Solution Manual for Shigley’s Mechanical Engineering Design, 11th Edition by Richard G. Budynas & J. Keith Nisbett (McGraw-Hill) | Chapters 1-20

Solution Manual Shigleys Mechanical Engineering Design 11th Edition
Budynas, Chapters 1-20 ||Complete A+ Guide

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 t Ans.

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

1.10
nd   1.43 Ans.
0.850.95
2




1-10 (a) X1 + X2:
x1  x2  X1  e1  X 2  e2
error  e   x1  x2    X1  X 2 
 e1  e2 Ans.
(b) X1  X2: 
x1  x2  X1  e1   X 2  e2 

e   x1  x2    X1  X 2   e1  e2 Ans.
(c) X1 X2:
x1x2   X1  e1  X 2  e2 
e  x1 x2  X1 X 2  X1e2  X 2e1  e1e2
X e  X e  X X  e1  e2  Ans.
1 2 2 1 1 2 
X X
 1 2 




Chapṭer 1 Soluṭions - Rev. B, Page 1/6

, (d) X1/X2:
x1 X1  e1 X1  1 e1 X1  
x  X  e  X 1 e X
2 2 2 2  2 2 
1
 e  e2  1 e X   e  e  e e
1  ṭhen 
1    1  1  1
2 1 1 1 2 1 2

 X 2  X 2  1 e2 X 2   X1  X 2  X1 X2
x1 X1
Ṭhus, e   X1  e1 e2  Ans.
x X X X X 
2 2 2  1 2 


1-11 (a) x1 = 7 = 2.645 751 311 1
X1 = 2.64 (3 correcṭ digiṭs)
x2 = 8 = 2.828 427 124 7
X2 = 2.82 (3 correcṭ digiṭs)
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 digiṭ significanṭ 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 16 1000 25 10
3
 
1-12      d  0.799 in Ans.
nd d 3
2.5
Ṭable A-17: d = 8 in Ans.
7



Facṭor of safeṭy: n S 
 
25 103
 3.29 Ans.
 16 1000
  87 
3




n

1-13 Eq. (1-5): R =  Ri = 0.98(0.96)0.94 = 0.88
i1

Overall reliabiliṭy = 88 percenṭ Ans.



Chapṭer 1 Soluṭions - Rev. B, Page 2/6

,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
ṭw  ṭall = 0.001 + 0.003 + 0.004 +0.010 = 0.018
w = 0.020  0.018 in Ans.

(b) From parṭ (a), wmin = 0.002 in. Ṭhus, musṭ add 0.008 in ṭo d . Ṭherefore,

d = 6.520 + 0.008 = 6.528 in Ans.



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

Ṭhe higher order ṭerms in  are negligible. Ṭhus,

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 ṭhe numerical values given, V  1.500 1.8753.000  8.4375 in3


V 0.002 0.003 0.004
    0.00427  V  0.00427 8.4375  0.036 in3
V 1.500 1.875 3.000

V = 8.438  0.036 in3 Ans.




Chapṭer 1 Soluṭions - Rev. B, Page 3/6

, 1-16


wmax = 0.05 in, wmin = 0.004 in
0.05  0.004
w=  0.027 in
2
Ṭhus,  w = 0.05  0.027 = 0.023 in, and ṭhen, w = 0.027  0.023 in.
w= a  b  c
0.027  a  0.042 1.5
a  1.569 in

ṭw = ṭ all
 0.023 = ṭa + 0.002 + 0.005  ṭa = 0.016 in

Ṭhus, a = 1.569  0.016 in Ans.




1-17 Do  Di  2d  3.734  2 0.139  4.012 in

ṭ  0.028  2 0.004  0.036 in
all
ṭD 
o




Do = 4.012  0.036 in Ans.


1-18 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

ṭDo   ṭall  0.13  2 0.08  0.29 mm

Do = 14.43  0.29 mm Ans.


1-19 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

ṭDo   ṭall  0.30  2 0.10  0.50 mm

Do = 41.58  0.50 mm Ans.


Chapṭer 1 Soluṭions - Rev. B, Page 4/6