Stability – I : Chapter 10

Ship’s wt	KG	VM
9840t	8.0m	78720tm
246t  ( shift)	4m	(-) 984tm

       Final W = 9840t                                                                                        Final VM = 77736tm

We know that:

Final KG = (Final VM/Final W)
Final KG = (77736/9840)
= 7.9m

GM (Solid) = (8.5 – 7.9) m
= 0.6m

FSC when tanks one not divided ;
FSC = (i di/ W)
= (LB³ x di )/ (12 x W )
= (12 x 10³ x 1.025)/(12x 9840)
= 0.104m

Total FSC = (1/n2)
= (1/2²) x (0.104 )
= 0.026m

GM fluid = (GM solid – FSC)
= (0.6 – 0.026)
= 0.574m

10. A ship displacing 10000t has KM 9. 9m. The following is her present condition:

Tank	KG (m)	1(m4)	contents	RD	Remarks
FP Tank	6.3	10	SW	1.025	Full
No 1 DBT	1.15	420	HFO	0.95	Slock
No 2 P or  S	0.65	720	HFO	0.95	Port slock  Stb empty
No 3p  or s	0.65	240	SW	1.025	Pord full  Stbd slack
No 3c	0.60	1200	HFO	0.95	Full
No 4p	0.70	300	FW	1.00	Both Slack
NO 5p	0.85	180	DO	0.88	Slack
NO 5s	0.85	100	HFO	0.95	Full
AP Tank	8.80	20	SW	1.025	Empty

If the final KG is 8.954 m, find the final GM fluid .

Solution:

Given:
W = 10000 t, KM = 9.9m & KG = 8.954m

FSM  NO.1  =  (  i x di )
= (420 x 0.95)
= 399

NO.2P   = ( i x di )
= (720 x 0.95)
= 684

NO.3S   = ( i x di)
= (240 x 1.025)
= 246

NO.4P   =  (  i x di )
= (300 x 1 )
= 300

NO.4S =   ( i x di)
= (300 x 1)
= 300

NO.5P =  ( i x di )
= ( 180 x 0.88 )
= 158.4

Total   FSM  =  2087.4 tm
FSC = (FSM / W)
= ( 2087.4 /10,000 )
= 0.208m

GM solid = (KM – KG)
=( 9.9 – 8.954 )
= 0.946m

GM fluid = (0.946 – 0.208)
= 0.738m.