# Screw and mating thread strength equations

by mc172
 P: 213 In addition to the friction between the mating threads (which you accounted for), you need to account for the friction between the head of the bolt and whatever surface it is coming in contact with.
 P: 3 Ah, yes. The nut doesn't come into contact with a mating face, instead the mating faces are stationary inside the nut using a slip-ring mechanism. Thanks, I must take the friction and effective radius into account. Is effective radius the mean diameter of the contact surfaces? Thanks.
P: 3
I'm English and was born after metrication so pounds are quite alien to me too. I assumed this forum was largely American hence the unit choice. And the thread is in inches, so it makes sense to use pounds.

Are you assuming two turns of the cross-section of the thread are effectively a beam with a uniformly distributed load? I don't understand how it is that simple, as the thread is attached to the nut...
Or is it a cantilever beam, 58.12mm long with a 1.268mm overhang, trapezoidal in section?

I also don't understand how you got this:
 Quote by WillemBouwer Thus σ_bending=3Ft⁄(bh^2 ) 505=3F(1)⁄((58.12*1.268^2 )
From this:
 Quote by WillemBouwer σ_bending=My⁄I with M = Ft/2; y = h/2 and I = bh^3/12
I try putting numbers back into My/I and get very different answers...

Thanks.
 P: 82 mc172, yes you treat it as a cantilever attached to the nut itself. Yes the threads will almost definitely not distribute the entire load evenly between them, mostly the firt two threads of your nut take the entire load, this is what I am assuming in my calcs. Remember this will be the peak bending stress in the thread. "σ_bending=My⁄I with M = Ft/2; y = h/2 and I = bh^3/12" By substituting we get σ_bending = (Ft/2)*(h/2)/bh^3/12 So σ_bending =(Fth/4)*(12/bh^3) hence σ_bending =3Ft⁄(bh^2) Hope you see this?

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