# Roof Wind Load



## PEin2010 (Sep 25, 2010)

Reference: Question #506, Structural depth, NCEES 2000 sample problems and solutions.

The question is to calculate the maximum roof diaphragm service load chord force (kips). the wind load is given as 400 lb/ft. The plan of the roof is 192' x 120'.

The solution calculates wl^2/8 = 400*192^2/8 = 1843200 lb-ft, 1843200/ 120' = 15360 lb = 15.4 kips is stated as the answer. What i don't understand is that why is the wind load applied perpendicular to the roof? will the wind ever act like that?

If this is of any help: the figure shows the elevation and plan view of a single - story bldg with precast, prestressed double-tee wall panels and double-tee roof elements supported on the wall panels on the outside and on an inverted tee beam on the inside.

Thank you.


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## MOOK (Sep 25, 2010)

PEin2010 said:


> Reference: Question #506, Structural depth, NCEES 2000 sample problems and solutions.
> The question is to calculate the maximum roof diaphragm service load chord force (kips). the wind load is given as 400 lb/ft. The plan of the roof is 192' x 120'.
> 
> The solution calculates wl^2/8 = 400*192^2/8 = 1843200 lb-ft, 1843200/ 120' = 15360 lb = 15.4 kips is stated as the answer. What i don't understand is that why is the wind load applied perpendicular to the roof? will the wind ever act like that?
> ...



This is the reaction from the wall to the roof due to wind load.


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## MA_PE (Sep 26, 2010)

Like Mook said they are providing a uniform wind load with no other information. The maximum forces in the roof are developed with the uniform load applied normal to the surface and over the full surface. Therefore that is the way you should apply it.


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