- Universiti Teknologi Malaysia

Transcription

[Type text]
SHIP HYDROSTATICS
AND STABILITY
A SYSTEMATIC APPROACH
Omar bin Yaakob
A systematic Approach
Table of Content
Table of Content ........................................................................................................................ii
Preface....................................................................................................................................... iv
Chapter 1 Ship Types, Basic Terms, Terminologies and Symbols........................................... 1
1.1
Introduction ................................................................................................................. 1
1.2
Types of ships.............................................................................................................. 1
1.3
Basic Terms, Terminologies and Symbols ................................................................. 7
1.3.1
Reference positions: ............................................................................................. 7
1.3.2
Linear Dimensions ............................................................................................... 8
1.3.3
Size of Ships ...................................................................................................... 10
1.3.4
Form Coefficients .............................................................................................. 12
1.3.5
Centroids ............................................................................................................ 13
1.4
Ship Lines Plan.......................................................................................................... 14
1.4.1
The importance of Ship Lines Plan .................................................................... 14
1.4.2
Body Plan ........................................................................................................... 16
1.4.3
Half Breadth Plan ............................................................................................... 18
1.4.4
Profile / Sheer Plan ............................................................................................ 20
1.4.5
Offsets Data ....................................................................................................... 21
1.5
Ship Geometry Coordinate System ........................................................................... 23
Chapter 2 Hydrostatics and Floatation ..................................................................................... 24
2.1
Archimedes Principles of Floatation ......................................................................... 24
2.2
Reduction of Weight of Immersed Objects ............................................................... 24
2.3
What makes a ship float? .......................................................................................... 27
2.4
Effect of Density ...................................................................................................... 27
2.5
Some Simple Problems ............................................................................................. 28
2.6
Tonne per centimeter immersion (TPC).................................................................... 31
2.7
Hydrostatics Particulars............................................................................................. 32
2.8
Hydrostatic Particulars of a Ship ............................................................................... 34
2.9
Using Hydrostatic Curves and Tables ....................................................................... 36
2.10
Bonjean Curves...................................................................................................... 37
2.11
Cross Sectional Area Curve ................................................................................... 38
2.12
Second Moments of Areas ..................................................................................... 38
Exercises .............................................................................................................................. 43
Chapter 3 Basic Stability Consideration .................................................................................. 45
3.1
Introduction ............................................................................................................... 45
3.2
What is stability? ....................................................................................................... 45
© Universiti Teknologi Malaysia, August 2012
ii
3.3
Longitudinal and Transverse Stability ...................................................................... 46
3.4
Basic Initial Stability: The role of GM...................................................................... 47
3.5
Determining the Centre of Gravity of ships after loading ......................................... 49
3.6
Effect of movement or addition of weights on centre of gravity .............................. 51
3.7
Hanging Weights, The Use Of Derricks And Cranes ............................................... 53
3.8
Free Surface Correction ............................................................................................ 54
3.8.1 The Effect of Free Surface on Ship Stability .......................................................... 54
3.8.2 Calculating Second Moment of Area ...................................................................... 55
Exercises .............................................................................................................................. 56
Chapter 4
Transverse Stability ........................................................................................... 58
4.1
List due to movement of weights onboard ................................................................ 58
4.2
Finding list after loading and unloading ................................................................... 61
4.3
Correcting Lists by moving or adding weight ........................................................... 63
Exercises .............................................................................................................................. 64
iii
Preface
Man has benefited from the sea in various ways. The sea has been the source of food,
ornaments as well as provided means of transportation. Conquests and defences
have been carried out on water. Water-based leisure activities are becoming more
common and varied. Lately, man has innovated the use of the sea beyond those
traditional applications. The sea has now become a primary source of petroleum,
gas and lately marine renewable energy.
To carry out various activities at sea, rivers and lakes, man uses various types of
marine structures, fixed and floating. The structures normally provide safe and
stable platforms upon which the activities are carried out. They must be designed
and built in various sizes, shapes and sophistication. Some of them are small and
simple such as a canoe or a raft while others are large and complicated such as an
aircraft carrier or a semi-submersible oil drilling platform.
Naval architecture is an engineering field covering the technology in design of ships
and floating structures ensuring that they can perform their stipulated functions and
missions effectively and safely. The persons having this expertise are called naval
architects. To build these structures, shipbuilders requires design plans and
guidelines prepared by naval architects. Knowledge in naval architecture is used to
carry out design calculation and to produce plans which can be used by the
shipyards.
Although man has been using marine transport for a long time, not all these vehicles
are designed and constructed using naval architecture knowledge. In fact the
discipline of knowledge on ship design and naval architecture only appeared in the
seventeenth century. Prior to that, shipbuilding is not based on science and
technology but rather on the skills of the master craftsmen.
This dependence on master craftsmen for shipbuilding can be traced back to the
earliest civilization of Egypt, Greek and China. Similarly the warships and
exploration vessels built by the Romans, Muslims as well as the European colonial
powers were not built using scientific methods.
By the seventeenth century a number of scientists and engineers tried to apply
science and mathematical methods in ship design. Among the earliest was Sir
Anthony Deane who wrote Doctrine of Naval Architecture in 1670. Among others, he
put forward a method to determine the draught of the ship before it was built, a
technique which formed the basis of we now understand as hydrostatics . Since
then, a number of scientists and engineers continued to study and document various
fields of naval architecture. In 1860, a professional body comprising of naval
architects was formed under the name Institution of Naval Architects. A hundred
years later the name was changed to Royal Institution of Naval Architects.
iv
A naval architect works to determine the size and shape of a ship tailored to its
intended use. In addition, he estimates its stability, propulsive power as well as
calculates the size and strength of its structure and the impact of waves on the
vessel. The types of machinery and equipment to be installed, materials to be used
and layout of ship are also determined based on naval architectural knowledge.
This book will cover only one aspect of naval architecture, that is ship hydrostatics
and stability since it is one of the most important subject in naval architecture. The
safety of ships, crew, passengers and cargo will be jeopardised if ships are not stable.
This book is written to enable readers to appreciate the basic terminologies, carry out
simple hydrostatics calculations and to equip them with basic tools to assess
stability of vessels.
Many books have been written on naval architecture, ship hydrostatics and stability.
Most books however are difficult to follow, due to the unsystematic way the
materials are presented.
This book differs from the rest in the sense that the content is not presented as
discrete topics, unrelated to each other. Instead, it is presented in a systematic and
logical manner.
It introduces the basic concepts and develops the understanding in a continuous
progression of knowledge acquisition as well as confidence building through handson calculations examples and exercises.
The Chapters are interrelated, as do the Sections. Readers can easily see the
relationships between each Chapters and Sections. Materials which are important
but not fit to be in the logical flow of the content are put in the appendices.
This book is also designed to be a self-teaching and self-explanatory book.
Knowledge is built in an incremental manner. Readers can read and understand the
step by step explanation and solutions to the problems and able to apply their
knowledge by solving the exercises provided.
In writing this book, I have depended on many persons. In particular I am thankful
to Ir Dr Mohamad Pauzi, Dr. Koh Kho King and Haji Yahya Samian who have
