Pin and Weld drill depth - potential problem?

[lysander]

I wouldn't even worry about it.

The stress even with that thin a wall will be around 50,000 psi. The yield strength of annealed barrel steel is round 55,000 psi. After heat treating that figure goes to around 90,000 to 100,000 psi.

ETA: If all a barrel had to do was contain the pressure, they could be very thin, the only reason barrel are as thick as they are is so they are stiff.

Yes and no.

The MIL-C-63989C says that M855 shall not be less than 12,000psi average at the port on an M16A2.

The bore pressure near the muzzle is nowhere near the same as the chamber pressure, so the pressure the pin and weld job would be far, far less then 50k PSI. See the chart below for bore pressures at the port for various length gas systems.

Attachment 59501

That being said, thin walls will expand and contract more as the bullet passes and attempts to push the barrel material out of the way so deformation, cracking, flaking and eventually material removal will happen relatively quickly at that spot. This could end up marring the bullet as it passes that point and will also lead to excessive erosion at the muzzle, both causing accuracy issues. Best not to run this way with a can.

I've had a smith accidentally drill all the way through into the bore on an MR556 pin an weld job nearly a decade ago. I had them counter bore the muzzle to 6mm ID and 5mm in to remove all contact and stress from the bullet, basically making a 90 degree crown set back by 5mm into the end of the muzzle.

Ended up being a tack driver and could print sub 1/2 MOA groups with the right ammo, trigger, and optics. Suppressed just fine as well.

You are confusing chamber pressure with hoop stress in the barrel. Nowhere in my post did I mention the chamber pressure.

"The stress even with that thin a wall will be around 50,000 psi."

"The yield strength of annealed barrel steel is round 55,000 psi."

The bullet diameter is .224", the groove diameter of the barrel is .224" so there is no bullet trying "to push the barrel material out of the way" in the grooves. In the lands, the engraving process takes place at the throat and the bullet is deformed to conform to the rifling long before it reaches the muzzle, again, no bullet is trying "to push the barrel material out of the way".

As to the dire prediction of expanding and contracting... it is a 0.125" length covering 8% of the perimeter, subjected to less than 10,000 psi pressure, that is where the hoop stress of 50,000 psi came from. You will be happy to know that the fatigue strength of 41xx barrel steels hover around the 90,000 to 110,000 psi range, so even if it does flex, it will not fatigue. I will leave it to you to calculated the exact amount of bow that small unsupported section of barrel with flex, but I will tell you that it ain't much, less than what can be reliably measured outside of a laboratory.

[lysander]

Will mine hold up for the long run, or will it slowly start giving me problems?

You will probably never notice it.

[GrumpyM4]

You are confusing chamber pressure with hoop stress in the barrel. Nowhere in my post did I mention the chamber pressure.

"The stress even with that thin a wall will be around 50,000 psi."

"The yield strength of annealed barrel steel is round 55,000 psi."

in the grooves. In the lands, the engraving process takes place at the throat and the bullet is deformed to conform to the rifling long before it reaches the muzzle, again, no bullet is trying "to push the barrel material out of the way".

As to the dire prediction of expanding and contracting... it is a 0.125" length covering 8% of the perimeter, subjected to less than 10,000 psi pressure, that is where the hoop stress of 50,000 psi came from. You will be happy to know that the fatigue strength of 41xx barrel steels hover around the 90,000 to 110,000 psi range, so even if it does flex, it will not fatigue. I will leave it to you to calculated the exact amount of bow that small unsupported section of barrel with flex, but I will tell you that it ain't much, less than what can be reliably measured outside of a laboratory.

Sure.

Except the yield strength based on wall thickness equals somewhere around 350psi,assuming a .004" wall thickness of barrel steel alone. The bore pressure at the muzzle is still well in excess of 350psi.

Now, given that the average thickness of chrome lining in a rifle barrel is somewhere between .002" and .004" thick, and Chrome is a harder and much more brittle material having a far smaller elasticity window under strain and virtually no plasticity prior to fracture, this makes the OPs predicament a bit more of an issue.

