A box shaped vessel is 80m long, 12m wide and floats at a draft of 4m. A full width midship compartment 15m long is bilged and this results in the draft increasing to 4.5m. Calculate each of the tollowing
(a) The permeability of the compartment (4)
(b) The change in metacentric height due to bilging (12)
Structured for DG Shipping MEO Class II examination scoring criteria.
Box-shaped vessel: length L = 80 m, breadth B = 12 m, draft d = 4 m.
A full-width midship compartment 15 m long is bilged, and the draft increases to 4.5 m.
When a compartment is bilged, the increase in draft is due to the loss of buoyancy of the compartment.
The volume of lost buoyancy = the volume of water entering the compartment = (permeability x compartment volume).
The increase in draft causes an increase in displacement equal to the weight of water entering.
Increase in draft = 4.5 - 4.0 = 0.5 m.
The increase in displacement = L x B x increase in draft x density = 80 x 12 x 0.5 x 1.025 = 492 tonne.
This equals the weight of water entering the compartment.
Weight of water entering = permeability x (compartment volume) x density = permeability x (15 x 12 x 4) x 1.025.
So 492 = permeability x 720 x 1.025 = permeability x 738.
Permeability = 492/738 = 0.667.
Answer: Permeability = 0.667 (66.7%).
The bilged compartment is full width, so the waterplane area is reduced by the area of the compartment at the waterline.
Original waterplane area = L x B = 80 x 12 = 960 m2.
The compartment is full width (12 m) and 15 m long, so the lost waterplane area = 15 x 12 = 180 m2.
Effective waterplane area after bilging = 960 - 180 = 780 m2.
Original displacement = L x B x d x density = 80 x 12 x 4 x 1.025 = 3936 tonne.
Original KB = d/2 = 4/2 = 2 m.
Original BM = I/V = (L x B^3/12)/(L x B x d) = B^2/(12 x d) = 144/(12 x 4) = 3 m.
Original KM = KB + BM = 2 + 3 = 5 m.
After bilging, the draft is 4.5 m.
New KB = 4.5/2 = 2.25 m.
New BM = I/V. The second moment of area of the effective waterplane = (L x B^3/12) - (15 x 12^3/12) = (80 x 1728/12) - (15 x 1728/12) = (80 x 144) - (15 x 144) = 11520 - 2160 = 9360 m4.
New volume of displacement = 3936/1.025 = 3840 m3 (unchanged, since the ship's weight is unchanged).
New BM = 9360/3840 = 2.4375 m.
New KM = new KB + new BM = 2.25 + 2.4375 = 4.6875 m.
The change in metacentric height = new KM - original KM = 4.6875 - 5.0 = -0.3125 m.
Answer: The metacentric height is reduced by 0.3125 m (GM decreases by 0.3125 m).