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Old January 25, 2012, 03:18 AM   #2
Scorch
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Join Date: February 13, 2006
Location: Washington state
Posts: 15,248
Bolt thrust is equivalent to the pressure times the area of the case head. For example:
Assume you have a loaded 30-06 that operates at 62,000 psi measured with a strain gauge.
30-06 case head diameter is .470".
30-06 case head area is .1734944 square inches.
Bolt thrust would be the chamber pressure X the case head area, i.e. 10,756 pounds of pressure.

In reality, it is a bit more complex than that. You have to take into account the duration of the pressure impulse, the degree of taper in the case, etc, etc, etc, to determine the bolt load, but this will give you a quick working number. Shear is the force you are concerned about. The type of metal, thickness of the bolt lugs and the receiver ring will give you the info needed for the critical strength calculations. If you have, say, a Spanish Modelo 1893 Mauser, designed for 52,000 psi cartridges, produced from carburized mild steel, shear yield strength may be fairly low, bolt thrust from 9,021 pounds of pressure will be safe for a long time and 10,756 pounds might weaken it very quickly. If you are using a Weatherby Mark V rifle, produced from 41XX alloy steel, pressure excursions into the neighborhood of 70,000 psi are expected, the shear forces were calculate before the action was made, and the rifle is expected to fare very well for many thousands of rounds.

The brass cartridge case is just a seal or a liner for the pressure vessel itself, and adds very little to the strength of the action assembly.
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