Page 195 - Rock Mechanics For Underground Mining
P. 195

CLOSED-FORM SOLUTIONS FOR SIMPLE EXCAVATION SHAPES

                                        on the point of interest. The various geometric parameters are defined as follows:

                                                                  (W + H)
                                                              e 0 =
                                                                  (W − H)
                                                                      2
                                                                          2
                                                                   4(x + z )
                                                                      1
                                                                          1
                                                              b =    2    2
                                                                  (W − H )
                                                                          2
                                                                      2
                                                                   8(x − z )
                                                                          1
                                                                      1
                                                              d =           − 1
                                                                  (W − H )
                                                                     2
                                                                          2
                                                                       e 0  2   1/2
                                                              u = b +    (b − d)
                                                                      |e 0 |
                                                                       e 0  2   1/2
                                                               e = u +   (u − 1)
                                                                      |e 0 |

                                                                           e+1   z 1
                                                                = arctan
                                                                          e − 1  x 1
                                                                       "            #
                                                                               
 2
                                                                           e+1    z 1
                                                              
 = arctan
                                                                          e − 1   x 1
                                                              C = 1 − ee 0
                                                                       2
                                                              J = 1 + e − 2e cos 2
                                        The stress components are given by
                                                  p(e 0 − e)        2    C
                                               ll =         (1 + K)(e − 1)
                                                     J 2                 2e 0
                                                             J

                                                  + (1 − K)    (e − e 0 ) + Ce cos 2(  +  ) − C cos 2
                                                             2
                                                  p          2
                                              mm =  {(1 + K)(e − 1) + 2(1 − K)e 0 [e cos 2(  +  ) − cos 2 ]}−   ll
                                                  J
                                                  p(e 0 − e)      Ce
                                              lm =          (1 + K)  sin 2  + (1 − K) e(e 0 + e) sin 2
                                                     J 2           e 0

                                                                    J          2
                                                  + e sin 2(  −  ) −  (e 0 + e) + e e 0 sin 2(  +  )  (6.21)
                                                                    2
                                        In applying these formulae, it should be noted that the angle 
, defining the orientation
                                        of the local reference axes l, m relative to the ellipse local axes x 1 , z 1 , is not selected
                                        arbitrarily. It is defined uniquely in terms of the ellipse shape and the point’s position
                                        co-ordinates.
                                          The boundary stresses around an elliptical opening with axes inclined to the field
                                        stress directions are obtained by selecting values of x 1 , z 1 which fall on the boundary
                                        contour. For this case, e = e 0 ,   ll =   lm = 0, and the l axis is directed normal to the
                                        boundary. For the problem geometry defined in Figure 6.4b, the boundary stress is
                                        given by
                                                         p              2        2
                                                      =    {(1 + K)[(1 + q ) + (1 − q ) cos 2(  −  )]
                                                        2q
                                                                       2             2
                                                        − (1 − K)[(1 + q) cos 2  + (1 − q ) cos 2 ]}  (6.22)
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