Solucionário Elementos de Máquinas de Shigley 10 edição

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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. 
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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 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
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
−
 + +
= =  
+ + 
      +
+ ≈ − ≈ + − ≈ + −      
+      
 
= − ≈ − 
 
 
______________________________________________________________________________ 
 
 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 ( ) ( )33 25 1032 1000 1.006 in .2.5d
S d Ans
n d
σ
pi
= ⇒ = ⇒ = 
 Table A-17: d = 141 in Ans. 
 Factor of safety: 
( )
( )
( )
3
3
25 10
4.79 .
32 1000
1.25
S
n Ans
σ
pi
= = =
 
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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 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 MED, 10th edition 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 
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 
Σ 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
=
−
 −
= = = 
− − 
∑
 
______________________________________________________________________________ 
 
1-15 122.9 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 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 
Σ 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
µ
σ
− − −
= = = 
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. 
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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 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)10
sy
T nd Ans
Spi pi
   
= = = =       
 
______________________________________________________________________________ 
 
1-19 Obtain the coefficients of variance for stress and strength 
 
ˆ ˆ 5 0.09231
65
PC
Pσ
σ
σ
σ σ
µ
= = = = 
 
 
ˆ
ˆ 6.59 0.06901
95.5
ySS
S
S y
C
S
σσ
µ
= = = =
 
 
 (a) 1.2n = 
 
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 Cσ
− −
= − = − = −
+ +
 
 Interpolating Table A-10, 
 1.61 0.0537 
 1.6127 Φ ⇒ Φ = 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
pi
σ pi pi pi
= = = ⇒ = = = 
 
 (b) 1.5n = 
 
 ( )2 2 2
1.5 1 3.605
1.5 0.06901 0.09231
z
−
= − = −
+
 
 
 3.6 0.000159 
 3.605 Φ ⇒ Φ = 0.00015645 
 3.7 0.000108 
 
 R = 1 − 0.00015645 = 0.9998 Ans. 
 
 
( )
( )
4 65 1.54 1.140 in .
95.5y
Pnd Ans
Spi pi
= = = 
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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= + = + = 
 
 
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
= = = = 
Shigley’s MED, 10th edition Chapter 1 Solutions, Page 8/12 
 
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 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 in 
 allt 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 to d . Therefore, 
 
 d = 6.520 + 0.008 = 6.528 in Ans. 
 
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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, ( ) 31.500 1.875 3.000 8.4375 inV = = 
 
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 yields 
8.4735
8.4015
V ≈ in, whereas, exact is 
8.473551..
8.401551..
V = in 
______________________________________________________________________________ 
 
1-23 
 
 
 wmax = 0.05 in, wmin = 0.004 in 
 
0.05 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. 
 
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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. 
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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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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. 
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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. 
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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. 
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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. 
 
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) 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. 
 
 (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) σ = M /Z = 1770/0.934 = 1895 psi = 1.90 kpsi Ans. 
 
 (b) σ = 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) θ = Tl /GJ = 9 740(9.85)/[11.3(106)(pi /32)1.004] = 8.648(10−2) rad = 4.95° Ans. 
 
______________________________________________________________________________ 
 
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. 
______________________________________________________________________________ 
 
1-33 
Shigley’s MED, 10th edition Chapter 1 Solutions, Page 12/12 
 (a) τ =F /A = 2 700/[pi (0.750)2/4] = 6110 psi = 6.11 kpsi Ans. 
 
 (b) σ = 32Fa/pi d 3 = 32(180)31.5/[pi (1.25)3] = 29 570 psi = 29.6 kpsi Ans. 
 
 (c) Z =pi (do4 − di4)/(32 do) = pi (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 
 
( )
( )
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
= = ⋅ 
______________________________________________________________________________ 
 
2-4 
 AISI 1018 CD steel: Table A-5 
 
( ) ( )
6
630.0 10 106 10 in .
0.282
E Ans
γ
= = 
 2011-T6 aluminum: Table A-5 
 
( ) ( )
6
610.4 10 106 10 in .
0.098
E Ans
γ
= = 
 
Shigley’s MED, 10th edition Chapter 2 Solutions, Page 2/22 
 Ti-6Al-6V titanium: Table A-5 
 
( ) ( )
6
616.5 10 103 10 in .
0.160
E Ans
γ
= =
 
 No. 40 cast iron: Table A-5 
 
( ) ( )
6
614.5 10 55.8 10 in .
0.260
E Ans
γ
= = 
______________________________________________________________________________ 
 
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
−
= = 
 The percent difference from the value in Table A-5 is 
 
 
0.208 0.211 0.0142 1.42 percent .
0.211
Ans− = − = − 
______________________________________________________________________________ 
 
2-6 (a) A0 = pi (0.503)2/4 = 0.1987 in2, σ = Pi / A0 
 
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 1l 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 
 
 σ = 30.5(106)∫ − 61 000 (1) 
 The equation for the line between data points 8 and 9 is 
 σ = 7.60(105)∫ + 42 900 (2) 
 Solving Eqs. (1) and (2) simultaneously yields σ = 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 0.1077100 100 45.8 % .
0.1987
fA AR Ans
A
−
−
= = = 
Data Point Pi ∆l, Ai ∫ σ 
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 
Shigley’s MED, 10th edition Chapter 2 Solutions, Page 4/22 
 
 (a) Linear range 
 
 
 (b) Offset yield 
 
 
 (c) Complete range 
y = 3.05E+07x - 1.06E+01
0
10000
20000
30000
40000
50000
0.000 0.001 0.001 0.002
S
tr
e
ss
 (
p
si
)
Strain
Series1
Linear (Series1)
0
5000
10000
15000
20000
25000
30000
35000
40000
45000
50000
0.000 0.002 0.004 0.006 0.008 0.010 0.012 0.014
S
tr
e
ss
 (
p
si
)
Strain
Y
0
10000
20000
30000
40000
50000
60000
70000
80000
90000
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
S
tr
e
ss
 (
p
si
)
Strain
U
Shigley’s MED, 10th edition Chapter 2 Solutions, Page 5/22 
 (c) The material is ductile since there is a large amount of deformation beyond yield. 
 
 (d) The closest material to the values of Sy, Sut, and R is SAE 1045 HR with Sy = 45 kpsi, 
Sut = 82 kpsi, and R = 40 %. Ans. 
______________________________________________________________________________ 
 
2-7 To plot σ true vs.ε, the following equations are applied to the data. 
 true
P
A
σ = 
 Eq. (2-4) 
 
0
0
ln for 0 0.0028 in (0 8400 lbf )
ln for 0.0028 in ( 8400 lbf )
l l P
l
A l P
A
ε
ε
= ≤ ∆ ≤ ≤ ≤
= ∆ > >
 
 where 
2
2
0
(0.503) 0.1987 in
4
A pi= =
 
 The results are summarized in the table below and plotted on the next page. The last 5 
points of data are used to plot log σ vs log ε 
 
 The curve fit gives m = 0.2306 
log σ0 = 5.1852 ⇒ σ0 = 153.2 kpsi Ans. 
 
