# Calculating, estimating shear capacity of a pin....

Awl --
Suppose I have square solid alum bar, with an accurate square collar that
slides along the bar.
Suppose I drill a 3/16" hole clear thru both, for a pin. The pin will now
be in pure shear at two points.
How would I estimate/calculate the load rating of the pin? What type of
steel would be best? I'd prefer SS. Seems to me like hardened dowels-type
mat'l would be near-indestructible. The pins will be threaded at one end,
for a handle of sorts. I would like not to have to harden the pin after
machining, but if nec, I will.
Would anyone KNOW the shear capacity of such a pin in various materials?
If worse comes to worse, I could try to determine this empirically, with
some kind of hydraulic setup/gauge. Or just load the collar with heavy
plates?! Hints on this would be useful, as well.
Just calculate the cross area of the pin x 2 x tensile strength of the material.
If the pin is steel and the surrounding material aluminum, won't the surrounding material fail before the pin shears? Then your failure mode will be some combination of bending and shear on the pin (and you won't get a nice clean break).
What is the material of the collar? If it's also Al unless it's a more exotic type than just run-of-the-mill, there's good chance the failure will be there before you'll shear a steel pin. And, of course, same query/caveat on the bar; there will be quite a lot of deformation there unless it's also hard alloy.
So, answer is, sure it's possible to calculate shear strength pretty accurately but need more info than given to have any meaningful chance...
Holy shit.... now I know why I didn't do better in school!!!
OK, for 303 SS cold drawn bar
I get a yield tensile strength of 60,200 psi, which gives 3,340 lbs as the "sustainable load"? Double this if I use the ultimate tensile strength. I presume the lower value is what OSHA and testing labs would use, correct?
Is cold drawn 303 SS what I would typically get from a supplier? Annealed? If annealed, the load is a little over half the above figure. I want something that is reasonably strong, machinable (groove-able in a manual lathe), that I don't have to heat treat. Better choices than 303?
A 3,000 lb load (or 1500+ lbs) for a 3/16 pin is pretty impressive, to me. Am I indeed in the ballpark??
Regarding the comments form Tim, dpb, altho the collar is approx. 1/8" mat'l, there is sig. area of alum around the 3/16" hole, so altho there will be some deformation, there is more "mat'l support around the hole" that does not apply to pin calcs..
ALSO: Even if there is deformation to the collar, if you think about it, it is deformation "on top" of the pin, still shearing downwards on the pin -- likely still in perfect shear, as the collar is not likely to buckle *away* from the solid alum. So altho there are some differences (the pin is now not symmetrically "contained" by the collar), it intuitively seems to me that the conditions of shear still largely prevail. Correct me if I"m wrong.... :)
With forces along the length of the bar, or rotary forces? The former should be a lot easier to calculate.
One consideration is the gap between the bar and the inside of the collar. With close to zero, it is pure shear, as the gap increases, the forces get more complex. (And punches are made with a gap proportional to the thickness of the workpiece to make the task easier.)
To complicate this -- since the bar at least is aluminum, it will deform under the loads. Is the collar aluminum or steel? The softer the aluminum, the more the deformation will increase the *apparent* gap, and thus weaken the assembly.
Personally, given the variables (material of bar and collar, gap, etc) I would say the empirical test would be the most reasonable way to go. I'm not even sure whether a scale model would scale the forces reasonably, even with the same materials.
Good Luck, DoN.
The shear of interest probably is along the length of the bar, because both the bar and the collar have square cross-sections. The whole bar would have to shear (or deform) to rotate the collar, so the pin would hardly be involved, at least at first.
That's right, pure shear, along the length of the bar, along which the collar slides Altho the collar is 1/8" wall, it has considerable overall area, so the pin would be the proverbial weaker link, I would think. As I mentioned, in this scenario, under load, the pure shear condition is likely to remain, as the collar probably will not pull away from the bar, but actually would get tighter.
Someone sent me this, which distinguishes between tensile strength and shear strength, and in fact tabulates shear psi separately -- not a big biggie, as both are in the same ballpark:
Just curious as to whether "shear psi" is a more accurate way to go, or if it is in fact itself derived from tensile psi. I did not see a "shear psi" on matweb, which leads me to believe that it is derived from tensile psi, or some other quantity.
Inded you are...
I found this on the net:
"With direct shear joints the shank of the bolts sustain the shear force directly giving rise to a shear stress in the bolt. The shear strength of a steel fastener is about 0.6 times the tensile strength. This ratio is largely independent of the tensile strength. The shear plane should go through the unthreaded shank of a bolt if not than the root area of the thread must be used in the calculation."
So basically the shear strength of a pin is 60% of its tensil strength. In your case 120% since you have a double-shear situation.
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Only if that material is, essentially, to get your conclusion undeformable.
The problem is, the Al is _much_ less strong than a SS pin so there's a good chance it will simply tear instead of the pin shearing. Of course, the end result is the same. :)
If you're trying to compute the loading that arrangement will take axially, I'd bet the limiting factor is the ability of that 1/8" thick section (1/4" if both edges do see same force but there will likely be deformation that prevents that at least to a degree) to withstand the load.
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Those are really *amazing* numbers for a 3/16 pin.... and very good news for moi, as it makes my contraption do-able.
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Yet again, _only_ if the supporting material _around_ the pin can transmit those forces to the pin to cause it to shear. The thinwall Al collar will not hold up nearly to the level of a steel pin. It's the weakest link that will fail...
Or in other words "Shaft fails, protecting shear pin".
OK, just ballpark, suppose you have an infinitely strong 3/16 pin in the middle of an infinitely wide/long 1/8" alum sheet, with the region around the hole (pin) well-supported so the sheet doesn't buckle.
What would the failure point of the hole in this sheet be? Or, what would I need to calculate it?
And how would this differ if the pin, hole were in the same alum, but only, say, 4" x 4".
...
Roughly it's the same failure (shear) of the Al plate -- the ratio of the tensile strengths of the two materials will be a first approximation reduced by roughly a factor of about another 20-25% for the propensity of Al to tear more easily than steel.
I don't recall a number for Al otomh; I'm guessing it may be only roughly half(?) of a run-of-the-mill SS so I'd expect you wouldn't get but about 40% of the pin shear strength out of the plate. There are, of course, some pretty good Al alloys but you'll have to go out of your way to get one; it won't be in the bin at the local Ace Hdwe.
Some of the metals guys here may have better handle on numbers than I do otomh or you can look stuff up as well as I, of course.
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And, of course, there's the yield bearing load on that thin edge (1/8") to consider as well among other failure modes/criteria that a real design would need to consider.
The point is all the axial load transmitted to the pin has to come across those two thin cross sections and if there's any eccentricity in application on one side preferentially to the other it basically boils down to one. The local bearing area there is only 1/8*3/16 = 3/128 = 0.023 in^2 per face or >50ksi on the one face which > yield strength of even many of the good Al alloys. Even if it is evenly distributed across the two it's a very sizable fraction and that means severe distortion is highly likely.
None of these very crude guesstimates has anything about the load concentration, either, what more if it happens to be an impact force instead of constant load.
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Suppose I had an infinitely tough knife edge, 3/16 thick with a 3/32 radius on its cutting edge, plowing through a 1/8" sheet of, say, 6061, held in such a way that it would not buckle or bend, but just tear. Sort of like a razor going thru a taut sheet of paper.
How much force would that take?
That is in essence what would have to happen to make the alum in my gadget fail.
Intuitively, I can't see that happening before the pin would start to shear or deform.
What would likely happen before the pin starts to shear is the aluminum will deform around the pin... That's because the pin is harder than the base material. And we don't yet have any real numbers to go by.
It's down to that. How much force are you talking about? How much area is that force distributed over? What are the material properties of the parts?
So, you want a shear pin in aluminum, a softer pin? (By a lot) Plastic maybe?
Here is a formal write up of all that is involved...

