In the attached report, safety factor is se to calculate from ultimate tensile stress (485 MPa), and it indicates 1.33 with stresses:
Von-Mises: 323 MPa..............SF = 1.5
1st principal: 303 MPa.............SF = 1.6
2nd principal: -181.2 MPa....... SF =2.67
So, how is inventor getting 1.33 ?
EDIT: if I set it to calculate from yield ( 260 MPa) it acurately reports SF = 260/323=0.8
what could be wrong?
EDIT: added part file
Solved! Go to Solution.
Solved by henderh. Go to Solution.
I couldn't get your files to download cleanly so I re-ran the simulation. My results are a bit different than yours BUT:
1/ when using yield strength the factor of safety is as expected
2/ when using ultimate strength there is a discrepancy in the factor of safety
3/ you can work around this by substituting the yield strength into the ultimate strength field but the problem remains.
Yeah thats a possible workaround but as you said the problem is still there , not to mention the report will list that you put ultimate tensile stress instead of yield, absolutely ruining the report for third parties :P.
Mb someone from autodesk can post about this issue?
Thanks
Hi merlob and rhinterhoeller,
I've confirmed the issue and forwarded it on to Development as 1444075. We'll do our best to address this for R2012_SP2.
Thank you for bringing this to our attention, and I apologize for the inconvenience this is causing.
Sincerely, -Hugh
[ps: I'd like to mention that the SF calculations based on either the YS or UTS are not always valid. The reason is that Stress Analysis in Inventor uses only the linear portion of the stress-strain curve. Therefore the stress results (and corresponding Safety and Failure factor results) will not be valid beyond the yield point of the material.]
Thanks for the response Hugh.
About the stress beyond the linear portion, is this statement right: Inventor assumes the material has a linear stress-strain curve past yield.
Would this mean that setting SF to calculate from UTS is valid only on fragile materials which are linear up to rupture?
In another words, every time I´m running a simulation and the stress calculated is past yield, i cannot determine whether the part will break or just deform plastically?
Hi rmerlob,
Yes, I agree your three statements are correct.
There are other assumptions of linear stress analysis we've outlined here: Stress analysis assumptions
If non-linear analysis is needed, we recommend using Autodesk Simulation Mechanical (formerly Algor). Here is a link to download a free trial.
Best regards! -Hugh
It turns out we are calculating the Safety Factor based on UTS correctly (the failure factor).
There are other considerations for brittle materials, so we need to calculate the FF based on prinicpal stresses, not just von Mises.
Computed FF=1.23 how do we get this value?
1) Find the location of the minimum FF
2) Add a probe to this location
3) Get the P1 and P3 values (325.5 and -67.8 MPa)
4) In this case, since P1>0 and P3<0 we use the equation SF=abs[UTS/(P1-P3)] => 1.23 which matches the displayed value
I apologize for any confusion I caused.
Thanks, -Hugh
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ps: If both P1 and P3 are positive, we use just P1 in the denominator. If P1 is negative, we use just P3 in the denominator.
Inventor is really helpful for HiTech FEA analysis and calculations of safety factors! ~ Sunny