Stability – I : Chapter 10

16. A ship of 5000t displacement has a DB tank 18m long and 12m wide , partly full of SW . Find the FSC in the following cases:

1. If the tank is undivided .
2. If the tank is divided into identical P and S watertight divisions and
   1. Both side are slack.
   2. only one side is slack.
3. If the tank is divided in to P, S and C identical watertight division and
   1. All three of these are slack .
   2. Only two of these are slack .
   3. Only one of these are slack .
4. If the tank is divided into four identical watertight division – Port inner ,  port outer ,Stbd  inner , Stbd outer, – and
   1. All four of these are slack .
   2. Any three of these are slack .
   3. Any two of these are slack .
   4. Any one of these is slack .

NOTE : The use of 1/n2 would be a very quick method of solving this question .

Solution :

a).  if the tank is undivided

We know that
FSC = (LB 3 x i )/  (12 x W)
= ( 123x  18 x 1.025)/ (12 x 5000)
So, FSC = 0.531m

b).  If Divided into identical P and S

1 ).   Both sides are slack
FSC = (1/n2)
= (1/22 ) x 0.531
= 0.133m

2.) If only one side is slack
We can calculate
FSC = (0.133 /2)
= 0.066m

C). If the tank is divided in to P , S and C  identical water tight division and

1.)  If all three of these are slack .
We know that
FSC = (1/n2)
= (1/32 ) x 0.531
= 0.059m

2.)  If only two of these are slack then
FSC   = (0.059 /2)
= 0.039m

3.) If only one of these is slack
= (0.059/3)
= 0.019m

d). If the tank is divided into four identical tanks

1).  All four of these are slack
We can calculate:
FSC = (1/n2) x 0.531
= (1/42 ) x 0.531
= 0.0332m.

2). Any three of these are slack
FSC would be   0.0247 m

3). Any two of these are slack
FSC would be (0.0332/2) m
= 0.0165m

4). Any one of these is slack = (8.25/1000)
= .0082m.