1477 Kojex 12 13. Laws of Buoyancy

8/6/2019 1477 Kojex 12 13. Laws of Buoyancy

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A.A. B. Dinariyana

JurusanTeknikSistem PerkapalanFakultasTeknologi Kelautan ITS SurabayaTA 2009-2010


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A basic requirement for any ship is that itmust have sufficient buoyancy to support theweight of the ship.

Therefore, ships must satisfy not only thebuoyancy condition but also a staticequilibrium condition.

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Archimedes principle states that when abody is wholly or partially immersed in a fluidit appears to suffer a loss in mass equal to themass of the fluid it displaces.

The mass density of fresh water is 1000 kgper m 3. Therefore, when a body is immersed

in fresh water it will appear to suffer a loss inmass of 1000 kg for every 1 m 3 of water itdisplaces.

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Since the actual mass of the box is not changed, there must be aforce acting vertically upwards to create the apparent loss of massof 1000 kg.

This force is called the force of buoyancy, and is considered to actvertically upwards through a point called the centre of buoyancy.The centre of buoyancy is the centre of gravity of the underwatervolume.


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The box has a mass of 4000 kg, but has a volume of 8 m 3.If totally immersed in fresh water it will displace 8 m 3 of water, and since 8 m 3 of fresh water has a mass of 8000 kg, there will be an upthrust or force of buoyancycausing an apparent loss of mass of 8000 kg. Theresultant apparent loss of mass is 4000 kg.When released, the box will rise until a state of equilibriumis reached


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Now consider the box to be floating in fresh water with half itsvolume immersed.

If a mass of 1000 kg be loaded on deck the new mass of the body

will be 5000 kg, and since this exceeds the buoyancy by 1000 kg, itwill move downwards.

The downwards motion will continue until buoyancy is equal tothe mass of the body. This will occur when the box is displacing 5m3 of water and the buoyancy is 5000 kg.


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W

B

G

G

B

A ship afloat in the water in static equilibrium.

The integration of upward components of hydrostatic pressuresover the surface of the body, or the buoyant force is equal to theweight of the displaced water ( ).

This force must be exactly balanced by the gravitational force of the bodys mass (W), directed downwards.

W


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The TPC for any draft is the mass which must beloaded or discharged to change a ship's meandraft in salt water by one centimeter.

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WPA = AWP


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Consider a ship floating in salt water at the waterline WL.Now let a mass of `w' tonnes be loaded so that the meandraft is increased by one centimeter. The ship thenfloats at the waterline W1 L1.Since the draft has been increased by one centimeter, themass loaded is equal to the TPC for this draft.The mass of water in the layer between WL and W1 L1 isalso equal to the TPC


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The draft at which a rectangular homogeneouslog will float may be found as follows:

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Find the distance between the centers of gravity andbuoyancy of a rectangular log 1.2m wide, 0.6 m deep,and of relative density 0.8 when floating in freshwater with two of its sides parallel to the waterline.

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It is well known that a ship floats at a deeper draughtin fresh water than it does in seawater. fresh wateris less dense than seawater, thus it requires a largevolume of fresh water to produce the same buoyantforce

The ship will also trim in order to keep LCB and LCG inthe same vertical line.

In order to determine the changes in draught whenmoving from seawater to fresh water we can use thetons per centimeter immersion concept.

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tAWP

F

B

t

The volume of the layer may be expressed as the differencebetween displacements in fresh water and seawater oralternatively by the waterplane area multiplied by the parallelsinkage. Equating the two equations

Since the weight of the ship remains constant

Substituting into the above equation

tA WPSF

F

SSFSSFF

1A

ttAF

S

WP

SWPS

F

SS


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A container ship, 161 meter long by 23.2 meterbeam, floats at a pier where the density of thewater is 1.010 t/m 3.It is desired to load this ship at the pier to adraught such that upon entering seawater thedraught will be 8.75 meter.For seawater at a draught of 8.75 meter, thedisplacement is 19420 tons and TPC is 27.62t/cm.To what draught should the ship be loaded atthe pier?

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Displacement volume

Waterplane area

Parallel sinkage

the draught at the pier may be 0.104 meter more than therequired seawater draught, so the ship may be loaded to a draught

of T at the pier : 8.75 + 0.104 = 8.854 m

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3

S

m18946025.1

19420

2

S

1WP m6.2694

025.1

62.27100T100A

m104.01010.1

025.1

6.2694

189461

At

F

S

WP


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A 12300 ton barge is 100 meters long and has constantcross section as shown below

At what draught will it float in seawater and infreshwater?

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8 m

6 m

16 m


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Displacement volume in sea water

Waterplane area consists of triangular prism and arectangular prism

The draught in sea water is T=10.5 . The parallel sinkage infresh water:

The draft in fresh water is T =10.5+0.1875=10.6875 m

3

S

m12000025.1

12300W

12000)6t(1610061610021

m1875.01

1

025.1

16100

120001

A

tF

S

WP


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Keel (K) :The base line reference point from which all otherreference point measurements are compared.

Centre of Buoyancy (B) : The geometric centre of the shipsunderwater hull body. It is the point at which all the forces of buoyancy may be considered to act in a vertically upwarddirection.

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B

G

M

K


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Centre of Gravity (G) : The point at which all forces of gravityacting on the ship can be considered to act. The position of thecentre of gravity depends upon the distribution of weights withinthe ship.Metacentre (M) : The intersection of the lines of buoyant forces asthe ship heels through small angles of heel.

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B

G

M

K


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Ship Stability for Masters and Mates, FourthEdition, Revised, D.R. Derrett, B-H Newnes,1990

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1477 Kojex 12 13. Laws of Buoyancy