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boundary integration in mechanics module
Posted 2010/06/11 8:43 GMT-4 4 Replies
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Hi all,
I'm wondering why in boundary integration with "predifined quantities" = "displecement" I have "Unit of integral" = "m^3" ?!?
As I understand it should be "m".
I tried to apply prescribed displacemebt on my sample and than I verified this displacemet using boundary integration... and I didn't get the same displacement...
Please tell me how this boundary integration works and why I have there "m^3"
Thanks in advance,
Alex.
I'm wondering why in boundary integration with "predifined quantities" = "displecement" I have "Unit of integral" = "m^3" ?!?
As I understand it should be "m".
I tried to apply prescribed displacemebt on my sample and than I verified this displacemet using boundary integration... and I didn't get the same displacement...
Please tell me how this boundary integration works and why I have there "m^3"
Thanks in advance,
Alex.
4 Replies Last Post 2010/06/21 13:04 GMT-4
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Posted:
1 decade ago
Hi
an integration
int(f(x,y))*dx*dy)
results in the product of the units of f(x,y), dx and dy (but in COMSOL the dx*dy are implicit, not shown, but still there)
as if dx[m], dy[m] and f(x,y)[m] you end up with [m^3].
if you are looking for the "average" then you must divide the results by the volume/surface/length (3D,2D,1D respectively) and you find back your units of f() (these are the integration of the value "1" one)
There are some exception though, not fully clear why
- in V3.5 the integration coupling variables do not port their units to the results, ending in "red" flags,
- in V4 if you ask for int(1)*dxdy its result is given without units, but it should in my view have the units of a surface or [m^2], this might results in some "red flags too" for the unit conversions later on
Hope I answered your question
Have fun Comsoling
Ivar
an integration
int(f(x,y))*dx*dy)
results in the product of the units of f(x,y), dx and dy (but in COMSOL the dx*dy are implicit, not shown, but still there)
as if dx[m], dy[m] and f(x,y)[m] you end up with [m^3].
if you are looking for the "average" then you must divide the results by the volume/surface/length (3D,2D,1D respectively) and you find back your units of f() (these are the integration of the value "1" one)
There are some exception though, not fully clear why
- in V3.5 the integration coupling variables do not port their units to the results, ending in "red" flags,
- in V4 if you ask for int(1)*dxdy its result is given without units, but it should in my view have the units of a surface or [m^2], this might results in some "red flags too" for the unit conversions later on
Hope I answered your question
Have fun Comsoling
Ivar
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Posted:
1 decade ago
Hi Ivar,
ok, but how about "reaction force" in boundary integration then?
Unit of integral in this case is [N] and not [N*m^2] so I'm confused..
If I need value of total reaction force wich is acting on my boundary... should I divide the results by the volume/surface/length or not???
Best regards,
Alex
ok, but how about "reaction force" in boundary integration then?
Unit of integral in this case is [N] and not [N*m^2] so I'm confused..
If I need value of total reaction force wich is acting on my boundary... should I divide the results by the volume/surface/length or not???
Best regards,
Alex
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Posted:
1 decade ago
Hi,
in order to obtain a force you integrate the pressure across your boundary.
Regards
Edgar
in order to obtain a force you integrate the pressure across your boundary.
Regards
Edgar
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Posted:
1 decade ago
Hi
You should check the different items/vaiables, some are global (integration aleady done) and some need to be integrated over the length/surface ...
That's when I find the units useful (apart that in V3.5 the integration coupling variables do not pass the units, and in V4 there are still a few cases not passing the variables correctly, these are to be fixed on next release have I been told)
So if you end up with a N*m^2 as result, then you need to divide by a Area=int(1) over the boundary, and in fact you want the average and not the integration ("aveop#()" operator in V4)
Hope this helps
Ivar
You should check the different items/vaiables, some are global (integration aleady done) and some need to be integrated over the length/surface ...
That's when I find the units useful (apart that in V3.5 the integration coupling variables do not pass the units, and in V4 there are still a few cases not passing the variables correctly, these are to be fixed on next release have I been told)
So if you end up with a N*m^2 as result, then you need to divide by a Area=int(1) over the boundary, and in fact you want the average and not the integration ("aveop#()" operator in V4)
Hope this helps
Ivar