Bearing Stress Simulation

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firasb posted this 29 November 2017

Hello,

I'm trying to perform bearing stress simulation to compare against the results of an actual experiment. 

In the experiment, the (aluminum) material began to yield at about 6000N of force. The bearing stress ( = force / bearing area ) was calculated to be 297Mpa ( 6000N / 0.0002016 m^2 ). However, the simulation produced some different numbers.

My question is: How do you find bearing stress in ANSYS Mechanical?

Below are some screenshots of my simulation so far. Please help.

 

Normal Stress in Y-direction

 

Elements selected in the bearing area

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peteroznewman posted this 29 November 2017

Verification of model results with experimental data can be challenging.

Please describe in more detail the experimental setup. It looks like you have an aluminum plate with a hole in it.
What material was the shaft that was in the hole?
What was the diameter of the hole and the diameter of the shaft?
How long is the shaft and how was the load applied to the shaft?
Was the plate and the shaft in a tensile testing machine that was measuring force and displacement?
How was the onset of yielding determined?  Was it the 0.2% offset on the stress-strain curve?
What is the area you are using to compute the stress from the force?

I would create a model of both the shaft and the hole and use frictional contact.
With that model, a small contact patch area will develop, but you will need a higher element density around the hole and a much higher element density near the contact point.

The model you have that used bearing loads is a simplified method of applying bearing loads to a structure and is not recommended for the detailed analysis of the contact stresses of a shaft to a hole.

 

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firasb posted this 30 November 2017

Thank you for your reply, Peter.

Here are the details of the experiment: An aluminum coupon (100mm long, 25.4mm wide, 3.175mm thick) was pulled on an Instron machine. The coupon was supported at the bottom by a large clamp and at the top by a 6.35mm diameter 38.1mm long steel alloy pin that goes through a 6.35mm hole. The hole was reamed and the diameter of it is within 0.025±mm of the pin diameter.

 

 

Here is the data output of the experiment (force/displacement chart):

From the chart I made the assumption that the 6061 aluminum started to yield at around 6000N.

The instron pulled the shaft straight up from both of its sides. After the experiment, there was almost no deflection/deformation in the shaft as it is much harder than the aluminum coupon.

To compute the bearing stress, I divided the force at yield onset  (6000N) by the bearing area which is thickness of coupon multiplied by the diameter of the hole (0.003175m x 0.00635m). Please let me know if you'd like me to provide any other information.

 

Regarding your suggestion: I think I can learn from tutorials how to model the shaft separately and create frictional contact between it and the hole. I can also increase the element density of the mesh in the areas you mentioned. However, how do I find the bearing stress? Do I average out the y-direction normal stresses for all the elements in the top half of the hole? Sorry for my confusion. Your help is highly appreciated.

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peteroznewman posted this 30 November 2017

Thanks for the detailed information!

Bearing Stress is calculated using the definition that you have shown and is useful for hand calculations for a first order analysis prior to an FEA model. 

The experiment is a physical test that creates some data which an FEA model can attempt to simulate.

An ANSYS Bearing Load applies forces to one half of a cylindrical face in a radial direction on the face with a magnitude proportional to the cosine of the angle of the face normal to the load direction. This means a factor of 1 on the top of the hole at a 0 degree angle to the load direction and a factor of 0.5 at a 60 degree angle and zero at 90 degrees and beyond. That force profile is much better than just putting equal parallel forces on all the nodes around the entire cylindrical face which is what a Force Load would do and creates tension on the opposite side of the hole!

A more accurate representation of the stress in the part than the Bearing Load is to model a cylindrical pin and use contact.

One missing dimension is from the edge of the hole to the top of the plate. Recommendations are 1.5 to 2 diameters, but looking at your images, it looks like you are at 1.0 diameters or 6.35 mm, so I created a model at that dimension.

You should always take advantage of symmetry in your models. I can slice this model down the center twice, and keep the left rear quarter, the other quarters have identical stress, however, a quarter model will only show a quarter of the load, so if you pull the pin 1 mm up, the model would show 3 kN not 12 kN that the experiment delivered.

You should also try to get brick-shaped hex elements, rather than tetrahedral elements because they deliver higher quality results for the same number of nodes. You should also have smaller elements where the stress changes rapidly. The image below is a quarter model of your plate. A pin of the same diameter is hidden in this view and is modeled as a rigid body.

The experiment included taking the material past yield and into plastic deformation. When you build a model you can add plasticity to the material model. There are several types of plasticity and I chose a simple Bilinear Kinematic Plasticity. The model displaces the pin up 1 mm and the resultant force required to do that is plotted below. Remember: 3 kN in the model = 12 kN in the test.

