174 4. RELIABILITY OF A COMPONENT UNDER STATIC LOAD
Table 4.9: Ultimate strength and yield strength of some stainless steels
Ultimate Strength S
u
(ksi) Yield Strength S
y
(ksi)
Materials Remarks
μ
S
u
σ
S
u
γ
S
u
Sample
size
μ
S
y
σ
S
y
γ
S
y
Sample
size
Type 202
Strip, austenitic
cool rolled 0.017-
0.022
˝
99.7 2.71 0.027 25 49.9 1.32 0.026 25
Type 301
Strip and sheet
annealed, 0.038–0.195
˝
105 5.68 0.054 100 46.8 4.7 0.100 100
Type 304
Round bars annealed,
0.50-4.625
˝
85 4.14 0.049 45 37.9 3.76 0.099 35
Type 403
Bar, ferritic
hot rolled
109.7 4.48 0.041 549 81.2 6.86 0.084 549
Type 347 Forging -100F 118 3.34 0.028 23 34.5 1.15 0.033 187
AM-350
Sheet
annealed, 0.025–0.125
˝
152.8 6.7 0.044 194 63.4 4.04 0.064 188
parameters are not available due to lack of sample data, we can use the following methods to
estimate their distribution parameters and then estimate the reliability of a component.
1. Estimate ultimate shear strength by using ultimate tensile strength.
Material mechanical properties from tensile tests are typically available from design hand-
books. We can use the following Equation (4.3) [4] to estimate ultimate shearing strength
by using ultimate tensile strength:
S
su
S
u
D
8
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
<
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
:
0:60 Aluminum alloy
0:75 Steel
0:90 Copper
0:90 Malleable iron
1:30 Cast iron;
(4.3)
where S
u
is ultimate tensile strength and S
su
is ultimate shear strength.
2. Estimate shear yield strength by using tensile yield strength.
4.5. ESTIMATION OF SOME DESIGN PARAMETERS 175
Table 4.10: Ultimate strength and yield strength of some irons
Ultimate Strength S
u
(ksi) Yield Strength S
y
(ksi)
Materials Remarks
μ
S
u
σ
S
u
γ
S
u
Sample
size
μ
S
y
σ
S
y
γ
S
y
Sample
size
Malleable
Ferrritic
foundary A
53.4 2.68 0.050 434 34.9 1.47 0.042 434
Iron grade
32510
Grade 32510
foundary B
56 1.41 0.025 785 39.1 1.22 0.031 785
Malleable
iron
Pearilitic 217-233
BHN spherodized
93.9 4.33 0.046 172 61.8 2.48 0.040 172
Nodular
iron grade
60-45-10
As cast
grade 60-45-10
73.2 5.37 0.073 412 54.5 4.65 0.085 412
Nodular
iron
As cast
auto crankshafts
99.3 6.14 0.062 125 74.9 4.62 0.062 125
e following Equation (4.4) [4] can be used to estimate shearing yield strength by using
tensile yield strength:
S
sy
S
y
D
(
0:5 based on the Maximum shear stress failure theory
0:577 based on distortion—energy failure theory;
(4.4)
where S
y
is tensile yield strength and S
sy
is shear yield strength.
3. Mean and standard deviation if the range of material mechanical properties are given.
If the range of a material mechanical property X is given, we can assume that X follows a
normal distribution with a mean
X
and a standard deviation
X
. We can use the following
equations to estimate the mean and the standard deviation of the material mechanical
property X :
X
D
X
min
C X
max
2
X
D
X
max
X
min
8
;
(4.5)
where X
min
and X
max
are the lower and upper limits of the material mechanical property
X, respectively.
176 4. RELIABILITY OF A COMPONENT UNDER STATIC LOAD
4. Mean and standard deviation if only the average of material mechanical property is given.
For a material mechanical strength such as yield strength or ultimate strength, if only an
average value is available, we assume that it follows a normal distribution and could use
the following approach to estimate its mean and standard deviation:
X
D X
average
X
D 0:1
X
D
X
X
D 0:1X
average
;
(4.6)
where X
average
is the average of the sample data of the material mechanical property.
X
is
the coefficient of variance.
X
and
X
is the mean and the standard deviation of its normal
distribution.
For Young’s modulus E, shear Young’s modulus G and the Poisson ratio , if their aver-
ages are known, we can assume that they follow a normal distribution and use following
equations [6] to estimate their mean and standard deviation.
For Young’s modulus E,
E
D 0:01
E
D E
average
E
D
E
E
D 0:01E
average
;
(4.7)
where
E
is the coefficient of variance of the material Young’s modulus. E
average
is the
average of sample data of the material Young’s modulus.
E
and
E
are the mean and the
standard deviation of normally distributed material Young’s modulus.
For shear Young’s modulus G,
G
D 0:025
G
D G
average
G
D
G
G
D 0:025 G
average
;
(4.8)
where,
G
is the coefficient of variance of the material shear Young’s modulus. G
average
is
the average of sample data of the material shear Young’s modulus.
G
and
G
are the mean
and the standard deviation of normally distributed material shear Young’s modulus.
For Poisson ratio ,
D 0:042
D
average
D
D 0:042
average
;
(4.9)
4.5. ESTIMATION OF SOME DESIGN PARAMETERS 177
where
is the coefficient of variance of the material Poisson ratio.
average
is the average
of sample data of the material Poisson ratio.
and
are the mean and the standard
deviation of normally distributed material Poisson ratio.
5. e stress concentration factor.
e stress concentration factor is a function of geometric shape and dimension. Since the
geometric dimension is a random variable, the stress concentration factor is also a random
variable and typically follows a normal distribution. We can use the following equations
to determine the mean and the standard deviation of stress concentration factor [6, 7]:
K
D 0:05
K
D K
Table
K
D
K
K
D 0:05K
Table
;
(4.10)
where
K
is the coefficient of variance of the stress concentration factor. K
Table
is the stress
concentration factor obtained from tables in current design handbooks or design books.
K
and
K
are the mean and the standard deviation of normally distributed stress con-
centration factor.
Example 4.6
A material with only three tensile tests has the following mechanical properties, as shown in
Table 4.11. If all mechanical properties are assumed to be normal distributions, estimate their
distribution parameters of this material mechanical properties.
Table 4.11: Average values from three tensile tests
S
y-average
(ksi) S
u-average
(ksi) E
average
(ksi)
ν
average
49.2 61.3 2.91 × 10
4
0.282
Solution:
We can use Equation (4.6) to estimate the mean and standard deviation of the yield strength and
the ultimate strength. Per Equations (4.7) and (4.9), we can estimate the mean and the standard
deviation of Young’s modulus and Poisson ratio, respectively.
We can use the relationship ration G D
E
2.1 C /
among Young’s modulus, shear Young’s
modulus, and Poisson ratio to calculate the shear Youngs modulus, and then use Equation (4.8)
to estimate the mean and standard deviation of the shear Young’s modulus.
We can use Equations (4.3) and (4.4) to estimate the ultimate shear strength and the shear
yield strength, respectively, and then use Equation (4.6) to estimate their mean and standard
deviation.
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