2" BASIC ULTRALIGHT WING CONSTRUTION ??'S

Discussion in 'Aircraft Design / Aerodynamics / New Technology' started by oldcrow, Jun 1, 2019.

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  1. Jun 1, 2019 #1

    oldcrow

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    Does any know or tried to figure out the wing load on a typical 2" tube spar ladder construction wing, using 2x.125 t6 tubing? I know most of the ultralights use 2x.058 but I want to build a low wing with the struts attaching above the fuse. (kind of like the cloud dancer)
    Or it would be ideal if someone could post the blueprints to build the cloud dancer wing.
    thanks all. Ken
     
  2. Jun 1, 2019 #2

    BJC

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    The Cloud Dancer wing failed in flight, so don’t copy it exactly.


    BJC
     
  3. Jun 2, 2019 #3

    proppastie

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    Might want to start with a proven set of plans before you jump into design and stress analysis.
     
  4. Jun 2, 2019 #4

    Winchester

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    The Cloud Dancer was an ultralight Motorglider with a span of 40 feet. A 30 foot wing will be stronger with the same wall thickness. Are you absolutely set on a low wing UL? They are usually harder to maintain and less forgiving on rough terrain. I would suggest forgoing .125 for a more conventional size and using an internal sleeve at the same thickness- Wing Root to several inches after the strut connection point.
     
  5. Jun 2, 2019 #5

    BBerson

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    Cloud dancer doesn't have struts.
     
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  6. Jun 2, 2019 #6

    Winchester

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    Last edited: Jun 2, 2019
  7. Jun 2, 2019 #7

    BBerson

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    The spar loads are same for low wing, just reversed. The strut loads are same also if same angle and length, but reversed to compression.
    But since a strut will buckle with minimal compression loads, all this must absolutely be understood. I don't think any are available to copy. So must engineer it from the start.
     
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  8. Jun 2, 2019 #8

    radfordc

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    It failed due to delamination of the aluminum sheet from the foam structure. Probably not applicable to this discussion.
     
  9. Jun 2, 2019 #9

    BJC

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    Yes, it did. From the OP
    Kind of an incongruous comparison, so I suggested “don’t copy it exactly.”


    BJC
     
  10. Jun 2, 2019 #10

    proppastie

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    If you do decide to become an aircraft designer please load test your finished product....much better to be surprised and disappointed on the ground than in the air.

    FAA Basic Glider Criteria has examples of load tests.. my proppastie site or google, free from FAA
    your tax dollars at work.
     
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  11. Jun 2, 2019 #11

    oldcrow

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    thank you all for the input so far. attached is a pic of where I am at so far, so you can get an idea of where I am going. The pilot will sit just behind the wing leading edge and the motor will sit just before the wing trailing edge. with the struts joining behind the pilots head centered. ( length 17.5 span 35 ) an open cockpit motor glider part 103 i'll post more pics as I get them drawn. SEA-1.jpg
     
  12. Jun 3, 2019 #12

    BoKu

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    First things first--Warning: I am not an engineer. This is not engineering advice. I sometimes do engineering under supervision. I always seek advice from an engineer or do actual static testing before throwing stuff into the air. What follows is just an example of how a non-engineer might approach things like this using common sense.

    What I generally do when starting a conceptual assessment is do a quickie estimate of the stresses at a couple of critical locations to make sure I'm at least within an order of magnitude of safe. After that I'll start breaking out some formal tools and refining the numbers.

    For the case at hand, I'd start with a guess that the critical issue is bending reaction in the wing main spar at the side of body (for a cantilever wing) or at the site of the strut connection (for an externally-braced wing).

    Step one is to make a reasonably safe estimate of the non-lifting mass, all the stuff that the wings have to lift and support. Usually this means the fuselage and tail and pilot and payload--basically the maximum gross weight minus the mass of the wings themselves. In the case at hand, I'd be inclined to take the maximum ultralight empty weight of 254 lbs, add about 220 lbs for pilot and fuel, which gets us to 474 lbs, and then assume that the wings weigh nothing. Which is probably conservative by 30 or 50 lbs, but a modest amount of conservatism is good.

