Nonlinear FEA with Von Misses Plasticity in 17-4 PH900 Stainless Steel 1

[JasonNicholson]

Software: Cosmos Advanced Professional
Purpose: To find the residual displacement of the tip.
Model: Plasticity Von Mises Model

Material: 17-4 PH900 Stainless Steel
Modulus of Elasticity: 28,500 ksi
Poisson's Ratio: .272
Density: .282 lbm/in^3
UTS: 210 ksi
.2% YS: 200 ksi

Approximate Stress Strain Curve:
TrueStrain TrueStress (psi)
0.00700 200000
0.00779 204101
0.00858 207878
0.00936 211383
0.01015 214658
0.01094 217733
0.01173 220633
0.01251 223380
0.01330 225990
0.01409 228478
0.01488 230855
0.01566 233133
0.01645 235320
0.01724 237424
0.01803 239451
0.01881 241408
0.01960 243299
0.02039 245130
0.02118 246905
0.02196 248627
0.02275 250300
0.02354 251926
0.02433 253509
0.02511 255051
0.02590 256554
0.02669 258021
0.02748 259453
0.02826 260852
0.02905 262220
0.02984 263559
0.03063 264869
0.03141 266152
0.03220 267409
0.03299 268642
0.03378 269851
0.03456 271037
0.03535 272202
0.03614 273346
0.03693 274470
0.03771 275574
0.03850 276660
0.03929 277728
0.04008 278780
0.04086 279814
0.04165 280832
0.04244 281835
0.04323 282823
0.04401 283797
0.04480 284757
0.04559 285703
0.04638 286635
0.04716 287555
0.04795 288463
0.04874 289359
0.04953 290243
0.05031 291116
0.05110 291978
0.05189 292829
0.05268 293670
0.05346 294500
0.05425 295321
0.05504 296132
0.05583 296934
0.05661 297727
0.05740 298511
0.05819 299286
0.05898 300052
0.05976 300811
0.06055 301561
0.06134 302304
0.06213 303039
0.06291 303766
0.06370 304486
0.06449 305199
0.06528 305905
0.06606 306604
0.06685 307296
0.06764 307982
0.06843 308661
0.06921 309334
0.07000 310001

Loading Conditions:
8 pound load at the tip is loaded and unloaded. The load is always vertical. Symmetry condition. Fixed at back end. See picture below.

Questions:
1. What are the difference between kinematic hardening and isotropic hardening? How does it affect the solution?
2. What is the proper control to use of these: Incremental Load Control Method, Incremental Displacement Control Method or Incremental Arc-Length Control Method?
3. Should I use large strain formulation along with large displacement formulation? Should I leave one out?
4. How do I troubleshoot when the model fails?

Currently I can't get the problem to solve with Incremental Load Control Method and the stress strain information above. We have tested this on exact problem on an Instron machine but I cannot get Cosmos to solve. I checked for buckling and it doesn't occur before yielding (load factor of 4). I am missing something. HELP!!!

-Jason Nicholson

 

[hondaknight1]

I do not know much about Cosmos any more, its been a long long time..

But I can say one thing, your figures do not quite match.
You give 0.2% YS at 200KSI do you mean 0.2% proof stress? Then you have the stress strain curve with 200ksi at 0.7% stress? Most FE packages like stuff to match up so if these numbers have been input then you might be confusing the solver.

 

[Jordonlaw]

Does this problem solve with linear-elastic properties ? You should do that first. Then I'd simplify the plasticity problem to a yield stress and post yield stiffness if Cosmos lets you do that. If this all works then in my experince its the input of the stress-strain curve that causes the problem (I have no experience of Cosmos by the way). The other thing to do is a simple cantilever with your material properties and take it post yield using a large strain formulation.

Hope this helps.

 

[corus]

Isotropic hardening is where the material yield surface changes uniformly in all drections. Kinematic hardening is used for cylic loading with a constant rate of hardening. A combination of the two is commonly used.
I'm not sure what the difference is between large strain and large displacment methods so how you can leave one out I have no idea. Use large displacement if the displacements are expected to be significant enough to change the load distribution, ie. the rod bends to such an extent that the bending moment will alter with displacement. I'd leave it out if possible as it adds a further non-linearity into the problem.
If you're applying a point load then you'll have problems getting a solution directly under the load position. Try and smooth it out a little. The stress strain curve you have isn't so non-linear as to cause problems in my opinion.

corus

 

[GBor]

The picture is too large to see the whole thing. You may want to "Red Flag" yourself and upload the file using the engineering.com attachment feature instead of embedding it into the post.

I thought isotropic hardening would only allow you to enter a post-yield modulus where kinematic hardening lets you put in the full stress-strain curve, but that may just be my software, which isn't COSMOS.

