Uniformly Loaded Beam Overhanging One Support

[Sedgehammer]

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;


For I his equation is correct;

MI for hollow rectangle beams = ((Width * Height³) - (Inside_width * Inside_height³)) / 12

E is 29,000,000
For x use the equation locating the point of zero shear in the middle graphic. for x¹  use a.

yes, that's a simple beam, but as you said, if with an overhang it acts as a deflection in the main support area. If beam passes simple beam, it passes the one with the small overhang
That's more math then I trust myself to do, to just get an idea what might work, to see if cost effective. simple math i'm better then most, but that wood take me to area's of my mind that haven't been explored in many years....... :laugh:
I'll also try and run it by my engineer, but he's not always conducive to answering things just to get an idea if

[Sedgehammer]

scratch all that. Just will use an H beam like your calcs allow, but heavier then needed. maybe a 14x43. Not that it is needed. just looks better i think. I can inlay the web w/ wood. Maybe drill 1" holes every 2' and use sq headed bolts and nuts to bolt it

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

[Gary_C]

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

A mighty big and well protected one because your life and entire structure may well depend on that single post. Better think about footing required too.  :

[Sedgehammer]

@Gary_C  yup. agreed

@Don P if you wood chime in on this please sir

according to this column toolbox and the numbers used, a SYP #1, 10x10 passes easily. But maybe my inputs are off, as it looks like a 2x2 wood pass, so somethings wrong i figure
I used the following inputs;
unbraced length = 106" ceiling is 10' - a 14' H beam = 106"
depth = 10"
width = 10"
compression = 7,130
elasticity = 1.79
load = 16,044

and what sized footing wood be recommended

[Gary_C]

Murphy's Law should be be telling you to look for another alternative. Putting all your eggs in one basket is risky. 

[Sedgehammer]

Eggs are for  digin1 .......  ;D

Looking at all options at this point, but I think a 10"x will suffice, but I'll see if Don or someone else can make tails or heads off my inputs were correct

I think a 3' sq x 1' deep footing will be more then adequate from what I have read

[Don P]

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.

Footing, the distance from edge of post horizontally to edge of footing is also how thick it should be. Close enough there. At least a # grid of rebar in the lower third, no steel closer than 3" to soil anywhere.

Footing size = load/allowable soil bearing capacity. Assuming 2,000psf allowable soil capacity x 9 sf of footing = 18,000 lbs allowable, plastic clays can be down around 1500 and muck can be worthless. Think about your wheel loads and whether you stick trucks. Too small is no bargain if its soft.

[Sedgehammer]

Thanks!

Stout indeed! Me bride wants to use one of our largest ERC logs that we cut down this spring. I see it's even stouter in that regard. I'm just having a hard time with right angled interior wood/metal w/ a round post...... does not compute, does not compute.....

[trimguy]

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

[Sedgehammer]

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

i've never seen a woman be so right about so many things and she's not arrogant about it. she's just smart. with that said, she's not always right.........

[Sedgehammer]

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.

If i twas to use a green syp post. say a 12x12 or even larger, what amount of end shrinkage wood I have? i know it wood take several years for it to completely dry, but is it an amount that wood have sheetrock not happy?

[Don P]

Very little shrinkage in length unless it is full of juvenile wood or twisted grain.