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144  Bending of thin plates

                   5.6.2  Potential energy of a transverse load

                  An element Sx x Sy of the transversely loaded plate of Fig. 5.8 supports a load qSxSy.
                  If the displacement of the element normal to the plate is w then the potential energy
                  SV of the load on the element referred to the undeflected plate position is
                                                SV = -wqSxSy                (See Section 4.2)
                  Therefore, the potential energy V of the total load on the plate is given by
                                                     ab
                                              V= -1 1 wqdxdy                         (5.39)
                                                     0 0


                   5.6.3  Potential energy of in-plane loads


                  We may consider each load N,,  Ny and Nxy in turn, then use the principle of super-
                   position  to  determine the  potential energy  of  the  loading  system when  they  act
                   simultaneously. Consider an elemental strip of  width Sy  along the length a of  the
                   plate in Fig. 5.15(a). The compressive load on this strip is N,Sy  and due to the bending
                   of the plate the horizontal length of the strip decreases by an amount A,  as shown in
                   Fig. 5.15(b). The potential energy SV, of the load NJy,  referred to the undeflected
                   position of the plate as the datum, is then

                                                SV, = -N,ASy                          (5.40)
                   From Fig. 5.15(b) the length of a small element Sa of the strip is

                                               Sa = (Sx2 + SW2)t
                                                             )’I
                   and since awlax is small then
                                            Sa  M sx [1+;  (g



                   Hence




                   giving
                                                     dl
                                            a=d+JoT(ax>  dx
                                                         dw
                   and

                                          A=a-d=j  afl -(-)  d~
                                                           dw
                                                       0  2  ax
                   Since

                                 jr  (g ) ’ dx  only differs from j:i  (g)*dx
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