Engineering Rock Mechanics

1 02 Fractures and hemispherical projection

0.1

and from this we find that the number of pieces longer than 0.2 m is

N0.2 = 0.057 Nt = 0.057 1922 = 109.5,

which rounds down to 109.
To compute the mean length of these pieces, we need to know their
overall length. This is found by evaluating the RQD for a threshold value
of 0.2 m,
RQDo.2 = lOO(14.344 - 0.2 + 1) exp(-14.344 - 0.2) = 22.0%,
and then computing the overall length as
LO.* = RQDo.2 L, = (22.0/100) , 134 = 29.4 m.
The mean length of the pieces is then

• Number of pieces - - -Overall length of pieces -- 0.100
  -Theoretical mean length
  O.Oo0

0 Computed mean length

Lo,* 29.4

20.2 = - = - = 0.27 m.

N0.2 109

This mean is much greater than the mean of all pieces (which is the

reciprocal of the frequency, or 0.07 m), and so it is interesting to see how

this mean length varies with the threshold we choose. The plot below

shows this.

1000 1 I 0.900

0.800

0.700

0.600 g

v

0.500 =

P

0.400 ;

8

0.300 z

a,

Q7.2 Based on a sample of 128 fracture spacing values which gave

a mean spacing of 0.215 m, estimate the range of the population

mean fracture spacing and frequency at the 80% confidence level.

How many fractures should be in the sample for an error of flO% at

the 90% and 95% confidence levels?