NOTA Large Angle Stability 2009

© Omar Yaakob 2009 1


WB 1

K

Z

M

G

B

W 1

W

ExternalHeelingmoment

M

G Z

L

L1

righting moment = x GZ

for small angles,

GZ = GMT

sin

where is the angle of heel.

Righting Lever and Moment

16. LARGE ANGLE STABILITY


T

TIBM

M’TM

GZ at Large Angles

The location of the metacentre is no longerstationary at large angles. This is due to thedifferent shapes of waterplane areas atsuccessive angle of heels. This differencesgive rise to different transverse secondmoments of area. The distance from thecentre of buoyancy to the metacentre isgiven as

At small angle of heel, IT can be assumedconstant and hence the metacentre can beconsidered stationary. However at largerangles of heels, the waterplane shapeschanges significantly leading to movementof the metacentre . Because of this reason,the accuracy of the expression GZ=GMT sindiminishes at large angles. In other wordsthere is no simple expression relating GZ toGMT .

B’’

B ’

G


GZ or Statical Stability or RightingArm CurvesShows relationship between GZand .

20

GZ

(m)

GZ value at 20

Angle of heel

Angle of Vanishing Stability

Range of Stability

Point of inflexion

At the design stage, the naval architectmust ensure that the curves arecalculated.The GZ data is produced at differentdisplacements. An assumed KG is used.Later, when the actual and KG at anyloading condition are known, actual valuesof GZ can be extracted from the data.

Various methods are used to calculatestatical stability curves. We do not havethe opportunity to deal with these methodsin this short course. Details can be foundin Principles of Naval Architecture (Vol 1).


2,000

5

80

15

50

30

40SN (m)

KS= 4.2m

Cross Curves of StabilityThe GZ values calculated at the designstage are presented either in tables or in

Cross Curves of Stability.The GZ values calculated are carried out atonly a few displacements and heel angles.To obtain values at other displacements or

angles, linear interpolation is used.The values of GZ will depend on the KG butthe value of KG changes from time to time.Since it is difficult to calculate GZ at manydisplacements, various angles of heel anddifferent values of KG, a fixed value of KG

is normally used in the design stage.This assumed position of G is designated S,

the height above keel being KS, thecorresponding point on the line of action ofbuoyancy is called N i.e. not Z. The valueof the righting lever is thus SN instead of

GZ.

2,500

Displacement ( tonnes)

1,500 3,000


N

K

M

G

B

Z

S

N

If S is above G ; GZ = SN + SG SinIf S is below G; GZ = SN - SG Sin

A better method is to put S at the keel i.e.KS=0. In this case SN = KN. The assumedrighting arm, KN values are plotted instead of

SN at various displacement. When G isknown for any condition, GZ can be

calculated:GZ = KN – KG Sin

It must be noted that the values of KG usedis the virtual or fluid KG i.e. taking into

consideration the FSC.


ExampleA ship with lightship displacement 1,700 tonnes, KG 3.5m is load ed with 1,800tonnes of cargo at Kg 3.8m. KMT after loading is 3.8m while KN values are asfollows. Plot the GZ curve and find the area under the curve up to 300.

3.263.252.922.201.540.774,000

3.263.192.842.161.500.753,000

756045302010Displacement(tonnes)

Angle of heel ()

Solution:i. Carry out loading calculation to obtain final displacement and KG.ii. Find KN values at that displacement.iii. Correct KN to obtain GZ using GZ = KN – KG siniv. Plot the curve (note that initial slope= GMT=0.15m)v. Use Simpson rule to find area under the curve up to 30 .



