Naval Architectural Calculation, Spring 2016, Myung-Il Roh 1
Ship Stability
Ch. 1 Introduction to Ship Stability
Spring 2016
Myung-Il Roh
Department of Naval Architecture and Ocean Engineering Seoul National University
Lecture Note of Naval Architectural Calculation
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Contents
Ch. 1 Introduction to Ship Stability
Ch. 2 Review of Fluid Mechanics
Ch. 3 Transverse Stability Due to Cargo Movement
Ch. 4 Initial Transverse Stability
Ch. 5 Initial Longitudinal Stability
Ch. 6 Free Surface Effect
Ch. 7 Inclining Test
Ch. 8 Curves of Stability and Stability Criteria
Ch. 9 Numerical Integration Method in Naval Architecture
Ch. 10 Hydrostatic Values and Curves
Ch. 11 Static Equilibrium State after Flooding Due to Damage
Ch. 12 Deterministic Damage Stability
Ch. 13 Probabilistic Damage Stability
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Ch. 1 Introduction to Ship Stability
1. Generals
2. Static Equilibrium
3. Restoring Moment and Restoring Arm 4. Ship Stability
5. Examples for Ship Stability
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1. Generals
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The force that enables a ship to float
It is directed upward.
It has a magnitude equal to the weight of the fluid which is displaced by the ship.
Water tank
How does a ship float? (1/3)
Ship
Water Ship
“Buoyant Force”
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How does a ship float? (2/3)
Archimedes’ Principle
The magnitude of the buoyant force acting on a floating body in the fluid is equal to the weight of the fluid which is displaced by the floating body.
The direction of the buoyant force is opposite to the gravitational force.
Equilibrium State (“Floating Condition”)
Buoyant force of the floating body
= Weight of the floating body
Displacement = Weight
G: Center of gravity B: Center of buoyancy W: Weight, : Displacement
: Density of fluid
V: Submerged volume of the floating body (Displacement volume, )
G
B W
= -W = - gV
Buoyant force of a floating body
= the weight of the fluid which is displaced by the floating body (“Displacement”)
Archimedes’ Principle
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How does a ship float? (3/3)
Displacement() = Buoyant Force = Weight(W)
Weight = Ship weight (Lightweight) + Cargo weight (Deadweight)
DWT LWT
W
C T B
L B
T: Draft C
B
: Block coefficient
: Density of sea water LWT: Lightweight DWT: Deadweight
Ship
Water Ship
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What is “Stability”?
Stability = Stable + Ability
F B
G B
F G
W L
℄ F B
G
F G
B
1W
1L
1B
℄ Inclining
(Heeling)
Restoring
F B G
F G
B
1B
Capsizing ℄
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What is a “Hull form”?
Hull form
Outer shape of the hull that is streamlined in order to satisfy requirements of a ship owner such as a deadweight, ship speed, and so on
Like a skin of human
Hull form design
Design task that designs the hull form Hull form of the VLCC(Very Large Crude oil Carrier)
Wireframe model Surface model
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What is a “Compartment”?
Compartment
Space to load cargos in the ship
It is divided by a bulkhead which is a diaphragm or peritoneum of human.
Compartment design (General arrangement design)
Compartment modeling + Ship calculation
Compartment modeling
Design task that divides the interior parts of a hull form into a number of compartments
Ship calculation (Naval architecture calculation)
Design task that evaluates whether the ship satisfies the required cargo capacity by a ship owner and, at the same time, the international regulations related to stability, such as MARPOL and SOLAS, or not
Compartment of the VLCC
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What is a “Hull structure”?
Hull structure
Frame of a ship comprising of a number of hull structural parts such as plates, stiffeners, brackets, and so on
Like a skeleton of human
Hull structural design
Design task that determines the specifications of the hull structural parts such as the size, material, and so on
Hull structure of the VLCC
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Principal Characteristics (1/2)
LOA (Length Over All) [m]: Maximum Length of Ship
LBP (Length Between Perpendiculars (A.P. ~ F.P.)) [m]
A.P.: After perpendicular (normally, center line of the rudder stock)
F.P.: Inter-section line between designed draft and fore side of the stem, which is perpendicular to the baseline
Lf (Freeboard Length) [m]: Basis of freeboard assignment, damage stability calculation
96% of Lwl at 0.85D or Lbp at 0.85D, whichever is greater
Rule Length (Scantling Length) [m]: Basis of structural design and equipment selection
Intermediate one among (0.96 Lwl at Ts, 0.97 Lwl at Ts, Lbp at Ts) Lwl
Loa
A.P. Lbp F.P.
