Page 426 - Aircraft Stuctures for Engineering Student
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10.4 Fuselage frames and wing ribs  407















                                                            H
                                 (a)                            (b)
               Fig. 10.41  Free body diagrams of stiffeners JK and HKD in the beam of  Example 10.13.


               concentrations are produced if a stiffener ends in the centre of a web panel. We note
               also that the web panels are only effective in shear so that the shear flow is constant
               throughout a particular web panel; the assumed directions of the shear flows are
               shown in Fig. 10.40.
                 It is instructive at this stage to examine the physical role of the different structural
               components in supporting the applied loads. Generally, stiffeners are assumed to
               withstand axial forces only so that  the horizontal component of the load at K  is
               equilibrated locally by  the axial load in the stiffener JK and not by the bending of
               stiffener HKD.  By  the same argument the vertical component of  the load at K is
               resisted  by  the  axial load  in  the  stiffener  HKD.  These  axial stiffener loads  are
               equilibrated in turn by the resultants of the shear flows q1 and q2 in the web panels
               CDKJ and JKHG. Thus we  see that the web panels resist the shear component of
               the externally applied load  and  at the  same time transmit  the bending and  axial
               load  of  the  externally applied load  to the  beam flanges; subsequently, the flange
               loads are reacted at the support points A and E.
                 Consider  the  free  body  diagrams  of  the  stiffeners  JK  and  HKD  shown  in
               Figs 10.41(a) and (b).
                 From the equilibrium of stiffener JK we have
                                 (ql - q2) x 250 = 4000 sin 60" = 3464.1 N           (i)
               and from the equilibrium of stiffener HKD

                                  200q1 + 100q2 = 4000 COS 60" = 2000 N              (ii)
               Solving Eqs (i) and (ii) we obtain
                                   q1  = 11.3 N/mm,   q2 = -2.6  N/mm
               The vertical shear force in the panel BCGF is equilibrated by the vertical resultant of
               the shear flow q3. Thus
                                      3ooq3 = 4000 COS 60" = 2000 N

               whence
                                             q3 = 6.7N/mm
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