---
title: "Thin cylindrical shell: change in length and diameter"
subject: "Strength of Materials"
chapter: 5
key: "semester_3/strength_of_materials:thin-cyl-shell-4m-25nmm2"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/thin-cyl-shell-4m-25nmm2
site: BCT Engineering
---

# Thin cylindrical shell: change in length and diameter

Strength of Materials · Chapter 5: Thin Walled Vessels

- Asked in: 2069 Asar, 6 marks
- Video: [THIN CYLINDER | DERIVATION | HOOP OR CIRCUMFERENTIAL STRAIN | LONGITUDINAL STRAIN | CHANGE IN VOLUME](https://www.youtube.com/watch?v=j3MuFDxRsMo) — Structural Funda by:- KheNiSiR
- 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-cyl-shell-4m-25nmm2
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-5

## Question

A thin cylindrical shell 4m long and thickness 1.5cm is of 1.5cm internal diameter. Calculate the change in length and diameter if the shell is subjected to an internal pressure of $25\ \text{N/mm}^2$. $E = 2.05\times10^5\ \text{N/mm}^2$ and poisson's ratio = 0.3.

## Short answer

Hoop stress 12.5 N/mm², longitudinal 6.25 N/mm². Hoop strain 5.18×10⁻⁵ gives δd = 0.00078 mm; longitudinal strain 1.22×10⁻⁵ gives δL = 0.049 mm (both increases).

## Answer

**Data (as given):** $L = 4000$ mm, $t = 15$ mm, $d = 15$ mm, $p = 25\ \text{N/mm}^2$, $E = 2.05\times10^5\ \text{N/mm}^2$, $\nu = 0.3$.

#### Stresses
$$
\sigma_c = \frac{pd}{2t} = \frac{25(15)}{2(15)} = 12.50\ \text{N/mm}^2, \qquad \sigma_l = \frac{pd}{4t} = 6.25\ \text{N/mm}^2
$$

#### Strains
$$
\varepsilon_c = \frac{\sigma_c-\nu\sigma_l}{E} = \frac{12.50-0.3(6.25)}{2.05\times10^5} = 5.1829\times10^{-5}
$$

$$
\varepsilon_l = \frac{\sigma_l-\nu\sigma_c}{E} = \frac{6.25-0.3(12.50)}{2.05\times10^5} = 1.2195\times10^{-5}
$$

#### Changes
$$
\delta d = \varepsilon_c d = 0.00078\ \text{mm}, \qquad \delta L = \varepsilon_l L = 0.0488\ \text{mm}
$$

**Answer (data as given): $\delta L \approx 0.049\ \text{mm}$ and $\delta d \approx 0.0008\ \text{mm}$ (both increases).**

*Note:* with $t = d$ the shell is not really thin, so the thin-shell formulas are only approximate here. If the internal diameter was meant to be 150 mm, the same method gives $\sigma_c = 125.0$, $\sigma_l = 62.5\ \text{N/mm}^2$, $\delta d = 0.078$ mm and $\delta L = 0.488$ mm.

## Check yourself

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

1. With d = t = 15 mm and p = 25 N/mm^2, what is the hoop stress?
   A. 25 N/mm^2
   B. 6.25 N/mm^2
   C. 50 N/mm^2
   D. 12.5 N/mm^2
   Answer: D. sigma_c = pd/2t = 25 x 15/30 = 12.5 N/mm^2. (Key point: Stresses)

2. How is the longitudinal strain found?
   A. (sigma_l - nu sigma_c)/E
   B. sigma_l/E only
   C. (sigma_l + nu sigma_c)/E
   D. (sigma_c - nu sigma_l)/E
   Answer: A. Hoop stress causes lateral contraction along the axis, reducing longitudinal strain. (Key point: Strain formula)

3. What change in length results?
   A. About 0.00078 mm
   B. About 0.049 mm increase
   C. About 0.49 mm
   D. About 0.012 mm
   Answer: B. eps_l = 1.22e-5 times 4000 mm = 0.0488 mm. (Key point: Change in length)

4. Why is the result only approximate?
   A. E is too small
   B. Poisson ratio is too large
   C. With t = d the shell is not thin (needs t < d/20)
   D. Pressure is too low
   Answer: C. Thin-shell formulas assume t much less than d. (Key point: Thin-shell validity)

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