7/30/2019 Formula Stability

1/25

Chapter1BasicHullFormRelatedCalculationFormula

CBLBTLBT

Cm

BT

Am

Cwp LB

Aw CP LAm

CVPwTA

TPC100

Aw

7/30/2019 Formula Stability

2/25

Simpson1st rule 1,4,1

Atleast

3offset

Oddnumberofequiwidthoffset

Simpson2nd rule 1,3,3,1

Atleast4offset

(n1)exactlydivisibleby3

7/30/2019 Formula Stability

3/25

Chapter1 Calculation

Example1:

Ashipdisplacement6000thasalengthof120m,breadth12manddraft5m.Calculate

theships

block

coefficient.

Example2:

Avessel

120

m

long

and

15

m

wide

has

adraught

of

7.5

m

in

SW

when

its

displacement

is10800tonne.Theimmersedmidship areais107m2andwaterplane areais1500m2.

Calculate(i)Cw ,(ii)Cm ,(iii)Cb &(iv)Cp

7/30/2019 Formula Stability

4/25

Chapter1 Calculation

Example3: Aship oflength240.0m,beam32.0m floatsatadraught7.20minsea

water. Atthisdraught,ithasaCB,CW andCM of 0.56,0.76and0.975 respectively.

Determineits

i) displacement, ii) TPCinseawater iii) CVP, and iv) CP

Example4: A

ship

of

length

240.0

m,

beam

32.0

m

floats

at

adraught

7.20

m

in

sea

water. Atthisdraught,ithasaCB,CW andCM of 0.56,0.76and0.975 respectively.

Determineits

i) displacement, ii) TPCinseawater iii) CVP

, iv) CPIf

the

ship

now

steams

upriver

for

one

hour

to

aport

where

the

water

density

is

1004

kg/m3,determineitsnewdraughtifthechangeinitsblockcoefficientisnegligible.

7/30/2019 Formula Stability

5/25

Example1.(WaterPlaneArea)

Aship120metreslongatthewaterlinehasequidistantlyspacedhalfordinates

commencingfromaftasfollows:

0.1,4.6,7.5,7.6,7.6,3.7,0metres,respectively.

FindtheareaofthewaterplaneandtheTPCatthisdraft.

Chapter1 Calculation

Example2.(LCF)

Ashipwaterplaneis120mlong.Halfbreadthsatequalintervalsfromaftare:

0.1,4.6,

7.5,

7.6,

7.6,

3.7

and

0metres,

respectively.

FindthepositionofCOF.

7/30/2019 Formula Stability

6/25

Chapter1 Calculation

Example3.(DisplacementwithWaterPlaneArea)

Thewaterplaneareasofashipatonemeterintervalsfromkeelupwardsare:

1730,1925,

2030,

2100,

2150

m2.

FindthedisplacementinSWat4mdraft.

Example4.(KBwithWaterPlaneArea)

Theareaof successivewaterplanesof ashipatintervalsof1.5mstartingfrom1m

waterplane are:

1531,1665,

1789

and

1900m2

.The

appendage

below

the

1m

waterplane has

a

displacementof1750tonneinseawaterandtheKBof0.55m.

CalculatethetotalvolumeofdisplacementandthevalueofKBtothe5.5mwaterline.

7/30/2019 Formula Stability

7/25

Chapter1 Calculation

Example5.(DisplacementwithTransversecrosssectionarea)

Findthevolumedisplacementofabarge48mlong,whoseunderwatertranseverse cross

sectionalareaare:

19.6,25,17.5,13,0m2

7/30/2019 Formula Stability

8/25

Chapter2CGSmallAngleStabilityFormula

KG

Verticalmomentofallshipmassesabt keel

Allship

masses

LCG

Longitudinalmomentofallshipmassesabt midship

Allship

masses

TCG

Transversemomentofallshipmassesabt centreline

Allshipmasses

KB

Allverticalmomentabt keel

Totalmoment

7/30/2019 Formula Stability

9/25

Effectonmassshifting

xmGoG1

ymGoG1

Effectonmassaddition/removal

f

xm

GoG1

f

ym

GoG1

7/30/2019 Formula Stability

10/25

Suspendedmasscalculation

Kg ofthe

suspended

mass

taking

from

the

point

ofsuspensiontokeel.