(ii) (iii) (v)

-0.2663.5260.9663.2675

0.0593.1610.8663.2260

1.549sum0.2992.5810.7072.8845

0.35510.3551.8250.5002.1830

0.81530.2721.2480.3421.5220

0.37930.1260.6340.1740.7610

0100000

fASMGZ(m)KG Sin Sin KN(m)

Area under the curve up to 30 ,

Area = 3 x 10 xx 1.549 = 0.1014 mrad8 180


Loading Calculation Table

Initial Ship ConditionΔ, KG, LCG

SHIP HULL FORM

New deadweightloading/unloadingw, Kg, lcg

Calculate DraughtsTA and TF

Use Δto read

Hydrostatics dataTmean, LCB, LCF,MCTC,

KMT

FINALΔ,KG, FSC LCG

Calculate GMT

HydrostaticsData

LOADINGCALCULATIONS


Loading Calculation Table

Initial Ship ConditionΔ, KG, LCG

SHIP HULL FORM

New deadweightloading/unloadingw, Kg, lcg

Calculate DraughtsTA and TF

Use Δto read

Hydrostatics dataTmean, LCB, LCF,MCTC,

KMT

FINALΔ,KG, FSC LCG

Calculate GMT

HydrostaticsData

SN or KN CrossCurvesStability

GZ CURVE

StabilityAssessment

KN/SN Valuesat correctΔ

STABILITYASSESSMENT PROCESS



© Omar Yaakob 2009 98Title: Small Ship Survey- Stability Dr. Omar Yaakob




Large Ship(IMO)Stability Criteria

0.15 mInitial GM6.

25.0 degAngle at Maximum GZ5.

0.20 mMaximum GZ4.

0.03 m radArea Under Curve 30o-40o

or up to f (flooding Angle)3.

0.090 mrad

Area Under Curve 0 o-40o orup to f (flooding Angle)

2.

0.055m.rad

Area Under Curve 0 o-30o1.


7.6.1 Load Line Criteria.

0.35 m0.15 m8. Initial GM

30 deg15 deg7. Angle at Maximum GZ

0.2 m0.2 m6. Maximum GZ

0.03 m.rad0.03 m.rad5. Area Under Curve 30o-40o orup to qf (flooding Angle)

N.A0.055 + 0.001(30-qmax) if max.GZ occur between 15 to 30 deg.4. Area Under Curve 15o-30o

0.089 m.radN.A3 Area Under Curve 0o-40o orup to qf (flooding Angle)

0.055 m.rad0.055 m.rad if max. GZ occur at30 deg.

2. Area Under Curve 0o-30o

N.A0.07 m.rad if max. GZ occur at15 to 30 deg.1. Area Under Curve 0o-15o

Fishing Vessel(IMO)

Small Craft -Passenger/Cargo(HSC Code)Stability Criteria


The stability is assessed at a number of loading conditions depe ndingon ship types, for example:

1. Lightship2. Homogenous Full Load Departure3. Homogenous Full Load Arrival4. Ballast Departure5. Ballast Arrival6. Etc.


ExampleA ship 144m LBP lightship 5200 tonnes, KG 7.4m, LCG 2.0m aft of amidships. Thehomogenous loaded departure condition are as follows:

----

44.0 F17.0 F22.0 A49.0 F

10.17.18.09.0

1700190018001330

Cargo No.1Cargo No.2Cargo No.3Cargo No.4

-14.0 F9.550Crew and Store

21060.0 F7.2170Fresh Water

270350

1.2 A8.0 A

1.53.2

1350210

Fuel Tank 1Fuel Tank 2

FSM(tm)

LCG fromamidship(m)

VCG(m)

Mass(t)

Item


Extracts of hydrostatics data at two relevant draughts are as follows:

8.0837.0883.00178.5613,9999.00

7.956.802.80170.012,9948.50

KMT

(m)LCF(m from)

LCB(m from)

MCTC(tm )

Displacement(tonnes)

T (m)

KN values at two displacements are as follows:

7.0716.5775.5594.0542.1241.4140.70713900

7.0676.5755.5554.0522.1221.4120.70513500

7560453015105o

Find the draughts at the perpendiculars, calculate GMT and check if the vessel pass theIMO merchant ship stability criteria.



