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Can not visualize factor of safety

4 REPLIES 4
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Message 1 of 5
ignacio_ivb
559 Views, 4 Replies

Can not visualize factor of safety

Hello, 

 

 

I´m still learning  simulation mechanical. I have a doubt that maybe it's a little too basic but I still can´t figure it out.

 

I will separate them in two questions:

 

First question:

 

I have made the analysis of a simple structure. All three parts are brick type element with steel A36 as material. Each part is 4x10x100cm and the load is distributed 80kgf in the Y axis. The problem is that I don´t know how to interpret the Von mises factor of safety. I attach a picture. 

 

Second question:

 

When I change the material of the same structure to wood, and I click Factor of Safety, I does not show anything, becouse It shows Maximum and minimum Value not available. If I click "set allawable stress values" and click Load Yield Stress or Ultimate Stress, It loads all as 0. If I change that to some value like (the tensile yield strenght of pine) 91N/mm2, it displays a factor of safety in the results analysis that I think is too big.

 

Thank you in advance

Smiley Happy

4 REPLIES 4
Message 2 of 5
John_Holtz
in reply to: ignacio_ivb

Hi ignacio_ivb. Welcome to Simulation Mechanical and the forum.

 

Are you looking at minimum value or the maximum value? The maximum value is not important. If there were some point in the structure where the actual stress were 0, then the factor of safety would be infinite.

 

From the Help documentation, the factor of safety = allowable stress/actual stress, so the minimum factor of safety is caused by the maximum actual stress.

 

For steel, your image shows 17.7488=248.2/actual stress. This gives the maximum actual stress=13.98

 

For wood, you have 6.6377=91/actual stress. This gives the maximum actual stress=13.71.

 

As I would expect, the maximum stress in the structure does not depend on the material. The maximum stress is based on the geometry, so it is the basically the same regardless of the material. (There is a small difference because the von Mises stress is based on the normal stresses and shear stress in all 6 directions, and some of those are affected by the Poisson's ratio.) But the largest stress component is due to the fact that you have two cantilever beams supporting the load, so the stress = bending moment/section modulus, and all of those are independent of the material properties.

 


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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 5
ignacio_ivb
in reply to: John_Holtz

Hi,

 

Thanks for the reply. Much clear now. There is something that I still don´t know and I couldn´t find it with detail.

 

For steel I choose the tensile yield strengh, but , as wood is not isotropic, it has more types of stregth. What should I use for the Allowable Stress in the analysis?

Because when I click "Load Yield Stress" it loads 0 Mpa.

 

Here is an example of the wood i´m using:

 

 

 

Message 4 of 5
John_Holtz
in reply to: ignacio_ivb

The Factor of Safety command is intended for isotropic materials that behave the same in tension and compression. As you point out, wood is orthotropic and has different allowable stresses in tension and compression and direction. Therefore, you will not be able to use the Factor of Safety command.

 

I think you will not be able to create one contour that shows the factor of safety for the entire model. So your options are

  1. Use the custom results ("Results Contours > Other Results > Custom Results") to calculate the factor of safety for parts in tension, another calculation for parts in compression, and so on.
  2. Display the stress and use probes or annotations to point to the nodes with the smallest factor of safety (which you will have to calculate manually)
  3. Export all of the results to a spreadsheet and do a more sophisticated analysis manually.

I do not know how wood responds to the loads, so I do not know if the principal stresses will be aligned to the grain or not. If they are aligned, then you may be able to use the principal stresses for your calculation. If they are not aligned, then you should use the stress tensors for your calculations.

 

Does anyone else have any ideas?



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 5
ignacio_ivb
in reply to: John_Holtz

I will try that. Thank you very much Smiley Happy

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