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 2 = 50/0.005

P = 100 parts Ans.
______________________________________________________________________________
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
0.85  0.95 
2
1.43
Ans.
______________________________________________________________________________
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    X 1  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 2e1  X 1 X 2  1  2 
 X1 X 2 
Shigley’s MED, 11th edition
Ans.
Chapter 1 Solutions, Page 1/12
(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 
X1  
X2 
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,
1-11 (a)
e
x1 = 7 = 2.645 751 311 1
X1 = 2.64
(3 correct digits)
x2 = 8 = 2.828 427 124 7
X2 = 2.82
(3 correct digits)
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
______________________________________________________________________________
S

nd
1-12

32  1000  25  10

d3
2.5
Table A-17: d =
1 14
in
3


d 1.006 in
Ans.
Ans.
25  103 
S
n 
4.79
 32  1000 
Ans.
  1.25
Factor of safety:
______________________________________________________________________________
3
Shigley’s MED, 11th edition
Chapter 1 Solutions, Page 2/12
1-13
(a)
x
60
70
80
90
100
110
120
130
140
150
160
170
180
190
200
210
fx
f x2
120
7200
70
4900
240
19200
450
40500
800
80000
1320
145200
720
86400
1300
169000
1120
156800
750
112500
320
51200
510
86700
360
64800
190
36100
0
0
210
44100
8480 1 104 600
f
2
1
3
5
8
12
6
10
8
5
2
3
2
1
0
1
69

x 
Eq. (1-6)
1
N
k
fx
i i
i 1

8480
122.9 kcycles
69
Eq. (1-7)
k
sx 
fx
2
i i
 N x2
i 1
N  1
 1 104 600  69(122.9) 2 


69  1


z115 
(b)
Eq. (1-5)
1/ 2
 30.3 kcycles Ans.
x  x
x  x 115  122.9
 115

  0.2607
ˆ x
sx
30.3
Interpolating from Table (A-10)
0.2600
0.2607
0.2700
0.3974
x
0.3936

N(0.2607) = 69 (0.3971) = 27.4  27
Shigley’s MED, 11th edition
x = 0.3971
Ans.
Chapter 1 Solutions, Page 3/12
From the data, the number of instances less than 115 kcycles is
2 + 1 + 3 + 5 + 8 + 12 = 31 (the data is not perfectly normal)
____________________________________________________________________________
1-14
x
174
182
190
198
206
214
222

x 
Eq. (1-6)
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 i
i 1
k
fx
2
i i

39 126
198.61 kpsi
197
 N x2
12
 7 789 204  197(198.61) 2 
sx 

 9.68 kpsi Ans.
N1
197  1


Eq. (1-7)
______________________________________________________________________________
i 1
1-15
L 122.9 kcycles and sL  30.3 kcycles
z10 
Eq. (1-5)
Thus,
x   x x10  L x10  122.9


sL
30.3
ˆ
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, 11th edition
Chapter 1 Solutions, Page 4/12
1-16
x
f
93
95
97
99
101
103
105
107
109
111

19
25
38
17
12
10
5
4
4
2
136
x 
Eq. (1-6)
1
N
k
fx
i i
fx
2
i i
13 364 / 136  98.26471 = 98.26 kpsi
 N x2
i 1
N1
Eq. (1-7)
fx2
164331
225625
357542
166617
122412
106090
55125
45796
47524
24642
1315704
i 1
k
sx 
fx
1767
2375
3686
1683
1212
1030
525
428
436
222
13364
 1 315 704  136(98.26471) 2 


136  1


12
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  x
x  x
x  98.26
z0.01 
 0.01
 0.01
sx
4.30
ˆ 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
 Ri
Eq. (1-9):
R = i 1 = 0.98(0.96)0.94 = 0.88
Overall reliability = 88 percent
Ans.
Shigley’s MED, 11th edition
Chapter 1 Solutions, Page 5/12
______________________________________________________________________________
1-18 Obtain the coefficients of variance for strength and stress
ˆ S
23.5
CS  sy 
 0.07532
S sy
312
C 
ˆ
ˆ
145
 T 
 0.09667

T
1 500
For R = 0.99, from Table A-10, z =  2.326.
Eq. (1-12):
n 
1  1   1  z 2CS2   1  z 2C2 
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 
From the given equation for stress,
 max 
S sy
n

Ans.
16T
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
  S 
  (312)10

sy 

______________________________________________________________________________
1-19
Obtain the coefficients of variance for stress and strength
ˆ
ˆ
5
C    P 
 0.09231

P
65
CS 
(a)
6.59
ˆ S ˆ S y


 0.06901
S
Sy
95.5
n 1.2
z 
Eq. (1-11):
Shigley’s MED, 11th edition
nd  1
nd2CS2  C2

1.2  1
1.22  0.069012   0.092312
  1.6127
Chapter 1 Solutions, Page 6/12
n 
Sy


(b)
Interpolating Table A-10,
1.61
0.0537
1.6127

1.62
0.0526

R = 1  0.0534 = 0.9466
Ans.
Sy
P / ( d / 4)
2

 d 2S y
4P
 d 
 = 0.0534
4  65  1.2
4 Pn

 1.020 in Ans.
 Sy
  95.5 
n 1.5
1.5  1
z 
1.52  0.069012   0.092312
3.6
3.605
3.7
0.000159

0.000108
R = 1  0.00015645 = 0.9998
  3.605

 = 0.00015645
Ans.
4  65  1.5
4Pn

 1.140 in Ans.
 Sy
  95.5 
d 
______________________________________________________________________________
1-20
 max   max   a   b  90  383  473 MPa
From footnote 9 of text,
ˆ max   ˆ2a  ˆ2b 
C max 
CS y 
n 
1/ 2
 (8.42  22.32 )1/ 2  23.83 MPa
ˆ max
ˆ
23.83
  max 
 0.0504
473
 max
 max
ˆ S y
S y
Sy
 max
Shigley’s MED, 11th edition


ˆ S y
Sy

42.7
 0.0772
553
553
 1.169  1.17 Ans.
473
Chapter 1 Solutions, Page 7/12
z 
Eq. (1-11):
From Table A-10,
nd  1
nd2 CS2  C2

1.169  1
1.1692  0.0772 2   0.0504 2
  1.635
( 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
V 0.002 0.003 0.004



0.004267
V
1.500 1.875 3.000
V = 8.4375  0.0360 in3
Shigley’s MED, 11th edition

V 0.004267  8.4375  0.0360 in 3
Ans.
Chapter 1 Solutions, Page 8/12
8.4735
8.473551..
V 
8.4015 in, whereas, exact is
8.401551.. in
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
a = 1.569  0.016 in

ta = 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.
______________________________________________________________________________
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
Shigley’s MED, 11th edition
Chapter 1 Solutions, Page 9/12
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
tDo  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)  = 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
(f) E = 145/6.89 = 21.0 Mpsi
(g) v = 75/1.61 = 46.6 mi/h
Ans.
Ans.
Ans.
Ans.
Ans.
(h) V = 1000/946 = 1.06 qt
Ans.
______________________________________________________________________________
1-30 From Table A-2,
Shigley’s MED, 11th edition
Chapter 1 Solutions, Page 10/12
(a) l = 5(0.305) = 1.53 m
Ans.
(b)  = 90(6.89) = 620 MPa
(c) p = 25(6.89) = 172 kPa
(d) Z =12(16.4) = 197 cm3
Ans.
Ans.
Ans.
(e) w = 0.208(175) = 36.4 N/m
Ans.
(f)  = 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)  = M /Z = 1770/0.934 = 1895 psi = 1.90 kpsi
(b)  = 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)  = Tl /GJ = 9 740(9.85)/[11.3(106)( /32)1.004] = 8.648(102) rad = 4.95
Ans.
______________________________________________________________________________
1-32
(a)  =F / wt = 1000/[25(5)] = 8 MPa
Ans.
(b) I = bh3 /12 = 10(25)3/12 = 13.0(103) mm4
Ans.
(c) I = d4/64 =  (25.4)4/64 = 20.4(103) mm4
Ans.
(d)  =16T / d 3 = 16(25)103/[ (12.7)3] = 62.2 MPa
Ans.
______________________________________________________________________________
1-33
(a)  =F /A = 2 700/[ (0.750)2/4] = 6110 psi = 6.11 kpsi
Ans.
(b)  = 32Fa/ d 3 = 32(180)31.5/[ (1.25)3] = 29 570 psi = 29.6 kpsi
(c) Z = (do4  di4)/(32 do) =  (1.504  1.004)/[32(1.50)] = 0.266 in3
Shigley’s MED, 11th edition
Ans.
Ans.
Chapter 1 Solutions, Page 11/12
(d) k = (d 4G)/(8D 3 N) = 0.062 54(11.3)106/[8(0.760)3 32] = 1.53 lbf/in
Ans.
______________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 1 Solutions, Page 12/12
Chapter 8
______________________________________________________________________________
8-1
(a) Thread depth= 2.5 mm Ans.
Width = 2.5 mm Ans.
dm = 25 - 1.25 - 1.25 = 22.5 mm
dr = 25 - 5 = 20 mm
l = p = 5 mm Ans.
(b) Thread depth = 2.5 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  1.226 869 p
d m  d  0.649 519 p
d  1.226 869 p  d  0.649 519 p
 d  0.938 194 p
d 
2
d 2

(d  0.938 194 p) 2
Ans.
4
4
______________________________________________________________________________
At 
8-3

From Eq. (c) of Sec. 8-2,
tan   f
PR  F
1  f tan 
Pd
Fd m tan   f
TR  R m 
2
2 1  f tan 
1  f tan 
T
Fl / (2 ) 1  f tan 
e 0 
 tan 
TR
Fd m / 2 tan   f
tan   f
Shigley’s MED, 11th edition
Ans.
Chap. 8 Solutions, Page 1/68
Using f = 0.08, form a table and plot the efficiency curve.
e
, 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, and dm = d  p/2 = 25  5/2 = 22.5 mm, the torque required to
raise the load is found using Eqs. (8-1) and (8-6)
TR 
5  22.5   5    0.09  22.5  5  0.06  45
 15.85 N  m



2
22.5
0.09
5
2







Ans.
The torque required to lower the load, from Eqs. (8-2) and (8-6) is
TL 
5  22.5     0.09  22.5  5  5  0.06  45
 7.83 N  m


2
22.5
0.09
5
2








Ans.
Since TL is positive, the thread is self-locking. From Eq.(8-4) the efficiency is
5 5
 0.251
Ans.
2 15.85 
______________________________________________________________________________
e
8-5
Collar (thrust) bearings, at the bottom of the screws, must bear on the collars. The bottom
segment of the screws must be in compression. Whereas, tension specimens and their
grips must be in tension. Both screws must be of the same-hand threads.
______________________________________________________________________________
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
Shigley’s MED, 11th edition
Chap. 8 Solutions, Page 2/68
(b) F = 2500 lbf/screw
V = 28.67(0.25) = 7.17 in/min
Ans.
d m  2  0.25 / 2  1.875 in
sec   1 / cos(29o / 2)  1.033
Eq. (8-5):
TR 
2500(1.875)  0.25   (0.05)(1.875)(1.033) 

  221.0 lbf · in
2
  (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
AISI 1006 CD steel. Table A-20: Sy = 41 kpsi.
(a) The handle has maximum bending moment where it enters the screw body.
M = (3.5  0.375) F = 3.125 F
32M
32(3.125) F
Sy   

 41 000
3
d
 (0.375)3
F = 67.9 lbf Ans.
(b) Using Fig. 8-3 for the Acme thread, p = l = 1/6 = 0.1667 in
d m  d  p / 2  0.75  1/12  0.6667 in , d r  d  p  0.75  1/ 6  0.5833 in
 = 29/2 = 14.5, sec 14.5 = 1.033
From Eqs. (8-5) and (8-6),
Fd  l   fd m sec   Ff c d c
Ttotal  m 

2   d m  fl sec  
2
F  0.6667   0.1667    0.15  0.6667 1.033   F  0.15 1
 0.1542 F


2
2
   0.6667   0.15  0.1667 1.033  
From part (a), Ttotal = 3.5 F = 3.5(67.9) = 237.7 lbf٠in
237.7
F
 1542 lbf
Ans.
0.1542
(c) Using Eqs. (8-11), (8-8), (8-7), and (8-12):
Bending stress in first thread, with the force on the first thread being 0.38F and nt = 1,
6(0.38 F )
6(0.38)(1542)
x 

 11 510 psi  11.5 kpsi
 d r nt p  (0.5833)(1)(0.1667)

Shigley’s MED, 11th edition
Chap. 8 Solutions, Page 3/68
Axial stress in body of screw,
4F
4(1542)
y  

 5770 psi  5.77 kpsi
2
 dr
 (0.5833) 2
Torsion in body of screw:
16T
16(237.7)
 yz 

 6100 psi  6.10 kpsi
3
 d r  (0.5833)3
The tangential shear stress given by Eq. (8-12) with one thread carrying 0.38 T:
4  0.38T 
4  0.38  237.7
 zx   2

  2030 psi  2.03 kpsi
2
 d r 1 p
  0.5833 1 0.1667
From Eq. (5-14),
1/2
2
1
2
2
2
11.5   5.77     5.77  0    0  11.5   6  6.10   6  2.03
 
2
 18.9 kpsi
Sy
41

 2.2 Ans.
ny 
  18.9
(d) The column has one end fixed and the other end pivoted in the swivel joint of the
anvil striker, so from Table 4-2, C = 1.2. We will use the root diameter of the screw body
to check buckling, neglecting the effect of the threads.
A = (0.58332)/4 = 0.267 in2, Sy = 41 kpsi, E = 30(106) psi, L = 8 in,


k I
 d 4 / 64

 d / 4  0.5833 / 4  0.1458 in , L/k = 8/0.1458 = 54.9
A
d2 / 4
From Eq. (4-45),
 2 2 1.2  30 106  
  131.7

41 000


Since 54.9 < 131.7, the J.B. Johnson formula is applicable. From Eq. (4-48), the critical
clamping force for buckling is
2

 Sy l  1 
Pc r  A  S y  



 2 k  CE 
2
3




41
10


1


(54.9) 
 0.267 41103   
 9995 lbf
Ans
6
2
1.2
30
10









P
9995
 6.5 Ans.
n  cr 
F 1542
It is confirmed that the weak link is the yielding of the handle.
______________________________________________________________________________
8-8
T = 8(3.5) = 28 lbf  in
1/ 2
2
 l   2 CE 


  
 k 1  S y 
dm 
1/2
3 1
  0.6667 in
4 12
Shigley’s MED, 11th edition
Chap. 8 Solutions, Page 4/68
l =
1
= 0.1667 in,
6
 =
290
= 14.50,
2
sec 14.50 = 1.033
From Eqs. (8-5) and (8-6)
Ttotal 
0.6667 F
2
 0.1667    0.15  0.6667 1.033  0.15 1 F
 0.1542 F


0.6667
0.15
0.1667
1.033
2









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),
2.2 103  (1.375)  0.5   (0.10)(1.375)  2.2 103  (0.15)(2.25)
TR 
  (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 

 
H 
T 
F 
36 F  8   (0.14)(36)  0.09(100) F

2   (36)  0.14(8) 
2
(3.831  4.5) F  8.33F N · m (F in kN)
2 n  2 (1)  2 rad/s
T
3000
H

 477 N · m
2

477
 57.3 kN Ans.
8.33
Shigley’s MED, 11th edition
Chap. 8 Solutions, Page 5/68
Fl
57.3(8)

 0.153 Ans.
2 T
2 (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 =  (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
km 
0.5774  207 14
0.5774 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 =  (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)
Shigley’s MED, 11th edition
Chap. 8 Solutions, Page 6/68
From Eq. (8-22)
km 
0.5774  207 14
0.5774 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
From Table 8-7, ld = L  LT = 40 34 = 6 mm, lt = l  ld = 22  6 = 16 mm
Ad =  (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
0.5774  207 14
0.5774 Ed

 3 624.4 MN/m
Ans.
 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  
______________________________________________________________________________
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.
km 
(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 =  (0.52)/4 = 0.1963 in2. From Table 8-2, At = 0.1419 in2. From Eq. (8-17)
kb 
(c)
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, 11th edition
Ans.
Chap. 8 Solutions, Page 7/68
Top steel frustum: t = 1.5 in, d = 0.5 in, D = 0.75 in, E = 30 Mpsi. From Eq. (8-20)
k1 
0.5774  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 =  (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

 1.705 Mlbf/in
Ad lt  At ld 0.1963  0.69   0.1419  2.5 
Shigley’s MED, 11th edition
Ans.
Chap. 8 Solutions, Page 8/68
(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 
0.5774  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
Ad =  (0.52)/4 = 0.1963 in2. From Table 8-2, At = 0.1419 in2. From Eq. (8-17)
Shigley’s MED, 11th edition
Chap. 8 Solutions, Page 9/68
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 
0.5774  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
From Table 8-7, ld = L  LT = 4.75  1.25 = 3.5 in, lt = l  ld = 4.19  3.5 = 0.69 in
Shigley’s MED, 11th edition
Chap. 8 Solutions, Page 10/68
Ad =  (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 
0.5774  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
km = (1/ km + 1/ km )1 = km /2 = 8.373/2 = 4.19 Mlbf/in
Ans
______________________________________________________________________________
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
From Table 8-7, ld = L  LT = 4.75  1.25 = 3.5 in, lt = l  ld = 4.19  3.5 = 0.69 in
Shigley’s MED, 11th edition
Chap. 8 Solutions, Page 11/68
Ad =  (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)
0.5774 10.3  0.5
 7.23 Mlbf/in
k1 
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.
(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 =  (102) / 4 = 78.54 mm2. From Table 8-1,
Shigley’s MED, 11th edition
Chap. 8 Solutions, Page 12/68
At = 58 mm2. From Eq. (8-17)
78.54  58.0  207
Ad At E
kb 

 292.1 MN/m
Ad lt  At ld 78.54 16   58.0  34 
Ans.
(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)
0.5774  7110
k1 
 1576 MN/m
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 
0.5774  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
Ans.
8-20 (a) From Table A-31, the nut height is H = 8.4 mm. L > l + H = 60 + 8.4 = 68.4 mm.
Shigley’s MED, 11th edition
Chap. 8 Solutions, Page 13/68
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 =  (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)
0.5774 10.3  71
k1 
 1576 MN/m
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
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
Shigley’s MED, 11th edition
Chap. 8 Solutions, Page 14/68
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 =  (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)
0.5774 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
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)
Shigley’s MED, 11th edition
Chap. 8 Solutions, Page 15/68
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 ), Sec. 8-7: C 
kb  k m
Ad At E
Eq. (8-17):
kb 
Ad lt  At ld
Eq. (8-22):
0.5774  207  d
km 
 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, mm 2, 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 ), Sec. 8-7: C 
kb  k m
Ad At E
Eq. (8-17):
kb 
Ad lt  At ld
Shigley’s MED, 11th edition
Chap. 8 Solutions, Page 16/68
For upper frustum, Eq. (8-20), with D = 1.5 d and t = 1.5 in:
k1 
0.5774  30  d
 1.155 1.5   0.5d   2.5d  
ln  

 1.155 1.5   2.5d   0.5d  

0.5774  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:
0.5774  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:
0.5774 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, in 2, Mlbf/in):
Shigley’s MED, 11th edition
Chap. 8 Solutions, Page 17/68
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
______________________________________________________________________________
kb
8-24 Equation (f ), Sec. 8-7: C 
kb  k m
Eq. (8-17):
kb 
Shigley’s MED, 11th edition
Ad At E
Ad lt  At ld
Chap. 8 Solutions, Page 18/68
Top frustum, Eq. (8-20), with E = 10.3Mpsi, D = 1.5 d, and t = l /2:
0.5774 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
0.5774 10.3  d
k2 
1.155  0.5  0.5l   0.5d  2  l  0.5  tan 30 0   2.5d  2  l  0.5  tan 30 0 
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
0.5774  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, in 2, 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, 11th edition
Chap. 8 Solutions, Page 19/68
8-25 For half of joint, Eq. (8-20): t = 20 mm, d = 14 mm, D = 21 mm, and E = 207 GPa
k1 
0.5774  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 
0.5774  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 = (0.752)/4 = 0.4418 in2, At = 0.373 in2 (Table 8-2)
Eq. (8-17),
0.4418  0.373  30
Ad At E

 1.296 Mlbf/in
kb 
Ans.
Ad lt  At ld 0.4418 1.25   0.373  8.75 
Eq. (4-4),
2
2
Am Em  / 4  1.125  0.75  30
km 
Ans.

 1.657 Mlbf/in
l
10
Eq. (f), Sec. 8-7,
Shigley’s MED, 11th edition
C = kb/(kb + km) = 1.296/(1.296 + 1.657) = 0.439
Ans.
Chap. 8 Solutions, Page 20/68
(b)
Let: Nt = no. of turns, p = pitch of thread (in), N = no. of threads per in = 1/p. Then,
 = b + m = Nt p = Nt / N
(1)
But, b = Fi / kb, and, m = 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
Ans.
 15 150 lbf
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 =  / 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

Nt   b
Ans.


360
 kb km  Kd
______________________________________________________________________________
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, 11th edition
Chap. 8 Solutions, Page 21/68
 5.21  8.95 
 k  km 
Nt   b
(14.4)(103 )11
 Fi N  
6
 5.218.95 10 
 kb km 
 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
120  0.141 9 
 1.10
CP  Fi 0.2 13.33  12.77
(b) From Eq. (8-29), the overload factor is
np 
nL 
S p At

S p At  Fi
CP

Ans.
120  0.141 9   12.77
 1.60
0.2 13.33
Ans.
(c) From Eq. (8-30), the joint separation factor of safety is
Fi
12.77

 1.20
Ans.
P 1  C  13.33 1  0.2 
______________________________________________________________________________
n0 
8-30
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
Eq. (8-27), T = KFi d = 0.2(13.8)103(0.5)/12 = 115 lbf  ft
Ans.
______________________________________________________________________________
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
kb
1
Eq. (f ), Sec. 8-7,
C

 0.278
kb  km 1  2.6
Eq. (8-28) with np = 1,
Shigley’s MED, 11th edition
Chap. 8 Solutions, Page 22/68
P
0.25S p At
S p At  Fi

C
C
Ptotal = NP = 8(6.869) = 55.0 kN

0.25  20.1 380 10 3 
Ans.
0.278
 6.869 kN
(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 ), Sec. 8-7,
C
Eq. (8-28) with np = 1,
 S A  Fi
Ptotal  N  p t
C

kb
4

 0.25
kb  k m 4  12
 0.25 NS p At

C

80  0.25 
PtotalC

 4.70
0.25S p At 0.25 120  0.141 9
Round to N = 5 bolts
Ans.
N
(b) Eq. (8-30) with n0 = 1,
 F 
Ptotal  N  i 
 1 C 
P 1  C  80 1  0.25 

 4.70
N  total
12.77
Fi
Round to N = 5 bolts
Ans.
______________________________________________________________________________
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 = (122)/4 = 113.1 mm2, Table 8-1: At = 84.3 mm2
Shigley’s MED, 11th edition
Chap. 8 Solutions, Page 23/68
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 
0.5774  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
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 = [6 (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):
Shigley’s MED, 11th edition
Chap. 8 Solutions, Page 24/68
Fi
41.10

 11.8
Ans.
P 1  C  4.712 1  0.263
______________________________________________________________________________
n0 
8-34
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 =  (0.52)/4 = 0.196 3 in2, Table 8-2: At = 0.141 9 in2
Eq. (8-17):
0.196 3  0.141 9  30
Ad At E
kb 

 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),
k1 
0.5774  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 = [1 500 (3.52)/4](103)/10 = 1.443 kips/bolt
Yielding factor of safety, Eq. (8-28):
S p At
85  0.141 9 

 1.27
np 
CP  Fi 0.299 1.443  9.05
Overload factor of safety, Eq. (8-29):
Shigley’s MED, 11th edition
Ans.
Chap. 8 Solutions, Page 25/68
nL 
S p At  Fi

CP
Separation factor of safety, Eq. (8-30):
85  0.141 9   9.05
0.299 1.443
 6.98
Ans.
Fi
9.05

 8.95
Ans.
P 1  C  1.443 1  0.299 
______________________________________________________________________________
8-35 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 = (102)/4 = 78.5 mm2, Table 8-1: At = 58.0 mm2
Eq. (8-17):
78.5  58.0  207
Ad At E
kb 

 320.8 MN/m
Ad lt  At ld 78.5 16   58.0  29 
n0 
Members: Steel cyl. head: t = 20 mm, d = 10 mm, D = 15 mm, E = 207 GPa. Eq. (8-20),
0.5774  207 10
k1 
 3503 MN/m
1.155  20   15  10  15  10 
ln
1.155  20   15  10  15  10 
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 = [550 (0.82)/4]/36 = 7.679 kN/bolt
Shigley’s MED, 11th edition
Chap. 8 Solutions, Page 26/68
Yielding factor of safety, Eq. (8-28):
830  58.0 103

 1.27
np 
CP  Fi 0.228  7.679   36.1
S p At
Ans.
Overload factor of safety, Eq. (8-29):
S p At  Fi 830  58.0 103  36.1
nL 

 6.88 Ans.
CP
0.228  7.679 
Separation factor of safety, Eq. (8-30):
Fi
36.1

 6.09
n0 
Ans.
P 1  C  7.679 1  0.228 
______________________________________________________________________________
8-36
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 =  (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
kb 

 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),
0.5774  30  0.4375
 31.40 Mlbf/in
k1 
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 
Cast iron: Has two frusta. Midpoint of complete joint is at (3/8 + 1/2)/2 =
0.4375 in.
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
Shigley’s MED, 11th edition
Chap. 8 Solutions, Page 27/68
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 = [1 200 (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):
S p At  Fi 120  0.106 3  9.57
nL 

 8.80 Ans.
CP
0.2911.244 
Separation factor of safety, Eq. (8-30):
Fi
9.57

 10.9
n0 
Ans.
P 1  C  1.244 1  0.291
______________________________________________________________________________
8-37
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
ld = L  LT = 40  20 = 10 mm
lt = l  ld = 26  10 = 16 mm
From Table 8-1, At = 84.3 mm2. Ad =  (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. 8-23, 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 
0.5774  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
Shigley’s MED, 11th edition
Chap. 8 Solutions, Page 28/68
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 = [6 (1002)/4](103)/10 = 4.712 kN/bolt
Yielding factor of safety, Eq. (8-28)
S p At
650  84.310 3

 1.29
np 
CP  Fi 0.275  4.712   41.1
Overload factor of safety, Eq. (8-29)
S p At  Fi 650  84.310 3  41.1
nL 

 10.7
CP
0.275  4.712 
Ans.
Ans.
Separation factor of safety, Eq. (8-30)
Fi
41.1

 12.0
n0 
Ans.
P 1  C  4.712 1  0.275
______________________________________________________________________________
8-38
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. 8-23, we have three frusta.
Top frusta, steel: t = l / 2 = 0.375 in, d = 0.5 in, D = 0.75 in, E = 30 Mpsi
0.5774  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
Shigley’s MED, 11th edition
Chap. 8 Solutions, Page 29/68
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 = [1 500 (3.52)/4](103)/10 = 1.443 kips/bolt
Yielding factor of safety, Eq. (8-28)
S p At
85  0.141 9 

 1.27
np 
Ans.
CP  Fi 0.296 1.443  9.05
Overload factor of safety, Eq. (8-29)
S p At  Fi 85  0.141 9   9.05
nL 

 7.05
CP
0.296 1.443
Ans.
Separation factor of safety, Eq. (8-30)
Fi
9.05

 8.91
n0 
Ans.
P 1  C  1.443 1  0.296 
______________________________________________________________________________
8-39
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 =  (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. 8-23, 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)
0.5774  207 10
k1 
 4 163 MN/m
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
Shigley’s MED, 11th edition
Chap. 8 Solutions, Page 30/68
P = [550 (0.82)/4]/36 = 7.679 kN/bolt
Yielding factor of safety, Eq. (8-28)
S p At
830  58.0 10 3

 1.27
np 
Ans.
CP  Fi 0.253  7.679   36.1
Overload factor of safety, Eq. (8-29)
S p At  Fi 830  58.0 103  36.1

 6.20
nL 
CP
0.253  7.679 
Ans.
Separation factor of safety, Eq. (8-30)
Fi
36.1

 6.29
n0 
Ans.
P 1  C  7.679 1  0.253
______________________________________________________________________________
8-40
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 =  (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
kb 

 5.724 Mlbf/in
Ad lt  At ld 0.150 3  0.46875   0.106 3  0.125 
Similar to Fig. 8-23, we have three frusta.
Top frusta, steel: t = l / 2 = 0.296875 in, d = 0.4375 in, D = 0.65625 in, E = 30 Mpsi
0.5774  30  0.4375
 35.52 Mlbf/in
k1 
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 
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
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
Shigley’s MED, 11th edition
Chap. 8 Solutions, Page 31/68
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 = [1 200 (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 
CP  Fi 0.318 1.244   9.57
Overload factor of safety, Eq. (8-29)
S p At  Fi 120  0.106 3  9.57
nL 

 8.05
CP
0.318 1.244 
Ans.
Ans.
Separation factor of safety, Eq. (8-30)
Fi
9.57

 11.3
n0 
Ans.
P 1  C  1.244 1  0.318 
______________________________________________________________________________
8-41
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 = (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), Sec. 8-7, C = 0.238.
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
 Db

  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.
Shigley’s MED, 11th edition
Chap. 8 Solutions, Page 32/68
5. The external load requirement per bolt is P = 1.15 pg Ag/N, where from Prob 8-33, pg =
6 MPa, and Ag =  (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, 11th edition
Chap. 8 Solutions, Page 33/68
8-42 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 = (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), Sec. 8-7, 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
 Db

  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 =  (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, 11th edition
Chap. 8 Solutions, Page 34/68
8-43 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 = (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), Sec. 8-7, 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
 Db
 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.
N

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 =  (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
Shigley’s MED, 11th edition
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
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 35/68
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 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 = (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), Sec. 8-7, 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
 Db

 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 8-36,
pg = 1 200 psi, and Ag =  (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.
d
km
H
L
0.375 7.42 0.3281 1.25
Shigley’s MED, 11th edition
LT
1
ld
lt
Ad
At
kb
0.25 0.625 0.1104 0.0775 2.905
Chap. 8 Solutions, Page 36/68
d
0.375
0.375
0.375
0.375
C
0.281
0.281
0.281
0.281
SAE
grade
1
2
4
5
N
13
13
13
13
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 
Half of the external moment is contributed by the line load in the interval 0 ≤  ≤ 

M
  FbR 2 sin  d 
0
2
M

 Fb,max R 2
2
2

0
Fb,max R 2 sin 2  d
M
 R2
from which Fb,max 
Fmax 

2

1
FbR sin  d 
M 2
M
R sin  d 
(cos 1 - cos  2 )
2 
R 1
R
Noting 1 = 75, 2 = 105,
Fmax 
(b)
12 000
(cos 75 - cos105 )  494 lbf
 (8 / 2)
Fmax  Fb,max R 
Fmax 
Ans.
M
2M
 2 
( R)   
2
 N
RN
R
2(12 000)
 500 lbf
(8 / 2)(12)
Ans.
(c) F = Fmax sin 
M = 2 Fmax R [(1) sin2 90 + 2 sin2 60 + 2 sin2 30 + (1) sin2 (0)] = 6FmaxR
from which,
Fmax 
Shigley’s MED, 11th edition
M
12 000

 500 lbf
6R
6(8 / 2)
Ans.
Chap. 8 Solutions, Page 37/68
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 
0.5774  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 =  (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, 11th edition
Chap. 8 Solutions, Page 38/68
Yield: From Eq. (8-28)
np 
S p At
CP  Fi

600  35310 3
0.266  4.5   190.6
 1.10 Ans.
Load factor: From Eq. (8-29)
nL 
S p At  Fi
CP

600  353103  190.6
0.266  4.5 
 17.7
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 =  (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, 11th edition
Chap. 8 Solutions, Page 39/68
km 
0.5774  207  20
0.5774 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 p At
600  245 103

 1.07 Ans.
np 
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
i  i 
 90.0 kpsi
At 0.141 9
CP 0.2 13.33
Eq. (8-39),
a 

 9.39 kpsi
2 At 2  0.141 9 
Eq. (8-41),
 m   a   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   i  23.2 150  90.0 

 0.856
n f  e ut
Ans.
 a  Sut  Se  9.39 150  23.2 
(b) Gerber Eq. (8-46)
1 
nf 
Sut Sut2  4 Se  S e   i   Sut2  2 i Se 

2 a Se 

1
150 150 2  4  23.2  23.2  90.0   150 2  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, 11th edition
Chap. 8 Solutions, Page 40/68
nf 


Se
S p S p2  Se2   i2   i Se
2
2
 a  S p  Se 

23.2
120 120 2  23.22  90 2  90  23.2    1.30

9.39 120 2  23.2 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  km 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 p At
380  20.110 3

 0.98
np 
Ans.
Yield, Eq. (8-28),
CP  Fi 0.278  7.5   5.73
S p At  Fi
380  20.1103  5.73
 0.915
0.278  7.5 
Ans.
(b) Overload, Eq. (8-29),
nL 
(c) Separation, Eq. (8-30),
n0 
(d) Goodman, Eq. (8-35),
C  Pb max  Pb min  0.278  7.5  2.5 103

 34.6 MPa
a 
2 At
2  20.1
CP

Fi
5.73

 1.06
P 1  C  7.5 1  0.278 
Ans.
C  Pb max  Pb min  Fi 0.278  7.5  2.5 103 5.73 10
Eq. (8-36),  m 
 

At
2 At
2  20.1
20.1
3
  354.2
MPa
Table 8-11, Sut = 520 MPa, i = 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. (6-10):
Se  0.5Sut  0.5  520   260 MPa .
Table 6-2:
Eq. (6-18):
a = 3.04, b =  0.217
k a  aSutb  3.04  5200.217   0.783
Eq. (6-19):
kb = 1
Eq. (6-25):
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, from
Eq. (6-17),
Se = ka kb kc S e / Kf = 0.783(1)0.85(260) / 2.2 = 78.7 MPa
Shigley’s MED, 11th edition
Chap. 8 Solutions, Page 41/68
Eq. (8-38),
nf 
78.7  520  285 
Se  Sut   i 

 0.789
Sut a  Se  m   i  520  34.6   78.7  354.2  285 
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  i = Fi /At = 0.75 Sp = 0.75(120) =90 kpsi
Eq. (8-35),
a 
Eq. (8-36),
m 
Eq. (8-38),
C  Pb max  Pb min  0.25  8  2 

 5.29 kpsi
2 At
2  0.141 9 
0.25  8  2 
C  Pb max  Pb min 
i 
 90  98.81 kpsi
2 At
2  0.141 9 
23.2 150  90 
Se  Sut   i 

 1.39
Ans.
Sut a  Se  m   i  150  5.29   23.2  98.81  90 
______________________________________________________________________________
8-51 Given: (8) M8  1.5 class 9.8 bolts, kb = 1.5 MN/mm, km = 3.9 MN/mm, reversing load
with Pmax = 50 kN.
Per bolt, P = Pmax/N = 50/8 = 6.25 kN
Table 8-1, At = 36.6 mm2, Table 8-11, Sut = 900 MPa, Sp = 650 MPa,
Table 8-17, Se = 140 MPa
Eq. (f), Sec. 8-7: C = kb / (kb + km) = 1.5/(1.5 + 3.9) = 0.278
Eqs. (8-31) and (8-32): Fi = 0.75 At Sp = 0.75(36.6)650(103) = 17.84 kN
i = Fi / At = 17.84(103)/36.6 = 487.5 MPa
Eq. (8-39): a = CP / (2At) = 0.278(6.25)103/[2(36.6)] = 23.74 MPa
(a) Goodman: Eq. (8-45):
S  S   i  140  900  487.5 
n f  e ut

 2.34 Ans.
 a  Sut  Se  23.74  900  140 
(b) Gerber: Eq. (8-46):
1 
nf 
Sut Sut2  4 Se  S e   i   Sut2  2 i S e 

2 a Se 
nf 
1
900 900 2  4 140 140  487.5   900 2  2  487.5    3.52 Ans.

2  23.74 140  
(c) Morrow: Use Goodman Eq. (8-45) replacing Sut with  f . Equation (6-44) gives

Shigley’s MED, 11th edition
Chap. 8 Solutions, Page 42/68
 f = Sut + 345 = 900 + 345 = 1 245MPa
nf 
Se  f   i 
 a  f  Se 

140 1245  487.5 
23.74 1245  140 
 3.23
Ans.
______________________________________________________________________________
8-52 From Prob. 8-33, C = 0.263, Pmax = 4.712 kN / bolt, Fi = 41.1 kN, Sp = 650 MPa, and
At = 84.3 mm2
i = 0.75 Sp = 0.75(650) = 487.5 MPa
3
CP 0.263  4.712 10
a 

 7.350 MPa
2 At
2  84.3
Eq. (8-39):
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   i  140  900  487.5 
Eq. (8-45):

 7.55
n f  e ut
Ans.
 a  Sut  Se  7.350  900  140 
(b) Gerber:
Eq. (8-46):
1 
nf 
Sut Sut2  4 Se  Se   i   Sut2  2 i Se 

2 a Se 
m 
Eq. (8-40)

1
900 9002  4 140 140  487.5   900 2  2  487.5 140  

2  7.350 140 
 11.4
Ans.
(c) ASME-elliptic:
Eq. (8-47):
nf 


Se
S p S p2  Se2   i2   i S e
2
2
 a  S p  Se 

140
650 6502  140 2  487.52  487.5 140    9.73
2
2 

7.350  650  140 
Ans.
______________________________________________________________________________
8-53 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
 i  0.75S p  0.75  85   63.75 kpsi
Shigley’s MED, 11th edition
Chap. 8 Solutions, Page 43/68
Eq. (8-37):
a 
CP 0.299 1.443

 1.520 kpsi
2 At
2  0.141 9 
Eq. (8-38)
m 
CP
  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   i  18.8 120  63.75 
Eq. (8-45):
n f  e ut
Ans.

 5.01
 a  Sut  Se  1.520 120  18.8 
(b) Gerber:
Eq. (8-46):
1 
nf 
Sut Sut2  4 Se  Se   i   Sut2  2 i S e 


2 a S e

1
120 1202  4 18.6 18.6  63.75   120 2  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   i2   i Se
2
2
 a  S p  Se 

18.6
85 852  18.6 2  63.752  63.75 18.6   6.22
2
2 

1.520  85  18.6 
Ans.
______________________________________________________________________________
8-54 From Prob. 8-35, C = 0.228, Pmax = 7.679 kN/bolt, Fi = 36.1 kN, Sp = 830 MPa, and
At = 58.0 mm2
i = 0.75 Sp = 0.75(830) = 622.5 MPa
3
CP 0.228  7.679 10
a 

 15.09 MPa
Eq. (8-37):
2 At
2  58.0 
CP
  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
Eq. (8-38)
Shigley’s MED, 11th edition
m 
Chap. 8 Solutions, Page 44/68
Eq. (8-45):
nf 
Se  Sut   i  162 1040  622.5 

 3.73
 a  Sut  S e  15.09 1040  162 
Ans.
(b) Gerber:
Eq. (8-46):
1 
nf 
Sut Sut2  4 Se  Se   i   Sut2  2 i Se 


2 a Se

1
1040 10402  4 162 162  622.5   1040 2  2  622.5 162  

2 15.09 162 
 5.74
Ans.
(c) ASME-elliptic:
Eq. (8-47):
nf 


Se
S p S p2  Se2   i2   i S e
 a  S p2  Se2 

162
830 8302  162 2  622.52  622.5 162    5.62
2
2 

15.09 830  162 
Ans.
______________________________________________________________________________
8-55 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
 i  0.75S p  0.75 120   90 kpsi
Eq. (8-37):
a 
CP 0.2911.244 

 1.703 kpsi
2 At
2  0.106 3
Eq. (8-38)
m 
CP
  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 i 
23.2 150  90 
Eq. (8-45):
n f  e ut
Ans.

 4.72
 a  Sut  Se  1.703 150  23.2 
(b) Gerber:
Eq. (8-46):
Shigley’s MED, 11th edition
Chap. 8 Solutions, Page 45/68
nf 

1
 S S 2  4 S  S     S 2  2 S 
ut
ut
e
e
i
ut
i e

2 a Se 
1
150 1502  4  23.2  23.2  90   150 2  2  90  23.2 

2 1.703 23.2 
 7.28
Ans.
(c) ASME-elliptic:
Eq. (8-47):
nf 


Se
S p S p2  Se2   i2   i Se
2
2
 a  S p  Se 

23.2
120 1202  23.22  90 2  90  23.2    7.24

1.703 1202  18.62  
Ans.
______________________________________________________________________________
8-56 From Prob. 8-52, C = 0.263, Se = 140 MPa, Sut = 900 MPa, At = 84.4 mm2, i = 487.5
MPa, and Pmax = 4.712 kN.
Pmin = Pmax / 2 = 4.712/2 = 2.356 kN
a 
Eq. (8-35):
Eq. (8-36):
m 
C  Pmax  Pmin  0.263  4.712  2.356 103

 3.675 MPa
2 At
2  84.3
C  Pmax  Pmin 
2 At
 i
0.263  4.712  2.356 103

 487.5  498.5 MPa
2  84.3
Eq. (8-38):
140  900  487.5 
Se  Sut   i 

 11.9
Ans.
Sut a  Se  m   i  900  3.675   140  498.5  487.5 
______________________________________________________________________________
nf 
Shigley’s MED, 11th edition
Chap. 8 Solutions, Page 46/68
8-57 From Prob. 8-53, C = 0.299, Se = 18.8 kpsi, Sut = 120 kpsi, At = 0.141 9 in2, i = 63.75
kpsi, and Pmax = 1.443 kips
Pmin = Pmax / 2 = 1.443/2 = 0.722 kips
a 
Eq. (8-35):
Eq. (8-36):
m 

C  Pmax  Pmin 
2 At

0.299 1.443  0.722 
2  0.141 9 
 0.760 kpsi
C  Pmax  Pmin 
i
2 At
0.299 1.443  0.722 
 63.75  66.03 kpsi
2  0.141 9 
Eq. (8-38):
Se  Sut   i 
18.8 120  63.75 
Ans.

 7.89
Sut a  Se  m   i  120  0.760   18.8  66.03  63.75 
______________________________________________________________________________
nf 
8-58 From Prob. 8-54, C = 0.228, Se = 162 MPa, Sut = 1040 MPa, At = 58.0 mm2, i = 622.5
MPa, and Pmax = 7.679 kN.
Pmin = Pmax / 2 = 7.679/2 = 3.840 kN
a 
Eq. (8-35):
Eq. (8-36):
m 
C  Pmax  Pmin  0.228  7.679  3.840 103

 7.546 MPa
2 At
2  58.0 
C  Pmax  Pmin 
2 At
i
0.228  7.679  3.840 103

 622.5  645.1 MPa
2  58.0 
Eq. (8-38):
Shigley’s MED, 11th edition
Chap. 8 Solutions, Page 47/68
162 1040  622.5 
Se  Sut   i 

 5.88
Ans.
Sut a  Se  m   i  1040  7.546   162  645.1  622.5 
______________________________________________________________________________
nf 
8-59 From Prob. 8-55, C = 0.291, Se = 23.2 kpsi, Sut = 150 kpsi, At = 0.106 3 in2, i = 90
kpsi, and Pmax = 1.244 kips
Pmin = Pmax / 2 = 1.244/2 = 0.622 kips
a 
Eq. (8-35):
Eq. (8-36):
m 

C  Pmax  Pmin 
2 At

0.2911.244  0.622 
2  0.106 3
 0.851 kpsi
C  Pmax  Pmin 
i
2 At
0.2911.244  0.622 
 90  92.55 kpsi
2  0.106 3
Eq. (8-38):
23.2 150  90 
Se  Sut   i 

 7.45
Ans.
Sut a  Se  m   i  150  0.851  23.2  92.55  90 
______________________________________________________________________________
nf 
8-60 Let the repeatedly-applied load be designated as P. From Table A-22, Sut = 93.7 kpsi.
Referring to the Figure of Prob. 3-136, 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, with R = 0.5 in,
R2
0.52
rn 

 1.968 246 in
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
 (12 ) / 4  0.7854 in 2
If P is the maximum load
Shigley’s MED, 11th edition
Chap. 8 Solutions, Page 48/68
M  Prc  2 P
2(0.468) 
P
rc 
P 
 i  1  c i  
1 
  26.29 P
A
eri  0.785 4 
0.031 754(1.5) 
26.294 P

 13.15P
a  m  i 
2
2
(a) Eye: Section AA,
Table 6-2:
a = 11.0 kpsi, b =  0.650
Eq. (6-18):
k a  11.0(93.7) 0.650  0.575
Eq. (6-23):
de = 0.370 d
Eq. (6-19):
Eq. (6-25):
Eq. (6-10):
Eq. (6-18):
0.107
 0.37 
kb  
 0.978

 0.30 
kc = 0.85
Se  0.5Sut  0.5  93.7   46.85 kpsi
Se = 0.575(0.978)0.85(46.85) = 22.4 kpsi
From Eq. 6-48, for Gerber,
2
2

 2 m S e  
1  Sut   a 
1  1  
nf  


2   m  Se 
Sut a  



With m = a,
2
2


 2Se   1
1 Sut2 
93.7 2
2(22.4)   1.616

 1  1  
1  1  
nf 
 
  
P
2  a Se 
 93.7  
 Sut   2 13.15P(22.4) 


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 ,  = P/At = P /0.663 = 1.51 P, a = m = /2 = 0.755 P
From Eq. 6-48, Gerber
2
2


 2S e   1
1 Sut2 
93.72
2(14.7)   19.01

 1  1  
1  1  
nf 
 
 
P
2  a Se 
 93.7  
 Sut   2 0.755P(14.7) 


Comparing 19.01/P with 1.616/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 section 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.
Shigley’s MED, 11th edition
Chap. 8 Solutions, Page 49/68
(c) For nf = 2
1.616 103 
 808 lbf, max. load Ans.
2
______________________________________________________________________________
P 
8-61 Member, Eq. (8-22) with E =16 Mpsi, d = 0.75 in, and l = 1.5 in
km 
0.5774 16  0.75
0.5774 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,
Eq. (8-17),
At = 0.373 in2
Ad = (0.752)/4 = 0.442 in2
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  km 8.09  13.32
Eq. (8-35),
a 
C  Pmax  Pmin  0.378  6  4 

 1.013 kpsi
2 At
2  0.373
m 
C  Pmax  Pmin  Fi 0.378  6  4 
25
 

 72.09 kpsi
At
2 At
2  0.373
0.373
Eq.(8-36),
(a) From Table 8-9, Sp = 85 kpsi, and Eq. (8-51), the yielding factor of safety is
np 
Shigley’s MED, 11th edition
Sp
m a

85
 1.16
72.09  1.013
Ans.
Chap. 8 Solutions, Page 50/68
(b) From Eq. (8-29), the overload factor of safety is
nL 
S p At  Fi
CPmax

85  0.373  25
 2.96
0.378  6 
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
i = Fi / At = 25/0.373 = 67.0 kpsi; and from Eq. (8-38)
Se  Sut   i 
18.6 120  67.0 

 4.56
Ans.
Sut a  S e  m   i  120 1.013  18.6  72.09  67.0 
______________________________________________________________________________
nf 
8-62 (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),
i = 0.75 Sp = 0.75(120) = 90 kpsi
kb
4

 0.2
kb  k m
4  16
CP
0.2 P
a 

 0.705P kpsi
2 At
2(0.141 9)
C 
Eq. (8-45) for the Goodman criterion,
S S  i 
23.2(150  90)
11.4


 2
n f  e ut
 a  Sut  Se  0.705 P(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),
S p At
120  0.141 9 

 1.22
np 
Ans.
CP  Fi 0.2  5.70   12.77
Load factor, Eq. (8-29),
S A - Fi 120(0.141 9)  12.77
nL  p t

 3.74 Ans.
CP
0.2(5.70)
Separation load factor, Eq. (8-30)
Shigley’s MED, 11th edition
Chap. 8 Solutions, Page 51/68
12.77
Fi

 2.80 Ans.
P(1 - C ) 5.70(1  0.2)
______________________________________________________________________________
n0 
8-63 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
i = 0.75 Sp = 0.75(74) = 55.5 kpsi
a 
CP
0.30 P

 0.155P kpsi
2 At
2(0.969)
Gerber, Eq. (8-46),
nf 
1
 S S 2  4 S  S     S 2  2 S 
ut
ut
e
e
i
ut
i e

2 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,
64.28
P 
 32.14 kip Ans.
2
Fine thread,
Fi = 0.75 At Sp = 0.75(1.073)74 = 59.55kips
i = 0.75 Sp = 0.75(74) = 55.5 kpsi
CP
0.32 P
a 

 0.149 P kpsi
2 At
2(1.073)

The only thing that changes in Eq. (8-46) is a. Thus,
0.155 64.28 66.87
nf 

 2  P  33.43 kips
Ans.
0.149 P
P
Percent improvement,
33.43  32.14
(100) ฀ 4% Ans.
32.14
______________________________________________________________________________
8-64 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
Shigley’s MED, 11th edition
Chap. 8 Solutions, Page 52/68
Eq. (8-31),
Fi = 0.75Fp = 0.75 At Sp
= 0.75(5610600(103) = 252.45 kN
i = 0.75 Sp = 0.75(600) = 450 MPa
Eq. (8-39),
CP 0.28  65 10
a 

 16.22 MPa
2 At
2  561
3
Gerber, Eq. (8-46),
nf 

1
 S S 2  4 S  S     S 2  2 S 
ut
ut
e
e
i
ut
i e

2 a S e 
1
830 830 2  4 129 129  450   830 2  2  450 129 

2 16.22 129 
 4.75
Ans.
The yielding factor of safety, from Eq. (8-28) is
600  561103

 1.24
np 
CP  Fi 0.28  65   252.45
S p At
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
Ans.

 5.39
P 1  C  65 1  0.28 
______________________________________________________________________________
8-65 (a) Table 8-2,
At = 0.077 5 in2
Table 8-9,
Sp = 85 kpsi, Sut = 120 kpsi
Table 8-17,
Se = 18.6 kpsi
Unthreaded grip,
n0 
Ad E
 (0.375)2 (30)
kb 

 0.245 Mlbf/in per bolt
l
4(13.5)
Am 

[( D  2t ) 2 - D 2 ] 

(4.752 - 4 2 )  5.154 in 2
4
4
Am E
5.154(30)  1 
km 

   2.148 Mlbf/in/bolt.
l
12
6
Shigley’s MED, 11th edition
Ans.
Ans.
Chap. 8 Solutions, Page 53/68
(b)
Fi = 0.75 At Sp = 0.75(0.0775) (85) = 4.94 kip
 i  0.75S p  0.75(85)  63.75 kpsi
2000  

(4) 2   4189 lbf/bolt

6 4

kb
0.245
C 

 0.102
kb  k m
0.245  2.148
CP
0.102(4.189)

 2.77 kpsi
a 
2 At
2(0.0775)
P  pA 
From Eq. (8-46) for Gerber fatigue criterion,
nf 
1
 S S 2  4 S  S     S 2  2 S 
ut
ut
e
e
i
ut
i e

2 a S e 
1
120 1202  4 18.6 18.6  63.75   120 2  2  63.75 18.6   4.09

2  2.77 18.6 
(c) Pressure causing joint separation from Eq. (8-30)

Ans.
Fi
1
P(1  C )
Fi
4.94

 5.50 kip
P 
1  C 1  0.102
P
5.50

p 
6  2.63 kpsi Ans.
A  (42 ) / 4
______________________________________________________________________________
n0 
8-66 From the solution of Prob. 8-64, At = 0.077 5 in2, Sut = 120 kpsi, Se = 18.6 kpsi, C = 0.102,
i = 63.75 kpsi
Pmax = pmaxA = 2  (42)/4 = 25.13 kpsi, Pmin = pminA = 1.2  (42)/4 = 15.08 kpsi,
C  Pmax  Pmin  0.102  25.13  15.08 
Eq. (8-35),

 6.61 kpsi
a 
2 At
2  0.077 5 
Eq. (8-36),
m 
C  Pmax  Pmin 
0.102  25.13  15.08 
i 
 63.75  90.21 kpsi
2 At
2  0.077 5 
Eq. (8-38),
Se  Sut   i 
18.6 120  63.75 
nf 

 0.814
Sut a  Se  m   i  120  6.61  18.6  90.21  63.75 
Ans.
This predicts a fatigue failure.
______________________________________________________________________________
8-67 Members: Sy = 57 kpsi, Ssy = 0.577(57) = 32.89 kpsi.
Shigley’s MED, 11th edition
Chap. 8 Solutions, Page 54/68
Bolts: SAE grade 5, Sy = 92 kpsi, Ssy = 0.577(92) = 53.08 kpsi
Shear in bolts,
  (0.252 ) 
 0.0982 in 2
As  2 

4


AS
0.0982(53.08)
Fs  s sy 
 2.61 kips
n
2
Bearing on bolts,
Ab = 2(0.25)0.25 = 0.125 in2
AS
0.125(92)
Fb  b yc 
 5.75 kips
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.
______________________________________________________________________________
8-68 Members, Table A-20, Sy = 42 kpsi
Bolts, Table 8-9, Sy = 130 kpsi, Ssy = 0.577(130) = 75.01 kpsi
Shear of bolts,
   5 /16 2 
2
As  2 
  0.1534 in
4



n
Fs
5

 32.6 kpsi
As 0.1534
S sy


75.01
 2.30
32.6
Ans.
Bearing on bolts,
Ab = 2(0.25) (5/16) = 0.1563 in2
5
b  
 32.0 kpsi
0.1563
Shigley’s MED, 11th edition
Chap. 8 Solutions, Page 55/68
n
Sy
b

130
 4.06
32.0
Bearing on members,
n
Tension of members,
Sy
b

Ans.
42
 1.31
32
Ans.
At = [2.375  2(5/16)] (1/4) = 0.4375 in2
t 
5
 11.4 kpsi
0.4375
Sy
42
 3.68
Ans.
 t 11.4
______________________________________________________________________________
n

8-69 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,
  (202 ) 
 628.3 mm 2
As  2 

 4 
AS
628.3(242.3)10 3
Fs  s sy 
 60.9 kN
n
2.5
Bearing on bolts,
Ab = 2(20)20 = 800 mm2
AS
800(420)10 3
Fb  b yc 
 134 kN
n
2.5
Bearing on member,
800(490)10 3
Fb 
 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.
______________________________________________________________________________
Shigley’s MED, 11th edition
Chap. 8 Solutions, Page 56/68
8-70 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 =  (202)/4 = 314.2 mm2
90 103 
s 
 95.48 MPa
3  314.2 
n
S sy
s

242.3
 2.54
95.48
Ans.
Bearing on bolt,
Ab = 3(20)15 = 900 mm2
90 103 
b  
 100 MPa
900
S
420
n y 
 4.2 Ans.
b
100
Bearing on members,
S
320
n y 
 3.2 Ans.
b
100
Tension on members,
90 103 
F
t 

 46.15 MPa
A 15[190  3  20 ]
S
320
n y 
 6.93 Ans.
46.15
t
______________________________________________________________________________
8-71 Members: Sy = 57 kpsi
Bolts: Sy = 100 kpsi, Ssy = 0.577(100) = 57.7 kpsi
Shear of bolts,
  1/ 4 2 
2
A  3
  0.1473 in
4


F
5
s 

 33.94 kpsi
As 0.1473
n
Bearing on bolts,
Shigley’s MED, 11th edition
S sy
s

57.7
 1.70
33.94
Ans.
Ab = 3(1/4) (5/16) = 0.2344 in2
Chap. 8 Solutions, Page 57/68
b  
n
Bearing on members,
Sy
b
5
F

 21.3 kpsi
Ab
0.2344

100
 4.69
21.3
Ans.
Ab = 0.2344 in2 (From bearing on bolts calculation)
b =  21.3 kpsi (From bearing on bolts calculation)
n
Sy
b

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
Sy
57
n

 6.68
Ans.
 t 8.534
______________________________________________________________________________
t 
8-72 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)  50RA  0

RA  8 kN
A
0
200(1.6)  50RB  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

Bolt shear,
As  (122 )  113.1 mm 2
4
Shigley’s MED, 11th edition
Chap. 8 Solutions, Page 58/68
Fmax
8(103 )
 

 70.73 MPa
As
113.1
S
242.3
n  sy 
 3.43

70.73
Ab = td = 10(12) = 120 mm2
8(103 )
b  
 66.67 MPa
120
S
370
 5.55
n y 
b
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
B  A 
(103 )  97.37 MPa
IA
102.7(103 )
S
370
n y 
 3.80
 A 97.37
At the center, call it point C,
MC = 1.6(350) = 560 N · m
1
(10)(503 )  104.2(103 ) mm 4
IC 
12
560(25)
M c
(103 )  134.4 MPa
C  C 
3
104.2(10 )
IC
S
370
n y 
 2.75  3.80 more critical at C
 C 134.4
n  min(3.04, 3.80, 2.75)  2.72 Ans.
______________________________________________________________________________
Shigley’s MED, 11th edition
Chap. 8 Solutions, Page 59/68
8-73 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 =  (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),
0.5774  30  0.5
0.5774 Ed
km 

 16.65 Mlbf/in
 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  
kb
4.574
C

 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
i = 0.75 Sp = 0.75(65) = 48.75 kips
Eq. (a):
Direct shear,
CP  Fi 0.216 1.071  6.918

 50.38 kpsi
At
0.141 9
F
3
s  s 
 21.14 kpsi
At 0.141 9
b 
Shigley’s MED, 11th edition
Chap. 8 Solutions, Page 60/68
von Mises stress, Eq. (5-15),
    b2  3 s2 
1/2
 50.382  3  21.142  
1/2
 62.3 kpsi
Sp
65

 1.04
Ans.
  62.3
______________________________________________________________________________
n
8-74
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 

(202 )  314.2 mm 2
4
Fi  0.75(245)(380)(103 )  69.83 kN
 i  0.75  380   285 MPa
CP  Fi  0.25(1.75)  69.83  3

 (10 )  287 MPa
245
At


3
7(10 )
F
  s 
 22.3 MPa
314.2
Ad
   [287 2  3(22.32 )]1/ 2  290 MPa
b 
Sp
380

 1.31
Ans.
  290
______________________________________________________________________________
n
8-75 Using the result of Prob. 5-80 for lubricated assembly (replace 0.2 with 0.18 per Table
8-15)
2 f T
Fx 
0.18d
With a design factor of nd gives
T 
or T/d = 716. Also,
Shigley’s MED, 11th edition
0.18nd Fxd
0.18(3)(1000)d

 716d
2 f
2 (0.12)
Chap. 8 Solutions, Page 61/68
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
3
8
At
T/d = 11 475At
n
0.0364
417.70 1.75
- 24
0.058
665.55 2.8
 24
0.0878
1007.50 4.23
where the factor of safety in the last column of the table comes from
n
Select a
3"
8
2 f (T / d )
2 (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
2 f T
2 (0.12)(400)

 4.47
0.18Fx d
0.18(1000)(0.375)
______________________________________________________________________________
n
8-76
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
Shigley’s MED, 11th edition
Chap. 8 Solutions, Page 62/68
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 = (102) / 4 = 78.54 mm2
Max. force,
Ssy = 0.577(420) = 242.3 KPa
 max 
FC S sy

As
n
From Eq. (1), FC = 2.343 F. Thus
F
Ssy  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.
S  A  170  64  3
F  y  b 

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: At C
I
1
12   503  103   1.24 105  mm 4
12
 max 
Shigley’s MED, 11th edition
Sy
n

Mc 151Fc

I
I
 F
Sy  I 


n  151 c 
Chap. 8 Solutions, Page 63/68
5
190 1.24 10   3

 10  3.12 kN
F
2.0  151 25  


So F = 2.32 kN based on bearing on channel.
Ans.
______________________________________________________________________________
8-77 Bolts, from Table 8-11, Sy = 420 MPa
Bracket, from Table A-20, Sy = 210 MPa
12
 4 kN; M  12(200)  2400 N · m
3
2400
 37.5 kN
FA  FB 
64
FA  FB  (4) 2  (37.5) 2  37.7 kN
FO  4 kN
F 
Bolt shear:
The shoulder bolt shear area, As = (122) / 4 = 113.1 mm2
Ssy = 0.577(420) = 242.3 KPa
37.7(10)3
 333 MPa
113
S
242.3
n  sy 
Ans.
 0.728

333
Bearing on bolts:
Ab  12(8)  96 mm 2
37.7(10)3
b  
 393 MPa
96
S
420
n  yc 
Ans.
 1.07
b
393
 
Bearing on member:
 b  393 MPa
S
210
n  yc 
 0.534
b
393
Shigley’s MED, 11th edition
Ans.
Chap. 8 Solutions, Page 64/68
Bending stress in plate:
 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) mm
Ans.
2400(68)
Mc
 
(10)3  110 MPa

6
1.48(10)
I
Sy
210

 1.91
n 
Ans.

110
I 
Failure is predicted for bolt shear and bearing on member.
______________________________________________________________________________
8-78
FA  1208  125  1083 lbf,
Bolt shear:
 3625 
FA  FB  
  1208 lbf
 3 
FB 1208  125  1333 lbf
As = ( / 4)(0.3752) = 0.1104 in2
 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
 max

57.7
 4.78
12.07
Ans.
Bearing on bolt: Bearing area is Ab = td = 0.375 (0.375) = 0.1406 in2.
Shigley’s MED, 11th edition
Chap. 8 Solutions, Page 65/68
1333
F

 9 481 psi
Ab
0.1406
b  
Sy
n
b

100
 10.55
9.481
Ans.
Bearing on member: From Table A-20, Sy = 54 kpsi. Bearing stress same as bolt
Sy
n
b

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  
Mc 3250 1

 13 080 psi
I
0.2484

Sy
54
 4.13
Ans.
 13.08
______________________________________________________________________________
n

8-79 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
Thus F 
 1000 lbf and the resultant bolt load is
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 =  (0.5)2/4 = 0.1963 in2
S
(0.577)(100)
n  sy 
 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
Shigley’s MED, 11th edition
100
 8.89
1.054 / 0.09375
Ans.
Chap. 8 Solutions, Page 66/68
Bearing on channel:
Bearing on plate:
n
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.
Strength of plate:
0.25(7.5)3 0.25(0.5)3

12
12
3
 0.25(0.5)

 2
  0.25  0.5  (2.5) 2   7.219 in 4
12


M  5000 lbf · in per plate
Mc 5000(3.75)
 

 2597 psi
I
7.219
42
 16.2 Ans.
n
2.597
______________________________________________________________________________
I 
A = 58 mm2. From Table A-9, beam 14,
R1 = R2 = F/2 = 10/2 = 5 kN,
M1 = M2 = Fl/8 = 10(0.4)103/8 = 500 N٠m.
 0.050 
M 1  500  4 F1 
  F1  5000 N
 2 
R 5000
F2  1 
 1250 N
4
4
Bolt A has the maximum force of
FA = F1 + F2 = 5000 + 1250 = 6250 N
F
6250
Ans.
 max   A  A 
 108 MPa
58
A
______________________________________________________________________________
8-81 Given: 250 hp at 600 rev/min, r1 = 2.5 in, r2 = 5 in.
63 025  250 
 26.26 103  lbf  in
Eq. (3-42): T 
600
F1 F2
r
5
(a)

 F2  2 F1 
F1  2 F1
2.5
r1
r2
r1
8-80
4 F1r1  4 F2 r2  T
 4 F1 (2.5)  4  2 F1  5  26.26 103 
Yields F1 = 525.2 lbf and F2 = 1050.4 lbf
 max 
4 F2
d2

d
From Table A-17 select d =
Shigley’s MED, 11th edition
4 F2
 max
5
16
in

4 1050.4 
 0.258 in
  20 103
Ans.
Chap. 8 Solutions, Page 67/68
(b) F 
 max d
2
4
nFr2 = T

20 103  

4
 
5
16
2
 1534 lbf
n (1534) 5 = 26.26 (103)

n = 3.4 bolts
Use four bolts without need of the inner bolts
Ans.
______________________________________________________________________________
8-82 to 8-84 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.
______________________________________________________________________________
8-85 Assume tensile forces to be proportional to deflections. Thus,
F1 F2


F2  9 F1
(1)
1
9
MA = 0 = 2 F1 (1) + 2 F2 (9)  1000 (10)
Substitute Eq. (1) in,
2 F1 (1) + 2 (9 F1 ) (9)  1000 (10) = 0
Yields F1 = 60.98 lbf, F2 = 548.8 lbf.
Combining the preload,
5000  548

 27.7 103 psi  27.7 kpsi
0.2
The shear forces are F3 =1000/4 = 250 lbf.
 
The shear stress is,
F 250
 3
 1250 psi  1.25 kpsi
A 0.2
The maximum tensile stress is,

2
2
27.7
 
 27.7 
2
    2 
 
Ans.
  1.25  27.8 kpsi
2
2
2
 2 
______________________________________________________________________________
 max 
Shigley’s MED, 11th edition
Chap. 8 Solutions, Page 68/68
Chapter 2
2-1 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.
______________________________________________________________________________
2-2 (a) Maximize yield strength: Q&T at 425C (800F) Ans.
(b) Maximize elongation: Q&T at 650C (1200F) Ans.
______________________________________________________________________________
2-3 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.
 76.5  102 
2011-T6 aluminum: Tables A-22 and A-5
3
S y 169  10 

62.3 kN m/kg
Ans.
 26.6  102 
Ti-6Al-4V titanium: Tables A-24c and A-5
3
S y 830  10 

187 kN m/kg
Ans.
 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

70.6  102 
______________________________________________________________________________
2-4 AISI 1018 CD steel: Table A-5
6
E 30.0  10 

106  106  in

0.282
2011-T6 aluminum: Table A-5
6
E 10.4  10 

106  106  in

0.098
Shigley’s MED, 11th edition
Ans.
Ans.
Chapter 2 Solutions, Page 1/22
Ti-6Al-6V titanium: Table A-5
6
E 16.5  10 

103  10 6  in Ans.

0.160
No. 40 cast iron: Table A-5
6
E 14.5  10 

55.8  106  in Ans.

0.260
______________________________________________________________________________
E  2G
2G
2-5
Using values for E and G from Table A-5,
30.0  2  11.5
v
0.304 Ans.
2  11.5 
Steel:
The percent difference from the value in Table A-5 is
2G (1  v) E

v
0.304  0.292
 0.0411  4.11 percent
0.292
10.4  2  3.90
v
0.333
2  3.90
Ans.
Ans.
Aluminum:
The percent difference from the value in Table A-5 is 0 percent Ans.
Beryllium copper:
18.0  2  7.0 
v
0.286
2  7.0 
Ans.
The percent difference from the value in Table A-5 is
0.286  0.285
 0.00351  0.351 percent
0.285
14.5  2  6.0 
v
0.208
2  6.0 
Ans.
Ans.
Gray cast iron:
The percent difference from the value in Table A-5 is
0.208  0.211
  0.0142   1.42 percent
Ans.
0.211
______________________________________________________________________________
2-6 (a) A0 =  (0.503)2/4 = 0.1987 in2,  = Pi / A0
Shigley’s MED, 11th edition
Chapter 2 Solutions, Page 2/22
For data in the elastic range, from Eq. (2–2),  = (l – l0) / l0 =  l / l0 =  l / 2
A  A
 0
A
For data in the plastic range, from Eq. (2–8),
On the next two pages, the data and plots are presented. Figure (a) shows the linear part of
the curve from data points 1 to 7. Figure (b) shows data points 1 to 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
 = 30.5(106)  61 000
(1)
The equation for the line between data points 8 and 9 is
 = 7.60(105) + 42 900
(2)
Solving Eqs. (1) and (2) simultaneously yields  = 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. (2-25) as
R
A0  Af
A0

0.1987  0.1077
0.458 45.8 %
0.1987
Data Point
Pi
l, Ai


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, 11th edition
Ans.
Chapter 2 Solutions, Page 3/22
Stress (psi)
45000
40000 f(x) = 30451576.81 x − 10.57
35000
30000
25000
20000
15000
10000
5000
0
0.000
0.001
0.001
0.002
Linear ()
Strain
Stress (psi)
(a) Linear range
50000
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
70000
Stress (psi)
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
Shigley’s MED, 11th edition
Chapter 2 Solutions, Page 4/22
(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  vs.  , the following equations are applied to the data.
P
 
A
Eq. (2-4)
l
 ln
for 0 l 0.0028 in (0 P 8 400 lbf )
l0
 ln
Eq. (2–9)
A0 
A0
A
for l  0.0028 in
(P  8400 lbf )
 (0.503) 2
0.1987 in 2
4
where
The results are summarized in the following table and plot. The last 5 points of data are
used to plot log  vs log  .
The curve fit gives
m = 0.2306
log 0 = 5.1852  0 = 153.2 kpsi
Ans.
The true strain corresponding to 20% cold work is given by Eq. (2–28) as
 1 
 1 
 ln 
 ln 
 0.2231
 1 W 
 1  0.2 
Eq. (2-30): S y  0 m 153.2(0.2231) 0.2306 108.4 kpsi
Ans.
Eq. (2-32), with Su 85.6 from Prob. 2-6,
S
85.6
Su  u 
107 kpsi
1  W 1  0.2
Shigley’s MED, 11th edition
Ans.
Chapter 2 Solutions, Page 5/22
l
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
0
0.000 4
0.000 6
0.001 0
0.001 3
0.002 3
0.002 8
Shigley’s MED, 11th edition
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

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
  
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
log 
-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 
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
Chapter 2 Solutions, Page 6/22
______________________________________________________________________________
2-8 Tangent modulus at  = 0 is
E

5000  0

25  106  psi
 0.2  10 3   0
Ans.
At  = 20 kpsi
Shigley’s MED, 11th edition
Chapter 2 Solutions, Page 7/22
 26  19   103 
E20 
 1.5  1  10 3 
 (10-3)
0
0.20
0.44
0.80
1.0
1.5
2.0
2.8
3.4
4.0
5.0
14.0  106  psi
Ans.
 (kpsi)
0
5
10
16
19
26
32
40
46
49
54
______________________________________________________________________________
2-9 W = 0.20,
(a) Before cold working: Annealed AISI 1018 steel. Table A-22, Sy = 32 kpsi, Su = 49.5

kpsi, 0 = 90.0 kpsi, m = 0.25, f = 1.05
m u 0.25
After cold working: Eq. (2-23),
 1 
 1 
20 ln 
 ln 
 0.223  u
 1 W 
 1  0.2 
Eq. (2-28):
Eq. (2-30):
S y  020m 90  0.223
Eq. (2-32):
S
49.5
Su   u 
61.9 kpsi
1  W 1  0.20
Su 49.5

1.55
Sy
32
0.25
61.8 kpsi
Ans.
93% increase
Ans.
Ans.
25% increase
Ans.
Su 61.9

1.002
S y 61.8
(b) Before:
After:
Ans.
Lost most of its ductility.
______________________________________________________________________________
2-10 W = 0.20,
(a) Before cold working: AISI 1212 HR steel. Table A-22, Sy = 28 kpsi, Su = 61.5 kpsi, 0

= 110 kpsi, m = 0.24, f = 0.85

After cold working: Eq. (2-23), m  u 0.24
Shigley’s MED, 11th edition
Chapter 2 Solutions, Page 8/22
Eq. (2-28):
 1
20 ln 
 1 W

 1 
 ln 
 0.223  u

 1  0.2 
Eq. (2-30):
S y  020m 110  0.223
Eq. (2-32):
S
61.5
Su   u 
76.9 kpsi
1  W 1  0.20
Su 61.5

2.20
Sy
28
0.24
76.7 kpsi
Ans.
174% increase
Ans.
Ans.
25% increase
Ans.
Su 76.9

1.002
S y 76.7
(b) Before:
After:
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, 0 = 100 kpsi, m = 0.15, f = 0.18

After cold working: Eq. (2-23), m  u 0.15
 1 
 1 
20 ln 
 ln 
 0.223   f
 1 W 
 1  0.2 
Eq. (2-28):
Material fractures.
Ans.
______________________________________________________________________________
2-12 For HB = 275, Eq. (2-36), Su = 3.4(275) = 935 MPa
Ans.
______________________________________________________________________________
2-13 Gray cast iron, HB = 200.
Eq. (2-37),
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. (2-36), 0.5HB = 100  HB = 200
Ans.
______________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 2 Solutions, Page 9/22
2-15 For the data given, converting HB to Su using Eq. (2-36)
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
Su =
Eq. (1-6)
Su 
S
u
N
1172
Su2 (kpsi)
13225
13456
13456
13689
13806.25
13806.25
13806.25
13924
13924
14280.25
Su2 = 137373
1172

117.2 117 kpsi Ans.
10
Eq. (1-7),
10
S
2
u
 NSu2
137373  10  117.2 
sSu 

1.27 kpsi
Ans.
N1
9
______________________________________________________________________________
2
i 1
2-16 For the data given, converting HB to Su using Eq. (2-37)
HB
230
232
232
234
235
235
235
236
236
239
Su =
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
414.12
Su2 = 17152.63
Eq. (1-6)
Shigley’s MED, 11th edition
Chapter 2 Solutions, Page 10/22
Su 
S
u
N

414.12
41.4 kpsi Ans.
10
Eq. (1-7),
10
S
2
u
 NSu2
17152.63  10  41.4 
sSu 

1.20
Ans.
N1
9
______________________________________________________________________________
2
i 1
uR 
2-17 (a) Eq. (2-16)
45.62
34.7 in lbf / in 3
2(30)
Ans.
(b) A0 = (0.5032)/4 = 0.19871 in2
l
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
(A0 / A) – 1
A
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
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

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
 = P/A0
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
Shigley’s MED, 11th edition
Ans.
Chapter 2 Solutions, Page 11/22
Shigley’s MED, 11th edition
Chapter 2 Solutions, Page 12/22
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:

For diameter,
F
F

S y
A   / 4 d 2

d
4F
 Sy
Weight/length = A, Cost/length = $/in = ($/lbf) Weight/length,
Deflection/length =  /L = F/(AE)
With F = 100 kips = 100(103) lbf,
Young's
Yield
Material Modulus Density Strength Cost/lbf Diameter
units
Mpsi
lbf/in3
kpsi
$/lbf
in
1020 HR
1020 CD
1040
4140
Al
Ti
30
30
30
30
10.4
16.5
0.282
0.282
0.282
0.282
0.098
0.16
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
Weight/
length
lbf/in
Cost/
length
$/in
0.9400
0.4947
0.3525
0.1709
0.1960
0.1333
0.25
0.15
0.12
0.14
0.22
$0.93
Deflection/
length
in/in
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
2
Al [ (1 in) / 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 A20, perform a Brinell hardness test, then use Eq. (2-36) to estimate an ultimate strength of
Shigley’s MED, 11th edition
Chapter 2 Solutions, Page 13/22
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
2
Al [ (1 in) / 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
W
9.0 lbf

0.318 lbf/in 3
Al [ (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   (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 This problem is for student research. No standard solution is provided.
______________________________________________________________________________
2-25 For strength,  = F/A = S  A = F/S
For mass, m = Al = (F/S) l
Shigley’s MED, 11th edition
Chapter 2 Solutions, Page 14/22
Thus,
f 3(M ) =  /S , and maximize S/
( = 1)
In Fig. (2-27), draw lines parallel to S/
The higher strength aluminum alloys have the greatest potential, as determined by
comparing each material’s bubble to the S/ guidelines.
Ans.
______________________________________________________________________________
2-26 For stiffness, k = AE/l  A = kl/E
For mass, m = Al = (kl/E) l =kl2  /E
Thus,
f 3(M) =  /E , and maximize E/
( = 1)
In Fig. (2-24), draw lines parallel to E/
Shigley’s MED, 11th edition
Chapter 2 Solutions, Page 15/22
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.
______________________________________________________________________________
2-27 For strength,
 = Fl/Z = S
(1)
where Fl is the bending moment and Z is the section modulus [see Eq. (3-26b)]. 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.
4  
For example, for a circular cross section, C =
Fl
S
CA3/2
Shigley’s MED, 11th edition

1
. Then, for strength, Eq. (1) is
 Fl 
A 

 CS 
2/3
(2)
Chapter 2 Solutions, Page 16/22
For mass,
Thus,
 Fl 
m  Al  

 CS 
2/3
F
l   
C
f 3(M) =  /S 2/3, and maximize S 2/3/
2/3
  
l 5/3  2/3 
S 
( = 2/3)
In Fig. (2-27), draw lines parallel to S 2/3/
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.
______________________________________________________________________________
2-28 Equation (2-41) 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. (2-42) becomes
Shigley’s MED, 11th edition
Chapter 2 Solutions, Page 17/22
 kl 3 
A 

 3CE 
and Eq. (2-44) becomes
1/2
 k 
m  Al  

 3C 
1/2
  
l 5/2  1/2 
E 
E1/2

M
f3  M   1/2
 . From Fig. (2-24)
E , or maximize
Thus, minimize
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.
______________________________________________________________________________
2-29 For stiffness, k = AE/l  A = kl/E
For mass, m = Al = (kl/E) l =kl2  /E
So,
f 3(M) =  /E, and maximize E/ . Thus,  = 1.
Ans.
______________________________________________________________________________
2-30 For strength,  = F/A = S  A = F/S
Shigley’s MED, 11th edition
Chapter 2 Solutions, Page 18/22
For mass, m = Al = (F/S) l
Ans.
So, f 3(M ) =  /S, and maximize S/ . Thus,  = 1.
______________________________________________________________________________
2-31 Equation (2-41) 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, C = (4)1
[see Eq. (2-41)]. 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. (2-42) becomes
 kl 3 
A 

 3CE 
and Eq. (2-44) becomes
1/2
 k 
m  Al  

 3C 
1/2
  
l 5/2  1/2 
E 
1/2
E

M
f3  M   1/2
 . Thus,  = 1/2. Ans.
E , or maximize
So, minimize
______________________________________________________________________________
2-32 For strength,
 = Fl/Z = S
(1)
where Fl is the bending moment and Z is the section modulus [see Eq. (3-26b)]. 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
4  
section, Z = d /(32)and the area is A = d /4. This leads to C =
3
2
1
. So, with
3/2
Z =C (A) , for strength, Eq. (1) is
Fl
S
CA3/2
For mass,
 Fl 
A 

 CS 

 Fl 
m  Al  

 CS 
2/3
F
l   
C
2/3
2/3
(2)
  
l 5/3  2/3 
S 
So, f 3(M) =  /S 2/3, and maximize S 2/3/. Thus,  = 2/3. Ans.
______________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 2 Solutions, Page 19/22
2-33 For stiffness, k=AE/l, or, A = kl/E.
Thus, m = Al = (kl/E )l = kl 2  /E. Then, M = E / and  = 1.
From Fig. 2-24, lines parallel to E / for ductile materials include steel, titanium,
molybdenum, aluminum alloys, and composites.
For strength, S = F/A, or, A = F/S.
Thus, m = Al = F/Sl = Fl  /S. Then, M = S/ and  = 1.
From Fig. 2-27, lines parallel to S/ 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.
______________________________________________________________________________
s
2-34 See Prob. 1-13 solution for x = 122.9 kcycles and x = 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.
ˆ
From Eq. (1-4), the probability density function (PDF), with   x and   sx , is
 1  x  x 2 
 1  x  122.9  2 
1
1
f ( x) 
exp   
exp   
  
 
sx 2
30.3 2
 2  sx  
 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.
Shigley’s MED, 11th edition
Chapter 2 Solutions, Page 20/22
Range
midpoint
(kcycles)
x
60
70
80
90
100
110
120
130
140
150
160
170
180
190
200
210
Frequency
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
0.02
0.02
0.01
0.01
Normal Distribution
Histogram
0.01
0.01
0.01
0
0
0
0 20 40 60 80 100 120 140 160 180 200 220
Shigley’s MED, 11th edition
Chapter 2 Solutions, Page 21/22
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).
Shigley’s MED, 11th edition
Chapter 2 Solutions, Page 22/22
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
100(KW-KB)/KW
1.1
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
 m Auscu  6.894757  25.4  Auscu ฀ 6.895  25.4  Auscu
kpsi in
Ans.
For music wire, from Table 10-4:
Auscu = 201 kpsiinm,
m = 0.145;
what is ASI?
ASI = 6.895(25.4)0.145 (201) = 2215 MPammm
Ans.
______________________________________________________________________________
10-3
Given: Music wire, d = 2.5 mm, OD = 31 mm, plain ground ends, Nt = 14 coils.
Shigley’s MED, 11th edition
Chapter 10 Solutions, Page 1/47
(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 
(c)
L0 
 L0 cr 
Ans.
A
2211

 1936 MPa
m
d
2.50.145
4C  2 4 11.4   2

 1.117
4C  3 4 11.4   3
 d 3 S sy
8K B D

  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
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.
Eq. (10-19): 3 ≤ Na ≤ 15 Na = 13
O.K.
F
167.9
Eq. (10-17):   s  1 
 1  0.29
F1
130
Shigley’s MED, 11th edition
Chapter 10 Solutions, Page 2/47
Eq. (10-20):   0.15,   0.29
From Eq. (10-7) for static service
O.K .
8(130)(28.5)
 8F1D 
 1.117
 674 MPa
3 
 (2.5)3
 d 
S
871.2
 1.29
n  sy 
1
674
1  K B 
Eq. (10-21):
ns ≥ 1.2, n = 1.29 O.K.
 167.9 
 167.9 
  674 
  870.5 MPa
 130 
 130 
S sy /  s  871.2 / 870.5 ฀ 1
 s  1 
Ssy/s ≥ (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
(b) Table 10-1:
Table 10-5:
Na = Nt  2 = 12  2 = 10 coils
G = 11.2 Mpsi
Eq. (10-9):
k
Ans.
0.24 11.2 106
d 4G

 28 lbf/in
8D3 N a
8  23 10
Fs = k ys = k (L0  Ls ) = 28(5  2.6) = 67.2 lbf
(c) Eq. (10-1):
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):
 s  KB
Table 10-4:
m = 0.187, A = 147 kpsiinm
Shigley’s MED, 11th edition
Ans.
8  67.2  2
8 Fs D
 1.135
 48.56 103  psi
3
3
d
  0.2 
Chapter 10 Solutions, Page 3/47
147
A

 198.6 kpsi
m
d
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.
 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
Eq. (10-9):
Table 10-1:
4
3
d 4G 4  77.2 10
Na 

 11.6 coils
8kD3 8  3.333 403
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):
 s  KB
Shigley’s MED, 11th edition
8  98.66  40
8 Fs D
 1.135
 178.2 MPa
3
d
  43 
Chapter 10 Solutions, Page 4/47
S sy
715.5
 4.02
Ans.
 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
140
A
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
KB 
Eq. (10-5):

 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
 d 3S sy / ns   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

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
Shigley’s MED, 11th edition
C = 10
O.K.
Chapter 10 Solutions, Page 5/47
Eq. (10-19):
3 ≤ Na ≤ 15
Na = 7
O.K.
Fs
18.78
1 
 1  0.14
F1
16.5
Eq. (10-20):   0.15,   0.14 not O.K . , but probably acceptable.
From Eq. (10-7) for static service
Eq. (10-17):
 
 8F D 
8(16.5)(0.8)
 74.5 103  psi  74.5 kpsi
 1  K B  1 3   1.135
3
d
(0.080)




n
Eq. (10-21):
Ssy
1

101.8
 1.37
74.5
ns ≥ 1.2, n = 1.37 O.K.
 18.78 
 18.78 
  74.5 
  84.8 kpsi
 16.5 
 16.5 
ns  S sy /  s  101.8 / 84.8  1.20
 s  1 
Eq. (10-21):
ns ≥ 1.2, ns = 1.2 It is solid-safe (basis of design). O.K.
Eq. (10-13) and Table 10-2: L0 ≤ (L0)cr
1.78 in  4.21 in O.K.
______________________________________________________________________________
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

 1.340
KB 
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
8D3 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
 s  K B s 3  1.340
 325.1103  psi
d
  0.0073 
(1)
A = 201 kpsiinm, m = 0.145
Shigley’s MED, 11th edition
Chapter 10 Solutions, Page 6/47
Eq. (10-14):
Table 10-6:
201
A

 412.7 kpsi
m
d
0.007 0.145
Ssy = 0.45 Sut = 0.45(412.7) = 185.7 kpsi
Sut 
s > Ssy, that is, 325.1 > 185.7 kpsi, the spring is not solid-safe. Return to Eq. (1) with
Fs = kys and s = Ssy /ns, and solve for ys, giving
 Ssy / ns   d 3 185.7 103  / 1.2   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

 1.169
KB 
4C  3 4  8.143  3
Na = Nt  2 = 16  2 = 14 coils
G = 6 Mpsi
0.014 4  6 106
d 4G
k

 1.389 lbf/in
8D3 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
 s  K B s 3  1.169
 47.42 103  psi
3
d
  0.014 
(1)
A = 145 kpsiinm, m = 0
145
A
 145 kpsi
Sut  m 
d
0.0140
Ssy = 0.35 Sut = 0.35(135) = 47.25 kpsi
s > Ssy, that is, 47.42 > 47.25 kpsi, the spring is not solid-safe. Return to Eq. (1) with
Fs = kys and s = Ssy /ns, and solve for ys, giving
 Ssy / ns   d 3 47.25 103  / 1.2   0.0143 
ys 

8 K B kD
The free length should be wound to
Shigley’s MED, 11th edition
8 1.169 1.389  0.114 
 0.229 in
Chapter 10 Solutions, Page 7/47
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

 1.385
KB 
4C  3 4  4   3
Na = Nt  2 = 11.2  2 = 9.2 coils
G = 10 Mpsi
0.050 4 10 106
d 4G
k

 106.1 lbf/in
8D3 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
 s  K B s 3  1.385
 71.8 103  psi
d
  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.
 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)
4C  2 4 13.32   2

 1.099
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
8D3 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
Shigley’s MED, 11th edition
Chapter 10 Solutions, Page 8/47
8  39.20 1.972
8 Fs D
 1.099
 66.7 103  psi
3
3
d
  0.148 
Eq. (10-7):
 s  KB
Table 10-4:
A = 140 kpsiinm, m = 0.190
140
A
 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
 1.36 Spring is solid-safe (ns > 1.2)
Ans.
 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. (10-1):
Eq. (10-5):
Table 10-1:
C = D/d = 0.782/0.138 = 5.667
4C  2 4  5.667   2

 1.254
KB 
4C  3 4  5.667   3
Na = Nt  2 = 12  2 = 10 coils
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
 s  K B s 3  1.254
 124.7 103  psi
d
  0.1383 
(1)
A = 147 kpsiinm, m = 0.187
A
147
 212.9 kpsi
Sut  m 
d
0.1380.187
Ssy = 0.50 Sut = 0.50(212.9) = 106.5 kpsi
s > Ssy, that is, 124.7 > 106.5 kpsi, the spring is not solid-safe. Return to Eq. (1) with
Fs = kys and s = Ssy /ns, and solve for ys, giving
 Ssy / ns   d 3 106.5 103  / 1.2   0.1383 
ys 
8 K B kD

8 1.254 109.0  0.782 
 0.857 in
The free length should be wound to
Shigley’s MED, 11th edition
Chapter 10 Solutions, Page 9/47
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)

4
13.86
2

  1.095
4C  2

KB 
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
 s  K B s 3  1.095
 110.1 103  psi
d
  0.1853 
(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 sy 112.2

 1.02 Spring is not solid-safe (ns < 1.2)
ns 
 s 110.1
Return to Eq. (1) with Fs = kys and s = Ssy /ns, and solve for ys, giving
3
3
S sy / ns   d 3 112.2 10  / 1.2    0.185 


 5.109 in
ys 
8 K 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)
4C  2 4  2.8   2
Eq. (10-5):

 1.610
KB 
4C  3 4  2.8   3
Shigley’s MED, 11th edition
Chapter 10 Solutions, Page 10/47
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  0.7 3  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
 s  K B s 3  1.610
 1303 MPa
d
  0.253 
(1)
Table 10-4 (dia. less than table):
A = 1867 MPammm, m = 0.146
1867
A
Eq. (10-14): Sut  m 
 2286 MPa
d
0.250.146
Table 10-6: Ssy = 0.35 Sut = 0.35(2286) = 734 MPa
s > Ssy, that is, 1303 > 734 MPa, the spring is not solid-safe. Return to Eq. (1) with
Fs = kys and s = Ssy /ns, and solve for ys, giving
 Ssy / ns   d 3  734 / 1.2    0.253 

 1.22 mm
8 K B kD
8 1.610  2.728  0.7 
The free length should be wound to
ys 
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):

 1.368
KB 
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, 11th edition
Chapter 10 Solutions, Page 11/47
8  60.03 5.3
8 Fs D
 1.368
 641.4 MPa
3
d
 1.23 
Eq. (10-7):
 s  KB
Table 10-4:
A = 2211 MPammm, m = 0.145
2211
A
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.
 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

 1.098
KB 
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
 s  K B s 3  1.098
 702.8 MPa
d
  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)
 s 702.8
Return to Eq. (1) with Fs = kys and s = Ssy /ns, and solve for ys, giving
S sy / ns   d 3
 734 / 1.2    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, 11th edition
Chapter 10 Solutions, Page 12/47
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

 1.192
KB 
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
 s  K B s 3  1.192
 165.2 MPa
d
  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 sy 299.5

 1.81 Spring is solid-safe (ns > 1.2) Ans.
ns 
 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, 11th edition
Chapter 10 Solutions, Page 13/47
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
 s  K B s 3  1.092
 832 MPa
d
  4.53 
(1)
A = 2005 MPammm, m = 0.168
2005
A
 1557 MPa
Sut  m 
d
4.50.168
Ssy = 0.50 Sut = 0.50(1557) = 779 MPa
s > Ssy, that is, 832 > 779 MPa, the spring is not solid-safe. Return to Eq. (1) with
Fs = kys and s = Ssy /ns, and solve for ys, giving
 Ssy / ns   d 3  779 / 1.2    4.53 

 139.5 mm
8 K B kD
8 1.092  2.357  64.7 
The free length should be wound to
ys 
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, 11th edition
Chapter 10 Solutions, Page 14/47
4(13.81)  2
 1.096
4(13.81)  3
8F D
8(18.2)(1.865)
 38.5 103  psi  38.5 kpsi
 s  K B s 3  1.096
3
d
  0.135 
KB 
Ans.
______________________________________________________________________________
10-21 Given: Plain end, hard drawn steel, 12 gauge W & M wire, OD = 0.75 in, Nt = 20 coils,
L0 = 3.75 in.
Table A-28: d = 0.1055 in
(a) D = OD  d = 0.75  0.1055 = 0.6445 in. C = D / d = 0.6445/0.1055 = 6.109 Ans.
(b) Table 10-1: Na = Nt = 20 coils,
p = (L0  d)/ Na = (3.75  0.1055)/20 = 0.1822 in/coil Ans.
(c) Table 10-1: Ls = d (Nt + 1) = 0.1055 (20 + 1) = 2.2155 in,
ys = L0  Ls = 3.75  2.2155 = 1.5345 in
Ans.
(d) Eq. (10-8):
0.10554 11.5 106 1.5345 
d 4Gys

 50.36 lbf
Fs 
Ans.
1 



3
1
8D N 1  2  8  0.6445 3  20  1 
2
 2C 
 2  6.109  
(e) Eq. (10-5): K B 
Eq. (10-7):
 s  KB
4C  2 4  6.109   2

 1.233
4C  3 4  6.109   3
8  50.36  0.6445
8 Fs D
 1.233
 86.8 103  psi  86.8 kpsi
3
3
d
  0.1055 
(f) Table 10-4 and Eq. (10-14): Sut 
Ans.
140
A

 214.6 kpsi
0.190
m
d
 0.1055 
Table 10-6: Ssy = 0.45 Sut = 0.45(214.6) = 96.57 kpsi.
S sy 96.57
ns 

 1.11
Ans.
s
86.8
(g) Exact, k = Fs / ys = 50.36/1.5345 = 32.82 lbf/in
Ans.
4
4
0.1055 11.5 106
d G
Approximate using Eq. (10-9): k 

 33.26 lbf/in
3
8D3 N
8  0.6445  20
Ans.
Approximate is 1.34 percent higher than the exact.
Ans.
______________________________________________________________________________
10-22 Given: Squared and ground, oil tempered steel, d = 3 mm, OD = 30 mm, Nt = 32 coils,
L0 = 240 mm.
Table A-28: d = 0.1055 in
(a) D = OD  d = 30  3 = 27 mm. C = D / d = 27/3 = 9
Ans.
(b) Table 10-1: Na = Nt  2 = 32  2 = 30 coils,
p = (L0  2d)/ Na = [240  2(3)] / 30 = 7.8 mm/coil
Ans.
(c) Table 10-1: Ls = d Nt = 3 (32) = 96 mm,
ys = L0  Ls = 240  96 = 144 mm
Ans.
Shigley’s MED, 11th edition
Chapter 10 Solutions, Page 15/47
(d) Eq. (10-8):
 0.003 77.2 109  0.144
d 4Gys

 189.45 N
Fs 
Ans.
1 



3
1
3
8D N 1 
2 
8  0.027   30  1 
2
 2C 
 2  9  
4C  2 4  9   2
(e) Eq. (10-5): K B 

 1.152
4C  3 4  9   3
Eq. (10-7):
8 189.45  0.027
8F D
 s  K B s 3  1.152
106   555.8 MPa
Ans.

