)
2
σ A, σ B =
15
 15 
±   + (−10) 2 = 20, − 5 kpsi
2
 2
−1
 20 −5 
(a) BCM: Eq. (5-32b), n =  −
 = 1.45 Ans.
 31 109 
S
31
MM: Eq. (5-32a), n = ut =
= 1.55 Ans.
σ A 20
(b), (c) The plot is shown in the solution to Prob. 5-21.
OG 1.48"
=
= 1.44 Ans.
OF 1.03"
OH 1.60"
MM: n =
=
= 1.55 Ans.
OF 1.03"
______________________________________________________________________________
BCM: n =
5-24
From Table A-24, Sut = 31 kpsi, Suc = 109 kpsi
BCM: Eq. (5-31), MM: Eqs. (5-32)
2
σ A, σ B =
−10 − 25
 −10 + 25 
2
± 
 + (−10) = −5, − 30 kpsi
2
2


Shigley’s MED, 10th edition
Chapter 5 Solutions, Page 18/52
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109
= 3.63 Ans.
−30
S
109
MM: Eq. (5-32c), n = − uc = −
= 3.63 Ans.
σB
−30
(b), (c) The plot is shown in the solution to Prob. 5-21.
(a) BCM: Eq. (5-31c), n = −
Suc
σB
−
OJ 5.53"
=
= 3.64 Ans.
OI 1.52"
______________________________________________________________________________
BCM and MM: n =
5-25
From Table A-24, Sut = 31 kpsi, Suc = 109 kpsi
BCM: Eq. (5-31), MM: Eqs. (5-32)
2
σ A, σ B =
−35 + 13
 −35 − 13 
2
± 
 + (−10) = 15, − 37 kpsi
2
2


−1
 15 −37 
(a) BCM: Eq. (5-31b), n =  −
 = 1.21 Ans.
 31 109 
−1
−1
 (109 − 31)15 −37 
 ( S − Sut ) σ A σ B 
MM: Eq. (5-32b), n =  uc
−
−
 = 1.46 Ans.
 =
Suc Sut
Suc 
109
31
109
(
)



(b), (c) The plot is shown in the solution to Prob. 5-21.
OL 2.42"
=
= 1.21 Ans.
OK 2.00"
OM 2.91"
MM: n =
=
= 1.46 Ans.
OK 2.00"
______________________________________________________________________________
BCM: n =
5-26
Sut = 36 kpsi, Suc = 35 kpsi
BCM: Eq. (5-31),
(a) σA = 15, σB = − 10 kpsi.
−1
 15 −10 
BCM: Eq. (5-31b), n =  −
 = 1.42 Ans.
 36 35 
Shigley’s MED, 10th edition
Chapter 5 Solutions, Page 19/52
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(b) The plot is shown below is for Probs. 5-26 to 5-30. Note: The drawing in this manual
may not be to the scale of original drawing. The measurements were taken from the
original drawing.
OB 1.28"
n=
=
= 1.42 Ans.
OA 0.90"
______________________________________________________________________________
5-27
Sut = 36 kpsi, Suc = 35 kpsi
BCM: Eq. (5-31),
(a) σA = 15, σB = − 15 kpsi.
−1
 10 −15 
BCM: Eq. (5-31b), n =  −
 = 1.42 Ans.
 36 35 
(b) The plot is shown in the solution to Prob. 5-26.
OD 1.28"
=
= 1.42 Ans.
OC 0.90"
______________________________________________________________________________
n=
5-28
Sut = 36 kpsi, Suc = 35 kpsi
BCM: Eq. (5-31),
2
12
 12 
(a) σ A , σ B = ±   + (−8)2 = 16, − 4 kpsi
2
 2
−1
 16 −4 
BCM: Eq. (5-31b), n =  −  = 1.79 Ans.
 36 35 
(b) The plot is shown in the solution to Prob. 5-26.
Shigley’s MED, 10th edition
Chapter 5 Solutions, Page 20/52
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OF 1.47"
=
= 1.79 Ans.
OE 0.82"
______________________________________________________________________________
n=
5-29
Sut = 36 kpsi, Suc = 35 kpsi
BCM: Eq. (5-31),
2
(a) σ A , σ B =
−10 − 15
 −10 + 15 
2
± 
 + 10 = −2.2, − 22.8 kpsi
2
2


BCM: Eq. (5-31c), n = −
35
= 1.54 Ans.
−22.8
(b) The plot is shown in the solution to Prob. 5-26.
OH 1.76"
=
= 1.53 Ans.
OG 1.15"
______________________________________________________________________________
n=
5-30
Sut = 36 kpsi, Suc = 35 kpsi
BCM: Eq. (5-31),
2
(a) σ A , σ B =
15 + 8
2
 15 − 8 
± 
 + ( −8 ) = 20.2, 2.8 kpsi
2
 2 
BCM: Eq. (5-31a), n =
36
= 1.78 Ans.
20.2
(b) The plot is shown in the solution to Prob. 5-26.
OJ 1.82"
=
= 1.78 Ans.
OI 1.02"
______________________________________________________________________________
n=
5-31
Sut = 36 kpsi, Suc = 35 kpsi. MM: Use Eq. (5-32). For this problem, MM reduces to the
MNS theory.
S
36
(a) σA = 15, σB = − 10 kpsi. Eq. (5-32a), n = ut =
= 2.4 Ans.
σ A 15
(b) The plot on the next page is for Probs. 5-31 to 5-35. Note: The drawing in this manual
may not be to the scale of original drawing. The measurements were taken from the
original drawing.
Shigley’s MED, 10th edition
Chapter 5 Solutions, Page 21/52
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n=
OB 2.16"
=
= 2.43 Ans.
OA 0.90"
______________________________________________________________________________
Sut = 36 kpsi, Suc = 35 kpsi. MM: Use Eq. (5-32).For this problem, MM reduces to the
MNS theory.
(a) σA = 10, σB = − 15 kpsi. Eq. (3-32b) is not valid and must use Eq, (3-32c),
S
35
n = − uc = −
= 2.33 Ans.
σB
−15
(b) The plot is shown in the solution to Prob. 5-31.
OD 2.10"
n=
=
= 2.33 Ans.
OC 0.90"
______________________________________________________________________________
5-32
5-33
Sut = 36 kpsi, Suc = 35 kpsi. MM: Use Eq. (5-32).For this problem, MM reduces to the
MNS theory.
2
12
 12 
(a) σ A , σ B = ±   + (−8)2 = 16, − 4 kpsi
2
 2
S
36
n = ut =
= 2.25 Ans.
σ A 16
(b) The plot is shown in the solution to Prob. 5-31.
OF 1.86"
=
= 2.27 Ans.
OE 0.82"
______________________________________________________________________________
n=
5-34
Sut = 36 kpsi, Suc = 35 kpsi. MM: Use Eq. (5-32).For this problem, MM reduces to the
MNS theory.
Shigley’s MED, 10th edition
Chapter 5 Solutions, Page 22/52
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2
−10 − 15
 −10 + 15 
2
(a) σ A , σ B =
± 
 + 10 = −2.2, − 22.8 kpsi
2
2


S
35
n = − uc = −
= 1.54 Ans.
σB
−22.8
(b) The plot is shown in the solution to Prob. 5-31.
OH 1.76"
=
= 1.53 Ans.
OG 1.15"
______________________________________________________________________________
n=
5-35
Sut = 36 kpsi, Suc = 35 kpsi. MM: Use Eq. (5-32).For this problem, MM reduces to the
MNS theory.
2
15 + 8
2
 15 − 8 
± 
 + ( −8 ) = 20.2, 2.8 kpsi
2
 2 
S
36
n = ut =
= 1.78 Ans.
σ A 20.2
(b) The plot is shown in the solution to Prob. 5-31.
(a) σ A , σ B =
OJ 1.82"
=
= 1.78 Ans.
OI 1.02"
______________________________________________________________________________
n=
5-36
Given: AISI 1006 CD steel, F = 0.55 kN, P = 4.0 kN, and T = 25 N⋅m. From Table A-20,
Sy =280 MPa. Apply the DE theory to stress elements A and B
4 ( 4 )103
4P
A:
σx = 2 =
= 22.6 (106 ) Pa = 22.6 MPa
2
πd
π ( 0.015 )
3
16 ( 25 )
16T 4 V
4  0.55 (10 ) 
 = 41.9 (106 ) Pa = 41.9 MPa
τ xy = 3 +
=
+ 
2
3
π d 3 A π ( 0.015 ) 3  (π / 4 ) 0.015 


σ ′ =  22.62 + 3 ( 41.92 )
n=
B:
280
= 3.68
76.0
1/2
= 76.0 MPa
Ans.
σx =
3
3
32 Fl 4 P 32 ( 0.55 )10 ( 0.1) 4 ( 4 )10
+
=
+
= 189 (106 ) Pa = 189 MPa
πd3 πd2
π ( 0.0153 )
π ( 0.0152 )
τ xy =
16 ( 25 )
16T
=
= 37.7 (106 ) Pa = 37.7 MPa
3
3
πd
π ( 0.015 )
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Chapter 5 Solutions, Page 23/52
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σ ′ = (σ x2 + 3τ xy2 )
1/2
= 1892 + 3 ( 37.72 ) 
1/2
= 200 MPa
S y 280
=
= 1.4
Ans.
σ ′ 200
______________________________________________________________________________
n=
5-37
From Prob. 3-44, the critical location is at the top of the beam at x = 27 in from the left
end, where there is only a bending stress of σ = − 7 456 psi. Thus, σ′ = 7 456 psi and
(Sy)min = nσ′ = 2(7 456) = 14 912 psi
Choose (Sy)min = 15 kpsi
Ans.
______________________________________________________________________________
5-38
From Table A-20 for 1020 CD steel, Sy = 57 kpsi. From Eq. (3-42), p.116,
T=
63 025H
n
(1)
where n is the shaft speed in rev/min. From Eq. (5-3), for the MSS theory,
τ max =
Sy
2nd
=
16T
πd3
(2)
where nd is the design factor. Substituting Eq. (1) into Eq. (2) and solving for d gives
 32 ( 63 025 ) Hnd 
d =

nπ S y


1/3
(3)
Substituting H = 20 hp, nd = 3, n = 1750 rev/min, and Sy = 57(103) psi results in
 32 ( 63 025 ) 20 ( 3) 
= 0.728 in
d min = 
Ans.
3 
 1750π ( 57 )10 
______________________________________________________________________________
1/3
5-39
From Table A-20, Sy = 54 kpsi. From the solution of Prob. 3-68, in the plane of analysis
σ1 = 16.5 kpsi, σ2 = − 1.19 kpsi, and τmax = 8.84 kpsi
The state of stress is plane stress. Thus, the three-dimensional principal stresses are
σ1 = 16.5 kpsi, σ2 = 0, and σ3 = − 1.19 kpsi
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MSS: From Eq. (5-3),
n=
Sy
σ1 − σ 3
=
54
= 3.05
16.5 − ( −1.19 )
Ans.
Note: Whenever the two principal stresses of a plane stress state are of opposite sign, the
maximum shear stress found in the analysis is the true maximum shear stress. Thus, the
factor of safety could have been found from
Sy
54
= 3.05 Ans.
2τ max 2 ( 8.84 )
DE: The von Mises stress can be found from the principal stresses or from the stresses
found in part (d) of Prob. 3-68. That is,
n=
=
Eqs. (5-13) and (5-19)
S
Sy
54
n= y =
=
1/2
1/2
σ ′ (σ A2 − σ Aσ B + σ B2 )
16.52 − 16.5 ( −1.19 ) + ( −1.19 ) 2 


= 3.15 Ans.
or, Eqs. (5-15) and (5-19) using the results of part (d) of Prob. 3-68
Sy
Sy
54
=
=
σ ′ (σ 2 + 3τ 2 )1/2 15.32 + 3 ( 4.432 )1/2


= 3.15 Ans.
______________________________________________________________________________
n=
5-40
From Table A-20, Sy = 370 MPa. From the solution of Prob. 3-69, in the plane of analysis
σ1 = 275 MPa, σ2 = − 12.1 MPa, and τmax = 144 MPa
The state of stress is plane stress. Thus, the three-dimensional principal stresses are
σ1 = 275 MPa, σ2 = 0, and σ3 = − 12.1 MPa
MSS: From Eq. (5-3), n =
Sy
σ1 − σ 3
=
370
= 1.29
275 − ( −12.1)
Ans.
DE: From Eqs. (5-13) and (5-19)
S
Sy
370
=
n= y =
1/2
1/2
2
2
′
2
σ (σ A − σ Aσ B + σ B )
 275 − 275 ( −12.1) + ( −12.1)2 


= 1.32
Ans.
______________________________________________________________________________
Shigley’s MED, 10th edition
Chapter 5 Solutions, Page 25/52
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5-41
From Table A-20, Sy = 54 kpsi. From the solution of Prob. 3-70, in the plane of analysis
σ1 = 22.6 kpsi, σ2 = − 1.14 kpsi, and τmax = 11.9 kpsi
The state of stress is plane stress. Thus, the three-dimensional principal stresses are
σ1 = 22.6 kpsi, σ2 = 0, and σ3 = − 1.14 kpsi
MSS: From Eq. (5-3), n =
Sy
σ1 − σ 3
=
54
= 2.27
22.6 − ( −1.14 )
Ans.
DE: From Eqs. (5-13) and (5-19)
Sy
Sy
54
n=
=
=
1/2
1/2
σ ′ (σ A2 − σ Aσ B + σ B2 )
 22.62 − 22.6 ( −1.14 ) + ( −1.14 ) 2 


= 2.33 Ans.
______________________________________________________________________________
5-42
From Table A-20, Sy = 370 MPa. From the solution of Prob. 3-71, in the plane of analysis
σ1 = 78.2 MPa, σ2 = − 5.27 MPa, and τmax = 41.7 MPa
The state of stress is plane stress. Thus, the three-dimensional principal stresses are
σ1 = 78.2 MPa, σ2 = 0, and σ3 = − 5.27 MPa
MSS: From Eq. (5-3), n =
Sy
σ1 − σ 3
=
370
= 4.43
78.2 − ( −5.27 )
Ans.
DE: From Eqs. (5-13) and (5-19)
S
Sy
370
n= y =
=
1/2
1/2
σ ′ (σ A2 − σ Aσ B + σ B2 )
78.22 − 78.2 ( −5.27 ) + ( −5.27 ) 2 


= 4.57
Ans.
______________________________________________________________________________
5-43
From Table A-20, Sy = 54 kpsi. From the solution of Prob. 3-72, in the plane of analysis
σ1 = 36.7 kpsi, σ2 = − 1.47 kpsi, and τmax = 19.1 kpsi
The state of stress is plane stress. Thus, the three-dimensional principal stresses are
σ1 = 36.7 kpsi, σ2 = 0, and σ3 = − 1.47 kpsi
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MSS: From Eq. (5-3), n =
Sy
σ1 − σ 3
=
54
= 1.41
36.7 − ( −1.47 )
Ans.
DE: From Eqs. (5-13) and (5-19)
S
Sy
54
n= y =
=
1/2
1/2
2
2
σ ′ (σ A − σ Aσ B + σ B )
36.7 2 − 36.7 ( −1.47 ) + ( −1.47 ) 2 


= 1.44
Ans.
______________________________________________________________________________
5-44
From Table A-20, Sy = 370 MPa. From the solution of Prob. 3-73, in the plane of analysis
σ1 = 376 MPa, σ2 = − 42.4 MPa, and τmax = 209 MPa
The state of stress is plane stress. Thus, the three-dimensional principal stresses are
σ1 = 376 MPa, σ2 = 0, and σ3 = − 42.4 MPa
MSS: From Eq. (5-3), n =
Sy
σ1 − σ 3
=
370
= 0.88
376 − ( −42.4 )
Ans.
DE: From Eqs. (5-13) and (5-19)
S
Sy
370
n= y =
=
1/2
1/2
σ ′ (σ A2 − σ Aσ B + σ B2 )
3762 − 376 ( −42.4 ) + ( −42.4 ) 2 


= 0.93 Ans.
______________________________________________________________________________
5-45
From Table A-20, Sy = 54 kpsi. From the solution of Prob. 3-74, in the plane of analysis
σ1 = 7.19 kpsi, σ2 = − 17.0 kpsi, and τmax = 12.1 kpsi
The state of stress is plane stress. Thus, the three-dimensional principal stresses are
σ1 = 7.19 kpsi, σ2 = 0, and σ3 = − 17.0 kpsi
MSS: From Eq. (5-3), n =
Sy
=
54
= 2.23
7.19 − ( −17.0 )
Ans.
σ1 − σ 3
DE: From Eqs. (5-13) and (5-19)
S
Sy
54
n= y =
=
1/2
1/2
2
2
σ ′ (σ A − σ Aσ B + σ B )
7.192 − 7.19 ( −17.0 ) + ( −17.0 ) 2 


= 2.51 Ans.
______________________________________________________________________________
Shigley’s MED, 10th edition
Chapter 5 Solutions, Page 27/52
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5-46
From Table A-20, Sy = 54 kpsi. From the solution of Prob. 3-76, in the plane of analysis
σ1 = 1.72 kpsi, σ2 = − 35.9 kpsi, and τmax = 18.8 kpsi
The state of stress is plane stress. Thus, the three-dimensional principal stresses are
σ1 = 1.72 kpsi, σ2 = 0, and σ3 = − 35.9 kpsi
MSS: From Eq. (5-3), n =
Sy
σ1 − σ 3
=
54
= 1.44
1.72 − ( −35.9 )
Ans.
DE: From Eqs. (5-13) and (5-19)
Sy
Sy
54
n=
=
=
1/2
1/2
σ ′ (σ A2 − σ Aσ B + σ B2 )
1.722 − 1.72 ( −35.9 ) + ( −35.9 )2 


= 1.47
Ans.
______________________________________________________________________________
5-47
From Table A-20, Sy = 370 MPa. From the solution of Prob. 3-77,
Bending: σB = 68.6 MPa, Torsion: τB = 37.7 MPa
For a plane stress analysis it was found that τmax = 51.0 MPa. With combined bending
and torsion, the plane stress analysis yields the true τmax.
S
370
MSS: From Eq. (5-3), n = y =
= 3.63
Ans.
2τ max 2 ( 51.0 )
DE: From Eqs. (5-15) and (5-19)
Sy
Sy
370
n=
=
=
1/2
1/2
σ ′ (σ B2 + 3τ B2 )
68.62 + 3 ( 37.7 2 ) 


= 3.91 Ans.
______________________________________________________________________________
5-48
From Table A-20, Sy = 54 kpsi. From the solution of Prob. 3-79,
Bending: σC = 3460 psi, Torsion: τC = 882 kpsi
For a plane stress analysis it was found that τmax = 1940 psi. With combined bending and
torsion, the plane stress analysis yields the true τmax.
54 (103 )
S
MSS: From Eq. (5-3), n = y =
= 13.9
Ans.
2τ max 2 (1940 )
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Chapter 5 Solutions, Page 28/52
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DE: From Eqs. (5-15) and (5-19)
54 (103 )
Sy
Sy
n=
=
=
σ ′ (σ C2 + 3τ C2 )1/2 34602 + 3 ( 8822 ) 1/2


= 14.3 Ans.
______________________________________________________________________________
5-49
From Table A-20, Sy = 54 kpsi. From the solution of Prob. 3-80, in the plane of analysis
σ1 = 17.8 kpsi, σ2 = − 1.46 kpsi, and τmax = 9.61 kpsi
The state of stress is plane stress. Thus, the three-dimensional principal stresses are
σ1 = 17.8 kpsi, σ2 = 0, and σ3 = − 1.46 kpsi
MSS: From Eq. (5-3), n =
Sy
σ1 − σ 3
=
54
= 2.80
17.8 − ( −1.46 )
Ans.
DE: From Eqs. (5-13) and (5-19)
S
Sy
54
n= y =
=
1/2
1/2
2
2
σ ′ (σ A − σ Aσ B + σ B )
17.82 − 17.8 ( −1.46 ) + ( −1.46 ) 2 


= 2.91 Ans.
______________________________________________________________________________
5-50
From Table A-20, Sy = 54 kpsi. From the solution of Prob. 3-81, in the plane of analysis
σ1 = 17.5 kpsi, σ2 = − 1.13 kpsi, and τmax = 9.33 kpsi
The state of stress is plane stress. Thus, the three-dimensional principal stresses are
σ1 = 17.5 kpsi, σ2 = 0, and σ3 = − 1.13 kpsi
MSS: From Eq. (5-3), n =
Sy
σ1 − σ 3
=
54
= 2.90
17.5 − ( −1.13)
Ans.
DE: From Eqs. (5-13) and (5-19)
S
Sy
54
=
n= y =
1/2
1/2
2
2
σ ′ (σ A − σ Aσ B + σ B )
17.52 − 17.5 ( −1.13) + ( −1.13) 2 


= 2.98 Ans.
______________________________________________________________________________
Shigley’s MED, 10th edition
Chapter 5 Solutions, Page 29/52
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5-51
From Table A-20, Sy = 54 kpsi. From the solution of Prob. 3-82, in the plane of analysis
σ1 = 21.5 kpsi, σ2 = − 1.20 kpsi, and τmax = 11.4 kpsi
The state of stress is plane stress. Thus, the three-dimensional principal stresses are
σ1 = 21.5 kpsi, σ2 = 0, and σ3 = − 1.20 kpsi
MSS: From Eq. (5-3), n =
Sy
σ1 − σ 3
=
54
= 2.38
21.5 − ( −1.20 )
Ans.
DE: From Eqs. (5-13) and (5-19)
Sy
Sy
54
n=
=
=
1/2
1/2
σ ′ (σ A2 − σ Aσ B + σ B2 )
 21.52 − 21.5 ( −1.20 ) + ( −1.20 ) 2 


= 2.44
Ans.
______________________________________________________________________________
5-52
From Table A-20, Sy = 54 kpsi. From the solution of Prob. 3-83, the concern was failure
due to twisting of the flat bar where it was found that τmax = 14.3 kpsi in the middle of the
longest side of the rectangular cross section. The bar is also in bending, but the bending
stress is zero where τmax exists.
S
54
MSS: From Eq. (5-3), n = y =
= 1.89
Ans.
2τ max 2 (14.3)
DE: From Eqs. (5-15) and (5-19)
S
Sy
54
n= y =
=
= 2.18
Ans.
1/2
1/2
2
′
σ ( 3τ max )
3 (14.32 ) 


______________________________________________________________________________
5-53
From Table A-20, Sy = 54 kpsi. From the solution of Prob. 3-84, in the plane of analysis
σ1 = 34.7 kpsi, σ2 = − 6.7 kpsi, and τmax = 20.7 kpsi
The state of stress is plane stress. Thus, the three-dimensional principal stresses are
σ1 = 34.7 kpsi, σ2 = 0, and σ3 = − 6.7 kpsi
MSS: From Eq. (5-3), n =
Sy
σ1 − σ 3
=
54
= 1.30
34.7 − ( −6.7 )
Shigley’s MED, 10th edition
Ans.
Chapter 5 Solutions, Page 30/52
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DE: From Eqs. (5-13) and (5-19)
Sy
Sy
54
n=
=
=
1/2
1/2
σ ′ (σ A2 − σ Aσ B + σ B2 )
34.7 2 − 34.7 ( −6.7 ) + ( −6.7 ) 2 


= 1.40
Ans.
______________________________________________________________________________
5-54
From Table A-20, Sy = 54 kpsi. From the solution of Prob. 3-85, in the plane of analysis
σ1 = 51.1 kpsi, σ2 = − 4.58 kpsi, and τmax = 27.8 kpsi
The state of stress is plane stress. Thus, the three-dimensional principal stresses are
σ1 = 51.1 kpsi, σ2 = 0, and σ3 = − 4.58 kpsi
MSS: From Eq. (5-3), n =
Sy
σ1 − σ 3
=
54
= 0.97
51.1 − ( −4.58 )
Ans.
DE: From Eqs. (5-13) and (5-19)
Sy
Sy
54
n=
=
=
1/2
1/2
σ ′ (σ A2 − σ Aσ B + σ B2 )
51.12 − 51.1( −4.58 ) + ( −4.58 ) 2 


= 1.01 Ans.
______________________________________________________________________________
5-55
From Table A-20, Sy = 54 kpsi. From the solution of Prob. 3-86, in the plane of analysis
σ1 = 59.7 kpsi, σ2 = − 3.92 kpsi, and τmax = 31.8 kpsi
The state of stress is plane stress. Thus, the three-dimensional principal stresses are
σ1 = 59.7 kpsi, σ2 = 0, and σ3 = − 3.92 kpsi
MSS: From Eq. (5-3), n =
Sy
σ1 − σ 3
=
54
= 0.85
59.7 − ( −3.92 )
Ans.
DE: From Eqs. (5-13) and (5-19)
Sy
Sy
54
n=
=
=
1/2
1/2
2
2
′
2
σ (σ A − σ Aσ B + σ B )
59.7 − 59.7 ( −3.92 ) + ( −3.92 ) 2 


= 0.87
Ans.
______________________________________________________________________________
Shigley’s MED, 10th edition
Chapter 5 Solutions, Page 31/52
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5-56
For Prob. 3-84, from Prob. 5-53 solution, with 1018 CD, DE theory yields, n = 1.40.
From Table A-21, for 4140 Q&T @400°F, Sy = 238 kpsi. From Prob. (3-87) solution
which considered stress concentrations for Prob. 3-84
σ1 = 53.0 kpsi, σ2 = − 8.48 kpsi, and τmax = 30.7 kpsi
DE: From Eqs. (5-13) and (5-19)
S
Sy
238
n= y =
=
1/2
1/2
σ ′ (σ A2 − σ Aσ B + σ B2 )
53.02 − 53.0 ( −8.48 ) + ( −8.48 ) 2 


= 4.12
Ans.
Using the 4140 versus the 1018 CD, the factor of safety increases by a factor of
4.12/1.40 = 2.94. Ans.
______________________________________________________________________________
5-57
Design Decisions Required:
•
•
•
•
Material and condition
Design factor
Failure model
Diameter of pin
Using F = 416 lbf from Ex. 5-3,
σ max =
32M
πd3
1
 32M  3
d =

 πσ max 
Decision 1: Select the same material and condition of Ex. 5-3 (AISI 1035 steel, Sy = 81
kpsi)
Decision 2: Since we prefer the pin to yield, set nd a little larger than 1. Further
explanation will follow.
Decision 3: Use the Distortion Energy static failure theory.
Decision 4: Initially set nd = 1
S
S
σ max = y = y = 81 000 psi
1
nd
1
 32(416)(15)  3
d =
 = 0.922 in
 π (81 000) 
Shigley’s MED, 10th edition
Chapter 5 Solutions, Page 32/52
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Choose preferred size of d = 1.000 in
π (1)3 (81 000)
F=
n=
32(15)
= 530 lbf
530
= 1.27
416
Set design factor to nd = 1.27
Adequacy Assessment:
σ max =
Sy
nd
=
81 000
= 63 800 psi
1.27
1
 32(416)(15)  3
d =
 = 1.00 in (OK)
 π (63 800) 
F=
π (1)3 (81 000)
= 530 lbf
32(15)
530
n=
= 1.27 (OK)
416
______________________________________________________________________________
5-58
From Table A-20, for a thin walled cylinder made of AISI 1020 CD steel, Syt = 57 kpsi,
Sut = 68 kpsi.
Since r/t = 7.5/0.0625 = 120 > 10, the shell can be considered thin-wall. From the
solution of Prob. 3-93 the principal stresses are
σ1 = σ 2 =
pd
p (15)
=
= 60 p,
4t 4(0.0625)
σ3 = − p
From Eq. (5-12)
1
2
1
=
2
σ′ =
1/ 2
(σ 1 − σ 2 ) 2 + (σ 2 − σ 3 ) 2 + (σ 3 − σ 1 ) 2 
1/ 2
(60 p − 60 p ) 2 + (60 p + p ) 2 + (− p − 60 p) 2 
= 61 p
For yield, σ′ = Sy ⇒ 61p = 57 (103) ⇒ p = 934 psi Ans.
For rupture, 61 p = 68 ⇒ p = 1.11 kpsi Ans.
________________________________________________________________________
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Chapter 5 Solutions, Page 33/52
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5-59
For AISI 1020 HR steel, from Tables A-5 and A-20, w = 0.282 lbf/in3, Sy = 30 kpsi, and
ν = 0.292. Then, ρ = w/g = 0.282/386 lbf⋅s2/in. For the problem, ri = 3 in, and ro = 5 in.
Substituting into Eqs. (3-55), p. 129, gives
σt =
9 ( 25 ) 1 + 3 ( 0.292 ) 2 
0.282 2  3 + 0.292  
ω 
r 
 9 + 25 + 2 −
r
386
8
3 + 0.292



225


= 3.006 (10−4 ) ω 2  34 + 2 − 0.5699r 2  = F ( r ) ω 2
r


225


σ r = 3.006 (10−4 ) ω 2  34 − 2 − r 2  = G ( r ) ω 2
r


(1)
(2)
For the distortion-energy theory, the von Mises stress will be
σ ′ = (σ t2 − σ tσ r + σ 22 )
1/2
= ω 2  F 2 (r ) − F (r )G (r ) + G 2 (r ) 
1/2
(3)
Although it was noted that the maximum radial stress occurs at r = (rori )1/2 we are more
interested as to where the von Mises stress is a maximum. One could take the derivative
of Eq. (3) and set it to zero to find where the maximum occurs. However, it is much
easier to plot σ′/ω 2 for 3 ≤ r ≤ 5 in. Plotting Eqs. (1) through (3) results in
70
60
50
σ'/ω2
40
tan
30
radial
20
von Mises
10
0
3
3.5
4
4.5
5
r (in)
It can be seen that there is no maxima, and the greatest value of the von Mises stress is
the tangential stress at r = ri. Substituting r = 3 in into Eq. (1) and setting σ′ = Sy gives




30 (103 )

ω=
 3.006 10−4  34 + 225 − 0.5699 32  
( ) 
( )  

32

1/2
= 1361 rad/s
Shigley’s MED, 10th edition
Chapter 5 Solutions, Page 34/52
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60ω 60(1361)
=
= 13 000 rev/min
Ans.
2π
2π
________________________________________________________________________
n=
5-60
Since r/t = 1.75/0.065 = 26.9 > 10, we can use thin-walled equations. From Eqs. (3-53)
and (3-54), p. 128,
di = 3.5 − 2(0.065) = 3.37 in
p(d i + t )
2t
500(3.37 + 0.065)
σt =
= 13 212 psi = 13.2 kpsi
2(0.065)
pd 500(3.37)
σl = i =
= 6481 psi = 6.48 kpsi
4t
4(0.065)
σ r = − pi = −500 psi = −0.5 kpsi
σt =
These are all principal stresses, thus, from Eq. (5-12),
{
}
1
2
2
2 1/2
(13.2 − 6.48 ) + [6.48 − (−0.5)] + ( −0.5 − 13.2 )
2
= 11.87 kpsi
Sy
46
n=
=
σ ′ 11.87
n = 3.88 Ans.
________________________________________________________________________
σ′ =
5-61
From Table A-20, S y = 320 MPa
With pi = 0, Eqs. (3-49), p. 127, are
 b2 
r2 p  r2 
σ t = − 2o o 2 1 + i 2  = c 1 + 2 
ro − ri  r 
 r 
σr = −
(1)
 b2 
ro2 po  ri 2 
1 − 2  = c 1 − 2 
2
2 
ro − ri  r 
 r 
For the distortion-energy theory, the von Mises stress is
σ ′ = (σ t2 − σ tσ r + σ
)
2 1/2
r
 b 2 2  b 2  b 2   b 2  2 
= c 1 + 2  − 1 + 2 1 − 2  + 1 − 2  
 r   r  r   r  
1/2
1/2

b4 
= c 1 + 3 4 
r 

Shigley’s MED, 10th edition
Chapter 5 Solutions, Page 35/52
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We see that the maximum von Mises stress occurs where r is a minimum at r = ri. Here,
σr = 0 and thus σ′ = − σt . Setting − σt = Sy = 320 MPa at r = 0.1 m in Eq. (1) results in
2ro2 po 2 ( 0.15 ) po
=
= 3.6 po = 320 ⇒ po = 88.9 MPa
Ans.
r = ri
ro2 − ri 2 0.152 − 0.12
________________________________________________________________________
2
−σ t
5-62
=
From Table A-24, Sut = 31 kpsi for grade 30 cast iron. From Table A-5, ν = 0.211 and
w = 0.260 lbf/in3. In Prob. 5-59, it was determined that the maximum stress was the
tangential stress at the inner radius, where the radial stress is zero. Thus at the inner
radius, Eq. (3-55), p. 129, gives
1 + 3v 2  0.260 2  3.211  
1 + 3(0.211) 2 
 3+ v  2
2
2
2
σ t = ρω 2 
ω 
3 
ri  =
  2ro + ri −
 2 (5 ) + 3 −
8
3
+
386
8
3.211
v






( )
= 0.01471ω 2 = 31 103
⇒
1452 rad/ sec
Ans.
n = 60(1452)/(2π ) = 13 870 rev/min
________________________________________________________________________
5-63
From Table A-20, for AISI 1035 CD, Sy = 67 kpsi.
From force and bending-moment equations, the ground reaction forces are found in two
planes as shown.
The maximum bending moment will be at B or C. Check which is larger. In the xy plane,
M B = 223(8) = 1784 lbf ⋅ in and M C = 127(6) = 762 lbf ⋅ in.
In the xz plane, M B = 123(8) = 984 lbf ⋅ in and M C = 328(6) = 1968 lbf ⋅ in.
Shigley’s MED, 10th edition
Chapter 5 Solutions, Page 36/52
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1
M B = [(1784)2 + (984) 2 ] 2 = 2037 lbf ⋅ in
1
M C = [(762) 2 + (1968) 2 ] 2 = 2110 lbf ⋅ in
So point C governs. The torque transmitted between B and C is T = (300 − 50)(4) = 1000
lbf·in. The stresses are
τ xz =
16T 16(1000) 5093
=
= 3 psi
d
πd3
πd3
σx =
32M C 32(2110) 21 492
=
=
psi
d3
πd3
πd3
For combined bending and torsion, the maximum shear stress is found from
1/2
τ max
 σ x  2 2 
=   + τ xz 
 2 

1/ 2
 21.49 2  5.09  2 
= 
+ 3  
3 
 2d   d  
=
11.89
kpsi
d3
Max Shear Stress theory is chosen as a conservative failure theory. From Eq. (5-3)
Sy
11.89
67
=
⇒
d = 0.892 in
Ans.
3
2n
d
2 ( 2)
________________________________________________________________________
τ max =
5-64
=
As in Prob. 5-63, we will assume this to be a statics problem. Since the proportions are
unchanged, the bearing reactions will be the same as in Prob. 5-63 and the bending moment
will still be a maximum at point C. Thus
xy plane: M C = 127(3) = 381 lbf ⋅ in
xz plane: M C = 328(3) = 984 lbf ⋅ in
So
1/ 2
M max =  (381) 2 + (984) 2  = 1055 lbf ⋅ in
32 M C 32(1055) 10 746
10.75
σx =
=
=
psi = 3 kpsi
3
3
3
πd
πd
d
d
Since the torsional stress is unchanged,
τ xz =
5.09
kpsi
d3
For combined bending and torsion, the maximum shear stress is found from
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1/ 2
τ max
 σ  2

=  x  + τ xz2 
 2 

1/ 2
 10.75  2  5.09 2 
= 
+ 3  
3 
 2d   d  
=
7.40
kpsi
d3
Using the MSS theory, as was used in Prob. 5-63, gives
S
7.40
67
τ max = y = 3 =
⇒
d = 0.762 in
Ans.
2n
d
2 ( 2)
________________________________________________________________________
5-65
For AISI 1018 HR, Table A-20 gives Sy = 32 kpsi. Transverse shear stress is a maximum
at the neutral axis, and zero at the outer radius. Bending stress is a maximum at the outer
radius, and zero at the neutral axis.
Model (c): From Prob. 3-40, at outer radius,
σ ′ = σ = 17.8 kpsi
Sy
32
=
= 1.80
σ ′ 17.8
At neutral axis,
n=
σ ′ = 3τ 2 = 3 ( 3.4 ) = 5.89 kpsi
2
Sy
32
=
= 5.43
′
σ 5.89
The bending stress at the outer radius dominates. n = 1.80
n=
Ans.
Model (d): Assume the bending stress at the outer radius will dominate, as in model
(c). From Prob. 3-40,
σ ′ = σ = 25.5 kpsi
n=
Sy
32
=
= 1.25
σ ′ 25.5
Ans.
Model (e): From Prob. 3-40,
σ ′ = σ = 17.8 kpsi
Sy
32
=
= 1.80
Ans.
σ ′ 17.8
Model (d) is the most conservative, thus safest, and requires the least modeling time.
Model (c) is probably the most accurate, but model (e) yields the same results with less
modeling effort.
________________________________________________________________________
n=
5-66
For AISI 1018 HR, from Table A-20, Sy = 32 kpsi. Model (d) yields the largest bending
moment, so designing to it is the most conservative approach. The bending moment is
M = 312.5 lbf⋅in. For this case, the principal stresses are
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σ1 =
32M
, σ2 = σ3 = 0
πd3
Using a conservative yielding failure theory use the MSS theory and Eq. (5-3)
σ1 − σ 3 =
Sy
S
32 M
= y
3
n
πd
⇒
n
1/3
⇒
 32 Mn 
d =
 π S 
y 

 32 ( 312.5 ) 2.5 
11
Thus, d = 
in
Ans.
 = 0.629 in ∴ Use d =
3
16
 π ( 32 )10 
________________________________________________________________________
1/3
5-67
When the ring is set, the hoop tension in the ring is equal to the screw tension.
σt =
ri 2 pi  ro2 
1 + 
ro2 − ri 2  r 2 
The differential hoop tension dF at r for the ring of width w, is dF = wσ t dr . Integration
yields
F=∫
ro
ri
wr 2 p
wσ t dr = 2 i 2i
ro − ri
ro
∫
ro
ri
 ro2 
wri 2 pi 
ro2 
1
dr
r
+
=
−


 = wri pi (1)
2 
ro2 − ri 2 
r r
 r 
i
The screw equation is
Fi =
T
0.2d
(2)
From Eqs. (1) and (2)
pi =
F
T
=
wri 0.2d wri
dFx = fpi ri dθ
Fx = ∫
2π
0
=
fpi wri dθ =
2π f T
0.2d
2π
f Tw
ri ∫ dθ
0.2d wri 0
Ans.
________________________________________________________________________
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5-68 T = 20 N⋅m, Sy = 450 MPa
(a) From Prob. 5-67, T = 0.2 Fi d
T
20
Fi =
=
= 16.7 (103 ) N = 16.7 kN
0.2d 0.2  6 (10−3 ) 


(b) From Prob. 5-67, F =wri pi
pi =
( )
Ans.
16.7 103
F
F
= i =
= 111.3 106 Pa = 111.3 MPa
−
−
3
3
wri wri 12 10  ( 25 / 2 ) 10 



(
)
(
(
( )
)
pi ri 2 + ro2
ri 2 pi  ro2 
σ t = 2 2 1 +  =
ro − ri 
r r =r
ro2 − ri 2
(c)
Ans.
)
i
=
(
111.3 0.01252 + 0.0252
0.025 − 0.0125
2
2
) = 185.5 MPa
Ans.
σ r = − pi = −111.3 MPa
(d)
τ max =
σ1 − σ 3
=
σt −σr
2
2
185.5 − (−111.3)
=
= 148.4 MPa
2
σ ' = (σ A2 − σ Aσ B + σ B2 )1/2
Ans.
= 185.52 − (185.5)( −111.3) + ( −111.3) 2 
= 259.7 MPa
1/2
Ans.
(e) Maximum Shear Stress Theory
n=
Sy
2τ max
=
450
= 1.52
2 (148.4 )
Ans.
Distortion Energy theory
Sy
450
=
= 1.73 Ans.
σ ′ 259.7
________________________________________________________________________
n=
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5-69
The moment about the center caused by
the force F is F re where re is the effective
radius. This is balanced by the moment
about the center caused by the tangential
(hoop) stress. For the ring of width w
ro
Fre = ∫ rσ t w dr
ri
=
re =
w pi ri 2
ro2 − ri 2
∫
ro
ri

ro2 
r
+

 dr
r 

ro 
w pi ri 2  ro2 − ri 2
2
+
r
ln


o
ri 
F ( ro2 − ri 2 )  2
From Prob. 5-67, F = wri pi. Therefore,
re =
ri  ro2 − ri 2
r 
+ ro2 ln o 
2
2 
ro − ri  2
ri 
For the conditions of Prob. 5-67, ri = 12.5 mm and ro = 25 mm
12.5  252 − 12.52
25 
+ 252 ln
Ans.
 = 17.8 mm
2
2 
25 − 12.5 
2
12.5 
________________________________________________________________________
re =
5-70 (a) The nominal radial interference is δnom = (2.002 − 2.001) /2 = 0.0005 in.
From Eq. (3-57), p. 130,
2
2
2
2
Eδ  ( ro − R )( R − ri ) 


p= 3
2R 
ro2 − ri 2


30 (106 ) 0.0005  (1.52 − 12 )(12 − 0.6252 ) 

 = 3072 psi Ans.
=
1.52 − 0.6252
2 (13 )


Inner member: pi = 0, po = p = 3072 psi. At fit surface r =R = 1 in,
Eq. (3-49), p. 127,
 12 + 0.6252 
R 2 + ri 2
σ t = − p 2 2 = −3072  2
= −7 010 psi
2 
R − ri
 1 − 0.625 
σr = − p = − 3072 psi
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Eq. (5-13)
σ ′ = (σ A2 − σ Aσ B + σ A2 )
1/2
1/2
2
= ( −7010 ) − ( −7010 )( −3072 ) + ( −3072 ) 


= 6086 psi
Ans.
Outer member: pi = p = 3072 psi, po = 0. At fit surface r =R = 1 in,
Eq. (3-49), p. 127,
σt = p
 1.52 + 12 
ro2 + R 2
=
3072
 2 2  = 7 987 psi
ro2 − R 2
 1.5 − 1 
σr = − p = − 3072 psi
Eq. (5-13)
σ ′ = (σ A2 − σ Aσ B + σ A2 )
1/2
= 7987 2 − 7987 ( −3072 ) + ( −3072 ) 
1/2
= 9888 psi
Ans.
(b) For a solid inner tube,
30 (106 ) 0.0005  (1.52 − 12 )(12 ) 

 = 4167 psi Ans.
p=
1.52
2 (13 )


Inner member: σt = σr = − p = − 4167 psi
1/2
2
2
σ ′ = ( −4167 ) − ( −4167 )( −4167 ) + ( −4167 ) 


= 4167 psi
Ans.
Outer member: pi = p = 4167 psi, po = 0. At fit surface r =R = 1 in,
Eq. (3-49), p. 127,
σt = p
 1.52 + 12 
ro2 + R 2
=
4167
 2 2  = 10834 psi
ro2 − R 2
 1.5 − 1 
σr = − p = − 4167 psi
Eq. (5-13)
σ ′ = (σ A2 − σ Aσ B + σ A2 )
1/2
= 10 8342 − 10 834 ( −4167 ) + ( −4167 ) 
1/2
= 13 410 psi
Ans.
________________________________________________________________________
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5-71 From Table A-5, E = 207 (103) MPa. The nominal radial interference is δnom = (40 −
39.98) /2 = 0.01 mm.
From Eq. (3-57), p. 130,
p=
2
2
2
2
Eδ  ( ro − R )( R − ri ) 


ro2 − ri 2
2R3 


207 (103 ) 0.01  ( 32.52 − 202 )( 202 − 102 ) 

 = 26.64 MPa Ans.
=
32.52 − 102
2 ( 203 )


Inner member: pi = 0, po = p = 26.64 MPa. At fit surface r =R = 20 mm,
Eq. (3-49), p. 127,
σt = − p
 202 + 102 
R 2 + ri 2
26.64
=
−
= −44.40 MPa
 2
2 
R 2 − ri 2
 20 − 10 
σr = − p = − 26.64 MPa
Eq. (5-13)
σ ′ = (σ A2 − σ Aσ B + σ A2 )
1/2
1/2
= ( −44.40 ) − ( −44.40 )( −26.64 ) + ( −26.64 ) 


2
= 38.71 MPa
Ans.
Outer member: pi = p = 26.64 MPa, po = 0. At fit surface r =R = 20 mm,
Eq. (3-49), p. 127,
σt = p
 32.52 + 202 
ro2 + R 2
26.64
=
= 59.12 MPa

2
2 
ro2 − R 2
 32.5 − 20 
σr = − p = − 26.64 MPa
Eq. (5-13)
σ ′ = (σ A2 − σ Aσ B + σ A2 )
1/2
= 59.122 − 59.12 ( −26.64 ) + ( −26.64 ) 
1/2
= 76.03 MPa
Ans.
________________________________________________________________________
5-72 From Table A-5, E = 207 (103) MPa. The nominal radial interference is δnom = (40.008 −
39.972) /2 = 0.018 mm.
From Eq. (3-57), p. 130,
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2
2
2
2
Eδ  ( ro − R )( R − ri ) 


p=
ro2 − ri 2
2R3 


207 (103 ) 0.018  ( 32.52 − 202 )( 202 − 102 ) 

 = 47.94 MPa Ans.
32.52 − 102
2 ( 203 )


Inner member: pi = 0, po = p = 47.94 MPa. At fit surface r =R = 20 mm,
=
Eq. (3-49), p. 127,
 202 + 102 
R 2 + ri 2
σ t = − p 2 2 = −47.94  2
= −79.90 MPa
2 
R − ri
 20 − 10 
σr = − p = − 47.94 MPa
Eq. (5-13)
σ ′ = (σ A2 − σ Aσ B + σ A2 )
1/2
1/2
2
= ( −79.90 ) − ( −79.90 )( −47.94 ) + ( −47.94 ) 


= 69.66 MPa
Ans.
Outer member: pi = p = 47.94 MPa, po = 0. At fit surface r =R = 20 mm,
Eq. (3-49), p. 127,
σt = p
 32.52 + 202 
ro2 + R 2
47.94
=
= 106.4 MPa

2
2 
ro2 − R 2
 32.5 − 20 
σr = − p = − 47.94 MPa
Eq. (5-13)
σ ′ = (σ A2 − σ Aσ B + σ A2 )
1/2
= 106.42 − 106.4 ( −47.94 ) + ( −47.94 ) 
1/2
= 136.8 MPa
Ans.
________________________________________________________________________
5-73 From Table A-5, for carbon steel, Es = 30 kpsi, and νs = 0.292. While for Eci = 14.5 Mpsi,
and νci = 0.211. For ASTM grade 20 cast iron, from Table A-24, Sut = 22 kpsi.
For midrange values, δ = (2.001 − 2.0002)/2 = 0.0004 in.
Eq. (3-50), p. 127,
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δ
p=
 1 r +R
 1  R 2 + ri 2

ν
R 
+
 +  2 2 −ν i  
o
2
 Ei  R − ri

 Eo  r − R
0.0004
=
= 2613 psi
2
2

 2 +1

 12

1
1
1
+ 0.211 +
− 0.292  
2
6  2
6  2
14.5 (10 )  2 − 1
 30 (10 )  1
 
2
o
2
o
2
At fit surface, with pi = p =2613 psi, and po = 0, from Eq. (3-50), p. 127
σt = p
 22 + 12 
ro2 + R 2
=
2613
 2 2  = 4355 psi
ro2 − R 2
 2 −1 
σr = − p = − 2613 psi
From Modified-Mohr theory, Eq. (5-32a), since σA > 0 > σB and | σB /σA | <1,
22
= 5.05
Ans.
σ A 4.355
________________________________________________________________________
n=
5-74
Sut
=
E = 207 GPa
Eq. (3-57), p. 130, can be written in terms of diameters,
( )
3
Eδ d  (d o2 − D 2 )( D 2 − d i2 )  207 10 (0.062)  (502 − 452 )(452 − 402 ) 
p=

=


3
2D3 
(d o2 − di2 )
(502 − 402 )
2 ( 45 )



p = 15.80 MPa
Outer member: From Eq. (3-50),
Outer radius: (σ t )o =
452 (15.80)
(2) = 134.7 MPa
502 − 452
(σ r ) o = 0
Inner radius: (σ t )i =
452 (15.80)  502
1 +
502 − 452  452

 = 150.5 MPa

(σ r )i = −15.80 MPa
Bending (no slipping):
I = (π /64)(504 − 404) = 181.1 (103) mm4
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At ro :
(σ x ) o = ±
Mc
675(0.05 / 2)
=±
= ±93.2(106 ) Pa = ±93.2 MPa
−9
I
181.1 10
At ri :
(σ x )i = ±
675(0.045 / 2)
= ±83.9 106 Pa = ±83.9 MPa
−9
181.1 10
(
(
)
( )
)
Torsion: J = 2I = 362.2 (103) mm4
At ro :
(τ )
=
Tc 900(0.05 / 2)
=
= 62.1 106 Pa = 62.1 MPa
J 362.2 10−9
At ri :
(τ )
=
900(0.045 / 2)
= 55.9 106 Pa = 55.9 MPa
−9
362.2 10
xy o
xy i
(
(
( )
)
( )
)
Outer radius, is plane stress. Since the tangential stress is positive the von Mises stress
will be a maximum with a negative bending stress. That is,
σ x = −93.2 MPa, σ y = 134.7 MPa, τ xy = 62.1 MPa
σ ′ = (σ x2 − σ xσ y + σ y2 + 3τ xy2 )
1/2
Eq. (5-15)
2
2
= ( −93.2 ) − ( −93.2 )(134.7 ) + 134.7 2 + 3 ( 62.1) 


S y 415
no =
=
= 1.84 Ans.
σ ′ 226
1/2
= 226 MPa
Inner radius, 3D state of stress
From Eq. (5-14) with τyz = τzx = 0 and σx = + 83.9 MPa
1/2
1
(83.9 − 150.5)2 + (150.5 + 15.8) 2 + (−15.8 − 83.9) 2 + 6(55.9) 2  = 174 MPa
σ′ =

2
With σx = − 83.9 MPa
1/2
1
(−83.9 − 150.5) 2 + (150.5 + 15.8) 2 + (−15.8 + 83.9) 2 + 6(55.9) 2  = 230 MPa
σ′ =
2
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S y 415
=
= 1.80 Ans.
σ ′ 230
________________________________________________________________________
5-75 From the solution of Prob. 5-74, p = 15.80 MPa
ni =
Inner member: From Eq. (3-50),
Outer radius: (σ t )o = −
ro2 + ri 2
452 + 402
(15.80) = −134.8 MPa
=
−
p
o
ro2 − ri 2
452 − 402
(σ r )o = − p = −15.80 MPa
Inner radius: (σ t )i
( )
2 452
2ro2
= − 2 2 po = − 2
(15.80) = −150.6 MPa
45 − 402
ro − ri
(σ r )i = 0
Bending (no slipping): I = (π /64)(504 − 404) = 181.1 (103) mm4
Mc
675(0.045 / 2)
= ±83.9(106 ) Pa = ±83.9 MPa
At ro :
(σ x )o = ± = ±
−9
I
181.1 10
(
At ri :
(σ x ) i = ±
)
( )
675(0.040 / 2)
= ±74.5 106 Pa = ±74.5 MPa
−9
181.1 10
(
)
Torsion: J = 2I = 362.2 (103) mm4
At ro :
(τ )
=
Tc 900(0.045 / 2)
=
= 55.9 106 Pa = 55.9 MPa
−9
J
362.2 10
At ri :
(τ )
=
900(0.040 / 2)
= 49.7 106 Pa = 49.7 MPa
−9
362.2 10
xy o
xy i
(
(
)
( )
)
( )
Outer radius, 3D state of stress
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From Eq. (5-14) with τyz = τzx = 0 and σx = + 83.9 MPa
1/2
1
(83.9 + 134.8) 2 + (−134.8 + 15.8) 2 + (−15.8 − 83.9) 2 + 6(55.9) 2  = 213 MPa
σ′ =
2
With σx = − 83.9 MPa
1/2
1
(−83.9 + 134.8) 2 + (−134.8 + 15.8) 2 + (−15.8 + 83.9) 2 + 6(55.9) 2  = 142 MPa
σ′ =

2
S y 415
no =
=
= 1.95 Ans.
σ ′ 213
Inner radius, plane stress. Worst case is when σx is positive
σ x = 74.5 MPa, σ y = −150.6 MPa, τ xy = 49.7 MPa
Eq. (5-15)
1/2
σ ′ = (σ x2 − σ xσ y + σ y2 + 3τ xy2 )
2 1/ 2
=  74.52 − 74.5 ( −150.6 ) + ( −150.6 ) + 3 ( 49.7 )  = 216 MPa


S y 415
ni =
=
= 1.92 Ans.
σ ′ 216
______________________________________________________________________________
2
5-76
For AISI 1040 HR, from Table A-20, Sy = 290 MPa.
From Prob. 3-110, pmax = 65.2 MPa. From Eq. (3-50) at the inner radius R of the outer
member,
r 2 + R2
502 + 252
=
= 108.7 MPa
σ t = p o2
65.2
ro − R 2
502 − 252
σ r = − p = −65.2 MPa
These are principal stresses. From Eq. (5-13)
σ o′ = (σ t2 − σ tσ r + σ r2 )
1/ 2
2
= 108.7 2 − 108.7 ( −65.2 ) + ( −65.2 ) 


1/ 2
= 152.2 MPa
Sy
290
=
= 1.91 Ans.
σ o′ 152.2
________________________________________________________________________
n=
5-77
For AISI 1040 HR, from Table A-20, Sy = 42 kpsi.
From Prob. 3-111, pmax = 9 kpsi. From Eq. (3-50) at the inner radius R of the outer
member,
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ro2 + R 2
22 + 12
9
=
= 15 kpsi
ro2 − R 2
22 − 12
σ r = − p = −9 kpsi
These are principal stresses. From Eq. (5-13)
σt = p
σ o′ = (σ t2 − σ tσ r + σ r2 )
1/2
2
= 152 − 15( −9) + ( −9 ) 


1/2
= 21 kpsi
S y 42
=
= 2 Ans.
σ o′ 21
________________________________________________________________________
n=
5-78
For AISI 1040 HR, from Table A-20, Sy = 290 MPa.
From Prob. 3-111, pmax = 91.6 MPa. From Eq. (3-50) at the inner radius R of the outer
member,
r 2 + R2
502 + 252
σ t = p o2
91.6
=
= 152.7 MPa
ro − R 2
502 − 252
σ r = − p = −91.6 MPa
These are principal stresses. From Eq. (5-13)
σ o′ = (σ t2 − σ tσ r + σ r2 )
1/ 2
2
= 152.7 2 − 152.7( −91.6) + ( −91.6 ) 


1/2
= 213.8 MPa
Sy
290
=
= 1.36 Ans.
σ o′ 213.8
________________________________________________________________________
n=
5-79
For AISI 1040 HR, from Table A-20, Sy = 42 kpsi.
From Prob. 3-111, pmax = 12.94 kpsi. From Eq. (3-50) at the inner radius R of the outer
member,
r 2 + R2
22 + 12
=
= 21.57 kpsi
σ t = p o2
12.94
ro − R 2
22 − 12
σ r = − p = −12.94 kpsi
These are principal stresses. From Eq. (5-13)
σ o′ = (σ t2 − σ tσ r + σ r2 )
1/2
2
=  21.57 2 − 21.57( −12.94) + ( −12.94 ) 


1/2
= 30.20 kpsi
Sy
42
=
= 1.39 Ans.
σ o′ 30.2
________________________________________________________________________
n=
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Chapter 5 Solutions, Page 49/52
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5-80
For AISI 1040 HR, from Table A-20, Sy = 290 MPa.
From Prob. 3-111, pmax = 134 MPa. From Eq. (3-50) at the inner radius R of the outer
member,
r 2 + R2
502 + 252
σ t = p o2
134
=
= 223.3 MPa
502 − 252
ro − R 2
σ r = − p = −134 MPa
These are principal stresses. From Eq. (5-13)
σ o′ = (σ t2 − σ tσ r + σ r2 )
1/2
2
=  223.32 − 223.3( −134) + ( −134 ) 


1/2
= 312.6 MPa
Sy
290
=
= 0.93 Ans.
σ o′ 312.6
________________________________________________________________________
n=
5-81
For AISI 1040 HR, from Table A-20, Sy = 42 kpsi.
From Prob. 3-111, pmax = 19.13 kpsi. From Eq. (3-50) at the inner radius R of the outer
member,
r 2 + R2
22 + 12
=
= 31.88 kpsi
σ t = p o2
19.13
ro − R 2
22 − 12
σ r = − p = −19.13 kpsi
These are principal stresses. From Eq. (5-13)
σ o′ = (σ t2 − σ tσ r + σ r2 )
1/ 2
2
= 31.882 − 31.88( −19.13) + ( −19.13 ) 


1/2
= 44.63 kpsi
Sy
42
=
= 0.94 Ans.
σ o′ 44.63
________________________________________________________________________
n=
5-82
1 K
θ   K
θ
θ
3θ 
σ p =  2 I cos  ±  I sin cos sin 
2
2   2π r
2
2
2 
2π r
2
1/2
2
3θ  
θ
θ
 KI
+
sin cos cos  
2
2
2  
 2π r
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1/2
 θ  2θ

2θ
2 3θ
2θ
2θ
2 3θ 
±
+
cos
sin
cos
sin
sin
cos
cos


 
2 
2
2
2
2
2
2  

θ
θ
θ
θ
θ
KI 
KI
=
cos 1 ± sin 
 cos ± sin cos  =
2
2
2
2
2
2π r 
2π r
=
KI
2π r
Plane stress: The third principal stress is zero and
KI
θ
θ
KI
θ
θ
cos 1 + sin  , σ 2 =
cos 1 − sin  , σ 3 = 0 Ans.
σ1 =
2
2
2
2
2π r
2π r
Plane strain: Equations for σ1 andσ2 are still valid,. However,
KI
θ
cos
Ans.
2
2π r
________________________________________________________________________
σ 3 = ν (σ 1 + σ 2 ) = 2ν
5-83
For θ = 0 and plane strain, the principal stress equations of Prob. 5-82 give
σ1 = σ 2 =
(a) DE: Eq. (5-18)
or,
KI
, σ 3 = 2ν
2π r
KI
= 2νσ 1
2π r
1 
2
2
2 1/2
σ 1 − σ 1 ) + (σ 1 − 2νσ 1 ) + ( 2νσ 1 − σ 1 )  = S y
(

2
σ1 − 2νσ1 = Sy
1
For ν = ,
3

 1 
1 − 2  3   σ 1 = S y
 

(a) MSS: Eq. (5-3) , with n =1
⇒
σ 1 = 3S y
σ1 − σ3 = Sy
ν=
1
3
⇒
⇒
Ans.
σ1 − 2νσ1 = Sy
σ 1 = 3S y
Ans.
2
3
σ 3 = σ1 − S y = 2S y ⇒ σ 3 = σ1
Radius of largest circle
1
2  σ
R = σ1 − σ1  = 1
2
3  6
________________________________________________________________________
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5-84
Given: a = 16 mm, KIc = 80 MPa⋅ m and S y = 950 MPa
(a) Ignoring stress concentration
F = SyA =950(100 − 16)(12) = 958(103) N = 958 kN
Ans.
(b) From Fig. 5-26: h/b = 1, a/b = 16/100 = 0.16, β = 1.3
Eq. (5-37)
K I = βσ π a
80 = 1.3
F
π (16)10−3
100(12)
F = 329.4(103) N = 329.4 kN
Ans.
________________________________________________________________________
5-85
Given: a = 0.5 in, KIc = 72 kpsi⋅ in and Sy = 170 kpsi, Sut = 192 kpsi
ro = 14/2 = 7 in, ri = (14 − 2)/2 = 6 in
0.5
a
=
= 0.5,
ro − ri 7 − 6
Fig. 5-30:
β B 2.4
Eq. (5-37):
K Ic = βσ π a
ri 6
= = 0.857
ro 7
⇒
72 = 2.4σ π ( 0.5 )
⇒
σ = 23.9 kpsi
Eq. (3-50), p. 127, at r = ro = 7 in:
r2 p
62 p
σ t = 2i i 2 ( 2 ) ⇒ 23.9 = 2 i 2 ( 2 ) ⇒ pi = 4.315 kpsi
Ans.
ro − ri
7 −6
________________________________________________________________________
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Chapter 6
6-1
Eq. (2-21):
Eq. (6-8):
Table 6-2:
Eq. (6-19):
Sut = 3.4 H B = 3.4(300) = 1020 MPa
Se′ = 0.5Sut = 0.5(1020) = 510 MPa
a = 1.58, b = −0.085
ka = aSutb = 1.58(1020) −0.085 = 0.877
kb = 1.24d −0.107 = 1.24(10) −0.107 = 0.969
Se = ka kb Se′ = (0.877)(0.969)(510) = 433 MPa Ans.
Eq. (6-18):
______________________________________________________________________________
Eq. (6-20):
6-2
(a) Table A-20:
Eq. (6-8):
Sut = 80 kpsi
Se′ = 0.5(80) = 40 kpsi
Ans.
(b) Table A-20:
Eq. (6-8):
Sut = 90 kpsi
Se′ = 0.5(90) = 45 kpsi
Ans.
(c) Aluminum has no endurance limit. Ans.
(d) Eq. (6-8):
Sut > 200 kpsi, Se′ = 100 kpsi
Ans.
______________________________________________________________________________
6-3
Sut = 120 kpsi, σ rev = 70 kpsi
Fig. 6-18:
f = 0.82
Eq. (6-8):
Se′ = Se = 0.5(120) = 60 kpsi
Eq. (6-14):
a=
Eq. (6-15):
 f Sut 
1
1
 0.82(120) 
b = − log 
 = − log 
 = −0.0716
3
3
60


 Se 
( f Sut ) 2 [ 0.82(120) ]
=
= 161.4 kpsi
60
Se
2
1
σ 
 70  −0.0716
Eq. (6-16):
N =  rev  = 
= 116 700 cycles Ans.

 161.4 
 a 
______________________________________________________________________________
1/ b
6-4
Sut = 1600 MPa, σ rev = 900 MPa
Fig. 6-18:
Sut = 1600 MPa = 232 kpsi. Off the graph, so estimate f = 0.77.
Eq. (6-8):
Sut > 1400 MPa, so Se = 700 MPa
Eq. (6-14):
( f Sut )2 [ 0.77(1600)]
a=
=
= 2168.3 MPa
700
Se
2
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Eq. (6-15):
 f Sut
1
b = − log 
3
 Se

1
 0.77(1600) 
 = − log 
 = −0.081838
3
700



1
σ 
 900  −0.081838
N =  rev  = 
= 46 400 cycles Ans.
Eq. (6-16):

 2168.3 
 a 
______________________________________________________________________________
1/ b
6-5
Sut = 230 kpsi, N = 150 000 cycles
Fig. 6-18, point is off the graph, so estimate:
f = 0.77
Eq. (6-8):
Sut > 200 kpsi, so Se′ = Se = 100 kpsi
Eq. (6-14):
a=
Eq. (6-15):
 f Sut
1
b = − log 
3
 Se
Eq. (6-13):
S f = aN b = 313.6(150 000)−0.08274 = 117.0 kpsi
( f Sut ) 2 [ 0.77(230) ]
=
= 313.6 kpsi
100
Se
2

1
 0.77(230) 
 = − log 
 = −0.08274
3
 100 

Ans.
______________________________________________________________________________
6-6
Sut = 1100 MPa = 160 kpsi
Fig. 6-18:
f = 0.79
Eq. (6-8):
Se′ = Se = 0.5(1100) = 550 MPa
Eq. (6-14):
( f Sut ) 2 [ 0.79(1100) ]
a=
=
= 1373 MPa
Se
550
Eq. (6-15):
 f Sut
1
b = − log 
3
 Se
Eq. (6-13):
S f = aN b = 1373(150 000)−0.06622 = 624 MPa
2

1
 0.79(1100) 
 = − log 
 = −0.06622
3
550



Ans.
______________________________________________________________________________
6-7
Sut = 150 kpsi, S yt = 135 kpsi, N = 500 cycles
Fig. 6-18:
f = 0.798
From Fig. 6-10, we note that below 103 cycles on the S-N diagram constitutes the lowcycle region, in which Eq. (6-17) is applicable.
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Eq. (6-17):
log( 0.798) /3
S f = Sut N ( log f ) /3 = 150 ( 500 ) 
= 122 kpsi
Ans.
The testing should be done at a completely reversed stress of 122 kpsi, which is below
the yield strength, so it is possible. Ans.
______________________________________________________________________________
6-8
The general equation for a line on a log Sf - log N scale is Sf = aNb, which is Eq. (6-13).
By taking the log of both sides, we can get the equation of the line in slope-intercept
form.
log S f = b log N + log a
Substitute the two known points to solve for unknowns a and b. Substituting point (1,
Sut),
log Sut = b log(1) + log a
From which a = Sut . Substituting point (103 , f Sut ) and a = Sut
log f Sut = b log103 + log Sut
From which b = (1/ 3) log f
∴ S f = Sut N (log f )/3 1 ≤ N ≤ 103
______________________________________________________________________________
6-9
Read from graph: (103 , 90 ) and (106 , 50). From S = aN b
log S1 = log a + b log N1
log S2 = log a + b log N 2
From which
log a =
log S1 log N 2 − log S2 log N1
log N 2 / N1
log 90 log106 − log 50 log103
log106 / 103
= 2.2095
=
a = 10log a = 102.2095 = 162.0 kpsi
log 50 / 90
= −0.0851
3
( S f ) ax = 162 N −0.0851 103 ≤ N ≤ 106 in kpsi
b=
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Ans.
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Check:
( S f ) ax  3 = 162(103 )−0.0851 = 90 kpsi
10
( S f ) ax  6 = 162(106 ) −0.0851 = 50 kpsi
10
The end points agree.
______________________________________________________________________________
6-10
d = 1.5 in, Sut = 110 kpsi
Se′ = 0.5(110) = 55 kpsi
Eq. (6-8):
Table 6-2:
a = 2.70, b = − 0.265
Eq. (6-19):
ka = aSut b = 2.70(110) −0.265 = 0.777
Eq. (6-20):
kb = 0.879 d −0.107= 0.879(1.5) −0.107 =0.842
Eq. (6-18):
Se = kakb Se′ = 0.777(0.842)(55) = 36.0 kpsi Ans.
______________________________________________________________________________
6-11
For AISI 4340 as-forged steel,
Eq. (6-8):
Table 6-2:
Eq. (6-19):
Se = 100 kpsi
a = 39.9, b = − 0.995
ka = 39.9(260)−0.995 = 0.158
Eq. (6-20):
 0.75 
kb = 

 0.30 
−0.107
= 0.907
Each of the other modifying factors is unity.
Se = 0.158(0.907)(100) = 14.3 kpsi
Ans.
For AISI 1040:
Se′ = 0.5(113) = 56.5 kpsi
ka = 39.9(113)−0.995 = 0.362
kb = 0.907 (same as 4340)
Each of the other modifying factors is unity
Se = 0.362(0.907)(56.5) = 18.6 kpsi
Ans.
Not only is AISI 1040 steel a contender, it has a superior endurance strength. Ans.
______________________________________________________________________________
6-12
D = 1 in, d = 0.8 in, T = 1800 lbf⋅in, f = 0.9, and from Table A-20 for AISI 1020 CD,
Sut = 68 kpsi, and Sy = 57 kpsi.
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(a) Fig. A-15-15:
r 0.1
D
1
=
= 0.125, =
= 1.25, K ts = 1.40
d 0.8
d 0.8
Get the notch sensitivity either from Fig. 6-21, or from the curve-fit Eqs. (6-34) and
(6-35b). Using the equations,
a = 0.190 − 2.51(10−3 ) ( 68 ) + 1.35 (10−5 ) ( 68) − 2.67 (10−8 )( 683 ) = 0.07335
2
1
qs =
a
r
1+
Eq. (6-32):
=
1
= 0.812
0.07335
1+
0.1
Kfs = 1 + qs (Kts − 1) = 1 + 0.812(1.40 − 1) = 1.32
For a purely reversing torque of T = 1800 lbf⋅in,
τ a = K fs
Tr K fs 16T 1.32(16)(1800)
=
=
= 23 635 psi = 23.6 kpsi
J
πd3
π (0.8)3
Eq. (6-8):
Se′ = 0.5(68) = 34 kpsi
Eq. (6-19):
ka = 2.70(68)−0.265 = 0.883
Eq. (6-20):
kb = 0.879(0.8)−0.107 = 0.900
Eq. (6-26):
kc = 0.59
Eq. (6-18) (labeling for shear):
Sse = 0.883(0.900)(0.59)(34) = 15.9 kpsi
For purely reversing torsion, use Eq. (6-54) for the ultimate strength in shear.
Eq. (6-54):
Ssu = 0.67 Sut = 0.67(68) = 45.6 kpsi
Adjusting the fatigue strength equations for shear,
Eq. (6-14):
Eq. (6-15):
( f Ssu )
a=
S se
2
[0.9(45.6)]
=
15.9
 f S su
1
b = − log 
3
 S se
1
2
= 105.9 kpsi

1
 0.9(45.6) 
 = − log 
 = −0.137 27
3
 15.9 

1
 τ  b  23.3  −0.137 27
Eq. (6-16):
= 61.7 103 cycles Ans.
N = a  =

 a   105.9 
(b) For an operating temperature of 750o F, the temperature modification factor,
from Table 6-4 is kd = 0.90.
( )
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Sse = 0.883(0.900)(0.59)(0.9)(34) = 14.3 kpsi
( f Ssu )
a=
2
[0.9(45.6)]
=
2
= 117.8 kpsi
14.3
S se
 f S su 
1
1
 0.9(45.6) 
b = − log 
 = − log 
 = −0.152 62
3
3
 14.3 
 S se 
1
1
 τ  b  23.3  −0.152 62
= 40.9 103 cycles
N = a  =
Ans.

a
117.8


 
______________________________________________________________________________
6-13
( )
L = 0.6 m, Fa = 2 kN, n = 1.5, N = 104 cycles, Sut = 770 MPa, S y = 420 MPa (Table A-20)
First evaluate the fatigue strength.
Se′ = 0.5(770) = 385 MPa
ka = 57.7(770) −0.718 = 0.488
Since the size is not yet known, assume a
typical value of kb = 0.85 and check later.
All other modifiers are equal to one.
Eq. (6-18):
Se = 0.488(0.85)(385) = 160 MPa
In kpsi, Sut = 770/6.89 = 112 kpsi
Fig. 6-18:
f = 0.83
Eq. (6-14):
a=
Eq. (6-15):
Eq. (6-13):
( f Sut )
Se
2
[0.83(770)]
2
=
160
= 2553 MPa
 f Sut 
1
1
 0.83(770) 
b = − log 
 = − log 
 = −0.2005
3
3
 160 
 Se 
S f = aN b = 2553(104 )−0.2005 = 403 MPa
Now evaluate the stress.
M max = (2000 N)(0.6 m) = 1200 N ⋅ m
σ a = σ max =
Mc M ( b / 2 ) 6 M 6 (1200 ) 7200
=
=
=
= 3 Pa, with b in m.
I
b(b3 ) /12 b3
b3
b
Compare strength to stress and solve for the necessary b.
403 (106 )
S
= 1.5
n= f =
σ a 7200 / b3
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b = 0.0299 m Select b = 30 mm.
Since the size factor was guessed, go back and check it now.
1/ 2
Eq. (6-25):
d e = 0.808 ( hb ) = 0.808b = 0.808 ( 30 ) = 24.24 mm
−0.107
 24.2 
kb = 
= 0.88

 7.62 
Our guess of 0.85 was slightly conservative, so we will accept the result of
Eq. (6-20):
b = 30 mm.
Ans.
Checking yield,
σ max =
7200
10−6 ) = 267 MPa
3 (
0.030
Sy
420
= 1.57
σ max 267
______________________________________________________________________________
ny =
6-14
=
Given: w =2.5 in, t = 3/8 in, d = 0.5 in, nd = 2. From Table A-20, for AISI 1020 CD,
Sut = 68 kpsi and Sy = 57 kpsi.
Eq. (6-8):
Se′ = 0.5(68) = 34 kpsi
Table 6-2:
Eq. (6-21):
Eq. (6-26):
ka = 2.70(68) −0.265 = 0.88
kb = 1 (axial loading)
kc = 0.85
Eq. (6-18):
Se = 0.88(1)(0.85)(34) = 25.4 kpsi
Table A-15-1: d / w = 0.5 / 2.5 = 0.2, K t = 2.5
Get the notch sensitivity either from Fig. 6-20, or from the curve-fit Eqs. (6-34) and
(6-35a). The relatively large radius is off the graph of Fig. 6-20, so we will assume the
curves continue according to the same trend and use the equations to estimate the notch
sensitivity.
a = 0.246 − 3.08 (10−3 ) ( 68 ) + 1.51(10−5 ) ( 68) − 2.67 (10−8 )( 683 ) = 0.09799
2
1
= 0.836
a 1 + 0.09799
1+
0.25
r
K f = 1 + q ( K t − 1) = 1 + 0.836(2.5 − 1) = 2.25
q=
Eq. (6-32):
1
σa = K f
=
Fa
2.25 Fa
=
= 3Fa
A (3 / 8)(2.5 − 0.5)
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Since a finite life was not mentioned, we’ll assume infinite life is desired, so the
completely reversed stress must stay below the endurance limit.
25.4
=2
σ a 3Fa
Fa = 4.23 kips Ans.
______________________________________________________________________________
nf =
6-15
Se
=
Given: D = 2 in, d = 1.8 in, r = 0.1 in, M max = 25 000 lbf ⋅ in, M min = 0.
From Table A-20, for AISI 1095 HR, Sut = 120 kpsi and Sy = 66 kpsi.
Eq. (6-8):
Se′ = 0.5Sut = 0.5 (120 ) = 60 kpsi
Eq. (6-19):
ka = aSutb = 2.70(120) −0.265 = 0.76
Eq. (6-24):
d e = 0.370d = 0.370(1.8) = 0.666 in
Eq. (6-20):
kb = 0.879d e −0.107 = 0.879(0.666) −0.107 = 0.92
Eq. (6-26):
kc = 1
Eq. (6-18):
Se = ka kb kc Se′ = (0.76)(0.92)(1)(60) = 42.0 kpsi
Fig. A-15-14: D / d = 2 /1.8 = 1.11,
r / d = 0.1 /1.8 = 0.056
⇒ K t = 2.1
Get the notch sensitivity either from Fig. 6-20, or from the curve-fit Eqs. (6-34) and
(6-35a). Using the equations,
a = 0.246 − 3.08 (10−3 ) (120 ) + 1.51(10−5 ) (120 ) − 2.67 (10−8 )(1203 ) = 0.04770
2
1
q=
1+
=
a
r
1
= 0.87
0.04770
1+
0.1
K f = 1 + q ( K t − 1) = 1 + 0.87(2.1 − 1) = 1.96
Eq. (6-32):
I = (π / 64)d 4 = (π / 64)(1.8)4 = 0.5153 in 4
Mc 25 000(1.8 / 2)
=
= 43 664 psi = 43.7 kpsi
I
0.5153
=0
σ max =
σ min
Eq. (6-36):
σm = K f
σ max + σ min
2
= (1.96 )
( 43.7 + 0 ) = 42.8 kpsi
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σa = K f
σ max − σ min
2
= (1.96 )
( 43.7 − 0 )
2
= 42.8 kpsi
1 σ a σ m 42.8 42.8
=
+
=
+
nf
Se Sut 42.0 120
Eq. (6-46):
n f = 0.73
Ans.
A factor of safety less than unity indicates a finite life.
Check for yielding. It is not necessary to include the stress concentration for static
yielding of a ductile material.
Sy
66
= 1.51 Ans.
σ max 43.7
______________________________________________________________________________
ny =
6-16
=
From a free-body diagram analysis, the bearing reaction forces are found to be 2.1 kN at
the left bearing and 3.9 kN at the right bearing. The critical location will be at the
shoulder fillet between the 35 mm and the 50 mm diameters, where the bending moment
is large, the diameter is smaller, and the stress concentration exists. The bending moment
at this point is M = 2.1(200) = 420 kN∙mm. With a rotating shaft, the bending stress will
be completely reversed.
σ rev =
Mc 420 (35 / 2)
=
= 0.09978 kN/mm 2 = 99.8 MPa
I
(π / 64)(35) 4
This stress is far below the yield strength of 390 MPa, so yielding is not predicted. Find
the stress concentration factor for the fatigue analysis.
Fig. A-15-9: r/d = 3/35 = 0.086, D/d = 50/35 = 1.43, Kt =1.7
Get the notch sensitivity either from Fig. 6-20, or from the curve-fit Eqs. (6-34) and
(6-35a). Using the equations, with Sut = 470 MPa = 68.2 kpsi and r = 3 mm = 0.118 in,
a = 0.246 − 3.08 (10−3 ) ( 68.2 ) + 1.51(10−5 ) ( 68.2 ) − 2.67 (10−8 ) ( 68.2 ) = 0.09771
2
Eq. (6-32):
1
= 0.78
0.09771
a 1+
1+
0.118
r
K f = 1 + q ( K t − 1) = 1 + 0.78(1.7 − 1) = 1.55
Eq. (6-8):
S e' = 0.5 Sut = 0.5(470) = 235 MPa
Eq. (6-19):
ka = aSutb = 4.51(470) −0.265 = 0.88
Eq. (6-24):
kb = 1.24d −0.107 = 1.24(35) −0.107 = 0.85
q=
1
3
=
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Eq. (6-26):
kc = 1
Eq. (6-18):
S e = k a kb kc S e' = (0.88)(0.85)(1)(235) = 176 MPa
176
= 1.14 Infinite life is predicted.
Ans.
K f σ rev 1.55 ( 99.8 )
______________________________________________________________________________
nf =
6-17
Se
=
From a free-body diagram analysis, the
bearing reaction forces are found to be
RA = 2000 lbf and RB = 1500 lbf. The shearforce and bending-moment diagrams are
shown. The critical location will be at the
shoulder fillet between the 1-5/8 in and the
1-7/8 in diameters, where the bending
moment is large, the diameter is smaller, and
the stress concentration exists.
M = 16 000 – 500 (2.5) = 14 750 lbf ∙ in
With a rotating shaft, the bending stress will
be completely reversed.
Mc 14 750(1.625 / 2)
=
= 35.0 kpsi
I
(π / 64)(1.625) 4
This stress is far below the yield strength of 71 kpsi, so yielding is not predicted.
σ rev =
Fig. A-15-9: r/d = 0.0625/1.625 = 0.04, D/d = 1.875/1.625 = 1.15, Kt =1.95
Get the notch sensitivity either from Fig. 6-20, or from the curve-fit Eqs. (6-34) and
(6-35a). Using the equations,
a = 0.246 − 3.08 (10−3 ) ( 85) + 1.51(10−5 ) ( 85 ) − 2.67 (10−8 ) ( 85 ) = 0.07690
2
1
q=
=
1
= 0.76 .
0.07690
1+
0.0625
Eq. (6-32):
a
r
K f = 1 + q ( K t − 1) = 1 + 0.76(1.95 − 1) = 1.72
Eq. (6-8):
S e' = 0.5Sut = 0.5(85) = 42.5 kpsi
Eq. (6-19):
ka = aSutb = 2.70(85) −0.265 = 0.832
Eq. (6-20):
Eq. (6-26):
kb = 0.879 d −0.107 = 0.879(1.625) −0.107 = 0.835
1+
3
kc = 1
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Eq. (6-18):
S e = ka kb kc S e' = (0.832)(0.835)(1)(42.5) = 29.5 kpsi
29.5
= 0.49
Ans.
K f σ rev 1.72 ( 35.0 )
Infinite life is not predicted. Use the S-N diagram to estimate the life.
nf =
Se
=
Fig. 6-18: f = 0.867
Eq. (6-14):
Eq. (6-15):
( f Sut )
a=
Se
2
[0.867(85)]
=
2
29.5
= 184.1
 f Sut 
1
1
 0.867(85) 
b = − log 
 = − log 
 = −0.1325
3
3
 29.5 
 Se 
1
1
 K f σ rev  b  (1.72)(35.0)  −0.1325
Eq. (6-16):
N =
= 4611 cycles
 =

 a   184.1 
N = 4600 cycles
Ans.
______________________________________________________________________________
6-18
From a free-body diagram analysis, the
bearing reaction forces are found to be RA =
1600 lbf and RB = 2000 lbf. The shear-force
and bending-moment diagrams are shown.
The critical location will be at the shoulder
fillet between the 1-5/8 in and the 1-7/8 in
diameters, where the bending moment is
large, the diameter is smaller, and the stress
concentration exists.
M = 12 800 + 400 (2.5) = 13 800 lbf ∙ in
With a rotating shaft, the bending stress will
be completely reversed.
Mc 13 800(1.625 / 2)
σ rev =
=
= 32.8 kpsi
I
(π / 64)(1.625)4
This stress is far below the yield strength of 71 kpsi, so yielding is not predicted.
Fig. A-15-9: r/d = 0.0625/1.625 = 0.04, D/d = 1.875/1.625 = 1.15, Kt =1.95
Get the notch sensitivity either from Fig. 6-20, or from the curve-fit Eqs. (6-34) and
(6-35a). Usingthe equations,
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a = 0.246 − 3.08 (10−3 ) ( 85 ) + 1.51(10−5 ) ( 85 ) − 2.67 (10−8 ) ( 85 ) = 0.07690
2
1
3
Eq. (6-32):
1
= 0.76
a 1 + 0.07690
1+
0.0625
r
K f = 1 + q ( K t − 1) = 1 + 0.76(1.95 − 1) = 1.72
Eq. (6-8):
S e' = 0.5Sut = 0.5(85) = 42.5 kpsi
Eq. (6-19):
ka = aSutb = 2.70(85) −0.265 = 0.832
Eq. (6-20):
kb = 0.879 d −0.107 = 0.879(1.625) −0.107 = 0.835
Eq. (6-26):
kc = 1
Eq. (6-18):
S e = ka kb kc S e' = (0.832)(0.835)(1)(42.5) = 29.5 kpsi
q=
=
29.5
= 0.52
Ans.
K f σ rev 1.72 ( 32.8 )
Infinite life is not predicted. Use the S-N diagram to estimate the life.
f = 0.867
Fig. 6-18:
Se
nf =
( f Sut )
2
[0.867(85)]
2
Eq. (6-14):
a=
Eq. (6-15):
 f Sut
1
b = − log 
3
 Se
Se
=
=
29.5
1
= 184.1

1
 0.867(85) 
 = − log 
 = −0.1325
3
 29.5 

1
 K f σ rev  b  (1.72)(32.8)  −0.1325
= 7527 cycles
N =
Eq. (6-16):
 =

 a   184.1 
N = 7500 cycles
Ans.
______________________________________________________________________________
6-19
Table A-20: Sut = 120 kpsi, S y = 66 kpsi
N = (950 rev/min)(10 hr)(60 min/hr) = 570 000 cycles
One approach is to guess a diameter and solve the problem as an iterative analysis
problem. Alternatively, we can estimate the few modifying parameters that are dependent
on the diameter and solve the stress equation for the diameter, then iterate to check the
estimates. We’ll use the second approach since it should require only one iteration, since
the estimates on the modifying parameters should be pretty close.
First, we will evaluate the stress. From a free-body diagram analysis, the reaction forces
at the bearings are R1 = 2 kips and R2 = 6 kips. The critical stress location is in the middle
of the span at the shoulder, where the bending moment is high, the shaft diameter is
smaller, and a stress concentration factor exists. If the critical location is not obvious,
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prepare a complete bending moment diagram and evaluate at any potentially critical
locations. Evaluating at the critical shoulder,
M = 2 kip (10 in ) = 20 kip ⋅ in
σ rev =
Mc M ( d / 2 ) 32 M 32 ( 20 ) 203.7
=
=
=
=
kpsi
I
π d 4 / 64 π d 3
πd3
d3
Now we will get the notch sensitivity and stress concentration factor. The notch
sensitivity depends on the fillet radius, which depends on the unknown diameter. For
now, let us estimate a value of q = 0.85 from observation of Fig. 6-20, and check it later.
Fig. A-15-9: D / d = 1.4d / d = 1.4,
Eq. (6-32):
r / d = 0.1d / d = 0.1,
K t = 1.65
K f = 1 + q ( K t − 1) = 1 + 0.85(1.65 − 1) = 1.55
Now, evaluate the fatigue strength.
Se' = 0.5(120) = 60 kpsi
ka = 2.70(120) −0.265 = 0.76
Since the diameter is not yet known, assume a typical value of kb = 0.85 and check later.
All other modifiers are equal to one.
Se = (0.76)(0.85)(60) = 38.8 kpsi
Determine the desired fatigue strength from the S-N diagram.
Fig. 6-18:
f = 0.82
Eq. (6-14):
( f Sut )
a=
2
[0.82(120)]
=
38.8
Se
2
= 249.6
Eq. (6-15):
 f Sut 
1
1
 0.82(120) 
b = − log 
 = − log 
 = −0.1347
3
3
 38.8 
 Se 
Eq. (6-13):
S f = aN b = 249.6(570 000) −0.1347 = 41.9 kpsi
Compare strength to stress and solve for the necessary d.
nf =
Sf
K f σ rev
=
41.9
= 1.6
(1.55 ) ( 203.7 / d 3 )
d = 2.29 in
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Since the size factor and notch sensitivity were guessed, go back and check them now.
Eq. (6-20):
kb = 0.91d −0.157 = 0.91( 2.29 )
−0.157
= 0.80
From Fig. 6-20 with r = d/10 = 0.229 in, we are off the graph, but it appears our guess for
q of 0.85 is low. Assuming the trend of the graph continues, we’ll choose q = 0.91 and
iterate the problem with the new values of kb and q.
Intermediate results are Se = 36.5 kpsi, Sf = 39.6 kpsi, and Kf = 1.59. This gives
nf =
Sf
K f σ rev
=
39.6
= 1.6
(1.59 ) ( 203.7 / d 3 )
d = 2.36 in
Ans.
A quick check of kb and q show that our estimates are still reasonable for this diameter.
______________________________________________________________________________
6-20
S e = 40 kpsi, S y = 60 kpsi, Sut = 80 kpsi, τ m = 15 kpsi, σ a = 25 kpsi, σ m = τ a = 0
Obtain von Mises stresses for the alternating, mid-range, and maximum stresses.
σ a′ = (σ a2 + 3τ a2 )
1/2
σ m′ = (σ m2 + 3τ m2 )
1/2
1/ 2
2
=  252 + 3 ( 0 ) 


= 25.00 kpsi
1/2
2
= 02 + 3 (15 ) 


2
2
′ = (σ max
σ max
+ 3τ max
)
1/2
( )
= 25.98 kpsi
1/2
2
2
= (σ a + σ m ) + 3 (τ a + τ m ) 


1/2
=  252 + 3 152  = 36.06 kpsi
Sy
60
ny =
=
= 1.66 Ans.
′
σ max
36.06
(a) Modified Goodman, Table 6-6
nf =
1
= 1.05
(25.00 / 40) + (25.98 / 80)
Ans.
(b) Gerber, Table 6-7
2
2

 2(25.98)(40)  
1  80   25.00  
nf = 
 = 1.31
 
 −1 + 1 + 
2  25.98   40  
 80(25.00)  


(c) ASME-Elliptic, Table 6-8
Ans.
1
= 1.32 Ans.
(25.00 / 40) + (25.98 / 60) 2
______________________________________________________________________________
nf =
2
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6-21
S e = 40 kpsi, S y = 60 kpsi, Sut = 80 kpsi, τ m = 20 kpsi, σ a = 10 kpsi, σ m = τ a = 0
Obtain von Mises stresses for the alternating, mid-range, and maximum stresses.
σ a′ = (σ a2 + 3τ a2 )
1/ 2
σ m′ = (σ m2 + 3τ m2 )
1/2
1/2
2
= 102 + 3 ( 0 ) 


= 10.00 kpsi
1/2
2
= 02 + 3 ( 20 ) 


2
2
′ = (σ max
σ max
+ 3τ max
)
1/2
( )
= 34.64 kpsi
1/2
2
2
= (σ a + σ m ) + 3 (τ a + τ m ) 


1/2
= 102 + 3 202  = 36.06 kpsi
Sy
60
ny =
=
= 1.66 Ans.
′
σ max
36.06
(a) Modified Goodman, Table 6-6
1
nf =
= 1.46
(10.00 / 40) + (34.64 / 80)
(b) Gerber, Table 6-7
Ans.
2 

2
 2(34.64)(40)  
1  80   10.00  
nf = 
  = 1.74
 
  −1 + 1 + 
2  34.64   40  
 80(10.00)  

Ans.
(c) ASME-Elliptic, Table 6-8
1
= 1.59 Ans.
(10.00 / 40) + (34.64 / 60) 2
______________________________________________________________________________
nf =
6-22
2
S e = 40 kpsi, S y = 60 kpsi, Sut = 80 kpsi, τ a = 10 kpsi, τ m = 15 kpsi, σ a = 12 kpsi, σ m = 0
Obtain von Mises stresses for the alternating, mid-range, and maximum stresses.
σ a′ = (σ a2 + 3τ a2 )
1/2
1/2
2
= 122 + 3 (10 ) 


σ m′ = (σ m2 + 3τ m2 )
1/2
1/ 2
2
=  02 + 3 (15 ) 


2
2
′ = (σ max
σ max
+ 3τ max
)
1/2
= 21.07 kpsi
= 25.98 kpsi
1/2
2
2
= (σ a + σ m ) + 3 (τ a + τ m ) 


2 1/2
= (12 + 0 ) + 3 (10 + 15 )  = 44.93 kpsi


Sy
60
ny =
=
= 1.34 Ans.
′
σ max
44.93
2
(a) Modified Goodman, Table 6-6
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nf =
1
= 1.17
(21.07 / 40) + (25.98 / 80)
Ans.
(b) Gerber, Table 6-7
2

2
 2(25.98)(40)  
1  80   21.07  
nf = 
  = 1.47
 
  −1 + 1 + 
2  25.98   40  
 80(21.07)  

(c) ASME-Elliptic, Table 6-8
Ans.
1
= 1.47 Ans.
(21.07 / 40) + (25.98 / 60) 2
______________________________________________________________________________
nf =
6-23
2
S e = 40 kpsi, S y = 60 kpsi, Sut = 80 kpsi, τ a = 30 kpsi, σ m = σ a = τ a = 0
Obtain von Mises stresses for the alternating, mid-range, and maximum stresses.
σ a′ = (σ a2 + 3τ a2 )
1/ 2
σ m′ = (σ m2 + 3τ m2 )
1/ 2
1/2
2
= 02 + 3 ( 30 ) 


= 51.96 kpsi
= 0 kpsi
2
2
′ = (σ max
+ 3τ max
σ max
)
1/2
1/2
2
2
= (σ a + σ m ) + 3 (τ a + τ m ) 


1/2
2
= 3 ( 30 )  = 51.96 kpsi


Sy
60
ny =
=
= 1.15 Ans.
′
σ max 51.96
(a) Modified Goodman, Table 6-6
nf =
1
= 0.77
(51.96 / 40)
Ans.
(b) Gerber criterion of Table 6-7 is only valid for σm > 0; therefore use Eq. (6-47).
σ′
S
40
nf a = 1 ⇒ nf = e =
= 0.77 Ans.
Se
σ a′ 51.96
(c) ASME-Elliptic, Table 6-8
nf =
1
= 0.77
(51.96 / 40) 2
Ans.
Since infinite life is not predicted, estimate a life from the S-N diagram. Since σ'm = 0,
the stress state is completely reversed and the S-N diagram is applicable for σ'a.
Fig. 6-18: f = 0.875
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Eq. (6-14):
( f Sut ) 2 [ 0.875(80) ]
a=
=
= 122.5
Se
40
Eq. (6-15):
 f Sut
1
b = − log 
3
 Se
2

1
 0.875(80) 
 = − log 
 = −0.08101
3
40



1
σ 
 51.96  −0.08101
N =  rev  = 
Eq. (6-16):
= 39 600 cycles Ans.

 122.5 
 a 
______________________________________________________________________________
1/ b
6-24
S e = 40 kpsi, S y = 60 kpsi, Sut = 80 kpsi, τ a = 15 kpsi, σ m = 15 kpsi, τ m = σ a = 0
Obtain von Mises stresses for the alternating, mid-range, and maximum stresses.
σ a′ = (σ a2 + 3τ a2 )
1/ 2
1/2
2
= 02 + 3 (15 ) 


σ m′ = (σ m2 + 3τ m2 )
= 25.98 kpsi
1/2
2
= 152 + 3 ( 0 ) 


1/2
2
2
′ = (σ max
+ 3τ max
σ max
)
1/2
= 15.00 kpsi
1/2
2
2
= (σ a + σ m ) + 3 (τ a + τ m ) 


2 1/2
= (15 ) + 3 (15 )  = 30.00 kpsi


Sy
60
ny =
=
= 2.00 Ans.
′
σ max
30
2
(a) Modified Goodman, Table 6-6
nf =
1
= 1.19
(25.98 / 40) + (15.00 / 80)
Ans.
(b) Gerber, Table 6-7
2 

2
 2(15.00)(40)  
1  80   25.98  
nf = 
  = 1.43
 
  −1 + 1 + 
2  15.00   40  
 80(25.98)  

(c) ASME-Elliptic, Table 6-8
Ans.
1
= 1.44 Ans.
(25.98 / 40) + (15.00 / 60) 2
______________________________________________________________________________
nf =
6-25
2
Given: Fmax = 28 kN, Fmin = −28 kN . From Table A-20, for AISI 1040 CD,
Sut = 590 MPa, S y = 490 MPa,
Check for yielding
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σ max =
Fmax
28 000
=
= 147.4 N/mm 2 = 147.4 MPa
A
10(25 − 6)
Sy
490
= 3.32 Ans.
σ max 147.4
Determine the fatigue factor of safety based on infinite life
ny =
=
Eq. (6-8):
S e' = 0.5(590) = 295 MPa
Eq. (6-19):
ka = aSutb = 4.51(590) −0.265 = 0.832
Eq. (6-21):
kb = 1 (axial)
Eq. (6-26):
kc = 0.85
Eq. (6-18):
S e = ka kb kc S e' = (0.832)(1)(0.85)(295) = 208.6 MPa
Fig. 6-20:
q = 0.83
Fig. A-15-1: d / w = 0.24, K t = 2.44
K f = 1 + q ( K t − 1) = 1 + 0.83(2.44 − 1) = 2.20
σa = K f
28 000 − ( −28 000 )
Fmax − Fmin
= 2.2
= 324.2 MPa
2A
2(10)(25 − 6)
Fmax + Fmin
=0
2A
Note, since σm = 0, the stress is completely reversing, and
σm = K f
208.6
= 0.64 Ans.
σ a 324.2
Since infinite life is not predicted, estimate the life from the S-N diagram. With σm = 0,
the stress state is completely reversed, and the S-N diagram is applicable for σa.
nf =
Se
=
Sut = 590/6.89 = 85.6 kpsi
Fig. 6-18:
f = 0.87
Eq. (6-14):
a=
Eq. (6-15):
 f Sut
1
b = − log 
3
 Se
( f Sut )2 [ 0.87(590)]
=
= 1263
208.6
Se
2

1
 0.87(590) 
 = − log 
 = −0.1304
3
 208.6 

1
σ 
 324.2  −0.1304
= 33 812 cycles
N =  rev  = 
Eq. (6-16):

 1263 
 a 
N = 34 000 cycles
Ans.
________________________________________________________________________
1/ b
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6-26
Sut = 590 MPa, S y = 490 MPa, Fmax = 28 kN, Fmin = 12 kN
Check for yielding
Fmax
28 000
=
= 147.4 N/mm 2 = 147.4 MPa
A
10(25 − 6)
Sy
490
ny =
=
= 3.32 Ans.
σ max 147.4
σ max =
Determine the fatigue factor of safety based on infinite life.
From Prob. 6-25: S e = 208.6 MPa, K f = 2.2
σa = K f
28 000 − (12 000 )
Fmax − Fmin
= 2.2
= 92.63 MPa
2A
2(10)(25 − 6)
σm = K f
 28 000 + 12 000 
Fmax + Fmin
= 2.2 
 = 231.6 MPa
2A
 2(10)(25 − 6) 
Modified Goodman criteria:
1 σ a σ m 92.63 231.6
=
+
=
+
n f Se Sut 208.6 590
n f = 1.20
Ans.
Gerber criteria:
2
2

 2σ m Se  
1  Sut  σ a 
−1 + 1 + 
nf = 


2  σ m  Se 
 Sutσ a  


2
2

 2(231.6)(208.6)  
1  590  92.63 
= 
−1 + 1 + 


2  231.6  208.6 
 590(92.63)  


n f = 1.49
Ans.
ASME-Elliptic criteria:
nf =
1
1
=
(σ a / Se ) 2 + (σ m / S y ) 2
(92.63 / 208.6)2 + (231.6 / 490) 2
= 1.54 Ans.
The results are consistent with Fig. 6-27, where for a mean stress that is about half of the
yield strength, the Modified Goodman line should predict failure significantly before the
other two.
______________________________________________________________________________
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6-27
Sut = 590 MPa, S y = 490 MPa
(a) Fmax = 28 kN, Fmin = 0 kN
Check for yielding
σ max =
ny =
Fmax
28 000
=
= 147.4 N/mm 2 = 147.4 MPa
A
10(25 − 6)
Sy
σ max
=
490
= 3.32
147.4
Ans.
From Prob. 6-25: S e = 208.6 MPa, K f = 2.2
σa = K f
Fmax − Fmin
28 000 − 0
= 2.2
= 162.1 MPa
2A
2(10)(25 − 6)
 28 000 + 0 
Fmax + Fmin
= 2.2 
 = 162.1 MPa
2A
 2(10)(25 − 6) 
For the modified Goodman criteria,
σm = K f
1 σ a σ m 162.1 162.1
=
+
=
+
n f Se Sut 208.6 590
n f = 0.95 Ans.
Since infinite life is not predicted, estimate a life from the S-N diagram. First, find an
equivalent completely reversed stress. For the modified Goodman criteria, see Ex. 6-12.
σ rev =
Fig. 6-18:
σa
162.1
=
= 223.5 MPa
1 − (σ m / Sut ) 1 − (162.1 / 590)
f = 0.87
Eq. (6-14):
( f Sut )2 [ 0.87(590)]
a=
=
= 1263
Se
208.6
Eq. (6-15):
 f Sut
1
b = − log 
3
 Se
Eq. (6-16):
σ 
N =  rev 
 a 
2
1/ b

1
 0.87(590) 
 = − log 
 = −0.1304
3
 208.6 

1
 223.5  −0.1304
=
= 586 000 cycles

 1263 
Ans.
(b) Fmax = 28 kN, Fmin = 12 kN
The maximum load is the same as in part (a), so
σ max = 147.4 MPa
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n y = 3.32
Ans.
Factor of safety based on infinite life:
Fmax − Fmin
28 000 − 12 000
= 2.2
= 92.63 MPa
2A
2(10)(25 − 6)
σa = K f
 28 000 + 12 000 
Fmax + Fmin
= 2.2 
 = 231.6 MPa
2A
 2(10)(25 − 6) 
1 σ a σ m 92.63 231.6
=
+
=
+
n f Se Sut 208.6 590
σm = K f
n f = 1.20
Ans.
(c) Fmax = 12 kN, Fmin = −28 kN
The compressive load is the largest, so check it for yielding.
σ min =
ny =
Fmin
−28 000
=
= −147.4 MPa
A
10(25 − 6)
S yc
=
σ min
−490
= 3.32
−147.4
Ans.
Factor of safety based on infinite life:
σa = K f
12 000 − ( −28 000 )
Fmax − Fmin
= 2.2
= 231.6 MPa
2A
2(10)(25 − 6)
σm = K f
12 000 + ( −28 000 ) 
Fmax + Fmin
= 2.2 
 = −92.63 MPa
2A
 2(10)(25 − 6) 
208.6
= 0.90 Ans.
σ a 231.6
Since infinite life is not predicted, estimate a life from the S-N diagram. For a negative
mean stress, we shall assume the equivalent completely reversed stress is the same as the
actual alternating stress. Get a and b from part (a).
For σm < 0,
nf =
Se
=
1
σ 
 231.6  −0.1304
Eq. (6-16):
N =  rev  = 
Ans.
= 446 000 cycles

 1263 
 a 
______________________________________________________________________________
1/ b
6-28
Eq. (2-21): Sut = 0.5(400) = 200 kpsi
Eq. (6-8):
S e' = 0.5(200) = 100 kpsi
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Eq. (6-19):
ka = aSutb = 14.4(200) −0.718 = 0.321
Eq. (6-25):
d e = 0.37 d = 0.37(0.375) = 0.1388 in
Eq. (6-20):
kb = 0.879d e −0.107 = 0.879(0.1388) −0.107 = 1.09
Since we have used the equivalent diameter method to get the size factor, and in doing so
introduced greater uncertainties, we will choose not to use a size factor greater than one.
Let kb = 1.
Eq. (6-18):
Se = (0.321)(1)(100) = 32.1 kpsi
40 − 20
40 + 20
Fa =
= 10 lb
Fm =
= 30 lb
2
2
32M a 32(10)(12)
σa =
=
= 23.18 kpsi
π d3
π (0.375)3
32 M m 32(30)(12)
=
= 69.54 kpsi
σm =
πd3
π (0.375)3
(a) Modified Goodman criterion
1 σ a σ m 23.18 69.54
=
+
=
+
nf
Se Sut
32.1
200
n f = 0.94
Ans.
Since infinite life is not predicted, estimate a life from the S-N diagram. First, find an
equivalent completely reversed stress (See Ex. 6-12).
σa
23.18
=
= 35.54 kpsi
1 − (σ m / Sut ) 1 − (69.54 / 200)
f = 0.775
σ rev =
Fig. 6-18:
Eq. (6-14):
( f Sut )2 [ 0.775(200) ]
a=
=
= 748.4
Se
32.1
Eq. (6-15):
 f Sut
1
b = − log 
3
 Se
Eq. (6-16):
σ 
N =  rev 
 a 
2
1/ b

1
 0.775(200) 
 = − log 
 = −0.228
3
32.1 


1
 35.54  −0.228
=
= 637 000 cycles

 748.4 
Ans.
(b) Gerber criterion, Table 6-7
Shigley’s MED, 10th edition
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2
2

 2σ m Se  
1  Sut  σ a 
nf = 
−1 + 1 + 


2  σ m  Se 
 Sutσ a  


2
2

 2(69.54)(32.1)  
1  200  23.18 
= 
−1 + 1 + 


2  69.54  32.1 
200(23.18)  



Ans.
= 1.16 Infinite life is predicted
______________________________________________________________________________
6-29
E = 207.0 GPa
1
(a) I = (20)(43 ) = 106.7 mm 4
12
Fl 3
3EIy
y=
⇒ F= 3
3EI
l
9
3(207)(10 )(106.7)(10−12 )(2)(10−3 )
Fmin =
= 48.3 N
1403 (10−9 )
Fmax =
3(207)(109 )(106.7)(10 −12 )(6)(10 −3 )
= 144.9 N
1403 (10 −9 )
Ans.
Ans.
(b) Get the fatigue strength information.
Sut = =3.4HB = 3.4(490) = 1666 MPa
Eq. (2-21):
From problem statement: Sy = 0.9Sut = 0.9(1666) = 1499 MPa
Eq. (6-8):
Se′ = 700 MPa
Eq. (6-19):
Eq. (6-25):
Eq. (6-20):
Eq. (6-18):
ka = 1.58(1666)-0.085 = 0.84
de = 0.808[20(4)]1/2 = 7.23 mm
kb = 1.24(7.23)-0.107 = 1.00
Se = 0.84(1)(700) = 588 MPa
This is a relatively thick curved beam, so
use the method in Sect. 3-18 to find the
stresses. The maximum bending moment
will be to the centroid of the section as
shown.
M = 142F N∙mm, A = 4(20) = 80 mm2, h =
4 mm, ri = 4 mm, ro = ri + h = 8 mm,
rc = ri + h/2 = 6 mm
Table 3-4:
rn =
h
4
=
= 5.7708 mm
ln( ro / ri ) ln(8 / 4)
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e = rc − rn = 6 − 5.7708 = 0.2292 mm
ci = rn − ri = 5.7708 − 4 = 1.7708 mm
co = ro − rn = 8 − 5.7708 = 2.2292 mm
Get the stresses at the inner and outer surfaces from Eq. (3-65) with the axial stresses
added. The signs have been set to account for tension and compression as appropriate.
σi = −
σo =
Mci F
(142 F )(1.7708) F
− =−
−
= −3.441F MPa
Aeri A
80(0.2292)(4) 80
Mco F (142 F )(2.2292) F
− =
−
= 2.145 F MPa
Aero A
80(0.2292)(8) 80
(σ i ) min = −3.441(144.9) = −498.6 MPa
(σ i ) max = −3.441(48.3) = −166.2 MPa
(σ o ) min = 2.145(48.3) = 103.6 MPa
(σ o ) max = 2.145(144.9) = 310.8 MPa
(σ i ) a =
(σ i )m =
−166.2 − ( −498.6 )
= 166.2 MPa
2
−166.2 + ( −498.6 )
= −332.4 MPa
2
310.8 − 103.6
(σ o ) a =
= 103.6 MPa
2
310.8 + 103.6
(σ o ) m =
= 207.2 MPa
2
To check for yielding, we note that the largest stress is –498.6 MPa (compression) on the
inner radius. This is considerably less than the estimated yield strength of 1499 MPa, so
yielding is not predicted.
Check for fatigue on both inner and outer radii since one has a compressive mean stress
and the other has a tensile mean stress.
Inner radius:
S
588
Since σm < 0, n f = e =
= 3.54
σ a 166.2
Outer radius:
Since σm > 0, using the Modified Goodman line,
σ σ
103.6 207.2
1/ n f = a + m =
+
Se Sut
588 1666
n f = 3.33
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Infinite life is predicted at both inner and outer radii. The outer radius is critical, with a
Ans.
fatigue factor of safety of nf = 3.33.
______________________________________________________________________________
6-30
From Table A-20, for AISI 1018 CD, Sut = 64 kpsi, S y = 54 kpsi
Eq. (6-8):
S e' = 0.5(64) = 32 kpsi
Eq. (6-19):
ka = 2.70(64) −0.265 = 0.897
Eq. (6-20):
kb = 1 (axial)
Eq. (6-26):
kc = 0.85
Eq. (6-18):
Se = (0.897)(1)(0.85)(32) = 24.4 kpsi
Fillet:
Fig. A-15-5: D / d = 3.5 / 3 = 1.17, r / d = 0.25 / 3 = 0.083, K t = 1.85
Use Fig. 6-20 or Eqs. (6-34) and (6-35a) for q. Estimate a little high since it is off the
graph. q = 0.85
K f = 1 + q ( K t − 1) = 1 + 0.85(1.85 − 1) = 1.72
σ max =
Fmax
5
=
= 3.33 kpsi
w2 h 3.0(0.5)
σ min =
−16
= −10.67 kpsi
3.0(0.5)
σa = K f
σ max − σ min
2
 σ max + σ min
2

σm = K f 
ny =
Sy
σ min
=
= 1.72
3.33 − (−10.67)
= 12.0 kpsi
2

 3.33 + (−10.67) 
 = −6.31 kpsi
 = 1.72 
2



54
= 5.06
−10.67
∴ Does not yield.
Since the midrange stress is negative,
nf =
Se
σa
=
24.4
= 2.03
12.0
Hole:
Fig. A-15-1: d / w1 = 0.4 / 3.5 = 0.11 ∴ K t = 2.68
Use Fig. 6-20 or Eqs. (6-34) and (6-35a) for q. Estimate a little high since it is off the
graph, q = 0.85
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K f = 1 + 0.85(2.68 − 1) = 2.43
σ max =
Fmax
5
=
= 3.226 kpsi
h ( w1 − d ) 0.5(3.5 − 0.4)
σ min =
Fmin
−16
=
= −10.32 kpsi
h ( w1 − d ) 0.5(3.5 − 0.4)
σa = K f
σ max − σ min
2
 σ max + σ min
2

σm = K f 
ny =
Sy
σ min
=
= 2.43
3.226 − (−10.32)
= 16.5 kpsi
2

 3.226 + (−10.32) 
 = −8.62 kpsi
 = 2.43 
2



54
= 5.23
−10.32
∴ does not yield
Since the midrange stress is negative,
nf =
Se
σa
=
24.4
= 1.48
16.5
Thus the design is controlled by the threat of fatigue at the hole with a minimum factor of
safety of n f = 1.48. Ans.
______________________________________________________________________________
6-31
Sut = 64 kpsi, S y = 54 kpsi
Eq. (6-8):
S e' = 0.5(64) = 32 kpsi
Eq. (6-19):
Eq. (6-20):
Eq. (6-26):
Eq. (6-18):
ka = 2.70(64) −0.265 = 0.897
kb = 1 (axial)
kc = 0.85
Se = (0.897)(1)(0.85)(32) = 24.4 kpsi
Fillet:
Fig. A-15-5: D / d = 2.5 / 1.5 = 1.67, r / d = 0.25 /1.5 = 0.17, K t ≈ 2.1
Use Fig. 6-20 or Eqs. (6-34) and (6-35a) for q. Estimate a little high since it is off the
graph. q = 0.85
K f = 1 + q ( K t − 1) = 1 + 0.85(2.1 − 1) = 1.94
σ max =
Fmax
16
=
= 21.3 kpsi
w2 h 1.5(0.5)
σ min =
−4
= −5.33 kpsi
1.5(0.5)
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σa = K f
σ max − σ min
2
 σ max + σ min
2

σm = K f 
ny =
Sy
σ max
=
= 1.94
21.3 − (−5.33)
= 25.8 kpsi
2

 21.3 + (−5.33) 
 = 15.5 kpsi
 = 1.94 
2



54
= 2.54
21.3
∴ Does not yield.
Using Modified Goodman criteria,
1 σ a σ m 25.8 15.5
=
+
=
+
nf
Se Sut 24.4 64
n f = 0.77
Hole:
Fig. A-15-1: d / w1 = 0.4 / 2.5 = 0.16 ∴ K t = 2.55
Use Fig. 6-20 or Eqs. (6-34) and (6-35a) for q. Estimate a little high since it is off the
graph. q = 0.85
K f = 1 + 0.85(2.55 − 1) = 2.32
σ max =
Fmax
16
=
= 15.2 kpsi
h ( w1 − d ) 0.5(2.5 − 0.4)
σ min =
−4
Fmin
=
= −3.81 kpsi
h ( w1 − d ) 0.5(2.5 − 0.4)
σ max − σ min
 15.2 − (−3.81) 
= 2.32 
 = 22.1 kpsi
2
2


 σ + σ min 
 15.2 + (−3.81) 
σ m = K f  max
 = 13.2 kpsi
 = 2.32 
2
2




Sy
54
ny =
=
= 3.55 ∴ Does not yield.
σ max 15.2
σa = K f
Using Modified Goodman criteria
1 σ a σ m 22.1 13.2
=
+
=
+
nf
Se Sut 24.4 64
n f = 0.90
Thus the design is controlled by the threat of fatigue at the fillet with a minimum factor
of safety of n f = 0.77
Ans.
______________________________________________________________________________
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6-32
Sut = 64 kpsi, S y = 54 kpsi
From Prob. 6-30, the fatigue factor of safety at the hole is nf = 1.48. To match this at the
fillet,
S
S
24.4
nf = e ⇒ σ a = e =
= 16.5 kpsi
n f 1.48
σa
where Se is unchanged from Prob. 6-30. The only aspect of σa that is affected by the fillet
radius is the fatigue stress concentration factor. Obtaining σa in terms of Kf,
σa = K f
σ max − σ min
2
= Kf
3.33 − ( −10.67)
= 7.00 K f
2
Equating to the desired stress, and solving for Kf,
σ a = 7.00 K f = 16.5
⇒
K f = 2.36
Assume since we are expecting to get a smaller fillet radius than the original, that q will
be back on the graph of Fig. 6-20, so we will estimate q = 0.8.
K f = 1 + 0.80( K t − 1) = 2.36
⇒
K t = 2.7
From Fig. A-15-5, with D / d = 3.5/3 = 1.17 and Kt = 2.6, find r / d. Choosing r / d =
0.03, and with d = w2 = 3.0,
r = 0.03w2 = 0.03 ( 3.0 ) = 0.09 in
At this small radius, our estimate for q is too high. From Fig. 6-20, with r = 0.09, q
should be about 0.75. Iterating, we get Kt = 2.8. This is at a difficult range on Fig. A-155 to read the graph with any confidence, but we’ll estimate r / d = 0.02, giving r = 0.06
in. This is a very rough estimate, but it clearly demonstrates that the fillet radius can be
relatively sharp to match the fatigue factor of safety of the hole.
Ans.
______________________________________________________________________________
6-33
S y = 60 kpsi, Sut = 110 kpsi
Inner fiber where rc = 3 / 4 in
3
3
ro = +
= 0.84375
4 16(2)
3 3
ri = −
= 0.65625
4 32
Table 3-4, p. 135,
rn =
h
3 /16
=
= 0.74608 in
ro
0.84375
ln
ln
0.65625
ri
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e = rc − rn = 0.75 − 0.74608 = 0.00392 in
ci = rn − ri = 0.74608 − 0.65625 = 0.08983
 3  3 
A =     = 0.035156 in 2
 16   16 
Eq. (3-65), p. 133,
σi =
Mci
−T (0.08983)
=
= −993.3T
Aeri (0.035156)(0.00392)(0.65625)
where T is in lbf∙in and σ i is in psi.
1
2
σ a = 496.7T
σ m = (−993.3)T = −496.7T
Eq. (6-8):
Se' = 0.5 (110 ) = 55 kpsi
Eq. (6-19):
ka = 2.70(110) −0.265 = 0.777
Eq. (6-25):
d e = 0.808 ( 3 / 16 )( 3 / 16 ) 
Eq. (6-20):
kb = 0.879 ( 0.1515 )
Eq. (6-19):
Se = (0.777)(1)(55) = 42.7 kpsi
1/ 2
−0.107
= 0.1515 in
= 1.08 (round to 1)
For a compressive midrange component, σ a = S e / n f . Thus,
42.7
3
T = 28.7 lbf ⋅ in
0.4967T =
Outer fiber where rc = 2.5 in
3
= 2.59375
32
3
ri = 2.5 −
= 2.40625
32
3 / 16
rn =
= 2.49883
2.59375
ln
2.40625
e = 2.5 − 2.49883 = 0.00117 in
co = 2.59375 − 2.49883 = 0.09492 in
Mco
T (0.09492)
σo =
=
= 889.7T psi
Aero (0.035156)(0.00117)(2.59375)
ro = 2.5 +
1
2
σ m = σ a = (889.7T ) = 444.9T psi
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(a) Using Eq. (6-46), for modified Goodman, we have
σa σm 1
+
=
Se Sut n
0.4449T 0.4449T 1
+
=
42.7
110
3
T = 23.0 lbf ⋅ in
Ans.
(b) Gerber, Eq. (6-47), at the outer fiber,
2
nσ a  nσ m 
+
 =1
Se  Sut 
2
3(0.4449T )  3(0.4449T ) 
+
 =1
42.7
110


