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Rock Mechanics.pdf - Mining and Blasting

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Figure 16.6 (a) Rectangular block<br />

geometry, <strong>and</strong> (b) assumed distribution<br />

of water pressure with depth, for<br />

limiting equilibrium analysis of chimney<br />

caving.<br />

MINING-INDUCED SURFACE SUBSIDENCE<br />

An example of wide applicability is that shown in Figure 16.6a. A block of width<br />

a <strong>and</strong> base length b has one pair of sides oriented in the direction of the strike of the<br />

orebody <strong>and</strong> the other pair in the dip direction. The dip is , the maximum height<br />

of the block is h, <strong>and</strong> the water table is a distance d below the horizontal ground<br />

surface. The groundwater pressure is assumed to be zero at the water table, to have an<br />

initially hydrostatic rate of increase with depth, to be zero at the stope hanging wall<br />

<strong>and</strong> to have an infinite rate of decrease with increasing depth at this point. The skewed<br />

parabolic water pressure distribution shown in Figure 16.6b satisfies these boundary<br />

conditions. For this distribution, the maximum water pressure is z ′ w/2, <strong>and</strong> the total<br />

water pressure force generated over a depth z ′ = z − d, isz ′2 w/3.<br />

In this case, it is necessary to evaluate the shear resistance developed on each of<br />

the four vertical faces (Figure 16.7). The total shear resistance is then given by<br />

Q = 2QBCGF + QDCGH + QABFE<br />

(16.5)<br />

Consider the side face BCGF when 0 ≤ d ≤ h − b sin , i.e. groundwater level intersects<br />

the up-dip face DCGH. If Coulomb’s shear strength criterion is used, equation<br />

16.3 applies. This gives<br />

b cos z<br />

QBCGF =<br />

0 0<br />

(c ′ + k z tan ′ )dz −<br />

(z − d)2<br />

w tan <br />

3<br />

′<br />

<br />

dx<br />

From Figure 16.7, z = h − x tan . Substitution for z <strong>and</strong> integrating gives<br />

QBCGF = Q1 − w tan ′ <br />

b cos h<br />

3<br />

2 − hb sin + b2<br />

3 sin2 <br />

<br />

−d[2h − b sin − d]<br />

where<br />

490<br />

Q1 =<br />

b cos <br />

2<br />

<br />

c(2h − b sin ) + k tan ′<br />

<br />

h 2 − bh sin + b2 sin 2 <br />

<br />

3

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