06 Transverse Righting Moment

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2009 Fall, Ship Stability

SDAL@Advanced Ship Design Automation Lab.http://asdal.snu.ac.kr Seoul National Univ.

N a v a

l A r c

h i t e c t u r e

& O c e a n

E n g

i n e e r i n g

SDAL@Advanced Ship Design Automation Lab.http://asdal.snu.ac.kr Seoul National Univ.2009 Fall, Ship Stability

- Ship Stability -

Part.1-II Righting Force and Moment

2009 Fall

Prof. Kyu-Yeul Lee

Department of Naval Architecture and Ocean Engineering,Seoul National University


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@



Righting Moment

Overview of Ship Stability

Force & Moment on a Floating Body

Newtons 2nd

LawEuler Equation

Stability Criteria

Damage Stability- MARPOL regulation

Pressure Integration Technique

Calculation Method to find GZwith respect to IMO regulation

sinGZ GM = , GM KB BM KG= +

sin L LGZ GM = , L LGM KB BM KG= +

- Overview of Ship Stability

BF GZ Transverse RightingMoment :

B LF GZ Longitudinal RightingMoment :

GZ Calculation

( )G BGZ y y= +( ) L G BGZ x x= +

Z

K

z

O

CL

y

z M

restoring

e

G

FG

B B1

N

FB

1 B y

G y

F B: Buoyancy force : Angle of Heel, : Angle of Trim( xG ,yG ,zG) : Center of mass in waterplane fixed frame( x B ,y B ,z B) : Center of buoyancy in waterplane fixed frame

y' G , y' B in body fixed frame

Rotational Transformation! yG , y B in waterplane fixed frame

Fundamental of Ship Stability

Properties which is related to hullform of the ship

Hydrostatic Values

Intact Stability- IMO Requirement (GZ)- Grain Stability- Floodable Length

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- Contents -Part.1-I Fundamentals of ship stability

Ch.1 Overview of Ship StabilityCh.2 Physics for Ship Stability

Ch.3 Hydrostatic Pressure, Force and Moment on a floating bodyCh.4 Concept of Righting MomentCh.5 Hydrostatic Values

Part.1-II Righting MomentCh.6 Transverse Righting Moment

Ch.7 Longitudinal Righting MomentCh.8 Heeling Moment caused by Fluid in Tanks

Part.1-III Stability CriteriaCh.9 Intact StabilityCh.10 Damage Stability

Part.1-IV Pressure Integration TechniqueCh.11 Calculation of Static Equilibrium PositionCh.12 Governing Equation of Force and Moment with Immersion, Heel and TrimCh.13 Partial derivatives of force and moments with immersion, heel, and trim

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N a v a

l A r c

h i t e c t u r e

& O c e a n

E n g

i n e e r i n g


Ch.6 Transverse Righting Moment- Sec.1 Calculation of Center of Buoyancy -

2009

Prof. Kyu-Yeul Lee


- Ship Stability -


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Sec.1 Calculation of Center of BuoyancySec.2 Calculation of BM, GZ in Wall Sided Ship

Sec.3 Inclining TestSec.4 Transverse Stability of ship (Unstable condition)Sec.5 Transverse Righting Moment due to Movement of CargoSec.6 Calculation of Heeling Angle due to Shift of Center of Mass

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Sec.1 Calculation of Center of Buoyancy- Rotational Transformation of Point and Frame- Example) Calculation of Center of Buoyancy of Ship with

Constant SectionMethod Direct calculating center of buoyancy in waterplane fixed frameMethod Transformation center of buoyancy from body fixed frame to

waterplane fixed frame

-

Calculation of Center of Buoyancy of Ship with Various Station

Shape 6/124


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(Review)Transverse Righting Moment

Z

K

z

O,O'

y

z

)(+

CL

y

z M

righting

e

G

FG

B B1

N

FB

1 B y

G y


O'x'y'z' : Body fixed frameOxyz : Waterplane fixed frame

Righting Moment : Moment toreturn the ship to the upright floatingposition (Righting moment, Momentof statical stability))

BGZ F = i Transverse Righting moment

1( )righting G B B y y F = + i

Righting arm

Righting Arm ( GZ )

1G BGZ y y= +From direct calculation

We should know yG , y B1 in waterplane fixed frame

From geometrical figure withassumption that M does not changewithin small angle of heel (about 10 )

sinGZ GM = GM is related to below equation bygeometrical figure

GM KB BM KG= +

How to calculate inwaterplane fixed frame ?

1 1, B B y z

x,x'

G: Center of mass B: Center of buoyancy B1: Changed center of buoyancyK : Keel F G : Weight of ship F B : Buoyant force acting on ship

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

M : The intersection of the line of buoyant force through B1 with the centerline of the ship7

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B

K

G

O,O'

y

z

Base Line

F B

F G

z

y

e

External moment ( e) is applied on theship in clockwise. A ship is heeled aboutorigin O through an angle of

- Transverse Righting Moment


Method 1. calculate center of the buoyancy(B 1) with respect to waterplane fixed frame

Question : How to calculate center of the buoyancy(B 1) with respect to waterplane fixed frame?

x,x'

y

z( )+

j

k

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dydzdA = = dA A= ydA M y A, = zdA M z A,

A, Mz, My (with respect to waterplane fixed frame)

Integral value(area and 1 st moment of area) haveto be calculated for every position when position of ship is changed.


O,O'

y

z

F B

F G

e

B

G

B11 B

y

1 B z

Center of buoyancy is changed from B to B1.


How to calculate center of buoyancy B1 withrespect to waterplane fixed frame?

Method 1. Calculate Center of buoyancy B1 withrespect to waterplane fixed frame directly


O'x'y'z' : Body fixed frame

Oxyz : Waterplane fixed frame

x,x'

y

z( )+

j

k

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Integral value(area and 1 st moment of area) haveto be calculated for every position when position of ship is changed.

y

z

B

G

B1

e

1 B y

1 B z

position change

y

z

Wetted surfaceat present

z

y

Area and moment of areahave to be calculated again

Wetted surfacewith changed position

=

A

M

A

M z y z A y A B B

,, ,),(11

Center of buoyancy with respect towaterplane fixed frame







A, Mz, My (with respect to waterplane fixed frame)



Center of buoyancy is changed from B to B1.


O,O'

x,x'

y

z( )+

j

k

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y

z


O

y

z

B

K

G

z

y B1

e


Method 2. Calculate center of buoyancy B1with respect to body fixed frame, thentransform B1 to waterplane fixed frame

with respect to body fixed frame, ,, , A y A z A M M

' 'dA dy dz= = dA A, ' ' A y M y dA= , ' ' A z M z dA=

Integral value could be used as it is exceptintersection region with waterplane area whenposition of ship is changed.

