Wind Load Calculation Formula for a Communications Cabinet?

From a pressure vessel handbook.

70 MPH = 13 psf (Pounds per Square Feet)

80 = 17

90 = 21

100 = 26

110 = 31

120 = 37

130 = 44

The above info was dated 1982

I hope this helps.

Thanks,

This jives fairly well with Willyap06's post #6 below which yields 16.4 PSF for 80 MPH.

jh

Greetings,

I hope you will be able to help me. I am a sales rep for a Solar company. We pre-assemble 3 panels and frame them, wire them, attach a microinverter, ready to go. We have sold a couple of jobs in the Miami Dade area, a 146 mph wind zone. I have submitted our module to a testing company however they would like to know what is the PSF would be for this and what the accepted amount of deflection is? Could you answer this or steer me in the right direction? Thanks,

Mike Kantor mikekantor@readysolar.com

6H x 5W x 3D.

Units? Cubit? span? Metre? Foot?

Inch foot by a thick wide ?

First sentence of 2nd pp begins "In feet the cabinet measures..."

Ah got it cheers..

Have u checked in CP 3, Chapter V, Part 2 for wind load calculation?

Thanks for posting.

Pardon my ignorance but I am not familiar with CP 3. I am EE by degree but have been pretty much a software weenie since '80 so maybe one of you 'civil' gentlemen could point me in the right direction.

Thanx,

jh

What is your location?

Who is the local "AHJ" (Authority Having Jurisdiction)?

What "Building Code" do they follow?

The structural section of the building code will usually state the required wind loading in terms of pounds per square foot or miles per hour. Depending on the type of building or occupancy, they may even assign an "Improtance Factor" multiplier to applly to the requirement.

Ried the "maniacal inginear"

Formula:

q=0.002558V²

where: q = pressure in psf, on plane normal in the direction of the wind.

V= velocity of wind in mph.

Apply corresponding horizontal cross section factor for different shapes.

Thanks.

All I need now is the cross section factor for a flat surface. I have seen 1.0 and 2.0 for flat surfaces. Does your formula account for negative pressure on the backside?

Thanx,

jh

The formula is the same for the windward and leeward side of the structure.

To determine the effective pressure, multiply q by the corresponding horizontal factor C.

For a more complex structure like building, there is also consideration of coefficients, say k, for vertical section configuration (e.g., 0.80 windward/0.40 leeward) multiplied by the horizontal cross section factor C. The effective pressure will then be C x k x q.

The negative factor is for suction effect of the wind.

I thaught wind resistance for a vehicle...( not exactly the same I know) was proportional to the cube of the speed...

Wouldn't this imply wind load should be proportional to cube of wind speed?

I'm not trying to be argumentative... just genuinely curious.. (hmmm we know what curiosity did..)

Del

I'm curious too!

Somebody in the CR4 community may help shed light on this!

HELP... please help.

Stress on bolts

Folks, sorry about the link given in my last post. After I actually looked at the site in detail, I noticed that the formula given for the stressed area is incorrect. I also noticed, even if the formula had been correct, there's no information given for elastic modulus of steel (or any other metal, for that matter).

Yield strength for steel ranges between 35 kips and 100 kips (bolts => 50,000 psi), and elastic modulus ranges from about 28 X 10⁶ to 30 X 10⁶ psi (bolts => 100,000 psi).

Since I could have sent you on a wild goose chase, the root diameters and stress areas for various sized bolts are: 1/2" => .422", .140 in²; 5/8" => .531", .222 in²; 3/4" => .656", .338 in²; 7/8" => .766", .461 in²; 1" => .875, .601 in².

Wind load, in psf, on a flat surface given by willyap06 is 0.00256v². If the velocity is 80 mph, then the total load is 16.4 psf. The total area facing wind is 30 sq. ft. maximum, yielding a maximum load of about 500 lb. A 500 lb load 3 feet up (half the total distance) amounts to 1,500 lb tensile load on two bolts on windward side and 1,500 lb compression load on two bolts on leeward side.

Stresses on various bolts: 1/2" => 5,360 psi; 5/8" => 3,380 psi; 3/4" => 2,220 psi; 7/8" => 1,630 psi; 1" => 1,250 psi

Please! Somebody help me get my head outta my a**!

Thanks . . . I needed that.

A communications cabinet is not very big really but your location might be a little bit problematic i.e. if it falls off the roof it might kill someone.

Given that your measurements are in feet, I am guessing that you are located in America. The advice to use CP3 (Code of Practice 3) is refering to the old British codes which have since been superseded by BS 6399 (British Standards) which are currently being superseded by Eurocode 1. However, some countries do still use a version of CP3 e.g. Singapore. They use the local wind speeds as measured over the past 50 years or so with a statistical measure of 3 second gust speed which then corresponds to the manner in which the maximum wind speed is calculated based on location of building (valley, side of hill etc), height of building, form of building (is it long rather than tall).

But let's say for the sake of argument that your maximum wind speed on top of your building is 80mph (corresponds to about 35m/s which is a hurricane on the Beaufort scale which is conservative unless you know otherwise). Your building is 10m or less in height and does actually have that wind speed on the top of it.

http://en.wikipedia.org/wiki/Beaufort_scale

The basic wind pressure is 0.613 x 35 x 35 = 750Pa = .75kN/m2

The coefficients for windward and leeward are respectively +0.8 amd -0.5 for rectangular objects = +1.3 overall.

1.3 x .75 = .957kN/m2 say 1.0kN/m2. That's your overall wind pressure.

From that you generally need to apply a load factor for stress related calcualtions. For wind in the british codes this is 1.4.

1.0 x 1.4 = 1.4kN/m2.

Now all you have to do is calculate the force in the bolts. Main difficulty is generally not the bolts, it is the substrate to which you are attaching them.

Thank you all for posting. I think the beast has now been sufficiently slayed.

Best Regards,

John