offset beams on catamaran genius or deathtrap?

Discussion in 'Boat Design' started by sailor182, Jun 1, 2013.

  1. sailor182
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    sailor182 Junior Member

    As to what style of catamaran I'm thinking something like a Stiletto 27
     

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  2. groper
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    groper Senior Member

    Yes, i thought so... big problem is the beam which the mast is stepped and the added compression from it. A convenient way of solving this might be with a dolphin striker arrangement, with turn buckles that can be tensioned after you slide the beams out. A problem might be dragging it through the water as the bridge clearance looks very low on those types of beach cats...

    As Adhoc suggests, you wont know if it can be done unless you run the numbers. Take a look through the veristar e-rules link i posted earlier. All the formulas to calculate the beam loads are in there. Then once you know the loads, its another problem again to work out how strong your beam sleeves need to be. I dont know how to work that part out however, would require some thought and some more googling no doubt...
     
  3. Ad Hoc
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    Ad Hoc Naval Architect

    It is not that onerous.

    You have worked out the bending moment, M. You size the smaller beam, for the span, side to side.

    The size of the beam at A-A being the same as at B-B. Whereas at C-C this is double. And assuming for simplicity the beam is an RHS, it is shown as per the sections.

    offset beam.jpg

    Where there is a break and a cross over, the fixings at each end must also accommodate this bending moment. So the couple to created by the resistance (to hold it) is simply the distance apart, d, and the force at each fixing = F. So the further apart the lower the force, the easier it is to "hold" it.

    From this we can calculate the shear area required in the beam, at C-C and then it must also be taken by the beams at A-A and B-B..since if the singular beam cannot take this shear load, but the double (at C-C) can, it’ll fail just pass the joint.

    Then the reaction loads, in this case as a UDL, into the lower flange and considered as built in beam.
     
  4. groper
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    groper Senior Member

    A quick google nets a stilletto 27 at 1250lbs, lets say 600kgs and be metric... So lets run some numbers, feel free to change them to suit your needs...

    Hulls, 200kg each = 400kgs
    Cross beams, center deck gear, mast and rigging, center outboard perhaps = 200kgs.
    Crew = 4 people @ 75kg each, = 300kgs all on the windward hull.
    All this whilst flying a hull...

    Lets assume a 4m beam - plucked the number out of thin air, but seems reasonable.

    Thus your cantilever, supported at one end, bending moment = 2m* 200kgs (crossdeck) + 4m * 700kgs (windward hull and crew)
    = (2 * 2000N) + (4*7000N)
    = 4kNm + 28kNm
    = 32kNm

    This would be divided by the 2 beams, so 16kNm per beam so far (torsion still to add) and this is the moment at the hull to beam connection. Now the center of the beam where the sleeve is;

    We will assume the central boat weight load is taken directly by the sleeve itself (the outer section) and so ignore the cross structure weight for simplicity. Thus the simple transverse bending moment would be the 700kgs*2m or 14,000Nm.

    If we have a 0.3m sleeve, then thats F=m/d = 14000/0.3 = 46,666Nm or 46.6kN - divided by 2 beams again, = 23.3kN per beam

    And the shear = 23.3kN/shear area - we dont know the section size of the beam so we cant calculate, we need to assume some sizes and see if they are upto the task.

    Is this correct so far?
     
  5. groper
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    groper Senior Member

    It just occured to me that if the above is correct, then if we increase the sleeve length so that the beam to hull connection = the sleeve force connection, then the sliding section size will be optimized for both ends. ie at the sleeve and at the hull....
     
  6. FAST FRED
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    FAST FRED Senior Member

    The stiffness of the tubes is related mostly to the tube diameter.

    A much higher web depth could be had by simply allowing a pivoting fold operation with a simple cable holding the hulls from moving fore and aft .

    For trailering one goes fwd , one goes aft , to sail line them up and clip the cables on.
     
  7. Ad Hoc
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    Ad Hoc Naval Architect

    The hull weight is 1 hull not both, since only one hull is "flying". So the moment is 4m x 500kg (1 hull [4x200] + crew [4x300])
     
  8. Richard Woods
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    Richard Woods Woods Designs

    OK Why is the hull flying in the first place? because it is being lifted up by the windward shroud. So what opposes that lift? The compression down the mast and onto the mast beam

    A dolphin striker under the mastbeam can take that load back to the hulls. Often the maststep is "floating" through the beam, the whole mast compression is taken by the dolphin striker. The mastbeam instead of being bent is actually then in compression.

    But I think that is academic. Your problem is the 12in of bury, it's not enough, even if you make a very strong beam. The problem is that the beam will flex in the socket - there has to be a gap between beam and socket to if you want to move it easily. So you need the longest possible lever to reduce that movement.

    The telescoping beam cats, like the Stiletto and International 23 are very narrow, because of the beam system. Ultimately you'll want a wider boat the first time you sail in any wind and you won't care about the fact it takes you a few minutes more to rig.

    But as always, it's not the hull assembly that takes the time, it's raising the mast and rigging the boat. So you need to consider those details as well

    Richard Woods of Woods Designs

    www.sailingcatamarans.com
     
  9. groper
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    groper Senior Member

    Yes Richard, i always wondered about that, even with a mast and rigging present, it seems to be ignored in the global transverse beam calculations when in reality it forms a giant kingpost and the shrouds end up forming a truss. Using the shrouds in tension effectively unloads the transverse beams. Are you aware of any class rules that account for this or is it always just ignored?
     
  10. Ad Hoc
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    Ad Hoc Naval Architect

    It is a "localised" load as such, not a Global load. Therefore it is ignored. Plus if the shrouds fail you've lost all your global strength. You may satisfy yourself that it works (especially in a 2D analysis), but you're placing far to much faith in several parts to work in unison and consistently without the need for any maintenance - as global structure should be. Not to mention the fact that the shrouds will only support tensile loads, not compression or shear/torsion.
     
  11. Richard Woods
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    Richard Woods Woods Designs

    Agreed, if you break a shroud you loose the hull lifting facility, but on the other hand your mast is lying under the boat, so its not the beam strength you are worrying about then, more the hull impact strength.

    Attached is a photo of a 35ft x 20 ft catamaran that is a successful racing boat. This photo was taken after finishing first in the 2008 Swiftsure race, beating a F40 over the line - just. The fastest speed we did was 22 knots. Quite exciting at 1am and going down a channel between two unlit rocks

    It has a 50ft mast (approx) and weighs maybe 1.8T. As you can see there are only 3 beams holding the boat together. the biggest is 6in the other two are maybe 4in. The beams haven't broken, but the forward beam mounting has. It is quite a flexible structure though, and very wet to sail at speed.

    Take the conventional cantilever calculation method and the beams are not strong enough. It is held together by the rig.

    Richard Woods of Woods Designs

    www.sailingcatamarans.com
     

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  12. Ad Hoc
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    Ad Hoc Naval Architect

    So, have you done your calc's yet and what results do you get for your 2 or 3 beams?
     

  13. sailor182
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    sailor182 Junior Member

    sorry no after thinking it over I think I'm gonna go in a different direction.
     
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