T = 28.2 lbf ⋅ in
Ans.
(c) To guard against yield, use T of part (b) and the inner stress.
Sy
60
= 2.14 Ans.
σ i 0.9933(28.2)
______________________________________________________________________________
ny =
6-34
=
From Prob. 6-33, S e = 42.7 kpsi, S y = 60 kpsi, and Sut = 110 kpsi
(a) Assuming the beam is straight,
σ max =
Goodman:
6T
Mc M ( h / 2 ) 6 M
=
= 2 =
= 910.2T
3
I
bh /12 bh
(3 /16)3
0.4551T 0.4551T 1
+
=
42.7
110
3
T = 22.5 lbf ⋅ in
Ans.
2
(b) Gerber:
3(0.4551T )  3(0.4551T ) 
+
 =1
42.7
110


T = 27.6 lbf ⋅ in Ans.
Sy
60
= 2.39 Ans.
σ max 0.9102(27.6)
______________________________________________________________________________
(c)
ny =
=
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6-35
K f ,bend = 1.4, K f ,axial = 1.1, K f ,tors = 2.0, S y = 300 MPa, Sut = 400 MPa, Se = 200 MPa
Bending:
Axial:
σ m = 0, σ a = 60 MPa
σ m = 20 MPa, σ a = 0
τ m = 25 MPa, τ a = 25 MPa
Torsion:
Eqs. (6-55) and (6-56):
σ a′ =
[1.4(60) + 0]
σ m′ =
[0 + 1.1(20)]
2
2
+ 3[ 2.0(25) ] = 120.6 MPa
2
+ 3[ 2.0(25)] = 89.35 MPa
2
Using Modified Goodman, Eq. (6-46),
1 σ a′ σ m′ 120.6 89.35
=
+
=
+
n f Se Sut
200
400
n f = 1.21
Ans.
′ = σ a′ + σ m′ ,
Check for yielding, using the conservative σ max
Sy
300
= 1.43 Ans.
120.6 + 89.35
______________________________________________________________________________
ny =
6-36
σ a′ + σ m′
=
K f ,bend = 1.4, K f ,tors = 2.0, S y = 300 MPa, Sut = 400 MPa, Se = 200 MPa
Bending: σ max = 150 MPa, σ min = −40 MPa, σ m = 55 MPa, σ a = 95 MPa
Torsion: τ m = 90 MPa, τ a = 9 MPa
Eqs. (6-55) and (6-56):
σ a′ =
[1.4(95)]
σ m′ =
[1.4(55)]
2
2
+ 3[ 2.0(9)] = 136.6 MPa
2
+ 3[ 2.0(90) ] = 321.1 MPa
2
Using Modified Goodman,
1 σ a′ σ m′ 136.6 321.1
=
+
=
+
n f Se Sut
200
400
n f = 0.67
Ans.
′ = σ a′ + σ m′ ,
Check for yielding, using the conservative σ max
ny =
Sy
σ a′ + σ m′
=
300
= 0.66
136.6 + 321.1
Ans.
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Since the conservative yield check indicates yielding, we will check more carefully with
′ obtained directly from the maximum stresses, using the distortion energy failure
σ max
theory, without stress concentrations. Note that this is exactly the method used for static
failure in Ch. 5.
′ =
σ max
(σ max )
+ 3 (τ max ) =
2
2
(150 )
2
+ 3 ( 90 + 9 ) = 227.8 MPa
2
Sy
300
=
= 1.32 Ans.
′
σ max
227.8
Since yielding is not predicted, and infinite life is not predicted, we would like to
estimate a life from the S-N diagram. First, find an equivalent completely reversed stress
(See Ex. 6-12).
ny =
σ a′
136.6
=
= 692.5 MPa
1 − (σ m′ / Sut ) 1 − (321.1 / 400)
This stress is much higher than the ultimate strength, rendering it impractical for the S-N
diagram. We must conclude that the stresses from the combination loading, when
increased by the stress concentration factors, produce such a high midrange stress that the
equivalent completely reversed stress method is not practical to use. Without testing, we
are unable to predict a life.
______________________________________________________________________________
σ rev =
6-37
Table A-20: S ut = 64 kpsi, S y = 54 kpsi
From Prob. 3-68, the critical stress element experiences σ = 15.3 kpsi and τ = 4.43 kpsi.
The bending is completely reversed due to the rotation, and the torsion is steady, giving
σa = 15.3 kpsi, σm = 0 kpsi, τa = 0 kpsi, τm = 4.43 kpsi. Obtain von Mises stresses for the
alternating, mid-range, and maximum stresses.
σ a′ = (σ a2 + 3τ a2 )
1/2
σ m′ = (σ m2 + 3τ m2 )
1/ 2
2
= 15.32 + 3 ( 0 ) 


1/2
2
=  0 2 + 3 ( 4.43) 


2
2
′ = (σ max
σ max
+ 3τ max
)
1/2
= 15.3 kpsi
1/2
= 7.67 kpsi
2
= 15.32 + 3 ( 4.43 ) 


1/ 2
= 17.11 kpsi
Check for yielding, using the distortion energy failure theory.
Sy
54
=
= 3.16
ny =
′
σ max 17.11
Obtain the modifying factors and endurance limit.
Eq. (6-8):
Se′ = 0.5 ( 64 ) = 32 kpsi
Eq. (6-19):
ka = 2.70(64) −0.265 = 0.90
Eq. (6-20):
kb = 0.879(1.25) −0.107 = 0.86
Shigley’s MED, 10th edition
Chapter 6 Solutions, Page 32/58
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Eq. (6-18):
Se = 0.90(0.86)(32) = 24.8 kpsi
Using Modified Goodman,
1 σ a′ σ m′ 15.3 7.67
=
+
=
+
nf
Se Sut 24.8 64
n f = 1.36 Ans.
______________________________________________________________________________
6-38
Table A-20: S ut = 440 MPa, S y = 370 MPa
From Prob. 3-69, the critical stress element experiences σ = 263 MPa and τ = 57.7 MPa.
The bending is completely reversed due to the rotation, and the torsion is steady, giving
σa = 263 MPa, σm = 0, τa = 0 MPa, τm = 57.7 MPa. Obtain von Mises stresses for the
alternating, mid-range, and maximum stresses.
σ a′ = (σ a2 + 3τ a2 )
1/2
σ m′ = (σ m2 + 3τ m2 )
1/ 2
1/2
2
=  2632 + 3 ( 0 ) 


= 263 MPa
1/ 2
2
= 02 + 3 ( 57.7 ) 


2
2
′ = (σ max
σ max
+ 3τ max
)
1/ 2
= 99.9 MPa
1/ 2
2
=  2632 + 3 ( 57.7 ) 


= 281 MPa
Check for yielding, using the distortion energy failure theory.
Sy
370
=
= 1.32
′
281
σ max
Obtain the modifying factors and endurance limit.
ny =
Eq. (6-8):
Se′ = 0.5 ( 440 ) = 220 MPa
Eq. (6-19):
ka = 4.51(440) −0.265 = 0.90
Eq. (6-20):
kb = 1.24(30) −0.107 = 0.86
Eq. (6-18):
Se = 0.90(0.86)(220) = 170 MPa
Using Modified Goodman,
1 σ a′ σ m′ 263 99.9
=
+
=
+
nf
Se Sut 170 440
n f = 0.56
Infinite life is not predicted.
Ans.
______________________________________________________________________________
6-39
Table A-20: S ut = 64 kpsi, S y = 54 kpsi
From Prob. 3-70, the critical stress element experiences σ = 21.5 kpsi and τ = 5.09 kpsi.
The bending is completely reversed due to the rotation, and the torsion is steady, giving
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σa = 21.5 kpsi, σm = 0 kpsi, τa = 0 kpsi, τm = 5.09 kpsi. Obtain von Mises stresses for the
alternating, mid-range, and maximum stresses.
σ a′ = (σ a2 + 3τ a2 )
1/2
σ m′ = (σ m2 + 3τ m2 )
1/2
2
=  21.52 + 3 ( 0 ) 


1/2
2
=  0 2 + 3 ( 5.09 ) 


2
2
′ = (σ max
σ max
+ 3τ max
)
1/ 2
1/ 2
= 21.5 kpsi
= 8.82 kpsi
2
=  21.52 + 3 ( 5.09 ) 


1/2
= 23.24 kpsi
Check for yielding, using the distortion energy failure theory.
ny =
Sy
54
=
= 2.32
′
σ max
23.24
Obtain the modifying factors and endurance limit.
ka = 2.70(64) −0.265 = 0.90
kb = 0.879(1) −0.107 = 0.88
Se = 0.90(0.88)(0.5)(64) = 25.3 kpsi
Using Modified Goodman,
1 σ a′ σ m′ 21.5 8.82
=
+
=
+
n f Se Sut 25.3 64
n f = 1.01 Ans.
______________________________________________________________________________
6-40
Table A-20: S ut = 440 MPa, S y = 370 MPa
From Prob. 3-71, the critical stress element experiences σ = 72.9 MPa and τ = 20.3 MPa.
The bending is completely reversed due to the rotation, and the torsion is steady, giving
σa = 72.9 MPa, σm = 0 MPa, τa = 0 MPa, τm = 20.3 MPa. Obtain von Mises stresses for
the alternating, mid-range, and maximum stresses.
σ a′ = (σ a2 + 3τ a2 )
1/2
σ m′ = (σ m2 + 3τ m2 )
1/ 2
2
=  72.92 + 3 ( 0 ) 


1/ 2
2
=  02 + 3 ( 20.3) 


2
2
′ = (σ max
σ max
+ 3τ max
)
1/2
1/2
= 72.9 MPa
= 35.2 MPa
2
=  72.92 + 3 ( 20.3) 


1/2
= 80.9 MPa
Check for yielding, using the distortion energy failure theory.
Sy
370
=
= 4.57
ny =
′
σ max 80.9
Shigley’s MED, 10th edition
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Obtain the modifying factors and endurance limit.
Eq. (6-8):
Se′ = 0.5 ( 440 ) = 220 MPa
Eq. (6-19):
ka = 4.51(440) −0.265 = 0.90
Eq. (6-20):
kb = 1.24(20) −0.107 = 0.90
Eq. (6-18):
Se = 0.90(0.90)(220) = 178.2 MPa
Using Modified Goodman,
1 σ a′ σ m′
72.9 35.2
=
+
=
+
nf
Se Sut 178.2 440
n f = 2.04 Ans.
______________________________________________________________________________
6-41
Table A-20: S ut = 64 kpsi, S y = 54 kpsi
From Prob. 3-72, the critical stress element experiences σ = 35.2 kpsi and τ = 7.35 kpsi.
The bending is completely reversed due to the rotation, and the torsion is steady, giving
σa = 35.2 kpsi, σm = 0 kpsi, τa = 0 kpsi, τm = 7.35 kpsi. Obtain von Mises stresses for the
alternating, mid-range, and maximum stresses.
σ a′ = (σ a2 + 3τ a2 )
1/2
σ m′ = (σ m2 + 3τ m2 )
1/2
2
= 35.2 2 + 3 ( 0 ) 


1/2
2
=  0 2 + 3 ( 7.35 ) 


2
2
′ = (σ max
+ 3τ max
σ max
)
1/2
1/2
= 35.2 kpsi
= 12.7 kpsi
2
= 35.2 2 + 3 ( 7.35 ) 


1/ 2
= 37.4 kpsi
Check for yielding, using the distortion energy failure theory.
ny =
Sy
54
=
= 1.44
′
σ max 37.4
Obtain the modifying factors and endurance limit.
Eq. (6-8):
Se′ = 0.5(64) = 32 kpsi
Eq. (6-19):
ka = 2.70(64) −0.265 = 0.90
Eq. (6-20):
kb = 0.879(1.25) −0.107 = 0.86
Se = 0.90(0.86)(32) = 24.8 kpsi
Eq. (6-18):
Using Modified Goodman,
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1 σ a′ σ m′ 35.2 12.7
=
+
=
+
nf
Se Sut 24.8 64
n f = 0.62 Infinite life is not predicted.
Ans.
______________________________________________________________________________
6-42
Table A-20: S ut = 440 MPa, S y = 370 MPa
From Prob. 3-73, the critical stress element experiences σ = 333.9 MPa and τ = 126.3
MPa. The bending is completely reversed due to the rotation, and the torsion is steady,
giving σa = 333.9 MPa, σm = 0 MPa, τa = 0 MPa, τm = 126.3 MPa. Obtain von Mises
stresses for the alternating, mid-range, and maximum stresses.
σ a′ = (σ a2 + 3τ a2 )
1/ 2
σ m′ = (σ m2 + 3τ m2 )
1/2
2
= 333.9 2 + 3 ( 0 ) 


1/2
2
=  0 2 + 3 (126.3) 


2
2
′ = (σ max
+ 3τ max
σ max
)
1/2
1/2
= 333.9 MPa
= 218.8 MPa
2
= 333.9 2 + 3 (126.3 ) 


1/ 2
= 399.2 MPa
Check for yielding, using the distortion energy failure theory.
ny =
Sy
370
=
= 0.93
′
σ max
399.2
The sample fails by yielding, infinite life is not predicted.
Ans.
The fatigue analysis will be continued only to obtain the requested fatigue factor of
safety, though the yielding failure will dictate the life.
Obtain the modifying factors and endurance limit.
Eq. (6-8):
Se′ = 0.5(440) = 220 MPa
Eq. (6-19):
ka = 4.51(440) −0.265 = 0.90
Eq. (6-20):
Eq. (6-18):
kb = 1.24(50) −0.107 = 0.82
Se = 0.90(0.82)(220) = 162.4 MPa
Using Modified Goodman,
1 σ a′ σ m′ 333.9 218.8
=
+
=
+
nf
Se Sut 162.4 440
n f = 0.39
Infinite life is not predicted.
Ans.
______________________________________________________________________________
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6-43
Table A-20: Sut = 64 kpsi, S y = 54 kpsi
From Prob. 3-74, the critical stress element experiences completely reversed bending
stress due to the rotation, and steady torsional and axial stresses.
σ a ,bend = 9.495 kpsi,
σ a ,axial = 0 kpsi,
τ a = 0 kpsi,
σ m,bend = 0 kpsi
σ m,axial = −0.362 kpsi
τ m = 11.07 kpsi
Obtain von Mises stresses for the alternating, mid-range, and maximum stresses.
σ a′ = (σ a2 + 3τ a2 )
1/ 2
σ m′ = (σ m2 + 3τ m2 )
1/2
2
2
= ( 9.495 ) + 3 ( 0 ) 


1/ 2
= 9.495 kpsi
2
2
= ( −0.362 ) + 3 (11.07 ) 


2
2
′ = (σ max
+ 3τ max
σ max
)
1/2
1/2
= 19.18 kpsi
2
2
= ( −9.495 − 0.362 ) + 3 (11.07 ) 


1/2
= 21.56 kpsi
Check for yielding, using the distortion energy failure theory.
Sy
54
ny =
=
= 2.50
′
σ max
21.56
Obtain the modifying factors and endurance limit.
Eq. (6-8):
Se′ = 0.5(64) = 32 kpsi
Eq. (6-19):
ka = 2.70(64) −0.265 = 0.90
Eq. (6-20):
Eq. (6-18):
kb = 0.879(1.13) −0.107 = 0.87
Se = 0.90(0.87)(32) = 25.1 kpsi
Using Modified Goodman,
1 σ a′ σ m′ 9.495 19.18
=
+
=
+
nf
Se Sut
25.1
64
n f = 1.47 Ans.
______________________________________________________________________________
6-44
Table A-20: S ut = 64 kpsi, S y = 54 kpsi
From Prob. 3-76, the critical stress element experiences completely reversed bending
stress due to the rotation, and steady torsional and axial stresses.
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σ a ,bend = 33.99 kpsi,
σ a ,axial = 0 kpsi,
τ a = 0 kpsi,
σ m,bend = 0 kpsi
σ m,axial = −0.153 kpsi
τ m = 7.847 kpsi
Obtain von Mises stresses for the alternating, mid-range, and maximum stresses.
σ a′ = (σ a2 + 3τ a2 )
1/ 2
σ m′ = (σ m2 + 3τ m2 )
1/2
2
2
= ( 33.99 ) + 3 ( 0 ) 


1/2
= 33.99 kpsi
2
2
= ( −0.153) + 3 ( 7.847 ) 


2
2
′ = (σ max
+ 3τ max
σ max
)
1/2
1/2
= 13.59 kpsi
2
2
= ( −33.99 − 0.153 ) + 3 ( 7.847 ) 


1/2
= 36.75 kpsi
Check for yielding, using the distortion energy failure theory.
ny =
Sy
54
=
= 1.47
′
σ max
36.75
Obtain the modifying factors and endurance limit.
Eq. (6-8):
Se′ = 0.5(64) = 32 kpsi
Eq. (6-19):
ka = 2.70(64) −0.265 = 0.90
Eq. (6-20):
kb = 0.879(0.88) −0.107 = 0.89
Eq. (6-18):
Se = 0.90(0.89)(32) = 25.6 kpsi
Using Modified Goodman,
1 σ a′ σ m′ 33.99 13.59
=
+
=
+
nf
Se Sut
25.6
64
n f = 0.65
Infinite life is not predicted.
Ans.
______________________________________________________________________________
6-45
Table A-20: S ut = 440 MPa, S y = 370 MPa
From Prob. 3-77, the critical stress element experiences σ = 68.6 MPa and τ = 37.7 MPa.
The bending is completely reversed due to the rotation, and the torsion is steady, giving
σa = 68.6 MPa, σm = 0 MPa, τa = 0 MPa, τm = 37.7 MPa. Obtain von Mises stresses for
the alternating, mid-range, and maximum stresses.
σ a′ = (σ a2 + 3τ a2 )
1/2
σ m′ = (σ m2 + 3τ m2 )
1/2
2
=  68.6 2 + 3 ( 0 ) 


1/ 2
2
=  02 + 3 ( 37.7 ) 


2
2
′ = (σ max
+ 3τ max
σ max
)
1/2
1/2
= 68.6 MPa
= 65.3 MPa
2
=  68.62 + 3 ( 37.7 ) 


Shigley’s MED, 10th edition
1/2
= 94.7 MPa
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Check for yielding, using the distortion energy failure theory.
ny =
Sy
370
=
= 3.91
′
σ max
94.7
Obtain the modifying factors and endurance limit.
Eq. (6-8):
Se′ = 0.5(440) = 220 MPa
Eq. (6-19):
ka = 4.51(440) −0.265 = 0.90
Eq. (6-20):
kb = 1.24(30) −0.107 = 0.86
Eq. (6-18):
Se = 0.90(0.86)(220) = 170 MPa
Using Modified Goodman,
1 σ a′ σ m′ 68.6 65.3
=
+
=
+
nf
Se Sut 170 440
n f = 1.81 Ans.
______________________________________________________________________________
6-46
Table A-20: Sut = 64 kpsi, S y = 54 kpsi
From Prob. 3-79, the critical stress element experiences σ = 3.46 kpsi and τ = 0.882 kpsi.
The bending is completely reversed due to the rotation, and the torsion is steady, giving
σa = 3.46 kpsi, σm = 0, τa = 0 kpsi, τm = 0.882 kpsi. Obtain von Mises stresses for the
alternating, mid-range, and maximum stresses.
σ a′ = (σ a2 + 3τ a2 )
1/ 2
σ m′ = (σ m2 + 3τ m2 )
1/ 2
1/2
2
= 3.462 + 3 ( 0 ) 


2
= 02 + 3 ( 0.882 ) 


2
2
′ = (σ max
σ max
+ 3τ max
)
1/ 2
= 3.46 kpsi
1/2
= 1.53 kpsi
1/ 2
2
= 3.46 2 + 3 ( 0.882 ) 


= 3.78 kpsi
Check for yielding, using the distortion energy failure theory.
ny =
Sy
54
=
= 14.3
′
σ max
3.78
Obtain the modifying factors and endurance limit.
Eq. (6-8):
Se′ = 0.5(64) = 32 kpsi
Eq. (6-19):
ka = 2.70(64) −0.265 = 0.90
Eq. (6-20):
kb = 0.879(1.375) −0.107 = 0.85
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Eq. (6-18):
Se = 0.90(0.85)(32) = 24.5 kpsi
Using Modified Goodman,
1 σ a′ σ m′ 3.46 1.53
=
+
=
+
nf
Se Sut 24.5 64
n f = 6.06
Ans.
______________________________________________________________________________
6-47
Table A-20: Sut = 64 kpsi, S y = 54 kpsi
From Prob. 3-80, the critical stress element experiences σ = 16.3 kpsi and τ = 5.09 kpsi.
Since the load is applied and released repeatedly, this gives σmax = 16.3 kpsi, σmin = 0
kpsi, τmax = 5.09 kpsi, τmin = 0 kpsi. Consequently, σm = σa = 8.15 kpsi, τm = τa = 2.55
kpsi.
For bending, from Eqs. (6-34) and (6-35a),
a = 0.246 − 3.08 (10−3 ) ( 64 ) + 1.51(10−5 ) ( 64 ) − 2.67 (10−8 ) ( 64 ) = 0.10373
2
1
1
= 0.75
a 1 + 0.10373
1+
0.1
r
K f = 1 + q ( K t − 1) = 1 + 0.75(1.5 − 1) = 1.38
q=
Eq. (6-32):
3
=
For torsion, from Eqs. (6-34) and (6-35b),
a = 0.190 − 2.51(10−3 ) ( 64 ) + 1.35 (10−5 ) ( 64 ) − 2.67 (10−8 ) ( 64 ) = 0.07800
2
1
1
= 0.80
0.07800
a 1+
1+
0.1
r
K fs = 1 + qs ( K ts − 1) = 1 + 0.80(2.1 − 1) = 1.88
q=
Eq. (6-32):
3
=
Obtain von Mises stresses for the alternating and mid-range stresses from Eqs. (6-55) and
(6-56).
{
σ a′ = (1.38 )( 8.15 )  + 3 (1.88 )( 2.55 ) 
2
}
2 1/2
= 13.98 kpsi
σ m′ = σ a′ = 13.98 kpsi
′ = σ a′ + σ m′ ,
Check for yielding, using the conservative σ max
Sy
54
=
= 1.93
ny =
σ a′ + σ m′ 13.98 + 13.98
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Obtain the modifying factors and endurance limit.
Eq. (6-8):
Se′ = 0.5(64) = 32 kpsi
Eq. (6-19):
ka = aSutb = 2.70(64) −0.265 = 0.90
Eq. (6-24):
de = 0.370d = 0.370 (1) = 0.370 in
Eq. (6-20):
kb = 0.879d e −0.107 = 0.879(0.370) −0.107 = 0.98
Eq. (6-18):
Se = (0.90)(0.98)(32) = 28.2 kpsi
Using Modified Goodman,
1 σ a′ σ m′ 13.98 13.98
=
+
=
+
nf
Se Sut
28.2
64
n f = 1.40
Ans.
______________________________________________________________________________
6-48
Table A-20: Sut = 64 kpsi, S y = 54 kpsi
From Prob. 3-81, the critical stress element experiences σ = 16.4 kpsi and τ = 4.46 kpsi.
Since the load is applied and released repeatedly, this gives σmax = 16.4 kpsi, σmin = 0
kpsi, τmax = 4.46 kpsi, τmin = 0 kpsi. Consequently, σm = σa = 8.20 kpsi, τm = τa = 2.23
kpsi.
For bending, from Eqs. (6-34) and (6-35a),
a = 0.246 − 3.08 (10−3 ) ( 64 ) + 1.51(10−5 ) ( 64 ) − 2.67 (10−8 ) ( 64 ) = 0.10373
2
1
= 0.75
0.10373
a 1+
1+
0.1
r
K f = 1 + q ( K t − 1) = 1 + 0.75(1.5 − 1) = 1.38
q=
Eq. (6-32):
1
3
=
For torsion, from Eqs. (6-34) and (6-35b),
a = 0.190 − 2.51(10−3 ) ( 64 ) + 1.35 (10−5 ) ( 64 ) − 2.67 (10−8 ) ( 64 ) = 0.07800
2
1
= 0.80
a 1 + 0.07800
1+
0.1
r
K fs = 1 + qs ( K ts − 1) = 1 + 0.80(2.1 − 1) = 1.88
q=
Eq. (6-32):
1
3
=
Obtain von Mises stresses for the alternating and mid-range stresses from Eqs. (6-55) and
(6-56).
Shigley’s MED, 10th edition
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{
σ a′ = (1.38 )( 8.20 )  + 3 (1.88 )( 2.23) 
2
}
2 1/2
= 13.45 kpsi
σ m′ = σ a′ = 13.45 kpsi
′ = σ a′ + σ m′ ,
Check for yielding, using the conservative σ max
Sy
54
ny =
=
= 2.01
σ a′ + σ m′ 13.45 + 13.45
Obtain the modifying factors and endurance limit.
Eq. (6-8):
Se′ = 0.5(64) = 32 kpsi
Eq. (6-19):
Eq. (6-24):
ka = aSutb = 2.70(64) −0.265 = 0.90
Eq. (6-20):
Eq. (6-18):
kb = 0.879d e −0.107 = 0.879(0.370) −0.107 = 0.98
d e = 0.370d = 0.370(1) = 0.370 in
Se = (0.90)(0.98)(32) = 28.2 kpsi
Using Modified Goodman,
1 σ a′ σ m′ 13.45 13.45
=
+
=
+
nf
S e Sut
28.2
64
n f = 1.46
Ans.
______________________________________________________________________________
6-49
Table A-20: Sut = 64 kpsi, S y = 54 kpsi
From Prob. 3-82, the critical stress element experiences repeatedly applied bending,
axial, and torsional stresses of σx,bend = 20.2 kpsi, σx,axial = 0.1 kpsi, and τ = 5.09 kpsi..
Since the axial stress is practically negligible compared to the bending stress, we will
simply combine the two and not treat the axial stress separately for stress concentration
factor and load factor. This gives σmax = 20.3 kpsi, σmin = 0 kpsi, τmax = 5.09 kpsi, τmin =
0 kpsi. Consequently, σm = σa = 10.15 kpsi, τm = τa = 2.55 kpsi.
For bending, from Eqs. (6-34) and (6-35a),
a = 0.246 − 3.08 (10−3 ) ( 64 ) + 1.51(10−5 ) ( 64 ) − 2.67 (10−8 ) ( 64 ) = 0.10373
2
1
= 0.75
a 1 + 0.10373
1+
0.1
r
K f = 1 + q ( K t − 1) = 1 + 0.75(1.5 − 1) = 1.38
q=
Eq. (6-32):
1
3
=
For torsion, from Eqs. (6-34) and (6-35b),
Shigley’s MED, 10th edition
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a = 0.190 − 2.51(10−3 ) ( 64 ) + 1.35 (10−5 ) ( 64 ) − 2.67 (10−8 ) ( 64 ) = 0.07800
2
1
1
= 0.80
a 1 + 0.07800
1+
0.1
r
K fs = 1 + qs ( K ts − 1) = 1 + 0.80(2.1 − 1) = 1.88
q=
Eq. (6-32):
3
=
Obtain von Mises stresses for the alternating and mid-range stresses from Eqs. (6-55) and
(6-56).
{
σ a′ = (1.38 )(10.15 ) + 3 (1.88 )( 2.55 ) 
2
}
2 1/2
= 16.28 kpsi
σ m′ = σ a′ = 16.28 kpsi
′ = σ a′ + σ m′ ,
Check for yielding, using the conservative σ max
Sy
54
ny =
=
= 1.66
σ a′ + σ m′ 16.28 + 16.28
Obtain the modifying factors and endurance limit.
Eq. (6-8):
Se′ = 0.5(64) = 32 kpsi
Eq. (6-19):
Eq. (6-24):
ka = aSutb = 2.70(64) −0.265 = 0.90
Eq. (6-20):
Eq. (6-18):
kb = 0.879d e −0.107 = 0.879(0.370) −0.107 = 0.98
d e = 0.370d = 0.370(1) = 0.370 in
Se = (0.90)(0.98)(32) = 28.2 kpsi
Using Modified Goodman,
1 σ a′ σ m′ 16.28 16.28
=
+
=
+
nf
S e Sut
28.2
64
n f = 1.20
Ans.
____________________________________________________________________________
6-50
Table A-20: Sut = 64 kpsi, S y = 54 kpsi
From Prob. 3-83, the critical stress element on the neutral axis in the middle of the
longest side of the rectangular cross section experiences a repeatedly applied shear stress
of τmax = 14.3 kpsi, τmin = 0 kpsi. Thus, τm = τa = 7.15 kpsi. Since the stress is entirely
shear, it is convenient to check for yielding using the standard Maximum Shear Stress
theory.
S / 2 54 / 2
ny = y
=
= 1.89
τ max
14.3
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Find the modifiers and endurance limit.
Eq. (6-8):
Se′ = 0.5(64) = 32 kpsi
Eq. (6-19):
ka = aSutb = 2.70(64) −0.265 = 0.90
The size factor for a rectangular cross section loaded in torsion is not readily available.
Following the procedure on p. 297, we need an equivalent diameter based on the 95
percent stress area. However, the stress situation in this case is nonlinear, as described on
p. 116. Noting that the maximum stress occurs at the middle of the longest side, or with a
radius from the center of the cross section equal to half of the shortest side, we will
simply choose an equivalent diameter equal to the length of the shortest side.
de = 0.25 in
Eq. (6-20):
kb = 0.879de −0.107 = 0.879(0.25)−0.107 = 1.02
We will round down to kb = 1.
Eq. (6-26):
kc = 0.59
Eq. (6-18):
S se = 0.9(1)(0.59)(32) = 17.0 kpsi
Since the stress is entirely shear, we choose to use a load factor kc = 0.59, and convert the
ultimate strength to a shear value rather than using the combination loading method of
Sec. 6-14. From Eq. (6-54), Ssu = 0.67Su = 0.67 (64) = 42.9 kpsi.
Using Modified Goodman,
1
1
=
= 1.70 Ans.
(τ a / S se ) + (τ m / S su ) (7.15 / 17.0) + (7.15 / 42.9)
______________________________________________________________________________
nf =
6-51
Table A-20: Sut = 64 kpsi, S y = 54 kpsi
From Prob. 3-84, the critical stress element experiences σ = 28.0 kpsi and τ = 15.3 kpsi.
Since the load is applied and released repeatedly, this gives σmax = 28.0 kpsi, σmin = 0
kpsi, τmax = 15.3 kpsi, τmin = 0 kpsi. Consequently, σm = σa = 14.0 kpsi, τm = τa = 7.65
kpsi. From Table A-15-8 and A-15-9,
D / d = 1.5 /1 = 1.5,
K t ,bend = 1.60,
r / d = 0.125 /1 = 0.125
K t ,tors = 1.39
Eqs. (6-34) and (6-35), or Figs. 6-20 and 6-21:
Eq. (6-32):
qbend = 0.78, qtors = 0.82
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K f ,bend = 1 + qbend ( Kt ,bend − 1) = 1 + 0.78 (1.60 − 1) = 1.47
K f ,tors = 1 + qtors ( Kt ,tors − 1) = 1 + 0.82 (1.39 − 1) = 1.32
Obtain von Mises stresses for the alternating and mid-range stresses from Eqs. (6-55) and
(6-56).
{
σ a′ = (1.47 )(14.0 )  + 3 (1.32 )( 7.65 )
2
}
2 1/2
= 27.0 kpsi
σ m′ = σ a′ = 27.0 kpsi
′ = σ a′ + σ m′ ,
Check for yielding, using the conservative σ max
Sy
54
ny =
=
= 1.00
σ a′ + σ m′ 27.0 + 27.0
Since stress concentrations are included in this quick yield check, the low factor of safety
is acceptable.
Eq. (6-8):
Se′ = 0.5(64) = 32 kpsi
Eq. (6-19):
ka = aSutb = 2.70(64) −0.265 = 0.897
Eq. (6-24):
Eq. (6-20):
Eq. (6-18):
de = 0.370d = 0.370 (1) = 0.370 in
kb = 0.879d e −0.107 = 0.879(0.370) −0.107 = 0.978
Se = (0.897)(0.978)(0.5)(64) = 28.1 kpsi
Using Modified Goodman,
1 σ a′ σ m′ 27.0 27.0
=
+
=
+
nf
Se Sut 28.1 64
n f = 0.72
Ans.
Since infinite life is not predicted, estimate a life from the S-N diagram. First, find an
equivalent completely reversed stress (See Ex. 6-12).
σ rev =
σ a′
27.0
=
= 46.7 kpsi
1 − (σ m′ / Sut ) 1 − (27.0 / 64)
Fig. 6-18:
f = 0.9
Eq. (6-14):
( f Sut ) 2 [ 0.9(64) ]
a=
=
= 118.07
Se
28.1
Eq. (6-15):
 f Sut
1
b = − log 
3
 Se
2

1
 0.9(64) 
 = − log 
 = −0.1039
3
 28.1 

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1
σ 
 46.7  −0.1039
Eq. (6-16):
N =  rev  = 
Ans.
= 7534 cycles 7500 cycles

 118.07 
 a 
______________________________________________________________________________
1/ b
6-52
Table A-20: Sut = 64 kpsi, S y = 54 kpsi
From Prob. 3-85, the critical stress element experiences σx,bend = 46.1 kpsi, σx,axial = 0.382
kpsi and τ = 15.3 kpsi. The axial load is practically negligible, but we’ll include it to
demonstrate the process. Since the load is applied and released repeatedly, this gives
σmax,bend = 46.1 kpsi, σmin,bend = 0 kpsi, σmax,axial = 0.382 kpsi, σmin,axial = 0 kpsi, τmax = 15.3
kpsi, τmin = 0 kpsi. Consequently, σm,bend = σa,bend = 23.05 kpsi, σm,axial = σa,axial = 0.191
kpsi, τm = τa = 7.65 kpsi. From Table A-15-7, A-15-8 and A-15-9,
D / d = 1.5 /1 = 1.5,
r / d = 0.125 /1 = 0.125
K t ,bend = 1.60,
K t ,tors = 1.39,
K t ,axial = 1.75
Eqs. (6-34) and (6-35), or Figs. 6-20 and 6-21: qbend = qaxial =0.78, qtors = 0.82
Eq. (6-32):
K f ,bend = 1 + qbend ( Kt ,bend − 1) = 1 + 0.78 (1.60 − 1) = 1.47
K f ,axial = 1 + qaxial ( Kt ,axial − 1) = 1 + 0.78 (1.75 − 1) = 1.59
K f ,tors = 1 + qtors ( Kt ,tors − 1) = 1 + 0.82 (1.39 − 1) = 1.32
Obtain von Mises stresses for the alternating and mid-range stresses from Eqs. (6-55) and
(6-56).
1/ 2
2
 

0.191) 
2
(


σ a′ =  (1.47 )( 23.05 ) + (1.59 )
3
1.32
7.65
+
(
)(
)


 
0.85 
 

{
σ m′ = (1.47 )( 23.05 ) + (1.59 )( 0.191) + 3 (1.32 )( 7.65 ) 
2
}
2 1/2
= 38.45 kpsi
= 38.40 kpsi
′ = σ a′ + σ m′ ,
Check for yielding, using the conservative σ max
Sy
54
ny =
=
= 0.70
σ a′ + σ m′ 38.45 + 38.40
Since the conservative yield check indicates yielding, we will check more carefully with
′ obtained directly from the maximum stresses, using the distortion energy
with σ max
failure theory, without stress concentrations. Note that this is exactly the method used for
static failure in Ch. 5.
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′ =
σ max
ny =
(σ
+ σ max,axial ) + 3 (τ max ) =
2
max,bend
Sy
54
=
= 1.01
′
53.5
σ max
2
( 46.1 + 0.382 )
2
+ 3 (15.3) = 53.5 kpsi
2
Ans.
This shows that yielding is imminent, and further analysis of fatigue life should not be
interpreted as a guarantee of more than one cycle of life.
Eq. (6-8):
Se′ = 0.5(64) = 32 kpsi
Eq. (6-19):
ka = aSutb = 2.70(64) −0.265 = 0.897
Eq. (6-24):
de = 0.370d = 0.370 (1) = 0.370 in
Eq. (6-20):
kb = 0.879d e −0.107 = 0.879(0.370) −0.107 = 0.978
Eq. (6-18):
Se = (0.897)(0.978)(0.5)(64) = 28.1 kpsi
Using Modified Goodman,
1 σ a′ σ m′ 38.45 38.40
=
+
=
+
nf
Se Sut
28.1
64
n f = 0.51
Ans.
Since infinite life is not predicted, estimate a life from the S-N diagram. First, find an
equivalent completely reversed stress (See Ex. 6-12).
σ rev =
σ a′
38.45
=
= 96.1 kpsi
1 − (σ m′ / Sut ) 1 − (38.40 / 64)
This stress is much higher than the ultimate strength, rendering it impractical for the S-N
diagram. We must conclude that the fluctuating stresses from the combination loading,
when increased by the stress concentration factors, are so far from the Goodman line that
the equivalent completely reversed stress method is not practical to use. Without testing,
we are unable to predict a life.
______________________________________________________________________________
6-53
Table A-20: Sut = 64 kpsi, S y = 54 kpsi
From Prob. 3-86, the critical stress element experiences σx,bend = 55.5 kpsi, σx,axial = 0.382
kpsi and τ = 15.3 kpsi. The axial load is practically negligible, but we’ll include it to
demonstrate the process. Since the load is applied and released repeatedly, this gives
σmax,bend = 55.5 kpsi, σmin,bend = 0 kpsi, σmax,axial = 0.382 kpsi, σmin,axial = 0 kpsi, τmax = 15.3
kpsi, τmin = 0 kpsi. Consequently, σm,bend = σa,bend = 27.75 kpsi, σm,axial = σa,axial = 0.191
kpsi, τm = τa = 7.65 kpsi. From Table A-15-7, A-15-8 and A-15-9,
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D / d = 1.5 /1 = 1.5,
K t ,bend = 1.60,
r / d = 0.125 /1 = 0.125
K t ,tors = 1.39,
K t ,axial = 1.75
Eqs. (6-34) and (6-35), or Figs. 6-20 and 6-21: qbend = qaxial =0.78, qtors = 0.82
Eq. (6-32):
K f ,bend = 1 + qbend ( Kt ,bend − 1) = 1 + 0.78 (1.60 − 1) = 1.47
K f ,axial = 1 + qaxial ( Kt ,axial − 1) = 1 + 0.78 (1.75 − 1) = 1.59
K f ,tors = 1 + qtors ( Kt ,tors − 1) = 1 + 0.82 (1.39 − 1) = 1.32
Obtain von Mises stresses for the alternating and mid-range stresses from Eqs. (6-55) and
(6-56).
1/2
2
 

0.191) 
2
(
σ a′ =  (1.47 )( 27.75 ) + (1.59 )
 + 3 (1.32 )( 7.65 )  
0.85 
 

{
σ m′ = (1.47 )( 27.75 ) + (1.59 )( 0.191)  + 3 (1.32 )( 7.65 ) 
2
}
2 1/ 2
= 44.71 kpsi
= 44.66 kpsi
Since these stresses are relatively high compared to the yield strength, we will go ahead
and check for yielding using the distortion energy failure theory.
′ =
σ max
ny =
(σ
+ σ max,axial ) + 3 (τ max ) =
2
max,bend
Sy
54
=
= 0.87
′
61.8
σ max
2
( 55.5 + 0.382 )
2
+ 3 (15.3) = 61.8 kpsi
2
Ans.
This shows that yielding is predicted. Further analysis of fatigue life is just to be able to
report the fatigue factor of safety, though the life will be dictated by the static yielding
failure, i.e. N = 1/2 cycle.
Ans.
Eq. (6-8):
Se′ = 0.5 ( 64 ) = 32 kpsi
Eq. (6-19):
ka = aSutb = 2.70(64) −0.265 = 0.897
Eq. (6-24):
de = 0.370d = 0.370 (1) = 0.370 in
Eq. (6-20):
kb = 0.879d e −0.107 = 0.879(0.370) −0.107 = 0.978
Eq. (6-18):
Se = (0.897)(0.978)(0.5)(64) = 28.1 kpsi
Using Modified Goodman,
1 σ a′ σ m′ 44.71 44.66
=
+
=
+
nf
Se Sut
28.1
64
n f = 0.44
Ans.
______________________________________________________________________________
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6-54
From Table A-20, for AISI 1040 CD, Sut = 85 kpsi and Sy = 71 kpsi. From the solution to
Prob. 6-17 we find the completely reversed stress at the critical shoulder fillet to be σrev =
35.0 kpsi, producing σa = 35.0 kpsi and σm = 0 kpsi. This problem adds a steady torque
which creates torsional stresses of
τm =
Tr 2500 (1.625 / 2 )
=
= 2967 psi = 2.97 kpsi, τ a = 0 kpsi
J
π (1.6254 ) / 32
From Table A-15-8 and A-15-9, r/d = 0.0625/1.625 = 0.04, D/d = 1.875/1.625 = 1.15,
Kt,bend =1.95, Kt,tors =1.60
Eqs. (6-34) and (6-35), or Figs. 6-20 and 6-21: qbend = 0.76, qtors = 0.81
Eq. (6-32):
K f ,bend = 1 + qbend ( Kt ,bend − 1) = 1 + 0.76 (1.95 − 1) = 1.72
K f ,tors = 1 + qtors ( Kt ,tors − 1) = 1 + 0.81(1.60 − 1) = 1.49
Obtain von Mises stresses for the alternating and mid-range stresses from Eqs. (6-55) and
(6-56).
{
}
σ ′ = {(1.72 )( 0 )  + 3 (1.49 )( 2.97 )  }
σ a′ = (1.72 )( 35.0 )  + 3 (1.49 )( 0 ) 
2 1/2
2
2 1/2
2
m
= 60.2 kpsi
= 7.66 kpsi
′ = σ a′ + σ m′ ,
Check for yielding, using the conservative σ max
Sy
71
ny =
=
= 1.05
′
′
σ a + σ m 60.2 + 7.66
From the solution to Prob. 6-17, Se = 29.5 kpsi. Using Modified Goodman,
1 σ a′ σ m′ 60.2 7.66
=
+
=
+
nf
Se Sut 29.5 85
n f = 0.47 Ans.
Since infinite life is not predicted, estimate a life from the S-N diagram. First, find an
equivalent completely reversed stress (See Ex. 6-12).
σ rev =
Fig. 6-18:
σ a′
60.2
=
= 66.2 kpsi
1 − (σ m′ / Sut ) 1 − (7.66 / 85)
f = 0.867
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( f Sut )
2
[0.867(85)]
2
Eq. (6-14):
a=
=
Eq. (6-15):
 f Sut
1
b = − log 
3
 Se
Eq. (6-16):
σ 
N =  rev 
 a 
Se
1/ b
29.5
= 184.1

1
 0.867(85) 
 = − log 
 = −0.1325
3
 29.5 

1
 66.2  −0.1325
=
= 2251 cycles

 184.1 
N = 2300 cycles
Ans.
______________________________________________________________________________
6-55
From the solution to Prob. 6-18 we find the completely reversed stress at the critical
shoulder fillet to be σrev = 32.8 kpsi, producing σa = 32.8 kpsi and σm = 0 kpsi. This
problem adds a steady torque which creates torsional stresses of
τm =
Tr 2200 (1.625 / 2 )
=
= 2611 psi = 2.61 kpsi,
J
π (1.6254 ) / 32
τ a = 0 kpsi
From Table A-15-8 and A-15-9, r/d = 0.0625/1.625 = 0.04, D/d = 1.875/1.625 = 1.15,
Kt,bend =1.95, Kt,tors =1.60
Eqs. (6-34) and (6-35), or Figs. 6-20 and 6-21: qbend = 0.76, qtors = 0.81
Eq. (6-32):
K f ,bend = 1 + qbend ( Kt ,bend − 1) = 1 + 0.76 (1.95 − 1) = 1.72
K f ,tors = 1 + qtors ( Kt ,tors − 1) = 1 + 0.81(1.60 − 1) = 1.49
Obtain von Mises stresses for the alternating and mid-range stresses from Eqs. (6-55) and
(6-56).
{
}
σ ′ = {(1.72 )( 0 )  + 3 (1.49 )( 2.61)  }
σ a′ = (1.72 )( 32.8 )  + 3 (1.49 )( 0 ) 
2 1/2
2
2 1/2
2
m
= 56.4 kpsi
= 6.74 kpsi
′ = σ a′ + σ m′ ,
Check for yielding, using the conservative σ max
Sy
71
=
= 1.12
ny =
σ a′ + σ m′ 56.4 + 6.74
From the solution to Prob. 6-18, Se = 29.5 kpsi. Using Modified Goodman,
1 σ a′ σ m′ 56.4 6.74
=
+
=
+
nf
Se Sut 29.5 85
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n f = 0.50 Ans.
Since infinite life is not predicted, estimate a life from the S-N diagram. First, find an
equivalent completely reversed stress (See Ex. 6-12).
σ rev =
σ a′
56.4
=
= 61.3 kpsi
′
1 − (σ m / Sut ) 1 − (6.74 / 85)
Fig. 6-18:
f = 0.867
Eq. (6-14):
a=
Eq. (6-15):
 f Sut
1
b = − log 
3
 Se
( f Sut )
Se
σ 
Eq. (6-16): N =  rev 
 a 
1/ b
2
[0.867(85)]
2
=
29.5
= 184.1

1
 0.867(85) 
 = − log 
 = −0.1325
3
 29.5 

1
 61.3  −0.1325
=
= 4022 cycles

 184.1 
N = 4000 cycles
Ans.
______________________________________________________________________________
6-56
Sut = 55 kpsi, S y = 30 kpsi, K ts = 1.6, L = 2 ft, Fmin = 150 lbf , Fmax = 500 lbf
Eqs. (6-34) and (6-35b), or Fig. 6-21: qs = 0.80
Eq. (6-32): K fs = 1 + qs ( Kts − 1) = 1 + 0.80 (1.6 − 1) = 1.48
Tmax = 500(2) = 1000 lbf ⋅ in, Tmin = 150(2) = 300 lbf ⋅ in
τ max =
16 K fsTmax
τ min =
16 K fsTmin
τm =
τa =
πd
3
πd
τ max + τ min
3
2
τ max − τ min
2
=
16(1.48)(1000)
= 11 251 psi = 11.25 kpsi
π (0.875)3
16(1.48)(300)
= 3375 psi = 3.38 kpsi
π (0.875)3
11.25 + 3.38
=
= 7.32 kpsi
2
11.25 − 3.38
=
= 3.94 kpsi
2
=
Since the stress is entirely shear, it is convenient to check for yielding using the standard
Maximum Shear Stress theory.
S / 2 30 / 2
ny = y
=
= 1.33
τ max 11.25
Find the modifiers and endurance limit.
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Eq. (6-8):
Se′ = 0.5(55) = 27.5 kpsi
Eq. (6-19):
Eq. (6-24):
ka = 14.4(55) −0.718 = 0.81
Eq. (6-20):
kb = 0.879(0.324) −0.107 = 0.99
Eq. (6-26):
kc = 0.59
Eq. (6-18):
S se = 0.81(0.99)(0.59)(27.5) = 13.0 kpsi
d e = 0.370(0.875) = 0.324 in
Since the stress is entirely shear, we will use a load factor kc = 0.59, and convert the
ultimate strength to a shear value rather than using the combination loading method of
Sec. 6-14. From Eq. (6-54), Ssu = 0.67Su = 0.67 (55) = 36.9 kpsi.
(a) Modified Goodman, Table 6-6
nf =
1
1
=
= 1.99
(τ a / S se ) + (τ m / S su ) (3.94 / 13.0) + (7.32 / 36.9)
Ans.
(b) Gerber, Table 6-7
2
2

 2τ m S se  
1  S su  τ a 
−1 + 1 + 
nf = 


2  τ m  S se 
 S suτ a  


2
2

 2(7.32)(13.0)  
1  36.9   3.94  
= 

 
 −1 + 1 + 
2  7.32   13.0  
 36.9(3.94)  


n f = 2.49 Ans.
______________________________________________________________________________
6-57
Sut = 145 kpsi, S y = 120 kpsi
From Eqs. (6-34) and (6-35a), or Fig. 6-20, with a notch radius of 0.1 in, q = 0.9. Thus,
with Kt = 3 from the problem statement,
K f = 1 + q ( K t − 1) = 1 + 0.9(3 − 1) = 2.80
4 P −2.80(4)( P)
=
= −2.476 P
πd2
π (1.2) 2
1
σ m = −σ a = (−2.476 P ) = −1.238 P
2
f P ( D + d ) 0.3P ( 6 + 1.2 )
Tmax =
=
= 0.54 P
4
4
σ max = − K f
From Eqs. (6-34) and (6-35b), or Fig. 6-21, with a notch radius of 0.1 in, qs = 0.92. Thus,
with Kts = 1.8 from the problem statement,
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K fs = 1 + qs ( K ts − 1) = 1 + 0.92(1.8 − 1) = 1.74
16 K fsT
16(1.74)(0.54 P)
= 2.769 P
πd
π (1.2)3
τ
2.769 P
τ a = τ m = max =
= 1.385 P
2
2
Eqs. (6-55) and (6-56):
τ max =
3
=
σ a′ = [(σ a / 0.85)2 + 3τ a2 ]1/2 = [(1.238P / 0.85) 2 + 3(1.385P )2 ]1/2 = 2.81P
σ m′ = [σ m 2 + 3τ m2 ]1/2 = [(−1.238P) 2 + 3(1.385P )2 ]1/2 = 2.70 P
Eq. (6-8):
Se′ = 0.5(145) = 72.5 kpsi
Eq. (6-19):
ka = 2.70(145)−0.265 = 0.722
Eq. (6-20):
Eq. (6-18):
kb = 0.879(1.2) −0.107 = 0.862
Se = (0.722)(0.862)(72.5) = 45.12 kpsi
Modified Goodman:
1 σ a′ σ m′ 2.81P 2.70 P 1
=
+
=
+
=
n f Se Sut 45.12
145
3
P = 4.12 kips
Ans.
Sy
120
= 5.29 Ans.
σ a′ + σ m′ (2.81)(4.12) + (2.70)(4.12)
______________________________________________________________________________
Yield (conservative): n y =
6-58
=
From Prob. 6-57, K f = 2.80, K f s = 1.74, S e = 45.12 kpsi
4 Pmax
4(18)
= −2.80
= −44.56 kpsi
2
πd
π (1.22 )
4P
4(4.5)
σ min = − K f min2 = −2.80
= −11.14 kpsi
πd
π (1.2) 2
 D+d 
 6 + 1.2 
Tmax = f Pmax 
 = 0.3(18) 
 = 9.72 kip ⋅ in
 4 
 4 
 D+d 
 6 + 1.2 
Tmin = f Pmin 
 = 0.3(4.5) 
 = 2.43 kip ⋅ in
4


 4 
16Tmax
16(9.72)
τ max = K f s
= 1.74
= 49.85 kpsi
3
πd
π (1.2)3
16Tmin
16(2.43)
τ min = K f s
= 1.74
= 12.46 kpsi
3
πd
π (1.2)3
−44.56 − ( −11.14)
σa =
= 16.71 kpsi
2
σ max = − K f
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−44.56 + ( −11.14)
= −27.85 kpsi
2
49.85 − 12.46
τa =
= 18.70 kpsi
2
49.85 + 12.46
τm =
= 31.16 kpsi
2
σm =
Eqs. (6-55) and (6-56):
σ a′ = [(σ a / 0.85)2 + 3τ a2 ]1/2 = [(16.71/ 0.85)2 + 3(18.70) 2 ]1/2 = 37.89 kpsi
σ m′ = [σ m 2 + 3τ m2 ]1/2 = [(−27.85)2 + 3(31.16) 2 ]1/ 2 = 60.73 kpsi
Modified Goodman:
1 σ a′ σ m′ 37.89 60.73
=
+
=
+
n f Se Sut 45.12 145
nf = 0.79
Since infinite life is not predicted, estimate a life from the S-N diagram. First, find an
equivalent completely reversed stress (See Ex. 6-12).
σ rev =
σ a′
37.89
=
= 65.2 kpsi
1 − (σ m′ / Sut ) 1 − (60.73 /145)
Fig. 6-18:
f = 0.8
Eq. (6-14):
( f Sut )
a=
2
[0.8(145)]
=
2
45.12
Se
= 298.2
Eq. (6-15):
 f Sut 
1
1
 0.8(145) 
b = − log 
 = − log 
 = −0.1367
3
3
 45.12 
 Se 
Eq. (6-16):
σ 
N =  rev 
 a 
1/ b
1
 65.2  −0.1367
=
= 67 607 cycles

 298.2 
N = 67 600 cycles
Ans.
______________________________________________________________________________
6-59
For AISI 1020 CD, From Table A-20, Sy = 390 MPa, Sut = 470 MPa. Given: Se = 175
MPa.
360 + 160
360 − 160
First Loading:
= 260 MPa,
= 100 MPa
(σ m )1 =
(σ a )1 =
2
2
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Goodman:
(σ a )1
1 − (σ m )1 / Sut
2
0.9 ( 470 ) 
(σ a )e1 =
a=
=
100
= 223.8 MPa > Se ∴ finite life
1 − 260 / 470
= 1022.5 MPa
175
0.9 ( 470 )
1
= −0.127 767
b = − log
3
175
−1/0.127 767
 223.8 
N =
= 145 920 cycles

 1022.5 
320 + ( −200 )
320 − ( −200 )
Second loading: (σ m ) 2 =
= 60 MPa,
= 260 MPa
(σ a ) 2 =
2
2
(σ a ) e 2 =
(a) Miner’s method:
n1 n2
+
=1
N1 N 2
260
= 298.0 MPa
1 − 60 / 470
 298.0 
N2 = 

 1022.5 
−1/0.127767
= 15 520 cycles
n2
80 000
+
=1
145 920 15 520
⇒
⇒
n2 = 7000 cycles Ans.
(b) Manson’s method: The number of cycles remaining after the first loading
Nremaining =145 920 − 80 000 = 65 920 cycles
Two data points: 0.9(470) MPa, 103 cycles
223.8 MPa, 65 920 cycles
a2 (103 )
0.9 ( 470 )
=
b
223.8
a2 ( 65 920 ) 2
b2
1.8901 = ( 0.015170 ) 2
b
log1.8901
= −0.151 997
log 0.015170
223.8
= 1208.7 MPa
a2 =
−0.151 997
( 65 920 )
b2 =
1/ −0.151 997
 298.0 
= 10 000 cycles
n2 = 
Ans.

 1208.7 
______________________________________________________________________________
6-60
Given: Se = 50 kpsi, Sut = 140 kpsi, f =0.8. Using Miner’s method,
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0.8 (140 ) 
a=
= 250.88 kpsi
50
0.8 (140 )
1
b = − log
= −0.116 749
3
50
1/ −0.116 749
 95 
= 4100 cycles
σ 1 = 95 kpsi,
N1 = 

 250.88 
2
1/ −0.116 749
σ 2 = 80 kpsi,
 80 
N2 = 

 250.88 
σ 3 = 65 kpsi,
 65 
N3 = 

 250.88 
= 17 850 cycles
1/ −0.116 749
= 105 700 cycles
0.2 N 0.5 N
0.3 N
+
+
= 1 ⇒ N = 12 600 cycles Ans.
4100 17 850 105 700
______________________________________________________________________________
6-61
Given: Sut = 530 MPa, Se = 210 MPa, and f = 0.9.
(a) Miner’s method
0.9 ( 530 ) 
a=
= 1083.47 MPa
210
0.9 ( 530 )
1
b = − log
= −0.118 766
3
210
2
1/ −0.118 766
 350 

 1083.47 
σ 1 = 350 MPa, N1 = 
= 13 550 cycles
1/ −0.118 766
 260 
σ 2 = 260 MPa, N 2 = 

 1083.47 

225
= 165 600 cycles
1/ −0.118 766

σ 3 = 225 MPa, N 3 = 

 1083.47 
= 559 400 cycles
n1 n2 n3
+
+
=1
N1 N 2 N 3
n3
5000
50 000
+
+
= 184 100 cycles
13 550 165 600 559 400
Ans.
(b) Manson’s method:
The life remaining after the first series of cycling is NR1 = 13 550 − 5000 = 8550
cycles. The two data points required to define S e′,1 are [0.9(530), 103] and (350, 8550).
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3
0.9 ( 530 ) a2 (10 )
=
b
350
a2 ( 8550 ) 2
b2
b2 =
a2 =
⇒
log (1.362 9 )
log ( 0.116 96 )
350
( 8550 )
−0.144 280
1.3629 = ( 0.11696 ) 2
b
= −0.144 280
= 1292.3 MPa
−1/0.144 280
 260 
N2 = 
= 67 090 cycles

 1292.3 
N R 2 = 67 090 − 50 000 = 17 090 cycles
0.9 ( 530 )
260
b3 =
=
a3 (103 )
b3
a3 (17 090 )
b3
⇒
1.834 6 = ( 0.058 514 ) 2
b
log (1.834 6 )
260
= −0.213 785, a3 =
= 2088.7 MPa
−0.213 785
log ( 0.058 514 )
(17 090 )
−1/0.213 785
 225 
= 33 610 cycles Ans.
N3 = 

 2088.7 
______________________________________________________________________________
6-62
Given: Se = 45 kpsi, Sut = 85 kpsi, f = 0.86, and σa = 35 kpsi and σm = 30 kpsi for 12 (103)
cycles.
σa
35
Gerber equivalent reversing stress: σ rev =
=
= 39.98 kpsi
2
2
1 − (σ m / Sut ) 1 − ( 30 / 85 )
(a) Miner’s method: σrev < Se. According to the method, this means that the endurance
limit has not been reduced and the new endurance limit is S e′ = 45 kpsi. Ans.
(b) Manson’s method: Again, σrev < Se. According to the method, this means that the
material has not been damaged and the endurance limit has not been reduced. Thus,
the new endurance limit is S e′ = 45 kpsi. Ans.
______________________________________________________________________________
6-63
Given: Se = 45 kpsi, Sut = 85 kpsi, f = 0.86, and σa = 35 kpsi and σm = 30 kpsi for 12 (103)
cycles.
σa
35
=
= 54.09 kpsi
Goodman equivalent reversing stress: σ rev =
1 − (σ m / Sut ) 1 − ( 30 / 85 )
Initial cycling
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0.86 ( 85 ) 
a=
= 116.00 kpsi
45
0.86 ( 85 )
1
b = − log
= −0.070 235
3
45
2
σ 1 = 54.09 kpsi,
1/ −0.070 235
 54.09 
N1 = 

 116.00 
= 52 190 cycles
(a) Miner’s method (see discussion on p. 333): The number of remaining cycles at 54.09
kpsi is Nremaining = 52 190 − 12 000 = 40 190 cycles. The new coefficients are b′ = b,
and a′ =Sf /Nb = 54.09/(40 190) − 0.070 235 = 113.89 kpsi. The new endurance limit is
Se′,1 = a′N eb′ = 113.89 (106 )
− 0.070 235
= 43.2 kpsi
Ans.
(b) Manson’s method (see discussion on p. 334): The number of remaining cycles at
54.09 kpsi is Nremaining = 52 190 − 12 000 = 40 190 cycles. At 103 cycles,
Sf = 0.86(85) = 73.1 kpsi. The new coefficients are
b′ = [log(73.1/54.09)]/log(103/40 190) = − 0.081 540 and a′ = σ1/ (Nremaining) b′ =
54.09/(40 190) − 0.081 540 = 128.39 kpsi. The new endurance limit is
Se′,1 = a′N eb′ = 128.39 (106 )
− 0.081 540
= 41.6 kpsi
Ans.
______________________________________________________________________________
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Chapter 7
7-1
(a) DE-Gerber, Eq. (7-10):
A = 4 ( K f M a ) + 3 ( K fsTa ) = 4 [ (2.2)(70) ] + 3 [ (1.8)(45) ] = 338.4 N ⋅ m
2
2
2
2
B = 4 ( K f M m ) + 3 ( K fsTm ) = 4 [ (2.2)(55) ] + 3 [ (1.8)(35) ] = 265.5 N ⋅ m
2
2
2

 
6
 8(2)(338.4)    2(265.5) ( 210 ) 10

+
+
d =
1
1
6 
6

 π ( 210 ) 10    338.4 ( 700 ) 10
 

d = 25.85 (10−3) m = 25.85 mm Ans.
( )
( )
( )
2



2 1/2




1/3





(b) DE-elliptic, Eq. (7-12) can be shown to be
1/3


2
2
 16n A
338.4 )
265.5 )
(
(
B 
 16(2)

+ 2  =
+
d =
2
2 
2
 π

6
6
π
S
S
( 210 ) 10 
( 560 ) 10  
e
y 





 

−3
d = 25.77 (10 ) m = 25.77 mm Ans.
2
1/3
2
( )
( )
(c) DE-Soderberg, Eq. (7-14) can be shown to be
1/3
16(2)  338.4
265.5

=
+
6

 π
560 106
 210 10

d = 27.70 (10−3) m = 27.70 mm Ans.
16n  A B  
d=
 + 
 π  Se S y  
( )
( )
1/3




(d) DE-Goodman: Eq. (7-8) can be shown to be
1/3
16(2)  338.4
265.5  



=
+
6

 π  210 106
700
10



d = 27.27 (10−3) m = 27.27 mm Ans.
________________________________________________________________________
Criterion
d (mm)
Compared to DE-Gerber
DE-Gerber
25.85
DE-Elliptic
25.77
0.31% Lower
Less conservative
DE-Soderberg
27.70
7.2% Higher
More conservative
DE-Goodman
27.27
5.5% Higher
More conservative
______________________________________________________________________________
1/3
16n  A B  
d=
 +

 π  Se Sut  
7-2
( )
( )
This problem has to be done by successive trials, since Se is a function of shaft size. The
material is SAE 2340 for which Sut = 175 kpsi, Sy = 160 kpsi, and HB ≥ 370.
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Eq. (6-19), p. 295:
ka = 2.70(175) −0.265 = 0.69
Trial #1: Choose dr = 0.75 in
Eq. (6-20), p. 296:
kb = 0.879(0.75) −0.107 = 0.91
Eq. (6-8), p.290:
Eq. (6-18), p. 295:
Se′ = 0.5Sut = 0.5 (175 ) = 87.5 kpsi
Se = 0.69 (0.91)(87.5) = 54.9 kpsi
d r = d − 2r = 0.75D − 2 D / 20 = 0.65 D
d
0.75
D= r =
= 1.15 in
0.65 0.65
D 1.15
r=
=
= 0.058 in
20
20
Fig. A-15-14:
d = d r + 2r = 0.75 + 2(0.058) = 0.808 in
d 0.808
=
= 1.08
0.75
dr
r 0.058
=
= 0.077
0.75
dr
Kt = 1.9
Fig. 6-20, p. 296:
r = 0.058 in, q = 0.90
Eq. (6-32), p. 303:
Kf = 1 + 0.90 (1.9 – 1) = 1.81
Fig. A-15-15:
Kts = 1.5
Fig. 6-21, p. 304:
r = 0.058 in, qs = 0.92
Eq. (6-32), p. 303:
Kfs = 1 + 0.92 (1.5 – 1) = 1.46
We select the DE-ASME Elliptic failure criteria, Eq. (7-12), with d as dr, and
Mm = Ta = 0,
1/3
2
2 1/2 

 




1.46(400)  
16(2.5)  1.81(600)
 + 3

4
dr = 

3

 160 103   
 π   54.9 10 




 

dr = 0.799 in
( )
( )
Trial #2: Choose dr = 0.799 in.
kb = 0.879(0.799) −0.107 = 0.90
Se = 0.69 (0.90)(0.5)(175) = 54.3 kpsi
d
0.799
D= r =
= 1.23 in
0.65 0.65
r = D / 20 = 1.23/20 = 0.062 in
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Figs. A-15-14 and A-15-15:
d = d r + 2r = 0.799 + 2(0.062) = 0.923 in
d 0.923
=
= 1.16
d r 0.799
r 0.062
=
= 0.078
d r 0.799
With these ratios only slightly different from the previous iteration, we are at the limit of
readability of the figures. We will keep the same values as before.
K t = 1.9, K ts = 1.5, q = 0.90, qs = 0.92
∴ K f = 1.81, K fs = 1.46
Using Eq. (7-12) produces dr = 0.802 in. Further iteration produces no change. With
dr = 0.802 in,
0.802
D=
= 1.23 in
0.65
d = 0.75(1.23) = 0.92 in
A look at a bearing catalog finds that the next available bore diameter is 0.9375 in. In
nominal sizes, we select d = 0.94 in, D = 1.25 in, r = 0.0625 in Ans.
______________________________________________________________________________
7-3
F cos 20°(d / 2) = TA, F = 2 TA / ( d cos 20°) = 2(340) / (0.150 cos 20°) = 4824 N.
The maximum bending moment will be at point C, with MC = 4824(0.100) = 482.4 N∙m.
Due to the rotation, the bending is completely reversed, while the torsion is constant.
Thus, Ma = 482.4 N∙m, Tm = 340 N∙m, Mm = Ta = 0.
For sharp fillet radii at the shoulders, from Table 7-1, Kt = 2.7, and Kts = 2.2. Examining
Figs. 6-20 and 6-21 (pp. 303 and 304 respectively) with Sut = 560 MPa, conservatively
estimate q = 0.8 and qs = 0.9. These estimates can be checked once a specific fillet radius
is determined.
Eq. (6-32):
K f = 1 + 0.8(2.7 − 1) = 2.4
K fs = 1 + 0.9(2.2 − 1) = 2.1
(a) We will choose to include fatigue stress concentration factors even for the static
analysis to avoid localized yielding.
1/ 2
Eq. (7-15):
′
σ max
2
 32 K f M a  2
 16 K fsTm  
= 
 + 3
 
3
3
 π d

 π d  
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Eq. (7-16):
n=
Sy
π d 3S y
=
′
16
σ max
 4 ( K M )2 + 3 ( K T )2 
f
a
fs m


−1/2
Solving for d,
1/3
 16n
1/2 

 4( K f M a ) 2 + 3( K fsTa )2  
d =
 π S y

 16(2.5)
=
 π ( 420 ) 106

( ){
4 [ (2.4)(482.4)] + 3[ (2.1)(340) ]
d = 0.0430 m = 43.0 mm
(b)
}
2 1/2
2
1/3




Ans.
ka = 4.51(560) −0.265 = 0.84
Assume kb = 0.85 for now. Check later once a diameter is known.
Se = 0.84(0.85)(0.5)(560) = 200 MPa
Selecting the DE-ASME Elliptic criteria, use Eq. (7-12) with M m = Ta = 0.
2


 2.1(340)
16(2.5)   2.4(482.4) 
 + 3
d =
4
6

 420 106
 π   200 10 



= 0.0534 m = 53.4 mm
( )
( )




2 1/2




1/3





With this diameter, we can refine our estimates for kb and q.
Eq. (6-20):
kb = 1.51d −0.157 = 1.51( 53.4 )
−0.157
= 0.81
Assuming a sharp fillet radius, from Table 7-1, r = 0.02d = 0.02 (53.4) = 1.07 mm.
Fig. (6-20):
Fig. (6-21):
q = 0.72
qs = 0.77
Iterating with these new estimates,
Eq. (6-32):
Eq. (6-18):
Eq. (7-12):
Kf = 1 + 0.72 (2.7 – 1) = 2.2
Kfs = 1 + 0.77 (2.2 – 1) = 1.9
Se = 0.84(0.81)(0.5)(560) = 191 MPa
d = 53 mm
Ans.
Further iteration does not change the results.
_____________________________________________________________________________
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7-4
We have a design task of identifying bending moment and torsion diagrams which are
preliminary to an industrial roller shaft design. Let point C represent the center of the
span of the roller.
FCy = 30(8) = 240 lbf
FCz = 0.4(240) = 96 lbf
T = FCz (2) = 96(2) = 192 lbf ⋅ in
T 192
FBz =
=
= 128 lbf
1.5 1.5
FBy = FBz tan 20o = 128 tan 20o = 46.6 lbf
(a) xy-plane
ΣM O = 240(5.75) − FAy (11.5) − 46.6(14.25) = 0
240(5.75) − 46.6(14.25)
FAy =
= 62.3 lbf
11.5
ΣM A = FOy (11.5) − 46.6(2.75) − 240(5.75) = 0
240(5.75) + 46.6(2.75)
FOy =
= 131.1 lbf
11.5
Bending moment diagram:
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xz-plane
ΣM O = 0 = 96(5.75) − FAz (11.5) + 128(14.25)
96(5.75) + 128(14.25)
FAz =
= 206.6 lbf
11.5
ΣM A = 0 = FOz (11.5) + 128(2.75) − 96(5.75)
96(5.75) − 128(2.75)
FOz =
= 17.4 lbf
11.5
Bending moment diagram:
M C = 100 2 + ( −754) 2 = 761 lbf ⋅ in
M A = ( −128) 2 + ( −352) 2 = 375 lbf ⋅ in
Torque: The torque is constant from C to B, with a magnitude previously obtained of 192
lbf∙in.
(b) xy-plane
2
2
M xy = −131.1x + 15 x − 1.75 − 15 x − 9.75 − 62.3 x − 11.5
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1
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Bending moment diagram:
Mmax = –516 lbf ∙ in and occurs at 6.12 in.
M C = 131.1(5.75) − 15(5.75 − 1.75) 2 = 514 lbf ⋅ in
This is reduced from 754 lbf ∙ in found in part (a). The maximum occurs at x = 6.12 in
rather than C, but it is close enough.
xz-plane
2
2
M xz = 17.4 x − 6 x − 1.75 + 6 x − 9.75 + 206.6 x − 11.5
1
Bending moment diagram:
Let M net = M xy2 + M xz2
Plot Mnet(x), 1.75 ≤ x ≤ 11.5 in
Mmax = 516 lbf ∙ in at x = 6.25 in
Torque: The torque rises from 0 to 192 lbf∙in linearly across the roller, then is constant to
B. Ans.
______________________________________________________________________________
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This is a design problem, which can have many acceptable designs. See the solution for
Prob. 7-17 for an example of the design process.
______________________________________________________________________________
7-5
7-6
If students have access to finite element or beam analysis software, have them model the
shaft to check deflections. If not, solve a simpler version of shaft for deflection. The 1 in
diameter sections will not affect the deflection results much, so model the 1 in diameter
as 1.25 in. Also, ignore the step in AB.
From Prob. 7-4, integrate Mxy and Mxz.
xy plane, with dy/dx = y'
( )
131.1 2
62.3
3
3
2
x + 5 x − 1.75 − 5 x − 9.75 −
x − 11.5 + C1
2
2
131.1 3 5
5
62.3
4
4
3
EIy = −
x + x − 1.75 − x − 9.75 −
x − 11.5 + C1 x + C2
6
4
4
6
EIy ′ = −
(1)
( )
y = 0 at x = 0
From (1),
⇒
C2 = 0
y = 0 at x = 11.5 ⇒ C1 = 1908.4 lbf ⋅ in 3
x = 0:
EIy' = 1908.4
x = 11.5:
EIy' = –2153.1
xz plane (treating z ↑ + )
( )
17.4 2
206.6
3
3
2
x − 2 x − 1.75 + 2 x − 9.75 +
x − 11.5 + C3
2
2
17.4 3 1
1
206.6
4
4
3
EIz =
x ) − x − 1.75 + x − 9.75 +
x − 11.5 + C3 x + C4
(
6
2
2
6
EIz ′ =
z = 0 at x = 0
From (2),
⇒
(2)
C4 = 0
z = 0 at x = 11.5 ⇒ C3 = 8.975 lbf ⋅ in 3
x = 0:
EIz' = 8.975
x = 11.5:
EIz' = –683.5
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At O:
EIθ = 1908.42 + 8.9752 = 1908.4 lbf ⋅ in 3
At A:
EIθ = ( −2153.1) 2 + ( −683.5) 2 = 2259.0 lbf ⋅ in 3 (dictates size)
θ=
2259
= 0.000 628 rad
30 10 (π / 64 ) 1.254
n=
0.001
= 1.59
0.000 628
( )
(
6
)
At gear mesh, B
xy plane
With I = I1 in section OCA,
y′A = −2153.1/ EI1
Since y'B/A is a cantilever, from Table A-9-1, with I = I 2 in section AB
Fx( x − 2l ) 46.6
y′B / A =
(2.75)[2.75 − 2(2.75)] = −176.2 / EI 2
=
2 EI 2
2 EI 2
2153.1
176.2
∴ y′B = y′A + y′B / A = −
−
6
4
6
30 10 (π / 64 ) 1.25
30 10 (π / 64 ) 0.8754
( )
(
)
( )
(
)
= –0.000 803 rad (magnitude greater than 0.0005 rad)
xz plane
683.5
z ′A = −
,
EI1
z ′B = −
z ′B / A = −
(
128 2.752
2 EI 2
) = − 484
EI 2
484
683.5
−
= −0.000 751 rad
30 10 (π / 64 ) 1.254 30 106 (π / 64 ) 0.8754
( )
6
(
)
( )
(
)
θ B = (−0.000 803) 2 + (−0.000 751) 2 = 0.00110 rad
Crowned teeth must be used.
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Finite element results:
θO = 5.47(10−4 ) rad
Error in simplified model
3.0%
θ A = 7.09(10−4 ) rad
11.4%
0.0%
θ B = 1.10(10−3 ) rad
The simplified model yielded reasonable results.
Sut = 72 kpsi, S y = 39.5 kpsi
Strength
At the shoulder at A, x = 10.75 in. From Prob. 7-4,
M xy = −209.3 lbf ⋅ in, M xz = −293.0 lbf ⋅ in, T = 192 lbf ⋅ in
M = ( −209.3) 2 + ( −293) 2 = 360.0 lbf ⋅ in
Se′ = 0.5(72) = 36 kpsi
ka = 2.70(72) −0.265 − 0.869
−0.107
 1 
= 0.879
kb = 

 0.3 
kc = k d = k e = k f = 1
Fig. A-15-8:
Fig. A-15-9:
Fig. 6-20:
Fig. 6-21:
Eq. (6-32):
Se = 0.869(0.879)(36) = 27.5 kpsi
D / d = 1.25, r / d = 0.03
Kts = 1.8
Kt = 2.3
q = 0.65
qs = 0.70
K f = 1 + 0.65(2.3 − 1) = 1.85
K fs = 1 + 0.70(1.8 − 1) = 1.56
Using DE-ASME Elliptic, Eq. (7-11) with M m = Ta = 0,
1/2
1
16
=
n π 13
( )
2
 1.85(360)  2
1.56(192)  
+
4
3
 

 39 500  

 
  27 500 
n = 3.91
Perform a similar analysis at the profile keyway under the gear.
The main problem with the design is the undersized shaft overhang with excessive slope
at the gear. The use of crowned-teeth in the gears will eliminate this problem.
______________________________________________________________________________
7-7 through 7-16
These are design problems, which can have many acceptable designs. See the solution for
Prob. 7-17 for an example of the design process.
______________________________________________________________________________
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7-17
(a) One possible shaft layout is shown in part (e). Both bearings and the gear will be
located against shoulders. The gear and the motor will transmit the torque through the
keys. The bearings can be lightly pressed onto the shaft. The left bearing will locate the
shaft in the housing, while the right bearing will float in the housing.
(b) From summing moments around the shaft axis, the tangential transmitted load
through the gear will be
Wt = T / (d / 2) = 2500 / (4 / 2) = 1250 lbf
The radial component of gear force is related by the pressure angle.
Wr = Wt tan φ = 1250 tan 20o = 455 lbf
(
W = Wr2 + Wt 2
)
1/ 2
(
= 4552 + 12502
)
1/2
= 1330 lbf
Reactions RA and RB , and the load W are all in the same plane. From force and moment
balance,
RA = 1330(2 / 11) = 242 lbf
RB = 1330(9 / 11) = 1088 lbf
M max = RA (9) = 242(9) = 2178 lbf ⋅ in
Shear force, bending moment, and torque diagrams can now be obtained.
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(c) Potential critical locations occur at each stress concentration (shoulders and keyways).
To be thorough, the stress at each potentially critical location should be evaluated. For
now, we will choose the most likely critical location, by observation of the loading
situation, to be in the keyway for the gear. At this point there is a large stress
concentration, a large bending moment, and the torque is present. The other locations
either have small bending moments, or no torque. The stress concentration for the
keyway is highest at the ends. For simplicity, and to be conservative, we will use the
maximum bending moment, even though it will have dropped off a little at the end of the
keyway.
(d) At the gear keyway, approximately 9 in from the left end of the shaft, the bending is
completely reversed and the torque is steady.
M a = 2178 lbf ⋅ in Tm = 2500 lbf ⋅ in M m = Ta = 0
From Table 7-1, p. 365, estimate stress concentrations for the end-milled keyseat to be Kt
= 2.14 and Kts = 3.0. For the relatively low strength steel specified (AISI 1020 CD),
roughly estimate notch sensitivities of q = 0.75 and qs = 0.80, obtained by observation of
Figs. 6-20 and 6-21, assuming a typical radius at the bottom of the keyseat of r / d = 0.02,
and a shaft diameter of up to 3 inches.
Eq. (6-32):
K f = 1 + 0.75(2.14 − 1) = 1.9
K fs = 1 + 0.8(3.0 − 1) = 2.6
Eq. (6-19):
ka = 2.70(68) −0.265 = 0.883
For estimating kb , guess d = 2 in.
Eq. (6-20)
kb = (2 / 0.3) −0.107 = 0.816
Eq. (6-18)
Se = 0.883(0.816)(0.5)(68) = 24.5 kpsi
Selecting the DE-Goodman criteria for a conservative first design,
1/3
Eq. (7-8):
1/ 2
2 1/2



3 ( K T ) 2   
16n   4 ( K f M a ) 
fs
m

 
+
d=


Se
Sut
 π 




1/3
1/ 2
2 1/2



3 ( 2.6 ⋅ 2500 )2  
4
1.9
⋅
2178
(
)
16(1.5)  
 +
 
d=


π 
24 500
68 000





d = 1.57 in
Ans.
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With this diameter, the estimates for notch sensitivity and size factor were conservative,
but close enough for a first iteration until deflections are checked. Check yielding with
this diameter.
1/ 2
Eq. (7-15):
′
σ max
2
 32 K f M a  2
 16 K fsTm  
= 
 + 3
 
3
3
 π d

 π d  
1/2
2
 32(1.9)(2178)  2
 16(2.6)(2500)  
′ = 
σ max
 + 3
 
3
3
 π (1.57)

 π (1.57)
 
′ = 57 /18.4 = 3.1 Ans.
n y = S y / σ max
= 18389 psi = 18.4 kpsi
(e) Now estimate other diameters to provide typical shoulder supports for the gear and
bearings (p. 364). Also, estimate the gear and bearing widths.
(f) Entering this shaft geometry into beam analysis software (or Finite Element software),
the following deflections are determined:
Left bearing slope:
0.000 532 rad
Right bearing slope:
− 0.000 850 rad
Gear slope:
− 0.000 545 rad
Right end of shaft slope:
− 0.000 850 rad
Gear deflection:
− 0.001 45 in
Right end of shaft deflection:
0.005 10 in
Comparing these deflections to the recommendations in Table 7-2, everything is within
typical range except the gear slope is a little high for an uncrowned gear.
(g) To use a non-crowned gear, the gear slope is recommended to be less than 0.0005 rad.
Since all other deflections are acceptable, we will target an increase in diameter only for
the long section between the left bearing and the gear. Increasing this diameter from the
proposed 1.56 in to 1.75 in, produces a gear slope of − 0.000 401 rad. All other
deflections are improved as well.
______________________________________________________________________________
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7-18
(a) Use the distortion-energy elliptic failure locus. The torque and moment loadings on
the shaft are shown in the solution to Prob. 7-17.
Candidate critical locations for strength:
• Left seat keyway
• Right bearing shoulder
• Right keyway
Table A-20 for 1030 HR: Sut = 68 kpsi, S y = 37.5 kpsi, H B = 137
Eq. (6-8):
Se′ = 0.5(68) = 34.0 kpsi
Eq. (6-19):
ka = 2.70(68) −0.265 = 0.883
kc = k d = ke = 1
Left keyway
See Table 7-1 for keyway stress concentration factors,
Kt = 2.14 
 Profile keyway
Kts = 3.0 
For an end-mill profile keyway cutter of 0.010 in radius, estimate notch sensitivities.
Fig. 6-20:
Fig. 6-21:
Eq. (6-32):
q = 0.51
qs = 0.57
K fs = 1 + qs ( K ts − 1) = 1 + 0.57(3.0 − 1) = 2.1
K f = 1 + 0.51(2.14 − 1) = 1.6
−0.107
Eq. (6-20):
Eq. (6-18):
 1.875 
kb = 
= 0.822

 0.30 
Se = 0.883(0.822)(34.0) = 24.7 kpsi
1
Eq. (7-11):
2 2
 1.6(2178)  2
 2.1(2500)  
1
16
=
4 
 + 3
 
n f π (1.8753 )   24 700 
 37 500  

Ans.
nf = 3.5
Right bearing shoulder
The text does not give minimum and maximum shoulder diameters for 03-series bearings
(roller). Use D = 1.75 in.
r 0.030
D 1.75
=
= 0.019,
=
= 1.11
d 1.574
d 1.574
Fig. A-15-9:
Fig. A-15-8:
Kt = 2.4
K ts = 1.6
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Fig. 6-20:
Fig. 6-21:
Eq. (6-32):
q = 0.65
qs = 0.70
K f = 1 + 0.65(2.4 − 1) = 1.91
K fs = 1 + 0.70(1.6 − 1) = 1.42
 0.453 
M = 2178 
 = 493 lbf ⋅ in
 2 
1/ 2
Eq. (7-11):
2
  1.91(493)  2
 1.42(2500)  
1
16
=
4 
 + 3
 
n f π (1.5743 )   24 700 
 37 500  

Ans.
nf = 4.2
Right keyway
Use the same stress concentration factors as for the left keyway. There is no bending
moment, thus Eq. (7-11) reduces to:
1 16 3K fsTm 16 3(2.1)(2500)
=
=
π d 3S y
nf
π (1.53 ) (37 500)
nf = 2.7
Ans.
Yielding
Check for yielding at the left keyway, where the completely reversed bending is
maximum, and the steady torque is present. Using Eq. (7-15), with Mm = Ta = 0,
1/2
′
σ max
2
 32 K f M a 2
 16 K fsTm  
= 
 + 3
 
3
3
 π d

 π d  
1/ 2
2
 32 1.6 2178 2
 16 ( 2.1)( 2500 )  
(
)(
)
 + 3
 
= 
 π (1.875 )3  
 π (1.875 )3 


 

= 8791 psi = 8.79 kpsi
S
37.5
Ans.
ny = y =
= 4.3
′
σ max
8.79
Check in smaller diameter at right end of shaft where only steady torsion exists.
1/2
′
σ max
  16 K fsTm  2 
= 3 
 
3
  π d  
1/2
  16 2.1 2500 2 
( )(
) 
= 3 
3
 
 
π (1.5 )
 
 
= 13 722 psi = 13.7 kpsi
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ny =
Sy
37.5
=
= 2.7
′
σ max
13.7
Ans.
(b) One could take pains to model this shaft exactly, using finite element software.
However, for the bearings and the gear, the shaft is basically of uniform diameter, 1.875
in. The reductions in diameter at the bearings will change the results insignificantly. Use
E = 30 Mpsi for steel.
To the left of the load, from Table A-9, case 6,
θ AB =
d y AB
Fb
1449(2)(3 x 2 + 2 2 − 112 )
=
(3 x 2 + b 2 − l 2 ) =
6 EIl
6(30)(106 )(π / 64)(1.8754 )(11)
dx
= 2.4124(10 −6 )(3 x 2 − 117)
At x = 0 in:
θ = −2.823(10−4 ) rad
θ = 3.040(10−4 ) rad
At x = 9 in:
To the right of the load, from Table A-9, case 6,
θ BC =
(
d y BC
Fa
=
−3 x 2 + 6 xl − 2l 2 − a 2
dx
6 EIl
)
At x = l = 11 in:
θ=
Fa 2
1449(9)(112 − 92 )
l − a2 =
= 4.342(10−4 ) rad
6 EIl
6(30)(106 )(π / 64)(1.8754 )(11)
(
)
Obtain allowable slopes from Table 7-2.
Left bearing:
n fs =
Allowable slope
0.001
=
= 3.5
Actual slope
0.000 282 3
n fs =
0.0008
= 1.8
0.000 434 2
Ans.
Right bearing:
Ans.
Gear mesh slope:
Table 7-2 recommends a minimum relative slope of 0.0005 rad. While we don’t know the
slope on the next shaft, we know that it will need to have a larger diameter and be stiffer.
At the moment we can say
0.0005
= 1.6
Ans.
0.000 304
______________________________________________________________________________
n fs <
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7-19
The most likely critical locations for fatigue are at locations where the bending moment is
high, the cross section is small, stress concentration exists, and torque exists. The twoplane bending moment diagrams, shown in the solution to Prob. 3-72, indicate decreasing
moments in both planes to the left of A and to the right of C, with combined values at A
and C of MA = 5324 lbf∙in and MC = 6750 lbf∙in. The torque is constant between A and
B, with T = 2819 lbf∙in. The most likely critical locations are at the stress concentrations
near A and C. The two shoulders near A can be eliminated since the shoulders near C
have the same geometry but a higher bending moment. We will consider the following
potentially critical locations:
• keyway at A
• shoulder to the left of C
• shoulder to the right of C
Table A-20:
Eq. (6-8):
Sut = 64 kpsi, Sy = 54 kpsi
Se′ = 0.5(64) = 32.0 kpsi
Eq. (6-19):
ka = 2.70(64) −0.265 = 0.897
kc = k d = ke = 1
Keyway at A
Assuming r / d = 0.02 for typical end-milled keyway cutter (p. 365), with d = 1.75 in,
r = 0.02d = 0.035 in.
Table 7-1:
Kt = 2.14, Kts = 3.0
Fig. 6-20:
q = 0.65
Fig. 6-21:
qs = 0.71
Eq. (6-32):
K f = 1 + q ( K t − 1) = 1 + 0.65(2.14 − 1) = 1.7
K fs = 1 + qs ( K ts − 1) = 1 + 0.71(3.0 − 1) = 2.4
−0.107
Eq. (6-20):
Eq. (6-18):
 1.75 
= 0.828
kb = 

 0.30 
Se = 0.897(0.828)(32) = 23.8 kpsi
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We will choose the DE-Gerber criteria since this is an analysis problem in which we
would like to evaluate typical expectations.
Using Eq. (7-9) with Mm = Ta = 0,
A = 4 ( K f M a ) = 4 (1.7 )( 5324 )  = 18102 lbf ⋅ in = 18.10 kip ⋅ in
2
2
B = 3 ( K fsTm ) = 3 ( 2.4 )( 2819 )  = 11 718 lbf ⋅ in = 11.72 kip ⋅ in
2
2
2 1/2 

1
8 A    2 BSe   
=
1 + 1 + 
  
n π d 3 Se    ASut   
 
 
2 1/ 2 
 
 


2
11.72
23.8
(
)(
)
 
1
1
=
+
+

  

π (1.753 ) ( 23.8 )    (18.10 )( 64 )   
 
 
n = 1.3
8 (18.10 )
Shoulder to the left of C
r / d = 0.0625 / 1.75 = 0.036, D / d = 2.5 / 1.75 = 1.43
Fig. A-15-9:
Fig. A-15-8:
Fig. 6-20:
Fig. 6-21:
Eq. (6-32):
Kt = 2.2
Kts = 1.8
q = 0.71
qs = 0.76
K f = 1 + q ( K t − 1) = 1 + 0.71(2.2 − 1) = 1.9
K fs = 1 + qs ( K ts − 1) = 1 + 0.76(1.8 − 1) = 1.6
−0.107
Eq. (6-20):
Eq. (6-18):
 1.75 
kb = 
= 0.828

 0.30 
Se = 0.897(0.828)(32) = 23.8 kpsi
For convenience, we will use the full value of the bending moment at C, even though it
will be slightly less at the shoulder. Using Eq. (7-9) with Mm = Ta = 0,
A = 4 ( K f M a ) = 4 (1.9 )( 6750 )  = 25 650 lbf ⋅ in = 25.65 kip ⋅ in
2
2
B = 3 ( K fsTm ) = 3 (1.6 )( 2819 )  = 7812 lbf ⋅ in = 7.812 kip ⋅ in
2
2
2 1/2 

1
8 A    2 BSe   
=
1 + 1 + 
  
n π d 3 Se    ASut   

 

2 1/2 
 
   2 ( 7.812 )( 23.8 )   
=

1 + 1 + 

π (1.753 ) ( 23.8 )    ( 25.65 )( 64 )   




8 ( 25.65 )
Shigley’s MED, 10th edition
Chapter 7 Solutions, Page 18/45
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n = 0.96
Shoulder to the right of C
r / d = 0.0625 / 1.3 = 0.048, D / d = 1.75 / 1.3 = 1.35
Fig. A-15-9:
Fig. A-15-8:
Fig. 6-20:
Fig. 6-21:
Eq. (6-32):
Kt = 2.0
Kts = 1.7
q = 0.71
qs = 0.76
K f = 1 + q ( K t − 1) = 1 + 0.71(2.0 − 1) = 1.7
K fs = 1 + qs ( K ts − 1) = 1 + 0.76(1.7 − 1) = 1.5
−0.107
 1.3 
= 0.855
kb = 

 0.30 
Eq. (6-18):
Se = 0.897(0.855)(32) = 24.5 kpsi
For convenience, we will use the full value of the bending moment at C, even though it
will be slightly less at the shoulder. Using Eq. (7-9) with Mm = Ta = 0,
Eq. (6-20):
A = 4 ( K f M a ) = 4 (1.7 )( 6750 )  = 22 950 lbf ⋅ in = 22.95 kip ⋅ in
2
2
B = 3 ( K fsTm ) = 3 (1.5 )( 2819 )  = 7324 lbf ⋅ in = 7.324 kip ⋅ in
2
2
2 1/2 

1
8 A    2 BSe   
=
1 + 1 + 
  
n π d 3 Se    ASut   

 

2 1/2 
 
   2 ( 7.324 )( 24.5 )   
=

1 + 1 + 

π (1.33 ) ( 24.5 )    ( 22.95 )( 64 )   
 
 
n = 0.45
8 ( 22.95 )
The critical location is at the shoulder to the right of C, where n = 0.45 and finite life is
predicted. Ans.
Though not explicitly called for in the problem statement, a static check for yielding is
especially warranted with such a low fatigue factor of safety. Using Eq. (7-15), with
Mm = Ta = 0,
1/ 2
′
σ max
2
 32 K f M a  2
 16 K fsTm  
= 
 + 3
 
3
3
 π d

 π d  
1/ 2
2
2


 16 (1.5 )( 2819 )  
32
1.7
6750
(
)(
)
 + 3
 
= 
3
3


 

π
π
1.3
1.3
(
)
(
)


 

Shigley’s MED, 10th edition
= 55 845 psi = 55.8 kpsi
Chapter 7 Solutions, Page 19/45
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n=
Sy
54
=
= 0.97
′
σ max
55.8
This indicates localized yielding is predicted at the stress-concentration, though after
localized cold-working it may not be a problem. The finite fatigue life is still likely to be
the failure mode that will dictate whether this shaft is acceptable.
It is interesting to note the impact of stress concentration on the acceptability of the
proposed design. This problem is linked with several previous problems (see Table 1-2,
p. 34) in which the shaft was considered to have a constant diameter of 1.25 in. In each of
the previous problems, the 1.25 in diameter was more than adequate for deflection, static,
and fatigue considerations. In this problem, even though practically the entire shaft has
diameters larger than 1.25 in, the stress concentrations significantly reduce the
anticipated fatigue life.
______________________________________________________________________________
7-20
For a shaft with significantly varying diameters over its length, we will choose to use
shaft analysis software or finite element software to calculate the deflections. Entering
the geometry from the shaft as defined in Prob. 7-19, and the loading as defined in Prob.
3-72, the following deflection magnitudes are determined:
Location
Left bearing O
Right bearing C
Left Gear A
Right Gear B
Slope Deflection
(rad)
(in)
0.00640 0.00000
0.00434 0.00000
0.00260 0.04839
0.01078 0.07517
Comparing these values to the recommended limits in Table 7-2, we find that they are all
out of the desired range. This is not unexpected since the stress analysis of Prob. 7-19
also indicated the shaft is undersized for infinite life. The slope at the right gear is the
most excessive, so we will attempt to increase all diameters to bring it into compliance.
Using Eq. (7-18) at the right gear,
d new nd ( dy / dx )old
=
d old
( slope )all
1/4
=
(1)(0.01078)
0.0005
1/4
= 2.15
Multiplying all diameters by 2.15, we obtain the following deflections:
Location
Left bearing O
Right bearing C
Left Gear A
Right Gear B
Slope
(rad)
0.00030
0.00020
0.00012
0.00050
Deflection
(in)
0.00000
0.00000
0.00225
0.00350
Shigley’s MED, 10th edition
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This brings the slope at the right gear just to the limit for an uncrowned gear, and all
other slopes well below the recommended limits. For the gear deflections, the values are
below recommended limits as long as the diametral pitch is less than 20.
______________________________________________________________________________
7-21
The most likely critical locations for fatigue are at locations where the bending moment is
high, the cross section is small, stress concentration exists, and torque exists. The twoplane bending moment diagrams, shown in the solution to Prob. 3-73, indicate both
planes have a maximum bending moment at B. At this location, the combined bending
moment from both planes is M = 4097 N∙m, and the torque is T = 3101 N∙m. The
shoulder to the right of B will be eliminated since its diameter is only slightly smaller,
and there is no torque. Comparing the shoulder to the left of B with the keyway at B, the
primary difference between the two is the stress concentration, since they both have
essentially the same bending moment, torque, and size. We will check the stress
concentration factors for both to determine which is critical.
Table A-20:
Sut = 440 MPa, Sy = 370 MPa
Keyway at A
Assuming r / d = 0.02 for typical end-milled keyway cutter (p. 365), with d = 50 mm,
r = 0.02d = 1 mm.
Table 7-1:
Fig. 6-20:
Fig. 6-21:
Eq. (6-32):
Kt = 2.14, Kts = 3.0
q = 0.66
qs = 0.72
K f = 1 + q ( K t − 1) = 1 + 0.66(2.14 − 1) = 1.8
K fs = 1 + qs ( K ts − 1) = 1 + 0.72(3.0 − 1) = 2.4
Shoulder to the left of B
r / d = 2 / 50 = 0.04, D / d = 75 / 50 = 1.5
Fig. A-15-9:
Kt = 2.2
Shigley’s MED, 10th edition
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Fig. A-15-8:
Fig. 6-20:
Fig. 6-21:
Eq. (6-32):
Kts = 1.8
q = 0.73
qs = 0.78
K f = 1 + q ( K t − 1) = 1 + 0.73(2.2 − 1) = 1.9
K fs = 1 + qs ( K ts − 1) = 1 + 0.78(1.8 − 1) = 1.6
Examination of the stress concentration factors indicates the keyway will be the critical
location.
Eq. (6-8):
Se′ = 0.5(440) = 220 MPa
Eq. (6-19):
ka = 4.51(440) −0.265 = 0.899
−0.107
Eq. (6-20):
Eq. (6-18):
 50 
= 0.818
kb = 

 7.62 
kc = k d = ke = 1
Se = 0.899(0.818)(220) = 162 MPa
We will choose the DE-Gerber criteria since this is an analysis problem in which we
would like to evaluate typical expectations. Using Eq. (7-9) with Mm = Ta = 0,
A = 4 ( K f M a ) = 4 (1.8 )( 4097 )  = 14 750 N ⋅ m
2
2
B = 3 ( K fsTm ) = 3 ( 2.4 )( 3101)  = 12 890 N ⋅ m
2
2
2 1/2 

   2 BSe   
1 + 1 + 
  
   ASut   
2 1/2 
 
6

8 (14 750 )
   2 (12 890 )(162 ) (10 )   


1
1
=
+
+


π ( 0.0503 ) (162 ) (106 )    (14 750 )( 440 ) (106 )   
 
 
n = 0.25
Infinite life is not predicted.
Ans.
1
8A
=
n π d 3 Se
Though not explicitly called for in the problem statement, a static check for yielding is
especially warranted with such a low fatigue factor of safety. Using Eq. (7-15), with
Mm = Ta = 0,
1/2
′
σ max
2
 32 K f M a 2
 16 K fsTm  
= 
 + 3
 
3
3
 π d

 π d  
1/ 2
2
 32 1.8 4097 2
 16 ( 2.4 )( 3101)  
(
)(
)
 + 3
 
= 
 π ( 0.050 )3  
 π ( 0.050 )3 


 

( )
= 7.98 108 Pa = 798 MPa
Shigley’s MED, 10th edition
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n=
Sy
370
=
= 0.46
′
σ max
798
This indicates localized yielding is predicted at the stress-concentration. Even without
the stress concentration effects, the static factor of safety turns out to be 0.93. Static
failure is predicted, rendering this proposed shaft design unacceptable.
This problem is linked with several previous problems (see Table 1-2, p. 34) in which the
shaft was considered to have a constant diameter of 50 mm. The results here are
consistent with the previous problems, in which the 50 mm diameter was found to
slightly undersized for static, and significantly undersized for fatigue. Though in the
current problem much of the shaft has larger than 50 mm diameter, the added
contribution of stress concentration limits the fatigue life.
______________________________________________________________________________
7-22
For a shaft with significantly varying diameters over its length, we will choose to use
shaft analysis software or finite element software to calculate the deflections. Entering
the geometry from the shaft as defined in Prob. 7-21, and the loading as defined in Prob.
3-73, the following deflection magnitudes are determined:
Location
Slope
(rad)
0.01445
0.01843
0.00358
0.00366
Left bearing O
Right bearing C
Left Gear A
Right Gear B
Deflection
(mm)
0.000
0.000
3.761
3.676
Comparing these values to the recommended limits in Table 7-2, we find that they are all
well out of the desired range. This is not unexpected since the stress analysis in Prob.
7-21 also indicated the shaft is undersized for infinite life. The transverse deflection at
the left gear is the most excessive, so we will attempt to increase all diameters to bring it
into compliance. Using Eq. (7-17) at the left gear, assuming from Table 7-2 an allowable
deflection of yall = 0.01 in = 0.254 mm,
d new nd yold
=
d old
yall
1/4
(1)(3.761)
=
0.254
1/4
= 1.96
Multiplying all diameters by 2, we obtain the following deflections:
Location
Left bearing O
Right bearing C
Left Gear A
Right Gear B
Slope
(rad)
0.00090
0.00115
0.00022
0.00023
Deflection
(mm)
0.000
0.000
0.235
0.230
Shigley’s MED, 10th edition
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This brings the deflection at the gears just within the limit for a spur gear (assuming P <
10 teeth/in), and all other deflections well below the recommended limits.
______________________________________________________________________________
7-23
(a) Label the approximate locations of the effective centers of the bearings as A and B,
the fan as C, and the gear as D, with axial dimensions as shown. Since there is only one
gear, we can combine the radial and tangential gear forces into a single resultant force
with an accompanying torque, and handle the statics problem in a single plane. From
statics, the resultant reactions at the bearings can be found to be RA = 209.9 lbf and RB =
464.5 lbf. The bending moment and torque diagrams are shown, with the maximum
bending moment at D of MD = 209.9(6.98) = 1459 lbf∙in and a torque transmitted from D
to C of T = 633 (8/2) = 2532 lbf∙in. Due to the shaft rotation, the bending stress on any
stress element will be completely reversed, while the torsional stress will be steady.
Since we do not have any information about the fan, we will ignore any axial load that it
would introduce. It would not likely contribute much compared to the bending anyway.
Potentially critical locations are identified as follows:
• Keyway at C, where the torque is high, the diameter is small, and the keyway creates
a stress concentration.
Shigley’s MED, 10th edition
Chapter 7 Solutions, Page 24/45
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•
•
•
•
Keyway at D, where the bending moment is maximum, the torque is high, and the
keyway creates a stress concentration.
Groove at E, where the diameter is smaller than at D, the bending moment is still
high, and the groove creates a stress concentration. There is no torque here, though.
Shoulder at F, where the diameter is smaller than at D or E, the bending moment is
still moderate, and the shoulder creates a stress concentration. There is no torque
here, though.
The shoulder to the left of D can be eliminated since the change in diameter is very
slight, so that the stress concentration will undoubtedly be much less than at D.
Table A-20:
Eq. (6-8):
Sut = 68 kpsi, Sy = 57 kpsi
Se′ = 0.5(68) = 34.0 kpsi
Eq. (6-19):
ka = 2.70(68) −0.265 = 0.883
Keyway at C
Since there is only steady torsion here, only a static check needs to be performed. We’ll
use the maximum shear stress theory.
τ=
Eq. (5-3):
Tr 2532 (1.00 / 2 )
=
= 12.9 kpsi
J
π (1.004 ) / 32
ny =
Sy / 2
τ
=
57 / 2
= 2.21
12.9
Keyway at D
Assuming r / d = 0.02 for typical end-milled keyway cutter (p. 365), with d = 1.75 in,
r = 0.02d = 0.035 in.
Table 7-1:
Fig. 6-20:
Fig. 6-21:
Eq. (6-32):
Kt = 2.14, Kts = 3.0
q = 0.66
qs = 0.72
K f = 1 + q ( K t − 1) = 1 + 0.66(2.14 − 1) = 1.8
K fs = 1 + qs ( K ts − 1) = 1 + 0.72(3.0 − 1) = 2.4
−0.107
Eq. (6-20):
Eq. (6-18):
 1.75 
= 0.828
kb = 

 0.30 
Se = 0.883(0.828)(34.0) = 24.9 kpsi
We will choose the DE-Gerber criteria since this is an analysis problem in which we
would like to evaluate typical expectations.
Using Eq. (7-9) with Mm = Ta = 0,
Shigley’s MED, 10th edition
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A = 4 ( K f M a ) = 4 (1.8 )(1459 )  = 5252 lbf ⋅ in = 5.252 kip ⋅ in
2
2
B = 3 ( K fsTm ) = 3 ( 2.4 )( 2532 )  = 10 525 lbf ⋅ in = 10.53 kip ⋅ in
2
1
8A
=
n π d 3 Se
2
2 1/2 
 
 2 BSe   

1 + 1 + 
  
   ASut   
2 1/ 2 
 
   2 (10.53)( 24.9 )   
=

1 + 1 + 

π (1.753 ) ( 24.9 )    ( 5.252 )( 68 )   




n = 3.59
Ans.
8 ( 5.252 )
Groove at E
We will assume Figs. A-15-14 is applicable since the 2 in diameter to the right of the
groove is relatively narrow and will likely not allow the stress flow to fully develop. (See
Fig.7-9 for the stress flow concept.)
r / d = 0.1 / 1.55 = 0.065,
D / d = 1.75 / 1.55 = 1.13
Fig. A-15-14: Kt = 2.1
q = 0.76
Fig. 6-20:
Eq. (6-32):
K f = 1 + q ( K t − 1) = 1 + 0.76(2.1 − 1) = 1.8
−0.107
Eq. (6-20):
Eq. (6-18):
 1.55 
= 0.839
kb = 

 0.30 
Se = 0.883(0.839)(34) = 25.2 kpsi
Using Eq. (7-9) with Mm = Ta = Tm = 0,
A = 4 ( K f M a ) = 4 (1.8 )(1115 )  = 4122 lbf ⋅ in = 4.122 kip ⋅ in
2
2
B=0
2 1/2 
 
 2 BSe   

1 + 1 + 
  
   ASut   
1/2
8 ( 4.122 )
1 + ( 0 ) 2 
=
+
1


π (1.553 ) ( 25.2 )
1
8A
=
n π d 3 Se
{
n = 4.47
}
Ans.
Shoulder at F
Fig. A-15-9:
r / d = 0.125 / 1.40 = 0.089,
Kt = 1.7
D / d = 2.0 / 1.40 = 1.43
Shigley’s MED, 10th edition
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Fig. 6-20:
Eq. (6-32):
q = 0.78
K f = 1 + q ( K t − 1) = 1 + 0.78(1.7 − 1) = 1.5
−0.107
Eq. (6-20):
Eq. (6-18):
 1.40 
= 0.848
kb = 

 0.30 
Se = 0.883(0.848)(34) = 25.5 kpsi
Using Eq. (7-9) with Mm = Ta = Tm = 0,
A = 4 ( K f M a ) = 4 (1.5 )( 845 )  = 2535 lbf ⋅ in = 2.535 kip ⋅ in
2
2
B=0
2 1/2 
 
 2 BSe   

1 + 1 + 
  
   ASut   
1/2
8 ( 2.535 )
1 + ( 0 ) 2 
1
=
+


π (1.403 ) ( 25.5 )
1
8A
=
n π d 3 Se
{
n = 5.42
}
Ans.
(b) The deflection will not be much affected by the details of fillet radii, grooves, and
keyways, so these can be ignored. Also, the slight diameter changes, as well as the
narrow 2.0 in diameter section, can be neglected. We will model the shaft with the
following three sections:
Section Diameter Length
(in)
(in)
1
1.00
2.90
2
1.70
7.77
3
1.40
2.20
The deflection problem can readily (though tediously) be solved with singularity
functions. For examples, see Ex. 4-7, p. 173, or the solution to Prob. 7-24. Alternatively,
shaft analysis software or finite element software may be used. Using any of the
methods, the results should be as follows:
Location
Left bearing A
Right bearing B
Fan C
Gear D
Slope
Deflection
(in)
(rad)
0.000290 0.000000
0.000400 0.000000
0.000290 0.000404
0.000146 0.000928
Shigley’s MED, 10th edition
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Comparing these values to the recommended limits in Table 7-2, we find that they are all
within the recommended range.
______________________________________________________________________________
7-24
Shaft analysis software or finite element software can be utilized if available. Here we
will demonstrate how the problem can be simplified and solved using singularity
functions.
Deflection: First we will ignore the steps near the bearings where the bending moments
are low. Thus let the 30 mm dia. be 35 mm. Secondly, the 55 mm dia. is very thin, 10
mm. The full bending stresses will not develop at the outer fibers so full stiffness will not
develop either. Thus, ignore this step and let the diameter be 45 mm.
Statics: Left support: R1 = 7(315 − 140) / 315 = 3.889 kN
Right support: R2 = 7(140) / 315 = 3.111 kN
Determine the bending moment at each step.
x(mm)
0
40
100
140
210
275
315
M(N · m) 0 155.56 388.89 544.44 326.67 124.44 0
I35 = (π/64)(0.0354) = 7.366(10-8) m4, I40 = 1.257(10-7) m4, I45 = 2.013(10-7) m4
Plot M/I as a function of x.
x(m)
0
0.04
0.04
0.1
0.1
0.14
0.14
0.21
0.21
0.275
0.275
0.315
M/I (109 N/m3)
0
2.112
1.2375
3.094
1.932
2.705
2.705
1.623
2.6
0.99
1.6894
0
Step
Slope
52.8
∆Slope
–0.8745
30.942
–21.86
–1.162
19.325
–11.617
0
–15.457
–34.78
0.977
-24.769
-9.312
0.6994
-42.235
-17.47
Shigley’s MED, 10th edition
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The steps and the change of slopes are evaluated in the table. From these, the function
M/I can be generated:
0
1
0
M / I = 52.8 x − 0.8745 x − 0.04 − 21.86 x − 0.04 − 1.162 x − 0.1

1
1
0
−11.617 x − 0.1 − 34.78 x − 0.14 + 0.977 x − 0.21
1
−9.312 x − 0.21 + 0.6994 x − 0.275
0
1
− 17.47 x − 0.275  109

Integrate twice:
E
dy 
1
2
1
= 26.4 x 2 − 0.8745 x − 0.04 − 10.93 x − 0.04 − 1.162 x − 0.1

dx
2
2
1
−5.81 x − 0.1 − 17.39 x − 0.14 + 0.977 x − 0.21
2
+ C1  109 (1)

2
− 0.581 x − 0.1
1
−4.655 x − 0.21 + 0.6994 x − 0.275 − 8.735 x − 0.275
2
3
Ey = 8.8x3 − 0.4373 x − 0.04 − 3.643 x − 0.04

3
3
−1.937 x − 0.1 − 5.797 x − 0.14 + 0.4885 x − 0.21
3
−1.552 x − 0.21 + 0.3497 x − 0.275
2
2
2
− 2.912 x − 0.275
3
+ C1x + C2  109

Boundary conditions: y = 0 at x = 0 yields C2 = 0;
y = 0 at x = 0.315 m yields C1 = –0.295 25 N/m2.
Equation (1) with C1 = –0.295 25 provides the slopes at the bearings and gear. The
following table gives the results in the second column. The third column gives the results
from a similar finite element model. The fourth column gives the results of a full model
which models the 35 and 55 mm diameter steps.
x (mm)
0
140
315
θ (rad)
–0.001 4260
–0.000 1466
0.001 3120
F.E. Model
–0.001 4270
–0.000 1467
0.001 3280
Shigley’s MED, 10th edition
Full F.E. Model
–0.001 4160
–0.000 1646
0.001 3150
Chapter 7 Solutions, Page 29/45
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The main discrepancy between the results is at the gear location (x = 140 mm). The larger
value in the full model is caused by the stiffer 55 mm diameter step. As was stated
earlier, this step is not as stiff as modeling implicates, so the exact answer is somewhere
between the full model and the simplified model which in any event is a small value. As
expected, modeling the 30 mm dia. as 35 mm does not affect the results much.
It can be seen that the allowable slopes at the bearings are exceeded. Thus, either the load
has to be reduced or the shaft “beefed” up. If the allowable slope is 0.001 rad, then the
maximum load should be Fmax = (0.001/0.001 426)7 = 4.91 kN. With a design factor this
would be reduced further.
To increase the stiffness of the shaft, apply Eq. (7-18) to the most offending deflection (at
x = 0) to determine a multiplier to be used for all diameters.
d new nd ( dy / dx )old
=
d old
( slope )all
1/4
=
(1)(0.0014260)
0.001
1/4
= 1.093
Form a table:
Old d, mm
20.00 30.00 35.00 40.00 45.00 55.00
New ideal d, mm
21.86 32.79 38.26 43.72 49.19 60.12
Rounded up d, mm 22.00 34.00 40.00 44.00 50.00 62.00
Repeating the full finite element model results in
x = 0:
θ = –9.30 × 10-4 rad
x = 140 mm: θ = –1.09 × 10-4 rad
x = 315 mm: θ = 8.65 × 10-4 rad
This is well within our goal. Have the students try a goal of 0.0005 rad at the gears.
Strength: Due to stress concentrations and reduced shaft diameters, there are a number of
locations to look at. A table of nominal stresses is given below. Note that torsion is only
to the right of the 7 kN load. Using σ = 32M/(πd3) and τ = 16T/(πd3),
x (mm)
σ (MPa)
τ (MPa)
σ ′(MPa)
40
100
110
140
210
275
300
330
0 15
0
0 22.0 37.0 61.9 47.8 60.9 52.0 39.6 17.6
0 0
0
0
0
6
8.5 12.7 20.2 68.1
0 22.0 37.0 61.9 47.8 61.8 53.1 45.3 39.2 118.0
Table A-20 for AISI 1020 CD steel: Sut = 470 MPa, Sy = 390 MPa
At x = 210 mm:
Eq. (6-19):
ka = 4.51(470) −0.265 = 0.883
Shigley’s MED, 10th edition
Chapter 7 Solutions, Page 30/45
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Eq. (6-20):
Eq. (6-18):
Fig. A-15-8:
Fig. A-15-9:
Fig. 6-20:
Fig. 6-21:
Eq. (6-32):
kb = (40 / 7.62) −0.107 = 0.837
Se = 0.883 (0.837)(0.5)(470) = 174 MPa
D / d = 45 / 40 = 1.125, r / d = 2 / 40 = 0.05
Kts = 1.4
Kt = 1.9
q = 0.75
qs = 0.79
Kf = 1 + 0.75(1.9 –1) = 1.68
Kf s = 1 + 0.79(1.4 – 1) = 1.32
Choosing DE-ASME Elliptic to inherently include the yield check, from Eq. (7-11), with
Mm = Ta = 0,
1/2
2
2
 
