Axial friction coefficient and radial friction coefficient 1

[naah]

Hello dear
I hope it is the correct forum to discus the below question
I insert a shaft to a hub according to image below, then I could calculat the friction coefficient from insertion force. This assmbly is based on the interference insertion.
After shaft insertion I need to calculate the highest torque that the shaft can withstand in the hub.
My question is if the friction coefficient from shaft insertion is valid to use for calculating the torque?

Thanks in advance for your time and your knowledge.

 

[GregLocock]

Yes and no. The dynamic friction as you press the shaft in is likely to be lower than the static friction coefficient resisting the torque, so it is a safe assumption.

Cheers

Greg Locock


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[Denial]

No and yes.

I'm almost certainly overthinking it here, but unless the two parts are infinitely rigid the act of pushing the shaft into the hub will induce a slight "cupping" deflection in the hub.[ ] This will cause compression in the (very slightly) concave surface of the hub, which will in turn cause that end of its hole to grip the shaft harder and thereby increase the required pushing force.[ ] At the same time the axial compression force in the shaft will tend (via the Poisson effect) to expand the shaft's diameter, which will further increase the increased gripping force.

This argument, if correct and producing an appreciable effect, works counter to Greg's perfectly valid effect.

 

[hydtools]

Friction is proportional to normal force at the interface surface. Normal force is the result of compression due to interference at the interface. That interference is consistent, whether pushing or twisting the shaft. The coefficient of friction is either static or dynamic. So is friction torque capacity the same as the insertion force? Yes and no. Insertion force is resisted by dynamic friction. Torque capacity before slipping is resisted by static friction. So yes and no.

Ted

 

[Tmoose]

If the shaft s installed in the hub using a hydraulic press, etc the surfaces will likely be buffed a little smoother and even change slightly as the highs spots are smoothed down.

The specs for ball and roller bearing housing and shaft fits reflect this. Ground finished are preferred to minimize the effect. If machined turned finishes are used the fits specified are a little tighter.

The hardness of the shaft and hub can have an effect on successful assembly of interference fitted parts too.

And it is also possible for a poorly detailed shaft to destroy the hub bore during that first installation.
All it takes is an un-broken leading shaft edge, or to present the shaft at a slight angle to turn the shaft into a broaching tool, or initiate galling.
Then the party is over.

Controlled Heating the hub and cooling the shaft ( shrink fitting ) is less risky .

 

[BrianE22]

I better test might be to insert the shaft and then try to remove it. Measure the force while removing. The peak force should represent the static friction. The steady state torque, the dynamic COF.

 

[desertfox]

Hi naah

There are standard formula for the calculation you are performing just do a Google search.
Yes the friction coefficient plays a part in the formula but also the elasticity of the materials you propose to use, it would choose a low friction coefficient to err on the safe side and then put the connection under test.


“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

 

[dvd]

Not the answer to the question being asked, but it sounds like a bit of reinventing the wheel -

1. established interference fits for non-keyed hub-to-shaft fits are documented
2. a shrink fit may be a better way to get there

https://en.wikipedia.org/wiki/Engineering_fit

 

[IRstuff]

I insert a shaft to a hub according to image below, then I could calculat the friction coefficient from insertion force. This assmbly is based on the interference insertion.
After shaft insertion I need to calculate the highest torque that the shaft can withstand in the hub.

This is backwards, you design the interference fit to meet a requirement; you do not design and then see if it works.

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert!

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[SnTMan]

...and, how fine you trying to chop it?

Regards,

Mike

The problem with sloppy work is that the supply FAR EXCEEDS the demand

 

[mfgenggear]

Like irstuff said this design would be a press fit. But instead the shaft would be freezed. And the hub would be heated. And it would drop in and then press in if needed.

 

[naah]

BrianE22 thanks for your answer.
I am wondering if I can perfrom the break-awy torque, and from the highest torque calculating the friction?