stability basic coarse chapter 3 a-density b-relative density

STABILITY

BASIC COARSE

CHAPTER 3

A-DENSITYB-RELATIVE DENSITY

A-DENSITY• DENSITY : DEFINDED AS THE MASS PER UNIT VOLUME MEASURED IN

KG/M3 OR TON/M3 .

MASS IN KG OR TONS

DENSITY = ---------------------------------- kg/m3 or T/m3 VOLUME IN M3

VOLUME = L * B * D (LENGTH * BREADTH * DEPTH )

B-RELATIVE DENSITY

RELATIVE DENSITY: DEFFINED AS THE RATIO BETWEEN THE DENSITY OF

ANY LIQUID TO THE DENSITY OF FRESH WATER.

R.D = DENSITY OF ANY LIQUID

DENSITY OF FRESH WATER

DENSITY OF FRESH WATER = 1000 KG/M3 OR 1.000 T/M3 DENSITY OF SALT WATER = 1025 KG/M3 OR 1.025 T/M3

CHAPTER 4

LAW OF FLOATATION

LAW OF FLOATATION• LAW OF FLOATATION THE MASS OF ANY SUBSTANCE IS EQUAL TO THE

MASS OF THE WATER THE SUBSTANCE DISPLACES.MASS OF SUBSTANCE = MASS OF WATER DISPLACED

AS THE SHIP MASS = DENSITY OF SHIP * SHIPS . VOLUME ( L * B * DEPTH)AND

AS THE WATER MASS = DENSITY OF THE WATER * WATER VOLUME DISPLACED BY THE PART UNDER WATER

( L * B * DRAFT )SO SHIPS MASS = WATER DISPLACED MASS DENSITY OF SHIP * DEPTH = DENSITY OF WATER * DRAFT

LAW OF FLOATATION• THE WEIGHT OF ANY SHAPE IS ACTING ONLY AT A CERTAIN POINT WHICH IS

CALLED CENTRE OF GRAVITY CENTRE OF GRAVITY : IS DEFINED AS A POINT WHERE THE SHIPS WEIGHT IS CONCENTRATED , THIS

FORCE IS ACTING DOWNWARD & THE POINT ALWAYS LIES AT ½ THE DEPTH OF THE SHAPE

KG = ½ DEPTH EXAMPLE DEPTH = 4m SO KG = 2m

DEPTH

W

G ₀

LAW OF FLOATATION• THE CENTRE OF BOUYANCY• IS DEFINED AS A POINT WHERE THE SHIP’S BOUYANCY IS CONCENTRATED,

THIS FORCE IS ACTING UPWARD ,AND ALWAYS CENTERED AT ½ THE DRAFT . KB = ½ DRAFT ,e.g; DRAFT = 4m , SO KB = 2m

B’

W L

B

DRAFT ₀

LAW OF FLOATATIONW

KG = ½ DEPTH

DEPTH

• G• B

DRAFT K KB = ½ DRAFT

B

LAW OF FLOATATIONKG DEFINED AS THE HEIGHT THAT LIES BETWEEN THE KEEL & THE CENTRE of

GRAVITY.

KB DEFINED AS THE HEIGHT THAT LIES BETWEEN THE KEEL & THE CENTRE OF

BOUYANCY. REMARK ( B FORCE , G FORCE ) BOTH FORCES ACTS AGAINEST EACH OTHER S , IF THE G FORCE

INCREASED OVER THE B FORCE THE SHIP STARTS TO GO DOWN ;INCREASING THE SHIPS DRAFT BY THE DIFFRENCE IN FORCES .

RESERVE BOUYANCY

DEFINED AS THE SPACE THAT LIES BETWEEN THE WATER SURFACE AND THE FIRST WATER TIGHT INTEGRITY ( MAIN DECK).

Volume under water Area under water

Reserve bouyancy

draft

depth

RESERVE BOUYANCY = DEPTH - DRAFTORRESERVE BOUYANCY = VOLUME OF SHIP - VOLUME UNDER WATER ORRESERVE BOUYANCY = AREA OF THE SHIP - AREA UNDER WATER

EFFECT OF DENSITY ON SHIP’S VOLUME & DISPLACEMENT

A- BOX SHAPE VESSELS B- SHIP SHAPE VESSELS

CHAPTER 5

A-BOX SHAPED VESSELES

1-EFFECT OF DENSITY ON SHIP’S VOLUME 2-EFFECT OF DENSITY ON SHIP’DISPLACEMENT


• ANY BOX SHAPED VESSEL SAILS FROM ONE PORT TO ANOTHER CERTAIN CHANGES OCCURES OVER THE SHIP, AS A RESULT OF THE EFFECT OF DENSITY ON SHIP’S VOLUME & DISPLACEMENT