 For 20% cold work, Eq. (2-14) and Eq. (2-17) give, 
 
A = A0 (1 – W) = 0.1987 (1 – 0.2) = 0.1590 in2 
 
 
0
0.2306
0
0.1987ln ln 0.2231
0.1590
Eq. (2-18): 153.2(0.2231) 108.4 kpsi .
Eq. (2-19), with 85.6 from Prob. 2-6, 
85.6 107 kpsi .
1 1 0.2
m
y
u
u
u
A
A
S Ans
S
SS Ans
W
ε
σ ε
′
= = =
= = =
=
′ = = =
− −
 
 
Shigley’s MED, 10th edition Chapter 2 Solutions, Page 6/22 
P ∆l A ε σtrue log ε log σtrue 
0 0 0.198 7 0 0
1 000 0.000 4 0.198 7 0.000 2 5 032.71 -3.699 3.702 
2 000 0.000 6 0.198 7 0.000 3 10 065.4 -3.523 4.003 
3 000 0.001 0 0.198 7 0.000 5 15 098.1 -3.301 4.179 
4 000 0.001 3 0.198 7 0.000 65 20 130.9 -3.187 4.304 
7 000 0.002 3 0.198 7 0.001 15 35 229 -2.940 4.547 
8 400 0.002 8 0.198 7 0.001 4 42 274.8 -2.854 4.626 
8 800 0.198 4 0.001 51 44 354.8 -2.821 4.647 
9 200 0.197 8 0.004 54 46 511.6 -2.343 4.668 
9 100 0.196 3 0.012 15 46 357.6 -1.915 4.666 
13 200 0.192 4 0.032 22 68 607.1 -1.492 4.836 
15 200 0.187 5 0.058 02 81 066.7 -1.236 4.909 
17 000 0.156 3 0.240 02 108 765 -0.620 5.036 
16 400 0.130 7 0.418 89 125 478 -0.378 5.099 
14 800 0.107 7 0.612 45 137 419 -0.213 5.138 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Shigley’s MED, 10th edition Chapter 2 Solutions, Page 7/22 
 
 
______________________________________________________________________________ 
 
2-8 Tangent modulus at σ = 0 is( ) ( )
6
3
5000 0 25 10 psi
0.2 10 0
E σ
−
∆ −
= ≈ =
∆
−
ò
 Ans. 
 
 
 At σ = 20 kpsi 
 
Shigley’s MED, 10th edition 
 
( )
( )(20
26 19 10
1.5 1 10
E
−
≈ =
−
 
∫ (10-3) σ (kpsi) 
0 0 
0.20 5 
0.44 10 
0.80 16 
1.0 19 
1.5 26 
2.0 32 
2.8 40 
3.4 46 
4.0 49 
5.0 54 
 
 
______________________________________________________________________________
 
2-9 W = 0.20, 
 (a) Before cold working: 
kpsi, σ0 = 90.0 kpsi, m 
 After cold working: Eq. (2
 Eq. (2-14), 0
1 1 0.20i
A
A W
= = =
− −
 Eq. (2-17), ln ln1.25 0.223i uε ε= = = <
 Eq. (2-18), y iS Ansσ ε′ = = =
 Eq. (2-19), 
1 1 0.20u
S Ans′ = = =
− −
 
 (b) Before: 49.5
32
u
y
S
S
= =
 Lost most of its ductility
______________________________________________________________________________
 
2-10 W = 0.20, 
 (a) Before cold working: AISI 1212 HR steel. Table A
σ0 = 110 kpsi, m = 0.24, 
 After cold working: Eq. (2
 Chapter 2
)( )
( ) ( )
3
6
3
26 19 10
14.0 10 psi
1.5 1 10−
≈ = Ans. 
 
______________________________________________________________________________
) Before cold working: Annealed AISI 1018 steel. Table A-22, Sy
 =
 
0.25, ∫f = 1.05 
After cold working: Eq. (2-16), ∫u = m = 0.25 
1 1 1.25
1 1 0.20A W
= = =
− −
 
0ln ln1.25 0.223i u
i
A
A
ε ε= = = < 
( )0.250 90 0.223 61.8 kpsi .my iS Ansσ ε= = = 93% increase 
49.5 61.9 kpsi .
1 1 0.20
uSS Ans
W
= = =
− −
 25% increase 
49.5 1.55
32
= =
 After: 61.9 1.00
61.8
u
y
S
S
′
= =
′
 
ductility. 
______________________________________________________________________________
) Before cold working: AISI 1212 HR steel. Table A-22, Sy = 28 kpsi, 
0.24, ∫f = 0.85 
q. (2-16), ∫u = m = 0.24 
Chapter 2 Solutions, Page 8/22 
______________________________________________________________________________ 
y = 32 kpsi, Su = 49.5 
93% increase Ans. 
25% increase Ans. 
 Ans. 
______________________________________________________________________________ 
= 28 kpsi, Su = 61.5 kpsi, 
Shigley’s MED, 10th edition Chapter 2 Solutions, Page 9/22 
 Eq. (2-14), 0 1 1 1.25
1 1 0.20i
A
A W
= = =
− −
 
 Eq. (2-17), 0ln ln1.25 0.223i u
i
A
A
ε ε= = = < 
 Eq. (2-18), ( )0.240 110 0.223 76.7 kpsi .my iS Ansσ ε′ = = = 174% increase Ans. 
 Eq. (2-19), 61.5 76.9 kpsi .
1 1 0.20
u
u
SS Ans
W
′ = = =
− −
 25% increase Ans. 
 
 (b) Before: 61.5 2.20
28
u
y
S
S
= =
 After: 76.9 1.00
76.7
u
y
S
S
′
= =
′
 Ans. 
 Lost most of its ductility. 
______________________________________________________________________________ 
 
2-11 W = 0.20, 
 (a) Before cold working: 2024-T4 aluminum alloy. Table A-22, Sy = 43.0 kpsi, Su = 
64.8 kpsi, σ0 = 100 kpsi, m = 0.15, ∫f = 0.18 
 After cold working: Eq. (2-16), ∫u = m = 0.15 
 Eq. (2-14), 0 1 1 1.25
1 1 0.20i
A
A W
= = =
− −
 
 Eq. (2-17), 0ln ln1.25 0.223i f
i
A
A
ε ε= = = > Material fractures. Ans. 
______________________________________________________________________________ 
 
2-12 For HB = 275, Eq. (2-21), Su = 3.4(275) = 935 MPa Ans. 
______________________________________________________________________________ 
 
2-13 Gray cast iron, HB = 200. 
 Eq. (2-22), Su = 0.23(200) − 12.5 = 33.5 kpsi Ans. 
 
 From Table A-24, this is probably ASTM No. 30 Gray cast iron Ans. 
______________________________________________________________________________ 
 
2-14 Eq. (2-21), 0.5HB = 100 ⇒ HB = 200 Ans. 
______________________________________________________________________________ 
 
 
Shigley’s MED, 10th edition Chapter 2 Solutions, Page 10/22 
2-15 For the data given, converting HB to Su using Eq. (2-21) 
 
HB Su (kpsi) Su2 (kpsi) 
230 115 
 
13225 
232 116 
 
13456 
232 116 
 
13456 
234 117 
 
13689 
235 117.5 
 
13806.25 
235 117.5 
 
13806.25 
235 117.5 
 
13806.25 
236 118 
 
13924 
236 118 
 
13924 
239 119.5 
 
14280.25 
ΣSu = 1172 ΣSu2 = 137373 
 
 Eq. (1-6) 
 
1172 117.2 117 kpsi .
10
u
u
S
S Ans
N
= = = ≈
∑
 
 
 Eq. (1-7), 
 
 
( )
10
2 2
2
1 137373 10 117.2 1.27 kpsi .
1 9u
u u
i
S
S NS
s Ans
N
=
−
−
= = =
−
∑
 
