Unformatted text preview: oduced,
GM
• Equilibrium is unstable,
Equilibrium unstable
and
• GM is regarded as
GM
negative
negative
• If M coincides with G,
If coincides
• Body is in neutral
Body
equilibrium
equilibrium W = mg
G
B R=W
θ θ
G G
M B1 Overturning
moment M B1 x
RW Determination of The
Metacentric Height • Metacentric height of a
vessel can be determined
if the angle of tilt θ cause
by moving a load P a
know distance x across
the deck is measured P • Overturning moment due to
movement of load P = Px
• If GM is the metacentric
height, and
• W = mg is the total weight of
the vessel including P
Righting moment = W x GM x θ θ
x
P P • For equilibrium in tilt position,
the righting moment must
equal to overturning moment,
So that,
W x GM x θ = Px
Metacentric height,
GM = Px/Wθ
The true metacentric height is
the value GM as θ
0 θ
x
P P Determination of The Position of
The Metacentre Relative To The
Centre of Buoyancy • Metacentric height GM
determine the stability of
a floating body
• For a vessel of known
shape and displacement,
• Position of the centre of
gravity G and buoyancy
B is known
• So GM can be found by
knowing the distance BM
GM = BM – BG O
Small
area a x
O
M
A
A’ D
D’ C’ O
G
B C
B’ Upthrust, R = W θ O • AC is the original waterline
plane, and
• B is the centre of buoyancy in
equilibrium position
• When the vessel is tilted
through a small angle θ,
• Centre of buoyancy will move
to B’ (result in alteration in the
shape of the displaced fluid) Small
area a x
O
M
A
A’ D
D’ C’ O
G
B C
B’ Upthrust, R = W θ O • A’C’ is the waterline plane in
the displaced position
• For small angle of tilt,
BM = BB’/θ Small
area a x
O • Movement of B to B’ is the
result of removal of fluid
volume AOA’ and addition of
COC’
• Total weight of fluid displaced
remains unchanged M
A
A’ D
D’ C’ O
G
B C
B’ Upthrust, R = W θ Weight of
wedge AOA’ = O Weight of
wedge COC’ • a is a small area in the
waterline plane at a distance x
from the axis of rotation OO
• It generates a volume when
the vessel is tilted Small
area a x
O
M
A
A’ Volume swept = DD’ x a = axθ
out by a D
D’ C’ O
G
B C
B’ Upthrust, R = W θ O Summing all such volumes and,
Multiply by the specific weight
ρg of the liquid,
Weight of
=
wedge AOA’ Small
area a x x = AO ∑ ρgaxθ O x =0 M
A Similarly, A’ x = CO Weight of
=
wedge COC’ ∑ ρgaxθ
x =0 D
D’ C’ O
G
B C
B’ Upthrust, R = W θ O Since there is no chan...
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 Fall '11
 LamersAKBP
 Buoyancy, Force, upthrust, floating bodies

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