AS WE KNOW THAT THE

A RELATION BETWEEN THE DENSITY & MASS WOULD BE ; DIRECT PROPORTION

DENSITY ∞ MASS ( DIRECT PROPORTION ) WHICH MEANS THAT

WHEN DENSITY DECREASES THE MASS DECREASES WHEN DENSITY INCREASES THE MASS INCREASES

DENSITY = MASS kg VOLUME


• A RELATION BETWEEN THE DENSITY & VOLUME WOULD BE ; INV. PROPORTION

DENSITY 1 / ∞ VOLUME ( INV. PROPORTION ) WHICH MEANS THAT

WHEN DENSITY DECREASES THE VOLUME INCREASES WHEN DENSITY INCREASES THE VOLUME DECREASES

THE VOLUME IS THE SUM OF L * B * DRAFT , THE L & B NEVER CHANGE FROM PORT TO ANOTHER SO THE ONLY

PARAMETER THAT CHANGES IS THE DRAFT ,THERFORE THE VOLUME CHANGES ASWELL

A-BOX SHAPED SHIPS

1-EFFECT OF DENSITY ON VOLUME

EFFECT OF DENSITY ON VOLUME

• LETS SAY A BOX SHAPED VESSEL DISPLACES 20,000 TONS SAILED FROM PORT A HAS WATER DENSITY 1.OOO

TO PORT B HAS WATER DENSITY 1.025 , ACCORDING TO THE RELATION BETWEEN DENSITY AND VOLUME

“INV.PROPORTIONS” , WE DISCOVERS THAT AT PORT B, THE VOLUME WILL DECREASES AS THE WATER DENSITY INCREASES ( 1.000 PORT A TO 1.025 PORT B ) ,

WHILE THE SHIP STILL DISPLACES THE SAME 20,000TONS

SINCE THE VOLUME = L * B * DRAFT , SO THE CHANGE IN THE VOLUME COMES FROM THE CHANGE IN THE

DRAFT

EFFECT OF DENSITY ON VOLUME

SHIP’S MASS AT PORT A = SHIP’S MASS AT PORT BWHERE THE MASS = DENSITY * VOLUME

( OLD DENSITY * OLD DRAFT ) = ( NEW DENSITY * NEW DRAFT )

A-BOX SHAPED SHIPS

2-EFFECT OF DENSITY ON DISPLACEMENT

EFFECT OF DENSITY ON DISPLACEMENT

• A BOX SHAPED VESSEL DISPLACES 20,000 TONS SAILED FROM PORT A OF WATER DENSITY 1.OOO & DRAFT 7.0 mtrs

TO PORT B OF WATER DENSITY 1.025 , AS SHE ARRIVED TO PORT B , THE SHIP’S DRAFT STAYED THE SAME 7.0 mtrs. DESPITE THE DENSITY IS ALREADY CHANGED FROM 1.000 TO 1.025 , THAT MEANS A CHANGE OCCURRED ON THE SHIPS DISPLACEMENT (MASS) YOU WILL FIND THE SHIP DISPLACEMENT BECAME 21,000 TONS AS EXAMPLE.

THE RELATION BETWEEN DENSITY & DISPLACEMENT (MASS) IS DIRECT PROPORTIONS ,AS A RESULT THE DISPLACEMENT INCREASED WHEN DENSITY INCREASED ( 1.000 TO 1.025)

EFFECT OF DENSITY ON DISPLACEMENT

• SHIP’S VOLUME AT PORT A = SHIP’S VOLUME AT PORT B

THE SHIP DISPLACES THE SAME VOLUME OF WATER IN BOTH PORTS A & BWHERE THE VOLUME =

OLD MASS NEW MASS

------------------------- = ----------------------

OLD DENSITY NEW DENSITY

B- SHIP SHAPED VESSELS


EFFECT OF DENSITY ON VOLUME & DISPLACEMENT

• INORDER TO UNDER STAND THE EFFECT WE SHOULD VERY WELL UNDERSTAND THE PLYMSOL MARK ( DRAFT MEASURES)

FREE BOARD

(RESERVE BOUYANCY )

54

WNAWinter

SummerFWA

Fresh

Tropical F

Tropical

230mm

300mm

540mm


• FWA ( FRESH WATER ALLOWANCE ) DEFINED AS THE NUMBER OF MM THAT INCREASES OR DECREASES IN SHIPS MEAN

DRAFT WHEN THE SHIP SAILS FROM SALT WATER TO FRESH WATER & VISE VERSA

• T P C ( TONS PER CENTIMETRE) DEFINED AS THE NUMBER OF TONS LOADED OR DISCHARGED INORDER TO CHANGE

SHIPS DRAFT 1 CM IN SALT WATER

FWA = DISPLACEMENT 4 * TPC


• DWA (DOCK WATER ALLOWANCE) DEFINED AS THE NUMBER OF MM THAT INCREASES OR DECREASES IN SHIPS MEAN

DRAFT WHEN THE SHIP SAILS FROM SALT WATER TO DOCK WATER & VISE VERSA.

