Ship Stability

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Naval Architectural Calculation, Spring 2018, Myung-Il Roh 1

Ship Stability

Ch. 1 Introduction to Ship Stability

Spring 2018

Myung-Il Roh

Department of Naval Architecture and Ocean Engineering Seoul National University

Lecture Note of Naval Architectural Calculation

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Contents

þ Ch. 1 Introduction to Ship Stability þ Ch. 2 Review of Fluid Mechanics

þ Ch. 3 Transverse Stability Due to Cargo Movement þ Ch. 4 Initial Transverse Stability

þ Ch. 5 Initial Longitudinal Stability þ Ch. 6 Free Surface Effect

þ Ch. 7 Inclining Test

þ Ch. 8 Curves of Stability and Stability Criteria

þ Ch. 9 Numerical Integration Method in Naval Architecture þ Ch. 10 Hydrostatic Values and Curves

þ Ch. 11 Static Equilibrium State after Flooding Due to Damage þ Ch. 12 Deterministic Damage Stability

þ 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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þ The force that enables a ship to float

n It is directed upward.

n 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

Æ “Buoyant Force”

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How does a ship float? (2/3)

þ Archimedes’ Principle

n 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.

n The direction of the buoyant force is opposite to the gravitational force.

þ Equilibrium State (“Floating Condition”) n Buoyant force of the floating body

= Weight of the floating body

\Displacement = Weight

G: Center of gravity B: Center of buoyancy W: Weight, D: Displacement r: Density of fluid

V: Submerged volume of the floating body (Displacement volume, Ñ)

G

B W

D

D = -W = - r gV

Buoyant force of a floating body

= the weight of the fluid which is displaced by the floating body (“Displacement”) Æ Archimedes’ Principle

Buoyant force of a floating body

= the weight of the fluid which is displaced by the floating body (“Displacement”)

Æ Archimedes’ Principle

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How does a ship float? (3/3)

þ Displacement(D) = Buoyant Force = Weight(W)

þ Weight = Ship weight (Lightweight) + Cargo weight (Deadweight)

DWT LWT

W

C T B

L _B

+

=

=

×

×

×

×

=

D r

T: Draft CB: Block coefficient r: 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

1

W

1

L

1

B

℄ Inclining (Heeling)

Restoring

F B

G

F G

B1 B

Capsizing ℄

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What is a “Hull form”?

þ Hull form

n 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

n Like a skin of human þ Hull form design

n 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”?

þ Compartment

n Space to load cargos in the ship

n It is divided by a bulkhead which is a diaphragm or peritoneum of human.

þ Compartment design (General arrangement design) n Compartment modeling + Ship calculation

þ Compartment modeling

n Design task that divides the interior parts of a hull form into a number of compartments

þ Ship calculation (Naval architecture calculation)

n 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”?

þ Hull structure

n Frame of a ship comprising of a number of hull structural parts such as plates, stiffeners, brackets, and so on

n Like a skeleton of human þ Hull structural design

n 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)

þ LOA (Length Over All) [m]: Maximum Length of Ship

þ LBP (Length Between Perpendiculars (A.P. ~ F.P.)) [m]

n A.P.: After perpendicular (normally, center line of the rudder stock)

n F.P.: Inter-section line between designed draft and fore side of the stem, which is perpendicular to the baseline

þ Lf (Freeboard Length) [m]: Basis of freeboard assignment, damage stability calculation n 96% of Lwl at 0.85D or Lbp at 0.85D, whichever is greater

þ Rule Length (Scantling Length) [m]: Basis of structural design and equipment selection n 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.

D ra ft

D e p th A ir D ra ft

¡ 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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6 DOF Motions of a Ship

Heave

Surge Sway

x z

y Roll

Pitch

Yaw

Translational motions

Rotational motions

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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 n y n z n or x y z Point E: Origin of the inertial frame(n-frame) b-frame: Body fixed frame x b y b z b or x’ y’ z’

Point O: Origin of the body fixed frame(b-frame)

x b

y n

z n

x n

z b

y b

E

O

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 2ndlaw is only valid for the inertial frame.

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System of Coordinates for a Ship

Stem, Bow

Stern yb

zb xb

(a)

AP LBP FP

BL SLWL AP: aft perpendicular

FP: fore perpendicular LBP: length between perpendiculars.

