Dry-docking-All About To Know
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D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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D. Dry Docking
Trim
Effect on G
Simpson Rules
Dry Docking
Simplified Stab
Statical Stab
Revisions Ex.
Inclining Inclini ng Test Test
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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LEARNING OBJECTIVES To understand the virtual loss of GM and the calculations. •
•
•
D. Dry Docking
To calculate the maximum trim allowed to maintain a minimum stated GM. To understand the safe requirements for a ship prior enter into dry dock.
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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LEARNING OBJECTIVES To understand the critical period during dry docking process. •
•
•
D. Dry Docking
To calculate the ship‟s drafts after the water level has fallen and after the ship has taken the block overall. Effect to stability when vessel has run aground (single point).
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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Anybody would like to share their experience during dry docking….? docking….?
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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Before enter into dry dock, vessel must have… 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
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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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
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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F
No effect on ship‟s Initial Stability… Stability…
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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When coming into Dry Dock: •
The rate of pumping will be reduced the ship's sternpost near the as block.
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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Sueing Point
F
Commence touching the ground… „Sueing Point‟ „Sueing Point‟ D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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When coming into Dry Dock: •
•
D. Dry Docking
Once the sternpost is touching the block, the UP-THRUST UP-THRUST forces forces start to act against the sternpost. At this moment part of ship's weight gets transferred to the keel blocks.
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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P F
P is the Upthrust Force acting Force acting at first point of touching the ground. Commence Critical Period… Period… D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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Sueing Point at „AP‟
D. Dry Docking
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P
K
P is the Upthrust Force acting Force acting at first point of touching the ground. Commence Critical Period… Period… D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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D. Dry Docking
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When coming into Dry Dock: •
•
When ship's weight gets transferred to the keel blocks, vessel will suffer loss on her GM. The time interval between the sternpost landing on the blocks and the ship taking the blocks overall is referred to as the CRITICAL PERIOD. PERIOD.
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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P
F
P force is increasing gradually as the trim change by Head…Vessel is still in Critical Period… Period…
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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When coming into Dry Dock: •
Vessel must have positive effective GM that to be maintained throughout the critical period.
•
If not vessel may heel over, slip off the blocks when there is an external force acting and heel the ship.
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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P
F
Vessel is fully rest on the blocks… End of Critical Period
D. Dry Docking
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D. Dry Docking
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M G1
Initial GM loss by GG1 after completed Period… the Critical Period…
G This is due to Upthrust Force or „P‟ Force… Force… B
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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P
F
What is the total P Force during during Critical Period __?___ tonnes “How much weight to be discharged in order to bring the ship from trim by stern to even-keel… even-keel…” ”
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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CALCULATION OF UPTHRUST FORCE AT THE STERNPOST STERNPOST - 'P' FORCE
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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w d
F
Weight discharged to even keel the draft… draft… Trimming Moment
=
w x d t-m by Head
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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F
After weight discharged… discharged… T M By Head
=
T M By Stern
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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P
d
F
P is the Upthrust Force or weight discharged discharged to the blocks… blocks… T.M
=
w x d
=
P x d t-m by Head
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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P
F
Vessel is fully rest on the blocks, Change of Trim by Head and finally vessel at even keel drafts… End of Critical Period… Period…
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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Change of Trim =
Trimming MCTCMoment (TM)
Whereby TM
=
w x d
Change of Trim =
P x d MCTC
P
=
=
P x d
COT x MCTC tonnes d
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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Exercise in classroom MV OneSuch, LBP 120m is going to dry dock at the following condition in sea water Draft forward is 3.5m and aft is 4.0m, distance sueing point (AP) to F is 57.5m. Her displacement is 4600 tonnes, MCTC is 86 t-m and TPC 15.45 Calculate i. Th The e am amou ount nt of up up-t -thr hrus ustt fo forc rce e (P) at the end of Critical Period? ii.. Fin ii Final al dra drafts fts fo forw rward ard an and d aft? aft?
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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Calculation of P force… force… P
P
=
COT x MCTC d
=
50 x 86 57.5
=
74.8 tonnes
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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Body rise
CODA =
=
=
P TPC
=
74.8 15.45
=
4.8cm
=
0.048m
57.5 x 50 120
24cm
COT
=
Pxd MCTC
=
74.8 x 57.5 86
=
50cm
CODF = =
50 – 24 26cm
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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Forward
Aft
Initial draft
3.500m
4.000m
Body rise
0.048m -
0.048m –
COD
0.260m +
0.240m –
Final draft
3.712m
3.712m
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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d
P
F
P is the Upthrust Force or weight discharged discharged to the blocks… blocks… T.M
=
w x d
=
P x d t-m by Stern
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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P
F
Vessel is fully rest on the blocks, Change of Trim by Stern Stern and finally vessel at even keel drafts… End of Critical Period… Period…
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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Change of Trim
=
Trimming Moment (TM) MCTC
Whereby
TM
=
w x d
Change of Trim
=
P x d MCTC
=
COT x MCTC tonnes d
P
=
P x d
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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Virtual Loss Of GM During Critical Period
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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•
Method 1 – GG1
•
Method 2 – MM1
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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Method 1 •
When the vessel comes in contact with it is therethe is ablocks, transfer of assumed weight 'P'that weight 'P' from the keel to the blocks.
