4. Chapter 2 11
2-5 Distribution is uniform in interval 0.5000 to 0.5008 in, range numbers are a = 0.5000,
b = 0.5008 in.
(a) Eq. (2-22) µx =
a + b
2
=
0.5000 + 0.5008
2
= 0.5004
Eq. (2-23) σx =
b − a
2
√
3
=
0.5008 − 0.5000
2
√
3
= 0.000 231
(b) PDF from Eq. (2-20)
f (x) =
1250 0.5000 ≤ x ≤ 0.5008 in
0 otherwise
(c) CDF from Eq. (2-21)
F(x) =
0 x < 0.5000
(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
µx =
0.5002 + 0.5008
2
= 0.5005 in
ˆσx =
0.5008 − 0.5002
2
√
3
= 0.000 173 in
f (x) =
1666.7 0.5002 ≤ x ≤ 0.5008
0 otherwise
F(x) =
0 x < 0.5002
1666.7(x − 0.5002) 0.5002 ≤ x ≤ 0.5008
1 x > 0.5008
2-6 Dimensions produced are due to tool dulling and wear. When parts are mixed, the distribution
is uniform. From Eqs. (2-22) and (2-23),
a = µx −
√
3s = 0.6241 −
√
3(0.000 581) = 0.6231 in
b = µx +
√
3s = 0.6241 +
√
3(0.000 581) = 0.6251 in
We suspect the dimension was
0.623
0.625
in Ans.
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5. 12 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
2-7 F(x) = 0.555x − 33 mm
(a) Since F(x) is linear, the distribution is uniform at x = a
F(a) = 0 = 0.555(a) − 33
∴ a = 59.46 mm. Therefore, at x = b
F(b) = 1 = 0.555b − 33
∴ b = 61.26 mm. Therefore,
F(x) =
0 x < 59.46 mm
0.555x − 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
0 otherwise
Ans.
From the range numbers,
µx =
59.46 + 61.26
2
= 60.36 mm Ans.
ˆσx =
61.26 − 59.46
2
√
3
= 0.520 mm Ans.
1
(b) σ is an uncorrelated quotient ¯F = 3600 lbf, ¯A = 0.112 in2
CF = 300/3600 = 0.083 33, CA = 0.001/0.112 = 0.008 929
From Table 2-6, for σ
¯σ =
µF
µA
=
3600
0.112
= 32 143 psi Ans.
ˆσσ = 32 143
(0.083332
+ 0.0089292
)
(1 + 0.0089292)
1/2
= 2694 psi Ans.
Cσ = 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.
shi20396_ch02.qxd 7/21/03 3:28 PM Page 12
6. Chapter 2 13
2-8 Cramer’s rule
a1 =
y x2
xy x3
x x2
x2
x3
=
y x3
− xy x2
x x3 − ( x2)2
Ans.
a2 =
x y
x2
xy
x x2
x2
x3
=
x xy − y x2
x x3 − ( x2)2
Ans.
Ϫ0.05
0
0.05
0.1
0.15
0.2
0.25
0.3
0 0.2 0.4 0.6 0.8 1
Data
Regression
x
y
x y x2
x3
xy
0 0.01 0 0 0
0.2 0.15 0.04 0.008 0.030
0.4 0.25 0.16 0.064 0.100
0.6 0.25 0.36 0.216 0.150
0.8 0.17 0.64 0.512 0.136
1.0 −0.01 1.00 1.000 −0.010
3.0 0.82 2.20 1.800 0.406
a1 = 1.040 714 a2 = −1.046 43 Ans.
Data Regression
x y y
0 0.01 0
0.2 0.15 0.166 286
0.4 0.25 0.248 857
0.6 0.25 0.247 714
0.8 0.17 0.162 857
1.0 −0.01 −0.005 71
shi20396_ch02.qxd 7/21/03 3:28 PM Page 13
10. Chapter 2 17
(b) Eq. (2-35)
s ˆm =
0.556
√
2.0333
= 0.3899 lbf/in
k = (9.7656, 0.3899) lbf/in Ans.
