Engineering Rock Mechanics

Questions and answers: underground excavation instability mechanisms 367

inclination of the long axis to the horizontal, and x is a parameter for

locating position on the boundary. These parameters are shown in the

sketch below, together with another location parameter 8.

For the case when B = 45", then cos 2(x - B) = cos(2x - 90) = sin 2% and
cos2B = cos90 = 0. Substituting these values into the equation above,
reduces it to

In order to determine the locations on the boundary where the circum-
ferential stress is a maximum or a minimum, we can differentiate this
expression with respect to x, set the result to zero and solve for x. Thus,
we have

and, by equating this to zero and rearranging, we find

(1 + k)(l - 9)

(1 -k)(l +q)'

tan2x = -

For this particular excavation, we have q = 2.5 and k = 25.5p.5 = 3,
which leads to

= -20.3".

The form of the tangent function means that both x = -20.3" and
x = 90"+(-20.3") = 69.7" satisfy this equation. If we substitute these into
the equation for 08 in order to find the magnitude of the circumferential
stress at these orientations, we obtain

8.5

• (1 - 3) [(I + 2.5)' cos(-40.6)]} = 104.2 MPa

00 = ~ { (1 + 3) [ (1 + 2.52) + (1 - 2.52) sin(-40.6")]

2. 2.5

and
8.5
ne = - { (1 + 3) [ (1 + 2S2) + (1 - 2S2) sin( 139.4")]
2. 2.5

• (1 - 3) [(l + 2.5)'cos(139.4")]} = -5.6 MPa.