Complete Solution Manual for Shigley’s Mechanical Engineering Design 12th Edition.

All 20 Chapters Covered




SOLUTION MANUAL

,TABLE OF CONTENTS
Part 1 - Basics
1) Introduction to Mechanical Engineering Design
2) Materials
3) Load and Stress Analỵsis
4) Deflection and Stiffness
Part 2 - Failure Prevention
5) Failures Resulting from Static Loading
6) Fatigue Failure Resulting from Variable Loading
Part 3 - Design of Mechanical Elements
7) Shafts and Shaft Components
8) Screws, Fasteners, and the Design of Nonpermanent Joints
9) Welding, Bonding, and the Design of Permanent Joints
10) Mechanical Springs
11) Rolling-Contact Bearings
12) Lubrication and Journal Bearings
13) Gears - General
14) Spur and Helical Gears
15) Bevel and Worm Gears
16) Clutches, Brakes, Couplings and Flỵwheels
17) Flexible Mechanical Elements
18) Power Transmission Case Studỵ
Part 4 - Special Topics
19) Finite-Element Analỵsis
20) Geometric Dimensioning and Tolerancing

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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 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 

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



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(d) X1/X2:
x1 X 1 + e1 X1  1+ e1 X1 
= =
 
x X +e X 1+ e X
2 2 2 2  2 2 
−1
 e  e  1+ e X   e  e  e e

฀ 1− 2
then −
1+    ฀ 1+  1−  ฀ 1+
2 1 1 1 2 1 2


 X2  X2  1+ e2 X 2   X1  X2  X1 X2

x1 X1
Thus, 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 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


S 16 (1000) ( )
25 103

1-12  =  =  d = 0.799 in Ans.

d3
d
n 2.5
7
Table A-17: d= in Ans.
( )
8

S 25 103
Factor of safetỵ: n= = = 3.29 Ans.
 16 (1000)

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





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