Question

In the given portal frame, horizontal sway will be $\alpha \frac{P{l}^3}{EI}.$ The value of $\alpha$ is $\text{\_\_\_\_}\times {10}^{-2}.$

• A. 3.7

Answer 1

The correct option is A 3.7

Assuming, sway in the right direction


So, horizontal sway force =

$\frac{24EI\delta }{l^3}+\frac{3EI\delta }{l^3}=\frac{27EI\delta }{l^3}$

Let, P = sway force

$\therefore P=\frac{27EI\delta }{l^3}$

$\Rightarrow \delta =\frac{P{l}^3}{27EI}$

$\therefore \alpha =\frac{1}{27}=0.037=3.7\times {10}^{-2}$