## Monday, February 25, 2013

### Checking My Math

It has been almost a year since I made plans to rebuild the big telescope into a half-Suerrier truss using carbon fiber composite tubes.  In the meantime, several unexpected expenses came up, including expanding ScopeRoller's working space, a snow plow so that UPS can deliver materials to ScopeRoller, etc. In addition to the costs, I was also seriously busy teaching classes and trying to save the American Republic, so even had I felt like buying the materials, I would not have had the time.

The time is now coming available, because I am only teaching one class this semester.  Because carbon fiber composite tubing is pretty darn expensive, I want to make sure that I have the math right.  If I order too small a set of tubes, the truss isn't going to be stiff enough; if the tubes are thicker than they need to be, it will be slightly heavier and much more expensive than it needs to be.

I have checked my math several times, but I still do not believe the results that I have getting.  Or perhaps carbon fiber composite really is magic!

To calculate deflection caused by a load on a hollow tube, I use the cantilever beam, end load formula:

∆ = PL^3/3EI

where P = force, L = length, E = Young's modulus, and I =area moment of inertia

In this case, the force of the upper cage assembly is 10 pounds, or 4.54 kg.  The length of each side of the truss is 62 inches, or 1.575 meters.  The area moment of inertia for hollowing tubing is calculated by:

Ix = π (do4 - di4) / 64         (5)
where
do = cylinder outside diameter
di = cylinder inside diameter

For a piece of tubing that 0.25" ID (.00635 meters) and .378" OD (.01 meters), that gives an area moment of inertia of .256.

There seems to be considerable difference of opinion about the Young's modulus of carbon fiber composite, but I am using 138 gigapascals, because a number of vendors make this claim.

Plugging this in says that a piece of carbon fiber composite tube that is incredibly small will have 0.00000 meters of deflection with a 4.54 kg weight at the end.  For anyone who experience with long pieces of small carbon fiber composite tubing -- does this seem even slightly plausible?  It seems outside the realm of my experience.

UPDATE: I my formula in the spreadsheet for area moment of inertia wrong, and yes, newtons, not kg.  This means that the deflection from a single tube is enormous. The truss calculations I did a while back indicate that I don't need anything this stiff (cumulative benefits of how this stuff works).  Does it seem plausible that a truss consisting of six tubes (three triangles) each of 0.553" ID, .0625" OD could be stiff enough to hold ten pounds of weight with deflection measured in thousandths of an inch?

The temptation is strong to build a very small scale model of this, perhaps using carbon fiber tubing, and see how well the math from TDT4WIN mentioned here actually models reality.

Anthony said...

P = 4.54 kg * 9.81 ms^-2 = 44.54 N
L^3 = 3.91 m^3
I = pi(10^-8 - 1.63e-9)/64 = 4.11e-10 m^4
E 138e+9 Nm^-2

d = 3.07m

(Checking in SMath Studio, a MathCad clone, I get 3.73m, because small variations in do and di make a *big* difference.)

Clayton Cramer said...

Thanks -- I had a divide instead of a minus in the calculation of moment of inertia, along with using kg instead of newtons.

This makes even a pretty large tube insufficient.

Rorschach said...

yeah, that did not sound right at all to me either. think fishing rod... ever see a fishing rod with a 10 pound fish on the end? how much does it deflect?

Anthony said...

Trusses are different. If the connections are pure hinges (free to rotate), then there is no bending on the individual elements, and all the forces are carried in compression or tension. For a simple rectangular truss 3 times as long as it is high, with a weight pulling in the vertical direction, the maximum stress will be 3x the weight, carried in the diagonal.

For your 10-pound weight, I get compression in the diagonal described above of .01 mm; the total deflection of the frame will be on the same order, provided that the ends of the rods are *completely* free to rotate, and carry *no* bending.

You may also be able to calculate the overall 'I' of the truss by calculating the 'I' at a cross section, remembering that displacing your circle from the neutral axis adds an Ar^2 to each member, where A is the cross-sectional area, and r is the distance from the member to the neutral axis.

Anthony said...

Checking a bit more: if the truss is approximated by a hexagon of rods, 1 meter in diameter, and 3 meters total length, I = 3e-5 m^4, and total deflection is 7e-5 m. I can send you a calculation sheet if you'd like.

Clayton Cramer said...

Anthony: is this calculation sheet something that SMath Studio uses? Perhaps I should look at getting this program for my calculations.