Example : FWA 200mm (0.2mtrs) , DW DENSITY = 1.015 SO DWA = 0.2 * ( 10 ) = 0.08 mtrs ( 80 mm )

25

(1.025 - DWD)DW A = FWA ----------------------

25


• IF THE SHIP SAILS FROM PORT A WHOSE WATER DENSITY IS 1.000 TO PORT B WHOSE WATER DENSITY IS 1.025 ( THE DENSITY INCREASED) , SO ACCORDING TO THE RELATION BETWEEN DENSITY & VOLUME.

DENSITY 1 / ∞ VOLUME ( INV. PROPORTION ) WHICH MEANS THAT WHEN DENSITY DECREASES THE VOLUME INCREASES WHEN DENSITY INCREASES THE VOLUME DECREASES THE SHIPS DRAFT WILL DECREASES , THE VALUE OF DRAFT DECREASING EQUALS

THE FWA.

Eg. SHIP SHAPE V/L SAILED FROM PORT A WITH DENSITY 1.000 TO PORT B WITH DENSITY 1.025 FWA 200MM .OLD DRAFT 7.0mtrs so the new draft will decrease to 7.0 mt - FWA 200MM ( 20CM, 0.2mt )

7 - 0.2 = 6.8 mt ( NEW DRAFT )


• EXAMPLE SHIP SHAPE V/L SAILED FROM PORT A WITH DENSITY 1.025 TO PORT B WITH

DENSITY 1.015 FWA 200MM .OLD DRAFT 7.0mtrs , DWA 200MM ,

SO THE NEW DRAFT WILL INCREASE “ACCORDING TO THE INV.

RELATION “ BY THE VALUE OF THE DWA ( FROM SALT WATER DENSITY TO DOCK WATER DENSITY ) ,

OLD DRAFT + DWA = NEW DRAFT 7.0 + 200mm( 0.2mtrs) = 7.2mtrs

STATIC STABILITY

CHAPTER 6

STATIC STABILITY• HEELING , IS THE ANGLE OCCURES WHEN IN THE SHIP WHEN HEELS TO ONE SIDE DUE TO

EXTERNAL FORCES (WIND,WAVES)• LIST, IS THE ANGLE OCCURES IN THE SHIP WHEN HEELS TO ONE SIDE DUE TO

INTERNAL FORCES , LIST PORTSIDE OR LIST STRB SIDE. ( BALLAST,CARGO)• TRIM, IS THE DIFFRENCE BETWEEN THE FORWARD DRAFT & THE AFT DRAFT. TRIM COULD BE BY FORE ( FORWARD DRAFT LARGER THAN AFT DRAFT) 10 M FORE - 8.0 M AFT = 2.0 M BY FORE ( TRIM ) TRIM COULD BE BY AFT ( AFT DRAFT LARGER THAN FORE DRAFT)

10 M FORE - 15 M AFT = 5.0 M BY AFT ( TRIM )

STATIC STABILITY

K

G

K

M

G

M

K

B B B

STATIC STABILITY• KM = KG + GM• KM = KB + BM• KG = KB + BG• KG = KM - GM• GM = KM - KG

• KB = ½ DRAFT , KG = ½ DEPTH

CENTRE OF BOUYANCY ALWAYS MOVES TO THE HEELED SIDE TO BE CENTERED IN ½ THE UNDER

WATER VOLUME

KB = ½ DRAFT , KG = ½ DEPTHKB = ½ DRAFT , KG = ½ DEPTH

STATIC STABILITY• KG DEFINED AS THE HEIGHT BETWEEN THE KEEL & CENTRE OF GRAVITY• KM DEFINED AS THE HEIGHT BETWEEN THE KEEL & METACENTRE .THE HEIGHT

OF METACENTRE• GM DEFINED AS THE HEIGHT BETWEEN CENTRE OF GRAVITY & METACENTRE .

CALLED ( METACENTRIC HEIGHT)• GM COULD BE +VE ( G BELOW M ) STABLE SHIP GM COULD BE -VE ( G ABOVE M ) UNSTABLE SHIP

G

M

M

+ VEGM -VE GM

GW L•

•

•

•

STATIC STABILITY

• METACENTRE POINT DEFINED AS THE POINT THAT EXISTS WHEN THE SHIP HEELS OR LISTS TO A SIDE ,

THIS POINT OCCURS WHEN THE LINE OF BOUYANCY THAT ACTS UPWARD INTERSECT WITH THE CENTRE LINE.

B

M

B’

K

WL

G

B

W

•

STATIC STABILITYEQUILIBRIUM

• STABLE SHIP STABLE SHIP MEANS THAT THE SHIP HAS A +VE GM . AND WHEN HEELS OR LISTS

A RIGHTING LEVER APPEARS , THE LEVER HAS A MOMENT TO RIGHTEN THE SHIP & BRINGS HER BACK TO THE UPRIGHT CONDOTION . THE STATICAL RIGHTENING MOMENT IS THE SUM OF THE RIGHTENIG LEVER & THE SHIPS DISPLACEMENT.

THE RIGHTENING LEVER IS REPRESENTED BY GZ. THE GZ THAT APPEARS , STARTS FROM THE G POINT TO THE LINE OF

BOUANCY MAKING A RIGHT ANGLE.

