Question

Let a steel bar of lengthl, breadth $b$ and depth $d$ be loaded at the center by a load $W$. Then the sag of bending of beam is (Y = Young's modulus of steel)

• A. $\frac{W l^{3}}{2 b d^{3} Y}$
• B. $\frac{W l^{3}}{4 b d^{3} Y}$
• C. $\frac{W l^{3}}{2 db^{3} Y}$
• D. $\frac{W l^{3}}{4 b d^{3} Y}$

Answer 1

Answer: (B)

The sag of bending of beam is given by
$\delta=\frac{W l^{3}}{48 Y I}$ Where $W$ is load, $L$ is length, $Y$ is young's modulus and $\operatorname{Iis} \frac{b d^{3}}{12}$ (area moment of inertia)
$\therefore \delta=\frac{W l^{3}}{48 Y} \times \frac{12}{b d^{3}}=\frac{W l^{3}}{4 b d^{3} Y}$