Civil Engineering Reference
In-Depth Information
[
]
{} {}
{
()
}
Pn
>
n Z
,
t
,
Rt
[16.16]
F
t
i
i
i
1
i
1
i
1
i
n
=
ˆ
{} {
}
{
( )
}
=
PY
≤ −
1
W
Z
,
t
,
Rt
t
t
t
i
i
i
1
i
1
i
1
i
i
i
i
1
Now, taking the expectation of the expression in Eq. (16.16) over the dis-
tributions of { Z t i }, { t i } and { R ( t i )}, we obtain
n
ˆ
=
(
) =
{} {}
{
()
}
Pn
>
n
E
PY
≤ −
1
W
Z
,
t
,
Rt
[16.17]
F
t
t
t
i
i
i
i
i
i
1
i
1
i
1
i
1
where E [
] is the expectation. Since Ŷ t i does not depend on { t i } and { R ( t i )},
and depends only on Z t i , we have
ˆ
{} {}
{
()
}
(
)
PY
≤−
1
W
Z
,
t
,
Rt
=
F
1
W Z
[16.18]
ˆ
t
t
t
i
i
t
t
i
1
i
1
i
1
YZ
i
i
i
i
i
From Eqs. (16.17) and (16.18) we have,
n
=
(
) =
(
)
Pn
>
n
E
F
1
W Z
[16.19]
ˆ
F
t
t
YZ
i
i
i
1
It must be noted that in Eq. (16.19), W t i is a function of { t i } and { R ( t i )} and
the expectation is obtained over the distribution of { Z t i } i ≥1 , { t t } i ≥1 and { R ( t i )}
i
1 . The expression in Eq. (16.19) is general and as a special case, if N ( t ) is
a Poisson process with rate v , then t i has a Gamma distribution (Ang and
Tang, 2007) with parameters v and i where the mean is i / v . In the case of
no deterioration, P ( n F > n )
=
(1
P f ) n , where P f is the probability of failure
given the occurrence of a load.
Probability distribution of t n F . Considering failures due to excessive
demand only, the distribution for time to failure t n F is given by:
)
(
Nt
ˆ
(
) =
{} {}
{
()
}
Pt
>
t
E
PY
≤ −
1
W
Z
,
t
,
Rt
[16.20]
n
t
t
t
i
i
F
i
i
i
i
1
i
1
i
1
i
=
1
Therefore,
()
Nt
(
) =
(
)
Pt
>
t
E
F
1
W Z
[16.21]
n
ˆ
t
t
F
YZ
i
i
i
=
1
In particular, if N ( t ) is a Poisson process with rate v , then
()
n
ν
t
n
ν
t
e
!
(
) =
∑∏
(
)
Pt
>
t
E
F
1
WZ
[16.22]
n
ˆ
t
t
F
YZ
i
i
n
i
=
1
n
=
0
Probability distribution of t F . Now considering the failures due to both
excessive demand and excessive deterioration, the distribution for time
to failure t F is estimated as follows:
 
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