Engineering Rock Mechanics

Questions and answers: underground excavation instabiliiy mechanisms 365

situ case can be computed using

and so for the roof we find

croof = 0.429 [4.59 - (-20.5)) + (-20.5) = -9.74 MPa

and for the side wall we find

csidewall = 0.429 [84.2 - 1041 + 104 = 95.5 MPa.
In addition, the magnitude of the vertical stress is
ov = yz = 0.028 x 750 = 21 MPa.
For an elliptical opening oriented with its major axis horizontal, the
circumferential stress induced in the boundary at the top and bottom of
the opening is

W2

where ptop = -

2H

Combining these equations and substituting q = W/H leads to

atop = 0, [f +k - I]

and from this, knowing that q = 4 for this particular opening, we find

that

2 .0.429

(Tto, = 0” (9 + k - 1) = 21 (

The stress induced at the ends of this elliptical opening is given by

where pend = H/2 because the opening is an ovaloid. These relations
lead to

@end = nu (1 - k + 2&)

from which we obtain

(Tend = (^21) (^1 - 0.429 + 2& 1 = 96.0 MPa.
These approximate solutions agree reasonably well with the results
found from the rigorous numerical analysis. The first graph below gives
a visual assessment of how the various computed stresses compare, and
shows that for the side walls the agreement is very good, although for
the roof and floor the agreement is not so good. The second graph below
shows how the stresses compare for values of k = 0 and k = 1 for a
range of values of q. With the exception of the case for stress induced in
the roof when k = 1, the agreement is very good. Taken as a whole, these
results show that the approximate solution can give a good insight into
the behaviour of the opening.