MX:
“the load will be fixed directly down the center of the pin” has a number of potential different meanings; down the pin lengthwise, at its center, or across the round shape of the pin, at its center span, for example. Draw every possible alternative that your wording might allow/suggest, and you might see why we could be unsure of what you really meant. Again, you are looking at it, but we can’t see it from here, so you must be very careful that your verbiage only allows one word picture to be drawn, out of many. There isn’t anything wrong with a free hand sketch, but proportions are important, so reasonable scale and dimensioning is important, and that almost always shows things better than our verbiage.
By your sketch, I would say, it appears that you mean ‘the pin spans btwn. two .5" thk. support plates, 1.5" clear distance apart (btwn. them); and is loaded at its mid span by a third .5" thk. plate, and is shearing the pin through its round area (.785 sq.in.), in double shear.’ With the proportions your system has, I would say you have a simple beam with a span length of about 2", that gives you the worst bending stress on the pin; and you did say fail safe, so be conservative. The support plates aren’t thick enough to fix the pin ends, and the .125" dia. clearance certainly won’t fix them. In its simplest form the shear stress is Tau = Fmax./(2 x .785sq.in.). Then you might combine these two stresses and compare then to ultimate strength. You might determine that yielding is failure. And, in any case you will have some factor of safety or margin of safety against yield or ultimate. That’s called engineering judgement and is dependent upon the specifics of the problem. Finally, the specs. on the materials involved will give you some min. yield and ultimate tensile strength, and some chemistry, % elongation and the like, which start to indicate toughness, hardness, etc., but the actual yield for a given pin will most likely be higher than the min. value.
Furthermore, you will get considerable yielding in bearing (a Hertz stress problem) btwn. the pin and the plates before you shear the pin. And, any imperfections in the pin can drastically influence the actual failure number and mode. Impact loading and fatigue loading are also very important. This is actually a fairly complex little problem, and what you are looking for is a safe and educated force value that your system can withstand and still allow you to sleep at night, not hurt anyone, or seriously damage any equipment. You are not looking for an exact force value.
