---
title: "Hoop, longitudinal and max shear stress in thin cylinder"
subject: "Strength of Materials"
chapter: 5
key: "semester_3/strength_of_materials:thin-cyl-hoop-longitudinal-shear-relations"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/thin-cyl-hoop-longitudinal-shear-relations
site: BCT Engineering
---

# Hoop, longitudinal and max shear stress in thin cylinder

Strength of Materials · Chapter 5: Thin Walled Vessels

- Asked in: 2067 Asar (old course), 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-cyl-hoop-longitudinal-shear-relations
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-5

## Question

For thin cylinders loaded with internal pressure p, obtain the relation for Hoop stress, longitudinal stress and maximum shear stress.

## Short answer

Hoop stress from axial cut: σ_c = pd/(2t). Longitudinal stress from transverse cut: σ_l = pd/(4t). Maximum shear stress τ_max = (σ_c−σ_l)/2 = pd/(8t) = σ_l/2.

## Answer

For a thin cylinder (internal diameter $d$, length $L$, thickness $t$, internal pressure $p$, $t < d/20$) the radial stress is negligible and the wall is in a state of biaxial tension.

#### Hoop (circumferential) stress
Cut the cylinder by a plane along its axis. The pressure acts on the projected area $d\times L$; two wall sections of area $t\times L$ resist it.

$$
p\,d\,L = 2\,\sigma_c\,t\,L \quad\Rightarrow\quad \sigma_c = \frac{pd}{2t}
$$

#### Longitudinal stress
Cut across the cylinder. The pressure on the end area $\frac{\pi}{4}d^2$ is resisted by the ring area $\pi d\,t$.

$$
p\,\frac{\pi}{4}d^2 = \sigma_l\,\pi d\,t \quad\Rightarrow\quad \sigma_l = \frac{pd}{4t}
$$

#### Maximum shear stress
$\sigma_c$ and $\sigma_l$ are principal stresses (no shear on these planes), so

$$
\tau_{max} = \frac{\sigma_c-\sigma_l}{2} = \frac{pd}{8t}
$$

#### Summary
| Quantity | Expression | Relation |
|---|---|---|
| Hoop stress | $\dfrac{pd}{2t}$ | $2\sigma_l$ |
| Longitudinal stress | $\dfrac{pd}{4t}$ | $\tfrac12\sigma_c$ |
| Max shear stress | $\dfrac{pd}{8t}$ | $\tfrac12\sigma_l$, $\tfrac14\sigma_c$ |

## Check yourself

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

1. In the hoop derivation, what force does the pressure give on a cut along the axis?
   A. p pi d^2/4, resisted by 2 sigma_c t
   B. p d L, resisted by sigma_c t L
   C. p d L, resisted by 2 sigma_c t L
   D. p d t, resisted by sigma_c d L
   Answer: C. Pressure acts on the projected area d x L; two wall sections t x L resist. (Key point: Hoop derivation)

2. What is the area resisting the end force?
   A. Area d x L
   B. Area pi d^2/4
   C. Area 2 t L
   D. Ring area pi d t
   Answer: D. The cut across the cylinder shows a ring of circumference pi d and thickness t. (Key point: Longitudinal derivation)

3. Which relation among the stresses holds?
   A. sigma_c = 2 sigma_l and tau_max = sigma_l/2
   B. sigma_c = sigma_l and tau_max = sigma_c
   C. sigma_l = 2 sigma_c and tau_max = sigma_c/2
   D. tau_max = sigma_c + sigma_l
   Answer: A. pd/2t, pd/4t, pd/8t give those ratios. (Key point: Relations)

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