Solution Manual for Shigley's Mechanical Engineering Design: 2024 Release ISE by Budynas | All 20 Chapters Covered

All 20 Chapters Covered




SOLUTION MANUAL

,TABLE OF CONTENTS
Part 1 - Basics
1) Introduction to Mechanical Engineering Design
2) Materials
3) Load and Stress Analysis
4) Deflection and Stiffness
Part 2 - Failure Prevention
5) Failures Resulting from Static Loading
6) Fatigue Failure Resulting from Variable Loading
Part 3 - Design of Mechanical Elements
7) Shafts and Shaft Components
8) Screws, Fasteners, and the Design of Nonpermanent Joints
9) Welding, Bonding, and the Design of Permanent Joints
10) Mechanical Springs
11) Rolling-Contact Bearings
12) Lubrication and Journal Bearings
13) Gears - General
14) Spur and Helical Gears
15) Bevel and Worm Gears
16) Clutches, Brakes, Couplings and Flywheels
17) Flexible Mechanical Elements
18) Power Transmission Case Study
Part 4 - Special Topics
19) Finite-Element Analysis
20) Geometric Dimensioning and Tolerancing

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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.
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1-8 C A = C B ,

10 + 0.8 P = 60 + 0.8 P  0.005 P 2

P 2 = 50/0.005  P = 100 parts Ans.
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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


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1-10 (a) X 1 + X 2 :
x1  x2  X 1  e1  X 2  e2
error  e   x1  x2    X 1  X 2 
 e1  e2 Ans.
(b) X 1  X 2 :
x1  x2  X 1  e1   X 2  e2 
e   x1  x2    X 1  X 2   e1  e2 Ans.
( c) X 1 X 2 :
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 




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



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(d) X 1 /X 2 :
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 e
1    1    1 
then  1    1  2
 X2  X2  1  e2 X 2   X1   X2  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 
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1-11 (a) x 1 = 7 = 2.645 751 311 1
X 1 = 2.64 (3 correct digits)
x 2 = 8 = 2.828 427 124 7
X 2 = 2.82 (3 correct digits)
x 1 + x 2 = 5.474 178 435 8
e 1 = x 1  X 1 = 0.005 751 311 1
e 2 = x 2  X 2 = 0.008 427 124 7
e = e 1 + e 2 = 0.014 178 435 8
Sum = x 1 + x 2 = X 1 + X 2 + e
= 2.64 + 2.82 + 0.014 178 435 8 = 5.474 178 435 8 Checks
(b) X 1 = 2.65, X 2 = 2.83 (3 digit significant numbers)
e 1 = x 1  X 1 =  0.004 248 688 9
e 2 = x 2  X 2 =  0.001 572 875 3
e = e 1 + e 2 =  0.005 821 564 2
Sum = x 1 + x 2 = X 1 + X 2 + e
= 2.65 +2.83  0.001 572 875 3 = 5.474 178 435 8 Checks
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16 1000  25 10 
3
S
1-12      d  0.799 in Ans.
nd d3 2.5
Table A-17: d = 78 in Ans.

S 25 103 
Factor of safety: n   3.29 Ans.
 16 1000 

3
 7
8

______________________________________________________________________________
n
1-13 Eq. (1-5): R =  Ri = 0.98(0.96)0.94 = 0.88
i 1
Overall reliability = 88 percent Ans.
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Chapter 1 Solutions - Rev. B, Page 2/6



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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), w min = 0.002 in. Thus, must add 0.008 in to d . Therefore,

d = 6.520 + 0.008 = 6.528 in Ans.

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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 in 3


V 0.002 0.003 0.004
    0.00427  V  0.00427  8.4375   0.036 in 3
V 1.500 1.875 3.000

V = 8.438  0.036 in3 Ans.
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Chapter 1 Solutions - Rev. B, Page 3/6



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1-16


w max = 0.05 in, w min = 0.004 in
0.05  0.004
w=  0.027 in
2
Thus,  w = 0.05  0.027 = 0.023 in, and then, w = 0.027  0.023 in.
w= a b c
0.027  a  0.042  1.5
a  1.569 in

tw = t all
 0.023 = t a + 0.002 + 0.005  t a = 0.016 in

Thus, a = 1.569  0.016 in Ans.