W.L.
B.L.
W.L.
B.L.
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Definitions for the Length of a Ship
Main deck
Structures above main deck
(Main) Hull
Molded line Wetted line
Length overall(L OA ) Length on waterline(L WL )
Design waterline
Length between perpendiculars(L BP )
AP FP
Stem t stem
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Breadth
B.L. B.L.
Draf t
Depth Air Draft
B (Breadth) [m]: Maximum breadth of the ship, measured amidships
- B
molded: excluding shell plate thickness - B
extreme: including shell plate thickness
D (Depth) [m]: Distance from the baseline to the deck side line
- D
molded: excluding keel plate thickness - D
extreme: including keel plate thickness
Td (Designed Draft) [m]: Main operating draft - In general, basis of ship’s deadweight and speed/power performance
Ts (Scantling Draft) [m]: Basis of structural design
Air Draft [m]: Distance (height above waterline only or including operating draft) restricted by the port facilities, navigating route, etc.
- Air draft from baseline to the top of the mast - Air draft from waterline to the top of the mast - Air draft from waterline to the top of hatch cover - …
Principal Characteristics (2/2)
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Freeboard 1/2 Molded breadth(B
,mld)
Scantling draft
Dead rise Baseline
Camber
Scantling waterline
C L
Molded depth(D
,mld)
Keel
Depth
Sheer after Sheer forward
Deck plating
Deck beam
Centerline
Definitions for the Breadth and Depth of a Ship
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2. Static Equilibrium
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Center Plane
Before defining the coordinate system of a ship, we first introduce three planes, which are all standing perpendicular to each other.
Generally, a ship is symmetrical about starboard and port.
The first plane is the vertical longitudinal plane of symmetry, or center plane.
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The second plane is the horizontal plane, containing the bottom of the ship, which is called base plane.
Base Plane
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Midship Section Plane
The third plane is the vertical transverse plane through the midship, which is called midship section plane.
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Centerline in
(a) Elevation view, (b) Plan view, and (c) Section view
℄
℄ Centerline
℄: Centerline (a)
(b)
(c) Centerline:
Intersection curve between center plane and hull form
Elevation view
Plan view
Section view
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Baseline in
(a) Elevation view, (b) Plan view, and (c) Section view
℄
Baseline (a)
(b)
(c)
BL BL
Baseline:
Intersection curve between base plane and hull form
Elevation view
Plan view
Section view
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System of Coordinates
n-frame: Inertial frame x
nynznor x y z Point E: Origin of the inertial frame(n-frame) b-frame: Body fixed frame x
bybzbor x’ y’ z’
Point O: Origin of the body fixed frame(b-frame)
x b
y n
z n
x
nz
by
bE
O
1) Body fixed coordinate system
The right handed coordinate system with the axis called x b (or x’), y b (or y’), and z b (or z’) is fixed to the object. This coordinate system is called body fixed coordinate system or body fixed reference frame (b-frame) .
2) Space fixed coordinate system
The right handed coordinate system with the axis called x n (or x), y n (or y) and z n (or z) is fixed to the space. This coordinate system is called space fixed coordinate system or space fixed reference frame or inertial frame (n-frame) .
In general, a change in the position and orientation of the object is described with respect to
the inertial frame. Moreover Newton’s 2 nd law is only valid for the inertial frame.
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System of Coordinates for a Ship
Stem, Bow
Stern
y b z b
x b
(a)
AP LBP FP
BL SLWL
AP: aft perpendicular FP: fore perpendicular LBP: length between perpendiculars.
BL: baseline
SLWL: summer load waterline
(b)
x n y n
z n
x b y b z b
: midship
Space fixed coordinate system (n-frame): Inertial frame x
nynznor x y z
Body fixed coordinate system (b-frame): Body fixed frame x
bybzbor x’ y’ z’
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Center of Buoyancy (B) and Center of Mass (G)
Center of buoyancy (B)
It is the point at which all the vertically upward forces of support (buoyant force) can be considered to act.
