Hi,
how can I associate cartersian coordinate 1 2 3
to that of the imported model 3d ?
and in the Post processing displacments are in mm ?
it seems a wrong result.....
help please.
Wrong (?) results
Re: Wrong (?) results
HI,
the first question i don't really understand.
As for the second, ElmerPost makes no unit conversions but displays the input file values asis.
ElmerSolver does not know much about units either, but expects a consistent set of units to be
used in the mesh, material & other input parameters.
Regards, Juha
the first question i don't really understand.
As for the second, ElmerPost makes no unit conversions but displays the input file values asis.
ElmerSolver does not know much about units either, but expects a consistent set of units to be
used in the mesh, material & other input parameters.
Regards, Juha

 Posts: 72
 Joined: 10 Nov 2009, 19:17
Re: Wrong (?) results
As to the first question, I think the answer that you're looking for is that coordinates 1,2 and 3 relate to x, y, and z.
Verification example
Hello all,
I tried to calculate displacement on the end of beam with geometry 0,1 x 0,1 x 1 m. I defined the boundary displacement on the first end (wall) 0 0 0. On the second end is force in y direction 1000 N.
By theoretical equation is displacement y = ( F * L^3 ) / (3 * E * J ).
As a material I choose carbon steel with E = 2.00e+11 Pa, J is for rectangular profile b * h^3. In my example b = h and J is a^4. So J = 1e04 m^4. After calculation y = 0.0002 meter.
I created model of beam in Salome (Beam of size 100 x 100 x 1000, this was reason for using menu Operation  Transformation  Scale Transform  Scale Factor 0,001 to get proper size of beam model). Mesh consists of quadrangle elements for obtain good results. Then I exported it via unv file into Elmer ( change . to , in unv file was necessary, I 'm using Ubuntu 10.04).
For check of proper size of beam I used in Elmers menu Model  Summary, where I find
BOUNDING BOX
Xcoordinate: [ 0 , 0.1 ]
Ycoordinate: [ 0 , 0.1 ]
Zcoordinate: [ 0 , 1 ]
It seems to be correct in SI units. After choosing Steel Carbon as a material of beam, which has Youngs modul 200e9, poisson ratio 0.285 I defined force and displacement.
After calculation Elmer gave that maximum displacement in y direction is 1,69e06.
I am confused. Everything what I defined to Elmer was in meters, newtons  SI units. And differences between theoretical equation and Elmer is in order of 100.
What I am doing in the wrong way?
I giving my beam as a Elmer project into attachment.
Have anyone already compared results from linear static with empiric equations?
Have a nice day.
Karel
I tried to calculate displacement on the end of beam with geometry 0,1 x 0,1 x 1 m. I defined the boundary displacement on the first end (wall) 0 0 0. On the second end is force in y direction 1000 N.
By theoretical equation is displacement y = ( F * L^3 ) / (3 * E * J ).
As a material I choose carbon steel with E = 2.00e+11 Pa, J is for rectangular profile b * h^3. In my example b = h and J is a^4. So J = 1e04 m^4. After calculation y = 0.0002 meter.
I created model of beam in Salome (Beam of size 100 x 100 x 1000, this was reason for using menu Operation  Transformation  Scale Transform  Scale Factor 0,001 to get proper size of beam model). Mesh consists of quadrangle elements for obtain good results. Then I exported it via unv file into Elmer ( change . to , in unv file was necessary, I 'm using Ubuntu 10.04).
For check of proper size of beam I used in Elmers menu Model  Summary, where I find
BOUNDING BOX
Xcoordinate: [ 0 , 0.1 ]
Ycoordinate: [ 0 , 0.1 ]
Zcoordinate: [ 0 , 1 ]
It seems to be correct in SI units. After choosing Steel Carbon as a material of beam, which has Youngs modul 200e9, poisson ratio 0.285 I defined force and displacement.
After calculation Elmer gave that maximum displacement in y direction is 1,69e06.
I am confused. Everything what I defined to Elmer was in meters, newtons  SI units. And differences between theoretical equation and Elmer is in order of 100.
What I am doing in the wrong way?
I giving my beam as a Elmer project into attachment.
Have anyone already compared results from linear static with empiric equations?
Have a nice day.
Karel
 Attachments

 beam.tar.gz
 beam
 (263.97 KiB) Downloaded 188 times

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Re: Wrong (?) results
Hi Karel
The SI unit of "Force 2" should be [F]/[A]=Pa, not N. The area of the beam end is 0.01 which should give you the factor ~100 compared to teh theoretical results.
Unfortunately the keywords do not always have the most natural interpretations. Pressure as a keyword was already reserved by the NavierStokes equation so Force was chosen for the Navier's equation.
Peter
The SI unit of "Force 2" should be [F]/[A]=Pa, not N. The area of the beam end is 0.01 which should give you the factor ~100 compared to teh theoretical results.
Unfortunately the keywords do not always have the most natural interpretations. Pressure as a keyword was already reserved by the NavierStokes equation so Force was chosen for the Navier's equation.
Peter
Re: Wrong (?) results
Than you very much Peter. It is very helpfull. Now I understand to definiction. Now it is clear. It will be more usefull to define NODAL FORCES. Menu for NODAL FORCES is not defined in standart Elmer graphical menu, isn't it? It must be defined via text script, as I seen in forum. It is big disadvantage that Elmer have in LINEAR STATIC no menu for NODAL FORCE and have no possibility to define NODAL FORCES to choosen NODE in graphical menu. Thank you one more time.
Karel
Karel

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 Posts: 3498
 Joined: 22 Aug 2009, 11:57
 Antispam: Yes
 Location: Espoo, Finland
 Contact:
Re: Wrong (?) results
Hi Karel
Elmer menu structure is fully programmable so you can add the nodal load to the menus even by yourself. Just edit the corresponding .xml file in edf directory and add "varname Load" to boundary condition menu definition.
Peter
Elmer menu structure is fully programmable so you can add the nodal load to the menus even by yourself. Just edit the corresponding .xml file in edf directory and add "varname Load" to boundary condition menu definition.
Peter
Re: Wrong (?) results
Great, thank you. I will try it.
Karel
Karel