Miscelleneous

Stability – I : Chapter 8

  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

As we know that  KB can be calculated as
= (d/2)
= (3.2/2)
= 1.6m

  1. 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.

  1. 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)

We know that:
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

As we know that 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.

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