I have a K-38 that has been destroyed by an overload (not by me). There is no bulge in the barrel so I suspect a double charge of a correct powder or a charge of a wrong powder. The chamber that the round was in and the two adjacent chambers are destroyed and the top strap is peeled upwards like a banana. I see from the 'net that .38 Special PSI is around 17,000 for regular loads and 20,000 for +P loads in modern handguns. As a guesstimate what level PSI would be required to do this sort of damage?
I have searched the net and cannot find what is the PSI that they use for proof loads for .38 Special. I have checked the SAAMI website and they don't specifically list the data. I would like to know what that PSI is since it would seem to be the level that won't destroy the revolver and then I could have a point to guesstimate what the PSI was for this event.
(This is similar to what can happen with a Webley that has been 'shaved' to accept .45 ACP rounds. There is a sticky thread about this in this forum which links to pictures on the British Militaria website. I have seen figures that the operating pressure for the .455 Webleys to be in the 12700 to 13200 psi range the latter specifically cited for the Mark VI, which along with the Mark V, was the only one designed for smokeless powder. The .45 ACP is stated to have a standard operating pressure of between 19000 to 21000 psi depending on the source I read. There are pictures on the 'net of .455 Webley cylinders which failed with .45 ACP rounds. So it seems that something in the range of six to seven thousand PSI beyond the standard operating pressure is enough to blow the cylinder. I have also read, can't find the source now, that the .45 ACP round pressure is higher than the proof load pressure for the Webleys. I acknowledge that the metallurgy in the early 1900s was not as good as it is now and that may contribute, to some degree, to the Webleys fracturing with the .45 ACP round.)