---
title: "Composite cylinder by shrink fitting"
subject: "Mechanics of Solids"
chapter: 8
key: "semester_5/mechanics_of_solids:compound-cylinder-shrink-fit"
source: https://www.bctengineer.com/q/semester_5/mechanics_of_solids/compound-cylinder-shrink-fit
site: BCT Engineering
---

# Composite cylinder by shrink fitting

Mechanics of Solids · Chapter 8: Thick- Wall Cylinders

- Asked in: year not given, 8 marks
- Video: [Compound Cylinder](https://www.youtube.com/watch?v=se6cMJedDcc) — Rathin Dutta
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/mechanics_of_solids/compound-cylinder-shrink-fit
- Whole chapter: https://www.bctengineer.com/md/semester_5/mechanics_of_solids/chapter-8

## Question

A steel tube of outer radius 50 mm and inner radius 30 mm is fitted inside another steel tube of inner radius 50 mm (before assembly) and outer radius 70 mm, with a diametral interference of 0.06 mm at the common surface. Take $E = 200$ GPa. (a) Calculate the contact pressure. (b) Find the hoop stresses at the inner and outer surfaces of each tube after assembly. (c) By how much must the outer tube be heated above the inner one to slip it on? Take $\alpha = 12\times10^{-6}$/$^\circ$C.

## Short answer

Contact pressure after shrink fit is 23.04 MPa. Hoop stresses: inner tube −72 MPa at bore (r=30) and −48.96 MPa at interface (r=50); outer tube 71.04 MPa at interface and 48 MPa at outer surface (r=70). The outer tube must be heated by at least 50 °C above the inner tube for assembly.

## Answer

#### Data

Inner tube: $a = 30$ mm, $r = 50$ mm. Outer tube: $r = 50$ mm, $b = 70$ mm. Radial interference $\delta = 0.06/2 = 0.03$ mm. Both tubes are steel, $E = 200\times10^3$ MPa.

```text
        outer tube (50 to 70 mm)
     .--------------------.
    /   .--------------.   \
   |   /  inner tube    \   |
   |  |   (30 to 50)     |  |
   |   \                /   |
    \   '--------------'   /
     '--------------------'
        contact pressure p at r = 50
```

#### Contact pressure

After assembly, a pressure $p$ acts on the contact surface. The inner tube has external pressure $p$ only, the outer tube has internal pressure $p$ only. The radial interference equals the sum of the outward movement of the outer tube's inner surface and the inward movement of the inner tube's outer surface.

For a tube with internal pressure $p$ (inner radius $r$, outer radius $b$): hoop stress at inner surface $= p\dfrac{b^2 + r^2}{b^2 - r^2}$, so

$$
u_{outer} = \frac{p\,r}{E}\left[\frac{b^2 + r^2}{b^2 - r^2} + \nu\right]
$$

For a tube with external pressure $p$ only (inner radius $a$, outer radius $r$): hoop stress at outer surface $= -p\dfrac{r^2 + a^2}{r^2 - a^2}$, so the inward movement is

$$
u_{inner} = \frac{p\,r}{E}\left[\frac{r^2 + a^2}{r^2 - a^2} - \nu\right]
$$

Sum (the Poisson terms cancel for the same material):

$$
\delta = \frac{p\,r}{E}\left[\frac{b^2 + r^2}{b^2 - r^2} + \frac{r^2 + a^2}{r^2 - a^2}\right]
$$

$$
\frac{b^2 + r^2}{b^2 - r^2} = \frac{4900 + 2500}{2400} = 3.0833,\qquad
\frac{r^2 + a^2}{r^2 - a^2} = \frac{2500 + 900}{1600} = 2.1250
$$

$$
p = \frac{\delta E}{r\,(3.0833 + 2.1250)} = \frac{0.03\times200\times10^3}{50\times5.2083} = 23.04\ \text{MPa}
$$

#### (b) Hoop stresses

**Outer tube** (internal pressure $p$):

$$
\sigma_\theta(r=50) = p\,\frac{b^2 + r^2}{b^2 - r^2} = 23.04\times3.0833 = 71.04\ \text{MPa (tension)}
$$

$$
\sigma_\theta(r=70) = \frac{2pr^2}{b^2 - r^2} = \frac{2\times23.04\times2500}{2400} = 48.00\ \text{MPa (tension)}
$$

**Inner tube** (external pressure $p$):

$$
\sigma_\theta(r=50) = -p\,\frac{r^2 + a^2}{r^2 - a^2} = -23.04\times2.125 = -48.96\ \text{MPa}
$$

$$
\sigma_\theta(r=30) = -\frac{2pr^2}{r^2 - a^2} = -\frac{2\times23.04\times2500}{1600} = -72.00\ \text{MPa (compression)}
$$

| Surface | Hoop stress (MPa) |
|---|---|
| Inner tube, r = 30 | -72.00 |
| Inner tube, r = 50 | -48.96 |
| Outer tube, r = 50 | 71.04 |
| Outer tube, r = 70 | 48.00 |

The radial stress is $-p = -23.04$ MPa at the contact surface and zero on the free surfaces.

#### (c) Heating of the outer tube

To slip the outer tube over, its bore must expand by the diametral interference $0.06$ mm (a small extra clearance is used in practice). The bore of 100 mm diameter expands by $\alpha\,d\,\Delta T$:

$$
\Delta T = \frac{0.06}{12\times10^{-6}\times100} = 50\ ^\circ\text{C}
$$

**Answer:** Contact pressure $= 23.04$ MPa; hoop stresses: outer tube $71.0$ MPa (bore) and $48.0$ MPa (outside); inner tube $-49.0$ MPa (outside) and $-72.0$ MPa (bore); outer tube to be heated by at least $50^\circ$C above the inner tube.

## Check yourself

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

1. The radial interference at the common surface equals
   A. The outward movement of the outer tube bore only
   B. The sum of the outward movement of the outer tube's inner surface and the inward movement of the inner tube's outer surface
   C. The difference of the two movements
   D. The diametral interference 0.06 mm
   Answer: B. Contact pressure p squeezes the inner tube and expands the outer; their sum equals the radial interference 0.03 mm. (Key point: Interference condition)

2. The contact pressure is about
   A. 11.5 MPa
   B. 23.0 MPa
   C. 46.1 MPa
   D. 71.0 MPa
   Answer: B. p = delta E / (r (3.0833 + 2.125)) = 0.03 x 200000 / (50 x 5.2083) = 23.04 MPa. (Key point: Contact pressure)

3. After assembly the hoop stress in the inner tube and outer tube is
   A. Both tensile
   B. Inner tube compressive (e.g. -72 MPa at bore), outer tube tensile (e.g. 71 MPa at contact)
   C. Inner tube tensile, outer compressive
   D. Both compressive
   Answer: B. The inner tube has external pressure only (compression) and the outer tube internal pressure only (tension). (Key point: Hoop stress signs)

4. The outer tube must be heated above the inner by about
   A. 25 degrees C
   B. 50 degrees C
   C. 100 degrees C
   D. 5 degrees C
   Answer: B. The bore must expand by 0.06 mm: dT = 0.06/(12e-6 x 100) = 50 degrees C. (Key point: Heating)

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