3
d
  0.003
(f) Table 10-4 and Eq. (10-14): Sut 
4
1855
A

 1510.5 MPa
0.187
m
d
 3
Table 10-5: Ssy = 0.45 Sut = 0.45(1510.5) = 679.7 MPa.
S sy 679.7
ns 

 1.22
Ans.
 s 555.8
(g) Exact, k = Fs / ys = 189.45/144 = 1.316 N/mm
Ans.
Approximate using Eq. (10-9):
4
 0.003 77.2 109  3
d 4G

k
10   1.324 N/mm
3
8D3 N
8  0.027  30
Ans.
Approximate is 0.61 percent higher than the exact.
Ans.
______________________________________________________________________________
10-23 y = 50 mm, F = 90 N, k = F / y = 90/50 = 1.8 N/mm Ans.
ys = 60 mm, Fs = k ys = 1.8(60) = 108 N.
A 2065
Sut  m  0.263
Eq. (10-14), Table 10-4, assume 2.5  d  5 mm:
d
d
2065 722.75
(1)
Table 10-6 (includes KB): S ys  0.35Sut  0.35 0.263  0.263
d
d
Eq. (10-7) (with ns = 1.2 but without KB):
8n F D 8n F C 8 1.2 108 10  3300
 max  s s3  s s2 
 2 MPa
(2)
d
d
d
d2
722.75 3300
3300
 d 1.737 
 d  2.40 mm
Equate Eqs. (1) and (2), 0.263  2
d
d
722.75
Since this is less than 2.5 mm, return to Eq. (10-14), Table 10-4, for 0.3  d  2.5 mm:
653.45
A 1867

(1)
Sut  m  0.146
S ys  0.146
d
d
d
Again, equate Eqs. (1) and (2),
653.45 3300
3300
 2
 d 1.854 
 d  2.40 mm
Ans.
0.146
d
d
653.45
The final factor of safety is
Shigley’s MED, 11th edition
Chapter 10 Solutions, Page 16/47
653.45 / d 0.146 653.45  2.40 
ns 


 1.20
Ans.
s
8 108 10
 8 Fs C 

2 
 d 
OD =Cd + d = (C + 1) d = 11(2.40) = 26.4 mm
Ans.
ID = (C  1) d = 9(2.40) = 21.6 mm
Ans.
k = 1.8 N/mm
Ans. (found earlier)
Table 10-5, G = 69.0 GPa, Eq. (10-9):
2.4 103  69 109 
d 4G
dG
Na 


 11.5 coils
8 D3k 8C 3 k
8 103 1.8 103 
1.854
S ys
Table 10-1:
Nt = Na + 2 = 13.5 coils
Ans.
Ls = d (Nt + 1) = 2.4(13.5 + 1) = 34.8 mm Ans.
L0 = Ls + ys = 34.8 + 60 = 94.8 mm
Ans.
10  2.4 
D
Eq. (10-13):   2.63  2.63
 0.666
L0
94.8
Stable if supported between fixed-fixed ends. Otherwise would need to be supported by
hole or rod.
______________________________________________________________________________
10-24 Phosphor-bronze, closed ends, C = 10, at y = 2 in F = 15 lbf, ys = 3 in, ns = 1.2.
k = F / y = 15/2 = 7.5 lbf/in, Fs = k ys = 7.5 (3) = 22.5 lbf.
Eq. (10-14) with Table 10-4 assuming 0.022  d  0.075 in, A = 121 kpsi٠inm and m =
0.028. Then, from Table 10-5, Sys = 0.45 Sut:
121 54.45
(1)
S ys  0.45 0.028  0.028
d
d
4C  2 4 10   2

 1.135
4C  3 4 10   3
Eq. (10-5): K B 
Eq. (10-7) with ns = 1.2:
8  22.5 10
8F C
0.7804
 max  ns K B s 2  1.2 1.135 
103  

2
d
d
d2
(2)
Where max is in kpsi. Equating (1) and (2) gives
54.45 0.7804

d 0.028
d2
 d 1.972 
0.7804
54.45

d  0.116 in
Returning to Table 10-4, use A =110 kpsi٠inm and m = 0.028 for 0.075  d  0.3 in,
S ys  0.45
110
49.5 0.7804
0.7804
 0.064 
 d 1.936 
 d  0.117 in
0.064
2
d
d
d
49.5
Table A-17, select the preferred size of:
Shigley’s MED, 11th edition
d = 0.12 in
Ans.
Chapter 10 Solutions, Page 17/47
Check ns, ns 
S ys
s

49.5 /  d 0.064 
 FC

K B  8 s 2 103 
 d


49.5 103   d 1.936
49.5 103    0.12 
(3)
8 K B Fs C
1.936
ns 
8 1.135  22.5 10 
 1.26
Ans.
OD = Cd + d = (C + 1) d = (10 + 1)0.12 = 1.32 in
Ans.
ID = (C  1) d = (10  1)0.12 = 1.08 in
Found earlier, k = 7.5 lbf/in
Ans.
Table 10-5, G = 6 Mpsi. Eq. (10-9): N a 
Table 10-1:
0.12  6 106
dG

 12 coils
8C 3 k
8 103  7.5
Nt = Na + 2 = 14 coils
Ans.
Ls = d (Nt + 1) = 0.12(14 + 1) = 1.8 in
L0 = Ls + ys = 1.8 + 3 = 4.8 in
Ans.
Ans.
Eq. (10-13):  = 2.63 D / L0 = 2.63 (10)0.12/4.8 = 0.658
Table 10-2: Stable if supported between fixed-fixed ends. Otherwise would need to be
supported by hole or rod.
______________________________________________________________________________
10-25 From Prob. 10-24, d = 0.12 in and from Eq. (3),
1.936
49.5 103   d 1.936 49.5 103    0.12 
C

 10.46
Ans.
8ns K B Fs
8 1.2 1.135  22.5 
OD = (C + 1) d = 11.46(0.12) = 1.375 in
Ans.
ID = (C  1) d = 9.46(0.12) = 1.135 in
Ans.
Table 10-5, G = 6 Mpsi, Eq. (10-9): N a 
Table 10-1:
0.12  6 106
dG

 10.5 coils
8C 3 k 8 10.463  7.5
Nt = Na + 2 = 12.5 coils
Ans.
Ls = d (Nt + 1) = 0.12(12.5 + 1) = 1.62 in
Ans.
L0 = Ls + ys = 1.62 + 3 = 4.62 in
Ans.
Eq. (10-13):  = 2.63 D / L0 = 2.63 (10.46)0.12/4.62 = 0.715
Shigley’s MED, 11th edition
Chapter 10 Solutions, Page 18/47
Table 10-2: Stable if supported between fixed-fixed ends, or one end on flat surface and
other end hinged. Otherwise would need to be supported by hole or rod.
______________________________________________________________________________
10-26 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 s, F = Fs = 20(1 + ) = 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
s
s
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
______________________________________________________________________________
10-27 In Prob. 10-26, 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. 10-26. For part (b), use
OD = D + d = 0.950 in.
(3)
Shigley’s MED, 11th edition
Chapter 10 Solutions, Page 19/47
and,
d
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)
s
81.167
95.223
ns = Ssy/s
ns
1.580
1.358
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)
s
93.643
ns = Ssy/s
ns
1.381
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. 10-26, this was due to the initial
values picked and not the approach.
______________________________________________________________________________
10-28 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
24 (79.3)103
d 4G
Na 

 15.9 coils
Eq. (10-9):
8kD 3 8(4.286)13.253
Assume squared and closed.
Table 10-1: Nt = Na + 2 = 15.9 + 2 = 17.9 coils
Ls = dNt = 2(17.9) =35.8 mm
Shigley’s MED, 11th edition
Chapter 10 Solutions, Page 20/47
Eq. (10-5):
ys = L0  Ls = 48  35.8 = 12.2 mm
Fs = kys = 4.286(12.2) = 52.29 N
4C  2 4  6.625   2

 1.213
KB 
4C  3 4  6.625   3
 8(52.29)13.25 
8Fs D
1.213


  267.5 MPa
d3
  23 


Eq. (10-7):
 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
s

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.
______________________________________________________________________________
10-29 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, 11th edition
Chapter 10 Solutions, Page 21/47
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/s
C
d
d
D
C
Na
Nt
Ls
ys
Fs
A
m
Sut
Ssy
KB
s
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. 10-28, is
that once d is calculated, the closest wire size must be selected. Iterating on d uses
available wire sizes from the beginning.
______________________________________________________________________________
10-30 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.
• It is better to familiarize yourself with vendor resources rather than invent them
yourself.
• Sample catalog pages can be given to students for study.
______________________________________________________________________________
10-31 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:
Table 10-5:
Eq. (10-9):
Shigley’s MED, 11th edition
Na = Nt  2 = 30  2 = 28 coils
G = 11.5 Mpsi
0.0667 4 11.5 106
d 4G

 3.424 lbf/in
k
8D3 N a
8  0.667 3  28
Ans.
Chapter 10 Solutions, Page 22/47
(d) Table 10-4:
Eq. (10-14):
Table 10-6:
Eq. (10-5):
A = 140 kpsiinm, m = 0.190
140
A
 234.2 kpsi
Sut  m 
d
0.0667 0.190
Ssy = 0.45 Sut = 0.45 (234.2) = 105.4 kpsi
Fs = kys = 3.424(3) = 10.27 lbf
4C  2 4 10   2

 1.135
KB 
4C  3 4 10   3
Eq. (10-7):
 s  KB
8 10.27  0.667
8Fs D
1.135

d3
  0.06673 
 66.72 103  psi  66.72 kpsi
S sy
105.4
 1.58
Ans.
 s 66.72
(e) a = m = 0.5s = 0.5(66.72) = 33.36 kpsi, r = a / m = 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 obtained using Eqs. (10-28)
and (10-29b).
S sa
35
S se 

 39.9 kpsi
2
2
1   S sm / S su  1   55 /156.9 
The Gerber fatigue criterion from Eq. (6-48), adapted for shear,
2
2

 2 m S se  
1  Ssu   a 
1  1  
nf  


2   m  S se 
Ssu a  



2

2
 2  33.36  39.9  
1  156.9  33.36 
 
1  1  


2  33.36  39.9 
156.9  33.36   



 1.13
Ans.
______________________________________________________________________________
10-32 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.
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:
Shigley’s MED, 11th edition
Na = Nt  2 = 9  2 = 7 coils
Chapter 10 Solutions, Page 23/47
(c) Table 10-5:
Eq. (10-9):
(d)
G = 11.75 Mpsi
k
0.14 11.75 10 6
d 4G

 40.98 lbf/in
8D3 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):
 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
8 Fs D

1.172
 195.7 103  psi  195.7 kpsi
3
3
d
  0.1 
201
A
 0.145  280.7 kpsi
m
d
0.1
S sy
s

126.3
 0.645
195.7
Ans.
(e) a = m = s /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 obtained using Eqs. (10-28)
and (10-29b).
S sa
35
S se 

 38.3 kpsi
2
2
1   S sm / S su  1   55 / 188.1
The Gerber fatigue criterion from Eq. (6-48), adapted for shear,
2
2

 2 m S se  
1  S su   a 
nf  
1  1  


S su a  
2   m  S se 



2

2
 2  97.85  38.3  
1  188.1  97.85 
 
1  1  


2  97.85  38.3 
188.1 97.85   



 0.38
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.
Shigley’s MED, 11th edition
Chapter 10 Solutions, Page 24/47
______________________________________________________________________________
10-33 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
y
1
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
147
A
 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):
 s  KB
(e) Table 10-4:
Eq. (10-14):
Fs 
Ans.
4C  2 4  7   2

 1.2
4C  3 4  7   3
8 Fs D
 S sy
d3
 d 3 S sy
8K B D

  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:
Shigley’s MED, 11th edition
Ls = Nt d = 8.46(0.2375) = 2.01 in
Chapter 10 Solutions, Page 25/47
L0 = Ls + ys = 2.01 + 1.69 = 3.70 in
Ans.
______________________________________________________________________________
10-34 For a coil radius given by:
R  R1 
R2  R1

2 N
The torsion of a section is T = PR where dL = R d
U
T
1
1 2 N 3

T
dL 
PR d

P
P
GJ
GJ 0
3
P 2 N 
R2  R1 

  d
 R1 
2 N
GJ 0 

P 
P

GJ

2 N
4
R2  R1  
 1   2 N  
 
  R1 
 
2 N
 4   R2  R1  
  0
 PN
2GJ ( R2
R
 R)
4
2
 R14  
 PN
( R1  R2 )  R12  R22 
2GJ
PN
16
J 
d4  p 
( R1  R2 )  R12  R22 
4
Gd
32
P
d 4G

k 
Ans.
 P 16 N ( R1  R2 )  R12  R22 

1
______________________________________________________________________________
10-35 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
Ssu = 0.67Sut = 163.7 kpsi,
Ssy = 0.35Sut = 85.5 kpsi
The Gerber ordinate intercept for the Zimmerli data is obtained using Eqs. (10-28) and
(10-29b).
S se 
S sa
35

 39.5 kpsi
2
1  (S sm / S su )
1  (55 / 163.7) 2
Shigley’s MED, 11th edition
Chapter 10 Solutions, Page 26/47
Let r = a /m = 7/11. The Gerber fatigue criterion from Eq. (6-48), adapted for shear,
2

 2S se  
1 S su2 r 
1  1  
nf 

2  m S se 
S su r  



Solving for m gives,
2
2
2


 2  39.5   
 2 S se   1 163.7   7 / 11 
1 S su2 r 
m 
1  1  
1  1  

 
S su r   2 1.5  39.5 
2 n f S se 
163.7  7 / 11  






 36.70 kpsi
But,
8 F C 4C  2  8 FmC 
m  KB m2 


4C  3   d 2 
d
8 11103
8F
Let  = m = 36.70 kpsi, and   m2 
 4.377 kpsi From Eq. (10-23),
2
d
  0.08 
2
 2    3
2  
C
 
 
4
 4  4

2  36.70   4.377
4  4.377 
2
 2  36.70   4.377  3  36.70 
 
 6.98
 
4  4.377 

 4  4.377 
D = Cd = 6.98(0.08) = 0.558 in
4C  2 4(6.98)  2

 1.201
4C  3
4(6.98)  3
 8(7)(0.558) 3 
 8F D 
 a  K B  a 3   1.201 
(10 )   23.3 kpsi
3
 d 
  (0.08 )

KB 
The Gerber fatigue criterion from Eq. (6-48), adapted for shear,
2
2

 2 m S se  
1  S su   a 
nf  
1  1  


2   m  S se 
 S su a  


2
2
1  163.7  23.3 
  11  39.5  
 
 1  1   2  

 
2  36.6  39.5 
  7  163.7  


 1.50
checks
4
Gd
10(106 )(0.08) 4
Na 

 31.02 coils
8kD3
8(9.5)(0.558)3
Shigley’s MED, 11th edition
Chapter 10 Solutions, Page 27/47
Nt = 31.02 + 2 = 33 coils
Ls = dNt = 0.08(33) = 2.64 in
ymax = Fmax / k = 18 / 9.5 = 1.895 in
ys = (1 + ) ymax = (1 + 0.15)(1.895) = 2.179 in
L0 = Ls + ymax = 2.64 + 2.179 = 4.819 in
(L0)cr = 2.63 D /  = 2.63(0.558) / 0.5 = 2.935 in
s = (1 + )( Fmax /Fa) a = 1.15(18/7)23.3 = 68.8 kpsi
ns = Ssy / s = 85.5/68.9 = 1.24
kg
9.5(386)

 109 Hz
2
2
 d DN a
 (0.08 )(0.558)(31.02)(0.283)
These steps are easily implemented on a spreadsheet, as shown below, for different
diameters.
f 
2
2
d
m
A
d1
d2
d3
d4
0.080
0.092
0.106
0.121
0.146
0.146
0.263
0.263
169.000 169.000 128.000 128.000
Sut
244.363 239.618 231.257 223.311
Ssu
163.723 160.544 154.942 149.618
Ssy
85.527
83.866
80.940
78.159
Ssa
35.000
35.000
35.000
35.000
Sse
39.452
39.654
40.046
40.469
m


C
D
36.667
36.667
4.377
6.977
0.558
36.667
36.667
3.346
9.603
0.879
36.667
36.667
2.517
13.244
1.397
36.667
36.667
1.929
17.702
2.133
KB
1.201
1.141
1.100
1.074
a
23.333
23.333
23.333
23.333
nf
1.500
1.500
1.500
1.500
Na
30.993
13.594
5.975
2.858
Nt
32.993
15.594
7.975
4.858
Ls
2.639
1.427
0.841
0.585
ys
2.179
2.179
2.179
2.179
Shigley’s MED, 11th edition
Chapter 10 Solutions, Page 28/47
L0
4.818
3.606
3.020
2.764
(L0)cr
2.936
4.622
7.350
11.220
69.000
69.000
69.000
69.000
1.240
1.215
1.173
1.133
s
ns
f,
(Hz)
108.895 114.578 118.863 121.775
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.
______________________________________________________________________________
10-36 The steps are the same as in Prob. 10-35 except that the Gerber-Zimmerli criterion is
replaced with the Goodman-Zimmerli relationship of Eq. (10-29a) :
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
below (see solution to Prob. 10-35 for additional details).
Iteration of d for the first trial
d
d1
0.080
d2
0.0915
d3
0.1055
d4
0.1205
d1
0.080
d2
0.0915
d3
0.1055
d4
0.1205
m
A
0.146
169.000
0.146
169.000
0.263
128.000
0.263
128.000
KB
a
1.151
29.008
1.108
29.040
1.078
29.090
1.058
29.127
Sut
Ssu
244.363
163.723
239.618
160.544
231.257
154.942
223.311
149.618
nf
Na
1.500
14.191
1.500
6.456
1.500
2.899
1.500
1.404
Ssy
Ssa
85.527
35.000
83.866
35.000
80.940
35.000
78.159
35.000
Nt
Ls
16.191
1.295
8.456
0.774
4.899
0.517
3.404
0.410
Sse
m


52.706
45.585
53.239
45.635
54.261
45.712
55.345
45.771
ys
L0
2.179
3.474
2.179
2.953
2.179
2.696
2.179
2.589
45.585
4.377
45.635
3.346
45.712
2.517
45.771
1.929
(L0)cr
3.809
85.782
5.924
85.876
9.354
86.022
14.219
86.133
C
D
9.052
0.724
12.309
1.126
16.856
1.778
22.433
2.703
ns
f, (Hz)
d
s
0.997
0.977
0.941
0.907
141.284 146.853 151.271 154.326
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.
Shigley’s MED, 11th edition
Chapter 10 Solutions, Page 29/47
Iteration of d for the second trial
d
m
d1
0.080
0.146
d2
0.0915
0.146
d3
0.1055
0.263
d4
0.1205
0.263
A
Sut
169.000
244.363
169.000
239.618
128.000
231.257
128.000
223.311
a
Ssu
Ssy
163.723
85.527
160.544
83.866
154.942
80.940
Ssa
Sse
35.000
52.706
35.000
53.239
m


34.188
34.188
4.377
6.395
0.512
C
D
d1
0.080
1.221
d2
0.0915
1.154
d3
0.1055
1.108
d4
0.1205
1.079
nf
21.756
2.000
21.780
2.000
21.817
2.000
21.845
2.000
149.618
78.159
Na
Nt
40.243
42.243
17.286
19.286
7.475
9.475
3.539
5.539
35.000
54.261
35.000
55.345
Ls
ys
3.379
2.179
1.765
2.179
1.000
2.179
0.667
2.179
34.226
34.226
34.284
34.284
34.329
34.329
L0
(L0)cr
5.558
2.691
3.944
4.266
3.179
6.821
2.846
10.449
3.346
8.864
0.811
2.517
12.292
1.297
1.929
16.485
1.986
s
d
KB
ns
f, (Hz)
64.336
1.329
99.816
64.407 64.517 64.600
1.302
1.255
1.210
105.759 110.312 113.408
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 = 3.944 in, and
.Nt = 19.3 turns. Ans.
______________________________________________________________________________
10-37 This is the same as Prob. 10-35 since Ssa = 35 kpsi. Therefore, 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.
______________________________________________________________________________
10-38 For the Gerber-Zimmerli fatigue-failure criterion, Ssu = 0.67Sut ,
S sa
,
S se 
1  (S sm / S su ) 2

1 S su2 r 
m 
1  1 
2 n f Sse 

2
 2S se  


 S su r  

See the process used in Prob. 10-36. The last 2 columns of diameters of Ex. 10-5 are
presented below with additional calculations.
Shigley’s MED, 11th edition
Chapter 10 Solutions, Page 30/47
d
0.105
0.112
d
0.105
0.112
Sut
278.691 276.096
Nt
10.915
8.190
Ssu
186.723 184.984
Ls
1.146
0.917
L0
3.446
3.217
Sse
38.325
38.394
m


38.508
38.502 (L0)cr
6.630
8.160
38.508
38.502
KB
1.111
1.095
2.887
2.538
a
23.105
23.101
C
12.004
13.851
nf
1.500
1.500
D
1.260
1.551
s
70.855
70.844
ID
1.155
1.439
ns
1.770
1.754
OD
1.365
1.663
fn
Na
8.915
6.190
fom
105.433 106.922
-0.973
-1.022
There are only slight changes in the results.
______________________________________________________________________________
10-39 As in Prob. 10-38, the basic change is Ssa.
S sa
1  (S sm / S su )
Recalculate m using Eq. (6-41) for shear. That is,
For Goodman, using Eq. (10-29a):

 
nf   a  m 
 S se S su 
1

S se 
S se S su
S se S su

 a S su   m S se  m  rS su  S se 
Where r = a / m. Thus,
m 
S se S su
n f  rS su  S se 
See the process used in Prob. 10-36. Calculations for the last 2 diameters of Ex. 10-5 are
given below.
Shigley’s MED, 11th edition
Chapter 10 Solutions, Page 31/47
d
0.105
0.112
d
0.105
0.112
Sut
278.691 276.096
Nt
11.153
8.353
Ssu
186.723 184.984
Ls
1.171
0.936
L0
3.471
3.236
Sse
49.614
49.810
m


38.207
38.201 (L0)cr
6.572
8.090
38.207
38.201
KB
1.112
1.096
2.887
2.538
a
22.924
22.920
C
11.899
13.732
nf
1.500
1.500
D
1.249
1.538
s
70.301
70.289
ID
1.144
1.426
ns
1.784
1.768
OD
1.354
1.650
fn
Na
9.153
6.353
fom
104.509 106.000
-0.986
-1.034
There are only slight differences in the results.
______________________________________________________________________________
10-40 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
 A  max2 [( K ) A (16C )  4]  y
ny
d
 d 2S y
4C 2  C  1
(16C )  4 
n y Fmax
4C (C  1)
  d 2S y

4C 2  C  1  (C  1) 
 1
 4ny Fmax



2
2


 d Sy
1
1   d Sy
C 2  1 
 1 C  
 2  0


4 
4n y Fmax
4  4n y Fmax


Shigley’s MED, 11th edition
Chapter 10 Solutions, Page 32/47
2


2
  d 2S y 
 d 2S y
1   d Sy
 

 2  take positive root
C 
 16n F 

2 16n y Fmax
4
n
F
y max 
y max



1   (0.067 2 )(175.5)(103 )
 
2
16(1.5)(18)
2

  (0.067) 2 (175.5)(103 ) 
 (0.067) 2 (175.5)(103)

2
 


  4.590

16(1.5)(18)
4(1.5)(18)



D  Cd  4.59  0.067   0.3075 in
Fi 
 d 3 i
8D

d3 
33 500
C  3 

 1000  4 


8D  exp(0.105C )
6.5  

Use the lowest Fi in the preferred range. This results in the best fom.
Fi 
 (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
d 4G
(0.067) 4 (11.5)(10 6 )
Na 

 45.28 turns
8kD3
8(22)(0.3075)3
G
11.5
Nb  N a 
 45.28 
 44.88 turns
E
28.6
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 
 max 
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

d3
 (0.067)3
(n y ) body 
S sy
 max

105.3
 1.70
62.1
r2  2d  2(0.067)  0.134 in,
(K )B 
C2 
4C2  1
4(4)  1

 1.25
4C2  4
4(4)  4
Shigley’s MED, 11th edition
2r2
2(0.134)

4
d
0.067
Chapter 10 Solutions, Page 33/47
 8(18)(0.3075)  3
 8Fmax D 
 1.25 
 (10 )  58.58 kpsi
3 
3
 d 
  (0.067) 
S
105.3
(ny ) B  sy 
 1.80
B
58.58
 B  (K ) B 
fom  (1)
 2d 2 ( N b  2) D
4

 2 (0.067) 2 (44.88  2)(0.3075)
4
 0.160
Several diameters, evaluated using a spreadsheet, are shown below.
d
Sut
Ssy
Sy
C
D
Fi (calc)
Fi (rd)
k
Na
Nb
L0
L18 lbf
KB
max
(ny)body
B
(ny)B
(ny)A
fom
0.067
233.97
7
105.29
0
175.48
3
4.589
0.307
6.505
7.0
22.000
45.29
44.89
3.556
4.056
1.326
62.118
1.695
58.576
1.797
1.500
0.160
0.072
230.79
9
103.86
0
173.10
0
5.412
0.390
5.773
6.0
24.000
27.20
26.80
2.637
3.137
1.268
60.686
1.711
59.820
1.736
1.500
0.144
0.076
228.441
102.798
171.331
6.099
0.463
5.257
5.5
25.000
19.27
18.86
2.285
2.785
1.234
59.707
1.722
60.495
1.699
1.500
-0.138
0.081
225.69
2
101.56
1
169.26
9
6.993
0.566
4.675
5.0
26.000
13.10
12.69
2.080
2.580
1.200
58.636
1.732
61.067
1.663
1.500
0.135
0.085
223.63
4
100.63
5
167.72
6
7.738
0.658
4.251
4.5
27.000
9.77
9.36
2.026
2.526
1.179
57.875
1.739
61.367
1.640
1.500
0.133
0.09
0.095
0.104
221.21 218.95 215.22
9
8
4
99.548 98.531 96.851
165.91
4
8.708
0.784
3.764
4.0
28.000
7.00
6.59
2.071
2.571
1.157
57.019
1.746
61.598
1.616
1.500
0.135
164.21
8
9.721
0.923
3.320
3.5
29.000
5.13
4.72
2.201
2.701
1.139
56.249
1.752
61.712
1.597
1.500
0.138
161.41
8
11.650
1.212
2.621
3.0
30.000
3.15
2.75
2.605
3.105
1.115
55.031
1.760
61.712
1.569
1.500
0.154
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.
______________________________________________________________________________
10-41 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
or
Shigley’s MED, 11th edition
Ans.
2d + L0 = 2(0.162) + 16.12 = 16.45 in overall
Chapter 10 Solutions, Page 34/47
D 1.338

 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 
 i  K B  i 3   1.166
 14 950 psi
 (0.162)3
 d 
(c) From Table 10-5 use: G = 11.4(106) psi and E = 28.5(106) psi
(b)
C 
11.4
G
 84 
 84.4 turns
28.5
E
(0.162) 4 (11.4)(106 )
d 4G
k 

 4.855 lbf/in
8D 3 N a
8(1.338)3 (84.4)
(d) Table 10-4:
A = 147 psi · inm ,
m = 0.187
147
Sut 
 207.1 kpsi
(0.162)0.187
S y  0.75(207.1)  155.3 kpsi
S sy  0.50(207.1)  103.5 kpsi
Ans.
N a  Nb 
Body
 d 3S sy
F 
 KBD
 (0.162)3 (103.5)(103 )

8(1.166)(1.338)
Ans.
 110.8 lbf
Torsional stress on hook point B
2r2
2(0.25  0.162 / 2)

 4.086
d
0.162
4C2  1
4(4.086)  1

 1.243
(K )B 
4C2  4 4(4.086)  4
 (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)
C1 
Shigley’s MED, 11th edition
Chapter 10 Solutions, Page 35/47
4 
16( K ) A D
S yt    F 

3
 d 2 
 d
155.3(103 )
F 
 85.8 lbf
16(1.099)(1.338) /  (0.162)3   4 /  (0.162)2 
 min(110.8, 103.9, 85.8)  85.8 lbf Ans.


(e) Eq. (10-48):
85.8  16
F  Fi

 14.4 in Ans.
k
4.855
______________________________________________________________________________
y 
10-42
Fmin = 9 lbf, Fmax = 18 lbf
18  9
18  9
Fa 
 4.5 lbf , Fm 
 13.5 lbf
2
2
A313 stainless:
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
Sut 
 243.9 kpsi
(0.081)0.146
S su  0.67 Sut  163.4 kpsi
S sy  0.35Sut  85.4 kpsi
S y  0.55Sut  134.2 kpsi
Table 10-8:
m = 0.146
m = 0.263
Sr = 0.45Sut = 109.8 kpsi
109.8 / 2
Sr / 2

 57.8 kpsi
Se 
2
1  [ Sr / (2Sut )]
1  [(109.8 / 2) / 243.9]2
r   a /  m  Fa / Fm  4.5 / 13.5
For Gerber, Eq. (6-48), solving for (m)A gives
2

 2Se  
1 Sut2 r 
1  1  
 m  A 

2 n f Se 
 Sut r  


2

2

 
2  57.8 
1 243.9  4.5 / 13.5 

1  1  
   63.3 kpsi
2
2  57.8 
 243.9  4.5 / 13.5   


Hook bending

16C
4  13.5  (4C 2  C  1)16C

( m ) A  Fm ( K ) A
4




 (1)
 d 2  d 2   d 2 
4C (C  1)


Shigley’s MED, 11th edition
Chapter 10 Solutions, Page 36/47
Let  
 a  A 103  d 2

13.5
Equation (1) reduces to
C2 

16
63.3 103    0.081
C
13.5
 8
16
2
 96.65
0
The useable root for C is
2
2
 1  96.647

1  
 
 96.647 



C
     8 
 

96.647

8

 4 8

4 8
8
8 






 4.91
F
4.5
63.3  21.1 kpsi
 a  A  a  m  A 
13.5
Fm
n 
f
A
2
2

 2 m Se  
1  Sut    a  
 
   1  1  

Sut a  
2   m   Se  



2
2

 2(63.3)(57.8)  
1  243.9   21.1  
 
  2.00
 
  1  1  
2  63.3   57.8  
243.9(21.1)  



checks
D = Cd = 0.398 in
Using Eq. (10-4) for i
C  3 
 d 3 i  d 3  33 500

Fi 

 1000  4 


8D
8D  exp(0.105C )
6.5  

Use the lowest Fi in the preferred range.
 (0.081)3 
33 500
4.91  3  

 1000  4 
Fi 


8(0.398)  exp[0.105(4.91)]
6.5  

 8.55 lbf
For simplicity we will round Fi up to next 1/4 integer. Let Fi = 8.75 lbf.
18  9
 36 lbf/in
0.25
d 4G
(0.081) 4 (10)(106 )
Na 

 23.7 turns
8kD3
8(36)(0.398)3
G
10
 23.7 
 23.3 turns
Nb  N a 
E
28
L0  (2C  1  N b )d  [2(4.91)  1  23.3](0.081)  2.602 in
k 
Lmax  L0  ( Fmax  Fi ) / k  2.602  (18  8.75) / 36  2.859 in
Shigley’s MED, 11th edition
Chapter 10 Solutions, Page 37/47
 a  A 
Body:
Fa
4.5
63.3  21.1 kpsi
 m  A 
13.5
Fm
4C  2 4(4.91)  2

 1.300
4C  3
4(4.91)  3
8(1.300)(4.5)(0.398) 3
(10 )  11.16 kpsi
 a body 
 (0.081)3
F
13.5
(11.16)  33.48 kpsi
 m body  m  a body 
Fa
4.5
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 
From Eq. (6-48),
73.17 / 2
 38.5 kpsi
1  [(73.17 / 2) / 163.4]2
2
2

 2 m S se  
1  S su    a  
(n f ) body  
 
 1  1  

2   m   S se  
S su a  



2
2

 2(33.47)(38.5)  
1  163.4   11.16  
 
  2.53
 
  1  1  
2  33.47   38.5  
163.4(11.16)  



Let r2 = 2d = 2(0.081) = 0.162
2r
4(4)  1
C2  2  4, ( K ) B 
 1.25
4(4)  4
d
(K )B
1.25
( a ) B 
a 
(11.16)  10.73 kpsi
1.30
KB
(K )B
1.25
m 
( m ) B 
(33.48)  32.19 kpsi
1.30
KB
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:
Shigley’s MED, 11th edition
Chapter 10 Solutions, Page 38/47

4 Fmax  (4C 2  C  1)
 1
2 
d 
C 1

2

4(18)  4(4.91)  4.91  1

 1 (10-3 )  84.4 kpsi
2 
4.91  1
 (0.081 ) 

134.2
(n y ) A 
 1.59
84.4
( A ) max 
Body:
 i  ( Fi / Fa ) a  (8.75 / 4.5)(11.16)  21.7 kpsi
r   a /( m   i )  11.16 / (33.47  21.7)  0.948
r
0.948
(S sy   i ) 
(85.4  21.7)  31.0 kpsi
r 1
0.948  1
(S )
31.0
 sa y 
 2.78
11.16
a
(S sa ) y 
(n y ) body
Hook shear:
S sy  0.3Sut  0.3(243.9)  73.2 kpsi
 max  ( a ) B  ( m ) B  10.73  32.18  42.9 kpsi
73.2
 1.71
(n y ) B 
42.9
7.6 2d 2 ( N b  2) D
7.6 2 (0.081) 2 (23.3  2)(0.398)

 1.239
fom  
4
4
A tabulation of several wire sizes follow
0.081
d
0.085
0.092
0.098
0.105
0.120
Sut
243.920 242.210 239.427 237.229 234.851 230.317
Ssu
163.427 162.281 160.416 158.943 157.350 154.312
Ssy
Sr
85.372
84.773
83.800
83.030
82.198
80.611
109.764 108.994 107.742 106.753 105.683 103.643
Se
57.809
(m)A
C
63.331 62.887 62.164 61.594 60.976 59.799
96.695 105.734 122.443 137.659 156.443 200.389
4.916
5.497
6.563
7.527
8.713 11.477
(a)A
21.110
20.962
20.721
20.531
20.325
19.933
(nf)A
D
OD
Fi
(calc)
2.000
0.398
0.479
2.000
0.467
0.552
2.000
0.604
0.696
2.000
0.738
0.836
2.000
0.915
1.020
2.000
1.377
1.497
8.537
7.842
6.769
5.960
5.117
3.618
Fi (rd)
8.750
8.750
8.750
8.750
8.750
8.750

Shigley’s MED, 11th edition
57.403
56.744
56.223
55.659
54.585
Chapter 10 Solutions, Page 39/47
k
36.000
36.000
36.000
36.000
36.000
36.000
Na
23.677
17.767
11.301
7.979
5.512
2.756
Nb
23.320
17.410
10.944
7.622
5.155
2.399
L0
2.604
2.329
2.122
2.124
2.266
2.922
Lmax
2.861
2.586
2.379
2.381
2.523
3.179
KB
1.300
1.263
1.215
1.184
1.157
1.117
(a)body
11.162
11.015
10.796
10.638
10.478
10.197
(m)body
33.486
33.044
32.388
31.913
31.433
30.591
Ssr
73.176
72.663
71.828
71.169
70.455
69.095
Sse
38.519
38.249
37.809
37.462
37.087
36.371
(nf)body
2.526
2.542
2.564
2.578
2.591
2.611
(KB)B
4.000
4.000
4.000
4.000
4.000
4.000
(a)B
10.732
10.898
11.106
11.226
11.320
11.416
(m)B
32.196
32.695
33.319
33.679
33.960
34.248
(Ssr)B
68.298
67.819
67.040
66.424
65.758
64.489
(Sse)B
35.708
35.458
35.050
34.728
34.380
33.717
(nf)B
2.512
2.457
2.383
2.336
2.293
2.230
Sy
(A)max
134.156 133.215 131.685 130.476 129.168 126.674
84.441
83.849
82.886
82.125
81.302
79.732
1.589
1.589
1.589
1.589
1.589
1.589
r
21.704
0.947
21.417
0.947
20.992
0.947
20.684
0.947
20.373
0.947
19.828
0.947
(Ssa)y
30.974
30.822
30.555
30.330
30.077
29.570
2.775
2.798
2.830
2.851
2.871
2.900
(Ssy)B
73.176
72.663
71.828
71.169
70.455
69.095
(B)max
42.928
43.594
44.426
44.905
45.280
45.664
(ny)B
fom
1.705
-1.240
1.667
-1.229
1.617
-1.240
1.585
-1.278
1.556
-1.353
1.513
-1.636
(ny)A
i
(ny)body
optimal fom
The shaded areas show the conditions not satisfied.
______________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 10 Solutions, Page 40/47
10-43 For the hook,
M = FR sin,
F 
1
EI

 /2
0
∂M/∂F = R sin
F  R sin   R d 
2
 FR3
2 EI
The total deflection of the body and the two hooks
  FR 3  8FD3 N b
8FD 3 N b
 F (D / 2)3



2


d 4G
d 4G
E ( / 64)(d 4 )
 2 EI 
8FD 3 
G  8FD3 N a
 4  Nb   
d G 
E
d 4G
G
Q.E.D.
 N a  Nb 
E
______________________________________________________________________________
 
10-44 Table 10-5 (d = 4 mm = 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
1783
A
Sut  m  0.190  1370 MPa
d
4
Sy = all = 0.78 Sut = 0.78(1370) = 1069 MPa
D = OD  d = 32  4 = 28 mm
C = D/d = 28/4 = 7
Eq. (10-43):
Ki 
2
4C 2  C  1 4  7   7  1

 1.119
4C (C  1)
4(7)(7  1)
32 Fr
d3
At yield, Fr = My ,  = Sy. Thus,
Eq. (10-44):
  Ki
My 
 d 3S y
32 Ki
Count the turns when M = 0
N  2.5 
Shigley’s MED, 11th edition

  43 1069 103 
32(1.119)
 6.00 N · m
My
k
Chapter 10 Solutions, Page 41/47
where from Eq. (10-51):
k
d 4E
10.8 DN
Thus,
N  2.5 
Solving for N gives
N 

My
4
d E / (10.8DN )
2.5
1  [10.8DM y / (d 4 E )]
2.5


1  10.8(28)(6.00) / 4 4 (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 
My

 87.3 N Ans.
Fmax 
r
68.75
______________________________________________________________________________
10-45 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 
(0.081) 4 (28.6)(106 )
d 4E

 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
0.078(360) = 28.1 ). The original configuration of the spring was
Shigley’s MED, 11th edition
Chapter 10 Solutions, Page 42/47
Ans.
(b)
D
0.419

 5.17
d
0.081
4C 2  C  1 4(5.17) 2  5.17  1

 1.168
Ki 
4C  C  1
4(5.17)(5.17  1)
C 
  Ki
 32(13.25) 
32M
 1.168 
 297 103  psi  297 kpsi
3
3
d
  (0.081) 
Ans.
To achieve this stress level, the spring had to have set removed.
______________________________________________________________________________
10-46 (a) Consider half and double results
Straight section:
M = 3FR,
M
 3R
F
Upper 180 section:
M  F [ R  R(1  cos  )]
M
 FR(2  cos  ),
 R(2  cos  )
F
Lower section:
M = FR sin ,
Considering bending only:
Shigley’s MED, 11th edition
M
 R sin 
F
Chapter 10 Solutions, Page 43/47

U
2  l /2

9 FR 2 dx   FR 2 (2  cos  ) 2 R d 

0
F
EI  0

2F  9 2


  

R l  R 3  4  4sin  0    R 3   

EI  2
2

 4 
2
2
2 FR  19
9  FR

R  l 
(19 R  18l )

EI  4
2  2 EI
 

 /2
0
F ( R sin  ) 2 R d 

The spring rate is
k
F


2 EI
R (19 R  18 l )
2
Ans.
(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   0.0024  /  64 
0.006 19  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,
i 
Eq. (3-65):
Axial:
a 
Combining,
Mci
3FRci

2
Aeri  d / 4  e  R  d / 2 
F
F

A d2 / 4
 max   i   a 
F
4F
d2


3Rci
 1  S y

 e  R  d / 2 
 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):
Shigley’s MED, 11th edition
Sut = A/dm = 1783/20.190 = 1563 MPa
Sy = 0.78 Sut = 0.78(1563) = 1219 MPa
Chapter 10 Solutions, Page 44/47
Table 3-4:
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):
  0.0022 1219 106 
 46.0 N
Ans.
 3  6  0.95804

 1
4

0.04196
6
1




______________________________________________________________________________
F
10-47 (a)
M
 x
F
M   Fx,
M  Fl  FR 1  cos   ,

0 xl
M
 l  R 1  cos  
F
0   / 2
 /2
2
1 l



(
)
Fx
x
dx
F l  R 1  cos    Rd


0
0
EI
F

4l 3  3R  2 l 2  4   2  l R   3  8  R 2 
12 EI
F 



The spring rate is
k
F
F

12 EI
4l  3R  2 l  4   2  l R   3  8  R 2 
Shigley’s MED, 11th edition
3
2
Ans.
Chapter 10 Solutions, Page 45/47
(b) Given: A313 stainless wire, d = 0.063 in, R = 0.625 in, and l = 0.5 in.
Table 10-5:
E = 28 Mpsi
I
k

64
d4 

64
 0.063   7.733 10  in
4
7
4
12  28 106  7.733107
4  0.53   3  0.625   2  0.52   4   2  0.5  0.625    3  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):
Ki 
2
4C 2  C  1 4  20.8   20.8  1

 1.037
4C  C  1
4  20.8  20.8  1
Eq. (10-44), setting  = Sy:
Ki
Solving for F yields
32 Fr
 Sy
d3

F = 3.25 lbf
1.037
32 F  0.5  0.625 
  0.063
3

 154.4 103 
Ans.
Try solving part (c) of this problem using curved beam theory. You should obtain the
same answer.
______________________________________________________________________________
10-48 (a)
M =  Fx

M
Fx
Fx

 2
I / c I / c bh / 6
Constant stress,
Shigley’s MED, 11th edition
Chapter 10 Solutions, Page 46/47
bh 2 Fx

6

6 Fx
b
h

(1)
Ans.
At x = l,
6 Fl
b
ho 
(b)

h  ho x / l
Ans.
M =  Fx,  M / F = x
l
y
0

l
l
 Fx   x 
M  M / F 
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
bho E
3 bho3 E
F bho3 E
 3
Ans.
y
8l
______________________________________________________________________________
k
10-49 Computer programs will vary.
______________________________________________________________________________
10-50 Computer programs will vary.
Shigley’s MED, 11th edition
Chapter 10 Solutions, Page 47/47
Chapter 4
For a torsion bar, kT = T/ = Fl/, and so  = Fl/kT. For a cantilever, kl = F/ , = F/kl. For
the assembly, k = F/y, or, y = F/k = l + 
Thus
F Fl 2 F
y 

k
kT kl
Solving for k
kk
1
 2l T
k 2
Ans.
l
1 kl l  kT

kT kl
______________________________________________________________________________
4-1
For a torsion bar, kT = T/ = Fl/, and so  = Fl/kT. For each cantilever, kl = F/l, l = F/kl,
and,L = F/kL. For the assembly, k = F/y, or, y = F/k = l + l +L.
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 kL
______________________________________________________________________________
4-2
4-3
(a) For a torsion bar, k =T/ =GJ/l.
Two springs in parallel, with J =di 4/32,
and d1 = d2 = d,
J G J G   d4 d4 
k  1  2  G 1  2 
x
l  x 32  x l  x 

1 
1
Gd 4  

32
 x lx
Deflection equation,


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 
Shigley’s MED, 11th edition
Chapter 4 Solutions, Page 1/96
lx
T2 
  T2  1500
 x 
T2  1500

x
l
Ans.
(3)
lx
Ans. (4)
l
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,  
3
3
 di
  0.5 
Substitute into Eq. (2) resulting in
T1  1500
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
1=2=
T1  4 

32
Or,
d14G

T2  6 

32
d 24G


32
T1  15 103  d14
 0.54  G

4T1 6T2
 4  60 103 
4
d1
d2
(1)
(2)
T2  10 103  d 24
Equal stress,  1   2 
750  5
(3)
T
T
16T1 16T2

 13  23
3
3
d1 d 2
 d1  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.
Shigley’s MED, 11th edition
Chapter 4 Solutions, Page 2/96
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
Ans.
Ans.
343  4 
T1l1

 0.053 18 rad
J1G   / 32   0.388 84 11.5 106 
Deflection of T is
1 
Spring constant is
k
The stress in d1 is
1 
16  343
16T1

 29.7 103  psi  29.7 kpsi
3
 d1   0.388 8 3
Ans.
The stress in d1 is
2 
16 1 157 
16T2

 29.7 103  psi  29.7 kpsi
3
3
 d 2   0.583 2 
Ans.
T
1

1500
 28.2 103  lbf  in
0.053 18
Ans.
______________________________________________________________________________
4-5
(a) Let the radii of the straight sections be r1 = d1 /2 and r2 = d2 /2. Let the angle of the
taper be  where tan  = (r2  r1)/2. Thus, the radius in the taper as a function of x is
r = r1 + x tan , and the area is A =  (r1 + x tan )2. The deflection of the tapered portion
is
l
l
1
F
F
dx
F
 

dx 
2

 E 0  r1  x tan  
 E  r1  x tan   tan 
AE
0


1 1
1
F  1
F



  
 E  r1 tan  tan   r1  l tan     E tan   r1 r2 

F
r2  r1
F
l tan 
Fl


 E tan  r1r2
 E tan  r1r2
 r1r2 E

4 Fl
 d1d 2 E
l
0
Ans.
(b) For section 1,
1 
4 Fl
4(1000)(2)
Fl


 3.40(10 4 ) in
2
2
6
AE  d1 E  (0.5 )(30)(10 )
For the tapered section,
4 Fl
4
1000(2)


 2.26(10 4 ) in
6
 d1d 2 E  (0.5)(0.75)(30)(10 )
For section 2,
Shigley’s MED, 11th edition
Ans.
Ans.
Chapter 4 Solutions, Page 3/96
4 Fl
4(1000)(2)
Fl


 1.51(10 4 ) in Ans.
2
2
6
AE  d1 E  (0.75 )(30)(10 )
______________________________________________________________________________
2 
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  where tan  = (r2  r1)/2. Thus, the radius in the taper as a function of x is
r = r1 + x tan , and the polar second area moment is J = ( /2) (r1 + x tan )4. The angular
deflection of the tapered portion is
l
l
T
dx
2T
1 2T
1
   dx 

4

GJ
3  G  r1  x tan  3 tan 
 G 0  r1  x tan  
0

l
0
 2
T 1 1
2 T  1
1


 3

  
3
3 G  r1 tan  tan   r1  l tan    3 G tan   r13 r23 
2
2
2
2 T  l  r23  r13
2 Tl  r1  r1r2  r2 
T
r23  r13





3 G tan  r13r23
3 G  r2  r1  r13 r23
3 G
r13r23

2
2
32 Tl  d1  d1d 2  d 2 
3 G
d13d 23
Ans.
(b) The deflections, in degrees, are:
Section 1,
Tl
GJ
1 
32(1500)(2)  180 
 180  32Tl  180 




  2.44 deg
4
4
6 
    d1 G     (0.5 )11.5(10 )   
Ans.
Tapered section,

32 Tl (d12  d1d 2  d 2 2 )  180 


Gd13 d 23
3
  
2
2
32 (1500)(2) 0.5  (0.5)(0.75)  0.75   180 


  1.14 deg
3
11.5(106 )(0.53 )(.753 )
  
Ans.
Section 2,
32(1500)(2)  180 
Tl  180  32Tl  180 
Ans.





  0.481 deg
4
GJ     d 2 G     (0.754 )11.5(106 )   
______________________________________________________________________________
2 
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  = 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 = (A)(lx)
Shigley’s MED, 11th edition
Chapter 4 Solutions, Page 4/96
c  
l
o
1 
 
 l 2 0.282 500(12) 
Fdx w l
  (l  x )dx   lx  x 2  

 0.169 in
AE E 0
E
2  0 2E
2(30)106
2
l
From the weight at the bottom of the cable,
4(5000) 500(12) 
Wl
4Wl


 5.093 in
AE  d 2 E
 (0.52 )30(106 )
   c  W  0.169  5.093  5.262 in
Ans.
W 
The percentage of total elongation due to the cable’s own weight
0.169
(100)  3.21%
Ans.
5.262
______________________________________________________________________________
4-8
Fy = 0 = R 1  F  R 1 = F
MA = 0 = M1  Fa  M1 = Fa
VAB = F, MAB =F (x  a ), VBC = MBC = 0
Section AB:

F  x2
1
 AB 
F
x

a
dx



  ax   C1

EI
EI  2

AB = 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:
 BC 
1
EI
  0  dx  0  C
3
From Eq. (1), at x = a (with C1 = 0),  
 BC  

F  a2
Fa 2
a
(
a
)



= C3. Thus,


EI  2
2 EI

Fa 2
2 EI
Shigley’s MED, 11th edition
Chapter 4 Solutions, Page 5/96
yBC  
Fa 2
Fa 2
dx


x  C4
2 EI 
2 EI
(3)
F  a3
a2 
Fa3
From Eq. (2), at x = a (with C2 = 0), y 
. Thus, from Eq. (3)
 a   
EI  6
2 
3EI

Fa 2
Fa 3
a  C4  
2 EI
3EI
yBC  
C4 

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
MC = 0 = F (l /2)  R1 l  R1 = F /2
Fy = 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:
 AB 
1
EI
Fx
F x2
dx

 C1
2
EI 4
2
l
F 
 2 C  0
From symmetry, AB = 0 at x = l /2 
1
4 EI
 AB 
F x 2 Fl 2
F


 4x2  l 2 
EI 4 16 EI 16 EI
Shigley’s MED, 11th edition

C1  
Fl 2
. Thus,
16 EI
(1)
Chapter 4 Solutions, Page 6/96
y AB 
F
16 EI
yAB = 0 at x = 0
y AB 
  4x
2
 l 2  dx 
F  4 x3 2 
 l x   C2

16 EI  3

 C2 = 0, and,
Fx
4 x 2  3l 2 

48 EI
(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 

2
4
3
l




  

48EI   2 
48EI

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
wl4
(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
9.5(106 )(71)
My

 163 MPa Ans.
I
4.14(106 )
______________________________________________________________________________
 max  
Shigley’s MED, 11th edition
Chapter 4 Solutions, Page 7/96
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
 15 
Z  2.28 in 3
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 
 max 
0.5 
450(72)(120)
300(2403 )
2
2
2



120
72
240

 48(30)(106 ) I
6(30)(106 ) I (240)
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
 5.70 kpsi
 max 
6.00
______________________________________________________________________________
4-12
I

(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(30)106 (0.2485)39 
(150 /12)(15)
 2(39)(152 )  153  393   0.0978 in

6
24(30)10 (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, 11th edition
Chapter 4 Solutions, Page 8/96
y
340(15)(19.5)
19.52  152  39 2 
6
6(30)(10 )(0.2485)(39)
(150 / 12)(19.5)
 2(39)(19.52 )  19.53  393   0.1027 in

24(30)(106 )(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  3 x 2 )
dx
6 EIl
dy AB
 A
dx
Fal 2 Fal
A 

6 EIl 6 EI
Fa 2l
yO   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)(500 2 )
yE 

 1.11 mm Ans.
l     
6 EIl 
24 207 103 16.384 103 
 2   24 EI
_____________________________________________________________________________
 4
4-14 I 
(2  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, 11th edition
Chapter 4 Solutions, Page 9/96
200  4(12)
300  2(12) 
FBl 3 FAa 2
yB  


(a  3l ) 
 2(12)  3(4)(12)
3EI 6 EI
3(10.4)106 (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)
______________________________________________________________________________
 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
FB l 3
Fa
1
  FB l 3  2 Fa(4a 2  3l 2 ) 
yB  

(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 mm4
d
4
64

I 
4
64

(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 
Ans.


6 EIl
d M
F (x  l)

   A x3  3lx 2  2l 2 x   ( x  l ) 
[( x  l ) 2  a (3x  l )]
6 EI
  x l
 dx  6 EIl
y AB 
yBC










M Al
F (x  l)
(x  l) 
[( x  l ) 2  a (3 x  l )]
6 EI
6 EI
(x  l)

 M Al  F  ( x  l ) 2  a (3 x  l ) 
Ans.
6 EI
______________________________________________________________________________


Shigley’s MED, 11th edition

Chapter 4 Solutions, Page 10/96
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
 Fy  0  2l  2l  a   R2  wa  R2  2l Ans.
wa
w
VAB  R1  wx =
 2l  a   wx =  2l  a  x   a 2  Ans.
2l
2l
2
wa
VBC   R2  
Ans.
2l

x2 
w 
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 
 AB   AB dx 

 alx  a x  lx   C3 


2
3 
EI
EI 2l
EI  2l 

M BC   VBC dx   
y AB    AB dx 
1 w 2 1 2 2 1 3

 alx  a x  lx   C3  dx


2
3 
EI  2l 

1 w 1 3 1 2 3 1 4

 alx  a x  lx   C3 x  C4 

6
12 
EI  2l  3

 0 at x  0  C4  0

y AB
 BC
M
1
  BC dx 
EI
EI

wa 2
1  wa 2 
1 2
 2l (l  x) dx  EI  2l  lx  2 x   C5 
 AB   BC at x  a 
1
EI

wa 3
w  2 1 4 1 3
 1  wa 2 
1 2








 C5
ala
a
la
C
la
a
C
C
3
5
3




 2l 
2
3 
2 
6
 
 EI  2l 

Shigley’s MED, 11th edition
(1)
Chapter 4 Solutions, Page 11/96


1  wa 2 
1 2
1  wa 2  1 2 1 3 
lx

x

C
dx

5


 lx  x   C5 x  C6 



2 
6 
EI  2l 
EI  2l  2


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    BC dx 
yBC
yBC
y AB  yBC at x  a 
1 5 1 4
w 1
wa 2  1 2 1 3 1 3 
3
ala

a

la

C
a

3


 la  a  l   C5 (a  l )
2l  3
6
12
2l  2
6
3 

C3 a 
wa 2
 3la 2  4l 3   C5 (a  l )
24l
Substituting (1) into (2) yields C5 
(2)
wa 2
a 2  4l 2  . Substituting this back into (2) gives

24l
wa 2
C3 
4al  a 2  4l 2  . Thus,

24l
w
y AB 
4alx 3  2a 2 x 3  lx 4  4a 3lx  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
y BC 
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  

24 EIl 
w 
2
2bx 3 (2l  b)  lx 4  b 2 x  2l  b 


24 EIl

yBC
  4alx 3  2a 2 x 3  lx 4  4a 2l 2 x  4a 3lx  a 4 x   l ( x  a ) 4 
Shigley’s MED, 11th edition
Ans.
Ans.
Chapter 4 Solutions, Page 12/96
w
 4blx3  2b 2 x3  lx 4  4b 2l 2 x  4b3lx  b 4 x  l ( x  b) 4 
24 EIl
w
 4alx 3  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
______________________________________________________________________________
4-20 Note to the instructor: See the note in the solution for Problem 4-18.
wa 2
wa
 Fy  0  RB  2l  wa  RB  2l  2l  a  Ans.
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
EI 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
wa 2l
yAB = 0 at x = l  C1 

12
w a 2 3 w a 2l
wa 2 x 2
wa 2 x 2
EIy AB  
x 
x
l  x 2   y AB 

 l  x2  Ans.
12l
12
12l
12 EIl
w
3
EI BC   M BC dx    l  a  x   C3
6
w
4
EIyBC    l  a  x   C3 x  C4
24
wa 4
wa 4
yBC = 0 at x = l  
 C3l  C4  0  C4 
 C3l
(1)
24
24
wa 2l wa 2l wa 3
wa 2
AB = BC 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 

1  w
wa 2
wa 4 wa 2l
4
yBC 
x l  a  

 l  a 
 l  a  x  
EI  24
6
24
6

yCD 
w 
4
Ans.
 l  a  x   4a 2  l  x  l  a   a 4 

24 EI
______________________________________________________________________________

Shigley’s MED, 11th edition
Chapter 4 Solutions, Page 13/96
4-21
Table A-9, beam 7,
w l 100(10)
R1  R2 

 500 lbf 
2
2
wx
100 x
 2(10) x 2  x3  103 
2lx 2  x 3  l 3  
y AB 

6
24 EI
24  30 10  0.05 
 2.7778 10 6  x  20 x 2  x3  1000 
Slope:
 AB 
d y AB
w

6lx 2  4 x3  l 3 

dx
24 EI
w
wl 3
2
3
3



6
l
l
4
l
l


x l
24 EI
24 EI
100 103 
w l3
x l 
x  l 
 x  10   2.7778 103   x  10 
6
x l
24 EI
24(30)10 (0.05)
At x = l,  AB
yBC   AB

From Prob. 4-20,
2
wa 2 100  4 
RA 

 80 lbf 
2l
2(10)
y AB
RB 
100  4 
wa
 2l  a  
 2(10)  4  480 lbf 
2l
2(10)
100  42  x
wa 2 x 2
2

10 2  x 2   8.8889 10 6  x 100  x 2 
l x 


6
12 EIl
12  30 10  0.05 
w 
4
 l  a  x   4a 2  l  x  l  a   a 4 

24 EI
100
10  4  x 4  4  4 2  10  x 10  4   44 

6

24  30 10  0.05 
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
yBC  2.7778 103
 x  20 x  x  1000   8.8889 10  x 100  x 
  x 10   2.7778 10  14  x   896 x  9216 
2
3
6
6
4
2
Ans.
Ans.
Ans.
The deflection equations can be simplified further. However, they are sufficient for
plotting.
Using a spreadsheet,
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, 11th edition
Chapter 4 Solutions, Page 14/96
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
12.5
0.001244 0.001419
0.002
13
0.001575
13.5
14
0.001722 0.001867
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 48EI
 3
y
l
3
kl 3 1800  36 
I

 0.05832 in 4
6
48 E 48(30)10
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
Shigley’s MED, 11th edition
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
Chapter 4 Solutions, Page 15/96
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
4 I 4(60)103 (0.05729)
Mc ( Fl / 4)c

 F

 1528 lbf

I
I
lc
 36  (0.25)
(1528)  36 
Fl 3
y

 0.864 in Ans.
48 EI 48(30)106 (0.05729)
______________________________________________________________________________
3
4-23 From the solutions to Prob. 3-79, T1  60 lbf and T2  400 lbf
I
d4
64

 (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
6(30)10 (0.1198)(40)
460(12)(10)

102  122  40 2   0.0332 in

6
6(30)10 (0.1198)(40)

Ans.
dz
Fb x
 d 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 (3x 2  b12  l 2 )  2 2 (3 x 2  b2 2  l 2 ) 
6 EIl
 6 EIl
 x 10in
 A  y   

(575)(30)
3 102   30 2  40 2 
6

6(30)10 (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. 3-80, T1  2880 N and T2  432 N
I
d4
64

 (30) 4
Shigley’s MED, 11th edition
64
 
 39.76 103 mm 4
Chapter 4 Solutions, Page 16/96
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   BC
C
 a2   beam6   y A beam10
(1)
 d  F1a1  l  x  2
 
 Fa

x  a12  2lx      1 1  6lx  3 x 2  a12  2l 2  


 x l
  x l  6 EIl
 dx  6 EIl
 BC C  
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

   3(207)103 (39.76)103  510  300 
6(207)103 (39.76)103 (510)
 7.99 mm
Ans.
yA 
The slope at A, relative to the z axis is
 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 (3x  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. 3-81, T1  392.16 lbf and T2  58.82 lbf
I
d4

64
From Table A-9, beam 6,
Shigley’s MED, 11th edition
 (1) 4
64
 0.049 09 in 4
Chapter 4 Solutions, Page 17/96
( 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  62  22 2   0.0428 in Ans.

6
6
EIl
6(30)10
(0.049
09)(22)

 x 8in
The displacement magnitude is   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
 A  z  

(350)(14)
3  82   14 2  22 2   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
 A  y   

(450.98)(6)
3  82   62  222   0.00356 rad
6


6(30)10 (0.04909)(22)
Ans.
The slope magnitude is  A  0.00242 2   0.00356   0.00430 rad Ans.
2
______________________________________________________________________________
4-26 From the solutions to Prob. 3-82, T1  250 N and T2  37.5 N
I
d4
64

 (20)4
64
 7 854 mm 4
345sin 45o  (550)(300)

 F1 y b1 x 2
2
2 
( x  b1  l ) 
3002  550 2  8502 
yA  


3
6(207)10 (7 854)(850)
 6 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
 345 cos 45  (550)(300)
o

 300
6(207)10 (7 854)(850)
2
3

 5502  850 2 
287.5(150)(300)
3002  150 2  8502   0.650 mm

3
6(207)10 (7 854)(850)
The displacement magnitude is  
Shigley’s MED, 11th edition
Ans.
y A2  z A2  1.602   0.650   1.73 mm
2
Ans.
Chapter 4 Solutions, Page 18/96
F1 y b1
 d  F1 y b1 x 2
 
d y
x  b12  l 2    
 
(3a12  b12  l 2 )


 d x  x a1  dx  6 EIl
  x a1 6 EIl
 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
 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  A  0.002432  0.0001912  0.00244 rad Ans.
______________________________________________________________________________
4-27 From the solutions to Prob. 3-83, FB  750 lbf
I
d4
64

 (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)
 0.0805 in
16
2
 14  30  
2
 750sin 20  (9)(16)
o
2
6(30)106 (0.119 8)(30)
 30
2
 162 
Ans.
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

16
6(30)10 (0.119 8)(30)
6
 0.1169 in
2
 14  30
2

 750 cos 20  (9)(16)

o
 30
6(30)10 (0.119 8)(30)
6
2
 162 
Ans.
The displacement magnitude is  
Shigley’s MED, 11th edition
2
y A2  z A2  0.08052   0.1169   0.142 in
2
Ans.
Chapter 4 Solutions, Page 19/96
F2 y a2 x 2
 
d y
 d  F1 y b1 x 2
 
x  b12  l 2  
l  x2  



6 EIl
 d x  x  a1  dx  6 EIl
  x  a1
 A  z  
F1 y b1
F2 y a2
 3a  b  l   6EIl  l  3a 
6 EIl
 300 cos 20  (14) 3 16  14  30 

 

6(30)10 (0.119 8)(30) 

2
1
2
1
2
2
2
1
o
2
2
2
6
 750sin 20  (9)
o

302  3 162    8.06 105  rad

6(30)10 (0.119 8)(30) 
6
Ans.
dz
F ax
 d F b x

    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
 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


302  3 162  

3 16   14  30  
6
6


6(30)10 (0.119 8)(30)
6(30)10 (0.119 8)(30) 

 0.00115 rad
Ans.
The slope magnitude is  A  8.06 10 5    0.001152  0.00115 rad Ans.
______________________________________________________________________________
2
4-28 From the solutions to Prob. 3-84, FB = 22.8 (103) N
4
 d 4   50 

 306.8 103  mm 4
I
64
64
From Table A-9, beam 6,
F2 y b2 x 2
 F1 y b1 x 2

yA  
( x  b12  l 2 ) 
( x  b2 2  l 2 ) 
6 EIl
 6 EIl
 x  400mm
11103  sin 20o  (650)(400)

4002  6502  10502 


3
3
6(207)10 (306.8)10 (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, 11th edition
Chapter 4 Solutions, Page 20/96
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 
6(207)103 (306.8)103 (1050)
 22.8 103  cos 25o  (300)(400)


400 2  3002  1050 2   1.791 mm

3
3
6(207)10 (306.8)10 (1050)
The displacement magnitude is  
y A2  z A2 
 3.735 
2
 1.7912  4.14 mm
Ans.
Ans.
d y
F bx
 d F b x

   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
 A  z  
F1 y b1
F2 y b2
 3a  b  l   6EIl  3a  b  l 
6 EIl
1110  sin 20  (650)

3  400   650
 
6(207)10 (306.8)10 (1050) 

2
1
3
2
1
2
2
1
2
2
o
2
3
2
3
2
 1050 2 
 22.8 103  sin 25o  (300)

3  4002   3002  1050 2 
 
3
3

6(207)10 (306.8)10 (1050) 
 0.00507 rad
Ans.
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
 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  400 2   650 2  1050 2 
 
3
3

6(207)10 (306.8)10 (1050) 

 22.8 103  cos 25o  (300)

3  4002   300 2  1050 2 
 
3
3

6(207)10 (306.8)10 (1050) 
Ans.
 0.00489 rad
The slope magnitude is  A 
 0.00507    0.00489 
2
2
 0.00704 rad Ans.
______________________________________________________________________________
4-29 From the solutions to Prob. 3-79, 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, 11th edition
Chapter 4 Solutions, Page 21/96
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
  1z 1  b12  l 2   2 z 2  b2 2  l 2   
 302  402 
6 EIl
6 EIl
6(30)106 (0.119 8)(40)
O  y   

460(12)
12 2  40 2   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  3x 2  a22  
6 EIl
 6 EIl
 x l
C  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. 3-80, 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
Fa x
 d Fb x

O  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 (3x 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
C  z  
Fa
Fa
Fal
 Fa

  1 1 (6lx  2l 2  3 x 2  a12 )  2 2 (l 2  3 x 2 )   1 1 (l 2  a12 )  2 2
6 EIl
3EI
 6 EIl
 x l 6 EIl
2 070(300)(510)
3 312(230)
(510 2  230 2 ) 

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. 3-81, 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, 11th edition
Chapter 4 Solutions, Page 22/96
F b
d y
 
 d  F1 y b1 x 2
x  b12  l 2     1 y 1 (b12  l 2 )

  
 d x  x  0  dx  6 EIl
  x 0 6 EIl
350(14)

142  22 2   0.00726 rad
Ans.

6
6(30)10 (0.04909)(22)
O  z  
dz
F b
 d  F2 z b2 x 2

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)
 0.00624 rad
Ans.
O  y   
The slope magnitude is O  0.007262   0.00624   0.00957 rad Ans.
2
 d  F1 y a1 (l  x) 2
 
d y
x  a12  2lx   

  
 d x  x l  dx  6 EIl
  x l
F1 y a1 2
 F1 y a1


6lx  2l 2  3x 2  a12   
(l  a12 )

 6 EIl
 x l 6 EIl
350(8)
222  82   0.00605 rad
Ans.


6
6(30)10 (0.0491)(22)
C  z  
dz
 d  F2 z a2 (l  x) 2

x  a22  2lx   

   
  x l
 dx  6 EIl
 d x  x l
C  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 C 
 0.00605 
2
Ans.
 0.008462  0.0104 rad Ans.
______________________________________________________________________________
4-32 From the solutions to Prob. 3-82, T1 =250 N and T1 =37.5 N, and Prob. 4-26, I = 7 854
mm4. From Table A-9, beam 6
F b
 
d y
 d  F1 y b1 x 2
x  b12  l 2     1 y 1 (b12  l 2 )

  
 d x  x  0  dx  6 EIl
  x 0 6 EIl
O  z  
 345sin 45o  (550)

5502  8502   0.00680 rad

3
6(207)10 (7 854)(850)
Shigley’s MED, 11th edition
Ans.
Chapter 4 Solutions, Page 23/96
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
O  y   
345 cos 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 O  0.00680 2  0.00316 2  0.00750 rad Ans.
 d  F1 y a1 (l  x ) 2
 
F a

d y
x  a12  2lx      1 y 1  6lx  2l 2  3 x 2  a12  

  
 d x  x l  dx  6 EIl
  x l  6 EIl
 x l
C  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
C  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.

3
6(207)10 (7 854)(850)
The slope magnitude is C 
 0.00558
2
 6.04 105    0.00558 rad Ans.
2
________________________________________________________________________
4-33 From the solutions to Prob. 3-83, 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
O  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
F a
F b
F al
F b

  1 y 1  3 x 2  b12  l 2   2 y 2  l 2  3x 2   1 y 1  b12  l 2   2 y 2
6 EIl
6 EI
 6 EIl
 x 0 6 EIl
 300 cos 20o  (14)
750sin 20o  (9)(30)
2
2

14  30   6(30)106 (0.119 8)  0.00751 rad
6(30)106 (0.119 8)(30)
Shigley’s MED, 11th edition
Ans.
Chapter 4 Solutions, Page 24/96
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
O  y   
F a
F b
F al
F b

   1z 1  3 x 2  b12  l 2   2 z 2  l 2  3 x 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 O  0.007512  0.01042  0.0128 rad Ans.
F ax
 
 dy 
 d  F1 y a1 (l  x) 2
x  a12  2lx   2 y 2  l 2  x 2   

  
6 EIl
 dx  x l  dx  6 EIl
  x l
F a
F a
F al
F a

  1 y 1  6lx  2l 2  3x 2  a12   2 y 2  l 2  3 x 2    1 y 1 (l 2  a12 )  2 y 2
6 EIl
3EI
 6 EIl
 x l 6 EIl
C  z  
750sin 20o  (9)(30)
 300 cos 20o  (16)
2
2

 30  16   3(30)106 (0.119 8)  0.0109 rad
6(30)106 (0.119 8)(30)
Ans.
dz
F a x
 d  F1z a1 (l  x) 2

x  a12  2lx   2 z 2  l 2  x 2   

   
6 EIl
  x l
 dx  6 EIl
 d x  x l
C  y   
F a
F a
F al
F a

   1z 1  6lx  2l 2  3x 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
1
6


 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 C 
2
Ans.
 0.0222 rad Ans.
______________________________________________________________________________
4-34 From the solutions to Prob. 3-84, FB = 22.8 kN, and Prob. 4-28, I = 306.8 (103) mm4.
From Table A-9, beam 6
F2 y b2 x 2
 d  F1 y b1 x 2
 
d y
x  b12  l 2  
x  b22  l 2  


  
6 EIl
 d x  x  0  dx  6 EIl
  x 0
O  z  
11103  sin 20o  (650)

b l 
b l   

6502  10502 



3
3
6 EIl
6 EIl
6(207)10 (306.8)10 (1050)
F1 y b1
2
1
2
F2 y b2
2
2
2
 22.8 103  sin 25o  (300)

 
300 2  10502   0.0115 rad

3
3
6(207)10 (306.8)10 (1050)
Shigley’s MED, 11th edition
Ans.
Chapter 4 Solutions, Page 25/96
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
F b
F b
  1z 1  b12  l 2   2 z 2  b22  l 2 
6 EIl
6 EIl
11103  cos 20o  (650)

 
 6502  10502 
6(207)103 (306.8)103 (1050)
O  y   
 22.8 103  cos 25o  (300)

300 2  1050 2   0.00427 rad


3
3
6(207)10 (306.8)10 (1050)
The slope magnitude is O 
 0.0115   0.00427 
2
2
Ans.
 0.0123 rad Ans.
F2 y a2 (l  x) 2
 
d y
 d  F1 y a1 (l  x) 2
x  a12  2lx  
x  a22  2lx   


  
6 EIl
 d x  x l  dx  6 EIl
  x l
F2 y a2
 F1 y a1

(6lx  2l 2  3x 2  a12 ) 
6lx  2l 2  3x 2  a22  


6 EIl
 6 EIl
 x l
C  z  
11103  sin 20o  (400)



 l  a   6EIl  l  a   6(207)10
10502  4002 
3
6 EIl
(306.8)103 (1050)
F1 y a1
2
2
1
F2 y a2
2
2
2
 22.8 103  sin 25o  (750)

10502  750 2   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
C  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


10502  4002 
l  a1  
l  a2    



3
3
6 EIl
6 EIl
6(207)10 (306.8)10 (1050)
 22.8 103  cos 25o  (750)


10502  7502   0.0112 rad

3
3
6(207)10 (306.8)10 (1050)
Ans.
The slope magnitude is C  0.01332  0.01122  0.0174 rad Ans.
______________________________________________________________________________
4-35 The required new slope in radians is  new = 0.06( /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 (O)y =  0.00468 rad.
Since is inversely proportional to I,
Shigley’s MED, 11th edition
Chapter 4 Solutions, Page 26/96
 new Inew =  old Iold

4
Inew =  d new
/64 =  old Iold / new
1/ 4
or,
d new
 64  old


I old 
 

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.
  0.00105

______________________________________________________________________________
4-36 The required new slope in radians is  new = 0.06( /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 (C)y =  0.0191 rad.
See the solution to Prob. 4-35 for the development of the equation
1/ 4
d new
 64  old


I old 
 

new


1/ 4
 64 0.0191

d new  
39.76 103    62.0 mm
Ans.
  0.00105

______________________________________________________________________________
4-37 The required new slope in radians is  new = 0.06( /180) = 0.00105 rad.
In Prob. 4-31, I = 0.0491 in4, and the maximum slope is C = 0.0104 rad.
See the solution to Prob. 4-35 for the development of the equation
1/ 4
d new
 64  old

I old 

 

new


1/ 4
 64 0.0104

d new  
0.0491  1.77 in
Ans.
  0.00105

______________________________________________________________________________
4-38 The required new slope in radians is  new = 0.06( /180) = 0.00105 rad.
In Prob. 4-32, I = 7 854 mm4, and the maximum slope is O = 0.00750 rad.
See the solution to Prob. 4-35 for the development of the equation
Shigley’s MED, 11th edition
Chapter 4 Solutions, Page 27/96
1/ 4
d new
 64  old


I old 
 

new


1/ 4
 64 0.00750

d new  
7 854   32.7 mm
Ans.
  0.00105

______________________________________________________________________________
4-39 The required new slope in radians is  new = 0.06( /180) = 0.00105 rad.
In Prob. 4-33, I = 0.119 8 in4, and the maximum slope  = 0.0222 rad.
See the solution to Prob. 4-35 for the development of the equation
1/ 4
d new
 64  old

I old 

 

new


1/ 4
 64 0.0222

d new  
0.119 8   2.68 in
Ans.
  0.00105

______________________________________________________________________________
4-40 The required new slope in radians is  new = 0.06( /180) = 0.00105 rad.
In Prob. 4-34, I = 306.8 (103) mm4, and the maximum slope is C = 0.0174 rad.
See the solution to Prob. 4-35 for the development of the equation
1/ 4
d new
 64  old

I old 

 

new


1/ 4
 64 0.0174

d new  
306.8 103    100.9 mm
Ans.
  0.00105

______________________________________________________________________________
4-41 IAB =  14/64 = 0.04909 in4, JAB = 2 IAB = 0.09818 in4, IBC = (0.25)(1.5)3/12 = 0.07031 in4,
ICD =  (3/4)4/64 = 0.01553 in4. For Eq. (3-41), b/c = 1.5/0.25 = 6   = 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, B1, due to the force and moment acting on
B (y2 = CD B1 = 2B1).
Shigley’s MED, 11th edition
Chapter 4 Solutions, Page 28/96
3. The vertical deflection due to the rotation at B, B2, due to the torsion acting at B (y3 =
BC B1 = 5B1).
4. The vertical deflection of C due to the force acting on C (y4).
5. The rotation at C, C, due to the torsion acting at C (y3 = CD C = 2C).
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  6 2 
200  63 
y1  

 0.01467 in
3  30 10 6  0.04909  2  30 106  0.04909 
2. From Table A-9, beams 1 and 4
 d  Fx 2
M x2  
Fx
M x
 B1   
 x  3l   B      3x  6l   B 
EI  x l
2 EI   x l  6 EI
 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)
 TL 
1000  6 
B2  
 0.005314 rad
 
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 10 6  0.07031
 0.00395 in
5. For twist of BC, from Eq. (3-41), with T = 2(200) = 400 lbfin
C 
400  5 
 0.02482 rad
0.299 1.5  0.253 11.5 10 6
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 
Shigley’s MED, 11th edition
 0.00114 in
Chapter 4 Solutions, Page 29/96
Summing the deflections results in
6
y D   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. 4-78.
______________________________________________________________________________
4-42 The deflection of D in the x direction due to Fz is from:
1. The deflection due to the slope at B, B1, due to the force and moment acting on B (x1 =
BC B1 = 5B1).
2. The deflection due to the moment acting on C (x2).
1. For AB, IAB =  14/64 = 0.04909 in4. From Table A-9, beams 1 and 4
M B x 2  
M x
 d  Fx 2
 Fx


x
l
3


 3x  6l   B 
 


EI  x l
2 EI   x l  6 EI
 dx  6 EI
 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
2  100  5
M Cl 2

 0.04267 in
x2 
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 =  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, B2, due to the moment acting on B (x1 = BC B2 =
5B2). With IAB = 0.04907 in4,
B2 
5  150  6
M Bl

 0.003056 rad
EI
30 106  0.04909
Shigley’s MED, 11th edition
Chapter 4 Solutions, Page 30/96
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   

6
3
3
30
10
0.001953
EI





 BC
6. The elongation of CD due to the tension. For CD, the area is A =  (0.752)/4 = 0.4418
in2
150  2 
 Fl 
x6  
 2.26 10 5  in
 
6
 AE CD 0.4418  30 10
Summing the deflections results in
6
xD   xi  0.01019  0.04267  3.82 105 
i 1
 0.01528  0.10667  2.26 105   0.1749 in Ans.
______________________________________________________________________________
4-43
JOA = JBC =  (1.54)/32 = 0.4970 in4, JAB =  (14)/32 = 0.09817 in4, IAB =  (14)/64 =
0.04909 in4, and ICD =  (0.754)/64 = 0.01553 in4.
T  lOA l AB lBC 
 Tl 
 Tl 
 Tl 
 



 
 
  
 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)
s 

GJ 11.5 106   0.09817 

Ans.
 s  0.0345 rad
Ans.
Simplified is 0.0345/0.0260 = 1.33 times greater Ans.
yD 
Fy lOC 3
3EI AB
  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, 11th edition
Chapter 4 Solutions, Page 31/96
y
 3000 / 12  x 2 25 x 2  x3   25 12  3
wx
2lx 2  x 3  l 3   
 

  
24 EI
24  30 106  485 

 7.159 1010  x  27 106   600 x 2  x 3 

Ans.
The maximum height occurs at x = 25(12)/2 = 150 in
ymax  7.159 10 10 150  27 106   600 150 2   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

3x2  b2  l 2 
dx 6 EIl
dyL
dx
Let  
dyL
dx

Fb  b 2  l 2 
6 EIl
x 0
and set I 
x 0
dL 
 d L4
64
. Thus,
32 Fb  b 2  l 2 
1/ 4
Ans.
3 El
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 
3 El
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,
Shigley’s MED, 11th edition
Chapter 4 Solutions, Page 32/96
 32nFb  l 2  b 2  

d 
3 El


d
4
32(1.28)(3000)(200)  300 2  200 2 
3 (207)103 (300)(0.001)
d  38.1 mm
I
1/4
  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

3 x 2  6  300  x  1002  2  3002    0  x 2  600 x  63333  0
x
1
600  600 2  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 =  (1.254)/64 = 0.1198 in4. From Table A-9, beam 6
2
 F a (l  x) 2
 F b x

  1 1
( x  a12  2lx)    2 2 ( x 2  b2 2  l 2 ) 
 6 EIl
  6 EIl

2
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.
______________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 4 Solutions, Page 33/96
4-48 I =  (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
6 EIl
6  30 10  0.1198  20 
 5.217 10 6  x  x 2  175 
5  x  20
y
F1a1  l  x 
6 EIl
x
2
(1)
 a12  2lx  
150  5  20  x 
 x 2  52  2  20  x 
6  30 10  0.1198  20 
6
 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  10 2  20 2 

6
6 EIl
6  30 10  0.1198  20 
 5.797 106  x  x 2  300 
(3)
10  x  20
F a l  x  2
250 10  20  x 
 x 2  102  2  20  x 
z 2 2
x  a22  2lx  

6 EIl
6  30 106  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
 = 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)
______________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 4 Solutions, Page 34/96
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

(3 x 2  3a 2  6al  2l 2 )
dx
6 EIl
y AB 
With I =  d 4/64, the slope at the left bearing is
dy AB
dx
x 0
 A 
MB
(3a 2  6al  2l 2 )
4
6 E  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

3 E Al
3 (30)10 (0.002)(10) 
Ans.
 0.461 in
______________________________________________________________________________
4-50 From Table A-5, E = 10.4 Mpsi
MO = 0 = 18 FBC  6(100)  FBC = 33.33 lbf
The cross sectional area of rod BC is A =  (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
 
6
 AE  BC  0.1963 30 10 
Ans.
______________________________________________________________________________
4-51 MO = 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
  0.5  / 4  30 10 
 AE  AC


By similar triangles the deflection at B due to the elongation of the rod AC is
y A y B1

6
18

y B1  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,
Shigley’s MED, 11th edition
Chapter 4 Solutions, Page 35/96
 dy
 Fa 2
 d  F (x  l)
Fa 2

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
2
3( x  l )  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, EA = 71.7 GPa, ES = 207 GPa.
MO = 0 = 450 FCD – 650(4000)  FCD = 5777.8 N
5 777.8  220 
 FL 
 0.2172 mm
yD   
 
 / 4  0.0062  207 109
 AE CD
The deflection of B due to yD is
(yB)1 = (650/450) ( 0.2172) =  0.3137 mm
Treat beam OADB as simply-supported at O and D. Use beam 10 of Table A-9 and use
the equation for yC,
2
Fa 2  l  a 
4000  0.2   0.450  0.2 

 yB 2  
3EI
3  71.7 109  0.012  0.0503 / 12 
 3.868 10 3  m  3.868 mm
Superposition:
yB = (yB)1 + (yB)2 =  0.3137  3.868 =  4.18 mm
Ans.
______________________________________________________________________________
4-53 From Table A-5, EA = 71.7 MPa, ES = 207 MPa
MO = 0 = 450 FCD – 150(4000)  FCD = 1333 N
1333  220 
 FL 
yD   
 0.0501 mm
 
 / 4  0.0062  207 109
 AE CD
The deflection of B due to yD is
(yB)1 = (650/450) ( 0.0501) =  0.0724 mm
Treat beam OADB as simply-supported at O and D. Find slope at B for beam 6 of Table
A-9,
Fa
  x 3  3lx 2   2l 2  a 2  x  la 2 
yBC 

6 EIl 
dy
Fa
 3 x 2  6lx  2l 2  a 2 
 BC  BC 
dx
6 EIl
For OADB,
Shigley’s MED, 11th edition
Chapter 4 Solutions, Page 36/96
4000  0.15   3  0.450   6  0.450  0.450  2  0.450   0.152 


D 
9
3
6  71.7 10  0.012   0.050 / 12  0.450
2
2
 0.004 463 rad
Deflection of B due to slope at D is
(yB)2 = 0.004 463(200) = 0.8926 mm
Superposition:
yB = (yB)1 + (yB)2 =  0.0725 + 0.8926 = 0.820 mm
Ans.
______________________________________________________________________________
4-54 From Table A-5, EA = 10.4 Mpsi, ES = 30 Mpsi
MO = 0 = 18 FCD – 6(100)  FCD = 33.33 lbf
33.33 12 
 FL 
 6.792 105  in
yB   
 
2
6
 / 4  0.5  30 10
 AE  BC
The deflection of A due to yB is
(yA)1 = (6/18) ( 6.792)105 =  2.264 (105) in
Treat beam OAB as simply-supported at O and D. Use beam 6 of Table A-9
 y A 2
100 12  6  62  122  182 
Fbx 2
2
2

 5.538 10 3  in
x b l   

6
3
6 EIl
6 10.4 10  0.25  2 / 12   18
Superposition:
yA = (yA)1 + (yA)2 =  2.264 (105)  5.538 (103) =  5.56 (103) in
Ans.
______________________________________________________________________________
4-55 From Table A-5, ES = 30 Mpsi
MO = 0 = 18 FAC – 11(100)  FAC = 183.3 lbf
183.3 12 
 FL 
 3.734 104  in
yA   
Ans.
 
2
6
AE

/
4
0.5
30
10





 AC
_____________________________________________________________________________________________________________________
4-56
From Table A-5, EA = 71.7 MPa, ES = 207 MPa
MO = 0 = 450 FCD – 650(4000)  FCD = 5777.8 N
5 777.8  220 
 FL 
yD   
 0.2172 mm
 
 / 4  0.0062  207 109
 AE CD
The deflection of A due to yD is
(yA)1 = (150/300) ( 0.2172) =  0.1086 mm
Treat beam OADB as simply-supported at O and D. Use beam 10 of Table A-9
4000  0.2  0.15  0.450 2  0.150 2 
Fax 2
2
 y A 2 
 l  x   6 71.7 109  0.012 0.0503 / 12  0.450
6 EIl
  

 
 0.8926 103  m  0.8926 mm
Superposition:
yA = (yA)1 + (yA)2 =  0.1086 + 0.8926 = 0.784 mm
Ans.
______________________________________________________________________________
4-57 From Table A-5, EA = 71.7 MPa, ES = 207 MPa
Shigley’s MED, 11th edition
Chapter 4 Solutions, Page 37/96
MO = 0 = 450 FCD – 150(4000)  FCD = 1333 N
1333  220 
 FL 
yD   
 0.0501 mm
 
 / 4  0.0062  207 109
 AE CD
The deflection of A due to yD is
(yA)1 = (150/300) ( 0.0501) =  0.02505 mm
Treat beam OADB as simply-supported at O and D. Use beam 6 of Table A-9
4000  0.3 0.15  0.152  0.32  0.452 
Fbx 2
2
2
 y A 2 
 x  b  l   6 71.7 109  0.012 0.0503 / 12  0.450
6 EIl
  

 
 0.6695 103  m  0.6695 mm
Superposition:
yA = (yA)1 + (yA)2 =  0.02505  0.6695 =  0.695 mm
Ans.
______________________________________________________________________________
4-58 From Table A-5, E = 207 GPa, and G = 79.3 GPa.
FlOC l AB 2
Fl AC l AB 2
Fl AB 3
Fl AB 3
 Tl 
 Tl 
yB  
l
l





 AB 
 AB
3EI AB G  dOC 4 / 32  G  d AC 4 / 32  3E  d 34 / 64 
 GJ OC
 GJ  AC

l
32 Fl AB 2  lOC
2l AB 
 AC 4 

4
4

 Gd OC Gd AC 3Ed AB 
The spring rate is k = F/ yB. Thus
 32l 2
k   AB
 
 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

 79.3 10 18 79.3 10 12 3  207 10 8   

 8.10 N/mm
Ans.
_____________________________________________________________________________
4-59 For the beam deflection, use beam 5 of Table A-9.
F
R1  R2 
2
F
F
1 
, and  2 
2k1
2k 2
y AB  1 
1   2
l
x
Fx
(4 x 2  3l 3 )
48 EI
 1 k2  k1

x

y AB  F  
x
(4 x 2  3l 3 ) 
48 EI
 2k1 2k1k2l

Shigley’s MED, 11th edition
Ans.
Chapter 4 Solutions, Page 38/96
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 Fk 2  Fk1

yBC 
x
 x   2lx 
2k1
2k1k 2l
4
EIl


 1 k2  k1
 l  x  4 x 2  l 2  8lx  Ans.
 F 

x


48 EI
 2k1 2k1k2l

______________________________________________________________________________
4-60
Fa
F
, and R2  (l  a )
l
l
Fa
F
, and  2 
(l  a )
1 
lk1
lk 2
R1 
y AB  1 
1   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

 
F (x  l)
( x  l ) 2  a (3 x  l ) 
yBC  1  1 2 x 
l
6 EI 
 a

x
(x  l)
( x  l ) 2  a (3 x  l )  
yBC  F  
Ans.
 k a  k1  l  a   
2  2
6 EI
 k1l k1k2l

______________________________________________________________________________
4-61 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  b2  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.
______________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 4 Solutions, Page 39/96
4-62
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
6
2
EI
 9.5 10  x  2 750 x   1250 x  2000  C1
6
dx
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  3000 2  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
 max  

 163 106  Pa  163MPa Ans.
4.14(106 )
I
The solutions are the same as Prob. 4-10.
______________________________________________________________________________
4-63 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, 11th edition
Chapter 4 Solutions, Page 40/96
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. 4-11.
______________________________________________________________________________
 max 
4-64 I =  (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
EIy  75.5 x 3  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 10 4  x 


EI
Evaluating at x = 15 in,
Shigley’s MED, 11th edition
Chapter 4 Solutions, Page 41/96
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 10 4  (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
4-65 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  C 2
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
Ans.
 6.667 107  70 x 3  4.167 x 4  163.3 x  10  2833 x 


The tabular results and plot are exactly the same as Prob. 4-21.
______________________________________________________________________________
4-66 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
Shigley’s MED, 11th edition
Chapter 4 Solutions, Page 42/96
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.
______________________________________________________________________________
4-67
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
1
dy 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

y = 0 at x = l

EIy 
y
C2 = 0
1
1
C1   R1l 2  M Al . Thus,
6
2
1
1
1
1
3
 1

R1 x3  M A x 2  R2 x  l    R1l 2  M Al  x
6
2
6
2
 6

1 
3
 M A  Fa  x 3  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  2 M 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, 11th edition
Chapter 4 Solutions, Page 43/96
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  x 3  3 x 2l   x  l   2 xl 2   F   ax 3   l  a  x  l   axl 2 





6 EIl
1
2

 M A  x  l  l 2  Fl  x  l   x  l   a  3 x  l  


6 EIl
 x  l   M l  F  x  l 2  a 3x  l 
Ans.

 

A

6 EI
yBC 






The solutions reduce to the same as Prob. 4-17.
______________________________________________________________________________
w b  a 
1


 R1 
4-68  M D  0  R1l  w  b  a  l  b   b  a  
 2l  b  a 
2
2l


w
w
2
2
M  R1 x 
xa 
xb
2
2
dy 1
w
w
3
3
EI
 R1 x 2 
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
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  1 w b  a
w
w
4
 2l  b  a  x3  x  a  x  b

2l
24
24
EI  6
4
1  1 w b  a 
w
w
4
4 

x 
 2l  b  a  l 3   l  a    l  b   
2l
24
24
l 6
 


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,
w
4
4
2  b  a  2l  b  a  x 3  x  2  b  a  2l  b  a  l 2   l  a    l  b  
y AB 



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

Shigley’s MED, 11th edition
Chapter 4 Solutions, Page 44/96
yBC 

w
4
2  b  a  2l  b  a  x3  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  x3  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.
______________________________________________________________________________
4-69 I1 =  (1.3754)/64 = 0.1755 in4, I2 =  (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
I2
 I1 I 2 
 I1 I 2 
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
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
Shigley’s MED, 11th edition
Chapter 4 Solutions, Page 45/96
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
-0.002
0
1
2
3
4
5
6
7
8
-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.
______________________________________________________________________________
4-70 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
M
F
Fl Fl
l

x

x
I
2 I1
2 I1 4 I1
2
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
E
dy F 2 Fl
Fl
l

x 
x
x
2 I1
4 I1
2
dx 4 I1
Shigley’s MED, 11th edition
1

F
l
x
4 I1
2
2
 C1
Chapter 4 Solutions, Page 46/96
dy/dx = 0 at x = 0
 C1 = 0
2
3
F 3 Fl 2 Fl
l
F
l
Ey 
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
2 x
y
 2 x  6lx  3l x 

24 EI1 
2
2 
3
2

5 Fl 3
F  l
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 
Ans.
 2  l   6l  l   3l  l    2  x     
24 EI1 
2  
16 EI1
 2

The answers are identical to Ex. 4-10.
______________________________________________________________________________
4-71 Place a fictitious 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
ymax
 1
 2
 EI
l/2

0
 M
M
 Q
 
2
 dx  
 Q  0 EI
l /2

0
 wl 
wx2   x 
x

 dx
 2 
2   2 
 
Note, after differentiating with respect to Q, it can be set to zero
l/2
l/2
w  x 3l x 4 
5w


ymax
x
l
x
dx
0   2EI  3  4   384 EI Ans.
0
______________________________________________________________________________
w

2 EI
2
4-72 Place a fictitious force Q pointing downwards at the end. Use the variable x originating at
the free end and positive to the left
wx2
M
M  Qx 
 x
2
Q
l
l
 1 l  M  
1  wx 2 
w
ymax    M 
x3dx

   x  dx 
 dx  


2 EI 0
 EI 0   Q   Q  0 EI 0  2 
wl 4
Ans.
8EI
______________________________________________________________________________

Shigley’s MED, 11th edition
Chapter 4 Solutions, Page 47/96
4-73 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
A 
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)(60 4 ) / 4

 0.182 in in the direction of the 150 lbf force
30(106 )(3.70)
 y A   0.182 in Ans.
______________________________________________________________________________
4-74
P a
RA   Q
2 l
P l  a 
RB  
Q
l
2
M a
P a 
M    Q x
 x
Q l
2 l 
l
l
M a
P a 

M    Q x  P x  
 xl
 x
Q l
2
2
2 l 

M
l  x  (l  a)
M  Q l  a  x 
 l  a  x
Q
We observe that for section BD, the only force is Q, which is zero, so there is no
contribution to the deflection at D from the strain energy in section BD.
0 x
l
2
l  Pa
 1 l  a M 
Pa 2 Pa  
1  l /2 Pa 2
yD  
M
dx
x
dx
x2 
x 
x  dx








l
0
0
/2
Q Q  0 EI 
l
2l
2  
 2l
 EI
The first two integrals can be combined from 0 to l,
3
2
1  Pal 3 Pa  3  l   Pa  2  l    Pal 2




yD 
l
l
Ans.



  
   
EI  6l
l 
 2   4 
 2    16 EI
(b) Table A-9, beam 5 with F = P,
dy
P
P
Slope:  
12 x 2  3l 2  


 4x2  l 2 
dx 48EI
16 EI
Shigley’s MED, 11th edition
Chapter 4 Solutions, Page 48/96
Pl 2
Pl 2
. By symmetry,  C 
16 EI
16 EI
2
Pal
yC  a C 
Ans.
This agrees with part (a)
16 EI
______________________________________________________________________________
At x = 0,  A  
4-75
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
The deflection at the tip is



k
U Tl AC T TlCO T



F GJ AC F GJ CO F
l AB

0
Tl l
Tl l
1
M M
dx  AC AB  CO AB 
EI 3 F
GJ AC
GJ CO
EI AB
l AB
 Fx dx
2
0
3
2
2
3
Tl AC l AB TlCO l AB
Fl AC l AB
FlCO l AB
Fl AB
Fl AB





4
4
4
GJ AC
GJ CO 3EI AB G  d AC
/ 32  G  dCO
/ 32  3E  d AB
/ 64 
2
 l AC
l
32 Fl AB
2l AB 
 CO4 

4
4 
  Gd AC Gd CO 3Ed AB

F



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  
______________________________________________________________________________

4-76 I1 =  (1.3754)/64 = 0.1755 in4, I2 =  (1.754)/64 = 0.4604 in4
Place a fictitious force Q pointing downwards at the midspan of the beam, x = 8 in
Shigley’s MED, 11th edition
Chapter 4 Solutions, Page 49/96
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
3
8
 8 M M 
1
1 
2
2
   2
dx  
900 x dx 
900 x  90  x  3  x dx



EI 2 3 
 0 EI Q Q 0 EI1 0
3
8
300 x 3
1 
1
9



300 x3  90( x 4  2 x 3  x 2 ) 

EI1 0 EI 2 
4
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.
______________________________________________________________________________
4-77 I =  (0.54)/64 = 3.068 (103) in4, J = 2 I = 6.136 (103) in4, A = (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 F x
M  4.2 F
Fa  0.6 F
M
 0.6 x
F
M
 4.2
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
Shigley’s MED, 11th edition
Chapter 4 Solutions, Page 50/96
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
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 10 6 
 B  F 
7
15
15  4.22 
15
2
0.6


x
d
x
dx


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.
*Note. Be careful, this is not the actual deflection of point B. For this, fictitious 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
B = 0.0831 i  0.2862 j  0.00770 k in
From this, the deflection of B in the direction of F is
 B  F  0.6  0.0831  0.8  0.2862   0.279 in
which agrees with our result.
______________________________________________________________________________
4-78 Strain energy. AB: Bending and torsion, BC: Bending and torsion, CD: Bending.
IAB =  (14)/64 = 0.04909 in4, JAB = 2 IAB = 0.09818 in4, IBC = 0.25(1.53)/12 = 0.07031 in4,
ICD =  (0.754)/64 = 0.01553 in4.
For the torsion of bar BC, Eq. (3-41) is in the form of  =TL/(JG), where the equivalent of
J is Jeq = bc 3. With b/c = 1.5/0.25 = 6, JBC = bc 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
Shigley’s MED, 11th edition
Chapter 4 Solutions, Page 51/96
Torsion
BC: Bending
Torsion
CD: Bending
T  5F
M Fx
T  2F
M Fx
T
5
F
M
x
F
T
2
F
M
x
F
U
M
1
Tl T

 M
dx
F
F
JG F
EI
6
5F  6 
2F 5
1
2



F  x  2 d x
5
2


6

3
6
6
0.09818 11.5 10
7.008 10 11.5 10 
30 10  0.04909 0
D 

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.
______________________________________________________________________________
4-79 AAB =  (12)/4 = 0.7854 in2, IAB =  (14)/64 = 0.04909 in4, IBC = 1.5 (0.253)/12 = 1.953
(103) in4, ACD =  (0.752)/4 = 0.4418 in2, IAB =  (0.754)/64 = 0.01553 in4. For (D )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:
0
M y  Fz x
F
Fa
1
Fa  F
F
M y
M y  F x  2 Fz
x
BC:
F
Fa
0
Fa  Fz
F
M y
5
M y  5 F  2 Fz  Fz x
AB:
F
Fa
Fa  F
1
F
Shigley’s MED, 11th edition
Chapter 4 Solutions, Page 52/96
 D  x
1
U  FL  Fa




F  AE CD F EI BC


F  2
1
EI AB
0.4418  30 10

6
5
  F x  2F  x d x
z
0
6
  5F  2 F
z
0
1 
 FL  Fa
 Fz x  5  d x  

 AE  AB F
1
3
F
 5   Fz  52 
3 
30 10 1.953 10   3

6
F  6
F
1


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
 D  x  8.135 104   150   5.286 10 4   100   0.1749 in
Ans.
______________________________________________________________________________
4-80 IOA = IBC =  (1.54)/64 = 0.2485 in4, JOA = JBC = 2 IOA = 0.4970 in4, IAB =  (14)/64 =
0.04909 in4, JAB = 2 IAB = 0.09818 in4, ICD =  (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,
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

EI
F
F
 JG OC F
The terms involving the torsion and bending moments in OC must be split up because of
the changing second-area moments.
 D  y 
Shigley’s MED, 11th edition
Chapter 4 Solutions, Page 53/96
 D  y 
2
12 F  4 
12 F  9 
1


12
12
F x 2d x


6 
6 

6
0.4970 11.5 10
0.09818 11.5 10
30 10  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,
 B  y 
13
12
12 F 13
1
1
2
12


F
x
d
x
F x 2d x

6 


6
6
0.09818 11.5 10
30 10  0.04909 0
30 10  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
Ans.
 0.848 in
Simplified is 0.848/0.706 = 1.20 times greater Ans.
______________________________________________________________________________
4-81 Place a fictitious 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
F  Q  33.33
F
1
Q
 FL  F 
 0  33.3312
 
1  6.79 105  in Ans.

