Question

At a draught of 1.0 m in sea water density 1025 kg/m3 the dispalcement of a ship is 900 tonne and the height of the centre of buoyancy above the keel (KB) is 0.6 m. Values of tonne per centimeter immersion (TPC) in sea water for a range of draught are given in the table.

(Table will be here soon)

(a) Calcualte EACH of the following for a draught of 6.0m in sea water

(i) The displacement

(ii) The height of the centre of buoyancy above the keel

(b) At a draught of 6.0m, the height of the longitudinal metacentre above the keel (KML) is 128m and the second moment of are of the waterplane above a transverse axis through midship is 996728m4. The centre of flotation is aft of midships. Calculate the distance of the centre of flotation (LCF) from midship.

Answer 1

At draught 1.0 m in sea water (density 1025 kg/m3): displacement = 900 tonne, KB = 0.6 m.

TPC values in sea water for a range of draughts are given in the table (not reproduced here). The standard method is as follows.

Part (a)

For a draught of 6.0 m in sea water:

(i) Displacement.

The displacement at 6.0 m is found by integrating the TPC curve (or the waterplane area) from 1.0 m to 6.0 m and adding the displacement at 1.0 m.
Displacement at 6.0 m = displacement at 1.0 m + integral of (TPC x 100) over the draught from 1.0 to 6.0 m.
Using Simpson's rule on the TPC values at the draughts given in the table, the added displacement = (h/3) x sum of (TPC x multiplier) x 100, where h is the draught interval in metres.
Displacement at 6.0 m = 900 + added displacement (tonne).

(ii) Height of the centre of buoyancy above the keel (KB) at 6.0 m.

KB is found from the moment of the displacement about the keel.
KB at 6.0 m = (moment of displacement about keel)/(total displacement).
The moment of the added displacement about the keel is found by integrating the TPC curve with the lever arm (draught) using Simpson's rule, and adding the moment of the initial 900 tonne at KB 0.6 m.
KB = (900 x 0.6 + moment of added displacement)/(total displacement at 6.0 m).

Part (b)

Distance of the centre of flotation (LCF) from midship at draught 6.0 m.

Given: KML = 128 m, second moment of area of the waterplane about a transverse axis through midship I = 996728 m4, centre of flotation is aft of midships.

The longitudinal metacentre above the keel: KML = KB + BML.

BML = I/V, where V is the volume of displacement = displacement/density = (displacement at 6.0 m)/1.025 m3.

So BML = 996728/V.

KML = KB + BML = 128 m, so BML = 128 - KB.

The distance of the LCF from midship is found from the relationship between the second moment of area about midship and about the centroid (LCF):

I about centroid = I about midship - A x (LCF distance)^2.

The second moment of area about the centroid (through the LCF) is related to BML by BML = I_about_LCF/V.

So I_about_LCF = BML x V.

Then (LCF distance)^2 = (I about midship - I about LCF)/A, where A is the waterplane area at 6.0 m (found from the TPC: A = TPC x 100/1.025).

LCF distance = sqrt((996728 - I_about_LCF)/A) metres, aft of midships.

The numerical values are obtained by substituting the displacement, KB, waterplane area and BML found in part (a) and from the given data. The LCF is aft of midships as stated.

✓ Verified Model Answer (Text Solution)

For a draught of 6.0 m in sea water:

Distance of the centre of flotation (LCF) from midship at draught 6.0 m.