1.32(107)  
1
16
 1.68(326.67) 
 +3
 
=
4 
6
n π 0.043   174 106 
390
10

 


 

n = 1.98
(
)
( )
( )
At x = 330 mm:
The von Mises stress is the highest but it comes from the steady torque only.
Fig. A-15-9:
Fig. 6-21:
Eq. (6-32):
Eq. (7-11):
D / d = 30 / 20 = 1.5, r / d = 2 / 20 = 0.1
Kts = 1.42
qs = 0.79
Kf s = 1 + 0.79(1.42 – 1) = 1.33
1
16
=
n π 0.023
(



() 3 ) 1.33(107)
390 (10 ) 

6

n = 2.49
Note that since there is only a steady torque, Eq. (7-11) reduces to essentially the
equivalent of the distortion energy failure theory.
Check the other locations.
If worse-case is at x = 210 mm, the changes discussed for the slope criterion will
improve the strength issue.
______________________________________________________________________________
7-25 and 7-26 With these design tasks each student will travel different paths and almost all
details will differ. The important points are
• The student gets a blank piece of paper, a statement of function, and some constraints
– explicit and implied. At this point in the course, this is a good experience.
• It is a good preparation for the capstone design course.
Shigley’s MED, 10th edition
Chapter 7 Solutions, Page 31/45
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•
The adequacy of their design must be demonstrated and possibly include a designer’s
notebook.
• Many of the fundaments of the course, based on this text and this course, are useful.
The student will find them useful and notice that he/she is doing it.
• Don’t let the students create a time sink for themselves. Tell them how far you want
them to go.
______________________________________________________________________________
7-27
This task was once given as a final exam problem. This problem is a learning experience.
Following the task statement, the following guidance was added.
•
•
•
Take the first half hour, resisting the temptation of putting pencil to paper, and decide
what the problem really is.
Take another twenty minutes to list several possible remedies.
Pick one, and show your instructor how you would implement it.
The students’ initial reaction is that he/she does not know much from the problem
statement. Then, slowly the realization sets in that they do know some important things
that the designer did not. They knew how it failed, where it failed, and that the design
wasn’t good enough; it was close, though.
Also, a fix at the bearing seat lead-in could transfer the problem to the shoulder fillet, and
the problem may not be solved.
To many students’ credit, they chose to keep the shaft geometry, and selected a new
material to realize about twice the Brinell hardness.
______________________________________________________________________________
7-28
In Eq. (7-22) set
πd4
πd2
I=
, A=
64
4
to obtain
 π   d  gE
ω =   
 l  4 γ
2
(1)
or
d=
4l 2ω
γ
π
gE
2
(2)
(a) From Eq. (1) and Table A-5
9
 π   0.025  9.81(207)(10 )
= 883 rad/s
 

76.5 103
 0.6   4 
2
ω =
( )
Shigley’s MED, 10th edition
Ans.
Chapter 7 Solutions, Page 32/45
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(b) From Eq. (1), we observe that the critical speed is linearly proportional to the
diameter. Thus, to double the critical speed, we should double the diameter to d = 50
mm.
Ans.
(c) From Eq. (2),
lω =
π2 d
gE
4 l
γ
Since d / l is the same regardless of the scale,
lω = constant = 0.6(883) = 529.8
529.8
ω=
= 1766 rad/s Ans.
0.3
Thus the first critical speed doubles.
______________________________________________________________________________
7-29
From Prob. 7-28, ω = 883 rad/s
A = 4.909 (10 −4 ) m 2 , I = 1.917 (10 −8 ) m 4 ,
E = 207(109 ) Pa,
(
)
γ = 7.65 (104 ) N/m 3
( )
w = Aγ l = 4.909 10 −4 7.65 10 4 (0.6) = 22.53 N
One element:
Eq. (7-24):
(
0.3(0.3) 0.62 − 0.32 − 0.32
)
( )
( ) (0.6)
y = w δ = 22.53(1.134) (10 ) = 2.555 (10 ) m
y = 6.528 (10 )
Σw y = 22.53(2.555) (10 ) = 5.756 (10 )
Σw y = 22.53(6.528) (10 ) = 1.471(10 )
δ11 =
( )
9
6(207) 10 (1.917) 10
= 1.134 10−6 m/N
−8
−6
1
−5
1 11
−10
2
1
−5
2
−4
−10
−8
(
(
)
)
5.756 10−4
Σw y
= 9.81
= 620 rad/s
ω1 = g
Σw y 2
1.471 10−8
(30% low)
Two elements:
Shigley’s MED, 10th edition
Chapter 7 Solutions, Page 33/45
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δ11 = δ 22 =
δ12 = δ 21 =
(
) = 6.379 10 m/N
( )
(0.6)
0.45(0.15) 0.62 − 0.452 − 0.152
( )
(
9
6(207) 10 (1.917) 10
−8
)
−7
0.15(0.15)(0.6 − 0.15 − 0.152 )
= 4.961 10−7 m/N
9
−8
6(207) 10 (1.917) 10 (0.6)
2
2
( )
(
(
)
(
)
)
(
)
(
)
y1 = y2 = w1δ11 + w2δ12 = 11.265(6.379) 10 −7 + 11.265(4.961) 10 −7 = 1.277 10−5 m
(
y12 = y22 = 1.632 10 −10
)
(
)
(
Σw y = 2(11.265)(1.277) 10 −5 = 2.877 10 −4
(
Σw y = 2(11.265)(1.632) 10
2
(
(
 2.877 10−4
ω1 = 9.81 
 3.677 10−9
−10
)
) = 3.677 (10 )
−9
)  = 876 rad/s
) 
(0.8% low)
Three elements:
δ11 = δ 33 =
(
0.5(0.1) 0.62 − 0.52 − 0.12
( )
(
9
6(207) 10 (1.917) 10
(
0.3(0.3) 0.62 − 0.32 − 0.32
δ 22 =
( )
(
9
6(207) 10 (1.917) 10
δ12 = δ 32 =
(
−8
−8
)
) (0.6)
) (0.6)
( )
(
6(207) 10 (1.917) 10
(
0.1(0.1) 0.62 − 0.12 − 0.12
−8
)
)
(
)
= 3.500 10 −7 m/N
(
)
= 1.134 10−6 m/N
0.3(0.1) 0.62 − 0.32 − 0.12
9
)
)
(0.6)
(
)
= 5.460 10−7 m/N
( )
( )
( ) (0.6)
y = 7.51 3.500 (10 ) + 5.460 (10 ) + 2.380 (10 )  = 8.516 (10 )
y = 7.51 5.460 (10 ) + 1.134 (10 ) + 5.460 (10 )  = 1.672 (10 )
y = 7.51  2.380 (10 ) + 5.460 (10 ) + 3.500 (10 )  = 8.516 (10 )
Σw y = 7.51 8.516 (10 ) + 1.672 (10 ) + 8.516 (10 )  = 2.535 (10 )
Σw y = 7.51{8.516 (10 )  + 1.672 (10 )  + 8.516 (10 )  } = 3.189 (10 )
δ13 =
9
6(207) 10 (1.917) 10
= 2.380 10−7 m/N
−8
−7
−7
−7
−6
−7
−6
−7
−5
−7
−7
−7
−6
1
2
3
−6
2
−5
−6
2
−6
−5
2
Shigley’s MED, 10th edition
−4
−6
2
−9
Chapter 7 Solutions, Page 34/45
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(
(
 2.535 10−4
ω1 = 9.81 
 3.189 10−9
)  = 883 rad/s
) 
The result is the same as in Prob. 7-28. The point was to show that convergence is rapid
using a static deflection beam equation. The method works because:
• If a deflection curve is chosen which meets the boundary conditions of momentfree and deflection-free ends, as in this problem, the strain energy is not very
sensitive to the equation used.
• Since the static bending equation is available, and meets the moment-free and
deflection-free ends, it works.
______________________________________________________________________________
7-30
(a) For two bodies, Eq. (7-26) is
(m1δ11 − 1/ ω 2 )
m2δ12
m1δ 21
(m2δ 22 − 1/ ω 2 )
=0
Expanding the determinant yields,
2
 1 
 1 
 2  − (m1δ11 + m2δ 22 )  2  + m1m2 (δ11δ 22 − δ12δ 21 ) = 0
ω 
 ω1 
(1)
Eq. (1) has two roots 1 / ω12 and 1 / ω22 . Thus
 1
1  1
1 
 2 − 2  2 − 2  = 0
 ω ω1   ω ω2 
or,
2
2
1  1   1  1 
 1   1
 2  +  2 + 2    +  2  2  = 0
 ω   ω1 ω2   ω   ω1   ω2 
(2)
Equate the third terms of Eqs. (1) and (2), which must be identical.
1 1
ω12 ω22
= m1m2 (δ11δ 22 − δ12δ 21 )
⇒
1
ω22
= ω12 m1m2 (δ11δ 22 − δ12δ 21 )
and it follows that
Shigley’s MED, 10th edition
Chapter 7 Solutions, Page 35/45
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ω2 =
g2
1
ω1 w1w2 (δ11δ 22 − δ12δ 21 )
Ans.
(b) In Ex. 7-5, part (b), the first critical speed of the two-disk shaft (w1 = 35 lbf,
w 2 = 55 lbf) is ω1 = 124.8 rad/s. From part (a), using influence coefficients,
1
3862
= 466 rad/s Ans.
124.8 35(55)  2.061(3.534) − 2.2342  10−8
______________________________________________________________________________
ω2 =
7-31
(
)
In Eq. (7-22), for ω1, the term I / A appears. For a hollow uniform diameter shaft,
ω1 ∝
(
(
)
)
(
)(
)
2
2
2
2
π do4 − di4 / 64
I
1 do + di do − di
1
=
=
=
d o2 + di2
2
2
2
2
A
d o − di
16
4
π d o − di / 4
This means that when a solid shaft is hollowed out, the critical speed increases beyond
that of the solid shaft of the same size. Ans.
By how much?
(1/ 4) d o2 + di2
(1/ 4) d o2
d 
= 1+  i 
 do 
2
The possible values of di are 0 ≤ di ≤ d o , so the range of the critical speeds is
ω1 1 + 0 to about ω1 1 + 1
or from ω1 to 2 ω1 .
For the specific case where the inner diameter is half of the outer diameter, the ratio of
the critical speeds is
2
1 
 do 
 di 
Ans.
1 +   = 1 +  2  = 1.12
 do 
 do 


______________________________________________________________________________
2
7-32
All steps will be modeled using singularity functions with a spreadsheet. Programming
Shigley’s MED, 10th edition
Chapter 7 Solutions, Page 36/45
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both loads will enable the user to first set the left load to 1, the right load to 0 and
calculate δ11 and δ21. Then set the left load to 0 and the right to 1 to get δ12 and δ22. The
spreadsheet shows the δ11 and δ21 calculation. A table for M / I vs. x is easy to make.
First, draw the bending-moment diagram as shown with the data.
x
M
0
0
1
2
3
0.875 1.75 1.625
4
1.5
5
6
7
1.375 1.25 1.125
8
1
M (lbf-in)
2
1.5
1
0.5
0
0
2
4
6
8
10
12
14
16
x (in)
x
M
9
10
11
0.875 0.75 0.625
12
0.5
13
14
15
0.375 0.25 0.125
16
0
The second-area moments are:
0 ≤ x ≤ 1 in and 15 ≤ x ≤ 16 in, I1 = π ( 2 4 ) / 64 = 0.7854 in 4
1 ≤ x ≤ 9 in , I 2 = π ( 2.4724 ) / 64 = 1.833 in 4
9 ≤ x ≤ 15 in , I 3 = π ( 2.7634 ) / 64 = 2.861 in 4
Divide M by I at the key points x = 0, 1, 2, 9, 14, 15, and 16 in and plot
x
M/I
x
M/I
0
0
1
1
2
2
3
4
5
6
7
8
1.1141 0.4774 0.9547 0.9547 0.8865 0.8183 0.7501 0.6819 0.6137 0.5456
9
9
10
11
12
13
14
14
15
15
0.4774 0.3058 0.2621 0.2185 0.1748 0.1311 0.0874 0.0874 0.0437 0.1592
Shigley’s MED, 10th edition
16
0
Chapter 7 Solutions, Page 37/45
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1.2
M/I (lbf/in3)
1
0.8
0.6
0.4
0.2
0
0
2
4
6
8
10
12
14
16
x (in)
From this diagram, one can see where changes in value (steps) and slope occur. Using a
spreadsheet, one can form a table of these changes. An example of a step is, at x = 1 in,
M/I goes from 0.875/0.7854 = 1.1141 lbf/in3 to 0.875/1.833 = 0.4774 lbf/in3, a step
change of 0.4774 − 1.1141 = − 0.6367 lbf/in3. A slope change also occurs at at x = 1 in.
The slope for 0 ≤ x ≤ 1 in is 1.1141/1 = 1.1141 lbf/in2, which changes to (0.9547 −
0.4774)/1 = 0.4774 lbf/in2, a change of 0.4774 − 1.1141 = − 0.6367 lbf/in2. Following
this approach, a table is made of all the changes. The table shown indicates the column
letters and row numbers for the spreadsheet.
A
x
1a
1b
2
2
9a
9b
14
14
15a
15b
16
1
2
3
4
5
6
7
8
9
10
11
12
B
M
0.875
0.875
1.75
1.75
0.875
0.875
0.25
0.25
0.125
0.125
0
C
M/I
1.114085
0.477358
0.954716
0.954716
0.477358
0.305854
0.087387
0.087387
0.043693
0.159155
0.000000
D
step
0.000000
-0.636727
0.000000
0.000000
0.000000
-0.171504
0.000000
0.000000
0.000000
0.115461
0.000000
E
Slope
1.114085
0.477358
0.477358
-0.068194
-0.068194
-0.043693
-0.043693
-0.043693
-0.043693
-0.159155
-0.159155
F
∆ Slope
0.000000
-0.636727
0.000000
-0.545552
0.000000
0.024501
0.000000
0.000000
0.000000
-0.115461
0.000000
The equation for M / I in terms of the spreadsheet cell locations is:
0
1
M / I = E2 ( x) + D3 x − 1 + F3 x − 1 + F5 x − 2
0
1
0
1
+ D7 x − 9 + F7 x − 9 + D11 x − 15 + F11 x − 15
Shigley’s MED, 10th edition
1
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Integrating twice gives the equation for Ey. Assume the shaft is steel. Boundary
conditions y = 0 at x = 0 and at x = 16 inches provide integration constants (C1 = − 4.906
lbf/in and C2 = 0). Substitution back into the deflection equation at x = 2 and 14 in
provides the δ ’s. The results are: δ11 = 2.917(10–7) and δ12 = 1.627(10–7). Repeat for
F1 = 0 and F2 = 1, resulting in δ21 = 1.627(10–7) and δ22 = 2.231(10–7). This can be
verified by finite element analysis.
y1
y2
y12
∑ wy
= 18(2.917)(10−7 ) + 32(1.627)(10−7 ) = 1.046(10−5 )
= 18(1.627)(10−7 ) + 32(2.231)(10−7 ) = 1.007(10−5 )
= 1.093(10−10 ), y22 = 1.014(10−10 )
= 5.105(10−4 ), ∑ w y 2 = 5.212(10−9 )
Neglecting the shaft, Eq. (7-23) gives
ω1 = 386
5.105(10−4 )
= 6149 rad/s or 58 720 rev/min
5.212(10−9 )
Ans.
Without the loads, we will model the shaft using 2 elements, one between 0 ≤ x ≤ 9 in,
and one between 0 ≤ x ≤ 16 in. As an approximation, we will place their weights at
x = 9/2 = 4.5 in, and x = 9 + (16 − 9)/2 = 12.5 in. From Table A-5, the weight density of
steel is γ = 0.282 lbf/in3. The weight of the left element is
π
π 
w1 = γ ∑ d 2l = 0.282    22 (1) + 2.4722 ( 8 )  = 11.7 lbf
4
4
The right element is
π 
w2 == 0.282    2.7632 ( 6 ) + 2 2 (1)  = 11.0 lbf
4
The spreadsheet can be easily modified to give
δ11 = 9.605 (10−7 ) , δ12 = δ 21 = 5.718 (10−7 ) , δ 22 = 5.472 (10−7 )
y1 = 1.753 (10 −5 ) , y2 = 1.271(10 −5 )
y12 = 3.072 (10−10 ) , y22 = 1.615 (10−10 )
∑ w y =3.449 (10 ) , ∑ w y
−4
2
=5.371 (10−9 )
Shigley’s MED, 10th edition
Chapter 7 Solutions, Page 39/45
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 3.449 (10−4 ) 
 = 4980 rad/s
−9
 5.371(10 ) 
ω1 = 386 
A finite element model of the exact shaft gives ω1 = 5340 rad/s. The simple model is
6.8% low.
Combination: Using Dunkerley’s equation, Eq. (7-32):
1
1
1
+
⇒ ω1 3870 rad/s Ans.
2
6149 49802
ω
______________________________________________________________________________
2
1
7-33
≈
We must not let the basis of the stress concentration factor, as presented, impose a viewpoint on the designer. Table A-16 shows Kts as a decreasing monotonic as a function of
a/D. All is not what it seems. Let us change the basis for data presentation to the full
section rather than the net section.
τ = K tsτ 0 = K ts′τ 0′
K ts =
32T
 32T 
= K ts′ 
3
3 
π AD
πD 
Therefore
K ts′ =
K ts
A
Form a table:
K ts′ has the following attributes:
• It exhibits a minimum;
• It changes little over a wide range;
•
Its minimum is a stationary point minimum at a / D B 0.100;
Shigley’s MED, 10th edition
Chapter 7 Solutions, Page 40/45
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•
Our knowledge of the minima location is
0.075 ≤ ( a / D ) ≤ 0.125
We can form a design rule: In torsion, the pin diameter should be about 1/10 of the shaft
diameter, for greatest shaft capacity, that is, a ≈ 0.10 D.
Ans.
However, it is not catastrophic if one forgets the rule.
______________________________________________________________________________
7-34
From the solution to Prob. 3-72, the torque to be transmitted through the key from the
gear to the shaft is T = 2819 lbf∙in. From Prob. 7-19, the nominal shaft diameter
supporting the gear is 1.00 in. From Table 7-6, a 0.25 in square key is appropriate for a
1.00 in shaft diameter. The force applied to the key is
F=
T
2819
=
= 5638 lbf
r 1.00 / 2
Selecting 1020 CD steel for the key, with Sy = 57 kpsi, and using the distortion-energy
theory, Ssy = 0.577 Sy = (0.577)(57) = 32.9 kpsi.
Failure by shear across the key:
F F
=
A tl
S
S
n = sy = sy
F / tl
τ
τ=
⇒
l=
1.1( 5638 )
nF
=
= 0.754 in
tS sy 0.25 ( 32 900 )
Failure by crushing:
σ=
S
Sy
F
F
=
n= y =
A (t / 2) l
σ 2 F / ( tl )
⇒
l=
2 Fn 2 ( 5638 )(1.1)
=
= 0.870 in
tS y
0.25 ( 57 ) (103 )
Select ¼-in square key, 7/8 in long, 1020 CD steel. Ans.
______________________________________________________________________________
7-35
From the solution to Prob. 3-73, the torque to be transmitted through the key from the
gear to the shaft is T = 3101 N∙m. From Prob. 7-21, the nominal shaft diameter
supporting the gear is 50 mm. To determine an appropriate key size for the shaft
diameter, we can either convert to inches and use Table 7-6, or we can look up standard
metric key sizes from the internet or a machine design handbook. It turns out that the
recommended metric key for a 50 mm shaft is 14 x 9 mm. Since the problem statement
specifies a square key, we will use a 14 x 14 mm key. For comparison, using Table 7-6
as a guide, for d = 50 mm = 1.97 in, a 0.5 in square key is appropriate. This is equivalent
to 12.7 mm. A 14 x 14 mm size is conservative, but reasonable after rounding up to
standard sizes.
Shigley’s MED, 10th edition
Chapter 7 Solutions, Page 41/45
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The force applied to the key is
F=
T
3101
=
= 124 (103 ) N
r 0.050 / 2
Selecting 1020 CD steel for the key, with Sy = 390 MPa, and using the distortion-energy
theory, Ssy = 0.577 Sy = 0.577(390) = 225 MPa.
Failure by shear across the key:
τ=
F F
=
A tl
n=
S sy
τ
=
S sy
F / ( tl )
⇒
1.1(124 ) (103 )
nF
=
= 0.0433 m = 43.3 mm
l=
tS sy ( 0.014 )( 225 ) (106 )
Failure by crushing:
n=
Sy
σ
σ=
F
F
=
A (t / 2) l
=
Sy
2 F / ( tl )
⇒
2 (124 ) (103 ) (1.1)
2 Fn
l=
=
= 0.0500 m = 50.0 mm
tS y ( 0.014 )( 390 ) (106 )
Select 14 mm square key, 50 mm long, 1020 CD steel. Ans.
______________________________________________________________________________
7-36
Choose basic size D = d = 15 mm. From Table 7-9, a locational clearance fit is
designated as 15H7/h6. From Table A-11, the tolerance grades are ∆D = 0.018 mm and
∆d = 0.011 mm. From Table A-12, the fundamental deviation is δF = 0 mm.
Hole:
Eq. (7-36):
Shaft:
Eq. (7-37):
7-37
Dmax = D + ∆D = 15 + 0.018 = 15.018 mm
Dmin = D = 15.000 mm
Ans.
Ans.
dmax = d + δF = 15.000 + 0 = 15.000 mm
Ans.
dmin = d + δF – ∆d = 15.000 + 0 – 0.011 = 14.989 mm
Ans.
Choose basic size D = d = 1.75 in. From Table 7-9, a medium drive fit is designated as
H7/s6. From Table A-13, the tolerance grades are ∆D = 0.0010 in and ∆d = 0.0006 in.
From Table A-14, the fundamental deviation is δF = 0.0017 in.
Hole:
Eq. (7-36):
Dmax = D + ∆D = 1.75 + 0.0010 = 1.7510 in
Shigley’s MED, 10th edition
Ans.
Chapter 7 Solutions, Page 42/45
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Dmin = D = 1.7500 in
Ans.
Shaft:
Eq. (7-38):
Ans.
dmin = d + δF = 1.75 + 0.0017 = 1.7517 in
dmax = d + δF + ∆d = 1.75 + 0.0017 + 0.0006 = 1.7523 in Ans.
______________________________________________________________________________
7-38
Choose basic size D = d = 45 mm. From Table 7-9, a sliding fit is designated as H7/g6.
From Table A-11, the tolerance grades are ∆D = 0.025 mm and ∆d = 0.016 mm. From
Table A-12, the fundamental deviation is δF = –0.009 mm.
Hole:
Eq. (7-36):
Dmax = D + ∆D = 45 + 0.025 = 45.025 mm
Dmin = D = 45.000 mm
Ans.
Ans.
Shaft:
Eq. (7-37):
dmax = d + δF = 45.000 + (–0.009) = 44.991 mm
Ans.
dmin = d + δF – ∆d = 45.000 + (–0.009) – 0.016 = 44.975 mm
Ans.
______________________________________________________________________________
7-39
Choose basic size D = d = 1.250 in. From Table 7-9, a close running fit is designated as
H8/f7. From Table A-13, the tolerance grades are ∆D = 0.0015 in and ∆d = 0.0010 in.
From Table A-14, the fundamental deviation is δF = –0.0010 in.
Hole:
Eq. (7-36):
Dmax = D + ∆D = 1.250 + 0.0015 = 1.2515 in
Dmin = D = 1.2500 in
Ans.
Ans.
Shaft:
Eq. (7-37):
Ans.
dmax = d + δF = 1.250 + (–0.0010) = 1.2490 in
dmin = d + δF – ∆d = 1.250 + (–0.0010) – 0.0010 = 1.2480 in
Ans.
______________________________________________________________________________
7-40
Choose basic size D = d = 35 mm. From Table 7-9, a locational interference fit is
designated as H7/p6. From Table A-11, the tolerance grades are ∆D = 0.025 mm and
∆d = 0.016 mm. From Table A-12, the fundamental deviation is δF = 0.026 mm.
Hole:
Eq. (7-36):
Dmax = D + ∆D = 35 + 0.025 = 35.025 mm
Dmin = D = 35.000 mm
The bearing bore specifications are within the hole specifications for a locational
interference fit. Now find the necessary shaft sizes.
Shaft:
Eq. (7-38):
dmin = d + δF = 35 + 0.026 = 35.026 mm
Shigley’s MED, 10th edition
Ans.
Chapter 7 Solutions, Page 43/45
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dmax = d + δF + ∆d = 35 + 0.026 + 0.016 = 35.042 mm
Ans.
______________________________________________________________________________
7-41
Choose basic size D = d = 1.5 in. From Table 7-9, a locational interference fit is
designated as H7/p6. From Table A-13, the tolerance grades are ∆D = 0.0010 in and
∆d = 0.0006 in. From Table A-14, the fundamental deviation is δF = 0.0010 in.
Hole:
Eq. (7-36):
Dmax = D + ∆D = 1.5000 + 0.0010 = 1.5010 in
Dmin = D = 1.5000 in
The bearing bore specifications exactly match the requirements for a locational
interference fit. Now check the shaft.
Shaft:
Eq. (7-38):
dmin = d + δF = 1.5000 + 0.0010 = 1.5010 in
dmax = d + δF + ∆d = 1.5000 + 0.0010 + 0.0006 = 1.5016 in
The shaft diameter of 1.5020 in is greater than the maximum allowable diameter of
1.5016 in, and therefore does not meet the specifications for the locational interference
fit. Ans.
______________________________________________________________________________
7-42
(a) Basic size is D = d = 35 mm.
Table 7-9:
H7/s6 is specified for medium drive fit.
Table A-11: Tolerance grades are ∆D = 0.025 mm and ∆d = 0.016 mm.
Table A-12: Fundamental deviation is δ F = +0.043 mm.
Eq. (7-36):
Dmax = D + ∆D = 35 + 0.025 = 35.025 mm
Dmin = D = 35.000 mm
Ans.
Eq. (7-38):
dmin = d + δF = 35 + 0.043 = 35.043 mm
Ans.
dmax = d + δF + ∆d = 35 + 0.043 + 0.016 = 35.059 mm
(b)
Eq. (7-42):
Eq. (7-43):
Eq. (7-40):
δ min = d min − Dmax = 35.043 − 35.025 = 0.018 mm
δ max = d max − Dmin = 35.059 − 35.000 = 0.059 mm
pmax
(
)(
)
2
2
2
2
Eδ max  d o − d d − di 


=
d o2 − di2
2d 3 


207 109 ( 0.059 )  602 − 352 352 − 0

=
602 − 0
2 353

( )
( )
(
)(
Shigley’s MED, 10th edition
)  = 115 MPa

Ans.
Chapter 7 Solutions, Page 44/45
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pmin
(
)(
)
2
2
2
2
Eδ min  d o − d d − di 


=
d o2 − di2
2d 3 


207 109 ( 0.018)  602 − 352 352 − 0

=
602 − 0
2 353

( )
( )
(
)(
)  = 35.1 MPa

Ans.
(c) For the shaft:
Eq. (7-44):
σ t ,shaft = − p = −115 MPa
Eq. (7-46):
σ r ,shaft = − p = −115 MPa
Eq. (5-13):
σ ′ = (σ 12 − σ 1σ 2 + σ 22 )
1/2
1/ 2
=  (−115) 2 − (−115)(−115) + (−115) 2 
n = S y / σ ′ = 390 /115 = 3.4
= 115 MPa
Ans.
For the hub:
 602 + 352 
do2 + d 2
=
= 234 MPa
115
 2
2 
do2 − d 2
 60 − 35 
Eq. (7-45):
σ t ,hub = p
Eq. (7-46):
σ r ,hub = − p = −115 MPa
Eq. (5-13):
σ ′ = (σ 12 − σ 1σ 2 + σ 22 )
1/2
=  (234) 2 − (234)( −115) + ( −115) 2 
1/ 2
= 308 MPa
n = S y / σ ′ = 600 / 308 = 1.9
Ans.
(d) A value for the static coefficient of friction for steel to steel can be obtained online or
from a physics textbook as approximately f = 0.8.
Eq. (7-49)
T = (π / 2) f pmin ld 2
( )
= (π / 2)(0.8)(35.1) 106 (0.050)(0.035) 2 = 2700 N ⋅ m
Ans.
______________________________________________________________________________
Shigley’s MED, 10th edition
Chapter 7 Solutions, Page 45/45
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Chapter 8
_________________________
______________________________________
____________________
Notes to the Instructor.
ugh 8-44 and 8-51 through 8-58. There is a sta
statement preceding
1. For Problems 8-33 through
Prob. 8-33 which covers a cha
hange which will be corrected in the second printing
pri
of the 10th
edition.
2. For Probs. 8-41 to 8-44. Th
These problems, as well as many others in this
is chapter are best
implemented using a spreadsh
sheet.
_________________________
______________________________________
____________________
8-1
mm Ans.
(a) Thread depth= 2.55 m
Width = 2.5 mm Ans.
ns.
dm = 25 - 1.25 - 1.25 = 22.5
2
mm
dr = 25 - 5 = 20 mm
l = p = 5 mm Ans.
(b) Thread depth = 2.55 m
mm Ans.
Width at pitch line = 2.5 mm Ans.
dm = 22.5 mm
dr = 20 mm
l = p = 5 mm Ans.
_________________________
______________________________________
____________________
8-2
From Table 8-1,
d r = d − 1.226
1.
869 p
d m = d − 0.649
0
519 p
d − 1.226
1.
869 p + d − 0.649 519 p
d =
= d − 0.938 194
19 p
2
πd 2 π
= (d − 0.938 194 p)2 Ans.
4
4
_________________________
______________________________________
____________________
At =
8-3
From Eq. (c) of Sec. 8--2,
tan λ + f
PR = F
1 − f tan λ
Shigley’s MED, 10th edition
Chap
ap. 8 Solutions, Page 1/70
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PRd m
Fd m tan λ + f
=
2
2 1 − f tan λ
T0
Fl / (2π ) 1 − f tan λ
1 − f tan λ
e=
=
= tan λ
TR
Fd m / 2 tan λ + f
tan λ + f
TR =
Ans.
Using f = 0.08, form a table and plot the efficiency curve.
e
λ, deg.
0
0
0
0.678
20
0.796
30
0.838
40
0.8517
45
0.8519
_________________________
______________________________________
____________________
8-4
Given F = 5 kN, l = 5 mm,
m and dm = d − p/2 = 25 − 5/2 = 22.5 mm,
m, the torque required to
raise the load is found uusing Eqs. (8-1) and (8-6)
TR =
5 ( 22.5 )  5 + π ( 0.09 ) 22.5  5 ( 0.06 ) 45
= 15.85 N ⋅ m

+
2
π
22.5
−
0.09
5
2
(
)
(
)


Ans.
The torque required too lower
l
the load, from Eqs. (8-2) and (8-6) is
TL =
5 ( 22.5 )  π ( 0.09 ) 22.5 − 5  5 ( 0.06 ) 45
= 7.83 N ⋅ m

+
2
π
22.5
+
0.09
5
2
(
)
(
)


Ans.
Since TL is positive, the
he thread is self-locking. From Eq.(8-4) the effi
fficiency is
5 ( 5)
= 0.251
Ans.
2π (15.85 )
_________________________
______________________________________
____________________
e=
Collar (thrust) bearings,
gs, at the bottom of the screws, must bear on th
the collars. The bottom
segment of the screwss must
m be in compression. Whereas, tension spe
pecimens and their
grips must be in tension
on. Both screws must be of the same-hand threa
reads.
_________________________
_______________________________________
____________________
8-5
Shigley’s MED, 10th edition
Chap
ap. 8 Solutions, Page 2/70
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8-6
Screws rotate at an angular rate of
n=
1720
= 28.67 rev/min
60
(a) The lead is 0.25 in, so the linear speed of the press head is
V = 28.67(0.25) = 7.17 in/min Ans.
(b) F = 2500 lbf/screw
d m = 2 − 0.25 / 2 = 1.875 in
sec α = 1 / cos(29o / 2) = 1.033
Eq. (8-5):
TR =
2500(1.875)  0.25 + π (0.05)(1.875)(1.033) 

 = 221.0 lbf · in
2
 π (1.875) − 0.05(0.25)(1.033) 
Eq. (8-6):
Tc = 2500(0.08)(3.5 / 2) = 350 lbf · in
Ttotal = 350 + 221.0 = 571 lbf · in/screw
571(2)
Tmotor =
= 20.04 lbf · in
60(0.95)
Tn
20.04(1720)
H =
=
= 0.547 hp Ans.
63 025
63 025
______________________________________________________________________________
8-7
L = 3.5 in, T = 3.5 F, M = (L − 3/8) F = (3.5 − 0.375) F = 3.125 F, Sy = 41 kpsi.
32M
32(3.125) F
σ = Sy =
=
= 41 000
3
πd
π (0.1875)3
F = 8.49 lbf
T = 3.5(8.49) = 29.7 lbf ∙ in
Ans.
(b) Eq. (8-5), 2α = 60° , l = 1/10 = 0.1 in, f = 0.15, sec α = 1.155, p = 0.1 in
3
d m = − 0.649 519 ( 0.1) = 0.6850 in
4
Fclamp (0.6850)  0.1 + π (0.15)(0.6850)(1.155) 
TR =


2
 π (0.6850) − 0.15(0.1)(1.155) 
TR = 0.075 86Fclamp
TR
29.7
Fclamp =
=
= 392 lbf Ans.
0.075 86 0.075 86
Shigley’s MED, 10th edition
Chap. 8 Solutions, Page 3/70
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(c) The column has one end fixed and the other end pivoted. Base the decision on the
mean diameter column. Input: C = 1.2, D = 0.685 in, A = π(0.6852)/4 = 0.369 in2, Sy = 41
kpsi, E = 30(106) psi, L = 6 in, k = D/4 =0.171 25 in, L/k = 35.04. From Eq. (4-45),
 2π 2 (1.2 ) 30 (106 ) 
 = 131.7
=
41 000


From Eq. (4-46), the limiting clamping force for buckling is
1/2
1/2
2
 l   2π CE 

  = 
 k 1  S y 
2

 Sy l  1 

Fclamp = Pc r = A  S y − 


 2π k  CE 
2


 41(103 )

1


3
= 0.369  41(10 ) − 
= 14.6 (103 ) lbf
35.04 
6
 2π
 1.2 ( 30 )10 


Ans
(d) This is a subject for class discussion.
______________________________________________________________________________
8-8
T = 8(3.5) = 28 lbf ⋅ in
dm =
3 1
− = 0.6667 in
4 12
l =
1
= 0.1667 in,
6
α =
290
= 14.50,
2
sec 14.50 = 1.033
From Eqs. (8-5) and (8-6)
Ttotal =
0.6667 F
2
 0.1667 + π ( 0.15 )( 0.6667 )(1.033)  0.15 (1) F
= 0.1542 F

+
2
 π ( 0.6667 ) − 0.15 ( 0.1667 )(1.033) 
28
= 182 lbf
Ans.
0.1542
_____________________________________________________________________________
F=
8-9
dm = 1.5 − 0.25/2 = 1.375 in, l = 2(0.25) = 0.5 in
From Eq. (8-1) and Eq. (8-6),
Shigley’s MED, 10th edition
Chap. 8 Solutions, Page 4/70
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2.2 (103 ) (1.375)  0.5 + π (0.10)(1.375)  2.2 (103 ) (0.15)(2.25)
TR =
 π (1.375) − 0.10(0.5)  +
2
2


= 330 + 371 = 701 lbf · in
Since n = V/l = 2/0.5 = 4 rev/s = 240 rev/min
so the power is
701( 240 )
Tn
=
= 2.67 hp
Ans.
63 025
63 025
______________________________________________________________________________
H=
8-10 dm = 40 − 4 = 36 mm, l = p = 8 mm
From Eqs. (8-1) and (8-6)
T =
=
ω =
H =
T =
F =
36 F  8 + π (0.14)(36)  0.09(100) F
+
2  π (36) − 0.14(8) 
2
(3.831 + 4.5) F = 8.33F N · m (F in kN)
2π n = 2π (1) = 2π rad/s
Tω
3000
H
=
= 477 N · m
2π
ω
477
= 57.3 kN Ans.
8.33
57.3(8)
Fl
=
= 0.153 Ans.
2π T
2π (477)
______________________________________________________________________________
e=
8-11 (a) Table A-31, nut height H = 12.8 mm. L ≥ l + H = 2(15) + 12.8 = 42.8 mm. Rounding
up,
L = 45 mm
Ans.
(b) From Eq. (8-14), LT = 2d + 6 = 2(14) +6 = 34 mm
From Table 8-7, ld = L − LT = 45− 34 = 11 mm, lt = l − ld = 2(15) − 11 = 19 mm,
Ad = π (142) / 4 = 153.9 mm2. From Table 8-1, At = 115 mm2. From Eq. (8-17)
kb =
153.9 (115 ) 207
Ad At E
=
= 874.6 MN/m
Ad lt + At ld 153.9 (19 ) + 115 (11)
Ans.
(c) From Eq. (8-22), with l = 2(15) = 30 mm
Shigley’s MED, 10th edition
Chap. 8 Solutions, Page 5/70
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km =
0.5774π ( 207 )14
0.5774π Ed
=
= 3 116.5 MN/m
 0.5774l + 0.5d 
 0.5774 ( 30 ) + 0.5 (14 ) 
2 ln  5
 2 ln 5

 0.5774l + 2.5d 
 0.5774 ( 30 ) + 2.5 (14 ) 
Ans.
8-12 (a) Table A-31, nut height H = 12.8 mm. Table A-33, washer thickness t = 3.5 mm. Thus,
the grip is l = 2(15) + 3.5 = 33.5 mm. L ≥ l + H = 33.5 + 12.8 = 46.3 mm. Rounding up
L = 50 mm
Ans.
(b) From Eq. (8-14), LT = 2d + 6 = 2(14) +6 = 34 mm
From Table 8-7, ld = L − LT = 50− 34 = 16 mm, lt = l − ld = 33.5 − 16 = 17.5 mm,
Ad = π (142) / 4 = 153.9 mm2. From Table 8-1, At = 115 mm2. From Eq. (8-17)
kb =
153.9 (115 ) 207
Ad At E
=
= 808.2 MN/m
Ad lt + At ld 153.9 (17.5 ) + 115 (16 )
Ans.
(c)
From Eq. (8-22)
km =
0.5774π ( 207 )14
0.5774π Ed
=
= 2 969 MN/m
 0.5774l + 0.5d 
 0.5774 ( 33.5 ) + 0.5 (14 ) 
2 ln  5
 2 ln 5

 0.5774l + 2.5d 
 0.5774 ( 33.5 ) + 2.5 (14 ) 
Ans.
______________________________________________________________________________
8-13 (a) Table 8-7, l = h + d /2 = 15 + 14/2 = 22 mm. L ≥ h + 1.5d = 36 mm. Rounding up
L = 40 mm
Ans.
(b) From Eq. (8-14), LT = 2d + 6 = 2(14) +6 = 34 mm
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From Table 8-7, ld = L − LT = 40− 34 = 6 mm, lt = l − ld = 22 − 6 = 16 mm
Ad = π (142) / 4 = 153.9 mm2. From Table 8-1, At = 115 mm2. From Eq. (8-17)
kb =
153.9 (115 ) 207
Ad At E
=
= 1 162.2 MN/m
Ad lt + At ld 153.9 (16 ) + 115 ( 6 )
Ans.
(c) From Eq. (8-22), with l = 22 mm
km =
0.5774π ( 207 )14
0.5774π Ed
=
= 3 624.4 MN/m
 0.5774l + 0.5d 
 0.5774 ( 22 ) + 0.5 (14 ) 
2 ln  5
 2 ln 5

 0.5774l + 2.5d 
 0.5774 ( 22 ) + 2.5 (14 ) 
Ans.
______________________________________________________________________________
8-14 (a) From Table A-31, the nut height is H = 7/16 in. L ≥ l + H = 2 + 1 + 7/16 = 3 7/16 in.
Rounding up, L = 3.5 in
Ans.
(b) From Eq. (8-13), LT = 2d + 1/4 = 2(0.5) + 0.25 = 1.25 in
From Table 8-7, ld = L − LT = 3.5 − 1.25 = 2.25 in, lt = l − ld = 3 − 2.25 = 0.75 in
Ad = π (0.52)/4 = 0.1963 in2. From Table 8-2, At = 0.1419 in2. From Eq. (8-17)
kb =
0.1963 ( 0.1419 ) 30
Ad At E
=
= 1.79 Mlbf/in
Ad lt + At ld 0.1963 ( 0.75 ) + 0.1419 ( 2.25 )
Shigley’s MED, 10th edition
Ans.
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(c)
Top steel frustum: t = 1.5 in, d = 0.5 in, D = 0.75 in, E = 30 Mpsi. From Eq. (8-20)
k1 =
0.5774π ( 30 ) 0.5
1.155 (1.5 ) + 0.75 − 0.5 ( 0.75 + 0.5 )
ln 
1.155 (1.5 ) + 0.75 + 0.5 ( 0.75 − 0.5 )
= 22.65 Mlbf/in
Lower steel frustum: t = 0.5 in, d = 0.5 in, D = 0.75 + 2(1) tan 30° = 1.905 in, E = 30
Mpsi. Eq. (8-20) ⇒ k2 = 210.7 Mlbf/in
Cast iron: t = 1 in, d = 0.5 in, D = 0.75 in, E = 14.5 Mpsi (Table 8-8). Eq. (8-20) ⇒
k3 = 12.27 Mlbf/in
From Eq. (8-18)
km = (1/k1 + 1/k2 +1/k3)−1 = (1/22.65 + 1/210.7 + 1/12.27)−1 = 7.67 Mlbf/in
Ans.
8-15 (a) From Table A-32, the washer thickness is 0.095 in. Thus, l = 2 + 1 + 2(0.095) = 3.19
in. From Table A-31, the nut height is H = 7/16 in. L ≥ l + H = 3.19 + 7/16 = 3.63 in.
Rounding up, L = 3.75 in
Ans.
(b) From Eq. (8-13), LT = 2d + 1/4 = 2(0.5) + 0.25 = 1.25 in
From Table 8-7, ld = L − LT = 3.75 − 1.25 = 2.5 in, lt = l − ld = 3.19 − 2.5 = 0.69 in
Ad = π (0.52)/4 = 0.1963 in2. From Table 8-2, At = 0.1419 in2. From Eq. (8-17)
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kb =
0.1963 ( 0.1419 ) 30
Ad At E
=
= 1.705 Mlbf/in
Ad lt + At ld 0.1963 ( 0.69 ) + 0.1419 ( 2.5 )
Ans.
(c)
Each steel washer frustum: t = 0.095 in, d = 0.531 in (Table A-32), D = 0.75 in, E = 30
Mpsi. From Eq. (8-20)
k1 =
0.5774π ( 30 ) 0.531
1.155 ( 0.095 ) + 0.75 − 0.531 ( 0.75 + 0.531)
ln 
1.155 ( 0.095 ) + 0.75 + 0.531 ( 0.75 − 0.531)
= 89.20 Mlbf/in
Top plate, top steel frustum: t = 1.5 in, d = 0.5 in, D = 0.75 + 2(0.095) tan 30° = 0.860 in,
E = 30 Mpsi. Eq. (8-20) ⇒ k2 = 28.99 Mlbf/in
Top plate, lower steel frustum: t = 0.5 in, d = 0.5 in, D = 0.860 + 2(1) tan 30° = 2.015 in,
E = 30 Mpsi. Eq. (8-20) ⇒ k3 = 234.08 Mlbf/in
Cast iron: t = 1 in, d = 0.5 in, D = 0.75 + 2(0.095) tan 30° = 0.860 in, E = 14.5 Mpsi
(Table 8-8). Eq. (8-20) ⇒ k4 = 15.99 Mlbf/in
From Eq. (8-18)
km = (2/k1 + 1/k2 +1/k3+1/k4)−1 = (2/89.20 + 1/28.99 + 1/234.08 + 1/15.99)−1
= 8.08 Mlbf/in Ans.
______________________________________________________________________________
8-16 (a) From Table 8-7, l = h + d /2 = 2 + 0.5/2 = 2.25 in.
L ≥ h + 1.5 d = 2 + 1.5(0.5) = 2.75 in Ans.
(b) From Table 8-7, LT = 2d + 1/4 = 2(0.5) + 0.25 = 1.25 in
ld = L − LT = 2.75 − 1.25 = 1.5 in, lt = l − ld = 2.25 − 1.5 = 0.75 in
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Ad = π (0.52)/4 = 0.1963 in2. From Table 8-2, At = 0.1419 in2. From Eq. (8-17)
kb =
0.1963 ( 0.1419 ) 30
Ad At E
=
= 2.321 Mlbf/in
Ad lt + At ld 0.1963 ( 0.75 ) + 0.1419 (1.5 )
Ans.
(c)
Top steel frustum: t = 1.125 in, d = 0.5 in, D = 0.75 in, E = 30 Mpsi. From Eq. (8-20)
k1 =
0.5774π ( 30 ) 0.5
1.155 (1.125 ) + 0.75 − 0.5 ( 0.75 + 0.5 )
ln 
1.155 (1.125 ) + 0.75 + 0.5 ( 0.75 − 0.5 )
= 24.48 Mlbf/in
Lower steel frustum: t = 0.875 in, d = 0.5 in, D = 0.75 + 2(0.25) tan 30° = 1.039 in, E =
30 Mpsi. Eq. (8-20) ⇒ k2 = 49.36 Mlbf/in
Cast iron: t = 0.25 in, d = 0.5 in, D = 0.75 in, E = 14.5 Mpsi (Table 8-8). Eq. (8-20) ⇒
k3 = 23.49 Mlbf/in
From Eq. (8-18)
km = (1/k1 + 1/k2 +1/k3)−1 = (1/24.48 + 1/49.36 + 1/23.49)−1 = 9.645 Mlbf/in Ans.
______________________________________________________________________________
8-17 a) Grip, l = 2(2 + 0.095) = 4.19 in. L ≥ 4.19 + 7/16 = 4.628 in.
Rounding up, L = 4.75 in
Ans.
(b) From Eq. (8-13), LT = 2d + 1/4 = 2(0.5) + 0.25 = 1.25 in
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From Table 8-7, ld = L − LT = 4.75 − 1.25 = 3.5 in, lt = l − ld = 4.19 − 3.5 = 0.69 in
Ad = π (0.52)/4 = 0.1963 in2. From Table 8-2, At = 0.1419 in2. From Eq. (8-17)
kb =
0.1963 ( 0.1419 ) 30
Ad At E
=
= 1.322 Mlbf/in
Ad lt + At ld 0.1963 ( 0.69 ) + 0.1419 ( 3.5 )
Ans.
(c)
Upper and lower halves are the same. For the upper half,
Steel frustum: t = 0.095 in, d = 0.531 in, D = 0.75 in, and E = 30 Mpsi. From Eq. (8-20)
k1 =
0.5774π ( 30 ) 0.531
1.155 ( 0.095 ) + 0.75 − 0.531 ( 0.75 + 0.531)
ln 
1.155 ( 0.095 ) + 0.75 + 0.531 ( 0.75 − 0.531)
= 89.20 Mlbf/in
Aluminum: t = 2 in, d = 0.5 in, D =0.75 + 2(0.095) tan 30° = 0.860 in, and E = 10.3
Mpsi. Eq. (8-20) ⇒ k2 = 9.24 Mlbf/in
For the top half, km′ = (1/k1 + 1/k2)−1 = (1/89.20 + 1/9.24)−1 = 8.373 Mlbf/in
Since the bottom half is the same, the overall stiffness is given by
Ans
km = (1/ km′ + 1/ km′ )−1 = km′ /2 = 8.373/2 = 4.19 Mlbf/in
______________________________________________________________________________
8-18 (a) Grip, l = 2(2 + 0.095) = 4.19 in. L ≥ 4.19 + 7/16 = 4.628 in.
Rounding up, L = 4.75 in
Ans.
(b) From Eq. (8-13), LT = 2d + 1/4 = 2(0.5) + 0.25 = 1.25 in
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From Table 8-7, ld = L − LT = 4.75 − 1.25 = 3.5 in, lt = l − ld = 4.19 − 3.5 = 0.69 in
Ad = π (0.52)/4 = 0.1963 in2. From Table 8-2, At = 0.1419 in2. From Eq. (8-17)
kb =
0.1963 ( 0.1419 ) 30
Ad At E
=
= 1.322 Mlbf/in
Ad lt + At ld 0.1963 ( 0.69 ) + 0.1419 ( 3.5 )
Ans.
(c)
Upper aluminum frustum: t = [4 + 2(0.095)] /2 = 2.095 in, d = 0.5 in, D = 0.75 in, and
E = 10.3 Mpsi. From Eq. (8-20)
0.5774π (10.3) 0.5
k1 =
= 7.23 Mlbf/in
1.155 ( 2.095 ) + 0.75 − 0.5 ( 0.75 + 0.5 )
ln
1.155 ( 2.095 ) + 0.75 + 0.5 ( 0.75 − 0.5 )
Lower aluminum frustum: t = 4 − 2.095 = 1.905 in, d = 0.5 in,
D = 0.75 +4(0.095) tan 30° = 0.969 in, and E = 10.3 Mpsi. Eq. (8-20) ⇒ k2 = 11.34
Mlbf/in
Steel washers frustum: t = 2(0.095) = 0.190 in, d = 0.531 in, D = 0.75 in, and E = 30 Mpsi.
Eq. (8-20) ⇒ k3 = 53.91 Mlbf/in
From Eq. (8-18)
km = (1/k1 + 1/k2 +1/k3)−1 = (1/7.23 + 1/11.34 + 1/53.91)−1 = 4.08 Mlbf/in Ans.
______________________________________________________________________________
8-19 (a) From Table A-31, the nut height is H = 8.4 mm. L > l + H = 50 + 8.4 = 58.4 mm.
Rounding up, L = 60 mm
Ans.
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(b) From Eq. (8-14), LT = 2d + 6 = 2(10) + 6 = 26 mm, ld = L − LT = 60 − 26 =
34 mm, lt = l − l = 50 − 34 = 16 mm. Ad = π (102) / 4 = 78.54 mm2. From Table 8-1,
At = 58 mm2. From Eq. (8-17)
78.54 ( 58.0 ) 207
Ad At E
kb =
=
= 292.1 MN/m
Ans.
Ad lt + At ld 78.54 (16 ) + 58.0 ( 34 )
(c)
Upper and lower frustums are the same. For the upper half,
Aluminum: t = 10 mm, d = 10 mm, D = 15 mm, and from Table 8-8, E = 71 GPa.
From Eq. (8-20)
0.5774π ( 71)10
= 1576 MN/m
k1 =
1.155 (10 ) + 15 − 10  (15 + 10 )
ln
1.155 (10 ) + 15 + 10  (15 − 10 )
Steel: t = 15 mm, d = 10 mm, D = 15 + 2(10) tan 30° = 26.55 mm, and E = 207
GPa. From Eq. (8-20)
k2 =
0.5774π ( 207 )10
1.155 (15 ) + 26.55 − 10  ( 26.55 + 10 )
ln 
1.155 (15 ) + 26.55 + 10  ( 26.55 − 10 )
= 11 440 MN/m
For the top half, km′ = (1/k1 + 1/k2)−1 = (1/1576 + 1/11 440)−1 = 1385 MN/m
Since the bottom half is the same, the overall stiffness is given by
km = (1/ km′ + 1/ km′ )−1 = km′ /2 = 1385/2 = 692.5 MN/m
Shigley’s MED, 10th edition
Ans.
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8-20 (a) From Table A-31, the nut height is H = 8.4 mm. L > l + H = 60 + 8.4 = 68.4 mm.
Rounding up, L = 70 mm
Ans.
(b) From Eq. (8-14), LT = 2d + 6 = 2(10) + 6 = 26 mm, ld = L − LT = 70 − 26 =
44 mm, lt = l − ld = 60 − 44 = 16 mm. Ad = π (102) / 4 = 78.54 mm2. From Table 8-1,
At = 58 mm2. From Eq. (8-17)
kb =
78.54 ( 58.0 ) 207
Ad At E
=
= 247.6 MN/m
Ad lt + At ld 78.54 (16 ) + 58.0 ( 44 )
Ans.
(c)
Upper aluminum frustum: t = 10 mm, d = 10 mm, D = 15 mm, and E = 71 GPa. From Eq.
(8-20)
0.5774π (10.3) 71
= 1576 MN/m
k1 =
1.155 ( 2.095 ) + 15 − 10  (15 + 10 )
ln
1.155 ( 2.095 ) + 15 + 10  (15 − 10 )
Lower aluminum frustum: t = 20 mm, d = 10 mm, D = 15 mm, and E = 71 GPa. Eq.
(8-20) ⇒ k2 = 1 201 MN/m
Top steel frustum: t = 20 mm, d = 10 mm, D = 15 + 2(10) tan 30° = 26.55 mm, and E =
207 GPa. Eq. (8-20) ⇒ k3 = 9 781 MN/m
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Lower steel frustum: t = 10 mm, d = 10 mm, D = 15 + 2(20) tan 30° = 38.09 mm, and E =
207 GPa. Eq. (8-20) ⇒ k4 = 29 070 MN/m
From Eq. (8-18)
km = (1/k1 + 1/k2 +1/k3+1/k4)−1 = (1/1 576 + 1/1 201 + 1/9 781 +1/29 070)−1
= 623.5 MN/m Ans.
______________________________________________________________________________
8-21 (a) From Table 8-7, l = h + d /2 = 10 + 30 + 10/2 = 45 mm. L ≥ h + 1.5 d =
10 + 30 + 1.5(10) = 55 mm Ans.
(b) From Eq. (8-14), LT = 2d + 6 = 2(10) + 6 = 26 mm, ld = L − LT = 55 − 26 =
29 mm, lt = l − ld = 45 − 29 = 16 mm. Ad = π (102) / 4 = 78.54 mm2. From Table 8-1,
At = 58 mm2. From Eq. (8-17)
78.54 ( 58.0 ) 207
Ad At E
=
= 320.9 MN/m
kb =
Ans.
Ad lt + At ld 78.54 (16 ) + 58.0 ( 29 )
(c)
Upper aluminum frustum: t = 10 mm, d = 10 mm, D = 15 mm, and E = 71 GPa. From Eq.
(8-20)
0.5774π (10.3) 71
= 1576 MN/m
k1 =
1.155 ( 2.095 ) + 15 − 10  (15 + 10 )
ln
1.155 ( 2.095 ) + 15 + 10  (15 − 10 )
Lower aluminum frustum: t = 5 mm, d = 10 mm, D = 15 mm, and E = 71 GPa. Eq.
(8-20) ⇒ k2 = 2 300 MN/m
Top steel frustum: t = 12.5 mm, d = 10 mm, D = 15 + 2(10) tan 30° = 26.55 mm, and E =
207 GPa. Eq. (8-20) ⇒ k3 = 12 759 MN/m
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Lower steel frustum: t = 17.5 mm, d = 10 mm, D = 15 + 2(5) tan 30° = 20.77 mm, and E
= 207 GPa. Eq. (8-20) ⇒ k4 = 6 806 MN/m
From Eq. (8-18)
km = (1/k1 + 1/k2 +1/k3+1/k4)−1 = (1/1 576 + 1/2 300 + 1/12 759 +1/6 806)−1
= 772.4 MN/m Ans.
______________________________________________________________________________
kb
8-22 Equation (f ), p. 428: C =
kb + k m
Ad At E
Eq. (8-17):
kb =
Ad lt + At ld
0.5774π ( 207 ) d
km =
Eq. (8-22):
 0.5774 ( 40 ) + 0.5d 
2 ln 5

 0.5774 ( 40 ) + 2.5d 
See Table 8-7 for other terms used.
Using a spreadsheet, with coarse-pitch bolts (units are mm, mm2, MN/m):
d
10
12
14
16
20
24
30
d
10
12
14
16
20
24
30
l
40
40
40
40
40
40
40
At
58
84.3
115
157
245
353
561
ld
24
25
21
17
14
11
4
lt
16
15
19
23
26
29
36
Ad
78.53982
113.0973
153.938
201.0619
314.1593
452.3893
706.8583
kb
356.0129
518.8172
686.2578
895.9182
1373.719
1944.24
2964.343
H
8.4
10.8
12.8
14.8
18
21.5
25.6
L>
48.4
50.8
52.8
54.8
58
61.5
65.6
km
1751.566
2235.192
2761.721
3330.796
4595.515
6027.684
8487.533
L
50
55
55
55
60
65
70
LT
26
30
34
38
46
54
66
C
0.16892
0.188386
0.199032
0.211966
0.230133
0.243886
0.258852
Use a M14 × 2 bolt, with length 55 mm.
Ans.
_____________________________________________________________________________
kb
8-23 Equation (f ), p. 428: C =
kb + k m
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Eq. (8-17):
kb =
Ad At E
Ad lt + At ld
For upper frustum, Eq. (8-20), with D = 1.5 d and t = 1.5 in:
k1 =
0.5774π ( 30 ) d
 1.155 (1.5 ) + 0.5d  ( 2.5d ) 
ln  

 1.155 (1.5 ) + 2.5d  ( 0.5d ) 
=
0.5774π ( 30 ) d
 (1.733 + 0.5d ) 
ln 5

 (1.733 + 2.5d ) 
Lower steel frustum, with D = 1.5d + 2(1) tan 30° = 1.5d + 1.155, and t = 0.5 in:
0.5774π ( 30 ) d
k2 =
 (1.733 + 0.5d )( 2.5d + 1.155 ) 
ln 

 (1.733 + 2.5d )( 0.5d + 1.155 ) 
For cast iron frustum, let E = 14. 5 Mpsi, and D = 1.5 d, and t = 1 in:
0.5774π (14.5 ) d
k3 =
 (1.155 + 0.5d ) 
ln 5

 (1.155 + 2.5d ) 
Overall,
km = (1/k1 +1/k2 +1/k3)−1
See Table 8-7 for other terms used.
Using a spreadsheet, with coarse-pitch bolts (units are in, in2, Mlbf/in):
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d
0.375
0.4375
0.5
0.5625
0.625
0.75
0.875
At
0.0775
0.1063
0.1419
0.182
0.226
0.334
0.462
d
0.375
0.4375
0.5
0.5625
0.625
0.75
0.875
ld
2.5
2.375
2.25
2.125
2.25
2
1.75
Ad
H
L>
0.110447 0.328125 3.328125
0.15033 0.375
3.375
0.19635 0.4375 3.4375
0.248505 0.484375 3.484375
0.306796 0.546875 3.546875
0.441786 0.640625 3.640625
0.60132
0.75
3.75
lt
0.5
0.625
0.75
0.875
0.75
1
1.25
L
3.5
3.5
3.5
3.5
3.75
3.75
3.75
LT
1
1.125
1.25
1.375
1.5
1.75
2
l
3
3
3
3
3
3
3
kb
k1
k2
k3
km
C
1.031389 15.94599 178.7801 8.461979 5.362481 0.161309
1.383882 19.21506 194.465 10.30557 6.484256 0.175884
1.791626 22.65332 210.6084 12.26874 7.668728 0.189383
2.245705 26.25931 227.2109 14.35052 8.915294 0.20121
2.816255 30.03179 244.2728 16.55009 10.22344 0.215976
3.988786 38.07191 279.7762 21.29991 13.02271 0.234476
5.341985 46.7663 317.1203 26.51374 16.06359 0.24956
Ans.
Use a 169 −12 UNC × 3.5 in long bolt
______________________________________________________________________________
8-24 Equation (f ), p. 428:
Eq. (8-17):
kb =
C=
kb
kb + k m
Ad At E
Ad lt + At ld
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Top frustum, Eq. (8-20), with E = 10.3Mpsi, D = 1.5 d, and t = l /2:
0.5774π (10.3) d
k1 =
 1.155 l / 2 + 0.5d 
ln 5

 1.155 l / 2 + 2.5d 
Middle frustum, with E = 10.3 Mpsi, D = 1.5d + 2(l − 0.5) tan 30°, and t = 0.5 − l /2
0.5774π (10.3) d
k2 =
1.155 ( 0.5 − 0.5l ) + 0.5d + 2 ( l − 0.5 ) tan 300   2.5d + 2 ( l − 0.5 ) tan 300 
ln
1.155 ( 0.5 − 0.5l ) + 2.5d + 2 ( l − 0.5 ) tan 300  0.5d + 2 ( l − 0.5 ) tan 300 
{
{
}
}
Lower frustum, with E = 30Mpsi, D = 1.5 d, t = l − 0.5
0.5774π ( 30 ) d
k3 =
 1.155 ( l − 0.5 ) + 0.5d  
ln 5  

 1.155 ( l − 0.5 ) + 2.5d  
See Table 8-7 for other terms used.
Using a spreadsheet, with coarse-pitch bolts (units are in, in2, Mlbf/in)
Size
1
2
3
4
5
6
8
10
d
0.073
0.086
0.099
0.112
0.125
0.138
0.164
0.19
At
0.00263
0.0037
0.00487
0.00604
0.00796
0.00909
0.014
0.0175
Ad
0.004185
0.005809
0.007698
0.009852
0.012272
0.014957
0.021124
0.028353
L>
0.6095
0.629
0.6485
0.668
0.6875
0.707
0.746
0.785
L
0.75
0.75
0.75
0.75
0.75
0.75
0.75
1
LT
0.396
0.422
0.448
0.474
0.5
0.526
0.578
0.63
l
0.5365
0.543
0.5495
0.556
0.5625
0.569
0.582
0.595
ld
0.354
0.328
0.302
0.276
0.25
0.224
0.172
0.37
Size
1
2
3
4
5
6
8
10
d
0.073
0.086
0.099
0.112
0.125
0.138
0.164
0.19
lt
0.1825
0.215
0.2475
0.28
0.3125
0.345
0.41
0.225
kb
0.194841
0.261839
0.333134
0.403377
0.503097
0.566787
0.801537
1.15799
k1
1.084468
1.321595
1.570439
1.830494
2.101297
2.382414
2.974009
3.602349
k2
1.954599
2.449694
2.993366
3.587564
4.234381
4.936066
6.513824
8.342138
k3
7.09432
8.357692
9.621064
10.88444
12.14781
13.41118
15.93792
18.46467
km
0.635049
0.778497
0.930427
1.090613
1.258846
1.434931
1.809923
2.214214
C
0.23478
0.251687
0.263647
0.27
0.285535
0.28315
0.306931
0.343393
Use a 2−56 UNC × 0.75 in long bolt.
Ans.
______________________________________________________________________________
Shigley’s MED, 10th edition
Chap. 8 Solutions, Page 19/70
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8-25 For half of joint, Eq. (8-20): t = 20 mm, d = 14 mm, D = 21 mm, and E = 207 GPa
k1 =
0.5774π ( 207 )14
1.155 ( 20 ) + 21 − 14  ( 21 + 14 )
ln 
1.155 ( 20 ) + 21 + 14  ( 21 − 14 )
= 5523 MN/m
km = (1/k1 + 1/k1)−1 = k1/2 = 5523/2 = 2762 MN/m
Ans.
From Eq. (8-22) with l = 40 mm
km =
0.5774π ( 207 )14
 0.5774 ( 40 ) + 0.5 (14 ) 
2 ln 5

 0.5774 ( 40 ) + 2.5 (14 ) 
= 2762 MN/m
Ans.
which agrees with the earlier calculation.
For Eq. (8-23), from Table 8-8, A = 0.787 15, B = 0.628 73
km = 207(14)(0.78 715) exp [0.628 73(14)/40] = 2843 MN/m
Ans.
This is 2.9% higher than the earlier calculations.
______________________________________________________________________________
8-26 (a) Grip, l = 10 in. Nut height, H = 41/64 in (Table A-31).
L ≥ l + H = 10 + 41/64 = 10.641 in. Let L = 10.75 in.
Table 8-7, LT = 2d + 0.5 = 2(0.75) + 0.5 = 2 in, ld = L − LT = 10.75 − 2 = 8.75 in,
lt = l − ld = 10 − 8.75 = 1.25 in
Ad = π(0.752)/4 = 0.4418 in2, At = 0.373 in2 (Table 8-2)
Eq. (8-17),
0.4418 ( 0.373) 30
Ad At E
kb =
Ans.
=
= 1.296 Mlbf/in
Ad lt + At ld 0.4418 (1.25 ) + 0.373 ( 8.75 )
Eq. (4-4), p. 163,
2
2
Am Em (π / 4 ) (1.125 − 0.75 ) 30
=
= 1.657 Mlbf/in
km =
Ans.
l
10
Eq. (f), p. 428, C = kb/(kb + km) = 1.296/(1.296 + 1.657) = 0.439
Shigley’s MED, 10th edition
Ans.
Chap. 8 Solutions, Page 20/70
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(b)
Let: Nt = no. of turns, p = pitch of thread (in), N = no. of threads per in = 1/p. Then,
δ = δb + δm = Nt p = Nt / N
(1)
But, δb = Fi / kb, and, δm = Fi / km. Substituting these into Eq. (1) and solving for Fi gives
Fi =
kb k m N t
kb + k m N
( 2)
1.296 (1.657 )106 1/ 3
= 15 150 lbf
Ans.
1.296 + 1.657 16
______________________________________________________________________________
=
8-27 Proof for the turn-of-nut equation is given in the solution of Prob. 8-26, Eq. (2), where
Nt = θ / 360°.
The relationship between the turn-of-nut method and the torque-wrench method is as
follows.
 k + km 
Nt =  b
 Fi N
 kb k m 
T = KFd
i
(turn-of-nut)
(torque-wrench)
Eliminate Fi
 k + km  NT
θ
=
Nt =  b
Ans.
o

k
k
Kd
360
 b m 
______________________________________________________________________________
8-28 (a) From Ex. 8-4, Fi = 14.4 kip, kb = 5.21(106) lbf/in, km = 8.95(106) lbf/in
Eq. (8-27):
T = kFid = 0.2(14.4)(103)(5/8) = 1800 lbf · in Ans.
From Prob. 8-27,
Shigley’s MED, 10th edition
Chap. 8 Solutions, Page 21/70
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 5.21 + 8.95 
 k + km 
Nt =  b
(14.4)(103 )11
 Fi N = 
6
k
k
5.21
8.95
10
( ) 
 b m 

o
= 0.0481 turns = 17.3
Ans.
Bolt group is (1.5) / (5/8) = 2.4 diameters. Answer is much lower than RB&W
recommendations.
______________________________________________________________________________
8-29 C = kb / (kb + km) = 3/(3+12) = 0.2, P = Ptotal/ N = 80/6 = 13.33 kips/bolt
Table 8-2, At = 0.141 9 in2; Table 8-9, Sp = 120 kpsi; Eqs. (8-31) and (8-32),
Fi = 0.75 At Sp = 0.75(0.141 9)(120) = 12.77 kips
(a) From Eq. (8-28), the factor of safety for yielding is
S p At
np =
120 ( 0.141 9 )
=
CP + Fi 0.2 (13.33) + 12.77
(b) From Eq. (8-29), the overload factor is
nL =
S p At − Fi
CP
=
= 1.10
120 ( 0.141 9 ) − 12.77
0.2 (13.33)
Ans.
= 1.60
Ans.
(c) From Eq. (803), the joint separation factor of safety is
Fi
12.77
=
= 1.20
Ans.
P (1 − C ) 13.33 (1 − 0.2 )
______________________________________________________________________________
n0 =
1/2 − 13 UNC Grade 8 bolt, K = 0.20
(a) Proof strength, Table 8-9, Sp = 120 kpsi
Table 8-2, At = 0.141 9 in2
Maximum, Fi = Sp At = 120(0.141 9) = 17.0 kips
Ans.
(b) From Prob. 8-29, C = 0.2, P = 13.33 kips
Joint separation, Eq. (8-30) with n0 = 1
Minimum Fi = P (1 − C) = 13.33(1 − 0.2) = 10.66 kips Ans.
(c) Fi = (17.0 + 10.66)/2 = 13.8 kips
Ans.
Eq. (8-27), T = KFi d = 0.2(13.8)103(0.5)/12 = 115 lbf ⋅ ft
______________________________________________________________________________
8-30
8-31 (a) Table 8-1, At = 20.1 mm2. Table 8-11, Sp = 380 MPa.
Eq. (8-31), Fi = 0.75 Fp = 0.75 At Sp = 0.75(20.1)380(10−3) = 5.73 kN
Eq. (f ), p. 428,
C=
kb
1
=
= 0.278
kb + k m 1 + 2.6
Shigley’s MED, 10th edition
Chap. 8 Solutions, Page 22/70
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Eq. (8-28) with np = 1,
−3
S p At − Fi 0.25S p At 0.25 ( 20.1) 380 (10 )
P=
=
=
= 6.869 kN
C
C
0.278
Ptotal = NP = 8(6.869) = 55.0 kN Ans.
(b) Eq. (8-30) with n0 = 1,
F
5.73
P= i =
= 7.94 kN
1 − C 1 − 0.278
Ptotal = NP = 8(7.94) = 63.5 kN Ans. Bolt stress would exceed proof strength
______________________________________________________________________________
8-32 (a) Table 8-2, At = 0.141 9 in2. Table 8-9, Sp = 120 kpsi.
Eq. (8-31), Fi = 0.75 Fp = 0.75 At Sp = 0.75(0.141 9)120 = 12.77 kips
Eq. (f ), p. 428,
C=
kb
4
=
= 0.25
kb + k m 4 + 12
Eq. (8-28) with np = 1,
 S A − Fi  0.25 NS p At
Ptotal = N  p t
=
C
C


80 ( 0.25 )
P C
N = total
=
= 4.70
0.25S p At 0.25 (120 ) 0.141 9
Round to N = 5 bolts Ans.
(b) Eq. (8-30) with n0 = 1,
 F 
Ptotal = N  i 
 1− C 
P (1 − C ) 80 (1 − 0.25 )
N = total
=
= 4.70
Fi
12.77
Round to N = 5 bolts Ans.
Shigley’s MED, 10th edition
Chap. 8 Solutions, Page 23/70
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______________________________________________________________________________
Problems 8-33 through 8-44 and 8-51 through 8-58. Note to the instructor: In previous
editions of this solutions manual, in Probs. 8-33 through 8-36 the cylinder diameter was
mistakenly used, instead of the effective gasket diameter, in determining the external force from
the cylinder pressure. To correct this, the second printing of the 10th edition changes the problem
statement so that the old value of the cylinder diameter is used for the effective gasket diameter.
Consequently, the solutions to these problems, as well as several others that reference these
problems remain numerically the same as before, with the diameter being redefined in the
problem statements.
______________________________________________________________________________
8-33 Bolts: From Table A-31, the nut height is H = 10.8 mm. L ≥ l +H = 40 + 10.8 = 50.8
mm.
Round up to L = 55 mm
Ans.
Eq. (8-14): LT = 2d + 6 = 2(12) + 6 = 30 mm
Table 8-7: ld = L − LT = 55 − 30 = 25 mm, lt = l −ld = 40 − 25 = 15 mm
Ad =π (122)/4 = 113.1 mm2, Table 8-1: At = 84.3 mm2
Eq. (8-17):
kb =
113.1( 84.3) 207
Ad At E
=
= 518.8 MN/m
Ad lt + At ld 113.1(15 ) + 84.3 ( 25 )
Members: Steel cyl. head: t = 20 mm, d = 12 mm, D = 18 mm, E = 207 GPa. Eq. (8-20),
k1 =
0.5774π ( 207 )12
1.155 ( 20 ) + 18 − 12  (18 + 12 )
ln 
1.155 ( 20 ) + 18 + 12  (18 − 12 )
= 4470 MN/m
Cast iron: t = 20 mm, d = 12 mm, D = 18 mm, E = 100 GPa (from
Table 8-8). The only difference from k1 is the material
k2 = (100/207)(4470) = 2159 MN/m
Eq. (8-18): km = (1/4470 + 1/2159)−1 = 1456 MN/m
C = kb / (kb + km) = 518.8/(518.8+1456) = 0.263
Table 8-11:
Sp = 650 MPa
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Chap. 8 Solutions, Page 24/70
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For a non-permanent connection, using Eqs. (8-31) and (8-32)
Fi = 0.75 At Sp = 0.75(84.3)(650)10−3 = 41.1 kN
The total external load is Ptotal = pg Ag, where Ag is the effective area of the cylinder, based
on the effective sealing diameter of 100 mm. The external load per bolt is P = Ptotal /N.
Thus
P = [6π (1002)/4](10−3)/10 = 4.712 kN/bolt
Yielding factor of safety, Eq. (8-28):
np =
S p At
CP + Fi
650 ( 84.3)10−3
= 1.29
0.263 ( 4.712 ) + 41.10
=
Ans.
Overload factor of safety, Eq. (8-29):
nL =
S p At − Fi
CP
=
650 ( 84.3)10−3 − 41.10
= 11.1
0.263 ( 4.712 )
Ans.
Separation factor of safety, Eq. (8-30):
Fi
41.10
=
= 11.8
Ans.
P (1 − C ) 4.712 (1 − 0.263)
______________________________________________________________________________
n0 =
8-34
Instructors: See note preceding Prob. 8-33.
Bolts: Grip, l = 1/2 + 5/8 = 1.125 in. From Table A-31, the nut height is H = 7/16 in.
L ≥ l + H = 1.125 + 7/16 = 1.563 in.
Round up to L = 1.75 in
Ans.
Eq. (8-13): LT = 2d + 0.25 = 2(0.5) + 0.25 = 1.25 in
Table 8-7: ld = L − LT = 1.75 − 1.25 = 0.5 in, lt = l −ld = 1.125 − 0.5 = 0.625 in
Ad = π (0.52)/4 = 0.196 3 in2, Table 8-2: At = 0.141 9 in2
Eq. (8-17):
kb =
0.196 3 ( 0.141 9 ) 30
Ad At E
=
= 4.316 Mlbf/in
Ad lt + At ld 0.196 3 ( 0.625 ) + 0.141 9 ( 0.5 )
Members: Steel cyl. head: t = 0.5 in, d = 0.5 in, D = 0.75 in, E = 30 Mpsi. Eq. (8-20),
Shigley’s MED, 10th edition
Chap. 8 Solutions, Page 25/70
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k1 =
0.5774π ( 30 ) 0.5
1.155 ( 0.5 ) + 0.75 − 0.5 ( 0.75 + 0.5 )
ln 
1.155 ( 0.5 ) + 0.75 + 0.5 ( 0.75 − 0.5 )
= 33.30 Mlbf/in
Cast iron: Has two frusta. Midpoint of complete joint is at (1/2 + 5/8)/2 =
0.5625 in.
Upper frustum, t = 0.5625 −0.5 = 0.0625 in, d = 0.5 in,
D = 0.75 + 2(0.5) tan 30° = 1.327 in, E = 14.5 Mpsi (from Table 8-8)
Eq. (8-20) ⇒ k2 = 292.7 Mlbf/in
Lower frustum, t = 0.5625 in, d = 0.5 in, D = 0.75 in, E = 14.5 Mpsi
Eq. (8-20) ⇒ k3 = 15.26 Mlbf/in
Eq. (8-18): km = (1/33.30 + 1/292.7 + 1/15.26)−1 = 10.10 Mlbf/in
C = kb / (kb + km) = 4.316/(4.316+10.10) = 0.299
Table 8-9: Sp = 85 kpsi
For a non-permanent connection, using Eqs. (8-31) and (8-32)
Fi = 0.75 At Sp = 0.75(0.141 9)(85) = 9.05 kips
The total external load is Ptotal = pg Ag, where Ag is the effective area of the cylinder, based
on the effective sealing diameter of 3.5 in. The external load per bolt is P = Ptotal /N. Thus,
P = [1 500π (3.52)/4](10−3)/10 = 1.443 kips/bolt
Yielding factor of safety, Eq. (8-28):
S A
85 ( 0.141 9 )
np = p t =
= 1.27
CP + Fi 0.299 (1.443) + 9.05
Ans.
Overload factor of safety, Eq. (8-29):
nL =
S p At − Fi
CP
=
85 ( 0.141 9 ) − 9.05
0.299 (1.443)
= 6.98
Ans.
Separation factor of safety, Eq. (8-30):
Fi
9.05
=
= 8.95
Ans.
P (1 − C ) 1.443 (1 − 0.299 )
______________________________________________________________________________
n0 =
Shigley’s MED, 10th edition
Chap. 8 Solutions, Page 26/70
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8-35 Instructors: See note preceding Prob. 8-33.
Bolts: Grip: l = 20 + 25 = 45 mm. From Table A-31, the nut height is H = 8.4 mm.
L ≥ l +H = 45 + 8.4 = 53.4 mm.
Round up to L = 55 mm
Ans.
Eq. (8-14): LT = 2d + 6 = 2(10) + 6 = 26 mm
Table 8-7: ld = L − LT = 55 − 26 = 29 mm, lt = l −ld = 45 − 29 = 16 mm
Ad =π (102)/4 = 78.5 mm2, Table 8-1: At = 58.0 mm2
Eq. (8-17):
kb =
78.5 ( 58.0 ) 207
Ad At E
=
= 320.8 MN/m
Ad lt + At ld 78.5 (16 ) + 58.0 ( 29 )
Members: Steel cyl. head: t = 20 mm, d = 10 mm, D = 15 mm, E = 207 GPa. Eq. (8-20),
k1 =
0.5774π ( 207 )10
1.155 ( 20 ) + 15 − 10  (15 + 10 )
ln 
1.155 ( 20 ) + 15 + 10  (15 − 10 )
= 3503 MN/m
Cast iron: Has two frusta. Midpoint of complete joint is at (20 + 25)/2 = 22.5 mm
Upper frustum, t = 22.5 − 20 = 2.5 mm, d = 10 mm,
D = 15 + 2(20) tan 30° = 38.09 mm, E = 100 GPa (from Table 8-8),
Eq. (8-20) ⇒ k2 = 45 880 MN/m
Lower frustum, t = 22.5 mm, d = 10 mm, D = 15 mm, E = 100 GPa
Eq. (8-20) ⇒ k3 = 1632 MN/m
Eq. (8-18): km = (1/3503 + 1/45 880 + 1/1632)−1 = 1087 MN/m
C = kb / (kb + km) = 320.8/(320.8+1087) = 0.228
Table 8-11:
Sp = 830 MPa
For a non-permanent connection, using Eqs. (8-31) and (8-32)
Fi = 0.75 At Sp = 0.75(58.0)(830)10−3 = 36.1 kN
The total external load is Ptotal = pg Ag, where Ag is the effective area of the cylinder, based
on the effective sealing diameter of 0.8 m. The external load per bolt is P = Ptotal /N. Thus,
P = [550π (0.82)/4]/36 = 7.679 kN/bolt
Shigley’s MED, 10th edition
Chap. 8 Solutions, Page 27/70
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Yielding factor of safety, Eq. (8-28):
830 ( 58.0 )10−3
np =
=
= 1.27
CP + Fi 0.228 ( 7.679 ) + 36.1
S p At
Ans.
Overload factor of safety, Eq. (8-29):
nL =
S p At − Fi
CP
=
830 ( 58.0 )10−3 − 36.1
0.228 ( 7.679 )
= 6.88 Ans.
Separation factor of safety, Eq. (8-30):
Fi
36.1
Ans.
=
= 6.09
P (1 − C ) 7.679 (1 − 0.228 )
______________________________________________________________________________
n0 =
8-36 Instructors: See note preceding Prob. 8-33.
Bolts: Grip, l = 3/8 + 1/2 = 0.875 in. From Table A-31, the nut height is H = 3/8 in.
L ≥ l + H = 0.875 + 3/8 = 1.25 in.
Let L = 1.25 in Ans.
Eq. (8-13): LT = 2d + 0.25 = 2(7/16) + 0.25 = 1.125 in
Table 8-7: ld = L − LT = 1.25 − 1.125 = 0.125 in, lt = l −ld = 0.875 − 0.125 =
0.75 in
Ad = π (7/16)2/4 = 0.150 3 in2, Table 8-2: At = 0.106 3 in2
Eq. (8-17),
kb =
0.150 3 ( 0.106 3) 30
Ad At E
=
= 3.804 Mlbf/in
Ad lt + At ld 0.150 3 ( 0.75 ) + 0.106 3 ( 0.125 )
Members: Steel cyl. head: t = 0.375 in, d = 0.4375 in, D = 0.65625 in, E = 30 Mpsi. Eq.
(8-20),
k1 =
0.5774π ( 30 ) 0.4375
1.155 ( 0.375 ) + 0.65625 − 0.4375 ( 0.65625 + 0.4375 )
ln 
1.155 ( 0.375 ) + 0.65625 + 0.4375 ( 0.65625 − 0.4375 )
= 31.40 Mlbf/in
Cast iron: Has two frusta. Midpoint of complete joint is at (3/8 + 1/2)/2 =
0.4375 in.
Shigley’s MED, 10th edition
Chap. 8 Solutions, Page 28/70
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Upper frustum, t = 0.4375 −0.375 = 0.0625 in, d = 0.4375 in,
D = 0.65625 + 2(0.375) tan 30° = 1.089 in, E = 14.5 Mpsi (from Table
8-8)
Eq. (8-20) ⇒ k2 = 195.5 Mlbf/in
Lower frustum, t = 0.4375 in, d = 0.4375 in, D = 0.65625 in, E = 14.5
Mpsi
Eq. (8-20) ⇒ k3 = 14.08 Mlbf/in
Eq. (8-18): km = (1/31.40 + 1/195.5 + 1/14.08)−1 = 9.261 Mlbf/in
C = kb / (kb + km) = 3.804/(3.804 + 9.261) = 0.291
Table 8-9: Sp = 120 kpsi
For a non-permanent connection, using Eqs. (8-31) and (8-32)
Fi = 0.75 At Sp = 0.75(0.106 3)(120) = 9.57 kips
The total external load is Ptotal = pg Ag, where Ag is the effective area of the cylinder, based
on the effective sealing diameter of 3.25 in. The external load per bolt is P = Ptotal /N.
Thus,
P = [1 200π (3.252)/4](10−3)/8 = 1.244 kips/bolt
Yielding factor of safety, Eq. (8-28):
S p At
120 ( 0.106 3)
=
= 1.28
np =
Ans.
CP + Fi 0.291(1.244 ) + 9.57
Overload factor of safety, Eq. (8-29):
nL =
S p At − Fi
CP
=
120 ( 0.106 3) − 9.57
= 8.80 Ans.
0.291(1.244 )
Separation factor of safety, Eq. (8-30):
n0 =
Fi
9.57
=
= 10.9
P (1 − C ) 1.244 (1 − 0.291)
Ans.
______________________________________________________________________________
8-37 Instructors: See note preceding Prob. 8-33.
From Table 8-7, h = t1 = 20 mm
For t2 > d, l = h + d /2 = 20 + 12/2 = 26 mm
L ≥ h + 1.5 d = 20 + 1.5(12) = 38 mm. Round up to L = 40 mm
LT = 2d + 6 = 2(12) + 6 = 30 mm
Shigley’s MED, 10th edition
Chap. 8 Solutions, Page 29/70
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ld = L − LT = 40 − 20 = 10 mm
lt = l − ld = 26 − 10 = 16 mm
From Table 8-1, At = 84.3 mm2. Ad = π (122)/4 = 113.1 mm2
Eq. (8-17),
113.1( 84.3) 207
Ad At E
kb =
=
= 744.0 MN/m
Ad lt + At ld 113.1(16 ) + 84.3 (10 )
Similar to Fig. 8-21, we have three frusta.
Top frusta, steel: t = l / 2 = 13 mm, d = 12 mm, D = 18 mm, E = 207 GPa. Eq. (8-20)
k1 =
0.5774π ( 207 )12
1.155 (13) + 18 − 12  (18 + 12 )
ln 
1.155 (13) + 18 + 12  (18 − 12 )
= 5 316 MN/m
Middle frusta, steel: t = 20 − 13 = 7 mm, d = 12 mm, D = 18 + 2(13 − 7) tan 30° = 24.93
mm, E = 207 GPa. Eq. (8-20) ⇒ k2 = 15 660 MN/m
Lower frusta, cast iron: t = 26 − 20 = 6 mm, d = 12 mm, D = 18 mm, E = 100 GPa (see
Table 8-8). Eq. (8-20) ⇒ k3 = 3 887 MN/m
Eq. (8-18),
km = (1/5 316 + 1/15 660 + 1/3 887)−1 = 1 964 MN/m
C = kb / (kb + km) = 744.0/(744.0 + 1 964) = 0.275
Table 8-11: Sp = 650 MPa. For a non-permanent connection, using Eqs. (8-31) and (8-32)
Fi = 0.75 At Sp = 0.75(84.3)(650)10−3 = 41.1 kN
The total external load is Ptotal = pg Ag, where Ag is the effective area of the cylinder, based
on the effective sealing diameter of 100 mm. The external load per bolt is P = Ptotal /N.
Thus
P = [6π (1002)/4](10−3)/10 = 4.712 kN/bolt
Yielding factor of safety, Eq. (8-28)
np =
S p At
CP + Fi
=
650 ( 84.3)10−3
0.275 ( 4.712 ) + 41.1
= 1.29
Ans.
Overload factor of safety, Eq. (8-29)
nL =
S p At − Fi
CP
=
650 ( 84.3)10−3 − 41.1
= 10.7
0.275 ( 4.712 )
Shigley’s MED, 10th edition
Ans.
Chap. 8 Solutions, Page 30/70
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Separation factor of safety, Eq. (8-30)
Fi
41.1
=
= 12.0
Ans.
P (1 − C ) 4.712 (1 − 0.275 )
______________________________________________________________________________
n0 =
8-38 Instructors: See note preceding Prob. 8-33.
From Table 8-7, h = t1 = 0.5 in
For t2 > d, l = h + d /2 = 0.5 + 0.5/2 = 0.75 in
L ≥ h + 1.5 d = 0.5 + 1.5(0.5) = 1.25 in. Let L = 1.25 in
LT = 2d + 0.25 = 2(0.5) + 0.25 = 1.25 in. All threaded.
From Table 8-1, At = 0.141 9 in2. The bolt stiffness is kb = At E / l = 0.141 9(30)/0.75 =
5.676 Mlbf/in
Similar to Fig. 8-21, we have three frusta.
Top frusta, steel: t = l / 2 = 0.375 in, d = 0.5 in, D = 0.75 in, E = 30 Mpsi
0.5774π ( 30 ) 0.5
k1 =
= 38.45 Mlbf/in
1.155 ( 0.375 ) + 0.75 − 0.5 ( 0.75 + 0.5 )
ln
1.155 ( 0.375 ) + 0.75 + 0.5 ( 0.75 − 0.5 )
Middle frusta, steel: t = 0.5 − 0.375 = 0.125 in, d = 0.5 in,
D = 0.75 + 2(0.75 − 0.5) tan 30° = 1.039 in, E = 30 Mpsi.
Eq. (8-20) ⇒ k2 = 184.3 Mlbf/in
Lower frusta, cast iron: t = 0.75 − 0.5 = 0.25 in, d = 0.5 in, D = 0.75 in, E = 14.5 Mpsi.
Eq. (8-20) ⇒ k3 = 23.49 Mlbf/in
Eq. (8-18),
km = (1/38.45 + 1/184.3 + 1/23.49)−1 = 13.51 Mlbf/in
C = kb / (kb + km) = 5.676 / (5.676 + 13.51) = 0.296
Table 8-9, Sp = 85 kpsi. For a non-permanent connection, using Eqs. (8-31) and (8-32)
Fi = 0.75 At Sp = 0.75(0.141 9)(85) = 9.05 kips
The total external load is Ptotal = pg Ag, where Ag is the effective area of the cylinder, based
on the effective sealing diameter of 3.5 in. The external load per bolt is P = Ptotal /N. Thus
P = [1 500π (3.52)/4](10−3)/10 = 1.443 kips/bolt
Yielding factor of safety, Eq. (8-28)
np =
S p At
CP + Fi
=
85 ( 0.141 9 )
= 1.27
0.296 (1.443) + 9.05
Ans.
Overload factor of safety, Eq. (8-29)
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Chap. 8 Solutions, Page 31/70
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nL =
S p At − Fi
CP
=
85 ( 0.141 9 ) − 9.05
0.296 (1.443)
= 7.05
Ans.
Separation factor of safety, Eq. (8-30)
Fi
9.05
Ans.
=
= 8.91
P (1 − C ) 1.443 (1 − 0.296 )
______________________________________________________________________________
n0 =
8-39 Instructors: See note preceding Prob. 8-33.
From Table 8-7, h = t1 = 20 mm
For t2 > d, l = h + d /2 = 20 + 10/2 = 25 mm
L ≥ h + 1.5 d = 20 + 1.5(10) = 35 mm. Let L = 35 mm
LT = 2d + 6 = 2(10) + 6 = 26 mm
ld = L − LT = 35 − 26 = 9 mm
lt = l − ld = 25 − 9 = 16 mm
From Table 8-1, At = 58.0 mm2. Ad = π (102)/4 = 78.5 mm2
Eq. (8-17),
78.5 ( 58.0 ) 207
Ad At E
kb =
=
= 530.1 MN/m
Ad lt + At ld 78.5 (16 ) + 58.0 ( 9 )
Similar to Fig. 8-21, we have three frusta.
Top frusta, steel: t = l / 2 = 12.5 mm, d = 10 mm, D = 15 mm, E = 207 GPa. Eq. (8-20)
0.5774π ( 207 )10
= 4 163 MN/m
k1 =
1.155 (12.5 ) + 15 − 10  (15 + 10 )
ln
1.155 (12.5 ) + 15 + 10  (15 − 10 )
Middle frusta, steel: t = 20 − 12.5 = 7.5 mm, d = 10 mm, D = 15 + 2(12.5 − 7.5) tan 30° =
20.77 mm, E = 207 GPa. Eq. (8-20) ⇒ k2 = 10 975 MN/m
Lower frusta, cast iron: t = 25 − 20 = 5 mm, d = 10 mm, D = 15 mm, E = 100 GPa (see
Table 8-8). Eq. (8-20) ⇒ k3 = 3 239 MN/m
Eq. (8-18),
km = (1/4 163 + 1/10 975 + 1/3 239)−1 = 1 562 MN/m
C = kb / (kb + km) = 530.1/(530.1 + 1 562) = 0.253
Table 8-11: Sp = 830 MPa. For a non-permanent connection, using Eqs. (8-31) and (8-32)
Fi = 0.75 At Sp = 0.75(58.0)(830)10−3 = 36.1 kN
The total external load is Ptotal = pg Ag, where Ag is the effective area of the cylinder, based
on the effective sealing diameter of 0.8 m. The external load per bolt is P = Ptotal /N. Thus
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P = [550π (0.82)/4]/36 = 7.679 kN/bolt
Yielding factor of safety, Eq. (8-28)
np =
S p At
CP + Fi
=
830 ( 58.0 )10−3
= 1.27
0.253 ( 7.679 ) + 36.1
Ans.
Overload factor of safety, Eq. (8-29)
nL =
S p At − Fi
CP
=
830 ( 58.0 )10−3 − 36.1
= 6.20
0.253 ( 7.679 )
Ans.
Separation factor of safety, Eq. (8-30)
Fi
36.1
Ans.
=
= 6.29
P (1 − C ) 7.679 (1 − 0.253)
______________________________________________________________________________
n0 =
8-40 Instructors: See note preceding Prob. 8-33.
From Table 8-7, h = t1 = 0.375 in
For t2 > d, l = h + d /2 = 0.375 + 0.4375/2 = 0.59375 in
L ≥ h + 1.5 d = 0.375 + 1.5(0.4375) = 1.031 in. Round up to L = 1.25 in
LT = 2d + 0.25 = 2(0.4375) + 0.25 = 1.125 in
ld = L − LT = 1.25 − 1.125 = 0.125
lt = l − ld = 0.59375 − 0.125 = 0.46875 in
Ad = π (7/16)2/4 = 0.150 3 in2, Table 8-2: At = 0.106 3 in2
Eq. (8-17),
0.150 3 ( 0.106 3) 30
Ad At E
=
= 5.724 Mlbf/in
Ad lt + At ld 0.150 3 ( 0.46875 ) + 0.106 3 ( 0.125 )
Similar to Fig. 8-21, we have three frusta.
Top frusta, steel: t = l / 2 = 0.296875 in, d = 0.4375 in, D = 0.65625 in, E = 30 Mpsi
kb =
k1 =
0.5774π ( 30 ) 0.4375
1.155 ( 0.296875 ) + 0.656255 − 0.4375 ( 0.75 + 0.656255 )
ln 
1.155 ( 0.296875 ) + 0.75 + 0.656255 ( 0.75 − 0.656255 )
= 35.52 Mlbf/in
Middle frusta, steel: t = 0.375 − 0.296875 = 0.078125 in, d = 0.4375 in,
D = 0.65625 + 2(0.59375 − 0.375) tan 30° = 0.9088 in, E = 30 Mpsi.
Eq. (8-20) ⇒ k2 = 215.8 Mlbf/in
Lower frusta, cast iron: t = 0.59375 − 0.375 = 0.21875 in, d = 0.4375 in, D = 0.65625 in,
E = 14.5 Mpsi. Eq. (8-20) ⇒ k3 = 20.55 Mlbf/in
Shigley’s MED, 10th edition
Chap. 8 Solutions, Page 33/70
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Eq. (8-18),
km = (1/35.52 + 1/215.8 + 1/20.55)−1 = 12.28 Mlbf/in
C = kb / (kb + km) = 5.724/(5.724 + 12.28) = 0.318
Table 8-9, Sp = 120 kpsi. For a non-permanent connection, using Eqs. (8-31) and (8-32)
Fi = 0.75 At Sp = 0.75(0.106 3)(120) = 9.57 kips
The total external load is Ptotal = pg Ag, where Ag is the effective area of the cylinder, based
on the effective sealing diameter of 3.25 in. The external load per bolt is P = Ptotal /N.
Thus
P = [1 200π (3.252)/4](10−3)/8 = 1.244 kips/bolt
Yielding factor of safety, Eq. (8-28)
np =
S p At
CP + Fi
=
120 ( 0.106 3)
= 1.28
0.318 (1.244 ) + 9.57
Ans.
Overload factor of safety, Eq. (8-29)
nL =
S p At − Fi
CP
=
120 ( 0.106 3) − 9.57
0.318 (1.244 )
= 8.05
Ans.
Separation factor of safety, Eq. (8-30)
Fi
9.57
=
= 11.3
Ans.
P (1 − C ) 1.244 (1 − 0.318 )
______________________________________________________________________________
n0 =
8-41 Instructors: See note preceding Prob. 8-33.
This is a design problem and there is no closed-form solution path or a unique solution.
What is presented here is one possible iterative approach. We will demonstrate this with
an example.
1. Select the diameter, d. For this example, let d = 10 mm. Using Eq. (8-20) on members,
and combining using Eq. (8-18), yields km = 1 141 MN/m (see Prob. 8-33 for method of
calculation.
2. Look up the nut height in Table A-31. For the example, H = 8.4 mm. From this, L is
rounded up from the calculation of l + H = 40 + 8.4 = 48.4 mm to 50 mm. Next,
calculations are made for LT = 2(10) + 6 = 26 mm, ld = 50 − 26 = 24 mm, lt = 40 − 24 = 16
mm. From step 1, Ad =π (102)/4 = 78.54 mm2. Next, from Table 8-1, At = 78.54 mm2.
From Eq. (8-17), kb = 356 MN/m. Finally, from Eq. (f), p. 428, C = 0.238.
Shigley’s MED, 10th edition
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3. From Prob. 8-33, the bolt circle diameter is E = 200 mm. Substituting this for Db in Eq.
(8-34), the number of bolts are
N=
π Db
=
π ( 200 )
4d
4 (10 )
Rounding this up gives N = 16.
= 15.7
4. Next, select a grade bolt. Based on the solution to Prob. 8-33, the strength of ISO 9.8
was so high to give very large factors of safety for overload and separation. Try ISO 4.6
with Sp = 225 MPa. From Eqs. (8-31) and (8-32) for a non-permanent connection, Fi =
9.79 kN.
5. The external load requirement per bolt is P = 1.15 pg Ag/N, where from Prob 8-33, pg =
6 MPa, and Ag = π (1002)/4. This gives P = 3.39 kN/bolt.
6. Using Eqs. (8-28) to (8-30) yield np = 1.23, nL = 4.05, and n0 = 3.79.
Steps 1 - 6 can be easily implemented on a spreadsheet with lookup tables for the tables
used from the text. The results for four bolt sizes are shown below. The dimension of each
term is consistent with the example given above.
d
8
10
12
14
km
854
1141
1456
1950
H
6.8
8.4
10.8
12.8
L
50
50
55
55
LT
22
26
30
34
ld
28
24
25
21
lt
12
16
15
19
d
8
10
12
14
C
0.215
0.238
0.263
0.276
N
20
16
13*
12
Sp
225
225
225
225
Fi
6.18
9.79
14.23
19.41
P
2.71
3.39
4.17
4.52
np
1.22
1.23
1.24
1.25
Ad
At
kb
50.26 36.6 233.9
78.54 58
356
113.1 84.3 518.8
153.9 115 686.3
nL
3.53
4.05
4.33
5.19
n0
2.90
3.79
4.63
5.94
*Rounded down from13.08997, so spacing is slightly greater than four diameters.
Any one of the solutions is acceptable. A decision-maker might be cost such as
N × cost/bolt, and/or N × cost per hole, etc.
________________________________________________________________________
Shigley’s MED, 10th edition
Chap. 8 Solutions, Page 35/70
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8-42 Instructors: See note preceding Prob. 8-33.
This is a design problem and there is no closed-form solution path or a unique solution.
What is presented here is one possible iterative approach. We will demonstrate this with
an example.
1. Select the diameter, d. For this example, let d = 0.5 in. Using Eq. (8-20) on three frusta
(see Prob. 8-34 solution), and combining using Eq. (8-19), yields km = 10.10 Mlbf/in.
2. Look up the nut height in Table A-31. For the example, H = 0.4375 in. From this, L is
rounded up from the calculation of l + H = 1.125 + 0.4375 = 1.5625 in to 1.75 in. Next,
calculations are made for LT = 2(0.5) + 0.25 = 1.25 in, ld = 1.75 − 1.25 = 0.5 in, lt = 1.125
− 0.5 = 0.625 in. From step 1, Ad =π (0.52)/4 = 0.1963 in2. Next, from Table 8-1, At =
0.141 9 in2. From Eq. (8-17), kb = 4.316 Mlbf/in. Finally, from Eq. (f), p. 428, C = 0.299.
3. From Prob. 8-34, the bolt circle diameter is E = 6 in. Substituting this for Db in Eq. (834), for the number of bolts
N=
π Db
=
π ( 6)
4d
4 ( 0.5 )
Rounding this up gives N = 10.
= 9.425
4. Next, select a grade bolt. Based on the solution to Prob. 8-34, the strength of SAE grade
5 was adequate. Use this with Sp = 85 kpsi. From Eqs. (8-31) and (8-32) for a nonpermanent connection, Fi = 9.046 kips.
5. The external load requirement per bolt is P = 1.15 pg Ag/N, where from Prob 8-34,
pg = 1 500 psi, and Ag = π (3.52)/4 . This gives P = 1.660 kips/bolt.
6. Using Eqs. (8-28) to (8-30) yield np = 1.26, nL = 6.07, and n0 = 7.78.
d
0.375
0.4375
0.5
0.5625
km
6.75
9.17
10.10
11.98
H
0.3281
0.375
0.4375
0.4844
d
0.375
0.4375
0.5
0.5625
C
0.261
0.273
0.299
0.308
N
13
11
10
9
L
LT
ld
1.5
1
0.5
1.5 1.125 0.375
1.75 1.25 0.5
1.75 1.375 0.375
Sp
85
85
85
85
Fi
4.941
6.777
9.046
11.6
P
1.277
1.509
1.660
1.844
lt
0.625
0.75
0.625
0.75
np
1.25
1.26
1.26
1.27
Ad
At
0.1104 0.0775
0.1503 0.1063
0.1963 0.1419
0.2485 0.182
nL
4.95
5.48
6.07
6.81
kb
2.383
3.141
4.316
5.329
n0
5.24
6.18
7.78
9.09
Any one of the solutions is acceptable. A decision-maker might be cost such as
N × cost/bolt, and/or N × cost per hole, etc.
________________________________________________________________________
Shigley’s MED, 10th edition
Chap. 8 Solutions, Page 36/70
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8-43 Instructors: See note preceding Prob. 8-33.
This is a design problem and there is no closed-form solution path or a unique solution.
What is presented here is one possible iterative approach. We will demonstrate this with
an example.
1. Select the diameter, d. For this example, let d = 10 mm. Using Eq. (8-20) on three frusta
(see Prob. 8-35 solution), and combining using Eq. (8-19), yields km = 1 087 MN/m.
2. Look up the nut height in Table A-31. For the example, H = 8.4 mm. From this, L is
rounded up from the calculation of l + H = 45 + 8.4 = 53.4 mm to 55 mm. Next,
calculations are made for LT = 2(10) + 6 = 26 mm, ld = 55 − 26 = 29 mm, lt = 45 − 29 = 16
mm. From step 1, Ad =π (102)/4 = 78.54 mm2. Next, from Table 8-1, At = 58.0 mm2. From
Eq. (8-17), kb = 320.9 MN/m. Finally, from Eq. (f), p. 428, C = 0.228.
3. From Prob. 8-35, the bolt circle diameter is E = 1000 mm. Substituting this for Db in
Eq. (8-34), for the number of bolts
N=
π Db
=
π (1000 )
= 78.5
4d
4 (10 )
Rounding this up gives N = 79. A rather large number, since the bolt circle diameter, E is
so large. Try larger bolts.
4. Next, select a grade bolt. Based on the solution to Prob. 8-35, the strength of ISO 9.8
was so high to give very large factors of safety for overload and separation. Try ISO 5.8
with Sp = 380 MPa. From Eqs. (8-31) and (8-32) for a non-permanent connection, Fi =
16.53 kN.
5. The external load requirement per bolt is P = 1.15 pg Ag/N, where from Prob 8-35, pg =
0.550 MPa, and Ag = π (8002)/4 . This gives P = 4.024 kN/bolt.
6. Using Eqs. (8-28) to (8-30) yield np = 1.26, nL = 6.01, and n0 = 5.32.
Steps 1 - 6 can be easily implemented on a spreadsheet with lookup tables for the tables
used from the text. The results for three bolt sizes are shown below. The dimension of
each term is consistent with the example given above.
d
10
20
36
km
1087
3055
6725
H
8.4
18
31
L
55
65
80
d
10
20
36
C
0.228
0.308
0.361
N
79
40
22
Sp
380
380
380
LT
26
46
78
ld
29
19
2
lt
16
26
43
Fi
P
np
16.53 4.024 1.26
69.83 7.948 1.29
232.8 14.45 1.3
Shigley’s MED, 10th edition
Ad
78.54
314.2
1018
At
58
245
817
nL
6.01
9.5
14.9
n0
5.32
12.7
25.2
kb
320.9
1242
3791
Chap. 8 Solutions, Page 37/70
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A large range is presented here. Any one of the solutions is acceptable. A decision-maker
might be cost such as N × cost/bolt, and/or N × cost per hole, etc.
________________________________________________________________________
8-44 Instructors: See note preceding Prob. 8-33.
This is a design problem and there is no closed-form solution path or a unique solution.
What is presented here is one possible iterative approach. We will demonstrate this with
an example.
1. Select the diameter, d. For this example, let d = 0.375 in. Using Eq. (8-20) on three
frusta (see Prob. 8-36 solution), and combining using Eq. (8-19), yields km = 7.42 Mlbf/in.
2. Look up the nut height in Table A-31. For the example, H = 0.3281 in. From this,
L ≥ l + H = 0.875 + 0.3281 = 1.2031 in. Rounding up, L = 1.25. Next, calculations are
made for LT = 2(0.375) + 0.25 = 1 in, ld = 1.25 − 1 = 0.25 in, lt = 0.875 − 0.25 = 0.625 in.
From step 1, Ad =π (0.3752)/4 = 0.1104 in2. Next, from Table 8-1, At = 0.0775 in2. From
Eq. (8-17), kb = 2.905 Mlbf/in. Finally, from Eq. (f), p. 428, C = 0.263.
3. From Prob. 8-36, the bolt circle diameter is E = 6 in. Substituting this for Db in Eq. (834), for the number of bolts
N=
π Db
=
π (6)
4d
4 ( 0.375 )
Rounding this up gives N = 13.
= 12.6
4. Next, select a grade bolt. Based on the solution to Prob. 8-36, the strength of SAE grade
8 seemed high for overload and separation. Try SAE grade 5 with Sp = 85 kpsi. From Eqs.
(8-31) and (8-32) for a non-permanent connection, Fi = 4.941 kips.
5. The external load requirement per bolt is P = 1.15 pg Ag/N, where from Prob 8-34,
pg = 1 200 psi, and Ag = π (3.252)/4. This gives P = 0.881 kips/bolt.
6. Using Eqs. (8-28) to (8-30) yield np = 1.27, nL = 6.65, and n0 = 7.81.
Steps 1 - 6 can be easily implemented on a spreadsheet with lookup tables for the tables
used from the text. For this solution we only looked at one bolt size, 83 − 16 , but evaluated
changing the bolt grade. The results for four bolt grades are shown below. The dimension
of each term is consistent with the example given above.
H
L
d
km
0.375 7.42 0.3281 1.25
LT
1
ld
lt
Ad
At
kb
0.25 0.625 0.1104 0.0775 2.905
Shigley’s MED, 10th edition
Chap. 8 Solutions, Page 38/70
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d
0.375
0.375
0.375
0.375
C
0.281
0.281
0.281
0.281
N
13
13
13
13
SAE
grade
1
2
4
5
Sp
33
55
65
85
Fi
1.918
3.197
3.778
4.941
P
0.881
0.881
0.881
0.881
np
1.18
1.24
1.25
1.27
nL
2.58
4.30
5.08
6.65
n0
3.03
5.05
5.97
7.81
Note that changing the bolt grade only affects Sp, Fi , np, nL, and n0. Any one of the
solutions is acceptable, especially the lowest grade bolt.
________________________________________________________________________
8-45 (a) Fb′ = RFb′,max sin θ
Half of the external moment is contributed by the line load in the interval 0 ≤ θ ≤ π
π
M
= ∫ Fb′R 2 sin θ dθ =
0
2
M
π
= Fb′,max R 2
2
2
π
0
Fb′,max R 2 sin 2 θ dθ
M
from which Fb′,max =
Fmax =
∫
π R2
φ2
∫φ
Fb′R sin θ dθ =
1
M φ2
M
(cos φ1 - cos φ2 )
R sin θ dθ =
2 ∫φ
πR 1
πR
Noting φ1 = 75°, φ2 = 105°,
Fmax =
12 000
(cos 75o - cos105o ) = 494 lbf
π (8 / 2)
Fmax = Fb′,max R∆φ =
(b)
Fmax =
Ans.
M
2M
 2π 
( R)   =
2
 N
RN
πR
2(12 000)
= 500 lbf
(8 / 2)(12)
Ans.
(c) F = Fmax sin θ
M = 2 Fmax R [(1) sin2 90° + 2 sin2 60° + 2 sin2 30° + (1) sin2 (0)] = 6FmaxR
from which,
Fmax =
M
12 000
=
= 500 lbf
6R
6(8 / 2)
Shigley’s MED, 10th edition
Ans.
Chap. 8 Solutions, Page 39/70
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The simple general equation resulted from part (b)
2M
RN
________________________________________________________________________
Fmax =
8-46
(a) From Table 8-11, Sp = 600 MPa. From Table 8-1, At = 353 mm2.
Eq. (8-31):
Table 8-15:
Fi = 0.9 At S p = 0.9 ( 353 )( 600 ) (10 −3 ) = 190.6 kN
K = 0.18
Eq. (8-27):
T = K Fi d = 0.18(190.6)(24) = 823 N⋅m
Ans.
(b) Washers: t = 4.6 mm, d = 24 mm, D = 1.5(24) = 36 mm, E = 207 GPa.
Eq. (8-20),
k1 =
0.5774π ( 207 ) 24
1.155 ( 4.6 ) + 36 − 24  ( 36 + 24 )
ln 
1.155 ( 4.6 ) + 36 + 24  ( 36 − 24 )
= 31 990 MN/m
Cast iron: t = 20 mm, d = 24 mm, D = 36 + 2(4.6) tan 30° = 41.31 mm, E = 135 GPa.
Eq. (8-20) ⇒ k2 = 10 785 MN/m
Steel joist: t = 20 mm, d = 24 mm, D = 41.31 mm, E = 207 GPa. Eq. (8-20) ⇒ k3 = 16
537 MN/m
Eq. (8-18):
km = (2 / 31 990 + 1 / 10 785 +1 / 16 537)−1 = 4 636 MN/m
Bolt: l = 2(4.6) + 2(20) = 49.2 mm. Nut, Table A-31, H = 21.5 mm. L > 49.2 + 21.5 = 70.7
mm. From Table A-17, use L = 80 mm. From Eq. (8-14)
LT = 2(24) + 6 = 54 mm, ld = 80 − 54 = 26 mm, lt = 49.2 − 26 = 23.2 mm
From Table (8-1), At = 353 mm2, Ad = π (242) / 4 = 452.4 mm2
Eq. (8-17):
kb =
452.4 ( 353) 207
Ad At E
=
= 1680 MN/m
Ad lt + At ld 452.4 ( 23.2 ) + 353 ( 26 )
C = kb / (kb + km) = 1680 / (1680 + 4636) = 0.266, Sp = 600 MPa, Fi = 190.6 kN,
P = Ptotal / N = 18/4 = 4.5 kN
Shigley’s MED, 10th edition
Chap. 8 Solutions, Page 40/70
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Yield: From Eq. (8-28)
np =
S p At
CP + Fi
=
600 ( 353)10−3
= 1.10 Ans.
0.266 ( 4.5 ) + 190.6
Load factor: From Eq. (8-29)
nL =
S p At − Fi
CP
=
600 ( 353)10−3 − 190.6
= 17.7
0.266 ( 4.5 )
Ans.
Separation: From Eq. (8-30)
n0 =
Fi
190.6
=
= 57.7
P (1 − C ) 4.5 (1 − 0.266 )
Ans.
As was stated in the text, bolts are typically preloaded such that the yielding factor of
safety is not much greater than unity which is the case for this problem. However, the
other load factors indicate that the bolts are oversized for the external load.
______________________________________________________________________________
8-47 (a) ISO M 20 × 2.5 grade 8.8 coarse pitch bolts, lubricated.
Table 8-2,
At = 245 mm2
Table 8-11,
Sp = 600 MPa
Fi = 0.90 At Sp = 0.90(245)600(10−3) = 132.3 kN
Table 8-15,
K = 0.18
Eq. (8-27),
T = KFi d = 0.18(132.3)20 = 476 N ⋅ m
Ans.
(b) Table A-31, H = 18 mm, L ≥ LG + H = 48 + 18 = 66 mm. Round up to L = 80 mm per
Table A-17.
LT = 2d + 6 = 2(20) + 6 = 46 mm
ld = L - LT = 80 − 46 = 34 mm
lt = l - ld = 48 − 34 = 14 mm
Ad = π (202) /4 = 314.2 mm2,
kb =
Ad At E
314.2(245)(207)
=
= 1251.9 MN/m
Ad lt + Alt d
314.2(14) + 245(34)
Members: Since all members are steel use Eq. (8-22) with E = 207 MPa, l = 48 mm, d =
20mm
Shigley’s MED, 10th edition
Chap. 8 Solutions, Page 41/70
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km =
0.5774π ( 207 ) 20
0.5774π Ed
=
= 4236 MN/m
 0.5774l + 0.5d 
 0.5774 ( 48 ) + 0.5 ( 20 ) 
2 ln  5
 2 ln 5

 0.5774l + 2.5d 
 0.5774 ( 48 ) + 2.5 ( 20 ) 
kb
1251.9
=
= 0.228
kb + km 1251.9 + 4236
P = Ptotal / N = 40/2 = 20 kN,
C=
Yield: From Eq. (8-28)
S A
600 ( 245 )10−3
np = p t =
= 1.07 Ans.
CP + Fi 0.228 ( 20 ) + 132.3
Load factor: From Eq. (8-29)
nL =
S p At − Fi
CP
=
600 ( 245 )10 −3 − 132.3
0.228 ( 20 )
= 3.22
Ans.
Separation: From Eq. (8-30)
Fi
132.3
=
= 8.57
Ans.
P (1 − C ) 20 (1 − 0.228 )
______________________________________________________________________________
n0 =
8-48 From Prob. 8-29 solution, Pmax =13.33 kips, C = 0.2, Fi = 12.77 kips, At = 0.141 9 in2
F
12.77
σi = i =
= 90.0 kpsi
At 0.141 9
CP 0.2 (13.33)
Eq. (8-39),
σa =
=
= 9.39 kpsi
2 At 2 ( 0.141 9 )
Eq. (8-41),
σ m = σ a + σ i = 9.39 + 90.0 = 99.39 kpsi
(a) Goodman Eq. (8-45) for grade 8 bolts, Se = 23.2 kpsi (Table 8-17), Sut = 150 kpsi
(Table 8-9)
S ( S − σ i ) 23.2 (150 − 90.0 )
=
= 0.856
n f = e ut
Ans.
σ a ( Sut + Se ) 9.39 (150 + 23.2 )
(b) Gerber Eq. (8-46)
1 
nf =
Sut Sut2 + 4 Se ( Se + σ i ) − Sut2 − 2σ i Se 

