Topic: Strength of Materials
A frame of two arms of equal length L is shown in the figure. The flexural rigidity of each arm of the frame is EI. The vertical deflection at the point of application of load P is
- A. $\frac{PL^{3}}{3EI}$
- B. $\frac{2PL^{3}}{3EI}$
- C. $\frac{4PL^{3}}{3EI}$
- D. None of the above
Answer: C. $\frac{4PL^{3}}{3EI}$
Explanation: Using Castigliano's theorem, we calculate the strain energy for both sections. For the vertical section AB, the moment is $M = P \times x$. The strain energy is $U_{AB} = \int_{0}^{L} \frac{M^{2}dx}{2EI} = \int_{0}^{L} \frac{(Px)^{2}dx}{2EI} = \frac{P^{2}}{2EI}[\frac{x^{3}}{3}]_{0}^{L} = \frac{P^{2}L^{3}}{6EI}$. For the horizontal section BC, the moment is constant at $M = P \times L$. The strain energy is $U_{BC} = \int_{0}^{L} \frac{(PL)^{2}dx}{2EI} = \frac{P^{2}L^{2}}{2EI}\int_{0}^{L} dx = \frac{P^{2}L^{3}}{2EI}$. The total vertical deflection is the partial derivative of the total strain energy with respect to P: $\frac{\partial U_{AB}}{\partial P} + \frac{\partial U_{BC}}{\partial P} = \frac{2PL^{3}}{6EI} + \frac{2PL^{3}}{2EI} = \frac{PL^{3}}{3EI} + \frac{PL^{3}}{EI} = \frac{4PL^{3}}{3EI}$.