contributed to some of the sections. There are many others who have assisted in
various ways.
v
Chapter 1 Ship Types, Basic Terms,
Terminologies and Symbols
1.1 Introduction
1.2 Types of ships
There are various ways of categorizing ships. Ship types can be classed according to
a number of criteria such as the number of hulls, hull form shapes, the way it is
supported in water, and its mission/function.
1. Number of Hulls
Ships can be categorized in terms of the number of hulls. Most ships have only a
single hull; these are called mono-hulls. Some ships have multiple hulls such as
catamaran and trimaran.
Figure 1 A mono hull Fishing Boat
1
Figure 2 A Catamaran
2. Shape of hull form
The shape of the hull are different from one ship and the other. Most large slow
ships have round-bilge hull form while smaller faster boats normally have chine
hulls. The fishing boat in Figure 1 has a round-bilge while the Catamaran in Figure 2
has chine hull form. Chine hulls consists of two or more almost flat surfaces, the line
connecting the surfaces is called the chine line. When the hull is made up of two
surfaces, then there is a single chine. Double chine vessels have three surfaces. Chine
hulls are also called V-shaped hulls while round-bilge hulls are called U-shaped
hulls.
Figure 3 Mono hull planing craft with single chine
3. How the body is supported in water
When a ship is in water, the total weight of the ship is being supported by various
forces, depending on the types of hullform. Round-bilge hull forms are normally
supported hydrostatically, i.e. all the weight of the vessel is supported by buoyancy
forces which equals the total weight of water displaced by the vessel. These are also
called displacement hull.
2
Figure 4 Marine Craft Support Triangle
Volker Bertram , Overview of High-Performance Marine Vehicles as Naval Platforms , Volume 38, Number 2 Summer 2008
Hydrofoils are an examples of vessels supported by the dynamic lift due to its flat
lower foils. At high speeds, the lift forces provided by the foils is enough to support
the ship, lifting it out of water. At these speeds, the hydrostatic buoyancy forces are
insignificant. By lifting the body above the water surface, the drag of water on the
hull is reduced and the vessel can travel at high speeds.
3
Figure 5 High speed Hydrofoil Ferry
Another kind of ships are the hovercrafts, which operates above the water surface.
Air cushion is provided by large fans which pump air to lift the vessel above the
water surface. The total weight of the vessel is supported by the air cushion,
sometimes referred to as aeropowered lift forces.
Figure 6 A Hovercraft Passenger Ferry
Many vesssels have combinations of support. For example, when a chine hull vessel
is stationary, it is hydrostatically supported. However, when it starts to move and
reaches a certain speed, water moving along the lower hull will lift the vessel,
reducing the hydrostatic buoyancy forces. In this case, at the cruising speeds the
vessel is supported both by a combination of hydrostatics and hydrodynamic forces.
Chine-hull vessels which operate at high speeds using partial hydrodynamic
support are called planing hull vessels, see Figure 3.
4
4. Its function/mission
Ships can also be categorized according to their functions i.e. how they are used for
the benefit of mankind. For example, some ships are meant for transport such as
crude oil tankers, bulk carriers, containerships, passenger ships, general cargo ships,
liquefied natural gas (LNG) carriers. The ships used in the navy may take various
size, dimensions and functions such as aircraft carrier, submarine, frigate, destroyer,
patrol craft, and minesweeper. Some ships are not meant to carry cargo but to carry
out certain service at sea. Examples of work or service vessels include tugs, supply
boats, crew boats, heavy lift, crane ships, fishing boats and rescue boats. Some other
boats are used for recreational purpose such as luxury yacht, cruise ships, tourists
boats.
Figure 7 This 396m long containership is one of the largest ship in the world
5
Figure 8 A crude oil tanker
6
1.3
Basic Terms, Terminologies and Symbols
The terms and terminologies used in naval architecture are unique. Proper and
uniform understanding is important since these terms and symbols will be used not
only in hydrostatics and stability calculations but also in various other naval
architecture calculations such as ship resistance and propulsion, ship structure and
ship hydrodynamics. These symbols and terms are used by those in the academic
world as well as by the practioners in the industry. Therefore proper understanding
is important to ensure smooth and efficient communications.
1.3.1 Reference positions:
Distances and relative locations are measured from certain reference positions on
ships.
Design Waterline (DWL)
Waterlines are lines of the water surface at which the ship is expected to float at.
DWL is the waterline at which the ship is expected to float at its fully loaded or
operational condition.
After perpendicular(AP)
AP is the line which is perpendicular to the intersection of the after side of the
rudder-post with the DWL. For some ships without rudder-post, the AP is taken as
the centre-line of the rudder stock or the intersection of the DWL with the transom.
Forward perpendicular (FP)
A line drawn perpendicular to the intersection of the DWL with the forward side of
the stem.
Amidships or Midships (
)
The point midway between the forward and after perpendiculars.
Base line
The lowest part of the ship, normally the underside of keel where a horizontal line is
drawn. This becomes a reference line for measurements in the vertical direction.
7
1.3.2 Linear Dimensions
Important linear dimensions of the ship are shown in Figure 9.
Figure 9 Linear Dimensions
Length between perpendiculars (LBP or LPP)
The horizontal distance between AP and FP. This is the most important length
measurement during ship design development stages. Most calculations such as
stability, propulsion, maneuvering use LBP.
8
Length on the design load water-line (LWL)
The length on the water-line of the ship when floating in still water at DWL. In many
cases, this is similar to LBP and also important during calculations in the deisgn
stages.
Length overall (LOA)
The length measured from the extreme point forward to the extreme point aft. This
length is an important measure during operational stage of the ship.
Breadth or Beam (B)
The maximum breadth or beam of the ship is usually measured at amidships. Some
ships have the largest breadth not at amidships.
Depth (D)
The vertical height of the uppermost continuous deck measured at the side
amidships from the base line.
Draught or Draft (T)
The depth of immersion from baseline to any waterline.
Freeboard
The height of the deck at side above the LWL. It is equal to the difference between
the depth and the load draught.
Trim
The difference between the draughts at AP and FP. If the draught forward is greater
than the draught aft it is called trim forward, by the head, or by bow. If the draught
aft is greater, it is called trim aft, by stern. Ships without trim are said to be level
keel or even keel. Trim are sometimes stated as trim angles, θ .
Moulded, Extreme and Displacement Dimensions
Moulded linear dimensions refer to measurements of inner dimensions of the ship,
i.e. the measurements neglect the thickness of plating. Subscripts mld are used.
Moulded breadth for example (Bmld) is the breadth measured between the inside
plating on the two sides of the ship, moulded depth (Dmld) and draughts (Tmld) are
measured from the top of the keel plate. Moulded dimensions are normally used
during ship construction process, especially when lofting is carried out. In
hydrostatics and stability calculations, the outer or extreme dimensions are used.
These are given the subscript ext, for example Text. These dimensions consider the
9
surfaces which are in contact with water. Unless otherwise stated, all dimensions
without subscripts are normally referred to extreme dimensions.
1.3.3 Size of Ships
The size of ships are normally stated in terms of displacement, deadweight or gross
registered ton (GRT). Some ships such are measured in terms of their carrying
capacity, for example TEUs (Twenty feet Equivalent Units) for containerships, the
number of cars for car carriers or the number of heads for livestock carriers.
In ship design and operation field, the term “weight” and “mass” are used
interchangeably and units of mass (tonnes/tons/pounds/kilograms) are normally
used. Although this is not strictly correct, the net impact is the same and as long as
consistency is maintained, there should not be any problem.
Volume Displacement (  )
A floating ship displaces water. The volume of water being displaced is the amount
of water ‘pushed aside’ by the ship. The volume of water is called the volume
displacement of the ship usually expressed in m3.
Mass Displacement (  )
When a ship is floating, it is displacing the volume of water whose weight or mass is
equivalent to its own weight or mass. The total weight of the vessel is the same as
the weight of water being displaced. So the term ‘mass displacement’ is the same as
the total weight or mass of the ship in sea water, normally expressed in units of
tonnes or kilogram.
This weight of water equals the volume displacement multiplied by the density of
water,
In fresh water = x 1000 kg/m³
In sea water = x 1025 kg/m³
Most of the time, the term ‘displacement’ refers to mass displacement.
Mass displacement or weight of the ship is equal to the sum of lightship weight and
deadweight. The operational displacement of the ship or its total weight will
actually vary from time to time. While the lightship weight is constant, the
deadweight and hence the displacement will vary from time to time during
operation depending on the loading conditions of the ship.
10
Lightship
Lightship weight is the weight of an empty ship, without cargo, crew, water, fuel
and other payload components. It is normally associated with a ship that has just
been built and ready to sail. It is the non-variable component of the mass
displacement.
Deadweight
The difference between the mass displacement and the lightship weight is called the
deadweight. This is the variable component of displacement and includes cargo,
fuel, crew, passengers, stores, etc, expressed in tonnes. The sizes of tankers and bulk
carriers are often quoted in terms of the deadweight tonnage, which is the maximum
deadweight the ship is designed to carry. Since the weight of the cargo make up
most of the deadweight, the deadweight tonnage is a good measure of the cargo
carrying capacity of the tankers and bulk carriers.
Displacement tonnage
The designed total weight of the ship is called ship displacement tonnage and this is
normally stated in the ship particulars. The sizes of non-cargo carrying ships such as
ships belonging to government agencies
are normally stated in terms of
displacement tonnage.
Gross tonnage (GRT)
Although the terms ‘tonnage’ and ‘tons’ are used, GRT is not a measure of weight.
Instead, gross tonnage is the total volume of enclosed spaces in a ship including the
