2017-01-24
Lecture Note of Naval Architectural Calculation
Ship Stability
Ch. 1 Introduction to Ship Stability
Spring 2016
Myung-Il Roh
Department of Naval Architecture and Ocean Engineering Seoul National University
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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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How does a ship float? (1/3)
The force that enables a ship to float “Buoyant Force” It is directed upward. It has a magnitude equal to the weight of the fluid which is displaced by the ship.
Ship
Ship Water tank Water
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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. Buoyant force of a floating body = the weight of the fluid which is displaced by the floating body (“Displacement”) Archimedes’ Principle Equilibrium State (“Floating Condition”) Buoyant force of the floating body W = -W = -gV = Weight of the floating body G Displacement = Weight
G: Center of gravity B: Center of buoyancy B W: Weight, : Displacement : Density of fluid V: Submerged volume of the floating body (Displacement volume, ) 6 Naval Architectural Calculation, Spring 2016, Myung-Il Roh
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How does a ship float? (3/3)
Displacement() = Buoyant Force = Weight(W)
L B T C T: Draft B CB: Block coefficient : Density of sea water W LWT DWT LWT: Lightweight DWT: Deadweight
Weight = Ship weight (Lightweight) + Cargo weight(Deadweight) Ship
Ship Water
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What is “Stability”? FG
G B Capsizing ℄ B1
FB FG FG Inclining W L WL(Heeling) 1 1 G G
B B B1 Restoring
FB ℄ ℄ FB Stability = Stable + Ability
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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 10 Naval Architectural Calculation, Spring 2016, Myung-Il Roh
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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
W.L. W.L.
B.L. B.L.
A.P. Lbp F.P. Lwl 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)
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Definitions for the Length of a Ship
Structures above main deck Main deck
(Main) Hull
Wetted line Molded line
Length overall(LOA)
Length on waterline(LWL)
Stem tstem Design waterline
Length between perpendiculars(L ) AP BP FP
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Principal Characteristics (2/2)
B (Breadth) [m]: Maximum breadth of the ship, measured amidships
-Bmolded: excluding shell plate thickness -Bextreme: including shell plate thickness Air Draft
D (Depth) [m]: Distance from the baseline to the deck side line
-Dmolded: excluding keel plate thickness -Dextreme: including keel plate thickness
Td (Designed Draft) [m]: Main operating draft
Depth - In general, basis of ship’s deadweight and speed/power performance Draft
Ts (Scantling Draft) [m]: Basis of structural design
B.L. B.L. Breadth
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 -…
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Definitions for the Breadth and Depth of a Ship
1/2 Molded breadth(B ) ,mld Deck plating
Camber Deck beam
Freeboard
Scantling waterline Molded depth(D,mld)
Scantling draft Centerline
Dead rise CL Baseline Keel Sheer after Sheer forward
Depth
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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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Base Plane
The second plane is the horizontal plane, containing the bottom of the ship, which is called 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: Intersection curve between center plane and hull form
Centerline Elevation view
Plan view (a) ℄ (c) Section view ℄ ℄: Centerline (b)
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Baseline in (a) Elevation view, (b) Plan view, and (c) Section view Baseline: Intersection curve between base plane and hull form
Elevation view
BL BL Plan view (a) ℄ (c) Section view
Baseline (b)
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System of Coordinates
zb zn yb O
n-frame: Inertial frame xn yn zn or x y z E yn xb Point E: Origin of the inertial frame(n-frame)
b-frame: Body fixed frame xb yb zb or x’ y’ z’ Point O: Origin of the body fixed frame(b-frame) xn 1) Body fixed coordinate system
The right handed coordinate system with the axis called xb(or x’), yb(or y’), and zb(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 xn(or x), yn(or y) and zn(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 2nd law is only valid for the inertial frame.
