huangyhg

load spreading in wood diaphragm
i'm doing a project where i've got a diaphragm as shown in the image... i am wondering if anyone knows of any study to find a spread angle for loading in wood sheathed diaphragm. my diaphragm is 19/32" shtg with 10d nails @ 6" o.c. edge, 12" o.c. field.
my thoughts are that the diaphragm shown will distribute the lateral wind loading in the north/south direction as follows... (assuming the south to north direction to be the worst case)
- line a will take half the distance between b and a
- line b will take half the distance between a and c and share its load with a.5 and b.5
- line c will take half the distance between b and d and share its load with b.5 and c.5
- line d will take half the distance between c and d
i know i can accomplish this by ibc code with providing sub-diaphragms with blocking, but am wondering if anyone knows a study or section in the code which allows a spreading of the load such as the 45 degree vertical load spreading allowed in masonry...
if you need further clarification, please ask questions.
thank you for any help you may offer.
tyler
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you're making it either more complicated or less complicated than you need to, depending on how you look at it. the following two paragraphs briefly outline the two commonly used approaches:
if the walls are relatively rigid compared to the roof diaphragm, then the approach is to simply distribute the lateral force to each line of resistance according to its tributary width.
if the walls are relatively flexible compared to the diaphragm, the analysis must take torsional effects into consideration, and the analysis becomes a bit more complex. the force introduced to each shearwall (moment frame, etc) is calculated according to two ratios: the wall's lateral rigidity compared to the total lateral rigidity, and the wall's torsional rigidity compared to the total torsional rigidity.
the building code governing your work will have limiting ratios of diaphragm deflection to story drift to determine whether the diaphragm is flexible or rigid. this will determine which approach is appropriate.
in either case, draw a shear diagram (just like you would do for a beam) of the diaphragm in each direction and check the calculated shear forces against the allowable values for your diaphragm. this analysis sometimes leads to the subdiaphragms you mentioned in your post.
examples can be found in design of wood structures by breyer and seismic design of building structures by lindeburg.
regarding the last paragraph in your post: diaphragms don't work like that. failure will occur where you try to introduce a point load into the diaphragm, before any 45° spread of load can occur. the load has to be distributed into the diaphragm through a collector (aka drag strut).
i believe what you are referring to with the 45 degrees is the reduced vertical load pattern to a cmu or concrete lintel when the depth of the lintel is greater than half its length.
i don't see how this concept for load reduction could be applied to a flexible wood diaphragm when the lintel is rigid.
nkt is correct. just keep it simple. i would be concerned with the two interior cmu walls at grids b and c being able to take 1/3 of the shear each, being so short. the drag forces are going to be high and the foundations large for the overturning unless you use grade beams, not to mention the shearing stresses to the walls.
too bad the atrchitect couldn't use two steps on the north side of the building too, aligning them with the south face. this would make things a whole lot simpler structurally. oh well, if we didn't have architects, they wouldn't need us, right?
mike mccann
mmc engineering
does he actually have cmu walls?? if so i would just try and take out all in the load in the two exterior walls. to me, it seems like there is no need to fool around trying to figure a contribution from the interior walls unless you need them. i guess the floor diaphragm capacity will govern your analysis.
you didn't mention where the shearwalls or braces were. could you, please do so?

this seems to me, as others have mentioned, to be an issue of whether your diaphragm is rigid or flexible, and where the lateral force resisting elements are located. that will determine the answer to your question.
and it's probably more appropriately called load distribution than load spreading.
ok... so to answer some questions.
- the diaphragm is wood and therefore flexible.
- the walls are wood 2x6 construction.
- all exterior walls are pretty much taken up by door and window openings. love this
- i've provided shearwalls at the steps in the building (i.e a.5, b, b.5, c, c.5) along with a perforated shearwall along line a and line d, blkg at lines a.5, b.5, and c.5 to create sub-diaphragms, and drag strut connections at lines b and c.
my initial intention was to try and get away without necessitating the blkg at lines a.5, b.5 and c.5...

in the situation you describe, "exterior walls pretty much taken up by door and window openings", the diaphragm may not be flexible. do the analysis and check it.
you will need a drag strut on each line where you use the walls as shearwalls, to avoid introducing a point load into the diaphragm. it is not necessary to use all of the available walls as shearwalls, just make sure to not build any unintended rigidity into the walls you don't intend to use as shearwalls.

i would check the roof diaphragm to check to see if it can span from end wall to end wall and just use your two end walls as shear walls. if it can not, i would next see what it takes to make the roof diaphragm rigid. using a rigid diapragm almost all your shear force will go into the end walls and everything should work fine. i would only use a flexable diaphragm using all of the walls as shear walls as a last resort design method. even if i did not use the short interior walls as shear walls, i would dowel and reinforce both ends as they will take some shear load in the real world.
i agree with nkt--if you are going to use any of those interior shear walls, you must have a truss lined up with the wall, and you must use a drag strut connector from said truss to the shear wall (simpson makes a nifty drag strut connector). without these drag struts, all the load will go to the outside shear walls.
daveatkins

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