. . . as was already mentioned earlier in this thread, an attempt to quantify the service life of a rubber component (in the present case a seal)one is confronted with several limitations concerning a proper methodology and lack of knowledge on the fatigue processes of rubbers.
Despite of these limitations I would suggest the following procedure to follow in order to receive a prognoses for service life of the o-ring in question:
Perform a non-linear finite element analysis of your sealing problem, where a typical cyclic loading should be simulated. Detect the areas of the cross section where strains are large. Get the full information on the strain tensor in those areas. Determine the components of the strain tensor which are in tension, because they may be the most dangerous ones. If you have the impression that one of the tension-strain components is much greater than the other ones (which often appears to be)you may be justified to handle that deformation mode for further considerations only. Now correlate the strain energy density found in the considered area of the cross section with the tear energy appropriate for the given deformation mode (see for example "How to design rubber components" by A. Gent). Thus you will get a relation dealing with a virtual crack you think about to be present in the considered area of the cross section of the o-ring. Then you may assume you would know the critical tearing energy ( ... and indeed one can assume a "most-lower" value - to be on the safe side). With all those assumption in mind you can state whether the current dynamical loading may be critical concerning a possible growth of some virtual cracks (flaws etc of certain crack-length). If you thus come to the conclusion that initial cracks ( always be present, independently of compound quality which may have a typical length of the order of several tens of micro-meters)subjected to the given loading condition would most probably grow, then you should be justified to claim that the seal may not survive. Otherwise you may of course claim the opposite.
Well, that procedure works well only if the assumptions made concerning the distribution of crack-length of pre-existing flaws and other imperfections in the bulk rubber is rather good.
But even if you do not know ( ...as is usually the case)the mentioned "distribution" you can get benefit out of the above described "calculus" when you for example state that you have in your rubber component a rather big number of small cracks with a crack-length of 50 micro-meters or so.