# Elasticity/Hertz contact

## The Hertz problem : rigid cylindrical punch

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

We have,

${\displaystyle {\frac {d^{2}u_{0}}{dx^{2}}}=-{\frac {1}{R}}}$

Hence,

${\displaystyle u_{0}=C_{0}-{\frac {x^{2}}{2R}}=C_{0}-{\frac {a^{2}\cos(2\phi )}{4R}}-{\frac {a^{2}}{4R}}}$

and

${\displaystyle {\frac {du_{0}}{d\phi }}=-{\frac {a^{2}\sin(2\phi )}{2R}}}$

Therefore,

${\displaystyle u_{1}=0~;~~u_{2}={\frac {a^{2}}{2R}}~;~~u_{n}=0~(n>2)}$

and

${\displaystyle p_{0}=-{\frac {F}{\pi a}}~;~~p_{1}=0~;~~p_{2}={\frac {2\mu a}{R(\kappa +1)}}~;~~p_{n}=0~(n>2)}$

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

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

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

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

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

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

### 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 ${\displaystyle a\,}$

${\displaystyle a={\sqrt {{\frac {FR_{1}R_{2}}{2\pi (R_{1}+R_{2})}}\left({\frac {\kappa _{1}+1}{\mu _{1}}}+{\frac {\kappa _{2}+1}{\mu _{2}}}\right)}}}$

and the force distribution ${\displaystyle p}$

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