Solutions for Shigley’s Mechanical Engineering Design, 11th Edition – (Budynas, 2020) | All 20 Chapters Covered

All 20 Chapters Covered




SOLUTION MANUAL

, www.konkur.in




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 ) − ( X1 − 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 2 e1 + e1e2
 e e 
≈ X 1e2 + X 2 e1 = X 1 X 2  1 + 2  Ans.
 X1 X 2 




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

, www.konkur.in




(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− then   ≈ 1 + 1 −  ≈ 1+ −
 X2  X2  1 + e2 X 2   X 1  X 2  X1 X 2
x1 X 1 X 1  e1 e2 
Thus, −e= ≈  −  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 σ = ⇒ = ⇒ d = 1.006 in Ans.
nd πd3 2.5
Table A-17: d = 1 14 in Ans.
S 25 (103 )
Factor of safety: n= = = 4.79 Ans.
σ 32 (1000 )
π (1.25 )
3


______________________________________________________________________________




Shigley’s MED, 10th edition Chapter 1 Solutions, Page 2/12

, www.konkur.in




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
Σ 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 − µx x − x 115 − 122.9
(b) Eq. (1-5) z115 = = 115 = = −0.2607
σˆ x sx 30.3

Interpolating from Table (A-10)

0.2600 0.3974
0.2607 x ⇒ x = 0.3971
0.2700 0.3936

NΦ(−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 3/12