______________________________________________________________________________ 
 
2-16 For the data given, converting HB to Su using Eq. (2-22) 
 
HB Su (kpsi) Su2 (kpsi) 
230 40.4 
 
1632.16 
232 40.86 
 
1669.54 
232 40.86 
 
1669.54 
234 41.32 
 
1707.342 
235 41.55 
 
1726.403 
235 41.55 
 
1726.403 
235 41.55 
 
1726.403 
236 41.78 
 
1745.568 
236 41.78 
 
1745.568 
239 42.47 
 
1803.701 
ΣSu = 414.12 ΣSu2 = 17152.63 
 
Shigley’s MED, 10th edition Chapter 2 Solutions, Page 11/22 
 Eq. (1-6) 
 
414.12 41.4 kpsi .
10
u
u
S
S Ans
N
= = =
∑
 
 
 Eq. (1-7), 
 
( )
10
2 2
2
1 17152.63 10 41.4 1.20 .
1 9u
u u
i
S
S NS
s Ans
N
=
−
−
= = =
−
∑
 
______________________________________________________________________________ 
 
2-17 (a) Eq. (2-9) 
2
345.6 34.7 in lbf / in .
2(30)Ru Ans≈ = ⋅ 
 
 (b) A0 = pi(0.5032)/4 = 0.19871 in2 
 
 
P ∆L A (A0 / A) – 1 ∫ σ = P/A0 
0 0 0 0 
1 000 0.000 4 0.000 2 5 032. 
2 000 0.000 6 0.000 3 10 070 
3 000 0.001 0 0.000 5 15 100 
4 000 0.001 3 0.000 65 20 130 
7 000 0.002 3 0.001 15 35 230 
8 400 0.002 8 0.001 4 42 270 
8 800 0.003 6 0.001 8 44 290 
9 200 0.008 9 0.004 45 46 300 
9 100 0.196 3 0.012 28 0.012 28 45 800 
13 200 0.192 4 0.032 80 0.032 80 66 430 
15 200 0.187 5 0.059 79 0.059 79 76 500 
17 000 0.156 3 0.271 34 0.271 34 85 550 
16 400 0.130 7 0.520 35 0.520 35 82 530 
14 800 0.107 7 0.845 03 0.845 03 74 480 
 
 
 From the figures on the next page, 
 
( )
( ) ( )
( )
5
1
3 3
1 (43 000)(0.001 5) 45 000(0.004 45 0.001 5)
2
1 45 000 76 500 (0.059 8 0.004 45)
2
81 000 0.4 0.059 8 80 000 0.845 0.4
66.7 10 in lbf/in .
T i
i
u A
Ans
=
≈ = + −
+ + −
+ − + −
≈ ⋅
∑
 
Shigley’s MED, 10th edition 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 Chapter 2
 
Chapter 2 Solutions, Page 12/22 
Shigley’s MED, 10th edition Chapter 2 Solutions, Page 13/22 
 2-18, 2-19 These problems are for student research. No standard solutions are provided. 
______________________________________________________________________________ 
 
2-20 Appropriate tables: Young’s modulus and Density (Table A-5)1020 HR and CD (Table A-
20), 1040 and 4140 (Table A-21), Aluminum (Table A-24), Titanium (Table A-24c) 
 
 Appropriate equations: 
 For diameter, ( ) 2
4
/ 4 y y
F F FS d
A d S
σ
pi pi
= = = ⇒ = 
 
 Weight/length = ρA, Cost/length = $/in = ($/lbf) Weight/length, 
 Deflection/length = δ /L = F/(AE) 
 
 With F = 100 kips = 100(103) lbf, 
 
 
Material 
Young's 
Modulus Density 
Yield 
Strength Cost/lbf Diameter 
Weight/ 
length 
Cost/ 
length 
Deflection/ 
length 
units Mpsi lbf/in3 kpsi $/lbf in lbf/in $/in in/in 
 
 
 
1020 HR 30 0.282 30 0.27 2.060 0.9400 0.25 1.000E-03 
1020 CD 30 0.282 57 0.30 1.495 0.4947 0.15 1.900E-03 
1040 30 0.282 80 0.35 1.262 0.3525 0.12 2.667E-03 
4140 30 0.282 165 0.80 0.878 0.1709 0.145.500E-03 
Al 10.4 0.098 50 1.10 1.596 0.1960 0.22 4.808E-03 
Ti 16.5 0.16 120 7.00 1.030 0.1333 $0.93 7.273E-03 
 
 The selected materials with minimum values are shaded in the table above. Ans. 
______________________________________________________________________________ 
 
2-21 First, try to find the broad category of material (such as in Table A-5). Visual, magnetic, 
and scratch tests are fast and inexpensive, so should all be done. Results from these three 
would favor steel, cast iron, or maybe a less common ferrous material. The expectation 
would likely be hot-rolled steel. If it is desired to confirm this, either a weight or bending 
test could be done to check density or modulus of elasticity. The weight test is faster. 
From the measured weight of 7.95 lbf, the unit weight is determined to be 
 
 
3
2
7.95 lbf 0.281 lbf/in[ (1 in) / 4](36 in)
W
Al pi
= = =w
 
 
which agrees well with the unit weight of 0.282 lbf/in3 reported in Table A-5 for carbon 
steel. Nickel steel and stainless steel have similar unit weights, but surface finish and 
darker coloring do not favor their selection. To select a likely specification from Table 
A-20, perform a Brinell hardness test, then use Eq. (2-21) to estimate an ultimate strength 
Shigley’s MED, 10th edition Chapter 2 Solutions, Page 14/22 
of 0.5 0.5(200) 100 kpsiu BS H= = = . Assuming the material is hot-rolled due to the 
rough surface finish, appropriate choices from Table A-20 would be one of the higher 
carbon steels, such as hot-rolled AISI 1050, 1060, or 1080. Ans. 
______________________________________________________________________________ 
 
2-22 First, try to find the broad category of material (such as in Table A-5). Visual, magnetic, 
and scratch tests are fast and inexpensive, so should all be done. Results from these three 
favor a softer, non-ferrous material like aluminum. If it is desired to confirm this, either a 
weight or bending test could be done to check density or modulus of elasticity. The 
weight test is faster. From the measured weight of 2.90 lbf, the unit weight is determined 
to be 
 
3
2
2.9 lbf 0.103 lbf/in[ (1 in) / 4](36 in)
W
Al pi
= = =w
 
 
which agrees reasonably well with the unit weight of 0.098 lbf/in3 reported in Table A-5 
for aluminum. No other materials come close to this unit weight, so the material is likely 
aluminum. Ans. 
______________________________________________________________________________ 
 
2-23 First, try to find the broad category of material (such as in Table A-5). Visual, magnetic, 
and scratch tests are fast and inexpensive, so should all be done. Results from these three 
favor a softer, non-ferrous copper-based material such as copper, brass, or bronze. To 
further distinguish the material, either a weight or bending test could be done to check 
density or modulus of elasticity. The weight test is faster. From the measured weight of 
9 lbf, the unit weight is determined to be 
 
 
3
2
9.0 lbf 0.318 lbf/in[ (1 in) / 4](36 in)
W
Al pi
= = =w
 
 
which agrees reasonably well with the unit weight of 0.322 lbf/in3 reported in Table A-5 
for copper. Brass is not far off (0.309 lbf/in3), so the deflection test could be used to gain 
additional insight. From the measured deflection and utilizing the deflection equation for 
an end-loaded cantilever beam from Table A-9, Young’s modulus is determined to be 
 
