I was wondering if anyone had stats on psi exerted by the hook on the pin upon firing.

A lot would depend on how effectively it was jointed (or re-jointed) on the circle and draw. Effective bearing surfaces there will distribute the stress and strain off the pin to some degree. Calling Pugwash ... he could answer better.

Let's see... how about 10,237 psi on the "hingh pin" ?

Of course it's a lot less outside of Calcutta.

Especially without US foreign aid.

Gough Thomas, who was a highly qualified engineer once wrote that the strain on the pin is about 1/3 of a ton, but did not elaborate on how he calculated it.

Calculating the strain would involve taking into account the Poisson effect of thick walled tubes contraction and expansion of the chambers, the force exerted by the recovery of the action body as well as the forward thrust of the charge going through the forcing cone etc.

The action body flexes back on firing, according to accepted theory. If it flexes back then it must recover and that recovery must by definition involve a rather powerful push on the barrels and through them on the cross pin.

More than calculations I would love to see a high speed film sequence of a double shotgun action during firing. That would give visual proof of what actually happens and not what is commonly thought happens.

Short and simple store is as follows. The maximum back-thrust on the action face would be akin to the bore area times the maximum pressure. Assume pi X (0.729/2)**2 X 10,000 psi = 0.4172 X 10,000 = 4,172 force pounds. The hook-pin contact on a typical W&S BLE measured about 90% of a 0.500" hinge pin and 0.350" wide hook. Circumference of the pin is pi times diameter = 3.14 X 0.500 = 1.57". The hook bares on 90% of the pin, so the effective bearing length is 1.413". The bearing area then is length times width = 1.413" X 0.350" = 0.495 square inches. Assuming equal pressure distribution (not really valid, but OK for this kind of swag), 4,172#/ 0.495" = 8,428 psi. That is well below the working strength of typical soft, low carbon steel. Nothing of the hook or pin is likely to yield! If a gun is worn off-face, the hook and pin go into a battering mode and that is a different story altogether.

If, the concern is wear, then we have to have a working idea of the action flex as wear requires relative motion. Clearly, there is little motion from firing. However, there is relatively a lot of motion when the barrel is cycled open and closed. Modern high pressure lubes will handle the above pressures nicely. However, grit will infect and taint the lube such that it will act as a grinding agent. Clean modern lube is, IMO, the key to longivity of gun bearing surfaces.

Questions, comments? As always, check my math and logic, please.

DDA

So, a small pin would be taking more psi vs a larger pin?

Good point!

Yes, a small pin would have less area (square inches) over which to distribute the force (pounds). A large bearing will carry more load than will a small one for equal strain on the components.

Questions?

DDA

RM: I think you better re-check your math and reasoning. I have never owned a gun with a hinge pin as large as 0.50". I would think that most of mine are 0.375" or less, and the bearing area cannot be wider than the hook. Also, the hook cannot bear on over half the pin diameter, or you wouldn't be able to remove the barrels without removing the pin. I don't know what other makers used, but the pin in the LC Smith gun is a taper pin, according to the drawings. Also, most guns that I am familiar with have a portion of the barrel lug that bears against the frame and will carry part of the load if properly fitted.

Clean modern lube is, IMO, the key to longivity of gun bearing surfaces.


Rocketman, would you mind sharing what brand of lube(s) you prefer?
Also does anyone have any photos exactly where to lube a box lock, 2 AH Foxes, and an AYA 453 ?
TIA

I do have an old H&A Single on which the lug has a hole rather than a hook so it is necessary to remove the pin to take it down. Even on this gun the forward half of the pin would bear no load upon firing. Assuming that you were meaning the hook covered 90% of "Half" the pin then that would be 45% of the circumference. Even using the .500" this would give .707 x .350 = .247 sq In for a load of more like 16,686 PSI on the hinge. It would of course vary from this depending upon actual diameter of the pin & width of the lug.
A question I have is are we as concerned so much about the actual area involved as to the shear strength of the pin. It would seem to me, that at least as far as the pin is concerned that once an adequate with was achieved making the lug wider would not increase its strength to any significant amount. The force require to deform or eventually shear it at the juncture with the frame sides would remain essentially the same regardless of its width.

See this pipe gun:

http://youtu.be/7Va87gB_4AI

Obviously there is NO hinge pin on this construction. Observe how both the barrel and the outer "breech" pipe recoil together, dragging the "breech" backwards.

I suspect that the expanding shell sticks to the "chamber" and barrel and shell casing act as one unit pushing back on to the "breech". If something similar happens in double guns then the strain on the cross pin is minimal. From this point of view the friction coefficient of the chamber would become a factor to consider.

