# Dutch Pin

Does anyone know of any existing formulii that can be used to calculate stress concentrations in the so called “dutch pin” that many teams advocate for attaching sprockets to shafts?

I am using this type of pin in a project and need to prove that it is stronger (and how much stronger) than a normal pin.

Thanks,
Patrick

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 ® 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 / (RDL)

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.

Actually it is called a “dutch key” or sometimes just a “round key”. Some mold makers use this for locating “pins” as well.

Joe’s formulas are correct. I just checked them because the simplification looked funny to me. I thought that maybe the great Dr Joe might have slipped up. But when I wrote it out it looked better, and I got the same answer. That’s what we get for using single line formulas.

According to Shigley, (Mechanical Engineering Design, 1977ed) Joe’s example pin is a little large for the shaft. He recommends the key diameter be 1/4 of the shaft diameter and that the key be the length of the hub. If you’re REALLY interested he provides stress concentration factors for the shaft, both keyed and pinned, as well.

I find that in practice Shigley is usually a little on the conservative side for our application. But then we don’t do millions of cycles either. I usually follow Shigley unless something looks totally out of line and I’ve never had anything break yet. Though we do have the occasional weight problem

Joe,

Thanks for the analysis! The “dutch pin” I had heard from other teams over the past couple years was actually the second example you listed, but I think I like the the first example even better. I think the stress in the second example would be very difficult to calculate… However the fact is that with your example you are distributing the force over a much greater area than with any other type of pin – so I think that will be best even though shear yield is typically about half of normal yield.

Thanks,
Patrick

*Originally posted by Joe Johnson *
**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.
**

The name of the second case that Joe J. mentioned is called a “Dutchman” pin. I talked about it last year extensively. I do not know the formula, nor am I inclined to figure it out right now - maybe some other time. I do know that it is a lot stronger than you might think by doing normal stress calculations or estimations. It has to do with the unique orientation of the pin in relation to the direction of applied forces (which is longitudinal to the pin).

If you ever try to abuse one of these joints and examine it, you will notice that it fails (but only after extraordinary forces) in a fashion that I would call “a wiping shear” or “skinning”. It does not shear in the normal sense or break (unless you use a brittle material). It just deforms into an undescribeable shape. In fact, you can only get the pin to fail when the material surrounding the pin yields to allow some freeplay in the joint. And in some cases with free play, rather than breaking the pin, it gets pushed out of the joint. So you need a very tight fit for this type of pin.

The first case described by Joe would also work quite well. It would behave somewhat like a clutch bearing where the rollers in the bearing get wedged in one direction of rotation. But in this case it wedges in either direction.

Raul,
I would describe the failure something like;
the mallable pin material is extruded through the very small opening that is created by excessive repeated wear. This failure is actually quite impressive as one wonders how’d that pin turn into that odd shape. Interesting!