Question

A ship of length 120m displaces 11750 tonne when floating in sea water of density 1025 kg/m3. The centre of gravity is 2m above the centre of buovanoy and the waterplane is defined by the following equidistant half breadths given in table below:

Section: AP 1 2 3 4 5 6 7 FP

Half-breadth (m) 3.3 6.8 7.6 8.1 8.1 8.0 6.6 2.8 0

Calculate EACH of the following:

(a) The area of the waterplane (3)

(b) The position of the centroid of the waterplane from midships (3)

(c) The second moment of area of the waterplane about a transverse axis through the centroid (5)

(d) The moment to change trim one centimetre (MCT1cm) (5)

Answer 1

Ship length L = 120 m, displacement = 11750 tonne, sea water density 1025 kg/m3.

KG - KB = 2 m (centre of gravity 2 m above centre of buoyancy).

Half-breadths at sections AP(0),1,2,3,4,5,6,7,FP(8):

Section: 0 1 2 3 4 5 6 7 8

Half-breadth (m): 3.3 6.8 7.6 8.1 8.1 8.0 6.6 2.8 0

Number of intervals n = 8, common interval h = 120/8 = 15 m.

Part (a)

Area of the waterplane.

Simpson's First Rule, multipliers 1,4,2,4,2,4,2,4,1:

Sum = 1x3.3 + 4x6.8 + 2x7.6 + 4x8.1 + 2x8.1 + 4x8.0 + 2x6.6 + 4x2.8 + 1x0

= 3.3 + 27.2 + 15.2 + 32.4 + 16.2 + 32.0 + 13.2 + 11.2 + 0 = 150.7.

Half waterplane area = (h/3) x sum = (15/3) x 150.7 = 5 x 150.7 = 753.5 m2.

Full waterplane area = 2 x 753.5 = 1507 m2.

Answer: Waterplane area = 1507 m2.

Part (b)

Position of the centroid of the waterplane from midships.

Midship is at station 4. First moment about station 4:

Distances in intervals: station 0 = -4, 1 = -3, 2 = -2, 3 = -1, 4 = 0, 5 = +1, 6 = +2, 7 = +3, 8 = +4.

Moment = 1x3.3x(-4) + 4x6.8x(-3) + 2x7.6x(-2) + 4x8.1x(-1) + 2x8.1x0 + 4x8.0x(+1) + 2x6.6x(+2) + 4x2.8x(+3) + 1x0x(+4)

= -13.2 - 81.6 - 30.4 - 32.4 + 0 + 32.0 + 26.4 + 33.6 + 0 = -65.6.

Distance of centroid from midship = (moment/sum) x h = (-65.6/150.7) x 15 = -0.4353 x 15 = -6.53 m.

Negative means forward of midship.

Answer: Centroid (LCF) is 6.53 m forward of midship.

Part (c)

Second moment of area of the waterplane about a transverse axis through the centroid.

Second moment about midship (station 4) using the product of (multiplier x half-breadth x distance^2):

Moment2 = 1x3.3x16 + 4x6.8x9 + 2x7.6x4 + 4x8.1x1 + 2x8.1x0 + 4x8.0x1 + 2x6.6x4 + 4x2.8x9 + 1x0x16

= 52.8 + 244.8 + 60.8 + 32.4 + 0 + 32.0 + 52.8 + 100.8 + 0 = 576.4.

Second moment of half waterplane about midship = (1/3) x h^3 x sum = (1/3) x 15^3 x 576.4 = (1/3) x 3375 x 576.4 = 1125 x 576.4 = 648450 m4.

Full waterplane second moment about midship = 2 x 648450 = 1296900 m4.

Using the parallel axis theorem to transfer to the centroid:

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

= 1296900 - 1507 x (6.53)^2 = 1296900 - 1507 x 42.64 = 1296900 - 64258 = 1232642 m4.

Answer: Second moment of area about the transverse axis through the centroid = 1.233 x 10^6 m4.

Part (d)

MCT 1 cm.

MCT 1 cm = (displacement x GML)/(100 x L), where GML is the longitudinal metacentric height.

GML = KML - KG. KML = KB + BML.

BML = I / Volume of displacement.

Volume of displacement = 11750/1.025 = 11463.4 m3.

BML = 1232642/11463.4 = 107.53 m.

KB is not given directly, but KG - KB = 2 m. We need KB.

Using the waterplane and displacement, KB can be estimated, but a standard approach:

KML = KB + BML. GML = KML - KG = (KB + BML) - (KB + 2) = BML - 2 = 107.53 - 2 = 105.53 m.

MCT 1 cm = (11750 x 105.53)/(100 x 120) = 1240000/12000 = 103.3 tonne-m per cm.

Answer: MCT 1 cm = 103.3 tonne-m per cm.

✓ Verified Model Answer (Text Solution)