Solution Manual For Shigley's Mechanical Engineering Design, 10th Edition

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Solution Manual For Shigley's Mechanical Engineering Design, 10th Edition offers a comprehensive guide to solving every question in your textbook, helping you master the material.

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Shigley’s MED, 10th edition Chapter 1 Solutions, Page 1/12
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( )
2
1.10 1.43 .
0.85 0.95
dn Ans= =
______________________________________________________________________________
1-10 (a) X1 + X2:( ) ( )
1 2 1 1 2 2
1 2 1 2
1 2
error
.
x x X e X e
e x x X X
e e Ans
+ = + + +
= = + − +
= +
(b) X1 − X2:( )
( ) ( )
1 2 1 1 2 2
1 2 1 2 1 2 .
x x X e X e
e x x X X e e Ans
− = + − +
= − − − = −
(c) X1 X2:( )( )
1 2 1 1 2 2
1 2 1 2 1 2 2 1 1 2
1 2
1 2 2 1 1 2
1 2
.
x x X e X e
e x x X X X e X e e e
e e
X e X e X X Ans
X X
= + +
= − = + +
 
 + = + 
 

Shigley’s MED, 10th edition Chapter 1 Solutions, Page 2/12
(d) X1/X2:1 1 1 1 1 1
2 2 2 2 2 2
1
2 2 1 1 1 2 1 2
2 2 2 2 1 2 1 2
1 1 1 1 2
2 2 2 1 2
1
1
1
1 1 then 1 1 1
1
Thus, .
x X e X e X
x X e X e X
e e e X e e e e
X X e X X X X X
x X X e e
e Ans
x X X X X
−
 + +
= =  
+ + 
      +
+  −  + −  + −      
+      
 
= −  − 
 
______________________________________________________________________________
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
______________________________________________________________________________
1-12( )
( )
3
3
25 1032 1000 1.006 in .
2.5d
S d Ans
n d


=  =  =
Table A-17: d =1
4
1 in Ans.
Factor of safety:( )
( )
( )
3
3
25 10 4.79 .
32 1000
1.25
S
n Ans
 
= = =
______________________________________________________________________________

Shigley’s MED, 10th edition Chapter 1 Solutions, Page 3/12
1-13 (a)
Eq. (1-6)1
1 8480 122.9 kcycles
69
k
i i
i
x f x
N =
= = =
Eq. (1-7)2 2 1/ 22
1 1 104 600 69(122.9) 30.3 kcycles .
1 69 1
k
i i
i
x
f x N x
s Ans
N
=
−  −
= = = − − 

(b) Eq. (1-5)115
115
115 122.9 0.2607
ˆ 30.3
x
x x
x x x
z s


− − −
= = = = −
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
x f f x 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

Shigley’s MED, 10th edition Chapter 1 Solutions, Page 4/12
2 + 1 + 3 + 5 + 8 + 12 = 31 (the data is not perfectly normal)
____________________________________________________________________________
1-14
x f f x 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
Eq. (1-6)1
1 39 126 198.61 kpsi
197
k
i i
i
x f x
N =
= = =
Eq. (1-7)2 2 1 22
1 7 789 204 197(198.61) 9.68 kpsi .
1 197 1
k
i i
i
x
f x N x
s Ans
N
=
−  −
= = = − − 

______________________________________________________________________________
1-15122.9 kcycles and 30.3 kcyclesLL s= =
Eq. (1-5)10 10
10
122.9
ˆ 30.3
x
L
x x L x
z s


− − −
= = =
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, 10th edition Chapter 1 Solutions, Page 5/12
1-16
x f fx fx2
93 19 1767 164331
95 25 2375 225625
97 38 3686 357542
99 17 1683 166617
101 12 1212 122412
103 10 1030 106090
105 5 525 55125
107 4 428 45796
109 4 436 47524
111 2 222 24642
 136 13364 1315704
Eq. (1-6)1
1 13 364 / 136 98.26471 = 98.26 kpsi
k
i i
i
x f x
N =
= = =
Eq. (1-7)2 2 1 22
1 1 315 704 136(98.26471) 4.30 kpsi
1 136 1
k
i i
i
x
f x N x
s N
=
−  −
= = = 
− − 