The plasticity model used a Yield Strength of 450 MPa and a Tangent Modulus of 300 MPa.
A multilinear plasticity model would do a better job of creating the more gradual curve shown in the experimental data.

You are welcome to have a copy of this model but it was built in ANSYS 17.2 so you would have to upgrade to open it.

Regards, Peter

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firasb posted this 30 November 2017

Peter, 

Sorry for the missing info. Hole center to edge distance is 12.7mm which puts me at a ratio of 2.

Thank you so much for all the information. I'll try to follow your method.

I'd like to use your model as a guide. How can I obtain your model? And is it ok to reach out for help in the next few days if needed?

peteroznewman posted this 30 November 2017

The ANSYS 17.2 archive file is attached, but you can't open that in 17.1.
The attached file has a .txt file extension. Use Save Target As to download, then change the file extension to .wbpz before you use File Restore Archive.

I didn't move the hole away from the edge, but you can change the dimension in DesignModeler. The ratio from the edge of the hole is 1.5 if the center is at 2 diameters from the edge.

Ask for any help you want. Others may comment also.

Regards, Peter

Attached Files

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firasb posted this 02 December 2017

Hi Peter,

I started fresh and created a pin geometry with diameter of 6.35mm and a hole 6.375mm in diameter. A frictional contact condition was created where I made the cylindrical face of the pin the contact and the hole wall the target. Friction coefficient was set to 0.2.

Boundary conditions twin forces in the Y-direction, one at each end face of the pin. And two fixed supports at the bottom of the coupon to simulate flat jaws of the clamp.

I tried very hard and long to get ansys to produce a solution but it keeps showing error messages. I've tried changing the mesh, geometry, solution type and other variables, but can't get past the error messages.

Could you please point me in the right direction?

Below are some screenshots of the mesh, contact and error messages..

 

firasb posted this 02 December 2017

Ok, I was finally able to get a solution but I had to go with a ridiculously large element size.. My machine usually doesn't have a problem processing a much larger number of elements/nodes than what's in this example.. so I'm stumped. Not sure where I went wrong. Your help is highly appreciated.

peteroznewman posted this 02 December 2017

Getting frictional contact to solve is difficult if you don't know all the steps to follow, and adding in plasticity makes it very difficult to solve to the full load. You can show your appreciation by clicking "Like" on the posts that are helpful.

1) Make sure the geometry is in contact

I bet that the pin and hole are concentric, which means when you made the hole larger than the pin, they are no longer touching. I calculate a radial clearance of 0.0125 mm. You must translate the pin up by that distance to get it to touch the top of the hole. That way, the solver can start with the mesh in initial contact.

2) Add a Contact Tool to the Connections Folder

When you Generate Initial Contact Results from the Contact Tool, it will tell you if there is initial contact or not, and the size of the gap or penetration. A very small penetration is good, a very small gap is a problem. If you have a very small gap, use Adjust to Touch.

3) Use Adjust to Touch

Go to the details for the Frictional Contact where in Geometry Modification, Interface Treatment you will find Adjust to Touch. This will take your very small gap and close it.

4) Turn On Auto Time Stepping instead of Program Controlled

In Analysis Settings, Step Controls, change Auto Time Stepping from Program Controlled to On. Define By Substeps. Set Initial Substeps to 50, Minimum Substeps to 10 and Maximum Substeps to 500. You don't want the solver trying to apply the full load in a single step, since the displacement can be so large that the pin ends up meters past the hole. You need to tell it to take 50 substeps as an initial step size. The solver will automatically increase the step size as it converges on each substep. Also, if you want to plot a curve on the way to the full load, you want to specify a minimum step size, so you are guaranteed at least 10 points on your curve.

5) Turn on Large Deflection.

Under Analysis Settings. This is essential when using a plasticity model that will see large strains.

6) Displacements are easier to converge on than Forces

You want a Force-Displacement curve, so you can either apply force and calculate displacement, or you can apply a displacement and calculate reaction force. It is much easier for the solver to do the later.

Applied displacement is required when using plasticity if you want to simulate beyond the point where the force-displacement curve approaches a maximum. 

You haven't used symmetry as I advised above and there is a potential problem in your model with forces applied at the two ends of a pin that extends so far from the plate. Is each end of the pin exactly the same distance from the face of the plate? If not, there will be a moment that tries to tilt the pin and you don't want that. Symmetry would fix that. However, I strongly recommend you delete the force BC on the ends of the pin and replace it with a displacement BC. You can specify a 1 or 2 mm Y displacement. You could either set X and Z to zero or leave them free. There is less chance of a convergence error if you set them to zero.