under-deck and the enclosed space in the superstructure of the ships. Due to its
history of its use, although its unit is tons, it is a measure of volume, not weight
where 1 ton is equivalent to 100 ft3. The sizes of most commercial vessels are stated
in terms of gross tonnage.
Net or register tonnage
GRT is the total volume of enclosed spaces onboard a ship, NRT is the net volume
after deductions of non-freight earning spaces such as engine room and crew
accommodation. Net tonnage are used when charges are levied for services are
provided for the commercial ship, for example for pilotage and port charges.
11
1.3.4 Form Coefficients
When comparing one ship's form with another, the naval architect makes use of a
number of coefficients. These coefficients of forms are used as a general term to
describe the fullness or the fineness of the ship hulls. These coefficients are
important and used in power, stability, strength and design calculations. Coefficient
of forms are stated in terms ratios between the actual area or volumes divided the
are or volume of the circumscribing box. The higher the value, the fuller the ship
form. However the values do not exceed 1.
Block coefficient (CB)
This is a measure of the fullness of the form of the ship and is the ratio of the volume
of displacement to a given water-line, and the volume of the circumscribing box
having the same length, breadth and draught as the ship.
ie: CB =  ÷ (L x B x T)
CB varies between ships. Typically slimmer, faster ships such as frigates and patrol
crafts have CB between 0.5-0.65 while slower full form ships such as tankers and bulk
carriers have CB 0f around 0.7-0.85.
Midship section area coefficient(CM)
This is the ratio of the midship section area to the area of the circumscribing
rectangle having a breadth equal to the breadth of the ship and a depth equal to the
draught.
ie: CM = AM ÷ (B x T)
Water-plane area coefficient(CWP)
This is the ratio of the area of the water-plane to the area of the circumscribing
rectangle having a length equal to the LPP and a breadth equal to B.
ie:
CWP = AW ÷ (L x B)
Prismatic coefficient (CP)
The ratio of the volume of displacement of the ship to the volume of the
circumscribing box having a constant section equal to the immersed midship section
area AM, and a length equal to the LwL
i.e. CP =  ÷ (AM x L)
12
The above is the most typical prismatic coefficient, sometimes called longitudinal
prismatic coefficient, because it is a measure of the longitudinal distribution of
displacement of the ship.
In certain cases, vertical prismatic coefficient CPV is calculated
CPV =  ÷ Awp x L
1.3.5 Centroids
The location of centroids of areas, volumes and weights are important in
hydrostatics and stability calculations.
Centre of flotation (F)
When a ship trims at small angles of trim, the ship is pivoting about a transverse
axis passing through the centre of floatation, F. When viewed from the side,
consecutive water-lines are assumed to be passing through the centre of floatation.
The centre of floatation coincides with the centre of the waterplane area at that
draught. The location of F is measured longitudinally from the references axes, either
amidships, AP or FP. This distance is called longitudinal centre of floation (LCF).
Centre of buoyancy (B)
The single buoyancy force representing the summation of all hydrostatic forces
acting on a ship is considered to act upwards through a single point called the centre
of buoyancy (B). This coincides with the centroid of the underwater volume of a
ship. Its position is defined by:
(a) Vertical centre of buoyancy (VCB) which indicates the location in the vertical
direction. The reference line must be stated. In normal practice, the keel line is used
as the reference line and in this case, this height is stated as KB which is the vertical
distance above keel.
(b) The longitudinal distance measured either from amidships or AP or FP is called
the longitudinal centre of buoyancy (LCB).
Centre of gravity (G)
This is the point through which the total weight of the ship may be assumed to act.
Similar to the centre of buoyancy, the location of centre of gavity also is defined by:
(a) Vertical centre of gravity (VCG) which indicates the location in the vertical
direction. In normal practice, KG is used where the keel line is used as the reference
line.
13
(b) Longitudinal centre of gravity (LCG) which is
measured either from amidships or AP or FP.
1.4
the longitudinal distance
Ship Lines Plan
1.4.1 The importance of Ship Lines Plan
Ship has a complex and unique hull shape due to its double curvature and nonhomogeneous cross sections. Unlike simple object like cylinder, box, and cone which
can be represented in simple orthographic drawing, ship hull require special way of
representing its unique and complex shape. Not only it require to be shown in three
different orthogonal views, more lines are also needed in order to represents its
shape at different cross sections or planes. For this reason, the ship hull drawing is
always called as Lines Plan Drawing.
Lines Plan Drawing is a lines drawing that represent the shape of the ship hull
looking from three orthogonal (perpendicular to each other) views i.e. front, side and
top views. The front view is termed as Body Plan, the side view is the Sheer Plan
and the top view is the Half Breadth Plan. Since all of these views represent the
same hull, they are interrelated to each other, thus the preparation of lines plan
drawing must follow certain standard procedure.
Lines plan drawing has always regarded by the naval architects as the most
important piece of information about the ship. This is due to two reasons i.e. the ship
performance and ship design process. On the performance of the ship, the shape of
the hull form determines the power required to drive the ship, thus reflect the ship
speed, its also determine the amount of pay load (capacity), comfort, habitability, etc.
On the ship design process, lines plan drawing is the first information that needs to
be made available. Without lines plan drawing, no calculation, design and analysis
works can be performed. Construction process also can only be commenced after the
lines plan drawing is completed.
Some samples of the various hull form are shown in Figure 10 to 12.
14
Figure 10: Body plan of a displacement hull (Container Ship)
Figure 11: Body plan of a planning hull (Vee hull with hard chine)
15
Figure 12: Body plan of a catamaran
1.4.2 Body Plan
Body Plan represents the shape of the ship hull when viewing from the front or rear
of the ship at every ship stations as shown in Figure 10 and 13. Station is a
transverse cross-section along the ship length which normally equally spaced. The
body plan concept can be better understood by referring to Figure 14. A ship is
normally divided into 11 or 21 stations from after perpendicular, AP (Sometimes
noted as station 0) until forward perpendicular, FP (or noted as station 10 0r 20). Half
or even quarter station may also be used especially at the region with high
curvature. Body plan is normally placed at the top right hand side of the drawing
although it can also be placed at the middle or on top of the sheer plan drawing
depending on the size and type of ship.
Since most ships have symmetrical shape for both port (left side looking from rear)
and starboard (right) sides, only one side is shown in the drawing. Therefore, it is
almost a standard practice to show the stations of the rear region of the ship at the
left side of body plan while the right hand side of the body plan represents the
stations at the forward region of the ship. The curve on the body plan is also call
station curve. The centre line of the body plan represents the centre line of the ship.
Apart from showing the station curves, the body plan also shows the waterlines and
the buttock lines grid. These grid lines are essential not only for reference lines but
also used for transferring and checking data from one plan to another.
16
Figure 13: Body plan
Figure 14: 3-Dimensional body plan
17
1.4.3 Half Breadth Plan
The same hull form if it is viewed from top will produce the plan view of the ship.
However since the hull shape is complex and unique, the plan view must be made at
several waterline planes. Thus Half Breadth Plan is a lines drawing that represents
the shape of the ship hull looking from top view at every waterlines of the ship.
Waterline is the horizontal plane that cut the ship along its vertical axis, thus
creating the waterlines curves as shown in Figure 15. Waterline is normally equally
spaced, although half waterline may also be used at the lower region of the ship.
Since the hull is symmetry about its centre line, only half of the hull is shown in this
plan as shown in Figure 16.
Apart from waterline curves, the deck line curve needs to be drawn on this plan. If
the ship has bulwark, chines or / and knuckles lines, these curves have also to be
shown in the drawing. In this plan, the grid lines shown are the stations and buttock
lines of the ship.
18
Figure 15: 3-Dimensional half-breadth plan
Figure 16: Half breadth plan
19
1.4.4 Profile / Sheer Plan
Sheer Plan which is usually placed at the top left hand side of the lines plan drawing
represent the shape of the ship hull looking from the side of ship at several buttock lines.
Buttock line is the vertical plane that cuts the ship along its length, creating the buttock line
curves as indicated in Figure 17. The middle buttock line (normally labeled as BL 0) is the
plane that cuts the ship along its centre line which creates the profile curve of the ship. Other
buttock lines are drawn outward (offsets) of ship’s centre line and normally at equally
spaced distance. The stations and waterlines grids are shown in this sheer plan drawing. A
typical sheer plan drawing is shown in Figure 18.
Plan B = Plan C
Figure 17: 3-Dimensional sheer plan
20
Figure 18: Profile / Sheer Plan
1.4.5 Offsets Data
Offsets data is the data that is extracted (measured) from the lines plan drawing and
considered the most important data for the design, calculation, analysis and construction of
the ship. As the name implied, Offset Data is the distance measured from the centre line of
the ship to the specific point on the curves (station or waterline curves). The offset data must
be measured at every intersection points on each stations and waterlines including deck line,
chines, knuckles and bulwarks (if any). Offset data also called as half breadth data, because
it represents the half breadth of the ship at every station and waterlines. A typical example
of offsets data is shown in Table 1 and the measurement of offsets data is illustrated in
Figure 19.
In the offsets Table, it is also a standard practice to indicate the data of height above based
for deck, chine, bulwark, and knuckles lines. The height above base of buttock lines may also
be included whenever necessary.
A sample of the complete lines plan drawing containing the body plan, profile, halfbreadth plan and offset are shown in Figure 20.
Table 1: Offsets table















