flyrotary@lancaironline.net Mailing List Archive

I sense that you’re well on your way to push-back accordion fold door; but here’s my 2-cents worth after using that type and a power-lift bifold, each for about 2 years.

The accordion fold doors had translucent fiberglass over the framing, which was very nice for light when the doors were closed. The lower track wasn’t drained, so it filled with water at every rain and the track and the rollers rusted in a couple of years. And, the doors were a bit of a chore opening and closing, although these were rather tall; 16-18 ft, so were a bit heavy.

The power-lift bi-fold is great. If I were doing it; I’d add some framing so it could have some translucent panels in the upper half for light when the door is closed.

Al G

-----Original Message-----
From: Rotary motors in aircraft [mailto:flyrotary@lancaironline.net] On Behalf Of Finn Lassen
Sent: Tuesday, December 28, 2010 9:46 PM
To: Rotary motors in aircraft
Subject: [FlyRotary] Off topic: Hangar doors

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?

$I_0 = \frac{bh^3}{12}$

(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.
$\pi \left(\frac{{r_2}+{r_1}}{2}\right)^3 \left({r_2}-{r_1}\right)$
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

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