Solution Manual - Shigley's Mechanical Engineering Design 11th Edition (Budynas, 2019) All Chapters Graded A+

Shigleys Mechanical Engineering Design 11th Edition Budynas
Solutions Manual

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 = C B,

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   



1 2 2 1
X X
 1 2 




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, (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
2   1
1 
then

1 1

 1 1
 1 2
  1 1
 2

 X 2  X 2  1 e2 2  
X X 1  X 2  X 1 X 2
x1 X1
Thus,  e    X1  e1 e2  Ans.

x X X  


 




2 2 2 
X 1 X 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 321000 25103 
1-12   
 
 
 d  1.006 in Ans.
nd d 3
2.5
Table A-17: d = 1 14 in Ans.

n
S 25103 
Factor of safety:   4.79 Ans.
 321000
 1.25 
3




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



k
1 8480
N
 fx
Eq. (1-6) x  i i   122.9 kcycles
i 1 69

Eq. (1-7)
k


sx 
fx 2
i i  N x2 1104 600  69(122.9)2 1/ 2
 
i 1   30.3 kcycles Ans.
N 1





 69  1 


x x x 115  122.9
(b) Eq. (1-5) z115  ˆ x  115s  30.3  0.2607
x x



Interpolating from Table (A-10)

0.2600 0.3974
0.2607 x  x = 0.3971
0.2700 0.3936

N(0.2607) = 69 (0.3971) = 27.4  27 Ans.

From the data, the number of instances less than 115 kcycles is


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, 2 + 1 + 3 + 5 + 8 + 12 = 31 (the data is not perfectly normal)



1-14
x f fx f x2
174 6 1044 181656
182 9 1638 298116
190 44 8360 1588400
198 67 13266 2626668
206 53 10918 2249108
214 12 2568 549552
222 6 1332 295704
 197 39126 7789204


k
1 39 126
N
 fx
Eq. (1-6) x  i i   198.61 kpsi
i 1 197

k

Eq. (1-7) sx 
f x i i
2
 N x2 12
  7 789 204 197(198.61)   9.68 kpsi Ans.
2
i1
197 1 
N 1  



1-15 L  122.9 kcycles and sL  30.3 kcycles

Eq. (1-5) x  x x10  L x10 122.9
z10   
ˆ sL 30.3

Thus, x10 = 122.9 + 30.3 z10 = L10

From Table A-10, for 10 percent failure, z10 = 1.282. Thus,

L10 = 122.9 + 30.3(1.282) = 84.1 kcycles Ans.




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