Wing weight

Discussion in 'General Experimental Aviation Questions' started by Dave Hodges, May 17, 2019.

Help Support HomeBuiltAirplanes Forum by donating:

  1. Jun 19, 2019 #61

    Dave Hodges

    Dave Hodges

    Dave Hodges

    Well-Known Member

    Jun 29, 2014
    Likes Received:
    Reidsville, GA
    Thank you very much for doing these calculations. I didn't know about the half-elipse formula. So I just divided the U into a L and a J. A piece of paper inside the U was 13/16 inch long. A piece of paper outside the U was 1-1/32 inch long. So averaging those two and multiplying by 0.041 gave me a cross-sectional area of 0.0382 inches∆2. You calculated 0.0337 inch∆2. We're in the same ball park there. Your minor and major radii are of the inside of the U. That makes sense, being conservative. Do you suppose that's why the actual load slightly exceeded the calculated load? But when you consider that cross-sectional area was about 0.038, rather than 0.0337 inches∆2, the actual load should have been greater than it was.
    Thank you for telling me that a long flange, with zigzags welded all along it, should be considered to be a series of pinned columns, rather than fixed columns. Applying that principle does indeed validate the Euler equation. But I suspect that if my joints had been bolted together, rather than welded, the truss would have been much weaker.

Share This Page