Question

A cantilever beam carries moment at the free end. The depth/breadth ratio of beam is 2 : 1. The deflection of beam is ${\delta }_1$. Now, the span is halved and the cross -section is reversed. The deflection of this new arrangement due to same loading as in previous case is ${\delta }_2$. The ratio of $\frac{{\delta }_2}{{\delta }_1}$ is:

• A. 1 : 1
• B. 1 : 2
• C. 2 : 1
• D. 1 : 4

Answer 1

The correct option is A 1 : 1

Deflection at free end of cantilever beam due to moment at free end
${\delta }_1=\frac{M{L}^2}{2EI}=\frac{M{L}^2}{2E\left(\frac{b{d}^3}{12}\right)}....\left(1\right)$
Now, for $L^{\prime }=\frac{L}{2}$ and reversed cross section
${\delta }_2=\frac{M(L/2{)}^2}{2E\left(\frac{d{b}^3}{12}\right)}.....\left(2\right)$
$\frac{{\delta }_2}{{\delta }_1}=\frac{\frac{M{\left(\frac{L}{2}\right)}^2}{2E\left(\frac{d{b}^3}{12}\right)}}{\frac{M{L}^2}{2E\left(\frac{b{d}^3}{12}\right)}}=\frac{1}{4}\times {\left(\frac{d}{b}\right)}^2$
$=\frac{1}{4}\times 2^2=1(\because \frac{d}{b}=2(\text{given}\left)\right)$