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Help me figure this out Pls.

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posted on Jun, 1 2011 @ 05:56 PM
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I have a small out building, it is 10 x 16 that is 14 feet tall with a 12/12 pitch roof. It weights about 2300 pounds and sits on the ground. I would like to figure out what wind speeds would it take to blow it over.

I know there are formulas to do this but I do not understand the math..

Can anyone help?

thanks




posted on Jun, 1 2011 @ 06:02 PM
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reply to post by 22-250
 


I think your missing some crucial data. What is it made of? Wood, steel, concrete? How are you anchoring it to the earth? Are you setting it in a foundation? is it a pole building? ect....



posted on Jun, 1 2011 @ 06:10 PM
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reply to post by 22-250
 


Hopefully you did not build it out of straw or twigs.

All it would take to blow it over is a "huff and a puff".



We would also need to know what it is resting on and how it is anchored as well.



posted on Jun, 1 2011 @ 06:10 PM
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It's wood construction with concrete fiber siding. It sits on a limestone base, With a grid of 4x4 making up the floor, other than that it is built ike a house. I was going to anchor it down but have not done it yet. Last summer it endured 70 MPH winds, lucky me, I am wondering just what it would take for it to move or blow over. The west wall is 16 feet long and 9 feet tall. Thats where the wind always comes from during bad weather..



posted on Jun, 1 2011 @ 06:13 PM
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70mph or just below 90 mph within that range. If it's sturdy enough or has something securing it into the ground it should hold to just above 90mph no more.



posted on Jun, 1 2011 @ 06:13 PM
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reply to post by highfreq
 


Actually, the only truly crucial info that's missing is the orientation of the roof, and whether or not the 14' includes the roof or not. So...

1) Does the 14-foot height include the roof? Or is the roof on top of that?
2) Are the slanted sides of the roof along the 16-foot or 10-foot side? That is, does the peak run parallel to the 16-foot side, or to the 10-foot side?

Unless, of course, someone figured it out intuitively, without actually calculating it.
edit on 1-6-2011 by CLPrime because: (no reason given)



posted on Jun, 1 2011 @ 06:14 PM
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Here's a couple of articles to help you determine that it's one article in two parts: www.nwda.net...

www.nwda.net...



posted on Jun, 1 2011 @ 06:16 PM
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The building is 14 feet to the peak of the roof which is 12/12, quite steep and runs the 16 feet.



posted on Jun, 1 2011 @ 06:22 PM
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Explanation: S&F!

How to convert windspeed to psi. [ehow.com]


Blowing wind exerts pressure on the objects that are in its way. The amount of pressure exerted by wind on an object depends on the wind's speed and density, and the object's shape. If you know these three variables, you can easily convert wind speed to pressure in pounds per square inch (psi). Before attempting this calculation, it is useful to know that the density of dry air at sea level is roughly 1.25kg per cubic meter and that every object has a drag coefficient (C) that can be estimated based on its shape.


Nuke blast damage and injury. [stevequayle.com]


The is a definite relationship between the overpressure and the dynamic pressure. The overpressure and dynamic pressure are equal at 70 psi, and the wind speed is 1.5 times the speed of sound. Below an overpressure of 70 psi, the dynamic pressure is less than the overpressure; above 70 psi it exceeds the overpressure. Since the relationship is fixed it is convenient to use the overpressure alone as a yardstick for measuring blast effects. At 20 psi overpressure the wind speed is still 500 mph, higher than any tornado wind.

As a general guide, city areas are completely destroyed (with massive loss of life) by overpressures of 5 psi, with heavy damage extending out at least to the 3 psi contour. The dynamic pressure is much less than the overpressure at blast intensities relevant for urban damage, although at 5 psi the wind speed is still 162 mph - close to the peak wind speeds of the most intense hurricanes.

Humans are actually quite resistant to the direct effect of overpressure. Pressures of over 40 psi are required before lethal effects are noted. This pressure resistance makes it possible for unprotected submarine crews to escape from emergency escape locks at depths as great as one hundred feet (the record for successful escape is actually an astonishing 600 feet, representing a pressure of 300 psi). Loss of eardrums can occur, but this is not a life threatening injury.

The danger from overpressure comes from the collapse of buildings that are generally not as resistant. The violent implosion of windows and walls creates a hail of deadly missiles, and the collapse of the structure above can crush or suffocate those caught inside.



