Calculating stability of vessel containing liquid

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

  1. Robert Wilkinson
    Joined: Jan 2018
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    Location: New Zealand

    Robert Wilkinson Junior Member

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

    A tank barge of unladen weight 1000 kN is 24m long, 4.5m wide and the centre of gravity of the structure is 1.2m above the keel. It is divided into two tanks, each 18m long and 1.8m wide, the base of the tanks being 0.3m above the keel. One tank is partially filled with 45,000 litres of oil of specific gravity 0.89, the other tank is empty. The draft under these conditions is 2.1m and the centre of buoyancy is then 0.9m below the surface.
    Take moment of inertia of waterline plane as 75 per cent of circumscribing rectangle.
    To what angle will the barge tilt?

    Answer from book is given as 16 degrees.

    I'm just wondering how you calculate second moment of inertia for the oil in the tank?
    Only one of the 2 tanks has oil, do you have to use parallel axis theorem, but how do you do that?

    I think I can workout the rest from here?
     
  2. TANSL
    Joined: Sep 2011
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    Location: Spain

    TANSL Senior Member

    If the tank is rectangular with, say, a length "l" and a beam "b" the moment of inertia with respect to the longitudinal axis of the tank is : Ix = l * b ^ 3/12.
    If you need the inertia of the tank with respect to the longitudinal axis of the vessel, you must apply Steiner's theorem.
     
  3. Robert Wilkinson
    Joined: Jan 2018
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    Location: New Zealand

    Robert Wilkinson Junior Member

    Normally you would added the two second moment of area together, but in this case it is a bit different. One of the tanks is empty and the other one is partly full with oil.
    Ix = 18 * 1.8^3 / 12 = 8.748m^4 for the oil,
    but does Ix = O for the empty tank

    So I am just wondering how you take this fact into account when you are calculating the inclination?
     
  4. Robert Wilkinson
    Joined: Jan 2018
    Posts: 11
    Likes: 0, Points: 1
    Location: New Zealand

    Robert Wilkinson Junior Member

    Normally you would added the two second moment of area together, but in this case it is a bit different. One of the tanks is empty and the other one is partly full with oil.
    Ix = 18 * 1.8^3 / 12 = 8.748m^4 for the oil,
    but does Ix = O for the empty tank

    So I am just wondering how you take this fact into account when you are calculating the inclination?
     

  5. TANSL
    Joined: Sep 2011
    Posts: 4,805
    Likes: 99, Points: 58, Legacy Rep: 300
    Location: Spain

    TANSL Senior Member

    Yes,of course.
    When defining the load condition, the CoG of all the weights, light ship, loads and liquids in tanks is calculated. This will result in a CoG displaced with respect to the center line and, therefore, to an initial heel.
     
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