RESOLUÇÃO ELEMENTOS DE MÁQUINAS 10 ED SHIGLEY

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Shigley’s Mechanical Engineering Design, 10th edition in SI Units 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
 
 
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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. 
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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 
 
 ( )2
1.10 1.43 .
0.85 0.95d
n Ans= = 
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 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 eX e X e X X Ans
X X
= + +
= − = + +
 
≈ + = + 
 
 
 
Shigley’s Mechanical Engineering Design, 10th edition in SI Units 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
11 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
−
 + +
= =  
+ + 
      +
+ ≈ − ≈ + − ≈ + −      
+      
 
= − ≈ − 
 
 
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 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 
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1-12 ( ) ( )63 172.4 1032 113 0.0052 m 5.2861 mm .2.5σ pi= ⇒ = ⇒ = =d
S d Ans
n d
 
 Table A-17: d = 5.5 mm Ans. 
 Factor of safety: 
( )
( )
6
33
172.4 10
2.82 .113
5.5 10
σ
pi −
= = =
×
S
n Ans 
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Shigley’s Mechanical Engineering Design, 10th edition in SI Units 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 1104 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) 115115
115 122.9 0.2607
ˆ 30.3
x
x x
x x x
z
s
µ
σ
− − −
= = = = − 
 
 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 
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 
Σ 69 8480 1 104 600 
Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 1 Solutions, Page 4/12 
 2 + 1 + 3 + 5 + 8 + 12 = 31 (the data is not perfectly normal) 
 
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1-14 
x f f x f x2 
1.2 6 7.2 8.64 
1.25 9 11.25 14.0625 
1.31 44 57.64 75.5084 
1.37 67 91.79 125.7523 
1.42 53 75.26 106.8692 
1.48 12 17.76 26.2848 
1.53 6 9.18 14.0454 
Σ 197 270.08 371.1626 
 
 
 
Eq. (1-6) 
1
1 270.08 1.37 GPa
197
=
= = =∑
k
i i
i
x f x
N
 
 
 Eq. (1-7) 
2 2
1 22
1 371.16 197(1.37) 0.067 GPa .
1 197 1
=
−
 −
= = = 
− − 
∑
k
i i
i
x
f x N x
s Ans
N
 
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1-15 122.9 kcycles and 30.3 kcyclesLL s= = 
 
 Eq. (1-5) 10 1010
122.9
ˆ 30.3
x
L
x x L x
z
s
µ
σ
− − −
= = =
 
 
 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. 
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Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 1 Solutions, Page 5/12 
1-16 
x f fx fx2 
0.64 19 12.16 7.78 
0.66 25 16.50 10.89 
0.64 38 24.32 15.56 
0.67 17 11.39 7.63 
0.7 12 8.40 5.88 
0.71 10 7.10 5.04 
0.72 5 3.60 2.59 
0.74 4 2.96 2.19 
0.75 4 3.00 2.25 
0.77 2 1.54 1.19 
Σ 136 90.97 61.01 
 
 Eq. (1-6) 
1
1 90.97 / 136 0.67 MPa
=
= = =∑
k
i i
i
x f x
N
 
 
 
 Eq. (1-7) 
2 2
1 22
1 61.01 136(0.67) 0.034 MPa
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
0.67
ˆ 0.034
µ
σ
− − −
= = =
x
x x
x x x x
z
s
 
Solving for the yield strength gives 
 
 x0.01 = 0.67 + 0.034 z0.01 
 
 From Table A-10, z0.01 = − 2.326. Thus 
 
 x0.01 = 0.67 + 0.034(− 2.326) = 0.60 MPa Ans. 
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Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 1 Solutions, Page 6/12 
1-17 
 
 
 
 
 
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1-18 Obtain the coefficients of variance for strength and stress 
 
ˆ 23.5 0.07532
312
syS
S
sy
C
S
σ
= = = 
 
 
ˆ ˆ 145 0.09667
1 500
TC
T
τ
σ
σ σ
τ
= = = = 
 
 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
σ+ − − −
=
−
   + − − − − −   
= = =
− −
 
 From the given equation for stress, 
 
 
max 3
16syS T
n d
τ
pi
= = 
 
 Solving for d gives 
 
 
1/3 1/ 3
6
16 16(1500)1.3229 0.0319 m 31.9 mm .(312)10sy
T nd Ans
Spi pi
   
= = = =       
 
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1-19 Obtain the coefficients of variance for stress and strength 
 
ˆ ˆ 22.24 0.07692
289.13
σ
σ
σ
σ σµ
= = = =
PC
P
 
 
Eq. (1-9) 
Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 1 Solutions, Page 7/12 
 
ˆ
ˆ 45.44 0.06901
658.45
σσ
µ
= = = =
ySS
S
S y
C
S
 
 
 (a) 1.2n = 
 
 Eq. (1-11): ( )2 2 2 2 2 2
1 1.2 1 1.7695
1.2 0.06901 0.07692σ
− −
= − = − = −
+ +
d
d S
n
z
n C C
 
 Interpolating Table A-10, 
 1.76 0.0392 
 1.7695 Φ ⇒ Φ = 0.03844 
 1.77 0.0384 
 
 R = 1 − 0.03844 = 0.96156 Ans. 
 
 
( )
( )
2
2
3
6
/ ( / 4) 4
4 289.13 10 1.24 0.0259m 25.9mm .
658.45 10
pi
σ pi
pi pi
= = =
×
⇒ = = = =
×
y y y
y
S S d S
n
P d P
Pnd Ans
S
 
 
 (b) 1.5n = 
 
 ( )2 2 2 2 2 2
1 1.5 1 3.8770
1.2 0.06901 0.07692σ
− −
= − = − = −
+ +
d
d S
n
z
n C C
 
 
 3.8 0.000072 
 3.8770 Φ ⇒ Φ = 0.00005367 
 3.9 0.000048 
 
 R = 1 − 0.00005367 = 0.99994633 Ans. 
 
 
( )
( )
3
6
4 289.13 10 1.54 0.029m 29mm .
658.45 10pi pi
×
= = = =
×y
Pnd Ans
S
 
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1-20 
max max
90 383 473 MPaa bσµ σ σ σ= = + = + = 
 
 From footnote 9, p. 25 of text, 
 
 ( )
max
1/ 22 2 2 2 1/ 2
ˆ ˆ ˆ (8.4 22.3 ) 23.83 MPa
a bσ σ σ
σ σ σ= + = + = 
Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 1 Solutions, Page 8/12 
 
 
max max
max
max max
ˆ ˆ 23.83 0.0504
473
C σ σσ
σ
σ σ
µ σ
= = = = 
 
 
ˆ ˆ 42.7 0.0772
553
y y
y
y
S S
S
S y
C
S
σ σ
µ
= = = =
 
 
max
553 1.169 1.17 .
473
yS
n Ans
σ
= = = = 
 
 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 Cσ
− −
= − = − = −
+ +
 
 
 From Table A-10, Φ(− 1.635) = 0.05105 
 
 R = 1 − 0.05105 = 0.94895 = 94.9 percent Ans. 
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1-21
 
 
 
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Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 1 Solutions, Page 9/12 
1-22
 
 
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1-23 
 
 
 
 
 
 
Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 1 Solutions, Page 10/12 
 
1-24 
 
 
 
 
 
 
 
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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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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. 
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1-27 
 
 
 
Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 1 Solutions, Page 11/12 
1-28
 
 
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1-29 From Table A-2, 
 
 (a) σ = 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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1-30 From Table A-2, 
 (a) l = 5(0.305) = 1.53 m Ans. 
 (b) σ = 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) δ = 0.001 89(25.4) = 0.048 0 mm Ans. 
 
Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 1 Solutions, Page 12/12 
 (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. 
 
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1-31 
 
 
 
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1-32 
 (a) σ =F / wt = 1000/[25(5)] = 8 MPa Ans. 
 
 (b) I = bh3 /12 = 10(25)3/12 = 13.0(103) mm4 Ans. 
 
 (c) I =pi d4/64 = pi (25.4)4/64 = 20.4(103) mm4 Ans. 
 
