Composite pocket boom: sharing calculations and asking for help

Discussion in 'Sailboats' started by RMA, Oct 19, 2022.

  1. RMA
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    RMA Junior Member

    For various reasons, (weight reduction, better reefing, easier stowing of mainsail) I am designing a composite pocket boom for my C&C 38-ii.


    I think I have most of the maths figured out already. See the attached spreadsheet. However, there are a few components of the design that I'm unsure of.

    The starting point has been the pocket booms made by GMT composites (Rhode Island, USA). They make booms for large yachts and construct their pocket booms from H80 foam and carbon fiber. My first iteration in the design uses 19mm H80 for the core and symmetrical laminates on each face of:
    [ 300g CF-45,45, 3(250g CF uni), 300g CF-45,45]

    Resulting in the total elastic modulus of the laminate at ~49 GPa (again, see spreadsheet, sheet 3 for details).

    My concerns are:

    1. At best, it seems like we can only roughly estimate the elastic modulus of the laminate schedule because mechanical properties for composite textiles are often given for the dry fiber only. In the resin matrix, these properties are likely to be much lower. I've applied a 70% reduction to the listed properties of the fabrics in my calculations of total elastic modulus (see sheet 3 in spreadsheet). Am I being too conservative or not enough?

    2. This is a fairly large structure and consequently, the moments of inertia (Iy & Ix) are an order of magnitude larger than traditional aluminum extrusions. And these values are important later on. For example, for the critcal load that a sandwich composite structure can withstand, Larsson & Eliasson give the formula:
    P = ( pi^2 * E I ) / L^2 ​
    Where E = elastic modulus of the laminate, I = moment of inertia for the structure, and L = length of structure. L&E write I as the moment of inertia of a rectangle. Can I simply swap my calculated I of the whole structure and still use this formula? It becomes:
    P = ( pi^2 * 49GPa * 8794cm4 ) / 378^2 = 2968421 N ​
    Which to me, seems spuriously large. (see spreadsheet, sheet 3 for details)


    3. The section modulus calculation is uncertain to me. I have the moment of inertia and the distances from the neutral axis to the outermost fibers, but I'm not sure this logic applies to this shape. Can anyone chime in?


    4. The section modulus is used to determine the maximum bending moment on the boom as:
    M = σf · SM​
    Where M = maximum bending moment, σf = normal stress (I believe this is the face stress: P/cross sectional area of laminate), and SM = section modulus. Which for my design becomes:
    M = 142575 N/cm2 * 761 cm3 = 1085332 Nm (vertical axis)​
    Again, this also seems spuriously high.

    5. Lastly, L&E do not give formulae for determining the loads on the boom, rather they state:
    "The bending forces (wind pressure on mainsail counteracted by the sheet and kicker) that the boom has to withstand result in requirements for minimum section modulus. This is the vertical SM, the horizontal SM is allowed to be half of the vertical SM."​
    They offer a formula for dimensioning the SM based on righting moment, vang position, heeling arm, and the yield strength of the boom (which is not explained), but this formula does not generate meaningful answers. See my deconstruction of their example in the spreadsheet, sheet 3.

    I suppose a workaround would be to use the force at the mainsheet attachment (dimensioned as the RM/heeling arm) to estimate the force at the vang (proportional to the fulcrum length of the mainsheet) and combine these two values as an approximate bending moment. I have no idea if this would be accurate but it seems logical to me.

    Any help or comments are greatly appreciated. Thanks!
     

    Attached Files:

  2. gonzo
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    gonzo Senior Member

    Are you laminating the part yourself? In that case, the first thing is to laminate samples and test them. Otherwise, you don't have any data about the material. It is not likely you will get the same quality as a factory.
     
  3. RMA
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    RMA Junior Member

    Yes, and may vacuum bag to up the fiber:resin content. Certainly. I would never expect to get factory quality and I will make some test pieces but I'm hoping to compensate with a slightly oversized laminate. If any of the numbers above are real, it seems like the boom as designed is already extremely overbuilt.
     
  4. gonzo
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    gonzo Senior Member

    You are guessing at the material properties. Until you actually test the material you can produce, there is no data to design to. I don't know what your experience with carbon fiber is, but it is much harder to work with than fiberglass. A pre-preg should give better results. One of the issues is that this is an asymmetrical shape with asymmetrical forces applied to it, and open on top. Besides the bending, there are compression and torsion forces. You need to add shock loads too. The forces will be concentrated at probably three small locations: gooseneck, vang attachment and sheet/clew attachment. Other things to consider are, for example, that the section will tend to open/close/skew when it bends/twists because it is open on top. Looking at GMT website, their booms are boxed in the bottom, and tapered. There is a FEA analysis image, that shows the complex design and laminate schedule.
     
  5. TANSL
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    TANSL Senior Member

    It is not necessary to test the material to have the mechanical properties of a composite material. Of course, if you have done that test, all the better, but most designers use formulas that give, with sufficient approximation, the mechanical properties of any laminate. No one, with sufficient knowledge, assumes anything. Perfectly contrasted formulas are applied.
    I think that putting a sandwich type laminate on the boom is not suitable. In addition, the cross section of the boom should be a closed tube, as a really resistant element, plus wings that were only used to collect and store the sail, but that would not form part of the resistant structure of the boom.
    Regarding the spreadsheet that you have shown us, the calculations seem correct if it is a question of obtaining the global resistance of the laminate, but you should complete them with the detailed analysis of each one of the layers of the laminate. One can get very unpleasant surprises if this analysis is not done.
     