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System of Coordinates for a Ship
Body fixed coordinate system (b-frame): Body fixed frame xb yb zb or x’ y’ z’
Space fixed coordinate system (n-frame): Inertial frame xn yn zn or x y z
Stem, Bow
zb zn zb SLWL yb yn xb x n BL yb AP LBP xb FP
AP: aft perpendicular : midship FP: fore perpendicular LBP: length between perpendiculars. Stern BL: baseline (a) SLWL: summer load waterline (b)
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K : keel Center of Buoyancy (B) LCB: longitudinal center of buoyancy LCG: longitudinal center of gravity VCB: vertical center of buoyancy VCG : vertical center of gravity and Center of Mass (G) TCB: transverse center of buoyancy TCG : transverse center of gravity z z Elevation view Section view y x x y G LCG VCG B G B LCB VCB K CL Plan view y z TCG z x G B G TCB LCB LCG B K CL ※ In the case that the shape of a ship is asymmetrical Center of buoyancy (B) with respect to the centerline. 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.
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Static Equilibrium (1/3)
Static Equilibrium
nd F ① Newton’s 2 law G ma F
FG G
m: mass of ship G: Center of mass
a: acceleration of ship FG: Gravitational force of ship
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Static Equilibrium (2/3)
Static Equilibrium
nd F ① Newton’s 2 law G ma F
FG FB G for the ship to be in static equilibrium 0,(0) Fa
FG FB B
FB
B: Center of buoyancy at upright position(center of volume of the submerged volume of the ship)
FB: Buoyant force acting on ship
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Static Equilibrium (3/3)
Static Equilibrium
nd F ① Newton’s 2 law G ma F
FG FB G for the ship to be in static equilibrium 0,(0) Fa F F Static Equilibrium G B B ②Euler equation I FB for the ship to be in static equilibrium 0,(0)
When the buoyant force (FB) lies on the same : Moment I: Mass moment of inertia line of action as the gravitational force (FG), : Angular velocity total summation of the moment becomes 0.
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What is “Stability”? FG
G B Capsizing ℄ B1
FB FG FG Inclining W L WL(Heeling) 1 1 G G
B B B1 Restoring
FB ℄ ℄ FB Stability = Stable + Ability
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Stability of a Floating Object
You have a torque on this object Static Equilibrium relative to any point that you choose. It does not matter where you pick a point. ① Newton’s 2nd law ma F The torque will only be zero when the F F buoyant force and the gravitational G B force are on one line. Then the torque for the ship to be in static equilibrium becomes zero. 0,(0) Fa
Rotate FG FB
②Euler equation I for the ship to be in static equilibrium 0,(0)
When the buoyant force (FB) lies on the same line of action as the gravitational force (FG), total summation of the moment becomes 0.
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Stability of a Ship
You have a torque on this object Static Equilibrium relative to any point that you choose. It does not matter where you pick a point. ① Newton’s 2nd law ma F The torque will only be zero when the F F buoyant force and the gravitational G B force are on one line. Then the torque for the ship to be in static equilibrium becomes zero. 0,(0) Fa
Rotate FG FB F G FG ②Euler equation I G G for the ship to be in static equilibrium B B 0,(0)
F When the buoyant force (FB) lies on the same B FB line of action as the gravitational force (FG), (a) (b) total summation of the moment becomes 0. Static Equilibrium 30 Naval Architectural Calculation, Spring 2016, Myung-Il Roh
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Interaction of Weight and Buoyancy of a Floating Body (1/2) Restoring Torque Moment (Heeling r Moment) F e FG G
WLW1 L1 G G
B B B1
FB ℄ ℄ FB (a) (b) Euler equation: I 0
Interaction of weight and buoyancy resulting in intermediate state
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Interaction of Weight and Buoyancy of a Floating Body (2/2)
Static Equilibrium Heeling Moment
e FG FG
WLW1 L1 G G
B B B1
FB ℄ ℄ FB (a) (b) Euler equation: I 0
Interaction of weight and buoyancy resulting in static equilibrium state
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Stability of a Floating Body (1/2)
Restoring Moment Inclined
B FB B FG
FB G FG G
(a) (b)
Floating body in stable state
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Stability of a Floating Body (2/2)
Overturning Moment Inclined F G G FG G
B B FB FB
(a) (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?
z Fz k F y P i z j F x y
Fx r (,x yz ,) O P PPP y
x MrP F
ijkTransverse moment Longitudinal moment Yaw moment xP yP z Pij()()() yF Pz zF Py xF Pz zF Px k xF Py yF Px M M M FFFxyz x y z 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. 35 Naval Architectural Calculation, Spring 2016, Myung-Il Roh
Equations for Static Equilibrium (1/3)
Suppose there is a floating ship. The force equilibrium states that the sum of total forces is zero.
