---
title: "Longitudinal stress is half of circumferential stress"
subject: "Strength of Materials"
chapter: 5
key: "semester_3/strength_of_materials:longitudinal-stress-half-circumferential"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/longitudinal-stress-half-circumferential
site: BCT Engineering
---

# Longitudinal stress is half of circumferential stress

Strength of Materials · Chapter 5: Thin Walled Vessels

- Asked in: 2074 Asoj, 4 marks; 2072 Chaitra, 8 marks (asked 2 times)
- Video: [PRESSURE VESSEL(THIN CYLINDER) - HOOP STRESS/ CIRCUMFERENTIAL STRESS](https://www.youtube.com/watch?v=UEmgT1JhMYs) — TECHNICAL CLASSES
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/longitudinal-stress-half-circumferential
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-5

## Question

Prove that longitudinal stress is half of the circumferential stress for the thin cylinder with neat sketch.

## Short answer

From transverse force balance, p(πd²/4)=σ_l(πdt) → σ_l=pd/(4t). From longitudinal cut, σ_c=pd/(2t). Thus σ_l=σ_c/2, longitudinal stress is half the circumferential stress.

## Answer

A thin cylinder of internal diameter $d$, wall thickness $t$ and internal pressure $p$ is subjected to two stresses: circumferential (hoop) stress $\sigma_c$ acting on longitudinal sections, and longitudinal stress $\sigma_l$ acting on circular cross-sections. We show that $\sigma_l = \tfrac12\sigma_c$.

```text
   Longitudinal section        Transverse section
   (hoop stress)               (longitudinal stress)

     ______________             _________
    |<----- d ---->|           /  p  p   \
  sc|   p  p  p    |sc        | p  p  p   |-> sl
    |______________|           \_________/
       length L                 area = pi*d*t
```

#### Circumferential stress (cut along a diameter)
Cut the cylinder by a longitudinal plane through its axis. The pressure on the projected area $d\times L$ is balanced by the tension in the two walls:

$$
\text{Bursting force} = p\times d\times L, \qquad \text{Resisting force} = 2\times\sigma_c\times L\times t
$$

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

#### Longitudinal stress (cut across the cylinder)
Cut the cylinder by a transverse plane. The pressure acts on the end area $\frac{\pi}{4}d^2$. This force is resisted by the metal ring of area $\pi d t$ (thin wall):

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

#### Relation
$$
\frac{\sigma_l}{\sigma_c} = \frac{pd/4t}{pd/2t} = \frac{1}{2} \quad\Rightarrow\quad \sigma_l = \frac{\sigma_c}{2}
$$

So the longitudinal stress is half of the circumferential stress. This is why a boiler shell fails along a longitudinal seam first, and why the longitudinal joint needs more strength than the circumferential joint.

## Check yourself

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

1. Cutting along a diameter plane, which balance gives the hoop stress?
   A. p x d x L = 2 sigma_c L t
   B. p x d x L = sigma_c L t
   C. p x (pi/4)d^2 = sigma_c pi d t
   D. p x d = sigma_c x t
   Answer: A. Bursting force on the projected area d x L is resisted by two wall sections each of area L x t. (Key point: Hoop stress equilibrium)

2. Cutting across the cylinder, what resists the pressure force on the end area?
   A. Two wall strips of area L t, giving pd/2t
   B. The end area pi d^2/4, giving pd/t
   C. The metal ring of area pi d t, giving sigma_l = pd/4t
   D. The ring of area d t, giving pd/2t
   Answer: C. p (pi/4) d^2 = sigma_l pi d t gives pd/4t. (Key point: Longitudinal stress equilibrium)

3. What follows from sigma_l = sigma_c / 2?
   A. The circumferential joint must be stronger
   B. A shell fails along a longitudinal seam first, so that seam needs more strength
   C. The shell fails at the end caps first
   D. Both seams fail together at equal pressure
   Answer: B. Hoop stress is double, so the longitudinal seam carries the larger stress. (Key point: Practical consequence)

## 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
- [Max shear stress and volumetric strain in thin cylinder](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Amax-shear-half-longitudinal-volumetric-strain) — Strength of Materials, chapter 5
- [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
- [Thin shell: stresses and dimension changes](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Athin-shell-5m-1m-20mm-5nmm2) — Strength of Materials, chapter 5
- [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×