Chapter 8 - Stability Calculation

1. A box shaped vessel of displacement 1640t is 50m long 10m wide and 8m high . Find her KB in SW, if she is one an even keel and upright.

Solution:


Displacement = 1640

Volume of box shape vessel = (L x B x H) = (50m x10m x 8m)

RD = 1.025

Displacement = (u/w volume) x(density of water displaced) 1640= ( L x B x d ) x (1.025) 1640 =(50 x 10 x d) x (1.025 )

Hence, d = 1640/ (50 x 10x 1.025) = 3.2m

KB can be calculated as = (d/2) = (3.2/2) = 1.6m

2. A box shaped vessel 60m x 10m x10m floats in DW of RD 1.020 at an even keel draft of 6m . Find her KB in DW of RD 1.004.

Solution :


Volume of box shape vessel = ( L X B XH) =(60m x 10m x 10m)

RD = 1.020, Draft = 6m

Displacement at RD of 1.020 = (u/w volume)x(density) =( L x B x d ) x 1.020 = (60 x 10 x 6) x 1.020 = 3672t

Now we have to calculate draft of vessel at RD of 1.004, keeping the displacement same = (L x B x d ) x (1.004) 3672 = (60 x 10 x d )x (1.004) d = 3672 /(60 x 10 x 1.004) = 6.095m

Hence, KB = (d/2 = (6.095 /2) = 3.047m.

3. A triangular shaped vessel of displacement 650t floats in DW RD 1.015. Her water plane is rectangle 30m x 8m . Find her KB .

Solution:


Displacement = 650t, RD = 1.95

Water plane Area =(L x B) = (30m x 8m) Volume of triangular vessel = ½ ( L x B x H)

Displacement = (u/w volume)x (Density) 650 = ½ (L x B x d ) x( density) =(½ x 30 x 8 x d) x(1.015) d = (650 x 2)/ (30 x 8 x 1.015) = 5.336m

centeriod of triangle is 2/3rd of the perpendicular bisector of the base .Hence centre of buoyancy will be always 2/3rd of perpendicular bisector of base.

KB =( 2/3 x height of perpendicular bisector of base) = (2/3 x 5.336) = 3.557m.

4. A triangular shaped vessel floats in SW. Her water plane is a rectangle 40m x 12m. If her KB is 3.6m, find her displacement .

Solution :


Water plane area = (L x B) = 40m x 12m

Initial KB =3.6m,

centeriod of triangle is 2/3rd of the perpendicular bisector of the base .hence centre of buoyancy will be always 2/3rd of perpendicular bisector of base.

So, KB = 2/3 x (Draft) 3.6 = 2/3 x d

So, Draft(d) =(3.6 x 3 / 2) = 5.4m Volume of triangular vessel = ½ ( L x B x H)

Displacement =(u/w volume) x (density) = ½ (L x B x d) x (1.025) = 1/2 (40 x 12 x 5.4) x (1.025 = 1328.4t .

5. A homogeneous log of wood 3m x 0.75m x 0.75m floats in SW with one face horizontal .If the RD of the log is 0.8m, calculate the vertical distance between its COG and its COB .

Solution:


Volume of log of wood = (L x B x H) = (3m x 0.75m x 0.75m)

RD of SW = 1.025 RD of the log = 0.8m

Mass of the log = (volume x density) = (3 x 0.75 x 0.75 ) x (0.8) = 1.35 t

Since, log floats freely so Mass = Displacement 1.35 = (u/w volume) x(density) 1.35 = (L x B x draft) x (1.025) 1.35 = ( 3x 0.75 x draft) x (1.025)

So, draft = 1.35/(3 x 0.75 x 1.025) Draft = 0.585 m COB = (1/2 x draft) = (0.585/2) = 0.2925m

COG = (1/2 x height) = (0.75 / 2) = 0.375m

Hence, Vertical distance between COB and COG = (0.375 – 0.295) = 0.083m

6. A homogeneous log of wood of 0.5m square section floats in water of RD 1.005 at a draft 0.4m with one of its faces horizontal . Find the vertical distance between its COG and its COB in water of RD 1.020.

Solution: