Solutions Manual for Shigley’s Mechanical Engineering Design 9th Edition 2026 Update

Shigleys mechanical engineering design 9th edition solutions manual Page 1 1




Chapter 1

Problems 1-1 through 1-6 are for student research. No standard solutions are provided.

1-7 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.
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1-8 C A = C B ,

10 + 0.8 P = 60 + 0.8 P  0.005 P 2

P 2 = 50/0.005  P = 100 parts Ans.
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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

1.10
nd   1.43 Ans.
0.85  0.95 
2


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1-10 (a) X 1 + X 2 :
x1  x2  X 1  e1  X 2  e2
error  e   x1  x2    X 1  X 2 
 e1  e2 Ans.
(b) X 1  X 2 :
x1  x2  X 1  e1   X 2  e2 
e   x1  x2    X 1  X 2   e1  e2 Ans.
( c) X 1 X 2 :
x1 x2   X 1  e1  X 2  e2 
e  x1 x2  X 1 X 2  X 1e2  X 2 e1  e1e2
 e e 
 X 1e2  X 2 e1  X 1 X 2  1  2  Ans.
 X1 X 2 




Chapter 1 Solutions - Rev. B, Page 1/6


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(d) X 1 /X 2 :
x1 X 1  e1 X 1  1  e1 X 1 
   
x2 X 2  e2 X 2  1  e2 X 2 
1
 e2   1  e1 X 1  
e2 e1   e2  e1 e
1    1    1 
then  1    1  2
 X2   1  e2 X 2  
X2 X1   X2  X1 X 2
x X X  e e 
Thus, e  1  1  1  1  2  Ans.
x2 X 2 X 2  X 1 X 2 
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1-11 (a) x 1 = 7 = 2.645 751 311 1
X 1 = 2.64 (3 correct digits)
x 2 = 8 = 2.828 427 124 7
X 2 = 2.82 (3 correct digits)
x 1 + x 2 = 5.474 178 435 8
e 1 = x 1  X 1 = 0.005 751 311 1
e 2 = x 2  X 2 = 0.008 427 124 7
e = e 1 + e 2 = 0.014 178 435 8
Sum = x 1 + x 2 = X 1 + X 2 + e
= 2.64 + 2.82 + 0.014 178 435 8 = 5.474 178 435 8 Checks
(b) X 1 = 2.65, X 2 = 2.83 (3 digit significant numbers)
e 1 = x 1  X 1 =  0.004 248 688 9
e 2 = x 2  X 2 =  0.001 572 875 3
e = e 1 + e 2 =  0.005 821 564 2
Sum = x 1 + x 2 = X 1 + X 2 + e
= 2.65 +2.83  0.001 572 875 3 = 5.474 178 435 8 Checks
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16 1000  25 10 
3
S
1-12      d  0.799 in Ans.
nd d3 2.5
Table A-17: d = 78 in Ans.

S 25 103 
Factor of safety: n   3.29 Ans.
 16 1000 

3
 7
8

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n
1-13 Eq. (1-5): R =  Ri = 0.98(0.96)0.94 = 0.88
i 1
Overall reliability = 88 percent Ans.
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Chapter 1 Solutions - Rev. B, Page 2/6


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1-14 a = 1.500  0.001 in
b = 2.000  0.003 in
c = 3.000  0.004 in
d = 6.520  0.010 in
(a) 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), w min = 0.002 in. Thus, must add 0.008 in to d . Therefore,

d = 6.520 + 0.008 = 6.528 in Ans.

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1-15 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

V bca  acb  abc a b c a b c
and,        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
    0.00427  V  0.00427  8.4375   0.036 in 3
V 1.500 1.875 3.000

V = 8.438  0.036 in3 Ans.
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Chapter 1 Solutions - Rev. B, Page 3/6


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1-16


w max = 0.05 in, w min = 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 = t all
 0.023 = t a + 0.002 + 0.005  t a = 0.016 in

Thus, a = 1.569  0.016 in Ans.

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1-17 Do  Di  2d  3.734  2  0.139   4.012 in

t Do   tall  0.028  2  0.004   0.036 in

D o = 4.012  0.036 in Ans.
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1-18 From O-Rings, Inc. (oringsusa.com), D i = 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

D o = 14.43  0.29 mm Ans.
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1-19 From O-Rings, Inc. (oringsusa.com), D i = 34.52  0.30 mm, d = 3.53  0.10 mm

Do  Di  2d  34.52  2  3.53  41.58 mm

t Do   tall  0.30  2  0.10   0.50 mm

D o = 41.58  0.50 mm Ans.
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Chapter 1 Solutions - Rev. B, Page 4/6


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