d1mc semester 1 / ship stability / march 2007 /capt. mrd. dry docking 1 1 to be a world class...

D1MC Semester 1 / Ship Stability / March 2007 /Capt. MRD. Dry Docking 11

To Be A World Class Maritime Academy





Revisions Ex.Revisions Ex.

Simplified StabSimplified Stab

Simpson RulesSimpson Rules

TrimTrim Effect on GEffect on G

Dry DockingDry Docking

Statical StabStatical Stab

Inclining TestInclining Test




LEARNING OBJECTIVESLEARNING OBJECTIVES

• To understand the virtual loss of GM and To understand the virtual loss of GM and the calculations. the calculations.

• To calculate the maximum trim allowed To calculate the maximum trim allowed to maintain a minimum stated GM. to maintain a minimum stated GM.

• To understand the safe requirements for To understand the safe requirements for a ship prior enter into dry dock. a ship prior enter into dry dock.




LEARNING OBJECTIVESLEARNING OBJECTIVES

• To understand the critical period during To understand the critical period during dry docking process. dry docking process.

• To calculate the ship’s drafts after the To calculate the ship’s drafts after the water level has fallen and after the ship water level has fallen and after the ship has taken the block overall. has taken the block overall.

• Effect to stability when vessel has run Effect to stability when vessel has run aground (single point). aground (single point).




Anybody would like to share their experience Anybody would like to share their experience during dry docking….?during dry docking….?




Before enter into dry dock, vessel must have…Before enter into dry dock, vessel must have…

• Positive initial GM (GM fluid)• Upright • Trim - if possible even-keel or

slight trim by stern • Double bottom tank kept either dry

or pressed up - reduced FSE• If initial GM is small - D.B. tank to

be pressed up to increase GM

• Positive initial GM (GM fluid)• Upright • Trim - if possible even-keel or

slight trim by stern • Double bottom tank kept either dry

or pressed up - reduced FSE• If initial GM is small - D.B. tank to

be pressed up to increase GM




When coming into Dry Dock: • The vessel will line-up with her

centerline vertically over the keel blocks

• Dock gate will be closed and commence pumping out water

When coming into Dry Dock: • The vessel will line-up with her

centerline vertically over the keel blocks

• Dock gate will be closed and commence pumping out water




FF

No effect on ship’s Initial Stability…No effect on ship’s Initial Stability…




When coming into Dry Dock: • The rate of pumping will be

reduced as the ship's sternpost near the block.

When coming into Dry Dock: • The rate of pumping will be

reduced as the ship's sternpost near the block.




Commence touching the ground… ‘Commence touching the ground… ‘Sueing PointSueing Point’’

Sueing PointSueing Point

FF




When coming into Dry Dock:

• Once the sternpost is touching the block, the UP-THRUST forces start to act against the sternpost.

• At this moment part of ship's weight gets transferred to the keel blocks.

When coming into Dry Dock:

• Once the sternpost is touching the block, the UP-THRUST forces start to act against the sternpost.

• At this moment part of ship's weight gets transferred to the keel blocks.




PP

P is the P is the Upthrust ForceUpthrust Force acting at first point of acting at first point of touching the ground. Commence touching the ground. Commence Critical PeriodCritical Period……

FF




Sueing PointSueing Point

at ‘at ‘APAP’’




PP

KKKK

P is the P is the Upthrust ForceUpthrust Force acting at first point of acting at first point of touching the ground. Commence touching the ground. Commence Critical PeriodCritical Period……




When coming into Dry Dock:When coming into Dry Dock:

• When ship's weight gets transferred to When ship's weight gets transferred to the keel blocks, vessel will suffer loss on the keel blocks, vessel will suffer loss on her GM. her GM.

• The time interval between the sternpost The time interval between the sternpost landing on the blocks and the ship taking landing on the blocks and the ship taking the blocks overall is referred to as the the blocks overall is referred to as the CRITICAL PERIODCRITICAL PERIOD. .




PP

P force is increasing gradually as the trim change P force is increasing gradually as the trim change by Head…Vessel is still in by Head…Vessel is still in Critical PeriodCritical Period……

FF




When coming into Dry Dock:When coming into Dry Dock:

• Vessel must have positive effective Vessel must have positive effective GM that to be maintained GM that to be maintained throughout the critical period. throughout the critical period.

• If not vessel may heel over, slip off If not vessel may heel over, slip off the blocks when there is an the blocks when there is an external force acting and heel the external force acting and heel the ship. ship.




P

Vessel is fully rest on the blocks… Vessel is fully rest on the blocks… End of Critical End of Critical PeriodPeriod

FF




<>




M

G1

G

B

Initial GM loss by Initial GM loss by GGGG11 after after completed the Critical completed the Critical Period…Period…

This is due to Upthrust This is due to Upthrust Force or ‘P’ Force…Force or ‘P’ Force…




PP

What is the total P Force during Critical Period __What is the total P Force during Critical Period __??___ tonnes___ tonnes

““How much weight to be discharged in order to bring the ship How much weight to be discharged in order to bring the ship from trim by stern to even-keel…from trim by stern to even-keel…””

FF




CALCULATION CALCULATION

OF UPTHRUST FORCE OF UPTHRUST FORCE

AT THE STERNPOSTAT THE STERNPOST

- 'P' FORCE - 'P' FORCE




ww

d

FF

Weight discharged to even keel the draft…

Trimming Moment = w x d t-m by Head

Weight discharged to even keel the draft…

Trimming Moment = w x d t-m by Head




FF

After weight discharged…

T M By Head = T M By Stern

After weight discharged…

T M By Head = T M By Stern




P

P is the Upthrust Force or weight dischargedP is the Upthrust Force or weight discharged to to the blocks…the blocks…

T.M T.M = = ww x d x d = = P P xx dd t-m by Head t-m by Head

d

FF




PP

Vessel is fully rest on the blocks, Change of Trim Vessel is fully rest on the blocks, Change of Trim by Head and finally vessel at even keel drafts… by Head and finally vessel at even keel drafts… End of End of Critical PeriodCritical Period……

FF




Change of TrimChange of Trim == Trimming Moment (TM)Trimming Moment (TM) MCTCMCTC

WherebyWhereby TM TM == w x dw x d

Change of TrimChange of Trim == PP x d x d MCTCMCTC

PP == COT x MCTCCOT x MCTC tonnestonnesdd


== PP x d x d




Exercise in classroomExercise in classroom

MV OneSuch, LBP 120m is going to dry dock MV OneSuch, LBP 120m is going to dry dock at the following condition in sea waterat the following condition in sea water

Draft forward is 3.5m and aft is 4.0m, Draft forward is 3.5m and aft is 4.0m, distance sueing point (AP) to F is 57.5m.distance sueing point (AP) to F is 57.5m.