It probably won't matter in this model, but the best practice is to put the contact side of the contact pair on the softer material and the target side of the contact pair on the harder material. So in your case you should flip the contact pair. If the two materials are similar modulus, then the contact side goes on the geometry with a smaller radius and the target side goes on the larger radius (or flat) side of the contact pair.

The reason the ridiculously large elements worked was because they "cut the corner" on the radial clearance to give you an initially closed contact.  Go back to smaller elements after you translate the pin to be in contact.

Good luck,

Peter

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firasb posted this 03 December 2017

Hi Peter,

I used symmetry initially as you suggested but switched back to full geometry because of the trouble I was having in trying to get ANSYS to produce a solution. I wasn't sure if it was causing a problem so I eliminated it in the beginning. Thank you for the guidelines and tips in your last post by the way. After following them, I was able to get the model to work. You were correct about the pin and hole being concentric. I shifted the pin up to touch the hole and now the model works!

The model's accuracy is another story though. I added a force reaction at the contact area and it showed a maximum force of 18180N ( 4545N x 4 ) in the y-direction at 2.5mm pin displacement. That's 37% higher than the measured force of 13300N which the instron recorded. Also the mesh might need some work. I played with several mesh features but can't seem to produce a mesh as neat as the one in your model. I think a better mesh and a properly set up bilinear or multilinear properties chart for aluminum will get me closer. Any tips are  appreciated although you've already helped a lot.

Below are some screenshots of the model.

Thank you for your patience and your time. It's helped me a lot.

-Firas

 

 

 

 

 

peteroznewman posted this 04 December 2017

Hi Firas,

Here is a video to show you how I made my mesh.

 

Regards,

Peter

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peteroznewman posted this 04 December 2017

One material property that is very important in a plasticity material model is Elongation at Break. For 6061-O Aluminum, Elongation at Break = 25%. That means in a tensile test sample, when Total Strain > 0.25 the sample fractures in two pieces.

In a model with 6061-O, where the load causes plastic deformation, the model is only valid up to the point when the maximum Equivalent Total Strain = 0.25.

In the hole shear model, the Equivalent Total Strain of 0.25 occurs at a pin displacement of 0.46 mm. After that point, the model is no longer simulating reality.

This is why your model is overestimating the experimental force at a pin displacement of 2.5 mm, it is way past the point with valid strain values. The image below shows the Total Equivalent Strain at 1 mm of pin displacement. The color purple and above shows the material that has exceeded 25% elongation.

In order for a model to simulate beyond the Elongation at Break value, it must include an element failure algorithm. In other words, when an element exceeds 0.25 Equivalent Total Strain, it is removed from the model, weakening the structure, which results in less reaction force needed to continue to displace the pin further. This capability is included in an ACT extension called EKILL, but that is not available for the Student license. However, the Explicit Dynamics solver does include the capability to include element failure.

Regards,

Peter

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firasb posted this 07 December 2017

Hi Peter, it's finals week but I'm looking forward to trying out your latest tips for the mesh and element failure algorithm as soon as I'm done next week. I'll post my progress. Thanks again!

peteroznewman posted this 10 December 2017

Hi Firas,

Good luck on the final exams!

I've heard that Explicit Dynamics can solve an intractable contact problem that the Static Structural solver fails to solve, but I had not tried it myself, so I decided to try it out on your problem.

Explicit Dynamics was setup to move the pin 2.5 mm in 0.25 seconds, ramping up the velocity from zero to 80 mm/s. I used a very coarse mesh to reduce the computational time, but it still took nearly 33 hours of computation on 15 cores. It took 9.77 million time steps to get to 0.25 seconds.

The material model included a Principal strain failure of 0.25 and 0.5 shear strain. This helps the elements to not fail in "compression."

The model is too coarse and the velocity too fast to be useful for quantitative results to represent your experimental data, but it is useful to understand the concept that as the material fails, those elements disappear and no longer carry any load, which is transferred to adjacent elements.

 

Best regards,

Peter

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firasb posted this 18 December 2017

Hi Peter,

 

Thanks for the wishes. I think I did alright.

I'm back to the simulation now. I tried following your mesh tutorial video, but when I went to generate the mesh for the coupon body it gave me some errors and only generated a mesh on the outer face of the coupon. I think the new mesh settings might be conflicting with my older settings and I forgot which settings I had modified originally. Is there a way to get the default mesh settings back? (I searched google but came up empty handed). Also, is there a way to edit the title of this thread? 

Below is what I have so far.

Thanks and best regards,

Firas

 

peteroznewman posted this 19 December 2017

Hi Firas,

To edit the Title, just edit your first post and you can change the title.