•
Hence there is a virtual rise of ship's G (discharged (discharged of weight below G) G)
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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P
d
F
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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P
d
F
Trimming Moment by… Head Head
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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P
K
P is the Upthrust Force acting at first point of touching the ground. Commence Critical Period…”weight Period…”weight discharged from the ship” ship”
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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M
G1
G
K
Reduction or or Loss Loss of GM = GM = GG1
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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M G1
Initial GM loss by GG1 during the Critical Period… Period…
G
B
This is due to Upthrust Upthrust Force Force or „P‟ Force… Force…
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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1 Method 1
GG1
=
P x KG W-P
in metres
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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P
M G1 G B
W
During Critical Period… part of ship body is still floating floating
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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P
W-P
External Force
M
G B B1
W
is force… inclined Vessel external force… to a small angle by an
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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P M
External Force
G1 G
K W-P
Method 1 Discharged 1 Discharged of weight, shift of GG1
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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P M X
X = KG1 Sin
G1
G
K W-P
Method 1 Discharged 1 Discharged of weight, shift of GG1
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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P M
Y
G1 G
K W-P
W
Method 1 Discharged 1 Discharged of weight, shift of GG1
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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Y G1
Y = GG1 Sin
G 1… … Discharged of weight, shift of GG1… Method 1
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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P M
Y
G1 X G
K W-P
W
Method 1 Discharged 1 Discharged of weight, shift of GG1
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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P
X
Y
G1 G
K
W
=G
G1
W D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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X =
KG1 x Sin PX
Y =
GG1 Sin
=
WY
P x KG1 x Sin
=
W x GG1 Sin
P x KG1
=
W x GG1
=
W x GG1
P x (KG + GG1)
(P x KG) + (P x GG1) =
W x GG1
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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(P x KG) + (P x GG1)
=
W x GG1
P x KG
=
(W x GG1) – (P x GG1)
P x KG
=
(W – P) x GG1
P x KG W – P
=
GG1
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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Therefore the formula is…
GG1
=
P x KG W-P
D. Dry Docking
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W-P External Force
M
G1
Z
B B1
W-P
Righting Moment at small angle of heel… heel…
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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W-P
M
Righting Moment = W x GZ = W x GM Sin In this case,
Righting Moment = (W – P) x G1M Sin
G1
Z W-P
D. Dry Docking
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•
Method 1 – GG1
•
Method 2 – MM1
D. Dry Docking
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Method 2 •
•
When the vessel comes in contact with the blocks, it is assumed that there is a transfer of buoyancy 'P' buoyancy 'P' to the keel blocks. Hence there is a reduction in KM while the weight and KG are remains constant.
D. Dry Docking
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P
d F Reduction in Buoyancy
D. Dry Docking
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P
d
Reduction in Buoyancy
F
D. Dry Docking
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P
K
P is the Upthrust Force acting at first point of touching the ground. Commence Critical Period…”buoyancy Period…” buoyancy reduction from the ship” ship”
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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P
Buoyancy Reduction
K
P is the Upthrust Force acting at first point of touching the ground. Commence Critical Period…”buoyancy Period…” buoyancy reduction from the ship” ship”
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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M M1
B B1 K
Reduction or or Loss Loss of GM = GM = MM1
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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Method 2 2
MM1
=
P x KM in metres W
D. Dry Docking
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M Initial GM loss by MM1 after M1 G
B
the Critical Period… Period… This is due to Upthrust Upthrust Force or „P‟ Force… Force…
D. Dry Docking
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P
M M1 G B
W During Critical Period, part of ship body is still floating
D. Dry Docking
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P
W-P
External Force
M
G B B1
W
Vessel is inclined to a small angle by an external force… force…
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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W-P
P W
M
Y
M 1 X G
K
W
2… Transferred of buoyancy, shift of MM1… Method 2…
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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P
X
W
W
W-P
M1 G
=
M
M1 Y
K
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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X = KM1 x Sin PX
Y =
MM1 Sin
=
(W – P) x Y
=
(W – P) x MM1 Sin
P x KM1
=
(W – P) x MM1
P x KM1
=
W x MM1 – P x MM1
P x KM1 x Sin
P x KM1 + P x MM1 =
W x MM1
P (KM1 + MM1)
W x MM1
=
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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P (KM1 + MM1) =
W x MM1
P x KM
=
W x MM1
P xWKM
=
MM1
Therefore the formula is… is…
MM1
=
P x KM W
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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W M External Force
M1
G
Z
B B1
W
Righting Moment at small angle of heel… heel… D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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W
M1
Righting Moment = W x GZ = W x GM Sin In this case,
Righting Moment = W x GM1 Sin
G
Z
W D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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SUMMARY… SUMMARY…
•
•
Method 1 – GG1 : Weight transferred Method 2 – MM1 : Buoyancy transferred
D. Dry Docking
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Exercise in classroom
…continued
MV OneSuch is going to dry dock at the following condition in sea water Draft forward is 3.5m and aft is 4.0m, distance sueing point (AP) to F is 57.5m. Her displacement is 4600 tonnes, MCTC is 86 t-m, Calculate the amount of up-thrust force (P) during Critical Period and the virtual loss of
GM if GM if KM is 8.0m and KG is 7.2m. 7.2m. D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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Calculation of P force… force… P
=
COT x MCTC d
P
=
50 x 86 57.5
=
74.8 tonnes
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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Virtual loss of GM ( (GG GG1) method… method… GG1 =
P x KG W-P
=
GG1 =
74.8 x 7.2 4600 – 74.8 0.119m
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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Virtual loss of GM ( (MM MM1) method… method… MM1 =
P x KM W
=
MM1 =
74.8 x 8.0 4600 0.130m
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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Comparison the Virtual loss of GM between (MM1) and (GG (GG1) method… method…
Different is… =
0.130 – 0.119
=
0.011m … ± 1cm
D. Dry Docking
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Effect of Trim Trim In Dry Docking
D. Dry Docking
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Change of Trim
=
Trimming Moment (TM) MCTC
Whereby
TM
=
w x d
Change of Trim
=
P x d MCTC
=
COT x MCTC tonnes d
P
=
P x d
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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Example 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?
D. Dry Docking
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Case 1 0 m / Even keel
Case 2 0.5 m by Stern
Case 3 5 m by Stern
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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Case 1 0 m / Even keel
Case 2 0.5 m by Stern
Case 3 5 m by Stern
„P‟… Calculate „P ‟…
Calculate „P „P‟… ‟…
Calculate „P „P‟… ‟…
P
P
P
= MCTC x trim d
= MCTC x trim d
= MCTC x trim d
D. Dry Docking
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Case 1 0 m / Even keel
Case 2 0.5 m by Stern
Case 3 5 m by Stern
„P‟… Calculate „P‟…
Calculate „P‟… „P‟…
Calculate „P‟… „P‟…
P
P
P
= MCTC x trim d
= MCTC x trim d
= 200 x 0
= 200 x 50
80
80
= MCTC x trim d = 200 x 500 80
D. Dry Docking
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Case 2 0.5 m by Stern
Case 1 0 m / Even keel
Case 3 5 m by Stern
„P‟… Calculate „P‟…
Calculate „P‟… „P‟…
Calculate „P‟… „P‟…
P
P
P
P
= MCTC x trim d
= MCTC x trim d
= 200 x 0
= 200 x 50
80
80
= 0 tonne
P = 125 tonnes
= MCTC x trim d = 200 x 500 80
P = 1250 tonnes tonnes
D. Dry Docking
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Case 1 0 m / Even keel
Case 2 0.5 m by Stern
Case 3 5 m by Stern
GM… Virtual Loss of GM…
Virtual Loss of GM… GM…
Virtual Loss of GM… GM…
D. Dry Docking
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Case 1 0 m / Even keel
Case 2 0.5 m by Stern
Case 3 5 m by Stern
GM… Virtual Loss of GM…
Virtual Loss of GM… GM…
Virtual Loss of GM… GM…
GG1
GG1
GG1
=
P
x KG W – P
=
P
x KG W – P
=
P
x KG W-P
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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Case 1 0 m / Even keel
Case 2 0.5 m by Stern
Case 3 5 m by Stern
GM… Virtual Loss of GM…
Virtual Loss of GM… GM…
Virtual Loss of GM… GM…
GG1
GG1
GG1
=
=
P
x KG W – P 0 x 6
5000 – 0
=
P
x KG W – P
= 125 125 x x 6 5000 –125 125
=
P
x KG W-P
= 1250 1250 x x 6 5000 –1250
D. Dry Docking
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Case 1 0 m / Even keel
Case 2 0.5 m by Stern
Case 3 5 m by Stern
GM… Virtual Loss of GM…
Virtual Loss of GM… GM…
Virtual Loss of GM… GM…
GG1
GG1
GG1
=
=
P
x KG W – P 0x6
=
= 0m
x KG W – P
=
0.154 m m
P
x KG W-P
= 1250 x 6
5000 –125 GG1
=
= 125 x 6
5000 – 0 GG1
P
5000 –1250 GG1
= 2.0 m
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
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Case 1 0 m / Even keel
Case 2 0.5 m by Stern
Case 3 5 m by Stern
Old GM = 1.0m
Old GM = 1.0m
Old GM = 1.0m
New GM… GM…
New GM… GM…
New GM… GM…
=
GM - GG1
=
GM - GG1
=
GM - GG1
=
1.0 – 0 0
=
1.0 – 0.154 0.154
=
1.0 – 2.0 2.0
=
1.0 m
=
0.846 m
=
- 1.0 m
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
91
To Be A World Class Maritime Academy
Residual GM MAX. TRIM…? TRIM…?
1.0
TRIM increased GM decreased
0.846
TRIM 0
- 1.0
0.5
5.0
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
92
To Be A World Class Maritime Academy
•
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. Maximum Trim is….? is….?
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
93
To Be A World Class Maritime Academy
During Critical Period… Period… 1.0m P force is …? Initial GM 1.0m
M
G1
Virtual Loss of GM = 1.0m
GG1 is Virtual Loss of GM GM
G
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
94
To Be A World Class Maritime Academy
During Critical Period… Period… 1.0m P force is …? Initial GM 1.0m
M
G1
Virtual Loss of GM = 1.0m
GG1 1.0 1.0
=
x KG W-P = Px6 5000 – P
5000 - P 5000 P
P
= 6P = 7P = 714.28 tonnes
GG1 is Virtual Loss of GM GM
G
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
95
To Be A World Class Maritime Academy
1.0m P force is …? Initial GM 1.0m
Maximum trim is …? …?
Virtual Loss of GM = 1.0m
P
= 714.28 tonnes
GG
P
Trim
= MCTC x trim d = Pxd MCTC = 714.28 x 80 200 = 285.7 cms
Trim
= 2.86 m by Stern
1
1.0
=
x KG W-P = Px6 5000 – P
5000 - P 5000 P
P
= 6P = 7P = 714.28 tonnes
Trim
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
96
To Be A World Class Maritime Academy
Residual GM 1.0
MAX. TRIM …… ……? 2.86m ?
0.846
TRIM 0
- 1.0
0.5
5.0
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
97
To Be A World Class Maritime Academy
CONCLUSION: •
The virtual loss of GM is NIL NIL as as vessel having zero trim.
•
•
The loss is increased as the trim increased. Maximum trim is depend upon the initial GM
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
98
To Be A World Class Maritime Academy
WORKED EXAMPLE 1 A ship of 140m 140m in in length, displacement 5000t 5000t and and upright is to enter dry dock with drafts forward 3.84m 3.84m,, aft 4.60m 4.60m.. Given the following hydrostatic particulars: TPC 20 tonnes 20 tonnes MCTC 150 t150 t- m CF 5m forward 5m forward of amidships KM 9.75m The blocks of the dry dock are horizontal horizontal.. i.
Calc Ca lcul ulat ate e the the draf drafts ts of of the the vess vessel el at th the e inst instan ants ts whe when n she she is taking the blocks forward and aft.
ii.
The shi ship's p's eff effect ectiv ive e GM at thi this s mome moment nt if th the e KG is 7.75m 7.75m
iii. The Right Righting ing Moment Moment at this this inst instant ant for an an angle angle of of heel heel 5º. 5º.
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR
99
To Be A World Class Maritime Academy
F 4.60m
Trim 76 cm by Stern
No effect on ship‟s Initial Stability… Stability…
3.84m
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 100
To Be A World Class Maritime Academy
P
F 4.60m
Trim 76 cm by Stern
3.84m
P is the Upthrust Force acting Force acting at first point of
touching the ground, commence Critical Period… Period… D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 101
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P
F
Change of Trim 76cms by Head
Even keel draft
What is the total P Force during Critical Period?
End of Critical Period… Period… D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 102
To Be A World Class Maritime Academy
Ship‟s trimmed i.
a.
=
4.60 – 3.84 = 0.76 m by Stern Stern
P
=
MCTC x trim d
P
=
152 tonnes tonnes
Bodily rise =
P
=
= 152
TPC b. Change of Trim
=
150 x 76 75
= 7.6 cms =
20 76 cms by Head Head
0.076 m m
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 103
To Be A World Class Maritime Academy
c. Ch Chan ange ge of dr draf aftt aft aft du due e COT COT =
l
x COT
L = 75 x 76 140 = 40.7cm = 0.407 m m
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 104
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d. Change of draft forward due COT = COT – Change of draft aft = 76 – 40.7 = 35.3cm = 0.353 m m
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 105
To Be A World Class Maritime Academy
e.
Fwd
Aft
Initial drafts Bodily rise Change of drafts
3.840 0.076 0.353 +
4.600 0.076 0.407 -
Final drafts
4.117 m m
4.117 m
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 106
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F 4.117m
4.117m
End of Critical Period, vessel is fully rested on
blocks, draft is at even keel D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 107
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i.
