Engineering Rock Mechanics

126 Rock masses: deformabilify, strength and failure

for a 2-D analysis with uniaxial loading.
The equation is for two orthogonal frac-
ture sets, where E is the modulus of the
intact rock, A, and A2 are the fracture fre-
quencies of the two sets, si , s12, sY1, and
si2 are the normal and shear compliances
for the fractures in each set, and a is the
angle between the applied normal stress
and the normal to the first set.
The equation reduces for one set (Le.
x2 = 0) to
11
Em E

• = - + AS: COS^^ + ~sJ:,cos~asin~u

and if we put si1 = s and si2 = ks, and then

rearrange, we obtain

+ AS cos2a (cos's + ksin2a)

_--^11

Em - E

from which we find

Em - = 1 - EmLs c0s2~ (COS~O~ + ksin2a).

Using E = 2.75 GPa, A = 2 rn-l and s = 0.05 m/GPa, the plots shown

to the right in Diagram 1 result.

For two sets with A1 = A2 = A, si1 = s:, = s and si2 = sz2 = ks, the

basic equation reduces to

E

_- Em - I - E,,,A~ (cos4a + 2k cos2a sin2a + sin4a).

E

Using E = 2.75 GPa, 1 = 2 m-l and s = 0.05 m/GPa, the plots in

Diagram 2 result.

Explain why terms such as c0s4a appear in the formula and

comment on any general principles that are apparent from these

illustrative plots.

A8.3 In Section 3.1, we saw that a cos2 term is required for resolving

stresses - both the force and the area have to be resolved - and

this leads to Mohr's circle representation of a 2-D stress state. Terms

such as c0s4ac appear in the formula above because four resolutions

are now required: two to account for the stress transformation; one to

resolve the fracture frequency (see A7.6); and one to resolve the fracture

displacements.

Although the plots in Diagrams 1 and 2 have been calculated for

specific circumstances, we can note the following general principles.

The rock mass modulus will always be less than the intact rock

modulus (the outer circle in the plots), except when the loading is

parallel to the fractures in a rock mass containing only one set.