( )
( )
33
4
100 24
17.7 Mpsi
3 3 (1) 64 (17 / 32)
FlE
Iy pi
= = =
 
 
which agrees better with the modulus for copper (17.2 Mpsi) than with brass (15.4 Mpsi). 
The conclusion is that the material is likely copper. Ans. 
______________________________________________________________________________ 
 
2-24 and 2-25 These problems are for student research. No standard solutions are provided. 
______________________________________________________________________________ 
 
2-26 For strength, σ = F/A = S ⇒ A = F/S 
Shigley’s MED, 10th edition Chapter 2 Solutions, Page 15/22 
 For mass, m = Alρ = (F/S) lρ 
 
 Thus, f 3(M ) = ρ /S , and maximize S/ρ (β = 1) 
 
 In Fig. (2-19), draw lines parallel to S/ρ 
 
 
 The higher strength aluminum alloys have the greatest potential, as determined by 
comparing each material’s bubble to the S/ρ guidelines. Ans. 
______________________________________________________________________________ 
 
2-27 For stiffness, k = AE/l ⇒ A = kl/E 
 For mass, m = Alρ = (kl/E) lρ =kl2 ρ /E 
 
 Thus, f 3(M) = ρ /E , and maximize E/ρ (β = 1) 
 
 In Fig. (2-16), draw lines parallel to E/ρ 
 
 
Shigley’s MED, 10th edition Chapter 2 Solutions, Page 16/22 
 
 From the list of materials given, tungsten carbide (WC) is best, closely followed by 
aluminum alloys. They are close enough that other factors, like cost or availability, would 
likely dictate the best choice. Polycarbonate polymer is clearly not a good choice 
compared to the other candidate materials. Ans. 
______________________________________________________________________________ 
 
2-28 For strength, 
 σ = Fl/Z = S (1) 
 
 where Fl is the bending moment and Z is the section modulus [see Eq. (3-26b), p. 104 ]. 
The section modulus is strictly a function of the dimensions of the cross section and has 
the units in3 (ips) or m3 (SI). Thus, for a given cross section, Z =C (A)3/2, where C is a 
number. For example, for a circular cross section, C = ( ) 14 pi − . Then, for strength, Eq. 
(1) is 
 
2/3
3/2
Fl FlS A
CA CS
 
= ⇒ =  
 
 (2) 
 
 
Shigley’s MED, 10th edition Chapter 2 Solutions, Page 17/22 
 For mass, 
2/3 2/3
5/3
2/3 
Fl F
m Al l l
CS C S
ρρ ρ     = = =     
     
 
 
 Thus, f 3(M) = ρ /S 2/3, and maximize S 2/3/ρ (β = 2/3) 
 
 In Fig. (2-19), draw lines parallel to S 2/3/ρ 
 
 
 
 
 
 From the list of materials given, a higher strength aluminum alloy has the greatest 
potential, followed closely by high carbon heat-treated steel. Tungsten carbide is clearly 
not a good choice compared to the other candidate materials. .Ans. 
______________________________________________________________________________ 
 
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 
Shigley’s MED, 10th edition Chapter 2 Solutions, Page 18/22 
 
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) 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 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. 
______________________________________________________________________________ 
 
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. 
______________________________________________________________________________ 
2-31 For strength, σ = F/A = S⇒ A = F/S 
Shigley’s MED, 10th edition Chapter 2 Solutions, Page 19/22 
 For mass, m = Alρ = (F/S) lρ 
 
 So, f 3(M ) = ρ /S, and maximize S/ρ . Thus, β = 1. Ans. 
______________________________________________________________________________ 
 
2-32 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 the circular cross section (see 
p.77), C = (4pi)−1. Another 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    = =    
   
 
 So, minimize ( )3 1/2f M E
ρ
= , or maximize
1/2EM
ρ
= . Thus, β = 1/2. Ans. 
______________________________________________________________________________ 
 
2-33 For strength, 
 σ = Fl/Z = S (1) 
 
 where Fl is the bending moment and Z is the section modulus [see Eq. (3-26b), p. 104 ]. 
The section modulus is strictly a function of the dimensions of the cross section and has 
the units in3 (ips) or m3 (SI). The area of the cross section has the units in2 or m2. Thus, for 
a given cross section, Z =C (A)3/2, where C is a number. For example, for a circular cross 
section, Z = pid 3/(32)and the area is A = pid 2/4. This leads to C = ( ) 14 pi − . So, with 
 Z =C (A)3/2, for strength, Eq. (1) is 
 
2/3
3/2
Fl FlS A
CA CS
 
= ⇒ =  
 
 (2) 
 
 
 For mass, 
2/3 2/3
5/3
2/3 
Fl F
m Al l l
CS C S
ρρ ρ     = = =     
     
 
 
 So, f 3(M) = ρ /S 2/3, and maximize S 2/3/ρ. Thus, β = 2/3. Ans. 
______________________________________________________________________________ 
 
Shigley’s MED, 10th edition Chapter 2 Solutions, Page 20/22 
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. 
______________________________________________________________________________ 
 
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 
 
 
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. 
 
Shigley’s MED, 10th edition Chapter 2 Solutions, Page 21/22 
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 0.004832507 
90 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. 
 
 
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 MED, 10th edition Chapter 2 Solutions, Page 22/22 
 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). 
Shigley’s MED, 10th edition Chapter 3 Solutions, Page 1/100 
Chapter 3 
 
 
 
3-1 
 
0OMΣ = 
 
18 6(100) 0BR − = 
 
33.3 lbf .BR Ans= 
 
 
0yFΣ = 
 
100 0O BR R+ − = 
 
66.7 lbf .OR Ans= 
 
33.3 lbf .C BR R Ans= = 
 
______________________________________________________________________________ 
 
3-2 
 Body AB: 
 
 
0xFΣ = Ax BxR R= 
 
 
0yFΣ = Ay ByR R= 
 
 
0BMΣ = (10) (10) 0Ay AxR R− = 
 
Ax AyR R= 
 
 Body OAC: 
 
 
0OMΣ = (10) 100(30) 0AyR − = 
 
300 lbf .AyR Ans= 
 
 
0xFΣ =
 
300 lbf .Ox AxR R Ans= − = − 
 
 
0yFΣ = 100 0Oy AyR R+ − = 
 
200 lbf .OyR Ans= − 
______________________________________________________________________________ 
 
 
 
 
 
 
 
 
 
Shigley’s MED, 10th edition 
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
= + =
= −
= + − =
= −
= + =
∑
∑
 
 
 Step 2: Find components of 
 
 
(
( )4
4
0
 400(4.5) 7.794 1.9 0
 305.4 N .
0 305.4 N
0 ( ) 400 N
C
D
x Cx
y Cy
M
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
 Chapter 3
1.39 kN .R Ans 
1.6 kN .R Ans 
______________________________________________________________________________
0 400cos30 0
 346.4 N
0 400 400sin 30 0
 200 N
346.4 200 400 N .R Ans
= + =
= + − =
= + =
o
o
 
Step 2: Find components of RC and RD on link 4 
)
4
4
 400(4.5) 7.794 1.9 0
 305.4 N .
0 305.4 N
0 ( ) 400 N
DR
R Ans
− − =
=
= − 
9 7.794(400cos30 )
4.5(400sin 30 ) 0− =
o
Chapter 3 Solutions, Page 2/100 
______________________________________________________________________________ 
Shigley’s MED, 10th edition 
Step 3: Find components of RC
 
( )
( )
( )
2
2
2
0
 305.4 346.4 0
 41 N
0
 200 N
x
Cx
Cx
y
Cy
F
R
R
F
R
=
+ − =
=
=
=
∑
∑
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
_____________________________________________________________________________________________________________________
 
 
 
 
 
 
 
 
 
 
 
 
 Chapter 3
C on link 2305.4 346.4 0+ − =
 
 Ans. 
_____________________________________________________________________________________________________________________
 
Chapter 3 Solutions, Page 3/100 
_____________________________________________________________________________________________________________________ 
Shigley’s MED, 10th edition 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 = − = 
_____________________________________________________________________________ 
 
3-6 
 
 
0yFΣ = 
 500 40(6) 740 lbf .OR Ans= + = 
 
0OMΣ = 
 500(8) 40(6)(17) 8080 lbf in .OM Ans= + = ⋅ 
 
 1 8080 740(8) 2160 lbf in .M Ans= − + = − ⋅ 
 
2 2160 240(6) 720 lbf in .M Ans= − + = − ⋅ 
 
3
1720 (240)(6) 0 checks!
2
M = − + = 
 
 
 
 
 
______________________________________________________________________________ 
 
Shigley’s MED, 10th edition 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 = − + = 
 
 
 
 
______________________________________________________________________________ 
 
3-8 
 Break at the hinge at B 
 
 Beam OB: 
 From symmetry, 
 1 200 lbf .BR V Ans= = 
 
 Beam BD: 
 0DMΣ = 
 2200(12) (10) 40(10)(5) 0R− + = 
 2 440 lbf .R Ans= 
 
 0yFΣ = 
 3200 440 40(10) 0R− + − + = 
 3 160 lbf .R Ans= 
 
 
 
 
 
Shigley’s MED, 10th edition Chapter 3 Solutions, Page 6/100 
 
 
 1 200(4) 800 lbf in .M Ans= = ⋅ 
 
2 800 200(4) 0 checks at hingeM = − = 
 
3 800 200(6) 400 lbf in .M Ans= − = − ⋅ 
 
4
1400 (240)(6) 320 lbf in .
2
M Ans= − + = ⋅ 
 
5
1320 (160)(4) 0 checks!
2
M = − = 
 
 
 
 
 
 
 
 
 
______________________________________________________________________________ 
 
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. 
______________________________________________________________________________ 
Shigley’s MED, 10th edition Chapter 3 Solutions, Page 7/100 
3-10 
1 2 1 0 0
1 0 1 1
1 2 2
500 8 40 14 40 20
500 8 40 14 40 20 (1)
500 8 20 14 20 20 (2)
at 20 in, 0, Eqs. (1) and (2) give
O O
O O
O O
q R x M x x x x
V R M x x x x
M R x M x x x
x V M
R
− − −
−
+
= − − − − − + −
= − − − − − + −
= − − − − − + −
= = =
( )
2
500 40 20 14 0 740 lbf .
(20) 500(20 8) 20(20 14) 0 8080 lbf in .
O O
O O O
R Ans
R M M Ans
− − − = ⇒ =
− − − − − = ⇒ = ⋅
 
 
0 8 : 740 lbf, 740 8080 lbf in
8 14 : 740 500 240 lbf
 740 8080 500( 8) 240 4080 lbf in
14 20 : 740 500 40( 14) 40 800 lbf
 740 8080
x V M x
x V
M x x x
x V x x
M x
≤ ≤ = = − ⋅
≤ ≤ = − =
= − − − = − ⋅
≤ ≤ = − − − = − +
= − −
2 2500( 8) 20( 14) 20 800 8000 lbf inx x x x− − − = − + − ⋅
 
 
 Plots of V and M are the same as in Prob. 3-6. 
______________________________________________________________________________ 
 
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. 
______________________________________________________________________________ 
 
Shigley’s MED, 10th edition Chapter 3 Solutions, Page 8/100 
3-12 
1 1 1 0 0 1
1 2 3
0 0 1 1 0
1 2 3
1 1 2 2 1
1 2 3
1
400 4 10 40 10 40 20 20
400 4 10 40 10 40 20 20 (1)
400 4 10 20 10 20 20 20 (2)
0 at 8 in 8 400(
q R x x R x x x R x
V R x R x x x R x
M R x x R x x x R x
M x R
− − − −
= − − + − − − + − + −
= − − + − − − + − + −
= − − + − − − + − + −
= = ∴ − 18 4) 0 200 lbf .R Ans− = ⇒ =
 
 
at x = 20+, V =M = 0. Applying Eqs. (1) and (2), 
2 3 2 3
2
2 2
200 400 40(10) 0 600
200(20) 400(16) (10) 20(10) 0 440 lbf .
 
R R R R
R R Ans
− + − + = ⇒ + =
− + − = ⇒ =
3 600 440 160 lbf .R Ans= − =
 
 
0 4 : 200 lbf, 200 lbf in
4 10 : 200 400 200 lbf,
 200 400( 4) 200 1600 lbf in
10 20 : 200 400 440 40( 10) 640 40 lbf
 200 400( 4)
x V M x
x V
M x x x
x V x x
M x x
≤ ≤ = = ⋅
≤ ≤ = − = −
= − − = − + ⋅
≤ ≤ = − + − − = −
= − − + ( )2 2440( 10) 20 10 20 640 4800 lbf inx x x x− − − = − + − ⋅
 
Plots of V and M are the same as in Prob. 3-8. 
______________________________________________________________________________ 
 
3-13 Solution depends upon the beam selected. 
______________________________________________________________________________ 
 
3-14 (a) Moment at center, 
 
( )
( )
2
2
2
2
2 2 2 2 4
c
c
l a
x
l l lM l a a
−
=
    
= − − = −    
     
w wl
 
At reaction, 2 2
r
M a= w 
 
a = 2.25, l = 10 in, w = 100 lbf/in 
 
( )2
100(10) 10 2.25 125 lbf in
2 4
100 2.25
253 lbf in .
2
c
r
M
M Ans
 
= − = ⋅ 
 
= = ⋅
 
(b) Optimal occurs when c rM M= 
Shigley’s MED, 10th edition 
 
2
2 4 2
l l a
a a al l − = ⇒ 
 
w w
Taking the positive root 
 
 
(2 21 4 0.25 2 1 0.207 .2 2a l l l l Ans = − + + = − =  
for l = 10 in, w = 100 lbf
 ( )min 100 2 2.07 214 lbfinM = = ⋅
______________________________________________________________________________
 
3-15 
(a) 
20 10 5 kpsi
2
C −= = 
20 10 15 kpsi
2
CD += =
2 215 8 17 kpsiR = + =
1 5 17 22 kpsiσ = + = 
2 5 17 12 kpsiσ = − = − 
11 8tan 14.04 cw
2 15p
φ −  = = 
 
1 17 kpsi
45 14.04 30.96 ccws
Rτ
φ
= =
= − =
o o o
 
 
(b) 
9 16 12.5 kpsi
2
C += = 
16 9 3.5 kpsi
2
CD −= = 
2 25 3.5 6.10 kpsiR = + =
1 12.5 6.1 18.6 kpsiσ = + =
2 12.5 6.1 6.4 kpsiσ = − =
11 5tan 27.5 ccw
2 3.5p
φ −  = = 
 
1 6.10 kpsi
45 27.5 17.5 cws
Rτ
φ
= =
= − =
o o o
 Chapter 3
2 20.25 0a a al l⇒ + − =
 
 
) ( )2 24 0.25 2 1 0.207 .2 2la l l l l Ans = − + + = − =  
 
= 100 lbf, a = 0.207(10) = 2.07 in 
2100 2 2.07 214 lbf in= = ⋅
 
______________________________________________________________________________
15 kpsi 
15 8 17 kpsi 
5 17 12 kpsi 
 
tan 14.04 cwo
 
45 14.04 30.96 ccwo o o
 
12.5 kpsi 
3.5 kpsi 
5 3.5 6.10 kpsi 
12.5 6.1 18.6 kpsi
 
12.5 6.1 6.4 kpsi 
tan 27.5 ccwo 
45 27.5 17.5 cwo o o 
Chapter 3 Solutions, Page 9/100 
______________________________________________________________________________ 
Shigley’s MED, 10th edition 
(c) 
 
2 2
1
2
24 10 17 kpsi
2
24 10 7 kpsi
2
7 6 9.22 kpsi
17 9.22 26.22 kpsi
17 9.22 7.78 kpsi
C
CD
R
σ
σ
+
= =
−
= =
= + =
= + =
= − =
11 790 tan 69.7 ccw
2 6p
φ −  = + =  
  
o o
1 9.22 kpsi
69.7 45 24.7 ccws
Rτ
φ
= =
= − =
o o o
 
 
 
(d) 
 
 
2 2
1
2
12 22 5 kpsi
2
12 22 17 kpsi
2
17 12 20.81 kpsi
5 20.81 25.81 kpsi
5 20.81 15.81 kpsi
C
CD
R
σ
σ
− +
= =
+
= =
= + =
= + =
= − = −
11 1790 tan 72.39 cw
2 12p
φ −  = + =  
  
o o
1 20.81 kpsi 
72.39 45 27.39 cws
Rτ
φ
= =
= − =
 
 
______________________________________________________________________________
 
 Chapter 3
17 kpsi
7 kpsi
7 6 9.22 kpsi
17 9.22 26.22 kpsi
17 9.22 7.78 kpsi
 
 
 
90 tan 69.7 ccw  = + =  
  
o o
69.7 45 24.7 ccwo o o
 
5 kpsi
17 kpsi
17 12 20.81 kpsi
5 20.81 25.81 kpsi
5 20.81 15.81 kpsi
 
1 1790 tan 72.39 cw
2 12
  
= + =  
  
o o
 72.39 45 27.39 cwo 
______________________________________________________________________________
 
Chapter 3 Solutions, Page 10/100 
______________________________________________________________________________ 
Shigley’s MED, 10th edition 
3-16 
(a) 
2 2
1
2
8 7 0.5 MPa
2
8 7 7.5 MPa
2
7.5 6 9.60 MPa
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 −  = = 
 
1 8.078 MPa
45 10.9 34.1 ccws
Rτ
φ
= =
= − =
o o o
 
 
 Chapter 3
0.5 MPa
7.5 MPa
7.5 6 9.60 MPa
9.60 0.5 9.10 MPa
0.5 9.6 10.1 Mpa
 
90 tan 70.67 cwo o
 
70.67 45 25.67 cwo o o
 
7.5 MPa
7.5 3 8.078 MPa
1.5 8.078 9.58 MPa
1.5 8.078 6.58 MPa
 
tan 10.9 cwo
 
45 10.9 34.1 ccwo o o
 
 
Chapter 3 Solutions, Page 11/100 
Shigley’s MED, 10th edition 
 (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 −  = = 
 
1 9.71 MPa
45 27.75 17.25 cws
Rτ
φ
= =
= − =
o o o
 
 
 
______________________________________________________________________________
 
 Chapter 3
8 7 10.63 MPa
4 10.63 14.63 MPa
4 10.63 6.63 MPa
 
90 tan 69.4 ccw  = + =  
  
o o
69.4 45 24.4 ccwo o o
 
5.5 MPa
5.5 8 9.71 MPa
0.5 9.71 10.21 MPa
0.5 9.71 9.21 MPa
 
tan 27.75 ccwo
 
45 27.75 17.25 cwo o o 
______________________________________________________________________________
 
Chapter 3 Solutions, Page 12/100 
______________________________________________________________________________ 
Shigley’s MED, 10th edition 
3-17 
(a) 
2 2
1
2
12 6 9 kpsi
2
12 6 3 kpsi
2
3 4 5 kpsi
5 9 14 kpsi
9 5 4 kpsi
C
CD
R
σ
σ
+
= =
−
= =
= + =
= + =
= − =
 
 
 
11 4tan 26.6 ccw
2 3p
φ −  = =  
1 5 kpsi
45 26.6 18.4 ccws
Rτ
φ
= =
= − =
o o o
 
 
(b) 
2 2
1
2
30 10 10 kpsi
2
30 10 20 kpsi
2
20 10 22.36 kpsi
10 22.36 32.36 kpsi
10 22.36 12.36 kpsi
C
CD
R
σ
σ
−
= =
+
= =
= + =
= + =
= − = −
 
 
11 10tan 13.28 ccw
2 20p
φ −  = = 
 
1 22.36 kpsi
45 13.28 31.72 cws
Rτ
φ
= =
= − =
o o o
 
 
 
 
 Chapter 3
3 4 5 kpsi
tan 26.6 ccwo
 
45 26.6 18.4 ccwo o o
 
20 kpsi
20 10 22.36 kpsi
10 22.36 32.36 kpsi
10 22.36 12.36 kpsi
 
tan 13.28 ccwo
 
45 13.28 31.72 cwo o o
 
 
Chapter 3 Solutions, Page 13/100 
Shigley’s MED, 10th edition 
(c) 
 
2 2
1
2
10 18 4 kpsi
2
10 18 14 kpsi
2
14 9 16.64 kpsi
4 16.64 20.64 kpsi
4 16.64 12.64 kpsi
C
CD
R
σ
σ
− +
= =
+
= =
= + =
= + =
= − = −
 
 
11 1490 tan 73.63 cw
2 9p
φ −  = + =  
  
o o
1 16.64 kpsi 
73.63 45 28.63 cws
Rτ
φ
= =
= − =
 
 
 (d) 
 
 
2 2
1
2
9 19 14 kpsi
2
19 9 5 kpsi
2
5 8 9.434 kpsi
14 9.43 23.43 kpsi
14 9.43 4.57 kpsi
C
CD
R
σ
σ
+
= =
−
= =
= + =
= + =
= − =
11 590 tan 61.0 cw
2 8p
φ −  = + =  
  
o o
1 9.34 kpsi
61 45 16 cws
Rτ
φ
= =
= − =
o o o
 
 
______________________________________________________________________________
 
 
 Chapter 3
4 kpsi
14 kpsi
14 9 16.64 kpsi
4 16.64 20.64 kpsi
4 16.64 12.64 kpsi
 
1 1490 tan 73.63 cw  = + =  
  
o o
 
16.64 kpsi 
73.63 45 28.63 cwo
 
5 8 9.434 kpsi
14 9.43 23.43 kpsi
14 9.43 4.57 kpsi
 
90 tan 61.0 cw  = + =  
  
o o
 
61 45 16 cw 
______________________________________________________________________________
 
Chapter 3 Solutions, Page 14/100 
______________________________________________________________________________ 
Shigley’s MED, 10th edition 
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 54.1MPa
2
39.1 19.6 MPa
2
69.1 34.6 MPa
2
τ
τ
τ
+
= =
= =
= =
 
