Question

An overhang beam of uniform EI is loaded as shown in the figure below. the deflection at the free end will be

• A. $\frac{P{L}^3}{81EI}$
• B. $\frac{P{L}^3}{27EI}$
• C. $\frac{4P{L}^3}{81EI}$
• D. $\frac{2P{L}^3}{27EI}$

Answer 1

The correct option is C $\frac{4P{L}^3}{81EI}$

The over hanging part of thr beam can be replaced by moment $\frac{PL}{3}$ and load P at point B. Slope at B

${\theta }_B=\left(\frac{PL}{3}\right)\frac{L}{3EI}=\frac{P{L}^2}{9EI}$



If there was no load on overhanging part the deflection of load point would have been ${\theta }_B\times \frac{L}{3}$. however dur to load the additional deflection $\frac{P(L/3{)}^3}{3EI}=\frac{P{L}^3}{81EI}$ is caused.

$\therefore Deflectionunderload$

${\Delta }_{Load}=\frac{P{L}^2}{9EI}\times \frac{L}{3}+\frac{P{L}^3}{81EI}=\frac{4P{L}^3}{81EI}$