- This topic has 47 replies, 7 voices, and was last updated 16 years, 2 months ago by
Adrian.
Adrian.-
Posts
-
Bruce R. Schaefer
Just let me know when you guys get around to flying it… 😈
Chris LaPanse
That’s why I mentioned starting with relatively thin plate, and then adding quite a few layers of T2T. That in essence makes the entire fin can one piece of CF, and that should allow for these kinds of loads.
Warren B. Musselman
I’m with Chris on this – dadoed slots in the airframe with relatively thin, low mass cores properly glued in, then 8 maybe 10 or even more layers of tip to tip, vacuum bagged and with a proper cure process.
W
Chris LaPanse
8 or 10 layers sounds like a bit much – you only need to build it up to roughly 1/8″ plate. Anyone know how many layers there are in 1/8″ plate CF?
Chris LaPanse
On a related note – it’s amazing how fast you forget things once you’re out of school for the summer. Things like how to properly read the table in the back of the aerodynamics textbook. (apparently, 1.9270 and 1.9270*10^-3 are not the same number…)
All my numbers above are off by several orders of magnitude. The new value seems more reasonable (something seemed wrong when performing that calculation – now I know why). The side force on the fins is now just 72.6 pounds of force. Therefore, flutter becomes the main concern (and I still believe that the flutter would not be a problem with 0.1-0.125 inch CF). I also made an error in the required fin thickness to support the amount of force listed above, though it’s irrelevant now that I’ve realized just how far off that value was.
😳
Adrian
Well, after some calculations…
Ah, the cool refreshing sound of real engineering.
Very interesting results. 150 ft/sec of instant shear is a pretty conservative worst case, but it shows the margin we have here. What did you assume for the distribution of the fiber direction through the fin thickness? All uni, or just the outer layers? We would need some layers in the other direction to give it some resistance to handling and landing dings.
86,000 Kpsi dynamic pressure? Yikes. I think that the leading edges of the fins may not hold up to stagnation pressure that high. I have heard of other people getting some erosion /delamination of the leading edges for their high speed flights. Wrappng the leading edges with aluminum may be worthwhile to reduce the risk of delamination.
Next you should figure out how much bending moment gets generated in the tube by 76000 lbs of fin force. It would probably snap most tubes like a toothpick. Honestly it seems too high to be reasonable, but that probably goes back to the assumption of 2+ degrees angle of attack at max Q. Here again uni carbon in the tube will hel bail us out. But first, figure out the lateral forces on the part of the tube that is unsupported by the motor. start with calculating the angular acceleration. I think you’ll find here that nose weight will make things somewhat worse by adding more pitch inertia. I would bet that the straight compressive load of the load from drag and inertia will be small compared to the bending load, but it would be interesting to see how it comes out.
Adrian
Ok, 76 lbs is a lot more reasonable than stacking 10 SUVs on the end of the tube. I think i’ll use rocksim to double check. It provides restoring torque as an optionin the graphs. I’ll see if it let’s me give it a super long tower that extends to max q.
What do your new calcs say about the necessary thickness?
Chris LaPanse
New calculations:
(these ones are right, I promise)
New dynamic pressure: 86.4 PSI
New lifting force on a single fin: 72.6lbs
New thickness required (ignoring flutter): 0.019 inches of top, aerospace grade CF.
New peak stress within 1/8″ thick fin in this scenario: 3500PSI (which also happens to be well within the maximum published specs by West System, Aeropoxy, and Pro-Set for bond strength, so bonding shouldn’t be a problem). I’m now inclined to think that wind shear isn’t as big of an issue as flutter.My problem above was because my aerodynamics textbook uses a standard atmosphere table that has a multiplier at the top of each column. I looked at it, and saw that the density at 7000 feet was 1.9 slugs/ft^3 (ahh, the joy of english units 🙄 ). I missed the fact that every number in that column should be multiplied by 10^-3 to be accurate.
Chris LaPanse
Well, I just ran it through rocksim, and rocksim seems to think the total force would peak at ~220lbf (that’s not too far from my value of ~72lbs on a single fin, considering all of the assumptions that I had to make to get there). It’s kind of a shame that you can’t just graph or predict dynamic pressure in rocksim though.
Adrian
Rocksim indeed lets you use a launch tower that is 5000 feet high, and enter 150 ft/sec of cross-wind. When I ran it, it predicted that such a situation would result in just under 300 N*m of torque when it left the tower at Mach 3 🙄 , given my fins with about 1.6 calibers of stability margin. With 3 fins, half of the torque would come from the fin perpendicular to line of action, so figure 150 N*m of torque, applied by one fin 16 inches (0.4m) behind the CG of the rocket. That makes about 375 lbs of side pressure on one fin.
If I ignore the center layers of the layup, and assume that the middle 6″ of the fin is taking all the stress, then with a 0.01″ thick layer of uni carbon on each side of the core, and 0.04″ between the midpoints of the layers, (for a total thickness of 0.050″) then I get 62500 psi stress at the root of the fins. The published strength data on the aircraft spruce & specialty website for a mostly-uni layer comes out to about 90 ksi. In other words, you could put a 430 lb weight 3″ from the root a 6″ long fin, and the fin would only have to be 60 mils thick, with 10 mil outer layers of uni-carbon, to withstand that force without breaking.
When I laid up my carbon plate for my fins, 10 plies resulted in about 36 mils thickness, so you could use just about anything to make a 0.035″ core, put on 3 layers of unidirectional tip-to-tip carbon, and you can be assured that wind shear won’t snap the fins off at the root. With a tip-to-tip layup, I think the only way the joint could be weaker than the fin itself might be with a debonding of the T2T layer from the fillet material. Even then though, I think the fin just above the fillet would be the weakest point.
For flutter, you’d need to calculate the resonant frequency for the bending/twisting mode shape you’d get with the flutter, and show that it’s faster than the flutter frequency that can be generated by airflow at the given speed. Definitely more than a back-of-the-envelope calculation.
Bruce R. Schaefer
Chris LaPanse
Warren B. Musselman
Adrian- You must be logged in to reply to this topic.