Calculating stability of vessel containing liquid

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

  1. Robert Wilkinson
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    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
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    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
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    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
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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?
     
  5. TANSL
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    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.
     
  6. Robert Wilkinson
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    Robert Wilkinson Junior Member

    Hi TANSL

    The other day I went through this question and this is what I came up with

    First calculate Depth of Oil in Tank

    45,000 litres is 45 m^3 and Tank is 18 m long by 1.8 m wide
    So Depth of Oil in Tank = Tank contents Volume / Surface Area = 45 / (18 * 1.8) = 1.389 m
    So Depth of Oil in Tank = 1.389 m
    Centre of Gravity of Oil in Tank = Height of Base of Tank above keel + (Depth of Oil in Tank / 2) = 0.3 + (1.389/2)= 0.994 m

    Now calculate the mass of Oil in the partially full Tank

    For 45,000 litres of oil of specific gravity of 0.89 Mass of Oil = 45,000 * 0.89 = 40,050 kg
    Tank Barge Mass = 1000 KN / 9.81 = 101.9367992 Tonnes
    Mass of Oil = 40.05 Tonnes
    Total Mass of Barge and Oil = 141.9867992 Tonnes

    Find Combined Centre of Gravity from bottom of Keel

    Let Height of Centre of Gravity of barge and Oil = HG
    And Height of Centre of Buoyancy of barge and oil = HB,
    Now HB = 2.1 - 0.9 m = 1.2m
    HG = [( 101.9368 * 1.2) + (40.05 * 0.99444)] / 141.9867992 = ( 122.324159 + 39.8275) /141.9867992
    HG = 1.142019258 m

    Effective Metacentric Height = NM
    Now NM = HB + BM - (HG + GN)

    BM = I / V and GN = GG' / Theta = (w1 * I1) / w * V

    Effective Metacentric Height
    NM = HB - HG + [I - {((w1 * I1) / w) / V} ]

    w1 = Specific Gravity of Oil = 0.89
    w = Specific Gravity of Water = 1
    I1 = Second Moment of Area of Oil in Tank I1 = 18 * 1.8^3 / 12 = 8.748 m^4
    I = Second Moment of Waterline Plane I = 0.75 * 24 * 4.5^3 / 12 = 136.6875 m^4

    NM = 1.2 - 1.142 + [ 182.25 - {((0.89 * 8.748) / 1) / 141.9867992} ]
    NM = 0.057980742 + ( 128.90178 / 141.9867992)
    NM = 0.965824153 m

    Centre of Oil Load is 0.9m from centre of Barge

    So Overturning Moment = Oil Load * Centre of Oil Load from centre of Barge
    So Overturning Moment = 40.05 * 0.9 Tonne m = 36.045 Tonne m

    Righting Moment = W * NM * Theta where W = Mass of Barge and Oil and Theta = Angle of Inclination
    Righting Moment = 141.987 * 0.9658 * Theta = 137.13428 Theta

    For Barge to be stable, Overturning Moment = Righting Moment

    36.045 Tonne m = 137.13428 Theta
    Theta = 36.045 / 137.13428 = 0.262844564 radians
    Now there are 2 Pi radians in one circle of 360 degrees so 1 radian = 360 / 2 * Pi = 57.29577951 degrees
    so Theta = 0.262844564 * 57.29577951 = 15.05988417 degrees

    The barge will tilt to 15.05988417 degrees

    The books answer is 16 degrees
    This is my best attempt at trying to get close to this answer.
    Do you think that the books answer may be a little bit wrong or do you think my answer is correct?
    If there is anything I might of missed in getting to the answer may you please let me know.
    Your help so far is much appreciated.
     
    Last edited: Feb 25, 2018

  7. TANSL
    Joined: Sep 2011
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    TANSL Senior Member

    I can not follow all your calculations in detail but they seem correct.
    Perhaps if you took the density of oil equal to 0.9 kg / dm3 and that of water equal to 1.025 kg / dm3 you would reach a figure closer to 16 degrees.
    Probably it is an error due to the rounding of decimals. ????
     
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