Questions and answers: in situ rock stress 49such as regions of thrust faults where the stress is being transmitted
through small areas of the fault;
the fractures are formed in a vertical plane, which can be checked
with an impression packer;
the rock is impermeable, which is acceptable for rapid tests in
granite, but we can incorporate permeability if required; and
the stress concentration around the borehole can be estimated by
the Kirsch equations (see ERM 1, Section 19.2); this depends on
the rock being linearly elastic, and is an acceptable assumption in
unweathered granite.
For a vertical fracture, a horizontal section through the borehole
is as in the sketch above. From the Kirsch equations for the stress
concentrations around a circular hole in an elastic material, the equation
for the breakdown pressure is found to be: PB = 30h - CTH +aI where PB is
the breakdown pressure, ah is the minor horizontal stress, OH is the major
horizontal stress, and at is the tensile strength of the rock. This equation
results because the water has to be pressurized to a value that will
overcome the stress concentration of three times the minor horizontal
principal stress and the tensile strength. However, because the maximum
horizontal principal stress is aiding the breakdown pressure, it appears
as a unit negative stress concentration in the equation.
Rearranging gives OH = 30h - PB + at. We know that a, is 10 MPa, and
we will assume that the density of the rock is 27 kN/m3.
Test 2 PB = 14.0 MPa, PS = 8.0 MPa, depth, z = 500 m, vertical
stress, a, = yz = 0.027 x 500 = 13.5 MPa, ah = the shut-in pressure,
Ps = 8.0 MPa
CTH = 3ah - PB + a, = 3 x 8.0 - 14.0 + 10.0, and hence CH = 20 MPa.
Thus, UH = 20 MPa > a, = 13.5 MPa > ah = 8 MPa, which would
produce a vertical fracture as assumed (because the fracture develops
perpendicular to the least principal stress 5).
The ratio, k, between the mean of the horizontal stresses and the
vertical stresses is :(oh + uH)/az = i(8.0 + 20.0)/13.5, giving k = 1.04.
Test2 PB = 24.5 MPa, PS = 16.0 MPa, depth, z = lo00 m, vertical
stress, a, = yz = 0.027 x lo00 = 27 ma, ah = the shut-in pressure,
Ps = 16.0MPa.
OH = 30h - PB f 0, = 3 X 16.0 - 24.5 + 10.0 = 33.5 MPa.
Thus, CTH = 33.5 MPa > 0, = 27 MPa > ah = 16 MPa, which again would
produce a vertical fracture, as assumed.
For this test, we have k = ;(ah + UH)/~~ = i(16.0 + 33.5)/27.0, giving
k = 0.92.
sNote that for this question, the principal stresses have already been assumed to be
vertical and horizontal. More advanced hydraulic fracturing methods using pre-existing
fractures are available in which the orientation of the principal stresses can be determined
from the measurements.