Page 236 - Rock Mechanics For Underground Mining
P. 236

EXCAVATION DESIGN IN MASSIVE ELASTIC ROCK












































              Figure 7.19 Problem geometry for  necessary to consider other mechanisms to explain the mode of action of support and
              determining the effect of support  reinforcement systems.
              load on the elastic stress distribution  The following example is intended to illustrate the main features of the function
              around an elliptical opening.
                                        of surface support on excavation peripheral rock, and is based on an analysis by
                                        Ladanyi (1974). It is not a basis for comprehensive description of the interaction
                                        between installed support and tunnel peripheral rock, which is a more complex,
                                        statically indeterminate problem. The problem geometry is shown in Figure 7.20a,
                                        with a circular opening excavated in a medium subject to hydrostatic stress. The field
                                        stresses and rock mass strength are such that an annulus of failed rock is generated in
                                        the excavation periphery. The main questions are the relation between the radius, r e ,
                                        of the failed zone, the applied support presure, p i , and the stress distribution in the
                                        fractured rock and elastic domains. It is assumed that the strength of the rock mass is
                                        described by a Mohr–Coulomb criterion, i.e.

                                                                   (1 + sin  )  2c cos
                                                               1 =   3       +
                                                                   (1 − sin  )  1 − sin
                                        or

                                                                     1 = b  3 + C 0                    (7.9)
                                        218
   231   232   233   234   235   236   237   238   239   240   241