Ouch......good luck!.....appears to be high risk for long-term decay since water will find way behind edge of slab (and there is exposure to inside from under slab also)......at very least, use treated-wood. However, I find hard to believe competent masonry contractor could not simply install...
Doug.........I continue to disagree about shape of pressure within contact zone, and doubt there is any sudden concentration of load at edge of contact zone....I have to believe pressure develops similar to "beam on elastic foundation" which has smooth transition.
You seemed to be claiming...
As noted above by others........relatively simple just analyze segment to left of hinge first to obtain point load to be applied to end of cantilever.........then analyze remainder of beam as separate element (to right of hinge only). Of course if you were to consider moving load on the...
Doug......I must disagree with your assumed distribution of reaction force..........so then, by the same logic, if the load on upper beam were highly irregular.....say, with completely non-uniform shape......would you say that shape of pressure within contact area must somehow be the same...
Deflection should be limited to code-specified limits.........even if only for your own (or your firms) liability.
Take full advantage of live load reduction allowed by code, which will likely reduce floor live load substantially (up to 50-percent), depending on total floor area being supported...
Doug......sounds like you are assuming uniform contact pressure, and then adding point loads (somewhat arbitrarily) to make up difference.
Although I have been "tripped up" on first-pass at logic with this analysis........it appears that distributed load should not be uniform, but should taper...
But wait!.......there's more.......oh geez........as close as I can get to "pure" contact for I-upper-24 (I-upper = 24-percent I-lower).......I end up with the following;
Reaction at midspan = 21.01 percent of total load
Reaction at 0.49 span (that is, one-percent of upper-beam-span to left of...
Wrapping this up, at least to a "point".........looks like I owe Doug a beer!
Using only point load reaction at midspan.......I calculate that "perfect" contact occurs only at midspan, down to I-upper-25 (that is I-upper = 25-percent of I-lower).......for which configuration the midspan...
Just some musings on fundamentals........though have not quite fitted all pieces into this puzzle.
For continuous beam.......moment for each beam segment is of course the same at interior support. Slope is also the same at end of each beam segment at this support (that is how segments are...
For first contact, ratio of 1/2.4 is same as the 41-2/3-percent I reported, so at least we agree on starting point!
However there is no obvious reason to then claim that point contact (only) continues for all I-upper values down to 25-percent of I-lower........or that distributed contact then...
Hazards of long posts.......one important correction;
There is difference between curvature and slope. Looking at this case for example.........at midspan of both beams, slope is the same (zero) yet curvature is not the same. We see this by looking at the total shape of each beam. This is one...
Doug......see your graphs again........quite informative
Upon comparing my "20-percent" solution (I-upper = 20-percent I-lower) to what is your graph of same configuration (you have ratio just expressed other way, as multiple of 5).......I get similar results, though one qualitative difference...
Alrighty then.....my bad......updating yet again......going to get this!
I-upper = 41-2/3 percent of I-lower
M-upper = 50-percent of M-lower (in area of contact, at and near midspan of both beams)
I-upper = 40-percent of I-lower
M-upper = 47-percent of M-lower
I-upper = 33-1/3 percent of...
Ah, clicking tad too quickly through formulas..........revised moments for Case 2;
I-upper = 41-2/3 percent of I-lower
M-upper = 50-percent of M-lower (in area of contact, at and near midspan of both beams)
I-upper = 40-percent of I-lower
M-upper = 49-percent of M-lower
I-upper = 33-1/3...
