Is there a general ROT for when a brace is needed for post?
I have seven 6×6 post supporting two roof structures. They all sit on concrete piers, and 4 of them are locked in at the bottom by deck framing. The other three support a roof over a drive way. Maximum height of any of them is 8′ from the Simpson post anchors to the bottom of the beams.
Dave
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Replies
Are you talking about diagonal bracing? Or a mechanical attachment from post top to beam?
Diagonal bracing at the top of the post.
What I have is four post on the outside of a 36' wide by 14' deep porch/deck. The two outside corner post carry double 14"LVL beams that support the 36' roof truss above the porch. The two inside post are not load bearing for the roof, just 2x10s let in above and the lower beams for the floor joist. The deck is essentialy at grade level, so all the post are not only anchored on the piers, but bottom braced by the 2x10s on the deck framing. I am not to concerned about the two inner post in the line but with the two corner post that are load bearing. From the top of the deck to the bottom of the LVL beams I have 7' of post. Do I need diagonal bracing at the the beam/post connection?
The other area is a small 18' wide by 10' roof over a driveway. Gable truss roof setting on three post and 10' beams to the building. Post are anchored to piers with Simpson post hardware. From the top of the piers to the bottom of the beams it is 7'- 3". Do I need diagonal bracing at the top of the two outside load bearing post ?
BTW roof is 4/12 on the building.
Dave
Like Dave said, diagonal bracing--trangles will not "rack" (what the second poster talked about where you rectangle suddely becomes a very short parallelogram, crushing stuff beneath it).
I gather that the porch roof is a shed roof.If that is the case the roof, after sheathing, will be the brace. The plywood/roof joist will can not move side ways.Not clear on the other one.
Neither roof is a shed roof. The 36' is spanned by a gable end truss and the18' is also a Gable end truss.
I understand the reason for knee braces on post. They are of particular importance on decks where there can be large live loads (literally moving masses of people). My roof trusses are designed for local ground snow loads and wind loads, so the only possible weak link in the load transfer to the piers and footings are at the post beam connections. The remainder of the roof is over an additional 33' of framed building on the 36'side and the small 18' roof tees into the side of the larger one.
I don't know that I need knee braces on the post. I will run it past my engineer in the next few days, but in the meantime I was wondering if there is a general rule for when they are needed.
Dave
ROT -- It took me ten minutes: Rule Of Thumb.
What you must see in your mind is the weakest link -- where the structure will fail. A toddler can build a tower of blocks on a table, but when somebody bumps the table, they come crashing down. Likewise, you can build your standing 6 X 6 columns and they will hold up for years until the next hurricane, tornado or earthquake. Or maybe time and rust will be your undoing.
There is something known in the iron workers trade as the "two nolt rule". ENR reported a couple of years ago about a tragic accident in which somebody got crushed to death. They installed two columns and then a beam across them. Each was secured by one bolt. This acted as a parallelogram. Something happened and it just folded over. You can prove it to yourself with some 1 X 1 sticks and some 3/16" screws.
Now adding two screws at each connection with give you a certain amount of rihidity to prevent this but not much. Think of the principle of the lever. Remember that Greek guy who said, "Give me an adequate fulcrum and a long enough lever, and I can shift the earth from its orbit". Those Simpson post bases have maybe 12" separation between bo;ys yet the lever arm is 96": an 8 to 1 ratio.
The surest way us to build a prototype and see where it fails. This is expensive. Or you could build a model and let your toddler ttry to destroy it. Also time consuming and expensive.
The engineer tries to eliminate all this experimenting with mathematics. Actually they start with experiments and take measurements. Then they derive a formula to replicate these results. The result looks like this: A^3-.0178-((SQRT B +/- d*C) * f).
In this example, A is something you already known such as the thickness of your beam. B is some sort of hypothetical number based on the species and grade and condition and age and sex of the wood you are ysing. You can look this up in a book of tables if you have it. C is a constant for which you need another book. The clincher to this nonsense is f which is the factor.
What the engineers do is multiply the result of their equation so that the resulting number in this particular instance matches the results of the experiment. For the above, I refer you to the movie, A Day at the Races in which Chico haros his wheelbarrow of books to Groucho.
That said and done, look at your proposed structure and consider where and how it is likely to fail.
The rule of thumb is to overbuild and always follow the advice of the clerk at Home de Pot.
~Peter
No incipet remarks necessary.