It is equal to the center of volume of the submerged volume of the ship. Also, It is equal to the first moment of the submerged volume of the ship about particular axis divided by the total buoyant force (displacement).
Center of mass or Center of gravity (G)
It is the point at which all the vertically downward forces of weight of the ship (gravitational force) can be considered to act.
It is equal to the first moment of the weight of the ship about particular axis divided by the total weight of the ship.
※ In the case that the shape of a ship is asymmetrical with respect to the centerline.
K
x
z z
LCB
LCB VCB
B
B
B
C
L
K z
B
TCB
C
L LCG
G LCG
G G TCG
VCG G
y
x y
z
x y
K
: keel
LCB
: longitudinal center of buoyancy
VCB
: vertical center of buoyancy
LCG: longitudinal center of gravity
VCG: vertical center of gravity
TCB: transverse center of buoyancy
TCG: transverse center of gravity Elevation view
Plan view
Section view
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① Newton’s 2
ndlaw
Static
Equilibrium (1/3)
Static Equilibrium
ma F G
F
GF
G
m: mass of ship
a: acceleration of ship G: Center of mass FG
: Gravitational force of ship
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① Newton’s 2
ndlaw ma F
F
G
Static Equilibrium (2/3)
Static Equilibrium
F
BB
B: Center of buoyancy at upright
position(center of volume of the submerged volume of the ship)
FB
: Buoyant force acting on ship
F
B
for the ship to be in static equilibrium
0 F ,( a 0)
G
F
B F G
F
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Static Equilibrium (3/3)
F
BB
I
② Euler equation
When the buoyant force ( F
B) lies on the same line of action as the gravitational force ( F
G), total summation of the moment becomes 0.
I: Mass moment of inertia
: Angular velocity
for the ship to be in static equilibrium
0 ,( 0) Static Equilibrium
: Moment
Static Equilibrium
① Newton’s 2
ndlaw ma F
F
G F
Bfor the ship to be in static equilibrium
0 F ,( a 0)
G
F
B F G
F
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What is “Stability”?
Stability = Stable + Ability
F B
G B
F G
W L
℄ F B
G
F G
B
1W
1L
1B
℄ Inclining
(Heeling)
Restoring
F B G
F G
B
1B
Capsizing ℄
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Stability of a Floating Object
You have a torque on this object relative to any point that you choose. It does not matter where you pick a point.
The torque will only be zero when the buoyant force and the gravitational force are on one line. Then the torque becomes zero.
I
② Euler equation
When the buoyant force ( F
B) lies on the same line of action as the gravitational force ( F
G), total summation of the moment becomes 0.
for the ship to be in static equilibrium
0 ,( 0) Static Equilibrium
① Newton’s 2
ndlaw ma F
F
G F
Bfor the ship to be in static equilibrium
0 F ,( a 0)
G
F
B F
Rotate
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Stability of a Ship
You have a torque on this object relative to any point that you choose. It does not matter where you pick a point.
The torque will only be zero when the buoyant force and the gravitational force are on one line. Then the torque becomes zero.
I
② Euler equation
When the buoyant force ( F
B) lies on the same line of action as the gravitational force ( F
G), total summation of the moment becomes 0.
for the ship to be in static equilibrium
0 ,( 0) Static Equilibrium
① Newton’s 2
ndlaw ma F
F
G F
Bfor the ship to be in static equilibrium
0 F ,( a 0)
G
F
B F
Static Equilibrium
(a) F B (b)
G B
F G Rotate
F B G
F G
B
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a Floating Body (1/2)
(a) (b)
F B
G B
F G
F B
G
F G
B
1W L W
1L
1B
℄ ℄
I
Euler equation: 0 Interaction of weight and buoyancy resulting in intermediate state
Restoring Moment
r Torque
(Heeling Moment)
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Interaction of Weight and Buoyancy of a Floating Body (2/2)
(a) (b)
F B
G B
F G
F B
G
F G
B
1W L W
1L
1B
℄ ℄
Interaction of weight and buoyancy resulting in static equilibrium state
Heeling Moment
eI
Euler equation: 0
Static Equilibrium
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Stability of a Floating Body (1/2)
Floating body in stable state
G B
F B
F G
(a) (b)
G B
Restoring Moment Inclined
F B
F G
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Stability of a Floating Body (2/2)
Overturning Moment Inclined
G
B
(a) (b)
F G
F B G
B
F G
F B
Floating body in unstable state
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Transverse, Longitudinal, and Yaw Moment
Question) If the force F is applied on the point of rectangle object, what is the moment?
