Bolted assembly, friction force, clamping force

[Ricky Bobby]

I think I have solved my problem for a friction force between a bolted assembly, or at least I think did. If a torque was given of 267.6 in-lbs and a constant of k=0.2 and the diameter(d) =0.5,
T=k*P*d should equal 267.6=0.2*P*0.5, resulting in P=2676lbs.
Then using the equation:
Tc=F*fc*dc/2
Where Tc=Torque
F= force =2676lbs
fc= collar friction coefficient =0.15
dc=mean collar diameter= dm+1.5dm/2=2.5dm/2=1.25dm: dc=dm*diameter=1.25*0.5=0.625
Tc=2676*0.15*0.625/2
Tc=125.438in-lbs
Then
Tc=ff*n
Tc=Torque=125.438in-lbs
ff=friction force
n=center of bolt hole to outer edge of part =0.46inches
125.438in-lbs=ff*0.46in
ff=272.69lbs?
Seems high for a frictional force for a 1/2 inch diameter bolt.
Any help is greatly appreciated. Thank you for any help you can lend.

 

[KataKoniK]

Hello,

Thank you for sharing your solution for the friction force between a bolted assembly. Your calculations and equations seem to be correct, and I agree that the calculated friction force of 272.69lbs does seem high for a 1/2 inch diameter bolt.

One possible explanation for this could be the assumption of a constant friction coefficient of 0.15 for the collar. In reality, the friction coefficient can vary depending on factors such as surface roughness, lubrication, and contact pressure. It is also important to note that the friction coefficient may not remain constant throughout the entire assembly, as it can change at different contact points.

Another factor that could contribute to the high calculated friction force is the use of a mean collar diameter. This assumes that the collar is perfectly round, which may not be the case in reality. The actual contact area between the collar and the bolt may be smaller, leading to a higher contact pressure and thus a higher friction force.

In conclusion, while your solution may be mathematically correct, it is important to consider the uncertainties and assumptions involved in calculating the friction force in a bolted assembly. Further analysis and testing may be needed to accurately determine the friction force in this scenario.

I hope this helps. Thank you for your contribution to the forum.