---
title: "Thin shell: stresses and dimension changes"
subject: "Strength of Materials"
chapter: 5
key: "semester_3/strength_of_materials:thin-shell-5m-1m-20mm-5nmm2"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/thin-shell-5m-1m-20mm-5nmm2
site: BCT Engineering
---

# Thin shell: stresses and dimension changes

Strength of Materials · Chapter 5: Thin Walled Vessels

- Asked in: 2076 Chaitra, 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/thin-shell-5m-1m-20mm-5nmm2
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-5

## Question

A thin cylindrical shell is 5m long and has 1m internal diameter and 20mm metal thickness. Calculate the maximum intensity of shear stress, longitudinal stress and circumferential stress induced, if subjected to an internal pressure of $5\ \text{N/mm}^2$. Also calculate change in diameter, length and volume of the shell. Take $E = 200\ \text{GPa}$ and poisons ratio = 0.3.

## Short answer

Stresses: circumferential 125 N/mm², longitudinal 62.5 N/mm², max shear 31.25 N/mm². Strains ε_c=5.3125×10⁻⁴, ε_l=1.25×10⁻⁴. Changes: diameter 0.531 mm, length 0.625 mm, volume 4.663×10⁶ mm³.

## Answer

**Data:** $d = 1000$ mm, $L = 5000$ mm, $t = 20$ mm, $p = 5\ \text{N/mm}^2$, $E = 200\ \text{GPa} = 2\times10^5\ \text{N/mm}^2$, $\nu = 0.3$.

#### Stresses
$$
\sigma_c = \frac{pd}{2t} = \frac{5(1000)}{2(20)} = 125.0\ \text{N/mm}^2, \qquad \sigma_l = \frac{pd}{4t} = 62.5\ \text{N/mm}^2
$$

$$
\tau_{max} = \frac{\sigma_c-\sigma_l}{2} = 31.25\ \text{N/mm}^2
$$

(maximum shear stress in the plane of the wall; it equals half the longitudinal stress).

#### Strains
$$
\varepsilon_c = \frac{125.0-0.3(62.5)}{2\times10^5} = 5.3125\times10^{-4}, \qquad \varepsilon_l = \frac{62.5-0.3(125.0)}{2\times10^5} = 1.2500\times10^{-4}
$$

$$
\varepsilon_v = 2\varepsilon_c + \varepsilon_l = 1.1875\times10^{-3}
$$

#### Changes in dimensions
$$
\delta d = \varepsilon_c d = 0.5312\ \text{mm}, \qquad \delta L = \varepsilon_l L = 0.6250\ \text{mm}
$$

$$
V = \frac{\pi}{4}(1000)^2(5000) = 3.9270\times10^{9}\ \text{mm}^3, \qquad \delta V = \varepsilon_v V = 4.6633\times10^{6}\ \text{mm}^3
$$

**Answer:** $\sigma_c = 125.0$, $\sigma_l = 62.5$, $\tau_{max} = 31.25$ N/mm²; $\delta d = 0.531$ mm, $\delta L = 0.625$ mm, $\delta V = 4.663\times10^6$ mm³ ($\approx 4663$ cm³).

## Check yourself

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

1. For d = 1000, t = 20, p = 5, what are sigma_c and sigma_l?
   A. 62.5 and 125 N/mm2
   B. 125 and 62.5 N/mm2
   C. 250 and 125 N/mm2
   D. 125 and 31.25 N/mm2
   Answer: B. pd/2t = 125; pd/4t = 62.5. (Key point: Stresses)

2. What is the maximum shear stress?
   A. 31.25 N/mm2, half the longitudinal stress
   B. 62.5 N/mm2
   C. 93.75 N/mm2
   D. 125 N/mm2
   Answer: A. (125 - 62.5)/2 = 31.25. (Key point: Maximum shear)

3. Which formula gives the change in length?
   A. delta L = e_c x L, 2.66 mm
   B. delta L = e_l x d, 0.125 mm
   C. delta L = e_l x L, e_l = (sigma_l - nu sigma_c)/E, 0.625 mm
   D. delta L = sigma_l L / E only, 1.56 mm
   Answer: C. e_l = (62.5 - 37.5)/2e5 = 1.25e-4, times 5000 = 0.625 mm. (Key point: Length change)

4. What is the change in volume?
   A. About 466 cm3
   B. About 2331 cm3
   C. About 46 630 cm3
   D. About 4663 cm3
   Answer: D. e_v = 2 x 5.3125e-4 + 1.25e-4 = 1.1875e-3 times 3.927e9 mm3. (Key point: Volume change)

## Related in Strength of Materials

- [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×
- [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
- [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
- [Hoop, longitudinal and max shear stress in thin cylinder](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Athin-cyl-hoop-longitudinal-shear-relations) — Strength of Materials, chapter 5
- [Longitudinal stress is half of circumferential stress](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Alongitudinal-stress-half-circumferential) — Strength of Materials, chapter 5, asked 2×
- [Steel spherical vessel: circumferential stress and diameter change](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Asteel-spherical-vessel-1000mm-1mpa) — Strength of Materials, chapter 5

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