Layup schedule for rudder

Discussion in 'Fiberglass and Composite Boat Building' started by erdben, Dec 15, 2016.

  1. erdben
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    erdben Junior Member

    Hi,

    I’d like to build a transom-hung cassette-type rudder for my boat (Santana2023R), and I’m trying to figure out the layup schedule.

    This is what I have so far:

    Rudder blade is 65”long, 20” of this is in the cassette, 45” submerged, cord is 12", lateral area is 3.6 sqft.

    I’d like to make it as light as possible without breaking the bank, and I’m planning to use EPS (blue/pink) foam for core with maybe a 2” wide strip of cedar at max thickness to form an I-beam with the skins.

    There are many boat building blogs describing the method, but there seems to be quite a big variation in what people put in the skin. I’ve been looking at i550 builds that are smaller boats than mine, but they can plane, so could reach higher speeds I guess.

    I figured the worse-case scenario for the rudder is broaching while surfing down a big wave at 10kts of speed with the rudder angled to provide max lift (that’s kind of how I busted my previous rudder).

    I got ~5600 N for max side force and 3136Nm for max bending moment (using arm = half the immersed length of the rudder).

    I played a little with VectorLam (http://www.vectorply.com/), and assumed that from a structural point of view, the blade is a foam- or cedar-cored laminate that has the thickness of the blade, and the width of about 50% of the cord (since beyond that the skins are too close to each other to work effectively). VectorLam spits out a bunch of parameters, and I thought I needed ultimate bending moment and flexural stiffness to judge whether a certain laminate 1) would not break under max force and 2) would not flex too much.

    Ultimate bending moment is given in Nm / m, and I figured I need to multiply this with the width of my laminate (the 6” assumed width) to get the max bending moment where it would break.

    Flexural stiffness is given in N x m^2 / m, and once again, I figured I need to multiply this with my hypothetical laminate width (6”) to get stiffness in N x m^2.

    Then, I used the equation for flex for the free end (Y) of a continuously loaded beam fixed on one side: Y = Force x Length^3 / 8 x E x I, and replaced the “E x I” with the flexural stiffness I got above.

    It did give believable numbers, but I was surprised to see that most of these i550 build blogs I found describe laminates that would either break under similar max loads or even if they have high enough ultim. bend. moment, they would bend pretty significantly.

    Here is one laminate I came up with using carbon uni for stiffness, glass +/-45 biax for torsion and 6oz glass on the outside to protect the carbon.:

    1.2” foam core +
    9 oz glass biax
    9 oz carbon uni
    9 oz glass biax
    9 oz carbon uni
    9 oz glass biax
    9 oz carbon uni
    9 oz carbon uni
    6oz glass cloth.

    VectorLam says the skin thickness would be about 1/8”.

    For this laminate, I got 5606Nm ultimate bending moment, and 3” deflection at the end of the 45” free span when under the expected max load (3136Nm). If I replace the foam core with cedar I get 6743Nm for ult. bend. moment and 2.5” for the flex for max expected load. (These flex numbers seem rather high...)

    Sorry this post got so long – does any of this actually make sense?:confused:

    If this is the right approach, what would be some target numbers for ultimate bending moment and acceptable flex at maximum load?

    Finally, is it a good idea to use glass biax and carbon uni as in my test laminate above, and are there any rules on the order of the various carbon and glass biax layers in a laminate?

    Thanks a lot!
     
  2. TANSL
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    TANSL Senior Member

    According to ISO 12215-8, when the rudder blade is constructed with a PRF coating on a foam core, the required minimum fiber weight per square meter, Wr, should be as follows:
    Wr = k5 * 1.1 * (0.115 * Lwl + 0.15) kg/m2
    Where :
    • k5 = 1,0 for glass fiber reinforcements containing up to 50% mat laminate by weight
    • k5 = 0.9 for continuous glass reinforcements (eg bi-axial, roving, unidirectional)
    • k5 = 0.7 for continuous reinforcements using aramid, carbon or a hybrid of both.
    The outer and inner part of the laminate are the ones with the highest tension/compression, with the minimum stress being in the vicinity of the neutral axis. So placing the sturdier material in the center of your scheme does not make much sense.
    You should use the bending moment, etc., to calculate the rudder shaft and its bearings.
     
  3. erdben
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    erdben Junior Member

    Thanks I'll check how that formula compares with what I did.

    Maybe I wasn't clear, but all that glass and carbon are in the skin (on both sides); the foam is in the middle. What I'd like to know is that once you determined how much carbon / glass you need, how do you determine in what order you put those in the skin. Do you alternate them or try to put all the uni close to the outside in one group?
     
  4. TANSL
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    TANSL Senior Member

    Yes, I understood you perfectly. Maybe it's me who has not been able to explain himself.
    You can place the layers as you wish, but if you make a cross section of the skin, in the center of that cut the tensions are smaller than in the outer and inner layers. Therefore, it is best to place the most resistant layers on the outside or inside the skin. However, so that the warp of the fabric is not visible from the outside, you should put some layers of light mat on the outside
     
  5. erdben
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    erdben Junior Member

    That's good to know, thanks.

    About that formula:

    Wr = k5 * 1.1 * (0.115 * Lwl + 0.15) kg/m2

    Isn't it for rudders with stainless steel or carbon rudder stock and internal structure? It gives a very light fiber weight. For my boat, it specifies 20.7 oz/sq yd carbon or 26.6 oz/sq yd continuous glass. That's significantly lighter than any laminate I've seen for these kind of rudders. It also doesn't factor in the aspect ratio of the rudder, which would greatly affect the bending moment.