ALTERNATIVE METHOD
a.
Mean draft
b.
True mean draft correction
=
4.220 m.
=
Dist. CF to amidships x trim LBP
=
5 x 0.76 140
=
0.027 m
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 108
To Be A World Class Maritime Academy
i.
ALTERNATIVE METHOD
c. True mean draft =
Mean draft – correction
= =
4.220 – 4.193 m 0.027
= = =
4.193 0.076 m m4.117 m even keel
d. Therefore: True meanrise draft Bodily Final drafts
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 109
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ii.
GG1
=
P x KG W – P
=
152 x 7.75 5000 – 152
=
1178
=
0.243 m
=
9.75 m – 7.75
4848 Initial GM
= =
Effective Effectiv e GM =
KM – KG 2.00 m
2.00 – 0.243 =
1.757 m
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 110
To Be A World Class Maritime Academy
M G1
Initial GM loss by GG1 after the Critical Period… Period…
G
B
This is due to Upthrust Force or „P‟ Force… Force…
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 111
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OR MM1 =
P x KM
=
W
Effective Effecti ve GM
152 x 9.75 5000
=
0.296 m
=
2.00 – 0.296
=
1.704 m
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 112
To Be A World Class Maritime Academy
M Initial GM loss by MM1 after M1 G
B
the Critical Period… Period… This is due to Upthrust Force or „P‟ Force… Force…
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 113
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W-P External Force
M
G1
Z
B B1
W-P
Righting Moment at small angle of heel… heel… D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 114
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W-P
M Righting Moment = W x GZ =
W x GM Sin
In this case, Righting Moment = (W – P) x G1M Sin
G1
Z
W-P D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 115
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W M External Force
M1
G
Z
B B1
W
Righting Moment at small angle of heel… heel… D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 116
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W
M1 Righting Moment = W x GZ =
W x GM Sin
In this case, Righting Moment = W x GM1 Sin
G
Z
W D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 117
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iii.
RM
=
(W – P) x G1M Sin
=
(5000 – 152) x 1.757 x Sin 5
=
742.4 t-m
=
W x GM1 Sin
=
5000 x 1.704 x Sin 5
=
742.6 t-m
OR RM
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 118
To Be A World Class Maritime Academy
WORKED EXAMPLE 2 A ship of length 165m 165m,, KG 7.30m 7.30m is floating in a graving dock with drafts forward 5.50m 5.50m,, aft 7.86m 7.86m in in water RD 1.025. At the aft perpendicular the keel is 0.24m above the top of the horizontal blocks. If the 0.24m water level has fallen in the dock by 1.22m 1.22m,, the ship‟s become ship‟s become unstable (GM (GM = 0m). 0m).
Calculate i. The drafts forward and aft at which it occurs ii. The original/initial GM Given Displacement for a hydrostatic mean draft of 6.65m 6.65m is 9151 9151 tonnes. tonnes. TPC 24 24,, MCTC 120 120 t-m t-m and CF 3.66 3.66 m m
abaft amidships. D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 119
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F 7.86m
5.50m
Clearance 24cm
No effect on ship‟s Initial Stability, Stability, initial trim is
2.36m by Stern D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 120
To Be A World Class Maritime Academy
F 5.50m Depth of water 7.86 + 0.24 = 8.10m
No effect on ship‟s Initial Stability… Stability…
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 121
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F 7.86m
Clearance 24cm
No effect on ship‟s Initial Stability… Stability…
5.50m
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 122
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F 7.86m
5.50m
Clearance 16cm
Drop of water level by 8cm 8cm.. No effect on ship‟s
Initial Stability. D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 123
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F 7.86m
Clearance 12cm
5.50m
Drop of water level by 12cm 12cm.. No effect on ship s Initial Stability. D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 124
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F 7.86m
Clearance 6cm
5.50m
Drop of water level by 18cm 18cm.. No effect on ship s Initial Stability. D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 125
To Be A World Class Maritime Academy
F 7.86m
5.50m
Drop of water level by 24cm 24cm… … stern post start to touch the block… block… D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 126
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P
F 7.86m
5.50m
P is the Upthrust Force acting Force acting at first point of
touching the block. Commence Critical Period… Period… D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 127
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P
F 6.88m
Drop of water level 98cm, Vessel become
unstable… Zero GM. Vessel is still in Critical Period… Period… D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 128
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7.86m
WL
Reduction : 98cms 6.88m
WL
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 129
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P d
F
Body rise & Trimming Moment by… Head Head
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 130
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7.86m A:
WL
Body rise
7.86m - Br
6.88m
WL
WL
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 131
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7.86m
7.86m - Br B :
WL
WL
Change of draft aft due to COT by Head
6.88m
WL
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 132
To Be A World Class Maritime Academy
7.86m A:
Body rise
WL Reduction : 98cm
7.86m - Br B :
WL
Change of draft aft due to COT by Head
6.88m
WL
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 133
To Be A World Class Maritime Academy
Reduction = A
+
B
where
A
Body Rise
B
Change of draft aft due to COT
REDUCTION =
Body rise + Change of draft aft due to COT
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 134
To Be A World Class Maritime Academy
Fallen of water level = where
A
+
A
Body Rise
B
Change of draft aft due to COT
Fallen of water level =
B
Body rise + Change of draft aft due to COT
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 135
To Be A World Class Maritime Academy
Fallen WL
Fallen WL
98
=
=
=
P TPC
+
P TPC
+
P 24
+
l
L
l
L
x
TM MCTC
x Pxd MCTC
78.84 165
x P x 78.84 120
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 136
To Be A World Class Maritime Academy
98
=
P 24
+
[78.84 165
98
=
P 24
+
0.313926545P 0.313926545 P 1
98
=
P
+ 24
7.534P 7.534P
2352 =
8.534 8.534P P
x
P x 78.84 ] 120
P
=
275.6 tonnes tonnes
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 137
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P
=
275.6 tonnes tonnes
If we calculate until vessel is FULLY REST REST,,
P
=
MCTC x trim d
P
=
359.2 tonnes
=
120 x 236 236 78.84
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 138
To Be A World Class Maritime Academy
To find the drafts forward and aft… aft…
i.