 
2
6

AE
Q


 BC

 Q 0   0.5  / 4  30 10 
Q 0
______________________________________________________________________________
B 
4-82
U
Q
IOB = 0.25(23)/12 = 0.1667 in4
AAC =  (0.52)/4 = 0.1963 in2
MO = 0 = 6 RC  11(100)  18 Q
Shigley’s MED, 11th edition
Chapter 4 Solutions, Page 54/96
RC = 3Q + 183.3
MA = 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
 2 x
Q
M    2Q  83.33 x
Axial in AC:
F  3Q  183.3
F
3
Q
 U 
 FL  F 
 1

  


 Q Q 0  AE  Q  Q 0  EI
B  
183.3 12 
1

3 
6  
0.1963  30 10
EI
 1.121103  
5
M
M 
dx 
Q Q 0
6
2
 100 x  7  x  d x  0 2 83.33 x dx
0
5
6


2


100
7
166.7
x
x
d
x
x
dx



0 
10.4 106  0.1667  0
1
 1.121103   5.768 107  100 129.2   166.7  72    0.0155 in
Ans.
______________________________________________________________________________
4-83 Table A-5, EA = 71.7 GPa, ES = 207 GPa, EI = 71.7(109)0.012(0.053)/12 = 8962.5 N٠m2
F =  4000 lbf
MO = 0 = 450 FAC + 650F  FAC =  1.444 F
F
FAC AC LAC
F
 yB 1 
EA
1.444  4000  1.444  220

207 109   / 4  0.006 2
 0.3135 mm
Shigley’s MED, 11th edition
Chapter 4 Solutions, Page 55/96
MA = 0 = – 450 RO + 200F  RO = 0.4444 F
0  x  0.450 m
M  0.4444 Fx
M / F  0.4444 x
450  x  0.650 m
M / F   0.650  x 
0.65
M
1  0.45
2
0.4444  Fx 2 dx   F  0.65  x  dx 
dx 


0
0.45

F
EI  0
2

4000   0.4444 
1
3 0.65

0.453   0.65  x 

  3.867 mm
0.45
8962.5 
3
3

yB = (yB)1 + (yB)2 =  0.3135  3.867 =  4.18 mm Ans.
______________________________________________________________________________
4-84 Table A-5, EA = 71.7 GPa, ES = 207 GPa, EI = 71.7(109)0.012(0.053)/12 = 8962.5 N٠m2
MO = 0 = 450 FCD + 650Q – 150(4000)  FCD = 1333  1.444 Q
F
FCD CD LCD
Q
Q 0
 yB 1 
EA
1333  1.444  220

207 109   / 4  0.0062
 yB 2 
1
EI
M  F  0.650  x 
l
M
 0.0724 mm
Fy = 0 = RO – 4000 + 1333 – 1.444 Q + Q  RO = 2667 + 0.444 Q
0  x  0.150 m
M   2667  0.444Q  x
M / Q  0.444 x
150  x  0.450 m
M   2667  0.444Q  x  4000  x  0.150    1333  0.444Q  x  600
0.450  x  0.650 m
M / Q  0.444 x
M   0.65  x  Q
 yB 2 
1
EI
l
M
0
M / Q   x
0.45
M
1  0.15
dx

2667 x  0.444 x  dx    1333 x  600  0.444 x  dx 

0.15

Q
EI  0
Q 0
 2667  0.444  0.153  1333  0.444 
600  0.444 
 
0.453  0.153  
0.452  0.152   




3
3
2


 
1

 7.996   8.922 104  m  0.8922 mm
8962.5
yB = (yB)1 + (yB)2 =  0.0724 + 0.8922 = 0.820 mm Ans.
______________________________________________________________________________
4-85 Table A-5, EA = 10.4 Mpsi, ES = 30 Mpsi, EI = 10.4(106)0.25(23)/12 = 1.733 (106) lbf٠in3
MO = 0 = 18 FBC + 6F  FBC =  F / 3

1
EI
Shigley’s MED, 11th edition
Chapter 4 Solutions, Page 56/96
 y A 1 
FBC
FBC
LBC
F
AE
 1/ 3  100 12

 / 4  0.52  30 106
2
 2.264 105  in
Fy = 0 = RO +F + FBC
2
0  x  6 in
M   Fx
3

RO =  2 F / 3
2
M / F   x
3
1
1
M  FBC x   F x
M / F   x
3
3
6
12
l
M
1
1 
2
2
dx 
2 / 3 Fx 2 dx    1/ 3  F x 2 d x 
 y A 2  0 M




0
0
EI
F
EI 
F  4 3 1 3
100

 96   5.540 103  in
 6  12  
6
EI  27
27
 1.733 10 
0  x  12in
yA = (yA)1 + (yA)2 =  2.264 (105)  5.540 (103) =  5.56 (103) in
Ans.
______________________________________________________________________________
4-86 Table A-5, EA = 10.4 Mpsi, ES = 30 Mpsi
MO = 0 =6(FAC +Q)  11(100)
FAC
FAC = 183.3  Q
 1
Q
 y A 1 
FAC
FAC
LAC
Q
Q 0
AE
183.3  112

 3.734 104  in
2
6
 / 4  0.5  30 10
Treating beam OADB as a simply-supported beam pinned at O and A we see that the
force Q does not induce any bending. Thus,
yA = (yA)1 =  3.734 (104) in
Ans.
______________________________________________________________________________
4-87
Table A-5, EA = 71.7 GPa,
k = P /  = 10 (103)/(103)
= 10 (106) N/m,
EI = 71.7 (109)0.005(0.033)/12
= 806.6 N٠m2
(a) OA: RO = F
M
M = RO x = Fx
x
x
Shigley’s MED, 11th edition
Chapter 4 Solutions, Page 57/96
 yB 1 
1
EI
l
M
0
1
M
dx 
EI
F

0.250
0
Fx 2 dx 
300
0.2503 103  1.937 mm

3  806.6 
Ans.
M
x
F
Since AB is the same length as OA, the integration is identical to part (a). Thus
(yB)2 =  1.937 mm
Ans.
(c) AC: Eq. (4-15): MO = 0 = 250 FAC + 500 F  FAC =  2F
F
FAC AC 2 F 2
2
FAC
   4  300   0.12 mm Ans.
U
F 
  y B 3 

U
2k
k
k
F
10 106 
(b) AB:
M Fx
(d)
yB = (yB)1 + (yB)2 + (yB)3 = 2( 1.937)  0.12 =  3.994 mm
Ans.
______________________________________________________________________________
4-88 Table A-5, E = 30 Mpsi, G = 11.5 Mpsi. (EI)AB = 30(106) (/64)14 = 1.473 (106)
lbf٠in2, (JG)AB = (/32)14 (11.5)106 = 1.129 (106) lbf٠in2, (EI)BD = 30(106) 0.25
(1.253)/12 = 1.221 (106) lbf٠in2. F = 200 lbf.
M y
(a) AB at xB: M y  500 xB
1
 EI  AB

l AB
0
0
No contribution from F
M z
 xB
F
M z   F  300  xB
 y D i 
F
Mz
6
 200  300  63
M z
1
2


dxB 
F
x
dx
300

 B B
F
3 1.473106
1.473 106  0
 4.888 10 3  in
Tx = 4F
Shigley’s MED, 11th edition
Tx
4
F
 yD ii
Tx
LAB 4 200 4 6
     0.01701 in
 F

1.129 106 
 JG  AB
Tx
Chapter 4 Solutions, Page 58/96
(yD)1 = (yD)i + (yD)ii =  4.888 (103) +0.01701 = 0.0121 in
(b) BC at xC has no F. Therefore, (yD)2 = 0
Ans.
(c) BD at xD:
M x
 xD
F
M x  FxD
 y D 3
1

 EI  BD

4
0
2
D
Fx dxD 
 3.494 103  in
(d) yD = (yD)1 + (yD)2 + (yD)3 = 0.0121 + 0 + 0.0035 = 0.0156 in
200  43 
3 1.221106
Ans.
Ans.
______________________________________________________________________________
4-89 Table A-5, E = 207 GPa, G = 79.3 GPa, (EI)AB = 207 (109)(/64) 0.0254 = 3.969 (103) m4,
(JG)AB = (/32) 0.0254 (79.3)109 = 3.041 (103) m4
(a) AB: Bending:
M y
0
M y  300 xB
Q
No contribution from Q.
M z
 xB
Q
M z   450  Q  xB
1
EI
 y D i 


M z
dxB
Q
Q 0
l
M
0
1
 EI  AB
z

0.150
0
450 xB2 dxB
450  0.1503 
3  3.969  103
Torsion:
 1.276 104  m  0.1275 mm
Tx  Q  0.1  450  0.125  0.1Q  56.25
 yD ii 
Tx
Tx
LAB
Q
Q 0
 JG  AB

56.25  0.1 0.150
3.041 10
3

Tx
 0.1
Q
 2.775 104  m  0.2775 mm
(yD)1 = (yD)i + (yD)ii = 0.1275  0.2775 =  0.150 mm
(b) BC: Break at xC has no Q in it. Thus, (yD)2 = 0
Ans.
Ans.
(c) BD: Axial has no Q. Thus no contribution. Bending only has Q and since Q = 0, no
contribution. Therefore, (yD)3 = 0
Ans.
Shigley’s MED, 11th edition
Chapter 4 Solutions, Page 59/96
(d)
yD = (yD)1 +(yD)2 +(yD)3 = 0.150 + 0 + 0 =  0.150 mm
Ans.
______________________________________________________________________________
4-90
Table A-5, E = 30 Mpsi, G = 11.5 Mpsi. (EI)AB = 30 (106) (/64) 14 = 1.473 (106) lbf٠in2.
(a) AB: Break at xB:
 z D i 

1
 EI  AB

l
0
My
500  63 
M y   Q  500  xB
M y
Q
M y
Q
 xB
dxB
Q 0
3 1.473106
 0.0244 in
No contribution of Q to Tx, Mz. Thus,
(zD)1 = (zD)i =  0.0244 in
Ans.
(b) BC: Break xC shows no contributions from Q. Thus, (zD)2 = 0
Ans.
(c) BD: Break xD shows no contributions from Q for Mx, My, or Tz. For axial,
F Q
F
 1 but setting Q = 0 gives nothing. Thus, (zD)3 = 0
Q
Ans.
(d) zD = (zD)1 =  0.0244 in Ans.
______________________________________________________________________________
4-91 Table A-5, E = 207 GPa, (EI)AB = 207 (109)(/64) 0.0254 = 3.969 (103) m4,
(EA)BD = 207(109) (/4) 0.0252 = 101.6 (106) m2. F =  300 N.
Shigley’s MED, 11th edition
Chapter 4 Solutions, Page 60/96
(a) AB: Break at xB shows only contribution to My from F
M y
M y  FxB
1
 zD 1 
EI
l
M
0
y
F
 xB
M y
1
dx 
F
 EI  AB

0.150
0
2
B
Fx dxB 
 8.50 105  m  0.0850 mm
300  0.1503 
3  3.969 103
Ans.
(b) BC: Break at xB shows no contribution from F. Thus, (zD)2 = 0
Ans.
(c) BD: Break at xD shows only contribution to axial force from F
Fz  F
 z D 3
Fz
1
F
Fz
LBD 300 1 0.100

F



 2.95 107  m   2.95 10 4  mm
6
EA
101.6
10
 BD
 
Fz
(d) zD =  0.0850 + 0  2.95 (104) =  0.0853 mm
Ans.
Ans.
______________________________________________________________________________
4-92
There is no bending in AB. Using the variable, rotating counterclockwise from B
M  PR sin 
Fr  P cos 
F  P sin 
MF
 2 PR sin 2 
P
M
 R sin 
P
Fr
 cos 
P
F
 sin 
P
A  6(4)  24 mm 2 , ro  40  12 (6)  43 mm, ri  40  12 (6)  37 mm,
From Table 3-4, 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
Shigley’s MED, 11th edition
Chapter 4 Solutions, Page 61/96
From Eq. (4-38)

M  M
  2
0 AeE  P

2
4-93
2

2
PR  sin  


1   MF 

2
2 CFr R  Fr
d


d






0 AE
0
P
AG  P


 d


2 PR sin 2 
2 CPR  cos  

d  
d  
d  
d
0
0
0
0
AeE
AE
AE
AG
EC 
 PR  R
 (10)(40)  40
(207 103 )(1.2) 
1
2
1
2












4 AE  e
G  4(24)(207 103 )  0.07511
79.3 103 
  0.0338 mm
Ans.
______________________________________________________________________________

P  R sin  


2 F R  F
d




0 AE  P


2
2

2
Place a fictitious 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
variable, rotating counterclockwise from B
M
M  PR sin   Q  R  R sin  
 R 1  sin  
Q
Fr
Fr   P  Q  cos 
 cos 
Q
F
F   P  Q  sin 
 sin 
Q
MF   PR sin   QR 1  sin     P  Q  sin 
MF
 PR sin 2    PR sin  1  sin    2QR sin  1  sin  
Q
But after differentiation, we can set Q = 0. Thus,
MF
 PR sin  1  2sin  
Q
A  6(4)  24 mm 2 , ro  40  12 (6)  43 mm, ri  40  12 (6)  37 mm,
From Table 3-4, 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),
Shigley’s MED, 11th edition
Chapter 4 Solutions, Page 62/96




1   MF 
2 F R  F 
2
2 CFr R  Fr 
d

d

d








 d



0
0 AE
0 AE
0
Q
AG  Q 

 Q 
PR 2 2
PR 2 2
PR 2
d
d









sin
1
sin
sin
sin  1  2sin   d


AeE 0
AE 0
AE 0
CPR 2

cos 2  d

0
AG
2
 CE 
R

 PR  PR  
 PR  CPR PR  


   1

   2
  1  2 

4 G 
 4  AeE 4 AE  4
 AE 4 AG AE  4  e
 

2
M  M

AeE  Q
3
 
 1.2  207 10 
 40


1
2





24  207 103  4  0.07511
4 79.3 103  


Ans.
 0.0766 mm
______________________________________________________________________________
4-94 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

10  40 
Section is equivalent to the “T” section of Table 3-4,
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

 
3
  (1.5)  2.25   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    /2
Shigley’s MED, 11th edition
Chapter 4 Solutions, Page 63/96
Fr
 cos  ,
F
Fr  F cos 
F  F sin 
F
 sin 
F
M
 (4  2.125sin  )
F
MF
MF  F (4  2.125sin  ) F sin 
 2 F (4  2.365sin  ) sin 
F
M  F (4  2.125sin  )
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
 /2
F (4)(1)
2 1 4 2
(4  2.125sin  ) 2
F
d

    Fx dx 
E I 0
3.375(G / E ) 0
3.375(0.329)
 / 2 2 F (4  2.125sin  ) sin 
F sin 2  (2.125)
d  
d
0
0
3.375
3.375
2
 / 2 (1) F cos  (2.125)

d
0
3.375(G / E )

 /2
Substitute
I = 2.689 in4, F = 6700 lbf, E = 30 (106) psi, G = 11.5 (106) psi
2  6700   43
  
4
1
  
16    17(1)  4.516   




6 
30 10   3  2.689  3.375(11.5 / 30) 3.375(0.329)   2 
 4 

2.125   
2 
2.125
  
   


4
1
2.125


 





3.375  4  3.375 
 4   3.375 11.5 / 30   4  
 0.0226 in
Ans.
______________________________________________________________________________
4-95 Since R/h = 35/4.5 = 7.78 use Eq. (4-38), integrate from 0
to  , and double the results
M
 R 1  cos 
M  FR 1  cos 
F
Fr
 sin 
Fr  F sin 
F
F
 cos 
F  F cos 
F
MF  F 2 Rcos 1  cos 
  MF 
F
 2 FRcos 1  cos 
From Eq. (4-38),
Shigley’s MED, 11th edition
Chapter 4 Solutions, Page 64/96
 FR 2
FR


(1  cos  ) 2 d 
cos 2  d
  2


0
0
AE
 AeE


2 FR 
1.2 FR  2

cos  1  cos   d 
sin  d 


AE 0
AG 0

E
2 FR  3 R 3

 0.6 

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,
4.5
h
rn 

 34.95173 mm
37.25
ro
ln
ln
32.75
ri
and e = R  rn = 35  34.95173 = 0.04827 mm. Thus,
2 F  35   3
35
3
207 


 0.6
  0.08583 F
3 
13.5  207 10  2 0.04827 2
79.3 
1
where F is in N. For  = 1 mm, F 
 11.65 N
Ans.
0.08583
Note: The first term in the equation for  dominates and this is from the bending moment.
Try Eq. (4-41), and compare the results.
______________________________________________________________________________
4-96 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 ( B ) H  0.
M 
FR
(1  cos  )  HR sin 
2
M
  R sin 
H
0  

2
By symmetry, we may consider only half of the wire form and use twice the strain energy
Eq. (4-41) then becomes,
( B ) H 

 /2
0
2
U

H EI
 /2

M 
  M H  Rd  0
0
 FR

 2 (1  cos  )  HR sin   (  R sin  ) R d  0
F F

F 30
  H 0  H  
 9.55 N
2 4
4
 
Shigley’s MED, 11th edition
Ans.
Chapter 4 Solutions, Page 65/96
The reaction at A is the same where H goes to the left. Substituting H into the moment
equation we get,
FR
M
R


[ (1  cos  )  2sin  ]
0  
 (1  cos  )  2 sin  
2
F 2
2
2

/2
FR
U
2  M 
2
P 

[ (1  cos  )  2sin  ]2 R d
M
Rd 
2

0
EI  F 
EI
P
4
3
 /2
FR
 2  ( 2   2 cos 2   4sin 2   2 2 cos   4 sin   4 sin  cos  ) d
2 EI 0

FR 3    
 
 
 2  2     2    4    2 2  4  2 
2 EI   2 
4
4

M 

(30)(403 )
(3 2  8  4) FR 3 (3 2  8  4)

 0.224 mm
EI
8
8
207 103    2 4  / 64 
Ans.
______________________________________________________________________________
4-97 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   Q ( R sin   l )
 R sin   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),

 A  


1  M 
1
M
Rd  
EI  Q 
 Q 0 EI
 /2
0
PR 2

EI

 /2
0

 /2
0
PR sin   R sin   l Rd
1(2.52 )
PR 2  



 R sin   l sin   d  EI  4 R  l  
 (2.5)  2 
6
4

30 10    0.125  / 64   4
2
 0.0689 in
Ans.
______________________________________________________________________________
4-98 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  )  l 
M BC
  R (1  cos  )  l 
P
From Eq. (4-41) with the addition of the bending strain energy in the straight portion of
the wire,
Shigley’s MED, 11th edition
Chapter 4 Solutions, Page 66/96
 /2 1 
M BC 
1 
M AB 
M AB
dx  


 M BC
Rd
0 EI
0
EI 
P 
P 

P l 2
PR  /2
2
x dx 
R (1  cos  )  l  d




EI 0
EI 0
3
Pl
PR  / 2 2
 R (1  2 cos   cos 2  )  2 Rl (1  cos  )  l 2 d



0
3EI EI
Pl 3 PR  / 2 2
 R cos 2    2 R 2  2 Rl  cos   ( R  l ) 2 d



3EI EI 0 

Pl 3 PR   2

R   2 R 2  2 Rl   ( R  l )2 



2
3EI EI  4

A  
l
P

EI

l3  3

2
2
  R  R  2 R  2 Rl   R( R  l ) 
2
3 4

 23 
1

2
3
2
 3  4 (2.5 )  2.5  2(2.5 )  2(2.5)(2)   2  2.5  2.5  2  
6
4
30 10    0.125  / 64 

 0.106 in Ans.
______________________________________________________________________________
4-99
R = 2.5 in, d = 0.125 in, l = 2 in, P = 1 lbf, E = 30 Mpsi.
EI = 30 (106) (/64) 0.1254 = 359.53 lbf٠in2.
Member ABC: 0  x  l / 2
M  Qx
M
x
Q
Since Q = 0, there is no contribution.
M  Qx  P  x  l / 2 
l/2 xl
 1
 EI
 A 1  
l
M
0
M
x
Q
l
M 
1 
0   P  x  l / 2  xdx 
dx  
l /2

Q  Q  0 EI 
l
3
2
P  x 3 x 2l 
P  l 3   l / 2   l 3  l / 2  l 

 
 

 
 
EI  3
4  l / 2 EI  3  3  4
4 


3
3
5 1 2
5 Pl


 2.32 103  in
48 EI 48  359.53 
Member CD:
Shigley’s MED, 11th edition
Chapter 4 Solutions, Page 67/96
M  Q l  R 1  cos     P l / 2  R 1  cos   
0  
M
 l  R 1  cos   
Q 
1 
P l / 2  R 1  cos    l  R 1  cos    Rd
EI 0 
PR   2
2
l / 2   3l / 2  R 1  cos    R 2 1  cos    d



EI 0 
PR  2
l / 2   3l / 2  R   3l / 2  R cos   R 2  2 R 2 cos   R 2 cos 2   d



0
EI
PR   2
 1  cos 2
l / 2   3l / 2  R  R 2   3l / 2  R cos   2 R 2 cos   R 2 



0
EI 
2

 A  2 

  d

PR  2
l / 2   3l / 2  R   3 / 2  R 2    3l / 2  R  2 R 2  cos   1 / 2  R 2 cos 2 d

0
EI
 1 2.5
 PR 2
l / 2   3l / 2  R   3 / 2  R 2  
22 / 2   3  2  / 2   2.5    3 / 2  2.52



359.53
EI






 0.4123 in
A = (A)1 +(A)2 = 2.32 (103) + 0.4123 = 0.4146 in Ans.
______________________________________________________________________________
4-100 E = 30 Mpsi, EI = 30 (106) (/64) (0.125)4 = 359.53 lbf٠in2.
0  1   / 2
M   P  Q  R cos 1
0  2  
M  QR cos  2
M
 R cos 1
Q
M
 R cos  2
Q
No contribution in 2 range since Q = 0. Thus
1
A 
EI


 /2
0
M
1

M
Rd1
Q
EI
Q 0
PR 3   sin 21
 
2 EI  2
2
 /2
0

 /2
0
PR 3
PR cos 1 Rd1 
2 EI
2
 /2
 1  cos 2 d
2
0
  PR 3  1 2.53

 0.0341 in
 
4
EI
4
359.53



1
1
Ans.
______________________________________________________________________________
4-101 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.
Shigley’s MED, 11th edition
Chapter 4 Solutions, Page 68/96

Place a fictitious 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  
Q
M  PR sin   QR 1  cos  
From Eq. (4-41)
  / 2 1  M
 A V   0
M
EI
 Q



PR 3
EI

 /2
0


1
Rd  

 Q  0 EI
 sin   sin  cos   d 
1 2.53 
2  30 106   0.1254  / 64 

 /2
0
PR sin   R 1  cos  Rd
PR 3  1  PR 3
1   
EI  2  2 EI
 0.0217 in 
Ans.
_____________________________________________________________________________
4-102 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 fictitious 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
  R sin 
Q
M  P  R 1  cos    l   QR sin 
From Eq. (4-41)
  / 2 1  M 

1
 A V   0
M
Rd  
EI  Q 

 Q  0 EI


PR 2
EI
 /2

0

 /2
0
P  R 1  cos    l    R sin   Rd
 R sin   R sin  cos   l sin   d  
1 2.52 
2  30 106   0.1254  / 64 
PR 2 
PR 2
1 
R
l
R




 R  2l 


EI 
2 
2 EI
 2.5  2  2    0.0565 in
Since the deflection is negative,  is in the opposite direction of Q. Thus the deflection is
  0.0565 in 
Ans.
_____________________________________________________________________________
4-103 EI =30(106) (/64)(0.125)4 = 359.53 lbf٠in2.
Shigley’s MED, 11th edition
Chapter 4 Solutions, Page 69/96
AB: Only Q and since Q = 0, no contribution to (A)V.
CD:
M  QR sin   P  l / 2   R 1  cos  
 A V

1
EI


0
M
M
 R sin 
Q
M
Rd
Q
Q 0
1 
P  l / 2   R 1  cos    R sin  Rd
EI 0 

PR 2
PR 2
   l / 2  cos   R cos    R / 2  sin 2   

l  2R 
0
EI
EI
1 2.52


 2  2  2.5    0.1217 in 
Ans.
359.53 

______________________________________________________________________________
4-104 EI =30(106) (/64)(0.125)4 = 359.53 lbf٠in2.
AB: Only Q. Since Q = 0, no contribution.
BC:
M
 R 1  cos  
Q
M  QR 1  cos    PR sin 
 A V

1
EI

 /2
0
M
M
Rd
Q

Q 0
1
EI
 /2
  PR sin   R 1  cos   d
0
 /2
1 2 
PR 3 
PR 3  1 
cos
sin








1  
EI 
EI  2 
2
0
3  2.53 
3PR 3


 0.0652 in 
2 EI
2  359.53
Ans.
________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 4 Solutions, Page 70/96
4-105 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  be counterclockwise originating at D
M
M  FR sin 
 R sin 
0  
F
Using Eqs. (4-29) and (4-41)
 Fl 
F

1 
M 
Fl

FR 3
AB
 

sin 2  d
1  0

M
Rd 
0
EI  F 
AE
EI
 AE  AB F
  403  
Fl  FR 3 F  l  R 3 
9.81 
80



  


AE 2 EI
E  A 2 I  207 103     2 2  / 4  2   24  / 64  




 6.067 mm
Ans.
_____________________________________________________________________________
4-106 AOA = 2(0.25) = 0.5 in2,
IOAB = 0.25(23)/12 = 0.1667 in4,
IAC =  (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
Bending M OA  2 Fx
AB: Bending
M AB   F x
M OA
 2 x
F
M AB
 x
F
AC: Isolating the upper curved section
M AC  3FR  sin   cos   1
Shigley’s MED, 11th edition
M AC
 3R  sin   cos   1
F
Chapter 4 Solutions, Page 71/96
10
1
F
 Fl 
1
OA

 
4 Fx 2 dx 



AE
F
EI
 OAB 0
 EI OAB

OA
9 FR3

 EI  AC

 /2
  sin   cos   1
0
3F 10 
0.5 10.4 106

 3 
9 F 103 
2
20
Fx
2
dx
0
d
4 F 103 
F  203 

3 10.4 106  0.1667  3 10.4 10 6  0.1667 
 /2
30 10  3.068 10
6
3
 sin

2
  2 sin  cos   2 sin   cos 2   2 cos   1 d
0



 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.
_____________________________________________________________________________
4-107 AOA = 2(0.25) = 0.5 in2,
IOAB = 0.25(23)/12 = 0.1667 in4,
IAC =  (0.54)/64 = 3.068 (10-3) in4
Applying a vertical fictitious force, Q, at
A, from statics the reactions are as
shown. The fictitious force is transmitted
through section OA and member AC.
FOA
1
OA: FOA  3F  Q
Q
AC:
M AC   3F  Q  R sin    3F  Q  R 1  cos  
M AC
 R  sin   cos   1
Q
 Fl   FOA   1   / 2

M AC
  
Rd 
     M AC
 
Q
 AE OA  Q   EI  AC 0
 Q 0


3FlOA
3FR 3

 AE OA  EI  AC
3 100 10
 /2
  sin   cos   1
2
d
0
3 100 103



  1  2   2    0.462 in
4
2
10.4 10  0.5 30 10  3.068 10   4
6

6
3
Ans.
_____________________________________________________________________________
4-108
I =  (64)/64 = 63.62 mm4
0 /2
Shigley’s MED, 11th edition
Chapter 4 Solutions, Page 72/96
M
 R sin 
F
T
T  FR (1  cos  )
 R (1  cos  )
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 
( A ) y  
U
1
 
F
 EI

 /2
0
F ( R sin  )2 R d 
1
GJ

 /2
0

F [ R (1  cos  )]2 R d 

Integrating and substituting J  2 I and G  E /  2 1   
FR3  
FR 3
 3













(1
)
2
4
8
(3
8)




EI  4
4 EI
 4

3
(250)(80)
 [4  8  (3  8)(0.29)]
  12.5 mm Ans.
4(200)103  63.62 
_____________________________________________________________________________
( A ) y  
4-109 The force applied to the copper and steel wire assembly is
Fc  Fs  400 lbf
(1)
Since the deflections are equal,  c   s
 Fl   Fl 

 

 AE c  AE  s
Fc l
Fs l

2
6
3( / 4)(0.1019) (17.2)10
( / 4)(0.1055) 2 (30)10 6
Yields, Fc = 1.6046 Fs. Substituting this into Eq. (1) gives
1.6046 Fs  Fs  2.6046Fs  400

Fs  153.6 lbf
Fc  1.6046 Fs  246.5 lbf
F
246.5
c  c 
 10 075 psi  10.1 kpsi Ans.
Ac 3( / 4)(0.1019)2
F
153.6
s  s 
 17 571 psi  17.6 kpsi Ans.
As ( / 4)(0.10552 )
 Fl 
153.6(100)(12)
 0.703 in
 
 
2
6
 AE  s ( / 4)(0.1055) (30)10
Ans.
_____________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 4 Solutions, Page 73/96
 b  0.75(65)  48.8 kpsi
4-110 (a) Bolt stress
Ans.
 
Fb  6 b Ab  6(48.8)   (0.52 )  57.5 kips
4
F
57.43
Cylinder stress
c   b 
 13.9 kpsi Ans.
Ac ( / 4)(5.52  52 )
(b) Force from pressure
 D2
 (52 )
(500)  9817 lbf  9.82 kip
P
p
4
4
Fx = 0
Total bolt force
Pb + Pc = 9.82
(1)
Since  c   b ,
Pc l
Pb l

2
2
( / 4)(5.5  5 ) E 6( / 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
 b  48.8 
 50.7 kpsi Ans.
6( / 4)(0.52 )
7.638
 c  13.9 
 12.0 kpsi Ans.
( / 4)(5.52  52 )
_____________________________________________________________________________
4-111
(1)
Tc + T s = T
c = s

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.
_____________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 4 Solutions, Page 74/96
4-112 RO + RB = W
(1)
OA = AB
 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 
A  
 0.0223 mm Ans.
 
3
 AE OA 10(60)(71.7)(10 )
______________________________________________________________________________
From Eq. (2)
4-113 See figure in Prob. 4-112 solution.
Procedure 1:
1. Let RB be the redundant reaction.
2. Statics. RO + RB = 4 000 N
3. Deflection of point B.  B 
 RO = 4 000  RB
RB  600 
AE

(1)
 RB  4000  400   0
AE
(2)
4. From Eq. (2), AE cancels and RB = 1 600 N
Ans.
and from Eq. (1), RO = 4 000  1 600 = 2 400 N
Ans.
2 400(400)
 Fl 
 0.0223 mm Ans.
 
3
 AE OA 10(60)(71.7)(10 )
_____________________________________________________________________________
A  
4-114 (a) Without the right-hand wall, the deflection of point C would be
5 103  8
2 103  5
Fl
C  


AE   / 4  0.752 10.4 106  / 4  0.52 10.4 106
 0.01360 in  0.005 in  Hits wall
Shigley’s MED, 11th edition
Ans.
Chapter 4 Solutions, Page 75/96
(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



 
C 
2
6
2
 / 4  0.75 10.4 10  / 4  0.5 10.4 106
 0.01360 
4 RC
5 
 8
 2   0.005
6 
2
 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.

 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
2 947  8 
RA  8 

 5.13 10 3  in  Ans.
AAB E   / 4  0.752 10.4 106
_____________________________________________________________________________
 B   AB 
4-115 Since OA = AB,
TOA (4) TAB (6)

JG
JG
Statics. TOA + TAB = 200 (2)

Substitute Eq. (1) into Eq. (2),
3
5
TAB  TAB  TAB  200

2
2
From Eq. (1)
3
TOA  TAB
2
TAB  80 lbf  in
(1)
Ans.
3
3
TOA  TAB  80  120 lbf  in
Ans.
2
2
80  6 
180
 TL 
A  
 0.390 0
 
4
6
 JG  AB  / 32  0.5 11.5 10 
 max 
16T
d3
 AB 

16 80 
  0.53 
 OA 
16 120 
  0.53 
Ans.
 4890 psi  4.89 kpsi
 3260 psi  3.26 kpsi
Ans.
Ans.
_____________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 4 Solutions, Page 76/96
4-116 Since OA = AB,
TOA (4)
TAB (6)

4
 / 32  0.5 G  / 32  0.754 G
TOA  0.2963 TAB

(1)
Statics. TOA + TAB = 200 (2)
Substitute Eq. (1) into Eq. (2),
0.2963TAB  TAB  1.2963TAB  200
TAB  154.3 lbf  in

TOA  0.2963TAB  0.2963 154.3  45.7 lbf  in
From Eq. (1)
A 
154.3  6 
180
 0.1480
4
6
 / 32  0.75 11.5 10 
 max 
16T
d3
 AB 

 OA 
16 154.3
  0.753 
16  45.7 
  0.53 
Ans.
Ans.
Ans.
 1862 psi 1.86 kpsi
 1862 psi 1.86 kpsi
Ans.
Ans.
_____________________________________________________________________________
4-117 Procedure 1
1. Arbitrarily, choose RC as a redundant reaction.
2. Statics. Fx = 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)
C  

 C
0
AE
4. The deflection equation simplifies to
 45 RC + 60(103) = 0
From Eq. (1),
AE
AE
 RC = 1 333 lbf = 1.33 kip
RO = 6(103)  1 333 = 4 667 lbf = 4.67 kip
Ans.
Ans.
FAB = FB + RC = 6 +1.333 = 7.333 kips compression
Shigley’s MED, 11th edition
Chapter 4 Solutions, Page 77/96
FAB 7.333

 14.7 kpsi Ans.
A
(0.5)(1)
Deflection of A. Since OA is in tension,
Rl
4 667(20)
 A   OA  O OA 
 0.00622 in Ans.
AE
(0.5)(1)(30)106
_____________________________________________________________________________
 AB 
4-118 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 
yB 
RB  l  a 
3
3EI

w l  a 
2
24 EI
 4l  l  a    l  a 2  6l 2   0


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
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
_____________________________________________________________________________
4-119 See figure in Prob. 4-118 solution.
MC 
Procedure 1
1. Choose RB as redundant reaction.
2. Statics. RC = wl  RB
(1)
1
M C  wl 2  RB  l  a 
2
Shigley’s MED, 11th edition
(2)
Chapter 4 Solutions, Page 78/96
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
yB 
1
l
1
U
M
M
dx


RB EI 0 RB
l
or,
M
 xa
RB
1
l
1
1
1  1


 w x 2  0  dx 
 w x 2  RB  x  a    x  a  dx  0



EI 0 2
EI a  2

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 
4-120 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 
Shigley’s MED, 11th edition
(1)
(2)
Chapter 4 Solutions, Page 79/96
1
M  R1 x  w x 2  M 1
2
1
dy 1
 R1 x 2  w x3  M 1 x  C1
EI
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
12.75 R2 
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
y x 12in 
1
1 500 4 1
1
3

554.59
12
12  1310.112 2


6 
24 12
2
25.5 10   6
10.2 1310.112  17  554.59  
 5.51103  in
Ans.
______________________________________________________________________________
4-121 Cable area, A 
Procedure 2

4
(0.52 )  0.1963 in 2
Shigley’s MED, 11th edition
Chapter 4 Solutions, Page 80/96
1. Statics. RA + FBE + FDF = 5(103)
(1)
3 FDF + FBE = 10(103)
(2)
2. Bending moment equation.
1
M  RA x  FBE x  16  5000 x  32
1
1
dy 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 
 6.453(106 ) FBE
yB   
 
6
AE
0.1963(30)10

 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
Simplifying gives 682.7 RA + 232.3 FBE + 16 C1 = 0
(5)
B.C. 2: At x = 48 in,
FDF (38)
 Fl 
yD   
 6.453(106 ) FDF
 
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
 
(6)
Equations (1), (2), (5) and (6) in matrix form are
1
1
0   RA   5000 
 1




1
3
0   FBE   10 000 
 0



0
 682.7 232.3
0
16   FDF  



6



3.413
10
C
18
432
5
461
232.3
48



 1  

Solve simultaneously or use software. The results are
RA =  970.5 lbf,
Shigley’s MED, 11th edition
FBE = 3956 lbf,
FDF = 2015 lbf, and C1 =  16 020 lbfin2.
Chapter 4 Solutions, Page 81/96
3956
2015
 20.2 kpsi,  DF 
 10.3 kpsi
0.1963
0.1963
EI = 30(106)(1.2) = 36(106) lbfin2
 BE 
y

Ans.
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
3
B: x = 16 in,
yC 
1
 161.8  323   659.3  32  16 3  16 020  32  
6 

36 10 
 0.0865 in
 
36 10
6

1
 161.8 163   16 020 16    0.0255 in
6 

36 10 
D: x = 48 in,
1
3
 833.3 x  32  16 020 x
yB 
C: x = 32 in,
yD 
3
Ans.
Ans.
 
 161.8 483  659.3  48  16 3  833.3  48  32 3  16 020  48  


 0.0131 in Ans.
______________________________________________________________________________
4-122 Beam: EI = 207(103)21(103)
= 4.347(109) Nmm2.
Rods: A = ( /4)82 = 50.27 mm2.
Procedure 2
1. Statics.
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
Shigley’s MED, 11th edition
Chapter 4 Solutions, Page 82/96
1
1
dy
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 
yB   
 4.805 106  FBE
 
3
50.27
207
10
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,
4.347 109   6.246 106  FDF   
Simplifying gives
1000
1
3
2253   FBE  225  75 

3
6
1
3
 RC  225  150   C1  225  C2
6
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
Shigley’s MED, 11th edition
Chapter 4 Solutions, Page 83/96
 2 103  
1
1
0 0
1



  RC  
3
1
2
0
0
0

6 10  

  FBE 





0
20.89 103 
0
75 1   F   140.6 106 
 

  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.
7
2
8
3
and C1 = 1.036 (10 ) Nmm , C2 =  7.243 (10 ) Nmm .
The bolt stresses are BE = 4189/50.27 = 83.3 MPa, DF =  189/50.27=  3.8 MPa Ans.
The deflections are
From Eq. (4)
yA 
1
 7.243 108   0.167 mm

4.347 109  
Ans.
For points B and D use the axial deflection equations*.
4189  50 
 Fl 
yB   
 0.0201 mm
 
50.27  207 103
 AE  BE
Ans.
189  65 
 Fl 
yD  
Ans.
 1.18 10 3  mm
 
3
AE
50.27
207
10
 

 DF
*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.
______________________________________________________________________________
4-123 Everything in Ex. 4-15 is the same except kB = 40 MN/m.
yB =  FB / kB =  FB / 40 (106) =  2.5 (108) FB
Equation (7) is replaced by
EI  2.5 108 FB  
FA
F
3
3
 0.2   B  0   0.2C1  C2
6
6
With EI = 1.25 (104) N٠m2, the equation reduces to
 1.3333 (103) FA + 3.125 (104) FB + 0.2 C1 + C2
Shigley’s MED, 11th edition
Chapter 4 Solutions, Page 84/96
The remaining equations come from Ex. 4-15
1
0
1
1


4
0
0
3

4

 4.7746 10 
0
0
0

 1.3333 103  3.125 104 
0
0.2

3
4
4

 7.1458 10  5.625 10  6.3662 10  0.35
0
 FA   0 
0   
 FB
0 



1   F   18.75

 C 



1  C1
0 

  
C
0





1  2  
Solving for the unknowns gives
 FA   2350 N

 F   5484 N

 B  


F
3134
N
 C 

 C   95.239 N  m 2 
 1 

C2  17.628 N  m3 
Equation (3): 1.25 10 4  y  
Ans.
2350 3 5484
3
x 
x  0.2  95.239 x  17.628
6
6
Which reduces to y =  0.03133 x3 + 0.07312 x  0.23 +7.619 (103) x  1.410 (103)
At x = 0, y = yA =  1.410 (103) m =  1.41 mm
Ans.
At x = 0.2 m, y = yB =  0.03133(0.2)3 +7.619(103)0.2  1.410(103)
=  1.37 (104) m =  0.137 mm
Ans.
At x = 0.35 m,
y = yC =  0.03133(0.35)3 + 0.07312 (0.35  0.2)3 +7.619(103)0.35  1.410(103)
= 1.60 (104) m =  0.160 mm
Ans.
______________________________________________________________________________
4-124 (a) The cross section at A does not rotate. Thus, for a single quadrant we have
U
0
M A
The bending moment at an angle  to the x axis is
FR
M
1
M  MA 
1  cos  
2
M A
The rotation at A is
 /2
U
M
1
A 

M
Rd  0

M A EI 0
M A
Shigley’s MED, 11th edition
Chapter 4 Solutions, Page 85/96
Thus,
1
EI
 /2
or,

  M
A

0
FR
1  cos    1 Rd  0
2


FR   FR

0
MA 
 
2 2
2

FR  2 
1  
2  
Substituting this into the equation for M gives
2
FR 
M
 cos    (1)

2 
The maximum occurs at B where  =  /2
MA 
M max  M B  
FR
Ans.

(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   
F

2
Thus,
U
4

D 
F EI
 /2

0
FR 3
M
M
Rd 
F
EI
 /2

0
2
FR3   2 
2

cos

d




  
 
EI  4  

3
FR
 2  8  Ans.

4 EI
______________________________________________________________________________

4-125
Pcr 
I
C 2 EI
l2

 D4  d 4  
 D4
1  K 
4
where K 
64
64
2
4

C E   D
1  K 4 
Pcr  2 

l
 64

d
D
1/4


64 Pcr l 2

D 3
Ans.
4
  CE 1  K  
______________________________________________________________________________
4-126 A 

D 2 1  K 2  ,
I

D 4 1  K 4  
4
64
The radius of gyration, k, is given by
Shigley’s MED, 11th edition

64
D 4 1  K 2 1  K 2  , where K = d / D.
Chapter 4 Solutions, Page 86/96
k2 
I D2

1 K 2 

A 16
From Eq. (4-46)
S y2l 2
S y2l 2
Pcr
S 
S 
 / 4  D 2 1  K 2  y 4 2 k 2CE y 4 2  D 2 / 16 1  K 2  CE
4 Pcr   D 1  K
2
2
S
y

 D 1  K  S y  4 Pcr 
2
2
4 S y2l 2 D 2 1  K 2 
 2 D 2 1  K 2  CE
4 S y2l 2 1  K 2 
 1  K 2  CE


4 S y2l 2 1  K 2 
4 Pcr


D
2
2
2
  S y 1  K   1  K  CE 1  K  S y 
1/ 2
1/2


S yl 2
Pcr

 2

Ans.
2
2
2
  S y 1  K   CE 1  K  
______________________________________________________________________________
0.9
4-127 (a) M 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   2 (207)(10 ) 

  157.4
  
6
 k 1  165(10 ) 
Since (l / k )1  (l / k ) use Johnson formula.
Try 25 mm x 12 mm,
2


165 106

1


6


(142.7)
 29.1 kN
Pcr  0.025(0.012) 165 10 
9 
 2
 1(207)10 



This is significantly greater than the design load of 5492 N found earlier. Check out-ofplane.
 
Shigley’s MED, 11th edition
 
Chapter 4 Solutions, Page 87/96
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
1.2 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
dh
0.012(0.011)
No, bearing stress is not significant. Ans.
______________________________________________________________________________
 
(b)  b  
4-128 (a) M A  0  800  750  
9
2
9  52
FBD  1373 N
(b)  
FBD  500 
Ans.
1373
F

 4.99 MPa
A 25 11
Ans.
(c) lBD  0.92  0.52  1.030 m
In plane, Table 4-2 
C=1
1/ 2
 bh /12 
k  I / A 
 bh 


3
 h / 12  0.025 / 12  0.007 217 m
l / k = 1.030/0.007 217 = 142.7.
1/ 2
2
 l   2 CE 
Eq. (4-45):    

 k 1  S y 
 2 2 1 207 109  


165 106 


1/ 2
 157.4
Since (l / k)1 > l / k, use Johnson formula, Eq. (4-48)
Shigley’s MED, 11th edition
Chapter 4 Solutions, Page 88/96
2
2


165 106 


 Sy l  1 
1


6


Pcr  A  S y  
  0.25  0.011 165 10  
142.7 

9 
 2
 1 207 10  

 2 k  CE 


 26 720 N  26.72 kN
n
Pcr
26.72

 19.5
FBD 1.373
Ans.
(d) Out of plane, Table 4-2, C = 1.2, k = 0.011/ 12 = 3.175 (103) m,
l / k = 1.030/3.175 (103) = 324.4. Since l / k > (l / k)1, use the Euler formula, Eq. (4-44)
 C 2 E 
1.2 2  207 109 
Pcr  A 
0.025
0.011





  6 407 N
2
324.42
  l / k  


n
Pcr 6 407

 4.67
FBD 1373
Ans.
______________________________________________________________________________
4-129 Out of plane bending, C =1.2.


82  4 2
M A  0  400  1200 
FBC 


8


FBC  1386 N
Ans.
k I/A
 bh
3
/12  /  bh   h / 12  0.010 / 12  2.887 103  m
l / k = 0.8/[2.887(103)] = 277.1
1/2
2
 l   2 CE 
Eq. (4-45):    

 k 1  S y 
 2 2 1.2  207 109  


200 106 


1/ 2
 156.6
Since l / k > (l / k)1, use the Euler formula, Eq. (4-44)
 C 2 E 
1.2 2  207 109 
Pcr  A 
 0.02  0.01 
  6 386 N
2 
277.12
  l / k  


Shigley’s MED, 11th edition
Chapter 4 Solutions, Page 89/96
n
Pcr 6 386

 4.6 Ans.
FBD 1386
______________________________________________________________________________
4-130 This is an open-ended design problem with no one distinct solution.
______________________________________________________________________________
F = 1500( /4)22 = 4712 lbf. From Table A-20, Sy = 37.5 kpsi
Pcr = nd F = 2.5(4712) = 11 780 lbf
4-131
(a) Assume Euler with C = 1
1/4
1/ 4
 64 11790  502 
 64 Pcr l 2 
 3



d  3

6
  1 30 10  
  CE 
Use d = 1.25 in. The radius of gyration, k = ( I / A)1/2 = d /4 = 0.3125 in
l
50

 160
k 0.3125

P l2
I  d 4  cr 2
64
C E
 2 2 (1)30 106  
  126

 37.5 103  


2
6
4
  30 10  / 64 1.25
 14194 lbf
Pcr 
502
 1.193 in
1/ 2
1/2
2
 l   2 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
  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   
 12 748 lbf
85.33
6
4
 2
 1 30 10 



2
Use d = 0.75 in.
(c)
Shigley’s MED, 11th edition
Chapter 4 Solutions, Page 90/96
n( a ) 
14194
 3.01
4 712
Ans.
12 748
Ans.
 2.71
4 712
______________________________________________________________________________
n(b ) 
4-132 From Table A-20, Sy = 180 MPa
4F sin = 2 943
735.8
sin 
In range of operation, F is maximum when  = 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
Try b = 5 mm. Out of plane, k = b / 12 = 5/ 12 = 1.443 mm
l
350

 242.6
k 1.443
2
9
 l   2 1.4  207 10  

  
180 106 
 k 1 

1/ 2
Pcr  A
C 2 E
l / k 
2
 5(30)
 178.3
1.4 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
1.4 2  207 103
C 2 E

6(30)
 12 605 N
Pcr  A
2
2
l / k 
 202.1
O.K. Use 25  6 mm bars
Ans. The factor of safety is
12 605
 4.43 Ans.
2843
______________________________________________________________________________
n
4-133 P = 1 500 + 9 000 = 10 500 lbf
Shigley’s MED, 11th edition
Ans.
Chapter 4 Solutions, Page 91/96
MA = 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
10 500  1.607  3 / 2  
P  ec 
1 
  17157 psi  17.16 kpsi Ans.
1  2   
A k 
2.160 
0.953 
______________________________________________________________________________
c  
4-134 This is a design problem which has no single distinct solution.
______________________________________________________________________________
4-135 Loss of potential energy of weight = W (h + )
1
Increase in potential energy of spring = k 2
2
1
W (h + ) = k 2
2
2
2
W
W
or,  2 

h  0 . W = 30 lbf, k = 100 lbf/in, h = 2 in yields
k
k
 2  0.6   1.2 = 0
Taking the positive root [see discussion after Eq. (b), Sec. 4-17]
1
 max  0.6  ( 0.6) 2  4(1.2)   1.436 in
Ans.

2
Fmax = k  max = 100 (1.436) = 143.6 lbf
Ans.
______________________________________________________________________________
4-136 The drop of weight W1 converts potential energy, W1 h, to kinetic energy
Equating these provides the velocity of W1 at impact with W2.
W1h 
Shigley’s MED, 11th edition
1 W1 2
v1
2 g

v1  2 gh
1 W1 2
v1 .
2 g
(1)
Chapter 4 Solutions, Page 92/96
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
W1  W2
W
v2  1 v1
g
g

v2 
W1
W1
v1 
W1  W2
W1  W2
2 gh
(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    k 2
2
2
g
Substituting in Eq. (1) and rearranging results in
W1  W2
W12 h
 2
0
(3)
k
W1  W2 k
Solving for the positive root [see discussion after Eq. (b), Sec. 4-17]
2 2
2
W1  W2 
W12 h 
1  W1  W2


  2
 4

8

2
k
k
W1  W2 k 




W1 = 40 N, W2 = 400 N, h = 200 mm, k = 32 kN/m = 32 N/mm.
(4)
2
1   40  400 
40  400 
402 200 

  29.06 mm
  2 
4
8





2   32 
40  400 32 
 32 


Ans.
Fmax = k = 32(29.06) = 930 N Ans.
______________________________________________________________________________
1
4-137 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
1
1
2
k1a 2  k1  a  y   k 2 y 2
2
2
2
Simplifying gives
Shigley’s MED, 11th edition
 k1  k2  y 2  2ak1 y  0
Chapter 4 Solutions, Page 93/96
2k1a
. Without damping the weight will vibrate between these
k1  k 2
2k1a
two limits. The maximum displacement is thus y max =
Ans.
k1  k 2
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 
Shigley’s MED, 11th edition
Chapter 4 Solutions, Page 94/96
Chapter 3
3-1
M O  0
18 RB  6(100)  0
RB  33.3 lbf
Ans.
Fy  0
RO  RB  100  0
RO  66.7 lbf
Ans.
RC  RB  33.3 lbf
Ans.
______________________________________________________________________________
3-2
Body AB:
Fx  0
RAx  RBx
Fy  0
RAy  RBy
M B  0
RAy (10)  RAx (10)  0
RAx  RAy
Body OAC:
M O  0
RAy (10)  100(30)  0
RAy  300 lbf
Ans.
Fx  0
ROx   RAx  300 lbf
Fy  0
ROy  RAy  100  0
ROy  200 lbf
Ans.
Ans.
______________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 1/114
3-3
0.8
 1.39 kN Ans.
tan 30
0.8
RA 
 1.6 kN Ans.
sin 30
RO 
______________________________________________________________________________
3-4
Step 1: Find RA & RE
4.5
 7.794 m
tan 30
M A  0
h
9 RE  7.794(400 cos 30 )
4.5(400sin 30 )  0
RE  400 N
F
x
F
y
0
Ans.
RAx  400 cos 30  0
RAx  346.4 N
0
RAy  400  400sin 30  0
RAy  200 N
RA  346.42  2002  400 N
Ans.
Step 2: Find components of RC and RD on link 4
M
C
0
400(4.5)   7.794  1.9  RD  0
RD  305.4 N
F
F
Ans.
x
0

 RCx 4  305.4 N
y
0

( RCy ) 4  400 N
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 2/114
Step 3: Find components of RC on link 2
 Fx  0
 RCx 2  305.4  346.4  0
 RCx 2  41 N
F
y
0
R 
Cy 2
 200 N
Ans.
_____________________________________________________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 3/114
3-5
M C  0
1500 R1  300(5)  1200(9)  0
R1  8.2 kN
Ans.
Fy  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
Fy  0
RO  500  40(6)  740 lbf
Ans.
M 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
______________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 4/114
3-7
M B  0
2.2 R1  1(2)  1(4)  0
R1  0.91 kN
Fy  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:
M D  0
200(12)  R2 (10)  40(10)(5)  0
R2  440 lbf
Ans.
Fy  0
200  440  40(10)  R3  0
R3  160 lbf
Ans.
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 5/114
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
0
M  R1 x  9 x  300  5 x  1200  R2 x  1500
1
(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
0  x  300 :

R1  8.2 kN
Ans.
Ans.
V  8.2 kN, M  8.2 x N  m
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.
______________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 6/114
3-10
q  RO x
1
2
 MO x
V  RO  M O x
1
 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.
0  x  1.2 :
V  0.91 kN, M  0.91x kN  m
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.
______________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 7/114
3-12
q  R1 x
1
 400 x  4
0
1
 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  8R1  400(8  4)  0

R1  200 lbf
1
(1)
1
(2)
Ans.
at x = 20+, V =M = 0. Applying Eqs. (1) and (2),
200  400  R2  40(10)  R3  0
R2  R3  600

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
Plots of V and M are the same as in Prob. 3-8.
______________________________________________________________________________
2
3-13 Solution depends upon the beam selected.
______________________________________________________________________________
3-14
(a) Moment at center,
 l  2a 
xc 
2
2
wl
 l   wl  l

2
M c   l  a      
 a
2  2
 2   2  4

At reaction, M r  wa 2 2
a = 2.25, l = 10 in, w = 100 lbf/in
Mc 
100(10)  10

  2.25   125 lbf  in
2  4

100  2.252 
 253 lbf  in
2
(b) Optimal occurs when M c  M r
Mr 
Shigley’s MED, 11th edition
Ans.
Chapter 3 Solutions, Page 8/114
2
wl  l
 wa
a


 a 2  al  0.25l 2  0


2 4
2

Taking the positive root
1
l
2  1  0.207 l
l  l 2  4  0.25l 2   


 2
2
for l = 10 in, w = 100 lbf, a = 0.207(10) = 2.07 in
M min  100 2  2.07 2  214 lbf  in

a

Ans.
______________________________________________________________________________
3-15
(a)
20  10
 5 kpsi
2
20  10
CD 
 15 kpsi
2
C
R  152  82  17 kpsi
 1  5  17  22 kpsi
 2  5  17  12 kpsi
1
8
 p  tan 1    14.04 cw
2
 15 
 1  R  17 kpsi
s  45  14.04  30.96 ccw
(b)
9  16
 12.5 kpsi
2
16  9
CD 
 3.5 kpsi
2
C
R  52  3.52  6.10 kpsi
 1  12.5  6.1  18.6 kpsi
 2  12.5  6.1  6.4 kpsi
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 9/114
1
 5 

tan 1 
  27.5 ccw
2
 3.5 
 1  R  6.10 kpsi
p 
s  45  27.5  17.5 cw
(c)
24  10
 17 kpsi
2
24  10
CD 
 7 kpsi
2
C
R  7 2  62  9.22 kpsi
 1  17  9.22  26.22 kpsi
 2  17  9.22  7.78 kpsi
1
 7 
 
 p  90  tan 1     69.7 ccw
2
6

 1  R  9.22 kpsi
s  69.7  45  24.7 ccw
(d)
12  22
 5 kpsi
2
12  22
CD 
 17 kpsi
2
C
R  17 2  122  20.81 kpsi
 1  5  20.81  25.81 kpsi
 2  5  20.81  15.81 kpsi
1
 17  
 

 1  R  20.81 kpsi
 p  90  tan 1     72.39 cw
2
12
s  72.39  45  27.39 cw
______________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 10/114
3-16
(a)
8  7
 0.5 MPa
2
87
 7.5 MPa
CD 
2
C
R  7.52  62  9.60 MPa
 1  9.60  0.5  9.10 MPa
 2  0.5  9.6  10.1 Mpa
1
 7.5  

 p  90  tan 1 
   70.67 cw
2
 6 
 1  R  9.60 MPa
s  70.67  45  25.67 cw
(b)
96
 1.5 MPa
2
96
 7.5 MPa
CD 
2
C
R  7.52  32  8.078 MPa
 1  1.5  8.078  9.58 MPa
 2  1.5  8.078  6.58 MPa
1
 3 

 p  tan 1 
  10.9 cw
2
 7.5 
 1  R  8.078 MPa
s  45  10.9  34.1 ccw
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 11/114
(c)
12  4
 4 MPa
2
12  4
 8 MPa
CD 
2
C
R  82  7 2  10.63 MPa
 1  4  10.63  14.63 MPa
 2  4  10.63  6.63 MPa
1
 8 
 

 1  R  10.63 MPa
 p  90  tan 1     69.4 ccw
2
7
s  69.4  45  24.4 ccw
(d)
65
 0.5 MPa
2
65
 5.5 MPa
CD 
2
C
R  5.52  82  9.71 MPa
 1  0.5  9.71  10.21 MPa
 2  0.5  9.71  9.21 MPa
1
 8 
 p  tan 1    27.75 ccw
2
 5.5 
 1  R  9.71 MPa
s  45  27.75  17.25 cw
______________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 12/114
3-17
(a)
12  6
 9 kpsi
2
12  6
CD 
 3 kpsi
2
C
R  32  42  5 kpsi
 1  5  9  14 kpsi
 2  9  5  4 kpsi
1
 4
tan 1    26.6 ccw
 3
2
 1  R  5 kpsi
p 
s  45  26.6  18.4 ccw
(b)
30  10
 10 kpsi
2
30  10
 20 kpsi
CD 
2
C
R  202  102  22.36 kpsi
 1  10  22.36  32.36 kpsi
 2  10  22.36  12.36 kpsi
1
 10 
 p  tan 1    13.28 ccw
2
 20 
 1  R  22.36 kpsi
s  45  13.28  31.72 cw
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 13/114
(c)
10  18
 4 kpsi
2
10  18
 14 kpsi
CD 
2
C
R  142  9 2  16.64 kpsi
 1  4  16.64  20.64 kpsi
 2  4  16.64  12.64 kpsi
1
 14  
 p  90  tan 1     73.63 cw
2
 9 
 1  R  16.64 kpsi
s  73.63  45  28.63 cw
(d)
9  19
 14 kpsi
2
19  9
 5 kpsi
CD 
2
C
R  52  82  9.434 kpsi
 1  14  9.43  23.43 kpsi
 2  14  9.43  4.57 kpsi
1
 5 
 
 p  90  tan 1     61.0 cw
2
8

 1  R  9.34 kpsi
s  61  45  16 cw
______________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 14/114
3-18
(a)
80  30
 55 MPa
2
80  30
 25 MPa
CD 
2
C
R  252  202  32.02 MPa
 1  0 MPa
 2  55  32.02  22.98  23.0 MPa
 3  55  32.0  87.0 MPa
1 2 
23
 11.5 MPa,
2
 2 3  32.0 MPa,
1 3 
87
 43.5 MPa
2
(b)
30  60
 15 MPa
2
60  30
 45 MPa
CD 
2
C
R  452  302  54.1 MPa
 1  15  54.1  39.1 MPa
 2  0 MPa
 3  15  54.1  69.1 MPa
39.1  69.1
 54.1 MPa
2
39.1
 19.6 MPa
1 2 
2
69.1
 34.6 MPa
2 3 
2
1 3 
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 15/114
(c)
40  0
 20 MPa
2
40  0
CD 
 20 MPa
2
C
R  202  202  28.3 MPa
 1  20  28.3  48.3 MPa
 2  20  28.3  8.3 MPa
 3   z  30 MPa
1 3 
48.3  30
 39.1 MPa,
2
 1 2  28.3 MPa,
2 3 
30  8.3
 10.9 MPa
2
(d)
50
 25 MPa
2
50
CD 
 25 MPa
2
C
R  252  302  39.1 MPa
 1  25  39.1  64.1 MPa
 2  25  39.1  14.1 MPa
 3   z  20 MPa
64.1  20
20  14.1
 42.1 MPa,  1 2  39.1 MPa,  2 3 
 2.95 MPa
2
2
______________________________________________________________________________
1 3 
3-19
(a)
Since there are no shear stresses on the
stress element, the stress element
already represents principal stresses.
 1   x  10 kpsi
 2  0 kpsi
 3   y  4 kpsi
10  ( 4)
 7 kpsi
2
10
 1 2   5 kpsi
2
0  (4)
 2 kpsi
2 3 
2
1 3 
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 16/114
(b)
0  10
 5 kpsi
2
10  0
 5 kpsi
CD 
2
C
R  52  4 2  6.40 kpsi
 1  5  6.40  11.40 kpsi
 2  0 kpsi,  3  5  6.40  1.40 kpsi
 1 3  R  6.40 kpsi,
1 2 
11.40
 5.70 kpsi,
2
3 
1.40
 0.70 kpsi
2
(c)
2  8
 5 kpsi
2
82
 3 kpsi
CD 
2
C
R  32  42  5 kpsi
 1  5  5  0 kpsi,  2  0 kpsi
 3  5  5  10 kpsi
1 3 
10
 5 kpsi,
2
 1 2  0 kpsi,
 2 3  5 kpsi
(d)
10  30
 10 kpsi
2
10  30
 20 kpsi
CD 
2
C
R  202  102  22.36 kpsi
 1  10  22.36  12.36 kpsi
 2  0 kpsi
 3  10  22.36  32.36 kpsi
12.36
32.36
 6.18 kpsi,  2 3 
 16.18 kpsi
2
2
______________________________________________________________________________
 1 3  22.36 kpsi,
Shigley’s MED, 11th edition
1 2 
Chapter 3 Solutions, Page 17/114
3-20
From Eq. (3-15),
 3  (6  18  12) 2   6(18)  (6)(12)  18(12)  92  62  ( 15)2  
  6(18)( 12)  2(9)(6)(15)  ( 6)(6) 2  18( 15) 2  ( 12)(9) 2   0
 3  594  3186  0
Roots are: 21.04, 5.67, –26.71 kpsi
Ans.
21.04  5.67
 7.69 kpsi
2
5.67  26.71
 16.19 kpsi
2 3 
2
21.04  26.71
 23.88 kpsi
 max   1 3 
2
1 2 
Ans.
_____________________________________________________________________________
3-21
From Eq. (3-15)


2
 3  (20  0  20) 2   20(0)  20(20)  0(20)  402  20 2  0 2  




  20(0)(20)  2(40) 20 2 (0)  20 20 2

 3  40 2  2 000  48 000  0
Roots are: 60, 20, –40 kpsi
60  20
 20 kpsi
2
20  40
 30 kpsi
2 3 
2
60  40
 50 kpsi
 max   1 3 
2

2

 0(0) 2  20(40) 2   0

Ans.
1 2 
Ans.
_____________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 18/114
3-22 From Eq. (3-15)
2
2
 3  (10  40  40) 2  10(40)  10(40)  40(40)  20 2   40    20   


 10(40)(40)  2(20)( 40)( 20)  10( 40)  40( 20)  40(20)   0
 3  90 2  0
2
Roots are: 90, 0, 0 MPa
2
2
Ans.
2 3  0
 1 2   1 3   max 
90
 45 MPa
2
Ans.
_____________________________________________________________________________
3-23

15000
F

 33 950 psi  34.0 kpsi
A  4   0.752 

60
FL
L
   33 950
 0.0679 in
AE
E
30 106 

0.0679
1  
 1130 106   1130
L
60
From Table A-5, v = 0.292
 2  v1  0.292(1130)  330 
Ans.
Ans.

Ans.
Ans.
d   2 d  330 106  (0.75)  248 10 6  in
Ans.
_____________________________________________________________________________
3-24

3000
F

 6790 psi  6.79 kpsi
A  4   0.752 

60
FL
L
   6790
 0.0392 in
AE
E
10.4 106 

0.0392
 653 106   653
L
60
From Table A-5, v = 0.333
1 

 2  v1  0.333(653)  217 
Ans.
Ans.
Ans.
Ans.
d   2 d  217 10 6  (0.75)  163 10 6  in
Ans.
_____________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 19/114
3-25
d 0.0001d

 0.0001
d
d
From Table A-5, v = 0.326, E = 119 GPa

0.0001
1  2 
 306.7 106 
v
0.326
FL
F
and   , so

AE
A
E
=
 1 E  306.7 10 6  (119) 109   36.5 MPa
L
2 
F   A  36.5 10
6

  0.03
2
Ans.
 25 800 N  25.8 kN
4
Sy = 70 MPa >  , so elastic deformation assumption is valid.
_____________________________________________________________________________
3-26

FL
L
8(12)
   20 000
 0.185 in
AE
E
10.4 106 
Ans.
_____________________________________________________________________________
3-27

FL
L
3
   140 106 
 0.00586 m  5.86 mm
AE
E
71.7 109 
Ans.
_____________________________________________________________________________
3-28

FL
L
10(12)
   15 000
 0.173 in
AE
E
10.4 106 
Ans.
_____________________________________________________________________________
3-29
With  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,
x 
E x
E y
v
1
1
v
v
1

E x  vE y
1  v2

E   x  v y 
1  v2
Likewise,
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 20/114
y 
E   y   x 
1  v2
From Table A-5, E = 207 GPa and ν = 0.292. Thus,
E   x  v y 
x 
2

207 109  0.0019  0.292  0.000 72  
2
1 v
1  0.292
9
207 10   0.000 72  0.292  0.0019  
10   382 MPa
6
Ans.
106   37.4 MPa Ans.
1  0.2922
_____________________________________________________________________________
y 
3-30
With  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,
x 
E x
E y
v
1
1
v
v
1

E x  vE y
1  v2

E   x  v y 
1  v2
Likewise,
y 
E   y   x 
1  v2
From Table A-5, E = 71.7 GPa and ν = 0.333. Thus,
E   x  v y  71.7 109   0.0019  0.333  0.000 72  
106   134 MPa Ans.

x 

2
2
1 v
1  0.333
9
71.7 10   0.000 72  0.333  0.0019  
y 
106   7.04 MPa Ans.
1  0.3332
_____________________________________________________________________________
3-31 For plane strain, z = 0. From the third equation of Eq. (3-19),
Ans.
 z    x   y 
First of Eq. (3-19),
1
 x   x   y    x   y 
E
1
 1  2   x  1    y 
E
1 
1    x  y 

E 
Similarly,

Shigley’s MED, 11th edition

Ans.
Chapter 3 Solutions, Page 21/114
1 
1    y  x 
Ans.
E 
______________________________________________________________________________
3-32
c
ac
F
(a) R1  F M max  R1a 
l
l
6M
6 ac
 bh 2 l
Ans.
FF
 2  2
bh
bh l
6ac
y 
   bm b  hm h   lm l1   1( s)( s) 2 (s)  s 2
F
(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.
_____________________________________________________________________________
3-33
wl
w l  l  wl 2
R1 
, M max x l /2 
(a)
l   
2
2 2 2
8
6M
6 wl 2 3Wl
 2  2

4bh 2
bh
bh 8
(b)

4  bh 2
W
3 l
Ans.
Wm ( m /  )(bm / b)(hm / h) 2 1( s)( s )2


 s2
W
lm / l
s
Ans.
w m lm
wm s 2
 s2 
  s Ans.
wl
w
s
For equal stress, the model load w varies linearly with the scale factor.
_____________________________________________________________________________
3-34
(a)
Can solve by iteration or derive
equations for the general case. Find
maximum moment under wheel W3 .
WT  W at centroid of W’s
l  x3  d 3
RA 
WT
l
Under wheel 3,
M 3  RA x3  W1a13  W2 a23 
For maximum,
 l  x3  d3  W
l
dM 3
W
 0   l  d3  2 x3  T
dx3
l
Substitute into M  M 3 
Shigley’s MED, 11th edition
 l  d3 
4l
T
x3  W1a13  W2 a23

x3 
l  d3
2
2
WT  W1a13  W2 a23
Chapter 3 Solutions, Page 22/114
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: W 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)  20128 kip  in
2
4(1200)
Wheel 2: d 2  238  84  154 in
Wheel 1: d1 
M2 
(1200  154)2
(104.4)  26.1(84)  21 605 kip  in  M max
4(1200)
Ans.
Check if all of the wheels are on the rail.
xmax  600  77  523 in Ans.
(b)
(c)
See above sketch.
(d)
Inner axles
_____________________________________________________________________________
3-35
(a) Let a = total area of entire envelope
Let b = area of side notch
A  a  2b  40(2)(37.5)  25  34   2150 mm2
1
1
3
3
 40 75   34 25
12
12
6
4
I  1.36 10  mm
Ans.
I  I a  2Ib 
(b)
Aa  0.375(1.875)  0.703 125 in 2
Dimensions in mm.
Ab  0.375(1.75)  0.656 25 in 2
A  2(0.703125)  0.656 25  2.0625 in 2
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 23/114
2(0.703 125)(0.9375)  0.656 25(0.6875)
 0.858 in Ans.
2.0625
0.375(1.875)3
 0.206 in 4
Ia 
12
1.75(0.375)3
 0.007 69 in 4
Ib 
12
I1  2 0.206  0.703 125(0.0795) 2   0.00769  0.656 25(0.1705) 2   0.448 in 4
y
Ans.
(c)
Use two negative areas.
Aa  625 mm 2 , Ab  5625 mm 2 , Ac  10 000 mm 2
A  10 000  5625  625  3750 mm 2 ;
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
 2.637 106  mm 4
Ib 
12
100(100)3
 8.333 106  in 4
Ic 
12
2
2
I1  8.333 106   10 000(7.29) 2    2.637 10 6   5625  7.29    8138  625  57.29  6.25  

 

Ia 
I1  4.29 106  in 4
Ans.
(d)
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 24/114
Aa  4  0.875   3.5 in 2
Ab  2.5  0.875   2.1875 in 2
A  Aa  Ab  5.6875 in 2
y
2.9375  3.5   1.25(2.1875)
5.6875
 2.288 in
Ans.
1
1
3
2
3
2
(4)  0.875   3.5  2.9375  2.288    0.875  2.5   2.1875  2.288  1.25 
12
12
4
I  5.20 in
Ans.
_____________________________________________________________________________
I
3-36
1
(20)(40)3  1.067 105  mm4
12
A  20(40)  800 mm 2
Mmax is at A. At the bottom of the section,
Mc 450 000(20)
 max 

 84.3 MPa Ans.
I
1.067 105 
I
Due to V, max is between A and B at y = 0.
3 V 3  3000 
 
 max 
  5.63 MPa Ans.
2 A 2  800 
__________________________________________________________________________
3-37
1
(1)(2)3  0.6667 in 4
12
A  1(2)  2 in 2
I
M 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,
 max 
Mc 3200(0.5)

 2400 psi
I
0.6667
Ans.
Due to V,  max is at A, at y = 0.
3 V 3  800 
 
  600 psi Ans.
2 A 2 2 
_____________________________________________________________________________
 max 
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 25/114
3-38
1
(0.75)(2)3  0.5 in 4
12
A  (0.75)(2)  1.5 in 2
I
M A  0
15 RB  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,
 max 
Mc 5000(1)

 10 000 psi
I
0.5
Ans.
Due to V,  max is between B and C at y = 0.
3 V 3  1000 
 
  1000 psi Ans.
2 A 2  1.5 
_____________________________________________________________________________
 max 
3-39
I
d4

 (50) 4
 
 306.796 103 mm 4
64
64
2
d
 (50) 2
A

 1963 mm 2
4
4
M 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,  max is at A, at y = 0.
 max 
Ans.
4 V 4  750 
2
 
  0.509 kN/mm  509 MPa Ans.
3 A 3  1963 
_____________________________________________________________________________
 max 
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 26/114
3-40
8 I
wl 2
wl 2 c
  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  60 2 
(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.
_____________________________________________________________________________
3-41
I

 0.5   3.068 10  in ,
64
4
3
4
A

4
 0.5   0.1963 in
2
2
Model (c)
500(0.5) 500(0.75 / 2)
M 

 218.75 lbf  in
2
2
Mc 218.75(0.25)


I
3.068 10 3 
  17 825 psi  17.8 kpsi
 max 
Ans.
4 V 4 500

 3400 psi  3.4 kpsi
3 A 3 0.1963
Shigley’s MED, 11th edition
Ans.
Chapter 3 Solutions, Page 27/114
Model (d)
M  500(0.625)  312.5 lbf  in

Mc 312.5(0.25)

I
3.068 103 
  25 464 psi  25.5 kpsi
 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

Mc 218.75(0.25)

I
3.068 103 
  17 825 psi  17.8 kpsi
 max 
Ans.
4 V 4 500

 3400 psi  3.4 kpsi
3 A 3 0.1963
Ans.
____________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 28/114
3-42
I


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)


1018
I
  88.4 N/mm 2  88.4 MPa Ans.
 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
  141.5 N/mm 2  141.5 MPa Ans.
4 V 4  2000 
2
 max 
 
  23.6 N/mm  23.6 MPa
3 A 3  113.1 

Ans.
Model (e)
M  2000(7.5)  15000 N  mm
Mc 15000(6)

I
1018
  88.4 N/mm 2  88.4 MPa Ans.
4 V 4  2000 
2
 max 
 
  23.6 N/mm  23.6 MPa
3 A 3  113.1 

Ans.
_____________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 29/114
3-43
(a)  
d
(b)  
d
(c)  
d
Mc M  d / 2  32M


I
 d 4 / 64  d 3
3
32 M


3
32(218.75)
 0.420 in Ans.
 (30 000)
V
V

A d2 / 4
4V

4(500)
 0.206 in
 (15000)

Ans.
4V 4
V

3 A 3  d 2 / 4 
4 4V

3 
4 4(500)
 0.238 in
3  (15000)
Ans.
_____________________________________________________________________________
3-44
p1  p2
1
x  l  terms for x  l  a
a
p  p2
1
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 
a
2a
2
pa
p  p2 3
6 F (l  a )
 F (l  a )  1  1
a 0 
2 p1  p2 
a2
2
6a
 F  p1a 
From (1) and (2)
Shigley’s MED, 11th edition
p1 
2F
(3l  2a),
a2
p2 
2F
(3l  a)
a2
(1)
(2)
(3)
Chapter 3 Solutions, Page 30/114
From similar triangles
b
a

p2 p1  p2

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
 p2
p1
p
  Fl  F ( a  2b)  ( a  2b) 2  1
(a  2b)3
2
6a
Normally Mmax =  Fl
M max   F (l  a  2b) 
The fractional increase 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(1500)
3 1.5  2(1.2)   14 375 lbf/in
1.22 
p2 
2(1500)
3 1.5  1.2  11 875 lbf/in
1.22 
b = 1.2(11 875)/(14 375 + 11 875) = 0.5429 in
Substituting into (5) yields
 = 0.036 89 or 3.7% higher than -Fl
_____________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 31/114
3-45
300(30) 40
 1800  6900 lbf
2
30
300(30) 10
R2 
 1800  3900 lbf
2
30
3900
a
 13 in
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,  x  
 3176 psi
8.5
18000(2.5)
At x  10 in, y  2.5 in,  x  
 5294 psi
8.5
25350( 1.5)
At x  27 in, y  1.5 in,  x  
 4474 psi
8.5
25350(2.5)
At x  27 in, y  2.5 in,  x  
 7456 psi
8.5
Ans.
Max tension  5294 psi
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
 max V
(c)
VQ 5100(3.125)

 1875 psi
Ans.
Ib
8.5(1)
There are three potentially critical locations for the maximum shear stress, all at

Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 32/114
x = 27 in: (i) at the top where the bending stress is maximum, (ii) at the neutral axis
where 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,

7456
 3728 psi
 max  max 
2
2
For (ii):
The bending stress is zero, and the transverse shear stress was found previously to be
1875 psi. Thus, max = 1875 psi.
For (iii):
The bending stress, at y = – 0.5 in, is
18000(0.5)
x  
 1059 psi
8.5
The transverse shear stress is
Q  yA  (1)(3)(1)  3.0 in 3

VQ 5100(3.0)

 1800 psi
Ib
8.5(1)
From Mohr’s circle,
2
 1059 
2
  1800  1876 psi
 2 
The critical location is at x = 27 in, at the top surface, where max = 3728 psi.
Ans.
_____________________________________________________________________________
3-46
 max  
(a)
I = bh3/12 = 50(503)/(12) = 520.83 (103) mm4
Element A:
2 000  200  50 / 2 
My
A   A  
 19.2 MPa
I
520.83 103 
A 
VQA
, QA  0   A  0
Ib
Shigley’s MED, 11th edition
Ans.
Chapter 3 Solutions, Page 33/114
Element B:
B  
B 
MyB
, yB  0   B  0
I
QB  yB A   25 / 2  25  50   15.625 103  mm3
3
VQB 2000 15.625 10

 1.20 MPa
Ib
520.83 103  50
Element C:
C  
2 000  200  25 / 2 
520.83 103 
Ans.
 9.60 MPa
QC   25 / 2  25 / 4  25 / 2  50  11.719 103  mm3
C 
2 000 11.719 103
520.83 103  50
 0.90 MPa
Ans.
(b) Point A:
 max 
A
2

19.2
 9.6 MPa
2
Ans.
Point B:
 max   B  1.20 MPa
Ans.
Point C:
2
 max
 
  C    C2
 2 
2
 9.60 
2
 
  0.9  4.88 MPa
 2 
(c) Point A is critical.
Shigley’s MED, 11th edition
Ans.
Ans.
Chapter 3 Solutions, Page 34/114
(d) Transverse shear does not change with respect to the length of the beam.
2 000 L  50 / 2 
 1.20  L  25.0 mm
Ans.
2
2  520.83 103
______________________________________________________________________________
3-47 I = (/64)404 = 125.66(103) mm4, J = 2I
(a) Point A:
3
M A c A 50 10 10  40 / 2 
A 

 79.6 MPa
125.66
I
Ans.
3
800
10
40
/
2




Tr
A  A A 
 63.7 MPa
2 125.66 103
J
 A max 
A

Point B:
B  0
3
4 V 4 50 10 
B 

 53.1 MPa
3 A 3   40 / 2 2
Ans.
Point C:
C  0
C   B 
Ans.
Tr
 53.1  63.7  116.8 MPa
J
(b) Point A:
 79.6 
2
2
 max  A  
  63.7  75.1 MPa
 2 
Ans.
Point B:
(max)B = 53.1 MPa
Shigley’s MED, 11th edition
Ans.
Chapter 3 Solutions, Page 35/114
Point C:
(max)C = 116.8 MPa
(c) max = 116.8 MPa at point C
Ans.
Ans.
______________________________________________________________________________
(a) L = 10 in. Element A:
3-48
My
(1000)(10)(0.5)

10 3  101.9 kpsi
4
I
( / 64)(1)
VQ
A 
, Q  0  A  0
Ib

A  
2
 max

2
 
 101.9 
2
  A    A2  
  (0)  50.9 kpsi
 2 
 2 
Ans.
Element B:
B  
My
, y0
I
 4r
Q  y A  
 3
B 
B  0

2
3
4  0.5 
   r  4r


 1/ 12 in 3


6
6
 2 
3
VQ (1000)(1/ 12)
103  1.698 kpsi

4
Ib ( / 64)(1) (1)

0

2
 max     1.6982  1.698 kpsi
2
Ans.
Element C:
C  
(1000)(10)(0.25)
My

103  50.93 kpsi
I
( / 64)(1) 4
Shigley’s MED, 11th edition


Chapter 3 Solutions, Page 36/114
r
r
r
y1
y1
y1


Q   ydA   y (2 x )dy   y 2 r 2  y 2 dy
2
  r 2  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
C 
VQ
(1000)(0.05413)
103  1.273 kpsi

4
Ib ( / 64)(1) (0.866)
 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.
 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.
2
0
 max     0 psi Ans.
2
 1.698  0 
% error  
 * (100)  100% Ans.
 1.698 
Element C:
2
 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.
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 37/114
Bending
stress,
kpsi)
Transverse
shear stress,
kpsi)
Max shear
stress,
max kpsi)
Max shear
stress,
neglecting 
max 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
L = 10 in
A
B
C
L = 4 in
A
B
C
L = 1 in
A
B
C
L = 0.1in
A
B
C
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.
_____________________________________________________________________________
3-49
c
F
l
c
M  Fx 0  x  a
l
6M 6  c l  Fx
 2 
bh
bh2
R1 
h
6 Fcx
lb max
0 xa
Ans.
_____________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 38/114
3-50
c
From Problem 3-49, R1  F  V , 0  x  a
l
3V
3 (c / l ) F
3 Fc
 max 

 h
2 bh 2 bh
2 lb max
Ans.
6 Fcx
.
lb max
Sub in x = e and equate to h above.
3 Fc
6 Fce

lb max
2 lb max
From Problem 3-49, h( x) 
3 Fc max
Ans.
2
8 lb max
_____________________________________________________________________________
e
3-51
(a)
x-z plane
M O  0  1.5(0.5)  2(1.5) sin(30 )(2.25)  R2 z (3)
R2 z  1.375 kN Ans.
Fz  0  R1z  1.5  2(1.5) sin(30 )  1.375
R1z  1.625 kN Ans.
x-y plane
M O  0  2(1.5) cos(30 )(2.25)  R2 y (3)
R2 y  1.949 kN
Ans.
Fy  0  R1 y  2(1.5) cos(30 )  1.949
R1 y  0.6491 kN
Ans.
(b)
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 39/114
(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
My  
 x 
V dx 
z
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) Vy (kN)
–1.625 0.6491
–1.625 0.6491
–0.1250 0.6491
0
0.4327
0.2500
0
1.375
–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
M
(kNm)
0
0.8749
1.352
1.406
1.427
0
(d) The bending stress is obtained from Eq. (3-27),
M z yA M y z A
x 

Iz
Iy
The maximum tensile bending stress will be at point A in the cross section of Prob. 3-35
(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
Iy  2
 2.67(105 ) mm 4  2.67(107 ) m 4

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
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 40/114
found at x = 1.77 m, where My = – 0.9408 kN·m, Mz = 1.075 kN·m, and x = 100.1 MPa.
Ans.
_____________________________________________________________________________
3-52
(a) x-z plane
3
600
M O  0  (1000)(4) 
(10)  M Oy
5
2
M Oy  1842.6 lbf  in Ans.
3
600
Fz  0  ROz  (1000) 
5
2
ROz  175.7 lbf Ans.
x-y plane
4
600
(10)  M Oz
M O  0   (1000)(4) 
5
2
M Oz  7442.5 lbf  in Ans.
4
600
Fy  0  ROy  (1000) 
5
2
ROy  1224.3 lbf Ans.
(b)
(c)
1/ 2
V ( x)  V y ( x ) 2  Vz ( x) 2 
1/ 2
M ( x)   M y ( x )2  M z ( x ) 2 
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 41/114
x (m)
0
4
10
Vz (kN)
–175.7
–175.7
424.3
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
3(2) (2.625)(1.625) 3
Iy 

 1.601 in 4
12
12
At x = 0, using Eq. (3-27),
M y M z
x   z  y
Iz
Iy
( 7442.6)(1.5) ( 1842.6)( 1)
x  

 6594 psi
2.051
1.601
Check at x = 4 in,
( 2545.4)(1.5) ( 2545.4)( 1)
x  

 2706 psi
2.051
1.601
The critical location is at x = 0, where x = 6594 psi. Ans.
_____________________________________________________________________________
3-53 (a) Moments at A: My = 300(0.050) = 15 N٠m,
Mz = 200(0.055) = 11 N٠m, Torque at A: Tx = 200(0.060) = 12 N٠m
M  M y2  M z2  152  112  18.601 N  m
 Mz 
 11 
 tan 1    36.25o

M 
 15 
 y
o
Critical point will be 90 from , i.e. 126.25o from the vertical y axis.
(b)
M c 32 M 32 18.601 6
 bend 


10  109.6 MPa
I
 d3
  0.0123 
  tan 1 
 axial 
Ans.
4  300 
F 4F
106  2.65 MPa


2
A d
  0.0122 
 total   bend   axial  109.6  2.65  112.3 MPa

16 12 
Tx c 16Tx
10 6  35.4 MPa


3
3
J
d
  0.012 
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 42/114
Ans.
(c)
Ans.
2
112.3
 112.3 
2
 56.2 MPa, R  
  35.4  66.4 MPa
2
 2 
 1  C  R  56.2  66.4  122.6 MPa,  2  0,
(d)
C
 3  C  R  56.2  66.4  10.2 MPa
 max  R  66.4 MPa
Ans.
______________________________________________________________________________
3-54 (a) Moments at A: My =  200(0.060) =  12 N٠m, Mz = 300(0.095) = 28.5 N٠m
Torque at A: Tx =  300(0.050) =  15 N٠m
M  M y2  M z2 
 Mz
M
 y
  tan 1 
 12
2
 28.52  30.92 N  m

1  28.5 
o
  tan 
  112.8
 12 

Critical point will be 90o from , i.e. 202.8o from the vertical y. axis.
M c 32 M 32  30.92  6
 bend 


10  182.3 MPa
 d3
I
  0.0123 
 axial 
Ans.
4  200 
F 4F


106  1.77 MPa
2
2
A d
  0.012 
 total   bend   axial  182.3  1.77  184.1 MPa

16 15 
Tx c 16Tx


10 6  44.2 MPa
3
3
d
J
  0.012 
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 43/114
Ans.
(c)
Ans.
(d)
2
184.1
 184.1 
2
C
 92.1 MPa, R  
  44.2  102.1 MPa
2
 2 
 1  C  R  92.1  102.1  194.2 MPa,  2  0,
 3  C  R  92.1  102.1  10 MPa
 max  R  102.1 MPa
Ans.
______________________________________________________________________________
3-55
(a) Moments at A: My = 60(5) + 200(5.5) = 1400 lbf٠in,
Mz =  (300 – 75)5.5 =  1238 lbf٠in,
Torque at A: Tx =  75(5) =  375 lbf٠in
M  M y2  M z2  14002   1238   1869 lbf  in
2
 Mz 
 1238 
o
 tan 1 

  318.5  41.5
M 
1400


 y
o
Critical point will be 90 from , i.e. 48.5o from the vertical y axis. Ans.
M c 32 M 32 1869  3
 bend 
10  45.13 kpsi


(b)
I
 d3
  0.753 
  tan 1 
 axial 
4  60 
F 4F


 136 psi
2
A d
  0.752 
 total   bend   axial  45.126  0.136  45.3 kpsi

Tx c 16Tx 16  375  3


10  4.53 kpsi
 d 3   0.753 
J
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 44/114
Ans.
(c)
Ans.
(d)
2
45.3
 45.3 
2
C
 22.7 kpsi, R  
  4.53  23.1 kpsi
2
 2 
 1  C  R  22.7  23.1  45.8 kpsi,  2  0,
 3  C  R  22.7  23.1  0.45 kpsi
 max  R  23.1 kpsi
Ans.
______________________________________________________________________________
3-56
Given: b = 3.6 in, c = 2.5 in, and l = 40 in.
From Table A-5, G = 11.5 Mpsi. From Table A-20, Sy = 42 kpsi.
(a) For the table for Eq. (3-40), b/c = 3.6/2.5 = 1.44
b/c_
_____
0.208
1

1.44
0.231
1.5
0.231  
1.5  1.44

 0.12    0.231  0.12  0.231  0.208   0.228
0.231  0.208
1.5  1
Equation (3-40):
30 103 
T
 max 

 5 850 psi  5.85 kpsi
Ans.
 bc 2 0.228  3.6  2.52
(b) For the table for Eq. (3-41), b/c = 3.6/2.5 = 1.44
_____
b/c_
0.141
1

1.44
0.196
1.5
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 45/114
0.196  
1.5  1.44

 0.12    0.196  0.12  0.196  0.141  0.189
0.196  0.141
1.5  1
Equation (3-41):
30 103  40
Tl


 9.815 10 3  rad  0.562o Ans.
3
3
6
 bc G 0.189  3.6  2.5 11.5 10
(c) Ssy = 0.5(42) = 21 kpsi. Yield factor of safety,
S sy
21
ny 

 3.59
Ans.
 max 5.85
______________________________________________________________________________
3-57 Given: 250 hp at 540 rev/min, allow = 15 kpsi.
63 025
63 025
Eq. (3-42):
T
H
250  29.178 103  lbf  in
n
540
Eq. (3-41) with table where for square cross-section, b/c =1
29.178 103 
T
 max 

 15 103   b  2.107 in
2
3
0.208bc
0.208b
2
From Table A-17, use b = 14 in
Ans.
______________________________________________________________________________
3-58 Given: T = 50 kN-m. OD = 300 mm, t = 2 mm, and l = 2 m.
J = (/32)(0.3004 – 0.2964) = 4.157(105) m4. From Table A-5, G = 79.3 GPa.
(a) Eq. (3-37):
3
Tr 50 10  0.15 6
 max  
10  180.4 MPa
Ans.
J
4.157 10 5 
(b) Eq. (3-45):
50 103 
T
 max 

106  179.2 MPa
2
2 Amt 2  / 4  0.298  0.002 
Answer is 0.67 percent lower than part (a).
(c) Eq. (3-35):
50 103  2
TL


 0.0303 rad  1.74o
JG 4.157 10 5  79.3 109 
Ans.
Ans.
(d) Eq. (3-46):
50 103    0.298  2
TLml

 0.303 rad  1.74o
  1l 
4GAm2 t 4  79.3109  / 4  0.2982  2 0.002


Ans.
Within the same accuracy, the answer is the same as part (c).
______________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 46/114
3-59
Given: Rectangular tube with inner dimensions 1.5 in  2.0 in, t =
1
8
in, 1035 CD steel,
n = 540 rev/min, and Ssy =0.5 Sy.
(a) Table A-20, Sy = 67 kpsi. Factor of safety against yield, ny = 2. For the table for Eq.
(3-40), with b/c = 2/1.5 = 1.333,
_____
b/c_
0.208
1

1.333
0.231
1.5
0.231  
1.5  1.333

 0.3333    0.231  0.3333  0.231  0.208   0.2233
0.231  0.208
1.5  1
Eq. (3-40):
 max 
Eq. (3-42):
0.5S y 0.5  67 103
T
T



 T  16.83 103  lbf  in
2
2
2
ny
 bc 0.2233  2 1.5
16.83 103  540
Tn
H

 144 hp
63 025
63 025
(b) Eq. (3-45):
 max 
0.5S y
ny

Ans.
T
2 Amt
0.5  67 103
T

 T  14.46 103  lbf  in
1
1
1
2


2  1.5  8 2.0  8  8
14.46 103  540
H
 124 hp
Ans.
63 025
______________________________________________________________________________
3-60 Outer dimensions 20  30 mm, t = 1 mm, l = 1 m, 1018 CD steel, Ssy = 0.5 Sy.
Table A-20, Sy = 370 MPa. Ssy = 0.5(370) = 185 MPa. Table A-5, G = 79.3 GPa.
(a) Am = 19(29) = 551 mm2.
Eq, (3-45):
T
 185 106   T  2(551)106  0.001185 10 6   204 N  m

Ans.
2 Amt
(b) Eq. (3-46):
T  2 19  29 10 3 1
TLml
  
  1l 
 3
 T  52.5 N  m Ans.

2
4GAm2 t
 180  4  79.3109  551 1012  0.001



______________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 47/114
3-61
(a) The area within the wall median line, Am, is
Square: Am  (b  t )2 . From Eq. (3-45)
Tsq  2 Amt all  2(b  t ) 2 t all
Round: Am   (b  t ) 2 / 4
Trd  2 (b  t ) 2 t all / 4
Ratio of Torques
Tsq
2(b  t )2 t all
4

  1.27 Ans.
2
Trd  (b  t ) t all / 2 
(b) Twist per unit length from Eq. (3-46) is
1 
TLm
 L
L
2 A t L
 m all2 m  all m  C m
2
4GAm t
4GAm t
2G Am
Am
Square:
sq  C
Round:
rd  C
4(b  t )
(b  t )2
 (b  t )
2
 (b  t ) / 4
C
4(b  t )
(b  t )2
 sq
 1 . Twists are the same. Ans.
 rd
_____________________________________________________________________________
3-62
(a) The area enclosed by the section median line is Am = (1 – 0.0625)2 = 0.8789 in2 and
the length of the section median line is Lm = 4(1  0.0625) = 3.75 in. From Eq. (3-45),
T  2 Am t  2(0.8789)(0.0625)(12 000)  1318 lbf  in
From Eq. (3-46),
  1l 
TLm l
4GAm2 t

(1318)(3.75)  36 
 
4 11.5  10 (0.8789)  0.0625 
6
2
Ans.
 0.0801 rad  4.59 Ans.
(b) The radius at the median line is rm = 0.125 + (0.5) (0.0625) = 0.15625 in. The area enclosed
by the section median line is Am = (1  0.0625)2 – 4(0.15625)2 + 4(π /4) (0.15625)2 = 0.8579
in2. The length of the section median line is Lm = 4[1 – 0.0625 – 2(0.15625)] + 2π (0.15625) =
3.482 in.
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 48/114
From Eq. (3-45),
T  2 Am t  2(0.8579)(0.0625)(12 000)  1287 lbf  in
Ans.
From Eq. (3-46),
(1287)(3.482)  36 
TLm l
  1l 

 0.0762 rad  4.37 Ans.
2
6
2
4GAmt 4 11.5  10 (0.8579)  0.0625 
 
_____________________________________________________________________________
3-63
1 
3Ti
Ti 

GLi ci3
T  T1  T2  T3 
1G
3
1GLi ci3
3
 Li ci3
i 1
3
Ans.
From Eq. (3-47), G1c
G and 1 are constant, therefore the largest shear stress occurs when c is a maximum.
 max  G1cmax
Ans.
_____________________________________________________________________________
3-64
(b) Solve part (b) first since the angle of twist per unit length is needed for part (a).
 max   allow  12  6.89   82.7 MPa
1 
 max
Gcmax

   0.348 rad/m
79.3 10  (0.003)
82.7 106
9
(a)
T1 
T2 
1GL1c13
3
 2GL2 c23
3


3GL3c33
 
0.348(79.3) 109 (0.020)(0.0023 )
3
0.348(79.3) 109  (0.030)(0.0033 )
3
9
0.348(79.3) 10  (0)(03 )
Ans.
 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.
_____________________________________________________________________________
T3 
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 49/114
3-65
(b) Solve part (b) first since the angle of twist per unit length is needed for part (a).

12 000
1  max 
 8.35 103 rad/in Ans.
6
Gcmax 11.5 10 (0.125)

 
(a)
T1 
1GL1c13
3

 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
3GL3c33  8.35  10  11.5  10   0.625   0.0625 
T3 

 4.88 lbf  in Ans.
3
3
T  T1  T2  T3  5.86  62.52  4.88  73.3 lbf  in Ans.
_____________________________________________________________________________
T2 

3
3-66
(b) Solve part (b) first since the angle of twist per unit length is needed for part (a).
 max   allow  12  6.89   82.7 MPa
1 
 max
Gcmax

   0.348 rad/m
79.3 10  (0.003)
82.7 106
9
(a)
T1 
T2 
1GL1c13
3
 2GL2 c23
3


3GL3c33
 
0.348(79.3) 109 (0.020)(0.0023 )
3
0.348(79.3) 109  (0.030)(0.0033 )
3
9
0.348(79.3) 10  (0.025)(0.0023 )
Ans.
 1.47 N  m
 7.45 N  m
Ans.
Ans.
Ans.

 1.84 N  m
3
3
T  T1  T2  T3  1.47  7.45  1.84  10.8 N  m Ans.
_____________________________________________________________________________
T3 
3-67
(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 
 max bc 2
Tmax  2(3.08)  6.16 N  m
Shigley’s MED, 11th edition
Ans.
Chapter 3 Solutions, Page 50/114
From the table for Eqs. (3-40) and (3-41), with b/c = 30/2 = 15, and has a value between
0.313 and 0.333. From Eq. (3-40),
1

 0.321
3  1.8 / (30 / 2)
From Eq. (3-41),
3.08(0.3)
Tl


 0.151 rad
3
 bc G 0.321 0.030  0.0023  79.3 109

kt 
T


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 
 max bc 2
Ans.
Interpolating from the table for Eqs. (3-40) and (3-41), with b/c = 30/4 = 7.5,
7.5  6

(0.307  0.299)  0.299  0.305
86
From Eq. (3-41)
Tl
11.9(0.3)


 0.0769 rad
3
 bc G 0.305  0.030  0.0043  79.3 109

kt 
T



11.9
 155 N  m
0.0769
 
Ans.
Ans.
(b) From Eq. (3-47), with two 2-mm strips,




 
2
6
Lc 2  0.030  0.002  80  10
T

 3.20 N  m
3
3
Tmax  2(3.20)  6.40 N  m Ans.
3Tl
3(3.20)(0.3)
 3 
 0.151 rad
Lc G  0.030  0.0023  79.3 109
 
kt  T   6.40 0.151  42.4 N  m
Ans.
Ans.
From Eq. (3-47), with a single 4-mm strip,
Tmax


 
2
6
Lc 2  0.030  0.004  80  10


 12.8 N  m
3
3
Shigley’s MED, 11th edition
Ans.
Chapter 3 Solutions, Page 51/114

3Tl
3(12.8)(0.3)

 0.0757 rad
3
Lc G  0.030  0.0043  79.3 109


 
kt  T   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.
_____________________________________________________________________________
3-68
(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
 max   3
J d
13
 16 152.8 

   0.0223 m  22.3 mm

6
   70  10 


H
(40000)
 1528 N  m
(b) T  9.55  9.55
n
250
13
 16T 
d 

  max 
 
Ans.
13


16(1528) 

d
 0.0481 m  48.1 mm Ans.
   70  106 


_____________________________________________________________________________
 
3-69
(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
 max   3
J d
13
13
 16T 
 16 1261 
d 
 
  0.685 in
  max 
  (20 000) 
63025H 63025(50)

 12610 lbf  in
(b) T 
n
250
Ans.
13
16(12 610) 
d 
  1.48 in Ans.
  (20 000) 
_____________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 52/114
3-70
 max
16T

d3
 max d 3
 50  106    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
_____________________________________________________________________________
3-71


16T
  3  T   d 3  110  106 0.0203  173 N  m
16
16
d
 

 0.0204  79.3 109 15

4
Tl
 d G
 180 
 l


32T
32(173)
JG
l  1.89 m Ans.
_____________________________________________________________________________

T
 
 

3-72


 


16T
 T   d 3   30 000  0.753  2485 lbf  in
3
16
16
d
Tl
32Tl
32(2485)(24)



 0.167 rad  9.57 Ans.
4
JG  d G  0.754 11.5 106





 
_____________________________________________________________________________
3-73
(a) Tsolid
J max  d o4 max


16d o
r
Thollow
J max  (d o4  d i4 ) max


16d o
r
 
 
364
Tsolid  Thollow
di4
%T 
(100%)  4 (100%) 
(100%)  65.6%
Tsolid
do
404
(b) Wsolid  kdo2 ,

Whollow  k do 2  di 2

Ans.
 
 
362
Wsolid  Whollow
di2
%W 
(100%)  2 (100%) 
(100%)  81.0%
Wsolid
do
402
Ans.
_____________________________________________________________________________
3-74
4
 4
J max  d 4 max
J max   d   xd    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, 11th edition
Chapter 3 Solutions, Page 53/114

Whollow  k d 2   xd 
(b) Wsolid  kd 2
2

2
%W 
Wsolid  Whollow
 xd  (100%)  x 2 (100%)
(100%) 
Wsolid
d2
Ans.
Plot %T and %W versus x.
Percent Reduction in
Torque and Weight
Percent Reduction
100
80
60
Weight
40
Torque
20
0
difference
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 .
_____________________________________________________________________________
3-75
Tc
(a)  
J
 
120 10

 
13
4200  d 2 
6

4
32   d 4   0.70d  



 
2.8149 104
d
3
 2.8149 104 
  6.17 102 m  61.7 mm
d 
6
 120(10 ) 


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  di 2 
4200  0.050 2 
Tc


 30.8 MPa
Ans.
4
J  32   d 4   d    32   0.080 4   0.050 4 
i




_____________________________________________________________________________

Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 54/114
3-76
T  9.55
H
(1500)
 9.55
 1433 N  m
n
10
13
 16 1433 

   0.045 m  45 mm
16T
 16T 
 3
 dC  
 = 
 dC
  
 80 106 
  

From Table A-17, select 50 mm. Ans.
16  2 1433
 117 106  Pa  117 MPa Ans.
(a)  start 
3
  0.050 
13
 
(b) Design activity
_____________________________________________________________________________
3-77
T
63 025 H 63 025(1)

 7880 lbf  in
n
8
16T
 3
 dC

13
 16T 
dC  

  
13
 16  7880  
=

  15 000  
 1.39 in
From Table A-17, select 1.40 in.
Ans.
_____________________________________________________________________________
3-78
For a square cross section with side length b, and a circular section with diameter d,


Asquare  Acircular  b 2  d 2
 b
d
4
2
From Eq. (3-40) with b = c,
 max square
3
1.8  T  1.8  T  2 
T 
T
 2 3
(4.8)  6.896 3
  3 3
 3

bc 
b/c b 
1  d   
d
For the circular cross section,
16T
T
 max circular  3  5.093 3
d
d
T
 max square 6.896 d 3

 1.354
 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 for Eq. (3-41), β = 0.141. From Eq. (3-41),
Tl
Tl
Tl
Tl
square 


 11.50 4
4
3
4
d G
 bc G  b G
  
0.141
d G
 2 
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 55/114
For the circular cross section,
Tl
Tl
Tl
 rd 

 10.19 4
4
GJ G  d 32 
d G
Tl
 sq 11.50 d 4G

 1.129
 rd 10.19 Tl
4
d G
The angle of twist in the square cross section is 12.9% greater.
Ans.
_____________________________________________________________________________
3-79 (a)
T1  0.15T2
 T  0  (500  75)(4)  T
2
1700  4.25T2  0

T1  0.15  400   60 lbf
(b)
M
O
T2  400 lbf
Ans.
Ans.
 0  575(10)  460(28)  RC (40)
RC  178.25 lbf
F 0 R
O
(c)
 T1  5   1700  T2  0.15T2  5 
Ans.
 575  460  178.25
RO  293.25 lbf
Ans.
(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, 11th edition
Chapter 3 Solutions, Page 56/114
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.
d3
 (1.25)3
I
Tr 16T 16(1700)
  3
 4433 psi = 4.43 kpsi
Ans.
 (1.25)3
J d

(e)
2
x
2
2
15.3
2
 
 15.3 
1,  2 
  x    xy  
 
   4.43
2
2
 2 
 2 
 1  16.5 kpsi
Ans.
 2  1.19 kpsi
Ans.
2
2
2
2
 
 15.3 
 max   x    xy   
Ans.
   4.43   8.84 kpsi
 2 
 2 
_____________________________________________________________________________
3-80 (a)
T2  0.15T1
 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
(c)
 0  RO  3312  1794  2070
RO  3036 N Ans.
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 57/114
(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.
3
3
d
 (0.030)
I
16(306)
Tr 16T
  3
 57.7 106 Pa  57.7MPa
Ans.
3
 (0.030)
J d

 
(e)
2
x
2
2
263
2
 
 263 
1,  2 
  x    xy  
 
   57.7 
2
2
 2 
 2 
 1  275 MPa
Ans.
 2  12.1 MPa
Ans.
2
2
2
2
 
 263 
 max   x    xy   
   57.7   144 MPa
 2 
 2 
Ans.
_____________________________________________________________________________
3-81
(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.
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 58/114
(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.
d3
 (1)3
I
Tr 16T 16(1000)
  3
 5093 psi = 5.09 kpsi
Ans.
J d
 (1)3

(e)
2
x
2
2
21.5
2
 
 21.5 
  x    xy  
 
1,  2 
   5.09 
2
2
 2 
 2 
Ans.
 1  22.6 kpsi
 2  1.14 kpsi
2
Ans.
2
2
2
 
 21.5 
Ans.
 max   x    xy   
   5.09   11.9 kpsi
 2 
 2 
_____________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 59/114
3-82
(a)
T2  0.15T1
 T  0   300  45 (125)  T
2
31 875  127.5T1  0
(b)

 T1  (150)  31 875   0.15T1  T1  (150)
T1  250 N  mm Ans.
T2  0.15  250   37.5 N  mm Ans.
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,
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 60/114
M
 47.37    32.16 
2
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 32 M 32  57.26 



 72.9 106  Pa  72.9 MPa
Ans.
 d 3  (0.020)3
I
Tr 16T 16(31.88)
  3
 20.3 106  Pa  20.3 MPa
Ans.
J d
 (0.020)3
(e)
2
x
2
2
72.9
2
 
 72.9 
1,  2 
  x    xy  
 
   20.3
2
2
 2 
 2 
 1  78.2 MPa
Ans.
 2  5.27 MPa
Ans.
2
2
2
2
 
 72.9 
 max   x    xy   
Ans.
   20.3   41.7 MPa
 2 
 2 
_____________________________________________________________________________
3-83
(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.
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 61/114
(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
d3
 (1.25)3
Tr 16T 16(2819)
  3
 7351 psi = 7.35 kpsi
Ans.
 (1.25)3
J d

(e)
2
x
2
2
35.2
2
 
 35.2 
1,  2 
  x    xy  
 
   7.35 
2
2
 2 
 2 
 1  36.7 kpsi
Ans.
 2  1.47 kpsi
2
Ans.
2
2
2
 
 35.2 
Ans.
 max   x    xy   
   7.35   19.1 kpsi
 2 
 2 
_____________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 62/114
3-84
(a)
 T  0   11 000(cos 20º )(300)  F (cos 25º )(150)
B
(b)
FB  22 810 N
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 )  22 810 sin(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, 11th edition
Chapter 3 Solutions, Page 63/114
Mc 32 M 32  4097 


 333.9 10 6  Pa  333.9 MPa
Ans.
3
3
d
 (0.050)
I
Tr 16T 16(3101)
  3
 126.3 106 Pa  126.3 MPa
Ans.
J d
 (0.050)3

 
(e)
2
x
2
2
333.9
2
 
 333.9 
  x    xy  
 
1,  2 
  126.3
2
2
 2 
 2 
Ans.
 1  376 MPa
 2  42.4 MPa
Ans.
2
2
2
2
 
 333.9 
Ans.
 max   x    xy   
  126.3   209 MPa
 2 
 2 
_____________________________________________________________________________
3-85
(a)
 M D  z  6.13Cx  3.8(92.8)  3.88(362.8)  0
C x  287.2 lbf
Ans.
 M C  z  6.13Dx  2.33(92.8)  3.88(362.8)  0
Dx  194.4 lbf
Ans.
3.8
(808)  500.9 lbf Ans.
6.13
2.33
(808)  307.1 lbf Ans.
 M C  x  0  Dz 
6.13
(b) For DQC, let x, y, z correspond to the original  y, x, z axes.
 M D  x  0
 Cz 
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 64/114
(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  1167 2  1345 lbf  in
16T 16(3135)
 3
 11 070 psi Ans.
d
 1.133 
b  
32M
32(1345)

  9495 psi
3
d
 1.133
Ans.
a  
362.8
F

 362 psi
A
( / 4) 1.132
Ans.




