---
title: "Spherical vessel 1.9 m, volume increase 400 cm3"
subject: "Strength of Materials"
chapter: 5
key: "semester_3/strength_of_materials:spherical-vessel-1-9m-400cm3"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/spherical-vessel-1-9m-400cm3
site: BCT Engineering
---

# Spherical vessel 1.9 m, volume increase 400 cm3

Strength of Materials · Chapter 5: Thin Walled Vessels

- Asked in: 2078 Bhadra, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/spherical-vessel-1-9m-400cm3
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-5

## Question

A seamless spherical vessel of 1.9 m internal diameter and 6 mm thick is filled with a fluid under pressure until its volume increases by $400\ \text{cm}^3$. Calculate the pressure exerted by the fluid in the vessel. Take $E = 2\times10^5\ \text{N/mm}^2$ and Poisson's ratio = 0.25.

## Short answer

Using ΔV/V=3pd(1-ν)/(4tE) with V=3.5914×10⁹ mm³, ΔV=4×10⁵ mm³ yields pressure p=0.125 N/mm².

## Answer

**Data:** $d = 1900$ mm, $t = 6$ mm, $\delta V = 400\ \text{cm}^3 = 4\times10^5\ \text{mm}^3$, $E = 2\times10^5\ \text{N/mm}^2$, $\nu = 0.25$.

For a thin sphere, $\sigma = \dfrac{pd}{4t}$ in all directions, the linear strain is $\varepsilon = \dfrac{\sigma}{E}(1-\nu)$ and

$$
\frac{\delta V}{V} = 3\varepsilon = \frac{3pd}{4tE}(1-\nu)
$$

#### Calculation
$$
V = \frac{\pi d^3}{6} = \frac{\pi(1900)^3}{6} = 3.5914\times10^{9}\ \text{mm}^3
$$

$$
\frac{4\times10^5}{3.5914\times10^{9}} = 1.1138\times10^{-4} = \frac{3p(1900)(0.75)}{4(6)(2\times10^5)} = 8.9063\times10^{-4}\,p
$$

$$
p = 0.1251\ \text{N/mm}^2
$$

**Answer: $p \approx 0.125\ \text{N/mm}^2$.**

## Check yourself

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

1. Which formula gives the volume used?
   A. V = pi d^2 / 4
   B. V = pi d^3 / 6
   C. V = 4 pi d^3 / 3
   D. V = pi d^3 / 12
   Answer: B. A sphere of diameter d has V = pi d^3 / 6. (Key point: Volume of sphere)

2. Which relation links pressure and volume change?
   A. dV/V = pd(1 - nu)/(4tE)
   B. dV/V = 3pd(1 + nu)/(4tE)
   C. dV/V = 3pd(1 - nu)/(4tE)
   D. dV/V = 3pd(1 - nu)/(2tE)
   Answer: C. Three equal linear strains of pd(1 - nu)/(4tE). (Key point: Volumetric strain relation)

3. What pressure is found?
   A. About 0.125 N/mm2
   B. About 1.25 N/mm2
   C. About 0.0125 N/mm2
   D. About 12.5 N/mm2
   Answer: A. 1.1138e-4 = 8.9063e-4 p gives p = 0.1251. (Key point: Result)

## Close variants set in other exams

- [Spherical vessel, volume increase 300 cm3, find pressure](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aspherical-vessel-300cm3) — Strength of Materials, chapter 5, asked 2×

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