Static coefficient of friction between copper or brass

[Platanus3]

Hello all

I'm trying to calculate a torque needed to tighten a threaded terminal. (Both male and female parts are copper alloy)

I'm using Motosh's equation which requires me to use the coefficient of friction.

Web sources tells me that the copper-copper coefficient of friction is 1.0.

The problem is that coefficient of friction value of 1.0 gives me a torque value that seems to be too high.

Can anyone advise me on this issue?

Thanks.

 

[bsheikho]

Whats your the pitch of your screw? Also the angle at which the contact is made between the 2 points, I imagine the force to calculate friction would be from the axial force created from turning the screw.

 

[JBA]

If you are using the generally accepted Motosh equation, I have verified the average coefficient of copper to equal 1.00, the result you are getting may not be what you want but it should be considered accurate.

 

[Platanus3]

Whats your the pitch of your screw? Also the angle at which the contact is made between the 2 points, I imagine the force to calculate friction would be from the axial force created from turning the screw.

Sorry for the late response.
Pitch is 0.0833 inch because the threads belong to unified fine thread (1.500-12UN-2A and -2B)
Thread angle is 60 degree (I believe a standard value), so I used alpha value of 30 degree in Motosh equation.
My machinery handbook tells me to calculate the yield clamping force and yes it does take the friction between threads into account. But the value is too high. I get like 2771 lb-ft of torque.

 

[Platanus3]

If you are using the generally accepted Motosh equation, I have verified the average coefficient of copper to equal 1.00, the result you are getting may not be what you want but it should be considered accurate.

OK. If then, what is the realistic safety factor I should apply?
The result seems too high.
For 1.500-12UN-2A (and 2B), I get like 2771 ft-lbs of torque.
Yield strength I used is 45,000 psi (Copper alloy)

 

[Mech_Engineer]

How much force are you trying to get out of the terminal once tightened?

 

[Platanus3]

How much force are you trying to get out of the terminal once tightened?

How much force are you trying to get out of the terminal once tightened?

I'm using yield clamping force which calculates to be 12315 lb force

 

[JBA]

It is always preferable to stretch a fastener beyond its required design contact or tensile loading but this can be achieved without actually designing for loading to yield. Are you sure that you really require a level of loading that high?

Is your fastener size based entirely upon getting a required design load or compensation for potential thermal expansion loosening or contact pressure loss; or is there some other factor that is causing you to use a fastener size greater than what would be required for sufficient fastener grip (i.e. for an electrical service it might be current carrying capacity) causing you to select a bolt loading higher than what is required.

 

[Mech_Engineer]

I'm using yield clamping force which calculates to be 12315 lb force

For that level of force and a frictional coefficient of 1.0, I'm not surprised you need 2700 ft-lbs. You will need to lower the coefficient of friction (maybe with a grease) if you want to reduce the torque requirement.