Tapering rod: total extension under load and self weight

The rod is a frustum of a cone, fixed at the top (d₁ = 80 mm) with the 45 kN load at the bottom (d₂ = 40 mm). The extension has two parts: that due to the axial load and that due to the rod's own weight.

Data: $L = 3000$ mm, $E = 210\times10^3\ \text{N/mm}^2$, $P = 45\,000$ N, $\rho = 2\times10^5\ \text{kg/m}^3$ (as given), $g = 9.81\ \text{m/s}^2$.

Extension due to the load

For a uniformly tapering circular bar:

$$\delta_P = \frac{4PL}{\pi E d_1 d_2} = \frac{4(45\,000)(3000)}{\pi(210\,000)(80)(40)} = 0.2558\ \text{mm}$$

Extension due to self weight

Volume of the frustum $V = \dfrac{\pi L}{12}(d_1^2 + d_1d_2 + d_2^2) = 0.008796\ \text{m}^3$, so the weight is

$$W = \rho g V = 2\times10^5(9.81)(0.008796) = 17.26\ \text{kN}$$

Let $y$ be measured up from the bottom, with diameter $d = d_2 + ky$ where $k = \dfrac{d_1-d_2}{L} = 0.01333$. The weight below that section is $\dfrac{\rho g\pi}{12k}(d^3-d_2^3)$, and the self-weight extension is

$$\delta_w = \int_0^L \frac{\text{(weight below)}}{E\,\frac{\pi}{4}d^2}\,dy = \frac{\rho g}{3k^2E}\left[\frac{d_1^2-d_2^2}{2} + d_2^3\left(\frac{1}{d_1}-\frac{1}{d_2}\right)\right] = 0.0280\ \text{mm}$$

Total extension

$$\delta = \delta_P + \delta_w = 0.2558 + 0.0280 = 0.2838\ \text{mm}$$

Answer: total extension $\approx 0.284\ \text{mm}$.