Introduction to Elasticity/Hertz contact

The Hertz Problem: Rigid Cylindrical Punch

Hertz indentation

• The contact length $a$ depends on the load $F\,$.
• There is no singularity at $x = \pm a$.
• The radius of the cylinder ($R\,$) is large.

We have,

$\frac{d^2 u_0}{dx^2} = -\frac{1}{R}$

Hence,

$u_0 = C_0 - \frac{x^2}{2R} = C_0 - \frac{a^2\cos(2\phi)}{4R}-\frac{a^2}{4R}$

and

$\frac{d u_0}{d\phi} = -\frac{a^2\sin(2\phi)}{2R}$

Therefore,

$u_1 = 0 ~;~~ u_2 = \frac{a^2}{2R} ~;~~ u_n = 0 ~(n > 2)$

and

$p_0 = -\frac{F}{\pi a} ~;~~ p_1 = 0 ~;~~ p_2 = \frac{2\mu a}{R(\kappa+1)} ~;~~ p_n = 0 ~(n > 2)$

Plug back into the expression for $p(\theta)$ to get

$p(\theta) = \left(-\frac{F}{\pi a} + \frac{2\mu a}{R(\kappa+1)}\cos(2\theta) \right)/\sin\theta$

This expression is singular at $\theta=0$ and $\theta=\pi$, unless we choose

$\frac{F}{\pi a} = \frac{2\mu a}{R(\kappa+1)} \Rightarrow a = \sqrt{\frac{F(\kappa+1)R}{2\pi\mu}}$

Plugging $a$ into the equation for $p(\theta)$,

$p(\theta) = -\frac{2F\sin\theta}{\pi a} \Rightarrow p(x) = -\frac{2F\sqrt{a^2-x^2}}{\pi a^2}$

Two deformable cylinders

If instead of the half-plane we have an cylinder; and instead of the rigid cylinder we have a deformable cylinder, then a similar approach can be used to obtain the contact length $a\,$

$a = \sqrt{\frac{FR_1R_2}{2\pi(R_1+R_2)}\left( \frac{\kappa_1+1}{\mu_1} + \frac{\kappa_2+1}{\mu_2}\right)}$

and the force distribution $p$

$p(x) = -\frac{2F\sqrt{a^2-x^2}}{\pi a^2}$