Enforced Displacement 3

[New2FE]

Hello anyone Im new to FEA and I am trying to work a few things out.

I understand that when you know what your applied loads are, and F=KX, the displacement can be calculated using KinverseF=X applied to a reduced stiffness matrix (where zero displacements are 'eliminated' from the stiffness matrix). Then once you know the displacements you calculate F=KX on the full stiffness matrix to get the reactions.

So how does one go about workind out what the forces are if you enforce a displacement. For example I have a simple cantilever with an enforced displacement at the tip.

 

[ESPcomposites]

You can partition the matrix into two separate matrices. You separate the two matrices by the ones with enforced displacement DOF and ones that do not. For the matrix with the enforced displacements, you can solve for the associated forces. You then enter those forces back into the other partitioned matrix to solve for the remaining displacements. You should now have all displacements plus the forces that it took to create the enforced displacement.

Do you have a FEM book? It may have an example.

Brian

http://www.espcomposites.com

 

[Elabbasi]

One way is to partition the matrix as Brian mentioned, but solve first the part without enforced displacements and then you will have all displacements. Then substitute in the enforced displacement part and get reaction forces. Oh, and when you partition remember that the "F" vector of the non-enforced part will be changed due to the enforced displacements values.

You can also keep the matrix as is, and add Lagrange Multipliers that enforce the displacements, and solve the expanded system of equations (for unknown displacements + Lagrange Multipliers). The multipliers are the reaction forces. No matrix partitioning there, but a few more equations to solve for.

You'll find details/examples in most FEM books.


Nagi Elabbasi

http://www.veryst.com

 

[New2FE]

Hello,

unfortutely my book doesnt have an enforced displacement example. I tried partitioning the matrix as you described, but I dont seem to get the right answer.

I am doing it wrong, I think, is it right to remove rows and columns associated with DOF that have no enforced displacement?

In my cantilever, one end of the the beam has zero displacement and rotation, while the other end has a negative displacement (of say for example - 0.1659mm) ie bending downwards. I am trying to find the force that would produce this displacement using this method (I know the load should be -35000N based on E = 200e11, length = 1.6m and I =1.44e-6m^4)

 

[New2FE]

Here attached is a spreadsheet I made to for this problem.
The first worksheet "Displacements" uses excel to invert the assembled stiffness matrix and solve for given load (-35000N).

I remove column and rows by replacing the stiffness terms (for zero displacement DOF) with zeros except for the diagonal, where it is =1. I highligted my displacement at the tip in yellow, and this is the value I want to use in the next problem, on the worksheet "Forces".

In that worksheet, I have my displacements of 0 at the end and -0.1659m at the tip. However when I multiply the reduced stiffness matrix by the known displacements I get a force too big (highlighted red. Obviosuly something is really worng in what im doing, what should my partitioned matrix look like?

 

[rb1957]

this sounds a lot like homework ... but it's good to solve FE by hand (so you know what the black box is doing).

on "displacements", rows 21-27 don't look right, the next matrix looks right, 'cause you're accounting for the reactions. not sure about yor sign convention, should the moment at the reaction be -ve ? (draw a free body diagram, solve the problem by hand to see if you're close, and to see the inaccuracy due to modelling the beam with only two elements ... maybe it's significant, maybe it isn't)

on "forces" ... i'm not sure what you're doing with the "reduced matrix" is valid. compare with the matricies on the 1st sheet.

also not particularly happy with what you're doing to the 1st couple of rows and columns. yes they don't affect the solution ('cause they're associated with zero displacement) but changing they should mess with the matrix inversion making it poorly conditioned (the matrix elements are different by 6 orders of magnitude).

 

[rb1957]

i wonder why your marix doesn't invert ?

 

[rb1957]

that'd be your element matricies ... rows 2-19.

rows 22-27 inverts.

i think your reduced matrix should be 4x4.

 

[New2FE]

Hi!

OK I worked it out. The modification to the force vector was tripping me up. Attached is my final spreadsheet. I was trying to basically represent the same problem two ways: applying a load to one end, or by enforcing a displacement that achives the same reactions.

The spreadsheet isnt very tidy but basically the shaded blue cells can be changed for different loads and boundary conditions.

It was just an exercise to try replicate a beam I've modelled in Nastran, I was just trying to replicate the results, which they now do thankfuly. My coordinate system is with rotations positive anticlockwise (ie pointing out of the screen)

Anyway thanks for the help guys.

 

[ESPcomposites]

Good job. I can see how this may not be immediately obvious, but it is a good example problem that will help you to better understand the "black box".

It should also force you to think about how DOF's are applied and perhaps to better understand the mechanics.

Side note: Some time ago, I became reasonably knowledgeable about FEM theory, but still did not have a solid enough mechanics background. So while I understood the black box, I did not necessarily understand the actual engineering problem. In my opinion, that is usually the greater cause of poor FEM results. In fact, it would seem that a good amount of posts on the FEM board are actually mechanics questions posed as FEM questions. This usually implies a situation like what I went through, which is fairly common in today's computer based engineering environment.

Brian

http://www.espcomposites.com