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4.7 Solution of statically indeterminate systems  97

               Hence




               Assuming that the fuselage frame is linearly elastic we have, from Eqs (ii) and (iii)




               Substituting from Eqs (iv) and (v) into Eqs (vi) gives two simultaneous equations
                                            PR
                                            27r   - MA + NAR                       (vii)
                                           7PR  -MA+$NAR                           (viii)
                                            87r
               These equations may be written in matrix form as follows




               so that





               or
                                                              -1/2
                                  { zf}  [ -i/R             { -7/8}
                                         =
               which gives
                                              PR         -3P
                                        MA  = -,    NA  = -
                                              4.rr        47r
               The bending moment distribution follows from Eq. (iv) and is
                                           PR
                                      M = -(1   - $cos0 - 0sin0)
                                           27r
               The solution of Eqs (ix) involves the inversion of the matrix


                                              [: 3RR/2]
               which may be  carried out using any of  the standard methods detailed in texts on
               matrix analysis. In this example Eqs (vi$  and (viii) are clearly most easily solved
               directly; however, the  matrix  approach illustrates the  technique and  serves as  a
               useful introduction to the more detailed discussion in Chapter 12.

               Example 4.7
               A two-cell fuselage has circular frames with a rigidly attached straight member across
               the middle. The bending stiffness of the lower half of the frame is 2EI, whilst that of
               the upper half and also the straight member is EI.
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