---
title: "Thin vessel with extra water: stresses"
subject: "Strength of Materials"
chapter: 5
key: "semester_3/strength_of_materials:thin-vessel-water-10nmm2-bulk-modulus"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/thin-vessel-water-10nmm2-bulk-modulus
site: BCT Engineering
---

# Thin vessel with extra water: stresses

Strength of Materials · Chapter 5: Thin Walled Vessels

- Asked in: 2066 Jestha (old course), 8 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-vessel-water-10nmm2-bulk-modulus
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-5

## Question

A thin walled cylindrical pressure vessel of 1m diameter and 8mm thick is filled with water at atmosphere pressure. Additional water is pumped and the internal pressure is raised to $10\ \text{N/mm}^2$. Find principal stresses, maximum shear stress in the wall of the vessel material, $E = 200{,}000\ \text{N/mm}^2$, $\nu = 0.3$. For water, $K = 2100\ \text{N/mm}^2$.

## Short answer

Hoop stress σ_c = 625.0 N/mm², longitudinal σ_l = 312.5 N/mm². Maximum shear stress τ_max = (625.0−312.5)/2 = 156.25 N/mm².

## Answer

**Data:** $d = 1000$ mm, $t = 8$ mm, $p = 10\ \text{N/mm}^2$ (the pressure raised above atmospheric), $E = 2\times10^5\ \text{N/mm}^2$, $\nu = 0.3$, $K_{water} = 2100\ \text{N/mm}^2$. (E, ν and K are needed only for volume change; the stresses depend only on $p$, $d$, $t$.)

#### Principal stresses
$$
\sigma_1 = \sigma_c = \frac{pd}{2t} = \frac{10(1000)}{2(8)} = 625.0\ \text{N/mm}^2, \qquad
\sigma_2 = \sigma_l = \frac{pd}{4t} = 312.5\ \text{N/mm}^2
$$

The radial stress (maximum $-p = -10$ N/mm² on the inner surface) is small compared with these and is neglected in thin-shell theory.

#### Maximum shear stress
$$
\tau_{max} = \frac{\sigma_1-\sigma_2}{2} = \frac{625.0-312.5}{2} = 156.25\ \text{N/mm}^2
$$

(in the plane of the wall, on planes at $45^\circ$ to the axis). If the radial stress is taken as zero, the absolute maximum shear stress, $\sigma_1/2 = 312.5$ N/mm², occurs on planes at $45^\circ$ to the wall surface.

**Answer:** $\sigma_1 = 625.0$ N/mm², $\sigma_2 = 312.5$ N/mm², $\tau_{max} = 156.25$ N/mm².

## Check yourself

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

1. What are the principal stresses at 10 N/mm^2?
   A. 312.5 and 156.25 N/mm^2
   B. 625 and 312.5 N/mm^2
   C. 1250 and 625 N/mm^2
   D. 625 and 10 N/mm^2
   Answer: B. pd/2t = 10 x 1000/16 = 625 and half of it is 312.5. (Key point: Principal stresses)

2. Are E, nu and K needed for the stresses?
   A. Yes, E is needed
   B. Yes, K is needed
   C. No; stresses depend only on p, d, t
   D. Yes, nu is needed
   Answer: C. They matter only for the volume change. (Key point: E, nu, K role)

3. What is the in-plane maximum shear stress?
   A. 312.5 N/mm^2
   B. 625 N/mm^2
   C. 468.75 N/mm^2
   D. 156.25 N/mm^2
   Answer: D. (625 - 312.5)/2 = 156.25. (Key point: Max shear)

4. What is the radial stress treatment?
   A. At most -p = -10 N/mm^2, neglected as small
   B. Equal to 625 N/mm^2
   C. Zero everywhere
   D. Equal to the longitudinal stress
   Answer: A. Radial stress is at most p at the inner surface, tiny relative to 625. (Key point: Radial stress)

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