( ) ( ) ( )
P
P P P P z P y P z P x P y P x
x y z
x y z y F z F x F z F x F y F
F F F
M r F i j k
i j k
M
xM
yM
zTransverse moment Longitudinal moment Yaw moment
The x-component of the moment, i.e., the bracket term of unit vector i, indicates the transverse moment, which is the moment caused by the force F acting on the point P about x axis. Whereas the y-component, the term of unit vector j, indicates the longitudinal moment about y axis, and the z-component, the last term k, represents the yaw moment about z axis.
z
x
O
P
zF
F
yy F
z j k i x
F
xy ( , , )
P
x y z
P P Pr
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Equations for Static Equilibrium (1/3)
Suppose there is a floating ship. The force equilibrium states that the sum of total forces is zero.
, where
F
G.zand F
B.zare the z component of the gravitational force vector and the buoyant force vector, respectively, and all other components of the vectors are zero.
, , 0
G z B z
F F F
Also the moment equilibrium must be satisfied, this means, the resultant moment should be also zero.
G B
τ M M 0
where M
Gis the moment due to the gravitational force and M
Bis the moment due to the buoyant
force.
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Equations for Static Equilibrium (2/3)
G B
τ M M 0
where M
Gis the moment due to the gravitational force and M
Bis the moment due to the buoyant force.
From the calculation of a moment we know that M
Gand M
Bcan be written as follows:
, , ,
, , , , , ,
( ) ( ) ( )
G G G
G G G
G x G y G z
G G z G G y G G z G G x G G y G G x
x y z
F F F
y F z F x F z F x F y F
M r F
i j k
i j k
, , ,
, , , , , ,
( ) ( ) ( )
B B B
B B B
B x B y B z
B B z B B y B B z B B x B B y B B x
x y z
F F F
y F z F x F z F x F y F
M r F
i j k
i j k
, , , , , ,
( ) ( ) and ( ) ( )
G
y
G F
G z z
G F
G y x F
G
G z B y F
B
B z z F
B
B y x F
B B zM i j M i j
, , , ,
( ) ( ) and ( ) ( )
G
y
G F
G z x F
G G z B y F
B
B z x F
B B zM i j M i j
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Equations for Static Equilibrium (3/3)
G B
τ M M 0
where M
Gis the moment due to the gravitational force and M
Bis the moment due to the buoyant force.
, , , ,
( ) ( )
G B y F G G z y F B B z x F G G z x F B B z
τ M M i j 0
, , 0
G G z B B z
y F y F
Substituting
G B 0
G B
y y y y
, ,
G z B z
F F (force equilibrium)
, , , ,
( ) ( ) and ( ) ( )
G y F G G z x F G G z B y F B B z x F B B z
M i j M i j
, ,
and x F G G z x F B B z 0
G B 0
G B
x x x x
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3. Restoring Moment and Restoring Arm
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Restoring Moment Acting on an Inclined Ship
B
1F
BB G F
GZ
B
G Z
r Heeling
Moment
eRestoring Moment
W
1L
1F B
B G F G
W L
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Restoring Arm (GZ, Righting Arm)
G: Center of mass K: Keel B: Center of buoyancy at upright position B
1: Changed center of buoyancy
F
G: Weight of ship F
B: Buoyant force acting on ship restoring F GZ B
• Transverse Restoring Moment B
1F
BB G F
GZ
B
G Z
r Heeling
Moment
eRestoring Moment
• The value of the restoring moment is found by multiplying the
buoyant force of the ship (displacement), , by the
perpendicular distance from G to the line of action of .
• It is customary to label as Z
the point of intersection of the line of action of and the parallel line to the waterline through G to it.
• This distance GZ is known as the
‘restoring arm’ or ‘righting arm’.