Momentofstatical stability=Rightingmoment

GZ

sinGMT

7/30/2019 Formula Stability

11/25

Angleoflist

Momentofstatical stability=Rightingmoment

GZ

sinGMT

L

T

101

GM

GGtan

7/30/2019 Formula Stability

12/25

FreesurfacecorrectionFSC

FSC

I

12bl 3

GMTeff FSCGMT

7/30/2019 Formula Stability

13/25

Chapter2 Calculation

Suspendedmasscalculation

LCGcalculation

MassshiftingcalculatenewKG

Massaddition/removalcalculatenewKG

Example

(from

Notes

Pg 3)

displacement,

GMT,

LCG

7/30/2019 Formula Stability

14/25

TutorialQuestion1:

Ashipwhichistobemodified,hasthefollowingoriginalparticulars:

Displacement 150.00t

GM 0.45m

KG 1.98m

KB 0.9m

TPC 2.00t

Draught 1.65m

Aftermodification,20tisaddedatKgat3.6m.CalculatetheGM,assuming

constantwaterplane areaoverthechangeindraught.

Chapter2 Calculation

TutorialQuestion2:

Ashiphasadisplacementof10000t,KG7m,GM0.2mandwaterplane area1600m.

Asumming KMremains

constant

throughout,

calculate

(i) ThemassofdeckcargothatcanbeaddedatKg13mwithoutmakingthe

shipunstable,&

(ii) Thechangeofdraughtiftheshipmovesintofreshwaterofrelativedensity

1.004after

loading

of

cargo

in

S.W.

2

7/30/2019 Formula Stability

15/25

Chapter2 Calculation

TutorialQuestion3:

Avessel,havingadisplacementof11900tonne andGM0.85m,hasalisttoportof

2.75degree.100tonne ofcargoisstilltobeloaded,andtwocompartmentsare

available,one

6.1m

off

centreline to

starboard

and

the

other

8.4m

off

centreline to

port.

AssumingKMisconstant,calculatethemassrequiredineachcompartmenttothe

vesselupright.

7/30/2019 Formula Stability

16/25

Chapter3LargeAngleStabilityCalculationFormula

GZ

GZf

sinKGKN

cosGGsinGGZG 211000

G0G1.sin

/G1G2/.cos

Signconvectionsapply!

7/30/2019 Formula Stability

17/25

Curve of statical stability

1. Range of stability

2. Angle of vanishing stability

3. Maximum righting lever

4. Initial GMT:

=0tov

=v

=GZmax

Drawtangenttocurveat =0,drawverticallineat =1radian(57.3o)totangentialline,verticaldistanceisinitial GMT.

7/30/2019 Formula Stability

18/25

Chapter3 Calculation

Inacertainconditionofloading,theshipdisplacementismadeupasfollows:

Afterloading,followingchangestookplaceagain,

2000tonneisremovedfromalocation,whichis5mfromkeel.

Additionally,2000tonnecargowasloweredby2mtowardskeel.

Item Mass (t) Centre of Gravityfrom Keel (m)

Lightship 5000 6.00

Cargo 6000 3.00

Fuel 2500 2.00

Store 1500 5.00

7/30/2019 Formula Stability

19/25

Chapter3 Calculation

Item Mass (tonne) VCG(m)

Lightship 4 200 6.0

Cargo 9 100 7.0

Fuel 1 500 1.1Stores 200 7.0

TutorialQ1:

TheangleofinclinationandcorrespondingrightingleverforavesselatanassumedKGof

6.5masgivenincrosscurvesare:

In

a

particular

loaded

condition

the

displacement

mass

is

made

up

as

follows:

Plotthecurveofstatical stabilityfortheinformationobtaineddirectlyfromthecross

curvesandalsothemodifiedcurverepresentingstatical stabilityattheactualloading

condition.Determine

the

initial

stability

and

the

range

of

stability

for

the

loaded

condition

given.