Note, for accuracy in the calculation given above,x needs to be of more significant
figures than the rounded value.
For a normal distribution, from Eq. (1-5), and a yield strength exceeded by 99 percent
(R = 0.99, pf = 0.01),0.01 0.01
0.01
98.26
ˆ 4.30
x
x x
x x x x
z s


− − −
= = =
Solving for the yield strength gives
x0.01 = 98.26 + 4.30 z0.01
From Table A-10, z0.01 = − 2.326. Thus
x0.01 = 98.26 + 4.30(− 2.326) = 88.3 kpsi Ans.
______________________________________________________________________________
1-17 Eq. (1-9): R =1
n
i
i
R
=
 = 0.98(0.96)0.94 = 0.88
Overall reliability = 88 percent Ans.
______________________________________________________________________________

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Shigley’s MED, 10th edition Chapter 1 Solutions, Page 6/12
1-18 Obtain the coefficients of variance for strength and stressˆ 23.5 0.07532
312
syS
S
sy
C S

= = =ˆ ˆ 145 0.09667
1 500
T
C T


 

= = = =
For R = 0.99, from Table A-10, z = − 2.326.
Eq. (1-12):( )( )
( ) ( ) ( ) ( )
( ) ( )
2 2 2 2
2 2
2 2 2 2
2 2
1 1 1 1
1
1 1 1 2.326 0.07532 1 2.326 0.09667
1.3229 1.32 .
1 2.326 0.07532
S
S
z C z C
n z C
Ans

+ − − −
= −
   + − − − − −
   = = =
− −
From the given equation for stress,max 3
16syS T
n d
 
= =
Solving for d gives1/ 3 1/ 3
6
16 16(1500)1.3229 0.0319 m 31.9 mm .
(312)10sy
T n
d Ans
S


   
= = = =       
______________________________________________________________________________
1-19 Obtain the coefficients of variance for stress and strengthˆ ˆ 5 0.09231
65
P
C P

 
 

= = = =ˆˆ 6.59 0.06901
95.5
ySS
S
S y
C S



= = = =
(a)1.2n =

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Shigley’s MED, 10th edition Chapter 1 Solutions, Page 7/12
Eq. (1-11):( )
2 2 2 2 2 2
1 1.2 1 1.6127
1.2 0.06901 0.09231
d
d S
n
z n C C

− −
= − = − = −
+ +
Interpolating Table A-10,
1.61 0.0537
1.6127   = 0.0534
1.62 0.0526
R = 1 − 0.0534 = 0.9466 Ans.( )
( )
2
2
4 65 1.24 1.020 in .
/ ( / 4) 4 95.5
y y y
y
S S d S Pn
n d Ans
P d P S





= = =  = = =
(b)1.5n =( )
2 2 2
1.5 1 3.605
1.5 0.06901 0.09231
z −
= − = −
+
3.6 0.000159
3.605   = 0.00015645
3.7 0.000108
R = 1 − 0.00015645 = 0.9998 Ans.( )
( )
4 65 1.54 1.140 in .
95.5y
Pn
d Ans
S


= = =
______________________________________________________________________________
1-20max max 90 383 473 MPaa b
   
= = + = + =
From footnote 9, p. 25 of text,( )
max
1/ 2
2 2 2 2 1/ 2
ˆ ˆ ˆ (8.4 22.3 ) 23.83 MPaa b



  
= + = + =max max
max
max max
ˆ ˆ 23.83 0.0504
473
C


 
 
 
= = = =ˆ ˆ 42.7 0.0772
553
y y
y
y
S S
S
S y
C S
 

= = = =

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Shigley’s MED, 10th edition Chapter 1 Solutions, Page 8/12max
553 1.169 1.17 .
473
yS
n Ans

= = = =
Eq. (1-11):( )
2 2 2 2 2 2
1 1.169 1 1.635
1.169 0.0772 0.0504
d
d S
n
z n C C

