Stability of Floating Bodies

Any floating body is subjected by two opposing vertical forces. One is the body's weight W which is downward, and the other is the buoyant force BF which is upward. The weight is acting at the center of gravity G and the buoyant force is acting at the center of buoyancy BO. W and BF are always equal and if these forces are collinear, the body will be in upright position as shown below.
 

floating_upright-pos.gif

 

The body may tilt from many causes like wind or wave action causing the center of buoyancy to shift to a new position BO' as shown below.
 

floating_inclined-pos.gif

 

Point M is the intersection of the axis of the body and the line of action of the buoyant force, it is called metacenter. If M is above G, BF and W will produce a righting moment RM which causes the body to return to its neutral position, thus the body is stable. If M is below G, the body becomes unstable because of the overturning moment OM made by W and BF. If M coincides with G, the body is said to be just stable which simply means critical. The value of righting moment or overturning moment is given by
 

$RM \, \text{ or } \, OM = Wx = W \, (MG \sin \theta)$

 

The distance MG is called metacentric height.
 

Metacentric height, $MG = MB_O \pm GB_O$

 

Use (-) if G is above BO and (+) if G is below BO. Note that M is always above BO.
 

Value of MBO
Assume that the body is rectangular at the top view and measures B by L at the waterline when in upright position. The moment due to the shifting of the buoyant force is equal to the moment due to shifting of wedge.
 

floating_stable-pos.gif

 

$BF \, z = Fs$

$\gamma V_D \, (MB_O \sin \theta) = (\gamma v)s$

$V_D MB_O \sin \theta = vs$

$MB_O = \dfrac{vs}{V_D \sin \theta}$

 

$MB_O = \dfrac{[ \, \frac{1}{2}(\frac{1}{2}B)(\frac{1}{2}B \tan \theta)L \,](\frac{2}{3}B)}{V_D \sin \theta}$

$MB_O = \dfrac{\frac{1}{12}LB^3 \tan \theta}{V_D \sin \theta}$
 

For small value of θ, tan θ ≡ sin θ and note that 1/12 LB3 = I, thus,

$MB_O = \dfrac{I \sin \theta}{V_D \sin \theta}$

$MB_O = \dfrac{I}{V_D}$

The formula above can be applied to any section.
 

Where
W = weight of the body
BF = buoyant force
M = metacenter
G = center of gravity of the body
BO = center of buoyancy in upright position
BO' = center of buoyancy in tilted position
MG = metacentric height or the distance from M to G
MBO = distance from M to BO
GO = distance from G to BO
v = volume of the wedge either immersion or emersion
s = horizontal distance between the center of gravity of the wedges
θ = angle of tilting
I = moment of inertia of the waterline section of the body
RM = righting moment
OM = overturning moment
 

For rectangular section

$MB_O = \dfrac{B^2}{12D}\left(1 + \dfrac{\tan^2 \theta}{2} \right)$

 

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