Consider a bar of circular cross section and uniform diameter throughout. Consider it to be suspended from a rigid support and its top end, such that it is in a hanging in a vertical position as shown in the figure.

Let,

A = Uniform cross sectional area of the bar

E = Young’s modulus for the bar

L = Length of the bar

ρ = Weight of the bar, per unit length, for the material of the bar

Consider an element of length ‘dy’ at a distance of ‘y’ from the bottom of the bar being elongated due to the force ‘P’, at section x-x, as shown in the figure.

Weight of the portion below x-x = P = ρ × A × y

Change in the length of the element ‘dy’ = $\frac{Pl}{AE} = \frac{ρ × (A × y) × dy}{AE}$

=$\frac{ρ × y × dy}{E}$

For total change in the length of the bar, we need to integrate along the length

Total change in length = $\int_0^L\frac{ρy.dy}{E}$

On integrating, we get,

$δL = \frac{ρL^2}{2E}$

This is the expression for the elongation of a uniform bar under self weight.