Her displacement is 4600 tonnes, MCTC is 86 Her displacement is 4600 tonnes, MCTC is 86 t-m and TPC 15.45t-m and TPC 15.45

Calculate Calculate i.i. The amount of up-thrust force (The amount of up-thrust force (PP) at ) at

the the end of Critical Period? end of Critical Period? ii.ii. Final drafts forward and aft?Final drafts forward and aft?




PP == COT x MCTCCOT x MCTC dd

== 50 x 8650 x 86 57.557.5

PP == 74.8 tonnes74.8 tonnes

Calculation of P force…Calculation of P force…




CODF CODF == 50 – 2450 – 24

== 26cm26cm

COT COT == P x dP x dMCTCMCTC

== 74.8 x 57.574.8 x 57.5 8686

== 50cm50cm

CODA CODA == 57.557.5 x 50 x 50120120

== 24cm24cm

Body rise Body rise == PPTPCTPC

== 74.874.815.4515.45

= = 4.8cm4.8cm

== 0.048m0.048m




ForwardForward AftAft

Initial draftInitial draft 3.500m3.500m 4.000m4.000m

Body riseBody rise 0.048m -0.048m - 0.048m –0.048m –

CODCOD 0.260m +0.260m + 0.240m –0.240m –

Final draftFinal draft 3.712m3.712m 3.712m3.712m




P

P is the Upthrust Force or weight dischargedP is the Upthrust Force or weight discharged to to the blocks…the blocks…

T.M T.M = = ww x d x d = = PP x d x d t-m by Stern t-m by Stern

d

FF




PP

Vessel is fully rest on the blocks, Change of Trim Vessel is fully rest on the blocks, Change of Trim by by SternStern and finally vessel at even keel drafts… and finally vessel at even keel drafts… End of Critical Period…End of Critical Period…

FF





WherebyWhereby TM TM == w x dw x d == PP x d x d







Virtual Loss Of GM

During

Critical Period

Virtual Loss Of GM

During

Critical Period




• Method 1 – GG1

• Method 2 – MM1






Method 1Method 1 • When the vessel comes in contact When the vessel comes in contact

with the blocks, it is assumed that with the blocks, it is assumed that there is a there is a transfer of weighttransfer of weight 'P' 'P' from the keel to the blocks.from the keel to the blocks.

• Hence there is a virtual rise of Hence there is a virtual rise of ship's G (ship's G (discharged of weight discharged of weight below Gbelow G))




PPd

FF




PP

dd

FF

Trimming Moment by… Trimming Moment by… HeadHead




PP

KKKK

P is the Upthrust Force acting at first point of touching P is the Upthrust Force acting at first point of touching the ground. Commence Critical Period…”the ground. Commence Critical Period…”weight weight discharged from the shipdischarged from the ship””




KKKK

GGGG

GG11

MM

Reduction Reduction oror Loss of GM Loss of GM = GG = GG11




M

G1

G

B

Initial GM loss by GGInitial GM loss by GG11 during during the Critical Period…the Critical Period…

This is due to This is due to UpthrustUpthrust Force Force or ‘P’ Force…or ‘P’ Force…




Method 1Method 1

GGGG11 == P x KGP x KG in metres in metres W - PW - P

GGGG11 == P x KGP x KG in metres in metres W - PW - P




During Critical Period… part of ship body is still floating During Critical Period… part of ship body is still floating

PP

BBBB

MM

WW

GGGG

GG11




Vessel is inclined to a small angle by an external force… Vessel is inclined to a small angle by an external force…

P

B1B1

BB

M

W - P

W

G

External Force




Method 1 Discharged of weight, shift of GG1Method 1 Discharged of weight, shift of GG1

PP

GG

MM

GG11

W - PW - P

KK

External Force




P

G

M

G1

W - PW - P

X

KK

X = KGX = KG11 Sin Sin





P

G

M

W

G1

W - PW - P

Y

KK





Method 1… Discharged of weight, shift of GG1…Method 1… Discharged of weight, shift of GG1…

G1

G

Y

Y = GGY = GG1 1 Sin Sin




P

G

M

W

G1

W - PW - PW - PW - P

X

Y

KK





=

GG11

P

KK

X GG11

GG

Y

GG

WWW




XX = = KGKG11 x Sin x Sin YY = = GGGG11 Sin Sin

PPXX = = WWYY

P x P x KGKG11 x Sin x Sin == W x W x GGGG11 Sin Sin

P x KGP x KG11 == W x GGW x GG11

P x (KG + GGP x (KG + GG11)) == W x GGW x GG11

(P x KG) + (P x GG(P x KG) + (P x GG11)) == W x GGW x GG11




(P x KG) + (P x GG(P x KG) + (P x GG11)) == W x GGW x GG11

P x KGP x KG == (W x GG(W x GG11) – (P x GG) – (P x GG11))

P x KGP x KG == (W – P) x GG(W – P) x GG11

P x KGP x KG == GGGG11

W – PW – P




Therefore the formula is… Therefore the formula is…

GGGG11 == P x KGP x KG W - PW - PGGGG11 == P x KGP x KG W - PW - P




Righting Moment at small angle of heel… Righting Moment at small angle of heel…

B1B1

B

G1

M

W - P

W - P

Z

External Force




G1 Z

MW - P

W - P

Righting Moment Righting Moment = W x GZ= W x GZ= W x GM Sin = W x GM Sin

In this case,In this case, Righting MomentRighting Moment

= = (W – P) x G(W – P) x G11M Sin M Sin







Method 2Method 2 • When the vessel comes in contact When the vessel comes in contact

with the blocks, it is assumed that with the blocks, it is assumed that there is a there is a transfer of buoyancytransfer of buoyancy 'P' 'P' to the keel blocks.to the keel blocks.

