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General questions about contact best practices, and trying to speed up simulations:

joshua.hill
#1) Is there a general convention of which body to make the i or j body in contact for best performance? i.e. the larger body should be i, smaller should be j.
 
#2) Regarding complex geometry, I noticed that it is suggested to split apart geometry to the region of contact. Does this mean that, say I have contact in an inner diameter of a larger part, that I should make the region around the contact area a separate part? i.e. make the single, larger part into two separate parts (connect with a fixed joint)? My understanding is that this is easier on the solver trying to locate points of contact
  • 0
    Hi Joshua,
     
    I am not aware of any benefits in selecting the larger body as the i-body. But the second note is definitely helpful if you have large and complex CAD files but you know that only part of the geometry is coming into contact. However, you do not need to create separate bodies after splitting up the CAD geometry. You can simply have two separate geometries under the same part BUT only reference the geometry of interest (e.g., the inner region in your example) in the contact object. Please note that the contact object references geometries rather than parts.
     
    I hope this helps.
     
    Thank you,
    Maziar Rostamian
    MSC Software
  • 0
    Hi Maziar,
    When I click on the link you provided, I get an "invalid page". Does the link still work for you?
     
    Chris
  • 0
    Hi Chris,
     
    It does work for me. I think I can help you with this issue. I will contact you offline.
     
    Thank you,
    Maziar
  • 0
    Some of the more important things I've learned over the years:
    • Download and install HertzWin. It helps you get a feel for the stiffness.
    • Between each pair of geometries, there is only one contact, at the point of maximum penetration. If you have complex surfaces that will contact at several points, only one point will have a force at a given time. And from this follows that a very small change in position or attitude can move the point of largest penetration from one area to another. This will cause an unstable model and solver problems. This is one case where breaking up a geometry in multiple smaller geometries really can help.
      • As a very simple example: Imagine a rubber brick on a table. In reality, the the block lays there very stable. You push at one corner and release and it is still stable. In Adams, maybe not. Pushing at one corner will move the point of biggest penetration to corner closest to the applied force. Releasing the force, the block will bounce up on that corner, then another corner will have the largest penetration. Cutting the block into four equal smaller blocks (can still belong to the same part), now have created a very stable system where you have four contact points when the block lies flat on the surface.
    • Be careful with units and exponent. Yes, the solver likes it better when the exp>1 and it is also agrees better with Hertzian contact theory. Exp=1.5 for a ball on flat surface for example. But your stiffness does not have the units of force/length, but force/length^exp. Read thus that I posted a few weeks ago:
    You have probably seen somewhere that a contact exponent = 2.2 is good for the solver. That typically works excellent if you work in millimeter. But in meter, that means that there is almost no contact force for small (<10 mm) penetration.
    Take for example a state where your penetration is 0.1 mm, about what you expect for a brake pad to disk.
    Modeling in mm, using k=1e5, you get a stiffness force of F=1e5*0.1^2.2 = 631 N. At the same state when model in meter, with k=1e8 the force becomes F=1e8*1e-4^2.2 = 0.158 N.
    Why does this happen? Because you believe that the unit for stiffness is F/L (N/mm or N/m) and convert accordingly.
    But the stiffness unit is really F/L^2.2 (N/mm^2.2 or N/m^2.2).
    To make it easier for reasoning, use the exponent = 2. Then you see that the stiffness is really N/mm^2 or N/m^2, i.e., stiffness per area unit, or pressure(!). and need to be converted that way.
    In your case, you should really use the stiffness k=1e5*1000^2.2 = 3.98e11.
    Then the force will be F=3.98e11*1e-4^2.2 = 1e5*1000^2.2*(0.1/1000)^2.2 = 1e5*1000^2.2 * 0.1^2.2 / 1000^2.2 = 1e5*0.1^2.2 = 621 N
     
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