I've been looking at various span charts for decks and I'm still not quite sure how to calculate it all when you take into account not just the joists, but the beams and posts as well.
What we currently have:
23' wide deck attached to house with ledger. All pretreated wood.
Extends 10' with 2×6 joists 16" OC.
Outside 'beam' is 2×10 with joist hangers. Beam supported by 3 4×4 posts.
I have a hunch this isn't to code…or…if it is…just barely. It's solid–but bouncy.
I'm not thrilled with the 2×6 joists, but I also am not thrilled with having to scrap the entire deck. Plus, the 2x6s do help with the headroom underneath the deck (which we plan on turning into a patio.
What I'd like to do:
1) replace the exterior 2×10 beam with an actual beam 1 or 2' in and under the joists (so the joists are cantilevered and rest upon the beam)
2) replace the 3 4×4 posts with either 3 6×6 or 4 4x6s.
My question is: how do I calculate what that new beam has to be given the existing joists and the desired post options?
Best Answer
First, your question
This is a bit long winded, but it is how we figure it out accurately.
Figure out a design load per square foot (PSF), we typically use 40 PSF live load and 10 PSF dead load for floors and decks. In a simple span, the beam carries the load halfway to the next support or tributary width; in this case 5'. Multiply the PSF x the tributary width to get the pounds per lineal foot (PLF) on the beam, in this case 50 PSF x 5 Ft = 250 PLF. Then, using software or load tables find a beam that can carry that load based on the span of the beam; allowable loads reduce as the span increases. In your case this requires (3) 2x10 DF #2.
I typically use Forte for beam sizing instead of load tables.
For the post sizes, calculate the tributary load which, in simple spans, is the load halfway between supports. In your case, this would be half the beam length between post multiplied by PLF. For your center post that is 250 PLF x 11.5 FT = 2875 LB. Then, using load tables or calculations find a post that can support that load based on height and bracing of the column. Take into account bearing area of the 2x10 at about 625 PSI perpendicular to the grain for a required bearing area of 4.6 Sq. In. In your case a 4x4 is plenty.
Try a column load calculator here
For multiple span beams, cantilevers, and anything outside of "simple span" things are a little different. For this reason I typically use the beam sizing program.
If this is too intense, maybe just be guided by the following.
If I read correctly, you have deck joist spanning about 10' and on one end a single 2x10 "beam" or rim joist that the joist hang into that spans approximately 11'-6" between (3) 4x4 post.
Based on this, your deck joist are structurally OK although maxed out at a 40/10 load, but are deflecting (bending) a good amount giving part of the bouncy feel. You can add 2x6 or larger joist down the middle of the existing joist to firm things up, or just live with a safe bouncy deck.
The 2x10 is definitely undersized and should be addressed. It also contributes to the bounciness. The simplest solution to this is to add (2) intermediate post mid span reducing the span of the 2x10 to below 6'. OR beef up the 2x10 by adding (2) more 2x10's to it nailing 3 16D nails per foot per beam lamination would easily transfer the load between beams.
Finally, the 4x4 post are more than adequate to support a load of 2,875lb. An 8'-0" length of 4x4 can support over 6,000 lb in wet service.
We applied over 10,000lb to a 7' 4x4 in jacking up a floor with no indication of possible buckling; we were compression limited on the member being jacked.