180 Testing techniques
Set1 Set2 Set3 Set4
Scanline 1 8.0 88.0 76.0 69.0
Scanline 2 48.2 41.5 125.1 76.6
Scanline3 66.2 112.6 105.2 131.0
Scanline4 46.9 126.8 53.4 87.7
Scanline5 87.6 111.7 17.3 51.9
Scanline 6 73.4 57.3 68.0 9.8and from this we obtain the matrix of I cos 0,i 1 values,
e=
0.990 0.035 0.242 0.358
0.667 0.749 0.576 0.232
0.403 0.384 0.262 0.656
0.683 0.599 0.597 0.040
0.041 0.370 0.955 0.617
0.286 0.540 0.375 0.985
As a result, the frequencies of the fracture sets are found to be
A = (Q'. Q)-'. (0'. A) =2.138 1.268 1.283 1.108 20.457-1[ 1.283 1.268 1.498 1.454 1.454 1.867 1.2221 1.374 [ Wi] = I:] rn-]
Knowing the fracture frequencies of the sets enables the calculation of
the fracture frequencies along lines of any orientation in a rock mass, as
required in the next question in which the fracture frequencies have been
slightly changed.QI 1.4 As part of a site investigation study, a rock mass was found to
contain four fracture sets with dip/dip direction and frequencies as
follows (see figure on next page):1.108 1.222 1.374 1.965Set 1 : 08/145,3.48/m
Set 2: 88/148,3.91 /m
Set 3: 76/021,3.58/m
Set 4: 69/087,3.26/m
In order to establish in which directions through the rock mass an
excavation will encounter the minimal and maximal numbers of
fractures, the fracture frequency in different directions through the
rock mass, A,, was calculated using the formula 1, = E:=, 11; cos&I
(see 47.6 and A7.6). The results are presented below on a hemi-
spherical projection, with the contouring corresponding to the frac-
ture frequency values in the different directions.
Explain from first principles why the directions of the minimal and
maximal frequencies occur where they do.