2σ a Se 
=
1
150 1502 + 4 ( 23.2 )( 23.2 + 90.0 ) − 1502 − 2 ( 90.0 ) 23.2  = 1.32

2 ( 9.39 ) 23.2 
Ans.
(c) ASME-elliptic Eq. (8-47) with Sp = 120 kpsi (Table 8-9)
Shigley’s MED, 10th edition
Chap. 8 Solutions, Page 42/70
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nf =
=
(
Se
S p S p2 + Se2 − σ i2 − σ i Se
2
2
σ a ( S p + Se )
)
23.2
120 1202 + 23.22 − 902 − 90 ( 23.2 )  = 1.30
2
2 

9.39 (120 + 23.2 )
Ans.
______________________________________________________________________________
8-49 Attention to the Instructor. Part (d) requires the determination of the endurance strength,
Se, of a class 5.8 bolt. Table 8-17 does not provide this and the student will be required to
estimate it by other means [see the solution of part (d)].
Per bolt, Pbmax = 60/8 = 7.5 kN, Pbmin = 20/8 = 2.5 kN
kb
1
C=
=
= 0.278
kb + k m 1 + 2.6
(a) Table 8-1, At = 20.1 mm2; Table 8-11, Sp = 380 MPa
Eqs. (8-31) and (8-32), Fi = 0.75 At Sp = 0.75(20.1)380(10−3) = 5.73 kN
S A
380 ( 20.1)10−3
np = p t =
= 0.98
Ans.
Yield, Eq. (8-28),
CP + Fi 0.278 ( 7.5) + 5.73
S p At − Fi
(b) Overload, Eq. (8-29),
nL =
(c) Separation, Eq. (8-30),
n0 =
(d) Goodman, Eq. (8-35),
σa =
Eq. (8-36), σ m =
C ( Pb max + Pb min )
2 At
CP
=
380 ( 20.1)10−3 − 5.73
0.278 ( 7.5 )
Fi
5.73
=
= 1.06
P (1 − C ) 7.5 (1 − 0.278)
+
C ( Pb max − Pb min )
2 At
=
= 0.915
Ans.
Ans.
0.278 ( 7.5 − 2.5 )103
2 ( 20.1)
3
Fi 0.278 ( 7.5 + 2.5 )10 5.73 (10
=
+
At
2 ( 20.1)
20.1
3
= 34.6 MPa
) = 354.2
MPa
Table 8-11, Sut = 520 MPa, σi = Fi /At = 5.73(103)/20.1 = 285 MPa
We have a problem for Se. Table 8-17 does not list Se for class 5.8 bolts. Here, we will
estimate Se using the methods of Chapter 6. Estimate S e′ from the,
Eq. (6-8), p. 290,
Se′ = 0.5Sut = 0.5 ( 520 ) = 260 MPa .
Table 6-2, p. 296,
Eq. (6-19), p. 295,
a = 4.51, b = − 0.265
ka = aSutb = 4.51( 520 −0.265 ) = 0.860
Eq. (6-21), p. 296,
kb = 1
Eq. (6-26), p. 298,
kc = 0.85
The fatigue stress-concentration factor, from Table 8-16, is Kf = 2.2. For simple axial
loading and infinite-life it is acceptable to reduce the endurance limit by Kf and use the
nominal stresses in the stress/strength/design factor equations. Thus,
Eq. (6-18), p. 295,
Se = ka kb kc S e′ / Kf = 0.86(1)0.85(260) / 2.2 = 86.4 MPa
Shigley’s MED, 10th edition
Chap. 8 Solutions, Page 43/70
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Eq. (8-38),
nf =
Se ( Sut − σ i )
Sutσ a + Se (σ m − σ i )
=
86.4 ( 520 − 285 )
520 ( 34.6 ) + 86.4 ( 354.2 − 285 )
= 0.847
Ans.
It is obvious from the various answers obtained, the bolted assembly is undersized. This
can be rectified by a one or more of the following: more bolts, larger bolts, higher class
bolts.
______________________________________________________________________________
8-50 Per bolt, Pbmax = Pmax /N = 80/10 = 8 kips, Pbmin = Pmin /N = 20/10 = 2 kips
C = kb / (kb + km) = 4/(4 + 12) = 0.25
(a) Table 8-2, At = 0.141 9 in2, Table 8-9, Sp = 120 kpsi and Sut = 150 kpsi
Table 8-17, Se = 23.2 kpsi
Eqs. (8-31) and (8-32), Fi = 0.75 At Sp ⇒ σi = Fi /At = 0.75 Sp = 0.75(120) =90 kpsi
Eq. (8-35),
σa =
Eq. (8-36),
σm =
Eq. (8-38),
nf =
C ( Pb max − Pb min )
2 At
C ( Pb max + Pb min )
2 At
Se ( Sut − σ i )
=
=
0.25 ( 8 − 2 )
2 ( 0.141 9 )
+σi =
= 5.29 kpsi
0.25 ( 8 + 2 )
2 ( 0.141 9 )
+ 90 = 98.81 kpsi
23.2 (150 − 90 )
= 1.39
Ans.
Sutσ a + Se (σ m − σ i ) 150 ( 5.29 ) + 23.2 ( 98.81 − 90 )
______________________________________________________________________________
8-51 Instructors: See note preceding Prob. 8-33.
From Prob. 8-33, C = 0.263, Pmax = 4.712 kN / bolt, Fi = 41.1 kN, Sp = 650 MPa, and
At = 84.3 mm2
σi = 0.75 Sp = 0.75(650) = 487.5 MPa
3
CP 0.263 ( 4.712 )10
σa =
=
= 7.350 MPa
Eq. (8-39):
2 At
2 ( 84.3)
Eq. (8-40)
σm =
CP Fi
+ = 7.350 + 487.5 = 494.9 MPa
2 At At
(a) Goodman: From Table 8-11, Sut = 900 MPa, and from Table 8-17, Se = 140 MPa
S ( S − σ i ) 140 ( 900 − 487.5 )
Eq. (8-45):
=
= 7.55
n f = e ut
Ans.
σ a ( Sut + Se ) 7.350 ( 900 + 140 )
(b) Gerber:
Eq. (8-46):
Shigley’s MED, 10th edition
Chap. 8 Solutions, Page 44/70
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nf =
=
1
 S S 2 + 4 S ( S + σ ) − S 2 − 2σ S 
ut
ut
e
e
i
ut
i e

2σ a Se 
1
900 9002 + 4 (140 )(140 + 487.5 ) − 9002 − 2 ( 487.5 )(140 ) 

2 ( 7.350 )140 
= 11.4
Ans.
(c) ASME-elliptic:
Eq. (8-47):
nf =
=
(
Se
S p S p2 + Se2 − σ i2 − σ i Se
2
2
σ a ( S p + Se )
)
140
650 650 2 + 1402 − 487.52 − 487.5 (140 )  = 9.73

7.350 ( 650 2 + 1402 ) 
Ans.
______________________________________________________________________________
8-52 Instructors: See note preceding Prob. 8-33.
From Prob. 8-34, C = 0.299, Pmax = 1.443 kips/bolt, Fi = 9.05 kips, Sp = 85 kpsi, and
At = 0.141 9 in2
σ i = 0.75S p = 0.75 (85) = 63.75 kpsi
Eq. (8-37):
σa =
CP 0.299 (1.443)
=
= 1.520 kpsi
2 At
2 ( 0.141 9 )
Eq. (8-38)
σm =
CP
+ σ i = 1.520 + 63.75 = 65.27 kpsi
2 At
(a) Goodman: From Table 8-9, Sut = 120 kpsi, and from Table 8-17, Se = 18.8 kpsi
S ( S − σ i ) 18.8 (120 − 63.75 )
Eq. (8-45):
n f = e ut
Ans.
=
= 5.01
σ a ( Sut + Se ) 1.520 (120 + 18.8 )
(b) Gerber:
Eq. (8-46):
Shigley’s MED, 10th edition
Chap. 8 Solutions, Page 45/70
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nf =
=
1
 S S 2 + 4 S ( S + σ ) − S 2 − 2σ S 
ut
ut
e
e
i
ut
i e

2σ a Se 
1
120 1202 + 4 (18.6 )(18.6 + 63.75 ) − 1202 − 2 ( 63.75 )(18.6 ) 

2 (1.520 )18.6 
= 7.45
Ans.
(c) ASME-elliptic:
Eq. (8-47):
nf =
=
(
Se
S p S p2 + S e2 − σ i2 − σ i Se
2
2
σ a ( S p + Se )
)
18.6
85 852 + 18.62 − 63.752 − 63.75 (18.6 )  = 6.22

1.520 ( 852 + 18.62 ) 
Ans.
______________________________________________________________________________
8-53 Instructors: See note preceding Prob. 8-33.
From Prob. 8-35, C = 0.228, Pmax = 7.679 kN/bolt, Fi = 36.1 kN, Sp = 830 MPa, and
At = 58.0 mm2
σi = 0.75 Sp = 0.75(830) = 622.5 MPa
3
CP 0.228 ( 7.679 )10
Eq. (8-37):
σa =
=
= 15.09 MPa
2 At
2 ( 58.0 )
CP
+ σ i = 15.09 + 622.5 = 637.6 MPa
2 At
(a) Goodman: From Table 8-11, Sut = 1040 MPa, and from Table 8-17, Se = 162 MPa
S ( S − σ i ) 162 (1040 − 622.5 )
Eq. (8-45):
=
= 3.73
n f = e ut
Ans.
σ a ( Sut + Se ) 15.09 (1040 + 162 )
Eq. (8-38)
σm =
Shigley’s MED, 10th edition
Chap. 8 Solutions, Page 46/70
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(b) Gerber:
Eq. (8-46):
1 
nf =
Sut Sut2 + 4 Se ( Se + σ i ) − Sut2 − 2σ i Se 

2σ a Se 
=
1
1040 10402 + 4 (162 )(162 + 622.5 ) − 10402 − 2 ( 622.5 )(162 ) 

2 (15.09 )162 
= 5.74
Ans.
(c) ASME-elliptic:
Eq. (8-47):
nf =
=
(
Se
S p S p2 + Se2 − σ i2 − σ i Se
2
2
σ a ( S p + Se )
)
162
830 8302 + 1622 − 622.52 − 622.5 (162 )  = 5.62
2
2 

15.09 ( 830 + 162 )
Ans.
______________________________________________________________________________
8-54 Instructors: See note preceding Prob. 8-33.
From Prob. 8-36, C = 0.291, Pmax = 1.244 kips/bolt, Fi = 9.57 kips, Sp = 120 kpsi, and
At = 0.106 3 in2
σ i = 0.75S p = 0.75 (120 ) = 90 kpsi
Eq. (8-37):
σa =
CP 0.291(1.244 )
=
= 1.703 kpsi
2 At
2 ( 0.106 3)
Eq. (8-38)
σm =
CP
+ σ i = 1.703 + 90 = 91.70 kpsi
2 At
(a) Goodman: From Table 8-9, Sut = 150 kpsi, and from Table 8-17, Se = 23.2 kpsi
S (S −σi )
23.2 (150 − 90 )
Eq. (8-45):
=
= 4.72
n f = e ut
Ans.
σ a ( Sut + Se ) 1.703 (150 + 23.2 )
(b) Gerber:
Eq. (8-46):
Shigley’s MED, 10th edition
Chap. 8 Solutions, Page 47/70
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nf =
=
1
 S S 2 + 4 S ( S + σ ) − S 2 − 2σ S 
ut
ut
e
e
i
ut
i e

2σ a Se 
1
150 1502 + 4 ( 23.2 )( 23.2 + 90 ) − 1502 − 2 ( 90 )( 23.2 ) 

2 (1.703) 23.2 
= 7.28
Ans.
(c) ASME-elliptic:
Eq. (8-47):
nf =
=
(
Se
S p S p2 + S e2 − σ i2 − σ i Se
2
2
σ a ( S p + Se )
)
23.2
120 1202 + 23.22 − 902 − 90 ( 23.2 )  = 7.24

1.703 (1202 + 18.62 ) 
Ans.
______________________________________________________________________________
8-55 Instructors: See note preceding Prob. 8-33.
From Prob. 8-51, C = 0.263, Se = 140 MPa, Sut = 900 MPa, At = 84.4 mm2, σi = 487.5
MPa, and Pmax = 4.712 kN.
Pmin = Pmax / 2 = 4.712/2 = 2.356 kN
σa =
Eq. (8-35):
Eq. (8-36):
σm =
C ( Pmax − Pmin )
2 At
=
0.263 ( 4.712 − 2.356 )103
2 ( 84.3)
= 3.675 MPa
C ( Pmax + Pmin )
+σi
2 At
0.263 ( 4.712 + 2.356 )103
=
+ 487.5 = 498.5 MPa
2 ( 84.3)
Eq. (8-38):
Se ( Sut − σ i )
140 ( 900 − 487.5 )
=
= 11.9
Ans.
Sutσ a + Se (σ m − σ i ) 900 ( 3.675 ) + 140 ( 498.5 − 487.5 )
______________________________________________________________________________
nf =
Shigley’s MED, 10th edition
Chap. 8 Solutions, Page 48/70
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8-56 Instructors: See note preceding Prob. 8-33.
From Prob. 8-52, C = 0.299, Se = 18.8 kpsi, Sut = 120 kpsi, At = 0.141 9 in2, σi = 63.75
kpsi, and Pmax = 1.443 kips
Pmin = Pmax / 2 = 1.443/2 = 0.722 kips
σa =
Eq. (8-35):
Eq. (8-36):
σm =
=
C ( Pmax − Pmin ) 0.299 (1.443 − 0.722 )
=
= 0.760 kpsi
2 At
2 ( 0.141 9 )
C ( Pmax + Pmin )
+σi
2 At
0.299 (1.443 + 0.722 )
+ 63.75 = 66.03 kpsi
2 ( 0.141 9 )
Eq. (8-38):
nf =
Se ( Sut − σ i )
=
18.8 (120 − 63.75 )
= 7.89
Ans.
Sutσ a + Se (σ m − σ i ) 120 ( 0.760 ) + 18.8 ( 66.03 − 63.75 )
______________________________________________________________________________
8-57 Instructors: See note preceding Prob. 8-33.
From Prob. 8-53, C = 0.228, Se = 162 MPa, Sut = 1040 MPa, At = 58.0 mm2, σi = 622.5
MPa, and Pmax = 7.679 kN.
Pmin = Pmax / 2 = 7.679/2 = 3.840 kN
σa =
Eq. (8-35):
Eq. (8-36):
σm =
=
C ( Pmax − Pmin )
2 At
=
0.228 ( 7.679 − 3.840 )103
2 ( 58.0 )
= 7.546 MPa
C ( Pmax + Pmin )
+σi
2 At
0.228 ( 7.679 + 3.840 )103
+ 622.5 = 645.1 MPa
2 ( 58.0 )
Eq. (8-38):
Shigley’s MED, 10th edition
Chap. 8 Solutions, Page 49/70
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Se ( Sut − σ i )
162 (1040 − 622.5 )
Ans.
=
= 5.88
Sutσ a + Se (σ m − σ i ) 1040 ( 7.546 ) + 162 ( 645.1 − 622.5 )
______________________________________________________________________________
nf =
8-58 Instructors: See note preceding Prob. 8-33.
From Prob. 8-54, C = 0.291, Se = 23.2 kpsi, Sut = 150 kpsi, At = 0.106 3 in2, σi = 90
kpsi, and Pmax = 1.244 kips
Pmin = Pmax / 2 = 1.244/2 = 0.622 kips
σa =
Eq. (8-35):
Eq. (8-36):
σm =
=
C ( Pmax − Pmin )
2 At
=
0.291(1.244 − 0.622 )
2 ( 0.106 3)
= 0.851 kpsi
C ( Pmax + Pmin )
+σi
2 At
0.291(1.244 + 0.622 )
+ 90 = 92.55 kpsi
2 ( 0.106 3)
Eq. (8-38):
Se ( Sut − σ i )
23.2 (150 − 90 )
=
= 7.45
Ans.
Sutσ a + Se (σ m − σ i ) 150 ( 0.851) + 23.2 ( 92.55 − 90 )
______________________________________________________________________________
nf =
8-59 Let the repeatedly-applied load be designated as P. From Table A-22, Sut = 93.7 kpsi.
Referring to the Figure of Prob. 3-122, the following notation will be used for the radii of
Section AA.
ri = 1.5 in, ro = 2.5 in, rc = 2.0 in
From Table 3-4, p. 135, with R = 0.5 in
R2
0.52
=
= 1.968 246 in
rn =
2 rc − rc2 − R 2
2 2 − 22 − 0.52
e
co
ci
A
=
=
=
=
(
)
(
)
rc − rn = 2.0 − 1.968 246 = 0.031 754 in
ro - rn = 2.5 − 1.968 246 = 0.531 754 in
rn - ri = 1.968 246 − 1.5 = 0.468 246 in
π (12 ) / 4 = 0.7854 in 2
If P is the maximum load
Shigley’s MED, 10th edition
Chap. 8 Solutions, Page 50/70
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M = Prc = 2 P
P
rc 
P 
2(0.468) 
σ i = 1 + c i  =
1 +
 = 26.29P
A
eri  0.785 4 
0.031 754(1.5) 
26.294 P
σ
σa = σm = i =
= 13.15P
2
2
(a) Eye: Section AA,
Table 6-2, p. 296, a = 14.4 kpsi, b = − 0.718
Eq. (6-19), p. 295,
ka = 14.4(93.7) −0.718 = 0.553
Eq. (6-23), p. 297,
de = 0.370 d
Eq. (6-20), p. 296,
 0.37 
kb = 

 0.30 
−0.107
= 0.978
Eq. (6-26), p. 298,
kc = 0.85
Eq. (6-8), p. 290,
Se′ = 0.5Sut = 0.5 ( 93.7 ) = 46.85 kpsi
Eq. (6-18) p. 295,
Se = 0.553(0.978)0.85(46.85) = 21.5 kpsi
From Table 6-7, p. 315, for Gerber
2
2

 2σ m Se  
1  Sut  σ a 
nf = 
−1 + 1 + 


2  σ m  Se 
 Sutσ a  

With σm = σa,
2
2


 2Se   1
1 Sut2 
93.7 2
 2(21.5)   1.557

nf =
−1 + 1 + 
−1 + 1 + 
 =
 =
P
2 σ a Se 
 93.7  
 Sut   2 13.15P(21.5) 



where P is in kips.
Thread: Die cut. Table 8-17 gives Se = 18.6 kpsi for rolled threads. Use Table 8-16 to find
Se for die cut threads
Se = 18.6(3.0/3.8) = 14.7 kpsi
2
Table 8-2, At = 0.663 in , σ = P/At = P /0.663 = 1.51 P, σa = σm =σ /2 = 0.755 P
From Table 6-7, Gerber
2
2


 2S e   1
1 Sut2 
93.7 2
 2(14.7)   19.01

1
1
nf =
−1 + 1 + 
=
−
+
+


 =
2 σ a Se 
Sut   2 0.755P(14.7) 
P
 93.7  





Shigley’s MED, 10th edition
Chap. 8 Solutions, Page 51/70
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Comparing 1910/P with 19 200/P, we conclude that the eye is weaker in fatigue. Ans.
(b) Strengthening steps can include heat treatment, cold forming, cross section change (a
round is a poor cross section for a curved bar in bending because the bulk of the material
is located where the stress is small). Ans.
(c) For nf = 2
P =
1.557 (103 )
= 779 lbf, max. load Ans.
2
______________________________________________________________________________
8-60 Member, Eq. (8-22) with E =16 Mpsi, d = 0.75 in, and l = 1.5 in
km =
0.5774π (16 ) 0.75
0.5774π Ed
=
= 13.32 Mlbf/in
 0.5774l + 0.5d 
 0.5774 (1.5 ) + 0.5 ( 0.75 ) 
2 ln  5
 2 ln 5

 0.5774l + 2.5d 
 0.5774 (1.5 ) + 2.5 ( 0.75 ) 
Bolt, Eq. (8-13),
LT = 2d + 0.25 = 2(0.75) + 0.25 = 1.75 in
l = 1.5 in
ld = L − LT = 2.5 − 1.75 = 0.75 in
lt = l − ld = 1.5 − 0.75 = 0.75 in
Table 8-2,
At = 0.373 in2
Ad = π(0.752)/4 = 0.442 in2
Eq. (8-17),
kb =
0.442 ( 0.373) 30
Ad At E
=
= 8.09 Mlbf/in
Ad lt + At ld 0.442 ( 0.75 ) + 0.373 ( 0.75 )
C=
kb
8.09
=
= 0.378
kb + k m 8.09 + 13.32
Eq. (8-35),
σa =
C ( Pmax − Pmin )
2 At
=
0.378 ( 6 − 4 )
2 ( 0.373)
= 1.013 kpsi
Eq.(8-36),
Shigley’s MED, 10th edition
Chap. 8 Solutions, Page 52/70
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σm =
C ( Pmax + Pmin ) Fi 0.378 ( 6 + 4 )
25
+ =
+
= 72.09 kpsi
2 At
2 ( 0.373)
0.373
At
(a) From Table 8-9, Sp = 85 kpsi, and Eq. (8-51), the yielding factor of safety is
np =
Sp
σm +σa
=
85
= 1.16
72.09 + 1.013
Ans.
(b) From Eq. (8-29), the overload factor of safety is
nL =
S p At − Fi
CPmax
=
85 ( 0.373) − 25
0.378 ( 6 )
= 2.96
Ans.
(c) From Eq. (8-30), the factor of safety based on joint separation is
n0 =
Fi
25
=
= 6.70
Pmax (1 − C ) 6 (1 − 0.378 )
Ans.
(d) From Table 8-17, Se = 18.6 kpsi; Table 8-9, Sut = 120 kps; the preload stress is
σi = Fi / At = 25/0.373 = 67.0 kpsi; and from Eq. (8-38)
nf =
Se ( Sut − σ i )
=
18.6 (120 − 67.0 )
= 4.56
Ans.
Sutσ a + Se (σ m − σ i ) 120 (1.013) + 18.6 ( 72.09 − 67.0 )
______________________________________________________________________________
8-61 (a) Table 8-2,
At = 0.1419 in2
Table 8-9,
Sp = 120 kpsi, Sut = 150 kpsi
Table 8-17,
Se = 23.2 kpsi
Eqs. (8-31) and (8-32),
σi = 0.75 Sp = 0.75(120) = 90 kpsi
kb
4
=
= 0.2
kb + k m
4 + 16
CP
0.2 P
σa =
=
= 0.705P kpsi
2 At
2(0.141 9)
C =
Eq. (8-45) for the Goodman criterion,
S (S − σi )
23.2(150 − 90)
11.4
=
=
= 2
n f = e ut
σ a ( Sut + Se ) 0.705P(150 + 23.2)
P
⇒ P = 5.70 kips
Ans.
(b) Fi = 0.75At Sp = 0.75(0.141 9)120 = 12.77 kips
Yield, Eq. (8-28),
Shigley’s MED, 10th edition
Chap. 8 Solutions, Page 53/70
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np =
S p At
=
120 ( 0.141 9 )
= 1.22
0.2 ( 5.70 ) + 12.77
CP + Fi
Load factor, Eq. (8-29),
S A - Fi 120(0.141 9) − 12.77
=
= 3.74
nL = p t
CP
0.2(5.70)
Ans.
Ans.
Separation load factor, Eq. (8-30)
Fi
12.77
=
= 2.80 Ans.
P(1 - C ) 5.70(1 − 0.2)
______________________________________________________________________________
n0 =
8-62 Table 8-2, At = 0.969 in2 (coarse), At = 1.073 in2 (fine)
Table 8-9,
Sp = 74 kpsi, Sut = 105 kpsi
Table 8-17,
Se = 16.3 kpsi
Coarse thread,
Fi = 0.75 At Sp = 0.75(0.969)74 = 53.78 kips
σi = 0.75 Sp = 0.75(74) = 55.5 kpsi
CP
0.30 P
σa =
=
= 0.155P kpsi
2 At
2(0.969)
Gerber, Eq. (8-46),
nf =
1
 S S 2 + 4 S ( S + σ ) − S 2 − 2σ S 
ut
ut
e
e
i
ut
i e

2σ a Se 
1
105 1052 + 4 (16.3)(16.3 + 55.5 ) − 1052 − 2 ( 55.5 )16.3 = 64.28


P
2 ( 0.155 P )16.3
With nf =2,
=
P =
64.28
= 32.14 kip
2
Ans.
Fine thread,
Fi = 0.75 At Sp = 0.75(1.073)74 = 59.55kips
σi = 0.75 Sp = 0.75(74) = 55.5 kpsi
CP
0.32 P
σa =
=
= 0.149 P kpsi
2 At
2(1.073)
The only thing that changes in Eq. (8-46) is σa. Thus,
0.155 64.28 66.87
nf =
=
= 2 ⇒ P = 33.43 kips
0.149 P
P
Shigley’s MED, 10th edition
Ans.
Chap. 8 Solutions, Page 54/70
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Percent improvement,
33.43 − 32.14
(100) 4% Ans.
32.14
______________________________________________________________________________
8-63 For an M 30 × 3.5 ISO 8.8 bolt with P = 65 kN/bolt and C = 0.28
Table 8-1,
Table 8-11,
Table 8-17,
At = 561 mm2
Sp = 600 MPa, Sut = 830 MPa
Se = 129 MPa
Eq. (8-31),
Fi = 0.75Fp = 0.75 At Sp
= 0.75(5610600(10−3) = 252.45 kN
σi = 0.75 Sp = 0.75(600) = 450 MPa
Eq. (8-39),
σa =
CP 0.28 ( 65)10
=
= 16.22 MPa
2 At
2 ( 561)
3
Gerber, Eq. (8-46),
nf =
=
1
 S S 2 + 4 S ( S + σ ) − S 2 − 2σ S 
ut
ut
e
e
i
ut
i e

2σ a Se 
1
830 8302 + 4 (129 )(129 + 450 ) − 8302 − 2 ( 450 )129 

2 (16.22 )129 
= 4.75
Ans.
The yielding factor of safety, from Eq. (8-28) is
np =
S p At
CP + Fi
=
600 ( 561)10−3
0.28 ( 65 ) + 252.45
= 1.24
Ans.
From Eq. (8-29), the load factor is
nL =
S p At − Fi
CP
=
600 ( 561)10−3 − 252.45
= 4.62
0.28 ( 65)
Ans.
The separation factor, from Eq. (8-30) is
Fi
252.45
=
= 5.39
Ans.
P (1 − C ) 65 (1 − 0.28 )
______________________________________________________________________________
n0 =
Shigley’s MED, 10th edition
Chap. 8 Solutions, Page 55/70
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At = 0.077 5 in2
Sp = 85 kpsi, Sut = 120 kpsi
Se = 18.6 kpsi
8-64 (a) Table 8-2,
Table 8-9,
Table 8-17,
Unthreaded grip,
kb =
Am =
Ad E
π (0.375) 2 (30)
=
= 0.245 Mlbf/in per bolt
l
4(13.5)
π
[( D + 2t ) 2 - D 2 ] =
π
(4.752 - 42 ) = 5.154 in 2
4
4
Am E
5.154(30)  1 
km =
=
  = 2.148 Mlbf/in/bolt.
l
12
6
(b)
Ans.
Ans.
Fi = 0.75 At Sp = 0.75(0.0775)(85) = 4.94 kip
σ i = 0.75S p = 0.75(85) = 63.75 kpsi
2000  π
2
P = pA =
(4)  = 4189 lbf/bolt
6  4

kb
0.245
C =
=
= 0.102
kb + k m
0.245 + 2.148
CP
0.102(4.189)
σa =
=
= 2.77 kpsi
2 At
2(0.0775)
From Eq. (8-46) for Gerber fatigue criterion,
nf =
=
1
 S S 2 + 4 S ( S + σ ) − S 2 − 2σ S 
ut
ut
e
e
i
ut
i e

2σ a Se 
1
120 1202 + 4 (18.6 )(18.6 + 63.75 ) − 1202 − 2 ( 63.75 )18.6  = 4.09

2 ( 2.77 )18.6 
Ans.
(c) Pressure causing joint separation from Eq. (8-30)
Fi
=1
P(1 − C )
4.94
Fi
=
= 5.50 kip
P=
1 − C 1 − 0.102
5.50
P
=
p =
6 = 2.63 kpsi Ans.
A π (42 ) / 4
______________________________________________________________________________
n0 =
8-65 From the solution of Prob. 8-64, At = 0.077 5 in2, Sut = 120 kpsi, Se = 18.6 kpsi, C = 0.102,
σi = 63.75 kpsi
Pmax = pmaxA = 2 π (42)/4 = 25.13 kpsi, Pmin = pminA = 1.2 π (42)/4 = 15.08 kpsi,
Shigley’s MED, 10th edition
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Eq. (8-35),
σa =
Eq. (8-36),
σm =
C ( Pmax − Pmin ) 0.102 ( 25.13 − 15.08 )
=
= 6.61 kpsi
2 At
2 ( 0.077 5 )
C ( Pmax + Pmin )
2 At
+σi =
0.102 ( 25.13 + 15.08 )
2 ( 0.077 5 )
+ 63.75 = 90.21 kpsi
Eq. (8-38),
Se ( Sut − σ i )
18.6 (120 − 63.75 )
nf =
=
= 0.814
Sutσ a + Se (σ m − σ i ) 120 ( 6.61) + 18.6 ( 90.21 − 63.75 )
Ans.
This predicts a fatigue failure.
______________________________________________________________________________
8-66 Members: Sy = 57 kpsi, Ssy = 0.577(57) = 32.89 kpsi.
Bolts: SAE grade 5, Sy = 92 kpsi, Ssy = 0.577(92) = 53.08 kpsi
Shear in bolts,
 π (0.252 ) 
2
As = 2 
 = 0.0982 in
4


As Ssy
0.0982(53.08)
=
= 2.61 kips
Fs =
n
2
Bearing on bolts,
Ab = 2(0.25)0.25 = 0.125 in2
AS
0.125(92)
= 5.75 kips
Fb = b yc =
n
2
Bearing on member,
0.125(57)
Fb =
= 3.56 kips
2
Tension of members,
At = (1.25 − 0.25)(0.25) = 0.25 in2
0.25(57)
= 7.13 kip
2
F = min(2.61, 5.75, 3.56, 7.13) = 2.61 kip
Ft =
Ans.
The shear in the bolts controls the design.
______________________________________________________________________________
8-67 Members, Table A-20, Sy = 42 kpsi
Bolts, Table 8-9, Sy = 130 kpsi, Ssy = 0.577(130) = 75.01 kpsi
Shear of bolts,
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 π ( 5 /16 )2 
As = 2 
 = 0.1534 in 2
4


τ=
Fs
5
=
= 32.6 kpsi
As 0.1534
n=
S sy
τ
=
75.01
= 2.30
32.6
Ans.
Bearing on bolts,
Ab = 2(0.25)(5/16) = 0.1563 in2
5
σb = −
= −32.0 kpsi
0.1563
S
130
n= y =
= 4.06
Ans.
σ b 32.0
Bearing on members,
n=
Tension of members,
Sy
σb
=
42
= 1.31
32
Ans.
At = [2.375 − 2(5/16)](1/4) = 0.4375 in2
σt =
5
= 11.4 kpsi
0.4375
Sy
42
= 3.68
Ans.
σ t 11.4
______________________________________________________________________________
n=
=
8-68 Members: Table A-20, Sy = 490 MPa, Ssy = 0.577(490) = 282.7 MPa
Bolts: Table 8-11, ISO class 5.8, Sy = 420 MPa, Ssy = 0.577(420) = 242.3 MPa
Shear in bolts,
 π (202 ) 
As = 2 
= 628.3 mm 2

 4 
AS
628.3(242.3)10−3
= 60.9 kN
Fs = s sy =
n
2.5
Bearing on bolts,
Ab = 2(20)20 = 800 mm2
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Fb =
Ab S yc
n
=
800(420)10−3
= 134 kN
2.5
Bearing on member,
Fb =
800(490)10−3
= 157 kN
2.5
Tension of members,
At = (80 − 20)(20) = 1 200 mm2
1 200(490)10−3
Ft =
= 235 kN
2.5
F = min(60.9, 134, 157, 235) = 60.9 kN
Ans.
The shear in the bolts controls the design.
______________________________________________________________________________
8-69 Members: Table A-20, Sy = 320 MPa
Bolts: Table 8-11, ISO class 5.8, Sy = 420 MPa, Ssy = 0.577(420) = 242.3 MPa
Shear of bolts,
As = π (202)/4 = 314.2 mm2
90 (103 )
τs =
= 95.48 MPa
3 ( 314.2 )
S
242.3
n = sy =
= 2.54 Ans.
τs
95.48
Bearing on bolt,
Ab = 3(20)15 = 900 mm2
90 (103 )
σb = −
= −100 MPa
900
S
420
n= y =
= 4.2 Ans.
σb
100
Bearing on members,
S
320
n= y =
= 3.2 Ans.
σb
100
Tension on members,
90 (103 )
F
σt =
=
= 46.15 MPa
A 15[190 − 3 ( 20 )]
S
320
n= y =
= 6.93 Ans.
46.15
σt
______________________________________________________________________________
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8-70 Members: Sy = 57 kpsi
Bolts: Sy = 100 kpsi, Ssy = 0.577(100) = 57.7 kpsi
Shear of bolts,
 π (1/ 4 ) 2 
A = 3
 = 0.1473 in 2
4


F
5
τs =
=
= 33.94 kpsi
As 0.1473
n=
Bearing on bolts,
S sy
τs
57.7
= 1.70
33.94
Ans.
Ab = 3(1/4)(5/16) = 0.2344 in2
σb = −
Sy
n=
Bearing on members,
=
σb
F
5
=−
= −21.3 kpsi
Ab
0.2344
=
100
= 4.69
21.3
Ans.
Ab = 0.2344 in2 (From bearing on bolts calculation)
σb = − 21.3 kpsi (From bearing on bolts calculation)
n=
Sy
σb
=
57
= 2.68
21.3
Ans.
Tension in members, failure across two bolts,
At =
5
 2.375 − 2 (1 / 4 )  = 0.5859 in 2
16
F
5
=
= 8.534 kpsi
At 0.5859
S
57
n= y =
= 6.68
Ans.
σ t 8.534
σt =
______________________________________________________________________________
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8-71 By symmetry, the reactions at each support is 1.6 kN. The free-body diagram for the left
member is
∑M
∑M
B
=0
1.6(250) − 50 RA = 0
⇒
RA = 8 kN
A
=0
200(1.6) − 50 RB = 0
⇒
RB = 6.4 kN
Members: Table A-20, Sy = 370 MPa
Bolts: Table 8-11, Sy = 420 MPa, Ssy = 0.577(420) = 242.3 MPa
π
Bolt shear,
As = (122 ) = 113.1 mm 2
4
F
8(103 )
τ = max =
= 70.73 MPa
113.1
As
S
242.3
= 3.43
n = sy =
τ
70.73
Ab = td = 10(12) = 120 mm2
8(103 )
σb = −
= −66.67 MPa
120
S
370
n= y =
= 5.55
σb
66.67
Bearing on member,
Strength of member. The bending moments at the hole locations are:
in the left member at A, MA = 1.6(200) = 320 N · m. In the right member at B, MB =
8(50) = 400 N · m. The bending moment is greater at B
1
I B = [10(503 ) − 10(123 )] = 102.7(103 ) mm 4
12
400(25)
M c
σB = A =
(103 ) = 97.37 MPa
102.7(103 )
IA
S
370
n= y =
= 3.80
σ A 97.37
At the center, call it point C,
MC = 1.6(350) = 560 N · m
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Chap. 8 Solutions, Page 61/70
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1
(10)(503 ) = 104.2(103 ) mm 4
12
560(25)
M c
(103 ) = 134.4 MPa
σC = C =
3
IC
104.2(10 )
Sy
370
n=
=
= 2.75 < 3.80 more critical at C
σ C 134.4
n = min(3.04, 3.80, 2.75) = 2.72 Ans.
______________________________________________________________________________
IC =
8-72 The free-body diagram of the bracket, assuming the upper bolt takes all the shear and
tensile load is
Fs = 2500 lbf
P=
2500 ( 3)
7
= 1071 lbf
Table A-31, H = 7/16 = 0.4375 in. Grip, l = 2(1/2) = 1 in. L ≥ l + H = 1.4375 in. Use 1.5
in bolts.
Eq. (8-13),
LT = 2d + 0.25 = 2(0.5) + 0.25 = 1.25 in
Table 8-7,
ld = L − LT = 1.5 − 1.25 = 0.25 in
lt = l − ld = 1 − 0.25 = 0.75 in
Table 8-2,
At = 0.141 9 in2
Ad = π (0.52) /4 = 0.196 3 in2
0.196 3 ( 0.141 9 ) 30
Ad At E
Eq. (8-17),
kb =
=
= 4.574 Mlbf/in
Ad lt + At ld 0.196 3 ( 0.75 ) + 0.141 9 ( 0.25 )
Eq. (8-22),
0.5774π ( 30 ) 0.5
0.5774π Ed
=
= 16.65 Mlbf/in
km =
 0.5774 l + 0.5d 
 0.5774 (1) + 0.5 ( 0.5 ) 
2 ln  5
 2 ln  5

 0.5774 l + 2.5d 
 0.5774 (1) + 2.5 ( 0.5 ) 
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kb
4.574
=
= 0.216
kb + km 4.574 + 16.65
Table 8-9,
Sp = 65 kpsi
Eqs. (8-31) and (8-32),
Fi = 0.75 At Sp = 0.75(0.141 9)65 = 6.918 kips
σi = 0.75 Sp = 0.75(65) = 48.75 kips
C=
CP + Fi 0.216 (1.071) + 6.918
=
= 50.38 kpsi
At
0.141 9
F
3
τs ≈ s =
= 21.14 kpsi
At 0.141 9
Eq. (a), p. 432, σ b =
Direct shear,
von Mises stress, Eq. (5-15), p. 237
σ ′ = (σ b2 + 3τ s2 )
1/2
= 50.382 + 3 ( 21.142 ) 
1/2
= 62.3 kpsi
Sp
65
=
= 1.04
Ans.
σ ′ 62.3
______________________________________________________________________________
n=
8-73
2P(200) = 14(50)
14(50)
P =
= 1.75 kN per bolt
2(200)
Fs = 7 kN/bolt
S p = 380 MPa
At = 245 mm 2 , Ad =
π
(202 ) = 314.2 mm 2
4
Fi = 0.75(245)(380)(10−3 ) = 69.83 kN
σ i = 0.75 ( 380 ) = 285 MPa
CP + Fi  0.25(1.75) + 69.83  3
=
 (10 ) = 287 MPa
245
At


7(103 )
F
τ = s =
= 22.3 MPa
314.2
Ad
σ ′ = [287 2 + 3(22.32 )]1/ 2 = 290 MPa
σb =
Sp
380
=
= 1.31
Ans.
σ ′ 290
______________________________________________________________________________
n=
Shigley’s MED, 10th edition
Chap. 8 Solutions, Page 63/70
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8-74 Using the result of Prob. 5-67 for lubricated assembly (replace 0.2 with 0.18 per Table
8-15)
2π f T
Fx =
0.18d
With a design factor of nd gives
T =
0.18nd Fx d
0.18(3)(1000)d
=
= 716d
2π f
2π (0.12)
or T/d = 716. Also,
T
= K (0.75S p At )
d
= 0.18(0.75)(85 000) At
= 11 475 At
Form a table
Size
1
4 - 28
5
16 - 24
3
8 − 24
At
T/d = 11 475At
n
417.70 1.75
0.0364
0.058
665.55 2.8
1007.50 4.23
0.0878
where the factor of safety in the last column of the table comes from
n=
Select a
3"
8
2π f (T / d )
2π (0.12)(T / d )
=
= 0.0042(T / d )
0.18Fx
0.18(1000)
- 24 UNF cap screw. The setting is given by
T = (11 475At )d = 1007.5(0.375) = 378 lbf · in
Given the coarse scale on a torque wrench, specify a torque wrench setting of 400 lbf · in.
Check the factor of safety
2π f T
2π (0.12)(400)
=
= 4.47
0.18Fx d
0.18(1000)(0.375)
______________________________________________________________________________
n=
Shigley’s MED, 10th edition
Chap. 8 Solutions, Page 64/70
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8-75
Bolts, from Table 8-11, Sy = 420 MPa
Channel, From Table A-20, Sy = 170 MPa. From Table A-7, t = 6.4 mm
Cantilever, from Table A-20, Sy = 190 MPa
F′A = F′B = F′C = F / 3
M = (50 + 26 + 125) F = 201 F
FA′′ = FC′′ =
201F
= 2.01 F
2 ( 50 )
1

(1)
FC = FC′ + FC′′ =  + 2.01 F = 2.343F
3

Shear on Bolts: The shoulder bolt shear area, As = π(102) / 4 = 78.54 mm2
Max. force,
Ssy = 0.577(420) = 242.3 KPa
τ max =
FC S sy
=
As
n
From Eq. (1), FC = 2.343 F. Thus
F=
S sy  As  242.3  78.54  −3
=

10 = 4.06 kN
n  2.343 
2.0  2.343 
Bearing on bolt: The bearing area is Ab = td = 6.4(10) = 64 mm2. Similar to shear
F=
S y  Ab  420  64  −3
=

10 = 5.74 kN
n  2.343  2.0  2.343 
Bearing on channel: Ab = 64 mm2, Sy = 170 MPa.
Shigley’s MED, 10th edition
Chap. 8 Solutions, Page 65/70
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F=
S y  Ab  170  64  −3
=

10 = 2.32 kN
n  2.343  2.0  2.343 
Bearing on cantilever:: Ab = 12(10) = 120 mm2, Sy = 190 MPa.
F=
S y  Ab  190  120  −3
=

10 = 4.87 kN
n  2.343  2.0  2.343 
Bending of cantilever:: A
At C
I=
1
(12 ) ( 503 − 103 ) = 1.24 (105 ) mm 4
12
σ max =
F=
Sy
n
=
Mc 151Fc
=
I
I
⇒ F=
Sy  I 


n  151
51c 
5
190 1.24 (10 )  −3

 10 = 3.12 kN
2.0  151( 25 ) 


So F = 2.32 kN basedd oon bearing on channel. Ans.
_________________________
______________________________________
____________________
1, Sy = 420 MPa
8-76 Bolts, from Table 8-11,
Bracket, from Table A--20, Sy = 210 MPa
12
= 4 kN;
k M = 12(200) = 2400 N · m
3
240
400
FA′′ = FB′′ =
= 37.5 kN
64
(4 2 + (37.5)2 = 37.7 kN
FA = FB = (4)
FO = 4 kN
F′ =
Bolt shear:
The shoulder bolt shear
ar area, As = π(122) / 4 = 113.1 mm2
420) = 242.3 KPa
Ssy = 0.577(42
37.7(10)3
= 333 MPa
113
S
242
42.3
n = sy =
= 0.728
33
333
τ
τ =
Ans.
Shigley’s MED, 10th edition
Chap.
p. 8 Solutions, Page 66/70
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Bearing on bolts:
Ab = 12(8) = 96
9 mm 2
37.7(10
10)3
σb = −
= −393 MPa
966
S
420
20
n = yc =
= 1.07
Ans.
393
93
σb
Bearing on member:
σ b = −393 MP
Pa
S
210
10
n = yc =
= 0.534
393
93
σb
Ans.
Bending stress in plate:
e:
 bd 3

bh3 bd 3
−
− 2
+ a 2bd 
12
12
 12

3
3
3
 8(12)

8(136)
8(12)
=
−
− 2
+ (32) 2 (8)(12) 
12
12
 12

6
4
= 1.48(10) m
mm
Ans.
Mc
24
2400(68)
σ =
=
(10)3 = 110 MPa
1.4
.48(10)6
I
S
210
10
n= y =
= 1.91
Ans.
110
10
σ
I =
Failure is predicted forr bolt shear and bearing on member.
_________________________
______________________________________
____________________
Shigley’s MED, 10th edition
Chap.
p. 8 Solutions, Page 67/70
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8-77
FA = 1208 − 125 = 1083 lbf,
Bolt shear:
 3625 
FA′′ = FB′′ = 
 = 1208 lbf
 3 
FB = 1208 + 125 = 1333 lbf
As = (π / 4)(0.3752) = 0.1104 in2
τ max =
Fmax
1333
=
= 12 070 psi
As
0.1104
From Table 8-10, Sy = 100 kpsi, Ssy = 0.577(100) = 57.7 kpsi
n=
S sy
τ max
=
57.7
= 4.78
12.07
Ans.
Bearing on bolt: Bearing area is Ab = td = 0.375 (0.375) = 0.1406 in2.
σb = −
n=
Sy
σb
F
1333
=−
= −9 481 psi
0.1406
Ab
=
100
= 10.55
9.481
Ans.
Bearing on member: From Table A-20, Sy = 54 kpsi. Bearing stress same as bolt
n=
Sy
σb
=
54
= 5.70
9.481
Ans.
Bending of member: At B, M = 250(13) = 3250 lbf⋅in
I=
3
1 3 3 3 
4
−
2
 
   = 0.2484 in
12  8  
8
  
Shigley’s MED, 10th edition
Chap. 8 Solutions, Page 68/70
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Mc 3250 (1)
=
= 13 080 psi
I
0.2484
σ=
Sy
54
= 4.13
Ans.
σ 13.08
______________________________________________________________________________
n=
=
8-78 The direct shear load per bolt is F′ = 2000/6 = 333.3 lbf. The moment is taken only by the
four outside bolts. This moment is M = 2000(5) = 10 000 lbf · in.
10 000
= 1000 lbf and the resultant bolt load is
Thus F ′′ =
2(5)
F =
(333.3) 2 + (1000) 2 = 1054 lbf
Bolt strength, Table 8-9, Sy = 100 kpsi; Channel and Plate strength, Sy = 42 kpsi
Shear of bolt: As = π (0.5)2/4 = 0.1963 in2
n=
Ssy
τ
=
(0.577)(100)
= 10.7
1.054 / 0.1963
Ans.
Bearing on bolt: Channel thickness is t = 3/16 in, Ab = 0.5(3/16) = 0.09375 in2
n=
Bearing on channel:
Bearing on plate:
n=
100
= 8.89
1.054 / 0.09375
42
= 3.74
1.054 / 0.09375
Ab = 0.5(0.25) = 0.125 in2
n=
42
= 4.98
1.054 / 0.125
Ans.
Ans.
Ans.
Strength of plate:
I =
0.25(7.5)3 0.25(0.5)3
−
12
12
 0.25(0.5)3

− 2
+ ( 0.25 )( 0.5 ) (2.5) 2  = 7.219 in 4
12


Shigley’s MED, 10th edition
Chap. 8 Solutions, Page 69/70
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M = 5000 lbf · in per plate
Mc 5000(3.75)
σ =
=
= 2597 psi
7.219
I
42
= 16.2 Ans.
n=
2.597
______________________________________________________________________________
8-79 to 8-81 Specifying bolts, screws, dowels and rivets is the way a student learns about such
components. However, choosing an array a priori is based on experience. Here is a chance
for students to build some experience.
Shigley’s MED, 10th edition
Chap. 8 Solutions, Page 70/70
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Chapter 9
Figure for Probs.
9-1 to 9-4
9-1
Given, b = 50 mm, d = 50 mm, h = 5 mm, τallow = 140 MPa.
F = 0.707 hlτallow = 0.707(5)[2(50)](140)(10−3) = 49.5 kN Ans.
______________________________________________________________________________
9-2
Given, b = 2 in, d = 2 in, h = 5/16 in, τallow = 25 kpsi.
Ans.
F = 0.707 hlτallow = 0.707(5/16)[2(2)](25) = 22.1 kip
______________________________________________________________________________
9-3
Given, b = 50 mm, d = 30 mm, h = 5 mm, τallow = 140 MPa.
F = 0.707 hlτallow = 0.707(5)[2(50)](140)(10−3) = 49.5 kN Ans.
______________________________________________________________________________
9-4
Given, b = 4 in, d = 2 in, h = 5/16 in, τallow = 25 kpsi.
F = 0.707 hlτallow = 0.707(5/16)[2(4)](25) = 44.2 kip
Ans.
______________________________________________________________________________
9-5
Prob. 9-1 with E7010 Electrode.
Table 9-6:
f = 14.85 h kip/in = 14.85 [5 mm/(25.4 mm/in)] = 2.923 kip/in
= 2.923(4.45/25.4) = 0.512 kN/mm
F = f l = 0.512[2(50)] = 51.2 kN
Ans.
______________________________________________________________________________
9-6
Prob. 9-2 with E6010 Electrode.
Table 9-6:
f = 14.85 h kip/in = 14.85(5/16) = 4.64 kip/in
Shigley’s MED, 10th edition
Chapter 9 Solutions, Page 1/35
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F = f l = 4.64[2(2)] = 18.6 kip
Ans.
______________________________________________________________________________
9-7
Prob. 9-3 with E7010 Electrode.
Table 9-6:
f = 14.85 h kip/in = 14.85 [5 mm/(25.4 mm/in)] = 2.923 kip/in
= 2.923(4.45/25.4) = 0.512 kN/mm
F = f l = 0.512[2(50)] = 51.2 kN
Ans.
______________________________________________________________________________
9-8
Prob. 9-4 with E6010 Electrode.
Table 9-6:
f = 14.85 h kip/in = 14.85(5/16) = 4.64 kip/in
F = f l = 4.64[2(4)] = 37.1 kip
Ans.
______________________________________________________________________________
9-9
Table A-20:
1018 CD: Sut = 440 MPa, Sy = 370 MPa
1018 HR: Sut = 400 MPa, Sy = 220 MPa
Cold-rolled properties degrade to hot-rolled properties in the neighborhood of the weld.
Table 9-4:
τ all = min(0.30Sut , 0.40S y )
= min[0.30(400), 0.40(220)]
= min(120, 88) = 88 MPa
for both materials.
Eq. (9-3):
F = 0.707hlτall = 0.707(5)[2(50)](88)(10−3) = 31.1 kN Ans.
______________________________________________________________________________
9-10
Table A-20:
1020 CD: Sut = 68 kpsi, Sy = 57 kpsi
1020 HR: Sut = 55 kpsi, Sy = 30 kpsi
Cold-rolled properties degrade to hot-rolled properties in the neighborhood of the weld.
Table 9-4:
τ all = min(0.30Sut , 0.40S y )
= min[0.30(55), 0.40(30)]
= min(16.5, 12.0) = 12.0 kpsi
for both materials.
Eq. (9-3):
F = 0.707hlτall = 0.707(5/16)[2(2)](12.0) = 10.6 kip Ans.
______________________________________________________________________________
Shigley’s MED, 10th edition
Chapter 9 Solutions, Page 2/35
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Table A-20:
1035 HR: Sut = 500 MPa, Sy = 270 MPa
1035 CD: Sut = 550 MPa, Sy = 460 MPa
Cold-rolled properties degrade to hot-rolled properties in the neighborhood of the weld.
Table 9-4:
τ all = min(0.30Sut , 0.40S y )
= min[0.30(500), 0.40(270)]
= min(150, 108) = 108 MPa
for both materials.
Eq. (9-3):
F = 0.707hlτall = 0.707(5)[2(50)](108)(10−3) = 38.2 kN Ans.
______________________________________________________________________________
9-11
Table A-20:
1035 HR: Sut = 72 kpsi, Sy = 39.5 kpsi
1020 CD: Sut = 68 kpsi, Sy = 57 kpsi, 1020 HR: Sut = 55 kpsi, Sy = 30 kpsi
Cold-rolled properties degrade to hot-rolled properties in the neighborhood of the weld.
Table 9-4:
τ all = min(0.30Sut , 0.40S y )
= min[0.30(55), 0.40(30)]
= min(16.5, 12.0) = 12.0 kpsi
for both materials.
Eq. (9-3):
F = 0.707hlτall = 0.707(5/16)[2(4)](12.0) = 21.2 kip Ans.
______________________________________________________________________________
9-12
9-13
2 (100 ) (103 )
2F
Eq. (9-3):
τ=
=
= 141 MPa
Ans.
hl
5  2 ( 50 + 50 ) 
______________________________________________________________________________
9-14
Eq. (9-3):
τ=
2 ( 40 )
2F
=
= 22.6 kpsi
hl
( 5 /16 )  2 ( 2 + 2 ) 
Ans.
______________________________________________________________________________
2 (100 ) (103 )
2F
9-15
Eq. (9-3): τ =
=
= 177 MPa
Ans.
hl
5  2 ( 50 + 30 ) 
______________________________________________________________________________
9-16
Eq. (9-3): τ =
2 ( 40 )
2F
=
= 15.1 kpsi
hl
( 5 / 16 )  2 ( 4 + 2 ) 
Ans.
______________________________________________________________________________
Shigley’s MED, 10th edition
Chapter 9 Solutions, Page 3/35
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9-17
b = d =50 mm, c = 150 mm, h = 5 mm, and τallow = 140 MPa.
(a) Primary shear, Table 9-1, Case 2 (Note: b and d are interchanged between problem
figure and table figure. Note, also, F in kN and τ in MPa):
F (103 )
V
= 2.829 F
τ ′y = =
A 1.414 ( 5)( 50 )
Secondary shear, Table 9-1:
Ju =
d ( 3b 2 + d 2 )
6
=
50 3 ( 502 ) + 502 
6
= 83.33 (103 ) mm3
J = 0.707 h Ju = 0.707(5)(83.33)(103) = 294.6(103) mm4
τ x′′ = τ ′′y =
Mry
J
=
175 F (103 ) ( 25 )
294.6 (103 )
= 14.85 F
τ max = τ x′′2 + (τ ′y + τ ′′y ) = F 14.852 + ( 2.829 + 14.85 ) = 23.1F (1)
2
F=
τ allow
23.1
=
2
140
= 6.06 kN Ans.
23.1
(b) For E7010 from Table 9-6, τallow = 21 kpsi = 21(6.89) = 145 MPa
1020 HR bar:
Sut = 380 MPa, Sy = 210 MPa
1015 HR support: Sut = 340 MPa, Sy = 190 MPa
Table 9-3, E7010 Electrode: Sut = 482 MPa, Sy = 393 MPa
The support controls the design.
Table 9-4:
τallow = min(0.30Sut, 0.40Sy ) =min[0.30(340), 0.40(190) = min(102, 76)
= 76 MPa
The allowable load, from Eq. (1) is
τ allow
76
=
= 3.29 kN
Ans.
23.1 23.1
______________________________________________________________________________
F=
9-18
b = d =2 in, c = 6 in, h = 5/16 in, and τallow = 25 kpsi.
Shigley’s MED, 10th edition
Chapter 9 Solutions, Page 4/35
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(a) Primary shear, Table 9-1(Note: b and d are interchanged between problem figure and
table figure. Note, also, F in kip and τ in kpsi):
V
F
=
= 1.132 F
A 1.414 ( 5 /16 )( 2 )
Secondary shear, Table 9-1:
2
2
d ( 3b 2 + d 2 ) 2 3 ( 2 ) + 2 
Ju =
=
= 5.333 in 3
6
6
τ ′y =
J = 0.707 h Ju = 0.707(5/16)(5.333) = 1.178 in4
τ x′′ = τ ′′y =
Mry
J
=
7 F (1)
1.178
= 5.942 F
τ max = τ x′′2 + (τ ′y + τ ′′y ) = F 5.9422 + (1.132 + 5.942 ) = 9.24 F (1)
2
F=
τ allow
9.24
=
2
25
= 2.71 kip Ans.
9.24
(b) For E7010 from Table 9-6, τallow = 21 kpsi
1020 HR bar:
Sut = 55 kpsi, Sy = 30 kpsi
1015 HR support: Sut = 50 kpsi, Sy = 27.5 kpsi
Table 9-3, E7010 Electrode: Sut = 70 kpsi, Sy = 57 kpsi
The support controls the design.
Table 9-4:
τallow = min(0.30Sut, 0.40Sy ) =min[0.30(50), 0.40(27.5) = min(15, 11)
= 11 kpsi
The allowable load, from Eq. (1) is
τ allow
11
= 1.19 kip
Ans.
9.24 9.24
______________________________________________________________________________
F=
9-19
=
b =50 mm, c = 150 mm, d = 30 mm, h = 5 mm, and τallow = 140 MPa.
(a) Primary shear, Table 9-1, Case 2 (Note: b and d are interchanged between problem
figure and table figure. Note, also, F in kN and τ in MPa):
Shigley’s MED, 10th edition
Chapter 9 Solutions, Page 5/35
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F (103 )
V
=
= 2.829 F
A 1.414 ( 5)( 50 )
τ ′y =
Secondary shear, Table 9-1:
Ju =
d ( 3b 2 + d 2 )
6
=
50 3 ( 302 ) + 502 
6
= 43.33 (103 ) mm3
J = 0.707 h Ju = 0.707(5)(43.33)(103) = 153.2(103) mm4
τ x′′ =
Mry
J
175F (103 ) (15 )
=
153.2 (103 )
= 17.13F
Mrx 175F (10 ) ( 25 )
=
= 28.55 F
J
153.2 (103 )
3
τ ′′y =
τ max = τ x′′2 + (τ ′y + τ ′′y ) = F 17.132 + ( 2.829 + 28.55 ) = 35.8 F (1)
2
F=
τ allow
35.8
=
2
140
= 3.91 kN Ans.
35.8
(b) For E7010 from Table 9-6, τallow = 21 kpsi = 21(6.89) = 145 MPa
1020 HR bar:
Sut = 380 MPa, Sy = 210 MPa
1015 HR support: Sut = 340 MPa, Sy = 190 MPa
Table 9-3, E7010 Electrode: Sut = 482 MPa, Sy = 393 MPa
The support controls the design.
Table 9-4:
τallow = min(0.30Sut, 0.40Sy ) =min[0.30(340), 0.40(190) = min(102, 76)
= 76 MPa
The allowable load, from Eq. (1) is
τ allow
76
= 2.12 kN
Ans.
35.8 35.8
______________________________________________________________________________
F=
9-20
=
b = 4 in, c = 6 in, d = 2 in, h = 5/16 in, and τallow = 25 kpsi.
Shigley’s MED, 10th edition
Chapter 9 Solutions, Page 6/35
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(a) Primary shear, Table 9-1(Note: b and d are interchanged between problem figure and
table figure. Note, also, F in kip and τ in kpsi):
V
F
=
= 0.5658F
A 1.414 ( 5 /16 )( 4 )
Secondary shear, Table 9-1:
τ ′y =
Ju =
d ( 3b2 + d 2 )
=
6
4 3 ( 22 ) + 42 
6
= 18.67 in 3
J = 0.707 h Ju = 0.707(5/16)(18.67) = 4.125 in4
τ x′′ =
τ ′′y =
Mry
J
8 F (1)
=
4.125
= 1.939 F
Mrx 8 F ( 2 )
=
= 3.879 F
J
4.125
τ max = τ x′′2 + (τ ′y + τ ′′y ) = F 1.9392 + ( 0.5658 + 3.879 ) = 4.85F (1)
2
F=
τ allow
4.85
=
2
25
= 5.15 kip Ans.
4.85
(b) For E7010 from Table 9-6, τallow = 21 kpsi
1020 HR bar:
Sut = 55 kpsi, Sy = 30 kpsi
1015 HR support: Sut = 50 kpsi, Sy = 27.5 kpsi
Table 9-3, E7010 Electrode: Sut = 70 kpsi, Sy = 57 kpsi
The support controls the design.
Table 9-4:
τallow = min(0.30Sut, 0.40Sy ) =min[0.30(50), 0.40(27.5) = min(15, 11)
= 11 kpsi
The allowable load, from Eq. (1) is
τ allow
11
= 2.27 kip
Ans.
4.85 4.85
______________________________________________________________________________
F=
=
Shigley’s MED, 10th edition
Chapter 9 Solutions, Page 7/35
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9-21
Given, b = 50 mm, c = 150 mm, d = 50 mm, h = 5 mm, τallow = 140 MPa.
Primary shear (F in kN, τ in MPa, A in mm2):
F (103 )
V
′
τy = =
= 1.414 F
A 1.414 ( 5 )( 50 + 50 )
Secondary shear:
Table 9-1:
(b + d )
=
3
( 50 + 50 )
=
3
= 166.7 (103 ) mm3
6
6
J = 0.707 h Ju = 0.707(5)166.7(103) = 589.2(103) mm4
Ju
τ x′′ = τ ′′y =
Mry
J
=
175 F (103 ) (25)
589.2 (103 )
= 7.425F
Maximum shear:
τ max = τ x′′2 + (τ ′y + τ ′′y ) = F 7.4252 + (1.414 + 7.425 ) = 11.54 F
2
τ allow
2
140
= 12.1 kN Ans.
11.54 11.54
______________________________________________________________________________
F=
9-22
=
Given, b = 2 in, c = 6 in, d = 2 in, h = 5/16 in, τallow = 25 kpsi.
Primary shear:
τ ′y =
V
F
=
= 0.5658 F
A 1.414 ( 5 /16 )( 2 + 2 )
Secondary shear:
Table 9-1:
(b + d )
=
3
( 2 + 2)
=
3
= 10.67 in 3
6
6
J = 0.707 h Ju = 0.707(5/16)10.67 = 2.357 in4
Ju
τ x′′ = τ ′′y =
Mry
J
=
7 F (1)
= 2.970 F
2.357
Maximum shear:
τ max = τ x′′2 + (τ ′y + τ ′′y ) = F 2.9702 + ( 0.566 + 2.970 ) = 4.618 F
2
τ allow
2
25
= 5.41 kip Ans.
4.618 4.618
______________________________________________________________________________
F=
9-23
=
Given, b = 50 mm, c = 150 mm, d = 30 mm, h = 5 mm, τallow = 140 MPa.
Shigley’s MED, 10th edition
Chapter 9 Solutions, Page 8/35
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Primary shear (F in kN, τ in MPa, A in mm2):
F (103 )
V
τ ′y = =
= 1.768 F
A 1.414 ( 5 )( 50 + 30 )
Secondary shear:
Table 9-1:
Ju =
(b + d )
3
=
( 50 + 30 )
3
= 85.33 (103 ) mm3
6
6
J = 0.707 h Ju = 0.707(5)85.33(103) = 301.6(103) mm4
τ x′′ =
τ ′′y =
Mry
J
=
175F (103 ) (15)
301.6 (103 )
= 8.704 F
3
Mrx 175F (10 ) (25)
=
= 14.51F
J
301.6 (103 )
Maximum shear:
τ max = τ x′′2 + (τ ′y + τ ′′y ) = F 8.7042 + (1.768 + 14.51) = 18.46 F
2
τ allow
2
140
= 7.58 kN Ans.
18.46 18.46
______________________________________________________________________________
F=
9-24
=
Given, b = 4 in, c = 6 in, d = 2 in, h = 5/16 in, τallow = 25 kpsi.
Primary shear:
τ ′y =
V
F
=
= 0.3772 F
A 1.414 ( 5 /16 )( 4 + 2 )
Secondary shear:
Table 9-1:
Ju =
(b + d )
3
=
( 4 + 2)
3
= 36 in 3
6
6
J = 0.707 h Ju = 0.707(5/16)36 = 7.954 in4
τ x′′ =
Mry
τ ′′y =
Mrx 8 F (2)
=
= 2.012 F
J
7.954
J
=
8 F (1)
= 1.006 F
7.954
Maximum shear:
τ max = τ x′′2 + (τ ′y + τ ′′y ) = F 1.0062 + ( 0.3772 + 2.012 ) = 2.592 F
2
Shigley’s MED, 10th edition
2
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τ allow
25
=
= 9.65 kip Ans.
2.592 2.592
______________________________________________________________________________
F=
9-25
Given, b = 50 mm, d = 50 mm, h = 5 mm, E6010 electrode.
A = 0.707(5)(50 +50 + 50) = 530.3 mm2
Member endurance limit: From Table A-20 for AISI 1010 HR, Sut = 320 MPa.
Eq. 6-19 and Table 6-2, pp. 295 and 296, respectively:
ka = 272(320)−0.995 = 0.875
kb = 1 (uniform shear), kc = 0.59 (torsion, shear), kd = 1
Eqs. (6-8) and (6-18):
Se = 0.875(1)(0.59)(1)(0.5)(320) = 82.6 MPa
Electrode endurance: E6010, Table 9-3,
Sut = 427 MPa
Eq. 6-19 and Table 6-2, pp. 295 and 296, respectively:
ka = 272(427)−0.995 = 0.657
As before, kb = 1 (direct shear), kc = 0.59 (torsion, shear), kd = 1
Se = 0.657(1)(0.59)(1)(0.5)(427) = 82.8 MPa
The members and electrode are basically of equal strength. We will use Se = 82.6 MPa.
For a factor of safety of 1, and with Kfs = 2.7 (Table 9-5)
τ A 82.6 ( 530.3)
F = allow =
= 16.2 (103 ) N = 16.2 kN
Ans.
K fs
2.7
______________________________________________________________________________
9-26
Given, b = 2 in, d = 2 in, h = 5/16 in, E6010 electrode.
A = 0.707(5/16)(2 +2 + 2) = 1.326 in2
Member endurance limit: From Table A-20 for AISI 1010 HR, Sut = 47 kpsi.
Eq. 6-19 and Table 6-2, pp. 295 and 296, respectively:
ka = 39.9(47)−0.995 = 0.865
kb = 1 (uniform shear), kc = 0.59 (torsion, shear), kd = 1
Eqs. (6-8) and (6-18):
Se = 0.865(1)(0.59)(1)(0.5)(47) = 12.0 kpsi
Electrode endurance: E6010, Table 9-3,
Sut = 62 kpsi
Eq. 6-19 and Table 6-2, pp. 295 and 296, respectively:
Shigley’s MED, 10th edition
ka = 39.9(62)−0.995 = 0.657
Chapter 9 Solutions, Page 10/35
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As before, kb = 1 (uniform shear), kc = 0.59 (torsion, shear), kd = 1
Se = 0.657(1)(0.59)(1)(0.5)(62) = 12.0 kpsi
Thus the members and electrode are of equal strength. For a factor of safety of 1, and
with Kfs = 2.7 (Table 9-5)
τ A 12.0 (1.326 )
F = allow =
= 5.89 kip
Ans.
K fs
2.7
______________________________________________________________________________
9-27
Given, b = 50 mm, d = 30 mm, h = 5 mm, E7010 electrode.
A = 0.707(5)(50 +50 + 30) = 459.6 mm2
Member endurance limit: From Table A-20 for AISI 1010 HR, Sut = 320 MPa.
Eq. 6-19 and Table 6-2, pp. 295 and 296, respectively:
ka = 272(320)−0.995 = 0.875
kb = 1 (direct shear), kc = 0.59 (torsion, shear), kd = 1
Eqs. (6-8) and (6-18):
Se = 0.875(1)(0.59)(1)(0.5)(320) = 82.6 MPa
Electrode endurance: E6010, Table 9-3,
Sut = 482 MPa
Eq. 6-19 and Table 6-2, pp. 295 and 296, respectively:
ka = 272(482)−0.995 = 0.582
As before, kb = 1 (direct shear), kc = 0.59 (torsion, shear), kd = 1
Se = 0.582(1)(0.59)(1)(0.5)(482) = 82.7 MPa
The members and electrode are basically of equal strength. We will use Se =82.6 MPa.
For a factor of safety of 1, and with Kfs = 2.7 (Table 9-5)
τ A 82.6 ( 459.6 )
F = allow =
= 14.1(103 ) N = 14.1 kN
Ans.
K fs
2.7
______________________________________________________________________________
9-28
Given, b = 4 in, d = 2 in, h = 5/16 in, E7010 electrode.
A = 0.707(5/16)(4 +4 + 2) = 2.209 in2
Member endurance limit: From Table A-20 for AISI 1010 HR, Sut = 47 kpsi.
Eq. 6-19 and Table 6-2, pp. 295 and 296, respectively:
ka = 39.9(47)−0.995 = 0.865
kb = 1 (direct shear), kc = 0.59 (torsion, shear), kd = 1
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Eqs. (6-8) and (6-18):
Se = 0.865(1)(0.59)(1)(0.5)(47) = 12.0 kpsi
Electrode endurance: E7010, Table 9-3,
Sut = 70 kpsi
Eq. 6-19 and Table 6-2, pp. 295 and 296, respectively:
ka = 39.9(70)−0.995 = 0.582
As before, kb = 1 (direct shear), kc = 0.59 (torsion, shear), kd = 1
Se = 0.582(1)(0.59)(1)(0.5)(70) = 12.0 kpsi
Thus the members and electrode are of equal strength. For a factor of safety of 1, and
with Kfs = 2.7 (Table 9-5)
τ A 12.0 ( 2.209 )
F = allow =
= 9.82 kip
Ans.
K fs
2.7
______________________________________________________________________________
9-29
τ′ = 0 (why?)
Primary shear:
Secondary shear:
Table 9-1: Ju = 2π r3 = 2π (1.5)3 = 21.21 in3
J = 0.707 h Ju = 0.707(1/4)(21.21) = 3.749 in4
2 welds:
Mr 8 F (1.5 )
=
= 1.600 F
2 J 2 ( 3.749 )
τ ′′ =
τ ′′ = τ allow
⇒ 1.600 F = 20 ⇒ F = 12.5 kip
Ans.
______________________________________________________________________________
9-30
l = 2 + 4 + 4 = 10 in
2 (1) + 4 ( 0 ) + 4 ( 2 )
= 1 in
10
2 ( 4) + 4 ( 2) + 4 ( 0)
y=
= 1.6 in
10
x=
M = FR = F(10 − 1) = 9 F
r1 =
(1 − 1) + ( 4 − 1.6 )
r3 =
( 2 − 1)
2
2
2
= 2.4 in,
r2 = 12 + ( 2 − 1.6 ) = 1.077 in
2
+ 1.62 = 1.887 in
Shigley’s MED, 10th edition
Chapter 9 Solutions, Page 12/35
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1
( 0.707 )( 5 /16 ) ( 23 ) = 0.1473 in 4
12
1
= J G3 = ( 0.707 )( 5 / 16 ) ( 43 ) = 1.178 in 4
12
J G1 =
J G2
3
(
J = ∑ J i + Ai rG2i
i =1
)
= 0.1473 + 0.707 ( 5 /16 )( 2 ) ( 2.42 ) + 1.178 + 0.707 ( 5 / 16 )( 4 ) (1.077 2 )
+ 1.178 + 0.707 ( 5 / 16 )( 4 ) (1.887 2 ) = 9.220 in 4
 1.6 
o
 = 28.07
−
4
1


α = tan −1 
r = 1.62 + ( 4 − 1) = 3.4 in
2
Primary shear (τ in kpsi, F in kip) :
τ′ =
Secondary shear:
τ ′′ =
V
F
=
= 0.4526 F
A 0.707 ( 5 /16 )(10 )
Mr 9 F ( 3.4 )
=
= 3.319 F
J
9.220
τ max =
( 3.319F sin 28.07 ) + ( 3.319F cos 28.07
o 2
o
+ 0.4526 F )
2
= 3.724 F
τmax = τallow ⇒ 3.724 F = 25 ⇒ F = 6.71 kip
Ans.
______________________________________________________________________________
Shigley’s MED, 10th edition
Chapter 9 Solutions, Page 13/35
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9-31
l = 30 + 50 + 50 = 130 mm
30 (15 ) + 50 ( 0 ) + 50 ( 25 )
= 13.08 mm
130
30 ( 50 ) + 50 ( 25 ) + 50 ( 0 )
y=
= 21.15 mm
130
x=
M = FR = F(200 − 13.08)
= 186.92 F (M in N⋅m, F in kN)
r1 =
(15 − 13.08) + ( 50 − 21.15)
r3 =
( 25 − 13.08)
2
2
= 28.92 mm, r2 = 13.082 + ( 25 − 21.15) = 13.63 mm
2
2
+ 21.152 = 24.28 mm
1
( 0.707 )( 5 ) ( 303 ) = 7.954 (103 ) mm 4
12
1
= J G3 = ( 0.707 )( 5 ) ( 503 ) = 36.82 (103 ) mm 4
12
J G1 =
J G2
3
(
J = ∑ J i + Ai rG2i
i =1
)
= 7.954 (103 ) + 0.707 ( 5 )( 30 ) ( 28.922 ) + 36.82 (103 ) + 0.707 ( 5 )( 50 ) (13.632 )
+ 36.82 (103 ) + 0.707 ( 5 )( 50 ) ( 24.282 ) = 307.3 (103 ) mm 4
 21.15 
o
 = 29.81
 50 − 13.08 
α = tan −1 
r = 21.152 + ( 50 − 13.08) = 42.55 mm
2
Primary shear (τ in MPa, F in kN) :
F (103 )
V
τ′ = =
= 2.176 F
A 0.707 ( 5 )(130 )
Secondary shear:
τ ′′ =
3
Mr 186.92 F (10 ) ( 42.55 )
=
= 25.88 F
J
307.3 (103 )
Shigley’s MED, 10th edition
Chapter 9 Solutions, Page 14/35
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τ max =
( 25.88F sin 29.81 ) + ( 25.88F cos 29.81
o 2
o
+ 2.176 F )
2
= 27.79 F
τmax = τallow ⇒ 27.79 F = 140 ⇒ F = 5.04 kN
Ans.
______________________________________________________________________________
9-32
Weld
Pattern
1.
Figure of merit
 a2 
J
a 3 /12 a 2
=
= 0.0833  
fom′ = u =
12h
lh
ah
 h 
a ( 3a 2 + a 2 )
3.
( 2a ) − 6a 2 a 2 = 5a 2 = 0.2083  a 2 
fom′ =
 
12 ( a + a ) 2ah 24h
 h 
4.
fom′ =
5.
( 2a )
fom′ =
6 ( 2a ) h
5
 a2 
a2
= 0.3333  
3h
 h 
2.
fom′ =
=
Rank______
1
4
4
 a2 
1  8a 3 + 6 a 3 + a 3
a4 
−

 = 0.3056  
3ah 
12
2a + a 
 h 
2
3
 a2 
1
8a 3
=
= 0.3333  
6h 4a 24ah
 h 
1
2π ( a / 2 )
 a2 
a3
3
=
= 0.25  
4ah
π ah
 h 
______________________________________________________________________________
3
6.
fom′ =
Shigley’s MED, 10th edition
Chapter 9 Solutions, Page 15/35
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9-33
Weld
Pattern
1.
2.
3.
4.*
5. & 7.
6. & 8.
Figure of merit
( a3 /12 ) = 0.0833  a 2 
I
fom′ = u =
 
lh
ah
 h 
( a3 / 6 ) = 0.0833  a 2 
fom′ =
 
2ah
 h 
2
( aa / 2 ) = 0.25  a 2 
fom′ =
 
2ah
 h 
2
( a / 12 ) ( 6a + a ) = 7a 2 = 0.1944  a 2 
fom′ =
 
3ah
36h
 h 
a
a2
a
x= , y=
=
2
a + 2a 3
2
3
2a
a3
a
2 a
Iu =
− 2a + ( a + 2a )   =
3
3
3
3
Rank______
fom′ =
5
3
 a2 
I u ( a / 3) 1  a 2 
=
=   = 0.1111 
lh
3ah
9 h 
 h 
2
( a / 6 ) ( 3a + a ) = 1  a 2  = 0.1667  a 2 
fom′ =
 
 
4ah
6 h 
 h 
π ( a / 2) a2
 a2 
=
= 0.125  
8h
π ah
 h 
6
6
1
2
3
3
9.
fom′ =
4
*Note. Because this section is not symmetric with the vertical axis, out-of-plane
deflection may occur unless special precautions are taken. See the topic of “shear center”
in books with more advanced treatments of mechanics of materials.
______________________________________________________________________________
9-34
Attachment and member (1018 HR), Sy = 220 MPa and Sut = 400 MPa.
The member and attachment are weak compared to the properties of the lowest electrode.
Decision Specify the E6010 electrode
Controlling property, Table 9-4: τall = min[0.3(400), 0.4(220)] = min(120, 88) = 88 MPa
For a static load, the parallel and transverse fillets are the same. Let the length of a bead
be l = 75 mm, and n be the number of beads.
Shigley’s MED, 10th edition
Chapter 9 Solutions, Page 16/35
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τ=
F
= τ all
n ( 0.707 ) hl
100 (103 )
F
=
= 21.43
nh =
0.707lτ all 0.707 ( 75 )( 88 )
where h is in millimeters. Make a table
Number of beads, n
1
2
3
4
Leg size, h (mm)
21.43
10.71
7.14
5.36 → 6 mm
Specify h = 6 mm on all four sides.
Decision
Weldment specification:
Pattern: All-around square, four beads each side, 75 mm long
Electrode: E6010
Leg size: h = 6 mm
______________________________________________________________________________
9-35
Decision: Choose a parallel fillet weldment pattern. By so-doing, we’ve chosen an
optimal pattern (see Prob. 9-32) and have thus reduced a synthesis problem to an analysis
problem:
Table 9-1, case 2, rotated 90°: A = 1.414hd = 1.414(h)(75) = 106.05h mm2
Primary shear
τ ′y =
12 (103 ) 113.2
V
=
=
A 106.05h
h
Secondary shear:
d (3b 2 + d 2 )
6
75[3(752 ) + 752 ]
=
= 281.3 (103 ) mm3
6
J = 0.707(h)(281.3) (103 ) = 198.8 (103 ) h mm 4
Ju =
With α = 45°,
Shigley’s MED, 10th edition
Chapter 9 Solutions, Page 17/35
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τ ′′x =
3
Mry 12 (10 ) (187.5)(37.5) 424.4
Mr cos 45o
=
=
=
= τ ′′y
J
J
h
198.8 (103 ) h
τ max = τ ′′x 2 + (τ ′y + τ ′′y ) =
2
1
684.9
424.42 + (113.2 + 424.4) 2 =
h
h
Attachment and member (1018 HR): Sy = 220 MPa, Sut = 400 MPa
Decision: Use E60XX electrode which is stronger
τ all = min[0.3(400), 0.4(220)] = 88 MPa
τ max = τ all =
h=
684.9
= 88 MPa
h
684.9
= 7.78 mm
88
Decision: Specify 8 mm leg size
Weldment Specifications:
Pattern: Parallel horizontal fillet welds
Electrode: E6010
Type: Fillet
Length of each bead: 75 mm
Leg size: 8 mm
______________________________________________________________________________
9-36
Problem 9-35 solves the problem using parallel horizontal fillet welds, each 75 mm long
obtaining a leg size rounded up to 8 mm.
For this problem, since the width of the plate is fixed and the length has not been
determined, we will explore reducing the leg size by using two vertical beads 75 mm long
and two horizontal beads such that the beads have a leg size of 6 mm.
Decision: Use a rectangular weld bead pattern with a leg size of 6 mm (case 5 of Table
9-1 with b unknown and d = 75 mm).
Materials:
Attachment and member (1018 HR): Sy = 220 MPa, Sut = 400 MPa
From Table 9-4, AISC welding code,
τall = min[0.3(400), 0.4(220)] = min(120, 88) = 88 MPa
Select a stronger electrode material from Table 9-3.
Decision: Specify E6010
Solving for b: In Prob. 9-35, every term was linear in the unknown h. This made solving
for h relatively easy. In this problem, the terms will not be linear in b, and so we will use
an iterative solution with a spreadsheet.
Throat area and other properties from Table 9-1:
A = 1.414(6)(b + 75) = 8.484(b + 75)
(1)
Shigley’s MED, 10th edition
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Ju
( b + 75 )
=
6
3
, J = 0.707 (6) Ju = 0.707(b +75)3
(2)
Primary shear (τ in MPa, h in mm):
τ ′y =
12 (103 )
V
=
A
A
(3)
Secondary shear (See Prob. 9-35 solution for the definition of α) :
Mr
τ ′′ =
J
3
Mry 12 (10 ) (150 + b / 2 ) (37.5)
Mr
cos α =
τ ′′x = τ ′′ cos α =
=
3
J
J
0.707 ( b + 75 )
τ ′′y = τ ′′ sin α =
3
Mr
Mrx 12 (10 ) (150 + b / 2 ) (b / 2)
sin α =
=
3
J
J
0.707 ( b + 75 )
τ max = τ ′y 2 + (τ ′′x + τ ′′y )
2
(4)
(5)
(6)
Enter Eqs. (1) to (6) into a spreadsheet and iterate for various values of b. A portion of
the spreadsheet is shown below.
b (mm) A (mm2) J (mm4)
41
984.144 1103553.5
42
992.628 1132340.4
43
1001.112 1161623.6
44
1009.596 1191407.4
τ'y (Mpa) τ"y (Mpa) τ"x (Mpa)
12.19334
12.08912
11.98667
11.88594
69.5254
67.9566
66.43718
64.96518
38.00722
38.05569
38.09065
38.11291
τmax
(Mpa)
90.12492
88.63156
87.18485 < 88 Mpa
85.7828
We see that b ≥ 43 mm meets the strength goal.
Weldment Specifications:
Pattern: Horizontal parallel weld tracks 43 mm long, vertical parallel weld tracks 75 mm
long
Electrode: E6010
Leg size: 6 mm
______________________________________________________________________________
9-37
Materials:
Member and attachment (1018 HR):
Table 9-4:
S y = 32 kpsi,
Sut = 58 kpsi
τ all = min[0.3(58), 0.4(32)] = 12.8 kpsi
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Decision: Use E6010 electrode. From Table 9-3: S y = 50 kpsi, Sut = 62 kpsi,
τ all = min[0.3(62), 0.4(50)] = 20 kpsi
Decision: Since 1018 HR is weaker than the E6010 electrode, use τ all = 12.8 kpsi
Decision: Use an all-around square weld bead track.
l1 = 6 + a = 6 + 6.25 = 12.25 in
Throat area and other properties from Table 9-1:
A = 1.414 h(b + d ) = 1.414( h )(6 + 6) = 16.97 h
Primary shear
τ ′y =
( )
3
1179
V F 20 10
psi
= =
=
A A 16.97h
h
Secondary shear
(b + d )3 (6 + 6)3
=
= 288 in 3
6
6
J = 0.707 h(288) = 203.6h in 4
Ju =
τ x′′ = τ ′′y =
τ max
Mry
=
( )
20 103 (6.25 + 3)(3)
=
2726
psi
h
203.6h
1
4762
= τ x′′2 + (τ ′y + τ ′′y ) 2 =
27262 + (1179 + 2726) 2 =
psi
h
h
J
Relate stress to strength
τ max = τ all
⇒
( )
4762
= 12.8 103
h
⇒
h=
4762
= 0.372 in
12.8 103
( )
Decision:
Specify 3 / 8 in leg size
Specifications:
Pattern: All-around square weld bead track
Electrode: E6010
Type of weld: Fillet
Weld bead length: 24 in
Leg size: 3 / 8 in
Attachment length: 12.25 in
______________________________________________________________________________
Shigley’s MED, 10th edition
Chapter 9 Solutions, Page 20/35
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9-38
This is a good analysis task to test a student’s understanding.
(1) Solicit information related to a priori decisions.
(2) Solicit design variables b and d.
(3) Find h and round and output all parameters on a single screen. Allow return to Step 1
or Step 2.
(4) When the iteration is complete, the final display can be the bulk of your adequacy
assessment.
Such a program can teach too.
______________________________________________________________________________
9-39
The objective of this design task is to have the students teach themselves that the weld
patterns of Table 9-2 can be added or subtracted to obtain the properties of a
contemplated weld pattern. The instructor can control the level of complication. We have
left the presentation of the drawing to you. Here is one possibility. Study the problem’s
opportunities, and then present this (or your sketch) with the problem assignment.
Use b1 as the design variable. Express properties as a function of b1. From Table 9-3,
case 3:
A = 1.414h(b − b1 )
bd 2 b1d 2 (b − b1 )d 2
−
=
2
2
2
I = 0.707hI u
V
F
τ′ = =
A 1.414h(b − b1 )
Mc Fa (d / 2)
τ ′′ =
=
I
0.707 hI u
Iu =
Parametric study
Let a = 10 in, b = 8 in, d = 8 in, b1 = 2 in, τ all = 12.8 kpsi, l = 2(8 − 2) = 12 in
Shigley’s MED, 10th edition
Chapter 9 Solutions, Page 21/35
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A = 1.414h(8 − 2) = 8.48h in 2
I u = (8 − 2)(82 / 2) = 192 in 3
I = 0.707(h)(192) = 135.7h in 4
10 000 1179
=
τ′ =
psi
8.48h
h
10 000(10)(8 / 2) 2948
=
τ ′′ =
psi
135.7h
h
1
3175
τ max =
11792 + 29482 =
= 12 800 psi
h
h
from which h = 0.248 in. Do not round off the leg size – something to learn.
Iu
192
=
= 64.5 in
hl 0.248(12)
A = 8.48(0.248) = 2.10 in 2
fom ' =
I = 135.7(0.248) = 33.65 in 4
h2
0.2482
l=
12 = 0.369 in 3
2
2
I
33.65
=
= 91.2 in
eff =
vol 0.369
1179
= 4754 psi
τ′ =
0.248
2948
= 11 887 psi
τ ′′ =
0.248
3175
τ max =
= 12 800 psi
0.248
vol =
Now consider the case of uninterrupted welds,
b1 = 0
A = 1.414( h )(8 − 0) = 11.31h
I u = (8 − 0)(82 / 2) = 256 in 3
I = 0.707(256)h = 181h in 4
10 000 884
τ′ =
=
11.31h
h
10 000(10)(8 / 2) 2210
τ ′′ =
=
181h
h
1
2380
τ max =
= τ all
8842 + 22102 =
h
h
τ
2380
= 0.186 in
h = max =
τ all 12 800
Do not round off h.
Shigley’s MED, 10th edition
Chapter 9 Solutions, Page 22/35
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A = 11.31(0.186) = 2.10 in 2
I = 181(0.186) = 33.67 in 4
884
0.1862
16 = 0.277 in 3
= 4753 psi, vol =
0.186
2
2210
τ ′′ =
= 11 882 psi
0.186
I
256
= 86.0 in
fom ' = u =
hl 0.186(16)
I
33.67
=
= 121.7 in
eff = 2
(h / 2)l (0.1862 / 2)16
τ′ =
Conclusions: To meet allowable stress limitations, I and A do not change, nor do τ and σ.
To meet the shortened bead length, h is increased proportionately. However, volume of
bead laid down increases as h2. The uninterrupted bead is superior. In this example, we
did not round h and as a result we learned something. Our measures of merit are also
sensitive to rounding. When the design decision is made, rounding to the next larger
standard weld fillet size will decrease the merit.
Had the weld bead gone around the corners, the situation would change. Here is a follow
up task analyzing an alternative weld pattern.
______________________________________________________________________________
9-40
From Table 9-2
For the box
A = 1.414h(b + d )
Subtracting b1 from b and d1 from d
A = 1.414h ( b − b1 + d − d1 )
Iu =
d2
d3 b d2 1
1
(3b + d ) − 1 − 1 = ( b − b1 ) d 2 + d 3 − d13
6
6
2
2
6
(
)
l = 2(b − b1 + d − d1 )
fom = I u / hl
______________________________________________________________________________
Length of bead
Shigley’s MED, 10th edition
Chapter 9 Solutions, Page 23/35
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9-41 Computer programs will vary.
______________________________________________________________________________
9-42
τall = 12 kpsi. Use Fig. 9-17(a) for general geometry, but employ
beads and then
beads.
Horizontal parallel weld bead pattern
b = 3 in, d = 6 in
Table 9-2:
A = 1.414hb = 1.414(h)(3) = 4.24h in 2
bd 2 3(6)2
=
= 54 in 3
2
2
I = 0.707 hI u = 0.707( h)(54) = 38.2h in 4
10
2.358
τ′ =
kpsi
=
4.24h
h
Mc 10(10)(6 / 2) 7.853
τ ′′ =
kpsi
=
=
I
38.2h
h
Iu =
τ max = τ ′2 + τ ′′2 =
1
8.199
2.3582 + 7.8532 =
kpsi
h
h
Equate the maximum and allowable shear stresses.
τ max = τ all =
8.199
= 12
h
from which h = 0.683 in. It follows that
I = 38.2(0.683) = 26.1 in 4
The volume of the weld metal is
vol =
h 2l (0.683) 2 (3 + 3)
=
= 1.40 in 3
2
2
The effectiveness, (eff)H, is
(eff) H =
I
26.1
=
= 18.6 in
vol 1.4
Ans.
Shigley’s MED, 10th edition
Chapter 9 Solutions, Page 24/35
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Vertical parallel weld beads
b = 3 in
d = 6 in
From Table 9-2, case 2
A = 1.414hd = 1.414(h)(6) = 8.48h in 2
d 3 63
= = 72 in 3
6
6
I = 0.707 hI u = 0.707(h)(72) = 50.9h
10
1.179
τ′ =
psi
=
8.48h
h
Mc 10(10)(6 / 2) 5.894
τ ′′ =
psi
=
=
I
50.9h
h
1
6.011
τ max = τ ′2 + τ ′′2 =
1.1792 + 5.8942 =
kpsi
h
h
Iu =
Equating τ max to τ all gives h = 0.501 in. It follows that
I = 50.9(0.501) = 25.5 in 4
h 2l 0.5012
(6 + 6) = 1.51 in 3
=
2
2
I
25.5
(eff ) V =
=
= 16.7 in
vol 1.51
vol =
Ans.
______________________________________________________________________________
9-43
F = 0, T = 15 kip⋅in.
Table 9-1:
Ju = 2π r 3 = 2π (1)3 = 6.283 in3, J = 0.707(1/4) 6.283 = 1.111 in4
Tr 15 (1)
=
= 13.5 kpsi Ans.
J 1.111
______________________________________________________________________________
τ max =
9-44
F = 2 kip, T = 0.
Table 9-2:
A = 1.414 π h r = 1.414 π (1/4)(1) = 1.111 in2
Iu = π r 3 = π (1)3 = 3.142 in3, I = 0.707(1/4) 3.142 = 0.5553 in4
Shigley’s MED, 10th edition
Chapter 9 Solutions, Page 25/35
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τ′ =
V
2
=
= 1.80 kpsi
A 1.111
τ ′′ =
Mr 2 ( 6 )(1)
=
= 21.6 kpsi
I
0.5553
τ max = (τ′ 2 + τ′′ 2)1/2 = (1.802 + 21.62)1/2 = 21.7 kpsi
Ans.
______________________________________________________________________________
9-45
F = 2 kip, T = 15 kip⋅in.
Bending:
Table 9-2: A = 1.414 π h r = 1.414 π (1/4)(1) = 1.111 in2
Iu = π r 3 = π (1)3 = 3.142 in3, I = 0.707(1/4) 3.142 = 0.5553 in4
τ′ =
V
2
=
= 1.80 kpsi
A 1.111
(τ ′′) M
=
Mr 2 ( 6 )(1)
=
= 21.6 kpsi
I
0.5553
Torsion:
Table 9-1:
Ju = 2π r 3 = 2π (1)3 = 6.283 in3, J = 0.707(1/4) 6.283 = 1.111 in4
(τ ′′)T
=
Tr 15 (1)
=
= 13.5 kpsi
J 1.111
τ max = τ ′2 + (τ ′′ )M + (τ ′′ )T = 1.802 + 21.62 + 13.52 = 25.5 kpsi
2
2
Ans.
______________________________________________________________________________
9-46
F = 2 kip, T = 15 kip⋅in.
Bending:
Table 9-2: A = 1.414 π h r = 1.414 π h (1) = 4.442h in2
Iu = π r 3 = π (1)3 = 3.142 in3, I = 0.707 h (3.142) = 2.221h in4
τ′ =
V
2
0.4502
kpsi
=
=
A 4.442h
h
Shigley’s MED, 10th edition
Chapter 9 Solutions, Page 26/35
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(τ ′′ )M
Mr 2 ( 6 )(1) 5.403
kpsi
=
=
I
2.221h
h
=
Torsion:
Ju = 2π r 3 = 2π (1)3 = 6.283 in3, J = 0.707 h (6.283) = 4.442 in4
Table 9-1:
(τ ′′)T
=
Tr 15 (1) 3.377
kpsi
=
=
J 4.442h
h
τ max = τ ′ + (τ ′′ ) M + (τ ′′ )T
2
2
τ max = τ all
2
2
6.387
= 20
h
⇒
2
2
6.387
 0.4502   5.403   3.377 
= 
kpsi
 +
 +
 =
h
 h   h   h 
Should specify a 83 in weld.
⇒
h = 0.319 in
Ans.
Ans.
______________________________________________________________________________
9-47
h = 9 mm,
d = 200 mm,
b = 25 mm
From Table 9-2, case 2:
A = 1.414(9)(200) = 2.545(103) mm2
d 3 2003
=
= 1.333 106 mm3
6
6
I = 0.707h Iu = 0.707(9)(1.333)(106) = 8.484(106) mm4
( )
Iu =
( )
25 103
F
τ′ = =
= 9.82 MPa
A 2.545(103 )
M = 25(150) = 3750 N⋅m
τ ′′ =
Mc 3750(100)
=
(103 ) = 44.20 MPa
I
8.484(106 )
τ max = τ ′2 + τ ′′2 = 9.822 + 44.202 = 45.3 MPa Ans.
______________________________________________________________________________
9-48
h = 0.25 in, b = 2.5 in, d = 5 in.
Table 9-2, case 5:
A = 0.707h (b +2d) = 0.707(0.25)[2.5 + 2(5)] = 2.209 in2
Shigley’s MED, 10th edition
Chapter 9 Solutions, Page 27/35
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d2
52
=
= 2 in
b + 2d 2.5 + 2 ( 5 )
y=
2d 3
− 2d 2 y + ( b + 2d ) y 2
3
2 ( 53 )
=
− 2 ( 52 ) ( 2 ) +  2.5 + 2 ( 5 )  ( 22 ) = 33.33 in 3
3
Iu =
I = 0.707 h Iu = 0.707(1/4)(33.33) = 5.891 in4
Primary shear:
F
2
=
= 0.905 kpsi
A 2.209
τ′ =
Secondary shear (the critical location is at the bottom of the bracket):
y = 5 − 2 = 3 in
τ ′′ =
My 2 ( 5 )( 3)
=
= 5.093 kpsi
I
5.891
τ max = τ ′2 + τ ′′2 = 0.9052 + 5.0932 = 5.173 kpsi
τ all
18
=
= 3.48
Ans.
τ max 5.173
______________________________________________________________________________
n=
9-49
The largest possible weld size is 1/16 in. This is a small weld and thus difficult to
accomplish. The bracket’s load-carrying capability is not known. There are geometry
problems associated with sheet metal folding, load-placement and location of the center
of twist. This is not available to us. We will identify the strongest possible weldment.
Use a rectangular, weld-all-around pattern – Table 9-2, case 6:
Shigley’s MED, 10th edition
Chapter 9 Solutions, Page 28/35
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A = 1.414 h(b + d ) = 1.414(1 / 16)(1 + 7.5)
= 0.7512 in 2
x = b / 2 = 0.5
.5 in
y = d / 2 = 7.5
.5 / 2 = 3.75 in
2
d
7.52
Iu =
(3b + d ) =
[3(1) + 7.5] = 98.44 in 3
6
6
I = 0.707hI u = 0.707(1 / 16)(98.44) = 4.350 in 4
M = (3.75 + 0.5)
.5)W = 4.25W
V
W
τ′ =
=
= 1.331W
A 0.7512
12
Mc
4.25
25W (7.5 / 2)
= 3.664W
=
τ ′′ =
4.350
I
τ max = τ ′2 + τ ′′2 = W 1.3312 + 3.6642 = 3.90W
T allowable stress given is low. Let’s demon
onstrate that.
Material properties: The
For the 1020 CD brack
cket, use HR properties of Sy = 30 kpsi and Sut = 55. The 1030 HR
support, Sy = 37.5 kpsi
si and Sut = 68. The E6010 electrode has streng
ngths of Sy = 50 and
Sut = 62 kpsi.
Allowable stresses:
1020 HR:
τall = min[0.3(55), 0.4(30)] = min(16.5, 12) = 12 kpsi
1020 HR:
τall = min[0.3(68), 0.4(37.5)] = min(20.4, 15)
5) = 15 kpsi
E6010:
τall = min[0.3(62), 0.4(50)] = min(18.6, 20) = 18.6 kpsi
Since Table 9-6 givess 118.0 kpsi as the allowable shear stress, use this
th lower value.
Therefore, the allowab
able shear stress is
τall = min(14.4, 12, 18.0) = 12 kpsi
However, the allowabl
ble stress in the problem statement is 1.5 kpsii which
w
is low from the
weldment perspective.
e. The load associated with this strength is
τ max = τ all = 3.90W = 1500
W =
1500
= 385 lbf
3.90
If the welding can bee aaccomplished (1/16 leg size is a small weld),
), the
t weld strength is
12 000 psi and the load
ad associated with this strength is W = 12 000/
0/3.90 = 3077 lbf. Can
the bracket carry suchh a load?
There are geometry problems
pro
associated with sheet metal folding.. Load
L
placement is
important and the cente
nter of twist has not been identified. Also, thee load-carrying
l
capability of the top be
bend is unknown.
Shigley’s MED, 10th edition
Chapter
ter 9 Solutions, Page 29/35
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These uncertainties ma
may require the use of a different weld pattern.
n. Our solution provides
the best weldment andd thus insight for comparing a welded joint to one which employs
screw fasteners.
_________________________
_______________________________________
____________________
9-50
F
FB
FBx
FBy
= 100 lbf , τ all = 3 kpsi
= 100(16 / 3) = 533.3 lbf
= −533.3cos 60o = −266.7 lbf
= −533.3cos 30o = −462 lbf
It follows that RAy = 56
562 lbf and RAx = 266.7 lbf, RA = 622 lbf
M = 100(16) = 1600 lbf · in
The OD of the tubess is 1 in. From Table 9-1, case 6:
A = 2 [1.414(π hr ) ] = 2(1.414)(π h)(1 / 2) = 4.442
2h in 2
J u = 2π r 3 = 2π (1 / 2)3 = 0.7854 in 3
J = 2(
2(0.707)hJ u = 1.414(0.7854)h = 1.111h in 4
The weld only carriess the
t torsional load between the handle and tub
ube A. Consequently,
the primary shear in the
th weld is zero, and the maximum shear stress
ss is comprised entirely
of the secondary shear.
ar.
Shigley’s MED, 10th edition
Chapter
ter 9 Solutions, Page 30/35
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Tc
Mc 1600(0.5) 720.1
psi
=
=
=
1.111h
J
J
h
720.1
=
= 3000
h
τ max = τ ′′ =
τ max = τ all
h=
720.1
= 0.240 → 1 / 4 in
3000
Decision: Use 1/4 in fillet welds
Ans.
______________________________________________________________________________
9-51
For the pattern in bending shown, find the centroid G of the weld group.
x=
75 ( 6 )(150 ) + 325 ( 9 )(150 )
( 6 )(150 ) + ( 9 )(150 )
I 6 mm = 2 ( I G + Ax 2 )
= 225 mm
6 mm
 0.707 ( 6 ) (1503 )

2
= 2
+ 0.707 ( 6 )(150 )( 225 − 75 )  = 31.02 (106 ) mm 4
12


I 9 mm
 0.707 ( 9 ) (1503 )

2
= 2
+ 0.707 ( 9 )(150 )(175 − 75 )  = 22.67 (106 ) mm4
12


I = I 6 mm + I 9 mm = (31.02 + 22.67)(106) = 53.69(106) mm4
The critical location is at B. With τ in MPa, and F in kN
F (103 )
V
τ′ = =
= 0.3143F
A 2 0.707 ( 6 + 9 )(150 ) 
3
Mc 200 F (10 ) ( 225 )
τ ′′ =
=
= 0.8381F
I
53.69 (106 )
Shigley’s MED, 10th edition
Chapter 9 Solutions, Page 31/35
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τ max = τ ′2 + τ ′′2 = F 0.31432 + 0.83812 = 0.8951F
Materials:
1015 HR (Table A-20): Sy = 190 MPa, E6010 Electrode(Table 9-3): Sy = 345 MPa
τall = 0.577(190) = 109.6 MPa
Eq. (5-21), p. 239
τ all / n
109.6 / 2
= 61.2 kN
Ans.
0.8951 0.8951
______________________________________________________________________________
F=
9-52
=
In the textbook, Fig. Problem 9-52b is a free-body diagram of the bracket. Forces and
moments that act on the welds are equal, but of opposite sense.
(a)
M = 1200(0.366) = 439 lbf · in Ans.
(b)
Fy = 1200 sin 30° = 600 lbf Ans.
(c)
Fx = 1200 cos 30° = 1039 lbf Ans.
(d) From Table 9-2, case 6:
A = 1.414(0.25)(0.25 + 2.5) = 0.972 in 2
d2
2.52
Iu =
(3b + d ) =
[3(0.25) + 2.5] = 3.39 in 3
6
6
The second area moment about an axis through G and parallel to z is
I = 0.707hI u = 0.707(0.25)(3.39) = 0.599 in 4 Ans.
(e) Refer to Fig. Problem 9-52b. The shear stress due to Fy is
τ1 =
Fy
A
=
600
= 617 psi
0.972
The shear stress along the throat due to Fx is
τ2 =
Fx
1039
=
= 1069 psi
0.972
A
The resultant of τ1 and τ2 is in the throat plane
τ ′ = τ12 + τ 22 = 6172 + 10692 = 1234 psi
The bending of the throat gives
τ ′′ =
Mc
439(1.25)
=
= 916 psi
I
0.599
Shigley’s MED, 10th edition
Chapter 9 Solutions, Page 32/35
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The maximum shear stress is
τ max = τ ′2 + τ ′′2 = 12342 + 9162 = 1537 psi
(f) Materials:
1018 HR Member:
E6010 Electrode:
n=
S sy
τ max
Ans.
Sy = 32 kpsi, Sut = 58 kpsi (Table A-20)
Sy = 50 kpsi (Table 9-3)
=
0.577 S y
τ max
=
0.577(32)
= 12.0
1.537
Ans.
(g) Bending in the attachment near the base. The cross-sectional area is approximately
equal to bh.
A1 bh = 0.25(2.5) = 0.625 in 2
F
1039
= 1662 psi
τ xy = x =
0.625
A1
0.25(2.5) 2
I
bd 2
=
=
= 0.260 in 3
c
6
6
At location A,
F
M
σy = y +
A1
I /c
600
439
+
= 2648 psi
σy =
0.625 0.260
The von Mises stress σ ′ is
σ ′ = σ y2 + 3τ xy2 =
26482 + 3(1662) 2 = 3912 psi
Thus, the factor of safety is,
S
32
n= y =
= 8.18
σ ′ 3.912
Ans.
The clip on the mooring line bears against the side of the 1/2-in hole. If the clip fills
the hole
F
−1200
σ =
=
= −9600 psi
td
0.25(0.50)
S
32(103 )
n=− y =−
= 3.33 Ans.
σ′
−9600
Further investigation of this situation requires more detail than is included in the task
statement.
(h) In shear fatigue, the weakest constituent of the weld melt is 1018 HR with Sut = 58
kpsi, Eq. (6-8), p. 290, gives
Shigley’s MED, 10th edition
Chapter 9 Solutions, Page 33/35
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Se′ = 0.5Sut = 0.5(58) = 29.0 kpsi
ka = 14.4(58)-0.718 = 0.780
Eq. (6-19), p. 295:
For uniform shear stress on the throat, assume kb = 1.
Eq.(6-26), p. 298: kc = 0.59
From Eq. (6-18), p. 295, the endurance strength in shear is
Sse = 0.780(1)(0.59)(29.0) = 13.35 kpsi
From Table 9-5, the shear stress-concentration factor is Kf s = 2.7. The loading is
repeatedly-applied
τ
1.537
τ a = τ m = K f s max = 2.7
= 2.07 kpsi
2
2
Table 6-7, p. 315: Gerber factor of safety nf, adjusted for shear, with Ssu = 0.67Sut
2
1  S su  τ a 
 −1 +
nf = 

2  τ m  S se 

 2τ S  
1 +  m se  
 S suτ a  
2
2

 2(2.07)(13.35)  
1  0.67(58)   2.07  
= 

 −1 + 1 + 
  = 5.83 Ans.
2  2.07   13.35  
 0.67(58)(2.07)  


Attachment metal should be checked for bending fatigue.
______________________________________________________________________________
9-53
(a) Use b = d = 4 in. Since h = 5/8 in, the primary shear is
τ′ =
F
= 0.2829 F
1.414(5 / 8)(4)
The secondary shear calculations, for a moment arm of 14 in give
4[3(42 ) + 42 ]
= 42.67 in 3
6
J = 0.707hJ u = 0.707(5 / 8)42.67 = 18.85 in 4
Mry 14F (2)
=
= 1.485F
τ ′′x = τ ′′y =
J
18.85
Ju =
Thus, the maximum shear and allowable load are:
Shigley’s MED, 10th edition
Chapter 9 Solutions, Page 34/35
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τ max = F 1.4852 + (0.2829 + 1.485) 2 = 2.309F
25
τ all
F =
=
= 10.8 kip Ans.
2.309 2.309
The load for part (a) has increased by a factor of 10.8/2.71 = 3.99
Ans.
(b) From Prob. 9-18b, τall = 11 kpsi
Fall =
τ all
2.309
=
11
= 4.76 kip
2.309
The allowable load in part (b) has increased by a factor of 4.76/1.19 = 4 Ans.
______________________________________________________________________________
Purchase the hook having the design shown in Fig. Problem 9-54b. Referring to text Fig.
9-29a, this design reduces peel stresses.
______________________________________________________________________________
9-54
9-55
(a)
l/2
1 l / 2 Pω cosh(ω x)
A
dx = A1 ∫ cosh(ω x) dx = 1 sinh(ω x)
τ = ∫
/
2
/
2
l
l
−
−
l
4b sinh(ωl / 2)
ω
−l / 2
A1
A1
= [sinh(ωl / 2) − sinh(−ωl / 2)] = [sinh(ωl / 2) − (− sinh(ωl / 2))]
l/2
ω
=
(b)
(c)
2 A1 sinh(ωl / 2)
ω
Pω
P
=
[2sinh(ωl / 2)] =
ω
4bl sinh(ωl / 2)
2bl
Pω cosh(ω l / 2)
Pω
τ (l / 2) =
=
Ans.
4b sinh(ω l / 2)
4b tanh(ω l / 2)
K =
Pω
τ (l / 2)
ωl / 2
 2bl 
=

