Ship Stability

(1)

Naval Architectural Calculation, Spring 2016, Myung-Il Roh 1

Ship Stability

Ch. 5 Initial Longitudinal Stability

Spring 2016

Myung-Il Roh

Department of Naval Architecture and Ocean Engineering Seoul National University

Lecture Note of Naval Architectural Calculation

 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. 5 Initial Longitudinal Stability

1. Longitudinal Stability

2. Longitudinal Stability in Case of Small Angle of Trim 3. Longitudinal Righting Moment Arm

4. Derivation of Longitudinal Metacentric Radius (BM

_L

) 5. Moment to Trim One Degree and Moment to Trim One Centimeter (MTC)

6. Example of Longitudinal Stability

1. Longitudinal Stability

(3)

L1

W1

F

G

F

B G

B

Static Equilibrium

I  ^   

② Euler equation

When the buoyant force (F_B)and the gravitational force(F_G) are on one line, the total momentabout the transverse axis through any pointbecomes 0.

It does not matter where the axis of rotation is selected if the two forces are on one line.

for the ship to be in static equilibrium

0    ,( ^  ^  0)

① Newton’s 2^ndlaw

ma   F

F

G

   F

_B

for the ship to be in static equilibrium

0   F ,( ^ a  0)

G

F

B

 F 

G: Center of mass of a ship F_G: Gravitational force of a ship B: Center of buoyancy at initial position FB: Buoyant force acting on a ship I: Mass moment of inertia

: Angular velocity

 : Moment m: Mass of a ship a: Acceleration of a ship

W₁L₁: Waterline at initial position

.

A P F P .

Longitudinal Stability - Stable Equilibrium

② Then, release the external moment by moving the weight to its original position.

③ Test whether it returns to its initial equilibrium position.

ZL: Intersection point of the vertical line to the waterline W2L2through the changed position of the center of buoyancy (B1) with the horizontal line parallel to the waterline W2L2through the center of mass of the ship (G)

Return to its initial equilibrium position

Stable

B₁:Changed position of the center of buoyancy after the ship has been trimmed

P

A. F.P

FG

FB

B

e

W₁L₁: Waterline at initial position W₂L₂: Waterline after trim

FG

FB

B1

① Produce an external trim moment by moving a weight in the longitudinal direction (e ).

W2

L2

ZL

G

A.P : after perpendicular, F.P : forward perpendicular

r

Longitudinal righting

moment (τ_r) GZLLongitudinal righting arm

L1

W1 G1



(4)

Reference Frames

FG

FB

B

e

FG

FB

B1

W2

L2

ZL

G

r

L1

W1



e

B1

r



B

G L1

W12 ^Z^L L₂

W

FB

FG

Rotation of the water plane fixed frame while the body (ship) fixed frame is fixed

Rotation of the body (ship) fixed frame with respect to the water plane fixed frame

Same in view of the Mechanics!!

FB

In this figure, the angle of inclination, , is negative, which means that the ship is trimmed by the stern.

 ,

O O x

z

Longitudinal Stability of a Ship - Stable Condition (1/3)

B

1

z

x

 F

G

, z

, x



t Trimmoment

F

B

G

B

① Apply an external trim moment to the ship.

② Then release the external moment.

③ Test whether it returns to its initial equilibrium position.

stem

stern

(5)

Longitudinal Stability of a Ship - Stable Condition (2/3)

B B

1

x

z

x z

, O O



t Trimmoment

F

B

r

B1

 r

G

Resultant moment about y-axis through point O ( ): τ

^e

F

G

r

_B₁

F

_B

τ

e

, ,

( )

_G _{G z} _G _{G y}

G G z G G x

G G y G G x

y F z F x F z F x F y F

  

    

   



i j k

1 1

, ,

( )

B B z B B y

B B z B B x

B B y B B x

y F z F

x F z F

x F y F

   

    

   

i j k

, , ,

G G G

G x G y z

G

G x y z

F F F

 

i j k

r F

, ,

( )

G G z G G y

G G z G G x

G G y G G x

y F z F x F z F x F y F

  

 

   

   

i j k

, ,

( x F_G _{G z}z F_G _{G x})



j

1 , 1 ,

( x_B F_{B z} z_B F_{B x})

 j   

, , ,

0 0

G x

G G y

G z

F F

F W

   

   

     

      

 

F

, , ,

0 0

B x

B B y

B z

F F F

   

   

     

      

  F

 

₁

(

 x_G W x_B

)



j  

 

₁

(

x_G  x_B

)



j     

(

x_G x_B1

)





j



 W If

F

G

r

_G



G

In this figure, the angle of inclination, , is negative, which means that the ship is trimmed by the stern.

