Question 27.10: The panel shown in Fig. P.27.10 is idealized into a combinat......
The panel shown in Fig. P.27.10 is idealized into a combination of direct stress carrying booms and shear stress carrying plates; the boom areas are shown and the plate thickness is t. Derive expressions for the distribution of a direct load in each boom and state how the load distributions are affected when A = B.
The assumed directions of shear flow are shown in Fig. S.27.10(a). For the equilibrium of an element of the top boom, Fig. S.27.10(b),
P_{1}+{\frac{\partial P_{1}}{\partial z}}\delta z-P_{1}+q_{1}\delta z=0from which
{\frac{\partial P_{1}}{\partial z}}=-q_{1} (i)
Similarly, for an element of the central boom,
{\frac{\partial P_{2}}{\partial z}}=q_{1}-q_{2} (ii)
For overall equilibrium of the panel, at any section z,
P_{1}+P_{2}+P_{3}=-6P (iii)
and taking moments about boom 3,
P_{1}2d+P_{2}d+3P2d+2P d=0so that
2P_{1}+P_{2}=-8P (iv)
The compatibility condition is shown in Fig. S.27.10(c) for an element of the top panel.
Then
i.e.,
{\frac{\mathrm{d}\gamma_{1}}{\mathrm{d}z}}={\frac{1}{d}}(\varepsilon_{2}-\varepsilon_{1})-{\frac{\partial^{2}\nu}{\partial z^{2}}} (v)
Similarly, for an element of the lower panel,
{\frac{\mathrm{d}\gamma_{2}}{\mathrm{d}z}}={\frac{1}{d}}(\varepsilon_{3}-\varepsilon_{2})-{\frac{\partial^{2}\nu}{\partial z^{2}}} (vi)
Subtracting Eq. (vi) from (v),
{\frac{\mathrm{d}\gamma_{1}}{\mathrm{d}z}}-{\frac{\mathrm{d}\gamma_{2}}{\mathrm{d}z}}={\frac{1}{d}}(2\varepsilon_{2}-\varepsilon_{1}-\varepsilon_{3}) (vii)
But
\gamma_{1}=\frac{q_{1}}{G t}\quad\gamma_{2}=\frac{q_{2}}{G t}\quad\varepsilon_{2}=\frac{P_{2}}{B E}\quad\varepsilon_{1}=\frac{P_{1}}{A E}\quad\varepsilon_{3}=\frac{P_{3}}{A E}Then, from Eq. (vii),
{\frac{\mathrm{d}q_{1}}{\mathrm{d}z}}-{\frac{\mathrm{d}q_{2}}{\mathrm{d}z}}={\frac{G t}{d E}}\left({\frac{2P_{2}}{B}}-{\frac{P_{1}}{A}}-{\frac{P_{3}}{A}}\right) (viii)
Substituting in Eq. (viii) for q_{1}-q_{2} from Eq. (ii) and for P_{1}{\mathrm{~and~}}P_{3} from Eqs (iv) and (iii) and rearranging,
{\frac{\partial^{2}P_{2}}{\partial z^{2}}}-\mu^{2}P_{2}={\frac{6G t P}{d E A}} (ix)
where
\mu^{2}{=}\frac{G t(2A+B)}{d E A B}The solution of Eq. (ix) is
P_{2}=C\cosh\mu z+D\sinh\mu z-{\frac{6P B}{2A+B}}\,When z=0,P_{2}=-2P, and when z=L,\ q_{1}=q_{2}=0 so that, from Eq. (ii), \partial P_{2}/\partial z=0. These give
C=4P\left({\frac{B-A}{2A+B}}\right)\quad D=-4P\left({\frac{B-A}{2A+B}}\right)\operatorname{tanh}\mu LThen
P_{2}={\frac{6P}{2A+B}}\left[-B+{\frac{2}{3}}(B-A){\frac{\cosh\mu(L-z)}{\cosh\mu L}}\right]From Eq. (iv),
P_{1}={\frac{6P}{2A+B}}\left[-\left({\frac{B+8A}{6}}\right)-{\frac{1}{3}}(B-A){\frac{\cosh\mu(L-z)}{\cosh\mu L}}\right]and from Eq. (iii),
P_{3}={\frac{6P}{2A+B}}\left[-\left({\frac{4A-B}{6}}\right)-{\frac{1}{3}}(B-A){\frac{\cosh\mu(L-z)}{\cosh\mu L}}\right]When A=B,
P_{1}=-3P\ \ P_{2}=-2P\ \ P_{3}=-Pand there is no shear lag effect.
