Calculating $W$ - $Z$ bosons mass difference Naively one might expect the W weak vector bosons to have slightly higher mass than the Z due to EM self-energy, while the opposite is true (80 vs. 91 Gev). Presumably due to the strong interactions and mediated by quarks since that is the only particle coupled to both interactions, but I can't find a discussion of this, even qualitative, anywhere.  Anyone aware of any?  
Plenty of recent papers relating to the ATLAS and CMS measurements of the W mass, but none seem to touch this point.  And of course the pions do agree with the naive assumption but the kaons don't, so perhaps all that can be said was said so long ago that it doesn't come up in todays' searches.  The percent difference is much greater in the vector boson case though, than for the mesons.       
 A: No, absolutely not! Naively, one might expect $M_W=M_Z ~ \cos \theta_W < M_Z$, where $\theta_W=   \arccos \frac{g}{\sqrt{g^2+g'^2}}$  is the weak mixing angle, at the heart of the logic of the  standard model! 
Geometrically, $M_W$ is the base and $M_Z$ the hypotenuse of the SM  coupling space right triangle,
g being the su(2) coupling and g' the hypercharge coupling. The hypotenuse is always larger than the base. In our world this Weinberg angle is almost 30°. 
In a hypothetical alternative world with very small Weinberg angle, the electric charge e would be essentially the hypercharge and then the Z would be only slightly heavier than the W. 
I'm not quite sure what you could have in mind w.r.t. EM self-energy of a neutral object, or with strong corrections; there are heavy quark loop and Higgs radiative corrections to both gauge boson propagators, but they are small! The dominant tree-level result here is what it is: an experimental number fixed by the EM charge e and the g implicit in the Fermi constant of β decay, $\theta_W=   \arcsin e/g$. Given these couplings, the mass ratio expected in the model is fixed, and, of course, validated just right by observation.