Longitudinal Stability of a Ship - Stable Condition (3/3)

B B

1

x z

, O O

F

B

r

B1

F

G

r

G

x

z



The moment arm induced by the

buoyant force and gravitational force is expressed by GZ_L, where ZLis the intersection point of the line of buoyant force() through the new position of the center of buoyancy(B₁) with a transversely parallel line to a waterline through the center of the ship’s mass(G).

r

GZ

L

  

•Longitudinal Righting Moment

Stable!!

 r

G

Resultant moment about y-axis through point O ( ): τ

^e

F

G

r

_B₁

F

_B

τ

e

GZ

L

    j

( x

_G

x

_B1

)



 j  

Z

L

x

G

B1

x

, z

, x



r Restoring moment

(6)

Position and Orientation of a Ship

with Respect to the Water Plane Fixed and Body(Ship) Fixed Frame

Rotation of the water plane fixed frame while the body (ship) fixed frame is fixed Rotation of the body (ship) fixed framewhile the water plane fixed frame is fixed

Same in view of the Mechanics!!

B B

1

x

z

x z

, O O

FB

FG



G

B

관계ID가 rId28인 이미 지 부분을 파일에서 찾 을 수 없습니다.

x

z

x z

, O O

FB

FG



G

Emerged volume

Submerged volume

The ship is rotated.

The water plane is rotated.

2. Longitudinal Stability

in Case of Small Angle of Trim

(7)

Assumptions for Small Angle of Trim

Assumptions

① Small angle of inclination

(3~5 for trim)

② The submerged volume and the emerged volume are to be the same.



P

A. F.P

FG

FB

Emerged volume Submerged volume

e

W1L1: Waterline at initial position W2L2 : Waterline after trim

FG

FB

L1

W1

W2

L2 L G

Z

B B1

r

Longitudinal Metacenter (M _L ) (1/3)

Longitudinal Metacenter (M

_L

)

The intersection pointof

a vertical line through the center of buoyancy at a previous position(B) with a vertical line through the center of buoyancy at the present position(B₁) after the ship has been trimmed.



P

A. F.P

FB

W1L1: Waterline at previous position W2L2 : Waterline after trim

FB

L1

W²1

W

L2

1

M

L

B B1

G

※ Metacenter “M” is valid for small angle of inclination.

(8)



P

A. F.P

FB

W2

L2

2

M

L

3 G W

L3

M_Lremains at the same positionfor small angle of trim, up to about 3~5 degrees.

FB

As the ship is inclined with a small trim angle, Bmoves on the arc of circle whose center is at ML. The BM_Lis longitudinal metacentric radius.

B2B₁B

The GMLis longitudinal metacentric height.

1

M

L

Longitudinal Metacenter (M _L ) (2/3)



P

A. F.P

3 G W

L3

FB

W4

L4

M_Ldoes not remain in the same positionfor large trim angles over 5 degrees.

FB

B3

B2B₁^B

Thus, the longitudinal metacenter, M

_L

, is only valid for a small trim angle.

3

M

L ^M^L¹^,^M^L²

Longitudinal Metacenter (M _L ) (3/3)

(9)

② The submerged volume and the emerged volume are to be the same.

Longitudinal Stability for a Box-Shaped Ship

① Apply an external trim moment (_e) which results in the ship to incline with a trim angle .

② For the submerged volume and the emerged volume are to be the same, the ship rotates about the transverse axis through the point O.

: Midship



P

A. F.P

FG

K FB

e

r

W1L1: Waterline at initial position W2L2: Waterline after a small angle of trim

FG

FB

L1

W1

W2

L2

O B B1

G

Assumption

①A small trim angle(3~5)

About which point a box-shaped ship rotates, while the submerged volume and the emerged volume are to be the same?

Longitudinal Stability for a General Ship

What will happen if the hull form of a ship is not symmetric about the transverse (midship section) plane through point O?

The submerged volume and the emerged volume are not same!

Assumption

①A small angle of inclination(3~5 for trim)

The hull form of a general ship is not symmetric about the transverse (midship section) plane.