 (d) τ =16T /pi d 3 = 16(25)103/[pi (12.7)3] = 62.2 MPa Ans. 
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1-33 
 
 
 
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Shigley’s Mechanical Engineering Design, 10th edition in SI Units 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. 
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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 
 
( )
( )
3370 10
47.4 kN m/kg .
76.5 102
yS Ans
ρ
= = ⋅ 
 2011-T6 aluminum: Tables A-22 and A-5 
 
( )
( )
3169 10
62.3 kN m/kg .
26.6 102
yS Ans
ρ
= = ⋅ 
 Ti-6Al-4V titanium: Tables A-24c and A-5 
 
( )
( )
3830 10
187 kN m/kg .
43.4 102
yS Ans
ρ
= = ⋅ 
 ASTM No. 40 cast iron: Tables A-24a and A-5.Does not have a yield strength. Using the 
ultimate strength in tension 
 
( )( )
( )
342.5 6.89 10
40.7 kN m/kg
70.6 102
utS Ans
ρ
= = ⋅ 
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Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 2 Solutions, Page 2/22 
 
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2-5 
22 (1 )
2
E GG 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 is 
 
 
0.304 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 is 
 
 
0.286 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
−
= = 
 Thepercent difference from the value in Table A-5 is 
 
 
0.208 0.211 0.0142 1.42 percent .
0.211
Ans− = − = − 
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Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 2 Solutions, Page 3/22 
 
 
Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 2 Solutions, Page 4/22 
 
 
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Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 2 Solutions, Page 9/22 
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2-12 For HB = 275, Eq. (2-21), Su = 3.4(275) = 935 MPa Ans. 
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Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 2 Solutions, Page 10/22 
 
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2-18, 2-19 These problems are for student research. No standard solutions are provided. 
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Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 2 Solutions, Page 15/22 
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2-24 and 2-25 These problems are for student research. No standard solutions are provided. 
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Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 2 Solutions, Page 17/22 
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Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 2 Solutions, Page 19/22 
Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 2 Solutions, Page 20/22 
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2-29 Eq. (2-26), p. 77, applies to a circular cross section. However, for any cross section shape 
it can be shown that I = CA 2, where C is a constant. For example, consider a rectangular 
section of height h and width b, where for a given scaled shape, h = cb, where c is a 
constant. The moment of inertia is I = bh 3/12, and the area is A = bh. Then I = h(bh2)/12 
= cb (bh2)/12 = (c/12)(bh)2 = CA 2, where C = c/12 (a constant). 
 Thus, Eq. (2-27) becomes 
 
1/23
3
klA
CE
 
=  
 
 
 and Eq. (2-29) becomes 
 
1/2
5/2
1/23
k
m Al l
C E
ρρ    = =    
   
 
 Thus, minimize ( )3 1/2f M E
ρ
= , or maximize
1/2EM
ρ
= . From Fig. (2-16) 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 2 Solutions, Page 21/22 
 
 From the list of materials given, aluminum alloys are clearly the best followed by steels 
and tungsten carbide. Polycarbonate polymer is not a good choice compared to the other 
candidate materials. Ans. 
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2-30 For stiffness, k = AE/l ⇒ A = kl/E 
 For mass, m = Alρ = (kl/E) lρ =kl2 ρ /E 
 
 So, f 3(M) = ρ /E, and maximize E/ρ . Thus, β = 1. Ans. 
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2-31 For strength, σ = F/A = S ⇒ A = F/S 
 For mass, m = Alρ = (F/S) lρ 
 
 So, f 3(M ) = ρ /S, and maximize S/ρ . Thus, β = 1. Ans. 
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Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 2 Solutions, Page 22/22 
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2-34 For stiffness, k=AE/l, or, A = kl/E. 
 
 Thus, m = ρAl =ρ (kl/E )l = kl 2 ρ /E. Then, M = E /ρ and β = 1. 
 
 From Fig. 2-16, lines parallel to E /ρ for ductile materials include steel, titanium, 
molybdenum, aluminum alloys, and composites. 
 
 For strength, S = F/A, or, A = F/S. 
 
 Thus, m = ρAl =ρ F/Sl = Fl ρ /S. Then, M = S/ρ and β = 1. 
 
 From Fig. 2-19, lines parallel to S/ρ give for ductile materials, steel, aluminum alloys, 
nickel alloys, titanium, and composites. 
 
 Common to both stiffness and strength are steel, titanium, aluminum alloys, and 
composites. Ans. 
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2-35 See Prob. 1-13 solution for x = 122.9 kcycles and xs = 30.3 kcycles. Also, in that solution 
it is observed that the number of instances less than 115 kcycles predicted by the normal 
distribution is 27; whereas, the data indicates the number to be 31. 
 From Eq. (1-4), the probability density function (PDF), with xµ = and ˆ xsσ = , is 
 
Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 2 Solutions, Page 23/22 
 
2 21 1 1 1 122.9( ) exp exp
2 2 30.32 30.3 2xx
x x xf x
ss pi pi
    
− −  = − = −    
        
 (1) 
 
 The discrete PDF is given by f /(Nw), where N = 69 and w = 10 kcycles. From the Eq. (1) 
and the data of Prob. 1-13, the following plots are obtained. 
 
Range 
midpoint 
(kcycles) Frequency 
Observed 
PDF 
Normal 
PDF 
x f f /(Nw) f (x) 
60 2 0.002898551 0.001526493 
70 1 0.001449275 0.002868043 
80 3 0.004347826 0.00483250790 5 0.007246377 0.007302224 
100 8 0.011594203 0.009895407 
110 12 0.017391304 0.012025636 
120 6 0.008695652 0.013106245 
130 10 0.014492754 0.012809861 
140 8 0.011594203 0.011228104 
150 5 0.007246377 0.008826008 
160 2 0.002898551 0.006221829 
170 3 0.004347826 0.003933396 
180 2 0.002898551 0.002230043 
190 1 0.001449275 0.001133847 
200 0 0 0.000517001 
210 1 0.001449275 0.00021141 
 
 Plots of the PDF’s are shown below. 
 
Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 2 Solutions, Page 24/22 
 
 It can be seen that the data is not perfectly normal and is skewed to the left indicating that 
the number of instances below 115 kcycles for the data (31) would be higher than the 
hypothetical normal distribution (27). 
______________________________________________________________________________ 
0
0,002
0,004
0,006
0,008
0,01
0,012
0,014
0,016
0,018
0,02
0 20 40 60 80 100 120 140 160 180 200 220
Normal Distribution
Histogram
L (kcycles)
P
ro
b
a
b
il
it
y
 D
e
n
si
ty
Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 1/100 
Chapter 3 
 
 
 
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Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 2/100 
 
3-3 
 
 
 
 
 
0.8 1.39 kN .
tan 30O
R Ans= =
o
 
 
0.8 1.6 kN .
sin 30A
R Ans= =
o
 
 
 
 
 
______________________________________________________________________________ 
 
3-4 
 Step 1: Find RA & RE 
 
 
 
 
 
 
 
 
 
 
2 2
0 400cos30 0
 346.4 N
0 400 400sin 30 0
 200 N
346.4 200 400 N .
x Ax
Ax
y Ay
Ay
A
F R
R
F R
R
R Ans
= + =
= −
= + − =
= −
= + =
∑
∑
o
o
 
 
 
 Step 2: Find components of RC and RD on link 4 
 
 
( )
( )4
4
0
 400(4.5) 7.794 1.9 0
 305.4 N .
0 305.4 N
0 ( ) 400 N
C
D
D
x Cx
y Cy
M
R
R Ans
F R
F R
=
− − =
=
= ⇒ =
= ⇒ = −
∑
∑
∑ 
4.5 7.794 m
tan 30
0
9 7.794(400cos30 )
4.5(400sin 30 ) 0
400 N .
A
E
E
h
M
R
R Ans
= =
Σ =
−
− =
=
o
o
o
Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 3/100 
 
Step 3: Find components of RC on link 2 
 
( )
( )
( )
2
2
2
0
 305.4 346.4 0
 41 N
0
 200 N
x
Cx
Cx
y
Cy
F
R
R
F
R
=
+ − =
=
=
=
∑
∑
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 Ans. 
_____________________________________________________________________________________________________________________ 
 
 
 
 
 
Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 4/100 
3-5 
 
0CMΣ = 
 11500 300(5) 1200(9) 0R− + + = 
 1 8.2 kN .R Ans= 
 
 
0yFΣ = 
 28.2 9 5 0R− − + = 2 5.8 kN .R Ans= 
 
 
 
 
 
 
 
 
 
1 8.2(300) 2460 N m .M Ans= = ⋅ 
 
2 2460 0.8(900) 1740 N m .M Ans= − = ⋅ 
 
3 1740 5.8(300) 0 checks!M = − = 
_____________________________________________________________________________ 
 
 
 
 
 
 
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Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 5/100 
3-7 
 
0BMΣ = 
 
12.2 1(2) 1(4) 0R− + − = 
 1 0.91 kN .R Ans= − 
 
0yFΣ = 
 
20.91 2 4 0R− − + − = 
 
2 6.91 kN .R Ans= 
 
 
 
 
 
1 0.91(1.2) 1.09 kN m .M Ans= − = − ⋅
 
 
2 1.09 2.91(1) 4 kN m .M Ans= − − = − ⋅
 
 3 4 4(1) 0 checks!M = − + = 
 
 
 
 
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Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 6/100 
 
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3-9 
1 1 1 1
1 2
0 0 0
1 2
1 1 1
1 2
9 300 5 1200 1500
9 300 5 1200 1500 (1)
9 300 5 1200 1500 (2)
q R x x x R x
V R x x R x
M R x x x R x
− − − −
= − − − − + −
= − − − − + −
= − − − − + −
 
 
At x = 1500+ V = M = 0. Applying Eqs. (1) and (2), 
 
1 2 1 29 5 0 14R R R R− − + = ⇒ + = 
 
1 11500 9(1500 300) 5(1500 1200) 0 8.2 kN .R R Ans− − − − = ⇒ =
 
 
 
2 14 8.2 5.8 kN .R Ans= − = 
 
0 300 : 8.2 kN, 8.2 N m
300 1200 : 8.2 9 0.8 kN
 8.2 9( 300) 0.8 2700 N m
1200 1500 : 8.2 9 5 5.8 kN
 8.2 9( 300
x V M x
x V
M x x x
x V
M x x
≤ ≤ = = ⋅
≤ ≤ = − = −
= − − = − + ⋅
≤ ≤ = − − = −
= − − ) 5( 1200) 5.8 8700 N mx x− − = − + ⋅
 
 
Plots of V and M are the same as in Prob. 3-5. 
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Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 7/100 
________________________________________________________________________ 
 
3-11 
1 1 1 1
1 2
0 0 0
1 2
1 1 1
1 2
2 1.2 2.2 4 3.2
2 1.2 2.2 4 3.2 (1)
2 1.2 2.2 4 3.2 (2)
q R x x R x x
V R x R x x
M R x x R x x
− − − −
= − − + − − −
= − − + − − −
= − − + − − −
 
at x = 3.2+, V = M = 0. Applying Eqs. (1) and (2), 
 
 
1 2 1 2
1 2 1 2
2 4 0 6 (3)
3.2 2(2) (1) 0 3.2 4 (4)
R R R R
R R R R
− + − = ⇒ + =
− + = ⇒ + =
 
 
Solving Eqs. (3) and (4) simultaneously, 
 R1 = -0.91 kN, R2 = 6.91 kN Ans. 
 
0 1.2 : 0.91 kN, 0.91 kN m
1.2 2.2 : 0.91 2 2.91 kN
 0.91 2( 1.2) 2.91 2.4 kN m
2.2 3.2 : 0.91 2 6.91 4 kN
 0.91 2(
x V M x
x V
M x x x
x V
M x x
≤ ≤ = − = − ⋅
≤ ≤ = − − = −
= − − − = − + ⋅
≤ ≤ = − − + =
= − − −1.2) 6.91( 2.2) 4 12.8 kN mx x+ − = − ⋅
 
 
Plots of V and M are the same as in Prob. 3-7. 
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Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 8/100 
 
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3-13 Solution depends upon the beam selected. 
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Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 9/100 
 
 
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Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 10/100 
 
 
Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 11/100 
 
 
Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 12/100 
 
 
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Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 13/100 
3-16 
(a) 
2 2
1
2
8 7 0.5 MPa
2
8 7 7.5 MPa
2
7.5 6 9.60MPa
9.60 0.5 9.10 MPa
0.5 9.6 10.1 Mpa
C
CD
R
σ
σ
− +
= = −
+
= =
= + =
= − =
= − − = −
 
 
11 7.590 tan 70.67 cw
2 6p
φ −  = + =  
  
o o
 
1 9.60 MPa
70.67 45 25.67 cws
Rτ
φ
= =
= − =
o o o
 
 
 
(b) 
2 2
1
2
9 6 1.5 MPa
2
9 6 7.5 MPa
2
7.5 3 8.078 MPa
1.5 8.078 9.58 MPa
1.5 8.078 6.58 MPa
C
CD
R
σ
σ
−
= =
+
= =
= + =
= + =
= − = −
 
 
 
 
 
11 3tan 10.9 cw
2 7.5p
φ −  = = 
 
o
 
1 8.078 MPa
45 10.9 34.1 ccws
Rτ
φ
= =
= − =
o o o
 
 
 
Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 14/100 
 (c) 
2 2
1
2
12 4 4 MPa
2
12 4 8 MPa
2
8 7 10.63 MPa
4 10.63 14.63 MPa
4 10.63 6.63 MPa
C
CD
R
σ
σ
−
= =
+
= =
= + =
= + =
= − = −
 
11 890 tan 69.4 ccw
2 7p
φ −  = + =  
  
o o
1 10.63 MPa
69.4 45 24.4 ccws
Rτ
φ
= =
= − =
o o o
 
 
 
 
 
 (d) 
2 2
1
2
6 5 0.5 MPa
2
6 5 5.5 MPa
2
5.5 8 9.71 MPa
0.5 9.71 10.21 MPa
0.5 9.71 9.21 MPa
C
CD
R
σ
σ
−
= =
+
= =
= + =
= + =
= − = −
 
 
 
 
11 8tan 27.75 ccw
2 5.5p
φ −  = = 
 
o
 
1 9.71 MPa
45 27.75 17.25 cws
Rτ
φ
= =
= − =
o o o
 
 
 
 
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Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 15/100 
Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 16/100 
 
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Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 17/100 
3-18 
(a) 
 
2 2
1
2
3
80 30 55 MPa
2
80 30 25 MPa
2
25 20 32.02 MPa
0 MPa
55 32.02 22.98 23.0 MPa
55 32.0 87.0 MPa
C
CD
R
σ
σ
σ
− −
= = −
−
= =
= + =
=
= − + = − = −
= − − = −
 
1 2 2 3 1 3
23 8711.5 MPa, 32.0 MPa, 43.5 MPa
2 2
τ τ τ= = = = =
 
 
(b) 
2 2
1
2
3
30 60 15 MPa
2
60 30 45 MPa
2
45 30 54.1 MPa
15 54.1 39.1 MPa
0 MPa
15 54.1 69.1 MPa
C
CD
R
σ
σ
σ
−
= = −
+
= =
= + =
= − + =
=
= − − = −
 
1 3
1 2
2 3
39.1 69.1 54.1 MPa
2
39.1 19.6 MPa
2
69.1 34.6 MPa
2
τ
τ
τ
+
= =
= =
= =
 
 
Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 18/100 
(c) 
2 2
1
2
3
40 0 20 MPa
2
40 0 20 MPa
2
20 20 28.3 MPa
20 28.3 48.3 MPa
20 28.3 8.3 MPa
30 MPaz
C
CD
R
σ
σ
σ σ
+
= =
−
= =
= + =
= + =
= − = −
= = −
 
1 3 1 2 2 3
48.3 30 30 8.339.1 MPa, 28.3 MPa, 10.9 MPa
2 2
τ τ τ
+ −
= = = = =
 
 
(d) 
2 2
1
2
3
50 25 MPa
2
50 25 MPa
2
25 30 39.1 MPa
25 39.1 64.1 MPa
25 39.1 14.1 MPa
20 MPaz
C
CD
R
σ
σ
σ σ
= =
= =
= + =
= + =
= − = −
= = −
 
1 3 1 2 2 3
64.1 20 20 14.142.1 MPa, 39.1 MPa, 2.95 MPa
2 2
τ τ τ
+ −
= = = = =
 
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Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 19/100 
Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 20/100 
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Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 21/100 
 
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3-22 From Eq. (3-15) 
 
( ) ( )2 23 2 2
2 2 2
3 2
(10 40 40) 10(40) 10(40) 40(40) 20 40 20
10(40)(40) 2(20)( 40)( 20) 10( 40) 40( 20) 40(20) 0
90 0
σ σ σ
σ σ
 
− + + + + + − − − − −
 
 − + − − − − − − − = 
− =
 
 
Roots are: 90, 0, 0 MPa Ans.
 