STATICAL RIGHTENIG MOMENT = RIGHTENING LEVER * DISPLACEMENT RM ( TON METER) = GZ (mtrs) * ∆ ( tons )

STATIC STABILITYSTABLE SHIP

• STABLE SHIP B W

w

k

B

G

M

B

W

B

B B’

G

M

K

Z

G

•

•

•

•

•

•

STATICAL RIGHTENING MOMENT = GZ * DISPLACEMENTA COUPLING IS SET TO BRING THE SHIP BACK TO UP RIGHT CONDOTION

STATIC STABILITYUNSTABLE SHIP

• UNSTABLE SHIP MEANS THAT THE SHIP HAS A -VE GM ,THERFORE A CAPSIZING LEVER WILL APPEARS ,WITH THE SHIP’S DISPLACEMENT A CAPSIZING MOMENT OCCURES; WHICH HEELS THE SHIP EVEN MORE TO THE HEELED OR THE LISTED SIDE.

STATICAL CAPSIZING MOMENT = - GZ * DISPLACEMENT - RM = - GZ * ∆

STATIC STABILITYUNSTABLE SHIP

• UNSTABLE SHIP

W

K

B

M

G

B

W

K

B B’

M

GZ

B

B

W

GZ

•

•

•

•

•

••

STATICAL CAPSIZING MOMENT = - GZ * DISPLACEMENTA COUPLING IS SET & INCREASES THE SHIPS HEEL OR LIST

STATIC STABILITYNEUTRAL SHIP

• NEUTRAL SHIP DEFINED AS A SHIP HAS HER G POINT COINSIDE WITH THE M POINT AS A RESULT NO LEVER APPEARS THERFORE NO MOMENT OCCURS ,& NO COUPLING ARISES .THE SHIP STAYES HEELED . UNABLE TO BE UPRIGHT.

THE

K

B

M G

B

W

B B’

K

G M

W

BB

W

•

• •

STATIC STABILITYTENDER & STIFF SHIPS

• TENDER SHIP A SHIP SAID TO BE TENDER WHEN SHEHAS A SMALL GM , WHEN SHE HEELS GZ SMALLCONSEQUNTLY STATICAL RIGHTENING MOMENT IS ALSO SMALL.THERFORE PERIOD OF ROLLING IS LONG EXAMPLE : PASSENGER SHIPS , CARGO SHIPS

K

G

M

STATIC STABILITYTENDER & STIFF SHIPS

• STIFF SHIPA SHIP SAID TO BE STIFF WHEN SHEHAS A LARGE GM , WHEN SHE HEELS GZ LARGECONSEQUNTLY STATICAL RIGHTENING MOMENT IS ALSO LARGE.THERFORE PERIODE OF ROLLING IS SHORT EXAMPLE : WAR SHIPS

K

G

M

STATIC STABILITYANGLE OF LOLL

• ANGLE OF LOLL

THE ANGLE THAT APPEARS WHEN THE SHIP HEELS TO A SIDE WHILE THE SHIP HAS A –VE GM . A CAPSIZING MOMENT CREATED INCREASES THE HEELING ,

BY THAT TIME THE CENTRE OF BOUYANCY B STARTS TO MOVE TO THE HEELED SIDE UNTILL B REACHES A POINT JUST BELOW THE LINE OF GRAVITY. THE ANGLE WHERE THAT HAPPENS IS CALLED ANGLE OF LOLL .

WE NOTICE THAT THE SHIP AT THE ANGLE OF LOLL , HAS NO GZ, NO GM, NO MOMENT AT ALL.AS A RESULT THE SHIP STAYES ON THIS CONDITION ( HEELED)


IF THE SHIP HEELED MORE CAUSE OF ANY REASON (WIND), THE CENTRE OF BOUYANCY B MOVES FAR FURTHER AWAY IN THE HEELED SIDE, AS A RESULT B IS NO MORE ACTING BELOW THE SAME LINE OF GRAVITY, AND

A RIGHTNING MOMENT CREATED TO BRING BACK THE SHIP NOT TO THE UPRIGHT CONDITION BUT TO THE ANGLE OF LOLL AGAIN. THE SHIP KEEPPS ROLLING AROUND THE ANGLE OF LOLL ,TILL THE PROBLEM IS SOLVED.


M

GZ

BB’

K

BB’

M G

B B’

G Z

M

B

W

B

W

B

W

CAPSIZING MOMENT

WINDWIND

WIND

RIGHTENING MOMENT

Fig.1 Fig.2

Fig. 3

LOLL

•

••

•

•

•

•

••

• •

•

STATIC STABILITYCORRECTING ANGLE OF LOLL

INORDER TO CORRECT < OF LOLL WE MUST LOWER THE G BELOW M , PUTTING INTO CONSIDERATION THE SEQUENCE.

1. FILLING THE ½ FULL BALLAST TANKS ( TO REMOVE FREE SURFACE)2. LOWERING DOWN ANY UPPER LOADS ( CRANES , TOPSIDES TODOUBLE

BOTTOM TANKS)3. FILLING THE D.B TANKS IN THE HEELED SIDE 4. THEN FILL THE D.B TANKS IN THE OTHER SIDE TO THE HEELED SIDE & THAT

SHOULD BE GRADUALLY.WHY THE HEELED SIDE FIREST ?AS FILLING THE TANKS IN THE HEELED SIDE THE G WILL MOVE UP SLOWLY

&INCREASING LOLL ANGLE ;DUE TO FREE SURFACS ,BUT EVENTUALLY AFTER A WHILE THE G STARTS TO MOVE DOWN ,ANGLE OF LOLL STARTS TO BE REDUCED GRADUALLY ,UNTILL IT DISAPPEARS . G RETURNS BELOW M TO THE + VE CONDITION CREATING A RIGHTENING MOMENT, MAKES THE SHIP BACK TO THE UPRIGHT CONDITION.

STATIC STABILITYCORRECTING ANGLE OF LOLL

• IF WE STARTS FILLING D.B TANKS IN THE HIGH SIDE , THE TANKS GETS FILLED GRADUALLY ,AND OFCOARSE FREE SURFACE WILL MAKES THE G MOVES MORE UP ,INCREASING THE HEEL;& ANGLE OF LOLL ; EVENTUALLY THE FREE SURFACE EFFECT STARTS TO DISAPPEAR & THE SHIP STARTS TO BE ADJUSTED & RETURNS TO THE UPRIGHT CONDITION CAUSE THE G STARTS TO MOVE DOWN ,ANGLE OF LOLL DECREASES GRADUALLY , & THEN DISAPPEARS , & G TURNS TO BE BELOW THE M (+VE GM),A RIGHTENING MOMENT IS CREATED BUT VERY STRONG ONE.

• UNFORTUNATLY ,THE GZ CREATED IS VERY LARGE , THE RETURN WILL BE VERY SEVERE ,STIFF AND IN A MATTER OF SECONDS; & LEADS TO A VERY DANGEROUS SITUATION TO THE SHIP.

FINAL KG

CHAPTER

FINAL KG• ANY SHIP DURING LOADING / DISCHARGING CARGO; THE CENTRE OF GRAVITY G

STARTS TO MOVE EITHER TOWARD OR AWAY FROM THE CENTRE OF GRAVITY g OF THE WEIGHTS LOADED / DISCHARGED .

• As WE SEE(fig.1) G MOVED TO G’ RELATED TO g of the weight• As WE SEE(fig.2) G MOVED TO G’ RELATED TO g of the weight

»

K K

G G

G’

g

g

G’

Fig. 1 Fig.2

FINAL KG• ACCORDING TO THE ILLUSTRATION , WE DISCOVER THAT THE G OF THE SHIP

KEEPS MOVING UP AND DOWN WITH THE g OF THE WEIGHTS LOADED /DISCHARGED ,UNTILL IT IS SET IN A FINAL POSITION AFTER FINISHING THE LOADING/DISCHARGING PROCESS.

• SO ,WE HAVE AN INITIAL KG , ENDS UP BY FINAL KG .

• THE FINAL KG LEADS TO THE FINAL GM.

FINAL GM = KM - FINAL KGFINAL GM = KM - FINAL KG

FINAL KG• INORDER TO GET THE FINAL KG , EVERY WEIGHT HAS ITS Kg , THE G MOVES BY THE EFFECT

OF THE MOMENT OCCURRED FROM THE Kg & w ,TILL G STOPS AT A FINAL POSITION ( KG )

FINAL KG’ = TOTAL MOMENT 2000 = FINAL KG’ TOTAL W 300

IF THE SHIP’S KM = 8 m so the final G’.M = KM - FINAL KG’ 8 - 6.6 = final G’M

w/tons Kg/m MOMENT/ ton m

100 10 1000

200 5.0 1000

Total w Total M

300 2000

6.6m

1.4m

FINAL KG

• GG’IS THE MOVE OF G TO G’ DURING LOAD/DISCH

LEADING TO THE FINAL KG, & FINAL GM

K

100 T

g

k

10m (kg)

200 T

g

k

5m (kg)

G

G’

Initial KG

FINAL KG

MFinal G’M

INITIAL GM

GZ CURVES

CHAPTER

GZ CURVES• GZ IS THE LEVER THAT OCCURES WHEN THE SHIP HEELS ,THE GZ LEVER IS

RESPONSIBLE FOR RETURNING THE SHIP BACK TO THE UP RIGHT CONDITION.• THE LENGTH OF GZ LEVER DEPENDS ON TWO PARAMETERS ,

GM & ANGLE OF HEEL.Ѳ

Ѳ heel

GZ = GM * SIN Ѳ

B

M

K

G

B’

Z

G Z

M B’

W

•

•

•

GZ CURVESGM

• AS THE Ѳ INCREASES , GZ INCREASE TILL REACHES THE MAX THEN DROP DOWN AGAIN TO REACH THE VANISHING ANGLE.

