---
title: "Thick cylinder stresses and failure criteria"
subject: "Mechanics of Solids"
chapter: 8
key: "semester_5/mechanics_of_solids:thick-cylinder-stresses-failure-numerical"
source: https://www.bctengineer.com/q/semester_5/mechanics_of_solids/thick-cylinder-stresses-failure-numerical
site: BCT Engineering
---

# Thick cylinder stresses and failure criteria

Mechanics of Solids · Chapter 8: Thick- Wall Cylinders

- Asked in: year not given, 8 marks
- Video: [Strength of Materials - Thick cylinder - Shrinkage allowance](https://www.youtube.com/watch?v=4EEuX9NMplk) — Engineering Design
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/mechanics_of_solids/thick-cylinder-stresses-failure-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_5/mechanics_of_solids/chapter-8

## Question

A thick steel cylinder has an inside diameter of 160 mm and an outside diameter of 240 mm. It is subjected to an internal pressure of 40 MPa, the outer surface is free. The ends are closed. Take $E = 200$ GPa, $\nu = 0.3$, yield stress $\sigma_y = 350$ MPa. (a) Find the hoop and radial stresses at the inner and outer surfaces and at mid-wall radius of 100 mm, and the axial stress. (b) Find the factor of safety against yielding by the maximum shear stress (Tresca) theory and the von Mises theory. (c) Find the hoop strain, axial strain and volumetric strain at the inner surface.

## Short answer

By Lamé equations: inner surface (r=80 mm) σθ=104 MPa, σr=−40 MPa; mid-wall (r=100 mm) σθ=78.1 MPa, σr=−14.1 MPa; outer surface (r=120 mm) σθ=64 MPa, σr=0; axial stress 32 MPa. Factor of safety: 2.43 (Tresca), 2.81 (von Mises). Hoop strain 5.32×10⁻⁴, axial strain 6.4×10⁻⁵, volumetric strain 1.92×10⁻⁴.

## Answer

#### Data

$a = 80$ mm, $b = 120$ mm, $p_i = 40$ MPa, $p_o = 0$.

$$
\frac{p_ia^2}{b^2 - a^2} = \frac{40\times6400}{14400 - 6400} = 32\ \text{MPa}
$$

#### (a) Lame's stresses

$$
\sigma_\theta = 32\left(1 + \frac{14400}{r^2}\right),\quad \sigma_r = 32\left(1 - \frac{14400}{r^2}\right)
$$

| Radius $r$ (mm) | $b^2/r^2$ | $\sigma_\theta$ (MPa) | $\sigma_r$ (MPa) |
|---|---|---|---|
| 80 (inner) | 2.25 | 104.00 | -40.00 |
| 100 | 1.44 | 78.08 | -14.08 |
| 120 (outer) | 1.00 | 64.00 | 0.00 |

Axial stress for closed ends: $\sigma_z = p_ia^2/(b^2 - a^2) = 32.0$ MPa.

Check: $\sigma_\theta + \sigma_r = 2A = 64$ MPa is the same at every radius, for example $104 - 40 = 64$ at the inner surface and $64 + 0 = 64$ at the outer surface. $\checkmark$

At the inner surface: $\sigma_\theta = 104$, $\sigma_r = -40$, $\sigma_z = 32$ MPa.

#### (b) Failure criteria

**Maximum shear stress (Tresca):** $\sigma_1 = \sigma_\theta = 104$, $\sigma_3 = \sigma_r = -40$ MPa (the axial stress, 32 MPa, lies between them).

$$
\sigma_1 - \sigma_3 = 104 - (-40) = 144\ \text{MPa},\qquad \tau_{max} = 72\ \text{MPa}
$$

$$
FS = \frac{\sigma_y}{\sigma_1 - \sigma_3} = \frac{350}{144} = 2.43
$$

**von Mises (distortion energy):**

$$
\begin{aligned}
\sigma_{vm} &= \sqrt{\tfrac12\left[(\sigma_\theta - \sigma_r)^2 + (\sigma_r - \sigma_z)^2 + (\sigma_z - \sigma_\theta)^2\right]} \\
&= \sqrt{\tfrac12\left[144^2 + 72^2 + 72^2\right]} = 124.7\ \text{MPa}
\end{aligned}
$$

$$
FS = \frac{350}{124.7} = 2.81
$$

Tresca is more conservative, which is why it is commonly used for thick cylinders.

