Perfect Solution Manual Shigleys Mechanical Engineering Design 11th Edition Budynas latest 2026

SOLUTION MANUAL FOR SHIGLEYS
MECHANICAL ENGINEERING DESIGN 11TH
EDITION LATEST EDITION 2026




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

,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 ANSWER:

1-8 C A = CB,

10 + 0.8 P = 60 + 0.8 P  0.005 P 2

P 2 = 50/0.005  P = 100 parts ANSWER:


1-9 Max. load = 1.10 P
Min. area = (0.95)2 A
Min. strength = 0.85 S
To offset the absolute uncertainties, the design factor, from Eq. (1-1) should be

1.10
n d   1.43 ANSWER:
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 ANSWER:
(b) X1  X2 : 
x 1  x 2  X1  e1   X 2  e2 

e   x1  x2    X1  X 2   e1  e2 ANSWER:
(c) X1 X2 :
x1x2   X1  e1  X 2  e2 
e  x1 x2  X1 X 2  X1 e2  X 2 e1  e1 e2
 X e  X e  X X  e1  e2  ANSWER:
1 2 2 1 1 2 
X X
 1 2 




Chapter 1 Solutions - Rev. B, Page 2/6

, (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  then 
  1   1
2 1 1 1 2 1 2
1    1
 X 2  X 2  1 e 2 X 2   X1  X 2  X1 X2
x1 X1
Thus, e    X1  e1 e2  ANSWER:
x X X  X  X 
2 2 2  1 2 



1-11 (a) x 1 = 7 = 2.645 751 311 1
X1 = 2.64 (3 correct digits)
x 2 = 8 = 2.828 427 124 7
X2 = 2.82 (3 correct digits)
x 1 + x 2 = 5.474 178 435 8
e1 = x 1  X1 = 0.005 751 311 1
e2 = x 2  X2 = 0.008 427 124 7
e = e1 + e2 = 0.014 178 435 8
Sum = x 1 + x 2 = 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 = x 1  X1 =  0.004 248 688 9
e2 = x 2  X2 =  0.001 572 875 3
e = e1 + e2 =  0.005 821 564 2
Sum = x 1 + x 2 = 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 ANSWER:
n d 3 2.5
7
Table A-17: d= in ANSWER:
 
8
25 10 3
Factor of safety: nS   3.29 ANSWER:
 16  
1000

 7
3




1-13 Eq. (1-5): R =  Ri = 0.98(0.96)0.94 = 0.88
i1
Overall reliability = 88 percent ANSWER:



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

, 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
tw  tall = 0.001 + 0.003 + 0.004 +0.010 = 0.018
w = 0.020  0.018 in ANSWER:

(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 ANSWER:



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,        ANSWER:
V abc a b c a b c

For the 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 ANSWER:




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