• THE RED LINE CALLED ARCHI . LINE ,FROM THIS LINE WE GET THE INITIAL GM OF THE SHIP. FROM Ѳ 57.3 ⁰ EXTEND UP A LINE TO CUT THE ARCHI .LINE AT A POINT. FROM THIS POINT WE EXTEND A HORIZONTAL LINE TO READ THE GM, ON THE GZ SCALE .THE ARCHI LINE DRAWN AS A TANGENT FROM 0 AND SLOPE OF THE CURVE AS SHOWN BELOW.

3.9m

57.3

Vanishing angle 91 ⁰

Max GZѲ 40⁰Max GZ

ARCHI LINE GZ

10 20 30 40 50 60 70 80 90

GM 1.1 m

4

3

2

1

0

GZ CURVESSTABLE SHIP

• MAX GZ = 4.0 m AT Ѳ 39.0⁰ RANGE OF STABILITY = 0—90 ⁰• INITIAL GM = 1.3 m AT Ѳ 57.3⁰ VANISHING ANGLE = 90⁰ GZ

GM GM

57,3

STABLE SHIP +VE GZ

10 20 30 40 50 60 70 80 90

4

1

2

0

3

1.3

GZ CURVESSTATICAL MOMENT

• IF THE SHIP DISPLACEMENT = 5000T THE MOMENT AT 25⁰ WOULD BE • GZ * W = MOMENT 3.0 * 5000 = 15000 Tm ( at 25⁰ ) GZ 4 3 2 GM 1

57,3 25 10 20 30 40 50 60 70 80 90

GZ CURVESUNSTABLE SHIP

GZ RANGE OF STABILITY 17 ⁰--- 83⁰ Ѳ LOLL 17⁰

MAX GZ 3.8m at Ѳ 43⁰ VANISHING Ѳ 83⁰ MAX GZ AT 43⁰

Ѳ LOLL 17⁰

43⁰

UNSTABLE SHIP –VE GZ CURVE

83⁰

RANGE OF UNSTABILITY 0⁰ --- 17⁰

< LOLL

GZ

10 20 30 40 50 60 70 80 90

0

-1

-2

1

2

3

4.0

GZ CURVESUNSTABLE SHIP

4_ 3_

2_ 1_ 0 | | | | | | | | | | -1

UNSTABLE SHIP -VE GZ

57.3

-2

-3

Ѳ LOLL 22⁰

GM – 3m

RANGE OF UNSTABILITY 0⁰--- 22⁰RANGE OF STABILITY 22⁰ -- 92⁰INITIAL GM - 3 m

GZ

10 20 30 40 50 60 70 80 90 100

FREE SURFACE

CHAPTER 7

FREE SURFACE

• FREE SURFACE IS DEFINED AS THE SURFACE THAT CAN MOVE FREELY FROM ONE SIDE TO

ANOTHER FREELY , EXAMPLE A TANK ½ FULL OF BALLAST .

THE FREE SURFACE HAS A NEGATIVE EFFECT OVER THE SHIP’S STABLE CONDITION, MORE CLEARLY THE FREE SURFACE LEADS TO LOSS IN THE G M , WHICH MEANS

THAT IT COULD REDUCES THE GM TO THE EXTENT OF CONVERTING THE +VE GM

TO -VE GM ( STABLE SHIP TO UNSTABLE SHIP ),SPECIALLY IF THE SHIP STARTED HER VOYAGE WITH A SMALL INITIAL G.M , AS A RESULT THE SHIP CAN EASILY CAPSIZE & SINKS.

FREE SURFACE• THE FREE SURFACE REDUCES THE SHIP RIGHTENING MOMENT BY REDUCING THE

GZ LEVER, THE LEVER WHICH USED TO BRING THE SHIP BACK TO THE UPRIGHT CONDITION .

• , THE FREE SURFACE MAKES AN EXTRA CAPSIZING MOMENT OVER THE SHIP, AS A RESULT OF THE EXTRA WEIGHT ADDED FROM THE LIQUID IN THE ½ FULL

TANK IN THE HEELED SIDE. g moved to g1 ALSO // G MOVED TO G’AS LIQUIDE HEELEDG’Z < GZ NEW MOMENY< OLD MOMENTNEW G1M < OLD GM GG1 = LOSS IN GM

M

G1Z1

GZ

BB’g

g

G’

FREE SURFACE

• CONSEQUENTLY IT IS OBVIOUS THAT THE EFFECT OF THE FREE SURFACE ON THE SHIP’S STABILITY IS SIMMILLAR AS SHIFTING A LOAD VERTICALLY UP.