2-12 The expression = δ/l is of the form x/y. Now δ = (0.0015, 0.000 092) in, unspecified
distribution; l = (2.000, 0.0081) in, unspecified distribution;
Cx = 0.000 092/0.0015 = 0.0613
Cy = 0.0081/2.000 = 0.000 75
From Table 2-6, ¯ = 0.0015/2.000 = 0.000 75
ˆσ = 0.000 75
0.06132
+ 0.004 052
1 + 0.004 052
1/2
= 4.607(10−5
) = 0.000 046
We can predict ¯ and ˆσ but not the distribution of .
2-13 σ = E
= (0.0005, 0.000 034) distribution unspecified; E = (29.5, 0.885) Mpsi, distribution
unspecified;
Cx = 0.000 034/0.0005 = 0.068,
Cy = 0.0885/29.5 = 0.030
σ is of the form x, y
Table 2-6
¯σ = ¯ ¯E = 0.0005(29.5)106
= 14 750 psi
ˆσσ = 14 750(0.0682
+ 0.0302
+ 0.0682
+ 0.0302
)1/2
= 1096.7 psi
Cσ = 1096.7/14 750 = 0.074 35
2-14
δ =
Fl
AE
F = (14.7, 1.3) kip, A = (0.226, 0.003)in2
, l = (1.5, 0.004) in, E = (29.5, 0.885) Mpsi dis-
tributions unspecified.
CF = 1.3/14.7 = 0.0884; CA = 0.003/0.226 = 0.0133; Cl = 0.004/1.5 = 0.00267;
CE = 0.885/29.5 = 0.03
Mean of δ:
δ =
Fl
AE
= Fl
1
A
1
E
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11. 18 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
From Table 2-6,
¯δ = ¯F ¯l(1/ ¯A)(1/ ¯E)
¯δ = 14 700(1.5)
1
0.226
1
29.5(106)
= 0.003 31 in Ans.
For the standard deviation, using the first-order terms in Table 2-6,
ˆσδ
.
=
¯F ¯l
¯A ¯E
C2
F + C2
l + C2
A + C2
E
1/2
= ¯δ C2
F + C2
l + C2
A + C2
E
1/2
ˆσδ = 0.003 31(0.08842
+ 0.002672
+ 0.01332
+ 0.032
)1/2
= 0.000 313 in Ans.
COV
Cδ = 0.000 313/0.003 31 = 0.0945 Ans.
Force COV dominates. There is no distributional information on δ.
2-15 M = (15000, 1350) lbf · in, distribution unspecified; d = (2.00, 0.005) in distribution
unspecified.
σ =
32M
πd3
, CM =
1350
15 000
= 0.09, Cd =
0.005
2.00
= 0.0025
σ is of the form x/y, Table 2-6.
Mean:
¯σ =
32 ¯M
πd3
.
=
32 ¯M
π ¯d3
=
32(15 000)
π(23)
= 19 099 psi Ans.
Standard Deviation:
ˆσσ = ¯σ C2
M + C2
d3 1 + C2
d3
1/2
From Table 2-6, Cd3
.
= 3Cd = 3(0.0025) = 0.0075
ˆσσ = ¯σ C2
M + (3Cd)2
(1 + (3Cd))2 1/2
= 19 099[(0.092
+ 0.00752
)/(1 + 0.00752
)]1/2
= 1725 psi Ans.
COV:
Cσ =
1725
19 099
= 0.0903 Ans.
