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3.2 St. Venant warping function solution  59

         from which (see Eq. (3.12))
                                            7ra3b3
                                       J=                                      (VI
                                           (a2 + b2)
           The shear stress distribution is obtained in terms of the torque by substituting for
         the product G dO/dz in Eq. (iii) from Eq. (iv) and then differentiating as indicated by
         the relationships of Eqs (3.2). Thus
                                         2TY         2  Tx
                                  rZx = -~      TZY =
                                        rub3 ’      7ra3b
                                                    ~
           So far we  have  solved for the stress distribution,  Eqs (vi), and the rate of twist,
         Eq. (iv). It remains to determine the warping distribution  w  over the cross-section.
         For this we return to Eqs (3.10) which become, on substituting from the above for
         rzx, rzy and dO/dz
                   dw  -   2Ty    T  (a2+b2)   dw - 2Tx      T  (a2 +h2)
                   -
                      - --
                                                   -
                   dx     7rab3G+G  m3b3 y7  %  =-E             7ra3b3  X
         or
                                                                              (vii)

         Integrating both of Eqs (vii)

                          T(b2 - a2)               T(b2 - a2)
                      W=           yx+fib),  w=             XY +f2 (XI
                           7ra3 b3 G                nu3 b3 G
         The warping displacement given by each of these equations must have the same value
         at identical points (x,y). It follows thatfi(y) =f2(x) = 0. Hence

                                          T(b2 - a2)
                                     W=            XY                        (viii)
                                           nu3 b3 G
         Lines of constant w therefore describe hyperbolas with the major and minor axes of
         the  elliptical  cross-section  as  asymptotes.  Further,  for  a  positive  (anticlockwise)
         torque the warping is negative in the first and third quadrants (a > b) and positive
         in the second and fourth.


         v-- --I
            3.2  St. Venant warping function-solution
         In formulating his stress function solution Prandtl made assumptions concerned with
         the stress distribution  in the bar. The alternative approach presented  by  St. Venant
         involves  assumptions  as  to  the  mode  of  displacement  of  the  bar;  namely,  that
         cross-sections of a bar subjected to torsion  maintain their original unloaded  shape
         although they  may  suffer warping  displacements  normal  to their  plane.  The first
         of  these  assumptions  leads  to  the  conclusion  that  cross-sections  rotate  as  rigid
         bodies about a centre of rotation  or twist. This fact was also found to derive from
         the  stress function  approach  of  Section  3.1 so that, referring  to Fig.  3.4 and Eq.
         (3.9), the components of displacement in the x and y  directions of a point  P in the
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