1078001286709990613710FINALDISPLACEMENT

830107800118270614268510DEADWEIGHT

---

7480032300

-

--

3960065170

44.0 F17.0 F22.0 A49.0 F

17170134901440011970

10.17.18.09.0

1700190018001330

Cargo No.1Cargo No.2Cargo No.3Cargo No.4

-1020014.0 F4759.550Crew and Store

210-1020060.0 F12247.2170Fresh Water

270350

--

16201680

1.2 A8.0 A

2025675

1.53.2

1350210


--104002.0 A384807.45200LIGHTSHIP

FSM(tm)

M F(tm)

MA(tm)

LCG from(m from )

M v(tm)

VCG(m)

Mass(t)

Item

KG = 7.287m LCG = 1.52m aft of amidships Free Surface Correction, F.S.C = 0.061m


GMSOLID

= mF.S.C = m

GMFluid

= m

Title: SHIP STABILITY – Dr. Omar Bin Yaakob

1299413999

1299413710

Enter hydrostatic Table at = 13710 tonnes, to obtain:

Tmean

= 8.50 + (9.00 – 8.50)

= 8.856m

and,MCTC = 176.1 t.m. ; KM

T= 8.045 m;

LCB = 2.949 m.A.LCF = 7.005 m.A.

For transverse stabilityKM = mKG = m


TRIM = by bow/stern

100)(

MCTCxxp

TA = T mean ± TA

TF = T mean ± TF

For longitudinal stability

TRIM =

TA= trim x (72 – )144

= m

TF=


GZ (m)

KGf sin

7.0696.5765.5574.0532.1231.4130.706KN (m)

7560453015105o

KGf = 7.287 + 0.061GZ = KN - KGf Sin Plot GZ curve and calculate and check criteria.

OK>0.15m0.697mGMT

OK0.03 m-rad0.072 m.radA40

– A30



OK0.2m @ 30 0.393m @ 37GZmax

EvaluationRequired by IMORule

ActualCriteria

i.e. fulfill all Load Line Rule Criteria

Values obtainedfrom KN Table atDisplacement =13,710 tonnes

KG fluid, takinginto account freesurfacecorrection


Determining Steady Angle of Heel due to Heeling Moment

Righting moment = Righting Lever x = GZ

Curves of displacement multiplied by GZ values at various angles of heel are called therighting moment curve.

When a ship is acted upon by an external or internal moments, the steady angle of heel canbe obtained as the point of equilibrium between the curves of heeling moment and the rightingmoment curve.

Heeling mmnt cos

Steady Angle of Heel

Moment

Heel angle

GZ

A

D

cDE

E

F


The same information can be represented using the righting arm (instead of rightingmoment) curve and curveHeeling moment cos

Heeling mmt cos

Steady Angle of Heel Heel angle

GZ

AD

cDE

E

F

Arm (m)We are not only concerned aboutthe resultant steady angle of heelbut also possibility ofovershooting.

US Navy Criteria:•For crowding of personnel and lifting of weight:•Angle at C for lifting of weight or crowding of personnel < 15 degrees•Heeling arm at C < 0.6 maximum•Area A1 > 0.4 total area under curve.

Area 1



During operation, the craft has always been expected toencounter the heeling arm or upsetting arm due to :

• wind,• lifting of weight,• movement of personnel and• high speed turning.

The effect of heeling arm need to be calculated and assessedagainst the specified criteria.

Presently, the stability of ship subjected to heeling arm isassessed statically that is by superimposing the heelingarm curve onto the GZ curve.

