NAVAL ARCHITECTURE 1
Class Notes
w tonnes
G
G1
d
d
G1
G
W tonnes
d
G1
G
w tonnes
Omar bin Yaakob
Naval Architecture Notes
Chapter 1
Introduction
Introduction
To carry out various activities at sea, rivers and lakes, man uses various types of
marine structures, fixed and floating. The structures 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. 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 war ships 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.
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.
A naval architects 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.
Ship hydrostatics and stability 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. In this book, readers will be able to appreciate the basic
terminologies, carry out simple hydrostatics calculations and will be equipped with
basic tools to assess stability of vessels.
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Naval Architecture Notes
Chapter 2
Ship Types, Basic Terms,
Terminologies and Symbols
1. Types of Ships
Ship types can be classed according to:
1. No of Hull
a) Monohull/Single hull
b) Multi-hull
Catamaran
Trimaran
Quadramaran
Pentamaran
2. Shape of hull form
a) Roundbilge
b) Chine
Single Chine
Multiple Chine
3. How the body is supported in water
a) Hydrostatic
b) Hydrodynamic
c) Aeropowered Lift
4. Its function/mission
a) Transport
Tanker
Bulk Carrier
Containership
Passenger ship
General Cargo
LNG Carrier
b) Navy
Aircrft Carrier
Submarine
Frigate
Destroyer
Patrol Craft
Minesweeper
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Naval Architecture Notes
c) Work/Service Vessels
Tugs
Supply boat
Crew Boats
Heavy Lift
Crane ships
Fuel Supply Ships
Fishing Boat
Fire Fighting Boats
Rescue Boats
d) Leisure Vessels
Cruise ships
Tourist Boats
Water Taxi
Boat Houses
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Naval Architecture Notes
2. Basic Terms, Terminologies and Symbols
After
perpendicular(AP)
BML
This is represented by a line which is perpendicular to the intersection of
the after edge of the rudder-post with the designed load water-line. This
is the case for both single- and twin-screw merchant ships. For some
classes of warships, and for merchant ships having no rudder-post, the
after perpendicular is taken as the centre-line of the rudder stock.
This is the point midway between the forward and after perpendiculars.
This is the maximum beam, or breadth, of the ship measured at
amidships.
This is the rounded plating at the lower corners between the vertical shell
plating and the outer bottom plating.
Longitudinal metacentric radius measured from centre of buoyancy
BMT
Transverse metacentric radius measured from centre of buoyancy
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 solid of constant rectangular cross-section having the
same length, breadth and draught as the ship.
ie: CB = ÷ (L x B x T)
The LPP is normally used in calculating the value of CB which varies with
the type of ship.
Amidships (
Breadth (B)
)
Bilge
Fast ships
0.50-0.65 (fine form)
Ordinary ships
0.65-0.75 (moderate form)
Slow ships
0.75-0.85 (full form)
Camber or round of This is the transverse curvature given to the decks, and is measured by
beam
the difference between the heights of the deck at side and centre. The
amount of camber amidships is often one-fiftieth of the beam of the ship.
Coefficients of
Form is used as a general term to describe the shape of the ship's hull;
form
and when comparing one ship's form with another, the naval architect
makes use of a number of coefficients. These coefficients are of great use
in power, stability, strength and design calculations. Examples are Cb,
Cp, Cw etc.
Centre of flotation This is the centre of the area, or centroid, of the water-plane of a ship.
(F)
For small angles of trim consecutive water-lines pass through F. The
location is normally on the centerline and longitudinally the distance
from AP or amidships is referred to as LCF
Centre of buoyancy This is the centroid of the underwater form of a ship, and is the point
(B)
through which the total force of buoyancy may be assumed to act. Its
position is defined by:
(a) KB the vertical distance above the base, sometimes referred to as VCB
(b) LCB the longitudinal distance measured either from amidships or AP
or FP.
Centre of gravity
This is the point through which the total weight of the ship may be
(G)
assumed to act. It also is defined by:
(a) KG the vertical distance above the base
(b) LCG the longitudinal distance measured either from amidships or AP
or FP
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Naval Architecture Notes
CP
Prismatic coefficient, CP =
AM x L
Depth (D)
This is the vertical distance between the base line and the top of the
uppermost continuous deck measured at the side amidships.
Draught (T)
This is the depth of immersion from the keel to any waterline.
Displacement
This equals the volume ( ) or weight ( ) of water displaced by the hull.
Displacement as a This is volume of water displaced by the ship. It can be imagined as the
volume ( )
volume of the hole in the water occupied by the ship measured in cubic
metres.
Displacement as a This is the weight of water displaced by the ship. It equals the volume
weight ( )
displaced multiplied by a constant representing the density of water, ie:
In fresh water = x 1000 kg/m³
In sea water = x 1025 kg/m³
Weight (or mass) displacement equals the total weight of the ship when
the ship is at rest in equilibrium in still water.
Deadweight
This is the difference between the weight displacement and the lightship
weight. This is the measure of a ship's capacity to carry cargo, fuel,
passengers, stores, etc, expressed in tonnes. The size of tankers is often
given in terms of deadweight tonnage, which is the design deadweight.
Ships are usually chartered on the deadweight tonnage.
Displacement
This represents the designed total weight of the ship. It is the sum of
tonnage
lightship weight and deadweight. The size of warshipsand government
ships is always given in terms of displacement tonnage.
Entrance and run
These are the shaped underwater portions of the ship forward and aft of
the parallel middle body.
Forward
This is represented by a line which is perpendicular to the intersection of
perpendicular (FP) the designed load water-line with the forward side of the stem.
This may be considered to be the height amidships, of the freeboard deck
Freeboard
at side above the normal summer load water-line.
GML
Longitudinal metacentric height measured from centre of gravity
GMT
Transverse metacentric height measured from centre of gravity
Gross tonnage
This is a measure of the total volume of enclosed spaces in a ship
(GRT)
including the under-deck, 'tween-deck spaces and enclosed spaces
above the upper deck. The size of most ordinary merchant ships is
quoted in terms of gross tonnage. Although it unit is tons, it must be
remembered that it is a measure of volume, not weight. 1 ton = 100 ft3.
This is the amount of inclination of the ship in the transverse direction,
Heel ()
and is usually measured in degrees.
IL
Longitudinal moment of inertia of waterplane about amidship
ILCF
Longitudinal moment of inertia of waterplane about F
IT
Transverse moment of inertia of waterplane about centreline
KML
Height of longitudinal metacentre above keel line
KMT
Height of transverse metacentre above keel line
Length between
This is the horizontal distance between the forward and after
perpendiculars (LPP) perpendiculars.
Length on the
This is the length, as measured on the water-line of the ship when
designed load
floating in still water in the loaded, or designed, condition.
water-line (LWL)
Length overall (LOA) This is the length measured from the extreme point forward to the
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Naval Architecture Notes
Lightship weight
MCT1CM
extreme point aft.
This equals the weight of an empty ship i.e. a ship without load. It is
fully equipped and ready to proceed to sea, but with no crew, passengers,
stores, fuel, water, or cargo on board. The boiler or boilers, however, are
filled with water to their working level.
Moment to change trim 1 cm, MCT1CM =
GML
BML
=
pIL
100 L
100L
100L
Midship section
This is the transverse section of the ship amidships.
Base line
This represents the lowest extremity of the ship. At the point where this
line cuts the midship section a horizontal line is drawn, and it is this line
which acts as the datum, or base line, for all hydrostatic calculations.
Normally, this is the underside of keel.
Midship section
This is the ratio of the immersed area of the midship section to the area
area coefficient(CM) 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)
CM values range from about 0.85 for fast ships to 0.99 for slow ships.
Net or register
This represents the tonnage of a ship after certain approved deductions,
tonnage
ie nonfreight earning spaces, have been made from gross tonnage. A
register ton represents 100 cubic feet of volume.
Parallel middle
This is the length over which the midship section remains unchanged.
body (LP)
Prismatic
This is the ratio of the volume of displacement of the ship to the volume
coefficient (CP)
of the circumscribing solid having a constant section equal to the
immersed midship section area AM, and a length equal to the LPP
i.e. CP = ÷ (AM x L)
The Cp is a measure of the longitudinal distribution of displacement of
the ship, and its value ranges from about 0.55 for fine ships to 0.85 for
full ships.
Rise of floor
This is the amount by which the line of the outer bottom plating
amidships rises above the base line, when continued to the moulded
breadth lines at each side.
Sheer
This is the curvature given to the decks in the longitudinal direction, and
is measured at any point by the difference between the height at side at
that point and the height at side amidships.
This is the difference between the draughts forward and aft. If the
Trim
draught forward is greater than the draught aft it is called trim by the
head, or bow. If the draught aft is greater, it is called trim by the stern.
This is the amount by which the midship section falls in from the halfTumble-home
breadth line at any particular depth.
Tonnes per
This is the mass which must be added to, or deducted from, a ship in
centimetre (TPC)
order to change its mean draught by 1 cm.
Water-plane area
This is the ratio of the area of the water-plane to the area of the
coefficient(CWP)
circumscribing rectangle having a length equal to the LPP and a breadth
equal to B.
ie: CWP = AW ÷ (L x B)
The range of values is from about 0.70 for a fine ship to 0.90 for a full
ship.
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Naval Architecture Notes
Courtesy http://www.dynagen.co.za/eugene/hulls/terms.html
Exercise:
Visit these
1.
2.
3.
websites and get acquainted with more ship terms:
http://www.midwestconnection.com/glshpng/glossary.htm
http://www.scribd.com/doc/18008262/Ship-Terms-Glossary
http://cruises.about.com/od/cruiseglossary/Cruise_Ship_and_Na
utical_Term_Glossary.htm
4. http://www.islandregister.com/terms.html
5. http://phrontistery.info/nautical.html
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Naval Architecture Notes
Chapter 3
Hydrostatics and Floatation
3.1 Archimedes Law of Floatation
Archimedes (born 287 B.C) Law states that
“An object immersed in a liquid experience a lift equivalent to the mass of
liquid the object displaces.”
A man immersed in water for example will feel a weight reduction because part of
the weight is supported by buoyancy. This buoyancy is equal to the weight of water
displaced by his immersed body.
3.2 Reduction of Weight of Immersed Objects
The maximum buoyancy is when the object is fully immersed and this equal the
total outside volume of the object multiplied by the density of the fluid. 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 weight reduction.
Example 3.1:
Consider a cuboid having dimensions 1m x 1m x 2m. 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.
? tonne
In this case, displaced volume = 1 x 1 x 2 =
2 m3
This is the volume of liquid pushed aside by
the cuboid.
Archimedes says that the weight of this
object in liquid is reduced due to the support
given by liquid on the object. The apparent
weight equals the weight in air minus the
reduction in weight of the object; or the
buoyancy i.e.
Buoyancy
=
=
=
=
=
=
2m3
volume diplacement x density of liquid
mass displacement
2m3 x 1000 kg/m3
2000 kg
2 tonnes
reduction in weight
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Naval Architecture Notes
Apparent weight = weight in air – buoyancy
Since the object weighs 3 tonne in air, it will apparently weigh only 1 tonne in
water.
Exercise 3.1
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?
3.3
What make a Ship Floats?
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
balances 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 provided by the outer shell of the ship,
the ship will sink.
3.4
Effect of Density
An object experiences buoyancy force equivalent to the weight of fluid it displaces.
For a particular object, the buoyancy force will depend on the density of the fluid,
since its volume is constant. This explains for example why a bather will feel more
buoyant while swimming at sea compared to in the river or lake. Also, a floating
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Naval Architecture Notes
object of constant weight will sink at a deeper draught in freshwater compared to in
seawater.
Total weight W = Buoyancy = Displaced volume x liquid
Since weight does not change, the buoyancy is also constant. So displaced volume
will be inversely proportional to the density of fluid. For floating object, this will
determine its level of sinkage or draught.
3.5
Some Simple Problems
The fact that a floating object displaces fluid equivalent to its weight can be used to
solve a number of problems.
= Displaced volume x
Total weight W = Buoyancy
water
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.
Just to understand the concept, consider a floating box of dimension L x B x D,
floating at a draught T.
CASE 1: We know its weight, we can find its draught
In this case, we know the weight of the object, we can find the displaced volume:
Displaced volume
=
W
water
i.e. for a box-shaped vessel:
Displaced volume = L x B x T
Hence draught T of the cuboid can be found.
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Naval Architecture Notes
Example 3.2
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
=LxBxT
Weight of displaced water = 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 3.2
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, 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 3.3
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 =
=LxBxT
Weight of barge =
Weight displacement,
W
=
= x salt water
= 100 x 20 x 5 x 1.025
= 10250 tonnes
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Naval Architecture Notes
Exercise 3.3
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 3.4
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.
Exercise 3.5
1.0m
A cylindrical container weighing 5 tonne floats
with its axis vertical. If the diameter is 1.0m, find
its draught in:
i.
sea water
ii.
oil of density 870 kg/ m3.
Exercise 3.6
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.
3.6
Hydrostatic Particulars
A floating object will be at a certain draught depending on the total weight of the
object, density of water and the shape of the object. For a ship, the shape of the
object has strong influence on the draught of the ship; the shape and draught have
to provide enough buoyancy to support the ship.
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Naval Architecture Notes
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, WSA
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 out of
water, and draught becomes zero, the particulars ceased to exist.
As long as draught and trim is 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. Once draught or trim changes, the particulars
will also change.
This change in draught will normally occur due to changes in total weight of the
ship, or if a force is applied to the ship to make it sink to a deeper draught.
Example 3.4
A box 2m x 1m (LxB) in sea water is floating at a draught of 0.3m.
Calculate its , , CB, CWP and TPC.
i. = L x B x T = 2 x 1 x 0.3 = 0.6m3
ii. = L x B x T x = x = 0.6 x 1.025 = 0.615 tonnes
iii. CB =
LxBxT
=
0.6 = 1.00
0.6
iv. CWP = Awp
2x1
=
LxB
2x1
v. TPC = Awp x
100
=
=
1.00
2 x 1 x 1.025
100
= 0.0205
Exercise 3.7
Calculate the particulars at draught of 0.4, 0.5, 0.6 and 0.7m.
Exercise 3.8
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 2300 tonne, what is its draught? If the barge is floating at a
draught of 4m, what is its CB?
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Naval Architecture Notes
It can be seen from Exercise 3.8 that for a box-shaped object at different draughts,
the waterplane areas are constant. Hence, many hydrostatics particulars remain
constant.
Exercise 3.9:
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.
iii. 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
3.7
D 4
64
.
Hydrostatic Particulars of a Ship
Hydrostatic particulars of a real ship will be different. Consider the ship whose lines
plan is shown below. At different draughts, the ship will have different waterplane
areas, volumes and centroids. Hence, the hydrostatic particularly will vary as the
draughts changes.
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Naval Architecture Notes
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.
Exercise 3.10
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
2
(m )
(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 ro
LBT
Aw
(LB)
Aw x ro
100
The particulars can be presented in two forms, either as a set of curves or in tabular
format. Table 3.1 shows a typical table of hydrostatic particulars while an example
of hydrostatic curves is shown on page 18.
Table 3.1 Hydrostatic Particulars of Bunga Kintan LBP 100m
Draught Displacement
(m)
(tones)
Cb
KB
(m)
BMT
(m)
BML
(m)
MCTC
(tonne-m)
LCB
LCF
(m from
(m from )
)
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
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Naval Architecture Notes
3.8
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 3.5
From MV Bulker hydrostatic Curves (pg18) 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 3.11
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?
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.
HOMEWORK 1:
By using the hydrostatic particulars of Bunga Kintan shown in Table 3.1:
i.
ii.
iii.
iv.
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?
Submission Date: _______________________
Exercise 3.11:
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.
© Omar bin Yaakob, July 2008
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Naval Architecture Notes
MV Bulker Hyrostatic Curves
© Omar bin Yaakob, July 2008
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Naval Architecture Notes
Chapter 4
Basic Stability Consideration
4.1
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. For the ship's master, he must be able to load
and stow cargo and handle the ship while ensuring that the ship will be stable and
safe.
4.2
What is stability?
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.
Figure
4.1
An
analog
y for
stability is often given of the marble. In Figure 1 (a), the marble in the bowl will
return to its original position at the bottom of the bowl is it is moved to the left or
the right. This marble is in a condition called positively stable. A slight push on
the marble which is put on an upside down bowl as in Figure 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.
