


Help!
Carbon wing joiner tube sizing
Hello guys,
Is there any folks in here able to tell me, calculate, carbon tube size i need for a wing joiner of 3m wingspan plane, TOW 15kg /2G max? I am about to prototype this plane: https://www.rcgroups.com/forums/show...2007762&page=8 Does a 3K carbon tube 20/18mm 1m length wing joiner + 4 screw M5 (sky hunter style) going through wing root and a carbon plate stuck into the fuselage would be enough? Does carbon 3K is a good choice, standard tube thickness is only 1mm...? would you recommend a carbon rod with lower section? confused: thanks for help you could bring, all inputs welcome!! STrato 

Last edited by stratocus; Feb 21, 2014 at 03:36 AM.





See attached. This is for a winched glider.
For your case, the winch load F would instead be weight*g_factor. You state 2 for the g_factor, which is way too low for an RC plane  a shallow dive and a blip of elevator will fold the wing. I'd go with a g_factor of at least 5 or even 10. 





Stratocus, I'm afraid that some of your translation isn't coming through very well either. It's hard to figure out what the details of your wing are so that we can suggest a joiner or possibly point out options in this thread and in your linked project thread.
You also typed "/2" which may have been meant to indicate 12G's. But a "/" does not equal a "1". If you did mean only 2G's then as pointed out above that won't even begin to get the job done. Generally a wing joiner does not need to be as long as 1 meter unless you need it to span out that far. Usually if the wing is designed to carry the load you only need the joiner to extend into the panel by about 25cm. But if you are making the wing so that the loads are spread out more and the joiner will not only join the wings together but also act as a sort of secondary spar then a 1meter long joiner does work. But you need to figure out if that is what you are doing or if you intend the wings to carry their own load and this is only a joiner. For a 3M model you will most certainly need something heavier than a 1mm wall thickness. That won't even come close to doing the job. A better option would be something up around a 2.5 to 3mm wall thickness for a model of this sort of size and weight. And as well it should have a blend of longitudinal fiber and a woven layer to bind it all together solidly. 



Quote:
I will take 5G with a safety margin of 2, 10G will be a better one ;) Could you confirm i will come close to the result if i change winch load by the actual weight of the model? Also, to make it clear, does sigma is a coefficient that i have to multiply with maximum G factor model will have to handle, in our case, i decided 10. Am i right? I came out with a pultruded carbon fiber rod of 16mm, sounds correct isn't it? Mark, thanks a 1000's for your quick and ultra efficient answer! BMatthews, thanks too, will take your input into consideration. 






Quote:





First a few notes to think about. Carbon is strong. But "strong" and "stiff" are two different things.
So, on a 3 meter model as I understand this to be I would suggest that the booms being made from single tubes are going to be too flexible for the model to hold proper pitch trim over the flight speed range. And if the tubes bend due to drag from the inverted V tail then your flight will become very exciting in a hurry. It would be better if you use two tubes of the 16/12 size stacked and glued into a figure "8" shape to increase the stiffness. The details of your wing are still not clear. Where are the joints between the panels? And I see you still show the 18/16mm tube as a spar. Or is that only as a slip tube? The tail booms will produce a lot of twisting load in the wings. You look like you will have a fairly strong and stiff main spar. But your rear spar is going to be very flexible compared to the main spar. The tail booms will produce a lot of twisting leverage as they lift up and down which will easily flex the rear spar. So your rear spar should be bigger and stiffer. Something like the same size tube or rod as the front spar. Tubes or rods make for poor spar design though. A better spar is made by using flat caps joined by vertical fill so you make a rectangular box for the spar. This puts more material out at the upper and lower extreme to withstand the bending loads to a much better degree. So you can do more with less. It is also a very bad idea to stop the spars at mid point, shift the loads through ribs to a separate outer spar and then only use one spar in the tips. The main spars should extend all the way from root to tip. Joiner rods are OK but you don't want to try to move the loads through ribs as you show. You are asking for trouble. It's a very bad weak point. Clearly you need to learn more about model aircraft structures. I suggest you search for plans you can download for various sailplanes. Study what they do for the wing spars and joiners and copy those methods. A good place to start is with the plans for Mark Drela's models that you can find at; http://www.charlesriverrc.org/newsite/index.html But there are lots of other plans to study as well. Google for "free model airplane plans download" to find some other web sites with plans. The only difference between these gliders and your twin boomer is that you will want to use two main spars to better withstand the twisting loads from the booms But the way the spars are designed and how the slip rods hold the wings together is still a method you should copy. 