P

A. F.P

FG

K FB

B

FG

FB

B1

L1

W1

W2

L2

G

So, the draftmustbe adjustedto maintain same displacement.

O

(10)

The intersection of the initial waterline (W1L1) with the adjusted waterline (W3L3) is a point F, on which the submerged volumeand the emerged volumeare supposed to be the same.

So, the draftmustbe adjustedto maintain same displacement.



P

A. K F. P

B F

G

F

B

1

L

1

W

1

^G

W

3

L

3

Submerged volume

Emerged volume

W

2

L

2

F O

Expanded view

What we want to find out is the point F. How can we find the point F?

W1L1: Waterline at initial position W2L2 : Waterline after a small angle of trim W3L3 : Adjusted waterline

x

W

1

x



Rotation Point (F)

= Emerged volume (vf) Submerged volume (va)

.

. ( tan ) ( tan )

F F P

A P y x dx F yx dx

 

. .

F F P

A Px y dx   F x y dx  

 

What does this equation mean?

( ) V x

( ) A x dx 





tan x y 



dx





 

L1

P F.

W1

W2

L2



P A.



K

F

tan



  

^y^ ^x^^tan

^ 

  

( ) A x

tan x y 



  

L

C

x

G

B1

B

Longitudinal Center of Floatation (LCF) (1/3)

: Longitudinal coordinate from the temporary origin F : Breadth of waterline W1L1at any position

F

O

From now on, the adjusted waterline is indicated as W₂L₂.

W1L1: Waterline at initial position W2L2 : Adjusted waterline

x

x z

y

z

(11)

Naval Architectural Calculation, Spring 2016, Myung-Il Roh 21 .

F A Px y dx  







That means the point Flieson the transverse axis through the centroid of the water plane, called longitudinal center of floatation (LCF).

a f

v v ^.

. ( tan ) ( tan )

F F P

A Py x



dx F y x



dx

 

. .

F F P

A Px y dx   F x y dx  

 

Longitudinal moment of the after water plane area from Fabout the transverse axis (y’) through the point F

dx y

x dA

F x

y

. F A Px dA



Longitudinal moment of the forward water plane area from Fabout the transverse axis (y’) through the point F _{F P}_.

F x dA







F^{F P}^.x y dx 

Since these moments are equal and opposite, the longitudinal moment of the entire water plane area about the transverse axis through pointFis zero.

Longitudinal Center of Floatation (LCF) (2/3)

Therefore, for the ship to incline under the condition that the submerged volume and emerged volume are to be the same,

the ship rotates about the transverse axis through the longitudinal center of floatation(LCF).

F

Longitudinal Center of Floatation(LCF) L1

P F.

W1

W2

L2

P A.

M

L



K

F

^G B1

B

Longitudinal Center of Floatation (LCF) (3/3)



L O Z

x

x z

y

(12)

3. Longitudinal Righting Moment Arm

B ML

Longitudinal Righting Moment Arm (GZ _L )

: Vertical center of buoyancy at initial position : Longitudinal metacentric height

: Longitudinal metacentric radius : Vertical center of mass of the ship K

: Keel

GM

L

 KB  BM

_L

 KG

KB ≒ 51~52% draft Vertical center of mass of the ship How can you get the

value of the BML?



B1

F

B

ZL

F

G

: Longitudinal Metacenter

F G



L B

 GZ F

Longitudinal Righting Moment

From geometrical configuration

L L

sin

GZ  GM  

with assumption that M_Lremains at the same position within a small angle of trim (about 3~5)

x

x z

(13)

4. Derivation of Longitudinal Metacentric Radius (BM _L )

Derivation of Longitudinal Metacentric Radius (BM

_L

) (1/8)

Because the triangle gg'g1is similar with BB'B1 1

1

BB BB gg gg

 



1 1

BB v gg



The distance BB1equals to v ₁

gg



 B  x

B

 z

B

 

B

, ,

( )

B v a v f

x

x x

  

   ( _, _,)

B v a v f

z

z z

  

 

x'_v,f, z'_v,f: longitudinal and vertical distance of the emerged volume from F x'_v,a, z'_v,a: longitudinal and vertical distance of

the submerged volume from F B₁: Changed position of the center of buoyancy

B: Center of buoyancy at initial position

: Displacement volume v: Submerged / Emerged volume

Let’s derive longitudinal metacentric radius BM_L when a ship is trimmed.

