Complete Solution Manual Shigleys Mechanical Engineering Design 11th Edition by Keith Nisbett Richard Budynas 2026 perfect solution

Complete Solution Manual Shigleys Mechanical
Engineering Design 11th Edition By Keith Nisbett
Richard Budynas 2026 Perfect Solution




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+  1 −   1+ −
2 1 1 1 2 1 2
 n 
 X2  X2  1+ X2   X1  X2  X1 X2
E2
X1 X1
Thus, E = − X1  E2 Ans:
 E1 −
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 10 3
1-12 =  =  D = 0.799 Ans:
In
D
d
N 2.5
3

Table A-17: D = 87 In Ans:

Factor N=
S
=
( )
25 10 3
= 3.29
Safety:Of Ans:
 16 (1000 )

 ( 78 )
3




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

, n

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