All 20 Chapters Covered
SOLUTION MANUAL
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Chapter 1
Problems 1-1 through 1-6 are for student research. No standard solutions are provided.
1-7 From Fig. 1-2, cost of grinding to 0.0005 in is 270%. Cost of turning to 0.003 in is 60%.
Relative cost of grinding vs. turning = 270/60 = 4.5 times Ans.
1-8 CA = CB,
10 + 0.8 P = 60 + 0.8 P 0.005 P 2
P 2 = 50/0.005 P = 100 parts Ans.
1-9 Max. load = 1.10 P Min.
area = (0.95)2A Min.
strength = 0.85 S
To offset the absolute uncertainties, the design factor, from Eq. (1-1) should be
nd 1.10 1.43
2 Ans.
0.85 0.95
1-10 (a) X1 + X2:
x1 x2 X1 e1 X 2 e2
error e x1 x2 X1 X 2
e1 e2 Ans.
(b) X1 X2:
x1 x2 X1 e1 X 2 e2
e x1 x2 X1 X 2 e1 e2 Ans.
(c) X1 X2: x1x2 X1 e1 X 2 e2
e x1x2 X1 X 2 X1e2 X 2e1 e1e2
Xe +X e X X e1 e2 Ans.
1 2 2 1 1 2
X1 X2
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(d) X1/X2:
x1 X1 e1 X1 1 e1 X1
x2 X 2 e2 X 2 1 e2 X 2
1 e then
e2 1 2 1 e1 X1 e1 e2 1 e1 e2
1 1 1 −
X2 X 1 e2 X 2 X1 X2 X1 X2
x1 X1 2
e X1 e1 e2 Ans.
Thus,
x2 X2 X2 X1 X2
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
S 32 1000 25 103
1-12 d 1.006 in Ans.
nd d3 2.5
Table A-17: d = 1 1 in Ans.
4 S
n 25 103
Factor of safety: 4.79 Ans.
32 1000
3
1.25
Shigley’s MED, 10th edition Chapter 1 Solutions, Page 2/12
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1-13 (a)
x f fx f x2
60 2 120 7200
70 1 70 4900
80 3 240 19200
90 5 450 40500
100 8 800 80000
110 12 1320 145200
120 6 720 86400
130 10 1300 169000
140 8 1120 156800
150 5 750 112500
160 2 320 51200
170 3 510 86700
180 2 360 64800
190 1 190 36100
200 0 0 0
210 1 210 44100
69 8480 1 104 600
1 k
fi xi 122.9 kcycles
x 8480
Eq. (1-6)
N i1 69
Eq. (1-7)
k fi xi
Nx
− 2 1/ 2
s i1 2 2 30.3 kcycles Ans.
x
1 104 600 69(122.9)
N 1 69 1
z115 x x
x115 x s 115 122.9 0.2607
(b) Eq. (1-5) ˆx x
30.3
Interpolating from Table (A-10)
0.2600 0.3974
0.2607 x x = 0.3971
0.2700 0.3936
N ( 0.2607) = 69 (0.3971) = 27.4 27 Ans.
From the data, the number of instances less than 115 kcycles is
Shigley’s MED, 10th edition Chapter 1 Solutions, Page 3/12
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2 + 1 + 3 + 5 + 8 + 12 = 31 (the data is not perfectly normal)
1-14
x f fx f x2
174 6 1044 181656
182 9 1638 298116
190 44 8360 1588400
198 67 13266 2626668
206 53 10918 2249108
214 12 2568 549552
222 6 1332 295704
197 39126 7789204
1
x k
39 126
Eq. (1-6) N f ix i 197 198.61 kpsi
i1
k fi xi N x 12
2 2
7 789 204 197(198.61) 2
9.68 kpsi
Eq. (1-7) sx i1 Ans.
N 1 197 1
1-15 L 122.9 kcycles and sL 30.3 kcycles
Eq. (1-5) x x10 L x10 122.9
x
z10
ˆ sL 30.3
Thus, x10 = 122.9 + 30.3 z10 = L10
From Table A-10, for 10 percent failure, z10 = 1.282. Thus,
L10 = 122.9 + 30.3( 1.282) = 84.1 kcycles Ans.