 
 
2 3
1 2 1 3 max
0
90 45 MPa .
2
Ans
τ
τ τ τ
=
= = = =
 
 
 
 
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Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 22/100 
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Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 23/100 
 
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______________________________________________________________________________ 
 
 
______________________________________________________________________________ 
 3-29 
 With 0,zσ = solve the first two equations of Eq. (3-19) simulatenously. Place E on the 
left-hand side of both equations, and using Cramer’s rule, 
 
( )
2 2
1
1 1 1
1
x
y x yx y
x
E v
E E vE vE
v v v
v
σ
−
++
= = =
−
− −
−
ò
ò ò òò ò
 
 Likewise, 
 
Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 24/100 
( )
21
y x
y
E
v
ν
σ
+
=
−
ò ò
 
 
 From Table A-5, E = 207 GPa and ν = 0.292. Thus, 
 
 
( ) ( ) ( ) ( )
( ) ( ) ( )
9
6
2 2
9
6
2
207 10 0.0019 0.292 0.000 72
10 382 MPa .
1 1 0.292
207 10 0.000 72 0.292 0.0019
10 37.4 MPa .
1 0.292
x y
x
y
E v
Ans
v
Ans
σ
σ
−
−
+ − +  
= = =
− −
− +  
= = −
−
ò ò
 
_____________________________________________________________________________ 
 
3-30 
 With 0,zσ = solve the first two equations of Eq. (3-19) simulatenously. Place E on the 
left-hand side of both equations, and using Cramer’s rule, 
 
( )
2 2
1
1 1 1
1
x
y x yx y
x
E v
E E vE vE
v v v
v
σ
−
++
= = =
−
− −
−
ò
ò ò òò ò
 
 Likewise, 
 
( )
21
y x
y
E
v
ν
σ
+
=
−
ò ò
 
 
From Table A-5, E = 71.7 GPa and ν = 0.333. Thus, 
( ) ( ) ( ) ( )
( ) ( ) ( )
9
6
2 2
9
6
2
71.7 10 0.0019 0.333 0.000 72
10 134 MPa .
1 1 0.333
71.7 10 0.000 72 0.333 0.0019
10 7.04 MPa .
1 0.333
x y
x
y
E v
Ans
v
Ans
σ
σ
−
−
+ − +  
= = =
− −
− +  
= = −
−
ò ò
 
_____________________________________________________________________________ 
 
3-31 
(a) 1 max 1 
c acR F M R a F
l l
= = =
 
 
2
2 2
6 6
6
M ac bh lF F
bh bh l ac
σ
σ = = ⇒ = Ans. 
(b) ( )( )( ) ( )( )( )
2 2
1 21( )( ) ( )
 .( )( )
m m m mm
m m
b b h h l lF s s s
s Ans
F a a c c s s
σ σ
= = =
 
 For equal stress, the model load varies by the square of the scale factor.
 
 
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Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 25/100 
3-32 
(a) 
2
1 max /2,2 2 2 2 8x l
l l l lR M l
=
 
= = − = 
 
w w w
 
2 2
2 2 2
6 6 3 4
 .
8 4 3
M l Wl bhW Ans
bh bh bh l
σ
σ = = = ⇒ =
w
 
 
(b) 
2 2
2( / )( / )( / ) 1( )( )
.
/
m m m m
m
W b b h h s s
s Ans
W l l sσ σ
= = =
 
2
2
 .
m m ml ss s Ans
l s
= ⇒ = =
w w
w w
 
 
 For equal stress, the model load w varies linearly with the scale factor. 
_____________________________________________________________________________ 
 
Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 26/100 
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Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 27/100 
3-35 
( )3 5 4
2
1 (20)(40) 1.067 10 mm
12
20(40) 800 mm
I
A
= =
= =
 
Mmax is at A. At the bottom of the section, 
( )max 5
450 000(20) 84.3 MPa .
1.067 10
Mc Ans
I
σ = = = 
Due to V, τmax is between A and B at y = 0. 
max
3 3 3000 5.63 MPa .
2 2 800
V Ans
A
τ  = = = 
 
 
 
 
 
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Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 28/100 
 
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Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 29/100 
 
 
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Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 30/100 
 
 
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Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 31/100 
 
3-41 
( ) ( )4 4 2 212 1018 mm , 12 113.1 mm64 4I A
pi pi
= = = = 
Model (c) 
2
2
max
2000(6) 2000(9) 15 000 N mm
2 2
15 000(6)
1018
88.4 N/mm 88.4 MPa .
4 4 2000 23.6 N/mm 23.6 MPa .
3 3 113.1
M
Mc
I
Ans
V Ans
A
σ
σ
τ
= + = ⋅
= =
= =
 
= = = = 
 
 
Model (d) 
 
2
2000(12) 24 000 N mm
24 000(6)
1018
141.5 N/mm 141.5 MPa .
M
Mc
I
Ans
σ
σ
= = ⋅
= =
= =
 
2
max
4 4 2000 23.6 N/mm 23.6 MPa .
3 3 113.1
V Ans
A
τ  = = = = 
 
 
 
 
 
 
 
Model (e) 
 
2
2000(7.5) 15000 N mm
15000(6)
1018
88.4 N/mm 88.4 MPa .
M
Mc
I
Ans
σ
σ
= = ⋅
= =
= =
 
2
max
4 4 2000 23.6 N/mm 23.6 MPa .
3 3 113.1
V Ans
A
τ
 
= = = = 
 
 
 
 
 
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Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 32/100 
 
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Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 33/100 
 
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Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 34/100 
 
 
Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 35/100 
 
 
Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 36/100 
 
 
 
 
 
 
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Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 37/100 
Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 38/100 
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Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 39/100 
 
 
3-46 
 
1
 0
cR F
l
cM Fx x a
l
=
= ≤ ≤
 
( )
2 2
max
66
6
 0 .
c l FxM
bh bh
Fcxh x a Ans
lb
σ
σ
= =
= ≤ ≤
 
_____________________________________________________________________________ 
 
3-47 
From Problem 3-46, 1 , 0
cR F V x a
l
= = ≤ ≤
 
max
max
3 3 ( / ) 3
 .
2 2 2
V c l F Fch Ans
bh bh lb
τ
τ
= = ⇒ =
 
From Problem 3-46, 
max
6( ) Fcxh x
lbσ
=
 . 
Sub in x = e and equate to h above. 
max max
max
2
max
3 6
2
3
 .
8
Fc Fce
lb lb
Fc
e Ans
lb
τ σ
σ
τ
=
= 
_____________________________________________________________________________ 
 
3-48 (a) 
x-z plane 
 
20 1.5(0.5) 2(1.5)sin(30 )(2.25) (3)O zM RΣ = = + −o 
 
2 1.375 kN .zR Ans= 
 
10 1.5 2(1.5)sin(30 ) 1.375z zF RΣ = = − − +o 
 
1 1.625 kN .zR Ans= 
x-y plane 
 
20 2(1.5)cos(30 )(2.25) (3)O yM RΣ = = − +o 
 
2 1.949 kN .yR Ans= 
 
10 2(1.5) cos(30 ) 1.949y yF RΣ = = − +o 
 
1 0.6491 kN .yR Ans= 
 
Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 40/100 
 
 
(b) 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
(c) The transverse shear and bending moments for most points of interest can readily be 
taken straight from the diagrams. For 1.5 < x < 3, the bending moment equations are 
parabolic, and are obtained by integrating the linear expressions for shear. For 
convenience, use a coordinate shift of x′ = x – 1.5. Then, for 0 < x′ < 1.5, 
( )
( )
2
2
0.125
0.125
2
At 0, 0.9375 0.5 0.125 0.9375
z
y z
y y
V x
x
M V dx x C
x M C M x x
′= −
′
′ ′= = − +
′ ′ ′= = = − ⇒ = − +
∫
 
( )
( )
2
2
1.949 0.6491 1.732 0.6491
1.125
1.732 0.6491
2
At 0, 0.9737 0.8662 0.125 0.9375
y
z
z z
V x x
M x x C
x M C M x x
′ ′= − + = − +
−
′ ′= + +
′ ′ ′= = = ⇒ = − − −
 
By programming these bending moment equations, we can find My, Mz, and their vector 
combination at any point along the beam. The maximum combined bending moment is 
found to be at x = 1.79 m, where M = 1.433 kN·m. The table below shows values at key 
locations on the shear and bending moment diagrams. 
 