• Hence there is a reduction in KM Hence there is a reduction in KM while the weight and KG are while the weight and KG are remains constant. remains constant.




Reduction in BuoyancyReduction in Buoyancy

PPd

FF




PP

KKKK

P is the Upthrust Force acting at first point of P is the Upthrust Force acting at first point of touching the ground. Commence Critical touching the ground. Commence Critical Period…”Period…”buoyancy reduction from the shipbuoyancy reduction from the ship””




PP

KKKK

Reduction Reduction Buoyancy Buoyancy

P is the Upthrust Force acting at first point of P is the Upthrust Force acting at first point of touching the ground. Commence Critical touching the ground. Commence Critical Period…”Period…”buoyancy reduction from the shipbuoyancy reduction from the ship””




KKKK

BB

MM11

MM

Reduction Reduction oror Loss of GM Loss of GM = MM = MM11

BB11BB11




Method 2Method 2

MMMM11 == P x KMP x KM in metres in metres WW

MMMM11 == P x KMP x KM in metres in metres WW




M

M1

G

B

Initial GM loss by MM1 after the Critical Period…

This is due to Upthrust Force or ‘P’ Force…






During Critical Period, part of ship body is still floatingDuring Critical Period, part of ship body is still floating

PP

BB

MM

W

GG

MM11MM11




Vessel is inclined to a small angle by an external Vessel is inclined to a small angle by an external force…force…

PP

B1B1

B

GG

MM

W - PW - P

W

External Force




Method 2Method 2…… Transferred of buoyancy, shift of MM Transferred of buoyancy, shift of MM11……

P

G

M

W

M1

W - P

X

Y

KK

W




==

MM11

P

KK

X MM

MM11 Y

GGGG

W W W - P




X X = = KMKM11 x Sin x Sin YY = = MMMM11 Sin Sin

PPXX == (W – P) x (W – P) x YY

P x P x KMKM11 x Sin x Sin == (W – P) x (W – P) x MMMM11 Sin Sin

P x KMP x KM11 == (W – P) x MM(W – P) x MM11

P x KMP x KM11 == W x MMW x MM11 – P x MM – P x MM11

P x KMP x KM11 + P x MM + P x MM11 = = W x MMW x MM11

P (KMP (KM11 + MM + MM11)) == W x MMW x MM11




P (KMP (KM11 + MM + MM11)) == W x MMW x MM11

P x KMP x KM == W x MMW x MM11

P x KMP x KM == MMMM11 W W

Therefore the formula is…Therefore the formula is…

MMMM11 == P x KMP x KM WW





Righting Moment at small angle of heel… Righting Moment at small angle of heel…

BB11BB11

BB

G

M1

W

W

Z

MExternal Force




G Z

M1

W

W



= = W x GMW x GM11 Sin Sin







• Method 1 – GG1 : Weight transferred

• Method 2 – MM1 : Buoyancy transferred

• Method 1 – GG1 : Weight transferred

• Method 2 – MM1 : Buoyancy transferred

SUMMARY…SUMMARY…




Exercise in classroom Exercise in classroom …continued…continued

MV OneSuch is going to dry dock at the MV OneSuch is going to dry dock at the following condition in sea waterfollowing condition in sea water

Draft forward is 3.5m and aft is 4.0m, Draft forward is 3.5m and aft is 4.0m, distance sueing point (AP) to F is 57.5m.distance sueing point (AP) to F is 57.5m.

Her displacement is 4600 tonnes, MCTC is 86 Her displacement is 4600 tonnes, MCTC is 86 t-m,t-m,

Calculate the amount of up-thrust force (P) Calculate the amount of up-thrust force (P) during Critical Period and the during Critical Period and the virtual loss of virtual loss of GMGM if KM is 8.0m and KG is 7.2m. if KM is 8.0m and KG is 7.2m.




PP == COT x MCTCCOT x MCTC dd

== 50 x 8650 x 86 57.557.5


Calculation of P force…Calculation of P force…




GGGG11 == P x KGP x KG W - PW - P

== 74.8 x 7.274.8 x 7.24600 – 74.84600 – 74.8

GGGG11 == 0.119m0.119m

Virtual loss of GM (Virtual loss of GM (GGGG11) method…) method…





== 74.8 x 8.074.8 x 8.0 46004600

MMMM11 == 0.130m0.130m

Virtual loss of GM (Virtual loss of GM (MMMM11) method…) method…




Comparison the Virtual loss of GM between Comparison the Virtual loss of GM between ((MMMM11) and () and (GGGG11) method…) method…

Different is… Different is…

== 0.130 – 0.119 0.130 – 0.119

= = 0.011m0.011m

… … ±± 1cm1cm




Effect of Trim

In

Dry Docking

Effect of Trim

In

Dry Docking





WherebyWhereby TM TM == w x dw x d == PP x d x d







Example

• Vessel displacement 5000 tonnes, distance

sueing point to CF is 80 m, MCTC 200 t-m,

KM 7.0 m and KG 6.0 m.

What will be the maximum trim allowed?

Example

• Vessel displacement 5000 tonnes, distance

sueing point to CF is 80 m, MCTC 200 t-m,

KM 7.0 m and KG 6.0 m.

What will be the maximum trim allowed?