FFGz,, F Bz 0 , where
FG.z and FB.z are the z component of the gravitational force vector and the buoyant force vector, respectively, and all other components of the vectors are zero.
Also the moment equilibrium must be satisfied, this means, the resultant moment should be also zero.
τMGB M 0
where MG is the moment due to the gravitational force and MB is the moment due to the buoyant force.
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Equations for Static Equilibrium (2/3)
τMGB M 0
where MG is the moment due to the gravitational force and MB is the moment due to the buoyant force.
From the calculation of a moment we know that MG and MB can be written as follows:
MrFGGG ijk xyzGGG FFFGx,,, Gy Gz
ij()()()yFGGzGGy,, zF xF GGzGGx ,, zF k xF GGyGGx , yF ,
MrFBBB ijk xyzBBB FFFBx,,, By Bz
ij()()()yFB BzBBy,, zF xF BBzBBx ,, zF k xF BByBBx , yF ,
MiG()()and()()yF G Gz,, zF G Gy j xF G Gz , Mi B yF B Bz ,, zF B By j xF B Bz ,
MiGGGzGGzBBBzBBz()()and()()yF,, j xF Mi yF ,, j xF
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Equations for Static Equilibrium (3/3)
τMGB M 0
where MG is the moment due to the gravitational force and MB is the moment due to the buoyant force.
MiGGGzGGzBBBzBBz()()and()()yF,, j xF Mi yF ,, j xF
τMGB M i()()yF GGzBBzGGzBBz ,, yF j xF ,, xF 0
yFGGzBBz,, yF 0 and xFGGzBBz,, xF 0
Substituting FGz,, F Bz (force equilibrium)
yyGB0 xxGB 0
yyGB xGB x
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3. Restoring Moment and Restoring Arm
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Restoring Moment Acting on an Inclined Ship
Restoring Heeling Moment Moment F r e G
FG Z G G Z WLW1 L1
B B B1
FB FB
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Restoring Arm (GZ, Righting Arm)
• The value of the restoring moment Heeling Restoring is found by multiplying the Moment Moment buoyant force of the ship e r (displacement), , by the perpendicular distance from G to FG G Z the line of action of . • It is customary to label as Z the point of intersection of the line B B of action of and the parallel line 1 to the waterline through G to it. • This distance GZ is known as the ‘restoring arm’ or ‘righting arm’. FB • Transverse Restoring Moment
restoring F B GZ
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
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• Restoring Moment F GZ Metacenter (M) restoring B
M Definition of M (Metacenter) • The intersection point of the vertical F e G line through the center of buoyancy at previous position (B) with the G Z vertical line through the center of
buoyancy at new position (B1) after inclination • The term meta was selected as a prefix for center B because its Greek meaning implies movement. The B1 metacenter therefore is a moving center. • GM Metacentric height
F • From the figure, GZ can be obtained B with assumption that M does not change within a small angle of r inclination (about 7 to 10), as below. Z: The intersection point of the line of buoyant force through B1 with the transverse line through G GZ GM sin
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Restoring Moment at Large Angle of Inclination (1/3)
M GZ GM sin F e For a small angle of inclination G (about 7 to 10)
G // Z • The use of metacentric height (GM) // as the restoring arm is not valid for a ship at a large angle of inclination.
B B1
To determine the restoring F arm ”GZ”, it is necessary to know B the positions of the center of mass G: Center of mass of a ship r (G) and the new position of the FG: Gravitational force of a ship B: Center of buoyancy in the previous state (before inclination) center of buoyancy (B1). FB: Buoyant force acting on a ship
B1: New position of center of buoyancy after the ship has been inclined 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)
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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 (Bi-1) with the vertical line through the center of buoyancy at present position (Bi) after inclination
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Restoring Moment at Large Angle of Inclination (3/3)
M: The intersection point of the vertical line through the center
of buoyancy at previous position (Bi-1) with the vertical line through the center of buoyancy at present position (Bi) after inclination
GZ GM 35sin 35 C35 C30
=35
G Z
FB,35
FB,30
L35 L30
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Stability of a Ship According to Relative Position between “G”, “B”, and “M” at Small Angle of Inclination
• 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.
