Complete Solution Manual Shigleys Mechanical Engineering Design 11th Edition by Keith Nisbett Richard Budynas 2024 Release|| Complete Guide A+

Solution Manual

Shigleys Mechanical Engineering Design
By Keith Nisbett Richard Budynas


11th edition




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

,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 = 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 2 1 1 2 
x x
 1 2 




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

, (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
1 − then −
1+      
2 1 1 1 2 1 2
1+ 1− 1+
 x2  x2  1+ e2 x 2   x1  x2  x1 x 2
x1 x1
Thus, e = − x1  e1 e2  ans.
x x x x −x 
2 2 2  1 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 16 (1000) ( )
25 103
1-12  =  =  d = 0.799 in ans.
Nd d 3 2.5
7
table a-17: d= in ans.
( )
8

S 25 103
Factor of safety: n= = = 3.29 ans.
 16 (1000)
 (7)
3

8


n

1-13 eq. (1-5): r =  ri = 0.98(0.96)0.94 = 0.88
I=1

Overall reliability = 88 percent ans.



Chapter 1 Solutions - Rev. B, Page 3/6

, 1-14 a = 1.500  0.001 in
B = 2.000  0.003 in
C = 3.000  0.004 in
D = 6.520  0.010 in
(a) W = d − a − b − c = 6.520 − 1.5 − 2 − 3 = 0.020 in
tw = tall = 0.001 + 0.003 + 0.004 +0.010 = 0.018
W = 0.020  0.018 in ans.

(b) from part (a), wmin = 0.002 in. Thus, must add 0.008 in to d . Therefore,

D = 6.520 + 0.008 = 6.528 in ans.



1-15 v = xyz, and x = a   a, y = b   b, z = c   c,

V = abc

V = (a  a)(b  b)(c  c)
= abc  bca  acb  abc  abc  bca  cab  abc

The higher order terms in  are negligible. Thus,

v bca + acb + abc

v bca + acb + abc a b c a b c
And, = + + = + + ans.
v abc a b c a b c

for the numerical values given, v = 1.500 (1.875)3.000 = 8.4375 in3


v 0.002 0.003 0.004
= + + = 0.00427  v = 0.00427 (8.4375) = 0.036 in3
V 1.500 1.875 3.000

V = 8.438  0.036 in3 ans.




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