Question 27.10: The panel shown in Fig. P.27.10 is idealized into a combinat......

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),

from 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,

so 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

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

The solution of Eq. (ix) is

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

Then

From Eq. (iv),

and from Eq. (iii),

When A=B,

and there is no shear lag effect.