(d) The critical stress element will be where the bending stress and axial stress are both in
compression.
 max  9495  362  9857 psi
2
 max
 9857 
2
 
  11 070  12 118 psi  12.1 kpsi
 2 
Ans.
2
9857
 9857 
2
1 , 2 
 
  11 070
2
2


 1  7189 psi  7.19 kpsi Ans.
 2  17 046 psi  17.0 kpsi
Ans.
_____________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 65/114
3-86
(a)
 M D  z  0
6.13Cx  3.8(46.6)  3.88(140)  0
C x  117.5 lbf
Ans.
 M C  z  0
6.13Dx  2.33(46.6)  3.88(140)  0
Dx  70.9 lbf
Ans.
3.8
(406)  251.7 lbf Ans.
6.13
2.33
(406)  154.3 lbf Ans.
 M C  x  0  Dz 
6.13
(b) For DQC, let x, y, z correspond to the original  y, x, z axes.
 M D  x  0
 Cz 
(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)
 3
 8021 psi Ans.
d
 13 
b  
32M
32(647.1)

  6591 psi
3
d
 13
Shigley’s MED, 11th edition
 
Ans.
Chapter 3 Solutions, Page 66/114
a  
F
140

 178.3 psi
A
( / 4) 12
 
Ans.
(d) The critical stress element will be where the bending stress and axial stress are both in
compression.
 max  6591  178.3  6769 psi
2
 max
 6769 
2
 
  8021  8706 psi  8.71 kpsi
 2 
Ans.
2
6769
 6769 
2
 
1, 2 
  8021
2
2


 1  5321 psi  5.32 kpsi Ans.
 2  12 090 psi  12.1 kpsi
Ans.
_____________________________________________________________________________
3-87
 M B  z  5.62(362.8)  1.3(92.8)  3 Ay  0
Ay  639.4 lbf
Ans.
B y  276.6 lbf
Ans.
 M A  z  2.62(362.8)  1.3(92.8)  3By  0
5.62
(808)  1513.7 lbf
3
2.62
 0  Bz 
(808)  705.7 lbf
3
 M B  y  0
 M A  y
 Az 
Shigley’s MED, 11th edition
Ans.
Ans.
Chapter 3 Solutions, Page 67/114
(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.
T  808(1.3)  1050 lbf  in
16(1050)

 7847 psi Ans.
 0.883


M  (829.8) 2  (2117) 2  2274 lbf  in
32M
32(2274)
b  

 33 990 psi
3
d
 0.883

a  

92.8
F

 153 psi
A
( / 4) 0.882


Ans.
Ans.
(d) The critical stress will occur when the bending stress and axial stress are both in
compression.
 max  33 990  153  34 143 psi
2
 max
 34 143 
2
 
  7847  18 789 psi  18.8 kpsi
2


Ans.
2
34143
 34143 
2
1 , 2 
 
  7847
2
2 

 1  1717 psi  1.72 kpsi Ans.
 2  35 860 psi  35.9 kpsi
Ans.
_____________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 68/114
3-88
Ft 
T
100

 1600 N
c / 2 0.125 / 2
  M A z  0
Fn  1600 tan 20  582.4 N
TC  Ft  b 2   1600  0.250 2   200 N  m 450 RDy  582.4(325)  2667(75)  0
P
TC
200

 2667 N
 a 2   0.150 2 
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
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 R A2 y  R A2z  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.
32M B 32(181.9)
B 

 68.6 106 Pa  68.6 MPa
3
3
d
 0.030

B 

 
Ans.
16TB
16(200)

 37.7 106 Pa  37.7 MPa
3
3
d
 0.030
 max 


 
2
B
2
68.6
 
 68.6 
2
  B    B2 
 
  37.7  85.3 MPa
2
2
 2 
 2 
2
Ans.
2
 
 68.6 
2
 max   B    B2  
  37.7  51.0 MPa Ans.
 2 
 2 
_____________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 69/114
3-89
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
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.
2
M B  AB RA2 y  RA2z  0.075 161.82   1778   133.9 N  m
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)
B 

 50.5 106 Pa  50.5 MPa
3
3
d
 0.030

B 

 
16TB
16(200)

 37.7 106 Pa  37.7 MPa
3
3
d
 0.030
Shigley’s MED, 11th edition


 
Chapter 3 Solutions, Page 70/114
 max
2
B
2
50.5
 
 50.5 
2

  B    B2 
 
  37.7  70.6 MPa
2
2
2
2




2
2
 
 50.5 
2
 max   B    B2  
  37.7  45.4 MPa
 2 
 2 
Ans.
Ans.
_____________________________________________________________________________
3-90
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
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  R A2z  4 140 2  542  600 lbf  in
M C  CD RD2 y  RD2 z  6 75.92  1262  883 lbf  in
The maximum stresses occur at C.
C 
C 
32M C
d
3
16TC
d
3


32(883)

 1.3753
16(450)

 1.3753
Shigley’s MED, 11th edition


Ans.
 3460 psi
 882 psi
Chapter 3 Solutions, Page 71/114
 max
2
C
2
3460
 
 3460 
2

  C    C2 
 
  882  3670 psi
2
2
2
2




2
2
 
 3460 
2
 max   C    C2  
  882  1940 psi
 2 
 2 
Ans.
Ans.
_____________________________________________________________________________
3-91
(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  32M 32  8  200 
x 



 16 297 psi  16.3 kpsi
3
I
 d 4 / 64  d 3
 1
 xz 
Tr T  d / 2  16T 16  5  200 



 5093 psi  5.09 kpsi
3
J  d 4 / 32  d 3
 1
(c)
2
x
2
16.3
2
2
 
 16.3 
  x    xz  
 
   5.09 
2
2
 2 
 2 
 1  17.8 kpsi
Ans.
1,  2 
 2  1.46 kpsi
2
Ans.
2
2
2
 
 16.3 
Ans.
 max   x    xz   
   5.09   9.61 kpsi
 2 
 2 
_____________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 72/114
3-92
(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
 1400 
 Mz 
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  32 M tot 32 1612 
 x  tot  tot 4


 16 420 psi  16.4 kpsi
3
I
 d / 64
 d3
 1
 = tan 1 

Tr T  d / 2  16T 16  5 175 



 4456 psi  4.46 kpsi
3
J  d 4 / 32  d 3
 1
(c)
2
x
2
16.4
2
 
 16.4 
1,  2 
  x   2 
 
   4.46 
2
2
 2 
 2 
 1  17.5 kpsi
Ans.
 2  1.13 kpsi
2
Ans.
2
2
 
 16.4 
 max   x    2  
   4.46   9.33 kpsi
 2 
 2 
Ans.
_____________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 73/114
3-93
(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
M


z


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  32 M tot 32 1985 
 x ,bend  tot  tot 4


 20 220 psi  20.2 kpsi
3
I
 d / 64
d3
 1
 = tan 1 
 x ,axial 
Fx
Fx
75


 95.5 psi  0.1 kpsi , which is essentially negligible
2
A  d / 4  12 / 4
 x   x ,axial   x ,bend  20 220  95.5  20 316 psi  20.3 kpsi

Tr 16T 16  5  200 


 5093 psi  5.09 kpsi
3
J d3
 1
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 74/114
(c)
2
x
2
20.3
2
 
 20.3 
1,  2 
  x   2 
 
   5.09 
2
2
 2 
 2 
 1  21.5 kpsi
Ans.
 2  1.20 kpsi
2
Ans.
2
2
 
 20.3 
 max   x    2  
Ans.
   5.09   11.4 kpsi
 2 
 2 
_____________________________________________________________________________
3-94
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 for Eq. (3-40), with b/c = 1.5/0.25 = 6,  =
0.299. From Eq. (3-40),
T
400
 max 

 14 270 psi  14.3 kpsi
Ans.
2
2
 bc
 0.299 1.5  0.25
____________________________________________________________________________
3-95
(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  32M 32 11 250 
x 



 28 011 psi  28.0 kpsi
3
I
 d 4 / 64  d 3
 1
 xz 
Tr T  d / 2  16T 16 12  250 



 15 279 psi  15.3 kpsi
3
J  d 4 / 32  d 3
 1
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 75/114
(c)
2
x
2
28.0
2
2
 
 28.0 
1,  2 
  x    xz  
 
  15.3
2
2
 2 
 2 
 1  34.7 kpsi
Ans.
 2  6.7 kpsi
Ans.
2
2
2
2
 
 28.0 
 max   x    xz   
Ans.
  15.3  20.7 kpsi
 2 
 2 
____________________________________________________________________________
3-96
(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

 = 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 
 x ,bend  tot  tot 4


 46 142 psi  46.1 kpsi
3
I
 d / 64
 d3
 1
 x ,axial 
Fx
Fx
300


 382 psi  0.382 kpsi
2
A  d / 4  12 / 4
 x   x ,axial   x ,bend  46 142  382  46 524 psi  46.5 kpsi

Tr 16T 16 12  250 


 15 279 psi  15.3 kpsi
3
J d3
 1
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 76/114
(c)
2
x
2
46.5
2
 
 46.5 
1,  2 
  x   2 
 
  15.3
2
2
 2 
 2 
 1  51.1 kpsi
Ans.
 2  4.58 kpsi
Ans.
2
2
2
 
 46.5 
 max   x    2  
  15.3  27.8 kpsi
 2 
 2 
Ans.
____________________________________________________________________________
3-97
(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

 = 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 
 x ,bend  tot  tot 4


 55 462 psi  55.5 kpsi
3
I
 d / 64
 d3
 1
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 77/114
 x ,axial 
Fx
Fx
300


 382 psi  0.382 kpsi
2
A  d / 4  12 / 4
 x   x ,axial   x ,bend  55 462  382  55 844 psi  55.8 kpsi

Tr 16T 16 12  250 


 15 279 psi  15.3 kpsi
3
J d3
 1
(c)
2
x
2
55.8
2
 
 55.8 
  x   2 
 
  15.3
2
2
 2 
 2 
 1  59.7 kpsi
Ans.
1,  2 
 2  3.92 kpsi
Ans.
2
 max
2
2
 
 55.8 
  x   2  
  15.3  31.8 kpsi
 2 
 2 
Ans.
____________________________________________________________________________
3-98
(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
Fig. A-15-9
K t ,bend  1.59
 x  K t ,bend
32 11 250 
Mc
32 M
 K t ,bend
 (1.59)
 44 538 psi  44.5 kpsi
3
3
d
I
 1
 xz  K t ,torsion
16 12  250 
Tr
16T
 K t ,torsion

(1.39)
 21 238 psi  21.2 kpsi
3
d3
J
 1
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 78/114
(c)
2
x
2
44.5
2
2
 
 44.5 
1,  2 
  x    xz  
 
   21.2 
2
2
 2 
 2 
 1  53.0 kpsi
Ans.
 2  8.48 kpsi
Ans.
2
2
2
2
 
 44.5 
 max   x    xz   
Ans.
   21.2   30.7 kpsi
 2 
 2 
____________________________________________________________________________
3-99
(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

 = 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
Shigley’s MED, 11th edition
Fig. A-15-8
Chapter 3 Solutions, Page 79/114
Fig. A-15-9
K t ,bend  1.59
 x ,bend  K t ,bend
32  4530 
Mc
32 M
 K t ,bend
 (1.59)
 73 366 psi  73.4 kpsi
3
3
d
I
 1
 x ,axial  K t ,axial
Fx
300
 1.75 
 668 psi  0.668 kpsi
2
A
 1 / 4
 x   x ,axial   x ,bend  73 366  668  74 034 psi  74.0 kpsi
  K t ,torsion
16 12  250 
Tr
16T
 K t ,torsion
 (1.39)
 21 238 psi  21.2 kpsi
3
3
d
J
 1
(c)
2
x
2
74.0
2
 
 74.0 
  x   2 
 
   21.2 
2
2
 2 
 2 
 1  79.6 kpsi
Ans.
1,  2 
 2  5.64 kpsi
2
 max
Ans.
2
2
 
 74.0 
  x   2  
   21.2   42.6 kpsi
 2 
 2 
Ans.
____________________________________________________________________________
3-100
(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.
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 80/114
My = (300)(12) – (–100)(11) = 4700 lbf·in
Mz = (250)(11) = 2750 lbf·in
M tot  M y2  M z2

 4700    2750 
2
2
 5445 lbf  in
 M
z
 My


 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)
r / d  0.125 /1  0.125
 = tan 1 
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
 x ,bend  K t ,bend
32  5445 
Mc
32 M
 K t ,bend

(1.59)
 88185 psi  88.2 kpsi
3
d3
I
 1
 x ,axial  K t ,axial
Fx
300
 1.75 
 668 psi  0.668 kpsi
2
A
 1 / 4
 x   x ,axial   x ,bend  88 185  668  88 853 psi  88.9 kpsi
  K t ,torsion
16 12  250 
Tr
16T
 K t ,torsion
 (1.39)
 21 238 psi  21.2 kpsi
3
3
d
J
 1
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 81/114
(c)
2
x
2
88.9
2
 
 88.9 
1,  2 
  x   2 
 
   21.2 
2
2
 2 
 2 
Ans.
 1  93.7 kpsi
 2  4.80 kpsi
Ans.
2
2
2
 
 88.9 
Ans.
 max   x    2  
   21.2   49.2 kpsi
 2 
 2 
____________________________________________________________________________
3-101 (a) (Eq. 3-42):
200  60 
Tn

 0.1904 hp
Ans.
63 025 63 025
(b) The output torque To = (i /o) Ti. But i r1 = o r2. Thus, To = (r2 / r1) Ti. So,
To = (2.5/1) 200 = 500 lbf٠in
Ans.
P
(c)
 Mx = 0, (FG cos 20o)(1) – 200 = 0
FG = 212.84 lbf
 (MB)y = 0, 2(FG cos 20o) – 3.5RAz = 0
RAz = 2(212.84 cos 20o)/3.5 = 114.29 lbf
 Fz = 0, RBz + 114.29 – 212.84 cos 20o = 0
RBz = 85.71 lbf
 (MB)z = 0, 2 (212.84 sin 20o) + 3.5 (FA)y = 0
RAy = – 41.60 lbf
 Fy = 0, RBy – 41.60 +212.84 sin 20o = 0  RBy = – 31.19 lbf
2
2
RA  RAy
 RAz

 41.60 
2
2
RB  RBy
 RBz

 31.19 
2
 114.292  121.6 lbf
2
 85.712  91.21 lbf
Ans.
Ans.
(d)
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 82/114
(e) (My)max = 2( 85.71) =  171.42 lbf-in, (Mz)max = 2( 31.19) =  62.38 lbf-in
 171.42    62.38  182.42
32 M 32 182.42  3


10  14.87 kpsi
3
d3
  0.5
16T 16  200  3
 3
10  8.15 kpsi
3
d
  0.5
2
M max 
2
lbf-in
Ans.
Ans.
(f)
2

2
14.87
 
 14.87 
2
 
1, 2       2 
  8.15
2
2
2
2
 


 18.47,  3.60 kpsi
Ans.
2
2
 
 14.87 
2
Ans.
 max      2  
  8.15  11.03 kpsi
2
 2 
______________________________________________________________________________
3-102 (a) (Eq. 3-42):
200  60 
Tn
P

 0.1904 hp
Ans.
63 025 63 025
(b) The output torque To = (i /o) Ti. But i r1 = o r2. Thus, To = (r2 / r1) Ti. So,
To = (2.5/1) 200 = 500 lbf٠in
Ans.
(c)
 Mx = 0 = (FG cos 20o)(1) – 200
FG = 212.84 lbf
 (MB)y = 0 = – 2(FG cos 20o) – 3.5RCz
RCz = – 2(212.84 cos 20o)/3.5 = – 114.29 lbf
 Fz = 0 = RDz – 114.29 + 212.84 cos 20o
RDz = – 85.71 lbf
 (MB)z = 0 = – 2 (212.84 sin 20o) + 3.5 RCy  RCy = 41.60 lbf
 Fy = 0 = RDy + 41.60 – 212.84 sin 20o  RDy = 31.19 lbf
2
2
RC  RCy
 RCz
 41.60 2   114.29   121.6 lbf
Ans.
2
2
RD  RDy
 RDz
 31.19 2   85.71  91.21 lbf
Ans.
2
2
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 83/114
(d)
(e) (My)max = 2(85.71) = 171.42 lbf٠in, (Mz)max = 2(31.19) = 62.38 lbf٠in
M max  171.42 2  62.382  182.42 lbf  in

32 M 32 182.42  3

10  14.87 kpsi
3
d3
  0.5

16T 16  500  3

10  20.37 kpsi
 d 3   0.5 3
Ans.
Ans.
(f)
2

2
14.87
 
 14.87 
2
    2 
 
  20.37
2
2
2
 2 
 29.1,  14.2 kpsi
Ans.
 1, 2 
2
2
 
 14.87 
2
Ans.
 max      2  
  20.37  21.7 kpsi
2
 2 
______________________________________________________________________________
3-103
(a) M = F(p / 4), c = p / 4, I = bh3 / 12, b =  dr nt, h = p / 2
 F  p / 4    p / 4 
Mc
Fp 2
b  



3
I
bh3 /12
16   d r nt  p / 2  /12
6F
Ans.
 d r nt p
F
F
4F
(b)  a     2
 2
Ans.
A
 dr / 4
 dr
Tr T  d r / 2  16T
t  

Ans.
 d r4 / 32  d r3
J
b  
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 84/114
(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.
(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
 x  b  
6 1500 
6F

  4584 psi =  4.584 kpsi
 d r nt p
 1.25  2  0.25
 y  a  
4 1500 
4F
=
=  1222 psi =  1.222 kpsi
2
 dr
 1.252 
 yz   t  
16  235 
16T
=
=  612.8 psi =  0.6128 kpsi
3
 dr
 1.253 
Use Eq. (3-15) for the three-dimensional stress element.
2
2
 3   4.584  1.222   2   4.584  1.222    0.6128        4.584  0.6128    0

  5.806  5.226  1.721  0
3



2
The roots are at 0.2543, – 4.584, and –1.476. Thus, the ordered principal stresses are
1 = 0.2543 kpsi, 2 = –1.476 kpsi, and 3 = – 4.584 kpsi.
Ans.
From Eq. (3-16), the principal shear stresses are
0.2543   1.476 
 
Ans.
 1/ 2  1 2 
 0.8652 kpsi
2
2
   3  1.476    4.584 
Ans.
 2/3  2

 1.554 kpsi
2
2
0.2543   4.584 
 
 1/3  1 3 
 2.419 kpsi
Ans.
2
2
____________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 85/114
3-104 As shown in Fig. 3-32, the maximum stresses occur at the inside fiber where r = ri.
Therefore, from Eq. (3-50)
 t ,max
ri2 pi  ro2 
 2 2 1  2 
ro  ri  ri 
 r2  r2 
 pi  o2 i2 
 r r 
 o i 
Ans.
ri2 pi  ro2 
1     pi Ans.
ro2  ri2  ri2 
______________________________________________________________________________
 r ,max 
3-105 If pi = 0, Eq. (3-49) becomes
t 
 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

 t ,max  
For σr, we have
r 
2 po ro2
Ans.
ro2  ri 2
 po ro2  ri 2 ro2 po / r 2
ro2  ri 2
po ro2  ri 2 
 2 2  2  1
ro  ri  r

So σr = 0 at r = ri. Thus at r = ro
po ro2  ri 2  ro2 
 r ,max  2 2  2    po Ans.
ro  ri  ro 
______________________________________________________________________________
3-106 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
( t ) av   1   2 
p  / 4  di2
 di t

pd i
4t
Ans.
The radial stress on the inner surface of the shell is, 3 =  p Ans.
______________________________________________________________________________
3-107 σt > σl > σr
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 86/114
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
 max  
ro2  ri 2
32  2.752
(10 000)  1597 psi Ans.
 pi 
 max 
ro 2
32
______________________________________________________________________________
3-108 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 
 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
( max  pi )
 max
 100
(25  4)106
 91.7 mm
25 106 
t  ro  ri  100  91.7  8.3 mm Ans.
______________________________________________________________________________
3-109 σt > σl > σr
τmax = (σt − σr)/2 at r = ri
ri 2 pi  ro2 
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  ri 2  ro2  ri 2
 max  
4 2 (500)
 4129 psi Ans.
42  3.752
______________________________________________________________________________

3-110 From Eq. (3-49) with pi = 0,
r2 p  r2 
 t   2o o 2 1  i 2 
ro  ri  r 
ro2 po  ri 2 
 r   2 2 1  2 
ro  ri  r 
σt > σl > σr, and since σt and σr are negative,
τmax = (σr − σt)/2 at r = ro
 max

ro2 po  ri 2  ri 2 po
1  ro2 po  ri 2  ro2 po  ri 2  
   2 2 1  2   2 2 1  2   2 2  2   2 2
2  ro  ri  ro  ro  ri  ro   ro  ri  ro  ro  ri
po 
ro2  ri 2
32  2.752
(10 000)  1900 psi Ans.


max
2.752
ri 2
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 87/114
______________________________________________________________________________
3-111 From Eq. (3-49) with pi = 0,
r2 p  r2 
 t   2o o 2 1  i 2 
ro  ri  r 
r  
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
ro2 po  ri2  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
 max   

ri  ro
 max
( max  po )
 100
25 106 
 25  4 106
 92.8 mm
t  ro  ri  100  92.8  7.2 mm Ans.
______________________________________________________________________________
3-112 From Eq. (3-49) with pi = 0,
t  
ro2 po  ri 2 
1  
ro2  ri 2  r 2 
r  
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
r2 p  r2  r2 p
1  r 2 p  r 2  r 2 p  r 2 
 max    2o o 2 1  i 2   2o o 2 1  i 2    2o o 2  i 2   2i o 2
2  ro  ri  ro  ro  ri  ro   ro  ri  ro  ro  ri
3.752 (500)
 2
 3629 psi Ans.
4  3.752
______________________________________________________________________________
3-113 From Table A-20, Sy=490 MPa
From Eq. (3-49) with pi = 0,
ro2 po  ri 2 
 t   2 2 1  2 
ro  ri  r 
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 88/114
Maximum will occur at r = ri
0.8(490)  25  19 
(r  r )

2r 2 p
 82.8 MPa Ans.
 t ,max   2 o o2  po   t ,max o2 i  
ro  ri
2ro
2(252 )
______________________________________________________________________________
2
2
2
2
3-114 From Table A-20, Sy = 71 kpsi
From Eq. (3-49) with pi = 0,
ro2 po  ri 2 
 t   2 2 1  2 
ro  ri  r 
Maximum will occur at r = ri
 t ,max  ro  ri 
 0.8(71)  1  0.75 
2ro2 po
 t ,max   2 2  po  


 12.4 kpsi Ans.
ro  ri
2ro2
2(12 )
______________________________________________________________________________
2
2
2
2
3-115 From Table A-20, Sy=490 MPa
From Eq. (3-50)
t 
ri 2 pi  ro2 
1  
ro2  ri 2  r 2 
Maximum will occur at r = ri
 t ,max
2
2
ri 2 pi  ro2  pi  ro  ri 
 2 2 1  2  
ro  ri  ri 
ro2  ri 2
 t ,max (ro2  ri 2 )
0.8(490)  (252  19 2 )


 105 MPa
Ans.
ro2  ri 2
(252  192 )
______________________________________________________________________________
3-116 From Table A-20, Sy =71 MPa
From Eq. (3-50)

t 
pi 
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 )
 t ,max  2i i 2 1  o2   i 2o 2i
ro  ri  ri 
ro  ri
 t ,max (ro2  ri 2 )
0.8(71)  (12  0.752 )  15.9 ksi
Ans.
ro2  ri 2
(12  0.752 )
______________________________________________________________________________

Shigley’s MED, 11th edition
pi 

Chapter 3 Solutions, Page 89/114
3-117 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 s = 0.282 lbf/in3. The area of the wall
is
Awall = ( /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 = (2 /3)(1803  179.253) = 152.0 (103) in3
The weight of the structure on the wall area at the tank bottom is
W = s Vtotal = 0.282(609.5 +152.0) (103) = 214.7(103) lbf
214.7 103 
W
l  

 254 psi
Awall
846.5
The maximum pressure will occur at the bottom of the tank, pi = water h. From Eq. (3-50)
with r  ri
t 
 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 
ri 2 pi  ro2 
 23.8 psi Ans.
 r  2 2 1  2    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.
Since 1  2  3, 1 = t = 5708 psi, 2 = r =  24 psi and3 = l =  254 psi. From
Eq. (3-16),
5708  254
 2981  2980 psi
2
5708  24
Ans.
 2866  2870 psi
1 2 
2
24  254
 115 psi
2 3 
2
______________________________________________________________________________
1 3 
3-118 Stresses from additional pressure are,
Eq. (3-52),
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 90/114
 l 50psi 
Eq. (3-50)
(r)50 psi
50 179.252 
180 2  179.252
=  50 psi
 5963 psi
1802  179.252
 11 975 psi
180 2  179.252
Adding these to the stresses found in Prob. 3-117 gives
 t 50psi  50
t = 5708 + 11 975 = 17683 psi = 17.7 kpsi Ans.
r =  23.8  50 =  73.8 psi Ans.
l =  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
1 3 
 8879 psi
2
17 683  5709
Ans.
 5987 psi
1 2 
2
5709  23.8
 2866 psi
2 3 
2
______________________________________________________________________________
3-119 (a) For shapes let  (3 +  )/8 = 1,  = 0.3, ro = 2, and ri = 1. Then,
1  3 1  3  0.3

 0.5758
3 
3  0.3
Thus, Eqs. (3-55) are
4
4
 t  1  4  2  0.5758 r 2 ,
r  1 4  2  r2
r
r
These equations are plotted:
Rotating Ring
10
8
6
4
2
0
1
1.2
1.4
1.6
tangential stress
1.8
2
radial stress
(b) The tangential stress, t is maximum at r = ri and is given by
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 91/114
3     2 2 ri 2 ro2 1  3 2 
ri 
  ri  ro  2 
3 
ri
 8 

 t max   2 

 3      3    1  3   2
2
  2 
 ri  2ro 
 
3 
 8   


 2
1   ri 2   3   ro2 
Ans.
4
r = 0 at r = ri and r = ro. (r)max occurs where dr / dr = 0.
2 2

d r
 3    2ri ro
4
2 2
  2 
  3  2r   0  r  ri ro  r  ri ro
dr
r
8



Thus,

3    2 2 ri 2 ro2
2
 3  
 ri ro    2 
Ans.
 r max   2 
  ri  ro 
  ro  ri 
ri ro
 8 
 8 

______________________________________________________________________________

3-120 Since σt and σr are both positive and σt > σr
 max   t  max 2
From Eq. (3-55), t is maximum at r = ri = 0.3125 in. The term
2
 0.282  2  5000    3  0.292 

 

  82.42 lbf/in
60
8
 386 





0.31252 2.752 1  3(0.292)
2
2
2 
 82.42 0.3125  2.75 

0.3125


3  0.292
0.31252


 1260 psi
 3 
 
 8
2
 t max
 max 

1260
 630 psi
2
Ans.

r 2r 2




Radial stress:

 r  k  ri2  ro2  i 2o  r 2 
r


Maxima:
 ri2 ro2

d r
 k  2 3  2r   0
 r

dr


Shigley’s MED, 11th edition
 r  ri ro  0.3125(2.75)  0.927 in
Chapter 3 Solutions, Page 92/114




0.31252 2.752
2
2
2

 0.927
 r max  82.42 0.3125  2.75 


0.927 2


 490 psi Ans.
______________________________________________________________________________
3-121  = 2 (2000)/60 = 209.4 rad/s,  = 3320 20 kg/m3,  = 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.
______________________________________________________________________________
 t  3320(209.4)2 
3-122 Eq (3-55):
 2 2 ri 2 ro2 r 2 
 2 2 ri 2 ro2

K
r
r
K
 t   i  o  2   (1),
and  r   ri  ro  2  r 2  (2)
2
r
r



2
Where K =  (3 +  )/8 and (1 + 3 )/(3 +  ) = [1 + 3(0.2)] / (3 + 0.2) = 1/2.
It can be seen that t is always positive. Check for maxima.
 r 2r 2

d t
 K  2 i 3o  r   0  r 4  2ri 2 ro2 No roots, no maxima. Check at extremes.
dr
r


At r = ri,

 t i  K  ri2  ro2  ro2 
 2 ri 2 
ri 2 

K

 2ro  
2
2


 2 2 2 ro2 
 2 ro2 
At r = ro,
 t i  K  ri  ro  ri    K  2ri  
2
2


Eq. (3) > (4). Thus, (t)max = (t)i .
For Eq. (2), r = 0 at r = ri and r = ro. Check for maxima.
 r 2r 2

d r
(5)
 K  2 i 3o  2r   0  r  ri ro
dr
r


Substitute Eq. (5) into (2),
(3)
(4)
 r max  K  ri2  ro2  ri ro  ri ro   K  ro2  2ri ro  ri2   K  ro  ri 
2
(6)

r2 
Comparing Eqs. (3) and (6) it is clear that K  2ro2  i   K  ro2  ri 2  2ri ro  . Thus, the
2

 3  
2
2
maximum stress is given by Eq. (3) simplified as,  max   2 
  4ro  ri  Ans.
 16 
2
2
3
(b) Vol = (/4)(5 – 0.75 )(1/16) = 1.1996 in .  = 2(12 000)/60 = 1256.6 rad/s.
5 /16
 6.749 104  lbf-s 2 /in 4

386 1.1996 
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 93/114
2
0.375 
2
 3  0.2  
 5 360 psi Ans.
Eq. (3):  t  max  6.749 104 1256.62 
  2  2.5  
2 
 8 
The factor of safety corresponding to fracture is:
S
12
n u 
 2.24
Ans.
 max 5.36
______________________________________________________________________________
3-123  = 2 (3500)/60 = 366.5 rad/s,
mass of blade = m = V = (0.282 / 386) [1.25(30)(0.125)] = 3.425(103) lbfs2/in
F = (m/2)  2r
= [3.425(103)/2]( 366.52)(7.5)
= 1725 lbf
Anom = (1.25  0.5)(1/8) = 0.093 75 in2
nom = 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.
______________________________________________________________________________
3-124  = 0.292, E = 207 GPa, ri = 0, R = 25 mm, ro = 50 mm
Eq. (3-57),
207(109 )  (0.052  0.0252 )(0.0252  0)  9
3

 10  3.105(10 )
2(0.025)3 
(0.052  0)

where p is in MPa and  is in mm.

p

(1)
Maximum interference,
1
 max  [50.042  50.000]  0.021 mm Ans.
2
Minimum interference,
1
 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.
______________________________________________________________________________
3-125  = 0.292, E = 30 Mpsi, ri = 0, R = 1 in, ro = 2 in
Eq. (3-57),
30(106 )  (2 2  12 )(12  0) 
7
p

  1.125(10 )
3
2
2(1 ) 
(2  0)

Shigley’s MED, 11th edition
(1)
Chapter 3 Solutions, Page 94/114
where p is in psi and  is in inches.
Maximum interference,
1
 max  [2.0016  2.0000]  0.0008 in
2
Minimum interference,
1
2
 min  [2.0010  2.0010]  0
Ans.
Ans.
From Eq. (1),
pmax = 1.125(107)(0.0008) = 9 000 psi
Ans.
pmin = 1.125(107)(0) = 0
Ans.
______________________________________________________________________________
3-126  = 0.292, E = 207 GPa, ri = 0, R = 25 mm, ro = 50 mm
Eq. (3-57),
207(109 )  (0.052  0.0252 )(0.0252  0)  9
3

 10  3.105(10 )
2(0.025)3 
(0.052  0)

where p is in MPa and  is in mm.

p

(1)
Maximum interference,
1
 max  [50.059  50.000]  0.0295 mm Ans.
2
Minimum interference,
1
 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.
______________________________________________________________________________
3-127  = 0.292, E = 30 Mpsi, ri = 0, R = 1 in, ro = 2 in
Eq. (3-57),
30(106 )  (2 2  12 )(12  0) 
7
p

  1.125(10 )
2(13 ) 
(22  0)

(1)
where p is in psi and  is in inches.
Maximum interference,
1
 max  [2.0023  2.0000]  0.00115 in
2
Shigley’s MED, 11th edition
Ans.
Chapter 3 Solutions, Page 95/114
Minimum interference,
1
2
 min  [2.0017  2.0010]  0.00035
Ans.
From Eq. (1),
pmax = 1.125(107)(0.00115) = 12 940 psi
pmin = 1.125(107)(0.00035) = 3 938
Ans.
Ans.
______________________________________________________________________________
3-128  = 0.292, E = 207 GPa, ri = 0, R = 25 mm, ro = 50 mm
Eq. (3-57),
207(109 )  (0.052  0.0252 )(0.0252  0)  9
3
 10  3.105(10 )
3 
2
2(0.025) 
(0.05  0)

where p is in MPa and  is in mm.

p

(1)
Maximum interference,
1
 max  [50.086  50.000]  0.043 mm Ans.
2
Minimum interference,
1
 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.
______________________________________________________________________________
3-129  = 0.292, E = 30 Mpsi, ri = 0, R = 1 in, ro = 2 in
Eq. (3-57),
30(106 )  (2 2  12 )(12  0) 
7
p

  1.125(10 )
3
2
2(1 ) 
(2  0)

(1)
where p is in psi and  is in inches.
Maximum interference,
1
 max  [2.0034  2.0000]  0.0017 in
2
Minimum interference,
1
2
 min  [2.0028  2.0010]  0.0009
Shigley’s MED, 11th edition
Ans.
Ans.
Chapter 3 Solutions, Page 96/114
From Eq. (1),
pmax = 1.125(107)(0.0017) = 19 130 psi
Ans.
pmin = 1.125(107)(0.0009) = 10 130
Ans.
______________________________________________________________________________
3-130 From Table A-5, Ei = Eo = 30 Mpsi, i o ri = 0, R = 1 in, ro = 1.5 in
1
The radial interference is    2.002  2.000   0.001 in Ans.
2
Eq. (3-57),


2
2
2
2
E  ro  R R  ri
p 3
ro2  ri 2
2R 

 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
( t )i r  R   p 2 i2  (8333) 2 2  8333 psi  8.33 kpsi Ans.
R  ri
1 0
ro2  R 2
1.52  12

 21 670 psi  21.7 kpsi Ans.
(8333)
rR
ro2  R 2
1.52  12
______________________________________________________________________________
( t )o
p
3-131 From Table A-5, Ei = 30 Mpsi, Eo =14.5 Mpsi, i o 

ri = 0, R = 1 in, ro = 1.5 in
1
The radial interference is    2.002  2.000   0.001 in Ans.
2
Eq. (3-56),
p
p

 1  r 2  R2
 1
 o  
R   o2
2
 Ei
 Eo  ro  R
 R 2  ri 2

 2 2  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  0 2
 4599 psi Ans.
( t )i r  R   p 2 i 2  (4599) 2
R  ri
1  02
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 97/114
ro2  R 2
1.52  12

 11 960 psi Ans.
(4599)
r R
ro2  R 2
1.52  12
______________________________________________________________________________
( t )o
p
3-132 From Table A-5, Ei = Eo = 30 Mpsi, i o ri = 0, R = 0.5 in, ro = 1 in
The minimum and maximum radial interferences are
1
 min  1.002  1.002   0.000 in Ans.
2
1
 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
E  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.
R 2  ri 2
0.52  02
( t )i r  R   p 2 2  (22 500) 2
 22 500 psi Ans.
R  ri
0.5  0 2
ro2  R 2
12  0.52
(22
500)

 37 500 psi Ans.
r R
ro2  R 2
12  0.52
______________________________________________________________________________
( t )o
p
3-133 From Table A-5, Ei = 10.4 Mpsi, Eo =30 Mpsi, i o 

ri = 0, R = 1 in, ro = 1.5 in
The minimum and maximum radial interferences are
1
 min  [2.003  2.002]  0.0005 in Ans.
2
1
 max  [2.006  2.000]  0.003 in Ans.
2
Eq. (3-56),
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 98/114
p
p

 1  r 2  R2
 1
R   o2
 o  
2
 Ei
 Eo  ro  R
 R 2  ri 2

 2 2  i  
 R  ri


 1
 1.52  12

 12  02

1
1

0.292


0.333


2
2
2
6 
6  2
 30 10  1.5  1
 10.4 10  1  0
 
 
p  6.229 10   psi
p  6.229 10  
p  6.229 10  
 
6
6
min
6
max
min
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
( t )i min   pmin 2 i 2  (3.11) 2 2  3.11 kpsi Ans.
R  ri
1 0
( t )o
min
 pmin
ro2  R 2
1.52  12
 (3.11) 2 2  8.09 kpsi
ro2  R 2
1.5  1
Ans.
Maximum interference:
R2  r 2
12  02
( t )i max   pmax 2 i2  (18.7) 2
 18.7 kpsi
R  ri
1  02
Ans.
ro2  R 2
1.52  12
(18.7)

 48.6 kpsi Ans.
max
ro2  R 2
1.52  12
______________________________________________________________________________
( t )o
 pmax
3-134 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 

10 2
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   d 2 / 4   (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,
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 99/114
4000
190 000(9.4677)
F Mci



 300 MPa
A Aeri 314.16 314.16(0.53228)(37.5)
i 
Ans.
4000
190 000(10.5323)
F Mco



 195 MPa Ans.
A Aero 314.16 314.16(0.53228)(57.5)
______________________________________________________________________________
o 
3-135 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   d 2 / 4   (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,
750
1218.8(0.35307)
F Mci



 37 230 psi  37.2 kpsi
A Aeri 0.44179 0.44179(0.02193)(1.25)
i 
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)
______________________________________________________________________________
o 
3-136 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   d 2 / 4   (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,
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 100/114
300
3900(2.82456)
F Mci



 233 MPa
A Aeri 28.2743 28.2743(0.17544)(10)
i 
Ans.
300
3900(3.17544)
F Mco



 145 MPa Ans.
A Aero 28.2743 28.2743(0.17544)(16)
______________________________________________________________________________
o 
3-137
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   d 2 / 4   (6) 2 / 4  28.2743 mm 2
The angle  of the line of radius centers is
 Rd /2 

1  10  6 / 2 
  sin 1 
  sin 
  30
 Rd R
 10  6  10 
M  F  R  d / 2  sin   300 10  6 / 2  sin 30  1950 N  mm
Using Eq. (3-65) for the bending stress, and combining with the axial stress,
F sin  Mci 300 sin 30
1950(2.82456)



 116 MPa
A
Aeri
28.2743
28.2743(0.17544)(10)
i 
Ans.
1950(3.17544)
F sin  Mco 300sin 30



 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.
______________________________________________________________________________
o 
3-138 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
2 0.625  0.625  0.125
2

 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
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 101/114
A   d 2 / 4   (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,
i 
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)
______________________________________________________________________________
3-139 d  0.25 in, ri  0.5 in, ro  0.75 in
o 
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   d 2 / 4   (0.25) 2 / 4  0.049087 in 2
The angle  of the line of radius centers is
 Rd /2 
1  0.5  0.25 / 2 

  sin 1 
  sin 
  30
 Rd R
 0.5  0.25  0.5 
M  F  R  d / 2  sin   75  0.5  0.25 / 2  sin 30  23.44 lbf  in
Using Eq. (3-65) for the bending stress, and combining with the axial stress,
i 
F sin  Mci 75sin 30
23.44(0.118686)



 18 716 psi  18.7 kpsi
A
Aeri 0.049087 0.049087(0.006314)(0.5)
Ans.
F sin  Mco 75sin 30
23.44(0.131314)



 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.
______________________________________________________________________________
o 
3-140
(a)   
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,
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 102/114
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. 3-35. Using
Eq. (3-65),
i 
Mci
12(0.049006)

 10 070 psi  10.1 kpsi
Aeri 0.08205(0.005694)(0.125)
o  
Mco
12(0.060394)

 6618 psi  6.62 kpsi
Aero
0.08205(0.005694)(0.2344)
 i 10.1

 1.26
 8.02

6.62
Ko  o 
 0.825
 8.02
(c) K i 
Ans.
Ans.
Ans.
Ans.
______________________________________________________________________________
3-141
(a)   
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. 3-35. Using
Eq. (3-65),
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 103/114
i 
Mci
12(0.061552)

 6406 psi  6.41 kpsi
Aeri 0.10545(0.008748)(0.125)
o  
Ans.
Mco
12(0.079048)

 3872 psi  3.87 kpsi
Aero
0.10545(0.008748)(0.2656)
Ans.
 i 6.41

 1.32
Ans.
 4.86

3.87
Ko  o 
 0.80
Ans.
 4.86
______________________________________________________________________________
(c) K i 
3-142
(a)   
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. 3-35. Using
Eq. (3-65),
i 
Mci
12(0.051398)

 9106 psi  9.11 kpsi
Aeri 0.08205(0.003302)(0.25)
o  
Ans.
Mco
12(0.058002)

 7148 psi  7.15 kpsi
Aero
0.08205(0.003302)(0.3594)
Ans.
 i 9.11

 1.14
Ans.
 8.02

7.15
Ko  o 
 0.89
Ans.
 8.02
______________________________________________________________________________
(c) K i 
3-143 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.
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 104/114
rn 
A
  dA / r 
For a rectangular area with constant width b, the denominator is
ro  bdr 
r
ri  r   b ln roi
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 mm 2
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 60  3.2 cos 0  4.40 kN
Fy  F1 y  F2 y  2.4sin 60  3.2sin 0  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
  tan 1
Fy
Fx
 tan 1
2.08
 25.3
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  2.4cos  60  25.3   1.97 kN
Fn  2.4sin  60  25.3   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,
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 105/114
i 
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)
______________________________________________________________________________
o 
3-144 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. 3-143,
dA
3.625
6
 0.75 ln
 0.75ln
 0.682 920 in
r
2
4.375
2.4375
A
rn 

 3.56923 in
0.682
920
(
/
)
dA
r


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,
i 
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)
______________________________________________________________________________
o 
3-145
ri = 12 in, ro = ri + h = 12 + 3 = 15 in, rc = 12 + 3/2 = 13.5 in


I  a 3b  (1.53 )(0.75)  1.988 in 4
4
4
A   ab   (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
o   o c 

 55.5 kpsi Ans.
A
Iro
3.534
1.988(15)
______________________________________________________________________________
3-146
ri = 1.25 in, ro = ri + h = 1.25 + 0.5 + 1 + 0.5 = 3.25 in
i 
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 106/114
rc = (ri + ro) / 2 = (1.25 + 3.25)/2 = 2.25 in
Ans.
ro
 dA 
For outer rectangle,  
  b ln
ri
 r 


 A 
r2

 , A   r2
 
For circle, 
O


2
2
   dA r  
r
r
r


2
c
 c
 O
O
 dA 
    2 (rc  rc2  r 2 )
 r O
Combine the integrals subtracting the circle from the rectangle




3.25
dA
 1.25 ln
 2 2.25  2.252  0.52  0.840 904 in
r
1.25
A  1.25(2)   (0.52 )  1.714 60 in 2
Ans.
A
1.714 60
rn 

 2.0390 in Ans.
0.840904
dA
r
(
/
)


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
i  


 20 720 psi = 20.7 kpsi Ans.
A Aeri 1.7146 1.7146(0.2110)(1.25)
2000
13 500(1.2110)
F Mco
o  


 12 738 psi  12.7 kpsi
Ans.
A Aero 1.7146 1.7146(0.2110)(3.25)
______________________________________________________________________________
3-147 Table A-5: Glass: EG = 46.2 GPa, G = 0.245, Steel: ES =207 GPa, S = 0.292
Eq. (3-68):
a
3
2
2
3F 1  1  / E1  1  2  / E2
8
1/ d1  1/ d 2
2
9
2
9
3  5  1  0.245  /  46.2 10   1  0.292  / 207 10  

8
1/ 0.030  1/ 
3
 1.1168 104  m  0.11168 mm
Eq. (3-69):
pmax 
3 5
3F

 191.4 MPa
2
2
2 a
2  0.11168
Ans.
(b) Eq. (3-70):
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 107/114


1 
1
z


1
 1   2   pmax 1  tan
 1   G  
2 
a
z / a 
2 1   z / a   






1
1
z
 

 1
 191.4  1  
 1  0.245 
 tan
2 
 z / 0.11168  
2 1   z / 0.11168   
   0.11168 


Eq. (3-71):
 pmax
191.4
3 

(2)
2
2
1   z / a  1   z / 0.11168 
The maximum shear stress is given by
 3
(3)
 max  1
2
(c) Plots of Eqs. (1) – (3) as absolute dimensionless stresses,stress / pmax, are given.
Note, at z = 0, 1 = 2 =  0.745 pmax.
(1)
1.0
0.9
0.8

Stress/pmax
0.7
Ans.
1,2
0.6
0.5
0.4
max
0.3
0.2
0.1
0.0
0.0
0.2
0.4
0.6
0.8
1.0
z/a
(d) max  0.323 pmax = 0.323(191.4) = 61.8 MPa at z = 0.05 mm
Ans.
(e) From Fig. 3-38, max = 0.3(191.4) = 57.4 MPa
Ans.
______________________________________________________________________________
3-148 From Eq. (3-68),
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 108/114
13
a  KF 1 3

2 1  2  E  
1 3  3  
 F  

2 1 d 
 8 

Use  = 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
2 a
2 ( KF )
2 K
2 (0.03685)
From Eq. (3-71), the maximum principal stress occurs on the surface where z = 0, and is
equal to – pmax.
 max   z   pmax  352 F 1/3 MPa
Ans.
From Fig. 3-38,
 max  0.3 pmax  106 F 1/3 MPa
Ans.
______________________________________________________________________________
3-149 From Eq. (3-68),
1  12  E1  1  22  E2

3
F


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

 487.2 MPa
pmax 
2
2 a
2  0.09902 
From Fig. 3-38, 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.


1
 1   2  487.2  1  0.48 tan 1 1/ 0.48   1  0.333 
 101.3 MPa

2 1  0.482  

487.2
3 
 396.0 MPa
1  0.482
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 109/114
From Eq. (3-72),
 101.3   396.0   147.4 MPa Ans.
 
 max  1 3 
2
2
Note that if a closer examination of the applicability of the depth assumption from Fig. 338 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 . For =
0.333 for aluminum, the maximum shear stress occurs at a depth of z = 0.492a with max
= 0.3025 pmax.
This gives max = 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.
______________________________________________________________________________
3-150 From the solution to Prob. 3-149, a = 0.0990 mm and pmax = 487.2 MPa. Assuming
applicability of Fig. 3-38, 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.

 1   2  487.2  1  0.48 tan 1 1/ 0.48   1  0.292  

3 
487.2
 396.0 MPa
1  0.482
From Eq. (3-72),
 92.09    396.0   152.0 MPa
 
 max  1 3 
2
2

1
  92.09 MPa
2 1  0.482  
Ans.
Note that if a closer examination of the applicability of the depth assumption from Fig. 338 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 . For
= 0.292 for steel, the maximum shear stress occurs at a depth of z = 0.478a with max =
0.3119 pmax.
______________________________________________________________________________
3-151 From Eq. (3-68),
2 1  2  E
F
3


a3

 8  1 d1  1 d 2
2
 3  20   2 1  0.292   207 000 
a3
 0.1258 mm

1 30  1 
 8 
From Eq. (3-69),
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 110/114
pmax 
3  20 
3F

 603.4 MPa
2
2 a
2 0.12582


From Fig. 3-38, the maximum shear stress occurs at a depth of
z  0.48a  0.48  0.1258   0.0604 mm
Ans.
Also from Fig. 3-38, the maximum shear stress is
 max  0.3 pmax  0.3(603.4)  181 MPa
Ans.
______________________________________________________________________________
3-152 Aluminum Plate-Ball interface: From Eq. (3-68),
2
2
 3F  1  1  E1  1  2  E2
a 

1 d1  1 d 2
 8 
3
2
6
2
6
 3F  1  0.292   30  10   1  0.333  10.4  10  
a 
 3.517 103  F 1/3 in

1 11 
 8 
3
From Eq. (3-69),
3F
3F

 3.860 104 F 1/3 psi
pmax 
2
2
2 a
2 3.517 103 F 1/3 
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,


1
 1  3.86 104  F 1/3  1  0.48 tan 1 1/ 0.48   1  0.333 
 8025 F 1/3 psi
2 
2 1  0.48  

3.86 104  F 1/3
3 
 3.137 104  F 1/3 psi
2
1  0.48

From Eq. (3-72),
 max 
1   3
 

 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),
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 111/114
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),
3F
3F

 4.369 10 4 F 1/3 psi
pmax 
2
2
3
1/3
2 a
2 3.306 10 F 
The steel ball has a higher Poisson’s ratio than the cast iron table, so it will dominate.

 


 1  4.369 10 4  F 1/3  1  0.48 tan 1 1/ 0.48  1  0.292  
3 

4.369 10 4  F 1/3
1  0.48
From Eq. (3-72),
 max 
1   3
2

1
 8258 F 1/3 psi
2 
2 1  0.48  
 3.55110 4  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.
______________________________________________________________________________
3-153 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
12
 2 1  0.3332  10.4 106    1  0.2112  14.5 106  




Kc  
  (2)

1 /1.25  1/  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. 3-40 applicable.
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 112/114
 max  0.3 pmax
pmax 
4000
 13 300 psi
0.3
From Eq. (3-74), pmax = 2F / (bl ), so we have
pmax
2F
2F 1 2


 lK c F 1 2  lK c
So,
  lK c pmax 
F 

2


2
  (2)(2.593) 10 4  (13 300) 




2


 117.4 lbf Ans.
______________________________________________________________________________
2
3-154
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):
12
 2(40) 2 1  0.2922  30 106   


b
  (0.75)

1/ 0.94  1/ 1.24


Eq. (3-74):
pmax 
 1.052 103  in
2  40 
2F

 32 275 psi  32.3 kpsi
 bl
 1.052  103   0.75 
Ans.
From Fig. 3-40,
 max  0.3 pmax  0.3  32 275 = 9682.5 psi  9.68 kpsi
Ans.
______________________________________________________________________________
3-155 Use Eqs. (3-73) through (3-77).
1/ 2
 2 F (1  12 ) / E1  (1  v22 ) / E2 
b

(1/ d1 )  (1/ d 2 )
 l

1/ 2
 2(600) (1  0.292 2 ) / (30(10 6 ))  (1  0.292 2 ) / (30(106 )) 


1/ 5  1/ 
  (2)

b  0.007 631 in
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 113/114
pmax 
2F
2(600)

 25 028 psi
 bl  (0.007 631)(2)

z2 z 
   2  0.292  25 028 
2

b
b 

Ans.
 7102 psi  7.10 kpsi
 x  2 pmax  1 

1  0.786 2  0.786



z2
 1  2 0.7862

 1 2 2



z
b


 y   pmax
2
 25 028
 2  0.786  
2
2



b
 1   0.786 
z
1





b2


 4 646 psi  4.65 kpsi
Ans.
 pmax
25 028
z 

 19 677 psi  19.7 kpsi
Ans.
z2
1  0.7862
1 2
b
   z 4 646   19 677 

 7 516 psi  7.52 kpsi
 max  y
Ans.
2
2
______________________________________________________________________________
3-156 Use Eqs. (3-73) through (3-77).
 2 F 1  12  E1  1  2 2  E2 

b
 l

1/ d1  1/ d 2


12
12
 2(2000) 1  0.2922   207 103   1  0.2112  100 103   





  (40)

1/150  1/ 


b  0.2583 mm
pmax 
2F
2(2000)

 123.2 MPa
 bl  (0.2583)(40)

z2 z 
   2  0.292 123.2  1  0.786 2  0.786
2

b
b 

Ans.
 35.0 MPa


z2
 1  2 0.7862

 1 2 2



z
b


2
 123.2
 2  0.786  
 y   pmax
2


2


b
z
 1   0.786 


 1 2

b


 22.9 MPa
Ans.
 x  2 pmax  1 
Shigley’s MED, 11th edition


Chapter 3 Solutions, Page 114/114
z 
 pmax
 max 

z2
1 2
b
 y  z
2

123.2
1  0.786 2
 96.9 MPa
22.9   96.9 
2
Ans.
 37.0 MPa
Ans.
______________________________________________________________________________
3-157
Use Eqs. (3-73) through (3-77).
 2 F 1  12  E1  1  2 2  E2 

b
 l

1/ d1  1/ d 2


12
12
 2(250) 1  0.2112  14.5 106    1  0.2112  14.5 106  





  (1.25)

1/ 3  1/ 


b  0.007 095 in
pmax 
2F
2(250)

 17 946 psi
 bl  (0.007 095)(1.25)

z2 z 
   2  0.21117 946 
2

b
b 

 3 680 psi  3.68 kpsi
Ans.
 x  2 pmax  1 

1  0.786 2  0.786



z2
 1  2 0.7862

 1 2 2



z
b


2
 17 946
 2  0.786  
 y   pmax
2


2


b
z
 1   0.786 
1





b2


Ans.
 3 332 psi  3.33 kpsi
 pmax
17 946
z 

 14 109 psi  14.1 kpsi
Ans.
z2
1  0.7862
1 2
b
   z 3 332   14 109 

 5 389 psi  5.39 kpsi
 max  y
Ans.
2
2
______________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 3 Solutions, Page 115/114
Chapter 13
13-1
d P  17 / 8  2.125 in
N
1120
dG  2 d P 
 2.125  4.375 in
N3
544
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   m  3 mm Ans.
13-2
C  3 15  60   2  112.5 mm
Ans.
Ans.
______________________________________________________________________________
NG  16  4   64 teeth
13-3
Ans.
dG  NG m  64  6   384 mm
d P  N P m  16  6   96 mm
C   384  96  / 2  240 mm
Ans.
Ans.
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   / P   / 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 20  6.578 in
Gear Base-Circle:
Ans.
d 2  N 2 / P  28 / 3  9.333 in
d 2b  9.333cos 20  8.770 in
Base pitch:
Ans.
pb  pc cos    / 3 cos 20  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, 11th edition
Chapter 13 Solutions, Page 1/36
______________________________________________________________________________
13-5
(a)
 14 / 6  2  32 / 6 2 
AO  
 
 
 2   2  
(b)
  tan 1 14 / 32   23.63
Ans.
  tan 1  32 /14   66.37
Ans.
(c)
d P  14 / 6  2.333 in
1/2
 2.910 in
Ans.
d G  32 / 6  5.333 in Ans.
(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
Shigley’s MED, 11th edition
Ans.
Ans.
Chapter 13 Solutions, Page 2/36
____________________________________________________________________
13-6
(a)
pn   / Pn   / 4  0.7854 in
pt  pn / cos  0.7854 / cos 30  0.9069 in
px  pt / tan  0.9069 / tan 30  1.571 in
(b)
Eq. (13-7):
(c)
Pt  Pn cos  4 cos 30  3.464 teeth/in
pnb  pn cos n  0.7854 cos 25  0.7380 in
Ans.
t  tan 1  tan n / cos   tan 1 (tan 25 / cos 30 )  28.3 Ans.
(d)
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 30
36
dG 
 10.39 in Ans.
4 cos 30
______________________________________________________________________________
13-7
N P  19 teeth, N G  57 teeth, n  20 , mn  2.5 mm
(a)
pn   mn    2.5  7.854 mm
Ans.
pn
7.854

 9.069 mm Ans.
cos cos 30
pt
9.069
px 

 15.71 mm Ans.
tan tan 30
mn
2.5
mt 

 2.887 mm Ans.
cos cos 30
pt 
(b)
Shigley’s MED, 11th edition
Chapter 13 Solutions, Page 3/36
 tan 20 
 22.80
 
cos
30


t  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
dG  57  2.887   164.6 mm
Ans.
Ans.
______________________________________________________________________________
13-8
(a) Using Eq. (13-11) with k = 1, = 20º, and m = 2,
NP 


2k
m  m 2  1  2m  sin 2 
1  2m  sin 2 
 2 
1  2  2   sin  20 
2 1
2

 2
2

   14.16 teeth
 1  2  2   sin 2 20
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, = 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, 11th edition
Chapter 13 Solutions, Page 4/36
13-9
Repeating the process shown in the solution to Prob. 13-8, except with = 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 = 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 
2
3sin 
2 1
1 
3sin 20
2
 12.32


1  3sin 2 20
 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 
2
1  2m  sin 
2 1
2.5 
1  2  2.5   sin 20
2
 14.64

2.52  1  2  2.5   sin 2 20

15 teeth


Ans.
The largest gear-tooth count possible to mesh with this pinion, from Eq. (13-12) is
N P2 sin 2   4k 2
NG 
4k  2 N P sin 2 

152 sin 2 20  4 1
2
4 1  2 15  sin 2 20
 45.49  45 teeth
Ans.
(c) The smallest pinion that will mesh with a rack, from Eq. (13-13),
Shigley’s MED, 11th edition
Chapter 13 Solutions, Page 5/36
2 1
2k

2
sin  sin 2 20
 17.097  18 teeth Ans.
______________________________________________________________________________
NP 
13-11 n  20 ,  30
From Eq. (13-19), t  tan 1  tan 20 / cos 30   22.80
(a) The smallest pinion tooth count that will run with itself, from Eq. (13-21) is
NP 


2k cos
1  1  3sin 2 t
3sin 2 t
2 1 cos 30
1 
3sin 22.80
2


1  3sin 2 22.80
 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 30
NP 
2.5  2.52  1  2  2.5   sin 2 22.80
2

1  2  2.5   sin 22.80
 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 t  4k 2 cos 2 
4k cos  2 N P sin 2 t
102 sin 2 22.80  4 1 cos 2 30
4 1 cos 2 30  2  20  sin 2 22.80
 26.08  26 teeth
Ans.
(c) The smallest pinion that will mesh with a rack, from Eq. (13-24) is

2k cos 2 1 cos 30
NP 

sin 2 t
sin 2 22.80
 11.53  12 teeth Ans.
______________________________________________________________________________
13-12 From Eq. (13-19),
Shigley’s MED, 11th edition

 tan n 
1  tan 20 
 22.796
t  tan 
  tan 
 
 cos 
 cos 30 
1
Chapter 13 Solutions, Page 6/36
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.
______________________________________________________________________________

 tan n 
1  tan 20 

 27.236
tan


 
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),
t  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 
Setting k = 1 for full depth teeth, NP = 9 teeth, and solving for ,
  sin 1
2 1
2k
 sin 1
 28.126
NP
9
Ans.
______________________________________________________________________________
13-15
(a)
(b)
Eq. (13-3):
pn   mn  3 mm
Eq. (13-16):
pt  pn / cos  3 / cos 25  10.40 mm
Eq. (13-17):
px  pt / tan  10.40 / tan 25  22.30 mm
Eq. (13-3):
mt  pt /   10.40 /   3.310 mm
Eq. (13-19):
Shigley’s MED, 11th edition
t  tan 1
Ans.
Ans.
Ans.
tan n
tan 20
 tan 1
 21.88

cos
cos 25

Ans.
Ans.
Chapter 13 Solutions, Page 7/36
(c)
Eq. (13-2):
dp = mt Np = 3.310 (18) = 59.58 mm
Ans.
Eq. (13-2):
dG = mt NG = 3.310 (32) = 105.92 mm
Ans.
______________________________________________________________________________
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 30  1.155 in
dG 3


 48 / 12 cos 30   4.619 in

1.155  4.619
 2.887 in
2
 16 /  8 cos 25   2.207 in
Cab 
dP4
Ans.

Ans.

d G 5  36 / 8 cos 25  4.965 in
Cbc  3.586 in Ans.
______________________________________________________________________________
20  8   20  4
   
40  17   60  51
4
nd   00   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
Shigley’s MED, 11th edition
Chapter 13 Solutions, Page 8/36
13-19 (a)
(b)
12 1
  540   162 rev/min
40 1



 Ans.
cw about x, as viewed from the positive x axis 



d P  12 /  8 cos 23   1.630 in
nc 


d G  40 / 8 cos 23  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, = 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, 11th edition
Chapter 13 Solutions, Page 9/36
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, = 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
N5 = 20 teeth
Ans.
______________________________________________________________________________
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, = 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
Shigley’s MED, 11th edition
Chapter 13 Solutions, Page 10/36
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, = 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, i = 2500 rev/min
Let ωo = 300 rev/min for minimal gear ratio to minimize gear size.
o 300
N N
1


 2 4
i 2500 8.333 N 3 N 5
Let
N2 N4
1
1



N3 N5
8.333 2.887
From Eq. (13-11) with k = 1, = 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, 11th edition
Chapter 13 Solutions, Page 11/36
 15   15 
o       2500   304.2
 43   43 
Too big. Try N3 = N5 = 44.
 15   15 
o       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  nA   12  nA
51
e
Shigley’s MED, 11th edition
Chapter 13 Solutions, Page 12/36
nA 
12
 17.49 rev/min
35 / 51
(negative indicates cw as viewed from the bottom of the figure)



 Ans.



______________________________________________________________________________
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  nA    85  nA 
51
 16 
 nA    n A  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  = 20o, P = 6 teeth/in. Since N  d, then,
N2 + N3 = N4 + N5
(1)
Hour hand moves 1/12 of minute hand. Thus,  5 /  2 = 121 . Now,
 5 /  4 = N 4 / N 5 ,  3 /  2 = N 2 / N 3, and  4 =  3. Thus,
2 N 3 N 5

 12  4  3
2 N 2 N 4
So, try N3 = 4 N2, and N5 = 3 N4. Substituting N5 = 3 N4 into Eq. (1) gives
N2 + N3 = 4 N4  N4 = (N2 + N3)/4. Let N2 = 1. Then, N3 = 4N2 = 4(1) = 4,
N4 = (N2 + N3)/4 = (1 + 4)/4 = 5/4, N5 = 3 N4 = 3(5/4) = 15/4. Teeth must be a multiple of
4. To avoid interference for the smaller pinion use Eq. (13-11) with m = N 3 / N 2 = 4:
2k
N2 
m  m 2  1  2m  sin 2 
2
1  2m  sin 



2 1
4 
1  2  4   sin 20
2
o

42  1  2  4   sin 2 20o  15.4 teeth
Use N2 = 16 teeth which is a multiple of 4. Then, N3 = 4(16) = 64 teeth, N4 = (5/4)16 = 20
teeth, and N5 = (15/4) 16 = 60 teeth. Thus,
Shigley’s MED, 11th edition
Chapter 13 Solutions, Page 13/36
N2 = 16 teeth, N3 = 64 teeth, N4 = 20 teeth, and N5 = 60 teeth
Ans.
______________________________________________________________________________
13-30 The geometry condition is d 5 / 2  d 2 / 2  d 3  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 )  N3 / 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  n A 
nA  
17
 0  nA 
3
3
 320   68.57 rev/min
14
 nA  68.57 rev/min cw Ans.
The negative sign indicates opposite of n2.
______________________________________________________________________________
13-31 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 
13-32 (a)
  2 n / 60
H  T   2 Tn / 60
Shigley’s MED, 11th edition
(T in N  m, H in W)
Chapter 13 Solutions, Page 14/36
 
60 H 103
So
T
So
F32t 
2 n
 9550 H / n
(H in kW, n in rev/min)
9550  75 
 398 N  m
Ta 
1800
mN 2 5 17 
r2 

 42.5 mm
2
2
Ta 398

 9.36 kN
r2 42.5
F3b   Fb3  2  9.36   18.73 kN in the positive x-direction.
(b)
Ans.
mN 4 5  51

 127.5 mm
2
2
Tc 4  9.36 127.5   1193 N  m ccw
r4 
T4c  1193 N  m cw
Ans.
Note: The solution is independent of the pressure angle.
______________________________________________________________________________
N N
13-33
d 
P 6
d 2  4 in, d 4  4 in, d5  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  n A 6
Shigley’s MED, 11th edition
Chapter 13 Solutions, Page 15/36
n A  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   2 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
hp
n A 200 rev/min 
  min   2 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 20  287 lbf
Wt 
Gear 4
FA4  2W t  2  788   1576 lbf
Gear 5
Shigley’s MED, 11th edition
Chapter 13 Solutions, Page 16/36
Arm
Tout  1576  9   1576  4   7880 lbf  in
Ans.
______________________________________________________________________________
13-34 (a)  = 20o, P = 6 teeth/in. Since N  d, then,
N2 + N3 = N4 + N5
(1)
e = 40 = 8  5. Then, let N 2 / N 3 = 8 and N 4 / N 5 = 5. Thus, N 2 = 8 N 3 and N 4 = 5 N 5.
Substituting N 4 = 5 N 5 into Eq. (1) gives N2 + N3 = 6N5  N5 = (N2 + N3)/6.
To avoid interference use Eq. (13-11) with m = N 2 / N 3 = 8:
2k
N2 
m  m 2  1  2m  sin 2 
2
1  2m  sin 


2 1


8  82  1  2  8   sin 2 20o  16.2 teeth
1  2  8   sin 2 20o
Try N 3 = 17 teeth  N 2 = 8 N 3 = 8(17) = 136 teeth, N5 = (136 + 17)/6 = 25.5 teeth
which is unacceptable. Next, try N 3 = 18 teeth  N 2 = 8 N 3 = 8(18) = 144 teeth,
N5 = (144 + 18)/6 = 27 teeth, and N 4 = 5 N5 = 5 (27) = 135 teeth. Thus,
N 2 = 144 teeth, N 3 = 18 teeth, N 4 = 135 teeth, and N5 = 27 teeth
Ans.
(b) Gear diameter is d = N / P (with P = 6 teeth/in), 0.5 in wall clearance on two sides,
addendum of each gear is 1/P = 1/6 in, and each wall thickness is 0.75 in. Thus,
N /2 N /2
N
1
 2  0.5   2    2  0.75 
Y 2 3  4
P
P
P
P
144 18 / 2 135 / 2
1



 2  0.5   2    2  0.75   39.6 in
Ans.
6
6
6
6
______________________________________________________________________________

13-35 (a)  = 25o, P = 6 teeth/in. Since N  d, then,
N2 + N3 = N4 + N5
(1)
e = 40 = 8  5. Let N 2 / N 3 = 8 and N 4 / N 5 = 5. Thus, N 2 = 8 N 3 and N 4 = 5 N 5.
Substituting N 4 = 5 N 5 into Eq. (1) gives N2 + N3 = 6N5  N5 = (N2 + N3)/6.
To avoid interference use Eq. (13-11) with m = N 2 / N 3 = 8:
2k
N2 
m  m 2  1  2m  sin 2 
2
1  2m  sin 


2 1


8  82  1  2  8  sin 2 25o  10.7 teeth
1  2  8   sin 2 25o
Try N 3 = 11 teeth  N 2 = 8 N 3 = 8(11) = 88 teeth, N5 = (88 + 11)/6 = 16.5 teeth which
is unacceptable. Next, try N 3 = 12 teeth  N 2 = 8 N 3 = 8(12) = 96 teeth,
N5 = (96 + 12)/6 = 18 teeth, and N 4 = 5 N5 = 5 (18) = 90 teeth. Thus,
N 2 = 96 teeth, N 3 = 12 teeth, N 4 = 90 teeth, and N5 = 18 teeth
Ans.

Shigley’s MED, 11th edition
Chapter 13 Solutions, Page 17/36
(b) Gear diameter is d = N / P (with P = 6 teeth/in), 0.5 in wall clearance on two sides,
addendum of each gear is 1/P = 1/6 in, and each wall thickness is 0.75 in. Thus,
N /2 N /2
N
1
 2  0.5   2    2  0.75 
Y 2 3  4
P
P
P
P
96 12 / 2 90 / 2
1



 2  0.5   2    2  0.75   27.3 in Ans.
6
6
6
6
______________________________________________________________________________
13-36 (a) n = 20o,  = 45o, P = 6 teeth/in. Since N  d, then,
N2 + N3 = N4 + N5
(1)
e = 40 = 8  5. Let N 2 / N 3 = 8 and N 4 / N 5 = 5. Thus N 2 = 8 N 3 and N 4 = 5 N 5.
Substituting N 4 = 5 N 5 into Eq. (1) gives N2 + N3 = 6N5  N5 = (N2 + N3)/6.
From Eq. (13-19),
o
 tan n 
1  tan 20 

 20o
tan
t  tan 1 


o 
 tan 
 tan 45 
Eq. (13-22):
2k cos
Np 
m  m 2  1  2m  sin 2 t
2
1  2m  sin t


2 1 cos 45o


8  82  1  2  8  sin 2 20o  11.5 teeth
1  2 8   sin 2 20o
Try N 3 = 12 teeth  N 2 = 8 N 3 = 8(12) = 96 teeth, N5 = (96 + 12)/6 = 18 teeth, and
N 4 = 5 N5 = 5 (18) = 90 teeth. Thus,
N 2 = 96 teeth, N 3 = 12 teeth, N 4 = 90 teeth, and N5 = 18 teeth
Ans.
(b) Gear diameter is d = N / P (with P = 6 teeth/in), 0.5 in wall clearance on two sides,
addendum of each gear is 1/P = 1/6 in, and each wall thickness is 0.75 in. Thus,
N /2 N /2
N
1
 2  0.5   2    2  0.75 
Y 2 3  4
P
P
P
P
96 12 / 2 90 / 2
1



 2  0.5   2    2  0.75   27.3 in Ans.
6
6
6
6
______________________________________________________________________________

13-37 e  40, P = 6 teeth/in,  = 20o.
(a) Minimum size is N 2 / N 3 = N 4 / N 5 =
(13-11) with m = 6.325:
N2 


40 = 6.325. To avoid interference use Eq.
2k
m  m 2  1  2m  sin 2 
2
m


1
2
sin


2 1
6.325 
1  2  6.325   sin 20
Shigley’s MED, 11th edition
2
o


6.2352  1  2  6.325   sin 2 20o  16.0 teeth
Chapter 13 Solutions, Page 18/36
Let N3 = 16 teeth. N2 = 16 40 = 101.2. Use N2 = 101 teeth, e = (101/16)2 = 39.85 which
is ok since it is between 38 and 42. Thus,
N 2 = N 4 = 101 teeth, and N 3 = N5 = 16 teeth
Ans.
(b) Gear diameter is d = N / P (with P = 6 teeth/in), 0.5 in wall clearance on two sides,
addendum of each gear is 1/P = 1/6 in, and each wall thickness is 0.75 in. Thus,
N /2 N /2
N
1
 2  0.5   2    2  0.75 
Y 2 3  4
P
P
P
P
101 116 / 2 101/ 2
1



 2  0.5   2    2  0.75   29.4 in
Ans.
6
6
6
6
Comparing this to Prob. 13-34 where Y = 39.6 in, we see a large reduction in gearbox
size.
______________________________________________________________________________
13-38 (a) P = 6 teeth/in,  = 20o. Since N  d, then,
N2
N
N
N
30
60 20
80
 N3  4  5  N 6  7 
 20 

 N6 
2
2
2
2
2
2
2
2
Solving for N6 yields
N6 = 15 teeth
Ans.
N
N N / 2 N 7 / 2 2 60 20 30 / 2 80 / 2 2

 
 

  22.83 in Ans.
(b) Y  4  3  2
P
P
P
P
P 6
6
6
6
6
(c) Gear 2, n2 = 300 rev/min,
 d n   N 2 / P  n2   30 / 6  300

 392.7 ft/min
Ans.
Eq, (13-34): V2  2 2 
12
12
12
H
4
Wt  33 000  33 000
 336.1 lbf
Ans.
(d) Eq. (13-35):
V
392.7
Ans,
(e)
Fr = Wt tan  = 336.1 tan20o = 122.3 lbf
63 025 H 63 025  4 
(f)

 70.0 lbf  in
Ti 
Ans.
12n
12  300 
(g)
e
N 2 N 3 N 5 N 6 N 2 N 5 30  20 


 0.125
N 3 N 4 N 6 N 7 N 4 N 7 60  80 
1
 1 
To  Ti    70.0 
Ans.
  560 lbf  ft
e
 0.125 
Ans.
(h)
o = ei = 0.125 (300) = 37.5 rev/min
(i) Assuming no losses,
Po = Pi = 4 hp
Ans.
______________________________________________________________________________
13-39 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
Shigley’s MED, 11th edition
Chapter 13 Solutions, Page 19/36
From Eq. (13-36),
60 000 150 
60 000 H

 7.368 kN
Wt 
 dn
  216 1800 
d 
 216 
Ta 2  Wt  2   7.368 
  795.7 N  m
 2 
 2 
W r  7.368 tan 20  2.682 kN
Gears 3 and 4
 384 
 216 
Wt 
  7.368
2
 2 
t
W  13.10 kN
W r  13.10 tan 20  4.768 kN
Ans.
______________________________________________________________________________
13-40 Given: P = 5 teeth/in, N2 = 18T, N3 = 45T,
n  20 , 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 20  226 lbf
dP 
Fat2  W32t  622 lbf,

Fa 2  622 2  2262
Shigley’s MED, 11th edition

1/ 2
Far2  W32r  226 lbf
 662 lbf
Chapter 13 Solutions, Page 20/36
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
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.
______________________________________________________________________________
13-41 Given: P = 4 teeth/in,
n  20 ,
NP = 20T,
n2 = 900 rev/min
N P 20

 5.000 in
P
4
63025  30  2 
Tin 
 4202 lbf  in
900
W32t  Tin /  d2 / 2   4202 /  5 / 2   1681 lbf
d2 
W32r  1681 tan 20  612 lbf
The motor mount resists the equivalent forces and torque.
The radial force due to torque is
Shigley’s MED, 11th edition
Chapter 13 Solutions, Page 21/36
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
M 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
For W32r ,
Shigley’s MED, 11th edition
Chapter 13 Solutions, Page 22/36
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:
2
2
2
2
FS 1  153  179  5.625 / 8.98    179  7 / 8.98  
At A and B, the shear forces are:
FS 2  153  179  5.625 / 8.98    179  7 / 8.98  
 145 lbf
The shear forces are independent of the rotational sense.
The bolt tensions and the shear forces for cw rotation are,
For ccw rotation,
______________________________________________________________________________
13-42 (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 
Shigley’s MED, 11th edition
Chapter 13 Solutions, Page 23/36
(b)
(c)
 d 2 n2
  2.5  2500 
 1636 ft/min Ans.
12
 d n   2.5   2500 15 / 44  
Vo  V4  V5  4 4 
 558 ft/min
12
12
Vi  V2  V3 
Input gears:
Wti  33000
12

H 33000  25 

 504.3 lbf  504 lbf
Vi
1636
Wri  Wti tan   504.3 tan 20  184 lbf
W
504.3
Wi  ti 
 537 lbf
Ans.
cos  cos 20
Output gears:
H 33000  25 
 1478 lbf
Wto  33000 
Vo
558
Ans.
Ans.
Ans.
Ans.
Wro  Wto tan   1478 tan 20  538 lbf Ans.
Wto
1478
Wo 

 1573 lbf Ans.

cos 20
cos 20
(d)
d
Ti  Wti  2
 2

 2.5 
  630 lbf  in
  504.3 

 2 
2
Ans.
2
 44 
 44 
To  Ti    630    5420 lbf  in Ans.
 15 
 15 
______________________________________________________________________________
(e)
13-43 H  35 hp, ni  1200 rev/min,  =20
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
Shigley’s MED, 11th edition
Ans.
N
P
16
 1.6 in
10
48
d3  d5 
 4.8 in
10
d2  d4 
Ans.
Ans.
Ans.
Chapter 13 Solutions, Page 24/36
 d 2 n2
 1.6 1200 
 502.7 ft/min Ans.
12
12
 d n  1.6  400 
Vo  V4  V5  4 4 
 167.6 ft/min Ans.
12
12
Vi  V2  V3 
(c)

H 33000  35 

 2298 lbf lbf
Vi
502.7
Wti  33000
Wri  Wti tan   2298 tan 20  836.4 lbf
W
2298
Wi  ti 
 2445 lbf
Ans.
cos  cos 20
H 33000  35 

 6891 lbf
Vo
167.6
Wto  33000
Ans.
Ans.
Ans.
Wro  Wto tan   6891tan 20  2508 lbf Ans.
Wto
6891
Wo 

 7333 lbf Ans.

cos 20
cos 20
(d)
d 
 1.6 
Ti  Wti  2   2298 
  1838 lbf  in
 2 
 2 
2
Ans.
2
 48 
 48 
(e)
To  Ti    1838    16 540 lbf  in Ans.
 16 
 16 
______________________________________________________________________________

2
13-44 (a)
For o  , from Eq. (13-11), with m = 2, k = 1,   20
i 1
NP 
2 1
2 
1  2  2   sin 20
So
N P min  15
(b)
P
2


22  1  2  2   sin 2 20  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  = 20°,
WtA = FA cos20° = 300 cos20° = 281.9 lbf
For A, with  = 25°, same transmitted load,
Shigley’s MED, 11th edition
Chapter 13 Solutions, Page 25/36
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
FB 

 777.6 lbf Ans.

cos 25
cos 25
______________________________________________________________________________
13-45 (a)
For
o 5
 , from Eq. (13-11), with m = 5, k = 1,   20
i 1
2 1
5  52  1  2  5   sin 2 25  10.4
NP 
2

1  2  5  sin 25

So
(b)
N P min  11
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  = 20°,
WtA = FA cos20° = 11 cos20° = 10.33 kN
For A, with  = 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 / 2   W  d A   11.40  600   22.80 kN
WtB  WtA A

 
 d B / 2  tA  d B 
 300 
WtB
22.80

 25.16 kN Ans.

cos 25
cos 25
______________________________________________________________________________
FB 
13-46 (a) Using Eq. (13-11) with k = 1, = 20º, and m = 2,
Shigley’s MED, 11th edition
Chapter 13 Solutions, Page 26/36
NP 


2k
m  m 2  1  2m  sin 2 
2
1  2m  sin 
 2 
1  2  2   sin  20 
2 1
2

 2
2

   14.16 teeth
 1  2  2   sin 2 20
Round up for the minimum integer number of teeth.
NF = 15 teeth, NC = 30 teeth
d 125

 8.33 mm/tooth Ans.
N 15
2 kW
H
 1000 W   rev   60 s 

 100 N  m
T
 191 rev/min  kW   2 rad   min 
(b)
m
(c)
(d)
Ans.
From Eq. (13-36),
Wt 
60 000  2 
60 000 H

 1.60 kN  1600 N
 dn
 125 191
Ans.
Or, we could have obtained Wt directly from the torque and radius,
Wt 
100
T

 1600 N
d / 2 0.125 / 2
Wr  Wt tan   1600 tan 20  583 N Ans.
Wt
1600
W

 1700 N Ans.
cos  cos 20
______________________________________________________________________________
13-47 (a) Using Eq. (13-11) with k = 1, = 20º, and m = 2,
NP 


2k
m  m 2  1  2m  sin 2 
2
1
2
sin


m


 2 
1  2  2   sin  20 
2 1
2

 2
2

   14.16 teeth
 1  2  2   sin 2 20
Round up for the minimum integer number of teeth.
NC = 15 teeth, NF = 30 teeth
Shigley’s MED, 11th edition
Ans.
Chapter 13 Solutions, Page 27/36
(b)
P
(c)
T
(d)
N 30

 3 teeth/in
d 10
Ans.
 550 lbf  ft/s   12 in   rev   60 s 
1 hp




 70 rev/min 
hp
  ft   2 rad   min 
T  900 lbf  in
Ans.
H

From Eqs. (13-34) and (13-35),
 dn
 10  70 
 183.3 ft/min
12
H 33000 1
 180 lbf Ans.
Wt  33000 
V
183.3
Wr  Wt tan   180 tan 20  65.5 lbf Ans.
Wt
180
W

 192 lbf Ans.
cos  cos 20
______________________________________________________________________________
N 
d 
 1.30 

13-48 (a)
Eq. (13-14):   tan 1  P   tan 1  P   tan 1 
Ans.
  18.5
N
d
3.88


 G
 G
V
12

 dn
  2 1.30  600 
(b)
Eq. (13-34):
V
(c)
Eq. (13-35):
Wt  33 000
12

12
 408.4 ft/min
Ans.
Eq. (13-38):
H
 10 
 33 000 
  808 lbf
V
 408.4 
Wr  Wt tan  cos   808 tan 20 cos18.5  279 lbf
Ans.
Eq. (13-38):
Wa  Wt tan  sin   808 tan 20 sin18.5  93.3 lbf
Ans.
Ans.
The tangential and axial forces agree with Prob. 3-85, but the radial force given in Prob.
3-85 is shown here to be incorrect. Ans.
______________________________________________________________________________
13-49
  tan 1  2 / 4   26.565
  tan 1  4 / 2   63.435
rav  2  1.5sin 26.5650  / 2  1.665 in
Tin  63 025H / n  63025  2.5  / 240  656.5 lbf  in
W t  T / rav  656.5 / 1.665  394.3 lbf


a  2  1.5 cos 26.565 / 2  2.671 in
W r  394.3 tan 20 cos 26.565  128.4 lbf
Shigley’s MED, 11th edition
Chapter 13 Solutions, Page 28/36
W a  394.3 tan 20 sin 26.565  64.2 lbf
W = 128.4i – 64.2j + 394.3k lbf
RAG = –1.665i + 5.171j,
RAB = 2.5j
M 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  2039i  656.5 j  557.1k
So,
 2039i  656.5 j  557.1k    2.5 FBz i  2.5 FBxk  Tj  0
FBz  2039 / 2.5  815.6 lbf
T  656.5 lbf  in
FBx  557.1 / 2.5  222.8 lbf
So,
FB = 222.8i 815.6k lbf
Ans.
1/2
2
2
FB   222.88    815.6  


 845.5 lbf
FA = – (FB + W)
= – (–222.8i – 815.6k + 128.4i – 64.2j + 394.3k)
= 94.4i + 64.2j + 421.3k Ans.

FA (radial)  94.4 2  421.32

1/ 2
 431.7 lbf
FA (thrust)  64.2 lbf
______________________________________________________________________________
13-50
d 2  18 / 10  1.8 in, d3  30 / 10  3.0 in
 d2 / 2 
1  0.9 

  tan 
  30.96
d
/
2
1.5


 3 
  tan 1 
  90    59.04
rav  3.0 / 2   0.5sin 59.04o  / 2  1.286 in
9
 0.5cos 59.04 / 2  0.6911 in
16
t
W  25 lbf
W r  25 tan 20 cos 59.04  4.681 lbf
DE 


W a  25 tan 20 sin 59.04  7.803 lbf
W = –4.681i – 7.803j +25k
Shigley’s MED, 11th edition
Chapter 13 Solutions, Page 29/36
RDG = 1.286i + 0.6911j
RDC = –0.625j
M D  R DG  W  R DC  FC  T  0
R DG  W  17.28i  32.15 j  6.800k
R DC  FC  0.625 FCz i  0.625 FCx k
17.28i  32.15 j  6.800k    0.625 FCz i  0.625 FCxk   Tj  0
T  32.15 lbf  in
Ans.
FC  10.88i  27.65k lbf

FC  10.882  27.652
F  0

1/ 2
Ans.
 29.7 lbf
Ans.
FD  6.20i  7.80 j  52.65k lbf
2 1/2
FD (radial)   6.20    52.65    53.0 lbf Ans.


a
FD (thrust)  W  7.80 lbf Ans.
______________________________________________________________________________
2
13-51
NOTE: The shaft forces exerted on the gears are not shown in the figures above.
Pt  Pn cos  5 cos 30  4.330 teeth/in
t  tan 1
dP 
tan n
tan 20
 tan 1
 22.80
cos
cos 30
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.
Shigley’s MED, 11th edition
Chapter 13 Solutions, Page 30/36
Pinion (Gear 2)
W r  W t tan t  800 tan 22.80  336 lbf
W a  W t tan  800 tan 30  462 lbf
W  336i  462 j  800k lbf Ans.
1/2
W   336    462   800 2 


2
Gear 3
2
 983 lbf
Ans.
W  336i  462 j  800k lbf Ans.
W  983 lbf Ans.
32
dG 
 7.390 in
4.330
TG  W t r  800  7.390   5912 lbf  in
______________________________________________________________________________

tan n
1 tan 20
t  tan
 tan
 22.80
13-52

cos
cos 30
Pinion (Gear 2)
W r  W t tan t  800 tan 22.80  336 lbf
1
W a  W t tan  800 tan 30  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.
Shigley’s MED, 11th edition
Chapter 13 Solutions, Page 31/36
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.
______________________________________________________________________________
13-53 Gear 3:
Pt  Pn cos  7 cos 30  6.062 teeth/in
tan t 
tan 20
 0.4203, t  22.8

cos 30
54
d3 
 8.908 in
6.062
W t  500 lbf
W a  500 tan 30  288.7 lbf
W r  500 tan 22.8  210.2 lbf
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 30  1114 lbf
W r  1929 tan 22.8  811 lbf
W4  811i  1114 j  1929k lbf Ans.
______________________________________________________________________________
13-54
Pt  6 cos 30  5.196 teeth/in
42
d3 
 8.083 in
5.196
t  22.8
16
d2 
 3.079 in
5.196
63025  25 
T2 
 916 lbf  in
1720
Shigley’s MED, 11th edition
Chapter 13 Solutions, Page 32/36
T
916

 595 lbf
r 3.079 / 2
W a  595 tan 30  344 lbf
W r  595 tan 22.8  250 lbf
Wt 
W  344i  250 j  595k lbf
R DC  6i,
(1)
R DG  3i  4.04 j
M D  R DC  FC  R DG  W  T  0
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
F  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, 11th edition
Chapter 13 Solutions, Page 33/36
13-55
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
 1600 rev/min  kW   2 rad   min 
35.81
T
Wt 

 1119 N
d 2 / 2 0.064 / 2
T
H

Shigley’s MED, 11th edition
Chapter 13 Solutions, Page 34/36
W r  W t tan t  1119 tan 20  407.3 N
W a  W t tan  1119 tan15  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.
F4c  407.3i  1119 j  299.8k N Ans.
______________________________________________________________________________
13-56
14
36
N

 2.021 in, d3 
 5.196 in

Pn cos 8 cos 30
8cos 30
15
45
d4 
 3.106 in, d 5 
 9.317 in

5 cos15
5cos15
d2 
For gears 2 and 3: t  tan 1  tan n / cos   tan 1  tan 20 / cos 30   22.8
For gears 4 and 5: t  tan 1  tan 20 / cos15   20.6
F23t  T2 / r2  1200 /  2.021/ 2   1188 lbf
5.196
 1987 lbf
3.106
F23r  F23t tan t  1188 tan 22.8  499 lbf
F54t  1188
F54r  1986 tan 20.6  746 lbf
F23a  F23t tan   1188 tan 30  686 lbf
F54a  1986 tan15  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, 11th edition
Chapter 13 Solutions, Page 35/36
R CG  1.553 j  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
M 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  
F  FC  F54  F23  FD  0
Substituting, gives
F
x
C

i  FCy j   1987i  746 j  532k    1188i  499 j  686k 


 1610i  425 j  FDz k  0
Solving gives
FCx  1987  1188  1610  1565 lbf
FCy  746  499  425  672 lbf
FDz  532  686  154 lbf
FC = 1565i + 672j lbf
Ans.
FD = 1610i  425j + 154k lbf
Ans.
______________________________________________________________________________
So,
Shigley’s MED, 11th edition
Chapter 13 Solutions, Page 36/36
13-57
VW 
WW t
 dW nW

  0.100  600 
60
60
H 2000


 637 N
VW

  m/s
L  px NW  25 1  25 mm
  tan 1
 tan 1
L
 dW
25
 4.550 lead angle
 100 
WW t
W
cos n sin   f cos 
V

 3.152 m/s
VS  W 
cos  cos 4.550
In ft/min: VS = 3.28(3.152) = 10.33 ft/s = 620 ft/min
Use f = 0.043 from curve A of Fig. 13-38. Then, from the first of Eq. (13-43)
W
637
 5323 N
cos14.5 sin 4.55  0.043cos 4.55



W y  W sin n  5323sin14.5  1333 N


W z  5323 cos14.5 cos 4.55  0.043sin 4.55   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
M 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
Shigley’s MED, 11th edition
Ans.
Chapter 13 Solutions, Page 37/36
1/2
2
2
FB   613   318.5  


F  FA  W  R B  0
Or
 691 N radial
FA    W  FB     637i  1333 j  5119k  318.5i  613j
 318.5i  1946 j  5119k
Radial
Ans.
FAr  318.5i  1946 j N
2 1/2
FAr   318.5    1946    1972 N


a
Thrust
FA  5119 N
______________________________________________________________________________
2
13-58 From Prob. 13-57,
WG  637i  1333 j  5119k N
pt  p x
So
dG 
N G px


48  25 

 382 mm
Bearing D takes the thrust load.
M 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,
F  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  637i N (thrust)
______________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 13 Solutions, Page 38/36
13-59
 1.5  600 
 235.7 ft/min
12
33000  0.75 
W x  WW t 
 105.0 lbf
235.7

pt  px   0.3927 in
8
L  0.3927  2   0.7854 in
VW 
0.7854
 9.46
 1.5 
105.0
W
 515.3 lbf

cos 20 sin 9.46  0.05 cos 9.46
W y  515.3sin 20  176.2 lbf
W z  515.3 cos 20  cos 9.46   0.05sin 9.46   473.4 lbf
  tan 1
So
W  105i  176 j  473k lbf
T  105  0.75  78.8 lbf  in
Ans.
Ans.
______________________________________________________________________________
13-60 Computer programs will vary.
Shigley’s MED, 11th edition
Chapter 13 Solutions, Page 39/36
Chapter 9
Figure for Probs.
9-1 to 9-4
9-1
Given, b = 50 mm, d = 50 mm, h = 5 mm, allow = 140 MPa.
F = 0.707 hlallow = 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, allow = 25 kpsi.
F = 0.707 hlallow = 0.707(5/16)[2(2)](25) = 22.1 kip
Ans.
______________________________________________________________________________
9-3
Given, b = 50 mm, d = 30 mm, h = 5 mm, allow = 140 MPa.
F = 0.707 hlallow = 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, allow = 25 kpsi.
F = 0.707 hlallow = 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, 11th edition
Chapter 9 Solutions, Page 1/39
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:
 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 = 0.707hlall = 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:
 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 = 0.707hlall = 0.707(5/16)[2(2)](12.0) = 10.6 kip Ans.
______________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 9 Solutions, Page 2/39
9-11
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:
 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 = 0.707hlall = 0.707(5)[2(50)](108)(103) = 38.2 kN Ans.
______________________________________________________________________________
9-12
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:
 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 = 0.707hlall = 0.707(5/16)[2(4)](12.0) = 21.2 kip Ans.
______________________________________________________________________________
9-13
2 100  103 
2F


 141 MPa
Ans.
Eq. (9-3):
hl
5  2  50  50  
______________________________________________________________________________
9-14
Eq. (9-3):

2  40 
2F

 22.6 kpsi
hl
 5 /16   2  2  2 
Ans.
______________________________________________________________________________
2 100  103 
2F

 177 MPa
Ans.
9-15
Eq. (9-3):  
hl
5  2  50  30  
______________________________________________________________________________
9-16
Eq. (9-3):  
2  40 
2F

 15.1 kpsi
hl
 5 /16   2  4  2 
Ans.
______________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 9 Solutions, Page 3/39
9-17
A = (throat area)(length) = 0.707h (b + d) Ans.
b
b2
x b  d   0  d   b 

x
2
2 b  d 
y b  d   0 b 
d
d 
2

x
d2
2 b  d 
Ans.
Ans.
For line b:
 J u b
2
 b
 b3
 b2

b3

2
  b   x   y    b   bx  x 2  y 2 
12

 4

 2
 12
 b2
 4b3  b  d 2  6b 4  b  d   3b5  3bd 4
b3
b3
b4
d4
 b 



2
2
2
12
12  b  d 
 4 2  b  d  4  b  d  4  b  d  
Similarly,
 J u d 
4d 3  b  d   6d 4  b  d   3d 5  3b 4 d
2
12  b  d 
J u   J u b   J u  d
2
4b3  b  d   6b 4  b  d   3b5  3bd 4  4d 3  b  d   6d 4  b  d   3d 5  3b 4 d
2

2
12  b  d 
2
4b3  b  d   6b 4  b  d   3b 4  b  d   3d 4  b  d   4d 3  b  d   6d 4  b  d 
2

2
12  b  d 
2
This reduces to
b 4  4b3 d  d 4  4bd 3
1
12  b  d 
Add and subtract 6b2d 2 to Eq. (1) giving
b 4  4b3d  6b 2 d 2  4bd 3  d 4   6b 2 d 2

Ju 
12  b  d 
Ju 
 b  d   6b 2 d 2

12  b  d 
4
which is the same as Table 9  1
Ans.
______________________________________________________________________________
9-18 A = (throat area) (length) = 0.707h (2b + d) Ans.
d
b2
x  2b  d   0  d    b 

x
Ans.
2
2b  d
d

d b  
2 d
d 
y  2b  d   0  b   d    d  b   y  
Ans.

2b  d
2
2
For lines b:
Shigley’s MED, 11th edition
Chapter 9 Solutions, Page 4/39
 J u b
2
 b
  b3
 b2

 b3

2 
 2   b   x   y     2b   bx  x 2  y 2 

 4

 2
  6
12
2b 4
2b5
2b 4
2b5
b3 b3
bd 2 2b3
bd 2
  






6 2 2b  d  2b  d 2
2
3 2b  d  2b  d 2
2
Line d:
d3
d3
b4 d
 J u d   d x 2  
12
12  2b  d  2
Ju = (Ju)b + (Ju)d
bd 2 d 3
b4 d
2b 3
2b 4
2b5
Ju 





3 2b  d  2b  d 2
2
12  2b  d 2
4
5
4
8b3  6bd 2  d 3 2b  2b  d   2b  b d


2
12
 2b  d 

4
4
8b3  6bd 2  d 3 2b  2b  d   b  2b  d 

2
12
 2b  d 
b4
8b3  6bd 2  d 3
Ans.

12
2b  d
______________________________________________________________________________

9-19
b = d =50 mm, c = 150 mm, h = 5 mm, and allow = 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  in MPa):
F 103 
V
 2.829 F
 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
 x   y 
Mry
J

175 F 103   25 
294.6 103 
 14.85 F
 max   x2   y   y   F 14.852   2.829  14.85   23.1F (1)
2
Shigley’s MED, 11th edition
2
Chapter 9 Solutions, Page 5/39
F
 allow
23.1
140
 6.06 kN Ans.
23.1

(b) For E7010 from Table 9-6, allow = 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:
allow = min(0.30Sut, 0.40Sy ) =min[0.30(340), 0.40(190) = min(102, 76)
= 76 MPa
The allowable load, from Eq. (1) is
 allow
76
 3.29 kN
Ans.
23.1 23.1
______________________________________________________________________________
F
9-20

b = d =2 in, c = 6 in, h = 5/16 in, and allow = 25 kpsi.
(a) Primary shear, Table 9-1(Note: b and d are interchanged between problem figure and
table figure. Note, also, F in kip and  in kpsi):
V
F

 1.132 F
A 1.414  5 /16  2 
Secondary shear, Table 9-1:
2
2
d  3b2  d 2  2 3  2   2 

 5.333 in 3
Ju 
6
6
 y 
J = 0.707 h Ju = 0.707(5/16)(5.333) = 1.178 in4
 x   y 
Mry
J

7 F 1
1.178
 5.942 F
 max   x2   y   y   F 5.942 2  1.132  5.942   9.24 F (1)
2
F
Shigley’s MED, 11th edition
 allow
9.24

2
25
 2.71 kip Ans.
9.24
Chapter 9 Solutions, Page 6/39
(b) For E7010 from Table 9-6, allow = 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:
allow = 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
 allow
11
 1.19 kip
Ans.
9.24 9.24
______________________________________________________________________________
F
9-21

b =50 mm, c = 150 mm, d = 30 mm, h = 5 mm, and allow = 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  in MPa):
 y 
F 103 
V

 2.829 F
A 1.414  5 50 
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
 x 
Mry
J

175 F 103  15 
153.2 103 
 17.13F
Mrx 175 F 10   25 

 28.55 F
J
153.2 103 
3
 y 
 max   x2   y   y   F 17.132   2.829  28.55   35.8 F (1)
2
Shigley’s MED, 11th edition
2
Chapter 9 Solutions, Page 7/39
F
 allow
35.8
140
 3.91 kN
35.8

Ans.
(b) For E7010 from Table 9-6, allow = 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:
allow = min(0.30Sut, 0.40Sy ) =min[0.30(340), 0.40(190) = min(102, 76)
= 76 MPa
The allowable load, from Eq. (1) is
 allow
76
 2.12 kN
Ans.
35.8 35.8
______________________________________________________________________________
F
9-22

b = 4 in, c = 6 in, d = 2 in, h = 5/16 in, and allow = 25 kpsi.
(a) Primary shear, Table 9-1(Note: b and d are interchanged between problem figure and
table figure. Note, also, F in kip and  in kpsi):
V
F

 0.5658 F
A 1.414  5 /16  4 
Secondary shear, Table 9-1:
 y 
Ju 
d  3b 2  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
 x 
 y 
Mry
J

8 F 1
4.125
 1.939 F
Mrx 8 F  2 

 3.879 F
J
4.125
 max   x2   y   y   F 1.9392   0.5658  3.879   4.85 F (1)
2
Shigley’s MED, 11th edition
2
Chapter 9 Solutions, Page 8/39
F
 allow
4.85

25
 5.15 kip Ans.
4.85
(b) For E7010 from Table 9-6, allow = 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:
allow = 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
 allow
11
 2.27 kip
Ans.
4.85 4.85
______________________________________________________________________________
F
9-23

A = (throat area)(length) = 0.707h (2b + d) Ans.
d
b2
x  2b  d   0  d    b 

x
Ans.
2
2b  d
d

d b  
2 d
d 
y  2b  d   0  b   d    d  b   y  
Ans.

