I think it is important to identify that one, it is a torsional load and two, it is distributed through all the spiders and through all the teeth engaged with the chain. There is no single point of the force vector from the crank or ring but a system of 5 force vectors distributed through the spiders which in turn impacting a distributed area of the ring and then through ring teeth. I would not be surprised if the variation of the force from the pedals was reasonably uniform at the ring bolts at any given time, if they are uniformly torqued. Then we get into the details of the tolerance of each of the holes and fastners relative to each spider and the resulting impact to the force distribution at the interface.

Nominally, the spiders see a cyclical load based on the force applied to the pedals in a given location, like a sine wave. Granted, the spider with the crack can only tolerate a load less than the others due to the failure on one of two spider sections, but the other four will pick up the difference while the remaining one will continue to work with the load until it fails. When it does the other four will have to pick up the difference. So initially the spiders distributed the load into 5 spider, with the complete failure of the 5th the load is distributed through 4. Note the resistance force vector is pointed counter-clockwise (your trying to get the ring to move clock-wise and the rear wheel and ground are resisting it), forcing a compressive force on the crack.

So squirtdad, experiencing the risk is your call. As little as we know of the design parameters of the crank and the load (s), what is the impact of the risk event of the second section of the failing spider? Are the remaining spiders up to the task of eduring your mashing? Is the failure of the second section truely catestrophic (one stroke) or will it propogate over time (n+1 strokes)? When it fails, what is the likelyhood of the ring folding with a distributed load around 180 degrees of teeth and being supported by the internal ring (would have to "fold" both rings)? Do you want to find out?