( ) ( )+

1 B y 1 B z




Method 2. calculate center of the buoyancy(B 1) with respect to body fixed framethen transform frame from body fixed frame to waterplane fixed frame


O,O' x,x'

y

z( )+

j

k

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y

z

B

K

G

z

y

e

y

zWetted surfaceat present

( ) ( )+ position change y

Only changed area andmoment of area have to calculated.


z ( )+( )

- Transverse Ri htin Moment










O,O' x,x'

y

z( )+

j

k

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y

z


( ) ( )+ y



z

y

z

B

K

G

z

y

e

position change( )+

( )position change

y

zWetted surfaceat present z

y


Is area invariant with respect to referenceframe?

Wetted surface area with respect to waterplane fixed frame

Area is invariant with respect toreference frame.











O,O' x,x'

y

z( )+

j

k

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y

z

B

K

G

z

y B1

e

y

z


( ) ( )+ position change y



1 B y 1 B z

1 1

, ' , '( ' , ' ) , A y A z B B M M

y z A A

=

Center of buoyancy in body fixed frame











( )+( )

1 1

1 1

'cos sin

sin cos ' B B

B B

y y

z z

=

Center of buoyancy in waterplane fixedframe : Rotational Transformation

O,O' x,x'

y

z( )+

j

k

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y

z

B

G

B1

1 B y

1 B z

O

y

z

B

K

G

z

y

B1

=

A M

A M z y z A y A B B

,, ,),(11

Center of buoyancy with respect towaterplane fixed frame


with respect to waterplane fixedframe

y z M M A ,,

1 1

, ' , '( ' , ' ) , A y A z B B M M

y z A A

=

1 1

1 1

'cos sinsin cos '

B B

B B

y y z z

=

Center of buoyancy with respect to waterplanefixed frame

Rotational transformation


Method 2. Calculate center of buoyancy B1 withrespect to body fixed frame, then transform B1 towaterplane fixed frame

Same

with respect to body fixed frame, ' , ', , A y A z A M M

' 'dA dy dz = , ' ' A y M y dA= , ' ' A z M z dA=

Convenient

1 B y1 B

z


1 B y

1 B z



Comparison between Method 1 and Method 2


O,O'

x,x'

y

z( )+

j

k

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Rotational Transformation of Point and Frame(A) Rotation of the point (B) Rotation of the frame

Given : Coordinate of P with respect to oyz frameFind : Coordinate of Q which is rotated coordinate of Pabout origin O in the yz plane through an angle of .

P

Pr

y

z

o

QQr

=P

P

Q

Q

z

y

z

y

cossinsincos

Given: Coordinate of P with respect to oyz frameFind: Coordinate of P with respect to oyz which is rotatedframe about origin O ' from o ' y' z ' through an angle of .

' yP y

P y

P z

P z

cos sinsin cos

P P

P P

y y

z z

=

Rotational transformation of pointRotational transformation of frame

P

Pr

y

z

,o o

' z


cos sinsin cos

P P

P P

y y

z z

=

...(1) ...(2 1)

...(2 2)

Rotational transformation of frame16

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Rotational Transformation of Point and Frame

Given : Coordinate of P with respect to oyz frameFind : Coordinate of Q which is rotated coordinate of Pabout origin O in the yz plane through an angle of .

P

Pr

y

z

o

QQr

=P

P

Q

Q

z y

z y

cossinsincos

P y

P z

Rotational transformation of point through an angle of equals to

rotational transformation of frame through an angle of - .

Rotational transformation of point

y

z

,o o


Given: Coordinate of P with respect to oyz frameFind: Coordinate of P with respect to oyz which is rotatedframe about origin O ' from o ' y' z ' through an angle of - .

(A) Rotation of the point (B) Rotation of the frame

cos sinsin cos

P P

P P

y y z z

= ...(2 1)

If we substitude - into

cos sinsin cos

P P

P P

y y

z z

=

...(2 1)

Pr

y

z

P y

P z P

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(Proof) Rotational Transformation of Point

sincoscossin)sin( +=+Summation formula of trigonometric function

sinsincoscos)cos( =+

Coordinate of point P, Q is expressed byan angle

)sin()cos(

+=+=

QQ

QQ

z y

rr

sincos

PP

PP

z y

rr

==

( ) ( )

sincos

sinsincoscos

sinsincoscos

)cos(

PP

PP

QQ

QQ

z y

y

==

=

+=

rr

rr

r

Let coordinate of Q be expressed by difference formulaof trigonometric function.

( ) ( )

sincos

sincoscossin

sincoscossin

)sin(

PP

PP

QQ

QQ

y z

z

+=+=

+=

+=

rr

rr

r

)(, QP rr =

)(, QP rr =

=

P

P

Q

Q

z

y

z

y

cossinsincos

In the matrix form

P

Pr

y

z

o

QQr


Given : Coordinate of P with respect to oyz frameFind : Coordinate of Q which is rotated coordinate of P about origin O in the yz plane through anangle of .

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(Proof) Rotational Transformation of Frame

sincoscossin)sin( +=+Summation formula of trigonometric function

sinsincoscos)cos( =+

Coordinate of point P is expressed by an angle

cos( )sin( )

P P

P P

y z

= + = +

rr

cossin

P P

P P

y z

==

rr

( ) ( )

cos( )cos cos sin sincos cos sin sin

cos sin

P P

P P

P P

P P

y

y z

= += =

=

rr rr r

Let coordinate of P be expressed by difference formula of trigonometric function.

( ) ( )

sin( )

sin cos cos sin

sin cos cos sin

cos sin

P P

P P

P P

P P

z

z y

= += += += +

r

r r

r r

cos sinsin cos

P P

P P

y y

z z

=

In the matrix form

yP y

P y

P zP z

P

Pr

y

z

o

z


Given: Coordinate of P with respect to oyz frameFind: Coordinate of P with respect to oyz which is rotated frame about origin O ' from o ' y' z ' throughan angle of - .