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1-17 Do  Di  2d  3.734  2  0.139   4.012 in

t Do   tall  0.028  2  0.004   0.036 in

D o = 4.012  0.036 in Ans.
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1-18 From O-Rings, Inc. (oringsusa.com), D i = 9.19  0.13 mm, d = 2.62  0.08 mm

Do  Di  2d  9.19  2  2.62   14.43 mm

t Do   tall  0.13  2  0.08   0.29 mm

D o = 14.43  0.29 mm Ans.
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1-19 From O-Rings, Inc. (oringsusa.com), D i = 34.52  0.30 mm, d = 3.53  0.10 mm

Do  Di  2d  34.52  2  3.53  41.58 mm

t Do   tall  0.30  2  0.10   0.50 mm

D o = 41.58  0.50 mm Ans.
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Chapter 1 Solutions - Rev. B, Page 4/6



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1-20 From O-Rings, Inc. (oringsusa.com), D i = 5.237  0.035 in, d = 0.103  0.003 in

Do  Di  2d  5.237  2  0.103  5.443 in

t Do   tall  0.035  2  0.003  0.041 in

D o = 5.443  0.041 in Ans.
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1-21 From O-Rings, Inc. (oringsusa.com), D i = 1.100  0.012 in, d = 0.210  0.005 in

Do  Di  2d  1.100  2  0.210   1.520 in

t Do   tall  0.012  2  0.005   0.022 in

D o = 1.520  0.022 in Ans.
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1-22 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.

(h) V = 1000/946 = 1.06 qt Ans.
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1-23 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.

Chapter 1 Solutions - Rev. B, Page 5/6



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(d) Z =12(16.4) = 197 cm3 Ans.

(e) w = 0.208(175) = 36.4 N/m Ans.

(f)  = 0.001 89(25.4) = 0.0480 mm Ans.

(g) v = 1200(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.
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1-24
(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 = 9740(9.85)/[11.3(106)( /32)1.004] = 8.648(102) rad = 4.95 Ans.

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1-25
(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.
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1-26
(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 = (d o 4  d i 4)/(32 d o ) =  (1.504  1.004)/[32(1.50)] = 0.266 in3 Ans.

(d) k = (d 4G)/(8D 3 N) = 0.06254(11.3)106/[8(0.760)3 32] = 1.53 lbf/in Ans.

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



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Chapter 2

2-1 From Tables A-20, A-21, A-22, and A-24c,
(a) UNS G10200 HR: S ut = 380 (55) MPa (kpsi), S yt = 210 (30) Mpa (kpsi) Ans.
(b) SAE 1050 CD: S ut = 690 (100) MPa (kpsi), S yt = 580 (84) Mpa (kpsi) Ans.
(c) AISI 1141 Q&T at 540C (1000F): S ut = 896 (130) MPa (kpsi), S yt = 765 (111)
Mpa (kpsi) Ans.
(d) 2024-T4: S ut = 446 (64.8) MPa (kpsi), S yt = 296 (43.0) Mpa (kpsi) Ans.
(e) Ti-6Al-4V annealed: S ut = 900 (130) MPa (kpsi), S yt = 830 (120) Mpa (kpsi) Ans.
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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.
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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
S y 370 10 
3

  47.4 kN  m/kg Ans.
 76.5 102 
2011-T6 aluminum: Tables A-22 and A-5
S y 169 10 
3

  62.3 kN  m/kg Ans.
 26.6 102 
Ti-6Al-4V titanium: Tables A-24c and A-5
Sy 830 103 
  187 kN  m/kg Ans.
 43.4 102 
ASTM No. 40 cast iron: Tables A-24a and A-5.Does not have a yield strength. Using the
ultimate strength in tension
Sut 42.5  6.89  10 
3

  40.7 kN  m/kg Ans
 70.6 102 
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2-4
AISI 1018 CD steel: Table A-5
E 30.0 10 
6

  106 106  in Ans.
 0.282
2011-T6 aluminum: Table A-5
E 10.4 10 
6

  106 106  in Ans.
 0.098
Ti-6Al-6V titanium: Table A-5

Chapter 2 - Rev. D, Page 1/19



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16.5 106 
 103 106  in
E
 Ans.
 0.160
No. 40 cast iron: Table A-5
E 14.5 10 
6

  55.8 106  in Ans.
 0.260
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2-5
E  2G
2G (1  v)  E  v
2G
From Table A-5
30.0  2 11.5 
Steel: v   0.304 Ans.
2 11.5 

10.4  2  3.90
Aluminum: v  0.333 Ans.
2  3.90

18.0  2  7.0 
Beryllium copper: v  0.286 Ans.
2  7.0 

14.5  2  6.0 
Gray cast iron: v
 0.208 Ans.
2  6.0 
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2-6 (a) A 0 =  (0.503)2/4,  = P i / A 0

For data in elastic range,  =  l / l 0 =  l / 2
l l  l0 l A
For data in plastic range,     1  0 1
l0 l0 l0 A
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 A 0 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)


Chapter 2 - Rev. D, Page 2/19



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