Shear Pin Stress Calculation

[JLBK56]

Hello all.

I've been doing some review on this forum and other sources related to stresses in shear pins and I've not come across a definite conclusion. As it relates to shear pins, what is the correct way to analyze for failure? Most of the sources I've seen have considered failure on the basis of average shear stress (V/A) compared with the shear strength of the pin material. This includes my undergraduate solid mechanics textbook. However, other sources I've seen recommend using the maximum shear stress in the pin, per beam theory (4V/3A). I'm not convinced that the beam theory calculation applies for shear pins, since I believe 4V/3A only applies for cases where L>>D. Interestingly, I've also done a quick FEA on a steel pin in single shear between two steel plates with no separation (linear) contact between the plate surfaces and the pin, and such a simulation does not match the beam theory hand calculation, but does match the average stress result. I'm interested to hear from the community what the theoretically correct approach is to analyzing shear pins, and the justification for that. Is there a way to determine the maximum stress in a shear pin cross section, or is there a reason that the average stress is usually sufficient? Thanks.

 

[Compositepro]

All shear pins that I have seen, which are designed to fail at a fairly precise load, have a machined groove with a concave cross-section. This controls all the factors that lead to failure that would otherwise also be involved, like sharpneness of corners and bearing stress.

 

[desertfox]

Hi JMP1098

Because strengths of materials vary even with the same grade of material it’s hard to define a set value for the shear pin strength.
I once had to design some vandal proof nuts for an electrical cabinet, the nuts were made from hexagonal shaped bar, which consisted of a spherical portion with a small thin hexagon section at one end for use by the installer. With the correct size of tapped hole going right through the component,I used the average shear stress theory to get the ball park failure torque for the nuts but because of possible work hardening of the material and material strength variation, I got the manufacturer to test one of the nuts to failure and then alter the machined groove between the hexagonal and spherical portion of my original design so the component would fail at the correct torque setting. As far as I know they worked well.
Because of the short lengths involved between the two shearing points in your application then the bending stress will be negligible and therefore using the shear stress due to bending is not really viable.

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

 

[Ng2020]

It depends on the overall configuration of the joint, but in my domain pin joints are usually sized by (conservative) bending strength checks. For more on where I'm coming from google "Melcon, Hoblitt, Cozzone", or the USAF stress manual, or the NASA atsronautic structures manual.

 

[SirPhobos]

The beam theory calculation is derived based on equations for bending. It is a result of both vertical shear and a bending moment causing internal shear along the length of the member. Thus, if we were to isolate a single point of material, the total shear force would be a vector at an angle, not perpendicular to the cross-section. This is why the resulting stress is not simply V/A, as with pure shear.

In the case of a shear pin, the force and support are acting at the same point along the length of the beam. There is no bending moment. Thus, the correct equation is V/A.

https://wp.optics.arizona.edu/optomech/wp-content/uploads/sites/53/2016/10/OPTI_222_W10.pdf

https://www.informit.com/articles/article.aspx?p=2982118&seqNum=7

-- SirPhobos

 

[Ng2020]

"the force and support are acting at the same point along the length of the beam. There is no bending moment." In theory perhaps.

The usual assumption in my industry is that the lugs locally yield, such that applied loads and reactions on the pin become distributed loads. Because of this there is a small moment arm between the resultant applied load and resultant reaction. Add to that any gaps in the joint and the moment arm grows. Although it's small, loads are typically high, so pin bending becomes an issue.

 

[rb1957]

I'd use V/A if I had to (and not a beam calc). AIUI a shear pin has little moment in it (at failure) as the shear forces tend towards the parting plane (at failure).
Our initial loading assumption might be uniformly distributed load (load applied on one lug, reacted on the other); this implies a fair amount of bending in the pin. As load increases the common assumption is to reduce the moment arm (between load and reaction) by assuming a triangular distribution ... the load and reaction are now closer to the parting plane and bending reduced. Ultimately you coudl say there is no offset.

Of course the interaction between shear and bending is small, but modelling the pin (particularly a shear pin) as a beam to gain allowable is "suspect".

Of course too, at the end of the day pins have an allowable shear load (which I'd use in preference to V/A).

another day in paradise, or is paradise one day closer ?