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Date: Wed, 29 Dec 2010 23:07:06 -0500
From: Finn Lassen <finn.lassen@verizon.net>
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To: Rotary motors in aircraft <flyrotary@lancaironline.net>
Subject: Re: [FlyRotary] Off topic: Hangar doors
References: <list-4657369@logan.com>
In-reply-to: <list-4657369@logan.com>
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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.

I = 0.2184

Maximum deflection at center:

5 * 450 * 120^3 / (384 * 30,000,000 * 0.2184) = 1.55".
At 110 mph wind pressure (30 lbs/sqft).
Supports spaced 1.5 ft apart = 450 pounds distributed load over each 10 
ft tube.

That should be more than acceptable.

Can anyone spot any error in these formulas or calculations?

Finn


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?
> 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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<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">
<html>
  <head>
    <meta content="text/html; charset=ISO-8859-1"
      http-equiv="Content-Type">
    <title></title>
  </head>
  <body text="#000000" bgcolor="#ffffff">
    A lurker on the list was kind enough to point me to the formula for
    I for a square tube:<br>
    <p align="left">I = <img
        src="cid:part1.00000702.07000709@verizon.net" width="62"
        height="51"><img src="cid:part2.01080407.08010609@verizon.net"
        width="150" border="0" height="150"><br>
    </p>
    Looks like 1/8" wall 1.5"x1.5" square tubing will be adequate.<br>
    <div align="center">
      <center>
        <p align="left">I = 0.2184<br>
        </p>
        <p align="left">Maximum deflection at center:<br>
          <img src="cid:part3.02040501.09020202@verizon.net" alt=""><br>
          5 * 450 * 120^3 / (384 * 30,000,000 * 0.2184) = 1.55".<br>
          At 110 mph wind pressure (30 lbs/sqft).<br>
          Supports spaced 1.5 ft apart = 450 pounds distributed load
          over each 10 ft tube.<br>
        </p>
        <p align="left">That should be more than acceptable.<br>
        </p>
        <p align="left">Can anyone spot any error in these formulas or
          calculations?<br>
        </p>
        <p align="left">Finn<br>
        </p>
        <br>
      </center>
    </div>
    On 12/29/2010 12:46 AM, Finn Lassen wrote:
    <blockquote cite="mid:list-4657369@logan.com" type="cite">
      <meta content="text/html; charset=ISO-8859-1"
        http-equiv="Content-Type">
      <title></title>
      Looking to design and build hangar doors.<br>
      <br>
      I kinda fancy vertical harmonica doors.<br>
      /\/\/\/\/\<br>
      The inner points carried by wheels in groove in concrete slab.<br>
      Wheels able to pivot in bottom of door frames.<br>
      Top supported by rollers in a steel U-channel, able to pivot in
      top of door frames.<br>
      I figure 3 feet wide sections hinged at edges.<br>
      The thinner the better, but will have to be able to withstand wind
      pressure without deforming.<br>
      <br>
      10 feet tall.<br>
      Max wind pressure 31 pounds/sq ft. (110 mph wind zone).<br>
      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,<br>
      <br>
      For calculations the vertical supports will be spaced 1.5 feet
      apart. (Actually 3 feet apart but doubled at each edge).<br>
      <br>
      I'm figuring maximum of 450 pounds of distributed weight (wind
      pressure) on each vertical support.<br>
      <br>
      I've seen a couple of different formulas for maximum deflection,
      for example:<br>
      5/384 * W * L^3 / (E * I).<br>
      <br>
      I assume I can safely use E = 30,000,000 for steel.<br>
      <br>
      Let's say I'll allow 3" maximum deflection:<br>
      I = 0.113?<br>
      <br>
      However, I'm having trouble arriving at values for I (area moment
      of inertia).<br>
      <br>
      What would I be for a 2"x2" 1/8" wall square steel tube?<br>
      <table class="wikitable">
        <tbody>
          <tr>
            <td><img class="tex" alt="I_0 = \frac{bh^3}{12}"
                src="cid:part4.00080305.01020704@verizon.net"></td>
            <td><br>
            </td>
            <td><sup id="cite_ref-rect_3-0" class="reference"><a
                  moz-do-not-send="true"><span></span></a></sup><br>
            </td>
          </tr>
        </tbody>
      </table>
      <br>
      (2 * 2^3) / 12 = 1.33? <br>
      Or would that only be for a solid 2x2" bar?<br>
      <br>
      For a 1.5" square tube?<br>
      I = 0.42?<br>
      <br>
      (I also need guard railing for my porch. Seems a1.5"D&nbsp; tube with
      1/8" wall is I =&nbsp; 0.1276.<br>
      <img class="tex" alt="\pi \left(\frac{{r_2}+{r_1}}{2}\right)^3
        \left({r_2}-{r_1}\right)"
        src="cid:part5.05040401.07040202@verizon.net"><br>
      Does that seem right?)<br>
      <br>
      Also, I guess I'm not so much concerned with deflection as with
      point of permanent deformation.<br>
      How would I arrive at that number?<br>
      <br>
      My objective is the thin hangar doors and the lightest guard
      rails. And of course cheapest materials.<br>
      <br>
      Sanity check appreciated.<br>
      <br>
      Finn<br>
    </blockquote>
    <br>
  </body>
</html>

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