− −
= − = − = −
+ +
From Table A-10, (− 1.635) = 0.05105
R = 1 − 0.05105 = 0.94895 = 94.9 percent Ans.
______________________________________________________________________________
1-21 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)d a b c= − − −w = 6.520 − 1.5 − 2 − 3 = 0.020 inallt t= w
= 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 tod . Therefore,d
= 6.520 + 0.008 = 6.528 in Ans.
______________________________________________________________________________
1-22 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   +  + 
and,.
V bc a ac b ab c a b c a b c Ans
V abc a b c a b c
  +  +       
 = + + = + +
For the numerical values given,( )
3
1.500 1.875 3.000 8.4375 inV = =

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Shigley’s MED, 10th edition Chapter 1 Solutions, Page 9/12( )
30.002 0.003 0.004 0.004267 0.004267 8.4375 0.0360 in
1.500 1.875 3.000
V V
V
  + + =    =
V = 8.4375  0.0360 in3 Ans.
This answer yields8.4735
8.4015
V  in, whereas, exact is8.473551..
8.401551..
V = in
______________________________________________________________________________
1-23
wmax = 0.05 in, wmin = 0.004 in0.05 0.004 0.027 in
2
+ =w =
Thus,  w = 0.05 − 0.027 = 0.023 in, and then, w = 0.027  0.023 in.0.027 0.042 1.5
1.569 in
a b c
a
a
− −
= − −
=
w =
tw =all
t  0.023 = ta + 0.002 + 0.005  ta = 0.016 in
Thus, a = 1.569  0.016 in Ans.
______________________________________________________________________________
1-24( )
2 3.734 2 0.139 4.012 ino iD D d= + = + =( )
all 0.028 2 0.004 0.036 inoDt t= = + =
Do = 4.012  0.036 in Ans.
______________________________________________________________________________
1-25 From O-Rings, Inc. (oringsusa.com), Di = 9.19  0.13 mm, d = 2.62  0.08 mm( )
2 9.19 2 2.62 14.43 mmo iD D d= + = + =( )
all 0.13 2 0.08 0.29 mmoDt t= = + =
Do = 14.43  0.29 mm Ans.
______________________________________________________________________________

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Shigley’s MED, 10th edition Chapter 1 Solutions, Page 10/12
1-26 From O-Rings, Inc. (oringsusa.com), Di = 34.52  0.30 mm, d = 3.53  0.10 mm( )
2 34.52 2 3.53 41.58 mmo iD D d= + = + =( )
all 0.30 2 0.10 0.50 mmoDt t= = + =
Do = 41.58  0.50 mm Ans.
______________________________________________________________________________
1-27 From O-Rings, Inc. (oringsusa.com), Di = 5.237  0.035 in, d = 0.103  0.003 in( )
2 5.237 2 0.103 5.443 ino iD D d= + = + =( )
all 0.035 2 0.003 0.041 inoDt t= = + =
Do = 5.443  0.041 in Ans.
______________________________________________________________________________
1-28 From O-Rings, Inc. (oringsusa.com), Di = 1.100  0.012 in, d = 0.210  0.005 in( )
2 1.100 2 0.210 1.520 ino iD D d= + = + =( )
all 0.012 2 0.005 0.022 inoDt t= = + =
Do = 1.520  0.022 in Ans.
______________________________________________________________________________
1-29 From Table A-2,
(a)
 = 150/6.89 = 21.8 kpsi Ans.
(b) F = 2 /4.45 = 0.449 kip = 449 lbf Ans.
(c) M = 150/0.113 = 1330 lbf  in = 1.33 kip  in Ans.
(d) A = 1500/ 25.42 = 2.33 in2 Ans.
(e) I = 750/2.544 = 18.0 in4 Ans.
(f) E = 145/6.89 = 21.0 Mpsi Ans.
(g) v = 75/1.61 = 46.6 mi/h Ans.

