Engineering Rock Mechanics

Question and answers: design of underground excavations 385

approach is required to solve the system. The iteration follows the steps
as shown in the diagram above.
Although it is easiest to determine the operating point of each pil-
lar graphically and perform the iteration manually, it is reasonably
straightforward to use a spreadsheet program to perform the iteration
automatically. In this case, calculations are required in order to find an
operating point in terms of the intersection of a ground characteristic
and the appropriate segment of the pillar characteristic.
An example of the solution, starting with an assumed pressure of
25 MPa at pillar A, is given in the table below. This particular starting
pressure has been selected simply for demonstration: a better initial
estimate could have been made on the basis of tributary area theory.

Itera- Pillar A Pillar B
tion
(MPa) (MPa) (MPa) (MPa) (MPa) (MPa)

m c pressure strain m c pressure strain

1 25.00

2 -1666 3.739 3.178 3.368 x

3 -1564 14.50 11.18 2.126 x

4 -1666 12.345 9.891 1.473 x

5 -1564 12.54 10.11 1.552 x

6 -1666 13.01 10.29 1.631 x

8 -1666 13.05 10.31 1.642 x

9 -1564 12.42 10.04 1.519 x

10 -1666 13.05 10.31 1.643 x lop3

11 -1564 12.42 10.04 1.519 x

12 -1666 13.05 10.31 1.643 x lop3

7 -1564 12.42 10.05 1.521 x 10-3

The operating conditions are therefore pillar pressures of 10.0 MPa and

10.3 MPa at A and B, respectively. The figures below show how the system

converges.

Ground and pillar characteristics at iterations 2 and 3:

14.0

12.0

10.0

a m

s 8.0

vi

??

c v, 6.0

4.0

2.0

0.0

O.OE+OO 2.OE-03 4.OE-03 6.OE-03 8.OE-03 1 .OE-02 1.2E-02 1.4E-02

strain