So is it accurate to say that the pin will fail in the areas of the reduced cross section (The 5mm grooves you describe)? Maybe elaborate on the loading scenario a bit for us.
It is pretty typical (for designing purposes) to assume that a cross section in shear will fail at about 70% of its UTS in tension. Now this is a pretty broad assumption considering the array of material properties that can affect these types of failures but it's a place to start.
Could you approach this problem with hand calcs first?
I've been able to establish a shear factor of 0.758 from a previous destructive test having used 0.7 initially. This was done using hand calcs.
The shear pin indeed did fracture at its area of smallest cross section. The shaft is part of a weak-link in a chain, with the two shear areas being pulled apart in equal and opposing directions.
Having played around quite abit, it seems the 2nd Principle Stress approaches the UTS at the failure load found using hand calculations. Does this have any significant and how would I have known this was the failure criteria?