Let a steel bar of length l,breadth b,and dep | vedclass

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(B) The sag (depression) $\delta$ of a beam supported at both ends and loaded at the centre is given by the formula: $\delta = \frac{W l^3}{48 Y I}$.H

Vedclass · Accepted Answer

$\frac{W l^3}{4 b d^3 Y}$

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(B) The sag (depression) $\delta$ of a beam supported at both ends and loaded at the centre is given by the formula: $\delta = \frac{W l^3}{48 Y I}$.Here,$W$ is the load,$l$ is the length,$Y$ is Young's modulus,and $I$ is the geometrical moment of inertia.For a rectangular cross-section of breadth $b$ and depth $d$,the moment of inertia is $I = \frac{b d^3}{12}$.Substituting the value of $I$ into the formula for $\delta$:$\delta = \frac{W l^3}{48 Y (\frac{b d^3}{12})}$$\delta = \frac{W l^3}{48 Y} 	imes \frac{12}{b d^3}$$\delta = \frac{W l^3}{4 b d^3 Y}$.

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

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