Elongation of tapered bar and self-weight

Tapered circular bar

Let $x$ be measured from the end of diameter $d_1$. The diameter at $x$ varies linearly:

$$d_x = d_1 + \frac{d_2 - d_1}{L}x = d_1 + kx, \qquad k = \frac{d_2-d_1}{L}$$

The axial force is the same ($P$) at every section, but the area changes. Elongation of a small length $dx$:

$$d\delta = \frac{P\,dx}{A_xE} = \frac{4P\,dx}{\pi E\,(d_1+kx)^2}$$ $$\delta = \frac{4P}{\pi E}\int_0^L \frac{dx}{(d_1+kx)^2} = \frac{4P}{\pi E}\left[-\frac{1}{k(d_1+kx)}\right]_0^L = \frac{4P}{\pi E k}\left(\frac{1}{d_1}-\frac{1}{d_2}\right)$$

Substituting $k = (d_2-d_1)/L$:

$$\boxed{\delta = \frac{4PL}{\pi E\,d_1d_2}}$$

Check: when $d_1 = d_2 = d$ this gives $\frac{4PL}{\pi E d^2} = \frac{PL}{AE}$, the prismatic bar result.

Bar hanging under its own weight

The axial force at any section equals the weight of the part of the bar below it. Measuring $y$ upward from the free end, $F_y = w A y$.

$$d\delta = \frac{F_y\,dy}{AE} = \frac{wAy\,dy}{AE} \;\Rightarrow\; \delta = \frac{w}{E}\int_0^L y\,dy = \frac{wL^2}{2E}$$

Since the total weight is $W = wAL$, $\delta = \dfrac{WL}{2AE}$: the elongation is the same as if half the weight acted at the free end.