## General Mechanical

#### Mohr-Coulomb material model

• krishnadondapati
Subscriber

Hi,

I am using workbench 18.1 to simulate 4 point loading test on rock with Mohr - Coulomb criterion, but even after reaching its tensile limit its not showing any material failure.

so can some one help me to solve it

Thanks

• peteroznewman
Subscriber

Please attach your project archive for examination.

• krishnadondapati
Subscriber

Hi

Here I'm attaching project file

• krishnadondapati
Subscriber

Sorry I tried to attach the file but its size is 250MB may that's the reason it wasn't uploaded

• maurya
Subscriber

Hello krishna

According to me loading roller must have frictional contact with stone.

I know that brittle material fails with principal stresses. So use the principal stresses.

In ansys if you want to see the failure you have to use the material non linearity, otherwise it interpolates the linear elastic curve.

For accurate answer wait for Peter sir reply.

Below file is attached with friction having zero plastic strain.

thanks

Deepak

• peteroznewman
Subscriber

Hello Krishna,

There are many students here who have used the Mohr-Coulomb material model. If you edit the Title of this discussion and use "Mohr-Coulomb material model" instead of "Dondapati", some of them may even read this thread!

Did you read the ANSYS documentation for Mohr-Coulomb material model? It's under Mechanical APDL > Material Reference > Geomechanics. There you will find this

The model defines yielding when the combination of pressure and shear stress reaches the cohesion of the material particles. Yielding occurs when the shear stress on any plane in the material reaches this criterion:

The value of c in your material is 19 MPa.

Did you read about the Mohr-Coulomb Stress Safety Tool in the documentation? It's under Mechanical Applications > Mechanical User's Guide > Using Results > Structural Results > Stress Tools.

Read the full page, not just this excerpt.

I recommend a few changes to your model.

Use Symmetry.
You have two planes of symmetry, half way along the length, and half way through the front-to-back depth. If you split the model on those two planes, you cut the volume to 1/4 and can therefore use smaller elements to get a more refined mesh that is still within the Student license limits. Another advantage is those planes take up 5 degrees of freedom on the body, so a displacement on the y coordinate is all that is needed to have a static model.

Replace Contact with Split Faces
Use a thin face to apply a displacement of Y=0 on the bottom and another thin face to apply a Y = -0.1 displacement on the top. It takes less time to solve if you have displacements than if you have contact.

Here is the result of the Stress Tool:

This shows the material has exceeded the failure threshold above about 40% of the applied displacement. However, this shows the failure is at the narrow face. Perhaps a wider face and smaller elements next to the face would be better for showing when the material on the center plane gets to the failure point.

• krishnadondapati
Subscriber

Hi Maurya,

I'm not able to open your file as yours is a future version. Can I know which version it was saved in?

Thanks

• peteroznewman
Subscriber

Hi Krishna,

Much trial and error has gone into the attached model which I am going to let run overnight. I see that it can be difficult to create a model that converges up to the point of failure. Take a look at the attached ANSYS 18.1 archive and I will report in the morning if it made it further than the models with larger element sizes that ran faster.

Regards,

Peter

• krishnadondapati
Subscriber

Thank you Peter

I'm going through your suggestion regarding M-C stress tool.

Regards

Krishna

• maurya
Subscriber

Hello krishna

it is ansys 19 file.

But you must see the Peter sir ATTACHED FILE.

he didn't use the friction at top contact as well as bottom.

But he is genius he removed the support by displacement (with 2 direction free)  and loading tool by surface displacement with free movement in other directions.

So there is no meaning to check my file

All contact removed problem became linear.

regards

Deepak

• peteroznewman
Subscriber

Hello Krishna,

The file I provided above ran slow on a computer with only 16 GB of RAM. The solver needed more than that to run in memory, therefore it ran for a much longer time using the slow HDD as storage. It did eventually stop solving but managed to converge past the point where failure is predicted.

Here is the Safety Factor Plot with a Probe showing SF < 1 near the center of the beam on the bottom face. This is a view of the bottom of the beam.

Failure at the support was predicted at much less displacement, but that failure could be delayed by spreading out the contact force over a larger area. The failure at the center of the span cannot be mitigated, so is mostly independent of the testing setup.

The model ran for 103 iterations before stopping due to a failure to converge.

Your model had an applied displacement of -0.1 mm on the center pair of faces. Where did that number come from?
My model was able to solve until -0.0454 mm then could not converge, but the model predicts failure at the center at this displacement if you ignore the failure at the support face. The force reaction for this displacement was 977 N for the quarter model or 3908 N for the full model.  Do you have experimental data to compare this with?

Regards,

Peter

• maurya
Subscriber

Hello peter sir

is this  linear or non-linear problem after removing contact.

Material and geometric non linearity is not there, i confirmed. is  SOF  of Mohr-coulumb criteria?

thank you

• peteroznewman
Subscriber

Hello Deepak,

This could have been a linear analysis, since I am looking for the point where the material reaches its yield point and it is elastic up to that point. I set Large Deflection to On but that was probably unnecessary.

I have learned that a combination of using SOLID65 elements and the CONC material model will allow cracking to be computed on an element basis.

Regards,

Peter

• maurya
Subscriber

Hello sir

in my fracture mechanic tutorial i created the cracks in apdl. Upto ansys 18 version only formation of crack is there without propagation,in ansys 19 they provide the benefits of propagation.

KSCON i know, concentrated keypoint about which all analyses occur.

Variation of stress concentration without moving the crack.

• krishnadondapati
Subscriber

Hi Peter

Thank you for your info, Sorry for the late reply as I'm working on another one.

In the experiments, failure is initiated in the middle of the beam and also want to compare the results with the experimental strain values.

can you please give me an insight on identifying the displacement at the last converging point. Also, one more thing about the selection of no of steps and step time.

Thanks & Regards

Krishna

• peteroznewman
Subscriber

Hi Krishna,

Sorry for the late reply also.

This post has some relevant insights on Step Control.

When the solver fails to converge and stops, the Deformation plots often show the Unconverged solution, which is not what I want to see, I want to see the last converged solution. To see that, go to the Tabular data table for that result and scroll the the end. In a 2 step solution, you will see the time is 2 on the last row but 1.235 on the second last row for example.  Right click on that second last row and select Retrieve this result.  Now you are looking at the last converged result. You can probe the displacement anywhere on that model.

You can do the same for strain, picking the second last row.  Since your strain gauge was aligned along the length of the beam, request Normal Strain and pick the appropriate axis for the strain component.  You can then probe to pick off the value at the location where the strain gauge was mounted to compare with the experimental result. You can select a node nearest the strain gauge and plot the time history of strain.

Kind regards,
Peter

Viewing 15 reply threads
• You must be logged in to reply to this topic.