(a)
4.3
(b)
(c)
Longitudinal and Transverse Stability
Ship initial stability can be seen from two aspects, longitudinally and transversely.
From longitudinal viewpoint, the effect of internal and external moments on ship's
trim is considered. 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. The trim angle is rarely taken into consideration.
Transverse stability calculation considers the ship stability in the port and
starboard direction. We are interested in the behaviour of the ship when external
statical moment is applied such as due to wind, waves or a fishing net hanging from
the side. The effect of internally generated moment such as movement of masses
on-board transversely is also studied. An important relationship considered is that
© Omar bin Yaakob, July 2008
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Naval Architecture Notes
between heeling and righting moments and the resulting angle of heel and its
consequence on the safety of the boat.
This Chapter will focus on basic transverse stability particularly the relationships
between the metacentre and the centre of gravity.
4.4
Basic Initial Stability: The role of GM
w
MT
MT
w
G
W
G
L
W
w1
B
W
B
L1
L
B1
K
W
K
(a)
(b)
Figure 4.2
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. When the ship is slightly disturbed from upright, the
centre of buoyancy being centre of underwater volume moves to the right. The line
of action of buoyancy vertically upward crosses 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 moment.
In this case, M was originally above G and we can see that the returning moment is
positive. If M was below G i.e. GM negative, the returning 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.
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Naval Architecture Notes
Consider the triangle shown below:
Righting moment = x GZ
MT
and
GZ = GMT sin
For any displacement, righting moment depends
on GZ.
G
And GZ depends on GM. The bigger GZ, the
bigger Righting Moment.
Z
MT
G
Relationships between K, B, G and MT are important.
B
KMT = KB + BMT
K
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. Therefore the distance GMT will
only depend on the height of centre of gravity. In other words, to ensure a large
GMT, we can only ‘control’ KG.
4.5
Determining Centre of Gravity, Areas or Volumes of Composite Bodies
The above section has shown that the relative position of M and G are important in
determining ship stability. Since M is constant for any particular draught, only G
will finally determine the value of GM.
Before we go into the details of stability calculations, we have to consider how to
determine the location of G. Consider a composite body consisting of two portions
shown in Figure 4.3.
Area A
ca
C
cb
Area B
xb
xa
Figure 4.3
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X
21
Naval Architecture Notes
Distance of Centre of Composite to the reference axis:
X = A x xa + B x xb
A+B
i.e.,
X = Total moment of area about the reference axis
Total area
If the composite consists of volumes,
Centre of Volume
X = Total moment of volume about the reference axis
Total volume
If the composite consists of weights,
Centre of Gravity
X = Total moment of weight about the reference axis
Total weight
Example 4.1
Find centre of area (from AP) for an object consisting of four components shown in
the figure below.
Component
Area
(m2)
1
2
3
4
TOTAL
Centroid from AP
© Omar bin Yaakob, July 2008
Distance from AP
(m)
-2.5
Moment of Area
about AP (m3)
= Total moment of area about AP
Total area
=
m
22
Naval Architecture Notes
Example 4.2
A trimaran has three hulls and the respective volume displacements, LCB and KB
are shown below. Find the total displacement, LCB and KB.
Hull
Side 1
Main
Side 2
Hull
Side 1
Main
Side 2
TOTAL
Volume
Displacemen
t (m3)
158.7
1045.8
158.7
Volume
Displace
ment
(m3)
158.7
1045.8
158.7
Lcb (m aft of Kb
amidships)
(m above keel)
13.0
2.0
13.0
2.5
2.0
2.5
lcb (m aft Moment about Kb
Moment
of
amidships (m4)
(m
above about keel
amidships)
keel)
(m4)
13.0
2.0
13.0
2.5
2.0
2.5
LCB= Total moment about amidships
Total Volume
=
4.56 m aft of amidships
KB= Total moment about keel
Total Volume
=
2.12 m
Example 4.3
A stack of weights consists of one 3kg weight and two 2kg weights. Find centre of
gravity of the stack above the floor:
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Naval Architecture Notes
Item
Weight (kg)
CG above floor
(cm)
Moment about Floor
(kgcm)
Wt A
Wt B
Wt C
JUMLAH
Final CG =
=
cm
Example 4.4
A ship has three parts and the respective weights and Kg are as follows. Find the
total weight and KG.
Part
Lightship
Cargo 1
Cargo 2
Weight
(tonnes)
2000
300
500
Kg
(m above keel)
5.5
7.6
2.5
Part
Weight
(tonnes)
Lightshi
p
Cargo 1
Cargo 2
TOTAL
2000
Kg
(m above
keel)
5.5
300
500
7.6
2.5
Moment
about keel
(tonne-m)
KG = Total moment about Keel =
Total weight
m
Example 4.5
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
500 tonnes, Kg 3.5 m
750 tonnes, Kg 9.0 m
The following cargo is then discharged:
450 tonnes of cargo Kg 0.6 m
And 800 tonnes of cargo Kg 3.0 m
Find the final GM.
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Naval Architecture Notes
Item
Weight (tonne)
Kg
Ship
Loaded
Cargo1
Cargo2
Cargo3
Unloaded
Cargo
6000
1000
500
750
6.0
2.5
3.5
9.0
-450
-800
0.6
3.0
Final KG
=
=
Final KG
=
Moment about keel
(tonne-m)
Final moment
Final displacement
________
Final KM
=
Final KG
=
Ans. Final GM =
m
m
m
m
Homework 2
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.
Submission Date: _______________
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Naval Architecture Notes
4.6
Movement of Centre of Areas, Volumes and Weights
When a portion is added or removed from an object, its centre moves.
Consider a homogenous object as shown below:
i. If weight is moved a distance y:
Centre of gravity moved x = GG’ = m x y
M
i.e. total moment divided by total weight
ii. If weight m is removed:
The remaining weight M-m
Movement of centre of gravity x = GG’ = m x d
M-m
i.e. total moment divided by remaining weight.
Example 4.6
A ship weighing 7000 tonnes is floating at the wharf. At that time, KM = 6.5 m and
GM 0.5m. Find new GM when a 30 tonnes box is loaded at Kg 10.0m. Assume no
change in KM.
Method 1:
Find rise in KG
Original KG = KM - GM =
m
Distance 30 tonnes box from original G =
GG’= 30 x 4.0 = 0.017m
7030
M
G’
m
G
B
KG’= KG+ GG’ =
m
10
m
K
KM does not change, therefore, GM =
© Omar bin Yaakob, July 2008
=
m
26
Naval Architecture Notes
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
4.7 Hanging Weights, The Use Of Derricks And Cranes
The use of cranes and derricks will make the weights suspended. Suspended
weights acts 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 by the derrick. The centre of gravity
KG will suddenly increase and since KM is constant, GM will reduce suddenly. If
the rise in KG is more than the original GM, the net GM will be negative, leading to
instability.
Example 4.7
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.
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
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Naval Architecture Notes
4.8 Free Surface Correction
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
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 being considered.
fluid
is the
Once the FSC is known, the new reduced GM called GMfluid is obtained
GM
fluid
= GMsolid - FSC
It is important that free surface be avoided or at least minimised.
Note also that KG in ships having free surface is
increased by FSC.
called KG fluid
and regarded
KGfluid = KGsolid + FSC
For tanks with a rectangular surface:
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 density of fluid
12
ship displacement
EXERCISE 4
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.
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Naval Architecture Notes
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?
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.
© Omar bin Yaakob, July 2008
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Naval Architecture Notes
Appendix A
Second Moments of Areas
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. Let dA in Figure A.1
represent an element of an area and let y be its distance from the axis AB
dA
y
A
B
Fig. A.1
The second moment of the element about AB is equal to dA x y2
2. To find the second moment of a rectangle about an axis parallel to one of its
sides and passing through the centroid
l
dx
x
b
G
B
A
Fig. A.2
In Figure A.2, l represents the length of the rectangle and b represents the breadth.
Let G be the centroid and let AB, an axis parallel to one of the sides, pass through
the centroid.
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Naval Architecture Notes
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
l. x
1AB
2
.dx
- b/2
b/2
1AB l
x 2 .dx
- b/2
b / 2
x3
l
3 b / 2
1AB
lb 3
12
3. To find the second moment of a rectangle about one of its sides.
l
dx
b
x
B
A
Fig. A.3
Consider the second moment (i) of the elementary strip shown in Figure A.3 about
the axis AB.
i= l dx x x2
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Naval Architecture Notes
Let IAB be the second moment of the rectangle about the axis AB. Then :b
1AB l. x 2 .dx
O
b
x3
l
3 O
or
1AB
lb 3
3
4. 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 axis parallel to the first reduced by the
product of the area and the square of the perpendicular distance between the
two axes. Thus, in Figure A.4, if G represents the centroid of the area (A) and
the axis OZ is parallel to AB then:-
I OZ I AB - Ay2
A
B
y
O
Z
G
Fig. A.4
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Naval Architecture Notes
5. Second moment of area of a circle
X
y
A
D
B
X
Fig. A.5
For circle, the second moment of area about an axis AB.
I AB
D4
64
What is IXX?
6. Applications.
Second moment of areas are used in calculations of BML and BMT :
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.
© Omar bin Yaakob, July 2008
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Naval Architecture Notes
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.
6m
4m
Diameter
1m.
If the total weight of the structure is 3 tonnes, floating in sea water calculate:
i)
ii)
iii)
draught
KMT
KML
Homework 3:
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
Submission Date: _______________
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Naval Architecture Notes
Chapter 4 Calculations of Ship
Hydrostatic Particulars
4.1
The Importance of Hydrostatic Particulars
In the previous chapter, we have seen the importance of knowing the hydrostatic
particulars of a vessel. If we have the hydrostatic particulars in the form of tables,
curves, or our own direct calculation, we can obtain details about the ship in any
particular condition. We can also determine or estimate what would happen when
ship condition changes such as due to addition or removal of weights.
To draw hydrostatic curves or to make the table, we need to calculate the
particulars. The hydrostatic particulars can be obtained only if we carry out
calculations of area, volumes and moment at various draughts or water plane area.
Using some known relationships, the particulars can be derived from areas,
volumes and moments,
If the body has a uniform shape, such as cuboids, cones, spheres or prisms,
calculation of areas, volumes and moments are easy. For example water plane
areas, block coefficients, TPC, MCTC, KB and LCB of such objects can be found
using simple formulae. We can easily obtain the particulars at any draught and if
necessary plot the curves.
However not all ships have simple and uniform shapes as above. In fact, most
ships have hull shapes which are varying in three directions. This makes it difficult
to calculate hydrostatic particulars.
4.2
Methods to Calculate Areas
Figure 4.1 Typical Ship Half-Breadth Plan
Consider the shape of the ship whose body plan is shown in Figure 4.1. If we want
to find the area of the section or water plane for example, we do not have simple
methods. Similarly to find volume displacement or LCF will not be easy.
If we want calculate the water plane area of the ship in Figure 4.1 at a particular
draught, we may use a few methods.
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Naval Architecture Notes
First is to plot the curve on a graph paper from where the area under the curve can
be obtained by counting the squares. To improve accuracy, smaller boxes or
triangles can be used. The method is tedious and it’s accuracy depends on the size
of the smallest grid. To use this method, we need to plot the curve first; a
disadvantage when sometimes we are only provided with offset data, i.e. halfbreadth at various stations.
The second method is to use an equipment called the planimeter. This equipment
can be used to measure the area of a shape drawn on paper. Again, this equipment
can only be used only when hard copy of the waterline drawing is available.
Moreover, similar to graphical method, planimeter requires a lot of man power.
Figure 3.2 A Planimeter
The third method is to use mathematical approximation. In this method, an
attempt is made to represent the curve or shape by a mathematical expression. By
using calculus, area and moments of the area bounded by the curve can be found
by integration.
Mathematical methods are normally preferred for a number of reasons. First there
is no need for a hard copy of the curves. Offset tables are normally available and
the data can be used directly in the calculations. A very important feature of
mathematical methods is the ability to make use of the technology offered by
computers. The use of mathematical methods also enable us to obtain not only
areas but all hydrostatic particulars. As we have seen in chapter 3, we need to
calculate not only areas but also volumes, positions of centroids of waterplanes
(LCF) and centroids of volumes (KB and LCB). In addition we require second
moments of areas for calculations of MCTC and metacentric heights. Unlike
graphical or planimeter methods, mathematical methods can easily be used to
calculate these particulars.
A very important caution should be noted when using mathematical methods. The
accuracy of the calculations will mainly depend on the degree of fit of the actual
curve to the mathematical expression representing it.
4.3
Mathematical Methods
offset
© Omar bin Yaakob, July 2008
ST1
ST2
36
h
ST3
ST4
ST5
Naval Architecture Notes
Figure 4.3 Waterline or Sectional Area Curve
Figure 4.3 shows a curve which may represent a half-waterplane area or a curve of
sectional areas. A waterplane curve is represented by offsets made up of halfbreadth at various stations. Stations are positions along the length of the ship and
normally separated by a common-interval, h. To cater for the fast changing slopes
of the curve at the stern and bow regions, half stations may be used.
To calculate the area, centroid and moment under such curve, its offsets and h are
required. By assuming that the curve can be represented by a certain
mathematical formulae, calculations can be made. A number of methods have been
developed for these purpose such as Newton-Cottes, Tchebycheff, Trapezoidal and
Simpson methods. In this course, we will concentrate on the two most popular
methods; Trapezoidal and Simpson methods.
4.4
Trapezoidal Method
When a curve can be assumed to be represented by a set of trapezoids, the area
under the curve can be calculated.
D
E
A
y3
y4
C
B
y2
y1
h
A
F
Figure 4.4 Waterline or Sectional Area Curve
In Figure 4.4, the area under the curve is the are area of trapezoid ABCDEF.
Area
=
=
1
y1 y 2 h 1 y 2 y3 h 1 y3 y 4
2
2
2
1
h y1 2 y 2 2 y3 y 4
2
Exercise
1. Find the Trapezoidal formulae for curves made up of
© Omar bin Yaakob, July 2008
37
Naval Architecture Notes
i)
ii)
iii)
6 offsets
9 offsets
n offsets
2. The midship section of a chine vessel has the following offsets:
Draught(m) 0
Half-breadth (m)
0
0.25
0.6
0.5
1.0
0.75
1.5
1.0
1.9
Calculate its midship section coefficient at draught of 1.00m.
3. Find the water plane area of a ship LBP = 10m made up of the following offsets:
Station
Half-breadth
(m)
0
0
1
0.3
2
1.0
3
1.2
4
1.1
Find its area, TPC and waterplane coefficient
4.5
Simpson Rules for Areas.
Simpson rule is the most popular method being used in ship calculations to
calculate volumes, second moments of areas and centroid. This is because it is
flexible, easy to use and its mathematical basis is easily understood.
Basically, the rule states that the ship waterlines or sectional area curves can be
represented by polynomials. By using calculus, the areas, volumes, centroids and
moments can be calculated. Since the separation between stations are constant,
the calculus has been simplified by using multiplying factors or multipliers.
There are three Simpson rules, depending on the number and locations of the
offsets.
4.5.1 Simpson First Rule
Simpson’s First Rule
D
C
B
y3
y2
y1
h
-h
A
O
E
A
Figure 4.5 Waterline or Sectional Area Curve with Three Offsets
© Omar bin Yaakob, July 2008
38
Naval Architecture Notes
Assume that the offsets are points on a polynomial curve of form
y a0 a1 x a2 x 2 a3 x 3
Then area ABCDE =
h
h
y.x
= 2a0 h+ 2a2h3
3
At x= -h y1 = a0 - a1h + a2h2 - a3h3
At x= 0 y2 = a0
At x= h y3 = a0 + a1h + a2h2 + a3h3
a0 y 2
Therefore,
a2
y1 y3 2 y 2
2h 2
Substituting these values into the above equation
Area ABCD =
h
y1 4 y 2 y3
3
First Rule is used when there is an odd number of offsets. The basic multiplier for
three offsets are 1,4,1. For more stations, the multipliers are developed as follows:
Station
1
2
3
4
5
6
7
Offset
a1
a2
a3
a4
a5
a6
a7
Multiplier
1
4
1
1
4
2
4
1
1
2
4
4
1
1
1
4
Area = 1/3 x h x (multiplier x offset)
Where h = common interval
Exercise
4. Use Simpson first rule to find the area of the midship section of the chine vessel
in exercise 2. Explain the difference in area.
