How Hertzian contact stress is worked out

A ball on its raceway, a wheel on a rail, a cam on its follower — this works out the stress where two curved surfaces meet. Treated as rigid bodies they would meet at a point or a line, and with zero area no stress can be found. In reality both deflect a little and meet over a finite area; Hertzian contact stress is that area and pressure derived from elasticity.

When it applies

It only applies where the following three hold. Outside them the value found here is not even a rough guide.

The contact area is small enough compared with the bodies themselves — the material around it has to be treatable as a half-space
The deformation stays within the elastic range. Once it yields, the formulas no longer hold
There is no sliding and no friction at the interface. Tangential forces are not considered

The lubricant film, fatigue under repeated load and surface roughness are not included either. Do not judge a problem where those dominate — rolling bearing life, for instance — from these formulas alone.

Combining the two bodies into one

Before the formulas, the shape and the material of the two bodies are each reduced to a single value. Once those two are fixed, the rest follows from whether the contact is spherical or cylindrical.

Equivalent radius of curvature

The shapes combined into one value, through the sum of the reciprocals.

\frac{1}{R_e} = \frac{1}{R_1} + \frac{1}{R_2} \qquad R_e = \frac{R_1 R_2}{R_1 + R_2}

Equivalent modulus of elasticity

The materials combined into one value. Poisson's ratio enters here.

\frac{1}{E_e} = \frac{1 - \nu_1^2}{E_1} + \frac{1 - \nu_2^2}{E_2} \qquad E_e = \frac{E_1 E_2}{E_2 (1 - \nu_1^2) + E_1 (1 - \nu_2^2)}

$E_e$ has the units of a modulus but is not the modulus of either material. It combines how readily the two bodies deflect, and belongs to this contact only.

Flat and concave surfaces

The formulas do not change with the shape. Only the way $R_e$ is found changes.

Against a flat surface

A flat surface is a curved one of infinite radius. Since $1 / R_2 \to 0$ , the equivalent radius is simply the radius of the convex side.

\frac{1}{R_e} = \frac{1}{R_1} + 0 \qquad R_e = R_1

Against a concave surface

A convex surface seated in a concave one, as a ball sits in its outer ring. The radius of the concave surface is taken as negative.

\frac{1}{R_e} = \frac{1}{R_1} + \frac{1}{-R_2} \qquad R_e = \frac{R_1 R_2}{R_2 - R_1}

$R_1 < R_2$ is required. If the convex surface is the larger it does not seat, $R_e$ turns negative and the formulas lose their meaning. The calculator rejects inputs that fail this condition.

A concave surface gives a larger $R_e$ than a flat or convex one, so the same load spreads over a wider area and the pressure falls. That is why bearing raceways are grooved.

Spherical contact (point contact)

The contact area is a circle. Write its radius as $a$ .

a = \sqrt[3]{\frac{3 F R_e}{4 E_e}}

How far the centres of the two bodies approach each other.

\delta = \sqrt[3]{\frac{9 F^2}{16 R_e E_e^2}}

The pressure over the contact is not uniform but semi-elliptical, highest at the centre and zero at the rim. Writing the distance from the centre as $r$ gives the following.

p(r) = p_{max} \sqrt{1 - \left(\frac{r}{a}\right)^2} \qquad p_{max} = \frac{3F}{2 \pi a^2}

$p_{max}$ is 1.5 times the mean pressure (the mean being $F / \pi a^2$ ). Checking with load divided by area therefore underestimates it.

Cylindrical contact (line contact)

Two cylinders with their axes parallel. The contact is not a circle but a strip; write its half-width as $b$ and the length along the axis as $L$ .

b = \sqrt{\frac{4 F R_e}{\pi E_e L}}

The pressure is semi-elliptical across the strip.

p(x) = p_{max} \sqrt{1 - \left(\frac{x}{b}\right)^2} \qquad p_{max} = \frac{2F}{\pi b L}

Here $p_{max}$ is 4/π ≒ 1.27 times the mean pressure.

The approach is not worked out. In line contact it depends on the shape outside the contact as well, so the Hertz formulas alone do not fix it. The calculator does not output it either.

Contact pressure is not the internal stress

$p_{max}$ is the greatest pressure acting on the surfaces in contact, not the greatest stress inside the material. Which quantity to look at depends on what you want to check.

To judge indentation or wear, use $p_{max}$ — the pressure the surface itself carries
To judge cracking or spalling, use the internal stress. Rolling contact fatigue starts not at the surface but a little below it

For spherical contact the internal stresses can be written in proportion to $p_{max}$ .

\sigma_{max} = \frac{1 - 2\nu}{3} \, p_{max}

Maximum tensile stress $\sigma_{max}$ — at the rim of the contact circle, on the surface, acting outwards
Maximum shear stress $\tau_{max}$ — directly below the centre of the contact, at a depth of about $0.48\,a$ . With $\nu = 0.3$ it comes to $\tau_{max} \fallingdotseq 0.31\,p_{max}$

The numbers change with $\nu$ . 0.31 and 0.48 are the figures for $\nu = 0.3$ , roughly the value for steel; another ratio gives other figures. The calculator does not output these values. It goes as far as $p_{max}$ .

Notes

$F$ is the force normal to the surface at the contact. Resolve self-weight and any external load into the component normal to the interface before entering it
For cylindrical contact, $L$ is the length actually in contact. Near the ends the pressure rises — edge loading — so conditions there are more severe than these formulas suggest
The calculator also outputs $R_e$ and $E_e$ : when checking against a hand calculation, get these two to agree first.