2b  d
2
2
2
d3
d2
d 
 2b   
Ans.
 6b  d 
12
 2  12
______________________________________________________________________________
9-24 A = (throat area)(length) = 0.707h (b + 2d) Ans.
b

b  d 
b
2
b
x  b  2d   0  d    b   b  d   x  
Ans.
b  2d
2
2
d
d2
y  b  2d   0  b    2 d  
y
Ans.
2
b  2d
Iu 
Shigley’s MED, 11th edition
Chapter 9 Solutions, Page 9/39
2
I u  by 2 
d3 d3
2d 3
d

 2d   y  

 b y 2  2d 2 y  2d y 2
12
6
2
2

2 3
d  2d 2 y   b  2 d  y 2
Ans.
3
______________________________________________________________________________

9-25
Given, b = 50 mm, c = 150 mm, d = 50 mm, h = 5 mm, allow = 140 MPa.
Primary shear (F in kN,  in MPa, A in mm2):
F 103 
V

 1.414 F
y  
A 1.414  5  50  50 
Secondary shear:
b  d 

3
Table 9-1:
 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
 x   y 
Mry
J

175 F 103  (25)
589.2 103 
 7.425 F
Maximum shear:
 max   x2   y   y   F 7.4252  1.414  7.425   11.54 F
2
 allow
2
140
 12.1 kN Ans.
11.54 11.54
______________________________________________________________________________
F
9-26

Given, b = 2 in, c = 6 in, d = 2 in, h = 5/16 in, allow = 25 kpsi.
Primary shear:
 y 
Secondary shear:
Table 9-1:
V
F

 0.5658 F
A 1.414  5 /16  2  2 
b  d 

Shigley’s MED, 11th edition
2  2

3
 10.67 in 3
6
6
J = 0.707 h Ju = 0.707(5/16)10.67 = 2.357 in4
Ju
 x   y 
Maximum shear:
3
Mry
J

7 F (1)
 2.970 F
2.357
Chapter 9 Solutions, Page 10/39
 max   x2   y   y   F 2.9702   0.566  2.970   4.618 F
2
 allow
2
25
 5.41 kip Ans.
4.618 4.618
______________________________________________________________________________
F
9-27

Given, b = 50 mm, c = 150 mm, d = 30 mm, h = 5 mm, allow = 140 MPa.
Primary shear (F in kN,  in MPa, A in mm2):
F 103 
V
 1.768 F
 y  
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
 x 
 y 
Mry
J

175 F 103  (15)
301.6 103 
 8.704 F
3
Mrx 175 F 10  (25)

 14.51F
J
301.6 103 
Maximum shear:
 max   x2   y   y   F 8.7042  1.768  14.51  18.46 F
2
 allow
2
140
 7.58 kN Ans.
18.46 18.46
______________________________________________________________________________
F
9-28

Given, b = 4 in, c = 6 in, d = 2 in, h = 5/16 in, allow = 25 kpsi.
Primary shear:
 y 
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
 x 
Shigley’s MED, 11th edition
V
F

 0.3772 F
A 1.414  5 /16  4  2 
Mry
J

8 F (1)
 1.006 F
7.954
Chapter 9 Solutions, Page 11/39
 y 
Maximum shear:
Mrx 8 F (2)

 2.012 F
J
7.954
 max   x2   y   y   F 1.006 2   0.3772  2.012   2.592 F
2
 allow
2
25
 9.65 kip Ans.
2.592 2.592
______________________________________________________________________________
9-29 Given: b = 50 mm, c = 150 mm, d = 50 mm, h = 5 mm, allow = 140 MPa.
F 103  sin 45o
V
Primary shear (F in kN):  x   y  
F
A 1.414  5  50  50 
F

Secondary shear: M = 0.707 F (103)(175  25) = 106.05(103) F
 
Table 9-1: J u   b  d  / 6   50  50  / 6  166.7 103 mm3
3
3
J  0.707 h J u  0.707  5 166.7 103   589.3 103  mm 4
Mry
 x   y 
J

106.5 103  F  25 
589.3 103 
 4.50 F
Upper right end of weld:
 max 
 x   x    y   y 
2
2
 2  F  4.5 F   7.778 F
2
140
 18.0 kN
Ans.
7.778
______________________________________________________________________________
9-30 Given: b = 2 in, c = 6 in, d = 2 in, h = 165 in, allow = 25 kpsi.
 max   allow
F

Primary shear (F in kip) :  x   y 
0.707 F
V

 0.4 F
A 1.414 5  2  2 
 
16
Secondary shear: M = 0.707 F (7  1) = 4.242 F
3
3
Table 9-1: J u   b  d  / 6   2  2  / 6  10.667 in 3
J  0.707 h J u  0.707
Mry
 x   y 
Upper right end of weld:
 max 
J

 10.667  2.357 in
5
16
4.242 F 1
2.357
 x   x    y   y 
2
2
4
 1.80 F
 2  0.4 F  1.80 F   3.111F
2
25
 8.04 kip
Ans.
3.111
______________________________________________________________________________
 max   allow
Shigley’s MED, 11th edition

F
Chapter 9 Solutions, Page 12/39
9-31
Given: b = 50 mm, c = 150 mm, d = 30 mm, h = 5 mm, allow = 140 MPa.
F 103  sin 45o
V
Primary shear (F in kN):  x   y  
 1.25 F
A 1.414  5  50  30 
Secondary shear: M = 0.707 F (103)(175  15) = 113.12 (103) F
 
Table 9-1: J u   b  d  / 6   50  30  / 6  85.33 103 mm3
3
3
J  0.707 h J u  0.707  5  85.33 103   301.65 103  mm 4
 x 
Mry

J
113.12 103  F 15 
301.65 103 
 5.625 F
3
Mrx 113.12 10  F  25
 y 

 9.375 F
J
301.65 103 
Upper right end of weld:
 x   x    y   y 
2
2
2
 F 1.25  5.625   1.25  9.375    12.66 F


140
 max   allow

F
 11.1 kN
Ans.
12.66
______________________________________________________________________________
9-32 Given: b = 4 in, c = 6 in, d = 2 in, h = 165 in, allow = 25 kpsi.
 max 
2
Primary shear (F in kip) :  x   y 
0.707 F
V

 0.2667 F
A 1.414 5  4  2 
 
16
Secondary shear: M = 0.707 F (8  1) = 4.949 F
3
3
Table 9-1: J u   b  d  / 6   4  2  / 6  36 in 3
J  0.707 h J u  0.707
  36  7.954 in
5
16
4
4.949 F 1
 0.6222 F
J
7.954
Mr 4.949 F  2 
 1.244 F
 y  x 
J
7.954
Upper right end of weld:
 x 
Mry

 x   x    y   y 
2
2
2
 F  0.2667  0.6222    0.2667  1.244    1.753F


25
 max   allow

F
 14.3 kip
Ans.
1.753
______________________________________________________________________________
 max 
9-33
2
Given, b = 50 mm, d = 50 mm, h = 5 mm, E6010 electrode.
Shigley’s MED, 11th edition
Chapter 9 Solutions, Page 13/39
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. (6-18) and Table 6-2:
ka = 54.9(320)0.758 = 0.693
kb = 1 (uniform shear), kc = 0.59 (torsion, shear), kd = 1
Eqs. (6-10) and (6-17):
Se = 0.693(1)(0.59)(1)(0.5)(320) = 65.4 MPa
Electrode endurance: E6010, Table 9-3,
Eq. (6-18) and Table 6-2:
Sut = 427 MPa
ka = 54.9(427)0.758 = 0.557
As before, kb = 1 (direct shear), kc = 0.59 (torsion, shear), kd = 1
Se = 0.557(1)(0.59)(1)(0.5)(427) = 70.2 MPa
The members and electrode are basically of equal strength. We will use Se = 65.4 MPa.
For a factor of safety of 1, and with Kfs = 2.7 (Table 9-5)
 A 65.4  530.3 
F  allow 
 12.8 103  N  12.8 kN
Ans.
K fs
2.7
______________________________________________________________________________
9-34
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. (6-18) and Table 6-2:
ka = 12.7(47)0.758 = 0.686
kb = 1 (uniform shear), kc = 0.59 (torsion, shear), kd = 1
Eqs. (6-10) and (6-17):
Se = 0.686(1)(0.59)(1)(0.5)(47) = 9.51 kpsi
Electrode endurance: E6010, Table 9-3,
Sut = 62 kpsi
Eq. (6-18) and Table 6-2:
ka = 12.7(62)0.758 = 0.556
As before, kb = 1 (uniform shear), kc = 0.59 (torsion, shear), kd = 1
Se = 0.556(1)(0.59)(1)(0.5)(62) = 10.2 kpsi
For a factor of safety of 1, with Kfs = 2.7 (Table 9-5), using Se = 9.51 kpsi
 A 9.511.326 
F  allow 
 4.67 kip
Ans.
K fs
2.7
______________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 9 Solutions, Page 14/39
9-35
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. (6-18) and Table 6-2:
ka = 54.9(320)0.758 = 0.693
kb = 1 (direct shear), kc = 0.59 (torsion, shear), kd = 1
Eqs. (6-10) and (6-17):
Se = 0.693(1)(0.59)(1)(0.5)(320) = 65.4 MPa
Electrode endurance: E6010, Table 9-3,
Eq. (6-18) and Table 6-2:
Sut = 482 MPa
ka = 54.9(482)0.758 = 0.508
As before, kb = 1 (direct shear), kc = 0.59 (torsion, shear), kd = 1
Se = 0.508(1)(0.59)(1)(0.5)(482) = 72.2 MPa
For a factor of safety of 1,with Kfs = 2.7 (Table 9-5), and using Se =65.4 MPa
 A 65.4  459.6 
F  allow 
 11.1103  N  11.1 kN
Ans.
K fs
2.7
______________________________________________________________________________
9-36
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. (6-18) and Table 6-2:
ka = 12.7(47)0.758 = 0.686
kb = 1 (direct shear), kc = 0.59 (torsion, shear), kd = 1
Eqs. (6-10) and (6-17):
Se = 0.686(1)(0.59)(1)(0.5)(47) = 9.51 kpsi
Electrode endurance: E7010, Table 9-3,
Eq. (6-18) and Table 6-2:
Sut = 70 kpsi
ka = 12.7(70)0.758 = 0.507
As before, kb = 1 (direct shear), kc = 0.59 (torsion, shear), kd = 1
Se = 0.507(1)(0.59)(1)(0.5)(70) = 10.5 kpsi
For a factor of safety of 1, with Kfs = 2.7 (Table 9-5), and using Se = 9.51 kpsi
Shigley’s MED, 11th edition
Chapter 9 Solutions, Page 15/39
9.51 2.209 
 7.78 kip
Ans.
K fs
2.7
______________________________________________________________________________
9-37 Primary shear:
 = 0 (why?)
Secondary shear:
F
 allow A

Table 9-1: Ju = 2 r3 = 2 (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 
    allow
 1.600 F  20  F  12.5 kip
Ans.
______________________________________________________________________________
9-38 Direct shear Vz = F, Torsion Tx = 8F, Bending My = 6F
B: Direct shear:
V
F
F

 0.6014 F
 V   z  z 
A 1.414 hr 1.414 1 1.5

4
Torsion, Table 9-1:
Ju = 2r3 = 2 (1.5)3 = 21.21 in3
J = 0.707h Ju = 0.707 ( 14 ) 21.21 = 3.749 in4
Tr
 T   z  x 
8 F 1.5 
 3.201F
J
3.749
 max   z   z  0.6014 F  3.201F  3.802 F

20
F  allow 
 5.26 kip
Ans.
3.802 3.802
C: Direct shear, torsion, and bending.
Bending, Table 9-2:
Iu = r3 =  (1.5)3 = 10.603 in3, I = 0.707h Iu = 0.707( 14 ) 10.603 =1.874 in4.
  M
 max 

Mc 6 F 1.5 

 4.803F
I
1.874
 V   T    M
2
2
2
 allow

 0.6014 F 
2
  3.201F    4.803F   5.803F
2
2
20
 3.45 kip
Ans.
5.803 5.803
______________________________________________________________________________
F
Shigley’s MED, 11th edition

Chapter 9 Solutions, Page 16/39
9-39
l = 2 + 4 + 4 = 10 in
2 1  4  0   4  2 
 1 in
10
2  4  4  2  4  0
 1.6 in
y
10
x
M = FR = F (10  1) = 9 F
1  1   4  1.6 
2
r1 
2
 2.4 in
r2  12   2  1.6   1.077 in
2
 2  1
r3 
2
 1.62  1.887 in
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   2.4 2   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 
  tan 1 
r  1.62   4  1  3.4 in
2
Primary shear ( in kpsi, F in kip) :
 
Secondary shear:
  
Shigley’s MED, 11th edition
V
F

 0.4526 F
A 0.707  5 /16 10 
Mr 9 F  3.4 

 3.319 F
J
9.220
Chapter 9 Solutions, Page 17/39
 max 
 3.319F sin 28.07    3.319F cos 28.07
o 2
o
 0.4526 F 
2
 3.724 F
Ans.
max = allow  3.724 F = 25  F = 6.71 kip
______________________________________________________________________________
9-40
l = 30 + 50 + 50 = 130 mm
30 15   50  0   50  25 
 13.08 mm
130
30  50   50  25   50  0 
 21.15 mm
y
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 
2
2
 28.92 mm
r2  13.082   25  21.15   13.63 mm
2
 25  13.08
r3 
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 
  tan 1 
r  21.152   50  13.08   42.55 mm
2
Shigley’s MED, 11th edition
Chapter 9 Solutions, Page 18/39
Primary shear ( in MPa, F in kN) :
F 103 
V
 2.176 F
  
A 0.707  5 130 
Secondary shear:
3
Mr 186.92 F 10   42.55 

 25.88 F
  
J
307.3 103 
 max 
 25.88F sin 29.81    25.88F cos 29.81
o 2
o
 2.176 F 
2
 27.79 F
Ans.
max = allow  27.79 F = 140  F = 5.04 kN
______________________________________________________________________________
9-41
Weld
Pattern
1.
2.
Figure of merit
 a2 
J u a 3 / 12 a 2





fom
0.0833  
lh
ah
12h
 h 
a  3a 2  a 2  a 2
 a2 
fom 

 0.3333  
6  2a  h
3h
 h 
3.
 2a   6a 2 a 2  5a 2  0.2083  a 2 
fom 
 
12  a  a  2ah 24h
 h 
4.
fom 
5.
 2a 
fom 
Rank______
5
1
4
4
 a2 
1  8a 3  6a 3  a 3
a4 
0.3056




 
3ah 
12
2a  a 
 h 
2
3
 a2 
1
8a 3

 0.3333  
6h 4a 24ah
 h 
1
2  a / 2 
 a2 
a3
6.
3
fom 

 0.25  
4ah
 ah
 h 
______________________________________________________________________________
3
Shigley’s MED, 11th edition
Chapter 9 Solutions, Page 19/39
9-42
Weld
Pattern
1.
2.
3.
4.*
5. & 7.
Figure of merit
3
 a2 
I u  a / 12 
 0.0833  
fom  
lh
ah
 h 
a3 / 6 

 a2 
 0.0833  
fom 
2ah
 h 
aa 2 / 2 

 a2 
fom 
 0.25  
2ah
 h 
a 2 /12   6a  a  7 a 2

 a2 

fom 

 0.1944  
3ah
36h
 h 
a
a2
a
x , y

2
a  2a 3
Rank______
6
6
1
2
2
a3
2a 3
a
2 a
Iu 
 2a   a  2a    
3
3
3
3
6. & 8.
3
 a2 
I u  a / 3 1  a 2 
    0.1111 
fom  
3ah
9 h 
lh
 h 
a 2 / 6   3a  a  1  a 2 

 a2 
fom 
    0.1667  
4ah
6 h 
 h 
  a / 2 a2
 a2 
fom 

 0.125  
8h
 ah
 h 
5
3
3
9.
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.
______________________________________________________________________________
9-43
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: all = 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, 11th edition
Chapter 9 Solutions, Page 20/39

F
  all
n  0.707  hl
100 103 
F

 21.43
nh 
0.707l all 0.707  75  88 
where h is in millimeters. Make a table
Number of beads, n
1
2
3
4
Decision
Leg size, h (mm)
21.43
10.71
7.14
5.36  6 mm
Specify h = 6 mm on all four sides.
Weldment specification:
Pattern: All-around square, four beads each side, 75 mm long
Electrode: E6010
Leg size: h = 6 mm
______________________________________________________________________________
9-44
Decision: Choose a parallel fillet weldment pattern. By so-doing, we’ve chosen an
optimal pattern (see Prob. 9-41) 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
12 103  113.2
V
 y 


A 106.05h
h
Secondary shear:
d (3b 2  d 2 )
6
75[3(752 )  752 ]

 281.3 103  mm 3
6
J  0.707(h)(281.3) 103   198.8 103  h mm 4
Ju 
With  = 45,
Shigley’s MED, 11th edition
Chapter 9 Solutions, Page 21/39
 x 
3
Mry 12 10  (187.5)(37.5) 424.4
Mr cos 45o



  y
J
J
h
198.8 103  h
 max   x 2   y   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
 all  min[0.3(400), 0.4(220)]  88 MPa
 max   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
______________________________________________________________________________
9-45
Problem 9-44 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,
all = 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. 9-44, 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)
Shigley’s MED, 11th edition
(1)
Chapter 9 Solutions, Page 22/39
Ju
 b  75 

6
3
, J = 0.707 (6) Ju = 0.707(b +75)3
(2)
Primary shear ( in MPa, h in mm):
 y 
12 103 
V

A
A
(3)
Secondary shear (See Prob. 9-44 solution for the definition of ) :
Mr
  
J
3
Mry 12 10  150  b / 2  (37.5)
Mr
 x    cos  

cos  
3
J
J
0.707  b  75 
3
Mr
Mrx 12 10  150  b / 2  (b / 2)

 y    sin  
sin  
3
J
J
0.707  b  75 
 max   y 2   x   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)
41
42
43
44
A (mm2)
984.144
992.628
1001.112
1009.596
J (mm4)
1103553.5
1132340.4
1161623.6
1191407.4
'y (Mpa) "y (Mpa) "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
max
(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
______________________________________________________________________________
9-46
Materials:
Member and attachment (1018 HR):
Table 9-4:
Shigley’s MED, 11th edition
S y  32 kpsi,
Sut  58 kpsi
 all  min[0.3(58), 0.4(32)]  12.8 kpsi
Chapter 9 Solutions, Page 23/39
Decision: Use E6010 electrode. From Table 9-3: S y  50 kpsi, Sut  62 kpsi,
 all  min[0.3(62), 0.4(50)]  20 kpsi
Decision: Since 1018 HR is weaker than the E6010 electrode, use  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  1.414h(b  d )  1.414( h)(6  6)  16.97 h
Primary shear
 
3
V F 20 10
1179

 y   
psi
A A 16.97 h
h
Secondary shear
(b  d )3 (6  6)3

 288 in 3
6
6
J  0.707 h(288)  203.6h in 4
Ju 
 x   y 
 max
Mry

 
20 103 (6.25  3)(3)

2726
psi
h
203.6h
1
4762
  x2  ( y   y ) 2 
27262  (1179  2726) 2 
psi
h
h
J
Relate stress to strength
 max   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, 11th edition
Chapter 9 Solutions, Page 24/39
9-47
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.
______________________________________________________________________________
9-48
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
Iu 


2
2
2
I  0.707 hI u
V
F
  
A 1.414h(b  b1 )
Mc Fa ( d / 2)

  
I
0.707hI u
Parametric study
Let a  10 in, b  8 in, d  8 in, b1  2 in,  all  12.8 kpsi, l  2(8  2)  12 in
Shigley’s MED, 11th edition
Chapter 9 Solutions, Page 25/39
A  1.414h(8  2)  8.48h in 2
I u  (8  2)(82 / 2)  192 in 3
I  0.707( h)(192)  135.7 h in 4
10 000 1179
 
psi

8.48h
h
10 000(10)(8 / 2) 2948

  
psi
135.7h
h
1
3175
 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
eff 

 91.2 in
vol 0.369
1179
 4754 psi
 
0.248
2948
 11 887 psi
  
0.248
3175
 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

 
11.31h
h
10 000(10)(8 / 2) 2210

  
181h
h
1
2380
 max 
  all
8842  2210 2 
h
h

2380
 0.186 in
h  max 
 all 12 800
Do not round off h.
Shigley’s MED, 11th edition
Chapter 9 Solutions, Page 26/39
A  11.31(0.186)  2.10 in 2
I  181(0.186)  33.67 in 4
884
0.186 2
 4753 psi, vol 
16  0.277 in 3
0.186
2
2210
 11882 psi
  
0.186
I
256
fom '  u 
 86.0 in
hl 0.186(16)
33.67
I
eff  2

 121.7 in
(h / 2)l (0.1862 / 2)16
 
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.
______________________________________________________________________________
9-49
From Table 9-2
For the box
A  1.414 h(b  d )
Subtracting b1 from b and d1 from d
A  1.414h  b  b1  d  d1 
Iu 
Length of bead
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
______________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 9 Solutions, Page 27/39
9-50 Computer programs will vary.
______________________________________________________________________________
9-51
all = 12 kpsi. Use Fig. 9-17(a) for general geometry, but employ
beads.
beads and then
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

 
kpsi
4.24h
h
Mc 10(10)(6 / 2) 7.853


  
kpsi
I
38.2h
h
Iu 
 max   2   2 
1
8.199
2.3582  7.8532 
kpsi
h
h
Equate the maximum and allowable shear stresses.
 max   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 
Shigley’s MED, 11th edition
26.1
I

 18.6 in
vol 1.4
Ans.
Chapter 9 Solutions, Page 28/39
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

 
psi
8.48h
h
Mc 10(10)(6 / 2) 5.894


  
psi
I
50.9h
h
1
6.011
 max   2   2 
1.1792  5.894 2 
kpsi
h
h
Iu 
Equating  max to  all gives h  0.501 in. It follows that
I  50.9(0.501)  25.5 in 4
h 2l 0.5012
vol 
(6  6)  1.51 in 3

2
2
I
25.5

 16.7 in
(eff ) V 
vol 1.51
Ans.
______________________________________________________________________________
9-52
F = 0, T = 15 kipin.
Table 9-1:
Ju = 2 r 3 = 2 (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
______________________________________________________________________________
 max 
9-53
F = 2 kip, T = 0.
Table 9-2:
A = 1.414  h r = 1.414  (1/4)(1) = 1.111 in2
Iu =  r 3 =  (1)3 = 3.142 in3, I = 0.707(1/4) 3.142 = 0.5553 in4
Shigley’s MED, 11th edition
Chapter 9 Solutions, Page 29/39
 
V
2

 1.80 kpsi
A 1.111
  
Mr 2  6 1

 21.6 kpsi
I
0.5553
 max = ( 2 +  2)1/2 = (1.802 + 21.62)1/2 = 21.7 kpsi
Ans.
______________________________________________________________________________
9-54
F = 2 kip, T = 15 kipin.
Bending:
Table 9-2: A = 1.414  h r = 1.414  (1/4)(1) = 1.111 in2
Iu =  r 3 =  (1)3 = 3.142 in3, I = 0.707(1/4) 3.142 = 0.5553 in4
 
V
2

 1.80 kpsi
A 1.111
 M

Mr 2  6 1

 21.6 kpsi
I
0.5553
Torsion:
Table 9-1:
Ju = 2 r 3 = 2 (1)3 = 6.283 in3, J = 0.707(1/4) 6.283 = 1.111 in4
 T

Tr 15 1

 13.5 kpsi
J 1.111
 max   2    M    T  1.802  21.62  13.52  25.5 kpsi
2
2
Ans.
______________________________________________________________________________
9-55
F = 2 kip, T = 15 kipin.
Bending:
Table 9-2: A = 1.414  h r = 1.414  h (1) = 4.442h in2
Iu =  r 3 =  (1)3 = 3.142 in3, I = 0.707 h (3.142) = 2.221h in4
 
Shigley’s MED, 11th edition
V
2
0.4502


kpsi
A 4.442h
h
Chapter 9 Solutions, Page 30/39
  M
Mr 2  6 1 5.403


kpsi
I
2.221h
h

Torsion:
Ju = 2 r 3 = 2 (1)3 = 6.283 in3, J = 0.707 h (6.283) = 4.442 in4
Table 9-1:
 T
 max

Tr 15 1 3.377


kpsi
J 4.442h
h
2
2
2
6.387
2
2
 0.4502   5.403   3.377 
kpsi
  2     M    T  
 
 
 
h
 h   h   h 
 max   all
6.387
 20
h

Should specify a 83 in weld.

h  0.319 in
Ans.
Ans.
______________________________________________________________________________
9-56
h  9 mm,
d  200 mm,
b  25 mm
From Table 9-2, case 2:
A = 1.414(9)(200) = 2.545(103) mm2
Iu 
d 3 2003

 1.333 106 mm3
6
6
 
I = 0.707h Iu = 0.707(9)(1.333)(106) = 8.484(106) mm4
 
25 103
F
  
 9.82 MPa
A 2.545(103 )
M = 25(150) = 3750 Nm
  
Mc 3750(100)

103  44.20 MPa
I
8.484(106 )
 
 max   2   2  9.822  44.202  45.3 MPa Ans.
______________________________________________________________________________
9-57
h = 0.25 in, b = 2.5 in, d = 5 in.
Table 9-2, case 5:
Shigley’s MED, 11th edition
A = 0.707h (b +2d) = 0.707(0.25)[2.5 + 2(5)] = 2.209 in2
Chapter 9 Solutions, Page 31/39
y
d2
52

 2 in
b  2d 2.5  2  5 
2d 3
 2d 2 y   b  2 d  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
Secondary shear (the critical location is at the bottom of the bracket):
y = 5  2 = 3 in
  
My 2  5  3 

 5.093 kpsi
I
5.891
 max   2   2  0.9052  5.0932  5.173 kpsi
 all
18

 3.48
Ans.
 max 5.173
______________________________________________________________________________
n
9-58
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, 11th edition
Chapter 9 Solutions, Page 32/39
A  1.414 h(b  d )  1.414(1 / 16)(1  7.5)
 0.7512 in 2
x  b / 2  0.5 in
y  d / 2  7.5 / 2  3.75 in
d2
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)W  4.25W
V
W

 1.331W
 
A 0.7512
Mc
4.25W (7.5 / 2)

 3.664W
  
I
4.350
 max   2   2  W 1.3312  3.664 2  3.90W
Material properties: The allowable stress given is low. Let’s demonstrate that.
For the 1020 CD bracket, use HR properties of Sy = 30 kpsi and Sut = 55. The 1030 HR
support, Sy = 37.5 kpsi and Sut = 68. The E6010 electrode has strengths of Sy = 50 and
Sut = 62 kpsi.
Allowable stresses:
1020 HR:
all = min[0.3(55), 0.4(30)] = min(16.5, 12) = 12 kpsi
1020 HR:
all = min[0.3(68), 0.4(37.5)] = min(20.4, 15) = 15 kpsi
E6010:
all = min[0.3(62), 0.4(50)] = min(18.6, 20) = 18.6 kpsi
Since Table 9-6 gives 18.0 kpsi as the allowable shear stress, use this lower value.
Therefore, the allowable shear stress is
all = min(14.4, 12, 18.0) = 12 kpsi
However, the allowable stress in the problem statement is 1.5 kpsi which is low from the
weldment perspective. The load associated with this strength is
 max   all  3.90W  1500
W 
1500
 385 lbf
3.90
If the welding can be accomplished (1/16 leg size is a small weld), the weld strength is
12 000 psi and the load associated with this strength is W = 12 000/3.90 = 3077 lbf. Can
the bracket carry such a load?
There are geometry problems associated with sheet metal folding. Load placement is
important and the center of twist has not been identified. Also, the load-carrying
capability of the top bend is unknown.
Shigley’s MED, 11th edition
Chapter 9 Solutions, Page 33/39
These uncertainties may require the use of a different weld pattern. Our solution provides
the best weldment and thus insight for comparing a welded joint to one which employs
screw fasteners.
______________________________________________________________________________
9-59
F
FB
FBx
FBy
 100 lbf ,  all  3 kpsi
 100(16 / 3)  533.3 lbf
 533.3cos 60  266.7 lbf
 533.3cos 30  462 lbf
It follows that RAy  562 lbf and RAx  266.7 lbf, RA = 622 lbf
M = 100(16) = 1600 lbf · in
The OD of the tubes is 1 in. From Table 9-1, case 6:
A  2 1.414( hr )   2(1.414)( h)(1 / 2)  4.442h in 2
J u  2 r 3  2 (1 / 2)3  0.7854 in 3
J  2(0.707)hJ u  1.414(0.7854)h  1.111h in 4
The weld only carries the torsional load between the handle and tube A. Consequently,
the primary shear in the weld is zero, and the maximum shear stress is comprised entirely
of the secondary shear.
Shigley’s MED, 11th edition
Chapter 9 Solutions, Page 34/39
Tc
Mc 1600(0.5) 720.1



psi
J
J
h
1.111h
720.1

 3000
h
 max    
 max   all
h
720.1
 0.240  1 / 4 in
3000
Decision: Use 1/4 in fillet welds
Ans.
______________________________________________________________________________
9-60
For the pattern in bending shown, find the centroid G of the weld group.
x
75  6 150   325  9 150 
 225 mm
 6 150    9 150 
I 6 mm  2  I G  Ax 2 
6mm
 0.707  6  1503 

2
 2
 0.707  6 150  225  75    31.02 10 6  mm 4
12


I 9mm
 0.707  9  1503 

2
 2
 0.707  9 150 175  75   22.67 106  mm 4
12


I = I 6 mm + I 9 mm = (31.02 + 22.67)(106) = 53.69(106) mm4
The critical location is at B. With  in MPa, and F in kN
F 103 
V
  
 0.3143F
A 2 0.707  6  9 150  
3
Mc 200 F 10   225 

 0.8381F
  
I
53.69 106 
Shigley’s MED, 11th edition
Chapter 9 Solutions, Page 35/39
 max   2   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
Eq. (5-21):
all = 0.577(190) = 109.6 MPa
 all / n
109.6 / 2
 61.2 kN
Ans.
0.8951 0.8951
______________________________________________________________________________
F
9-61

In the textbook, Fig. Problem 9-61b 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 9-61b. The shear stress due to Fy is
1 
Fy
A

600
 617 psi
0.972
The shear stress along the throat due to Fx is
2 
Fx
1039

 1069 psi
A
0.972
The resultant of 1 and 2 is in the throat plane
    12   22  617 2  10692  1234 psi
The bending of the throat gives
  
Shigley’s MED, 11th edition
Mc
439(1.25)

 916 psi
I
0.599
Chapter 9 Solutions, Page 36/39
The maximum shear stress is
 max   2   2  1234 2  9162  1537 psi
(f) Materials:
1018 HR Member:
E6010 Electrode:
n
S sy
 max
Ans.
Sy = 32 kpsi, Sut = 58 kpsi (Table A-20)
Sy = 50 kpsi (Table 9-3)

0.577 S y
 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
1039
F
 1662 psi
 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
y  y 
A1 I / c
600
439
y 

 2648 psi
0.625 0.260
The von Mises stress   is
    y2  3 xy2 
26482  3(1662) 2  3912 psi
Thus, the factor of safety is,
S
32
n y 
 8.18
  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
 

 9600 psi
td
0.25(0.50)
S
32(103 )
 3.33 Ans.
n y 

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. (6-10), gives
Shigley’s MED, 11th edition
Chapter 9 Solutions, Page 37/39
Se  0.5Sut  0.5(58)  29.0 kpsi
ka = 12.7(58)-0.758 = 0.585
Eq. (6-18):
For uniform shear stress on the throat, assume kb = 1.
Eq.(6-25):
kc = 0.59
From Eq. (6-17), the endurance strength in shear is
Sse = 0.585(1)(0.59)(29.0) = 10.0 kpsi
From Table 9-5, the shear stress-concentration factor is Kf s = 2.7. The loading is
repeatedly-applied

1.537
 a   m  K f s max  2.7
 2.07 kpsi
2
2
Eq. (6-48): Gerber factor of safety nf, adjusted for shear, with Ssu = 0.67Sut
2
1  S su   a 
 1 
nf  

2   m  S se 

 2 S  
1   m se  
 S su a  
2
2

 2(2.07)(10.0)  
1  0.67(58)   2.07  
 

  1  1  
   4.55 Ans.
2  2.07   10.0  
 0.67(58)(2.07)  


Attachment metal should be checked for bending fatigue.
______________________________________________________________________________
9-62
(a) Use b = d = 4 in. Since h = 5/8 in, the primary shear is
 
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 14 F (2)
 x   y 

 1.485F
J
18.85
Ju 
Thus, the maximum shear and allowable load are:
Shigley’s MED, 11th edition
Chapter 9 Solutions, Page 38/39
 max  F 1.4852  (0.2829  1.485) 2  2.309 F
 all
25

 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
F 
Ans.
(b) From Prob. 9-20b, all = 11 kpsi
Fall 
 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.
______________________________________________________________________________
9-63
Purchase the hook having the design shown in Fig. Problem 9-63b. Referring to text Fig.
9-29a, this design reduces peel stresses.
______________________________________________________________________________
9-64
(a)
l/2
l /2
A
1 l / 2 P cosh( x)
dx  A1  cosh( x) dx  1 sinh( x)
  
l
/
2
l
/
2


l

4b sinh(l / 2)
l / 2
A1
A1
 [sinh(l / 2)  sinh(l / 2)]  [sinh(l / 2)  ( sinh(l / 2))]


(b)
(c)
2 A1 sinh(l / 2)

P
P
[2sinh(l / 2)] 
4bl sinh(l / 2)
2bl

P cosh( l / 2)
P
 (l / 2) 

Ans.
4b sinh( l / 2)
4b tanh( l / 2)
K 

 (l / 2)
P
l / 2
 2bl 




4b tanh(l / 2)  P  tanh( l / 2)
Ans.
Ans.
For computer programming, it can be useful to express the hyperbolic tangent in
terms of exponentials:
l exp(l / 2)  exp(l / 2)
K 
Ans.
2 exp(l / 2)  exp(l / 2)
______________________________________________________________________________
9-65
This is a computer programming exercise. All programs will vary.
Shigley’s MED, 11th edition
Chapter 9 Solutions, Page 39/39
Chapter 5
5-1
Sy = 350 MPa.
MSS: 1  3 = Sy /n
DE:
n

    A2   A B   B2 
1/ 2
n
Sy
 1   3 
  x2   x y   y2  3 xy2 
1/ 2
Sy

(a) MSS:
1 = 100 MPa,2 = 100 MPa,3 = 0
n
DE:
(b) MSS:
Ans.
   (100 2  100(100)  1002 )1/ 2  100 MPa,
350
 3.5
100  0
350
 3.5
100
Ans.
350
 4.04
86.6
Ans.
n
1 = 100 MPa,2 = 50 MPa,3 = 0
n
DE:
350
 3.5
100  0
Ans.
   (100 2  100(50)  50 2 )1/ 2  86.6 MPa,
n
2
100
 100 
2
(c)  A ,  B 
 
  (75)  140,  40 MPa
2
2


 1  140,  2  0,  3  40 MPa
350
 1.94 Ans.
MSS: n 
140  ( 40)
DE:
   100 2  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


 1  0,  2  11.0,  3  114.0 MPa
350
MSS:
n
 3.07 Ans.
0  (114.0)
DE:
   [(50) 2  ( 50)(75)  (75) 2  3( 50) 2 ]1/2  109.0 MPa
350
n
 3.21 Ans.
109.0
(d)  A ,  B 
(e)  A ,  B 
2
100  20
2
 100  20 
 
   20   104.7, 15.3 MPa
2
2


Shigley’s MED, 11th edition
Chapter 5 Solutions, Page 1/58
 1  104.7,  2  15.3,  3  0 MPa
350
 3.34
104.7  0
MSS:
n
DE:
2
   1002  100(20)  20 2  3  20  
Ans.
1/2
 98.0 MPa


350
n
 3.57 Ans.
98.0
______________________________________________________________________________
5-2
Sy = 350 MPa.
MSS: 1  3 = Sy /n
DE:
n

    A2   A B   B2 
(a) MSS:
DE:
(b) MSS:
DE:
1/ 2

Sy
 1   3 
Sy
n

n
 1  100 MPa,  3  0  n 
n
Sy

350
 3.5
100  0
350
 3.5
[100  (100)(100)  1002 ]1/ 2
2
 1  100 ,  3  100 MPa 
n
Ans.
Ans.
350
 1.75
100  ( 100)
Ans.
350
 2.02 Ans.
1/2
1002  (100)( 100)   100  2 


350
(c) MSS:
 1  100 MPa,  3  0  n 
 3.5 Ans.
100  0
350
DE:
n
 4.04 Ans.
1/ 2
1002  (100)(50)  50 2 
350
 1  100,  3  50 MPa  n 
 2.33 Ans.
(d) MSS:
100  ( 50)
350
DE:
n
 2.65
Ans.
1/2
1002  (100)( 50)   50  2 


350
(e) MSS:  1  0,  3  100 MPa  n 
 3.5 Ans.
0  (100)
350
DE:
n
 4.04
Ans.
1/2
 50 2  ( 50)(100)   100 2 


______________________________________________________________________________
5-3
n
From Table A-20, Sy = 37.5 kpsi
Shigley’s MED, 11th edition
Chapter 5 Solutions, Page 2/58
MSS: 1  3 = Sy /n
n

    A2   A B   B2 
1/ 2
DE:
n
Sy
 1   3 
  x2   x y   y2  3 xy2 
1/ 2
Sy

(a) MSS:
 1  25 kpsi,  3  0
DE:
n

37.5
 252  (25)(15)  152 
1/ 2
n
 1.72
 1  15 kpsi,  3  15 
(b) MSS:
DE:
n
37.5
 1.5
25  0
n
37.5
1/ 2
152  (15)(15)   15 2 


Ans.
Ans.
37.5
 1.25
15  (15)
 1.44
Ans.
2
20
 20 
    ( 10)2  24.1,  4.1 kpsi
2
 2 
 1  24.1,  2  0,  3  4.1 kpsi
37.5
n
 1.33 Ans.
MSS:
24.1  ( 4.1)
(c)  A ,  B 
    202  3  102  
DE:
Ans.
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


 1  17.7,  2  0,  3  14.7 kpsi
37.5
MSS:
n
 1.16 Ans.
17.7  ( 14.7)
(d)  A ,  B 
1/ 2
   (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 
 
   15   9,  39 kpsi
2
2


 1  0,  2  9,  3  39 kpsi
37.5
MSS:
n
 0.96 Ans.
0   39 
(e)  A ,  B 
Shigley’s MED, 11th edition
Chapter 5 Solutions, Page 3/58
DE:
2
2
2
    24    24  24    24   3  15  
1/ 2
 35.4 kpsi


37.5
n
 1.06 Ans.
35.4
______________________________________________________________________________
5-4
From Table A-20, Sy = 47 kpsi.
MSS: 1  3 = Sy /n
n

Sy
 1   3 
Sy
n

47
(a) MSS:
 1  30 kpsi,  3  0  n 
 1.57 Ans.
30  0
47
n
 1.57 Ans.
DE:
2
[30  (30)(30)  302 ]1/ 2
47
(b) MSS:
 1  30 ,  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:
 1  30 kpsi,  3  0  n 
 1.57 Ans.
30  0
47
DE:
n
 1.81 Ans.
1/ 2
2
30  (30)(15)  152 
47
 1  0,  3  30 kpsi  n 
 1.57 Ans.
(d) MSS:
0  (30)
47
DE:
n
 1.81
Ans.
1/ 2
2
 30   ( 30)(15)   15 2 


47
 0.78 Ans.
(e) MSS:  1  10,  3  50 kpsi  n 
10  (50)
47
DE:
n
 0.84
Ans.
1/2
2
 50   ( 50)(10)  10 2 


______________________________________________________________________________
DE:
    A2   A B   B2 
Shigley’s MED, 11th edition
1/ 2

Sy

n
Chapter 5 Solutions, Page 4/58
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, 11th edition
Chapter 5 Solutions, Page 5/58
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) A = 25 kpsi, B = 15 kpsi
MSS:
n
DE:
OB 4.37"

 1.50
OA 2.92"
Ans.
OC 5.02"

 1.72 Ans.
OA 2.92"
(b) A = 15 kpsi, B =  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:
(c)  A ,  B 
MSS: n 
2
20
 20 
    ( 10) 2  24.1,  4.1 kpsi
2
 2 
OH 3.25"

 1.34
OG 2.43"
Ans. DE: n 
OI 3.46"

 1.42
OG 2.43"
Ans.
2
12  15
 12  15 
2
(d)  A ,  B 
 
  ( 9)  17.7, 14.7 MPa
2
2


MSS: n 
OK 2.67"

 1.16
OJ 2.30"
(e)  A ,  B 
Ans. DE: n 
OL 3.06"

 1.33
OJ 2.30"
Ans.
2
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, 11th edition
Chapter 5 Solutions, Page 6/58
5-7
Sy = 295 MPa, A = 75 MPa, B =  35 MPa,
(a) n 

Sy
2
A
  A B  

2 1/ 2
B

295
752  75  35    35  2 


1/ 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, 11th edition
Chapter 5 Solutions, Page 7/58
5-8
Sy = 295 MPa, A = 30 MPa, B =  100 MPa,
(a) n 

Sy
2
A
  A B  

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, 11th edition
Chapter 5 Solutions, Page 8/58
5-9
2
100
 100 
2
Sy = 295 MPa,  A ,  B 
 
  (25)  105.9,  5.9 MPa
2
 2 
Sy
295

 2.71
(a) n 
1/2
2
2 1/ 2
 A   A B   B  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, 11th edition
Chapter 5 Solutions, Page 9/58
5-10
2
30  65
 30  65 
2
Sy = 295 MPa,  A ,  B 
 
  40  3.8, 91.2 MPa
2
2


Sy
295
(a) n 

 3.30
1/ 2
2
2 1/ 2
2
 A   A B   B   3.8   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, 11th edition
Chapter 5 Solutions, Page 10/58
5-11
2
80  30
2
 80  30 
Sy = 295 MPa,  A ,  B 
 
   10   30.9, 80.9 MPa
2
2


Sy
295
(a) n 

 2.95 Ans.
1/ 2
2
2 1/ 2
 A   A B   B  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

 
n 1  3 
S

 yt S yc 
(a)  1  25 kpsi,  3  0
(b)  1  15,  3  15 kpsi
Shigley’s MED, 11th edition
1
1
 25 0 
n      2.40
 60 75 

Ans.
1

 15 15 
n 
  2.22
 60 75 
Ans.
Chapter 5 Solutions, Page 11/58
2
20
 20 
(c)  A ,  B 
    ( 10) 2  24.1,  4.1 kpsi ,
2
 2 
 1  24.1,  2  0,  3  4.1 kpsi 
(d)  A ,  B 
Ans.
2
12  15
 12  15 
2
 
  ( 9)  17.7, 14.7 kpsi
2
2


 1  17.7,  2  0,  3  14.7 kpsi
(e)  A ,  B 
1
 24.1 4.1 

n
  2.19
75 
 60
1

 17.7 14.7 
n

  2.04
75 
 60
Ans.
2
24  24
2
 24  24 
 
   15   9,  39 kpsi
2
2


1
 0 39 
 1  0,  2  9,  3  39 kpsi  n   
Ans.
  1.92
 60 75 
______________________________________________________________________________
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)  A  25,  B  15 kpsi
n
OB 3.49"

 2.39 Ans.
OA 1.46"
(b)  A  15,  B  15 kpsi
n
(c)
OD 2.36"

 2.23 Ans.
OC 1.06"
2
20
 20 
 A ,  B      ( 10) 2  24.1,  4.1 kpsi
2
 2 
n
(d)
OF 2.67"

 2.19 Ans.
OE 1.22"
2
12  15
 12  15 
2
 A , B 
 
  ( 9)  17.7, 14.7 kpsi
2
2


Shigley’s MED, 11th edition
Chapter 5 Solutions, Page 12/58
n
(e)
OH 2.34"

 2.03 Ans.
OG 1.15"
2
24  24
2
 24  24 
 A , B 
 
   15   9,  39 kpsi
2
2


OJ 3.85"
n

 1.93 Ans.
OI 2.00"
______________________________________________________________________________
5-14
Since f > 0.05, and Syt  Syc, the Coulomb-Mohr theory for ductile materials will be
used.
(a) From Eq. (5-26),
1
1

 
 150 50 
n 1  3  

  1.23 Ans.
S

S
235
285


yt
yc


(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

 
 50 150 
n 1  3  

  1.35 Ans.
S

S
235
285


yc 
 yt
Shigley’s MED, 11th edition
Chapter 5 Solutions, Page 13/58
(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
5-16
2
125
 125 
2
 A,  B 
 
  (75)  160,  35 kpsi
2
2


(a) From Eq. (5-26),
1
1
 1  3 
 160 35 
n


  
  1.24 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.
OF 2.04"

 1.24 Ans.
OE 1.64"
______________________________________________________________________________
n
5-17
2
80  125
 80  125 
2
 
 A,  B 
  50  47.7,  157.3 kpsi
2
2


(a) From Eq. (5-26),
1
1
 1  3 
157.3 
 0
n


  
  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
5-18
2
125  80
 125  80 
2
 
 A,  B 
  ( 75)  180.8, 24.2 kpsi
2
2


(a) From Eq. (5-26),
Shigley’s MED, 11th edition
Chapter 5 Solutions, Page 14/58
1
1
 1  3 
0 
 180.8
n


  
  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), MM: Eqs. (5-32)
(a) A = 25 kpsi, B = 15 kpsi
BCM : Eq. (5-31a), n 
MM: Eq. (5-32a), n 
Sut
A
Sut


A
(b) A = 15 kpsi, B = 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: A  0  B, and B /A   1, Eq. (5-32a), n 
Sut
A

30
 2.0 Ans.
15

30
 1.24 Ans.
24.14
2
20
 20 
(c)  A ,  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: A  0  B, and B /A   1, Eq. (5-32a), n 
(d)  A ,  B 
Sut
A
2
15  10
 15  10 
2
 
  ( 15)  17.03,  22.03 kpsi
2
2


Shigley’s MED, 11th edition
Chapter 5 Solutions, Page 15/58
1
 17.03 22.03 
BCM: Eq. (5-31b), n  

  1.23 Ans.
90 
 30
MM: A  0  B, and B /A   1, Eq. (5-32b),
1
1
  90  30 17.03 22.03 
  S  Sut   A  B 
n   uc


  1.60 Ans.
 
Suc Sut
Suc 
90  30 
90 


(e)  A ,  B 
2
20  20
 20  20 
2
 
  ( 15)  5,  35 kpsi
2
2


BCM: Eq. (5-31c), n  
Suc
B

90
 2.57 Ans.
35
90
 2.57 Ans.
B
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) A = 25, B = 15 kpsi
BCM & MM:
OB 1.74"
n

 1.19 Ans.
OA 1.46"
(b) A = 15, B =  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)  A ,  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, 11th edition
Chapter 5 Solutions, Page 16/58
2
15  10
 15  10 
2
(d)  A ,  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
(e)  A ,  B 
 
  ( 15)  5,  35 kpsi
2
2


OM 4.55"
BCM and MM: n 

 2.57 Ans.
OL 1.77"
______________________________________________________________________________
5-21
From Table A-24, Sut = 31 kpsi, Suc = 109 kpsi
BCM: Eqs. (5-31), MM: Eqs. (5-32)
(a) A = 15, B = 10 kpsi.
31
 2.07 Ans.
 A 15
S
31
MM: Eq. (5-32a), n  ut 
 2.07 Ans.
 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"
______________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 5 Solutions, Page 17/58
5-22
Sut = 31 kpsi, Suc = 109 kpsi
BCM: Eq. (5-31), MM: Eqs. (5-32)
(a) A = 15, B =  50 kpsi, B /A  > 1
1
 15 50 
BCM: Eq. (5-31b), n   
  1.06 Ans.
 31 109 
1
1
 109  3115 50 
  S  Sut   A  B 
MM: Eq. (5-32b), n   uc


  1.24 Ans.
 
Suc Sut
Suc 
109 

 109  31
(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)
 A,  B 
2
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.
 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)
 A,  B 
2
10  25
 10  25 
2
 
  ( 10)  5,  30 kpsi
2
2


Shigley’s MED, 11th edition
Chapter 5 Solutions, Page 18/58
109
 3.63 Ans.
 B 30
S
109
MM: Eq. (5-32c), n   uc  
 3.63 Ans.
30
B
(b), (c) The plot is shown in the solution to Prob. 5-21.
(a) BCM: Eq. (5-31c), n  
Suc

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
35  13
 35  13 
2
 
 A,  B 
  ( 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   A  B 
MM: Eq. (5-32b), n   uc


  1.46 Ans.
 
Suc Sut
Suc 
109 

 109  31
(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) A = 15, B =  10 kpsi.
1
 15 10 
BCM: Eq. (5-31b), n   
  1.42 Ans.
 36 35 
Shigley’s MED, 11th edition
Chapter 5 Solutions, Page 19/58
(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) A = 15, B =  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)  A ,  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, 11th edition
Chapter 5 Solutions, Page 20/58
OF 1.47"

 1.79 Ans.
OE 0.82"
______________________________________________________________________________
n
5-29
Sut = 36 kpsi, Suc = 35 kpsi
BCM: Eq. (5-31),
2
10  15
 10  15 
2
(a)  A ,  B 
 
  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
15  8
2
 15  8 
(a)  A ,  B 
 
   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) A = 15, B =  10 kpsi. Eq. (5-32a), n  ut 
 2.4 Ans.
 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, 11th edition
Chapter 5 Solutions, Page 21/58
n
OB 2.16"

 2.43 Ans.
OA 0.90"
______________________________________________________________________________
5-32
Sut = 36 kpsi, Suc = 35 kpsi. MM: Use Eq. (5-32).For this problem, MM reduces to the
MNS theory.
(a) A = 10, B =  15 kpsi. Eq. (5-32b) is not valid and must use Eq, (5-32c),
S
35
n   uc  
 2.33 Ans.
B
15
(b) The plot is shown in the solution to Prob. 5-31.
OD 2.10"
n

 2.33 Ans.
OC 0.90"
______________________________________________________________________________
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)  A ,  B      (8) 2  16,  4 kpsi
2
 2
S
36
n  ut 
 2.25 Ans.
 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, 11th edition
Chapter 5 Solutions, Page 22/58
2
10  15
 10  15 
2
(a)  A ,  B 
 
  10  2.2,  22.8 kpsi
2
2


S
35
n   uc  
 1.54 Ans.
22.8
B
(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.
 A 20.2
(b) The plot is shown in the solution to Prob. 5-31.
(a)  A ,  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:
x 

 22.6 106  Pa  22.6 MPa
2
2
d
  0.015 
3
16  25 
16T 4 V
4  0.55 10  

  33.6 106  Pa  33.6 MPa


 xy  3 
2
3
 d 3 A   0.015  3   / 4  0.015 


    22.62  3  33.62  
1/ 2
n
B:
280
 4.49
62.4
 62.4 MPa
Ans.
3
3
32 Fl 4 P 32  0.55 10  0.1 4  4 10
x 



 189 106  Pa  189 MPa
d3 d2
  0.0153 
  0.0152 
 xy 
16  25 
16T

 37.7 106  Pa  37.7 MPa
3
3
d
  0.015 
Shigley’s MED, 11th edition
Chapter 5 Solutions, Page 23/58
    x2  3 xy2 
1/2
 1892  3  37.7 2  
1/ 2
 200 MPa
S y 280

 1.4
Ans.
  200
______________________________________________________________________________
n
5-37
From Prob. 3-45, 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  =  7 456 psi. Thus,  = 7 456 psi and
(Sy)min = n = 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)
T
63 025H
n
(1)
where n is the shaft speed in rev/min. From Eq. (5-3), for the MSS theory,
 max 
Sy
2nd

16T
 d3
(2)
where nd is the design factor. Substituting Eq. (1) into Eq. (2) and solving for d gives
1/3
 32  63 025  Hnd 
d 

n S y


(3)
Substituting H = 20 hp, nd = 3, n = 1750 rev/min, and Sy = 57(103) psi results in
1/3
 32  63 025  20  3 
d min  
Ans.
 0.728 in
3 
 1750  57 10 
______________________________________________________________________________
5-39
Given: d = 30 mm, AISI 1018 steel, H = 10 kW, n = 200 rev/min.
Table A-20, Sy = 220 MPa
Eq. (3-44): T = 9.55 H/n = 9.55(10)103/200 = 477.5 N٠m
16T 16  477.5  6
 max  3 
10  90.07 MPa
3
d
  0.030 
(a) Eq. (5-3): n 
Shigley’s MED, 11th edition
Sy
2 max

220
 1.22
2  90.07 
Ans.
Chapter 5 Solutions, Page 24/58
2
  3 max
 3  90.07   156.0 MPa
(b) From Eq. (5-13),  max
Sy
220

 1.41
Ans.

 max
156
______________________________________________________________________________
Eq. (5-19): n 
5-40
Given: d = 20 mm, AISC 1035 HR steel, ns = 400 rev/min, ny = 1.5.
Table A-20, Sy = 270 MPa
S
270
(a) Eq. (5-30):  max  y 
 90 MPa
2n y 2 1.5 
Substituting Eq. (3-44) in the equation for max gives
 max
16T
16 
H
9.55 


3
3 
d
d 
ns 
H
 d 3 ns max
16  9.55 

  0.023  400  90 106
16  9.55 
 5.92 103  W  5.92 kW
  3  max 
(b) From Eq. (5-13):  max
Thus,
H

 d 3 ns max
Sy
ny

270
1.5
  0.023  400 103.9 106

 max 
Ans.
270
 103.9 MPa
1.5 3
 6.84 103  W  6.84 kW
Ans
16  9.55 
16  9.55 
______________________________________________________________________________
5-41

Table A-20 for AISI 1040 CD steel, Sy = 490 MPa.
From Prob. 3-47,
2
2
 
 79.6 
2
A: x =79.6 MPa, xy = 63.7 MPa.  max      xy2  
  63.7  75.1 MPa.
2
 2 
B: max = zx = 53.1 MPa.
C: max = zx = 116.8 MPa.
Critical case is at point C.
S
490
(a) MSS Theory: n y  y 
 2.1
Ans.
2 max 2 116.8 
Sy
490
 2.4
Ans.
3 max
3 116.8 
______________________________________________________________________________
(b) DE Theory: n 
5-42

Table A-20 for AISI 1040 CD steel, Sy = 490 MPa
From Prob. 3-53,  1  122.6 MPa,  2  0,  3  10.2 MPa,  max  R  66.4 MPa
Sy
490
(a) MSS Theory:
ny 

 3.69
Ans.
2 max 2  66.4 
2
(b) DE Theory, Eq. (5-13):    122.62  122.6  10.2    10.2  


Shigley’s MED, 11th edition
1/ 2
 128 MPa
Chapter 5 Solutions, Page 25/58
S y 490

 3.83
Ans.
  128
______________________________________________________________________________
ny 
5-43
Table A-20 for AISI 1020 CD steel, Sy = 390 MPa
From Prob. 3-54,  1  194.2 MPa,  2  0,  3  10 MPa,  max  102.1 MPa
Sy
490
(a) MSS Theory:
ny 

 1.91
Ans.
2 max 2 102.1
2 1/ 2
(b) DE Theory, Eq. (5-13):    194.22  194.2  10    10    199.4 MPa


Sy
490
ny 

 1.96
Ans.
  194.2
______________________________________________________________________________
5-44
Table A-20 for AISI 1035 CD steel, Sy = 67 kpsi
From Prob. 3-55,  1  45.8 kpsi,  2  0,  3  0.45 kpsi,  max  23.1 kpsi
Sy
67
(a) MSS Theory:
ny 

 1.45
Ans.
2 max 2  23.1
2 1/ 2
(b) DE Theory, Eq. (5-13):     45.82  45.8  0.45    0.45    46.0 MPa


S y 67
ny 

 1.46
Ans.
  46
______________________________________________________________________________
5-45
Table A-20 for AISI 1040 CD steel, Sy = 71 kpsi
From Prob. 3-101,  1  18.47 kpsi,  2  3.60 kpsi,  max  11.03 kpsi
Sy
71
(a) MSS Theory:
ny 

 3.22
Ans.
2 max 2 11.03 
2 1/2
(b) DE Theory, Eq. (5-13):    18.47 2  18.47  3.60    3.60    20.51 MPa


Sy
71
ny 

 3.46
Ans.
  20.51
______________________________________________________________________________
5-46
Table A-20 for AISI 1040 CD steel, Sy = 71 kpsi
From Prob. 3-102,  1  29.1 kpsi,  2  14.2 kpsi,  max  21.7 kpsi
Sy
71
(a) MSS Theory:
ny 

 1.64
Ans.
2 max 2  21.7 
2
(b) DE Theory, Eq. (5-13):     29.12  29.1 14.2    14.2  


Shigley’s MED, 11th edition
1/ 2
 38.2 MPa
Chapter 5 Solutions, Page 26/58
Sy
71

 1.86
Ans.
  38.2
______________________________________________________________________________
ny 
5-47
Table A-21 for AISI 4140 steel Q & T 400o F , Sy = 238 kpsi, F = 15 kip.
MA = 0 = 3 RD  2 F  RD = 2 (15)/3 = 10 kip,
Fy = 0 = RA + RD  F  RA = 15  10 = 5 kip
Critical sections are at points B and C where the areas are minimal.
B: dB = 1.1 in, MB = RA (1) = 5 kip٠in, VB = RA = 5 kip, TB = 7 kip٠in
AB = (/4) 1.12 = 0.9503 in2,
C: dC = 1.3 in, MC = RD (1) = 10 kip٠in, VC = RD = 10 kip, TC = 7 kip٠in
AC = (/4) 1.32 = 1.327 in2,
Critical locations are at the outer surfaces where bending stresses are maximum, and at the
center planes where the transverse shear stresses are maximum. In both cases, there exists
the torsional shear stresses.
B: Outer surface:
32  5 
16  7 
32 M B
16TB
B 

 38.26 kpsi,  B 

 26.78 kpsi
3
3
3
3
d
d
 1.1
 1.1
 B max
2
2
 
 38.26 
2
  B    B2  
  26.78  32.9 kpsi
2
2




 B   B2  3 B2  38.262  3  26.78   60.1 kpsi
2
Center plane:
 B V

4 5
4VB

 7.02 kpsi
3 AB 3  0.9503
 B max   B   B V  26.78  7.02  33.8
 B  3  B max  3  33.8   58.5 kpsi
kpsi
C: Outer surface:
16  7 
32 M C 32 10 
16TC
C 

 46.36 kpsi,  C 

 16.23 kpsi
3
3
3
3
d
d
 1.3
 1.3
 C max
2
2
 
 46.36 
2
  C    C2  
  16.23  28.30 kpsi
 2 
 2 
 C   C2  3 C2  46.362  3 16.23  54.2 kpsi
2
Center plane:
 C V

4 10 
4VC

 10.05 kpsi
3 AC 3 1.327 
 C max   C   C V  16.23  10.05  26.28
 C  3  C max  3  26.28   45.5 kpsi
Shigley’s MED, 11th edition
kpsi
Chapter 5 Solutions, Page 27/58
(a) MSS Theory: Critical location is at point B at the center plane:
Sy
238
ny 

 3.52
Ans.
2 max 2  33.8 
(b) DE Theory: Critical location is at point B at the outer surface:
S y 238
ny 

 3.96
Ans.
 B 60.1
______________________________________________________________________________
5-48
MO = 0 = 40 RC  30(575) +12(460)
RC = 293.25 lbf
Fy = 0 = RO + 293.25 +460  575
RO =  178.25 lbf
Mmax = 2.9325 kip٠in
32 M max 32  2.9325 


 29.87 kpsi
d3
 13 

16T 16 1.5 

 7.64 kpsi
 d 3  13 
2
 29.87 
2
(a)  max  
  7.64  16.78 kpsi
2


Sy
50
ny 

 1.49
2 max 2 16.78 
Ans.
(b) Eq. (5-15): ’ = [29.872 +3 (7.64)2]1/2 = 32.67 kpsi
Sy
50
ny 

 1.53
Ans.
  32.67
______________________________________________________________________________
5-49
Given: AISI 1010 HR, ny = 2, L = 0.5 m, F = 150 N, T = 25 N٠m
Table A-20, Sy = 180 MPa.
Mz = FL = 150 (0.5) = 75 N٠m
32 M z 32  75  2400



 d3
d3
d3
16T 16  25  400
 3

d
d3
d3
6
2
2
2
402.63 S y 180 10 
 
 1200   400 
2
(a)  max       
 3  


3 
d3
2n
2  2
2
 d  d 
 4  402.63 

d 
6
 180 10  
Shigley’s MED, 11th edition
1/3
 0.0208 m  20.8 mm
Ans.
Chapter 5 Solutions, Page 28/58
(b)      3
2

2 1/ 2
2
 2400  2
 400  
 
 3 3  
3 
  d  
  d 
1/2
6
795.14 S y 180 10 



2
d3
ny
1/3
 2  795.14  
  0.0207 m  20.7 mm
d 
Ans.
6
 180 10  
______________________________________________________________________________
5-50
From Table A-20, Sy = 54 kpsi. From the solution of Prob. 3-79, in the plane of analysis
1 = 16.5 kpsi, 2 =  1.19 kpsi, and max = 8.84 kpsi
The state of stress is plane stress. Thus, the three-dimensional principal stresses are
1 = 16.5 kpsi, 2 = 0, and 3 =  1.19 kpsi
MSS: From Eq. (5-3),
n
Sy
1   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.
2 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. 3-79. That is,
n

Eqs. (5-13) and (5-19)
S
Sy
54

n y 
1/2
1/2
   A2   A B   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. 3-79
Sy
Sy
54


1/
2
1/2
   2  3 2 
15.32  3  4.432 


 3.15 Ans.
______________________________________________________________________________
n
5-51
From Table A-20, Sy = 370 MPa. From the solution of Prob. 3-80, in the plane of analysis
Shigley’s MED, 11th edition
Chapter 5 Solutions, Page 29/58
1 = 275 MPa, 2 =  12.1 MPa, and max = 144 MPa
The state of stress is plane stress. Thus, the three-dimensional principal stresses are
1 = 275 MPa, 2 = 0, and 3 =  12.1 MPa
MSS: From Eq. (5-3), n 
Sy
1   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
   A2   A B   B2 
 2752  275  12.1   12.1 2 


 1.32
Ans.
______________________________________________________________________________
5-52
From Table A-20, Sy = 54 kpsi. From the solution of Prob. 3-81, in the plane of analysis
1 = 22.6 kpsi, 2 =  1.14 kpsi, and max = 11.9 kpsi
The state of stress is plane stress. Thus, the three-dimensional principal stresses are
1 = 22.6 kpsi, 2 = 0, and 3 =  1.14 kpsi
MSS: From Eq. (5-3), n 
Sy
1   3

54
 2.27
22.6   1.14 
Ans.
DE: From Eqs. (5-13) and (5-19)
S
Sy
54
n y 

1/2
1/2
2
2
   A   A B   B 
 22.6 2  22.6  1.14    1.14  2 


 2.33 Ans.
______________________________________________________________________________
5-53
From Table A-20, Sy = 370 MPa. From the solution of Prob. 3-82, in the plane of analysis
1 = 78.2 MPa, 2 =  5.27 MPa, and max = 41.7 MPa
The state of stress is plane stress. Thus, the three-dimensional principal stresses are
1 = 78.2 MPa, 2 = 0, and 3 =  5.27 MPa
Shigley’s MED, 11th edition
Chapter 5 Solutions, Page 30/58
MSS: From Eq. (5-3), n 
Sy
1   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
2
2
   A   A B   B 
78.22  78.2  5.27    5.27  2 


 4.57
Ans.
______________________________________________________________________________
5-54
From Table A-20, Sy = 54 kpsi. From the solution of Prob. 3-83, in the plane of analysis
1 = 36.7 kpsi, 2 =  1.47 kpsi, and max = 19.1 kpsi
The state of stress is plane stress. Thus, the three-dimensional principal stresses are
1 = 36.7 kpsi, 2 = 0, and 3 =  1.47 kpsi
MSS: From Eq. (5-3), n 
Sy
1   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
   A2   A B   B2 
36.7 2  36.7  1.47    1.47  2 


 1.44
Ans.
______________________________________________________________________________
5-55
From Table A-20, Sy = 370 MPa. From the solution of Prob. 3-84, in the plane of analysis
1 = 376 MPa, 2 =  42.4 MPa, and max = 209 MPa
The state of stress is plane stress. Thus, the three-dimensional principal stresses are
1 = 376 MPa, 2 = 0, and 3 =  42.4 MPa
MSS: From Eq. (5-3), n 
Shigley’s MED, 11th edition
Sy
1   3

370
 0.88
376   42.4 
Ans.
Chapter 5 Solutions, Page 31/58
DE: From Eqs. (5-13) and (5-19)
S
Sy
370
n y 

1/2
1/2
   A2   A B   B2 
3762  376  42.4    42.4 2 


 0.93 Ans.
______________________________________________________________________________
5-56
From Table A-20, Sy = 54 kpsi. From the solution of Prob. 3-85, in the plane of analysis
1 = 7.19 kpsi, 2 =  17.0 kpsi, and max = 12.1 kpsi
The state of stress is plane stress. Thus, the three-dimensional principal stresses are
1 = 7.19 kpsi, 2 = 0, and 3 =  17.0 kpsi
MSS: From Eq. (5-3), n 
Sy

54
 2.23
7.19   17.0 
Ans.
1   3
DE: From Eqs. (5-13) and (5-19)
S
Sy
54
n y 

1/2
1/2
   A2   A B   B2 
7.192  7.19  17.0    17.0  2 


 2.51 Ans.
______________________________________________________________________________
5-57 From Table A-20, Sy = 54 kpsi. From the solution of Prob. 3-87, in the plane of analysis
1 = 1.72 kpsi, 2 =  35.9 kpsi, and max = 18.8 kpsi
The state of stress is plane stress. Thus, the three-dimensional principal stresses are
1 = 1.72 kpsi, 2 = 0, and 3 =  35.9 kpsi
MSS: From Eq. (5-3), n 
Sy
1   3

54
 1.44
1.72   35.9 
Ans.
DE: From Eqs. (5-13) and (5-19)
S
Sy
54
n y 

1/2
1/2
   A2   A B   B2 
1.722  1.72  35.9    35.9 2 


 1.47
Ans.
______________________________________________________________________________
5-58
From Table A-20, Sy = 370 MPa. From the solution of Prob. 3-88,
Bending: B = 68.6 MPa, Torsion: B = 37.7 MPa
Shigley’s MED, 11th edition
Chapter 5 Solutions, Page 32/58
For a plane stress analysis it was found that max = 51.0 MPa. With combined bending
and torsion, the plane stress analysis yields the true max.
S
370
MSS: From Eq. (5-3), n  y 
 3.63
Ans.
2 max 2  51.0 
DE: From Eqs. (5-15) and (5-19)
S
Sy
370

n y 
1/
2
1/ 2
   B2  3 B2 
68.62  3  37.7 2  


 3.91 Ans.
______________________________________________________________________________
5-59
From Table A-20, Sy = 54 kpsi. From the solution of Prob. 3-90,
Bending: C = 3460 psi, Torsion: C = 882 kpsi
For a plane stress analysis it was found that max = 1940 psi. With combined bending and
torsion, the plane stress analysis yields the true max.
54 103 
Sy
MSS: From Eq. (5-3), n 

 13.9
Ans.
2 max 2 1940 
DE: From Eqs. (5-15) and (5-19)
54 103 
Sy
Sy


n
   C2  3 C2 1/ 2 34602  3  8822  1/2


 14.3
Ans.
______________________________________________________________________________
5-60
From Table A-20, Sy = 54 kpsi. From the solution of Prob. 3-91, in the plane of analysis
1 = 17.8 kpsi, 2 =  1.46 kpsi, and max = 9.61 kpsi
The state of stress is plane stress. Thus, the three-dimensional principal stresses are
1 = 17.8 kpsi, 2 = 0, and 3 =  1.46 kpsi
MSS: From Eq. (5-3), n 
Shigley’s MED, 11th edition
Sy
1   3

54
 2.80
17.8   1.46 
Ans.
Chapter 5 Solutions, Page 33/58
DE: From Eqs. (5-13) and (5-19)
S
Sy
54
n y 

1/2
1/2
   A2   A B   B2 
17.82  17.8  1.46    1.46  2 


 2.91 Ans.
______________________________________________________________________________
5-61
From Table A-20, Sy = 54 kpsi. From the solution of Prob. 3-92, in the plane of analysis
1 = 17.5 kpsi, 2 =  1.13 kpsi, and max = 9.33 kpsi
The state of stress is plane stress. Thus, the three-dimensional principal stresses are
1 = 17.5 kpsi, 2 = 0, and 3 =  1.13 kpsi
MSS: From Eq. (5-3), n 
Sy
1   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
   A   A B   B 
17.52  17.5  1.13   1.13 2 


 2.98 Ans.
______________________________________________________________________________
5-62
From Table A-20, Sy = 54 kpsi. From the solution of Prob. 3-93, in the plane of analysis
1 = 21.5 kpsi, 2 =  1.20 kpsi, and max = 11.4 kpsi
The state of stress is plane stress. Thus, the three-dimensional principal stresses are
1 = 21.5 kpsi, 2 = 0, and 3 =  1.20 kpsi
MSS: From Eq. (5-3), n 
Sy
1   3

54
 2.38
21.5   1.20 
Ans.
DE: From Eqs. (5-13) and (5-19)
S
Sy
54
n y 

1/2
1/2
   A2   A B   B2 
 21.52  21.5  1.20    1.20  2 


 2.44
Ans.
______________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 5 Solutions, Page 34/58
5-63
From Table A-20, Sy = 54 kpsi. From the solution of Prob. 3-94, the concern was failure
due to twisting of the flat bar where it was found that max = 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 max exists.
S
54
MSS: From Eq. (5-3), n  y 
 1.89
Ans.
2 max 2 14.3
DE: From Eqs. (5-15) and (5-19)
Sy
Sy
54
n


 2.18
Ans.
1/
2
1/ 2
2
2
   3 max


3
14.3
   
______________________________________________________________________________
5-64
From Table A-20, Sy = 54 kpsi. From the solution of Prob. 3-95, in the plane of analysis
1 = 34.7 kpsi, 2 =  6.7 kpsi, and max = 20.7 kpsi
The state of stress is plane stress. Thus, the three-dimensional principal stresses are
1 = 34.7 kpsi, 2 = 0, and 3 =  6.7 kpsi
MSS: From Eq. (5-3), n 
Sy
1   3

54
 1.30
34.7   6.7 
Ans.
DE: From Eqs. (5-13) and (5-19)
S
Sy
54

n y 
1/2
1/2
2
2
   A   A B   B 
34.7 2  34.7  6.7    6.7 2 


 1.40
Ans.
______________________________________________________________________________
5-65
From Table A-20, Sy = 54 kpsi. From the solution of Prob. 3-96, in the plane of analysis
1 = 51.1 kpsi, 2 =  4.58 kpsi, and max = 27.8 kpsi
The state of stress is plane stress. Thus, the three-dimensional principal stresses are
1 = 51.1 kpsi, 2 = 0, and 3 =  4.58 kpsi
MSS: From Eq. (5-3), n 
Shigley’s MED, 11th edition
Sy
1   3

54
 0.97
51.1   4.58 
Ans.
Chapter 5 Solutions, Page 35/58
DE: From Eqs. (5-13) and (5-19)
S
Sy
54
n y 

1/2
1/2
   A2   A B   B2 
51.12  51.1 4.58    4.58  2 


 1.01 Ans.
______________________________________________________________________________
5-66
From Table A-20, Sy = 54 kpsi. From the solution of Prob. 3-97, in the plane of analysis
1 = 59.7 kpsi, 2 =  3.92 kpsi, and max = 31.8 kpsi
The state of stress is plane stress. Thus, the three-dimensional principal stresses are
1 = 59.7 kpsi, 2 = 0, and 3 =  3.92 kpsi
MSS: From Eq. (5-3), n 
Sy
1   3

54
 0.85
59.7   3.92 
Ans.
DE: From Eqs. (5-13) and (5-19)
S
Sy
54
n y 

1/2
1/2
2
2
   A   A B   B 
59.7 2  59.7  3.92    3.92  2 


 0.87
Ans.
______________________________________________________________________________
5-67
For Prob. 3-95, from Prob. 5-64 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. 3-98 solution which
considered stress concentrations for Prob. 3-95
1 = 53.0 kpsi, 2 =  8.48 kpsi, and max = 30.7 kpsi
DE: From Eqs. (5-13) and (5-19)
S
Sy
238

n y 
1/2
1/2
   A2   A B   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.
______________________________________________________________________________
5-68
Design Decisions Required:
Shigley’s MED, 11th edition
Chapter 5 Solutions, Page 36/58




Material and condition
Design factor
Failure model
Diameter of pin
Using F = 416 lbf from Ex. 5-3,
 max 
32M
d3
1
 32M  3
d 

  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
 max  y  y  81 000 psi
nd
1
1
 32(416)(15)  3
d 
  0.922 in
  (81 000) 
Choose preferred size of d  1.000 in
F
n
 (1)3 (81 000)
32(15)
 530 lbf
530
 1.27
416
Set design factor to nd  1.27
Adequacy Assessment:
 max 
Sy
nd

81 000
 63 800 psi
1.27
1
 32(416)(15)  3
d 
  1.00 in (OK)
  (63 800) 
Shigley’s MED, 11th edition
Chapter 5 Solutions, Page 37/58
 (1)3 (81 000)
 530 lbf
32(15)
530
n
 1.27 (OK)
416
______________________________________________________________________________
F
5-69
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. 3-106 the principal stresses are
1   2 
pd
p(15)

 60 p,
4t 4(0.0625)
3   p
From Eq. (5-12)
1
2
1

2
 
1/ 2
( 1   2 )2  ( 2   3 ) 2  ( 3   1 ) 2 
1/ 2
(60 p  60 p ) 2  (60 p  p )2  (  p  60 p) 2 
 61 p
For yield,  = Sy  61p = 57 (103)  p = 934 psi Ans.
For rupture, 61 p  68  p  1.11 kpsi Ans.
________________________________________________________________________
5-70
Given: AISI CD 1040 steel, ny = 2, OD = 50 mm, ID = 42 mm, L = 150 mm.
Table A-20, Sy = 490 MPa
At r = ri = 21 mm, Eq. (3-51) gives
ro2  ri 2
252  212
 t max  pi 2 2  pi 2 2  5.793 pi   1
25  21
ro  ri
 r max   pi   3
Closed end, Eq. (3-52) gives
p  21
p r2
 l  2 i i 2  i2
 2.397 pi   2
25  212
ro  ri
2
(a)  max 
1   3
2
 max 

Sy
2n y
(b) Eq. (5-12):
Shigley’s MED, 11th edition
5.793 pi    pi 
2

 3.397 pi
3.397 pi 
490
2  2

pi  36.1 MPa
Ans.
Chapter 5 Solutions, Page 38/58
1/ 2
  1   2  2    2   3  2    3   1  2 
  

2


1/ 2
  5.793  2.397 2   2.397  12   1  5.7932 

 pi  5.883 pi
2


S
490
  y
5.883 pi 
pi  41.6 MPa
Ans.


ny
2
______________________________________________________________________________
5-71
Given: AISI 1040 CD steel, OD = 50 mm, ID = 42 mm, L = 150 mm, pi = 40 MPa
Table A-20, Sy = 490 MPa
At r = ri = 21 mm, Eq. (3-51) gives
r2  r2
252  212
 t max  pi o2 i2  40 2 2  231.74 MPa   1
ro  ri
25  21
 r max   pi  40
MPa   3
Closed end, Eq. (3-52) gives
40  21
pi ri 2
l  2 2  2
 95.87 MPa   2
ro  ri
25  212
2
(a)  max 
1   3
ny 
2
Sy
2 max
(b) Eq. (5-12):
231.74   40 
 135.87 MPa
2
490

 1.80
Ans.
2 135.87 

1/ 2
  1   2  2    2   3  2    3   1  2 
  

2


1/ 2
  231.74  95.87 2   95.87  40 2   40  231.74 2 


2


 235.3 MPa
Sy
490

 2.08
Ans.
  235.3
_____________________________________________________________________________
ny 
5-72
For AISI 1020 HR steel, from Tables A-5 and A-20, w = 0.282 lbf/in3, Sy = 30 kpsi, and
 = 0.292. Then,  = w/g = 0.282/386 lbfs2/in. For the problem, ri = 3 in, and ro = 5 in.
Substituting into Eqs. (3-55), gives
Shigley’s MED, 11th edition
Chapter 5 Solutions, Page 39/58
t 
9  25  1  3  0.292  2 
0.282 2  3  0.292  
 
r 
 9  25  2 
386
8
3  0.292
r



225


 3.006 104   2  34  2  0.5699r 2   F  r   2
r


225


 r  3.006 104   2  34  2  r 2   G  r   2
r


(1)
(2)
For the distortion-energy theory, the von Mises stress will be
    t2   t r   22 
1/ 2
1/2
  2  F 2 (r )  F (r )G (r )  G 2 ( r ) 
(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 / 2 for 3  r  5 in. Plotting Eqs. (1) through (3) results in
70
60
50
'/2
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  = Sy gives
1/2




30 103 


 3.006 104  34  225  0.5699 32  
   
  

32

 1361 rad/s
60 60(1361)

 13 000 rev/min
Ans.
2
2
________________________________________________________________________
n
5-73
Since r/t = 1.75/0.065 = 26.9 > 10, we can use thin-walled equations. From Eqs. (3-53)
and (3-54),
Shigley’s MED, 11th edition
Chapter 5 Solutions, Page 40/58
di  3.5  2(0.065)  3.37 in
p(di  t )
2t
500(3.37  0.065)
t 
 13 212 psi  13.2 kpsi
2(0.065)
pd 500(3.37)
 6481 psi  6.48 kpsi
l  i 
4t
4(0.065)
 r   pi  500 psi  0.5 kpsi
t 
These are all principal stresses, thus, from Eq. (5-12),


1/ 2
1
2
2
2
13.2  6.48   6.48  (0.5)   0.5  13.2 
2
 11.87 kpsi
Sy
46
n

  11.87
n  3.88 Ans.
________________________________________________________________________
 
5-74
From Table A-20, S y  320 MPa
With pi = 0, Eqs. (3-49) are
 b2 
r2 p  r2 
 t   2o o 2 1  i 2   c 1  2 
ro  ri  r 
 r 
(1)
 b2 
ro2 po  ri 2 
 r   2 2 1  2   c 1  2 
ro  ri  r 
 r 
For the distortion-energy theory, the von Mises stress is
    t2   t r  

2 1/ 2
r
1/ 2
 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

b4 
 c 1  3 4 
r 

We see that the maximum von Mises stress occurs where r is a minimum at r = ri. Here,
r = 0 and thus  =  t . Setting  t = Sy = 320 MPa at r = 0.1 m in Eq. (1) results in
2ro2 po 2  0.15  po
Ans.
 t r  r  2 2 
 3.6 po  320  po  88.9 MPa
i
ro  ri
0.152  0.12
________________________________________________________________________
2
Shigley’s MED, 11th edition
Chapter 5 Solutions, Page 41/58
5-75
From Table A-24, Sut = 31 kpsi for grade 30 cast iron. From Table A-5,  = 0.211 and
w = 0.260 lbf/in3. In Prob. 5-72, 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) gives
1  3v 2  0.260 2  3.211  
1  3(0.211) 2 
 3 v  2
2
2
2
ri  
 t   2 
 
3 
  2ro  ri 
 2 5   3 
3  v  386
3.211
 8 
 8 

 
 0.01471 2  31 103

1452 rad/ sec
Ans.
n = 60(1452)/(2 ) = 13 870 rev/min
________________________________________________________________________
5-76
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.
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
Shigley’s MED, 11th edition
Chapter 5 Solutions, Page 42/58
 xz 
16T 16(1000) 5093

 3 psi
d
d3
d3
x 
32 M C 32(2110) 21 492


psi
d3
 d3
d3
For combined bending and torsion, the maximum shear stress is found from
1/2
 max
  x  2 2 
     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
________________________________________________________________________
5-77 As in Prob. 5-76, we will assume this to be a statics problem. Since the proportions are
unchanged, the bearing reactions will be the same as in Prob. 5-76 and the bending moment
will still be a maximum at point C. Thus
 max 

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 2(1055) 10 746
10.75
x 


psi 
kpsi
3
3
3
d
d
d
d3
Since the torsional stress is unchanged,
 xz 
5.09
kpsi
d3
For combined bending and torsion, the maximum shear stress is found from
1/2
 max
   2

  x    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. 5-76, gives
S
7.40
67
 max  y  3 

d  0.762 in
Ans.
2n
d
2  2
________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 5 Solutions, Page 43/58
5-78
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. 3-41, at outer radius,
     17.8 kpsi
Sy
32

 1.80
  17.8
At neutral axis,
n
   3 2  3  3.4   5.89 kpsi
2
Sy
32

 5.43
  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. 3-41,
     25.5 kpsi
n
Sy
32

 1.25
  25.5
Ans.
Model (e): From Prob. 3-41,
     17.8 kpsi
Sy
32

 1.80
Ans.
  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
5-79
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
32 M
1 
,  2  3  0
d3
Using a conservative yielding failure theory use the MSS theory and Eq. (5-3)
1   3 
Sy
n
S
32 M
 y
3
d
n

1/3

 32 Mn 
d 
  S 
y 

1/3
 32  312.5  2.5 
11
Thus, d  
in
Ans.
  0.629 in  Use d 
3

32
10
16




________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 5 Solutions, Page 44/58
5-80
When the ring is set, the hoop tension in the ring is equal to the screw tension.
t 
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  w t dr . Integration
yields
F 
ro
ri
wr 2 p
w t dr  2 i 2i
ro  ri

ro
ri
ro
 ro2 
ro2 
wri 2 pi 

dr

r

1


  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 d
Fx  
2
0

fpi wri d 
2 f T
0.2d
2
f Tw
ri  d
0
0.2d wri
Ans.
________________________________________________________________________
5-81 T = 20 Nm, Sy = 450 MPa
(a) From Prob. 5-80, T = 0.2 Fi d
20
T
Fi 

 16.7 103  N  16.7 kN
3

0.2d 0.2  6 10  


Ans.
(b) From Prob. 5-80, F =wri pi
16.7 103 
Fi
F
pi 


 111.3 106  Pa  111.3 MPa
wri wri 12 10 3    25 / 2  103  



Shigley’s MED, 11th edition
Ans.
Chapter 5 Solutions, Page 45/58


pi ri 2  ro2
ri 2 pi  ro2 
 t  2 2 1   
ro  ri 
r r r
ro2  ri 2
(c)
i


111.3 0.01252  0.0252
2
0.025  0.0125
2
  185.5 MPa
Ans.
 r   pi  111.3 MPa
(d)
 max 
1   3

t r
2
2
185.5  ( 111.3)

 148.4 MPa
2
 '  ( A2   A B   B2 )1/ 2
Ans.
1/ 2
 185.52  (185.5)(111.3)  ( 111.3) 2 
 259.7 MPa
Ans.
(e) Maximum Shear Stress Theory
n
Sy
2 max

450
 1.52
2 148.4 
Ans.
Distortion Energy theory
Sy
450

 1.73 Ans.
  259.7
________________________________________________________________________
n
Shigley’s MED, 11th edition
Chapter 5 Solutions, Page 46/58
5-82
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   r t w dr
ri

re 
w pi ri 2
ro2  ri 2

ro
ri

ro2 
r


 dr
r 

r 
w pi ri 2  ro2  ri 2
 ro2 ln o 
2
2 
ri 
F  ro  ri   2
From Prob. 5-80, F = wri pi. Therefore,
re 
ri  ro2  ri 2
r 
 ro2 ln o 
2
2 
ro  ri  2
ri 
For the conditions of Prob. 5-80, ri = 12.5 mm and ro = 25 mm
 252  12.52
12.5
25 
 252 ln
Ans.
  17.8 mm
2
2 
25  12.5 
2
12.5 
________________________________________________________________________
re 
5-83 (a) The nominal radial interference is nom = (2.002  2.001) /2 = 0.0005 in.
From Eq. (3-57),
2
2
2
2
E   ro  R  R  ri  


p
2R3 
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):
 12  0.6252 
R 2  ri 2
 t   p 2 2  3072  2
 7 010 psi
2 
R  ri
 1  0.625 
r =  p =  3072 psi
Shigley’s MED, 11th edition
Chapter 5 Solutions, Page 47/58
Eq. (5-13):
    A2   A B   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,
t  p
Eq. (3-49):
 1.52  12 
ro2  R 2
3072

 2 2   7 987 psi
ro2  R 2
 1.5  1 
r =  p =  3072 psi
Eq. (5-13):
    A2   A B   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: t = r =  p =  4167 psi
1/2
2
2
    4167    4167  4167    4167  


 4167 psi
Ans.
Outer member: pi = p = 4167 psi, po = 0. At fit surface r =R = 1 in,
 1.52  12 
ro2  R 2
 4167  2 2   10834 psi
t  p 2
ro  R 2
 1.5  1 
Eq. (3-49):
r =  p =  4167 psi
Eq. (5-13):
    A2   A B   A2 
1/ 2
1/ 2
 10 8342  10 834  4167    4167  
 13 410 psi
Ans.
________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 5 Solutions, Page 48/58
5-84 From Table A-5, E = 207 (103) MPa. The nominal radial interference is nom = (40 
39.98) /2 = 0.01 mm.
From Eq. (3-57),
2
2
2
2
E   ro  R  R  ri  


p
2 R3 
ro2  ri 2


207 103  0.01   32.52  202  20 2  10 2  

  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,
t   p
Eq. (3-49):
 202  102 
R 2  ri 2


 44.40 MPa
26.64
 2
2 
R 2  ri 2
 20  10 
r =  p =  26.64 MPa
Eq. (5-13):
    A2   A B   A2 
1/ 2
1/ 2
2
  44.40    44.40  26.64    26.64  


 38.71 MPa
Ans.
Outer member: pi = p = 26.64 MPa, po = 0. At fit surface r =R = 20 mm,
t  p
Eq. (3-49):
 32.52  202 
ro2  R 2
26.64

 59.12 MPa

2
2 
ro2  R 2
 32.5  20 
r =  p =  26.64 MPa
Eq. (5-13):
    A2   A B   A2 
1/ 2
1/ 2
 59.12 2  59.12  26.64    26.64  
 76.03 MPa
Ans.
________________________________________________________________________
5-85 From Table A-5, E = 207 (103) MPa. The nominal radial interference is nom = (40.008 
39.972) /2 = 0.018 mm.
From Eq. (3-57),
Shigley’s MED, 11th edition
Chapter 5 Solutions, Page 49/58
2
2
2
2
E   ro  R  R  ri  


p
2R3 
ro2  ri 2


207 103  0.018   32.52  20 2  202  10 2  

  47.94 MPa Ans.

32.52  10 2
2  203 


Inner member: pi = 0, po = p = 47.94 MPa. At fit surface r =R = 20 mm,
t   p
Eq. (3-49):
 202  102 
R 2  ri 2
47.94


 79.90 MPa
 2
2 
20
10
R 2  ri 2



r =  p =  47.94 MPa
Eq. (5-13):
    A2   A B   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,
t  p
Eq. (3-49):
 32.52  202 
ro2  R 2

 106.4 MPa
47.94

2
2 
ro2  R 2
 32.5  20 
r =  p =  47.94 MPa
Eq. (5-13):
    A2   A B   A2 
1/ 2
 106.4 2  106.4  47.94    47.94  
1/2
 136.8 MPa
Ans.
________________________________________________________________________
5-86 From Table A-5, for carbon steel, Es = 30 kpsi, and s = 0.292. While for Eci = 14.5 Mpsi,
and ci = 0.211. For ASTM grade 20 cast iron, from Table A-24, Sut = 22 kpsi.
For midrange values,  = (2.001  2.0002)/2 = 0.0004 in.
Eq. (3-50):
Shigley’s MED, 11th edition
Chapter 5 Solutions, Page 50/58
p


1
R
 Eo
 ro2  R 2
 1  R 2  ri 2



 2
 2 2  i  
o 
2
 ro  R
 Ei  R  ri

0.0004

 22  12

 12

1
1
1
0.211
0.292







2
2
2
6
6
14.5 10   2  1
 30 10   1
 
 2613 psi
At fit surface, with pi = p =2613 psi, and po = 0, from Eq. (3-50)
t  p
 22  12 
ro2  R 2
2613

 2 2   4355 psi
ro2  R 2
 2 1 
r =  p =  2613 psi
From Modified-Mohr theory, Eq. (5-32a), since A > 0 > B and  B /A  <1,
22
 5.05
Ans.
 A 4.355
________________________________________________________________________
n
5-87
Sut

E = 207 GPa
Eq. (3-57) can be written in terms of diameters,
E d
p
2 D3
 
3
 (d o2  D 2 )( D 2  di2 )  207 10 (0.062)  (502  452 )(452  40 2 ) 




3
( d o2  di2 )
(502  40 2 )
2  45 




p  15.80 MPa
Outer member: From Eq. (3-50),
Outer radius:  t o 
452 (15.80)
(2)  134.7 MPa
50 2  452
 r  o  0
Inner radius:  t i 
452 (15.80)  502 
1 
  150.5 MPa
502  452  452 
 r i  15.80 MPa
Bending (no slipping):
Shigley’s MED, 11th edition
I = ( /64)(504  404) = 181.1 (103) mm4
Chapter 5 Solutions, Page 51/58
At ro :
  x o  
Mc
75(0.05 / 2)

 93.2(106 ) Pa  93.2 MPa
9
I
181.1 10
At ri :
 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 :
 

Tc 900(0.05 / 2)

 62.1 106 Pa  62.1 MPa
J 362.2 109
At ri :
 

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,
 x  93.2 MPa,  y  134.7 MPa,  xy  62.1 MPa
    x2   x y   y2  3 xy2 
1/ 2
Eq. (5-15)
1/ 2
2
2
  93.2    93.2 134.7   134.7 2  3  62.1 


S y 415
no 

 1.84 Ans.
  226
 226 MPa
Inner radius, 3D state of stress
From Eq. (5-14) with yz = zx = 0 and x = + 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
 

2
With x =  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
 
2
Shigley’s MED, 11th edition
Chapter 5 Solutions, Page 52/58
S y 415

 1.80 Ans.
  230
________________________________________________________________________
5-88 From the solution of Prob. 5-87, p = 15.80 MPa
ni 
Inner member: From Eq. (3-50),
Outer radius:  t o  
ro2  ri 2
452  402


p
(15.80)  134.8 MPa
o
ro2  ri 2
452  40 2
 r o   p  15.80 MPa
Inner radius:  t i
 
2 452
2ro2
  2 2 po   2
(15.80)  150.6 MPa
ro  ri
45  402
 r i  0
Bending (no slipping): I = ( /64)(504  404) = 181.1 (103) mm4
Mc
75(0.045 / 2)
 83.9(106 ) Pa  83.9 MPa
At ro :
  x o    
9
I
181.1 10

At ri :
  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 :
 

Tc 900(0.045 / 2)

 55.9 106 Pa  55.9 MPa
9
J
362.2 10
At ri :
 

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
Shigley’s MED, 11th edition
Chapter 5 Solutions, Page 53/58
From Eq. (5-14) with yz = zx = 0 and x = + 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

2
With x =  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
 

2
S y 415
no 

 1.95 Ans.
  213
Inner radius, plane stress. Worst case is when x is positive
 x  74.5 MPa,  y  150.6 MPa,  xy  49.7 MPa
Eq. (5-15)
    x2   x y   y2  3 xy2 
1/ 2
2 1/2
  74.52  74.5  150.6    150.6   3  49.7    216 MPa


S y 415
ni 

 1.92 Ans.
  216
______________________________________________________________________________
2
5-89
For AISI 1040 HR, from Table A-20, Sy = 290 MPa.
From Prob. 3-124, 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
 t  p o2
65.2
ro  R 2
502  252
 r   p  65.2 MPa
These are principal stresses. From Eq. (5-13)
 o   t2   t r   r2 
1/ 2
1/ 2
2
 108.7 2  108.7  65.2    65.2  


 152.2 MPa
Sy
290

 1.91 Ans.
 o 152.2
________________________________________________________________________
n
5-90
For AISI 1040 HR, from Table A-20, Sy = 42 kpsi.
Shigley’s MED, 11th edition
Chapter 5 Solutions, Page 54/58
From Prob. 3-125, pmax = 9 kpsi. From Eq. (3-50) at the inner radius R of the outer
member,
ro2  R 2
22  12
t  p 2
 9 2 2  15 kpsi
ro  R 2
2 1
 r   p  9 kpsi
These are principal stresses. From Eq. (5-13)
 o   t2   t r   r2 
1/ 2
2
 152  15( 9)   9  


1/ 2
 21 kpsi
S y 42

 2 Ans.
 o 21
________________________________________________________________________
n
5-91
For AISI 1040 HR, from Table A-20, Sy = 290 MPa.
From Prob. 3-126, pmax = 91.6 MPa. From Eq. (3-50) at the inner radius R of the outer
member,
r 2  R2
502  252
 t  p o2
91.6

 152.7 MPa
ro  R 2
50 2  252
 r   p  91.6 MPa
These are principal stresses. From Eq. (5-13)
 o   t2   t r   r2 
1/2
2
 152.7 2  152.7( 91.6)   91.6  


1/ 2
 213.8 MPa
Sy
290

 1.36 Ans.
 o 213.8
________________________________________________________________________
n
5-92
For AISI 1040 HR, from Table A-20, Sy = 42 kpsi.
From Prob. 3-127, pmax = 12.94 kpsi. From Eq. (3-50) at the inner radius R of the outer
member,
ro2  R 2
22  12
t  p 2
 12.94 2 2  21.57 kpsi
ro  R 2
2 1
 r   p  12.94 kpsi
These are principal stresses. From Eq. (5-13)
 o   t2   t r   r2 
1/ 2
Shigley’s MED, 11th edition
2
  21.57 2  21.57( 12.94)   12.94  


1/ 2
 30.20 kpsi
Chapter 5 Solutions, Page 55/58
Sy
42

 1.39 Ans.
 o 30.2
________________________________________________________________________
5-93 For AISI 1040 HR, from Table A-20, Sy = 290 MPa.
n
From Prob. 3-128, pmax = 134 MPa. From Eq. (3-50) at the inner radius R of the outer
member,
r 2  R2
502  252
 t  p o2
134

 223.3 MPa
ro  R 2
502  252
 r   p  134 MPa
These are principal stresses. From Eq. (5-13)
 o   t2   t r   r2 
1/ 2
1/ 2
2
  223.32  223.3(134)   134  


 312.6 MPa
Sy
290

 0.93 Ans.
 o 312.6
________________________________________________________________________
n
5-94
For AISI 1040 HR, from Table A-20, Sy = 42 kpsi.
From Prob. 3-129, pmax = 19.13 kpsi. From Eq. (3-50) at the inner radius R of the outer
member,
ro2  R 2
22  12
t  p 2
 19.13 2 2  31.88 kpsi
ro  R 2
2 1
 r   p  19.13 kpsi
These are principal stresses. From Eq. (5-13)
 o   t2   t r   r2 
1/2
2
 31.882  31.88(19.13)   19.13 


1/ 2
 44.63 kpsi
Sy
42

 0.94 Ans.
 o 44.63
________________________________________________________________________
n
5-95
   K


1 K
3 
 p   2 I cos    I sin cos sin 
2
2   2 r
2
2
2 
2 r
2
2


3  
 KI
sin cos cos  

2
2
2  
 2 r
Shigley’s MED, 11th edition
1/2
Chapter 5 Solutions, Page 56/58
1/2
   2

2
2 3
2
2
2 3 


cos
sin
cos
sin
sin
cos
cos


 
2 
2
2
2
2
2
2  






KI 
KI

cos 1  sin 
 cos  sin cos  
2
2
2
2
2
2 r 
2 r

KI
2 r
Plane stress: The third principal stress is zero and
KI
KI




cos 1  sin  ,  2 
cos 1  sin  ,  3  0 Ans.
1 
2
2
2
2
2 r
2 r
Plane strain: Equations for 1 and2 are still valid,. However,
KI

Ans.
cos
2
2 r
________________________________________________________________________
 3    1   2   2
5-96
For  = 0 and plane strain, the principal stress equations of Prob. 5-95 give
1   2 
(a) DE: Eq. (5-18)
or,
KI
KI
,  3  2
 2 1
2 r
2 r
1 
2
2
2 1/ 2
 1   1    1  2 1    2 1   1    S y


2
1  21 = Sy
1
For   ,
3

 1 
1  2  3    1  S y
 

(a) MSS: Eq. (5-3) , with n =1

 1  3S y
1  3 = Sy

1
3


Ans.
1  21 = Sy
 1  3S y
Ans.
2
3
 3   1  S y  2S y   3   1
Radius of largest circle
1
2  
R  1  1   1
2
3  6
________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 5 Solutions, Page 57/58
5-97
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,  = 1.3
K I    a
Eq. (5-37)
80  1.3
F
 (16)103
100(12)
F = 329.4(103) N = 329.4 kN
Ans.
________________________________________________________________________
5-98
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:
Eq. (5-37):
  2.4
K Ic    a
ri 6
  0.857
ro 7

72  2.4   0.5

  23.9 kpsi
Eq. (3-50) at r = ro = 7 in:
ri 2 pi
62 pi
 t  2 2  2   23.9  2 2  2   pi  4.315 kpsi
Ans.
ro  ri
7 6
________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 5 Solutions, Page 58/58
Chapter 7
7-1
Eq. (7-6)
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
2
(a) DE-Goodman, Eq. (7-8):
 16(2)  338.4
265.5  




   210  106  700  10 6   



3
d = 27.27 (10 ) m = 27.27 mm Ans.
 16n  A B
d 
 
   Se Sut


 
1/3
1/3
(b) DE-Morrow, Eq. (7-10):
 16(2)  338.4
265.5  




   210  106  1045  106   



3
d = 26.68 (10 ) m = 26.68 mm Ans.
 16n  A B  
d 
 

   Se  f  
1/3
1/3
(c) DE-Gerber, Eq. (7-12):
1/3
2 1/ 2  


 8nA    2 BSe    
1 1 
d 
  
  Se    ASut    



 
6
 8(2)(338.4)    2(265.5)  210  10

1 1 
6 
6

   210  10    338.4  700  10


3
d = 25.85 (10 ) m = 25.85 mm Ans.
  
