---
title: "Cylindrical vessel 3 m long: change in volume"
subject: "Strength of Materials"
chapter: 5
key: "semester_3/strength_of_materials:cyl-vessel-3m-volume-change"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/cyl-vessel-3m-volume-change
site: BCT Engineering
---

# Cylindrical vessel 3 m long: change in volume

Strength of Materials · Chapter 5: Thin Walled Vessels

- Asked in: 2079 Bhadra, 4 marks
- Video: [Change in volume of cylindrical vessel due to internal pressure](https://www.youtube.com/watch?v=7ieNTRolAwE) — Pavika fluid says Tawi
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/cyl-vessel-3m-volume-change
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-5

## Question

A cylindrical vessel 3 m long and 600 mm diameter with 10 mm thick plates is subjected to an internal pressure of 3 MPa. Calculate the change in volume of the vessel. Take $E = 200\ \text{GPa}$ and Poisson's ratio = 0.3 for the vessel material.

## Short answer

Circumferential stress 90 MPa, longitudinal 45 MPa. Volumetric strain 8.55×10⁻⁴, initial volume 8.4823×10⁸ mm³. Change in volume ≈7.252×10⁵ mm³ or 725.2 cm³ (increase).

## Answer

**Data:** $d = 600$ mm, $L = 3000$ mm, $t = 10$ mm, $p = 3\ \text{MPa}$, $E = 200\ \text{GPa} = 2\times10^5\ \text{MPa}$, $\nu = 0.3$.

#### Stresses
$$
\sigma_c = \frac{pd}{2t} = \frac{3(600)}{2(10)} = 90.0\ \text{MPa}, \qquad \sigma_l = \frac{pd}{4t} = 45.0\ \text{MPa}
$$

#### Strains
$$
\varepsilon_c = \frac{\sigma_c-\nu\sigma_l}{E} = \frac{90.0 - 0.3(45.0)}{2\times10^5} = 3.825\times10^{-4}, \qquad
\varepsilon_l = \frac{\sigma_l-\nu\sigma_c}{E} = 9.000\times10^{-5}
$$

$$
\varepsilon_v = 2\varepsilon_c + \varepsilon_l = 8.550\times10^{-4}
$$

#### Change in volume
$$
V = \frac{\pi}{4}(600)^2(3000) = 8.4823\times10^{8}\ \text{mm}^3, \qquad \delta V = \varepsilon_vV = 7.2524\times10^{5}\ \text{mm}^3
$$

**Answer: $\delta V \approx 725.2\ \text{cm}^3$ (725,237 mm³, an increase).**

## Check yourself

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

1. For d = 600 mm, t = 10 mm, p = 3 MPa, what are the hoop and longitudinal stresses?
   A. 90 and 45 MPa
   B. 45 and 90 MPa
   C. 180 and 90 MPa
   D. 90 and 22.5 MPa
   Answer: A. pd/2t = 90 and pd/4t = 45. (Key point: Stresses)

2. How is volumetric strain obtained?
   A. e_c + 2 e_l = 5.9e-4
   B. 2 e_c + e_l = 8.55e-4
   C. e_c + e_l = 4.7e-4
   D. 3 e_c = 1.15e-3
   Answer: B. e_c = 3.825e-4, e_l = 9e-5. (Key point: Volumetric strain)

3. What is the volume change?
   A. About 7.25 cm3 increase
   B. About 72 500 cm3 increase
   C. About 725 cm3 increase
   D. About 725 cm3 decrease
   Answer: C. 8.55e-4 x 8.4823e8 mm3 = 7.25e5 mm3. (Key point: Change in volume)

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