---
title: "Composite bar: total change in length and stresses"
subject: "Strength of Materials"
chapter: 3
key: "semester_3/strength_of_materials:composite-bar-aluminium-copper-steel"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/composite-bar-aluminium-copper-steel
site: BCT Engineering
---

# Composite bar: total change in length and stresses

Strength of Materials · Chapter 3: Simple Stress and Strain

- Asked in: 2080 Baisakh, 8 marks
- Video: [Som:stresses in varying bars how to find loads for each bar & elongation of the bar](https://www.youtube.com/watch?v=I77E5RkzBd4) — ENGINEERING TRICKS
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/composite-bar-aluminium-copper-steel
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-3

## Question

The composite section of aluminum, copper and steel bar is rigidly held by supports on left end. Find the total change in length and stresses in each bar, if the composite bar is subjected to loads at section as shown in figure. Take $E_{copper} = 100\ \text{kN/mm}^2$, $E_{aluminum} = 70\ \text{kN/mm}^2$ and $E_{steel} = 200\ \text{kN/mm}^2$. [Figure: bar fixed at the left end; aluminium bar $105\ \text{mm}\ \phi$, length 300 mm, with a 50 kN load acting to the left at its right end; copper bar $85\ \text{mm}\ \phi$, length 250 mm, with a 25 kN load acting to the left at its right end; steel bar $50\ \text{mm}\ \phi$, length 100 mm, with a 100 kN load acting to the right at its free end.]

## Short answer

Axial forces are 25, 75, and 100 kN in aluminum, copper, steel respectively (all tension). Stresses: σ_al =2.89, σ_cu =13.22, σ_st =50.93 N/mm²; total elongation =0.0709 mm.

## Answer

Order from the fixed wall: aluminium (300 mm), copper (250 mm), steel (100 mm). Loads: 50 kN (←) at the aluminium/copper junction, 25 kN (←) at the copper/steel junction and 100 kN (→) at the free end. Check: 100 kN to the right balances 50 + 25 = 75 kN to the left plus the wall reaction of 25 kN (←), so the wall pulls on the bar with 25 kN.

#### Axial force in each part (tension +ve, from the free end)
```text
|=== Al ===|=== Cu ===|== St ==>  100 kN
Wall   <-50 kN     <-25 kN
```
- Steel: $N_s = +100$ kN (tension)
- Copper: $N_c = 100 - 25 = +75$ kN (tension)
- Aluminium: $N_a = 100 - 25 - 50 = +25$ kN (tension)

#### Areas
$$
A_a = \frac{\pi}{4}(105)^2 = 8659.0,\quad A_c = \frac{\pi}{4}(85)^2 = 5674.5,\quad A_s = \frac{\pi}{4}(50)^2 = 1963.5\ \text{mm}^2
$$

#### Stresses
$$
\begin{aligned}
\sigma_a &= \frac{25\times10^3}{8659.0} = 2.887\ \text{N/mm}^2\\
\sigma_c &= \frac{75\times10^3}{5674.5} = 13.217\ \text{N/mm}^2\\
\sigma_s &= \frac{100\times10^3}{1963.5} = 50.930\ \text{N/mm}^2
\end{aligned}
$$

#### Change in length ($\delta = NL/AE$)
$$
\begin{aligned}
\delta_a &= \frac{25\times10^3\times300}{8659.0\times70\times10^3} = 0.0124\ \text{mm}\\
\delta_c &= \frac{75\times10^3\times250}{5674.5\times100\times10^3} = 0.0330\ \text{mm}\\
\delta_s &= \frac{100\times10^3\times100}{1963.5\times200\times10^3} = 0.0255\ \text{mm}\\
\delta_{total} &= 0.0709\ \text{mm}
\end{aligned}
$$

**Answer: $\sigma_{Al} = 2.89$, $\sigma_{Cu} = 13.22$, $\sigma_{St} = 50.93$ N/mm² (all tensile); total elongation = 0.0709 mm.**

## Check yourself

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

1. How is the axial force in each part found?
   A. Section method: sum of external loads on one side of the cut, starting from the free end
   B. Divide the total load equally among the three parts
   C. Use the same force 100 kN in all parts
   D. Add all the loads together
   Answer: A. Cutting the bar and balancing loads to one side gives 100, 75 and 25 kN in steel, copper and aluminium. (Key point: Axial force method)

2. What is the wall reaction on the bar?
   A. 100 kN to the right
   B. 25 kN pulling the bar to the left
   C. 75 kN to the left
   D. Zero
   Answer: B. 100 kN right balances 50 + 25 = 75 kN left plus a wall reaction of 25 kN left. (Key point: Wall reaction)

3. Which stress belongs to the copper part?
   A. 2.89 N/mm^2
   B. 50.93 N/mm^2
   C. 13.22 N/mm^2 (75 kN / 5674.5 mm^2)
   D. 13.22 N/mm^2 compressive
   Answer: C. sigma = N/A = 75000/5674.5 = 13.22 N/mm^2, tensile. (Key point: Stress formula)

4. How is the total change in length obtained?
   A. N L/(A E) of the steel part only
   B. Average of the three elongations
   C. Total load x total length / sum of areas
   D. Sum of N L/(A E) of the three parts
   Answer: D. Each part has its own N, A, E and L; their elongations add to 0.0709 mm. (Key point: Total elongation)

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