 
 
 
 
 
 
 
Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 41/100 
x (m) Vz (kN) Vy (kN) V (kN) 
My 
(kN⋅m) 
Mz 
(kN⋅m) 
M 
(kN⋅m) 
0 –1.625 0.6491 1.750 0 0 0 
0.5− –1.625 0.6491 1.750 –0.8125 0.3246 0.8749 
1.5 –0.1250 0.6491 0.6610 0.9375 0.9737 1.352 
1.625 0 0.4327 0.4327 –0.9453 1.041 1.406 
1.875 0.2500 0 0.2500 –0.9141 1.095 1.427 
3− 1.375 –1.949 2.385 0 0 0 
 
 (d) The bending stress is obtained from Eq. (3-27), 
 
y Az A
x
z y
M zM y
I I
σ
−
= + 
 The maximum tensile bending stress will be at point A in the cross section of Prob. 3-34 
(a), where distances from the neutral axes for both bending moments will be maximum. 
At A, for Mz, yA = –37.5 mm, and for My, zA = –20 mm. 
3 3
6 4 6 440(75) 34(25) 1.36(10 ) mm 1.36(10 ) m
12 12z
I −= − = = 
3 3
5 4 7 425(40) 25(6)2 2.67(10 ) mm 2.67(10 ) m
12 12y
I −
 
= + = = 
 
 
It is apparent the maximum bending moment, and thus the maximum stress, will be in the 
parabolic section of the bending moment diagrams. Programming Eq. (3-27) with the 
bending moment equations previously derived, the maximum tensile bending stress is 
found at x = 1.77 m, where My = – 0.9408 kN·m, Mz = 1.075 kN·m, and σx = 100.1 MPa. 
Ans. 
_____________________________________________________________________________Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 42/100 
 
Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 43/100 
______________________________________________________________________ 
Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 44/100 
___________________________________________________________________ 
 
Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 45/100 
___________________________________________________________________________
 
 
3-52 
 
3
1
1 3
3
 
3
i i i
i
i i
T GL cT
GL c
θθ = ⇒ =
 
 
3 31
1 2 3
1
 .
3 i ii
GT T T T L c Ansθ
=
= + + = ∑
 
From Eq. (3-47), τ = Gθ1c 
G and θ1 are constant, therefore the largest shear stress occurs when c is a maximum. 
max 1 max .G c Ansτ θ=
 
_____________________________________________________________________________
 3-53 
(b) Solve part (b) first since the angle of twist per unit length is needed for part (a). 
 
( )max allow 12 6.89 82.7 MPaτ τ= = =
 
 
( )
( )
6
max
1 9
max
82.7 10
0.348 rad/m .
79.3 10 (0.003)
Ans
Gc
τθ = = =
 
(a) 
Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 46/100 
( )9 331 1 1
1
0.348(79.3) 10 (0.020)(0.002 )
1.47 N m .
3 3
GL cT Ansθ= = = ⋅
 ( )
( )
9 33
2 2 2
2
9 33
3 3 3
3
1 2 3
0.348(79.3) 10 (0.030)(0.003 )
7.45 N m .
3 3
0.348(79.3) 10 (0)(0 )
0 .
3 3
1.47 7.45 0 8.92 N m .
GL cT Ans
GL cT Ans
T T T T Ans
θ
θ
= = = ⋅
= = =
= + + = + + = ⋅
 
_____________________________________________________________________________
 
 
_____________________________________________________________________________
 3-55 
(b) Solve part (b) first since the angle of twist per unit length is needed for part (a). 
 
( )max allow 12 6.89 82.7 MPaτ τ= = =
 
 
( )
( )
6
max
1 9
max
82.7 10
0.348 rad/m .
79.3 10 (0.003)
Ans
Gc
τθ = = =
 
(a) 
( )9 331 1 1
1
0.348(79.3) 10 (0.020)(0.002 )
1.47 N m .
3 3
GL cT Ansθ= = = ⋅
 ( )
( )
9 33
2 2 2
2
9 33
3 3 3
3
1 2 3
0.348(79.3) 10 (0.030)(0.003 )
7.45 N m .
3 3
0.348(79.3) 10 (0.025)(0.002 )
1.84 N m .
3 3
1.47 7.45 1.84 10.8 N m .
GL cT Ans
GL cT Ans
T T T T Ans
θ
θ
= = = ⋅
= = = ⋅
= + + = + + = ⋅
 
Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 47/100 
_____________________________________________________________________________
 3-56 
(a) From Eq. (3-40), with two 2-mm strips,
 
( )( )( )( )
( )
6 22
max
max
80 10 0.030 0.002
3.08 N m
3 1.8 / ( / ) 3 1.8 / 0.030 / 0.002
2(3.08) 6.16 N m .
bcT
b c
T Ans
τ
= = = ⋅
+ +
= = ⋅
 
From the table on p. 116, with b/c = 30/2 = 15, α = β and has a value between 0.313 and 0.333. 
From Eq. (3-40), 
1 0.321
3 1.8 / (30 / 2)α ≈ =+
 
From Eq. (3-41), 
 
( )( )( )( )3 3 9
3.08(0.3) 0.151 rad .
0.321 0.030 0.002 79.3 10
6.16 40.8 N m .
0.151t
Tl Ans
bc G
Tk Ans
θ β
θ
= = =
= = = ⋅ 
 
From Eq. (3-40), with a single 4-mm strip, 
( ) ( )( )( )
( )
6 22
max
max
80 10 0.030 0.004
11.9 N m .
3 1.8 / ( / ) 3 1.8 / 0.030 / 0.004
bcT Ans
b c
τ
= = = ⋅
+ +
 
Interpolating from the table on p. 116, with b/c = 30/4 = 7.5,
 
 
7.5 6 (0.307 0.299) 0.299 0.305
8 6
β −= − + =
−
 
From Eq. (3-41) 
( )( )( )( )3 3 9
11.9(0.3) 0.0769 rad .
0.305 0.030 0.004 79.3 10
11.9 155 N m .
0.0769t
Tl Ans
bc G
Tk Ans
θ β
θ
= = =
= = = ⋅
 
(b) From Eq. (3-47), with two 2-mm strips, 
 
( )( )( )( )2 62
max
0.030 0.002 80 10
3.20 N m
3 3
2(3.20) 6.40 N m .
LcT
T Ans
τ
= = = ⋅
= = ⋅
 
Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 48/100 
( )( )( )( )3 3 9
3 3(3.20)(0.3) 0.151 rad .
0.030 0.002 79.3 10
6.40 0.151 42.4 N m .t
Tl Ans
Lc G
k T Ans
θ
θ
= = =
= = = ⋅
 
 
 From Eq. (3-47), with a single 4-mm strip, 
( ) ( ) ( ) ( )2 62
max
0.030 0.004 80 10
12.8 N m .
3 3
LcT Ansτ= = = ⋅
 
( )( )( )( )3 3 9
3 3(12.8)(0.3) 0.0757 rad .
0.030 0.004 79.3 10
12.8 0.0757 169 N m .t
Tl Ans
Lc G
k T Ans
θ
θ
= = =
= = = ⋅
 
The results for the spring constants when using Eq. (3-47) are slightly larger than when using 
Eq. (3-40) and Eq. (3-41) because the strips are not infinitesimally thin (i.e. b/c does not equal 
infinity). The spring constants when considering one solid strip are significantly larger (almost 
four times larger) than when considering two thin strips because two thin strips would be able 
to slip along the center plane. 
_____________________________________________________________________________
 3-57 
(a) Obtain the torque from the given power and speed using Eq. (3-44). 
 
