Derivation of Extension for Tapering Rod

This solution explains how to find the extension ($\Delta$) of a uniformly tapering rod under an axial tensile load (P).

Problem Setup

We are given a rod with:

• Length: $L$
• Diameter at one end: $d_1$
• Diameter at the other end: $d_2$
• Axial tensile load: $P$
• Modulus of elasticity: $E$

We need to find the total extension $\Delta$.

Finding the Diameter at any Section

For a uniformly tapering rod, the diameter $d$ at a distance $x$ from the end with diameter $d_1$ varies linearly. The formula is:

$d(x) = d_1 + \frac{(d_2 - d_1)x}{L}$

Differential Extension Calculation

Consider a small element of the rod of length $dx$ at position $x$.

The cross-sectional area $A(x)$ at this section is:

$A(x) = \frac{\pi [d(x)]^2}{4} = \frac{\pi}{4} \left( d_1 + \frac{(d_2 - d_1)x}{L} \right)^2$

The extension $d\Delta$ of this small element $dx$ under load $P$ is given by the formula:

$d\Delta = \frac{P dx}{A(x) E}$

Substituting the expression for $A(x)$:

$d\Delta = \frac{P dx}{\frac{\pi}{4} \left( d_1 + \frac{(d_2 - d_1)x}{L} \right)^2 E} = \frac{4P dx}{\pi E \left( d_1 + \frac{(d_2 - d_1)x}{L} \right)^2}$

Total Extension Integration

To find the total extension $\Delta$, we integrate $d\Delta$ from $x=0$ to $x=L$:

$\Delta = \int_{0}^{L} \frac{4P dx}{\pi E \left( d_1 + \frac{(d_2 - d_1)x}{L} \right)^2}$

Let $u = d_1 + \frac{(d_2 - d_1)x}{L}$. Then $du = \frac{d_2 - d_1}{L} dx$, which means $dx = \frac{L}{d_2 - d_1} du$.

The limits of integration change: when $x=0$, $u=d_1$; when $x=L$, $u=d_2$.

The integral becomes:

$\Delta = \int_{d_1}^{d_2} \frac{4P}{\pi E} \left( \frac{L}{d_2 - d_1} \right) \frac{du}{u^2}$

$\Delta = \frac{4PL}{\pi E (d_2 - d_1)} \int_{d_1}^{d_2} u^{-2} du$

$\Delta = \frac{4PL}{\pi E (d_2 - d_1)} \left[ -\frac{1}{u} \right]_{d_1}^{d_2}$

$\Delta = \frac{4PL}{\pi E (d_2 - d_1)} \left( -\frac{1}{d_2} - (-\frac{1}{d_1}) \right)$

$\Delta = \frac{4PL}{\pi E (d_2 - d_1)} \left( \frac{1}{d_1} - \frac{1}{d_2} \right)$

$\Delta = \frac{4PL}{\pi E (d_2 - d_1)} \left( \frac{d_2 - d_1}{d_1 d_2} \right)$

$\Delta = \frac{4PL}{\pi E d_1 d_2}$

Final Formula and Answer

The extension of the uniformly tapering rod is:

$\Delta = \frac{4PL}{\pi E d_1 d_2}$

This matches Option A.