Page 349 - Aircraft Stuctures for Engineering Student
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330  Open and closed, thin-walled beams


                                                                        -
                                                           1
                                                                        2
                                                                        v            3
                                         1.5 mm    200 mm        -           -       t
                                                                        -
                                          2mm                           -
                                                                        5            4
                      6                                    6

                 Fig. 9.48  Idealization of a wing section.
                 plane the direct stress at any point in the actual wing section is directly proportional
                 to its distance from the horizontal axis of  symmetry. Further, the distribution of
                 direct stress in all the panels will be linear so that either of Eqs (9.70)  or (9.71)  may
                 be used. We note that, in addition to contributions from adjacent panels, the boom
                 areas include the existing spar flanges. Hence




                                                          2.0 x 600 (  ;;l)
                 or
                                            6
                               B1  = 300 + 3.0x400(2-l)+     6     2+-
                 which gives

                                            B1(= B6) = 1050mm2
                 Also




                 i.e.
                                                                               )
                                                                                 M

                                                                             )
                                                   +
                   B2  = 2 x 300 +  2.0 x 600 (  2M)  2.5 x  300  (2-1)+   1.5 x  600 (  K
                                                                               2+-
                                          2+-
                                                                         6
                                    6                   6
                 from which
                                           B2(=  Bs) = 1791.7~
                                                               2
                 Finally
                 i.e.                    1.5 x 600 (
                               B3  = 300 +             k%)   +  2.0 x  200
                                             6     2+-                (2 - 1)
                 so that
                                            B3(= B4) = 891.7mm 2
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