I'm a new user of FEA and have gone through a few examples. I was trying to do an analysis on a part and am wondering why the scale on the contour view doesn't change when I changed the load. Whether the load is 10 or 500 the scale is from 1060 to 45.13. My first guess is that scale is a force multiplication factor, ie the max stress for a load of 10 is 1060*10 and the max stress at 500 is 1060*500. Is this a correct interpretation? Or am I doing something wrong (My account settings won't let me attach a model.)
Also, why does the animated deformation view go beyond the unloaded position? For example, I have loaded a beam in the -y direction. I would expect the animated view to simply toggle between the unloaded position and loaded position. However, the extremes in the animated view go from the deformed view to a position that seems to be as if the beam was loaded in the +y direction. Is this by design, or am I doing something wrong?
Finally, is there documentation on how to interpret the deformation scaling? A scale factor of 10% seems to produce a larger than expected deformation.
Dear Chris, 1.- If you change the load and run the solver again then results will change for sure!. In linear static analysis if for a load of 10 N you have a maximum vonMises stress of say 1060 MPa, if you edit the load an enter a value of 500 N, then stress results will change in the ratio of 1060 x 500/10 = 53000 MPa. This is simply linear static analysis, where the material is linear & static (not plasticity exist) and small displacement exist. If the stress results are bigger than the material yield stress, or large displacements effect exist, then the linear static analysis results are useless ... 2.- Click on POST-OPTIONS and in the menu you can set OFF "ANIMATION - Load and Unload". Animation is simply a way to see the deformed shape, choose the option that fits you better, but remember that is pure graphics, to see the trend of deformation.
3.- Click again on POST-OPTIONS. In top of the menu you have:
Scale Deformation: here you can specify the value of Scale deformation. Actual Deformation: When checked, actual deformation at scale 1:1 is being shown. When unchecked, scaled deformation is being shown.
Hi Blas, Thank you very much for the response and explanations. While I wish the contour scale would update with the max/min stresses of each analysis, I can make the calculation you outlined fairly easily to find the true stresses in the part. However, it is not so easy with the deformations. I am still having the issue that when I modify the load and rerun the analysis, the "old" deformation is still displayed. Deleting the analysis and creating a new one doesn't work either! Is there something in the procedure that I am missing? Using the example "Analyzing Buckling for a Bracket", if I went through all the steps, and at the end, wanted to change the load to 200, how would I go about that in such a way to have the deformed view show the deformation associated with the new load? Do I have to start over entirely? Also, is it possible to measure the distance between a point on the undeformed model and the deformed model (viewed with a 1:1 scale.) I see that you can measure the deformation between nodes on the deformed model, but I can't seem to find how to measure the translation of the deformed model relative to the original. For example, if you have a cantilevered beam under load, I'm trying to measure the deflection between a point at the end of the beam. In other words how do a measure the effect of the strains within the beam on a point at the end of the beam? I wish I could post attachments, I bet that would help! Either way, thanks for your help. Chris
Dear Chris, 1.- If the deformation is the same, not matter you apply 100 N or 1000 N, then you need to set correctly the postprocessing options. Click F6 Then on POSTPROCESSING play with the different options: _ 2.- Regarding meassuring distance between nodes, please note that resultant displacements is just the real deformation that experience a node in the model, then simply LIST displacements and you are done!. You will get TX,TY,TZ and rotations, as well as RESULTANT displacements. You have command TOOLS > MEASURE > DISTANCE BETWEEN NODES that allows to measure distance between deformed nodes:
And also you can update your FE model with the deformed shape of the structure based in the resultant displacements, go to "CUSTOM TOOLS > POSTPROCESSINGS > NODES MOVE BY DEFORM WITH OPTIONS". In summary, plenty of options ... And please note, do not get confused with STRAIN & DISPLACEMENTS: strains are unitless values, is just DISPLACEMENTS divided by length. Best regards, Blas.
I didn't have as much success with your previous post as I did the first! I've attached the file I am working with, if you wanted to see why I can't get the deformed view to change when I change the load. As for measuring the deflection, I did not explain myself very well. Now that I'm able to post attachments, I am trying to measure the distance between the circle points/nodes.
Dear Chris, If you activate ACTUAL DEFORMATION and then set SCALE DEFORMATION = 1 as explained in my first post, ie, real life deformation, then the deformed shape will vary according the applied load. The first picture is for loading FY=-1, then you get displacements = 0.158, and the second picture is the resultant displacement for loading FY=-10, then the resultant displacement will be 10 times bigger, ie, 1.58, and the deformed shape will change as well.