FG • Stable Condition ( G < M ) • Neutral Condition ( G = M ) • Unstable Condition ( G > M ) F Z G G G, Z, M M FG M
G Z
F B B1 B B1 G B B1 FG G M FG M K K K G M F G Z F F B B B B B B
F F B B FB 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 B1 with the transverse line through G
M: The intersection of the line of buoyant force through B1 with the centerline of the ship
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Importance of Transverse Stability
FG F GFG FG FG FG G G
e e
B B2 B1 B1 B
F B0 F FF B F BF0BB 1 B2 1 2
The ship is inclined further from it. The ship is inclined further from it. 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.
One of the most important factors of stability is the breadth. So, we usually consider that transverse stability is more important than longitudinal
stability. - Overview of Ship Stability 47 Naval Architectural Calculation, Spring 2016, Myung-Il Roh
4. Ship Stability
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Summary of Static Stability of a Ship (1/3)
e 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 G1, off the centerline. Because the buoyant force and the F F G1 G1 gravitational force are not on one line, the
G G1 forces induces a moment to incline the ship. * We have a moment on this object relative to any point that we choose. B It does not matter where we pick a point.
FB
G: Center of mass of a ship G1: New position of center of mass after the object on the deck moves to the right side 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 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) 49 Naval Architectural Calculation, Spring 2016, Myung-Il Roh
Summary of Static Stability of a Ship (2/3)
e
F F G FG1 G1 GG G1 G1
BB B1
The total moment will only be zero FB FF when the buoyant force and the BB gravitational force are on one line. If the moment becomes zero, the ship is in static equilibrium state. r G: Center of mass of a ship G1: New position of center of mass after the object on the deck moves to the right side 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 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) 50 Naval Architectural Calculation, Spring 2016, Myung-Il Roh
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Summary of Static Stability of a Ship (3/3)
When the object on the deck returns to the e initial position in the centerline, the center of mass of the ship returns to the initial point G. Then, because the buoyant force and the gravitational force are not on one line, the forces induces a restoring moment to FG return the ship to the initial position. G Z G1 ※ Naval architects refer to the restoring moment as “righting moment”. The moment arm of the buoyant force and gravitational force about G is B B1 expressed by GZ, where Z is defined as the intersection point of the line of buoyant force(F ) through the new position of the F B B center of buoyancy(B1) with the transversely parallel line to the waterline through the center of mass of the ship (G). r G: Center of mass of a ship • Transverse Righting Moment G1: New position of center of mass after the object on the deck moves to the right side F : G Gravitational force of a ship righting FGZ B 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 By the restoring moment, the ship Z F : The intersection of a line of buoyant force( B) through the new position returns to the initial position. of the center of buoyancy (B1) with the transversely parallel line to the waterline through the center of mass of a ship(G) 51 Naval Architectural Calculation, Spring 2016, Myung-Il Roh
Evaluation of Stability : Merchant Ship Stability Criteria – IMO Regulations for Intact Stability (IMO Res.A-749(18) ch.3.1) IMO recommendation on intact stability for passenger and cargo ships.
Righting arm = const. Area A: Area under the righting arm curve (GZ) between the heel angle of 0 and 30 (: displacement) 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 GM m: Heel angle of maximum righting arm 57.3 ※ After receiving approval of A B calculation of IMO regulation Angle of heel from Owner and Classification ( []) 0304010 20 50 60 70 80 Society, ship construction can m f proceed. IMO Regulations for Intact Stability (a) Area A ≥ 0.055 m-rad (b) Area A + B ≥ 0.09 m-rad The work and energy (c) Area B ≥ 0.030 m-rad considerations (dynamic stability) (d) GZ ≥ 0.20 m at an angle of heel equal to or greater than 30
(e) GZmax should occur at an angle of heel preferably exceeding 30 but not less than 25. Static considerations (f) The initial metacentric height GMo should not be less than 0.15 m.
