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## Q. 10.18

Figure 10.33 shows a compound beam loaded at its free end. If the flexural rigidity is constant throughout the beam, calculate the total strain energy stored. Using this strain energy, find the deflection at point E.

## Verified Solution

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

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