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Chapter 20
20-1
(a)
12
10
8
6
4
2
0
60
70
80
90
100
110
120
130
140
150
160
170
180
190 200
210
(b) f
( N
x )
=
f
/
(69
·
10)
=
f
/
690
x
f
f x
fx 2
f /
( N x )
60
2
120
7200
0.0029
70
1
70
4900
0.0015
80
3
240
19 200
0.0043
90
5
450
40 500
0.0072
100
8
800
80 000
0.0116
110
12
1320
145 200
0.0174
120
6
720
86 400
0.0087
130
10
1300
169 000
0.0145
140
8
1120
156 800
0.0116
150
5
750
112 500
0.0174
160
2
320
51 200
0.0029
170
3
510
86 700
0.0043
180
2
360
64 800
0.0029
190
1
190
36 100
0.0015
200
0
0
0
0
210
1
210
44 100
0.0015
69
8480
1 104 600
Eq. (20-9)
x
=
8480
69
=
122
.
9 kcycles
1 104 600 − 8480 2
1 / 2
/ 69
Eq. (20-10)
s x =
69
−
1
=
30.3 kcycles Ans.
/
¯
672654812.005.png 672654812.006.png
2
Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
20-2 Data represents a 7-class histogram with N
=
197.
x
f
f x
fx 2
174
6
1044
181 656
182
9
1638
298 116
190
44
8360
1 588 400
198
67
13 266
2 626 688
206
53
10 918
2 249 108
214
12
2568
549 552
220
6
1320
290 400
197
39 114
7 789 900
x =
39 114
197 =
198
.
55 kpsi Ans.
7 783 900
1 / 2
−
39 114 2
/
197
s x =
197
−
1
=
9
.
55 kpsi Ans.
20-3
Form a table:
x
f
f x
fx 2
64
2
128
8192
68
6
408
27 744
72
6
432
31 104
76
9
684
51 984
80
19
1520
121 600
84
10
840
70 560
88
4
352
30 976
92
2
184
16 928
58
4548
359 088
x =
4548
58 =
78
.
4 kpsi
359 088
−
4548 2
/
58
1 / 2
s x =
=
6
.
57 kpsi
58
−
1
From Eq. (20-14)
exp
2
1
1
2
x
−
78
.
4
f ( x )
=
57 √ 2
−
6
.
π
6
.
57
672654812.007.png 672654812.008.png
Chapter 20
3
20-4 (a)
y
f
f y
fy 2
y
f
/
( N
w
)
f ( y )
g ( y )
5.625
1
5.625
31.640 63
5.625
0.072 727
0.001 262
0.000 295
5.875
0
0
0
5.875
0
0.008 586
0.004 088
6.125
0
0
0
6.125
0
0.042 038
0.031 194
6.375
3
19.125
121.9219
6.375
0.218 182
0.148 106
0.140 262
6.625
3
19.875
131.6719
6.625
0.218 182
0.375 493
0.393 667
6.875
6
41.25
283.5938
6.875
0.436 364
0.685 057
0.725 002
7.125
14
99.75
710.7188
7.125
1.018 182
0.899 389
0.915 128
7.375
15
110.625
815.8594
7.375
1.090 909
0.849 697
0.822 462
7.625
10
76.25
581.4063
7.625
0.727 273
0.577 665
0.544 251
7.875
2
15.75
124.0313
7.875
0.145 455
0.282 608
0.273 138
8.125
1
8.125
66.015 63
8.125
0.072 727
0.099 492
0.106 72
55
396.375
2866.859
For a normal distribution,
2866
1 / 2
.
859
−
(396
.
375 2
/
55)
y =
396
.
375
/
55
=
7
.
207,
s y =
=
0
.
4358
55
−
1
exp
2
1
1
2
x
−
7
.
207
f ( y )
=
4358 √ 2
−
0
.
π
0
.
4358
For a lognormal distribution,
ln √ 1 + 0 . 060 474 2
s x = ln √ 1 + 0 . 060 474 2
x =
ln 7
.
206 818
−
= 1 . 9732,
= 0 . 0604
exp
2
1
1
2
ln x
−
1
.
