Interference Fit in ANSYS Workbench 17.1

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firasb posted this 2 weeks ago

Hello,

I would like to run a bearing stiffness simulation. 

I have a CFRP coupon and a pin modeled as separate bodies, and the pin must be pressed into a hole in the CFRP coupon as shown in the screenshots below. The pin is 6.35 mm in diameter all the way, and the hole in the coupon is 6.35102 mm on the pin side and 6.3373 mm at the other side. I looked up tutorials to learn how to perform an interference fit and they all instruct to to right-click on "Geometry" then select "Import Geometry". Unfortunately, that option does not show up for me when I right-click on "Geometry" and I'm not sure how to proceed. 

Please help me press the pin into the coupon.

 

Iso-view of pin and coupon:

Side-view of pin and hole profile:

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peteroznewman posted this 2 weeks ago

In Workbench, create a Static Structural model, which has a Geometry cell. When you open DesignModeler from the Geometry cell, you can construct the geometry you show above.

Draw the pin inserted in the hole, not outside the hole. The pin solid will interfere with the hole.  Construct the center of the pin in the center of the hole so that it interferes equally in the X direction and the Y direction.

Close DesignModeler and double click on the Model item which will open Mechanical. The geometry you created in DesignModeler will automatically show up in Mechanical. Create a frictionless contact between the hole and the pin and add a fixed support to the pin.

In Mechanical, click on Analysis Settings and in Step Controls, type 3 for number of steps.  Step 1 is when the interference is resolved and there are no other loads. Step 2 is when you pull on the plate (or pin) with a unit force in the X direction, and step 3 is when you pull on the coupon in the Y direction.

Add a force to pull the coupon in the X direction.  The value is 0 for step 1, 100 for step 2 and 0 for step 3. Add a force to pull the coupon in the Y direction. The value is 0 for step 1 and 2, and 100 for step 3.

The solution shows the displacement. Look at the solution at time step 2. Divide the force by the displacement in X to get the bearing stiffness in the X direction. Look at the solution at time step 3. Divide the force by the displacement in Y to get the bearing stiffness in the Y direction.

If you have more questions, follow these directions to attach a project archive and remind us what version of ANSYS you are using (17.1).

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firasb posted this 2 weeks ago

Peter,

I want to compare the results of the simulation to experimental results where there is only force in the Y direction. So are 3 steps necessary? Or should I just do 2 steps where step 1 is 0 and step 2 is the force in the Y direction from the experiment?

Also, the pin is made of alloy steel and the coupon is CFRP. When I drew the pin inserted in the hole the software clumped them together as one body but I want them to be separate (because of their different properties). How can I create them as two separate bodies or entities?

Attached is the project archive as you instructed. It is in version 17.1

Attached Files

peteroznewman posted this 2 weeks ago

firasb,

To create a separate body for the pin, set Extrude 6 to Add Frozen rather than Add Material. Then you will have two separate bodies.

Use the YZ plane to Slice All bodies to cut the pin in half, then suppress the half of the pin that is not needed.

If you are only interested in the Y direction, then you only need 2 steps, not 3.

 

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peteroznewman posted this 2 weeks ago

firasb,

I did the step mentioned above, and noticed a very small round protruding from the hole. Was this supposed to be removing material on the edge to make a blend, rather than a protrusion?  Either way, it is a mistake to include such a small feature when calculating bearing stiffness.

When you say "the hole in the coupon is 6.35102 mm on the pin side and 6.3373 mm at the other side" you mean that the hole has a taper. I edited your Sketch 2 that makes the hole to have those two radial dimensions. Now there is clearance on one side and interference on the other side.

I also suggest you make the two sides of the pin the Fixed Support, and apply a Force to the -Y face of the coupon.

Under Analysis Settings, here is the two step settings for step 1. Note that Auto Time Stepping is On and Initial Substeps is 100.

The Force is zero in step 1 then 100 N in step 2.

The first time I tried to solve this, the solver had difficulty. I edited the contact in the Geometry Modification to offset the contact surface by -0.006 mm, which is most of the 0.00635 of the interference, and ramp the -0.006 to zero in substeps. 

That was not sufficient. I added a displacement BC to the Y face of the coupon and set displacement to Free, Free, 0 to make sure that the coupon would not slide off the pin, since this is a frictionless contact. Adding friction is another way to prevent this problem.

When it nearly passed step 2, I added the APDL command neqit,50 so that it would not give up after 26 tries.

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firasb posted this 2 weeks ago

Peter, 

Thank you so very much for the help so far!

To answer your question regarding the different hole diameters, yes the hole is tapered. The round protrusion you found was a mistake. I was trying to put a small radius on the edges of the hole as to avoid unrealistically high stress concentrations in those regions. I went ahead and removed the round features per your instruction.

