Q8 (10 Marks) Ship Stability
SC&S • Written Exam

(a) Explain why the bilging of empty double-bottom or deep tanks below the waterline leads to an increase in GM. (6)

(b) A ship of 22000 tonne displacement is 160 m long and MCT1cm 280 tonne m, waterplane area 3060 m2 centre of buoyancy 1 m aft of midships and centre of flotation 4 m aft of midships. It floats in water of 1.007 t/ m3 at draughts of 8.15 m forward and 8.75 m aft. Calculate the new draughts if the vessel moves into sea water of 1.026 t/ m2. (10)

Appeared In: Nov 2023

Verified Model Answer (Text Solution)

Structured for DG Shipping MEO Class II examination scoring criteria.

Exam Ready
Part (a)

Why bilging empty double-bottom or deep tanks below the waterline increases GM.

When an empty tank below the waterline is bilged (flooded), the water that enters is at the same level as the sea and, being below the waterline, the added weight of water is exactly balanced by the added buoyancy of the extra volume displaced (the ship sinks slightly). The net effect is that the centre of gravity of the added water is low (in the tank, near the bottom of the ship), while the added buoyancy acts at the centre of buoyancy of the flooded volume, which is also low. Because the added weight is low, the overall centre of gravity KG is lowered, and because the added buoyancy is low, the centre of buoyancy KB is also lowered but by less than the weight effect. The result is that GM increases (the ship becomes stiffer). Additionally, once the tank is full there is no free surface, so there is no free-surface loss of GM. Hence bilging an empty low tank increases GM and improves stability (though it increases displacement and draft).

Part (b)

New draughts when the vessel moves into sea water.

Ship 22,000 t displacement, 160 m long, MCT1cm = 280 t-m, waterplane area 3060 m2, centre of buoyancy 1 m aft of midships, centre of flotation 4 m aft of midships. It floats in water of 1.007 t/m3 at draughts of 8.15 m forward and 8.75 m aft. Calculate the new draughts if the vessel moves into sea water of 1.026 t/m3.

Step 1 - change of mean draught.

Volume in water of 1.007 = 22000/1.007 = 21847 m3; volume in sea water (1.026) = 22000/1.026 = 21442 m3. Decrease in volume = 405 m3.

Decrease in mean draught = 405/3060 = 0.132 m. New mean draught = 8.45 - 0.132 = 8.318 m.

Step 2 - change of trim.

The decrease in buoyancy (405 t) acts at the centre of flotation (4 m aft of midships), while the centre of buoyancy is 1 m aft of midships. The moment about the centre of flotation = 405 x (4 - 1) = 405 x 3 = 1215 t-m.

Change of trim = moment/MCT1cm = 1215/280 = 4.34 cm. Since the buoyancy is removed aft of the centre of buoyancy, the stern rises, reducing the stern trim. Original trim = 0.6 m (8.75 - 8.15) by the stern. New trim = 0.6 - 0.0434 = 0.5566 m by the stern.

Step 3 - new draughts.

The change of trim of 0.0434 m (4.34 cm) is distributed about the centre of flotation (4 m aft of midships). The length is 160 m, so the aft portion is 80 - 4 = 76 m and the forward portion is 80 + 4 = 84 m.

Change in aft draught = 0.0434 x 76/160 = 0.0206 m (stern rises, so subtract).

Change in forward draught = 0.0434 x 84/160 = 0.0228 m (bow rises, so subtract).

New aft draught = 8.75 - 0.132 - 0.0206 = 8.597 m.

New forward draught = 8.15 - 0.132 - 0.0228 = 7.995 m.

Answer: new draughts are about 8.0 m forward and 8.6 m aft.

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