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- Calculation of Center of Buoyancy of Ship with Various Station

Shape 20/124


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Example) Calculation of Center of Buoyancy of Ship with Constant Section: Method Direct calculating center of buoyancy in waterplane fixed frame

G: Center of mass K :Keel B: Center of buoyancy B1 : Changed center of buoyancy

Given : Breadth ( B):20, Depth ( D):20, Draft(T ) 10, Angle of Heel( ) : -30 Find : Center of buoyancy ( y B , z B) in Waterplane fixed frame

z

y

O

30 B

K

B1

2020 z

y P

Q

R

S

Question) A ship with a breadth 20m, depth 20m, draft 10m is heel about origin O through anangle of -30 . Calculate center of buoyancy with respect to waterplane fixed frame

z,z

y,y O

B

K

20

10

20

QP

R S


Section view

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2009 Fall, Ship StabilitySDAL@Advanced Ship Design Automation Lab.http://asdal.snu.ac.kr Seoul National Univ.


z

yO

30 B

K

B1

2020

z

y P

Q

R

SQuestion) A ship with a breadth 20m, depth 20m, draft 10m is heel about origin Othrough an angle of -30 . Calculate center of buoyancy with respect to waterplanefixed frame

Sol.) P, Q, R,S with respect to waterplanefixed frame

( , )P PP x y is calculated by rotational transformation from

Q( xQ ,yQ), R( x R ,y R), S ( xS ,yS ) are calculated in the same way.

=

A

M

A

M z y z A y A B B

,, ,),(11

cos(30) sin(30) 10 3.66sin(30) cos(30) 10 13.66

Q

Q

x y = =

cos(30) sin(30) 10 13.66

sin(30) cos(30) 10 3.66 R

R

x

y

= =

cos(30) sin(30) 10 3.66

sin(30) cos(30) 10 13.66S

S

x

y

= =

cos(30) sin(30)sin(30) cos(30)

0.866 0.5 10 13.660.5 0.866 10 3.66

P P

P P

x x

y y

= = =



Given : Breadth ( B):20, Depth ( D):20, Draft(T ) 10, Angle of Heel( ) : -30 Find : Center of buoyancy in Waterplane fixed frame1 1( , ) B B y z

( , )P PP x y

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Sol.) Area

y

z

=

A

M

A

M z y z A y A B B

,, ,),(11

P

Q

R

S

2 11 1

2 1

( ) z z

z z y y y y

=

Intersection point between straight line and y

axis have to be calculated in order to know area.Equation of straight line P1S1 have to becalculated in order to know intersection point.

Equation of straight line through two points isas follows.

Substituting two points of P1, S1 into equation of straight line

13.66 ( 3.66)( 3.66) ( ( 13.66))

( 3.66) ( 13.66) z y

=

(-13.66,-3.66)

(-3.66, 13.66)

(13.66,3.66)

(3.66,-13.66)

1.732 20 z y = +

T

Intersection point T ( 11.55,0)T

Equation of straight line of Q1R 1 is obtained in the same way

1.732 20 z y =

U

Intersection point U (11.55,0)U



z

yO

30 B

K

B1

2020

z

y P

Q

R

S


Question) A ship with a breadth 20m, depth 20m, draft 10m is heel about origin Othrough an angle of -30 . Calculate center of buoyancy with respect to waterplanefixed frame


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l ) l l f f f h h


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Sol.) Area

y

z

=

A

M

A

M z y z A y A B B

,, ,),(11

P

Q

R

S

Divide area into 4 part , A1,A

2,A

3,A

4.

Then calculate area of A1.

(-13.66,-3.66)

(-3.66, 13.66)

(13.66,3.66)

(3.66,-13.66)

T (-11.55,0) U(11.55,0)A1 A2

A3A4

1

111.55 ( 13.66) 3.66

23.867

A Area = =

Area of A2, A3, A4 can be calculated in the same way.

2 55.66, A Area =3

86.6 A Area =

453.87 A Area =



z

yO

30 B

K

B1

2020 z

y P

Q

R

S




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E l ) C l l i f C f B f Shi i h C S i


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Sol.) Centroid

y

z

=

A

M

A

M z y z A y A B B

,, ,),(11

P

Q

R

S

(-13.66,-3.66)

(-3.66, 13.66)

(13.66,3.66)

(3.66,-13.66)

T (-11.547,0) U(11.547,0)A1 A2

A3A4

1 _

211.54 ( 11.54 ( 13.66))

312.25

c A x = +

=

1 _

23.66 ( ) 2.44

3c A y = =

Centroid of A1 can be calculated as follows1 1 _ _ ( , )c A c A x y

Centroids of A2, A3, A4 are calculated in the same way.

2 2 _ _ ( , ) ( 3.96, 1.83)c A c A x y =

3 3 _ _ ( , ) ( 2.11, 6.99)c A c A x y =

4 4 _ _ ( , ) (6.29, 4.55)c A c A x y =



z

yO

30 B

K

B1

2020 z

y P

Q

R

S




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Sol.) 1st moment of area

=

A

M

A

M z y z A y A B B

,, ,),(11

1st

moments of area are calculated with areasand centroids which are calculated in previous.

(13.66,3.66)

y

z

P

Q

R

S

(-13.66,- 3.66)

(-3.66, 13.66)

(3.66,-13.66)

T (-11.547,0) U(11.547,0)A1 A2

A3A4Area y c zc Area yc Area zc

A1 3.87 -12.25 -2.44 -47.38 -9.44

A2 55.66 -3.96 -1.83 -220.24 -101.85

A3 86.60 -2.11 -6.99 -183.01 -605.62

A4 53.87 6.29 -4.55 338.78 -245.28

Sum 200.00 -111.85 -962.19

Centoid of total area is calculated as follows.

( )1 1

, , 111.85 962.( , ) , , 0.56, 4.81200 200

A y A z B B

M M y z

A A

= = =

, A y M , A z M



z

yO

30 B

K

B1

2020 z

y P

Q

R

S




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l ) l l f f f h h


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2009 Fall, Ship StabilitySDAL@Advanced Ship Design Automation Lab.http://asdal.snu.ac.kr Seoul National Univ.2009 Fall, Ship Stability

Sol.) Centroid

Centroid of total area before heel

_ _ ( , ) (0, 5)c A c A y z =

Centroids of changed area after heel

1 1 _ _

2 1( , ) ( 10, 10 tan 30 )

3 3

( 6.67, 1.92)

c A c A y z =

=

2 2 _ _ 2 1

( , ) ( 10, 10 tan 30 )3 3

(6.67,1.92)

c A c A y z = =


Example) Calculation of Center of Buoyancy of Ship with Constant Section: Method Transformation of center of buoyancy from body fixed frame to waterplane fixed frame

z

y

y

z

O 30

B

K

B1

20

20

QP

R S

A1

A2

A

z

y

y

z

O 30

B

K

B1

20

20

QP

R S




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Sol.) 1st moment of area

= A

M

A

M z y z A y A B B

',', ','

)','(11

1st moments of area are calculated with areas

and centroids which are calculated in previous.Area yc zc Area y Area z

A 200.00 0.00 -5.00 0.00 -1000.00A1 -28.87 -6.67 -1.92 192.45 55.56A2 28.87 6.67 1.92 192.45 55.56

Sum 200.00 384.90 -888.89

, A y M , A z M

Centroid of total area after heel with respect to body fixed frame is as follows( )

