Engineering Rock Mechanics

Questions and answers: fractures and hemispherical projection 1 03

A7.2 Using the negative exponential distribution, for which the mean
and standard deviation are equal, the standard form of the error band
around the mean, X f OE, can be simplified to X(1 f E). Here, o is the
standard deviation, not stress, and E is the proportional error, not the
strain!
At the 80% confidence level, the standard noma1 variable, z, is found
from tables: Cp(z) = 0.80 + z = 1.282. This is the value of the normally
distributed random variable, z, which has a mean of 0 and a standard
deviation of 1, with 80% of the population within fz.
Hence
E=-=-- z 1.282 - 0.113.
.JNm
The range for the mean spacing value is then given by X(1 f E) =
0.215(1 f 0.113). This gives 0.191 m 5 x 5 0.239 m, with the range of fre-
quency being the reciprocal of these figures: 4.184 m- 5 h 5 5.236 m-l.
Note that the mean of these two frequency values is 4.710 m-, whereas
the frequency given by the mean spacing is 1 = I/X = 1/0.215 =
4.651 m-l. The correct value for the mean frequency is the latter one
(because the reciprocal of the average of two reciprocal values is not the
same as the average of the two values).
To find the sample size required at the 90% and 95% confidence levels,
we use the same technique to find N given E.
Thus, at the 90% confidence level, we have
4 (z) = 0.90 j z = 1.645
and hence

This value has to be rounded up to the next integer (rounding down

will give a sample size that is too small to reach the 90% confidence

level), giving 271 fractures.

At the 95% confidence level, we have

4 (2) = 0.95 j z = 1.960

and hence

1.960 '

N= (:12 - = ( - o.lo ) = 384.2 or 385 fractures.

This shows how the size of the sample increases significantly as the

required Confidence level rises.

47.3 The mean fracture frequency in a vertical direction in a sand-

stone rock mass is 1.22 m-', and a total of 500 vertical 3 m long

rockbolts are to be installed to stabilize the roof of an underground

excavation in this rock mass. How many rockbolts would you expect

to:

intersect no fractures;

intersect less than 3 fractures; and,

intersect more than 4 fractures?