Adding each ring's moment of inertia would be conservative. Since the moment of inertia for a simple bar is I = b*h^3 / 12, if you take each bar individually you lose out on the h^3 advantage. If they are truly back to back and well connected to the shell - as in continuously welded, not intermitent (aka stich or skip) welded - then I'd calc the I as if the bars were a single continuous stiffener. What may complicate the issue a bit is if you need to account for corrosion of the inner stiffener in which case you'd have to deal with the situation by adding the individual I's then calculating their center of mass and the combined center of mass and doing the Ad^2 component for each part. Its not too hard, but best to look it up in a book or chat face to face with a Mech Engr or Structural Engr to get it figured out.
