Question

$\text{The rotation of the overhanging end 'A' for the beam loaded as shown in figure is}\frac{KML}{EI}.$
Then the magnitude of K is

• A. 0.67

Answer 1

The correct option is A 0.67

$\sum M_x=0$

$\Rightarrow R_B\times L+M=0$
$\Rightarrow R_B=\frac{M}{L}$

$and{R}_C=\frac{M}{L}$

$\begin{array}{cccc}\text{Member}& M_x& \frac{\partial M_x}{\partial M}& \text{Range}\\ \text{AB}& \text{M}& \text{1}& \text{0 to}\frac{L}{3}\\ \text{CB}& \frac{M}{L}x& \frac{x}{L}& \text{0 to L}\end{array}$

$\therefore {\theta }_A=\frac{\partial U}{\partial M}=\int M_x\frac{\partial M_x}{\partial M}\frac{dx}{EI}$

$={\int }_0^{L/3}M.1\frac{dx}{EI}+{\int }_0^L\frac{M}{L}x.\frac{x}{L}\frac{dx}{EI}$

$=\frac{ML}{3EI}+\frac{ML}{3EI}=\frac{2ML}{3EI}\left(clockwise\right)$