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Metacenter (M) • Restoring Moment
restoring F GZ
B
rG Z
B
1B
Z:
The intersection point of the line of buoyant force through B1with the transverse line through GF
GF
B
eDefinition of M (Metacenter)
• The intersection point of the vertical line through the center of buoyancy at previous position (B) with the vertical line through the center of buoyancy at new position (B
1) after inclination
• GM Metacentric height
M
• The term meta was selected as a prefix for center because its Greek meaning implies movement. The metacenter therefore is a moving center.
• From the figure, GZ can be obtained with assumption that M does not change within a small angle of inclination (about 7 to 10), as below.
sin
GZ GM
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Restoring Moment at Large Angle of Inclination (1/3)
G: Center of mass of a ship FG: Gravitational force of a ship
B: Center of buoyancy in the previous state (before inclination) FB: Buoyant force acting on a ship
B1: New position of center of buoyancy after the ship has been inclined
To determine the restoring arm ”GZ”, it is necessary to know the positions of the center of mass (G) and the new position of the center of buoyancy (B 1 ).
• The use of metacentric height (GM) as the restoring arm is not valid for a ship at a large angle of inclination.
Z: The intersection point of a vertical line through the new position of the center of buoyancy(B1) with the transversely parallel line to a waterline through the center of mass(G)
F
G
eG
B B
1
rZ
F
BM
//
//
sin GZ GM
For a small angle of inclination (about 7 to 10)
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Restoring Moment at Large Angle of Inclination (2/3)
M: The intersection point of the vertical line through the center
of buoyancy at previous position ( B
i-1) with the vertical line
through the center of buoyancy at present position ( B
i) after
inclination
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Restoring Moment at Large Angle of Inclination (3/3)
G
L
35 ,35F
B ,30F
BC
35=35
Z
35 sin 35
GZ GM
L
30C
30M: The intersection point of the vertical line through the center of buoyancy at previous position ( B
i-1) with the vertical line through the center of buoyancy at present position ( B
i) after inclination
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• Righting (Restoring) Moment: Moment to return the ship to the upright floating position
• Stable / Neutral / Unstable Condition: Relative height of G with respect to M is one measure of stability.
Stability of a Ship According to
Relative Position between “G”, “B”, and “M” at Small Angle of Inclination
• Stable Condition ( G < M ) • Neutral Condition ( G = M ) • Unstable Condition ( G > M )
FB FG
Z G
B M
FB FG
G M
B
FB FG
M G B
B K
F B
B
1F G
G, Z, M
G: Center of mass K: Keel
B: Center of buoyancy at upright position B1: Changed center of buoyancy FG: Weight of ship FB: Buoyant force acting on ship
Z: The intersection of the line of buoyant force through B1with the transverse line through G M: The intersection of the line of buoyant force through B1with the centerline of the ship
B K
F B
B
1F G
Z G
M
B K
F B
B
1F G
Z G
M
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F
GB0
F
eG
B
F
GB0
F
eG
1
B
B B
2B
1F
GF
GF
GF
GB1
F
B2
F F
B1B2
F
The ship is inclined further from it. The ship is inclined further from it.
- Overview of Ship Stability
One of the most important factors of stability is the breadth.
So, we usually consider that transverse stability is more important than longitudinal stability.
Importance of Transverse Stability
The ship is in static equilibrium state. Because of the limit of the breadth, “B” can not move further. the ship will capsize.
As the ship is inclined, the position of the center of buoyancy “B” is changed.
Also the position of the center of mass “G” relative to inertial frame is changed.
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4. Ship Stability
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G F
GSummary of Static Stability of a Ship (1/3)
When an object on the deck moves to the right side of a ship, the total center of mass of the ship moves to the point G
1, off the centerline.
B
e Because the buoyant force and the gravitational force are not on one line, the forces induces a moment to incline the ship.
* We have a moment on this object relative to any point that we choose.
It does not matter where we pick a point.
