Question

The strength of a beam varies as the product of its breadth $b$ and square of its depth $d$. A beam cut out of a circular log of radius $r$ would be stiffest when

• A. $b=d=\frac{r}{\sqrt{2}}$
• B. $b^2=\frac{r}{\sqrt{2}}=d$
• C. $d=\sqrt{2}b=2\sqrt{\frac{2}{3}}r$
• D. $d=\sqrt{3}b=2\sqrt{\frac{2}{3}}r$

Answer 1

The correct option is C $d=\sqrt{2}b=2\sqrt{\frac{2}{3}}r$

Given that $s\propto b{d}^2$
$s=kb{d}^2$
and $r^2=b^2+d^2$
$\Rightarrow S=kb(r^2-b^2)$
for max. stiffness $\frac{ds}{db}=k(r^2-3b^2)=0$
$\Rightarrow r=\sqrt{3}b$
now, $\frac{d^{2}s}{d{b}^2}=-6kbr<0$ at $r=\sqrt{3}b$
Therefore, $s$ is max. at $r=\sqrt{3}b$