Page 391 - Aircraft Stuctures for Engineering Student
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372  Stress analysis of aircraft components

                    In this example the stringer areas do not vary along the length of the beam but the
                  method of solution is identical.
                    We are required to find the shear flow distribution at a section 2 m from the built-in
                  end of the beam. We therefore calculate the boom loads at sections, say 0.1 m either
                  side of this section. Thus, at a distance 2.1 m from the built-in end
                                        M, = -100  x  1.9 = -190kNm
                  The dimensions of this section are easily found by proportion and are width = 1.18 m,
                  depth = 0.59 m. Thus the second moment of area is
                             I,,  = 4 x 900 x 295’  + 2 x 1200 x 295’  = 5.22 x  lo8 1ll111~
                  and
                                              -190  x  lo6
                                        0z:r =            = -0.364~~
                                              5.22 x  lo8 Yr
                  Hence
                             Pi  = P3  = -P4  = -P6  = -0.364  X 295 X 900 = -96642N
                  and

                                 P2   -Ps  = -0.364  x 295 x  1200 = -128  856N
                  At a section 1.9 m from the built-in end
                                        M, = -100  x 2.1 = -2lOkNm
                  and the section dimensions are width = 1.22m, depth = 0.61 m so that
                             I,,  = 4 x 900 x  305’ + 2 x  1200 x  305’  = 5.58 x lo8 mm4
                  and

                                              -210  x  lo6
                                        0z>r =            = -0.376~~
                                              5.58 x  lo8 Yr
                  Hence
                            Pi  = P3  = -P4  = -P6  = -0.376  x  305 x 900 = -103212N

                  and
                                 P2  = -Ps  = -0.376  x 305 x  1200 = -137616N
                  Thus, there is an increase in compressive load of 103 212 - 96 642 = 6570 N in booms 1
                  and 3 and an increase in tensile load of 6570 N in booms 4 and 6 between the two sections.
                  Also, the compressive load in boom 2 increases by  137 616 - 128 856 = 8760N while
                  the tensile load in boom 5  increases by  8760N. Therefore, the change in boom load
                  per unit length is given by
                                                              6570
                                 AP1  = AP3 = -AP4  = -AP6  = -
                                                                     32.85 N
                                                               200
                  and
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