---
title: "Max shear stress and volumetric strain in thin cylinder"
subject: "Strength of Materials"
chapter: 5
key: "semester_3/strength_of_materials:max-shear-half-longitudinal-volumetric-strain"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/max-shear-half-longitudinal-volumetric-strain
site: BCT Engineering
---

# Max shear stress and volumetric strain in thin cylinder

Strength of Materials · Chapter 5: Thin Walled Vessels

- Asked in: 2068 Chaitra, 2+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/max-shear-half-longitudinal-volumetric-strain
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-5

## Question

Prove that maximum shear stress in a thin cylinder is half of the longitudinal stress. Also derive an expression for volumetric strain for thin cylinder.

## Short answer

Principal stresses: hoop σ_c = pd/(2t), longitudinal σ_l = pd/(4t). Max shear τ_max = (σ_c−σ_l)/2 = pd/(8t) = σ_l/2. Volumetric strain ε_v = 2ε_c+ε_l = pd(5−4ν)/(4tE).

## Answer

#### Maximum shear stress is half of the longitudinal stress
In a thin cylinder of diameter $d$, thickness $t$ and pressure $p$, the two principal stresses are the hoop stress and the longitudinal stress (the radial stress is neglected):

$$
\sigma_1 = \sigma_c = \frac{pd}{2t}, \qquad \sigma_2 = \sigma_l = \frac{pd}{4t}
$$

The maximum shear stress (in the plane of the wall) is half the difference of the principal stresses:

$$
\tau_{max} = \frac{\sigma_1-\sigma_2}{2} = \frac{1}{2}\left(\frac{pd}{2t} - \frac{pd}{4t}\right) = \frac{pd}{8t}
$$

Since $\sigma_l = \dfrac{pd}{4t}$:

$$
\tau_{max} = \frac{pd}{8t} = \frac{1}{2}\,\sigma_l
$$

Hence the maximum shear stress is half of the longitudinal stress (and one quarter of the hoop stress). It acts on planes at $45^\circ$ to the axis.

#### Volumetric strain
Volume $V = \dfrac{\pi}{4}d^2L$. Taking the differential,

$$
\frac{\delta V}{V} = 2\frac{\delta d}{d} + \frac{\delta L}{L} = 2\varepsilon_c + \varepsilon_l
$$

The strains (including Poisson's effect) are

$$
\varepsilon_c = \frac{\sigma_c}{E} - \nu\frac{\sigma_l}{E} = \frac{pd}{2tE}\left(1-\frac{\nu}{2}\right), \qquad \varepsilon_l = \frac{\sigma_l}{E} - \nu\frac{\sigma_c}{E} = \frac{pd}{2tE}\left(\frac{1}{2}-\nu\right)
$$

$$
\varepsilon_v = \frac{pd}{2tE}\left[2\left(1-\frac{\nu}{2}\right) + \frac{1}{2} - \nu\right] = \frac{pd}{2tE}\left(\frac{5}{2}-2\nu\right) = \frac{pd}{4tE}\left(5-4\nu\right)
$$

so the change in volume is $\delta V = \dfrac{pd}{4tE}(5-4\nu)\,V$.

## Check yourself

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

1. Which two stresses are principal in a thin cylinder?
   A. Hoop and radial
   B. Longitudinal and shear
   C. Hoop pd/2t and longitudinal pd/4t
   D. Radial and shear
   Answer: C. Radial stress is neglected, leaving hoop and longitudinal. (Key point: Principal stresses)

2. Why is tau_max equal to half of sigma_l?
   A. tau_max = sigma_c/2 = sigma_l
   B. tau_max = sigma_c + sigma_l
   C. tau_max = sigma_l
   D. tau_max = (pd/2t - pd/4t)/2 = pd/8t = sigma_l/2
   Answer: D. Half the difference of principal stresses is pd/8t, which is half of pd/4t. (Key point: Proof)

3. How is volumetric strain written in terms of linear strains?
   A. eps_v = 2 eps_c + eps_l
   B. eps_v = eps_c + 2 eps_l
   C. eps_v = eps_c + eps_l
   D. eps_v = 3 eps_c
   Answer: A. From V = (pi/4) d^2 L, dV/V = 2 dd/d + dL/L. (Key point: Volumetric strain)

4. What is the final volumetric strain?
   A. pd(5 - 2 nu)/(4tE)
   B. pd(5 - 4 nu)/(4tE)
   C. pd(3 - 4 nu)/(4tE)
   D. pd(1 - nu)/(2tE)
   Answer: B. eps_v = pd/(2tE)[2(1 - nu/2) + 1/2 - nu] = pd/(2tE)(5/2 - 2nu). (Key point: Final result)

## Related in Strength of Materials

- [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
- [Volumetric strain is three times longitudinal strain (cube)](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Avolumetric-strain-three-times-longitudinal) — Strength of Materials, chapter 3
- [Volumetric strain of thin cylindrical vessel](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Avolumetric-strain-thin-cylinder-derive) — Strength of Materials, chapter 5, asked 2×
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