The Von Mises failure citerion is as follows :
Sy = sqrt{((S1-S2)^2+(S2-S3)^2+(S1-S3)^2)/2) where S1,S2,S3 are the three pricipal stresses and Sy is the yield strength in a uniaxially loaded bar - in other words if you calculate the "Von Mises Stress", and it is greater than the tensile yield strength that you can usually look up, failure will occur. The "Von Mises stresses" given by your FEA program are calculated from this equation. This criterion is generally accepted as being the best "theory of failure" for ductile materials - such as a piece of mild steel. If you have brittle behavior you should use maximum normal stress theory. (I should have mentioned this in my all too brief original answer).
In ductile materials, it is the shear stresses that cause failure. If the three priciplal stresses are not all the same - you've got shear. If, for example, you have pure hydrostatic compression, nothing will happen, however high the pressure becomes - all the principal stresses will be the same and the Von Mises stresses will be zero. The same thing would be true in tension - which would cause the criterion to fail in the extreme, because the atoms would eventually have to be pulled directly apart - but I don't think anyone has ever figured out how to do this in practice, and it would involve an enormous stress.
In your case - it appears that S2 and S3 are zero - in other words you have pure tension - and the Von Mises stress is the same as the first principal stress S1 (study the equation). So in this particular case, you can compare your tensile stress directly with that present in a normal tensile test. Hardly worth doing an FEA was it ? You can gain a better understanding of all this by looking at the Mohr stress circle. Look in a book about elementary strength of materials - it should help you. Since you are doing an FEA, you should not need to consider stress concentration factors - but you may need to consider fatigue failure if you have cyclic loading. As a rough guide, you can look up the endurance limit (assuming the material has one - not all materials do).You also need a factor of safety in all designs - but from what you say you may be OK in that respect.