Sounds about right. I would imagine the brass would flow if anything deflected significantly so I would use the OD of the cartridge/ID of the cylinder.

Now think about the stress on the thinnest part of the cylinder closest to the OD.... The frame is in bending because of its cross section, so you would really need FEA to determine the peak stresses. The same for the cylinder because the wall thickness is varying all around the chamber.

Big revolvers are quite a feat of engineering....

Quote Originally Posted by Fuel Fire Desire View Post
I’m doing some basic high school math. I’m curious what peak frame stress my revolver is encountering during firing (just for fun).

The .454 Casull produces 53,000 psi.

The .454 Casull bullet is 0.452”

Pi x r squared is 0.16 square inches of area.

53,000 psi x 0.16 square inches is 8,480 ponds of force.


Is my math correct on thinking that the frame of my revolver is withstanding 4.24 tons of stretching force upon firing? Just bored and curious. Pretty impressive as well.