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Rock Mechanics.pdf - Mining and Blasting

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Figure 7.6 A plane of weakness,<br />

oriented perpendicular to the major<br />

principal stress, intersecting a circular<br />

opening along the horizontal diameter.<br />

EXCAVATION DESIGN IN MASSIVE ELASTIC ROCK<br />

For the special case of a zone of influence defined by a 5% departure from the field<br />

stresses, A is set equal to 10 in the preceding expressions.<br />

7.3 Effect of planes of weakness on elastic stress distribution<br />

In excavation design problems where major discontinuities penetrate the prospective<br />

location of the opening, questions arise concerning the validity of elastic analysis in<br />

the design process <strong>and</strong> the potential effect of the discontinuity on the behaviour of the<br />

excavation periphery. It is now shown that, in some cases, an elastic analysis presents<br />

a perfectly valid basis for design in a discontinuous rock mass, <strong>and</strong> in others, provides<br />

a basis for judgement of the engineering significance of a discontinuity. The following<br />

discussion takes account of the low shear strengths of discontinuities compared with<br />

that of the intact rock. It assumes that a discontinuity has zero tensile strength, <strong>and</strong> is<br />

non-dilatant in shear, with a shear strength defined by<br />

= n tan (7.3)<br />

As observed earlier, although the following discussion is based on a circular opening,<br />

for purposes of illustration, the principles apply to an opening of arbitrary shape. In<br />

the latter case, a computational method of stress analysis would be used to determine<br />

the stress distribution around the opening.<br />

Case 1. (Figure 7.6) From the Kirsch equations (equations 6.18), for = 0, the<br />

shear stress component r = 0, for all r. Thus rr, are the principal stresses xx,<br />

yy <strong>and</strong> xy is zero. The shear stress on the plane of weakness is zero, <strong>and</strong> there is no<br />

tendency for slip on it. The plane of weakness therefore has no effect on the elastic<br />

stress distribution.<br />

Case 2. (Figure 7.7a) Equations 6.18, with = /2, indicate that no shear stress<br />

is mobilised on the plane of weakness, <strong>and</strong> thus the elastic stress distribution is not<br />

204

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