The bullet diameter is .224", the groove diameter of the barrel is .224" so there is no bullet trying "to push the barrel material out of the way"

Also, Yes, the bullet does try to push the barrel out of the way. Simply physics still apply. Bullet passage under pressure does in fact bell out the barrel during passage. It's at such a small level that it's difficult to detect, but it does happen. Were your belief to be completely true, the jacket of the bullet would wear away quickly enough to allow freefloating of the bullet well before it reached the muzzle. Action/reaction. Pressure slightly enlarges the bore as the bullet passes, yet due to the elasticity of the steel, it returns to original form because the pressure never exceeds yield strength and heads into plasticity territory. By your belief, we could shoot all the .22 lead rounds we wanted with a .22 conversion kit in a standard DI AR without worrying about lead buildup in the gas port, nor would there ever be a squib round stuck in the barrel past the throat. Two issues that we know for a fact happen. Lead without a copper jacket shaves off into a gas port and builds up over time and squib rounds get stuck in all sorts of places in barrels.

You're trying to dumb down the process and only focus on a few factors as well as ignore simple facts such as material thickness affecting yield strength under pressure. Determining the actual PSI yield strength of a pipe based on wall thickness, OD, and it's safety rating (which is based on the materials overall given yield strength) is not that difficult. You're also ignoring thermal input regarding the steel, bullet deformation in the bore due to pressure as well as heat from both burning propellant and friction. There is a LOT happening on many different levels every time we pull a trigger.

[lysander]

Sure.

Except the yield strength based on wall thickness equals somewhere around 350psi,assuming a .004" wall thickness of barrel steel alone. The bore pressure at the muzzle is still well in excess of 350psi.

?????

First - Yield strength is a mechanical property of the barrel steel, it does not vary with thickness. For AISI 4150 steel in the annealed condition is around 55,000 psi, and on the heated treated condition around 90,000 to 100,000 depending on the temper.

Second - You obviously do not know how to calculate hoop stress. So, I will just refer you here where a computer will do all the math for you.

Enter the following data:

15,000 = inside pressure (MPa, psi)

0 = outside pressure (MPa, psi)

.112 = inside radius (mm, in)

.152 = outside radius (mm, in)

.112 = radius to point in the wall (mm, in)

Inside pressure is the pressure inside the barrel over the pin location. this is way high but hey let's have a safety factor.

Outside pressure is actually 14.6, but it's so much less than the inside pressure it will have no influence on the magnitude of the resulting stress.

Inside radius is half of .224

Outside radius is the inside radius plus the 0.040 the OP said was how close he came to breaking through ("A little math shows I have roughly .040 between the barrel ID and drill")

Radius to point in the wall is where you want to calculated the stress, it will be highest on the inside surface, so the inside radius is used.

Results - Circumferential (Hoop) Stress: 50636 (MPa, psi)

Now, compare the resulting hoop stress to the yield stress of the steel, and the fatigue limit (both around 100,000 psi). And then remember this stress is if we had cut the entire outside diameter of the barrel down to a 0.040" wall thickness. What we actually have is a 8% section of the barrel this thin.....

Now, given that the average thickness of chrome lining in a rifle barrel is somewhere between .002" and .004" thick, and Chrome is a harder and much more brittle material having a far smaller elasticity window under strain and virtually no plasticity prior to fracture, this makes the OPs predicament a bit more of an issue.

The chromium plating thickness in the bore is between 0.0004" and 0.0015".

