Engineering Rock Mechanics

395

I

I

I

(where mi is the value of m for intact rock) and substituting a1 = 110 MPa,
a3 = 4 MPa and s = 1, we find that

I

,_____________________L_______--__---______~__-________-_______- I I

mi = [ (v)2 -s] = [ (x)2 110 - 4 - l] 100 = 3.09.

4

Using this value of mi together with RMR = 64, we obtain

RMR- 100 64 - 100

m = mi exp (

28

) = 3.09 exp ( 28 ) = 3.09 x 0.277 = 0.86