Stress COV dominates. No information of distribution of σ.
shi20396_ch02.qxd 7/21/03 3:28 PM Page 18
12. Chapter 2 19
2-16
Fraction discarded is α + β. The area under the PDF was unity. Having discarded α + β
fraction, the ordinates to the truncated PDF are multiplied by a.
a =
1
1 − (α + β)
New PDF, g(x), is given by
g(x) =
f (x)/[1 − (α + β)] x1 ≤ x ≤ x2
0 otherwise
More formal proof: g(x) has the property
1 =
x2
x1
g(x) dx = a
x2
x1
f (x) dx
1 = a
∞
−∞
f (x) dx −
x1
0
f (x) dx −
∞
x2
f (x) dx
1 = a {1 − F(x1) − [1 − F(x2)]}
a =
1
F(x2) − F(x1)
=
1
(1 − β) − α
=
1
1 − (α + β)
2-17
(a) d = U[0.748, 0.751]
µd =
0.751 + 0.748
2
= 0.7495 in
ˆσd =
0.751 − 0.748
2
√
3
= 0.000 866 in
f (x) =
1
b − a
=
1
0.751 − 0.748
= 333.3 in−1
F(x) =
x − 0.748
0.751 − 0.748
= 333.3(x − 0.748)
x1
f(x)
x
x2
␣ 
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13. 20 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
(b) F(x1) = F(0.748) = 0
F(x2) = (0.750 − 0.748)333.3 = 0.6667
If g(x) is truncated, PDF becomes
g(x) =
f (x)
F(x2) − F(x1)
=
333.3
0.6667 − 0
= 500 in−1
µx =
a + b
2
=
0.748 + 0.750
2
= 0.749 in
ˆσx =
b − a
2
√
3
=
0.750 − 0.748
2
√
3
= 0.000 577 in
2-18 From Table A-10, 8.1% corresponds to z1 = −1.4 and 5.5% corresponds to z2 = +1.6.
k1 = µ + z1 ˆσ
k2 = µ + z2 ˆσ
From which
µ =
z2k1 − z1k2
z2 − z1
=
1.6(9) − (−1.4)11
1.6 − (−1.4)
= 9.933
ˆσ =
k2 − k1
z2 − z1
=
11 − 9
1.6 − (−1.4)
= 0.6667
The original density function is
f (k) =
1
0.6667
√
2π
exp −
1
2
k − 9.933
0.6667
2
Ans.
2-19 From Prob. 2-1, µ = 122.9 kcycles and ˆσ = 30.3 kcycles.
z10 =
x10 − µ
ˆσ
=
x10 − 122.9
30.3
x10 = 122.9 + 30.3z10
From Table A-10, for 10 percent failure, z10 = −1.282
x10 = 122.9 + 30.3(−1.282)
= 84.1 kcycles Ans.
0.748
g(x) ϭ 500
x
f(x) ϭ 333.3
0.749 0.750 0.751
shi20396_ch02.qxd 7/21/03 3:28 PM Page 20
17. 24 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
For no yield, m = Sy − σ ≥ 0
z =
m − µm
ˆσm
=
0 − µm
ˆσm
= −
µm
ˆσm
µm = ¯Sy − ¯σ = 27.47 kpsi,
ˆσm = ˆσ2
σ + ˆσ2
Sy
1/2
= 12.32 kpsi
z =
−27.47
12.32
= −2.230
From Table A-10, pf = 0.0129
R = 1 − pf = 1 − 0.0129 = 0.987 Ans.
