Question

The vertical deflection of point ${}^{\prime }{B}^{\prime }$ for the beam $AB$ shown in figure is given as ${\delta }_B=\frac{P{l}^3}{\alpha EI}$. (correct upto two decimals) is

• A. 24

Answer 1

The correct option is A 24
Let the moment reaction developed at ${}^{\prime }{B}^{\prime }$ is ${}^{\prime }{M}^{\prime }$


To find $M$, put slope at ${}^{\prime }{B}^{\prime }$ zero

${\theta }_B=\frac{P(l/2{)}^2}{2EI}-\frac{Ml}{EI}=0$

$\Rightarrow M=\frac{Pl}{8}$

Deflection of end ${}^{\prime }{B}^{\prime }$

${\delta }_B=\frac{P(l/2{)}^3}{3EI}+\frac{P(l/2{)}^2}{2EI}\times \left(\frac{l}{2}\right)-\frac{M{l}^2}{2EI}$

${\delta }_B=\frac{P{l}^3}{24EI}+\frac{P{l}^3}{16EI}-\frac{P{l}^3}{16EI}=\frac{P{l}^3}{24EI}$