This is a fascinating subject and really could benefit from some high speed video so we can see what really happens to the breech during firing.

In my previous posting I mentioned the Poisson effect. Thick walled cylinders expand radially under pressure and contract axially. Obviously after the pressure is reduced the cylinder returns to its original dimensions and this recovery must involve some force on the breech face and cross pin. As far as I know there has been no analysis of this phenomenon. Most authorities focus on the breech flexing, assuming that there is no dynamic processes going on in the barrels. The authorities consider that the pressure on the breech face is via the shell alone, while the pipe gun tends to indicate that barrel and shell together press the breech face. Just looking at the imprints left on the breech face of double guns seems to indicate that there is some pressure exerted by the barrels too.

To add to the above. How much strain is there on the barrel of an auto which is held via a shallow thread on the magazine tube? If the strain is as great as calculated above those shallow threads should show some serious wear. That they do not tends to indicate that there is little forward thrust via the barrel on the threads.

Note that "the strain is as great as calculated above" is NOT forward on the bbls but rearward on the breech. You do bring up some interesting points, but that the forward thrust on the barrel is not equal to the rearward thrust on the breech is really obvious, by the fact that when the two are bolted together the gun recoils "Backward".
A barrel would have some forward component of thrust on it from the friction of the load being pushed through it, but even in the case of the autoloader the breech bolt is locked to the barrel at the moment of firing thus actually exerting a larger rearward force to the barrel than its forward component.

Poisson effect? Something downright fishy there.

Piper you have a point re the frictional force in the barrel.

Overall the locking of double guns seems to be a result of empirically derived choices rather than analysis.

Over the years I have measured many guns. Some have cross pins of 8 millimeters and are solid, others have 10 millimeter cross pins and rattle. Obviously it is hard to isolate the causes of the looseness, or to prioratise them, ie does friction contribute more or less than strain from firing, etc.

The impression left after much measuring and comparing is that the fewer locking elements there are, and more generously dimensioned they are, the more robust and long lasting the lockup. Prime example is the Winchester Model 21 which as far as I can determine has a two point lockup- the cross pin and the round locking bolt. The French Manufrance Robust is another solid locker with two lock up points, so is the Beretta 626.

Shotgunlover;
There are several American guns of renown which utilize Two Point bolting. The L C Smith, Ithaca NID & Fox guns all depend upon the hinge pin & the top rotary bolt. Lefever Arms Co guns use the ball hinge & a bolted square shouldered dolls head. D M Lefever guns use the ball hinge & top Cross Bolt, very much akin to Greener's. Parkers used the pin & Single underbolt but unlike the Win 21 had a square shouldered, unbolted, Doll's Head. Baker Bativia Models as well as many Stevens/Savage models used the hinge pin & a bolted straight rib extension. For multiple bolts to all be effective they must be extremely well fitted so they all bear simultaneously, Something unfeasible on many lower priced guns one sees advertising a multitude of bolts. Even then it is not unusual to see guns advertised as being triple or even quadruple bolted with all the bolts restricting the rotational motion of the barrels & only the hinge pin containing axial forces. Such in my opinion is purely advertisement Eye Wash.

Thanks, Tom. 2-p made a good catch on the 90% of half = 45% of the circumference. His math is right --- see below.

I don't think the pin taper, if any, will make much difference.

I agree that circle jointing, doll's heads, etc. complicate the issue. I was shooting for a SWAG.

DDA

2-p, you are correct on catching the half circumference. That is what I was thinking, but forgot to divide by two in me equation. Shame, Shame on me!!! Thanks for your keen eye. I always appreciate your review.

Even the possible 16,860 psi is still well within the working limits of steel. The above number is interesting in terms of relative rotation between the hook and pin. However, I think the more relevant number is the shear stress on the pin. That is the stress required to yield the pin (in two places) across its diameter. I measured the hook of the gun considered above and got a pin of very near 0.50". So, the section area of a 1/2" rod is pi X radius squared = 1/2 divided by 2 quantity times itself times 3.14 = 1/4 X 1/4 X 3.14 = 0.0625 X 3.14 = 0.196 square inches. Shear on the pin in two places means the shear area is 0.392 square inches. So, we have a shear stress of 4276# divided by 0.392 square inches = 10,908 psi shear on the pin. This is still well within the usual working stress for low carbon steel.

For whoever above observed that most of these designs seem to be emperical, I agree.

