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[julian1998]

Hello, 

I'm doing a simple simulation of an interference fit. I have defined a step where the hub is pulled away from the shaft, so I can determine the force that is needed to do so. With the force reaction function im getting the following values and graph: (The second step is where its pulled away)

Now, my question is: what value is the force that is needed to pull the hub from the shaft? is it the 1205.9 N or is it the difference from 442.69 N to -1205.9 N? 

What value is the 442.69 N? (In the first step, the intereference fit itself, without any other loads is simulated) 

Thank you very much for the help

[Akshay Maniyar]

 

Hi Julian1998,

Thank you for your post. Can you share some more details about the loadsteps and how load are applied?Also, where are you calculating force reaction like for displacement boundary condition or at contact region? This will help everyone to understand your model and answer your query properly.

Meanwhile please check the Ansys How to video on INTERFERENCE FIT.

How to Setup and Solve an Interference Fit using Ansys Mechanical

Thank you,

Akshay Maniyar

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[julian1998]

Hello Akshay Maniyar,

thank you for your reply.

The loadsteps are similar to the Ansys How To Video you have send me.

The first loadstep is just the interference fit without any external load. I have applied a frictional contact where the interference of shaft and hub lays. On the second load step a displacement of 5 mm is applied so that the hub is pulled away from the shaft (external displacement "A"). The shaft is fixed on one side (External displacement "B") on both steps. 

External displacement B:

External Displacement A: 

The force reaction graph in the post is from the displacement "A"

[Akshay Maniyar]

Hi Julien1998,

The reaction you are getting in the first load step is because we have constrained the hub in the axial direction and when the interference is solved there will be some axial displacement of the hub but it is restricted. Then the reaction in the second load step is the force required to pull out the hub by 5 mm.  

Thank you,

Akshay Maniyar

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