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

Answer 1

Answer: (B)

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