
[Sponsors] 
September 25, 2012, 20:16 
Find where force is applied

#1 
Senior Member

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: functionsIt 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
__________________
 Julien de Charentenay 

September 26, 2012, 02:53 

#2  
Senior Member
Join Date: Aug 2010
Location: Groningen, The Netherlands
Posts: 216
Rep Power: 9 
Dear Julien,
I hope I got your question right. But actually you posted your answer already: Quote:
used for the calculations of the moments So in your case you chose the origin. I hope I could contribute regards Colin 

September 26, 2012, 06:46 

#3 
Senior Member

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 ydirection. Let me know if it is unclear. julien
__________________
 Julien de Charentenay 

September 28, 2012, 08:38 

#4 
Senior Member
Vieri Abolaffio
Join Date: Jul 2010
Location: Always on the move.
Posts: 308
Rep Power: 7 
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 

September 30, 2012, 20:53 

#5 
Senior Member

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 xcomponent 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 "workaround" using ParaView (http://www.cfdonline.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
__________________
 Julien de Charentenay Last edited by julien.decharentenay; September 30, 2012 at 21:20. Reason: Just wanted to clarify. 

Thread Tools  
Display Modes  


Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
OpenFOAM 1.6ext git installation on Ubuntu 11.10 x64  Attesz  OpenFOAM Installation  45  January 13, 2012 13:38 
Problem Building OF on Centos cluster (no admin rights)  CKH  OpenFOAM Installation  5  November 13, 2011 07:32 
Timeaveraged force in CFX  rjmcsherry  CFX  2  October 21, 2010 10:34 
Body force at the cell face  Souviktor  Fluent UDF and Scheme Programming  0  March 31, 2009 08:54 
plotting force magnitude  force  CDadapco  6  July 11, 2008 04:19 