120 3. COMPUTATIONAL METHODS FOR THE RELIABILITY OF A COMPONENT
max
in the critical section. erefore, the limit state function of this component will be:
g
.
S
u
;
max
/
D S
u
max
D
8
ˆ
ˆ
<
ˆ
ˆ
:
> 0 Safe
D 0 Limit state
< 0 Failure:
(a)
Since both S
u
and
max
are normally distributed random variables and the limit state function
g
.
S
u
;
max
/
is a linear function of S
u
and
max
, the limit state function will be a normally
distributed random variable too. e mean
g
and the standard deviation
g
can be calculated
by Equations (3.7) and (3.8):
g
D
S
u
max
D 62:5 30:3 D 32:2 .ksi/ (b)
g
D
q
2
S
u
C
2
max
D
p
5:8
2
C 15:4
2
D 16:456: (c)
e reliability of the component can be calculated by Equations (3.9), (3.10), or (3.11). Use
Equation (3.11) to run the calculation. In this example, the reliability index ˇ per Equa-
tion (3.12) is:
ˇ D
g
g
D
32:2
16:456
D 1:9567: (d)
Per Equation (3.11) and Table 2.10, the reliability of the component is:
R D ˆ
.
ˇ
/
D ˆ
.
1:9567
/
D 0:9748:
3.3.3 COMPUTATION OF RELIABILITY WHEN BOTH ARE
LOG-NORMAL DISTRIBUTIONS
When both component strength index S and component stress index Q follow lognormal dis-
tributions, the event .S > Q/ is the same as the event .ln S > ln Q/ because both S and Q are
always positive. erefore, for this case, the limit state function of a component can be expressed
as
g
.
S; Q
/
D ln S ln Q D
8
ˆ
ˆ
<
ˆ
ˆ
:
> 0 Safe
D 0 Limit state
< 0 Failure:
(3.13)
Now, since ln S and ln Q are normal distributions, the limit state function g
.
S; Q
/
will be a
normal distribution too. Per Equations (3.7) and (3.8), the mean and standard deviation of the
limit state function in this case are:
g
D
ln S
ln Q
(3.14)
3.3. RELIABILITY OF A COMPONENT WITH TWO RANDOM VARIABLES 121
g
D
q
2
ln S
C
2
ln Q
; (3.15)
where
ln S
and
ln S
are the log-mean and log-standard deviation of the log-normally dis-
tributed component strength index S.
ln Q
and
ln Q
are the log-mean and log-standard devi-
ation of the log-normally distributed component stress index Q.
Equations (3.9), (3.10), or (3.11) can be used to calculate the reliability of a component
when both are log-normal distributions.
Example 3.6
e failure mode of a component will be a fatigue failure. According to the design information,
at the specified cyclic stress level, both the number of cycles to failure N
f
of the component and
the service number of cycles n
s
are log-normal distributions. e log mean and log standard de-
viation of N
f
are
ln N
f
D 13:305 and
ln N
f
D 0:121. e log mean and log standard deviation
of n
s
are
ln n
s
D 11:886 and
ln n
s
D 0:654. Calculate the reliability of this component.
Solution:
Per the provided information, the component strength index S will be the number of cycles
to failure N
f
of the component at the specified cyclic stress level. e component stress index
Q will be the service number of cycles n
s
at the specified cyclic stress level. So, the limit state
function of the component is:
g
N
f
; n
s
D N
f
n
s
D
8
ˆ
ˆ
<
ˆ
ˆ
:
> 0 Safe
D 0 Limit state
< 0 Failure:
(a)
Since both N
f
and n
s
are positive variables, the limit state function of this component can be
expressed equivalently by the following equation. We will use the following equation to solve
this example:
g
N
f
; n
s
D ln.N
f
/ ln.n
s
/ D
8
ˆ
ˆ
<
ˆ
ˆ
:
> 0 Safe
D 0 Limit state
< 0 Failure:
(b)
Per Equations (3.14) and (3.15), we have the mean
g
and standard deviation
g
of the limit
state function:
g
D
ln
.
N
f
/
ln
.
n
s
/
D 13:305 11:886 D 1:419 (c)
g
D
q
2
ln
.
N
f
/
C
2
ln
.
n
s
/
D
p
0:121
2
C 0:654
2
D 0:6651: (d)
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