Engineering Rock Mechanics

37% Design of underground excavations

Compute the water pressure required to open the critical fracture

Using these results, predict the flow regime and overall stability of

Does this affect the layout of the boreholes and operation of the

set (1) 250 m above, (2) 250 m below, and (3) at the target horizon.

the rock mass when water is injected at the target horizon.

system, and if so, how?

A20.3 The relations given for the in situ stress magnitudes are

a, = 262, ah = 6 + 122 and OH = 15 + 282.

The unit weight of rock is usually about 25 kN/m3, and so it appears
from these equations that the depth, z, has units of kilometres, with the
constant of proportionality having units of MPa/km. The constant of
proportionality in the relation for a, is comparable in magnitude to the
unit weight of rock, and so shows that these are total stress relations. To
convert them into effective stress relations, we need to subtract the unit
weight of water, giving

ai = (26 - 1O)z = 162, a,!, =^6 + (12 - 1O)z =^6 + 22, and

~fr = 15 + (28 - 10)~ = 15 + 182.

At the target horizon, which is situated at a depth of 2 km, the in situ

effective stresses are then ai = 32 MPa, a; = 10 MPa and a; = 51 MPa.

In order to determine which of the three fracture sets is most suscept-

ible to being jacked open by the pressurized water, we need to determine

the effective normal stress acting on each of the sets. For the horizontal

set, this is the vertical stress, giving aLl = 32 MPa. For the two vertical

sets, we need to apply a stress transformation in the horizontal plane

to determine the normal stresses. The geometry of the problem in the

horizontal plane is as shown below.

= 51

I set 3

This diagram shows that the angle between the major principal stress com-

ponent in the plane, a;, and the normal to set 2 is 130" - (75" + 90") = -35",