Hello,

Length of Bar = L

Radius of Broad end = r1. Hence, Diameter = d1.

Radius of smaller end = r2. Hence, Diameter = d2.

Let, the force due to self weight = P

Now, consider a small element of length dx of the bar at distance x from the broad end. Hence, diameter of bar at x from broad end = d1 - ( d1 - d2 ). x / L

Let, d1 - d2 / L = k

Hence, diameter of bar at distance x = d1 - kx

So, Cross sectional area at x = π /4 x ( d1 - kx )^2

We know that,

Stress = Force / Area

Hence, Stress = P / [ π / 4 x ( d1 - kx )^2 ]

So, Stress = 4P / π ( d1 - kx )^2

Also, Stress / Strain = E

Hence, Strain = Stress / E

So, Stress = 4P / [ π ( d1 - kx )^2 ] E

Hence, Elongation of elementary length = { 4P / [ π ( d1 - kx )^2 ] E }.dx

Let, Total elongation = dL

So,  dL = lim. ( 0 to L ) ∫ { 4P / [ π ( d1 - kx )^2 ] E }.dx

Solving the above integration, we get,

dL = 4P / πEk [ ( 1/ d1 - kl ) - 1/d1 ]

Resubstituting  k = d1 - d2 / L

So, dL = 4PL / πEd2.d1

Hence, the total deformation due to the self weight of bar is 4PL / πEd1d2

Best Wishes.