© Omar bin Yaakob, July 2008
39
Naval Architecture Notes
© Omar bin Yaakob, July 2008
40
Naval Architecture Notes
Example 1
Find the waterplane coefficient for the waterplane of a 27m LBP boat represented by
the following offsets:
Station
0
1
2
3
4
5
6
Half-breadth (m) 1.1 2.7 4.0 5.1 6.1 6.9 7.7
.
.
Simpson
Multiplier
1
4
2
4
2
4
1
Offset
Station
0
1
2
3
4
5
6
1.1
2.7
4.0
5.1
6.1
6.9
7.7
Area= 1/3
x h x ∑ Product
Area
Product
Area
= ___________ m2
Cw = _______
4.5.2 Simpson Second Rules
D
E
A
y3
y4
C
B
y2
y1
h
A
-3h
2
-h
2
0 h
2
F
3h
2
Figure 4.6 Waterline or Sectional Area Curve with Four offsets
Assume that the offsets are points on a polynomial curve of form
y a0 a1 x a2 x 2 a3 x 3
© Omar bin Yaakob, July 2008
41
Naval Architecture Notes
Then area ABCD
=
3h / 2
y.x
3 h / 2
a1 x 2 a 2 x 3 a3 x 4
a 0
2
3
4
=
=
3a0
But
3a1 h 9a 2 h 2 27a3 h 3
2
4
8
3h
x
2
y1 a0
h
x
2
a1 h a 2 h 2 a3 h 3
y 2 a0
2
4
8
a1 h a 2 h 2 a3 h 3
y3 a0
2
4
8
3a1 h 9a 2 h 2 27a3 h 3
y 4 a0
2
4
8
h
x
3h
2
x
2
Adding (2)
2a 0
(1)
3
a2 h 2
4
(2)
and (4)
a2 h 2
y1 y 2
2
Adding (2) and (5)
2a 0
9a 2 h 2
y1 y 4
2
4a2 h 2 y1 y 4 y 2 y3
a2
y 1 y 4 y 2 y3
4h 2
a0
y 2 y3 a2 h 2
2
2
9 y 2 9 y 3 y1 y 4
16
16 10 10
Then area ABCD
=
3
h y1 3 y 2 3 y3 y 4
8
The basic multipliers are thus 1,3,3,1 and Area = 3/8 x h x (multiplier x offset)
The rule can only be used when number of offsets = 3N +1
© Omar bin Yaakob, July 2008
42
Naval Architecture Notes
4.5.3 Simpson Third Rule
Simpson third rule is used when we have three offsets and we require the area
between two of the offsets.
Example 2
7.02
5.98
1.06
1.06
Figure 4.7 Midship Section Curve with Three offsets
A midship section curve has halfbreadth 1.06, 5.98 and 7.02 m spaced at 9.0m
draught interval. Find the area between the first two draughts.
½ Breadth
1.06
5.98
7.02
Multiplier
5
8
-1
Product
5.30
47.84
-7.02
46.12
Area= 1/12 x 9 x 46.12 x 2= 69.18 m2
If we require the area between two upper draughts, the calculations are as follows:
½ Breadth
7.02
5.98
1.06
Multiplier
Product
5
8
-1
35.10
47.84
-1.06
81.88
Area = 1/12 x 9 x 81.88 x 2 = 122.82 m2
Total Area = 192.0 m2
Exercise
5.
Find the total area under the curve using Simpson first rule and compare.
© Omar bin Yaakob, July 2008
43
Naval Architecture Notes
5.6
Obtaining Volume
Volumes and hence displacement of a ship at any draught can be calculated if we
know either:
i)
ii)
Waterplane areas at various waterlines up to the required draught
Sectional areas up to the required draught at various stations.
Example 3
Sectional areas of a 180m LBP ship up to 5m draught in sea water at constant
interval along the length are as follows. Find its volume displacement, mass
displacement and prismatic coefficient.
Station
Area
(m2)
0
5
1
2
3
4
5
6
7
8
9
10
118 233 291 303 304 304 302 283 171 0
Figure 3.8 Section Shapes at
Various Stations
Statio
n
0
1
2
3
4
5
6
7
8
9
© Omar bin Yaakob, July 2008
Section Area
Simpso
n
Multipli
er
Produc
t Vol
5
118
233
291
303
304
304
302
283
171
44
Naval Architecture Notes
10
0
∑f vol
Volume
=
1
h fvol
3
=
____________ m3
Similarly if we have waterplane areas, we can use Simpson rules to integrate the
areas to obtain volume. In this case the common interval is the waterline spacing.
4.7
Considering Half and Quarter Stations
Rapidly changing curvature at both ends of the ship necessitates the use of half and
quarter stations. To take this into consideration, Simpson Multipliers are also
divided as follows:
ST
1
½
1
1½
2
3…
Figure 4.9 Shape with Half Stations
If we consider full stations only:
1
4
2
4…
1
2
1
4
1
1 1/2
4
1
If we consider half stations:
1
2
1/
4
2
1
2
1
2
2
2
4
2
1
2
Example 4
A waterplane for a 120 m LBP ship has the following offsets:
Station
1/2 ord
0
0.6
1/4
1/2
3/4
2.8
4.0
5.2
1
6.2
2
9.0
3
9.8
4
8.4
5
4.8
51/2
2.2
6
0.0
51/2
6
Find the waterplane area, waterplane coefficient and TPC for
the waterplane.
Station
0
1/
4
© Omar bin Yaakob, July 2008
1/
2
3/
4
1
2
3
4
5
45
Naval Architecture Notes
Multiplie
r
1/4
1/
4
1
1/4
1
1/
4
1/
2
© Omar bin Yaakob, July 2008
1
1
1
4
1
1/
4
11/4
4
1
2
4
4
1/2
2
1/2
1
11/2
2
1/
2
46
Naval Architecture Notes
Statio
n
1/2
SM
Product
ord
0
0.6
1/
4
1/4
2.8
1
1/2
4.0
1/2
3/4
5.2
1
1
6.2
2
9.0
4
3
9.8
2
4
8.4
4
5
4.8
5.5
2.2
2
6
0.0
1/2
. .Area= 1/3
Area
Lever
Product
Product
1st moment
2nd mmt
11/4
11/2
x h x ∑ Product
Area
= ___________ m2
Cw = _______
TPC = ________
4.8
Obtaining LCF, LCB and Longitudinal Second moment of Area
dx
y
x
A
© Omar bin Yaakob, July 2008
47
Naval Architecture Notes
Area
=
1st moment
=
ydx
ydx x
x . ydx
=
2st moment
=
=
y
dx x 2
x
ydx
2
=> 1/3 x productA x h
=> 1/3 x product1stmmt x h x h
=> 1/3 x product2ndmmt x h x h2
Values of x are given in multiples of h, the common interval.
If the product for area is multiplied by multiples of h, called levers, the sum of
products can be used to find the first moment and hence the longitudinal position
of the centroid.
LCF
1
h h product 1st moment
3
=
1
h product area
3
product 1st moment x h
=
product area
If the offsets are half-breadths, the centroid is LCF. If the offsets are sectional
areas, the centroid is centre of volume i.e. LCB. The LCF is measured from the axis
where levers are taken.
For second moment, Simpson’s product for areas are multiplied twice with levers.
Again, the second moment are taken about the axis from where levers are taken.
IL
1
h h h product 2nd moment
3
Example 5
Find the area, LCF , second moment of area about amidships, transverse second
moment of area about centreline for the waterplane of a ship LBP 180m with the
following ordinates.
Stesen
1/2ord (m)
AP
0
1/2
5
© Omar bin Yaakob, July 2008
1
8
2
10.5
3
12.5
4
13.5
5
13.5
6
12.5
7
11.0
8
7.5
9
3.0
91/2
1.0
FP
0
48
Naval Architecture Notes
AP
½
ordinat
e
0
½
5.0
2
10.0
+4 ½
+45.0
4½
+202.5
1
8.0
1½
12.0
+4
+48.0
+4
+192.0
2
10.5
4
42.0
+3
+126.0
+3
+378.0
3
12.5
2
25.0
+2
+50.0
+2
+100.0
4
13.5
4
54.0
+1
+54.0
+1
+54.0
5
13.5
2
27.0
0
Sum_aft
0
-
Station
SM
Product
Area
Lever
Product
1stmmt
Lever
Product
2ndmmt
½
-
+5
-
+5
-
+323.0
6
12.5
4
50.0
-1
-50.0
-1
+50.0
7
11.0
2
22.0
-2
-44.0
-2
+88.0
8
7.5
4
30.0
-3
-90.0
-3
+270.0
9
3.0
1½
4.5
-4
-18.0
-4
+72.0
9½
1.0
2
2.0
-4 ½
-9.0
-4½
+40.5
FP
0
½
-
-5
-
-5
-
278.5
Sum_fwd
1447.0
-211.0
Waterplane Area =
LCF =
1 180
278.5 2 3342.0 m 2
3 10
product 1st moment x h
product area
LCF =
IL
(323 211) 180
7.24 m aft of amidships
278.5
10
1
h h h product 2nd moment
3
IL = 2 x 1 x 183x 1447.0 = 5,625, 936 m4.
3
© Omar bin Yaakob, July 2008
49
Naval Architecture Notes
Exercise 6:
1. Repeat the Example 5 but this time,
Calculate LCF from AP and 2nd Moment of area about AP.
Check that the answers are identical.
2. Calculate the centroid of the midship section in Example 2 measured from
the top-most waterline (page 8).
3. Calculate LCB of the vessel in on Example 3.
4. Calculate LCF from amidship and longitudinal second moment of area about
amidship of the ship in Example 4 on page 11.
4.9
Obtaining Second Moment Of Area About The Centreline
dx
y
x
A
If the shaded area is a rectangle, second moment of area about the x-axis is
i
=
1
dxy 3
3
for the whole area :
IT
1 3
y dx
3
If the ordinates are cubed and Simpson multipliers are applied,
1 1
I T h product 2nd moment
3 3
1
h product 2nd moment
9
© Omar bin Yaakob, July 2008
50
Naval Architecture Notes
Example
Find BMT for a waterplane of a ship LBP = 100m with the following half breadths.
At this draught the ship has a displacement of 11275 tonnes in sea water.
AP
0
½
5
1
8
2
10.5
Station
3
12.5
4
13.5
5
13.5
6
12.5
7
11
8
7.5
(½
ordinate)3
SM
AP
½
ordinat
e
0
-
½
Product for
Second
Moment T
-
½
5.0
125.0
2
250.0
1
8.0
512.0
1½
768.0
2
10.5
1157.6
4
4630.4
3
12.5
1953.1
2
3906.2
4
13.5
2460.4
4
9841.6
5
13.5
2460.4
2
4920.8
6
12.5
1953.1
4
7812.4
7
11.0
1331.0
2
2662.0
8
7.5
421.9
4
1687.6
9
3.0
27.0
1½
40.5
9½
1.0
1.0
2
2.0
FP
0
-
½
-
9
3
91/2
1
FP
0
36521.5
2nd Moment = 1/3 x
81158.9 m4
about amidships
1/3
x h x product mmt x 2 =
Volume Displacement
BMT
= 11275 = 11000 m3
1.025
= 81158.9 = 7.38m
11000
4.10 Appendages
Appendages are the portion of the hull which is protruding from the main body. It
may be part of underwater volume such as a skeg or keel or parts of a waterplane
area which is not suitable to be integrated with the main area due to its abrupt
change in area.
Areas, volumes and moment are calculated separately for the appendages and later
incorporated using composite body method explained in Chapter 4 of NA1 notes.
© Omar bin Yaakob, July 2008
51
Naval Architecture Notes
Example
A ship length 150m, breadth 22m has the following areas at the various draft.
Draught (m)
Area of
Waterplane(m2 )
2
1800
4
2000
6
2130
8
2250
10
2370
There is an appendage (between waterline 0 and 2m) with displacement 2600 tonne
in sea water and Kb of 1.2m. Find the total displacement, KB and Cb of the ship at
10m draught.
Solution:
Draught
(m)
Aw (m2 )
Multiplier
2
4
6
8
10
1800
2000
2130
2250
2370
1
4
2
4
1
Product
for
Volume
1800
8000
4260
9000
2370
25,430
Lever
0
1
2
3
4
Product
for 1st
Moment
0
8000
8520
27000
9480
53,000
Volume Displacement = 1/3 x 2x 25430 = 16960 m3
Mass = 16960 x =17380 tonne
Centre of Buoyancy = 53000 x 2 = 4.16m above 2m WL.
25430
Composite Table
Portion
Main(2m10m)
Appendage
Total
Displacemen
t
(tonnes)
17380
2600
19980
KB
Moment
6.16
107,000
1.20
3,120
110,120
KB = 110,120 = 5.51m
19980
CB =
19980
150 x 22 x 10 x 1.025
© Omar bin Yaakob, July 2008
= 0.59
52
Naval Architecture Notes
© Omar bin Yaakob, July 2008
53
Naval Architecture Notes
4.11 Simpson Rules for Radial Integration
C
dθ
r
A
B
Strip Area
=
1 2
r dθ
2
Total Area
=
2r
=
1
r 2d θ
2
1
2
dθ
In Simpson terms, if first rule is used;
Total area
=
1 1
h fA
2 3
in radians
Example
e.g.
A figure is bounded by two radii at right angles to each other and a plane
curve. The polar coordinates of the curve at equal interval of angle are 10,9,8,7,6,5
and 4 meters respectively. Find the area of the figure and its centroid from the 10m
radius.
© Omar bin Yaakob, July 2008
54
Naval Architecture Notes
Angle
r
r2
SM
0
15
30
45
60
75
90
10
9
8
7
6
5
4
100
81
64
49
36
25
16
1
4
2
4
2
4
1
Area
r3
Sin
angle
r3xsinxSM
3157.95
1
h
fA
2
3
Product
for
Area
100
324
128
196
72
100
16
936
1 1 90
936
2 3 6 180
= 40.6 Sq. metres
Centroid is measured perpendicular from one boundary
1st moment about AB =
1
r 3 sin θ d θ
3
moment
area
Centroid `x
=
=
=
1 1
h fmmt
3 3
1 1
h f
2 3
2 fmmt
3
f
Centroid from 10m boundary = 2 x 3157.95 = 2.25m
3 x 936
Exercise 7
Find area bounded by a plane curve and two radii 900 apart, if the lengths of the
radii at equal angle intervals are 2,3,5,8, and 10 metres respectively. Also find the
distance of the centroid of the figure from the 2m radius.
© Omar bin Yaakob, July 2008
55
Naval Architecture Notes
4.12 Tchebycheff’s Rule
When y2 is the middle ordinate and y1 and y3 are located 0.7071l to the left and
right of y2,
Area = C (y1 +y2+y3) where C= L/number of ordinates and l is 0.5L
Ordinates are not equally spaced and their positions in the length depend on
number of ordinates, n.
n
2
3
4
5
6
10
Position of ordinates from centre of length
expressed as fraction of half length
0.5773
0
0.7071
0.1876
0.7947
0
0.3745
0.9325
0.2666
0.4225
0.8662
0.0838
0.3127
0.50 0.6873
0.9162
Example
Find area of a 200m waterplane if the half breadth at Tchebycheff stations are as
follows:
1.2, 5.0, 8.4, 10.5, 11.7, 11.8, 11.1, 9.6, 7.4, 3.8
C= L/10 = 200/10
Sum of y =
Area = 3220 m2
© Omar bin Yaakob, July 2008
56
Naval Architecture Notes
EXERCISES 4
Question 1
A cargo ship 120m, breadth 25m and depth 16m is floating at 8.5m draught
in sea water. The area of sections at various stations are shown in the
following table:
Statio
n
As
(m2)
AP
1
2
3
4
5
6
7
8
9
FP
12.8
64.5
100.
0
120.
6
154.
2
166.
8
140.
7
125.
9
97.6
43.2
0.0
Calculate
i.
ii.
iii.
iv.
v.
Mass Displacement
Longitudinal Centre of Buoyancy (LCB) from amidships.
Block Coefficient (CB)
Midship Section Coefficient (CM)
Prismatic Coefficient (Cp)
COPYING (zero marks), UNTIDY (minus up to 1 mark)
Question 2
a. At a draught of 4m, the waterplane of Containership Bunga Bawang
(LBP=88m) has the following offsets.
