Does a conversion factor exist that tranforms wind speed to pressure? For example, what would be the value in psi exerted by a 100 mph wind?

Hi Mark,

Fortunately, there is a formula.

it is:
pressure = 1.25/2 * speed^2
[speed in meter per second - m/s, result is in pascals - Pa]

Anyway, you wanted it in psi and mph. I hope I got it right...
pressure = 1/55201 * speed^2
[speed in mph, result is in psi]

good luck
Michal
www.tfdesign.cz

There is no easy answer when it comes to wind speed. Although the equation listed above will get you in the ball-park, and probably be slightly conservative, which is what you want unless all the details are known.

To get the exact answer, you need to know the site conditions, the building size and shape, the local governing code, and what you are trying to determine with the pressure.

Hi,

I'm not very familiar with your codes over sea but the very first thing we do is turning speed into pressure. Other factors, like those listed above, come in play after that.
Anyway, the question was clear.

Best
Michal

Michal,
Thanks alot. I'm sure there's more here than meets the eye! I'll multiply the result by two just for safety

Hi Mark & Michael,

I just checked my windmilling book which has a formula to calculate windpressure on a flat plate with the wind applied normal to that plate :-

Pressure = (mph)[squared] x 0.005 lb/sq ft.

This would yield a result twice that given by Michael's formula.

Which one is correct ?

Regards

Ken Hume

Gentlemen,
I am so far over my head it's not even funny. But that's never stopped me before... Anyway, a look at Simplified Engineering for Architects and Builders gives static pressure as:
Qs (psf)=0.00256*V*V (mph).
That happens to agree with Michal. A friend who designed towers for the telephone used factors of 20 psf for 70 mph and 40 psf for 100 mph. That also agrees with Qs with a added safety factor of approx. 50%. Anyway it would be nice to see examples of how Qs is actually used. For example what would be the uplift on a 20'x20' pavilion with a given slope? And if I wanted to get even deeper over my head, I might ask if the uplift came from the wind moving over the top of the roof or from below or even some combination.

Hi,

To make it clear, one of my formula was taken from EuroCode 1 and apply to plain perpendicular to the wind. The other one was just my derivation which should work for you better because of units.

The exact given formula was:
g = 1/2 ro * v^2

where ro is an air density and can be safely set to value 1.25 (kg/m^3 - kilograms per cubic meter), so we are getting the formula stated above (pressure = 1.25/2 * speed^2)

This is just basic value and you have to use other formulas to get required/expected figures for designing your roofs and walls.
It is usually something like:
needed_value = g * building_height_factor * shape_geometry_factor * safety_factor
and can be found in your local code.

Michal

Accurately applying windloads are not for the faint of heart. There are entire books dedicated to the subject on how to apply them accurately, and specialty software programs and the like. I've outlined the basic engineering procedure for determining wind load pressures below:

According to ACSE-7 (American Society of Civil Engineers Minimum Design Loads for Buildings and Other Structures), which is THE source for engineers in the US, the wind pressure is determined by:

qz = 0.00256 * Kz * Kzt * Kd * V * V * I

where

Kz is a factor ranging from 0.68 to 1.89 based on the building height and exposure (open field, in a woods, etc.)

Kzt is a factor relating to the surrounding terrain. (Is the structure on a hill, edge of a cliff, etc.) Wind speeds up going up and over a hill.

Kd is a directionality factor, based on the type of structure. For most structures, this is 0.85.

V is the wind velocity in mph, based on a three-second gust windspeed commonly seen in the new IRC and IBC building codes, not the "Fastest Mile" windspeed seen in the UBC/BOCA.

I in an importance factor. For most structures this is 1, agricultural buildings less than 1, and for hospitals/shelters, greater than 1.

After evaluating the above equation, we get the wind velocity pressure for a particular building at a paticular height.

To determine the pressure to apply to the structure, you use a series of different equations, based on the structure type, but is in the order of:

p = q * G * Cp - qi * (GCPi)

where

q is the pressure developed above (qz). (qz is based on a particular height. Different pressures can be applied to the building at different heights.)

G is a gust factor based on the rigidity of the structure (again, usually 0.85)

Cp is the external pressure coefficient, which can be a positive or negative number, depending on building shape, windward or leeward side, etc.

qi and GCPi are similiar, but are the internal pressure coefficients.

So, Mark, assuming a 20' x 20' open pavillion, in a 100 mph wind zone, making some general assumptions, qz is 18.5 psf. For a 10 degree roof slope with the ridge perpendicular to the wind direction, the uplift (design pressure, p) is 24 psf on the windward side, and 13 psf on the leeward side. For a 12/12 pitch, with would switch up to a 5.6 psf DOWN pressure on the windward side and a 11.1 psf uplift on the leeward side. And, with open structures, the uplift comes from air moving on both sides of the roof (above and below). Air must move faster over the top, since it has to travel up and down the roof, which causes a lower pressure, pulling/pushing the roof up. Essentially, same as an airplane wing.

Hi Joe,

Thanks for the in depth explanation and though at first sight it appears that the simple windmilling formula is out by a factor of 2 it might also just be that it is correct for windmills when the appropriate factors are selected !

Since windmills are tall and slim (50 feet not including sails) and generally set on a very exposed position on top of a hill or upward hill slope to deliberatley catch the wind then the various factors that you have quoted above could just about get you to the values quoted in the old timer windmilling formula to which I made reference above.

I think that the principals you outline above are much the same as used in Europe today for general building design and your point is well made that it might pay to get professional help on such matters.

Regards

Ken Hume