# Multilinear Kinematic Plasticity created from a stress-strain graph

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peteroznewman posted this 30 October 2017

I was sent a stress-strain graph of a brittle metal tested in an Instron machine. It looked like this.

I have matlab and downloaded Grabit, a utility that makes digitizing data from a graph very easy. Here is what that looked like:

Now that I had the curve digitized, I could convert the Stress and Strain into True Stress and Plastic Strain so I could enter values in a material model using Multilinear Kinematic Plasticity. The formulas to convert are shown below:

E = 13,000 ksi

True Strain = ln(1+Strain)   where ln is the natural logarithm function.

True Stress = Stress(1+Strain)

Elastic Strain = True Stress/E

Plastic Strain = True Strain - Elastic Strain

I used a spreadsheet to correct the offset in the origin of the graph to put it back on zero.

The first two columns came from the matlab Grabit utility.

The last two columns are copy paste into the Multlinear Plasticity material model, except for the first line.

The first first line is the yield stress. This was the material model used for the thread fracture analysis.

Fabricio.Urquhart posted this 09 July 2018

Peter, do you have any information like this post for strucutre steel materials?

- ASTM A572 Steel, Grade 50 (beams and columns)

- ASTM 36 (plates)

- ASTM A325 (bolts)

I found in matweb.com. But plasticity properties I did not find.

Thank you!!

peteroznewman posted this 09 July 2018

I can borrow the Atlas of Stress-Strain Curves from my Engineering Library.

Curve 6 is ASTM 36.

Fabricio.Urquhart posted this 09 July 2018

Thank you, very very much Peter.

For the bolts do you have any curve?

Attached Files

peteroznewman posted this 10 July 2018

Fabricio,

When I search astm A325 stress strain curve on Google, it gives me this.

Figure 7 is the stress strain curve for ASTM A325.

You should try Google for yourself, it is really very good!

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Fabricio.Urquhart posted this 10 July 2018

I have found a curve for bolts in this article whcih I have attached. If would like to add in your engineering library, I shared with you.

Fabricio.Urquhart posted this 10 July 2018

Peter I am using multilinear isotropic hardening plasticiy, but I could not see the nonlinearity.

The model is a beam, fixed in both extreme, but I modeled only the half, and use a plane of symmetry,

I could not see the nonlinear curve in equivalent total strain neither nonlinearity in the equivalent stress:

In spite of reach the yeld and rupture stress, the curve is linear, I am not understanding why.

Attached Files

Fabricio.Urquhart posted this 10 July 2018

You are right!!

Attached Files

Fabricio.Urquhart posted this 10 July 2018

Peter, I solved. The problem was the load, which was normal to surface. So with component, the plasticity occure easier.

mekafime posted this 10 August 2018

Peter

How you get the first and second column?

peteroznewman posted this 10 August 2018

First and second columns were digitized from the plot of Engineering Stress vs Engineering Strain.

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mekafime posted this 10 August 2018

I used Engauge Digitizier, open source, Thanks!

SandeepMedikonda posted this 10 August 2018

For digitizing data from research articles, WebPlotDigitizer is a very good app that I used extensively in my research.

It works as an App on your Chrome browser.

Regards,

Sandeep

mekafime posted this 11 August 2018

Hi,

the third column (Strain) , how you get?

SandeepMedikonda posted this 11 August 2018

Third column in a Stress vs Strain plot, Can you explain or post a picture?

You can digitize any no. of curves from a 2-D plot using WebPlot Digitizer.

~Sandeep

Fabricio.Urquhart posted this 11 August 2018

I did not understand, too. Which column??First and second column?Third column? I did not understand nothing.

Can you explain?

Fabricio

peteroznewman posted this 11 August 2018

This post has a table of numbers in an Excel spreadsheet, the first few lines are below.

Column 1 is Offset Strain and is the value of strain read off the plot (using digitizing sofware) in that same post.

Column 2 is the Stress read off the plot

Column 3 is Strain = Offset Strain - 0.0003    (which is the first value in the table).

So Columns 2 and 3 are Engineering Stress and Engineering Strain (which start at 0,0) that get converted to True Stress and True Strain and the components of True Strain, which are Elastic Strain and Plastic Strain using the formulas listed in that post.

Published sources of Stress-Stain data usually start at (0,0), but raw data straight from the Instron can have offsets that need to be corrected. I hope that clarifies things for everyone.

Regards,

Peter

Fabricio.Urquhart posted this 12 August 2018

Peter, it clarifies.

But the column that I have to input in Ansys is the last (True Stress) and one before (plastic strain), right?

Another point, your first column is offstet strain. In the picture below in the X axis, I have the strain, and in your material graph you have the strain, too. Why did you use offset strain? I did not understand.

I am with difficult, to input the ASTM A325 curve. Because I did not find the engineering stress. Only the true stress, and I think that is not reliable.

I think that is not reliable, because I was reading about ASTM A325 curve, and it has a yield landing as the other two types of steel.

Can you help me with it?

Attached Files

peteroznewman posted this 12 August 2018

I created a column called offset strain because the plot was clearly not going through zero strain, which was at the red cursor location. The plot went through the point 0.0003 strain instead of zero due to some slack in the Instron testing setup. Therefore, the whole curve needed to shift to the left by 0.0003 to pass through zero. The column where the offset was subtracted off was called Strain, which was the Engineering Strain.

Fabricio.Urquhart posted this 12 August 2018

Now I understood.

But I have problems with the ASTM A325 curve. Because this, the connection does not have the stiffness that it should have.

The model is attached.

Attached Files

peteroznewman posted this 12 August 2018

Fabricio,

I opened the spreadsheet from above and looked at the A325 tab. Here are the columns:

You have left two columns off: Elastic Strain and Plastic Strain, and you don't have a cell for E, the Young's Modulus.