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5. Examples for Ship Stability
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[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. 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) Assumption) (1) Gravitational acceleration = 10 [m/s2], Density of sea water = 1.025 [ton/m3] (2) When the ship will be in the static equilibrium finally, the deck will not be immersed and the bottom will not emerge.
FkNG 205
0.3m
5m 3m 0.4m Baseline 10m : Location of the center of
gravity of the ship 54 Naval Architectural Calculation, Spring 2016,AP Myung-Il Roh FP
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F 205 kN Solution) G (1) Static Equilibrium (1/3) 0.3m
5m 3m
0.4m Baseline When the ship is floating in sea water, the requirement 10m A FP : Location of the center of mass of the ship for ship to be in static equilibrium state is derived from P Newton’s 2nd law and Euler equation as follows. (1-1) Newton’s 2nd Law: Force Equilibrium The resultant force should be zero to be in static equilibrium. nn n FF Gz,, F Bz 0 , where n FG.z : zn-coordinate of the gravitational force n FB.z : zn-coordinate of the buoyant force (1-2) Euler Equation: Moment Equilibrium
The resultant moment should be zero to be in static equilibrium. nn n τM GB M0 , where n MG : the moment due to the gravitational force n MB : the moment due to the buoyant force.
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F 205 kN Solution) G (1) Static Equilibrium (2/3) 0.3m
5m 3m
0.4m Baseline 10m
AP FP The first step is to satisfy the Newton- : Location of the center of mass of the ship Euler equation which requires that the sum of total forces and moments acting z z' on the ship is zero. FG As described earlier, in order to satisfy a stable equilibrium, the buoyant force and yG G gravitational force should act on the same vertical line, therefore, the moment x,x' O,E arm of the buoyant force and ф˚ y gravitational force must be same. B y' yB K yyGB FB
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F 205 kN Solution) G (1) Static Equilibrium (3/3) 0.3m
5m 3m
0.4m Baseline yyGB 10m AP FP : Location of the center of mass of the ship z z'
yyGGcos sin yyB cos sin B FG zzGGsin cos zzB sin cos B yG G By representing yG and y B with yzy GG ,, B , and z B ,
we can get x,x' O,E ф˚ y y' yzyzGGcos sin BB cos sin B1 yB K
FB In this equation, we suppose that y'G and z'G are already given, and y'B and z'B can be geometrically calculated.
Body fixed coordinate system(b-frame): Body fixed frame x’ y’ z’ Space fixed coordinate system(n-frame): Inertial frame x y z
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yy Solution) GB
(2-1) Changed Center of Buoyancy, B1, with Respect to the Body Fixed Frame
z z'
The centroid of A with respect to the FG body fixed frame:
yG G
M Az, M Ay, yzCA__,, CA x,x' O,E ф˚ AAAA y y' , where y B1 A B AA : the area of A K st MA,z’ : 1 moment of area of A about z’ axis st FB MA,y’ : 1 moment of area of A about y’ axis.
To obtain the centroid of A, the followings are required. -The area of A -1st moment of area of A about z’ axis -1st moment of area of A about y’ axis
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yy Solution) GB (2-2) Center of Buoyancy and Center of Gravity with Respect to the Body Fixed Frame (1/2)
1) Center of buoyancy, B , with respect to the body The centroid of A with respect 1 to the body fixed frame: fixed frame M Az, M Ay, yzCA__,, CA AA To calculate the centroid of A using the geometrical AA
relations, we use the areas, A1, A2, and A3.
2a
R 2b S 0 ф˚ A2 A3 R0 t A A1 A2 A3 S A A1 Q P
To describe the values of A1, A2, and A3 using the geometrical parameters (a, t, and ), y’ and z’ coordinate of the points P, Q, R, R0, S, S0 with respect to the body fixed frame is used, which are given as follows.
