Re: Possibly interesting balsa test

From: valkyrie83 <gbower_at_gmail.com>
Date: Tue, 01 Aug 2006 04:24:33 -0000

Hey guys,

Its been a while. School, work, and constantly moving between the two
have really kept me busy the past couple of years. I've still got lots
of indoor ideas kicking around that I need to try at some point. I'm
still following this forum among others. Anyways, I thought I might be
able to add some insight to this discussion...

For three and up equally spaced boron, the orientation doesn't matter.
Unless you don't want to space them evenly in order to tailor the
stiffness (which in my opinion is advantageous in most cases). For
example, with 4 boron it's probably more advantageous to put the boron
at 1,5,7,11 for an unbraced stick. That will give very good stiffness
in the plane of bending induced by the motor tension. It will also
give some stiffness in the perpendicular plane to prevent buckling,
but not as much.

The bending stiffness is proportional to the material stiffness
(young's modulus) and the area moment of inertia. For a rolled tube
fuselage, a majority of the bending stiffness comes from the boron
(assmuing it doesn't locally buckle away from the tube). You can
generally ignore the balsa in these calculations, since it has a few
orders of magnitudes lower young's modulus. The balsa tube is very
important when it comes to the torsional stiffness though, in which
case the boron does essentially nothing. The area moment of inertia
can be computed by taking the area of the boron cross section times
the distance from neutral axis squared. This is valid since the boron
is so compact and is away from the neutral axis. For a rectangle (say
a balsa spar) the area moment of inertia is 1/12*width*height^3.

Say the boron has an area of 1 unit squared, and the tube diameter is
10 units. Lets look at a couple examples. In each example the neutral
axis is perpendicular to the bending plane and passes through the
center of the tube.

Boron at 12 and 6 o'clock:
Relative bending stiffness in plane of rubber bending = 1 unit^2 *(5
unit)^2 + 1 unit^2 * (5 unit)^2 = 50 unit^4
in the other plane, the boron is on the neutral axis, so it
contributes essentially no bending stiffness.

Boron at 3,6,9,12:
Bending stiffness in the plane of rubber bending is unchanged since
the 3 and 9 o'clock boron are on its neutral axis. == 50 unit^4
Similarly for bending in the perpendicular axis the 3 and 9 provide 50
unit^4 of stiffness and the 6 and 12 provide none.

Now 3 boron spaced at 4,8,12:
Plane of rubber bending = 1 unit^2 * (5 unit)^2 + 1 unit^2*(5*sin(30
deg))^2 + 1 unit^2 *(5 unit*sin(30 deg))^2 = 25 + 25/4 + 25/4 = 37.5
unit^4
Perpendicular plane = 1 unit^2 * 0^2 + 1 unit* (5 unit*cos(30))^2 + 1
unit * (5 unit*cos(30))^2 = 37.5 unit^4

4 boron at 1,5,7,11
Plane of rubber bending = 4 * 1 unit^2 *(5 unit*cos(30))^2 = 75 unit^4
perpendicular plane = 4 * 1 unit^2 * (5 unit*sin(30))^2 = 25 unit^4

I hope this clears some stuff up, and doesn't confuse you all too
much. A diagram of some sort would probably helpful. It might be
worthwile to draw a circlular tube and the boron locations to try and
see how the calculations work, then they can be repeated for any boron
locations you want (for random locations, the neutral axis is
essentially the center of area of the boron fibers).

[Side note: This last point allows you to see why one boron fiber
doesn't work well at all. Its young's modulus is so high, that the
neutral axis moves very close to the fiber even when the effects of
the balsa tube are included. Then when computing the moment of inertia
the distance from the neutral axis to the fiber is very small, so you
get very little bending stiffness. Hence the huge jump in stiffness
from going from 1 boron to 2 or more.]

Geoff

Currently in Thousand Oaks, CA




--- In Indoor_Construction_at_yahoogroups.com, "John Kagan"
<john_kagan_at_...> wrote:
>
> --- In Indoor_Construction_at_yahoogroups.com, "akihirodanjo" <adanjo-
> 373_at_> wrote:
> > Does it also apply to 2 borons (for instance, 12&6 o'clock and 3&9
> > o'clock) ?
>
> I believe it requires 3 or more boron. It is pretty apparent that the
> theory doesn't work with 2. If Ray is on maybe he can add more
> clarity :)
>
Received on Mon Jul 31 2006 - 21:25:46 CEST