X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Received: from vms173019pub.verizon.net ([206.46.173.19] verified) by logan.com (CommuniGate Pro SMTP 5.3.11) with ESMTP id 4657368 for flyrotary@lancaironline.net; Wed, 29 Dec 2010 00:46:44 -0500 Received-SPF: pass receiver=logan.com; client-ip=206.46.173.19; envelope-from=finn.lassen@verizon.net Received: from [192.168.1.100] ([unknown] [71.98.181.96]) by vms173019.mailsrvcs.net (Sun Java(tm) System Messaging Server 7u2-7.02 32bit (built Apr 16 2009)) with ESMTPA id <0LE600DR5DCVJA80@vms173019.mailsrvcs.net> for flyrotary@lancaironline.net; Tue, 28 Dec 2010 23:46:09 -0600 (CST) Message-id: <4D1ACB1F.2040203@verizon.net> Date: Wed, 29 Dec 2010 00:46:07 -0500 From: Finn Lassen User-Agent: Mozilla/5.0 (Windows; U; Windows NT 5.1; en-US; rv:1.9.2.13) Gecko/20101207 Thunderbird/3.1.7 MIME-version: 1.0 To: Rotary motors in aircraft Subject: Off topic: Hangar doors References: In-reply-to: Content-type: multipart/alternative; boundary=------------040401080801020806090902 This is a multi-part message in MIME format. --------------040401080801020806090902 Content-Type: text/plain; charset=ISO-8859-1; format=flowed Content-Transfer-Encoding: 7bit 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 --------------040401080801020806090902 Content-Type: multipart/related; boundary="------------040209050600010608050508" --------------040209050600010608050508 Content-Type: text/html; charset=ISO-8859-1 Content-Transfer-Encoding: 7bit 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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