300 Surkrce excavation instability mechanisms
- Lower bound theorem. If any stress distribution throughout the struc-
ture can be found which is everywhere in equilibrium internally and
balances certain external loads and at the same time does not violate the
yield condition, those loads will be carried safely by the structure.
An upper bound solution results from an analysis in which a geometry of
discrete blocks is postulated and the associated forces then determined, and
a lower bound solution results from an analysis in which the sustainability
of a stress distribution is analysed.
At the surface of a rock mass, the applied and in situ stresses are generally
so low as to prevent ductile behaviour and plasticity theorems will be
inapplicable. However, the concepts can be applied usefully to rock
foundations by:
(a) using the upper bound analysis in the study of foundations where the
instability is governed by the movement of rigid blocks along pre-
existing discontinuities; and
(b) using lower bound analysis in the study of foundations where the
instability is governed by a general yielding of the rock material, which
could occur for highly loaded weak rocks.
Discontinuurn analysis. In Fig. 17.14, there is a cross-section through a
uniform line loading of width D on a rock foundation containing three
discontinuities. For simplicity in this analysis, the discontinuities are
assumed to have some cohesion but zero angle of friction, although the
analysis can easily take account of a non-zero angle of friction. Application
of the equations of static equilibrium to the forces shown acting on the free-
body diagrams of the two wedges (also shown in the figure) permits
calculation of the applied load which will cause instability of the system
and, for the geometry and discontinuity strength shown, this is p = 6c.
The analysis refers to the problem of a discontinuous rock and the
solution is mechanically correct. However, if the analysis were being
considered as part of a plastic analysis of a continuum, then this solution
BFy = O,i+ ve, block 11:
Geometry W + cDV2/.\/2 + CD-N~IV~ = 0
=-N, = (W + 2cD)d2
Total load = Dp
ZF, = 0, -. +ve, block 11:
N2-N,1d2-~D~2/d2 = 0 -N2= W + 3cD
Z
N1 ZF, = 0,d + ve, block I: N, = (W + 4cD).\/2Free body diagrams ZF, = O,i+ ve, block I: &
Figure 17.14 Equilibrium analysis of a foundation on discontinuous rock.