It might be more instructive to transform the cross sectional area of
the boron to its balsa equivalent.
Multiply the area of the boron fiber by modular ratio of the two
materials (E boron/E balsa) to get the transformed area. It will
become readily apparent as to the relative effectiveness of the two
materials and where best to place the boron.
What is the Elastic Modulus for boron?
Steve Smith
Castle Rock, CO
--- In Indoor_Construction_at_yahoogroups.com, "valkyrie83" <gbower@...>
wrote:
>
> 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 Wed Aug 02 2006 - 05:44:29 CEST