The load each beam carries is porportional the I of each beam. Then each beam can be reviewed with its load and the stresses determined. If done properly, the deflections will be equal.
I would follow the first post above. You're assuming that the deflection is the same for both, and solving for the load that each beam carries based on that.
Lutfi is correct. The problem is the same as for a loaded welded connection when it can't be assumed that the load will pass through some point of contact, but only through the weld.
If the beams are initially in contact prior to loading, then eureka is correct in that as load is applied to the top beam, the two beams will each deflect equally, with the interface between the two beams sliding (horizontal shear effect).
If two unequal beams span the same amount, and deflect the same amount, the the bending moments and shears in each will be proportional to the relative stiffnesses of the two beams. So if beam A has an I = 100 and beam B has an I = 200, then beam B will take 2/3 (200/(100+200)) of the load and beam A will take 1/3.
If they aren't in initial contact, (a gap) then the top beam will take all the load until they come into contact, then the remaining load will be shared by the two as above.
All of these calculations are based on two distinct beams, each with their own Areas, moments of intertia, etc.
If you connect the two beams with adequate welds or bolts, then you prohibit the horizontal sliding I mentioned above and you've created a combined section - where you must calculate the total Area and Moment of Inertia for the two shapes made one.
I agree with eureka and JStephen if the initial conditions are as follows:
1. Both beams are the same length.
2. The bottom beam has simple supports. The top beam has no supports (other than the bottom beam).
3. Before any load is applied, the beams are in full contact along their entire lengths.
4. Assume no friction between the beams.
5. Vertical (down) load is applied to the top beam.
The calculated answer would be a reasonable and conservative approximation (most likely for several reasons). Probably the most important being that in "real life" there will be some friction between the beams. The actual deflection should be slightly less than calculated, but the defelection of both beams should still be equal.
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