Question 27.12: The idealized cantilever beam shown in Fig. P.27.12 carries ......
The idealized cantilever beam shown in Fig. P.27.12 carries a uniformly distributed load of intensity w. Assuming that all direct stresses are carried by the booms while the panels are effective only in shear, determine the distribution of direct load in the central boom in the top cover. Young’s modulus for the booms is E, and the shear modulus of the walls is G.
The forces acting on the top cover of the box are shown in Fig. 27.12(a). Then, for the equilibrium of the element \delta z of the edge boom shown in Fig. S.27.12(b),
\frac{\partial P_{B}}{\partial z}=-q+\frac{w z}{2h} (i)
Similarly, for the central boom,
{\frac{\partial P_{\mathrm{A}}}{\partial z}}=2q (ii)
For the equilibrium of a length z of the cover,
2P_{\mathrm{B}}+P_{\mathrm{A}}-2{\frac{w z^{2}}{4h}}=0 (iii)
The compatibility of displacement condition is shown in Fig. S.27.12(c).
Then
which gives
{\frac{\partial\gamma}{\partial z}}={\frac{1}{d}}(\varepsilon_{\mathrm{A}}-\varepsilon_{\mathrm{B}}) (iv)
But
\gamma={\frac{q}{G t}},\ \ \varepsilon_{A}{=}{\frac{P_{\mathrm{A}}}{A E}},\ \ \varepsilon_{B}{=}{\frac{P_{\mathrm{B}}}{B E}}Substituting in Eq. (iv),
{\frac{\mathrm{d}q}{\mathrm{d}z}}={\frac{G t}{\mathrm{d}E}}{\Biggl(}{\frac{P_{\mathrm{A}}}{A}}-{\frac{P_{\mathrm{B}}}{B}}{\Biggr)}Substituting for q from Eq. (ii) and P_{\mathrm{{B}}} from Eq. (iii) and rearranging,
\frac{\partial^{2}P_{A}}{\partial z^{2}}-\mu^{2}P_{\mathrm{A}}=-\frac{G t w}{2d E h B}z^{2} (v)
where
\mu^{2}{=}\frac{G t(2B+A)}{d E A B}The solution of Eq. (v) is
P_{\mathrm{A}}=C\cosh\mu z+D\sinh\mu z+{\frac{w}{2h(2B+A)}}\left({\frac{2}{\mu^{2}}}+z^{2}\right)When z=0,P_{\mathrm{A}}=0, which gives
C=-{\frac{\it w A}{h(2B+A)\mu^{2}}}When z=L,\ \partial P_{\mathrm{A}}/\partial z=0, since q=0 at z=L. This gives
D=-{\frac{w A}{\mu h(2B+A)\cosh\mu h}}\biggl(L+{\frac{\sinh\mu L}{\mu}}\biggr)Hence,
P_{\mathrm{A}}=-{\frac{w A}{h(2B+A)}}\left[{\frac{\cosh\mu z}{\mu^{2}}}+\left({\frac{\mu L+\sinh\mu L}{\mu^{2}\cosh\mu L}}\right)\sinh\mu z-{\frac{1}{\mu^{2}}}-{\frac{z^{2}}{2}}\right]