  
 
2 1/ 2




1/3





(d) DE-SWT, Eq. (7-14):
1/3
1/2 
 16  2 
1/2 
 16n 2
 338.4  2  338.4  265.5   
d 
A  AB   

6 
 
   210  10
  Se


d = 27.99 (103) m = 27.99 mm Ans.
________________________________________________________________________
Criterion
d (mm)
Compared to DE-Goodman
1/3
DE-Goodman
DE-Morrow
DE-Gerber
Shigley’s MED, 11th edition
27.27
26.68
25.85
2.2% Lower
5.2% Lower
Less conservative
Less conservative
Chapter 7 Solutions, Page 1/48
DE-SWT
27.99
2.6% Higher
More conservative
______________________________________________________________________________
7-2
Given: AISI 1050 CD steel, Ma = 650 lbf ۰ in, Mm = 500 lbf ۰ in, Ta = 400 lbf ۰ in, Tm =
300
lbf ۰ in, Se = 30 kpsi, Kf = 2.3, Kfs = 1.9, ny = 2.5. Table A-20, Sy = 77 kpsi, Sut = 100 kpsi.
Eq. (7-6):
A  4  K f M a   3  K fsTa   4  2.3  650    3  1.9  400   3266.9 lbf in
2
2
2
2
B  4  K f M m   3  K fsTm   4  2.3  500    3  1.9  300   2502.9 lbf in
(a) DE-Goodman criterion, Eq. (7-8):
2
 16n  A B  
d 
 

   Se Sut  
1/3
 16n  A B  
d 
 

 
   Se  f  
2
2
 16(2.5)  3266.9
2502.9  





   30  106  100  106   



(b) DE-Morrow criterion,
1/3
2
 f
1/3
0.119 in
Ans.
= 50 + 100 = 150 kpsi, Eq. (7-10):
 16(2.5)  3266.9
2502.9  





   30  106  150  106   



1/3
0.117 in
Ans.
(c) DE-Gerber criterion, Eq. (7-12):
1/3
2 1/ 2  


 8nA    2 BSe    
d 
1 1 
  
  Se    ASut    


1/3

2 1/2 
 
6



 8  2.5  3266.9    2  2502.9  30  10    


1  1  
 0.113 in
6
6
   30  10    3266.9  100  10   

 
 
(d) DE-SWT criterion, Eq. (7-14):
1/2 
 16n 2
d 
A  AB  

  Se

Ans.
1/3
1/3
 16  2.5 

1/2 
 3266.92  3266.9  2502.9    0.123 in

Ans.
6 
   30  10

______________________________________________________________________________
7-3
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.
Shigley’s MED, 11th edition
Chapter 7 Solutions, Page 2/48
Eq. (6-18):
ka 2.00(175)  0.217 0.65
Trial #1: Choose dr = 0.75 in
Eq. (6-19
):
Eq. (6-10):
Eq. (6-17):
kb 0.879(0.75)  0.107 0.91
Se 0.5Sut 0.5  175  87.5 kpsi
Se = 0.65 (0.91)(87.5) = 51.8 kpsi
d r d  2r 0.75 D  2 D / 20 0.65D
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
dr
0.75
r 0.058

0.077
dr
0.75
Kt = 1.9
Fig. 6-26:
r = 0.058 in, q = 0.90
Eq. (6-32):
Kf = 1 + 0.90 (1.9 – 1) = 1.81
Fig. A-15-15:
Kts = 1.5
Fig. 6-27:
r = 0.058 in, qs = 0.92
Eq. (6-32):
Kfs = 1 + 0.92 (1.5 – 1) = 1.46
For the DE-Goodman criteria, Eqs. (7-6) and (7-8), with d as dr, Ma = 600 lbf ۰ in,
Tm = 400 lbf ۰ in, and Mm = Ta = 0,
 16n  1
d r 
  S
 e

 16  2.5 

 

0.847 in
2 1/ 2
 4 K M  
f
a


1

Sut
2 1/2
 3 K T  
fs m



 

1/3


1
1
2 1/ 2
2 1/2 




4
1.81

600

3
1.46

400

 

  

3 
3 

51.8
10
175
10

 
 
1/3
 
Ans.
Trial #2: Choose dr = 0.85 in.
kb 0.879(0.85)  0.107 0.894
Shigley’s MED, 11th edition
Chapter 7 Solutions, Page 3/48
Se = 0.65 (0.894)(0.5)(175) = 50.8 kpsi
d
0.85
D r 
1.31 in
0.65 0.65
r = D / 20 = 1.31/20 = 0.0655 in
Figs. A-15-14 and A-15-15:
d d r  2r 0.85  2(0.0655) 0.981 in
d 0.981

1.15
dr
0.85
r 0.0655

0.077
dr
0.85
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,
 K f 1.81, K fs 1.46
qs 0.92
Using Eq. (7-8) produces dr = 0.852 in. Further iteration produces no change. With
dr = 0.835 in,
d
0.852
D r 
1.31 in
0.65 0.65
d 0.75 D 0.75(1.31) 0.983 in Ans.
______________________________________________________________________________
7-4
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. 6-26 and 6-27 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.
Eq. (7-15):
2
  32 K f M a  2
 16 K fsTm  
  
 max
  3
 
3
3
   d

  d  
Shigley’s MED, 11th edition
1/ 2
Chapter 7 Solutions, Page 4/48
Eq. (7-16):
Solving for d,
Sy
 d 3S y
n


 max
16
 4  K M  2  3 K T  2 
f
a
fs m


 1/2
1/3
 16n
1/2 

 4( K f M a ) 2  3( K fsTa ) 2  
d 
  S y

 16(2.5)

   420  106

 
4  (2.4)(482.4)   3  (2.1)(340) 
2
d = 0.0430 m = 43.0 mm

2 1/2




1/3
Ans.
ka 3.04(560) 0.217 0.77
(b) Eq. (6-18):
Assume kb = 0.85 for now. Check later once a diameter is known.
Se = 0.77(0.85)(0.5)(560) = 183 MPa
Selecting the DE-Goodman criteria, use Eqs. (7-6) and (7-8) with M m Ta 0.
 16n  1
d 
  S
 e

4 K M 2
  f a  
1/2
1

Sut
3 K T 2
  fs m  
1/2
1/3

 

 16  2.5  

1
1
2 1/2
2 1/2 








4  2.4 482.4 

3  2.1 340 

6 
6 

 
 
183
10
560
10

 

0.0574 m 57.4 mm
Ans.
 
1/3
 
With this diameter, we can refine our estimates for kb and q.
Eq. (6-19
):
kb 1.51d  0.157 1.51 57.4 
 0.157
0.80
Assuming a sharp fillet radius, from Table 7-1, r = 0.02d = 0.02 (57.4) = 1.15 mm.
Fig. (6-26):
Fig. (6-27):
q = 0.72
qs = 0.77
Iterating with these new estimates,
Eq. (6-32):
Eq. (6-18):
Eq. (7-8):
Kf = 1 + 0.72 (2.7 – 1) = 2.2
Kfs = 1 + 0.77(2.2 – 1) = 1.9
Se = 0.77(0.80)(0.5)(560) = 172.5 MPa
d = 57 mm
Ans.
Shigley’s MED, 11th edition
Chapter 7 Solutions, Page 5/48
Further iteration does not change the results.
_____________________________________________________________________________
7-5
Given: AISI 1045 CD steel, Se = 40 kpsi, Kf = 2.1, Kfs = 1.7, nf = 2.5.
Table A-20, Sy = 77 kpsi, Sut = 91 kpsi.
(MO)y = 0 =  18 RCz + 12(300)
RCz = 200 lbf
(MO)z = 0 =  18 RCy + 6(600)
RCy = 200 lbf
(MB)y =  6(200) =  1200 lbf ۰ in,
(MB)z = 6(200) = 1200 lbf ۰ in
MB 
  1200 
2
 12002 1697 lbf in
Mm = 0, Ma = 1697 lbf ۰ in, Tm = Ta = 1800/2 = 900 lbf ۰ in
(a) DE-Gerber criterion, Eqs. (7-6) and Eq. (7-12):
A  4  2.1 1697    3 1.7  900   7604 lbf in
2
2
B  3 1.7  900   2650 lbf in
2
1/3

2 1/2 
 
3



 8  2.5  7604    2  2650  40  10    

d 
1 1 
 1.353 in
3 
3
   40  10    7604  91 10   

 
 
(b) DE-Goodman criterion, Eq. (7-6) and Eq. (7-8):
Ans.
1/3
 16  2.5   7604
2650  

  1.408 in
d 

Ans.
 40  103  91 103   
 



______________________________________________________________________________
7-6
(a) Let Kfb = (Kf)bending, Kfa = (Kf)axial, Kfs = (Kfs)torsion.
Eqs. (6-66) and (6-67):
2

32M a
4 Fa 
16Ta

 a   K fb

K
 3  K fs
fa
3
2 
d
d 
d3

 


 
1/2
1/2
2

32M m
4Fm 
16Tm  

 m   K fb

K
 3  K fs

fa
3
2 
d
d 
 d 3  

 
For the DE-Goodman criterion, substitute stresses into Eq. (6-41),
2
2 1/2
1 


8
K
M

K
F
d

48
K
T




fb
a
fa
a
fs
a



 d 3  Se 

nf 


1/2
2
2
4  1 


 8K fb M m  K fa Fm d   48  K fsTm   
 Sut 

(b) For the free-body diagram in the problem statement,
 Fx 0  FCx 173 lbf
Shigley’s MED, 11th edition
1
Ans.
Chapter 7 Solutions, Page 6/48
  M  0 7 F  3.75  173  3.5  58  F 121.68 lbf
  M  0  7 F  3.5  500   F 250 lbf
  T  0 3.75  500   T  T 1875 lbf in
A y
A z
z
C
z
C
y
C
y
C
A x
M m 0,
M a M B 3.5 121.682  2502 973.1 lbf in
Tm = 1875 lbf ۰ in, Ta = 0, Fm = 173 lbf, Fa = 0
From part (a), with d = 1.5 in,

2 1/2
1

8
2
973.1

0









3
  1.53   40  10 
nf 

2
2
1
4 

 2.3  173 1.5   48  1.5  1875  
3
 90  10 

(c) Without axial force, use Eq. (7-7):




2
1
16 
1

4  2  973.1 

3 
3
n   1.5   40  10 


1/ 2

2
1
3 1.5  1875  

3
90  10 


 1



 4.38
1/2 




Ans.


 4.38


Ans.
1/2
Axial force is too small to add any appreciable difference.
______________________________________________________________________________
7-7
From Prob. 7-6, Ma = 973.1 lbf ۰ in, Tm = 1875 lbf ۰ in, Ta = 0. From Eqs. (7-6) and (7-8):
1/3

 16  2.5  
2 1/2
2 1/2 
1
1

  1.245 in Ans.
d 
4  2  973.1 
3  1.5  1875  


   40  103  
90  103  


______________________________________________________________________________
7-8
From Chap. 13, Eq. (13-1), with N = number of gear teeth and d = gear pitch diameter,
the gear pitch is P = N/d = 90/6 = 15 teeth/in.
Table 7-2, yall = 0.005 in. Eq. (7-17),
1/4
1/4
1.5  0.007 
nd yold
d new dold
0.75
0.903 in
Ans.
yall
0.005
______________________________________________________________________________

7-9



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
Shigley’s MED, 11th edition
Chapter 7 Solutions, Page 7/48
T FCz (2) 96(2) 192 lbf in
T 192
FBz 

128 lbf
1.5 1.5
FBy FBz tan 20 128 tan 20 46.6 lbf
(a) xy-plane
M 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
M 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:
xz-plane
M O 0 96(5.75)  FAz (11.5)  128(14.25)
96(5.75) 128(14.25)
FAz 
206.6 lbf
11.5
M A 0 FOz (11.5)  128(2.75)  96(5.75)
Shigley’s MED, 11th edition
Chapter 7 Solutions, Page 8/48
96(5.75)  128(2.75)
17.4 lbf
11.5
Bending moment diagram:
FOz 
M C  1002  ( 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
Bending moment diagram:
1
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:
Shigley’s MED, 11th edition
Chapter 7 Solutions, Page 9/48
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.
______________________________________________________________________________
7-10
This is a design problem, which can have many acceptable designs. See the solution for
Prob. 7-22 for an example of the design process.
______________________________________________________________________________
7-11
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. 7-9, 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  
Shigley’s MED, 11th edition
(1)
Chapter 7 Solutions, Page 10/48
From (1),
y 0 at x 0

C2 0
y 0 at x 11.5

C1 1908.4 lbf in 3
x = 0:
x = 11.5:
EIy' = 1908.4
EIy' = –2153.1
xz plane (treating z   )
 
17.4 2
206.6
3
3
2
EIz  
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
(2)
 
z 0 at x 0

C4 0
From (2),
z 0 at x 11.5 
C3 8.975 lbf in 3
x = 0:
EIz' = 8.975
x = 11.5:
EIz' = –683.5
At O:
EI  1908.42  8.9752 1908.4 lbf in 3
At A:
EI  ( 2153.1) 2  ( 683.5) 2 2259.0 lbf in 3

n
(dictates size)
2259
0.000 628 rad
30  10    / 64   1.254 
6
0.001
1.59
0.000 628
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
yB / A 

(2.75)[2.75  2(2.75)]  176.2 / EI 2
2 EI 2
2 EI 2
2153.1
176.2
 yB  yA  yB / A 

6
4
6
30 10   / 64  1.25
30 10   / 64  0.8754
 


 


= –0.000 803 rad (magnitude greater than 0.0005 rad)
Shigley’s MED, 11th edition
Chapter 7 Solutions, Page 11/48
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   / 64  1.254 30 106   / 64  0.8754
 
6


 


 B  ( 0.000 803) 2  ( 0.000 751) 2 0.00110 rad
Crowned teeth must be used.
Finite element results:
O 5.47(10 4 ) rad
Error in simplified model
3.0%
4
 A 7.09(10 ) rad
11.4%
3
 B 1.10(10 ) rad
0.0%
The simplified model yielded reasonable results.
Strength
Sut 72 kpsi, S y 39.5 kpsi
At the shoulder at A, x 10.75 in. From Prob. 7-9,
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 2.00(72)  0.217  0.791
 0.107
 1 
kb 
0.879

 0.3 
kc kd ke k f 1
Se 0.791(0.879)(36) 25.0 kpsi
Fig. A-15-8:
Fig. A-15-9:
Fig. 6-26:
D / d = 1.25, r / d = 0.03
Kts = 1.8
Kt = 2.3
q = 0.65
Shigley’s MED, 11th edition
Chapter 7 Solutions, Page 12/48
Fig. 6-27:
Eq. (6-32):
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-Goodman, Eqs. (7-6) and (7-7) with M m Ta 0,


B  3 1.56  192   
A  4  1.85  360  
2 1/2
2 1/2
1332 lbf in
518.8 lbf in
  13   1332
518.8 


n

16  25  103  72  103  


1
n = 3.25
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.
______________________________________________________________________________
7-12 through 7-21
These are design problems, which can have many acceptable designs. See the solution for
Prob. 7-22 for an example of the design process.
______________________________________________________________________________
7-22 (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  1250 tan 20 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
Shigley’s MED, 11th edition
Chapter 7 Solutions, Page 13/48
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.
(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, 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. 6Shigley’s MED, 11th edition
Chapter 7 Solutions, Page 14/48
26 and 6-27, 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. (6-18):
ka 2.00(68)  0.217 0.801
Eq. (6-20)
For estimating kb , guess d 2 in.
kb (2 / 0.3) 0.107 0.816
Eq. (6-17)
Se 0.801(0.816)(0.5)(68) 22.2 kpsi
Selecting the DE-Goodman criteria for a conservative first design,
Eq. (7-8) with Eq. (7-6):
1/2
2 1/ 2



 3 K T  2   
 16n   4  K f M a  
fs m

  
d 



Se
Sut
  
 


 


1/3

  4 1.9 2178 2  1/2  3 2.6 2500 2  1/2  
 
  
 16(1.5)   
 
d 



3
3

22.2
10
68
10







 
d 1.60 in
1/3
 
Ans.
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.
Eq. (7-15):
2
  32 K f M a  2
 16 K fsTm  
  
 max
  3
 
3
3
d

   d


 
1/2
2
  32(1.9)(2178)  2
 16(2.6)(2500)  
  
 max
  3
 
3
3
   (1.60)

  (1.60)
 
 57 /17.4 3.3 Ans.
n y S y /  max
1/2
17 374 psi 17.4 kpsi
(e) Now estimate other diameters to provide typical shoulder supports for the gear and
bearings. Also, estimate the gear and bearing widths.
Shigley’s MED, 11th edition
Chapter 7 Solutions, Page 15/48
(f) Entering this shaft geometry into beam analysis software (or Finite Element software),
the following deflections are determined:
Left bearing slope:
0.000 493 rad
Right bearing slope:
 0.000 788 rad
Gear slope:
 0.000 505 rad
Right end of shaft slope:
 0.000 788 rad
Gear deflection:
 0.000 927 in
Right end of shaft deflection:
0.005 73 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 1.60 in to 1.75 in, produces a gear slope of  0.000 353 rad. All other deflections
are improved as well.
______________________________________________________________________________
7-23
(a) Use the DE-Goodman fatigue criterion. The torque and moment loadings on the shaft
are shown in the solution to Prob. 7-22.
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. (6-10):
Se 0.5(68) 34.0 kpsi
Eq. (6-18):
ka 2.00(68)  0.217 0.801
kc k d ke 1
Left keyway
See Table 7-1 for keyway stress concentration factors,
Shigley’s MED, 11th edition
Chapter 7 Solutions, Page 16/48
K t 2.14 
 Profile keyway
K ts 3.0 
For an end-mill profile keyway cutter of 0.010 in radius, estimate notch sensitivities.
Fig. 6-26:
q 0.51
Fig. 6-27:
qs 0.57
Eq. (6-32):
K fs 1  qs ( Kts  1) 1  0.57(3.0  1) 2.1
K f 1  0.51(2.14  1) 1.6
 0.107
Eq. (6-19):
 1.875 
kb 

 0.30 
Eq. (6-17):
Se 0.801(0.822)(34.0) 22.4 kpsi
0.822
A  4  (1.6)(2178)  6969.6
2
Eq. (7-6):
B  3  (2.1)(2500)  9093.3
2
Eq. (7-7):
 (1.8753 )  6969.6
9093.3 


nf 

 22.4  103  68  103  
16


nf = 2.9
1
Ans.
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:
Kt 2.4
Fig. A-15-8:
Kts 1.6
Fig. 6-26:
q 0.65
Fig. 6-27:
qs 0.70
Eq. (6-32):
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 
A  4  (1.9)(493) 1873.4
2
Eq. (7-6):
Shigley’s MED, 11th edition
Chapter 7 Solutions, Page 17/48
B  3  (1.42)(2500)  6148.8
2
Eq. (7-7):
 (1.5743 )  1873.4
6148.8 


nf 

 22.4  103  68  103  
16


nf = 4.4
Ans.
1
Right keyway
Use the same stress concentration factors as for the left keyway. There is no bending
moment, so there is no alternating stress, so fatigue is not predicted at this location
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,
2
  32 K f M a  2
 16 K fsTm  
  
 max
  3
 
3
3
   d

  d  
1/ 2
2
  32 1.6 2178  2
 16  2.1  2500   




  3
 
 
   1.875  3  
    1.875  3 


 

1/ 2
8791 psi 8.79 kpsi
S
37.5
ny  y 
4.3

 max
8.79
Ans.
Check in smaller diameter at right end of shaft where only steady torsion exists.
  16 K fsTm  2 
  3 
 max
 
3
   d  
1/2
  16 2.1 2500  2 
 
 
 3 
3
 
 
  1.5 
 
 
13 722 psi 13.7 kpsi
S
37.5
ny  y 
2.7

 max
13.7
1/2
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,
Shigley’s MED, 11th edition
Chapter 7 Solutions, Page 18/48
 AB 
d y AB
Fb
1449(2)(3 x 2  2 2  112 )

(3 x 2  b 2  l 2 ) 
dx
6 EIl
6(30)(106 )( / 64)(1.8754 )(11)
2.4124(10 6 )(3 x 2  117)
At x = 0 in:
  2.823(10 4 ) rad
4
At x = 9 in:  3.040(10 ) rad
To the right of the load, from Table A-9, case 6,
 BC 

d yBC
Fa

 3 x 2  6 xl  2l 2  a 2
dx
6 EIl

At x = l = 11 in:
Fa 2
1449(9)(112  9 2 )

l  a2 
4.342(10 4 ) rad
6
4
6 EIl
6(30)(10 )( / 64)(1.875 )(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 
7-24
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. 3-83, 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
Shigley’s MED, 11th edition
Chapter 7 Solutions, Page 19/48
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. (6-10):
Sut = 64 kpsi, Sy = 54 kpsi
Se 0.5(64) 32.0 kpsi
Eq. (6-18):
ka 2.00(64)  0.217 0.811
kc k d ke 1
Keyway at A
Assuming r / d = 0.02 for typical end-milled keyway cutter (see discussion prior to Table
7-1), with d = 1.75 in, r = 0.02d = 0.035 in.
Table 7-1:
Kt = 2.14, Kts = 3.0
Fig. 6-26:
q = 0.65
Fig. 6-27:
qs = 0.71
K f 1  q  K t  1 1  0.65(2.14  1) 1.7
Eq. (6-32):
K fs 1  qs ( K ts  1) 1  0.71(3.0  1) 2.4
 0.107
Eq. (6-19):
 1.75 
kb 

 0.30 
Eq. (6-17):
Se 0.811(0.828)(32) 21.5 kpsi
0.828
We will choose the DE-Gerber criteria since this is an analysis problem in which we
would like to evaluate typical expectations.
Using Eqs. (7-6) and (7-11) 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
Shigley’s MED, 11th edition
2
Chapter 7 Solutions, Page 20/48
2 1/2 

1
8 A    2 BSe   

1   1  
  
n  d 3 Se    ASut   
 
 
2 1/2 
 
8  18.10 
   2  11.72   21.5    


1  1  

  1.753   21.5      18.10   64    




n = 1.2
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. 6-26:
Fig. 6-27:
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
Eq. (6-32):
K fs 1  qs ( K ts  1) 1  0.76(1.8  1) 1.6
 0.107
Eq. (6-19):
 1.75 
kb 

 0.30 
Eq. (6-17):
Se 0.811(0.828)(32) 21.5 kpsi
0.828
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 Eqs. (7-6) and (7-11) 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 
 
 2 BSe   

1   1  
  
AS

ut

  
 

2 1/2 
 
 


8  25.65 
2
7.812
21.5




 



1
1

  

  1.753   21.5      25.65  64    
 
 
1
8A
 3
n  d Se
n = 0.87
Shoulder to the right of C
r / d = 0.625 / 1.3 = 0.048, D / d = 1.75 / 1.3 = 1.35
Fig. A-15-9:
Kt = 2.0
Shigley’s MED, 11th edition
Chapter 7 Solutions, Page 21/48
Fig. A-15-8:
Fig. 6-26:
Fig. 6-27:
Kts = 1.7
q = 0.71
qs = 0.76
K f 1  q  K t  1 1  0.71(2.0  1) 1.7
Eq. (6-32):
K fs 1  qs ( K ts  1) 1  0.76(1.7  1) 1.5
 0.107
Eq. (6-19):
 1.3 
kb 
0.855

 0.30 
Se 0.811(0.855)(32) 22.2 kpsi
Eq. (6-17):
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 Eqs. (7-6) and (7-11) with Mm = Ta = 0,
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  d 3 Se    ASut   
 
 
2 1/2 

   2  7.324   22.2    


1  1  

  1.33   22.2      22.95   64    
 
 
8  22.95 
n = 0.41
The critical location is at the shoulder to the right of C, where n = 0.41 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,
2
  32 K f M a  2
 16 K fsTm  
  
 max
  3
 
3
3
   d

  d  
1/2
2
  32 1.7 6750  2
 16  1.5   2819   




  3
 
 
3
3


 

  1.3
  1.3


 

1/2
55 845 psi 55.8 kpsi
S
54
n y 
0.97

 max
55.8
Shigley’s MED, 11th edition
Chapter 7 Solutions, Page 22/48
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)
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.
______________________________________________________________________________
7-25
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. 7-24, and the loading as defined in Prob.
3-83, the following deflection magnitudes are determined:
Location
Left bearing O
Right bearing C
Left Gear A
Right Gear B
Slope
(rad)
0.0064
0
0.0043
4
0.0026
0
0.0107
8
Deflectio
n
(in)
0.00000
0.00000
0.04839
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. 7-24
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:
Shigley’s MED, 11th edition
Location
Slope
(rad)
Left bearing O
Right bearing C
0.00030
0.00020
Deflectio
n
(in)
0.00000
0.00000
Chapter 7 Solutions, Page 23/48
Left Gear A
0.00012
0.00225
Right Gear B
0.00050
0.00350
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.
______________________________________________________________________________
7-26
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. 3-84, 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 (see discussion prior to Table
7-1) with d = 50 mm, r = 0.02d = 1 mm.
Table 7-1:
Fig. 6-26:
Fig. 6-27:
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
Eq. (6-32):
K fs 1  qs ( K ts  1) 1  0.72(3.0  1) 2.4
Shoulder to the left of B
Shigley’s MED, 11th edition
Chapter 7 Solutions, Page 24/48
r / d = 2 / 50 = 0.04, D / d = 75 / 50 = 1.5
Fig. A-15-9:
Fig. A-15-8:
Fig. 6-26:
Fig. 6-27:
Kt = 2.2
Kts = 1.8
q = 0.73
qs = 0.78
K f 1  q  K t  1 1  0.73(2.2  1) 1.9
Eq. (6-32):
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. (6-10):
Se 0.5(440) 220 MPa
Eq. (6-18):
ka 3.04(440) 0.217 0.811
 0.107
Eq. (6-19):
Eq. (6-17):
 50 
kb 
0.818

 7.62 
kc k d ke 1
Se 0.811(0.818)(220) 150 MPa
We will choose the DE-Gerber criteria since this is an analysis problem in which we
would like to evaluate typical expectations. Using Eqs. (7-6) and (7-11) 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
BS
1
8A 
e
 3 1   1  
  
n  d Se    ASut   
 
 
2 1/2 
 
6



2
12
890
150
10




8  14 750 


 
 



1

1

3
6 
6




  0.050   150   10  
  14 750   440   10    



n = 0.23
Infinite life is not 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,
Shigley’s MED, 11th edition
Chapter 7 Solutions, Page 25/48
2
  32 K f M a  2
 16 K fsTm  
  
 max
  3
 
3
3
   d

  d  
1/2
2
  32 1.8 4097  2
 16  2.4   3101  




  3
 
 
   0.050  3  
    0.050  3 


 

S
370
n y 
0.46

 max
798
1/ 2
 
7.98 108 Pa 798 MPa
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) 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.
______________________________________________________________________________
7-27
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. 7-26, and the loading as defined in Prob.
3-84, 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.
7-26 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
Shigley’s MED, 11th edition
1/4
(1)(3.761)

0.254
1/4
1.96
Chapter 7 Solutions, Page 26/48
Multiplying all diameters by 2, we obtain the following deflections:
Location
Slope
Deflection
(rad)
(mm)
Left bearing O
0.00090
0.000
Right bearing C
0.00115
0.000
Left Gear A
0.00022
0.235
Right Gear B
0.00023
0.230
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.
______________________________________________________________________________
7-28
(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.
Shigley’s MED, 11th edition
Chapter 7 Solutions, Page 27/48
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.
 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. (6-10):
Sut = 68 kpsi, Sy = 57 kpsi
Se 0.5(68) 34.0 kpsi
Eq. (6-18):
ka 2.00(68)  0.217 0.801
Keyway at C
Shigley’s MED, 11th edition
Chapter 7 Solutions, Page 28/48
Since there is only steady torsion here, only a static check needs to be performed. We’ll
use the maximum shear stress theory.

Eq. (5-3):
Tr 2532  1.00 / 2 

12.9 kpsi
J
  1.004  / 32
ny 
Sy / 2


57 / 2
2.21
12.9
Keyway at D
Assuming r / d = 0.02 for typical end-milled keyway cutter (see discussion prior to Table
7-1), with d = 1.75 in, r = 0.02d = 0.035 in.
Table 7-1:
Fig. 6-26:
Fig. 6-27:
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
Eq. (6-32):
K fs 1  qs ( K ts  1) 1  0.72(3.0  1) 2.4
 0.107
Eq. (6-19):
 1.75 
kb 

 0.30 
Eq. (6-17):
Se 0.801(0.828)(34.0) 22.5 kpsi
0.828
We will choose the DE-Gerber criteria since this is an analysis problem in which we
would like to evaluate typical expectations.
Using Eqs. (7-6) and (7-11) with Mm = Ta = 0,
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
2
2 1/2 

1
8 A    2 BSe   

1   1  
  
n  d 3 Se    ASut   
 
 
2 1/2 
 
8  5.252 
   2  10.53  22.5    


1  1  

  1.753   22.5      5.252   68    




n = 3.39
Ans.
Groove at E
Shigley’s MED, 11th edition
Chapter 7 Solutions, Page 29/48
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
Fig. 6-26:
q = 0.76
K f 1  q  K t  1 1  0.76(2.1  1) 1.8
Eq. (6-32):
 0.107
 1.55 
kb 
0.839

 0.30 
Eq. (6-19):
Eq. (6-17):
Se 0.801(0.839)(34) 22.8 kpsi
Using Eqs. (7-6) and (7-11) with Mm = Ta = Tm = 0,
A  4  K f M a   4   1.8   1115   4122 lbf in 4.122 kip in
B=0
2 1/2 

1
8 A    2 BSe   

1   1  
  
n  d 3 Se    ASut   
 
 
1/2
8  4.122 
1   0 2 

1



 1.553  22.8 
2

2


n = 4.04

Ans.
Shoulder at F
Fig. A-15-9:
Fig. 6-26:
Eq. (6-32):
r / d = 0.125 / 1.40 = 0.089,
D / d = 2.0 / 1.40 = 1.43
Kt = 1.7
q = 0.78
K f 1  q  K t  1 1  0.78(1.7  1) 1.5
 0.107
Eq. (6-19):
Eq. (6-17):
 1.40 
kb 
0.848

 0.30 
Se 0.801(0.848)(34) 23.4 kpsi
Using Eqs. (7-6) and (7-11) with Mm = Ta = Tm = 0,
A  4  K f M a   4   1.5   845   2535 lbf in 2.535 kip in
B=0
2
Shigley’s MED, 11th edition
2
Chapter 7 Solutions, Page 30/48
2 1/2 

1
8 A    2 BSe   

1   1  
  
n  d 3 Se    ASut   
 
 
1/2
8  2.535 
1   0  2 

1



  1.403   23.4 

n = 4.97

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:
Sectio
n
1
2
3
Diamete
r
(in)
1.00
1.70
1.40
Lengt
h
(in)
2.90
7.77
2.20
The deflection problem can readily (though tediously) be solved with singularity
functions. For examples, see Ex. 4-7, or the solution to Prob. 7-29. 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
(rad)
0.00029
0
0.00040
0
0.00029
0
0.00014
6
Deflectio
n
(in)
0.000000
0.000000
0.000404
0.000928
Comparing these values to the recommended limits in Table 7-2, we find that they are all
within the recommended range.
______________________________________________________________________________
7-29
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.
Shigley’s MED, 11th edition
Chapter 7 Solutions, Page 31/48
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
M(N ∙ m) 0 155.56 388.89 544.44 326.67 124.44
31
5
0
I35 = (/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
The steps and the change of slopes are evaluated in the table. From these, the function
M/I can be generated:
Shigley’s MED, 11th edition
Chapter 7 Solutions, Page 32/48
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
0
1
1
 9.312 x  0.21  0.6994 x  0.275
0
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
1
 C1  109 (1)

2
 0.581 x  0.1
 4.655 x  0.21  0.6994 x  0.275  8.735 x  0.275
Ey   8.8 x3  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
2
3
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
 (rad)
F.E. Model
Full F.E. Model
–0.001 4260
–0.000 1466
0.001 3120
–0.001 4270
–0.000 1467
0.001 3280
–0.001 4160
–0.000 1646
0.001 3150
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
Shigley’s MED, 11th edition
Chapter 7 Solutions, Page 33/48
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
New ideal d, mm
Rounded up d,
mm
20.00 30.00 35.00 40.00 45.00 55.00
21.86 32.79 38.26 43.72 49.19 60.12
22.00 34.00 40.00 44.00 50.00 62.00
Repeating the full finite element model results in
x = 0:
 = –9.30  10-4 rad
x = 140 mm:  = –1.09  10-4 rad
x = 315 mm:  = 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  = 32M/(d3) and  = 16T/(d3),
x (mm)
 (MPa)
 (MPa)
 (MPa)
0 15
40
100
110
140
210
275
300
330
0 22.0 37.0 61.9 47.8 60.9 52.0 39.6 17.6
0
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:
ka 3.04(470)  0.217 0.800
Eq. (6-18):
Eq. (6-19):
Eq. (6-17):
Fig. A-15-8:
Fig. A-15-9:
Fig. 6-26:
Fig. 6-27:
Eq. (6-32):
kb (40 / 7.62)  0.107 0.837
Se = 0.800 (0.837)(0.5)(470) = 157 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
Shigley’s MED, 11th edition
Chapter 7 Solutions, Page 34/48
Using DE-Goodman, from Eqs. (7-6) and (7-7), with Mm = Ta = 0,
A  4   1.68   326.67   1097.6
2
B  3   1.32   107   244.6
2
  0.043   1097.6
244.6 


n

 157  106  390  106  
16


n 1.65
1
At x = 330 mm:
The von Mises stress is the highest but it comes from the steady torque only. So we will
do a check on yield, using Eqs. (7-15) and (7-16).
  16(1.32)(107)  2 

 max  3 
 
3
   (0.02 )  
1/2
155.7 MPa
S
390
n y 
2.5

 max
155.7
Check the other locations.
If worse-case is at x = 210 mm, the changes discussed for the slope criterion will
improve the strength issue.
______________________________________________________________________________
7-30 and 7-31 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.
 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.
______________________________________________________________________________
7-32
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.
Shigley’s MED, 11th edition
Chapter 7 Solutions, Page 35/48

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.
______________________________________________________________________________
7-33
In Eq. (7-22) set
d4
d2
I
, A
64
4
to obtain
2
    d  gE
    
 l   4 
(1)
or
d
4l 2
2

gE
(2)
(a) From Eq. (1) and Table A-5
2
9
    0.025  9.81(207)(10 )
 
883 rad/s
 

76.5  103 
 0.6   4 
Ans.
(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),
2 d
l 
4 l
gE

Since d / l is the same regardless of the scale,
l constant 0.6(883) 529.8
529.8

1766 rad/s Ans.
0.3
Shigley’s MED, 11th edition
Chapter 7 Solutions, Page 36/48
Thus the first critical speed doubles.
______________________________________________________________________________
7-34
From Prob. 7-33,  883 rad/s
A 4.909  10 4  m 2 , I 1.917  10  8  m 4 ,
E 207(109 ) Pa,
 7.65  10 4  N/m3
w  Al 4.909  10  4  7.65  10 4  (0.6) 22.53 N
One element:
Eq. (7-24):
11 
0.3(0.3)  0.62  0.32  0.32 
6(207)  10  (1.917)  10
9
8

y1 w111 22.53(1.134) 10 6
y12 6.528  10 10 

1.134  10 6  m/N
 (0.6)
 2.555  10  m
5


wy 22.53(2.555) 10  5 5.756 10  4

wy 2 22.53(6.528)  10  10  1.471 10 8 
5.756  10 4 
wy
1  g
 9.81
620 rad/s
wy 2
1.471 10 8 
(30% low)
Two elements:
11  22 
12  21 
0.45(0.15)  0.62  0.452  0.152 
6(207)  10  (1.917)  10
9
8
 (0.6)
6.379  10 7  m/N
0.15(0.15)(0.62  0.152  0.152 )
4.961 10 7 m/N
9
8
6(207) 10 (1.917) 10 (0.6)
 










y1  y2 w111  w212 11.265(6.379) 10 7  11.265(4.961) 10 7 1.277 10 5 m
y12  y22 1.632  10 10 
wy 2(11.265)(1.277) 10 5 2.877 10 4

wy 2(11.265)(1.632)  10
9
2
Shigley’s MED, 11th edition



 10
 3.677  10 
Chapter 7 Solutions, Page 37/48


 2.877 10 4
1  9.81 
 3.677 10 9
  876 rad/s
 
(0.8% low)
Three elements:
11  33 
 22 

0.5(0.1) 0.62  0.52  0.12

6(207) 10 (1.917) 10
8
 (0.6)
0.3(0.3)  0.62  0.32  0.32 
6(207)  10  (1.917)  10
9
12  32 
13 
 
9
8

 (0.6)
 

6(207) 10 (1.917) 10
8
6(207)  10  (1.917)  10

 (0.6)
0.1(0.1)  0.62  0.12  0.12 


3.500 10 7 m/N
1.134  10 6  m/N
0.3(0.1) 0.62  0.32  0.12
9



5.460 10 7 m/N
2.380  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 
wy 7.51  8.516  10   1.672  10   8.516  10   2.535  10 
wy 7.51  8.516  10     1.672  10     8.516  10    3.189  10 
9
8
7
7
7
6
7
6
7
5
7
7
7
6
1
2
3
6
5
6
2


 2.535 10 4
1  9.81 
 3.189 10 9
2
6
5
2
4
6
2
9
  883 rad/s
 
The result is the same as in Prob. 7-33. 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.
______________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 7 Solutions, Page 38/48
7-35
(a) For two bodies, Eq. (7-26) is
( m111  1/  2 )
m212
m1 21
( m2 22  1/  2 )
0
Expanding the determinant yields,
2
 1 
 1 
 2   (m111  m2 22 )  2   m1m2 (11 22  12 21 ) 0
 
 1 
(1)
2
2
Eq. (1) has two roots 1/ 1 and 1/ 2 . Thus
 1
1  1
1 
 2  2   2  2  0
  1    2 
or,
2
2
1  1   1  1 
 1   1
 2    2  2      2   2  0
    1 2      1   2 
(2)
Equate the third terms of Eqs. (1) and (2), which must be identical.
1 1
m1m2 (11 22  12 21 )
12 22

1
12 m1m2 (11 22  12 21 )
22
and it follows that
1
2 
1
g2
w1w2 (11 22  12 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 1 = 124.8 rad/s. From part (a), using influence coefficients,
1
3862
2 
466 rad/s Ans.
124.8 35(55)  2.061(3.534)  2.234 2   10  8 
______________________________________________________________________________
7-36
In Eq. (7-22), for 1, the term I / A appears. For a hollow uniform diameter shaft,
Shigley’s MED, 11th edition
Chapter 7 Solutions, Page 39/48
1 
2
2
2
2
  d o4  di4  / 64
I
1  d o  di   do  di  1


 d o2  d i2
2
2
2
2
A
16
d o  di
4
  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 d i are 0 d i d o , so the range of the critical speeds is
1 1  0 to about 1 1  1
or from 1 to 2 1 .
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 
1     1   2  1.12
 do 
 do 


2
Ans.
______________________________________________________________________________
7-37
All steps will be modeled using singularity functions with a spreadsheet. Programming
both loads will enable the user to first set the left load to 1, the right load to 0 and
calculate 11 and 21. Then set the left load to 0 and the right to 1 to get 12 and 22. The
spreadsheet shows the 11 and 21 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
x
M
0
0
1
2
3
0.875 1.75 1.625
9
10
11
0.875 0.75 0.625
Shigley’s MED, 11th edition
12
0.5
4
1.5
5
6
7
1.375 1.25 1.125
13
14
15
0.375 0.25 0.125
8
1
16
0
Chapter 7 Solutions, Page 40/48
2
M (lbf-in)
1.5
1
0.5
0
0
2
4
6
8
10
12
14
16
x (in)
The second-area moments are:
0  x 1 in and 15  x 16 in, I1   2 4  / 64 0.7854 in 4
1 x 9 in , I 2   2.472 4  / 64 1.833 in 4
9 x 15 in , I 3   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
16
0
1.2
1
M/I (lbf/in3)
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
Shigley’s MED, 11th edition
Chapter 7 Solutions, Page 41/48
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
1
0
D7 x  9  F7 x  9  D11 x  15  F11 x  15
1
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  ’s. The results are: 11 = 2.917(10–7) and 12 = 1.627(10–7). Repeat for
F1 = 0 and F2 = 1, resulting in 21 = 1.627(10–7) and 22 = 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 ),  wy 2  5.212(10 9 )
Neglecting the shaft, Eq. (7-23) gives
Shigley’s MED, 11th edition
Chapter 7 Solutions, Page 42/48
1  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  = 0.282 lbf/in3. The weight of the left element is

 
w1   d 2l 0.282    22  1  2.472 2  8   11.7 lbf
4
 4
The right element is
 
w2 0.282    2.7632  6   22  1  11.0 lbf
 4
The spreadsheet can be easily modified to give
11 9.605  10 7  , 12  21 5.718  10  7  ,  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 
 wy 3.449  10  ,  wy
4
2
5.371 10 9 
 3.449  10  4  
 4980 rad/s
1  386 
9
 5.371 10  
A finite element model of the exact shaft gives 1 = 5340 rad/s. The simple model is
6.8% low.
Combination: Using Dunkerley’s equation, Eq. (7-32):
1
1
1


 1 3870 rad/s Ans.
2
2
1 6149 4980 2
______________________________________________________________________________
7-38
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
Shigley’s MED, 11th edition
Chapter 7 Solutions, Page 43/48
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.
 K ts 0 K ts 0
32T
 32T 
K ts 
K ts 
3
3 
 AD
D 
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  0.100;
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.
______________________________________________________________________________
7-39
From the solution to Prob. 3-83, the torque to be transmitted through the key from the
gear to the shaft is T = 2819 lbf∙in. From Prob. 7-24, 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
T
2819
F 
5638 lbf
r 1.00 / 2
Shigley’s MED, 11th edition
Chapter 7 Solutions, Page 44/48
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



l
1.1 5638 
nF

0.754 in
tS sy 0.25  32 900 
Failure by crushing:

F
F n Sy  Sy

 2 F /  tl 
A  t / 2 l

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.
______________________________________________________________________________
7-40
From the solution to Prob. 3-84, the torque to be transmitted through the key from the
gear to the shaft is T = 3101 N∙m. From Prob. 7-26, 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.
The force applied to the key is
T
3101
F 
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:

n
F F

A tl
S sy


S sy
F /  tl 
Shigley’s MED, 11th edition

1.1 124   103 
nF
l

0.0433 m 43.3 mm
tS sy  0.014   225   10 6 
Chapter 7 Solutions, Page 45/48
Failure by crushing:

n
Sy


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.
______________________________________________________________________________
7-41
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 F = 0 mm.
Hole:
Eq. (7-36)
:
Shaft:
Eq. (7-37):
7-42
Dmax = D + D = 15 + 0.018 = 15.018 mm
Ans.
Dmin = D = 15.000 mm
Ans.
Ans.
dmax = d + F = 15.000 + 0 = 15.000 mm
dmin = d + F – 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 F = 0.0017 in.
Hole:
Eq. (7-36
):
Dmax = D + D = 1.75 + 0.0010 = 1.7510 in
Ans.
Dmin = D = 1.7500 in
Ans.
Shaft:
Eq. (7-38):
dmin = d + F = 1.75 + 0.0017 = 1.7517 in
Ans.
dmax = d + F + d = 1.75 + 0.0017 + 0.0006 = 1.7523 in
Ans.
______________________________________________________________________________
7-43
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 F = –0.009 mm.
Hole:
Eq. (7-36):
Dmax = D + D = 45 + 0.025 = 45.025 mm
Ans.
Dmin = D = 45.000 mm
Ans.
Shaft:
Shigley’s MED, 11th edition
Chapter 7 Solutions, Page 46/48
dmax = d + F = 45.000 + (–0.009) = 44.991 mm
Ans.
dmin = d + F – d = 45.000 + (–0.009) – 0.016 = 44.975 mm
Ans.
______________________________________________________________________________
Eq. (7-37):
7-44
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 F = –0.0010 in.
Hole:
Eq. (7-36):
Dmax = D + D = 1.250 + 0.0015 = 1.2515 in Ans.
Dmin = D = 1.2500 in
Ans.
Shaft:
Eq. (7-37):
dmax = d + F = 1.250 + (–0.0010) = 1.2490 in
Ans.
dmin = d + F – d = 1.250 + (–0.0010) – 0.0010 = 1.2480 in Ans.
______________________________________________________________________________
7-45
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 F = 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 + F = 35 + 0.026 = 35.026 mm
Ans.
dmax = d + F + d = 35 + 0.026 + 0.016 = 35.042 mm
Ans.
______________________________________________________________________________
7-46 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 F = 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 + F = 1.5000 + 0.0010 = 1.5010 in
dmax = d + F + d = 1.5000 + 0.0010 + 0.0006 = 1.5016 in
Shigley’s MED, 11th edition
Chapter 7 Solutions, Page 47/48
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.
______________________________________________________________________________
7-47
(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.
 0.043 mm.
Table A-12: Fundamental deviation is F
Dmax = D + D = 35 + 0.025 = 35.025 mm
Eq. (7-36):
Dmin = D = 35.000 mm
Eq. (7-38):
dmin = d + F = 35 + 0.043 = 35.043 mm
Ans.
dmax = d + F + d = 35 + 0.043 + 0.016 = 35.059 mm
Ans.
(b)
Eq. (7-42):
 min d min  Dmax 35.043  35.025 0.018 mm
Eq. (7-43):
 max d max  Dmin 35.059  35.000 0.059 mm
pmax
Eq. (7-40):


2
2
2
2
E max  d o  d d  di


2d 3 
d o2  di2

 
 
 




207 109  0.059   602  352 352  0 

 115 MPa

602  0
2 353


pmin



2
2
2
2
E min  d o  d d  di 



d o2  di2
2d 3 


207 109  0.018   602  352 352  0 

 35.1 MPa

602  0
2 353


 
 
Ans.



Ans.
(c) For the shaft:
 t ,shaft  p  115 MPa
Eq. (7-44):
 r ,shaft  p  115 MPa
Eq. (7-46):
Eq. (5-13):
    12   1 2   22 
1/2
 ( 115) 2  ( 115)(  115)  (  115) 2 
n S y /   390 /115 3.4
1/2
115 MPa
Ans.
For the hub:
Shigley’s MED, 11th edition
Chapter 7 Solutions, Page 48/48
 t ,hub  p
Eq. (7-45):
 602  352 
d o2  d 2

234 MPa
115
 2
2 
d o2  d 2
 60  35 
Eq. (7-46):
 r ,hub  p  115 MPa
Eq. (5-13):
    12   1 2   22 
1/2
 (234) 2  (234)( 115)  ( 115) 2 
n S y /   600 / 308 1.9
1/ 2
308 MPa
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.
T ( / 2) f pmin ld 2
Eq. (7-49)
( / 2)(0.8)(35.1)  106  (0.050)(0.035) 2 2700 N m
Ans.
______________________________________________________________________________
7-48 Basic size D = 2.5 in, L = 3 in, OD = 4 in, ID = 2.5 in, f = 0.7.
(a) Table 7-9: medium drive fit  H7/S6
Table A-13: D = 0.0012 in, d = 0.0007 in
Table A-14: F = 0.0021 in
Eq. (7-36): Dmax = D + D = 2.5 + 0.0012 = 2.5012 in
Ans.
Dmin = D = 2.5000 in
Ans.
Eq. (7-38): dmax = d +F + d = 2.5 + 0.0021 + 0.0007 = 2.5028 in
Ans.
dmin = d +F = 2.5 + 0.0021 = 2.5021 in
Ans.
(b) Eq. (7-42): min = dmin  Dmax = 2.5021  2.5012 = 0.0009 in
Eq. (7-40):
2
2
2
2
6
2
2
2
2
E   d o  d   d  di   30  10  0.0009   4  2.5   2.5  0  


 3291 psi
p 3
3
2d 
d o2  d i2
42  0 2
2
2.5








Eq. (7-49):


Tmax  fpLd 2   0.7  3291 3 2.52 67 850 lbf in 67.85 kip in
2
2
(c) (i) Eq. (7-43): max = dmax  Dmin = 2.5028  2.5 = 0.0028 in
Eq. (7-40):
30  106  0.0028   4 2  2.52   2.52  0 2  

 10 240 psi =10.24 kpsi
p
3
4 2  02
2  2.5 


(ii) Eq. (7-45):
d2 d2
42  2.52
23.4 kpsi
  t  hub  p o2 2 10.24 2
do  d
4  2.52
t 
(iii)
 p  10.24 kpsi
Ans.
Ans.
hub
    12   1 2   22 
Shigley’s MED, 11th edition
Ans.
1/2
2
 23.42  23.4   10.2     10.2  


1/2
29.8 kpsi
Chapter 7 Solutions, Page 49/48
S y 100

3.36
Ans.
  29.8
______________________________________________________________________________
n
Shigley’s MED, 11th edition
Chapter 7 Solutions, Page 50/48
Chapter 6
6-1
Eq. (2-36):
Eq. (6-10):
Table 6-2:
Eq. (6-18):
Sut  3.4 H B  3.4(300)  1020 MPa
Se  0.5Sut  0.5(1020)  510 MPa
a  1.38, b  0.067
k a  aSutb  1.38(1020) 0.067  0.868
Eq. (6-19):
kb  1.24 d 0.107  1.24(10) 0.107  0.969
Se  k a kb Se  (0.868)(0.969)(510)  429 MPa Ans.
Eq. (6-17):
______________________________________________________________________________
6-2
(a) Table A-20:
Eq. (6-10):
Sut = 80 kpsi
Se  0.5(80)  40 kpsi
Ans.
(b) Table A-20:
Eq. (6-10):
Sut = 90 kpsi
Se  0.5(90)  45 kpsi
Ans.
(c) Aluminum has no endurance limit. Ans.
(d) Eq. (6-10):
Sut > 200 kpsi, Se  100 kpsi
Ans.
______________________________________________________________________________
6-3
Sut  120 kpsi,  ar  70 kpsi
Fig. 6-23:
f  0.82
Eq. (6-10):
Se  S e  0.5(120)  60 kpsi
Eq. (6-13):
( f Sut ) 2  0.82(120) 
a

 161.4 kpsi
Se
60
Eq. (6-14):
 f Sut
1
b   log 
3
 Se
2
1/ b

1
 0.82(120) 
   log 
  0.0716
3
60



1
 
 70  0.0716
N   ar   
Eq. (6-15):
 117 000 cycles Ans.

 161.4 
 a 
______________________________________________________________________________
6-4
Sut  1600 MPa,  ar  900 MPa
Fig. 6-23:
Sut = 1600 MPa. Off the graph, so estimate f = 0.77.
Eq. (6-10):
Sut > 1400 MPa, so Se = 700 MPa
Eq. (6-13):
( f Sut ) 2  0.77(1600) 
a

 2168.3 MPa
Se
700
2
Shigley’s MED, 11th edition
Chapter 6 Solutions, Page 1/58
Eq. (6-14):
 f Sut
1
b   log 
3
 Se

1
 0.77(1600) 
   log 
  0.081838
3
700



1
1/ b
 
 900  0.081838
N   ar   
Eq. (6-15):
 46 400 cycles Ans.

 2168.3 
 a 
______________________________________________________________________________
6-5
Sut  230 kpsi, N  150 000 cycles
Fig. 6-23, point is off the graph, so estimate:
f = 0.77
Eq. (6-10):
Sut > 200 kpsi, so Se  Se  100 kpsi
Eq. (6-13):
( f Sut ) 2  0.77(230) 
a

 313.6 kpsi
Se
100
Eq. (6-14):
 f Sut
1
b   log 
3
 Se
Eq. (6-12):
S f  aN b  313.6(150 000) 0.08274  117.0 kpsi
2

1
 0.77(230) 
   log 
  0.08274
3
 100 

Ans.
______________________________________________________________________________
6-6
Sut  1100 MPa = 160 kpsi
Fig. 6-23:
f = 0.79
Eq. (6-10):
Se  S e  0.5(1100)  550 MPa
Eq. (6-13):
( f Sut ) 2  0.79(1100) 
a

 1373 MPa
Se
550
Eq. (6-14):
 f Sut
1
b   log 
3
 Se
Eq. (6-12):
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. 6-23:
f = 0.80
From Fig. 6-21, we note that below 103 cycles on the S-N diagram constitutes the lowcycle region. The stress-life approach is not very reliable in this region, but for a rough
Shigley’s MED, 11th edition
Chapter 6 Solutions, Page 2/58
response to this question, we can write an equation in log-log scale for the line between
(100, Sut) and (103, fSut) as
S f  Sut N 
log f  /3
log  0.80   /3
 150  500 
 123 kpsi
Ans.
The testing should be done at a completely reversed stress of 123 kpsi, which is below
the yield strength, so it is possible. Ans.
______________________________________________________________________________
6-8
d = 1.5 in, Sut = 110 kpsi
Se  0.5(110)  55 kpsi
Eq. (6-10):
Table 6-2:
a = 2.00, b =  0.217
Eq. (6-18):
k a  aSut b  2.00(110) 0.217  0.721
Eq. (6-19):
kb = 0.879 d 0.107= 0.879(1.5) 0.107 =0.842
Eq. (6-17):
Se = kakb S e = 0.721(0.842)(55) = 33.4 kpsi Ans.
______________________________________________________________________________
6-9
For AISI 4340 as-forged steel,
Eq. (6-10):
Table 6-2:
Eq. (6-18):
Se = 100 kpsi
a = 12.7, b =  0.758
ka = 12.7(260)0.758 = 0.188
Eq. (6-19):
 0.75 
kb  

 0.30 
0.107
 0.907
Each of the other modifying factors is unity.
Se = 0.188(0.907)(100) = 17.1 kpsi
Ans.
For AISI 1040:
Se  0.5(113)  56.5 kpsi
ka  12.7(113)0.758  0.353
kb  0.907 (same as 4340)
Each of the other modifying factors is unity
Se  0.353(0.907)(56.5)  18.1 kpsi
Ans.
Not only is AISI 1040 steel a contender, it has a superior endurance strength. Ans.
______________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 6 Solutions, Page 3/58
6-10
From Table A-20, Sut = 570 MPa, Sy = 310 MPa.
From a free-body diagram analysis, the bearing
reaction forces are found to be R1 = 3.25 kN and
R2 = 9.75 kN. The shear-force and bendingmoment diagrams are shown. The critical
location is at the section where the bending
moment is maximum, on the outer surface where
the bending stress is maximum. With a rotating
shaft, the bending stress will be completely
reversed.
Mc 487 500(25 / 2)

 317.8 MPa
I
( / 64)(25) 4
Sy
310

 0.98
Ans.
ny 
 max 317.8
 max   ar 
(a)
Yielding is predicted, on the outer surface. For
some applications, this might not prevent the part
from being used, so we will continue checking for
fatigue.
(b) Eq. (6-10):
Se'  0.5S ut  0.5(570)  285 MPa
Eq. (6-18):
k a  aSutb  3.04(570) 0.217  0.767
Eq. (6-19):
kb  1.24d 0.107  1.24(25) 0.107  0.879
Eq. (6-25):
kc  1
Ans.
Eq. (6-17):
Se  ka kb kc S e'  (0.767)(0.879)(1)(285)  192 MPa
(c) For completely reversed stress, the fatigue factor of safety can be assessed as the ratio
of the endurance limit to the completely reversed stress.
S
192
nf  e 
 0.60
Ans.
 ar 317.8
Infinite life is not predicted. Use the S-N diagram to estimate the life.
(d)
Fig. 6-23, or Eq. (6-11): f = 0.87
 f Sut 
2
 0.87(570)

2
Eq. (6-13):
a
Eq. (6-14):
 f Sut 
1
1
 0.87(570) 
b   log 
   log 
  0.1374
3
3
 192 
 Se 
192
Se
1
Eq. (6-15):
 1280.8
1
   b  317.8  0.1374
 25444
N   ar   

 a   1280.8 
N = 25000 cycles
Ans.
______________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 6 Solutions, Page 4/58
6-11
From Table A-20, Sut = 400 MPa, Sy = 220 MPa.
Free-body, shear-force, and bending-moment
diagrams are shown.
32 M 32(45000)
 max 

 458366 / d 3
3
3
d
d
The load is repeatedly applied and released, so from
Eqs. (6-8) and (6-9),
 m   a   max / 2  229183 / d 3
Be sure to confirm that the units are legitimate for
stress in MPa and d in mm.
For yielding,
n y  1.5 
Sy
 max

220
458366 / d 3
d  14.62 mm
Now check fatigue, opting for the linear Goodman criterion for simplicity of solving for
the diameter. First, determine the adjusted endurance limit.
Eq. (6-10):
Se'  0.5Sut  0.5(400)  200 MPa
Eq. (6-18):
k a  aSutb  3.04(400) 0.217  0.828
Estimate the size factor from the diameter determined for yielding. It can be adjusted
later.
Eq. (6-19):
kb  1.24 d 0.107  1.24(15) 0.107  0.93
Eq. (6-25):
kc  1
Ans.
Eq. (6-17):
Se  ka kb kc S e'  (0.767)(0.879)(1)(285)  192 MPa
(c) For completely reversed stress, the fatigue factor of safety can be assessed as the ratio
of the endurance limit to the completely reversed stress.
S
192
nf  e 
 0.60
Ans.
 ar 317.8
Infinite life is not predicted. Use the S-N diagram to estimate the life.
(d)
Fig. 6-23, or Eq. (6-11): f = 0.87
 f Sut 
2
 0.87(570)

Eq. (6-13):
a
Eq. (6-14):
 f Sut
1
b   log 
3
 Se
192
Se
1
Eq. (6-15):
2
 1280.8

1
 0.87(570) 
   log 
  0.1374
3
192



1
   b  317.8  0.1374
N   ar   
 25444

 a   1280.8 
N = 25000 cycles
Ans.
Shigley’s MED, 11th edition
Chapter 6 Solutions, Page 5/58
6-12
D = 1 in, d = 0.8 in, T = 1800 lbfin, and from Table A-20 for AISI 1020 CD,
Sut = 68 kpsi, and Sy = 57 kpsi.
r 0.1
D
1
(a) Fig. A-15-15:

 0.125, 
 1.25, K ts  1.40
d 0.8
d 0.8
Get the notch sensitivity either from Fig. 6-27, or from the curve-fit Eqs. (6-33) and
(6-36). Using the equations,
a  0.190  2.51103   68   1.35 105   68   2.67 10 8  683   0.07335
2
qs 
Eq. (6-32):
1
1

a
r
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,
Tr K fs 16T 1.32(16)(1800)


 23 635 psi  23.6 kpsi
d3
 (0.8)3
J
Se  0.5(68)  34 kpsi
 a  K fs
Eq. (6-10):
Eq. (6-18):
ka = 2.00(68)0.217 = 0.80
Eq. (6-19):
kb = 0.879(0.8)0.107 = 0.90
Eq. (6-25):
kc = 0.59
Eq. (6-17) (labeling for shear):
Sse = 0.80(0.90)(0.59)(34) = 14.4 kpsi
For purely reversing torsion, use Eq. (6-58) for the ultimate strength in shear.
Eq. (6-58):
Ssu = 0.67 Sut = 0.67(68) = 45.6 kpsi
Fig. 6-23:
f = 0.9
Adjusting the fatigue strength equations for shear,
 0.9(45.6)

Eq. (6-13):
 f S su 
a
Eq. (6-14):
 f S su
1
b   log 
3
 S se
S se
2
1
2
14.4
 117.0 kpsi

1
 0.9(45.6) 
   log 
  0.151 61
3
 14.4 

1
   b  23.6  0.151 61
 38.5 103 cycles Ans.
Eq. (6-15):
N  a  

a
117.0

  
(b) Estimate the ultimate strength at the operating temperature.
Eq. (6-26):
( ST S RT )750  0.98  3.5(10 4 )(750)  6.3(10 7 )750 2  0.89
Thus,
 Sut 750   ST
Shigley’s MED, 11th edition

 
S RT 750  Sut 70  0.89(68)  60.5 kpsi
Chapter 6 Solutions, Page 6/58
Eq. (6-10):
Eq. (6-17):
 Se 750
 0.5  Sut 750  0.5(60.5)  30.3 kpsi
Sse = 0.80(0.90)(0.59)(30.3) = 12.9 kpsi

Note that we use kd = 1 since the ultimate strength has been adjusted for the operating
temperature.
S su  0.67  Sut 750  0.67  60.5  40.5 kpsi
Eq. (6-58):
 f S su 
a
2
S se
0.9(40.5)

2
 103.0 kpsi
12.9
 f S su
1
b   log 
3
 S se

1
 0.9(40.5) 
   log 
  0.150 37
3
 12.9 

1
1
   b  23.6  0.150 37
N  a  
Ans.
 18.0 103 cycles

 a   103.0 
______________________________________________________________________________
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
k a  38.6(770) 0.650  0.51
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. (6-17):
Fig. 6-23:
Se = 0.51(0.85)(385) = 167 MPa
f = 0.83
Eq. (6-13):
a
Eq. (6-14):
Eq. (6-12):
 f Sut 
Se
2
 0.83(770) 

2
167
 2446 MPa
 f Sut 
1
1
 0.83(770) 
b   log 
   log 
  0.1943
3
3
 167 
 Se 
S f  aN b  2446(104 ) 0.1943  409 MPa
Now evaluate the stress.
M max  (2000 N)(0.6 m)  1200 N  m
 a   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.
Shigley’s MED, 11th edition
Chapter 6 Solutions, Page 7/58
Sf
409 106 
 1.5
7200 / b3
b = 0.0298 m Select b = 30 mm.
n
a

Since the size factor was guessed, go back and check it now.
1/ 2
Eq. (6-24):
d e  0.808  hb   0.808b  0.808  30   24.2 mm
0.107
 24.2 
Eq. (6-19):
kb  
 0.88

 7.62 
Our guess of 0.85 was slightly conservative, so we will accept the result of
b = 30 mm.
Checking yield,
 max 
Ans.
7200
106   267 MPa
0.0303
Sy
420
 1.57
 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. (6-10):
Se  0.5(68)  34 kpsi
Table 6-2:
Eq. (6-20):
Eq. (6-25):
ka  2.00(68) 0.217  0.80
kb = 1 (axial loading)
kc = 0.85
Eq. (6-17):
Se = 0.80(1)(0.85)(34) = 23.1 kpsi
Table A-15-1: d / w  0.5 / 2.5  0.2, K t  2.5
Get the notch sensitivity either from Fig. 6-26, or from the curve-fit Eqs. (6-33) and
(6-35). The relatively large radius is off the graph of Fig. 6-26, 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):
Shigley’s MED, 11th edition
1

Chapter 6 Solutions, Page 8/58
Fa
2.25 Fa
=
 3Fa
A (3 / 8)(2.5  0.5)
a  K f
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.
23.1
2
 a 3Fa
Fa  3.85 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. (6-10):
Se  0.5Sut  0.5 120   60 kpsi
Eq. (6-18):
k a  aSutb  2.00(120) 0.217  0.71
Eq. (6-23):
d e  0.370d  0.370(1.8)  0.666 in
Eq. (6-19):
kb  0.879d e 0.107  0.879(0.666) 0.107  0.92
Eq. (6-25):
kc  1
Eq. (6-17):
Se  ka kb kc Se  (0.71)(0.92)(1)(60)  39.2 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. 6-26, or from the curve-fit Eqs. (6-33) and
(6-35). 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
Eq. (6-32):
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
I  ( / 64) d 4  ( / 64)(1.8) 4  0.5153 in 4
Mc 25 000(1.8 / 2)

 43 664 psi  43.7 kpsi
I
0.5153
0
 max 
 min
Shigley’s MED, 11th edition
Chapter 6 Solutions, Page 9/58
Eqs. (6-8) and (6-9):  m  K f
a  K f
 43.7  0   42.8 kpsi
 max   min
 1.96 
 max   min
 1.96 
2
2
2
 43.7  0 
1
2
  
 42.8 42.8 
nf   a  m   


S
S
39.2
120 

e
ut


Eq. (6-41):
n f  0.69
 42.8 kpsi
1
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.
 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.
 ar 
Mc 420 (35 / 2)

 0.09978 kN/mm 2  99.8 MPa
4
I
( / 64)(35)
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. 6-26, or from the curve-fit Eqs. (6-33) and
(6-35). Using the equations, with Sut = 470 MPa and r = 3 mm,
a  1.24  2.25 103  (470)  1.60 10 6  (470) 2  4.11 10 10  (470)3  0.4933
1
q
1
Shigley’s MED, 11th edition
a
r

1
 0.78
0.4933
1
3
Chapter 6 Solutions, Page 10/58
Eq. (6-32):
K f  1  q ( K t  1)  1  0.78(1.7  1)  1.55
Eq. (6-10):
Se'  0.5Sut  0.5(470)  235 MPa
Eq. (6-18):
k a  aSutb  3.04(470) 0.217  0.80
Eq. (6-19):
kb  1.24 d 0.107  1.24(35) 0.107  0.85
Eq. (6-25):
kc  1
Eq. (6-17):
Se  ka kb kc S e'  (0.80)(0.85)(1)(235)  160 MPa
160
 1.03 Infinite life is predicted.
Ans.
K f  ar 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
( / 64)(1.625) 4
This stress is far below the yield strength of 71 kpsi, so yielding is not predicted.
 ar 
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. 6-26, or from the curve-fit Eqs. (6-33) and
(6-35). 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
a
r
K f  1  q ( K t  1)  1  0.76(1.95  1)  1.72
1
Eq. (6-32):

3
Shigley’s MED, 11th edition
Chapter 6 Solutions, Page 11/58
Eq. (6-10):
Se'  0.5S ut  0.5(85)  42.5 kpsi
Eq. (6-18):
k a  aSutb  2.00(85) 0.217  0.76
Eq. (6-19):
kb  0.879d 0.107  0.879(1.625) 0.107  0.835
Eq. (6-25):
kc  1
Eq. (6-17):
Se  ka kb kc S e'  (0.76)(0.835)(1)(42.5)  27.0 kpsi
27.0
 0.45
Ans.
K f  ar 1.72  35.0 
Infinite life is not predicted. Use the S-N diagram to estimate the life.
nf 
Fig. 6-23:
Se
f = 0.87

 0.87(85) 

Eq. (6-13):
 f Sut 
a
Eq. (6-14):
 f Sut 
1
1
 0.87(85) 
b   log 
   log 
  0.1459
3
3
 27.0 
 Se 
Se
2
2
27.0
 202.5
1
1
 K f  ar  b  (1.72)(35.0)  0.1459
 4082 cycles
N 
Eq. (6-15):
 

 a   202.5 
N = 4100 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)
 ar 

 32.8 kpsi
I
( / 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
Shigley’s MED, 11th edition
Chapter 6 Solutions, Page 12/58
Get the notch sensitivity either from Fig. 6-26, or from the curve-fit Eqs. (6-33) and
(6-35). Using the equations,
a  0.246  3.08 103   85   1.51105  85   2.67 108   85   0.07690
2
Eq. (6-32):
1
 0.76
0.07690
a 1
1
0.0625
r
K f  1  q ( K t  1)  1  0.76(1.95  1)  1.72
Eq. (6-10):
Se'  0.5S ut  0.5(85)  42.5 kpsi
Eq. (6-18):
k a  aSutb  2.00(85) 0.217  0.76
Eq. (6-19):
kb  0.879d 0.107  0.879(1.625) 0.107  0.835
Eq. (6-25):
kc  1
Eq. (6-17):
Se  ka kb kc S e'  (0.76)(0.835)(1)(42.5)  27.0 kpsi
q
1
3

27.0
 0.48
Ans.
K f  ar 1.72  32.8 
Infinite life is not predicted. Use the S-N diagram to estimate the life.
Fig. 6-23:
f = 0.87
2
2
 f Sut   0.87(85) 
a
Eq. (6-13):

 202.5
Se
27.0
nf 
Eq. (6-14):
Se

 f Sut 
1
1
 0.87(85) 
b   log 
   log 
  0.1459
3
3
 27.0 
 Se 
1
1
 K f  ar  b  (1.72)(32.8)  0.1459
 6370 cycles
N 
Eq. (6-15):
 

 a   202.5 
N = 6400 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.
Shigley’s MED, 11th edition
Chapter 6 Solutions, Page 13/58
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,
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
 ar 
Mc M  d / 2  32M 32  20  203.7




kpsi
I
d3
 d 4 / 64  d 3
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. 6-26, 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  2.00(120) 0.217  0.71
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 = (0.71)(0.85)(60) = 36.2 kpsi
Determine the desired fatigue strength from the S-N diagram.
Fig. 6-23:
f = 0.82
Eq. (6-13):
a
Eq. (6-14):
 f Sut
1
b   log 
3
 Se
Eq. (6-12):
S f  aN b  267.5(570 000) 0.1448  39.3 kpsi
 f Sut 
Se
2
 0.82(120)

2
36.2
 267.5

1
 0.82(120) 
   log 
  0.1448
3
 36.2 

Compare strength to stress and solve for the necessary d.
Shigley’s MED, 11th edition
Chapter 6 Solutions, Page 14/58
nf 
Sf
K f  ar

39.3
 1.6
1.55   203.7 / d 3 
d = 2.34 in
Since the size factor and notch sensitivity were guessed, go back and check them now.
Eq. (6-19):
kb  0.91d 0.157  0.91 2.34 
 0.80
This is a little lower than our initial guess.
From Fig. 6-26 with r = d/10 = 0.234 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 = 34.1 kpsi, Sf = 37.2 kpsi, and Kf = 1.59. This gives
0.157
nf 
Sf
K f  ar

37.2
 1.6
1.59   203.7 / d 3 
d = 2.41 in
Ans.
A quick check of kb and q show that our estimates are still reasonable for this diameter.
______________________________________________________________________________
6-20
Se  40 kpsi, S y  60 kpsi, Sut  80 kpsi,  m  15 kpsi,  a  25 kpsi,  m   a  0
Obtain von Mises stresses for the alternating, mean, and maximum stresses.
 a   a2  3 a2 
1/ 2
 m   m2  3 m2 
1/ 2
2
  252  3  0  


1/ 2
2
  0 2  3 15  


2
2
   max
 3 max
 max

1/2
 
1/ 2
 25.00 kpsi
 25.98 kpsi
1/2
2
2
  a   m   3  a   m  


1/2
  252  3 152   36.06 kpsi
Sy
60
ny 

 1.66 Ans.

 max
36.06
(a) Goodman, Equation (6-41)
nf 
1
 1.05
(25.00 / 40)  (25.98 / 80)
Ans.
(b) Gerber, Equation (6-48)
2 
2

 2(25.98)(40)  
1  80   25.00  
nf  
  1.31
 
 1  1  
2  25.98   40  
 80(25.00)  


Shigley’s MED, 11th edition
Ans.
Chapter 6 Solutions, Page 15/58
(c) Morrow
Estimate the fatigue strength coefficient.
Eq. (6-44):  f  Sut  50  80  50  130 kpsi
Eq. (6-46): n f 
1
 1.21
(25.00 / 40)  (25.98 / 130)
Ans.
______________________________________________________________________________
6-21 Se  40 kpsi, S y  60 kpsi, Sut  80 kpsi,  m  20 kpsi,  a  10 kpsi,  m   a  0
Obtain von Mises stresses for the alternating, mean, and maximum stresses.
 a   a2  3 a2 
1/ 2
 m   m2  3 m2 
1/2
2
 10 2  3  0  


1/ 2
2
  02  3  20  


2
2
   max
 3 max
 max

1/2
 
 10.00 kpsi
1/ 2
 34.64 kpsi
1/2
2
2
  a   m   3  a   m  


1/2
 102  3 202   36.06 kpsi
Sy
60
ny 

 1.66 Ans.

 max
36.06
(a) Goodman, Equation (6-41)
1
 1.46
(10.00 / 40)  (34.64 / 80)
(b) Gerber, Equation (6-48)
nf 
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) Morrow
Estimate the fatigue strength coefficient.
Eq. (6-44):  f  Sut  50  80  50  130 kpsi
1
 1.94 Ans.
(10.00 / 40)  (34.64 / 130)
______________________________________________________________________________
Eq. (6-46): n f 
6-22
Se  40 kpsi, S y  60 kpsi, Sut  80 kpsi,  a  10 kpsi,  m  15 kpsi,  a  12 kpsi,  m  0
Obtain von Mises stresses for the alternating, mean, and maximum stresses.
 a   a2  3 a2 
1/ 2
 m   m2  3 m2 
1/ 2
Shigley’s MED, 11th edition
2
 12 2  3 10  


2
  0 2  3 15  


1/ 2
1/ 2
 21.07 kpsi
 25.98 kpsi
Chapter 6 Solutions, Page 16/58
2
2
   max
 3 max
 max

1/ 2
1/ 2
2
2
  a   m   3  a   m  


1/ 2
2
2
 12  0   3 10  15    44.93 kpsi


Sy
60
ny 

 1.34 Ans.

 max
44.93
(a) Goodman, Equation (6-41)
nf 
1
 1.17
(21.07 / 40)  (25.98 / 80)
Ans.
(b) Gerber, Equation (6-48)
2

2
 2(25.98)(40)  
1  80   21.07  
nf  
   1.47
 
  1  1  
2  25.98   40  
 80(21.07)  


Ans.
(c) Morrow
Estimate the fatigue strength coefficient.
Eq. (6-44):  f  Sut  50  80  50  130 kpsi
1
 1.38 Ans.
(21.07 / 40)  (25.98 / 130)
______________________________________________________________________________
Eq. (6-46): n f 
6-23
Se  40 kpsi, S y  60 kpsi, Sut  80 kpsi,  a  30 kpsi,  m   a   a  0
Obtain von Mises stresses for the alternating, mean, and maximum stresses.
 a   a2  3 a2 
1/ 2
 m   m2  3 m2 
1/ 2
1/2
2
 02  3  30  


 51.96 kpsi
 0 kpsi
2
2
   max
 3 max
 max

1/ 2
1/ 2
2
2
  a   m   3  a   m  


1/ 2
2
 3  30    51.96 kpsi


Sy
60
ny 

 1.15 Ans.

 max
51.96
(a) through (c)
With a mean stress of zero, the Goodman, Gerber, and Morrow criteria all simplify to the
same simple comparison of the alternating stress to the endurance limit,
nf 
Shigley’s MED, 11th edition
Se
40

 0.77
 a 51.96
Ans.
Chapter 6 Solutions, Page 17/58
Since infinite life is not predicted, estimate a life from the S-N diagram. Since 'm = 0,
the stress state is completely reversed and the S-N diagram is applicable for 'a.
Fig. 6-23:
f = 0.875
Eq. (6-13):
( f Sut ) 2  0.875(80) 
a

 122.5
Se
40
Eq. (6-14):
 f Sut 
1
1
 0.875(80) 
b   log 
   log 
  0.08101
3
3
40


 Se 
2
1
1/ b
 
 51.96  0.08101
Eq. (6-15):
N   ar   
 39 600 cycles Ans.

 122.5 
 a 
______________________________________________________________________________
6-24
Se  40 kpsi, S y  60 kpsi, Sut  80 kpsi,  a  15 kpsi,  m  15 kpsi,  m   a  0
Obtain von Mises stresses for the alternating, mean, and maximum stresses.
 a   a2  3 a2 
1/2
 m   m2  3 m2 
1/2
1/2
2
 02  3 15  


 25.98 kpsi
1/2
2
 152  3  0  


2
2
   max
 3 max
 max

1/ 2
 15.00 kpsi
1/ 2
2
2
  a   m   3  a   m  


2 1/ 2
 15   3 15    30.00 kpsi


Sy
60
ny 

 2.00 Ans.

 max
30
2
(a) Goodman, Eq. (6-41)
nf 
1
 1.19
(25.98 / 40)  (15.00 / 80)
Ans.
(b) Gerber, Eq. (6-48)
2

2
 2(15.00)(40)  
1  80   25.98  
nf  
   1.43
 
  1  1  
2  15.00   40  
 80(25.98)  


Ans.
(c) Morrow
Estimate the fatigue strength coefficient.
Eq. (6-44):  f  Sut  50  80  50  130 kpsi
Eq. (6-46): n f 
1
 1.31
(25.98 / 40)  (15.00 /130)
Ans.
______________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 6 Solutions, Page 18/58
6-25
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
F
28 000
 max  max 
 147.4 N/mm 2  147.4 MPa
A
10(25  6)
Sy
490
 3.32 Ans.
 max 147.4
Determine the fatigue factor of safety based on infinite life
ny 

Eq. (6-10):
Se'  0.5(590)  295 MPa
Eq. (6-18):
k a  aSutb  3.04(590) 0.217  0.76
Eq. (6-20):
kb  1 (axial)
Eq. (6-25):
kc  0.85
Eq. (6-17):
Se  ka kb kc S e'  (0.76)(1)(0.85)(295)  190.6 MPa
Fig. 6-26:
Fig. A-15-1:
q = 0.83
d / w  0.24, K t  2.44
K f  1  q ( K t  1)  1  0.83(2.44  1)  2.20
a  K f
28 000   28 000 
Fmax  Fmin
 2.2
 324.2 MPa
2A
2(10)(25  6)
Fmax  Fmin
0
2A
Note, since m = 0, the stress is completely reversing, and
m  K f
190.6
 0.59 Ans.
 a 324.2
Since infinite life is not predicted, estimate the life from the S-N diagram. With m = 0,
the stress state is completely reversed, and the S-N diagram is applicable for a.
nf 
Fig. 6-23:
Se

f = 0.87
Eq. (6-13):
( f Sut ) 2  0.87(590) 
a

 1382
Se
190.6
Eq. (6-14):
 f Sut 
1
1
 0.87(590) 
b   log 
   log 
  0.1434
3
3
 190.6 
 Se 
Eq. (6-15):
 
N   ar 
 a 
2
1/ b
Shigley’s MED, 11th edition
1
 324.2  0.1434

 24 613 cycles

 1382 
Chapter 6 Solutions, Page 19/58
N = 25 000 cycles
Ans.
________________________________________________________________________
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.
 max 147.4
 max 
Determine the fatigue factor of safety based on infinite life.
From Prob. 6-25: Se  190.6 MPa, K f  2.2
a  K f
28 000  12 000 
Fmax  Fmin
 2.2
 92.63 MPa
2A
2(10)(25  6)
m  K f
 28 000  12 000 
Fmax  Fmin
 2.2 
  231.6 MPa
2A
 2(10)(25  6) 
Goodman criteria, Equation (6-41):
1
  
 92.63 231.6 
nf   a  m   


 190.6 590 
 Se Sut 
n f  1.14
1
Ans.
Gerber criteria, Equation (6-48):
2
2

 2 m Se  
1  Sut   a 
1  1  
nf  


2   m  Se 
 Sut a  


2
2

 2(231.6)(190.6)  
1  590  92.63 
 
1  1  


2  231.6  190.6 
590(92.63)  



n f  1.42
Ans.
Morrow criteria:
Estimate the fatigue strength coefficient.
Eq. (6-44):
 f  Sut  345  590  345  935 MPa
1
Eq. (6-46):
1
  
 92.63 231.6 
nf   a  m   


 Se  f 
 190.6 935 


n f  1.36
Shigley’s MED, 11th edition
Ans.
Chapter 6 Solutions, Page 20/58
The results are consistent with Fig. 6-36, where for a mean stress that is about half of the
yield strength, the Goodman line should predict failure significantly before the other two.
______________________________________________________________________________
6-27 Sut  590 MPa, S y  490 MPa
From Prob. 6-25: Se  190.6 MPa, K f  2.2
(a) Fmax  28 kN, Fmin  0 kN
Check for yielding
 max 
ny 
Fmax
28 000

 147.4 N/mm 2  147.4 MPa
A
10(25  6)
Sy
 max
a  K f

490
 3.32
147.4
Ans.
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 Goodman criteria, Eq. (6-41):
m  K f
1
1
  
 162.1 162.1 

nf   a  m   
  0.89
 190.6 590 
 Se Sut 
Ans.
Since infinite life is not predicted, estimate a life from the S-N diagram. First, find an
equivalent completely reversed stress. Using the Goodman criterion,
a
162.1
Eq. (6-58):
 ar 

 223.5 MPa
1  ( m / Sut ) 1  (162.1 / 590)
Fig. 6-23:
f = 0.87
Eq. (6-13):
( f Sut ) 2  0.87(590) 
a

 1382
Se
190.6
Eq. (6-14):
 f Sut 
1
1
 0.87(590) 
b   log 
   log 
  0.1434
3
3
 190.6 
 Se 
Eq. (6-15):
 
N   ar 
 a 
2
1/ b
1
 223.5  0.1434

 329 000 cycles

 1382 
Ans.
(b) Fmax  28 kN, Fmin  12 kN
Shigley’s MED, 11th edition
Chapter 6 Solutions, Page 21/58
The maximum load is the same as in part (a), so
 max  147.4 MPa
n y  3.32 Ans.
Factor of safety based on infinite life:
a  K f
Fmax  Fmin
28 000  12 000
 2.2
 92.63 MPa
2A
2(10)(25  6)
m  K f
 28 000  12 000 
Fmax  Fmin
 2.2 
  231.6 MPa
2A
 2(10)(25  6) 
1
Eq. (6-41):
1
  
 92.63 231.6 

nf   a  m   
  1.14
 190.6 590 
 Se Sut 
Ans.
(c) Fmax  12 kN, Fmin  28 kN
The compressive load is the largest, so check it for yielding.
 min 
ny 
28 000
Fmin

 147.4 MPa
A
10(25  6)
S yc
 min

490
 3.32
147.4
Ans.
Factor of safety based on infinite life:
a  K f
12 000   28 000 
Fmax  Fmin
 2.2
 231.6 MPa
2A
2(10)(25  6)
m  K f
12 000   28 000  
Fmax  Fmin
 2.2 
  92.63 MPa
2A
 2(10)(25  6) 
190.6
 0.82 Ans.
 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, consistent with the horizontal fatigue line in Fig. 6-34. Get a
and b from part (a).
For m < 0, Eq. (6-42):
nf 
Se
1/ b

1
 
 231.6  0.1434
Eq. (6-15):
N   ar   
Ans.
 257 000 cycles

 1382 
 a 
______________________________________________________________________________
6-28
Eq. (2-36): Sut = 0.5(400) = 200 kpsi
Shigley’s MED, 11th edition
Chapter 6 Solutions, Page 22/58
Eq. (6-10):
Se'  0.5(200)  100 kpsi
Eq. (6-18):
k a  aSutb  11.0(200) 0.650  0.35
Eq. (6-24):
d e  0.37 d  0.37(0.375)  0.1388 in
Eq. (6-19):
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. (6-17):
Se  (0.35)(1)(100)  35.0 kpsi
40  20
40  20
Fa 
 10 lb
Fm 
 30 lb
2
2
32M a 32(10)(12)
a 

 23.18 kpsi
d3
 (0.375)3
m 
32M m 32(30)(12)

 69.54 kpsi
d3
 (0.375)3
(a) Goodman criterion, Eq. (6-41):
1  a  m 23.18 69.54




nf
Se Sut
35.0
200
n f  0.99
Ans.
Since infinite life is not predicted, estimate a life from the S-N diagram. First, find an
equivalent completely reversed stress. Using the Goodman criterion,
a
23.18
Eq. (6-58):
 ar 

 35.54 kpsi
1  ( m / Sut ) 1  (69.54 / 200)
Fig. 6-23:
f = 0.78
Eq. (6-13):
( f Sut ) 2  0.78(200) 
a

 695.3
Se
35
Eq. (6-14):
 f Sut 
1
1
 0.78(200) 
b   log 
   log 
  0.2164
3
3
 35.0 
 Se 
Eq. (6-15):
 
N   ar 
 a 
2
1/ b
1
 35.54  0.2164

 929 000 cycles

 695.3 
Ans.
(b) Gerber criterion, Eq. (6-48):
Shigley’s MED, 11th edition
Chapter 6 Solutions, Page 23/58
2
2 

 2 m Se  
1  Sut   a 
nf  
1  1  


2   m  Se 
Sut a  



2
2

 2(69.54)(35.0)  
1  200  23.18 
 
1  1  


2  69.54  35.0 
 200(23.18)  


Ans.
 1.23 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.
Eq. (2-36):
Sut = =3.4HB = 3.4(490) = 1666 MPa
From problem statement: Sy = 0.9Sut = 0.9(1666) = 1499 MPa
Se  700 MPa
Eq. (6-10):
Eq. (6-18):
Eq. (6-24):
Eq. (6-19):
Eq. (6-17):
ka = 1.38(1666)-0.067 = 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 
Shigley’s MED, 11th edition
h
4

 5.7708 mm
ln(ro / ri ) ln(8 / 4)
Chapter 6 Solutions, Page 24/58
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. (3-76) with the axial stresses
added. The signs have been set to account for tension and compression as appropriate.
i  
o 
Mci F
(142 F )(1.7708) F
 

 3.441F MPa
80(0.2292)(4) 80
Aeri A
Mco F (142 F )(2.2292) F
 

 2.145 F MPa
80(0.2292)(8) 80
Aero A
( i ) min  3.441(144.9)  498.6 MPa
( i ) max  3.441(48.3)  166.2 MPa
( o ) min  2.145(48.3)  103.6 MPa
( o ) max  2.145(144.9)  310.8 MPa
( i ) a 
( i )m 
166.2   498.6 
 166.2 MPa
2
166.2   498.6 
 332.4 MPa
2
310.8  103.6
 103.6 MPa
( o ) a 
2
310.8  103.6
( 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 m < 0, Eq. (6-42): n f  e 
 3.54
 a 166.2
Outer radius:
Since m > 0, using the Goodman line, Eq. (6-41),
Shigley’s MED, 11th edition
Chapter 6 Solutions, Page 25/58
1
  
 103.6 207.2 

nf   a  m   

 588 1666 
 Se Sut 
n f  3.33
1
Infinite life is predicted at both inner and outer radii. The outer radius is critical, with a
fatigue factor of safety of nf = 3.33.
Ans.
______________________________________________________________________________
6-30
From Table A-20, for AISI 1018 CD, Sut  64 kpsi, S y  54 kpsi
Eq. (6-10):
Se'  0.5(64)  32 kpsi
Eq. (6-18):
k a  2.00(64) 0.217  0.81
Eq. (6-19):
kb  1 (axial)
Eq. (6-25):
kc  0.85
Eq. (6-17):
Se  (0.81)(1)(0.85)(32)  22.0 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. 6-26 or Eqs. (6-33) and (6-35) 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
 max 
Fmax
5

 3.33 kpsi
w2 h 3.0(0.5)
 min 
16
 10.67 kpsi
3.0(0.5)
a  K f
 max   min
2
  max   min
2

m  K f 
ny 
Sy
 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 mean stress is negative, use Eq. (6-42).
nf 
Se
a

22.0
 1.83
12.0
Hole:
Shigley’s MED, 11th edition
Chapter 6 Solutions, Page 26/58
Fig. A-15-1: d / w1  0.4 / 3.5  0.11  K t  2.68
Use Fig. 6-26 or Eqs. (6-33) and (6-35) for q. Estimate a little high since it is off the
graph, q = 0.85
K f  1  0.85(2.68  1)  2.43
 max 
Fmax
5

 3.226 kpsi
h  w1  d  0.5(3.5  0.4)
 min 
Fmin
16

 10.32 kpsi
h  w1  d  0.5(3.5  0.4)
 max   min
a  K f
2
  max   min
2

m  Kf 
ny 
Sy
 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 mean stress is negative, use Eq. (6-42).
nf 
Se
a

22.0
 1.33
16.5
Thus the design is controlled by the threat of fatigue at the hole with a minimum factor of
safety of n f  1.33. Ans.
______________________________________________________________________________
6-31
Sut  64 kpsi, S y  54 kpsi
Eq. (6-10):
Se'  0.5(64)  32 kpsi
Eq. (6-18):
k a  2.00(64) 0.217  0.81
Eq. (6-19):
kb  1 (axial)
Eq. (6-25):
kc  0.85
Eq. (6-17):
Se  (0.81)(1)(0.85)(32)  22.0 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. 6-26 or Eqs. (6-33) and (6-35) 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
Shigley’s MED, 11th edition
Chapter 6 Solutions, Page 27/58
 max 
Fmax
16

 21.3 kpsi
w2 h 1.5(0.5)
4
 5.33 kpsi
1.5(0.5)
   min
21.3  ( 5.33)
 1.94
 25.8 kpsi
 a  K f max
2
2
 min 
  max   min
2

m  K f 
ny 
Sy
 max


 21.3  ( 5.33) 
  15.5 kpsi
  1.94 
2



54
 2.54
21.3
 Does not yield.
Using Goodman criteria, Eq. (6-41),
1
1
  
 25.8 15.5 
nf   a  m   

  0.71
 22.0 64 
 Se Sut 
Hole:
Fig. A-15-1: d / w1  0.4 / 2.5  0.16  K t  2.55
Use Fig. 6-26 or Eqs. (6-33) and (6-35) 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
 max 
Fmax
16

 15.2 kpsi
h  w1  d  0.5(2.5  0.4)
 min 
4
Fmin

 3.81 kpsi
h  w1  d  0.5(2.5  0.4)
 max   min
 15.2  ( 3.81) 
 2.32 
  22.1 kpsi
2
2


    min 
 15.2  ( 3.81) 
 m  K f  max
  13.2 kpsi
  2.32 
2
2




Sy
54
ny 

 3.55  Does not yield.
 max 15.2
a  Kf
Using Goodman criteria, Eq. (6-41),
1
1
  
 22.1 13.2 
nf   a  m   

  0.83
 22.0 64 
 Se Sut 
Shigley’s MED, 11th edition
Chapter 6 Solutions, Page 28/58
Thus the design is controlled by the threat of fatigue at the fillet with a minimum factor
of safety of n f  0.71 Ans.
______________________________________________________________________________
6-32
Sut  64 kpsi, S y  54 kpsi
From Prob. 6-30, the fatigue factor of safety at the hole is nf = 1.33. To match this at the
fillet,
S
S
22.0
nf  e  a  e 
 16.5 kpsi
n f 1.33
a
where Se is unchanged from Prob. 6-30. The only aspect of a that is affected by the fillet
radius is the fatigue stress concentration factor. Obtaining a in terms of Kf,
a  K f
 max   min
2
 Kf
3.33  ( 10.67)
 7.00 K f
2
Equating to the desired stress, and solving for Kf,
 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. 6-26, 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. 6-26, 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, S ut  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,
Shigley’s MED, 11th edition
Chapter 6 Solutions, Page 29/58
rn 
3 / 16
h

 0.74608 in
0.84375
ro
ln
ln
0.65625
ri
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),
i 
Mci
T (0.08983)

 993.3T
Aeri (0.035156)(0.00392)(0.65625)
where T is in lbf∙in and  i is in psi.
1
2
 a  496.7T
 m  (993.3)T  496.7T
Eq. (6-10):
Se'  0.5 110   55 kpsi
Eq. (6-18):
k a  2.00(110) 0.217  0.72
Eq. (6-24):
d e  0.808  3 /16  3 /16  
Eq. (6-19):
kb  0.879  0.1515 
Eq. (6-18):
Se  (0.72)(1)(55)  39.6 kpsi
1/ 2
0.107
 0.1515 in
 1.08 (round to 1)
For a compressive mean component, from Eq. (6-42),  a  S e / n f . Thus,
39.6
3
T  26.6 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
ro  2.5 
Shigley’s MED, 11th edition
Chapter 6 Solutions, Page 30/58
o 
Mco
T (0.09492)

 889.7T psi
Aero (0.035156)(0.00117)(2.59375)
1
2
(a) Using Eq. (6-41), for Goodman, we have
 m   a  (889.7T )  444.9T psi
1
1

 
 0.4449T 0.4449T 

nf   a  m   
 3
39.6
110
S
S


ut 
 e
T  21.8 lbf  in
Ans.
(b) For Morrow, estimate the fatigue strength coefficient from Eq. (6-44),
 f  Sut  50  110  50  160 kpsi
1
Eq. (6-46):
1
a m 
 0.4449T 0.4449T 



nf  
 3
 S   
160 
 39.6
f 
 e
T  23.8 lbf  in
Ans.
(c) To guard against yield, use T of part (b) and the inner stress.
Sy
60
 2.54 Ans.
 i 0.9933(23.8)
______________________________________________________________________________
ny 
6-34

From Prob. 6-33, Se  39.6 kpsi, S y  60 kpsi, and Sut  110 kpsi
(a) Assuming the beam is straight,
 max 
Mc M  h / 2  6 M
6T

 2 
 910.2T
3
I
bh /12
bh
(3 / 16)3
Using Eq. (6-41), for Goodman, we have
1
1

 
 0.4551T 0.4551T 
nf   a  m   

 3
110 
 39.6
 Se Sut 
T  21.3 lbf  in
Ans.
(b) ) For Morrow, estimate the fatigue strength coefficient from Eq. (6-44),
 f  Sut  50  110  50  160 kpsi
1
Eq. (6-46):
1
a m 
 0.4551T 0.4551T 



nf  
 3
 S   
160 
 39.6
f 
 e
T  23.3 lbf  in
Shigley’s MED, 11th edition
Ans.
Chapter 6 Solutions, Page 31/58
Sy
60
 2.83 Ans.
 max 0.9102(23.3)
______________________________________________________________________________
(c)
6-35
ny 

K f ,bend  1.4, K f ,axial  1.1, K f ,tors  2.0, S y  300 MPa, Sut  400 MPa, Se  160 MPa
Bending:
Axial:
 m  0,  a  60 MPa
 m  20 MPa,  a  0
 m  35 MPa,  a  35 MPa
Torsion:
Eqs. (6-66) and (6-67):
 a 
1.4(60)  0
 m 
 0  1.1(20)
2
2
 3 2.0(35)   147.5 MPa
2
 3 2.0(35)   123.2 MPa
2
   a   m ,
Check for yielding, using the conservative  max
ny 
Sy

 a   m
300
 1.11
147.5  123.2
Yielding is not predicted. Ans.
Using Goodman, Eq. (6-41),
1
  
 147.5 123.2 
nf   a  m   


S
S
160
400 

ut 
 e
n f  0.81
1
Ans.
Finite life is predicted. To use the Walker criterion for estimating an equivalent
completely reversed stress, estimate the material fitting parameter for steels with Eq. (657).
  0.0002 Sut  0.8818  0.0002(400)  0.8818  0.8018
Eq. (6-61):
Fig. 6-23:
 ar   m   a   a  123.2  147.5 
Off the chart, so use f = 0.9
1 
1 0.8018
Eq. (6-13):
( f Sut ) 2  0.9(400) 
a

 810
Se
160
Eq. (6-14):
 f Sut
1
b   log 
3
 Se
Eq. (6-15):
 
N   ar 
 a 
147.50.8018  166.4 MPa
2
1/ b
Shigley’s MED, 11th edition

1
 0.9(400) 
   log 
  0.1174
3
 160 

1
 166.4  0.1174

 716 000 cycles

 810 
Ans.
Chapter 6 Solutions, Page 32/58
______________________________________________________________________________
6-36
K f ,bend  1.4, K f ,tors  2.0, S y  300 MPa, S ut  400 MPa, Se  160 MPa
Bending:  max  150 MPa,  min  40 MPa,  m  55 MPa,  a  95 MPa
Torsion:  m  90 MPa,  a  9 MPa
Eqs. (6-66) and (6-67):
 a 
1.4(95) 
 m 
1.4(55) 
 3 2.0(9)   136.6 MPa
2
2
2
 3 2.0(90)   321.1 MPa
2
Using Goodman, Eq. (6-41),
1
1
  
 136.6 321.1 
nf   a  m   

  0.60
400 
 160
 Se Sut 
Ans.
   a   m ,
Check for yielding, using the conservative  max
ny 
Sy
 a   m

300
 0.66
136.6  321.1
Ans.
Since the conservative yield check indicates yielding, we will check more carefully with
 obtained directly from the maximum stresses, using the distortion energy failure
 max
theory, without stress concentrations. Note that this is exactly the method used for static
failure in Ch. 5.
 
 max
 max 
2
 3  max  
2
150 
2
 3  90  9   227.8 MPa
2
Sy
300

 1.32 Ans.

227.8
 max
Since yielding is not predicted, and infinite life is not predicted, we would like to
estimate a life from the S-N diagram.
ny 
To use the Walker criterion for estimating an equivalent completely reversed stress,
estimate the material fitting parameter for steels with Eq. (6-57).
  0.0002 Sut  0.8818  0.0002(400)  0.8818  0.8018
Eq. (6-61):
Fig. 6-23:
 ar   m   a   a   321.1  136.6 
Off the chart, so use f = 0.9
Eq. (6-13):
( f Sut ) 2  0.9(400) 
a

 810
Se
160
1 
1 0.8018
136.60.8018  173.6 MPa
2
Shigley’s MED, 11th edition
Chapter 6 Solutions, Page 33/58
Eq. (6-14):
 f Sut
1
b   log 
3
 Se

1
 0.9(400) 
   log 
  0.1174
3
 160 

1
1/ b
 
 173.6  0.1174
N   ar   
Ans.
Eq. (6-15):
 499 000 cycles

 810 
 a 
_____________________________________________________________________________
6-37
Table A-20: S ut  64 kpsi, S y  54 kpsi
From Prob. 3-79, the critical stress element experiences  = 15.3 kpsi and  = 4.43 kpsi.
The bending is completely reversed due to the rotation, and the torsion is steady, giving
a = 15.3 kpsi, m = 0 kpsi, a = 0 kpsi, m = 4.43 kpsi. Obtain von Mises stresses for the
alternating, mean, and maximum stresses.
 a   a2  3 a2 
1/ 2
 m   m2  3 m2 
1/2
1/ 2
2
 15.32  3  0  


 15.3 kpsi
1/ 2
2
  0 2  3  4.43  


2
2
   max
 max
 3 max

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
ny 

 3.16

 max
17.11
Obtain the modifying factors and endurance limit.
Eq. (6-10):
Se  0.5  64   32 kpsi
Eq. (6-18):
ka  2.00(64) 0.217  0.81
Eq. (6-19):
kb  0.879(1.25) 0.107  0.86
Se  0.81(0.86)(32)  22.3 kpsi
Eq. (6-17):
Using Goodman, Eq. (6-41),
1
1
  
 15.3 7.67 
nf   a  m   

Ans.
  1.24
 22.3 64 
 Se Sut 
______________________________________________________________________________
6-38
Table A-20: S ut  440 MPa, S y  370 MPa
From Prob. 3-80, the critical stress element experiences  = 263 MPa and  = 57.7 MPa.
The bending is completely reversed due to the rotation, and the torsion is steady, giving
a = 263 MPa, m = 0, a = 0 MPa, m = 57.7 MPa. Obtain von Mises stresses for the
alternating, mean, and maximum stresses.
Shigley’s MED, 11th edition
Chapter 6 Solutions, Page 34/58
 a   a2  3 a2 
1/ 2
 m   m2  3 m2 
1/ 2
1/ 2
2
  2632  3  0  


 263 MPa
1/ 2
2
 0 2  3  57.7  


2
2
   max
 3 max
 max

1/ 2
 99.9 MPa
2
  2632  3  57.7  


1/ 2
 281 MPa
Check for yielding, using the distortion energy failure theory.
Sy
370

 1.32

281
 max
Obtain the modifying factors and endurance limit.
ny 
Eq. (6-10):
Se  0.5  440   220 MPa
Eq. (6-18):
k a  3.04(440) 0.217  0.81
Eq. (6-19):
kb  1.24(30) 0.107  0.86
Eq. (6-17):
Se  0.81(0.86)(220)  153 MPa
Using Goodman,
Eq. (6-41):
1
  
 263 99.9 

nf   a  m   

S
S
153
440 

e
ut


n f  0.51
1
Infinite life is not predicted.
Ans.
______________________________________________________________________________
6-39
Table A-20: Sut  64 kpsi, S y  54 kpsi
From Prob. 3-81, the critical stress element experiences  = 21.5 kpsi and  = 5.09 kpsi.
The bending is completely reversed due to the rotation, and the torsion is steady, giving
a = 21.5 kpsi, m = 0 kpsi, a = 0 kpsi, m = 5.09 kpsi. Obtain von Mises stresses for the
alternating, mean, and maximum stresses.
 a   a2  3 a2 
1/ 2
 m   m2  3 m2 
1/ 2
1/ 2
2
  21.52  3  0  


1/ 2
2
  0 2  3  5.09  


2
2
   max
 max
 3 max

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 
Shigley’s MED, 11th edition
Sy
54

 2.32

 max
23.24
Chapter 6 Solutions, Page 35/58
Obtain the modifying factors and endurance limit.
Se  0.5  64   32 kpsi
Eq. (6-10):
Eq. (6-18):
ka  2.00(64) 0.217  0.81
Eq. (6-19):
kb  0.879(1) 0.107  0.88
Eq. (6-17):
Se  0.81(0.88)(32)  22.8 kpsi
Using Goodman, Eq. (6-41),
1
1
  
 21.5 8.82 
nf   a  m   

Ans.
  0.93
 22.8 64 
 Se Sut 
______________________________________________________________________________
6-40
Table A-20: S ut  440 MPa, S y  370 MPa
From Prob. 3-82, the critical stress element experiences  = 72.9 MPa and  = 20.3 MPa.
The bending is completely reversed due to the rotation, and the torsion is steady, giving
a = 72.9 MPa, m = 0 MPa, a = 0 MPa, m = 20.3 MPa. Obtain von Mises stresses for
the alternating, mean, and maximum stresses.
 a   a2  3 a2 
1/ 2
 m   m2  3 m2 
1/2
1/ 2
2
 72.92  3  0  


1/ 2
2
  0 2  3  20.3  


2
2
   max
 max
 3 max

1/ 2
 72.9 MPa
 35.2 MPa
2
  72.9 2  3  20.3 


1/ 2
 80.9 MPa
Check for yielding, using the distortion energy failure theory.
Sy
370
ny 

 4.57

80.9
 max
Obtain the modifying factors and endurance limit.
Eq. (6-10):
Se  0.5  440   220 MPa
Eq. (6-18):
ka  3.04(440) 0.217  0.81
Eq. (6-19):
kb  1.24(20) 0.107  0.90
Eq. (6-17):
Se  0.81(0.90)(220)  160.4 MPa
Using Goodman, Eq. (6-41),
1
1
  
 72.9 35.2 
nf   a  m   

Ans.
  1.87
 160.4 440 
 Se Sut 
______________________________________________________________________________
6-41
Table A-20: S ut  64 kpsi, S y  54 kpsi
Shigley’s MED, 11th edition
Chapter 6 Solutions, Page 36/58
From Prob. 3-83, the critical stress element experiences  = 35.2 kpsi and  = 7.35 kpsi.
The bending is completely reversed due to the rotation, and the torsion is steady, giving
a = 35.2 kpsi, m = 0 kpsi, a = 0 kpsi, m = 7.35 kpsi. Obtain von Mises stresses for the
alternating, mean, and maximum stresses.
 a   a2  3 a2 
1/ 2
 m   m2  3 m2 
1/2
1/ 2
2
 35.2 2  3  0  


 35.2 kpsi
1/ 2
2
  0 2  3  7.35  


2
2
   max
 3 max
 max

1/ 2
 12.7 kpsi
2
 35.22  3  7.35  


1/ 2
 37.4 kpsi
Check for yielding, using the distortion energy failure theory.
ny 
Sy
54

 1.44

37.4
 max
Obtain the modifying factors and endurance limit.
Eq. (6-10):
Se  0.5(64)  32 kpsi
Eq. (6-18):
ka  2.00(64) 0.217  0.81
Eq. (6-19):
kb  0.879(1.25) 0.107  0.86
Eq. (6-17):
Se  0.81(0.86)(32)  22.3 kpsi
Using Goodman, Eq. (6-41),
1
1
  
 35.2 12.7 
nf   a  m   

Ans.
  0.56
 22.3 64 
 Se Sut 
______________________________________________________________________________
6-42
Table A-20: S ut  440 MPa, S y  370 MPa
From Prob. 3-84, the critical stress element experiences  = 333.9 MPa and  = 126.3
MPa. The bending is completely reversed due to the rotation, and the torsion is steady,
giving a = 333.9 MPa, m = 0 MPa, a = 0 MPa, m = 126.3 MPa. Obtain von Mises
stresses for the alternating, mean, and maximum stresses.
 a   a2  3 a2 
1/ 2
 m   m2  3 m2 
1/ 2
2
 333.9 2  3  0  


1/ 2
2
  0 2  3 126.3  


2
2
   max
 max
 3 max

1/ 2
1/ 2
 333.9 MPa
 218.8 MPa
1/2
2
 333.92  3 126.3  


 399.2 MPa
Check for yielding, using the distortion energy failure theory.
Shigley’s MED, 11th edition
Chapter 6 Solutions, Page 37/58
ny 
Sy
370

 0.93

 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. (6-10):
Se  0.5(440)  220 MPa
Eq. (6-18):
ka  3.04(440) 0.217  0.81
Eq. (6-19):
kb  1.24(50) 0.107  0.82
Eq. (6-17):
Se  0.81(0.82)(220)  146.1 MPa
Using Goodman, Eq. (6-41),
1
1
  
 333.9 218.8 
nf   a  m   

Ans.
  0.36
 146.1 440 
 Se Sut 
______________________________________________________________________________
6-43
Table A-20: Sut  64 kpsi, S y  54 kpsi
From Prob. 3-85, the critical stress element experiences completely reversed bending
stress due to the rotation, and steady torsional and axial stresses.
 a ,bend  9.495 kpsi,
 a ,axial  0 kpsi,
 a  0 kpsi,
 m,bend  0 kpsi
 m,axial  0.362 kpsi
 m  11.07 kpsi
Obtain von Mises stresses for the alternating, mean, and maximum stresses.
 a   a2  3 a2 
1/ 2
 m   m2  3 m2 
1/ 2
2
2
  9.495   3  0  


1/ 2
 9.495 kpsi
2
2
  0.362   3 11.07  


2
2
   max
 max
 3 max

1/ 2
1/ 2
 19.18 kpsi
1/ 2
2
2
  9.495  0.362   3 11.07  


 21.56 kpsi
Check for yielding, using the distortion energy failure theory.
Sy
54
ny 

 2.50

 max
21.56
Obtain the modifying factors and endurance limit.
Shigley’s MED, 11th edition
Chapter 6 Solutions, Page 38/58
Eq. (6-10):
Se  0.5(64)  32 kpsi
Eq. (6-18):
ka  2.00(64) 0.217  0.81
Eq. (6-19):
kb  0.879(1.13) 0.107  0.87
Eq. (6-17):
Se  0.81(0.87)(32)  22.6 kpsi
Using Goodman, Eq. (6-41),
1
1
  
 9.495 19.18 
nf   a  m   

Ans.
  1.39
64 
 22.6
 Se Sut 
______________________________________________________________________________
6-44
Table A-20: S ut  64 kpsi, S y  54 kpsi
From Prob. 3-87, the critical stress element experiences completely reversed bending
stress due to the rotation, and steady torsional and axial stresses.
 a ,bend  33.99 kpsi,  m ,bend  0 kpsi
 a ,axial  0 kpsi,
 a  0 kpsi,
 m,axial  0.153 kpsi
 m  7.847 kpsi
Obtain von Mises stresses for the alternating, mean, and maximum stresses.
 a   a2  3 a2 
1/ 2
 m   m2  3 m2 
1/ 2
2
2
  33.99   3  0  


1/ 2
 33.99 kpsi
2
2
  0.153  3  7.847  


2
2
   max
 max
 3 max

1/ 2
1/ 2
 13.59 kpsi
1/ 2
2
2
  33.99  0.153   3  7.847  


 36.75 kpsi
Check for yielding, using the distortion energy failure theory.
ny 
Sy
54

 1.47

 max
36.75
Obtain the modifying factors and endurance limit.
Eq. (6-10):
Se  0.5(64)  32 kpsi
Eq. (6-18):
ka  2.00(64) 0.217  0.81
Eq. (6-19):
kb  0.879(0.88) 0.107  0.89
Eq. (6-17):
Se  0.81(0.89)(32)  23.1 kpsi
Using Goodman, Eq. (6-41),
1
1
  
 33.99 13.59 
nf   a  m   

  0.59
64 
 23.1
 Se Sut 
Shigley’s MED, 11th edition
Ans.
Chapter 6 Solutions, Page 39/58
______________________________________________________________________________
6-45
Table A-20: S ut  440 MPa, S y  370 MPa
From Prob. 3-88, the critical stress element experiences  = 68.6 MPa and  = 37.7 MPa.
The bending is completely reversed due to the rotation, and the torsion is steady, giving
a = 68.6 MPa, m = 0 MPa, a = 0 MPa, m = 37.7 MPa. Obtain von Mises stresses for
the alternating, mean, and maximum stresses.
 a   a2  3 a2 
1/ 2
 m   m2  3 m2 
1/ 2
1/ 2
2
  68.62  3  0  


1/ 2
2
  0 2  3  37.7  


 68.6 MPa
 65.3 MPa
2 1/ 2
2
2
   max
 3 max
 max

1/ 2
  68.62  3  37.7    94.7 MPa


Check for yielding, using the distortion energy failure theory.
ny 
Sy
370

 3.91

 max
94.7
Obtain the modifying factors and endurance limit.
Eq. (6-10):
Se  0.5(440)  220 MPa
Eq. (6-18):
ka  3.04(440) 0.217  0.81
Eq. (6-19):
kb  1.24(30) 0.107  0.86
Eq. (6-17):
Se  0.81(0.86)(220)  153 MPa
Using Goodman, Eq. (6-41),
1
1
  
 68.6 65.3 
nf   a  m   

Ans.
  1.68
S
S
153
440


ut 
 e
______________________________________________________________________________
6-46
Table A-20: Sut  64 kpsi, S y  54 kpsi
From Prob. 3-90, the critical stress element experiences  = 3.46 kpsi and  = 0.882 kpsi.
The bending is completely reversed due to the rotation, and the torsion is steady, giving
a = 3.46 kpsi, m = 0, a = 0 kpsi, m = 0.882 kpsi. Obtain von Mises stresses for the
alternating, mean, and maximum stresses.
 a   a2  3 a2 
1/ 2
 m   m2  3 m2 
1/ 2
1/ 2
2
 3.462  3  0  


1/ 2
2
 02  3  0.882  


2
2
   max
 max
 3 max

1/ 2
Shigley’s MED, 11th edition
 3.46 kpsi
 1.53 kpsi
1/ 2
2
 3.46 2  3  0.882  


 3.78 kpsi
Chapter 6 Solutions, Page 40/58
Check for yielding, using the distortion energy failure theory.
ny 
Sy
54

 14.3

 max
3.78
Obtain the modifying factors and endurance limit.
Eq. (6-10):
Se  0.5(64)  32 kpsi
Eq. (6-18):
k a  2.00(64) 0.217  0.81
Eq. (6-19):
kb  0.879(1.375) 0.107  0.85
Eq. (6-17):
Se  0.81(0.85)(32)  22.0 kpsi
Using Goodman,
1
1
  
 3.46 1.53 
nf   a  m   

Eq. (6-41):

 22.0 64 
 Se Sut 
Ans.
n f  5.5
______________________________________________________________________________
6-47
Table A-20: Sut  64 kpsi, S y  54 kpsi
From Prob. 3-91, the critical stress element experiences  = 16.3 kpsi and  = 5.09 kpsi.
Since the load is applied and released repeatedly, this gives max = 16.3 kpsi, min = 0
kpsi, max = 5.09 kpsi, min = 0 kpsi. Consequently,m = a = 8.15 kpsi, m = a = 2.55
kpsi.
For bending, from Eqs. (6-33) and (6-35),
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. (6-33) and (6-36),
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):
Shigley’s MED, 11th edition
1
3

Chapter 6 Solutions, Page 41/58
Obtain von Mises stresses for the alternating and mean stresses from Eqs. (6-66) and (667).

