Engineering Rock Mechanics

Questions and answers: rock masses 123

is written out,

tl +t2+t3+...

• tl + - t2 + -.. t3..
  El E2 E3

Fractured strata
Although the total thickness of the rock mass is the sum of the thickness
of each stratum, L = xi ti, because the fractures are now present the
equation should also include a term to account for the thickness (i.e.
aperture) of each fracture. However, this is assumed to be negligible
compared to the thickness of the intact rock and has been ignored here.
As before, the strain in the intact rock of each unit is ~i = 8i/ti = a/Ei,
and so the displacement of each unit due to straining of the intact rock is
Si = ti (a/ Ei). For any fracture, its normal displacement is 8d = a/ Ed, and
SO, for n fractures in unit i, the displacement is 8di = ni8d = hiti(a/Edi).
The total normal displacement is the sum of the displacements due to
the intact rock and the fractures and is thus 8~ = xi 8i + xi 8di which, on
substitution of the expressions for ai and 8di, becomes

a a

8T = Eli- + xkiti- =

i Ei Edi i

The total strain is thus

8T

L

&=-I^1 J

ti ti

i i

and the rock mass modulus is

ti

The influences of the fracture frequency and fracture deformation
moduli values on this composite rock mass modulus, E,, are illustrated
below.
Note that in the diagram below for the conditions assumed, the frac-
ture deformation modulus is more important than the fracture frequency.
The value of E, does not reduce much when many fractures are present
providing the fracture deformation modulus is high (the Ed = loo0 GPa
curve in the diagram below). On the other hand, just one or two frac-
tures every few metres will have a significant effect on E, if Ed is, say
10 GPa/m or less. This diagram also shows that it is critical to make an
assessment of fracture normal stiffness in order to be able to understand
the rock mass modulus.