Engineering Rock Mechanics

384 Design of underground excavations

Three-pillar system
For the three-pillar layout, we have the following three ground charac-
teristics (i.e. one for each pillar)

8A = (56.20 - 3.304~~ - 1.121PJ3 - 0.533~~) x iop3

= (66.52 - 1.121PA - 3.602~~ - 1.121pC) x iop3

6c = (56.20 - 0.533~~ - 1.121PB - 3.304~~) X iop3.

As the layout is symmetric, we can put pc = pA to give

SA = (56.20 - 3.837~~ - 1.121pB) x lop3

SB = (66.52 - 2.242~~ - 3.602pB) x lop3

and from these we find (by using Eq. (20.4))

EA = (9.867 - 0.6395~~ - 0.1868~s) x

ER = (1 1.587 - 0.3737~~ - 0.6003~~) x lop3

Rearranging these equations into the usual format of y = mx + c gives

lo00 9.867 - 0.1868~~

PA = -- (20.5)

0.6395 EA + 0.6395

pB = -- (20.6)
0.6003 E’ + 0.6003
In this case we have four unknowns (EA, EB, PA, pB) and four equations
(the two ground characteristics above and a pillar characteristic for each
of the two pillars). If we had linear equations for the pillar characteristics,
we would be able to solve the system of equations directly. However,
as the pillar characteristics are nonlinear this means that an iterative

lo00 11.587 - 0.3737~~

Assume an initial value for pa

Derive a pillar characteristic formula for pillar B in terms of pe and 6,

I

Find the operating point of pillar B and hence determine ps and 6,

Use the new value of p, to derive a pillar characteristic formula for pillar A in termsofp, and6,

Compare the new values of

pn andp, with the previous

values:are the changes

( Yes:stop]