    I couldn't find any scantling formula for a "foam core and skin only, no internal structure" rudder, that's why I tried to come up with numbers from ultimate bending moment and stiffness.
     
  6. TANSL
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    TANSL Senior Member

    You could consult ISO 12215-8. It deals with the problem of all types of rudder.
    If you break the rudder blade (PRF-coated foam core) into strips of, say, an inch wide, each strip will withstand forces, due to water pressure and blows from the waves, which are not very important. Therefore, they do not need to be very thick. But the sum of these forces is very important and generate bending moments that must be supported by the rudder stock and its bearings. Here it is necessary to take into account, logically, the situation of the center of pressure.
    That is my interpretation of the facts. Maybe some expert could shed more light on us.
     
  7. erdben
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    erdben Junior Member

    That's the thing, these rudders don't have rudder stock, they are basically like a dagger board that can be raised-lowered in a transom-hung cassette. So the core+skin have to be strong enough to support the bending moments.
     
  8. TANSL
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    TANSL Senior Member

    I'm sorry but I can not conceive of a stockless rudder. I am quite sure that your rudder has a stock.
    If the rudder blade was hollow, the skin may have some problem, but being solid, the blade will withstand hydrodynamic and wave pressures perfectly. All the forces applied on the blade have a resultant, applied in the center of pressure, which is the one that creates a distribution of bending moments along the blade and the stock. If you want to do a direct calculation, instead of being guided by the norm ISO, find out the law of distribution of bending moments. With it you will be able to know the tension in each zone of the blade and you will be able to check if the material of the core, plus the skin, are able to withstand the stresses that each corresponds to bear. Although not easy, you should use some procedure to know the maximum allowable stress in the core and skin.
    I do not recommend direct calculation.
    Read what the ISO 12215-8 standard says and apply its formulas. If you are not convinced about what the ISO standard determines, apply a safety coefficient of 2 to the results obtained with it.
     
  9. erdben
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    erdben Junior Member

    I guess I should have done earlier, but here is a picture of a cassette-type rudder, similar to what I'd like to build.

    [​IMG]

    Here is another rudder of that type

    [​IMG]

    You can lift / lower the blade in the cassette.

    Currently I have a kick-up rudder, and the blade is built from some heavy filler as core + heavy fiberglass skin. The blade is sandwiched between aluminum cheek plates. The whole thing is heavy, ugly and as it turned out still not strong enough, since during the summer I managed to bend the cheek plates and break the bolt holding the rudder blade during a spectacular broach.

    Since I have to fix it anyway, I figured I would rather build a cassette rudder, which would be nicer, and would allow me to steer the boat in shallow water or lift the blade partially for better performance in light air.

    I tried to find formulas for the layup of transom-hung rudders, I don't think any of those have a rudder stock, but couldn't come up with anything. That's why I tried to use VectorLam. I'm pretty confident in my load calculation, but I'm not sure I'm using the numbers from VectorLam correctly.
     
  10. TANSL
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    TANSL Senior Member

    If you give me your boat and rudder data, I can calculate the blade and shaft according to ISO. (In Europe, pleasure boats must comply with this standard)
     
  11. PAR
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    PAR Yacht Designer/Builder

    A cassette style of rudder doesn't need a shaft TANSL.

    I question the need or desire for the biax in the laminate schedule. I'd use a finishing cloth on the exterior, as a way to clean up the layers of unidi carbon, but I wouldn't intermix in biax. A relatively recent lay up I did for a blade of this size (44"), was a straight 9 layer schedule of unidi carbon, with a 6 ounce cloth cover to making fairing it up easier. This was over an 8 pound foam with a modest aluminum armature for the pivot pin (kickup) and root of the blade. With a bigger (longer) armature, I could have decreased the laminate requirements a bit. I ended up with about a fat 1/8" skin.

    This is the crux of the deal with carbon blades - how much money do you want to spend on carbon, when a less costly armature and/or core arrangement can come to play.
     
  12. TANSL
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    TANSL Senior Member

    Well, what I want to express is that for the rudder to turn it needs a turning axis, a shaft. It can be formed by the rudder stock or by the pintles or both.
     
  13. erdben
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    erdben Junior Member

    Hi PAR,

    that's great info, thanks. Can I ask what was the weight of the carbon unis? When you come up with a schedule like this, is it just based on experience or is there actually a way to calculate how much reinforcement you need based on expected loads? Does the method I described in my first post make any sense?
     
  14. PAR
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    PAR Yacht Designer/Builder

    Yes, you can calculate the loads and develop the schedule, but there are lots of variables to alter the schedule, such as the foam and armature. My fabric was 8.8 ounce. I'd done this size foil previously, so no calculations were necessary, as he was just a weekend buoy racer. It was lighter (by a lot) than the previous all 'glass blade and stiffer by a significant amount too. If you want to carve the foil down to it's ultimate weight savings/stiffness benefits, then the math is necessary. If you can live within 10% - 15% of these figures, your math will do the job, though I'd substitute the 'glass biax with 12 ounce carbon biax (45/45).
     

  15. erdben
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    erdben Junior Member

    I can certainly live with a little heavier rudder to err on the side of safety. So basically you recommend modifying my original layup so it would have 4 layers of 9oz carbon uni and 3 layers of 12 oz carbon +/-45 biax? Does the order of the various layers make any difference?

    Thanks again!
     
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