Elasticity/Hertz contact

The Hertz problem : rigid cylindrical punch

 * 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} $$