---
title: "Elongation of uniform bar under its self weight"
subject: "Strength of Materials"
chapter: 3
key: "semester_3/strength_of_materials:bar-self-weight-elongation"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/bar-self-weight-elongation
site: BCT Engineering
---

# Elongation of uniform bar under its self weight

Strength of Materials · Chapter 3: Simple Stress and Strain

- Asked in: 2075 Asoj, 5 marks; 2068 Baisakh (old course), 4 marks (asked 2 times)
- Video: [Elongation of a bar due to self weight | simple stress & strain13 | Engg Mechanics | Sumit Dwivedi |](https://www.youtube.com/watch?v=WWYjVEWLCB8) — Learning with Sumit
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/bar-self-weight-elongation
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-3

## Question

Derive the expression for the total elongation of a uniform bar of length L and cross section area A under its self weight.

## Short answer

Elongation δ = γL²/(2E) = WL/(2AE), where γ is unit weight, W total weight. Derived by considering stress γx at distance x from free end and integrating strain over length.

## Answer

Consider a uniform vertical bar of length $L$, area $A$, Young's modulus $E$ and unit weight (weight density) $\gamma = \rho g$, hanging from a fixed support at the top and carrying only its own weight.

```text
 ═══════  fixed support
   │ ↑ x measured from the free (bottom) end
   │
   │  dx
   │
   ○  free end
```

#### Derivation
Take a section at distance $x$ from the free end. The load it carries is the weight of the bar below it:

$$
P_x = \gamma A x
$$

The stress at the section is $\sigma_x = \dfrac{P_x}{A} = \gamma x$, which varies linearly from zero at the free end to $\gamma L$ at the support.

The elongation of a small element of length $dx$:

$$
d\delta = \frac{\sigma_x}{E}dx = \frac{\gamma x}{E}dx
$$

Total elongation:

$$
\delta = \int_0^L\frac{\gamma x}{E}dx = \frac{\gamma L^2}{2E}
$$

With the total weight $W = \gamma A L$:

$$
\boxed{\delta = \frac{\gamma L^2}{2E} = \frac{WL}{2AE}}
$$

**Meaning:** the elongation due to self-weight is half the elongation that the same weight $W$ would cause if it were applied as a point load at the free end ($\delta = WL/AE$). Maximum stress $\sigma_{max} = \gamma L$ occurs at the support.

**Example:** a steel bar of $L = 100$ m with $\gamma = 77$ kN/m³ and $E = 200$ GPa: $\delta = \dfrac{77\times10^{-6}\ \text{N/mm}^3\times(10^5)^2}{2\times200\times10^3} = 1.925$ mm, with $\sigma_{max} = 7.7$ N/mm².

## Check yourself

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

1. What axial load acts on a section at distance x from the free end?
   A. γAL
   B. γA(L - x)
   C. P = γAx (weight of the part below)
   D. W
   Answer: C. Only the portion below the section is carried. (Key point: Load at section)

2. How does stress vary along the bar and where is it maximum?
   A. Constant
   B. Linearly from zero at the free end to γL at the support
   C. Maximum at the free end
   D. Parabolic
   Answer: B. σx = γx, so it peaks at the support. (Key point: Stress)

3. What is the total elongation?
   A. γL^2/2E = WL/2AE
   B. WL/AE
   C. γL^2/E
   D. WL/4AE
   Answer: A. Integrating γx/E from 0 to L gives γL^2/2E. (Key point: Result)

4. How does it compare with the same weight W applied as a point load at the end?
   A. Equal
   B. Twice
   C. Four times
   D. Half as much
   Answer: D. WL/2AE is half of WL/AE, because the load builds up gradually. (Key point: Meaning)

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