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)