PyPP,, z a t,,, Qy QQ z a t Ry(,)(,tan), z aa R( y , z )(,0) a RR0 R00 R Sy(,)(, z a a tan), S( y , z )(,0) a SS0 S00 S
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yy Solution) GB (2-2) Center of Buoyancy and Center of Gravity with Respect to the Body Fixed Frame (2/2)
2a
R 2b S 0 ф˚ A2 A3 R0 t A A1 A2 A3 S A A1 Q P 1 Area: aatan 2 z 21 a tan Centroid: yzCC,,tan a a Cy CC, z 33 A 1/3a tan 2 Moment of area about z’ axis: a y 121 2/3a Area y a atan a a3 tan C 233 Moment of area about y’ axis:
11132 Area zC a atan a tan a tan 23660 Naval Architectural Calculation, Spring 2016, Myung-Il Roh
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yy Solution) GB (2-3) Center of Buoyancy and Center of Gravity with Respect to the Body Fixed Frame (1/2)
1) Center of buoyancy, B1, with respect to the body The centroid of A with respect fixed frame to the body fixed frame: M M yz,, Az, Ay, CA__ CA AAAA A A1 A2 A3 2a The table blow summarizes the results of the area, centroid with st R respect to the body fixed frame and 1 moment of area with 2b respect to the body fixed frame of A , A , A , and A. S 1 2 3 0 ф˚ A2 A R Area Centroid Moment of area Moment of area 3 0 t (,)yz S A AA CC about z'-axis about y'-axis A1 ()yA ()z A Q C C P A 2at t 0 at 2 1 0, 2 A 1 2tanaa a3 tan a3 tan 2 2 aatan , 2 33 3 6 A 1 2tanaa a3 tan a3 tan 2 3 aatan , 2 33 3 6 3 2 A 2at - 2tana a3 tan at2 3 (=A1+A2-A3) 3
The center of buoyancy, B1, with respect to the body fixed frame is 2 2 2 M Az, M Ay,' at tan a tan yzBB,, , AA326 t t AA 61 Naval Architectural Calculation, Spring 2016, Myung-Il Roh
yy Solution) GB (2-3) Center of Buoyancy and Center of Gravity with Respect to the Body Fixed Frame (2/2)
2 at2 tan a2 tan yz,, 2) Center of gravity, G, with respect to the body BB 326tt fixed frame FG z’ The center of gravity, G, with respect to z 2a the body fixed frame is given by G geometrical relations as shown in the figure, which is 2b d
y yz', ' dbt ,2 GG xn, xb O,E ф˚ y’ t
B B1
K FB
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Solution) FG
(3) Comparison between the Figure Describing the Ship Inclined zb zn and the Figure Describing the Water Plane Inclined (1/2) 2a G
Let us calculate the center of buoyancy, B1, and 2b d the center of gravity, G, using the Fig. (b). yn
The center of buoyancy, B1, and the center of gravity, G, x ,x O,E ф˚ n b y with respect to the body fixed frame t b B B1 2 at2 tan a2 tan yz,, K BB 326tt F B
yz', ' dbt ,2 GG (b)
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.
yzyzGGcos sin BB cos sin
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Solution) (3) Comparison between the Figure Describing the Ship Inclined and the Figure Describing the Water Plane Inclined (2/2)
yzyzGGcos sin BB cos sin
2 at2 tan a2 tan yz,, BB 326tt
yz',GG ' dbt ,2
32taa222 tansin2 dbtcos (2 ) sin 6t Substituting a=2.5m, b=1.5m, t=0.4m, d=0.3m into this equation and rearranging
15.025 15.625 2 2.6 sin 0.3 cos sin tan 36 tan 0.159 [rad] 9.019 [deg]
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[Example] Equilibrium Position and Orientation of a Box-shaped Ship Question 2) The center of mass is moved to 2 [m] in the direction of the forward perpendicular.
A box-shaped ship of 10 meter length, 5 meter breadth and 3 meter height weighs 205 [kN]. The center of mass is moved to 2 [m] in the direction of the forward perpendicular. When the ship is in static equilibrium state, determine the equilibrium position and orientation of the ship. Assumption) (1) Gravitational acceleration = 10 [m/s2], Density of sea water = 1.025 [ton/m3] (2) When the ship will be in the static equilibrium finally, the deck will not be immersed and the bottom will emerge.
