Predictions OF natural in situ stress stutes based on elasticity theory 57as the depth below the ground surface increases, due to the weight of the
overburden. As rules of thumb, taking the typical density of rock into
account:
1 MPa is induced by 40 m of overlying rock, or
1 psi is induced by 1 ft of overlying rock.More generally, we should use the expression
induced vertical stress, 0, = p MPa
where z is the depth, measured in metres, below the ground surface and
yis the unit weight, measured in MN/m3. Examples of yare:
y = 0.01 MN/m3 for some coals,
= 0.023 MN/m3 for some shales,
= 0.03 MN/m3 for gabbro.
This approach is always used as an estimate of the vertical stress
component unless, of course, the stress determination programme does
include direct measurement of the vertical stress. We have seen, for
example, that during the course of data reduction in hydraulic fracturing,
the vertical stress component is estimated by this technique. Conversely,
using the CSIRO gauge, the complete stress tensor is determined and so it
is not necessary to estimate the vertical stress component. We will be
discussing later in this chapter whether the measured stress states do
correspond to such preconceived notions.
4.6.2 The horizontar/ stress components
We now consider the magnitudes of the horizontal stress components.
Given that the vertical stress has a particular magnitude at a point in a rock
mass, we might expect that a horizontal stress would be induced as a result
of the vertical compression of the rock. To provide an initial estimate of this
stress, based on elasticity theory and assuming isotropic rock, we must
introduce the parameters Young’s modulus and Poisson’s ratio (a more
detailed treatment of the elastic constants is given in Chapter 5 and a
discussion of the validity of elasticity theory itself is given in Chapter 10).
In Fig. 4.13, an illustration of an element of rock being uniaxially stressed
is given-the applied axial stress is 0, and the resulting axial strain is E,.
There is also a lateral strain induced, E~, because the element expands
laterally as it is being axially compressed. From these values, we define the
Young’s modulus and Poisson’s ratio as:
axial stress - oa
axialstrain E,Poisson’s ratio,~ = lateral strain - _- E,
axialstrain E,Young‘s modulus, E = _-
Utilizing these parameters, we can derive expressions for the skain along
any axis for the small cube at depth in a rock mass illustrated in Fig. 4.13(c).