Screw and mating thread strength equations

by mc172
 P: 209 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: 82

Screw and mating thread strength equations

I'll try to give some input from stress principals.
First of all I'm a South African so imperial units confuse the hell out of me, but that is a discussion for another day, haha...

Obviously you did some homework...
I'm gonna use a M10 TR GR8.8 bolt say with a S30400 nut... with the nut being weaker...
Nut material: BS 970 PART 4 GR 304 S15. It is assumed that the following properties apply to the material:
Minimum ultimate tensile strength: Sut = 505 MPa
Tensile yield strength: Syt = 205 MPa
Shear ultimate strength: Ssu = 0.6x505 MPa = 303 MPa
Shear yield strength: Ssy = 0.577x205 MPa = 118 MPa
Modulus of elasticity: E =193 GPa

Nut thickness: a = 8 mm
Depth of thread: t = 1 mm
Pitch: p = 2 mm
Thread length: b = ∏*d_m*n = ∏(9.25)(2) = 58.12 mm

Use weaker thread as design criteria:
Please note that yielding will first take place but if you want to know the force at the point where the thread will rupture use rupture tensile strength:
Bending: σ_bending=My⁄I
with M = Ft/2; y = h/2 and I = bh^3/12
Thus σ_bending=3Ft⁄(bh^2 )
505=3F(1)⁄((58.12*1.268^2 )
F = 15.73 kN
Shear: tau_max=1.5F⁄bh
303=1.5F⁄(58.12*1.268)
F = 14.89 kN
Because bending is involved, the shear stress distribution is not uniform; it occurs at the centre of the thread and is 50% higher than the average stress. At this point the bending stress is zero because maximum bending stress occurs at the outer fibres. And this stresses will not be combined:

Hence rupture will occur due to shear at 14.89 kN and this translates to roughly 29 Nm, didn't do the exact calculations here.

Note this is only for the threads, you can do some calcs on the bolt aswell but seeing that the material of the bolt is stronger it can be thought logical that it will not break before the nut threads...

Remember that if you want safe working load than use the yielding stresses and check the resulting thrusts and torques.

Hoped this helped a bit, not an expert, just the type of calculations I would do for this problem, so just use your parameters. Hope I did not give you the direct answer as I will get some stick from some of the PF police guys, haha...
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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