Rock Mechanics.pdf - Mining and Blasting

LONGWALL AND CAVING MINING METHODS
(c) Using equation 10.84, calculate the mining-induced elastic displacements of the
upper and lower surfaces at the midpoint of each of the excavations if G =
16 GPa and = 0.20.
2 A 1.0 m thick, flat-lying, metalliferous orebody of large areal extent is to be extracted
by a longwall method. The in situ vertical stress at the mining depth is 85 MPa,
and Young’s modulus and Poisson’s ratio of the rock are 48 GPa and 0.20, respectively.
(a) Calculate the critical span of a longwall face under these conditions. What total
amount of energy is released per unit length of stope and per unit volume of
extraction, for this critical span? What values do these quantities approach as
the span increases above the critical?
(b) Partial extraction is to be used to control the energy release rates in this case.
Using equations 10.88 and 15.1, write down an expression for the total energy
released per unit volume of extraction when the orebody is mined in a number of
parallel panels of span, l, spaced on centres of S as in Figure 15.5a. Determine
values of l and S that will restrict this energy release rate to 15 MJ m −3 and
maintain an extraction ratio of at least 0.80. In order to ensure the long-term
stability of the pillars it is necessary that they have width to height ratios of at
least 20.
3 A 2.5 m thick horizontal coal seam at a depth of 230 m is to be extracted by a
series of parallel longwall panels with a single row of pillars between them as shown
in Figure 15.7. The roadways are to be 4.5 m wide and the weighted unit weight of
the overburden materials is = 25 kN m −3 . The distribution of vertical stress with
distance, x, from the pillar rib side is of the type illustrated in Figure 15.9 with the
width of the yield zone xb = 10 m and the peak stress yy = 12 MPa. The vertical
stress, zz, decays with distance into the solid coal away from the yield zone according
to the equation
where v is the in situ vertical stress.
482
(zz − v) = (yy − v) exp{(xb − x)/12}