Question 11.CA.2: A load P is supported at B by two rods of the same material ......

Using the forces F_{B C} \text { and } F_{B D} in members BC and BD and recalling Eq. (11.14), the strain energy of the system is

U=\frac{P^2 L}{2 A E}            (11.14)

U=\frac{F_{B C}^2(B C)}{2 A E}+\frac{F_{B D}^2(B D)}{2 A E}            (1)

From Fig. 11.12a,

BC = 0.6l          BD = 0.8l

From the free-body diagram of pin B and the corresponding force triangle (Fig. 11.12b),

F_{B C}=+0.6 P \quad F_{B D}=-0.8 P

Substituting into Eq. (1) gives

U=\frac{P^2 l\left[(0.6)^3+(0.8)^3\right]}{2 A E}=0.364 \frac{P^2 l}{A E}