Meshing

First check that your geometry is sweepable. Right click on Mesh and Show Sweepable bodies. If your coupon doesn't light up in green, then it is not sweepable. Change to Multizone method instead of Sweep, or edit the geometry to make it sweepable. That means the same topology on the top and bottom faces.

Second, Clear Generated Data on Mesh, then try to Generate Mesh.

Defaults

I think the 17.2 default Mesh Sizing function is Curvature, but you can try different sizing functions and see what it does to your mesh.

If you Archive your model, you can attach the .wbpz file to your post and I can download it and take a closer look. I can upload a file, but only in 17.2 or 18.2 so you can't open that without doing an upgrade. If you upgrade, I have a 17.2 and 18.2 archive I can attach.

Peter

 

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firasb posted this 20 December 2017

Hey Peter,

I made the top and bottom surface topology the same by removing my original fixed support. I just chopped it off in the Geometry Modeler. After doing that, I was able to generate the mesh! It looks great, thank you.

I want to switch my coupon material to CFRP and the hole geometry from cylindrical to tapered. That should conclude my simulation.

To switch to CFRP, will I need all these material properties below? Or should I be able to get close with only some of them?

Also, I wasn't able to get the Explicit Dynamics model to work and I looked for Elongation at Break in the material properties but couldn't find it. Could you elaborate a little more on the process to set that up if you would please?

Unfortunately, we're stuck here at school with the 17.1 and 18.1 versions. IT techs will not upgrade our software. Otherwise, file sharing would have been a great way to go as you suggested.

As always, thank you for the help.

-Firas

peteroznewman posted this 20 December 2017

Hi Firas,

Carbon Fiber Reinforced Polymer (CFRP) is an orthotropic material, so you will need all the Othotropic Elasticity properties. The Stress and Strain Limits are optional as they are only used to calculate factors of safety, not to calculate stress and strain. Note that Young's Modulus is 22 times larger in the X direction than the Y or Z directions, since that is the direction the carbon fibers are running. That means you have to know what direction the fibers are running in the coupon being tested. Let's assume the fibers are running along the coupon from the clamp to the hole. 

Using an orthotropic material in ANSYS means you have to define an element coordinate system where the X axis of the new coordinate system points in the direction of the global Y axis, since the coupon in your model is along the global Y axis. This post has instructions on how to do that.

Elongation at Break is the strain when rupture occurs. Looking at the table you provided, the strain in the X direction when rupture occurs is 0.0092 or 0.92 % This is consistent with the data in Figure 4 of this paper. However for your test, the material in contact with the top of the pin will be in compression and you can see that the values for failure in compression (either stress or strain) in the X direction are about half that for tension, so this is where the coupon would begin to fail.

I will post a whole new discussion soon on Explicit Dynamics using your geometry and aluminum material and will circle back and provide you a link.

Finally, if you add taper to the hole, you will no longer be able to use symmetry through the thickness. You can still use symmetry through the width.

Best regards,

Peter

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firasb posted this 04 January 2018

Hi Peter, Happy holidays to you, I hope you had a good break!

I'm trying to finish the final leg of the simulation. I started the CFRP simulation using everything I've learned from this thread so far.

A few issues I've run into.. 

- Mesh: The Aluminum coupon mesh settings are not working as well for the CFRP coupon. I suspect this is due to the tapered hole throwing things off..

- Sliding: With a Y-displacement of only 1mm (just to test the model), the coupon slides on the pin in the direction of the open taper to keep the angle of the hole wall as close to parallel as possible with the pin surface. 

Below are photos of what I have so far. What can be done to correct those issues?

By the way, I'm planning to conduct the CFRP test within the next two weeks, so I'll have some data for comparison. After that I can fine tune the model.

I won't be able to run an Explicit Dynamics analysis because of the massive processing power required unfortunately.. but the video you posted above is very interesting. Thank you for sharing that.

Peter, thank you as always for your help.. and Happy New Year!

Best regards,

Firas]

 

Mesh (Z-direction)

 

Mesh (X-direction)

 

Tapered hole:

 

Coupon sliding and element triads:

peteroznewman posted this 04 January 2018

Hi Firas,

Good progress!

You can have a sweepable body that includes a tapered hole, so the mesh methods should work.

Frictionless contact between the tapered hole and the pin provides no lateral force to oppose the bending force that contact on one side of the hole with the pin generates. You can change from frictionless to frictional contact and that should provide an adequate frictional force to oppose the bending forces. If the coefficient of friction is low enough on your physical sample between the hole and the pin, the sample will slide just like your simulation.

I played around with Explicit Dynamics to crush a concrete cylinder, with and without steel rebar, as well as a 2D simulation of a lattice. See the three videos.

Best regards,

Peter

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