Community
Inventor Forum
Welcome to Autodesk’s Inventor Forums. Share your knowledge, ask questions, and explore popular Inventor topics.
cancel
Showing results for 
Show  only  | Search instead for 
Did you mean: 

FEA analysis stress value not correct

45 REPLIES 45
Reply
Message 1 of 46
manish
3566 Views, 45 Replies

FEA analysis stress value not correct

Hi,

 

I have some issues with the stress analysis. I have attached two picture files. One is with support and another without support. All the setttings for stress analysis are made already. All the factors are same for both files. You can just simulate directly.

 

Practically stress must be less in the case which has supprot while here in the software its opposite. When I add support, stress increases which is not quite promising.

 

thanks,

Manish

Manish Jangir
Tags (2)
45 REPLIES 45
Message 41 of 46
manish
in reply to: conklinjm


@conklinjm wrote:


 

Was that change made before you did the pictures of your analysis that you provided in message 19 or did you use a different yield strength for that analysis for the Safety Factor image seems inconsistent with either a material with 207MPa or 330MPa yield strength.  .

 

HTH




I don't know there is something wrong in safety factor calculator,

Values given,

 

Yield strength: 330 MPa

Tensile strength: 520 MPa

 

Von mises stress: 322.1 MPa (same)

Safety Factor: 1.47(different),

 

Please check report attached. Last time S.F was 0.97 with same values. and before that, 1.04 which was reasonable based on stress.

Manish Jangir
Message 42 of 46
henderh
in reply to: manish

Hi Manish,

 

  In the part / simulation attached to posts 23 and 24, I noticed that the the Safety Factor in the Assign Materials dialog is specified based on Ultimate Tensile Strength (as opposed to the default Yield strength).  This is why the SF is 0.97 in the part / simulation (material = steel, UTS = 345MPa) and the SF is 1.47 in the report (material = GS, UTS = 520 MPa).

 

  Also, when the Safety Factor based on UTS, we calculate it differently according to failure critieria in brittle materials where it's calculated using the principal stresses, not the Von mises stress.  It's described in more detail here.

 

  I hope this may explain why you see the difference from expected.

 

Best regards,

-Hugh



Hugh Henderson
QA Engineer (Fusion Simulation)
Message 43 of 46
rmerlob
in reply to: manish

Ah the old "difference in calculating the safety factor for brittle materials" answer (you already explained this to me once, thanks Hugh)

Maybe Autodesk should put up a reminder/warning when you set it to UTS I'm pretty sure lots of people don't know that. (Based on how you're the only one that comes up with that answer when it's needed)
Message 44 of 46
henderh
in reply to: rmerlob

Hi rmerlob,

 

  Thank you for improvment suggestion for a reminder/warning when the user switches the Safety Factor to UTS.  I have authored and sent an internal proposal to include this in a future release.

 

  In the meantime, we do have a blog that also goes into the details with an example.

 

Best regards,

 

 



Hugh Henderson
QA Engineer (Fusion Simulation)
Message 45 of 46
manish
in reply to: henderh


@henderh wrote:

Hi Manish,

 

  In the part / simulation attached to posts 23 and 24, I noticed that the the Safety Factor in the Assign Materials dialog is specified based on Ultimate Tensile Strength (as opposed to the default Yield strength).  This is why the SF is 0.97 in the part / simulation (material = steel, UTS = 345MPa) and the SF is 1.47 in the report (material = GS, UTS = 520 MPa).

 

  Also, when the Safety Factor based on UTS, we calculate it differently according to failure critieria in brittle materials where it's calculated using the principal stresses, not the Von mises stress.  It's described in more detail here.

 

  I hope this may explain why you see the difference from expected.

 

Best regards,

-Hugh


Hi Hugh,

 

Yes you are right. Now I changed the simulation back to yield strength and got safety factor 1.03( approx. same as before).

 

But still my question is the same, safety factor is much lower than what we expect?

 

regards,

Manish

 

 

 

Manish Jangir
Message 46 of 46
rmerlob
in reply to: manish

Hi manish,

 

Attached is the simulation I did with your file, I used the simmetry to simulate only half (with half the load) to be able to see the actual depth of the highly stressed area.

 

Like we´ve all mentioned anything past yield is not valid with inventor models, what you can get from this simulation is that at 88 tons, the part yields, this yielding is localized though and strain causes steel to harden, the strain hardening problem is much more complex that what Inventors simulation can handle, I would say that this part won´t break under that load but definitely is not safe in the traditional sense of design (yield/SF = maximum stress), sadly there are some areas of engineering that force us to design into the plastic zone, that´s were testing or more complex FEA tools come into play.

 

As for your concern of the part not breaking under a lot more force that that of the simulation, just look at how much of the material is very well into the safe zone (SF over 5) this coupled with the localized strain hardening of the critical portions mean that the part is a long way off from breaking, but to predict exactly at what load it will break is well outside of Inventor´s simulation capabilities.

 

 

BTW: I´ve used FEA to design manhole covers, and there is no way you can design a weight/cost efficient manhole cover that resists the standard (EN 124) required load without allowing some plastic deformation at testing (even the standard allows for some), so maybe your case is similar, you just can´t allow yourself to have a huge safety factor over yield because 1. budget, 2. you´ve done physical testing and you know it holds.

 

Warning: even though it holds, it doesnt mean it will always hold, over the yield point I´m guessing you have to double-concern yourself if fatigue is involved. In this case we could use a bit more info as to how this works in real life.

 

Regards,

 

RM

 

 

Can't find what you're looking for? Ask the community or share your knowledge.

Post to forums  

Autodesk Design & Make Report