---
title: "Cylindrical shell with spherical ends: volume change"
subject: "Strength of Materials"
chapter: 5
key: "semester_3/strength_of_materials:cyl-shell-spherical-ends-volume-change"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/cyl-shell-spherical-ends-volume-change
site: BCT Engineering
---

# Cylindrical shell with spherical ends: volume change

Strength of Materials · Chapter 5: Thin Walled Vessels

- Asked in: 2080 Bhadra, 6 marks
- Video: [Thin-Walled PRESSURE VESSELS in 8 MINUTES - Mechanics of Materials](https://www.youtube.com/watch?v=ic8LmMUr8js) — Less Boring Lectures
- 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-shell-spherical-ends-volume-change
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-5

## Question

The cylindrical shell is 1 m long and has diameter 200 mm. The metal thickness is 6 mm, it has spherical ends. Calculate the change in volume, if the internal pressure is raised to $1.5\ \text{N/mm}^2$. Take $E = 2\times10^5\ \text{N/mm}^2$ and Poisson's ratio is 0.25.

## Short answer

Cylinder hoop stress 25 N/mm², longitudinal 12.5 N/mm², volumetric strain 2.5×10⁻⁴ gives ΔV_cyl=7854 mm³. Sphere stress 12.5 N/mm², strain 4.6875×10⁻⁵ gives ΔV_s=589 mm³. Total volume increase ≈8.44 cm³.

## Answer

**Assumption:** the 1 m length is the cylindrical portion; the two spherical (hemispherical) ends have the same diameter, 200 mm, and thickness, 6 mm. Total change in volume = change in the cylindrical part + change in the two ends (which together make one full sphere).

**Data:** $d = 200$ mm, $t = 6$ mm, $L = 1000$ mm, $p = 1.5\ \text{N/mm}^2$, $E = 2\times10^5\ \text{N/mm}^2$, $\nu = 0.25$.

#### Cylindrical part
$$
\sigma_c = \frac{pd}{2t} = 25.00\ \text{N/mm}^2, \qquad \sigma_l = \frac{pd}{4t} = 12.50\ \text{N/mm}^2
$$

$$
\varepsilon_c = \frac{25.00 - 0.25(12.50)}{2\times10^5} = 1.0938\times10^{-4}, \qquad \varepsilon_l = \frac{12.50 - 0.25(25.00)}{2\times10^5} = 3.1250\times10^{-5}
$$

$$
\varepsilon_v = 2\varepsilon_c + \varepsilon_l = 2.5000\times10^{-4}, \quad V_{cyl} = \frac{\pi}{4}(200)^2(1000) = 3.1416\times10^{7}\ \text{mm}^3
$$

$$
\delta V_{cyl} = 2.5000\times10^{-4}\times3.1416\times10^{7} = 7854.0\ \text{mm}^3
$$

#### Spherical ends (one complete sphere)
$$
\sigma = \frac{pd}{4t} = 12.50\ \text{N/mm}^2, \qquad \varepsilon = \frac{\sigma}{E}(1-\nu) = 4.6875\times10^{-5}
$$

$$
V_s = \frac{\pi d^3}{6} = 4.1888\times10^{6}\ \text{mm}^3, \qquad \delta V_s = 3\varepsilon V_s = 589.0\ \text{mm}^3
$$

#### Total
$$
\delta V = 7854.0 + 589.0 = 8443.0\ \text{mm}^3 = 8.44\ \text{cm}^3
$$

**Answer: increase in volume $\approx 8.44\ \text{cm}^3$.**

## Check yourself

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

1. How is the total change in volume found for a cylinder with spherical ends?
   A. Cylindrical part change plus change of two ends together as one sphere
   B. Cylinder alone, ignoring the ends
   C. Only the change of the spherical ends
   D. The cylinder change minus the sphere change
   Answer: A. The ends are two hemispheres which together make a complete sphere of the same diameter and thickness. (Key point: Total volume change composition)

2. How is the cylinder's volumetric strain obtained?
   A. e_c + 2 e_l
   B. 2 e_c + e_l = 2.5e-4
   C. 3 e_c
   D. e_c + e_l
   Answer: B. Hoop strain is counted twice because volume depends on d^2. (Key point: Cylinder volumetric strain)

3. What is the spherical stress and strain used?
   A. sigma = pd/2t = 25; strain = sigma/E
   B. sigma = pd/4t = 12.5; strain = sigma(1 + nu)/E
   C. sigma = pd/4t = 12.5; strain = sigma(1 - nu)/E; delta V = 3 strain x V_sphere
   D. sigma = pd/2t = 25; delta V = 2 strain x V
   Answer: C. Sphere stress is pd/4t in all directions, giving delta V = 589 mm3. (Key point: Spherical ends)

4. What is the final increase in volume?
   A. About 8.44 cm3
   B. About 7.85 cm3
   C. About 0.59 cm3
   D. About 84.4 cm3
   Answer: A. 7854 + 589 = 8443 mm3. (Key point: Total)

## Related in Strength of Materials

- [Thin cylindrical shell: change in length and diameter](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Athin-cyl-shell-4m-25nmm2) — Strength of Materials, chapter 5
- [Cylindrical vessel 3 m long: change in volume](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Acyl-vessel-3m-volume-change) — Strength of Materials, chapter 5
- [Cylindrical shell: stresses, max shear and volume change](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Acyl-shell-4m-1m-2nmm2-8mm) — Strength of Materials, chapter 5
- [Cylindrical shell with joint efficiencies: change in dimensions](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Acyl-shell-joint-efficiency) — Strength of Materials, chapter 5
- [Cylindrical shell: volume increase, 6 N/mm2 pressure](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Acyl-shell-4m-300mm-6nmm2-volume) — Strength of Materials, chapter 5
- [Thin cylindrical shell: max shear, dimension change](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Athin-shell-shear-dimension-change) — Strength of Materials, chapter 5, asked 2×

## Related in other subjects

- [Diameter to thickness ratio for stress limit](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Apipe-diameter-thickness-ratio-hoop-stress) — Hydraulics, chapter 4
- [Economic diameter; penstock thickness (1.25 m)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Apenstock-economic-diameter-thickness-1-25m) — Hydropower Engineering, chapter 6
- [Pressure change from displacement thickness in duct](https://www.bctengineer.com/question.md?key=semester_3%2Ffluid_mechanics%3Aduct-displacement-thickness-pressure) — Fluid Mechanics, chapter 8
- [Economic diameter factors; penstock wall thickness (3 m)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Aeconomic-diameter-and-wall-thickness-factors) — Hydropower Engineering, chapter 6