Thread: Dutch Pin
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Unread 26-02-2002, 17:13
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First order estimate:

I am not sure what you mean by a "dutch pin" because there is some confusion on this subject in my mind.

If I understand it correctly, you drill a hole that is parallel to the axis of rotation and is centered on the circle that is interface between the driver member and the driven member.

I have also heard a hole that is 90 degrees from axis of rotation but offset the same amount so that the center of the pin is tangent to the cylinder that is the interface between the driver member and the driven member.

I will analyze the first case, not the second.


But, before I do that, I have to analyze the baseline case: A standard shear pin.

Given a radius (R) for the interface between the driven and driver member, a torque (T) between the two members and a diameter (D) of the shear pin, the stress (S) on the shear pin is then

S = F / A
(F is force, A is area)

S = (T/(2R)) / (Pi*(D/2)^2)

Simplifying:
S = (2/Pi)*(T/ (R * D^2))

For the dutch pin case:
Given all the above assumptions and a length of the shear pin (L) we get:

S = F / A

S = (T/R) / (D*L)

Simplifying:
S = T / (R*D*L)


Plugging in sum numbers makes the situation clear:
Given T of 150 in-lbs (the Globe motor for example)
Given D of (1/8) in
Given R of (3/16) in
Given L of (1/2) in

Standard shear pin:
S = 33,000 psi

Dutch Pin:
S = 13,000 psi

There are simplifications galore in the above, but it is a reasonable first order estimate.

Hope this helps.

Joe J.