352 Underground excavation instability mechanisms
the lower mass of rock is replaced with a statically equivalent force p,A,
where A is the area of the imaginary plane under consideration. If one
now considers the case after the tunnel has been excavated, as shown in
Fig. 19.12(c), it can be seen that the portion of rock mass excavated no
longer transmits any stress, but the statically equivalent force must remain
the same as in Fig. 19.12@) to preserve equilibrium. However, this force
now acts over a smaller area than before (because the rock which is in the
location of the tunnel has been removed) and so the stress across the
horizontal plane must be increased.
The load no longer carried by the rock removed from the tunnel
is p, x 2, per unit length of tunnel, and, in order to preserve equilibrium,
this must equal the load re-distributed into the surrounding rock (Fig.
19.12(d)). This can be demonstrated by using the Kirsch equations for a
uniaxial stress state (i.e. k = 0), and determining the vertical stress across
the horizontal plane in question by taking the expression for tangential
stress with^6 = 0,
To obtain the total redistributed load, subtract p, from this expression and
integrate between the limits of a and 00 (the Kirsch equations are defined
for an infinite volume of rock, which implies the area A of the horizontal
plane is also infinite) for both sides of the tunnel,redistributed load = 2pzr( -$ + $)di
which reduces to 2pza, the load bearing capability lost by tunnel
excavation.
By integrating between limits of a and 3a, or a and 5u, the percentage of
the load is obtained which is re-distributed within one tunnel diameter and
three tunnel diameters, respectively, on either side of the excavated tunnel.
These are 81.5% and 89.6%, showing that it is the region close to the tunnel
in which the load re-distribution occurs. For this example, the loadTunnel diameters into the rock
Figure 19.13 Redistribution of vertical load adjacent to a circular tunnel.