Looks like UTS should have been listed as 310ksi, but I think that is just a typo.

Like JordanLaw says, you need to run it through a linear analysis first and see how much displacement you get. If it is large compared to the dimensions of the part, you need to use a LaGrangian technique (Large Displacement) formulation. Not sure what a "Large Strain" Formulation is mathematically.

What error are you getting when the model fails? You say you "can't get it to solve", but what error are you getting?

 

[prost]

I think you can think of isotropic and kinematic hardening as part of the constitutive law that describes the stress-strain behavior of the material; examples of constitutive laws are 'linear elasticity', 'elastic-plastic,' 'Mooney Rivlin rubber'. There are an infinite number, since many laws are merely curve fits of experimental data.

The decision to use isotropic or kinematic hardening (for a plastically deforming material) depends on which (isotr. or kinem.) appears to fit the data best. How could you find out? For aluminums and steels, one way might be to run a simple tension-compression test, one cycle--take a simple dogbone (ASTM standard) coupon, watch the stress-strain curve as you load the coupon up in tension until the stress-strain curve starts 'bends' over--the material can be said be deforming plastically or 'becomes plastic.' Then load in compression with the same load. Now the question is how to interpret what you observed? If the stress when the coupon becomes plastic is greater in tension than compression (that is, the magnitude of the stress when it goes plastic in tension is greater than the magnitude of the stress when it goes plastic in compression), then the material is undergoing kinematic hardening. In isotropic hardening material, you might see a couple of behaviors--perhaps this stress in compression when the curve starts to deviate from a straight line is the same as tension, but the slope of the stress-strain curve after this stress is higher than the corresponding tension side. When someone says "isotropic hardening is when the yield surface expands the same amount in all directions'--this increased slope in the stress strain curve corresponds to this expanding yield surface.

The 'isotropic vs. kinematic' hardening decision will have great influence on your finite element results. As a rule of thumb, isotropic hardening rules work very well for single loadings. Kinematic hardening is generally needed when you for multiple (or cyclic) loadings.

There is something called the 'Bauschinger effect' which is a kinematic hardening related term. That's probably a whole 'nuther thread, which I am not qualified to talk about, since I have little experience with the Bauschinger effect. If I am not mistaken, you can observe the Bauschinger effect with some materials that undergo kinematic hardening, or perhaps you don't observe it.

Please feel free to chime in to correct me. I am just getting started with this, so my understanding is perhaps more meager than the plasticity experts on this panel.

 

[prost]

To use large strain formulations in FE software, you're going to need a material constitutive relation that uses compatible stress-strain measures. If you've ever had a continuum mechanics class, you might recognize these terms for large strain measures: Eulerian, Cauchy, Almansi, Lagragian, Green, St. Venant. Similarly, you have corresponding stress measures: First Piola-Kirchoff, 2nd Piola-Kirchoff, and Cauchy. What's the difference between all of them? Coordinate systems! Are your strains (or deformation gradients) in terms of a global (spatial) coordinate system, in which the coordinate system is stationary, or are strains in terms of a material coordinate system, in which the coordinate system follows the deformation of the material? These strain and stress measures are valid for arbitrarily large strains and displacements.

The key here is what stress and strain measures does the constitutive law of your material use? If you have a Mooney Rivlin rubber, the strain energy is function of two parameters, and two deformation tensor invariants. Stresses are derivatives of this strain energy function relative to the various deformation tensor invariants. This is all quite complicated if you've never seen it before, but the bottom line is if you have large strains and displacements, you need to use a nonlinear material in your FE with the appropriate constitutive law. For instance, I used ADINA circa 1994, had my own constitutive law to use, but had to use a large strain measure that ADINA could give me--I had to program a Fortran subroutine that returned 2nd Piola-Kirchoff stress stress, given a Lagrange strain--these both used a material coordinate system, so were completely compatible (consistent? I forget the precise term for stress and strain measures that use the same coordinate system).

 

[crashengineer]

1. What are the difference between kinematic hardening and isotropic hardening? How does it affect the solution?
2. What is the proper control to use of these: Incremental Load Control Method, Incremental Displacement Control Method or Incremental Arc-Length Control Method?
3. Should I use large strain formulation along with large displacement formulation? Should I leave one out?
4. How do I troubleshoot when the model fails?

Jason
1. Use Isotrpoic hardening. Kinematic is used for cyclical loading. In your case both will give similar results.
2. Use incremental disp control. Whenever you are trying to find load-deflection curve in the non-linear range apply displacement. Load can decrease displacement cannot.
3. start out with small-strain and small-displacement. You dont need large disp formulation. A good example of large displacement is a "fishing rod" when displaced at the tip it has a very large curvature.
4. troubleshoot :?? change variables slightly to see the difference in answers. For example : loading direction, loading location....