Station
½ Breadth
(m)
0(AP)
1
2
4
6
7
8(FP)
2.20
4.48
6.22
7.10
5.02
2.53
0
Calculate area of waterplane, waterplane coefficient, TPC and LCF from
amidship.
b. The waterplane areas of Bunga Bawang at other draughts are as follows:
Draughts
Area (m2)
1m
2m
3m
520
690
830
Between the keel and 1m waterplane, there is an appendage with volume
420 m3 and centroid 0.60m above keel.
Use all the information to calculate for the ship at draught of 4m, the total
mass displacement in sea water, its block coefficient and centre of buoyancy
above keel.
© Omar bin Yaakob, July 2008
57
Naval Architecture Notes
Question 3
The 2m waterplane of a catamaran boat LBP 20m is shown in Figure 1. The
half-breadths of one hull is shown in the following table:
Station
0 (AP)
1
2
3
4
½ lebar
(m)
2.20
2.18
2.16
2.14
2.12
Station
½ lebar
(m)
6
2.0
0
7
1.8
0
8
1.6
0
8.5
1.2
0
9
0.9
0
9.5
0.4
0
5
(amidships)
2.10
10 (FP)
0.00
Calculate for the total waterplane:
area of waterplane, TPC, LCF and second moment of area about the
centreline
20m
CL
3m
Figure 1 Catamaran Waterplane
Question 4
A ship LBP 90m, lebar 17.2 m is floating in seawater. At 5m draught, the
waterplane has the following offsets.
Stn.
½
breadth
(m)
AP
0.0
1
5.5
2
8.0
3
8.4
4
8.5
5
8.6
6
8.5
7
8.0
8
7.0
9
4.5
FP
0.0
The ship has the following waterplane area at other draughts:
Draught (m)
Waterplane
Area(m2)
0.0
10
© Omar bin Yaakob, July 2008
0.5
500
1.0
800
2.0
1100
3.0
1200
4.0
1260
58
Naval Architecture Notes
Using all the information, calculate for draught of 5.0m:
i. Waterplane Area
ii. Second moment of area about the centreline.
iii. Mass displacement.
iv. Block Coefficient Cb
v. Height of Metacentre, KMT
© Omar bin Yaakob, July 2008
59
Naval Architecture Notes
Question 5
The cross-section of a tank can be represented by a plane curve and two radii 90 0 apart as
shown in Figure 2. The lengths of the radii at equal angle intervals are 12,14, 16,18, and 20
metres respectively. Calculate the area of the cross-section.
20m
12m
Figure 2
Question 6
a. Sebuah lengkung dinyatakan seperti berikut:
y = 2 +3x +4x2
Tentukan luas di bawah lengkung yang disempadani oleh x = 0 to x = 4
dan paksi x menggunakan kaedah:
i)
ii)
iii)
iv)
Simpsons Pertama
Simpsons Kedua
Trapezoid
Pengamiran
Berikan komentar terhadap keputusan yang diperolehi.
b. Ofset bagi sebuah kapal LBP 60m adalah seperti berikut:
Stesen
Separuh
Lebar
(m)
Kirakan :
i)
ii)
iii)
iv)
0
(AP)
0.5
1
2
2
1.4
2.6
4.3
½
3
4
5
(FP)
5.4
6.6
7.0
Luas Satahair
LCF dari peminggang
Momen luas kedua melintang pada garis tengah.
Momen luas kedua membujur pada pusat keapungan.
© Omar bin Yaakob, July 2008
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Naval Architecture Notes
© Omar bin Yaakob, July 2008
61
Naval Architecture Notes
Chapter 5 Transverse Stability
Consider a ship floating upright as shown in Figure 5.1. The centres of gravity and
buoyancy are on the centre line. The resultant force acting on the ship is zero, and
the resultant moment about the centre of gravity is zero.
W
MW
L
G
B-
K
W
Figure. 5.1
Now let a weight already on board the ship be shifted transversely such that G
moves to G1 as in Figure 5.2. This will produce a listing moment of W X GG 1 and
the ship will start to list until GI and the centre of buoyancy are in the same vertical
line as in Figure 5.3.
W
MW
L
G
G1
B-
K
W
Figure. 5.2
© Omar bin Yaakob, July 2008
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Naval Architecture Notes
W
W
M
W1
L1
M
Z
G1
G
B
B1
L
K
W
Figure. 5.3
G
G1
In this position G1 will also lie vertically under M so long as the angle of list is small.
Therefore, if the final positions of the metacentre and the centre of gravity are
known, the final list can be found, using trigonometry, in the triangle GG1M which
is right-angled at G.
In triangle 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 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. ‘
Note. It will be found more convenient in calculations, when taking moments, to
consider the ship to be upright throughout the operation.
© Omar bin Yaakob, July 2008
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Naval Architecture Notes
Example I
A ship of 6,000 tonnes displacement has KM = 7.3 m, and KG = 6.7 m, and is
floating upright. A weight of 60 tonnes already on board is shifted 12 m
transversely.
Find the resultant list.
Figure 5.4(a) shows the initial position of G before the weight was shifted and Figure
5.4(b) shows the final position of G after the weight has been shifted.
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 ‘
W
W
M-
M-
W
L
W
L
G-
G
B-
B-
K
K
G1
W
W
(a)
Figure. 5.4
(b)
Example 2
A ship of 8,000 tonnes displacement has KM = 8.7 m, and KG = 7.6 m The following
weights are then loaded and discharged:
Load 250 tonnes cargo KG 6.1 m and centre of gravity 7.6 m to starboard of
the centre line.
Load 300 tonnes fuel oil KG 0.6 m and centre of gravity 6.1 m to port of the
centre line.
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Naval Architecture Notes
Discharge 50 tonnes of ballast KG 1.2 m and centre of gravity 4.6 m to port
of the centre line.
Find the final list.
Note. In this type of problem find the final KG by taking moments about the keel,
and the final listing moment by taking moments about the centre line.
(1) Moments about the keel
Final KG
Weight
KG
8000
250
300
-50
8500
7.6
6.1
0.6
1.2
=
about
Final moment
Final displacement
=
Final KG
Moment
keel
60800
1525
180
-60
62445
62.445
8500
=
KM
Final KG
=
=
8.7 m.
7.34 m.
Final GM
=
1.36 m.
7.34 m
M-
M-
50 G - 7.6 m
t
4.6 m
K
250t
G
G1
B-
6.1 m
K
W
300t
(a)
(b)
Figure 5.5
Moments about the centre line
W
.d
Listing
moment
(tonne-m)
to port
© Omar bin Yaakob, July 2008
to
65
Naval Architecture Notes
starboard
250 7.6
-50 4.6
300 6.1
1900
-230
1830
1600
1900
Net listing moment
300
Since the final position of the centre of gravity must lie vertically under M, it follows
that the ship will list degrees to starboard.
Tan
=
Listing Moment
W x GM
=
300
8500 x 1.36
=
1 29 ½’
Ans. Final list
= 129 ½’ to starboard
Example 4
A ship of 13,750 tonnes displacement, GM = 0.75 m, is listed 2.5 degrees to
starboard and has yet to load 250 tonnes of cargo. There is space available in each
side of No.3 ‘tween deck (centre of gravity, 6.1 m out from the centre line). Find
how much cargo to load on each side if the ship is to be upright on completion of
loading.
M-
2 ½
G1
G
w
6.1
m
6.1
m
w
250-w
K
Figure 5.6
Tan = listing mmt
W x GM
1. Find listing moment that is initialy listing the ship to starboard;
listing moment = W x GM x tan = 13750 x 0.75 x tan 2.5 = 450.25 tonnem S clockwise
2. Load ‘w’ tonnes to port and (250-w tonnes) to starboard.
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Naval Architecture Notes
Net moment of the new load (anticlockwise) must counterbalance the original
moment (clockwise):
[New moment to port]– [New moment to starboard] = 450.25
[w x 6.1] – [(250 - w) x 6.1] = 450.25
w = 161.9 tonne to port
88.1 tonne to starboard
Example 5
A ship of 9,900 tonnes displacement has KM = 7.3 m and KG = 6.4 m. She has yet
to load two 50 tonne boxes with her own gear 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 as shown in Figure 5.7.
12m
6m
o
G1
W
15m
50t
50t
G2
9m
G-
L
W
K
Figure 5.7
(1) Moments about the keel
© Omar bin Yaakob, July 2008
weight
KG
Moment
9900
6.4
63,360
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Naval Architecture Notes
50
50
1000
0
Final KG
=
9.0
15.0
450
750
64,560
Final moment
Final displacement
=
=
64,560
10000
6.456m
(2) Moments about the centre line
Listing moment
W
50
50
.d
12
6
to port
to starboard
-
600
300
900
i.e. listing moment =
900 tonnes metres
(3)
New GM
tan
=
=
=
=
Ans. Maximum list
© Omar bin Yaakob, July 2008
=
7.3- 6.456
0.844 m.
Listing moment
W x GM
900
10000 x 0.844
6 6’
68
Naval Architecture 2 Notes
Exercise 5
1.
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.
2.
A ship of 5,000 tonnes displacement has KG 4.2 m, KM 4.5m and is listed
5 degrees to port. Assuming that KM remains constant, find the final list if
80 tonne of bunker is loaded in No 2 starboard tank whose centre of
gravity is 1 meter above the keel and 4 metre out from the centre line. (6
deg 3 min)
3.
A ship of 4,515 tonnes displacement is upright and has KG 5.4 m and KM
5.8 m It is required to list the ship 2 degrees to starboard and a weight of
15 tonnes is to be shifted transversely for this purpose. Find the distance
through which it must be shifted. (4.2m)
4.
A ship of 7,800 tonnes displacement has a mean draft of 6.8 m and is to be
loaded to a mean draft of 7 metres. GM 0.7 m, TPC 20 tonnes. The ship is
at present listed 4 degrees to starboard. How much more cargo can be
shipped in the port and starboard ‘tween decks, centres of gravity 6 m and
5 m respectively from the centre line, for the ship to complete loading and
finish upright. (216.5 tonnes Port, 183.5 tonnes Stb)
5.
A ship of 1,500 tonnes displacement has KG 2.7 m, and KM 3.1 m and is
floating upright in salt water. Find final list if a weight of 10 tonnes is
shifted transversely across the deck through a distance of 10 metres. (9.5
deg)
6.
A weight of 12 tonnes when moved transversely across the deck through a
distance of 12 m, causes a ship of 4,000 tonnes displacement to list 3.8
degrees to starboard. If KM 6 m, find the KG. (5.46m)
7.
A quantity of grain, estimated at 100 tonnes, shifts 10 m horizontally and
1.5 m vertically in a ship of 9,000 tonnes displacement. If the ship’s
original GM was 0.5 m, find the resulting list. (13 deg)
8.
A ship of 7,500 tonnes displacement has KM 8.6 m, KG 7.8 m and 20 m
beam. A quantity of deck cargo is lost from the starboard side (KG 12 m
and centre of gravity 6 m in from the rail). If the resulting list is 3 degrees
20 minutes to port, find how much deck cargo was lost. (XX.X tonnes)
9.
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)
10.
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 deck on the starboard side (centre of gravity 7.5 m from the centre
line), and a tender of 40 tonnes. Find how far from the centre line the
tender must be placed if the ship is to complete loading upright, and also
find the final GM. (KG of the deck cargo is 7 m.)
© Omar bin Yaakob, July 2006
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Naval Architecture 2 Notes
11.
A ship of 9,500 tonnes displacement is listed 3.5 degrees to starboard and
has KM 9.5 m and KG 9.3 m. She loads 300 tonnes of bunkers in No.3
double-bottom tank port side (KG 0.6 m and centre of gravity 6 m from the
centre line), and discharges two parcels of cargo each of 50 tonnes from the
port side of No.2 Shelter Deck (KG 11 m and centre of gravity 5 m from the
centre line). Find the final list. (14 deg)
12.
A ship of 6,500 tonnes displacement is floating upright and has GM 0.15 m.
A weight of 50 tonnes. already on board, is moved 1.5 m vertically
downwards and 5m transversely to starboard. Find the list. (13 deg)
13.
A ship of 5,600 tonnes displacement is floating upright and has KG 5.5 m,
and GM 0.5 m. A weight of 30 tonnes is lifted from the port side of No.2
‘tween deck to the starboard side of No.2 shelter deck (10 m horizontally
and 3 m vertically). Find the weight of water to be transferred in No.3
double-bottom tank from starboard to port to keep the ship upright. The
distance between the centres of gravity of the tanks is 6 metres.
© Omar bin Yaakob, July 2006
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Naval Architecture 2 Notes
Chapter 6 Longitudinal Stability
6.1 Introduction
In the previous chapter, we looked at transverse stability i.e. we considered
the resultant angles of heel when heeling moments are applied to the ship. In
this chapter, we will consider stability in the longitudinal direction. In this
case we will look at the effect of change in trimming moments on the trim of
the ship and the resulting draughts at the perpendiculars.
6.2 Trim due to movement of weights
ML
d
w
G’
G
F
B’
B
Figure 6.1
Consider the ship as in Figure 6.1. If the weight w is moved a distance d
meter, G will move to G’ parallel to the direction of movement of w.
G G’
=
wxd
The shift in weight results in a trimming moment wd and the ship will trim
until G and B are in line. LCF, the centre of floatation is the centre of area
of the water plane. For small trim, the ship is assumed to be trimming about
LCF.
The trimming moment causes change in trim and hence change in draughts
at AP and FP.
Change Trim =
trimming moment
MCTC
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Naval Architecture 2 Notes
Changes in draught forward, TF and aft, TA can be obtained by dividing
trim in proportion to the distance from LCF to the positions where the
draughts are measured, normally AP and FP.
x
TA
F
TA
TF
trim
TF
LBP
Figure 6.2
trim = TA + TF
LBP
- x
2
LBP
TA = trim x
LBP
+ x
2
TF = trim x
LBP
Example 6.1
A ship LBP 100 m has MCTC 125 tonne-m while its LCF is 2.0 m aft of
amidships. Its original draughts are 4.5 m at AP and 4.45 m at FP.
Find new draughts when a 100 tonne weight already on board is moved 50m
aft.
Change Trim =
trimming moment
MCTC
Change in trim =
100 50
125
= 40 cm = 0.4 m by stern
50 2.0
TF 0.4
100
= - 0.208 m
(since ship trims aft, forward draught is reduced i.e. negative)
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Naval Architecture 2 Notes
and similarly,
TA
=
+ 0.192 m, positive due to increase in draught aft.
Final draught table;
Original
T
New
TA
4.5
+ 0.192
4.692 m
TF
4.45
- 0.208
4.242 m
Exercise:
A ship LBP 50m is floating at Ta= 5.1m and Tf =5.3m. In this condition its MCTC is
30 tonne-m, LCF 5m fwd of amidships.
Find new draughts at Ap and FP when 50 tonne weight is moved 15m forward.
6.3 Small Weight Changes
If a small weight w is added or removed from a ship, the draught of the ship
will change as follows:
i. Parallel sinkage/rise
ii. Change in trim =
w
TPC
trimming moment
w x distance to LCF
=
MCTC
MCTC
Once trim is obtained, the changes TF and TA can be calculated and the
final draughts will include the parallel rise/sinkage and TF & TA.
© Omar bin Yaakob, July 2006
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Naval Architecture 2 Notes
Example 6.2
A ship LBP 100 m has LCF 3 m aft of amidships and floats at 3.2 m and 4.4
m at FP and AP respectively. Its TPC is 10 tonne while MCTC 100 tonne-m.
50 tonne cargo is removed from 20 m forward of amidships while 30 tonne is
unloaded from cargo hold 15 m aft of amidships. Find the final draughts at
the perpendiculars.
When cargo is removed, draught reduces i.e. the ship rise.
Parallel rise = 80 = 8 cm = 0.08 m
10
At the same time, the ship trims because there is a change of moment about
LCF.
Change in Moment
(aft)
Change in trim =
790
100
= sum of moment of weight about LCF
= 50 x (20 + 3) – 30 x (15 – 3)
=
790 tonne-m
= 0.079 m
TA
=
0.079
TF
=
0.079
Original
Rise
T
100
3
2
=
100
100
3
2
=
100
+ 0.037 m
- 0.042 m
TA
4.4 m
- 0.08
+ 0.037
4.357 m
TF
3.2 m
- 0.08
- 0.041
3.089 m
6.4 Effect of Large Changes in Mass
The effects discussed in Section 6.3 are related to small changes in weights.
