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derek.s
Posts: 40
Registered: ‎02-21-2005
Message 1 of 3 (359 Views)

FEA confusion

359 Views, 2 Replies
05-25-2010 09:59 AM
I am trying to understand how fixed constraint works using Vector Components. It shows inches input, not ft/lbs. What I am seeing is if I put down 1 in in the Y insert and it shows the X and Z are unconstrained. What I am trying to is put on a fixed constraint to one direction which is Y. When I use 1 in for Y, it looks like it is using force of some kind. Can anyone explain how this Vector Components work?

Thanks,
Derek
Employee
henderh
Posts: 649
Registered: ‎06-07-2007
Message 2 of 3 (360 Views)

Re: FEA confusion

05-25-2010 11:59 AM in reply to: derek.s
Hi Derek,

The fixed constraint is used to remove all DOFs (translations and rotations) from the selected entities (faces, edges or vertices).

However, Fixed constraint can also induce a stress / load by applying a fixed displacement. This is the inverse of fixing a face and applying a load. For example, if you had a spring, fixed one end and applied a force on the other end, you can calculate the displacement based on the stiffness, etc. Conversely, you can calculate the force required to achieve a certain displacement (stretching or compressing) of the spring. This is one way of using vector components in the Fixed constraint.

If you use the vector components, you can define a displacement in x,y,z. This is done by using the checkbox(es) for the corresponding direction to enable the input. If the checkbox remains unchecked, this DOF remains free.

In your example, by using 1 inch in the y displacement, that means the face, edge, etc will move 1 inch in the positive y-direction. If there is another fixed constraint, this will induce stress in most cases.

Say that you have a simple block part. You completely fix one face (no vector components). Apply a tensile load to the face on the opposite side of the block from the fixed one. You will get a positive strain, positive (tensile) stress, and some displacement.

To archive the same stress result using fixed constraint, instead of applying the load, apply the displacement value result (from the applied load simulation) as a fixed displacement. You can further confirm this by RMB on the fixed constraint node in the browser and view the reaction forces and moments.

I Hope this helped clear things up. Please let us know if you have any additional questions, comments or suggestions.

Best regards, -Hugh (Autodesk)

[Edit: Perhaps what you're looking for is the use of the Frictionless constraint. If you'd like to restrict the displacement to primarily the y-direction, you could apply a frictionless constraint to faces that have normals perpendicular to the y-direction.] Edited by: henderh on May 25, 2010 9:05 PM


Hugh Henderson
Simulation QA Engineer
DLS
Contributor
salariua
Posts: 12
Registered: ‎08-05-2005
Message 3 of 3 (186 Views)

Re: FEA confusion

04-17-2013 02:09 AM in reply to: henderh

Hugh,

 

I was hoping for a different result. I have modeled a large assembly with coarse mesh to determine displacement as a general value and I was hopping I can apply the displacement values on a component modeled with a finer mesh.

 

Can this be achieved? I know I can have local mesh control but selecting multiple faces in a part (shrinkwrap of asm) is impossible without a window select feature not available yet in Inventor.

Regards,
ADS

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