  


21















 
 



Figure 20: Offset data relation to lines plan
1.5 Ship Geometry Coordinate System
Sometimes, there is a need to define the locations and positions on the ship hull using
coordinates. In such cases, the following system is used:
Z axis for vertical direction, positive upwards
X axis for longitudinal direction, positive forward
Y axis in the transverse direction, positive to starboard.
See Figure 21.
Y
The origin (x=0, y=0, z=0) is normally taken at amidships, the centre line and the
baseline. Sometimes, the origin in the longitudinal direction i.e. x = 0 is taken at AP
or FP.
A Systematic Approach
Chapter 2 Hydrostatics and Floatation
2.1
Archimedes Principles of Floatation
Archimedes’ principle states that “An object immersed in a fluid experiences a lift
equivalent to the mass of fluid the object displaces.”
It means that when an object is inside a fluid, there is a force acting vertically upwards upon
it, the magnitude of which is equivalent to the mass of fluid displaced by the object. This
force is called buoyancy.
A man immersed in water for example will feel a weight reduction because part of his weight
is supported by buoyancy. This buoyancy is equal to the weight of water displaced by the
immersed parts of his body.
2.2
Reduction of Weight of Immersed Objects
The maximum buoyancy force exist when the object is fully immersed and this equals the
weight of the fluid being displaced. This in turn equals the total volume of the space
occupied by the object multiplied by the density of the fluid. If the weight of the immersed
object is less that the weight of fluid being displaced, the object experience a net positive
force upwards i.e. lift.
For example, a helium balloon is ‘lifted up’ by the buoyancy force equivalent to the
displaced mass of air. Although the volume of helium in the balloon is similar to the volume
of air that the balloon displaces (disregarding the thin skin of the balloon), the weight of
helium in the balloon is less than the weight of air being displaced due to the lower density
of helium. Therefore, the net force is positively up, i.e. the balloon floats up in the air.
This also explains the strong force required to keep a ball to stay underwater. The weight of
water being displaced by the ball is more than the weight of the ball itself, and hence the ball
will experience a strong net buoyancy force upwards.
© Omar bin Yaakob and Mohamad Pauzi Abdul Ghani, July 2006
24
W
Δ> W, giving net lift pushing the ball upwards
equilibirum
Δ1> W, ball floats in
If the ball is slowly brought to the surface, the ball will be pushed upwards until an
equilibrium is reached such that the ball floats with only a certain portion of it being
immersed in water. In this case, the smaller weight of water being displaced now equals to
the weight of the ball.
When maximum available buoyancy is less than the weight of the object, the object will sink.
That is why an anchor will sink to the bottom. However the object will still feel the effect of
weight reduction.
25
Example 2.1:
Consider a cuboid having dimensions 1m x 1m x 2m in Figure 3.1. If it weighs 3 tonnes in
air, what is its apparent weight in water density 1000 kg/m3?
If the object is immersed in liquid, it will
displace liquid around it equivalent to its
external volume.
In this case, displaced volume = 1 x 1 x 2 = 2
m3
This is the volume of liquid displaced or
pushed aside by the cuboid.
According to Archimedes Principle, the
weight of this object in liquid is reduced due
to the support given by the liquid on the
object. The apparent weight is equal to the
weight in air minus the reduction in weight
of the object or the buoyancy i.e.
Buoyancy
= weight of fluid displaced
= volume displacement x density of liquid
= 2m3 x 1000 kg/m3
= 2000 kg
= 2 tonnes
= reduction in weight
Apparent weight = weight in air – buoyancy
Since the object weighs 3 tonne in air, it will apparently weigh only 1 tonne in water.
26
Exercise 2.1
Do similar calculations to find out the apparent weight in oil (density 0.85
tonne/m3) and muddy water (density 1.3 tonne/m3) and mercury (density 13,000
kg/m3)
Fluid
Density
(
)
Fluid Support
(
)
Apparent Weight
(
)
Oil
Fresh Water
Muddy Water
Mercury
What can be concluded about relationships between buoyancy of objects and
the densities of fluids in which they are immersed?
2.3
What makes a ship float?
As can be seen in Section 3.1, when the maximum available buoyancy is more than the
weight of the object, the object will rise to the surface. It will rise to the surface until the
weight of the object is equal to the buoyancy provided by its immersed portions. When the
object is floating, its buoyancy is just enough to support its weight. At that point:
Total weight W = Buoyancy = Displaced volume x  liquid
This principle explains why a steel or concrete ship can float. As long as the outer shell of the
ship can provide enough volume to displace the surrounding water exceeding the actual
weight of the ship, the ship will float. A floating ship is such that the total weight of its hull,
machinery and deadweight equals to the weight of water displaced by its outer shell. If,
while it is floating weights are added until the total weight exceeds the maximum buoyancy
That can be provided by the outer shell of the ship, the ship will sink.
2.4
Effect of Density
An object experiences buoyancy force equivalent to the weight of fluid it displaces. As seen
in Example 3.1, for a particular object, the buoyancy force will depend on the density of the
fluid, since its volume is constant. The basic relationship can be written as follows:
Total weight W = Buoyancy = Displaced volume x  liquid
(3.1)
27
From Equation (3.1), the displaced volume is inversely proportional to the density of fluid.
For a floating object, this will determine its level of sinkage or draught.
This explains for example why a bather will feel more buoyant while swimming at sea
compared to the river or lake. Also, a floating object of constant weight will sink at a deeper
draught in freshwater compared to in seawater.
Consider a boat moving from the sea to a river. There is a change of density from the more
dense seawater to freshwater which has a lower density. Since the weight of the boat does
not change, the buoyancy force to support the boat is also constant. With the lower density
of fluid, the boat need to increase its displaced volume. To ensure this, the boat will need to
sink deeper i.e. increase its draught.
.
2.5
Some Simple Problems
The fact that a floating object displaces fluid equivalent to its weight as shown in Equation
3.1 can be used to solve a number of problems.
From this equation, we can obtain the weight of the object if we know the volume of water
displaced. On the other hand, if we know its weight, we can work out its displaced volume.
To illustrate the concept, consider a floating box of dimension L x B x D, floating at a draught
T, shown in Figure 3.2.
CASE 1: If we know its weight, we can find its draught
In this case, we know the weight of the object, so we can find the displaced volume:
Displaced volume,  =
W
water
Since it is a box-shaped vessel:
Displaced volume = L x B x T
Hence draught T of the cuboid can be found.
28
Example 2.2
A cuboid shaped wooden block (L x B x D) 1.45m x 0.5m x 0.25m floats in water. If the block
weighs 0.154 tonnes, find its draught if it floats in freshwater density 1.00 tonne/m3.
Solution:
The weight of the block of 0.154 tonnes must be supported by displaced water i.e. the block
must displace 0.154 tonnes of water:
In fresh water,
Volume of displaced water
Weight of displaced water
=LxBxT
 =  x  FW
 = 1.45 x 0.5 x T x FW
This must equal 0.154 tonne
1.45 x 0.5 x T x fw = 0.154 tonnes
T = 0.212 m
Exercise 2.2
Do similar calculations for salt water (density 1025 kg/m3 and oil density 0.85 tonne/m3)
CASE 2: If we know its draught, we can know its volume displacement and we can find its
weight
If we know the draught of the cuboid, we can find its volume displacement and hence the
weight of the object;
Say if we know its draught T, volume displacement = L x B x T
Weight = Buoyancy = Volume Displacement x  water
Weight = L x B x T x  water
Example 2.3
A box barge length 100m breadth 20m floats at a draught of 5m in sea water 1.025 tonne/m3.
Find its weight.
Solution
While floating in sea water density 1.025 tonne/m3:
Volume Displacement =  = L x B x T
Weight of barge =
Weight displacement, 
W
=

