2-p, you are correct on catching the half circumference. That is what I was thinking, but forgot to divide by two in me equation. Shame, Shame on me!!! Thanks for your keen eye. I always appreciate your review.
Even the possible 16,860 psi is still well within the working limits of steel. The above number is interesting in terms of relative rotation between the hook and pin. However, I think the more relevant number is the shear stress on the pin. That is the stress required to yield the pin (in two places) across its diameter. I measured the hook of the gun considered above and got a pin of very near 0.50". So, the section area of a 1/2" rod is pi X radius squared = 1/2 divided by 2 quantity times itself times 3.14 = 1/4 X 1/4 X 3.14 = 0.0625 X 3.14 = 0.196 square inches. Shear on the pin in two places means the shear area is 0.392 square inches. So, we have a shear stress of 4276# divided by 0.392 square inches = 10,908 psi shear on the pin. This is still well within the usual working stress for low carbon steel.
For whoever above observed that most of these designs seem to be emperical, I agree.
DDA
