Page 192 - Rock Mechanics For Underground Mining
P. 192

METHODS OF STRESS ANALYSIS




























              Figure 6.3 Problem geometry, co-  to Kirsch (1898), are
              ordinate system and nomenclature for
              specifying the stress and displacement  p           a 2  
             a 2  3a 4
              distribution around a circular excava-    rr =  (1 + K) 1 −  2  − (1 − K) 1 − 4  2  +  4  cos 2
              tion in a biaxial stress field.        2             r                  r    r
                                                    p             a 2  
            3a 4
                                                

 =  (1 + K) 1 +     + (1 − K) 1 +      cos 2
                                                    2             r 2                r 4
                                                    p             2a 2  3a 4
                                                r
 =  (1 − K) 1 +     −      sin 2
                   (6.18)
                                                    2              r  2  r  4
                                                     pa 2                            a 2
                                               u r =−     (1 + K) − (1 − K) 4(1 − 	) −  cos 2
                                                     4Gr                             r 2
                                                     pa 2                    a 2
                                               u 
 =−     (1 − K) 2(1 − 2	) +    sin 2
                                                     4Gr                     r 2
                                          In these expressions u r , u 
 are displacements induced by excavation, while
                                          rr ,   

 ,   r
 are total stresses after generation of the opening.
                                          By putting r = a in equation 6.18, the stresses on the excavation boundary are
                                        given as

                                                             

 = p[(1 + K) + 2(1 − K) cos 2
]
                                                             rr = 0                                   (6.19)
                                                             r
 = 0

                                        Equations 6.19 confirm that the solutions satisfy the imposed condition that the ex-
                                        cavation boundary is traction free. Similarly, for 
 = 0, and r large, the stress com-
                                        ponents are given by

                                                              rr = Kp,    

 = p,    r
 = 0
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