---
title: "Contact stress between two steel cylinders"
subject: "Mechanics of Solids"
chapter: 11
key: "semester_5/mechanics_of_solids:steel-cylinders-hertz-numerical"
source: https://www.bctengineer.com/q/semester_5/mechanics_of_solids/steel-cylinders-hertz-numerical
site: BCT Engineering
---

# Contact stress between two steel cylinders

Mechanics of Solids · Chapter 11: Contact Stresses

- Asked in: year not given, 6 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/mechanics_of_solids/steel-cylinders-hertz-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_5/mechanics_of_solids/chapter-11

## Question

Two parallel steel cylinders of radii 50 mm and 150 mm and length 40 mm are pressed together externally with a force of 10 kN. For both, $E = 200$ GPa and $\nu = 0.3$. Using Hertz theory find the half-width of the contact band, the maximum contact pressure, the principal stresses at the surface along the load axis, and the maximum shear stress with its depth.

## Short answer

Equivalent radius R = 37.5 mm, E* = 109.9 GPa. Half-width b = √(4FR/(πlE*)) = 0.330 mm. Maximum pressure p_max = 2F/(πbl) = 483 MPa. Surface principal stresses: -483, -483, -290 MPa. Maximum shear stress 145 MPa at depth 0.26 mm.

## Answer

#### Data

$R_1 = 50$ mm, $R_2 = 150$ mm, $l = 40$ mm, $F = 10$ kN, $E = 200$ GPa, $\nu = 0.3$.

Equivalent radius:

$$
\frac1R = \frac1{50} + \frac1{150} = 0.02667\ \text{mm}^{-1}\ \Rightarrow\ R = 37.5\ \text{mm}
$$

Equivalent modulus (same materials):

$$
\frac1{E^*} = \frac{2(1-\nu^2)}{E} \Rightarrow E^* = \frac{200\times10^3}{2\times0.91} = 109890\ \text{MPa}
$$

#### Half-width of the contact band

$$
\begin{aligned}
b &= \sqrt{\frac{4FR}{\pi l E^*}} = \sqrt{\frac{4\times10^4\times37.5}{\pi\times40\times109890}} \\
&= \sqrt{\frac{1.5\times10^6}{1.381\times10^7}} = 0.3296\ \text{mm}
\end{aligned}
$$

(Equivalently $b = \sqrt{8F(1-\nu^2)/[\pi lE(1/R_1 + 1/R_2)]}$.)

#### Maximum contact pressure

$$
p_{max} = \frac{2F}{\pi b\,l} = \frac{2\times10^4}{\pi\times0.3296\times40} = 482.9\ \text{MPa}
$$

#### Surface stresses on the load axis

With $x$ along the cylinder axis, $y$ across the contact width and $z$ in depth:

$$
\sigma_z = \sigma_y = -p_{max} = -482.9\ \text{MPa},\qquad \sigma_x = -2\nu p_{max} = -289.7\ \text{MPa}
$$

All three are compressive.

#### Maximum shear stress

$$
\tau_{max} \approx 0.300\,p_{max} = 0.300\times482.9 = 144.9\ \text{MPa},\quad z \approx 0.786\,b = 0.786\times0.3296 = 0.259\ \text{mm}
$$

It occurs about 0.26 mm below the surface, so the material beneath the surface yields first. This depth is of the order of the width of the contact band.

**Answer:** $b = 0.330$ mm (band width $0.659$ mm); $p_{max} = 483$ MPa; surface principal stresses $-483$, $-483$ and $-290$ MPa; $\tau_{max} = 145$ MPa at $0.26$ mm depth.

## Check yourself

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

1. How is the equivalent radius R found for the 50 mm and 150 mm cylinders pressed externally?
   A. R = (50 + 150)/2 = 100 mm
   B. 1/R = 1/50 + 1/150, so R = 37.5 mm
   C. R = 150 - 50 = 100 mm
   D. R = 50 x 150 = 7500 mm
   Answer: B. For external contact curvatures add: 1/R = 1/R1 + 1/R2 = 0.02667, so R = 37.5 mm. (Key point: Equivalent radius)

2. For two identical steels, what is E*?
   A. E/(2(1 - nu^2)) = 109890 MPa
   B. E/2 = 100 GPa
   C. E(1 - nu^2) = 182 GPa
   D. 2E = 400 GPa
   Answer: A. 1/E* = 2(1 - nu^2)/E, so E* = 200000/(2 x 0.91) = 109890 MPa. (Key point: Equivalent modulus)

3. What are the surface principal stresses on the load axis?
   A. -483, -483 and -290 MPa, all compressive
   B. +483, +483 and +290 MPa
   C. -483, 0 and 0 MPa
   D. -483, -483 and -483 MPa
   Answer: A. sigma_z = sigma_y = -p_max = -482.9 MPa and sigma_x = -2 nu p_max = -289.7 MPa. (Key point: Surface stresses on axis)

4. Where does the maximum shear stress of about 145 MPa occur?
   A. On the surface at the contact centre
   B. About 0.26 mm below the surface (0.786b)
   C. At 3 mm depth
   D. At the contact edge on the surface
   Answer: B. tau_max = 0.30 p_max at z = 0.786b = 0.259 mm, so subsurface material yields first. (Key point: Maximum shear depth)

## Related in Mechanics of Solids

- [Contact stresses and Hertz assumptions](https://www.bctengineer.com/question.md?key=semester_5%2Fmechanics_of_solids%3Ahertz-contact-stress-notes) — Mechanics of Solids, chapter 11

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