=
4b tanh(ωl / 2)  P  tanh(ωl / 2)
τ
Ans.
Ans.
For computer programming, it can be useful to express the hyperbolic tangent in
terms of exponentials:
ωl exp(ωl / 2) − exp(−ωl / 2)
K =
Ans.
2 exp(ωl / 2) + exp(−ωl / 2)
______________________________________________________________________________
9-56
This is a computer programming exercise. All programs will vary.
Shigley’s MED, 10th edition
Chapter 9 Solutions, Page 35/35
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Chapter 10
10-1
From Eqs. (10-4) and (10-5)
KW − K B =
4C − 1 0.615 4C + 2
+
−
4C − 4
C
4C − 3
Plot 100(KW − KB)/ KW vs. C for 4 ≤ C ≤ 12 obtaining
1.4
1.3
1.2
1.1
100(KW-KB)/KW
1
0.9
0.8
0.7
4
6
8
10
12
C
We see the maximum and minimum occur at C = 4 and 12 respectively where
Maximum = 1.36 % Ans.,
and Minimum = 0.743 % Ans.
______________________________________________________________________________
A = Sdm
dim(Auscu) = [dim (S) dim(d m)]uscu = kpsi⋅inm
10-2
dim(ASI) = [dim (S) dim(d m)]SI = MPa⋅mmm
ASI =
MPa mm m
m
⋅ m Auscu = 6.894 757 ( 25.4 ) Auscu
kpsi in
6.895 ( 25.4 ) Auscu
m
Ans.
For music wire, from Table 10-4:
Auscu = 201 kpsi⋅inm,
m = 0.145;
what is ASI?
Ans.
ASI = 6.895(25.4)0.145 (201) = 2215 MPa⋅mmm
______________________________________________________________________________
10-3
Given: Music wire, d = 2.5 mm, OD = 31 mm, plain ground ends, Nt = 14 coils.
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Chapter 10 Solutions, Page 1/41
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(a) Table 10-1:
Na = Nt − 1 = 14 − 1 = 13 coils
D = OD − d = 31− 2.5 = 28.5 mm
C = D/d = 28.5/2.5 = 11.4
Table 10-5:
d = 2.5/25.4 = 0.098 in
Eq. (10-9):
k=
(b) Table 10-1:
G = 81.0(103) MPa
⇒
2.54 ( 81)103
d 4G
=
= 1.314 N / mm
8 D 3 N a 8 ( 28.53 )13
Ls = d Nt = 2.5(14) = 35 mm
A = 2211 MPa⋅mmm
Table 10-4:
m = 0.145,
Eq. (10-14):
Sut =
Table 10-6:
Ssy = 0.45(1936) = 871.2 MPa
Eq. (10-5):
KB =
Eq. (10-7):
Fs =
L0 =
(c)
Ans.
A
2211
=
= 1936 MPa
m
d
2.50.145
4C + 2 4 (11.4 ) + 2
=
= 1.117
4C − 3 4 (11.4 ) − 3
π d 3 S sy
8K B D
=
π ( 2.53 ) 871.2
8 (1.117 ) 28.5
= 167.9 N
Fs
167.9
+ Ls =
+ 35 = 162.8 mm
k
1.314
Ans.
Ans.
2.63 ( 28.5 )
= 149.9 mm . Spring needs to be supported. Ans.
0.5
______________________________________________________________________________
(d)
10-4
( L0 )cr =
Given: Design load, F1 = 130 N.
Referring to Prob. 10-3 solution, C = 11.4, Na = 13 coils, Ssy = 871.2 MPa, Fs = 167.9 N,
L0 = 162.8 mm and (L0)cr = 149.9 mm.
Eq. (10-18): 4 ≤ C ≤ 12
C = 11.4
O.K.
O.K.
Eq. (10-19): 3 ≤ Na ≤ 15 Na = 13
Fs
167.9
Eq. (10-17): ξ =
−1 =
− 1 = 0.29
130
F1
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Eq. (10-20): ξ ≥ 0.15, ξ = 0.29 O.K .
From Eq. (10-7) for static service
 8F D 
τ 1 = K B  1 3  = 1.117
 πd 
n=
Eq. (10-21):
S sy
τ1
=
8(130)(28.5)
= 674 MPa
π (2.5)3
871.2
= 1.29
674
ns ≥ 1.2, n = 1.29 O.K.
 167.9 
 167.9 
 = 674 
 = 870.5 MPa
 130 
 130 
S sy / τ s = 871.2 / 870.5 1
τ s = τ1 
Ssy/τs ≥ (ns )d : Not solid-safe (but was the basis of the design). Not O.K.
L0 ≤ (L0)cr: 162.8 ≥ 149.9 Not O.K.
Design is unsatisfactory. Operate over a rod? Ans.
______________________________________________________________________________
10-5
Given: Oil-tempered wire, d = 0.2 in, D = 2 in, Nt = 12 coils, L0 = 5 in, squared ends.
(a) Table 10-1:
Ls = d (Nt + 1) = 0.2(12 + 1) = 2.6 in Ans.
(b) Table 10-1:
Table 10-5:
Na = Nt − 2 = 12 − 2 = 10 coils
G = 11.2 Mpsi
0.2 (11.2 )10
d 4G
=
= 28 lbf/in
3
8D N a
8 ( 23 )10
4
Eq. (10-9):
k=
6
Fs = k ys = k (L0 − Ls ) = 28(5 − 2.6) = 67.2 lbf
(c) Eq. (10-1):
Ans.
C = D/d = 2/0.2 = 10
4C + 2 4 (10 ) + 2
=
= 1.135
4C − 3 4 (10 ) − 3
Eq. (10-5):
KB =
Eq. (10-7):
τ s = KB
Table 10-4:
m = 0.187, A = 147 kpsi⋅inm
8 ( 67.2 ) 2
8 Fs D
= 1.135
= 48.56 (103 ) psi
3
3
πd
π ( 0.2 )
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Chapter 10 Solutions, Page 3/41
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A
147
=
= 198.6 kpsi
d m 0.20.187
Eq. (10-14):
Sut =
Table 10-6:
Ssy = 0.50 Sut = 0.50(198.6) = 99.3 kpsi
S sy
99.3
= 2.04
Ans.
τ s 48.56
______________________________________________________________________________
ns =
10-6
=
Given: Oil-tempered wire, d = 4 mm, C = 10, plain ends, L0 = 80 mm, and at F = 50 N,
y = 15 mm.
(a)
k = F/y = 50/15 = 3.333 N/mm
(b)
D = Cd = 10(4) = 40 mm
Ans.
OD = D + d = 40 + 4 = 44 mm
Ans.
(c) From Table 10-5, G = 77.2 GPa
4
3
d 4G 4 ( 77.2 )10
=
= 11.6 coils
8kD3 8 ( 3.333) 403
Eq. (10-9):
Na =
Table 10-1:
Nt = Na = 11.6 coils
Ans.
(d) Table 10-1:
Ls = d (Nt + 1) = 4(11.6 + 1) = 50.4 mm
(e) Table 10-4:
m = 0.187, A = 1855 MPa⋅mmm
Ans.
A 1855
=
= 1431 MPa
d m 40.187
Eq. (10-14):
Sut =
Table 10-6:
Ssy = 0.50 Sut = 0.50(1431) = 715.5 MPa
ys = L0 − Ls = 80 − 50.4 = 29.6 mm
Fs = k ys = 3.333(29.6) = 98.66 N
4C + 2 4(10) + 2
=
= 1.135
4C − 3 4(10) − 3
Eq. (10-5):
KB =
Eq. (10-7):
τ s = KB
8 ( 98.66 ) 40
8 Fs D
= 1.135
= 178.2 MPa
3
πd
π ( 43 )
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Chapter 10 Solutions, Page 4/41
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S sy
715.5
= 4.02
Ans.
τ s 178.2
______________________________________________________________________________
ns =
10-7
=
Static service spring with: HD steel wire, d = 0.080 in, OD = 0.880 in, Nt = 8 coils, plain
and ground ends.
Preliminaries
Table 10-5: A = 140 kpsi · inm, m = 0.190
A
140
Eq. (10-14): Sut = m =
= 226.2 kpsi
d
0.0800.190
Table 10-6: Ssy = 0.45(226.2) = 101.8 kpsi
Then,
D = OD − d = 0.880 − 0.080 = 0.8 in
Eq. (10-1):
C = D/d = 0.8/0.08 = 10
4C + 2 4(10) + 2
Eq. (10-5):
KB =
=
= 1.135
4C − 3
4(10) − 3
Table 10-1: Na = Nt − 1 = 8 − 1 = 7 coils
Ls = dNt = 0.08(8) = 0.64 in
Eq. (10-7) For solid-safe, ns = 1.2 :
3
3
π d 3S sy / ns π ( 0.08 ) 101.8 (10 ) / 1.2
Fs =
=
= 18.78 lbf
8K B D
8(1.135)(0.8)
Eq. (10-9):
k =
0.084 (11.5 )106
d 4G
=
= 16.43 lbf/in
8D 3 N a
8 ( 0.83 ) 7
ys =
Fs 18.78
=
= 1.14 in
k 16.43
(a) L0 = ys + Ls = 1.14 + 0.64 = 1.78 in Ans.
L
1.78
(b) Table 10-1: p = 0 =
= 0.223 in Ans.
Nt
8
(c) From above: Fs = 18.78 lbf Ans.
(d) From above: k = 16.43 lbf/in Ans.
2.63D
2.63(0.8)
(e) Table 10-2 and Eq. (10-13):
( L0 )c r =
=
= 4.21 in
α
0.5
Since L0 < (L0)cr, buckling is unlikely Ans.
______________________________________________________________________________
10-8
Given: Design load, F1 = 16.5 lbf.
Referring to Prob. 10-7 solution, C = 10, Na = 7 coils, Ssy = 101.8 kpsi, Fs = 18.78 lbf,
ys = 1.14 in, L0 = 1.78 in, and (L0)cr = 4.21 in.
Eq. (10-18):
4 ≤ C ≤ 12
C = 10
O.K.
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Eq. (10-19):
3 ≤ Na ≤ 15
Na = 7
O.K.
Fs
18.78
−1 =
− 1 = 0.14
16.5
F1
Eq. (10-20): ξ ≥ 0.15, ξ = 0.14 not O.K . , but probably acceptable.
From Eq. (10-7) for static service
Eq. (10-17):
ξ =
 8F D 
= 74.5 (103 ) psi = 74.5 kpsi
τ 1 = K B  1 3  = 1.135
3
d
(0.080)
π
π


n=
S sy
τ1
=
8(16.5)(0.8)
101.8
= 1.37
74.5
Eq. (10-21):
ns ≥ 1.2, n = 1.37 O.K.
Eq. (10-21):
 18.78 
 18.78 
 = 74.5 
 = 84.8 kpsi
 16.5 
 16.5 
ns = S sy / τ s = 101.8 / 84.8 = 1.20
ns ≥ 1.2, ns = 1.2 It is solid-safe (basis of design). O.K.
τ s = τ1 
1.78 in ≤ 4.21 in O.K.
Eq. (10-13) and Table 10-2: L0 ≤ (L0)cr
______________________________________________________________________________
10-9
Given: A228 music wire, squared and ground ends, d = 0.007 in, OD = 0.038 in,
L0 = 0.58 in,
Nt = 38 coils.
Eq. (10-1):
Eq. (10-5):
Table 10-1:
Table 10-5:
Eq. (10-9):
Table 10-1:
Eq. (10-7):
Table 10-4:
D = OD − d = 0.038 − 0.007 = 0.031 in
C = D/d = 0.031/0.007 = 4.429
4C + 2 4 ( 4.429 ) + 2
KB =
=
= 1.340
4C − 3 4 ( 4.429 ) − 3
Na = Nt − 2 = 38 − 2 = 36 coils
(high)
G = 12.0 Mpsi
0.007 4 (12.0 )106
d 4G
k=
=
= 3.358 lbf/in
8D 3 N a
8 ( 0.0313 ) 36
Ls = dNt = 0.007(38) = 0.266 in
ys = L0 − Ls = 0.58 − 0.266 = 0.314 in
Fs = kys = 3.358(0.314) = 1.054 lbf
8 (1.054 ) 0.031
8F D
τ s = K B s 3 = 1.340
= 325.1(103 ) psi
3
πd
π ( 0.007 )
(1)
A = 201 kpsi⋅inm, m = 0.145
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Eq. (10-14):
Table 10-6:
A
201
=
= 412.7 kpsi
d m 0.007 0.145
Ssy = 0.45 Sut = 0.45(412.7) = 185.7 kpsi
Sut =
τs > Ssy, that is, 325.1 > 185.7 kpsi, the spring is not solid-safe. Return to Eq. (1) with
Fs = kys and τs = Ssy /ns, and solve for ys, giving
( Ssy / ns ) π d 3 185.7 (103 ) / 1.2 π ( 0.0073 )
ys =
=
8 K B kD
The free length should be wound to
8 (1.340 ) 3.358 ( 0.031)
L0 = Ls + ys = 0.266 + 0.149 = 0.415 in
= 0.149 in
Ans.
This only addresses the solid-safe criteria. There are additional problems.
______________________________________________________________________________
10-10 Given: B159 phosphor-bronze, squared and ground. ends, d = 0.014 in, OD = 0.128 in,
L0 = 0.50 in, Nt = 16 coils.
Eq. (10-1):
Eq. (10-5):
Table 10-1:
Table 10-5:
Eq. (10-9):
Table 10-1:
Eq. (10-7):
Table 10-4:
Eq. (10-14):
Table 10-6:
D = OD − d = 0.128 − 0.014 = 0.114 in
C = D/d = 0.114/0.014 = 8.143
4C + 2 4 ( 8.143) + 2
KB =
=
= 1.169
4C − 3 4 ( 8.143) − 3
Na = Nt − 2 = 16 − 2 = 14 coils
G = 6 Mpsi
0.0144 ( 6 )106
d 4G
k=
=
= 1.389 lbf/in
8D 3 N a 8 ( 0.1143 )14
Ls = dNt = 0.014(16) = 0.224 in
ys = L0 − Ls = 0.50 − 0.224 = 0.276 in
Fs = kys = 1.389(0.276) = 0.3834 lbf
8 ( 0.3834 ) 0.114
8F D
τ s = K B s 3 = 1.169
= 47.42 (103 ) psi
3
πd
π ( 0.014 )
(1)
A = 145 kpsi⋅inm, m = 0
A
145
Sut = m =
= 145 kpsi
d
0.0140
Ssy = 0.35 Sut = 0.35(135) = 47.25 kpsi
τs > Ssy, that is, 47.42 > 47.25 kpsi, the spring is not solid-safe. Return to Eq. (1) with
Fs = kys and τs = Ssy /ns, and solve for ys, giving
( Ssy / ns ) π d 3 47.25 (103 ) /1.2 π ( 0.0143 )
ys =
=
8 K B kD
The free length should be wound to
8 (1.169 )1.389 ( 0.114 )
Shigley’s MED, 10th edition
= 0.229 in
Chapter 10 Solutions, Page 7/41
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L0 = Ls + ys = 0.224 + 0.229 = 0.453 in
Ans.
______________________________________________________________________________
10-11 Given: A313 stainless steel, squared and ground ends, d = 0.050 in, OD = 0.250 in,
L0 = 0.68 in, Nt = 11.2 coils.
Eq. (10-1):
Eq. (10-5):
Table 10-1:
Table 10-5:
Eq. (10-9):
Table 10-1:
Eq. (10-7):
Table 10-4:
Eq. (10-14):
Table 10-6:
D = OD − d = 0.250 − 0.050 = 0.200 in
C = D/d = 0.200/0.050 = 4
4C + 2 4 ( 4 ) + 2
KB =
=
= 1.385
4C − 3 4 ( 4 ) − 3
Na = Nt − 2 = 11.2 − 2 = 9.2 coils
G = 10 Mpsi
0.0504 (10 )106
d 4G
k=
=
= 106.1 lbf/in
8D 3 N a
8 ( 0.23 ) 9.2
Ls = dNt = 0.050(11.2) = 0.56 in
ys = L0 − Ls = 0.68 − 0.56 = 0.12 in
Fs = kys = 106.1(0.12) = 12.73 lbf
8 (12.73) 0.2
8F D
τ s = K B s 3 = 1.385
= 71.8 (103 ) psi
πd
π ( 0.0503 )
A = 169 kpsi⋅inm, m = 0.146
A
169
= 261.7 kpsi
Sut = m =
d
0.0500.146
Ssy = 0.35 Sut = 0.35(261.7) = 91.6 kpsi
S sy
91.6
= 1.28 Spring is solid-safe (ns > 1.2) Ans.
τ s 71.8
______________________________________________________________________________
ns =
=
10-12 Given: A227 hard-drawn wire, squared and ground ends, d = 0.148 in, OD = 2.12 in,
L0 = 2.5 in, Nt = 5.75 coils.
Eq. (10-1):
Eq. (10-5):
Table 10-1:
Table 10-5:
Eq. (10-9):
Table 10-1:
D = OD − d = 2.12 − 0.148 = 1.972 in
C = D/d = 1.972/0.148 = 13.32
(high)
+
4
13.32
2
(
) = 1.099
4C + 2
KB =
=
4C − 3 4 (13.32 ) − 3
Na = Nt − 2 = 5.75 − 2 = 3.75 coils
G = 11.4 Mpsi
0.1484 (11.4 )106
d 4G
k=
=
= 23.77 lbf/in
8D 3 N a
8 (1.9723 ) 3.75
Ls = dNt = 0.148(5.75) = 0.851 in
ys = L0 − Ls = 2.5 − 0.851 = 1.649 in
Fs = kys = 23.77(1.649) = 39.20 lbf
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Chapter 10 Solutions, Page 8/41
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8 ( 39.20 )1.972
8Fs D
= 1.099
= 66.7 (103 ) psi
3
πd
π ( 0.1483 )
Eq. (10-7):
τ s = KB
Table 10-4:
A = 140 kpsi⋅inm, m = 0.190
A
140
= 201.3 kpsi
Sut = m =
d
0.1480.190
Ssy = 0.35 Sut = 0.45(201.3) = 90.6 kpsi
Eq. (10-14):
Table 10-6:
S sy
90.6
Ans.
= 1.36 Spring is solid-safe (ns > 1.2)
τ s 66.7
______________________________________________________________________________
ns =
=
10-13 Given: A229 OQ&T steel, squared and ground ends, d = 0.138 in, OD = 0.92 in,
L0 = 2.86 in, Nt = 12 coils.
D = OD − d = 0.92 − 0.138 = 0.782 in
Eq. (10-5):
C = D/d = 0.782/0.138 = 5.667
4C + 2 4 ( 5.667 ) + 2
KB =
=
= 1.254
4C − 3 4 ( 5.667 ) − 3
Table 10-1:
Na = Nt − 2 = 12 − 2 = 10 coils
Eq. (10-1):
A229 OQ&T steel is not given in Table 10-5. From Table A-5, for carbon steels,
G = 11.5 Mpsi.
0.1384 (11.5 )106
d 4G
Eq. (10-9):
k=
=
= 109.0 lbf/in
8D3 N a
8 ( 0.7823 )10
Table 10-1:
Eq. (10-7):
Table 10-4:
Eq. (10-14):
Table 10-6:
Ls = dNt = 0.138(12) = 1.656 in
ys = L0 − Ls = 2.86 − 1.656 = 1.204 in
Fs = kys = 109.0(1.204) = 131.2 lbf
8 (131.2 ) 0.782
8F D
τ s = K B s 3 = 1.254
= 124.7 (103 ) psi
πd
π ( 0.1383 )
(1)
A = 147 kpsi⋅inm, m = 0.187
A
147
Sut = m =
= 212.9 kpsi
d
0.1380.187
Ssy = 0.50 Sut = 0.50(212.9) = 106.5 kpsi
τs > Ssy, that is, 124.7 > 106.5 kpsi, the spring is not solid-safe. Return to Eq. (1) with
Fs = kys and τs = Ssy /ns, and solve for ys, giving
3
3
S sy / ns ) π d 3 106.5 (10 ) /1.2  π ( 0.138 )
(
ys =
=
= 0.857 in
8K B kD
8 (1.254 )109.0 ( 0.782 )
The free length should be wound to
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L0 = Ls + ys = 1.656 + 0.857 = 2.51 in
Ans.
______________________________________________________________________________
10-14 Given: A232 chrome-vanadium steel, squared and ground ends, d = 0.185 in, OD = 2.75
in, L0 = 7.5 in, Nt = 8 coils.
Eq. (10-1):
Eq. (10-5):
D = OD − d = 2.75 − 0.185 = 2.565 in
C = D/d = 2.565/0.185 = 13.86
(high)
4C + 2 4 (13.86 ) + 2
KB =
=
= 1.095
4C − 3 4 (13.86 ) − 3
Table 10-1:
Na = Nt − 2 = 8 − 2 = 6 coils
Table 10-5:
G = 11.2 Mpsi.
0.1854 (11.2 )106
d 4G
k=
=
= 16.20 lbf/in
8D3 N a
8 ( 2.5653 ) 6
Eq. (10-9):
Table 10-1:
Eq. (10-7):
Ls = dNt = 0.185(8) = 1.48 in
ys = L0 − Ls = 7.5 − 1.48 = 6.02 in
Fs = kys = 16.20(6.02) = 97.5 lbf
8 ( 97.5 ) 2.565
8F D
τ s = K B s 3 = 1.095
= 110.1(103 ) psi
3
πd
π ( 0.185 )
(1)
A = 169 kpsi⋅inm, m = 0.168
A
169
Eq. (10-14): Sut = m =
= 224.4 kpsi
d
0.1850.168
Table 10-6: Ssy = 0.50 Sut = 0.50(224.4) = 112.2 kpsi
S
112.2
ns = sy =
= 1.02 Spring is not solid-safe (ns < 1.2)
τ s 110.1
Return to Eq. (1) with Fs = kys and τs = Ssy /ns, and solve for ys, giving
( Ssy / ns ) π d 3 = 112.2 (103 ) / 1.2 π ( 0.1853 ) = 5.109 in
ys =
8K B kD
8 (1.095 )16.20 ( 2.565 )
The free length should be wound to
Table 10-4:
L0 = Ls + ys = 1.48 + 5.109 = 6.59 in
Ans.
______________________________________________________________________________
10-15 Given: A313 stainless steel, squared and ground ends, d = 0.25 mm, OD = 0.95 mm,
L0 = 12.1 mm, Nt = 38 coils.
D = OD − d = 0.95 − 0.25 = 0.7 mm
Eq. (10-1):
C = D/d = 0.7/0.25 = 2.8
(low)
4
2.8
2
+
( ) = 1.610
4C + 2
Eq. (10-5):
KB =
=
4C − 3 4 ( 2.8 ) − 3
Shigley’s MED, 10th edition
Chapter 10 Solutions, Page 10/41
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Table 10-1:
Na = Nt − 2 = 38 − 2 = 36 coils
Table 10-5:
G = 69.0(103) MPa.
0.254 ( 69.0 )103
d 4G
k=
=
= 2.728 N/mm
8D3 N a
8 ( 0.73 ) 36
Eq. (10-9):
Table 10-1:
Eq. (10-7):
(high)
Ls = dNt = 0.25(38) = 9.5 mm
ys = L0 − Ls = 12.1 − 9.5 = 2.6 mm
Fs = kys = 2.728(2.6) = 7.093 N
8 ( 7.093) 0.7
8F D
τ s = K B s 3 = 1.610
= 1303 MPa
πd
π ( 0.253 )
(1)
Table 10-4 (dia. less than table):
A = 1867 MPa⋅mmm, m = 0.146
A
1867
Eq. (10-14): Sut = m =
= 2286 MPa
d
0.250.146
Table 10-6: Ssy = 0.35 Sut = 0.35(2286) = 734 MPa
τs > Ssy, that is, 1303 > 734 MPa, the spring is not solid-safe. Return to Eq. (1) with
Fs = kys and τs = Ssy /ns, and solve for ys, giving
S sy / ns ) π d 3 ( 734 / 1.2 ) π ( 0.253 )
(
ys =
=
= 1.22 mm
8 K B kD
8 (1.610 ) 2.728 ( 0.7 )
The free length should be wound to
L0 = Ls + ys = 9.5 + 1.22 = 10.72 mm
Ans.
This only addresses the solid-safe criteria. There are additional problems.
______________________________________________________________________________
10-16 Given: A228 music wire, squared and ground ends, d = 1.2 mm, OD = 6.5 mm, L0 = 15.7
mm, Nt = 10.2 coils.
D = OD − d = 6.5 − 1.2 = 5.3 mm
Eq. (10-1):
C = D/d = 5.3/1.2 = 4.417
4C + 2 4 ( 4.417 ) + 2
Eq. (10-5):
KB =
=
= 1.368
4C − 3 4 ( 4.417 ) − 3
Table (10-1): Na = Nt − 2 = 10.2 − 2 = 8.2 coils
Table 10-5 (d = 1.2/25.4 = 0.0472 in):
G = 81.7(103) MPa.
1.24 ( 81.7 )103
d 4G
Eq. (10-9):
k=
=
= 17.35 N/mm
8D3 N a
8 ( 5.33 ) 8.2
Table 10-1:
Ls = dNt = 1.2(10.2) = 12.24 mm
ys = L0 − Ls = 15.7 − 12.24 = 3.46 mm
Fs = kys = 17.35(3.46) = 60.03 N
Shigley’s MED, 10th edition
Chapter 10 Solutions, Page 11/41
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8 ( 60.03) 5.3
8 Fs D
= 1.368
= 641.4 MPa
3
πd
π (1.23 )
Eq. (10-7):
τ s = KB
Table 10-4:
A = 2211 MPa⋅mmm, m = 0.145
A
2211
Sut = m = 0.145 = 2153 MPa
d
1.2
Ssy = 0.45 Sut = 0.45(2153) = 969 MPa
Eq. (10-14):
Table 10-6:
S sy
(1)
969
= 1.51 Spring is solid-safe (ns > 1.2) Ans.
τ s 641.4
______________________________________________________________________________
ns =
=
10-17 Given: A229 OQ&T steel, squared and ground ends, d = 3.5 mm, OD = 50.6 mm,
L0 = 75.5 mm, Nt = 5.5 coils.
Eq. (10-1):
Eq. (10-5):
Table 10-1:
D = OD − d = 50.6 − 3.5 = 47.1 mm
C = D/d = 47.1/3.5 = 13.46 (high)
4C + 2 4 (13.46 ) + 2
KB =
=
= 1.098
4C − 3 4 (13.46 ) − 3
Na = Nt − 2 = 5.5 − 2 = 3.5 coils
A229 OQ&T steel is not given in Table 10-5. From Table A-5, for carbon steels,
G = 79.3(103) MPa.
3.54 ( 79.3)103
d 4G
Eq. (10-9):
k=
=
= 4.067 N/mm
8D3 N a
8 ( 47.13 ) 3.5
Table 10-1:
Eq. (10-7):
Ls = dNt = 3.5(5.5) = 19.25 mm
ys = L0 − Ls = 75.5 − 19.25 = 56.25 mm
Fs = kys = 4.067(56.25) = 228.8 N
8 ( 228.8 ) 47.1
8F D
τ s = K B s 3 = 1.098
= 702.8 MPa
πd
π ( 3.53 )
(1)
A = 1855 MPa⋅mmm, m = 0.187
A
1855
Eq. (10-14): Sut = m =
= 1468 MPa
d
3.50.187
Table 10-6: Ssy = 0.50 Sut = 0.50(1468) = 734 MPa
S sy
734
ns =
=
= 1.04 Spring is not solid-safe (ns < 1.2)
τ s 702.8
Return to Eq. (1) with Fs = kys and τs = Ssy /ns, and solve for ys, giving
( S sy / ns ) π d 3 = ( 734 / 1.2 ) π ( 3.53 ) = 48.96 mm
ys =
8 K B kD
8 (1.098 ) 4.067 ( 47.1)
The free length should be wound to
Table 10-4:
Shigley’s MED, 10th edition
Chapter 10 Solutions, Page 12/41
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L0 = Ls + ys = 19.25 + 48.96 = 68.2 mm
Ans.
______________________________________________________________________________
10-18 Given: B159 phosphor-bronze, squared and ground ends, d = 3.8 mm, OD = 31.4 mm,
L0 = 71.4 mm, Nt = 12.8 coils.
Eq. (10-1):
Eq. (10-5):
D = OD − d = 31.4 − 3.8 = 27.6 mm
C = D/d = 27.6/3.8 = 7.263
4C + 2 4 ( 7.263) + 2
KB =
=
= 1.192
4C − 3 4 ( 7.263) − 3
Table 10-1:
Na = Nt − 2 = 12.8 − 2 = 10.8 coils
Table 10-5:
G = 41.4(103) MPa.
3.84 ( 41.4 )103
d 4G
k=
=
= 4.752 N/mm
8 D 3 N a 8 ( 27.63 )10.8
Eq. (10-9):
Table 10-1:
Eq. (10-7):
Ls = dNt = 3.8(12.8) = 48.64 mm
ys = L0 − Ls = 71.4 − 48.64 = 22.76 mm
Fs = kys = 4.752(22.76) = 108.2 N
8 (108.2 ) 27.6
8F D
τ s = K B s 3 = 1.192
= 165.2 MPa
πd
π ( 3.83 )
(1)
Table 10-4 (d = 3.8/25.4 = 0.150 in):
A = 932 MPa⋅mmm, m = 0.064
A
932
Eq. (10-14): Sut = m =
= 855.7 MPa
d
3.80.064
Table 10-6: Ssy = 0.35 Sut = 0.35(855.7) = 299.5 MPa
S
299.5
ns = sy =
= 1.81 Spring is solid-safe (ns > 1.2) Ans.
τ s 165.2
______________________________________________________________________________
10-19 Given: A232 chrome-vanadium steel, squared and ground ends, d = 4.5 mm, OD = 69.2
mm, L0 = 215.6 mm, Nt = 8.2 coils.
Eq. (10-1):
Eq. (10-5):
D = OD − d = 69.2 − 4.5 = 64.7 mm
C = D/d = 64.7/4.5 = 14.38 (high)
4C + 2 4 (14.38 ) + 2
=
= 1.092
KB =
4C − 3 4 (14.38 ) − 3
Table 10-1:
Na = Nt − 2 = 8.2 − 2 = 6.2 coils
Table 10-5:
G = 77.2(103) MPa.
4.54 ( 77.2 )103
d 4G
k=
=
= 2.357 N/mm
8D3 N a
8 ( 64.73 ) 6.2
Eq. (10-9):
Table 10-1:
Ls = dNt = 4.5(8.2) = 36.9 mm
Shigley’s MED, 10th edition
Chapter 10 Solutions, Page 13/41
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Eq. (10-7):
Table 10-4:
Eq. (10-14):
Table 10-6:
ys = L0 − Ls = 215.6 − 36.9 = 178.7 mm
Fs = kys = 2.357(178.7) = 421.2 N
8 ( 421.2 ) 64.7
8F D
τ s = K B s 3 = 1.092
= 832 MPa
πd
π ( 4.53 )
(1)
A = 2005 MPa⋅mmm, m = 0.168
A
2005
Sut = m =
= 1557 MPa
d
4.50.168
Ssy = 0.50 Sut = 0.50(1557) = 779 MPa
τs > Ssy, that is, 832 > 779 MPa, the spring is not solid-safe. Return to Eq. (1) with
Fs = kys and τs = Ssy /ns, and solve for ys, giving
( S sy / ns ) π d 3 ( 779 / 1.2 ) π ( 4.53 )
ys =
=
= 139.5 mm
8 K B kD
8 (1.092 ) 2.357 ( 64.7 )
The free length should be wound to
L0 = Ls + ys = 36.9 + 139.5 = 176.4 mm
Ans.
This only addresses the solid-safe criteria. There are additional problems.
______________________________________________________________________________
10-20 Given: A227 HD steel.
From the figure: L0 = 4.75 in, OD = 2 in, and d = 0.135 in. Thus
D = OD − d = 2 − 0.135 = 1.865 in
(a) By counting, Nt = 12.5 coils. Since the ends are squared along 1/4 turn on each end,
N a = 12.5 − 0.5 = 12 turns Ans.
p = 4.75 / 12 = 0.396 in Ans.
The solid stack is 13 wire diameters
Ls = 13(0.135) = 1.755 in
Ans.
(b) From Table 10-5, G = 11.4 Mpsi
0.1354 (11.4) (106 )
d 4G
k =
=
= 6.08 lbf/in
8D 3 N a
8 (1.8653 ) (12)
(c) Fs = k(L0 - Ls ) = 6.08(4.75 − 1.755) = 18.2 lbf
Ans.
Ans.
(d) C = D/d = 1.865/0.135 = 13.81
Shigley’s MED, 10th edition
Chapter 10 Solutions, Page 14/41
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4(13.81) + 2
= 1.096
4(13.81) − 3
8F D
8(18.2)(1.865)
τ s = K B s 3 = 1.096
= 38.5 (103 ) psi = 38.5 kpsi
πd
π ( 0.1353 )
KB =
Ans.
______________________________________________________________________________
10-21 For the wire diameter analyzed, G = 11.75 Mpsi per Table 10-5. Use squared and ground
ends. The following is a spread-sheet study using Fig. 10-3 for parts (a) and (b). For Na,
k = Fmax /y = 20/2 = 10 lbf/in. For τs, F = Fs = 20(1 + ξ) = 20(1 + 0.15) = 23 lbf.
Source
Eq. (10-1)
Eq. (10-9)
Table 10-1
Table 10-1
1.15y + Ls
Eq. (10-13)
Table 10-4
Table 10-4
Eq. (10-14)
Table 10-6
Eq. (10-5)
Eq. (10-7)
Eq. (10-3)
Eq. (10-22)
(a) Spring over a Rod
(b) Spring in a Hole
Parameter Values
Source
Parameter Values
d
0.075
0.080
0.085
d
0.075
0.080
ID
0.800
0.800
0.800
OD
0.950
0.950
D
0.875
0.880
0.885
D
0.875
0.870
C
11.667
11.000
10.412 Eq. (10-1)
C
11.667
10.875
Na
6.937
8.828
11.061 Eq. (10-9)
Na
6.937
9.136
Nt
8.937
10.828
13.061 Table 10-1
Nt
8.937
11.136
Ls
0.670
0.866
1.110
Table 10-1
Ls
0.670
0.891
L0
2.970
3.166
3.410
1.15y + Ls
L0
2.970
3.191
(L0)cr
4.603
4.629
4.655
Eq. (10-13) (L0)cr
4.603
4.576
A
201.000 201.000 201.000 Table 10-4
A
201.000
201.000
m
0.145
0.145
0.145
Table 10-4
m
0.145
0.145
Sut
292.626 289.900 287.363 Eq. (10-14)
Sut
292.626
289.900
Ssy
131.681 130.455 129.313 Table 10-6
Ssy
131.681
130.455
KB
1.115
1.122
1.129
Eq. (10-5)
KB
1.115
1.123
135.335 112.948 95.293 Eq. (10-7)
135.335
111.787
τs
τs
ns
0.973
1.155
1.357
Eq. (10-3)
ns
0.973
1.167
fom
−0.282
−0.391
−0.536 Eq. (10-22) fom
−0.282
−0.398
For ns ≥ 1.2, the optimal size is d = 0.085 in for both cases.
0.085
0.950
0.865
10.176
11.846
13.846
1.177
3.477
4.550
201.000
0.145
287.363
129.313
1.133
93.434
1.384
−0.555
______________________________________________________________________________
10-22 In Prob. 10-21, there is an advantage of first selecting d as one can select from the
available sizes (Table A-28). Selecting C first requires a calculation of d where then a
size must be selected from Table A-28.
Consider part (a) of the problem. It is required that
ID = D − d = 0.800 in.
(1)
From Eq. (10-1), D = Cd. Substituting this into the first equation yields
d=
0.800
C −1
(2)
Starting with C = 10, from Eq. (2) we find that d = 0.089 in. From Table A-28, the closest
diameter is d = 0.090 in. Substituting this back into Eq. (1) gives D = 0.890 in, with C =
0.890/0.090 = 9.889, which are acceptable. From this point the solution is the same as
Prob. 10-21. For part (b), use
Shigley’s MED, 10th edition
Chapter 10 Solutions, Page 15/41
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OD = D + d = 0.950 in.
d=
and,
(3)
0.800
C −1
(a) Spring over a rod
Parameter Values
C
10.000
10.5
Eq. (2)
d
0.089
0.084
Table A-28
d
0.090
0.085
Eq. (1)
D
0.890
0.885
Eq. (10-1)
C
9.889
10.412
Eq. (10-9)
Na
13.669
11.061
Table 10-1
Nt
15.669
13.061
Table 10-1
Ls
1.410
1.110
1.15y + Ls
L0
3.710
3.410
Eq. (10-13) (L0)cr
4.681
4.655
Table 10-4
A
201.000 201.000
Table 10-4
m
0.145
0.145
Eq. (10-14)
Sut
284.991 287.363
Table 10-6
Ssy
128.246 129.313
Eq. (10-5)
KB
1.135
1.128
Eq. (10-7)
τs
81.167
95.223
ns
1.580
1.358
ns = Ssy/τs
Eq. (10-22) fom
-0.725
-0.536
Source
(4)
(b) Spring in a Hole
Source Parameter Values
C
10.000
Eq. (4)
d
0.086
Table A-28
d
0.085
Eq. (3)
D
0.865
Eq. (10-1)
C
10.176
Eq. (10-9)
Na
11.846
Table 10-1
Nt
13.846
Table 10-1
Ls
1.177
1.15y + Ls
L0
3.477
Eq. (10-13)
(L0)cr
4.550
Table 10-4
A
201.000
Table 10-4
m
0.145
Eq. (10-14)
Sut
287.363
Table 10-6
Ssy
129.313
Eq. (10-5)
KB
1.135
Eq. (10-7)
τs
93.643
ns
1.381
ns = Ssy/τs
Eq. (10-22)
fom
-0.555
Again, for ns ≥ 1.2, the optimal size is = 0.085 in.
Although this approach used less iterations than in Prob. 10-21, this was due to the initial
values picked and not the approach.
______________________________________________________________________________
10-23 One approach is to select A227 HD steel for its low cost. Try L0 = 48 mm, then for
y = 48 − 37.5 = 10.5 mm when F = 45 N. The spring rate is k = F/y = 45/10.5 = 4.286
N/mm.
For a clearance of 1.25 mm with screw, ID = 10 + 1.25 = 11.25 mm. Starting with
d = 2 mm,
D = ID + d = 11.25 + 2 = 13.25 mm
C = D/d = 13.25/2 = 6.625
(acceptable)
Table 10-5 (d = 2/25.4 = 0.0787 in): G = 79.3 GPa
d 4G
24 (79.3)103
Eq. (10-9):
Na =
=
= 15.9 coils
8kD 3 8(4.286)13.253
Assume squared and closed.
Shigley’s MED, 10th edition
Chapter 10 Solutions, Page 16/41
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Table 10-1:
Eq. (10-5):
Nt = Na + 2 = 15.9 + 2 = 17.9 coils
Ls = dNt = 2(17.9) =35.8 mm
ys = L0 − Ls = 48 − 35.8 = 12.2 mm
Fs = kys = 4.286(12.2) = 52.29 N
4C + 2 4 ( 6.625 ) + 2
KB =
=
= 1.213
4C − 3 4 ( 6.625 ) − 3
 8(52.29)13.25 
8Fs D
1.213
=

 = 267.5 MPa
πd3
π ( 23 )


Eq. (10-7):
τ s = KB
Table 10-4:
A = 1783 MPa · mmm, m = 0.190
A 1783
Sut = m = 0.190 = 1563 MPa
d
2
Ssy = 0.45Sut = 0.45(1563) = 703.3 MPa
Eq. (10-14):
Table 10-6:
ns =
S sy
τs
=
703.3
= 2.63 >1.2
267.5
O.K .
No other diameters in the given range work. So specify
A227-47 HD steel, d = 2 mm, D = 13.25 mm, ID = 11.25 mm, OD = 15.25 mm, squared
and closed, Nt = 17.9 coils, Na = 15.9 coils, k = 4.286 N/mm, Ls = 35.8 mm, and L0 = 48
mm.
Ans.
______________________________________________________________________________
10-24 Select A227 HD steel for its low cost. Try L0 = 48 mm, then for y = 48 − 37.5 = 10.5 mm
when F = 45 N. The spring rate is k = F/y = 45/10.5 = 4.286 N/mm.
For a clearance of 1.25 mm with screw, ID = 10 + 1.25 = 11.25 mm.
and,
D − d = 11.25
(1)
D =Cd
(2)
Starting with C = 8, gives D = 8d. Substitute into Eq. (1) resulting in d = 1.607 mm.
Selecting the nearest diameter in the given range, d = 1.6 mm. From this point, the
calculations are shown in the third column of the spreadsheet output shown. We see that
for d = 1.6 mm, the spring is not solid safe. Iterating on C we find that C = 6.5 provides
acceptable results with the specifications
A227-47 HD steel, d = 2 mm, D = 13.25 mm, ID = 11.25 mm, OD = 15.25 mm, squared
and closed, Nt = 17.9 coils, Na = 15.9 coils, k = 4.286 N/mm, Ls = 35.8 mm, and L0 = 48
mm.
Ans.
Shigley’s MED, 10th edition
Chapter 10 Solutions, Page 17/41
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Source
Eq. (2)
Table A-28
Eq. (1)
Eq. (10-1)
Eq. (10-9)
Table 10-1
Table 10-1
L0 − Ls
Fs = kys
Table 10-4
Table 10-4
Eq. (10-14)
Table 10-6
Eq. (10-5)
Eq. (10-7)
ns = Ssy/τs
C
d
d
D
C
Na
Nt
Ls
ys
Fs
A
m
Sut
Ssy
KB
τs
ns
Parameter Values
8.000
7
6.500
1.607
1.875
2.045
1.600
1.800
2.000
12.850
13.050
13.250
8.031
7.250
6.625
7.206
10.924
15.908
9.206
12.924
17.908
14.730
23.264
35.815
33.270
24.736
12.185
142.594
106.020
52.224
1783.000 1783.000 1783.000
0.190
0.190
0.190
1630.679 1594.592 1562.988
733.806
717.566
703.345
1.172
1.200
1.217
1335.568 724.943
268.145
0.549
0.990
2.623
The only difference between selecting C first rather than d as was done in Prob. 10-23, is
that once d is calculated, the closest wire size must be selected. Iterating on d uses
available wire sizes from the beginning.
______________________________________________________________________________
10-25 A stock spring catalog may have over two hundred pages of compression springs with up
to 80 springs per page listed.
• Students should be made aware that such catalogs exist.
• Many springs are selected from catalogs rather than designed.
• The wire size you want may not be listed.
• Catalogs may also be available on disk or the web through search routines. For
example, disks are available from Century Spring at
1 - (800) - 237 - 5225
www.centuryspring.com
• It is better to familiarize yourself with vendor resources rather than invent them
yourself.
• Sample catalog pages can be given to students for study.
______________________________________________________________________________
10-26 Given: ID = 0.6 in, C = 10, L0 = 5 in, Ls = 5 − 3 = 2 in, sq. & grd ends, unpeened, HD
A227 wire.
(a) With ID = D − d = 0.6 in and C = D/d = 10 ⇒10 d − d = 0.6 ⇒ d = 0.0667 in Ans.,
and D = 0.667 in.
(b) Table 10-1:
Ls = dNt = 2 in ⇒ Nt = 2/0.0667 = 30 coils Ans.
(c) Table 10-1:
Na = Nt − 2 = 30 − 2 = 28 coils
Shigley’s MED, 10th edition
Chapter 10 Solutions, Page 18/41
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Table 10-5:
G = 11.5 Mpsi
0.0667 4 (11.5 )106
d 4G
k=
=
= 3.424 lbf/in
8D3 N a
8 ( 0.6673 ) 28
Eq. (10-9):
(d) Table 10-4:
Eq. (10-14):
Ans.
A = 140 kpsi⋅inm, m = 0.190
A
140
Sut = m =
= 234.2 kpsi
d
0.0667 0.190
Table 10-6:
Ssy = 0.45 Sut = 0.45 (234.2) = 105.4 kpsi
Eq. (10-5):
Fs = kys = 3.424(3) = 10.27 lbf
4C + 2 4 (10 ) + 2
=
= 1.135
KB =
4C − 3 4 (10 ) − 3
Eq. (10-7):
τ s = KB
8 (10.27 ) 0.667
8 Fs D
= 1.135
3
πd
π ( 0.06673 )
= 66.72 (103 ) psi = 66.72 kpsi
S sy
105.4
= 1.58
Ans.
τ s 66.72
(e) τa = τm = 0.5τs = 0.5(66.72) = 33.36 kpsi, r = τa / τm = 1. Using the Gerber fatigue
failure criterion with Zimmerli data,
ns =
Eq. (10-30):
=
Ssu = 0.67 Sut = 0.67(234.2) = 156.9 kpsi
The Gerber ordinate intercept for the Zimmerli data is
S sa
35
S se =
=
= 39.9 kpsi
2
2
1 − ( S sm / S su ) 1 − ( 55 / 156.9 )
Table 6-7, p. 315,
2

 2 S se  
r 2 S su2 
−1 + 1 + 
S sa =

2 S se 
rS su  



2
12 (156.92 ) 
 2 ( 39.9 )  
=
 −1 + 1 + 
  = 37.61 kpsi
2 ( 39.9 ) 
1(156.9 )  



nf =
S sa
=
37.61
= 1.13
Ans.
τ a 33.36
______________________________________________________________________________
10-27 Given: OD ≤ 0.9 in, C = 8, L0 = 3 in, Ls = 1 in, ys = 3 − 1 = 2 in, sq. ends, unpeened,
music wire.
(a) Try OD = D + d = 0.9 in, C = D/d = 8 ⇒ D = 8d ⇒ 9d = 0.9 ⇒ d = 0.1 Ans.
Shigley’s MED, 10th edition
Chapter 10 Solutions, Page 19/41
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D = 8(0.1) = 0.8 in
(b) Table 10-1:
Ls = d (Nt + 1) ⇒ Nt = Ls / d − 1 = 1/0.1− 1 = 9 coils Ans.
Table 10-1:
(c) Table 10-5:
Eq. (10-9):
(d)
Na = Nt − 2 = 9 − 2 = 7 coils
G = 11.75 Mpsi
k=
0.14 (11.75 )106
d 4G
=
= 40.98 lbf/in
8D 3 N a
8 ( 0.83 ) 7
Ans.
Fs = kys = 40.98(2) = 81.96 lbf
4C + 2 4 ( 8 ) + 2
=
= 1.172
4C − 3 4 ( 8 ) − 3
Eq. (10-5):
KB =
Eq. (10-7):
τ s = KB
Table 10-4:
A = 201 kpsi⋅inm, m = 0.145
Eq. (10-14):
Sut =
Table 10-6:
Ssy = 0.45 Sut = 0.45(280.7) = 126.3 kpsi
ns =
8 ( 81.96 ) 0.8
8Fs D
= 1.172
= 195.7 (103 ) psi = 195.7 kpsi
3
3
πd
π ( 0.1 )
A
201
=
= 280.7 kpsi
m
d
0.10.145
S sy
τs
=
126.3
= 0.645
195.7
Ans.
(e) τa = τm = τs /2 = 195.7/2 = 97.85 kpsi. Using the Gerber fatigue failure criterion with
Zimmerli data,
Eq. (10-30):
Ssu = 0.67 Sut = 0.67(280.7) = 188.1 kpsi
The Gerber ordinate intercept for the Zimmerli data is
S sa
35
S se =
=
= 38.3 kpsi
2
2
1 − ( S sm / S su ) 1 − ( 55 /188.1)
Table 6-7, p. 315,
2

 2 S se  
r 2 S su2 
−1 + 1 + 
S sa =

2 S se 
rS su  



2
12 (188.12 ) 
 2 ( 38.3)  
=
 −1 + 1 + 
  = 36.83 kpsi
2 ( 38.3) 
1(188.1)  



Shigley’s MED, 10th edition
Chapter 10 Solutions, Page 20/41
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nf =
S sa
τa
=
36.83
= 0.376
97.85
Ans.
Obviously, the spring is severely under designed and will fail statically and in fatigue.
Increasing C would improve matters. Try C = 12. This yields ns = 1.83 and nf = 1.00.
______________________________________________________________________________
10-28 Given: Fmax = 300 lbf, Fmin = 150 lbf, ∆y = 1 in, OD = 2.1 − 0.2 = 1.9 in, C = 7,
unpeened, squared & ground, oil-tempered wire.
(a)
D = OD − d = 1.9 − d
(1)
C = D/d = 7 ⇒ D = 7d
(2)
Substitute Eq. (2) into (1)
7d = 1.9 − d ⇒ d = 1.9/8 = 0.2375 in
(b) From Eq. (2):
D = 7d = 7(0.2375) = 1.663 in
(c)
k=
(d) Table 10-5:
G = 11.6 Mpsi
Ans.
∆F 300 − 150
=
= 150 lbf/in
1
∆y
Ans.
4
6
d 4G 0.2375 (11.6 )10
=
= 6.69 coils
8D 3k
8 (1.6633 )150
Eq. (10-9):
Na =
Table 10-1:
Nt = Na + 2 = 8.69 coils
Ans.
Table 10-6:
A = 147 kpsi⋅inm, m = 0.187
A
147
= 192.3 kpsi
Sut = m =
d
0.23750.187
Ssy = 0.5 Sut = 0.5(192.3) = 96.15 kpsi
Eq. (10-5):
KB =
Eq. (10-7):
τ s = KB
(e) Table 10-4:
Eq. (10-14):
Ans.
4C + 2 4 ( 7 ) + 2
=
= 1.2
4C − 3 4 ( 7 ) − 3
8 Fs D
= S sy
π d3
Shigley’s MED, 10th edition
Chapter 10 Solutions, Page 21/41
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Fs =
π d 3 S sy
8K B D
=
π ( 0.23753 ) 96.15 (103 )
8 (1.2 )1.663
= 253.5 lbf
ys = Fs / k = 253.5/150 = 1.69 in
Table 10-1:
Ls = Nt d = 8.46(0.2375) = 2.01 in
L0 = Ls + ys = 2.01 + 1.69 = 3.70 in
Ans.
______________________________________________________________________________
10-29 For a coil radius given by:
R = R1 +
R2 − R1
θ
2π N
The torsion of a section is T = PR where dL = R dθ
∂U
1
∂T
1 2π N 3
=
T
dL =
PR dθ
∫
∂P
∂P
GJ
GJ ∫0
3
P 2π N 
R2 − R1 
=
θ  dθ
 R1 +
GJ ∫0 
2π N

δP =
2π N
4
P  1   2π N  
R2 − R1  
=
+
θ
R

  1
 
 
2π N
GJ  4   R2 − R1  
  0
π PN
π PN
=
( R1 + R2 ) ( R12 + R22 )
R24 − R14 ) =
(
2GJ ( R2 − R1)
2GJ
π 4
16 PN
∴ δp =
J =
d
( R1 + R2 ) ( R12 + R22 )
Gd 4
32
P
d 4G
=
k =
Ans.
δ P 16 N ( R1 + R2 ) ( R12 + R22 )
______________________________________________________________________________
10-30 Given: Fmin = 4 lbf, Fmax = 18 lbf, k = 9.5 lbf/in, OD ≤ 2.5 in, nf = 1.5.
For a food service machinery application select A313 Stainless wire.
Table 10-5:
G = 10(106) psi
Note that for
0.013 ≤ d ≤ 0.10 in A = 169,
m = 0.146
0.10 < d ≤ 0.20 in
A = 128,
m = 0.263
18 − 4
18 + 4
Fa =
= 7 lbf , Fm =
= 11 lbf , r = 7 / 11
2
2
169
Try, d = 0.080 in, Sut =
= 244.4 kpsi
(0.08)0.146
Shigley’s MED, 10th edition
Chapter 10 Solutions, Page 22/41
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Ssu = 0.67Sut = 163.7 kpsi,
Ssy = 0.35Sut = 85.5 kpsi
Try unpeened using Zimmerli’s endurance data: Ssa = 35 kpsi, Ssm = 55 kpsi
S sa
35
Gerber:
S se =
=
= 39.5 kpsi
2
1 − (S sm / S su )
1 − (55 / 163.7) 2
2


 
(7 / 11) 2 (163.7)2 
2(39.5)
S sa =
 −1 + 1 + 
  = 35.0 kpsi
2(39.5)
 (7 / 11)(163.7)  


α = S sa / n f = 35.0 / 1.5 = 23.3 kpsi
β =
 8(7)  −3
8Fa
(10−3 ) = 
(10 ) = 2.785 kpsi
2
2 
πd
 π (0.08 ) 
2
C =
D=
KB =
τa =
nf =
 2(23.3) − 2.785 
2(23.3) − 2.785
3(23.3)
−
= 6.97
+ 

4(2.785)
4(2.785)
4(2.785)


Cd = 6.97(0.08) = 0.558 in
4C + 2 4(6.97) + 2
=
= 1.201
4C − 3
4(6.97) − 3
 8(7)(0.558) −3 
 8F D 
K B  a 3  = 1.201 
(10 )  = 23.3 kpsi
3
 πd 
 π (0.08 )

35 / 23.3 = 1.50 checks
Gd 4
10(106 )(0.08) 4
= 31.02 coils
=
8kD 3
8(9.5)(0.558)3
= 31.02 + 2 = 33 coils, Ls = dN t = 0.08(33) = 2.64 in
= Fmax / k = 18 / 9.5 = 1.895 in
= (1 + ξ ) ymax = (1 + 0.15)(1.895) = 2.179 in
= 2.64 + 2.179 = 4.819 in
D
2.63(0.558)
= 2.63 =
= 2.935 in
0.5
α
= 1.15(18 / 7)τ a = 1.15(18 / 7)(23.3) = 68.9 kpsi
= S sy / τ s = 85.5 / 68.9 = 1.24
Na =
Nt
ymax
ys
L0
( L0 )cr
τs
ns
f =
kg
=
π d DN aγ
2
2
9.5(386)
= 109 Hz
π (0.08 )(0.558)(31.02)(0.283)
2
2
These steps are easily implemented on a spreadsheet, as shown below, for different
diameters.
Shigley’s MED, 10th edition
Chapter 10 Solutions, Page 23/41
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d1
d
0.080
m
0.146
A
169.000
Sut
244.363
Ssu
163.723
Ssy
85.527
Sse
39.452
Ssa
35.000
23.333
α
2.785
β
C
6.977
D
0.558
KB
1.201
23.333
τa
nf
1.500
Na
30.993
Nt
32.993
LS
2.639
ys
2.179
L0
4.818
(L0)cr
2.936
τs
69.000
ns
1.240
f,(Hz) 108.895
d2
0.0915
0.146
169.000
239.618
160.544
83.866
39.654
35.000
23.333
2.129
9.603
0.879
1.141
23.333
1.500
13.594
15.594
1.427
2.179
3.606
4.622
69.000
1.215
114.578
d3
0.1055
0.263
128
231.257
154.942
80.940
40.046
35.000
23.333
1.602
13.244
1.397
1.100
23.333
1.500
5.975
7.975
0.841
2.179
3.020
7.350
69.000
1.173
118.863
d4
0.1205
0.263
128
223.311
149.618
78.159
40.469
35.000
23.333
1.228
17.702
2.133
1.074
23.333
1.500
2.858
4.858
0.585
2.179
2.764
11.220
69.000
1.133
121.775
The shaded areas depict conditions outside the recommended design conditions. Thus,
one spring is satisfactory. The specifications are: A313 stainless wire, unpeened, squared
and ground, d = 0.0915 in, OD = 0.879 + 0.092 = 0.971 in, L0 = 3.606 in, and Nt = 15.59
turns
Ans.
______________________________________________________________________________
10-31 The steps are the same as in Prob. 10-30 except that the Gerber-Zimmerli criterion is
replaced with Goodman-Zimmerli:
S se =
S sa
1 − ( S sm S su )
The problem then proceeds as in Prob. 10-30. The results for the wire sizes are shown
Shigley’s MED, 10th edition
Chapter 10 Solutions, Page 24/41
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below (see solution to Prob. 10-30 for additional details).
d
m
A
Sut
Ssu
Ssy
Sse
Ssa
α
β
C
D
Iteration of d for the first trial
d1
d2
d3
d4
d1
d2
d3
d4
0.080
0.0915 0.1055 0.1205 d
0.080 0.0915 0.1055 0.1205
0.146
0.146
0.263
0.263
KB
1.151
1.108
1.078
1.058
29.008 29.040 29.090 29.127
169.000 169.000 128.000 128.000 τa
244.363 239.618 231.257 223.311 nf
1.500
1.500
1.500
1.500
163.723 160.544 154.942 149.618 Na
14.191
6.456
2.899
1.404
85.527 83.866 80.940 78.159 Nt
16.191
8.456
4.899
3.404
52.706 53.239 54.261 55.345 Ls
1.295
0.774
0.517
0.410
43.513 43.560 43.634 43.691 ymax
2.875
2.875
2.875
2.875
29.008 29.040 29.090 29.127 L0
4.170
3.649
3.392
3.285
2.785
2.129
1.602
1.228
(L0)cr
3.809
5.924
9.354 14.219
85.782 85.876 86.022 86.133
9.052
12.309 16.856 22.433 τs
0.724
1.126
1.778
2.703
ns
0.997
0.977
0.941
0.907
f (Hz) 140.040 145.559 149.938 152.966
Without checking all of the design conditions, it is obvious that none of the wire sizes
satisfy ns ≥ 1.2. Also, the Gerber line is closer to the yield line than the Goodman. Setting
nf = 1.5 for Goodman makes it impossible to reach the yield line (ns < 1) . The table
below uses nf = 2.
d
m
A
Sut
Ssu
Ssy
Sse
Ssa
α
β
C
D
Iteration of d for the second trial
d1
d2
d3
d4
d1
0.080 0.0915 0.1055 0.1205 d
0.080
0.146
0.146
0.263
0.263 KB
1.221
169.000 169.000 128.000 128.000 τa
21.756
244.363 239.618 231.257 223.311 nf
2.000
163.723 160.544 154.942 149.618 Na
40.243
85.527 83.866 80.940 78.159 Nt
42.243
52.706 53.239 54.261 55.345 Ls
3.379
43.513 43.560 43.634 43.691 ymax
2.875
6.254
21.756 21.780 21.817 21.845 L0
2.785
2.129
1.602
1.228 (L0)cr
2.691
64.336
6.395
8.864 12.292 16.485 τs
0.512
0.811
1.297
1.986 ns
1.329
f (Hz) 98.936
d2
d3
d4
0.0915 0.1055 0.1205
1.154
1.108
1.079
21.780 21.817 21.845
2.000
2.000
2.000
17.286
7.475
3.539
19.286
9.475
5.539
1.765
1.000
0.667
2.875
2.875
2.875
4.640
3.875
3.542
4.266
6.821 10.449
64.407 64.517 64.600
1.302
1.255
1.210
104.827 109.340 112.409
The satisfactory spring has design specifications of: A313 stainless wire, unpeened,
squared and ground, d = 0.0915 in, OD = 0.811 + 0.092 = 0.903 in, L0 = 4.640 in, and
.Nt = 19.3 turns. Ans.
______________________________________________________________________________
10-32 This is the same as Prob. 10-30 since Ssa = 35 kpsi. Therefore, the specifications are:
A313 stainless wire, unpeened, squared and ground, d = 0.0915 in, OD = 0.879 + 0.092 =
Shigley’s MED, 10th edition
Chapter 10 Solutions, Page 25/41
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0.971 in, L0 = 3.606 in, and Nt = 15.59 turns
Ans.
______________________________________________________________________________
10-33 For the Gerber-Zimmerli fatigue-failure criterion, Ssu = 0.67Sut ,
S sa
S se =
,
1 − (S sm / S su ) 2
S sa

r 2S su2 
=
−1 + 1 +
2S se 

2
 2S se  


 rS su  
The equation for Ssa is the basic difference. The last 2 columns of diameters of Ex. 10-5
are presented below with additional calculations.
0.105
0.112
d
0.105
0.112
278.691 276.096 Na
8.915
6.190
186.723 184.984 Ls
1.146
0.917
38.325 38.394 L0
3.446
3.217
125.411 124.243 (L0)cr
6.630
8.160
34.658 34.652 KB
1.111
1.095
23.105 23.101 τa
23.105 23.101
α
1.732
1.523 nf
1.500
1.500
β
C
12.004 13.851 τs
70.855 70.844
D
1.260
1.551 ns
1.770
1.754
ID
1.155
1.439 fn
105.433 106.922
OD
1.365
1.663 fom
−0.973 −1.022
d
Sut
Ssu
Sse
Ssy
Ssa
There are only slight changes in the results.
______________________________________________________________________________
10-34 As in Prob. 10-34, the basic change is Ssa.
S sa
For Goodman,
S se =
1 − ( S sm / S su )
Recalculate Ssa with
rS se S su
S sa =
rS su + S se
Calculations for the last 2 diameters of Ex. 10-5 are given below.
Shigley’s MED, 10th edition
Chapter 10 Solutions, Page 26/41
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0.105
278.691
186.723
49.614
125.411
34.386
22.924
α
1.732
β
C
11.899
D
1.249
ID
1.144
OD
1.354
d
Sut
Ssu
Sse
Ssy
Ssa
0.112
276.096
184.984
49.810
124.243
34.380
22.920
1.523
13.732
1.538
1.426
1.650
d
0.105
0.112
Na
9.153
6.353
Ls
1.171
0.936
L0
3.471
3.236
(L0)cr
6.572
8.090
KB
1.112
1.096
22.924 22.920
τa
nf
1.500
1.500
70.301 70.289
τs
ns
1.784
1.768
fn
104.509 106.000
fom
−0.986 −1.034
There are only slight differences in the results.
______________________________________________________________________________
10-35 Use: E = 28.6 Mpsi, G = 11.5 Mpsi, A = 140 kpsi · inm , m = 0.190, rel cost = 1.
140
Try
d = 0.067 in, Sut =
= 234.0 kpsi
(0.067)0.190
Table 10-6: Ssy = 0.45Sut = 105.3 kpsi
Table 10-7: Sy = 0.75Sut = 175.5 kpsi
Eq. (10-34) with D/d = C and C1 = C
S
F
σ A = max2 [( K ) A (16C ) + 4] = y
ny
πd
π d 2S y
4C 2 − C − 1
(16C ) + 4 =
n y Fmax
4C (C − 1)
 π d 2S y

4C 2 − C − 1 = (C − 1) 
− 1
 4n F

 y max

2


π d 2S y
1
1  π d Sy
− 1 C + 
− 2 = 0
C 2 − 1 +




4
4n y Fmax
4  4n y Fmax


2


2
 π d 2S y 
π d 2S y
1  π d Sy

± 
−
+ 2 take positive root
C =
 16n F 

2 16n y Fmax
4n y Fmax
y max 



1  π (0.067 2 )(175.5)(103 )
= 
2
16(1.5)(18)
+
2

 π (0.067) 2 (175.5)(103 ) 
π (0.067) 2 (175.5)(103 )

−
+ 2  = 4.590


16(1.5)(18)
4(1.5)(18)



Shigley’s MED, 10th edition
Chapter 10 Solutions, Page 27/41
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D = Cd = 4.59 ( 0.067 ) = 0.3075 in
Fi =
π d 3τ i
8D
=
πd3 
C − 3 
33 500

± 1000  4 −


8D  exp(0.105C )
6.5  

Use the lowest Fi in the preferred range. This results in the best fom.
Fi =
π (0.067)3 
33 500
4.590 − 3  

− 1000  4 −

  = 6.505 lbf
8(0.3075)  exp[0.105(4.590)]
6.5  

For simplicity, we will round up to the next integer or half integer. Therefore, use Fi = 7
lbf
18 − 7
k =
= 22 lbf/in
0.5
(0.067) 4 (11.5)(106 )
d 4G
Na =
=
= 45.28 turns
8kD3
8(22)(0.3075)3
G
11.5
Nb = N a −
= 45.28 −
= 44.88 turns
28.6
E
L0 = (2C − 1 + N b )d = [2(4.590) − 1 + 44.88](0.067) = 3.555 in
L18 lbf = 3.555 + 0.5 = 4.055 in
Body: K B =
τ max
(n y ) body
r2
(K )B
τB
(n y ) B
fom
4C + 2 4(4.590) + 2
=
= 1.326
4C − 3
4(4.590) − 3
8K B Fmax D 8(1.326)(18)(0.3075) −3
=
=
(10 ) = 62.1 kpsi
π d3
π (0.067)3
S
105.3
= sy =
= 1.70
τ max
62.1
2r
2(0.134)
= 2d = 2(0.067) = 0.134 in, C2 = 2 =
=4
0.067
d
4C2 − 1
4(4) − 1
=
=
= 1.25
4C2 − 4 4(4) − 4
 8(18)(0.3075)  −3
 8F D 
= ( K ) B  max3  = 1.25 
 (10 ) = 58.58 kpsi
3
 πd 
 π (0.067) 
S
105.3
= sy =
= 1.80
τB
58.58
π 2d 2 ( Nb + 2) D
π 2 (0.067) 2 (44.88 + 2)(0.3075)
= −(1)
=−
= −0.160
4
4
Several diameters, evaluated using a spreadsheet, are shown below.
Shigley’s MED, 10th edition
Chapter 10 Solutions, Page 28/41
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0.072
0.076
0.081
0.085
0.09
0.095
0.104
d
0.067
Sut
233.977 230.799 228.441 225.692 223.634 221.219 218.958 215.224
Ssy
105.290 103.860 102.798 101.561 100.635 99.548 98.531 96.851
Sy
175.483 173.100 171.331 169.269 167.726 165.914 164.218 161.418
C
4.589
5.412
6.099
6.993
7.738
8.708
9.721 11.650
D
0.307
0.390
0.463
0.566
0.658
0.784
0.923
1.212
Fi (calc)
6.505
5.773
5.257
4.675
4.251
3.764
3.320
2.621
Fi (rd)
7.0
6.0
5.5
5.0
4.5
4.0
3.5
3.0
k
22.000 24.000 25.000 26.000 27.000 28.000 29.000 30.000
Na
45.29
27.20
19.27
13.10
9.77
7.00
5.13
3.15
Nb
44.89
26.80
18.86
12.69
9.36
6.59
4.72
2.75
L0
3.556
2.637
2.285
2.080
2.026
2.071
2.201
2.605
L18 lbf
4.056
3.137
2.785
2.580
2.526
2.571
2.701
3.105
KB
1.326
1.268
1.234
1.200
1.179
1.157
1.139
1.115
62.118 60.686 59.707 58.636 57.875 57.019 56.249 55.031
τmax
(ny)body
1.695
1.711
1.722
1.732
1.739
1.746
1.752
1.760
58.576 59.820 60.495 61.067 61.367 61.598 61.712 61.712
τB
(ny)B
1.797
1.736
1.699
1.663
1.640
1.616
1.597
1.569
(ny)A
1.500
1.500
1.500
1.500
1.500
1.500
1.500
1.500
fom
-0.138 −0.135 −0.133 −0.135 −0.138 −0.154
−0.160 −0.144
Except for the 0.067 in wire, all springs satisfy the requirements of length and number of
coils. The 0.085 in wire has the highest fom.
______________________________________________________________________________
10-36 Given: Nb = 84 coils, Fi = 16 lbf, OQ&T steel, OD = 1.5 in, d = 0.162 in.
D = OD − d = 1.5 − 0.162 = 1.338 in
(a) Eq. (10-39):
L0 = 2(D − d) + (Nb + 1)d
= 2(1.338 − 0.162) + (84 + 1)(0.162) = 16.12 in
2d + L0 = 2(0.162) + 16.12 = 16.45 in overall
D 1.338
(b)
C =
=
= 8.26
d
0.162
4C + 2 4(8.26) + 2
KB =
=
= 1.166
4C − 3
4(8.26) − 3
8(16)(1.338)
 8F D 
τ i = K B  i 3  = 1.166
= 14 950 psi
π (0.162)3
 πd 
(c) From Table 10-5 use: G = 11.4(106) psi and E = 28.5(106) psi
Ans.
or
G
11.4
= 84 +
= 84.4 turns
E
28.5
d 4G
(0.162)4 (11.4)(106 )
=
= 4.855 lbf/in
k =
8D3 N a
8(1.338)3 (84.4)
Ans.
Na = Nb +
Shigley’s MED, 10th edition
Ans.
Chapter 10 Solutions, Page 29/41
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A = 147 psi · inm ,
m = 0.187
147
= 207.1 kpsi
Sut =
(0.162)0.187
S y = 0.75(207.1) = 155.3 kpsi
S sy = 0.50(207.1) = 103.5 kpsi
(d) Table 10-4:
Body
π d 3S sy
F =
π KBD
π (0.162)3 (103.5)(103 )
=
8(1.166)(1.338)
= 110.8 lbf
Torsional stress on hook point B
2r2
2(0.25 + 0.162 / 2)
=
= 4.086
0.162
d
4C2 − 1
4(4.086) − 1
(K )B =
=
= 1.243
4C2 − 4 4(4.086) − 4
π (0.162)3 (103.5)(103 )
F =
= 103.9 lbf
8(1.243)(1.338)
C2 =
Normal stress on hook point A
2r1 1.338
=
= 8.26
d
0.162
4C12 − C1 − 1 4(8.26) 2 − 8.26 − 1
=
= 1.099
(K ) A =
4C1(C1 − 1)
4(8.26)(8.26 − 1)
4 
16( K ) A D
S yt = σ = F 
+
3
π d 2 
 πd
155.3(103 )
F =
= 85.8 lbf
[16(1.099)(1.338)] / π (0.162)3  + 4 / π (0.162)2 
C1 =
= min(110.8, 103.9, 85.8) = 85.8 lbf
{
}
Ans.
(e) Eq. (10-48):
F − Fi
85.8 − 16
=
= 14.4 in Ans.
k
4.855
______________________________________________________________________________
y =
10-37
Fmin = 9 lbf, Fmax = 18 lbf
18 − 9
18 + 9
Fa =
= 4.5 lbf , Fm =
= 13.5 lbf
2
2
Shigley’s MED, 10th edition
Chapter 10 Solutions, Page 30/41
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0.013 ≤ d ≤ 0.1
A = 169 kpsi · inm ,
0.1 ≤ d ≤ 0.2
A = 128 kpsi · inm ,
E = 28 Mpsi, G = 10 Gpsi
Try d = 0.081 in and refer to the discussion following Ex. 10-7
169
= 243.9 kpsi
Sut =
(0.081)0.146
S su = 0.67Sut = 163.4 kpsi
S sy = 0.35Sut = 85.4 kpsi
S y = 0.55Sut = 134.2 kpsi
A313 stainless:
m = 0.146
m = 0.263
Table 10-8:
Sr = 0.45Sut = 109.8 kpsi
Sr / 2
109.8 / 2
=
= 57.8 kpsi
Se =
2
1 − [Sr / (2Sut )]
1 − [(109.8 / 2) / 243.9]2
r = Fa / Fm = 4.5 / 13.5 = 0.333

r 2 Sut2 
Table 6-7:
−1 +
Sa =
2 Se 


(0.333) 2 (243.92 ) 
−1 + 1 +
Sa =

2(57.8)

2
 2S e  
1+

 rSut  
2
 2(57.8)  
 0.333(243.9)   = 42.2 kpsi



Hook bending
16C
4 
Sa
S

+
=
= a
(σ a ) A = Fa ( K ) A
2
2
2
πd
π d  (n f ) A

 S
4.5  (4C 2 − C − 1)16C
+ 4 = a
2 
4C (C − 1)
2
πd 

This equation reduces to a quadratic in C (see Prob. 10-35). The useable root for C is
2
 2
 π d 2Sa 
π d Sa
C = 0.5 
+ 
 −
 144
 144 


 π (0.081)2 (42.2)(103 )
= 0.5 
+
144

= 4.91
π d 2Sa
36

+ 2


2

 π (0.081) 2 (42.2)(103 ) 
π (0.081)2 (42.2)(103 )

−
+
2



144
36



Shigley’s MED, 10th edition
Chapter 10 Solutions, Page 31/41
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D = Cd = 0.398 in
C − 3 
π d 3τ i π d 3  33 500

Fi =
=
± 1000  4 −


8D
8D  exp(0.105C )
6.5  

Use the lowest Fi in the preferred range.
π (0.081)3 
33 500
4.91 − 3  

Fi =
− 1000  4 −


8(0.398)  exp[0.105(4.91)]
6.5  

= 8.55 lbf
For simplicity we will round up to next 1/4 integer.
Fi = 8.75 lbf
18 − 9
k =
= 36 lbf/in
0.25
(0.081) 4 (10)(106 )
d 4G
Na =
=
= 23.7 turns
8kD3
8(36)(0.398)3
10
G
Nb = N a −
= 23.7 −
= 23.3 turns
28
E
L0 = (2C − 1 + N b )d = [2(4.91) − 1 + 23.3](0.081) = 2.602 in
Lmax = L0 + ( Fmax − Fi ) / k = 2.602 + (18 − 8.75) / 36 = 2.859 in

4.5(4)  4C 2 − C − 1
+ 1
(σ a ) A =
2 
πd  C − 1


18(10-3 )  4(4.912 ) − 4.91 − 1
+ 1 = 21.1 kpsi
2 
π (0.081 ) 
4.91 − 1

Sa
42.2
=
= 2 checks
(n f ) A =
(σ a ) A
21.1
=
Body:
4C + 2 4(4.91) + 2
=
= 1.300
4C − 3
4(4.91) − 3
8(1.300)(4.5)(0.398) −3
τa =
(10 ) = 11.16 kpsi
π (0.081)3
13.5
F
(11.16) = 33.47 kpsi
τm = m τa =
4.5
Fa
KB =
The repeating allowable stress from Table 10-8 is
Ssr = 0.30Sut = 0.30(243.9) = 73.17 kpsi
The Gerber intercept is given by Eq. (10-42) as
S se =
73.17 / 2
= 38.5 kpsi
1 − [(73.17 / 2) / 163.4]2
Shigley’s MED, 10th edition
Chapter 10 Solutions, Page 32/41
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From Table 6-7,
2
2

 2(33.47)(38.5)  
1  163.4   11.16  
−
+
+
1
1

 

 163.4(11.16)   = 2.53
2  33.47   38.5  

 


Let r2 = 2d = 2(0.081) = 0.162
2r
4(4) − 1
C2 = 2 = 4, ( K ) B =
= 1.25
d
4(4) − 4
(K )B
1.25
(τ a ) B =
τa =
(11.16) = 10.73 kpsi
KB
1.30
(K )B
1.25
(τ m ) B =
(33.47) = 32.18 kpsi
τm =
KB
1.30
(n f )body =
Table 10-8: (Ssr )B = 0.28Sut = 0.28(243.9) = 68.3 kpsi
68.3 / 2
= 35.7 kpsi
(S se ) B =
1 − [(68.3 / 2) / 163.4]2
2
2

 2(32.18)(35.7)  
1  163.4   10.73  
(n f ) B = 
 
  −1 + 1 + 
  = 2.51
2  32.18   35.7  
 163.4(10.73)  


Yield
Bending:

4 Fmax  (4C 2 − C − 1)
+ 1
(σ A ) max =
2 
C −1
πd 


4(18)  4(4.91) 2 − 4.91 − 1
+ 1 (10-3 ) = 84.4 kpsi
2 
4.91 − 1
π (0.081 ) 

134.2
(n y ) A =
= 1.59
84.4
=
Body:
τ i = ( Fi / Fa )τ a = (8.75 / 4.5)(11.16) = 21.7 kpsi
r = τ a /(τ m − τ i ) = 11.16 / (33.47 − 21.7) = 0.948
0.948
r
(S sy − τ i ) =
(85.4 − 21.7) = 31.0 kpsi
0.948 + 1
r +1
(S )
31.0
= sa y =
= 2.78
11.16
τa
(S sa ) y =
(n y ) body
Hook shear:
Ssy = 0.3Sut = 0.3(243.9) = 73.2 kpsi
τ max = (τ a ) B + (τ m ) B = 10.73 + 32.18 = 42.9 kpsi
73.2
= 1.71
(n y ) B =
42.9
7.6π 2d 2 ( N b + 2) D
7.6π 2 (0.081)2 (23.3 + 2)(0.398)
=−
= −1.239
fom = −
4
4
Shigley’s MED, 10th edition
Chapter 10 Solutions, Page 33/41
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A tabulation of several wire sizes follow
d
Sut
Ssu
Sr
Se
Sa
C
D
OD
Fi (calc)
Fi (rd)
k
Na
Nb
L0
L18 lbf
(σa)A
(nf)A
KB
(τa)body
(τm)body
Ssr
Sse
(nf)body
(K)B
(τa)B
(τm)B
(Ssr)B
(Sse)B
(nf)B
Sy
(σA)max
(ny)A
τi
r
(Ssy)body
(Ssa)y
(ny)body
(Ssy)B
(τB)max
(ny)B
fom
0.081
243.920
163.427
109.764
57.809
42.136
4.903
0.397
0.478
8.572
8.75
36.000
23.86
23.50
2.617
2.874
21.068
2.000
1.301
11.141
33.424
73.176
38.519
2.531
1.250
10.705
32.114
68.298
35.708
2.519
134.156
84.273
1.592
21.663
0.945
85.372
30.958
2.779
73.176
42.819
1.709
−1.246
0.085
242.210
162.281
108.994
57.403
41.841
5.484
0.466
0.551
7.874
9.75
36.000
17.90
17.54
2.338
2.567
20.920
2.000
1.264
10.994
32.982
72.663
38.249
2.547
1.250
10.872
32.615
67.819
35.458
2.463
133.215
83.682
1.592
23.820
1.157
84.773
32.688
2.973
72.663
43.486
1.671
−1.234
0.092
239.427
160.416
107.742
56.744
41.360
6.547
0.602
0.694
6.798
10.75
36.000
11.38
11.02
2.127
2.328
20.680
2.000
1.216
10.775
32.326
71.828
37.809
2.569
1.250
11.080
33.240
67.040
35.050
2.388
131.685
82.720
1.592
25.741
1.444
83.800
34.302
3.183
71.828
44.321
1.621
−1.245
0.098
237.229
158.943
106.753
56.223
40.980
7.510
0.736
0.834
5.987
11.75
36.000
8.03
7.68
2.126
2.300
20.490
2.000
1.185
10.617
31.852
71.169
37.462
2.583
1.250
11.200
33.601
66.424
34.728
2.341
130.476
81.961
1.592
27.723
1.942
83.030
36.507
3.438
71.169
44.801
1.589
−1.283
0.105
234.851
157.350
105.683
55.659
40.570
8.693
0.913
1.018
5.141
12.75
36.000
5.55
5.19
2.266
2.412
20.285
2.000
1.157
10.457
31.372
70.455
37.087
2.596
1.250
11.294
33.883
65.758
34.380
2.298
129.168
81.139
1.592
29.629
2.906
82.198
39.109
3.740
70.455
45.177
1.560
−1.357
0.12
230.317
154.312
103.643
54.585
39.786
11.451
1.374
1.494
3.637
13.75
36.000
2.77
2.42
2.918
3.036
19.893
2.000
1.117
10.177
30.532
69.095
36.371
2.616
1.250
11.391
34.173
64.489
33.717
2.235
126.674
79.573
1.592
31.097
4.703
80.611
40.832
4.012
69.095
45.564
1.516
−1.639
optimal fom
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Chapter 10 Solutions, Page 34/41
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The shaded areas show
ow the conditions not satisfied.
_________________________
______________________________________
____________________
10-38 For the hook,
M = FR sinθ, ∂M/∂F
∂F = R sinθ
δF =
1
EI
∫
π /2
0
F ( R sin
n θ ) R dθ =
2
π FR 3
2 EI
The total deflection of the body and the two hooks
 π FR 3  8FD 3 N b
8FD
D3Nb
π F ( D / 2)3
+
2
=
+


d 4G
d 4G
E (π / 64)((d 4 )
 2 EI 
8FD
D3 
G  8FD 3 N a
= 4  Nb +  =
dG 
E
d 4G
G
∴ N a = Nb +
Q.E.D.
E
_________________________
______________________________________
____________________
δ =
10-39 Table 10-5 (d = 4 mm
m = 0.1575 in): E = 196.5 GPa
Table 10-4 for A227:
Eq. (10-14):
Eq. (10-57):
A = 1783 MPa · mmm,
m = 0.190
A
1783
Sut = m = 0.190 = 1370 MPa
d
4
Sy = σall = 0.78 Sut = 0.78(1370) = 1069 MPaa
D = OD − d = 32 − 4 = 28 mm
C = D/d = 28/4 = 7
Eq. (10-43):
2
4C 2 − C − 1 4 ( 7 ) − 7 − 1
19
=
= 1.119
Ki =
4C (C − 1)
4(7)(7 − 1)
32Fr
π d3
At yield, Fr = My , σ = Sy. Thus,
Eq. (10-44):
σ = Ki
My =
π d 3S y π ( 43 )1069 (10−3 )
=
= 6.00 N · m
32Ki
32(1.119)
Shigley’s MED, 10th edition
Chapterr 110 Solutions, Page 35/41
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Count the turns when M = 0
N = 2.5 −
My
k=
where from Eq. (10-51):
k
d 4E
10.8 DN
Thus,
N = 2.5 −
My
4
d E / (10.8DN )
Solving for N gives
N =
=
2.5
1 + [10.8DM y / (d 4 E )]
2.5
{
}
1 + [10.8(28)(6.00) ] /  44 (196.5) 
This means (2.5 − 2.413)(360°) or 31.3° from closed.
= 2.413 turns
Ans.
Treating the hand force as in the middle of the grip,
87.5
r = 112.5 − 87.5 +
= 68.75 mm
2
6.00 (103 )
M
Fmax = y =
= 87.3 N Ans.
r
68.75
______________________________________________________________________________
10-40 The spring material and condition are unknown. Given d = 0.081 in and OD = 0.500,
(a) D = 0.500 − 0.081 = 0.419 in
Using E = 28.6 Mpsi for an estimate
k′ =
d 4E
(0.081)4 (28.6)(106 )
=
= 24.7 lbf · in/turn
10.8DN
10.8(0.419)(11)
for each spring. The moment corresponding to a force of 8 lbf
Fr = (8/2)(3.3125) = 13.25 lbf · in/spring
The fraction windup turn is
n =
Fr 13.25
=
= 0.536 turns
k′
24.7
The arm swings through an arc of slightly less than 180°, say 165°. This uses up
165/360 or 0.458 turns. So n = 0.536 − 0.458 = 0.078 turns are left (or
Shigley’s MED, 10th edition
Chapter 10 Solutions, Page 36/41
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0.078(360°) = 28.11° ). The original configuration of the spring
g was
w
Ans.
(b)
D 0.419
19
=
= 5.17
0.081
81
d
4C 2 − C − 1 4(5.17) 2 − 5.17 − 1
Ki =
=
= 1.168
4C ( C − 1)
4(5.17)(5.17 − 1)
C =
σ = Ki
 32(13.25) 
32M
= 1.168 
= 297 (103 ) psi = 297 kps
psi
3
3
πd
 π (0.081) 
Ans.
To achieve this stre
tress level, the spring had to have set removed.
d.
_________________________
______________________________________
____________________
10-41 (a) Consider half and
d double
d
results
Straight section:
M = 3FR,
∂M
= 3R
∂F
on:
Upper 180° section
M = F [ R + R(1 − cos φ )]
∂M
= FR(2 − cos φ ),
= R(2 − cos φ )
∂F
Lower section:
M = FR sin θ,
∂M
= R sin θ
∂F
Considering bendin
ding only:
Shigley’s MED, 10th edition
Chapterr 110 Solutions, Page 37/41
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π
∂U
2  l/2
9 FR 2 dx + ∫ FR 2 (2 − cos φ ) 2 R dφ +
=
∫
0
∂F
EI  0
2F  9 2
π
π

 π 
=
R l + R 3  4π − 4sin φ 0 +  + R 3   

2
EI  2

 4 
2
2
2FR  19π
9  FR
(19π R + 18l )
=
R + l =

2  2 EI
EI  4
δ =
∫
π /2
0
F ( R sin θ ) 2 R dθ 

The spring rate is
k=
F
δ
=
2 EI
R (19π R + 18 l )
Ans.
2
(b) Given: A227 HD wire, d = 2 mm, R = 6 mm, and l = 25 mm.
Table 10-5 (d = 2 mm = 0.0787 in):
k=
E = 197.2 GPa
2 (197.2 )109 π ( 0.0024 ) / ( 64 )
0.006 19π ( 0.006 ) + 18 ( 0.025 ) 
2
= 10.65 (103 ) N/m = 10.65 N/mm
Ans.
(c) The maximum stress will occur at the bottom of the top hook where the bendingmoment is 3FR and the axial fore is F. Using curved beam theory for bending,
σi =
Eq. (3-65), p. 133:
Axial:
σa =
Combining,
Mci
3FRci
=
2
Aeri (π d / 4 ) e ( R − d / 2 )
F
F
=
A πd2 / 4
σ max = σ i + σ a =
F=
4F
πd2


3Rci
+ 1 = S y

 e ( R − d / 2) 
π d 2Sy


3Rci
4
+ 1
 e ( R − d / 2) 
(1)
Ans.
For the clip in part (b),
Eq. (10-14) and Table 10-4:
Eq. (10-57):
Sut = A/dm = 1783/20.190 = 1563 MPa
Sy = 0.78 Sut = 0.78(1563) = 1219 MPa
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Chapter 10 Solutions, Page 38/41
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Table 3-4, p. 135:
rn =
12
(
2 6 − 62 − 12
)
= 5.95804 mm
e = rc − rn = 6 − 5.95804 = 0.04196 mm
ci = rn − (R − d /2) = 5.95804 − (6 − 2/2) = 0.95804 mm
Eq. (1):
F=
π ( 0.0022 )1219 (106 )
= 46.0 N
Ans.
 3 ( 6 ) 0.95804

+ 1
4
 0.04196 ( 6 − 1) 
______________________________________________________________________________
10-42 (a)
∂M
= −x
∂F
M = − Fx,
M = Fl + FR (1 − cos θ ) ,
{
0≤ x≤l
∂M
= l + R (1 − cos θ )
∂F
0 ≤θ ≤π / 2
π /2
2
1 l
− Fx(− x)dx + ∫ F l + R (1 − cos θ )  Rdθ
∫
0
EI 0
F
4l 3 + 3R  2π l 2 + 4 ( π − 2 ) l R + ( 3π − 8 ) R 2 
=
12 EI
δF =
{
}
}
The spring rate is
k=
F
δF
=
12 EI
4l + 3R  2π l + 4 (π − 2 ) l R + ( 3π − 8 ) R 2 
3
2
Ans.
(b) Given: A313 stainless wire, d = 0.063 in, R = 0.625 in, and l = 0.5 in.
Table 10-5:
E = 28 Mpsi
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Chapter 10 Solutions, Page 39/41
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I=
k=
π
64
d4 =
π
64
( 0.063 ) = 7.733 (10 ) in
−7
4
4
12 ( 28 )106 ( 7.733)10−7
4 ( 0.53 ) + 3 ( 0.625 )  2π ( 0.52 ) + 4 (π − 2 ) 0.5 ( 0.625 ) + ( 3π − 8 ) ( 0.6252 ) 
= 36.3 lbf/in
(c) Table 10-4:
Ans.
A = 169 kpsi⋅inm, m = 0.146
Eq. (10-14):
Sut = A/ d m = 169/0.0630.146 = 253.0 kpsi
Eq. (10-57):
Sy = 0.61 Sut = 0.61(253.0) = 154.4 kpsi
One can use curved beam theory as in the solution for Prob. 10-41. However, the
equations developed in Sec. 10-12 are equally valid.
C = D/d = 2(0.625 + 0.063/2)/0.063 = 20.8
Eq. (10-43):
2
4C 2 − C − 1 4 ( 20.8 ) − 20.8 − 1
=
= 1.037
Ki =
4C ( C − 1)
4 ( 20.8 )( 20.8 − 1)
Eq. (10-44), setting σ = Sy:
Ki
Solving for F yields
32 Fr
= Sy
πd3
⇒
F = 3.25 lbf
1.037
32 F ( 0.5 + 0.625 )
π ( 0.063
3
)
= 154.4 (103 )
Ans.
Try solving part (c) of this problem using curved beam theory. You should obtain the
same answer.
______________________________________________________________________________
10-43 (a)
M = − Fx
σ=
M
Fx
Fx
=
= 2
I / c I / c bh / 6
Constant stress,
bh 2 Fx
=
⇒
6
σ
h=
6 Fx
bσ
(1)
Ans.
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Chapter 10 Solutions, Page 40/41
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At x = l,
6 Fl
bσ
ho =
(b)
⇒
h = ho x / l
Ans.
M = − Fx, ∂ M /∂ F =− x
l
y=∫
0
=
l
l
M ( ∂M / ∂F )
− Fx ( − x )
1
12 Fl 3/ 2 1/2
=
dx = ∫ 1 3
dx
x dx
EI
E 0 12 bho ( x / l )3/2
bho3 E ∫0
2 12 Fl 3/2 3/2 8 Fl 3
l = 3
3 bho3 E
bho E
F bho3 E
= 3
Ans.
y
8l
______________________________________________________________________________
k=
10-44 Computer programs will vary.
______________________________________________________________________________
10-45 Computer programs will vary.
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Chapter 10 Solutions, Page 41/41
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Chapter 11
11-1
For the deep-groove 02-series ball bearing with R = 0.90, the design life xD, in multiples
of rating life, is
60L D nD 60 ( 25000 ) 350
L
=
= 525 Ans.
xD = D =
LR
L10
106
The design radial load is
FD = 1.2 ( 2.5) = 3.0 kN
1/3
Eq. (11-9):


525


C10 = 3.0 
1/1.483 
 0.02 + ( 4.459 − 0.02 ) ln (1/ 0.9 ) 

C10 = 24.3 kN
Table 11-2:
Ans.
Choose an 02-35 mm bearing with C10 = 25.5 kN.
Ans.
1.483
3
 

 525 ( 3 / 25.5 ) − 0.02  
Eq. (11-21): R = exp − 
Ans.
  = 0.920
4.459 − 0.02
 
 

______________________________________________________________________________
11-2
For the angular-contact 02-series ball bearing as described, the rating life multiple is
60L D nD 60 ( 40 000 ) 520
L
xD = D =
=
= 1248
106
LR
L10
The design radial load is
FD = 1.4 ( 725 ) = 1015 lbf = 4.52 kN
Eq. (11-9):
1/3


1248


C10 = 1015 
1/1.483 
 0.02 + ( 4.459 − 0.02 ) ln (1/ 0.9 ) 

= 10 930 lbf = 48.6 kN
Select an 02-60 mm bearing with C10 = 55.9 kN. Ans.
1.483
3
 

 1248 ( 4.52 / 55.9 ) − 0.02  
Eq. (11-21): R = exp − 
Ans.
  = 0.945
4.439
 
 

______________________________________________________________________________
Table 11-2:
Shigley’s MED, 10th edition
Chapter 11 Solutions, Page 1/28
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11-3
For the straight-roller 03-series bearing selection, xD = 1248 rating lives from Prob. 11-2
solution.
FD = 1.4 ( 2235) = 3129 lbf = 13.92 kN
 1248 
C10 = 13.92 

 1 
Table 11-3:
3/10
= 118 kN
Select an 03-60 mm bearing with C10 = 123 kN.
Ans.
1.483
10/3
 

 1248 (13.92 / 123) − 0.02  
  = 0.917 Ans.
Eq. (11-21): R = exp − 
4.459 − 0.02
 
 

______________________________________________________________________________
11-4
The combined reliability of the two bearings selected in Probs. 11-2 and 11-3 is
R = ( 0.945)( 0.917 ) = 0.867
Ans.
We can choose a reliability goal of 0.90 = 0.95 for each bearing. We make the
selections, find the existing reliabilities, multiply them together, and observe that the
reliability goal is exceeded due to the roundup of capacity upon table entry.
Another possibility is to use the reliability of one bearing, say R1. Then set the reliability
goal of the second as
R2 =
0.90
R1
or vice versa. This gives three pairs of selections to compare in terms of cost, geometry
implications, etc.
______________________________________________________________________________
11-5
Establish a reliability goal of
contact ball bearing,
0.90 = 0.95 for each bearing. For an 02-series angular
1/3


1248


C10 = 1015 
1/1.483 
 0.02 + 4.439 ln (1/ 0.95 ) 

= 12822 lbf = 57.1 kN
Select an 02-65 mm angular-contact bearing with C10 = 63.7 kN.
1.483
3
 

 1248 ( 4.52 / 63.7 ) − 0.02  
RA = exp − 
  = 0.962
4.439
 
 

Shigley’s MED, 10th edition
Chapter 11 Solutions, Page 2/28
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For an 03-series straight roller bearing,
3/10


1248


C10 = 13.92 
1/1.483 
 0.02 + 4.439 ln (1/ 0.95 ) 

= 136.5 kN
Select an 03-65 mm straight-roller bearing with C10 = 138 kN.
1.483
10/3
 

 1248 (13.92 /138 ) − 0.02  
RB = exp − 
  = 0.953
4.439
 
 

The overall reliability is R = (0.962)(0.953) = 0.917, which exceeds the goal.
______________________________________________________________________________
11-6
For the straight cylindrical roller bearing specified with a service factor of 1, R = 0.95 and
FR = 20 kN.
60L D nD 60 ( 8000 ) 950
L
xD = D =
=
= 456
106
LR
L10
3/10


456


C10 = 20 
Ans.
= 145 kN
1/1.483 
 0.02 + 4.439 ln (1/ 0.95 ) 

______________________________________________________________________________
11-7
Both bearings need to be rated in terms of the same catalog rating system in order to
compare them. Using a rating life of one million revolutions, both bearings can be rated
in terms of a Basic Load Rating.
1/ a
Eq. (11-3):
L 
C A = FA  A 
 LR 
= 8.96 kN
1/ a
 L n 60 
= FA  A A 
 LR 
 ( 3000 )( 500 )( 60 ) 
= 2.0 

106


1/3
Bearing B already is rated at one million revolutions, so CB = 7.0 kN. Since CA > CB,
bearing A can carry the larger load.
Ans.
______________________________________________________________________________
11-8
FD = 2 kN, LD = 109 rev, R = 0.90
1/ a
1/3
L 
 109 
C10 = FD  D  = 2  6  = 20 kN Ans.
 10 
 LR 
______________________________________________________________________________
Eq. (11-3):
Shigley’s MED, 10th edition
Chapter 11 Solutions, Page 3/28
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11-9
FD = 800 lbf, ΛD = 12 000 hours, nD = 350 rev/min, R = 0.90
 12 000 ( 350 )( 60 ) 
 L n 60 
Eq. (11-3):
C10 = FD  D D  = 800 
 = 5050 lbf Ans
106
 LR 


______________________________________________________________________________
1/3
1/ a
11-10 FD = 4 kN, ΛD = 8 000 hours, nD = 500 rev/min, R = 0.90
 8 000 ( 500 )( 60 ) 
 L n 60 
Eq. (11-3):
C10 = FD  D D  = 4 
 = 24.9 kN Ans
106
 LR 


______________________________________________________________________________
1/ a
1/3
11-11 FD = 650 lbf, nD = 400 rev/min, R = 0.95
ΛD = (5 years)(40 h/week)(52 week/year) = 10 400 hours
Assume an application factor of one. The multiple of rating life is
xD =
LD (10 400 )( 400 )( 60 )
=
= 249.6
LR
106
1/3


249.6


Eq. (11-9):
C10 = (1)( 650 ) 
1/1.483 
 0.02 + 4.439 ln (1/ 0.95 ) 

= 4800 lbf Ans.
______________________________________________________________________________
11-12 FD = 9 kN, LD = 108 rev, R = 0.99
Assume an application factor of one. The multiple of rating life is
xD =
LD 108
=
= 100
LR 106
1/3


100


C10 = (1)( 9 ) 
Eq. (11-9):
1/1.483 
 0.02 + 4.439 ln (1/ 0.99 ) 

= 69.2 kN Ans.
______________________________________________________________________________
11-13 FD = 11 kips, ΛD = 20 000 hours, nD = 200 rev/min, R = 0.99
Assume an application factor of one. Use the Weibull parameters for Manufacturer 2 in
Table 11-6.
The multiple of rating life is
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Chapter 11 Solutions, Page 4/28
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xD =
LD ( 20 000 )( 200 )( 60 )
=
= 240
LR
106
1/3


240


C10 = (1)(11) 
Eq. (11-9):
1/1.483 
 0.02 + 4.439 ln (1/ 0.99 ) 

= 113 kips Ans.
______________________________________________________________________________
11-14 From the solution to Prob. 3-68, the ground reaction force carried by the bearing at C is
RC = FD = 178 lbf. Use the Weibull parameters for Manufacturer 2 in Table 11-6.
xD =
LD 15000 (1200 )( 60 )
=
= 1080
106
LR
1/ a
Eq. (11-10):


xD
C10 = a f FD 

1/ b
 x0 + (θ − x0 )(1 − RD ) 
1/3


1080
C10 = 1.2 (178 ) 

1/1.483
 0.02 + ( 4.459 − 0.02 )(1 − 0.95 )

= 2590 lbf Ans.
______________________________________________________________________________
11-15 From the solution to Prob. 3-69, the ground reaction force carried by the bearing at C is
RC = FD = 1.794 kN. Use the Weibull parameters for Manufacturer 2 in Table 11-6.
xD =
LD 15000 (1200 )( 60 )
=
= 1080
106
LR
1/ a
Eq. (11-10):


xD
C10 = a f FD 

1/ b
 x0 + (θ − x0 )(1 − RD ) 
1/3


1080
C10 = 1.2 (1.794 ) 

1/1.483
 0.02 + ( 4.459 − 0.02 )(1 − 0.95 )

= 26.1 kN Ans.
______________________________________________________________________________
11-16 From the solution to Prob. 3-70, RCz = –327.99 lbf, RCy = –127.27 lbf
1/2
2
2
RC = FD = ( −327.99 ) + ( −127.27 )  = 351.8 lbf


Use the Weibull parameters for Manufacturer 2 in Table 11-6.
15000 (1200 )( 60 )
L
xD = D =
= 1080
LR
106
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Chapter 11 Solutions, Page 5/28
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1/ a
Eq. (11-10):


xD
C10 = a f FD 

1/ b
 xo + (θ − xo )(1 − RD ) 
1/3


1080
C10 = 1.2 ( 351.8 ) 

1/1.483
 0.02 + ( 4.459 − 0.02 )(1 − 0.95 )

= 5110 lbf Ans.
______________________________________________________________________________
11-17 From the solution to Prob. 3-71, RCz = –150.7 N, RCy = –86.10 N
1/ 2
2
2
RC = FD = ( −150.7 ) + ( −86.10 )  = 173.6 N


Use the Weibull parameters for Manufacturer 2 in Table 11-6.
xD =
LD 15000 (1200 )( 60 )
=
= 1080
106
LR
1/ a
Eq. (11-10):


xD
C10 = a f FD 

1/ b
 x0 + (θ − x0 )(1 − RD ) 
1/3


1080
C10 = 1.2 (173.6 ) 

1/1.483
 0.02 + ( 4.459 − 0.02 )(1 − 0.95 )

= 2520 N Ans.
______________________________________________________________________________
11-18 From the solution to Prob. 3-77, RAz = 444 N, RAy = 2384 N
(
RA = FD = 4442 + 23842
)
1/2
= 2425 N = 2.425 kN
Use the Weibull parameters for Manufacturer 2 in Table 11-6. The design speed is equal
to the speed of shaft AD,
d
125
nD = F ni =
(191) = 95.5 rev/min
250
dC
xD =
LD 12 000 ( 95.5 )( 60 )
=
= 68.76
LR
106
1/ a
Eq. (11-10):


xD
C10 = a f FD 

1/ b
 x0 + (θ − x0 )(1 − RD ) 
1/3


68.76
C10 = (1)( 2.425 ) 

1/1.483
 0.02 + ( 4.459 − 0.02 )(1 − 0.95 )

= 11.7 kN Ans.
______________________________________________________________________________
Shigley’s MED, 10th edition
Chapter 11 Solutions, Page 6/28
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11-19 From the solution to Prob. 3-79, RAz = 54.0 lbf, RAy = 140 lbf
(
RA = FD = 54.02 + 1402
)
1/ 2
= 150.1 lbf
Use the Weibull parameters for Manufacturer 2 in Table 11-6. The design speed is equal
to the speed of shaft AD,
d
10
nD = F ni = ( 280 ) = 560 rev/min
5
dC
xD =
LD 14 000 ( 560 )( 60 )
=
= 470.4
LR
106
1/ a
Eq. (11-10):


xD
C10 = a f FD 

1/ b
 x0 + (θ − x0 )(1 − RD ) 
3/10


470.4
C10 = (1)(150.1) 

1/1.483
 0.02 + ( 4.459 − 0.02 )(1 − 0.98 )

= 1320 lbf Ans.
______________________________________________________________________________
11-20 (a) Fa = 3 kN, Fr = 7 kN, nD = 500 rev/min, V = 1.2
From Table 11-2, with a 65 mm bore, C0 = 34.0 kN.
Fa / C0 = 3 / 34 = 0.088
From Table 11-1, 0.28 ≤ e ≤ 3.0.
Fa
3
=
= 0.357
VFr (1.2 )( 7 )
Since this is greater than e, interpolating Table 11-1 with Fa / C0 = 0.088, we obtain
X2 = 0.56 and Y2 = 1.53.
Eq. (11-12):
Fe = X iVFr + Yi Fa = ( 0.56 )(1.2 )( 7 ) + (1.53)( 3) = 9.29 kN
Fe > Fr so use Fe.
Ans.
(b) Use Eq. (11-10) to determine the necessary rated load the bearing should have to
carry the equivalent radial load for the desired life and reliability. Use the Weibull
parameters for Manufacturer 2 in Table 11-6.
10 000 ( 500 )( 60 )
L
xD = D =
= 300
106
LR
Shigley’s MED, 10th edition
Chapter 11 Solutions, Page 7/28
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1/ a
Eq. (11-10):


xD
C10 = a f FD 

1/ b
 x0 + (θ − x0 )(1 − RD ) 
1/3


300
C10 = (1)( 9.29 ) 

1/1.483
 0.02 + ( 4.459 − 0.02 )(1 − 0.95 )

= 73.4 kN
From Table 11-2, the 65 mm bearing is rated for 55.9 kN, which is less than the
necessary rating to meet the specifications. This bearing should not be expected to meet
the load, life, and reliability goals.
Ans.
______________________________________________________________________________
11-21 (a) Fa = 2 kN, Fr = 5 kN, nD = 400 rev/min, V = 1
From Table 11-2, 30 mm bore, C10 = 19.5 kN, C0 = 10.0 kN
Fa / C0 = 2 / 10 = 0.2
From Table 11-1, 0.34 ≤ e ≤ 0.38.
Fa
2
=
= 0.4
VFr (1)( 5 )
Since this is greater than e, interpolating Table 11-1, with Fa / C0 = 0.2, we obtain X2 =
0.56 and Y2 = 1.27.
Eq. (11-12): Fe = X iVFr + Yi Fa = ( 0.56 )(1)( 5) + (1.27 )( 2 ) = 5.34 kN Ans.
Fe > Fr so use Fe.
(b) Solve Eq. (11-10) for xD.
 C
xD =  10

 a f FD
a

1/ b
  x0 + (θ − x0 )(1 − RD ) 

3
 19.5 
 0.02 + ( 4.459 − 0.02 )(1 − 0.99 )1/1.483 
xD = 
 (1)( 5.34 )  



xD = 10.66
xD =
LD =
LD LD nD ( 60 )
=
LR
106
( ) = 10.66 (10 ) = 444 h
xD 106
nD ( 60 )
6
( 400 )( 60 )
Ans.
Shigley’s MED, 10th edition
Chapter 11 Solutions, Page 8/28
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______________________________________________________________________________
11-22 Fr = 8 kN, R = 0.9, LD = 109 rev
1/ a
Eq. (11-3):
L 
C10 = FD  D 
 LR 
1/3
 109 
= 8 6 
 10 
= 80 kN
From Table 11-2, select the 85 mm bore. Ans.
______________________________________________________________________________
11-23 Fr = 8 kN, Fa = 2 kN, V = 1, R = 0.99
Use the Weibull parameters for Manufacturer 2 in Table 11-6.
xD =
LD 10 000 ( 400 )( 60 )
=
= 240
106
LR
First guess: Choose from middle of Table 11-1, X = 0.56, Y = 1.63
Eq. (11-12):
Fe = 0.56 (1)( 8) + 1.63 ( 2 ) = 7.74 kN
Fe < Fr, so just use Fr as the design load.
Eq. (11-10):


xD
C10 = a f FD 

1/ b
 xo + (θ − xo )(1 − RD ) 
1/ a
1/3


240
C10 = (1)( 8 ) 

1/1.483
 0.02 + ( 4.459 − 0.02 )(1 − 0.99 )

= 82.5 kN
From Table 11-2, try 85 mm bore with C10 = 83.2 kN, C0 = 53.0 kN
Iterate the previous process:
Fa / C0 = 2 / 53 = 0.038
Table 11-1: 0.22 ≤ e ≤ 0.24
Fa
2
=
= 0.25 > e
VFr 1( 8)
Interpolate Table 11-1 with Fa / C0 = 0.038 to obtain X2 = 0.56 and Y2 = 1.89.
Eq. (11-12):
Fe = 0.56(1)8 + 1.89(2) = 8.26 > Fr
1/3
Eq. (11-10):
Table 11-2:
Iterate again:


240
C10 = (1)( 8.26 ) 
= 85.2 kN

1/1.483
 0.02 + ( 4.459 − 0.02 )(1 − 0.99 )

Move up to the 90 mm bore with C10 = 95.6 kN, C0 = 62.0 kN.
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Fa / C0 = 2 / 62 = 0.032
Again, 0.22 ≤ e ≤ 0.24
Fa
2
=
= 0.25 > e
VFr 1( 8 )
Interpolate Table 11-1 with Fa / C0 = 0.032 to obtain X2 = 0.56 and Y2 = 1.95.
Table 11-1:
Eq. (11-12):
Fe = 0.56(1)8 + 1.95(2) = 8.38 > Fr
1/3


240
Eq. (11-10): C10 = (1)( 8.38 ) 
 = 86.4 kN
1/1.483
 0.02 + ( 4.459 − 0.02 )(1 − 0.99 )

The 90 mm bore is acceptable. Ans.
______________________________________________________________________________
11-24 Fr = 8 kN, Fa = 3 kN, V = 1.2, R = 0.9, LD = 108 rev
First guess: Choose from middle of Table 11-1, X = 0.56, Y = 1.63
Eq. (11-12):
Fe = 0.56 (1.2 )( 8) + 1.63 ( 3) = 10.3 kN
Fe > Fr
1/ a
Eq. (11-3):
L 
C10 = Fe  D 
 LR 
1/3
 108 
= 10.3  6 
 10 
= 47.8 kN
From Table 11-2, try 60 mm with C10 = 47.5 kN, C0 = 28.0 kN
Iterate the previous process:
Fa / C0 = 3 / 28 = 0.107
0.28 ≤ e ≤ 0.30
Fa
3
=
= 0.313 > e
VFr 1.2 ( 8)
Interpolate Table 11-1 with Fa / C0 = 0.107 to obtain X2 = 0.56 and Y2 = 1.46
Table 11-1:
Eq. (11-12):
Fe = 0.56 (1.2 )( 8) + 1.46 ( 3) = 9.76 kN > Fr
1/3
 108 
Eq. (11-3):
C10 = 9.76  6  = 45.3 kN
 10 
From Table 11-2, we have converged on the 60 mm bearing. Ans.
______________________________________________________________________________
11-25 Fr = 10 kN, Fa = 5 kN, V = 1, R = 0.95
Use the Weibull parameters for Manufacturer 2 in Table 11-6.
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xD =
LD 12 000 ( 300 )( 60 )
=
= 216
LR
106
First guess: Choose from middle of Table 11-1, X = 0.56, Y = 1.63
Eq. (11-12):
Fe = 0.56 (1)(10 ) + 1.63 ( 5) = 13.75 kN
Fe > Fr, so use Fe as the design load.
Eq. (11-10):


xD
C10 = a f FD 

1/ b
 x0 + (θ − x0 )(1 − RD ) 
1/ a
1/3


216
C10 = (1)(13.75 ) 

1/1.483
 0.02 + ( 4.459 − 0.02 )(1 − 0.95 )

= 97.4 kN
From Table 11-2, try 95 mm bore with C10 = 108 kN, C0 = 69.5 kN
Iterate the previous process:
Fa / C0 = 5 / 69.5 = 0.072
Table 11-1:
0.27 ≤ e ≤ 0.28
Fa
5
=
= 0.5 > e
VFr 1(10 )
Interpolate Table 11-1 with Fa / C0 = 0.072 to obtain X2 = 0.56 and Y2 = 1.62 B 1.63
Since this is where we started, we will converge back to the same bearing. The 95 mm
bore meets the requirements. Ans.
______________________________________________________________________________
11-26 Fr = 9 kN, Fa = 3 kN, V = 1.2, R = 0.99
Use the Weibull parameters for Manufacturer 2 in Table 11-6.
xD =
LD 108
=
= 100
LR 106
First guess: Choose from middle of Table 11-1, X = 0.56, Y = 1.63
Eq. (11-12): Fe = 0.56 (1.2 )( 9 ) + 1.63 ( 3) = 10.9 kN
Fe > Fr, so use Fe as the design load.
1/ a
Eq. (11-10):


xD
C10 = a f FD 

1/ b
 x0 + (θ − x0 )(1 − RD ) 
Shigley’s MED, 10th edition
Chapter 11 Solutions, Page 11/28
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1/3


100
C10 = (1)(10.9 ) 

1/1.483
 0.02 + ( 4.459 − 0.02 )(1 − 0.99 )

= 83.9 kN
From Table 11-2, try 90 mm bore with C10 = 95.6 kN, C0 = 62.0 kN. Try this bearing.
Iterate the previous process:
Fa / C0 = 3 / 62 = 0.048
0.24 ≤ e ≤ 0.26
Fa
3
=
= 0.278 > e
VFr 1.2 ( 9 )
Interpolate Table 11-1 with Fa / C0 = 0.048 to obtain X2 = 0.56 and Y2 = 1.79
Table 11-1:
Eq. (11-12):
Fe = 0.56 (1.2 )( 9 ) + 1.79 ( 3) = 11.4 kN > Fr
11.4
83.9 = 87.7 kN
10.9
From Table 11-2, this converges back to the same bearing. The 90 mm bore meets the
requirements. Ans.
______________________________________________________________________________
C10 =
11-27 (a) nD = 1200 rev/min, LD = 15 kh, R = 0.95, a f = 1.2
From Prob. 3-72, RCy = 183.1 lbf, RCz = –861.5 lbf.
1/2
2
RC = FD = 183.12 + ( −861.5 )  = 881 lbf


15000 (1200 )( 60 )
L
xD = D =
= 1080
106
LR
1/3
Eq. (11-10):


1080
C10 = 1.2 ( 881) 

1/1.483
 0.02 + 4.439 (1 − 0.95 )

= 12800 lbf = 12.8 kips
Ans.
(b) Results will vary depending on the specific bearing manufacturer selected. A general
engineering components search site such as www.globalspec.com might be useful as
a starting point.
______________________________________________________________________________
11-28 (a) nD = 1200 rev/min, LD = 15 kh, R = 0.95, a f = 1.2
From Prob. 3-72, ROy = –208.5 lbf, ROz = 259.3 lbf.
2
RC = FD =  259.32 + ( −208.5 ) 


1/2
= 333 lbf
Shigley’s MED, 10th edition
Chapter 11 Solutions, Page 12/28
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xD =
LD 15000 (1200 )( 60 )
=
= 1080
LR
106
1/3


1080
Eq. (11-10):
C10 = 1.2 ( 333) 

1/1.483
 0.02 + 4.439 (1 − 0.95 )

= 4837 lbf = 4.84 kips Ans.
(b) Results will vary depending on the specific bearing manufacturer selected. A general
engineering components search site such as www.globalspec.com might be useful as
a starting point.
______________________________________________________________________________
11-29 (a) nD = 900 rev/min, LD = 12 kh, R = 0.98, a f = 1.2
From Prob. 3-73, RCy = 8.319 kN, RCz = –10.830 kN.
1/2
RC = FD = 8.319 2 + ( −10.830 )  = 13.7 kN


12 000 ( 900 )( 60 )
L
xD = D =
= 648
106
LR
2
1/3


648
Eq. (11-10): C10 = 1.2 (13.7 ) 
 = 204 kN Ans.
1/1.483
 0.02 + 4.439 (1 − 0.98 )

(b) Results will vary depending on the specific bearing manufacturer selected. A general
engineering components search site such as www.globalspec.com might be useful as
a starting point.
______________________________________________________________________________
11-30 (a) nD = 900 rev/min, LD = 12 kh, R = 0.98, a f = 1.2
From Prob. 3-73, ROy = 5083 N, ROz = 494 N.
(
RC = FD = 50832 + 4942
xD =
)
1/2
= 5106 N = 5.1 kN
LD 12 000 ( 900 )( 60 )
=
= 648
LR
106
1/3


648
Eq. (11-10): C10 = 1.2 ( 5.1) 
 = 76.1 kN Ans.
1/1.483
 0.02 + 4.439 (1 − 0.98 )

(b) Results will vary depending on the specific bearing manufacturer selected. A general
engineering components search site such as www.globalspec.com might be useful as
a starting point.
______________________________________
________________________________
________
11-31 Assume concentrated forces as shown.
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Pz = 8 ( 28) = 224 lbf
Py = 8 ( 35) = 280 lbf
T = 224 ( 2 ) = 448 lbf ⋅ in
ΣT x = −448 + 1.5 F cos 20o = 0
448
F=
= 318 lbf
1.5 ( 0.940 )
ΣM Oz = 5.75Py + 11.5RAy − 14.25 F sin 20o = 0
5.75 ( 280 ) + 11.5 RAy − 14.25 ( 318 )( 0.342 ) = 0
RAy = −5.24 lbf
ΣM Oy = −5.75 Pz − 11.5 RAz − 14.25 F cos 20o = 0
−5.75 ( 224 ) − 11.5 RAz − 14.25 ( 318 )( 0.940 ) = 0
1/2
2
2
RA = ( −482 ) + ( −5.24 ) 


ΣF z = ROz + Pz + RAz + F cos 20o = 0
RAz = −482 lbf;
= 482 lbf
ROz + 224 − 482 + 318 ( 0.940 ) = 0
ROz = −40.9 lbf
ΣF y = ROy + Py + RAy − F sin 20o = 0
ROy + 280 − 5.24 − 318 ( 0.342 ) = 0
ROy = −166 lbf
1/2
2
2
RO = ( −40.9 ) + ( −166 ) 


= 171 lbf
So the reaction at A governs.
Reliability Goal: 0.92 = 0.96
FD = 1.2 ( 482 ) = 578 lbf
xD = 35 000 ( 350 )( 60 ) / 106 = 735
1/3


735


C10 = 578 
1/1.483 
 0.02 + ( 4.459 − 0.02 ) ln (1/ 0.96 ) 

= 6431 lbf = 28.6 kN
From Table 11-2, a 40 mm bore angular contact bearing is sufficient with a rating of
31.9 kN.
Ans.
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Chapter 11 Solutions, Page 14/28
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______________________________________________________________________________
11-32 For a combined reliability goal of 0.95, use 0.95 = 0.975 for the individual bearings.
xD =
40 000 ( 420 )( 60 )
106
= 1008
The resultants of the given forces are
RO = [(–387)2 + 4672]1/2 = 607 lbf
RB = [3162 + (–1615)2]1/2 = 1646 lbf
At O:
1/3
Eq. (11-9):


1008


C10 = 1.2 ( 607 ) 
1/1.483 
 0.02 + ( 4.459 − 0.02 ) ln (1/ 0.975 ) 

= 9978 lbf = 44.4 kN
From Table 11-2, select an 02-55 mm angular-contact ball bearing with a basic load
rating of 46.2 kN. Ans.
At B:
3/10
Eq. (11-9):


1008


C10 = 1.2 (1646 ) 
1/1.483 
 0.02 + ( 4.459 − 0.02 ) ln (1/ 0.975 ) 

= 20827 lbf = 92.7 kN
From Table 11-3, select an 02-75 mm or 03-55 mm cylindrical roller.
______________________________________
________________________________
________
Ans.
11-33 The reliability of the individual
bearings is R = 0.98 = 0.9899
From statics,
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Chapter 11 Solutions, Page 15/28
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T = (270 − 50) = (P1 − P2)125
= (P1 − 0.15 P1)125
P1 = 310.6 N,
P2 = 0.15 (310.6) = 46.6 N
P1 + P2 = 357.2 N FAy = 357.2 sin 45o = 252.6 N = FAz
∑M
∑F
∑M
∑F
= 850REy + 300(252.6) = 0 ⇒ REy = −89.2 N
z
O
y
= 252.6 − 89.2 + ROy = 0 ⇒ ROy = −163.4 N
= −850REz − 700(320) + 300(252.6) = 0 ⇒ REz = −174.4 N
y
O
z
= −174.4 + 320 − 252.6 + ROz = 0 ⇒ ROz = 107 N
RO =
( −163.4 )
2
RE =
( −89.2 )
+ ( −174.4 ) = 196 N
2
+ 107 2 = 195 N
2
The radial loads are nearly the same at O and E. We can use the same bearing at both
locations.
xD =
60 000 (1500 )( 60 )
106
= 5400
1/3
Eq. (11-9):


5400


C10 = 1( 0.196 ) 
1/1.483 
 0.02 + 4.439 ln (1/ 0.9899 ) 

= 5.7 kN
From Table 11-2, select an 02-12 mm deep-groove ball bearing with a basic load rating
of 6.89 kN. Ans.
______________________________________________________________________________
11-34
R = 0.96 = 0.980
T = 12(240 cos 20o ) = 2706 lbf ⋅ in
F=
2706
= 498 lbf
6 cos 25o
In xy-plane:
ΣM Oz = −16(82.1) − 30(210) + 42 RCy = 0
RCy = 181 lbf
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ROy = 82.1 + 210 − 181 = 111.1 lbf
In xz-plane:
ΣM Oy = 16(226) − 30(451) − 42 RCz = 0
RCz = −236 lbf
ROz = 226 − 451 + 236 = 11 lbf
(
= (181
RO = 111.12 + 112
RC
xD =
2
+ 2362
)
)
1/ 2
= 112 lbf
Ans.
1/ 2
= 297 lbf
Ans.
50 000 ( 300 )( 60 )
106
= 900
1/3
( C10 )O


900


= 1.2 (112 ) 
1/1.483 
 0.02 + 4.439 ln (1/ 0.980 ) 

= 1860 lbf = 8.28 kN
1/3
( C10 )C


900


= 1.2 ( 297 ) 
1/1.483 
 0.02 + 4.439 ln (1/ 0.980 ) 

= 4932 lbf = 21.9 kN
Bearing at O: Choose a deep-groove 02-17 mm. Ans.
Bearing at C: Choose a deep-groove 02-35 mm. Ans.
______________________________________________________________________________
11-35 Shafts subjected to thrust can be constrained by bearings, one of which supports the
thrust. The shaft floats within the endplay of the second (roller) bearing. Since the thrust
force here is larger than any radial load, the bearing absorbing the thrust (bearing A) is
heavily loaded compared to bearing B. Bearing B is thus likely to be oversized and may
not contribute measurably to the chance of failure. If this is the case, we may be able to
obtain the desired combined reliability with bearing A having a reliability near 0.99 and
bearing B having a reliability near 1. This would allow for bearing A to have a lower
capacity than if it needed to achieve a reliability of 0.99 . To determine if this is the
case, we will start with bearing B.
Bearing B (straight roller bearing)
30 000 ( 500 )( 60 )
xD =
= 900
106
(
Fr = 36 2 + 67 2
)
1/ 2
= 76.1 lbf = 0.339 kN
Try a reliability of 1 to see if it is readily obtainable with the available bearings.
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3/10
Eq. (11-9):


900


C10 = 1.2 ( 0.339 ) 
1/1.483 
 0.02 + 4.439  ln (1/ 1.0 ) 

