Wind Base Shear vs. Wind Trib. to Diaphragm 1

[medeek]

When looking at the wind tributary to a diaphragm in order to calculate the shear wall capacities I normally resolve the wind forces to the roof and half the wall height (single story, four ext. walls for simplicity sake). This force is then applied to the shearwalls at the roof eave height.

However for overturning calculations it would seem more appropriate to include the forces on the lower half of the wall as well as resolve the roof forces at a height of the "mean roof height".

Does this seem logical? I've seen different calculations and spreadsheets that show the wind base shear as both of these numbers, one that includes the forces on the lower half of the windward wall and the other that is only the total forces tributary to the roof diaphragm. My feeling is that the wind base shear should be the total lateral force.

 

[jdgengineer]

When looking at the foundation, it would make sense to include the total wind shear (which is what our office does for foundation sliding). I don't believe overturning matters though as your force doubles but your moment arm gets cut in half.

Also you should visualize how the force gets into the system. The perpendicular walls take the shear force from the studs and the studs span between the roof and the foundation. Your shearwalls are then designed for the roof reaction. It's not necessary to design them for the foundation reaction as the studs transfer that load directly without the use of a shearwall.

 

[CELinOttawa]

The rule of setting aside the lower half of the wall is due to the way in which the load is transfered. An element (stud, wall section, tilt up panel, etc) that spans vertically and transfers reactions to the floor plate above and floor plate below does not have the bottom half contribute to lateral shear loads if at ground floor. These "bottom half of base wall" forces do not contribute to overturning either.

 

[medeek]

For simplicity sake lets ignore the roof for now. The picture below hopefully clarifies my question.

CEinOttawa you are correct in that the overturning and the shear loads are not affected by the loads on the bottom half of the wall.

However, if you consider foundation sliding then the full lateral load of "HW" is in play as Jdgengineer pointed out.

I guess it is just a matter of terminology, that I want to make sure I am getting correct. When we talk about the Wind Base Shear which is the correct value?

 

[CELinOttawa]

It still matter where you mean. In a side on of a whole house, your sketch is misleading. You cannot have the forces you show, since they are not on the part you have drawn.

If you have a solid block, your sketch is correct. For a building or house, your HW/2 is close, but the lower of the two is on a perpendicular section of foundation wall into (or out of) the page.

For your shear design, lateral analysis, detailing of connections, etc, etc, etc (all but the relatively obscure check of foundation sliding) this lower force is an out of plane force only.

 

[medeek]

Redrawing the Sketch, I still pose the same questions about the Wind Base Shear:

 

[CELinOttawa]

The total base shear as a global action is indeed HWL, but it is meaningless for design with only rare exceptions. The actions that matter are the top half of the wall except for one or two checks in narrow circumstances such as a few hot climates and very narrow designs which don't need depth due to other causes (tornado/cyclonic, soil conditions, etc, etc, etc).

Re your sketches you're missing the next box, where your blue HW/2 is resolved into the foundation wall parallel to the UDL shown, and the base shear that matters is the HWL/4 for each wall element. In terms of overturning those forces are present, but on a lever arm at or near zero.

Your building's base shear only matters if your building is little more than a structure on shallow strip footings. Even for slab on grade this is never going to govern any part of the design. You can call this the base shear until the cows come home, and you're technically right, but it is still never going to matter especially when this gets extended to a full size structure and multiple walls.

The reality is already complicated enough with the components of shear that come from leeward walls, windward and leeward roofs, etc.

 

[medeek]

Based on the preceding conversation and considering the following load pattern on a gable roof with wind in the transverse direction as shown below:

I then get the following loads applied to the roof diaphragm (plf):

From this we should be able to calculate the reactions on the diaphragm from the exterior shear walls (no interior shearwalls for this derivation).

We will then check the minimum load case where 8 psf and 16 psf are applied to the roof and walls respectively and also the case where the roof loads are neglected and only the wall loads are considered.

In this particular situation the roof zones 2 and 2E are negative so the load case where the roof loads are neglected will govern.

 

[medeek]

In summary, the Wind Base Shear @ Load Case A (transverse) would be the summation of all of these distributed loads multiplied by their length of application.

 

[medeek]

Interesting to note that for a gable roof only the wind forces on the walls contribute any lateral forces to the roof diaphragm and then to the shear walls for the longitudinal direction (wind parallel to ridge).

The breakdown of the forces would then be:

Transverse:

Longitudinal:

One could argue that there is a certain amount of wind force on the edge of overhangs but I'm thinking its insignificant.

 

[CELinOttawa]

The overhangs have negligible facing loads. Much more important would be to not forget drag forces which, to be fair, are also typically negligible until you get a long structure. But they are more important than any contribution of the overhang, and they should be included.

 

[medeek]

How do you consider the drag forces?

