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 {du_{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_{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 $p$

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

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