...
That's essentially what a tool does and they don't have any problems doing so.
Your intuition is simply wrong if you think the harder pin will cleanly shear before the softer surrounding plate deforms and yields...whether it will tear clear out is dependent on many factors and there's insufficient data to have any way to come close to calculating it--we don't have dimension drawings, just a general description.
Again, if you do use a harder Al, it will take more axial load before deformation is excessive (DOH!). As above, the thin profile means small x-sectional area which means high force amplification into quantitative stress levels. You can't get away from the previous basic force balance. Making the plate thicker helps (quite a lot, 'cuz even a 32nd over 1/8" is another 25% (5/4) which raises the area so lowers the stress).
As the model in the paper the other poster shows, even w/ small gaps the real world is that you can't have pure shear but bending moment comes into play and as his model shows the real answer is nonlinear even though the design manuals linearize the solutions. Making the clearances smaller helps there at the cost of more precise machining and perhaps assembly/disassembly if this is supposed to be able to be removed.
To do more than just generalities would require precise drawings, materials, loading profiles, etc.,etc.,etc., ...
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That was my point. I"m curious, in the above scenario, what the forces on the tool would have to be to begin to plough through 1/8" alum.
Oh, I figgered there would be "mutual deformation", just curious as to approx. relative amounts. Ergo my proposal for the "tool" above.

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