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.
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Pressure Loading due to Wind
- Thread starter Anthonyr
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What israelkk is refering to is sometimes called reference pressure. But wind velocity is measured in m/s not m/s^2, that was probably a typo.
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
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
The codes vary, and most of them are not intended for 20mph.
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.
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.
1prsplmps1
Structural
Just use the typical 0.003*V^2 to obtain your result in psf.
0.003*v^2, that was a simple formula but what is i based on?
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
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
1prsplmps1
Structural
ThomasH: yes, that would depend on how precise AnthonyR wants to be. As you know the 0.003 value really is 0.00256 that came from the density of air at roughly 60 deg. F. at standard atmospheric pressure. Then this density (0.00237… slugs/ft^3) is multiplied by (.5)[(5280 ft/mi)/(3600 sec/hr)]^2. Therefore, that "constant" can be altered by the temperature and pressure..
Cheers!!
1prsplmps1
Cheers!!
1prsplmps1
I guess I don't know why you would even bother worrying about a 20 mph wind - man - that's nothing - 1.2 psf - jeez that's a waste of time.
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.
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.
1prsplmps1:
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
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
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