Question

A uniform beam (EI=constant ) PQ in the form of a quarter-circle of radius R is fixed at end P and free at the end Q, where a load W is applied as shown. The vertical downward displacement, ${\delta }_q$, at the loaded point Q is given by:
${\delta }_q=\beta \left(\frac{w{R}^3}{EI}\right)$
Find the value of $\beta$ (correct to 4-decimal places).

• A. 0.7854

Answer 1

The correct option is A 0.7854
${\delta }_B=\beta \left(\frac{W{R}^3}{EI}\right)$
We need to find $\beta$
$M_{xx}=-WRsin\theta$
$\frac{\partial M_{xx}}{\partial w}=-Rsin\theta$
$\frac{\partial U}{\partial w}=\int M_{xx}\frac{\left(\partial M_{xx}/\partial w\right)dx}{EI}$

$={\int }_0^{\pi /2}\frac{\left(w{R}^{2}si{n}^2\theta \right)Rd\theta }{EI}$
$\Rightarrow \frac{\partial U}{\partial w}=\frac{w{R}^3}{2EI}{\int }_0^{\pi /2}2si{n}^2\theta d\theta$
$=\frac{w{R}^3}{2EI}{\int }_0^{\pi /2}(1-cos2\theta )d\theta$
$\Rightarrow \frac{\partial U}{\partial w}={\delta }_B=\frac{\pi }{4}\left(\frac{w{R}^3}{EI}\right)$
Thus $\beta =\frac{\pi }{4}=0.7854$