2-24 For a lognormal distribution,
Eq. (2-18) µy = ln µx − ln 1 + C2
x
Eq. (2-19) ˆσy = ln 1 + C2
x
From Prob. (2-23)
µm = ¯Sy − ¯σ = µx
µy = ln ¯Sy − ln 1 + C2
Sy
− ln ¯σ − ln 1 + C2
σ
= ln
¯Sy
¯σ
1 + C2
σ
1 + C2
Sy
ˆσy = ln 1 + C2
Sy
+ ln 1 + C2
σ
1/2
= ln 1 + C2
Sy
1 + C2
σ
z = −
µ
ˆσ
= −
ln
¯Sy
¯σ
1 + C2
σ
1 + C2
Sy
ln 1 + C2
Sy
1 + C2
σ
¯σ =
4 ¯P
πd2
=
4(30)
π(12)
= 38.197 kpsi
ˆσσ =
4 ˆσP
πd2
=
4(5.1)
π(12)
= 6.494 kpsi
Cσ =
6.494
38.197
= 0.1700
CSy
=
3.81
49.6
= 0.076 81
0
m
shi20396_ch02.qxd 7/21/03 3:28 PM Page 24
18. Chapter 2 25
z = −
ln
49.6
38.197
1 + 0.1702
1 + 0.076 812
ln (1 + 0.076 812)(1 + 0.1702)
= −1.470
From Table A-10
pf = 0.0708
R = 1 − pf = 0.929 Ans.
2-25
(a) a = 1.000 ± 0.001 in
b = 2.000 ± 0.003 in
c = 3.000 ± 0.005 in
d = 6.020 ± 0.006 in
¯w = d − a − b − c = 6.020 − 1 − 2 − 3 = 0.020 in
tw = tall = 0.001 + 0.003 + 0.005 + 0.006
= 0.015 in
w = 0.020 ± 0.015 in Ans.
(b) ¯w = 0.020
ˆσw = ˆσ2
all =
0.001
√
3
2
+
0.003
√
3
2
+
0.005
√
3
2
+
0.006
√
3
2
= 0.004 86 → 0.005 in (uniform)
w = 0.020 ± 0.005 in Ans.
2-26
V + V = (a + a)(b + b)(c + c)
V + V = abc + bc a + ac b + ab c + small higher order terms
V
¯V
.
=
a
a
+
b
b
+
c
c
Ans.
¯V = ¯a ¯b¯c = 1.25(1.875)(2.75) = 6.4453 in3
V
¯V
=
0.001
1.250
+
0.002
1.875
+
0.003
2.750
= 0.00296
V =
V
¯V
¯V = 0.00296(6.4453) = 0.0191 in3
Lower range number:
¯V − V = 6.4453 − 0.0191 = 6.4262 in3
Ans.
Upper range number:
¯V + V = 6.4453 + 0.0191 = 6.4644 in3
Ans.
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19. 26 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
2-27
(a)
wmax = 0.014 in, wmin = 0.004 in
¯w = (0.014 + 0.004)/2 = 0.009 in
w = 0.009 ± 0.005 in
¯w = ¯x − ¯y = ¯a − ¯b − ¯c
0.009 = ¯a − 0.042 − 1.000
¯a = 1.051 in
tw = tall
0.005 = ta + 0.002 + 0.002
ta = 0.005 − 0.002 − 0.002 = 0.001 in
a = 1.051 ± 0.001 in Ans.
(b) ˆσw = ˆσ2
all = ˆσ2
a + ˆσ2
b + ˆσ2
c
ˆσ2
a = ˆσ2
w − ˆσ2
b − ˆσ2
c
=
0.005
√
3
2
−
0.002
√
3
2
−
0.002
√
3
2
ˆσ2
a = 5.667(10−6
)
ˆσa = 5.667(10−6) = 0.00238 in
¯a = 1.051 in, ˆσa = 0.00238 in Ans.
2-28 Choose 15 mm as basic size, D, d. Table 2-8: fit is designated as 15H7/h6. From
Table A-11, the tolerance grades are D = 0.018 mm and d = 0.011 mm.
Hole: Eq. (2-38)
Dmax = D + D = 15 + 0.018 = 15.018 mm Ans.
Dmin = D = 15.000 mm Ans.
Shaft: From Table A-12, fundamental deviation δF = 0. From Eq. (2-39)
dmax = d + δF = 15.000 + 0 = 15.000 mm Ans.
dmin = d + δR − d = 15.000 + 0 − 0.011 = 14.989 mm Ans.
