Page 253 - Rock Mechanics For Underground Mining

ROOF BEAM ANALYSIS FOR LARGE VERTICAL DEFLECTION

potential for compressive failure of the roof beam should be based on the minimum
value for c for the transversely isotropic rock unit.
The possibility of shear failure of the roof beam by slip at the abutments can be
assessed directly from the calculated beam loads. Referring to Figure 8.7b, the lateral
thrust, T , at the abutment must mobilise a frictional resistance to slip sufficient to
provide the abutment shear force, V . The maximum shear force that can be mobilised
is

1
F = T tan = f c nt tan
2
and the abutment shear force is given by
1
V = st
2
The factor of safety against abutment shear failure is therefore defined by
f c n
FoS slip = tan (8.12)
s
To establish an engineering estimate for the yield threshold of the buckling limit, a
relation between the midspan deflection, , and the beam thickness, t, was proposed.
In various numerical studies, it was found that ultimate failure corresponded to a
displacement equivalent to approximately 0.25 t. The onset of non-linear behaviour
wasobservedtooccuratadeflectionequivalenttoabout0.1 t.Thatvaluewasproposed
as an allowable yield limit in roof design.

8.5 Roof beam analysis for large vertical deflection

The preceding analysis was based on the implicit assumption that the mechanics of
a voussoir beam were controlled by the central vertical crack and involved small
vertical deflections of the roof. The problem could then be solved from simple statics
using the free body diagram of Figure 8.7. In a more exhaustive analysis, Sofianos
(1996) allowed for the possibility of large vertical deflections of the roof beam. He
showed that the assumption by Evans of n = 0.5 was inadequate, and that the iterative
method proposed by Brady and Brown (1985) was appropriate for small defections
only. Starting from the free body diagram of Figure 8.7, and using a slightly different
notation with xx in place of f c , the moment equilibrium equation is

1
2
1
M A = ts = M R = xx ntz (8.13)
8 2
or
1 s 2
xx = (8.14)
4 nz
The problem is to find the equilibrium values of xx , n and z. Sofianos proposed that
to provide an efficient and quick analysis, the value of n could best be determined
by independent numerical methods. From the finite element studies of Wright (1972)
and the UDEC analysis of Sofianos et al. (1995), a graphical relation was established
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