Question

The thickness of a rectangular steel girder equals $h$. Using the equation $N\left(x\right)=EI\frac{d^{2}y}{d{x}^2}$, find the deflection $\lambda$ caused by the weight of the girder when the girder of length $2l$ rests freely on two supports (figure shown above):

• A. $\lambda =\frac{5\rho g{l}^4}{2E{h}^2}$
• B. $\lambda =\frac{7\rho g{l}^4}{2E{h}^2}$
• C. $\lambda =\frac{3\rho g{l}^3}{2E{h}^2}$
• D. $\lambda =\frac{5\rho g{l}^5}{2E{h}^3}$

Answer 1

The correct option is A $\lambda =\frac{5\rho g{l}^4}{2E{h}^2}$

$\begin{array}{c}\text{This bending momeut is clearly}\\ \begin{array}{cc}N=& {\int }_x^{2\ell }\omega d\epsilon (\epsilon -x)-\mathrm{wl}(2l-x)\\ & =\omega (2l^2-2xl+\frac{x^2}{2})\omega \ell (2\ell -x)\\ & =\omega (\frac{x^2}{2}-xl)\end{array}\end{array}$

$\begin{array}{c}\text{Here}\omega =\rho gbh\text{is weight of the bean perunit length}\\ \text{Now integrating}El\frac{dy}{dx}=\omega (\frac{x^3}{6}-\frac{x^{2}l}{2})+c_0\\ \text{or since}\frac{dy}{dx}=0\text{for}x=l,c_0=\frac{\omega l^3}{3}\end{array}$

$\begin{array}{cc}\text{As}y=0& \text{for}x=0,c_1=0\text{from this we find.}\\ \lambda & =y(x-1)=\frac{5w{l}^2}{24}/EI=\frac{5\rho g14}{2E{h}^2}\end{array}$

$\therefore \text{option}A\text{is correct}$