=  x salt water
29
= 100 x 20 x 5 x 1.025
= 10250 tonnes
Exercise 2.3
A block of wood length 5m, breadth 0.5m and depth 0.2m is floating in seawater at
a draught of 0.1m. Find the weight of the block.
Exercise 2.4
Find the new draught of the box in example 3.3 when it goes into river, water
density 1.000 tonne/m3. Also find a new draught if it is in sea water with density
1.100 tonne/m3.
1.0m
Exercise 2.5
A cylindrical container weighing 5 tonne floats with its
axis vertical. If the diameter is 1.0m, find its draught
in:
i.
ii.
sea water
oil of density 870 kg/ m3.
Exercise 2.6
A cylindrical tank diameter 0.6m and mass 200kg floats with its axis vertical.
Find its present draught in oil ( = 0.95 tonne/m3).
Find the weight of cargo to be added to ensure it will float at a draught of
0.85m.
30
Exercise 2.7
A boat with constant triangular crossection is floating with its apex down. Its LBP is
10m, breadth 2m and depth 0.7m. If it floats in sea water ( = 1025 kg/m3). at a
draught of 0.5m, what is its weight?
2.6
Tonne per centimeter immersion (TPC)
The weight required to increase or reduce trim by one centimeter ia called Tonne per
centimeter immersion (TPC)
For the ship having area of waterplane Awp, an increase in draught of one
centimeter will require an addition of weight equivalent to the additional volume of
displacement multiplied by the density.
Additional weight = Awp x 1cm
100
Example 2.4
A ship with TPC of 30 floats at a draught of 6.5m. What is the new mean draught
ahen 150 tonne is unloaded from the ship.
31
2.7
Hydrostatics Particulars
A floating object will be floating at a certain draught depending on the total weight of the
object, the density of water and the shape of the object. For a ship, the shape of the object has
a strong influence on the draught of the ship; the shape and draught have to provide enough
buoyancy to support the ship.
When a ship is floating at a certain draught, we can find the mass displacement and weight
of the ship if we can find its displaced volume . Also we can know its waterplane area,
calculate its TPC, KB, Cb etc. These particulars which are properties of the immersed part of
the ship are called hydrostatic particulars. Examples of hydrostatic particulars are:
,
, KB, LCB, Aw, BMT, BML, TPC, CB, CP, CM, CW, LCF, MCTC
These particulars describe the characteristics of the underwater portion of ship at a particular
draught. It is related to volumes, areas, centroids of volumes and areas and moments of
volumes and areas of the immersed portion. If the ship is taken out of water, and draught
becomes zero, the particulars ceased to exist.
As long as draught and trim are maintained, the size and shape of the underwater immersed
parts of the ship remains the same. The volumes, areas and moments of areas and volumes
remain the same and consequently, the hydrostatics particulars do not change. Once draught
or trim changes, the particulars will also change.
These changes in draught and trim will normally occur due to changes in the total weight of
the ship, movement of weights onboard or if external forces are applied.
Exercise 2.8
A box 2m x 1m (LxB) is floating in sea water.
Calculate its , , CB, CWP and TPC at draughts of 0.3 and 0.5m
32
Exercise 2.9
Find hydrostatic particulars in sea water (, ,Awp,LCB, LCF,TPC) of a box
barge with dimension L=100m, B=20m, at draughts of 1.0m, 3.0m, 5.0m, 7.0m,
9.0m. If the barge weighs 2,300 tonne, what is its draught? If the barge is
floating at a draught of 4m, what is its CB?
Exercise 2.10
Calculate
, , KB , LCB, Aw, TPC, CB, CP, CM, CW, LCF of a cylinder radius
1m floating with axis vertical at draughts of 1.0, 1.5, 2.0 and 2.5m.
It can be seen from Exercise 2.9 and 2.10 that for cuboids or cylinders, the waterplane areas
are constant at different draughts. Hence, many hydrostatics particulars which depend on
waterplane areas also remain constant.
Exercise 2.11
An empty cylindrical shaped tank is floating in sea water (density 1.025 t/m3) at a
draught of 8.0 m with its axis vertical. The external diameter of the tank is 12.0 m,
internal diameter 11.0 m, thickness of base 1.0 m and the overall height is 16.0 meter.
Its centre of gravity is 6 meter above its inner base.
Calculate:
.
i. Find Hydrostatic particulars , Awp, LCB, Cb, Cp, TPC, WSA
at T=1, 2, 4, 6, 8m.
ii. Plot hydrostatic curves similar to page 19 showing all data.
ii. Final draught of the tank after 500 m3 diesel oil (density 850 kg/m3) is poured
into the tank.
The second moment of area of a circle about its diameter is
D 4
.
64
33
2.8
Hydrostatic Particulars of a Ship
The objects considered in the previous sections are simple, uniform objects such as cuboids,
cylinders and prisms. The formulae involved in these cases are simple and familiar. In some
cases, the particulars such as block coefficient and waterplane areas remain constant at
various draughts. By using these objects, the calculations are simplified and can be used to
show the main concepts.
Those formulae used for the simple objects are no longer applicable for real ships, although
the concepts remain relevant. Unlike those simple objects considered earlier, real ships
rarely have uniform cross sectional areas or waterplane areas. The waterplanes are no longer
made up of straight lines. Therefore, there is no simple formulae to calculate their areas,
volumes and moments and hence hydrostatic particulars.
Consider the ship whose lines plan is shown in Figure 3.2 . At different draughts, the ship
will have different waterplane areas, sectional areas, volumes and centroids. Hence, the
hydrostatic particulars will also vary as the draught changes.
The methods to calculate areas, volumes, moments, centroids of the waterplanes and sections
of ships will be covered in Chapter 3.
If areas, volumes, moments, centroids of the waterplanes and sections of the ships can be
calculated, hydrostatic particulars of a ship can be obtained. These are calculated at the
design stage, once the shape and size of the ship has been decided.
The particulars can be presented in two forms, either as a set of curves or in tabular format.
Table 2.1 shows a typical table of hydrostatic particulars while an example of hydrostatic
curves is shown on Figure 2.3.
34
Table 2.1 Hydrostatic Particulars of Bunga Kintan LBP 100m
LCB
LCF
MCTC
(tonne-m) (m from ) (m from )
Draught
(m)
Displacement
(tones)
Cb
KB
(m)
BMT
(m)
BML
(m)
8.00
14820.00
0.72
4.07
3.66
180.00
190.00
2.50
2.00
7.50
13140.00
0.71
3.67
3.98
195.00
183.00
2.30
1.50
7.00
11480.00
0.70
3.26
4.46
219.00
180.00
2.00
0.70
6.50
9870.00
0.69
2.85
5.02
244.00
172.00
1.80
-0.06
6.00
8280.00
0.67
2.44
5.66
279.00
165.00
1.50
-1.00
5.50
6730.00
0.66
2.04
6.67
327.00
157.00
1.10
-2.00
5.00
5220.00
0.64
1.63
8.06
392.00
146.00
0.00
-3.00
Exercise 2.12
A ship with length 100m, breadth 22m has the following volumes and areas at
different waterlines. Calculate its , CB, CW and TPC in saltwater density
1.025tonnes/m3.
Draught
(m)
Aw
(m2 )
 (m3)
2
1800.0
3168.0
4
2000.0
6547.2
6
2100.0
10137.6
8
2120.0
13728.0
10
2130.0
17424.0

(tonnes)
Cb
Cw
TPC
x 

LBT
Aw
(LB)
Aw x 
100
35
2.9
Using Hydrostatic Curves and Tables
Hydrostatic curves and tables can be used to obtain all hydrostatic particulars of a ship once
the draught or any one of the particulars is known.
Example 2.5
From MV Bulker hydrostatic Curves (Figure 3.3 ) at a draught of 7m, we can obtain
displacement  = 31,000 tonnes, LCF = 2.0m forward of amidships and MCTC = 465 tonne-m
etc. Also if we know the ship weighs 40,000 tonnes, its draught, TPC, MCTC, LCF and LCB
can be obtained.
Exercise 2.13
Using MV Bulker Hydrostatic Curves, find displacement, LCB, LCF, TPC at
draught of 9.5m. If 1500 tonnes is added to the ship, what is its new draught?
36
Hydrostatic tables can be used in a similar manner to obtain hydrostatic particulars once
draught is known or to obtain draught and other particulars once the displacement or
another particular is known. There is however a need to interpolate the table to obtain
intermediate values.
Exercise 2.14
By using the hydrostatic particulars of Bunga Kintan shown in Table 3.1:
i.
ii.
iii.
iv.
2.10
Draw full hydrostatic curves of the ship
Find values of displacement
, KB , LCB, BMT, BML, MCTC,
CB, LCF of the ship if it is floating at a draught of 6.75m.
Find values of T, KB , LCB, BMT, BML, MCTC, CB, LCF of ship
if the ship weighs 11,480 tonnes.
When the ship is floating at a draught of 5.5m, 3000 tonne cargo
was added. What is its new draught?
Bonjean Curves
One important hydrostatic particular is the area of a section. For a particular transverse
section, the sectional area can be calculated up to a particular waterline.
This can be done for all waterlines at each station.
If these areas are plotted against draughts at all station positions along the ship, the resulting
diagram is called Bonjean Curves.
The curves are frequently drawn on the ships profile at the displacement stations or on a
centre line with those for stations in the fore body on the right hand side and for the after body
on the left hand side. They enable the displacement and LCB to be calculated for any
waterline, trimmed or even keel.
37
The plot for every stations are superimposed on the profile of the ship as follows:
2.11 Cross Sectional Area Curve
The shape of the hull can be defined by a curve representing the distribution of the crosssectional area (CSA) of each section at the respective stations up to a particular waterline,
normally DWL. The curve is called Sectional Area curve.
2.12 Second Moments of Areas
2.12.1 The Basic Formulae
The second moment of area or moment of inertia are used in the calculations of metacentric
heights. Therefore is important to have a firm understanding of this important quantity.
By definition, second moment of area is the product of area multiplied by square of the
distance of the centre of that area to the reference axis.
The second moment of an element of an area about an axis is equal to the product of the area
and the square of its distance from the axis
38
Consider the rectangle length l and breadth b. A small segment area length I and breadth dy
will have an area l x dy.
The second moment of the rectangle about an axis parallel to one of its sides and passing
through the centroid (XX) is
Ixx = Area x y2
Ixx = l x y2 x dy
= Consider the elementary strip which is shown shaded in the figure.
The second moment (i) of the strip about the axis AB is given by the equation:i= l dx x x2
Let I AB be the second moment of the whole rectangle about the axis AB then: b/2
IXX 
 l. y
2
.dy
- b/2
 b/2
1XX  l

y 2 .dy
- b/2
b / 2
 y3 
 l 
 3  b / 2
1XX 
lb 3
12
1AB 
lb 3
12
The second moment of a rectangle about one of its sides ( axis AB):b
1AB   l. x 2 .dx
O
39
b
x 
 l 
 3 O
3
1AB 
lb 3
3
According to the Theorem of Parallel Axes the second moment of an area about an
axis through the centroid is equal to the second moment about any other parallel
axis minus by the product of the area and the square of the perpendicular distance
between the two axes.
Thus, in Figure A.4,
I XX  I AB - Area  y 2
Fig. A.4
40
Second moment of area of a circle
X
X
Fig. A.5
For circle, the second moment of area about an axis AB.
I AB 
 D4
64
Exercise
Determine what is IXX?
2.12.2 Calculations of Metacentric Heights
Second moment of areas are used in calculations of the distances from ship centre
of buoyancy to the longitudinal and transverse metacentres, BML and BMT
respectively:
BM L 
IF

