Event simulation for vertical load test

Event simulation for vertical load test

bergloo
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Message 1 of 4

Event simulation for vertical load test

bergloo
Contributor
Contributor

Hi friend of autodesk
I have a question.
I have to make a vertical load test with these parameters:image001.jpg

For this I am using event simulation.

-I'm just using the container top.

-The material is high-density polypropylene

- Fixed them at the bottom.

- and finally a prescribed translation at the top.

 

 

 

Vertical Load Simulation Galon v25.pngVertical Load Simulation Galon v253.pngerror.pngVertical Load Simulation Galon v254.png

 

 

 

My big doubt and I believe that where I have the error, is in configuring the multiple curve. Taking into account a constant speed of 510 mm / sec and achieve a displacement of 6.35 mm.?

Another question in the manager - result output, how can I activate applied loads?

can you help me? I did not achieve the desired results

 

Thank you in advance

 

 

 

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Message 2 of 4

innovatenate
Autodesk Support
Autodesk Support

A 12 second simulation may be too long of a simulation window for Event Simulation. It's much better at handling much faster simulations. Please see the quickstart guide posted in this forum thread for more detail.

https://forums.autodesk.com/t5/design-validate-document/quick-start-for-event-simulation/td-p/668065...

 

Can you share your design with us to have a closer look?

 

If you look at the Solver Data

Screen Shot 2017-05-02 at 5.31.44 PM.png

 

... and the Solver Output

Screen Shot 2017-05-02 at 5.31.53 PM.png

 

 

 

You can see the External Work and what the value is at the time that the solver outputs to the results.

 

 

           Time   Procedure    Total    Kinetic   Internal  Viscous   External   Total    Elapsed     Remaining
    Inc    Step     Time       Time     Energy    Energy    Energy     Work      Energy   Wall Time   Wall Time
      0  2.50e-08  0.00e+00  0.00e+00  0.00e+00  0.00e+00  0.00e+00  0.00e+00  0.00e+00  0 00:00:00         n/a
     50  2.50e-08  1.25e-06  1.25e-06  2.66e-02  2.58e-02  7.15e-06  5.24e-02  1.47e-05  0 00:00:00  0 00:10:45
    100  2.50e-08  2.50e-06  2.50e-06  5.53e-02  5.02e-02  3.29e-06  1.06e-01  1.08e-05  0 00:00:01  0 00:10:47
....
....
....
>>>>>> Writing FEMAP output for time = 0.000900022 <<<<<<<
  36300  2.49e-08  9.01e-04  9.01e-04  6.28e+00  2.89e+01  1.40e-04  3.52e+01  7.38e-04  0 00:10:06  0 00:01:06
  36350  2.49e-08  9.03e-04  9.03e-04  6.29e+00  2.89e+01  1.44e-04  3.52e+01  7.22e-04  0 00:10:07  0 00:01:05
....
....
  40200  2.49e-08  9.98e-04  9.98e-04  6.41e+00  2.89e+01  1.04e-04  3.53e+01  4.84e-04  0 00:11:11  0 00:00:01
  40250  2.49e-08  1.00e-03  1.00e-03  6.49e+00  2.89e+01  1.07e-04  3.53e+01  4.61e-04  0 00:11:12  0 00:00:00
>>>>>> Writing FEMAP output for time = 0.00100002 <<<<<<<
  40269  2.49e-08  1.00e-03  1.00e-03  6.45e+00  2.89e+01  1.05e-04  3.53e+01  4.87e-04  0 00:11:12  0 00:00:00

 

 

In this case, there should be a conservation of energy for each step.

Total Energy = Kinetic Energy + Internal Energy + Viscous Energy - External Work

 

If you keep the value of Kinetic Energy very low (meaning there's not a lot of inertial forces), the solution will be quasi-static (I don't think I did a great job of this in the above example). 

 

I don't believe there is a way to get the Reaction Force of a constraint at each time step at this time. I will have to ask around and circle back.

 

However, you may be able to use the External Work value above to understand the forces acting on your bottle.

 

I hope this helps.

 

 




Nathan Chandler
Principal Specialist
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Message 3 of 4

lee.taylor
Alumni
Alumni

You should use the prescribed velocity boundary condition (instead of prescribed displacement) to move the grid points on the container top. You can set the curve to be a constant velocity of 510 mm/min. In this case your desired total displacement of 6.35 mm will require a total duration for the event of 0.747 sec (not 12 sec). Note that this constant velocity boundary condition will impart a shock to the container at time zero as the grid points on the container will see a instantaneous acceleration at the initial time. A 0.747 sec duration is still relatively long for Event Simulation and will result in a very large number of time increments for the analysis (probably on the order of tens of thousands of time increments). But, not to worry, Event Simulation will happily do the problem.

 

Your big problem here is extracting the reaction force from the results. Fusion Simulation does not offer an easy way to sum up all the reaction forces on the set of grid points on the top of the container. You will have to manually go around the top of the container and pick the nodes one at a time and plot the reaction force history. For each curve you can record the peak value and then add them all up.

 

Hope this helps.

 

Message 4 of 4

innovatenate
Autodesk Support
Autodesk Support

Thanks @lee.taylor

 

Before you run the simulation, you should verify that you have enabled the results output.

 

Screen Shot 2017-05-03 at 6.52.47 AM.png

 

Screen Shot 2017-05-03 at 9.01.55 AM.png

 

I hope that helps!

 

Thanks,

 

 




Nathan Chandler
Principal Specialist
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