1 1

, , 384.90 888.89( , ) , , 1.92, 4.44200 200

A y A z B B

M M y z

A A

= = =

Rotationaltransformation

1 1

1 1

cos sin 1.92 0.56cos(30 ) sin(30 )sin cos 4.44 4.81sin(30 ) cos(30 )

B B

B B

y y

z z

= = =

Transform coordinate of centroid of total area with respect to body fixed frame tocentroid with respect to waterplane fixed frame by rotational transformation

Same result



z

y

y

z

O 30

B

K

B1

20

20

QP

R S

A1

A2

A

z

y

y

z

O 30

B

K

B1

20

20

QP

R S




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1

1

0.56

4.81 B

B

y

z

=

1

1

0.56( )

4.81 B

B

y

z

=

Same

Calculation Result

Calculation Result



z

yO

30 B

K

B1

2020 z

y P

Q

R

S

z

y

y

z

O 30

B

K

B1

20

20

QP

R S

Method Direct calculating center of buoyancy inwaterplane fixed frame

Method Transformation of center of buoyancyfrom body fixed frame to waterplane fixed frame

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- Calculation of Center of Buoyancy of Ship with Various StationShape

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Calculation of Center of Buoyancy of Ship with Various Station Shape


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Calculation of Center of Buoyancy of Ship with Various Station Shape- Introduction

1 Br 1 B z

1 B y

' y

y

' z z

G

B

K 1 B

Lc

O O

-

1' B y

1' B z1 B

z

1 B y

1 Br

' z z

G

B

K

' y

y

Lc

dA

Method Direct calculating center of buoyancy in waterplane fixed frame

Method Transformation center of buoyancy from body fixed frameto waterplane fixed frame

How to calculate center of buoyancy of ship with various sections ?


How to calculate inwaterplane fixed frame ?

1 1, B B y z

Righting Moment : Moment toreturn the ship to the upright floatingposition (Restoring moment, Momentof statical stability))

BGZ F = i Transverse Restoring moment


Righting arm

Righting Arm ( GZ )

1G BGZ y y= +

From direct calculation

We should know yG , y B1 in waterplane fixed f rame


sinGZ GM = GM is related to below equation bygeometrical figure

GM KB BM KG= +

Z

K

z

O,O'

y

z

)(+

CL

y

z M

righting

e

G

FG

B B1

N

FB

1 B y

G y


x,x'

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VCB(Vertical Center of Buoyancy)


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VCB(Vertical Center of Buoyancy)Step Area, 1 st Moment of Area

z

y

z'

dA

y'

dz'

dy'

c y

c z G

1

n

ii

A dA dy dz

A=

= =

=

Area , A

dA dy dz=

Differential element of area, dA

, A y

c

M y dA y dy dz

y A

= =

=

1st

moment of area with respect to z axis, M A,x

, A z

c

M z dA z dy dz

z A

= =

=

1st moment of area with respect to y axis, M A,y

( ), ,, , A y A z c c M M

y z A A

= =

G

Centroid G

(Ai : Area of i-th element)

( , )c c y z : center of A

1) James M. Gere, Mechanics of materials, THOMSON, pp.828-850, 6 th Edition

- Transverse Righting Moment33

/124



34/124


VCB(Vertical Center of Buoyancy)Step Sectional Area (A M), Displacement Volume

Displacement volume

( ) ' A x dx =

== '' dzdydA ASectional Area

L

formhullr under wate

B

T

A

x y

z

( ) ''''''

dxdzdy

dzdydxdV

===

( ) A x

After calculation of each station area,displacement volume can be calculated byintegral of section area over the length of ship

Each stationarea Integral

in longitudinaldirection

Volume

A

Area of this Curve Volume( )

A.P x L

A

A1

A2

A3 An..



/124



35/124


VCB(Vertical Center of Buoyancy)Step Vertical Moment of Volume, Vertical Center of Buoyancy(VCB)

Vertical Center of Buoyancy

, , ( ) z A z M M x dx =

, z M VCB =

Vertical Moment of Volume

( )

, z M z dV

z dx dy dz

z dy dz dx

=

=

=

, ( ) A z M x M A,z : Vertical moment of area about y axis

After calculation of each vertical moment of stationarea about the y axis( M A,z), vertical moment of

displaced volume can be calculated by integral of vertical moment of section area over the length of ship

L

formhull

r under wate

B

T

x y

z A

vertical momentof station areaabout the y axis

Integralin longitudinaldirection

Verticalmoment of volume

, ' A z M , ' z M

Area of this Curve 1st Moment of Volume( )

, z M

A.P x L M A1, z

M A2, z

M A3, z

M An, z .

, A z M



z

35/124

1) Zill,C. Advanced Engineering Mathematics, p319, Jones and Bartlett, 3 rd


36/124


Area of Triangle by Vector

0 0( , ) x y

1 1( , ) x y

2 2( , ) x y

x

y

1 0r r

2 0r r

o

2r

1r

0r

1 0 2 01( ) ( ) ( )2

Area = r r r r r

1 0 2 0 2 0 1 01 ( )( ) ( )( )2

x x y y x x y y=

1 0 1 0

2 0 2 0

0

0

x x y y

x x y y

=

i j k

1)Cross Product

( )sin =a b a b n: Area of parellelogram sinh = ba

b

b

a

: Area of triangle with side a and b

A = a b

12

A = a b

Given : Position vector r 0, r 1, r 2, of vertex of triangle Find : Area of triangle


/124

Area of Triangle



37/124


Area of Triangleby Rotational Transformation

)'')(''()'')(''(21

01020201 y y x x y y x x =

1 0 2 0 2 0 1 0

1( )( ) ( )( )2 x x y y x x y y =

1 0 2 01( ) ( ) ( )2

Area = r r r r r

1 0 2 01( ') ( ' ' ) ( ' ' )2

Area = r r r r r

[ ] [ ]0 1 2 0 1 2cos sin' ' ' sin cos = r r r r r r

=210

210

210

210

cossinsincos

''''''

y y y x x x

y y y x x x

' = r R r

1 1 0 0 2 2 0 0

2 2 0 0 1 1 0 0

1 (( cos sin ) ( cos sin ))(( sin cos ) ( sin cos ))2

(( cos sin ) ( cos sin ))(( sin cos ) ( sin cos ))

x y x y x y x y

x y x y x y x y

= + +

+ +

)cos)(sin))((sin)(cos)((

)cos)(sin))((sin)(cos)((21

01010202

02020101

y y x x y y x x

y y x x y y x x

+

+=

2 21 0 2 0 1 0 2 0 2 0 1 0 1 0 2 0

2 21 0 2 0 2 0 1 0 1 0 2 0 1 0 2 0

1(( )( ) cos sin ( )( ) cos ( )( ) sin ( )( ) sin cos )