BL: baseline SLWL: summer load waterline(b)

ynxn zn ybxb zb : midship Space fixed coordinate system (n-frame): Inertial frame xn yn zn or x y z Body fixed coordinate system (b-frame): Body fixed frame xb yb zb or 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 y x

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 26

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① Newton’s 2nd law Static Equilibrium (1/3)

Static Equilibrium

ma = å F G

F G

F 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 ^nd law ma = å F

F G

= -

Static Equilibrium (2/3)

Static Equilibrium

F B

B

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 = å F , ( ^Q a = 0) G

F G

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Static Equilibrium (3/3)

F B

B

I w ^& = å t

② Euler equation When the buoyant force (FB)lies on the same

lineof action as the gravitational force (FG), total summation of the moment becomes 0.

I: Mass moment of inertia w: Angular velocity

for the ship to be in static equilibrium

0 = å t , ( ^Q w ^& = 0)

Static Equilibrium

t: Moment Static Equilibrium

① Newton’s 2ndlaw

ma = å F F G

= - + F _B

for the ship to be in static equilibrium

0 = å F , ( ^Q a = 0) G

F G

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What is “Stability”?

Stability = Stable + Ability

F B

G B

F G

W L

℄ F B

G

F G

B

1

W

1

L

1

B

℄ Inclining (Heeling)

Restoring

F B

G

F G

B1 B

Capsizing ℄

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Stability of a Floating Object l You have a torque on this object relative to any point that you choose. It does not matter where you pick a point.

l The torque will only be zero when the buoyant force and the gravitational force are on one line. Then the torque becomes zero.

I w ^& = å t

② Euler equation When the buoyant force (FB)lies on the same

lineof action as the gravitational force(FG), total summation of the moment becomes 0.

for the ship to be in static equilibrium

0 = å t , ( ^Q w ^& = 0)

Static Equilibrium

① Newton’s 2ndlaw

ma = å F F G

= - + F _B

for the ship to be in static equilibrium

0 = å F , ( ^Q a = 0)

G F B

F

\ =

Rotate

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Stability of a Ship

l You have a torque on this object relative to any point that you choose. It does not matter where you pick a point.

l The torque will only be zero when the buoyant force and the gravitational force are on one line. Then the torque becomes zero.

I w ^& = å t

② Euler equation When the buoyant force (FB)lies on the same

lineof action as the gravitational force (FG), total summation of the moment becomes 0.

for the ship to be in static equilibrium

0 = å t , ( ^Q w ^& = 0)

Static Equilibrium

① Newton’s 2ndlaw

ma = å F F G

= - + F _B

for the ship to be in static equilibrium

0 = å F , ( ^Q a = 0)

G F B

F

\ =

Static Equilibrium

(a) (b)

F B

G B

F G

Rotate

F B

G

F G

B

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Interaction of Weight and Buoyancy of a Floating Body (1/2)

(a) (b)

F B

G B

F G

F B

G

F G

B

1

W L W

1

L

1

B

℄ ℄

I w ^& = å t

Euler equation:

w & ¹ 0

Æ Interaction of weight and buoyancy resulting in intermediate state

Restoring Moment

t r

Torque (Heeling Moment)

t e

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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

1

W L W

1

L

1

B

℄ ℄

Interaction of weight and buoyancy resulting in static equilibrium state

Heeling Moment

t e

I w ^& = å t

Euler equation:

w & = 0

Æ Static Equilibrium

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Stability of a Floating Body (1/2)

Floating body in stablestate 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 unstablestate

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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

= ´

é ù

ê ú

= ê ú = × - × + - × + × + × - ×

ê ú

ë û

M r F

i j k

i j k

M x M _y M _z

Transverse momentLongitudinal 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 F z

F y

y F

z

j k i x

F x

y ( , , )

P x P y z P P

r

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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.z and F _B.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.

, , 0

G z B z

F = F + F =

å

Also the moment equilibrium must be satisfied, this means, the resultant moment should be also zero.

G B

= + =

å ^{τ M M 0}

where MGis the moment due to the gravitational force and MBis the moment due to the buoyant force.

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Equations for Static Equilibrium (2/3)

G B

= + =

å ^{τ M M 0}

where MGis the moment due to the gravitational force and MBis the moment due to the buoyant force.

From the calculation of a moment we know that MGand MBcan 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

= ´

é ù

ê ú

= ê ú

ê ú

ë û

= × - × + - × + × + × - ×

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

= ´

é ù

ê ú

= ê ú

ê ú

ë û

= × - × + - × + × + × - ×

M r F

i j k

i j k

, , , , , ,

( ) ( ) and ( ) ( )

G = y G × F G z - z G × F G y + - x G × F G z B = y B × F B z - z B × F B y + - x B × F B z

M i j M i j

, , , ,

( ) ( ) and ( ) ( )

G = y G × F G z + - x G × F G z B = y B × F B z + - x B × F B z

M i j M i j

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Equations for Static Equilibrium (3/3)

G B

= + =

å ^{τ M M 0}

where MGis the moment due to the gravitational force and MBis the moment due to the buoyant force.