 Chapter 3
55 MPa
25 MPa
25 20 32.02 MPa
55 32.02 22.98 23.0 MPa
55 32.0 87.0 MPa
= − + = − = −
 
1 2 2 3 1 3
23 8711.5 MPa, 32.0 MPa, 43.5 MPa
2 2
τ τ τ= = = = =
 
15 MPa
45 MPa
45 30 54.1 MPa
15 54.1 39.1 MPa
15 54.1 69.1 MPa
 
54.1 MPa
19.6 MPa
34.6 MPa
 
Chapter 3 Solutions, Page 15/100 
Shigley’s MED, 10th edition 
(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
+ −
= = = = =
______________________________________________________________________________
 
3-19 
(a) 
Since there are no shear stresses on the 
stress element, the stress element 
already represents principal stresses.
1
2
3
10 kpsi
0 kpsi
4 kpsi
x
y
σ σ
σ
σ σ
= =
=
= = −
 
1 3
1 2
2 3
10 ( 4) 7 kpsi
2
10 5 kpsi
2
0 ( 4) 2 kpsi
2
τ
τ
τ
− −
= =
= =
− −
= =
 Chapter 3
20 MPa
20 20 28.3 MPa
20 28.3 48.3 MPa
20 28.3 8.3 MPa
 
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
+ −
= = = = =
25 30 39.1 MPa
25 39.1 64.1 MPa
25 39.1 14.1 MPa
 
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
+ −
= = = = =
______________________________________________________________________________
Since there are no shear stresses on the 
stress element, the stress element 
already represents principal stresses. 
7 kpsi
2 kpsi
 
Chapter 3 Solutions, Page 16/100 
39.1 MPa, 28.3 MPa, 10.9 MPa
 
42.1 MPa, 39.1 MPa, 2.95 MPa
 
______________________________________________________________________________ 
Shigley’s MED, 10th edition 
(b) 
2 2
1
2 3
0 10 5 kpsi
2
10 0 5 kpsi
2
5 4 6.40 kpsi
5 6.40 11.40 kpsi
0 kpsi, 5 6.40 1.40 kpsi
C
CD
R
σ
σ σ
+
= =
−
= =
= + =
= + =
= = − = −
1 3 1 2 36.40 kpsi, 5.70 kpsi, 0.70 kpsiRτ τ τ= = = = = =
 
(c) 
2 2
1 2
3
2 8 5 kpsi
2
8 2 3 kpsi
2
3 4 5 kpsi
5 5 0 kpsi, 0 kpsi
5 5 10 kpsi
C
CD
R
σ σ
σ
− −
= = −
−
= =
= + =
= − + = =
= − − = −
 
1 3 1 2 2 3
10 5 kpsi, 0 kpsi, 5 kpsi
2
τ τ τ= = = =
 
(d) 
2 2
1
2
3
10 30 10 kpsi
2
10 30 20 kpsi
2
20 10 22.36 kpsi
10 22.36 12.36 kpsi
0 kpsi
10 22.36 32.36 kpsi
C
CD
R
σ
σ
σ
−
= = −
+
= =
= + =
= − + =
=
= − − = −
1 3 1 2 2 322.36 kpsi, 6.18 kpsi, 16.18 kpsτ τ τ= = = = =
______________________________________________________________________________
 
 
 
 
 Chapter 3
5 4 6.40 kpsi
5 6.40 11.40 kpsi
0 kpsi, 5 6.40 1.40 kpsi= = − = −
 
1 3 1 2 3
11.40 1.406.40 kpsi, 5.70 kpsi, 0.70 kpsi
2 2
τ τ τ= = = = = =
1 2
3 4 5 kpsi
5 5 0 kpsi, 0 kpsi
5 5 10 kpsi
σ σ= − + = = 
1 3 1 2 2 35 kpsi, 0 kpsi, 5 kpsiτ τ τ= = = =
 
10 kpsi
20 kpsi
20 10 22.36 kpsi
10 22.36 12.36 kpsi
10 22.36 32.36 kpsi
 
1 3 1 2 2 3
12.36 32.3622.36 kpsi, 6.18 kpsi, 16.18 kps
2 2
τ τ τ= = = = =
______________________________________________________________________________
Chapter 3 Solutions, Page 17/100 
6.40 kpsi, 5.70 kpsi, 0.70 kpsi
 
22.36 kpsi, 6.18 kpsi, 16.18 kpsi
 
______________________________________________________________________________ 
Shigley’s MED, 10th edition Chapter 3 Solutions, Page 18/100 
3-20 From Eq. (3-15), 
 
 
3 2 2 2 2
2 2 2
3
( 6 18 12) 6(18) ( 6)( 12) 18( 12) 9 6 ( 15)
 6(18)( 12) 2(9)(6)( 15) ( 6)(6) 18( 15) ( 12)(9) 0
594 3186 0
σ σ σ
σ σ
 − − + − + − + − − + − − − − − 
 − − − + − − − − − − − = 
− + =
 
 
Roots are: 21.04, 5.67, –26.71 kpsi Ans. 
 
1 2
2 3
max 1 3
21.04 5.67 7.69 kpsi
2
5.67 26.71 16.19 kpsi
2
21.04 26.71
 23.88 kpsi .
2
Ans
τ
τ
τ τ
−
= =
+
= =
+
= = =
 
 
 
 
_____________________________________________________________________________ 
 
3-21 
 From Eq. (3-15) 
 
( )
( ) ( )
23 2 2 2
2 2 2
3 2
(20 0 20) 20(0) 20(20) 0(20) 40 20 2 0
 20(0)(20) 2(40) 20 2 (0) 20 20 2 0(0) 20(40) 0
40 2 000 48 000 0
σ σ σ
σ σ σ
 
− + + + + + − − − −
  
 
− + − − − − − =
  
− − + = 
 
Roots are: 60, 20, –40 kpsi Ans. 
 
1 2
2 3
max 1 3
60 20 20 kpsi
2
20 40 30 kpsi
2
60 40 50 kpsi .
2
Ans
τ
τ
τ τ
−
= =
+
= =
+
= = =
 
 
 
_____________________________________________________________________________ 
 
Shigley’s MED, 10th edition Chapter 3 Solutions, Page 19/100 
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
=
= = = =
 
 
 
 
_____________________________________________________________________________ 
 
3-23 
( )( )
( )
( )
2
6
6
1
15000 33 950 psi 34.0 kpsi .
4 0.75
6033 950 0.0679 in .
30 10
0.0679 1130 10 1130 .
60
F Ans
A
FL L Ans
AE E
Ans
L
σ
pi
δ σ
δ µ−
= = = =
= = = =
= = = =ò
 
From Table A-5, v = 0.292 
 
( ) ( )
2 1
6 6
2
0.292(1130) 330 .
330 10 (0.75) 248 10 in .
v Ans
d d Ans
µ
− −
= − = − = −
∆ = = − = −
ò ò
ò
 
_____________________________________________________________________________ 
 
3-24 
( )( )
( )
( )
2
6
6
1
3000 6790 psi 6.79 kpsi .
4 0.75
606790 0.0392 in .
10.4 10
0.0392 653 10 653 .
60
F Ans
A
FL L Ans
AE E
Ans
L
σ
pi
δ σ
δ µ−
= = = =
= = = =
= = = =ò
 
From Table A-5, v = 0.333 
 
( ) ( )
2 1
6 6
2
0.333(653) 217 .
217 10 (0.75) 163 10 in .
v Ans
d d Ans
µ
− −
= − = − = −
∆ = = − = −
ò ò
ò
 
