Now, think of the action as the spring. Many times over the amount of metal in the core as in the case shell!! Post if you don't follow that logic or disagree.
Suppose we have a beam supported at each end and a heavy load midway. As that load was placed on the beam and the beam deflected the bottom of the beam would be placed into tension and the top of the beam into compression. If one were to graph the tension and compression in a cross section of that beam from top to bottom there would be a point where it is in neither tension nor compression. That point is only in shear. So that zero tension point contributes very little to the yield strength of the beam. And as you go to the top or bottom of the beam the compression and tension go up until they reach the top or the bottom. If the beam were an I beam the flanges on the I beam add strength exactly where the most tension and the most compression occur in the beam. The beam will take more load (pounds) before it yields (permanently bends) compared to another beam that was the same height and cross sectional area but with a rectangular cross section. And I vaguely remember that the distance from the flange to that point in the beam which is neither in tension nor compression is a major component of designing a beam.
And in a receiver the case hardened outer metal increases the the total yield point of the receiver. Pounds per square inch tension is not the only factor, it is how far that tension is from the zero stress point of the beam.
And the closer to that zero tension point in the beam the less the strain (stretching or compression) and the further that point is from it's yield point. And the less that point contributes to the resistance of the load because of the spring rate.
So it is not a simple case of calculating the percentage of the cross sectional area that is case hardened and pro-rating it's increased yield strength into the total strength of the action.
Thanks,
Mike
