Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Metacentric height correction due to free surface effect

  1. Sep 23, 2004 #1

    ar

    User Avatar

    Hi
    regarding the above subject there is a conflict of opinion between me and a collegue. So i need your help
    In order to calculate the corrected metacentric height of a floating structure that have a open compartment, filled with water, that is located eccentrically i have to use
    GMnew = GM - (density of tank fluid * accel. of gravity * second moment of inertia / structure displacement)
    Which is the second moment of inertia that i have to use?
    My opinion : is the moment of the free water surface with respect to it's axis of transverse rotation
    My collegues opinion : is the moment of the free water surface with respect to the structure's axis of transverse rotation
    I will appreciate your reply
    Thanks
    ar
     
  2. jcsd
  3. Sep 23, 2004 #2

    russ_watters

    User Avatar

    Staff: Mentor

    I'm a little ways away from my last statics/dynamics class and I'm not sure of all of your terms, so I'm not quite sure what you are asking, but perhaps drawing a simplified version of your problem would help:

    A 2d, square-cross section ship with 4 compartments floats 3/4 of the way out of the water. If the bottom-right compartment is opened, it fills halfway and the moment generated on the ship can be calculated by using a point-force at the cog of the water in relation to the cog of the ship.
     
  4. Sep 24, 2004 #3

    ar

    User Avatar

    Dear russ_waters
    thank you for your reply
    I will try to make the problem more specific.
    Let's say we have concrete box structure with plan dimensions 10m (transverse) & 20m(longitudinal) that is closed at the bottom with a slab and open on the top. The structure is divided in two symmetrical parts by a wall running on the longitudinal axis. So we have the concrete box structure with two empty tanks (20 by 5) that we are going to float (assume that height is sufficient, let's say 15m). The wall & slab thickness is, let's say 50cm.
    Now we add 1m of water inside the two tanks.
    Checking the transverse stability of the structure
    Weight = 1880 tons (1440 ton walls + 240 tons slab +200 ton water)
    Draft should be 1880/(20*10)= 9.4 m, for displaced volume = 1880 m3
    Center of gravity located at 5.7m
    Center of boyancy located at 4.7m

    Now, let's calculate the following values
    2nd moment of inertia of waterplane area Jw= 20 x 10^3/12 = 1667 m4

    2nd moment of inertia of each tank free surface with respect to tank centerline Jf1 = Jf2 = 20 x 5^3/12 = 208 m4

    2nd moment of inertia of each tank free surface with respect to structure's centerline Jf1' = Jf2' = (20 x 5^3/12) + (( 20 x 5) x 2.5^2)= 833 m4

    Metacentric height without considering the free water surface effect will be
    GM0=(J of waterplane area / displaced vol) + (5.7-4.7) = (1667/1880)+1 = 1.88 m

    NOW, MY QUESTION IS
    In order to consider the effect of free surfaces to the GM0 we have to reduce GMo by a value equal
    a) (Jf1+Jf2)/displ volume = 2 x 208/1880 = 0.22m so the GMcorrect=1.66m
    or
    b) (Jf1'+Jf2')/dipl volume = 2 x 833/1880 = 0.87, so the GMcorrect=1.01 m

    Thank you again for your interest to my inquiry
    ar
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Metacentric height correction due to free surface effect
  1. Unflattening a surface (Replies: 3)

Loading...