Page 186 - Rock Mechanics For Underground Mining
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METHODS OF STRESS ANALYSIS

                                          It is instructive to follow the procedure developed by Airy (1862) and described by
                                        Timoshenko and Goodier (1970), in establishing a particular form of the field equation
                                        for isotropic elasticity and plane strain. The differential equations of equilibrium in
                                        two dimensions for zero body forces are

                                                                  ∂  xx  ∂  xy
                                                                       +      = 0                      (6.1)
                                                                   ∂x     ∂y
                                                                  ∂  xy  ∂  yy
                                                                       +      = 0
                                                                   ∂x     ∂y
                                        or
                                                               2        2         2
                                                              ∂   xy   ∂   xx   ∂   yy
                                                                   =−        =−                        (6.2)
                                                              ∂x∂y      ∂x  2    ∂y 2
                                          For plane strain conditions and isotropic elasticity, strains are defined by
                                                                      1

                                                                ε xx =  (  xx − 	   yy )
                                                                      E
                                                                      1

                                                                ε yy =  (  yy − 	   xx )               (6.3)
                                                                      E
                                                                      1
                                                                  xy =    xy
                                                                      G

                                                                      2(1 + 	 )
                                                                   =           xy
                                                                         E
                                        where
                                                                           E

                                                                    E =
                                                                         1 − 	 2


                                                                     	 =
                                                                         1 −
                                          The strain compatibility equation in two dimensions is given by
                                                                 2      2       2
                                                                ∂ ε yy  ∂ ε xx  ∂   xy
                                                                     +      =                          (6.4)
                                                                 ∂x 2   ∂y 2   ∂x∂y
                                        Substituting the expressions for the strain components, (equations 6.3) in equation
                                        6.4, and then equations 6.2 in the resultant expression yields
                                                                   2
                                                                            2
                                                                                             2
                                             2
                                                      2
                                         1     ∂   yy  ∂   xx  
  1     ∂   xx  ∂   yy  
  (1 + 	 ) ∂   xy

                                                 − 	        +          − 	       = 2
                                        E     ∂x 2    ∂x 2    E    ∂y 2     ∂y 2       E    ∂x ∂y
                                                                                               2
                                                                                                      2
                                                                                     (1 + 	 )     ∂   xx  ∂   yy

                                                                                 =−                +
                                                                                       E      ∂x 2    ∂y 2
                                        which becomes, on simplification,
                                                           2       2      2       2
                                                          ∂   xx  ∂   xx  ∂   yy  ∂   yy
                                                               +       +      +       = 0
                                                           ∂x 2    ∂y 2   ∂x 2   ∂y 2
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