doublegunshop.com - home Forums General Discussion DoubleGun BBS @ doublegunshop.com Question For Chuck or Rocketman (Hoop Stress)

A while back I think one of you posted a formula for calculating hoop stress. I remember bore dia. as being a variable, thus smaller bores had less hoop stress when compared to equal pressure in a larger bore. Would it be possible to post that formula again?

Thanks

Given title I figured you were from Ohio

Ha! It took me a minute to figure that one out (not much of a B-ball fan). Besides, I'm much closer to FL than OH. I'm an Auburn grad - and we were the ONLY team to beat the Nat'l Champ FL football team this year (it felt good, but then AR and GA destroyed us). Oh, well. I'm left wondering if OH residents will ever go to the beach in FL again.

"The Elastic Strength of Guns" 1916, Alger, as used by the USNA in class, p28-- for a simple hollow cylinder, R=outer radius,r=inner radius, maximum internal pressure allowable=P.Then P= 3 times the quanity of(R squared-r squared)divided by (4 times R squared +2 times r squared).All then multiplied by the elastic limit of the material under tension.

Anothe formula sometimes quoted is known as Lamé's formula. Burrard quotes this as the "Gunmaker's Formula". Using t (tension) as the allowable internal pressure the formula is;
P = t x (R²-r²)/R²+r²). He states this gives the pressure necesary to stress the "Internal Layer" of the wall.
He quotes a different version of Alger's formula as stressing the entire wall as follows;
P = t x 3(R-r)/R+2r
Note that with either of these it is not actually necessary to obtain the radius, but diameter can be used by simply substituting D & d for R & r as long as you do so for the entire formula.
It is also, I think, worthy of note that actual experiments have been performed which by these figures "Should" have resulted in a burst bbl, but didn't. Greener reported on a 12ga chamber wall turned extremely thin & Hatcher reported on an '03 springfield bbl turned to .060" over the chamber as I recall. I can only in my limited ability account for this as similar to the LUP pressure difference from tranducer pressure, ie the time lag for the very short period of applied pressure, simply did not allow full deformation to take place. These formulas were, I think worked out for a continous load.

I have run figures for two different scenarios with all three above formula,s on steel having assumed Elastic Point of 50K psi (approx mill run 1035). First "chamber end" with dia of .800" & .100" wall (OD 1.000"). Next I ran bore dia of .730" with .030" wall (OD .790").
For allowable P in chamber I get;
Lamé's = 10,975psi
Alger's = 11,538psi
Alfer's = 19,852psi (as quoted by william)
For allowable bore P I get:
Lamé's = 3,941psi
Alger's = 4,000
Alger's = 9,562 (WEA)
Note; It appears, to me at least, the formula given by Mr Apperson looks to be much closer to the situation we expect to find in a gun bbl than the others. I thank you William for this formula. With wrought iron having an elastic limit of approx 25K psi these figures would be cut in half (rest of the formula would be identical for same dimensions, so P would be proprotionate to t for same dimensions) which can also explain why some fairly shoddy old guns have withstood firing with loads far stronger than were ever intended for. They are undoubtably running on the "Ragged Edge" but many have endured. With 4140 even in a fairly soft state running to 85K psi it is obvious the strength advantage to modern steels. All figures I have cited are also elastic limit (bulge point) not Ultimate Tensile which is much higher (Burst Point)

Thanks to both 2-P and Mr. A for the formuale. I'm trying to determine if these formulae will show whether a small bore (say a 20 ga.) will or will not withstand more pressure than a larger bore (say 12 ga.) WITH THE SAME WALL THICKNESS. I'm at work right now, so I'll have to plug-in later. Thanks again.

Mr. A - USNA = U. S. Naval Academy?

That is the beginning.By the last page(100),we learn how to calculate a built-up gun, the Mark 7,mod 3, 12".As designed in 1914, the tube was nickel steel tensile-90,000;elastic-55,000,the jacket was nickel steel tensile 90,000;elastic 60,000, and the hoops were nickel steel tensile 95,000;elastic 65,000.Max pressure 37,500 velocity 2900,projectile weight 870 lbs;weight of charge 336 lbs.Although my favorite is the 16" 50 caliber as on the Iowa class.However, you will notice that in 1914 some pretty good steel was available.

You are correct on USNA; however, from birth,under-graduate ,and graduate, I am of The University. It is located in Charlottesville and was founded by Mr. Jefferson.

William,
Having toured the battleship USS Alabama on display in Mobile Bay, I have to agree with your admiration for the 16 in. gun. The projectiles are on display in the magazine, and weigh around 2000 lbs. each, if I remember correctly. Couple that with a range of many miles (can't remember how many, but an astounding number) and incredible accuracy and one's jaw begins to slacken. I recall them explaining how the projectiles had to be moved around by hand, at times, and I pictured what it would be like to wrestle with a 5 ft., round, one ton chunk in rough seas trying to roll you over. Oh, yeah - people are trying to shell you, too, at the same time. Also, I can't begin to imagine the scene in the magazine of the Iowa (I believe it was) when the demonstration for the politician went horribly catastrophic.