Case 1 Case 1

0 m / Even keel0 m / Even keelCase 2Case 2

0.5 m by Stern0.5 m by SternCase 3 Case 3

5 m by Stern5 m by Stern




Case 1 Case 1




Calculate ‘Calculate ‘PP’…’…

P = P = MCTC x trimMCTC x trim

dd



dd



dd




Case 1 Case 1




Calculate ‘P’…Calculate ‘P’…


dd

= = 200 x 200 x 00

8080



dd

= = 200 x 200 x 5050

8080



dd

= = 200 x 200 x 500500

8080




Case 1 Case 1






dd

= = 200 x 200 x 00

8080

P = P = 0 tonne0 tonne



dd

= = 200 x 200 x 5050

8080

P = P = 125 tonnes125 tonnes



dd

= = 200 x 200 x 500500

8080

P = P = 1250 tonnes1250 tonnes




Case 1 Case 1




Virtual Loss of GM…Virtual Loss of GM… Virtual Loss of GM…Virtual Loss of GM… Virtual Loss of GM…Virtual Loss of GM…




Case 1 Case 1




Virtual Loss of GM…Virtual Loss of GM…

GGGG11 = = P x KGP x KG

W – PW – P



W – PW – P



W - PW - P




Case 1 Case 1






W – PW – P

= = 00 x 6 x 6

5000 – 5000 – 00



W – PW – P

= = 125125 x 6 x 6

5000 –5000 –125125



W - PW - P

= = 12501250 x 6 x 6

5000 –5000 –12501250




Case 1 Case 1






W – PW – P

= = 0 x 60 x 6

5000 – 05000 – 0

GGGG11 = = 0 m0 m



W – PW – P

= = 125 x 6125 x 6

5000 –125 5000 –125

GGGG11 = = 0.154 m0.154 m



W - PW - P

= = 1250 x 61250 x 6

5000 –12505000 –1250

GGGG11 = = 2.0 m2.0 m




Case 1 Case 1




Old GM = 1.0mOld GM = 1.0m

New GM…New GM…

= GM - GG= GM - GG11

= 1.0 – 0= 1.0 – 0

= = 1.0 m 1.0 m


New GM…New GM…


= 1.0 – 0.154= 1.0 – 0.154

= = 0.846 m 0.846 m


New GM…New GM…


= 1.0 – 2.0= 1.0 – 2.0

= = - 1.0 m - 1.0 m




Residual GMResidual GM

TRIMTRIM00

1.01.0

0.50.5

0.8460.846

- 1.0- 1.0

TRIM increasedTRIM increasedGM decreasedGM decreased

MAX. TRIM…?MAX. TRIM…?

5.05.0




• Vessel displacement 5000 tonnes, Vessel displacement 5000 tonnes, distance sueing point to CF is 80 distance sueing point to CF is 80 m, MCTC 200 t-m, KM 7.0 m and m, MCTC 200 t-m, KM 7.0 m and KG 6.0 m. KG 6.0 m.

Maximum Trim is….?Maximum Trim is….?




P force is …? Initial GM 1.0mP force is …? Initial GM 1.0m

Virtual Loss of GM = Virtual Loss of GM = 1.0m1.0m

GG

GG11MM

GG

GGGG11 is Virtual is Virtual Loss of GMLoss of GM

During Critical Period…During Critical Period…




P force is …? Initial GM 1.0mP force is …? Initial GM 1.0m

Virtual Loss of GM = Virtual Loss of GM = 1.0m1.0m


W - PW - P

1.01.0 = = P x 6P x 6

5000 – P5000 – P

5000 - P5000 - P = 6P = 6P

5000 = 7P5000 = 7P

P = P = 714.28 tonnes714.28 tonnes GG

GG11MM

GGGG11 is Virtual is Virtual Loss of GMLoss of GM

During Critical Period…During Critical Period…




P force is P force is …? Initial GM 1.0m…? Initial GM 1.0m Maximum trim is …?Maximum trim is …?

Virtual Loss of GM = 1.0mVirtual Loss of GM = 1.0m


W - PW - P

1.0 = 1.0 = P x 6P x 6

5000 – P5000 – P

5000 - P5000 - P = 6P = 6P

5000 = 7P5000 = 7P

P = 714.28 tonnesP = 714.28 tonnes

P = P = 714.28 tonnes714.28 tonnes


dd

Trim = Trim = P x dP x d

MCTCMCTC

= = 714.28 x 80714.28 x 80

200200

Trim = 285.7 cmsTrim = 285.7 cms

Trim = Trim = 2.86 m by Stern2.86 m by Stern




Residual GM

TRIM00

1.01.0

0.50.5

0.8460.846

- 1.0- 1.0

5.0

MAX. TRIM MAX. TRIM 2.86m2.86mMAX. TRIM MAX. TRIM …………??




CONCLUSION: CONCLUSION:

• The virtual loss of GM is The virtual loss of GM is NILNIL as vessel as vessel having zero trim. having zero trim.

• The loss is increased as the trim increased. The loss is increased as the trim increased.

• Maximum trim is depend upon the initial Maximum trim is depend upon the initial GMGM




WORKED EXAMPLE 1WORKED EXAMPLE 1

A ship of A ship of 140m140m in length, displacement in length, displacement 5000t5000t and upright is and upright is to enter dry dock with drafts forward to enter dry dock with drafts forward 3.84m3.84m, aft , aft 4.60m4.60m. Given . Given the following hydrostatic particulars: the following hydrostatic particulars:

TPCTPC 2020 tonnes tonnesMCTCMCTC 150150 t- m t- mCFCF 5m5m forward of amidships forward of amidshipsKMKM 9.75m9.75m

The blocks of the dry dock are horizontal. The blocks of the dry dock are horizontal.

i.i. Calculate the drafts of the vessel at the instants when she is Calculate the drafts of the vessel at the instants when she is taking the blocks forward and aft.taking the blocks forward and aft.

ii.ii. The ship's effective GM at this moment if the KG is The ship's effective GM at this moment if the KG is 7.75m7.75m

iii.iii. The Righting Moment at this instant for an angle of heel The Righting Moment at this instant for an angle of heel 5º.5º.




