Bilging
Bilging
Bilging
In the event of BILGING, the ship loses buoyancy of that particular compartment and
their effect on ship stability is as follows:
In the event of BILGING, there are two methods for finding ship’s stability:
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As per our syllabus, we will be discussing for stability of box shaped vessel in
bilged condition. Stability of ship shaped vessel in Bilged condition is beyond the
scope of our syllabus.
Now,
Let us consider, a box shaped vessel is floating in water @ WL.
v = d (A-a)
d =
Page 2 of 24
As per our syllabus, we will be discussing Bilging of midship, end and
intermediate compartments.
Worked Examples:-
It has amidship compartment of length 10m & extending side to side. The
midship compartment is bilged below WL. Find the Draft in bilged condition.
Solution:
= =
= = 0.556m
= 5.556m
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2. A box shaped vessel (Length 100m x Breadth 20m) is floating in water at an
EVEN KEEL Draft of 5.0m.
It has a midship compartment of length 10m & extending side to side.
The compartment has a W/T flat, 3m above the keel. Find the F&A drafts in the
event of the compartment getting bilged below the W/T flat.
Solution:
In this case,
Now,
Mean sinkage (d) =
= = = 0.3m
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The compartment has a W/T flat, 3m above the keel. Find the F&A drafts in the
event of the compartment getting bilged above the W/T flat.
Solution:
Now,
Mean sinkage (d) =
= = = 0.222m
Solution:
Now,
Mean sinkage (d) =
= = = 0.5m
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The compartment has a W/T flat, 5.2m above the keel. Find the F&A drafts in
the event of the compartment getting bilged below the water line.
Solution:
100 x 20 x = 10 x 20 x 5 – 90 x 20 x 0.2
= =
= 0.32 m
Therefore,
The above is not always true. As the compartment, may contain cargo. In those case,
in the event of bilging, volume of lost buoyancy will be equal to empty space in the
compartment bilged.
Permeability = x 100
Now, let a compartment is loaded with a cargo of relative density 0.6 t/ m3 with
stowage factor 2m3/t. find permeability ( ) ?
Solution:
0.6 t 1m3
1t = 1.67 m3
= x 100 = x 100
= 16.5%
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It has a midship compartment of length 10m & extending side to side. The
compartment is loaded with cargo such that permeability is 40%.
Find the Drafts F&A in the event of midship compartment getting bilged.
Solution:
= 400 m3
= 2000 – 80 = 1920 m2
Mean sinkage = =
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Calculate the draft and stability of the vessel in bilged condition.
Solution:
Mean sinkage =
= = 0.556m
Page 10 of 24
I = - = m4
V = 100 x 20 x 5m3
BM = = =
New BM = 6.0m
KB = (+) 2.778m
____________
New KM = 8.778m
KG =(-) 5.30m
____________
Final GM = 3.478m
It has a midship compartment of length 10m & extending side to side. KG = 5.3
The compartment has a W/T flat, 3m above the keel. KG of the ship 4.7m.
Solution:
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Volume of lost Buoyancy = 10 x 20 x 3m3
Mean sinkage =
= = 0.3m
New KB =
Page 12 of 24
BM = =
= =
= 6.667 m
New BM = 6.667m
New KM = 9.386m
KG =(-) 4.70 m
____________
New GM = 4.686m
Solution:
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Volume of lost Buoyancy = 10 x 20 x 2m3
Mean sinkage =
= = = 0.222m
New KB =
= =
BM = =
= = 6.0m
= 6.0 m
New KM = 8.544m
KG =(-) 4.70 m
____________
New GM = 3.844m
10. A box shaped vessel (Length 100m x Breadth 20m) is floating in water at an
EVEN KEEL Draft of 5.0m.
Page 14 of 24
It has a midship compartment of length 10m & extending side to side. The
compartment is loaded with cargo such that permeability is 40% KG. The ship is
4.7m. The compartment is bilged below water line.
Find the drafts and GM of the vessel in bilged condition.
Solution:
= 2000 – 80 = 1920m2
Mean sinkage =
= = 0.208m
= 2.604m
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BM = =
BM = = 6.4m
= 6.4m
KB = (+) 2.604m
____________
KM = 9.004m
KG = (-) 4.700
___________
GM = 4.304 m
Now, we will study, stability of vessel with end compartment getting bilged,
apart from metacentric height (GM) getting changed, a trimming moment is
created, which changes the Trim of the vessel.
11. A box shaped vessel (Length 100m x Breadth 20m) is floating in water at an
EVEN KEEL Draft of 5.0m.
The vessel has an END compartment of length 10m, located right forward and
extending side to side. KG of the ship 4.6m.
Solution:
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Mean sinkage =
= = = 0.556m
= 2.778m
MCTC =
BML = =
= =
BML = 121.5m
KML = 124.278m
Page 17 of 24
KG =(-) 4.60 m
______________
GML = 119.678m
Therefore,
MCTC = Always full length
MCTC = 122.67
T.M. = 51250 tm
Change of trim = =
12. A box shaped vessel (Length 100m x Breadth 20m) is floating in water at an
EVEN KEEL Draft of 5.0m.
It has an End compartment located right forward, of length 10m and extending
side to side. The compartment is loaded with cargo such that permeability 40%.
KG the ship 4.6m.
Solution:
Page 18 of 24
Mean sinkage =
= = 0.208m
= 2.604m
COF will shift away from the bilged compartment. New COF will be found by
taking moments about midship.
Let new COF shifts by m away from the original posn (midship)
= = - 1.875m
New BML = =
= 1673698.167m4
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ICF of Bilged water Plane Area= x + x(20 x 10) (46.875)2
= 666.667 + 175781.25
= 176447.917m4
= 1497250.25 m4
BML =
BML = 149.7m
KB = 2.604
______________
KML = 152.304m
KML = 152.304m
KG = (-) 4.6 m
_______________
GML = 147.704m
MCTC =
= 19218.75 tm
Draft Calculations:
F A
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BILGING OF INTERMEDIATE COMPARTMENT
The intermediate compartment is bilged and we will study, how ICF, BML, GML
MCTC and Trimming moments are calculated to find the change of trim and final
draft of the vessel in bilged condition.
13. A box shaped vessel (Length 100m x Breadth 20m) is floating in water at an
EVEN KEEL Draft of 5.0m.
The vessel has an intermediate compartment of length 10m and extending side
to side, located such that it’s forward transverse bulkhead is 15m behind the
forward most bulkhead. KG of the ship 4.7m.
The compartment gets bilged. Find the drafts F&A in bilged condition.
Solution:
Mean sinkage =
= = 0.556
Page 22 of 24
Mean Draft after Bilging = 5.556m
= 2.778m
= =
= - 3.333m
Now BML = =
= 1688884.445m4
= 1666.667 + 222177.78
= 223844.446m4
Now BML = =
Page 23 of 24
GML = 144.578m
MCTC =
MCTC = 148.2
= 34163.25 tm
= = 230.5 cms
Draft Calculations:
F A
*********
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