OK! I cannot stand this any longer - since we are building airplanes,
I doubt that we are using solid bars for this mount - for a rectangular
tube, or a hat section, the major working areas are the top and bottom
"flanges" so for a given wall thickness strength only goes up slightly
faster than linear with increased depth. dig out your old copies of
Rourke, and compute some I over C's for the real sections you are proposing.
Vance

I did Vance. A hat section with wide flanges is equivalent in "I" to a rectangular
tube with the lower wall split in half and the parts bent outwards.

For a  one inch high hat section with .0625 wall:

I = (bd^3-b1d1^3)/12

I1 = (2 X 1)^3 = 8    I2 = (2 X 0.875) ^3 = 5.359

I  = 8 - 5.358 = 2.641

Height of beam doubled to two inches high wall .0625:

I1 = (2 X 2)^3 = 64  I2 = (2 X 1.875)^3 = 52.73

Therefor I now equals 64 - 52.73 or 11.27 which is a whole bunch higher
than 2.641!!!!!!

Max Stress = MaxMoment/(I/c) where c is half the height of the flange.


BTW you computer clock is off :)

Paul Lamar

Paul, one difference between a rectangular tube and a hat section is that with the hat section, the edges are free, rather than constrained by being joined at the bottom of the rectangular tube, so aren't they more free to buckle outwards under load than when constrained by being in a tube format?

That said, Jerry's testing is the proof of the pudding in my estimation.

You mentioned stopping by on the way to Oshkosh, we will be leaving to go north to OSH on the Saturday before Oshkosh, so would you be able to stop by before that? Would like to have you visit.

Bill Schertz
KIS Cruiser # 4045

Yes you are correct. I few scrap straps as you suggested should  stabilize the flanges
and solve that problem.

Sure. My itinerary is flexible. I hope to have my Linux laptop up and running
so I can continue the newsletter while I am out of town.

Paul Lamar


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