Plastic strain to linear stress relation 2

[msii]

Hi All,
i received a stress analysis done in abaqus and have a question as using Inventor Professional 2022 to do similar analysis. In the received report instead of the stresses it talks about residual plastic strains in the material. If i want to do a linear static analysis in Autodesk inventor software, is there anyway that i can find what linear stress is equivalent to the strains I’m getting from residual plastic strain? The plastic strain reported that under load the strain is 0.07 and material wont collapse. the material is aluminium with Young modulus is 70 GPa. Is it correct that I say if i draw a line with the angle equivalent to young modulus on stress-strain curve and crossing the strain of 0.07, it would be my linear stress equivalent to plastic analysis?
actually, i m looking to find what maximum stress i need to aim to in my linear analysis to have the same strain in the material as was done in abaqus analysis. thanks in advance,

 

[rb1957]

what stress calc are you talking about ? plastic bending ??

why would you try to validate a plastic stress analysis using linear elastic analysis ??

So you want the linear elastic stress that results from a defined strain ? do you mean ... stress = E*strain ?

"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

 

[SWComposites]

You need to go back and review you strength of materials text and class notes for 5he details of plastic behavior (hopefully you had one).

What you are trying to do is completely incorrect. And why do you say “the material won’t collapse”?

 

[rb1957]

plastic bending uses a fictitious elastic bending stress (like form factor * Ftu)

"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

 

[Stress_Eng]

Sounds like you want to do the reverse of taking a linear stress and determining what the plastic stress and strain is. One way to do this is to use the Neuber method. From a plastic strain, determine the plastic stress and then use the method to calculate your elastic stress.

 

[Stress_Eng]

Plastic bending is well documented. Keep in mind however, if you’ve also got axial loading, your plastic bending capacity will reduce.

 

[LiftDivergence]

First of all, I don't know what the stress report you are looking at means by "collapse" in this context. Plastic collapse is a term that describes net tension failure across an entire cross section due to redistribution around local plasticity or a flaw. If you are just looking at a localized detail and writing a strain margin because a non-linear analysis has been performed, the check actually being made would be material "rupture", not "collapse"... two different things, I wonder if they understood that.

Furthermore, when doing a non-linear analysis, it should be required to check more than just material rupture. Usually it is appropriate to impose an MoS criteria on the plastic strain at ultimate, e'u, and depending on the geometry, certain triaxiality effects should be accounted for. I hope they did more than just finding an equivalent total strain from a FEM and writing a rupture margin. And if you are re-creating this work, I would strongly suggest that you don't simply follow along with what someone else did. You've basically told me nothing and I can already make guesses about the quality of this report.

Second, as I understand it:
1. You have a stress report which did non-linear analysis and reports a value of strain at peak load.
2. You are trying to re-create this, but you don't want to do a non-linear analysis.

Is it correct that I say if i draw a line with the angle equivalent to young modulus on stress-strain curve and crossing the strain of 0.07, it would be my linear stress equivalent to plastic analysis?

No, not at all. See the figure below. If you follow that method, all you will do is figure out the plastic portion of the strain. Remember, that value of 0.07 in/in is the TOTAL strain (both elastic and plastic). If you take the Young's modulus and intersect as you say, the abscissa will simply read off the plastic portion of the total strain. I also don't understand how you intend to read off a stress using that technique.

One way to do this is to use the Neuber method.

It is true that the Neuber technique assumes the elastic stress-strain product is equal to the plastic stress-strain product. This is an assumption - hence why it is called a "rule", not a theory. Generally you could indeed follow the Neuber parabola in either direction provided you know the correct Ramberg-Osgood parameters. However, I would NOT recommend doing that in this case.

Nueber is a pseudo-nonlinear technique that is hyper-localized based on notch root strain. It does NOT account for non-linear behavior of the entire component. So if the report you are referencing has done a complete FEM analysis with non-linear material based on a Ramberg-Osgood curve, it won't inherently show correlation with Neuber. Bascially, mixing and matching non-linear techniques...not recommended.

You haven't mentioned what this report is being used for... static analysis? Is this load cyclic? If so, the material could cyclically harden or soften and then this whole discussion changes.

I'm still confused why you even need the other report. You are saying you're tasked with doing something similar... just do your own analysis. If you want to use a linear model, fine. Probably better actually since it will be much easier to validate. Figure out where you have issues with yielding and then disposition them independently with plasticity corrections, etc.

Be very careful looking at substantiations that rely on non-linear analysis.


Keep em' Flying
//Fight Corrosion!

 

[Stress_Eng]

It is worth just becoming familiar with elastic plastic stress stain correction methods such as Neuber and Glinka (both based on equating strain energy densities) and their applications. As said, these methods are applicable to a point stress only.