Shear Pin Design 1

[ddelaiarro]

I've never really dealt with this before and want to make sure my thought process on this.

Setup
A shaft (D[sub]i[/sub]) feeds into a receiving hub (D[sub]o[/sub]). A shear pin of diameter, d, will go through a radial hole that is drilled through both the shaft and the receiving hub. The length of the shear pin will equal D[sub]o[/sub].

Main Questions
- Is the 'Double Shear Strenght' listed in most catalogs equal to the following?

P = T/D[sub]o[/sub] (where T is the torque 'seen' at this coupling)

If this isn't the way to figure out what Double Shear Strength I need, could someone please give me a hint here. I'm completely lost and it's frustrated the hell out of me. Maybe it's a case of sitting too close to the screen for two long :shrug:

Thanks in advance,
Dan

 

[Cockroach]

Double shear simply means that you have a force cutting across TWO shear planes similtaneously. In your case, you have a common pin between a shaft and hub, the pin has the opportunity to shear across ONE plane only, hence not double shear.

Therefore, if the pin of diameter "d" sits on bolt centre diameter "D", given the torque which is a moment:

F = T / (D/2) = 2T/D,
S = F / A = (2T/D) / (pi d^2 / 4) = 8T/(pi D d^2),
= (N m) / m^3 = N / m^2 = Pa

for F=shear force, S=shear stress, T=torsion. The last line simply validates dimensional consistancy.

Should this of been double shear or F cutting across two faces of the pin at the same time, then simply put we have twice the planar pin area or S = F / 2A. This means that the shear stress is essentially half that of single shear.

Hope this helps you out somewhat.

Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada

 

[Cockroach]

Just a quick correction, the above calculation assumes the pin is not completely through the linked piece. After looking at my post I realized the pin may indeed be completely through the piece, supporting your claim of double shear.

The analysis stands correct, S = F / 2A = 4T/(pi D d^2), or half that of single shear.

Sorry for the confusion.

Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada

 

[ddelaiarro]

Thanks Cockroach. Your second analysis was dead on as the pin does go all the way through. Actually posting it (and thinking about it in words instead of just equations) let me kind of work it out as well. I can see how it works now. It just didn't make sense to me (at the time) that the force acting on the pin (in two places) would be equal to the torque at that point divided by the diameter that it was acting on.

Thanks again.