Application of the subset simulation approach to spatially varying soils - Risk and Reliability in Geotechnical Engineering - page 559

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
order as shown in Table 15.5 . Then, the failure threshold was selected as the [( N s × p 0 ) + 1]
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
C 6 = −0.0936) as shown in Table 15.5 . Theoretically, the last failure threshold should be
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
function is negative and the chosen number N s of samples (i.e., 10). According to Table 15.5 ,
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-
lated in Table 15.5 is not accurate due to the small value of N s . For an accurate computation
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
in Table 15.6 . The thresholds corresponding to each N s value were calculated and shown
in this table. Table 15.6 indicates (as was shown before when N s = 10) that for the different
values of N s , the failure threshold decreases with the successive levels until reaching a nega-
tive value at the last level.
Figure 15.3a shows the effect of N s on the failure probability. It shows that for small val-
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
it may be seen from Table 15.6 and (ii) the corresponding final P f values are too close (they
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 (%)