(40000)9.55 9.55 152.8 N m
2500
HT
n
= = = ⋅
 
max 3
16Tr T
J d
τ
pi
= =
 
( )
( ) ( )
1 31 3
6
max
16 152.816 0.0223 m 22.3 mm .
70 10
Td Ans
piτ pi
    = = = = 
     
 (b) (40000)9.55 9.55 1528 N m
250
HT
n
= = = ⋅
 
( )( )
1 3
6
16(1528) 0.0481 m 48.1 mm .
70 10
d Ans
pi
 
 = = =
 
  
_____________________________________________________________________________
 
Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 49/100 
___________________________________________________________________________
 3-59 
( )( ) ( )6 33
max
max 3
50 10 0.0316
 265 N m
16 16
dT T
d
piτ pi
τ
pi
= ⇒ = = = ⋅
 
 
Eq. (3-44), ( )3265(2000) 55.5 10 W 55.5 kW .9.55 9.55TnH Ans= = = =
 
_____________________________________________________________________________
 3-60 
( )( )( )
( )( ) ( )
3 6 3
3
4 9
4
16
 110 10 0.020 173 N m
16 16
0.020 79.3 10 15
180
 
32 32(173)
1.89 m .
T T d
d
Tl d Gl
JG T
l Ans
pi pi
τ τ
pi
pi
pi
pi θθ
= ⇒ = = = ⋅
 
 
 
= ⇒ = =
=
 
_____________________________________________________________________________
 
___________________________________________________________________________
 3-62 
Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 50/100 
(a) 
4 4 4
max max max max
solid hollow
( )
 
16 16
o o i
o o
J d J d d
T T
r d r d
τ pi τ τ pi τ−
= = = =
 ( )
( )
44
solid hollow
4 4
solid
36
% (100%) (100%) (100%) 65.6% .
40
i
o
T T dT Ans
T d
−∆ = = = =
 
(b) ( )2 2 2solid hollow, o o iW kd W k d d= = −
 ( )
( )
22
solid hollow
2 2
solid
36
% (100%) (100%) (100%) 81.0% .
40
i
o
W W dW Ans
W d
−∆ = = = =
 
_____________________________________________________________________________
 3-63 
(a) 
( )444 max
max max max
solid hollow 16 16
d xdJ d JT T
r d r d
pi ττ pi τ τ
 
−
 
= = = =
 
4
4solid hollow
4
soli
( )% (100%) (100%) (100%) .
d
T T xdT x Ans
T d
−∆ = = =
 
(b) ( )( )22 2solid hollow W kd W k d xd= = −
 
( )2 2solid hollow
2
solid
% (100%) (100%) (100%) .xdW WW x Ans
W d
−∆ = = =
 
 Plot %∆T and %∆W versus x. 
 
 
The value of greatest difference in percent reduction of weight and torque is 25% andoccurs at 2 2x = . 
_____________________________________________________________________________
 
0
20
40
60
80
100
0 0,2 0,4 0,6 0,8 1
P
e
rc
e
n
t 
R
e
d
u
ct
io
n
x
Percent Reduction in 
Torque and Weight
Weight
Torque
difference
Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 51/100 
3-64 
(a) ( ) ( )( ) ( )
( )46
344
2.8149 104200 2
 120 10
32 0.70
dTc
J dd d
τ
pi
= ⇒ = =
 
−
  
( ) ( )
1 34
2
6
2.8149 10
6.17 10 m 61.7 mm
120(10 )
d −
 
 
= = =
 
 
 
 From Table A-17, the next preferred size is d = 80 mm. Ans. 
 di = 0.7d = 56 mm. The next preferred size smaller is di = 50 mm Ans. 
 
 (b) 
 
( )
( ) ( )
( )
( ) ( ) ( )4 4 44
4200 2 4200 0.050 2
30.8 MPa .
32 0.080 0.05032
i
i
dTc Ans
J d d
τ
pipi
= = = =
   
−−
  
 
_____________________________________________________________________________
 
 
3-65 
 
(1500)9.55 9.55 1433 N m
10
HT
n
= = = ⋅
 
( )
( )( )
1 3
1 3
3 6
16 143316 16
 = 0.045 m 45 mm 
80 10CC
T Td
d
τ
piτpi pi
 
   = ⇒ = = =    
 
 
From Table A-17, select 50 mm. Ans. 
(a) ( )( )( ) ( )
6
start 3
16 2 1433
117 10 Pa 117 MPa .
0.050
Ansτ
pi
= = =
 
 (b) Design activity 
_____________________________________________________________________________
 
___________________________________________________________________________
 3-67 For a square cross section with side length b, and a circular section with diameter d, 
2 2
square circular 4 2
A A b d b dpi pi= ⇒ = ⇒ =
 
Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 52/100 
 From Eq. (3-40) with b = c, 
( )
3
max 2 3 3 3square
1.8 1.8 23 3 (4.8) 6.896
/ 1
T T T T
bc b c b d d
τ
pi
    
= + = + = =     
     
 
 For the circular cross section, 
( )max 3 3circular 16 5.093T Td dτ pi= =
 
( )
( )
3max square
max circular 3
6.896
1.354
5.093
T
d
T
d
τ
τ
= =
 
 The shear stress in the square cross section is 35.4% greater. Ans. 
 (b) For the square cross section, from the table on p. 116, β = 0.141. From Eq. (3-41), 
square 43 4 411.50
0.141
2
Tl Tl Tl Tl
bc G b G d G
d G
θ β β pi
= = = =
 
 
 
 
 For the circular cross section, 
( ) 44 10.1932rd
Tl Tl Tl
GJ d GG d
θ
pi
= = =
 
4
4
11.50
1.129
10.19
sq
rd
Tl
d G
Tl
d G
θ
θ
= =
 
 The angle of twist in the square cross section is 12.9% greater. Ans. 
_____________________________________________________________________________
 
Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 53/100 
 
Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 54/100 
 
_________________________________________________________________________
 3-69 (a) 
 
( ) ( ) ( ) ( )
( )
( )
2 1
3
2 1 1 1
3
1 1
2
0.15
0 1800 270 (200) (125) 306 10 125 0.15
306 10 106.25 0 2880 N .
0.15 2880 432 N .
T T
T T T T T
T T Ans
T Ans
=
= = − + − = + −
− = ⇒ =
= =
∑
 
(b) 
 
0 3312(230) (510) 2070(810)
1794 N .
0 3312 1794 2070
3036 N .
O C
C
y O
O
M R
R Ans
F R
R Ans
= = + −
=
= = + + −
= −
∑
∑
 
(c) 
 
 
 
 
 
 
 
 
 
Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 55/100 
 
 
 
 
 (d) The maximum bending moment is at x = 230 mm, and is M = –698.3 N·m. Since the 
shaft rotates, each stress element will experience both positive and negative bending 
stress as it moves from tension to compression. The torque transmitted through the shaft 
from A to B is T = (1800 − 270)(0.200) = 306 N·m. For a stress element on the outer 
surface where the bending stress and the torsional stress are both maximum, 
 
 
( ) ( )33 332 698.332 263 10 Pa 263 MPa .(0.030)
Mc M Ans
I d
σ
pi pi
= = = = = 
( )63 316 16(306) 57.7 10 Pa 57.7MPa .(0.030)
Tr T Ans
J d
τ
pi pi
= = = = =
 
 (e) 
 
( ) ( )
( ) ( )
2 2
2 2
1 2
1
2
2 2
2 2
max
263 263
, 57.7
2 2 2 2
275 MPa .
12.1 MPa .
263 57.7 144 MPa .
2 2
x x
xy
x
xy
Ans
Ans
Ans
σ σ
σ σ τ
σ
σ
σ
τ τ
   
= ± + = ± +  
  
=
= −
   
= + = + =  
  
 
_____________________________________________________________________________
 
 
 
Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 56/100 
Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 57/100 
 
 
 
 
Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 58/100 
3-71 (a) 
 
( ) ( ) ( )
( )
2 1
2 1 1 1
1 1
2
0.15
0 300 45 (125) (150) 31 875 0.15 (150)
31 875 127.5 0 250 N mm .
0.15 250 37.5 N mm .
T T
T T T T T
T T Ans
T Ans
=
= = − + − = + −
− = ⇒ = ⋅
= = ⋅
∑
 
(b) 
 
o
o
o
o
0 345sin 45 (300) 287.5(700) (850)
150.7 N .
0 345cos 45 287.5 150.7
107.2 N .
0 345sin 45 (300) (850)
86.10 N .
0 345cos 45 86.10
O y C z
C z
z O z
O z
O z C y
C y
y O y
O y
M R
R Ans
F R
R Ans
M R
R Ans
F R
R
= = − −
= −
= = − + −
=
= = +
= −
= = + −
∑
∑
∑
∑
157.9 N .Ans= −
 
 (c) 
 
 
 
(d) From the bending moment diagrams, it is clear that the critical location is at A where 
both planes have the maximum bending moment. Combining the bending moments from 
the two planes, 
Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 59/100 
( ) ( )2 247.37 32.16 57.26 N mM = − + − = ⋅ 
The torque transmitted through the shaft from A to B is T = (300 − 45)(0.125) = 31.88 
N·m. For a stress element on the outer surface where the bending stress and the torsional 
stress are both maximum, 
 
( ) ( )63 332 57.2632 72.9 10 Pa 72.9 MPa .(0.020)
Mc M Ans
I d
σ
pi pi
= = = = = 
( )63 316 16(31.88) 20.3 10 Pa 20.3 MPa .(0.020)
Tr T Ans
J d
τ
pi pi
= = = = =
 
 (e) 
 
( ) ( )
( ) ( )
2 2
2 2
1 2
1
2
2 2
2 2
max
72.9 72.9
, 20.3
2 2 2 2
78.2 MPa .
5.27 MPa .
72.9 20.3 41.7 MPa .
2 2
x x
xy
x
xy
Ans
Ans
Ans
σ σ
σ σ τ
σ
σ
σ
τ τ
   
= ± + = ± +  
  
=
= −
   
= + = + =  
  
 
_____________________________________________________________________________
 
 
Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 60/100 
 
 
_____________________________________________________________________________
 
 
Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 61/100 
3-73 
(a) 
 
0 11000(cos 20º )(300) (cos 25º )(150)
22 810 N .
B
B
T F
F Ans
= = − +
=
∑
 
(b) 
 
0 11000(sin 20º )(400) 22 810(sin 25º )(750) (1050)
8319 N .
O z C y
C y
M R
R Ans
= = − − +
=
∑
 
 
0 11000(sin 20º ) 22 810sin(25º ) 8319
5083 N .
0 11 000(cos 20º )(400) 22 810(cos 25º )(750) (1050)
10 830 N .
0 11 000(cos 20º ) 22 810(cos25º ) 10 830
494 N .
y O y
O y
O y C z
C z
z O z
O z
F R
R Ans
M R
R Ans
F R
R Ans
= = − − +
=
= = − −
= −
= = − + −
=
∑
∑
∑
 
(c) 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
(d) From the bending moment diagrams, it is clear that the critical location is at B where 
both planes have the maximum bending moment. Combining the bending moments from 
the two planes, 
( ) ( )2 22496 3249 4097 N mM = + = ⋅ 
The torque transmitted through the shaft from A to B is 
 ( )( )11000cos 20º 0.3 3101 N mT = = ⋅ . 
For a stress element on the outer surface where the bending stress and the torsional stress 
are both maximum, 
Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 62/100 
 
( ) ( )63 332 409732 333.9 10 Pa 333.9 MPa .(0.050)
Mc M Ans
I d
σ
pi pi
= = = = = 
( )63 316 16(3101) 126.3 10 Pa 126.3 MPa .(0.050)
Tr T Ans
J d
τ
pi pi
= = = = =
 
 (e) 
 
( ) ( )
( ) ( )
2 2
2 2
1 2
1
2
2 2
2 2
max
333.9 333.9
, 126.3
2 2 2 2
376 MPa .
42.4 MPa .
333.9 126.3 209 MPa .
2 2
x x
xy
x
xy
Ans
Ans
Ans
σ σ
σ σ τ
σ
σ
σ
τ τ
   
= ± + = ± +  
  
=
= −
   
= + = + =  
  
 
_____________________________________________________________________________
 
__
Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 63/100 
______________________________________________________________________
 
Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 64/100 
 
 
Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 65/100 
 
 
_____________________________________________________________________________
 
 
Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 66/100 
Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 67/100 
_________________________________________________________________________
 
 
3-77 
 
100 1600 N
/ 2 0.125 / 2t
TF
c
= = = 
 
( ) ( )
( ) ( )
1600 tan 20 582.4 N
2 1600 0.250 2 200 N m
200 2667 N
2 0.150 2
n
C t
C
F
T F b
T
P
a
= =
= = = ⋅
= = =
( ) 0
450 582.4(325) 2667(75) 0
865.1 N
A z
Dy
Dy
M
R
R
=
− − =
=
∑
 
( ) 0 450 1600(325) 1156 NA Dz DzyM R R= = − + ⇒ =∑ 
0 865.1 582.4 2667 2384 Ny Ay AyF R R= = + − − ⇒ =∑
0 1156 1600 444 Nz Az AzF R R= = + − ⇒ =∑ 
 
Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 68/100 
AB The maximum bending moment will either be at B or C. If this is not obvious, sketch 
the shear and bending moment diagrams. We will directly obtain the combined moments 
from each plane. 
 
2 2 2 2
2 2 2 2
0.075 2384 444 181.9 N m
0.125 865.1 1156 180.5 N m
y z
y z
B A A
C D D
M AB R R
M CD R R
= + = + = ⋅
= + = + = ⋅
 
The stresses at B and C are almost identical, but the maximum stresses occur at B. Ans. 
 
( ) ( )
( ) ( )
6
3 3
6
3 3
32 32(181.9) 68.6 10 Pa 68.6 MPa
0.030
16 16(200) 37.7 10 Pa 37.7 MPa
0.030
B
B
B
B
M
d
T
d
σ
pi pi
τ
pi pi
= = = =
= = = =
 
 
2 2
2 2
max
68.6 68.6 37.7 85.3 MPa .
2 2 2 2
B B
B Ans
σ σ
σ τ
   
= + + = + + =  
  
 
 
2 2
2 2
max
68.6 37.7 51.0 MPa .
2 2
B
B Ans
σ
τ τ
   
= + = + =  
  
 
_____________________________________________________________________________
 
 
3-78 
 
100 1600 N
/ 2 0.125 / 2t
TF
c
= = = 
 
( ) ( )
( ) ( )
1600 tan 20 582.4 N
2 1600 0.250 2 200 N m
200 2667 N
2 0.150 2
n
C t
C
F
T F b
T
P
a
= =
= = = ⋅
= = =
 
 
 
 
 
( )
( )
0 450 582.4(325) 420.6 N
0 450 1600(325) 2667(75) 711.1 N
0 420.6 582.4 161.8 N
0 711.1 1600 2667 
A Dy Dyz
A Dz Dzy
y Ay Ay
z Az
M R R
M R R
F R R
F R
= = − ⇒ =
= = − + − ⇒ =
= = + − ⇒ =
= = + − +
∑
∑
∑
∑ 1778 NAzR⇒ = −
 
 
Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 69/100 
The maximum bending moment will either be at B or C. If this is not obvious, sketch 
shear and bending moment diagrams. We will directly obtain the combined moments 
from each plane. 
 
( )22 2 2
2 2 2 2
0.075 161.8 1778 133.9 N m
0.125 420.6 711.1 103.3 N m
y z
y z
B A A
C D D
M AB R R
M CD R R
= + = + − = ⋅
= + = + = ⋅
 
The maximum stresses occur at B. Ans. 
 