Please note this is LINEAR STATIC analysis, where displacements are proportional to loadings, and also small displacements is assumed. If resultant displaments are of order of the thickness of the element, then the problem behaviour should be considered as NONLINEAR. If I measure the thickness of the cantilever plate I note is 0.062 units, then applying a vertical load cause a displacement of 0.158, THIS MEANS THAT LINEAR STATIC ANALYSIS RESULST ARE MEANINGLESS!!, not accurate at all, understood??. At that loading level the problem behaves as nonlinear, at least by geometry, and real life displacements will be different as obtained running a linear static analysis, independtly of other "important" factors like mesh density, quality of mesh, or element type used. Best regards, Blas.
And regarding the distance value requestd, this is what you have comnputed by solving the linear static analysis, you get resultant displacements, that is the realtive displacements from the undeformed shape and deformed shape. Simply in the selector set NODES and activate SHOW TOOLTIPS, and pick in the node, then in the screen you will see displacement results.
Or list results, or use command LIST > OUTPUT > QUERY, etc... Best regards, Blas.
Thank you for your help (again!) I was following your instructions regarding displaying the actual deformation, however I wasn't activating the right result set. When I saw your screen shot, I saw that you had exploded the results tab and realized that each analysis creates a result set (even when I delete and redo the analysis with different constraints.)
For some reason I associated nonlinear analysis with plastic deformation, knowing that my part wasn't subject to that, I didn't think that it applied. Your explanation of displacement vs part thickness makes perfect sense though, so I will work through the nonlinear analysis example.
Even though the results are meaningless, I did have a question on how to constrain the part. First off, I have divided the part from the original design which was:
The part is supposed to be a handheld plastic clip that is squeeze as shown by the red arrows, and I am having a hard time constraining it correctly. I originally placed a fixed constraint along the symmetry plane, but these results had no deformation there, when in reality it should be significant. I think I need to to constrain the neutral axis of the part where it intersects with the XZ plane in the X and Y directions (the neutral axis is not at the central axis of the cross section.) Even though it won't be accurate, I selected all the nodes on the interior of this surface and fixed them as a test, but the analysis failed.
Do you have any advice on how to properly constrain this part?
Dear Chris, You have the following options: 1.- PLANE STRAIN This is a simply 2-D SOLID PLAIN STRAIN problem, then forgot at all to use "horrible" tetraedral meshing, you can mesh with QUAD 2-D plain strain elements using say four elements in the thickness of the plate. You can study any internal cross section of the plate, assuming you are far away of the edges. Remember when selecting the PLANE STRAIN property type to click on FORMULATION and change the NASTRAN formulation to CPLSTN4, if not the plate elemenmts will remains as SHELL CQUAD4. Alternatively, after meshing you can issue MODIFY > UPDATE ELEMENTS > FORMULATION. Plane strain elements are two-dimensional elements with membrane only stiffness and in-plane loading. The plane strain element idealization has zero strain in the thickness direction. These elements represent structures that are very thick relative to their lateral dimensions. The entries CPLSTN4, CPLSTN8, CPLSTN3, and CPLSTN6 are plane strain elements.
They must be defined in either the XZ or the XY plane of the basic coordinate system. They must be in the XZ plane for SOL 601. They are supported in all linear solutions expect cyclic solutions 114, 115 ,116, 118 and aero solutions 144, 145, 146. They are supported in the nonlinear solution 601. They behave linearly in solutions 106 and 129. All loads must be in-plane. The PLOADE1 entry can be used to define edge loads. Properties are defined on the PPLANE entry when linear materials are applied, or the PLPLANE entry for hyperelastic (SOL 601 only).
In the edge TY=0 (symmetry constraint), and also TX=0. By default, DOF TZ=0 in the full model (this is a 2-D problem).
Regarding loading, if you define a thickness of the element = 1.0, then loading will be per element thickness, ie, if the total load in the 3-D model is FY=-1.0 and the thickness of the solid = 2.5, then the load to prescribe to the 2-D model would be FY=-1/2.5=0.4.
2.- 3-D SOLID Alternatively, you can study the problem in 3-D using HEXAEDRAL elements CHEXA 8-nodes. Split the geometry to the half in the Z-axis (you have symmetry of loads & geometry) and prescribe TZ=0 in the cutting plane. Use a load FY=-0.5, and use the same constrainst as the previous case (TX=TY=0 in the surface). Results will be comparable between 2-D & 3-D models, but 2-D one is very fast!!