198 4. RELIABILITY OF A COMPONENT UNDER STATIC LOAD
Table 3.2 in Section 3.8. e estimated reliability of this component R is:
R D 0:9840: (c)
e estimated probability of the component failure F is
F D 1 R D 1 0:9840 D 0:016: (d)
e relative error of the probability of the failure is
"
D
0:0124:
(
e)
So, the range of the probability of the component failure with a 95% confidence level will
be:
F D 0:016 ˙ 0:016 0:0124 D 0:016 ˙ 0:0002: (f)
erefore, the range of the reliability of the pin with a 95% confidence level will be:
R D 1 F D 0:9840 ˙ 0:0002: (g)
4.8 RELIABILITY OF A SHAFT UNDER TORSION
4.8.1 RELIABILITY OF A SHAFT UNDER TORSION FOR A STRENGTH
ISSUE
A circular shaft under torsion T will have maximum shear stress on the outer layer of the shaft.
e following equation will calculate the maximum shear stress:
max
D
T d
o
=2
J
; (4.26)
where
max
is the maximum shear stress on a cross-section. T is resultant internal torque on the
cross-section of a shaft. d
o
is the outer diameter of a shaft on the cross-section. J is the polar
moment inertia of the shaft cross-section.
When the maximum shear stress of a shaft under torsion exceeds shaft material’s yield
strength, the shaft is treated as a failure. e limit state function of a solid shaft with circular
cross-section under torsion will be:
g
S
sy
; K
s
; d
o
; T
D S
sy
K
s
T d
o
=2
J
D S
sy
K
s
T d
o
=2
d
4
o
=32
D S
sy
K
s
16T
d
3
o
D
8
ˆ
<
ˆ
:
> 0 Safe
D 0 Limit state
< 0 Failure;
(4.27)
4.8. RELIABILITY OF A SHAFT UNDER TORSION 199
where S
sy
is material shear yield strength. K
s
is the shear stress concentration factor under tor-
sion. T and d
o
have the same meaning as those in Equation (4.26).
e limit state function of a hollow shaft with circular cross-section under torsion will be:
g
S
sy
; K
s
; d
o
; d
i
; T
D S
sy
K
s
T d
o
=2
J
D S
sy
K
s
T d
o
=2
d
4
o
d
4
i
=32
D S
sy
K
s
16T d
o
d
4
o
d
4
i
D
8
ˆ
<
ˆ
:
> 0 Safe
D 0 Limit state
< 0 Failure;
(4.28)
where d
i
is the inner diameter of a hollow shaft. e rests of parameters in Equation (4.28) have
the same meaning as those in Equation (
4.27).
When the component’s torsion is described by a PMF, we can use the total probability
theorem Equation (2.24) in Chapter 2 to calculate the reliability of the components. In this case,
the torque is expressed by PMF as:
P
.
T D T
i
/
D p
i
i D 1; 2; : : : ; n; (4.29)
where p
i
is the PMF when the torque T is equal to T
i
. Since it is a PMF, we have:
R D
n
X
iD1
.
p
i
R
i
/
; (4.30)
where R is the reliability of a shaft under such torsion. R
i
is the reliability of a shaft when the
torque T is equal to T
i
. Even T
i
can be a constant value, we still need to use Equations (4.27)
or (4.28) to calculate R
i
because other variables in the limit state function are still random
variables.
Equations (4.27), (4.28), and (4.30) can be used to calculate the reliability of a shaft by the
H-L method, R-F method, or Monte Carlo method which have been discussed in Chapter 3.
Two examples will demonstrate how to calculate the reliability of a shaft under torsion.
Example 4.14
e stepped shaft is with a smaller shaft diameter d
1
D 0:750 ˙ 0:005
00
and a larger diameter
d
2
D 1:000 ˙ 0:005
00
as shown in Figure 4.12. e fillet radius is 1=16
00
. e stepped shaft is
subjected to a torque T D 1000 ˙ 160 (lb/in). e shear yield strength of the shaft material
follows a normal distribution with a mean
S
sy
D 32,200 (psi) and the standard deviation
S
sy
D
3630 (ksi). Use the H-L method to calculate the reliability of the shaft under the torsion.
200 4. RELIABILITY OF A COMPONENT UNDER STATIC LOAD
T
T
d
2
= 1.000 ± 0.005˝
d
1
= 0.750 ± 0.005˝
r =
1
˝
16
Figure 4.12: A stepped-shaft under a torsion.