9732
g ( y )
=
0604)( √ 2
−
x (0
.
π
)
0
.
0604
(b) Histogram
f
1.2
Data
N
LN
1
0.8
0.6
0.4
0.2
0
5.63 5.88 6.13 6.38 6.63 6.88
log N
7.13 7.38 7.63 7.88 8.13
672654812.001.png 672654812.002.png
4
Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
20-5 Distribution is uniform in interval 0.5000 to 0.5008 in, range numbers are a
=
0.5000,
b
=
0.5008 in.
(a) Eq. (20-22)
µ x =
a
+
b
=
0
.
5000
+
0
.
5008
=
0
.
5004
2
2
Eq. (20-23)
σ x =
b
a
2 √ 3 =
−
0
.
5008
2 √ 3
−
0
.
5000
=
0
.
000 231
(b) PDF from Eq. (20-20)
f ( x )
=
1250 0
.
5000
≤
x
≤
0
.
5008 in
0
otherwise
(c) CDF from Eq. (20-21)
0
x
<
0
.
5000
F ( x )
=
( x
−
0
.
5)
/
0
.
0008 0
.
5000
≤
x
≤
0
.
5008
1
x
>
0
.
5008
If all smaller diameters are removed by inspection, a
=
0.5002, b
=
0.5008
µ x =
0
.
5002
+
0
.
5008
=
0
.
5005 in
2
σ x =
0
.
5008
2 √ 3
−
0
.
5002
=
0
.
000 173 in
1666
f ( x )
=
.
70
.
5002
≤
x
≤
0
.
5008
0
otherwise
0
x
<
0
.
5002
F
(
x
) =
1666
.
7
(
x
−
0
.
5002
)
0
.
5002
≤
x
≤
0
.
5008
1
x
>
0
.
5008
20-6 Dimensions produced are due to tool dulling and wear. When parts are mixed, the distrib-
ution is uniform. From Eqs. (20-22) and (20-23),
a
= µ x + √ 3 s
√ 3 s
=
0
.
6241
+ √ 3(0
√ 3(0
.
000 581)
=
0
.
6231 in
b
=
0
.
6241
.
000 581)
=
0
.
6251 in
We suspect the dimension was
0
.
623
in Ans.
0
.
625
= µ x −
−
672654812.003.png
Chapter 20
5
33 mm
(a) Since F ( x ) is linear, the distribution is uniform at x
=
0
.
555 x
−
=
a
F ( a )
=
0
=
0
.
555( a )
−
33
∴
a
=
59.46 mm. Therefore, at x
=
b
F ( b )
=
1
=
0
.
555 b
−
33
∴
b
=
61.26 mm. Therefore,
0
x
<
59
.
46 mm
F ( x )
=
0
.
555 x
−
33 59
.
46
≤
x
≤
61
.
26 mm
1
x
>
61
.
26 mm
The PDF is dF
/
dx , thus the range numbers are:
f ( x )
=
0
.
555 59
.
46
≤
x
≤
61
.
26 mm
Ans.
0
otherwise
From the range numbers,
µ x =
59
.
46
+
61
.
26
=
60
.
36 mm Ans.
2
σ
=
61
.
26
2 √ 3
−
59
.
46
=
0
.
520 mm Ans.
x
(b)
σ
is an uncorrelated quotient F
=
3600 lbf,
A
=
0
.
112 in 2
C F =
300
/
3600
=
0
.
083 33,
C A =
0
.
001
/
0
.
112
=
0
.
008 929
From Table 20-6, for
σ
σ = µ F
µ A =
3600
0
.
112 =
32 143 psi Ans.
32 143 (0
.
08333 2
+
0
.
008929 2 )
1 / 2
σ σ =
=
2694 psi Ans.
(1
+
0
.
008929 2 )
C σ =
2694
/
32 143
=
0
.
0838 Ans.
Since F and A are lognormal, division is closed and
σ
is lognormal too.
σ =
LN (32 143, 2694) psi Ans.
20-7 F ( x )
672654812.004.png

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