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562  Elementary aeroelasticity

                  where V(z) is the amplitude of the vibration at any section z. Substituting for TJ from
                  Eq. (13.48) in Eq. (13.47) yields
                                              d4 V  PAW’  V=O                       ( 13.49)
                                              d#     EI
                  Equation (13.49) is a fourth-order differential equation of standard form having the
                  general solution
                                  V = Bsin Xz + Ccos XZ + D sinh XZ + F cosh Xz     (1 3 .SO)
                  where

                                                      PAJ
                                                  4  =-
                                                       EI
                  and B, C, D and F are unknown constants which are determined from the boundary
                  conditions of the beam. The ends of the beam may be:

                    (1)  simply supported or  pinned,  in  which  case the  displacement  and  bending
                  moment are zero, and therefore in terms of  the function  V(z) we have  V = 0 and
                  d2V/&  = 0;
                    (2)  fixed, giving zero displacement and slope, that is V = 0 and dV/dz = 0;
                    (3)  free,  for  which  the  bending  moment  and  shear  force  are  zero,  hence
                  d2V/d2 = 0 and, from Eq. (13.43), d3V/dz3 = 0.

                  Example 13.4
                  Determine the first three normal modes of vibration and the corresponding natural
                  frequencies of the uniform, simply supported beam shown in Fig. 13.16.

                    Since both ends of the beam are simply supported, V = 0 and d2 V/d2 = 0 at z = 0
                  and z = L. From the first of these conditions and Eq. (13.50) we have
                                                 O=C+F                                  6)
                  and from the second
                                               0 = -X2C + X2F                          (ii)
                  Hence C = F = 0. Applying the above boundary conditions at z = L gives

                                           0 = BsinXL+DsinhXL                          (iii)
                  and
                                         0 = -X2B sin XL + X2D sinh XL                 6.1











                  Fig. 13.16 Beam of Example 13.4.
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