    So, we know that the wings have to carry about 474 lbs at 1g. Step 2 is to account for vertical acceleration to accommodate maneuvering, gusts, etc. For a carefully-operated ultralight, let's say that it's 3.8g limit load with a 150% safety factor for 5.7g ultimate. So the wings should be able to carry 474 * 5.7 = 2702 lbs, at least for a few seconds. And it stands to reason that each wing has to be able to carry half that, or 1351 lbs.

    Step 3 is to find the bending moment due to the lift at the place where the bending moment is the greatest. We already know that each wing has to be able to carry 1351 lbs, now we need to know the arm at which it is carried. For a cantilever wing, the usual thing is to assume that the lift distribution is somewhat elliptical, and that the center of lift is at about 42% semispan. For the 35-foot wing contemplated here, that's 17.5 * 0.42 = 7.35 feet outboard of the root. So at the root our cantilever wing has to be able to react a bending moment of 1351 lbs * 7.35 feet = 9930 ft-lbs. In the interest of making later calculations easier, let's convert that into inch-lbs by multiplying by 12 (the number of inches per foot): 119158 in-lbs. Let's round it up to 120,000 in-lbs for convenience sake.

    Okay, that's for a cantilever wing like the original Cloud Dancer. For a strut-braced wing, the maximum bending moment is probably at the site of the strut attachment. For this example, let's assume that the struts attach 6 feet outboard of the wing root, so there's 17.5 - 6 = 11.5 feet of wing outboard of the strut. The proper thing to do is some calculus to evaluate some integrals regarding the areas and centroids of portions of ellipses to figure out the bending moment at a point 6 feet out from the root. The next best thing is to make an Excel spreadsheet that does that for you. For a quickie estimate, I would just assume that the outboard 11.5 feet of a 17.5 foot wing creates 11.5 / 17.5 = 66% of the lift, or 892 lbs. For the arm, I'd assume again that the center of the area outboard of the strut is 42% of the way between there and the tip, or 11.5 * 0.42 = 4.83 feet. So our rough estimate for the bending moment at the strut would be something like 892 * 4.83 = 4308 ft-lbs or 51700 in-lbs (let's call it 52,000 in-lbs).

    So now we have some ballpark guesstimates of the bending moment in the wing at the critical location for a couple of cases. Are these guesstimates accurate enough to start cutting metal? No way! Are they good enough to guide some conceptual thinking about what is possible and practical? Maybe. Everything you read here might be wrong. Govern yourself accordingly.

    The next thing we need to do is figure out whether our proposed wing spar designs (aluminum tubes) have the capacity of reacting bending moments of 120,000 in-lbs for the cantilever case or 52,000 in-lbs for our 6-foot struts. In order to do that, we need to know the yield and ultimate tensile strengths of the aluminum and also the moment of inertia of the cross-section of the tube. The former we get from standard tables for the standard alloys, and the latter we get from relatively simple calculations shown in standard guides like Machinery's Handbook. However, since I don't have those resources handy, I just look them up in a cheat sheet that I drew up years ago using Excel:

    For 2" OD, 0.058" wall tubing:
    6061-T6 yield: 6670 in-lbs
    6061-T6 ultimate: 7500 in-lbs

    For 2" OD, 0.125" wall round tubing:
    6061-T6 yield: 12,300 in-lbs
    6061-T6 ultimate 14,600 in-lbs

    Even if we say that each of the two spars in each wing reacts half the bending load (It's really more like 60% front, 40% rear), 2" tubing with 0.058" wall or even 0.125" wall doesn't get us anywhere near the bending reaction we need for safe flight, even with the 6-foot wing struts.

    So, what to do? The usual compromises are some combination of:

    * Reduce the wingspan. A wingspan of 35' is a lot for a tube-ladder wing structure.

    * Attach the wing struts further out. This is probably only realistic if you make it high-wing and put the struts underneath so they're in tension during normal flight. The magic location for puddle-jumping airplanes seems to be around 60% semispan. As the strut attachments move outboard, the bending moment at the attachment point decreases, and the bending moment inboard of it increases. It gets into an area where you need to plot the moment on graph paper to understand it properly. Excel is great for stuff like that.