Elastic Strain = True Stress/E

Plastic Strain = True Strain - Elastic Strain

Since the input data for multilinear plasticity is Plastic Strain and True Stress, you haven't caluculated those numbers yet. Please add those to your spreadsheet.

Regards,

Peter

peteroznewman posted this 12 August 2018

Fabricio,

I opened the spreadsheet from above and looked at the A36 tab. There are errors in the formula in many columns.

After you fix the errors, there is still some data cleanup required due to noise in the data. For example, some negative plastic strains are calculated. These are in the elastic range and those rows won't even be used in the ANSYS material table, which has as its first row, the yield stress. The first entry in the material table could be 206.2 MPa as the yield stress, and call the plastic strain there zero, though it shows as 0.00008. However, that puts a sudden slope change in the data.

ANSYS multilinear materials require that the slope in the multilinear data is positive.  I added a column to compute the slope. To make sure this requirement is met, delete rows 7 and 10 from the spreadsheet. Now the positive slope requirement is met. Hide rows 2-4.  I would type over the 206.2 and make it 251 and type over the .00008 and make it 0. Now the Plastic Strain and True Stress columns have the data cleaned up for use in the material model, and the yield stress is 251 MPa.

The nonlinear algorithm will have a much easier time with clean data like this.

Regards,

Peter.

Fabricio.Urquhart posted this 13 August 2018

Thank you Peter. My error was that the curve strain X stress of ASTM A325 steel, is juts the true strain X stress, as was in the reference attached.

So this is the curve that I have to input in Ansys, isn't it?

Regards,

Fabricio

Attached Files

peteroznewman posted this 13 August 2018

Fabricio,

Are you using Isotropic or Kinematic Hardening?

I am reading the 18.2 help, 4.4.3.2.2.2. Specifying the Constants. The entry for Kinematic hardening says "No segment slope can be larger than the slope of the previous segment" while Isotropic hardening has no such restriction. See 4.4.2.2.2. Multilinear Isotropic Hardening

That means these curves of True Stress vs. Plastic Strain with an increase in slope can only be used by Isotropic Hardening.

If you want to use Kinematic hardening, then you would have to draw a line from the yield point to the tangent of the curve, and not use any of the points that I have crossed out below in a plot for A36 of True Stress (MPa) vs. Plastic Strain.

I would be interested to hear from any experts in plasticity if I have understood the manual correctly.

Regards,

Peter.

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Fabricio.Urquhart posted this 19 August 2018

Hello Peter.

I am using Isotropic Hardening. A point that I would like to ask if you agree. Is in the curve of true stress X plastic strain, when the rupture stress is exceeded, I shall use a value of true stress tending to inifinite, do you agree?

I have watched a lesson video, I do not remember who has sent to me, if you or Sandeep, it is teached in this lesson video, but I was not using this.

What do you think about it?

SandeepMedikonda posted this 19 August 2018

Fabricio,

Make sure that the last point on the stress-strain curve entered is well beyond any numerical strains anticipated since,  if you "run off the end of the input curve" the slope of the stress-strain curve is zero.  I usually include a data point of at least 100% strain to assure that this does not occur.  Postprocessing can be used to determine if the material ultimate limit has been reached.

Regards,

Sandeep

peteroznewman posted this 19 August 2018

ANSYS will use the last value of stress for any strains that exceed the final value of plastic strain, which means the slope of the stress-stain plot is flat. Below is the stress-strain data for Fabricio's A36 material.

If I create a one element model that has a zero displacement condition on three sides, and a non-zero displacement on a fourth face to stretch the body, I can plot the stress vs total strain.

Notice that the strain has gone past the end of the last value of strain in the table, and that the stress is constant.

If I suppress the non-zero displacement and instead use a force to pull on that same face, the solver can't get past a low value of stress and strain, well within the ranges defined in the material.  This is why I always choose displacement BCs over a force BC whenever possible. It is much more likely to converge to the end.

The solver is not able to make progress beyond the sharp knee in the material curve at 255 MPa. If it is not possible to change the model from a force input to a displacement input, one approach to allow the model to converge to higher values of stress and strain is to smooth out the material curve.

Here is a version of that material without the sharp knee.

Now the Force applied to the single element can make some progress, but convergence fails beyond 0.11 strain.

Below, I make an extrapolation in the material model as Sandeep suggests.

Now the Force input can solve to then end.

Bottom line: Displacement BCs are far more robust for convergence than Force BCs.

Regards,

Peter

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shazmi posted this 22 May 2019

Hi all,

Apologize, maybe my question is not related to this topic. By the way, may I know how to input stress/strain curve in workbench for multilinear kinematic hardening material model. Should I include elastic region in the curve or I just need to define plastic strain region only. Kindly need your help.

I actually have tried both ways (full stressxstrain curve and stressxplastic strain curve) and both gives same result. Is there any explanation behind it?

Thanks.

gurkanonal posted this 3 days ago

Hi Peter,

As you mentioned, first line of multilinear plasticity definition is the yield stress. I confused at this point. Since yield point on stress-strain curve consists of both plastic strain and elastic strain. Plastic strain is 0.002 according to 0.02% offset method + elastic strain must be equal to total strain. It means, at first line of multilinear plasticity definition plastic strain comprises of 0.002 plastic strain and yield stress but it is not allowed to enter 0.002 plastic strain. It must be 0. What is point that i miss?

Regards

peteroznewman posted this 3 days ago

Offset yield strength is an arbitrary approximation of a material's elastic limit. It is the stress that corresponds to a point at the intersection of a stress-strain curve and a line which is parallel to a specified modulus of elasticity line.

Multilinear Plasticity requires the true (non-offset) elastic limit definition of yield strength.

Instron has some nice graphics.

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