Snow drift on sloped roof

[vmirat]

We're designing an addition to an existing building. The existing building has a flat roof. The addition is a lower roof which is sloped (shed roof). The lower roof slope is 4 in 12. The difference in height between the two roofs is 12 feet. The calculated drift height is 3.5 feet. The drift width is 14 feet.

My question concerns the drift width. Is the width a horizontal projected distance or does it follow the slope of the roof?

 

[Teguci]

drift width is projected horizontal. But, that should only be used for the flat roof. The sloped roof will "shed" the snow load. What am I missing?

 

[SKIAK]

The first sentence of ASCE 7-05 section 7.4 says "Snow loads acting on a sloping surface shall be assumed to act on the horizontal projection to that surface." I usally don't account for the roof shedding snow unless I'm 100% certain that the roofing will be forever smooth and allow the snow to shed.

 

[vmirat]

The roof will have snow clips at the eave, so clearing of the roof is not possible.

I also read 7.4 about sloped surfaces. In the past, I've always used project area, but then I read the commentary which seems to imply that sloped surfaces have a clearing effect due to wind, hence the ability to use the horizontal projected load. However, because this is a lower shed roof, you won't have the same clearing effect, so that's why I asked the question. I'll probably still use the projected area though.

Thanks for the responses.

 

[SKIAK]

The ability of the roof to clear snow is definatly in the realm of "engineering judgement." If you believe that this roof will be less likely to clear then you might want to consider using a value closer to the flat roof snow load.

It should be conservative to use the snow load along the slanted axis of the beam. The figures 7-3, 7-4, and 7-5 all show the snow load presented on a horizontal beam even for curved and other roofs.

 

[Teguci]

yup - my fault. I'd still use the horizontal projection for the drift width.

 

[vmirat]

OK, I'm going to tag onto this thread for a different, but somewhat related, question. I'm dealing with a snow drift which is a triangular load that only partially covers the length of the beam. I can't find a beam formula for moment and deflection for this condition. I can find one that completely covers the beam, but not partial length. I'd rather not have to put this into my computer program if there is a simple plug/chug beam formula.

 

[SKIAK]

Eh... I don't have it handy but I'm sure "Roark's Formulas for Stress and Strain" would have it; although I'm not sure it is a "simple" plug-and-chug.

 

[strguy11]

In a related matter, what if the lower roof was steep and you could reduce it based on sliding. How do you take the drift and sliding into account?? Ch you reduce the drift load?

 

[vmirat]

SKIAK, I've looked through Roark and didn't find it. In fact, Roark doesn't have very many practical beam situations like the steel manual.

strguy11, ASCE 7 says that the drift load is superimposed on the balanced snow load. If your roof has a slope of 70 degrees or more, then the Cs factor is 0 and you have no balanced snow load, hence no drift load. Up to that point, I would assume that the code wants the drift load to be the full amount, not reduced by slope. However, you bring up an interesting question, which brings me back to my original post. It seems that it would be more realistic to apply the drift width parallel with the roof slope versus a horizontal projected area. This would act as a reduction of the drift load based on roof slope.

 

[vmirat]

I did find one formula in Roark, Table 8.1, condition 2.0, that looks like it could be reduced for a horizontal simply supported beam without end moments where "wa" is 0. However, you still have to pick an "x" value for the location of the moment and deflection desired. It would be nicer to have a formula already developed for Mmax and Dmax.

 

[SKIAK]

OIC, well that was my suggestion; I don't have any other references with information like that. I like Roark because it has a lot of those other non-practical (and by practical I mean typical) conditions. If you didn't want to use a computer I'm sure you could figure out a solution for the max moment by hand... if you had the time.