Bodily rise =
P = TPC
=
11.5 cms
=
0.115 m m
275.6 24
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 139
To Be A World Class Maritime Academy
To find the drafts forward and aft… aft…
ii.
COT =
Pxd MCTC
=
275.6 x 78.84 120
=
181cm by Head Head
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 140
To Be A World Class Maritime Academy
iii.
COT Change of draft aft due COT =
l x
COT
L =
86.5cm
=
m 0.865 m
=
78.84 165
x 181
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 141
To Be A World Class Maritime Academy
iv.
Forward Change of draft Forward =
COT –
Change of draft aft
=
181 –
86.5
=
94.5cm
=
m 0.945 m
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 142
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Fwd(m)
Aft(m)
Initial drafts Bodily rise Change of drafts
5.500 0.115 0.945 +
7.860 0.115 0.865 -
Final drafts drafts
6.330 6.330
6.880 6.880
v.
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 143
To Be A World Class Maritime Academy
GM… To find the initial GM… Mean draft
= 6.680m. Trim = 2.36m by stern
CF is 3.66m abaft amidships. TMD Correction Correction
=
Dist. CF to Amidships x Trim LBP
=
3.66 x 2.36 = 165
0.052 m
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 144
To Be A World Class Maritime Academy
GM… To find the initial GM… True Mean Draft (TMD) = = =
Mean draft + TMD Correction 6.680 + 0.052 6.732 m m
Diff of TMD = 6.732 – 6.650 = 0.082 m
=
8.2 cm cm
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 145
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To find the initial GM… GM… Therefore additional displacement displacement = =
8.2 cm x TPC (24) 196.8 t t
Displacement for TMD 6.732 m m = =
9151 + 196.8 9347.8 t t
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 146
To Be A World Class Maritime Academy
GM… To find the initial GM… When the ship become unstable, the GM = 0 m, therefore loss of GM must be equal to initial GM. GM.
GG1 = =
P x KG W – P
=
275.5 x 7.3 9347.8 – 275.5
2011.15 9072.3
=
0.222 m
Initial GM D. Dry Docking
=
0.222 m m D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 147
To Be A World Class Maritime Academy
Then… what will be the MAXIMUM TRIM TRIM allowed, safely docked if the initial GM is 0.222 m….? m….?
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 148
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GM 0.222
MAX. TRIM…? TRIM…?
2.36m 0 -ve
TRIM
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 149
To Be A World Class Maritime Academy
Initial GM 0.222m P force is …?
Virtual Loss of GM = 0.222m
P = 275.5 tonnes
Maximum trim is …? …?
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 150
To Be A World Class Maritime Academy
Initial GM 0.222m P force is …?
Virtual Loss of GM = 0.222m
Maximum trim is …? …? P
= 275.5 tonnes
P
= MCTC x trim d = Pxd MCTC
Trim P = 275.5 tonnes
=
Trim
275.5 x 78.84 120 = 181cm
Trim
D. Dry Docking
= 1.81m by Stern
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 151
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GM 0.222
MAX. TRIM 1.81m
2.36m 0 -ve
TRIM
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 152
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Trim 1.81m by Stern
Virtual loss of GM…?
P = MCTC x trim
GG1
= P x KG
d
W – P
P = 120 x 181
=
78.84 P = 275.5 tonnes
275.5 x 7.3 9347.8 – 275.5
GG1
= 0.222 0.222
Residual GM = 0.222 – 0.222 Residual GM = 0.000 D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 153
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Then… what will be the final drafts… drafts… if the initial GM is 0.222 m and trim now is is 1.81m by stern….? stern….?