These small changes lead to small changes in draughts and the hydrostatic
particulars are assumed unchanged.
If the weight changes are big, draughts will change significantly. Hydrostatic
data such as TPC and MCTC will also change and therefore the simple
formula used in Section 6.3 can no longer be used.
© Omar bin Yaakob, July 2006
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Naval Architecture 2 Notes
G
(original)
G (final)
Weight W
p
Buoyancy
B (original)
Figure 6.3 Relative Positions of G and B
Consider the relative positions of centre of gravity and centre of gravity as
shown in Figure 6.3. Before addition or removal of weight, B is directly
under G and in equilibrium, buoyancy equals weight W.
When weights are added, the draught of the ship can be considered to
change as follows:
i.)
ii.)
The effect of additional weight will cause the vessel to sink to new
draught. But since there is a big change in draught, TPC cannot be
used. Instead, the mean draught of the ship must be obtained from
hydrostatic particulars. At this mean draught, also obtain LCB, MCTC
and LCF.
From the way the vessel is loaded, the final LCG can be calculated.
The new location of G is no longer directly under the new LCB of the
ship i.e. a trimming moment is created which will trim the vessel. If
the longitudinal separation between G and B is p, then trim can be
calculated:
Trim (m) =
xp
MCTC x 100
LCG, LCB, MCTC and in this formula are for the final condition of
equilibrium.
This formula is important for two cases:
i)
ii)
If we know the condition of the ship in terms of its weight and LCG as
well as its hydrostatic particulars (LCB, LCF, MCTC) we can find its
trim and the draughts at the perpendiculars.
If we know ships draughts, we can find trim and hydrostatic
particulars (LCB, MCTC, LCF, displacement).
Using the above
formula, we can find LCG.
Direction of trim (by stern or bow) can be derived by considering the relative
position of LCG and LCB giving the direction of the trimming moment.
G
© Omar bin Yaakob,
) July 2006
G
)
75
Naval Architecture 2 Notes
Trim by Stern
Trim by Bow
Figure 6.4
6.5 Loading Calculations
Loading calculation is an important step in the ship design process.
Through this calculation, we are able to calculate the final displacement of
the ship, its KG, LCG and final draughts at the perpendiculars. Also, the
value of GM can be estimated. The calculation is done for a number of ship
loading conditions.
A loading table is used where the weights of the ship and any additional
deadweight and the respective kg and lcg are tabulated. Finally the total
weights, KG and LCG can be obtained. By using hydrostatic table, values of
mean draught, MCTC, LCF, KMT and LCB are obtained. The formula above
is used to calculate ships draughts at AP and FP. Also, since KG and KMT
are known, the value of GM can be obtained.
The above calculations are carried out while the ship is still on the drawing
board.
Loading calculation is also important for the ship's master to know the
current condition of his ship. Calculations can be done to determine the
current level of stability and draughts when some loading and unloading are
done on the ship.
To undertake loading calculations, the following items are required:
a) Hydrostatic data, table or particulars.
b) The weight, KG and LCG of the original ship whether lightship or already
loaded.
c) A list of loads to be added or removed from the ship; their masses, kg and
lcg.
Example 6.3
© Omar bin Yaakob, July 2006
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Naval Architecture 2 Notes
A ship LBP 125m having lightship mass 4000 tonne, LCG 1.6m aft of
amidships is loaded with the following:
8500 tonne cargo lcg 3.9m forward of amidships
1200 tonne fuel lcg 3.1m aft of amidships
200 tonne water lcg 7.6m aft of amidships.
100 tonne store lcg 30.5m fwd of amidships
Hydrostatic particulars indicate that at 14000 tonne displacement, mean
draught is 7.8m, MCTC 160 tonne-m, LCB 2.00m forward of amidships and
LCF 1.5m forward of amidships.
Find the final draughts at the
perpendiculars.
Item
Mass
(tonnes)
Lightship
Cargo
Fuel
Water
Store
TOTAL
4000
8500
1200
200
100
14000
LCG
(m from
amidship
s)
1.6A
3.9F
3.1A
7.6A
30.5F
Fwd Moment
about
amidships
(tonne-m)
Aft
Moment
about
amidships
(tonne-m)
6400
33150
3720
1520
3050
36200
11640
Forward Excess Moment
= 36200 - 11640
= 24560 tonne m
LCG from amidships
= 24560
14000
= 1.754m fwd of amidships
From hydrostatics, LCB 2.00m forward of amidships
Trim =
=
0.246 x 14000
160 x100
0.215m aft
+
125
- 1.5
2
= +0.11m
125
TA =
0.215 x
-
125
+ 1.5
= - 0.105m
0.215 x 2
125
TF =
Aft
© Omar bin Yaakob, July 2006
Fwd
77
Naval Architecture 2 Notes
T
(Original)
T
T (Final)
7.8m
7.8m
0.11
7.91m
-0.105
7.695m
6.6 Finding LCG of a Ship by measuring draughts
If we have the hydrostatic particulars of the vessel and we can measure the
draughts of the floating ship, we can know the ships weight and LCG.
The steps are as follows:
Read TF, TA
Calculate trim = | TF -TA |
Calculate Tmean = TF + TA
2
Use Tmean to obtain , MCTC and LCB from hydrostatic table.
Use trim formula to calculate the value of p i.e.
p =
MCTC x100 x Trim
since p= Distance (LCB ~ LCG), the location of LCG can be determined.
Note that the value of p is the actual distance between B and G in the
longitudinal direction. The actual position of G will depend on the direction
of trim as discussed earlier, see Figure 6.4.
6.7 Lightship survey or Inclining Experiment
The process to obtain the actual mass, KG and LCG of a ship is called
lightship survey and inclining experiment is the part in which the ship is
inclined to obtain KG.
It consists of taking a set of measurements and conducting analysis to
obtain the required objectives. Although sometimes the whole process is
called an inclining experiment, inclining the ship is only a part of the whole
procedure and will achieve only one aspect of the whole objectives. Inclining
experiment itself is very important and required to be done on every ship
under the Merchant Shipping Ordinance 1952.
The main objectives of the lightship survey are to determine for the ship in
as inclined condition and lightship condition:
i.
ii.
iii.
Displacement
KG
LCG
© Omar bin Yaakob, July 2006
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Naval Architecture 2 Notes
When the three particulars are known for a lightship condition, displacement
and centre of gravity for other conditions can be determined by using loading
calculation as explained in Section 6.5.
Preparation
The measurements are carried out when the ship is completed or nearly
complete. The vessel should be floating freely, not touching the bottom.
Gangways and ladders should be removed. Any loose cables and equipment
must be secured while tanks should either be fully pressed or emptied to
reduce free surface effects.
The draughts are measured at six locations around the ship. Mean draught
is calculated and is used to enter the hydrostatic tables to obtain the
hydrostatic particulars of , MCTC, KMT and LCB. The density of water is
also measured.
The inclining experiment itself is carried out by moving weights across the
ship. The weights are chosen such that the total weight on one side will give
about two degrees of heel. The angles of heel are usually measured using
three pendulums. If other devices are used, one pendulum must still be
used. To increase accuracy, the pendulum should be the longest possible
and to facilitate pendulum deflection reading, the pendulum bob may be
immersed in oil.
Movement of weights
Weights are moved one by one across the deck and after each move
pendulum reading is taken. When all three weights have been moved across,
readings are again taken each time the weights are returned to the original
position.
B
A
D
C
E
d F
Figure 6.5
Processing Results
The results are first processed for the ship in the condition at which the
measurements are taken. This is known as the ‘ as inclined condition’.
This condition is different from the final lightship conditions and therefore
corrections will have to be made later.
© Omar bin Yaakob, July 2006
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Naval Architecture 2 Notes
a. To obtain and other hydrostatic particulars in as inclined condition
From six draughts, obtain Tmean.
Use T mean to read KMT, LCB, MCTC, and LCF from hydrostatic
tables.
Hydrostatic tables are prepared normally using the assumed density of
1.025 tonnes/m3. Correction must be made to and MCTC for density
difference.
For example:
actual = table ρ actual
ρ table
b. To obtain KG as inclined
Based on the formula:
wd = GM tan
By plotting
6.6.
wd vs tan ,
a straight line is expected as shown in Figure
wd/
0.04
0.03
0.02
0.01
0
-0.4
-0.2
-0.01 0
-0.02
0.2
0.4
tan
-0.03
-0.04
Figure 6.6
Since KM is known from hydrostatic data, the height of centre of gravity
above keel,
KG= KM - GM
c. To obtain LCG as inclined
© Omar bin Yaakob, July 2006
80
Naval Architecture 2 Notes
The position of LCG is obtained using methods described in Section 6.6
i.e. by using the trim formula to calculate the value of p i.e.
p==
MCTC x100 x Trim
since p= Distance (LCB ~ LCG), the location of LCG can be determined.
d. Obtaining final , KG and LCG for lightship
A detailed record must be made of items that have yet to be installed
on the ship as well as items that are not part of the ship but present
on the ship during the lightship survey. Examples of the former are
deck equipment, cranes or any other equipment yet to be installed.
Items to be excluded include personnel, inclining weights and other
equipment.
The following format may be used:
Item
Mass
(tonnes)
Kg
(m)
Vertica
l
Momen
t
(tonnem)
Lcg
(m from
amidshi
ps)
Fwd
Moment
(tonne-m)
Aft
Moment
(tonne-m)
Ship
as
inclined
Items
to
remove
Incl.weights
Personnel
-12
-0.7
12
12.
5
+5
17
+2
1.3
Items to add
Deck
cranes
Generator
LIGHTSHIP
Example 6.4
Hydrostatic Particulars of MV Penyu LBP 50m is given below:
Drauf Displacement Cb
MCTC
LCB
LCF
tonnes
(tonne-m) (m from O)
(m dari O)
4.00
5000
0.75
100.00
-2.00
3.0
5.00
6000
0.76
110.00
-1.5
2.0
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Naval Architecture 2 Notes
6.00
7.00
7000
8000
0.77
0.78
120.00
130.00
0.0
0.0
0.0
0.0
The ship M.V. Penyu is floating at level keel draught of 4.50m.
Cargo are loaded as follows:
500
500
500
500
tonnes
tonnes
tonnes
tonnes
at
at
at
at
lcg
lcg
lcg
lcg
10m
10m
20m
15m
Aft of amidships
Fwd of amidships
Fwd of amidships
Aft of amidships
Find its final draughts at the perpendiculars.
SOLUTION:
i.
ii.
Find displacement (5500)and LCB (-1.75) of original ship. Since
ship is level keel, trim = 0, p= 0 therefore LCG= LCB.
Construct table to find final displacement and LCG.
Item
Weight
(tonnes)
Ori.
ship
Cargo 1
Cargo 2
Cargo 3
Cargo 4
TOTAL
5500
Lcg
(m from
amidships)
-1.75
Moment
about
amidship (tonnem)
LCG= Total moment amidship = 0.95 m aft of amidship
Total weight
iii. Use total weight to go into Hydrostatic Table and get
Tmean, MCTC, LCB, LCF
6.50m, 125tm, 0, 0
iv. find distance p where p is longitudinal distance between LCB
and LCG.
p = 0.95m
v. Use formula
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Naval Architecture 2 Notes
trim= displacement x p
MCTC x 100
Trim= 0.57m
vi. Use normal formula to calculate delta Ta and delta Tf. Since LCF is
at amidships, delta Ta= delta Tf = trim/2 =
vii. Make final table:
Tmean
Delta T
Final
Aft
Fwd
6.785
6.215
Exercise 6
1. A ship LBP 60m has lightship 500 tonnes, KG 3.7m and LCG 2.0m aft of
amidships. The following are loaded:
Item
Fuel Tank
Fresh Water
Cargo
Crew and Store
Mass
(tonnes)
50
10
735
5
Kg (m)
0.7
3.6
3.2
5.5
LCG from
(m)
1.5A
26.2F
1.5F
20.0A
Find Final KG and LCG
2. A ship LBP 100 m has LCF 3 m aft of amidships and floats at 3.2 m and
4.4 m at FP and AP respectively. Its TPC is 10 tonne while MCTC 100 tonnem.
The following cargo is added and removed:
UNLOAD
50 tonne cargo from 20 m forward of amidships
30 tonne cargo from 15 m aft of amidships.
10 tonne cargo at amidships.
LOAD
20 tonne cargo at 10m fwd of amidships
5 tonne fuel at 10m aft of amidships
Find the final draughts at the perpendiculars.
Item
Weight
Lcg (from D (m from LCF)
amidships)
Cargo1
Cargo2
Cargo3
© Omar bin Yaakob, July 2006
Moment
about
LCF (tonne-m)
83
Naval Architecture 2 Notes
Cargo4
Fuel
TOTAL
Use TPC, net weight reduced Rise =
Net moment = tm aft
Trim =
Find deltaTa and deltaTf
TA
TF
=
=
m
m
Make final table:
Original
Rise
T
FINAL
TA
4.4
TF
2. A ship is being loaded in port. At one point, its draughts are 10.5m A and
12.2m F, MCTC 200tonne-m and LCF 2m fwd of amidships. A further 5000
tonnes cargo is to be loaded at locations 10m fwd of amidships and 10m aft
of amidships. Determine how the cargo should be distributed to complete
loading with an even keel.
3. A ship arrives in port trimmed 25cms by stern. The centre of floatation is
amidships, MCTC 100 tm. A total of 1000 tonnes is to be discharged from
No1 hold (lcg 50m fwd of LCF) and No 4 Hold (lcg 45m aft of LCF). Find how
much to be discharged from each hold for the ship to complete loading on
even keel.
{Solution guide: Find trimming moment currently trimming ship using
Trim (cm) = trimming mmt/MCTC
Remove weights such that net moment from removal will counteract the
trimming moment.}
4. A ship is floating at draughts of 6.1m F and 6.7m A. The following cargo
is then loaded:
20
45
60
30
tonnes with lcg
30m forward of amidships.
tonnes
25m forward of amidships
tonnes
15m aft of amidships
tonnes
3m aft of amidships
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Naval Architecture 2 Notes
If LCF is at amidships, MCTC 200 tonne-m and TPC 35 tonne, find final
draughts.
5. An oil tanker 150m long, displacement 12,500 tonnes, LCF 1m aft of
amidships, MCTC 200 tonnes-m leaves port with draughts 7.2m F and 7.4m
A. There are 550 tonnes of fuel oil in the forward deep tank (centre of gravity
70m forward of LCF) and 600 tonnes in the after deep tank (centre of gravity
60m aft of LCF). During the sea passage, 450 tonnes of oil is consumed from
the aft tank. Find how much oil must be transferred from the forward to the
aft tank if the ship is to arrive on an even keel.
6. A ship arrives in port trimmed 0.3 m by the stern and is to discharge
4,600 tonnes of cargo from 4 holds. 1,800 tonnes of cargo is to be
discharged from No.2 and 800 tonnes from No.3 hold. Centre of floatation is
amidships. MCTC=250 tonne-m.
Centre of gravities of the holds No 1, 2, 3 and 4 from amidships are
45m forward, 25m forward, 20m aft and 50m aft respectively.
Find the amount of cargo to be discharged from Nos. 1 and 4 holds if the
ship is to sail on an even keel.
7. MV Bulker LBP 100 m is floating at a level keel draught of 7m. Its LCG is
4m fwd of amidships.
The following cargo are loaded:
2000 tonne lcg 10m aft of amidships
2000 tonne lcg 10m fwd of amidships
1000 tonne lcg
5m aft of amidships
Find the final draughts at the perpendiculars. Use the provided hydrostatic
curves.
8. MV bulker is floating at draughts of 7.8m F
displacement and LCG.
and
7.1m A. Find its
9. Bunga Kintan is floating at draught of 6.2mF and 5.8mA. Its GM was
measured and found to be 0.15m. Find its displacement , LCG and KG.
10. Some weights (w=200 tonnes,lcg= 2m fwd of amidships, kg=5.0m) are
taken off. What is its final displacement , LCG and KG ?
© Omar bin Yaakob, July 2006
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Naval Architecture 2 Notes
11. Pemerhatian berikut diperolehi daripada ujikaji sendeng. Kirakan sesaran
serta kedudukan pusat graviti membujur (LCG) bagi keadaan kapal kosong.