= 10.1 kN
The smallest capacity bearing from Table 11-3 has a rated capacity of 16.8 kN.
Therefore, we select the 02-25 mm straight cylindrical roller bearing. Ans.
Bearing at A (angular-contact ball)
With a reliability of 1 for bearing B, we can achieve the combined reliability goal of 0.99
if bearing A has a reliability of 0.99.
(
Fr = 36 2 + 212 2
)
1/ 2
= 215 lbf = 0.957 kN
Fa = 555 lbf = 2.47 kN
Trial #1:
Tentatively select an 02-85 mm angular-contact with C10 = 90.4 kN and C0 = 63.0 kN.
Fa 2.47
=
= 0.0392
C0 63.0
30 000 ( 500 )( 60 )
xD =
= 900
106
Table 11-1:
Interpolating, X2 = 0.56, Y2 = 1.88
Eq. (11-12):
Fe = 0.56 ( 0.957 ) + 1.88 ( 2.47 ) = 5.18 kN
1/3
Eq. (11-9):


900


C10 = 1.2 ( 5.18 ) 
1/1.483 
 0.02 + 4.439 ln (1/ 0.99 ) 

= 99.54 kN > 90.4 kN
Trial #2:
Tentatively select a 02-90 mm angular-contact ball with C10 = 106 kN and C0 = 73.5 kN.
Fa 2.47
=
= 0.0336
C0 73.5
Table 11-1:
Interpolating, X2 = 0.56, Y2 = 1.93
Fe = 0.56 ( 0.957 ) + 1.93 ( 2.47 ) = 5.30 kN
1/3


900


C10 = 1.2 ( 5.30 ) 
1/1.483 
 0.02 + 4.439 ln (1/ 0.99 ) 

Shigley’s MED, 10th edition
= 102 kN < 106 kN O.K.
Chapter 11 Solutions, Page 18/28
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Select an 02-90 mm angular-contact ball bearing. Ans.
______________________________________________________________________________
11-36 We have some data. Let’s estimate parameters b and θ from it. In Fig. 11-5, we will use
line AB. In this case, B is to the right of A.
( x )1 =
For F = 18 kN,
115 ( 2000 )( 60 )
106
= 13.8
This establishes point 1 on the R = 0.90 line.
The R = 0.20 locus is above and parallel to the R = 0.90 locus. For the two-parameter
Weibull distribution, x0 = 0 and points A and B are related by [see Eq. (11-8)]:
x A = θ ln (1/ 0.90 ) 
1/ b
(1)
xB = θ ln (1/ 0.20 ) 
1/ b
and xB/xA is in the same ratio as 600/115. Eliminating θ,
b=
ln ln (1/ 0.20 ) / ln (1/ 0.90 ) 
ln ( 600 /115 )
= 1.65
Ans.
Solving for θ in Eq. (1),
θ=
xA
ln (1/ RA ) 
1/1.65
=
1
 ln (1/ 0.90 ) 
1/1.65
Shigley’s MED, 10th edition
= 3.91
Ans.
Chapter 11 Solutions, Page 19/28
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Therefore, for the data at hand,
  x 
R = exp  − 

  3.91 
1.65



Check R at point B: xB = (600/115) = 5.217
  5.217 1.65 
R = exp  − 
  = 0.20
  3.91  
Note also, for point 2 on the R = 0.20 line,
log ( 5.217 ) − log (1) = log ( xm )2 − log (13.8)
( xm )2 = 72
______________________________________________________________________________
11-37 This problem is rich in useful variations. Here is one.
Decision: Make straight roller bearings identical on a given shaft. Use a reliability goal of
(0.99)1/6 = 0.9983.
Shaft a
(
= ( 502
FAr = 239 2 + 1112
FBr
2
)
1/2
= 264 lbf = 1.175 kN
)
1/2
+ 10752
= 1186 lbf = 5.28 kN
Thus the bearing at B controls.
xD =
10 000 (1200 )( 60 )
106
= 720
0.02 + 4.439 ln (1/ 0.9983) 
1/1.483
 720 
C10 = 1.2 ( 5.28 ) 

 0.080 26 
Select an 02-80 mm with C10 = 106 kN.
Shaft b
FCr = ( 8742 + 22742 )
1/2
(
FDr = 3932 + 657 2
)
1/2
= 0.080 26
0.3
= 97.2 kN
Ans.
= 2436 lbf
= 766 lbf
or
or
10.84 kN
3.41 kN
The bearing at C controls.
Shigley’s MED, 10th edition
Chapter 11 Solutions, Page 20/28
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xD =
10 000 ( 240 )( 60 )
106
= 144
 144 
C10 = 1.2 (10.84 ) 

 0.080 26 
0.3
Select an 02-90 mm with C10 = 142 kN.
Shaft c
(
= ( 417
FEr = 11132 + 23852
FFr
+ 895
2
)
1/2
)
2 1/ 2
= 123 kN
Ans.
= 2632 lbf
= 987 lbf
or
or
11.71 kN
4.39 kN
The bearing at E controls.
xD =
10 000 ( 80 )( 60 )
106
= 48
 48 
C10 = 1.2 (11.71) 

 0.080 26 
0.3
= 95.7 kN
Select an 02-80 mm with C10 = 106 kN.
Ans.
______________________________________________________________________________
11-38 Express Eq. (11-1) as
F1a L1 = C10a L10 = K
For a ball bearing, a = 3 and for an 02-30 mm angular contact bearing, C10 = 20.3 kN.
( )
( )
K = ( 20.3) 106 = 8.365 109
3
At a load of 18 kN, life L1 is given by:
( )
9
K 8.365 10
=
= 1.434 106 rev
3
a
18
F1
For a load of 30 kN, life L2 is:
L1 =
L2 =
( )
( ) = 0.310 10 rev
( )
30
8.365 109
6
3
In this case, Eq. (6-57) – the Palmgren-Miner cycle-ratio summation rule – can be
expressed as
Shigley’s MED, 10th edition
Chapter 11 Solutions, Page 21/28
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l1 l2
+
=1
L1 L2
Substituting,
l2
200 000
+
=1
6
1.434 10
0.310 106
l2
( )
( )
= 0.267 (10 ) rev Ans.
6
______________________________________________________________________________
11-39 Total life in revolutions
Let:
l = total turns
f1 = fraction of turns at F1
f2 = fraction of turns at F2
From the solution of Prob. 11-38, L1 = 1.434(106) rev and L2 = 0.310(106) rev.
Palmgren-Miner rule:
l1 l2
fl f l
+
= 1 + 2 =1
L1 L2 L1 L2
from which
l=
l=
1
f1 / L1 + f 2 / L2
1
{0.40 / 1.434 (10 )} + {0.60 / 0.310 (10 )}
6
= 451 585 rev
6
Ans.
Total life in loading cycles
4 min at 2000 rev/min = 8000 rev/cycle
6 min at 2000 rev/min = 12 000 rev/cycle
Total rev/cycle = 8000 + 12 000 = 20 000
451 585 rev
= 22.58 cycles
20 000 rev/cycle
Ans.
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Total life in hours
 min   22.58 cycles 
10

 = 3.76 h Ans.
 cycle   60 min/h 
______________________________________________________________________________
FrA = 560 lbf
11-40
FrB = 1095 lbf
Fae = 200 lbf
xD =
LD 40 000 ( 400 )( 60 )
=
= 10.67
LR
90 106
( )
R = 0.90 = 0.949
Eq. (11-18):
Eq. (11-18):
0.47 FrA 0.47 ( 560 )
=
= 175.5 lbf
KA
1.5
0.47 FrB 0.47 (1095 )
FiB =
=
= 343.1 lbf
1.5
KB
FiA =
FiA ≤ ? ≥ ( FiB + Fae )
175.5 lbf ≤ ( 343.1 + 200 ) = 543.1 lbf, so Eq. (11-19) applies.
We will size bearing B first since its induced load will affect bearing A, but is not itself
affected by the induced load from bearing A [see Eq. (11-19)].
From Eq. (11-19b), FeB = FrB = 1095 lbf.
Eq. (11-10):
FRB


10.67

= 1.4 (1095 ) 
1/1.5
 4.48 (1 − 0.949 ) 


3/10
= 3607 lbf
Ans.
Select cone 32305, cup 32305, with 0.9843 in bore, and rated at 3910 lbf with
K = 1.95. Ans.
With bearing B selected, we re-evaluate the induced load from bearing B using the actual
value for K.
0.47 FrB 0.47 (1095 )
Eq. (11-18): FiB =
=
= 263.9 lbf
1.95
KB
Find the equivalent radial load for bearing A from Eq. (11-19), which still applies.
Eq. (11-19a): FeA = 0.4 FrA + K A ( FiB + Fae )
FeA = 0.4 ( 560 ) + 1.5 ( 263.9 + 200 ) = 920 lbf
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FeA > FrA
Eq. (11-10):


10.67

FRA = 1.4 ( 920 ) 
1/1.5
 4.48 (1 − 0.949 ) 


3/10
= 3030 lbf
Tentatively select cone M86643, cup M86610, with 1 in bore, and rated at 3250 lbf
with K = 1.07. Iterating with the new value for K, we get FeA = 702 lbf and FrA = 2312
lbf. Ans.
By using a bearing with a lower K, the rated load decreased significantly, providing a
higher than requested reliability. Further examination with different combinations of
bearing choices could yield additional acceptable solutions.
______________________________________________________________________________
11-41 The thrust load on shaft CD is from the axial component of the force transmitted through
the bevel gear, and is directed toward bearing C. By observation of Fig. 11-14, direct
mounted bearings would allow bearing C to carry the thrust load. Ans.
From the solution to Prob. 3-74, the axial thrust load is Fae = 362.8 lbf, and the bearing
radial forces are FCx = 287.2 lbf, FCz = 500.9 lbf, FDx = 194.4 lbf, and FDz = 307.1 lbf.
Thus, the radial forces are
FrC = 287.2 2 + 500.9 2 = 577 lbf
FrD = 194.42 + 307.12 = 363 lbf
The induced loads are
0.47 FrC 0.47 ( 577 )
Eq. (11-18): FiC =
=
= 181 lbf
1.5
KC
0.47 FrD 0.47 ( 363)
Eq. (11-18): FiD =
=
= 114 lbf
KD
1.5
Check the condition on whether to apply Eq. (11-19) or Eq. (11-20), where bearings C
and D are substituted, respectively, for labels A and B in the equations.
FiC ≤ ? ≥ FiD + Fae
181 lbf < 114 + 362.8 = 476.8 lbf, so Eq.(11-19) applies
Eq. (11-19a): FeC = 0.4 FrC + KC ( FiD + Fae )
= 0.4 ( 577 ) + 1.5 (114 + 362.8 ) = 946 lbf > FrC , so use FeC
Assume for tapered roller bearings that the specifications for Manufacturer 1 in Table 116 are applicable.
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xD =
LD
108
=
= 1.11
LR 90 (106 )
R = 0.90 = 0.949


1.11

Eq. (11-10): FRC = 1( 946 ) 
1/1.5
 4.48 (1 − 0.949 ) 


Eq. (11-19b): FeD = FrD = 363 lbf
3/10
= 1130 lbf
Ans.
3/10


1.11
 = 433 lbf Ans.
Eq. (11-10): FRD = 1( 363) 
1/1.5
 4.48 (1 − 0.949 ) 


______________________________________________________________________________
11-42 The thrust load on shaft AB is from the axial component of the force transmitted through
the bevel gear, and is directed to the right. By observation of Fig. 11-14, indirect
mounted bearings would allow bearing A to carry the thrust load. Ans.
From the solution to Prob. 3-76, the axial thrust load is Fae = 92.8 lbf, and the bearing
radial forces are FAy = 639.4 lbf, FAz = 1513.7 lbf, FBy = 276.6 lbf, and FBz = 705.7 lbf.
Thus, the radial forces are
FrA = 639.4 2 + 1513.7 2 = 1643 lbf
FrB = 276.62 + 705.7 2 = 758 lbf
The induced loads are
0.47 FrA 0.47 (1643)
Eq. (11-18): FiA =
=
= 515 lbf
KA
1.5
0.47 FrB 0.47 ( 758 )
=
= 238 lbf
Eq. (11-18): FiB =
1.5
KB
Check the condition on whether to apply Eq. (11-19) or Eq. (11-20).
FiA ≤ ? ≥ FiB + Fae
515 lbf > 238 + 92.8 = 330.8 lbf, so Eq.(11-20) applies
Notice that the induced load from bearing A is sufficiently large to cause a net axial force
to the left, which must be supported by bearing B.
Eq. (11-20a): FeB = 0.4 FrB + K B ( FiA − Fae )
= 0.4 ( 758 ) + 1.5 ( 515 − 92.8) = 937 lbf > FrB, so use FeB
Assume for tapered roller bearings that the specifications for Manufacturer 1 in Table
11-6 are applicable.
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6
LD 500 (10 )
xD =
=
= 5.56
LR
90 (106 )
R = 0.90 = 0.949


5.56

Eq. (11-10): FRB = 1( 937 ) 
1/1.5
 4.48 (1 − 0.949 ) 


Eq. (11-19b): FeA = FrA = 1643 lbf
3/10
= 1810 lbf
Ans.
3/10


5.56
 = 3180 lbf Ans.
Eq. (11-10): FRA = 1(1643) 
 4.48 (1 − 0.949 )1/1.5 


______________________________________________________________________________
11-43 The lower bearing is compressed by the axial load, so it is designated as bearing A.
FrA = 25 kN
FrB = 12 kN
Fae = 5 kN
0.47 FrA 0.47 ( 25 )
=
= 7.83 kN
KA
1.5
0.47 FrB 0.47 (12 )
Eq. (11-18): FiB =
=
= 3.76 kN
1.5
KB
Check the condition on whether to apply Eq. (11-19) or Eq. (11-20)
Eq. (11-18):
FiA =
FiA ≤ ? ≥ FiB + Fae
7.83 kN < 3.76 + 5 = 8.76 kN, so Eq.(11-19) applies
Eq. (11-19a): FeA = 0.4 FrA + K A ( FiB + Fae )
= 0.4 ( 25 ) + 1.5 ( 3.76 + 5 ) = 23.1 kN < FrA, so use FrA
 60 min   8 hr   5 day   52 weeks 
LD = ( 250 rev/min ) 

 ( 5 yrs )


yr
 hr   day   week  

( )
= 156 106 rev
Assume for tapered roller bearings that the specifications for Manufacturer 1 in Table
11-6 are applicable.
Eq. (11-3):
L 
FRA = a f FD  D 
 LR 
3/10
( ) 
( ) 
156 106
= 1.2 ( 25 ) 
 90 106
3/10
= 35.4 kN
Ans.
Eq. (11-19b): FeB = FrB = 12 kN
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Chapter 11 Solutions, Page 26/28
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3/10
156 
FRB = 1.2 (12 ) 
= 17.0 kN Ans.
 90 
______________________________________________________________________________
Eq. (11-3):
11-44 The left bearing is compressed by the axial load, so it is properly designated as bearing A.
FrA = 875 lbf
FrB = 625 lbf
Fae = 250 lbf
Assume K = 1.5 for each bearing for the first iteration. Obtain the induced loads.
Eq. (11-18):
Eq. (11-18):
0.47 FrA 0.47 ( 875 )
=
= 274 lbf
1.5
KA
0.47 FrB 0.47 ( 625 )
FiB =
=
= 196 lbf
KB
1.5
FiA =
Check the condition on whether to apply Eq. (11-19) or Eq. (11-20).
FiA ≤ ? ≥ FiB + Fae
274 lbf < 196 + 250 lbf, so Eq.(11-19) applies
We will size bearing B first since its induced load will affect bearing A, but it is not
affected by the induced load from bearing A [see Eq. (11-19)].
From Eq. (11-19b), FeB = FrB = 625 lbf.
Eq. (11-3):
L 
FRB = a f FD  D 
 LR 
3/10
 90 000 (150 )( 60 ) 

= 1( 625 ) 
90 106


3/10
( )
FRB = 1208 lbf
Select cone 07100, cup 07196, with 1 in bore, and rated at 1570 lbf with K = 1.45. Ans.
With bearing B selected, we re-evaluate the induced load from bearing B using the actual
value for K.
0.47 FrB 0.47 ( 625 )
Eq. (11-18): FiB =
=
= 203 lbf
KB
1.45
Find the equivalent radial load for bearing A from Eq. (11-19), which still applies.
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Eq. (11-19a): FeA = 0.4 FrA + K A ( FiB + Fae )
= 0.4 ( 875) + 1.5 ( 203 + 250 ) = 1030 lbf
FeA > FrA
Eq. (11-3):
L 
FRA = a f FD  D 
 LR 
3/10
 90 000 (150 )( 60 ) 

= 1(1030 ) 
90 106


3/10
( )
FRA = 1990 lbf
Any of the bearings with 1-1/8 in bore are more than adequate. Select cone 15590, cup
15520, rated at 2480 lbf with K = 1.69. Iterating with the new value for K, we get FeA =
1120 lbf and FrA = 2160 lbf. The selected bearing is still adequate. Ans.
______________________________________________________________________________
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Chapter 12
12-1
Given: dmax = 25 mm, bmin = 25.03 mm, l/d = 1/2, W = 1.2 kN, µ = 55 mPa⋅s, and N =
1100 rev/min.
b − d max 25.03 − 25
cmin = min
=
= 0.015 mm
2
2
r 25/2 = 12.5 mm
r/c = 12.5/0.015 = 833.3
N = 1100/60 = 18.33 rev/s
P = W/ (ld) = 1200/ [12.5(25)] = 3.84 N/mm2 = 3.84 MPa
Eq. (12-7):
2
 55 (10−3 )18.33 
 r  µN
2
 = 0.182
= 833.3 
S = 
6
c P
 3.84 (10 ) 
Fig. 12-16:
h0 /c = 0.3
⇒
h0 = 0.3(0.015) = 0.0045 mm
Fig. 12-18:
f r/c = 5.4
⇒
f = 5.4/833.3 = 0.006 48
Ans.
T =f Wr = 0.006 48(1200)12.5(10−3) = 0.0972 N⋅m
Hloss = 2π TN = 2π (0.0972)18.33 = 11.2 W
Fig. 12-19:
Q/(rcNl) = 5.1 ⇒
Ans.
Q = 5.1(12.5)0.015(18.33)12.5 = 219 mm3/s
Qs = 0.81(219) = 177 mm3/s Ans.
Fig. 12-20:
Qs /Q = 0.81 ⇒
______________________________________________________________________________
12-2
Given: dmax = 32 mm, bmin = 32.05 mm, l = 64 mm, W = 1.75 kN, µ = 55 mPa⋅s, and N =
900 rev/min.
b − d max 32.05 − 32
=
= 0.025 mm
cmin = min
2
2
r ≈ 32/2 = 16 mm
r/c = 16/0.025 = 640
N = 900/60 = 15 rev/s
P = W/ (ld) = 1750/ [32(64)] = 0.854 MPa
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l/d = 64/32 = 2
Eq. (12-7):
2
 55 (10−3 )15 
 r  µN
 = 0.797
S = 
= 6402 
c P
 0.854 
Eq. (12-16), Figs. 12-16, 12-19, and 12-21
l/d
y∞
y1
y1/2
y1/4
yl/d
h0/c
2
0.98
0.83
0.61
0.36
0.92
P/pmax
Q/rcNl
2
2
0.84
3.1
0.54
3.45
0.45
4.2
0.31
5.08
0.65
3.20
h0 = 0.92 c = 0.92(0.025) = 0.023 mm
Ans.
pmax = P / 0.65 = 0.854/0.65 = 1.31 MPa
Ans.
Ans.
Q = 3.20 rcNl = 3.20(16)0.025(15)64 = 1.23 (103) mm3/s
______________________________________________________________________________
12-3
Given: dmax = 3.000 in, bmin = 3.005 in, l = 1.5 in, W = 800 lbf, N = 600 rev/min, and SAE
10 and SAE 40 at 150°F.
b − d max
3.005 − 3.000
=
= 0.0025 in
cmin = min
2
2
r ≈ 3.000 / 2 = 1.500 in
l / d = 1.5 / 3 = 0.5
r / c = 1.5 / 0.0025 = 600
N = 600 / 60 = 10 rev/s
W
800
=
= 177.78 psi
P=
ld 1.5(3)
Fig. 12-12: SAE 10 at 150°F, µ ′ = 1.75 µ reyn
1.75(10− 6 )(10) 
 r  µN
S = 
= 6002 
 = 0.0354
c P
 177.78

2
Figs. 12-16 and 12-21: h0/c = 0.11 and P/pmax = 0.21
h0 = 0.11(0.0025) = 0.000 275 in Ans.
pmax = 177.78 / 0.21 = 847 psi Ans.
Fig. 12-12: SAE 40 at 150°F, µ ′ = 4.5 µ reyn
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 4.5 
S = 0.0354 
 = 0.0910
 1.75 
h0 / c = 0.19, P / pmax = 0.275
h0 = 0.19(0.0025) = 0.000 475 in Ans.
pmax = 177.78 / 0.275 = 646 psi Ans.
______________________________________________________________________________
12-4
Given: dmax = 3.250 in, bmin = 3.256 in, l = 3 in, W = 800 lbf, and N = 1000 rev/min.
b − d max
3.256 − 3.250
=
= 0.003
cmin = min
2
2
r ≈ 3.250 / 2 = 1.625 in
l / d = 3 / 3.250 = 0.923
r / c = 1.625 / 0.003 = 542
N = 1000 / 60 = 16.67 rev/s
W
800
=
= 82.05 psi
P=
ld
3(3.25)
Fig. 12-12: SAE 20W at 150°F, µ′ = 2.40 µ reyn
Note to instructors: Some students may obtain a higher value of viscosity (2.85) from
Fig. 12-14. The value from Fig. 12-12 is used here as the preferred value since this figure
is specifically for single-viscosity oils.
 2.40(10− 6 )(16.67) 
 r  µN
= 5422 
S = 
 = 0.143
82.05
c P


2
From Eq. (12-16), and Figs. 12-16 and 12-21:
l/d
y∞
y1
y1/2
y1/4
yl/d
h0/c
0.923
0.82
0.44
0.26
0.14
0.42
P/pmax
0.923
0.83
0.44
0.31
0.21
0.42
h0 = 0.42c = 0.42(0.003) = 0.001 26 in
P
82.05
pmax =
=
= 195 psi Ans.
0.42
0.42
Ans.
Fig. 12-14: SAE 20W-40 at 150°F, µ′ = 4.4 µ reyn
S = 5422
4.4(10− 6 )(16.67)
= 0.263
82.05
From Eq. (12-16), and Figs. 12-16 and 12-21:
Shigley’s MED, 10th edition
Chapter 12 Solutions, Page 3/26
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l/d
y∞
y1
y1/2
y1/4
yl/d
h0/c
0.923
0.91
0.6
0.38
0.2
0.58
P/pmax
0.923
0.83
0.48
0.35
0.24
0.46
h0 = 0.58c = 0.58(0.003) = 0.001 74 in Ans.
P
82.05
pmax =
=
= 178 psi Ans.
0.46
0.46
______________________________________________________________________________
12-5
Given: dmax = 2.000 in, bmin = 2.0024 in, l = 1 in, W = 600 lbf, N = 800 rev/min, and SAE
20 at 130°F.
2.0024 − 2
b − d max
=
= 0.0012 in
cmin = min
2
2
d
2
= = 1 in, l / d = 1 / 2 = 0.50
r ≈
2
2
r / c = 1 / 0.0012 = 833
N = 800 / 60 = 13.33 rev/s
W
600
=
= 300 psi
P =
ld
2(1)
Fig. 12-12: SAE 20 at 130°F, µ ′ = 3.75 µ reyn
 3.75(10− 6 )(13.3) 
 r  µN
S = 
= 8332 
 = 0.115
300
c P


2
From Figs. 12-16, 12-18 and 12-19:
h0 / c = 0.23, r f / c = 3.8, Q / (rcNl ) = 5.3
h0 = 0.23(0.0012) = 0.000 276 in Ans.
3.8
f =
= 0.004 56
833
The power loss due to friction is
2π f WrN
2π (0.004 56)(600)(1)(13.33)
=
778(12)
778(12)
= 0.0245 Btu/s Ans.
Q = 5.3rcNl
= 5.3(1)(0.0012)(13.33)(1)
= 0.0848 in 3 / s Ans.
______________________________________________________________________________
H =
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Chapter 12 Solutions, Page 4/26
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12-6
Given: dmax = 25 mm, bmin = 25.04 mm, l/d = 1, W = 1.25 kN, µ = 50 mPa⋅s, and N =
1200 rev/min.
25.04 − 25
b − d max
cmin = min
=
= 0.02 mm
2
2
r ≈ d / 2 = 25 / 2 = 12.5 mm, l / d = 1
r / c = 12.5 / 0.02 = 625
N = 1200 / 60 = 20 rev/s
W
1250
P =
=
= 2 MPa
ld
252
 50(10− 3 )(20) 
 r  µN
S = 
= 6252 
 = 0.195
6
c P
 2(10 ) 
From Figs. 12-16, 12-18 and 12-20:
2
For µ = 50 mPa · s,
h0 / c = 0.52, f r / c = 4.5, Qs / Q = 0.57
h0 = 0.52(0.02) = 0.0104 mm Ans.
4.5
f =
= 0.0072
625
T = f Wr = 0.0072(1.25)(12.5) = 0.1125 N · m
The power loss due to friction is
H = 2πT N = 2π (0.1125)(20) = 14.14 W
Ans.
Qs = 0.57Q The side flow is 57% of Q Ans.
______________________________________________________________________________
12-7
Given: dmax = 1.25 in, bmin = 1.252 in, l = 2 in, W = 620 lbf, µ′ = 8.5 µ reyn, and N =
1120 rev/min.
b − d max 1.252 − 1.25
=
= 0.001 in
cmin = min
2
2
r = d / 2 = 1.25 / 2 = 0.625 in
r / c = 0.625 / 0.001 = 625
N = 1120 / 60 = 18.67 rev/s
620
W
=
= 248 psi
P =
ld 1.25(2)
 8.5(10− 6 )(18.67) 
 r  µN
= 6252 
S = 
 = 0.250
248
c P


l / d = 2 / 1.25 = 1.6
2
From Eq. (12-16), and Figs. 12-16, 12-18, and 12-19
Shigley’s MED, 10th edition
Chapter 12 Solutions, Page 5/26
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l/d
y∞
y1
y1/2
y1/4
yl/d
1.6
1.6
1.6
0.9
4.5
3
0.58
5.3
3.98
0.36
6.5
4.97
0.185
8
5.6
0.69
4.92
3.59
h0 = 0.69 c = 0.69(0.001) =0.000 69 in
Ans.
h0/c
fr/c
Q/rcNl
f = 4.92/(r/c) = 4.92/625 = 0.007 87
Ans.
Q = 1.6 rcNl = 1.6(0.625) 0.001(18.57) 2 = 0.0833 in3/s
Ans.
______________________________________________________________________________
12-8
Given: dmax = 75.00 mm, bmin = 75.10 mm, l = 36 mm, W = 2 kN, N = 720 rev/min, and
SAE 20 and SAE 40 at 60°C.
bmin − d max
75.10 − 75
=
= 0.05 mm
2
2
l / d = 36 / 75 = 0.48 0.5 (close enough)
r = d / 2 = 75 / 2 = 37.5 mm
r / c = 37.5 / 0.05 = 750
N = 720 / 60 = 12 rev/s
W
2000
=
= 0.741 MPa
P =
ld
75(36)
cmin =
Fig. 12-13: SAE 20 at 60°C, µ = 18.5 mPa · s
18.5(10− 3 )(12) 
 r  µN
S = 
= 7502 
 = 0.169
6
c P
 0.741(10 ) 
2
From Figures 12-16, 12-18 and 12-21:
h0 / c = 0.29, f r / c = 5.1, P / pmax = 0.315
h0 = 0.29(0.05) = 0.0145 mm Ans.
f = 5.1 / 750 = 0.0068
T = f Wr = 0.0068(2)(37.5) = 0.51 N · m
The heat loss rate equals the rate of work on the film
Hloss = 2πT N = 2π(0.51)(12) = 38.5 W
pmax = 0.741/0.315 = 2.35 MPa Ans.
Ans.
Fig. 12-13: SAE 40 at 60°C, µ = 37 mPa · s
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S = 0.169(37)/18.5 = 0.338
From Figures 12-16, 12-18 and 12-21:
h0 / c = 0.42, f r / c = 8.5, P / pmax = 0.38
h0 = 0.42(0.05) = 0.021 mm Ans.
f = 8.5 / 750 = 0.0113
T = f Wr = 0.0113(2)(37.5) = 0.85 N · m
H loss = 2π TN = 2π (0.85)(12) = 64 W Ans.
pmax = 0.741 / 0.38 = 1.95 MPa Ans.
_____________________________________________________________________________
12-9
Given: dmax = 56.00 mm, bmin = 56.05 mm, l = 28 mm, W = 2.4 kN, N = 900 rev/min, and
SAE 40 at 65°C.
b − d max
56.05 − 56
cmin = min
=
= 0.025 mm
2
2
r = d / 2 = 56 / 2 = 28 mm
r / c = 28 / 0.025 = 1120
l / d = 28 / 56 = 0.5, N = 900 / 60 = 15 rev/s
2400
P =
= 1.53 MPa
28(56)
Fig. 12-13: SAE 40 at 65°C, µ = 30 mPa · s
2
 30(10− 3 )(15) 
 r  µN
S = 
= 11202 
 = 0.369
6
c P
 1.53(10 ) 
From Figures 12-16, 12-18, 12-19 and 12-20:
h0 / c = 0.44, f r / c = 8.5, Qs / Q = 0.71, Q / (rcNl ) = 4.85
h0 = 0.44(0.025) = 0.011 mm Ans.
f = 8.5 / 1000 = 0.007 59
T = f Wr = 0.007 59(2.4)(28) = 0.51 N · m
H = 2π TN = 2π (0.51)(15) = 48.1 W Ans.
Q = 4.85 rcNl = 4.85(28)(0.025)(15)(28) = 1426 mm3 /s
Qs = 0.71(1426) = 1012 mm3 /s Ans.
_____________________________________________________________________________
12-10 Consider the bearings as specified by
minimum f :
d −+t0d , b−+0tb
maximum W:
d −′+td0 , b−+0tb
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and differing only in d and d′ .
Preliminaries:
l /d =1
P = W / (ld ) = 700 / (1.252 ) = 448 psi
N = 3600 / 60 = 60 rev/s
Fig. 12-16:
minimum f :
maximum W:
S ≈ 0.08
S ≈ 0.20
Fig. 12-12:
µ = 1.38(10−6) reyn
µN/P = 1.38(10−6)(60/448) = 0.185(10−6)
Eq. (12-7):
r
=
c
S
µN / P
For minimum f :
r
0.08
=
= 658
c
0.185(10− 6 )
c = 0.625 / 658 = 0.000 950
If this is cmin,
0.001 in
b − d = 2(0.001) = 0.002 in
The median clearance is
c = cmin +
t d + tb
t + tb
= 0.001 + d
2
2
and the clearance range for this bearing is
t + tb
∆c = d
2
which is a function only of the tolerances.
For maximum W:
r
0.2
=
= 1040
0.185(10− 6 )
c
c = 0.625 / 1040 = 0.000 600
0.0005 in
If this is cmin
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b − d ′ = 2cmin = 2(0.0005) = 0.001 in
t + tb
t + tb
= 0.0005 + d
c = cmin + d
2
2
t d + tb
∆c =
2
The difference (mean) in clearance between the two clearance ranges, crange, is
td + tb 
t + tb 
−  0.0005 + d

2
2 

= 0.0005 in
crange = 0.001 +
For the minimum f bearing
or
b − d = 0.002 in
d = b − 0.002 in
For the maximum W bearing
d′ = b − 0.001 in
For the same b, tb and td, we need to change the journal diameter by 0.001 in.
d ′ − d = b − 0.001 − (b − 0.002)
= 0.001 in
Increasing d of the minimum friction bearing by 0.001 in, defines d′ of the maximum
load bearing. Thus, the clearance range provides for bearing dimensions which are
attainable in manufacturing. Ans.
_____________________________________________________________________________
12-11 Given: SAE 40, N = 10 rev/s, Ts = 140°F, l/d = 1, d = 3.000 in, b = 3.003 in, W = 675
lbf.
b − d max
3.003 − 3
cmin = min
=
= 0.0015 in
2
2
r = d / 2 = 3 / 2 = 1.5 in
r / c = 1.5 / 0.0015 = 1000
675
W
P =
=
= 75 psi
3(3)
ld
Trial #1: From Figure 12-12 for T = 160°F, µ = 3.5 µ reyn,
∆T = 2(160 − 140) = 40°F
 3.5(10− 6 )(10) 
 r  µN
= 10002 
S = 
 = 0.4667
75
c P


2
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Chapter 12 Solutions, Page 9/26
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From Fig. 12-24,
9.70∆T
= 0.349 109 + 6.009 40(0.4667) + 0.047 467(0.4667) 2 = 3.16
P
P
75
∆T = 3.16
= 3.16
= 24.4°F
9.70
9.70
Discrepancy = 40 − 24.4 = 15.6°F
Trial #2: T = 150°F, µ = 4.5 µ reyn,
∆T = 2(150 − 140) = 20° F
 4.5 (10− 6 )10 
2
S = 1000 
 = 0.6
75


From Fig. 12-24,
9.70∆T
= 0.349 109 + 6.009 40(0.6) + 0.047 467(0.6) 2 = 3.97
P
P
75
∆T = 3.97
= 3.97
= 30.7°F
9.70
9.70
Discrepancy = 20 − 30.7 = − 10.7°F
Trial #3: T = 154°F, µ = 4 µ reyn,
∆T = 2(154 − 140) = 28°F
 4 (10− 6 )10 
S = 10002 
 = 0.533
75


From Fig. 12-24,
9.70∆T
= 0.349 109 + 6.009 40(0.533) + 0.047 467(0.533) 2 = 3.57
P
75
P
∆T = 3.57
= 3.57
= 27.6°F
9.70
9.70
Discrepancy = 28 − 27.6 = 0.4°F
O.K.
Tav = 140 +28/2 = 154°F Ans.
T1 = Tav − ∆T / 2 = 154 − (28 / 2) = 140°F
T2 = Tav + ∆T / 2 = 154 + (28 / 2) = 168°F
S = 0.4
From Figures 12-16, 12-18, to 12-20:
Shigley’s MED, 10th edition
Chapter 12 Solutions, Page 10/26
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h0
fr
Q
Qs
= 0.75,
= 11,
= 3.6,
= 0.33
c
c
rcN l
Q
h0 = 0.75(0.0015) = 0.001 13 in Ans.
11
f =
= 0.011
1000
T = f Wr = 0.0075(3)(40) = 0.9 N · m
2π ( 0.011) 675 (1.5 )10
2π f WrN
H loss =
=
= 0.075 Btu/s
778 (12 )
778 (12 )
Ans.
Q = 3.6rcN l = 3.6(1.5)0.0015(10)3 = 0.243 in 3 /s
Ans.
3
Qs = 0.33(0.243) = 0.0802 in /s Ans.
_____________________________________________________________________________
12-12 Given: d = 2.5 in, b = 2.504 in, cmin = 0.002 in, W = 1200 lbf, SAE = 20, Ts = 110°F,
N = 1120 rev/min, and l = 2.5 in.
P = W/(ld) = 1200/(2.5)2 = 192 psi,
N = 1120/60 = 18.67 rev/s
For a trial film temperature, let Tf = 150°F
Table 12-1:
Eq. (12-7):
µ′ = 0.0136 exp[1271.6/(150 + 95)] = 2.441 µ reyn
−6
2
2
 r  µ N  2.5 / 2  2.441(10 )18.67
S = 
=
= 0.0927

192
 c  P  0.002 
Fig. 12-24:
192
 0.349 109 + 6.009 40 ( 0.0927 ) + 0.047 467 ( 0.0927 2 ) 

9.70 
= 17.9°F
∆T =
∆T
17.9
= 110 +
= 119.0°F
2
2
T f − Tav = 150 − 119.0 = 31.0°F
Tav = Ts +
which is not 0.1 or less, therefore try averaging for the new trial film temperature, let
(T f ) new =
150 + 119.0
= 134.5°F
2
Proceed with additional trials using a spreadsheet (table also shows the first trial)
Shigley’s MED, 10th edition
Chapter 12 Solutions, Page 11/26
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Trial
Tf
µ'
S
∆T
Tav
Tf −Tav
New
Tf
150.0
134.5
127.9
125.3
124.3
124.0
123.8
2.441
3.466
4.084
4.369
4.485
4.521
4.545
0.0927
0.1317
0.1551
0.1659
0.1704
0.1717
0.1726
17.9
22.6
25.4
26.7
27.2
27.4
27.5
119.0
121.3
122.7
123.3
123.6
123.7
123.7
31.0
13.2
5.2
2.0
0.7
0.3
0.1
134.5
127.9
125.3
124.3
124.0
123.8
123.8
Note that the convergence begins rapidly. There are ways to speed this, but at this point
they would only add complexity.
(a)
µ ′ = 4.545(10−6 ), S = 0.1726
From Fig. 12-16:
h0
= 0.482, h0 = 0.482(0.002) = 0.000 964 in
c
From Fig. 12-17: φ = 56°
Ans.
(b) e = c − h0 = 0.002 − 0.000 964 = 0.001 04 in Ans.
f r
(c) From Fig. 12-18:
= 4.10, f = 4.10(0.002 /1.25) = 0.006 56
c
Ans.
(d) T = f Wr = 0.006 56(1200)(1.25) = 9.84 lbf · in
H =
(e) From Fig. 12-19:
2π T N
2π (9.84)(1120 / 60)
=
= 0.124 Btu/s
778(12)
778(12)
Ans.
Q
= 4.16
rcNl
 1120 
3
Q = 4.16(1.25)(0.002) 
 (2.5) = 0.485 in /s
 60 
Q
From Fig. 12-20: s = 0.6, Qs = 0.6(0.485) = 0.291 in 3/s
Q
Ans.
Ans.
W / ( ld ) 1200 / 2.52
P
= 0.45, pmax =
=
= 427 psi
pmax
0.45
0.45
From Fig. 12-22: φ pmax = 16° Ans.
(f) From Fig. 12-21:
Shigley’s MED, 10th edition
Ans.
Chapter 12 Solutions, Page 12/26
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(g) From Fig. 12-22: φ p0 = 82° Ans.
(h) From the trial table, Tf = 123.8°F Ans.
(i) With ∆T = 27.5°F from the trial table, Ts + ∆T = 110 + 27.5 = 137.5°F Ans.
_____________________________________________________________________________
12-13 Given: d = 1.250 in, td = 0.001 in, b = 1.252 in, tb = 0.003 in, l = 1.25 in, W = 250 lbf,
N = 1750 rev/min, SAE 10 lubricant, sump temperature Ts = 120°F.
P = W/(ld) = 250/1.252 = 160 psi,
N = 1750/60 = 29.17 rev/s
For the clearance, c = 0.002 ± 0.001 in. Thus, cmin = 0.001 in, cmedian = 0.002 in, and
cmax = 0.003 in.
For cmin = 0.001 in, start with a trial film temperature of Tf = 135°F
Table 12-1:
Eq. (12-7):
µ′ = 0.0158 exp[1157.5/(135 + 95)] = 2.423 µ reyn
−6
2
2
 r  µ N  1.25 / 2  2.423 (10 ) 29.17
=
= 0.1725
S = 

160
 c  P  0.001 
Fig. 12-24:
160 
0.349 109 + 6.009 40 ( 0.1725 ) + 0.047 467 ( 0.17252 ) 
9.70 
= 22.9°F
∆T =
∆T
22.9
= 120 +
= 131.4°F
2
2
T f − Tav = 135 − 131.4 = 3.6°F
Tav = Ts +
which is not 0.1 or less, therefore try averaging for the new trial film temperature, let
135 + 131.4
= 133.2°F
2
Proceed with additional trials using a spreadsheet (table also shows the first trial)
(T f ) new =
Trial
Tf
µ'
S
∆T
Tav
Tf−Tav
New
Tf
135.0
133.2
132.5
132.2
132.1
2.423
2.521
2.560
2.578
2.583
0.1725
0.1795
0.1823
0.1836
0.1840
22.9
23.6
23.9
24.0
24.0
131.4
131.8
131.9
132.0
132.0
3.6
1.4
0.6
0.2
0.1
133.2
132.5
132.2
132.1
132.1
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With Tf = 132.1°F, ∆T = 24.0°F, µ′ = 2.583 µ reyn, S = 0.1840,
Tmax = Ts + ∆T = 120 + 24.0 = 144.0°F
Fig. 12-16:
h0/c = 0.50, h0 = 0.50(0.001) = 0.000 50 in
∫ = 1 − h0/c = 1 − 0.50 = 0.05 in
Fig. 12-18:
r f /c = 4.25, f = 4.25/(0.625/0.001) = 0.006 8
Fig. 12-19:
Q/(rcNl) = 4.13, Q = 4.13(0.625)0.001(29.17)1.25 = 0.0941 in3/s
Fig. 12-20:
Qs/Q = 0.58, Qs = 0.58(0.0941) = 0.0546 in3/s
The above can be repeated for cmedian = 0.002 in, and cmax = 0.003 in. The results are
shown below.
cmin
0.001 in
cmedian
0.002 in
cmax
0.003 in
132.1
2.583
0.184
24.0
125.6
3.002
0.0534
11.1
124.1
3.112
0.0246
8.2
h0/c
144.0
0.5
131.1
0.23
128.2
0.125
h0
0.00050
0.00069
0.00038
∫
0.50
4.25
0.0068
4.13
0.0941
0.77
1.8
0.0058
4.55
0.207
0.88
1.22
0.0059
4.7
0.321
0.58
0.82
0.90
0.0546
0.170
0.289
Tf
µ′
S
∆Τ
Tmax
fr/c
f
Q/(rcNl)
Q
Qs /Q
Qs
_____________________________________________________________________________
12-14 Computer programs will vary.
_____________________________________________________________________________
12-15 In a step-by-step fashion, we are building a skill for natural circulation bearings.
• Given the average film temperature, establish the bearing properties.
• Given a sump temperature, find the average film temperature, then, establish the
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bearing properties.
• Now we acknowledge the environmental temperature’s role in establishing the sump
temperature. Sec. 12-9 and Ex. 12-5 address this problem.
Given: dmax = 2.500 in, bmin = 2.504 in, l/d = 1, N = 1120 rev/min, SAE 20 lubricant, W =
300 lbf, A = 60 in2, T∞ = 70°F, and α = 1.
600 lbf load with minimal clearance: We will start by using W = 600 lbf (nd = 2). The
task is to iteratively find the average film temperature, Tf , which makes Hgen and Hloss
equal.
b − d max
2.504 − 2.500
=
= 0.002 in
cmin = min
2
2
N = 1120/60 = 18.67 rev/s
P =
W
600
=
= 96 psi
ld
2.52
−6
2
2
 r  µ N  1.25  µ ′ (10 )18.67
S = 
=
= 0.0760µ ′

96
c P
 0.002 
µ′ = 0.0136 exp[1271.6/(Tf + 95)]
Table 12-1:
2545
fr
 f r  2545
WNc 
( 600 )18.67 ( 0.002 )
=
1050
c
 c  1050
fr
= 54.3
c
H gen =
H loss =
2.7 ( 60 / 144 )
h CR A
T f − T∞ ) =
(
(T f − 70)
α +1
1+1
= 0.5625 (T f − 70 )
Start with trial values of Tf of 220 and 240°F.
Trial Tf
220
240
µ′
0.770
0.605
S
0.059
0.046
f r/c
1.9
1.7
Hgen
103.2
92.3
Hloss
84.4
95.6
As a linear approximation, let Hgen = mTf + b. Substituting the two sets of values of Tf
and Hgen we find that Hgen = − 0.545 Tf +223.1. Setting this equal to Hloss and solving
for Tf gives Tf = 237°F.
Trial Tf
237
µ′
0.627
S
0.048
f r/c
1.73
Shigley’s MED, 10th edition
Hgen
93.9
Hloss
94.0
Chapter 12 Solutions, Page 15/26
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which is satisfactory.
Table 12-16:
h0/c = 0.21,
h0 = 0.21 (0.002) = 000 42 in
Fig. 12-24:
∆T =
96
0.349 109 + 6.009 4 ( 0.048 ) + 0.047 467 ( 0.0482 ) 

9.7 
= 6.31° F
T1 = Ts = Tf − ∆T = 237 − 6.31/2 = 233.8°F
Tmax = T1 + ∆T = 233.8 + 6.31 = 240.1°F
Trumpler’s design criteria:
0.002 + 0.000 04d = 0.002 + 0.000 04(2.5) = 0.000 30 in < h0
O.K.
Tmax = 240.1°F < 250°F O.K.
Wst
300
=
= 48 psi < 300 psi
ld
2.52
nd = 2 (assessed at W = 600 lbf)
O.K .
O.K.
We see that the design passes Trumpler’s criteria and is deemed acceptable.
For an operating load of W = 300 lbf, it can be shown that Tf = 219.3°F, µ′ = 0.78, S =
0.118, f r/c = 3.09, Hgen = Hloss = 84 Btu/h, h0 = , ∆T = 10.5°F, T1 = 224.6°F, and Tmax =
235.1°F.
_____________________________________________________________________________
+0.005
12-16 Given: d = 3.500 +−0.000
0.001 in, b = 3.505 −0.000 in , SAE 30, Ts = 120°F, ps = 50 psi,
N = 2000/60 = 33.33 rev/s, W = 4600 lbf, bearing length = 2 in, groove width = 0.250 in,
and Hloss ≤ 5000 Btu/hr.
b − d max
3.505 − 3.500
=
= 0.0025 in
cmin = min
2
2
r = d/ 2 = 3.500/2 = 1.750 in
r / c = 1.750/0.0025 = 700
l′ = (2 − 0.25)/2 = 0.875 in
l′ / d = 0.875/3.500 = 0.25
W
4600
P =
=
= 751 psi
4rl′ 4 (1.750 ) 0.875
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Trial #1: Choose (Tf )1 = 150°F. From Table 12-1,
µ′ = 0.0141 exp[1360.0/(150 + 95)] = 3.63 µ reyn
 3.63(10−6 )(33.33) 
 r  µN
S = 
= 7002 
 = 0.0789
751
c P


∫ = 0.9, f r/ c = 3.6
From Figs. 12-16 and 12-18:
2
From Eq. (12-24),
∆T =
=
0.0123( f r / c)SW 2
(1 + 1.5 ò2 ) ps r 4
0.0123 ( 3.6 ) 0.0789 ( 46002 )
1 + 1.5(0.9) 2  50 (1.7504 )
= 71.2°F
Tav = Ts + ∆T / 2 = 120 + 71.2/2 = 155.6°F
Trial #2: Choose (Tf )2 = 160°F. From Table 12-1
µ′ = 0.0141 exp[1360.0/(160 + 95)] = 2.92 µ reyn
 2.92 
S = 0.0789 
 = 0.0635
 3.63 
From Figs. 12-16 and 12-18:
∫ = 0.915, f r/ c =3
0.0123 ( 3) 0.0635 ( 46002 )
∆T =
= 46.9°F
1 + 1.5 ( 0.9152 )  50 (1.7504 )


Tav = 120 + 46.9/2 = 143.5°F
Trial #3: Thus, the plot gives (Tf )3 = 152.5°F. From Table 12-1
µ′ = 0.0141 exp[1360.0/(152.5 + 95)] = 3.43 µ reyn
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 3.43 
S = 0.0789 
 = 0.0746
 3.63 
From Figs. 12-16 and 12-18:
∫ = 0.905, f r/ c =3.4
∆T =
Tav
0.0123 ( 3.4 ) 0.0746 ( 46002 )
1 + 1.5 ( 0.9052 )  50 (1.7504 )


= 120 + 63.2/2 = 151.6°F
Result is close. Choose T f =
= 63.2°F
152.5 + 151.6
= 152.1°F
2
Try 152°F
µ′ = 0.0141 exp[1360.0/(152 + 95)] = 3.47 µ reyn
Table 12-1:
 3.47 
S = 0.0789 
 = 0.0754
 3.63 
f r
h0
= 3.4, ò = 0.902,
= 0.098
c
c
0.0123 ( 3.4 ) 0.0754 ( 46002 )
∆T =
= 64.1°F
1 + 1.5 ( 0.9022 )  50 (1.7504 )


Tav = 120 + 64.1 / 2 = 152.1°F O.K.
h0 = 0.098(0.0025) = 0.000 245 in
Tmax = Ts + ∆T = 120 + 64.1 = 184.1°F
Eq. (12-22):
π ( 50 )1.750 ( 0.00253 )
π ps rc3
2
Qs =
(1 + 1.5ò ) = 3 ( 3.47 )10− 6 ( 0.875) 1 + 1.5 ( 0.9022 )
3µ l′
= 1.047 in 3 /s
Hloss = ρ CpQs ∆T = 0.0311(0.42)1.047(64.1) = 0.877 Btu/s
= 0.877(602) = 3160 Btu/h O.K.
Trumpler’s design criteria:
0.0002 + 0.000 04(3.5) = 0.000 34 in > 0.000 245 Not O.K.
Tmax = 184.1°F < 250°F O.K.
Pst = 751 psi > 300 psi Not O.K.
n = 1, as done Not O.K.
_____________________________________________________________________________
+0.010
12-17 Given: d = 50.00 +−0.00
0.05 mm, b = 50.084 −0.000 mm , SAE 30, Ts = 55°C, ps = 200 kPa,
N = 2880/60 = 48 rev/s, W = 10 kN, bearing length = 55 mm, groove width = 5 mm, and
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Hloss ≤ 300 W.
bmin − d max
50.084 − 50
=
= 0.042 mm
2
2
r = d/ 2 = 50/2 = 25 mm
cmin =
r / c = 25/0.042 = 595
l′ = (55 − 5)/2 = 25 mm
l′ / d = 25/50 = 0.5
10 (103 )
W
P =
=
= 4 MPa
4rl′ 4 ( 25 ) 25
Trial #1: Choose (Tf )1 = 79°C. From Fig. 12-13, µ = 13 mPa · s.
13(10−3 )(48) 
 r  µN
= 5952 
S = 
 = 0.0552
6
c P
 4(10 ) 
From Figs. 12-16 and 12-18:
∫ = 0.85, f r/ c = 2.3
From Eq. (12-25),
978(106 ) ( f r / c)SW 2
∆T =
1 + 1.5 ò2
ps r 4
978(106 )  2.3(0.0552)(10 2 ) 
=
 = 76.3°C
1 + 1.5(0.85) 2 
200(25) 4

2
Tav = Ts + ∆T / 2 = 55 + 76.3/2 = 93.2°C
Trial #2: Choose (Tf )2 = 100°C. From Fig. 12-13, µ = 7 mPa · s.
7
S = 0.0552   = 0.0297
 13 
∫ = 0.90, f r/ c =1.6
From Figs. 12-16 and 12-18:
6
978(10 ) 1.6(0.0297)(102 ) 
∆T =
 = 26.9°C
1 + 1.5(0.9)2 
200(25) 4

Tav = 55 + 26.9/2 = 68.5°C
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Trial #3: Thus, the plot gives (Tf )3 = 85.5°C. From Fig. 12-13, µ = 10.5 mPa · s.
 10.5 
S = 0.0552 
 = 0.0446
 13 
From Figs. 12-16 and 12-18:
∫ = 0.87, f r/ c =2.2
∆T =
978(106 )  2.2(0.0457)(102 ) 
 = 58.9°C
1 + 1.5(0.87 2 ) 
200(25) 4

Tav = 55 + 58.9/2 = 84.5°C
Result is close. Choose T f =
Fig. 12-13:
85.5 + 84.5
= 85°C
2
µ = 10.5 mPa · s
 10.5 
S = 0.0552 
 = 0.0446
 13 
f r
h0
= 2.2,
= 0.13
ò = 0.87,
c
c
978(106 )  2.2(0.0457)(102 ) 
∆T =
 = 58.9°C
1 + 1.5(0.87 2 ) 
200(254 )

Tav = 55 + 58.9 / 2 = 84.5°C O.K.
or
138°F
From Eq. (12-22)
h0 = 0.13(0.042) = 0.005 46 mm or 0.000 215 in
Tmax = Ts + ∆T = 55 + 58.9 = 113.9°C
or
237°F
 π ( 200 ) 25 ( 0.0423 ) 
2


Qs = (1 + 1.5 ò )
= 1 + 1.5 ( 0.87 )  

−6
3µl′
 3 (10.5 )10 ( 25 ) 
= 3156 mm3 /s = 3156 ( 25.4− 3 ) = 0.193 in 3 /s
2
π ps rc3
Hloss = ρ CpQs ∆T = 0.0311(0.42)0.193(138) = 0.348 Btu/s
= 1.05(0.348) = 0.365 kW = 365 W not O.K.
Trumpler’s design criteria:
Not O.K.
0.0002 + 0.000 04(50/25.4) = 0.000 279 in > h0
Tmax = 237°F O.K.
Pst = 4000 kPa or 581 psi > 300 psi Not O.K.
n = 1, as done Not O.K.
_____________________________________________________________________________
12-18 So far, we have performed elements of the design task. Now let’s do it more completely.
The values of the unilateral tolerances, tb and td , reflect the routine capabilities of the
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bushing vendor and the in-house capabilities. While the designer has to live with these,
his approach should not depend on them. They can be incorporated later.
First we shall find the minimum size of the journal which satisfies Trumpler’s constraint
of Pst ≤ 300 psi.
W
Pst =
≤ 300
2dl ′
W
W
≤ 300 ⇒ d ≥
2
2d l ′ / d
600(l ′ / d )
d min =
900
= 1.73 in
2(300)(0.5)
In this problem we will take journal diameter as the nominal value and the bushing bore
as a variable. In the next problem, we will take the bushing bore as nominal and the
journal diameter as free.
To determine where the constraints are, we will set tb = td = 0, and thereby shrink the
design window to a point.
We set
d = 2.000 in
b = d + 2cmin = d + 2c
nd = 2 (This makes Trumpler’s nd ≤ 2 tight)
and construct a table.
c
b
d
Tf *
Tmax
h0 Pst Tmax n
0.0010
0.0011
0.0012
0.0013
0.0014
0.0015
0.0016
0.0017
0.0018
0.0019
0.0020
2.0020
2.0022
2.0024
2.0026
2.0028
2.0030
2.0032
2.0034
2.0036
2.0038
2.0040
2
2
2
2
2
2
2
2
2
2
2
215.50
206.75
198.50
191.40
185.23
179.80
175.00
171.13
166.92
163.50
160.40
312.0
293.0
277.0
262.8
250.4
239.6
230.1
220.3
213.9
206.9
200.6
×
×
×
×
×
×
×
×
√
×
√
√
√
√
√
√
√
√
√
√
√
√
√
√
√
√
√
√
√
√
√
√
√
√
√
√
√
√
√
√
√
√
√
√
fom
−5.74
−6.06
−6.37
−6.66
−6.94
−7.20
−7.45
−7.65
−7.91
−8.12
−8.32
*Sample calculation for the first entry of this column.
Iteration yields:
T f = 215.5°F
With T f = 215.5°F , from Table 12-1
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µ = 0.0136(10− 6 ) exp[1271.6 / (215.5 + 95)] = 0.817(10− 6 ) reyn
900
= 225 psi
N = 3000 / 60 = 50 rev/s, P =
4
2
−6
 1   0.817(10 )(50) 
S =
 
 = 0.182
225
 0.001  

From Figs. 12-16 and 12-18:
Eq. (12–24):
∫ = 0.7, f r/c = 5.5
0.0123(5.5)(0.182)(9002 )
= 191.6°F
[1 + 1.5(0.7 2 )](30)(14 )
191.6°F
= 215.8°F ≈ 215.5°F
Tav = 120°F +
2
∆TF =
For the nominal 2-in bearing, the various clearances show that we have been in contact
with the recurving of (h0)min. The figure of merit (the parasitic friction torque plus the
pumping torque negated) is best at c = 0.0018 in. For the nominal 2-in bearing, we will
place the top of the design window at cmin = 0.002 in, and b = d + 2(0.002) = 2.004 in. At
this point, add the b and d unilateral tolerances:
d = 2.000 +−0.000
b = 2.004 +−0.003
0.001 in,
0.000 in
Now we can check the performance at cmin , c , and cmax . Of immediate interest is the
fom of the median clearance assembly, − 9.82, as compared to any other satisfactory
bearing ensemble.
If a nominal 1.875 in bearing is possible, construct another table with tb = 0 and td = 0.
c
0.0020
0.0030
0.0035
0.0040
0.0050
0.0055
0.0060
b
1.879
1.881
1.882
1.883
1.885
1.886
1.887
d
1.875
1.875
1.875
1.875
1.875
1.875
1.875
Tf
157.2
138.6
133.5
130.0
125.7
124.4
123.4
Tmax h0 Pst Tmax n fom
√
√ − 7.36
194.30 × √
√
√ − 8.64
157.10 √ √
√
√
√
√ − 9.05
147.10
√
√ − 9.32
140.10 √ √
√
√
√
√ − 9.59
131.45
√
√ − 9.63
128.80 √ √
√
√
√ − 9.64
126.80 ×
The range of clearance is 0.0030 < c < 0.0055 in. That is enough room to fit in our
design window.
d = 1.875+−0.000
0.001 in,
b = 1.881+−0.003
0.000 in
The ensemble median assembly has a fom = − 9.31.
We just had room to fit in a design window based upon the (h0)min constraint. Further
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reduction in nominal diameter will preclude any smaller bearings. A table constructed for
a d = 1.750 in journal will prove this.
We choose the nominal 1.875-in bearing ensemble because it has the largest figure of
merit. Ans.
_____________________________________________________________________________
12-19 This is the same as Prob. 12-18 but uses design variables of nominal bushing bore b and
radial clearance c.
The approach is similar to that of Prob. 12-18 and the tables will change slightly. In the
table for a nominal b = 1.875 in, note that at c = 0.003 in the constraints are “loose.” Set
b = 1.875 in
d = 1.875 − 2(0.003) = 1.869 in
For the ensemble
b = 1.875+−0.003
0.001 in,
d = 1.869 +−0.000
0.001 in
Analyze at cmin = 0.003 in, c = 0.004 in and cmax = 0.005 in
At cmin = 0.003 in: T f = 138.4°F, µ′ = 3.160 µ reyn, S = 0.0297, H loss = 1035 Btu/h
and the Trumpler conditions are met.
At c = 0.004 in: T f = 130°F, µ′ = 3.872 µ reyn, S = 0.0205, Hloss = 1106 Btu/h, fom =
−9.246 and the Trumpler conditions are O.K.
At cmax = 0.005 in: T f = 125.68°F, µ′ = 4.325µ reyn, S = 0.014 66, Hloss = 1129 Btu/h
and the Trumpler conditions are O.K.
The ensemble figure of merit is slightly better; this bearing is slightly smaller. The
lubricant cooler has sufficient capacity.
_____________________________________________________________________________
12-20 Table 12-1:
µ (µ reyn) = µ 0 (106) exp [b / (T + 95)]
b and T in °F
The conversion from µ reyn to mPa⋅s is given on p. 612. For a temperature of C degrees
Celsius, T = 1.8 C + 32. Substituting into the above equation gives
µ (mPa⋅s) = 6.89 µ 0 (106) exp [b / (1.8 C + 32+ 95)]
= 6.89 µ 0 (106) exp [b / (1.8 C + 127)]
Ans.
For SAE 50 oil at 70°C, from Table 12-1, µ 0 = 0.0170 (10−6) reyn, and b = 1509.6°F.
From the equation,
µ = 6.89(0.0170) 10−6(106) exp {1509.6/[1.8(70) + 127]}
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= 45.7 mPa⋅s
Ans.
From Fig. 12-13, µ = 39 mPa⋅s
Ans.
The figure gives a value of about 15 % lower than the equation.
_____________________________________________________________________________
12-21 Originally
d = 2.000 +−0.000
b = 2.005+−0.003
0.001 in,
0.000 in
Doubled,
d = 4.000 +−0.000
b = 4.010 +−0.006
0.002 in,
0.000 in
The radial load quadrupled to 3600 lbf when the analyses for parts (a) and (b) were
carried out. Some of the results are:
Part
c
µ′
S
Tf
f r/c
Qs
h0 /c
∫
(a)
(b)
0.007
0.0035
3.416
3.416
0.0310
0.0310
135.1
135.1
0.1612
0.1612
6.56
0.870
0.1032
0.1032
0.897
0.897
Hloss
9898
1237
h0
0.000 722
0.000 361
Trumpler
h0
0.000 360
0.000 280
f
0.005 67
0.005 67
The side flow Qs differs because there is a c3 term and consequently an 8-fold increase.
Hloss is related by a 9898/1237 or an 8-fold increase. The existing h0 is related by a 2-fold
increase. Trumpler’s (h0)min is related by a 1.286-fold increase.
_____________________________________________________________________________
12-22 Given: Oiles SP 500 alloy brass bushing, L = 0.75 in, D = 0.75 in, T∞ = 70°F, F = 400 lbf,
N = 250 rev/min, and w = 0.004 in.
Table 12-8:
K = 0.6(10−10) in3⋅min/(lbf⋅ft⋅h)
P = F/ (DL) = 400/ [0.75(0.75)] = 711 psi
V = πDN/ 12 = π (0.75)250/12 = 49.1 ft/min
From Table 12-10, interpolation gives
V
33
49.1
100
f1
1.3
f1
1.8
= > f1 = 1.42
Table 12-11: f 2 = 1.0
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Table 12-12: PVmax = 46 700 psi⋅ft/min, Pmax = 3560 psi, Vmax = 100 ft/min
Pmax =
4 F
4 400
=
= 905 psi < 3560 psi O.K .
π DL π 0.752
PV = 711 (49.1) = 34 910 psi⋅ft/min < 46 700 psi⋅ft/min
O.K.
Eq. (12-32) can be written as
w = f1 f 2 K
4 F
Vt
π DL
Solving for t,
t =
π DLw
4 f1 f 2 KVF
=
π ( 0.75 ) 0.75 ( 0.004 )
4 (1.42 )1.0 ( 0.6 )10−10 ( 49.1) 400
= 1056 h = 1056 ( 60 ) = 63 400 min
Cycles = Nt = 250 (63 400) = 15.9 (106) cycles
Ans.
_____________________________________________________________________________
12-23 Given: Oiles SP 500 alloy brass bushing, wmax = 0.002 in for 1000 h, N = 200 rev/min,
F = 100 lbf, hCR = 2.7 Btu/(h∙ft2∙oF), Tmax = 300 oF, f s = 0.03, and nd = 2.
Using Eq. (12-38) with nd F for F, fs = 0.03 from Table 12-9, and hCR = 2.7
Btu/(h ⋅ ft2 ⋅ οF), gives
L≥
720 f s nd FN
720(0.03)2(100)200
=
= 1.79 in
778(2.7)(300 − 70)
J h CR (T f − T∞ )
From Table 12-13, the smallest available bushing has an ID = 1 in, OD = 1 83 in, and L =
2 in. With L/D = 2/1 = 2, this is inside of the recommendations of Eq. (12-33). Thus, for
the first trial, try the bushing with ID = 1 in, OD = 1 83 in, and L = 2 in. Thus,
Eq. (12-31):
Pmax =
P=
4 nd F
4 2(100)
=
= 127.3psi < 3560 psi (OK)
π DL
π 1(2)
nd F
2(100)
=
= 100 psi
1(2)
DL
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Eq. (12-29):
V =
π DN
12
=
π (1) 200
12
= 52.4 ft/min < 100 ft/min (OK)
PV = 100(52.4) = 5240 psi ⋅ ft/min < 46 700 psi ⋅ ft/min
(OK)
From Table 12-10, interpolation gives
V
33
52.4
100
f1
1.3
f1
1.8
= > f1 = 1.445
Eq. (12-32), with Tables 12-8 and 12-10:
w=
Answer
−11
f1 f 2 Knd FN t 1.445(1) 6 (10 ) 2 (100) 200 (1000)
=
= 0.000 578in < 0.001 in (OK)
3L
3(2)
Select ID = 1 in, OD = 1 83 in, and L = 2 in.
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Chapter 13
d P = 17 / 8 = 2.125 in
N
1120
dG = 2 d P =
( 2.125) = 4.375 in
544
N3
13-1
NG = PdG = 8 ( 4.375) = 35 teeth
Ans.
C = ( 2.125 + 4.375) / 2 = 3.25 in Ans.
______________________________________________________________________________
nG = 1600 (15 / 60 ) = 400 rev/min
p = π m = 3π mm Ans.
13-2
Ans.
C = 3 (15 + 60 )  2 = 112.5 mm Ans.
______________________________________________________________________________
NG = 16 ( 4 ) = 64 teeth
13-3
Ans.
dG = N G m = 64 ( 6 ) = 384 mm
d P = N P m = 16 ( 6 ) = 96 mm
Ans.
Ans.
C = ( 384 + 96 ) / 2 = 240 mm Ans.
______________________________________________________________________________
13-4
Mesh:
a = 1/ P = 1/ 3 = 0.3333 in Ans.
b = 1.25 / P = 1.25 / 3 = 0.4167 in Ans.
c = b − a = 0.0834 in Ans.
p = π / P = π / 3 = 1.047 in Ans.
t = p / 2 = 1.047 / 2 = 0.523 in Ans.
Pinion Base-Circle:
d1 = N1 / P = 21/ 3 = 7 in
d1b = 7 cos 20o = 6.578 in
Gear Base-Circle:
Ans.
d 2 = N 2 / P = 28 / 3 = 9.333 in
d 2b = 9.333cos 20o = 8.770 in
Base pitch:
Ans.
pb = pc cos φ = (π / 3) cos 20o = 0.984 in
Ans.
mc = Lab / pb = 1.53 / 0.984 = 1.55 Ans.
Contact Ratio:
See the following figure for a drawing of the gears and the arc lengths.
Shigley’s MED, 10th edition
Chapter 13 Solutions, Page 1/36
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______________________________________________________________________________
13-5
1/2
(a)
 14 / 6 2  32 / 6 2 
AO = 
 +
 
 2   2  
(b)
γ = tan −1 (14 / 32 ) = 23.63o
= 2.910 in
Ans.
Ans.
Γ = tan −1 ( 32 /14 ) = 66.37o
(c)
d P = 14 / 6 = 2.333 in
Ans.
dG = 32 / 6 = 5.333 in
Ans.
Ans.
Shigley’s MED, 10th edition
Chapter 13 Solutions, Page 2/36
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(d) From Table 13-3, 0.3AO = 0.3(2.910) = 0.873 in and 10/P = 10/6 = 1.67
0.873 < 1.67 ∴ F = 0.873 in Ans.
______________________________________________________________________________
13-6
(a)
pn = π / Pn = π / 4 = 0.7854 in
pt = pn / cosψ = 0.7854 / cos 30o = 0.9069 in
p x = pt / tanψ = 0.9069 / tan 30o = 1.571 in
pnb = pn cos φn = 0.7854 cos 25o = 0.7380 in
(b)
Eq. (13-7):
(c)
Pt = Pn cosψ = 4 cos 30o = 3.464 teeth/in
Ans.
φt = tan −1 ( tan φn / cosψ ) = tan −1 (tan 25o / cos30o ) = 28.3o Ans.
Table 13-4:
a = 1/ 4 = 0.250 in Ans.
b = 1.25 / 4 = 0.3125 in Ans.
20
dP =
= 5.774 in Ans.
4 cos 30o
36
dG =
= 10.39 in Ans.
4 cos 30o
______________________________________________________________________________
(d)
13-7
N P = 19 teeth, N G = 57 teeth, φn = 20o , mn = 2.5 mm
(a)
pn = π mn = π ( 2.5) = 7.854 mm
Ans.
pn
7.854
=
= 9.069 mm Ans.
cosψ cos 30o
pt
9.069
px =
=
= 15.71 mm Ans.
tanψ tan 30o
mn
2.5
mt =
=
= 2.887 mm Ans.
cosψ cos 30o
pt =
(b)
Shigley’s MED, 10th edition
Chapter 13 Solutions, Page 3/36
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 tan 20o 
= 22.80o
o 
cos
30


φt = tan −1 
(c)
a = mn = 2.5 mm
Ans.
Ans.
b = 1.25mn = 1.25 ( 2.5) = 3.125 mm
dP =
Ans.
N
= Nmt = 19 ( 2.887 ) =54.85 mm
Pt
Ans.
dG = 57 ( 2.887 ) = 164.6 mm Ans.
______________________________________________________________________________
13-8
(a) Using Eq. (13-11) with k = 1, φ = 20º, and m = 2,
NP =
=
(
2k
m + m 2 + (1 + 2m ) sin 2 φ
(1 + 2m ) sin 2 φ
{( 2) +
1 + 2 ( 2 )  sin ( 20 )
2 (1)
2
o
( 2)
2
)
( )} = 14.16 teeth
+ 1 + 2 ( 2 )  sin 2 20o
Round up for the minimum integer number of teeth.
NP = 15 teeth
Ans.
(b) Repeating (a) with m = 3, NP = 14.98 teeth. Rounding up, NP = 15 teeth. Ans.
(c) Repeating (a) with m = 4, NP = 15.44 teeth. Rounding up, NP = 16 teeth. Ans.
(d) Repeating (a) with m = 5, NP = 15.74 teeth. Rounding up, NP = 16 teeth. Ans.
Alternatively, a useful table can be generated to determine the largest gear that can mesh
with a specified pinion, and thus also the maximum gear ratio with a specified pinion.
The Max NG column was generated using Eq. (13-12) with k = 1, φ = 20º, and rounding
up to the next integer.
Min NP
13
14
15
16
17
18
Max NG
16
26
45
101
1309
unlimited
Max m = Max NG / Min NP
1.23
1.86
3.00
6.31
77.00
unlimited
With this table, we can readily see that gear ratios up to 3 can be obtained with a
minimum NP of 15 teeth, and gear ratios up to 6.31 can be obtained with a minimum NP
of 16 teeth. This is consistent with the results previously obtained.
______________________________________________________________________________
Shigley’s MED, 10th edition
Chapter 13 Solutions, Page 4/36
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13-9
Repeating the process shown in the solution to Prob. 13-8, except with φ = 25º, we obtain
the following results.
(a) For m = 2, NP = 9.43 teeth. Rounding up, NP = 10 teeth. Ans.
(b) For m = 3, NP = 9.92 teeth. Rounding up, NP = 10 teeth. Ans.
(c) For m = 4, NP = 10.20 teeth. Rounding up, NP = 11 teeth. Ans.
(d) For m = 5, NP = 10.38 teeth. Rounding up, NP = 11 teeth. Ans.
For convenient reference, we will also generate the table from Eq. (13-12) for φ = 25º.
Min NP
Max NG
Max m = Max NG / Min NP
9
13
1.44
10
32
3.20
11
249
22.64
12
unlimited
unlimited
______________________________________________________________________________
13-10 (a) The smallest pinion tooth count that will run with itself is found from Eq. (13-10).
NP ≥
≥
(
2k
1 + 1 + 3sin 2 φ
2
3sin φ
2 (1)
)
)
(
1 + 1 + 3sin 2 20o
3sin 2 20o
≥ 12.32 → 13 teeth Ans.
(b) The smallest pinion that will mesh with a gear ratio of mG = 2.5, from Eq. (13-11) is
NP ≥
≥
(
2k
m + m 2 + (1 + 2m ) sin 2 φ
2
(1 + 2m ) sin φ
2 (1)
{2.5 +
1 + 2 ( 2.5 )  sin 20
2
≥ 14.64
→
2.52 + 1 + 2 ( 2.5 )  sin 2 20o
o
15 teeth
)
}
Ans.
The largest gear-tooth count possible to mesh with this pinion, from Eq. (13-12) is
NG ≤
≤
N P2 sin 2 φ − 4k 2
4k − 2 N P sin 2 φ
152 sin 2 20o − 4 (1)
2
4 (1) − 2 (15 ) sin 2 20o
≤ 45.49 → 45 teeth
Ans.
(c) The smallest pinion that will mesh with a rack, from Eq. (13-13),
Shigley’s MED, 10th edition
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2 (1)
2k
=
2
sin φ sin 2 20o
≥ 17.097 → 18 teeth Ans.
______________________________________________________________________________
NP ≥
13-11 φn = 20o ,ψ = 30o
From Eq. (13-19), φt = tan −1 ( tan 20o / cos 30o ) = 22.80o
(a) The smallest pinion tooth count that will run with itself, from Eq. (13-21) is
NP ≥
≥
(
2k cosψ
1 + 1 + 3sin 2 φt
2
3sin φt
2 (1) cos 30o
(1 +
3sin 22.80
2
o
)
1 + 3sin 2 22.80o
≥ 8.48 → 9 teeth
)
Ans.
(b) The smallest pinion that will mesh with a gear ratio of m = 2.5, from Eq. (13-22) is
2 (1) cos 30o
NP ≥
2.5 + 2.52 + 1 + 2 ( 2.5 )  sin 2 22.80o
1 + 2 ( 2.5 )  sin 2 22.80o
≥ 9.95 → 10 teeth Ans.
{
}
The largest gear-tooth count possible to mesh with this pinion, from Eq. (13-23) is
NG ≤
≤
N P2 sin 2 φt − 4k 2 cos 2 ψ
4k cosψ − 2 N P sin 2 φt
102 sin 2 22.80o − 4 (1) cos 2 30o
4 (1) cos 2 30o − 2 ( 20 ) sin 2 22.80o
≤ 26.08 → 26 teeth
Ans.
(c) The smallest pinion that will mesh with a rack, from Eq. (13-24) is
NP ≥
o
2k cosψ 2 (1) cos 30
=
sin 2 φt
sin 2 22.80o
≥ 11.53 → 12 teeth Ans.
______________________________________________________________________________
13-12 From Eq. (13-19),
 tan φn
 cosψ
φt = tan −1 
o

−1  tan 20 
=
= 22.796o
tan


o 
cos
30



Shigley’s MED, 10th edition
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Program Eq. (13-23) on a computer using a spreadsheet or code, and increment NP. The
first value of NP that can be doubled is NP = 10 teeth, where NG ≤ 26.01 teeth. So NG = 20
teeth will work. Higher tooth counts will work also, for example 11:22, 12:24, etc.
Use NP = 10 teeth, NG = 20 teeth
Ans.
Note that the given diametral pitch (tooth size) is not relevant to the interference problem.
______________________________________________________________________________
o
 tan φn 
−1  tan 20 
=
= 27.236o
tan


o 
ψ
cos
cos
45




Program Eq. (13-23) on a computer using a spreadsheet or code, and increment NP. The
first value of NP that can be doubled is NP = 6 teeth, where NG ≤ 17.6 teeth. So NG = 12
teeth will work. Higher tooth counts will work also, for example 7:14, 8:16, etc.
13-13 From Eq. (13-19),
φt = tan −1 
Use NP = 6 teeth, NG = 12 teeth Ans.
______________________________________________________________________________
13-14 The smallest pinion that will operate with a rack without interference is given by Eq. (1313).
2k
NP =
sin 2 φ
Setting k = 1 for full depth teeth, NP = 9 teeth, and solving for φ,
2 (1)
2k
= sin −1
= 28.126o
Ans.
NP
9
______________________________________________________________________________
φ = sin −1
13-15
(a)
(b)
Eq. (13-3):
pn = π mn = 3π mm
Eq. (13-16):
pt = pn / cosψ = 3π / cos 25o = 10.40 mm
Eq. (13-17):
p x = pt / tanψ = 10.40 / tan 25 = 22.30 mm
Eq. (13-3):
mt = pt / π = 10.40 / π = 3.310 mm
Eq. (13-19):
Ans.
o
φt = tan −1
Ans.
Ans.
Ans.
tan φn
tan 20
= tan −1
= 21.88o
cosψ
cos 25o
o
Shigley’s MED, 10th edition
Ans.
Chapter 13 Solutions, Page 7/36
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Eq. (13-2):
dp = mt Np = 3.310 (18) = 59.58 mm
Ans.
Ans.
Eq. (13-2):
dG = mt NG = 3.310 (32) = 105.92 mm
______________________________________________________________________________
(c)
13-16 (a) Sketches of the figures are shown to determine
the axial forces by inspection.
The axial force of gear 2 on shaft a is in the
negative z-direction. The axial force of gear 3 on
shaft b is in the positive z-direction. Ans.
The axial force of gear 4 on shaft b is in the
positive z-direction. The axial force of gear 5 on
shaft c is in the negative z-direction. Ans.
12  16 
  ( 700 ) = +77.78 rev/min ccw
48  36 
(b)
nc = n5 =
(c)
d P 2 = 12 / 12 cos 30o = 1.155 in
dG 3
Ans.
(
)
= 48 / (12 cos 30 ) = 4.619 in
o
1.155 + 4.619
= 2.887 in Ans.
2
= 16 / ( 8 cos 25o ) = 2.207 in
Cab =
dP4
(
)
d G 5 = 36 / 8 cos 25o = 4.965 in
Cbc = 3.586 in Ans.
______________________________________________________________________________
20  8   20  4
   =
40  17   60  51
4
nd = ( 600 ) = 47.06 rev/min cw Ans.
51
______________________________________________________________________________
13-17
e=
6  18   20   3 
3
    =
10  38   48   36  304
3
n9 =
(1200 ) = 11.84 rev/min cw Ans.
304
______________________________________________________________________________
13-18
e=
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13-19 (a)



 Ans.
cw about x, as viewed from the positive x axis 


o
d P = 12 / ( 8 cos 23 ) = 1.630 in
nc =
(b)
12 1
⋅ ( 540 ) = 162 rev/min
40 1
(
)
d G = 40 / 8 cos 23o = 5.432 in
d P + dG
= 3.531 in
2
Ans.
N 32
=
= 8 in
Ans.
P
4
______________________________________________________________________________
(c)
d=
13-20 Applying Eq. (13-30), e = (N2 / N3) (N4 / N5) = 45. For an exact ratio, we will choose to
factor the train value into integers, such that
N2 / N3 = 9
N4 / N5 = 5
(1)
(2)
Assuming a constant diametral pitch in both stages, the geometry condition to satisfy the
in-line requirement of the compound reverted configuration is
N2 + N3 = N4 + N5
(3)
With three equations and four unknowns, one free choice is available. It is necessary that
all of the unknowns be integers. We will use a normalized approach to find the minimum
free choice to guarantee integers; that is, set the smallest gear of the largest stage to unity,
thus N3 = 1. From (1), N2 = 9. From (3),
N2 + N3 = 9 + 1 = 10 = N4 + N5
Substituting N4 = 5 N5 from (2) gives
10 = 5 N5 + N5 = 6 N5
N5 = 10 / 6 = 5 / 3
To eliminate this fraction, we need to multiply the original free choice by a multiple of 3.
In addition, the smallest gear needs to have sufficient teeth to avoid interference. From
Eq. (13-11) with k = 1, φ = 20°, and m = 9, the minimum number of teeth on the pinion to
avoid interference is 17. Therefore, the smallest multiple of 3 greater than 17 is 18.
Setting N3 = 18 and repeating the solution of equations (1), (2), and (3) yields
Shigley’s MED, 10th edition
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N2 = 162 teeth
N3 = 18 teeth
N4 = 150 teeth
N5 = 30 teeth
Ans.
______________________________________________________________________________
13-21 The solution to Prob. 13-20 applies up to the point of determining the minimum number
of teeth to avoid interference. From Eq. (13-11), with k = 1, φ = 25°, and m = 9, the
minimum number of teeth on the pinion to avoid interference is 11. Therefore, the
smallest multiple of 3 greater than 11 is 12. Setting N3 = 12 and repeating the solution of
equations (1), (2), and (3) of Prob. 13-20 yields
N2 = 108 teeth
N3 = 12 teeth
N4 = 100 teeth
Ans.
N5 = 20 teeth
______________________________________________________________________________
13-22 Applying Eq. (13-30), e = (N2 / N3) (N4 / N5) = 30. For an exact ratio, we will choose to
factor the train value into integers, such that
N2 / N3 = 6
N4 / N5 = 5
(1)
(2)
Assuming a constant diametral pitch in both stages, the geometry condition to satisfy the
in-line requirement of the compound reverted configuration is
N2 + N3 = N4 + N5
(3)
With three equations and four unknowns, one free choice is available. It is necessary that
all of the unknowns be integers. We will use a normalized approach to find the minimum
free choice to guarantee integers; that is, set the smallest gear of the largest stage to unity,
thus N3 = 1. From (1), N2 = 6. From (3),
N2 + N3 = 6 + 1 = 7 = N4 + N5
Substituting N4 = 5 N5 from (2) gives
7 = 5 N5 + N5 = 6 N5
N5 = 7 / 6
To eliminate this fraction, we need to multiply the original free choice by a multiple of 6.
In addition, the smallest gear needs to have sufficient teeth to avoid interference. From
Eq. (13-11) with k = 1, φ = 20°, and m = 6, the minimum number of teeth on the pinion to
avoid interference is 16. Therefore, the smallest multiple of 6 greater than 16 is 18.
Setting N3 = 18 and repeating the solution of equations (1), (2), and (3) yields
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Chapter 13 Solutions, Page 10/36
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N2 = 108 teeth
N3 = 18 teeth
N4 = 105 teeth
N5 = 21 teeth
Ans.
______________________________________________________________________________
13-23 Applying Eq. (13-30), e = (N2 / N3) (N4 / N5) = 45. For an approximate ratio, we will
choose to factor the train value into two equal stages, such that
N 2 / N3 = N 4 / N5 = 45
If we choose identical pinions such that interference is avoided, both stages will be
identical and the in-line geometry condition will automatically be satisfied. From Eq.
(13-11) with k = 1, φ = 20°, and m = 45 , the minimum number of teeth on the pinions
to avoid interference is 17. Setting N3 = N5 = 17, we get
N 2 = N 4 = 17 45 = 114.04 teeth
Rounding to the nearest integer, we obtain
N2 = N4 = 114 teeth
N3 = N5 = 17 teeth
Ans.
Checking, the overall train value is e = (114 / 17) (114 / 17) = 44.97.
______________________________________________________________________________
13-24 H = 25 hp, ωi = 2500 rev/min
Let ωo = 300 rev/min for minimal gear ratio to minimize gear size.
ωo 300
1
N N
=
=
= 2 4
ωi 2500 8.333 N 3 N 5
Let
N2 N4
1
1
=
=
=
8.333 2.887
N3 N5
From Eq. (13-11) with k = 1, φ = 20°, and m = 2.887, the minimum number of teeth on
the pinions to avoid interference is 15.
Let
N2 = N4 = 15 teeth
N3 = N5 = 2.887(15) = 43.31 teeth
Try N3 = N5 = 43 teeth.
Shigley’s MED, 10th edition
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 15   15 
   ( 2500 ) = 304.2
 43   43 
ωo = 
Too big. Try N3 = N5 = 44.
 15   15 
ωo =     ( 2500 ) = 290.55 rev/min
 44   44 
N2 = N4 = 15 teeth, N3 = N5 = 44 teeth Ans.
______________________________________________________________________________
13-25 (a) The planet gears act as keys and the wheel speeds are the same as that of the ring
gear. Thus,
nA = n3 = 900 (16 / 48 ) = 300 rev/min Ans.
(b)
nF = n5 = 0, nL = n6 , e = −1
n − 300
−1 = 6
0 − 300
300 = n6 − 300
n6 = 600 rev/min
Ans.
(c) The wheel spins freely on icy surfaces, leaving no traction for the other wheel.
The car is stalled. Ans.
______________________________________________________________________________
13-26 (a) The motive power is divided equally among four wheels instead of two.
(b) Locking the center differential causes 50 percent of the power to be applied to the
rear wheels and 50 percent to the front wheels. If one of the rear wheels rests on a
slippery surface such as ice, the other rear wheel has no traction. But the front wheels
still provide traction, and so you have two-wheel drive. However, if the rear
differential is locked, you have 3-wheel drive because the rear-wheel power is now
distributed 50-50.
______________________________________________________________________________
13-27 Let gear 2 be first, then nF = n2 = 0. Let gear 6 be last, then nL = n6 = –12 rev/min.
20  16  16 nL − nA
=
 =
30  34  51 nF − nA
16
( 0 − n A ) = −12 − nA
51
e=
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−12
= −17.49 rev/min
35 / 51



 Ans.
(negative indicates cw as viewed from the bottom of the figure) 


nA =
______________________________________________________________________________
13-28 Let gear 2 be first, then nF = n2 = 0 rev/min. Let gear 6 be last, then nL = n6 = 85 rev/min.
20  16  16 nL − nA
=
 =
30  34  51 nF − nA
16
( 0 − n A ) = ( 85 − nA )
51
 16 
− nA   + nA = 85
 51 
 16 
nA  1 −  = 85
 51 
85
nA =
= 123.9 rev/min
16
1−
51
e=
The positive sign indicates the same direction as n6. ∴ nA = 123.9 rev/min ccw Ans.
______________________________________________________________________________
13-29 The geometry condition is d5 / 2 = d 2 / 2 + d3 + d 4 . Since all the gears are meshed, they
will all have the same diametral pitch. Applying d = N / P,
N 5 / (2 P) = N 2 / (2 P) + N 3 / P + N 4 / P
N5 = N 2 + 2 N3 + 2 N 4 = 12 + 2 (16 ) + 2 (12 ) = 68 teeth
Ans.
Let gear 2 be first, nF = n2 = 320 rev/min. Let gear 5 be last, nL = n5 = 0 rev/min.
e=
12  16  12  3 nL − nA
   = =
16  12  68  17 nF − nA
320 − nA =
nA = −
17
( 0 − nA )
3
3
( 320 ) = −68.57 rev/min
14
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The negative sign indicates opposite of n2.
∴ nA = 68.57 rev/min cw Ans.
______________________________________________________________________________
13-30 Let nF = n2, then nL = n7 = 0.
e=−
nL − n5
20  16   36 
    = −0.5217 =
16  30   46 
nF − n5
0 − n5
= −0.5217
10 − n5
−0.5217 (10 − n5 ) = −n5
−5.217 + 0.5217 n5 + n5 = 0
n5 (1 + 0.5217 ) = 5.217
5.217
1.5217
n5 = nb = 3.428 turns in same direction
Ans.
______________________________________________________________________________
n5 =
13-31 (a)
ω = 2π n / 60
H = T ω = 2π Tn / 60
(T in N ⋅ m, H in W)
( )
60 H 103
So
T=
So
F32t =
2π n
(H in kW, n in rev/min)
= 9550 H / n
9550 ( 75 )
Ta =
= 398 N ⋅ m
1800
mN 2 5 (17 )
r2 =
=
= 42.5 mm
2
2
Ta 398
=
= 9.36 kN
r2 42.5
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F3b = − Fb 3 = 2 ( 9.36 ) = 18.73 kN in the positive x-direction.
Ans.
mN 4 5 ( 51)
=
= 127.5 mm
2
2
Tc 4 = 9.36 (127.5) = 1193 N ⋅ m ccw
r4 =
(b)
∴T4c = 1193 N ⋅ m cw
Ans.
Note: The solution is independent of the pressure angle.
______________________________________________________________________________
N N
13-32
d= =
P 6
d 2 = 4 in, d 4 = 4 in, d 5 = 6 in, d 6 = 24 in
 24   24   36 
e =  − −  +
 =1/ 6
 24   36   144 
nF = n2 = 1000 rev/min
nL = n6 = 0
n − nA
0 − nA
1
e= L
=
=
nF − nA 1000 − nA 6
nA = −200 rev/min
Noting that power equals torque times angular velocity, the input torque is
T2 =
 550 lbf ⋅ ft/s   60 s  1 rev  12 in 
H
25 hp
=




 = 1576 lbf ⋅ in
hp
n2 1000 rev/min 
  min  2π rad  ft 
For 100 percent gear efficiency, the output power equals the input power, so
Tarm =
H
25 hp  550 lbf ⋅ ft/s   60 s   1 rev  12 in 
=




 = 7878 lbf ⋅ in
nA 200 rev/min 
hp
  min   2π rad  ft 
Next, we’ll confirm the output torque as we work through the force analysis and
complete the free body diagrams.
Gear 2
1576
= 788 lbf
2
F32r = 788 tan 20o = 287 lbf
Wt =
Shigley’s MED, 10th edition
Chapter 13 Solutions, Page 15/36
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Gear 4
FA4 = 2W t = 2 ( 788) = 1576 lbf
Gear 5
Arm
Tout = 1576 ( 9 ) − 1576 ( 4 ) = 7880 lbf ⋅ in
Ans.
______________________________________________________________________________
13-33 Given: m = 12 mm, nP = 1800 rev/min cw,
N2 = 18T, N3 = 32T, N4 = 18T, N5 = 48T
Pitch Diameters:
d2 = 18(12) = 216 mm, d3 = 32(12) = 384 mm,
d4 = 18(12) = 216 mm, d5 = 48(12) = 576 mm
Gear 2
From Eq. (13-36),
60 000 (150 )
60 000 H
Wt =
=
= 7.368 kN
π dn
π ( 216 )(1800 )
d 
 216 
Ta 2 = Wt  2  = 7.368 
 = 795.7 N ⋅ m
 2 
 2 
W r = 7.368 tan 20o = 2.682 kN
Shigley’s MED, 10th edition
Chapter 13 Solutions, Page 16/36
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Gears 3 and 4
( 384)
 216 
Wt 
 = 7.368
2
 2 
t
W = 13.10 kN
W r = 13.10 tan 20o = 4.768 kN
Ans.
______________________________________________________________________________
13-34 Given: P = 5 teeth/in, N2 = 18T, N3 = 45T,
φn = 20o , H = 32 hp, n2 = 1800 rev/min
Gear 2
Tin =
63025 ( 32 )
1800
= 1120 lbf ⋅ in
18
= 3.600 in
5
45
dG =
= 9.000 in
5
1120
W32t =
= 622 lbf
3.6 / 2
W32r = 622 tan 20o = 226 lbf
dP =
Fat2 = W32t = 622 lbf,
(
Fa 2 = 622 2 + 226 2
)
1/2
Far2 = W32r = 226 lbf
= 662 lbf
Each bearing on shaft a has the same radial load of RA = RB = 662/2 = 331 lbf.
Gear 3
W23t = W32t = 622 lbf
W23r = W32r = 226 lbf
Fb 3 = Fb 2 = 662 lbf
RC = RD = 662 / 2 = 331 lbf
Shigley’s MED, 10th edition
Chapter 13 Solutions, Page 17/36
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Each bearing on shaft b has the same radial load which is equal to the radial load of
bearings A and B. Thus, all four bearings have the same radial load of 331 lbf. Ans.
______________________________________________________________________________
13-35 Given: P = 4 teeth/in,
φn = 20o ,
NP = 20T,
n2 = 900 rev/min
N P 20
=
= 5.000 in
P
4
63025 ( 30 )( 2 )
Tin =
= 4202 lbf ⋅ in
900
W32t = Tin / ( d 2 / 2 ) = 4202 / ( 5 / 2 ) = 1681 lbf
d2 =
W32r = 1681 tan 20o = 612 lbf
The motor mount resists the equivalent forces and torque.
The radial force due to torque is
Fr =
4202
= 150 lbf
14 ( 2 )
Forces reverse with rotational sense as
torque reverses.
The compressive loads at A and D are absorbed by the base plate, not the bolts. For W32t ,
the tensions in C and D are
Shigley’s MED, 10th edition
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ΣM AB = 0
1681( 4.875 + 15.25) − 2 F (15.25) = 0
F = 1109 lbf
If W32t reverses, 15.25 in changes to 13.25 in, 4.815 in changes to 2.875 in, and the forces
change direction. For A and B,
1681( 2.875) − 2 F1 (13.25 ) = 0 ⇒ F1 = 182.4 lbf
r
32
For W ,
M = 612 ( 4.875 + 11.25 / 2 ) = 6426 lbf ⋅ in
a=
(14 / 2 ) + (11.25 / 2 )
F2 =
6426
= 179 lbf
4 ( 8.98)
2
2
= 8.98 in
At C and D, the shear forces are:
FS 1 = 153 + 179 ( 5.625 / 8.98 )  + 179 ( 7 / 8.98 ) 
2
2
At A and B, the shear forces are:
Shigley’s MED, 10th edition
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FS 2 = 153 − 179 ( 5.625 / 8.98 )  + 179 ( 7 / 8.98 ) 
= 145 lbf
2
2
The shear forces are independent of the rotational sense.
The bolt tensions and the shear forces for cw rotation are,
For ccw rotation,
______________________________________________________________________________
13-36 (a)
N2 = N4 = 15 teeth,
P=
N
d
⇒ d=
N3 = N5 = 44 teeth
N
P
15
= 2.5 in Ans.
6
44
d3 = d5 =
= 7.33 in Ans.
6
d2 = d4 =
(b)
(c)
Vi = V2 = V3 =
π d 2 n2
π ( 2.5 )( 2500 )
= 1636 ft/min Ans.
12
π d n π ( 2.5) ( 2500 )(15 / 44 )
= 558 ft/min
Vo = V4 = V5 = 4 4 =
12
12
Input gears:
Wti = 33000
12
=
H 33000 ( 25 )
=
= 504.3 lbf = 504 lbf
Vi
1636
Wri = Wti tan φ = 504.3 tan 20o = 184 lbf
W
504.3
Wi = ti =
= 537 lbf
Ans.
cos φ cos 20o
Ans.
Ans.
Ans.
Shigley’s MED, 10th edition
Chapter 13 Solutions, Page 20/36
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Output gears:
H 33000 ( 25 )
= 1478 lbf
Wto = 33000 =
Vo
558
Ans.
Wro = Wto tan φ = 1478 tan 20o = 538 lbf Ans.
Wto
1478
Wo =
=
= 1573 lbf Ans.
cos 20o cos 20o
(d)
d 
 2.5 
Ti = Wti  2  = 504.3 
 = 630 lbf ⋅ in
 2 
 2 
2
Ans.
2
 44 
 44 
To = Ti   = 630   = 5420 lbf ⋅ in Ans.
 15 
 15 
______________________________________________________________________________
(e)
13-37 H = 35 hp, ni = 1200 rev/min, φ =20o
N 2 = N 4 = 16 teeth, N 3 = N 5 = 48 teeth, P = 10 teeth/in
N
16
(a)
nintermediate = n3 = n4 = 2 ni = (1200 ) = 400 rev/min
N3
48
(b)
no =
N2 N4
16  16 
ni =   (1200 ) = 133.3 rev/min
N3 N5
48  48 
P=
N
d
⇒ d=
Ans.
Ans.
N
P
16
= 1.6 in Ans.
10
48
d3 = d5 =
= 4.8 in Ans.
10
π d n π (1.6 )(1200 )
Vi = V2 = V3 = 2 2 =
= 502.7 ft/min Ans.
12
12
π d n π (1.6 )( 400 )
Vo = V4 = V5 = 4 4 =
= 167.6 ft/min Ans.
12
12
d2 = d4 =
(c)
Wti = 33000
H 33000 ( 35 )
=
= 2298 lbf lbf
Vi
502.7
Wri = Wti tan φ = 2298 tan 20o = 836.4 lbf
W
2298
Wi = ti =
= 2445 lbf
Ans.
cos φ cos 20o
Ans.
Ans.
Shigley’s MED, 10th edition
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H 33000 ( 35 )
=
= 6891 lbf
Vo
167.6
Wto = 33000
Ans.
Wro = Wto tan φ = 6891tan 20o = 2508 lbf Ans.
Wto
6891
Wo =
=
= 7333 lbf Ans.
o
cos 20
cos 20o
(d)
d 
 1.6 
Ti = Wti  2  = 2298 
 = 1838 lbf ⋅ in
 2 
 2 
2
Ans.
2
 48 
 48 
To = Ti   = 1838   = 16 540 lbf ⋅ in Ans.
 16 
 16 
______________________________________________________________________________
ω
2
13-38 (a)
For o = , from Eq. (13-11), with m = 2, k = 1, φ = 20o
ωi 1
(e)
NP =
2 (1)
{2 +
1 + 2 ( 2 )  sin 20
So
N P min = 15
(b)
P=
2
o
}
22 + 1 + 2 ( 2 )  sin 2 20o = 14.16
Ans.
N 15
=
= 1.875 teeth/in
d
8
Ans.
(c) To transmit the same power with no change in pitch diameters, the speed and
transmitted force must remain the same.
For A, with φ = 20°,
WtA = FA cos20° = 300 cos20° = 281.9 lbf
For A, with φ = 25°, same transmitted load,
FA = WtA/cos25° = 281.9/cos25° = 311.0 lbf
Ans.
Summing the torque about the shaft axis,
d 
d 
WtA  A  = WtB  B 
 2 
 2 
( d / 2 ) = W  d A  = 281.9  20  = 704.75 lbf
WtB = WtA A
(
) 
( d B / 2 ) tA  d B 
 8 
WtB
704.75
=
= 777.6 lbf Ans.
o
cos 25
cos 25o
______________________________________________________________________________
FB =
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13-39 (a)
For
ωo 5
= , from Eq. (13-11), with m = 5, k = 1, φ = 20o
ωi 1
2 (1)
NP =
5 + 52 + 1 + 2 ( 5)  sin 2 25o = 10.4
2
o
1 + 2 ( 5)  sin 25
{
So
N P min = 11
(b)
m=
}
Ans.
d 300
=
= 27.3 mm/tooth
N
11
Ans.
(c) To transmit the same power with no change in pitch diameters, the speed and
transmitted force must remain the same.
For A, with φ = 20°,
WtA = FA cos20° = 11 cos20° = 10.33 kN
For A, with φ = 25°, same transmitted load,
FA = WtA/cos25° = 10.33 / cos 25° = 11.40 kN
Ans.
Summing the torque about the shaft axis,
d 
d 
WtA  A  = WtB  B 
2
 
 2 
d 
( d / 2)
 600 
WtB = WtA A
= WtA  A  = (11.40 ) 
 = 22.80 kN
( dB / 2)
 300 
 dB 
WtB
22.80
FB =
=
= 25.16 kN Ans.
o
cos 25 cos 25o
______________________________________________________________________________
13-40 (a) Using Eq. (13-11) with k = 1, φ = 20º, and m = 2,
NP =
=
(
2k
m + m 2 + (1 + 2m ) sin 2 φ
2
1
2
m
sin
+
φ
(
)
{( 2) +
1 + 2 ( 2 )  sin ( 20 )
2 (1)
2
o
( 2)
2
)
( )} = 14.16 teeth
+ 1 + 2 ( 2 )  sin 2 20o
Round up for the minimum integer number of teeth.
NF = 15 teeth, NC = 30 teeth
Ans.
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d 125
=
= 8.33 mm/tooth Ans.
N 15
2 kW
H
 1000 W   rev   60 s 
=
= 100 N ⋅ m
T=
ω 191 rev/min  kW   2π rad   min 
m=
(b)
(c)
(d)
From Eq. (13-36),
Wt =
60 000 ( 2 )
60 000 H
=
= 1.60 kN = 1600 N
π dn
π (125 )(191)
Ans.
Or, we could have obtained Wt directly from the torque and radius,
Wt =
T
100
=
= 1600 N
d / 2 0.125 / 2
Wr = Wt tan φ = 1600 tan 20o = 583 N Ans.
Wt
1600
W=
=
= 1700 N Ans.
cos φ cos 20o
______________________________________________________________________________
13-41 (a) Using Eq. (13-11) with k = 1, φ = 20º, and m = 2,
NP =
=
(
2k
m + m 2 + (1 + 2m ) sin 2 φ
(1 + 2m ) sin 2 φ
{( 2) +
1 + 2 ( 2 )  sin ( 20 )
2 (1)
2
o
( 2)
2
)
( )} = 14.16 teeth
+ 1 + 2 ( 2 )  sin 2 20o
Round up for the minimum integer number of teeth.
NC = 15 teeth, NF = 30 teeth
(b)
P=
(c)
T=
(d)
N 30
=
= 3 teeth/in
d 10
Ans.
Ans.
 550 lbf ⋅ ft/s   12 in  rev  60 s 
1 hp





ω 70 rev/min 
hp
  ft  2π rad  min 
T = 900 lbf ⋅ in
Ans.
H
=
From Eqs. (13-34) and (13-35),
Shigley’s MED, 10th edition
Chapter 13 Solutions, Page 24/36
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V=
π dn
π (10 )( 70 )
= 183.3 ft/min
12
H 33000 (1)
Wt = 33000 =
= 180 lbf Ans.
V
183.3
Wr = Wt tan φ = 180 tan 20o = 65.5 lbf Ans.
W
180
W= t =
= 192 lbf Ans.
cos φ cos 20o
______________________________________________________________________________
N 
d 
 1.30 
o
13-42 (a)
Eq. (13-14): γ = tan −1  P  = tan −1  P  = tan −1 
Ans.
 = 18.5
N
d
3.88


 G
 G
12
=
π dn
Eq. (13-34):
V=
(c)
Eq. (13-35):
Wt = 33 000
12
=
π ( 2 )(1.30 )( 600 )
(b)
12
= 408.4 ft/min
Eq. (13-38):
H
 10 
= 33 000 
 = 808 lbf
V
 408.4 
Wr = Wt tan φ cos γ = 808 tan 20o cos18.5o = 279 lbf
Eq. (13-38):
Wa = Wt tan φ sin γ = 808 tan 20o sin18.5o = 93.3 lbf
Ans.
Ans.
Ans.
Ans.
The tangential and axial forces agree with Prob. 3-74, but the radial force given in Prob.
3-74 is shown here to be incorrect. Ans.
______________________________________________________________________________
13-43
Tin = 63 025H / n = 63025 ( 2.5) / 240 = 656.5 lbf ⋅ in
W t = T / r = 656.5 / 2 = 328.3 lbf
γ = tan −1 ( 2 / 4 ) = 26.565o
Γ = tan −1 ( 4 / 2 ) = 63.435o
(
)
a = 2 + 1.5 cos 26.565o / 2 = 2.67 in
W r = 328.3 tan 20o cos 26.565o = 106.9 lbf
W a = 328.3 tan 20o sin 26.565o = 53.4 lbf
W = 106.9i – 53.4j + 328.3k lbf
RAG = –2i + 5.17j,
RAB = 2.5j
ΣM 4 = R AG × W + R AB × FB + T = 0
Solving gives
R AB × FB = 2.5 FBz i − 2.5 FBx k
R AG × W = 1697i + 656.6 j − 445.9k
Shigley’s MED, 10th edition
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So,
(1697i + 656.6 j − 445.9k ) + ( 2.5FBz i − 2.5FBx k + Tj) = 0
FBz = −1697 / 2.5 = −678.8 lbf
T = −656.6 lbf ⋅ in
FBx = −445.9 / 2.5 = −178.4 lbf
So,
FB = −178.4i −678.8k lbf
Ans.
2 1/2
FB = ( −678.8 ) + ( −178.4 ) 


2
= 702 lbf
FA = – (FB + W)
= – (–178.8i – 678.8k + 106.9i – 53.4j + 328.3k)
= 71.5i + 53.4j + 350.5k Ans.
(
FA (radial) = 71.52 + 350.52
)
1/2
= 358 lbf
FA (thrust) = 53.4 lbf
______________________________________________________________________________
13-44
d 2 = 18 / 10 = 1.8 in, d3 = 30 / 10 = 3.0 in
 d2 / 2 
−1  0.9 
o
 = tan 
 = 30.96
 1.5 
 d3 / 2 
γ = tan −1 
Γ = 180o − γ = 59.04o
9
DE = + 0.5 cos 59.04o = 0.8197 in
16
t
W = 25 lbf
W r = 25 tan 20o cos 59.04o = 4.681 lbf
W a = 25 tan 20o sin 59.04o = 7.803 lbf
W = –4.681i – 7.803j +25k
RDG = 0.8197j + 1.25i
RDC = –0.625j
ΣM D = R DG × W + R DC × FC + T = 0
R DG × W = 20.49i − 31.25 j − 5.917k
R DC × FC = −0.625 FCz i + 0.625 FCx k
( 20.49i − 31.25 j − 5.917k ) + ( −0.625FCz i + 0.625FCxk ) + Tj = 0
T = 31.25 lbf ⋅ in
Ans.
FC = 9.47i + 32.8k lbf
Ans.
Shigley’s MED, 10th edition
Chapter 13 Solutions, Page 26/36
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(
FC = 9.47 2 + 32.82
ΣF = 0
)
1/2
= 34.1 lbf
Ans.
FD = −4.79i + 7.80 j − 57.8k lbf
2 1/2
FD (radial) = ( −4.79 ) + ( −57.8)  = 58.0 lbf Ans.


a
FD (thrust) = W = 7.80 lbf Ans.
______________________________________________________________________________
2
13-45
NOTE: The shaft forces exerted on the gears are not shown in the figures above.
Pt = Pn cosψ = 5 cos 30o = 4.330 teeth/in
φt = tan −1
dP =
tan φn
tan 20o
= tan −1
= 22.80o
cosψ
cos 30o
18
= 4.157 in
4.330
The forces on the shafts will be equal to the forces transmitted to the gears
through the meshing teeth.
Pinion (Gear 2)
W r = W t tan φt = 800 tan 22.80o = 336 lbf
W a = W t tanψ = 800 tan 30o = 462 lbf
W = −336i − 462 j + 800k lbf Ans.
1/2
W = ( −336 ) + ( −462 ) + 8002 


2
Gear 3
2
W = 336i + 462 j − 800k lbf
W = 983 lbf Ans.
32
dG =
= 7.390 in
4.330
= 983 lbf
Ans.
Ans.
Shigley’s MED, 10th edition
Chapter 13 Solutions, Page 27/36
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TG = W t r = 800 ( 7.390 ) = 5912 lbf ⋅ in
______________________________________________________________________________
13-46
φt = tan −1
tan φn
tan 20o
= tan −1
= 22.80o
cosψ
cos 30o
Pinion (Gear 2)
W r = W t tan φt = 800 tan 22.80o = 336 lbf
W a = W t tanψ = 800 tan 30o = 462 lbf
W = −336i − 462 j − 800k lbf Ans.
1/2
2
2
2
W = ( −336 ) + ( −462 ) + ( −800 ) 


= 983 lbf
Ans.
Idler (Gear 3)
From the diagram for the idler, noting that the radial and axial forces from gears 2 and 4
cancel each other, the force acting on the shaft is
W = +1600k lbf Ans.
Output gear (Gear 4)
W = 336i − 462 j − 800k lbf
2 1/2
2
2
W = ( −336 ) + ( −462 ) + ( −800 ) 


Ans.
= 983 lbf
Ans.
NOTE: For simplicity, the above figures only show the gear contact forces.
Also, notice that the idler shaft reaction contains a couple tending to turn the shaft endover-end. Also the idler teeth are bent both ways. Idlers are more severely loaded than
other gears, belying their name. Thus, be cautious.
______________________________________________________________________________
Shigley’s MED, 10th edition
Chapter 13 Solutions, Page 28/36
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13-47 Gear 3:
Pt = Pn cosψ = 7 cos 30o = 6.062 teeth/in
tan 20o
= 0.4203, φt = 22.8o
cos 30o
54
d3 =
= 8.908 in
6.062
W t = 500 lbf
W a = 500 tan 30o = 288.7 lbf
W r = 500 tan 22.8o = 210.2 lbf
tan φt =
W3 = 210.2i − 288.7 j − 500k lbf Ans.
Gear 4:
14
d4 =
= 2.309 in
6.062
8.908
W t = 500
= 1929 lbf
2.309
W a = 1929 tan 30o = 1114 lbf
W r = 1929 tan 22.8o = 811 lbf
W4 = −811i + 1114 j − 1929k lbf Ans.
______________________________________________________________________________
13-48
Pt = 6 cos 30o = 5.196 teeth/in
42
d3 =
= 8.083 in
5.196
φt = 22.8o
16
d2 =
= 3.079 in
5.196
63025 ( 25 )
T2 =
= 916 lbf ⋅ in
1720
Shigley’s MED, 10th edition
Chapter 13 Solutions, Page 29/36
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T
916
=
= 595 lbf
r 3.079 / 2
W a = 595 tan 30o = 344 lbf
W r = 595 tan 22.8o = 250 lbf
Wt =
W = 344i + 250 j + 595k lbf
R DC = 6i,
R DG = 3i − 4.04 j
ΣM D = R DC × FC + R DG × W + T = 0
(1)
R DG × W = −2404i − 1785 j + 2140k
R DC × FC = −6 FCz j + 6 FCy k
Substituting and solving Eq. (1) gives
T = 2404i lbf ⋅ in
FCz = −297.5 lbf
FCy = −365.7 lbf
ΣF = FD + FC + W = 0
Substituting and solving gives
FCx = −344 lbf
FDy = 106.7 lbf
FDz = −297.5 lbf
FC = −344i − 356.7 j − 297.5k lbf
Ans.
FD = 106.7 j − 297.5k lbf Ans.
______________________________________________________________________________
Shigley’s MED, 10th edition
Chapter 13 Solutions, Page 30/36
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13-49
Since the transverse pressure angle is specified, we will assume the given module is also
in terms of the transverse orientation.
d 2 = mN 2 = 4 (16 ) = 64 mm
d3 = mN3 = 4 ( 36 ) = 144 mm
d 4 = mN 4 = 4 ( 28) = 112 mm
6 kW
 1000 W   rev   60 s 
= 35.81 N ⋅ m
ω 1600 rev/min  kW   2π rad   min 
T
35.81
Wt =
=
= 1119 N
d 2 / 2 0.064 / 2
T=
H
=
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Chapter 13 Solutions, Page 31/36
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W r = W t tan φt = 1119 tan 20o = 407.3 N
W a = W t tanψ = 1119 tan15o = 299.8 N
F2 a = −1119i − 407.3 j − 299.8k N Ans.
F3b = (1119 − 407.3) i − (1119 − 407.3) j
= 711.7i − 711.7 j N Ans.
F4 c = 407.3i + 1119 j + 299.8k N Ans.
______________________________________________________________________________
13-50
N
14
36
=
= 2.021 in, d3 =
= 5.196 in
o
8cos 30o
Pn cosψ 8 cos 30
15
45
d4 =
= 3.106 in, d5 =
= 9.317 in
o
5cos15
5cos15o
d2 =
For gears 2 and 3: φt = tan −1 ( tan φn / cosψ ) = tan −1 ( tan 20o / cos 30o ) = 22.8o
For gears 4 and 5: φt = tan −1 ( tan 20o / cos15o ) = 20.6o
F23t = T2 / r2 = 1200 / ( 2.021/ 2 ) = 1188 lbf
5.196
= 1987 lbf
3.106
F23r = F23t tan φt = 1188 tan 22.8o = 499 lbf
F54t = 1188
F54r = 1986 tan 20.6o = 746 lbf
F23a = F23t tanψ = 1188 tan 30o = 686 lbf
F54a = 1986 tan15o = 532 lbf
Next, designate the points of action on gears 4 and 3, respectively, as points G and H, as
shown. Position vectors are
Shigley’s MED, 10th edition
Chapter 13 Solutions, Page 32/36
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R CG = 1.553j − 3k
R CH = −2.598 j − 6.5k
R CD = −8.5k
Force vectors are
F54 = −1986i − 748 j + 532k
F23 = −1188i + 500 j − 686k
FC = FCx i + FCy j
FD = FDx i + FDy j + FDz k
Now, a summation of moments about bearing C gives
ΣM C = R CG × F54 + R CH × F23 + R CD × FD = 0
The terms for this equation are found to be
R CG × F54 = −1412i + 5961j + 3086k
R CH × F23 = 5026i + 7722 j − 3086k
R CD × FD = 8.5 FDy i − 8.5 FDx j
When these terms are placed back into the moment equation, the k terms, representing
the shaft torque, cancel. The i and j terms give
3614
= −425 lbf
8.5
13683
FDx =
= 1610 lbf
8.5
Next, we sum the forces to zero.
FDy = −
ΣF = FC + F54 + F23 + FD = 0
Substituting, gives
( F i + F j) + ( −1987i − 746 j + 532k ) + ( −1188i + 499 j − 686k )
+ (1610i − 425 j + F k ) = 0
x
C
y
C
z
D
Solving gives
FCx = 1987 + 1188 − 1610 = 1565 lbf
FCy = 746 − 499 + 425 = 672 lbf
FDz = −532 + 686 = 154 lbf
So,
FC = 1565i + 672j lbf Ans.
Ans.
FD = 1610i − 425j + 154k lbf
______________________________________________________________________________
Shigley’s MED, 10th edition
Chapter 13 Solutions, Page 33/36
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VW =
13-51
WW t
π dW nW
=
π ( 0.100 )( 600 )
60
60
H 2000
=
=
= 637 N
VW
π
= π m/s
L = px NW = 25 (1) = 25 mm
λ = tan −1
= tan −1
L
π dW
25
= 4.550o lead angle
π (100 )
WW t
W=
cos φn sin λ + f cos λ
V
π
= 3.152 m/s
VS = W =
cos λ cos 4.550o
In ft/min: VS = 3.28(3.152) = 10.33 ft/s = 620 ft/min
Use f = 0.043 from curve A of Fig. 13-42. Then, from the first of Eq. (13-43)
W=
637
= 5323 N
cos14.5 sin 4.55o + 0.043cos 4.55o
o
(
)
W y = W sin φn = 5323sin14.5o = 1333 N
(
)
W z = 5323 cos14.5o cos 4.55o − 0.043sin 4.55o  = 5119 N
The force acting against the worm is
W = −637i + 1333 j + 5119k N
Thus, A is the thrust bearing.
Ans.
R AG = −0.05 j − 0.10k m, R AB = −0.20k m
ΣM A = R AG × W + R AB × FB + T = 0
R AG × W = −122.6i + 63.7 j − 31.85k N ⋅ m
R AB × FB = 0.2 FBy i − 0.2 FBx j
Substituting and solving gives
T = 31.85 N ⋅ m Ans.
FBx = 318.5 N, FBy = 613 N
So
FB = 318.5i + 613 j N
Ans.
Shigley’s MED, 10th edition
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1/2
2
2
FB = ( 613) + ( 318.5 ) 