_____________________________________________________________________________ 
Shigley’s MED, 10th edition Chapter 3 Solutions, Page 20/100 
3-25 
2
0.0001 0.0001d d
d d
∆ −
= = = −ò
 
 From Table A-5, v = 0.326, E = 119 GPa 
 
( )
( ) ( )
( ) ( )
62
1
6 9
1
2
6
0.0001 306.7 10
0.326
 and , so
= 306.7 10 (119) 10 36.5 MPa
0.03
36.5 10 25 800 N 25.8 kN .
4
v
FL F
AE A
E E
L
F A Ans
δ σ
δ
σ
pi
σ
−
−
− −
= = =
= =
= = =
= = = =
òò
ò
 
 Sy = 70 MPa > σ , so elastic deformation assumption is valid. 
_____________________________________________________________________________ 
 
3-26 
( )6
8(12)20 000 0.185 in .
10.4 10
FL L Ans
AE E
δ σ= = = =
 
_____________________________________________________________________________ 
 
3-27 
( ) ( )
6
9
3140 10 0.00586 m 5.86 mm .
71.7 10
FL L Ans
AE E
δ σ= = = = =
 
_____________________________________________________________________________3-28 
( )6
10(12)15 000 0.173 in .
10.4 10
FL L Ans
AE E
δ σ= = = =
 
_____________________________________________________________________________ 
 
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 MED, 10th edition Chapter 3 Solutions, Page 21/100 
( )
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.
 
 
_____________________________________________________________________________ 
Shigley’s MED, 10th edition Chapter 3 Solutions, Page 22/100 
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 s
σ 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. 
_____________________________________________________________________________ 
 
3-33 
(a) Can solve by iteration or derive 
equations for the general case. Find 
maximum moment under wheel 3W . 
 
TW W= Σ at centroid of W’s 
3 3
A T
l x dR W
l
− −
= 
 
 Under wheel 3, 
( )3 3
3 3 1 13 2 23 3 1 13 2 23A T
l x d
M R x W a W a W x W a W a
l
− −
= − − = − − 
 For maximum, ( )3 33 3 3
3
0 2
2
TdM l dWl d x x
dx l
−
= = − − ⇒ = 
 Substitute into ( )
2
3
3 1 13 2 234 T
l d
M M W W a W a
l
−
⇒ = − − 
 This means the midpoint of 3d intersects the midpoint of the beam. 
 For wheel i, ( )
2 1
1
,
2 4
i
ii
i i T j ji
j
l dl d
x M W W a
l
−
=
−
−
= = −∑ 
 
 Note for wheel 1: 0j jiW aΣ = 
1 2 3 4
104.4104.4, 26.1 kips
4T
W W W W W= = = = = = 
 
 Wheel 1: 
2
1 1
476 (1200 238)238 in, (104.4) 20128 kip in
2 4(1200)d M
−
= = = = ⋅ 
 Wheel 2: 2 238 84 154 ind = − = 
Shigley’s MED, 10th edition 
2
2 max
(1200 154)
4(1200)M M Ans
−
= − = ⋅ =
 
 Check if all of the wheels are on the rail
 
 
(b) max 600 77 523 in .x Ans= − =
(c) See above sketch.
(d) Inner axles 
_____________________________________________________________________________
 
3-34 
(a) Let a = total area of entire envelope
 Let b = area of side notch
 
(
( )6 4
2 40(2)(37.5) 25 34 2150 mm
1 12 40 75 34 25
12 12
1.36 10 mm .
a b
A a b
I I I
I Ans
= − = − =
= − = −
=
 
(b) 
0.375(1.875) 0.703 125 in
0.375(1.75) 0.656 25 in
2(0.703125) 0.656 25 2.0625 in
a
b
A
A
A
= =
= =
= + =
 
 
 
 
 
(c) 
Use two negative areas.
2 2 2625 mm , 5625 mm , 10 000 mm
10 000 5625 625 3750 mm ;
a b cA A A
A
= = =
= − − =
 
3
3
1
2(0.703 125)(0.9375) 0.656 25(0.6875)
2.0625
0.375(1.875) 0.206 in
12
1.75(0.375) 0.007 69 in
12
2 0.206 0.703 125(0.0795) 0.00769 0.65625(0.1
a
b
y Ans
I
I
I Ans
= =
= =
= =
   = + + + =   
 Chapter 3
2
2 max
(1200 154) (104.4) 26.1(84) 21605 kip in .M M Ans= − = ⋅ =
wheels are on the rail. 
600 77 523 in .x Ans= − = 
See above sketch. 
_____________________________________________________________________________
of entire envelope 
= area of side notch 
( )
)( ) ( )( )
2
3 3
6 4
2 40(2)(37.5) 25 34 2150 mm
1 12 40 75 34 25
12 12
1.36 10 mm .I Ans
= − = − =
= − = −
 
Dimensions in mm.
2
2
2
0.375(1.875) 0.703 125 in
0.375(1.75) 0.656 25 in
2(0.703125) 0.656 25 2.0625 in= + =
 
Use two negative areas. 
2 2 2
2
625 mm , 5625 mm , 10 000 mm
10 000 5625 625 3750 mm ;
a b cA A A= = =
= − − =
4
4
2 2 4
2(0.703 125)(0.9375) 0.656 25(0.6875) 0.858 in .
2.0625
0.206 in
0.007 69 in
2 0.206 0.703 125(0.0795) 0.00769 0.65625(0.1705) 0.448 in .
y Ans
I Ans
+
= =
   = + + + =   
Chapter 3 Solutions, Page 23/100 
(104.4) 26.1(84) 21605 kip in .M M Ans 
_____________________________________________________________________________ 
Dimensions in mm. 
2 2 4705) 0.448 in .I Ans   = + + + =   
Shigley’s MED, 10th edition Chapter 3 Solutions, Page 24/100 
( )
( )
1
3
4
3
6 4
3
6 4
6.25 mm, 50 mm, 50 mm
10 000(50) 5625(50) 625(6.25) 57.29 mm .
3750
100 57.29 42.71 mm .
50(12.5) 8138 mm
12
75(75) 2.637 10 mm
12
100(100) 8.333 10 in
12
a b c
a
b
c
y y y
y Ans
c Ans
I
I
I
= = =
− −
= =
= − =
= =
= =
= =
 
( ) ( ) ( ) ( )
( )
2 26 2 6
1
6 4
1
8.333 10 10000(7.29) 2.637 10 5625 7.29 8138 625 57.29 6.25
4.29 10 in .
I
I Ans
    = + − + − + −     
=
 
 
(d) 
( )
( )
( )
2
2
2
4 0.875 3.5 in
2.5 0.875 2.1875 in
5.6875 in
2.9375 3.5 1.25(2.1875)
2.288 in .
5.6875
a
b
a b
A
A
A A A
y Ans
= =
= =
= + =
+
= =
( ) ( ) ( )( ) ( )3 2 3 2
4
1 1(4) 0.875 3.5 2.9375 2.288 0.875 2.5 2.1875 2.288 1.25
12 12
5.20 in .
I
I Ans
= + − + + −
=
 
_____________________________________________________________________________ 
 
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  = = = 
 
 
 
 
 
_____________________________________________________________________________ 
 
Shigley’s MED, 10th edition Chapter 3 Solutions, Page 25/100 
3-36 
3 41 (1)(2) 0.6667 in
12
I = = 
21(2) 2 inA = = 
 
0OMΣ = 
 8 100(8)(12) 0AR − = 
 1200