Page 187 - Rock Mechanics For Underground Mining
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PRINCIPLES OF CLASSICAL STRESS ANALYSIS

                                        or

                                                                   2     2
                                                                  ∂     ∂
                                                                     +      (  xx +   yy ) = 0         (6.5)
                                                                  ∂x 2  ∂y 2
                                          Equation 6.5 demonstrates that the two-dimensional stress distribution for isotropic
                                        elasticity is independent of the elastic properties of the medium, and that the stress
                                        distribution is the same for plane strain as for plane stress. The latter point validates
                                        the use of photoelastic plane-stress models in estimating the stress distribution in
                                        bodies subject to loading in plane strain. Also, as noted in section 6.1, equation 6.5
                                        demonstrates that the sum of the plane normal stresses,   xx +   yy , satisfies the Laplace
                                        equation.
                                          The problem is to solve equations 6.1 and 6.5, subject to the imposed boundary
                                        conditions. The method suggested by Airy introduces a new function U(x, y), in
                                        terms of which the stress components are defined by
                                                                          2
                                                                         ∂ U
                                                                     xx =
                                                                         ∂y 2
                                                                          2
                                                                         ∂ U
                                                                     yy =                              (6.6)
                                                                         ∂x 2
                                                                            2
                                                                           ∂ U
                                                                     xy =−
                                                                           ∂x ∂y
                                        These expressions for the stress components satisfy the equilibrium equations 6.1,
                                        identically. Introducing them in equation 6.5 gives

                                                                       4
                                                                      ∇ U = 0
                                        where

                                                                             ∂  2  ∂ 2
                                                                         2
                                                                       ∇ =     2  +  2                 (6.7)
                                                                             ∂x   ∂y
                                        Equation 6.7 is called the biharmonic equation.
                                          Several methods may be used to obtain solutions to particular problems in terms
                                        of an Airy stress function. Timoshenko and Goodier (1970) transform equations
                                        6.5 and 6.6 to cylindrical polar co-ordinates, and illustrate a solution procedure by
                                        reference to a thick-walled cylinder subject to internal and external pressure, as shown
                                        in Figure 6.2. For this axisymmetric problem, the biharmonic equation assumes the
                                        form
                                                                           2
                                                           4
                                                                   3
                                                          d U    2d U   1 d U    1 dU
                                                              +       −       +       = 0
                                                                                  3
                                                                         2
                                                          dr 4   rdr 3  r dr 2   r dr
                                        for which a general solution for U is given by
                                                                                   2
                                                                          2
                                                           U = A  nr + Br  nr + Cr + D
                                        In this expression, the constants A, B, C, D are determined by considering both the
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