Re Mark Drela's Sizing of Wing Joiners Formula:
Rod Radius=(loadfactor X winchload X span / 2 X pi X max material stress)0.333pwr I am trying to size a pin for a140" plane but I decided to run the numbers for a Sagitta 900 first just to see if I had the formula right and it says the beefy 5/16" (.3125) Sagitta pin is only good for 70lbs winch load and that just doesn't seem right! Rod Radius=(.8488 X 70Lbs X 100"span / 2 X 3.14 X 250,000)0.333pwr=.156 .156X2=.312" dia. (got exp calc from rapidtables.com) However, if I take "span" in the formula to mean the span of half a wing then the numbers seem to work out I get 150lbs winch load. Should I be using half span in this formula where it says span or is there just a really big safety margin built into the formula? 

Last edited by dbatey; Mar 05, 2014 at 03:45 PM.




get a tube which will easily fit then add a vertical web of CF  doubles the bending moment IF YOU DO IT RIGHT






The Saggita was popular back when real steel was the wing rod of choice. Namely springy music wire or hardened and spring tempered drill rod. Carbon was super exotic and not readily available at that time. On top of that the really serious ping launches where folks simply stand on the winch the whole way and rocket off the end of the launch was just starting to come into vogue. So once again the Saggita design wasn't rated for full pedal zoom launches. Given that the 70lb rating might not be out of place at all. I also wonder what the rating of the original spring steel would look like. So all in all don't be surprised if your calculations come out looking wrong.
Also with many or most calculations for the carbon rod you want to use the specs for the actual rod product and not for carbon fibers in general. The spec for the actual rod's yield point will include the factors based on resin flex and possible failure occurring in the resin before the carbon itself lets go. So I have to ask where you got your max stress value of 250K from? Is that for the actual rod as provided by some supplier? Or is that some value for carbon fiber in general? 



Hi Bruce,
Thanks for the input. In order to see if I was getting the formula right I ran the numbers on an actual Sagitta 900 (something I know about). That's a steel pin. I used Drela's number for hardened steel. If I had used Drela's number for music wire (which probably would have been more accurate) the max winch load would have turned out to be even less than 70Lbsat least according to the way I calculated the formula. Having flown this model in F3B in the 80s I know I could break 100Lb test line without permanently bending the pin. So, I am thinking I may be missing something in my interpretation/application of the formula. 

Last edited by dbatey; Mar 06, 2014 at 10:25 AM.




Hmmm.... I'd have to agree. Either that or the formula isn't correctly dealing with the situation. Is that a formula for simple curving? Because of the way we use the rods they are only seeing a point load bending force at the two joints. The portion within the wing is supported by the wing spar system and isn't free to curve evenly. That leaves us with the portion that extends across the fuselage to do all the bending. Or if that portion is bound with a tube set in glue and bulkheads we have two small areas of bending occurring right at the join lines of the wing panels.
This gets even worse, or better, with something like the original joint on a Bird of Time where the two panels slip together than get rubber banded to the fuselage. Now all the bending forces are limited to that small area of the rod. What that does to the load on the rod and how it affects the material's springing modulus is the question. By inference from your experience of being able to snap 100lb test lines at will it suggests that the equation is very pessimistic. But then let's recall that the condition of the 100lb test line degraded quickly in the extreme use of winch duty as well. I recall my club at the time had to swap the line roughly twice a year when it started breaking all too frequently. So again that 100 lb line might have been snapping at around 70lbs depending on when you flew during the life cycle of the line. 



Dbatey, i will try to give you my point of view.
I believe this formula, is not applicable the way you used it. You try to determine by reciprocity the Max winch load. This is a more complex calculation... To many influent parameters as per given by BMathews and many more actually... In my opinion even if it's correct from a mathematical point view, this formula has to be used as an "empirical calculation" of wing joiner size. Actually you should take this formula the other way around, the way it has been planed to be used. EG: I did your calculation again with a 100lbs inch load, iso the 70 lbs you used, it gives only0.9 mm more on the diameter, according to the calculation you would need a joiner of 8.9 mm diameter. Not such a big difference on the diameter... So what it says? It only says that if you have a 8mm on your sagitta and you use a 70lbs tow line then you are safe, and you proof it by your experience... That's only what this formula says and that's only thing we need isn't it? Up to you knowing "margin" error to trim +/your joiner size. Sorry for my english, hope you will get what I mean. 

Last edited by stratocus; Mar 06, 2014 at 12:33 PM.

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