Drauf:
FP
4.92m
Peminggang 5.50m
AP
6.08m
Ketumpatan air laut 1025 kg/m3. Jumlah beban (termasuk pemberat dan air
balas) yang perlu dikeluarkan bagi mendapatkan kapal kosong ialah 354 tonne
pada lokasi 4.0m di depan peminggang dan 10.5m di atas lunas. Panjang kapal
ialah 125m dan butiran hidrostatiknya di dalam air berketumpatan 1025
Kg/m3 adalah seperti berikut;
Drauf
(m)
Sesaran
(tonne)
KMT
(m)
MCTC
(tonne-m)
6.00
5.00
10300
8200
8.4
9.0
141.0
131.0
© Omar bin Yaakob, July 2006
LCB
(m dari
)
3.8m Fwd
4.0m Fwd
LCF
(m dari
)
0.0 m
0.0 m
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Naval Architecture 2 Notes
HOMEWORK
1.
A ship LBP 60m has lightship 500 tonnes, KG 3.7m and LCG 2.0m aft of
amidships.
The following are loaded:
Item
Mass
(tonnes)
Kg (m)
Fuel Tank
Fresh Water
Cargo
Crew and Store
50
10
735
5
0.7
3.6
3.2
5.5
LCG
from
(m)
1.5A
26.2F
1.5F
20.0A
FSM
(tonnem)
40
15
Extracts from Hydrostatic Table as follows:
Drauf
(m)
3.6
3.8
1.
2.
3.
4.
Sesaran
(tonnes)
1280
1320
MCTC
(tonne-m)
14.5
14.7
LCB dari
(m)
0.30A
0.20A
LCF from
(m)
0.0
0.0
KMT (m)
4.29
4.15
Find the final Displacement, KG, LCG and GMT
Find Trim
Find new Draughts at AP and FP.
A 50 tonne weight is moved 6 m across. Find the resulting angle of heel.
© Omar bin Yaakob, July 2006
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Naval Architecture 2 Notes
Chapter 7 Large Angle Stability
7.1 Righting Lever and Moment
In Chapter 4, we have seen the importance of righting moment to ship stability.
For a ship being acted upon by external moments for example, the righting moment
will balance the external moment giving a steady angle of heel. If there is
inadequate righting moment, the vessel will capsize.
We have also seen that righting moment = x GZ
Where GZ is the righting lever or righting arm, the perpendicular connecting the
lines of action of buoyancy to the line of action of weight.
External
Heeling
moment
W
W
M
Z
G
W1
L1
L
B
W
K
B1
M
G
Z
B
Figure 7.1 The righting Lever
GZ
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Naval Architecture 2 Notes
In any loading condition, displacement does not change. Therefore the righting
moment in any loading condition will only depend on the righting arm GZ.
If we can determine the GZ values in that particular loading condition, we can know
its righting moment and hence the stability of the vessel.
As can be seen from Figure 7.1, GZ is the perpendicular distance between the two
lines of actions; weight through centre of gravity and buoyancy through centre of
buoyancy. Earlier, we have seen that at small angles, the metacentre is considered
stationary. From Figure 7.1, for small angles,
GZ = GMT sin
where is the angle of heel.
In any loading condition, GMT is constant.
directly proportional to sin .
Therefore the righting lever GZ is
PLOT GZ VS ANGLE OF HEEL
7.2 GZ at Large Angles
The location of the metacentre is no longer stationary at large angles. This is due to
the different shapes of waterplane areas at successive angle of heels. This
differences give rise to different transverse second moments of area. The distance
from the centre of buoyancy to the metacentre is given as
BMT
IT
At small angle of heel, IT can be assumed constant and hence the metacentre can be
considered stationary. However at larger angles of heels, the waterplane shapes
changes significantly leading to movement of the metacentre. Because of this
reason, the accuracy of the expression GZ=GMT sin diminishes at large angles. In
other words there is no simple expression relating GZ to GMT .
M’T
M
G
B’’
B’’
B B’
Figure 7.2 Changing metacentre at large angles
© Omar bin Yaakob, July 2006
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Naval Architecture 2 Notes
In general, the actual relationship between GZ and is shown in Figure 7.3. At
small angles, GZ can be considered as directly proportional to Sin . The slope
starts to increase until a point of inflexion is reached. This point marks the angle of
deck-edge immersion. The slope decreases until the maximum is reached. Then
the slope becomes negative and GZ reduces until the ship loses stability at the
angle of vanishing stability.
GZ curve is also called righting arm or statical stability curves. The curve is very
important in stability assessment. Load Line Rule requires GZ curves to be
calculated and the overall stability of the vessel assessed based on the curves.
Point of inflexion
G
Z
(
m
)
Angle of Vanishing Stability
GZ value at 20
2
0
Range of Stability
Figure 7.3
Angle of heel
GZ Curve
7.3 Obtaining Curves of Statical Stability
Curves of statical stability show the variation of GZ at various angles of heels. At
the design stage, the naval architect must ensure that the curves are calculated.
The GZ data is produced at different displacements. Also since KG is required in
the calculation, an assumed KG is used. Later, when the actual and KG at any
loading condition are known, actual values of GZ can be extracted from the data.
Various methods are used to calculate statical stability curves.
For a wall-sided vessel, GZ can be calculated using a simplified formula.
GZ = ( GM + ½ BM tan
2
) sin
Wall-sided formula is only valid for ships which have straight parallel sides. The
formula is valid as long as the deck is not immersed or the bilge is not raised out of
water.
For other types of (real) hullforms, other methods must be used. Details can be
found in Principles of Naval Architecture (Vol 1).
© Omar bin Yaakob, July 2006
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Naval Architecture 2 Notes
7.4 Cross Curves of Stability
The GZ values calculated at the design stage are presented either in tables or in
Cross Curves of Stability. The GZ values calculated are carried out at only a few
displacements and heel angles. Tables and curves are therefore presented at those
displacements and angles. To obtain values at other displacements or angles, linear
interpolation is used. A typical set of curves are shown in Fig. 7.4.
The value of KG of a ship is not fixed. It changes throughout the ship’s life
depending on how the ship is loaded. The values of GZ will depend on the KG.
However it is difficult to calculate GZ at many displacements, various angles of heel
and different values of KG.
For that purpose, a fixed value of KG is normally used in the design stage. This
assumed position of G is designated S, the height above keel being KS. Since S is
only an assumed centre of gravity, the corresponding point on the line of action of
buoyancy is called N i.e. not Z. The value of the righting lever is thus SN instead of
GZ.
40
SN (m)
50
30
80
KS= 4.2m
15
5
Displacement (tonnes)
Figure 7.4 Cross Curves of Stability
In other words, the cross curves of stability will show plots of the assumed righting
arm, SN rather than the actual righting arm, GZ. To obtain the actual GZ, SN
values at various angles are read at the right displacement, and then corrected for
the correct KG based on Figure 7.5.
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Naval Architecture 2 Notes
M
If S is above G ; GZ = SN + SG Sin
If S is below G; GZ = SN - SG Sin
N
S
G
Z
B
K
Figure 7.5
A better method is to put S at the keel i.e. KS=0. In this case SN = KN. The
assumed righting arm, KN values are plotted instead of SN at various displacement.
When G is known for any condition, GZ can be calculated:
GZ = KN – KG Sin
It must be noted that the values of KG used is the virtual or fluid KG i.e. taking into
consideration the FSC.
7.5 Initial Slope of GZ Curve
At small angles, the expression GZ = GMT Sin is valid. Also, at small angle, GZ
GMT i.e GZ approximates to a straight line with gradient GMT, as shown in
Figure 7.6.
In any loading condition, KMT is constant. Therefore the initial slope will solely
depend on KG.
G
Z
GM
57.3
© Omar bin Yaakob, July 2006
sudut sendeng
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Naval Architecture 2 Notes
Figure 7.6 GM as initial slope
© Omar bin Yaakob, July 2006
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Naval Architecture 2 Notes
Example 7.1
A ship with lightship displacement 1,700 tonnes, KG 3.5m is loaded with 1,800
tonnes of cargo at Kg 3.8m. KMT after loading is 3.8m while KN values are as
follows.
Displacemen
t
(tonnes)
3,000
4,000
10
20
0.75
0.77
1.50
1.54
Angle of heel ()
30
45
2.16
2.20
2.84
2.92
60
75
3.19
3.25
3.26
3.26
Plot the GZ curve and find the area under the curve up to 300.
Solution:
i)
Carry out loading calculation to obtain final displacement and KG. (3,500
tonnes, 3.65m)
Find KN values at that displacement.
Correct KN to obtain GZ using GZ = KN – KG sin
Plot the curve (note that initial slope= GMT=0.15m)
Use Simpson rule to find area under the curve up to 30 .
Does the ship pass IMO stability criteria?
ii)
iii)
iv)
v)
vi)
0
10
20
30
45
60
75
(ii)
KN(m)
0
0.76
1.52
2.18
2.88
3.22
3.26
0
0.174
0.342
0.500
0.707
0.866
0.966
0
0.634
1.248
1.825
2.581
3.161
3.526
(iii)
GZ(m)
0
0.126
0.272
0.355
0.299
0.059
-0.266
(v)
SM
1
3
3
1
sum
fA
0
0.379
0.815
0.355
1.549
Area under the curve up to 30 ,
Area = 3 x 10 x x 1.549 = 0.1014
8
180
© Omar bin Yaakob, July 2006
mrad
94
Naval Architecture 2 Notes
7.6 STABILITY ASSESSMENT
New deadweight
loading/unloading
w, Kg, lcg
Initial Ship Condition
Δ, KG, LCG
Loading Calculation Table
Use Δ to read
Hydrostatics
Data
SN or KN Curves
Curves of Statical
Stability
FINAL Δ,
KG, FSC
LCG
Hydrostatics data
Tmean, LCB, LCF,
MCTC,
KMT
KN/SN Values
at correct Δ
Calculate Draughts TA
and TF
Calculate GMT
Adjust for actual KG
GZ CURVE
SHIP HULL FORM
Stability
Assessment
IMO Load Line
Criteria
Wind Heeling
Heeling angle
due to heeling
moment
GZ CURVES OF SHIP
WITH NEGATIVE GM
© Omar bin Yaakob, July 2006
DYNAMIC STABILITY
95
Naval Architecture 2 Notes
7.6.1 Load Line Criteria.
Stability Criteria
1. Area Under Curve 0o15o
2. Area Under Curve 0o30o
3 Area Under Curve 0o-40o
or up to f (flooding Angle)
Large Ship
(IMO)
N.A
0.055
m.rad
0.090 m
rad
Small Craft Passenger/Cargo
(HSC Code)
0.07 m.rad if max. GZ
occur at 15 to 30 deg.
0.055 m.rad if max. GZ
occur at 30 deg.
Fishing Vessel
(IMO)
N.A
0.055 m.rad
N.A
0.089 m.rad
4. Area Under Curve 15o30o
N.A
0.055 + 0.001 (30 max) if max. GZ occur
between 15 to 30 deg.
N.A
5. Area Under Curve 30o40o or up to f (flooding
Angle)
0.03 m rad
0.03 m.rad
0.03 m.rad
6. Maximum GZ
0.20 m
0.2 m
0.2 m
7. Angle at Maximum GZ
30.0 deg
15 deg
30 deg
8. Initial GM
0.15 m
0.15 m
0.35 m
Table 1: Stability Criteria for Three Types of Vessel (courtesy
Yahya Samian)
The ship is assessed at a number of loading conditions, for example:
i.
ii.
iii.
iv.
v.
vi.
Lightship
Homogenous Full Load Departure
Homogenous Full Load Arrival
Ballast Departure
Ballast Arrival
Etc.
© Omar bin Yaakob, July 2006
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Naval Architecture 2 Notes
Example 7.2
A ship lightship 5200 tonnes, KG 7.4m, LCG 2.0m aft of amidships.
homogenous loaded Departure condition are as follows:
Item
Mass
(t)
VCG
(m)
Fuel Tank 1
Fuel Tank 2
Fresh Water
Crew and Store
Cargo No.1
Cargo No.2
Cargo No.3
Cargo No.4
1350
210
170
50
1700
1900
1800
1330
1.5
3.2
7.2
9.5
10.1
7.1
8.0
9.0
LCG from
amidship(m
)
1.2 A
8.0 A
60.0 F
14.0 F
44.0 F
17.0 F
22.0 A
49.0 F
The
FSM
(tm)
270
350
210
-
Extracts of hydrostatics data at two relevant draughts are as follows:
T
(m)
Displacement
(tonnes)
MCTC
(tm )
8.50
9.00
12,994
13,999
170.0
178.56
LCB
(m from
)
2.80
3.00
LCF
(m from
)
6.80
7.088
KMT (m)
7.95
8.083
KN values at two displacements are as follows:
o
5
10
15
30
45
60
75
13500
0.705 1.412 2.122 4.052 5.555 6.575 7.067
13900
0.707 1.414 2.124 4.054 5.559 6.577 7.071
Find the draughts at the perpendiculars, calculate GMT and check if the vessel pass
the IMO merchant ship stability criteria.
© Omar bin Yaakob, July 2006
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Naval Architecture 2 Notes
Solution:
Create the final loading table :
Item
Mass
(t)
VCG
(m)
Mv
(tm)
LCG
from
(m)
MA
(tm)
MF
(tm)
FSM
(tm)
LIGHTSHIP
5200
7.4
38480
2.0 A
10400
-
-
Fuel Tank 1
Fuel Tank 2
1350
210
1.5
3.2
2025
675
1.2 A
8.0 A
1620
1680
-
270
350
Fresh Water
170
7.2
1224
60.0 F
10200
-
210
Crew and
Store
50
9.5
475
14.0 F
10200
-
1700
1900
1800
1330
10.1
7.1
8.0
9.0
17170
13490
14400
11970
44.0 F
17.0 F
22.0 A
49.0 F
39600
65170
74800
32300
-
-
118270
107800
830
128670
107800
830
Cargo
Cargo
Cargo
Cargo
No.1
No.2
No.3
No.4
DEADWEIGHT
8510
FINAL
DISPLACEMEN
T
13710
F.S.C
61426
7.287
99906
1.522 A
830 = 0.0605m
13710
Enter hydrostatic Table at = 13710 tonnes, to obtain:
TLCF
= 8.50 +
= 8.50 +
13710 12994
13999 12994
716
x 0.5
1005
(9.00 – 8.50)
=
8.856m
and,
© Omar bin Yaakob, July 2006
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Naval Architecture 2 Notes
MCTC = 176.1 t.m. ; KMT = 8.045 m;
LCB = 2.949 m.A.
LCF = 7.005 m.A.
© Omar bin Yaakob, July 2006
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Naval Architecture 2 Notes
For transverse stability
KM
KG
= 8.045 m
= 7.287
GMSOLID
F.S.C
= 0.758 m
= 0.061
GMFluid
= 0.697m
For longitudinal stability
TRIM =
( p) x
MCTCx100
= (2.949 - 1.522) X 13710
176.2 x 100
= 1.111 m ke hadapan
TA
. TA = 8.856
-0.501
8.355 m
= 1.111 x (72 – 7.005)
144
TF = 8.956
+0.610
9.466 m
= - 0.501 m;
TF
1.111 x (72 7.005)
144
= + 0.610 m;
o
5
10
15
30
45
60
75
KN (m)
0.706
1.413
2.123
4.053
5.557
6.576
7.069
KGf sin
0.640
1.276
1.902
3.674
5.196
6.364
7.098
GZ (m)
0.066
0.137
0.221
0.379
0.361
0.212
-0.029
KGf = 7.287 + 0.061
GZ = KN - KGf Sin
© Omar bin Yaakob, July 2006
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Naval Architecture 2 Notes
Plot GZ curve and calculate and check criteria.
Criteria
GZ max
A30
A40
A40 – A30
GMT
Actual
0.393m @ 37
0.115 m.rad
0.187 m.rad
0.072 m.rad
0.697m
Required
0.2m @ 30
0.055 m-rad
0.09 m-rad
0.03 m-rad
>0.15m
Evaluation
OK
OK
OK
OK
OK
i.e. fulfill all Load Line Rule Criteria
7.6.2 Determining Steady Angle of Heel due to Heeling Moment
When a ship is acted upon by an external or internal moments, the steady angle of
heel can be obtained as the point of equilibrium between the external moment and
the righting moment curve.
