October 30, 2017 at 12:48 ampeteroznewmanSubscriber
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.
July 9, 2018 at 12:57 pmFabricio.UrquhartSubscriber
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.
July 9, 2018 at 5:11 pm
July 9, 2018 at 8:36 pmFabricio.UrquhartSubscriber
Thank you, very very much Peter.
For the bolts do you have any curve?
July 10, 2018 at 12:43 ampeteroznewmanSubscriber
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!
July 10, 2018 at 12:44 amFabricio.UrquhartSubscriber
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.
July 10, 2018 at 1:05 amFabricio.UrquhartSubscriber
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.
July 10, 2018 at 1:14 amFabricio.UrquhartSubscriber
You are right!!
July 10, 2018 at 2:02 amFabricio.UrquhartSubscriber
Peter, I solved. The problem was the load, which was normal to surface. So with component, the plasticity occure easier.
August 10, 2018 at 4:01 pmmekafimeSubscriber
How you get the first and second column?
August 10, 2018 at 5:27 pmpeteroznewmanSubscriber
First and second columns were digitized from the plot of Engineering Stress vs Engineering Strain.
August 10, 2018 at 5:56 pmmekafimeSubscriber
I used Engauge Digitizier, open source, Thanks!
August 10, 2018 at 8:29 pmSandeep MedikondaAnsys Employee
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.
August 11, 2018 at 1:22 pmmekafimeSubscriber
the third column (Strain) , how you get?
August 11, 2018 at 1:43 pmSandeep MedikondaAnsys Employee
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.
August 11, 2018 at 2:19 pmFabricio.UrquhartSubscriber
I did not understand, too. Which column??First and second column?Third column? I did not understand nothing.
Can you explain?
August 11, 2018 at 2:30 pmpeteroznewmanSubscriber
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.
August 12, 2018 at 2:10 amFabricio.UrquhartSubscriber
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?
August 12, 2018 at 2:36 ampeteroznewmanSubscriber
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.
August 12, 2018 at 2:00 pmFabricio.UrquhartSubscriber
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.
August 12, 2018 at 3:36 pm
August 12, 2018 at 4:13 pmpeteroznewmanSubscriber
August 13, 2018 at 1:39 amFabricio.UrquhartSubscriber
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?
August 13, 2018 at 2:28 ampeteroznewmanSubscriber
Are you using Isotropic or Kinematic Hardening?
I am reading the 18.2 help, 18.104.22.168.2.2. 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 22.214.171.124.2. 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.
August 19, 2018 at 2:45 amFabricio.UrquhartSubscriber
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?
August 19, 2018 at 1:48 pmSandeep MedikondaAnsys Employee
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.
August 19, 2018 at 5:21 pmpeteroznewmanSubscriber
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.
May 22, 2019 at 6:21 amshazmiSubscriber
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?
September 20, 2019 at 10:14 amgurkanonalSubscriber
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?
September 20, 2019 at 10:46 ampeteroznewmanSubscriber
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.
February 14, 2020 at 5:06 pmpeteroznewmanSubscriber
You haven't labeled the Strain axis with any numbers so I can't see how much of the data is plotted.
The red line in the first figure shows True Stress going down. That is not correct for a plasticity material model.
While it is true that the force went down, that was because the material changed from uniformly stretching to necking. You can't use any of the force-displacement data to build plasticity material models after the necking begins.
February 18, 2020 at 12:00 pmpeteroznewmanSubscriber
2) The real curve is decreasing because Engineering Stress and Strain are calculated on the original cross-sectional area of the sample. The True Stress and True Strain calculations update these values to estimate the actual, current cross-sectional area of the sample. A basic assumption about plasticity is that the material is flowing. If you have a volume of material, say 2x2x20mm = 80 mm^3 and you stretch that 20 mm out to 30 mm (strain = 0.5), the cross section doesn't remain at 4 mm^2 as the Engineering stress uses. It would be reduced to keep the volume constant; the cross sectional area would be 80/30 = 2.67 mm^2. If you divide the force by 4 you get the Engineering Stress. If you divide the force by 2.67 you get the True Stress.
February 21, 2020 at 9:45 pmpeteroznewmanSubscriber
In fig x, the True Stress is decreasing at high levels of strain. That is not correct for a ductile material. It is an artifact of the sample necking during the test. Once necking begins, you can't convert Stress-Strain to True-Stress-True-Strain.