, g: Center of the emerged volume , g₁: Center of the submerged volume

The center of buoyancy at initial position (B) moves parallel to gg1. ^L

M



B

1

P F.



P A.

W2

L2

M

L

B1



K

g

L1

W1

F

r

g

^{v f}^,

z

,

z

v a

,

x

v f ,

x

v a

1 1

BB B B gg g g

 



, ,

( ) ...(1)

B v a v f

x v x x

     



, ,

( ) ...(2)

B v a v f

z v z z

     



: Longitudinal translation of B to B1

: Vertical translation of B to B1

B B

1 O

g x

x z

y

F z

y

x

Assumption

①A small trim angle(3~5)

(14)

Derivation of Longitudinal Metacentric Radius (BM

_L

) (2/8)

tan 2

L B

BM 



B

2 L tan BM BB

 

2 2

BB BBB B

 2

1 tan BB B B

  

 

 

1 tan

tan x^B z^B 

 ^ ^

 

, , , ,

1 ( ) ( ) tan

tan ^{v a} ^{v f} ^{v a} ^{v f}

v v

x x z z 

^ ^ ^ ^ ^ ^

      



, , , ,



1 ( ) tan

tan ^{v a} ^{v f} ^{v a} ^{v f}

L v x v x v

BM z v z 

    

       

  Find!

, ,

( ) ...(1)

B v a v f

x v x x

     



, ,

( ) ...(2)

B v a v f

z v z z

     



Substituting (1) and (2) into and

 x

B

  z

B

 

B M

L



B

1 2

B

B

P F.



P A.

W2

L2

ML



K

L1

W1 F

r

g ^{v f}^,

z

,

xv f ,

xv a

B¹2

B B B

O

g g1

,

zv a

F z

y

x



, , , ,



1 ( ) tan

tan ^{v f} ^{v a} ^{v f} ^{v a}

L v x v x v

BM z v z 

 ^ ^ ^ ^

       

  (A)

(A) : Momentof the emerged volume

about y’-z’ plane P F.



P A.

W2

L2

M

L



K

g

1 L1

W1

r

g

^{v f}^,

z

,

z

v a

,

x

v f ,

x

v a

x

F

(B) (C) (D)

B2¹

B B B

,

v x 

v f

. ( )

F P F A x x dx 







 

. 2

tan ^{F P}

F x y dx

   







. tan

F P

F x y   x dx 





  

y: Breadth of waterline W1L1 at any distance x from point F[m]

x

x z

y

F

tan

 ^

  

^y^^ ^x^^^tan

^ 

( ) A x

tan x y 



  

L

C

Derivation of Longitudinal

Metacentric Radius (BM

_L

) (3/8)

(15)

x

x



, , , ,



1 ( ) tan

tan ^{v f} ^{v a} ^{v f} ^{v a}

L v x v x v

BM z v z 

 ^ ^ ^ ^

       

  (A)

,

v x 

v a

. ( )

F A PA x x dx 







 

²

tan ^F.

A Px y dx

   







 

. tan

F

A Px y   x dx 





  

P F.



P A.

W2

L2

M

L



K

g

1 L1

W1

r

g

^{v f}^,

z

,

z

v a

,

x

v f ,

x

v a

x

F

(B) (C) (D)

(B) : Momentof the submerged volume

about y’-z’ plane

tan



  

y x tan

 

  

( ) A x

  

^y ^x^tan



  

L F

C

B2¹

B B B

z

y

Derivation of Longitudinal Metacentric Radius (BM

_L

) (4/8)

x

x z

F

tan

 ^

  

^y^ ^ ^x^^^tan

^ 

( ) A x

tan x y 



  

,

v z 

v f (C)

. ( )

F P F A x z dx







 

. tan tan

2

F P F

x y   ^x^ ^dx





   

2 .

 

2

tan 2

F P

F x y dx

 _ _{ }







: Momentof the emerged volume about x’-y’ plane

1 tan z2x



L

C

P F.



P A.

W2

L2

M

L



K

g

1 L1

W1

r

g

^{v f}^,

z

,

z

v a

,

x

v f ,

x

v a

F



, , , ,



1 ( ) tan

tan ^{v f} ^{v a} ^{v f} ^{v a}

L v x v x v

BM z v z 

 ^ ^ ^ ^

       

  (A) (B) (C) (D)

2

x

B¹ B

B B y

Derivation of Longitudinal

Metacentric Radius (BM

_L

) (5/8)

(16)

x

x z

Derivation of Longitudinal Metacentric Radius (BM

_L

) (6/8)

P F.