FF

No effect on ship’s No effect on ship’s Initial StabilityInitial Stability……

4.60m4.60m3.84m3.84m

Trim 76 cm by SternTrim 76 cm by Stern




FF

PP

P is the P is the Upthrust ForceUpthrust Force acting at first point of acting at first point of touching the ground, commence touching the ground, commence Critical Critical PeriodPeriod……

4.60m4.60m3.84m3.84mTrim 76 cm by SternTrim 76 cm by Stern




PP

What is the total P Force during Critical What is the total P Force during Critical Period? End of Period? End of Critical PeriodCritical Period……

FFEven keel draftEven keel draftEven keel draftEven keel draft

Change of Trim 76cms by HeadChange of Trim 76cms by Head




Ship’s trimmed Ship’s trimmed = = 4.60 – 3.84 = 4.60 – 3.84 = 0.76 m by Stern0.76 m by Stern

i.i. P P = = MCTC x trimMCTC x trim = = 150 x 76150 x 76 dd 7575

PP = = 152 tonnes152 tonnes

a.a. Bodily riseBodily rise= = P P = = 152152 = 7.6 cms = = 7.6 cms = 0.076 m0.076 m TPCTPC 20 20

b.b. Change of TrimChange of Trim == 76 cms by Head76 cms by Head




c.c. Change of draft aft due COTChange of draft aft due COT

= = ll x COT x COT LL

= = 7575 x 76 x 76140140

= = 40.7cm 40.7cm

= = 0.407 m0.407 m




d.d. Change of draft forward due COTChange of draft forward due COT

= = COT – Change of draft aft COT – Change of draft aft

= = 76 – 40.776 – 40.7

= = 35.3cm 35.3cm

= = 0.353 m0.353 m




e.e. Fwd Fwd Aft Aft

Initial draftsInitial drafts 3.840 3.840 4.600 4.600 Bodily riseBodily rise 0.076 -0.076 - 0.076 -0.076 -Change of drafts Change of drafts 0.353 +0.353 + 0.407 -0.407 -

Final draftsFinal drafts 4.117 m4.117 m 4.117 m4.117 m




FF4.117m4.117m 4.117m4.117m

End of Critical Period, vessel is fully rested on End of Critical Period, vessel is fully rested on blocks, draft is at even keelblocks, draft is at even keel




i.i. ALTERNATIVE METHODALTERNATIVE METHOD

a.a. Mean draftMean draft = = 4.220 m.4.220 m.

b.b. True mean draft correction True mean draft correction

== Dist. CF to amidships x trimDist. CF to amidships x trimLBPLBP

= = 5 x 0.765 x 0.76 140140

= = 0.027 m0.027 m




i.i. ALTERNATIVE METHODALTERNATIVE METHOD

c.c. True mean draft True mean draft = = Mean draft – Mean draft –

correction correction = = 4.220 – 0.027 4.220 – 0.027 = = 4.193 m4.193 m

d.d. Therefore:Therefore:

True mean draftTrue mean draft = = 4.193 m4.193 mBodily riseBodily rise = = 0.076 m -0.076 m -Final drafts Final drafts == 4.117 m4.117 m even keel even keel




ii.ii. GGGG11 = = P x KGP x KG = = 152 x 7.75152 x 7.75 W – P W – P 5000 – 1525000 – 152

= = 11781178 = = 0.243 m0.243 m 48484848

Initial GM Initial GM == KM – KG KM – KG == 9.75 m – 7.75 9.75 m – 7.75 = = 2.00 m2.00 m

Effective GM =Effective GM = 2.00 – 0.2432.00 – 0.243 == 1.757 m1.757 m




M

G1

G

B

Initial GM loss by GGInitial GM loss by GG11 after after the Critical Period…the Critical Period…

This is due to Upthrust This is due to Upthrust Force or ‘P’ Force…Force or ‘P’ Force…




OROR

MMMM11 = = P x KMP x KM = = 152 x 9.75152 x 9.75 W W 5000 5000

= = 0.296 m0.296 m

Effective GM Effective GM = = 2.00 – 0.296 2.00 – 0.296

= = 1.704 m1.704 m




MM

M1

GG

BB






Righting Moment at small angle of heel…Righting Moment at small angle of heel…

B1B1

BB

G1

M

W - PW - P

W - P

Z

External Force




G1 Z

MW - P

W - P










Righting Moment at small angle of heel…Righting Moment at small angle of heel…

B1B1

BB

G

M1

W

W

Z

MExternal Force




G Z

M1

W

W










iii.iii. RM RM = = (W – P) x G(W – P) x G11M Sin M Sin

= = (5000 – 152) x 1.757 x Sin 5(5000 – 152) x 1.757 x Sin 5

= = 742.4 t-m742.4 t-mOROR

RM RM = = W x GMW x GM11 Sin Sin

= = 5000 x 1.704 x Sin 55000 x 1.704 x Sin 5

= = 742.6 t-m742.6 t-m




WORKED EXAMPLE 2WORKED EXAMPLE 2 A ship of length A ship of length 165m165m, KG , KG 7.30m7.30m is floating in a is floating in a graving dock with drafts forward graving dock with drafts forward 5.50m5.50m, aft , aft 7.86m7.86m in in water RD 1.025. At the aft perpendicular the keel is water RD 1.025. At the aft perpendicular the keel is 0.24m0.24m above the top of the horizontal blocks. If the above the top of the horizontal blocks. If the water level has fallen in the dock by water level has fallen in the dock by 1.22m1.22m, the , the ship’s become unstable (ship’s become unstable (GM = 0mGM = 0m). ).

Calculate Calculate i.i. The drafts forward and aft at which it occursThe drafts forward and aft at which it occursii.ii. The original/initial GM The original/initial GM

Given Given Displacement for a hydrostatic mean draft of Displacement for a hydrostatic mean draft of 6.65m6.65m is is 91519151 tonnes. TPC tonnes. TPC 2424, MCTC , MCTC 120120 t-m and CF t-m and CF 3.663.66 m m abaft amidships.abaft amidships.