  6. TANSL
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    TANSL Senior Member

    Bending always gives rise, as a consequence of the bending moment, to tension or compression in the resistant element. It is not "besides to". Shear stresses, interlaminar stresses and in the plane of the section are also produced. FEA is not required to calculate them. Things can be much simpler and realistic enough.
    One thing that I had not taken into account is the torsional forces on the boom and I would like to know the loads that give rise to those torsional forces and with respect to which axis that torsion occurs.
     
  7. RMA
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    RMA Junior Member

    The critical importance of having real world data is not lost on me. But we need a starting point before even building test samples, hence the use of reported material proteries and formulae to estimate laminate properties.

    My experience with carbon is that it is just harder to tell when it's fully wet out. Over saturating and fin rolling before vacuum bagging would help alleviate this challenge. What other challenges are you referring too?

    Yes, I think we all can intuit that the gooseneck, vang, sheet, and reefing block attachments will be heavily loaded and will require solid laminate (or marine ply core) and extra reinforcement in those areas. This is really all that I gather from the FEA analysis. Clearly some areas are not heavily loaded while others are, but we didn't need FEA to tell us this, right? ;) Not all of their pocket booms are tapered, I suspect this is mostly because they use a single 40' long tapered mold for pocket booms of all sizes. Starting at the widest end for the largest booms, ends in a more tapered section. The smaller booms start further down the mold and show less taper. Keep in mind, many of the booms shown on their site are on yachts 50-90'. Here, we are talking about a 12' boom on a 38' crusing sailboat with a relatively small mainsail (250 sqft or 23sqm). I think the loads on my boom should be relatively easy to overcome with proper design. Can you point me to some literature or formulae for calculating shock loads? Thanks!
     
    Last edited: Oct 20, 2022
  8. RMA
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    RMA Junior Member

    Thanks for your thoughts, TANSL. Yes, classically, the closed tube has proven a very good design and I had originally considered just adding a pocket shape on top of the original boom, but that simply adds weight, plus, where is the fun in that kind of design/build?

    Clearly, the open U shape can be done, and GMT builds them in H80 foam and uni and biax standard modulus carbon fiber. See the short article on their build process featured in CompositesWorld: Pocket Boom improves mainsail handling https://www.compositesworld.com/articles/pocket-boom-improves-mainsail-handling

    Thanks for checking over the spreadsheet. Yes, the initial goal is to design for global loads. Once those are sorted, we can zoom in to local loadings. I'm not sure what you mean about a detailed analysis of each layer in the laminate? I haven't come across this kind of analysis in L&E (I need to get more composites books!). They only discuss calculating the total elastic modulus of the laminate as a function of the moduli and thicknesses of each contributing layer. Can you point me in the right direction for this kind of analysis?
     
  9. RMA
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    RMA Junior Member

    How would I go about calculating these shear and interlaminar stress in the section plane?

    I would like to know about these torsional forces too.
     
  10. TANSL
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    TANSL Senior Member

    I enclose a sample of the analysis by layers that I have referred to in my previous post, with details of what is calculated (tension/compression in each layer, shear stress in the plane of the slice and interlaminar) and how to do it. The current study of composite materials makes this analysis essential.
     

    Attached Files:

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  11. Ad Hoc
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    Ad Hoc Naval Architect

    It seems you are over thinking this.
    The following shown HERE, will aid you in this quest.
     
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  12. Ad Hoc
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    Ad Hoc Naval Architect

    Don't forget...since you are using a low modulus material, deflection, NOT stress, is your criterion to use for "allowable".
    To establish what is an allowable deflection, design the member using steel and then using aluminium.
    Check the deflections..and see what you consider to be "appropriate" for the intended use.

    Then use THAT value, as your criterion using composite.
     
  13. TANSL
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    TANSL Senior Member

    @RMA , take a look at what others are doing. Perhaps they adopt a "less fun" shape of the cross-section of the boom, but for some reason, perhaps not fun at all, they decide that the resistant section should be closed, even if they incorporate some wings forming a box, with an open side, containing the furled main sail.
    [​IMG]
     
  14. tlouth7
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    tlouth7 Senior Member

    Section modulus of a symmetrical set of several rectangles is easy to determine! You will need to freshen up your integration from high school.

    However you cannot use the classic beam bending formulae because you don't have a constant stiffness value for all the parts of your section. Stress in your material will not vary linearly with distance from the neutral axis as per the beam bending equation. There are formulae for composite sections (generally stiff outer members with a less stiff core i.e. a sandwich beam) which you could probably generalise. You would treat the core, the 45 degree and 0 degree laminates as having different stiffnesses.

    Finally, for an open section like this you need to consider the buckling modes. In horizontal bending your side walls will deflect and change the section, which will allow for buckling at lower load than if the section stayed constant. If have always used standard sections which have lookup tables for buckling, or gone straight to FEA, so I don't know whether it is possible to find these results through hand calculations for arbitrary sections.

    Alternatively, you could approximate the section as a rectangular U section (take two extreme cases at the maximum and minimum widths of the section), and show that those easier to calculate sections have massive margins of safety on all failure modes.
     
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  15. gonzo
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    gonzo Senior Member

    You need to consider that this is also tapered end to end and has point loads.
     
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