I set up the boundary conditions so that a 4 kN force pulls on the edges of the pin in the +Y and the bottom face of the coupon was set as a fixed support. I worked through your steps above up to the "Analysis Settings" step. After that, I managed to run a quick solution using a very coarse/raw/simple mesh. The deformation solution is very much inline with the experimental results (which can be seen below). Experimental results show that 4 kN of tensile force applied translate to approximately 0.11 mm of displacement, which is almost exactly the result of the simulation. I'm going to try to refine the mesh by following your instructions in an older thread and then attempt to run the solution again. 

What do you recommend to improve upon this? Is my process sound in your opinion?

Experimental results:

 

Simple unrefined mesh:

 

"Top of hole" Displacement with 4 kN force applied:

peteroznewman posted this 2 weeks ago

This looks promising, but did you multiply the force on the half model by 2 to represent the force on the full model? 

Yes, you need many elements through the thickness. You can fit 12 elements and stay under the Student license limits.

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firasb posted this 2 weeks ago

Multiply the force by 2, or divide? Shouldn't the input force be proportional with the geometry? Half the model, half the force?

Thanks,

Firas

firasb posted this 2 weeks ago

I changed the force to

1.5 kN and got 0.04929 mm y-disp

6.0 kN and got 0.17304 mm y-disp

Off by quite a lot in both cases.. 

peteroznewman posted this 2 weeks ago

Experimentally, if you pull with 1 kN and you get 0.1 mm of y-disp then the stiffness is 10 kN/mm.

In a half model, you pull with 0.5 kN and you get 0.1 mm of y-disp.  What is the stiffness? 

You have to double to force before you divide by the displacement to calculate the stiffness of a full model to compare with the experiment.

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firasb posted this 2 weeks ago

If you pull a half model with 0.5 kN and you get 0.1 mm displacement then your stiffness is 5 kN/mm.

To get the stiffness of the full model, multiply by 2 to get 10 kN/mm. So the key is to multiply the force of the half model by 2. 

Thank you for leading me there. I thought you meant I need to double the input force if I'm working with a half model. I misunderstood, my apologies.

So pulling the half model at 1.5 kN yields 0.04929 mm, which means the full model is pulled at 3 kN to achieve the same displacement. This yields a numerical result of 60.9 kN/mm versus the experimental result of 36.2 kN/mm. 

What in your experience/opinion might be causing this? Incorrect material properties possibly?

peteroznewman posted this 2 weeks ago

Having too few elements in a region of high stress gradient also contributes to excessive stiffness. That is why a Mesh Convergence Study is essential, it will tell you if you have sufficiently small elements. I am currently running your model with a much finer mesh and 12 elements through the thickness. Unfortunately, you can't run that on the Student license as there are too many nodes and elements.

Solid elements can develop a condition during the solution called shear locking and volumetric locking that make them excessively stiff. One change that can reduce these tendencies is to change from Full Integration to Reduced Integration. I used that in a model I sent you above (I think).

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firasb posted this 2 weeks ago

Ok, I'll try to run 11 elements and see if it takes it. I'll report back with findings. Thank you Peter.

peteroznewman posted this 2 weeks ago

Another source of error between the experimental data and the simulation result is how the experimental setup recorded the relative displacement between the pin and coupon.

The simulation might hold the ends of a short pin fixed, pull on the bottom of the coupon and measure a displacement of that bottom edge to compute the stiffness.

The experimental fixture might have a much longer pin supported on a pair of vee-blocks. The model could lengthen the pin and use a simple support on the ends instead of fixed support, to allow the pin to bend more. This will result in larger displacements for the same force in the simulation and therefore lower stiffness values.

The wrong way to measure displacement on the experimental fixture is to use the cross-head displacement. This measurement includes all the flexibility of the frame, so a larger displacement is recorded for the same force.

The right way to measure displacement is with an extensometer that goes between the pin and the bottom of the coupon. The extensometer displacement is unaffected by the flexibility of the machine frame.

If the length of the coupon in the model is shorter than the length of coupon in the test fixture, that is another source of error.

peteroznewman posted this 2 weeks ago

I ran the model with very small values of force to look at the initial stretching of the hole. Due to the interference fit, the hole is pressing on the pin all the way around. It takes a little force to separate the hole from the pin on the tension side of the pin. 

The plot below shows how the system acts like there is a preload and the force has to ramp up before it enters a linear range. The slope of the red line is 25 N/micron for the half model so the stiffness of a full model would be 50 N/micron or 50 kN/mm. At much higher values of force, the slope may change again. See above for reasons why this could be so different from experimental results.

The plot starts with an initial deformation because the face at the end of the coupon moves slightly when the interference is resolved.

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