Solution:
1. e limit state function of the shaft on the stepped cross-section.
e critical section for this shaft will be on the stepped section with a fillet because of
stress concentration effect. e limit state function of the shaft for this problem is:
g
S
sy
; K
s
; d
o
; T
D S
sy
K
s
16T
d
3
1
D
8
ˆ
<
ˆ
:
> 0 Safe
0 Limit state
< 0 Failure:
(a)
ere are four random variables in the limit state function (a). e shear stress concen-
tration factor K
s
can be treated as a normal distribution. Its mean can be obtained from
current design handbook or some websites about the stress concentration factor. In this ex-
ample, the mean of the shear stress concentration factor K
s
is 1.92. Use Equation (4.10) to
estimate the standard deviations. e shaft diameter d
1
D 0:750 ˙ 0:005
00
can be treated as
a normal distribution. Its mean and standard deviation can be calculated by Equation (4.1).
e torque T D 1000 ˙ 160 (lb/in) cab be simplified as a normal distribution. Its mean
and standard deviation can be calculated by Equation (4.2). e distribution parameters
of all four random variables in the limit state function (a) are listed in Table 4.29.
Table 4.29: e distribution parameters of random variables for Equation (a)
S
sy
(psi)
Normal distribution
K
s
Normal distribution
d
1
(in)
Normal distribution
T (lb/in)
Normal distribution
μ
S
sy
σ
S
sy
μ
K
s
σ
K
s
μ
d
1
σ
d
1
μ
T
σ
T
32220 3630 1.92 0.096 0.750 0.00125 1000 40
4.8. RELIABILITY OF A SHAFT UNDER TORSION 201
2. e reliability of the shaft by the H-L method.
e limit state function (a) contains four normally distributed random variables and is
a nonlinear function. We will follow the H-L method discussed in Section 3.6 and the
program flowchart in Figure 3.6 to create a MATLAB program. e iterative results are
listed in Table 4.30. From the iterative results, the reliability index ˇ and corresponding
reliability R of the shaft in this example are:
ˇ D 2:2999 R D ˆ
.
2:2999
/
D 0:9893:
Table 4.30: e iterative results of Example 4.14 by the M-L method
Iterative#
S
y
*
K
s
*
d
1
*
T
*
β
*
|∆β
*
|
1
32220 1.92 0.75 1390.074 2.215268
2
25067.94 2.003943 0.749891 1035.755 2.300594 0.085326
3
24537.88 1.987212 0.749909 1022.464 2.299889 0.000705
4
24527.04 1.986438 0.74991 1022.418 2.299888 1.39E-07
Example 4.15
A solid shaft with a constant cross-section is subjected to a torque. e diameter of the shaft is
d
o
D 1:000 ˙ 0:005
00
. e torque can be described by a PMF:
P
.
T D T
i
/
D
(
0:45 T D 3500 (lb/in)
0:55 T D 4500 (lb/in):
e shear yield strength of the shaft material follows a normal distribution with a mean
S
sy
D
32,200 (psi) and a standard deviation
S
sy
D 3630 (ksi). Use the Monte Carlo method to calcu-
late the reliability of the shaft.
Solution:
1. Limit state function of the shaft under the torsion.
Since the torque in this example is described by a PMF, we will have two limit state func-
tions for this shaft.
202 4. RELIABILITY OF A COMPONENT UNDER STATIC LOAD
For T D 3500 (lb/in),
g
S
sy
; d
o
D S
sy
16 3500
d
3
o
D
8
ˆ
<
ˆ
:
> 0 Safe
0 Limit state
< 0 Failure:
(a)
For T D 4500 (lb/in),
g
S
sy
; d
o
D S
sy
16
4500
d
3
o
D
8
ˆ
<
ˆ
:
> 0 Safe
0 Limit state
< 0 Failure:
(
b)
ere are two random variables in the limit state functions. e shaft diameter d
o
D
1:000 ˙ 0:005
00
can be treated as a normal distribution. We can use Equation (4.1) to
calculate its mean and standard deviation. eir distribution parameters in this example
are listed in Table 4.31.
Table 4.31: e distribution parameters of random variables for Equations (a) and (b)
S
sy
(psi) Normal distribution d
0
(in) Normal distribution
μ
S
sy
σ
S
sy
μ
d
0
σ
d
0
32220 3630 1.000 0.00125
2. Reliability of the shaft.
e reliability of the shaft in this example can be calculated by Equation (4.30)
R D p
1
R
1
C p
2
R
2
; (c)
where p
1
D 0:45 for T D 2200 (lb/in). R
1
will be determined by the limit state func-
tion (a). p
2
D 0:55 for T D 2500 (lb/in). R
2
will be determined by the limit state func-
tion (b).
We will use the Monte Carlo method to calculate the reliability R
1
and R
2
of this example.
e Monte Carlo simulation method has been discussed in Section 3.8. We can follow the
Monte Carlo method and the program flowchart in Figure 3.8 to create a MATLAB pro-
gram. Since this problem is not complicated, we can use the trial number N D 1,598,400
for a key component from Table 3.2 in Section 3.8. e estimated reliability of this shaft
R is:
R D p
1
R
1
C p
2
R
2
D 0:55 0:99997 C 0:45 0:99466 D 0:99758:
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