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Rain Load ASCE 7
Rain Load ASCE 7
I am questioning my understanding of the applied rain load per ASCE 7...
Let's say you have a roof that was designed for 25psf snow load. You now want to know how far the main drain can be located away from your scupper (or secondary drain).
ASCE 7 says that the rain load is 5.2psf/in(h.s + h.d). Where:
d.s = depth of water on undeflected roof up to the inlet of the secondary drainage system.
d.h = additional depth of water above the secondary inlet (undeflected roof) at its design flow (hydraulic head)
Let's say you have a roof that was designed for 25psf snow load. You now want to know how far the main drain can be located away from your scupper (or secondary drain).
ASCE 7 says that the rain load is 5.2psf/in(h.s + h.d). Where:
d.s = depth of water on undeflected roof up to the inlet of the secondary drainage system.
d.h = additional depth of water above the secondary inlet (undeflected roof) at its design flow (hydraulic head)
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It appears as though this is applied as a uniform load when viewing the examples in the commentary. However this seems odd in that the slope of the roof is likely constant and thus the load would be triangular. So then I look again at the definition of d.s and it is stated as the depth of water on the undeflected roof... So I suppose this could be interpreted to vary based on the location/distance from the secondary drain. Which do you suppose is correct?Loader By Rain 3 0 Free
Loader By Rain 3 0 Torrent
Back to the example:If we assume that the load is uniform, the hydraulic head is 1' and the slope is 1/4' per foot then...
h.max = (25psf - 1' * 5.2)/(5.2) = 3.8in and then L.max = 4' / 0.25 in/ft = 15.2ft
If the load is varying then that changes things more of an iterative process maybe?
I am questioning my understanding of the applied rain load per ASCE 7...
Let's say you have a roof that was designed for 25psf snow load. You now want to know how far the main drain can be located away from your scupper (or secondary drain).
ASCE 7 says that the rain load is 5.2psf/in(h.s + h.d). Where:
d.s = depth of water on undeflected roof up to the inlet of the secondary drainage system.
d.h = additional depth of water above the secondary inlet (undeflected roof) at its design flow (hydraulic head)
If we assume that the load is uniform, the hydraulic head is 1' and the slope is 1/4' per foot then...
h.max = (25psf - 1' * 5.2)/(5.2) = 3.8in and then L.max = 4' / 0.25 in/ft = 15.2ft
If the load is varying then that changes things more of an iterative process maybe?
Thanks!
Let's say you have a roof that was designed for 25psf snow load. You now want to know how far the main drain can be located away from your scupper (or secondary drain).
ASCE 7 says that the rain load is 5.2psf/in(h.s + h.d). Where:
d.s = depth of water on undeflected roof up to the inlet of the secondary drainage system.
d.h = additional depth of water above the secondary inlet (undeflected roof) at its design flow (hydraulic head)
Loader By Rain 3 0 Pillows For Women Reviews
It appears as though this is applied as a uniform load when viewing the examples in the commentary. However this seems odd in that the slope of the roof is likely constant and thus the load would be triangular. So then I look again at the definition of d.s and it is stated as the depth of water on the undeflected roof... So I suppose this could be interpreted to vary based on the location/distance from the secondary drain. Which do you suppose is correct?Loader By Rain 3 0 Free
Loader By Rain 3 0 Torrent
Back to the example:If we assume that the load is uniform, the hydraulic head is 1' and the slope is 1/4' per foot then...
h.max = (25psf - 1' * 5.2)/(5.2) = 3.8in and then L.max = 4' / 0.25 in/ft = 15.2ft
If the load is varying then that changes things more of an iterative process maybe?
Thanks!
EIT
www.HowToEngineer.com
