---
title: "Thin cylindrical shell: max shear, dimension change"
subject: "Strength of Materials"
chapter: 5
key: "semester_3/strength_of_materials:thin-shell-shear-dimension-change"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/thin-shell-shear-dimension-change
site: BCT Engineering
---

# Thin cylindrical shell: max shear, dimension change

Strength of Materials · Chapter 5: Thin Walled Vessels

- Asked in: 2081 Bhadra, 6 marks; 2068 Baisakh (old course), 10 marks (asked 2 times)
- Video: [Mechanics of Materials: Lesson 41 - Thin Walled Pressure Vessel Basics](https://www.youtube.com/watch?v=0AbW_guBj5Y) — Jeff Hanson
- 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-shear-dimension-change
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-5

## Question

A thin cylindrical shell is 4 m long and has 1 m internal diameter and 12 mm metal thickness. Calculate the maximum intensity of shear produced and change in dimension of shell if it is subjected to an internal pressure of $2\ \text{N/mm}^2$. Take $E = 2\times10^5\ \text{N/mm}^2$ and $\nu = 0.3$.

## Short answer

Hoop stress 83.33, longitudinal 41.67 N/mm²; maximum shear τ_max=20.83 N/mm². Using Hooke's law, diameter increase 0.354 mm, length increase 0.333 mm, volume change 2487 cm³.

## Answer

**Data:** $d = 1000$ mm, $L = 4000$ mm, $t = 12$ mm, $p = 2\ \text{N/mm}^2$, $E = 2\times10^5\ \text{N/mm}^2$, $\nu = 0.3$.

#### Stresses
$$
\sigma_c = \frac{pd}{2t} = \frac{2(1000)}{2(12)} = 83.33\ \text{N/mm}^2, \qquad \sigma_l = \frac{pd}{4t} = 41.67\ \text{N/mm}^2
$$

#### Maximum shear stress
The principal stresses are $\sigma_c$ and $\sigma_l$ (the radial stress is neglected), so

$$
\tau_{max} = \frac{\sigma_c - \sigma_l}{2} = \frac{83.33 - 41.67}{2} = 20.83\ \text{N/mm}^2
$$

(this is the maximum in the plane of the wall; it equals $\sigma_l/2$ and $\sigma_c/4$).

#### Change in dimensions
$$
\varepsilon_c = \frac{\sigma_c - \nu\sigma_l}{E} = \frac{83.33 - 0.3(41.67)}{2\times10^5} = 3.542\times10^{-4}
$$

$$
\varepsilon_l = \frac{\sigma_l - \nu\sigma_c}{E} = \frac{41.67 - 0.3(83.33)}{2\times10^5} = 8.333\times10^{-5}
$$

$$
\delta d = \varepsilon_c\,d = 0.3542\ \text{mm}, \qquad \delta L = \varepsilon_l\,L = 0.3333\ \text{mm}
$$

Volumetric strain $\varepsilon_v = 2\varepsilon_c + \varepsilon_l = 7.917\times10^{-4}$, with $V = \frac{\pi}{4}d^2L = 3.1416\times10^{9}\ \text{mm}^3$:

$$
\delta V = \varepsilon_v V = 2.4871\times10^{6}\ \text{mm}^3 \;(2487.1\ \text{cm}^3)
$$

**Answer:** $\tau_{max} = 20.83$ N/mm²; $\delta d = 0.354$ mm (increase), $\delta L = 0.333$ mm (increase), $\delta V = 2487$ cm³.

## Check yourself

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

1. For d = 1000 mm, t = 12 mm, p = 2 N/mm2, what are the hoop and longitudinal stresses?
   A. 83.33 and 41.67 N/mm2
   B. 41.67 and 83.33 N/mm2
   C. 166.7 and 83.33 N/mm2
   D. 83.33 and 20.83 N/mm2
   Answer: A. sigma_c = pd/2t = 83.33; sigma_l = pd/4t = 41.67. (Key point: Stresses in the shell)

2. How is the maximum shear stress found?
   A. (sigma_c + sigma_l)/2 = 62.5 N/mm2
   B. (sigma_c - sigma_l)/2 = 20.83 N/mm2, with radial stress neglected
   C. sigma_c - sigma_l = 41.67 N/mm2
   D. sigma_c/2 = 41.67 N/mm2
   Answer: B. Principal stresses are the hoop and longitudinal; shear is half their difference. (Key point: Maximum shear)

3. Which formula gives the circumferential strain?
   A. (sigma_c + nu x sigma_l)/E
   B. (sigma_l - nu x sigma_c)/E
   C. (sigma_c - nu x sigma_l)/E
   D. sigma_c/E only
   Answer: C. Each strain includes the lateral effect of the other principal stress, with a negative sign. (Key point: Hoop strain)

4. What is the change in diameter?
   A. 0.333 mm, from epsilon_l x d
   B. 0.354 mm, from epsilon_c x L
   C. 3.54 mm, from epsilon_c x d
   D. 0.354 mm, from epsilon_c x d
   Answer: D. delta d = 3.5417e-4 x 1000 = 0.354 mm; delta L = 8.33e-5 x 4000 = 0.333 mm. (Key point: Change in diameter)

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