Mining - Deflection of Mine Roof Supports

Any design of a mine roof in bedded deposits which ignores differential deflections at the supports can quickly lead to dangerous overstressing. As illustrated by the typical case presented on page 1029, if one support deflects only 0.1 in. more than its neighbor, stresses 50 pct higher than anticipated will result. Previous studies1-3 have assumed that the mine roof acts like a beam clamped at both ends, with no deflection or slope' at the supports (Fig. 1). Essentially this assumes:5* ?A = ?, = ?b and dA = dB = 0. However, a host of conditions can cause differential sag or deflection at one support, and even an apparently trivial deviation can have major consequences in the beam: variations in spacing, size, and material of supports; local settlement due to floor heaving or lower mine openings; non-uniformity in wedging or bolt tensioning.8 In one particular instance concrete pillars, poured in place, shrank away from the roof upon setting.' When understood and controlled, differential sag becomes a definite advantage, permitting the transfer of loads from one support to another. Where one support is a bolt or prop and the other a pillar or rib, the load on the former can be reduced while the load on the pillar or rib is increased. This means two things: 1) savings can be effected by using smaller artificial supports such as props and bolts and 2) loads can be increased in the rib to induce eventual failure for longwall caving. In this last respect, another factor underlying coal bumps becomes understandable. The design of a mine roof by beam theory assumes that when a bedded formation is homogeneous, iso-tropic, and elastic, it will act like a uniformly loaded beam, fixed or clamped at both ends,6 s in Fig. 1.'" These are important assumptions. The uniform load is the weight of the beam itself or its body load. For a beam of unit width, this load becomes per foot: w = yhb. w = uniform load y = specific weight, lb per ft3 h = depth or thickness of beam, ft b = width of beam, ft Since b = 1, w - yh [ 1 ] A beam fixed at both supports is statically indeterminate to the second degree, that is, two equations in addition to those for static equilibrium are required to determine the reactions at the supports. These equations result from the geometry. From statistics: ^F,= 0;RA + RB — wL = 0 [2] wU S,M0 = 0 ; MB H------------MA — RBL = 0 [3] it Using superposition in which the left support is kept intact and the right one resolved into an active force and moment, we obtain from Fig. 2:" ?b = ?m + ? b2 + ?B3 = 0 dB = dB1 + dB2 + dB3 / wU Ma RBL \ V d°=(-r+-2—r)-zr [5] Where: Subscripts refer to position on axis of beam R = force, lb M = moment, ft-lb w = uniformly distributed load, lb per ft L = span, ft 0 = slope, radians d = deflection, ft E = Young's Modulus, lb per ft" 1 = Moment of inertia about the neutral axis, ft4 Solving Eqs. 2 through 5 for the four unknown reactions, we obtain: 2 L3 Fig. 3 is obtained by plotting MA and M, vs d, and then plotting RA and R, vs d, and adjusting their scales to coincide with MA and M, vs d,. It should be noted in Eqs. 6 and 7 that as dB increases, MA increases the same amount that Mn decreases. Eqs. 8 and 9 show a similar relationship between RA and R,. Proof that MA is the maximum absolute moment occurring in the beam is as follows: where M, = moment at x wx-When where M,., = maximum positive moment Eqs. 6 and 8 into above: For small dn: MA> M,,,