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196  Structural instability



















                  Fig. 6.30  Effect of taper on diagonal tension field beam calculations.

                  field beams since, for the latter case, k = 1 giving the results of Eqs (6.106), (6.107)
                  and (6.94).
                    In some cases beams taper along their lengths, in which case the flange loads are no
                  longer horizontal but have vertical components which reduce the shear load carried
                  by  the web.  Thus, in Fig.  6.30 where d  is  the depth of  the  beam  at the section
                  considered, we have, resolving forces vertically

                                    w- (~T+Fg)Sin@-at(dCOSa)Sina=O                  (6.117)
                  For horizontal equilibrium
                                        (FT - FB) COS /3  - gttd COS'   = 0         (6.118)
                  Taking moments about B
                                        wz - FTd COS @ + $gttd2 Cos2 a = 0          (6.1 19)
                  Solving Eqs (6.117), (6.118) and (6.119) for q, FT and FB

                                     at  = td 2w 2a (1  -$tan@)                     (6.120)
                                            sin
                                                                      )]
                                            W
                                     FT=-[z+?(              1 --tan@
                                          d cos @                                   (6.121)
                                            W
                                     FB=-[z-F(                        )]            (6.122)
                                          d cos @           1 - -tan@
                   Equation (6.102) becomes
                                                                                     (6.123)


                  Also the shear force S at any section of the beam is, from Fig. 6.30
                                            s  w - (FT + &j) sinp
                   or, substituting for FT  and FB  from Eqs (6.121) and (6.122)
                                                   G)
                                             S= W  1 --tan/3                         (6.124)
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