# Introduction to Elasticity/Flat punch indentation

## Indentation due to a frictionless rigid flat punch

 Indentation by a plat rigid punch
• Start with uneven surface profile ${\displaystyle u_{0}(x_{1})\,}$.
• Unsymmetric load ${\displaystyle F\,}$, but sufficient for complete contact over the area ${\displaystyle A\,}$.

Displacement in ${\displaystyle x_{2}\,}$ direction is

${\displaystyle u_{2}=-u_{0}(x_{1})+C_{1}x_{1}+C_{0}\,}$

where ${\displaystyle C_{0}\,}$ is a rigid body translation and ${\displaystyle C_{1}x_{1}\,}$ is a rigid body rotation.

Rigid body motions can be determined using a statically equivalent set of forces and moments

{\displaystyle {\begin{aligned}\int _{A}p(\xi )~d\xi &=-F\\\int _{A}p(\xi )\xi ~d\xi &=-F~d\end{aligned}}}

${\displaystyle -{\frac {du_{0}}{dx_{1}}}+C_{1}=-{\frac {(\kappa +1)}{4\pi \mu }}\int _{-a}^{a}{\frac {p(\xi )}{x-\xi }}~d\xi ~;~~-a

Integral is a Cauchy Singular Integral that appears often and very naturally when the problem is solved using complex variable methods.

Note that the only thing we are interested in is the distribution of contact forces ${\displaystyle p(\xi )\,}$.If we change the variables so that

${\displaystyle x=a\cos \phi \,}$ and ${\displaystyle \xi =a\cos \theta \,}$, then

${\displaystyle {\frac {1}{a\sin \phi }}{\frac {du_{0}}{d\phi }}+C_{1}=-{\frac {(\kappa +1)}{4\pi \mu }}\int _{0}^{\pi }{\frac {p(\theta )\sin \theta }{\cos \phi -\cos \theta }}~d\theta ~;~~0<\phi <\pi }$

If we write ${\displaystyle p(\theta )\,}$ and ${\displaystyle du_{0}/d\phi \,}$ as

{\displaystyle {\begin{aligned}p(\theta )&=\sum _{0}^{\infty }{\frac {p_{n}\cos(n\theta )}{\sin \theta }}\\{\frac {du_{0}}{d\phi }}&=\sum _{1}^{\infty }u_{n}\sin(n\phi )\end{aligned}}}

and do some algebra, we get

{\displaystyle {\begin{aligned}p_{0}&=-{\frac {F}{\pi a}}\\p_{1}&=-{\frac {Fd}{\pi a^{2}}}\\p_{n}&=-{\frac {4\mu u_{n}}{(\kappa +1)a}}~;~~n>1\end{aligned}}}

### Flat punch with symmetric load : ${\displaystyle u_{0}=C\,}$

In this case,

${\displaystyle {\frac {du_{0}}{d\phi }}=0\Rightarrow u_{n}=0~;~~n=1{\infty }}$

Also, ${\displaystyle d=0\,}$ (origin at the center of ${\displaystyle A\,}$), hence ${\displaystyle p_{1}=0\,}$. Therefore,

${\displaystyle p(x)={\frac {p_{0}}{\sin \phi }}=-{\frac {F}{\pi {\sqrt {a^{2}-x^{2}}}}}}$

At ${\displaystyle x=\pm a}$, the load is infinite, i.e. there is a singularity.