 

[KootK]

I've only considered drag forces for open structures with flat roofs. In those situations, standard wind load procedures give very small diaphragm loads so it seems prudent to bump them up a bit. Gas station canopies are the best example that I've encountered. I'm curious to know what the procedure is for handling drag loads myself. I've been simply loading these flat roofs as though they were mono sloped at 15 degrees. And that's just me making stuff up. I imagine the answer would depend on surface roughness and the behaviour of the stagnant air layer separating the surface from the fast moving air.

The greatest trick that bond stress ever pulled was convincing the world it didn't exist.

 

[CELinOttawa]

While they are always present, and I always consider them if I feel the structure warrants them, strictly speaking drag forces only need to be addressed to meet the legal minimum of competent practice if they are sighted or included by the code you're applying. They matter more in areas that get high winds, but are present regardless.

Note that drag forces build up to the point of being more than negligible when the length (either dimension) of the structure is four or more times the mean height. That makes them reasonably rare.

The code I am most familiar with for "frictional drag" is AS/NZS 1170.2 "Wind Actions". In that code the procedure is:

Fz = fz * Az

where Az = Area of surface along which drag is calculated

fz = 0.5ρ_air[V_des,Θ]²C_figC_dyn​

C_dyn is always 1.0, except where the structure is considered aerodynamically sensitive (fabric structures, for example)
C_fig for drag is equal to C_r, the frictional drag coefficient.

C_r = 0.04 (for surfaces with ribs across the wind direction)

= 0.02 (for surfaces with corrugations across the wind direction)​

= 0.01 (for smooth surfaces without corrugations or ribs or with corrugations or ribs parallel to the wind direction)​

Note that I have no idea what the US equivalent would, however the AS/NZS procedures are compatible with the Canadian codes and I do apply drag forces in my Canadian work where I believe they are warranted.

 

[CELinOttawa]

P.S. All that to say that Kootk's instincts were right... Lol.

Medeek: If you are writing the kind of all-encompassing software you seem to want to be producing, I believe you should include drag forces and make note of where these are coming from in your code comments, or release documentation (etc).

 

[KootK]

Thanks for taking the time to set this out for us CEL. Is there an actual requirement in the Canadian code for drag? I'll confess that it's been some time since I've gone through it myself in detail.

4 x h will encompass a fair number of one story buildings in at least one direction I'd think. Also, based on the coefficients that you've quoted, it sounds as though the "stagnant boundary layer" business that I was taught is mostly BS.

One of the projects on my "probably never get around to it" list is to compare all of the English speaking codes. I sense that each code (us/can/au/NZ/bs) contains some pearls that aren't found elsewhere. Kinda wish I could read German too. They seem to kicking ass when it comes to wood and glass.

The greatest trick that bond stress ever pulled was convincing the world it didn't exist.

 

[CELinOttawa]

No mention of drag in NBCC or OBC, at least not that I've found. The only minor nod is under free standing walls, where there is a transverse component to the loading.

 

[medeek]

I don't see anything for drag in the ASCE 7-10, at least not in Ch. 28 Part 1 Envelope Procedure. I think I will add a footnote to the longitudinal lateral load diagram explaining that if the building length (L) is longer than 4 times the mean roof height that wind loading due to drag should be further investigated. There will obviously be some limitations of the software I'm developing since I can't program every conceivable configuration. I am trying to make this calculator comply with the US standards (ASCE 7-10, IBC 2012).

 

[medeek]

Wow, I think I have it actually calculating the base shear for both transverse and longitudinal directions (gable roof type only). The overhangs are so messy to calculate, much simpler without them. It is actually quite interesting to include them though so one can get a better sense at how much they actually contribute to the lateral forces on the building.

I think the next step is to apply the loading calculated and generate the reactions at each end of the diaphragm for the shearwalls.

 

[medeek]

Shearwall reactions are now complete, assuming of course that only the exterior walls are acting as shear walls. This assumption is pretty useless for building with dimensions exceeding a certain limit since they will probably involve steel moment frames or multiple internal shear walls. I'm wondering if it might be useful to add some sort of advanced option which lets one specify an internal shearwall spacing or even a internal shearwall layout. The programming would get pretty complex.

I also added the ability to calculate truss/rafter uplift and horizontal load. Not entirely sure how useful this is either but I've personally ran into it enough times so I thought it might be of some benefit. The funny thing is after writing the code and creating the image for this section (3) I happened to look at a truss manufacturer's output for a garage I was designing a while back. I quickly noticed that the horizontal reactions and uplift were listed on the document so if you've already taken your design to get the trusses quoted you probably don't need this information calculated. Just out of curiosity I used the same parameters as the truss manufacturer used for their wind loads and after adjusting for the TC and BC dead loads both my horizontal reactions and uplift were within 0.5 lbs of their values. Nothing like a check nailing it so perfectly, that is why I love this stuff.