---
title: "Cylindrical shell: volume increase, 6 N/mm2 pressure"
subject: "Strength of Materials"
chapter: 5
key: "semester_3/strength_of_materials:cyl-shell-4m-300mm-6nmm2-volume"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/cyl-shell-4m-300mm-6nmm2-volume
site: BCT Engineering
---

# Cylindrical shell: volume increase, 6 N/mm2 pressure

Strength of Materials · Chapter 5: Thin Walled Vessels

- Asked in: 2076 Asoj, 6 marks
- Video: [Problem on Thin Cylindrical Shells (Pressure Vessels), Mechanics of Solids,( SOM)](https://www.youtube.com/watch?v=fVGKa0Z5nDM) — Reference Book
- 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-4m-300mm-6nmm2-volume
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-5

## Question

A cylindrical shell of length 4 m internal diameter 300 mm and wall thickness of 12 mm is initially filled with water at atmospheric pressure. Find the increase in volume if the water is pumped to increase the internal pressure to $6\ \text{N/mm}^2$. Take $E = 2.10\times10^5\ \text{N/mm}^2$, $\nu = 0.3$ and $K = 2100\ \text{N/mm}^2$.

## Short answer

Shell expansion ΔV_shell=1.919×10⁵ mm³, water compression ΔV_w=8.078×10⁵ mm³. Additional water required = total 9.997×10⁵ mm³ ≈999.7 cm³.

## Answer

The extra water pumped in must fill two things: the increase in the volume of the shell (it expands) and the reduction in the volume of the water already inside (water is compressed).

**Data:** $d = 300$ mm, $L = 4000$ mm, $t = 12$ mm, $p = 6\ \text{N/mm}^2$, $E = 2.1\times10^5\ \text{N/mm}^2$, $\nu = 0.3$, $K = 2100\ \text{N/mm}^2$.

#### Increase in shell volume
$$
\sigma_c = \frac{pd}{2t} = 75.0\ \text{N/mm}^2, \qquad \sigma_l = \frac{pd}{4t} = 37.50\ \text{N/mm}^2
$$

$$
\varepsilon_v = \frac{pd}{4tE}(5-4\nu) = \frac{6(300)}{4(12)(2.1\times10^5)}(5-1.2) = 6.7857\times10^{-4}
$$

$$
V = \frac{\pi}{4}(300)^2(4000) = 2.8274\times10^{8}\ \text{mm}^3, \qquad \delta V_{shell} = 1.9186\times10^{5}\ \text{mm}^3
$$

#### Compression of the water
$$
K = \frac{p}{\delta V_w/V} \Rightarrow \delta V_w = \frac{pV}{K} = \frac{6\times2.8274\times10^{8}}{2100} = 8.0784\times10^{5}\ \text{mm}^3
$$

#### Additional water required
$$
\delta V = \delta V_{shell} + \delta V_w = 1.9186\times10^{5} + 8.0784\times10^{5} = 9.9970\times10^{5}\ \text{mm}^3
$$

**Answer: additional volume of water $\approx 999.7\ \text{cm}^3$ (of which 191.9 cm³ is the expansion of the shell).**

## Check yourself

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

1. Why is the extra water volume the sum of two terms?
   A. It fills the shell's expansion and makes up water's compression
   B. It fills only the shell expansion
   C. It fills only the water compression
   D. It fills the shell expansion minus the water compression
   Answer: A. The shell expands while the water compresses, both needing extra water. (Key point: Two contributions)

2. Which expression gives the shell volumetric strain?
   A. pd(5 - 4 nu)/(2tE)
   B. pd(5 - 4 nu)/(4tE) = 6.79e-4
   C. pd(1 - nu)/(4tE)
   D. pd(5 + 4 nu)/(4tE)
   Answer: B. 2e_c + e_l = pd/(4tE) x (5 - 4 nu). (Key point: Shell volumetric strain)

3. How is water's compression found?
   A. delta V_w = KV/p
   B. delta V_w = pK/V
   C. delta V_w = pV/K
   D. delta V_w = p/(KV)
   Answer: C. K = p / (delta V / V), so delta V = pV / K. (Key point: Water compression)

4. What is the additional water needed?
   A. About 999.7 cm3, mostly from water compression
   B. About 191.9 cm3
   C. About 807.8 cm3 from the shell
   D. About 9997 cm3
   Answer: A. 191.9 + 807.8 = 999.7 cm3. (Key point: Total)

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