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5.3 Distributed transverse load  135

               Navier (1820) showed that these conditions are satisfied by representing the deflection
               w as an infinite trigonometrical or Fourier series
                                          m   w        mrx    nry
                                     w=      EA,,,,sin-   a   sin                 (5.27)
                                         m=l  n=l
               in  which  m  represents the  number  of  half  waves  in  the  x  direction  and  n  the
               corresponding number  in the y  direction. Further,  A,,,  are unknown  coefficients
               which must satisfy the above differential equation and may be determined as follows.
                 We may also represent the load q(x, y) by a Fourier series, thus
                                             m  w        mrx    nry
                                   q(x,y) =     Camnsin-sinb                      (5.28)
                                            m=l  n=l      a
               A particular coefficient amtnt is calculated by first multiplying both sides of Eq. (5.28)
               by  sin(m'rx/a) sin(n'ry/b)  and integrating with respect to x from 0 to a and with
               respect to y from 0 to b. Thus

                                      m'rx  . n'ry
                             q(x,y)sin-sin-      dxdy
                                       a      b
                                               .  mrx   m'rx   nry   n'ry
                             = m=l 2 n=l T/:jIamsin-sin- a   a   sin  - sin - dx dy
                                                                      b
                                                                b
                               ab
                             -
                             -
                                4
               since




                                                    -  a
                                                    _-  when  m=m'
                                                      2
               and
                                 so"  sin-sin-   b   dy = 0  when  n # n'
                                    .  nry  .  n'ry
                                       b
                                                      b
                                                    _-  when  n=n'
                                                    -
                                                      2
               It follows that
                                       4  a  b        m'rx   n'ry
                                am!,#  = - 1 1 q(x, y) sin - sin-   dx dy         (5.29)
                                       ab  o  o         a      b
               Substituting now for w and q(x, y) from Eqs (5.27) and (5.28) into the differential
               equation for w we have
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