128 Rock masses: deformability, strength and failureDiagram 2 -two orthogonal fracture sets.The formula E, = 2RMR - 100 gives E, = 124 - 100 = 24 GPa.
The formula E, = 107 gives E, = 107 = 1040 = 20 GPa.
RMR-IO 62-10 52
~The formula E, = 610- gives E, = E1Ov = 9GPa.
In the last formula, the square root factor has been incorporated by Hoek
and Brown (see the reference in Footnote 4) to reflect the increasing role
of the intact rock as it becomes weaker.
These empirical methods can provide surprisingly good estimates and
they are often the only practical way of estimating the in situ deformation
modulus, but they should be used with caution. An understanding of
the roles of the intact rock modulus and fracture stiffness contributions
assists in evaluating the reliability of empirical estimates for given
circumstances. We include further aspects of rock mass classification
systems in Chapter 12.
Q8.5 To study the effect of a fracture on the rock strength, plot a
graph of (al - a3) vs. flw using the single plane of weakness formula
included in Section 8.1. Assume a3 = 10 MPa, c = 0 and &, = 35O,