Engineering Rock Mechanics

350 Underground excavation instabiliiy mechanisms

OJD.. I Boundarv stresses at e = 0" and 180"

1 t = 90" and 2fO": (To = ~~(3k-l)

0 k

-1

Figure 19.11 Stress concentration factors due to a circular opening.

The highlights of this diagram are:

(a) under all stress fields, the opening alters the pre-existing state of stress,
i.e. the opening produces stress concentrations;
(b) there is a linear variation with k of the stress concentrations at the points
A and B (and, indeed, everywhere on the boundary);
(c ) in a uniaxial stress field (k = 0), the maximum stress concentration is
3 (i.e. compressive ), and the minimum stress concentration is -1 (i.e.
tensile);
(d) for a hydrostatic stress field (k = l), the stress concentration is 2
everywhere on the boundary (note that this may be demonstrated using
the information in (c) above, where the superposition of two orthogonal
and equal uniaxial stress fields results in the stress concentration of
3 + -1 = 2);
(e) tension on the boundary can only occur if k < Y3.

In the hydrostatic case (k = l), the stress concentration around the
excavation boundary is always 2p,. The solution for stresses anywhere
within the rock mass for this stress state is similarly simplified because there
are no shear stresses: the terms (1 - k) are all zero. Hence the equations for
radial and tangential stress reduce to

or = p,{l- (az/?)> and or = p,{l + (a2/?)>.

For many practical applications, it is useful to superpose the solution for
the stresses induced in the rock by a uniform internal pressure, p, with such
a pressure being due to either fluid pressure (water or mud for boreholes)
or support pressure (for tunnels and shafts). The contributions made by an
internal pressure to the radial stress, tangential stress, radial displacement
and tangential displacement are, respectively,

or = p(uz/?), oe = -p(az/?), u, = pa2/2Gr and ue=O.

If, again, we consider the case when k = 1, but now the opening is inter-
nally pressurized, the superposition of the above solutions gives

or = pz - (pz - p)(a2/?) and 00 = p, + (p, - p)(az/?).