Pressure Loading due to Wind
Pressure Loading due to Wind
(OP)
I am looking for an expression to determine the effective pressure on a rectangular surface due to wind. I have an application where the wind speed can reach a maximum speed of 20 mph and need to determine the effective pressure. Thanks in advance for your help.






RE: Pressure Loading due to Wind
RE: Pressure Loading due to Wind
RE: Pressure Loading due to Wind
F=0.5*Rho*V*V*A
Where :
Rho = Air specfic gravity ususlly 1.205 kg/m^3
V = wind velocity m/(sec^2)
A = Surface area m^2
RE: Pressure Loading due to Wind
RE: Pressure Loading due to Wind
RE: Pressure Loading due to Wind
What you might need is a design pressure. That means that you have to find the shape factor for your structure and also include the dynamic properties of the wind. If your structure is flexible you also have to include the dynamic behaviour of the structure.
Building codes usually cover this for structures regardless if they are buildings or not. However, your post does not include any information regarding how you will use the information.
As for the 20 mph, compared to any code requirements it's a very low value.
Regards
Thomas
RE: Pressure Loading due to Wind
Thanks for the correction.
Sure, it should be m/sec.
RE: Pressure Loading due to Wind
If I am doing structural calculations then I would use a much higher value as listed in ASCE 7.
Big difference in application.
RE: Pressure Loading due to Wind
RE: Pressure Loading due to Wind
For my applications (tanks), typical pressure on a flat surface is taken as 30 PSF at 100mph, and applied with allowable stress design. And you multiply by (V/100)^2 for other speeds, which will not give you much pressure.
The assumption made is that pressure is proportional to velocity squared, which assumes the drag coefficient for the object is a constant. In reality it isn't, and when you use a design wind speed vastly different from what the codes are intended for, you might get fairly meaningless results. So if you take building code formulas and plug in 400 MPH, don't expect to get real meaningful results.
The codes differ also as to whether the wind speed is a gust speed, or 3 sec average, at what height, etc. The pressures generated by the building codes are intended to applied with certain multipliers or stress adjustments, and if you don't know what those are, the pressure doesn't mean that much. (For example, the 30 PSF above is for allowable service stress with 1/3 increase).
Let us know a bit more about what you're doing and some of the folks here can get you more meaningful answers.
RE: Pressure Loading due to Wind
RE: Pressure Loading due to Wind
20 mph gives 0.003*20^2 = 1.2 psf = 57.5 Pa
20 mph = 8.94 m/2
Formula given by israelkk:
0.5 * 1.25 (kg/m^3) * (8.94 (m/s))^2 = 50 Pa.
And this is a very ideal situation. So I would be very careful using 0.003*v^2 without background information. It might be ok for low windspeeds like 20 mph. But it can also a oversimplification.
Regards
Thomas
RE: Pressure Loading due to Wind
Cheers!!
1prsplmps1
RE: Pressure Loading due to Wind
All mechanical equipment etc. that fall under most building codes are required to be designed for the code mandated wind speed for that region - which in most parts of the US is 90 mph. In most parts of the world its 90 mph except along coasts and in mountain regions where its even higher.
RE: Pressure Loading due to Wind
That's what I dislike about those simplified formulas. Yyou loose so much information "in the process" that the final result doesn't make sense.
I have such formulas myself but then I know the background, as I do now with your formula
JAE:
From a atructural point of view I definetly agree, but we don't know the application. Or do we?
Regards
Thomas
RE: Pressure Loading due to Wind