Have I got my spar bending / shear diagrams right?

Discussion in 'Aircraft Design / Aerodynamics / New Technology' started by Foundationer, Jun 20, 2019.

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  1. Jun 20, 2019 #1

    Foundationer

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    HIYA. I think I've got a handle on how the bending / shear loads work out in a two piece wing with sailplane style pinned overlapping spars. But there's a good chance I'm wrong so if I am could someone let me know and try and point me in the right direction please!

    Structure overview.
    Cantilever wing, overlapping spars with two connecting pins, lift loads transferred into the fuselage via two 'lift pins' in each root rib. Sailplane style.

    Shear.
    Shear rises from the tip to the fuselage attach where it's reacted into the fuselage. There is no lift related shear between the lift pins.

    Bending moment
    B.M rises till the root, continues level till the first main pin then decreases to zero at the second main pin.

    For the other wing half it's just a mirror image. Is that about right or am I missing something?
     

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  2. Jun 20, 2019 #2

    wsimpso1

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    Not there yet.

    The shear at any point between tip and root is the total lift outboard of that point. Shear accumulates to the root, then levels off until you get to the right pin, where shear in the beam changes by the reaction at that pin. If your shear is positive right of the pin, it is negative between the pins. The shear is flat from right pin to left pin, where the other reaction comes off and shear goes to zero.

    Moment builds much as you show it, and goes from from a rising rate curve to a straight line at the same slope as the moment curve had at the root. This up sloped line continues until you get to the right pin, then becomes a straight line between the right pin and zero at the left pin.

    If you know the moment at the right pin, the reaction at the left pin is that moment divided by the distance between pins. The reaction at the right pin is the sum of the lift on the wing and the reaction on the left pin. Nominally, with lift being positive on the wing, the reaction on the left pin is up, the right pin reaction is down. Reactions are the hieght of the shear step at each pin...

    More later.

    Bill
     
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  3. Jun 20, 2019 #3

    mcrae0104

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    I always find a free body diagram helpful, showing only the forces on an individual spar (the air load and a reaction on each pin). It makes it easy to visualize which reaction is pointing which way.
     
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  4. Jun 20, 2019 #4

    wsimpso1

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    Here is a run example.

    Half span is 180", fuselage wall is at BL24", lift is elliptically distributed, and is carried into the spar only until BL24". Inboard of the fuselage wall, the lift is carried by the fuselage. Lift is reacted at two pins one at BL20" and one at BL-20". I used the rigorous mathematical relationships in columns B through E. Columns G and H show a simple numerical integration done at 6" intervals.

    The shear at any point is just the accumulated lift outboard of that spot;
    The bending moment is the first integral of shear;
    Lift only accumulates between BL24 and BL180 - we can know what shear is to BL24;
    The shear goes flat for BL24 to BL20;
    Moment has accumulated to BL24. Moment between BL24 and BL20 is the moment at BL24 plus the shear times the distance to BL24;
    Now we have to do some solving:
    Moment at left pin is ZERO:
    Reaction at left pin is moment at right pin divided by distance between the pins;
    Reaction at right pin is opposite sense from lift and left pin reaction, and is the sum of left pin reaction and lift outboard of right pin;
    The hieght of the shear change at BL-20 is the reaction at -20;
    The hieght of the shear change at BL20 is the reaction at 20.

    You can review the attached file with a worked example. The worked example also has a numerical integration from for comparison.

    Billski
     

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    Last edited: Jun 21, 2019
  5. Jun 24, 2019 #5

    Foundationer

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    Thanks for this! The blanks have been filled in and I think I get it now.
     
  6. Jun 24, 2019 #6

    Foundationer

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    The shear force between my pins is 44 times greater than outboard of them! Wowsers. It didn't occur to me there'd be such a step at that location. I seem to need 7 layers of +-45 degree 280gsm bi-ax carbon whereas two will do outboard of them. Tomorrow I will scratch my head and check I've not made a SNAFU.

    In that spreadsheet example you're passing the wing lift into the fuselage at the main pins aren't you Billski? My only difference is using separate pins to react that load leaving the pins only linking the spars and dealing with pure bending.
     
  7. Jun 25, 2019 #7

    Bank angle

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    Hi Foundationer,

    For the structure you describe, wh
     
  8. Jun 25, 2019 #8

    Bank angle

    Bank angle

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    Oops.

    Hi Foundationer,

    For the structure you describe, which is typical for modern composite glider construction, the Shear Force increases, as you have drawn, to a peak at the root rib. Let's call this positive for the sake of orientation.

    This lift force is reacted at the root rib into the fuselage via the lift pins so the S.F. diagram drops to zero and remains zero until we reach the first main pin. The maximum wing bending moment is also at the root rib and is reacted into the other wing spar stub by equal and opposite vertical loads on the main pins. This load is the B.M. divided by the distance between the main pins. So the S.F. diagram now goes "negative" for this load and remains constant until it reaches the other main pin where it comes back up to the axis.

