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Estimate stopping force of roof versus tornado

  1. Feb 25, 2015 #1
    Problem statement

    In the not-so distant past, authorities recommended leaving the windows to a house open in a tornado to equalize the pressure and avoid blowing off the roof. We can estimate the validity of the advice. Assume the pressure inside the house is one atmosphere. Approximate Densities of roofing materials:

    Doug Fir 2x6: 1 kg/ft
    Plywood 5/8": 0.85 kg/ft^2
    Gravel roofing 2.5: kg/ft^2

    For a 40ft x 60f. area of roof, there are 40x20 ft of 2x6 (1.5 ft spacing between 2x6's) and 2400 ft^2 of plywood and gravel roofing. Assume 6 screws per 2x6 with a holding force of 500N each.

    Using the maximum pressure difference across the roof and the above data, can the roof blow off because of the pressure drop during a tornado, or is this not a valid concern?

    Relevant equations

    ##F=PA=\rho g h?##

    Work done so far

    Plywood and roofing: 40ft x 60ft = 12.192m x 18.288m = 731.52m^2
    Joists: 40ft x 20ft = 12.192m x 6.096m = 243.84m^2
    Screws: 6 screws x 40 joists = 240 x 500N = 120kN

    I’m afraid I do not know how to start this problem. It seems a little abstract from what we have already learned.
     
  2. jcsd
  3. Feb 25, 2015 #2

    Bystander

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    One step at a time: what's the mass of the roof; what additional downward forces are furnished; what's the lifting force due to pressure drops?
     
  4. Feb 25, 2015 #3
    ##m=V\rho##

    Douglas fir mass = (40 joists * 20 ft) * 1kg/ft = 2400 ft * 1kg/ft = 2400 kg
    Plywood mass = 2400 ft^2 * 0.85 kg/ft^2 = 2040 kg
    Roofing mass = 2400 ft^2 * 2.5 kg/ft^2 = 6000 kg

    Total mass = 2400kg + 2040kg + 6000kg = 10440kg

    ##f = mg##
    Downward force = 10440kg * 9.81m/s^2 = 102.416kN

    Pressure drop? I’m not sure how to calculate this from the given data.
     
  5. Feb 25, 2015 #4

    Bystander

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    From the problem statement (Check it carefully to see that it IS complete), the tornado is pulling a perfect vacuum. You've yet to calculate a total hold down force, by the way.
     
  6. Feb 25, 2015 #5
    So we are assuming equal pressure? So the pressure drop would be zero? I'm afraid we haven't really discussed pressure drop.

    Total downward force = 120kN + 102.416kN = 222.416kN
     
  7. Feb 25, 2015 #6

    Bystander

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    Pressure is defined as force per unit area. What's the area of the roof?
     
  8. Feb 25, 2015 #7
    Area = 40ft X 60ft = 12.192m X 18.288m =222.97m^2

    P = F/Area = 222.416kN/222.97m^2 = 997.52 N/m^2
     
  9. Feb 25, 2015 #8

    Bystander

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    Now you have a "pressure" holding the roof down, or a pressure difference that the roof will resist as far as lifting.
    Would you prefer Bernoulli, or just grab some rough numbers from the Uniform Building Code? Atmospheric pressure in SI units is ~ 100 kPa. Or, are you comfortable taking it from here?
     
  10. Feb 25, 2015 #9
    Going with Bernoulli, would the lift force be 1/2 * air_density * wind_speed^2 * roof_area? I am confused about the units and also confused how this ties in with opening the windows or not.
     
    Last edited: Feb 25, 2015
  11. Feb 25, 2015 #10

    haruspex

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    You don't have enough information to apply Bernoulli. Either look up some facts about tornadoes online or (I would think this is the intended approach) consider the maximum the pressure difference could possibly be.
     
  12. Feb 25, 2015 #11
    Why can't I apply Bernoulli. I have air density at 1.225 kg/m^3 and tornado wind speed varying by 70 to 300mph. And wouldn't the maximum pressure difference simply be the strongest tornado possible?
     
  13. Feb 25, 2015 #12

    haruspex

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    No wind speed was given. As I said, you could look up facts on the net. Alternatively, let the wind speed be arbitrarily great.
     
  14. Feb 26, 2015 #13
    Assuming an average tornado at 150 mph = 67.056 m/s, the dynamic pressure would be .5 * 1.225 * 67.056^2 = 2754 N/m^3. Taking this pressure vs. the pressure holding the roof down at 950 N/m^2, it is much greater. But I still don't see how this ties in with the windows being opened or not. Either way, the wind will blow the roof off. If the windows are open, the high pressure under the roof will blow it off and with the windows shut, the stagnant pressure against the house would still cause the roof to be blown off. Am I missing something here?
     
  15. Feb 26, 2015 #14

    Bystander

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    Does this address the original problem statement?
     
  16. Feb 26, 2015 #15
    I guess I wasn't sure of what the original problem statement was because this teacher is funny like that. I know from past experience he is looking for one thing that is not altogether clear from the problem statement sometimes.

    "Using the maximum pressure difference across the roof and the above data, can the roof blow off because of the pressure drop during a tornado, or is this not a valid concern?"

    So I would "estimate" a very big tornado at wind speeds of 350mph to assume the maximum pressure difference across the roof. And if I crunched the numbers, it would obviously blow the roof off because the upward force would be much greater than the downward force on the roof.

    I guess I was just looking for a better presentation of the evidence for the irrelevancy of leaving the windows open or shut.
     
  17. Feb 26, 2015 #16

    Bystander

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    The original statements of "open the windows" included that they be opened on the side of the house opposite the direction of the storm's approach. Given the low pressure in the "lee" of a bluff body like a house, the idea perhaps may have been to offset the Bernoulli lift of wind over the roof. You've found the roof mechanically able to resist ~0.01 atmosphere pressure difference. You've got some estimates of pressure differences generated by high winds. That's about all you can do.

    Don't know whether my expansion on the "folk wisdom" helps with the classroom problem. Been there --- wish you luck.
     
  18. Mar 2, 2015 #17
    As it turns out, I think my teacher is looking for something. I overheard him explain something to another student about this problem. He said if the windows are open the pressure drop, theoretically, is 0 because the pressure is equalized. But if the windows are open than the "maximum pressure drop" would be 1 atmosphere. It wouldn't get more than 1 so we are to use the value of "1 atmosphere" for our calculation. Does this make any sense to you?
     
  19. Mar 2, 2015 #18

    Bystander

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    If he's looking for "free" flow through a house with windows open as part of the answer, things have just left the realm of "real world problems." If there are four walls that are "all" window, and all open, and no "ground effect" difference between the lower and upper side of a flat roof (that's at least three impossible "ifs") one can say equal pressure.

    Play it by ear, but don't take the result of this particular exercise, too terribly seriously.
     
  20. Mar 2, 2015 #19
    What formula would I use in this crazy scenerio?
     
  21. Mar 2, 2015 #20

    Bystander

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    That would be Bernoulli both above and below the roof, same wind speeds parallel to the roof surface --- essentially treating the roof as a vane oriented in the direction of the airflow. I concur with your assessment,
     
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