In fig y, the True stress is only increasing, which is correct.
April 29, 2020 at 6:15 pmGmattosSubscriber
This has been informative. I have always defined my multi-linear isotropic hardening curves starting from the yield stress (as you outline above). However, I have received a model that has defined the multi-linear isotropic hardening curve starting from 1 psi stress, 0 strain (where the initial stress v strain curve represents the elastic portion). Are you able to clarify how the solver would interpret these two inputs given otherwise equally defined properties (i.e. elastic moduli and yield strength)? Would you expect the same result?
April 29, 2020 at 6:36 pmpeteroznewmanSubscriber
The results would be different in a two-step tensile test coupon simulation.
Step 1 is stretched by a force that results in a strain of 0.01
Step 2 that force is set to zero.
For the material properly defined with a yield stress of 10 ksi, the axial deformation at the end of step 2 will be numerically zero. The material remained elastic during the simulation and the coupon returns to its original length when the force is removed.
For the material defined with a yield stress of 1 psi, the axial deformation at the end of step 2 will be nonzero. The material immediately goes plastic during the simulation and the coupon will not return to its original length when the force is removed.
April 29, 2020 at 7:11 pmGmattosSubscriber
Would the force required to achieve 0.01 total strain at the end of step 1 be equivalent between the two models? I assume not...
Basically, the solver will default to using the curve as soon as the material reaches a stress defined within the curve (which should, in practice generally be the yield stress). Is that correct? I guess my question is, When does the solver use the information provided in the curve vs using the E and yield strength values?
April 29, 2020 at 7:52 pmpeteroznewmanSubscriber
Yes, the force will be very different. Look up the stress for each material at a strain of 0.01 and multiply by the cross-sectional area to get the force for the simple tensile test coupon described above.
When the stress is less than the yield stress (the first value in the table), only the value of E is used to compute stress and strain.
Once the stress exceeds the value of yield stress, then the data in the table is used to compute plastic strain and stress.
May 26, 2020 at 12:49 pmNeelkolekarSubscriber
I was looking for this tread thanks you very much for a detail explanation above.
I have few doubts on the workbench multi-linear material modelling.
Below I have shown a table with stress-strain data, where 221 MPa is material yield limit.
From column 5, it is clear that materiel has entered in plastic region (with positive values of plastic strain much before the theoretical yield stress (221 MPa).
When I put stress strain data in ansys from yield stress [3rd and 4th column] I have to make 1st strain point as zero plastic strain.This creates a not so good stress strain graph shown by red line in second image as it drags the 0.2% strain to 0% value).
So my question is should I start the stress strain data from a point where actual plastic strain (stress=140 MPa) instead of yield limit, is it acceptable and does it make any difference.
Need some advice here.
And when you say we should make sure that the last point on the stress-strain curve entered is well beyond any numerical strains anticipated since, what stress value we should use for a data point of 100% strain.
Thanks in advance,
May 26, 2020 at 6:17 pmpeteroznewmanSubscriber
Yes, use 140 MPa as the multilinear yield strength and the first entry in the stress column at zero plastic strain.
That is the point at which the linear part of the elastic stress-strain curve ends and the nonlinear part begins.
The 0.2% offset value of Yield Strength is irrelevant in this model. It is all about the linear to nonlinear parts of the True Stress - True Strain curve.
The last point on the curve can be the True Stress at the Plastic Strain when the material fractures. However, elements in the simulation can be stretched beyond the point at which material failure occurs. The material just becomes perfectly plastic after the last point. You have to compare total strain with elongation at break to decide if fracture has occurred.
June 6, 2020 at 8:58 pmAndreytestonSubscriber
April 25, 2021 at 8:46 pmMechEngineer2019SubscriberThanks for this post, Peter. It's been helpful in understanding the basics of nonlinear material definitions. nI've marked up the Instron graphic that was referenced, with the black modulus line representing a linear material model, the black then blue dashed line representing a bi-linear model and then the red SS curve can approximate a multi-linear model.nMy understanding is that for a given strain ϵ2, the linear model would generally show unrealistically high levels of stress (pt , the bi-linear model would show stress values over 0.2% yield (pt C), and the multi-linear model could show stress values below yield (pt D). More generally, any areas over yield in a linear model will also be over yield in a bi-linear model, but may very well be under yield in a multilinear model.nIs my understanding correct? I suppose the multi-linear material model would be less conservative in this sense.nThanks.n