Engineering Rock Mechanics

138 Rock masses: deformabilify, strength and failure

for the strength parameters of fracture set B to find

c^ = [OS x 0.0 + (1 - 0.5) x 16.31 = 8.15 MPa

and

tan3 = [OS x tan35 + (1 - 0.5) x tan431 = 0.8164 + f = 39.2"

Having now found the appropriate strength parameters for the inter-

mittent fracture set, the analysis proceeds as in A8.6 and A8.7 with the

results shown below.

Inclination ot discontinuity normal to major Fracture set A:

principal axis CA = 100kPaandf#A = 20

120

p 80

60

aJ

0 30 60 90

Inclination of discontinuity normal to major Fracture set B:

principal axis i? = 8.15 MPa and f = 39.2"

These diagrams are for each set in isolation. In the rock mass we have

& = 45" when PA = O", and so, using set A as the reference, a -45"

shift must be applied to 6 to bring the two sets to a common co-ordinate

system. The resulting plot is as shown on the next page with the heavier

line indicating the composite rock mass strength.

Note that in the second and fourth quadrants, i.e. 90-180" and 270-

360", the strength of the rock mass is only controlled by set A.

8.3 Additional points

We have emphasized that the fractures are the dominant feature gov-

erning rock mass behaviour. These fractures dominate the rock mass

geometry, deformation modulus, strength, failure behaviour, permeabil-

ity, and even the local magnitudes and directions of the in situ stress field.

There is, however, considerable complexity inherent in the geometrical

and mechanical properties of rock mass fracture systems, which should

be taken into account in direct modelling of the specific fractures. For