From Section 7.1,
Righting moment = Righting Lever x
= GZ
Curves of displacement multiplied by GZ values at various angles of heel are called
the righting moment curve, as shown in Figure 7.7.
For example, if a weight w already on board is shifted horizontally from port to
starboard a distance d, the heeling moment is wd Cos θ. The point of equilibrium
where the heeling moment equals the righting moment is the position at which the
ship will heel at a steady angle.
This analysis will also applicable to crane operation. However, in addition to
heeling moment crane operation is also associated with increase in KG i.e.
reduction in GM. The KN correction factor becomes more while the slope becomes
less. Hence the shape of the heeling moment curve reduces , leading to worse
situation.
E
M
G
o
Z
m
wdcos
A
e
C
F
n
D
t
c
O
Steady Angle of Heel
Figure 7.7
© Omar bin Yaakob, July 2006
E
D
Heel angle
Curve of Righting and Heeling Moments
101
Naval Architecture 2 Notes
US Navy Criteria:
For crowding of personnel and lifting of weight:
Angle at C for lifting of weight or crowding of personnel < 15 degrees
Heeling mmt at C < 0.6 maximum
Area A1 > 0.4 total area under curve.
7.6.3 Wind Heeling Moment
Wind heeling moment = k AV2 ℓ Cos2
Where A
V
l
buoyancy.
k
=
=
=
Projected area facing wind
Wind speed in knots
distance from centroid of wind pressure to centre of
=
Constant
V
knot
s
l
By plotting wind heeling moment and righting moment curves, the steady angle of
heel can be obtained.
Moment
(tonne-m)
A
E
k AV2 1 cos2
O
Figure 7.8
x GZ
C
D
c
D
Steady angle of heel due to constant wind moment
© Omar bin Yaakob, July 2006
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Naval Architecture 2 Notes
e.g US Navy, stability in wind satisfactory if:
Heeling moment at intersection of heeling mmt and rignting moment curves <
0.6 maximum.
Area A1 > 1.4 A2 where A2 extends 25 degress from C to direction of wind.
7.6.4 Determining Safety in Dynamic Mode.
The area under the righting moment curve is the dynamic stability of the vessel.
Dynamic stability defines the amount of energy that the vessel can absorb up to the
heeling angle in question.
When an inclining moment is applied, the baseline of the curve changes as shown
as broken line in Figure 7.7 and Figure 7.8.
With the change in baseline, the stability characteristics of the vessel also changes.
C is the new point of equilibrium (compared to 0º when there is no heeling moment)
where the steady heeling angle is obtained. The ship will return to this angle
whenever additional temporary moment is applied to port or starboard. The new
range of stability to starboard is from C to D. D is the new angle of vanishing
stability, which is less than the original.
Considering Figure 7.7, the area bounded by OAC is the work done by the the
heeling moment to heel the vessel to the steady angle of heel, C. In other words,
this is the potential energy available to heel the vessel from upright to C. If the
moment is applied quasistatically, the vessel will slowly heel to C. However, if it is
applied suddenly, the potential energy will be released suddenly causing the vessel
to overshoot C and reach E. The area OAC equals CEF, and the vessel will
oscillate about C. It is important to ensure that E does not exceed D, the new
angle of vanishing stability.
In the case of wind heeling, similar situation will happen if the wind moment is
applied suddenly as in a gust. Again, from Figure 7.8, the ship will capsize if the
potential energy (area OAC) could not be absorbed fully by the area CDE bounded
by the wind heeling moment curve and the righting moment curves, i.e. E exceeds
D.
7.6.5 Stability of ships with negative GM
A ship with negative GM will not stay upright. It will heel to one side until the
waterplane area is such that metacentric height exceeds the centre of gravity. It will
then stay at this new point of equilibrium, called the angle of loll. In other words a
ship with negative GM will not necessarily capsize.
If the negative GM ship is slowly released from upright ( = 0), it will slowly
approach the angle of loll. However, if the ship is released suddenly, which is
normally the case when GM becomes negative suddenly, the vessel will quickly heel
over by-passing the angle of loll to another angle called the angle of lurch.
© Omar bin Yaakob, July 2006
103
Naval Architecture 2 Notes
Area B
G
Z Angle of vanishing
stability
Angle of Loll
Heel Angle
Area A
Angle of Lurch
Figure 7.9 Angle of loll and lurch
The position of angle of lurch will depend on the energy available between upright (
= 0)and angle of loll (Area A) and the ability of the ship to absorb the energy
beyond the angle of loll (Area B). If the ability to absorb is low i.e. low righting
moment values, the angle of lurch will be big and if it exceeds angle of vanishing
stability, the ship will capsize.
© Omar bin Yaakob, July 2006
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Naval Architecture 2 Notes
EXERCISE 7 LARGE ANGLE STABILITY
1.
For one loading condition, a ship LBP 70m has displacement 1500 tonnes, KG
4.0 m and KM 4.5 m. SN values (with KS = 4.2m) are given in the table below.
10
0.23
Angle ()
SN (m)
15
0.35
20
0.44
30
0.44
45
0.35
60
0.19
75
0.01
a.
b.
c.
d.
Plot the GZ curve of the vessel.
Find area under the curve up to 30 degrees.
Check if the ship passes Load Line Rule stability criteria.
Check stability condition if a 50 tonne weight already onboard is moved 10
m across.
i.e. check steady angle of heel, area A1 and righting arm at steady angle of
heel.
2.
A fishing boat began its trip with displacement of 340 tonnes, KG 3.5 m and KM
3.75 m. After one day journey, the following events occurred :
20 tonne fish loaded (kg 3.0m)
6 tonne fuel consumed (kg 1.0m)
4 tonne fresh water consumed (kg 2.0m)
KM was found to be 3.8 m. The KN values are as follows:
Displaceme
nt
(tonne)
300
400
Angle of Heel
10
20
0.72
0
0.71
0
30
45
60
90
1.525
2.40
3.07
3.37
3.55
1.521
2.28
2.99
3.29
3.45
a. Plot its GZ and righting moment curve
b. In this condition the boat then used the power block to pull the fishing net.
The total weight of net and fish was 10 tonne and the boom of the power
block was 4 m from the boat’s centreline on the starboard side and 5 m above
keel. Plot roughly (but do not calculate) the new righting moment curve and
estimate its steady angle of heel.
c. When at that steady angle of heel, a gust suddenly hit the ship from port. If
the total wind moment is 110 tonne-m, comment on the safety of the vessel.
3.
a) Using wall sided formula, show that a wall-sided vessel will loll at an
angle of tan -1 √(2GM/BM).
b) Show the effect of separately increasing beam, increasing KG and increasing
freeboard
on the statical stability curve.
4.
i.
ii.
iii.
iv.
With the help of diagrams, explain:
What is GZ and why is it important in stability calculations?
Prove that for small angle of heel, GZ = GM sin .
Why is the above formulae not suitable for large heel angles?
What are SN, KN and GZ curves? Write down relationships between them.
© Omar bin Yaakob, July 2006
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Naval Architecture 2 Notes
v.
Sketch a GZ curve and indicate range of stability, angle of vanishing stability,
initial GM and maximum GZ.
© Omar bin Yaakob, July 2006
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Naval Architecture 2 Notes
5.
A ship 150m has lightship displacement 6500 tonnes, KG 8.0 m, LCG 1.5
m aft of amidships. The ship is loaded as follows :
Item
Mass
(t)
VCG
(m)
Fuel Tank 1
Fuel Tank 2
Fresh water Tank
Crew and Store
Cargo Hold No.1
Cargo Hold No.2
Cargo Hold No.3
1000
300
200
50
1800
2000
1600
1.5
3.2
7.2
9.5
10.1
7.1
8.0
LCG from
amidships
(m)
1.2 A
8.0 A
60.0 F
14.0 F
44.0 F
17.0 F
22.0 A
* free surface effect is neglected
Extract of hydrostatics data at relevant draughts as follows:
T
(m)
9.50
10.00
Displacement
(tonnes)
13,400
13,500
MCTC
(tm )
176.6
180.5
LCB
(m dari )
2.85F
3.01F
LCF
(m dari )
1.80A
2.30A
KMT (m)
7.95
8.08
KN values in meters at two displacements are as follows:
o
(tonnes)
13,400
13,500
a.
b.
c.
d.
5
10
15
30
45
60
75
0.700
0.706
1.400
1.413
2.023
2.123
4.000
4.053
5.268
5.557
6.346
6.576
6.873
7.069
Find Draughts at the perpendiculars
Find GMT
Check whether the vessel passes stability criteria.
Plot the righting arm (GZ) curve of the ship. Estimate its angle of heel when
the ship is lifting a 200 tonne cargo on the port side using its own derrick.
The derrick head is 15 m from the centreline and 15 m above keel.
e.
f. While lifting the cargo, the cable snapped. Find the maximum angle of roll on
the starboard side.
© Omar bin Yaakob, July 2006
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Naval Architecture 2 Notes
Chapter 8 Flooding
Stability
and
Damage
8.1 Introduction
Damage stability can be defined as the minimum adequate stability of a ship when
some part of the ship is damaged or opened to the sea.
All types of ships and boats are subject to the risk of sinking if they lose their
watertight integrity whether by collision, grounding or internal accident such as an
explosion. The most effective protection is provided by internal subdivision by
means of watertight transverse and/or longitudinal bulkheads and by some
horizontal subdivision – double bottom in commercial ships and watertight flats in
naval vessels.
A compartment, which has been opened to the sea, is said to have been bilged. It is
necessary to isolate the flooded volume in order to
restrict loss of transverse buoyancy or build up of such upsetting moment
that capsizing takes place
restrict change of trim i.e. lost of longitudinal stability
restrict loss of reserve buoyancy
restrict the damage to cargo
It is essential to have a standard of subdivision such that there is a reasonable
chance that the ship will remain afloat under such an emergency.
The requirements for subdivision have been fixed by legislature based on the
recommendations of various International Conferences on Safety of Life at Sea.
There are no Government requirements for the number of transverse bulkheads to
be fitted in a cargo ship. However, classification society rules specify the number of
bulkheads required and this is governed by the length of the ship.
Passenger ships – those that carry more than 12 passengers – must comply with
certain standards of subdivision. The method adopted is to determine a line beyond
which the ship should not sink and then ascertain the position and length of the
compartment which when flooded will cause sinkage to that line.This line beyond
which the ship should not sink is known as the Margin Line.
In introduction these concepts, it is necessary to establish certain basic definitions.
The most important and relevant of these are
Bulkhead Deck
The bulkhead deck is the uppermost deck to which the transverse watertight
bulkheads extend (usually the main deck).
Margin Line
The margin line is a line drawn parallel to, and three inches (76 mm) below, the
bulkhead deck at the side.
© Omar bin Yaakob and Mohamad Pauzi Abdul Ghani, July 2006
108
Naval Architecture 2 Notes
Permeability
The percentage volume of a space that can be flooded is known as the permeability.
available volume
total volume
When a compartment is flooded, it is rare for the total volume of this compartment
to becompletely filled with water. This is because the compartment will already
contain certain equipment or stores depending upon its use.
The table below from “Basic Ship Theory - 4th Edition” by Rawson & Tupper lists
some typical ship compartment permeabilities.
Permeability (%)
Watertight Compartment ( Warship)
97
Watertight Compartment ( Merchant
ship)
95
Accommodation Spaces
95
Machinery Compartments
85
Stores or Cargo Holds
60
Dry Cargo Spaces
70
Floodable Length
The maximum length of a compartment, which can be flooded so as to bring a
damaged ship to float at a waterline tangential to the margin line.
Curve of Floodable Length
This is a curve, which, at every point in its length, has an ordinate representing the
length of the ship, which may be flooded with the centre of the length at that point
and without the margin line being submerged.
Bonjean Curves
The area of the transverse section of a ship to successive waterlines can be
calculated and plotted as a curve showing the variation of sectional area with
draught. 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.
© Omar bin Yaakob and Mohamad Pauzi Abdul Ghani, July 2006
109
Naval Architecture 2 Notes
8. 2 The Effects of Damaged Compartments
A damaged ship could be lost in one of several ways.
If the ship is left with inadequate maximum righting moment or dynamical
stability, it could simply be overwhelmed by the seaway and the weather.
If the angle of list or trim is too great, placing non-watertight parts of the ship
underwater, then additional flooding will occur. In this case the ship could
lose
transverse stability, roll over and capsize
Longitudinal stability could also be lost in a similar manner causing the ship
to plunge (go down bow or stern first). One of the most notable examples of
plunging is the Titanic.
A ship may be lost even if stability is not compromised. It may simply sink.
This is called foundering.
Table 1 describes the damage scenarios and their effect to the ship
TABLE 1 : Damage Scenarios and Their Effect to the Ship
Case
Location of
Damage
Effect to the Ship
I
Long : At Centre of
Floatation, LCF
Trans : Symmetry about
Centre Line
1. Parallel singkage – Increase Draft
2. Increase or reduce GM depending on
Waterplane Area.
3. Reduce Freeboard parallel to the keel.
II
Long : Frd or Aft of LCF
Trans : Symmetry about
Centre Line
1. Parallel Singkage – Increase Draft
2. Trim by the Bow or Aft of Ship
3. Reduce Freeboard significantly at bow or
aft.
III
Long : At Centre of
Floatation, LCF
Trans : Port or Starboard
Side
1.
2.
3.
4.
Parallel Singkage – Increase Draft
Heeled to one side
Reduce GM and GZ
Reduce Freeboard Significantly at one side
© Omar bin Yaakob and Mohamad Pauzi Abdul Ghani, July 2006
110
Naval Architecture 2 Notes
1.
2.
3.
4.
5.
Long : Frd or Aft of LCF
Trans : Port or Starboard
Side
(Worst Scenario)
IV
8.3
Parallel Singkage – Increase Draft
Trim by the Bow or Aft of Ship
Heeled to one side
Reduce GM and GZ
Reduce Freeboard Significantly at bow or
aft and worst at one side.
Methods of Calculations
This section discusses the fundamental behavior of a damaged ship and introduces
two techniques that allow its analysis
The Added Weight Method
The Lost Buoyancy Method.
3.1 Flooding Calculation Methods
Firstly, define the degree of flooding to be examined (often given by regulations).
Then find
the waterline, trim and heel for the damaged floating condition (is there
sufficient reserve buoyancy left to prevent foudering?)
the damaged stability ( is there sufficient stability left to prevent capsizing
and plunging?)
Two techniques for flooding calculation explained on a rectangular vessel with a
damaged central compartment.
8.3.1 The Added Mass Method
Wo
A
B
L
W
E
WL
WoLo
ABFE
WoWLLo
Lo
F
Intact waterline
Damaged waterline
Added mass due to flooding
Additional buoyancy required
In order to calculate the added mass it is necessary to guess the damaged draught
AE and verify (trial and error)
© Omar bin Yaakob and Mohamad Pauzi Abdul Ghani, July 2006
111
Naval Architecture 2 Notes
8.3.2 The Lost Buoyancy Method
Wo
A
W
B
D
C
E
WL
WoLo
CDFE
Lo
L
F
Intact waterline
Damaged waterline
Lost of buoyancy i.e. must be made up by the buoyancies of WoACW
and BLoLD
The lost of CDFE can be calculated exactly. The additional buoyancy up to WoLo
can be found from parallel immersion/sinkage of the damaged waterplane
(excluding the portion AB).
8.3.3 Determining Draughts after Damage
The effects of bilging a mid-length compartment may be shown most simply by
considering a box barge of length L, breadth B and draught d having a mid-length
compartment of length l, permeability .
L
L
B
T
l
If this compartment is bilged, buoyancy is lost and must be replaced by
increasing the draught. The volume of buoyancy lost is the volume of the
compartment up to waterline WL, less the volume of water excluded by the
cargo in the compartment.
Volume of lost buoyancy =
l BT
© Omar bin Yaakob and Mohamad Pauzi Abdul Ghani, July 2006
112
Naval Architecture 2 Notes
This is replaced by the increase in draught multiplied by the area of the intact
part of the waterplane, i.e. the area of waterplane on each side of the bilged
compartment plus the area of cargo which projects through the waterplane in
the bilged compartment.
