Page 209 - Aircraft Stuctures for Engineering Student
P. 209

6.1 3 Tension field beams  193









                  400mm








                                                   1200 mm
                                                                                -I
               Fig. 6.28  Beam of Example 6.3.

               so that

                                               Q!  = 42.6"
                 The maximum flange stress will occur in the top flange at the built-in end where the
               bending moment on the beam is greatest and the stresses due to bending and diagonal
               tension are additive. Thus, from Eq. (6.98)
                                            5 x  1200     5
                                       FT  =
                                               400   -k 2 tan 42.6"
               i.e.
                                             FT = 17.7 kN
               Hence the direct stress in the top flange produced by the externally applied bending
               moment and the diagonal tension is  17.7 x  103/350 = 50.7N/mm2. In addition to
               this  uniform  compressive stress,  local  bending  of  the  type  shown  in  Fig.  6.27
               occurs. The local bending moment in the top flange at the  built-in end is found
               using Eq. (6.104), i.e.
                                     5 x lo3 x 3002 tan42.6"
                             Mnax =                      = 8.6 x 104Nmm
                                           12 x 400
               The maximum compressive stress corresponding to this bending moment occurs at
               the  lower extremity of  the  flange and  is  8.6 x  104/750 = 114.9N/mm2. Thus the
               maximum stress in a flange occurs on the inside of the top flange at the built-in end
               of the beam, is compressive and equal to 114.9 + 50.7 = 165.6N/mm2.
                 The compressive load in a stiffener is obtained using Eq. (6.102), i.e.
                                          5 x 300 tan 42.6"
                                      P=                 = 3.4 kN
                                               400
               Since, in this case, b < 1.5d, the equivalent length of a stiffener as a column is given by
               the first of Eqs (6.103). Thus

                                   1,  = 400/d4 - 2 x 300/400 = 253 mm
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