Solutions for Shigley's Mechanical Engineering Design, 11th Edition by Budynas & Nisbett

All 20 Chapters Covered
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SOLUTION MANUAL
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Chapter 1 gf




Problems 1-1 through 1-6 are for student research. No standard solutions are provided.
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1-7 From Fig. 1- gf gf



2, cost of grinding to  0.0005 in is 270%. Cost of turning to  0.003 in is 60%.
gf gf gf gf gf gf gf gf gf gf gf gf gf gf gf gf gf gf



Relative cost of grinding vs. turning = 270/60 = 4.5 times Ans. gf gf gf gf gf g f gf gf g f gf




1-8 CA = CB, gf g f




10 + 0.8 P = 60 + 0.8 P − 0.005 P 2
gf gf gf gf gf gf gf gf gf gf gf gf




P 2 = 50/0.005
gf gf gf  P = 100 parts Ans.
gf gf gf g f




1-9 Max. load = 1.10 P Migf gf gf gf gf



n. area = (0.95)2A Min
gf gf gf gf



. strength = 0.85 S
gf gf gf gf



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




1.10
= =1.43
gf

n g f gf gf
Ans.
0.85(0.95)
2
d




1-10 (a) X1 + X2:g f gf gf



x1 + x2 = X1 + e1 + X2 + e2
gf
gf
g f
gf
gf
gf
gf
gf gf
g f
gf




error = e = (x1 + x2 )−(X1 + X2) gf gf gf gf gf gf gf gf gf gf gf gf gf gf




= e1 + e2 gf gf gf Ans.
(b) X1 − X2: gf gf




x1 − x2 = X1 + e1 −(X2 + e2 )
gf gf gf gf gf gf gf gf gf gf gf gf gf




e = (x1 − x2)−(X1 − X2)= e1 −e2
gf gf gf gf gf gf gf gf gf gf gf gf gf gf gf gf gf Ans.
(c) X1 X2: gf




x1x2 = (X1 + e1)(X2 + e2)gf gf gf gf gf gf gf gf gf gf gf




e = x1x2 − X1 X2 = X1e2 + X2e1 + e1e2
gf gf
g f
gf
gf
gf
g f
gf
g f
gf gf
gf
gf



 e e 
 X e +X = X
g f g f g f




e X
gf

 
2
1
+ Ans.
gf gf g f gf g f g f g f
g f gf
gf
g f gf g f g f
gf


1 2 2 1 1 2
 X1 X2 
g f g f g f g f

gf
g f




Shigley’s MED, 10th edition
gf gf gf Chapter 1 Solutions, Page 1/12
gf gf gf gf

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(d) X1/X2:
X +e x X  1+ e X  g f gf g f gf gf g f g f g f g f



= 1
g f 1
=
1gfg f 1

1
gf
g f g f gfg f
gf
g f g f gfgf 1g f gf

x2 X2 + e2 X2 1+ e2 X2 
gf

gf gf gf
gf gf g f
g f g f gf
−1
 e  e2  1+ e X g f g f gf   e  e  e e
 −
gf g f g f g f g f g f g f



 1+ 2
 1 then
gf

g f g f 1 1 g f gf gf gf gfg f g
gf
g f gf

  1+
gf
1

gf1− 2
 +
 1 g f g f
gf
g f gf g f g f
gf
g f gf

gf
gf gf
g f g f 1g f gf
− g f g f
gf
2g f


f




 X2  X2 1+ e2 X2 gf
g f
gf
gf
gf
gf
g f   X1  X2 X1 g f
gf
g f
X2 gf




x X X  e e  g f  
Thus, e = 1 − 1
 1

1 g f gfg f g f g f gfg f g f g f gfgf g f g f g
Ans.
− 2 
gf gf gf gf
gf
f gf g f g f g f gf
gf


x2 X2 X2  X X2  gf
gf
g f gf
gf
g f



1




1-11 (a) x1 = 7 = 2.645 751 311 1
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X1 = 2.64 (3 correct digits)
gf gf gf gf




x2 = 8 = 2.828 427 124 7
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X2 = 2.82 (3 correct digits)
gf gf gf gf



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



e1 = x1 − X1 = 0.005 751 311 1
gf gf gf g f gf gf gf gf gf



e2 = x2 − X2 = 0.008 427 124 7
gf gf gf g f gf gf gf gf gf



e = e1 + e2 = 0.014 178 435 8 Su
gf gf gf gf gf gf gf gf gf gf



m = x1 + x2 = X1 + X2 + e
gf gf gf gf gf gf gf gf gf gf



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



e1 = x1 − X1 = − 0.004 248 688 9
gf gf gf g f gf gf gf gf gf gf



e2 = x2 − X2 = − 0.001 572 875 3
gf gf gf g f gf gf gf gf gf gf



e = e1 + e2 = − 0.005 821 564 2
gf gf gf gf gf gf gf gf gf gf gf



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



= 2.65 +2.83 − 0.001 572 875 3 = 5.474 178 435 8
gf gf gf gf gf gf gf gf gf gf gf gf Checks


S 32(1000) gf 25 103 ( ) gf


1-12  =gf  =  d =1.006 i gf gf g f Ans.
n
nd d3 gf 2.5
1
Table A-17:
gf d = 1 4 in
gf gf gf g f Ans.

Factor of safety:
gf gf
S
n= =
25 103
gf = 4.79
( ) gf

gf Ans.
 32(1000) gf




 (1.25)
3
gf




Shigley’s MED, 10th edition
gf gf gf Chapter 1 Solutions, Page 1/12
gf gf gf gf

, www.konkur.in




1-13 (a)
x f fx gf f x2 gf




60 2 120 7200
70 1 70 4900
80 3 240 19200
90 5 450 40500
100 8 800 80000
110 12 1320 145200
120 6 720 86400
130 10 1300 169000
140 8 1120 156800
150 5 750 112500
160 2 320 51200
170 3 510 86700
180 2 360 64800
190 1 190 36100
200 0 0 0
210 1 210 44100
 69 8480 1 104 600
gf gf




k 480
fi x i = = 122
g f


x = 1
N
gf g f g f gf

Eq. (1-6) gf g f
g f

8 g f .9 kcycles g f

g f i=1 69

Eq. (1-7) gf




fx − Nx
1/gf2
sx = = 1104 600 −69(122.9)2 
= 30.3 kcycles
gf g f g f g f gf

g f g f

gf gf gf gf
Ans.
i =1
69 − 1
g f gf
g f gf



N −1
gf gf

g f gf
 

x − x = x115 − = 115 − 122.9
= −0.2607
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(b) Eq. (1-5) z115 gf gf


ˆ
g f
g f g f
gf
g f

x 30.3 g f




=
f

s
x x



Interpolating from Table (A-10) gf gf gf




0.2600 0.3974
0.2607 x 
x = 0.397
g f g f



1
0.2700 0.3936



Shigley’s MED, 10th edition
gf gf gf Chapter 1 Solutions, Page 1/12
gf gf gf gf