Engineering Rock Mechanics

Questions and answers: design of surface excavations^33 1

Standard normal randoms Random values

0.353 1.084 -0.232 0.60 51.8 63.5

2.271 0.021 -0.003 0.89 61.4 82.7

0.310 -0.448 0.643 0.60 51.6 63.1

-0.361 0.741 0.414 0.50 48.2 56.4

-0.746 0.269 0.834 0.44 46.3 52.5

0.321 -1.190 0.429 0.60 51.6 63.2

0.975 -1.552 1.076 0.70 54.9 69.8

1.332 -2.155 0.310 0.75 56.7 73.3

-1.607 -2.746 0.186 0.31 42.0 43.9

2.140 -1.312 1.109 0.87 60.7 81.4

-1.140 0.665 -0.193 0.38 44.3 48.6

-0.962 -0.337 -0.306 0.41 45.2 50.4

1.306 -1.412 -1.343 0.75 56.5 73.1

Factor of safety

1.28

0.65

1.30

1.91

7.48

2.83

1.29

4.16

1.01

0.89

0.90

0.68

-4.30

Means: 0.57 50.5 61.0

Notice that some of the factors of safety are negative. This is due to the
combination of random values used, and such a factor of safety is clearly
wrong. In rigorous Monte Carlo simulation they have to be accounted
for, but here we will simply ignore them.
The factor of safety computed using the means of the random values
(shown at the foot of the table) is 1.43, whereas if the actual mean values
quoted at the head of the table are used the factor of safety is 1.51. This
discrepancy is simply due to the limited number of trials performed.
However, if the mean of all of the positive factors of safety is taken, the
result is 1.86, which is a very different value. This indicates that the mean
factor of safety taken over all of the trials is not the same as the factor
of safety computed using the mean values of the random variables, and
shows why Monte Carlo simulation should be performed.
The factors of safety are plotted as a histogram below.
0.25 T proportion

Even if we ignore the number of trials that result in a negative factor of

safety, there is a substantial proportion with a factor of safety less than one.

Indeed, for these results we have N(O < F) = 32, N(O < F < 1.0) = 7,