---
title: "Steel spherical vessel: circumferential stress and diameter change"
subject: "Strength of Materials"
chapter: 5
key: "semester_3/strength_of_materials:steel-spherical-vessel-1000mm-1mpa"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/steel-spherical-vessel-1000mm-1mpa
site: BCT Engineering
---

# Steel spherical vessel: circumferential stress and diameter change

Strength of Materials · Chapter 5: Thin Walled Vessels

- Asked in: 2066 Chaitra (old course), 8 marks
- Video: [Understanding Pressure Vessels](https://www.youtube.com/watch?v=uaWfZTLNPmA) — The Efficient Engineer
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/steel-spherical-vessel-1000mm-1mpa
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-5

## Question

A steel spherical pressure vessel of radius 1000mm having wall thickness of 10mm is filled with a fluid and the internal pressure is raised to 1 MPa. Calculate circumferential stresses and change in diameter. Take $E = 200\ \text{GPa}$ and $\nu = 0.25$.

## Short answer

Circumferential stress σ = pd/(4t) = 1×2000/(4×10) = 50 MPa. Change in diameter δd = (σ/E)(1-ν)d = 50×0.75×2000/200000 = 0.375 mm (increase).

## Answer

**Data:** radius $r = 1000$ mm, so $d = 2000$ mm; $t = 10$ mm, $p = 1\ \text{MPa}$, $E = 200\ \text{GPa} = 2\times10^5\ \text{MPa}$, $\nu = 0.25$.

#### Stress
In a thin sphere the stress in the wall is the same in every direction (tangential stress). Balancing the pressure force on a diametral section, $p\frac{\pi}{4}d^2$, with the wall force $\sigma\pi d t$:

$$
\sigma = \frac{pd}{4t} = \frac{1(2000)}{4(10)} = 50.0\ \text{MPa}
$$

#### Change in diameter
$$
\varepsilon = \frac{\sigma}{E} - \nu\frac{\sigma}{E} = \frac{\sigma}{E}(1-\nu) = \frac{50.0(0.75)}{2\times10^5} = 1.875\times10^{-4}
$$

$$
\delta d = \varepsilon\,d = 1.875\times10^{-4}\times2000 = 0.3750\ \text{mm}
$$

**Answer:** circumferential stress $= 50.0$ MPa; change in diameter $= 0.375$ mm (increase).

## Check yourself

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

1. What is the stress formula for a thin sphere?
   A. sigma = pd/2t
   B. sigma = pd/4t
   C. sigma = pd/t
   D. sigma = pd/8t
   Answer: B. Balancing p pi d^2/4 against sigma pi d t gives pd/4t. (Key point: Sphere stress)

2. With radius 1000 mm, which d is used?
   A. 1000 mm
   B. 500 mm
   C. 2000 mm
   D. 4000 mm
   Answer: C. d is twice the radius. (Key point: Diameter use)

3. What is the strain in the sphere wall?
   A. sigma/E
   B. (sigma/E)(1 + nu)
   C. (sigma/E)(1 - 2 nu)
   D. (sigma/E)(1 - nu)
   Answer: D. Stress is equal in two perpendicular directions, so each strain is sigma/E - nu sigma/E. (Key point: Strain)

4. Change in diameter?
   A. 0.375 mm increase
   B. 0.5 mm
   C. 0.0375 mm
   D. 3.75 mm
   Answer: A. 1.875e-4 x 2000 = 0.375 mm. (Key point: Result)

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