DDA

I like white lithium grease and, especially, Mobil 28 grease. Any of the modern gun oils will do fine; I use Rem-Oil mostly due to availability and cost. The trick is to wipe down bearing surfaces and relube frequently; like after every outting.

DDA

I don't think you can do a simple contact area /force calculation to get specific loading on the circumference of the pin. The pressure won't be the same as the surface changes from 90º to the force as it is around the curvature of the pin. I think its a very complex problem to solve. But I also think the surface pressures aren't the limiting factor.

Thanks Rocketman!

Re the shear stress on the cross pin. We are all assuming that the firing force is directed equally onto the breech face by the cartridge and onto the pin.

But, the pin is rarely unsupported in a double gun. A fair amount of metal is in FRONT of the pin, the knuckle, in full contact, thus any force exerted on the pin will be channeled to the action body. The same goes for OUs with bifurcated pins, they are supported at the front via protrusions of the action body metal. The pin undergoes deformation, more than shear stress. It is worth pondering whether this deformation comes from the forces of firing or via the recovery phase of the action body. If the action flexes back then it must spring back to its original form, squeezing the barrels between breech face and cross pin.

Personally I disagree with the assumption (fostered by Greener, Burrard and Thomas) that the cartridge head acts alone in pressing against the breech face. Whatever happens during firing is, I believe, far more complex than a simple thrust on the breech face by the cartridge head, and an equally direct and simple thrust on the cross pin. Over the years I have seen some strange damage to double guns but have yet to see a sheared cross pin, even in folding shotguns that have totally unsupported cross pins.

Good points, SL.



You never will, either. We are talking about double shear strength here, not single shear. For the hinge pin to shear it would have to shear at both ends, simultaneously. Ain't gonna happen from the forces generated by firing a shotshell.

I am about as far from an engineer as one man can be, but I deal with double shear situations daily on the farm. We pull subsoilers as deep as 18" in compacted soil. It is being held in place, until it hits an obstruction (stump, rock) by a Gr5 1/2" dia. bolt that does not have any tension on it from a nut, i.e. it is not being "squeezed together" to increase the shear strength. Gr5 is very similar to 4140 in strength, according to the charts I have looked at. Using a simple analogy of a 3/8" cross pin, or hinge pin, it doesn't take an engineer to see that there will never be a sheared doublegun hinge pin caused by any force that can be contained in the chamber and fired off a man's shoulder.

Chuck, I absolutely agree. I was not trying for a correct number. Rather, I was working on a ballpark SWAG. Nothing more.

DDA

SL, I think you have a point on how the barrel acts. I spent some time talking to a group of engineering students (one ME senior with fresh mech. of materials) about the likely behaviour of the chamber area under firing pressure and decay. They agree that it will not react strictly radially. However, we did not come up with any estimate of the magnitude of axial force. So, we all need to do a bit of research on this issue. Good thinkin' on this issue!!

DDA

Given that paper and tape shims work, if only as a temporary fix, I would guess there is not be much stress on the pin.

I think the force on the hook is mainly due to the mass of the barrels accelerating during recoil.

The barrel tubes want to basically stay motionless during firing. The reaction on the breech face pushes action and stock into your shoulder. As the action and stock go back, the barrels must hang on for the ride. This acceleration and puts force on the pin. For the same load, a heavy set of barrels will stress the pin more than a lighter set.

So I guess we could calculate the recoil energy, weight the various components of a gun, and by conservation of energy determine the force on the connections holding all the parts together.

If I am correct, the distance over which recoil is resisted will greatly affect the acceleration force on the pin. A lead sled contraption that sharply halts recoil will put more stress on the pin.

I say this is a lot like the forces put on rifle scope mounts.

Update. I ran a stress analysis for a typical 12 bore chamber with an inside radius of 0.405", outside radius of 0.600", and a radius within the wall of 0.500" at a pressure of 10,000 psi. The hoop stress is 20,400 psi, the axial stress is 8370 psi, and the radial stress (within the wall) is -3680 psi. All the above are well within the working stress levels for low carbon steel.

Gotta think about these numbers. What are they going to tell us?? Knowledge and/or opinions welcome.

DDA

Don,
There's some good S-N stuff on this link.

http://www.mscsoftware.com/training_vide...heory.15.3.html

The first chart on page 43 of this link is nice and straight forward for discussion.

http://books.google.com/books?id=Hbo8dI4...eel&f=false



I like the explanation of fatigue at this link.
http://www.sv.vt.edu/classes/MSE2094_NoteBook/97ClassProj/anal/kelly/fatigue.html#S_Ncurve