BM T 
IT

and
Where IF is longitudinal second moment of area of the waterplane about the centre of
floatation, IT is transverse second moment of area about the centreline and  is volume
displacement.
41
In hydrostatic calculations, the longitudinal second moment of area is initially calculated
about amidships. To obtain IF, theorem of parallel must be used:
I F  I   WPA LCF 2
Where WPA is the waterplane area and LCF is the centre of floatation measured from
amidship.
BML and BMT are important hydrostatic particulars which are required in the
calculations of ship stability.
a) Calculations of Moment to Change Trim one Centimeter (MCTC)
MCTC is the moment required to change trim one centimeter. ``````
The actual formula for MCTC is
MCTC 
  GML
tonne-m
100  L
Where GML is the distance from the centre of gravity to longitudinal metacentre and L is
LBP.
Usually, GML is large and can be approximated by BML.
Therefore the formula becomes:
MCTC 
Since
  IF
tonne-m
100  L  
  
MCTC 
  IF
100  L
42
Exercises
1. Find BML and BMT of a box shaped barge 120m x 20m x 10m floating at a draught of
7m.
2. A cylinder of radius r = 10m is floating upright at draught of 6m in fresh water. Find
its KML and KMT.
3. A fish cage consists of a wooden platform placed on used oil drums with the
following dimensions.
If the total weight of the structure is 3 tonnes, floating in sea water calculate:
i)
ii)
iii)
draught
KMT
KML
4. A catamaran consists of two box-shaped hulls spaced 5m apart, centreline to
centreline. Each hull measures (L x B x D) 10m x 0.5m x 1m. If its draught is
0.3m, find its :
i)
ii)
iii)
iv)
 and 
KB
BMT
Maximum allowable KG if GM minimum is 0.2m
43
5. A ship has the following characteristics at a draught of 2m:
LBP = 220m
Displacement = 153 tonnes
BML = 24.3 m
LCF = 0.97m aft of amidships
Calculate:
i.
ii.
iii.
MCTC in sea water density 1.025 tonne/m3.
longitudinal second moment of area of the waterplane about the centre
of floatation
longitudinal second moment of area of the waterplane about the centre
of amidships
44
Chapter 3 Basic Stability Consideration
3.1
Introduction
One of the factor threatening the safety of the ship, cargo and crew is the lost or lack of
stability of the vessel. Stability calculation is an important step in the design of the ship and
during its operation. While designing the ship, the designers must be able to estimate or
calculate to check whether the ship will be stable when constructed and ready to operate.
During operation, the ship's officers must be able to load and stow cargo and handle the ship
while ensuring that the ship will be stable and safe.
3.2
What is stability?
Stability is the tendency or ability of a system to return to its original condition when
disturbed or displaced from its normal equilibrium condition. Ship stability is the
tendency or ability of the ship to return to upright when displaced from the upright
position. A ship with a strong tendency to return to upright is regarded as a stable
vessel. On the other hand, a vessel is said to be not stable when it has little or no
ability to return to the upright condition. In fact, an unstable ship may require just a
small external force or moment to cause it to capsize.
A ship is normally stable to a certain degree of heel, after which it will capsize. Some
have a large or strong tendency to return to upright while others have a smaller
returning or righting moment. Some ships have a large range of stability, up to 90
degrees and beyond while others capsize when heeled beyond small angles of say, 20
degrees.
An analogy for stability is often given of the marble. In Figure 3.1 (a), the marble in
the bowl will return to its original position at the bottom of the bowl if it is moved to
the left or to the right. This marble is in a condition called positively stable.
Figure 3.1
45
A slight push on the marble which is put on an upside down bowl as in Figure 3.1 (b)
will cause it to roll off, a condition equivalent to instability. A neutrally stable ship is
analogous to a marble put on a flat surface, it will neither return nor roll any further.
It can also be seen there are various degrees of stability. For example a bowl with
very steep sides will have larger stability than a bowl with less sloping sides.
As we will see later, there are also varying degrees of ship stability, from ships with a
highly positive stability to those having negative stability.
3.3
Longitudinal and Transverse Stability
Ship initial stability can be seen from two aspects, longitudinally and transversely.
From longitudinal viewpoint, the effect of application internal and external moments
on ship's trim is considered. Examples of application of moment is the movement of
weights on board in the forward-aft direction or the addition or removal of weights
to/from the ship. Important parameters to be calculated are trim and the final
draughts at the perpendiculars of the ship. In any state, there is a definite relationship
between trim, draughts and the respective locations of the centres of buoyancy and
centre of gravity. Sometimes, the trim angle  is also calculated.
Transverse stability calculation considers the ship stability in the port and starboard
direction. We are interested in the behaviour of the ship when external moments are
applied such as due to wind, waves or a fishing net hanging from the side. The effect
of internally generated moments such as movement of weights
on-board
transversely is also studied. This includes the shifting of weights or crowding of
passengers on one side of the boat.
Important relationships considered are those between heeling, listing and righting
moments, as well as the resulting angle of heel or list  and its consequence on the
safety of the boat. Note that the words ‘list’ and ‘heel’ have similar meanings.
However, heel has a more general meaning and normally used for effects of outside
moments which are temporary such as due to wind, while list is normally used for
the effects of internally generated moment, such as movement of weights on-board.
This Chapter will focus on basic transverse stability particularly the relationship
between the metacentre and the centre of gravity and its effect on ship stability.
Further transverse stability calculation will be dealt with in the next Chapter while
the details regarding longitudinal stability will be covered in the Chapter 7.
46
3.4
Basic Initial Stability: The role of GM
Figure 3.2
Consider the ship floats upright in equilibrium as in the above Figure 4.2 (a). The
weight of the ship equals its displacement and the centre of buoyancy is directly
below the centre of gravity. The points B, G and M are centre of buoyancy, centre of
gravity and metacentre respectively.
When the ship is slightly disturbed from upright for example due to wind blowing
from port, the centre of buoyancy (which is also the centre of immersed or displaced
volume) moves to the right. The line of action of buoyancy points vertically upward
crossing the original centreline at the metacentre, M. Since G does not move, a
moment is generated to turn the ship back to its original position. This moment is
called the returning or righting moment.
In this case, M was originally above G and we can see that the righting moment is
positive i.e. the ship is stable. If M was below G i.e. GM negative, the righting
moment will be negative hence the ship is unstable. If M is at G, then the ship is
neutrally stable.
Righting moment is the real indication of stability i.e. the ability of the ship to return
to oppose any capsizing moment and return the ship to upright position.
The larger the righting moment, the better stability is.
47
Consider the triangle shown below:
Righting moment =  x GZ
and
GZ = GMT sin 
For any displacement, righting moment depends
on GZ.
And GZ depends on GM. The bigger the value of
GZ, the bigger will be the righting moment. Since
GZ is a function of GM, then bigger GM will lead to
larger GZ, bigger righting moment and hence better
stability.
It is very essential that ships have large enough GM
to ensure that it has enough righting moment to
overcome external moments.
The relationships between K, B, G and MT are important.
The relationships can be used to guide us how to increase GM.
KMT = KB + BMT
KMT = KG + GMT
For any particular draught or displacement at low angle of heel, keel K and
metacentre M are fixed. Therefore the values of KB, BM and hence KM are fixed, as
can be obtained from hydrostatic particulars and the distance GMT will only
depend on the height of the centre of gravity. In other words, if we can ‘control’
KG, we can also control GM and hence ship stability. The lower the centre of
gravity, the larger will be GM, and conversely a high value of KG may lead to small
or even negative GM, which means an unstable ship.
48
3.5
Determining the Centre of Gravity of ships after loading
The ability to pin-point the location of the ship’s centre of gravity is important
because its distance from keel will affect GM and hence the ship’s stability. The
centre of gravity will remain unchanged except under two conditions:
i.
ii.
Weights are added to or removed from the ship at locations not
coinciding with the original centre of gravity.
Weights already onboard are shifted, causing changes in moments
about the centre of gravity.
The effect of addition and removal of weights from ships on the centre of gravity can
be calculated by using the fundamental concepts of centroids of composite bodies.
The concept is given in Appendix 1. The tabular methods described in Appendix 1
can be applied directly. When considering the net centre of gravity after loading and
unloading, three groups of items are normally considered; the original ship. Their
respective weights and centres of gravity become the input data from where the final
KG is calculated.
Example 3.1
A ship originally weighs 2000 tonnes with KG= 5.5m. One cargo of 300 tonne was
unloaded from Kg=7.6m and another 500 tonne was loaded at Kg=2.5m. Find the
final KG.
Item
Lightship
Unload
Cargo 1
Load
Cargo 2
TOTAL
Weight
(tonnes)
2000
Kg
(m above keel)
5.5
-300
7.6
500
2.5
KG = Total moment about Keel =
Total weight
Moment
about
keel (tonne-m)
m
49
Example 3.2
A ship of 6,000 tonnes displacement has KG = 6 m and KM = 7.33 m. The following
cargo is loaded:
1000 tonnes, Kg 2.5 m
The following cargo is then discharged:
450 tonnes of cargo Kg 0.6 m
And 800 tonnes of cargo Kg 3.0 m
If KM on completion of loading is 7.3, what is the final GM.
Item
Weight (tonne)
Kg
Ship
Loaded
Cargo1
Cargo2
Cargo3
Unloaded
Cargo 4
Cargo 5
6000
1000
500
750
6.0
2.5
3.5
9.0
-450
-800
0.6
3.0
Final KG
=
=
Final KG
=
Final KM
=
Final KG
=
Ans. Final GM =
Moment about keel
(tonne-m)
Final moment
Final displacement
________
m
m
m
m
50
Exercise
A box-shaped barge is floating in sea water at a draught of 5m. The extreme
dimensions of the barge (L x B x D) are 12m x 11m x 10m. The wall and floor are
0.5m thick. Its centre of gravity is 4m above keel.
Calculate:
i.
ii.
iii.
The displacement and GMT of the empty barge.
The barge is to be used to carry mud (density1500 kg/m3). If the draught of
the barge cannot exceed 7.5m, find the maximum volume of mud that can be
loaded into the barge.
For the barge loaded as in (ii), find its GMT.
3.6 Effect of movement or addition of weights on centre of gravity
When portions of weights or areas or volumes are moved, added or removed
the overall centroid of the object will also move. The basic concepts to
determine the movement of centroids due to the movement or
addition/removal of weights, areas or volumes is described in Appendix 2.
The concepts given in Appendix 2 can be directly applied to ships. When