Area of intact waterplane
Increase in draught
=
=
(L- l
l B(1-)
LB- l
l B- l B
=
(L- l)B
=
volume of lost buoyancy
Area of intact waterplane
=
l BT
(L- l )B
=
lT
L- l
l may be regarded as the effective length of the bilged compartment.
Example 1
A box barge 30m long and 8 m beam floats at a level keel draught of 3m and
has a mid-length compartment 6m long. Calculate the new draught if this
compartment is bilged:
a) with =100%
b) with = 75%
(a) Volume of lost buoyancy =
Area of intact waterplane
6 x 8 x 3m3
=
(30-6) x 8m2
Increase in draught = 6 x 8 x 3 m3
24 x 8
New draught
= 0.75m
= 3.75m
(b) Volume of lost buoyancy =
Area of intact waterplane =
Increase in draught =
0.75 x 6 x 8 x 3m3
(30-0.75 x 6) x 8 m3
0.75 x 6 x 8 x 3
25.5 x 8
=
0.529m
=
=
3 + 0.529
3.529 m
New draught
© Omar bin Yaakob and Mohamad Pauzi Abdul Ghani, July 2006
113
Naval Architecture 2 Notes
8. 4 STABILITY AFTER FLOODING OF AMIDSHIPS COMPARTMENT
Stability is initially assessed by calculating metacentric height, GM T.
Example 8.2
A box-shaped vessel LBP = 30m, B =10 m floats at level keel draught of 2 m in
saltwater. Its KG = 3m. An amidship compartment, length 15m is damaged and
flooded. Calculate its final draught and transverse GM.
ORIGINALLY
= L x B x T x density = 30 x 10 x 2 x 1.025
=
615 tonnes
GM = KM-KG =(KB + BMT) - KG= (1.0 + 4.17) -3.0 = 2.17m
COMPARTMENT NOW BILGED (FLOODED)
Lost Buoyancy = 15 x 10 x 2
=
300 m3
Complete Waterplane Area
=
300 m2
Lost Waterplane Area
=
150 m2
Intact Waterplane area
=
150m2
Parallel Sinkage i.e. T
=
lost buoyancy
= 300
Intact waterplane 150
Hence New Draught
= 2.00 m
= 4.00 m
Two methods are normally used for calculation of metacentric heights during
flooding calculations. The methods uses two different basis but finally yield same
answers on the effect of flooding.
a. USING ADDED MASS BASIS
This method deals with the problem as if the hull is intact and an amount of water
is poured into the hull.
=
30 x 10 x 4 x 1.025
=
1230 tonne
KB
=
1
4.00
2
=
2.00 m
BM
=
100
12 4.00
=
2.083 m
The mass of water added (added mass) =
KG
=
615(3 2)
1230
=
0.500 m
© Omar bin Yaakob and Mohamad Pauzi Abdul Ghani, July 2006
114
1230 – 615
= 615 tonne
Naval Architecture 2 Notes
FSC
15 103
30 10 4 12
=
Final KGf
GMf
=
=
1.042 m
=
3.542 m
4.083 – 3.542
=
0.541 m
b. USING LOST BUOYANCY BASIS
This method considers the damaged and flooded compartment to be part of the sea
and has nothing to do with the intact vessel.
KB
=
BM
=
T
2
=
2.00 m
KM
(30-15) x 103
=
12 x 15 x 10 x 4
=
2.083 m (actual volume of displacement unchanged)
KG
=
3.000 m (unchanged since water not part of content of
=
1.083 m (note a different GM obtained)
4.083 m
the ship)
GM
Apparently, there are two different GM . However, actual measure of stability is
righting moments.
For example, check righting moment at 3 heel.
a) Added Mass method, =1230 tonnes and GM =0.541m
Righting Moment = x GM sin
= 1230 x 0.541 x sin
= 34.83 tonne-m
b) Lost Buoyancy method, = 615 tonnes and GM =1.083m
However the righting moment
=
x GM sin
=
615 x 1.083 x sin 3 = 34.83 tonne-m
i.e. similar righting moment, both methods give similar measure of stability
Exercise:
A box-shaped vessel LBP = 120m, B =20 m floats at level keel draught of 8 m in
saltwater. Its KG = 7 m. An amidship compartment (μ = 0.8) , length 25m is
damaged and flooded. Calculate its final draught and transverse GM using lost
buoyancy and added mass methods.
Show that both methods give the same assessment of the vessel’s initial stability.
© Omar bin Yaakob and Mohamad Pauzi Abdul Ghani, July 2006
115
Naval Architecture 2 Notes
8.5 FLOODING OF END COMPARTMENTS AND CHANGE OF DRAUGHTS
Flooding of end compartments will induce parallel sinkage as well as trim,
leading to large changes to draughts at the perpendiculars.
Example 3
A box shaped vessel 140m long, 20m wide floats in sea water at a draught of 4.5m.
A compartment extending the full width of the vessel with bulkheads at locations
15m and 40m forward of AP is flooded. The area and volume permeability for the
compartment are 75% and KG is 8.75m.
Calculate draughts and GM.
140m
140
m
20m
4.5
m
40m
15m
Lost buoyancy 25 20 4.5 0.75 1687.5 m 3
Intact WPA 140 20 - 25 20 0.75 2425m 2
1687.5
Parallel sinkage i.e. T
0.696m
2425
Shift in LCF, p
a q
Intact WPA
where a lost WPA
6.57 m fwd
q distance of lost WPA to original centroid
Trim
lost buoyancy Distance LCFnew to lost WPA centroid L
I Fnew
© Omar bin Yaakob and Mohamad Pauzi Abdul Ghani, July 2006
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Naval Architecture 2 Notes
Original second moment of area about old LCF,
I Fold
B L3
4,573,333 m 4
12
New second moment of area about new LCF,
I Fnew I Fold WPA p 2 i a p q
2
where i 2 nd moment of area of lost WPA and a is its area
I Fnew
m4
Trim
1687.5 49.07 140
TA
m by stern
76.57
1.36 m
140
TF Tp
Tf (m)
4.5
0.696
Original
Parallel
sinkage
T
FINAL
63.43
140
m
Ta (m)
4.5
0.696
To calculate GM,
Lost buoyancy Basis;
5.196
2.598 m
2
1/12140 - 25 20 3
BM
6.08 m
140 20 4.5
GM KB BM - KG 8.678 - 8.75 - 0.072 m
KB
Example 5
A ship LBP 150m, breadth 36m was involved in a collision and one of its
compartment ( = 0.95) length 36m with its centre 37m fwd of amidships is
damaged and open to sea. Just before collision, draughts of 3.1m and 7.3m were
recorded at forward and aft respectively. With these draughts the following data
were obtained from table of hydrostatics particulars.
AW
= 22166 m3
= 4820 m2
BML
LCF
= 320 m
= 2.5 m Aft of amidship
© Omar bin Yaakob and Mohamad Pauzi Abdul Ghani, July 2006
117
Naval Architecture 2 Notes
CB
= 0.8
At centre of hold, original draught = 3.1
Lost buoyancy
38
(7.3 3.1) 4.2 m
150
= 36 x 36 x 0.95 x 0.8 x 4.2
Lost wpa
= 36 x 36 x 0.95
Intact wpa
= 4820 – 1231
= 4137 m3
= 1231 m2
= 3589 m2
Parallel sinkage
=
4137
1.2 m
3589
Shift in LCF = lost wpA x dist to orig. F
Intact wpa
p
=
1231 (37 2.5)
13.55 m aftwards
3589
Trimming moment = lost buoyancy x dist. to new F
= 4137 x (13.55 + 2.5 + 37)
= 219468 m4
Original second moment of are about LCF,
IFold = BML x = 320 x 22166 = 7,093,120 m4
IFnew = {IFold + WPA x p2}- {i + a (p + q)2 }
where i = 2nd moment of area of lost waterplane and a is its
area.
IFnew = 4, 380, 145 m4
Trim =
Trimming moment x L =
IFnew
TF
=
7.52
91
4.56 m
150
TA
=
7.52
59
2.96 m
150
Original
Parallel
sinkage
T
FINAL
Tf (m)
3.1
1.2
Ta (m)
7.3
1.2
4.56
8.86
-2.96
5.54
© Omar bin Yaakob and Mohamad Pauzi Abdul Ghani, July 2006
118
219468
150 7.52 m by head
4380145
Naval Architecture 2 Notes
© Omar bin Yaakob and Mohamad Pauzi Abdul Ghani, July 2006
119
Naval Architecture 2 Notes
8.6 Mass and Centroid of Water entering ship bringing it to a particular
Waterline
Example 6
A ship of length 158 m has a displacement of 18,500 tonnes and the centre of
buoyancy 2.8 m abaft amidships. At a waterline tangential to the margin line the
areas of the immersed sections are as follows:
Stn
Area
AP
35
1/2
80
1
115
11/2
150
2
179
3
213
4
218
5
215
6
210
7
190
8
133
81/2
94
9
56
91/2
23
FP
0
Determine the mass of water that has entered the ship and the distanceof its
centroid from amidships.
Solution
Stn
AP
½
1
11/2
2
3
4
5
6
7
8
81/2
9
91/2
FP
Area
35
80
115
150
179
213
218
215
210
190
133
94
56
23
0
SM
½
2
1
2
11/2
4
2
4
2
4
11/2
2
1
2
½
f(A)
18
160
115
300
269
852
436
860
420
760
200
188
56
46
0
4680
Lever
5
41/2
4
31/2
3
2
1
0
1
2
3
31/2
4
41/2
5
Difference
1
Displacement 15.8 4680 1.025 25,200 tonnes
3
1638
LCB
15.8 5.53 m aft amidships
4680
Final Condition
Initial Condition
Difference
Displacement
(tones)
25,200
18,500
6,700
Lcb
Moment
5.53 A
2.80 A
139,350
51,800
87,500
© Omar bin Yaakob and Mohamad Pauzi Abdul Ghani, July 2006
120
fm(A)
90
720
460
1050
807
1704
436
5267 Aft
420
1520
600
658
224
207
0
3629 Fwd
1638 Aft
Naval Architecture 2 Notes
87550
13.07 m aft amidships
6700
mass of water that has entered the ship 6,700 tonnes
Centroid
centroid of water
13.07 m aft amidships
8.7 Location of Damaged Compartment
Waterline
bringing it to a particular
Example 7
A ship of length 128 m and displacement 12,200 tonnes with the centre of
buoyancy 0.77 m forward of amidships is brought, as the result of damage,to a
waterline at which the displacement is 14,200 tonnes and the centre of buoyancy is
7 m forward of amidships. The damage opens to the sea a compartment bounded by
transverse bulkheads and with a permeability of 80%. The areas of the immersed
sections for the fore body, in the damaged condition, at equidistant stations
commencing at amidships, are 139,148,158, 162, 139, 84 and 0 m2.
Determine the length and position of the damaged compartment.
Solution
Initial condition
Damaged condition
Water in compartment
Displacement (tonnes)
12,200
14,200
2,000
LCB (m)
0.77 F
7.00 F
45m F
Moment
9390
99400
90010
2000 0.975
2435 m 3
0.8
Assume a mean immersed sectional area of 120 m 2
Volume of compartment
length of compartment
2435
20.3 m
120
Assume bulkheads say 9 m aft and 11.3 m forward of centroid of flooded
compartment.
© Omar bin Yaakob and Mohamad Pauzi Abdul Ghani, July 2006
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Naval Architecture 2 Notes
2
Immersed sectional area (m )
100
AP
45 m fwd amidships
Aft End
Fwd End
Immersed area, As
158
124
65
SM
1
4
1
f(As)
158
496
65
719
Lever
0
1
2
fm(As)
0
496
130
626
Volume of compartment 1/3 719 10.15 2432 m 3
626
10.15 8.9 m fwd of aft end of compartment;
719
this more or less satisfies the requiremen ts
Centroid of compartment
Length of compartment 20.3 m
Aft end of compartment 45 - 9 36 m fwd of amidships
Fwd end of compartment 36 20.3 56.3 m fwd of amidships
8.8 Direct Flooding Calculations and Floodable Length
Δo
Δ1
L1
W1
Wo
g
G
B1
Bo
b
© Omar bin Yaakob and Mohamad Pauzi Abdul Ghani, July 2006
122
x
w
Lo
Naval Architecture 2 Notes
Figure shows the profile of a ship with waterline W1L1 tangential to the margin line.
The purpose of the calculation is to determine the extent and position of the
flooding which will bring the ship from waterline WoLo to the waterline W1L1.
Let
WoLo
waterline of undamaged
displacement and
centre of buoyancy
W1L1
g
G
ship
Δo
and
Bo
the
corresponding
waterline tangential to the margin line with Δ1 and B1 the
corresponding displacement and centre of buoyancy
centroid of lost buoyancy w
centre of gravity of ship for both conditions
Then
Mass of water gaining access to ship or the lost of buoyancy, w = Δ1 – Δo
Taking moments about Bo
Δ1 . b = w . x
x
1 b
w
Consequently the extent of the lost buoyancy or of the added weight to the4
waterline W1L1 and the position of the centroid can be determined. Thus, the
volume of water admitted is 0.975δΔ and the total volume of the compartment is
given by
v
0.975w
The length of the compartment is derived from a curve of areas of immersed
sections as follows:
l1
A
Portion of curve of immersed area
Figure shows a portion of such a curve derived from the Bonjean curves at the
waterline W1L1 which is tangential to the margin line. The centroid of the added
weight or of the lost buoyancy is on the ordinate at A. It is then necessary to
© Omar bin Yaakob and Mohamad Pauzi Abdul Ghani, July 2006
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Naval Architecture 2 Notes
determine an area under the curve, which will have its centroid on this ordinate at
A and also represent the volume
v
0.975w
The process is one of trial and error.
The procedure is to estimate a mean ordinate generally less than the ordinate at A,
say A1 and the first approximation to the length l1 of the compartment is given by
l1
v
0.975w 100
A1
A1
where is expressed as a percentage.
This length should be laid off so that the middle is on one side or the order of the
ordinate at A according to the shape of the curve. The volume and position of the
centroid corresponding to the length l1 can be determined by Simpson’s rule; this is
used as the basis for a second approximation. Normally the correct length and
position are obtained at the second attempt.
The length of the compartment so determined is known as the Floodable Length as
it is the length in the region considered which may be flooded without making the
ship sink beyond the margin line.
By calculations as indicated above for a series of waterlines which are tangential to
the margin line at different points throughtout the length of the ship it is possible to
determine a series of values for the plotting of a set of curves of floodable length as
shown below.
G
B
F
E
Aft
space
C
Engine
room
A
Fwd
space
D
Length of ship
Floodable Length Curve
In the above Figure, the point A is the mid-length of a compartment of length
represented by the ordinate AB. The vertical and horizontal scales used for plotting
the curves of floodable length are the same. Thus, CA = AD = 1/2AB and the
tangent of the angles BCA and BDA is 2.
© Omar bin Yaakob and Mohamad Pauzi Abdul Ghani, July 2006
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Naval Architecture 2 Notes
It is thus possible to ascertain whether any chosen length of compartment at any
position exceed the floodable length, by plotting the isosceles triangle with the
length of the compartment as the base.
Thus the length of the compartment represented by EF exceeds the floodable length
since the apex G lies above the curve.
Exercise 8
1. A box barge 100m long, depth 10m and 20 m beam floats at a level
keel draught of 8m and has a mid-length compartment 20m long. If
KG=8.4m, comment on stability of this vessel if the compartment is
bilged?
2. A box shaped vessel 150m long, 14m wide floats in sea water at a
draught of 7.0m. A compartment extending the full width of the vessel
with bulkheads at locations 35m and 65m forward of amidship is
flooded. The area and volume permeability for the compartment are
75% and KB can be assumed as equal to KG.
Determine whether the bow of the
calculate GM using lost buoyancy basis.
vessel will be submerged and
3. A ship of length 150 m has a displacement of 20,000 tonnes and the
centre of buoyancy 1.0 m aft of amidships.
At one waterline tangential to the margin line the areas of the
immersed sections are as follows:
Stn
Area
AP
0
1
2
3
4
5
6
7
8
9
60 135 200 220 220 215 205 175 110
FP
30
Find its floodable length and its centroid, assuming a permeability of
80%.
© Omar bin Yaakob and Mohamad Pauzi Abdul Ghani, July 2006
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Naval Architecture 2 Notes
© Omar bin Yaakob and Mohamad Pauzi Abdul Ghani, July 2006
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