Question 10.18: Figure 10.33 shows a compound beam loaded at its free end. I...

Let us draw the free-body diagram of the beams as shown in Figure 10.34.

From Figure 10.34(b) above

$\sum M_{ C }=0 \Rightarrow R_{ D }=2 P(\uparrow)$

Therefore,

$\sum F_y=0 \Rightarrow R_{ C }=P(\downarrow)$

Again from Figure 10.34(a)

$\sum M_{ A }=0 \Rightarrow R_{ B }=2 R_{ C }=2 P(\downarrow)$

and            $\sum F_y=0 \Rightarrow R_{ A }=P(\uparrow)$

So, both beams are symmetrically and identically loaded. Thus, the total strain energy of the system is

$U_{\text {bending }}=2\left[\left\lgroup \frac{1}{2 E I} \right\rgroup \int_0^a M_x^2 d x\right]_{ AC }+2\left[\left\lgroup \frac{1}{2 E I} \right\rgroup \int_0^a M_x^2 d x\right]_{ CE }$

$=\left\lgroup \frac{2}{E I} \right\rgroup \int_0^a M_x^2 d x$

$=\frac{2}{E I} \int_0^a P^2 x^2 d x=\frac{2}{3} \frac{P^2 a^2}{E I}$

The total strain energy of the system is $2 P^2 a^3 / 3 E I$

To calculate the deflection at point E, we can apply Castigliano’s second theorem [refer Eq. (10.66)].

$\frac{\partial U}{\partial Q_i}=\Delta_i ; \quad 1 \leq i \leq n$            (10.66)
Therefore,

$\Delta_{ E }=\frac{\partial U}{\partial P}=\frac{4 P a^3}{3 E I}$

Thus, vertical deflection at E is

$\Delta_{ E }=\frac{4 P a^3}{3 E I}(\downarrow)$