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#### gratefulgary

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I'm building a woodshed. It will sit on concrete pilings at the corners. It will be post and beam, that portion of the structure, walls, roof will not be supported by the beams in question.

The floor will be 12ftx12ft, with joists, 3/4 in plywood floor. It will need to support ~36,500 lb of wood. What dimension should the beams that form the perimeter of the floor be to safely support that loading? And how does one calculate that?

Gary

#### kwikfishron

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What do you have 6-7 cords of green Oak stacked 8’ high?

4x4 ground contact PT 16”oc for your 16” pieces should be fine.

#### Ron6519

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With that weight, I'd pour intermediate supports if this structure is off the ground.
Ron

#### dtsman

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In perspective,
36,500 lbs = 9 - Ford F150's.
On a 12x12 floor they would be 3 wide and stacked 3 high.

I would suggest a 6" slab based on the given info.

Bo

Remember,
If the women don't find you handsome,
they should at least find you handy.
(Red Green)

#### jcrack_corn

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oak is 35-55 ish lbs/cubic foot (wide range there, i know)

if your room is 12x12x10 thats enough room to stack about 44,000 pounds of oak if you use every single cubic inch. (including NO air between each piece of wood, lol). (using the lower range, ~40lbs/ft3)

are you sure you have 36000 pounds of wood, and if so, are you sure it will fit in a 12x12 room?

#### gratefulgary

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I used the "log cord weight calculator" here ( http://www.csgnetwork.com/logweight.html ). To give myself a little head-room I went on the side of hard maple, which is a goodly percentage of the wood I use. Using some simple arithmatic, with a height of 6.5 ft, and a weight of 5000lb/cord, that shed would hold 7.3 cord, or 36,562lb.

I cannot use a slab. I may be able to pour intermediate pilings. Nevertheless, I'm looking for a simple explanation of the load parameters so I can calculate the minimum beam size.

It's my understanding I only need the formula for the beams perpendicular the joists. The joists solution is a different formula and is relatively straight forward, I think.

And please, spare me the pick-up truck humor.

I didn't really follow what kwikfishron meant.

Thanks guys,

Gary

#### Ron6519

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I didn't really follow what kwikfishron meant.

Thanks guys,

Gary
Nobody ever does. He's a big Sci fi fan if that helps.
Ron

#### dtsman

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http://www.awc.org/pdf/wsdd/c2b.pdf

Here is a link to beam calculations. If the load is uniform, then it is just as straight forward as the joist. If you want to use 3 beams then you need 3 that will each hold 12000lbs, or 4 that will hold 9000lbs etc. The chart shows you the max spans for your intermediate footers.

Hope that helps.

Bo

Remember,
If the women don't find you handsome,
they should at least find you handy.
(Red Green)

#### bob22

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I didn't really follow what kwikfishron meant.

He meant, I think, place your floor joists on the earth so the earth is your beam(don't need a beam since the ground is there to support).

#### kwikfishron

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I didn't really follow what kwikfishron meant.

He meant, I think, place your floor joists on the earth so the earth is your beam(don't need a beam since the ground is there to support).
You got it Bob, the OP question was vague at best. I assumed it was firewood and wondered why you would want to build a floor system to support that kind of weight when you could simply cut in some 4x or 6x level into the ground and stack the wood.

I can’t remember what Planet I learned that technique on.

#### gratefulgary

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I thought it was rather specific, actually. If one doesn't know the answer, that doesn't make them a bad person. I need to construct it the way I described for several reasons, none of which are relevant. In a perfect world, I wouldn't have the constraints that exist and I could pour an f'ing slab. But that's not the case.

Nevertheless, thanks dtsman. Those tables are a big help.

G

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