Learning Forum

[momidor]

Hello, 

In the snapshot is the very thin crack in the metal. At the tip of is the ~ 500 MPa which is over the ultimate value which is for this particular material is 360 MPa. 

The question is how to interpret such issues ? 

Thanks in advance 

 

[Sandeep Medikonda]

Hi Momidor,

  Please check out this similar discussion with Rashi, where she was seeing a similar problem.

Regards,

Sandeep

[Ireneusz Malec]

@Sandeep could you provide the link from your post dated on 21.08.2018. The content is not more available unter the above link. Thank you.

 

[momidor]

 Hi Sandeep, 

Thanks but  regarding commands. Should I just write ERESX, NO under Static Structural tree ? Like below ? 

[peteroznewman]

 Hi momidor,

I was glad to learn about ERESX, NO from Sandeep.  Before I knew about that, my approach was to reduce the element size around the peak stress area so that the nodal values and integration point values were not so far apart.

You don't show the element size around your crack tip.

You don't say what the material model is. I assume you have added plasticity, but if the material was only linear elastic, we know that yield and ultimate strength values in the material database are not used to prevent stresses from exceeding those values during the simulation.

Please reply with details on your mesh and the material model.

Regards,

Peter

 

[Sandeep Medikonda]

Yes, this would show you the stresses at the integration points instead of the extrapolated values. Also, note that this command snippet needs to placed under the Static Structural branch/folder and not the Solution branch.

Regards,

Sandeep

[momidor]

Hi Peter Hi Sandeep

That is the model of welded joint with discontinuity. Material is standard stainless steel.

 

[peteroznewman]

Hi momidor,

Stainless steel has plasticity (or ductility), which means when the stress exceeds the yield strength, the material doesn't elastically stretch any further but instead flows and the part will not return to its original shape when the load is removed.  

A crack creates a stress concentration in the part. Under load, the stress gets concentrated at the crack tip and the material plastically flows to redistribute the stress, effectively blunting the sharp crack tip.

That behavior can be simulated in ANSYS by adding Plasticity. There is a lot to learn about plasticity. A good video to watch is at this link.  Once you have added plasticity, the stress will not exceed the ultimate strength of the material the way it does when you have only linear elastic materials in the model.

There are many Plasticity models to choose from. The simplest is the Bilinear Isotropic Hardening.

You only need to supply two numbers.

Type 0 for the Tangent Modulus to create an Elastic, Perfectly-Plastic material, which you may have read about in Engineering textbooks.

Regards,

Peter

[Sandeep Medikonda]

Momidor,

  You don't have any plasticity data in there. I am not really surprised to see a high stress...You would have to include some bi-linear plasticity data in there...The ultimate strength that is provided there is mostly for factor of safety calculations.

Regards,

Sandeep

[momidor]

 Hi 

Thanks for answers but let me ask.

Hereunder is the graph of the C-Mn carbon  steel just from the lab. 

1. If the "ERESX, NO" command is used, the maximum result is yield.  But I can imagine simulation where load creates stress near ultimate (Rm). Does the "eresx" sets a limit too low, i.e on yield instead of near ultimate ?

2. What is the full syntax ERESX, NO ? Should I type this command in ADPL, shouldn't I ? 

 

[Sandeep Medikonda]

In FEA, we are calculating the stresses at the integration points due to the numerical integration scheme used (Gaussian integration for example) and then extrapolating these values from the integration points to the nodes. When you put the ERESX, NO command in and evaluate the results, this will copy the integration point results to the nodes.

When the elements are too big, the distance between the nodes and the integration points can be high and you could see a significant difference. However, when you use this command snippet we are telling the code to not extrapolate. The values calculated at the integrations points are always the most accurate ones. Just right-click on Static Structural and insert Commands (APDL), this is just a text editor. In this, type in ERESX, NO as shown in this post.

In your case, the problem doesn't seem to be due to this but rather due to the lack of plasticity included.

Regards,

Sandeep

 

[momidor]

Thanks Sandeep, 

Works fine. Yield is set as 250 MPa. 

Another story is how to force to converge the simulations when bilinear isotropic hardening is set, because is at structural static is hardly works.  

All the best

Works fine. 

[Sandeep Medikonda]

Momidor,

  There are several best-practices in this forum and also outside on obtaining convergence. Check out some of these resources:

Resource 1

Resource 2

Resource 3

P.S: If a certain post answers your original question, please mark it as a solution so that it might help someone else in the future. Just as ERESX from the other post helped you.

Regards,

Sandeep

 

 

[momidor]

Hi Sandeep, 

 

One more issue before I press "solved".

Here is the steel bar axial stretching by 180 MPa. Although I used ERSX, at the narrow tip of the crack is the still crazy value... 577 MPa. Bilinear isotropic hardening was not set. How to approach to such results?

[peteroznewman]

Momidor,

This is not a crazy value, the stress is expected to be high at a crack tip. For simple geometries, there are even analytic equations to compute the stress. The linear elastic material model does not prevent the stress from exceeding yield. It's confusing that the Yield Strength is listed in the Engineering Data, but that is only there to compute Factor of Safety in post processing. It is not used during solving.

If you use large elements and ERESX, NO to prevent extrapolation, you can miss the true high value of stress for the linear elastic material. But as you use smaller and smaller elements and the mesh follows closely around that small curved crack tip, the result will converge on the true linear elastic material stress, which could be much higher even than you have shown.

To repeat what has been said before, these stresses cannot be achieved in a real material, or in a simulation that includes plasticity. Please add the bilinear isotropic hardening that includes the yield strength of the material and let plasticity distribute the stress around the crack tip.

Of course, there are convergence problems to solve once you do that, so you may have some new questions while resolving those problems.

Kind regards,

Peter