---
title: "Elongation of tapered bar and self-weight"
subject: "Strength of Materials"
chapter: 2
key: "semester_4/strength_of_materials_mechanical:tapered-bar-elongation-derivation"
source: https://www.bctengineer.com/q/semester_4/strength_of_materials_mechanical/tapered-bar-elongation-derivation
site: BCT Engineering
---

# Elongation of tapered bar and self-weight

Strength of Materials · Chapter 2: Stress and strain – axial load

- Asked in: year not given, 4+2 marks
- Video: [Deformation of a uniformily tapering circular bar](https://www.youtube.com/watch?v=viRLTivxhQY) — Make Mechanical Simple
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/strength_of_materials_mechanical/tapered-bar-elongation-derivation
- Whole chapter: https://www.bctengineer.com/md/semester_4/strength_of_materials_mechanical/chapter-2

## Question

Derive an expression for the total elongation of a circular bar of length L whose diameter varies uniformly from d1 at one end to d2 at the other, under an axial pull P. Also derive the elongation of a vertical prismatic bar of length L and weight density w (N/m^3) hanging under its own weight.

## Short answer

For a tapered circular bar with end diameters d1, d2: total elongation δ = 4PL/(π E d1 d2). For a prismatic bar hanging under own weight density w: axial force at distance y from free end is w A y, giving δ = wL²/(2E) = WL/(2AE).

## Answer

#### 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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is the total elongation of the tapered bar derived?
   A. Use PL/AE with the average diameter
   B. Integrate d(delta) = P dx / (A_x E) along the length with A_x varying
   C. Use PL/AE with the smallest area
   D. Multiply the end elongations
   Answer: B. Force P is constant but the area changes with x, so an element dx is integrated: delta = (4P/pi E) integral dx/(d1+kx)^2. (Key point: Tapered bar method)

2. What is the result for the tapered circular bar?
   A. delta = 4PL / (pi E (d1 + d2)^2/4)
   B. delta = 4PL / (pi E d1 d2)
   C. delta = 4PL / (pi E d1^2)
   D. delta = 2PL / (pi E d1 d2)
   Answer: B. The integral gives delta = 4PL/(pi E d1 d2), which reduces to PL/AE when d1 = d2. (Key point: Tapered bar result)

3. A vertical prismatic bar of weight density w hangs under its own weight. What is its elongation?
   A. w L^2 / E
   B. w L^2 / (2E)
   C. w L / (2E)
   D. w L^3 / (2E)
   Answer: B. The force at distance y from the free end is wAy, so delta = (w/E) integral y dy = wL^2/(2E), the same as half the weight acting at the free end. (Key point: Self-weight elongation)

## Related in Strength of Materials

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- [Thermal stress in a restrained bar](https://www.bctengineer.com/question.md?key=semester_4%2Fstrength_of_materials_mechanical%3Abar-between-rigid-walls-temperature) — Strength of Materials, chapter 2

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