Contact Stresses

Contact stresses

When two curved elastic bodies are pressed together, they touch over a very small area, and high compressive stresses (contact or Hertz stresses) develop at and below the surface. They govern the life of ball and roller bearings, gears, cams, wheels on rails, and rollers. Failure is by pitting, spalling or surface cracking, usually starting below the surface.

Assumptions (Hertz)

1. The materials are homogeneous, isotropic and linearly elastic.
2. The contact area is very small compared with the dimensions of the bodies and their radii of curvature.
3. Surfaces are smooth, so only normal pressure acts (no friction).
4. Loads are static and act normal to the contact area.
5. Deformations are small, so the contact area is elliptical (circular for spheres, a narrow strip for cylinders).

Point contact: two spheres

With radii $R_1, R_2$, load $F$, and $\dfrac{1}{E^*} = \dfrac{1-\nu_1^2}{E_1} + \dfrac{1-\nu_2^2}{E_2}$, $\dfrac{1}{R} = \dfrac1{R_1} + \dfrac1{R_2}$:

$$a = \left(\frac{3FR}{4E^*}\right)^{1/3},\qquad p_{max} = \frac{3F}{2\pi a^2}$$

The contact area is a circle of radius $a$, with a hemispherical pressure distribution.

Line contact: two parallel cylinders (length $l$)

$$b = \sqrt{\frac{4FR}{\pi l\,E^*}},\qquad p_{max} = \frac{2F}{\pi b\,l}$$

The contact is a strip of width $2b$ with a semi-elliptical pressure distribution.

For a cylinder on a plane, $R = R_1$.

Stresses

The principal stresses ($\sigma_x, \sigma_y, \sigma_z$) are compressive and maximum on the load axis. At the surface they equal about $-p_{max}$ ($-2\nu p_{max}$ for the axial direction in line contact). They fall off rapidly with depth. The maximum shear stress is not on the surface but at a depth below it: $\tau_{max} \approx 0.30\,p_{max}$ at $z \approx 0.78b$ for line contact, and $\approx 0.31\,p_{max}$ at $z \approx 0.48a$ for spheres ($\nu = 0.3$). This subsurface shear explains why fatigue cracks start below the surface.

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.