Elasticity/Hertz contact

The Hertz problem : rigid cylindrical punch

File:Hertz indentation.png 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,

$\frac{d^2{u}_0}{d{x}^2}=-\frac{1}{R}$

Hence,

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

and

$\frac{d{u}_0}{d\varphi }=-\frac{a^2\sin (2\varphi )}{2R}$

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

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

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 $a$

$a=\sqrt{\frac{F{R}_1{R}_2}{2\pi (R_1+R_2)}(\frac{{\kappa }_1+1}{{\mu }_1}+\frac{{\kappa }_2+1}{{\mu }_2})}$

and the force distribution $p$

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

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