2-29 Choose 45 mm as basic size. Table 2-8 designates fit as 45H7/s6. From Table A-11, the
tolerance grades are D = 0.025 mm and d = 0.016 mm
Hole: Eq. (2-38)
Dmax = D + D = 45.000 + 0.025 = 45.025 mm Ans.
Dmin = D = 45.000 mm Ans.
a
c b
w
shi20396_ch02.qxd 7/21/03 3:28 PM Page 26
20. Chapter 2 27
Shaft: From Table A-12, fundamental deviation δF = +0.043 mm. From Eq. (2-40)
dmin = d + δF = 45.000 + 0.043 = 45.043 mm Ans.
dmax = d + δF + d = 45.000 + 0.043 + 0.016 = 45.059 mm Ans.
2-30 Choose 50 mm as basic size. From Table 2-8 fit is 50H7/g6. From Table A-11, the tolerance
grades are D = 0.025 mm and d = 0.016 mm.
Hole:
Dmax = D + D = 50 + 0.025 = 50.025 mm Ans.
Dmin = D = 50.000 mm Ans.
Shaft: From Table A-12 fundamental deviation = −0.009 mm
dmax = d + δF = 50.000 + (−0.009) = 49.991 mm Ans.
dmin = d + δF − d
= 50.000 + (−0.009) − 0.016
= 49.975 mm
2-31 Choose the basic size as 1.000 in. From Table 2-8, for 1.0 in, the fit is H8/f7. From
Table A-13, the tolerance grades are D = 0.0013 in and d = 0.0008 in.
Hole:
Dmax = D + ( D)hole = 1.000 + 0.0013 = 1.0013 in Ans.
Dmin = D = 1.0000 in Ans.
Shaft: From Table A-14: Fundamental deviation = −0.0008 in
dmax = d + δF = 1.0000 + (−0.0008) = 0.9992 in Ans.
dmin = d + δF − d = 1.0000 + (−0.0008) − 0.0008 = 0.9984 in Ans.
Alternatively,
dmin = dmax − d = 0.9992 − 0.0008 = 0.9984 in. Ans.
2-32
Do = W + Di + W
¯Do = ¯W + ¯Di + ¯W
= 0.139 + 3.734 + 0.139 = 4.012 in
tDo
= tall = 0.004 + 0.028 + 0.004
= 0.036 in
Do = 4.012 ± 0.036 in Ans.
Do
WDiW
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21. 28 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
2-33
Do = Di + 2W
¯Do = ¯Di + 2 ¯W = 208.92 + 2(5.33)
= 219.58 mm
tDo
=
all
t = tDi
+ 2tw
= 1.30 + 2(0.13) = 1.56 mm
Do = 219.58 ± 1.56 mm Ans.
2-34
Do = Di + 2W
¯Do = ¯Di + 2 ¯W = 3.734 + 2(0.139)
= 4.012 mm
tDo
=
all
t2 = t2
Do
+ (2 tw)2 1/2
= [0.0282
+ (2)2
(0.004)2
]1/2
= 0.029 in
Do = 4.012 ± 0.029 in Ans.
2-35
Do = Di + 2W
¯Do = ¯Di + 2 ¯W = 208.92 + 2(5.33)
= 219.58 mm
tDo
=
all
t2 = [1.302
+ (2)2
(0.13)2
]1/2
= 1.33 mm
Do = 219.58 ± 1.33 mm Ans.
2-36
(a) w = F − W
¯w = ¯F − ¯W = 0.106 − 0.139
= −0.033 in
tw =
all
t = 0.003 + 0.004
tw = 0.007 in
wmax = ¯w + tw = −0.033 + 0.007 = −0.026 in
wmin = ¯w − tw = −0.033 − 0.007 = −0.040 in
The minimum “squeeze” is 0.026 in. Ans.
w
W
F
shi20396_ch02.qxd 7/21/03 3:28 PM Page 28
22. Chapter 2 29
(b)
Y = 3.992 ± 0.020 in
Do + w − Y = 0
w = Y − ¯Do
¯w = ¯Y − ¯Do = 3.992 − 4.012 = −0.020 in
tw =
all
t = tY + tDo
= 0.020 + 0.036 = 0.056 in
w = −0.020 ± 0.056 in
wmax = 0.036 in
wmin = −0.076 in
O-ring is more likely compressed than free prior to assembly of the
end plate.
