Environmental Engineering Reference
In-Depth Information
the failure threshold becomes negative were provided herein for illustration.
Table 15.5
indi-
cates that the failure threshold decreases with the successive levels until reaching a negative
value at the last level. This means that the samples generated by the SS successfully progress
toward the limit state surface
G
= 0. In order to select the failure threshold of a given level,
the calculated values of the performance function of this level were arranged in an increasing
th value of the arranged values of the performance function. Since
N
s
= 10 and
p
0
= 0.2, the
failure threshold is equal to the third value of the arranged values of the performance func-
tion. The SS computation continues until reaching a negative value (or a value of zero) of the
failure threshold. In this example, the negative value was reached in the sixth level (where
equal to zero. For this reason,
C
6
was set to zero. This means that the last conditional failure
probability
P
(
F
6
|
F
5
) is not equal to
p
0
. In this case, the last conditional failure probability
P
(
F
6
|
F
5
) is calculated as the ratio between the number of samples for which the performance
P
(
F
6
|
F
5
) is equal to 3/10 = 0.3. Thus, the failure probability of the footing under consideration
is equal to 0.2
5
× 0.3 = 9.6 × 10
−5
. It should be emphasized that the failure probability calcu-
of the failure probability,
N
s
should be increased. This number should be greater than 100
to provide a small bias in the calculated
P
f
value (see
Chapter 4
by Honjo in Phoon 2008).
In order to determine the optimal number of samples
N
s
to be used per level, different
values of
N
s
were considered to calculate
P
f
and its coefficient of variation COV
Pf
as shown
values of
N
s
, the failure threshold decreases with the successive levels until reaching a nega-
tive value at the last level.
ues of
N
s
, the failure probability largely changes with
N
s
. However, for high values of
N
s
,
the failure probability converges to an almost constant value.
Figure 15.3a
indicates that
2200 samples per level are required to accurately calculate the failure probability. This is
because (i) the
C
j
values corresponding to
N
s
= 2200 and 2400 samples are quasi-similar as
are respectively equal to 2.60 × 10
−3
and 2.63 × 10
−3
).
Table 15.6
Evolution of the failure threshold
C
j
with the different levels
j
and with the number
of realizations
N
s
when
p
0
=
0.2
Failure
threshold
C
j
for each
level j
Number of samples N
s
per level
10
100
200
1000
2000
2200
2400
C
1
1.4875
0.9397
1.0071
1.0638
1.0532
1.0466
1.0803
C
2
0.9740
0.4157
0.3969
0.4916
0.4467
0.4466
0.4942
C
3
0.7391
0.1011
0.1016
0.1513
0.1434
0.1347
0.1549
C
4
0.4007
−0.0491
−0.0437
−0.0307
−0.0616
−0.0536
−0.0564
C
5
0.1573
-
-
-
-
-
-
C
6
−0.0937
-
-
-
-
-
-
p
f
9.60
×
10
−5
2.80
×
10
−3
2.72
×
10
−3
2.20
×
10
−3
2.80
×
10
−3
2.60
×
10
−3
2.63
×
10
−3
221.4
57.9
42.1
18.7
13.3
12.8
12.4
COV
P
f
(%)
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