, , , ,

( ) ( )

G B y G F G z y B F B z x G F G z x B F B z

= + = × + × + - × - × =

å ^{τ M M i j 0}

, , 0

G G z B B z

y × F + y × F =

Substituting 0

G B

G B

y y y y

- =

\ =

, ,

G z B z

F = - F

(force equilibrium)

, , , ,

( ) ( ) and ( ) ( )

G = y G × F G z + - x G × F G z B = y B × F B z + - x B × F B z

M i j M i j

, ,

and - x _G × F _{G z} - x _B × F _{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 1

F B

B G F G

Z

B

G Z

t r

Heeling Moment

t e

Restoring Moment

W1 L1

F 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 B1: Changed center of buoyancy FG: Weight of ship FB: Buoyant force acting on ship

• Transverse Restoring Moment

B 1

F B

B G F G

Z

B

G Z

t r

Heeling Moment

t e

Restoring 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) ^{t restoring = F GZ B ×}

• Restoring Moment

t r

G Z

B 1

B

Z: The intersection point of the line of buoyant force through B1 with the transverse line through G

F G

F B

t e

Definition 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 (B1) after inclination

• GM ÆMetacentric height

M

• The term meta was selected as a prefix for center because its Greek meaning implies movement. The metacentertherefore is a moving center.

• From the figure, GZcan be obtained with assumption that M does not change within a small angle of

inclination(about 7° to 10°), as below.

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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 (B1).

• 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 ^{t e}

G

B B ₁

t r

Z

F B

M

//

//

sin GZ » GM × f

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

,35

F B ,30

F B

C35 f=35°

Z

35 sin 35

GZ ¹ GM × f

L30 C30

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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• 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 )

F B F G

Z G

B M

F B F G

G M

B

F _B F G

M G B

B K

F _B

B ₁

F _G

G, Z, M

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

Z: The intersection of the line of buoyant force through B 1 with the transverse line through G M: The intersection of the line of buoyant force through B 1 with the centerline of the ship

B K

F _B

B ₁

F _G

Z G

M

B K

F _B

B ₁

F _G

Z G M

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F G

B 0

F t e

G

B

F G

B 0

F t e

G

1 B

B B ² B ₁

F G

F G

F G

F G

B 1

F

B 2

F F B 1

B 2

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 theposition of the center of mass“G” relative toinertial frameis changed.

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4. Ship Stability

Naval Architectural Calculation, Spring 2018, Myung-Il Roh 50

G F G

Summary of Static Stability of a Ship (1/3)

l When an object on the deck moves to the right side of a ship, the total center of massof the ship moves to the point G1, off

the centerline.

B t e

l Because the buoyant force and the gravitational force are not on one line, the forces induce 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 G 1

F G 1 G 1

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 (2/3)

l 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.

t e

G F G

B

F B

G 1

F B

B 1

F B G 1

F

t r

B 1

F B

B G

t r

B G

G 1

t e

G 1

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

52 Naval Architectural Calculation, Spring 2018, Myung-Il Roh

Summary of Static Stability of a Ship (3/3)

l When the object on the deck returns to the initial positionin the centerline, the center of mass of the ship returns to the initial point G.

B 1

l Then, because the buoyant force and the gravitational force are not on one line, the forces induces a restoring momentto return the ship to the initial position.

l By the restoring moment, the ship returns to the initial position.

F B

G 1

B G

t r

t e

F G

Z

B

G Z

l 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(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 the ship (G).

righting F GZ B

t = ×

• 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

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Naval Architectural Calculation, Spring 2018, Myung-Il Roh 53

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 max should occur at an angle of heel preferably exceeding 30° but not less than 25°.

(f) The initial metacentric height GM o should 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

þ 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 )

※ f f : Heel angle at which openings in the hull f m : Heel angle of maximum righting arm

B 10

0 20 30 40 50 60 70 80

Angle of heel (f [°]) Righting arm

(GZ)

A

f _m

D = const.

(D: displacement)

f f

GM

57.3°

54 Naval Architectural Calculation, Spring 2018, Myung-Il Roh

5. Examples for Ship Stability

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Naval Architectural Calculation, Spring 2018, Myung-Il Roh 55

[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 ( f ) 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

G 205 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)

56 Naval Architectural Calculation, Spring 2018, Myung-Il Roh

Solution)

(1) Static Equilibrium (1/3)

: Location of the center of mass of the ship 0.4m 3m

Baseline

APFP10m

5m

G

205 F = - kN

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 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.

