mooose29 - 8-19-2011 at 11:04 AM
The recent storm in Mulege/Punta Chivato got my Dad and I thinking about wind loads on things like garage doors. We got into a debate about just how
much wind force is placed on objects at different wind speeds. I am not an engineer, but I did stay at a holiday inn express last night so use these
numbers at your own risk 
Also this is a basic formula for wind force over a
surface area, if you want to get really specific you can do a lot more.
The basic formula for wind force on an object is
L (load)=A (area) X .00256 X V (velocity) X V (velocity) X cd (drag coefficient)
I will use our garage doors as an example.
Area, are doors are 16 ft high by 12 feet wide so the area is 192 sq feet
Velocity, I will use Russ's average of about 45 MPH during the last storm
cd, = drag coefficient, for a flat object like a door it is 2, for rounder objects like a light pole it is 1.2, you can go online and find the cd for
the object you are looking for.
So for our doors for the last storm here was the average
L=192 X .00256 X 45mph X 45 mph X 2cd
L=1,989.12 pounds of force on the doors on avrerage
I believe Russ recorded a max wind speed of 67 mph which would equal a force on our doors of over 4,400 pounds.
That would be like standing a surburban on end and leaning it against the doors
dtbushpilot - 8-19-2011 at 01:46 PM
I think it would be more like turning the garage on end and parking the suburban on the door 
.....dt
durrelllrobert - 8-19-2011 at 03:50 PM
The .002046 in your equation is the coefficient of exposure (K) for flat areas with obstructions under 30 feet. If there is nothing between you and
the water then K = .003052:
1.Select the category of terrain for the structure. Choose category "A" for city centers with other structures nearby over 70 feet . Choose "B" for
wooded or urban areas with structures under 70 feet. Choose "C" for flat areas with obstructions under 30 feet in height. Choose "D" for flat,
unobstructed areas.
2 Use the following to find the coefficient of exposure (K) using the terrain category. For exposure "A" use .000307. For exposure "B" use .000940.
For exposure "C" use .002046. For exposure group "D" use .003052.
Next you need to use this formula to estimate wind pressure (q) on a structure: q = K x V^2 = coefficient of exposure x basic wind velocity SQUARED.
(ie 45 mph x 45mph = 2025) and 2025 x 003052 = 7.126 lbs/ sq ft wind pressure. If you retain your K = .002046 coefficent factor the wind pressure =
4.143 lb/sq ft
Since(most) garage doors are 7 ft high x 16ft wide the square area would be 112 sq ft and therefore, the wind pressure = 112 x 7.126 = 798 lbs. In the
case of your 192 sq ft door and using wind pressure = 4.143 lb/sq ft =the load would be 795 lb.
Read more: How to Calculate Wind Load on a Structure | eHow.com http://www.ehow.com/how_6587081_calculate-wind-load-structur...
[Edited on 8-19-2011 by durrelllrobert]
Cypress - 8-19-2011 at 04:09 PM
Been thru several hurricanes. Just imagine standing up in the back of a pickup truck going 75 miles/hr. Neat trick! Now, imagine someone has a high
pressure water hose spraying you. And the water is rising a foot every few minutes. You really can't, can you? 