These many different effects make it difficult to provide a simple rule of thumb for assessing the magnitude of harm produced by different blast intensities. A general guide is given below:

1 psi Window glass shatters
Light injuries from fragments occur.
3 psi Residential structures collapse.
Serious injuries are common, fatalities may occur.
5 psi Most buildings collapse.
Injuries are universal, fatalities are widespread.
10 psi Reinforced concrete buildings are severely damaged or demolished.
Most people are killed.
20 psi Heavily built concrete buildings are severely damaged or demolished.
Fatalities approach 100%.


Personal Disclosure: With the information provided so far I would confidently say between 70mph and 162mph.
I would hazzard a guess at maybe 120mph give or take 10mph.
I am not an expert.



posted on Jun, 1 2011 @ 06:31 PM
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Here is something else that maybe forgotten and provide some other insight as well.
Mythbusters.

Either way if you believe their science, it is still entertaining.
Grab some popcorn and enjoy.



posted on Jun, 1 2011 @ 06:57 PM
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reply to post by g146541
 


Thanks for that video, I can tell you after being in 28 hurricanes once the wind gets in it's game over...



posted on Jun, 1 2011 @ 07:06 PM
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I love it when math and experience coincide.

After working through the math, which happened to include a bit of calculus, as well as some fancy drawings, I have an answer of about 80 miles per hour (78.4 mph, to be exact...but it's hard to be that exact with something like this).



posted on Jun, 2 2011 @ 03:06 AM
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reply to post by CLPrime
 

You're not going to show your work?

I noticed one previous poster just referred to PSI on the side of the building. But I think you were on the right track by asking about the roof, the slope and height of the roof.

And, since the building is not anchored, did you calculate lift of the roof as well as the pressure on the roof? Mobile home roofs are really bad, they are shaped almost like the top of an airplane wing so they generate a lot of lift, so if they aren't anchored well, they can be blown over a lot more easily than just sidewall pressure calculations would suggest.

It's not a simple calculation. And that article someone posted called "Determining Wind Loads On Buildings" was pretty useless.



posted on Jun, 2 2011 @ 07:12 AM
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Thanks for the input guys.

Yeah lets see your work.



posted on Jun, 2 2011 @ 11:16 AM
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I was hoping I wouldn't have to (I can't even just post a photo of my work, 'cause it's too jumbled with wrong calculations and whatnot), but, if I must...

- Taking into account the profile of the building perpendicular to the wind, the center of gravity can be estimated to be just a bit above the middle of the square wall (it's off-center due to the addition of the roof on top, of course).
- From this, the angle the building needs to tip before it falls over can be found, as the point at which the center of gravity passes over the corner that remains on the ground. This angle turns out to be about 30 degrees.
- And this was where the calculus came in. The average force of gravity that needs to be overcome over the 30 degree tilt is given by:

9.81 * (3/pi) * [S(a=pi/6, b=pi/2) 1-sinx dx]

where S is the integral, taken over the 30-degree tilt.

- This works out to 1.69713 m/s^2
- The force required to move the 2300-lb (that is, 1043.26245-kg) building against this acceleration is. then, 1770.552 N.
- Now, pressure is force/surface area. However, the wind doesn't act to tilt the building over the entire side facing the wind. First, it actually pushes down a bit on the roof, since it's slanted into the wind. Also, it pushes sideways below the center of gravity, so this has to be eliminated, as well as a negligible force just above the center of gravity (a portion of which is due to the wind pushing down on the roof). So, after taking all of this into account, we're left with the wind acting to tilt the building against only about the top sixth of the facing wall.
- The surface area of this region of the facing wall (a sixth of it) is 2.23 m^2. The pressure, then, acting on this region is 794 N/m^2, or about 16.583 lb/ft^2.
- Finally, the method for converting from pressure (lb/ft^2) to wind speed (mph): divide by 0.0027 and square-root it...

sqrt(16.583/0.0027) = 78.37 miles per hour

Of course, there are plenty of sources of error with this: estimating the center of gravity, the tilt angle, and the region the winds acts on to tilt the building, as well as neglecting friction and the fact that the wind doesn't exactly push up against gravity (though, these last two should roughly cancel each other out). But, having a lot of experience with compensating for such errors, I feel comfortable with the result.
It should take about 80 mph to tip the building.



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