Folding system loads

Discussion in 'Multihulls' started by tamas, Nov 11, 2013.

  1. hump101
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    hump101 Senior Member

    See my response to Tamas, when calculating stresses in the strut, you can't use the strut total cross-section, you need to use the cross section remaining where the bar or bolt passes through, with a stress concentration factor of about 1.5 to account for the effect of the hole (the stress will not be evenly distributed across the remaining section).

    In respect of aluminium alloy, there are grades with yields more than twice those you've quoted, such as 6061-T6, which has a minimum yield of 245MPa, and more typically 275MPa. The 110MPa you quote is for 6061-T4.
     
  2. bruceb
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    bruceb Senior Member

    holes

    Does that mean you assume the "hole" to be 1.5 its actual diameter?
    B
     
  3. hump101
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    hump101 Senior Member

    You multiply the calculated stress by 1.5, which is equivalent to dividing the cross-sectional area of the plate either side of the hole by 1.5, but is not necessarily the same as multiplying the hole diameter by 1.5.

    Using Tamas' example, with his 60x12 plates, each with a 25mm+tolerance hole (say 26), then the remaining CSA is (60-26) x 12 = 408mm^2, so the max stress will be (max force in N) x 1.5 / (408).

    Note that the value of 1.5 is typical, but depends on your particular geometry, quality of the finish in the hole, tolerance of the hole around the pin, etc., so may not be valid for your application. We would normally do an FE analysis for a given geometry to determine a suitable SCF.
     
  4. bruceb
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    bruceb Senior Member

    Holes

    Thanks! So, on my lower attachment point the 304 stainless is 1/2"x 2 1/2" with 3/4" holes. .5 x 2.5 - .75 x .5 = .875 Divided by 1.5 = 0.58" cross section for calculating purposes? (about:))
    B
     
  5. hump101
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    hump101 Senior Member

    Yes, though being stickler for units it should be 0.58 "^2, not just ".

    As noted earlier, you should check the validity of the 1.5 SCF for your application. There is plenty of data on line to allow you to select a suitable SCF based on your specific geometry/hole tolerance, etc.
     
  6. ChineTri
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    ChineTri Junior Member

    This probably is the reason for the tapering at the end of the f-22 struts, which unfortunately I didn't measure. I would guess that yield at the location of the bolt is about the same as the rest of the strut.
    Jan Peter
     

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  7. hump101
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    hump101 Senior Member

    Yes, certainly. An optimised design will be tapered.

    However, these struts can also take compressive loads occasionally, so whilst you can taper to some extent, you must not compromise the compressive buckling strength of the section. I very much doubt that this is design-critical, due to the short span and the relatively small compressive load, but should be checked.
     
  8. ChineTri
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    ChineTri Junior Member

    Thanks catsketcher for your answer. Your post inspired me to apply this to some drawings i'm working on. The design has a float displacement of 1300 kg resulting in an upward force of 12753 N when completely submerged. In this case the trimaran will be heeled at 20 degrees. I've drawn this situation with the sizes of beam and strut to scale, and calculated the resulting forces, see attachment. I used the total displacement of the float for calculation of forces in 1 beam, where in reality 2 beams will distribute the load.
    I wonder if I made a mistake? For example you had a right angle in the triangle, while in my sketch there isn't...

    {edit}
    Re-reading catsketchers' PDF and assuming the beam would have to withstand a compression force of 200000 N. Combining this with some compression strength numbers I found here.
    For CF it gives a compression strength Xc of 570 MPa.
    Am I correct in thinking that a beam made of 50/50 carbon fiber and resin, with a cross sectional area of (200000/570) = 351 mm^2 should be able to withstand the compression force?
    {/edit}
    Jan Peter
     

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  9. catsketcher
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    catsketcher Senior Member

    Gday Jan Peter

    The right angle triangle is not necessary, in the old days engineers would often draw their forces on engineering graph paper, you are just making a force polygon and can still solve it either graphically or by using trig.

    Can you post your calculations of the moment? Your question of compression loading becomes interesting and in my mind quite tricky. The beam is not a simple compression strut as it is actually a cantilever beyond the strut and there is a grey area around the strut where it is undergoing bending and compression.

