Solution Manual for Shigleys Mechanical Engineering Design 11th Edition Budynas / All Chapters 1 - 20 / Full Complete 2025!!!

Chapter 1 t




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




1-7 From Fig. 1-2, cost of grinding to  0.0005 in is 270%. Cost of turning to  0.003 in is
t t t t t t t t t t t t t t t t t t t


60%.
t


Relative cost of grinding vs. turning = 270/60 = 4.5 times Ans.
t t t t t t t t t t




1-8 CA = CB,
t t




10 + 0.8 P = 60 + 0.8 P  0.005 P2
t t t t t t t t t t t t




P2 = 50/0.005
t t t  P = 100 parts Ans.
t t t t




1-9 Max. load = 1.10 P
t t t t


Min. area = (0.95)2A
t t t t


Min. strength = 0.85 S
t t t t t


To offset the absolute uncertainties, the design factor, from Eq. (1-1) should be
t t t t t t t t t t t t




1.10
nd  1.43 Ans.
0.850.95
t
t 2




1-10 (a) X1 + X2:
t t t


x1  x2  X1  e1  X2  e2
t
t
t
t
t
t
t
t t
t
t



error  e  x1  x2X1  X2 t t t t t t t t t t t t t t



 e1  e2 t
t
t Ans.
(b) X1  X2: t t 
x1  x2  X1 e1 X2 e2

t t t t t t t t t t t t t




e  x1  x2X1  X2 e1 e2
t t t t t t t t t t t t t t t t t Ans.
(c) X1 X2: t


x1x2  X1  e1X2  e2 t t t t t t t t t t t



e  x1x2  X1 X2  X1e2  X2e1  e1e2
t t
t t
t
t
t
t
t
t
t t
t
t
t


X e  Xe XX  e2 Ans.
e1  
t t t
t t


 
t

2 
t
1 2 2 1 1 t t t t t t
X X
 1 2 
t t

t




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

, (d) X1/X2:
x1 X 1  e1 X  X1 
1 e1  1
t
t t t t t

 
t t
t t

X e
t
x X 1 e X t t t t t t


2 2 2 2  2 2  t t
1
 e  e  1 e X   e  t e t e e t t t t t t t t t

1  2
then
1    1  1  1  2
2 t t 1 1 1t 2 t 1 t t t t t tt t t t t t t t t t t t t t
t t t t t t t
t t t

 X 2  X 2  1 e2 X 2 
t
 X 1 t
X 2  X 1 X2 t
t
t t
t
t
t
t



 t
x1 X1
Thus, e   X1  e Ans.
t

 2
t t t
t
t
e1 t
t


X X 
t

x X X
t


2  1 2 
t
2 2 t t




1-11 (a) x1 = 7 = 2.645 751 311 1
t t t t t t


X1 = 2.64 t(3 correct digits)
t t t



x2 = 8 = 2.828 427 124 7
t t t t t t


X2 = 2.82 t(3 correct digits)
t t t


x1 + x2 = 5.474 178 435 8
t t t t t t t


e1 = x1  X1 = 0.005 751 311 1
t t t t t t t t t


e2 = x2  X2 = 0.008 427 124 7
t t t t t t t t t


e = e1 + e2 = 0.014 178 435 8
t t t t t t t t t


Sum = x1 + x2 = X1 + X2 + e
t t t t t t t t t t t


= 2.64 + 2.82 + 0.014 178 435 8 = 5.474 178 435 8
t t t t t t t t t t t t t Checks
(b) X1 = 2.65, X2 = 2.83 (3 digit significant numbers)
t t t t t t t t t


e1 = x1  X1 =  0.004 248 688 9
t t t t t t t t t t


e2 = x2  X2 =  0.001 572 875 3
t t t t t t t t t t


e = e1 + e2 =  0.005 821 564 2
t t t t t t t t t t


Sum = x1 + x2 = X1 + X2 + e
t t t t t t t t t t t


= 2.65 +2.83  0.001 572 875 3 = 5.474 178 435 8
t t t t t t t t t t t t Checks


S 161000 25 103   t

1-12  
t

t    d  0.799 in t t t Ans.
n d 3
t 2.5
7
Table A-17: t d= t t
t
in Ans.
 
8

25 103
n S   3.29
t
Factor of safety:t t t t t t Ans.
 161000 t




 7 
3
t t t




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

i1

Overall reliability = 88 percent
t t t t Ans.
Chapter 1 Solutions - Rev. B, Page 2/6
t t t t t t t

,Chapter 1 Solutions - Rev. B, Page 3/6
t t t t t t t

, 1-14 a = 1.500  0.001 in
t t t t t


b = 2.000  0.003 in
t t t t t


c = 3.000  0.004 in
t t t t t


d = 6.520  0.010 in
t t t t t



(a) w  d a b c = 6.520  1.5  2  3 = 0.020 in t t t t t t t t t t t t t t t t t t t



tw  tall = 0.001 + 0.003 + 0.004 +0.010 = 0.018
t
t
t
t t t t t t t t



w = 0.020  0.018 in
t Ans.
t t t t




(b) From part (a), wmin = 0.002 in. Thus, must add 0.008 in to d . Therefore,
t t t t t t t t t t t t t t t t




d = 6.520 + 0.008 = 6.528 in
t t t t t t t Ans.



1-15 V = xyz, and x = a   a, y = b   b, z = c   c,
t t t t t t t t t t t t t t t t t t t t t t t




V  abc
t t




V  a  ab bc c
t t t t t t t t



 abc bca  acb  abc  abc bca cab  abc
t t t t t t t t t t t t t t t




The higher order terms in  are negligible. Thus,
t t t t t t t t




V bcaacbabc
t t t t t t




V bca  acb abc a b c a b c
and,      
t t t t t t t t t t t t
Ans. t t t t t t t

V abc a b c a b c

For the numerical values given, V 1.5001.8753.000  8.4375 in3
t t t t t t t t t t




V 0.002 0.003 0.004
    0.00427  V  0.004278.4375 0.036 in3
t t t t
t t t t t t t t t t

V 1.500 1.875 3.000

V = 8.438  0.036 in3
t t t t t Ans.




Chapter 1 Solutions - Rev. B, Page 4/6
t t t t t t t