F

No effect on ship’s Initial Stability,No effect on ship’s Initial Stability, initial trim is initial trim is 2.36m by Stern2.36m by Stern

7.86m7.86m5.50m5.50m

Clearance 24cmClearance 24cm





5.50m5.50m

Depth of water 7.86 + 0.24 = 8.10mDepth of water 7.86 + 0.24 = 8.10m

F





7.86m7.86m5.50m5.50m

Clearance 24cm Clearance 24cm

F




Drop of water level by Drop of water level by 8cm8cm. No effect on ship’s . No effect on ship’s Initial Stability.Initial Stability.

7.86m7.86m 5.50m5.50m


F




7.86m7.86m5.50m5.50m


F





7.86m7.86m5.50m5.50m

Clearance 6cm Clearance 6cm Clearance 6cm Clearance 6cm

F





5.50m5.50m7.86m7.86m

F

Drop of water level by Drop of water level by 24cm24cm… stern post start … stern post start to touch the block…to touch the block…




5.50m5.50m

PP

7.86m7.86m

P is the P is the Upthrust ForceUpthrust Force acting at first point of acting at first point of touching the block. Commence touching the block. Commence Critical PeriodCritical Period……

F




P

Drop of water level Drop of water level 98cm, 98cm, Vessel become Vessel become unstable… unstable… Zero GM. Zero GM. Vessel is still in Critical Vessel is still in Critical Period…Period…

6.88m6.88m

F




7.86m7.86m

WLWL

WLWL

6.88m6.88m

Reduction : 98cmsReduction : 98cms




PP

dd

FF

Body rise & Trimming Moment by… Head Body rise & Trimming Moment by… Head




7.86m7.86m

WLWL

WLWL

6.88m6.88m

A : A : Body riseBody rise

WLWL7.86m 7.86m -- BrBr




7.86m7.86m

WLWL

WLWL

6.88m6.88m


BB : : Change of draft aft due to COT by Change of draft aft due to COT by HeadHead




7.86m7.86m

WLWL

WLWL

6.88m6.88m



A : A : Body riseBody rise Reduction : 98cmReduction : 98cm




Reduction = Reduction = AA ++ BB

wherewhere AA Body RiseBody RiseB B Change of draft aftChange of draft aft

due to COTdue to COT

REDUCTIONREDUCTION

== Body rise Body rise ++ Change of draft aft Change of draft aft due to COT due to COT




Fallen of water level = Fallen of water level = AA ++ BB



Fallen of water level Fallen of water level

== Body rise Body rise + + Change of draft aft Change of draft aft due to COT due to COT




Fallen WLFallen WL = = PP ++ ll x TM x TM TPCTPC L MCTCL MCTC

Fallen WL Fallen WL == PP ++ ll x P x d x P x d TPCTPC L MCTCL MCTC

9898 == PP ++ 78.8478.84 x x P x 78.84P x 78.842424 165165 120120




98 98 == PP ++ [[78.8478.84 x x PP x x 78.8478.84 ]] 2424 165165 120 120

9898 == PP ++ 0.3139265450.313926545PP

2424 11

9898 == PP ++ 7.5347.534PP2424

23522352 == 8.5348.534PP

P P = = 275.6 tonnes275.6 tonnes




P P = = 275.6 tonnes275.6 tonnes If we calculate until vessel is If we calculate until vessel is FULLY RESTFULLY REST,,

PP == MCTC x trimMCTC x trim == 120 x 120 x 236236dd 78.84 78.84





To find the drafts forward and aft…To find the drafts forward and aft…

i.i. Bodily riseBodily rise= = P P = = 275.6275.6 TPCTPC 24 24

= = 11.5 cms 11.5 cms

= = 0.115 m0.115 m





ii.ii. COT COT = = P x dP x d MCTCMCTC

= = 275.6 x 78.84275.6 x 78.84120120

= = 181cm by Head181cm by Head




iii.iii. Change of draft aft due COTChange of draft aft due COT

= = ll x COT x COT = = 78.8478.84 x 181 x 181 LL 165 165

= = 86.5cm 86.5cm

= = 0.865 m0.865 m




iv.iv. Change of draft ForwardChange of draft Forward

= = COTCOT – – Change of draft aftChange of draft aft

== 181 181 – – 86.586.5

= = 94.5cm94.5cm

= = 0.945 m0.945 m




v. v. Fwd(m)Fwd(m) Aft(m)Aft(m)

Initial draftsInitial drafts 5.500 5.500 7.860 7.860 Bodily riseBodily rise 0.115 -0.115 - 0.115 -0.115 -Change of draftsChange of drafts 0.9450.945 + + 0.8650.865 - -

Final draftsFinal drafts 6.3306.330 6.8806.880




To find the initial GM…To find the initial GM…

Mean draft Mean draft = 6.680m. = 6.680m. Trim =Trim = 2.36m by stern2.36m by stern

CF is 3.66m abaft amidships.CF is 3.66m abaft amidships.

TMD CorrectionTMD Correction = = Dist. CF to Amidships x TrimDist. CF to Amidships x Trim

LBPLBP = = 3.66 x 2.363.66 x 2.36 = = 0.052 m0.052 m

165 165





True Mean Draft (TMD)True Mean Draft (TMD)

= = Mean draft + TMD Correction Mean draft + TMD Correction = = 6.680 + 0.052 6.680 + 0.052 = = 6.732 m6.732 m

Diff of TMD Diff of TMD = = 6.732 – 6.650 6.732 – 6.650 = = 0.082 m 0.082 m = = 8.2 cm8.2 cm





Therefore additional displacementTherefore additional displacement

= = 8.2 cm x TPC (24) 8.2 cm x TPC (24) = = 196.8 t196.8 t

Displacement for TMD 6.732 mDisplacement for TMD 6.732 m

= = 9151 + 196.8 9151 + 196.8 = = 9347.8 t9347.8 t





When the ship become unstable, the GM = 0 m, When the ship become unstable, the GM = 0 m, therefore loss of GM must be equal to initial GM.therefore loss of GM must be equal to initial GM.

