Hertzian contact

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Contact of two cylinders with axes parallel

If two circular cylinders with radii $R_1$ and $R_2$ are pressed together by a force per unit length of magnitude $F$ with their axes parallel, then the contact patch will be of half-width $b$ such that \begin{equation} b = \sqrt{\frac{2FR}{\pi E^*}} \end{equation} where $R$ and $E^*$ are the reduced radius of contact and the contact modulus defined by \begin{equation} \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}, \end{equation} \begin{equation} \frac{1}{E^*} = \frac{1-\nu_1^2}{E_1} + \frac{1-\nu_2^2}{E_2}. \end{equation}