Also, Yes, the bullet does try to push the barrel out of the way. Simply physics still apply. Bullet passage under pressure does in fact bell out the barrel during passage. It's at such a small level that it's difficult to detect, but it does happen. Were your belief to be completely true, the jacket of the bullet would wear away quickly enough to allow freefloating of the bullet well before it reached the muzzle. Action/reaction. Pressure slightly enlarges the bore as the bullet passes, yet due to the elasticity of the steel, it returns to original form because the pressure never exceeds yield strength and heads into plasticity territory. By your belief, we could shoot all the .22 lead rounds we wanted with a .22 conversion kit in a standard DI AR without worrying about lead buildup in the gas port, nor would there ever be a squib round stuck in the barrel past the throat. Two issues that we know for a fact happen. Lead without a copper jacket shaves off into a gas port and builds up over time and squib rounds get stuck in all sorts of places in barrels.

Expansion of the barrel does happen, but it is due to the internal pressure, not a bullet pushing a wave in front of it. But, like I stated before, the barrel stress stays below the fatigue limit and it can expand and contract infinitely.

Gas port leading is due to shaving, and you can have shaving with zero clearance. And, zero clearance is not a slip fit, you can get very high friction with a zero clearance fit.

You're trying to dumb down the process and only focus on a few factors as well as ignore simple facts such as material thickness affecting yield strength under pressure. Determining the actual PSI yield strength of a pipe based on wall thickness, OD, and it's safety rating (which is based on the materials overall given yield strength) is not that difficult. You're also ignoring thermal input regarding the steel, bullet deformation in the bore due to pressure as well as heat from both burning propellant and friction. There is a LOT happening on many different levels every time we pull a trigger.

You obviously do not have the engineering background necessary to understand what is important, what is not that important, and what assumptions you can make.

According to you every Government Model 1911 is on the verge to exploding every time you shoot a +P load. (23,000 psi chamber pressure, with a wall thickness of 0.062")

[GrumpyM4]

Ouch,

My bad. I added an extra zero (.004 rather than .04 for wall thickness) and added the safety factor for calculating bursting/yield pressure, making the expected safe operating range of the material thickness far lower than it should have been. I was also using Barlow's Formula for calculating bursting pressure.

I calculated yield/bursting pressure of the barrel using the typical calculations for pressure piping being wall thickness times yield strength times 2, divided by OD times safety factor added in to account for other variables such as pressure spikes, heat, brittleness, etc.. I'll show my (faulty) work.

In this case, given the known variables, it came out to .004 x 105,000 = 420 x 2 = 840 divided by OD x safety factor (.232 x 10.5 = 2.436) = 344.83psi for a safe constant pressure over time.

Removing safety factor puts that at 3,620.68psi, still well below the actual bore pressure at the muzzle, but all of that is irrelevant due to the misplaced zero that shouldn't have been there.

Proper bursting pressure (not factoring in effects of heat and ignoring safety factor) should have been .04 x 95,000 = 4,200 x 2 = 8,400/.232 = 36,206.9psi.

The further example you gave of the 1911 barrel only accounts for the barrel after the chamber where pressures have dropped significantly.

Nominal chamber thickness where the initial pressure is highest is .123", not .062, which would make the bursting pressure (assuming it's manufactured from modern 4140 steel with a yield pressure of 95k psi rather than the original barrel steel which I believe was only rated for 17k psi) .123 x 95,000 = 11,685 x 2 / (OD) .696 = 33,578psi, which is well above the 23k psi expected of a SAAMI spec'd +p .45ACP.

At the thinnest part of the barrel the calculation would look more like .063 (wall thickness according to blue print and factoring in maximum tolerance variation of both the OD of the barrel and ID of the bore to create the thinnest wall possible) x 95,000 = 5985 x 2 = 11,970 / .580 (OD) = 20,709psi bursting pressure.

Lastly, yes, the barrel does expand as the bullet passes through the bore. It's basic physics explained through complex calculations, although you are correct that it does not exceed fatigue strength of the steel in question. See he link below, "Stress Analysis of Gun Barrel Subjected to Dynamic Pressure".

http://article.sciencepublishinggrou...150304.14.html

I'll take my lumps for adding the extra zero.

?????