GGGG11 == P x KGP x KG = = 275.5 x 7.3275.5 x 7.3

W – PW – P 9347.8 – 275.59347.8 – 275.5

= = 2011.152011.15 = = 0.222 m0.222 m 9072.39072.3

Initial GM Initial GM = = 0.222 m0.222 m




Then… what will be the Then… what will be the MAXIMUM TRIMMAXIMUM TRIM

allowed, safely dockedallowed, safely docked

if the initial GM is if the initial GM is 0.222 m0.222 m….?….?




TRIMTRIM

MAX. TRIM…MAX. TRIM…??

00

0.2220.222

GMGM

2.36m2.36m

-ve-ve




Initial GM 0.222mInitial GM 0.222m

P force is …? P force is …? Maximum trim is …?Maximum trim is …?






Initial GM 0.222mInitial GM 0.222m

P force is …? P force is …? Maximum trim is …?Maximum trim is …?





dd

Trim = Trim = P x dP x d

MCTCMCTC

= = 275.5 x 78.84275.5 x 78.84

120120

Trim = 181cmTrim = 181cm

Trim = Trim = 1.81m by Stern1.81m by Stern




TRIMTRIM

MAX. TRIM MAX. TRIM 1.81m1.81m

00

0.2220.222

GMGM

2.36m2.36m

-ve-ve




Trim 1.81m by SternTrim 1.81m by Stern Virtual loss of GM…?Virtual loss of GM…?


dd

P = P = 120 x 181120 x 181

78.8478.84



W – PW – P

= = 275.5 x 7.3275.5 x 7.3

9347.8 – 275.59347.8 – 275.5

GGGG11 = = 0.2220.222

Residual GM = 0.222 – 0.222Residual GM = 0.222 – 0.222

Residual GM = Residual GM = 0.0000.000




Then… what will be the Then… what will be the final draftsfinal drafts……

if the initial GM is if the initial GM is 0.222 m 0.222 m and trim now isand trim now is

1.81m 1.81m by stern….?by stern….?





i.i. Bodily riseBodily rise= = P P = = 275.5275.5 TPCTPC 24 24

= = 11.5 cm 11.5 cm

= = 0.115 m0.115 m





ii.ii. COT COT = = P x dP x d = = 275.5 x 78.84275.5 x 78.84MCTCMCTC 120 120






= = l l x COT x COT = = 78.8478.84 x 181 x 181 LL 165 165

= = 86.5cm 86.5cm = = 0.865 m0.865 m






== 181 – 181 – 86.586.5 = 94.5cm= 94.5cm

= = 0.945 m0.945 m





Initial draftsInitial drafts 6.0506.050 7.860 7.860 Bodily riseBodily rise 0.115 -0.115 - 0.115 -0.115 -Change of draftsChange of drafts 0.9450.945 + + 0.8650.865 - -


Assuming aft draft maintain at 7.86m, new trim is 1.81m Assuming aft draft maintain at 7.86m, new trim is 1.81m by astern, therefore forward draft now is 6.05m…by astern, therefore forward draft now is 6.05m…




Worked Example 3Worked Example 3

Your vessel is going to dry dock with the following Your vessel is going to dry dock with the following conditions:conditions:

Draft forward 8.00 m and aft 9.00 m. Her displacement is Draft forward 8.00 m and aft 9.00 m. Her displacement is 30 000 tonnes. KM is 11.50 m, KG 10.90 m. MCTC 400 t-30 000 tonnes. KM is 11.50 m, KG 10.90 m. MCTC 400 t-m. TPC 38. LCF is 1.5 m abaft the amidships and LBP is m. TPC 38. LCF is 1.5 m abaft the amidships and LBP is 160 m.160 m.

The depth of water in the dock is initially The depth of water in the dock is initially 9.50m9.50m..

i.i. Find the effective GM and her new draft after water Find the effective GM and her new draft after water level level has fallen by has fallen by 95cm95cm in the dock. in the dock.

ii.ii. How much will be the further drop of water level so How much will be the further drop of water level so that vessel will take the blocks overall?that vessel will take the blocks overall?






9.0m9.0m

FF

9.5m9.5m9.5m9.5m




FF

Drop of water level by Drop of water level by 50cm50cm, No effect on ship’s , No effect on ship’s Initial Stability…Initial Stability…

9.0m9.0m9.0m9.0m

50cm drop of water level50cm drop of water level

9.0m9.0m9.0m9.0m

PP




8.55m8.55m

FF 45cm drop of water level45cm drop of water level

PP

Drop of water level by Drop of water level by 45cm45cm, effect on ship’s , effect on ship’s Initial Stability…Initial Stability…




9.00m9.00m

WLWL

WLWL

8.55m8.55m

Reduction : 45cmReduction : 45cm




9.00m9.00m

WLWL

WLWL

8.55m8.55m






9.00m9.00m

WLWL

WLWL

8.55m8.55m



A : A : Body riseBody rise Reduction : 45cmReduction : 45cm




Reduction = Reduction = AA ++ BB



REDUCTIONREDUCTION

== Body rise Body rise ++ Change of draft aft Change of draft aft due to COT due to COT




Fallen of water level = Fallen of water level = AA ++ BB



Fallen of water level Fallen of water level

== Body rise Body rise + + Change of draft aft Change of draft aft due to COT due to COT




Fallen WLFallen WL = = PP ++ ll x TM x TM TPCTPC L MCTCL MCTC

Fallen WL Fallen WL == PP ++ ll x P x d x P x d TPCTPC L MCTCL MCTC

4545 == PP ++ 78.578.5 x x P x 78.5P x 78.53838 160160 400 400




45 45 == PP ++ [ [ 78.578.5 x x PP x x 78.578.5 ]] 3838 160160 400 400

4545 == PP ++ 0.0962851560.096285156PP

3838 11

4545 == PP ++ 3.6593.659PP3838

17101710 == 4.6594.659PP





GGGG11 = = P x KGP x KG = = 367.05 x 10.9367.05 x 10.9 W – P W – P 30 000 – 367.030 000 – 367.0

