Solution Manual for Shigleys Mechanical Engineering Design 11th Edition Budynas / All Chapters 1 - 20 / Full Complete 2025!!!

Solution Manual for Shigleys Mechanical Engineering Design 11th Edition
Budynas ,Chapters Fully Complete 2025!!



Chapter 1 v




Problems 1-1 through 1-6 are for student research. No standard solutions are provided.
v v v v v v v v v v v v




1-7 From Fig. 1-2, cost of grinding to  0.0005 in is 270%. Cost of turning to  0.003 in is
v v v v v v v v v v v v v v v v v v v


60%.
v


Relative cost of grinding vs. turning = 270/60 = 4.5 timesv Ans. v v v v v v v v v




1-8 CA = CB,
v v




10 + 0.8 P = 60 + 0.8 P  0.005 P 2
v v v v v v v v v v v v




P 2 = 50/0.005
v v v  P = 100 parts Ans.
v v v v




1-9 Max. load = 1.10 P
v v v v


Min. area = (0.95)2A
v v v v


Min. strength = 0.85 S
v v v v v


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




1.10
nd   1.43 Ans.
0.850.95
v
v 2




1-10 (a) X1 +v v


vX2: x1  x2  X1  e1  X 2  e2
v
v
v
v
v
v
v
v v
v
v



error  e  x1  x2  X1  X2  v v v v v v v v v v v v v v



 e1  e2 v
v
v Ans.
(b) X1  X2: v v 
x1  x2  X1  e1  X2  e2 

v v v v v v v v v v v v v




e  x1  x2 X1  X2  e1  e2
v v v v v v v v v v v v v v v v v Ans.
(c) X1 X2: v


x1x2  X1  e1 X2  e2 
v v v v v v v v v v v



e  x1 x2  X1 X2  X1e2  X2e1  e1e2
v v
v v
v
v
v
v
v
v
v v
v
v
v


X e  Xe XX  e2 Ans.
e1  
v v v
v v


 
v

2 
v
1 2 2 1 1 v v v v v v
X X v v


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

, 1 2 v 




Chapter 1 Solutions - Rev. B, Page 2/6
v v v v v v v

, (d) X1/X2:
x1 X 1  e1 X  X1 
1 e1  1
v
v v v v v

 
v v
v v

X  e X 1 e X
v
x v v v v v v


2 2 2 2  2 2  v v
1
 e  e2  1 e X   e  e 
v v e v v v v v v v v v e
1  then
1    1  1  1 
2 v v 1 v1 1 v 2 v v v v v vv v v v v v v v v 1 v v v v 2 v
v v v v v v v
v v v

 X2  X2  1 e2 X 2   X1
v
v
X2 X1 v
v
v v
v
v X2 v



  v v
x1 X1
Thus, e   X1  e Ans.
v

 2
v v v
v
v
e1 v
v


X X 
v

x X X
v


2  1 2 
v
2 2 v v




1-11 (a) x1 = 7 = 2.645 751 311 1
v v v v v v


X1 = 2.64 (3 correct digits)
v v v v



x2 = 8 = 2.828 427 124 7
v v v v v v


X2 = 2.82 (3 correct digits)
v v v v


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


e1 = x1  X1 = 0.005 751 311 1
v v v v v v v v v


e2 = x2  X2 = 0.008 427 124 7
v v v v v v v v v


e = e1 + e2 = 0.014 178 435 8
v v v v v v v v v


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


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


e1 = x1  X1 =  0.004 248 688 9
v v v v v v v v v v


e2 = x2  X2 =  0.001 572 875 3
v v v v v v v v v v


e = e1 + e2 =  0.005 821 564 2
v v v v v v v v v v


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


= 2.65 +2.83  0.001 572 875 3 = 5.474 178 435 8
v v v v v v v v v v v v Checks


S 16 1000 25 103   v

1-12  
v

v    d  0.799 v v Ans.
in v


d 3
d
n 2.5 v

7
Table A-17: v d = 8 in Ans.
v v
v




Factor of safety:v v n S 
25 103
v  3.29 v v
  v
v Ans.
 16 1000 v




  78 
3
v v v




n

1-13 v v Eq. (1-5):
v R = Ri = 0.98(0.96)0.94 = 0.88
v v
v v
v v v

i1

Overall reliability = 88 percent
v v v v Ans.
Chapter 1 Solutions - Rev. B, Page 3/6
v v v v v v v

, Chapter 1 Solutions - Rev. B, Page 4/6
v v v v v v v