What's the formula for weight support?

Hi, folks. I'm planning on building a porch -- that is, a shed-type roof that will come off the back of the house and shelter an existing flagstone patio that is mortared to an eight-inch concrete slab.
What I'm trying to figure is how many posts I need.
Dimensions will be about eight by 24 feet. Construction will be post-and-beam with 6-by-6 posts and beams. The underside of the roof will be open, so the posts, beams, joists and decking will all be visible. Rather than plywood, the decking will be tongue-and-groove planking to provide a beaded effect.
I'm thinking three posts should be enough. That is, one post at each corner and one in the middle. So there will be support every 12 feet. Am I OK here?
I don't want to clutter the area unnecessarily, so the fewer posts the better.
Thanks
Mark
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Mark wrote>Hi, folks. I'm planning on building a porch -- that is, a shed-type

about 6-7 feet, for human support. You'll be fine with 3 posts. Tom Someday, it'll all be over....
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One more might help. A 6x6 with modulus S = bd^2/6 = 5.5x5.5^2/6 = 27.7 in^3 and max bending moment M = fS = 1000S = 27.7K in-lb = WL/8 might support WL = 222K in-lb, ie a total weight of W pounds over an L inch span.
With a 100 psf floor load and L in inches, an 8'xL" deck section supports 8x100L/12 pounds. If the building supports half of that, the beam along the outer edge of the deck supports W = 400L/12 lb, and 400L^2/12 = 222K makes L = 82 inches, so an 8' span might work.
Nick
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I had understood him to mean that he was building a roof over an existing surface, not anything that was intended to support foot-traffic. That being the case, the total load on the roof will be around, 30# x 192 sqft, = 5760, of which half will be supported on the outer edge, = 2880. Since that's presumeably the long edge, he needs a horizontal member that will support about 120 pounds per linear foot. IIRC, you can cantilever a beam about 1/4 of the distance it would span if it were supported at both ends, so shifting each post in 4' gives a 16' span and a 4' cantilever. Using L/240 for the maximum deflection, that requires a built up beam of two 2x10s. (Note: I wouldn't use 6x6s for the horizontal beam.)
If he wants the thing to look good, he's going to want diagonal braces anyway. How big those are will depend on how high the roof is, but assuming they start at 2' below the post, he can move the posts another foot in, and the diagonals buy him another 2' for a total clear span of only 10'. *THAT*, he can do with a 6x6..
(Note that, if you're going to use the diagonals as structure and not decoration, you can't just tack them in place using 16d nails.. A purist will use a big-ass shouldered mortice and tennon, with pins. A modern contractor will use 3/8" iron plates, and 1/2" bolts..)
--Goedjn
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Why 30 lbs? Is he in Colorado snow country or south Florida? Wind?
Mike
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