F
B G1F G
1 G1F
G1: New position of center of mass after the object on the deck moves to the right side
Z: The intersection of a line of buoyant force(FB) through the new position of the center of buoyancy (B1) with the transversely parallel line to the waterline through the center of mass of a ship(G)
G: Center of mass of a ship
FG: Gravitational force of a ship B: Center of buoyancy at initial position FB: Buoyant force acting on a ship
B1: New position of center of buoyancy after the ship has been inclined
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Summary of Static Stability of a Ship (2/3)
The total moment will only be zero when the buoyant force and the gravitational force are on one line. If the moment becomes zero, the ship is in static equilibrium state.
eG F
GB
F
BG
1F
BB
1F
B G1F
rB
1F
BB G
rB
G G
1
eG1
F
G1: New position of center of mass after the object on the deck moves to the right side
Z: The intersection of a line of buoyant force(FB) through the new position of the center of buoyancy (B1) with the transversely parallel line to the waterline through the center of mass of a ship(G)
G: Center of mass of a ship
FG: Gravitational force of a ship B: Center of buoyancy at initial position FB: Buoyant force acting on a ship
B1: New position of center of buoyancy after the ship has been inclined
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Summary of Static Stability of a Ship (3/3)
When the object on the deck returns to the initial position in the centerline, the center of mass of the ship returns to the initial point G.
B
1 Then, because the buoyant force and the gravitational force are not on one line, the forces induces a restoring moment to return the ship to the initial position.
By the restoring moment, the ship returns to the initial position.
F
BG
1B G
r
eF
GZ
B
G Z
The moment arm of the buoyant force and gravitational force about G is
expressed by GZ, where Z is defined as the intersection point of the line of buoyant force(F
B) through the new position of the center of buoyancy(B
1) with the
transversely parallel line to the waterline through the center of mass of the ship (G).
righting F GZ B
• Transverse Righting Moment
※ Naval architects refer to the restoring moment as “righting moment”.
G1: New position of center of mass after the object on the deck moves to the right side
Z: The intersection of a line of buoyant force(FB) through the new position of the center of buoyancy (B1) with the transversely parallel line to the waterline through the center of mass of a ship(G)
G: Center of mass of a ship
FG: Gravitational force of a ship B: Center of buoyancy at initial position FB: Buoyant force acting on a ship
B1: New position of center of buoyancy after the ship has been inclined
Naval Architectural Calculation, Spring 2016, Myung-Il Roh 52
Evaluation of Stability
: Merchant Ship Stability Criteria – IMO Regulations for Intact Stability
(a) Area A ≥ 0.055 m-rad
(c) Area B ≥ 0.030 m-rad
(d) GZ ≥ 0.20 m at an angle of heel equal to or greater than 30
(b) Area A + B ≥ 0.09 m-rad
(e) GZ
maxshould occur at an angle of heel preferably exceeding 30 but not less than 25.
(f) The initial metacentric height GM
oshould not be less than 0.15 m.
(IMO Res.A-749(18) ch.3.1)
※ After receiving approval of calculation of IMO regulation from Owner and Classification Society, ship construction can proceed.
IMO Regulations for Intact Stability
IMO recommendation on intact stability for passenger and cargo ships.
Static considerations The work and energy
considerations (dynamic stability) Area A: Area under the righting arm curve
between the heel angle of 0 and 30
Area B: Area under the righting arm curve
between the heel angle of 30 and min(40,
f)
※
f: Heel angle at which openings in the hull
m: Heel angle of maximum righting arm
B 10
0 20 30 40 50 60 70 80
Angle of heel ( []) Righting arm
(GZ)
A
m
= const.
(: displacement)
f
GM
57.3
Naval Architectural Calculation, Spring 2016, Myung-Il Roh 53
5. Examples for Ship Stability
Naval Architectural Calculation, Spring 2016, Myung-Il Roh 54
[Example] Equilibrium Position and Orientation of a Box-shaped Ship
Question 1) The center of mass is moved to 0.3 [m] in the direction of the starboard side.
A box-shaped ship of 10 meter length, 5 meter breadth and 3 meter height weighs 205 [kN].
The center of mass is moved 0.3 [m] to the left side of the center of the deck.
When the ship is in static equilibrium state, determine the angle of heel ( ) of the ship.
Assumption)
(1) Gravitational acceleration = 10 [m/s
2], Density of sea water = 1.025 [ton/m
3]
(2) When the ship will be in the static equilibrium finally, the deck will not be immersed and the bottom will not emerge.
