Elongation of tapering circular bar (derivation)

Consider a circular bar of length $L$ whose diameter varies uniformly from $d_1$ at one end to $d_2$ at the other, subjected to an axial pull $P$. Young's modulus is $E$.

                                       ____
                          ______----''''   |
             _____----''''                 |
   P <---   |d1            x -->         d2|   ---> P
             '''''----______               |
                          ''''''----____   |
                                       ''''
            |<------------- L ------------>|

Derivation

Measure $x$ from the small end (diameter $d_1$, taking $d_2 > d_1$). The diameter at distance $x$ is

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

Cross-sectional area: $A_x = \dfrac{\pi}{4}d_x^2$, so the stress is $\sigma_x = \dfrac{4P}{\pi d_x^2}$.

Elongation of a small length $dx$:

$$d\delta = \frac{\sigma_x}{E}dx = \frac{4P}{\pi E d_x^2}dx$$

Total elongation:

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

Since $d_1 + kL = d_2$, and $\dfrac{1}{d_1} - \dfrac{1}{d_2} = \dfrac{d_2 - d_1}{d_1d_2} = \dfrac{kL}{d_1d_2}$:

$$\boxed{\delta = \frac{4PL}{\pi E\,d_1\,d_2}}$$

Checks

• If $d_1 = d_2 = d$, then $\delta = \dfrac{4PL}{\pi Ed^2} = \dfrac{PL}{AE}$, the formula for a prismatic bar.
• The formula holds only for a gradual taper, away from the ends, where the stress is assumed uniform over each section (Saint-Venant's principle).

Example: $P = 50$ kN, $L = 1000$ mm, $d_1 = 20$ mm, $d_2 = 40$ mm, $E = 200$ GPa gives $\delta = \dfrac{4(50\times10^3)(1000)}{\pi(200\times10^3)(20)(40)} = 0.398$ mm.