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366  Stress analysis of aircraft components






                                                                    53.8







                  Fig. 10.3  Shear flow (N/mm) distribution at Section AA in Example 10.1

                  10.1.2  Open and closed section beams


                  We shall now consider the more general case of a beam tapered in two directions along
                  its length and comprising an arrangement of booms and skin. Practical examples of
                  such a beam are complete wings and fuselages. The beam may be of open or closed
                  section; the effects of taper are determined in an identical manner in either case.
                    Figure 10.4(a) shows a short length 6z of a beam carrying shear loads S, and Sy at
                  the section z; Sx and Sy are positive when acting in the directions shown. Note that if
                  the beam were of  open cross-section the shear loads would be applied through its
                  shear centre so that no twisting of the beam occurred. In addition to shear loads
                  the beam is subjected to bending moments Mx and My which produce direct stresses
                  a, in the booms and skin. Suppose that in the rth boom the direct stress in a direction
                  parallel to the z axis is ai,r, which may be found using either Eq. (9.6) or Eq. (9.7). The
                  component P,:, of the axial load P, in the rth boom is then given by
                                                 Pqr  = ai,rBr                       (10.8)

                  where B, is the cross-sectional area of the rth boom.
                    From Fig. 10.4(b)

                                                                                     (10.9)

                  Further, from Fig. 10.4(c)




                  or, substituting for Py,, from Eq. (10.9)

                                                                                    (10.10)
                  The axial load Pr is then given by

                                           Pr  =    + p',r + pI,r)1'2               (10.11)
                  or, alternatively

                                                                                    (10.12)
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