# Internal pressure

1. Oct 17, 2009

### tuoni

Does anyone have a quick way of estimating the amount of internal pressure a sealed tube would be able to handle without failing?

2. Oct 17, 2009

### FredGarvin

First: How do you define "failing?"

Second: It is easy, but the freshman year easy method does not account for stresses near the ends of the tube, only "far away" from them.

Are you attempting to use very high pressures? I hope not.

3. Oct 17, 2009

### tuoni

This is only a theoretical experiment. Too dangerous.

The reason I ask is because I became quite interested when I read about the Russian 7.62x42 mm cartridge a while back. It is a true silent cartridge, the only sound is from the action of the firearm. It utilises a special steel "capsule" that traps the escaping gases and unburned propellant within the casing.

The attached file "pressure.png" depicts a crude simplification of the setup.

The inner tube is able to move forward, providing additional space for the ejecta and propelling the bullet. I just thought this was a very interesting concept, and thought of doing some simple calculations just too see the outcome. At first for a fixed tube, how it would be able to handle various pressures, and then the telescoped variant.

Failing would mean not being able to safely trap the ejecta, slight deformation would be acceptable. The breech and chamber provide support for the rear and sides of the tube, so the critical part would be the front. In the first example, the front o the tube would have to stay intact. In the second example, the projecting tube would have to remain intact, and be held in place by the outer tube. The illustration doesn't include any such system, and I don't really know the exact method used, but let's assume that the tube only moves forward 3/4, the rest of the rear is somehow used to prevent the inner tube from being propelled along with the bullet.

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