2-37
(a) Figure defines w as gap.
The O-ring is “squeezed” at least 0.75 mm.
(b)
From the figure, the stochastic equation is:
Do + w = Y
or, w = Y − Do
¯w = ¯Y − ¯Do = 218.48 − 219.58 = −1.10 mm
tw =
all
t = tY + tDo
= 1.10 + 0.34 = 1.44 mm
wmax = ¯w + tw = −1.10 + 1.44 = 0.34 mm
wmin = ¯w − tw = −1.10 − 1.44 = −2.54 mm
The O-ring is more likely to be circumferentially compressed than free prior to as-
sembly of the end plate.
Ymax = ¯Do = 219.58 mm
Ymin = max[0.99 ¯Do, ¯Do − 1.52]
= max[0.99(219.58, 219.58 − 1.52)]
= 217.38 mm
Y = 218.48 ± 1.10 mm
Y
Do
w
w = F − W
¯w = ¯F − ¯W
= 4.32 − 5.33 = −1.01 mm
tw =
all
t = tF + tW = 0.13 + 0.13 = 0.26 mm
wmax = ¯w + tw = −1.01 + 0.26 = −0.75 mm
wmin = ¯w − tw = −1.01 − 0.26 = −1.27 mm
w
W
F
Ymax = ¯Do = 4.012 in
Ymin = max[0.99 ¯Do, ¯Do − 0.06]
= max[3.9719, 3.952] = 3.972 in
Y
Do
w
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23. 30 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
2-38
wmax = −0.020 in, wmin = −0.040 in
¯w =
1
2
(−0.020 + (−0.040)) = −0.030 in
tw =
1
2
(−0.020 − (−0.040)) = 0.010 in
b = 0.750 ± 0.001 in
c = 0.120 ± 0.005 in
d = 0.875 ± 0.001 in
¯w = ¯a − ¯b − ¯c − ¯d
−0.030 = ¯a − 0.875 − 0.120 − 0.750
¯a = 0.875 + 0.120 + 0.750 − 0.030
¯a = 1.715 in
Absolute:
tw =
all
t = 0.010 = ta + 0.001 + 0.005 + 0.001
ta = 0.010 − 0.001 − 0.005 − 0.001
= 0.003 in
a = 1.715 ± 0.003 in Ans.
Statistical: For a normal distribution of dimensions
t2
w =
all
t2
= t2
a + t2
b + t2
c + t2
d
ta = t2
w − t2
b − t2
c − t2
d
1/2
= (0.0102
− 0.0012
− 0.0052
− 0.0012
)1/2
= 0.0085
a = 1.715 ± 0.0085 in Ans.
2-39
x n nx nx2
93 19 1767 164 311
95 25 2375 225 625
97 38 3685 357 542
99 17 1683 166 617
101 12 1212 122 412
103 10 1030 106 090
105 5 525 55 125
107 4 428 45 796
109 4 436 47 524
111 2 222 24 624
136 13364 1315 704
¯x = 13 364/136 = 98.26 kpsi
sx =
1 315 704 − 13 3642
/136
135
1/2
= 4.30 kpsi
b c
w
d
a
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24. Chapter 2 31
Under normal hypothesis,
z0.01 = (x0.01 − 98.26)/4.30
x0.01 = 98.26 + 4.30z0.01
= 98.26 + 4.30(−2.3267)
= 88.26
.
= 88.3 kpsi Ans.