2

( )( ) cos sin ( )( ) cos ( )( ) sin ( )( ) sin cos )

x x x x x x y y x x y y y y y y

x x x x x x y y x x y y y y y y

= +

+ +

{ }{ }2 2 1 0 2 0 1 0 2 01

cos sin ( )( ) ( )( )2 x x y y y y y y = +

Is a area changed after rotation aboutorigin O? Given : Area before rotation Find : Area after rotation

0r

1r

2r

x0'r

1'r

2'r

y


,(R: )

o

37/124

Area of Triangle1) Zill,C. Advanced Engineering Mathematics, p319, Jones and Bartlett, 3 rd


38/124


Area of Triangleby Rotational Transformation

)'')(''()'')(''(21

01020201 y y x x y y x x =

1 0 2 0 2 0 1 0

1( )( ) ( )( )2 x x y y x x y y =

( )( ) ( )

Area

Area Area

= =

Rr

r r( R : Rotationaltransformation matrix)

1 0 2 01( ) ( ) ( )2

Area = r r r r r

1 0 2 01( ') ( ' ' ) ( ' ' )2

Area = r r r r r

{ }{ }2 2 1 0 2 0

1 0 2 0 1 0 2 0

1 0 2 0

1cos sin ( )( ) ( )(

1( )( )

)

(

2

)( )2

x x y y y y y

x x y y y y y y

y = +

=

Area is invariant with respect to frame.

Given : Find :

0r

1r

2r

x0'r

1'r

2'r

y


( )( ) a Area Are = rr

o

38/124

1st Moment of Area1) Zill,C. Advanced Engineering Mathematics, p319, Jones and Bartlett, 3 rd


39/124


1 Moment of Areaby Rotational Transformation

1 0 2 01( ) ( ) ( )

2

Area = r r r r r

( ) C A= M r r

0 0( , ) x y

1 1( , ) x y

2 2( , ) x y

x

y

C C r ( , ) x y

1 2 31

( )3C

= + +r r r r

A

B C

O

G

Position vector of centroid oftriangle ABC.

a

b

c

g

Let cent of side BC as D

( )12OD b c= +

Now, G is the point of internal division with ratio of 2 to 1

2 1 22 1 3

OD OA OD OAOG

+ += =+

( )1 123 2g b c a = + + =

( )13 a b c+ +

2

1

1 2 3

3 x x x x xdA xdxdy A+ += = = M

1 2 3

3 y y y y

ydA ydxdy A+ += = = M

Given : Position vector of vertex of triangle Find : 1 st moment of area of triangle

Position vector of centroid of triangle r 1r 2r 2

Area of triangle r 1r 2r 2

1st moment of area of triangle r 1r 2r 2 with respect to x andy axis

(*) ( ) , 10- ,40th ,2005, ,pp.15- Transverse Righting Moment

o

2r

0r1r

D39

/124


1st Moment of Area


40/124


1 2 31 ( ' ' ' )3C = + +r r r r

1 2 31 ( )3

= + +R r R r R r

1 2 31 ( )3

= + +R r r r

C A=R r( ) Moment = R r

( ) C Moment A=r r1 2 3

1 ( )3C

= + +r r r r

( ') 'C Moment A=r r

1 0 2 01( ) ( ) ( )2

Area = r r r r r

1st moment of area of triangle before rotation

( ') ( ) , ( ) Area Area A A= =r r1st moment of area of triangle after rotation

Given : Position vector of triangle, angle of rotation Find : 1st moment of area of triangle after rotationabout origin O

C = R rC C =r R r

( )

C

Moment

A=r

r[ ] [ ]0 1 2 0 1 2

cos sin' ' '

sin cos

= r r r r r r

=210

210

210

210

cossinsincos

''''''

y y y x x x

y y y x x x

' = r R r

0r

1r

2r

x 0'r

1'r

2'r

y

C

'C C r



C r

o

40/124

1) Zill,C. Advanced Engineering Mathematics, p319, Jones and Bartlett, 3 rd 1st Moment of Area


41/124


C C =r R r

C A= R r( ) Moment = R r

( ) ( )

C

Moment Moment

A

==

R r R r

R r( R : Rotationaltransformation matrix)

0r

1r

2r

x 0'r

1'r

2'r

y

C

'C C r

( ) C Moment A=r r1 2 3

1 ( )3C

= + +r r r r

( ') 'C oment A=M r r

1 0 2 01( ) ( ) ( )2

Area = r r r r r

1st moment of area of triangle before rotation

( ') ( ) , ( ) Area Area A A= =r r1st moment of area of triangle after rotation

1st moment of area is invariant withrespect to frame.

[ ] [ ]0 1 2 0 1 2cos sin

' ' 'sin cos

= r r r r r r

=210

210

210

210

cossinsincos

''''''

y y y x x x

y y y x x x

' = r R r

Given : Position vector of triangle, angle of rotation Find : 1st moment of area of triangle after rotationabout origin O



C r

o

41/124


42/124





- Calculation of Center of Buoyancy of Ship with Various StationShape

42/124

Example) Calculation of Center of Buoyancy of Ship with Various Station


43/124


a p e) Ca cu at o o Ce te o uoya cy o S p w t Va ous Stat oShape

Given : Length(L) : 50, Breadth(B) : 20, Depth(D) : 20, Draft(T):10, Angle of Heel( ) : -30 Find : Center of buoyancy ( y , c , z , c ) after heel in waterplane fixed frame

Problem) There is ship (L x B x D : 50 x 20 x 20) with various station shape. When thisship is heeled about x axis in counter-clock wise through an angle of 30 compulsorily ,calculate y and z coordinates of center of buoyancy with respect to waterplane fixed

frame.

20

CL

20

10

O,O ' y,y' z,z'

x,x'

10

10

20 20

10

20

20

20

20

20

2010


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O y z

A1

Example) Calculation of Center of Buoyancy of Ship


44/124


y dz dy dx

dz dy dx

=

y dz dy dx =

dz dy dx =

O y

x

A3

A2


Sectional area of each section(Body fixed frame)

( ) A x dA dy dz = =

1st moment of area of each section(Body fixed frame)

, A y M y dy dz

=

, A z M z dy dz =

Displacement volume(Body fixed frame)

dV =

1st moment of displacement volume(Body fixed frame)

, y M

, z M

.

.( )

F P

A P A x dx =

Center of buoyancy(Body fixed frame )

,c y

.

.( ( ))

F P

c A P y A x dx =

, ,( , )c c y z : Center of displaced volume

.

