Engineering Rock Mechanics

Question and answers: design of underground excavations 383

To determine the initial strain, we use the general form of Hooke’s

Ei = - [ai - U(Uj + 04. (20.3)

Now, as these pillars can be considered to be very long out of the plane

of the cross-section, we can assume that they are in a state of plane

strain. Taking the y-direction to be out of the plane of the cross-section,

this allows us to write

law, i.e.

1

E

1

E

O=-[ Cy - v(az + ax)]

and hence discover that

ay = v(a, + a,)

which, upon substitution into Eq. (20.3) leads to

[ (1 - v)a, - va,].

l+v

E, = -

E

Substituting v = 0.2 and E = 9.12 GPa, together with the initial stress

state of a, = 1 MPa and a, = 5 MPa, gives the initial strain in the vertical

direction as 0.5 x Eq. (20.2) then becomes

s

6

Substituting Eq. (20.1)) into Eq. (20.4) leads to

E = 0.5 x 10-3 + -_ (20.4)

6000 59.20

3.837 E + 3.837

p = -~ = -15646 + 15.43

which is the ground characteristic in terms of strain. This can be plotted
along with the pillar characteristic to find the operating point, as shown
in the diagram below. The operating point is then found to be at a pillar
pressure of 11.19 MPa but, much more importantly, is beyond the peak
strength of the pillar. As the figure shows, the system is only marginally
stable.

::::pe ground characteristic

O.OE+OO 2.OE-03 4.OE-03 6.OE-03 8.OE-03 1 .OE-02 1.2E-02 1.4E-02

strain