Question

For the same cross-section area and for a given load, the ratio of depression for the beam of a square cross-section and circular cross-section is

• A. $3:\pi$
• B. $\pi :3$
• C. $1:\pi$
• D. $\pi :1$

Answer 1

Answer: (A)

$3:\pi$

$\delta =\frac{W{l}^3}{3YI}$, where $W=$load$,l=$length of beam$,I=$moment of inertia$=\frac{b{d}^3}{12}$ for rectangular beam, and for square beam$,b=d.$ Thus, $I_1=\frac{b^4}{12}$
Now, for circular cross section, $I_2=\left[\frac{\pi r^4}{4}\right]$
$\therefore {\delta }_1=\frac{W{l}^3\times 12}{3Y{b}^4}=\frac{4W{l}^3}{Y{b}^4}$
and ${\delta }_2=\frac{W{l}^3}{3Y\left(\pi r^{\$}/4\right)}=\frac{4W{l}^3}{3Y\left(\pi r^4\right)}$
Thus, $\frac{{\delta }_1}{{\delta }_2}=\frac{3\pi r^4}{b^4}=\frac{3\pi r^4}{(\pi r^2{)}^2}=\frac{3}{\pi }$
$(\because b^2=\pi r^2$ as they have same cross sectional area)