when I am trying to do an NX Nastran Analysis using user Defined symmetric boundary condition it fails. But when I use the same Boundary condition in I-DEAS, I can get the output. when I check the f06 file it looks there is something missing in constraint. Please refer to the attached simulation file and advise on this.
Dear Logesh, This is not a question of using NX NASTRAN or I-DEAS solver, simply you have a rigid body motion in your model, it is free to move in the Z-axis!!, not any constraint defined, then if I-DEAS is able to solve the model ... wrong!!, revise your input ...
Please be aware that you are using 3-D solid CHEXA 8-nodos elements, then not rotations available, then to define symmetry constraints in one plane (for instance, Z-Y plane) simply do TX=0, the same for the others, etc..
Also, do not create LOCAL coordinate systems to apply constraints, better use global coordinate systems, try to do things as much simply as possible.
Thanks a lot for your advise.... I have create two boundary condition as shown in the image file using global coordinates. Is it correct? As Tet mess got Rotational DOF, If I use it, Is the same BC hold's good? please advise.
Dear Logesh, CTETRA elements as well as any other solid 3-D elements do not have rotational DOF, then is useless to constraint rotational DOF on solid elements, the NX NASTRAN solver will remove automatically ALL rotational DOF when solving the model, the PARAM,AUTOSPC,YES is included in the nastran input file.
Then please remember you need ALWAYS to constraint in any way the X, Y & Z translational directions in your model in order to avoid rigid body movements ...
Dear Logesh, No, I-DEAS is not wrong, is using the "general" rule when defining symmetry constraints. This is valid for ANY element, independently of the number of DOF of the element, and you can run this way and never will make things wrong.
For instance, if you want to apply a symmetry boundary condition to a Y-Z plane the rule is to constraint translational displacement in X-axis (normal to Y-Z plane) and rotational displacements in Y and Z axis (rotation in Y-Z plane is not permitted).
Now to your problem: 3-D solids elements do not have rotational DOF, then is useless to constraints rotations, you can do it, not problem, the results will be exactly the same, what I try to explain you is the reason why you can ignore rotations, OK?.
Of course, use the "general rule" when meshing with SHELL or BEAM elements, these are elements with six DOF!!. As you see in FEA, it is very importat to know very well the capabiities and features of the finite element library that support the solver!!.
Good Day. Belated Merry Christmas and Happy new year. I need some advice on the Linear static Analysis. I have done a setup on NX7.5 for Linear static analysis for a Thin ( 0.4mm Thick ) Plastic tray with a load of 5 N at the center and with a with a fixed boundary condition at the bottom surface ( Assuming that the stress strain relationship is linear). I have done the mess with a element size of 1.05mm. The Displacement result which i have got is 7.2mm. But my counter part who have done the physical test with 5N load found the Displacement is only 2.95mm. They also tested with Cosmos Express and got a result of 3.6mm. Could you please help me check weather my Pre processing setup is correct or not? The file is 16MB, So I have uploaded to GTAC scratch Area. The file name is Nastran.zip
Please advise on this Issue
Dear Logesh, I see the model, as you say the thickness is 0.4 mm, and global dimensions around 310x130mm, then ..... this is all but LINEAR, your results are meaningless!!.
Please note when you solve a model and displacement results are equal or higher than thickness, then this is a nonlinear problem "by the geometry", you need to solve the problem as nonlinear because the relation between load & displacement is all but linear, surely you will experience geometric nonlinearities behavior like stress softening or stress stiffening, depending how load is acting in compression or tension.
Regarding mesh, please note: if you mesh with a coarse mesh, your model will be stiffer than in reality, that means the displacements results will be smaller than real ones. If you refine mesh, you will realize that displacements will increase.
Said the above, please note your problem is not a "toy": you have a complex geometry, is not a simply plate, is not a thick solid, but a combination of two, then 3-D solid meshing is required + nonlinear analysis activation of "large displacements" effect.
Please note SOLID 3-D meshing required TO MESH with at least two elements in the thickness, three elements preferred!!!. Then you will have millions of nodes & elements (Esize=0.2 mm), not possible to afford for nonlinear analysis unless you have a powerful workstation with many, many GB of RAM memory, then please consider to study 1/4 problem -- Good luck!!.
Good Day. Thanks a lot for your valuable feedback. For the 3D Tet mesh the lowest i have tried is 0.8 mm on this part which has generated around 1.2 million Elements. I could not refine below as my workstation resources is not enough to handle this case. The statement which you have given on Mess size to deflection value is absolutely true. When I use 0.8 mm mesh size with Linear static the deflection value is 7.2mm. If I use 3mm Mesh size the deflection value is 4.8mm. ( Now I can guess why Cosmos give so lower value ( Almost nearer to Actual test) it is only 3.6 mm, As they use 5mm Mesh size).
So taking your advised I intended to go for Non Linear Study. I will simplify the model with mid surface model and apply 2D mesh. For this Geometric Non linear study Do i need to use NLSTATIC 106 or NLSTATIC-Multi constraint ADVNL 601,106, or ADVNL 601, 129?
Please help me to give a simple clarification on the difference between this solution types.