Solutions Manual for Shigley's Mechanical Engineering Design 11th Edition By Richard Budynas

Solutions Manual for Shigley's Mechanical Engineering Design, 11th Edition By Richard Budynas

Chapter 1-17, 20. No Solutions Files for Chapter 18 and 19 Provided by Author.
All Solutions Files Download link at the end of this PDF file.
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 = X 1 + e1 + X 2 + e2
error = e = ( x1 + x2 ) − ( X 1 + X 2 )
= e1 + e2 Ans.
(b) X1 − X2:
x1 − x2 = X 1 + e1 − ( X 2 + e2 )
e = ( x1 − x2 ) − ( X 1 − X 2 ) = e1 − e2 Ans.
(c) X1 X2:
x1 x2 = ( X 1 + e1 )( X 2 + e2 )
e = x1 x2 − X 1 X 2 = X 1e2 + X 2e1 + e1e2
 e e 
 X 1e2 + X 2 e1 = X 1 X 2  1 + 2  Ans.
 X1 X 2 




Shigley’s MED, 11th edition Chapter 1 Solutions, Page 1/12

, (d) X1/X2:
x1 X 1 + e1 X 1  1 + e1 X 1 
= =  
x2 X 2 + e2 X 2  1 + e2 X 2 
−1
 e2  e2  1 + e1 X 1   e1  e2  e1 e2
1 +   1−    1 +
then 1 −   1+ −
 X2  X2  1 + e2 X 2   X 1  X 2  X1 X 2
x X X e e 
Thus, e = 1 − 1  1  1 − 2  Ans.
x2 X 2 X 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
______________________________________________________________________________

32 (1000 ) 25 (10 )
3
S
1-12  =  =  d = 1.006 in Ans.
nd d3 2.5
Table A-17: d = 1 14 in Ans.
S 25 (103 )
Factor of safety: n= = = 4.79 Ans.
 32 (1000 )
 (1.25 )
3


______________________________________________________________________________




Shigley’s MED, 11th edition Chapter 1 Solutions, Page 2/12

,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
Eq. (1-6) x =
N
fx
i =1
i i =
69
= 122.9 kcycles


Eq. (1-7)
k

fx 2
i i − N x2
1 104 600 − 69(122.9) 2 
1/ 2

sx = i =1
=  = 30.3 kcycles Ans.
N −1  69 − 1 

x − x x − x 115 − 122.9
(b) Eq. (1-5) z115 = = 115 = = −0.2607
ˆ x sx 30.3

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, 11th edition Chapter 1 Solutions, Page 3/12

, 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
Eq. (1-6) x =
N
fx
i =1
i i =
197
= 198.61 kpsi


k

fx 2
− N x2
 7 789 204 − 197(198.61) 2 
i i 12

Eq. (1-7) sx = =
i =1
 = 9.68 kpsi Ans.
N −1  197 − 1 
______________________________________________________________________________

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

x −  x x10 − L x10 − 122.9
Eq. (1-5) 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.
___________________________________________________________________________




Shigley’s MED, 11th edition Chapter 1 Solutions, Page 4/12