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Shigley’s MED, 10th edition Chapter 1 Solutions, Page 11/12
(h) V = 1000/946 = 1.06 qt Ans.
______________________________________________________________________________
1-30 From Table A-2,
(a) l = 5(0.305) = 1.53 m Ans.
(b)
 = 90(6.89) = 620 MPa Ans.
(c) p = 25(6.89) = 172 kPa Ans.
(d) Z =12(16.4) = 197 cm3 Ans.
(e) w = 0.208(175) = 36.4 N/m Ans.
(f)
 = 0.001 89(25.4) = 0.048 0 mm Ans.
(g) v = 1 200(0.0051) = 6.12 m/s Ans.
(h)  = 0.002 15(1) = 0.002 15 mm/mm Ans.
(i) V = 1830(25.43) = 30.0 (106) mm3 Ans.
______________________________________________________________________________
1-31
(a)
 = M /Z = 1770/0.934 = 1895 psi = 1.90 kpsi Ans.
(b)
 = F /A = 9440/23.8 = 397 psi Ans.
(c) y =Fl3/3EI = 270(31.5)3/[3(30)106(0.154)] = 0.609 in Ans.
(d)
 = Tl /GJ = 9 740(9.85)/[11.3(106)(
 /32)1.004] = 8.648(10−2) rad = 4.95 Ans.
______________________________________________________________________________
1-32
(a)
 =F / wt = 1000/[25(5)] = 8 MPa Ans.
(b) I = bh3 /12 = 10(25)3/12 = 13.0(103) mm4 Ans.
(c) I =
 d4/64 =
 (25.4)4/64 = 20.4(103) mm4 Ans.
(d)
 =16T /
 d 3 = 16(25)103/[
 (12.7)3] = 62.2 MPa Ans.
______________________________________________________________________________
1-33

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Shigley’s MED, 10th edition Chapter 1 Solutions, Page 12/12
(a)
 =F /A = 2 700/[
 (0.750)2/4] = 6110 psi = 6.11 kpsi Ans.
(b)
 = 32Fa/
 d 3 = 32(180)31.5/[
 (1.25)3] = 29 570 psi = 29.6 kpsi Ans.
(c) Z =
 (do4 − di4)/(32 do) =
 (1.504 − 1.004)/[32(1.50)] = 0.266 in3 Ans.
(d) k = (d 4G)/(8D 3 N) = 0.062 54(11.3)106/[8(0.760)3 32] = 1.53 lbf/in Ans.
______________________________________________________________________________

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Shigley’s MED, 10th edition Chapter 2 Solutions, Page 1/22
Chapter 2
2-1 From Tables A-20, A-21, A-22, and A-24c,
(a) UNS G10200 HR: Sut = 380 (55) MPa (kpsi), Syt = 210 (30) MPa (kpsi) Ans.
(b) SAE 1050 CD: Sut = 690 (100) MPa (kpsi), Syt = 580 (84) MPa (kpsi) Ans.
(c) AISI 1141 Q&T at 540C (1000F): Sut = 896 (130) MPa (kpsi), Syt = 765 (111)
MPa (kpsi) Ans.
(d) 2024-T4: Sut = 446 (64.8) MPa (kpsi), Syt = 296 (43.0) MPa (kpsi) Ans.
(e) Ti-6Al-4V annealed: Sut = 900 (130) MPa (kpsi), Syt = 830 (120) MPa (kpsi) Ans.
______________________________________________________________________________
2-2 (a) Maximize yield strength: Q&T at 425C (800F) Ans.
(b) Maximize elongation: Q&T at 650C (1200F) Ans.
______________________________________________________________________________
2-3 Conversion of kN/m3 to kg/ m3 multiply by 1(103) / 9.81 = 102
AISI 1018 CD steel: Tables A-20 and A-5( )
( )
3
370 10 47.4 kN m/kg .
76.5 102
yS Ans

= = 
2011-T6 aluminum: Tables A-22 and A-5( )
( )
3
169 10 62.3 kN m/kg .
26.6 102
yS Ans

= = 
Ti-6Al-4V titanium: Tables A-24c and A-5( )
( )
3
830 10 187 kN m/kg .
43.4 102
yS Ans

