Question 27.6: The idealized cross-section of a beam is shown in Fig. P.27.......
The idealized cross-section of a beam is shown in Fig. P.27.6. The beam is of length L and is attached to a flexible support at one end, which only partially prevents warping of the cross-section; at its free end, the beam carries a concentrated torque T. Assuming that the warping at the built-in end is directly proportional to the free warping, that is, w=k w_{0}, derive an expression for the distribution of direct stress along the top right-hand corner boom. State the conditions corresponding to the values k = 0 and k = 1.
The warping distribution is given by Eq. (27.16), i.e.,
w=C\cosh\mu z+D\sinh\mu z+\frac{T}{8a b G}\left(\frac{b}{t_{b}}-\frac{a}{t_{a}}\right) (i)
in which the last term is the free warping, w_{0}, of the section. Eq. (i) may therefore be written
w=C\cosh\mu z+D\sinh\mu z+w_{0} (ii)
When z=0,\,w=k w_{0} so that, from Eq. (ii),
C=w_{0}(k-1)When z=L, the direct stress is zero. Then, from Chapter 1,
\sigma=E{\frac{\partial w}{\partial z}}=0so that
0=\mu C\sinh\mu L+\mu D\cosh\mu Lwhich gives
D=-w_{0}(k-1)\operatorname{tanh}\mu LEq. (ii) then becomes
w=w_{0}\left[1+(k-1){\frac{\cosh\mu(L-z)}{\cosh\mu L}}\right] (iii)
Then
\sigma=E{\frac{\partial w}{\partial z}}so that
\sigma=-\mu E w_{0}(k-1)\frac{\sinh\mu(L-z)}{\cosh\mu L}When k = 0,
\sigma=\mu E w_{0}{\frac{\sinh\mu(L-z)}{\cosh\mu L}}i.e., a rigid foundation.
When k = 1,
i.e., free warping (also from Eq. (iii)).