Indentation by a plat rigid punch
|
- Start with uneven surface profile
.
- Unsymmetric load
, but sufficient for complete contact over the area
.
Displacement in
direction is
![{\displaystyle u_{2}=-u_{0}(x_{1})+C_{1}x_{1}+C_{0}\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a073e4ade33565709b76903282807444162b49c0)
where
is a rigid body translation and
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b2b7a759579147a88122f2546ef479767d23b0d9)
The displacement gradient is
![{\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<x<a}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f76e45e13ebdbe2dacee5abc374fc454d4c1a702)
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
.If we change the variables so that
and
, 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 }](https://wikimedia.org/api/rest_v1/media/math/render/svg/c356d189f74252835d9db0233af73ff5f7fed2d1)
If we write
and
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dd37018e4936e93eaf9f0e4ca6e12fc4c00d9321)
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f64dce14d222ef4a139bcddab2f426bb0e3b4899)
![{\displaystyle u_{0}=C\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1a721267d24c896ba10eabeb5d300786366c8483)
In this case,
![{\displaystyle {\frac {du_{0}}{d\phi }}=0\Rightarrow u_{n}=0~;~~n=1{\infty }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1e47eddf150ac5a8e93c28af5b718966505a5d69)
Also,
(origin at the center of
), hence
. Therefore,
![{\displaystyle p(x)={\frac {p_{0}}{\sin \phi }}=-{\frac {F}{\pi {\sqrt {a^{2}-x^{2}}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/48e9735c0fd6267bdabe329f7414adac44f8f2ed)
At
, the load is infinite, i.e. there is a singularity.