= = 
ASTM No. 40 cast iron: Tables A-24a and A-5.Does not have a yield strength. Using the
ultimate strength in tension( )
( )
( )
3
42.5 6.89 10 40.7 kN m/kg
70.6 102
utS Ans

= = 
______________________________________________________________________________
2-4
AISI 1018 CD steel: Table A-5( ) ( )
6
6
30.0 10 106 10 in .
0.282
E Ans

= =
2011-T6 aluminum: Table A-5( ) ( )
6
6
10.4 10 106 10 in .
0.098
E Ans

= =

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Shigley’s MED, 10th edition Chapter 2 Solutions, Page 2/22
Ti-6Al-6V titanium: Table A-5( ) ( )
6
6
16.5 10 103 10 in .
0.160
E Ans

= =
No. 40 cast iron: Table A-5( ) ( )
6
6
14.5 10 55.8 10 in .
0.260
E Ans

= =
______________________________________________________________________________
2-52
2 (1 ) 2
E G
G v E v G
−
+ =  =
Using values for E and G from Table A-5,
Steel:( )
( )
30.0 2 11.5 0.304 .
2 11.5
v Ans
−
= =
The percent difference from the value in Table A-5 is0.304 0.292 0.0411 4.11 percent .
0.292 Ans
− = =
Aluminum:( )
( )
10.4 2 3.90 0.333 .
2 3.90
v Ans
−
= =
The percent difference from the value in Table A-5 is 0 percent Ans.
Beryllium copper:( )
( )
18.0 2 7.0 0.286 .
2 7.0
v Ans
−
= =
The percent difference from the value in Table A-5 is0.286 0.285 0.00351 0.351 percent .
0.285 Ans
− = =
Gray cast iron:( )
( )
14.5 2 6.0 0.208 .
2 6.0
v Ans
−
= =
The percent difference from the value in Table A-5 is0.208 0.211 0.0142 1.42 percent .
0.211 Ans
− = − = −
______________________________________________________________________________
2-6 (a) A0 =
 (0.503)2/4 = 0.1987 in2,
 = Pi / A0

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Shigley’s MED, 10th edition Chapter 2 Solutions, Page 3/22
For data in elastic range,
 =  l / l0 =  l / 2
For data in plastic range,0 0
0 0 0
1 1
l l Al l
l l l A
−
= = = − = −ò
On the next two pages, the data and plots are presented. Figure (a) shows the linear part of
the curve from data points 1-7. Figure (b) shows data points 1-12. Figure (c) shows the
complete range. Note: The exact value of A0 is used without rounding off.
(b) From Fig. (a) the slope of the line from a linear regression is E = 30.5 Mpsi Ans.
From Fig. (b) the equation for the dotted offset line is found to be
 = 30.5(106)
 − 61 000 (1)
The equation for the line between data points 8 and 9 is
 = 7.60(105)
 + 42 900 (2)
Solving Eqs. (1) and (2) simultaneously yields
 = 45.6 kpsi which is the 0.2 percent
offset yield strength. Thus, Sy = 45.6 kpsi Ans.
The ultimate strength from Figure (c) is Su = 85.6 kpsi Ans.
The reduction in area is given by Eq. (2-12) is( ) ( )
0
0
0.1987 0.1077
100 100 45.8 % .
0.1987
fA A
R Ans
A
− −
= = =
Data Point Pi l, Ai


1 0 0 0 0
2 1000 0.0004 0.00020 5032
3 2000 0.0006 0.00030 10065
4 3000 0.001 0.00050 15097
5 4000 0.0013 0.00065 20130
6 7000 0.0023 0.00115 35227
7 8400 0.0028 0.00140 42272
8 8800 0.0036 0.00180 44285
9 9200 0.0089 0.00445 46298
10 8800 0.1984 0.00158 44285
11 9200 0.1978 0.00461 46298
12 9100 0.1963 0.01229 45795
13 13200 0.1924 0.03281 66428
14 15200 0.1875 0.05980 76492
15 17000 0.1563 0.27136 85551
16 16400 0.1307 0.52037 82531
17 14800 0.1077 0.84506 74479

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Solution Manual For Shigley's Mechanical Engineering Design, 10th Edition

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