Shear Pin Design
Shear Pin Design
(OP)
I've never really dealt with this before and want to make sure my thought process on this.
Setup
A shaft (Di) feeds into a receiving hub (Do). 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 Do.
Main Questions
- Is the 'Double Shear Strenght' listed in most catalogs equal to the following?
P = T/Do (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
Setup
A shaft (Di) feeds into a receiving hub (Do). 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 Do.
Main Questions
- Is the 'Double Shear Strenght' listed in most catalogs equal to the following?
P = T/Do (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





RE: Shear Pin Design
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),
[S] = (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
RE: Shear Pin Design
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
RE: Shear Pin Design
Thanks again.