40 In situ rock stress
In the first case, the six components of the stress state are the six
components of the stress tensor with respect to the reference axes. In
the second case, the six components are given by the values of the three
principal stresses, 01, a2, and 03, plus three values providing the orient-
ation information needed to specify the three principal stress directions
relative to the reference axes; for example, the trend and plunge' of the
o1 direction and the trend of the 02 direction, remembering that 01, 02,
and 03 are mutually orthogonal.
In situ stress states, as estimated from geological information or as
determined from measurements during a site investigation, are usually
reported using the principal stresses. An example stress state, plotted on
a lower hemisphere projection, is shown below.
The diagram indicates a stress state
u1 = 16 MPa acting horizontally north-02 = 10 MPa acting vertically, and
o3 = 7 MPa acting horizontally east-The use of hemispherical projection is
explained in Appendix B of ERM 1 and
in this book in Section 7.1.south,7 MPa west.To establish why there is a natural stress state in the rock, we must
consider the geological circumstances. The stress state is caused by three
main factors.
(a) The weight of the rock. In Imperial units, a cube of rock 1 ft x 1 ft
x 1 ft has a base area of 12 in x 12 in = 144 in2. Assume that the cube
is on the floor and that it weighs 144 lbf, a representative value. The
cube will then exert a stress of 144 lbf/144 in2 = 1 lbf/in2 = 1 psi on
the floor. Thus, a useful rule of thumb is that in a rock mass the vertical
stress component will increase by about 1 psi for every foot of depth.
The equivalent rule in SI units is that the vertical stress will increase by
1 MPa for every 40 m of depth, because a representative value for rock
unit weight is 25 kN/m3. The exact values will depend on the local rock
density.
The pressure in a car tyre, say 25 psi, is equivalent to a rock depth of
25 ft or about 8 m. At a depth of 2000 ft, or about 615 m, the vertical
stress component is around 2000 psi or 15 MPa. This is equivalent to the
weight of a car on every square inch of rock, or the weight of 1500 cars
on every square metre.' The terms 'dip direction' and 'dip angle' are used for the orientation of a plane: the
'dip angle' is the angle between the steepest line in the plane and the horizontal; the
'dip direction' is the compass bearing or azimuth of the dip line. The words 'trend and
'plunge' are used for the orientation of a line: the 'trend' is the compass bearing of the
line; the 'plunge' is the angle between the line and the horizontal. Thus, the orientation
of a fracture will be specified by the dip direction and dip angle, whereas the orientation
of the normal to a fradure, or a borehole axis or a principal stress, will be specified by
the trend and plunge.