Question

A cantilever beam PQ of uniform flexural rigidity (EI) is subjected to a concentrated moment M at R as shown in the figure


The deflection at the free end Q is

• A. $\frac{3M{L}^2}{4EI}$
• B. $\frac{3M{L}^2}{8EI}$
• C. $\frac{M{L}^2}{4EI}$
• D. $\frac{M{L}^2}{6EI}$

Answer 1

The correct option is B $\frac{3M{L}^2}{8EI}$

We know that

$\Delta Q=\Delta Q_1+\Delta Q_2...\left(1\right)$
$\Delta Q1=\Delta R=\frac{M(L/2{)}^2}{2EI}=\frac{M{L}^2}{8EI}...\left(2\right)$
$And\Delta Q_2={\theta }_R.\left(\frac{L}{2}\right)$
$=\frac{M\left(\frac{L}{2}\right)}{EI}\times \left(\frac{L}{2}\right)$
$=\frac{M{L}^2}{4EI}...\left(3\right)$
From (1), (2) & (3) we can write
$\Delta Q=\frac{M{L}^2}{8EI}+\frac{M{L}^2}{4EI}$
$=\frac{(1+2)M{L}^2}{8EI}$
$\Rightarrow \Delta Q=\frac{3M{L}^2}{8EI}$