Engineering Rock Mechanics

386

Ground and pillar characteristics at iterations 11 and 12:

14.0

12.0

10.0

8.0

6.0

4.0

2.0

0.0

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

strain

From the graph of the final state, we can see along which segment of

the pillar characteristic the operating point lies. This allows us to demon-

strate how the problem could be solved were each pillar characteristic a

linear equation.

For both pillars the equation of the appropriate segment of the pillar

characteristic is

PA = 2200EA + 6.700 (20.7)

PB = 2200EB f 6.700 (20.8)

and so Eq. (20.5) though to Eq. (20.8) can be written as the following

system of linear equations:

1. PA + 0.2921~~ + 1563~~ + 0. EB = 15.43

0.6225~~ + 1. PB + 0. EA + 1666~~ = 19.30

1 ' PA +^0 ' PB - 2200E.4 + 0. &B = 6.70'

0. PA + 1. PB + 0. EA - 2200~ = 6.70
   In matrix form these are
   0.2921 1563 0
   0 1666

and the solution of this matrix equation is

1.519 x 10

1.643 x 10

-3

-3

PA

PB

EA

EB

15.43

19.30

6.70

6.70