- This topic has 47 replies, 7 voices, and was last updated 16 years, 2 months ago by
Adrian.
Adrian.-
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Adrian
After thinking about it some more, I realized that just compressing the fuel grains wouldn’t be a very representative test, considering the higher temperatures, internal pressure, and reduction in mass and volume of the fuel during the burn. It’d be pretty hard to test without flying it.
In my sims so far, I was somewhat surprised to see that this rocket doesn’t benefit from any additional mass beyond the minimum airframe + motor. The Cd vs. velocity for Rocksim’s nosecones are pretty hokey though, so when I get a chance I’ll sim it with my own Cd vs. velocity curve that I measured with my I flight to see the effect of mass. If we use the same nosecone shape, I think we should be able to get about the same overall Cd that I got on Violent Agreement; it will just be scaled up to a larger base diameter.
bryans
Disclaimer: From a person with almost zero MD experience, who has absolutely no formal training, and horrible at math.
Going backward in the thread a bit, it doesnt seem like this passes a sanity check…
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.
Assume that Chris’ data seems to match up somehat.
Also assume that the revised 76lb number is correct force that you have to deal with.
Now, i can buy the above if say, said material is a plane, and we have 6 inch wide fin, 6″ of which is hanging off of an infinite slab with perfect material creation and an infinitely strong joint. You place 76lb weight 3″ from the root. Assuming CF layup has zero deformation before its breaking point, you then have even load distribution over that 3 inch wide area; 60 mils still seems mighty thin, but lets give CF the benefit of the doubt.
Now do the exercise again, except where at the root, the layup starts curving away at 60 degrees over 1/2″. You have just created a lever with a very short load arm and very long effort arm. CF wont deform to spread out the load.
To put it a different way:
It seems like you are imparting the strength of the material in a planar configuration to the airframe-fin joint. Thats a moment arm, and it seems to me that the load distribution is all within a few millimeters of the joint.60 mils of CF in that configuration with a 6 inch joint is going to hold 76lb standing 3 inches from the root? Go look at something of similar construction and ask yourself what happens if you sit on it (two fins, 120 degrees apart, each bearing half of a 150-200lb load). Then think about how much less of a lever you have created if you have a four fin configuration. Not saying that four fins is a good idea, just trying to illustrate that moment arm.
Disclaimer: read disclaimer at top of post 🙂
Chris LaPanse
Keep in mind that Adrian was assuming that CF could withstand 62,500 pounds per square inch without trouble. That would require near perfect CF work – probably with a high pressure, high temperature cure and top grade epoxy. Realistically, most amateur carbon work could not stand up to anywhere near that (which is why that strength seems so incredible). I’m pretty sure the math works out though. Of course, you’d want them thicker than that for flutter reasons anyways.
Oh, and a fillet should reduce the stress. I don’t really follow your statement about the 60 degree curve – as long as the tube is fairly buckle-resistant (which it should be, given that the motor casing would be supporting it at that region), there shouldn’t be any problem with the strength at all. Oh, and wouldn’t the curve be 90 degrees, regardless of the fin count?
Adrian
Good question. I’m glad you asked. The quickie analysis I gave was for the flat part of the fin, where the two face sheets are held apart by 0.04″ of something, where that something could be G10 or CF or even balsa. It just has to be strong enough to keep the face sheets parallel with each other and keep them from buckling. That allows the face sheets to do all the work of resisting the bending. In both cases the main stress in the material is due to bending, but the flat part of the fin next to the root has has the worst of it, because its lever arm is only 0.04″ from center to center.
As you pointed out, the fin root is also in bending. And for a tip-to-tip layup, it resists bending the same way. Starting with the flat part of the fin closest to the tube, as you get closer to tube, the distance between the face sheets gets farther and farther apart, so the face sheets get more and more leverage to resist the bending moment. Now instead of just the 0.04″ core, you get into the fin root fillet material, and as long as it can keep the facesheets from buckling, the joint will get stronger as you get closer to the tube. On the tension side of the fillet, the fillet material has to keep the curved facesheet from straightening and pulling away from the fillet. On the compression side, the fillet material has to resist getting pushed inward by the facesheet. It is possible that if the fillet material is too wimpy, it will let the facesheets buckle or bend, and you can get a failure there. But the stress inward and outward is just a small fraction of what the facesheets see, so it’s usually not a problem.
In your example, assuming a relatively low friction surface, and a 160 lb load, then each fin would get 80 lbs at a 30 degree angle from pure bending. It would be equivalent to 40 lbs in the plane of the fin, and 70 lbs of bending. So you chose a good example, that’s pretty similar to the situation I was looking at. In that case, I would be confident that the fins would hold if I stood on the tube.
edward
I know all the discussion here has been on composite fincans but I wanted to give input about metal ones. I built one for my L3 – it was 2.5″ diameter with a 7″ root and 4″ span. I brazed the aluminum fins to an aluminum cylinder. It survived great except the part about landing in a ravine, one fin being wedged between two rocks while a 10′ long rocket fell and created a huge moment arm to bend the fin. It didn’t break, just bent.
I made v2 of my fincan tonight out of 1/16″ tube and 3/32 aluminum fins. I made two of them and destroyed one. The aluminum fin is the weak part because it becomes very malleable – but only at very high stresses. It took about 175 pounds at the very tip to start the fin bending and I had to get to 325 pounds to have it break the brazing off of the tube. I was impressed.
I have 3 hours into the fabrication of both fincans so they are very easy and not that time consuming to make.
Edward
Adrian
Cool. It sounds like brazing is a valuable skill to have. How much does the fincan weigh? What’s the 1/16″ tube for?
edward
You have to have something to braze the fins to – you cannot do it to the motor case. This slides right over the motor. The whole assembly (with cosmetic fillets) weighs just under 450 grams.
Edward
Adrian
Oh, I thought you meant diameter. Now I get it that you mean 1/16″ thickness.
Adrian
bryans
Chris LaPanse
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