Page 306 - Rock Mechanics For Underground Mining

ENERGY, MINE STABILITY, MINE SEISMICITY AND ROCKBURSTS

that the excavation boundary be traction free, for t > 0. Simplifying the expressions
for the case = 0, equations 10.58 and 10.59 yield
3
A =−a p/4G
a p/4G
B =−(1 − 2 ) 1/2 3
The expression for f (equation 10.57) then becomes

3
3
f = (−a p/4G) exp(− 0 T )[cos 0 T + (1 − 2 ) 1/2 sin 0 T ] + (a p/4G)
(10.60)
The various partial derivatives of the function f required to establish displacement
and strain components around the spherical opening are given by
2

f =−(1 − 2 ) 1/2 (a p/2G) exp(− 0 T ) sin 0 T (10.61)

f = [(1 − 2 )/(1 − )](ap/2G) exp(− 0 T )[cos 0 T − (1 − 2 ) 1/2 sin 0 T ]
(10.62)

Since the radial displacement is given by
2

u r = (1/r) f − (1/r ) f
the strain components are given, for the spherically symmetric problem, by
2
3

ε rr =−∂u r /∂r = (2/r ) f − (1/r) f − (2/r ) f (10.63)

2
3
ε

=−u r /r = (1/r ) f − (1/r ) f (10.64)
Thetotalstressesinthemediumareobtaineddirectlyfromtheinducedstrains,through
application of the stress–strain relations and superposition of the field stresses; i.e.
rr = + 2Gε rr + p (10.65)

= + 2Gε

+ p (10.66)
where

= ε rr + 2ε
Introduction of the expressions for ε rr and ε

in equations 10.65 and 10.66, and sub-
sequent substitution of the expressions for the function f and its derivatives (equations
10.60, 10.61 and 10.62), produce, after some simplification

a a a a 2a 1/2
3
3 2
− 0 T
rr = pe − cos 0 T + + − (1 − 2 ) sin 0 T
r 3 r r r 3 r 2
3
a
+ p 1 − (10.67)
r 3
a a a a a
3
3 2

= pe − 0 T − + cos 0 T + − +
(1 − ) r 2r 3 (1 − ) r 2r 3 r 2
#
a
3
∗ 1/2
(1 − 2 ) sin 0 T + p 1 + (10.68)
2r 3
288