Engineering Rock Mechanics

Question and answers: design of underground excavations 379

and the angle to the normal to set 3 is 130n - (155" - 90") = 65", reckoning

anticlockwise as positive. The Mohr circle diagram associated with this

geometry is shown below.

From this diagram we can see that set 3 has the lowest normal stress
acting on it, and will therefore be the fracture set that is first jacked open
by the pressurized water. By inspection of the diagram we see that the
normal stress is

an3 = - uH + u' +- - u' cos 130 = 30.5 + 20.5 cos 130 = 17.3 MPa.
2 2
As a consequence, the boreholes will need to be drilled such that
they intersect these fractures as near to perpendicular as possible; the
required borehole orientation is then either 065/00 or 245/00.
We can now examine the behaviour of these fractures at the three
required depths of 1.75 km, 2.25 km and 2.00 km. We start by comput-
ing the normal and shear stress on the fractures, and then determine
the fluid pressure required to induce shearing on the fracture. These
calculations are shown in the table below, and in these the angle 6J is
130". The limiting normal stress for a given induced shear stress is cal-
culated from the geometry of the strength criterion for the fractures, i.e.
t = a;, tan@ = o;, tan44.

Depth Induced Induced Limiting Critical
normal shear normal fluid
stress, stress, stress, pressure,
.;I u; c=- u& +a; r = a' Hh -u' c+rcose t=rsino ulirn=Z u-ulim
(km) (MPa) (MPa) (MPa) (MPa) (MPa) (MPa) (MPa) (ma)
1.75 46.5 9.5 28.0 18.5 16.11 14.17 14.68 1.43
2.00 51.0 10.0 30.5 20.5 17.32 15.70 16.26 1.06
2.25 55.5 10.5 33.0 22.5 18.54 17.24 17.85 0.69

2 2 tan 44

From this table, we see that the critical fluid pressure decreases with
depth. This means that, once the water pressure at a given horizon ap-