Contact stress between two steel cylinders

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.