2-40 From Prob. 2-39, µx = 98.26 kpsi, and ˆσx = 4.30 kpsi.
Cx = ˆσx/µx = 4.30/98.26 = 0.043 76
From Eqs. (2-18) and (2-19),
µy = ln(98.26) − 0.043 762
/2 = 4.587
ˆσy = ln(1 + 0.043 762) = 0.043 74
For a yield strength exceeded by 99% of the population,
z0.01 = (ln x0.01 − µy)/ˆσy ⇒ ln x0.01 = µy + ˆσyz0.01
From Table A-10, for 1% failure, z0.01 = −2.326. Thus,
ln x0.01 = 4.587 + 0.043 74(−2.326) = 4.485
x0.01 = 88.7 kpsi Ans.
The normal PDF is given by Eq. (2-14) as
f (x) =
1
4.30
√
2π
exp −
1
2
x − 98.26
4.30
2
For the lognormal distribution, from Eq. (2-17), defining g(x),
g(x) =
1
x(0.043 74)
√
2π
exp −
1
2
ln x − 4.587
0.043 74
2
x (kpsi) f/(Nw) f (x) g(x) x (kpsi) f/(Nw) f (x) g(x)
92 0.00000 0.03215 0.03263 102 0.03676 0.06356 0.06134
92 0.06985 0.03215 0.03263 104 0.03676 0.03806 0.03708
94 0.06985 0.05680 0.05890 104 0.01838 0.03806 0.03708
94 0.09191 0.05680 0.05890 106 0.01838 0.01836 0.01869
96 0.09191 0.08081 0.08308 106 0.01471 0.01836 0.01869
96 0.13971 0.08081 0.08308 108 0.01471 0.00713 0.00793
98 0.13971 0.09261 0.09297 108 0.01471 0.00713 0.00793
98 0.06250 0.09261 0.09297 110 0.01471 0.00223 0.00286
100 0.06250 0.08548 0.08367 110 0.00735 0.00223 0.00286
100 0.04412 0.08548 0.08367 112 0.00735 0.00056 0.00089
102 0.04412 0.06356 0.06134 112 0.00000 0.00056 0.00089
Note: rows are repeated to draw histogram
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25. 32 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
The normal and lognormal are almost the same. However the data is quite skewed and
perhaps a Weibull distribution should be explored. For a method of establishing the
Weibull parameters see Shigley, J. E., and C. R. Mischke, Mechanical Engineering Design,
McGraw-Hill, 5th ed., 1989, Sec. 4-12.
2-41 Let x = (S fe)104
x0 = 79 kpsi, θ = 86.2 kpsi, b = 2.6
Eq. (2-28)
¯x = x0 + (θ − x0) (1 + 1/b)
¯x = 79 + (86.2 − 79) (1 + 1/2.6)
= 79 + 7.2 (1.38)
From Table A-34, (1.38) = 0.88854
¯x = 79 + 7.2(0.888 54) = 85.4 kpsi Ans.
Eq. (2-29)
ˆσx = (θ − x0)[ (1 + 2/b) − 2
(1 + 1/b)]1/2
= (86.2 − 79)[ (1 + 2/2.6) − 2
(1 + 1/2.6)]1/2
= 7.2[0.923 76 − 0.888 542
]1/2
= 2.64 kpsi Ans.
Cx =
ˆσx
¯x
=
2.64
85.4
= 0.031 Ans.
2-42
x = Sut
x0 = 27.7, θ = 46.2, b = 4.38
µx = 27.7 + (46.2 − 27.7) (1 + 1/4.38)
= 27.7 + 18.5 (1.23)
= 27.7 + 18.5(0.910 75)
= 44.55 kpsi Ans.
f(x)
g(x)
Histogram
0
0.02
0.04
0.06
0.08
0.1
0.12
0.14
0.16
90 92 94 96 98 100 102 104 106 108
x (kpsi)
Probabilitydensity
110 112
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