Why bolt torque is T = (Friction factor) x F x D and not the radius? 2

[Abhijeet242]

We know that the torque is Force x moment arm. Here in this case why the moment arm is not radius. The real problem for me is that, I am calculating friction torque required by the rotating disc to overcome.
Here the I have done the calculations with below 2 methods:

1. In close approximation with Bolt torque :

Its a plain disc in Martensitic SS rotating on the fixed Bronze plate and acting as a bearing. The axial load is 500 kg including weight of the rotor. Hence like in case of bolt torque, we can say, the Axial force = 500 x 10 = 5000 N ( assuming g = 10 m/s2).
Now, T = 0.3 x 5000 x Avg. dia of the disc ( ( Inner dia. + Outer dia.)/2)

T = 0.3 x 5000 x ((0.119+0.0767)/2)
= 146.77 N-m
2. Considering simple approximation,

T = F x r = 0.3*5000*Avg. radius
= 73.38 N-m

So which one is correct? 73 or 146 Nm of torque?

 

[rb1957]

confusing to title the thread with something like the bolt preload equation ?

another day in paradise, or is paradise one day closer ?

 

[Abhijeet242]

Dear community, thanks for the overwhelming response. Sorry for the late reply. I see people asking about why am i confusing the torque required to rotate the disc with the Bolt torque.
Well, answer from my point of view, is that, both of the things are accomplishing quiet same results only the difference is that, in case of the Bolt, there is the axial elongation force involved and in case of rotating disc, weight of the rotor is involved. I think answer 74 N-m shall be correct for the specific query. Thanks again for the replies.

Now, speaking about the bolt, what i understood is that, Torque universally will always be T = F x r. In case of bolt now, previously what i was missing is, there are 2 surface friction bolt and nut experiences.
1. Friction between the threads
2. Friction between the Nut bottom surface and the surface of the component on which it is being tightened.
Here, the K factor which we use while calculating tightening torque is actually considering both these frictions and during the derivation "/2" has been taken care (i.e. r = d/2). Hence the simplified form what we see is T = K.F.d
where, d = bolt diameter.

 

[BrianE22]

I think you've found the help you needed but I'm curious, where did you see T = KFd? If you are using a nut and the nut is not spinning then there would not be a friction torque between it and the object it is tightening against.

Usually there are two components causing friction torque - friction between bolt head and surface and friction between bolt threads and mating part.

Maybe the source that supplied the T = KFd formula combined the two. That's quite the simplification though. At any rate, the numbers are all washed out by the choice of K.