---
title: "Compound tube: stresses and load share (1000 kN)"
subject: "Strength of Materials"
chapter: 3
key: "semester_3/strength_of_materials:compound-tube-steel-brass-1000kn"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/compound-tube-steel-brass-1000kn
site: BCT Engineering
---

# Compound tube: stresses and load share (1000 kN)

Strength of Materials · Chapter 3: Simple Stress and Strain

- Asked in: 2066 Chaitra (old course), 10 marks
- Video: [COMPOSITE/COMPOUND BARS](https://www.youtube.com/watch?v=eSLLf4fi6fk) — EASY SCIENCE PREP
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/compound-tube-steel-brass-1000kn
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-3

## Question

A compound tube consists of a steel tube 15cm internal diameter and 1cm thickness and an outer brass tube of 17cm internal diameter and 1cm thickness. Two tubes are of the same length. The compound tube carries an axial load of 1000 kN. Find the stresses and the load carried by each tube. Take $E_{steel} = 2\times10^7\ \text{N/cm}^2$ and $E_{brass} = 1\times10^7\ \text{N/cm}^2$

## Short answer

Equal strain and total load 1000 kN, with A_s=50.27 cm², A_b=56.55 cm², give σ_s = 2σ_b = 12732.4 N/cm² (127.32 MPa) and σ_b = 6366.2 N/cm² (63.66 MPa). Loads: steel 640.0 kN, brass 360.0 kN.

## Answer

Both tubes carry the load together and have the same length, so their strains are equal and $P_s + P_b = 1000$ kN $= 10^6$ N.

#### Areas
Steel: inner 15 cm, outer 17 cm. Brass: inner 17 cm, outer 19 cm.

$$
A_s = \frac{\pi}{4}(17^2 - 15^2) = 50.27\ \text{cm}^2, \qquad A_b = \frac{\pi}{4}(19^2 - 17^2) = 56.55\ \text{cm}^2
$$

#### Stresses
$$
\frac{\sigma_s}{E_s} = \frac{\sigma_b}{E_b} \Rightarrow \sigma_s = \frac{2\times10^7}{10^7}\sigma_b = 2\sigma_b
$$

$$
10^6 = \sigma_b(2A_s + A_b) = \sigma_b(2\times50.27 + 56.55) \Rightarrow \sigma_b = 6366.2\ \text{N/cm}^2
$$

$$
\sigma_s = 12732.4\ \text{N/cm}^2
$$

#### Loads
$$
P_s = \sigma_sA_s = 640.00\ \text{kN}, \qquad P_b = \sigma_bA_b = 360.00\ \text{kN}
$$

Check: $640.00 + 360.00 = 1000.00$ kN.

**Answer:** steel: $12732.4\ \text{N/cm}^2$ ($127.32$ MPa), load $640.0$ kN; brass: $6366.2\ \text{N/cm}^2$ ($63.66$ MPa), load $360.0$ kN.

## Check yourself

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

1. What are the steel and brass areas' diameters?
   A. Steel 15-16 cm, brass 17-18 cm
   B. Steel 17-19 cm, brass 15-17 cm
   C. Steel 15-17 cm, brass 17-19 cm
   D. Steel 13-15 cm, brass 15-17 cm
   Answer: C. Thickness 1 cm adds 2 cm to the diameter: steel outer 17 cm, brass inner 17 cm and outer 19 cm. (Key point: Outer diameters)

2. Why is sigma_s = 2 sigma_b?
   A. Strains are equal and E_s = 2 E_b
   B. Steel area is twice the brass area
   C. The loads are equal
   D. Brass is stiffer than steel
   Answer: A. sigma_s/E_s = sigma_b/E_b with E_s/E_b = 2. (Key point: Stress relation)

3. Which equation finds sigma_b?
   A. 10^6 = sigma_b (A_s + A_b)
   B. 10^6 = sigma_b (A_s - A_b)
   C. 10^6 = sigma_b (2 A_s + A_b)
   D. 10^6 = sigma_b A_b / 2
   Answer: C. Total load = sigma_s A_s + sigma_b A_b = sigma_b(2A_s + A_b). (Key point: Equilibrium equation)

4. What loads are carried?
   A. Steel 640 kN, brass 360 kN
   B. Steel 360 kN, brass 640 kN
   C. Steel 500 kN, brass 500 kN
   D. Steel 727 kN, brass 273 kN
   Answer: A. sigma_s A_s = 12732.4 x 50.27 = 640 kN and sigma_b A_b = 6366.2 x 56.55 = 360 kN. (Key point: Results)

## Close variants set in other exams

- [Compound tube: stresses, load share and shortening (750 kN)](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Acompound-tube-steel-brass-750kn) — Strength of Materials, chapter 3, asked 2×

## Related in Strength of Materials

- [Steel rod with two brass rods: stresses](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Asteel-brass-rods-60kn) — Strength of Materials, chapter 3