# Metacentric height correction due to free surface effect

1. Sep 23, 2004

### ar

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
Thanks
ar

2. Sep 23, 2004

### 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.

3. Sep 24, 2004

### ar

Dear russ_waters
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