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126  Bending of thin plates











                  Y






                                      (a 1                                  (b)

                  Fig. 5.5  (a)  Plate subjected to bending and twisting;  (b) tangential and normal moments on an arbitraty
                  plane.

                  the tangential components Mxy and MYx (again these are moments per unit length)
                  produce twisting of the plate about axes parallel to the x  and y axes. The system of
                  sufhes and the sign convention for these twisting moments must be clearly under-
                  stood to avoid confusion. Mxy is a twisting moment intensity in a vertical x plane
                  parallel to the y axis, while Myx is a twisting moment intensity in a vertical y plane
                  parallel to the x  axis.  Note  that  the first suffix  gives the direction of  the axis of
                  the twisting moment. We also define positive twisting moments as being clockwise
                  when  viewed  along their  axes in  directions parallel  to  the positive  directions of
                  the  corresponding x  or y  axis. In  Fig.  5.4,  therefore, all moment intensities are
                  positive.
                    Since the twisting moments are tangential moments or torques they are resisted by
                  a system of horizontal shear stresses  T,?,  as shown in Fig. 5.6. From a consideration of
                  complementary shear stresses (see Fig. 5.6) Mxy = -My,,  so that we may represent a
                  general moment application to the plate in terms of M,,  My and Mxy as shown in
                  Fig. 5.5(a). These moments produce tangential and normal moments, Mt and M,,
                  on an arbitrarily chosen diagonal plane FD. We may express these moment intensities
                  (in an analogous fashion to the complex stress systems of Section 1.6) in terms of M,,
                  My and MXy. Thus, for equilibrium of the triangular element ABC of Fig. 5.5(b) in a
                  plane perpendicular to AC
                         M,AC  = MxAB cos a + MyBC sin Q - MxyAB sin a - MXyBC cos a
                  giving

                                    M, = M, cos2 a + My sin2 a - Mxy sin 2a          (5.10)
                  Similarly for equilibrium in a plane parallel to CA
                         MtAC = M,AB  sin a - MYBC cos Q + MxyAB cos Q - MxyBC sin Q
                  or

                                           (Mx  -
                                      Mt =           sin 2a + Mxy cos 2a             (5.11)
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