---
title: "Elongation of a stepped steel bar"
subject: "Strength of Materials"
chapter: 2
key: "semester_4/strength_of_materials_mechanical:stepped-bar-axial-elongation"
source: https://www.bctengineer.com/q/semester_4/strength_of_materials_mechanical/stepped-bar-axial-elongation
site: BCT Engineering
---

# Elongation of a stepped steel bar

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

- Asked in: year not given, 8 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_4/strength_of_materials_mechanical/stepped-bar-axial-elongation
- Whole chapter: https://www.bctengineer.com/md/semester_4/strength_of_materials_mechanical/chapter-2

## Question

A steel bar ABCD is fixed at end A and free at end D. The segments are AB: 1.0 m long, 30 mm diameter; BC: 0.8 m long, 25 mm diameter; CD: 0.6 m long, 20 mm diameter. Axial loads act as follows: 60 kN pulling to the right at B, 40 kN pushing to the left at C, and 20 kN pulling to the right at D. Take E = 200 GPa. Find (a) the axial force and stress in each segment, (b) the total change in length of the bar, (c) the reaction at A.

## Short answer

By equilibrium, axial forces: AB = +40 kN, BC = -20 kN, CD = +20 kN. Stresses: AB 56.6 MPa (T), BC -40.7 MPa (C), CD 63.7 MPa (T). Total elongation = Σ(FL/AE) = 0.311 mm. Reaction at A = 40 kN.

## Answer

#### Free-body and axial forces
```text
 |A-----30mm-----B----25mm----C---20mm---D
 |<--- 1.0 m --->|<-- 0.8 m ->|<- 0.6 m ->
 R_A <---        60kN->   <-40kN     20kN->
```

Take forces to the right as positive. Cut each segment and balance the loads on the free side (right-hand side):

| Segment | Axial force $F$ | Nature |
|---|---|---|
| CD | $+20$ kN | Tension |
| BC | $+20 - 40 = -20$ kN | Compression |
| AB | $-20 + 60 = +40$ kN | Tension |

#### (a) Stresses
$A_{AB} = \frac{\pi}{4}(30)^2 = 706.9\ \text{mm}^2$, $A_{BC} = 490.9\ \text{mm}^2$, $A_{CD} = 314.2\ \text{mm}^2$.

| Segment | $\sigma = F/A$ (MPa) |
|---|---|
| AB | $40000/706.9 = +56.6$ |
| BC | $-20000/490.9 = -40.7$ |
| CD | $20000/314.2 = +63.7$ |

#### (b) Total change in length
$$
\delta = \sum \frac{F L}{A E}
$$

$$
\begin{aligned}
\delta_{AB} &= \frac{40000\times1000}{706.9\times200000} = +0.283\ \text{mm}\\
\delta_{BC} &= \frac{-20000\times800}{490.9\times200000} = -0.163\ \text{mm}\\
\delta_{CD} &= \frac{20000\times600}{314.2\times200000} = +0.191\ \text{mm}
\end{aligned}
$$

$$
\delta = 0.283 - 0.163 + 0.191 = +0.311\ \text{mm (elongation)}
$$

#### (c) Reaction at A
Equilibrium of the whole bar: $-R_A + 60 - 40 + 20 = 0$, so $R_A = 40$ kN acting to the left (equal to the force in AB).

**Answer: forces $+40$, $-20$, $+20$ kN; stresses $56.6$, $-40.7$, $63.7$ MPa; total elongation $= 0.311$ mm; $R_A = 40$ kN.**

## Check yourself

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

1. How are the axial forces in the segments found?
   A. Cut each segment and balance the loads on the free side
   B. Divide each load by the number of segments
   C. Add all loads and use the same force everywhere
   D. Use the reaction at D
   Answer: A. Cutting each segment and balancing the loads on the free (right) side gives +20, -20 and +40 kN for CD, BC and AB. (Key point: Method of sections)

2. What is the axial force in segment BC?
   A. +20 kN tension
   B. -20 kN compression
   C. +40 kN tension
   D. -40 kN compression
   Answer: B. BC: +20 (at D) - 40 (at C) = -20 kN, so it is in compression. (Key point: Force in BC)

3. How is the total change in length obtained for the stepped bar?
   A. Sum of F L / (A E) for the segments, with signs
   B. Total load times total length / (smallest A x E)
   C. Largest segment's elongation only
   D. Average stress times total length
   Answer: A. Each segment has its own F, L and A, so delta = sum of FL/AE = 0.283 - 0.163 + 0.191 = 0.311 mm. (Key point: Total elongation)

4. What is the reaction at A and why?
   A. 20 kN, equal to the force in CD
   B. 60 kN, the largest applied load
   C. 40 kN, from overall equilibrium (and equal to the force in AB)
   D. 120 kN, the sum of the magnitudes of all loads
   Answer: C. -R_A + 60 - 40 + 20 = 0 gives R_A = 40 kN, the same as the force in AB. (Key point: Reaction at A)

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