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10.1 Tapered beams  365

              2 mm and is fully effective in resisting direct stress. The beam tapers symmetrically
              about its horizontal centroidal axis and the cross-sectional area  of  each flange is
              400 mm2.

                The internal bending moment and shear load at the section AA produced by the
              externally applied load are, respectively
                                 Mx = 20 x 1 = 20kNm,  S,  = -2OkN

              The direct stresses parallel to the z axis in the flanges at this section are obtained either
              from Eq. (9.6) or Eq. (9.7) in which M,, = 0 and Zx,  = 0. Thus, from Eq. (9.6)

                                                   MXY
                                              uz = -
                                                   In
              in which
                                   Ixx = 2 x 400 x  1502 + 2 x 3003/12
              i.e.

                                         Zxx = 22.5  x  106m4
              Hence




              The components parallel to the z axis of the axial loads in the flanges are therefore
                                  PI = -Pz-2  = 133.3 x 400 = 53 320 N
                                    1
                                   -7
              The shear load resisted by the beam web is then, from Eq. (10.5)
                                                       6Yl        6Y2
                                S,.,," = -20  x lo3 - 53 320-  + 53 320-
                                                       6Z         6Z
              in which, from Figs 10.1 and 10.2, we see that
                              6y1 - -100   - -0.05,   6Y2 -  100   - 0.05
                                                    - - ---
                               sz   2 x  103         6,.  2 x  103
              Hence
                       S,.:w, = -20  x lo3 + 53 320 x 0.05 + 53 320 x 0.05 = -14668N
              The shear flow distribution in the web follows either from Eq. (10.6) or Eq. (10.7) and
              is (see Fig. 10.2(b))

                              412 = 22.5 x lo6 ([q150-s)ds+400   x 150
                                     14'''
              i.e.

                                 412 = 6.52 x        + 300s + 60000)                (ii)
              The maximum value of q12 occurs when s = 150mm and q12 (max) = 53.8 N/mm.
              The values of shear flow at points 1 (s = 0) and 2 (s = 300mm) are q1 = 39.1 N/mm
              and q2 = 39.1 N/mm; the complete distribution is shown in Fig. 10.3.
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