Re: [ATM] Pressure per square mm

["Mutalib Abdallah" <omegatroid@hotmail.com>]

Consider that the force is a constant plus a value proportional to distance 
from a "neutral axis".   You need to calculate the position of that axis, 
and then find the constant of proportionality.  This can be done easily if 
you know integral calculus using the idea that the moment on the top piece 
must be zero.   If you are not so good at integration over intersecting 
circular regions, you can do numerical integration by GQ, Newton-Cotes, or 
any of the other usual methods.   Use just a few terms, and you'll have a 
reasonable approximation.


>From: Woodchuck <djv@bedford.net>
>To: James Lerch <jlerch1@tampabay.rr.com>
>CC: ATM <atm@atmlist.net>
>Subject: Re: [ATM] Pressure per square mm
>Date: Wed, 17 Mar 2004 14:29:48 -0500 (EST)
>
>On Wed, 17 Mar 2004, James Lerch wrote:
>
> > Greetings All,
> >
> > I'm still working on the erosion simulator.  I need to solve for the 
>pressure
> > (aka normal force) for each square millimeter of the contact area, and 
>I'm not
> > sure how to do that, nor what 'topic' I'd search for information on the
> > procedure. :0
> >
> > Currently I'm applying the KISS principal, with the following 
>assumptions: (all
> > of which can be addressed as progress is made)
> >
> > 1) the contact area is flat
> > 2) the tool and mirror won't bend / flex
> > 3) the tool and mirror have uniform density
> > 4) the slurry is a uniform thickness
> >
> > I've created a picture that represents the problem I am trying to solve:
> > http://lerch.no-ip.com/atm/Pressure_per_unit_area.gif
> >
> > The image shows two situations.  The top situation has a rather straight 
>forward
> > solution, but things get complicated for the bottom situation.  Common 
>sense
> > says that the right side of the contact area should have a higher 
>pressure/mm
> > than the left side of the contact area.  I just don't know how to 
>calculate the
> > distribution..
>
>You will need to solve a few simultaneous integral equations.
>Off-list, I have sent a solution for a simplified problem, a
>cantilevered bar (not a disk).  It should demonstrate a way to
>proceed.
>
>This is not, in general, an easy problem.  "In general" means
>including the forces of the driving arm.
>
>Dave
>
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