    The B.M. diagram reaches a maximum at the root rib, as you've drawn, remains constant for the short distance too the first main pin then decreases linearly too zero at the second main pin.

    As you have observed, the shear loads in the spar between the main pins will be very high, particularly for single seat or tandem gliders where the distance between the main pins is relatively small.

    Take a look at the Slingsby Glass Fibre Repair Manual on the BGA website where it shows the wing of the Kestrel with the single spar stub having eleven layers of 92125 glass 280 gms/m2 on each side. ( pg 77/92.) From memory the web outboard of the root rib has two layers each side.

    The extra thickness, will of course, be useful in reacting the large bearing loads from the main pin bushes.

    I think beefing up this area would be prudent!
     
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  9. Jun 25, 2019 #9

    wsimpso1

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    Foundationeer,

    Send your spreadsheet to me on a PM, and I will look at your results for you...

    Billski
     
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  10. Jun 25, 2019 #10

    Foundationer

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    I've gone and downloaded that off the BGA website - will make some good bedtime reading.

    Thanks for further explanation re. the spars.
     
  11. Jun 26, 2019 #11

    Foundationer

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    So many SNAFUs! Turns out I can screw up my units even when I'm using SI: Newton Meters for bending and Kilograms rather than Newtons for shear so it's actually 4 times greater shear between the pins!

    Thanks for the offer Billski - will send over once I've tidied it up.
     
  12. Jun 27, 2019 #12

    Foundationer

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    This version has Shear in N and not Kg and everything (!).
     

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  13. Jun 27, 2019 #13

    wsimpso1

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    The shapes look roughly correct.

    Do your own status check;

    How much lift is one wing supposed to be supplying? That should be the amount of shear at the fuselage wall. Remember that the lift does not stop at the fuselage wall, lift continues over the fuselage and the fuselage is carrying it, so the fuselage lift is subtracted from the total, and we are looking at one half;

    Lessee. 750 pound airplane at 4 g is 3000 pounds. One side carries half of that, so we are at 1500 pounds. 90% of that is 1350 pounds. FOS is 2 for composites, so each wing at the pin will carry 2700 pounds or 12000 N - does your bird have a 750 pound gross?

    How much moment is supposed to be at the fuselage wall? Rough number for an elliptical lift distribution is 0.4 * (span-fuselage width)/2 * lift. Lift of an elliptical wing is centered about 40% out. For uniform distribution (it can not do this, but it is an upper bound), the leading coefficient is 0.5 not 0.4;

    Moment = 0.40 * 12000 N * (4.1-0.4)m = 17760 Nm. OK let's try out uniform distribution of lift. 0.5*12000*3.7 = 22000Nm. Hmmm, you are showing 25000 Nm.

    Something may not be correct.

    Billski
     
  14. Jun 27, 2019 #14

    Foundationer

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    Ok. Idiot check time! And finally I don't seem to be one...

    606lb airplane at 6g (Why not pick the bigger number?!) is 3636 lbs. So 1818 lbs / semi span. 90% of that is 1636 lbs. FOS 2 so 3,272lbs or 15,892N.

    .4 * (8.2-.7)/2* 15892 = 23,838N

    It's a rectangular wing so the Schrenk analysis pushes the centre of lift out beyond that ideal ellipse and 25,000N seems plausible I think?

    It also passes the lift = shear test but only when I remembered the wings have mass(!) and therefore 'inertia relief'.
     
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  15. Jun 27, 2019 #15

    David Lewis

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    To keep the load diagram on the conservative side, usually advisable to assume fuselage is not as efficient at producing lift as the wing.
     
  16. Jun 27, 2019 #16

    wsimpso1

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    Nobody is talking efficiency here, which pertains to drag created.

    We are talking lift produced by the wing. You have two choices - you either carry the lift distribution through the fuselage like the wing is continuous or make adjustments. Continuous wing is the way it is taught in the text books because we are supposed to minimize interactions between wing and fuselage. If you say that the fuselage is so disruptive of lift that significantly less lift is made than from the standard distribution, reduce the lift in the fuselage and increase it everywhere else.

    One thing not added in is that the horizontal tail is usually pulling down in flight, so that for the airplane to pull X g's, the wing has to lift X g's plus the downforce from the tail. So you need some idea of what your worse case tail download is too...

    Billski
     
  17. Jun 27, 2019 #17

    wsimpso1

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    Looks like you are in the ballpark. My offer still stands.

    Billski
     
  18. Jun 27, 2019 #18

    David Lewis

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    Right. That's the way it's taught, but that method underestimates the load on the wing.
     
  19. Jun 27, 2019 #19

    Lendo

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    David L./ Billski/ Foundationer,
    Sadraey suggests the fuselage provides roughly 45% of the lift of the estimated wing through the fuselage - that's what I've been using.
    George
     
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  20. Jun 29, 2019 #20

    wsimpso1

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    45% is mighty low, Sadraey's data must be for a fuselage with dreadful interactions with the wing...

    On the plus side, using that approximation, your stall speed will have more margin over requirements and the wing will be a little more conservatively designed too.

    Billski
     

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