---
title: "Lame's equations for thick cylinders"
subject: "Mechanics of Solids"
chapter: 8
key: "semester_5/mechanics_of_solids:lame-equations-thick-cylinder"
source: https://www.bctengineer.com/q/semester_5/mechanics_of_solids/lame-equations-thick-cylinder
site: BCT Engineering
---

# Lame's equations for thick cylinders

Mechanics of Solids · Chapter 8: Thick- Wall Cylinders

- Asked in: year not given, 8 marks
- Video: [Lame's Equation for Thick Cylinders—Derivation #mechanicsofsolids #strengthofmaterial #gate](https://www.youtube.com/watch?v=mBcgN7QbDpw) — Dr. Ravi's Academy
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/mechanics_of_solids/lame-equations-thick-cylinder
- Whole chapter: https://www.bctengineer.com/md/semester_5/mechanics_of_solids/chapter-8

## Question

Derive Lame's equations for the radial and hoop stresses in a thick-walled cylinder subjected to internal pressure $p_i$ and external pressure $p_o$. Sketch the variation of the stresses across the wall for internal pressure only, and state the axial stress for closed and open ends.

## Short answer

Lame's equations are sigma_r = A - B/r^2, sigma_theta = A + B/r^2 with constants from boundary conditions. For internal pressure only, hoop stress peaks at the inner surface and radial stress varies from -p_i to zero. Axial stress is p_i a^2/(b^2 - a^2) for closed ends, zero for open ends.

## Answer

#### Assumptions

Material is homogeneous, isotropic and linear elastic; the cylinder is long, and the section remains plane (axial strain uniform); the stress is symmetric about the axis, so $\sigma_r$ and $\sigma_\theta$ depend on $r$ only.

#### Equilibrium

Consider an element between radii $r$ and $r+dr$ subtending $d\theta$. Radial force balance:

$$
(\sigma_r + d\sigma_r)(r+dr)\,d\theta - \sigma_r\,r\,d\theta - 2\sigma_\theta\,dr\,\sin\frac{d\theta}{2} = 0
$$

For small angles ($\sin(d\theta/2) \approx d\theta/2$) and neglecting higher-order terms:

$$
\frac{d\sigma_r}{dr} + \frac{\sigma_r - \sigma_\theta}{r} = 0 \quad\text{or}\quad \sigma_\theta = \sigma_r + r\frac{d\sigma_r}{dr}
$$

#### Compatibility

With radial displacement $u$: $\varepsilon_\theta = u/r$ and $\varepsilon_r = du/dr$, so

$$
\frac{d\varepsilon_\theta}{dr} = \frac{\varepsilon_r - \varepsilon_\theta}{r}
$$

With Hooke's law and a constant axial stress $\sigma_z$ (equivalently constant $\varepsilon_z$), $\varepsilon_\theta = \dfrac1E[\sigma_\theta - \nu(\sigma_r + \sigma_z)]$ and $\varepsilon_r = \dfrac1E[\sigma_r - \nu(\sigma_\theta + \sigma_z)]$. Substituting:

$$
\frac{d}{dr}(\sigma_\theta - \nu\sigma_r) + \frac{1+\nu}{r}(\sigma_\theta - \sigma_r) = 0
$$

Put $\sigma_\theta = \sigma_r + r\sigma_r'$ from equilibrium:

$$
\sigma_r'' + \frac{3}{r}\sigma_r' = 0 \ \Rightarrow\ \sigma_r' = \frac{2B}{r^3}
$$

Integrating:

$$
\boxed{\sigma_r = A - \frac{B}{r^2},\qquad \sigma_\theta = A + \frac{B}{r^2}}
$$

Note that $\sigma_r + \sigma_\theta = 2A$ is constant across the wall.

#### Boundary conditions

At $r = a$ (inner): $\sigma_r = -p_i$. At $r = b$ (outer): $\sigma_r = -p_o$.

$$
A = \frac{p_ia^2 - p_ob^2}{b^2 - a^2},\qquad B = \frac{a^2b^2(p_i - p_o)}{b^2 - a^2}
$$

For $p_o = 0$:

$$
\sigma_r = \frac{p_ia^2}{b^2 - a^2}\left(1 - \frac{b^2}{r^2}\right),\qquad
\sigma_\theta = \frac{p_ia^2}{b^2 - a^2}\left(1 + \frac{b^2}{r^2}\right)
$$

#### Distribution

```text
 stress
   ^ sigma_theta (max at r=a, tension)
   |\
   | \_____
   |       ----__ sigma_theta (at r=b)
 --+-------------------> r
   | ____----
   |/  sigma_r (compressive; -p_i at r=a, 0 at r=b)
   a                   b
```

The hoop stress is highest at the inner surface, $\sigma_{\theta,max} = p_i\dfrac{b^2 + a^2}{b^2 - a^2}$, and always exceeds $p_i$, however thick the wall.

#### Axial stress

- Closed ends: $\sigma_z = \dfrac{p_ia^2}{b^2 - a^2}$ (uniform, from end force balance).
- Open ends (no end load): $\sigma_z = 0$.

## Check yourself

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

1. Radial equilibrium of an element of a thick cylinder gives
   A. dsigma_r/dr = sigma_theta
   B. sigma_theta = sigma_r + r dsigma_r/dr
   C. sigma_r = sigma_theta + r dsigma_theta/dr
   D. sigma_r + sigma_theta = 0
   Answer: B. Neglecting higher-order terms gives dsigma_r/dr + (sigma_r - sigma_theta)/r = 0. (Key point: Equilibrium equation)

2. The Lame solution has the form
   A. sigma_r = A - B/r^2, sigma_theta = A + B/r^2
   B. sigma_r = A + Br, sigma_theta = A - Br
   C. sigma_r = A/r, sigma_theta = B/r
   D. Both stresses are constant
   Answer: A. Compatibility gives sigma_r'' + 3 sigma_r'/r = 0, whose integration yields these forms; sigma_r + sigma_theta = 2A is constant. (Key point: Form of Lame solution)

3. For internal pressure only, the hoop stress is
   A. Maximum at the outer surface and equal to p_i
   B. Maximum at the inner surface, p_i (b^2 + a^2)/(b^2 - a^2), always greater than p_i
   C. Uniform across the wall
   D. Zero at the inner surface
   Answer: B. Hoop stress decreases outwards and is largest at r = a, always exceeding p_i however thick the wall. (Key point: Hoop stress under internal pressure)

4. The axial stress in a thick cylinder with closed ends and with open ends is
   A. p_i a^2/(b^2 - a^2) for closed, zero for open
   B. Zero for closed, p_i for open
   C. p_i a^2/(b^2 - a^2) for both
   D. Equal to hoop stress for closed ends
   Answer: A. Closed ends carry the end force giving a uniform axial stress; open ends have no end load so it is zero. (Key point: Axial stress)

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