rudder design - gudgeon size

Discussion in 'Multihulls' started by seandepagnier, Mar 20, 2022.

  1. seandepagnier
    Joined: Oct 2020
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    seandepagnier Junior Member

    I am in the processes of designing/building a kick-up rudder or my kraken 33' trimaran.

    I am now at the point of trying to figure out the right rudder hinge. Similar size monohulls all have 1.5" solid bronze shafts, however.. it seems this is more to do with the torque on the shaft than the forces holding the rudder? Otherwise, is there some kind of calculation for rudder surface area to shaft diameter?

    My initial thought is to use lashings for the rudder hinge, but I am concerned that these will loosen, and with the limited range (no skeg) will experience a lot of torque. This may still be workable and preferable. Thoughts?

    The other option is to use gudgeons, but I calculate I would need 1 inch diameter pins, which is quite large and I have not seen ones this size. Perhaps a hollow pipe is viable with an even larger diameter? In any case, it would be quite heavy and difficult to work with. What is the calculation for gudgeon size?
     
  2. oldmulti
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    oldmulti Senior Member

    This may help. Or look at Multihull Structure Thoughts tread index (on page 1) and look for the Buccaneer 28 plans or Command 10 plans.
     

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  3. seandepagnier
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    seandepagnier Junior Member

    Many thanks for your reply. The drawing is very helpful as that boat is very similar to my own. Any idea how he he determined a 5/8" diameter pin is sufficient?
     
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  4. jehardiman
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    jehardiman Senior Member

    Normally it is done using statics, assuming the rudder is put hard over (90 degrees unless there is a stop) at maximum speed. The maximum force is applied at the center of pressure and the reaction forces are calculated (note this also gives you the bending forces in the rudder blade and the tiller forces/moments). A safety factor is added and the necessary shear area (i.e. diameter) is calculated. Generally bearing area is set/assumed to be at least 1.25 diameter ( old ABS rules).
    Where did you get 1" from?
     
  5. seandepagnier
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    seandepagnier Junior Member

    A few things.. I don't know what "maximum speed" is, but I do know the dynamic forces of waves can be huge.

    Most keel hung rudders on boats similar size have 1.5 inch solid bronze shafts. This is more than 5 times the cross sectional area of 5/8" and also that rudder is supported at the bottom. I realize it's not possible to make a direct comparison, but it makes me wonder. As for 1" this is just scaling up based on rudder area from 20ft boats which use 1/2" pins. Furthermore, it seems most of these smaller boats are not intended for offshore work, and would work-harden and eventually break even a 1/2" size.

    So.. basically.. the more I try to learn, the less I know. If crowther says 5/8" is enough: it probably is, but it seems very light compared to other boats.
     
  6. oldmulti
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    oldmulti Senior Member

    Seandepagnier. Join the rest of us. If your Kurt Hughes you design rudder shafts that can handle almost any loads as he had a large cat bend the rudder shafts from a propeller blast when they gunned there big diesel engines. If you were Lock Crowther you designed rudders that would bend or break before the rudder did damage to the hull. (Lock and I had a discussion on the topic when he designed a cat for me). Result understand the designers ideas first then work from there. In the cat designed for me if I threw the helm hard over surfing down a wave front at 25 knots, bent rudder shafts would be the result. If Lock says 5/8" pins I would trust his design.
     
  7. seandepagnier
    Joined: Oct 2020
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    seandepagnier Junior Member

    An update.. I have mostly completed the rudder. I ended up using a similar design to the buccaneer kickup (thanks oldmulti)

    My blade ended up weighing about 22lbs, and the cheeks about the same, plus the hinges it is probably about 60 pounds. I think I overbuilt it. It is coosa board (26lb density foam) and 10 layers of carbon fiber 6oz cloth. This is to have a safety factor (so it wont even really weaken over time) and to account for my imperfect layup.

    I was reading that at 20knots of boat speed and 10 degrees of rudder there is more than 1 ton of force and this is what had me concerned.. now I am wondering how much lighter it could have been. Perhaps 10lb foam would do? Less carbon? How to calculate it? I also should have tried to obtain unidirectional carbon and could have used less layers.. but how many?
     
  8. rnlock
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    rnlock Senior Member

    If we can believe the calculator linked to below, at 20 knots, the dynamic pressure of fresh water is about 1120 lbs per square foot. A little more for salt water. This is what you can expect at a lift coefficient of 1. However, the Reynolds numbers are pretty high, so with some foils and strong, stubborn helmsmen, you might get somewhat more. Also in cases where you have a fixed skeg and a rudder, but I don't know how much of the load would fall on the rudder and how much on the skeg. With a lousy rudder shape, you might have a lift coefficient of less than 1. That might be annoying when trying to maneuver at slow speeds.

    If you want to verify the strength of the rudder in port instead of at sea, mount some identical gudgeons on something really solid, with the axis horizontal and the tiller lashed straight. Put the rudder on the gudgeons and weight the rudder, uniformly, to a bit more than 1120 lbs per square foot, depending on the foil, etc. That's not going to be easy, though. Maybe put on a reinforcing plate of some sort and then push down on it at a point about 40 percent of the way from the lower gudgeon to the tip, at 25 percent MAC. (The internet can tell you about MAC.) That ought to be a pretty good test, unless you were just starting to damage the rudder with that load. The solution, if you want to call it that, is to load it up until it actually breaks, and build another one. Or build a really ugly rudder to test in the first place.

