Question

The slope at point 'C' and deflection at point B in the beam as shown in figure are respectively

• A. $\frac{5}{4}\frac{P{l}^2}{EI},\frac{25P{l}^3}{12EI}$
• B. $\frac{3}{2}\frac{P{l}^2}{EI},\frac{25P{l}^3}{12EI}$
• C. $\frac{5}{4}\frac{P{l}^2}{EI},\frac{33P{l}^3}{12EI}$
• D. $\frac{3}{2}\frac{P{l}^2}{EI},\frac{26P{l}^3}{15EI}$

Answer 1

The correct option is C $\frac{5}{4}\frac{P{l}^2}{EI},\frac{33P{l}^3}{12EI}$

$\text{Slope at C}=\frac{P{l}^2}{2E\left(2I\right)}+\frac{2Pl\times l}{E\left(2I\right)}$

$=\frac{P{l}^2}{4EI}+\frac{2P{l}^2}{2EI}$

${\theta }_c=\frac{5P{l}^2}{4EI}$

$\text{Deflection at C}={\Delta }_c=\frac{P{l}^3}{3\times \left(2I\right)\times E}+\frac{2Pl\times (l{)}^2}{2\times (2I\times E)}$

$=\frac{P{l}^3}{EI}(\frac{1}{6}+\frac{1}{2})=\frac{4}{6}\frac{P{l}^3}{EI}$

Deflection at 'B'

${\Delta }_B={\theta }_C\times l+\frac{4P{l}^3}{6EI}+\frac{P{l}^3}{3EI}+\frac{Pl\times l^2}{2EI}$

$=\frac{5P{l}^3}{4EI}+\frac{4P{l}^3}{6EI}+\frac{P{l}^3}{3EI}+\frac{P{l}^3}{2EI}$

$=\frac{15P{l}^3+8P{l}^3+4P{l}^3+6P{l}^3}{12\times EI}$

$=\frac{33P{l}^3}{12EI}$