Engineering Rock Mechanics

Questions and answers: design of surface excavations 3 1 7

From these values we find that the modulus for the rock mass is

ti

E, =

= 2.11 GPa.

• 
  
  
  
  • 3+5+2
  
  
  
  

(b) The calculated value is greater than the value measured in the

If we consider greater thicknesses of the rock mass in turn, then we

plate-loading test of 1.0 GPa.

find the following:

(^1) -I
Em(Om-3m) = = 1 .OO GPa
18.0
= 1.76 GPa.
’ 3+5
13.48
Em(Om-8m) =
(& + ?ET) 3 + (A + E) * 5
This shows that the plate-loading test has determined the modulus
of only the uppermost 3 m of the rock mass. This is a feature of small
diameter plate loading tests, as a consideration of the vertical stress
induced on the plate axis shows.
For a loaded area (rather than a rigid plate, for which there is no
unique analytical solution) the magnitude of this stress is given by
where z is the depth below the surface and a is the radius of the loaded
area. A plot of ov/p against z/a shows how the induced stress rapidly
reduces with depth, with the 5% influence depth being 5.4~. For the case
of a 0.5 m diameter plate, this represents a depth of 1.35 m.
proportion of applied stress
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0