Elasto Plastic Curve and Multilinear Isotropic Hardening

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  • Last Post 05 August 2018
Fabricio.Urquhart posted this 11 July 2018

Hello, now I am trying to understand the application of plasticity in Ansys, so I am modeling a simple bar, fixed on both sides. But I am suing a plane of symmetry in the middle, because is the same plane of symmetry which I am using in my master thesis.

The question is about the multilinear isotropic hardening. I am using some curves from the literature, and in the reality we have a gap between the end of elasticity and the start of plasticity, where the stress is constant and there is a variation of the strain. I considered it in the multilinear material curve:

I would like some opinion. Because I have read some articles, and the people do not consider this "constant" gap. I think the because is difficult to know the strain when this gap starts. What do you think about it?

 

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SandeepMedikonda posted this 12 July 2018

Hi Fabricio,

  It looks like this is the engineering stress, is that correct? If so, you would have to convert this to true stress-strain data. Also, it looks like you don't have enough data points there. The reason why it is not recommended is since stiffness is zero it will introduce instability into your analysis.

  Check out this article, you might find useful.

Regards,

Sandeep

Fabricio.Urquhart posted this 17 July 2018

Thank you Sandeep. I have read the articles you said. It was very useful. 

The model attached is not converging. Do you think that the reason is that I am using engineering stress?

 

Attached Files

SandeepMedikonda posted this 17 July 2018

Hi Fabricio,

  It's possible, but more importantly, the metal plasticity material models are anticipating a true stress-strain data. So when you input Engineering stress-strain data you are getting a wrong result.

  Some suggestions:

  • Keep an eye on the plastic strain in your simulation is it may be exceeding the max. value you might have provided.
  • Remove the initial points in your simulation (0 & 0.0001) and just start from 0.02 plastic strain and see if it helps.
  • Increase the maximum sub-steps to about 1000. 

~Sandeep

 

Fabricio.Urquhart posted this 17 July 2018

Sandeep,

- I will keep an eye on the plastic strain

- Why remove?It is the elastic strain. And if I start with 0,02 strain, Ansys does not accept Mthe multilinear tabular. I did not agree with it.

- I will increase

 

Thank you

SandeepMedikonda posted this 17 July 2018

Fabricio,

When we define the data for Multilinear Isotropic Hardening, we are defining the plastic strain vs true stress right? But from looking at your data it looks like you had the complete engineering stress-strain profile input. So, I was suggesting you to separate the elastic strain data as discussed here.

Also, check out section 4.4.2.2.2 in the manual.

~Sandeep

Fabricio.Urquhart posted this 22 July 2018

Thank you Sandeep. I was reading about which element is better to multilinear isotropic hardening material. Some people say SOLID185, others sayd SOLID186. I have read here, but it doesnt say nothing about multilinear isotropic hardening.

 

What is your opinion about it, the element to use with this material data?

Regards

Fabricio

 

peteroznewman posted this 22 July 2018

 Fabricio,

I am interested to read Sandeep's reply and read any articles he links to, but I will give you my opinion now.

The SOLID185 is the linear element and the SOLID 186 is the quadratic element. Since you have a well structured mesh, you will have more detail in the results with the SOLID186 quadratic element because it can represent curvature in the stress results along its edge, even when the edge is a straight section. Better yet, since you have some cylindrical surfaces in your bodies (holes and shafts), the SOLID186 element can represent the geometric curvature of the geometry, while SOLID185 has straight edges so the edge of a hole will show facets with SOLID185 and you will need a finer mesh than if you use SOLID186 which can have curved edges.

Both elements support Rate Independent Plasticity, which includes Multilinear Isotropic Hardening.

Regards,

Peter

SandeepMedikonda posted this 22 July 2018

Hi Fabricio,

As far as the implementation is concerned both should work fine. However, since we expect higher deformations in the elements due to plasticity. I would recommend using SOLID 186 whenever possible. As you might already know, higher order elements refer to the use of higher-order shape functions. The shape function represents assumed behavior for a given element and how well each element's shape function matches the true behavior directly affects the accuracy of the solution. As Peter said, these elements can represent curved edges and surfaces more accurately and are not as sensitive to element distortion. They have also been reported to predict highly accurate stress and give better results than linear elements in many cases even with fewer elements. In my experience, this difference between different ordered elements is less severe for shells in comparison to solids.

Here is a link to a really good article that I had bookmarked a while ago.

Now the additional aspect that affects the accuracy of the solution in addition to the element order is the number of integration points, as the manual here says a quadratic element has no more integration points than a linear element.

Stresses (and strains) are calculated at the integration points, not at the nodes, and the variation of stress across an element is determined by the number of integration points. Having one integration point for a particular stress yields a constant value through the thickness while having two (i.e., 2x2x2 in 3D) yields a linear variation. The number of nodes does not determine the number of integration points, and different element formulations may use different number of integration points. 

(1) Uniform reduced integration uses one integration point. You will get a constant value for stress/strain within an element. 

(2) Enhanced Strain or Simplified Enhanced Strain have 2x2x2=8 (in 3D) for both volumetric and deviatoric terms. You will get a linear variation of any stress quantity requested.

Regards,

Sandeep

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Fabricio.Urquhart posted this 05 August 2018

Sandeep

I converted engineering stress to true stress using this article, and the graph is below:

When you said: "it looks like you don't have enough data points there". Are saying about the yield landing?

I took the engineering curves  (StressXStrain) from  ASM - Atlas of stress strain curves-2002. And apply the formulation from the article.

Am I doing something wrong?Because I was not waiting for the last point of true stress was smaller than the point before. Do you understand?

Thank you. 

 

SandeepMedikonda posted this 05 August 2018

Hi Fabricio, 

Yes, I meant for the yield data as according to my experience that is never flat. If you have confirmation of the data, you can ignore that.

Now, why is there an inconsistency between the units on the stress and strain? I understand that strain is dimensionless but I am wondering since it's so small, this could be the reason why you are seeing this behavior.

~Sandeep

peteroznewman posted this 05 August 2018

 Fabricio,

It is not permitted to have a negative slope on the True Stress vs Plastic Strain curve in a multilinear plasticity material model.

 

Also, the article you used to convert Engineering Stress and Strain to True Stress and Stain failed to mention that the simple equations that do this conversion are only valid until necking begins. Then those simple equations are not valid and you can't convert any more data on the Engineering Stress curve past the Ultimate Strength, as that is typically where necking begins. Perhaps this explains why your plot has a negative slope on the last segment of True Stress and Strain.

The webpage Sandeep provided explains that equation 4 to calculate true strain from strain is not valid after necking and gives an alternate equation using true area on the bottom of the page, but that data is rarely available.

Peter

Fabricio.Urquhart posted this 05 August 2018

Thank you Peter. I agree and understand that it is not permited to a negative slope.

Sandeep, I took the engineering stress X strain from ASM book. The picture is below:

So, after the ultimate stress, I use the ultimate stress, consequently the true stress grow up. Now I think that it is correct. I will watch the lesson that Peter sent.

Thank you!!!

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