Calculating Metacentric Height hollow cylinder

Discussion in 'Stability' started by Robert Wilkinson, Feb 11, 2018.

  1. Robert Wilkinson
    Joined: Jan 2018
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    Robert Wilkinson Junior Member

    I am trying to workout how to arrive at the answer for the following question.

    A hollow cylinder 2.4m high consists of two concentric cylindrical shells, each 6mm thick, of 2.4m and 3.0m diameter respectively. The top is open and the bottom is totally closed by a plate 25mm thick. The cylinder floats with its axis vertical and the density of the material is 7,700kg/ m^3.
    Determine the metacentric height when in this condition?
    Also find the new metacentric height when there is water in the inner chamber of 2.4m diameter to a depth of 0.6m?

    Answer from book is given as 0.725m and 0.298m

    I'm just wondering how you calculate the centre of gravity in both situations when it is a hollow cylinder and has concentric cylindrical shells?
  2. TANSL
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    TANSL Senior Member

    Forgive my way of expressing myself and, please, do not be offended. Iwant to help. Are you an amateur trying to learn more about naval architecture or are you a student of some branch related to naval architecture that tries to get someone to solve his problems? In the first case, I will be happy to help you. In the second case, the best advice I can give you is to study a bit.
  3. Robert Wilkinson
    Joined: Jan 2018
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    Robert Wilkinson Junior Member

    I'm an amateur trying to learn about naval architecture and I'd appreciate your help.
  4. TANSL
    Joined: Sep 2011
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    TANSL Senior Member

    Could you show a drawing of the two cylinders ?. The description you give does not allow me to know how they are and how they float.
    Meanwhile, maybe the following will help you:

    metacentric height : GM = KB + BM - KG
    KG = height of the center of gravity of the set on the base
    KB = center height of hull on the base
    BM = distance between the center of buoyancy and the metacenter = I / V

    I = second moment of inertia of the waterplane
    V = submerged volume of the hull​
  5. Ad Hoc
    Joined: Oct 2008
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    Ad Hoc Naval Architect

    I did a quickie calc...i get slightly different values, probably owing to rounding errors....and whether you have the disc which can be either butts onto to the outer cylinder, ie has a diameter of 2.4m, fits inside the outer cylinder, as a tight fitting disc...since it is not clear which is does, and, whether SW or FW. Both make a difference!!

    Well....assume the cylinders sit on top of the disc (butting onto it), the KG is taken about the this case...image the flat closing disc, which is 25mm thick....its KG is 12.5mm above the base....then KG of the cylinders are 1.20m (half their length) + disc depth of 25mm = 1.2+0.025 = 1.225m.

    When with water, the water's half its depth, i.e 300mm.

    That's it.

  6. Heimfried
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    Heimfried Senior Member

    The center of gravity (CoG) of any cylinder (faces rectangular to its axis and out of homogenius material) sits at its axis in its half height (if upright). The same applys to a hollow cylinder.
    I assume the height of the complete cylinder as 2,4 m, so the shell part without bottom plate is h(s) = 2.4 m - 0.025 m = 2.375 m
    The bottom surface of the bottom plate is at K = 0.0 m

    So the two shells of the cylinder without bottom plate have the same CoG(s) at 2.375 m / 2 = 1.1875 m, KG(s) = CoG(s) + thickness (bottom plate) = 1.1875 m + 0.025 m = 1.2125 m .
    The corresponding mass is m(s) = (D(in) + D(out)) * pi * h(s) * thickness (shell) * rho = (2.4 m + 3 m) * pi * 2,375 m * 0.006 m * 7,700 kg/m³ = 1,861 kg

    Bottom plate: KG(bp) = 0.0125 m (also a cylinder, so CoG at half its thickness). Mass(bp) = [D(out) / 2]^2 * pi * thickness(bp) * rho = 1.5 m * 1.5 m * pi * 0.025 m * 7,700 kg/m³ = 1,361 kg

    The joint KG = (KG(s) * m(s) + KG(bp) * m(bp)) / (m(s) + m(bp)) = (1.2125 m * 1,861 kg + 0.0125 m * 1,361 kg) / 3,222 kg = 0.706 m

    Including the water means a further cylinder to calculate with its KG(w) and its m(w) and joining in the equivalent way as done above.

    (Sorry, it is a bit rediculous to mix numbers with an exactness of half a millimeter with much more imprecise values, but the wording of the question is a bit weak regarding that point.)
    Last edited: Feb 13, 2018 at 4:23 AM
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