Using Calculix to model a rotatable joint

gwillisgwillis
I'm hoping someone can help me with modelling a rotatable joint in calculix. Imagine 2 bars bolted together in a cross shape, with the bars free to rotate on the bolt. That is the type of joint I am trying to model. Anything I have tried seems to give the same result as for a rigid joint. Thanks in advance. I hope this is a suitable forum to ask this question.

Comments

  • MishalMishal
    Hi gwillis

    Can you please show an image or a model as it is difficult to tell what you are aiming for. A cross shape for two bars would have 5 possible locations for bolting which of these should be free to rotate?
  • disladisla
    edited April 2021
    Hi,

    I would draw independent beams not sharing any node and impose the link through constrain equations on the vertical displacement for example.

    See attached file. 1 uz3 - 1 uz8 = 0

    Where 3 and 8 are the corresponding nodes at the connection point.

    This way you can impose to have the same rotation or displacement separatley of the pair of nodes . Be careful because the internal solver does not take torsion into account when computing stresses.

    Each solver has its own limitations regarding beam shape or further data extraction.

    ccx do not have equation to impose nodes same rotation.

    Victor suggest using internal solver .
  • gwillisgwillis
    edited April 2021
    Thanks for your comments. The simplest case I am examining is 2 poles connected with a bar between. The connections A and B can be bolted as a rigid connection, or bolted so the bar and pole can rotate relative to each other. I hope that is clear. I am assessing deflection at pole tip when a load is applied to the tip.image
  • gwillisgwillis
    edited April 2021

    A real example
  • gwillisgwillis
    image
  • disladisla
    Find example with arbitrary loads on the beam and free to rotate.
    Just need to keep in mind that Pinned means same node displacement and rigid same displacement plus same rotation.
    In this case, for large deflections I would go to nonlinear analisys which I'm not sure if it accepts this kind of constrain ecuations.
  • gwillisgwillis
    edited April 2021
    Thanks, I'll see if I can understand what your example file is doing.
  • disladisla
    It is a Mecway file .

    The file has 21 nodes so you can run it in Mecway I I'm not wrong.

    Victor can confirm that.
  • JohnMJohnM
    This might not be what you are looking for, but once you realize that solid elements do not have rotational degrees of freedom, merging two nodes across solid components makes a pin, which can be handy in some applications.



  • gwillisgwillis
    edited April 2021
    disla, I was able to open your file in mecway then save it as an abaqas file so I could examine it (I am using calculix because I need to programmatically generate the input file). Have I got this right? The equations describe a relationship between 2 nodes, in your sample file you are saying the displacements in x,y,z at those 2 nodes are the same - eg one node moves in +x direction meaning the connected node moves the same distance in the -x direction.

    If the other degrees of freedom are not specified (4-6? rotations around x,y,z axes?) as you have done, does that mean the members are free to rotate ie pinned? Just want to make sure I understand what the different specifications mean.

    JohnM, thanks for your sample. I have not had a chance to look at it yet.

    Thanks for all your help. This is great. I have been stuck on this for a while.
  • gwillisgwillis
    JohnM, thanks for your example. I am pretty new to FEA. I am using beams to model the poles and cross bar so I think your example is not applicable?
  • disladisla
    Not sure I understand your question,.. :/

    If two beams share a node, the software automatically consider the connection as rigid. (Same displacement and same rotation of both beams at that point)

    If you separate the beams without any node in common , you are more free to define the relationship between them by means of a pair of nodes that are close starting from completely independent (0 equations in common) , some degree of displacements in common (my example) up to finishing with all three rotations related (rigid connection like if they were fully welded). You are free to define the connection completely.

    You can also use "contact" but that’s another subject.

    Regarding your question , nodes 4 and 6 are not connected with anything. They are just part of the beam and will move and rotate in the space as far as the stiffness of the beam allow them.

    If you want to fix the displacement or rotation of a node with respect the exterior it can be done using the BC and imposing displacement and/or rotations =0 . By default, the software does not impose any BC with the exterior (pole base for example, nodes 1 and 2 has been set by me as clamped to the ground).

  • disladisla
    edited April 2021
    One thing that it is important gwillis,

    When you say: "If the other degrees of freedom are not specified (4-6? rotations around x,y,z axes?) as you have done...."

    -I have not specify anything around x,y,z axes. I have stablished a relationship between the degrees of freedom of two nodes. In other words, I have fixed their distance not to change. They must move together but they still can rotate one with respect the other. That means pinned connection.

    - If you impose same rotation, you will make the connection rigid.

    -Pined or rigid , there is no restriction on how the pair of nodes will move or rotate in the space. You should differentiate.

    This is kind of like they are married. Do what you want, but you are forced to do it together. :'( :D
  • gwillisgwillis
    Thanks, I'll read over that and see how I go.
  • gwillisgwillis
    I think I understand it now, thanks for your help. If I have more questions I'll ask again!

    Maybe if more marriages worked together society would be going better...
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