FkNG 205
Starboard
2m Port 5m 3m 0.4m Baseline 10m
: Location of the center of AP FP 65 Naval Architectural Calculation, Spring 2016, Myung-Il Roh mass of the ship
Solution)
FG 205 kN
Starboard
2m Port z 5m n Force Equilibrium 3m 0.4m Baseline 10m FF GB F 0 : Location of the center of AP FP mass of the ship
FG 250 FG 205 kN
FgVB
zb 1 1.025 10 ab 5 2 25.625 ab xb
FFGB F 250 25.625 ab b a 0 yynb, xn
FB O ab8
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Solution)
FG z z z z x
x x OE, yy, b Side view x a yy, (Profile view) FB OE, Instead of rotating the ship, we can consider the waterline rotated with an angle of while keeping the ship constant. z
FG z
b F x B OE, yy, a
x 67 Naval Architectural Calculation, Spring 2016, Myung-Il Roh
Solution)
FG 205 kN Starboard Moment Equilibrium 2m Port 5m zn 3m 0.4m MM M 0 Baseline GB 10m
: Location of the center of AP FP mass of the ship The centers of buoyancy B and gravity G should be in the same vertical line. FG
xG xB zb xG 3cos 3sin n xG x b 3cos 3sin
n x B b a yynb, xn
FB O
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z Solution) n
FG
zb n x G Moment Equilibrium x b 3cos 3sin
n x B b MM M 0 a yy, GB nb xn
FB O The centers of buoyancy B and gravity G z should be in the same vertical line. n zb n n xG xB
n xG 3cos 3sin ab n x cos sin B 33
n x ab B 3cos 3sin cos sin a /3 33 b /3 b a cos a b 3 sin 3
FB O
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Solution)
ab 3cos 3sin cos sin 33 dividing the both side of equation by cos ab 33tan tan 33 b babb tan 33 a aa33 multiplying3 a to the both side of equation 99abab22 9ab abab
From the force equilibrium if a b ab22 Unstable ab8 if a b a 8 Stable From the moment equilibrium 9ab abab b 1
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Solution)
Why is the ship unstable, when ab 22 ? Horizontal displacement of center of mass
zn xG 32sin 32 4.242 FG 32 xb
z 32 b
FB a 22 ynb, y b 22 x n O xB wedge 1 x wedge x x B b b total total 1 22 tan( ) 2tan( ) wedge 2 2tan( ) 4 x 2222 2 4 B x 0.66 4, x 2( ) 43 B total b 71 Naval Architectural Calculation, Spring 2016, Myung-Il Roh 233
Solution)
Why is the ship unstable, when ab 22 ? Horizontal displacement of center of mass
zn xG 32sin 32 32 4.242 F xb G
zb
a 22 ynb, y b 22 F x B n O xB wedge 1 x wedge x x B b b total total 1 Unstable 22 tan( ) 2tan( ) wedge 2 2tan( ) 4 x 2222 2 4 B x 0.66 4, x 2( ) 43 B total b 72 Naval Architectural Calculation, Spring 2016, Myung-Il Roh 233
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• Given: KB, KG, IT, Heeling moment Mh • Find: Angle of heel φ [Example] Heel Angle Caused by Movement • GZ of wall sided ship
1 2 GZ GM BM tan sin of Passengers in Ferry (1/2) 2 Question) Emergency circumstance happens in Ferry with displacement (mass) 102.5 ton. Heeling moment of 8 ton·m occurs due to passengers moving to the right of the ship. What will be an angle of heel? 4 Assume that wall sided ship with KB=0.6m, KG=2.4m, IT=200m . Solution) If it is in static equilibrium at an angle of heel
Righting moment in wall sided ship (Mr)= Heeling moment (Mh)
1 2 GM BM tan sin = 8ton m 2 ①CalculationofBM 102.5 ton /1.025 100 m3 I 200 BMmT 2 100 ②CalculationofGM GM KB BM KG 0.6 2 2.4 0.2 m
2 8 0.2 tan sin Nonlinear equation 102.5 about ? 예제5.2
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• Given: KB, KG, IT, Heeling moment Mh • Find: Angle of heel φ [Example] Heel Angle Caused by Movement • GZ of wall sided ship
1 2 GZ GM BM tan sin of Passengers in Ferry (2/2) 2 Question) Emergency circumstance happens in Ferry with displacement (mass) 102.5 ton. Heeling moment of 8 ton·m occurs due to passengers moving to the right of the ship. What will be an angle of heel? 4 Assume that wall sided ship with KB=0.6m, KG=2.4m, IT=200m . Solution) If it is in static equilibrium at an angle of heel