    * Use larger spar tubes. Most tube-ladder wings use 2-3/8" to 3" OD aluminum tubes. The strength of a bending beam scales with the cube of its depth, a little bit extra height gives you a lot more strength and stiffness. According to my cheat sheet, 2.5" OD, .058" wall 6061-T6 yields around 10,600 in-lbs; combined with other compromises, that might be enough.

    * Use internal sleeves in the tubes near the areas of maximum bending reaction.

    * Change to an I-beam spar located at the chordwise station of maximum depth. I-beams are much more effective application of material than round tubes in terms of strength and stiffness per unit mass.

    * Reduce the mass that the wings have to support.

    * Reduce the flight envelope and limit load--be very careful about this.

    --Bob K.
     
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  13. Jun 4, 2019 #13

    BoKu

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  14. Jun 4, 2019 #14

    proppastie

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    As to testing....there should be no permanent deformation (kinks, wrinkles,tears,) at limit load. I believe one should static test the structure to design limit load. I believe one should design the structure in aluminum or steel to 3x limit load. (limit x 1.5 for ultimate load x 1.5 FOS factor of safety) There are gust limit loads which are not the same as "G" limit loads. I believe the minimum limit loads should be "utility class" as one might find in a certified aircraft. FAA Glider Criteria has charts to help you with gust loading.
     
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  15. Jun 4, 2019 #15

    dino

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    Isn't 3X limit a bit conservative? I thought 1.5 FOS for aluminum structures was often used because it's approximately the relation of yield to rupture. In practice you load the structure to limit and it should return to its original shape. Increasing the load beyond yield will result in the structure taking a permanent set up to the point of failure at ultimate load hopefully allowing a safe but sorry landing.
     
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  16. Jun 4, 2019 #16

    oldcrow

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    THANK YOU BOB that was a lot of good information. As you can tell I am trying to build as simple as possible (not cheap just simple) so it looks like I will have to redesign the wing to an internal spar design. or maybe just scrub this and do an AirBike style fuse. and thanks for the link to Sandlins design methods. SEA-1.jpg
     
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  17. Jun 5, 2019 #17

    proppastie

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    There should be no catastrophic failure at Ultimate loading....some deformation is allowed. Were you an aircraft company you would test to destruction and that should not happen below the Ultimate load you designed your aircraft for. So unless your are a perfect designer and can design exactly to Ultimate load I believe the sizing of the material should be 1.5 FOS from Ultimate load.....and yes the the relation between yield to rupture for aluminum is aprox. 1.5..... If we go backwards with these numbers one would want to have a limit load of 1/3 the design sizing of the structure.

    Another way to look at it is..... material yield is a failure. The design must have a limit load less than the yield of the material.
     
  18. Jun 5, 2019 #18

    BoKu

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    Yes, I'd go with that. But I think that's covered by designing to limit load with a 1.5x FOS. Designing to ultimate with a 1.5x FOS results in a heavier structure with poorer performance and higher stall speed. The combination of those factors is likely result in such an increase operational risk factors as to more than compensate for the risk reduction of the increased structural margins. And I think that fulfills the technical definition of irony.
     
  19. Jun 5, 2019 #19

    proppastie

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    When you run the numbers it really is not that much weight.....I think it was both the AirBus and New Boeing that both failed ultimate load test....bummer if you are a little off at limit x 1.5 (or 2 for plastic)

    Edit :....ran some quick and dirty I-Beam spar numbers for a 12" thk., 36 ft wing rectangular platform. The difference in weight between the 8 g spar and 12 g spar was 8 lb. 30 lb vs 22 lb.

    400 lb gross, UL....1" caps.

    Percentage wise it is large, but 8 lb is not much......

    If your new design is the next RV hit....you will want the extra safety factor....it will cut down on the lawsuits.
     
    Last edited: Jun 5, 2019
  20. Jun 5, 2019 #20

    BJC

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    There is more difference between an 8 g airplane and a 12 g airplane than just the spar.

    The ideal safety factor is just above 1.0.


    BJC
     

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