Engineering Rock Mechanics

250 Excavation principles

is the energy required to com-

pletely destroy the rock speci- MPa

men. Express your answer in

joules.

(b) For how long would a do-

mestic 100-W light bulb have

cleared, the broken rock is removed, the next series of blastholes are
drilled, and then charged with explosive, the explosives are detonated
again, and so on. A typical cycle time is about 8 h.
Using a machine for excavation, say a bulldozer with a ripping attach-
ment or a partial face or full-face tunnel boring machine (Fig. 15.3b), the
energy is input to the rock at a much lower rate, but essentially con-
tinuously when the machine is cutting. Although neither the breakage of
rock by explosives nor the breakage of rock by rock cutters is fully un-
derstood, we can apply basic mechanics and empirical rules remarkably
effectively.

f

I

A 15. I (a) The area under the complete stress-strain curve is equivalent
to the product of the stress at failure and the strain at failure, because
the two triangles have the same area as the dashed square. Similarly, the
area under the complete load-displacement curve is the product of the
load at failure and the displacement at failure.
The load at failure = stress at failure x cross-sectional area =
200000000 x n x (50/2 x 1000)2 = 392699 N. The displacement at failure
= strain at failure x specimen length = 0.001 x 0.1 = 0.0001 m. Thus, the
energy used 392699 x 0.0001 = 3.93 x lo5 = 39 Nm. A joule is defined as
the energy expended by 1 N moving through 1 m, or 1 J = 1 Nm, so the
energy required = 39 J.
(b) To establish for how long a domestic 100-W light bulb has to be
illuminated in order to expend the same amount of energy, we need to