P A.

W2

L2

M

L



K

g

1 L1

W1

r

g

^{v f}^,

z

,

z

v a

,

x

v f ,

x

v a

F



, , , ,



1 ( ) tan

tan ^{v f} ^{v a} ^{v f} ^{v a}

L v x v x v

BM z v z 

 ^ ^ ^ ^

       

  (A) (B) (C) (D)

,

v z 

v a

. ( )

F A PA x z dx  







 

. tan tan

2

F A P

x y   ^x^ ^dx

    

 



2

 

2 .

tan 2

F

A P x y dx

 _ _{ }







(D) : Momentof the submerged volume

about x’-y’ plane

x

tan



  

^y^ ^x^ ^tan

^ 

  

( ) A x

  

^y ^x^tan



  

L F

C

1 tan z2x



B1

B B B2

y

Derivation of Longitudinal Metacentric Radius (BM

_L

) (7/8)

1 1 3

tan .tan

tan  I^L 2  I^L



 

      

1 2

1 tan

2

L

BMLI ^  ^

By substituting (A), (B), (C), and (D) into the BMLequation

,

v x v a ^^tan^



_AP^F ^x^²^^{y dx}^{ } (B)

,

v x v f^tan



_F^FP ^x²^{y dx}  (A)

(D) v z _{v a}_, ^tan²  ²

2

F

APx y dx

 _ _{ }







(C) v z v f, ^tan²  ²

2

FP

F x y dx

 _ _{ }







   



^. ² . ²



³



^.

^{ }

² .

^{ }

²



1 tan tan

tan 2

F P F F P F

F x y dx A Px y dx  F x y dx A P x y dx

 

            

  



 



 



 



 



, , , ,



1 ( ) tan

tan ^{v f} ^{v a} ^{v f} ^{v a}

L v x v x v

BM z v z 

 ^ ^ ^ ^

       

  (A) (B) (C) (D)

   



^. ² . ²



, _L ^{F P} ^F

F A P

I 



x y dx 



x y dx 

I_L: Moment of inertia of the entire water plane about transverse axis through its centroid F

   

²

 

²

 

. 2 2 . 2 2

. .

1 tan tan tan tan tan

tan 2 2

F P F F P F

F x y dx A P x y dx  F x y dx  A P x y dx

  



              

   



  



 



 



  

P F.



P A.

W2

L2

ML



K

g

g1 L1

W1 F

r

g ^{v f}^,

z

,

zv a xv a, x^{v f}^,

B2¹

B B B

O

x

x z

(17)

Derivation of Longitudinal Metacentric Radius (BM

_L

) (8/8)

L L

BM I



which is generally known as BM_L.

1

2

1 tan 2

L

BML

 I   ^    ^  

Ifθis small, tan²θ θ²=0

The BMLdoes not considerthe change of center of buoyancyin vertical direction.

In order to distinguish between them, the two will be indicated as follows:

2 0

(1 1tan ) 2

L L

BM I 





L L

BM I



(Consideringthe change of center of buoyancy in vertical direction)

(Without consideringthe change of center of buoyancy in vertical direction) P F.



P A.

W2

L2

ML



K

g

g1 L1

W1 F

r

g ^{v f}^,

z

,

zv a

,

xv f ,

xv a

B2¹

B B B

O z

x

x

5. Moment to Trim One Degree and Moment to Trim One Centimeter

(MTC)

(18)

Moment to Heel One Degree

B

1

M



e

F

B

Definition

It is the moment of external forcesrequired to produce one degree heel.

Assumption

Small inclination where the metacenter is not changed (about 7~10)



When the ship is heeled and in static equilibrium,

B

sin

F GM   



Moment to heel = Righting moment

Moment to heel one degree  F GM

_B

  sin1

^



r ^B

 F GZ 

(at small angle )

 Z

  1

^

Substituting G: Center of mass

B: Center of buoyancy at initial position F_G: Gravitational force of a ship FB: Buoyant force acting on a ship

B₁: New position of center of buoyancy after the ship has been inclined Z : The intersection of vertical line through the center of buoyancy with the transversally parallel line to a waterline through center of mass M: The intersection point of a vertical line through the center of

buoyancy at initial position(B) with a vertical line through the new position of the center of buoyancy(B₁) after the ship has been inclined transversally through a small angle

F

G

LCF: Longitudinal center of floatation

Z_L: Intersection point of the vertical line to the water surface through the changed center of buoyancy with the horizontal line parallel to the water surface through the center of mass

M_L: Intersection point of the vertical line to the water surface through the center of buoyancy at previous position(B) with the vertical line to the water surface through the changed position of the center of buoyancy(B1) after the ship has been trimmed.

