2021 It's PIGROAST TIME!

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For this it isn't a tipping, see-saw, problem. The "control" is bending failure in the main 20' span rather than in the 4' overhang. I would not use tile in the bathroom on that kind of beam span or joist span and spacing. Experiment with making the overhang longer and watch what it does to the main span bending moment and point of max moment, the see-saw is on that end.

"Plan view"And that is why we use drawings, that is nothing like what the word pictures had conjured up in my head. Don't try to describe stuff, draw it and draw it well.Tributary area supported by the beam is highlighted within the red box;(Image hidden from quote, click to view.)Trib width would be 16' x 50psf (40LL+10DL) = 800 lbs per lineal foot load on the beam.This calc;https://forestryforum.com/members/donp/oerhangbm.htmYou're cutting it mighty fine in #1SYP

MI for hollow rectangle beams = ((Width * Height^{3}) - (Inside_width * Inside_height^{3})) / 12

That is a simple beam calc, a beam between 2 supports. What we've been talking about is a beam with an overhang. That is a different set of equations, the load on the overhang is reducing the deflection. I also do not recognize the equation he is using, not saying I've run it against the standard equation to see if he is getting the same result by another path but based on the spelling and notational peculiarities I'm a little leery. Might try engineers edge for a calc.It would take more rewriting than I have time to modify the wood calc for this. This is the equations you need, it comes from the AISC steel construction manual. The bottom 2 equations are deflection;(Image hidden from quote, click to view.)For I his equation is correct;QuoteMI for hollow rectangle beams = ((Width * Height^{3}) - (Inside_width * Inside_height^{3})) / 12E is 29,000,000For x use the equation locating the point of zero shear in the middle graphic. for x^{1} use a.

Now what post does one use......

It does not have to compute, just accept it . You know she right.😁

I could look at the calc but with those dimensions there's another way to look at it. If the post is 1" thick in least dimension for each foot of height... that is if the post is a 10x10 and is 10' tall, it isn't going to fail by buckling. Old engineers called that ratio a short column. The only check you need to do is compression, it will fail by crushing. Take the 100 square inch post top and multiply it by the allowable Fc parallel to grain, shoot low, use 525 psi (there's your error above) =52,500 lbs allowable load. It is a pretty stout basket.

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