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 154
To Be A World Class Maritime Academy
To find the drafts forward and aft… aft…
i.
Bodily rise =
P = TPC
=
11.5 cm
=
0.115 m m
275.5 24
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 155
To Be A World Class Maritime Academy
To find the drafts forward and aft… aft…
ii.
COT
=
Pxd = MCTC
275.5 x 78.84 120
=
Head 181cm by Head
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 156
To Be A World Class Maritime Academy
iii.
Change of draft aft due COT COT =
x COT
=
78.84 165
86.5cm
=
m 0.865 m
l
L
=
x 181
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 157
To Be A World Class Maritime Academy
iv.
Change of draft Forward Forward =
COT
–
=
181 – 86.5
=
0.945 m m
Change of draft aft =
94.5cm
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 158
To Be A World Class Maritime Academy
Assuming aft draft maintain at 7.86m, new trim is 1.81m by astern, therefore forward draft now is 6.05m… 6.05m…
Fwd(m)
Aft(m)
Initial drafts Bodily rise Change of drafts
6.050 6.050 0.115 0.945 +
7.860 0.115 0.865 -
drafts Final drafts
6.880 6.880
6.880 6.880
v.
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 159
To Be A World Class Maritime Academy
Worked Example 3 Your vessel is going to dry dock with the following conditions: 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 tm. TPC 38. LCF is 1.5 m abaft the amidships and LBP is 160 m. The depth of water in the dock is initially 9.50m 9.50m.. i.
Find the effective GM and her new draft after water level has fallen by 95cm 95cm in in the dock.
ii. How much will be the further drop of water level so that vessel will take the blocks overall? D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 160
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F 9.0m 9.5m Clearance 50cm
No effect on ship‟s Initial Stability… Stability…
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 161
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P
F 9.0m
9.0m
50cm drop of water level
Drop of water level by 50cm 50cm,, No effect on ship‟s Initial Stability… Stability… D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 162
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P F 8.55m
45cm drop of water level
Drop of water level by 45cm 45cm,, effect on ship‟s Initial Stability… Stability… D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 163
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WL
9.00m
Reduction : 45cm
8.55m
D. Dry Docking
WL
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 164
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9.00m A:
WL
Body rise
9.00m - Br
WL
WL
8.55m
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 165
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9.00m
9.00m - Br B :
Change of draft aft due to COT by Head
WL
WL
WL
8.55m
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 166
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WL
9.00m A:
Body rise
Reduction : 45cm
9.00m - Br B :
Change of draft aft due to COT by Head
WL
WL
8.55m
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 167
To Be A World Class Maritime Academy
Reduction = A
+
B
where
A
Body Rise
B
Change of draft aft due to COT
REDUCTION =
Body rise + Change of draft aft
due to COT
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 168
To Be A World Class Maritime Academy
Fallen of water level = where
A
+
A
Body Rise
B
Change of draft aft due to COT
Fallen of water level =
B
Body rise + Change of draft aft
due to COT
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 169
To Be A World Class Maritime Academy
Fallen WL
Fallen WL
45
=
P TPC
+
l
x
L
=
P TPC
+
=
P 38
+
l
L
TM MCTC
x MCTC Pxd
78.5 160
x Px 78.5 400
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 170
To Be A World Class Maritime Academy
45
=
P 38
+
[ 78.5 160
45
=
P
+
0.096285156P 0.096285156P
38 45
=
1710 =
P
x
1 + 38
4.659 4.659P P
3.659P 3.659P
P x 78.5 ] 400
P
=
367.0 tonnes tonnes
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 171
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GG1
=
P x KG W – P
GG1
=
0.135 m
=
367.05 x 10.9 30 000 – 367.0
Initial GM
=
0.600 m
Effective GM
=
0.600 – 0.135
=
0.465 m
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 172
To Be A World Class Maritime Academy
OR
MM1 =
P x KM W
=
MM1 =
Effective GM
367.05 x 11.5 30 000 0.141 m
=
0.600 – 0.141
=
0.459 m
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 173
To Be A World Class Maritime Academy
aft… To find the drafts forward and aft…
i.
Bodily rise
=
P TPC
=
367.0 38
=
9.66cm
=
0.097 m m
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 174
To Be A World Class Maritime Academy
To find the drafts forward and aft… aft…
ii.
COT
=
Pxd = MCTC =
367.0 x 78.5 400 Head 72cm by Head
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 175
To Be A World Class Maritime Academy
iii.
Change of draft aft due COT COT =
l
x COT
=
78.5 160
=
35.3cm
=
m 0.353 m
L
x 72
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 176
To Be A World Class Maritime Academy
iv.
