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-   -   Find where force is applied (http://www.cfd-online.com/Forums/openfoam/107379-find-where-force-applied.html)

julien.decharentenay September 25, 2012 20:16

Find where force is applied
 
Hi,

I am using simpleFoam (of v2.1.1). I am trying to find where the forces are applied and have been outputing the forces and moments using the following in the controlDict:
functions
(
forces
{
type forces;
functionObjectLibs ( "libforces.so" );
outputControl timeStep;
outputInterval 1;
patches ( "Volume_0_Wheel-1" "Volume_0_Wheel-2" "Volume_0_Wheel-3" "Volume_0_Wheel-4" "Volume_0_Car" );
rhoName rhoInf;
rhoInf 1.205;
pName p;
UName U;
log true;
CofR (0.0 0.0 0.0);
}
);

It worked great and I got all the outputs in the forces\0\forces.dat file:

# Time forces( pressure, viscous ) moment( pressure, viscous )
997 ( ( ( -25.6521 1575.07 -2489 ) ( 0.0440792 73.7995 8.74441 ) ) ( ( -5215.72 15.0189 86.2846 ) ( -15.0971 -0.0198099 -0.124274 ) ) )
998 ( ( ( -25.6488 1575.07 -2488.99 ) ( 0.0442406 73.8003 8.74387 ) ) ( ( -5215.69 15.0191 86.2908 ) ( -15.0984 -0.0198765 -0.124476 ) ) )
999 ( ( ( -25.6506 1575.07 -2489.01 ) ( 0.0442343 73.8004 8.74362 ) ) ( ( -5215.72 15.0171 86.289 ) ( -15.0978 -0.02023 -0.124946 ) ) )

My problem is that I can't work out how to derive the x,y,z coordinates where the forces are applied to obtain the moment. I have been struggling for a couple of hours and have to admit that geometry is one of my many weak points... Any help would be appreciated.

Thanks in advance
Julien

colinB September 26, 2012 02:53

Dear Julien,

I hope I got your question right.
But actually you posted your answer already:

Quote:

...
CofR (0.0 0.0 0.0);
...
In my 'old' files (from OF 1.7.x) this was the reference point
used for the calculations of the moments
So in your case you chose the origin.

I hope I could contribute
regards
Colin

julien.decharentenay September 26, 2012 06:46

Hi Colin,

Thanks a lot for your answer. I probably need to clarify a bit. I am trying to know the distance - or vector - that define where the force is applied to generate the moment - I am just after the distance in the y-direction.

Let me know if it is unclear.

julien

sail September 28, 2012 08:38

you have the moments referred to a defined point, in this case 0 0 0.

moment is force time distance.

to find the distance from your reference ponint do moment/force

julien.decharentenay September 30, 2012 20:53

Hi,

Thanks a lot for your answer and comments.

My understanding is that:
- Forces are the integral over surface of p*Face_normal;
- Moment are the integral over the surface of p*Vector(CofR,Centroid)^Face_normal.

Based on the above, I was trying to inverse a system to get the y coordinate:
Mx = yFz - zFy
My = zFx - xFz
Mz = xFy - yFx

Just doing y = Mx/Fz was neglecting the effect of drag on the x-component of moment. I am unsure if the above is correct as it implies that MxFx + MyFy + MzFz = 0 (ie the moment is to be perpendicular to the Force), but when I calculate the expression based on the output from openFoam this is not true. So I am assuming that my interpretation is incorrect.

I am using a "work-around" using ParaView (http://www.cfd-online.com/Forums/ope...-parafoam.html) where I construct two expressions:

1) The forces: surface integral of p*Normals
2) The x moment of the z force: surface integral of coordsY*Forces_Z
The center of application of the z force is calculated as (2)/(1).

I noticed some differences in the forces calculated (order of 0.2% so negligeable). This seems to be doing what I need.

Kind regards,
Julien


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