---
title: "Total elongation of stepped-load steel bar"
subject: "Strength of Materials"
chapter: 3
key: "semester_3/strength_of_materials:steel-bar-total-elongation-abcd"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/steel-bar-total-elongation-abcd
site: BCT Engineering
---

# Total elongation of stepped-load steel bar

Strength of Materials · Chapter 3: Simple Stress and Strain

- Asked in: 2076 Chaitra, 6 marks
- Video: [Mechanics of Materials: Lesson 17 - Axial Elongation Due to Axial Load Example](https://www.youtube.com/watch?v=EypXa4lMQhA) — Jeff Hanson
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/steel-bar-total-elongation-abcd
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-3

## Question

Find the total elongation in the bar. Take E for the material as the 200 GPa. A Steel bar of $600\ \text{mm}^2$ cross-sectional area is carrying loads as shown in the figure given below. [Figure: bar A-B-C-D; 90 kN acting to the left at A, 20 kN acting to the right at B, 30 kN acting to the right at C, 40 kN acting to the right at D; A-B 750 mm, B-C 1000 mm, C-D 750 mm.]

## Short answer

Determine axial forces: AB 90 kN, BC 70 kN, CD 40 kN. Sum segment elongations Δ = NL/AE using E = 200 GPa and A = 600 mm². Total elongation = 1.396 mm.

## Answer

Loads: 90 kN (←) at A, 20 kN (→) at B, 30 kN (→) at C, 40 kN (→) at D. Check: 20 + 30 + 40 = 90 kN, so the bar is in equilibrium.

#### Axial force in each portion (tension +ve)
```text
 90 kN <--A-----750-----B--------1000--------C----750----D--> 40 kN
                        |                    |
                     20 kN -->            30 kN -->
```
Using the section method from the left end:

| Portion | Length (mm) | Axial force |
|---|---|---|
| AB | 750 | $+90$ kN (tension) |
| BC | 1000 | $90 - 20 = +70$ kN (tension) |
| CD | 750 | $70 - 30 = +40$ kN (tension) |

#### Elongation
$$
\begin{aligned}
\delta &= \frac{1}{AE}\sum N_iL_i = \frac{90\times10^3(750)+70\times10^3(1000)+40\times10^3(750)}{600\times200\times10^3}
= \frac{167500000}{120000000} = 1.3958\ \text{mm}
\end{aligned}
$$

Portion by portion: $\delta_{AB} = 0.5625$ mm, $\delta_{BC} = 0.5833$ mm, $\delta_{CD} = 0.2500$ mm, total $= 1.3958$ mm.

**Answer: total elongation = 1.396 mm.**

## Check yourself

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

1. How do you know the bar is in equilibrium?
   A. Loads to the right (20+30+40) equal 90 kN to the left
   B. The loads are all equal
   C. The bar is fixed
   D. Total of areas is zero
   Answer: A. 20 + 30 + 40 = 90 kN. (Key point: Equilibrium check)

2. What is the axial force in BC?
   A. 90 kN tension
   B. 70 kN tension
   C. 20 kN tension
   D. 50 kN compression
   Answer: B. Section from the left: 90 - 20 = 70 kN. (Key point: Axial force)

3. Which formula gives the total elongation?
   A. N_max L_total / A E
   B. Average N x L_total / A E
   C. Sum of N L over A E for each portion
   D. Sum of N / A E
   Answer: C. Each portion has its own force; elongations add to 1.396 mm. (Key point: Elongation)

4. What is the total elongation?
   A. 1.396 mm
   B. 0.5625 mm
   C. 2.0 mm
   D. 0.14 mm
   Answer: A. 0.5625 + 0.5833 + 0.2500 = 1.396 mm. (Key point: Result)

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