Righting moment in wall sided ship (Mr)= Heeling moment (Mh)
1 2 GM BM tan sin = 8ton m 2
2 0.2 tan sin 0.078 Righting Because of nonlinear equation, solve arm it by numerical method. 0.0858 Result of calculation is about =16.0˚. 0.0778 φ LHS RHS (Righting arm) (Heeling arm) Heeling 0.0703 arm 15˚ 0.0703 0.0780
16˚ 0.0778 0.0780
17˚ 0.0858 0.0780 15 17 In static equilibrium
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[Example] Heel Angle Caused by Movement of Cargo
Question) A cargo carrier of 10,000 ton displacement is floating. KB=4.0m, BM=2.5m, KG=5.0m. Cargo in hold of cargo carrier is shifted in vertical direction through a 10 meter, and shifted in transverse direction through a 20 meters. Find an angle of heel. • Given : displacement (), KB, BM, KG, weight of cargo(w) and moving distance • Find : angle of heel φ z
y
d=20.0 m
G 200 ton
h=10.0 m 5.0m 4.0m B Base Line CL 예제5.3
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[Example] Change of Center Caused by Movement of Cargo
Question) As below cases partial weight w of the ship is shifted. What is the shift distance of center of mass of the ship?
Case 1) Vertical shift of the partial weight Case 2) Horizontal shift of the partial weight
b
h
G1 G G1 G Base Base Line Line CL CL
예제5.5
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[Example] Calculation of Deadweight of Barge
Question) A barge is 40m length, 10m breadth, 5m depth, and is floating at 1 m draft. The vertical center of mass of the ship 20m is located in 2 m from the baseline. 40m C A cargo is supposed to be loaded in center of the deck. Find the maximum 5m loadable weight that keeps the 3 5m stability of ship. =1.0ton/m Base Line Hint) GM should be positive and deck CL should not submerged.
Problem to calculate the maximum load for the stability.
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[Example] Calculation of Position of Ship when Cargo is Moved by Crane
Question) 16.0 m A cargo carrier of 18,000 ton displacement is afloat and has GM = 1.5m. And we want to transfer the cargo of 200 ton weight from bottom of the ship to land. 27.0 m 200 A lifting height of cargo is 27.0 m from ton the original position. After lifting the cargo, turn the cargo to the right through a distance of 16.0 m from the centerline. Base What will be the angle of heel of the ship? Line CL Hint) Use the Moment to Heel One Degree and the heeling moment caused by the movement of the cargo. Moment to heel one degreeFGMB sin1 Problem to calculate the equilibrium angle of the ship when external force are applied.
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[Example] Calculation of Center of Buoyancy of Ship with Constant Section Example) A ship is inclined about x-axis through origin O with an angle of -30. Calculate center of buoyancy with respect to the water plane fixed frame. • Given: Breadth(B) 20m, Depth(D) 20m, Draft(T) 10m, Angle of Heel() -30˚
• Find: Center of buoyancy(yB,zB) G: Center of mass K:Keel
B: Center of buoyancy B1 : Changed center of buoyancy
Section view S z z,z‘ 20 z‘ S R 20 20
20 R O O y,y‘ -30˚ y B B 10 P B1 y‘ K P K Q Q
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[Example] Calculation of Center of Buoyancy of Ship with Various Station Shapes A ship with three varied section shape is given. When this ship is inclined about x axis with an angle of -30, calculate y and z coordinates of the center of buoyancy (with respect to the water plane fixed frame). • Given: Length(L) 50m, Breadth(B) 20m, Depth(D) 20m, Draft(T) 10m, Angle of Heel() -30˚
• Find: Center of buoyancy(y∇,c,z∇, c) after heeling 20
z,z' O,O' y,y' 20 20 20 x,x' 1 10 1 0 0 20 20 20 20 10
20 20 10
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