2
 a  1.38  8.15   3 1.88  2.55 

2 1/ 2
 13.98 kpsi
 m   a  13.98 kpsi
   a   m ,
Check for yielding, using the conservative  max
Sy
54

 1.93
ny 
 a   m 13.98  13.98
Obtain the modifying factors and endurance limit.
Eq. (6-10):
Se  0.5(64)  32 kpsi
Eq. (6-18):
k a  aSutb  2.00(64) 0.217  0.81
Eq. (6-23):
de  0.370d  0.370 1  0.370 in
Eq. (6-19):
kb  0.879d e 0.107  0.879(0.370) 0.107  0.98
Eq. (6-17):
Se  (0.81)(0.98)(32)  25.4 kpsi
Using Goodman,
Eq. (6-41):
1
  
 13.98 13.98 
nf   a  m   


64 
 25.4
 Se Sut 
n f  1.3
1
Ans.
______________________________________________________________________________
6-48
Table A-20: Sut  64 kpsi, S y  54 kpsi
From Prob. 3-92, the critical stress element experiences  = 16.4 kpsi and  = 4.46 kpsi.
Since the load is applied and released repeatedly, this gives max = 16.4 kpsi, min = 0
kpsi, max = 4.46 kpsi, min = 0 kpsi. Consequently,m = a = 8.20 kpsi, m = a = 2.23
kpsi.
For bending, from Eqs. (6-33) and (6-35),
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. (6-33) and (6-36),
Shigley’s MED, 11th edition
Chapter 6 Solutions, Page 42/58
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 mean stresses from Eqs. (6-66) and (667).

2
 a  1.38  8.20    3 1.88  2.23 

2 1/ 2
 13.45 kpsi
 m   a  13.45 kpsi
   a   m ,
Check for yielding, using the conservative  max
Sy
54

 2.01
ny 
 a   m 13.45  13.45
Obtain the modifying factors and endurance limit.
Eq. (6-10):
Se  0.5(64)  32 kpsi
Eq. (6-18):
k a  aSutb  2.00(64) 0.217  0.81
Eq. (6-23):
d e  0.370d  0.370(1)  0.370 in
Eq. (6-19):
kb  0.879d e 0.107  0.879(0.370) 0.107  0.98
Eq. (6-17):
Se  (0.81)(0.98)(32)  25.4 kpsi
Using Goodman,
Eq. (6-41):
1
  
 13.45 13.45 
nf   a  m   


S
S
25.4
64 

ut 
 e
n f  1.35
1
Ans.
______________________________________________________________________________
6-49
Table A-20: Sut  64 kpsi, S y  54 kpsi
From Prob. 3-93, the critical stress element experiences repeatedly applied bending,
axial, and torsional stresses of x,bend = 20.2 kpsi, x,axial = 0.1 kpsi, and  = 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 max = 20.3 kpsi, min = 0 kpsi, max = 5.09 kpsi, min =
0 kpsi. Consequently,m = a = 10.15 kpsi, m = a = 2.55 kpsi.
For bending, from Eqs. (6-33) and (6-35),
Shigley’s MED, 11th edition
Chapter 6 Solutions, Page 43/58
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
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):
3

For torsion, from Eqs. (6-33) and (6-36),
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 mean stresses from Eqs. (6-66) and (667).

2
 a  1.38 10.15    3 1.88  2.55 

2 1/ 2
 16.28 kpsi
 m   a  16.28 kpsi
   a   m ,
Check for yielding, using the conservative  max
Sy
54

 1.66
ny 
 a   m 16.28  16.28
Obtain the modifying factors and endurance limit.
Eq. (6-10):
Se  0.5(64)  32 kpsi
Eq. (6-18):
k a  aSutb  2.00(64) 0.217  0.81
Eq. (6-23):
d e  0.370d  0.370(1)  0.370 in
Eq. (6-19):
kb  0.879d e 0.107  0.879(0.370) 0.107  0.98
Eq. (6-17):
Se  (0.81)(0.98)(32)  25.4 kpsi
Using Goodman,
1
1
  
 16.28 16.28 
nf   a  m   

Eq. (6-41):

S
S
25.4
64 

ut 
 e
n f  1.12 Ans.
____________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 6 Solutions, Page 44/58
6-50
Table A-20: Sut  64 kpsi, S y  54 kpsi
From Prob. 3-94, 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 max = 14.3 kpsi, min = 0 kpsi. Thus, m = a = 7.15 kpsi. Since the stress is entirely
shear, it is convenient to check for yielding using the standard Maximum Shear Stress
theory.
S y / 2 54 / 2

 1.89
ny 
 max
14.3
Find the modifiers and endurance limit.
Eq. (6-10):
Se  0.5(64)  32 kpsi
Eq. (6-18):
k a  aSutb  2.00(64) 0.217  0.81
The size factor for a rectangular cross section loaded in torsion is not readily available.
An equivalent diameter based on the 95 percent stress area is not readily obtained, since
the stress situation in this case is nonlinear, as described in Section 3-12. 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.
d e  0.25 in
Eq. (6-19):
kb  0.879de 0.107  0.879(0.25) 0.107  1.02
We will round down to kb = 1.
Eq. (6-25):
kc  0.59
Eq. (6-17):
S se  0.81(1)(0.59)(32)  15.3 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. 6-16. From Eq. (6-58), Ssu = 0.67Su = 0.67 (64) = 42.9 kpsi.
Using Goodman,
1
1

 
 7.15 7.15 
nf   a  m   

Eq. (6-41):
Ans.
  1.58
S
S
15.3
42.9


se
su


______________________________________________________________________________
6-51
Table A-20: Sut  64 kpsi, S y  54 kpsi
From Prob. 3-95, the critical stress element experiences  = 28.0 kpsi and  = 15.3 kpsi.
Since the load is applied and released repeatedly, this gives max = 28.0 kpsi, min = 0
Shigley’s MED, 11th edition
Chapter 6 Solutions, Page 45/58
kpsi, max = 15.3 kpsi, min = 0 kpsi. Consequently,m = a = 14.0 kpsi, m = a = 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
Figs. 6-26 and 6-27: qbend = 0.78, qtors = 0.82
Eq. (6-32):
K f ,bend  1  qbend  K t ,bend  1  1  0.78 1.60  1  1.47
K f ,tors  1  qtors  K t ,tors  1  1  0.82 1.39  1  1.32
Obtain von Mises stresses for the alternating and mean stresses from Eqs. (6-66) and (667).

2
 a  1.47 14.0   3 1.32  7.65  

2 1/ 2
 27.0 kpsi
 m   a  27.0 kpsi
   a   m ,
Check for yielding, using the conservative  max
Sy
54
ny 

 1.00
 a   m 27.0  27.0
Since stress concentrations are included in this quick yield check, the low factor of safety
is acceptable.
Eq. (6-10):
Se  0.5(64)  32 kpsi
Eq. (6-18):
k a  aSutb  2.00(64) 0.217  0.81
Eq. (6-23):
de  0.370d  0.370 1  0.370 in
Eq. (6-19):
kb  0.879d e 0.107  0.879(0.370) 0.107  0.98
Eq. (6-17):
Se  (0.81)(0.98)(0.5)(64)  25.4 kpsi
For the Morrow criterion, estimate the fatigue strength coefficient for steel.
Eq. (6-44):
 f  Sut  50  64  50  114 kpsi
1
Eq. (6-46):
1
  a  m 
 27.0 27.0 
nf  




 S   
 25.4 114 
f 
 e
n f  0.77 Ans.
Since infinite life is not predicted, estimate a life from the S-N diagram. First, find an
equivalent completely reversed stress, again using Morrow.
Shigley’s MED, 11th edition
Chapter 6 Solutions, Page 46/58
Eq. (6-59):
Fig. 6-23:
 a
27.0

 35.4 kpsi
1  ( m /  f ) 1  (27.0 / 114)
Off the chart, so use f = 0.9
 ar 
Eq. (6-13):
( f Sut ) 2  0.9(64) 
a

 130.6
Se
25.4
Eq. (6-14):
 f Sut
1
b   log 
3
 Se
2

1
 0.9(64) 
   log 
  0.1185
3
 25.4 

1
1/ b
  ar 
 35.4  0.1185
N 
Ans.
Eq. (6-15):

 60856 cycles  61 000 cycles


 130.6 
 a 
______________________________________________________________________________
6-52
Table A-20: Sut  64 kpsi, S y  54 kpsi
From Prob. 3-96, the critical stress element experiences x,bend = 46.1 kpsi, x,axial = 0.382
kpsi and  = 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
max,bend = 46.1 kpsi, min,bend = 0 kpsi, max,axial = 0.382 kpsi, min,axial = 0 kpsi, max = 15.3
kpsi, min = 0 kpsi. Consequently,m,bend = a,bend = 23.05 kpsi, m,axial = a,axial = 0.191
kpsi, m = a = 7.65 kpsi. From Table A-15-7, 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,
K t ,axial  1.75
Eqs. (6-33), (6-35), and (6-36), or Figs. 6-26 and 6-27: 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 mean stresses from Eqs. (6-66) and (667).


   1.47  23.05  1.59  0.191   3 1.32  7.65  
2
2 1/ 2
2
2 1/2
 a  1.47  23.05   1.59  0.191   3 1.32  7.65 
m
 38.4 kpsi
 38.4 kpsi
   a   m ,
Check for yielding, using the conservative  max
Sy
54
ny 

 0.70
 a   m 38.4  38.4
Shigley’s MED, 11th edition
Chapter 6 Solutions, Page 47/58
Since the conservative yield check indicates yielding, we will check more carefully with
 obtained directly from the maximum stresses, using the distortion energy failure
 max
theory, without stress concentrations. Note that this is exactly the method used for static
failure in Ch. 5.
 
 max
ny 

  max,axial   3  max  
2
max,bend
Sy
54

 1.01

53.5
 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. (6-10):
Se  0.5(64)  32 kpsi
Eq. (6-18):
k a  aSutb  2.00(64) 0.217  0.81
Eq. (6-23):
de  0.370d  0.370 1  0.370 in
Eq. (6-19):
kb  0.879d e 0.107  0.879(0.370) 0.107  0.98
Eq. (6-17):
Se  (0.81)(0.98)(0.5)(64)  25.4 kpsi
For the Morrow criterion, estimate the fatigue strength coefficient for steel.
Eq. (6-44):
 f  Sut  50  64  50  114 kpsi
1
Eq. (6-46):
1
  
 38.4 38.4 

nf   a  m   

 S  
 25.4 114 
f 
 e
n f  0.54 Ans.
Since infinite life is not predicted, estimate a life from the S-N diagram. First, find an
equivalent completely reversed stress, again using Morrow.
Eq. (6-59):
Fig. 6-23:
 a
38.4

 57.9 kpsi
1  ( m /  f ) 1  (38.4 / 114)
Off the chart, so use f = 0.9
 ar 
Eq. (6-13):
( f Sut ) 2  0.9(64) 
a

 130.6
Se
25.4
Eq. (6-14):
 f Sut
1
b   log 
3
 Se
2
1/ b

1
 0.9(64) 
   log 
  0.1185
3
25.4



1
 
 57.9  0.1185
N   ar   
Ans.
Eq. (6-15):
 960 cycles

 130.6 
 a 
______________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 6 Solutions, Page 48/58
6-53
Table A-20: Sut  64 kpsi, S y  54 kpsi
From Prob. 3-97, the critical stress element experiences x,bend = 55.5 kpsi, x,axial = 0.382
kpsi and  = 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
max,bend = 55.5 kpsi, min,bend = 0 kpsi, max,axial = 0.382 kpsi, min,axial = 0 kpsi, max = 15.3
kpsi, min = 0 kpsi. Consequently,m,bend = a,bend = 27.75 kpsi, m,axial = a,axial = 0.191
kpsi, m = a = 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. (6-33), (6-35), and (6-36), or Figs. 6-26 and 6-27: 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 mean stresses from Eqs. (6-66) and (667).


   1.47  27.75  1.59  0.191   3 1.32  7.65   
2
2 1/2
2
2 1/2
 a  1.47  27.75  1.59  0.191   3 1.32  7.65  
m
 44.66 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.
 
 max
ny 

  max,axial   3  max  
2
max,bend
Sy
54

 0.87

 max
61.8
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. (6-10):
Se  0.5  64   32 kpsi
Eq. (6-18):
k a  aSutb  2.00(64) 0.217  0.81
Eq. (6-23):
de  0.370d  0.370 1  0.370 in
Eq. (6-19):
kb  0.879d e 0.107  0.879(0.370) 0.107  0.98
Eq. (6-17):
Se  (0.81)(0.98)(0.5)(64)  25.4 kpsi
For the Morrow criterion, estimate the fatigue strength coefficient for steel.
Shigley’s MED, 11th edition
Chapter 6 Solutions, Page 49/58
Eq. (6-44):
 f  Sut  50  64  50  114 kpsi
1
Eq. (6-46):
1
  a  m 
 44.66 44.66 
nf  




 S   
114 
 25.4
f 
 e
n f  0.47 Ans.
______________________________________________________________________________
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 ar =
35.0 kpsi, producing a = 35.0 kpsi and m = 0 kpsi. This problem adds a steady torque
which creates torsional stresses of
m 
Tr 2500 1.625 / 2 

 2967 psi  2.97 kpsi,
J
 1.6254  / 32
 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. (6-33), (6-35) and (6-36), or Figs. 6-26 and 6-27: qbend = 0.76, qtors = 0.81
Eq. (6-32):
K f ,bend  1  qbend  K t ,bend  1  1  0.76 1.95  1  1.72
K f ,tors  1  qtors  K t ,tors  1  1  0.811.60  1  1.49
Obtain von Mises stresses for the alternating and mean stresses from Eqs. (6-66) and (667).


   1.72  0    3 1.49  2.97   
2
 a  1.72  35.0    3 1.49  0  
2 1/2
2 1/ 2
2
m
 60.2 kpsi
 7.66 kpsi
   a   m ,
Check for yielding, using the conservative  max
Sy
71
ny 

 1.05
 a   m 60.2  7.66
From the solution to Prob. 6-17, Se = 27.0 kpsi. Using Goodman,
1
Eq. (6-41):
  
 60.2 7.66 
nf   a  m   


 27.0 85 
 Se Sut 
n f  0.43
1
Ans.
Since infinite life is not predicted, estimate a life from the S-N diagram. First, find an
equivalent completely reversed stress. Choosing the Goodman criterion,
Shigley’s MED, 11th edition
Chapter 6 Solutions, Page 50/58
Eq. (6-58):
Fig. 6-23:
Eq. (6-13):
 a
60.2

 66.2 kpsi
1  ( m / Sut ) 1  (7.66 / 85)
f = 0.867
2
2
 f Sut   0.867(85) 
a

 201.1
Se
27.0
 ar 
Eq. (6-14):
 f Sut
1
b   log 
3
 Se
Eq. (6-15):
 
N   ar 
 a 
1/ b

1
 0.867(85) 
   log 
  0.1454
3
 27.0 

1
 66.2  0.1454

 2084 cycles

 201.1 
N = 2100 cycles
Ans.
______________________________________________________________________________
6-55
From the solution to Prob. 6-18 we find the completely reversed stress at the critical
shoulder fillet to be rev = 32.8 kpsi, producing a = 32.8 kpsi and m = 0 kpsi. This
problem adds a steady torque which creates torsional stresses of
m 
Tr 2200 1.625 / 2 

 2611 psi  2.61 kpsi,
J
 1.6254  / 32
 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. (6-33), (6-35) and (6-36), or Figs. 6-26 and 6-27: qbend = 0.76, qtors = 0.81
Eq. (6-32):
K f ,bend  1  qbend  K t ,bend  1  1  0.76 1.95  1  1.72
K f ,tors  1  qtors  K t ,tors  1  1  0.811.60  1  1.49
Obtain von Mises stresses for the alternating and mean stresses from Eqs. (6-66) and (667).


   1.72  0    3 1.49  2.61  
2
 a  1.72  32.8   3 1.49  0  
2
m
2 1/ 2
2 1/2
 56.4 kpsi
 6.74 kpsi
   a   m ,
Check for yielding, using the conservative  max
Sy
71
ny 

 1.12
 a   m 56.4  6.74
From the solution to Prob. 6-18, Se = 27.0 kpsi. Using Goodman,
Shigley’s MED, 11th edition
Chapter 6 Solutions, Page 51/58
1
1
  
 56.4 6.74 

nf   a  m   
Eq. (6-41):

 27.0 85 
 Se Sut 
n f  0.46 Ans.
Since infinite life is not predicted, estimate a life from the S-N diagram. First, find an
equivalent completely reversed stress. Choosing the Goodman criterion,
 a
56.4

 61.3 kpsi
1  ( m / Sut ) 1  (6.74 / 85)
f = 0.867
2
2
f Sut 
0.867(85) 


a

 201.1
Se
27.0
 ar 
Eq. (6-58):
Fig. 6-23:
Eq. (6-13):
Eq. (6-14):
 f Sut
1
b   log 
3
 Se
Eq. (6-15):
 
N   ar 
 a 
1/ b

1
 0.867(85) 
   log 
  0.1454
3
 27.0 

1
 61.3  0.1454

 3536 cycles

 201.1 
N = 3500 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. (6-33) and (6-36), or Fig. 6-27: 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
 max 
 min 
m 
16 K fsTmax
d
3
16 K fsTmin
d
 max   min
3
2
 
 a  max min
2

16(1.48)(1000)
 11 251 psi  11.25 kpsi
 (0.875)3
16(1.48)(300)
 3375 psi  3.38 kpsi
 (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 y / 2 30 / 2
ny 

 1.33
 max 11.25
Shigley’s MED, 11th edition
Chapter 6 Solutions, Page 52/58
Find the modifiers and endurance limit.
Se  0.5(55)  27.5 kpsi
Eq. (6-10):
Eq. (6-18):
k a  11.0(55) 0.650  0.81
Eq. (6-23):
d e  0.370(0.875)  0.324 in
Eq. (6-19):
kb  0.879(0.324) 0.107  0.99
Eq. (6-25):
kc  0.59
Eq. (6-17):
S se  0.81(0.99)(0.59)(27.5)  13.0 kpsi
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. 6-16. From Eq. (6-58), Ssu = 0.67Su = 0.67 (55) = 36.9 kpsi.
(a) Goodman,
Eq. (6-41):
1
1

 
 3.94 7.32 
nf   a  m   

  1.99
 13.0 36.9 
 S se S su 
Ans.
(b) Gerber
2
2

 2 m S se  
1  S su   a 
1  1  
nf  
Eq. (6-48):


2   m  S se 
 S su a  


2
2

 2(7.32)(13.0)  
1  36.9   3.94  
nf  

 
 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. (6-33) and (6-35), or Fig. 6-26, 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
d2
 (1.2) 2
1
 m   a  (2.476 P )  1.238 P
2
f P  D  d  0.3P  6  1.2 

 0.54 P
Tmax 
4
4
 max   K f
Shigley’s MED, 11th edition
Chapter 6 Solutions, Page 53/58
From Eqs. (6-33) and (6-36), or Fig. 6-27, with a notch radius of 0.1 in, qs  0.92. Thus,
with Kts = 1.8 from the problem statement,
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
d
 (1.2)3

2.769 P
 a   m  max 
 1.385 P
2
2
Eqs. (6-66) and (6-67):
 max 
3

 a  [ a 2  3 a2 ]1/2  [(1.238 P)2  3(1.385P) 2 ]1/ 2  2.70 P
 m  [ m 2  3 m2 ]1/ 2  [(1.238P) 2  3(1.385P ) 2 ]1/2  2.70 P
Eq. (6-10):
Se  0.5(145)  72.5 kpsi
Eq. (6-18):
ka  2.00(145)0.217  0.68
Eq. (6-19):
kb  0.879(1.2) 0.107  0.862
Eq. (6-17):
Se  (0.68)(0.862)(72.5)  42.5 kpsi
1
Eq. (6-41):
1
  
 2.70 P 2.70 P 
nf   a  m   

 3
145 
 42.5
 Se Sut 
P  4.1 kips Ans.
Yield (conservative):
Sy
120
ny 

 5.4 Yielding is not predicted. Ans.
 a   m (2.70)(4.1)  (2.70)(4.1)
______________________________________________________________________________
6-58
From Prob. 6-57, K f  2.80, K f s  1.74, S e  42.5 kpsi
4 Pmax
4(18)
 2.80
 44.56 kpsi
2
d
 (1.2 2 )
4P
4(4.5)
 min   K f min2  2.80
 11.14 kpsi
d
 (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)
 max  K f s
 1.74
 49.85 kpsi
3
d
 (1.2)3
16Tmin
16(2.43)
 min  K f s
 1.74
 12.46 kpsi
3
d
 (1.2)3
 max   K f
Shigley’s MED, 11th edition
Chapter 6 Solutions, Page 54/58
44.56  ( 11.14)
 16.71 kpsi
2
44.56  ( 11.14)
m 
 27.85 kpsi
2
49.85  12.46
a 
 18.70 kpsi
2
49.85  12.46
m 
 31.16 kpsi
2
a 
Eqs. (6-66) and (6-67):
 a  [( a / 0.85)2  3 a2 ]1/2  [(16.71/ 0.85) 2  3(18.70) 2 ]1/ 2  37.89 kpsi
 m  [ m 2  3 m2 ]1/2  [(27.85)2  3(31.16) 2 ]1/2  60.73 kpsi
Goodman:
Eq. (6-41):
1
  
 37.89 60.73 

nf   a  m   

145 
 42.5
 Se Sut 
nf = 0.76
1
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).
Choosing the Goodman criterion,
Eq. (6-58):
Fig. 6-23:
 a
37.89

 65.2 kpsi
1  ( m / Sut ) 1  (60.73 / 145)
f = 0.8
 ar 
 f Sut 
2
0.8(145)

Eq. (6-13):
a
Eq. (6-14):
 f Sut
1
b   log 
3
 Se
42.5
Se
1/ b
Eq. (6-15):
  ar 
N 

 a 
2
 316.6

1
 0.8(145) 
   log 
  0.1454
3
 42.5 

1
 65.2  0.1454

 52 460 cycles

 316.6 
N = 52 500 cycles
Ans.
______________________________________________________________________________
6-59
For AISI 1020 CD, From Table A-20, Sy = 390 MPa, Sut = 470 MPa. Given: Se = 175
MPa.
Shigley’s MED, 11th edition
Chapter 6 Solutions, Page 55/58
 m 1 
First Loading:
360  160
 260 MPa,
2
 a 1 
360  160
 100 MPa
2
Goodman, Eq. (6-58):
 a e1 
Fig. 6-23:
 a 1
1   m 1 / Sut
100
 223.8 MPa  Se  finite life
1  260 / 470
Off the graph, so let f = 0.9.

2
0.9  470  
 1022.5 MPa
a
175
0.9  470 
1
 0.127 767
b   log
3
175
1/0.127767
 223.8 
 145 920 cycles
N 

 1022.5 
320   200 
320   200 
 60 MPa,
 260 MPa
Second loading:  m 2 
 a  2 
2
2
  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
Shigley’s MED, 11th edition
Chapter 6 Solutions, Page 56/58
0.9  470 
223.8

a2 103 
b2
a2  65 920  2
b
1.8901   0.015170  2
b
log1.8901
 0.151 997
log 0.015170
223.8
a2 
 1208.7 MPa
0.151 997
 65 920 
b2 
1/ 0.151 997
 298.0 
n2  
Ans.
 10 000 cycles

 1208.7 
______________________________________________________________________________
6-60
Given: Se = 50 kpsi, Sut = 140 kpsi, f =0.8. Using Miner’s method,
2
0.8 140  
 250.88 kpsi
a
50
0.8 140 
1
 0.116 749
b   log
3
50
1/ 0.116 749
 95 
 1  95 kpsi,
N1  
 4100 cycles

 250.88 
1/ 0.116 749
 2  80 kpsi,
 80 
N2  

 250.88 
 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
2
0.9  530  
 1083.47 MPa
a 
210
0.9  530 
1
 0.118 766
b   log
3
210
1/ 0.118 766
 350 
 1  350 MPa, N1  

 1083.47 
Shigley’s MED, 11th edition
 13 550 cycles
Chapter 6 Solutions, Page 57/58
1/ 0.118 766
 260 
 2  260 MPa, N 2  

 1083.47 
225

 165 600 cycles
1/ 0.118 766

 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 Se,1 are [0.9(530), 103] and (350, 8550).
0.9  530 
350

a2 103 
b2
a2  8550 
b2 
a2 

b2
1.3629   0.11696  2
b
log 1.362 9 
 0.144 280
log  0.116 96 
350
 8550 
0.144 280
 1292.3 MPa
1/0.144 280
 260 
 67 090 cycles
N2  

 1292.3 
N R 2  67 090  50 000  17 090 cycles
0.9  530 
260
b3 

a3 103 
b3
a3 17 090 
log 1.834 6 
log  0.058 514 
b3

1.834 6   0.058 514  2
 0.213 785, a3 
b
260
17 090 
0.213 785
 2088.7 MPa
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 a = 35 kpsi and m = 30 kpsi for 12 (103)
cycles.
Goodman equivalent reversing stress, Eq. (6-58):
Shigley’s MED, 11th edition
Chapter 6 Solutions, Page 58/58
 ar 
Initial cycling
a
35

 54.09 kpsi
1   m / Sut  1   30 / 85
2
0.86  85  
 116.00 kpsi
a 
45
0.86  85 
1
 0.070 235
b   log
3
45
 1  54.09 kpsi,
1/ 0.070 235
 54.09 
N1  

 116.00 
 52 190 cycles
(a) Miner’s method: 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: 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 = 1/ (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.
______________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 6 Solutions, Page 59/58
Chapter 11
11-1
Eq. (b), Sec. 11-3: L10 = 60 L R nR = 60(3000)500 = 90 (106) rev
Eq. (11-3):
 10  10  1800  60  
 L n 60 
 6.30 kN
C10  FD  D D  2.75 
Ans.
90  106 


 L R nR 60 
______________________________________________________________________________
1/ a
11-2
1/3
3
For the deep-groove 02-series ball bearing with R = 0.90, the design life xD, in multiples
of rating life, is
L
60L D nD 60  25000  350
xD  D 

525 Ans.
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.
  525 3 / 25.5 3  0.02  1.483 




R exp   
Ans.

 0.920
4.459  0.02

 

Eq. (11-21):
______________________________________________________________________________
11-3
For the angular-contact 02-series ball bearing as described, the rating life multiple is
L
60L D nD 60  40 000  520
xD  D 

1248
LR
L10
106
The design radial load is
FD 1.4  725  1015 lbf 4.52 kN
Eq. (11-9):
Shigley’s MED, 11th edition
Chapter 11 Solutions, Page 1/31
1/3


1248


C10 1015 
1/1.483 
 0.02   4.459  0.02   ln  1 / 0.9  

10 930 lbf 48.6 kN
Table 11-2:
Select an 02-60 mm bearing with C10 = 55.9 kN. Ans.
  1248 4.52 / 55.9 3  0.02  1.483 




R exp   
Ans.

 0.945
4.439

 

Eq. (11-21):
______________________________________________________________________________
11-4 For the straight-roller 03-series bearing selection, xD = 1248 rating lives from Prob. 11-3
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 

R exp  

 0.917 Ans.
4.459  0.02

 

Eq. (11-21):
______________________________________________________________________________
11-5
The combined reliability of the two bearings selected in Probs. 11-3 and 11-4 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.
______________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 11 Solutions, Page 2/31
11-6
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.


RA exp 

 1248  4.52 / 63.7  3  0.02 


4.439


1.483


 0.962

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.


RB exp 

 1248  13.92 /138  10/3  0.02 


4.439


1.483


 0.953

The overall reliability is R = (0.962)(0.953) = 0.917, which exceeds the goal.
______________________________________________________________________________
11-7
Given: R = 96 percent,  f = 1.2, x0 = 0,  = 4.48, b = 1.5. From Prob. 11-1,
FD = 2.75 kN, LR = L10 = 60 L R nR = 60(3000)500 = 90 (106) rev, and
LD = 60 L D nD = 60(10)103(1800) = 1.08 (109) rev
The dimensionless multiple of rating life is
9
LD 1.08  10 
xD  
12
LR
90  106 
Eq. (11-10):
1/ a
1/3




xD
12
C10 a f FD 

1.2
2.75
9.37 kN Ans.




1/ b
1/1.5 
 x0     x0   1  RD  
 4.48  1  0.96  
______________________________________________________________________________
11-8
For the straight cylindrical roller bearing specified with a service factor of 1, R = 0.95 and
FR = 20 kN.
Shigley’s MED, 11th edition
Chapter 11 Solutions, Page 3/31
xD 
LD 60L D nD 60  8000  950


456
LR
L10
106
3/10


456


C10 20 
145 kN
Ans.
1/1.483 
 0.02  4.439  ln  1 / 0.95  

______________________________________________________________________________
11-9
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.
Eq. (11-3):
L 
C A FA  A 
 LR 
8.96 kN
1/ a
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.
______________________________________________________________________________
11-10 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):
______________________________________________________________________________
11-11 FD = 800 lbf, D = 12 000 hours, nD = 350 rev/min, R = 0.90
 12 000  350   60  
 L n 60 
C10 FD  D D  800 
 5050 lbf Ans
LR 
106



Eq. (11-3):
______________________________________________________________________________
1/3
1/ a
11-12 FD = 4 kN, D = 8 000 hours, nD = 500 rev/min, R = 0.90
 8 000  500   60  
 L n 60 
C10 FD  D D  4 
 24.9 kN Ans
LR 
106



Eq. (11-3):
______________________________________________________________________________
1/ a
1/3
11-13 FD = 650 lbf, nD = 400 rev/min, R = 0.95
D = (5 years)(40 h/week)(52 week/year) = 10 400 hours
Assume an application factor of one. The multiple of rating life is
Shigley’s MED, 11th edition
Chapter 11 Solutions, Page 4/31
xD 
LD  10 400   400   60 

249.6
LR
106
1/3
Eq. (11-9):


249.6


C10  1  650  
1/1.483 
 0.02  4.439  ln  1 / 0.95  

4800 lbf Ans.
______________________________________________________________________________
11-14 FD = 9 kN, LD = 108 rev, R = 0.99
Assume an application factor of one. The multiple of rating life is
LD 108
xD   6 100
LR 10
1/3
Eq. (11-9):


100


C10  1  9  
1/1.483 
 0.02  4.439  ln  1 / 0.99  

69.2 kN Ans.
______________________________________________________________________________
11-15 FD = 11 kips, D = 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
xD 
LD  20 000   200   60 

240
LR
106
1/3
Eq. (11-9):


240


C10  1  11 
1/1.483 
 0.02  4.439  ln  1/ 0.99  

113 kips Ans.
______________________________________________________________________________
11-16 From the solution to Prob. 3-79, 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 
Shigley’s MED, 11th edition
LD 15000  1200   60 

1080
LR
106
Chapter 11 Solutions, Page 5/31
Eq. (11-10):


xD
C10 a f FD 

1/ b
 x0     x0   1  RD  
1/ a
1/3


1080
C10 1.2  178 

1/1.483
 0.02   4.459  0.02   1  0.95 

2590 lbf Ans.
______________________________________________________________________________
11-17 From the solution to Prob. 3-80, 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 
Eq. (11-10):
LD 15000  1200   60 

1080
LR
106


xD
C10 a f FD 

1/ b
 x0     x0   1  RD  
1/ a
1/3


1080
C10 1.2  1.794  

1/1.483
 0.02   4.459  0.02   1  0.95 

26.1 kN Ans.
______________________________________________________________________________
11-18 From the solution to Prob. 3-81, 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
Eq. (11-10):


xD
C10 a f FD 

1/ b
 xo     xo   1  RD  
1/ a
1/3


1080
C10 1.2  351.8  

1/1.483
 0.02   4.459  0.02   1  0.95 

5110 lbf Ans.
______________________________________________________________________________
11-19 From the solution to Prob. 3-82, 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.
Shigley’s MED, 11th edition
Chapter 11 Solutions, Page 6/31
xD 
Eq. (11-10):
LD 15000  1200   60 

1080
LR
106


xD
C10 a f FD 

1/ b
 x0     x0   1  RD  
1/ a
1/3


1080
C10 1.2  173.6  

1/1.483
 0.02   4.459  0.02   1  0.95 

2520 N Ans.
______________________________________________________________________________
11-20 From the solution to Prob. 3-88, 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
dC
250
xD 
Eq. (11-10):
LD 12 000  95.5   60 

68.76
LR
106


xD
C10 a f FD 

1/ b
 x0     x0   1  RD  
1/ a
1/3


68.76
C10  1  2.425  

1/1.483
 0.02   4.459  0.02   1  0.95 

11.7 kN Ans.
______________________________________________________________________________
11-21 From the solution to Prob. 3-90, 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
dC
5
xD 
Shigley’s MED, 11th edition
LD 14 000  560   60 

470.4
LR
106
Chapter 11 Solutions, Page 7/31
Eq. (11-10):


xD
C10 a f FD 

1/ b
 x0     x0   1  RD  
1/ a
3/10


470.4
C10  1  150.1 

1/1.483
 0.02   4.459  0.02   1  0.98 

1320 lbf Ans.
______________________________________________________________________________
11-22
(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
Ans.
Fe > Fr so use Fe.
(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
LR
106
Eq. (11-10):


xD
C10 a f FD 

1/ b
 x0     x0   1  RD  
1/ a


300
C10  1  9.29  

1/1.483
 0.02   4.459  0.02   1  0.95 

73.4 kN
1/3
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.
______________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 11 Solutions, Page 8/31
11-23 (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.
F  X iVFr  Yi Fa  0.56   1  5    1.27   2  5.34 kN
Ans.
Eq. (11-12): e
Fe > Fr so use Fe.
(b) Solve Eq. (11-10) for xD.
 C
xD  10
 a f FD

a

1/ b
  x0     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 L D nD  60 

LR
106
LD 
  10.66  10  444 h
xD 106
6
Ans.
nD  60   400   60 
______________________________________________________________________________
9
11-24 Fr 8 kN, R 0.9, LD 10 rev
Eq. (11-3):
L 
C10 FD  D 
 LR 
1/ a
 109 
8  6 
 10 
1/3
80 kN
From Table 11-2, select the 85 mm bore. Ans.
______________________________________________________________________________
Fr 8 kN, Fa 2 kN, V 1, R 0.99
11-25
Use the Weibull parameters for Manufacturer 2 in Table 11-6.
Shigley’s MED, 11th edition
Chapter 11 Solutions, Page 9/31
xD 
LD 10 000  400   60 

240
LR
106
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     xo   1  RD  
1/ a
1/3


240
C10  1  8  
82.5 kN

1/1.483
 0.02   4.459  0.02   1  0.99 

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.
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):
Eq. (11-10):
Fe 0.56(1)8  1.95(2) 8.38 > Fr


240
C10  1  8.38 

1/1.483
 0.02   4.459  0.02   1  0.99 

Shigley’s MED, 11th edition
1/3
86.4 kN
Chapter 11 Solutions, Page 10/31
The 90 mm bore is acceptable. Ans.
______________________________________________________________________________
F 8 kN, Fa 3 kN, V 1.2, R 0.9, LD 108 rev
11-26 r
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
Eq. (11-3):
L 
C10 Fe  D 
 LR 
1/ a
 108 
10.3  6 
 10 
1/3
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 
C10 9.76  6  45.3 kN
 10 
Eq. (11-3):
From Table 11-2, we have converged on the 60 mm bearing. Ans.
______________________________________________________________________________
Fr 10 kN, Fa 5 kN, V 1, R 0.95
11-27
Use the Weibull parameters for Manufacturer 2 in Table 11-6.
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     x0   1  RD  
Shigley’s MED, 11th edition
1/ a
Chapter 11 Solutions, Page 11/31


216
C10  1  13.75  

1/1.483
 0.02   4.459  0.02   1  0.95 

From Table 11-2, try 95 mm bore with C10 = 108 kN, C0 = 69.5 kN
Iterate the previous process:
1/3
97.4 kN
Fa / C0 = 5 / 69.5 = 0.072
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  1.63
Table 11-1:
Since this is where we started, we will converge back to the same bearing. The 95 mm
bore meets the requirements. Ans.
______________________________________________________________________________
Fr 9 kN, Fa 3 kN, V 1.2, R 0.99
11-28
Use the Weibull parameters for Manufacturer 2 in Table 11-6.
LD 108
xD   6 100
LR 10
First guess: Choose from middle of Table 11-1, X = 0.56, Y = 1.63
Fe 0.56  1.2   9   1.63  3 10.9 kN
Eq. (11-12):
Fe > Fr, so use Fe as the design load.
Eq. (11-10):


xD
C10 a f FD 

1/ b
 x0     x0   1  RD  
1/ a


100
C10  1  10.9  

1/1.483
 0.02   4.459  0.02   1  0.99 

1/3
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:
Shigley’s MED, 11th edition
Chapter 11 Solutions, Page 12/31
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 
11-29 (a)
nD 1200 rev/min, LD 15 kh, R 0.95, a f 1.2
From Prob. 3-83, 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
LR
106
Eq. (11-10):


1080
C10 1.2  881 

1/1.483
 0.02  4.439  1  0.95 

12800 lbf 12.8 kips Ans.
1/3
(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 1200 rev/min, LD 15 kh, R 0.95, a f 1.2
From Prob. 3-83, ROy = –208.5 lbf, ROz = 259.3 lbf.
1/ 2
2
RC FD  259.32    208.5   333 lbf


15000  1200   60 
L
xD  D 
1080
LR
106
1/3


1080
C10 1.2  333 

1/1.483
 0.02  4.439  1  0.95 

Eq. (11-10):
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.
______________________________________________________________________________
11-31 (a)
nD 900 rev/min, LD 12 kh, R 0.98, a f 1.2
From Prob. 3-84, RCy = 8.319 kN, RCz = –10.830 kN.
Shigley’s MED, 11th edition
Chapter 11 Solutions, Page 13/31
1/2
2
RC FD  8.3192    10.830   13.7 kN


12 000  900   60 
L
xD  D 
648
LR
106
1/3


648
C10 1.2  13.7  
 204 kN Ans.
1/1.483
 0.02  4.439  1  0.98 

Eq. (11-10):
(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.
______________________________________________________________________________
11-32 (a)
nD 900 rev/min, LD 12 kh, R 0.98, a f 1.2
From Prob. 3-84, ROy = 5083 N, ROz = 494 N.
RC FD  50832  494 2 
xD 
1/2
5106 N 5.1 kN
LD 12 000  900   60 

648
LR
106
1/3


648
C10 1.2  5.1 
 76.1 kN Ans.
1/1.483
 0.02  4.439  1  0.98 

Eq. (11-10):
(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.
______________________________________________________________________________
11-33 Assume concentrated forces as shown.
Pz 8  28  224 lbf
Py 8  35 280 lbf
T 224  2  448 lbf in
T x  448  1.5 F cos 20 0
448
F
318 lbf
1.5  0.940 
M Oz 5.75Py  11.5RAy  14.25 F sin 20 0
5.75  280   11.5 RAy  14.25  318   0.342  0
RAy  5.24 lbf
Shigley’s MED, 11th edition
Chapter 11 Solutions, Page 14/31
M Oy  5.75Pz  11.5 RAz  14.25 F cos 20 0
 5.75  224   11.5 RAz  14.25  318   0.940  0
2
2
RA    482     5.24  


z
z
z

F RO  Pz  RA  F cos 20 0
RAz  482 lbf;
1/2
482 lbf
ROz  224  482  318  0.940  0
ROz  40.9 lbf
F y ROy  Py  RAy  F sin 20 0
ROy  280  5.24  318  0.342  0
ROy  166 lbf
2
2
RO    40.9     166  


1/2
171 lbf
So the reaction at A governs.
Reliability Goal: 0.92 0.96
FD 1.2  482  578 lbf
xD 35000  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.
______________________________________________________________________________
11-34 For a combined reliability goal of 0.95, use 0.95 0.975 for the individual bearings.
xD 
40 000  420   60 
106
Shigley’s MED, 11th edition
1008
Chapter 11 Solutions, Page 15/31
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.
______________________________________________________________________________
11-35 The reliability of the individual
bearings is R  0.98 0.9899
From statics,
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  357.2sin 45  252.6 N  FAz
M
F
M
F
z
O
y
y
O
z
Shigley’s MED, 11th edition
 850REy  300(252.6)  0  REy   89.2 N
 252.6  89.2  ROy  0  ROy   163.4 N
  850 REz  700(320)  300(252.6)  0  REz   174.4 N
  174.4  320  252.6  ROz  0  ROz 107 N
Chapter 11 Solutions, Page 16/31
RO 
  163.4
RE 
  89.2 
2
2
 107 2  195 N
   174.4  196 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.
______________________________________________________________________________
11-36
R  0.96 0.980
T 12(240 cos 20 ) 2706 lbf in
2706
F
498 lbf
6 cos 25
In xy-plane:
RCy 181 lbf
ROy 82.1  210  181 111.1 lbf
In xz-plane:
M Oy 16(226)  30(451)  42 RCz 0
RCz  236 lbf
ROz 226  451  236 11 lbf
RO  111.12  112 

RC  1812  2362
xD 

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
Shigley’s MED, 11th edition
Chapter 11 Solutions, Page 17/31
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.
______________________________________________________________________________
11-37 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  362  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.
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  362  2122

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.
Shigley’s MED, 11th edition
Chapter 11 Solutions, Page 18/31
Fa 2.47

0.0392
C0 63.0
xD 
30 000  500   60 
106
900
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


900


C10 1.2  5.18  
1/1.483 
 0.02  4.439  ln  1/ 0.99  

99.54 kN  90.4 kN
Eq. (11-9):
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  
102 kN < 106 kN O.K.
1/1.483 


0.02

4.439
ln
1
/
0.99






Select an 02-90 mm angular-contact ball bearing. Ans.
______________________________________________________________________________
11-38 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.
115  2000   60 
For F = 18 kN,
 x 1 
106
13.8
This establishes point 1 on the R = 0.90 line.
Shigley’s MED, 11th edition
Chapter 11 Solutions, Page 19/31
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   ln  1/ 0.90  
1/ b
xB   ln  1/ 0.20  
1/ b
(1)
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),

xA
 ln  1/ RA  
1/1.65

1
 ln  1/ 0.90  
1/1.65
3.91
Ans.
Therefore, for the data at hand,

R exp  

 x 


 3.91 
1.65



Check R at point B: xB = (600/115) = 5.217
  5.217 1.65 
R exp   
  0.20
  3.91  
Shigley’s MED, 11th edition
Chapter 11 Solutions, Page 20/31
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
______________________________________________________________________________
11-39 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
FAr  2392 1112 
1/2


FBr  5022  10752
264 lbf 1.175 kN
1/ 2
1186 lbf 5.28 kN
Thus the bearing at B controls.
10 000  1200   60 
xD 
720
106
0.02  4.439  ln  1/ 0.9983  
 720 
C10 1.2  5.28  

 0.080 26 
1/1.483
0.3
97.2 kN
Select an 02-80 mm with C10 = 106 kN.
Shaft b

 393
FCr  8742  22742
FDr
2
 657 2 

1/2
1/2
0.080 26
Ans.
2436 lbf
766 lbf
or
or
10.84 kN
3.41 kN
The bearing at C controls.
10 000  240   60 
xD 
144
106
 144 
C10 1.2  10.84  

 0.080 26 
Select an 02-90 mm with C10 = 142 kN.
0.3
123 kN
Ans.
Shaft c
FEr  11132  23852 
Shigley’s MED, 11th edition
1/2
2632 lbf
or
11.71 kN
Chapter 11 Solutions, Page 21/31
FFr  417 2  8952 
1/2
987 lbf
or
4.39 kN
The bearing at E controls.
10 000  80   60 
xD 
48
106
 48 
C10 1.2  11.71 

 0.080 26 
0.3
95.7 kN
Select an 02-80 mm with C10 = 106 kN.
Ans.
______________________________________________________________________________
4
11-40 (a) Eq. (1-9):
(b)
R  Ri

Ri  4 R  4 0.92 0.9794 97.94 percent
Ans.
i 1
 Mx = 0 = (FG cos 20o)(1) – 200
FG = 212.84 lbf
 (MB)y = 0 = 2(FG cos 20o) – 3.5RAz
RAz = 2(212.84 cos 20o)/3.5 = 114.29 lbf
 (MB)z = 0, 2 (212.84 sin 20o) + 3.5 (FA)y = 0  RAy = – 41.60 lbf
2
2
RA  RAy
 RAz

  41.60 
2
 114.292 121.6 lbf
Ans.
(c) L10 = 1 (106) revolutions, LD = 30(103)60(60) = 108(106) revolutions.
6
LD 108  10 
xD 

108
L10
1 106 
Eq. (11-10) with af = 1, xD = 108, a = 3, FD = 121.6 lbf, RD = 0.9794, x0 = 0.02,
 = 4.459, and b =1.483:


xD
C10 a f FD 
1/ b 
 x0     x0   1  RD  
1/ a
1/3


108
1 121.6  
826.6 lbf Ans.
1/1.483 
 0.02   4.459  0.02   1  0.9794 

______________________________________________________________________________
11-41 For the output shaft: N1 / N2 = 2 /1  2 = (N1 / N2)2 = (30/60) 200 = 100 rev/min.
Shigley’s MED, 11th edition
Chapter 11 Solutions, Page 22/31
Service time in minutes:
 52 weeks   5 days   8 hours   60 minutes 
3
6 years 



 748.8  10  minutes
hour

 year   week   day  
3
For bearings C and D: LD = 748.8(10 )100 = 74.88(106) revolutions
74.88  10 6 
L
xD  D 
74.88
L10
1 106 
(a) R = RA RB RC RD  RD = R / (RA RB RC) = 0.90 / [0.97(0.96)0.98] = 0.986
Ans.
(b) Given Fr = 9 kN, Fa = 5 kN. Eq. (11-12) with V = 1 (inner ring rotates):
F e = X i F r + Yi F a
Iteration will be necessary. First try. From the middle of Table 11-1, X2 = 0.56, Y2 = 1.63
Fe = 0.56(9) + 1.63(5) = 13.19 kN
Eq. (11-10):
1/ a
1/3




xD
74.88

C10 a f FD 
1
13.19




1/ b
1/1.483 
 x0     x0   1  RD  
 0.02   4.459  0.02   1  0.986 

86.06 kN Ans.
From Table 11-2 we see that C10 exceeds the 83.2 kN rating for an 85 mm bore bearing.
To see if this bearing is possible, we need to determine Xi and Yi. For the 85 mm bearing:
C0 = 53.0 kN, Fa / C0 = 5/53 = 0.0943
Fa / C0
e
0084
0.28
0.0943
e
0.110
0.30
0.110  0.0943
0.3  e
0.3  e

 0.604 

e 0.29
0.110  0.084 0.3  0.28
0.02
Fa
5

0.556
VFr 1 9 
. Since 0.556 > 0.29, use X2 = 0.56, Y2 = 1.50 (interpolated)
Fe = 0.56(9) + 1.50(5) = 12.54 kN
12.54
C10 
86.06 81.8 kN
13.19
Since 81.8 kN < 83.2 kN it is acceptable to use the bearing with the 85 mm bore. Ans.
______________________________________________________________________________
11-42 From Table 11-3, for a 03-30 mm cylindrical roller bearing,
C10 =36.9 kN = 36.9/4.45 = 8.292 kips.
K C10a L10 8.29210/3  106  1.154  109 
Eq. (a), Sec. 11-7:
9
K 1.154  10 
L1  a 
5.399  106  rev
10/3
F1
5
At 5 kips:
Shigley’s MED, 11th edition
Chapter 11 Solutions, Page 23/31
9
K 1.154  10 
L2  a 
1.127  106  rev
10/3
F2
8
At 8 kips:
l1 l2
l2
300 000
 1


1
6
L1 L2
5.399  10  1.127  106 
Eq. (11-16):

300 000 
 1.06  106  rev
l2 1.127  106  1 
Ans.
6
 5.399  10  
______________________________________________________________________________
K C10a L10
and
11-43 From Table 11-3, select a 25 mm trial bearing. Using the equations
a
Li K / Fi
Size
25 mm
C10a L10
C10
28.6 kN
28.610/3  106 
7.154(1010)
L1
7.154  1010 
L2
7.154  1010 
L3
7.154  1010 
2010/3
3.294(106)
2510/3
1.566(106)
3010/3
852.7(103)
1


 f1 f 2 f 3 
0.3
0.5
0.2

l     


6
6
3
 3.294  10  1.566  10  852.7  10  
 L1 L2 L3 
Using a spreadsheet following the method above yields
1
1.55  106  rev
Size
l
25 mm 1.55(106) rev
30 mm 3.63(106) rev
35 mm 6.82(106) rev greater than 5 (106) rev. Thus, 35 mm
Ans.
______________________________________________________________________________
11-44 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 
L1  a 
1.434  106  rev
3
F1
18
For a load of 30 kN, life L2 is:
Shigley’s MED, 11th edition
Chapter 11 Solutions, Page 24/31
L2 
  0.310 10 rev
 
30
8.365 109
6
3
In this case, Eq. (6-68) – the Palmgren-Miner cycle-ratio summation rule – can be
expressed as
l1 l2

1
L1 L2
Substituting,
200 000
l2

1
6
1.434 10
0.310 106
l2
 
0.267  10 
6
 
rev
Ans.
______________________________________________________________________________
11-45 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. 11-44, 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
1
f1 / L1  f 2 / L2
l
1
 0.40 / 1.434  10      0.60 /  0.310  10   
451 585 rev
6
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
Shigley’s MED, 11th edition
Chapter 11 Solutions, Page 25/31
Total rev/cycle = 8000 + 12 000 = 20 000
451 585 rev
22.58 cycles
20 000 rev/cycle
Ans.
Total life in hours

min   22.58 cycles 
 10

 3.76 h Ans.
 cycle   60 min/h 
______________________________________________________________________________
FrA 560 lbf
11-46
FrB 1095 lbf
Fae 200 lbf
xD 
LD 40 000  400   60 

10.67
LR
90  106 
R  0.90 0.949
FiA 
0.47 FrA 0.47  560 

175.5 lbf
KA
1.5
FiB 
0.47 FrB 0.47  1095 

343.1 lbf
KB
1.5
Eq. (11-18):
Eq. (11-18):
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.
FRB
Eq. (11-10):


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.
Shigley’s MED, 11th edition
Chapter 11 Solutions, Page 26/31
FiB 
0.47 FrB 0.47  1095 

263.9 lbf
KB
1.95
Eq. (11-18):
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
FeA  FrA
Eq. (11-10):


10.67

FRA 1.4  920  
 4.48  1  0.949  1/1.5 


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.
______________________________________________________________________________
11-47 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. 3-85, 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 
FiC 

181 lbf
KC
1.5
Eq. (11-18):
FiD 
0.47 FrD 0.47  363

114 lbf
KD
1.5
Eq. (11-18):
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
Shigley’s MED, 11th edition
Chapter 11 Solutions, Page 27/31
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.
xD 
LD
108

1.11
LR 90  106 
R  0.90 0.949
FRC
Eq. (11-10):


1.11

1 946  
 4.48  1  0.949  1/1.5 


3/10
1130 lbf
Ans.
Eq. (11-19b): FeD FrD 363 lbf
3/10


1.11
 433 lbf Ans.
FRD 1 363 
1/1.5
 4.48  1  0.949 



Eq. (11-10):
______________________________________________________________________________
11-48 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. 3-87, 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  639.42  1513.7 2 1643 lbf
FrB  276.62  705.7 2 758 lbf
The induced loads are
0.47 FrA 0.47  1643
FiA 

515 lbf
K
1.5
A
Eq. (11-18):
FiB 
0.47 FrB 0.47  758 

238 lbf
KB
1.5
Eq. (11-18):
Check the condition on whether to apply Eq. (11-19) or Eq. (11-20).
FiA ? FiB  Fae
Shigley’s MED, 11th edition
Chapter 11 Solutions, Page 28/31
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.
6
LD 500  10 
xD  
5.56
LR
90  106 
R  0.90 0.949
Eq. (11-10):


5.56

FRB 1 937  
1/1.5
 4.48  1  0.949 



3/10
1810 lbf
Ans.
Eq. (11-19b): FeA FrA 1643 lbf
3/10


5.56
 3180 lbf Ans.
FRA 1 1643 
 4.48  1  0.949  1/1.5 


Eq. (11-10):
______________________________________________________________________________
11-49 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
FiA 
0.47 FrA 0.47  25

7.83 kN
KA
1.5
FiB 
0.47 FrB 0.47  12 

3.76 kN
KB
1.5
Eq. (11-18):
Eq. (11-18):
Check the condition on whether to apply Eq. (11-19) or Eq. (11-20)
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 
Shigley’s MED, 11th edition
Chapter 11 Solutions, Page 29/31
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
Eq. (11-19b): FeB FrB 12 kN
3/10
 156 
FRB 1.2  12  
17.0 kN

90


Eq. (11-3):
3/10
35.4 kN
Ans.
Ans.
______________________________________________________________________________
11-50 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.
FiA 
0.47 FrA 0.47  875 

274 lbf
KA
1.5
FiB 
0.47 FrB 0.47  625 

196 lbf
KB
1.5
Eq. (11-18):
Eq. (11-18):
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.
Shigley’s MED, 11th edition
Chapter 11 Solutions, Page 30/31
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 
FiB 

203 lbf
K
1.45
B
Eq. (11-18):
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 
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.
______________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 11 Solutions, Page 31/31
Shigley’s MED, 11th edition
Chapter 11 Solutions, Page 32/31
Chapter 14
d 
14-1
N
22

 3.667 in
P
6
Table 14-2:
Y = 0.331
 dn  (3.667)(1200)
V 

 1152 ft/min
12
12
Eq. (13-34):
1200  1152
Kv 
1.96
1200
Eq. (14-4b):
H
15
W t  33 000
 33 000
 429.7 lbf
V
1152
Eq. (13-35) :
K W t P 1.96(429.7)(6)
  v

 7633 psi  7.63 kpsi
FY
2(0.331)
Eq. (14-7):
Ans.
________________________________________________________________________
N
18

 1.8 in
P
10
Table 14-2: Y = 0.309
 dn  (1.8)(600)
V 

 282.7 ft/min
12
12
Eq. (13-34):
1200  282.7
Kv 
 1.236
1200
Eq. (14-4b):
H
2
W t  33 000
 33 000
 233.5 lbf
V
282.7
Eq. (13-35) :
K W t P 1.236(233.5)(10)
  v

 9340 psi  9.34 kpsi
FY
1.0(0.309)
Eq. (14-7):
d 
14-2
Ans.
________________________________________________________________________
d  mN  1.25(18)  22.5 mm
14-3
Table 14-2:
Eq. (14-6b):
Eq. (13-36):
Y = 0.309
 dn  (22.5)(10 3 )(1800)
V 

 2.121 m/s
60
60
6.1  2.121
Kv 
 1.348
6.1
60 000H
60 000(0.5)
Wt 

 0.2358 kN  235.8 N
 dn
 (22.5)(1800)
Shigley’s MED, 11th edition
Chapter 14 Solutions, Page 1/40
 
K vW t
1.348(235.8)

 68.6 MPa
FmY
12(1.25)(0.309)
Ans.
Eq. (14-8):
________________________________________________________________________
d  mN  8(16)  128 mm
14-4
Table 14-2:
Eq. (14-6b):
Eq. (13-36):
Y = 0.296
 dn  (128)(10 3 )(150)
V 

 1.0053 m/s
60
60
6.1  1.0053
Kv 
 1.165
6.1
60 000 H
60 000(6)
Wt 

 5.968 kN  5968 N
 dn
 (128)(150)
K vW t
1.165(5968)
 

 32.6 MPa
FmY
90(8)(0.296)
Ans.
Eq. (14-8):
________________________________________________________________________
d  mN  1(16)  16 mm
14-5
Table 14-2:
Eq. (14-6b):
Eq. (13-36):
Y = 0.296
 dn  (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
 dn
 (16)(400)
F 
Eq. (14-8):
K vW t
1.055(447.6)

 10.6 mm
 mY
150(1)(0.296)
From Table 13-2, use F = 11 mm or 12 mm, depending on availability. Ans.
________________________________________________________________________
d  mN  2(20)  40 mm
14-6
Table 14-2:
Eq. (14-6b):
Eq. (13-36):
Y = 0.322
 dn  (40)(10 3 )(200)
V 

 0.419 m/s
60
60
6.1  0.419
Kv 
 1.069
6.1
60 000H
60 000(0.5)
Wt 

 1.194 kN  1194 N
 dn
 (40)(200)
Shigley’s MED, 11th edition
Chapter 14 Solutions, Page 2/40
F 
Eq. (14-8):
K vW t
1.069(1194)

 26.4 mm
 mY
75(2.0)(0.322)
From Table 13-2, use F = 28 mm. Ans.
________________________________________________________________________
N
24

 4.8 in
P
5
Table 14-2: Y = 0.337
 dn  (4.8)(50)

 62.83 ft/min
V 
12
12
Eq. (13-34):
1200  62.83
Kv 
 1.052
1200
Eq. (14-4b):
H
6
W t  33 000
 33 000
 3151 lbf
V
62.83
Eq. (13-35) :
K W t P 1.052(3151)(5)
F  v

 2.46 in
3

Y
20(10
)(0.337)
Eq. (14-7):
d 
14-7
Use F = 2.5 in Ans.
________________________________________________________________________
N
16

 4.0 in
P
4
Table 14-2: Y = 0.296
 dn  (4.0)(400)
V 

 418.9 ft/min
12
12
Eq. (13-34):
1200  418.9
Kv 
 1.349
1200
Eq. (14-4b):
H
20
W t  33 000
 33 000
1575.6 lbf
V
418.9
Eq. (13-35) :
K W t P 1.349(1575.6)(4)
F  v

 2.39 in
3

Y
12(10
)(0.296)
Eq. (14-7):
d 
14-8
Use F = 2.5 in Ans.
________________________________________________________________________
14-9
Try P = 8 which gives d = 18/8 = 2.25 in and Y = 0.309.
V 
Eq. (13-34):
Shigley’s MED, 11th edition
 dn  (2.25)(600)

 353.4 ft/min
12
12
Chapter 14 Solutions, Page 3/40
Eq. (14-4b):
Eq. (13-35):
1200  353.4
 1.295
1200
H
2.5
W t  33 000
 33 000
 233.4 lbf
V
353.4
Kv 
F 
Eq. (14-7):
K vW t P 1.295(233.4)(8)

 0.783 in
Y
10(103 )(0.309)
Using coarse integer pitches from Table 13-2, the following table is formed.
d
9.000
2
6.000
3
4.500
4
3.000
6
V
Kv
1413.717
2.178
942.478
1.785
706.858
1.589
471.239
1.393
2.250
8
353.429
1.295
1.800
10
282.743
1.236
1.500
12
235.619
1.196
1.125
16
176.715
1.147
Wt F
58.356
0.082
87.535
0.152
116.713
0.240
175.06
0.473
9
233.42
0.782
6
291.78
1.167
2
350.13
1.627
9
466.85
2.773
2
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 
Eq. (14-6b):
Eq. (13-36):
 dn  (36)(10 3 )(900)

 1.696 m/s
60
60
6.1  1.696
1.278
6.1
60 000H
60 000(1.5)
Wt 

 0.884 kN  884 N
 dn
 (36)(900)
Kv 
F 
1.278(884)
 24.4 mm
75(2)(0.309)
Eq. (14-8):
Using the preferred module sizes from Table 13-2:
m
d
1.00
1.25
Shigley’s MED, 11th edition
V
18.0
22.5
Kv
Wt
F
0.848 1.139 1768.388 86.917
1.060 1.174 1414.711 57.324
Chapter 14 Solutions, Page 4/40
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
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
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
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
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 Given: Sc = 2.22 HB +200 MPa, HB = 200, Pinion NP = 16 teeth,  = 20o, m = 8 mm,
F = 90 mm, nP = 150 rev/min, W t = 6 kN, gear ratio 4:1. NG = 16(4) = 64 teeth.
Eq. (13-2): dP = m NP = 8(16) = 128 mm, dG = m NG = 8(64) = 512 mm
Pinion and Gear steel,
Eq. (14-13):


E

C p 
2
 2  1    
V
Eq. (14-12):
Eq. (14-14):
 207  109  


2
 2  1  0.292  
1/ 2
189.8  103  Pa
2 d P nP 2  0.128 150

1.005 m/s
2  60 
2  60 
Kv 
Eq. (14-6b):
1/2
6.1  1.005
1.165
6.1
rP sin  128sin 20o
r1 

21.89 mm,
2
2
 K W t  1 1 
 C  C p  v    
 F cos   r1 r2  
r2 4r1 87.56 mm
1/2
 1.165  6  103 
1
1

 189.8  10  



o
 0.090 cos 20  0.02189 0.08756  
1/2
3
 412.3  106  Pa  412.3 MPa
Sc = 2.22 HB +200 = 2.22(200) + 200 = 644 MPa
Shigley’s MED, 11th edition
Chapter 14 Solutions, Page 5/40
2
2
 S   644 
n  c  
Ans.
 2.44
412.3



 C
______________________________________________________________________________
14-12
NP
20
N
50

 2.5 in, d G  G 
 6.25 in
P
8
P
8
 (2.5)(1200)
V 
 785.4 ft/min
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
Table 14-8:
[Note: Using Eq. (14-13) can result in wide variation in
Cp due to wide variation in cast iron properties.]
2.5sin 20
6.25sin 20
r1 
 0.4275 in, r2 
1.069 in
2
2
Eq. (14-12):
Eq. (14-14):
 K W t  1 1 
C   Cp  v
  
F
cos

 r1 r2  

1/ 2
1/ 2
 1.655(504.2)  1
1 
  2100 



 1.5cos 20  0.4275 1.069  
  92.5(103 ) psi   92.5 kpsi Ans.
________________________________________________________________________
14-13
16
48
 1.333 in, d G 
 4 in
12
12
 (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
Kv 
C p  2100 psi
Table 14-8:
[Note: Using Eq. (14-13) can result in wide variation
in Cp due to wide variation in cast iron properties.]
Shigley’s MED, 11th edition
Chapter 14 Solutions, Page 6/40
Eq. (14-12):
Eq. (14-14):
r1 
1.333sin 20
 0.228 in,
2
r2 
 1.204(202.6)  1
1 
 C   2100 



 F cos 20  0.228 0.684  
4 sin 20
 0.684 in
2
1/ 2
  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, d G 5(48) 240 mm
14-14
V
Eq. (14-6a):
 (120)(10 3 )(50)
0.3142 m/s
60
Kv 
3.05  0.3142
1.103
3.05
60 000H
60(103 ) H

 3.183H
 dn
 (120)(50)
where H is in kW and Wt is in kN
Wt 
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].
120sin 20
240sin 20
r1 
 20.52 mm, r2 
 41.04 mm
2
2
Eq. (14-12):
Eq. (14-14):
 1.103(3.183)  103  H  1
1 
 690   163 



60 cos 20o
 20.52 41.04  

1/ 2
H 3.94 kW Ans.
________________________________________________________________________
14-15
Eq. (14-6a):
d P  4(20)  80 mm, dG  4(32)  128 mm
 (80)(10 3 )(1000)
V 
 4.189 m/s
60
3.05  4.189
Kv 
 2.373
3.05
Wt 
Shigley’s MED, 11th edition
60(10)(103 )
 2.387 kN  2387 N
 (80)(1000)
Chapter 14 Solutions, Page 7/40
C p  163 MPa
Table 14-8:
[Note: Using Eq. (14-13) can result in wide variation in
Cp due to wide variation in cast iron properties.]
80sin 20
128sin 20
r1 
 13.68 mm, r2 
 21.89 mm
2
2
Eq. (14-12):
1/ 2
 2.373(2387)  1
1 
 C   163 


    617 MPa Ans.
50
cos
20
13.68
21.89




Eq. (14-14):
________________________________________________________________________
14-16 The pinion controls the design.
Bending
YP = 0.303,
YG = 0.359
17
30
 1.417 in, dG 
 2.500 in
12
12
 d n  (1.417)(525)
V  P 
 194.8 ft/min
12
12
1200  194.8
Kv 
 1.162
1200
Se  0.5Sut  0.5(76)  38.0 kpsi
dP 
Eq. (14-4b):
Eq. (6-10):
Eq. (6-18):
Eq. (14-3):
ka = aSutb = 2.00(76)–0.217 = 0.781
2.25 2.25
l 

 0.1875 in
Pd
12
3Y
3(0.303)
x P 
 0.0379 in
2P
2(12)
t  4lx  4(0.1875)(0.0379)  0.1686 in
Eq. (b), Sec. 14-1:
d e  0.808 hb  0.808 0.875(0.1686)  0.310 in
Eq. (6-24):
 0.107
 0.107
 d 
 0.310 
kb  

 0.996


 0.3 
 0.3 
Eq. (6-19):
kc = kd = ke = 1
Account for one-way bending with kf = 1.66. (See Ex. 14-2.)
Eq. (6-17):
Se = 0.781(0.996)(1)(1)(1)(1.66)(38.0) = 49.07 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,
Shigley’s MED, 11th edition
Chapter 14 Solutions, Page 8/40
r
r
0.025
 f 
 0.148
d
t
0.1686
Approximate D/d = ∞ with D/d = 3; from Fig. A-15-6, Kt = 1.68.
From Fig. 6-26, with Sut = 76 kpsi and r = 0.025 in, q = 0.62.
Eq. (6-32):
Kf = 1 + q( Kt – 1) = 1 + 0.62 (1.68 – 1) = 1.42
S
49.07
 all  e 
15.36 psi
K f nd
1.42(2.25)
FY 
0.875(0.303)(15 360)
W t  P all 
 292.0 lbf
K vPd
1.162(12)
W tV
292.0(194.8)
H 

 1.72 hp Ans.
33 000
33 000
Wear
 1 =  2 = 0.292,
Eq. (14-13):
Eq. (14-12):
E1 = E2 = 30(106) psi


 






1
1
 
Cp  


2
2
2
  1   P 1  G  
  1  0.292  

 
   2 

EG  
 30  10 6   
  EP

 
1/ 2
 2285 psi
dP
1.417
sin  
sin 20  0.242 in
2
2
d
2.500
r2  G sin  
sin 20  0.428 in
2
2
1 1
1
1



 6.469 in  1
r1 r2
0.242 0.428
r1 
(SC )108  0.4H B  10 kpsi  [0.4(149)  10](103 )  49 600 psi
Eq. (6-79):
From the discussion and unnumbered equation immediately following Eq. (6-80),
 S  8  49 600
 C ,all   C 10 
  33 067 psi
n
2.25
From Eq. (14-14):


2

2
  
F cos     33 067   0.875cos 20 
W t   C  

 22.6 lbf
 C p    1 1    2285   1.162(6.469) 

 K
 
v
  r1 r2  
Shigley’s MED, 11th edition
Chapter 14 Solutions, Page 9/40
W tV
22.6(194.8)

 0.133 hp
33 000
33 000
Rating power (pinion controls):
H 
Ans.
H1 = 1.72 hp
H2 = 0.133 hp
Hall = (min 1.72, 0.133) = 0.133 hp Ans.
________________________________________________________________________
14-17 See Prob. 14-16 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   d P n / 12   (6.667)(870) / 12 1519 ft/min
K v  (1200  1519) / 1200  2.266
Se  0.5(113)  56.5 kpsi
ka  2.00(113)  0.217  0.717
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  0.717(0.873)(1)(1)(1)(1.66)(56.5)  58.7 kpsi
rf  0.300 / 3  0.100 in
r
r
0.100
 f 
 0.144
d
t
0.695
Kt = 1.75, q = 0.85, Kf = 1.64
S
58.7
 all  e 
 23.9 kpsi
K f nd
1.64(1.5)
FY 
2.5(0.322)(23 900)
W t  P all 
 2830 lbf
K vPd
2.266(3)
H  W tV / 33 000  2830(1519) / 33 000  130 hp
Ans.
Wear
Shigley’s MED, 11th edition
Chapter 14 Solutions, Page 10/40
Eq. (14-13):






1
Cp  

2
  1  0.292  
 2  30  106   

 
1/ 2
 2285 psi
Eq. (14-12):
r1 = (6.667/2) sin 20° = 1.140 in
r2 = (33.333/2) sin 20° = 5.700 in
Eq. (6-68):
SC = [0.4(262) – 10](103) = 94 800 psi
 C ,all   SC / nd   94 800 / 1.5   77 400 psi
2

 F cos  

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 130 hp
H 2  52.0 hp
H rated  min(130, 52.0)  52.0 hp Ans.
________________________________________________________________________
14-18 See Prob. 14-16 solution for equation numbers.
Given:  = 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
d G  6(30)  180 mm
 d Pn  (96)(10 3 )(1145)
V 

 5.76 m/s
60
(60)
6.1  5.76
Kv 
 1.944
6.1
Shigley’s MED, 11th edition
Chapter 14 Solutions, Page 11/40
Se  0.5(900)  450 MPa
ka  3.04(900)  0.217  0.695
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
d e  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  0.695(0.884)(1)(1)(1)(1.66)(450)  458.9 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
S
458.9
 all  e 
 223.4 MPa
K f nd 1.58  1.3
Wt 
Eq. (14-8):
Eq. (13-36):
FYm all
75(0.296)(6)(223.4)

 15 310 N
Kv
1.944
W t dn 15.31 (96)(1145)
H 

 88.1 kW
60 000
60 000
Ans.
Wear: Pinion and gear
Eq. (14-12):
Eq. (14-13):
Eq. (6-79):
r1 = (96/2) sin 20 = 16.42 mm
r2 = (180/2) sin 20 = 30.78 mm






1
Cp  

2
  1  0.292  
 2  207  103   

 
190 MPa
SC = 6.89[0.4(260) – 10] = 647.7 MPa
 647.7
 C ,all   SC / nd 
  568 MPa
1.3


W   C ,all 
 Cp 


t
Eq. (14-14):
1/ 2
Shigley’s MED, 11th edition
2

F cos  
1


K v  1 / r1  1 / r2 
Chapter 14 Solutions, Page 12/40
2
o
1
  568   75cos 20  



 
  3469 N
 190   1.944   1 / 16.42  1 / 30.78 
W t dn 3.469 (96)(1145)
H 

 20.0 kW
60 000
60 000
Eq. (13-36):
Thus, wear controls the gearset power rating; H = 20.0 kW. Ans.
________________________________________________________________________
NP = 17 teeth, NG = 51 teeth
N 17
dP  
 2.833 in
P
6
51
dG 
 8.500 in
6
V   d P n / 12   (2.833)(1120) / 12  830.7 ft/min
14-19
Eq. (14-4b):
Kv = (1200 + 830.7)/1200 = 1.692
S
90 000
 all  y 
 45 000 psi
nd
2
Table 14-2:
YP = 0.303, YG = 0.410
Wt 
Eq. (14-7):
H 
Eq. (13-35):
FYP all
2(0.303)(45 000)

 2686 lbf
K vP
1.692(6)
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):
Eq. (6-10):
Sut = 0.5 HB = 0.5(232) = 116 kpsi
Se  0.5Sut  0.5(116)  58 kpsi
Eq. (6-18):
ka  aSutb  2.00(116)  0.217  0.713
l 
1
1.25 2.25 2.25



 0.375 in
Pd
Pd
Pd
6
x
3YP
3(0.303)

 0.0758 in
2P
2(6)
Table 13-1:
Eq. (14-3):
Shigley’s MED, 11th edition
Chapter 14 Solutions, Page 13/40
Eq. (b), Sec. 14-1:
t  4lx  4(0.375)(0.0758)  0.337 in
d e  0.808 F t  0.808 2(0.337)  0.663 in
Eq. (6-24):
 0.107
 d 
 0.663 
kb  



 0.3 
 0.3 
kc = kd = ke = 1
Eq. (6-19):
 0.107
 0.919
Account for one-way bending with kf = 1.66. (See Ex. 14-2.)
Eq. (6-17): Se  0.713(0.919)(1)(1)(1)(1.66)(58)  63.1 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
 f 
 0.148
0.338
d
t
Fig. A-15-6:
Estimate D/d = ∞ by setting D/d = 3, Kt = 1.68.
Fig. 6-26:
q = 0.86
K f 1  q  K t  1 1  (0.86)(1.68  1) 1.58
Eq. (6-32):
S
63.1
 all  e 
 20.0 kpsi
K f nd 1.58(2)
FY 
2(0.303)(20 000)
W t  P all 
 1194 lbf
K vPd
1.692(6)
W tV
1194(830.7)
H 

 30.1 hp
33 000
33 000
Ans.
(b) Pinion fatigue
Wear
1/ 2
Eq. (14-13):
Eq. (14-12):
Shigley’s MED, 11th edition


1
Cp  
2
6 
 2 [(1 - 0.292 ) / 30(10 )] 
 2285 psi
dP
2.833
sin  
sin 20o  0.485 in
2
2
d
8.500
r2  G sin  
sin 20o  1.454 in
2
2
1 1
1
1

 2.750 in
   
r
r
0.485
1.454
2 
 1
r1 
Chapter 14 Solutions, Page 14/40
Eq. (6-79):
(SC )108  0.4 H B  10 kpsi
In terms of gear notation
C = [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
nd  2
. [See unnumbered equation immediately
to the load Wt by dividing by
after Eq. (6-80)].

82 800
  58 548 psi
 C ,all   c  
2
2
Solve Eq. (14-14) for Wt:
2
o
  58 548   2 cos 20 
W 
 
  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.
(c) Gear fatigue due to bending and wear
t
Bending
x
Eq. (14-3):
Eq. (b), Sec. 14-1:
Eq. (6-24):
3YG
3(0.4103)

 0.1026 in
2P
2(6)
t  4 lx  4(0.375)(0.1026)  0.392 in 
d e  0.808 F t  0.808 2(0.392)  0.715 in
 0.107
 0.715 
kb  
 0.911

0.30


Eq. (6-19):
kc = kd = ke = 1
kf = 1.66. (See Ex. 14-2.)
S

0.713(0.911)(1)(1)(1)(1.66)(58)
 62.5 kpsi
Eq. (6-17): e
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. 6-26: q = 0.82
K 1  q  K t  1 1  (0.82)(1.80  1) 1.66
Eq. (6-32): f
S
62.5
 all  e 
 18.8 kpsi
K f nd
1.66(2)
Shigley’s MED, 11th edition
Chapter 14 Solutions, Page 15/40
Wt 
FYP all
2(0.4103)(18 800)

 1520 lbf
K vPd
1.692(6)
W tV
1520(830.7)

 38.3 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 = 30.1 hp
Pinion wear:
H2 = 6.67 hp
Gear bending:
H3 = 38.3 hp
Gear wear:
H4 = 6.67 hp
Power rating of the gear set is thus
Hrated = min(30.1, 6.67, 38.3, 6.67) = 6.67 hp Ans.
________________________________________________________________________
14-20 Given: H = 5 hp, nP = 300 rev/min,  = 20o, HB = 200, Qv = 6, L = 108 cycles with R = 90
percent.
(a) Let F = 2 in, dP = NP/P = 16/6 = 2.667 in, dG = 48/6 = 8 in
d n
 (2.667)(300)
V  P P 
 209.5 ft/min
12
12
Eq. (13-34):
H
5
W t  33 000
 33 000
 787.6 lbf
V
209.5
Eq. (13-35):
Assuming uniform loading, Ko = 1. Qv = 6,
Eq. (14-28): B = 0.25(12  Qv) = 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  209.5 
K v  
  

A 
59.77



0.8255
 1.196
YP  0.296
Table 14-2:
From Eq. (a), Sec. 14-10, with F = 2 in
Eq. (14-31):
Eq. (14-32):
F Y
 2 0.296 
( K s ) P 1.192 
 1.192 

6
 P 


Cmc = 1
Shigley’s MED, 11th edition
0.0535
1.088
Chapter 14 Solutions, Page 16/40
F
2
 0.0375  0.0125F 
 0.0375  0.0125(2)  0.0625
10d P
10(2.667)
Eq.(4-33) assuming gears are centered between bearings: Cpm = 1
Fig. 14-11: Cma = 0.16
Eq. (14-35): Ce = 1
K 1  Cmc  C pf C pm  Cma Ce  1  1  0.0625  1  0.16  1  1.2225
Eq. (14-30): m
Eq. (14-40) assuming constant gear thickness: KB = 1
8  0.0178
 0.977
For the pinion, N = 108 cycles, and from Fig. 14-14: (YN ) P 1.3558(10 )
J P  0.27
Fig. 14-6:
Table 14-10, for R = 0.90:
KR = 0.85
Figs. 14-17 and 14-18:
KT = Cf = 1
Eq. (14-22): mG = NG/NP = 48/16 = 3
cos t sin t mG
cos 20o sin 20o  3 
I 


  0.1205
2  mN  mG  1
2(1)
3  1

Eq. (14-23):
Cp f 
C p  2300 psi
Table 14-8:
Strength: Grade 1 steel with HBP = HBG = 200
Fig. 14-2:
(St)P = 77.3(200) + 12 800 = 28 260 psi
Fig. 14-5:
(Sc)P = 322(200) + 29 100 = 93 500 psi
Fig. 14-15:
(ZN )P = 1.4488(108)–0.023 = 0.948
Sec. 14-12: HBP/HBG = 1  CH = 1
Pinion tooth bending
Eq. (14-15):
P K K
 6   (1.2207)(1) 
( ) P  W t K o K vK s d m B  787.8(1)(1.196)(1.088)   
F J
 2   0.27 
13 904 psi Ans.
Eq. (14-17):
S YN
28 260  0.977 
 all  t

Ans.

 32 482 psi
S F KT K R
1  1(0.85) 
28 260(0.977) / [(1)(0.85)]
 S Y / ( KT K R ) 
(S F ) P   t N

 2.34


13 904


Eq. (14-41):
Ans
Pinion tooth wear
Eq. (14-16):
1/ 2

K C 
( c ) P  C p  W t K o K vK s m f 
dPF I  P


 1.2207   1  
 2300  787.8(1)(1.196)(1.088) 


 2.667(2)   0.1205  

Shigley’s MED, 11th edition
1/ 2
101 485 psi
Ans.
Chapter 14 Solutions, Page 17/40
 all ,c 
Eq. (14-18):
Eq. (14-42):
Sc Z N CH 93 500  0.948(1) 


 104 280 psi
S H KT K R
1  1(0.85) 
Ans.
 S Z /  KT K R  
93 500  0.948  / [1 0.85  ]
(S H ) P   c N
 1.03
 
c
101 485

P
Ans.
Ans.
2
143 521
101 485 
F
F
(b) Only check wear. From Eq. (14-16) in part (a):
Sc ZN / (KT KR) = 93 500 (0.948)/[1(0.85)] = 104 280 psi
 S Z /  KT K R  
104 280
2
 SH  P   c N
 
c

 P 143 521 / F
c  P 
 143 521 
F 2 
 2.75 in
 104 280 
p = /6 = 0.5235 in, 3 p = 1.57 in and 5 p = 2.62. Fails to satisfy requirements since both
are less than 2.75 in.
Try P = 5 teeth/in. 3 p = 3/5 = 1.885 in and 5 p = 5/5 = 3.142 in.
dP = 16/5 = 3.2 in, V =  (3.2) 300/12 = 251.3 ft/min
5
W t 33 000
656.6 lbf
251.3
 59.77  251.3 
K v  

59.77


0.8255
1.214
Keep Km = 1.2207 and check later. KR = 0.85, I = 0.1205, Cp = 2300 psi , (Sc)P =
93 500 psi, (ZN)P = 0.948, and Sc ZN / (KT KR) = 104.28(103) psi.
1/2
115.5  103 

 1.2207  1 
  c  P 2300  656.6  1 1.214  1 

 
F
 3.2 F  0.1205 

 S Z /  KT K R  
104 280 F
SH )P   c N
2
 
c
115.5  103 

P
 115.5  103  
 2.22 in
F 2 
 104 280 


Use F = 2.5 in
Ans.
Rechecking Km.
F
2.5
Cp f 
 0.0375  0.0125 F 
 0.0375  0.0125(2.5)  0.0719
10d P
10(3.2)
Cma = 0.17
K m 1  Cmc  C pf C pm  Cma Ce  1  1  0.0719  1  0.17  1  1.2419
c  P
3
3
1/2
 1.2419  115.5  10  116.5  10 



F
F
 1.2207 
Shigley’s MED, 11th edition
Chapter 14 Solutions, Page 18/40
 116.5  103  
 2.24 in
F 2 
 104 280 


F = 2.5 in
Acceptable. Ans.
________________________________________________________________________
14-21
dP = 2.5(20) = 50 mm, dG = 2.5(36) = 90 mm
d n
 (50)(10 3 )(100)
V  P P 
 0.2618 m/s
60
60
60(120)
Wt 
 458.4 N
 (50)(10 3 )(100)
With no specific information given to indicate otherwise, assume
KB = Ko = Y = 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:
Eq. (14-38):
(YJ )P = 0.33, (YJ )G = 0.38
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 
Eq. (14-23):
Shigley’s MED, 11th edition
Chapter 14 Solutions, Page 19/40
Table 14-8:
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
 ZW  CH  1
Pinion tooth bending
Eq. (14-15):

1 KH KB 
( ) P   W t K o K vK s

bmt YJ  P

Eq. (14-41) for SI:
 1   1.27(1) 
 458.4(1)(1.099)(1) 

  43.08 MPa
 18(2.5)   0.33 
S Y 
194.9  0.977 
(S F ) P   t N  
 1(0.885)   4.99 Ans.

Y
Y
43.08


 Z P

Ans.
Gear tooth bending
 1   1.27(1) 
( )G  458.4(1)(1.099)(1) 
  0.38   37.42 MPa
18(2.5)



194.9  0.987 
( S F )G 
 5.81 Ans.
37.42  1(0.885) 
Ans.
Pinion tooth wear
Eq. (14-16):

K Z 
( c ) P   Z E W t K o K vK s H R 

d w1b Z I  P

Eq. (14-42) for SI:
 1.27   1 
191 458.4(1)(1.099)(1) 

  501.8 MPa
 50(18)   0.103 
S Z Z 
644  0.948(1) 
(S H ) P   c N W  

 1.37 Ans.
  c Y YZ  P 501.8  1(0.885) 
Ans.
Gear tooth wear
 (K ) 
( c )G   s G 
 (K s ) P 
( S H )G 
Shigley’s MED, 11th edition
1/ 2
 1
( c ) P   
 1
644 0.961(1)
1.39
501.8 1(0.885)
1/ 2
(501.8)  501.8 MPa
Ans.
Ans.
Chapter 14 Solutions, Page 20/40
________________________________________________________________________
14-22
Pt = Pn cos  = 6cos 30o = 5.196 teeth/in
16
48
dP 
 3.079 in, d G  (3.079)  9.238 in
5.196
16
d n
 (3.079)(300)
V  P P 
 241.8 ft/min
12
12
Eq. (13-34):
H
5
W t  33 000
 33 000
 682.3 lbf
V
241.8
Eq. (13-35):
Eq. (14-28): B = 0.25(12  Qv) = 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  241.8 
K v  
  

A 
59.77



From Prob. 14-20:
YP  0.296,
K R  0.85,
0.8255
 1.210
( K s ) P 1.088, K B  1,
mG  3, (YN ) P  0.977
(St ) P  28 260 psi, CH 1, ( Sc ) P  93 500 psi
( Z N ) P  0.948,
C p  2300 psi
The pressure angle is:
 tan 20 
t  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
1/ 2
2
2
2
2
Z    rP  a    rb  P     rG  a    rb  G    rP  rG  sin t




1/ 2
1/ 2
2
2
  3.079

  9.238



2
2
 
 0.167   1.419    
 0.167   4.258 


  2

  2

 3.079 9.238 


 sin 22.8
2 
 2
 0.9479  2.1852  2.3865  0.7466 Conditions O.K . for use
pN  pn cos n 
mN 
Eq. (14-21):
Shigley’s MED, 11th edition

cos 20  0.4920 in
6
pN
0.492

 0.6937
0.95Z
0.95(0.7466)
Chapter 14 Solutions, Page 21/40
 sin 22.8 cos 22.8   3 
cos t sin t
mG


  3  1   0.193
2mN
mG  1 
2(0.6937)


JP  0.45
I 
Eq. (14-23):
Fig. 14-7:
Fig. 14-8: Correction is 0.94. Thus, JP = 0.45(0.94) = 0.423.
Eq. (14-31): Cmc = 1
F
2
C pf 
 0.0375  0.0125 F 
 0.0375  0.0125(2) 0.0525
10d P
10(3.079)
Eq. (14-32:
Eq. (14-33) assuming gears are centered between bearings: Cpm = 1
Fig. 14-11:
Cma = 0.16
Eq. (14-35): Ce = 1
K 1  Cmc  C pf C pm  Cma Ce  1  1  0.0525  1  0.16  1  1.213
Eq. (14-30): m
Pinion tooth bending
P K K
 5.196  1.213(1)
( ) P  W t K o K vK s d m B  682.3(1)(1.21)(1.088) 

F J
 2  0.423
 6692 psi Ans.
Eq. (14-15):
(S F ) P 
Eq. (14-41):
Pinion tooth wear
Eq. (14-16):
StYN /  KT K R 
28 260(0.977) / [1(0.85)]

 4.85
6692
  P
Ans.
1/ 2

K C 
( c ) P  C p  W t K o K vK s m f 
dPF I  P

1/ 2

 1.213   1  
 2300 682.3(1)(1.21)(1.088) 
   69 640 psi Ans.

