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92  Energy methods of structural analysis

                 Table 4.5  (Tension positive)
                 0             0            0          @
                 Member        Length       Area       F

                 AB            =I2          AB         -R/2        -1 12        ~   1  4   ~   ~
                 BC            Ll2          AB         -R/2        -112         ~   1  4   ~   ~
                 CD            LI2          A           R           1           RIA
                 DE            LIZ          A           R           1           RIA
                 BD            LIZ          A          -R          -1           RIA
                 EB            LI2          A          -R          -1           RIA
                 A€            LI2          A           R           1           RIA

                 The two terms in Eq. (iii) may be evaluated separately, bearing in mind that only the
                 beam ABC contributes to the first term while the complete structure contributes to the
                  second. Evaluating the summation term by a tabular process we have Table 4.5.
                   Summation of column @ in Table 4.5 gives



                                        r=l
                 The bending moment at any section of the beam between A and F is
                                        3 d 3               dM      d3
                                   M=-Pz--Rz,         hence -=    --
                                        4      2            dR       2z
                  between F and B
                                      P         d3            dM      d3
                                 M  = - (L - z) - -Rz,  hence - = - -2
                                      4          2            dR       2
                  and between B and C
                                 P         fi                 dM      d3
                            M=-(L-z)  --R(L-z),         hence -=    --(L-z)
                                 4          2                  dR      2
                  Thus
                                &=A{











                  giving
                                                    -ll&PL3      RL3
                                                dz=           +-
                                                      768EI     16EI
                  Substituting from Eqs (iv) and (v) into Eq. (iii)
                                                m3
                                     11d3~~~  RL  A+IOAB
                                   -         +-
                                      768EI    16EI+4E (  ABA  ) =O
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