Solution Naval Numerical 24

Numerical 24: A ship of 8,000 tonnes displacement takes the ground on a sand bank on a falling tide at an even keel draft of 5.2 metres. KG 4.0 metres. The predicted depth of water over the sand bank at the following low water is 3.2 metres. Calculate the GM at this time assuming that the KM will then be 5.0 metres and that mean TPC is 15 tonne.

Solution: Theory of grounding

Given:

△ = 8000 t

initial draught= 5.2m

KG=4.0m

depth of water =3.2m

KM=5.0m

TPC =15cm

Upward force of sea bed;

P = TPC x fall in the water level.

P=15 x (5.2-3.2)x100= 3000 t.

$New GM = Old GM-Loss in GM$ 

or

$New GM = Old GM-\frac{P\times KM}{W}$

loss in GM = $\frac{P\times KM}{W}$ 

loss in GM = $\frac{3000\times 5}{8000}$ 

loss in GM = 1.88m

Old GM =KM - KG =5.0 - 4.0 =1.0m

New GM = 1.0 - 1.88

= -0.88m