Moment to Trim One Degree

FG

G B τe

F



B1

F

B

F

G

M

L

FB

Definition

It is the moment of external forces required to produce one degree trim.

Assumption

Small trim where the metacenter is not changed (about 3~5)

When the ship is trimmed and in static equilibrium,

B L

sin

F GM   



Moment to trim = Righting moment

Moment to trim one degree  F GM

_B



_L

 sin1

^

B L

 F GZ 

(at small angle )



  1

^

Substituting τr



ZL

(19)

F

G

G B τe

F B1

F

B

B L sin1 F GM ^

  

Moment to trim one degree

M

L

Often, we are more interested in the changes in draft produced by a trim moment than the changes in trim angle.

1

B L

100

BP

MTC F GM

   L



MTC: Moment to trim 1 cm

sin tan

BP

t







L Trim (t): d_a– d_f,

L

_BP

: Length between perpendiculars [m]

^d^a: Draft after df: Draft forward

P P F.

A. LBP

L2

(Ifis small) 1^

LBP

A.P F.P

t da

df



B.L F

LBP



Moment to Trim One Centimeter (MTC) (1/2)

1 t cm

Moment to Trim One Centimeter (MTC) (2/2)

1

B L 100 MTC F GM

  LBP



L L

GM KB BM KG

As practical matter, KB and KG are usually so small compared to GM

_L

that BM

_L

can be substituted for GM

_L

.

1 1

100 100

B L B L

BP BP

MTC F GM F BM

L L

     

 

1 100

L BP

MTC I

 L

   

 

100

L BP

I L

 

 

G B τe

F B1

F

B

M

L

IL: Moment of inertia of the water plane area about y’ axis B

,

F   

L ^L

BM  I

Substituting



K ²

L

F

G

P A.

1^

LBP F.P

(20)

6. Example of Longitudinal Stability

W1

L1

[Example] Calculation of Draft Change Due to Fuel Consumption (1/4)

During a voyage, a cargo ship uses up 320ton of consumable stores (H.F.O: Heavy Fuel Oil), located 88m forward of the midships.

Before the voyage, the forward draft marks at forward perpendicular recorded 5.46m, and the after marks at the after perpendicular, recorded 5.85m.

At the mean draft between forward and after perpendicular, the hydrostatic data show the ship to have LCF after of midship = 3m, Breadth = 10.47m, moment of inertia of the water plane area about transverse axis through point F = 6,469,478m

⁴

, Cwp = 0.8.

Calculate the draft mark the readings at the end of the voyage, assuming that there is no change in water density (=1.0ton/m

³

).

P P F.

A.

F

BP

195 L  m

3m

88 m

Consumable

Stores 5.46m

5.85m

Ship Stability

Ship Stability

Ch. 5 Initial Longitudinal Stability

Spring 2016

Myung-Il Roh

Department of Naval Architecture and Ocean Engineering Seoul National University

Lecture Note of Naval Architectural Calculation

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

Ch. 5 Initial Longitudinal Stability

1. Longitudinal Stability

2. Longitudinal Stability in Case of Small Angle of Trim 3. Longitudinal Righting Moment Arm

4. Derivation of Longitudinal Metacentric Radius (BM

) 5. Moment to Trim One Degree and Moment to Trim One Centimeter (MTC)

6. Example of Longitudinal Stability

1. Longitudinal Stability

F

F

Static Equilibrium

Static Equilibrium

I     

0    ,(     0)

ma   F

F

   F

0   F ,(  a  0)

F

 F 

.

A P F P .

Longitudinal Stability - Stable Equilibrium

Stable



Reference Frames





 ,

O O x

z

Longitudinal Stability of a Ship - Stable Condition (1/3)

B

z

x

 F

, z

, x



F

G

B

Longitudinal Stability of a Ship - Stable Condition (2/3)

B B

x

z

x z

, O O



F

r

 r

Resultant moment about y-axis through point O ( ): τ

F

r

F

τ





0 0

I  ^   

0    ,( ^  ^  0)

0   F ,( ^ a  0)