THE RIGHTENING MOMENT IS AFFECTED FROM THE FREE SURFACE ,AS THE G MOVES HORIZONTALLY TO G’ & PARALLEL TO g g1 , THAT MEANS THE GZ WILL BE REDUCED TO G’Z AND CONSEQUENTLY THE RIGHTENING MOMENT WILL ALSO BE REDUCED . RM = GZ * W

IN PRESENCE OF FREE SURFACE ,THE EFFECT RM = G’Z *W

• AS THE G ALSO MOVES UP VERTICALLY TO G1 , GM REDUCED BY THE VALUE OF THE MOVE OF G TO G1 & THAT IS CALLED THE LOSS IN GM (LOSS IN STABILITY) , THE NEW IS G1M

FREE SURFACE

• SUMMARY1. FREE SURFACE COMES FROM ½ FULL TANKS2. FREE SURFACE LEADS TO LOSS IN SHIPS STABILITY (LOSS IN GM)3. FREE SURFACE REDUCES THE SHIPS RIGHTENING MOMENT4. FREE SURFACE REDUCES THE GZ5. FREE SURFACE EFFECT ON SHIPS STABILITY IS EQUIVILANT TO THE EFFECT OF

SHIFTING A LOAD VERTICALLY UPWARD .6. FREE SURFACE MAKES THE LIQUID IN TANK TO LEAN TO THE HEELED SIDE , & ADDS

AN EXTRA HEELING MOMENT(CAPSIZING) ,I.E” REDUCES THE RIGHTENING MOMENT “WHICH MAKES THE SHIP TO HEEL WITH A LARGER Ѳ

TRANSVERSE STABILITYLIST

CHAPTER


• LIST IS THE ANGLE THAT OCCURES WHEN THE SHIP LEAN TO EITHER SIDE

PORT OR STRB AS ARESULT OF THE EFFECT OF AN INTERNAL FORCE SUCH AS BALLAST TANKS , CARGO DISTRIBUTION / SHIFTING .

• DURING LOADING /DISCHARGING A SHIP, THE WEIGHTS ADDED/REMOVED FROM THE SHIPS SIDES LEADS TO LIST HER TO EITHER SIDE.

• THE LIST THAT OCCURES DEPENDS ON THE MOMENT THAT EXISTS FROM THE SUM OF WEIGHTS ADDED /REMOVED & THERE DISTANCE FROM THE CENTRE LINE.

LIST MOMENT = W * d ( distance from centre line)


• The IDEA IS EQUIVILANT FROM THE point of VIEW OF A SIMPLE BALANCE .

2OO

1001OO

3OO3OO

5O

d d

Fig .1•AS THE Fig . 1 SHOWS, EVERY WEIGHT IS FAR FROM THE CENTRE BY ‘d ‘ ,INORDER TO KNOW WHICH SIDE IS HEAVIER AND LEADS THE BALANCE TO LEAN ,WE SHOULD GET THE TOTAL MOMENT PORT & TOTAL MOMENT STRB , MOMENT = W * D


• The SHIP LIST IS VERY SIMILLAR TO THE LAST EXAMPLE CONCEPT.

STBPORT

d d

dd

d d

dd

d d

100 50

200

100

150

300

200150

50

300

SO ,EACH WEIGHT IN THE SHIP IS FAR FROM THE CENTRE LINE BY DISTANCE “d”The SHIP WILL LEAN TO ONE SIDE ACCORDING TO THE MOMENT OF EACH SIDE. MOMENT = W * D


• A DEEPER VIEW TOWARD THE EFFECT OVER THE SHIP’S STBILITY “GM”THE G MOVES TO THE WEIGHT g FINALLY THE SHIP’S GGETS OUT OF THE CENTRELINE TO THE SIDE WHICH HAS THE BIGGER MOMENT;AS A RESULT THE SHIP LEANSTO THAT SIDE, & STOPS WHEN THE B’COMES JUST UNDER THE G’ ,AND ACTS ON THE SAME LINE OF WORK. SO THE SHIP’S G , SETTELED AT G’ , TAN Ѳ = GG ‘ GM

Ѳ IS THE LISTING ANGLE

K

G G’

M

Ѳ

BB’

W

B

G G’

MѲ


w D ( gg’)Distance from centre line

Momentport

MomentStrb

50 10 500

200 20 4000

150 10 1500

300 5 1500

100 5 500

100 10 1000

200 5 1000

150 10 1500

50 5 250

300 10 3000

1600 6750 8000

1600ton FINAL GG’ 1250 strb


• LISTING MOMENT = 1250 STRB• TOTAL WEIGHT = 1600 TON

• FINAL GG’ = TOTAL MOMENT 1250 = 0.781 mtrs.• TOTAL WEIGHT 1600• IF THE FINAL GM = 5.5 mtrs

TAN Ѳ = GG’ 0.781 = 8⁰ strb GM 5.50

G G’

M

0.781

5.5

8⁰

LONGITUDINAL STABILITYTRIM

CHAPTER


• TRIM IS THE DIFFERENCE BETWEEN THE AFT DRAFT & THE FORE DRAFT. TRIM COULD BE BY AFT OR BY FORE.

• IF THE FOR & AFT DRAFT WERE EQUAL & HAD NO DIFFERENCE ,THEN THE SHIP SAID TO BE ON AN EVEN KEEL.