    You can find estimates for the strength of composite layups on line, and figure out how much material you need, but accounting for local stresses where the pintles bolt on may be a bit trickier. As you get further down the rudder, you need fewer and fewer layers of cloth, because the bending stress is much less. On the other hand, the chance of local damage when you hit something on the bottom goes up. If I was building a rudder like yours, I'd put in some kind of insert or layup where hardware attached, with the fibers running parallel to the bolts, to take the compressive load from the bolts. The layup on the surface of the rudder would be thickest at the hardware, then get successively thinner as the distance to the hardware increases. Kind of like those layers you sometimes see in sails near the clue. Hypothetically, if you were never going to hit anything, you could use just one layer at the tip. On the other hand, 22 lb blades don't sound so bad (except for your budget). 38 pounds of hinges sounds appalling, but I'll admit I haven't run any numbers on this. I don't know how expensive it is, but pre-cured, pultruded carbon is much stronger than hand layups. Unidirectional hand layups should be stronger than cloth, at least in that one direction. Stitched carbon (biaxial?) should be stronger and stiffer than cloth. Nevertheless, it sounds to me like your hinges are where the extra weight is. All you have to do is find a transonic airplane with about the same size control surfaces, cantilevered the same amount, and copy the hinges. ;-) The load per square foot should be similar.

    Dynamic Pressure https://www.engineeringtoolbox.com/dynamic-pressure-d_1037.html
     
  9. seandepagnier
    Joined: Oct 2020
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    seandepagnier Junior Member

    Thanks for the reply.

    The hinges are only about 8 pounds. they are the only parts with bolts as they are bolted to the boat I used s-glass.

    The box the blade goes into is the remainder of the weight. I calculated 4000 lbs should be max force on this.. so I think it's "ok"

    As for pultruded carbon.. As far as I can tell, the strength of carbon in tension is irrelevant here as it will never fail that way. It will fail in compression correct? This means the compressed side of the blade will buckle, but since it wraps around the front everywhere, it is much stronger than just a foam sheet with carbon on each side... Some people build structures of carbon into the blade, but this seems to make it more complicated to build. Maybe it can be lighter? Maybe a skeleton of carbon fiber, inside of some kind of flexible "meat" with a skin over that?

    As for single layer etc... I put 2 layers full length with less and less, but by about halfway there are 6 or more layers... so maybe the tip can break off if it hits rocks or something, but so far this carbon blade I dropped a few times on the trailing edge and it does not even ding it or make any visible damage.

    This also leads me to believe that using coosa board 26lb foam with 900psi compression strength is overkill, and a much ligher 10lb foam would be still more than 4 strong enough?
     
  10. rnlock
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    rnlock Senior Member

    Pultruded carbon, since the fibers are much straighter, will be much stronger than hand layup carbon in compression as well. Keep in mind that uni and pultruded carbon are much stiffer than cloth and will take most of the stress on themselves, so they'd better be strong enough for the load in their direction without much help from the cloth.

    I think that, if the foam isn't going to get squashed by the pintle bolts, it's better to put any additional material where it will do the most good, i.e. on the outside. Sounds like the box is where the real weight problem is, though. And the stresses might be a bit tougher to figure out.

    4,000 lbs of force is fine, but the moment probably matters more. That is, 4,000 lbs times, say, 40 percent of the distance from the lower gudgeon to the tip of the rudder, assuming most of that's in the water. It's possible to refine that 40 percent, depending on the shape of the blade.

    If you are halfway from the tip to the gudgeon, you probably only need two or three layers, but the load builds up quickly as you get higher. You could go by the square of the distance from the tip of the rudder, but that will still overestimate the load close to the tip.

    I imagine that two layers of 6 oz. on that heavy foam is quite tough.
     
  11. seandepagnier
    Joined: Oct 2020
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    seandepagnier Junior Member

    I used all cloth which is the weakest way to build this it seems. Some of the cloth was 75% 25%, but only for 2 of the layers; the rest was biaxial

    I realize I should have gotten uni to save a few layers but I did not find a good price. I used 10 layers of 200gsm tampered with fewer and fewer layers until only 2 layers to the tip.

    I build the box the same overall strengh so it has 5 layers of 200gsm carbon everywhere inside and out. it weighed about 20 pounds.. The 60 pounds total includes the parts bolted to the boat and all the metal fasteners.
     

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    Last edited: Aug 9, 2022
  12. rnlock
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    rnlock Senior Member

    You've built something, and I'm sure it works FAR better than any of my imaginary rudders, at least on a real boat!
     

  13. seandepagnier
    Joined: Oct 2020
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    seandepagnier Junior Member

    I wanted to add, that it works better than the previous rudder enabling short tacking, but underway at 8-10 knots moving it only 2 degrees the boat turns significant... so it is very slight movements to keep a straight course.
     
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