Given your thread, 6.125 - 12 UN-3G, the geometry suggests a maximum shear area per unit length of 13.965 in^2/in for the Nut (3B) and 12.087 in^2/in on the Pin (3A). This means that the Pin should fail before the Nut simply on the basis of shear strength between geometries.
You would now multiply these numbers by 0.4331 inches respectively, to obtain shear area on both the Nut and Pin. Multiplying this again by the strength of your material will give you the force needed to send the thread into plastic deformation. Using tensile strength of the steel would give you the lower limit to maximum sustainable load needed to break these threads.
Note that thread shear is not the only mode of failure. You could strip the threads off the core of the steel bar by application of a normal load, the threads could also go by initiation of a crack from the root of the thread. You would need to study the fracture and determine the mode of failure from experience.
Hope this helps. Sorry I couldn't be more conclusive.
Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada