A lurker on the list was kind enough to point me to the formula for I for a square tube:

I =

Looks like 1/8" wall 1.5"x1.5" square tubing will be adequate.
On 12/29/2010 12:46 AM, Finn Lassen wrote:

Looking to design and build hangar doors.

I kinda fancy vertical harmonica doors.
/\/\/\/\/\
The inner points carried by wheels in groove in concrete slab.
Wheels able to pivot in bottom of door frames.
Top supported by rollers in a steel U-channel, able to pivot in top of door frames.
I figure 3 feet wide sections hinged at edges.
The thinner the better, but will have to be able to withstand wind pressure without deforming.

10 feet tall.
Max wind pressure 31 pounds/sq ft. (110 mph wind zone).
Obviously that's on the high side. There are trees in the vicinity. So 20 pounds/sq ft may be a more realistic number. But it doesn't hurt to be on the safe side,

For calculations the vertical supports will be spaced 1.5 feet apart. (Actually 3 feet apart but doubled at each edge).

I'm figuring maximum of 450 pounds of distributed weight (wind pressure) on each vertical support.

I've seen a couple of different formulas for maximum deflection, for example:
5/384 * W * L^3 / (E * I).

I assume I can safely use E = 30,000,000 for steel.

Let's say I'll allow 3" maximum deflection:
I = 0.113?

However, I'm having trouble arriving at values for I (area moment of inertia).

What would I be for a 2"x2" 1/8" wall square steel tube?

(2 * 2^3) / 12 = 1.33?
Or would that only be for a solid 2x2" bar?

For a 1.5" square tube?
I = 0.42?

(I also need guard railing for my porch. Seems a1.5"D  tube with 1/8" wall is I =  0.1276.

Does that seem right?)

Also, I guess I'm not so much concerned with deflection as with point of permanent deformation.
How would I arrive at that number?

My objective is the thin hangar doors and the lightest guard rails. And of course cheapest materials.

Sanity check appreciated.

Finn