Engineering Rock Mechanics

362 Underground excavation instability mechanisms

For the left-hand tunnel and point A, the parameters required for the
Kirsch solution are k = 1.0, a,, = 11.0 MPa, a = 1.5 m and r = 5 m. The
induced stresses at point A are then

ae = !a” 2 [(l +k) (1 + $) + (1 - k) (1 +33 cos28]

= a,, (1 + $) = 11.99 MPa

and

a--~,, (1+k) 1-- -(l-k) 1-4-+3- cos28

r-i [ ( ::) ( a2 r2 a4) r4 1

= a,, (1 - $) = 10.01 ma.

For the right-hand tunnel and point A, the parameters required for the

Kirsch solution are identical to those used above. Furthermore, as the in

situ stress state is isotropic, then the stresses induced at point A due to

the right-hand tunnel are the same as those due to the left-hand tunnel,

i.e. = 11.99 MPa and = 10.01 MPa.

In order to superpose the above results, we must first ensure that the

stress states are referred to the same co-ordinate system. Later, we are

asked to consider the stability of a horizontal fault through point A,

and on this basis it is appropriate to use the system of axes shown. As

a result, we must transform the stresses computed for the right-hand

tunnel. Here for simplicity we will use stress transformation equations

based on Mohr’s circle.

At point A and for the right-hand tunnel, the angle to be turned

through to rotate from the direction of the major principal stress (i.e. a@)

to the direction that is normal to the fault (i.e. vertical) is

(Y = arctan ($) = 53.1”.

The transformed stresses are then

11.99 + 10.01 11.99 - 10.01

cos(106.2) = 10.72 MPa

2 +2

00 +a, 00 - 0,

2 2

+- cos 2(a! + 90)

11.99 + 10.01 11.99 - 10.01

cos(286.2) = 1 1.28 MPa

2 +2

ne - 11.99 - 10.01

T,, = -- sin(2a) = - sin(106.2) = -0.95 ma

2 2

When we superpose stress states computed using the Kirsch solution,

we must remember to subtract the additional field stresses that are