Engineering Rock Mechanics

Questions and answers: rock musses 137

Weathered granite: dashed Intact granite: solid lines lines 8.00

t

t t

---._ ----..

-- friction angle

------s-*

___---- --

______------

___--- ______------

_---- ___ -- - - - - 1.00

0 -r*- I 0.00

__-*-

0.0 2.0 4.0 6.0 8.0 10.0

Normal stress. MPa

1

5.00

4.00 E

3.00 5

3

• m
  +-
  2.00 =

48.10 To study the influence of fracture persistence, consider a rock

mass containing two fracture sets, A and 6, mutually inclined at

clay infilling, while fracture set B is rough, clean and intermittent
with a 'two-dimensional proportional extent of fracturing' of 0.5. A
Plot the expected peak strength of this rock mass as a function of 07
the orientation of the principal stresses to the fracture directions,
for a minor principal stress of 5 MPa and given the following peak

45O, as shown to the right. Fracture set A is continuous and contains B

e

03

A
strength characteristics: to,
Intact rock: u1 = 75 + 5.29~3
Fracture set A: c = 100 kPa, 9 = 209
Fracture set B: c = 0, Q = 35'

A8.10 The key to answering this question is replacing the shear strength
of fracture set B with an effective shear strength that is a weighted
average of the intact rock and joint parameters

= [XCB + (1 - X) ci] + 0" [X taneB + (1 - XI tanei]

where x is a fracture : intact rock proportionality constant. In this case,
we have x = 0.5.
We require the shear strength parameters of the intact rock in B--t
space. The strength criterion has been given in principal stress space in
the question, but is easily converted using the equation
01 = (2ccos#)/(l - sine) + (I + sin#)/(l - sin#)a3.
Comparing this relation with the given strength criterion, we see that
5.29 = (1 + sin#i)/(l - sin&) + sin#i = 4.2916.29 +- +i = 43" and hence

2ci cos 43

1 - sin43

75 (1 - sin43)

2 cos 43

75 = 3 ci = = 16.3MPa

To find the strength parameters for the intermittent fracture set, we
combine the values obtained for ci and #+ together with the given values