Section 1 Basic Principles.pdf

SECTION

1 -BASIC

PRINCIPLES

INTRODUCTION

This section introduces the laws governing flotation and will help in the understanding of why shipsfloat.

It will form the basic level of understanding necessary to complete this learning program.

Learning Objectives

On completion of this section the learner will achieve the following:

Understand the terms Density, Mass and Volume and be able to complete simple

calculations relating to these terms.2.

Understand the laws governing flotation.3.

Understand the change in draught/freeboard that will occur when a box-shaped vessel

moves between water of different densities.4.

Applies (2) and (3) to calculations based on the flotation of box-shaped vessels.

CLASS 2/1 STABILITY -SECTION

1 Basic Principles

MASS AND VOLUME

1.1.1 Density

The density of any given substance is its mass per unit volume.

This can be expressed as:

r DENSiTY = -_~_=l

VOLUME I

For ship stability purposes the units commonly used are:

mass:

volume:

density:

tonnes (t)

cubic metres (m3)

tonnes per cubic metre

(t/m3)

Rearrangingthe aboveformulagives:

I VOLUME =.M8.§§.

l",n",D~~J

and:

Example 1

A piece of steel measures

0.1 m x 2.2 m x 6.0 m and has a density of 7.80 t/m3. Calculate its mass.

Solution

Mass = Volume x Density

Mass = (0.1 x 2.2 x 6.0) x 7.80

Mass = 10.296 tonnes

Exam/Jle 2

A block of aluminium measures 0.8 m x 0.6 m x 0.3 m and has a mass of 0.389 tonnes. Calculate

the density of the aluminium.

Solution

Mass = Volume

x Density

Density

ME.§..§..

Volume

Therefore:

(0.8 x 0.6 x 0.3)

= 2.701t/m3

Density =

0.144

Exam/Jle 3

A rectangular ballast tank is 12 m long, 8 m wide and has a depth of 4 m. Calculate the mass of

salt water ballast, density 1.025 t/m3, that can be loaded into the tank.

Solution

Mass = Volume x Density

Mass = (12 x 8 x 4) x 1.025

Mass

= 393.6 tonnes

Example 4

A fuel oil tank has length 4.2 m, breadth 3.4 m and a depth of 6.0 m. If 50 tonnes of fuel oil (density

0.84 t/m3)is loaded what will be the sounding (level) of oil in the tank?

Mass = Volume x Density

50 = (4.2 x 3.4 x sounding)

Solution

x 0.84

4.2 x 3.4 x 0.84

Sounding =

= 4.168 m

CLASS 2/1 STABILITY -SECTION

1 Basic Principles

1.1

DENSITY,

1.1.2 Relative Density (RD)

Quite otten the Relative

Density

(RD) of a substance

is quoted instead of Density.

This is simply a

ratio of the density of the substance

in question to that of Fresh Water.

The density of fresh water is 1.000 t/m3.

In the previous example the density of the oil was 0.84 t/m3. The relative density of the oil was

0.84; in other words, the density of the oil is 0.84 times that of fresh water!

1.1.3 Density of water in which a ship typically floats

A ship is presumed to always float in water that lies in the following density range:

Fresh water (FW):

Salt water (SW):

1.000 t/m3

1.025 t/m3

(RO 1.000)

(RO 1.025)

Water that lies between these two extremes

is termed Dock Water (OW).

If a question states that a ship is floating in salt water (SW) then it can be always assumed that the

water density is 1.025 Um3.

if in fresh water (FW) then a density of 1.000 t/m3 can be assumed.

CLASS 2/1 STABILITY -SECTION

1 Basic Principles

Similarly,

1.2

TH E LAWS OF FLOTATION

1.2.1 Archimedes'principle

This states that when a body is wholly or partially immersed

in a liquid, it experiences

an upthrust

(apparent

loss of mass -termed

Buoyancy force (Bf)), equal to the mass of liquid displaced.

