---
title: "Thin copper shell: pressure from extra 50 cc liquid"
subject: "Strength of Materials"
chapter: 5
key: "semester_3/strength_of_materials:thin-copper-shell-extra-liquid-50cc"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/thin-copper-shell-extra-liquid-50cc
site: BCT Engineering
---

# Thin copper shell: pressure from extra 50 cc liquid

Strength of Materials · Chapter 5: Thin Walled Vessels

- Asked in: 2069 Chaitra, 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/thin-copper-shell-extra-liquid-50cc
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-5

## Question

A thin walled cylindrical shell made up of copper plate has been filled with a liquid at atmospheric pressure. An additional 50 c.c. of liquid is then pumped in to 2m cylinder whose internal diameter is 25 cm and wall thickness is 12 mm. Find the values of pressure developed on the wall of cylinder due to this extra liquid. Take poission ratio = 0.34 and modules of Elasticity = $10^6\ \text{kg/cm}^2$.

## Short answer

Cylinder volume 98175 cm³, volumetric strain 50/V. From ε_v = pd(5−4ν)/(4tE) with ν=0.34, E=10⁶ kg/cm², pressure p = 26.86 kg/cm² (2.63 MPa).

## Answer

The extra liquid occupies the additional space made by the expansion of the shell (the liquid is taken as incompressible), so $\delta V = 50\ \text{cm}^3$.

**Data:** $L = 200$ cm, $d = 25$ cm, $t = 1.2$ cm, $\nu = 0.34$, $E = 10^6\ \text{kg/cm}^2$.

#### Relation between pressure and volume change
For a thin cylinder, $\sigma_c = \dfrac{pd}{2t}$, $\sigma_l = \dfrac{pd}{4t}$, and

$$
\varepsilon_v = 2\varepsilon_c + \varepsilon_l = \frac{pd}{4tE}(5-4\nu)
$$

$$
V = \frac{\pi}{4}(25)^2(200) = 98174.77\ \text{cm}^3
$$

$$
\frac{\delta V}{V} = \frac{50}{98174.77} = 5.0930\times10^{-4} = \frac{p(25)}{4(1.2)(10^6)}(5-4\times0.34) = 1.8958\times10^{-5}\,p
$$

$$
p = 26.864\ \text{kg/cm}^2 \approx 2.634\ \text{N/mm}^2
$$

**Answer: pressure developed $\approx 26.86\ \text{kg/cm}^2$ (2.63 MPa).**

## Check yourself

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

1. What does the pumped extra 50 cc represent?
   A. Compression of the liquid
   B. The shell's original volume
   C. Decrease in length
   D. The volume increase of the shell
   Answer: D. The liquid is taken as incompressible, so it fills only the expansion space. (Key point: Extra liquid)

2. Which relation connects delta V / V to p?
   A. dV/V = pd(5 - 4 nu)/(4 t E)
   B. dV/V = pd/(2tE)
   C. dV/V = pd(5 - 4 nu)/(8tE)
   D. dV/V = 3pd/(4tE)
   Answer: A. It comes from eps_v = 2 eps_c + eps_l. (Key point: Formula)

3. What pressure results?
   A. About 2.69 kg/cm^2
   B. About 26.9 kg/cm^2 (2.63 MPa)
   C. About 269 kg/cm^2
   D. About 13.4 kg/cm^2
   Answer: B. 5.093e-4 = 1.8958e-5 p gives p = 26.86 kg/cm^2. (Key point: Result)

## Close variants set in other exams

- [Thin copper shell: pressure from extra 80 cc liquid](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Athin-copper-shell-extra-liquid-80cc) — Strength of Materials, chapter 5, asked 2×