Test Bank: Conv Solutions for Shigley's Mechanical Engineering Design (11th Edition) - Graded A+ Comprehensive E...

STUDY NOTES & REFERENCE GUIDE




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SOLUTION MANUAL




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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
n = = 1.43 Ans.
0.85(0.95)
2
d




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 − X2 ) = e1 − e2 Ans.
(c) X1 X2:
x1x2 = ( X1 + e1 )( X 2 + e2 )
e = x1x2 − X1 X 2 = X1e2 + X 2e1 + e1e2
e e 
Xe +X e =X X 1
+ 2 Ans.
1 2 2 1 1 2 
 X1 X2 




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




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(d) X1/X2:
x X +e X  1+ e X 
= =
···




1 1 1 1 1 1
 
x2 X 2 + e2 X 2  1+ e2 X2 
−1
 e  e  1+ e X   e  e  e e
  1−
2
1+   1+  1−   1+ −
···




2 then 
1 1 1 2 1 2

X2   1+ e2 X 2   X1  X2 
··




 X2 X1 X2
x X X e e 
Thus, e= 1
− 1
 1

1
− 2
 Ans.
x2 X2 X 2  X1 X2 


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 1 in4 Ans.
Factor of safety: n=
S
=
25 103( ) = 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 8 480
Eq. (1-6) x = 1  fi xi = = 122
.9 kcycles
N i=1 69

Eq. (1-7)

fx − Nx
1/ 2
= 1104 600 − 69(122.9) 
2
sx = = 30.3 kcycles Ans.
i =1 69 − 1
N −1  

x − x = x115 − x =
115 − 122.9
(b) Eq. (1-5) z115 = ˆ s 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


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




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