Consider

a block of steel measuring

2 m x 2 m x 2 m that has a

density of 7.84 Um 3.

Example 5 (a)

If the block were to be suspended by a ship's crane that has a very

accurate load gauge, what mass would register on the gauge if the

block were suspended over the ship's side in air?

Solution (a)

The block is suspended

Fig. 1.1

in air!

Since:

Mass = Volume x Density;

Mass of the block = (2m x 2m x 2m) x 7.84 t/m3

= 62.72 t

The crane driver now lowers the block so that it becomes

half

submerged

in the dock water that has a density of 1.020 t/m3.

Example 5 (b)

What mass will the gauge now indicate?

OQpKWA~

DENSIT¥ ..

1,~2I)tI'"

Solution (b)

The block is now displacing

Fig 1.2

a volume of water

where: Volume

of water displaced

= (2m x 2m x 1 m) = 4m3

Mass of water displaced = Volume x Density of the dock water;

= 4 m3 x 1.020 t/m3

= 4.08 t which represents

the upthrust due to

buoyancy force (Bt) created by the displaced water.

Mass of block

62.72 t

Upthrust

due to Bf

4.08 t

t2:J

GauGe readinG

58.64 t

Example 5 (c)

What mass will the gauge indicate if the crane driver now

lowers

the block

so that it is completely

submerged

the dock water?

OOCKWATeR

DENSITY

1,CWiJrn$

Solution (c)

The block is now displacing a volume of water where:

Volume of water displaced

= (2m x 2m x 2m)= 8m3

Fig. 1.4

CLASS 2/1 STABILITY -SECTION

1 Basic Principles

]

Mass of water displaced

= Volume x Density of the dock water;

= 8 m3 x 1.020 t/m3

= 8.16 t which represents

the upthrust of the buoyancy force (Bt)

created

by the displaced

water.

(.."

-~""---

Mass of block

62.72 t

Upthrust due to Bf

8. 16 t

Gauqe readinq

54.56 t

Oem1TY

tt!lfJJ

"'._,"~ "","-"",,(~.".*

Fig.1.5

~~"' I ~ ~ ~~ ~, ~!<""

1.2.2 Law of flotation

This states that every floating body displaces it's own mass of the liquid in which it floats.

The displacement

of a ship (or any floating

object) is defined

as the number

of tonnes

of water it

displaces.

It is usual to consider

a ship displacing

salt water of density 1.025 tfm3, however,

fresh

water values

of displacement

(1.000 tfm3) are often quoted in ship's hydrostatic data.

The volume of displacement is the underwater volume of a ship afloat i.e. the volume below the

waterline.

To calculate the displacement (W) of a ship the following needs to be known:

The volume of displacement (V)

The density of the water in which it floats (p)

Since:

MASS = VOLUME x DENSITY

the mass, or displacement, of a ship is calculated by:

DISPLACEMENT = VOLUME OF DISPLACEMENTx WATER DENSITY

i.e.

~~~~J

1.2.3 Draught and Freeboard

Consider the ship shown.

1'1

""1,,,[=

~~FP

~=:t~"":

=~=~~ ~_..

Fig. 1.6

Draught is the distance from the keel to the waterline (WL), as measured at the forward and aft

ends of the ship. (More precisely the draught readings are taken as those read at the Forward and

Aft Perpendiculars -these terms are defined in Section 12). It is expressed in metres. If the

draughts forward and aft are the same then the ship is said to be on an even keel (as shown).

Freeboard is the distance between the waterline (WL) and the top of the uppermost continuousdeck.

It is usually expressed in millimetres and is measured amidships.

1.2.4 Reserve Buoyancy

This is the volume of the enclosed spaces above the waterline. Because reserve buoyancy is a

very important factor in determining a ship's seaworthiness minimum freeboards are assigned to a

ship to ensure that there is adequate reserve buoyancy at all times.

CLASS 2/1 STABILITY -SECTION

1 Basic Principles

:=:1

.-?1

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