Effect of Heeling Arm


Heeling Arm Curve Due to Turning at High Speed,Movement of Personnel and Lifting of Cargo


Heeling Arm Curve Due to Wind and Wave


Stability Criteria For Small Craft Under Heeling Arm

A1 > 0.4 Area UnderGZ Curve

A1 > 0.4 Area UnderGZ Curve

6. Area Under Curve For HighSpeedTurning

A1 > 1.4 A2A1 > 1.4 A25. Area Under Curve For BeamWind

0.6 of GZ max0.6 of GZ max4. GZ at Intersection Point (c)

15deg10 deg

3 Max Angle of Heel atIntersectionPoint (c) For Movement ofPersonnel

15deg8 deg

2. Max Angle of Heel atIntersectionPoint (c) For High SpeedTurning

15deg10 deg

1. Max Angle of Heel atIntersectionPoint (c) For Beam Wind

Fishing Vessel(IMO)

Small Craft -Passenger/Cargo(IMO and HSCCode)

Stability Criteria


EXERCISES

1. For one loading condition, a ship LBP 70m has displacement 1500 tonnes, KG 4.0 mand KM 4.5 m. SN values calculated at KS=4.2m are given in the table below.

0.010.190.350.440.440.350.23SN (m)

75604530201510Angle()

a.Plot the GZ curve of the vessel.b.Check if the ship passes Load Line Rule stability criteria.c.Find the angle of heel if a 50 tonne weight already onboard is moved 10 m across.


GZ (m)

SG sin

SN (m)

7560453015105o

pass>0.15m0.5mGMT

pass0.03 m-rad0.09A40 – A30

pass0.09 m-rad0.28A40

pass0.055 m-rad0.188A30

pass0.2m @ 25 0.54m at 30 degGZmax

EvaluationRequiredActualCriteria

i.e. fulfill all Load Line Rule Criteria

Since KS > KG, S is above G ; so GZ = SN + SG Sin



2. A fishing boat began its trip with displacement of 340 tonnes , KG 3.5 m and KM 3.75m. After one day journey, the following events occurred :

20 tonne fish loaded (kg 3.0m)6 tonne fuel consumed (kg 1.0m)4 tonne fresh water consumed (kg 2.0m)

KM was found to be 3.8 m. The KN values are as follows:

a. Plot its GZ Curveb. In this condition the boat then used the power block to lower a small boat to sea.

The weight of smal boat was 10 tonne and the boom of the power block was 4from the boat’s centreline on the starboard side and 5 m above keel. Plotroughly (but do not calculate) the new righting arm curve and estimate itssteady angle of heel.

3.453.292.992.281.5210.710400

3.553.373.072.401.5250.720300

906045302010Displacement(tonne)

Angle of Heel


GZ (m)

KGf

sin

KN (m)

o

KG= 3.53mGZ = KN - KG Sin .

Item Weight(tonnes)

KG (m) Moment about keel (tonne-m)

TOTAL


3. A ship 150m LBP has lightship displacement 6500 tonnes, KG 8.0 m, LCG 1.5 m aftof amidships. The ship is loaded as follows :

44.0 F17.0 F22.0 A

10.17.18.0

180020001600

Cargo Hold No.1Cargo Hold No.2Cargo Hold No.3

14.0 F9.550Crew and Store

60.0 F7.2200Fresh water Tank

1.2 A8.0 A

1.53.2

1000300


LCG from amidships(m)

VCG(m)

Mass(t)

Item

* free surface effect is neglectedExtract of hydrostatics data at relevant draughts as follows:

8.082.30A3.01F180.513,50010.00

7.951.80A2.85F176.613,4009.50

KMT (m)LCF(m dari)

LCB(m dari)

MCTC (tm )Displacement (tonnes)T (m)


7.0696.5765.5574.0532.1231.4130.70613,500

6.8736.3465.2684.0002.0231.4000.70013,400

7560453015105

o

(tonnes)

KN values in meters at two displacements are as follows:

a. Find Draughts at the perpendicularsb. Find GM T.c. Calculate and plot the righting arm curve of the ship. Check whe ther

the vessel passes stability criteria for merchant ship.d. Estimate its angle of heel when the ship is lifting a 200 tonne cargo

on the port side using its own derrick. The derrick head is 15 m fromthe centreline and 15 m above keel.

e. It is desired to reduce this heel angle to 5 degrees. How muchballast water must be transferred from port tank to starboard ta nk, adistance 6 meters?

NOTA Large Angle Stability 2009

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