#### (c) Strains at the inner surface

$$
\begin{aligned}
\varepsilon_\theta &= \frac{1}{E}\left[\sigma_\theta - \nu(\sigma_r + \sigma_z)\right] = \frac{104 - 0.3(-40 + 32)}{200\times10^3} = 5.32\times10^{-4} \\
\varepsilon_z &= \frac{1}{E}\left[\sigma_z - \nu(\sigma_\theta + \sigma_r)\right] = \frac{32 - 0.3(64)}{200\times10^3} = 6.4\times10^{-5} \\
\varepsilon_r &= \frac{1}{E}\left[\sigma_r - \nu(\sigma_\theta + \sigma_z)\right] = \frac{-40 - 0.3(136)}{200\times10^3} = -4.04\times10^{-4}
\end{aligned}
$$

Volumetric strain:

$$
\varepsilon_v = \varepsilon_\theta + \varepsilon_r + \varepsilon_z = \frac{1-2\nu}{E}(\sigma_\theta + \sigma_r + \sigma_z) = \frac{0.4\times96}{200\times10^3} = 1.92\times10^{-4}
$$

The inner radius increases by $u = \varepsilon_\theta\,a = 5.32\times10^{-4}\times80 = 0.0426$ mm.

**Answer:** (a) At $r = 80$: $\sigma_\theta = 104$ MPa, $\sigma_r = -40$ MPa; at $r = 100$: $\sigma_\theta = 78.1$ MPa, $\sigma_r = -14.1$ MPa; at $r = 120$: $\sigma_\theta = 64$ MPa, $\sigma_r = 0$; $\sigma_z = 32$ MPa. (b) FS $= 2.43$ (Tresca), $2.81$ (von Mises). (c) $\varepsilon_\theta = 5.32\times10^{-4}$, $\varepsilon_z = 6.4\times10^{-5}$, $\varepsilon_v = 1.92\times10^{-4}$.

## Check yourself

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

1. For a = 80 mm, b = 120 mm, p_i = 40 MPa, the inner-surface hoop and radial stresses are
   A. 104 MPa and -40 MPa
   B. 64 MPa and 0
   C. 78 MPa and -14 MPa
   D. 40 MPa and -40 MPa
   Answer: A. With p_i a^2/(b^2-a^2) = 32 MPa, sigma_theta = 32(1 + 2.25) = 104 and sigma_r = 32(1 - 2.25) = -40 MPa. (Key point: Lame constant)

2. The axial stress for the closed-ended cylinder is
   A. 64 MPa
   B. 32 MPa
   C. 40 MPa
   D. 0
   Answer: B. sigma_z = p_i a^2/(b^2 - a^2) = 32 MPa. (Key point: Axial stress for closed ends)

3. Tresca factor of safety is based on
   A. sigma_y / (sigma_theta - sigma_r) = 350/144 = 2.43
   B. sigma_y / sigma_theta = 3.37
   C. sigma_y / sigma_vm = 2.81
   D. sigma_y / (sigma_theta + sigma_r)
   Answer: A. Tresca uses the largest minus smallest principal stress, 104 - (-40) = 144 MPa. (Key point: Tresca factor of safety)

4. The hoop strain at the inner surface uses
   A. sigma_theta / E only = 5.2e-4
   B. [sigma_theta - nu(sigma_r + sigma_z)]/E = 5.32e-4
   C. [sigma_theta - nu sigma_r]/E
   D. [sigma_theta + nu(sigma_r + sigma_z)]/E
   Answer: B. Three-dimensional Hooke's law: (104 - 0.3(-40+32))/200000 = 5.32e-4. (Key point: Hoop strain)

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