( ) ( )
( ) ( )
6
3 3
6
3 3
32 32(133.9) 50.5 10 Pa 50.5 MPa
0.030
16 16(200) 37.7 10 Pa 37.7 MPa
0.030
B
B
B
B
M
d
T
d
σ
pi pi
τ
pi pi
= = = =
= = = =
 
 
2 2
2 2
max
50.5 50.5 37.7 70.6 MPa .
2 2 2 2
B B
B Ans
σ σ
σ τ
   
= + + = + + =  
  
 
 
2 2
2 2
max
50.5 37.7 45.4 MPa .
2 2
B
B Ans
σ
τ τ
   
= + = + =  
  
 
_____________________________________________________________________________
 
 
Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 70/100 
 
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Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 71/100 
 
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Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 73/100 
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Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 76/100 
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Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 77/100 
 
 
Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 78/100 
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Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 79/100 
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Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 80/100 
 
 
 
Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 81/100 
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Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 82/100 
Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions,Page 83/100 
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Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 84/100 
 
 
Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 85/100 
 
__________________________________________________________________________ 
 
3-91 As shown in Fig. 3-32, the maximum stresses occur at the inside fiber where r = ri. 
Therefore, from Eq. (3-50) 
 
 
2 2
,max 2 2 2
2 2
2 2
2 2
,max 2 2 2
1
 .
1 .
i i o
t
o i i
o i
i
o i
i i o
r i
o i i
r p r
r r r
r r
p Ans
r r
r p r
p Ans
r r r
σ
σ
 
= +  
−  
 +
=   
− 
 
= − = −  
−  
 
______________________________________________________________________________ 
 
3-92 If pi = 0, Eq. (3-49) becomes 
 
 
2 2 2 2
2 2
2 2
2 2 2
/
1
o o i o o
t
o i
o o i
o i
p r r r p r
r r
p r r
r r r
σ
− −
=
−
 
= − + 
−  
 
 The maximum tangential stress occurs at r = ri. So 
 
 
2
,max 2 2
2
 .
o o
t
o i
p r Ans
r r
σ = −
−
 
 For σr, we have 
 
2 2 2 2
2 2
2 2
2 2 2
/
1
o o i o o
r
o i
o o i
o i
p r r r p r
r r
p r r
r r r
σ
− −
=
−
 
= − 
−  
 
 So σr = 0 at r = ri. Thus at r = ro 
Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 86/100 
 
2 2 2
,max 2 2 2 .
o o i o
r o
o i o
p r r r p Ans
r r r
σ
 
−
= = − 
−  
 
______________________________________________________________________________ 
3-93 The force due to the pressure on half of the sphere is resisted by the stress that is 
distributed around the center plane of the sphere. All planes are the same, so 
 
 
( ) 2
av 1 2
/ 4( ) .
4
i i
t
i
p d pd Ans
d t t
pi
σ σ σ
pi
= = = =
 
 The radial stress on the inner surface of the shell is, σ3 = − p Ans.
 
______________________________________________________________________________ 
 
________________________________________________________________________ 
 
3-95 σt > σl > σr 
 τmax = (σt − σr)/2 at r = ri 
 
( )
2 2 2 2 2 2 2
max 2 2 2 2 2 2 2 2 2 2 2
6
max
6
max
1 1 1
2
( ) (25 4)10100 91.7 mm
25 10
100 91.7 8.3 mm .
i i o i i o i i o o i
o i i o i i o i i o i
i
i o
o i
r p r r p r r p r r p
r r r r r r r r r r r
p
r r
t r r Ans
τ
τ
τ
      
= + − − = =      
− − − −      
− −
⇒ = = =
= − = − =
 
______________________________________________________________________________ 
________________________________________________________________________ 
 
Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 87/100 
 
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3-98 From Eq. (3-49) with pi = 0, 
 
2 2
2 2 2
2 2
2 2 2
1
1
o o i
t
o i
o o i
r
o i
r p r
r r r
r p r
r r r
σ
σ
 
= − + 
−  
 
= − − 
−  
 
 
 σt > σl > σr, and since σt and σr are negative, 
 τmax = (σr − σt)/2 at r = ro 
 
 
( )
( )
2 2 2 2 2 2 2
max 2 2 2 2 2 2 2 2 2 2 2
6
max
6
max
1 1 1
2
25 10
100 92.8 mm( ) 25 4 10
100 92.8 7.2 mm .
o o i o o i o o i i o
o i o o i o o i o o i
i o
o
o i
r p r r p r r p r r p
r r r r r r r r r r r
r r
p
t r r Ans
τ
τ
τ
      
= − − + + = =      
− − − −      
⇒ = = =
+ +
= − = − =
 
______________________________________________________________________________ 
 
Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 88/100 
 
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3-100 From Table A-20, Sy=490 MPa 
 From Eq. (3-49) with pi = 0, 
 
 
2 2
2 2 21
o o i
t
o i
r p r
r r r
σ
 
= − + 
−  
 
 
 Maximum will occur at r = ri 
 
 
[ ]( )2 22 22
,max
,max 2 2 2 2
0.8( 490) 25 19( )2 82.8 MPa .
2 2(25 )
t o io o
t o
o i o
r rr p p Ans
r r r
σ
σ
− −
−
= − ⇒ = − = − =
−
 
______________________________________________________________________________ 
 
Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 89/100 
________________________________________________________________________ 
 
3-102 From Table A-20, Sy=490 MPa 
 From Eq. (3-50) 
 
 
2 2
2 2 21
i i o
t
o i
r p r
r r r
σ
 
= + 
−  
 
 
 Maximum will occur at r = ri 
 
 
( )
[ ]
2 22 2
,max 2 2 2 2 2
2 2 2 2
,max
2 2 2 2
1
( ) 0.8(490) (25 19 )
105 MPa .(25 19 )
i o ii i o
t
o i i o i
t o i
i
o i
p r rr p r
r r r r r
r r
p Ans
r r
σ
σ
+ 
= + = 
− − 
− −
⇒ = = =
+ +
 
______________________________________________________________________________ 
________________________________________________________________________ 
 
Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 90/100 
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Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 91/100 
__
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__
__ 
 
 
Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 92/100 
_________________________________________________________________ 
 
3-107 ω = 2pi (2000)/60 = 209.4 rad/s, ρ = 3320 20 kg/m3, ν = 0.24, ri = 0.01 m, ro = 0.125 m 
 
 Using Eq. (3-55) 
 
( ) ( )2 22 2 2 63 0.24 1 3(0.24)3320(209.4) 0.01 (0.125) (0.125) 0.01 (10)
8 3 0.24
1.85 MPa .
t
Ans
σ −
+ +   
= + + −   +   
=
 
______________________________________________________________________________ 
 
Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 93/100 
________________________________________________________________________ 
________________________________________________________________________ 
 
3-110 ν = 0.292, E = 207 GPa, ri = 0, R = 25 mm, ro = 50 mm 
 
 Eq. (3-57), 
 ( )9 2 2 2 9 33 2207(10 ) (0.05 0.025 )(0.025 0) 10 3.105(10 ) (1)2(0.025) (0.05 0)p
δ δ− − −= = 
− 
 
 where p is in MPa and δ is in mm. 
 
 Maximum interference, 
 
max
1 [50.042 50.000] 0.021 mm .
2
Ansδ = − = 
 Minimum interference, 
 
min
1 [50.026 50.025] 0.0005 mm .
2
Ansδ = − = 
 
 From Eq. (1) 
 pmax = 3.105(103)(0.021) = 65.2 MPa Ans. 
 
Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 94/100 
 pmin = 3.105(103)(0.0005) = 1.55 MPa Ans. 
______________________________________________________________________________ 
 
________________________________________________________________________ 
 
3-112 ν = 0.292, E = 207 GPa, ri = 0, R = 25 mm, ro = 50 mm 
 
 Eq. (3-57), 
 ( )9 2 2 2 9 33 2207(10 ) (0.05 0.025 )(0.025 0) 10 3.105(10 ) (1)2(0.025) (0.05 0)p
δ δ− − −= = 
− 
 
 where p is in MPa and δ is in mm. 
 
 Maximum interference, 
 
max
1 [50.059 50.000] 0.0295 mm .
2
Ansδ = − = 
 Minimum interference, 
 
min
1 [50.043 50.025] 0.009 mm .
2
Ansδ = − =