How to determine steady state of multi-step reaction with both reversible and irreversible steps?
Can anyone point me in the right direction for how to determine the relative concentrations at equilibrium?


This system will not reach equilibrium as long as there are irreversible reactions removing U2. If someone claims to have found an equilibrium state, the rates associated with k1 and k2 would be equal, but the [U2] would decrease because of those irreversible reactions.


  I know that kt=3kd must be true for the whole system to be in balance, but I have been struggling to figure out how to relate the other rate constants.


Usually, lowercase k is used for rate constants, with the rate of reaction given by a rate law that involves k, but also concentrations of the reactants and perhaps other species. The way you are talking about it, all reactions would be zero order. For zero order reactions, there is no equilibrium (a melting process has no equilibrium - if the temperature is above the melting point, everything melts, and if it is below the melting point, everything freezes). 


  I have read that the Keq for multi-step reactions is the product of each single equilibrium constant, but I'm not sure how to apply this with the irreversible reactions in the system.


You can't. There is the concept of steady state, though, where concentrations of intermediates remain constant while the forward and reverse rate of a given reaction are different. In this case, the rates increasing the concentration of an intermediate are equal to the rates decreasing the concentration of that same intermediate, but there is a net reaction. This is typical for living organisms (they can, however, reach equilibrium after they die).