Stability – I : Chapter 8

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)

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

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