Engineering Rock Mechanics

382 Design of underground excavations

6~ = (56.20 - 3.304~~ - 0.533~~) x 1 0-3

6c = (56.20 - 0.533~~ - 3.304~~) x 1 0-3

6~ = (56.20 - 3.304~~ - 1.1 21p~ - 0.533pc) x 1 0-3

6B=(66.52-1.121pA-3.602ps-1.121pc) X io-3

6c = (56.20 - 0.533~~ - 1.121~~ - 3.304~~) x

Analysis of these geometries has been undertaken using a CHILE
boundary element program, in order to determine the relation
between displacement and support pressure at the various pillar
locations. These relations are given below the sketches with 6 rep-
resenting displacement and p representing support pressure.
The support pressure is to be supplied by the rock pillars. Perform
a rock-support interaction analysis for each of the two geometries in
order to determine which is the preferred design. The stress-strain
characteristics of the two pillars are as given below:

u (MPa) 5.0 8.0 10.0 11.1 11.4 10.9 10.0 8.7 3.0 0.5

E x 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 7.0 12.0

A20.5 Two-pillar system
For the two-pillar layout, we have the following two ground character-
istics (i.e. one for each pillar):

6A = (56.20 - 3.304P.4 - 0.533~~) x
6c = (56.20 - 0.533~~ - 3.304~~) x
However, as the layout of the rooms is symmetrical, we can put
PA = pc = P and 6A = 6c = 6 to obtain one equation
6 = (56.20 - 3.837~) x lop3. (20.1)
This equation for the displacement of the boundary now has to be
converted into an equation for strain in the pillar. This strain has two
components: an initial strain due to the effect of the in situ stress
before the excavation was made; and a subsequent strain induced by
displacement of the boundary. It is therefore given by

(20.2)

6

h

& = &i +&d = Ei + -

where h, the height of the pillar, is 6 m in this case.