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[quote user="AnthonyV"] Great discussion guys.We can get a sense as to how much force the pin can transmit.  It appears that the pin is loaded in double shear, what makes it a pretty straightforward calculation.  What is the diameter of a typical pin, what type of steel is used, and what is the rated force for a coupler?For example, assuming an ultimate shear stress of the steel of 100,000 psi, a 1" diameter pin could transmit about 150,000 lb before shearing.  However, working stresses in practice are much lower due to fatique considerations, impact loads, etc.  As a result, actual working stresses are propably in the range of 25,000 psi, which would reduce the allowable load on the pin to only around 40,000 lbs.This is much lower than the maximum coupler force, which is at least 250,000 lbs.The result of this simple analysis suggests that the coupler load is not transmitted through the pin.ThanksAnthony V. [/quote]AnthonyV,Cordon has done a great job of photographing and diagramming the coupler, but it seems that the one unresolved question is whether or not the knuckle pivot pin bears any force when the coupler is mated, locked, and pulling a train.  In a few posts up, Nick explains that the pin does not bear a load, and refers to an alternate type of pivot pin made from plastic.  I reply that I looked at a coupler where the pivot pin does appear to bear a load when the closed, locked knuckle is pulled, as it would be when pulling a train.  I am not refuting Nick's explanation, but I cannot reconcile it with what I observed.But in the middle of last night, something occurred to me.  The four sets of force bearing ridges that cordon calls out in the photos follow a circular pattern as segments of an arc that is concentric to the pivot centerline of the knuckle.  The top and bottom sets of ridges that are very close to the knuckle pivot pin are performing the same function as the pivot pin.  Each set consists of a circular boss rotating in a circular bore.  There is one difference between the mated circular features near that pivot pin, and the combination of the pivot pin passing through the bores of the knuckle and the coupler body.  That difference is that, in the mated circular features, the circular bore is not a complete circle.  It is a segment of an arc.  It cannot be a complete circle because there would be no way to separate the knuckle from the coupler housing if it were.  There would also be no way to assemble it.  When viewing a coupler from the knuckle end, the train pulling force on a closed and locked knuckle would tend to rotate the knuckle to bear against the right side of the pivot pin.  In other words, the outer end or engaging mass of the knuckle would move from right to left.  From one of the photos showing the knuckle, it appears that the arc segment of the circular bore is positioned to bear the direction of this right-to-left loading.  So if the knuckle pin had a lose enough fit through its bore in the knuckle, the mating features of the circular boss and arc segment bore would bear all of the knuckle reaction side force, leaving the pivot pin completely free of any shear load.  However, as the mating features of the circular boss and arc segment bore wear, the pivot pin will begin to pick up some of the knuckle reaction side force.  I speculate that the coupler I inspected had suffered sufficient wear to cause this loading of the pin, which I did observe.  I also speculate that this is what causes pivot pin breakage of both steel pins, and especially plastic pins. This also explains the rationale of the patent for the plastic pins, which is that they will bend without breaking, as opposed to steel pins that break without bending.  If the pin is not intended to bear a load as the coupler is designed, but begins to bear a load as the coupler wears, it happens as a forcing surface gradually approaches the side of the pin during the wear period.  So once that forcing surface begins to load the pin, if the pin could move further away from the forcing surface even slightly, it would completely relieve itself from the load.  Therefore, a plastic pin could bend to accommodate this gradual encroachment of the forcing surface, and avoid loading, whereas a steel pin will not bend, thus quickly assuming the full load of the encroaching forcing surface.  Apparently the plastic pins are often forced to bend to a limit where they break.  And apparently, the steel pins are strong enough to carry the load of the forcing surface as it encroaches during the wear cycle, at least during the early development of that loading.  Apparently the steel pins are not strong enough to carry this shear load during repeated cycles, once it has developed to its maximum force.  When I say that the steel pins would not bend, I only mean as a comparison to the plastic pins.  Like the plastic pins, the steel pins would bend to graduate their assumption of the shear load, but just not as much as the plastic pins would bend.  In other words, the steel pins would assume far more load than the plastic pins for any amount of bend applied equally to both types of pins.
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