weights are shifted transversely on-board a ship, the moments change in the
port-starboard direction, causing the centre of gravity to shift.
Similarly, when weights are added or removed, the net effect is the centre of
gravity will shift. When a weight is added, the centre of gravity will move in
the direction towards the point at which added. On the other hand, when a
weight is removed, the centre of gravity will shift in the direction away from
the point at which the weight is removed.
Example 3.3
A boat displacement 150 tonnes has KG of 1.25m, GM 1.7m and floating upright in salt
water. A weight of 2 tonne already onboard is moved from port to starboard. Find the shift
of the centre of gravity.
51
Example 3.4
A ship weighing 7000 tonnes is floating at the wharf. At that time, KM = 6.5 m and
GM 0.5m. A 30 tonnes box is loaded at a distance 10.0m above keel. By considering
the shift in the centre of gravity, find new GM. Assume no change in KM.
Solution:
Find rise in KG
Original KG = KM - GM =
m
Distance 30 tonnes box from original G =
GG’= 30 x 4.0 = 0.017m
7030
KG’= KG+ GG’ =
m
10m
m
KM does not change, therefore, GM =
=
m
Exercise: Find the new KG and GM of the ship in Example 3.3 using
Method 2:
Find final KG using table of moment about keel
Portion
Mass (m)
Kg (m)
Ori. Ship
Box
Total
7000
30
7030
6.0
10.0
Moment about keel
(tonne-m)
KG = Sum of moment
Sum of weight
KG =
m
GM = KM - KG
KM - KG =
m
52
3.7 Hanging Weights, The Use Of Derricks And Cranes
The use of cranes and derricks will make the weights suspended. Suspended
weights are assumed to act at the point of suspension. Therefore a weight that was
initially located on the lower deck for example will instantly be transferred to the
point of suspension at the instant the weight is lifted off the deck by the derrick. The
centre of gravity KG will suddenly increase and because KM is constant, GM will
suddenly reduce. If the rise in KG is more than the original GM, the net GM will be
negative, leading to instability.
Example 3.5
A ship of 7,500 tonnes displacement is upright and has GM 0.20m and KM 6.5 m. A
heavy cargo of 100 tonnes already on the lower deck (kg=2m) is to be unloaded using
the ship’s crane. When lifting the cargo crane head is 15 m above keel. What is GM
during lifting. Comment of the safety of the operation.
Treat as if the weight is suddenly transferred from lower deck to the point of
suspension, a distance of 15 meters. The KG will rise, and since KM constant, GM
will be reduced.
53
Original KG = KM-GM= 6.5 – 0.2 = 6.3m
Rise in KG = 100 x 13
7,500
=0.173m
KG during lifting = KG + Rise = 6.473m
GM during lifting = KM- Kgnew = 6.5- 6.473 = 0.027m
3.8
Free Surface Correction
3.8.1 The Effect of Free Surface on Ship Stability
When free surface exists on board the ship, stability of ship is affected. The free
surface gives rise to free surface moment which in effect reduce GM. The reduction
is called Free Surface Correction (F.S.C).
FSC is calculated from the second moment of area of the surface of the fluid;
FSC = Free surface moment
Ship displacement
54
Free Surface Moment (FSM) = i x ρ fluid
Where i the second moment of area of the surface of the fluid and ρ
density of the fluid in the tank.
fluid
is the
Once the FSC is known, the new reduced GM called GMfluid is obtained
GM fluid = GMsolid - FSC
Note also that KG in ships having free surface is
increased by FSC.
called KGfluid
and regarded
KGfluid = KGsolid + FSC
As can be seen, the presence of liquid free surface effectively reduces GM or increase
KG. It is thus very important to ensure that free surface be avoided or at least
minimised during the design stages and while operating the ship.
3.8.2 Calculating Second Moment of Area
The method of calculating the second moment of area, described in Section 2.12 can
be used to calculate i . For free surface with rectangular and circular shapes, the
following formulae can be used.
For tanks with a rectangular surface area:
Free surface moment = 1 x tank length x tank breadth3 x density of fluid
12
Free surface correction = 1 x tank length x tank breadth3 x ρ fluid
12
Δ
For tanks with Circular surface:
Free surface moment = π x Diameter4
64
Free surface correction = π x Diameter4 x ρ fluid
64
Δ
1 x tank length x tank breadth3 x density of fluid
12
ship displacement
55
Exercises
1.
Bunga Kintan (Hydrostatic data given on page 12) is floating at draught of
6.5m. If its KG is 6.8m, what is its GM?
2.
A ship has a displacement of 1,800 tonnes and KG = 3m. She loads 3,400
tonnes of cargo (KG = 2.5 m) and 400 tonnes of bunkers (KG = 5.0m). Find the
final KG. 2.84m
3.
A ship sails with displacement 3,420 tonnes and KG = 3.75 m. During the
voyage bunkers were consumed as follows: 66 tonnes (KG = 0.45 m) and 64
tonnes (KG =1 m). Find the KG at the end of the voyage.
4.
A ship has displacement 2,000 tonnes and KG = 4m. She loads 1,500 tonnes of
cargo (KG = 6m), 3,500 tonnes of cargo (KG = 5m), and 1,520 tonnes of bunkers
(KG = 1m). She then discharges 2,000 tonnes of cargo (KG = 2.5 m) and
consumes 900 tonnes of oil fuel (KG = 0.5 m.) during the voyage. If KM=
5.5m, find the final GM on arrival at the port of destination.
5.
A ship arrives in port with displacement 6,000 tonnes and KG 6 m. She then
discharges and loads the following quantities:
Discharge
1250 tonnes of cargo
KG 4.5 metres
675 tonnes of cargo
KG 3.5 metres
420 tonnes of cargo
KG 9.0 metres
Load
980 tonnes of cargo
KG 4.25 metres
550 tonnes of cargo
KG 6.0 metres
700 tonnes of bunkers KG 1.0 metre
70 tonnes of FW KG 12.0 metres
During the stay in port 30 tonnes of oil (KG 1 m.) are consumed. If the final
KM is 6.8 m., find the GM on departure.
6.
A ship of 9,500 tonnes displacement has KM 9.5 m and KG 9.3 m. A load 300
tonnes on the lower deck (Kg 0.6 m) is lifted to the upper deck (Kg 11 m). Find
the final GM.
7.
A ship of 4,515 tonnes displacement is upright and has KG 5.4 m and KM 5.5
m. It is required to increase GM to 0.25m. A weight of 50 tonnes is to be
shifted vertically for this purpose. Find the height through which it must be
shifted.
8.
A ship of 7,500 tonnes displacement has KG 5.8 m. and GM 0.5 m. A weight of
50 tonnes is added to the ship, location Kg = 11m and 7m from centreline to
the starboard side. Find final location of G above keel and from the centreline.
What is its new GM?
56
9.
A ship has a displacement of 3,200 tonnes (KG = 3 m. and KM = 5.5 m.). She
then loads 5,200 tonnes of cargo (KG = 5.2 m.). Find how much deck cargo
having a KG = 10 m. may now be loaded if the ship is to complete loading
with a positive GM of 0.3 metres.
10.
A ship of 4,500 tonnes displacement is upright and has KG 5.4 m and KM 5.5
m. It is required to move a weight of 50 tonnes already on the deck (kg=6m)
using the ship’s derrick. The derrick head is 13 m above keel. Is it safe to do
so?
A ship of 9,500 tonnes displacement and has KM 9.5 m and KG 9.3 m. The ship
has two fuel tanks in double bottoms, rectangular shape each 20 x 5m
containing bunker density 900 kg/m3. Find GMfluid when free surface exists in
the tank.
Find Gmfluid for the ship in question 11 but with one tank only, length 20m
breadth 10m.
What happens to i when there are three tanks with b = 3.33m in question 11.
11.
12.
13.
57
Chapter 4 Transverse Stability
4.1 List due to movement of weights onboard
Normal ships are meant to be upright. However under certain conditions such as
when they are improperly loaded, out of balance moments may cause the ships to
list. Movement of weights onboard ships will effectively change the position of the
centre of gravity as described in Section 4.x. The shift in the centre of gravity will
cause a net moment which will cause the ship to list, either to port or starboard. The
net result is, the ship will heel to one side until it reaches an equilibrium i.e. a balance
of forces and moments and finally settles at a steady angle of list.
Consider a ship floating upright as shown in Figure 4.1. The centres of gravity and
buoyancy are on the centre line. The weight of the ship W is supported by an equal
and opposite force, the buoyancy, Δ. The resultant force acting on the ship is zero,
and the resultant moment about the centre of gravity is zero.
Figure 4.1
If a weight already on board the ship be shifted transversely from port to starboard,
the centre of gravity G moves to G1 as in Figure 4.2. The net effect will give rise to a
listing moment of W X GG1 in the clockwise direction and the ship will start to heel
in the clockwise direction (to starboard) until GI and the centre of buoyancy are in
the same vertical line as in Figure 4.3. The ship will settle at a steady angle of list,  .
58
Figure 4.2
Figure 4.3
59
In this position, for small angles of list is small, the buoyancy force will act vertically
upwards passing through the metacentre, M. G1 will also lie vertically under M. If
the positions of the metacentre and the centre of gravity are known, the final list
angle can be found:
Using trigonometry, in the triangle GG1M which is right-angled at G,
GG1M:
GG1 = w x d
W
Tan  =
GG1
GM
Tan  =
wxd
W x GM
The formula can be restated as:
Tan  =
listing moment
W x GM
It can be seen that GM plays a big role in determining angle of list. The bigger the
value of GM, the less the angle of list and vice-versa.
The final position of the centre of gravity and hence GM is found by taking moments
about the keel and about the centre line as discussed in Chapter 4. ‘
Example 5.1
A ship of 7,000 tonnes displacement is floating upright. It has KM = 7.3 m, and KG =
6.7 m, and is floating upright. Find the resultant list when a weight of 50 tonnes
already on board is shifted 12 m from port to starboard.
Solution:
When the weight is shifted transversely the ship’s centre of gravity will also shift
transversely, from G to G1. The ship will then list  degrees to bring G1 vertically
under M the metacentre
GM = KM - KG = 0.6m
Listing Moment = 60 x 12 tonne-m
Tan  =
60 x 12
6000 x 0.6
Tan  =
Ans. List
=
0.2
11 18 ½ ‘
60
(a)
(b)
Figure 4.4
4.2 Finding list after loading and unloading
As can be seen in Chapter 4 (See Section 4,x), loading or unloading a weight onboard
a ship not at the ship’s centre of gravity will cause a change in the height of the
centre of gravity, KG. In addition, if the location of the added or removed weight is
not along the centreline, there will be a change of moments in the starboard or port
direction (clockwise or anti-clockwise). The net moment will cause the ship to
change its list. To find the angle of list, calculations are carried out using three steps:
i)
Find the final KG (see Section 4.x)
ii)
Find the net final listing moment (port or starboard) about the centre line.
iii)
Use the Tan  formula to find angle of list.
Example 2
A ship of 5,000 tonnes displacement, KG = 7.6 m is loaded as follows:
 Load 100 tonnes cargo KG 6.1 m and centre of gravity 7.6 m to starboard of the
centre line.
 Load 200 tonnes fuel oil KG 0.6 m and centre of gravity 6.1 m to port of the
centre line.
 Discharge 70 tonnes of ballast KG 1.2 m and centre of gravity 4.6 m to port of
the centre line.
61
If the final KM is 8.0m, find the final list.
(1) Find Centre of Gravity after loading and unloading
Weigh
t
8000
250
300
-50
8500
Final KG
Final KG
KG
Moment
keel
60800
1525
180
-60
62445
7.6
6.1
0.6
1.2
about
=
Final moment
Final displacement
KM
Final KG
=
=
8.7 m.
7.34 m.
=
62.445
8500
Final GM
=
1.36 m.
=
7.34 m
Find moments about the centre line (set –ve value for port)
W
.d
Listing
(tonne-m)
to port
250
-50
300
7.6
4.6
6.1
moment
to starboard
1900
-230
1830
1600
Net listing moment
1900
300
Since the net moment is to starboard, the ship will list  degrees to starboard.
Tan 
=
Listing Moment
W x GM
=
300
8500 x 1.36