7/30/2019 Formula Stability

20/25

Chapter3 Calculation

Ashiphavingadisplacementof9500thasthefollowingGZcurve

0 10 15 30 45 60 75 90G

0Z(mm) 0 210 401 822 918 802 410 -430

Followingchangesinloadingtakesplace

a)60tonnes ofcargoshifted horizontallythroughadistanceof13m

b)120

tonnes of

cargo

shifted

vertically

downwards

adistance

of

5m

Draw(1)theoriginalGZcurve,

(2)theeffectonGZcurveof(a)alone

(3)theeffectonGZcurveof(b)aloneand

(4)theeffectonGZcurveofboth(a)and(b)simultaneously.

7/30/2019 Formula Stability

21/25

Chapter3 Calculation

Tutorial Q4For a particular ship of 1500 tonne displacement the GZ curve is defined by

the following relationships:

o 0 15 30 45 60 75 90

GZ(mm) 0 52 195 310 172 -24 -370

During a storm 100t of cargo shifts horizontally 5.0m and 2.5m verticallydownwards. Determine,

a) the angle of list and

b) the range of stability.

7/30/2019 Formula Stability

22/25

Chapter5LongitudinalStabilityFormula

t AF TT tc of tt

TM tM ii dm Massshifting

TM tM LCFlcgm ii Massaddition/removal

MCTCL100

GMLMCTC100M ttc

TPC100mdTp

TFf TFf

TFf TFf

LCF

2

L

L

tT cF0

LCF

2

L

L

tT cF0

LCF

2

L

L

tdTT cpF0

LCF

2

L

L

tdTT cpF0

7/30/2019 Formula Stability

23/25

Chapter5 Calculation

Example1

A160mshipfloatsatanevenkeeldraughtof8.20m. Itdisplaces17600tandhasa

KGof12.0m. Atthisdisplacement,KMLis157.0mandthecentreofflotationis1.80

maft

of

amidships.

To

have

agreater

immersion

of

the

propeller,

three

containers

of

masses16.0,14.0and12.0tareshiftedaftwards throughlongitudinaldistancesof

90.0,80.0and70.0respectively.

Calculatethefinalforwardandaftdraughts.

Example2

Aship,160.0mlong,displaces3600twithforwardandaftdraughtsof1.50mand

3.60mrespectivelyinthelightshipcondition. Atthisdisplacement,thecentreof

flotationis2.40maftofamidships,TPC12.0andMCTC120.0tm.

Ahelideck,ofmass60.0t,istobefitted65.0mforwardofamidships. Determinethe

forwardandaftdraughtsinthelightshipconditionaftertheinstallationofthe

helideck.

7/30/2019 Formula Stability

24/25

Chapter5 Calculation

Example3

Theforwardandaftdraftofavessel,76.2minlengthare5.18mand5.64mrespectively.

Themaximumpermissibledrafttocrossabaris5.5m.Calculate,usingthedatabelow,

themasstobedischargedfromaposition26maftofamidshipstoenablevesseltocross

thebar.

TPC:12 MCTC:91.5tm LCF:1.52maftmidships

Example4

Avessel

of

200m

length,

floats

at

draft

of

7.5m

aft

and

7.3m

forward.

Subsequently

followingloading/unloadingtakesplaceataport:

Loaded: a)390tat15mfwd ofmidship

b)450tat3maftofmidship

Unloaded: c)130tat

16m

forward

of

midship

d)100tat5maftofmidship

Baseyourcalculationsonthefollowingavailableshipshydrostaticdataforthisloading

condition: LCF=2.9maftofmidships TPC=22t MCTC=199tm

Calculatethe

final

forward

and

aft

drafts.

7/30/2019 Formula Stability

25/25

Chapter5 Calculation

Example5

Avessel146mlongfloatsatdraftsforwardandaftof5.14mand5.2mrespectivelyin

waterof

desity 1025kg/m3.

Thefollowingcargoisloaded: 400tat12mfwd ofmidship

600tat18maftofmidship

Usingthehydrostaticdatagivenbelow,caculate thedistanceanddirectioninwhich60t

ofballast

must

be

moved

to

give

afinal

trim

of

0.36m

by

stern,

and

the

final

drafts.

Hydrostaticdata:

LCF=3maftmidship TPC=20 MCTC=150tm