 3.079(2)   0.193  

 S Z /  KT K R  
93 500(0.948) / [(1)(0.85)]
(S H ) P   c N
1.50
 

69
640
c

P
(Eq. 14-43):
Ans.
______________________________________________________________________________
14-23 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
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
Shigley’s MED, 11th edition
(St )107  77.3H B  12 800
 77.3(232)  12 800  30 734 psi
Chapter 14 Solutions, Page 22/40
Fig. 14-14:
YN = 1.6831(108)–0.0323 = 0.928
V   d P n / 12   (2.833)(1120 / 12)  830.7 ft/min
KT  K R  1, S F  2, S H  2
30 734(0.928)
 all 
 14 261 psi
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
Cma  0.127  0.0158(2)  0.093(10 4 )(22 )  0.1586
Ce  1
Eq. (14-15):
K m  1  1[0.0581(1)  0.1586(1)]  1.217
K B 1
FJ P all
Wt 
K o K vK 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
Fig. 14-15:
ZN = 2.466N–0.056 = 2.466(108)–0.056 = 0.879
mG  51 / 17  3
I 
Eq. (14-23):
Shigley’s MED, 11th edition
cos 20o sin 20 o  3 

  1.205,
2
 3  1
CH  1
Chapter 14 Solutions, Page 23/40
Fig. 14-5:
0.99
( Sc )107  322 H B  29 100
 322(232)  29 100  103 804 psi
103 804(0.879)
 c,all 
 64 519 psi
2(1)(1)
2
Eq. (14-16):
 
Fd P I
W t   c,all 
 C  KK KK C
 p  o v s m f
2
 64 519   2(2.833)(0.1205) 

 2300   1(1.472)(1)(1.2167)(1) 
 300 lbf
300(830.7)
W tV
H 

 7.55 hp
33 000
33 000
The pinion controls, therefore Hrated = 7.55 hp Ans.
________________________________________________________________________
14-24
l = 2.25/ Pd,
x = 3Y / 2Pd
 2.25   3Y 
3.674
t  4 lx  4 
Y

 
Pd
 Pd   2 Pd 
 3.674 
F Y
d e  0.808 F t  0.808 F 
 Y  1.5487
Pd
 Pd 
 1.5487 F Y / P 
d

kb  


0.30


F Y
1
K s   1.192 
kb
 Pd
 0.107
F Y
 0.8389 
 Pd



 0.0535
0.0535

Ans.


________________________________________________________________________
14-25 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
 (5.5)(1145)
V 
1649 ft/min
12
Pinion bending
0.99 ( St )107  77.3H B  12 800  77.3(250)  12 800  32 125 psi
YN  1.6831[3(109 )] 0.0323  0.832
Shigley’s MED, 11th edition
Chapter 14 Solutions, Page 24/40
Eq. (14-17):
  all  P 
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
 59.77  1649 
K v  

59.77


K s  1, Cm  1
0.8255
1.534
F
 0.0375  0.0125 F
10d
3.25

 0.0375  0.0125(3.25)  0.0622
10(5.5)
Cmc 
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,
KT  1
W1t 
26 728(3.25)(0.345)
 3151 lbf
1.25(1.534)(1)(4)(1.240)
H1 
3151(1649)
 157.5 hp
33 000
Eq. (14-15):
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 ( S c )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
( c,all ) P 
109 600(0.727)
 79 679 psi
1(1)(1)
2
 
Fd P I
W   c,all 
 C p  K o K vK s K mC f


t
3
Shigley’s MED, 11th edition
Chapter 14 Solutions, Page 25/40
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?
________________________________________________________________________
14-26 From Prob. 14-25:
W1t  3151 lbf , W2t  3861 lbf ,
W3t  1061 lbf , W4t  1182 lbf
33 000 K o H
33 000(1.25)(40)
Wt 

 1000 lbf
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:
n3 
W3t
1061

 1.06
1000 1000
( S H ) P  1.06  1.03
Ans.
Gear wear
based on load:
based on stress:
Shigley’s MED, 11th edition
n4 
W4t
1182

 1.18
1000 1000
(S H )G  1.18  1.09
Ans.
Chapter 14 Solutions, Page 26/40
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.
________________________________________________________________________
14-27 Solution summary from Prob. 14-25: 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
( all ) P
( all )G
W1t
W2t
 26 728 psi
 27 546 psi
 3151 lbf ,
 3861 lbf ,
( St ) P  32 125 psi
( St ) G  32 125 psi
H 1  157.5 hp
H 2  192.9 hp
Wear
  20o ,
I  0.1176,
(Z N )G  0.769,
(Z N ) P  0.727
CP  2300 psi
( Sc ) P  Sc  322(390)  29 100  154 680 psi
154 680(0.727)
112 450 psi
1(1)(1)
154 680(0.769)
( c,all )G 
 118 950 psi
1(1)(1)
( c,all ) P 
2
 112 450 
W3t  
 (1061)  2113 lbf ,
 79 679 
H3 
2113(1649)
 105.6 hp
33 000
2


118 950
W 
 (1182)  2354 lbf ,
 109 600(0.769) 
t
4
H4 
2354(1649)
 117.6 hp
33 000
Rated power
Shigley’s MED, 11th edition
Chapter 14 Solutions, Page 27/40
Hrated = min(157.5, 192.9, 105.6, 117.6) = 105.6 hp
Ans.
Prob. 14-25:
Hrated = min(157.5, 192.9, 53.0, 59.0) = 53 hp
The rated power approximately doubled.
________________________________________________________________________
14-28 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
( all ) P  45 657 psi,
Similarly for (YN )G = 0.859
( all )G  47 161 psi,
W1t  4569 lbf ,
W2t  5668 lbf ,
From Table 14-8,
0.99
and
H1  228 hp
H 2  283 hp
C p  2300 psi.
(Sc )107
Also, from Table 14-6:
 180 000 psi
Modification of Sc by YN produces
( c,all ) P  130 525 psi
( 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.
________________________________________________________________________
14-29 Grade 2, 9310 carburized and case-hardened to 285 core and 580 case in Prob. 14-28.
Shigley’s MED, 11th edition
Chapter 14 Solutions, Page 28/40
Summary:
Table 14-3:
0.99
(St )107  65 000 psi
( all ) P  53 959 psi
( all )G  55 736 psi
and it follows that
W1t  5400 lbf ,
W2t  6699 lbf ,
H1  270 hp
H 2  335 hp
C p  2300 psi.
From Table 14-8,
Also, from Table 14-6:
Sc  225 000 psi
( c,all ) P  181 285 psi
( 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.
________________________________________________________________________
14-30 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
( all ) P  26 668 psi
( 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
For wear
Shigley’s MED, 11th edition
Chapter 14 Solutions, Page 29/40
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. 14-25, 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. 14-25
t
Wrated
 1061 lbf
Thus
304
1
1
 0.2865 
, not
Ans.
1061
3.49
4
________________________________________________________________________
14-31 SP = SH = 1,
Pd = 4,
JP = 0.345,
JG = 0.410,
Ko = 1.25
Bending
Table 14-4:
0.99
(St )107  13 000 psi
13 000(1)
 13 000 psi
1(1)(1)
 all FJ P
13 000(3.25)(0.345)


 1533 lbf
K o K vK 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
( all ) P  ( all )G 
W1t
H1
W2t
H2
Wear
Table 14-8:
Table 14-7:
C p  1960 psi
0.99
(Sc )107  75 000 psi  ( c,all ) P  ( c,all )G
Shigley’s MED, 11th edition
Chapter 14 Solutions, Page 30/40
2
 ( ) 
Fd p I
W3t   c,all P 
 C p  K o K vK s K mC f
2
 75 000  3.25(5.5)(0.1176)
 1295 lbf
W 

 1960  1.25(1.534)(1)(1.24)(1)
W4t  W3t  1295 lbf
1295(1649)
 64.7 hp
H 4  H3 
33 000
t
3
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.
________________________________________________________________________
14-32 From Table A-24a, Eav = 11.8(106) Mpsi
For  = 14.5 and HB = 156
SC 
1.4(81)
 51 693 psi
2sin14.5 / [11.8(106 )]
For  = 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 
14-33 Programs will vary.
________________________________________________________________________
14-34
(YN ) P  0.977,
(YN )G  0.996
(St ) P  (St )G  82.3(250)  12 150  32 725 psi
32 725(0.977)
 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
( all ) P 
Shigley’s MED, 11th edition
Chapter 14 Solutions, Page 31/40
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
( all )G 
Table 14-6:
0.99
(Sc )107  150 000 psi
 0.948(1) 
( 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
( c,allow )G 
(167 294) 171 706 psi
0.948
2
 171 706   1.963(1.5)(0.195) 
t
W4  
 
  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
14-35
(YN )G 1.6831(108 / 3.059)  0.0323  0.962
Table 14-3:
St = 55 000 psi
55 000(0.928)
( 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
( all )G 
(60 047)  62 247 psi
0.928
Shigley’s MED, 11th edition
Chapter 14 Solutions, Page 32/40
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
( c,all ) P 
180 000(0.8790)
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
( c,all )G 
(186 141) 198 169 psi
0.8790
2
 198 169   1.043 
t
W4  
 
 (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,
14-36
Table 14-3:
(YN )G = 0.962
(See Prob. 14-34)
St = 65 000 psi
65 000(0.928)
( 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)
( 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
Shigley’s MED, 11th edition
Chapter 14 Solutions, Page 33/40
Table 14-6:
Sc = 225 000 psi
(Z N ) P  0.8790,
( c,all ) P 
(Z N )G  0.9358
225 000(0.879)
 232 676 psi
1(0.85)
2
 232 676   1.963(1.5)(0.195) 
W 
 
  4013 lbf
 2300   1(1.404)(1.043) 
4013(925)
H3 
 112.5 hp
33 000
0.9358
( c,all )G 
(232 676)  247 711 psi
0.8790
2
 247 711   1.043 
t
W4  
 
 (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.
t
3
The bending of the pinion is the controlling factor.
________________________________________________________________________
P = 2 teeth/in, d = 8 in, N = dP = 8 (2) = 16 teeth
 
 
F 4 p 4   4   2
 P
2
 M x 0 10(300) cos 20  4 FB cos20
14-37
FB = 750 lbf
W t FB cos 20 750cos 20 705 lbf
n = 2400 / 2 = 1200 rev/min
 dn  (8)(1200)

 2513 ft/min
V 
12
12
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):
C p  2300 psi
Table 14-8:
Assume a typical quality number of 6.
Shigley’s MED, 11th edition
Chapter 14 Solutions, Page 34/40
Eq. (14-28):
B  0.25(12  Qv) 2 / 3  0.25(12  6) 2 /3  0.8255
Eq. (14-27):
A V 
 59.77  2513 
K v  
  

A 
59.77



A  50  56(1  B)  50  56(1  0.8255)  59.77
B
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
 2 0.296 
K s  1.192 
 1.192 
 1.23


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)
2
Cp f 
 0.0375  0.0125(2 )  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  0.0759 ln  1  0.95  0.885
With no specific information given to indicate otherwise, assume Ko = KB = KT = Cf = 1
Tooth bending
Eq. (14-15):
  W t K o K vK s
Pd K m K B
F J
 2
 705(1)(1.65)(1.23) 
 2
Eq. (14-41):
  (1.36)(1) 
 2294 psi

  0.27 
 S Y / ( KT K R ) 
SF   t N



Shigley’s MED, 11th edition
Chapter 14 Solutions, Page 35/40

47 000(0.9) / [(1)(0.885)]
 20.8
2294
Ans.
Tooth wear
Eq. (14-16):

K C 
 c  C p  W t K o K vK s m f 
dPF I 

1/ 2
1/ 2

 1.36   1  
 2300  705(1)(1.65)(1.23) 


 8(2 )   0.107  

 43 750 psi
Since gear B is a pinion, CH is not used in Eq. (14-42) (see Fig. 14-18), where
SH 
Sc Z N / ( KT K R )
c
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   m  4  18.75  236 mm
14-38
M
x
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
 dn  (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):
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
Table 14-8:
Z E 191 MPa
Assume a typical quality number of 6.
B  0.25(12  Qv) 2 / 3  0.25(12  6)2 / 3  0.8255
Eq. (14-28):
A  50  56(1  B)  50  56(1  0.8255)  59.77
Shigley’s MED, 11th edition
Chapter 14 Solutions, Page 36/40
B
Eq. (14-27):
 59.77  200(14.14) 
 A  200V 
K v  

  
A
59.77




0.8255
1.69
To estimate a size factor, get the Lewis Form Factor from Table 14-2, Y = 0.296.
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  0.0759 ln  1  0.98  0.955
With no specific information given to indicate otherwise, assume Ko = KB = KT = ZR = 1.
Tooth bending
  W t K o K vK s
Eq. (14-15):
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



Shigley’s MED, 11th edition
Chapter 14 Solutions, Page 37/40

324(0.9) / [(1)(0.955)]
 5.66
53.9
Ans.
Tooth wear
Eq. (14-16):

K Z 
 c  Z E  W t K o K vK s H R 
d w1b Z I 

1/ 2

 1.44   1  
 191  20 670(1)(1.69)(1.28) 


 300(236)   0.134  

 498 MPa
1/ 2
Since gear B is a pinion, CH is not used in Eq. (14-42) (see Fig. 14-18), where
Sc Z N / ( KT K R )
c
960(0.85) / [(1)(0.955)]

1.72 Ans
498
________________________________________________________________________
SH 
14-39 From the solution to Prob. 13-46, n = 191 rev/min, Wt = 1600 N, d = 125 mm,
N = 15 teeth, m = 8.33 mm/tooth.
F b 4 p 4   m  4  8.33 105 mm
V 
 dn  (0.125)(191)

 1.25 m/s
60
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):
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
Table 14-8:
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
Eq. (14-27):
 59.77  200(1.25) 
 A  200V 
K v  

  
A
59.77




A  50  56(1  B)  50  56(1  0.8255)  59.77
B
Shigley’s MED, 11th edition
0.8255
 1.21
Chapter 14 Solutions, Page 38/40
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
equations. d = 125/25.4 = 4.92 in, F = 105/25.4 = 4.13 in
Eq. (14-31): Cmc = 1 (uncrowned teeth)
4.13
C pf 
 0.0375  0.0125(4.13)  0.0981
10(4.92)
Eq. (14-32):
Eq. (14-33):
Fig. 14-11:
Eq. (14-35):
Eq. (14-30):
Cpm = 1
Cma = 0.32 (open gearing)
Ce = 1
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  0.0759 ln  1  0.95  0.885
Eq. 14-38:
With no specific information given to indicate otherwise, assume Ko = KB = KT = ZR = 1.
Tooth bending
1 KH KB
  W t K o K vK s
bmt YJ
Eq. (14-15):

  (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), where
 S Y / ( KT K R ) 
SF   t N



448(0.95) / [(1)(0.885)]

 32.7
14.7
Ans.
Tooth wear
Eq. (14-16):

K Z 
 c  Z E  W t K o K vK s H R 
d w1b Z I 

Shigley’s MED, 11th edition
1/ 2
Chapter 14 Solutions, Page 39/40

 1.42   1  
 191  1600(1)(1.21)(1.17) 


 125(105)   0.107  

 289 MPa
Eq. (14-42):
1/ 2
 S Z / ( KT K R ) 
SH   c N

c


1550(0.88) / [(1)(0.885)] 

  5.33 Ans
289


________________________________________________________________________
14-40 From the solution to Prob. 13-47, n = 2(70) = 140 rev/min, Wt = 180 lbf, d = 5 in
N = 15 teeth, P = 3 teeth/in.
 
 
F 4 p 4   4   4.2 in
 P
 3
 dn  (5)(140)
V 

 183.3 ft/min
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):
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
Table 14-8:
Assume a typical quality number of 6.
B  0.25(12  Qv) 2 / 3  0.25(12  6) 2 /3  0.8255
Eq. (14-28):
A  50  56(1  B)  50  56(1  0.8255)  59.77
B
Eq. (14-27):
A V 
 59.77  183.3 
K v  
  

A 
59.77



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):
F Y 
 4.2 0.290 
K s  1.192 
1.192 


3
 P 


Cmc = 1 (uncrowned teeth)
4.2
C pf 
 0.0375  0.0125(4.2)  0.099
10(5)
Shigley’s MED, 11th edition
0.0535
 1.17
Chapter 14 Solutions, Page 40/40
Eq. (14-33):
Fig. 14-11:
Eq. (14-35):
Eq. (14-30):
Cpm = 1
Cma = 0.32 (Open gearing)
Ce = 1
K m  1  1[0.099(1)  0.32(1)]  1.42
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  0.0759 ln  1  0.98  0.955
Eq. 14-38:
With no specific information given to indicate otherwise, assume Ko = KB = KT = Cf = 1.
Tooth bending
P K K
  W t K o K vK s d m B
F J
Eq. (14-15):
 3   (1.42)(1) 
 180(1)(1.18)(1.17) 
1010 psi

 4.2   0.25 
Eq. (14-41):
 S Y / ( KT K R ) 
SF   t N



65 000(0.95) / [(1)(0.955)]

 64.0
1010
Ans.
Tooth wear
Eq. (14-16):
1/ 2


 1.42   1  
K C 
 c  C p  W t K o K vK s m f   2300  180(1)(1.18)(1.17) 


dPF I 
 5(4.2)   0.107  


 28 800 psi
Since gear B is a pinion, CH is not used in Eq. (14-42), where
1/ 2
 S Z / ( KT K R ) 
SH   c N

c


 225 000(0.88) / [(1)(0.955)] 

  7.28 Ans
28 800


________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 14 Solutions, Page 41/40
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 = (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
0.8255
Eq. (15-5):
 59.77  785.3 
K v  

59.77


Eq. (15-8):
vt,max = [59.77 + (6 – 3)]2 = 3940 ft/min
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
Shigley’s MED, 11th edition
Chapter 15 Solutions, Page 1/20
Bending
Fig. 15-13:
0.99
St  sat  44(300)  2100  15 300 psi
( all ) P  swt 
Eq. (15-4):
WPt 
Eq. (15-3):
sat K L
S F KT K R

15 300(0.862)
10 551 psi
1(1)(1.25)
( all ) P FK x J P
Pd K o K vK 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

( all )G 
Eq. (15-4):
15 300(0.893)
 10 930 psi
1(1)(1.25)
10 930(1.25)(1)(0.216)
 620 lbf
6(1)(1.374)(0.5222)(1.106)
620(785.3)
H2 
 14.8 hp Ans.
33 000
WGt 
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)
( c,all ) P 
 112 630 psi
1(1)(1.118)
( 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
Shigley’s MED, 11th edition
Chapter 15 Solutions, Page 2/20
sac (C L )G CH
S H KT CR
125 920(1.0685)(1.001 38)
( c,all )G 
 120 511 psi
1(1)(1.118)
( 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):
 ( ) 
Fd P I
W   c,all P 
 C
 K K K CC
p

 o v m s xc
2
t
P
2

1.25(3.333)(0.083)
 112 630  



 2290   1(1.374)(1.106)(0.5937)(2) 
 464 lbf
464(785.3)
H3 
 11.0 hp
33 000
2

1.25(3.333)(0.083)
 120 511 
t
WG  


 2290   1(1.374)(1.106)(0.593 75)(2) 
 531 lbf
531(785.3)
H4 
 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,
n = 20, one gear straddle-mounted, Ko = 1, SF = 2, S H  2.
Fig. 15-7:
JP = 0.268, JG = 0.228
Shigley’s MED, 11th edition
Chapter 15 Solutions, Page 3/20
Mesh
dP = 30/6 = 5.000 in, dG = 60/6 = 10.000 in
vt =  (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
s K
4500(1)
swt  at L 
 2250 psi
S F KT K R
2(1)(1)
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
 
 v
I /c
FY
We expect the ratio CI/steel to be
 CI
( K v)CI
ECI


 steel ( K v)steel
Esteel
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
Shigley’s MED, 11th edition
0.8255
1.454
(all)P = swt = 2250 psi
Chapter 15 Solutions, Page 4/20
From Prob. 15-1, Kx = 1, Km = 1.106, Ks = 0.5222
( ) FK J
WPt  all P x P
Pd K o K vK s K m
Eq. (15-3):
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

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
swt   c,all 
 35 355 psi
2
2
Eq. (15-1):
 
Fd P I
W   c,all 
 C p  K o K vK mCsC xc


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

1.25(5.000)(0.086)
 35 355  
W 
 91.6 lbf



 1960   1(1.454)(1.106)(0.59375)(2) 
91.6(1178)
H3  H 4 
 3.27 hp
33 000
Ans.
t
Thus,
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.
________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 15 Solutions, Page 5/20
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,
S  S F  1,
n = 20, shaft angle 90°, n1 = 1800 rev/min, SF = 1, H
F = b = 25
C  190 MPa .
mm, Ko = KA = KT = K = 1 and p
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 = 1.25 (neither member straddle-mounted),
Km = KH = 1.25 + 5.6(10–6)(252) = 1.2535
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
(uncrowned)
Eq. (15-19):
KR = YZ = 0.50 – 0.25 log (1 – 0.999) = 1.25
CR  Z Z  YZ  1.25  1.118
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:
H lim = 2.35HB + 162.89
= 2.35(180) + 162.89 = 585.9 MPa
Fig. 15-6:
I = ZI = 0.066
Shigley’s MED, 11th edition
Chapter 15 Solutions, Page 6/20
( H ) P 
( H
Eq. (15-2):

) (Z NT ) P ZW
S H K Z Z
lim P
585.9(1)(1)
 524.1 MPa
1(1)(1.118)
2
 
bd e1Z I
W  H 
 C  1000 K K K Z Z
A v H  x xc
 p
Eq. (15-1):
t
The constant 1000 expresses W in kN.
t
P
 524.1 
W 

 190 
t
 dnW
1
H3 
60 000
t
P
Eq. (13-36):
Wear of Gear
2


25(88)(0.066)
 1000(1)(1.663)(1.2535)(0.56)(2)   0.473 kN


 (88)(1800)(0.473)

 3.92 kW
60 000
H lim = 585.9 MPa
585.9(1.0054)
( H )G 
 526.9 MPa
1(1)(1.118)
( )
 526.9 
WGt  WPt H G  0.473 
  0.476 kN
( H ) P
 524.1 
 (88)(1800)(0.476)
 3.95 kW
H4 
60 000
Thus in wear, the pinion controls the power rating; H = 3.92 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:
F lim = 0.30HB + 14.48
= 0.30(180) + 14.48 = 68.5 MPa
Eq. (15-13):
Kx = Y = 1
From Prob. 15-5:
Shigley’s MED, 11th edition
YZ = 1.25,
vet = 8.29 m/s,
Chapter 15 Solutions, Page 7/20
K A 1, K v 1.663, K 1,
Yx  0.52, K H  1.2535, YJ 1  0.23
( F ) P 
Eq. (5-4):
WPt 
Eq. (5-3):
 F limYNT
68.5(0.862)

 47.2 MPa
S F K YZ
1(1)(1.25)
( F ) P bmetY YJ 1
1000 K A K vYx K H 
47.2(25)(4)(1)(0.23)
 1.00 kN
1000(1)(1.663)(0.52)(1.2535)
  88   1800   1.00 
H1 
 8.29 kW
60 000
Bending of Gear

 F lim  68.5 MPa
68.5(0.864)
 47.3 MPa
1(1)(1.25)
47.3(25)(4)(1)(0.205)
WGt 
 0.895 kN
1000(1)(1.663)(0.52)(1.2535)
  88   1800   0.895 
H2 
 7.42 kW
60 000
Rating of mesh is
Hrating = min(8.29, 7.42, 3.92, 3.95) = 3.92 kW Ans.
with pinion wear controlling.
________________________________________________________________________
( F ) G 
15-7
(a)
 
 
(S F ) P   all   (S F )G   all 
  P
  G
(sat K L / KT K R ) P
( s K / KT K R )G
 t at L
(W Pd K o K vK s K m / FK x J ) P
(W Pd K o K vK s K m / FK x J )G
t
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 mG
( K L )G J G
JG
where  = – 0.0178 or  = – 0.0323 as appropriate. This equation is the same as
Shigley’s MED, 11th edition
Chapter 15 Solutions, Page 8/20
Eq. (14-44).
Ans.
(b) In bending


 sat K L

FK x J
FK x J
W t   all
 

 S F Pd K o K vK s K m  11  S F KT K R Pd K o K vK s K m 11
(1)
In wear
1/ 2
 sacCLCU 
 W t K o K vK mCsCxc 

C



p
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 vK mCsCxc  22
(2)
Equating the right-hand sides of Eqs. (1) and (2) and canceling terms, and
C  KR
recognizing that 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)
 
 
(S H ) P  ( S H )G   c,all    c,all 
 c  P  c  G
Substituting in the right-hand equality gives
[sacCL / (CR KT )]P
 C W t K K K C C / ( Fd I ) 
o v m s xc
P
 p
P

[ sacCLCH / (CR KT )]G
 C W t K K K C C / (Fd I ) 
o v m s xc
P
 p
G
Denominators cancel, leaving
Shigley’s MED, 11th edition
Chapter 15 Solutions, Page 9/20
(sac)P(CL)P = (sac)G(CL)GCH
Solving for (sac)P gives,
(sac ) P  (sac )G
(CL )G
CH
(CL ) P
(1)
 C  3.4822 N L 0.0602 and  CL  G 3.4822  N L / mG 
From Eq. (15-14), L P
Thus,
 0.0602
CH  sac  G mG0.0602CH
 sac  P  sac  G  1 mG 
Ans.
 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
mG
15 300 
 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 
Pinion
150 811  23 600
 373 Brinell
341
Core
(HB)11 = 300
Ans.
Case
(HB)12 = 373
A
n
s
.
(HB)21 = 339 (HB)22 = 345
_______________________________________________________________________
15-9
Pinion core
Shigley’s MED, 11th edition
Chapter 15 Solutions, Page 10/20
(sat ) P  44  300   2 100 15 300 psi
( all ) P 
Wt 
Shigley’s MED, 11th edition
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 
Chapter 15 Solutions, Page 11/20
Gear core
(sat )G  44(339)  2100 17 016 psi
17 016(0.893)
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)
( all )G 
Pinion case
(sac ) P  341(373)  23 620 150 813 psi
( c,all ) P 
150 813(1)
134 895 psi
1(1)(1.118)
 134 895 
W 

 2290 
2
t


1.25(3.333)(0.086)
 1(1.374)(1.106)(0.593 75)(2)   689.0 lbf


Gear case
(sac )G  341(345)  23 620 141 265 psi
( c,all )G 
141 265(1.0685)(1)
 135 010 psi
1(1)(1.118)
 135 010 
W 

 2290 
t
2


1.25(3.333)(0.086)
 1(1.1374)(1.106)(0.593 75)(2)   690.1 lbf


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, n = 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,
d G  40 / 10  4.000 in
 d P nP  (2)(1200)

 628.3 ft/min
12
12
K o  1, S F  1, S H  1
vt 
Eq. (15-6):
Eq. (15-5):
B = 0.25(12 – 5)2/3 = 0.9148
A = 50 + 56(1 – 0.9148) = 54.77
 54.77  628.3 
K v  

54.77


Shigley’s MED, 11th edition
0.9148
1.412
Chapter 15 Solutions, Page 12/20
Eq. (15-10):
Eq. (15-11):
Ks = 0.4867 + 0.2132/10 = 0.508
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
( swt ) P 
Eq. (15-4):
Wt 
Eq. (15-3):
15 300(1.040)
15 912 psi
1(1)(1)
(swt ) P FK x J P
Pd K o K vK 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

Gear:
Eq. (15-4):
(sat )G = 15 300 psi
15 300(1.053)
( swt )G 
16 111 psi
1(1)(1)
Wt 
16 111(0.71)(1)(0.201)
 256.0 lbf
10(1)(1.412)(0.508)(1.252)
H2 
256.0(628.3)
 4.9 hp
33 000
Eq. (15-3):
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):
C xc  2
(sac)P = 341(300) + 23 620 = 125 920 psi
125 920(1.419)(1)
( c,all ) P 
178 680 psi
1(1)(1)
Shigley’s MED, 11th edition
Chapter 15 Solutions, Page 13/20
2
 ( ) 
Fd P I
W   c,all P 
 C p  K o K vK mCsCxc
t
Eq. (15-1):
 178 680 


 2290 
 362.4 lbf
H3 
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
Gear:
(sac )G 125 920 psi
( c,all ) 
125 920(1.479)(1)
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 CsC xc
(sac )G 
(CL )G CH S F
N P IK s
Substituting (sat)P from above and the values of the remaining terms from
Ex. 15-1,
Shigley’s MED, 11th edition
Chapter 15 Solutions, Page 14/20
2290 1.52  10 020(1)(1)(0.216)(1)(0.575)(2) 
(sac )G 


1.32(1) 1.5 
25(0.065)(0.529)

 114 331 psi
114 331  23 620
( H B ) 22 
 266 Brinell
341
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
( H B )12 
 266 Brinell
341
Pinion
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 of the text. The decision set can be organized as follows:
Shigley’s MED, 11th edition
Chapter 15 Solutions, Page 15/20
A priori decisions:
• Function: H, Ko, rpm, mG, temp., NL, R
S  nd
• Design factor: nd (SF = nd , H
)
• Tooth system: Involute, Straight Teeth, Crowning, n
• 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, 11th edition
Chapter 15 Solutions, Page 16/20
Pinion Bending
Load-induced
stress (Allowable
stress)
Tabulated
strength
Associated
hardness
st 
W t PK o K vK m K s
 s11
FK x J P
( sat ) P 
Factor of
safety
Note:
st 
W t PK o K vK 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 t K o K vCsC xc 
 c  Cp 
  s12
Fd P I


s S KC
(sac ) P  12 H T R
(CL ) P (CH ) P
s22 = s12
(sac )G 
s22 S H KT CR
(CL )G (CH )G
 (sat ) P  2100

44
Bhn  
 (sat ) P  5980

48
 (sat )G  2100

44
Bhn  
 (sat )G  5980

48
 ( sac ) P  23 620

341
Bhn  
 ( sac ) P  29 560

363.6
 (sac ) P  23 620

341
Bhn  
 (sac ) 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
Chosen
hardness
New tabulated
strength
s11S F KT K R
(K L )P
Gear Bending
n11 
 all
(s ) ( K )
 at1 P L P
s11KT K R

n21 
(sat1)G ( K L )G
s21KT K R
 (s ) (C ) (C ) 
n12   ac1 P L P H P 
s12 KT CR


S F  nd , S H  S F
Shigley’s MED, 11th edition
Chapter 15 Solutions, Page 17/20
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, n = 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 = / 8 = 0.3927 in, C = 1.5 + 7 = 8.5 in
Eq. (15-39):
a = px / = 0.3927 / = 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):
Eq. (13-27):
Eq. (13-28):
Eq. (15-62):
( FW ) max  2 2  7   0.125   2.646 in
VW   (1.5)(1725 /12)  677.4 ft/min
 (7)(1725 / 56)
VG 
 56.45 ft/min
12
L  px NW  0.3927 in
 0.3927 
o
  tan  1 
  4.764

(1.5)


P
8
Pn  t 
 8.028
cos 
cos 4.764

pn 
 0.3913 in
Pn
 (1.5)(1725)
Vs 
12 cos 4.764
 679.8 ft/min
(b)
Eq. (15-38):
f  0.103exp   0.110(679.8)0.450   0.012  0.0250
Eq. (15-54):
cos n  f tan 
cos 20  0.0250 tan 4.764
e

 0.7563
cos n  f cot 
cos 20  0.0250 cot 4.764
Shigley’s MED, 11th edition
Ans.
Chapter 15 Solutions, Page 18/20
WGt 
Eq. (15-58):
33 000nd H o K a
33 000(1)(1)(1.25)

 966 lbf
VGe
56.45(0.7563)
WWt  WGt
Eq. (15-57):
Ans.
cos n sin   f cos 
cos n cos   f sin 
 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  562  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 ,
Eq. (15-61):
Eq. (15-63):
the mesh will survive at least 25 000 h.
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   8 / cos 4.764o  8.028
pn   / 8.028  0.3913 in
G 
966
 39 500 psi
0.3913(0.5)(0.125)
The stress is high. At the rated horsepower,
G 
Shigley’s MED, 11th edition
1
39 500  23 940 psi
1.65
acceptable
Chapter 15 Solutions, Page 19/20
(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,
Eq. (15-50):
1725
 0.13  0.568 ft · lbf/(min · in 2 · oF)
3939
17 530
ts  70 
 88.2o F Ans.
0.568(1700)
 CR 
Eq. (15-51):
________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 15 Solutions, Page 20/20
15-15 to 15-22
Problem statement values of 25 hp, 1125 rev/min, mG = 10, Ka = 1.25, nd = 1.1,
n = 20°, ta = 70°F are not referenced in the table. The first four parameters listed
in the table were selected as design decisions.
15-15 15-16 15-17 15-18 15-19 15-20 15-21 15-22
px
1.75
1.75
1.75
1.75
1.75
1.75
1.75
1.75
dW
3.60
3.60
3.60
3.60
3.60
4.10
3.60
3.60
FG
2.40
1.68
1.43
1.69
2.40
2.25
2.4
2.4
A
2000
2000
2000
2000
2000
2000
2500
2600
FAN
FAN
HW
38.2
38.2
38.2
38.2
38.2
38.0
41.2
41.2
HG
36.2
36.2
36.2
36.2
36.2
36.1
37.7
37.7
Hf
1.87
1.47
1.97
1.97
1.97
1.85
3.59
3.59
NW
3
3
3
3
3
3
3
3
NG
30
30
30
30
30
30
30
30
KW
125
80
50
115
185
Cs
607
854
1000
Cm
0.759
0.759
0.759
Cv
0.236
0.236
0.236
VG
492
492
492
492
492
563
492
492
t
WG
2430
2430
2430
2430
2430
2120
2524
2524
WWt
f
e
(Pt)G
Pn
C-to-C
ts
L

G
dG
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
Shigley’s MED, 11th edition
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
1284
0.0183 0.034
0.951
0.913
1.571
1.795
1.732
1.979
11.6
10.156
171
179.6
6.0
5.25
24.98
24.9
4158
5301
19.099
16.7
1284
0.034
0.913
1.795
1.979
10.156
179.6
5.25
24.9
5301
16.71
Chapter 15 Solutions, Page 21/20
Chapter 16
16-1
Given: r = 300/2 = 150 mm, a = R = 125 mm, b = 40 mm, f = 0.28, F = 2.2 kN, 1 = 0,
2 = 120, and a = 90.
From which, sina = sin90 = 1.
Eq. (1
0.28 pa (0.040)(0.150) 120
0 sin  (0.150  0.125cos  ) d
1
 2.993  10 4  pa N · m
Mf 
Eq. (16-3):
MN 
pa (0.040)(0.150)(0.125) 120 2
4
0 sin  d  9.478  10  pa N · m
1
c = 2(0.125 cos 30) = 0.2165 m
F 
9.478  10 4  pa  2.993  10 4  pa
Eq. (16-4):
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
2200 
Eq. (16-7):
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.
Ans.
(b) RH shoe:
Eq. (16-6):
0.28(734.5)103 (0.040)0.1502 (cos 0o  cos120o )
 277.6 N · m
1
LH shoe:
381.9
TL  277.6
144.4 N · m Ans.
734.5
Ttotal = 277.6 + 144.4 = 422 N · m Ans.
TR 
Shigley’s MED, 11th edition
Ans.
Chapter 16 Solutions, Page 1/27
(c)
RH shoe:
Fx = 2200 sin 30° = 1100 N,
Eqs. (16-8):
1

A   sin 2  
2
 0o
2 / 3 rad
120o
Rx 
Eqs. (16-9):
Ry 
Fy = 2200 cos 30° = 1905 N
 0.375,
 1

B    sin 2 
2 4
0
734.5  103  0.040(0.150)
1
1.264
[0.375  0.28(1.264)]  1100   1007 N
734.5  103  0.04(0.150)
R  [  1007 
LH shoe:
Eqs. (16-10):
2
[1.264  0.28(0.375)]  1905  4128 N
1
 41282 ]1/ 2  4249 N Ans.
Fx = 1100 N,
Rx 
Fy = 1905 N
381.9  103  0.040(0.150)
Ry 
1
[0.375  0.28(1.264)]  1100  570 N
381.9  103  0.040(0.150)
1
R   597 2  7512 
1/ 2
[1.264  0.28(0.375)]  1905  751 N
 959 N
Ans.
______________________________________________________________________________
16-2
Given: r = 300/2 = 150 mm, a = R = 125 mm, b = 40 mm, f = 0.28, F = 2.2 kN, 1 = 15,
2 = 105, and a = 90.
From which, sina = sin90 = 1.
Eq. (16-2):
0.28 pa (0.040)(0.150) 105
Mf 
sin  (0.150  0.125cos  ) d  2.177  10  4  pa

15

1
Shigley’s MED, 11th edition
Chapter 16 Solutions, Page 2/27
MN 
Eq. (16-3):
pa (0.040)(0.150)(0.125) 105 2
sin  d  7.765  10 4  pa

15

1
c = 2(0.125) cos 30° = 0.2165 m
F 
7.765  10 4  pa  2.177  10 4  pa
0.2165
Eq. (16-4):
 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)(0.150 2 )(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: 1 = 0°, 2 = 120°, a = 90°, sin a = 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 1 = 0:

f pabr 2
f pabr 
a 2 
Mf 
sin   r  a cos   d 
r
(1

cos

)

sin  2 
2

sin  a 1
sin  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 1 = 0:

Shigley’s MED, 11th edition
Chapter 16 Solutions, Page 3/27

MN
p bra 2 2
p bra   2 1

sin  d  a
 a
 sin 2 2 


sin  a 1
sin  a  2
4


pa (1.25)5.5(3.5)  120    1

  sin 2(120 ) 

1
 2  180  4

 30.41 pa lbf · in

 180o   2 
o
c  2r cos 
  2(5.5) cos 30  9.526 in
2


30.41 pa  14.31 pa
F  225 
 1.690 pa
9.526
pa  225 / 1.690 133.1 psi
Eq. (16-6):
f pabr 2 (cos 1  cos  2 ) 0.30(133.1)1.25(5.52 )

[1  ( 0.5)]
sin  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: 1 = 10°, 2 = 75°, a = 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:

2
2
2
2
1 2 


A  r  sin  d  a  sin  cos  d  r   cos     a  sin  
 1

1
1
2
 1
75
1

 200   cos   10  150  sin 2    77.5 mm
2
 10
75 /180 rad
2
 1

2
B   sin  d    sin 2 
 0.528
1
2 4
 10 /180 rad
75
2
C   sin  cos  d  0.4514
1
Now converting to Pascals and meters, we have from Eq. (16-2),
Shigley’s MED, 11th edition
Chapter 16 Solutions, Page 4/27
0.24  106  (0.075)(0.200)
f pabr
Mf 
A
(0.0775)  289 N · m
sin  a
sin 75
From Eq. (16-3),
MN 
pabra
106 (0.075)(0.200)(0.150)
B 
(0.528) 1230 N · m
sin  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 1  cos  2 )
sin  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
(1230 / 106 ) pa  (289 / 106 ) pa
5.70 
, solving gives
165
MN 
pa  619  103  Pa
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:
Shigley’s MED, 11th edition
Chapter 16 Solutions, Page 5/27
pabr
 C  f B   Fx
sin  a
106 (0.075)0.200

[0.4514  0.24(0.528)](10  3 )  5.70   0.658 kN
sin 75
p br
Ry  a ( B  f C )  Fy
sin  a
106 (0.075)0.200

[0.528  0.24(0.4514)]  10 3   0  9.88 kN
sin 75
Rx 
Secondary shoes:
Rx 

pabr
(C  f B)  Fx
sin  a
0.619  106  0.075(0.200)
sin 75
  0.143 kN
p br
Ry  a ( B  f C )  Fy
sin  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 that +y for secondary shoe is opposite to
+y for primary shoe.
Combining horizontal and vertical components,
RH   0.658  0.143   0.801 kN
RV  9.88  4.03  5.85 kN
R  ( 0.801) 2  5.852
 5.90 kN Ans.
______________________________________________________________________________
16-5
Given: Face width b = 1.25 in, F = 90 lbf, f = 0.25.
Preliminaries: 1 = 45°  tan1(6/8) = 8.13°, 2 = 98.13°, a = 90°,
a = (62 + 82)1/2 = 10 in
Eq. (16-2):

Mf 
f pabr 2
0.25 pa (1.25)6
sin   r  a cos   d 

sin  a 1
1
98.13
 sin   6  10 cos   d
8.13
 3.728 pa lbf · in
Shigley’s MED, 11th edition
Chapter 16 Solutions, Page 6/27
Eq. (16-3):

MN
p bra 2 2
p (1.25)6(10)
 a
sin  d  a

sin  a 1
1
98.13
 sin
2
 d
8.13
 69.405 pa lbf · in
Eq. (16-4): Using Fc = MN  Mf , we obtain
90(20)  (69.405  3.728) pa

pa  27.4 psi
Ans.
Eq. (16-6):
0.25(27.4)1.25  6 2   cos8.13  cos 98.13 
fpabr 2  cos 1  cos  2 
T 

sin  a
1
 348.7 lbf · in Ans.
______________________________________________________________________________
16-6
For
3ˆ f :
f  f  3ˆ 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 
Similarly, for
Ans.
 3ˆ f :
f  f  3ˆ f  0.25  3(0.025)  0.175
M f  (0.175 / 0.25) 3.728 pa  2.610 pa
90(20) = (69.405  2.610) pa
 pa = 26.95 psi
 0.175   26.95 
T 

 348.7  240.1 lbf · in Ans.
 0.25   27.4 
______________________________________________________________________________
16-7
Preliminaries: 2 = 180°  30°  tan1(3/12) = 136°, 1 = 20°  tan1(3/12) = 6°,
a = 90, sina = 1, a = (32 + 122)1/2 = 12.37 in, r = 10 in, f = 0.30, b = 2 in, pa = 150 psi.
Shigley’s MED, 11th edition
Chapter 16 Solutions, Page 7/27
Eq. (16-2):
Eq. (16-3):
Mf 
0.30(150)(2)(10) 136o
6 sin  (10  12.37 cos  ) d  12 800 lbf · in
sin 90
MN 
150(2)(10)(12.37) 136 2
6 sin  d  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,
Eq. (16-6):
FL 
53 300  12 800
 1446 lbf
28
TL 
0.30(150)(2)(10) 2 (cos 6  cos136 )
15 420 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,
cR = (24  2 tan 14°) cos 14° = 22.8 in
Fact  FL sin14  361 lbf Ans.
FR  FL / cos14 1491 lbf
1491 
Thus,
Then,
TR 
355.3  85.3
pa  pa  77.2 psi
22.8
0.30(77.2)(2)(10) 2 (cos 6  cos136 )
 7940 lbf · in
sin 90
Ttotal = 15 420 + 7940 = 23 400 lbf · in Ans.
______________________________________________________________________________
16-8
Shigley’s MED, 11th edition
Chapter 16 Solutions, Page 8/27
2
M f  2 ( fdN )(acos   r )
0
where dN  pbr d
2
 2 fpbr  (acos   r ) d  0
0
From which
2
2
0
0
a cos  d  r  d
r 2
r (60 )( / 180)
a 

 1.209r
sin  2
sin 60
Ans.
Eq. (16-15):
a 
4r sin 60
 1.170r
2(60)( / 180)  sin[2(60)]
Ans.
a differs with a  by 100(1.170 1.209)/1.209 =  3.23 % Ans.
______________________________________________________________________________
(a) Counter-clockwise rotation, 2 =  / 4 rad, r = 13.5/2 = 6.75 in
Eq. (16-15):
4r sin  2
4(6.75)sin( / 4)
a 

 7.426 in
2 2  sin 2 2
2 / 4  sin(2 / 4)
16-9
e  2a  2(7.426) 14.85 in
Ans.
(b)

M
F
x
R
= tan1(3/14.85) = 11.4°
 0  3F x  6.375P
 0   F x  Rx


F x  2.125P
R x  F x  2.125P
F y  F x tan11.4o  0.428P
 Fy   P  F y  R y
R y  P  0.428P  1.428P
Shigley’s MED, 11th edition
Chapter 16 Solutions, Page 9/27
Left shoe lever.
M
 0  7.78S x  15.28F x
15.28
Sx 
(2.125P)  4.174P
7.78
S y  f S x  0.30(4.174P)  1.252P
 Fy  0  R y  S y  F y
R
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  4.174P  2.125P  2.049P
Shigley’s MED, 11th edition
Chapter 16 Solutions, Page 10/27
(c) The direction of brake pulley rotation affects the sense of Sy, which has no effect on
the brake shoe lever moment and hence, no effect on Sx or the brake torque.
The brake shoe levers carry identical bending moments but the left lever carries a
tension while the right carries compression (column loading). The right lever is
designed and used as a left lever, producing interchangeable levers (identical levers).
But do not infer from these identical loadings.
______________________________________________________________________________
16-10 r = 13.5/2 = 6.75 in,
b = 6 in,
2 = 45° =  / 4 rad.
From Table 16-3 for a rigid, molded non-asbestos lining use a conservative estimate of
pa = 100 psi, f = 0.33.
Equation (16-16) gives the horizontal brake hinge pin reaction which corresponds to Sx in
Prob. 16-9. Thus,
p br
100(6)6.75
N  S x  a  2 2  sin 2 2  
2   / 4   sin  2  45  
2
2
 5206 lbf


which, from Prob. 16-9 is 4.174 P. Therefore,
4.174 P = 5206

P = 1250 lbf = 1.25 kip
Ans.
Applying Eq. (16-18) for two shoes, where from Prob. 16-9, a = 7.426 in
T  2a f N  2(7.426)0.33(5206)
 25 520 lbf · in  25.52 kip · in Ans.
______________________________________________________________________________
16-11 Given: D = 350 mm, b = 100 mm, pa = 620 kPa, f = 0.30,  = 270.
Eq. (16-22):
Shigley’s MED, 11th edition
Chapter 16 Solutions, Page 11/27
P1 
pabD
620(0.100)0.350

 10.85 kN
2
2
Ans.
f   0.30(270 )( / 180 ) 1.414
Eq. (16-19):
P2 = P1 exp( f  ) = 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,
f = 0.28,
pa 
Eq. (16-22):
b = 3.25 in,
 = 270°, P1 = 1800 lbf.
2 P1
2(1800)

 92.3 psi
bD 3.25(12)
Ans.
f   0.28(270o )( / 180 o )  1.319
P2  P1 exp( f  )  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
 MO = 0 = 100 P2  325 F  P2 = 325(300)/100 = 975 N
 100 
  cos  1 
  51.32
 160 
  270  51.32  218.7
f   0.30(218.7)   / 180  1.145
P1  P2 exp( f  )  975exp(1.145)  3064 N
Ans.
T   P1  P2  (D / 2)  (3064  975)(200 / 2)
 209  103  N · mm  209 N · m
Ans.
Ans.
______________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 16 Solutions, Page 12/27
16-14 (a) D = 16 in, b = 3 in
n = 200 rev/min
f = 0.20, pa = 70 psi
Eq. (16-22):
p bD
70(3)(16)
P1  a

 1680 lbf
2
2
f   0.20(3 / 2)  0.942
Eq. (16-14):
P2  P1 exp( f  )  1680 exp( 0.942)  655 lbf
D
16
 (1680  655)
2
2
 8200 lbf · in Ans.
Tn
8200(200)
H 

 26.0 hp
63 025
63 025
3P
3(1680)
P  1 
 504 lbf Ans.
10
10
T  ( P1  P2 )
Ans.
(b) Force of belt on the drum:
R = (16802 + 6552)1/2 = 1803 lbf
Force of shaft on the drum: 1680 and 655 lbf
TP1  1680(8)  13 440 lbf · in
TP2  655(8)  5240 lbf · in
Net torque on drum due to brake band:
Shigley’s MED, 11th edition
Chapter 16 Solutions, Page 13/27
T  TP1  TP2
 13 440  5240
 8200 lbf · in
The radial load on the bearing pair is 1803 lbf. If the bearing is straddle mounted with
the drum at center span, the bearing radial load is 1803/2 = 901 lbf.
Shigley’s MED, 11th edition
Chapter 16 Solutions, Page 14/27
(c) Eq. (16-21):
2P
bD
2 P1
2(1680)


 70 psi
3(16)
3(16)
p 
p
 0
Ans.
2 P2
2(655)

 27.3 psi Ans.
3(16)
3(16)
______________________________________________________________________________
p
 270

16-15 Given:  = 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  )  exp[0.2(3 / 2)]  2.566
If friction is not fully developed,
P1/P2 ≤ exp( f  )
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
Shigley’s MED, 11th edition
Chapter 16 Solutions, Page 15/27
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
P1
c
2.25
 2 
 2.25
P2
c1
1
ln( P1 / P2 ) ln 2.25
f 

 0.172
3 / 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)
p 1 
 89.6 psi Ans.
bD
2.125(8.25)
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.
Shigley’s MED, 11th edition
Chapter 16 Solutions, Page 16/27
(a) From Eq. (16-23),
2  4000 
2F
pa 

 0.194 N/mm 2 194 kPa
Ans.
 d ( D  d )  (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),
pa 
4F
4(4000)

 0.159 N/mm 2  159 kPa
2
2
 ( D  d )  (2502  1752 )
Ans.
Eq. (16-27):
3


f pa ( D 3  d 3 )  (0.30)159  103   2503  1753   10  3 
12
12
 128 N · m Ans.
______________________________________________________________________________
T 
16-17 Given: OD = 6.5 in, ID = 4 in, f = 0.24, pa = 120 psi.
(a) Eq. (16-23):
pd
 (120)(4)
F  a (D  d ) 
(6.5  4)  1885 lbf Ans.
2
2
Eq. (16-24) with N sliding planes:
 fpa d 2
 (0.24)(120)(4)
T 
(D  d 2 ) N 
(6.52  42 )(6)
8
8
 7125 lbf · in Ans.
T 
(b)
 (0.24)(120d )
(6.52  d 2 )(6)
8
d, in
2
3
4
5
6
T, lbf · in
5191
6769
7125
5853
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:
Shigley’s MED, 11th edition
Chapter 16 Solutions, Page 17/27
T 
 f pa d ( D 2  d 2 ) N  f pa N 2

D d  d3

8
8
Differentiating with respect to d and equating to zero gives
dT
 f pa N 2

 D  3d 2   0
dd
8
D
d* 
Ans.
3
d 2T
3 f pa N
 f pa N
 6
d 
d
2
dd
8
4
which is negative for all positive d. We have a stationary point maximum.
d* 
(b)
6.5
 3.75 in
3
Ans.
Eq. (16-24):
T* 


 (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
0.45   0.80
D
(d) Consider:
Multiply through by D,
0.45D  d  0.80 D
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 
  tan  1 
 12.53
 2.25 
Shigley’s MED, 11th edition
Chapter 16 Solutions, Page 18/27
Uniform wear
Eq. (16-45):
 f pa d 2
 D  d2
8sin 
 (0.28) pa (11) 2
1800 
12  112  128.2 pa

8sin12.53
1800
pa 
 14.04 psi Ans.
128.2
T 
Eq. (16-44):
F 
 pa d
 (14.04)11
(D  d ) 
(12  11)  243 lbf
2
2
Ans.
Uniform pressure
Eq. (16-48):
 f pa
 D3  d 3 
12sin 
 (0.28) pa
1800 
123  113  134.1 pa

12sin12.53
1800
pa 
 13.42 psi
Ans.
134.1
T 
Eq. (16-47):
 pa 2
 (13.42) 2
(D  d 2 ) 
12  112   242 lbf Ans.

4
4
______________________________________________________________________________
F 
16-20 Uniform wear
Eq. (16-34):
Eq. (16-33):
Thus,
1
T  ( 2  1) f pari  ro2  ri 2 
2
F = (2  1) pari (ro  ri)
(1 / 2)( 2  1) f pa ri  ro2  ri 2 
T

f FD
f ( 2  1) pa ri (ro  ri )( D)
r  ri
D / 2  d / 2 1
d
 o

  1   O.K .
2D
2D
4
D
Ans.
Uniform pressure
Shigley’s MED, 11th edition
Chapter 16 Solutions, Page 19/27
1
T  ( 2  1) f pa  ro3  ri3 
3
1
F  ( 2  1) pa  ro2  ri 2 
2
Eq. (16-38):
Eq. (16-37):
Thus,
(1 / 3)( 2  1) f pa  ro3  ri 3 
T
2  ( D / 2) 3  (d / 2) 3 




f FD (1 / 2) f ( 2  1) 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
  2 n / 60  2 500 / 60  52.4 rad/s
H
2(103 )
T 

 38.2 N · m
52.4

Key:
T
38.2

 3.18 kN
r
12
Average shear stress in key is
3.18(103 )
 
 13.2 MPa Ans.
6(40)
Average bearing stress is
F
3.18(103 )
b  

  26.5 MPa
Ab
3(40)
Let one jaw carry the entire load.
F 
Ans.
1  26 45 
rav  
   17.75 mm
2 2
2
T
38.2
F 

 2.15 kN
rav 17.75
The bearing and shear stress estimates are
b 
 2.15  103 
  22.6 MPa Ans.
10(22.5  13)
2.15(103 )
 
 0.869 MPa Ans.
10  0.25 (17.75) 2 
______________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 16 Solutions, Page 20/27
Shigley’s MED, 11th edition
Chapter 16 Solutions, Page 21/27
16-22
1  2 n / 60  2 (1600) / 60 167.6 rad/s
2  0
From Eq. (16-51),
I1I 2
Tt1
2800(8)


 133.7 lbf · in · s 2
I1  I 2
1  2 167.6  0
Eq. (16-52):
E 
I1I 2
133.7
2
(167.6  0) 2  1.877  106  lbf in
 1  2  
2  I1  I 2 
2
H = E / 9336 = 1.877(106) / 9336 = 201 Btu
In Btu, Eq. (16-53):
Eq. (16-54):
T 
H
201

 41.9 F
C pW
0.12(40)
Ans.
______________________________________________________________________________
16-23
n1  n2
260  240

 250 rev/min
2
2
Cs = ( 2   1) /  = (n2  n1) / n = (260  240) / 250 = 0.08
n
Eq. (16-62):
Ans.
 = 2 (250) / 60 = 26.18 rad/s
From Eq. (16-64):
6.75  103 
E2  E1
I 

 123.1 N · m · s 2
2
2
Cs
0.08(26.18)
I 
m 2
d o  di2 

8

m 
8I
8(123.1)
 2
 233.9 kg
2
d  di
1.5  1.4 2
Table A-5, cast iron unit weight = 70.6 kN/m3
Volume:
2
o
  = 70.6(103) / 9.81 = 7197 kg / m3.
V = m /  = 233.9 / 7197 = 0.0325 m3
V   t  d o2  d i2  / 4   t  1.52  1.42  / 4  0.2278t
Equating the expressions for volume and solving for t,
0.0325
t 
 0.143 m  143 mm Ans.
0.2278
______________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 16 Solutions, Page 22/27
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
2 n
2 (540)
 

 56.5 rad/s
60
60
Cs = 0.10
E  E
8.70(103 )
I  2 21 
 27.25 N · m · s 2
2
Cs
0.10(56.5)
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.
4
E2  E1  3(3531)  10 590 in · lbf
E  E
10 590
I  2 21 
 0.560 in · lbf · s 2 Ans.
2
Cs
0.30(251 )
______________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 16 Solutions, Page 23/27
16-26 (a)
(1)
(T2 )1   F21rP  
T2
T
rP  2
rG
n
Ans.
Equivalent energy
(2)
(1 / 2) I 222  (1 / 2)( I 2 )1  12 
( I 2 )1 
I
22
I  22
2 2
n
1
2
2
2
I G  rG   mG   rG   rG 
  

 n4
IP
 rP   mP   rP   rP 
(3)
From (2)
(b)
Ans.
( I 2 )1 
Ie  I M  I P  n2I P 
IL
n2
IG
n4I P

 n2I P
2
2
n
n
Ans.
Ans.
______________________________________________________________________________
16-27 (a) Reflect IL, IG2 to the center shaft
Shigley’s MED, 11th edition
Chapter 16 Solutions, Page 24/27
Reflect the center shaft to the motor shaft
I e  I M  I P  n 2I P 
(b) For R = constant = nm,
I P m2
I
 2 I P  2L 2
2
n
n
mn
I e  I M  I P  n2 I P 
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
n5
(c) For R = 10, n
n6  n2  200 = 0
From which
n*  2.430 Ans.
10
m* 
 4.115
2.430
Ans.
Notice that n*and m* are independent of IL.
______________________________________________________________________________
16-28 From Prob. 16-27,
IP
R 2I P
I

 L2
2
4
n
n
R
1 100(1) 100
10  1  n 2 (1)  2 
 2
n
n4
10
1
100
12  n 2  2  4
n
n
Ie  I M  I P  n2I P 
Shigley’s MED, 11th edition
Chapter 16 Solutions, Page 25/27
Optimizing the partitioning 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 includes the two additional
gears.
______________________________________________________________________________
16-29 Figure 16-29 applies,
t2  10 s, t1  0.5 s
t2 - t1 10  0.5

 19
t1
0.5
The load torque, as seen by the motor shaft (Rule 1, Prob. 16-26), is
TL 
1300(12)
 1560 lbf · in
10
The rated motor torque Tr is
Tr 
63 025(3)
 168.07 lbf · in
1125
For Eqs. (16-65):
2
(1125) 117.81 rad/s
60
2
s 
(1200) 125.66 rad/s
60
 Tr
168.07
a 

  21.41 lbf in s/rad
s  r
125.66  117.81
T
168.07(125.66)
b r s 
 2690.4 lbf · in
s  r 125.66  117.81
r 
Shigley’s MED, 11th edition
Chapter 16 Solutions, Page 26/27
The linear portion of the squirrel-cage motor characteristic can now be expressed as
TM = 21.41 + 2690.4 lbf · in
Eq. (16-68):
 1560  168.07 
T2  168.07 

 1560  T2 
19
One root is 168.07 which is for infinite time. The root for 10 s is desired. Use a
successive substitution method
19.30
24.40
26.00
26.50
Continue until convergence to
New T2
19.30
24.40
26.00
26.50
26.67
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
 
a
T  b 26.771  2690.4
max  2

 124.41 rad/s Ans.
 21.41
a
min 117.81 rad/s Ans.
124.41  117.81
 121.11 rad/s
 
2
124.41  117.81
max  min

 0.0545 Ans.
Cs 
(max  min ) / 2 (124.41  117.81) / 2
1
1
E1  I r2  (110.72)(117.81)2  768 352 in · lbf
2
2
1 2 1
E2  I 2  (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  2  0.0545(110.72)(121.11) 2
 88 508 in · lbf, close enough Ans.
During the punch
Shigley’s MED, 11th edition
Chapter 16 Solutions, Page 27/27
63 025H
n
TL (60 / 2 ) 1560(121.11)(60 / 2 )
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  di2
8
8g
8 gI
8(386)(110.72)
W  2

2
d o  di
d o2  di2
I 

If a mean diameter of the flywheel rim of 30 in is acceptable, try a rim thickness of 4 in
di  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 
l 2
l
d o  d i2   (322  282 )  188.5l

4
4
where l is the rim width as shown in Table A-18. The specific weight of cast iron is
 = 0.260 lbf / in3, therefore the volume of cast iron is
V 
W
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
Shigley’s MED, 11th edition
Chapter 16 Solutions, Page 28/27
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
r  117.81 / 10  11.781 rad/s
s 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   2141c  26 903.5 lbf · in
 15 600  1680.5 
T2  1680.6 

 15 600  T2 
19
The root is 10(26.67) = 266.7 lbf · in
 121.11 / 10 12.111 rad/s
Cs  0.0549 (same)
max 121.11 / 10 12.111 rad/s Ans.
min 117.81 / 10 11.781 rad/s Ans.
E1, E2, E and peak power are the same. From Table A-18
6
8 gI
8(386)(11 072) 34.19  10 
W  2

 2
d o  di2
d o2  di2
d o  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
Shigley’s MED, 11th edition
Chapter 16 Solutions, Page 29/27
W 
34.19  106 
 3026 lbf
802  702
W
3026
V 

 11 638 in 3

0.260

V  l (80 2  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, 11th edition
Chapter 16 Solutions, Page 30/27
Chapter 17
17-1
Begin with Eq. (17-10),
F1  Fc  Fi
2 exp( f  )
exp( f  )  1
Introduce Eq. (17-9):
 exp( f  )  1   2 exp( f  ) 
2T  exp( f  )
F1  Fc  d 
 Fc 



d  exp( f  ) 
 exp( f  )  1   exp( f  )  1 
exp( f  )
F1  Fc  F
exp( f  )  1

1 
 exp( f  ) 
Fc 

Now add and subtract  exp( f  )  1
 exp( f  ) 
 exp( f  ) 
 exp( f  ) 
F1  Fc  Fc 
 F 
 Fc 



 exp( f  )  1 
 exp( f  )  1 
 exp( f  )  1 
 exp( f  ) 
 exp( f  ) 
 ( Fc  F ) 
 Fc  Fc 


 exp( f  )  1 
 exp( f  )  1 
 exp( f  ) 
Fc
 ( Fc  F ) 


 exp( f  )  1  exp( f  )  1
( F  F ) exp( f  )  Fc
 c
Q.E.D.
exp( f  )  1
From Ex. 17-2: d = 3.037 rad, F = 664 lbf, exp( f ) = 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  F  Fc 
1
 802  73.4 
f   ln  1
ln 
 0.80
 
 d  F2  Fc  3.037  138  73.4 
F1 
Ans.
______________________________________________________________________________
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
Shigley’s MED, 11th edition
Chapter 17 Solutions, Page 1/39
V =  d n / 12 =  (2)(1750) / 12 = 916.3 ft/min
Eq. (17-1):
D = d / vel ratio = 2 / 0.5 = 4 in
 4 2
D d
 d    2sin  1
   2sin  1 
  3.123 rad
2C
 2(108) 
Table 17-2:
t = 0.05 in, dmin = 1.0 in, Fa = 35 lbf/in,  = 0.035 lbf/in3, f = 0.5
w = 12 bt = 12(0.035)6(0.05) = 0.126 lbf/ft
2
2
w V 
0.126  916.3 
Fc    

  0.913 lbf
g  60 
32.17  60 
(a) Eq. (e), Sec. 17-2:
Ans.
63 025H nom K snd
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
Ans.
F2 = (F1)a  [(F1)a  F2] = 147  90 = 57 lbf
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
Eq. (i), Sec. 17-2:
Using Eq. (17-7) solved for f  (see step 8, just after Table 17-8),
1  ( F )  Fc 
1
 147  0.913 
f   ln  1 a
ln 
 0.307
 
 d  F2  Fc  3.123  57  0.913 
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)
H 

 2.5 hp Ans.
33 000
33 000
Eq. (j), Sec. 17-2:
H
2.5
nf s 

1
H nom K s
2(1.25)
 D    2sin  1
Eq. (17-1):
Shigley’s MED, 11th edition
 4 2
D d
   2sin  1 
  3.160 rad
2C
 2(108) 
Chapter 17 Solutions, Page 2/39
L = [4C2  (D  d)2]1/2 + (DD + dd)/2
Eq. (17-2):
= [4(108)2  (4  2)2]1/2 + [4(3.160) + 2(3.123)]/2 = 225.4 in
(c) Eq. (17-13):
3C 2 w 3(108 / 12) 2 (0.126)
dip 

 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  )  Fc
exp( f  )  1
F2  F1  F
F  F2
Fi  1
 Fc
2
1  F  Fc 
f   ln  1

 d  F2  Fc 
F1 
dip 
3C 2 w
2Fi
which in this case, d = 3.123 rad, exp(f ) = 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
Shigley’s MED, 11th edition
Chapter 17 Solutions, Page 3/39
1
 114.8  0.913 
ln 
  0.50
3.123  24.8  0.913 
2
3  108 / 12  0.126
dip 
 0.222 in
2(68.9)
f 
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 =  d n / 12 =  (4)(1750) / 12 = 1833 ft/min
Eq. (17-1):
D = d / vel ratio = 4 / 0.5 = 8 in
 8 4
D d
 d    2sin  1
   2sin  1 
  3.123 rad
2C
 2(216) 
Table 17-2:
t = 0.11 in, dmin = 2.4 in, Fa = 60 lbf/in,  = 0.037 lbf/in3, f = 0.8
w = 12 bt = 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), Sec. 17-2:
Ans.
63 025H nom K snd
63 025(2)(1.25)(1)

 90.0 lbf · in
n
1750
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
Eq. (i), Sec. 17-2:
Eq. (17-9):
Shigley’s MED, 11th edition
Fi 
 F1  a
 F2
2
 Fc 
Ans.
Ans.
525.6  480.6
 17.0  486.1 lbf
2
Ans.
Chapter 17 Solutions, Page 4/39
1  ( F1) a  Fc 
1
 525.6  17.0 
ln 
ln 
 0.0297
 
 d  F2  Fc  3.123  480.6  17.0 
The friction is thus underdeveloped.
f 
(b) The transmitted horsepower is, with F = (F1)a  F2 = 45 lbf,
(F )V
45(1833)

 2.5 hp
33 000
33 000
H
2.5


1
H nom K s
2(1.25)
H 
nf s
 D    2sin  1
Eq. (17-1):
Eq. (17-2):
Ans.
 8 4 
D d
   2sin  1 
  3.160 rad
2C
 2(216) 
L = [4C2  (D  d)2]1/2 + (DD + dd)/2
= [4(216)2  (8  4)2]1/2 + [8(3.160) + 4(3.123)]/2 = 450.9 in
dip 
3C 2 w 3(216 / 12) 2 (0.586)

 0.586 in
2 Fi
2(486.1)
Ans.
Ans.
(c) Eq. (17-13):
______________________________________________________________________________
17-4
As a design task, the decision set just before Ex. 17-2 is useful.
A priori decisions:
• Function: Hnom = 60 hp, n = 380 rev/min, C = 192 in, Ks = 1.1
• Design factor: nd = 1
• Initial tension: Catenary
• Belt material. Table 17-2: Polyamide A-3, Fa = 100 lbf/in,  = 0.042 lbf/in3, f = 0.8
• Drive geometry: d = D = 48 in
• Belt thickness: t = 0.13 in
Design variable: Belt width.
Use a method of trials. Initially, choose b = 6 in
Shigley’s MED, 11th edition
Chapter 17 Solutions, Page 5/39
 dn  (48)(380)

 4775 ft/min
12
12
w  12bt  12(0.042)(6)(0.13)  0.393 lbf/ft
wV 2
0.393(4775 / 60) 2
Fc 

 77.4 lbf
g
32.17
63 025H nom K snd
63 025(60)(1.1)(1)

10 946 lbf · in
T 
n
380
2T
2(10 946)
F 

 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
V 
Transmitted power H
F (V )
456.1(4775)

 66 hp
33 000
33 000
F  F2
600  143.9
Fi  1
 Fc 
 77.4  294.6 lbf
2
2
1
F  Fc
1  600  77.4 
f   ln 1
 ln 
 0.656
 d F2  Fc   143.9  77.4 
H 
Eq. (17-2):
L = [4(192)2  (48  48)2]1/2 + [48() + 48()] / 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    Fc
exp  f    1
exp  f    exp(0.80 ) 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.
Shigley’s MED, 11th edition
Chapter 17 Solutions, Page 6/39
f 
1
F  Fc
1  573.7  77.4 
ln 1
 ln 
 0.80
 d F2  Fc   117.6  77.4 
3C 2 w 3(192 / 12) 2 (0.393)

 0.562 in
2 Fi
2(268.3)
______________________________________________________________________________
dip 
Shigley’s MED, 11th edition
Chapter 17 Solutions, Page 7/39
17-5
From the last equation given in the problem statement,
exp  f   
1
1   2T / [d (a0  a2 )b]


2T
1 
 exp  f    1

d
(
a
a
)
b
0
2




2T

 exp  f    exp  f    1
 d (a0  a2 )b 
b
1  2T

a0  a2  d
  exp  f  

  exp  f   


1 
But 2T/d = 33 000Hd/V. Thus,
1  33 000 H d
b

a0  a2 
V
  exp  f   
 Q.E.D.

  exp  f    1 
______________________________________________________________________________
17-6
Refer to Ex. 17-1 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  )  1  742.8  11.17  1 

 

  148.1 lbf
d  exp( f  )  1 
6  11.17  1 
Ans.
(b) See Prob. 17-4 statement. The final relation can be written
bmin 

 33 000 H a exp  f   
1


FaC pCv  (12t / 32.174)(V / 60) 2  V [exp  f    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
Shigley’s MED, 11th edition
Chapter 17 Solutions, Page 8/39
round up to an available width.
Eq. (17-1):
D d
 1  18  6 
 d    2sin  1 
    2sin 

 2C 
 2(96) 
 3.016 511 rad
D d
 1  18  6 
 D    2sin  1 
    2sin 

 2C 
 2(96) 
 3.266 674 rad
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  12bt  12(0.042)4.13(0.130)  0.271 lbf/ft
F 
(c)
2
2
w V 
0.271  2749 
  

 17.7 lbf
g  60 
32.17  60 
F  F2
289.1  41.5
Fi  1
 Fc 
 17.7 147.6 lbf
2
2
Fc 
Transmitted belt power H
F (V )
247.6(2749)

 20.6 hp
33 000
33 000
H
20.6
n fs 

 1.1
H nom K s
15(1.25)
H 
dip 
Dip:
2
3C 2w 3(96 / 12)  0.271

 0.176 in
2 Fi
2(147.6)
(d) If you only change the belt width, the parameters in the following table change as
shown.
Shigley’s MED, 11th edition
Chapter 17 Solutions, Page 9/39
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)
n-Fold
b = 6 in b = 12 in Change
25.65
51.3
2
270.35
664.9
2.46
420
840
2
a
F2
Ha
nfs
f
dip
172.4
20.62
1.1
0.139
0.328
592.4
20.62
1.1
0.125
0.114
3.44
1
1
0.90
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
F1
297.6
323.2
1.09
F2
50
75.6
1.51
f
0.80
0.80
1
dip 0.255
0.503
2
______________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 17 Solutions, Page 10/39
17-8
Find the resultant of F1 and F2:
D d
2C
D d
sin  
2C
2
1 D  d 
cos  1  

2  2C 
  sin  1
2

1 D  d  
R x  F1 cos   F2 cos   ( F1  F2 )  1  
 
2  2C  

D d
R y  F1 sin   F2 sin   ( F1  F2 )
Ans.
2C
Ans.
From Ex. 17-2, d = 16 in, D = 36 in, C = 16(12) = 192 in, F1 = 940 lbf, F2 = 276 lbf
 36  16 
o
  sin  1 
  2.9855
2(192)


2

1  36  16  
x
R  (940  276)  1  
   1214.4 lbf
2  2(192)  


 36  16 
R y  (940  276) 
  34.6 lbf
 2(192) 
d
 16 
T  ( F1  F2 )    (940  276)    5312 lbf · in
 2
 2
______________________________________________________________________________
17-9
This is a good class project. Form four groups, each with a belt to design. Once each
group agrees internally, all four should report their designs including the forces and
torques on the line shaft. If you give them the pulley locations, they could design the line
shaft.
______________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 17 Solutions, Page 11/39
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 snd
Ans.
eff
______________________________________________________________________________
Hd 
17-12 Some perspective on the size of Fc can be obtained from
2
w V 
12bt  V 
Fc    
 
g  60 
g  60 
2
An approximate comparison of non-metal and metal belts is presented in the table below.
Non-metal
Metal
0.04 0.280
,
l
b
f
/
i
n
3
b, in
t, in
5.00 1.000
0.20 0.005
The ratio w / wm is
w
12(0.04)(5)(0.2)

 29
wm 12(0.28)(1)(0.005)
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):
Shigley’s MED, 11th edition
Chapter 17 Solutions, Page 12/39
exp( f  )  1
exp( f  )  1
 F1
exp( f  )
exp( f  )
Assuming negligible centrifugal force and setting F1 = ab from step 3, just before Ex. 173,
F  F1  F2  ( F1  Fc )
F exp( f  )
a exp( f  )  1
(F )V
H d  H nom K snd 
33 000
33 000 H nom K s nd
F 
V
bmin 
Also,
(1)
1  33 000 H d  exp( f  )
bmin  

a
V
exp( f  )  1
Ans.
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,  = 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 snd 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.
Shigley’s MED, 11th edition
Chapter 17 Solutions, Page 13/39
 4 2 
 d    2sin  1 
  3.010 rad
 2(15.254) 
 4 2 
 D    2sin  1 
  3.273 rad
 2(15.274) 
For full friction development
exp( f  d ) exp[0.35(3.010)]  2.868
 dn  (2)(1750)
V 

 916.3 ft/min
12
12
Sy 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, just before Ex. 17-3,



Et
28(106 )(0.003) 
3
a Sf 
t

51.212(10
)

(0.003)

(1   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  d )
a exp( f  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
F  F2
74.25  28.87
Fi  1

 51.56 lbf
2
2
Existing friction
Shigley’s MED, 11th edition
Chapter 17 Solutions, Page 14/39
1  F1 
1
 74.25 
ln   
ln 
 0.314
 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 
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.
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 d and D .
______________________________________________________________________________
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,  = 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 snd  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 
 d    2sin  1 
  2.845 rad
 2(20.3) 
 9 3 
 D    2sin  1 
  3.438 rad
 2(20.3) 
For full friction development:
Shigley’s MED, 11th edition
Chapter 17 Solutions, Page 15/39
exp( f  d )  exp[0.32(2.845)]  2.485
 dn  (3)(1125)
V 

 883.6 ft/min
12
12
S f  56.67 kpsi
From selection step 3, just before Ex. 17-3,



Et
17(106 )(0.003) 
3
a Sf 
t   56.67(10 ) 
(0.003) 116.4 lbf/in
(1   2 )d 
(1  0.22 2 )(3) 


2T
2(385.2)
F 

 256.8 lbf
d
3
F  exp( f  d ) 
256.8  2.485 
bmin 

 

  3.69 in
a  exp( f  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
Fi  1

 337.3 lbf
2
2
Existing friction
1  F1 
1
 465.6 
ln   
ln 
  0.282
 d  F2 
2.845  208.8 
(F )V
256.8(883.6)
H 

 6.88 hp
33 000
33 000
H
6.88
n fs 

 1.1
5(1.25) 5(1.25)
f 
Fi can be reduced only to the point at which f   f  0.32. From Eq. (17-9)
Fi 
T  exp( f  d )  1  385.2  2.485  1 

 

  301.3 lbf
d  exp( f  d )  1 
3  2.485  1 
Eq. (17-10):
 2 exp( f  d ) 
 2(2.485) 
F1  Fi 
 429.7 lbf
  301.3 
 2.485  1 
 exp( f  d )  1 
F2  F1  F  429.7  256.8  172.9 lbf
Shigley’s MED, 11th edition
Chapter 17 Solutions, Page 16/39
f   f  0.32
and
______________________________________________________________________________
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.
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.
CD
0.002
4.000
20.300
109.700
3.000
9.000
310.600
0.003
3.500
20.300
131.900
3.000
9.000
333.300
439.000
461.700
182.200
209.000
1.100
60.000
0.309
301.200
1.100
60.000
0.285
301.200
429.600
429.600
172.800
172.800
0.320
0.320
t, in
0.005
4.000
20.300
110.900
3.000
9.000
315.20
0
443.60
0
186.80
0
1.100
60.000
0.304
301.20
0
429.60
0
172.80
0
0.320
0.008
1.500
18.700
194.900
5.000
15.000
215.300
0.010
1.500
20.200
221.800
6.000
18.000
268.500
292.300
332.700
138.200
204.300
1.100
70.000
0.288
195.700
1.100
80.000
0.192
166.600
272.700
230.800
118.700
102.400
0.320
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
Shigley’s MED, 11th edition
Chapter 17 Solutions, Page 17/39
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):




C  0.25   91.8  (12  6.2)  
2

 
 31.47 in
2




2
91.8
(12
6.2)
2(12
6.2)







2

 12  6.2 
 d    2sin -1 
  2.9570 rad
 2(31.47) 
exp( f  d )  exp[0.5123(2.9570)]  4.5489
 dn  (6.2)(3100)
V 

 5031.8 ft/min
12
12
Table 17-13:
Angle   d
180
 180 
 (2.957 rad) 
 169.42
  

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.
Eq. (17-17):
Ha = K1K2Htab = 0.9767(1)(4) = 3.91 hp
The allowable Fa is given by
Fa 
63 025H a
63 025(3.91)

 25.6 lbf
n(d / 2)
3100(6.2 / 2)
The allowable torque Ta is
Shigley’s MED, 11th edition
Chapter 17 Solutions, Page 18/39
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  )  1   79.4   4.5489  1 
Fi    

  20.0 lbf
 
 d   exp( f  )  1   6.2   4.5489  1 
Eq. (17-10):
 2 exp( f  ) 
 2(4.5489) 
F1  Fc  Fi 
 24.4  20 
 57.2 lbf

 4.5489  1 
 exp( f  )  1 
Thus, F2 = F1  Fa = 57.2  25.6 = 31.6 lbf
n fs 
Eq. (17-26):
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
576
 57.2 
 150.1 lbf
d
6.2
K
576
T2  F1  ( Fb ) 2  F1  b  57.2 
 105.2 lbf
D
12
T1  F1  ( Fb )1  F1 
From Table 17-17, K = 1193, b = 10.926, and from Eq. (17-27), the number of belt passes
is:
  K  b  K  b 
NP        
  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,
Shigley’s MED, 11th edition
Chapter 17 Solutions, Page 19/39
109 (91.8)
t 

720V
720(5031.8)
t  25 340 h Ans.
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
NH
NH
2(3.91)
n fs  b a  b a 
 2.0
Hd
H nom K s
3(1.3)
Also,
F1 = 40.8 lbf/belt,
Fi  9.99 lbf/belt,
( Fb )1  92.9 lbf/belt,
T1  133.7 lbf/belt,
N P  2.39(1010 ) passes,
F2 = 28.0 lbf/belt
Fc  24.4 lbf/belt
( Fb ) 2  48 lbf/belt
T2  88.8 lbf/belt
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):
 


C  0.25   86.8  (16  5.4)  
2



 26.05 in Ans.
2




2
86.8
(16
5.4)
2(16
5.4)







2

Eq. (17-1) in degrees:
 16  5.4 
 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:
Shigley’s MED, 11th edition
K2 = 1
Chapter 17 Solutions, Page 20/39
V 
Belt speed:
 (5.4)(1200)
 1696 ft/min
12
Use Table 17-12 to interpolate for Htab.
 2.62  1.59 
H tab 1.59  
 (1696  1000)  2.31 hp/belt
 2000  1000 
two belts:
Eq. (1717) for
H a  K1K 2 Nb 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,
Ha  Hd
4.36  1.25H nom
4.36
H nom 
 3.49 hp Ans.
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):
 


C  0.25   363.3  (26  26)  
2

 
140.8 in (nearly 144 in)
2




2
363.3

(26

26)

2(26

26)



2

 d   ,  D   , exp[0.5123 ]  5.0,
 dn  (26)(400)
V 

 2722.7 ft/min
12
12
Table 17-13: For  = 180°, K1 = 1
Table 17-14: For D360, K2 = 1.10
Table 17-12: Htab = 16.94 hp by interpolation
Shigley’s MED, 11th edition
Chapter 17 Solutions, Page 21/39
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
Ha
18.63
Round up to five belts. It is left to the reader to repeat the above for belts such as C360
and E360.
63 025H a
63 025(18.63)

 225.8 lbf/belt
n(d / 2)
400(26 / 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  )  1 
2935  5  1 
 exp( f  )  1   26  5  1   169.3 lbf/belt




 2 exp( f  ) 
 2(5) 
F1  Fc  Fi 
 25.9  169.3 
 308.1 lbf/belt

 5  1 
 exp( f  )  1 
F2  F1  Fa  308.1  225.8  82.3 lbf/belt
18.63  5
H N
nf s  a b 
1.109 Ans.
Hd
84
From Table 17-16,
T1  T2  F1 
Kb
5 680
 308.1 
 526.6 lbf
d
26
Eq. (17-27):
  K  b  K  b 
NP        
  T1 
 T2  
Thus,
NP > 109 passes
Shigley’s MED, 11th edition
1
 5.28  10 9  passes
Ans.
Chapter 17 Solutions, Page 22/39
t 
Eq. (17-28):
N P Lp
720V

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.
Decision: Use d = 11 in, C270 belts
Table 17-11: Lp = 270 + 2.9 = 272.9 in
Eq. (17-16b):
 


C  0.25   272.9  (60  11)  
2

 
 76.78 in
2




2
 272.9  2 (60  11)   2(60  11) 

This fits in the range
D  C  3( D  d )  60  C  3(60  11)  60 in  C  213 in
60  11
 d    2sin  1
 2.492 rad  142.8
2(76.78)
60  11
 D    2sin  1
 3.791 rad
2(76.78)
exp(f d) = exp[0.5123(2.492)] = 3.5846
For the flat on flywheel, f = 0.13 [see discussion just before Eq. (17-18)], exp(f D) =
exp[0.13(3.791)] = 1.637.
The belt speed is
V 
 dn  (11)(875)

 2520 ft/min
12
12
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
Shigley’s MED, 11th edition
Chapter 17 Solutions, Page 23/39
be entry for d = 11.65 in and n = 3000 ft/min. Then,
x  6.74
7.17  6.74


11.65  11
12  11
x  7.01 hp at 2000 ft/min
8.11  y
8.84  8.11


11.65  11
12  11
y  8.58 hp at 3000 ft/min
and
Interpolating these for 2520 ft/min gives
Eq. (17-17):
8.58  H tab
3000  2520

 H tab  7.83 hp/belt
8.58  7.01 3000  2000
Ha = K1K2Htab = 0.903(1.15)(7.83) = 8.13 hp
Eq. (17-19):
Hd = HnomKsnd = 50(1.2)(1.1) = 66 hp
H
66
Nb  d 
 8.1 belts
H
8.13
a
Eq. (17-20):
Decision: Use 9 belts. On a per belt basis,
63 025H a
63 025(8.13)

106.5 lbf/belt
n(d / 2)
875(11 / 2)
F d 106.5(11)
Ta  a 
 586.8 lbf · in per belt
2
2
Table 17-16: Kc = 1.716
2
2
 V 
 2520 
Fc  1.716 
  1.716 
  10.9 lbf/belt
1000
1000




Eq. (17-21):
Fa 
At fully developed friction, Eq. (17-9) gives
Fi 
T  exp( f  d )  1  586.9  3.5846  1 
 94.6 lbf/belt

 
d  exp( f  d )  1 
11  3.5846  1 
Eq. (17-10):
 2 exp( f  d ) 
 2(3.5846) 
F1  Fc  Fi 
158.8 lbf/belt
 10.9  94.6 
 3.5846  1 
 exp( f  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:
Shigley’s MED, 11th edition
Chapter 17 Solutions, Page 24/39
 Fb  1  Kb / 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 
NP           




 185.5 
  T1 
  304.3 

 T2  
9
9
 1.68  10  passes  10 passes
Ans.
1
Since NP is greater than 109 passes and is out of the range of Table 17-17, life from Eq.
(17-27) is
N L
109 (272.9)
t  P p 
 150  103  h
720V
720(2520)
Remember:
(Fi)drive = 9(94.6) = 851.4 lbf
Table 17-9: C-section belts are 7/8 in wide. Check sheave groove spacing to see if 14 in
width is accommodating.
(b) The fully developed friction torque on the flywheel using the flats of the V-belts, from
Eq. (17-9), is
 exp( f  )  1 
 1.637  1 
Tflat  Fi D 
 94.6(60) 
 1371 lbf · in per belt

 1.637  1 
 exp( f  )  1 
The flywheel torque should be
Tfly = mGTa = 5.147(586.9) = 3021 lbf · in per belt
but it is not. There are applications, however, in which it will work. For example,
make the flywheel controlling. Yes. Ans.
______________________________________________________________________________
17-21
(a)
S is the spliced-in string segment length
De is the equatorial diameter
D is the spliced string diameter
 is the radial clearance
S + De =  D = (De + 2) = De + 2
Shigley’s MED, 11th edition
Chapter 17 Solutions, Page 25/39
From which
 
S
2
The radial clearance is thus independent of De.
 
12(6)
 11.5 in
2
Ans.
This is true whether the sphere is the earth, the moon or a marble. Thinking in terms of a
radial or diametral increment removes the basic size from the problem.
Shigley’s MED, 11th edition
Chapter 17 Solutions, Page 26/39
(b) and (c)
Table 17-9: For an E210 belt, the thickness is 1 in.
210  4.5 210 4.5





4.5
2 

4.5
 0.716 in
 
2
d P  di 
The pitch diameter of the flywheel is
DP  2  D 
DP  D  2  60  2(0.716)  61.43 in
We could make a table:
Diametral
Growth A
1.3
2

B
1.8

Section
C
D
2.9 3.3


E
4.5

The velocity ratio for the D-section belt of Prob. 17-20 is
mG 
D  2
60  3.3 / 

 5.55
d
11
Ans.
for the V-flat drive as compared to ma = 60/11 = 5.455 for the VV drive.
The pitch diameter of the pulley is still d = 11 in, so the new angle of wrap, d, is
D  2  d
Ans.
2C
D  2  d
 D    2sin  1
Ans.
2C
Equations (17-16a) and (17-16b) are modified as follows
 d    2sin  1
Shigley’s MED, 11th edition
Chapter 17 Solutions, Page 27/39

( D    d )2
Lp  2C  ( D  2  d ) 
2
4C



C p  0.25   Lp  ( D  2  d ) 
2


Ans.
2




  L p  ( D  2  d )   2( D  2  d ) 2  Ans.
2



The changes are small, but if you are writing a computer code for a V-flat drive,
remember that d and D 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
K1
0.877
0.869
K2
1.05
1.15
d, in
3.0
4.2
D, in
22
30
d, rad
2.333
2.287
V, ft/min
1350.9
1891
3.304
3.2266
exp(fd )
Lp, in
91.3
101.3
C, in
24.1
31
Htab, uncorr.
0.783
1.662
NbHtab, uncorr.
3.13
3.326
Ta, lbf · in
26.45(105.8)
60.87(121.7)
Fa, lbf
17.6(70.4)
29.0(58)
Ha, hp
0.721(2.88)
1.667(3.33)
nf s
1.192
1.372
F1, lbf
26.28(105.2)
44(88)
F2, lbf
8.67(34.7)
15(30)
(Fb)1, lbf
73.3(293.2)
52.4(109.8)
(Fb)2, lbf
10(40)
7.33(14.7)
Fc, lbf
1.024
2.0
Fi, lbf
16.45(65.8)
27.5(55)
T1, lbf · in
99.2
96.4
T2, lbf · in
36.3
57.4
N, passes
1.61(109)
2.3(109)
t>h
93 869
89 080
Shigley’s MED, 11th edition
Chapter 17 Solutions, Page 28/39
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)
H2 
1000 K r N11.5 p 0.8
n11.5
Eq. (17-33):
Equating and solving for n1 gives
 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, 11th edition
Chapter 17 Solutions, Page 29/39
13  52
 82   49.5
2
2
p
52  13  

2
C   49.5  49.5  8 
   23.95 p
4
2  



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
F  
 744 lbf Ans.
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 (103) hr.
(a) From Table 17-19, pitch is p = 0.500 in. With C = 20 in,
Eq. (17-34):
L
2C N1  N 2  N1  N 2 



p
p
2
4 2C / p
2(20) 21  84
(84  21) 2



135 pitches (or links)
0.5
2
4 2 (20 / 0.5)
L = 135p = 135(0.500) = 67.5 in Ans.
2
(b) Table 17-20: No.40 chain, at 2000 rpm, Htab = 7.72 hp
Table 17-22: Assuming for now the “post-extreme”, that is, failure by fatigue of the
roller,
1.5
1.5
 N1 
 21 
K1       1.37
 17 
 17 
Table 17-23: Four strands,
K2 = 3.3
Eq. (17-37):
Ha = K1K2Htab = 1.37(3.3)7.72 = 34.9 hp
Ans.
Note that this table value does not account for the differences from the assumptions for
chain length and life.
(c) Eq. (17-32):
Shigley’s MED, 11th edition
Chapter 17 Solutions, Page 30/39
H1  0.004 N11.08n10.9 p (3  0.07 p)  0.004(21)1.08 (2000) 0.9 (0.5) 3  0.07(.5) 12.84 hp
To compare with the tabulated value from part (b), adjust with K2 for the number of
strands. It is not necessary to adjust with K1, since the number of teeth is included in Eq.
(17-37).
H a K 2 H1 3.3(12.84) 42.4 hp
This is larger than the table value from part (a), probably indicating that H2 is the limiting
power in the table.
H2 
1000K r N11.5 p 0.8 1000(17)(21)1.5 0.50.8

10.5 hp
n11.5
(2000)1.5
Eq. (17-33):
Adjusting for the number of strands,
Ha = K2H2 = 3.3(10.5) = 34.7 hp
Ans.
This compares pretty well to the adjusted table value from part (b) of 34.9 hp. Note that
this still doesn’t account for the desired chain length or life being different from the table
assumptions.
(d) Eq. (17-39):
0.4
  N 1.5
L p   15000  0.4 
0.8 
1
H 2 1000  K r 
 p 
 
 
  n1 
 100   h  
  21 1.5 0.8  135  0.4  15000  0.4 
1000  17 
 0.5 
 
  10.56 hp
 100   20000  
  2000 
Adjusting for the 4 strands,
H a K 2 H 2 3.3(10.56) 34.8 hp
Ans.
This is only slightly different from the 34.7 hp from part (c). Note that accounting for the
longer chain length increased the power rating slightly, while accounting for the longer
life decreased the power rating slightly.
N1 pn
21(0.5)(2000)

 1750 ft/min
12
12
(e) Eq. (17-30):
(f) From the standard relationship of power, torque, and speed, H=T
V 
T
Ans.
H
34.8 hp  550 ft-lbf/s 
1




  60 s/min  
 91.4 lbf-ft
 2000 rev/min 
hp
 2 rad/rev 

D
(g) Eq. (17-29):
Shigley’s MED, 11th edition
Ans.
p
0.5

3.355 in

sin(180 / N ) sin(180 / 21)
Chapter 17 Solutions, Page 31/39
F1 T / ( D / 2) 91.4 lbf-ft  12 in/ft  /   3.355 / 2  in 
654 lbf
Ans.
______________________________________________________________________________
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:
Strands
1
2
3
4
Shigley’s MED, 11th edition
Chain
H

tab
K2
No.
1.0 41.3 100
1.7 24.3 80
2.5 16.5 80
3.3 12.5 60
Htab
nf s Type
59.4 1.58 B
31.0 1.40 B
31.0 2.07 B
13.3 1.17 B
Chapter 17 Solutions, Page 32/39
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
2C N1  N 2 ( N 2  N1 ) 2



p
p
2
4 2C / p
17  84
(84  17) 2
 2(30 / 0.75) 

2
4 2 (30 / 0.75)
133.3
L = 0.75(133.3) = 100 in (no need to round)
A
N1  N 2 L 17  84 100



  82.83
2
p
2
0.75
Eq. (17-36) with p = 0.75 in:
Eq. (17-35):
2
p
N 2  N1  

2
C    A  A  8
 
4
2  



2

0.75
84  17  
2
    82.83    82.83  8 
  30.0 in

4 
2  



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,
Shigley’s MED, 11th edition
Chapter 17 Solutions, Page 33/39
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
 N1 
 19 
K1        1.18
 17 
 17 
Table 17-23:
K2 = 1, 1.7, 2.5, 3.3, 3.9, 4.6, 6.0
Decision variables for 15 000 h life goal:
K snd H nom 1.2(1.1)(50) 55.9


K1K 2
1.18K 2
K2
KK H
1.18K 2 H tab
 1 2 tab 
 0.0197K 2 H tab
K s H nom
1.2(50)
 
H tab
nf
s
(1)
Form a table for a 15 000 h life goal using these equations.
K2
1
1.7
2.5
3.3
3.9
4.6
6
H'tab
55.90
32.90
22.40
16.90
14.30
12.20
9.32
Chain #
120
120
120
120
80
60
60
Htab
21.6
21.6
21.6
21.6
15.6
12.4
12.4
nf s
0.423
0.923
1.064
1.404
1.106
1.126
1.416
Lub
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:
 ) 2.550 000
( H tab ) 2.515 000  ( H tab
 15 000

  
 )2.5 
H tab
( H tab
 50 000

Substituting from (1),
1/ 2.5

 0.618 H tab
 55.9 
34.5
H tab

  0.618 

K2
 K2 
The H  notation is only necessary because we constructed the first table, which we
normally would not do.
Shigley’s MED, 11th edition
Chapter 17 Solutions, Page 34/39
nf
s

 )
K1K 2 H tab
K K (0.618H tab
 1 2
 0.618[(0.0197) K 2 H tab ]
K s H nom
K s H nom
 0.0122 K 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
4 2C / p
19  38 (38  19) 2
 2(40) 

2
4 2 (40)
 108.7  110 even integer Ans.
Eq. (17-36):
A
Eq. (17-35):
N1  N 2 L 19  38 110



  81.5
2
p
2
1
C
1
 38  19 
   ( 81.5)  ( 81.5) 2  8 

p
4
 2 

2

  40.64


C = p(C/p) = 1.0(40.64/1.0) = 40.64 in (for reference) Ans.
______________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 17 Solutions, Page 35/39
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
  (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  
 wl   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
( p / Su )  0.0014
Su  240 kpsi, p. 920
0.0014(240)(2)(72)
Ff 
 24.2 kip
2
Ans.
(b) Factors of safety
Static, no bending:
n
Shigley’s MED, 11th edition
Fu
333

 28.3
Ft
11.76
Ans.
Chapter 17 Solutions, Page 36/39
Static, with bending:
ns 
Eq. (17-46):
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.
______________________________________________________________________________
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
Shigley’s MED, 11th edition
Chapter 17 Solutions, Page 37/39
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  
 wl   1    
 144d 2   1 

g  m
32.2 
m


5620

 162d 2 lbf, each wire
m
From Fig. 17-21 for 105 cycles, p/Su = 0.004. For plow steel, let Su = 240 kpsi.
Eq. (17-44):
Ff 
( p / Su )Su Dd
0.004(240 000)(30d )

14 400d lbf each wire
2
2
Eq. (17-41) and Table 17-27:
6
2
Er d w Am 12  10  0.067d  0.4d 
Fb 

 10 720d 3 lbf, each wire
D
30
Eq. (17-45):
F f  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* 
Shigley’s MED, 11th edition
14 400
 0.669 in
3(10 720)
Chapter 17 Solutions, Page 38/39
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.
 d 2 
Fu  (Su ) nom Anom  106 000 
 83 252d 2 lbf, each wire

 4 
2
F
83 452(0.5)
n u 
 7.32
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.
______________________________________________________________________________
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 (plow steel); Fig. 17-21: (p/Su)106 =
0.0014
Eq. (17-44):
Ff 
Table 17-24:
Eq. (17-47):
 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

a
Ft  (W  wl )  1  
g

2 

 (8  3.2d 2 )  1 


32.2 
 8.5  3.4d 2 kip
Shigley’s MED, 11th edition
Chapter 17 Solutions, Page 39/39
Note that bending is not included.
F
12.1d
nf  f 
8.5  3.4d 2
Ft
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
Ans.
1.120
1.095
(b) Try m = 4 strands
2 
8

Ft    3.2d 2   1 

32.2 
4

 2.12  3.4d 2 kip
F f  12.1d kip
12.1d
nf 
2.12  3.4d 2
d, in
0.5000
0.5625
0.6520
0.7500
nf
2.037
2.130
2.193
2.250 ← maximum nf
Ans.
0.8750 2.242
1.0000 2.192
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
dd
(b / m  cd 2 ) 2
nf 
From which
d* 
nf * 
Shigley’s MED, 11th edition
b
mc
Ans.
a b / (mc)
a m

(b / m)  c [b / (mc)] 2 bc
Ans.
Chapter 17 Solutions, Page 40/39
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
m
(1)
ad / n1  cd 2
  ad / n1   ad 2  (0)  b   a / n1   2cd 
dm
0  
2
dd
  ad / n1   cd 2 
a
d* 
Ans.
2
cn
1
From which
Substituting this result for d into Eq. (1) gives
4bcn1
Ans.
a2
______________________________________________________________________________
m* 
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(2 2 )  6.4 lbf/ft
wl  6.4(480)  3072 lbf
  wl /  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
l 2

AE 2 E
(1000  7000)(480)(12) 0.333(480 2 )12 2


1.6(12)(106 )
2(12)(106 )
1 
 2.4  0.460  2.860 in
Ans.
______________________________________________________________________________
17-35 to 17-38
Computer programs will vary.
Shigley’s MED, 11th edition
Chapter 17 Solutions, Page 41/39
______________________________________________________________________________
Shigley’s MED, 11th edition
Chapter 17 Solutions, Page 42/39
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.
______________________________________________________________________________
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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.
______________________________________________________________________________
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20–8
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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.
______________________________________________________________________________
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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.
______________________________________________________________________________
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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______________________________________________________________________________
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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.
sh
(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.
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(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.
______________________________________________________________________________
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.
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(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.
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(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.
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(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.
______________________________________________________________________________
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.
______________________________________________________________________________
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.
______________________________________________________________________________
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.
______________________________________________________________________________
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.
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(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.
______________________________________________________________________________
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.
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(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.
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(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.
______________________________________________________________________________
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.
______________________________________________________________________________
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.
______________________________________________________________________________
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.
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(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.
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(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.
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(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.
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(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.
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(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.
0.3, 0.4
Ans.
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(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.
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(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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20–28 (a) A datum feature is the physical surface that will be used to define the theoretical
datum. The datum B tag is associated with a control frame that is pointing to the datum
feature. The datum feature is best described by answer iv The outer surface of the
protruding cylinder. Ans.
(b) 10.1 Ans.
(c) Maximum material condition for a hole is when it is at its smallest allowed diameter
(maximum material), thus 10.0. Ans.
(d) The tolerance block indicates zero tolerance at maximum material condition, plus a
bonus of any amount that the hole is larger than maximum material condition. At a
diameter of 10.1, the hole is 0.1 larger than maximum material condition. Thus the
diameter of the tolerance zone is 0.1. Ans.
(e) The tolerance block indicates a 0.3 diameter tolerance zone, regardless of size. Thus,
the tolerance zone has a diameter of 0.3. Ans.
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(f) The universal profile tolerance of note at the bottom of the drawing applies to all
basic dimensions that aren’t otherwise toleranced. Thus, a profile tolerance of 0.2 is
centered on the basic dimension of 90. The diameter may vary between a minimum of
89.9 and a maximum of 90.1. Ans.
(g) The tolerance zone is located ideally by basic dimensions. The non-ideal center axis
of the hole must lie entirely within the tolerance zone. The center axis of the hole is
defined by its actual mating envelope. This is best describe by answer i, the largest gauge
pin that fits into the hole. Ans.
(h) There is not an explicit cylindricity control given, but the surface of the hole,
including its cylindricity, must stay within the ideal cylinder defined by the hole’s
maximum material condition. This is best described by answer iii. The envelope of a
perfect cylinder with diameter of 19.95. Ans.
(i) The tolerance for the holes is defined with the modifier for maximum material
condition. Therefore, the center axes of the bolt holes (defined by their actual mating
envelope) must be perfectly located if the holes are manufactured at maximum material
condition. If the holes are manufactured larger than maximum material condition, then
the tolerance zone size increases and the holes do not need to be perfectly located. Ans.
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