Engineering Rock Mechanics

3 1 6 Design of surface excavations

418.1 A foundation is proposed on a rock mass consisting of a
series of horizontal strata. The results of a site investigation and
associated laboratory testing programme are shown in the table
below.

Depth Description

(m) (%) elastic deformation

RQD Intact rock Fracture

modulus modulus

0 Moderately weathered sandstone 61 4.0 18.0

8 Slightly weathered limestone 80 39 115

10 Fresh, massive limestone 100 40 120

End of hole at 12 m

3 Slightly weathered sandstone 98 5.0 20.0

(a) Calculate the effective elastic modulus for the uppermost 10 m

of the rock mass.

(b) A plate-loading test using a 0.5 m diameter rigid plate at

the ground surface has indicated an elastic modulus of 1 .O GPa. By

considering the moduli of the various strata, explain why this might

be. What value of modulus should be used in the design?

(Aide memoire: For a rock mass consisting of i strata, each of which

contains a single set of planar, parallel fractures, oriented parallel

to the strata boundaries, the effective elastic modulus normal to the

fractures, Em, may be calculated using the expression

where, for each of the i strata, t; = thickness, €; =Young's modulus of

the intact rock, ).; = fracture frequency, Edi = modulus of deformation

of the fractures.)

A18.1 (a) If we assume a negative exponential distribution for discon-

tinuity spacings, then the relation between RQD and fracture frequency

is

RQD = 100 (At + 1) exp (-At)

For the moderately weathered sandstone, the RQD of which is 61%

(using the standard threshold value of t = 0.1 m), the corresponding

fracture frequency is 13.48 m-'. For the slightly weathered sandstone,

for which RQD = 98%, the corresponding frequency is 2.15 m-'. The

slightly weathered limestone has an RQD of 80% and hence a frequency

of 8.24 m-l. The frequency of the fresh, massive limestone is not re-

quired, as we only need to compute the modulus for the uppermost

10 m.