1 04 Fractures and hemispherical projectionWhat length should the rockbolts be if 95% of them are required to
intersect at least 3 fractures, i.e. extend to the fourth rock block back
into the rock mass?
A7.3 If we assume that fracture occurrence is a stochastic process, then
we can use the Poisson process which states that the probability of k
events occurring in an interval, n, is given by P(k, x) = e-Ax(Ax)k/k!. In
this question, x = 3 m (the length of a rockbolt), A = 1.22 m-*, and we
have to determine the probability for different values of k.
Intersect no fractures
Here we have k = 0 (i.e. no frachms over the length of the rockbolt), and
so P(0,3) = e-1.22x3(l.22 x 3)O/O! = 0.026 (O! is conventionally taken as 1).
The number of bolts is then 500 x 0.026 = 12.9, which rounds down to
12 bolts. We round downbecause we cannot have a fractional part of a bolt.
Intersect less than 3fractures
Less than 3 fractures includes 0,l and 2 fractures, and so we have
P(< 3,3) = P (0,3) + P (1,3) + P (2,3) = 0.026 + 0.094 + 0.172 = 0.292.
The number of bolts is 500 x 0.292 = 146.1, which rounds down to 246
bolts.Intersect more than 4fractures
Theoretically a bolt can intersect any number of fractures between zero
and infinity. The probability of it doing so is 1 (i.e. certainty). This tells us
that the probability of intersecting more than 4 fractures can be computed
from the Probability of intersecting 4 or fewer fractures asP(> 4,3) = 1 - P(5 4,3) which expands as
=
=
=1 - [P (0,3) + P (1,3) + P (2,3) + P (3,3) + P (4,3)1
1 - [0.026 + 0.094 + 0.172 + 0.210 + 0.1921
1 - 0.695 = 0.305
The number of bolts = 500 x 0.305 = 152.5, which rounds down to 152
bolts.What length of rockbolt is required?
The answer is given by the solution to the equation 0.95 = P(? 3, x).
Expanding and rearranging
0.95 = P(L3,x)=l-P(52,x)
= 1 - P (0, x) - P (1, x) - P (2, x).
Substituting w = Ax and expanding further gives= 1 -e-w (1 +w + f).