ΣF = FA + W + R B = 0
Or
= 691 N radial
FA = − ( W + FB ) = − ( −637i + 1333j + 5119k + 318.5i + 613j)
= 318.5i − 1946 j − 5119k
Ans.
FAr = 318.5i − 1946 j N
Radial
2 1/ 2
FAr = ( 318.5) + ( −1946 )  = 1972 N


Thrust
FAa = −5119 N
______________________________________________________________________________
2
13-52 From Prob. 13-51,
WG = 637i − 1333 j − 5119k N
pt = px
So
dG =
N G px
π
=
48 ( 25 )
π
= 382 mm
Bearing D takes the thrust load.
ΣM D = R DG × WG + R DC × FC + T = 0
R DG = −0.0725i + 0.191j m
R DC = −0.1075i m
R DG × WG = −977.7i − 371.1j − 25.02k N ⋅ m
R DC × FC = 0.1075 FCz j − 0.1075 FCy k N ⋅ m
Putting it together and solving,
T = 977.7 N ⋅ m Ans.
FC = −233 j + 3450k N,
ΣF = FC + WG + FD = 0
FC = 3460 N
Ans.
FD = − ( FC + WG ) = −637i + 1566 j + 1669k N
Ans.
Radial FDr = 1566 j + 1669k N
Or
(
FDr = 1566 2 + 1669 2
)
1/2
= 2289 N (total radial)
FDt = −637 i N (thrust)
______________________________________________________________________________
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Chapter 13 Solutions, Page 35/36
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13-53
VW =
π (1.5 )( 600 )
= 235.7 ft/min
12
33000 ( 0.75 )
W x = WW t =
= 105.0 lbf
235.7
π
pt = px = = 0.3927 in
8
L = 0.3927 ( 2 ) = 0.7854 in
0.7854
= 9.46o
π (1.5)
105.0
W=
= 515.3 lbf
o
cos 20 sin 9.46o + 0.05 cos 9.46o
W y = 515.3sin 20o = 176.2 lbf
W z = 515.3 cos 20o ( cos 9.46o ) − 0.05sin 9.46o  = 473.4 lbf
λ = tan −1
So
W = 105i + 176 j + 473k lbf
Ans.
T = 105 ( 0.75) = 78.8 lbf ⋅ in Ans.
______________________________________________________________________________
13-54 Computer programs will vary.
Shigley’s MED, 10th edition
Chapter 13 Solutions, Page 36/36
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Chapter 14
d =
14-1
N
22
=
= 3.667 in
P
6
Table 14-2:
Y = 0.331
π dn π (3.667)(1200)
Eq. (13-34): V =
=
= 1152 ft/min
12
12
1200 + 1152
Eq. (14-4b): K v =
= 1.96
1200
H
15
Eq. (13-35) : W t = 33 000
= 33 000
= 429.7 lbf
V
1152
K W t P 1.96(429.7)(6)
Eq. (14-7):
σ = v
=
= 7633 psi = 7.63 kpsi Ans.
FY
2(0.331)
________________________________________________________________________
N
18
=
= 1.8 in
P
10
Table 14-2: Y = 0.309
π dn π (1.8)(600)
Eq. (13-34): V =
=
= 282.7 ft/min
12
12
1200 + 282.7
Eq. (14-4b): K v =
= 1.236
1200
H
2
Eq. (13-35) : W t = 33 000
= 33 000
= 233.5 lbf
V
282.7
K W t P 1.236(233.5)(10)
Eq. (14-7):
σ = v
=
= 9340 psi = 9.34 kpsi Ans.
FY
1.0(0.309)
________________________________________________________________________
14-2
d =
14-3
d = mN = 1.25(18) = 22.5 mm
Y = 0.309
π dn π (22.5)(10−3 )(1800)
=
= 2.121 m/s
V =
60
60
6.1 + 2.121
Kv =
= 1.348
6.1
60 000 H
60 000(0.5)
=
= 0.2358 kN = 235.8 N
Wt =
π dn
π (22.5)(1800)
Table 14-2:
Eq. (14-6b):
Eq. (13-36):
K vW t
1.348(235.8)
=
= 68.6 MPa Ans.
FmY
12(1.25)(0.309)
________________________________________________________________________
Eq. (14-8):
σ =
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Chapter 14 Solutions, Page 1/39
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14-4
Table 14-2:
Eq. (14-6b):
Eq. (13-36):
d = mN = 8(16) = 128 mm
Y = 0.296
π dn π (128)(10−3 )(150)
=
= 1.0053 m/s
V =
60
60
6.1 + 1.0053
Kv =
= 1.165
6.1
60 000 H
60 000(6)
Wt =
=
= 5.968 kN = 5968 N
π dn
π (128)(150)
K vW t
1.165(5968)
=
= 32.6 MPa Ans.
FmY
90(8)(0.296)
________________________________________________________________________
Eq. (14-8):
14-5
Table 14-2:
Eq. (14-6b):
Eq. (13-36):
Eq. (14-8):
σ =
d = mN = 1(16) = 16 mm
Y = 0.296
π dn π (16)(10−3 )(400)
V =
=
= 0.335 m/s
60
60
6.1 + 0.335
Kv =
= 1.055
6.1
60 000H
60 000(0.15)
Wt =
=
= 0.4476 kN = 447.6 N
π dn
π (16)(400)
F =
K vW t
1.055(447.6)
=
= 10.6 mm
σ mY 150(1)(0.296)
From Table 13-2, use F = 11 mm or 12 mm, depending on availability. Ans.
________________________________________________________________________
14-6
Table 14-2:
Eq. (14-6b):
Eq. (13-36):
Eq. (14-8):
d = mN = 2(20) = 40 mm
Y = 0.322
π dn π (40)(10−3 )(200)
V =
=
= 0.419 m/s
60
60
6.1 + 0.419
Kv =
= 1.069
6.1
60 000 H
60 000(0.5)
Wt =
=
= 1.194 kN = 1194 N
π dn
π (40)(200)
F =
K vW t
1.069(1194)
=
= 26.4 mm
σ mY
75(2.0)(0.322)
From Table 13-2, use F = 28 mm. Ans.
________________________________________________________________________
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14-7
N
24
=
= 4.8 in
P
5
Table 14-2: Y = 0.337
π dn π (4.8)(50)
Eq. (13-34): V =
=
= 62.83 ft/min
12
12
1200 + 62.83
Eq. (14-4b): K v =
= 1.052
1200
H
6
Eq. (13-35) : W t = 33 000
= 33 000
= 3151 lbf
V
62.83
K W t P 1.052(3151)(5)
F = v
Eq. (14-7):
=
= 2.46 in
σY
20(103 )(0.337)
d =
Use F = 2.5 in Ans.
________________________________________________________________________
N
16
=
= 4.0 in
P
4
Table 14-2: Y = 0.296
π dn π (4.0)(400)
Eq. (13-34): V =
=
= 418.9 ft/min
12
12
1200 + 418.9
Eq. (14-4b): K v =
= 1.349
1200
H
20
Eq. (13-35) : W t = 33 000
= 33 000
= 1575.6 lbf
V
418.9
K vW t P 1.349(1575.6)(4)
Eq. (14-7):
=
= 2.39 in
F =
σY
12(103 )(0.296)
Use F = 2.5 in Ans.
________________________________________________________________________
d =
14-8
14-9
Try P = 8 which gives d = 18/8 = 2.25 in and Y = 0.309.
Eq. (13-34):
Eq. (14-4b):
Eq. (13-35):
Eq. (14-7):
V =
π dn π (2.25)(600)
=
= 353.4 ft/min
12
12
1200 + 353.4
= 1.295
1200
H
2.5
W t = 33 000
= 33 000
= 233.4 lbf
V
353.4
K W t P 1.295(233.4)(8)
=
= 0.783 in
F = v
σY
10(103 )(0.309)
Kv =
Using coarse integer pitches from Table 13-2, the following table is formed.
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d
V
F
Wt
Kv
9.000 1413.717 2.178 58.356 0.082
6.000 942.478 1.785 87.535 0.152
4.500 706.858 1.589 116.713 0.240
3.000 471.239 1.393 175.069 0.473
2.250 353.429 1.295 233.426 0.782
1.800 282.743 1.236 291.782 1.167
1.500 235.619 1.196 350.139 1.627
1.125 176.715 1.147 466.852 2.773
P
2
3
4
6
8
10
12
16
Other considerations may dictate the selection. Good candidates are P = 8
(F = 7/8 in) and P =10 (F = 1.25 in). Ans.
________________________________________________________________________
14-10 Try m = 2 mm which gives d = 2(18) = 36 mm and Y = 0.309.
V =
π dn
π (36)(10−3 )(900)
= 1.696 m/s
60
60
6.1 + 1.696
Eq. (14-6b): K v =
= 1.278
6.1
60 000H
60 000(1.5)
Eq. (13-36): W t =
=
= 0.884 kN = 884 N
π dn
π (36)(900)
1.278(884)
Eq. (14-8):
F =
= 24.4 mm
75(2)(0.309)
Using the preferred module sizes from Table 13-2:
m
1.00
1.25
1.50
2.00
3.00
4.00
5.00
6.00
8.00
10.00
12.00
16.00
20.00
25.00
32.00
40.00
50.00
=
d
18.0
22.5
27.0
36.0
54.0
72.0
90.0
108.0
144.0
180.0
216.0
288.0
360.0
450.0
576.0
720.0
900.0
V
0.848
1.060
1.272
1.696
2.545
3.393
4.241
5.089
6.786
8.482
10.179
13.572
16.965
21.206
27.143
33.929
42.412
Wt
F
Kv
1.139 1768.388 86.917
1.174 1414.711 57.324
1.209 1178.926 40.987
1.278 884.194 24.382
1.417 589.463 12.015
1.556 442.097 7.422
1.695 353.678 5.174
1.834 294.731 3.888
2.112 221.049 2.519
2.391 176.839 1.824
2.669 147.366 1.414
3.225 110.524 0.961
3.781
88.419 0.721
4.476
70.736 0.547
5.450
55.262 0.406
6.562
44.210 0.313
7.953
35.368 0.243
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Chapter 14 Solutions, Page 4/39
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Other design considerations may dictate the size selection. For the present design,
m = 2 mm (F = 25 mm) is a good selection. Ans.
________________________________________________________________________
14-11
NP
20
N
50
=
= 2.5 in, d G = G =
= 6.25 in
8
8
P
P
π (2.5)(1200)
= 785.4 ft/min
V =
12
1200 + 785.4
Kv =
= 1.655
1200
H
12
W t = 33 000
= 33 000
= 504.2 lbf
V
785.4
dP =
Eq. (14-4b):
Eq. (13-36):
C p = 2100 psi [Note: Using Eq. (14-13) can result in wide variation in
Table 14-8:
Cp due to wide variation in cast iron properties.]
2.5sin 20°
= 0.4275 in,
2
Eq. (14-12):
r1 =
r2 =
Eq. (14-14):
 K W t  1 1 
σ C = −C p  v
 + 
cos
F
φ
 r1 r2  

6.25 sin 20°
= 1.069 in
2
1/ 2
1/ 2
1.655(504.2)  1
1 
= −2100 
+


 1.5 cos 20°  0.4275 1.069  
= −92.5(103 ) psi = −92.5 kpsi Ans.
________________________________________________________________________
14-12
16
48
= 1.333 in, dG =
= 4 in
12
12
π (1.333)(700)
V =
= 244.3 ft/min
12
dP =
Eq. (14-4b):
Eq. (13-36):
1200 + 244.3
= 1.204
1200
H
1.5
Wt = 33 000
= 33 000
= 202.6 lbf
V
244.3
C p = 2100 psi [Note: Using Eq. (14-13) can result in wide variation
Kv =
Table 14-8:
in Cp due to wide variation in cast iron properties.]
Eq. (14-12):
r1 =
1.333sin 20°
= 0.228 in,
2
r2 =
4 sin 20°
= 0.684 in
2
Eq. (14-14):
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Chapter 14 Solutions, Page 5/39
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1.204(202.6)  1
1 
+


°
F
cos
20
0.228
0.684



1/ 2
σ C = −2100 
= −100(103 )
2
 2100  1.204(202.6)   1
1 
+
F =

 = 0.669 in
3  

 100(10 )   cos 20°   0.228 0.684 
Use F = 0.75 in Ans.
________________________________________________________________________
d p = 5(24) = 120 mm, dG = 5(48) = 240 mm
14-13
V=
π (120)(10−3 )(50)
= 0.3142 m/s
60
3.05 + 0.3142
Eq. (14-6a): K v =
= 1.103
3.05
60 000H
60(103 ) H
Wt =
=
= 3.183H
π dn
π (120)(50)
where H is in kW and Wt is in kN
Table 14-8: C p = 163 MPa [Note: Using Eq. (14-13) can result in wide variation in
Cp due to wide variation in cast iron properties].
120 sin 20°
240 sin 20°
Eq. (14-12): r1 =
= 20.52 mm, r2 =
= 41.04 mm
2
2
1.103(3.183) (103 ) H  1
1 
−690 = −163 
+


60 cos 20o
 20.52 41.04  

1/ 2
Eq. (14-14):
H = 3.94 kW Ans.
________________________________________________________________________
14-14
Eq. (14-6a):
d P = 4(20) = 80 mm, dG = 4(32) = 128 mm
π (80)(10−3 )(1000)
= 4.189 m/s
V =
60
3.05 + 4.189
Kv =
= 2.373
3.05
60(10)(103 )
Wt =
= 2.387 kN = 2387 N
π (80)(1000)
C p = 163 MPa [Note: Using Eq. (14-13) can result in wide variation in
Table 14-8:
Cp due to wide variation in cast iron properties.]
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Eq. (14-12):
r1 =
80 sin 20°
= 13.68 mm,
2
r2 =
128sin 20°
= 21.89 mm
2
1/ 2
 2.373(2387)  1
1 
+
Eq. (14-14): σ C = −163 

  = −617 MPa Ans.
 50 cos 20°  13.68 21.89  
________________________________________________________________________
14-15 The pinion controls the design.
Bending
YP = 0.303,
YG = 0.359
17
30
= 1.417 in, dG =
= 2.500 in
12
12
π d P n π (1.417)(525)
=
= 194.8 ft/min
V =
12
12
1200 + 194.8
Eq. (14-4b): K v =
= 1.162
1200
Se′ = 0.5(76) = 38.0 kpsi
Eq. (6-8), p. 290:
Eq. (6-19), p. 295:
ka = 2.70(76)–0.265 = 0.857
2.25 2.25
l =
=
= 0.1875 in
12
Pd
3Y
3(0.303)
Eq. (14-3):
x= P =
= 0.0379 in
2P
2(12)
dP =
Eq. (b), p. 729: t =
Eq. (6-25), p. 297:
4lx =
4(0.1875)(0.0379) = 0.1686 in
de = 0.808 hb = 0.808 0.875(0.1686) = 0.310 in
−0.107
 0.310 
kb = 
= 0.996

 0.3 
kc = kd = ke = 1
Account for one-way bending with kf = 1.66. (See Ex. 14-2.)
Eq. (6-20), p. 296:
Eq. (6-18), p. 295:
Se = 0.857(0.996)(1)(1)(1)(1.66)(38.0) = 53.84 kpsi
For stress concentration, find the radius of the root fillet (See Ex. 14-2).
rf =
0.300
0.300
=
= 0.025 in
P
12
From Fig. A-15-6,
r
0.025
r
= f =
= 0.148
d
t
0.1686
Approximate D/d = ∞ with D/d = 3; from Fig. A-15-6, Kt = 1.68.
From Fig. 6-20, with Sut = 76 kpsi and r = 0.025 in, q = 0.62.
Shigley’s MED, 10th edition
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Eq. (6-32):
Kf = 1 + 0.62 (1.68 – 1) = 1.42
Se
53.84
σ all =
=
= 16.85 psi
1.42(2.25)
K f nd
0.875(0.303)(16 850)
FYPσ all
Wt =
=
= 320.4 lbf
1.162(12)
K v Pd
320.4(194.8)
W tV
H =
=
= 1.89 hp Ans.
33 000
33 000
Wear
ν 1 = ν 2 = 0.292, E1 = E2 = 30(106) psi
1/ 2
Eq. (14-13):






1
Cp = 

2
  1 − 0.292  
 2π  30 (106 )  

 
Eq. (14-12):
r1 =
= 2285 psi
dP
1.417
sin φ =
sin 20° = 0.242 in
2
2
d
2.500
r2 = G sin φ =
sin 20° = 0.428 in
2
2
1 1
1
1
+
=
+
= 6.469 in −1
r1 r2
0.242 0.428
(SC )108 = 0.4H B − 10 kpsi = [0.4(149) − 10](103 ) = 49 600 psi
Eq. (6-68), p. 337:
From the discussion and equation developed on the bottom of p. 337,
( S ) 8 −49 600
σ C ,all = − C 10 =
= −33 067 psi
2.25
n
2
 −33 067   0.875cos 20° 
Eq. (14-14): W = 
 
 = 22.6 lbf
 2285   1.162(6.469) 
W tV
22.6(194.8)
H =
=
= 0.133 hp Ans.
33 000
33 000
Rating power (pinion controls):
t
H1 = 1.89 hp
H2 = 0.133 hp
Hall = (min 1.89, 0.133) = 0.133 hp Ans.
________________________________________________________________________
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14-16 See Prob. 14-15 solution for equation numbers.
Pinion controls:
YP = 0.322, YG = 0.447
Bending
dP = 20/3 = 6.667 in, dG = 100/3 = 33.333 in
V = π d P n / 12 = π (6.667)(870) / 12 = 1519 ft/min
K v = (1200 + 1519) / 1200 = 2.266
Se′ = 0.5(113) = 56.5 kpsi
ka = 2.70(113) −0.265 = 0.771
l = 2.25 / Pd = 2.25 / 3 = 0.75 in
x = 3(0.322) / [2(3)] = 0.161 in
t = 4(0.75)(0.161) = 0.695 in
d e = 0.808 2.5(0.695) = 1.065 in
kb = (1.065 / 0.30) −0.107 = 0.873
kc = k d = k e = 1
kf = 1.66 (See Ex. 14-2.)
Se = 0.771(0.873)(1)(1)(1)(1.66)(56.5) = 63.1 kpsi
rf = 0.300 / 3 = 0.100 in
r
0.100
r
= f =
= 0.144
d
t
0.695
Kt = 1.75, q = 0.85, Kf = 1.64
Se
63.1
σ all =
=
= 25.7 kpsi
K f nd 1.64(1.5)
FYPσ all
2.5(0.322)(25 700)
=
= 3043 lbf
Wt =
K v Pd
2.266(3)
H = W tV / 33 000 = 3043(1519) / 33 000 = 140 hp
Ans.
Wear
1/ 2
Eq. (14-13):
Eq. (14-12):






1
Cp = 

2
  1 − 0.292  
 2π  30 (106 )  

 
= 2285 psi
r1 = (6.667/2) sin 20° = 1.140 in
r2 = (33.333/2) sin 20° = 5.700 in
Eq. (6-68), p. 337:
σ C,all
SC = [0.4(262) – 10](103) = 94 800 psi
= −SC / nd = −94 800 / 1.5 = −77 400 psi
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2
σ
 F cos φ 

1
W =  C ,all 


 Cp 
K v  1 / r1 + 1 / r2 


2
1
 −77 400   2.5cos 20°  

=
 


 2285   2.266   1 / 1.140 + 1 / 5.700 
= 1130 lbf
W tV
1130(1519)
H =
=
= 52.0 hp Ans.
33 000
33 000
For 108 cycles (revolutions of the pinion), the power based on wear is 52.0 hp.
Rating power (pinion controls):
t
H1 = 140 hp
H 2 = 52.0 hp
H rated = min(140, 52.0) = 52.0 hp Ans.
________________________________________________________________________
14-17 See Prob. 14-15 solution for equation numbers.
Given: φ = 20°, n = 1145 rev/min, m = 6 mm, F = 75 mm, NP = 16 milled teeth,
NG = 30T, Sut = 900 MPa, HB = 260, nd = 3, YP = 0.296, and YG = 0.359.
Pinion bending
d P = mN P = 6(16) = 96 mm
dG = 6(30) = 180 mm
π d P n π (96)(10−3 )(1145)
V =
=
= 5.76 m/s
60
(60)
6.1 + 5.76
Kv =
= 1.944
6.1
Se′ = 0.5(900) = 450 MPa
ka = 4.51(900) −0.265 = 0.744
l = 2.25m = 2.25(6) = 13.5 mm
x = 3Ym / 2 = 3(0.296)6 / 2 = 2.664 mm
t =
4lx =
4(13.5)(2.664) = 12.0 mm
de = 0.808 75(12.0) = 24.23 mm
−0.107
 24.23 
kb = 
= 0.884

 7.62 
kc = k d = ke = 1
kf = 1.66 (See Ex. 14-2)
Se = 0.744(0.884)(1)(1)(1)(1.66)(450) = 491.3 MPa
rf = 0.300m = 0.300(6) = 1.8 mm
r/d = rf /t = 1.8/12 = 0.15, Kt = 1.68, q = 0.86, Kf = 1.58
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σ all =
Se
491.3
=
= 239.2 MPa
K f nd 1.58 (1.3)
Eq. (14-8):
Wt =
FYmσ all
75(0.296)(6)(239.2)
=
= 16 390 N
1.944
Kv
Eq. (13-36):
H =
W tπ dn 16.39π (96)(1145)
=
= 94.3 kW
60 000
60 000
Ans.
Wear: Pinion and gear
Eq. (14-12):
r1 = (96/2) sin 20° = 16.42 mm
r2 = (180/2) sin 20° = 30.78 mm
1/ 2






1
Eq. (14-13): C p = 
 = 190 MPa
2
  1 − 0.292  
 2π  207 (103 )  

 
Eq. (6-68), p. 337:
SC = 6.89[0.4(260) – 10] = 647.7 MPa
−647.7
σ C ,all = −SC / nd =
= −568 MPa
1.3
Eq. (14-14):
σ

W =  C ,all 
 Cp 


2
t

F cos φ 
1


K v  1 / r1 + 1 / r2 
2
Eq. (13-36):
o
1
 −568   75cos 20  

=

 
 = 3469 N
 190   1.944   1 / 16.42 + 1 / 30.78 
W tπ dn 3.469π (96)(1145)
=
= 20.0 kW
H =
60 000
60 000
Thus, wear controls the gearset power rating; H = 20.0 kW. Ans.
________________________________________________________________________
NP = 17 teeth, NG = 51 teeth
N 17
=
= 2.833 in
dP =
6
P
51
dG =
= 8.500 in
6
V = π d P n / 12 = π (2.833)(1120) / 12 = 830.7 ft/min
14-18
Eq. (14-4b):
Kv = (1200 + 830.7)/1200 = 1.692
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σ all =
Sy
nd
=
90 000
= 45 000 psi
2
Table 14-2:
YP = 0.303, YG = 0.410
Eq. (14-7):
Wt =
Eq. (13-35):
H =
FYPσ all
2(0.303)(45 000)
=
= 2686 lbf
1.692(6)
KvP
W tV
2686(830.7)
=
= 67.6 hp
33 000
33 000
Based on yielding in bending, the power is 67.6 hp.
(a) Pinion fatigue
Bending
Eq. (2-21), p. 52: Sut = 0.5 HB = 0.5(232) = 116 kpsi
Eq. (6-8), p. 290: Se′ = 0.5Sut = 0.5(116) = 58 kpsi
Eq. (6-19), p. 295: ka = 2.70(116)−0.265 = 0.766
Table 13-1, p. 688: l =
Eq. (14-3): x =
1
1.25
2.25
2.25
+
=
=
= 0.375 in
6
Pd
Pd
Pd
3YP
3(0.303)
=
= 0.0758 in
2P
2(6)
Eq. (b), p. 729: t =
4lx =
4(0.375)(0.0758) = 0.337 in
Eq. (6-25), p. 297: de = 0.808 F t = 0.808 2(0.337) = 0.663 in
 0.663 
Eq. (6-20), p. 296: kb = 

 0.30 
kc = kd = ke = 1
−0.107
= 0.919
Account for one-way bending with kf = 1.66. (See Ex. 14-2.)
Eq. (6-18): Se = 0.766(0.919)(1)(1)(1)(1.66)(58) = 67.8 kpsi
For stress concentration, find the radius of the root fillet (See Ex. 14-2).
0.300
0.300
rf =
=
= 0.050 in
P
6
r
r
0.05
Fig. A-15-6:
= f =
= 0.148
0.338
d
t
Estimate D/d = ∞ by setting D/d = 3, Kt = 1.68.
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Fig. 6-20, p. 303: q = 0.86
Eq. (6-32), p. 303: K f = 1 + (0.86)(1.68 − 1) = 1.58
Se
67.8
=
= 21.5 kpsi
1.58(2)
K f nd
FYPσ all
2(0.303)(21 500)
=
= 1283 lbf
Wt =
1.692(6)
K v Pd
σ all =
H =
W tV
1283(830.7)
=
= 32.3 hp
33 000
33 000
Ans.
(b) Pinion fatigue
Wear
1/ 2
Eq. (14-13):
Eq. (14-12):
Eq. (6-68):


1
Cp = 
2
6 
 2π[(1 - 0.292 ) / 30(10 )] 
= 2285 psi
dP
2.833
sin φ =
sin 20 o = 0.485 in
2
2
d
8.500
r2 = G sin φ =
sin 20 o = 1.454 in
2
2
1 1
1
1
+
= 2.750 in
 + =
 r1 r2  0.485 1.454
r1 =
(SC )108 = 0.4H B − 10 kpsi
In terms of gear notation
σC = [0.4(232) – 10]103 = 82 800 psi
We will introduce the design factor of nd = 2 and because it is a contact stress apply it
to the load Wt by dividing by nd = 2 . (See p. 337.)
σ C ,all = −
σc
2
=−
82 800
= −58 548 psi
2
Solve Eq. (14-14) for Wt:
2
o
 −58 548   2 cos 20 
Wt = 
 
 = 265 lbf
 2285  1.692(2.750) 
W tV
265(830.7)
H all =
=
= 6.67 hp Ans.
33 000
33 000
For 108 cycles (turns of pinion), the allowable power is 6.67 hp.
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(c) Gear fatigue due to bending and wear
Bending
x=
Eq. (14-3):
Eq. (b), p. 729: t =
3YG
3(0.4103)
=
= 0.1026 in
2P
2(6)
4 lx =
4(0.375)(0.1026) = 0.392 in ±
Eq. (6-25): de = 0.808 F t = 0.808 2(0.392) = 0.715 in
−0.107
 0.715 
Eq. (6-20): kb = 
= 0.911

 0.30 
kc = kd = ke = 1
kf = 1.66. (See Ex. 14-2.)
Eq. (6-18): Se = 0.766(0.911)(1)(1)(1)(1.66)(58) = 67.2 kpsi
r
r
0.050
= f =
= 0.128
d
t
0.392
Approximate D/d = ∞ by setting D/d = 3 for Fig. A-15-6; Kt = 1.80.
Fig. 6-20: q = 0.82
Eq. (6-32): K f = 1 + (0.82)(1.80 − 1) = 1.66
Se
67.2
=
= 20.2 kpsi
K f nd
1.66(2)
FYPσ all
2(0.4103)(20 200)
Wt =
=
= 1633 lbf
1.692(6)
K v Pd
σ all =
W tV
1633(830.7)
=
= 41.1 hp
33 000
33 000
The gear is thus stronger than the pinion in bending.
H all =
Ans.
Wear
Since the material of the pinion and the gear are the same, and the contact stresses are
the same, the allowable power transmission of both is the same. Thus, Hall = 6.67 hp
for 108 revolutions of each. As yet, we have no way to establish SC for 108/3
revolutions.
(d)
Pinion bending:
H1 = 32.3 hp
Pinion wear:
H2 = 6.67 hp
Gear bending:
H3 = 41.1 hp
Gear wear:
H4 = 6.67 hp
Power rating of the gear set is thus
Hrated = min(32.3, 6.67, 41.1, 6.67) = 6.67 hp Ans.
________________________________________________________________________
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dP = 16/6 = 2.667 in, dG = 48/6 = 8 in
π (2.667)(300)
V =
= 209.4 ft/min
12
33 000(5)
Wt =
= 787.8 lbf
209.4
14-19
Assuming uniform loading, Ko = 1.
Eq. (14-28): Qv = 6, B = 0.25(12 − 6)2 / 3 = 0.8255
A = 50 + 56(1 − 0.8255) = 59.77
0.8255
 59.77 + 209.4 
Eq. (14-27): K v = 

59.77


Table 14-2: YP = 0.296, YG = 0.4056
From Eq. (a), Sec. 14-10 with F = 2 in
 2 0.296 
( K s ) P = 1.192 

6


= 1.196
0.0535
= 1.088
0.0535
 2 0.4056 
( K s )G = 1.192 
= 1.097

6


From Eq. (14-30) with Cmc = 1
2
Cp f =
− 0.0375 + 0.0125(2) = 0.0625
10(2.667)
C p m = 1, Cma = 0.093 (Fig. 14 - 11), Ce = 1
K m = 1 + 1[0.0625(1) + 0.093(1)] = 1.156
Assuming constant thickness of the gears → KB = 1
mG = NG/NP = 48/16 = 3
With N (pinion) = 108 cycles and N (gear) = 108/3, Fig. 14-14 provides the relations:
(YN ) P = 1.3558(108 ) −0.0178 = 0.977
(YN )G = 1.3558(108 / 3)−0.0178 = 0.996
J P = 0.27,
Fig. 14-6:
Table 14-10: KR = 0.85
KT = C f = 1
Eq. (14-23):
Table 14-8:
I =
J G o 0.38
cos 20o sin 20o  3 

 = 0.1205
2(1)
 3 + 1
C p = 2300 psi
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Strength: Grade 1 steel with HBP = HBG = 200
Fig. 14-2:
(St)P = (St)G = 77.3(200) + 12 800 = 28 260 psi
Fig. 14-5:
(Sc)P = (Sc)G = 322(200) + 29 100 = 93 500 psi
Fig. 14-15:
(ZN )P = 1.4488(108)–0.023 = 0.948
(ZN )G = 1.4488(108/3)–0.023 = 0.973
Sec. 14-12:
HBP/HBG = 1
∴ CH = 1
Pinion tooth bending
Eq. (14-15):
(σ ) P = W t K o K v K s
Pd K m K B
F J
 6   (1.156)(1) 
= 787.8(1)(1.196)(1.088)   
 2   0.27 
= 13 170 psi Ans.
Eq. (14-41):
 S Y / ( KT K R ) 
(S F ) P =  t N

σ

28 260(0.977) / [(1)(0.85)]
=
= 2.47
13 170
Ans.
Gear tooth bending
Eq. (14-15):
Eq. (14-41):
 6   (1.156)(1) 
(σ )G = 787.8(1)(1.196)(1.097)   
= 9433 psi
 2   0.38 
28 260(0.996) / [(1)(0.85)]
( S F )G =
= 3.51 Ans.
9433
Ans.
Pinion tooth wear
1/ 2
Eq. (14-16):

K C 
(σ c ) P = C p  W t K o K v K s m f 
dP F I P

1/ 2

 1.156   1  
= 2300 787.8(1)(1.196)(1.088) 


 2.667(2)   0.1205  

= 98 760 psi Ans.
Eq. (14-42):
 S Z /( KT K R ) 
 93 500(0.948) /[(1)(0.85)] 
(S H ) P =  c N
 = 1.06
 =
σc
98 760


P 
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Ans.
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Gear tooth wear
1/ 2
1/ 2
 (K ) 
 1.097 
(σ c )G =  s G  (σ c ) P = 
(98 760) = 99 170 psi
 1.088 
 (K s )P 
93 500(0.973)(1) /[(1)(0.85)]
( S H )G =
= 1.08 Ans.
99 170
Ans.
The hardness of the pinion and the gear should be increased.
________________________________________________________________________
14-20
dP = 2.5(20) = 50 mm, dG = 2.5(36) = 90 mm
π d PnP π (50)(10−3 )(100)
V =
=
= 0.2618 m/s
60
60
60(120)
Wt =
= 458.4 N
π (50)(10−3 )(100)
With no specific information given to indicate otherwise, assume
KB = Ko = Yθ = ZR = 1
Eq. (14-28):
Eq. (14-27):
Table 14-2:
Qv = 6, B = 0.25(12 – 6)2/3 = 0.8255
A = 50 + 56(1 – 0.8255) = 59.77
 59.77 + 200(0.2618) 
Kv = 

59.77


YP = 0.322, YG = 0.3775
0.8255
= 1.099
Similar to Eq. (a) of Sec. 14-10 but for SI units:
Ks =
(
1
= 0.8433 mF Y
kb
)
0.0535
( K s ) P = 0.8433  2.5(18) 0.322 
0.0535
( K s )G = 0.8433  2.5(18) 0.3775 
Cmc = Ce = C pm = 1
= 1.003
0.0535
=1.007
use 1
use 1
18
− 0.025 = 0.011
10(50)
= 0.247 + 0.0167(0.709) − 0.765(10−4 )(0.7092 ) = 0.259
F = 18 / 25.4 = 0.709 in, C pf =
Cma
K H = 1 + 1[0.011(1) + 0.259(1)] = 1.27
Fig. 14-14:
(YN )P = 1.3558(108)–0.0178 = 0.977
(YN )G = 1.3558(108/1.8)–0.0178 = 0.987
Fig. 14-6:
(YJ )P = 0.33, (YJ )G = 0.38
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Eq. (14-38):
Table 14-8:
YZ = 0.658 – 0.0759 ln(1 – 0.95) = 0.885
cos 20o sin 20o  1.8 
ZI =

 = 0.103
2(1)
 1.8 + 1 
Z E = 191 MPa
Strength
Grade 1 steel, HBP = HBG = 200
Fig. 14-2:
Fig. 14-5:
Fig. 14-15:
(St)P = (St)G = 0.533(200) + 88.3 = 194.9 MPa
(Sc)P = (Sc)G = 2.22(200) + 200 = 644 MPa
(ZN )P = 1.4488(108)–0.023 = 0.948
(Z N )G = 1.4488(108 / 1.8) −0.023 = 0.961
Fig. 14-12:
H BP / H BG = 1
Eq. (14-23):
∴ ZW = CH = 1
Pinion tooth bending
Eq. (14-15):

1 KH KB 
(σ ) P =  W t K o K v K s

bmt YJ  P

 1  1.27(1) 
= 458.4(1)(1.099)(1) 

 = 43.08 MPa
18(2.5)   0.33 
S Y 
194.9  0.977 
Eq. (14-41) for SI: ( S F ) P =  t N  =
= 4.99
43.08 1(0.885) 
 σ Yθ YZ  P
Ans.
Ans.
Gear tooth bending
 1  1.27(1) 
(σ )G = 458.4(1)(1.099)(1) 

 = 37.42 MPa
18(2.5)   0.38 
194.9  0.987 
= 5.81 Ans.
( S F )G =
37.42 1(0.885) 
Ans.
Pinion tooth wear
Eq. (14-16):

K Z 
(σ c ) P =  Z E W t K o K v K s H R 

d w1b Z I  P

 1.27   1 
= 191 458.4(1)(1.099)(1) 

 = 501.8 MPa
 50(18)   0.103 
S Z Z 
644  0.948(1) 
Eq. (14-42) for SI: ( S H ) P =  c N W  =
= 1.37 Ans.
501.8 1(0.885) 
 σ c Yθ YZ  P
Ans.
Gear tooth wear
1/ 2
 (K ) 
(σ c )G =  s G 
 (K s )P 
1/ 2
1
(σ c ) P =  
1
(501.8) = 501.8 MPa
Shigley’s MED, 10th edition
Ans.
Chapter 14 Solutions, Page 18/39
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644 0.961(1)
= 1.39 Ans.
501.8 1(0.885)
________________________________________________________________________
( S H )G =
14-21
Pt = Pn cosψ = 6 cos 30° = 5.196 teeth/in
16
48
(3.079) = 9.238 in
dP =
= 3.079 in, dG =
5.196
16
π (3.079)(300)
V =
= 241.8 ft/min
12
0.8255
 59.77 + 241.8 
33 000(5)
t
W =
= 682.3 lbf , K v = 
= 1.210

241.8
59.77


From Prob. 14-19:
YP = 0.296, YG = 0.4056
( K s ) P = 1.088, ( K s )G = 1.097, K B = 1
mG = 3, (YN ) P = 0.977, (YN )G = 0.996, K R = 0.85
(St ) P = (St )G = 28 260 psi, CH = 1, (Sc ) P = (Sc )G = 93 500 psi
(Z N ) P = 0.948,
(Z N )G = 0.973,
C p = 2300 psi
The pressure angle is:
 tan 20° 
φt = tan −1 
 = 22.80°
 cos 30° 
3.079
cos 22.8° = 1.419 in,
2
a = 1 / Pn = 1 / 6 = 0.167 in
(rb ) P =
(rb )G = 3(rb ) P = 4.258 in
Eq. (14-25):
1/ 2
2
2
 3.079

 9.238



+ 0.167  − 1.4192  + 
+ 0.167  − 4.2582 
Z = 


 2

 2

 3.079 9.238 
−
+
 sin 22.8°
2 
 2
= 0.9479 + 2.1852 − 2.3865 = 0.7466 Conditions O.K . for use
p N = pn cos φ n =
Eq. (14-21):
mN =
π
6
1/ 2
cos 20° = 0.4920 in
pN
0.492
=
= 0.6937
0.95Z
0.95(0.7466)
Shigley’s MED, 10th edition
Chapter 14 Solutions, Page 19/39
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Eq. (14-23):
 sin 22.8° cos 22.8°   3 
I =
  3 + 1  = 0.193
2(0.6937)



Fig. 14-7:
J P′ o 0.45,
J G′ o 0.54
Fig. 14-8: Corrections are 0.94 and 0.98.
J P = 0.45(0.94) = 0.423, J G = 0.54(0.98) = 0.529
2
Cmc = 1, C pf =
− 0.0375 + 0.0125(2) = 0.0525
10(3.079)
C pm = 1, Cma = 0.093, Ce = 1
K m = 1 + (1)[0.0525(1) + 0.093(1)] = 1.146
Pinion tooth bending
 5.196  1.146(1)
= 6323 psi
(σ ) P = 682.3(1)(1.21)(1.088) 

 2  0.423
28 260(0.977) / [1(0.85)]
(S F ) P =
= 5.14 Ans.
6323
Ans.
Gear tooth bending
 5.196  1.146(1)
(σ )G = 682.3(1)(1.21)(1.097) 
= 5097 psi

 2  0.529
28 260(0.996) / [1(0.85)]
( S F )G =
= 6.50 Ans.
5097
Ans.
Pinion tooth wear
1/ 2

 1.146   1  
(σ c ) P = 2300 682.3(1)(1.21)(1.088) 


 3.078(2)   0.193  

93 500(0.948) / [(1)(0.85)]
= 1.54 Ans.
(S H ) P =
67 700
= 67 700 psi
Ans.
Gear tooth wear
1/ 2
1.097 
(σ c )G = 
(67 700) = 67 980 psi Ans.
1.088 
93 500(0.973) /[(1)(0.85)]
= 1.57 Ans.
( S H )G =
67 980
________________________________________________________________________
14-22 Given: R = 0.99 at 108 cycles, HB = 232 through-hardening Grade 1, core and case, both
gears. NP = 17T, NG = 51T,
Table 14-2: YP = 0.303, YG = 0.4103
Shigley’s MED, 10th edition
Chapter 14 Solutions, Page 20/39
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Fig. 14-6:
JP = 0.292, JG = 0.396
dP = NP / P = 17 / 6 = 2.833 in, dG = 51 / 6 = 8.500 in.
Pinion bending
From Fig. 14-2:
0.99
Fig. 14-14:
(St )107 = 77.3H B + 12 800
= 77.3(232) + 12 800 = 30 734 psi
YN = 1.6831(108)–0.0323 = 0.928
V = π d P n / 12 = π (2.833)(1120 / 12) = 830.7 ft/min
KT = K R = 1, S F = 2, S H = 2
30 734(0.928)
= 14 261 psi
σ all =
2(1)(1)
Qv = 5, B = 0.25(12 − 5)2 / 3 = 0.9148
A = 50 + 56(1 − 0.9148) = 54.77
 54.77 + 830.7 
K v = 

54.77


0.9148
= 1.472
0.0535
 2 0.303 
K s = 1.192 
= 1.089 ⇒ use 1

6


K m = Cm f = 1 + Cmc (C p f C p m + CmaCe )
Cmc = 1
F
− 0.0375 + 0.0125F
10d
2
=
− 0.0375 + 0.0125(2) = 0.0581
10(2.833)
=1
C pf =
C pm
Eq. (14-15):
Cma = 0.127 + 0.0158(2) − 0.093(10−4 )(22 ) = 0.1586
Ce = 1
K m = 1 + 1[0.0581(1) + 0.1586(1)] = 1.217
KB = 1
FJ Pσ all
Wt =
K o K v K s Pd K m K B
2(0.292)(14 261)
=
= 775 lbf
1(1.472)(1)(6)(1.217)(1)
W tV
775(830.7)
H =
=
= 19.5 hp
33 000
33 000
Pinion wear
Shigley’s MED, 10th edition
Chapter 14 Solutions, Page 21/39
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Fig. 14-15:
ZN = 2.466N–0.056 = 2.466(108)–0.056 = 0.879
mG = 51 / 17 = 3
Eq. (14-23):
I =
Fig. 14-5:
(Sc )107 = 322H B + 29 100
= 322(232) + 29 100 = 103 804 psi
103 804(0.879)
σ c,all =
= 64 519 psi
2(1)(1)
cos 20o sin 20o  3 

 = 1.205,
2
 3 + 1
CH = 1
0.99
2
Eq. (14-16):
σ 
Fd P I
W =  c,all 
 C p  K o K v K s K mC f


2
 64 519   2(2.833)(0.1205) 
=
 2300  1(1.472)(1)(1.2167)(1) 
= 300 lbf
300(830.7)
W tV
=
= 7.55 hp
H =
33 000
33 000
t
The pinion controls, therefore Hrated = 7.55 hp Ans.
________________________________________________________________________
14-23
l = 2.25/ Pd,
t =
4 lx =
x = 3Y / 2Pd
 2.25  3Y  3.674
4
Y

 =
Pd
 Pd  2Pd 
 3.674 
F Y
de = 0.808 F t = 0.808 F 
 Y = 1.5487
Pd
 Pd 
−0.107
−0.0535
 1.5487 F Y / P 
F Y 
d

= 0.8389 
kb = 



0.30
 Pd 


0.0535
F Y 
1
= 1.192 
Ks =
Ans.

kb
P
d


________________________________________________________________________
14-24 YP = 0.331, YG = 0.422, JP = 0.345, JG = 0.410, Ko = 1.25. The service conditions are
adequately described by Ko. Set SF = SH = 1.
dP = 22 / 4 = 5.500 in
dG = 60 / 4 = 15.000 in
Shigley’s MED, 10th edition
Chapter 14 Solutions, Page 22/39
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V =
Pinion bending
0.99
π (5.5)(1145)
12
= 1649 ft/min
(St )107 = 77.3H B + 12 800 = 77.3(250) + 12 800 = 32 125 psi
YN = 1.6831[3(109 )]−0.0323 = 0.832
Eq. (14-17):
32 125(0.832)
= 26 728 psi
1(1)(1)
B = 0.25(12 − 6)2 / 3 = 0.8255
A = 50 + 56(1 − 0.8255) = 59.77
(σ all ) P =
0.8255
 59.77 + 1649 
K v = 
= 1.534

59.77


K s = 1, Cm = 1
F
Cmc =
− 0.0375 + 0.0125 F
10d
3.25
=
− 0.0375 + 0.0125(3.25) = 0.0622
10(5.5)
Cma = 0.127 + 0.0158(3.25) − 0.093(10−4 )(3.252 ) = 0.178
Ce = 1
K m = Cm f = 1 + (1)[0.0622(1) + 0.178(1)] = 1.240
K B = 1,
Eq. (14-15):
KT = 1
26 728(3.25)(0.345)
= 3151 lbf
1.25(1.534)(1)(4)(1.240)
3151(1649)
H1 =
= 157.5 hp
33 000
W1t =
Gear bending By similar reasoning, W2t = 3861 lbf and H 2 = 192.9 hp
Pinion wear
mG = 60 / 22 = 2.727
cos 20o sin 20o  2.727 

 = 0.1176
2
 1 + 2.727 
0.99 ( Sc )107 = 322(250) + 29 100 = 109 600 psi
I =
(Z N ) P = 2.466[3(109 )]−0.056 = 0.727
(Z N )G = 2.466[3(109 ) / 2.727]−0.056 = 0.769
109 600(0.727)
(σ c,all ) P =
= 79 679 psi
1(1)(1)
Shigley’s MED, 10th edition
Chapter 14 Solutions, Page 23/39
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2
σ 
Fd P I
W =  c,all 
 C p  K o K v K s K mC f


t
3
2
 79 679   3.25(5.5)(0.1176) 
=
 
 = 1061 lbf
 2300  1.25(1.534)(1)(1.24)(1) 
1061(1649)
H3 =
= 53.0 hp
33 000
Gear wear
Similarly,
W4t = 1182 lbf ,
H 4 = 59.0 hp
Rating
H rated = min( H1, H 2 , H 3, H 4 )
= min(157.5, 192.9, 53, 59) = 53 hp
Ans.
Note differing capacities. Can these be equalized?
________________________________________________________________________
14-25 From Prob. 14-24:
W1t = 3151 lbf , W2t = 3861 lbf ,
W3t = 1061 lbf , W4t = 1182 lbf
33 000 Ko H
33 000(1.25)(40)
=
= 1000 lbf
Wt =
V
1649
Pinion bending: The factor of safety, based on load and stress, is
(S F ) P =
W1t
3151
=
= 3.15
1000 1000
Ans.
Gear bending based on load and stress
( S F )G =
W2t
3861
=
= 3.86
1000 1000
Ans.
Pinion wear
based on load:
based on stress:
W3t
1061
=
= 1.06
1000 1000
(S H ) P = 1.06 = 1.03
n3 =
Ans.
Gear wear
based on load:
n4 =
W4t
1182
=
= 1.18
1000 1000
Shigley’s MED, 10th edition
Chapter 14 Solutions, Page 24/39
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(SH )G = 1.18 = 1.09
based on stress:
Ans.
Factors of safety are used to assess the relative threat of loss of function 3.15, 3.86, 1.06,
1.18 where the threat is from pinion wear. By comparison, the AGMA safety factors
(SF)P, (SF)G, (SH)P, (SH)G
are
3.15, 3.86, 1.03, 1.09 or 3.15, 3.86, 1.061/2, 1.181/2
and the threat is again from pinion wear. Depending on the magnitude of the numbers,
using SF and SH as defined by AGMA, does not necessarily lead to the same conclusion
concerning threat. Therefore be cautious.
________________________________________________________________________
14-26 Solution summary from Prob. 14-24: n = 1145 rev/min, Ko = 1.25, Grade 1 materials,
NP = 22T, NG = 60T, mG = 2.727, YP = 0.331,YG = 0.422, JP = 0.345, JG = 0.410, Pd =
4T /in, F = 3.25 in, Qv = 6, (Nc)P = 3(109), R = 0.99, Km = 1.240, KT = 1, KB = 1,
dP = 5.500 in, dG = 15.000 in, V = 1649 ft/min, Kv = 1.534, (Ks )P = (Ks )G = 1, (YN )P =
0.832, (YN )G = 0.859, KR = 1
Pinion HB: 250 core, 390 case
Gear HB: 250 core, 390 case
Bending
(σ all ) P
(σ all )G
W1t
W2t
=
=
=
=
26 728 psi
27 546 psi
3151 lbf ,
3861 lbf ,
(St ) P = 32 125 psi
(St )G = 32 125 psi
H1 = 157.5 hp
H 2 = 192.9 hp
Wear
φ = 20o ,
I = 0.1176,
(Z N ) P = 0.727
(Z N )G = 0.769, CP = 2300 psi
(Sc ) P = Sc = 322(390) + 29 100 = 154 680 psi
154 680(0.727)
(σ c,all ) P =
= 112 450 psi
1(1)(1)
154 680(0.769)
(σ c,all )G =
= 118 950 psi
1(1)(1)
2
 112 450 
W =
 (1061) = 2113 lbf ,
 79 679 
t
3
H3 =
2113(1649)
= 105.6 hp
33 000
2


118 950
W4t = 
 (1182) = 2354 lbf ,
 109 600(0.769) 
H4 =
Shigley’s MED, 10th edition
2354(1649)
= 117.6 hp
33 000
Chapter 14 Solutions, Page 25/39
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Rated power
Hrated = min(157.5, 192.9, 105.6, 117.6) = 105.6 hp
Ans.
Prob. 14-24:
Hrated = min(157.5, 192.9, 53.0, 59.0) = 53 hp
The rated power approximately doubled.
________________________________________________________________________
14-27 The gear and the pinion are 9310 grade 1, carburized and case-hardened to obtain Brinell
285 core and Brinell 580–600 case.
Table 14-3:
0.99
(St )107 = 55 000 psi
Modification of St by (YN )P = 0.832 produces
(σ all ) P = 45 657 psi,
Similarly for (YN )G = 0.859
(σ all )G = 47 161 psi,
W = 4569 lbf ,
W = 5668 lbf ,
t
1
t
2
and
H1 = 228 hp
H 2 = 283 hp
From Table 14-8, C p = 2300 psi. Also, from Table 14-6:
0.99
(Sc )107 = 180 000 psi
Modification of Sc by YN produces
(σ c,all ) P = 130 525 psi
(σ c,all )G = 138 069 psi
and
W3t = 2489 lbf ,
W4t = 2767 lbf,
H 3 = 124.3 hp
H 4 = 138.2 hp
Rating
Hrated = min(228, 283, 124, 138) = 124 hp Ans.
________________________________________________________________________
Shigley’s MED, 10th edition
Chapter 14 Solutions, Page 26/39
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14-28 Grade 2, 9310 carburized and case-hardened to 285 core and 580 case in Prob. 14-27.
Summary:
Table 14-3:
0.99
(St )107 = 65 000 psi
(σ all ) P = 53 959 psi
(σ all )G = 55 736 psi
and it follows that
W1t = 5400 lbf ,
W2t = 6699 lbf ,
H1 = 270 hp
H 2 = 335 hp
From Table 14-8, C p = 2300 psi. Also, from Table 14-6:
Sc = 225 000 psi
(σ c,all ) P = 181 285 psi
(σ c,all )G = 191 762 psi
Consequently,
W3t = 4801 lbf ,
W4t = 5337 lbf ,
H 3 = 240 hp
H 4 = 267 hp
Rating
Hrated = min(270, 335, 240, 267) = 240 hp. Ans.
________________________________________________________________________
14-29 Given: n = 1145 rev/min, Ko = 1.25, NP = 22T, NG = 60T, mG = 2.727, dP = 2.75 in, dG =
7.5 in, YP = 0.331,YG = 0.422, JP = 0.335, JG = 0.405, P = 8T /in, F = 1.625 in, HB = 250,
case and core, both gears. Cm = 1, F/dP = 0.0591, Cf = 0.0419, Cpm = 1, Cma = 0.152,
Ce = 1, Km = 1.1942, KT = 1, KB = 1, Ks = 1,V = 824 ft/min, (YN )P = 0.8318, (YN )G =
0.859, KR = 1, I = 0.117 58
(St )107 = 32 125 psi
(σ all ) P = 26 668 psi
(σ all )G = 27 546 psi
0.99
and it follows that
W1t = 879.3 lbf ,
W2t = 1098 lbf ,
H1 = 21.97 hp
H 2 = 27.4 hp
Shigley’s MED, 10th edition
Chapter 14 Solutions, Page 27/39
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For wear
W3t = 304 lbf ,
W4t = 340 lbf ,
H 3 = 7.59 hp
H 4 = 8.50 hp
Rating
Hrated = min(21.97, 27.4, 7.59, 8.50) = 7.59 hp
In Prob. 14-24, Hrated = 53 hp. Thus,
7.59
1
= 0.1432 =
,
53.0
6.98
not
1
8
Ans.
The transmitted load rating is
t
Wrated
= min(879.3, 1098, 304, 340) = 304 lbf
In Prob. 14-24
t
Wrated
= 1061 lbf
Thus
304
1
1
= 0.2865 =
, not
Ans.
1061
3.49
4
________________________________________________________________________
14-30 SP = SH = 1,
Bending
Table 14-4:
Pd = 4,
0.99
JP = 0.345,
JG = 0.410, Ko = 1.25
(St )107 = 13 000 psi
13 000(1)
= 13 000 psi
1(1)(1)
σ all FJ P
13 000(3.25)(0.345)
=
=
= 1533 lbf
K o K v K s Pd K m K B 1.25(1.534)(1)(4)(1.24)(1)
1533(1649)
=
= 76.6 hp
33 000
= W1t J G / J P = 1533(0.410) / 0.345 = 1822 lbf
= H1J G / J P = 76.6(0.410) / 0.345 = 91.0 hp
(σ all ) P = (σ all )G =
W1t
H1
W2t
H2
Wear
Table 14-8:
Table 14-7:
C p = 1960 psi
0.99
(Sc )107 = 75 000 psi = (σ c,all ) P = (σ c,all )G
Shigley’s MED, 10th edition
Chapter 14 Solutions, Page 28/39
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2
 (σ ) 
Fd p I
W =  c,all P 
 C p  K o K v K s K mC f
t
3
2
 75 000  3.25(5.5)(0.1176)
W3t = 
= 1295 lbf

 1960  1.25(1.534)(1)(1.24)(1)
W4t = W3t = 1295 lbf
1295(1649)
H4 = H3 =
= 64.7 hp
33 000
Rating
Hrated = min(76.7, 94.4, 64.7, 64.7) = 64.7 hp
Ans.
Notice that the balance between bending and wear power is improved due to CI’s more
favorable Sc/St ratio. Also note that the life is 107 pinion revolutions which is (1/300) of
3(109). Longer life goals require power de-rating.
________________________________________________________________________
14-31 From Table A-24a, Eav = 11.8(106) Mpsi
For φ = 14.5° and HB = 156
SC =
1.4(81)
= 51 693 psi
2sin14.5° / [11.8(106 )]
For φ = 20°
1.4(112)
= 52 008 psi
2sin 20° / [11.8(106 )]
SC = 0.32(156) = 49.9 kpsi
The first two calculations were approximately 4 percent higher.
________________________________________________________________________
SC =
14-32 Programs will vary.
________________________________________________________________________
14-33
(YN ) P = 0.977,
(YN )G = 0.996
(St ) P = (St )G = 82.3(250) + 12 150 = 32 725 psi
32 725(0.977)
(σ all ) P =
= 37 615 psi
1(0.85)
37 615(1.5)(0.423)
W1t =
= 1558 lbf
1(1.404)(1.043)(8.66)(1.208)(1)
1558(925)
H1 =
= 43.7 hp
33 000
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32 725(0.996)
= 38 346 psi
1(0.85)
38 346(1.5)(0.5346)
W2t =
= 2007 lbf
1(1.404)(1.043)(8.66)(1.208)(1)
2007(925)
H2 =
= 56.3 hp
33 000
(Z N ) P = 0.948, (Z N )G = 0.973
(σ all )G =
Table 14-6:
0.99
(Sc )107 = 150 000 psi
 0.948(1) 
(σ c,allow ) P = 150 000 
 = 167 294 psi
 1(0.85) 
2
 167 294  1.963(1.5)(0.195) 
W3t = 
 
 = 2074 lbf
 2300   1(1.404)(1.043) 
2074(925)
H3 =
= 58.1 hp
33 000
0.973
(σ c ,allow )G =
(167 294) = 171 706 psi
0.948
2
 171 706  1.963(1.5)(0.195) 
W4t = 
 
 = 2167 lbf
 2300   1(1.404)(1.052) 
2167(925)
H4 =
= 60.7 hp
33 000
H rated = min(43.7, 56.3, 58.1, 60.7) = 43.7 hp Ans.
Pinion bending is controlling.
________________________________________________________________________
(YN ) P = 1.6831(108 ) −0.0323 = 0.928
14-34
(YN )G = 1.6831(108 / 3.059) −0.0323 = 0.962
Table 14-3:
St = 55 000 psi
55 000(0.928)
(σ all ) P =
= 60 047 psi
1(0.85)
60 047(1.5)(0.423)
W1t =
= 2487 lbf
1(1.404)(1.043)(8.66)(1.208)(1)
2487(925)
H1 =
= 69.7 hp
33 000
0.962
(σ all )G =
(60 047) = 62 247 psi
0.928
Shigley’s MED, 10th edition
Chapter 14 Solutions, Page 30/39
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62 247  0.5346 

 (2487) = 3258 lbf
60 047  0.423 
3258
H2 =
(69.7) = 91.3 hp
2487
W2t =
Table 14-6:
Sc = 180 000 psi
(Z N ) P = 2.466(108 )−0.056 = 0.8790
(Z N )G = 2.466(108 / 3.059)−0.056 = 0.9358
180 000(0.8790)
(σ c,all ) P =
= 186 141 psi
1(0.85)
2
 186 141  1.963(1.5)(0.195) 
W3t = 
 
 = 2568 lbf
 2300   1(1.404)(1.043) 
2568(925)
H3 =
= 72.0 hp
33 000
0.9358
(σ c ,all )G =
(186 141) = 198 169 psi
0.8790
2
 198 169   1.043 
W4t = 
 
 (2568) = 2886 lbf
 186 141   1.052 
2886(925)
H4 =
= 80.9 hp
33 000
H rated = min(69.7, 91.3, 72, 80.9) = 69.7 hp Ans.
Pinion bending controlling
________________________________________________________________________
(YN)P = 0.928, (YN )G = 0.962
14-35
Table 14-3:
(See Prob. 14-34)
St = 65 000 psi
65 000(0.928)
(σ all ) P =
= 70 965 psi
1(0.85)
70 965(1.5)(0.423)
W1t =
= 2939 lbf
1(1.404)(1.043)(8.66)(1.208)
2939(925)
H1 =
= 82.4 hp
33 000
65 000(0.962)
(σ all )G =
= 73 565 psi
1(0.85)
73 565  0.5346 
W2t =

 (2939) = 3850 lbf
70 965  0.423 
3850
H2 =
(82.4) = 108 hp
2939
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Table 14-6:
Sc = 225 000 psi
(Z N ) P = 0.8790, (Z N )G = 0.9358
225 000(0.879)
(σ c,all ) P =
= 232 676 psi
1(0.85)
2
 232 676  1.963(1.5)(0.195) 
W3t = 
 
 = 4013 lbf
 2300   1(1.404)(1.043) 
4013(925)
H3 =
= 112.5 hp
33 000
0.9358
(σ c ,all )G =
(232 676) = 247 711 psi
0.8790
2
 247 711   1.043 
W4t = 
 
 (4013) = 4509 lbf
 232 676   1.052 
4509(925)
H4 =
= 126 hp
33 000
H rated = min(82.4, 108, 112.5, 126) = 82.4 hp Ans.
The bending of the pinion is the controlling factor.
________________________________________________________________________
P = 2 teeth/in, d = 8 in, N = dP = 8 (2) = 16 teeth
π 
π 
F = 4 p = 4   = 4   = 2π
P
2
∑ M x = 0 = 10(300) cos 20 − 4FB cos20
FB = 750 lbf
W t = FB cos 20 = 750cos 20 = 705 lbf
n = 2400 / 2 = 1200 rev/min
π dn π (8)(1200)
V =
=
= 2513 ft/min
12
12
14-36
We will obtain all of the needed factors, roughly in the order presented in the textbook.
Fig. 14-2:
St = 102(300) + 16 400 = 47 000 psi
Fig. 14-5:
Fig. 14-6:
Sc = 349(300) + 34 300 = 139 000 psi
J = 0.27
cos 20o sin 20o  2 
I =

 = 0.107
2(1)
 2 + 1
Eq. (14-23):
Table 14-8:
C p = 2300 psi
Assume a typical quality number of 6.
Eq. (14-28): B = 0.25(12 − Qv )2 / 3 = 0.25(12 − 6)2/ 3 = 0.8255
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A = 50 + 56(1 − B) = 50 + 56(1 − 0.8255) = 59.77
B
Eq. (14-27):
A+ V 
 59.77 + 2513 
K v = 
 = 

A 
59.77



0.8255
= 1.65
To estimate a size factor, get the Lewis Form Factor from Table 14-2, Y = 0.296.
From Eq. (a), Sec. 14-10,
0.0535
0.0535
F Y 
 2π 0.296 
= 1.192 
= 1.23
K s = 1.192 


2
 P 


The load distribution factor is applicable for straddle-mounted gears, which is not the
case here since the gear is mounted outboard of the bearings. Lacking anything better,
we will use the load distribution factor as a rough estimate.
Eq. (14-31):
Eq. (14-32):
Eq. (14-33):
Fig. 14-11:
Eq. (14-35):
Eq. (14-30):
Cmc = 1 (uncrowned teeth)
2π
Cp f =
− 0.0375 + 0.0125(2π ) = 0.1196
10(8)
Cpm = 1.1
Cma = 0.23 (commercial enclosed gear unit)
Ce = 1
K m = 1 + 1[0.1196(1.1) + 0.23(1)] = 1.36
For the stress-cycle factors, we need the desired number of load cycles.
Fig. 14-14:
Fig. 14-15:
Eq. 14-38:
N = 15 000 h (1200 rev/min)(60 min/h) = 1.1 (109) rev
YN = 0.9
ZN = 0.8
K R = 0.658 − 0.0759 ln (1 − R ) = 0.658 − 0.0759ln (1 − 0.95) = 0.885
With no specific information given to indicate otherwise, assume Ko = KB = KT = Cf = 1
Tooth bending
Eq. (14-15):
Eq. (14-41):
Pd K m K B
F J
 2   (1.36)(1) 
= 705(1)(1.65)(1.23) 
= 2294 psi

 2π   0.27 
σ = W t KoKvKs
 S Y / ( KT K R ) 
SF =  t N

σ

47 000(0.9) / [(1)(0.885)]
=
= 20.8
2294
Shigley’s MED, 10th edition
Ans.
Chapter 14 Solutions, Page 33/39
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Tooth wear
1/ 2
Eq. (14-16):

K C 
σ c = C p W t Ko Kv Ks m f 
dPF I 

1/ 2

 1.36   1  
= 2300 705(1)(1.65)(1.23) 


 8(2π )   0.107  

= 43 750 psi
Since gear B is a pinion, CH is not used in Eq. (14-42) (see p. 753), where
Sc Z N / ( KT K R )
SH =
σc
139 000(0.8) / [(1)(0.885)] 
=
 = 2.9 Ans
43 750


________________________________________________________________________
m = 18.75 mm/tooth, d = 300 mm
N = d/m = 300 / 18.75 = 16 teeth
F = b = 4 p = 4 (π m ) = 4π (18.75) = 236 mm
14-37
∑M
= 0 = 300(11) cos 20 − 150 FB cos25
FB = 22.81 kN
W t = FB cos 25 = 22.81cos 25 = 20.67 kN
n = 1800 / 2 = 900 rev/min
π dn π (0.300)(900)
V =
=
= 14.14 m/s
60
60
We will obtain all of the needed factors, roughly in the order presented in the textbook.
Fig. 14-2:
Fig. 14-5:
Fig. 14-6:
Eq. (14-23):
Table 14-8:
x
St = 0.703(300) + 113 = 324 MPa
Sc = 2.41(300) + 237 = 960 MPa
J = YJ = 0.27
cos 20o sin 20o  5 
I = ZI =

 = 0.134
2(1)
 5 + 1
Z E = 191 MPa
Assume a typical quality number of 6.
Eq. (14-28): B = 0.25(12 − Qv ) 2/ 3 = 0.25(12 − 6)2 / 3 = 0.8255
A = 50 + 56(1 − B) = 50 + 56(1 − 0.8255) = 59.77
B
Eq. (14-27):
 59.77 + 200(14.14) 
 A + 200V 
K v = 

 = 

59.77
A




0.8255
= 1.69
To estimate a size factor, get the Lewis Form Factor from Table 14-2, Y = 0.296.
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Similar to Eq. (a) of Sec. 14-10 but for SI units:
Ks =
(
1
= 0.8433 mF Y
kb
)
0.0535
K s = 0.8433 18.75(236) 0.296 
0.0535
= 1.28
Convert the diameter and facewidth to inches for use in the load-distribution factor
equations. d = 300/25.4 = 11.81 in, F = 236/25.4 = 9.29 in
Eq. (14-31):
Eq. (14-32):
Eq. (14-33):
Fig. 14-11:
Eq. (14-35):
Eq. (14-30):
Cmc = 1 (uncrowned teeth)
9.29
C pf =
− 0.0375 + 0.0125(9.29) = 0.1573
10(11.81)
Cpm = 1.1
Cma = 0.27 (commercial enclosed gear unit)
Ce = 1
K m = K H = 1 + 1[0.1573(1.1) + 0.27(1)] = 1.44
For the stress-cycle factors, we need the desired number of load cycles.
N = 12 000 h (900 rev/min)(60 min/h) = 6.48 (108) rev
Fig. 14-14:
Fig. 14-15:
Eq. 14-38:
YN = 0.9
ZN = 0.85
K R = 0.658 − 0.0759 ln (1 − R ) = 0.658 − 0.0759ln (1 − 0.98) = 0.955
With no specific information given to indicate otherwise, assume Ko = KB = KT = ZR = 1.
Tooth bending
Eq. (14-15):
σ = W t KoKvKs
1 KH KB
bmt YJ

  (1.44)(1) 
1
= 20 670(1)(1.69)(1.28) 


 = 53.9 MPa
 236(18.75)   0.27 
Eq. (14-41):
 S Y / ( KT K R ) 
SF =  t N

σ

324(0.9) / [(1)(0.955)]
=
= 5.66
53.9
Ans.
Tooth wear
1/ 2
Eq. (14-16):

K Z 
σ c = Z E W t KoKvKs H R 
d w1b Z I 

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1/ 2

 1.44   1  
= 191  20 670(1)(1.69)(1.28) 


 300(236)   0.134  

= 498 MPa
Since gear B is a pinion, CH is not used in Eq. (14-42) (see p. 757), where
SH =
Sc Z N / ( KT K R )
σc
960(0.85) / [(1)(0.955)]
=
= 1.72 Ans
498
________________________________________________________________________
14-38 From the solution to Prob. 13-40, n = 191 rev/min, Wt = 1600 N, d = 125 mm,
N = 15 teeth, m = 8.33 mm/tooth.
F = b = 4 p = 4 (π m ) = 4π ( 8.33) = 105 mm
V =
π dn
60
=
π (0.125)(191)
= 1.25 m/s
60
We will obtain all of the needed factors, roughly in the order presented in the textbook.
Table 14-3:
Table 14-6:
Fig. 14-6:
Eq. (14-23):
Table 14-8:
St = 65 kpsi = 448 MPa
Sc = 225 kpsi = 1550 MPa
J = YJ = 0.25
cos 20o sin 20o  2 
I = ZI =

 = 0.107
2(1)
 2 + 1
Z E = 191 MPa
Assume a typical quality number of 6.
Eq. (14-28):
Eq. (14-27):
B = 0.25(12 − Qv )2 / 3 = 0.25(12 − 6)2/ 3 = 0.8255
A = 50 + 56(1 − B) = 50 + 56(1 − 0.8255) = 59.77
 A + 200V
K v = 
A

B
 59.77 + 200(1.25) 


 = 
59.77



0.8255
= 1.21
To estimate a size factor, get the Lewis Form Factor from Table 14-2, Y = 0.290.
Similar to Eq. (a) of Sec. 14-10 but for SI units:
Ks =
(
1
= 0.8433 mF Y
kb
)
0.0535
0.0535
K s = 0.8433 8.33(105) 0.290 
= 1.17
Convert the diameter and facewidth to inches for use in the load-distribution factor
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equations. d = 125/25.4 = 4.92 in, F = 105/25.4 = 4.13 in
Eq. (14-31): Cmc = 1 (uncrowned teeth)
4.13
Eq. (14-32): C pf =
− 0.0375 + 0.0125(4.13) = 0.0981
10(4.92)
Eq. (14-33): Cpm = 1
Fig. 14-11:
Cma = 0.32 (open gearing)
Eq. (14-35): Ce = 1
Eq. (14-30): K m = K H = 1 + 1[0.0981(1) + 0.32(1)] = 1.42
For the stress-cycle factors, we need the desired number of load cycles.
N = 12 000 h (191 rev/min)(60 min/h) = 1.4 (108) rev
Fig. 14-14:
YN = 0.95
Fig. 14-15:
ZN = 0.88
K R = 0.658 − 0.0759 ln (1 − R ) = 0.658 − 0.0759ln (1 − 0.95) = 0.885
Eq. 14-38:
With no specific information given to indicate otherwise, assume Ko = KB = KT = ZR = 1.
Tooth bending
Eq. (14-15):
σ = W t KoKvKs
1 KH KB
bmt YJ

  (1.42)(1) 
1
= 1600(1)(1.21)(1.17) 

 = 14.7 MPa
105(8.33)   0.25 
Since gear is a pinion, CH is not used in Eq. (14-42) (see p. 753), where
 S Y / ( KT K R ) 
SF =  t N

σ

448(0.95) / [(1)(0.885)]
=
= 32.7
14.7
Ans.
Tooth wear
1/ 2
Eq. (14-16):

K Z 
σ c = Z E W t KoKvKs H R 
d w1b Z I 

1/ 2

 1.42   1  
= 191 1600(1)(1.21)(1.17) 


 125(105)   0.107  

= 289 MPa
 S Z / ( KT K R ) 
SH =  c N

σc


1550(0.88) / [(1)(0.885)] 
=
 = 5.33 Ans
289


________________________________________________________________________
Eq. (14-42):
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14-39 From the solution to Prob. 13-41, n = 2(70) = 140 rev/min, Wt = 180 lbf, d = 5 in
N = 15 teeth, P = 3 teeth/in.
π 
π 
F = 4 p = 4   = 4   = 4.2 in
P
3
π dn π (5)(140)
=
= 183.3 ft/min
V =
12
12
We will obtain all of the needed factors, roughly in the order presented in the textbook.
Table 14-3:
Table 14-6:
Fig. 14-6:
Eq. (14-23):
Table 14-8:
St = 65 kpsi
Sc = 225 kpsi
J = 0.25
cos 20o sin 20o  2 
I =

 = 0.107
2(1)
 2 + 1
C p = 2300 psi
Assume a typical quality number of 6.
Eq. (14-28):
B = 0.25(12 − Qv )2 / 3 = 0.25(12 − 6)2/ 3 = 0.8255
A = 50 + 56(1 − B) = 50 + 56(1 − 0.8255) = 59.77
B
Eq. (14-27):
A+ V 
 59.77 + 183.3 
K v = 
 = 

59.77
A 



0.8255
= 1.18
To estimate a size factor, get the Lewis Form Factor from Table 14-2, Y = 0.290.
From Eq. (a), Sec. 14-10,
0.0535
Eq. (14-31):
Eq. (14-32):
Eq. (14-33):
Fig. 14-11:
Eq. (14-35):
Eq. (14-30):
F Y 
 4.2 0.290 
= 1.192 
K s = 1.192 


3
 P 


Cmc = 1 (uncrowned teeth)
4.2
C pf =
− 0.0375 + 0.0125(4.2) = 0.099
10(5)
Cpm = 1
Cma = 0.32 (Open gearing)
Ce = 1
K m = 1 + 1[0.099(1) + 0.32(1)] = 1.42
0.0535
= 1.17
For the stress-cycle factors, we need the desired number of load cycles.
N = 14 000 h (140 rev/min)(60 min/h) = 1.2 (108) rev
Fig. 14-14:
YN = 0.95
Fig. 14-15:
ZN = 0.88
K R = 0.658 − 0.0759 ln (1 − R ) = 0.658 − 0.0759ln (1 − 0.98) = 0.955
Eq. 14-38:
With no specific information given to indicate otherwise, assume Ko = KB = KT = Cf = 1.
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Tooth bending
Eq. (14-15):
Eq. (14-41):
Pd K m K B
F J
 3   (1.42)(1) 
= 180(1)(1.18)(1.17) 
= 1010 psi

 4.2   0.25 
σ = W t KoKvKs
 S Y / ( KT K R ) 
SF =  t N

σ

65 000(0.95) / [(1)(0.955)]
=
= 64.0
1010
Ans.
Tooth wear
1/ 2
Eq. (14-16):

K C 
σ c = C p W t Ko Kv Ks m f 
dPF I 

1/ 2

 1.42   1  
= 2300 180(1)(1.18)(1.17) 


 5(4.2)   0.107  

= 28 800 psi
Since gear B is a pinion, CH is not used in Eq. (14-42) (see p. 753), where
 S Z / ( KT K R ) 
SH =  c N

σc


 225 000(0.88) / [(1)(0.955)] 
=
 = 7.28 Ans
28 800


________________________________________________________________________
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Chapter 15
15-1
Given: Uncrowned, through-hardened 300 Brinell core and case, Grade 1, NC =
109 rev of pinion at R = 0.999, NP = 20 teeth, NG = 60 teeth, Qv = 6, Pd = 6
teeth/in, normal pressure angle = 20°, shaft angle = 90°, np = 900 rev/min, F =
1.25 in, SF = SH = 1, Ko = 1.
Fig. 15-7:
JP = 0.249, JG = 0.216
Mesh
dP = 20/6 = 3.333 in, dG = 60/6 = 10.000 in
Eq. (15-7):
vt = π(3.333)(900/12) = 785.3 ft/min
Eq. (15-6):
B = 0.25(12 – 6)2/3 = 0.8255
A = 50 + 56(1 – 0.8255) = 59.77
Eq. (15-5):
 59.77 + 785.3 
K v = 

59.77


Eq. (15-8):
vt,max = [59.77 + (6 – 3)]2 = 3940 ft/min
0.8255
= 1.374
Since 785.3 < 3904, Kv = 1.374 is valid. The size factor for bending is:
Eq. (15-10):
Ks = 0.4867 + 0.2132 / 6 = 0.5222
For one gear straddle-mounted, the load-distribution factor is:
Eq. (15-11):
Km = 1.10 + 0.0036 (1.25)2 = 1.106
For KL, the more conservative “critical” equation will be used here. Though it is
acceptable, according to AGMA, to use the less conservative “general” equation
for general applications.
Eq. (15-15): (KL)P = 1.6831(109)–0.0323 = 0.862
(KL)G = 1.6831(109 / 3)–0.0323 = 0.893
Eq. (15-14):
(CL)P = 3.4822(109)–0.0602 = 1
(CL)G = 3.4822(109 / 3)–0.0602 = 1.069
Eq. (15-19):
KR = 0.50 – 0.25 log(1 – 0.999) = 1.25 (or Table 15-3)
CR = K R = 1.25 = 1.118
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Bending
St = sat = 44(300) + 2100 = 15 300 psi
Fig. 15-13:
0.99
Eq. (15-4):
(σ all ) P = sw t =
Eq. (15-3):
Eq. (15-4):
sa t K L
=
15 300(0.862)
= 10 551 psi
1(1)(1.25)
S F KT K R
(σ all ) P FK x J P
WPt =
Pd K o K v K s K m
10 551(1.25)(1)(0.249)
=
= 690 lbf
6(1)(1.374)(0.5222)(1.106)
690(785.3)
H1 =
= 16.4 hp
33 000
15 300(0.893)
= 10 930 psi
1(1)(1.25)
10 930(1.25)(1)(0.216)
WGt =
= 620 lbf
6(1)(1.374)(0.5222)(1.106)
620(785.3)
H2 =
= 14.8 hp Ans.
33 000
(σ all )G =
The gear controls the bending rating.
________________________________________________________________________
15-2
Refer to Prob. 15-1 for the gearset specifications.
Wear
Fig. 15-12:
sac = 341(300) + 23 620 = 125 920 psi
For the pinion, CH = 1. From Prob. 15-1, CR = 1.118. Thus, from Eq. (15-2):
sac (CL ) P CH
S H KT C R
125 920(1)(1)
= 112 630 psi
(σ c,all ) P =
1(1)(1.118)
(σ c,all ) P =
For the gear, from Eq. (15-16),
B1 = 0.008 98(300 / 300) − 0.008 29 = 0.000 69
CH = 1 + 0.000 69(3 − 1) = 1.001 38
From Prob. 15-1, (CL)G = 1.0685. Equation (15-2) thus gives
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sac (CL )G CH
S H KT C R
125 920(1.0685)(1.001 38)
= 120 511 psi
(σ c,all )G =
1(1)(1.118)
(σ c,all )G =
For steel:
C p = 2290 psi
Eq. (15-9):
Cs = 0.125(1.25) + 0.4375 = 0.593 75
Fig. 15-6:
I = 0.083
Eq. (15-12):
Cxc = 2
Eq. (15-1):
 (σ ) 
Fd P I
W =  c ,all P 
 C p  K o K v K mCsC xc


2
t
P
2
=
=
H3 =
WGt =
=
H4 =

1.25(3.333)(0.083)
 112 630  

 

2290
1(1.374)(1.106)(0.5937)(2) 
464 lbf
464(785.3)
= 11.0 hp
33 000
2

1.25(3.333)(0.083)
 120 511 

 

2290
1(1.374)(1.106)(0.593 75)(2) 
531 lbf
531(785.3)
= 12.6 hp
33 000
The pinion controls wear:
H = 11.0 hp
Ans.
The power rating of the mesh, considering the power ratings found in Prob. 15-1,
is
H = min(16.4, 14.8, 11.0, 12.6) = 11.0 hp Ans.
________________________________________________________________________
15-3
AGMA 2003-B97 does not fully address cast iron gears. However, approximate
comparisons can be useful. This problem is similar to Prob. 15-1, but not
identical. We will organize the method. A follow-up could consist of completing
Probs. 15-1 and 15-2 with identical pinions, and cast iron gears.
Given: Uncrowned, straight teeth, Pd = 6 teeth/in, NP = 30 teeth, NG = 60 teeth,
ASTM 30 cast iron, material Grade 1, shaft angle 90°, F = 1.25, nP = 900 rev/min,
φn = 20°, one gear straddle-mounted, Ko = 1, SF = 2, S H = 2.
Fig. 15-7:
JP = 0.268, JG = 0.228
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dP = 30/6 = 5.000 in, dG = 60/6 = 10.000 in
Mesh
vt = π (5)(900 / 12) = 1178 ft/min
Set NL = 107 cycles for the pinion. For R = 0.99,
Table 15-7:
Table 15-5:
Eq. (15-4):
sat = 4500 psi
sac = 50 000 psi
sat K L
4500(1)
swt =
=
= 2250 psi
2(1)(1)
S F KT K R
The velocity factor Kv represents stress augmentation due to mislocation of tooth
profiles along the pitch surface and the resulting “falling” of teeth into
engagement. Equation (5-67) shows that the induced bending moment in a
cantilever (tooth) varies directly with E of the tooth material. If only the
material varies (cast iron vs. steel) in the same geometry, I is the same. From the
Lewis equation of Section 14-1,
M
K W tP
σ =
= v
I /c
FY
We expect the ratio σCI/σsteel to be
σ CI
( K v )CI
ECI
=
=
Esteel
σ steel ( K v )steel
In the case of ASTM class 30, from Table A-24(a)
(ECI)av = (13 + 16.2)/2 = 14.7 kpsi
( K v )CI =
Then,
14.7
( K v )steel = 0.7( K v )steel
30
Our modeling is rough, but it convinces us that (Kv)CI < (Kv)steel, but we are not
sure of the value of (Kv)CI. We will use Kv for steel as a basis for a conservative
rating.
Eq. (15-6):
B = 0.25(12 – 6)2/3 = 0.8255
A = 50 + 56(1 – 0.8255) = 59.77
Eq. (15-5):
 59.77 + 1178 
K v = 

59.77


Pinion bending
0.8255
= 1.454
(σall)P = swt = 2250 psi
From Prob. 15-1, Kx = 1, Km = 1.106, Ks = 0.5222
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Eq. (15-3):
(σ all ) P FK x J P
Pd K o K v K s K m
2250(1.25)(1)(0.268)
=
= 149.6 lbf
6(1)(1.454)(0.5222)(1.106)
149.6(1178)
H1 =
= 5.34 hp
33 000
WPt =
Gear bending
JG
 0.228 
= 149.6 
= 127.3 lbf
 0.268 
JP
127.3(1178)
H2 =
= 4.54 hp
33 000
WGt = WPt
The gear controls in bending fatigue. H = 4.54 hp Ans.
________________________________________________________________________
15-4
Continuing Prob. 15-3,
Table 15-5:
sac = 50 000 psi
50 000
sw t = σ c,all =
= 35 355 psi
2
2
Eq. (15-1):
σ 
Fd P I
W =  c,all 
 C p  K o K v K mCsCxc


Fig. 15-6:
I = 0.86
t
From Probs. 15-1 and 15-2: Cs = 0.593 75, Ks = 0.5222, Km = 1.106, Cxc = 2
C p = 1960 psi
From Table 14-8:
2
Thus,

1.25(5.000)(0.086)
 35 355  
W =
= 91.6 lbf


 1960  1(1.454)(1.106)(0.59375)(2) 
91.6(1178)
Ans.
H3 = H 4 =
= 3.27 hp
33 000
t
Rating
Side note: Based on results of Probs. 15-3 and 15-4,
H = min(5.34, 4.54, 3.27, 3.27) = 3.27 hp
The mesh is weakest in wear fatigue.
________________________________________________________________________
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15-5
Uncrowned, through-hardened to 180 Brinell (core and case), Grade 1, 109 rev of
pinion at R = 0.999, NP = z1 = 22 teeth, NG = z2 = 24 teeth, Qv = 5, met = 4 mm,
φn = 20°, shaft angle 90°, n1 = 1800 rev/min, SF = 1, S H = S F = 1, F = b = 25
mm, Ko = KA = KT = Kθ = 1 and C p = 190 MPa .
Fig. 15-7:
JP = YJ1 = 0.23, JG = YJ2 = 0.205
Mesh
dP = de1 = mz1 = 4(22) = 88 mm,
Eq. (15-7):
vet = 5.236(10–5)(88)(1800) = 8.29 m/s
Eq. (15-6):
B = 0.25(12 – 5)2/3 = 0.9148
A = 50 + 56(1 – 0.9148) = 54.77
dG = met z2 = 4(24) = 96 mm
0.9148
Eq. (15-5):
Eq. (15-10):
 54.77 + 200(8.29) 
Kv = 
= 1.663


54.77


Ks = Yx = 0.4867 + 0.008 339(4) = 0.520
Eq. (15-11): with Kmb = 1 (both straddle-mounted),
Km = KHβ = 1 + 5.6(10–6)(252) = 1.0035
From Fig. 15-8,
(CL ) P = (Z NT ) P = 3.4822(109 ) −0.0602 = 1.00
(CL )G = (Z NT )G = 3.4822[109 (22 / 24)]−0.0602 = 1.0054
Eq. (15-12):
Cxc = Zxc = 2
Eq. (15-19):
KR = YZ = 0.50 – 0.25 log (1 – 0.999) = 1.25
CR = Z Z = YZ = 1.25 = 1.118
(uncrowned)
From Fig. 15-10, CH = Zw = 1
Eq. (15-9):
Zx = 0.004 92(25) + 0.4375 = 0.560
Wear of Pinion
Fig. 15-12:
σH lim = 2.35HB + 162.89
= 2.35(180) + 162.89 = 585.9 MPa
Fig. 15-6:
I = ZI = 0.066
Eq. (15-2):
(σ H ) P =
(σ H
) (Z NT ) P ZW
S H Kθ Z Z
lim P
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=
585.9(1)(1)
= 524.1 MPa
1(1)(1.118)
2
σ 
bd e1Z I
Eq. (15-1):
W = H
 C p  1000K A K v K H β Z x Z xc


The constant 1000 expresses Wt in kN.
t
P
2

25(88)(0.066)
 524.1  
W =
= 0.591 kN
 

 190  1000(1)(1.663)(1.0035)(0.56)(2) 
t
π dnW
π (88)(1800)(0.591)
1
=
= 4.90 kW
H3 =
60 000
60 000
t
P
Eq. (13-36):
Wear of Gear
σH lim = 585.9 MPa
(σ H )G =
585.9(1.0054)
= 526.9 MPa
1(1)(1.118)
(σ H )G
 526.9 
= 0.591
 = 0.594 kN
(σ H ) P
 524.1 
π (88)(1800)(0.594)
H4 =
= 4.93 kW
60 000
WGt = WPt
Thus in wear, the pinion controls the power rating; H = 4.90 kW
Ans.
We will rate the gear set after solving Prob. 15-6.
________________________________________________________________________
15-6
Refer to Prob. 15-5 for terms not defined below.
Bending of Pinion
( K L ) P = (YNT ) P = 1.6831(109 ) −0.0323 = 0.862
( K L )G = (YNT )G = 1.6831[109 (22 / 24)]−0.0323 = 0.864
Fig. 15-13:
σF lim = 0.30HB + 14.48
= 0.30(180) + 14.48 = 68.5 MPa
Eq. (15-13):
Kx = Yβ = 1
From Prob. 15-5:
YZ = 1.25, vet = 8.29 m/s,
K A = 1, K v = 1.663, Kθ = 1,
Yx = 0.52, K H β = 1.0035, YJ 1 = 0.23
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σ F limYNT
68.5(0.862)
=
= 47.2 MPa
1(1)(1.25)
S F Kθ YZ
(σ F ) P bmetYβ YJ 1
Eq. (5-4):
(σ F ) P =
Eq. (5-3):
WPt =
1000 K A K vYx K H β
47.2(25)(4)(1)(0.23)
= 1.25 kN
1000(1)(1.663)(0.52)(1.0035)
π ( 88 )(1800 )(1.25 )
H1 =
= 10.37 kW
60 000
Bending of Gear
=
σ F lim = 68.5 MPa
68.5(0.864)
= 47.3 MPa
1(1)(1.25)
47.3(25)(4)(1)(0.205)
WGt =
= 1.12 kN
1000(1)(1.663)(0.52)(1.0035)
π ( 88 )(1800 )(1.12 )
H2 =
= 9.29 kW
60 000
Rating of mesh is
Hrating = min(10.37, 9.29, 4.90, 4.93) = 4.90 kW Ans.
with pinion wear controlling.
________________________________________________________________________
(σ F )G =
15-7
(a)
σ 
σ 
(S F ) P =  all  = (S F )G =  all 
 σ P
 σ G
( sat K L / KT K R ) P
( sat K L / KT K R )G
=
t
t
(W Pd K o K v K s K m / FK x J ) P
(W Pd K o K v K s K m / FK x J )G
All terms cancel except for sat , KL , and J,
(sat)P(KL)P JP = (sat )G(KL)G JG
From which
(sat )G =
(sat ) P ( K L ) P J P
J
= (sat ) P P mGβ
( K L )G J G
JG
where β = – 0.0178 or β = – 0.0323 as appropriate. This equation is the same as
Eq. (14-44). Ans.
(b) In bending
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σ

 sat K L

FK x J
FK x J
W t =  all
 =

 S F Pd K o K v K s K m 11  S F KT K R Pd K o K v K s K m 11
(1)
In wear
1/ 2
 sacCLCU 
 W t K o K v K mCsCxc 

 = Cp 

Fd P I
 S H KT CR  22

 22
Squaring and solving for Wt gives

 s 2 C 2C 2  
Fd P I
W t =  2 ac 2L 2H 2  

 S H KT CRCP  22  K o K v K mCsC xc 22
(2)
Equating the right-hand sides of Eqs. (1) and (2) and canceling terms, and
recognizing that CR = K R and PddP = NP, we obtain
(sac )22 =
S H2 (sat )11( K L )11 K x J11KT CsCxc
SF
CH2 N P K s I
Cp
(CL )22
For equal Wt in bending and wear
S H2
=
SF
(
SF
)
2
SF
=1
So we get
(sac )G =
Cp
(CL )G CH
(sat ) P ( K L ) P J P K x KT CsCxc
N P IK s
Ans.
(c)
σ 
σ 
(S H ) P = (S H )G =  c,all  =  c,all 
 σc  P  σc  G
Substituting in the right-hand equality gives
[sacCL / (CR KT )]P
[sacCLCH / (CR KT )]G
=
C W t K K K C C / ( Fd I ) 
C W t K K K C C / ( Fd I ) 
o v m s xc
P
o v m s xc
P
 p
P
 p
G
Denominators cancel, leaving
(sac)P(CL)P = (sac)G(CL)GCH
Solving for (sac)P gives,
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( sac ) P = ( sac )G
(CL )G
CH
(CL ) P
(1)
From Eq. (15-14), ( C L ) P = 3.4822 N L−0.0602 and ( C L )G = 3.4822 ( N L / mG )
Thus,
−0.0602
Ans.
C H = ( sac )G mG0.0602C H
( sac ) P = ( sac )G (1 mG )
−0.0602
.
This equation is the transpose of Eq. (14-45).
________________________________________________________________________
15-8
Given (HB)11 = 300 Brinell
Eq. (15-23):
(sat )P = 44(300) + 2100 = 15 300 psi
J P −0.0323
 0.249  −0.0323
= 15 300 
mG
) = 17 023 psi
 (3
JG
 0.216 
17 023 − 2100
( H B ) 21 =
= 339 Brinell Ans.
44
2290
15 300(0.862)(0.249)(1)(0.593 25)(2)
(sac )G =
1.0685(1)
20(0.086)(0.5222)
= 141 160 psi
141 160 − 23 600
( H B ) 22 =
= 345 Brinell Ans.
341
(sac ) P = (sac )G mG0.0602CH ≈ 141 160(30.0602 ) (1) = 150 811 psi
(sat )G = (sat ) P
( H B )12 =
150 811 − 23 600
= 373 Brinell
341
Ans.
Core
Case
Pinion (HB)11 = 300 (HB)12 = 373
Ans.
Gear
(HB)21 = 339 (HB)22 = 345
_______________________________________________________________________
15-9
Pinion core
(sat ) P = 44 ( 300 ) + 2 100 = 15 300 psi
(σ all ) P =
Wt =
15 300 ( 0.862 )
= 10 551 psi
1(1)(1.25 )
10 551(1.25 )( 0.249 )
= 689.7 lbf
6 (1)(1.374 )( 0.5222 )(1.106 )
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Gear core
(sat )G = 44(339) + 2100 = 17 016 psi
17 016(0.893)
(σ all )G =
= 12 156 psi
1(1)(1.25)
12 156(1.25)(0.216)
Wt =
= 689.3 lbf
6(1)(1.374)(0.5222)(1.106)
Pinion case
(sac ) P = 341(373) + 23 620 = 150 813 psi
150 813(1)
(σ c,all ) P =
= 134 895 psi
1(1)(1.118)
2

1.25(3.333)(0.086)
 134 895  
Wt = 
= 689.0 lbf
 

 2290  1(1.374)(1.106)(0.593 75)(2) 
Gear case
(sac )G = 341(345) + 23 620 = 141 265 psi
141 265(1.0685)(1)
(σ c,all )G =
= 135 010 psi
1(1)(1.118)
2

1.25(3.333)(0.086)
 135 010  
Wt = 
= 690.1 lbf
 

 2290  1(1.1374)(1.106)(0.593 75)(2) 
All four transmitted loads are essentially equal, so the equations developed within
Prob. 15-7 are effective. Ans.
________________________________________________________________________
15-10 The catalog rating is 5.2 hp at 1200 rev/min for a straight bevel gearset. Also
given: NP = 20 teeth, NG = 40 teeth, φn = 20°, F = 0.71 in, JP = 0.241, JG = 0.201,
Pd = 10 teeth/in, through-hardened to 300 Brinell-General Industrial Service, and
Qv = 5 uncrowned.
Mesh
d P = 20 / 10 = 2.000 in, dG = 40 / 10 = 4.000 in
π d P nP π (2)(1200)
vt =
=
= 628.3 ft/min
12
12
K o = 1, S F = 1, S H = 1
Eq. (15-6):
B = 0.25(12 – 5)2/3 = 0.9148
A = 50 + 56(1 – 0.9148) = 54.77
0.9148
Eq. (15-5):
Eq. (15-10):
 54.77 + 628.3 
K v = 
= 1.412

54.77


Ks = 0.4867 + 0.2132/10 = 0.508
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Eq. (15-11):
Km = 1.25 + 0.0036(0.71)2 = 1.252, where Kmb = 1.25
Eq. (15-15):
(KL)P = 1.3558(3∙106)–0.0178 = 1.040
(KL)G = 1.3558 (3∙106/2)–0.0178 = 1.053
Eq. (15-14):
(CL)P = 3.4822(3∙106)–0.0602 = 1.419
(CL)G = 3.4822(3∙106/2)–0.0602 = 1.479
Eq. (15-19):
KR = 0.50 – 0.25 log(1 – 0.99) = 1.0
CR = K R = 1.0
Bending
Pinion:
Eq. (15-23):
(sat )P = 44(300) + 2100 = 15 300 psi
Eq. (15-4):
(swt ) P =
Eq. (15-3):
Wt =
Gear:
(sat )G = 15 300 psi
15 300(1.053)
( sw t ) G =
= 16 111 psi
1(1)(1)
16 111(0.71)(1)(0.201)
Wt =
= 256.0 lbf
10(1)(1.412)(0.508)(1.252)
256.0(628.3)
H2 =
= 4.9 hp
33 000
Eq. (15-4):
Eq. (15-3):
15 300(1.040)
= 15 912 psi
1(1)(1)
(sw t ) P FK x J P
Pd K o K v K s K m
15 912(0.71)(1)(0.241)
=
= 303.2 lbf
10(1)(1.412)(0.508)(1.252)
303.2(628.3)
H1 =
= 5.8 hp
33 000
Wear
Pinion:
(CH )G = 1, I = 0.078, C p = 2290 psi,
Cs = 0.125(0.71) + 0.4375 = 0.526 25
Eq. (15-22):
Eq. (15-2):
Cxc = 2
(sac)P = 341(300) + 23 620 = 125 920 psi
125 920(1.419)(1)
(σ c,all ) P =
= 178 680 psi
1(1)(1)
Shigley’s MED, 10th edition
Chapter 15 Solutions, Page 12/20
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2
Eq. (15-1):
 (σ ) 
Fd P I
W =  c,all P 
 C p  K o K v K mCsC xc
t
 178 680 
=

 2290 
= 362.4 lbf
H3 =
Gear:
2


0.71(2.000)(0.078)
1(1.412)(1.252)(0.526 25)(2) 


362.4(628.3)
= 6.9 hp
33 000
(sac )G = 125 920 psi
125 920(1.479)(1)
(σ c,all ) =
= 186 236 psi
1(1)(1)
2

0.71(2.000)(0.078)
 186 236  
W =
 
 = 393.7 lbf
 2290  1(1.412)(1.252)(0.526 25)(2) 
393.7(628.3)
H4 =
= 7.5 hp
33 000
t
Rating:
H = min(5.8, 4.9, 6.9, 7.5) = 4.9 hp
Pinion wear controls the power rating. The catalog rating of 5.2 hp is slightly
higher than predicted here, but seems reasonable.
Ans.
________________________________________________________________________
15-11 From Ex. 15-1, the core hardness of both the pinion and gear is 180 Brinell. So
(HB)11 and (HB)21 are 180 Brinell and the bending stress numbers are:
(sat ) P = 44(180) + 2100 = 10 020 psi
(sat )G = 10 020 psi
The contact strength of the gear case, based upon the equation derived in Prob.
15-7, is
Cp
S H2 (sat ) P ( K L ) P K x J P KT CsCxc
(sac )G =
N P IK s
(CL )G CH S F
Substituting (sat)P from above and the values of the remaining terms from
Ex. 15-1,
Shigley’s MED, 10th edition
Chapter 15 Solutions, Page 13/20
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2290 1.52  10 020(1)(1)(0.216)(1)(0.575)(2) 


1.32(1) 1.5 
25(0.065)(0.529)

= 114 331 psi
114 331 − 23 620
( H B ) 22 =
= 266 Brinell
341
( sac )G =
The pinion contact strength is found using the relation from Prob. 15-7:
(sac ) P = (sac )G mG0.0602CH = 114 331(1)0.0602 (1) = 114 331 psi
114 331 − 23 600
= 266 Brinell
( H B )12 =
341
Core Case
Pinion 180 266
Gear
180 266
Realization of hardnesses
The response of students to this part of the question would be a function of the
extent to which heat-treatment procedures were covered in their materials and
manufacturing prerequisites, and how quantitative it was. The most important
thing is to have the student think about it.
The instructor can comment in class when students’ curiosity is heightened.
Options that will surface may include:
(a) Select a through-hardening steel which will meet or exceed core hardness in
the hot-rolled condition, then heat-treating to gain the additional 86 points of
Brinell hardness by bath-quenching, then tempering, then generating the teeth in
the blank.
(b) Flame or induction hardening are possibilities.
(c) The hardness goal for the case is sufficiently modest that carburizing and case
hardening may be too costly. In this case the material selection will be different.
(d)The initial step in a nitriding process brings the core hardness to 33–38
Rockwell C-scale (about 300–350 Brinell), which is too much.
________________________________________________________________________
15-12 Computer programs will vary.
________________________________________________________________________
15-13 A design program would ask the user to make the a priori decisions, as indicated
in Sec. 15-5, p. 798, of the text. The decision set can be organized as follows:
Shigley’s MED, 10th edition
Chapter 15 Solutions, Page 14/20
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A priori decisions:
• Function: H, Ko, rpm, mG, temp., NL, R
• Design factor: nd (SF = nd , S H = nd )
• Tooth system: Involute, Straight Teeth, Crowning, φn
• Straddling: Kmb
• Tooth count: NP (NG = mGNP)
Design decisions:
• Pitch and Face: Pd , F
• Quality number: Qv
• Pinion hardness: (HB)1, (HB)3
• Gear hardness: (HB)2, (HB)4
First, gather all of the equations one needs, then arrange them before coding. Find
the required hardnesses, express the consequences of the chosen hardnesses, and
allow for revisions as appropriate.
Shigley’s MED, 10th edition
Chapter 15 Solutions, Page 15/20
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Pinion Bending
Load-induced
stress (Allowable
stress)
Tabulated
strength
Associated
hardness
st =
W t PK o K v K m K s
= s11
FK x J P
( sat ) P =
Factor of
safety
st =
W t PK o K v K m K s
= s21
FK x J G
( sat )G =
s21S F KT K R
( K L )G
Pinion Wear
Gear Wear
1/ 2
 W K o K vCsC xc 

Fd P I


σ c = Cp 
t
( sac ) P =
= s12
s22 = s12
s12 S H KT CR
(CL ) P (CH ) P
( sac )G =
s22 S H KT CR
(CL )G (CH )G
 (sat ) P − 2100

44
Bhn = 
s
(
)
 at P − 5980

48
 (sat )G − 2100

44
Bhn = 
s
(
)
 at G − 5980

48
 (sac ) P − 23 620

341
Bhn = 
(
)
s
 ac P − 29 560

363.6
 (sac ) P − 23 620

341
Bhn = 
(
)
s
 ac P − 29 560

363.6
(HB)11
(HB)21
(HB)12
(HB)22
 44( H B )11 + 2100
(sat1) P = 
 48( H B )11 + 5980
44( H B ) 21 + 2100
(sat1)G = 
 48( H B )21 + 5980
341( H B )12 + 23 620
( sac1) P = 
363.6( H B )12 + 29 560
341( H B ) 22 + 23 620
(sac1)G = 
363.6( H B ) 22 + 29 560
σ
(s ) ( K )
n11 = all = at1 P L P
s11KT K R
σ
(s ) ( K )
n21 = at1 G L G
s21KT K R
 (s ) (C ) (C ) 
n12 =  ac1 P L P H P 
s12 KT CR


Chosen
hardness
New tabulated
strength
s11S F KT K R
(K L )P
Gear Bending
Note: S F = nd , S H =
Shigley’s MED, 10th edition
SF
Chapter 15 Solutions, Page 16/20
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2
n22
 (s ) (C ) (C ) 
=  ac1 G L G H G 
s22 KT CR


2
15-14 NW = 1, NG = 56, Pt = 8 teeth/in, d = 1.5 in, Ho = 1hp, φn = 20°, ta = 70°F,
Ka = 1.25, nd = 1, Fe = 2 in, A = 850 in2
(a)
mG = NG/NW = 56, dG = NG/Pt = 56/8 = 7.0 in
px = π / 8 = 0.3927 in, C = 1.5 + 7 = 8.5 in
Eq. (15-39):
a = px / π = 0.3927 / π = 0.125 in
Eq. (15-40):
b = 0.3683 px = 0.1446 in
Eq. (15-41):
ht = 0.6866 px = 0.2696 in
Eq. (15-42):
do = 1.5 + 2(0.125) = 1.75 in
Eq. (15-43):
dr = 3 – 2(0.1446) = 2.711 in
Eq. (15-44):
Dt = 7 + 2(0.125) = 7.25 in
Eq. (15-45):
Dr = 7 – 2(0.1446) = 6.711 in
Eq. (15-46):
c = 0.1446 – 0.125 = 0.0196 in
Eq. (15-47):
( FW ) max = 2 2 ( 7 )( 0.125 ) = 2.646 in
Eq. (13-27):
VW = π (1.5)(1725 /12) = 677.4 ft/min
π (7)(1725 / 56)
VG =
= 56.45 ft/min
12
L = px NW = 0.3927 in
Eq. (13-28):
 0.3927 
o
λ = tan −1 
 = 4.764
(1.5)
π


Pn =
pn =
Eq. (15-62):
(b)
Eq. (15-38):
Pt
8
=
= 8.028
cos λ
cos 4.764°
π
= 0.3913 in
Pn
π (1.5)(1725)
Vs =
= 679.8 ft/min
12 cos 4.764°
f = 0.103exp  −0.110(679.8) 0.450  + 0.012 = 0.0250
Eq. (15-54):
cos φn − f tan λ
cos 20° − 0.0250 tan 4.764°
e=
=
= 0.7563
cos φn + f cot λ
cos 20° + 0.0250 cot 4.764°
Shigley’s MED, 10th edition
Ans.
Chapter 15 Solutions, Page 17/20
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Eq. (15-58):
Eq. (15-57):
33 000nd H o K a
33 000(1)(1)(1.25)
=
= 966 lbf
VG e
56.45(0.7563)
cos φn sin λ + f cos λ
WWt = WGt
cos φn cos λ − f sin λ
WGt =
Ans.
 cos 20° sin 4.764° + 0.025cos 4.764° 
= 966 

 cos 20° cos 4.764° − 0.025sin 4.764° 
= 106.4 lbf Ans.
(c)
Eq. (15-33):
Cs = 1190 – 477 log 7.0 = 787
Eq. (15-36):
Cm = 0.0107 −56 2 + 56(56) + 5145 = 0.767
Eq. (15-37):
Cv = 0.659 exp[−0.0011(679.8)] = 0.312
Eq. (15-38):
(Wt)all = 787(7)0.8(2)(0.767)(0.312) = 1787 lbf
Since WGt < (W t )all , the mesh will survive at least 25 000 h.
Eq. (15-61):
Eq. (15-63):
0.025(966)
= −29.5 lbf
0.025sin 4.764° − cos 20° cos 4.764°
29.5(679.8)
Hf =
= 0.608 hp
33 000
106.4(677.4)
HW =
= 2.18 hp
33 000
966(56.45)
HG =
= 1.65 hp
33 000
Wf =
The mesh is sufficient
Ans.
Pn = Pt / cos λ = 8 / cos 4.764o = 8.028
pn = π / 8.028 = 0.3913 in
966
σG =
= 39 500 psi
0.3913(0.5)(0.125)
The stress is high. At the rated horsepower,
σG =
1
39 500 = 23 940 psi
1.65
acceptable
Shigley’s MED, 10th edition
Chapter 15 Solutions, Page 18/20
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(d)
Eq. (15-52):
Amin = 43.2(8.5)1.7 = 1642 in2 < 1700 in2
Eq. (15-49):
Hloss = 33 000(1 – 0.7563)(2.18) = 17 530 ft · lbf/min
Assuming a fan exists on the worm shaft,
1725
+ 0.13 = 0.568 ft · lbf/(min · in 2 · o F)
3939
17 530
Eq. (15-51): ts = 70 +
= 88.2o F Ans.
0.568(1700)
________________________________________________________________________
Eq. (15-50):
CR
=
Shigley’s MED, 10th edition
Chapter 15 Solutions, Page 19/20
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15-15 to 15-22
Problem statement values of 25 hp, 1125 rev/min, mG = 10, Ka = 1.25, nd = 1.1,
φn = 20°, ta = 70°F are not referenced in the table. The first four parameters listed
in the table were selected as design decisions.
px
dW
FG
A
HW
HG
Hf
NW
NG
KW
Cs
Cm
Cv
VG
WGt
t
W
W
f
e
(Pt)G
Pn
C-to-C
ts
L
λ
σG
dG
15-15
1.75
3.60
2.40
2000
15-16
1.75
3.60
1.68
2000
15-17
1.75
3.60
1.43
2000
15-18
1.75
3.60
1.69
2000
15-19
1.75
3.60
2.40
2000
15-20
1.75
4.10
2.25
2000
38.0
36.1
1.85
3
30
50
15-21
1.75
3.60
2.4
2500
FAN
41.2
37.7
3.59
3
30
115
15-22
1.75
3.60
2.4
2600
FAN
41.2
37.7
3.59
3
30
185
38.2
36.2
1.87
3
30
38.2
36.2
1.47
3
30
38.2
36.2
1.97
3
30
38.2
36.2
1.97
3
30
125
38.2
36.2
1.97
3
30
80
607
0.759
0.236
492
854
0.759
0.236
492
1000
0.759
0.236
492
492
492
563
492
492
2430
2430
2430
2430
2430
2120
2524
2524
1189
0.0193
0.948
1.795
1.979
10.156
177
5.25
24.9
5103
16.71
1189
0.0193
0.948
1.795
1.979
10.156
177
5.25
24.9
7290
16.71
1189
0.0193
0.948
1.795
1.979
10.156
177
5.25
24.9
8565
16.71
1189
0.0193
0.948
1.795
1.979
10.156
177
5.25
24.9
7247
16.71
1189
0.0193
0.948
1.795
1.979
10.156
177
5.25
24.9
5103
16.71
1038
0.0183
0.951
1.571
1.732
11.6
171
6.0
24.98
4158
19.099
1284
0.034
0.913
1.795
1.979
10.156
179.6
5.25
24.9
5301
16.7
1284
0.034
0.913
1.795
1.979
10.156
179.6
5.25
24.9
5301
16.71
Shigley’s MED, 10th edition
Chapter 15 Solutions, Page 20/20
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Chapter 16
16-1
Given: r = 300/2 = 150 mm, a = R = 125 mm, b = 40 mm, f = 0.28, F = 2.2 kN, θ1 = 0°,
θ2 = 120°, and θa = 90°.
From which, sinθa = sin90° = 1.
Eq. (16-2):
0.28 pa (0.040)(0.150) 120°
∫0° sin θ (0.150 − 0.125 cos θ ) dθ
1
= 2.993 (10− 4 ) pa N · m
Mf =
Eq. (16-3):
pa (0.040)(0.150)(0.125) 120° 2
sin θ dθ = 9.478 (10−4 ) pa N · m
∫
0
°
1
MN =
c = 2(0.125 cos 30°) = 0.2165 m
Eq. (16-4):
F =
9.478 (10− 4 ) pa − 2.993 (10− 4 ) pa
0.2165
= 2.995 (10− 3 ) pa
pa = F/ [2.995(10−3)] = 2200/ [2.995(10−3)]
= 734.5(103) Pa for cw rotation
Eq. (16-7):
2200 =
9.478 (10− 4 ) pa + 2.993 (10 − 4 ) pa
0.2165
pa = 381.9(103) Pa for ccw rotation
A maximum pressure of 734.5 kPa occurs on the RH shoe for cw rotation.
(b) RH shoe:
Eq. (16-6):
0.28(734.5)103 (0.040)0.1502 (cos 0o − cos120o )
= 277.6 N · m
TR =
1
LH shoe:
381.9
TL = 277.6
= 144.4 N · m Ans.
734.5
Ttotal = 277.6 + 144.4 = 422 N · m Ans.
Shigley’s MED, 10th edition
Ans.
Ans.
Chapter 16 Solutions, Page 1/27
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(c)
RH shoe:
Fx = 22200 sin 30° = 1100 N,
Eqs. (16-8):
1

A =  sin 2 θ 
2

0o
Eqs. (16-9):
Rx =
2π / 3 rad
120o
Ry =
= 0.375,
θ 1

B =  − sin 2θ 
2
4

0
7
734.5
(103 ) 0.040(0.150)
1
3
7
734.5
(10 ) 0.04(0.150)
R = [( −1007 )
LH shoe:
Fy = 2200 cos 30° = 1905
1
N
2
= 1.264
[0.375 − 0.28(1.264)]
)] − 1100 = −1007 N
[1.264 + 0.28(0.375)]] − 1905 = 4128 N
1
+ 41282 ]1/ 2 = 4249 N Ans.
11 N, Fy = 1905 N
Fx = 1100
Eqs. (16-10): Rx =
38 (103 ) 0.040(0.150)
381.9
[0.375 + 0.28(1.264)]
)] − 1100 = 570 N
1
3
381.9
103 ) 0.040(0.150)
(
Ry =
[1.264 − 0.28(0.375)]
)] − 1905 = 751 N
1
1/ 2
R = ( 597
59 2 + 7512 ) = 959 N Ans.
_________________________
______________________________________
____________________
16-2
Given: r = 300/2 = 150
50 mm, a = R = 125 mm, b = 40 mm, f = 0.28,
8, F = 2.2 kN, θ1 = 15°,
θ2 = 105°, and θa = 900°.
From which, sinθa = sin90° = 1.
Eq. (16-2):
0.28 pa (0.040)(0
)(0.150) 105°
Mf =
2.
(10− 4 ) pa
∫15° sin θ (0.150 − 0.125cos θ ) dθ = 2.177
1
Shigley’s MED, 10th edition
Chapter
ter 16 Solutions, Page 2/27
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MN =
Eq. (16-3):
pa (0.040)(0.150)(0.125) 105° 2
−4
∫15° sin θ dθ = 7.765 (10 ) pa
1
c = 2(0.125) cos 30° = 0.2165 m
F =
Eq. (16-4):
7.765 (10− 4 ) pa − 2.177 (10− 4 ) pa
0.2165
= 2.581(10− 3 ) pa
pa = 2200/ [2.581(10− 3)] = 852.4 (103) Pa
= 852.4 kPa on RH shoe for cw rotation
RH shoe:
TR =
Eq. (16-6):
LH shoe:
2200 =
Ans.
0.28(852.4)103 (0.040)(0.1502 )(cos15° − cos105°)
= 263 N · m
1
7.765 (10 − 4 ) pa + 2.177 (10 − 4 ) pa
0.2165
pa = 479.1(103 ) Pa = 479.1 kPa on LH shoe for cw rotation
Ans.
0.28(479.1)103 (0.040)(0.1502 )(cos15° − cos105°)
= 148 N · m
1
= 263 + 148 = 411 N · m Ans.
TL =
Ttotal
Comparing this result with that of Prob. 16-1, a 2.6% reduction in torque is obtained by
using 25% less braking material.
______________________________________________________________________________
16-3
Given: θ1 = 0°, θ2 = 120°, θa = 90°, sin θa = 1, a = R = 3.5 in, b = 1.25 in, f = 0.30,
F = 225 lbf, r = 11/2 = 5.5 in, counter-clockwise rotation.
LH shoe:
Eq. (16-2), with θ1 = 0:
θ
f pabr 2
f pabr 
a

sin θ ( r − a cos θ ) dθ =
Mf =
r (1 − cos θ 2 ) − sin 2 θ 2 
∫

sin θ a θ1
sin θ a 
2

0.30 pa (1.25)5.5 
3.5 2

5.5(1 − cos120o ) −
sin 120°

1
2


= 14.31 pa lbf · in
Eq. (16-3), with θ1 = 0:
θ
p bra 2 2
p bra θ 2 1

− sin 2θ 2 
MN = a
sin θ dθ = a
∫
sin θ a θ1
sin θ a  2
4

=

pa (1.25)5.5(3.5) 120°  π  1

 − sin 2(120°) 

1
 2  180°  4

= 30.41 pa lbf · in
=
Shigley’s MED, 10th edition
Chapter 16 Solutions, Page 3/27
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 180o − θ 2 
o
c = 2r cos 
 = 2(5.5) cos 30 = 9.526 in
2


30.41 pa − 14.31 pa
= 1.690 pa
F = 225 =
9.526
pa = 225 / 1.690 = 133.1 psi
Eq. (16-6):
f pabr 2 (cos θ1 − cos θ 2 ) 0.30(133.1)1.25(5.52 )
[1 − (−0.5)]
=
sin θ a
1
= 2265 lbf · in = 2.265 kip · in Ans.
TL =
RH shoe:
30.41 pa + 14.31 pa
= 4.694 pa
9.526
pa = 225 / 4.694 = 47.93 psi
47.93
TR =
2265 = 816 lbf ·in = 0.816 kip·in
133.1
F = 225 =
Ttotal = 2.27 + 0.82 = 3.09 kip ⋅ in
Ans.
______________________________________________________________________________
16-4
(a) Given: θ1 = 10°, θ2 = 75°, θa = 75°, pa = 106 Pa, f = 0.24, b = 0.075 m (shoe width),
a = 0.150 m, r = 0.200 m, d = 0.050 m, c = 0.165 m.
Some of the terms needed are evaluated here:
θ
2
θ2
θ2
θ
1