Shigley’s MED, 10th edition Chapter 1 Solutions, Page 4/12
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1-16
x f fx fx2
93 19 1767 164331
95 25 2375 225625
97 38 3686 357542
99 17 1683 166617
101 12 1212 122412
103 10 1030 106090
105 5 525 55125
107 4 428 45796
109 4 436 47524
111 2 222 24642
136 13364 1315704
1 k
fi xi 13 98.26471 = 98.26 kpsi
x
Eq. (1-6) N i1
k fi xi − N x
2 12
Eq. (1-7) s i1 2 2 4.30 kpsi
x
1 315 704 136(98.26471)
N 1 136 1
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),
z0.01 x x x0.01 x s x 98.26
ˆx x
0.01
4.30
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.
n
1-17 Eq. (1-9): R = Ri = 0.98(0.96)0.94 = 0.88
i1
Overall reliability = 88 percent Ans.
Shigley’s MED, 10th edition Chapter 1 Solutions, Page 5/12
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1-18 Obtain the coefficients of variance for strength and stress
ˆ Ssy 23.5 0.07532
CS
S sy 312
ˆ 0.09667
C ˆT 145
T 1 500
For R = 0.99, from Table A-10, z = 2.326. Eq. (1-
12):
1 1 1 z2 C 2 1 z 2 C 2
S
n
2 2
1 zC S
2 2 2 2
1 1 1 2.326 0.07532 1 2.326 0.09667
1.3229 1.32 Ans.
2 2
1 2.326 0.07532
From the given equation for stress,
max S sy 16T
n d3
Solving for d gives
1/ 3 1/ 3 0.0319 m 31.9 mm Ans.
16T n 16(1500)1.3229
d
S sy (312)106
1-19 Obtain the coefficients of variance for stress and strength
ˆ 5
C ˆP 0.09231
P 65
ˆS ˆS 6.59
y
C 0.06901
S
S Sy 95.5
(a) n 1.2
Shigley’s MED, 10th edition Chapter 1 Solutions, Page 6/12
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nd 1 1.2 1 1.6127
Eq. (1-11): z
2 2 2
nC C 2
1.2 0.069012 0.092312
d S
Interpolating Table A-10,
1.61 0.0537
1.6127 = 0.0534
1.62 0.0526
R = 1 0.0534 = 0.9466 Ans.
Sy Sy d 2S y 4Pn 4 ( 65)1.2
n d 1.020 in Ans.
P / ( d / 4)2
4P Sy (95.5)
(b) n 1.5
z 1.5 1 3.605
1.52 0.069012 0.092312
3.6 0.000159
3.605 = 0.00015645
3.7 0.000108
R = 1 0.00015645 = 0.9998 Ans.
d 4Pn 4 ( 65)1.5 1.140 in Ans.
Sy (95.5)
1-20 90 383 473 MPa
max max a b
From footnote 9, p. 25 of text,
1/ 2
ˆ ˆ2 ˆ2 (8.42 22.32 )1/ 2 23.83 MPa
max a b
ˆ ˆ 23.83
C max max 0.0504
max
max max 473
ˆS ˆS y
42.7
0.0772
CS y
y
S 553
Sy y
Shigley’s MED, 10th edition Chapter 1 Solutions, Page 7/12
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Sy
n 553
473 Ans.
max 1.169 1.17
nd 1 1.169 1 1.635
Eq. (1-11): z
2 2 2
n C C 1.169 0.07722 0.05042
2
d S
From Table A-10, ( 1.635) = 0.05105
R = 1 0.05105 = 0.94895 = 94.9 percent Ans.
1-21 a = 1.500 0.001 in
b = 2.000 0.003 in
c = 3.000 0.004 in
d = 6.520 0.010 in
(a) w d a b c = 6.520 1.5 2 3 = 0.020 in
tw tall = 0.001 + 0.003 + 0.004 +0.010 = 0.018
w = 0.020 0.018 in Ans.
(b) From part (a), wmin = 0.002 in. Thus, must add 0.008 in to d . Therefore,
d = 6.520 + 0.008 = 6.528 in Ans.
1-22 V = xyz, and x = a a, y = b b, z = c c,
V abc
V a a b b c c
abc bc a ac b ab c a b c b c a c a b a b c
The higher order terms in are negligible. Thus,
V bc a ac b ab c
and, V bc a ac b ab c a b c a b c
Ans.
V abc a b c a b c
For the numerical values given, V 1.500 1.875 3.000 8.4375 in3
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SOLUTION MANUAL
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Chapter 1
Problems 1-1 through 1-6 are for student research. No standard solutions are provided.
