No! A bike is a fairly simple mechanism that is light loaded. How many newton-meters does a cyclist exert? About 750N (~75kg weight) * crank length (0.1725m)=129Nm
The chainring has 39T (for example), that is z*modulus=Diameter, modulus of tooth for a bike is ~4.05mm => diameter of chainring is 0.157m, radius is 0.0635m
The tangential force at that point is: 129/0.0635=2031N (around 200kg)
Actually your calculation gives a higher tensil load than FBinNY's. He estimated about 300 lbf and yours is about 440 lbf.
A chain that holds 200kg is not a fuss. A comparation with motorcycle chain or industrial chain is not possible.
Why not? Bicycle chains are chosen to be flexible to allow sideways bending for shifting, for light weight and to be narrow enough to fit between the adjacent cogs. No industrial designer would ever use such a small, narrow chain for such high loads if they expected typical industrial service life.
And FBinNY, a chain is not calculated at psi load on the surface, and the area of contact is on multiple teeth, it's not that simple, such a calculus is made to evaluate the stress on a tooth of the sprocket, not on the chain. Of course, a fairly documented project should contain verification of bending the pins, and "scissoring" the pins, elongating the plates, etc.
An overall information about a chain as a whole is it's maximum tension before breaking, but this is also hard, the chain does not tear the plates or the pins, it only uncouples itself.
The "area of contact" pressure calculation is based on the tension in the upper run of the chain where the load is the greatest. The number of links supported by the chainring and cog teeth isn't germain.
Note to the OP: See what your seemingly innocent question started?