A =  r ∫ sin θ dθ − a ∫ sin θ cos θ dθ  = r [ − cos θ ]θ2 − a  sin 2 θ 

1
θ1
 θ1
2
θ1
75°
1

= 200 [ − cos θ ]10° − 150  sin 2 θ  = 77.5 mm
2
10°
75π /180 rad
θ2
θ 1

B = ∫ sin 2 θ dθ =  − sin 2θ 
= 0.528
θ1
2 4
10π /180 rad
75°
C =
θ2
∫θ
sin θ cos θ dθ = 0.4514
1
Now converting to Pascals and meters, we have from Eq. (16-2),
Mf =
0.24 (106 ) (0.075)(0.200)
f pabr
(0.0775) = 289 N · m
A=
sin θ a
sin 75°
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From Eq. (16-3),
MN =
pabra
106 (0.075)(0.200)(0.150)
(0.528) = 1230 N · m
B =
sin θ a
sin 75°
Finally, using Eq. (16-4), we have
F =
MN − M f
c
=
1230 − 289
= 5.70 kN
165
Ans.
(b) Use Eq. (16-6) for the primary shoe.
T =
=
fpabr 2 (cos θ1 − cos θ 2 )
sin θ a
0.24 (106 ) (0.075)(0.200) 2 (cos 10° − cos 75°)
sin 75°
= 541 N · m
For the secondary shoe, we must first find pa. Substituting
1230
289
pa and M f = 6 pa into Eq. (16 - 7),
6
10
10
6
(1230 / 10 ) pa + (289 / 106 ) pa
5.70 =
, solving gives pa = 619 (103 ) Pa
165
MN =
Then
T =
0.24 619 (103 )  0.075 ( 0.2002 ) ( cos 10° − cos 75° )
sin 75°
= 335 N · m
so the braking capacity is Ttotal = 2(541) + 2(335) = 1750 N · m
Ans.
(c) Primary shoes:
pabr
( C − f B ) − Fx
sin θ a
106 (0.075)0.200
=
[0.4514 − 0.24(0.528)](10− 3 ) − 5.70 = −0.658 kN
sin 75°
pabr
( B + f C ) − Fy
Ry =
sin θ a
106 (0.075)0.200
=
[0.528 + 0.24(0.4514)] (10− 3 ) − 0 = 9.88 kN
sin 75°
Rx =
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Secondary shoes:
Rx =
=
pabr
(C + f B) − Fx
sin θ a
0.619 (106 ) 0.075(0.200)
sin 75°
= −0.143 kN
p br
Ry = a ( B − f C ) − Fy
sin θ a
=
0.619 (106 ) 0.075(0.200)
= 4.03 kN
sin 75°
[0.4514 + 0.24(0.528)] (10− 3 ) − 5.70
[0.528 − 0.24(0.4514)] (10− 3 ) − 0
Note from figure th
that +y for secondary shoe is opposite to
+y for primary shoe
oe.
Combining horizon
ontal and vertical components,
RH = −0.658 − 0.143 = −0.801 kN
RV = 9.88 − 4.03
4
= 5.85 kN
R = (−0.801
01)2 + 5.852
= 5.90 kN
N Ans.
_________________________
______________________________________
____________________
16-5
Given: Face width b = 1.25 in, F = 90 lbf, f = 0.25.
5° − tan−1(6/8) = 8.13°, θ2 = 98.13°, θa = 90°,
Preliminaries: θ1 = 45°
2
2 1/2
a = (6 + 8 ) = 10 in
Eq. (16-2):
θ
f pabr 2
0.25 pa (1.25)6
Mf =
sin θ ( r − a cos θ ) dθ =
∫
sin θ a θ1
1
98.13°
∫
sinn θ ( 6 − 10 cos θ ) dθ
8.13°
= 3.728 pa lbf
bf · in
Eq. (16-3):
θ
MN =
pabra 2 2
p (1.25)6(10)
sin
in θ dθ = a
∫
sin θ a θ1
1
98.13°
∫
sin 2 θ dθ
8.13°
= 69.405 pa lb
lbf · in
Eq. (16-4): Using Fc = MN − Mf , we obtain
90(20) = (69.4055 − 3.728) pa
⇒
pa = 27.4 psi
Shigley’s MED, 10th edition
Ans.
Chapter
ter 16 Solutions, Page 6/27
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Eq. (16-6):
fp br 2 ( cos θ1 − cos θ 2 ) 0.25(27.4)1.25 ( 6 ) ( cos8.13° − cos 98.13° )
T = a
=
sin θ a
1
= 348.7 lbf · in Ans.
______________________________________________________________________________
2
16-6
For +3σˆ f :
f = f + 3σˆ f = 0.25 + 3(0.025) = 0.325
From Prob. 16-5, with f = 0.25, M f = 3.728 pa. Thus, M f = (0.325/0.25) 3.728 pa =
4.846 pa. From Prob. 16-5, M N = 69.405 pa.
Eq. (16-4): Using Fc = MN − Mf , we obtain
90(20) = (69.405 − 4.846) pa
⇒
pa = 27.88 psi
Ans.
From Prob. 16-5, pa = 27.4 psi and T = 348.7 lbf⋅in. Thus,
 0.325   27.88 
T =

 348.7 = 461.3 lbf ·in
 0.25   27.4 
Ans.
Similarly, for −3σˆ f :
f = f − 3σˆ f = 0.25 − 3(0.025) = 0.175
M f = (0.175 / 0.25) 3.728 pa = 2.610 pa
⇒ pa = 26.95 psi
90(20) = (69.405 − 2.610) pa
0.175
26.95



T =

 348.7 = 240.1 lbf · in Ans.
0.25
27.4



______________________________________________________________________________
16-7
Preliminaries: θ2 = 180° − 30° − tan−1(3/12) = 136°, θ1 = 20° − tan−1(3/12) = 6°,
θa = 90°, sinθa = 1, a = (32 + 122)1/2 = 12.37 in, r = 10 in, f = 0.30, b = 2 in, pa = 150 psi.
Eq. (16-2):
Mf =
0.30(150)(2)(10) 136o
∫6° sin θ (10 − 12.37 cos θ ) dθ = 12 800 lbf · in
sin 90°
Eq. (16-3):
MN =
150(2)(10)(12.37) 136° 2
∫6° sin θ dθ = 53 300 lbf · in
sin 90°
LH shoe:
cL = 12 + 12 + 4 = 28 in
Now note that Mf is cw and MN is ccw. Thus,
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Eq. (16-6):
FL =
53 300 − 12 800
= 1446 lbf
28
TL =
0.30(150)(2)(10) 2 (cos 6° − cos136°)
0.
= 15 4200 lbf · in
sin 90°
RH shoe:
M N = 53 300
pa
= 355.3 pa ,
150
M f = 12 800
pa
= 85.3 pa
150
On this shoe, both MN and Mf are ccw. Also,
24 − 2 tan 14°) cos 14° = 22.8 in
cR = (24
Fact = FL sin14° = 361 lbf Ans.
FR = FL / cos14° = 1491 lbf
Thus,
1491 =
Then,
TR =
355.3 + 85.3
pa ⇒ pa = 77.2 psi
22.8
0.30(77.2)(2)(10)2 (cos 6° − cos136°)
0.
= 7940
0 lbf
l · in
sin 90°
Ttotal = 15 420 + 7940 = 23 400 lbf · in Ans.
_________________________
______________________________________
____________________
16-8
θ2
M f = 2∫ ( fdN )(a′ cos θ − r )
0
where dN = pbr
br dθ
θ2
= 2 fpbr ∫ (a′ cos θ − r ) dθ = 0
0
From which
θ2
θ2
0
0
a′∫ cos θ dθ = r ∫ dθ
a′ =
rθ 2
r (60°)(π / 180)
=
= 1.209r
sin θ 2
sin 60°
Ans.
Eq. (16-15):
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a =
4r sin 60°
= 1.170r
2(60)(π / 180) + sin[2(60)]
2(6
Ans.
fers with a ′ by 100(1.170 −1.209)/1.209 = − 3.23
3 %
Ans.
a differ
_________________________
______________________________________
____________________
16-9
se rotation, θ2 = π / 4 rad, r = 13.5/2 = 6.75 in
(a) Counter-clockwise
Eq. (16-15):
4r sin θ 2
4(6.75) sin(π / 4)
a =
=
= 7.426 in
2θ 2 + sin 2θ 2
2π / 4 + sin(2π / 4)
e = 2a = 2(7.426) = 14.85 in
Ans.
(b)
α = tan−1(3/14.85)
5) = 11.4°
∑M
∑F
x
R
= 0 = 3F x − 6.375P
= 0 = −F x + R x
⇒ F x = 2.125P
⇒ R x = F x = 2.125P
F y = F x tan11.4o = 0.4
.428P
y
∑ Fy = −P − F + R y
R y = P + 0.428P = 1.42
.428P
Left shoe lever.
∑ M R = 0 = 7.78S x − 15.28F x
15.28
Sx =
(2.125P) = 4.174 P
7.78
S y = f S x = 0.30(4.174 P) = 1.252P
∑ Fy = 0 = R y + S y + F y
R y = − F y − S y = −0.428P − 1.252P = −1.68P
∑ Fx = 0 = R x − S x + F x
R x = S x − F x = 4.174 P − 2.125P = 2.049
2
P
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(c) The direction of brake
bra pulley rotation affects the sense of Sy, whhich has no effect on
the brake shoe leve
ver moment and hence, no effect on Sx or the brake
b
torque.
The brake shoe lev
evers carry identical bending moments but thee lleft lever carries a
tension while the ri
right carries compression (column loading). The
T right lever is
designed and usedd as a left lever, producing interchangeable lev
evers (identical levers).
But do not infer fro
from these identical loadings.
_________________________
______________________________________
____________________
16-10 r = 13.5/2 = 6.75 in,
b = 6 in,
θ2 = 45° = π / 4 rad.
From Table 16-3 for a rigid, molded non-asbestos lining use a conse
nservative estimate of
pa = 100 psi, f = 0.33..
es the horizontal brake hinge pin reaction whic
ich corresponds to Sx in
Equation (16-16) gives
Prob. 16-9. Thus,
p brr
100(6)6.75
N = S x = a ( 2θ 2 + sin 2θ 2 ) =
2 (π / 4 ) + sinn  2 ( 45° ) 
2
2
= 5206 lbf
{
}
which, from Prob. 6-99 is 4.174 P. Therefore,
4.174 P = 5206
⇒ P = 1250 lbf = 1.25 kip
A
Ans.
Applying Eq. (16-18)) for
f two shoes, where from Prob. 16-9, a = 7.426
7.4 in
T = 2a f N = 2(7.426)0.33(5206)
= 25 520 lbf · in = 25.52 kip · in Ans.
_________________________
______________________________________
____________________
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16-11 Given: D = 350 mm, b = 100 mm, pa = 620 kPa, f = 0.30, φ = 270°.
Eq. (16-22):
p bD
620(0.100)0.350
P1 = a
=
= 10.85 kN Ans.
2
2
f φ = 0.30(270°)(π / 180°) = 1.414
Eq. (16-19):
P2 = P1 exp(− f φ ) = 10.85 exp(− 1.414) = 2.64 kN
Ans.
T = ( P1 − P2 )( D / 2) = (10.85 − 2.64)(0.350 / 2) = 1.437 kN · m Ans.
______________________________________________________________________________
16-12 Given: D = 12 in,
Eq. (16-22):
f = 0.28,
pa =
b = 3.25 in,
φ = 270°, P1 = 1800 lbf.
2 P1
2(1800)
=
= 92.3 psi
bD
3.25(12)
Ans.
f φ = 0.28(270o )(π / 180o ) = 1.319
P2 = P1 exp(− f φ ) = 1800 exp(−1.319) = 481 lbf
T = ( P1 − P2 )( D / 2) = (1800 − 481)(12 / 2)
= 7910 lbf · in = 7.91 kip · in Ans.
______________________________________________________________________________
16-13
Σ MO = 0 = 100 P2 − 325 F ⇒ P2 = 325(300)/100 = 975 N
Shigley’s MED, 10th edition
Ans.
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 100 
 = 51.32°
 160 
φ = 270° − 51.32° = 218.7°
f φ = 0.30(218.7)
0
(π / 180° ) = 1.145
P1 = P2 exp( f φ ) = 975exp(1.145) = 3064 N
T = ( P1 − P2 ) ( D / 2) = (3064 − 975)(200 / 2)
α = co
cos −1 
= 209
20 (103 ) N · mm = 209 N · m
An
Ans.
Ans.
_________________________
______________________________________
____________________
16-14 (a) D = 16 in, b = 3 in
n = 200 rev/min
f = 0.20, pa = 70
0 ppsi
Eq. (16-22):
p bD
70(3
(3)(16)
P1 = a
=
= 1680 lbf
2
2
f φ = 0.20(3π / 2) = 0.942
Eq. (16-14):
P2 = P1 exp(− f φ ) = 1680 exp(−0.942) = 655
65 lbf
D
16
T = ( P1 − P2 ) = (1680 − 655)
2
2
= 8200 lbf · in Ans.
8200(200)
Tn
H =
ns.
=
= 26.0 hp Ans
63 025
63 025
3P
3(1680)
P = 1 =
= 504 lbf Ans.
10
10
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(b) Force of belt on the
th drum:
R = (16802 + 6552)1/2 = 1803 lbf
Force of shaft on th
the drum: 1680 and 655 lbf
TP1 = 1680(8) = 13 440 lbf · in
240 lbf · in
TP2 = 655(8) = 524
Net torque on drum
um due to brake band:
T = TP1 − TP2
40
= 13 440 − 5240
= 8200 lbf · in
The radial load onn the bearing pair is 1803 lbf. If the bearing is sstraddle mounted with
the drum at centerr span,
s
the bearing radial load is 1803/2 = 901
1 lbf.
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(c) Eq. (16-21):
2P
bD
2 P1
2(1680)
=
=
= 70 psi
3(16)
3(16)
p =
p
θ = 0°
Ans.
2 P2
2(655)
=
= 27.3 psi Ans.
3(16)
3(16)
______________________________________________________________________________
p
θ = 270°
=
16-15 Given: φ = 270°, b = 2.125 in, f = 0.20, T =150 lbf · ft, D = 8.25 in, c2 = 2.25 in (see
figure). Notice that the pivoting rocker is not located on the vertical centerline of the
drum.
(a) To have the band tighten for ccw rotation, it is necessary to have c1 < c2 . When
friction is fully developed,
P1 / P2 = exp( f φ ) = exp[0.2(3π / 2)] = 2.566
If friction is not fully developed,
P1/P2 ≤ exp( f φ )
To help visualize what is going on let’s add a force W parallel to P1, at a lever arm of
c3. Now sum moments about the rocker pivot.
∑M
= 0 = c3W + c1P1 − c2 P2
From which
c2 P2 − c1P1
c3
The device is self locking for ccw rotation if W is no longer needed, that is, W ≤ 0.
It follows from the equation above
W =
P1
c
≥ 2
P2
c1
When friction is fully developed
2.566 = 2.25 / c1
2.25
c1 =
= 0.877 in
2.566
When P1/P2 is less than 2.566, friction is not fully developed. Suppose P1/P2 = 2.25,
then
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c1 =
2.25
= 1 in
2.25
We don’t want to be at the point of slip, and we need the band to tighten.
c2
≤ c1 ≤ c2
P1 / P2
When the developed friction is very small, P1/P2 → 1 and c1 → c2
Ans.
(b) Rocker has c1 = 1 in
2.25
P1
c
= 2 =
= 2.25
1
P2
c1
ln( P1 / P2 ) ln 2.25
=
= 0.172
f =
φ
3π / 2
Friction is not fully developed, no slip.
T = ( P1 − P2 )
P
D
D
= P2  1 − 1
2
 P2
2
Solve for P2
2T
2(150)(12)
=
= 349 lbf
[( P1 / P2 ) − 1]D (2.25 − 1)(8.25)
P1 = 2.25P2 = 2.25(349) = 785 lbf
2P
2(785)
= 89.6 psi Ans.
p = 1 =
2.125(8.25)
bD
P2 =
(c) The torque ratio is 150(12)/100 or 18-fold.
349
P2 =
= 19.4 lbf
18
P1 = 2.25P2 = 2.25(19.4) = 43.6 lbf
89.6
p =
= 4.98 psi Ans.
18
Comment:
As the torque opposed by the locked brake increases, P2 and P1 increase (although
ratio is still 2.25), then p follows. The brake can self-destruct. Protection could be
provided by a shear key.
______________________________________________________________________________
16-16 Given: OD = 250 mm, ID = 175 mm, f = 0.30, F = 4 kN.
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(a) From Eq. (16-23),
2 ( 4000 )
2F
pa =
=
= 0.194 N/mm2 = 194 kPa
Ans.
π d ( D − d ) π (175)(250 − 175)
Eq. (16-25):
Ff
4000(0.30)
T =
(D + d ) =
(250 + 175)10− 3 = 127.5 N · m Ans.
4
4
(b) From Eq. (16-26),
4F
4(4000)
=
= 0.159 N/mm 2 = 159 kPa
2
2
π ( D − d ) π (2502 − 1752 )
pa =
Eq. (16-27):
T =
π
f pa ( D3 − d 3 ) =
π
(0.30)159 (103 )( 2503 − 1753 )(10− 3 )
Ans.
3
12
12
= 128 N · m Ans.
______________________________________________________________________________
16-17 Given: OD = 6.5 in, ID = 4 in, f = 0.24, pa = 120 psi.
(a) Eq. (16-23):
π pa d
π (120)(4)
F =
(D − d ) =
(6.5 − 4) = 1885 lbf Ans.
2
2
Eq. (16-24) with N sliding planes:
π fpa d 2
π (0.24)(120)(4)
T =
(D − d 2) N =
(6.52 − 42 )(6)
8
8
= 7125 lbf · in Ans.
(b)
T =
π (0.24)(120d )
8
(6.52 − d 2 )(6)
d, in T, lbf · in
2
5191
3
6769
4
7125
Ans.
5
5853
6
2545
(c) The torque-diameter curve exhibits a stationary point maximum in the range of
diameter d. The clutch has nearly optimal proportions.
______________________________________________________________________________
16-18 (a) Eq. (16-24) with N sliding planes:
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π f pa d ( D 2 − d 2 ) N
T =
8
=
π f pa N
8
(D d − d )
2
3
Differentiating with respect to d and equating to zero gives
dT
π f pa N 2
=
( D − 3d 2 ) = 0
dd
8
D
d* =
Ans.
3
3π f pa N
d 2T
π f pa N
= −6
d =−
d
8
4
dd 2
which is negative for all positive d. We have a stationary point maximum.
(b)
d* =
6.5
= 3.75 in
3
Eq. (16-24):
T* =
(
Ans.
)
π (0.24)(120) 6.5 / 3  2
2
6.5 − 6.5 / 3  (6) = 7173 lbf · in


8
(
)
(c) The table indicates a maximum within the range: 3 ≤ d ≤ 5 in
d
(d) Consider:
= 0.80
0.45 =
D
Multiply through by D,
0.45D ≤ d ≤ 0.80D
0.45(6.5) ≤ d ≤ 0.80(6.5)
2.925 ≤ d ≤ 5.2 in
*
1
d
= 0.577
  = d * /D =
3
D
which lies within the common range of clutches.
Yes. Ans.
______________________________________________________________________________
16-19 Given: d = 11 in,
l = 2.25 in,
T = 1800 lbf · in,
D = 12 in,
f = 0.28.
 0.5 
 = 12.53°
 2.25 
α = tan −1 
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Uniform wear
Eq. (16-45):
π f pa d 2
(D − d2)
8sin α
π (0.28) pa (11) 2
1800 =
(12 − 112 ) = 128.2 pa
8sin12.53°
1800
= 14.04 psi Ans.
pa =
128.2
T =
Eq. (16-44):
π pa d
F =
2
(D − d ) =
π (14.04)11
2
(12 − 11) = 243 lbf
Ans.
Uniform pressure
Eq. (16-48):
π f pa
D3 − d 3 )
(
12sin α
π (0.28) pa
1800 =
123 − 113 ) = 134.1 pa
(
12sin12.53°
1800
pa =
Ans.
= 13.42 psi
134.1
T =
Eq. (16-47):
π pa
π (13.42)
(122 − 112 ) = 242 lbf Ans.
4
4
______________________________________________________________________________
F =
(D 2 − d 2 ) =
16-20 Uniform wear
Eq. (16-34):
Eq. (16-33):
Thus,
1
(θ 2 − θ1) f pa ri ( ro2 − ri 2 )
2
F = (θ2 − θ1) pari (ro − ri)
T =
(1 / 2)(θ 2 − θ1) f pa ri ( ro2 − ri 2 )
T
=
f FD
f (θ 2 − θ1) pa ri (ro − ri )( D)
r + ri
D / 2 + d / 2 1
d
= o
=
= 1 +  O.K .
2D
2D
4
D
Ans.
Uniform pressure
Eq. (16-38):
T =
1
(θ 2 − θ1 ) f pa ( ro3 − ri3 )
3
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F =
Eq. (16-37):
1
(θ 2 − θ1) pa ( ro2 − ri 2 )
2
Thus,
(1 / 3)(θ 2 − θ1) f pa ( ro3 − ri3 )
T
2  ( D / 2)3 − (d / 2)3 
=
=


f FD (1 / 2) f (θ 2 − θ1) pa ( ro2 − ri 2 ) D
3  ( D / 2) 2 − (d / 2) 2 D  
2( D / 2)3 1 − (d / D)3 
1  1 − (d / D ) 3 
=
=

 O.K . Ans.
3( D / 2) 2 1 − (d / D) 2  D 3 1 − (d / D) 2 
______________________________________________________________________________
16-21
ω = 2π n / 60 = 2π 500 / 60 = 52.4 rad/s
T =
H
ω
=
2(103 )
= 38.2 N · m
52.4
Key:
T
38.2
=
= 3.18 kN
r
12
Average shear stress in key is
3.18(103 )
τ =
= 13.2 MPa Ans.
6(40)
Average bearing stress is
F
3.18(103 )
σb = −
=−
= −26.5 MPa
Ab
3(40)
Let one jaw carry the entire load.
F =
Ans.
1  26 45 
+

 = 17.75 mm
2 2
2
T
38.2
=
= 2.15 kN
F =
rav 17.75
rav =
The bearing and shear stress estimates are
σb =
−2.15 (103 )
= −22.6 MPa Ans.
10(22.5 − 13)
2.15(103 )
= 0.869 MPa Ans.
τ =
10 0.25π (17.75) 2 
______________________________________________________________________________
Shigley’s MED, 10th edition
Chapter 16 Solutions, Page 19/27
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16-22
ω1 = 2π n / 60 = 2π (1600) / 60 = 167.6 rad/s
ω2 = 0
From Eq. (16-51),
2800(8)
I1I 2
Tt1
=
=
= 133.7 lbf · in · s 2
I1 + I 2
ω1 − ω2 167.6 − 0
Eq. (16-52):
E =
I1I 2
133.7
2
(167.6 − 0) 2 = 1.877 (106 ) lbf ⋅ in
(ω1 − ω2 ) =
2 ( I1 + I 2 )
2
H = E / 9336 = 1.877(106) / 9336 = 201 Btu
In Btu, Eq. (16-53):
Eq. (16-54):
H
201
=
= 41.9°F Ans.
C pW
0.12(40)
______________________________________________________________________________
∆T =
16-23
n1 + n2
260 + 240
=
= 250 rev/min
2
2
Cs = (ω 2 − ω 1) / ω = (n2 − n1) / n = (260 − 240) / 250 = 0.08
n=
Eq. (16-62):
Ans.
ω = 2π (250) / 60 = 26.18 rad/s
From Eq. (16-64):
6.75 (103 )
E2 − E1
I =
=
= 123.1 N · m · s 2
2
2
Csω
0.08(26.18)
I =
m 2
( do + di2 )
8
⇒
m=
8I
8(123.1)
=
= 233.9 kg
2
1.52 + 1.42
d + di
Table A-5, cast iron unit weight = 70.6 kN/m3
Volume:
2
o
⇒ ρ = 70.6(103) / 9.81 = 7197 kg / m3.
V = m / ρ = 233.9 / 7197 = 0.0325 m3
V = π t ( d o2 − d i2 ) / 4 = π t (1.52 − 1.4 2 ) / 4 = 0.2278t
Equating the expressions for volume and solving for t,
0.0325
t =
= 0.143 m = 143 mm Ans.
0.2278
______________________________________________________________________________
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Chapter 16 Solutions, Page 20/27
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16-24 (a) The useful work performed in one revolution of the crank shaft is
U = 320 (103) 200 (10−3) 0.15 = 9.6 (103) J
Accounting for friction, the total work done in one revolution is
U = 9.6(103) / (1 − 0.20) = 12.0(103) J
Since 15% of the crank shaft stroke accounts for 7.5% of a crank shaft revolution, the
energy fluctuation is
E2 − E1 = 9.6(103) − 12.0(103)(0.075) = 8.70(103) J
(b) For the flywheel,
Since
Eq. (16-64):
Ans.
n = 6(90) = 540 rev/min
2π n
2π (540)
ω =
=
= 56.5 rad/s
60
60
Cs = 0.10
E −E
8.70(103 )
I = 2 21 =
= 27.25 N · m · s 2
0.10(56.5)2
Csω
Assuming all the mass is concentrated at the effective diameter, d,
md 2
4
4I
4(27.25)
m= 2 =
= 75.7 kg Ans.
d
1.22
______________________________________________________________________________
I = mr 2 =
16-25 Use Ex. 16-6 and Table 16-6 data for one cylinder of a 3-cylinder engine.
Cs = 0.30
n = 2400 rev/min or 251 rad/s
3(3368)
Tm =
= 804 lbf · in
Ans.
4π
E2 − E1 = 3(3531) = 10 590 in · lbf
10 590
E −E
I = 2 21 =
= 0.560 in · lbf · s 2 Ans.
Csω
0.30(2512 )
______________________________________________________________________________
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Chapter 16 Solutions, Page 21/27
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16-26 (a)
(1)
(T2 )1 = − F21rP = −
T2
T
rP = 2
rG
−n
Ans.
Equivalent energy
(2)
(1 / 2) I 2ω22 = (1 / 2)( I 2 )1 (ω12 )
( I 2 )1 =
I
ω22
I = 22
2 2
ω1
n
Anss.
2
2
2
I G  rG   mG   rG   rG 
=  
=
= n4
IP
 rP   mP   rP   rP 
(3)
From (2)
( I 2 )1 =
(b) I e = I M + I P + n 2 I P +
IL
n2
IG
n4I P
=
= n2I P
n2
n2
Ans.
Ans.
_________________________
______________________________________
____________________
16-27 (a) Reflect IL, IG2 to the center shaft
Shigley’s MED, 10th edition
Chapterr 116 Solutions, Page 22/27
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Reflect the center shaft
aft to the motor shaft
Ie = I M + I P + n2I P +
I P m2
I
+ 2 I P + 2L 2
2
n
n
mn
(b) For R = constant = nm, I e = I M + I P + n 2 I P +
(c) For R = 10,
Ans.
IP
R2I P
I
+
+ L2
2
4
n
n
R
Ans.
∂I e
2(1) 4(102 )(1)
= 0 + 0 + 2n(1) − 3 −
+0=0
∂n
n
n5
n6 − n2 − 200 = 0
From which
n* = 2.430 Ans.
10
m* =
= 4.115
2.430
Ans.
a independent of IL.
Notice that n*and m** are
_________________________
______________________________________
____________________
16-28 From Prob. 16-27,
IP
R2I P
I
+
+ L2
2
4
n
n
R
1 100(1) 100
2
= 10 + 1 + n (1) + 2 +
+ 2
n
n4
10
1 100
2
= 12 + n + 2 + 4
n
n
Ie = I M + I P + n2I P +
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Chapterr 116 Solutions, Page 23/27
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Optimizing the partitio
tioning of a double reduction lowered the gear--train inertia to
20.9/112 = 0.187, or to 19% of that of a single reduction. This inclu
ludes the two additional
gears.
_________________________
______________________________________
____________________
16-29 Figure 16-29 applies,
t2 = 10 s, t1 = 0.5 s
t2 - t1 10 − 0.5
=
= 19
0.5
t1
The load torque, as see
een by the motor shaft (Rule 1, Prob. 16-26),, is
TL =
1300(12)
13
= 1560 lbf · in
10
The rated motor torque
ue Tr is
Tr =
63 025(3)
= 168.07 lbf · in
1125
For Eqs. (16-65):
2π
(1125) = 117.81 rad/s
6
60
2π
ωs =
(1200) = 125.66 rad/s
6
60
−Tr
168.07
a =
=−
= −21.41 lbf ⋅ in ⋅ s/rad
ω s − ωr
125.66 − 117.81
168.07(125.66)
Trωs
=
= 2690.4 lbf · inn
b=
ωs − ωr 125.66 − 117.81
ωr =
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The linear portion of the squirrel-cage motor characteristic can now be expressed as
TM = −21.41ω + 2690.4 lbf · in
Eq. (16-68):
19
 1560 − 168.07 
T2 = 168.07 

 1560 − T2 
One root is 168.07 which is for infinite time. The root for 10 s is desired. Use a
successive substitution method
T2
0.00
19.30
24.40
26.00
26.50
New T2
19.30
24.40
26.00
26.50
26.67
Continue until convergence to
T2 = 26.771 lbf ⋅ in
Eq. (16-69):
a ( t2 − t1 )
−21.41(10 − 0.5)
=
= 110.72 lbf · in · s 2
ln (T2 / Tr ) ln(26.771 / 168.07)
T −b
ω =
a
T − b 26.771 − 2690.4
ωmax = 2
=
= 124.41 rad/s Ans.
−21.41
a
ωmin = 117.81 rad/s Ans.
124.41 + 117.81
ω =
= 121.11 rad/s
2
124.41 − 117.81
ωmax − ωmin
=
= 0.0545 Ans.
Cs =
(ωmax + ωmin ) / 2 (124.41 + 117.81) / 2
1
1
E1 = I ωr2 = (110.72)(117.81) 2 = 768 352 in · lbf
2
2
1 2 1
E2 = I ω2 = (110.72)(124.41) 2 = 856 854 in · lbf
2
2
∆E = E2 − E1 = 856 854 − 768 352 = 88 502 in · lbf
I =
Eq. (16-64):
∆E = Cs I ω 2 = 0.0545(110.72)(121.11) 2
= 88 508 in · lbf, close enough Ans.
During the punch
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Chapter 16 Solutions, Page 25/27
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63 025H
n
TLω (60 / 2π ) 1560(121.11)(60 / 2π )
H =
=
= 28.6 hp
63 025
63 025
T =
The gear train has to be sized for 28.6 hp under shock conditions since the flywheel is on
the motor shaft. From Table A-18,
(
(
)
m 2
W 2
d o + di2 =
d o + d i2
8
8g
8 gI
8(386)(110.72)
W = 2
=
d o + di2
d o2 + di2
I =
)
If a mean diameter of the flywheel rim of 30 in is acceptable, try a rim thickness of 4 in
d i = 30 − (4 / 2) = 28 in
d o = 30 + (4 / 2) = 32 in
8(386)(110.72)
W =
= 189.1 lbf
322 + 282
Rim volume V is given by
V =
πl
(d
4
2
o
− di2 ) =
πl
4
(322 − 282 ) = 188.5l
where l is the rim width as shown in Table A-18. The specific weight of cast iron is
γ = 0.260 lbf / in3, therefore the volume of cast iron is
V =
W
γ
=
189.1
= 727.3 in 3
0.260
Equating the volumes,
188.5 l = 727.3
727.3
l =
= 3.86 in wide
188.5
Proportions can be varied.
______________________________________________________________________________
16-30 Prob. 16-29 solution has I for the motor shaft flywheel as
I = 110.72 lbf · in · s2
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Chapter 16 Solutions, Page 26/27
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A flywheel located on the crank shaft needs an inertia of 102 I (Prob. 16-26, rule 2)
I = 102(110.72) = 11 072 lbf · in · s2
A 100-fold inertia increase. On the other hand, the gear train has to transmit 3 hp under
shock conditions.
Stating the problem is most of the solution. Satisfy yourself that on the crankshaft:
TL = 1300(12) = 15 600 lbf · in
Tr = 10(168.07) = 1680.7 lbf · in
ωr = 117.81 / 10 = 11.781 rad/s
ωs = 125.66 / 10 = 12.566 rad/s
a = −21.41(100) = −2141 lbf · in · s/rad
b = 2690.35(10) = 26903.5 lbf · in
TM = −2141ωc + 26 903.5 lbf · in
19
 15 600 − 1680.5 
T2 = 1680.6 

 15 600 − T2 
The root is 10(26.67) = 266.7 lbf · in
ω = 121.11 / 10 = 12.111 rad/s
Cs = 0.0549 (same)
ωmax = 121.11 / 10 = 12.111 rad/s Ans.
ωmin = 117.81 / 10 = 11.781 rad/s Ans.
E1, E2, ∆E and peak power are the same. From Table A-18
W =
6
8 gI
8(386)(11 072) 34.19 (10 )
=
=
d o2 + di2
d o2 + di2
d o2 + di2
Scaling will affect do and di , but the gear ratio changed I. Scale up the flywheel in the
Prob. 16-29 solution by a factor of 2.5. Thickness becomes 4(2.5) = 10 in.
d = 30(2.5) = 75 in
d o = 75 + (10 / 2) = 80 in
di = 75 − (10 / 2) = 70 in
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Chapter 16 Solutions, Page 27/27
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W =
34.19 (106 )
= 3026 lbf
802 + 702
W
3026
V =
=
= 11 638 in 3
γ
0.260
V =
π
l (802 − 702 ) = 1178 l
4
11 638
l =
= 9.88 in
1178
Proportions can be varied. The weight has increased 3026/189.1 or about 16-fold while
the moment of inertia I increased 100-fold. The gear train transmits a steady 3 hp. But the
motor armature has its inertia magnified 100-fold, and during the punch there are
deceleration stresses in the train. With no motor armature information, we cannot
comment.
______________________________________________________________________________
16-31 This can be the basis for a class discussion.
Shigley’s MED, 10th edition
Chapter 16 Solutions, Page 28/27
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Chapter 17
17-1
Begin with Eq. (17-10),
F1 = Fc + Fi
2 exp( f θ )
exp( f θ ) − 1
Introduce Eq. (17-9):
 exp( f θ ) + 1   2 exp( f θ ) 
2T  exp( f θ ) 
F1 = Fc + d 
= Fc +



d  exp( f θ ) − 1 
 exp( f θ ) − 1   exp( f θ ) + 1 
exp( f θ )
F1 = Fc + ∆F
exp( f θ ) − 1
 exp( f θ ) 
Now add and subtract Fc 

 exp( f θ ) − 1 
 exp( f θ ) 
 exp( f θ ) 
 exp( f θ ) 
+ ∆F 
− Fc 
F1 = Fc + Fc 



 exp( f θ ) − 1 
 exp( f θ ) − 1 
 exp( f θ ) − 1 
 exp( f θ ) 
 exp( f θ ) 
= ( Fc + ∆F ) 
+ Fc − Fc 


 exp( f θ ) − 1 
 exp( f θ ) − 1 
 exp( f θ ) 
Fc
= ( Fc + ∆F ) 
−

 exp( f θ ) − 1  exp( f θ ) − 1
( F + ∆F ) exp( f θ ) − Fc
Q.E.D.
= c
exp( f θ ) − 1
From Ex. 17-2: θd = 3.037 rad, ∆F = 664 lbf, exp( fθ ) = exp[0.80(3.037)] = 11.35, and
Fc = 73.4 lbf.
(73.4 + 664)11.35 − 73.4
= 802 lbf
(11.35 − 1)
F2 = F1 − ∆F = 802 − 664 = 138 lbf
802 + 138
Fi =
− 73.4 = 396.6 lbf
2
1  F1 − Fc 
1
 802 − 73.4 
= 0.80 Ans.
f′ =
ln 
ln 
=
θ d  F2 − Fc  3.037  138 − 73.4 
______________________________________________________________________________
F1 =
17-2
Given: F-1 Polyamide, b = 6 in, d = 2 in with n = 1750 rev/min, Hnom = 2 hp, C = 9(12) =
108 in, velocity ratio = 0.5, Ks = 1.25, nd = 1
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V = π d n / 12 = π (2)(1750) / 12 = 916.3 ft/min
Eq. (17-1):
Table 17-2:
D = d / vel ratio = 2 / 0.5 = 4 in
4−2
D−d
= π − 2sin −1 
θ d = π − 2sin −1
 = 3.123 rad
2C
 2(108) 
t = 0.05 in, dmin = 1.0 in, Fa = 35 lbf/in, γ = 0.035 lbf/in3, f = 0.5
w = 12 γbt = 12(0.035)6(0.05) = 0.126 lbf/ft
2
(a) Eq. (e), p. 877:
Fc =
2
0.126  916.3 
w V 
  =

 = 0.913 lbf
g  60 
32.17  60 
Ans.
63 025H nom K s nd
63 025(2)(1.25)(1)
=
= 90.0 lbf · in
n
1750
2T
2(90.0)
∆F = ( F1 )a − F2 =
=
= 90.0 lbf
d
2
T =
Table 17-4:
Cp = 0.70
Eq. (17-12):
(F1)a = bFaCpCv = 6(35)(0.70)(1) = 147 lbf
F2 = (F1)a − [(F1)a − F2] = 147 − 90 = 57 lbf
Ans.
Ans.
Do not use Eq. (17-9) because we do not yet know f ′
( F1 )a
+ F2
147 + 57
− 0.913 = 101.1 lbf
2
2
Using Eq. (17-7) solved for f ′ (see step 8, p.888),
1  ( F1) a − Fc 
1
 147 − 0.913 
ln 
ln 
= 0.307
f′ =
 =
θ d  F2 − Fc  3.123  57 − 0.913 
Eq. (i), p. 878:
Fi =
− Fc =
Ans.
The friction is thus underdeveloped.
(b) The transmitted horsepower is, with ∆F = (F1)a − F2 = 90 lbf,
(∆F )V
90(916.3)
Eq. (j), p. 879:
H =
=
= 2.5 hp Ans.
33 000
33 000
H
2.5
nf s =
=
=1
H nom K s
2(1.25)
Eq. (17-1):
Eq. (17-2):
4−2
D−d
= π + 2sin −1 
 = 3.160 rad
2C
 2(108) 
L = [4C2 − (D − d)2]1/2 + (DθD + dθd)/2
θ D = π + 2sin −1
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= [4(108)2 − (4 − 2)2]1/2 + [4(3.160) + 2(3.123)]/2 = 225.4 in
dip =
(c) Eq. (17-13):
3C 2 w 3(108 / 12) 2 (0.126)
=
= 0.151 in
2Fi
2(101.1)
Ans.
Ans.
Comment: The solution of the problem is finished; however, a note concerning the design
is presented here.
The friction is under-developed. Narrowing the belt width to 5 in (if size is available) will
increase f ′ . The limit of narrowing is bmin = 4.680 in, whence
w = 0.0983 lbf/ft
Fc = 0.713 lbf
T = 90 lbf · in (same)
∆F = ( F1) a − F2 = 90 lbf
Fi = 68.9 lbf
( F1) a = 114.7 lbf
F2 = 24.7 lbf
f ′ = f = 0.50
dip = 0.173 in
Longer life can be obtained with a 6-inch wide belt by reducing Fi to attain f ′ = 0.50.
Prob. 17-8 develops an equation we can use here
(∆F + Fc ) exp( f θ ) − Fc
exp( f θ ) − 1
F2 = F1 − ∆F
F + F2
Fi = 1
− Fc
2
1  F1 − Fc 
f′ =
ln 

θ d  F2 − Fc 
F1 =
dip =
3C 2w
2 Fi
which in this case, θd = 3.123 rad, exp(f θ) = exp[0.5(3.123)] = 4.766, w = 0.126 lbf/ft,
∆F = 90.0 lbf, Fc = 0.913 lbf, and gives
( 0.913 + 90 ) 4.766 − 0.913 = 114.8 lbf
F1 =
4.766 − 1
F2 = 114.8 − 90 = 24.8 lbf
Fi = (114.8 + 24.8)/ 2 − 0.913 = 68.9 lbf
f′ =
1
 114.8 − 0.913 
ln 
 = 0.50
3.123  24.8 − 0.913 
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3 (108 / 12 ) 0.126
= 0.222 in
2(68.9)
2
dip =
So, reducing Fi from 101.1 lbf to 68.9 lbf will bring the undeveloped friction up to 0.50,
with a corresponding dip of 0.222 in. Having reduced F1 and F2, the endurance of the belt
is improved. Power, service factor and design factor have remained intact.
______________________________________________________________________________
17-3
Double the dimensions of Prob. 17-2.
In Prob. 17-2, F-1 Polyamide was used with a thickness of 0.05 in. With what is available
in Table 17-2 we will select the Polyamide A-2 belt with a thickness of 0.11 in. Also, let
b = 12 in, d = 4 in with n = 1750 rev/min, Hnom = 2 hp, C = 18(12) = 216 in, velocity
ratio = 0.5, Ks = 1.25, nd = 1.
V = π d n / 12 = π (4)(1750) / 12 = 1833 ft/min
Eq. (17-1):
Table 17-2:
D = d / vel ratio = 4 / 0.5 = 8 in
8−4 
D−d
θ d = π − 2sin −1
= π − 2sin −1 
 = 3.123 rad
2C
 2(216) 
t = 0.11 in, dmin = 2.4 in, Fa = 60 lbf/in, γ = 0.037 lbf/in3, f = 0.8
w = 12 γbt = 12(0.037)12(0.11) = 0.586 lbf/ft
2
2
w V 
0.586  1833 
Fc =   =

 = 17.0 lbf
g  60 
32.17  60 
(a) Eq. (e), p. 877:
Ans.
63 025H nom K s nd
63 025(2)(1.25)(1)
=
= 90.0 lbf · in
1750
n
2T
2(90.0)
∆F = ( F1 )a − F2 =
=
= 45.0 lbf
d
4
T =
Table 17-4:
Cp = 0.73
Eq. (17-12):
(F1)a = bFaCpCv = 12(60)(0.73)(1) = 525.6 lbf
F2 = (F1)a − [(F1)a − F2] = 525.6 − 45 = 480.6 lbf
Fi =
Eq. (i), p. 878:
( F1 )a
2
+ F2
− Fc =
Ans.
Ans.
525.6 + 480.6
− 17.0 = 486.1 lbf
2
Ans.
Eq. (17-9):
f′ =
 ( F ) − Fc 
1
 525.6 − 17.0 
= 0.0297
ln  1 a
ln 
 =
θ d  F2 − Fc  3.123  480.6 − 17.0 
1
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The friction is thus
us underdeveloped.
ho
is, with ∆F = (F1)a − F2 = 45 lbf,
(b) The transmitted horsepower
(∆F )V
45(1833)
=
= 2.5 hp
3 000
33
33 000
H
2.5
=
=
=1
H nom K s
2(1.25)
H =
nf s
θ D = π + 2sin −1
Eq. (17-1):
Ans.
8−4 
D−d
= π + 2sin −1 
= 3.160 rad
2C
16) 
 2(216
L = [4C2 − (D − d)2]1/2 + (DθD + dθd)/2
Eq. (17-2):
= [4(216)2 − (8 − 4)2]1/2 + [8(3.160) + 4(3.123
23)]/2 = 450.9 in
Ans.
3C 2 w 3(216 / 12) 2 (0.586)
=
= 0.586
6 in Ans.
2Fi
2(486.1)
_________________________
______________________________________
____________________
dip =
(c) Eq. (17-13):
17-4
decision set on p. 885 is useful.
As a design task, the de
A priori decisions:
• Function: Hnom = 60 hp, n = 380 rev/min, C = 192 in, Ks = 1.1
• Design factor: nd = 1
• Initial tension: Caten
tenary
• Belt material. Tablee 117-2: Polyamide A-3, Fa = 100 lbf/in, γ = 0.042
0.
lbf/in3, f = 0.8
• Drive geometry: d = D = 48 in
• Belt thickness: t = 0.13
0
in
Design variable: Beltt width.
w
Use a method of trials.
ls. Initially, choose b = 6 in
V =
π dn
=
π (48)(380)
= 4775 ft/min
12
12
w = 12γ bt = 12(0.042)(6)(0.13)
1
= 0.393 lbf/ft
2
2
wV
0.393(4775 / 60)
=
= 77.4 lbf
Fc =
g
32.17
63 025H nom
63 025(60)(1.1)(1)
n K s nd
T =
=
= 10 946 lbf · in
n
380
Shigley’s MED, 10th edition
Chapter
ter 17 Solutions, Page 5/39
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2T
2(10 946)
=
= 456.1 lbf
d
48
F1 = ( F1)a = bFaC pCv = 6(100)(1)(1) = 600 lbf
F2 = F1 − ∆F = 600 − 456.1 = 143.9 lbf
∆F =
Transmitted power H
∆F (V )
=
33 000
F + F2
Fi = 1
−
2
F −
1
ln 1
f′ =
θ d F2 −
H =
Eq. (17-2):
456.1(4775)
= 66 hp
33 000
600 + 143.9
Fc =
− 77.4 = 294.6 lbf
2
Fc
1  600 − 77.4 
= ln 
= 0.656
Fc
π  143.9 − 77.4 
L = [4(192)2 − (48 − 48)2]1/2 + [48(π) + 48(π)] / 2 = 534.8 in
Friction is not fully developed, so bmin is just a little smaller than 6 in (5.7 in). Not having
a figure of merit, we choose the most narrow belt available (6 in). We can improve the
design by reducing the initial tension, which reduces F1 and F2, thereby increasing belt
life (see the result of Prob. 17-8). This will bring f ′ to 0.80
F1 =
( ∆F
+ Fc ) exp ( f θ ) − Fc
exp ( f θ ) − 1
exp ( f θ ) = exp(0.80π ) = 12.345
Therefore
(456.1 + 77.4)(12.345) − 77.4
= 573.7 lbf
12.345 − 1
F2 = F1 − ∆F = 573.7 − 456.1 = 117.6 lbf
F + F2
573.7 + 117.6
Fi = 1
− Fc =
− 77.4 = 268.3 lbf
2
2
F1 =
These are small reductions since f ′ is close to f , but improvements nevertheless.
1 F1 − Fc
1  573.7 − 77.4 
f′ =
= ln 
= 0.80
ln
θ d F2 − Fc π  117.6 − 77.4 
3C 2w 3(192 / 12)2 (0.393)
=
= 0.562 in
2 Fi
2(268.3)
______________________________________________________________________________
dip =
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17-5
From the last equation given in the problem statement,
exp ( f φ ) =
1
1 − {2T / [d (a0 − a2 )b]}


2T
1 −
 exp ( f φ ) = 1
d (a0 − a2 )b 



2T

 exp ( f φ ) = exp ( f φ ) − 1
 d (a0 − a2 )b 
b=
1  2T

a0 − a2  d
  exp ( f φ ) 


  exp ( f φ ) − 1 
But 2T/d = 33 000Hd/V. Thus,
  exp ( f φ ) 
 Q.E.D.

  exp ( f φ ) − 1 
______________________________________________________________________________
b=
17-6
1  33 000 H d

a0 − a2 
V
Refer to Ex. 17-1 on p. 882 for the values used below.
(a) The maximum torque prior to slip is,
T =
63 025H nom K s nd
63 025(15)(1.25)(1.1)
=
= 742.8 lbf · in
n
1750
Ans.
The corresponding initial tension, from Eq. (17-9), is,
Fi =
T  exp( f θ ) + 1  742.8  11.17 + 1 

=

 = 148.1 lbf
d  exp( f θ ) − 1 
6  11.17 − 1 
Ans.
(b) See Prob. 17-4 statement. The final relation can be written
bmin =
=
 33 000H a exp ( f θ ) 
1

2 
FaC pCv − (12γ t / 32.174)(V / 60)  V [exp ( f θ ) − 1] 
 33 000(20.6)(11.17) 
1
2 
100(0.7)(1) − {[12(0.042)(0.13)] / 32.174} (2749 / 60)  2749(11.17 − 1) 
= 4.13 in
Ans.
This is the minimum belt width since the belt is at the point of slip. The design must
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round up to an available width.
Eq. (17-1):
D −d
−1 18 − 6 
 = π − 2sin 

 2C 
 2(96) 
= 3.016 511 rad
D − d
−1 18 − 6 
θ D = π + 2sin −1 
 = π + 2sin 

 2C 
 2(96) 
= 3.266 674 rad
θ d = π − 2sin −1 
Eq. (17-2):
L = [4(96)2 − (18 − 6) 2 ]1/ 2 +
= 230.074 in
1
[18(3.266 674) + 6(3.016 511)]
2
Ans.
2T
2(742.8)
=
= 247.6 lbf
d
6
( F1 )a = bFaC pCv = F1 = 4.13(100)(0.70)(1) = 289.1 lbf
F2 = F1 − ∆F = 289.1 − 247.6 = 41.5 lbf
w = 12γ bt = 12(0.042)4.13(0.130) = 0.271 lbf/ft
∆F =
(c)
2
2
w V 
0.271  2749 
Fc =   =

 = 17.7 lbf
g  60 
32.17  60 
F + F2
289.1 + 41.5
− Fc =
− 17.7 = 147.6 lbf
Fi = 1
2
2
Transmitted belt power H
∆F (V )
247.6(2749)
=
= 20.6 hp
33 000
33 000
20.6
H
n fs =
=
= 1.1
H nom K s 15(1.25)
H =
Dip:
2
3C 2 w 3(96 / 12) ( 0.271)
=
= 0.176 in
dip =
2Fi
2(147.6)
(d) If you only change the belt width, the parameters in the following table change as
shown.
Shigley’s MED, 10th edition
Chapter 17 Solutions, Page 8/39
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b
w
Fc
(F1)a
F2
Fi
f′
dip
Ex. 17-1 This Problem
6.00
4.13
0.393
0.271
25.6
17.7
420
289
172.4
41.5
270.6
147.6
0.33*
0.80**
0.139
0.176
*Friction underdeveloped
**Friction fully developed
______________________________________________________________________________
17-7
The transmitted power is the same.
Fc
Fi
(F1)a
F2
Ha
nfs
f′
dip
n-Fold
b = 6 in b = 12 in Change
25.65
51.3
2
270.35
664.9
2.46
420
840
2
172.4
592.4
3.44
20.62
20.62
1
1.1
1.1
1
0.139
0.125
0.90
0.328
0.114
0.34
If we relax Fi to develop full friction (f = 0.80) and obtain longer life, then
n-Fold
b = 6 in b = 12 in Change
Fc
25.6
51.3
2
Fi
148.1
148.1
1
297.6
323.2
1.09
F1
F2
50
75.6
1.51
0.80
0.80
1
f′
0.503
2
dip 0.255
______________________________________________________________________________
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17-8
Find the resultant of F1 and F2:
D−d
2C
D−d
sin α =
2C
2
1D − d 
cos α 1 − 

2  2C 
α = sin −1
2

1 D − d  
R = F1 coss α + F2 cos α = ( F1 + F2 ) 1 − 
 
2  2C  

D−d
R y = F1 sin α − F2 sin α = ( F1 − F2 )
Ans.
2C
x
Ans.
From Ex. 17-2, d = 16
6 in, D = 36 in, C = 16(12) = 192 in, F1 = 940
0 lbf, F2 = 276 lbf
 36 − 16 
= 2.9855o

 2(192) 
α = sin
in −1 
2

1  36 − 16  
R = (940
40 + 276) 1 − 
  = 1214.4 lbf
2  2(192)  


 36 − 16 
R y = (940
40 − 276) 
 = 34.6 lbf
 2(192) 
d 
 16 
T = ( F1 − F2 )   = (940 − 276)   = 5312 lbf · in
i
2
 2
_________________________
______________________________________
____________________
x
This is a good class project.
pro
Form four groups, each with a belt to ddesign. Once each
group agrees internally
lly, all four should report their designs includin
ing the forces and
torques on the line shaf
haft. If you give them the pulley locations, they
ey could design the line
shaft.
_________________________
______________________________________
____________________
17-9
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17-10 If you have the students implement a computer program, the design problem selections
may differ, and the students will be able to explore them. For Ks = 1.25, nd = 1.1, d = 14
in and D = 28 in, a polyamide A-5 belt, 8 inches wide, will do (bmin = 6.58 in)
______________________________________________________________________________
17-11 An efficiency of less than unity lowers the output for a given input. Since the object of
the drive is the output, the efficiency must be incorporated such that the belt’s capacity is
increased. The design power would thus be expressed as
H nom K s nd
Ans.
eff
______________________________________________________________________________
Hd =
17-12 Some perspective on the size of Fc can be obtained from
w V 
12γ bt  V 
  =
 
g  60 
g  60 
2
Fc =
2
An approximate comparison of non-metal and metal belts is presented in the table below.
γ, lbf/in
b, in
t, in
3
Non-metal
0.04
5.00
0.20
Metal
0.280
1.000
0.005
The ratio w / wm is
12(0.04)(5)(0.2)
w
=
wm 12(0.28)(1)(0.005)
29
The second contribution to Fc is the belt peripheral velocity which tends to be low in
metal belts used in instrument, printer, plotter and similar drives. The velocity ratio
squared influences any Fc / (Fc)m ratio.
It is common for engineers to treat Fc as negligible compared to other tensions in the
belting problem. However, when developing a computer code, one should include Fc.
______________________________________________________________________________
17-13 Eq. (17-8):
exp( f θ ) − 1
exp( f θ ) − 1
F1
exp( f θ )
exp( f θ )
Assuming negligible centrifugal force and setting F1 = ab from step 3, p. 889,
∆F = F1 − F2 = ( F1 − Fc )
bmin =
∆F exp( f θ )
a exp( f θ ) − 1
Shigley’s MED, 10th edition
(1)
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(∆F )V
33 000
33 000 H nom K s nd
∆F =
V
H d = H nom K s nd =
Also,
1  33 000 H d  exp( f θ )
Ans.

 exp( f θ ) − 1
a
V
______________________________________________________________________________
bmin =
Substituting into Eq. (1),
17-14 The decision set for the friction metal flat-belt drive is:
A priori decisions
• Function: Hnom = 1 hp, n = 1750 rev/min, VR = 2 , C 15 in, Ks = 1.2 ,
Np = 106 belt passes.
• Design factor: nd = 1.05
• Belt material and properties: 301/302 stainless steel
Table 17-8:
Sy = 175 kpsi, E = 28 Mpsi, ν = 0.285
• Drive geometry: d = 2 in, D = 4 in
• Belt thickness: t = 0.003 in
Design variables:
• Belt width, b
• Belt loop periphery
Preliminaries
H d = H nom K s nd = 1(1.2)(1.05) = 1.26 hp
63 025(1.26)
T =
= 45.38 lbf · in
1750
A 15 in center-to-center distance corresponds to a belt loop periphery of 39.5 in. The
40 in loop available corresponds to a 15.254 in center distance.
 4−2 
θ d = π − 2sin −1 
 = 3.010 rad
 2(15.254) 
 4−2 
θ D = π + 2sin −1 
 = 3.273 rad
 2(15.274) 
For full friction development
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exp( f θ d ) = exp[0.35(3.010)] = 2.868
π dn π (2)(1750)
V =
=
= 916.3 ft/s
12
12
S y = 175 kpsi
Eq. (17-15):
S y = 14.17 (106 ) N p−0.407 = 14.17 (106 )(106 )
−0.407
= 51.212 (103 ) psi
From selection step 3, p. 889,



Et
28(106 )(0.003) 
3
=
−
51.212(10
)
(0.003)
a = S f −
t

(1 − ν 2 )d 
(1 − 0.2852 )(2) 


= 16.50 lbf/in of belt width
( F1)a = ab = 16.50b
For full friction development, from Prob. 17-13,
bmin =
∆F exp( f θ d )
a exp( f θ d ) − 1
∆F =
2T
2(45.38)
=
= 45.38 lbf
d
2
bmin =
45.38  2.868 

 = 4.23 in
16.50  2.868 − 1 
So
Decision #1: b = 4.5 in
F1 = ( F1) a = ab = 16.5(4.5) = 74.25 lbf
F2 = F1 − ∆F = 74.25 − 45.38 = 28.87 lbf
74.25 + 28.87
F + F2
Fi = 1
=
= 51.56 lbf
2
2
Existing friction
F
1
 74.25 
= 0.314
ln  1  =
ln 
θ d  F2  3.010  28.87 
(∆F )V
45.38(916.3)
Ht =
=
= 1.26 hp
33 000
33 000
Ht
1.26
n fs =
=
= 1.05
H nom K s 1(1.2)
f′ =
1
This is a non-trivial point. The methodology preserved the factor of safety corresponding
to nd = 1.1 even as we rounded bmin up to b.
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Decision #2 was taken care of with the adjustment of the center-to-center distance to
accommodate the belt loop. Use Eq. (17-2) as is and solve for C to assist in this.
Remember to subsequently recalculate θd and θD .
______________________________________________________________________________
17-15 Decision set:
A priori decisions
• Function: Hnom = 5 hp, N = 1125 rev/min, VR = 3, C 20 in, Ks = 1.25,
Np = 106 belt passes
• Design factor: nd = 1.1
• Belt material: BeCu, Sy = 170 kpsi, E = 17 Mpsi, ν = 0.220
• Belt geometry: d = 3 in, D = 9 in
• Belt thickness: t = 0.003 in
Design decisions
• Belt loop periphery
• Belt width b
Preliminaries:
H d = H nom K s nd = 5(1.25)(1.1) = 6.875 hp
63 025(6.875)
T =
= 385.2 lbf · in
1125
Decision #1: Choose a 60-in belt loop with a center-to-center distance of 20.3 in.
 9−3 
θ d = π − 2sin −1 
 = 2.845 rad
 2(20.3) 
 9−3 
θ D = π + 2sin −1 
 = 3.438 rad
 2(20.3) 
For full friction development:
exp( f θ d ) = exp[0.32(2.845)] = 2.485
π dn π (3)(1125)
=
= 883.6 ft/min
V =
12
12
S f = 56.67 kpsi
From selection step 3, p. 889,
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


Et
17(106 )(0.003) 
3
a = S f −
t
=
−
56.67(10
)
(0.003) = 116.4 lbf/in

(1 − ν 2 )d 
(1 − 0.222 )(3) 


2T
2(385.2)
∆F =
=
= 256.8 lbf
d
3
∆F  exp( f θ d )  256.8  2.485 
bmin =

 =

 = 3.69 in
a  exp( f θ d ) − 1  116.4  2.485 − 1 
Decision #2: b = 4 in
F1 = ( F1) a = ab = 116.4(4) = 465.6 lbf
F2 = F1 − ∆F = 465.6 − 256.8 = 208.8 lbf
F + F2
465.6 + 208.8
=
= 337.3 lbf
Fi = 1
2
2
Existing friction
F
1
 465.6 
ln  1  =
ln 
= 0.282
θ d  F2  2.845  208.8 
(∆F )V
256.8(883.6)
=
= 6.88 hp
H =
33 000
33 000
6.88
H
=
= 1.1
n fs =
5(1.25) 5(1.25)
f′ =
1
Fi can be reduced only to the point at which f ′ = f = 0.32. From Eq. (17-9)
Fi =
T
d
 exp( f θ d ) + 1  385.2  2.485 + 1 

 =

 = 301.3 lbf
3  2.485 − 1 
 exp( f θ d ) − 1 
Eq. (17-10):
 2 exp( f θ d ) 
 2(2.485) 
= 429.7 lbf
F1 = Fi 
 = 301.3 
f
θ
exp(
)
+
1
 2.485 + 1 
d


F2 = F1 − ∆F = 429.7 − 256.8 = 172.9 lbf
and
f ′ = f = 0.32
______________________________________________________________________________
17-16 This solution is the result of a series of five design tasks involving different belt
thicknesses. The results are to be compared as a matter of perspective. These design tasks
are accomplished in the same manner as in Probs. 17-14 and 17-15 solutions.
The details will not be presented here, but the table is provided as a means of learning.
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Five groups of students could each be assigned a belt thickness. You can form a table
from their results or use the table given here.
0.002
4.000
20.300
109.700
3.000
9.000
310.600
439.000
182.200
1.100
60.000
0.309
301.200
429.600
172.800
0.320
b
CD
a
d
D
Fi
F1
F2
nf s
L
f′
Fi
F1
F2
f
0.003
3.500
20.300
131.900
3.000
9.000
333.300
461.700
209.000
1.100
60.000
0.285
301.200
429.600
172.800
0.320
t, in
0.005
4.000
20.300
110.900
3.000
9.000
315.200
443.600
186.800
1.100
60.000
0.304
301.200
429.600
172.800
0.320
0.008
1.500
18.700
194.900
5.000
15.000
215.300
292.300
138.200
1.100
70.000
0.288
195.700
272.700
118.700
0.320
0.010
1.500
20.200
221.800
6.000
18.000
268.500
332.700
204.300
1.100
80.000
0.192
166.600
230.800
102.400
0.320
The first three thicknesses result in the same adjusted Fi, F1 and F2 (why?). We have no
figure of merit, but the costs of the belt and pulleys are about the same for these three
thicknesses. Since the same power is transmitted and the belts are widening, belt forces
are lessening.
______________________________________________________________________________
17-17 This is a design task. The decision variables would be belt length and belt section, which
could be combined into one, such as B90. The number of belts is not an issue.
We have no figure of merit, which is not practical in a text for this application. It is
suggested that you gather sheave dimensions and costs and V-belt costs from a principal
vendor and construct a figure of merit based on the costs. Here is one trial.
Preliminaries: For a single V-belt drive with Hnom = 3 hp, n = 3100 rev/min, D = 12 in,
and d = 6.2 in, choose a B90 belt, Ks = 1.3 and nd = 1. From Table 17-10, select a
circumference of 90 in. From Table 17-11, add 1.8 in giving
Lp = 90 + 1.8 = 91.8 in
Eq. (17-16b):

π


C = 0.25  91.8 − (12 + 6.2)  +
2

 
= 31.47 in
2

π


2
91.8
(12
6.2)
2(12
6.2)
−
+
−
−



2

Shigley’s MED, 10th edition
Chapter 17 Solutions, Page 16/39
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 12 − 6.2 
θ d = π − 2sin -1 
 = 2.9570 rad
 2(31.47) 
exp( f θ d ) = exp[0.5123(2.9570)] = 4.5489
π dn π (6.2)(3100)
V =
12
=
12
= 5031.8 ft/min
Table 17-13:
Angleθ = θ d
180°
π
 180° 
= (2.957 rad) 
= 169.42°
 π 
The footnote regression equation of Table 17-13 gives K1 without interpolation:
K1 = 0.143 543 + 0.007 468(169.42°) − 0.000 015 052(169.42°)2 = 0.9767
The design power is
Hd = HnomKsnd = 3(1.3)(1) = 3.9 hp
From Table 17-14 for B90, K2 = 1. From Table 17-12 take a marginal entry of Htab = 4,
although extrapolation would give a slightly lower Htab.
Ha = K1K2Htab = 0.9767(1)(4) = 3.91 hp
Eq. (17-17):
The allowable ∆Fa is given by
∆Fa =
63 025H a
63 025(3.91)
=
= 25.6 lbf
3100(6.2 / 2)
n(d / 2)
The allowable torque Ta is
Ta =
∆Fa d
25.6(6.2)
=
= 79.4 lbf · in
2
2
From Table 17-16, Kc = 0.965. Thus, Eq. (17-21) gives,
2
2
 V 
 5031.8 
Fc = K c 
 = 0.965 
 = 24.4 lbf
 1000 
 1000 
At incipient slip, Eq. (17-9) provides:
 T   exp( f θ ) + 1   79.4  4.5489 + 1 
Fi =   

 = 20.0 lbf
 =
 d   exp( f θ ) − 1   6.2  4.5489 − 1 
Eq. (17-10):
Shigley’s MED, 10th edition
Chapter 17 Solutions, Page 17/39
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 2 exp( f θ ) 
 2(4.5489) 
F1 = Fc + Fi 
= 24.4 + 20 
= 57.2 lbf

 4.5489 + 1 
 exp( f θ ) + 1 
Thus, F2 = F1 − ∆Fa = 57.2 − 25.6 = 31.6 lbf
Eq. (17-26):
n fs =
H a Nb
(3.91)(1)
=
= 1.003
Hd
3.9
Ans.
If we had extrapolated for Htab, the factor of safety would have been slightly less than
one.
Life
Use Table 17-16 to find equivalent tensions T1 and T2 .
Kb
= 57.2 +
d
K
T2 = F1 + ( Fb ) 2 = F1 + b = 57.2 +
D
T1 = F1 + ( Fb )1 = F1 +
576
= 150.1 lbf
6.2
576
= 105.2 lbf
12
From Table 17-17, K = 1193, b = 10.926, and from Eq. (17-27), the number of belt passes
is:
 K  − b  K  − b 
N P =   +   
 T1 
 T2  
−1
 1193  −10.926  1193  −10.926 
= 
+



 105.2 
 150.1 

−1
= 6.72(109 ) passes
From Eq. (17-28) for NP > 109,
109 (91.8)
720V
720(5031.8)
t > 25 340 h Ans.
t =
N P Lp
>
Suppose nf s was too small. Compare these results with a 2-belt solution.
H tab = 4 hp/belt, Ta = 39.6 lbf · in/belt,
∆Fa = 12.8 lbf/belt, H a = 3.91 hp/belt
2(3.91)
NH
Nb H a
=
= 2.0
n fs = b a =
3(1.3)
Hd
H nom K s
Also,
F1 = 40.8 lbf/belt,
F2 = 28.0 lbf/belt
Shigley’s MED, 10th edition
Chapter 17 Solutions, Page 18/39
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Fi = 9.99 lbf/belt,
Fc = 24.4 lbf/belt
( Fb )1 = 92.9 lbf/belt,
( Fb )2 = 48 lbf/belt
T1 = 133.7 lbf/belt,
T2 = 88.8 lbf/belt
10
N P = 2.39(10 ) passes,
t > 605 600 h
Initial tension of the drive:
(Fi)drive = NbFi = 2(9.99) = 20 lbf
______________________________________________________________________________
17-18 Given: two B85 V-belts with d = 5.4 in, D = 16 in, n = 1200 rev/min, and Ks = 1.25
Table 17-11: Lp = 85 + 1.8 = 86.8 in
Eq. (17-17b):

π


C = 0.25  86.8 − (16 + 5.4)  +
2

 
= 26.05 in Ans.
2

π


2
86.8
−
(16
+
5.4)
−
2(16
−
5.4)



2

Eq. (17-1) in degrees:
 16 − 5.4 
θ d = 180° − 2sin −1 
 = 156.5°
 2(26.05) 
From Table 17-13 footnote:
K1 = 0.143 543 + 0.007 468(156.5°) − 0.000 015 052(156.5°)2 = 0.944
Table 17-14:
K2 = 1
Belt speed:
V =
π (5.4)(1200)
12
= 1696 ft/min
Use Table 17-12 to interpolate for Htab.
 2.62 − 1.59 
H tab = 1.59 + 
 (1696 − 1000) = 2.31 hp/belt
 2000 − 1000 
Eq. (17-17) for two belts:
H a = K1K 2 N b H tab = 0.944(1)(2)(2.31) = 4.36 hp
Assuming nd = 1,
Hd = KsHnomnd = 1.25(1)Hnom
For a factor of safety of one,
Shigley’s MED, 10th edition
Chapter 17 Solutions, Page 19/39
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Ha = Hd
4.36 = 1.25H nom
4.36
= 3.49 hp Ans.
H nom =
1.25
______________________________________________________________________________
17-19 Given: Hnom = 60 hp, n = 400 rev/min, d = D = 26 in on 12 ft centers.
From Table 17-15, a conservative service factor of KS = 1.4 is selected.
Design task: specify V-belt and number of strands (belts). Tentative decision: Use D360
belts.
Table 17-11: Lp = 360 + 3.3 = 363.3 in
Eq. (17-16b):

π


C = 0.25  363.3 − (26 + 26)  +
2

 
= 140.8 in (nearly 144 in)
2

π


2
−
+
−
−
363.3
(26
26)
2(26
26)



2

θ d = π , θ D = π , exp[0.5123π ] = 5.0,
π dn π (26)(400)
V =
12
=
12
= 2722.7 ft/min
Table 17-13: For θ = 180°, K1 = 1
Table 17-14: For D360, K2 = 1.10
Table 17-12: Htab = 16.94 hp by interpolation
Thus,
Ha = K1K2Htab = 1(1.1)(16.94) = 18.63 hp / belt
Eq. (17-19):
Hd = HnomKs nd = 60(1.4)(1) = 84 hp
Number of belts, Nb
Nb =
Hd
84
=
= 4.51
18.63
Ha
Round up to five belts. It is left to the reader to repeat the above for belts such as C360
and E360.
Shigley’s MED, 10th edition
Chapter 17 Solutions, Page 20/39
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63 025H a
63 025(18.63)
=
= 225.8 lbf/belt
400(26 / 2)
n(d / 2)
(∆Fa )d
225.8(26)
Ta =
=
= 2935 lbf · in/belt
2
2
∆Fa =
Eq. (17-21):
2
2
 V 
 2722.7 
Fc = 3.498 
= 3.498 
= 25.9 lbf/belt
 1000 
 1000 
At fully developed friction, Eq. (17-9) gives
Fi =
Eq. (17-10):
Life
T
d
 exp( f θ ) + 1  2935  5 + 1 
 exp( f θ ) − 1  = 26  5 − 1  = 169.3 lbf/belt




 2 exp( f θ ) 
 2(5) 
= 25.9 + 169.3 
= 308.1 lbf/belt
F1 = Fc + Fi 

 5 + 1 
 exp( f θ ) + 1 
F2 = F1 − ∆Fa = 308.1 − 225.8 = 82.3 lbf/belt
18.63 ( 5)
H N
= 1.109 Ans.
nf s = a b =
84
Hd
From Table 17-16,
T1 = T2 = F1 +
Kb
5 680
= 308.1 +
= 526.6 lbf
d
26
Eq. (17-27):
  K  −b  K  − b 
N P =   +   
 T1 
 T2  
Thus,
NP > 10−9 passes
Eq. (17-28):
t =
N P Lp
720V
−1
= 5.28 (10−9 ) passes
Ans.
>
109 (363.3)
720(2722.7)
Thus,
t > 185 320 h Ans.
______________________________________________________________________________
17-20 Preliminaries: D 60 in, 14-in wide rim, Hnom = 50 hp, n = 875 rev/min, Ks = 1.2,
nd = 1.1, mG = 875/170 = 5.147, d 60 / 5.147 = 11.65 in
(a) From Table 17-9, an 11-in sheave exceeds C-section minimum diameter and
precludes D- and E-section V-belts.
Shigley’s MED, 10th edition
Chapter 17 Solutions, Page 21/39
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Decision: Use d = 11 in, C270 belts
Table 17-11: Lp = 270 + 2.9 = 272.9 in
Eq. (17-16b):

π


C = 0.25   272.9 − (60 + 11)  +
2

 
= 76.78 in
2

π


2
272.9
−
(60
+
11)
−
2(60
−
11)



2

This fits in the range
D < C < 3( D + d ) ⇒ 60 < C < 3(60 + 11) ⇒ 60 in < C < 213 in
60 − 11
θ d = π − 2sin −1
= 2.492 rad = 142.8°
2(76.78)
60 − 11
θ D = π + 2sin −1
= 3.791 rad
2(76.78)
exp(f θd) = exp[0.5123(2.492)] = 3.5846
For the flat on flywheel, f = 0.13 (see p. 892), exp(f θD) = exp[0.13(3.791)] = 1.637.
The belt speed is
V =
π dn
12
=
π (11)(875)
12
= 2520 ft/min
Table 17-13:
K1 = 0.143 543 + 0.007 468(142.8°) − 0.000 015 052(142.8°)2 = 0.903
Table 17-14: K2 = 1.15
For interpolation of Table 17-12, let x be entry for d = 11.65 in and n = 2000 ft/min, and y
be entry for d = 11.65 in and n = 3000 ft/min. Then,
x − 6.74
7.17 − 6.74
=
⇒ x = 7.01 hp at 2000 ft/min
11.65 − 11
12 − 11
and
8.11 − y
8.84 − 8.11
=
⇒ y = 8.58 hp at 3000 ft/min
11.65 − 11
12 − 11
Interpolating these for 2520 ft/min gives
8.58 − H tab
3000 − 2520
=
⇒ H tab = 7.83 hp/belt
8.58 − 7.01 3000 − 2000
Shigley’s MED, 10th edition
Chapter 17 Solutions, Page 22/39
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Eq. (17-17):
Ha = K1K2Htab = 0.903(1.15)(7.83) = 8.13 hp
Hd = HnomKsnd = 50(1.2)(1.1) = 66 hp
H
66
Eq. (17-20): N b = d =
= 8.1 belts
8.13
Ha
Eq. (17-19):
Decision: Use 9 belts. On a per belt basis,
63 025H a
63 025(8.13)
=
= 106.5 lbf/belt
875(11 / 2)
n(d / 2)
∆Fa d 106.5(11)
=
= 586.8 lbf · in per belt
Ta =
2
2
Table 17-16: Kc = 1.716
∆Fa =
2
Eq. (17-21):
2
 V 
 2520 
Fc = 1.716 
 = 1.716 
 = 10.9 lbf/belt
 1000 
 1000 
At fully developed friction, Eq. (17-9) gives
Fi =
T
d
 exp( f θ d ) + 1  586.9  3.5846 + 1 
= 94.6 lbf/belt

 =
11  3.5846 − 1 
 exp( f θ d ) − 1 
Eq. (17-10):
 2 exp( f θ d ) 
 2(3.5846) 
F1 = Fc + Fi 
= 158.8 lbf/belt
 = 10.9 + 94.6 
 3.5846 + 1 
 exp( f θ d ) + 1 
F2 = F1 − ∆Fa = 158.8 − 106.7 = 52.1 lbf/belt
NH
9(8.13)
nf s = b a =
= 1.11 O.K . Ans.
Hd
66
Durability:
( Fb )1 = K b / d = 1600 / 11 = 145.5 lbf/belt
( Fb )2 = Kb / D = 1600 / 60 = 26.7 lbf/belt
T1 = F1 + ( Fb )1 = 158.8 + 145.5 = 304.3 lbf/belt
T2 = F1 + ( Fb ) 2 = 158.8 + 26.7 = 185.5 lbf/belt
Eq. (17-27) with Table 17-17:
−1
 K  − b  K  − b 
 2038 −11.173  2038 −11.173 
N P =   +    = 
+



 185.5 
 T1 
 304.3 

 T2  
9
9
Ans.
= 1.68 (10 ) passes > 10 passes
Shigley’s MED, 10th edition
−1
Chapter 17 Solutions, Page 23/39
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Since NP is greater than
han 109 passes and is out of the range of Tablee 17-17,
1
life from Eq.
(17-27) is
N L
109 (272.9)
t = P p >
= 150 (103 ) h
720
20V
720(2520)
(Fi)drive = 9(94.6) = 851.4 lbf
Remember:
Table 17-9: C-sectionn belts
b
are 7/8 in wide. Check sheave groove spacing
sp
to see if 14 in
width is accommodatin
ting.
(b) The fully develope
ped friction torque on the flywheel using the flats
fla of the V-belts,
from Eq. (17-9), is
 exp( f θ ) − 1 
 1.637 − 1 
Tflat = Fi D 
= 94.6(60) 
 = 1371 lbf · in per belt

 1.637 + 1 
 exp( f θ ) + 1 
The flywheel torque should
sh
be
Tfly = mGTa = 5.147(586.9) = 3021 lbf · in per belt
but it is not. There are
re applications, however, in which it will work.
k. For example,
make the flywheel con
ontrolling. Yes. Ans.
_________________________
______________________________________
____________________
17-21
(a)
S is the spliced-in string segmentt length
l
De is the equatorial diameter
D′ is the spliced string diameter
δ is the radial clearance
S + πDe = π D′ = π(De + 2δ) = πDe + 2πδ
From which
δ =
S
2π
The radial clearance is thus independent of De.
δ =
12(
2(6)
= 11.5 in
2π
Ans.
This is true whether the sphere is the earth, the moon or a marble.. T
Thinking in terms of a
radial or diametral incr
crement removes the basic size from the proble
blem.
Shigley’s MED, 10th edition
Chapterr 117 Solutions, Page 24/39
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(b) and (c)
210 belt, the thickness is 1 in.
Table 17-9: For an E21
d P − di =
2δ =
210 + 4.5
π
−
210
π
=
4.5
π
4.5
π
4.5
= 0.716 in
δ =
2π
The pitch diameter off tthe flywheel is
DP − 2δ = D ⇒ DP = D + 2δ = 60 + 2(0.716) = 61.
1.43 in
We could make a table
le:
Dia
iametral
Gro
rowth
2δ
A
1.3
π
B
1.8
π
Section
C
D
2.9 3.3
π
π
E
4.5
π
t D-section belt of Prob. 17-20 is
The velocity ratio for the
m′G′ =
D + 2δ
60 + 3.3 / π
=
= 5.55
d
11
Ans.
for the V-flat drive ass compared
c
to ma = 60/11 = 5.455 for the VV drive.
d
The pitch diameter off tthe pulley is still d = 11 in, so the new anglee oof wrap, θd, is
D + 2δ − d
Ans.
2C
D + 2δ − d
θ D = π + 2sin −1
Ans.
2C
Equations (17-16a) and (17-16b) are modified as follows
θ d = π − 2sin −1
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(D + δ − d )2
2
4C

π

C p = 0.25   L p − ( D + 2δ + d ) 
2


L p = 2C +
π
( D + 2δ + d ) +
Ans.
2

π


+  L p − ( D + 2δ + d )  − 2( D + 2δ − d ) 2  Ans.
2



The changes are small, but if you are writing a computer code for a V-flat drive,
remember that θd and θD changes are exponential.
______________________________________________________________________________
17-22 This design task involves specifying a drive to couple an electric motor running at 1720
rev/min to a blower running at 240 rev/min, transmitting two horsepower with a center
distance of at least 22 inches. Instead of focusing on the steps, we will display two
different designs side-by-side for study. Parameters are in a “per belt” basis with per
drive quantities shown along side, where helpful.
Parameter
Four A-90 Belts Two A-120 Belts
mG
7.33
7.142
Ks
1.1
1.1
nd
1.1
1.1
0.877
0.869
K1
1.05
1.15
K2
d, in
3.0
4.2
D, in
22
30
2.333
2.287
θd, rad
V, ft/min
1350.9
1891
3.304
3.2266
exp(fθd )
Lp, in
91.3
101.3
C, in
24.1
31
0.783
1.662
Htab, uncorr.
NbHtab, uncorr.
3.13
3.326
26.45(105.8)
60.87(121.7)
Ta, lbf · in
17.6(70.4)
29.0(58)
∆Fa, lbf
0.721(2.88)
1.667(3.33)
Ha, hp
nf s
1.192
1.372
26.28(105.2)
44(88)
F1, lbf
8.67(34.7)
15(30)
F2, lbf
(Fb)1, lbf
73.3(293.2)
52.4(109.8)
10(40)
7.33(14.7)
(Fb)2, lbf
Fc, lbf
1.024
2.0
Fi, lbf
16.45(65.8)
27.5(55)
99.2
96.4
T1, lbf · in
T2, lbf · in
36.3
57.4
2.3(109)
1.61(109)
N ′, passes
t>h
93 869
89 080
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Chapter 17 Solutions, Page 26/39
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Conclusions:
• Smaller sheaves lead to more belts.
• Larger sheaves lead to larger D and larger V.
• Larger sheaves lead to larger tabulated power.
• The discrete numbers of belts obscures some of the variation. The factors of safety
exceed the design factor by differing amounts.
______________________________________________________________________________
17-23 In Ex. 17-5 the selected chain was 140-3, making the pitch of this 140 chain14/8 = 1.75
in. Table 17-19 confirms.
______________________________________________________________________________
17-24 (a) Eq. (17-32): H1 = 0.004 N11.08n10.9 p (3 − 0.07 p )
1000K r N11.5 p 0.8
n11.5
Equating and solving for n1 gives
Eq. (17-33):
H2 =
 0.25(106 ) K r N10.42 
n1 = 

p (2.2 − 0.07 p )


1/ 2.4
Ans.
(b) For a No. 60 chain, p = 6/8 = 0.75 in, N1 = 17, Kr = 17
1/ 2.4
 0.25(106 )(17)(17)0.42 
n1 = 

[2.2 − 0.07(0.75)]
 0.75

= 1227 rev/min
Ans.
Table 17-20 confirms that this point occurs at 1200 ± 200 rev/min.
(c) Life predictions using Eq. (17-40) are possible at speeds greater than 1227 rev/min.
Ans.
______________________________________________________________________________
17-25 Given: a double strand No. 60 roller chain with p = 0.75 in, N1 = 13 teeth at 300 rev/min,
N2 = 52 teeth.
(a) Table 17-20:
Table 17-22:
Table 17-23:
Use
Eq. (17-37):
Htab = 6.20 hp
K1 = 0.75
K2 = 1.7
Ks = 1
Ha = K1K2Htab = 0.75(1.7)(6.20) = 7.91 hp
Ans.
(b) Eqs. (17-35) and (17-36) with L/p = 82
Shigley’s MED, 10th edition
Chapter 17 Solutions, Page 27/39
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13 + 52
− 82 = − 49.5
2
2
p
 52 − 13  
C =  49.5 + 49.52 − 8 
 = 23.95 p
4
 2π  


C = 23.95(0.75) = 17.96 in, round up to 18 in Ans.
A=
(c) For 30 percent less power transmission,
H = 0.7(7.91) = 5.54 hp
63 025(5.54)
T =
= 1164 lbf · in
300
Ans.
Eq. (17-29):
0.75
= 3.13 in
sin(180o /13)
T
1164
=
= 744 lbf Ans.
F =
r
3.13 / 2
______________________________________________________________________________
D =
17-26 Given: No. 40-4 chain, N1 = 21 teeth for n = 2000 rev/min, N2 = 84 teeth, h = 20 000
hours.
(a) Chain pitch is p = 4/8 = 0.500 in and C 20 in.
Eq. (17-34):
2C N1 + N 2 ( N1 − N 2 )
+
+
p
2
4π 2C / p
2(20) 21 + 84
(84 − 21) 2
=
+
+
= 135 pitches (or links)
0.5
2
4π 2 (20 / 0.5)
L = 135(0.500) = 67.5 in Ans.
2
L
p
(b) Table 17-20:
Htab = 7.72 hp (post-extreme power)
Eq. (17-40): Since K1 is required, the N13.75 term is omitted (see p. 906).
( 7.72 ) (15 000)
2.5
constant =
= 18 399
135
1/ 2.5
18 399(135) 

′ = 
= 6.88 hp Ans.
H tab
 20 000 
(c) Table 17-22:
 21
K1 =  
 17 
1.5
= 1.37
Shigley’s MED, 10th edition
Chapter 17 Solutions, Page 28/39
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Table 17-23:
K2 = 3.3
H a = K1K 2 H tab
′ = 1.37(3.3)(6.88) = 31.1 hp
Ans.
N1 pn
21(0.5)(2000)
=
= 1750 ft/min
12
12
33 000(31.1)
F1 =
= 586 lbf Ans.
1750
______________________________________________________________________________
(d)
V =
17-27 This is our first design/selection task for chain drives. A possible decision set:
A priori decisions
• Function: Hnom, n1, space, life, Ks
• Design factor: nd
• Sprockets: Tooth counts N1 and N2, factors K1 and K2
Decision variables
• Chain number
• Strand count
• Lubrication type
• Chain length in pitches
Function: Motor with Hnom = 25 hp at n = 700 rev/min; pump at n = 140 rev/min;
mG = 700/140 = 5
Design Factor: nd = 1.1
Sprockets: Tooth count N2 = mGN1 = 5(17) = 85 teeth–odd and unavailable. Choose
84 teeth. Decision: N1 = 17, N2 = 84
Evaluate K1 and K2
Eq. (17-38):
Eq. (17-37):
Hd = HnomKsnd
Ha = K1K2Htab
Equate Hd to Ha and solve for Htab :
KnH
H tab = s d nom
K1K 2
Table 17-22:
K1 = 1
Table 17-23:
K2 = 1, 1.7, 2.5, 3.3 for 1 through 4 strands
H tab
′ =
1.5(1.1)(25)
41.25
=
(1) K 2
K2
Prepare a table to help with the design decisions:
Shigley’s MED, 10th edition
Chapter 17 Solutions, Page 29/39
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Strands
1
2
3
4
K2
1.0
1.7
2.5
3.3
Chain
No.
100
80
80
60
H tab
′
41.3
24.3
16.5
12.5
Htab
59.4
31.0
31.0
13.3
Lub.
nf s Type
1.58
B
1.40
B
2.07
B
1.17
B
Design Decisions:
We need a figure of merit to help with the choice. If the best was 4 strands of No. 60
chain, then
Decision #1 and #2: Choose four strand No. 60 roller chain with nf s = 1.17.
n fs =
K1K 2 H tab 1(3.3)(13.3)
=
= 1.17
K s H nom
1.5(25)
Decision #3: Choose Type B lubrication
Analysis:
Table 17-20:
Table 17-19:
Htab = 13.3 hp
p = 0.75 in
Try C = 30 in in Eq. (17-34):
L
p
2C N1 + N 2 ( N 2 − N1)2
+
+
p
2
4π 2C / p
17 + 84
(84 − 17) 2
= 2(30 / 0.75) +
+
2
4π 2 (30 / 0.75)
= 133.3
L = 0.75(133.3) = 100 in (no need to round)
Eq. (17-36) with p = 0.75 in: A =
N1 + N 2 L 17 + 84 100
−
=
−
= −82.83
2
2
0.75
p
Eq. (17-35):
p
 N − N1  
C =  − A + A2 − 8  2

4
 2π  

2
0.75 
84 − 17  
2
 − ( −82.83) + ( −82.83) − 8 
=
  = 30.0 in
4 
2π  



2
Shigley’s MED, 10th edition
Chapter 17 Solutions, Page 30/39
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Decision #4: Choose C = 30.0 in.
______________________________________________________________________________
17-28 Follow the decision set outlined in Prob. 17-27 solution. We will form two tables, the
first for a 15 000 h life goal, and a second for a 50 000 h life goal. The comparison is
useful.
Function: Hnom = 50 hp at n = 1800 rev/min, npump = 900 rev/min, mG = 1800/900 = 2,
Ks = 1.2, life = 15 000 h, then repeat with life = 50 000 h
Design factor: nd = 1.1
Sprockets: N1 = 19 teeth, N2 = 38 teeth
Table 17-22 (post extreme):
1.5
1.5
N 
 19 
K1 =  1  =   = 1.18
 17 
 17 
K2 = 1, 1.7, 2.5, 3.3, 3.9, 4.6, 6.0
Table 17-23:
Decision variables for 15 000 h life goal:
K s nd H nom 1.2(1.1)(50) 55.9
=
=
1.18K 2
K1K 2
K2
K1K 2 H tab 1.18K 2 H tab
=
=
= 0.0197 K 2 H tab
1.2(50)
K s H nom
′ =
H tab
nf
s
(1)
Form a table for a 15 000 h life goal using these equations.
K2
H'tab
1
1.7
2.5
3.3
3.9
4.6
6
55.90
32.90
22.40
16.90
14.30
12.20
9.32
Chain #
120
120
120
120
80
60
60
Htab
nf s
Lub
21.6
21.6
21.6
21.6
15.6
12.4
12.4
0.423
0.923
1.064
1.404
1.106
1.126
1.416
C'
C'
C'
C'
C'
C'
C'
There are 4 possibilities where nf s ≥ 1.1
Decision variables for 50 000 h life goal
From Eq. (17-40), the power-life tradeoff is:
′ )2.515 000 = ( H tab
′′ ) 2.5 50 000
( H tab
1/ 2.5
 15 000

′′ = 
′ )2.5 
H tab
( H tab
 50 000

= 0.618 H ′tab
Shigley’s MED, 10th edition
Chapter 17 Solutions, Page 31/39
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Substituting from (1),
 55.9  34.5
H tab
=
′′ = 0.618 
K2
 K 2 
The H ′′ notation is only necessary because we constructed the first table, which we
normally would not do.
′ )
K K H ′′
K K (0.618H tab
n f s = 1 2 tab = 1 2
= 0.618[(0.0197) K 2 H tab ]
K s H nom
K s H nom
= 0.0122K 2 H tab
Form a table for a 50 000 h life goal.
K2
1
1.7
2.5
3.3
3.9
4.6
6
H''tab
34.50
20.30
13.80
10.50
8.85
7.60
5.80
Chain #
120
120
120
120
120
120
80
Htab
21.6
21.6
21.6
21.6
21.6
21.6
15.6
nf s
0.264
0.448
0.656
0.870
1.028
1.210
1.140
Lub
C'
C'
C'
C'
C'
C'
C'
There are two possibilities in the second table with nf s ≥ 1.1. (The tables allow for the
identification of a longer life of the outcomes.) We need a figure of merit to help with
the choice; costs of sprockets and chains are thus needed, but is more information than
we have.
Decision #1: #80 Chain (smaller installation) Ans.
nf s = 0.0122K2Htab = 0.0122(8.0)(15.6) = 1.14 O.K.
Decision #2: 8-Strand, No. 80 Ans.
Decision #3: Type C′ Lubrication Ans.
Decision #4: p = 1.0 in, C is in midrange of 40 pitches
L
2C N1 + N 2 ( N 2 − N1)2
≈
+
+
p
p
2
4π 2C / p
19 + 38 (38 − 19) 2
= 2(40) +
+
2
4π 2 (40)
= 108.7 ⇒ 110 even integer Ans.
Eq. (17-36):
A=
N1 + N 2 L 19 + 38 110
−
=
−
= −81.5
2
2
1
p
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Eq. (17-35):
C
1
=  −(−81.5) +
p
4

 38 − 19 
(−81.5) 2 − 8 

 2π 
2

 = 40.64


C = p(C/p) = 1.0(40.64/1.0) = 40.64 in (for reference) Ans.
______________________________________________________________________________
17-29 The objective of the problem is to explore factors of safety in wire rope. We will express
strengths as tensions.
(a) Monitor steel 2-in 6 × 19 rope, 480 ft long.
Table 17-2: Minimum diameter of a sheave is 30d = 30(2) = 60 in, preferably
45(2) = 90 in. The hoist abuses the wire when it is bent around a sheave. Table 17-24
gives the nominal tensile strength as 106 kpsi. The ultimate load is
 π (2)2 
Fu = (Su ) nom Anom = 106 
 = 333 kip
 4 
Ans.
The tensile loading of the wire is given by Eq. (17-47)
a
W

Ft = 
+ w l  1 + 
g
m

W = 4(2) = 8 kip, m = 1
Table (17-24):
wl = 1.60d 2 l = 1.60(22)(480) = 3072 lbf = 3.072 kip
Therefore,
2 

Ft = (8 + 3.072) 1 +
 = 11.76 kip

32.2 
Eq. (17-41):
Ed A
Fb = r w m
D
and for the 72-in drum
Fb =
Ans.
12(106 )(2 / 13)(0.38)(22 )(10−3 )
= 39 kip
72
Ans.
For use in Eq. (17-44), from Fig. 17-21
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Chapter 17 Solutions, Page 33/39
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( p / Su ) = 0.0014
Su = 240 kpsi, p. 920
0.0014(240)(2)(72)
= 24.2 kip
Ff =
2
Ans.
(b) Factors of safety
Static, no bending:
n=
Static, with bending:
Eq. (17-46):
ns =
Fu
333
=
= 28.3
Ft
11.76
Ans.
Fu − Fb
333 − 39
=
= 25.0
Ft
11.76
Fatigue without bending:
F
24.2
nf = f =
= 2.06
Ft
11.76
Ans.
Ans.
Fatigue, with bending: For a life of 0.1(106) cycles, from Fig. 17-21
( p / Su ) = 4 / 1000 = 0.004
0.004(240)(2)(72)
Ff =
= 69.1 kip
2
Eq. (17-45):
nf =
69.1 − 39
= 2.56
11.76
Ans.
If we were to use the endurance strength at 106 cycles (Ff = 24.2 kip) the factor of
safety would be less than 1 indicating 106 cycle life impossible.
Comments:
• There are a number of factors of safety used in wire rope analysis. They are different,
with different meanings. There is no substitute for knowing exactly which factor
of safety is written or spoken.
• Static performance of a rope in tension is impressive.
• In this problem, at the drum, we have a finite life.
• The remedy for fatigue is the use of smaller diameter ropes, with multiple ropes
supporting the load. See Ex. 17-6 for the effectiveness of this approach. It will also
be used in Prob. 17-30.
• Remind students that wire ropes do not fail suddenly due to fatigue. The outer
wires gradually show wear and breaks; such ropes should be retired. Periodic
inspections prevent fatigue failures by parting of the rope.
______________________________________________________________________________
Shigley’s MED, 10th edition
Chapter 17 Solutions, Page 34/39
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17-30 Since this is a design task, a decision set is useful.
A priori decisions
• Function: load, height, acceleration, velocity, life goal
• Design Factor: nd
• Material: IPS, PS, MPS or other
• Rope: Lay, number of strands, number of wires per strand
Decision variables:
• Nominal wire size: d
• Number of load-supporting wires: m
From experience with Prob. 17-29, a 1-in diameter rope is not likely to have much of a
life, so approach the problem with the d and m decisions open.
Function: 5000 lbf load, 90 foot lift, acceleration = 4 ft/s2, velocity = 2 ft/s, life
goal = 105 cycles
Design Factor: nd = 2
Material: IPS
Rope: Regular lay, 1-in plow-steel 6 × 19 hoisting
Design variables
Choose 30-in Dmin. Table 17-27: w = 1.60d 2 lbf/ft
wl = 1.60d 2l = 1.60d 2(90) = 144d 2 lbf, each
Eq. (17-47):
a   5000
4 
W


Ft = 
+ w l  1 +  = 
+ 144d 2 1 +

g  m
32.2 
m


5620
=
+ 162d 2 lbf, each wire
m
Eq. (17-44):
Ff =
( p / Su ) Su Dd
2
From Fig. 17-21 for 105 cycles, p/Su = 0.004. From p. 912, Su = 240 kpsi, based on
metal area.
0.004(240 000)(30d )
Ff =
= 14 400d lbf each wire
2
Eq. (17-41) and Table 17-27:
Fb =
6
2
Er d w Am 12 (10 ) 0.067d ( 0.4d )
=
= 10 720d 3 lbf, each wire
D
30
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Chapter 17 Solutions, Page 35/39
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Eq. (17-45):
Ff − Fb
14 400d − 10 720d 3
Ft
(5620 / m) + 162d 2
We could use a computer program to build a table similar to that of Ex. 17-6.
Alternatively, we could recognize that 162 d 2 is small compared to 5620 / m, and
therefore eliminate the 162d 2 term.
nf =
nf
=
14 400d − 10 720d 3
m
=
(14 400d − 10 720d 3 )
5620 / m
5620
Maximize nf ,
∂n f
∂d
=0=
m
[14 400 − 3(10 720)d 2 ]
5620
From which
d* =
14 400
= 0.669 in
3(10 720)
Back-substituting
nf =
m
[14 400(0.669) − 10 720(0.6693 )] = 1.14 m
5620
Thus nf = 1.14, 2.28, 3.42, 4.56 for m=1, 2, 3, 4 respectively. If we choose d = 0.50 in,
then m = 2.
14 400(0.5) − 10 720(0.53 )
nf =
= 2.06
(5620 / 2) + 162(0.5) 2
This exceeds nd = 2
Decision #1: d = 1/2 in
Decision #2: m = 2 ropes supporting load. Rope should be inspected weekly for any
signs of fatigue (broken outer wires).
Comment: Table 17-25 gives n for freight elevators in terms of velocity.
 πd2 
= 83 252d 2 lbf, each wire
Fu = (Su )nom Anom = 106 000 

 4 
2
F
83 452(0.5)
= 7.32
n= u =
Ft
(5620 / 2) + 162(0.5) 2
By comparison, interpolation for 120 ft/min gives 7.08 - close. The category of
construction hoists is not addressed in Table 17-25. We should investigate this before
proceeding further.
______________________________________________________________________________
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Chapter 17 Solutions, Page 36/39
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17-31 Given: 2000 ft lift, 72 in drum, 6 × 19 MS rope, cage and load 8000 lbf, accel. = 2 ft/s2.
(a) Table 17-24: (Su)nom = 106 kpsi; Su = 240 kpsi (p. 912); Fig. 17-21: (p/Su)106 = 0.0014
Eq. (17-44):
Ff =
( p / Su ) Su dD
2
=
0.0014(240)d (72)
= 12.1 d kip
2
wl = 1.6d 2 2000(10−3) = 3.2d 2 kip
Table 17-24:

a
Ft = (W + wl ) 1 + 
g

2 

= (8 + 3.2d 2 ) 1 +


32.2 
= 8.5 + 3.4d 2 kip
Note that bending is not included.
F
12.1d
nf = f =
Ft
8.5 + 3.4d 2
Eq. (17-47):
d, in
0.500
1.000
1.500
1.625
1.750
2.000
nf
0.650
1.020
1.124
1.125 ← maximum nf
1.120
1.095
Ans.
(b) Try m = 4 strands
2 
8

Ft =  + 3.2d 2 1 +

32.2 
4

= 2.12 + 3.4d 2 kip
Ff = 12.1d kip
12.1d
nf =
2.12 + 3.4d 2
d, in
0.5000
0.5625
0.6520
0.7500
0.8750
1.0000
nf
2.037
2.130
2.193
2.250 ← maximum nf
2.242
2.192
Ans.
Shigley’s MED, 10th edition
Chapter 17 Solutions, Page 37/39
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Comparing tables, multiple ropes supporting the load increases the factor of safety,
and reduces the corresponding wire rope diameter, a useful perspective.
______________________________________________________________________________
17-32
ad
b / m + cd 2
dn f
(b / m + cd 2 )a − ad (2cd )
=
=0
(b / m + cd 2 )2
dd
nf =
From which
b
mc
d* =
nf * =
Ans.
a b / (mc)
a m
=
(b / m) + c [b / (mc)] 2 bc
Ans.
These results agree closely with the Prob. 17-31 solution. The small differences are due
to rounding in Prob. 17-31.
______________________________________________________________________________
17-33 From Prob. 17-32 solution:
n1 =
ad
b / m + cd 2
Solve the above equation for m
b
(1)
ad / n1 − cd 2
( ad / n1 ) − ad 2  (0) − b ( a / n1 ) − 2cd 
dm
=0= 
2
dd
( ad / n1 ) − cd 2 
m=
From which
d* =
a
2cn1
Ans.
Substituting this result for d into Eq. (1) gives
4bcn1
Ans.
a2
______________________________________________________________________________
m* =
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Chapter 17 Solutions, Page 38/39
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17-34 Table 17-27:
Am = 0.40d 2 = 0.40(22 ) = 1.6 in 2
Er = 12 Mpsi, w = 1.6d 2 = 1.6(22 ) = 6.4 lbf/ft
w l = 6.4(480) = 3072 lbf
γ w l / ( Aml ) = 3072 / [1.6(480)12] = 0.333 lbf/in 3
Treat the rest of the system as rigid, so that all of the stretch is due to the load of 7000 lbf,
the cage weighing 1000 lbf, and the wire’s weight. From the solution of Prob. 4-7,
Wl
γ l2
+
AE 2E
(1000 + 7000)(480)(12) 0.333(4802 )122
=
+
1.6(12)(106 )
2(12)(106 )
δ1 =
= 2.4 + 0.460 = 2.860 in
Ans.
______________________________________________________________________________
17-35 to 17-38 Computer programs will vary.
Shigley’s MED, 10th edition
Chapter 17 Solutions, Page 39/39
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Chapter 20
Refer to the first paragraph of Sec. 20–1, p. 970.
iv
Ans.
______________________________________________________________________________
20–1
Refer to the first paragraph on p. 974, and the first paragraph in Sec. 20–4 on p. 981.
The size dimension of a feature of size should be directly toleranced.
Ans.
______________________________________________________________________________
20–2
Refer to the last paragraph on p. 972.
ii
Ans.
______________________________________________________________________________
20–3
Refer to the first paragraph on p. 973.
Feature of size
Ans.
______________________________________________________________________________
20–4
Refer to the first paragraph on p. 973.
size, location, orientation, and form
Ans.
______________________________________________________________________________
20–5
Refer to the Form Controls sub-section on p. 985.
straightness, flatness, circularity, and cylindricity
Ans.
______________________________________________________________________________
20–6
Refer to the Orientation Controls sub-section on p. 987.
angularity, parallelism, and perpendicularity
Ans.
______________________________________________________________________________
20–7
Refer to the Location Controls sub-section on p. 990.
position, concentricity, symmetry, position
Ans.
The position control is the most prominently used. Ans.
______________________________________________________________________________
20–8
Refer to the Basic Dimensions sub-section on p. 983.
ii
Ans.
______________________________________________________________________________
20–9
20–10 All of the features with a +/– tolerance are features of size. Ans.
These include boss, hole, plate thickness, plate width, and plate height. The combined
thickness of the plate and the boss (70 +/– 0.1) technically satisfies the definition, though
it is not a typical feature of size.
______________________________________________________________________________
Shigley’s MED, 10th edition
Chapter 20 Solutions, Page 1/8
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20–11 The datum features are the physical surfaces of the plate. The datums are the
theoretically perfect planes corresponding to the datum feature simulators of the physical
surfaces. The datum reference frame is the xyz coordinate axis corresponding to the three
datum planes.
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20–12 For the first two questions, refer to the first two paragraphs in the Position sub-section on
p. 990. Also refer to step 4 of Ex. 20–1, p. 998.
iv
Ans.
iii
Ans.
For the third question, refer to the first full paragraph on p. 977.
v
Ans.
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20–13 (a) The boss is a feature of size, so it is directly toleranced.
The maximum and minimum diameters allowed are 50.3 and 49.7, respectively.
Ans.
(b) The position tolerance affects the location and orientation of the center axis of the
boss, but does not affect the diameter of the boss.
No effect.
Ans.
(c) The position tolerance always defines the allowed location and orientation of an axis,
center line, or center plane of a feature of size. Refer to the Position sub-section on p.
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990. For a strict definition of the axis, refer to the Actual Mating Envelopes sub-section
on p. 979.
The center axis of the boss, as determined by the related actual mating envelope, must be
within the tolerance zone.
Ans.
(d) Note that the position tolerance is specified at maximum material condition. If the
boss is produced at a diameter of 50.3, which is the maximum material condition, then
the specified tolerance applies.
The tolerance zone diameter is 0.2. Ans.
(e) If the boss is produced at a diameter of 49.7, which is 0.6 less than the maximum
material condition, the tolerance zone diameter is the sum of this 0.6 and the specified
tolerance of 0.2.
The tolerance zone diameter is 0.8. Ans.
(f) The part is stabilized with respect to datums A, B, and C, in that order, then the central
axis of the tolerance zone is oriented perpendicular to datum A, and located from datums
B and C by the basic dimensions.
Ans.
(g) The position control defines a tolerance zone that defines the allowed location and
orientation of the axis of the boss. Refer to the first paragraph of the Position sub-section
on p. 990. Depending on the amount of deviation from the maximum material condition,
the tolerance varies from 0.2 at MMC to 0.8 at LMC.
ii
Ans.
(h) Cylindricity is a characteristic of form. Since a specific cylindricity form control is
not specified, Rule #1 applies as the default form control for a feature of size. Refer to
the second full paragraph on p. 982.
iii
Ans.
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20–14 (a) The hole is a feature of size, so it is directly toleranced.
The maximum and minimum diameters allowed are 30.1 and 29.9, respectively.
Ans.
(b) The position tolerance affects the location and orientation of the center axis of the
hole, but does not affect the diameter of the boss.
No effect.
Ans.
(c) The position tolerance always defines the allowed location and orientation of an axis,
center line, or center plane of a feature of size. Refer to the Position sub-section on p.
990. For a strict definition of the axis, refer to the Actual Mating Envelopes sub-section
on p. 979.
The center axis of the hole, as determined by the related actual mating envelope, must be
within the tolerance zone.
Ans.
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(d) Note that the position tolerance is specified at maximum material condition, which for
the hole is 29.9. If the hole is produced at a diameter of 30.1, which is 0.2 greater than
the maximum material condition, the tolerance zone is the sum of this 0.2 and the
specified tolerance of 0.3.
The tolerance zone diameter is 0.5. Ans.
(e) If the hole is produced at a diameter of 29.9, which is the maximum material
condition, then the specified tolerance applies.
The tolerance zone diameter is 0.3. Ans.
(f) The part is stabilized with respect to datums A, B, and C, in that order, then the central
axis of the tolerance zone is oriented perpendicular to datum A, and located from datums
B and C by the basic dimensions.
Ans.
(g) The position control defines a tolerance zone that defines the allowed location and
orientation of the axis of the boss. Refer to the first paragraph of the Position sub-section
on p. 990. Depending on the amount of deviation from the maximum material condition,
the tolerance varies from 0.3 at MMC to 0.5 at LMC.
ii
Ans.
(h) Cylindricity is a characteristic of form. Since a specific cylindricity form control is
not specified, Rule #1 applies as the default form control for a feature of size. Refer to
the second full paragraph on p. 982.
iii
Ans.
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20–15 Refer to the Actual Mating Envelopes sub-section on p. 979.
An actual mating envelope is determined by fitting the largest gauge pin (or an expanding
mandrel) into the hole, just touching the high points of the hole. The center axis of the
gauge pin is defined to be the center axis of the hole.
Ans.
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20–16 Refer to the second full paragraph on p. 982.
iii
Ans.
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20–17 Refer to the first full paragraph on p. 982.
Shaft diameter at MMC = 20.2
Ans.
Shaft diameter at LMC = 19.8
Ans.
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20–18 Refer to the first full paragraph on p. 982.
Hole diameter at MMC = 19.8
Ans.
Hole diameter at LMC = 20.2
Ans.
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20–19 Refer to the second full paragraph on p. 982. The surface of the hole is limited by an
envelope of perfect form at MMC. There is no limiting envelope defined at LMC.
The diameter of the limiting envelope is 19.8.
Ans.
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20–20 Refer to the second full paragraph on p. 982. The surface of the shaft is limited by an
envelope of perfect form at MMC. There is no limiting envelope defined at LMC.
The diameter of the limiting envelope is 20.2.
Ans.
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20–21 Refer to Sec. 20–6, p. 994.
(a) Since no material condition is specified, the tolerance zone diameter is the specified
0.1 at all produced diameters of the boss.
0.1, 0.1, 0.1 Ans.
(b) Since the maximum material condition is specified, the tolerance zone diameter is the
specified 0.1 when the boss is produced at the MMC of 25.2, and increases by the amount
of deviation from the MMC.
0.5, 0.3, 0.1 Ans.
(c) Since the least material condition is specified, the tolerance zone diameter is the
specified 0.1 when the boss is produced at the LMC of 24.8, and increases by the amount
of deviation from the LMC.
0.1, 0.3, 0.5 Ans.
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20–22 Refer to Sec. 20–6, p. 994.
(a) Since no material condition is specified, the tolerance zone diameter is the specified
0.3 at all produced diameters of the hole.
0.3, 0.3, 0.3 Ans.
(b) Since the maximum material condition is specified, the tolerance zone diameter is the
specified 0.3 when the hole is produced at the MMC of 32.0, and increases by the amount
of deviation from the MMC.
0.3, 0.5, 0.7 Ans.
(c) Since the least material condition is specified, the tolerance zone diameter is the
specified 0.3 when the hole is produced at the LMC of 32.4, and increases by the amount
of deviation from the LMC.
0.7, 0.5, 0.3 Ans.
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20–23 Refer to the Cylindricity sub-section on p. 987.
cylindricity Ans.
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20–24 Refer to the third paragraph on p. 989.
profile of a surface Ans.
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20–25 Refer to the Form Controls sub-section on p. 985.
The form controls (straightness, flatness, circularity, and cylindricity) never reference
datums. They only relate to the form of the individual feature, and are independent of the
feature’s location.
Ans.
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20–26 (a) Refer to the first paragraph of Sec. 20–6 on p. 994 for the two primary modifiers, and
the last paragraph on p. 995 for the default material condition.
MMC, LMC, and RFS
Ans.
(b) Refer to the last paragraph on p. 995.
RFS
Ans.
(c) Refer to the last paragraph on p. 994 and the third paragraph on p. 995.
MMC and LMC
Ans.
(d) Refer to the last sentence on p. 994.
iv
Ans.
(e) Refer to the first paragraph of Sec. 20–6.
iii
Ans.
(f) Refer to the first paragraph on p. 995.
MMC
Ans.
(g) Refer to the first partial paragraph on p. 996.
RFS
Ans.
(h) Refer to the third paragraph on p. 995.
LMC
Ans.
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20–27 (a) The countersink diameter is specified as 12 ± 0.2.
The minimum countersink diameter is 11.8.
Ans.
(b) The countersink depth is specified as 3 ± 0.1.
The maximum countersink depth is 3.1.
Ans.
(c) The bolt holes are specified as 6 +0.1/–0.0. The maximum material condition for a
hole is the smallest hole diameter.
The diameter of the bolt holes at MMC is 6.0.
Ans.
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(d) Refer to the second paragraph on p. 973.
The features of size are all hole diameters and counterbore diameters, the slot width, the
base widths, the base thickness, and the center protrusion width. Ans.
(e) The default surface profile tolerance in the note provides a tolerance zone around all
surfaces of 0.5, centered on the surface. This gives a tolerance of 0.25 outside the base
dimension on each edge, for a total of 0.5. Therefore, the minimum and maximum
allowed base dimensions are 59.5 and 60.5. Ans.
(f) Datum feature A is the bottom surface. Its datum is a theoretical plane corresponding
to a datum feature simulator (such as a machine table). Datum feature B is the surface of
the hole. Its datum is the center axis corresponding to a datum feature simulator, such as
the largest gauge pin just touching the high points of the hole, while being held
perpendicular to datum A. Datum feature C is the surface of the slot. Its datum is the
center plane corresponding to a datum feature simulator, such as the largest gauge block
just touching the high points of the slot, while being held perpendicular to datum A.
Ans.
The datum reference frame is made of three mutually perpendicular planes, with the
origin of the coordinate axes at the bottom center of the datum B hole.
Ans.
The part is stabilized for manufacture or inspection by applying the datums in the
following sequence: Constrain the base to be in contact with a datum feature simulator
(an essentially flat surface) corresponding to datum A; constrain the translation of the
part on datum A with a gauge pin in the datum feature B hole; constrain the final rotation
of the part (around datum B) with a gauge pin or block in the groove.
Ans.
The purpose of the fixture is to locate and orient the large bore. This bore is located with
respect to the bottom surface (datum A), the pin hole (datum B), and the groove (datum
C). The edges of the base will not be touching anything for alignment, and can
consequently have much looser tolerances. It is usually easier and cheaper to precisely
locate a part with respect to locating pins than with respect to the edges of the part.
Ans.
(g) Since no material condition modifier is specified for the tolerance zone associated
with the hole, its tolerance zone is RFS (regardless of feature size). The specified
tolerance zone diameter is the specified 0.05 whether the hole diameter is 6.00, or 6.05,
or anything between.
0.05, 0.05
Ans.
(h) The top line of the position control of the bolt holes specifies the PLTZF tolerance of
0.3 at MMC (maximum material condition). For the bolt holes, when manufactured at
the MMC of 6.0, the diameter of the PLTZF tolerance zone is 0.3. When manufactured
at the LMC of 6.1, the diameter of the PLTZF tolerance zone is 0.4, i.e. the sum of the
specified tolerance plus the “bonus” tolerance of 0.1 due to the deviation of the hole from
its MMC.
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0.3, 0.4
Ans.
(i) The second line of the position control of the bolt holes specifies the FRTZF tolerance
of 0.1 at MMC (maximum material condition). For the bolt holes, when manufactured at
the MMC of 6.0, the diameter of the FRTZF tolerance zone is 0.1. When manufactured
at the LMC of 6.1, the diameter of the FRTZF tolerance zone is 0.2, i.e. the sum of the
specified tolerance plus the “bonus” tolerance of 0.1 due to the deviation of the hole from
its MMC.
0.1, 0.2
Ans.
(j) MMC ensures a minimum clearance for the bolts in their holes, but allows greater
deviations from ideal if the holes are produced larger.
Ans.
(k) Orientation and straightness of the axis are controlled by the position tolerance zone
which is specified in the feature control frame directly under the diameter dimension for
the bore. The position tolerance specifies a cylindrical tolerance zone that the axis of the
hole must be within, thus controlling its orientation and straightness.
Ans.
Cylindricity is a control of the surface of the bore, so is not controlled by the position
tolerance zone which controls the axis of the bore. Since there is not an explicit control
of the cylindricity specified, it could be controlled either by the default profile of a
surface control, specified with the note at the bottom of the drawing, or by Rule #1. In
this case, Rule #1 has the tighter tolerance, so it controls the cylindricity of the bore.
Rule #1 provides for the cylindricity of the surface of the bore to be within an envelope
of a perfect cylinder at MMC diameter of 14.9.
Ans.
(l) All exterior surfaces except for the bottom are controlled only by the default surface
profile tolerance, so as-cast is sufficient for them. Ans.
If the edges of the base were used as datum features, these surfaces would need to be
controlled more tightly in order to locate the critical holes from them. The edges of the
base could not be left as-cast. Ans.
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