LBP

фL1L2

LBP IS THE LENGTH BETWEEN PERPENDICULAR ф MIDSHIP L1 DISTANCE FROM AFT B. TO MID SHIP ,CF L2 DISTANCE FROM FORE B. TO MID SHIP,CF


• IF ANY LOADS ADDED OR REMOVED FROM THE SHIP ,THERE WILL BE AN EFFECT ON THE SHIPS DRAFTS & CONSEQUENTLY ON THE TRIM.

• THE LOADS WILL CHANGE THE DRAFTS AFT & FORE BY THE SAME VALUE,THAT ONLY HAPPENS IF THE CENTRE OF FLOATATION IS AMIDSHIP,IF NOT ,THE CHANGE WILL DEPEND ON THE CHANGE IN TRIM OCCURRED.& L1 ,L2 & L.

LBP

ф L1L2

DRAFTFORE

DRAFT AFTCF

L


• WHEN A LOAD IS ADDED ,THE G WILL MOVE TOWARD THE g of the weight,making THE SHIP TO LEAN FORWARD .THE SHIP STOPS LEANING FORWARD ONCE B MOVES & REACH JUST BELOW THE G’ , WHICH MEANS BOTH G ‘& B’ ACTS AGAIN ON THE SAME LINE OF WORK. THE FINAL GG’ ( DISTANCE BETWEEN G &G’) COULD BE CALCULATED FROM THE FINAL MOMENTS OF THE WEIGHTS & TOTAL WEIGHTS.

фW

GG’BB’

GML


• CENTRE OF FLOATATION IS THE CENTRE WHERE THE LINES OF WATER INTERSECTS . THE SHIP TRIM LONGITUDINALY AROUND THIS POINT. THE DRAFT AT THIS POINT IS CONSTANT.

LBP

ф

L1L2

CFNEW DRAFT AFTNEW

DRAFT FORE


• IF A LOAD IS ADDED AFT ,THE SHIPS DRAFT AFT WILL BE INCREASED WHILE THE SHIPS DRAFT FORE DECREASES, AS SHOWN IN THE fig. 1 BELOW. THE EFFECT OF THE WEIGHT OVER THE SHIP’S TRIM COMES FROM THE MOMENT IT MAKES.

• TRIMMING MOMENT IS THE MOMENT TO CHANGE THE SHIP’S TRIM ,& IT IS THE SUM OF THE W & DISTANCE OF W FROM CF.

• trimming moment = _w * d MEASURED IN TON METER W

LBP

ф

L1L2

CFNEW DRAFT AFTNEW

DRAFT FORE

W

Fig.1

d


• TRIMMING MOMENT = w * d MEASURED IN TON METER W MCTC : IS THE MOMENT THAT CHANGE THE TRIM BY 1 CM . CHANGE OF TRIM IS THE TOTAL CHANGE IN THE SHIPS TRIM FROM THE RATIO

BETWEEN THE MOMENTS OCCURRED & THE MCTC. MEASURED IN CM = TRIMMING MOMENT

MCTCLBP

ф

L1L2

CFNEW DRAFT AFTNEW

DRAFT FORE

W

Fig.1

d


• THE TOTAL CHANGE IN TRIM IN CM ,WILL BE DISTRIBUTED BETWEEN THE DRAFTS FORE & AFT. IF THE CF OF THE SHIP IS COINSIDE WITH THE MID SHIP POINT ,THE CHANGE IN TRIM WILL BE DIVIDED EQUALLY ON BOTH DRAFTS.

• EXAMPLE . CHANGE IN TRIM = 6 CM CF MID SHIP• SO DRAFT AFT = +3 CM DRAFT FORE = - 3 CM

LBP

ф L1L2

CF W

Fig.1

d


• THE TOTAL CHANGE IN TRIM IN CM ,WILL BE DISTRIBUTED BETWEEN THE DRAFTS FORE & AFT. IF THE CF OF THE SHIP IS NOT IN THE MID ,THE CHANGE IN TRIM WILL BE DISTRIBUTED BETWEEN THE DRAFTS BY THE FOLLOWING.

• DRAFT FORE = L2 * CHANGE OF TRIM (L2 DIST FROM CF TO FORE B ) L ( L1 DIST FROM CF TO AFT B )

DRAFT AFT = L1_ * CHANGE OF TRIM ( L IS THE LBP ) L

L

ф

L1L2

CFNEW DRAFT AFTNEW

DRAFT FORE

W

Fig.1

d


THE ADDED /DISCHARGED WEIGHT ALSO HAS AN EFFECT OVER THE SHIP , THE EFFECT

APPEARS OVER THE SHIPS MEAN DRAFT CALLED BODILY SINKAGE/RISE ,THIS CHANGE ADDED OR REMOVED TO BOTH DRAFTS FORE & AFT.

IF A WEIGHT ADDED THE EFFECT CALLED BODILY SINKAGE = _W _IF A WEIGHT DISCH. THE EFFECT CALLED BODILY RISE TPC

L

ф

L1L2

CFNEW DRAFT AFTNEW

DRAFT FORE

W

Fig.1

d

stability basic coarse chapter 3 a-density b-relative density

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