First - Yield strength is a mechanical property of the barrel steel, it does not vary with thickness. For AISI 4150 steel in the annealed condition is around 55,000 psi, and on the heated treated condition around 90,000 to 100,000 depending on the temper.

Second - You obviously do not know how to calculate hoop stress. So, I will just refer you here where a computer will do all the math for you.

Enter the following data:

15,000 = inside pressure (MPa, psi)

0 = outside pressure (MPa, psi)

.112 = inside radius (mm, in)

.152 = outside radius (mm, in)

.112 = radius to point in the wall (mm, in)

Inside pressure is the pressure inside the barrel over the pin location. this is way high but hey let's have a safety factor.

Outside pressure is actually 14.6, but it's so much less than the inside pressure it will have no influence on the magnitude of the resulting stress.

Inside radius is half of .224

Outside radius is the inside radius plus the 0.040 the OP said was how close he came to breaking through ("A little math shows I have roughly .040 between the barrel ID and drill")

Radius to point in the wall is where you want to calculated the stress, it will be highest on the inside surface, so the inside radius is used.

Results - Circumferential (Hoop) Stress: 50636 (MPa, psi)

Now, compare the resulting hoop stress to the yield stress of the steel, and the fatigue limit (both around 100,000 psi). And then remember this stress is if we had cut the entire outside diameter of the barrel down to a 0.040" wall thickness. What we actually have is a 8% section of the barrel this thin.....


The chromium plating thickness in the bore is between 0.0004" and 0.0015".


Expansion of the barrel does happen, but it is due to the internal pressure, not a bullet pushing a wave in front of it. But, like I stated before, the barrel stress stays below the fatigue limit and it can expand and contract infinitely.

Gas port leading is due to shaving, and you can have shaving with zero clearance. And, zero clearance is not a slip fit, you can get very high friction with a zero clearance fit.


You obviously do not have the engineering background necessary to understand what is important, what is not that important, and what assumptions you can make.

According to you every Government Model 1911 is on the verge to exploding every time you shoot a +P load. (23,000 psi chamber pressure, with a wall thickness of 0.062")

[lysander]

The further example you gave of the 1911 barrel only accounts for the barrel after the chamber where pressures have dropped significantly.

You are forgetting the fact that there are locking cuts in the barrel where the thickness is around 0.062" - .0.066" The first cut starts 0.400" the the case mouth. At that point the pressure from a +P load is still around 15,000 to 16,000 psi.



Even in a low pressure load, the pressure at 0.4 from the case mouth is pretty high:


EDIT:
I have been through these calculations before when trying to establish the minimum thickness you could make a gas block.

[JediGuy]

I have been through these calculations before when trying to establish the minimum thickness you could make a gas block.

I’m curious about what you reached on that question. Or, perhaps, have you ever had concerns about any small gas blocks on the market?

[lysander]

I’m curious about what you reached on that question. Or, perhaps, have you ever had concerns about any small gas blocks on the market?

I wanted to make the thinnest lightest gas block possible.

I forget the actual dimension I arrived at, but it was so thin that figuring a way to hold it during machining became a problem.

[GrumpyM4]

And by using the simplified Barlow's formula those portions still maintain around a 20,522psi capability.

I suspect this small window of variation is why some 1911 manufacturers do not recommend using +p ammunition in factory guns as it doesn't give enough leeway for a proper safety factor to account for pressure variations, heat, or other possibilities.

You are forgetting the fact that there are locking cuts in the barrel where the thickness is around 0.062" - .0.066" The first cut starts 0.400" the the case mouth. At that point the pressure from a +P load is still around 15,000 to 16,000 psi.



Even in a low pressure load, the pressure at 0.4 from the case mouth is pretty high:


EDIT:
I have been through these calculations before when trying to establish the minimum thickness you could make a gas block.

[lysander]

So.....

I take it we are in agreement that he doesn't have to worry about his near miss with the hole?

Besides, even if he drilled all the way through, he'd just have an extra gas port to nowhere....