Engineering Rock Mechanics

Questions and answers: underground excavation instability mechanisms 359

distance of the 5% zone of influence for the hydrostatic case and examine
the tunnel spacings on this basis.
For the first tunnel we have a = 1.5 m, from which we find that
r(5%) = ad% = 1.5m = 6.7 m. For the second tunnel, for which
a = 3.0 m, we find r(5%) = 13.4 m. The centre-to-centre spacing of
the tunnels is 10 m. Thus, on the basis of the 5% zone of influence,
we see that the first tunnel is affected by the second, but the second
is not affected by the first. This means that the stresses induced on the
boundary of the second tunnel are as computed above, but for the first
tunnel we need to use an approximation. We do this by computing the
stress state induced by the second tunnel at the position of the centre of
the first tunnel, and then using this computed stress state to determine
the stresses induced on the boundary of the first tunnel.
As the centre-to-centre distance of the tunnels is 10 m, and as the two
tunnels are at the same elevation, we use r = 10 m and 0 = 0” in the
Kirsch equations. As the second tunnel has a radius a = 3 m, for the case
of k = 0.3 we find 08 = 12.48 MPa and a, = 4.20 MPa. The shear stress is
zero. This means that, for the first tunnel, we have a: = 12.48 MPa and
k’ = 4.20/12.48 = 0.336. The stress induced in the crown and invert is
then found to be

08 = 0; [ (1 + k’) + 2( 1 - k’) COS 281

= 12.48 [(1 + 0.336) + 2(1 - 0.336) cos 1801 = 0.12 MPa

and the stress induced in the sidewalls is

08 = 0; [(I + k’) + 2(1 - k’) cos 281

= 12.48 [(l + 0.336) + 2(1 - 0.336) COSO] = 33.25 MPa.

For the case of k = 2.5 the stresses induced at the centre of the first
tunnel by the second tunnel are = 13.33 MPa and a, = 24.46 MPa,
which means that, for the first tunnel, we have 0; = 13.33 MPa and
k’ = 24.46/13.33 = 1.84. The stress induced in the crown and invert is
then found to be

00 = 0; [ (1 + k’) + 2( 1 - k’) cos 281

= 13.33 [(l + 1.84) + 2(1 - 1.84) cos 1801 = 60.05 MPa

and the stress induced in the sidewalls is

= 0; [ (1 + k’) + 2( 1 - k’) cos 281

= 13.33 [(l + 1.84) + 2(1 - 1.84)cosO] = 15.53 MPa.

As a check of the accuracy of the approximate method, we can analyse
the problem using a computational method. For the case of a CHILE
material under plane strain, a simple 2D boundary element method
is ideal for computing the induced stresses. Here, we have used the
program BOUND, as presented by Beer and Watson (1992) *.

Beer G. and Watson J. 0. (1992) Introduction to Finite Element and Boundary Element
Methods for Engineers. Wiley, Qlichester. Note that the program BOUND is a simple
educational tool and ideal for investigating problems such as the one in Q19.7.