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Deflection of a scaled down aircraft wing

  1. Dec 3, 2015 #1
    1. The problem statement, all variables and given/known data
    As a project, I have had to design and build a scaled down version of a glider wing. The actual glider wing would be made of aluminium and have a length of 7.5m and have a uniformly distributed load of 30kg/m^2, the scaled down version is 0.9m and made of balsa wood. I want to work out the theoretical maximum deflection of my scaled down wing and then test it to compare results.
    For balsa wood E=16GPa, for aluminium E= 69GPa.
    second moment of inertia for the scaled wing is 5.72×10-10m^4
    2. Relevant equations
    dmax=(UDLxL^4)/(8EI)

    3. The attempt at a solution
    dmax=(UDLx0.9^4)/(8x16x5.72×10-10)

    dmax=(30x7.5^4)/(8x69xI)

    How do I scale down the UDL?
     
    Last edited: Dec 3, 2015
  2. jcsd
  3. Dec 3, 2015 #2

    SteamKing

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    You might want to double check the second moment of area for the scaled-down wing. 0.572 m4 is pretty large. That's the second moment of area of a solid square cross section which measures 1.62 m on a side. Make sure that you haven't omitted a ×10-something
     
  4. Dec 3, 2015 #3
    Changed, thanks
     
  5. Dec 3, 2015 #4

    rcgldr

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    It would help if you can the "polars" of both wings noting the difference in Reynolds number in both cases. XFOIL is a program that can calculate polars if you can't find existing data. The scaled down version will have a smaller wing chord and travel at slower speed, so the Reynolds number will be less. The speed will be related to the wing loading more than the scale factor. If the scale ratio is 1/r, then to get the speed to scale down by 1/r, the mass of the smaller model would need to be (1/r)4 of the full size glider. More on this at this web site;

    http://www.charlesriverrc.org/articles/design/ibtherkelsen_scalespeed.htm
     
  6. Dec 3, 2015 #5
    The scaled down wing has a length 8.3 time smaller than the original, so the UDL of 30kg/m^2 would become 30/8.3^4 6.3x10^-3kg/m^2
    giving: dmax=(6.3x10^-3x0.9^4)/(8x16x5.72x10^-10) = 56455.283m. (Obviously too big)
    The area of the scaled down wing is 0.08m^2, therefore the mass I need to apply in experiment is: (0.08)(6.3x10^-3)=5.04x10^-4kg
    (This seems a lot too small)
     
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