    My answer is always reverse engineer something that works. The easiest beam to reverse engineer would be Farriers Tramp design with its alloy beams. They are often heavier than designed and have a reputation for being solid. Work out the loads from the boat and then size the strut and beam. That would give you very solid data to work from.

    Compression loading is something that I am interested in and it involves an equation derived by Euler.
    P = (Pi^2.E(Youngs modulus).I(moment of area))/Length ^2.Fixity(probably more than 1 here) You have to be careful when designing compression struts that you don't make a section with walls that are too thin. Sandwich structures often have more trouble than you would think in compression which is why carbon masts rarely have cores.

    http://en.wikipedia.org/wiki/Buckling

    Are you designing this or are the plans hard to follow?

    cheers

    Phil
     
  10. ChineTri
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    ChineTri Junior Member

    Hi Phil,
    I'm not sure which moment you mean? Perhaps the moment of the windforce trying to push the boat over? Or more probably the bending or torsional moment on the beams?
    Assuming the second option: I haven't calculated them (did some trials with approximate values) because the folding geometry is not final. Currently checking if the float in folded position fits beneath the main hull while folded width is no more than 2.5 m, and the float is not too deeply submerged. If it doesn't fit I'll have to rework the float and perhaps main hull. If it does fit I'll start designing length and overall shape of struts and beams.
    As for the sectional area of the beams I was thinking about a trapezoid shape, with the wide side on top. The sides would then point inward and down to the smaller bottom side, giving some downward spray deflection, and a wide walking area on top. But I haven't yet discovered how to calculate the moment of inertia of such a shape. Would it be sound to calculate the I of the trapezoid outside of the beam, and subtract the I of the trapezoid hollow inside of the beam? Perhaps a trapezoid shape is for some reason not a good choice?
    That is basically what I do. My boat will be a multi-chine trimaran based on the overall geometry of the F-22. (I hope multi-chine will shorten building time). I'm using what I know of the F-22 dimensions as a guideline in my design, more or less a way to double check outcomes...

    Unfortunately I live in a part of the world where very little Farriers are found, especially the older types.

    I guess I'll have use the compression loading calculation on the intended trapezoid shape when designing the beams. I have no idea what this fixity factor is?

    I'm designing, have a look at my website:https://sites.google.com/a/tammingas.nl/trimaran-design/design-considerations

    JP
     
  11. bruceb
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    bruceb Senior Member

    Buc 33 fold

    I have a basic drawing of my first full size folding prototype. The measurements are not exact- I used two different tapes for metric and inches, and they vary some. The geometry works well and should retract the float into a usable position for docking at around 14' beam, and 11.5' for transport. The beam is a fiberglass "top hat" section, with the bottom open. I assume the B-33 to be less than 6500 lbs all up, and float displacement around 6000 lbs.
    B
     

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  12. bruceb
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    bruceb Senior Member

    loading?

    H101, if I am understanding the loads correctly in CT's diagram, the load on the bottom pivot is about 6 times the float displacement? I don't trust my metric conversions very well, but that seems to be in the "expected" range based on other boats.
    B
     
  13. hump101
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    hump101 Senior Member

    I've had a quick look at CT's diagram and run his numbers. I get close to his numbers, but not identical (see attached spreadsheet to see the workings). In his case the horizontal offset (1.325m) is 6.02 times the vertical offset (0.22m), and hence the forces at the beam mounts on the centre hull will be about 6 times the force on the end of the beam. The effect of the angle of the beam elements and the heel angle change the relative compressive and tensile components in the inner beam elements, as per my spreadsheet.

    Need to remember that the vertical force on the float does not seem to include any dynamic element, so this would need to be reviewed before the values are used for design, but this gives the idea of the mechanics involved.
     

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  14. bruceb
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    bruceb Senior Member

    Numbers

    Thanks, that is a little easier to understand.
    B
     

  15. hump101
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    hump101 Senior Member

    No problem. I'll do a sheet for your geometry. Do you have dimensions for the first figure, in particular the offsets of the plates along the pins, so we can calculate the bending in the pins?
     
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