..

.

( )

( )

F P

c A PF P

A P

A x y dx

A x dx

=

, y M =

,c z , z M =

z dz dy dx

dz dy dx

=

.

..

.

( )

( )

F P

c A PF P

A P

A x z dx

A x dx

=

, ,

, ,

cos sinsin cos

c c

c c

y y

z z

=

Center of Buoyancy(Waterplane fixed frame )

z dz dy dx = .

.( ( ))

F P

c A P z A x dx =

( )c y A x =

( )c z A x =

( ) A x : Sectional area at x( , )c c y z : Centroid of section at x'

p ) y y pwith Various Station Shape

44/124



45/124


O,O ' y,y' z,z'

x,x'

)

y'

y

z' z

< Section A 1 >

< Section A 2 >

< Section A 3 >


Problem) There is ship (L x B x D : 50 x 20 x 20) with various station shape. When thisship is heeled about x axis in counter-clock wise through an angle of 30 compulsorily ,calculate y and z coordinates of center of buoyancy with respect to waterplane fixedframe.

2

CL

y

z

20

10

O y z

x

10

10

20 20

10

y'

y

z' z

y'

y

z' z



45/124



46/124

2009 Fall, Ship Stability SDAL@Advanced Ship Design Automation Lab.http://asdal.snu.ac.kr Seoul National Univ.

) In the same way of previous example, calculate center of buoyancy with respect to bodyfixed frame at first, then calculate center of buoyancy with respect to waterplane fixedframe by rotational transformation.

Coordinate of P 1, Q1 is known as (-5,0), (5,0) by geometric shape.Coordinate of P 1, P2, Q 1, Q 2 of section A3

< Section A 3 >

O y z

x

A3

A2

A1

z

y

y

z

O30

B

K

B1

20

20 Q2

Q1

P1P2

QP

Calculate equations of straight line PK, KQ in order to know P 2,Q2

The equation of straight line PK 2 10 z y =

The equation of straight line KQ 2 10 z y =

The equation of line of waterplane with respect to body fixed frame is as follows,

because waterplane is inclined through an angle of 30 .tan 30 0.5774 z y y = =

Intersection point P 2,Q2 betwwen waterplane and straight line PK, KQ can be calculated as followsP2(-3.88, -2.24), Q 2(7.03, 4.06)

20

CL

y

z

20

10

O y z

x

10

10

20 20

10


Area below waterplane can be calculated also by Gaussian Quadrature.




46/124



47/124


) -A3 : Sectional area of A 3O y z

x

A3

A2

A1

z

y

y

z

O30

B

K

B1

20

20 Q2

Q1

P1P2

QP

A3_2 A3_1

A3_0

3_ 0 0.5 10 10 50 A = =

3_1 1 2

1( ) ( )

2

0 5 0

0 3.88 2.245.60

A =

=

=

P O P O

i j k

3_ 2 1 2

1( ) ( )

2

0 5 0

0 7.03 4.0610.15

A =

=

=

Q O Q O

i j k

Total sectional area before heel

Changed area after heel

Total sectional area after heel

3 50 5.60 10.15 54.55 A = + =

If a ship is not a wall sided ship,area below waterplane is different.

P2(-3.88, -2.24), Q 2(7.03, 4.06)P1(-5,0), Q 1(5,0)

20

CL

y

z

20

10

O y z

x

10

10

20 20

10


( )

( )i

A x dA

dy dz

A x

= =

=

( )k A x: Sectional area at x

_ ( )k i A x : Partial area of section at x'




< Section A 3 >

47/124



48/124


) Centroid of A3O y z

x

A3

A2

A1

z

y

y

z

O30

B

K

B1

20

20 Q2

Q1

P1P2

QP

A3_2 A3_1

A3_0

P2(-3.88, -2.24), Q 2(7.03, 4.06)P1(-5,0), Q 1(5,0)

3_ 0

1Cetroid of (0, ( 10))3

10(0, )3

A =

=

3_10 5 3.88 0 0 2.24Cetroid of ( , )

3 3

( 2.96, 0.75)

A + =

=

3_ 20 7.03 5 0 0 4.06Cetroid of ( , )

3 3(4.01, 1.35)

A+ + + +=

=

20

CL

y

z

20

10

O y z

x

10

10

20 20

10





< Section A 3 >

48/124

Example) Calculation of Center


49/124


)

Calculate 1 st moment of area in order to know centroid of section A 3 with respect to body fixed frame

Area Area*y'c_3_i Area*z'c_3_i A3_0 50.00 0.00 -3.33 0.00 -166.67 A3_1 5.60 -2.96 -0.75 -16.57 -4.18 A3_2 10.15 4.01 1.35 40.69 13.73

- + 54.55 57.26 -148.76

- A3 : 1st moment of area of section A 3

Centroid of section A 3 with respect to body fixed frame is calculated as follows

( )

3 3

3 3

, , _ 3 _ 3( , ) ,

57.26 148.76, 1.05, 2.73

54.55 54.55

A y A zc c

A A

M M y z

Area Area =

= =

3 , A y M 3, A z M

3, A y

M : 1

st moment of area of section A 3 about z ' axis 3

, A z M : 1st moment of area of

section A 3 about y ' axis

O y z

x

A3

A2

A1

y

yO30

B

K

B1

20 Q2

Q1

P1P2

QP

A3_2 A3_1

A3_0

20

CL

y

z

20

10

O y z

x

10

10

20 20

10

z 20


,k A y M y dy dz

=

_ _ _ ( )c k i k i y A x = _ ( )c k k y A x =

of Buoyancy of Ship with Various Station Shape

_3 _ c i y _ 3_ c i z



< Section A 3 >

49/124



50/124


)

Calculate 1st

moment of area in order to know centroid of section A 2 with respect to body fixed frame

Area Area*y'c_2_i Area*z' c_2_i A3_0 200.00 0.00 -5.00 0.00 -1000.00 A3_1 -28.87 -6.67 -1.92 192.45 55.56 A3_2 28.87 6.67 1.92 192.45 55.56Sum 200.00 384.90 -888.89

-A2, - A2 : A2 1


( )3 32 2

, , _ 2 _ 2

384.90 888.89( , ) , , 1.92, 4.44

200 200 A y A z

c c A A

M M y z

Area Area = = =

y'

z'

y

z

A2_2 A2_1

A2_0

2 , A y M

: 1st moment of area of section A 2 about z ' axis

2 , A z M

: 1st moment of area of section A 2 about y ' axis

O y z

x

A3

A2

A1

20

CL

y

z

20

10

O y z

x

10

10

20 20

10



2 , A y M

2, A z M

_ 2 _ c i y _ 2_ c i z

,k A y M y dy dz

=

_ _ _ ( )c k i k i y A x = _ ( )c k k y A x =



< Section A 2 >

50/124

zExample) Calculation of Center of Buoyancy of Ship


51/124


)

Coordinate of R 1, S1 is known as (-7.5,0), (7.5,0) by geometric shape.

