Question 2: Extension of Tapered Wire

Solution

Consider an element of thickness $dx$ at a distance $x$ from the smaller end (radius $a$).

The radius $r$ at distance $x$ varies linearly:

$$ r(x) = a + \frac{b-a}{L}x $$

The extension $dl$ of this small element is given by Hooke’s Law:

$$ dl = \frac{F \cdot dx}{A(x) \cdot Y} = \frac{F \cdot dx}{\pi [r(x)]^2 Y} $$

Total extension $\Delta l$ is the integral from $x=0$ to $x=L$:

$$ \Delta l = \int_{0}^{L} \frac{F}{\pi Y} \frac{dx}{(a + \frac{b-a}{L}x)^2} $$

Let $u = a + \frac{b-a}{L}x$, then $du = \frac{b-a}{L}dx$. Limits: $a \to b$.

$$ \Delta l = \frac{F}{\pi Y} \cdot \frac{L}{b-a} \int_{a}^{b} \frac{du}{u^2} $$ $$ \Delta l = \frac{FL}{\pi Y (b-a)} \left[ -\frac{1}{u} \right]_{a}^{b} $$ $$ \Delta l = \frac{FL}{\pi Y (b-a)} \left( \frac{1}{a} – \frac{1}{b} \right) = \frac{FL}{\pi Y (b-a)} \left( \frac{b-a}{ab} \right) $$