, , 0

n n n

G z B z

F = F + F =

å

, where nFG.z: zn-coordinate of the gravitational force nFB.z: zn-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

= + =

å ^{τ M M 0}

, where nMG: the moment due to the gravitational force nMB: the moment due to the buoyant force.

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Solution)

(1) Static Equilibrium (2/3)

: Location of the center of mass of the ship 0.4m

3m

Baseline

APFP 10m G

205

F = - kN

0.3m

G B

y = y

z

y

O,E

K B z'

y' x,x'

F G

F B

ф˚

G

y G

y B

The 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.

58 Naval Architectural Calculation, Spring 2018, Myung-Il Roh

Solution)

(1) Static Equilibrium (3/3)

: Location of the center of mass of the ship 0.4m 3m

Baseline

APFP10m

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 G

y B

cos sin sin cos

G G

G G

y y

z z

f f

f f

é ù é ù é ¢ ù

ê ú = ê ë - ú û ê ¢ ú

ë û ë û

cos sin sin cos

B B

B B

y y

z z

f f

f f

é ù é ù é ¢ ù

ê ú = ê ë - ú û ê ¢ ú

ë û ë û

By representing and with , and , we can get

y G y _B y _G ¢ , z _G ¢ , y _B ¢ z¢ B

cos sin cos sin

G G B B

y ¢ × f + z ¢ × f = y ¢ × f + z ¢ × f

In this equation, we suppose that y'G and z'Gare already given, and y'Band 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

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Naval Architectural Calculation, Spring 2018, Myung-Il Roh 59

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 M y z

A A

¢

æ ¢ ö

¢ ¢ = ç ÷

è ø

, where AA: the area of A MA,z’: 1stmoment of area of A about z’ axis MA,y’: 1stmoment of area of A about y’ axis.

To obtain the centroid of A, the followings are required.

- The area of A - 1stmoment of area of A about z’ axis - 1stmoment of area of A about y’ axis z

y O,E

K B z'

y' x,x'

F G

F B

ф˚

G

y G

y B

A ¹

60 Naval Architectural Calculation, Spring 2018, Myung-Il Roh

Solution)

(2-2) 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 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

¢

æ ¢ ö

¢ ¢ = ç ÷

è ø

To calculate the centroid of A using the geometrical relations, we use the areas, A1, A2, and A3.

To describe the values of A1, A2, and A3 using the geometrical parameters (a, t, and f), 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.

( ) ( ) ( ) ( )

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

f f

¢ ¢ = - - ¢ ¢ = -

¢ ¢ = × ¢ ¢ =

¢ ¢ = - - × ¢ ¢ = -

R0 ф˚

2a

2b

P Q

R

tS

S0

A1 A2 A3

A

= ⁺ -

A A3

A2 A1

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Naval Architectural Calculation, Spring 2018, Myung-Il Roh 61

Solution)

(2-2) Center of Buoyancy and Center of Gravity with Respect to the Body Fixed Frame (2/2)

R 0

ф˚

2a

2b

P Q

R

t S

S 0

A 1

A ₂ A 3

A

= ⁺ -

A A3

A2 A1

Area:

1 tan

2 a a × × f

Centroid:

( ^, ) ^{2 , 1 tan}

3 3

C C

y ¢ z ¢ = a a f

Moment of area about z’ axis: 3

1 2 1

tan tan

2 3 3

Area y ´ C ^¢ = a a × × f ´ a = a f

Moment of area about y’ axis:

3 2

1 1 1

tan tan tan

2 3 6

Area z ´ C ¢ = a a × × f ´ a × f = a f

A

2

y¢

z¢

( _C , _C )

C y ¢ z ¢ a × tan f 1 / 3 × × a tan f a

2 / 3 a ×

G B

y = y

62 Naval Architectural Calculation, Spring 2018, Myung-Il Roh

Solution)

(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

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

¢

æ ¢ ö

¢ ¢ = ç ÷

è ø

The table blow summarizes the results of the area, centroid with respect to the body fixed frame and 1st moment of area with respect to the body fixed frame of A1, A2, A3, and A.

