Re: Possibly interesting balsa test

From: valkyrie83 <gbower_at_gmail.com>
Date: Wed, 02 Aug 2006 19:44:11 -0000

Boron Fiber Modulus = 58,000,000 = 400 GPa
Average Balsa (12 lb/ft^3) Modulus ~ 6300 psi = 0.0434 GPa
The difference is a factor of order 10^4

So a .004" diameter boron fiber w/ a 0.001" diameter tungsten core has
the same effective area as a 0.33" x 0.33" piece of balsa, if both
were located the same distance from a neutral axis. So an I-beam with
1 boron fiber on each end of the web would have the same stiffness as
a similar thickness I-beam w/ 0.33" x 0.33" balsa on each end of the web.

Keep in mind this is stiffness, not strength...

Geoff



--- In Indoor_Construction_at_yahoogroups.com, "Steve Smith"
<ssmith9831_at_...> wrote:
>
> 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 - 12:45:40 CEST