Solutions for Shigley’s Mechanical Engineering Design, 11th Edition – (Budynas, 2025/2026) | All 20 Chapters Covered

SOLUTION MANUAL
FOR SHIGLEYS
MECHANICAL
ENGINEERING DESIGN
TH
9 EDITION LATEST
EDITION 2025/2026




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

, 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.

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 Ans.


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
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  Ans.
(c) X1 X2 : E2

X1x2   X1  E1  X 2  E2 
E  X1 X2  X1 X 2  X1 e2  X 2 e1  E1 e2
XE X E  X X  E2 Ans.
E1  
1 2 2 1 1 2  
X X
 1 2 




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

, (d) X1 /X2 :
X1 X1  E1 X  1 X1 
E1   1
 
X X E X 1 E X
2 2 2 2  2 2 
1
 E   1 E X   E  E  E E
E
1 2 The 
  1   1
2 1 1 1 2 1 2
1  n   1
 X2  X2  1 X 2   X1  X 2  X1 X2
E2
X1 X1
Thus, E    X1  E Ans.
E1  2

X X X X X

2 2 2  1 2 



1-11 (A) 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



1-12 
S

16 1000 25 10

3
   D  0.799 Ans.
In
N D 2.5
3

Table A-17: D = 87 In Ans.

Factor Of N S 

25 10 3   3.29 Ans.
Safety:
 16 1000

 7
3




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

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

Overall Reliability = 88 Percent Ans.




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