Engineering Rock Mechanics

Questions and answers: rock masses 129

• 140-
  4 120.1
  r
  E:

5 100

L E 80..

60-

40-

a x 20-

0-

Y

and that the intact rock strength is given by u1 = 75 + 5.29~3. Also,

explain the form of the resulting graph.

..

I I

j (90 + @w)/2 = 62.5'

I. I

A8.5 The plot is shown below, Initially, as BW is increased from 0", i.e.

from when the fracture is perpendicular to the major principal stress

and parallel to the minor principal stress, the fracture does not induce

failure and the rock strength is constant, as the strength of the intact

rock: 129.7 MPa.

Failure of intact rock

48.6 If a rock mass contains more than one fracture set, we can

apply the single plane of weakness theory to each set, and superim-

pose the results to find a lowest-bound envelope of strength.

(a) Plot the 2-D variation in strength for a rock mass containing

two orthogonal sets of fractures, A and 6, the strengths of which

are CA = 100 kPa, (PA = 20° and CB = Oo, (PB = 355 when the minor

principal stress has the value 10 MPa. The intact rock strength is

again given by 61 = 75 + 5.29~3.

(b) Howwould this strength variation change if the minor principal

stress were reduced to zero?

A8.6 (a) For fracture set A only, the locus of rock strengths (see next
page) has been generated using the single plane of weakness formula in
Section 8.1.
The strength locus for fracture set B only, is similar, although the
minimal strength is higher than that for fracture set A.
To obtain the strength locus for a rock mass containing both sets of
fractures, we superimpose the strength loci for the two fracture sets
and take the minimal strength at each Bw angle, i.e. the envelope of