Coordinate of R 1, R 2, S1, S2 of section A 1O y z

x

A3

A2

A1

Calculate equations of straight line RR 3, SS 3 in order to know R 2,S2

The equation of straight line RR 3 4 30 z y =

The equation of straight line SS 3 4 30 z y =

The equation of line of waterplane with respect to body fixed frame is asfollows, because waterplane is inclined through an angle of 30 .tan 30 0.5774 z y y = =

Intersection point R 3,S3 between waterplane and straight line RR 3, SS 3 can be calculated as followsP2(-6.55,-3.78), Q 2(8.77,5.06)

z

y

y

z

O30

B

K

B1

20

20

SR

S2

S1

R 1

R 2

R 3 S3

10

20

CL

y

z

20

10

O y z

x

10

10

20 20

10


with Various Station Shape



< Section A 1 >

51/124

zExample) Calculation of Center of Buoyancy of Ship


52/124


) - A1 : Sectional area of section A 1

A1_2 A1_1

A1_0

1_ 0 0.5 (15 10) 10 125 A = + =

1_1 1 2

1( ) ( )

2

0 7.5 0

0 6.55 3.7814.19

A =

=

=

P O P O

i j k

1_ 2 1 2

1( ) ( )

2

0 7.5 0

0 8.77 5.0618.98

A =

=

=

Q O Q O

i j k

1 125 14.19 18.98 129.79 A = + =If a ship is not a wall sided ship,area below waterplane is different.

P2(-6.55,-3.78), Q 2(8.77,5.06)P1(-7.5,0), Q 1(7.5,0)

O y z

x

A3

A2

A1

z

y

y

z

O30

B

K

B1

20

20

SR

S2

S1

R 1

R 2

R 3 S3

10

20

CL

y

z

20

10

O y z

x

10

10

20 20

10


_

( )

( )

k

k ii

A x dAdy dz

A x

= ==




< Section A 1 >

Total sectional area before heel

Changed area after heel

Total sectional area after heel

52/124

OzExample) Calculation of Center of Buoyancy of Ship


53/124



) Centroid of section A 1

3_ 0 (15 10) ( 5) (0.5 2.5 10) 10 (2 / 3)Cetroid of (0, )125(0, 4.67)

A + =

=

3_10 7.5 6.55 0 0 3.78Cetroid of ( , )

3 3( 4.68, 1.26)

A + =

=

3_ 20 7.5 8.77 0 0 5.06Cetroid of ( , )

3 3(5.42, 1.69)

A+ + + +=

=

P2(-6.55,-3.78), Q 2(8.77,5.06)P1(-7.5,0), Q 1(7.5,0)

A1_2 A1_1

A1_0

O y z

x

A3

A2

A1

z

y

y

z

O30

B

K

B1

20

20

SR

S2

S1

R 1

R 2

R 3 S3

10

20

CL

y

z

20

10

O y z

x

10

10

20 20

10





< Section A 1 >

53/124

Oz



54/124



Calculate 1 st moment of area in order to know centroid of section A 1 with respect to body fixed frame

Area Area*y'c_1_iArea*z'c_1_i A1_0 125.00 0.00 -4.67 0.00 -583.34 A1_1 14.19 -4.68 -1.26 -66.48 -17.90 A1_2 18.98 5.42 1.69 102.90 32.02

- + 129.79 169.38 -533.42

- A1 : 1st moment of area of section A 1


1 , A y M

: 1st moment of area of section A 1 about z ' axis 1 , A z

M

: 1st moment of area of section A 1 about y ' axis

)

A1_2 A1_1

A1_0

O y z

x

A3

A2

A1

z

y

y

z

O30

B

K

B1

20

20

SR

S2

S1

R 1

R 2

R 3 S3

10

20

CL

y

z

20

10

O y z

x

10

10

20 20

10


, A y M y dy dz

=

_ ( )c i i y A x = ( )c y A x =


1 , A y M 1, A z M

_1_ c i y _1_ c i z

( )

1 1

1 1

, , _1 _1( , ) ,

169.38 533.42, 1.31, 4.11

129.79 129.79

A y A zc c

A A

M M y z

Area Area =

= =



< Section A 1 >

54/124


O yz


55/124




)

Area of this curve Volume( )

Displacement volume can be calculated by

integral of sectional area in longitudinal direction50

0( ) 7,304 A x dx = =

Displacement Volume

x' 0

Area

A1 A2 A3

54.55

200

127.79

25 50

' Adx =

O y z

x

A3

A2

A1

20

CL

y

z

20

10

O y z

x

10

10

20 20

10


dV dx dy dz = = dz dy dx =

.

.( )

F P

A P A x dx =



55/124


O y z


56/124




)

Area of this curve Transverse moment

of volume

x'

169.38

25 50

384.9

57.26

0After calculation of each transverse moment of sectional area about the z ' axis( M A,y' ), transverse moment of displaced volume can be calculated byintegral of transverse moment of section area over the length of ship

, A y M

-1 : Transverse moment of displaced volume

50

, ,012,455 y A y M M dx = =, A yM : Transverse moment of area about z ' axis

, A z M

: Vertical moment of area about y ' axis

1 , A y M

2 , A y M

3 , A y M

, ' , ' ' y A y M M dx=

O y z

x

A3

A2

A1

20

CL

y

z

20

10

O y

x

10

10

20 20

10


y dy dz dx = , y M .

.( ( ))

F P

c A P y A x dx =

,c y=

(Body fixed frame)



56/124


O y z


57/124




)-2 : Vertical moment of displaced volume

, A z M

Area of this curve Vertical moment

of volume

After calculation of each vertical moment of sectional area about the yaxis( M A,z), vertical moment of displaced volume can be calculated byintegral of vertical moment of section area over the length of ship

50

, ,030,794 z A z M M dx = =

x'

-148.76

-888.89

-533.42

25 50

1 , A z M

2 , A z M

3 , A z M

, ' , ' ' z A z M M dx=

O y z

x

A3

A2

A1

20

CL

y

z

20

10

O y

x

10

10

20 20

10


, z M z dy dz dx = .

.( ( ))

F P

c A P z A x dx =

,c z=

(Body fixed frame)

0



, A yM

: Transverse moment of area about z ' axis

, A z M

: Vertical moment of area about y ' axis

57/124

Example) Calculation of Center of Buoyancy of Shiph h O y

z


58/124




)Center of buoyancy (Body fixed frame)

Center of buoyancy can be calculated if we divide transverse, vertical moment of displaced volume by displaced volume.

