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40                  Mechanics and analysis of composite materials
             2.7.  Compatibility equations

               Consider  strain-displacement  equations,  Eqs. (2.22),  and  try  to  determine
             displacements u,,  u.,  and u,  in terms of strains E.~, E~,,  and yx,,,  yxz, y,=.  As can be
             seen,  there  are  six  equations  containing  only  three  unknown  displacements.  In
             the  general case,  such set of  equations is not  consistent,  and some compatibility
             conditions should be imposed on strains to provide the existence of the solution. To
             derive these conditions, decompose derivatives of the displacements as follows


                                                                               (2.32)


             Here
                 0 -’(”.-%)

                  ‘-2  ax           (x,y,z)                                    (2.33)

             is  the  angle  of  rotation  of  a  body  element  (such  as  the  cubic  element  shown
             in  Fig. 2.1)  around  the  z-axis.  Three  Eqs. (2.32)  including  one  and  the  same
             displacement  u,  allow  us  to  construct  three  couples  of  mixed  second-order
             derivatives of  u, with respect to x and y  or y and x, x and z  or z and x, y and z or
             z and y. Because the sequence of  differentiation  does not influence the result and
             since there  are two  other  groups  of  equations  in  Eqs. (2.32), we  arrive  at nine
             compatibility conditions that can be presented as





                                                                               (2.34)





             These equations are similar to Eqs. (2.32), Le., t..ey  allow us to determine rotation
             angles only if  some compatibility  conditions  are valid. These conditions compose
             the set of compatibility equations for strains and have the following final form



             where
                          aZE,  a%,   a2yxy
                 krJ&,y) =-+---
                                ax2  axay
                           ay2                 Y,4 7                           (2.36)
                               :
                              x

                 Tx(&,7)  ==----     -+-- ayJj         (x,y,z) .
                                                -
                          ayaz  i (aZ  ay
                          aZEx
                                                ax
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