1-7 From Fig. 1-2, cost of grinding to 0.0005 in is 270%. Cost of turning to 0.003 in is 60%.
Relative cost of grinding vs. turning = 270/60 = 4.5 times Ans.
1-8 CA = CB,
10 + 0.8 P = 60 + 0.8 P 0.005 P 2
P 2 = 50/0.005 P = 100 parts Ans.
1-9 Max. load = 1.10 P Min.
area = (0.95)2A Min.
strength = 0.85 S
To offset the absolute uncertainties, the design factor, from Eq. (1-1) should be
nd 1.10 1.43
2 Ans.
0.85 0.95
1-10 (a) X1 + X2:
x1 x2 X1 e1 X 2 e2
error e x1 x2 X1 X 2
e1 e2 Ans.
(b) X1 X2:
x1 x2 X1 e1 X 2 e2
e x1 x2 X1 X 2 e1 e2 Ans.
(c) X1 X2: x1x2 X1 e1 X 2 e2
e x1x2 X1 X 2 X1e2 X 2e1 e1e2
Xe +X e X X e1 e2 Ans.
1 2 2 1 1 2
X1 X2
Shigley’s MED, 10th edition Chapter 1 Solutions, Page 1/12
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(d) X1/X2:
x1 X1 e1 X1 1 e1 X1
x2 X 2 e2 X 2 1 e2 X 2
1 e then
e2 1 2 1 e1 X1 e1 e2 1 e1 e2
1 1 1 −
X2 X 1 e2 X 2 X1 X2 X1 X2
x1 X1 2
e X1 e1 e2 Ans.
Thus,
x2 X2 X2 X1 X2
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
S 32 1000 25 103
1-12 d 1.006 in Ans.
nd d3 2.5
Table A-17: d = 1 1 in Ans.
4 S
n 25 103
Factor of safety: 4.79 Ans.
32 1000
3
1.25
Shigley’s MED, 10th edition Chapter 1 Solutions, Page 2/12
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1-13 (a)
x f fx f x2
60 2 120 7200
70 1 70 4900
80 3 240 19200
90 5 450 40500
100 8 800 80000
110 12 1320 145200
120 6 720 86400
130 10 1300 169000
140 8 1120 156800
150 5 750 112500
160 2 320 51200
170 3 510 86700
180 2 360 64800
190 1 190 36100
200 0 0 0
210 1 210 44100
69 8480 1 104 600
1 k
fi xi 122.9 kcycles
x 8480
Eq. (1-6)
N i1 69
Eq. (1-7)
k fi xi
Nx
− 2 1/ 2
s i1 2 2 30.3 kcycles Ans.
x
1 104 600 69(122.9)
N 1 69 1
z115 x x
x115 x s 115 122.9 0.2607
(b) Eq. (1-5) ˆx x
30.3
Interpolating from Table (A-10)
0.2600 0.3974
0.2607 x x = 0.3971
0.2700 0.3936
N ( 0.2607) = 69 (0.3971) = 27.4 27 Ans.
From the data, the number of instances less than 115 kcycles is
Shigley’s MED, 10th edition Chapter 1 Solutions, Page 3/12
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2 + 1 + 3 + 5 + 8 + 12 = 31 (the data is not perfectly normal)
1-14
x f fx f x2
174 6 1044 181656
182 9 1638 298116
190 44 8360 1588400
198 67 13266 2626668
206 53 10918 2249108
214 12 2568 549552
222 6 1332 295704
197 39126 7789204
1
x k
39 126
Eq. (1-6) N f ix i 197 198.61 kpsi
i1
k fi xi N x 12
2 2
7 789 204 197(198.61) 2
9.68 kpsi
Eq. (1-7) sx i1 Ans.
N 1 197 1
1-15 L 122.9 kcycles and sL 30.3 kcycles
Eq. (1-5) x x10 L x10 122.9
x
z10
ˆ sL 30.3
Thus, x10 = 122.9 + 30.3 z10 = L10
From Table A-10, for 10 percent failure, z10 = 1.282. Thus,
L10 = 122.9 + 30.3( 1.282) = 84.1 kcycles Ans.