The center of buoyancy, B1, with respect to the body fixed frame is

( ) , , ' ² tan 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

f f

¢ æ × ö

æ ö ×

¢ ¢ = ç è ÷ ø = ç ç è - + ÷ ÷ ø

Area

( A A )

Centroid

( y C ¢ , z C ¢ )

Moment of area about z'-axis

( y _C ¢ × A )

Moment of area about y'-axis

( z _C ¢ × A )

A1

2a t × _0,

2 t

æ ö

ç - ÷

è ø

0 - × a t ²

A2

1 tan

2 × × × a a f ^{2 , tan}

3 3

a a × f

æ ö

ç ÷

è ø

3 tan 3

a × f ³ ( tan ) ²

6 a × f

A3

1 tan

2 × × × a a f ² , ^tan

3 3

a a × f

æ ö

- -

ç ÷

è ø

3 tan 3 a × f

- ³ ( tan ) ²

6 a × f

-

A (=A1+A2-A3)

2a t ×

^-

^{2 3 tan}

3

a × f 2 ³ ( tan ) ²

3 a t a × f - × +

R0 ф˚

2a 2b

P Q

R

tS S0 A1 A2

A3A

= ⁺ -

A A3

A2 A1

G B

y = y

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Naval Architectural Calculation, Spring 2018, Myung-Il Roh 63

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 ^{, 2} b t - )

( ) ² tan ² ( tan ) ²

, ,

3 2 6

B B

a t a

y z

t t

f f

æ × × ö

¢ ¢ = ç - + ÷

ç ÷

è ø

G B

y = y

y’

B

K O,E

2a

y

zz’

ф˚

F G

F _B

G

x n, x

b B1 2b

t d

64 Naval Architectural Calculation, Spring 2018, Myung-Il Roh

yb B K O,E

2a

yn znzb

ф˚

F G

F _B

G

xn,xb B1 2b

t d

(b) Solution)

(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, B1, and the center of gravity, G, using the Fig. (b).

l The center of buoyancy, B1, and the center of gravity, G, with respect to the body fixed frame

( ) ² tan ² ( ^tan ) ²

, ,

3 2 6

B B

a t a

y z

t t

f f

æ × × ö

¢ ¢ = ç - + ÷

ç ÷

è ø

( y ^{' , '} G z G ) ( = d ^{, 2} b t - )

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 ¢ × f + z ¢ × f = y ¢ × f + z ¢ × f

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Naval Architectural Calculation, Spring 2018, Myung-Il Roh 65

Solution)

(3) Comparison between the Figure Describing the Ship Inclined and the Figure Describing the Water Plane Inclined (2/2)

( )

{ ^{3 2 2 2 2 tan 2} } ^sin

cos (2 ) sin

6

t a a

d b t

t

f f

f f ^{- + + × ×}

× + - × =

( ) ²

15.025 15.625

2.6 sin 0.3 cos sin tan

3 6

f f f ^æ f ^ö

× + × = ç + ÷

è ø

Substituting a=2.5m, b=1.5m, t=0.4m, d=0.3m into this equation and rearranging

tan f = 0.159 [rad]

( ) ² tan ² ( tan ) ²

, ,

3 2 6

B B

a t a

y z

t t

f f

æ × × ö

¢ ¢ = ç - + ÷

ç ÷

è ø

( y ^{' , '} G z G ) ( = d ^{, 2} b t - )

cos sin cos sin

G G B B

y ¢ × f + z ¢ × f = y ¢ × f + z ¢ × f

66 Naval Architectural Calculation, Spring 2018, Myung-Il Roh

[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/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 emerge.

0.4m 3m

Baseline

AP FP

10m 5m

: Location of the center of mass of the ship Starboard

Port

G 205 F = - kN

2m

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Naval Architectural Calculation, Spring 2018, Myung-Il Roh 67 0.4m

3m

Baseline

APFP10m

5m

: Location of the center of mass of the ship

G

205 F = - kN

2m

Force Equilibrium

x n ^{y y n , b}

z n

a

z b

x b

G 205 F = - kN

a b

F B

G B 0 F = F + F =

å

G 250 F = -

1.025 10 1 5

2 25.625 F B g V

a b a b

= - × × r

æ ö

= × × ç × × × ÷

è ø

= × ×

250 25.625 0

G B

F F F

a b

= +

= - + × ×

=

å

8 a b

\ × =

O

Solution)

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Naval Architectural Calculation, Spring 2018, Myung-Il Roh

x

, y y¢

z

q

z¢

x¢

F G

b

a

F B O E ,

x ^{y y¢ ,}

z

q

z¢

x¢

F

G

a b

F

B

O E ,

x O E ,

, y y¢

z z¢

x¢

Side view (Profile view) Instead of rotating the ship, we can consider the

waterline rotated with an angle of  while keeping the ship constant.

Solution)