I do have an old H&A Single on which the lug has a hole rather than a hook so it is necessary to remove the pin to take it down. Even on this gun the forward half of the pin would bear no load upon firing. Assuming that you were meaning the hook covered 90% of "Half" the pin then that would be 45% of the circumference. Even using the .500" this would give .707 x .350 = .247 sq In for a load of more like 16,686 PSI on the hinge. It would of course vary from this depending upon actual diameter of the pin & width of the lug.
A question I have is are we as concerned so much about the actual area involved as to the shear strength of the pin. It would seem to me, that at least as far as the pin is concerned that once an adequate with was achieved making the lug wider would not increase its strength to any significant amount. The force require to deform or eventually shear it at the juncture with the frame sides would remain essentially the same regardless of its width.
