Aluminum tunnel hull floor and transom

Discussion in 'Metal Boat Building' started by Alumination, Nov 4, 2015.

  1. Mr Efficiency
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    Mr Efficiency Senior Member

    Key thing here,imo, is the torsional stresses which are greater in the tunnel hull than a mono, which is more one thing acting uniformly rather that two things joined together, each doing its own thing in a different setting. Welders are kept busy repairing some alloy cats that are subject to hard use. They'd probably be the ones that didn't allow for the different stresses involved, at the design stage.
     
  2. Alumination
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    Alumination Junior Member

    The torsional twisting is something I've been discussing with friends and co-workers. The floor won't really do much to add strength but bulkheads will. Also if I were to run a round or square tube down the center the entire length of the floor, a torque tube so to speak, connecting the transom to the bow and connected to the floor the entire length. I believe round is stronger in this type of application but square would be lots easier to weld into place.
     
  3. Barry
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    Barry Senior Member

    The moment of inertia is relevant to stress calculations in beams and not limited to compression.


    There are many loads that can create stresses in a beam but for a simple discussion let's assume that this beam is symmetrical around the x and y axis and the load is at 90 degrees to the x axis.
    Due to symmetry, the middle of the beam is called the neutral axis. Below the NA, the beam is in tension starting with 0 stress which then increases to a maximum tensile stress at the outer fibre. Above the NA, the beam is in compression, and has 0 stress at the NA going to a maximum compressive stress at the outer fibre.

    We will ignore shear flow stresses at the NA for now. We will say that the beam will have failed when it strains (stretches) permanently when the load is removed. Ie hits its yield stress

    The idea is that if you can get material further away from the NA then you can reduce stresses at the outer fibre for a given load.

    M is the bending moment due to the load, C is the distance from the neutral axis to the point where you want to calculate the stress, and I is the moment of inertia, and sigma is the stress in that spot that you are calculating the value of

    The formula is then
    Sigma = M {C divided by I}

    Multiply both sides by I\C and divide both sides by sigma you then get

    I\C = M\sigma
    As the maximum stress occurs at the outer fibre, a distance C from the neutral axis, this is the spot that failure will occur first in tension

    For a given load which determines the bending moment which you need to know when designing a beam and entering in an allowable maximum stress, you can now calculate the I\C value

    I\C is called the section modulus, the moment of inertia divided by the distance from the neutral axis to the outer fibre.

    If a beam has the same section modulus as another, the beams have the same strength

    The beams do not have the same strength if the moment of inertia is the same

    So when you know the bending moment and allowable stress, which gives you the section modulus you can just go to tables that give the section modulus and pick a profile that will suit the size that you want to use

    This is extremely simplified as there are many other considerations.

    To Alumination, your question as to use tubing or I beam
    The advantage of tubing is that if a load is not applied perpendicular to the x axis the tubings section modulus does not change as drastically as compared to I beam or channel

    This is a very simple explanation as there will be other loads imposed upon the outer fibre which sees the maximum bending stresses as others have pointed out
    Torsion of the floor members, vertical and horizontal shear, stress concentrations due to fastening the floor beams into the sponsons and of course dynamic loads introduced when pounding through waves.
     
  4. Ad Hoc
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    Ad Hoc Naval Architect

    I can't speak for Gonzo, but if one designs high speed vessel it is perfectly correct.

    One 12m HSC i designed displacement = 12 tonne ( 118kN), speed = 40knots.

    The bottom slamming pressure is 74kN/m^2. Given the weight of the boat is 118kN, the area required to support the slam load is simply 1.60m^2, or a panel that is 1.25 x 1.25m. Not much as that can easily be on just one side of the chine to centreline.

    This is consistent for just about every high speed vessel.
     
  5. TANSL
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    TANSL Senior Member

    When things are explained properly, I can not but agree.
    In short, I've only raised two issues:
    - Not good confuse pressure and force.
    - How Gonzo had come to the statement he made?
     
  6. rwatson
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    rwatson Senior Member

    Sketchup is a poor solution for boat design - very primitive. Solidworks is the best way
     

  7. lance linked
    Joined: Sep 2015
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    lance linked Junior Member

    I would agree that dynamically loaded structures can be difficult to figure accurately, so work in shorthand. In any case, I was considering a junior series beam in my example.

    More importantly, I make the point that designs often overemphasize the importance of shapes, and underemphasize the importance of the connections.
     
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