0.4m 3m
Baseline
AP FP
10m
5m 0.3m
:
Location of the center of gravity of the ship
G205
F kN
Given: Length ( L ): 10m, Breadth ( B ): 5m, Depth ( D ): 3m, Weight ( W ): 205kN,
Location of the Center of Gravity: 0.3m to the left side of the center of the deck
Find: Angle of Heel(ϕ)
Naval Architectural Calculation, Spring 2016, Myung-Il Roh 55
(1) Static Equilibrium (1/3)
: Location of the center of mass of the ship 0.4m
3m
Baseline
AP FP
10m
5m 0.3m
When the ship is floating in sea water, the requirement for ship to be in static equilibrium state is derived from Newton’s 2 nd law and Euler equation as follows.
(1-1) Newton’s 2 nd Law: Force Equilibrium
The resultant force should be zero to be in static equilibrium.
, , 0
n n n
G z B z
F F F
, where
n
F
G.z: z
n-coordinate of the gravitational force
n
F
B.z: z
n-coordinate of the buoyant force
(1-2) Euler Equation: Moment Equilibrium
The resultant moment should be zero to be in static equilibrium.
n n n
G B
τ M M 0
, where
n
M
G: the moment due to the gravitational force
n
M
B: the moment due to the buoyant force.
Naval Architectural Calculation, Spring 2016, Myung-Il Roh 56
Solution)
(1) Static Equilibrium (2/3)
: Location of the center of mass of the ship 0.4m
3m
Baseline
AP FP
10m
5m G 205
F kN 0.3m
G B
y y
z
O,E y K
B z'
y' x,x'
F G
F B ф˚
G y
Gy
BThe first step is to satisfy the Newton- Euler equation which requires that the sum of total forces and moments acting on the ship is zero.
As described earlier, in order to satisfy a stable equilibrium, the buoyant force and gravitational force should act on the same vertical line, therefore, the moment arm of the buoyant force and
gravitational force must be same.
Naval Architectural Calculation, Spring 2016, Myung-Il Roh 57
Solution)
(1) Static Equilibrium (3/3)
: Location of the center of mass of the ship 0.4m
3m
Baseline
AP FP
10m
5m G 205
F kN 0.3m
G B
y y
z
O,E y K
B z'
y' x,x'
F
GF
Bф˚
G y
Gy
Bcos sin sin cos
G G
G G
y y
z z
cos sin sin cos
B B
B B
y y
z z
By representing and with , and , we can get
Gy y
By z y
G , ,
G Bz
Bcos sin cos sin
G G B B
y z y z
In this equation, we suppose that y' G and z' G are already given, and y' B and z' B can be geometrically calculated.
Space fixed coordinate system(n-frame): Inertial frame x y z Body fixed coordinate system(b-frame): Body fixed frame x’ y’ z’
1
Naval Architectural Calculation, Spring 2016, Myung-Il Roh 58
Solution)
(2-1) Changed Center of Buoyancy, B
1, with Respect to the Body Fixed Frame
G B
y y
The centroid of A with respect to the body fixed frame:
C A_,
C A_
A z,,
A y,A A
M y z M
A A
, where
A A : the area of A
M A,z’ : 1 st moment of area of A about z’ axis M A,y’ : 1 st moment of area of A about y’ axis.
To obtain the centroid of A, the followings are required.
- The area of A
- 1 st moment of area of A about z’ axis - 1 st moment of area of A about y’ axis
z
O,E y K
B z'
y' x,x'
F
GF
Bф˚
G y
Gy
BA
1Naval Architectural Calculation, Spring 2016, Myung-Il Roh 59
(2-2) Center of Buoyancy and Center of Gravity with Respect to the Body Fixed Frame (1/2) 1) Center of buoyancy, B 1 , with respect to the body
fixed frame
The centroid of A with respect to the body fixed frame:
C A_,
C A_
A z,,
A y,A A
M y z M
A A
To calculate the centroid of A using the geometrical relations, we use the areas, A 1 , A 2 , and A 3 .
To describe the values of A 1 , A 2 , and A 3 using the geometrical parameters (a, t, and ), y’ and z’ coordinate of the points P, Q, R, R 0 , S, S 0 with respect to the body fixed frame is used, which are given as follows.