Change of draft Forward Forward =
COT
–
Change of draft aft
=
72.0
–
35.3
=
m 0.367 m
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 177
To Be A World Class Maritime Academy
Fwd(m)
Aft(m)
Initial Bodily drafts rise Change of drafts
8.000 0.097 0.367 +
9.000 0.097 0.353 -
drafts Final drafts
8.270 8.270
8.550 8.550
v.
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 178
To Be A World Class Maritime Academy
New Trim = 8.55 – 8.27
=
0.28m by Stern assuming „F‟ constant constant
P
=
400 x 28 78.5
=
142.7 tonnes
=
MCTC x T d P
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 179
To Be A World Class Maritime Academy
Further drop vessel fully rest
=
P TPC
+
=
142.7 38
+
=
17.5cm
l
L
x
Pxd MCTC
78.5 x 142.7 x 78.5 160 400
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 180
To Be A World Class Maritime Academy
SINGLE POINT
GROUNDING D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 181
To Be A World Class Maritime Academy
SINGLE POINT GROUNDING GROUNDING A vessel floating at drafts forward 8.70 m, aft 9.40 m grounds at a point 30 m aft of the forward perpendicular. Estimate the drafts of the vessel and the GM after the tide has fallen by 70cm. 70cm. MCTC 340 t-m, TPC 28, KG 7.60 m, KM 8.40
m, LBP 162 m. LCF 78 m forward of Aft Perpendicular and displacement is 29 000 tonnes. D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 182
To Be A World Class Maritime Academy
P 30m
F 9.40m
8.70m
Rock
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 183
To Be A World Class Maritime Academy
Tide fallen by 70cms… 70cms…
P = ?
F Fwd…? Aft…?
Rock
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 184
To Be A World Class Maritime Academy
Draft at P
Fallen of tide by 70cm
WL
New draft at P
D. Dry Docking
WL
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 185
To Be A World Class Maritime Academy
Draft at P
A:
WL
Body rise
Draft at P - Br
WL
WL
New draft at P
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 186
To Be A World Class Maritime Academy
Draft at P
Draft at P - Br
B :
Change of draft at P due to COT by Stern
WL
WL
WL
New draft at P
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 187
To Be A World Class Maritime Academy
WL
Draft at P
A:
Body rise
Fallen of tide by 70cm Draft at P - Br
B :
Change of draft at P due to COT by Stern
WL
WL
New draft at P
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 188
To Be A World Class Maritime Academy
Fallen of tide where
= A B
A
+
B
Body Rise Change of draft at P due to COT by Stern
Fallen of tide =
Body rise + Change of draft at P due to COT Stern
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 189
To Be A World Class Maritime Academy
Fallen of tide
Fallen of tide
70
=
l
P TPC
+
=
P TPC
+
l
=
P 28
+
54 x P x 54 54 162 340
L
L
x
TM MCTC
x Pxd MCTC
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 190
To Be A World Class Maritime Academy
P 30m 54m
F 9.40m
8.70m
Rock
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 191
To Be A World Class Maritime Academy
70
=
P 28
+
[
70
=
P
+
0.052941176P 0.052941176P
28 70
=
1960 =
P
54 x 162
1 + 28
2.482 2.482P P
1.482P 1.482P
P x 54 ] 340
P
=
789.7 tonnes tonnes
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 192
To Be A World Class Maritime Academy
aft… To find the drafts forward and aft…
i.
Bodily rise
=
P TPC
=
789.7 28
=
28cm
=
0.280 m
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 193
To Be A World Class Maritime Academy
aft… To find the drafts forward and aft…
ii.
COT
=
Pxd = MCTC
=
789.7 x 54 340
125.4cm by Stern
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 194
To Be A World Class Maritime Academy
iii.
COT Change of draft aft due COT =
l
x COT
=
78 x 162
=
60.4 cm
=
0.604 m
L
125.4
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 195
To Be A World Class Maritime Academy
iv.
Change of draft Forward Forward =
COT
–
=
125.4 – 60.4
=
65cm
=
0.650 m
Change of draft aft
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 196
To Be A World Class Maritime Academy
Fwd(m)
Aft(m)
Initial drafts Bodily rise Change of drafts
8.700 0.280 0.650 -
9.400 0.280 0.604 +
drafts Final drafts
7.770 m
9.724 m
v.
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 197
To Be A World Class Maritime Academy
ii.
Estimated GM
GG1
Initial GM
=
=
P x KG W – P
=
789.7 x 7.60 29000 – 789.7
=
0.213 m
8.40 m – 7.60
=
0.80 m
Effective GM =
0.800
0.213
D. Dry Docking
=
0.587 m
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 198
To Be A World Class Maritime Academy
ii.
Estimated GM MM1
Effective GM
=
P x KM W
=
0.80 – 0.229
=
0.571 m
=
789.7 x 8.40 29000
=
0.229 m
D. Dry Docking
D1MC Semester 1 / Ship Stability / March 2007 /Capt. MR 199
To Be A World Class Maritime Academy
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