Regarded as gm negative negative if m coincides with

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online