0 0
0 0
0 0
, , , , ,
( , ) ( , tan ), ( , ) ( , 0) ( , ) ( , tan ), ( , ) ( , 0)
P P Q Q
R R R R
S S S S
P y z a t Q y z a t
R y z a a R y z a
S y z a a S y z a
R
0ф˚
2a
2b
P Q
R
t S
S
0A
1A
2A
3A
A A1 A2 A3
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Solution)
(2-2) Center of Buoyancy and Center of Gravity with Respect to the Body Fixed Frame (2/2)
R
0ф˚
2a
2b
P Q
R
t S
S
0A
1A
2A
3A
A A1 A2 A3
Area: 1 tan 2 a a
Centroid: y z
C ,
C 2 3 a , 1 3 a tan
Moment of area about z’ axis:
1 2 1 3
tan tan
2 3 3
Area y
C a a a a Moment of area about y’ axis:
3 2
1 1 1
tan tan tan
2 3 6
Area z
C a a a a
A 2
y
z
C,
C
C y z a tan 1/ 3 a tan a
2 / 3 a
G B
y y
Naval Architectural Calculation, Spring 2016, Myung-Il Roh 61
Solution)
(2-3) Center of Buoyancy and Center of Gravity with Respect to the Body Fixed Frame (1/2) 1) Center of buoyancy, B 1 , with respect to the body
fixed frame The centroid of A with respect
to the body fixed frame:
C A_,
C A_
A z,,
A y,A A
M y z M
A A
The table blow summarizes the results of the area, centroid with respect to the body fixed frame and 1
stmoment of area with respect to the body fixed frame of A
1, A
2, A
3, and A.
The center of buoyancy, B 1 , with respect to the body fixed frame is
, , ' 2tan
2 tan
2, , ,
3 2 6
A y A z
B B
A A
M
M a t a
y z A A t t
Area
A
ACentroid
( , ) y z
C
CMoment of area about z'-axis ( y A
C )
Moment of area about y'-axis
( z A
C )
A
12a t 0,
2
t
0 a t
2A
21 tan
2 a a 2 , tan
3 3
a a
3
tan 3
a
3 tan
26
a
A
31 tan
2 a a 2 , tan
3 3
a a
3
tan 3
a
3 tan
26
a
A
(=A
1+A
2-A
3)
2a t - 2
3tan
3
a
2 3 tan
23
a t a
R0 ф˚
2a
2b
P Q
R
t S
S0
A1 A2 A3
A
A A1 A2 A3
G B
y y
Naval Architectural Calculation, Spring 2016, Myung-Il Roh 62
Solution)
(2-3) Center of Buoyancy and Center of Gravity with Respect to the Body Fixed Frame (2/2) 2) Center of gravity, G, with respect to the body
fixed frame
The center of gravity, G, with respect to the body fixed frame is given by geometrical relations as shown in the figure, which is
y ' , ' G z G d b t , 2
2tan
2 tan
2, ,
3 2 6
B B
a t a
y z t t
G B
y y
y’
B
K
O,E2a
y
z z’
ф˚
F G
F B
G
x
n, x
bB
12b
t
d
Naval Architectural Calculation, Spring 2016, Myung-Il Roh 63
yb
B K
O,E 2a
yn zn zb
ф˚
F
B Gx
n,x
b B12b
t
d
(b) (3) Comparison between the Figure Describing the Ship Inclined
and the Figure Describing the Water Plane Inclined (1/2)
Let us calculate the center of buoyancy, B 1 , and the center of gravity, G, using the Fig. (b).
The center of buoyancy, B 1 , and the center of gravity, G, with respect to the body fixed frame
2tan
2 tan
2, ,
3 2 6
B B
a t a
y z t t
y ' , '
Gz
G d b t , 2
Next, we use the condition that the moment arm of the buoyant force and gravitational force must be same and substitute the coordinates of the center of gravity and buoyancy with respect to the body fixed frame into the following equation.
cos sin cos sin
G G B B
y z y z
Naval Architectural Calculation, Spring 2016, Myung-Il Roh 64
Solution)
(3) Comparison between the Figure Describing the Ship Inclined and the Figure Describing the Water Plane Inclined (2/2)
3
22
2 2tan
2 sin
cos (2 ) sin
6 t a a
d b t
t
215.025 15.625
2.6 sin 0.3 cos sin tan
3 6
Substituting a=2.5m, b=1.5m, t=0.4m, d=0.3m into this equation and rearranging
tan 0.159 [rad] 9.019 [deg]
2tan
2 tan
2, ,
3 2 6
B B