doublegunshop.com - home Forums General Discussion DoubleGun BBS @ doublegunshop.com Blow Up EUREKA

I appreciate that you're looking where the facts seem to take you. On the surface of it, there would not seem to be a requirement for a very large pressure spike to cause this particular failure.

Maybe the lab folks would confirm your calculation and comment on possible adjustments to the formula because of the rapid rate of the pressure change. Maybe they'd comment on how the tensile test sample failures compared to barrel blowout.

You've mentioned recommending loads as intended for the original gun. Probably a good guideline and possibly easy to run questionable pressures without obstructions. Really looking forward to the article, nice job.

Drew;
My knowledge of this is somewhat limited, but my understanding is that Barlow's formula is for high Pressure tubing under a constant load. Firing a shotgun does not produce that constant load, but only very briefly reaches its peak pressure. It is my understanding it takes a considerably higher pressure to bulge or burst a barrel than that indicated by Barlow's formula. This is what I was hoping some of our engineering Friends would chime in on, as to just what formula is most applicable for this purpose.

As does Miller, I'd very much like to hear from someone who understands this stuff.

I did get some help from Adam at METL who is a metallurgical engineer and not a mechanical engineer.

Barlow's refers to a pipe capped at both ends with a static pressure. The barrels aren't designed to be pressure vessels as one end is open. This is why an obstruction could cause a rupture at a pressure below proof pressure. 12,205 psi may be the right value for stress the barrel can sustain if it were capped at both ends.

Drew;
I think that 12,205 psi would be the static pressure it would contain if capped at both ends & if it were of the uniform diameter at which you calculated it. Obviously if it were capped at both ends & a static pressure applied it could be no higher than the weakest point in the barrel could contain. Since in firing a shotgun we are dealing with a pressure curve rather than a uniform static pressure the entire barrel does not have to be able to contain the Max pressure. The forward, thinner portion can be burst at a pressure which is lower than the max chamber pressure. Indeed under controlled test conditions barrels have been fitted with the normal pressure sensor at the 1" mark & obstructed with the barrel being burst at the obstruction without recording any increase of pressure in the chamber.

It might be noted though that pressure is recorded in a gun barrel even though one end is not capped, and chances are the part of the tube where the sensor is located is subject to that force however brief. Maybe for an obstruction that causes a burst, enough pressure formed to cause the burst even if a lower pressure were recorded at a different location.

If you run Barlow's Formula for a standard .729" bore at a forward location having an .030" wall you get a burst point of 3,645PSI with that same 54,000 PSI steel. When the shot hits an obstruction a localized pressure builds up, at that point. If that local pressure exceeds the yield strength of the steel a bulge will occur, if it exceeds the ultimate strength a burst occurs. When the barrel bursts, the gasses are fred so pressure falls rapidly, thus never reach the pressure recorded at the breech upon firing. The load in the bore of course makes it a Sealed Chamber, But the load is Movable & Not totally fixed, thus not a True Closed Cell pressure.

As suggested by others, reference should be made to volume 3 of the Modern Shotgun,"The gun and cartridge", chapter 16 Bursts!.In this chapter Burrard reviews steel strength, yield versus ultimate strength,together with comments on performance applicable to Damascus. Formulae relating to barrel design/strength are also given. The effects of wave pressure caused by barrel obstructions are reviewed in detail. Perhaps a person interested in this issue with a degree in mathematics/physics could comment on the impact of wave pressure on the barrel failure as seen in the samples under review.

Roy;
I presume this is probably a lot of the same mat'l as found in Burrard's The Modern Shotgun. He does in this work show a barrel which received three bulges from one shot. The obstruction was not quite severe enough to cause a burst, but as it moved down the barrel the Wave pressure bounced back to the breech, forward again until it caugth back up, creating the second bulge & eventually the third. Each successive bulge was smaller than the previous one.
One often hears how "OutDated" Burrard's work is, but after having studied it for many years I am of the conclusion that except for the substitution of Plastic for paper & felt, not much has truly changed.

2-piper
First thank you. I have correct, my post it should have read, The Modern Shot Gun. Yet another of my Seniors moments.
In my opinion Burrard's research on bursts has not been surpassed. Today it can be enhanced by adopting current methods for evaluating materials chemistry, stress measurement and none destructive test methods,

I've got a mechanical engineer on this, but here are some formula for a thick wall (wall thickness greater than 1/10 - 1/20 ID) pressure vessel (again a closed system under static pressure)

Our subject Remington 1894 barrel:
Wall thickness at the end of the chamber was .119"
ID at the end of the chamber = .815�
OD = ID + wall thickness X 2 = 1.053�. This does not include the brazed flats which obviously provide significant additional metal/strength to the breech, but the barrel blew out laterally.

Barlow's formula P = 2 x S x t / D
P=Bursting pressure in psi.
S=Tensile strength of material in tube wall.
t=Wall thickness in inches.
D=Outside diameter in inches.

Burst pressure = 2 x 54000 x .119" divided by 1.053" = 12,205 psi

Burrard quotes Alger Burst Formula
Burst pressure = Ultimate tensile strength x 3(OD � ID) / OD + 2xID

54000 x 3(1.053 - .815) / 1.053 + 2 x .815 = 14,370 psi

Lame Formula
Burst pressure psi = Ultimate Tensile strength x (OD squared � ID squared) / OD squared + ID squared

54000 x .446 / 1.774 = 13,576 psi


Short version of wave pressure per Burrard is that it can be more than 200% of shell pressure depending on the load, location, and degree of obstruction