Hello,

Let,

A = Uniform cross sectional area of the bar

E = Young’s modulus for the bar

L = Length of the bar

w = Specific wt. of bar material

W = weight 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’

So, weight of bar for length y is w.A.y....... ( Since, Sp. Wt. = Weight / Area )

As elongation is taking place due to own weight, P = W

We know that,

Elongation = PL / AE

So, dL = Wdy / AE

So, dL = ( w.A.y ) x dy / A.E

So, dL = ( w / E ) x y.dy

Now, for total elongation, Intergrating above eqn with respect to y

So, dL = ( w / E ) x ∫ y.dy

So, dL = ( w / E ) x [ ( lim 0 to L ) y^2 / 2 ]

So, dL = ( wL ^ 2 ) / 2E

Since, W = wAL for full length of bar,

w = W / AL

So, dL = WL / 2AE

Hence, the elongation is WL / 2AE .

Best Wishes.