=
1 29 ½’
Ans. Final list
= 129 ½’ to starboard
62
4.3 Correcting Lists by moving or adding weight
When a ship is found to have an initial list, it means that it has an initial listing
moment.
Using
Tan  =
listing moment
W x GM
The formula can be restated as:
listing moment =
Tan  x W x GM
To bring a listing ship to upright:
1. Find listing moment that is initialy listing the ship.
2. Apply an equal and opposite moment such that the net moment is zero, i.e.
the applied moment must counterbalance the original moment.
Example
A ship of 1,500 tonnes displacement has KG 2.8 m, and KM 3.0 m and is
floating with a list of 7degrees to starboard. Find how much weight must
shifted transversely across the deck through a distance of 7 metres to bring
the ship to upright condition.
GM = KM-KG = 0.2m
Initial listing moment
= Tan 
x W x GM
= xxxxx tonne-m to starboard.
This moment must be neutralized or counteracted by an equal and opposite
moment to port.
W x 7 = xxxxx tonne-m to port
W = tonne
63
Exercises
1.
Carry out similar calculation as in Example 5.1, but with KG=7.0.
2.
A ship of 3,500 tonnes displacement (KG 2.7 m, and KM 3.1 m) is floating
upright. Find the listing angle when a weight of 12 tonnes is shifted 8 metres
across.
3.
A ship of 5,000 tonnes displacement is upright and has KG 6.0 m and KM
6.4 m. Find the distance through which a 15 tonnes weight already onboard
must be shifted to obtain a list of 2.5 degrees.
4.
During an inclining experiment of a 3,000 tonnes displacement ship, a
weight of 10 tonnes was moved transversely across the deck through a
distance of 12 m, causing the ship of to list 3 degrees to starboard. If KM 6
m, find the KG.
5.
A ship of 6,000 tonnes displacement has KM = 7.3 m, and KG = 6.7 m, and is
floating at a list of 11.3 degrees to starboard. Find how much water to be
transferred from starboard to port tanks, a distance of 5 meters to bring the
ship to upright.
6.
A ship of 10,000 tonnes displacement, is listed 2 degrees to starboard and has
yet to load 200 tonnes of cargo. There is space available (centre of gravity, 6
m to port and 3m to starboard from the centre line). Find how much cargo
to load on each side if the ship is to be upright on completion of loading.
7.
A ship of 5,000 tonnes displacement has KM = 5.5 m and KG = 5.0 m. She
has yet to load two 20 tonne boxes with her own derrick and the first box ls
to be placed on deck on the inshore side (KG 9 m and centre of gravity 6 m
out from the centre line). When the derrick plumbs the quay its head is 15 m
above the keel and 12m out from the centre line. Calculate maximum list
during operation. Note: The maximum list is obviously occur when the first
box is in place on the deck and the second box is suspended over the quay.
8.
A ship of 12,500 tonnes displacement, KM 7 m, KG 6.4 m, has a 3 degree list
to starboard and has yet to load 500 tonnes of cargo. There is space available
in the ‘tween decks, centres of gravity 6 m each side of the centre line. Find
how much cargo to load on each side if the ship is to complete loading
upright. (282.75 tonnes P)
9.
A ship is listed 2.5 degrees to port. The displacement is 8,500 tonnes KM 5.5
m, and KG 4.6 m. The ship has yet to load a locomotive of 90 tonnes mass on
64

- Universiti Teknologi Malaysia

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