Topic: Strength of Materials
A force P is applied at a distance 'X' from the end of the beam as shown in the figure. What would be value of 'X' so that the displacement at 'A' is equal to zero?
- A. 0.5 L
- B. 0.25 L
- C. 0.33 L
- D. 0.66 L
Answer: D. 0.66 L
Explanation: The moment applied at the support due to the offset load is $M = P(2L - X)$. The upward deflection at point A due to this moment is $\delta_{M} = \frac{M(2L)^{2}}{2EI} = \frac{4P(2L-x)L^{2}}{2EI}$. The downward deflection at point A due to the point load P at the end is $\delta_{P} = \frac{P(2L)^{3}}{3EI} = \frac{8PL^{3}}{3EI}$. Equating them for zero net displacement: $\frac{4P(2L-x)L^{2}}{2EI} = \frac{8PL^{3}}{3EI} \Rightarrow 2L-x = \frac{4}{3}L \Rightarrow x = 2L - \frac{4}{3}L = \frac{2}{3}L = 0.66L$.