GGGG11 = = 0.135 m0.135 m

Initial GM Initial GM == 0.600 m0.600 m

Effective GM Effective GM == 0.600 – 0.1350.600 – 0.135

== 0.465 m0.465 m




OROR

MMMM11 = = P x KMP x KM = = 367.05 x 11.5367.05 x 11.5 W W 30 000 30 000

MMMM11 = = 0.141 m0.141 m


= = 0.459 m0.459 m





i.i. Bodily riseBodily rise = = P P = = 367.0367.0 TPCTPC 38 38

= = 9.66cm9.66cm

= = 0.097 m0.097 m





ii.ii. COT COT = = P x dP x d = = 367.0 x 78.5367.0 x 78.5MCTCMCTC 400 400






= = l l x COT x COT = = 78.578.5 x 72 x 72 LL 160160

= = 35.3cm 35.3cm

= = 0.353 m0.353 m






== 72.0 – 72.0 – 35.335.3

= = 0.367 m0.367 m





Initial draftsInitial drafts 8.000 8.000 9.000 9.000 Bodily riseBodily rise 0.097 -0.097 - 0.097 -0.097 -Change of draftsChange of drafts 0.3670.367 + + 0.3530.353 - -





New Trim = 8.55 – 8.27New Trim = 8.55 – 8.27 == 0.28m by Stern 0.28m by Stern assuming ‘F’ constantassuming ‘F’ constant

PP == MCTC x TMCTC x T == 400 x 28400 x 28 dd 78.5 78.5





Further drop Further drop = = PP + + ll x x P x dP x d vessel fully restvessel fully rest TPC TPC L L MCTC MCTC

== 142.7142.7 + + 78.578.5 x x 142.7 x 78.5 142.7 x 78.5

38 38 160160 400 400

== 17.5cm17.5cm




SINGLE POINTSINGLE POINT

GROUNDING GROUNDING




SINGLE POINT GROUNDINGSINGLE POINT GROUNDING

A vessel floating at drafts forward 8.70 m, A vessel floating at drafts forward 8.70 m, aft 9.40 m grounds at a point 30 m aft of the aft 9.40 m grounds at a point 30 m aft of the forward perpendicular. forward perpendicular.

Estimate the drafts of the vessel and the GM Estimate the drafts of the vessel and the GM after the tide has fallen by 70cm.after the tide has fallen by 70cm.

MCTC 340 t-m, TPC 28, KG 7.60 m, KM 8.40 MCTC 340 t-m, TPC 28, KG 7.60 m, KM 8.40 m, LBP 162 m. LCF 78 m forward of Aft m, LBP 162 m. LCF 78 m forward of Aft Perpendicular and displacement is 29 000 Perpendicular and displacement is 29 000 tonnes. tonnes.




FF

9.40m9.40m 8.70m8.70m

PP

RockRock

30m




PP = = ??

RockRock

FF

Tide fallen by 70cms…Tide fallen by 70cms…

Aft…?Aft…?Fwd…?Fwd…?




Draft at PDraft at P

WLWL

WLWL

New draft at PNew draft at P

Fallen of tide by 70cmFallen of tide by 70cm





WLWL

WLWL



Draft at P Draft at P -- BrBrWLWL





WLWL

WLWL



BB : : Change of draft at Change of draft at PP due to COT by Stern due to COT by Stern




Fallen of tide by 70cmFallen of tide by 70cm


WLWL

WLWL



BB : : Change of draft at Change of draft at PP due to COT by Stern due to COT by Stern





Fallen of tide = Fallen of tide = AA ++ BB

wherewhere AA Body RiseBody RiseB B Change of draft at Change of draft at PP

due to COT by Sterndue to COT by Stern

Fallen of tide Fallen of tide

== Body rise Body rise + + Change of draft at Change of draft at PP due to COT Stern due to COT Stern




Fallen of tideFallen of tide = = PP ++ ll x TM x TM TPCTPC L MCTCL MCTC

Fallen of tideFallen of tide == PP ++ ll x P x d x P x d TPCTPC L MCTCL MCTC

7070 == PP ++ 5454 x x P x 54P x 542828 162162 340 340




FF

9.40m9.40m 8.70m8.70m

PP

RockRock

30m54m54m




70 70 == PP ++ [ [ 5454 x x PP x x 5454 ]] 2828 162162 340 340

7070 == PP ++ 0.0529411760.052941176PP

2828 11

7070 == PP ++ 1.4821.482PP2828

19601960 == 2.4822.482PP






i.i. Bodily riseBodily rise = = P P = = 789.7 789.7 TPCTPC 28 28

= = 28cm28cm

= = 0.280 m 0.280 m





ii.ii. COT COT = = P x dP x d = = 789.7 x 54 789.7 x 54 MCTCMCTC 340340

= = 125.4cm by Stern 125.4cm by Stern





= = l l x COT x COT = = 7878 x 125.4 x 125.4 LL 116262

= = 60.4 cm 60.4 cm

= = 0.604 m 0.604 m






== 125.4 – 60.4125.4 – 60.4

= = 65cm 65cm

= = 0.650 m 0.650 m





Initial draftsInitial drafts 8.700 8.700 9.400 9.400

Bodily riseBodily rise 0.280 -0.280 - 0.280 -0.280 -

Change of draftsChange of drafts 0.650 -0.650 - 0.604 +0.604 +

Final draftsFinal drafts 7.770 m7.770 m 9.724 m9.724 m




ii.ii. Estimated GMEstimated GM

GGGG11 = = P x KGP x KG = = 789.7 x 7.60789.7 x 7.60 W – P W – P 29000 – 789.729000 – 789.7

= = 0.213 m0.213 m

Initial GM Initial GM = = 8.40 m – 7.608.40 m – 7.60 == 0.80 m0.80 m

Effective GM = Effective GM = 0.800 – 0.213 0.800 – 0.213 = = 0.587 m0.587 m




ii.ii. Estimated GMEstimated GM

MMMM11 = = P x KMP x KM = = 789.7 x 8.40789.7 x 8.40 W W 29000 29000

= = 0.229 m0.229 m


= = 0.571 m0.571 m




Thank you…Thank you…

d1mc semester 1 / ship stability / march 2007 /capt. mrd. dry docking 1 1 to be a world class...

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