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Contact of two cylinders with axes parallel

Normal contact

If two circular cylinders with radii $R_1$ and $R_2$ are pressed together by a force per unit length of magnitude $P$ with their axes parallel, then the contact patch will be of half-width $b$ such that \begin{equation} b = \sqrt{\frac{2PR}{\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}

The resulting pressure distribution $p(x)$ is semielliptical, i.e., of the form \begin{equation} p(x) = p_0 \sqrt{1-\frac{x^2}{b^2}} \end{equation} where the peak pressure \begin{equation} p_0 = \sqrt{\frac{PE^*}{\pi R}}. \end{equation}

The coordinate $x$ is measured perpendicular to that of the cylinder axes.

For the case of nominal contact between cylinders closed from analytical solutions are available. \end{equation}

FreeFem++ numerical solution