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Press-fit load calculation always takes friction into consideration and gives you the load expression as follows,

$$F = µ\cdot PA$$

where µ is the friction coefficient, P (Interference Pressure) is the pressure developed due to interference and A being the total surface area in contact (after pressing).

So, if the friction coefficient is 0, does it mean the load required to create the press fit is 0?

Intuitively it doesn't seem like that, because there should be a minimum force to at least deform the hub/shaft. But that case isn't included in the formula mentioned above.

I'm asking this just to know the upper bound/lower bound of the press load.

Also, the formula doesn't take into account the entry chamfer of the hub. Without the chamfer, the case would be similar to that of punching sheet metals and not press-fit.

I have worked out the case that includes the chamfer, with no friction acting on the shaft. I'm not sure if it is a correct approach, but there's that. Any feedback would be greatly appreciated.

Press-fit force calculation without friction

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    $\begingroup$ Don't need a chamfer if you heat (or cool) one and do the opposite to the other. I fitted new valve guides by putting the new guides in the freezer and the head in the oven. $\endgroup$
    – Solar Mike
    Commented Feb 4, 2022 at 12:01
  • $\begingroup$ "So, if the friction coefficient is 0, does it mean the load required to create the press fit is 0?" - There is zero loads required to create the fit, but to maintain the fit by the friction on the contact faces. Without friction, the block will fall through. $\endgroup$
    – r13
    Commented Feb 4, 2022 at 15:55
  • $\begingroup$ @r13 I agree that without friction, the block will simply just fall. But don't you have to apply force for the deformation at least. How can the shaft enter the hub if there is a physical barrier and you are not providing any external load? I'm a bit confused here. $\endgroup$
    – Kaushik
    Commented Feb 4, 2022 at 17:40
  • $\begingroup$ @SolarMike Yes, you are right. You don't need the chamfer if you are just heating the hub, or cooling the shaft. But my question here is about direct insertion with the help of a servo/hydraulic press. $\endgroup$
    – Kaushik
    Commented Feb 4, 2022 at 17:43
  • $\begingroup$ My understanding of this equation of force is for the "best fit", not punch., if friction is zero on the contact faces, the element will simply fall by its own weight. $\endgroup$
    – r13
    Commented Feb 4, 2022 at 20:00

2 Answers 2

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Consider the initial situation shown in the image below. We are interested in the area in yellow.

enter image description here

After the load is applied, the deformed system will have the form shown below. Note that both components have deformed.

enter image description here

Because of the complications in the deformation, an approximation is typically made. A model of the deformation is shown below. An assumption made in the model is that the load needed to overcome the bump at the end is much smaller than that needed to expand the annular sleeve and contract the cylinder. So you are left with the need to solve a thick cylinder elasticity problem.

enter image description here

You can always play with the initial shapes to reduce unwanted effects and lip formation.

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  • $\begingroup$ Thanks for the effort. That was a good explanation and I also had thought the same with regards to the simultaneous deformation of shaft and hub. But my model takes into account the forces at any given instant, assuming that the deformation is uniform (as in, deformation is in the form of chamfer in both mating parts along with a straight radial compression and expansion). And since the force acting on both the mating parts is same, I just considered the shaft for easy calculation of compressive force. $\endgroup$
    – Kaushik
    Commented Feb 15, 2022 at 10:30
  • $\begingroup$ @Kaushik the problem with your model is that you dont know what the value of theta will ultimately be. So your back to the same problem as the friction formulation. You have just replaced mu with tan theta and you still dont know the value of the greek letter. See theta inside the object is nolonger the same as in the chamfer. $\endgroup$
    – joojaa
    Commented Feb 15, 2022 at 16:53
  • $\begingroup$ The good thing with the friction formulation is that its easy to measure the value in the workshop. $\endgroup$
    – joojaa
    Commented Feb 15, 2022 at 16:59
  • $\begingroup$ Ok, so let's take an example where the friction coefficient for a certain material combination is 0.01 for a given interference. This makes the value of force very little. Which is very counterintuitive. $\endgroup$
    – Kaushik
    Commented Feb 16, 2022 at 2:55
  • $\begingroup$ @joojaa: Both chamfers and friction are second-order effects. Chamfers ease the initial mating of parts and remove stress concentrations. Friction is dissipative. The first-order problem needs to be solved first. The standard mechanics way to solve the first-order problem is energy balance: to balance the work done by the applied load with the deformation energy created in the system. After that, friction and chamfers can be added to make the system more realistic. $\endgroup$
    – Biswajit Banerjee
    Commented Feb 16, 2022 at 18:56
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Friction coeficient is not one quantity and one effect. It is an experimenttaly derived value that includes multiple different physical effects lumped into one.

Therefore, all deformation and all other effects in that formula is rolled into the friction coeficient. This includes surface roughness peak distribution, lubrication status, deformation, wear, etc. You could make a model that is more complicated offcourse that would account for the individual physical processes better but it would be more complicated

But yes for purposes of of the formula you present a friction of 0 would indicate that no deformation or surface contact happens.

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  • $\begingroup$ I agree that friction coefficient will include all surface parameters. But would a zero friction really mean zero deformation? I have my doubts. $\endgroup$
    – Kaushik
    Commented Feb 15, 2022 at 7:18
  • $\begingroup$ @Kaushik for the model you use yes. If you want a model that has parts of friction taken out of the friction lump you need to derive a different model. So its not generally true but it is true for this particular formulation. $\endgroup$
    – joojaa
    Commented Feb 15, 2022 at 7:24
  • $\begingroup$ can you give an example for the model. I'm not sure how to begin with. $\endgroup$
    – Kaushik
    Commented Feb 15, 2022 at 9:11

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