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Two Questions - Surface Contact and Displacement Stress

12 REPLIES 12
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Message 1 of 13
tom
Participant
101432 Views, 12 Replies

Two Questions - Surface Contact and Displacement Stress

Hi there! I have two questions regarding boundary conditions and loading when using Autodesk Simulation.

 

1. I primarily design components that rest on the top surface of a piece of machinery and can end up being rather complex. Due to my assemblies being located on the top surface, I've been creating large solid fixed blocks below my part and using surface contact between them to prevent negative Y movements. However, this is computationally heavy and it has led me to wonder if there is such a thing as one-directional boundary conditions. In other words, is there a way to fix the bottom surface of my machinery to be free to move in all directions except in the negative Y direction?

 

2. If I know how much displacement is applied to a part, can Simulation calculate the resulting stress? For example, if I have a cantilevered beam that I know has deflected 1" on the free end, is that a load case I can apply and have displayed resulting stress?

 

Many thanks!

12 REPLIES 12
Message 2 of 13
John_Holtz
in reply to: tom

Hi Tom,

 

We do not have one-way boundary conditions yet, but we are discussing it for a future release.

 

One thing you can do to reduce the analysis size is to fix all of the nodes in the fixed blocks. The fix block can be as thin as reasonable; it will not deflect if all of the nodes are fixed.

 

You did not mention whether you were doing a Linear static stress analysis or an MES analysis.

  • If doing a linear stress analysis, there is not much you can do conveniently to make your own one-way constraints. If you want a manual procedure involving drawing gap elements, let us know.
  • If doing an MES analysis, and for all of the fixed blocks that are at the same elevation, draw one big element (on a new part number) that replaces the fixed blocks. (After meshing the model, just use the "Draw > Draw > Lines" command and connect the 8 corners. Then suppress the fixed blocks that came from the CAD model.) The surface-to-surface contact between the machine and the one element is sufficient.

For question 2, you can apply a prescribed displacement ("Setup > Constraints > Prescribed Displacement"). These are available for both Linear and MES.

 



John Holtz, P.E.

Global Product Support
Autodesk, Inc.


If not provided already, be sure to indicate the version of Inventor Nastran you are using!

"The knowledge you seek is at knowledge.autodesk.com" - Confucius 😉
Message 3 of 13
tom
Participant
in reply to: John_Holtz

Thanks John!

 

As far as the one-way boundary conditions go, if I could put my vote in, I'd say you should make that a priority for the next release. 🙂

 

For the Displacement initiated stress, I think I need a bit more guidance. I've attached a simplified version of my file if it helps. Essentially what's happening is I fix one end (cantilevered) and I put a presribed displacement on the other of .5" and a steel stiffness (30e6). Yet, the results show zero deflection or load. Can you help point me in the right direction?

 

Thanks! 

Message 4 of 13
John_Holtz
in reply to: tom

Hi Tom,

 

You made the same mistake that I make all of the time: you forgot to attach the file :-). Or maybe it was too large to attach to the forum. No problem. My guess is that your analysis type is Linear Static Stress. The prescribe displacements (displaced boundary elements in older versions) are multiplied by an input, but the default value is zero.

 

Go to "Setup > Model Setup > Parameters" ("Analysis > Parameters" in older versions). Enter the multiplier in the "Boundary" column of the grid on the "Multipliers" tab. That should do it.

 

The purpose of the grid is to let the user run the same model with different loads. So if you wanted displacements of 0.5, 1, and 3, you could set the magnitude of the prescribe displacement to 1 and enter the multipliers as 0.5, 1, and 3 in the grid.

 

 



John Holtz, P.E.

Global Product Support
Autodesk, Inc.


If not provided already, be sure to indicate the version of Inventor Nastran you are using!

"The knowledge you seek is at knowledge.autodesk.com" - Confucius 😉
Message 5 of 13
tom
Participant
in reply to: John_Holtz

Thanks! It worked perfectly!

 

Now about making that one-way boundary condition a reality... 

Message 6 of 13
tom
Participant
in reply to: John_Holtz

John-

 

Your answer helped get me to the place I need to be. However, now that I'm here, I have another question!

 

If I place my displaced boundary condition at the end of my cantilevered bar, is there a way to easily see how much load it took to make that happen? I see the Reactions button under Other Results, but my "Reaction Force Y Component" is unexpectedly low (only 10k lbs versus 200k expected from hand calcs). Am I doing something wrong, or misinterpreting what the Reactions button does?

 

Thanks!

Message 7 of 13
John_Holtz
in reply to: tom

Hi Tom,

 

Do you happen to have about 20 nodes on the end of your bar? Or stated another way, I just wanted to check that you summed the reaction forces for all of the nodes.

 

One of the reaction force results (probably "Reactions > Applied Force > ") should give the result you need. You can also use "Results Contours > Other Results > Element Forces > Axial Force" to see the load at each node.

 

Also check the displacements. If they are smaller than you expected, then the stiffness of the displaced elements is too low. I like to think of these constraints as springs that connect the model to the ground. The displacement you specify is how much the ground moves. So if the stiffness is too low, the model will not deflect as much as the ground moves.

 



John Holtz, P.E.

Global Product Support
Autodesk, Inc.


If not provided already, be sure to indicate the version of Inventor Nastran you are using!

"The knowledge you seek is at knowledge.autodesk.com" - Confucius 😉
Message 8 of 13
tom
Participant
in reply to: John_Holtz

John-

 

A bit of help please. I just cannot get this to make sense according to what my hand calculations and free body diagrams say the results should be.

 

I've attached the file so you can know what I'm talking about. I have a bar of steel that I need to bend into the shape dictated by the various displacement boundaries. I need to know the force at each reaction zone required to make the bar take this shape (I'm using hydraulic cylinders to force the bar into this shape, I'm sizing the cylinders). Yet, when I load a small surface (.125" wide) and sum the nodes, I get loads of 377,000lbf. If I force a cross-sectional line (instead of rectangle) into position, I get loads approximately 55,000lbf. Same stiffness in both. How can I better model what I am attempting to accomplish?

 

Also, when I look at the reaction arrows, they are all pointing down (negative Y), yet my free body diagram says at least a FEW on the left side need to be pointing up (positive Y) to get the shape dictated. Am I expecting too much of this software?


Please help.

 

Tom

 

Edit: Not sure how to attach. It's a simple bar, but at 21mb compressed, it's too big to upload. I've used adrive.com, an offsite ftp site. Here's the link to download, let me know if there's a better way to upload directly to this site.

http://www.adrive.com/public/fb1c2e6f1ba21fb457844fdf222ae3dd3ee9a8e482e13431b56b7bbb6ad3dfe9.html

Message 9 of 13
John_Holtz
in reply to: tom

Tom gets the "gold star" for doing the hand calculations to check the FEA results and noticing that "something was wrong". Congratulations Tom!

 

Unfortunately, I misread the post and spent most of my time running analyses with beam elements. Only when I was about to post a reply did I comprehend what you had written. The inaccurate results are related to the type of solver that was chosen. (Not your fault; the default is "Automatic", so the software decided which solver to use.) The analyses used an iterative solver, and I think that only the boundary conditions (constraints) are used to determine if the model is statically stable, not the boundary conditions and prescribe displacements together. With just the "pinned" constraints along one edge, the model is theoretically unstable, but mathematically with roundoff it was "just stable enough" to give a solution.

 

The key to catching that the results are inaccurate is to view the Summary File (accessible from the Report environment, bottom of the browser). When any of the "Sum of unfixed direction residuals" or "Largest unfixed direction residuals" are large, the solution should not be used.

 

The solution to the problem is to go to the "Setup" tab, "Model Setup > Parameters > Solution" and change the "Type of solver" to "Sparse". See the following pages in the documentation if you want a more indepth understanding.

 

  • Help > Autodesk Simulation > Setting Up and Performing the Analysis > Set Up Analyses Part 3 > Solvers in Finite Element Analysis
  • Help > Autodesk Simulation > Setting Up and Performing the Analysis > Set Up Analyses > Linear > Analyses Parameters > Static Stress with Linear Material Models

 

Tom, if you want me to go over the beam model with you, please send an email to me directly: John Holtz. If others on the forum are interested, I will start a new thread.

 

 



 



John Holtz, P.E.

Global Product Support
Autodesk, Inc.


If not provided already, be sure to indicate the version of Inventor Nastran you are using!

"The knowledge you seek is at knowledge.autodesk.com" - Confucius 😉
Message 10 of 13
bannecke
in reply to: John_Holtz

I have surface contacts in nearly all of my calculations.

So far I always had trouble getting decent results without spending exszessive time.

In my opinion the surface contact algorithm leaves some room for optimization 😉

I would strongly support tom´s suggestion:

Now about making that one-way boundary condition a reality... 

 

Regards

Boris

Message 11 of 13
tom
Participant
in reply to: John_Holtz

Aww, shucks, the Gold Star. Now I'm blushing. 🙂

 

Thank you, a ton John, for your help. I'm going to dive into this head-long this morning and read through the more information you linked. If I still have questions, I'll be sure to email you.

 

Again, thanks a ton for your help!

 

Tom

 

PS: Made any progress on that one-way boundary condition yet? Boris and I are waiting. 😉

Message 12 of 13
dharhay
in reply to: tom

John, what would you call "large".  Thanks.

Dave H
Message 13 of 13
John_Holtz
in reply to: dharhay

First, I need to make a partial retraction of my statement about checking the largest unfixed residual. I thought I understood it, but there is something happening when there are MPCs (multi-point constraints and smart bonding) and boundary elements in the model. (Boundary elements are "Displacement Boundary", "Elastic Boundary", and "Rigid Boundary" in 2011 and older, and "Prescribed Displacement", "1D Spring Support", and "3D Spring Support" respectively in 2012.) I will need to confir with the developers to understand how they affect the reactions.

 

So if the model does not contain any MPCs or "boundary elements", then the largest unfixed residuals should be small. (There are 3 translations and 3 rotations.) The unfixed residuals can be viewed as extra forces (or lack of forces) in the solution.

 

Each model is different. At the risk of making a general statement, I would say that the unfixed residuals should be less than a few percent of the applied loads.

 



John Holtz, P.E.

Global Product Support
Autodesk, Inc.


If not provided already, be sure to indicate the version of Inventor Nastran you are using!

"The knowledge you seek is at knowledge.autodesk.com" - Confucius 😉

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