( )

, ' , ', ,( , ) ,

12, 455 30,794, 1.71, 4.21

7,304 7,304

y zc c

M M y z TCB VCB

= = = = =

20

CL

y

z

20

10

O y

x

10

10

20 20

10


y dy dz dx

dx dy dz

=

,c

y , y M =

,c z , z M

=

z dy dz dx

dx dy dz

= Center of buoyancy with respect to waterplane fixed frame have

to be calculated. Rotational transformation



58/124

Example) Calculation of Center of Buoyancy of Shipi h V i S i Sh O y

z


59/124




)

20

CL

y

z

20

10

O y

x

10

10

20 20

10


Center of buoyancy (Waterplane fixed frame)

,

,

cos sin 1.71sin cos 4.21

cos(30) sin(30) 1.71 0.63sin(30) cos(30) 4.21 4.50

c

c

y

z

=

= =

, ,

, ,

cos sinsin cos

c c

c c

y y

z z

=

Center of buoyancy with respect to waterplane fixed frame have to be calculated. Rotational transformation



59/124


60/124



N a v a

l A r c

h i t e c t u r e

& O c e a n

E n g

i n e e r i n g


Ch.6 Transverse Righting Moment- Sec.2 Calculating BM, GZ in Wall Sided Ship -

2009

Prof. Kyu-Yeul Lee


- Ship Stability -



61/124



Sec.1 Calculation of Center of BuoyancySec.2 Calculation of BM, GZ in Wall Sided ShipSec.3 Inclining TestSec.4 Transverse Stability of ship (Unstable condition)Sec.5 Transverse Righting Moment due to Movement of CargoSec.6 Calculation of Heeling Angle due to Shift of Center of Mass

61/124

Transverse Righting Moment

Righting Moment : Moment toreturn the ship to the upright floating


62/124



Transverse Righting Moment

Z

K

z

O,O'

y

z

)(+

CL

y

z M

righting

e

G

FG

B B1

N

FB

1 B y

G y



return the ship to the upright floatingposition (Righting moment, Momentof statical stability))

BGZ F = i Transverse Righting moment


Righting arm

Righting Arm ( GZ )

1G BGZ y y= +

From direct calculation

We should know yG , y B1 in waterplane fixed frame


sinGZ GM =

GM is related to below equation bygeometrical figure

GM KB BM KG= +

x,x'

G: Center of mass B: Center of buoyancy B1: Changed center of buoyancyK : Keel F G : Weight of ship F B : Buoyant force acting on ship

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

M : The intersection of the line of buoyant force through B1 with the centerline of the ship

BM KB : Vertical center

of buoyancy BM : Transverse Metacenter Radius

KG : Vertical center of mass of shipis determined by the shape of ship

,KB BM KG is determined by loading condition of cargo

restoring BGZ F = sinGZ GM =

GM KB BM KG= + BM

62/124

Calculation of BM (1)(BM T M i R di )


63/124


SDAL@Advanced Ship Design Automation Lab.http://asdal.snu.ac.kr Seoul National Univ.2009 Fall, Ship Stability - Transverse Righting Moment M : The intersection of the line of buoyant force through B1 with the centerline of the ship B2: The intersection of the line of buoyant force through B1 with the transverse line through B

M

Z

B

G

z

y

z

K CL

B1=

== yW 2

W 1

L2

L1

g1g

B2

restoring

F B

(BM : Transverse Metacentric Radius)

yv,c


B1: Changed center of buoyancy

y

z( )+ j

k

Z : The intersection of the line of buoyant forcethrough B1 with the transverse line throughG

G: Center of mass B: Center of buoyancyF G : Weight of ship (=W )F B : Buoyancy (= g )

Wall sided ship: When a ship is in upright position, aship which have perpendicular side shellto waterplane is called wall sided ship .

Assumption

1. Wall sided ship Submerged volume is same with

emerged volume when the ship is

heeled. ( A ship is heel without changeof displacement volume )

2. A main deck is not flooded.

3. Center of rotation is not changed.

( M is not changed)

F G

Lets derive BM in case of simple section likea wall sided ship.

O,O' x,x'

63/124

Calculation of BM (2)(BM T M i R di )


64/124



M

Z

B

G

z

y

z

K CL

B1=

== yW 2

W 1

L2

L1

g1g

B2

restoring

F B


yv,c

Relation between moving distance of centerof changed displacement volume andmoving distance of center of center of buoyancy is as follows.

y

z( )+ j

k


1 1g v

BB ggg

=

1 12v

BB Og=

The shape of displacement volume is

changed as a ship is heeled.

F G

: Displacement volumev : Changed displacement volume

BB1: Moving distance of center of buoyancy

gg 1 : Moving distance of center of changed displacement volume

1 1g BB g v gg =

1 1( 2 )gg Og=

M : The intersection of the line of buoyant force through B1 with the centerline of the ship B2: The intersection of the line of buoyant force through B1 with the transverse line through B


B1: Changed center of buoyancy

Z : The intersection of the line of buoyant forcethrough B1 with the transverse line throughG

G: Center of mass B: Center of buoyancyF G : Weight of ship (=W )F B : Buoyancy (= g )

O,O' x,x'

64/124


65/124

dz dy dx = dV = ( ) A x dA dy dz = = 2 ,2 v c BB v y=

Calculation of BM (4)(BM T M t t i R di )


66/124



(R.H.S)


,2

v cv y

: y coordinate of center of changed displacement volume

: Changed displacement volume

: Displacement volume

,v c yv

.

.( )

F P

A P A x dx =

, ,

, ,

cos sinsin cos

v c v c

v c v c

y y

z z

=

, , ,cos sinv c v c v c y y z =

Represent center of buoyancy with respect to waterplanefixed frame as one with respect to body fixed frame., ,( , )v c v c y z

, ,co si2

s n( )v c v c y zv =

, ,

2 2cos sinv c v cv y v z =

Transverse moment of volume withrespect to body fixed frame.

Vertical moment of volume withrespect to body fixed frame.

y

yg1

L1

L2

z z

,cos

v c y

, sinv c z ,v c y


O,O'

x, x'

66/124

dz dy dx = dV = ( ) A x dA dy dz = =

Calculation of BM (5)(BM T M t t i R di )

2 ,2 v c BB v y=


67/124



(R.H.S)

,2

v cv y

.

.( )

F P

A P A x dx =

, ,

2 2cos sinv c v cv y v z =

Transverse moment of volumewith respect to body fixed frame.

Vertical moment of volume withrespect to body fixed frame.

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06 Transverse Righting Moment

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