Shigley’s MED, 10th edition Chapter 1 Solutions, Page 4/12
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1-16
x f fx fx2
93 19 1767 164331
95 25 2375 225625
97 38 3686 357542
99 17 1683 166617
101 12 1212 122412
103 10 1030 106090
105 5 525 55125
107 4 428 45796
109 4 436 47524
111 2 222 24642
136 13364 1315704
1 k
fi xi 13 98.26471 = 98.26 kpsi
x
Eq. (1-6) N i1
k fi xi − N x
2 12
Eq. (1-7) s i1 2 2 4.30 kpsi
x
1 315 704 136(98.26471)
N 1 136 1
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),
z0.01 x x x0.01 x s x 98.26
ˆx x
0.01
4.30
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.
n
1-17 Eq. (1-9): R = Ri = 0.98(0.96)0.94 = 0.88
i1
Overall reliability = 88 percent Ans.
Shigley’s MED, 10th edition Chapter 1 Solutions, Page 5/12
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1-18 Obtain the coefficients of variance for strength and stress
ˆ Ssy 23.5 0.07532
CS
S sy 312
ˆ 0.09667
C ˆT 145
T 1 500
For R = 0.99, from Table A-10, z = 2.326. Eq. (1-
12):
1 1 1 z2 C 2 1 z 2 C 2
S
n
2 2
1 zC S
2 2 2 2
1 1 1 2.326 0.07532 1 2.326 0.09667
1.3229 1.32 Ans.
2 2
1 2.326 0.07532
From the given equation for stress,
max S sy 16T
n d3
Solving for d gives
1/ 3 1/ 3 0.0319 m 31.9 mm Ans.
16T n 16(1500)1.3229
d
S sy (312)106
1-19 Obtain the coefficients of variance for stress and strength
ˆ 5
C ˆP 0.09231
P 65
ˆS ˆS 6.59
y
C 0.06901
S
S Sy 95.5
(a) n 1.2
Shigley’s MED, 10th edition Chapter 1 Solutions, Page 6/12
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nd 1 1.2 1 1.6127
Eq. (1-11): z
2 2 2
nC C 2
1.2 0.069012 0.092312
d S
Interpolating Table A-10,
1.61 0.0537
1.6127 = 0.0534
1.62 0.0526
R = 1 0.0534 = 0.9466 Ans.
Sy Sy d 2S y 4Pn 4 ( 65)1.2
n d 1.020 in Ans.
P / ( d / 4)2
4P Sy (95.5)
(b) n 1.5
z 1.5 1 3.605
1.52 0.069012 0.092312
3.6 0.000159
3.605 = 0.00015645
3.7 0.000108
R = 1 0.00015645 = 0.9998 Ans.
d 4Pn 4 ( 65)1.5 1.140 in Ans.
Sy (95.5)
1-20 90 383 473 MPa
max max a b
From footnote 9, p. 25 of text,
1/ 2
ˆ ˆ2 ˆ2 (8.42 22.32 )1/ 2 23.83 MPa
max a b
ˆ ˆ 23.83
C max max 0.0504
max
max max 473
ˆS ˆS y
42.7
0.0772
CS y
y
S 553
Sy y
Shigley’s MED, 10th edition Chapter 1 Solutions, Page 7/12
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Sy
n 553
473 Ans.
max 1.169 1.17
nd 1 1.169 1 1.635
Eq. (1-11): z
2 2 2
n C C 1.169 0.07722 0.05042
2
d S
From Table A-10, ( 1.635) = 0.05105
R = 1 0.05105 = 0.94895 = 94.9 percent Ans.
1-21 a = 1.500 0.001 in
b = 2.000 0.003 in
c = 3.000 0.004 in
d = 6.520 0.010 in
(a) w d a b c = 6.520 1.5 2 3 = 0.020 in
tw tall = 0.001 + 0.003 + 0.004 +0.010 = 0.018
w = 0.020 0.018 in Ans.
(b) From part (a), wmin = 0.002 in. Thus, must add 0.008 in to d . Therefore,
d = 6.520 + 0.008 = 6.528 in Ans.
1-22 V = xyz, and x = a a, y = b b, z = c c,
V abc
V a a b b c c
abc bc a ac b ab c a b c b c a c a b a b c
The higher order terms in are negligible. Thus,
V bc a ac b ab c
and, V bc a ac b ab c a b c a b c
Ans.
V abc a b c a b c
For the numerical values given, V 1.500 1.875 3.000 8.4375 in3
Shigley’s MED, 10th edition Chapter 1 Solutions, Page 8/12
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CLICK ON THE L.I.N.K
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