Stability – I : Chapter 11

22. On a ship 8000t displacement , 50t is shifted transversely by 4m. Find the list if the total FSM is 1216tm, KM 7.0m , KG 6.4m.

Solution :

W = 8,000t ,
w = 50t ,
transverse list =4m
FSM = 1216t ,
KM = 7.0m ,
KG = 6.4m

GM  = 7 – 6.4
= 0.6m

LM   =  W x D
= 50 x 4
= 200t

FSC = FSM / W
= 1216 / 8000
= 0.152m

Fluid GM = 0.6 – 0.152
= 0.448m

tanθ  = 200 / 8000 x 0.448
= 3.2deg11.6minut

23. A ship has W 10000 t, KM 7.8m , KG 7.075m, and is upright . NO3 port and stbd DB tanks are full of HFO RD 0.95. Each tank is rectangular , 15m long ,12 wide and 2m deep . calculate the list when HFO is consumed from NO3 stbd until the sounding is 1.2m.

Solution :

W = 10000t,
KM = 7.8 ,
KG = 7.075,
GM = 7.8 – 7.075 = 0.725m ,
LM = 0
RD 0f HFO = 0.95

We know that, L x B x D = 15 x 12 x 2
Volume of stbd tank = 15 x 12 x 2
= 360m³

Wt of consumed HFO = 15 x 12 x 0.8 x 0.95
= 136.8t

LM = W x D = ? doubt why ‘d’ taken as (6m)

FSC = LB cub x di / 12W
= 15 x 12 x 0.95 / 12 x 9863.2
=  0.208 m

Fluid GM = 0.649 – 0 .208
= 0.441m

 

24. A vessel displacing 9000t has KM 8.02m, KG 7.5m , and is up right . she loads 250t KG 12m, 3m to stbd of the centre line ; loads 1000t KG 3m, 1m to port of the centre line; discharges 250t KG 8m, 2m to stbd of the centreline.100t of cargo is then shifted transversly 3m to stbd . If the total FSM is 1200tm , Calculate the final list.

Solution :

Ship’s wt	KG	VM	D	LM
9000t	7.5m	67500		0
+ 250t	12m	+ 3,000	3s	750s
+ 1000t	3m	+ 3,000	1p	1000p
-250t	8m	– 2,000	2 (s)	500p
100	–	–	3s	300s

FW    = 10,000                 FKG = FVM / FW              71.500                          450p

FKG   = 71500 / 10,000
FKG = 7.15m

Final solid GM  = 8.02 – 7.15
= 0.87m

FSC  = FSM / FW
= 1200 / 10000
= 0.12m

Fluid GM   =  0.87 – 0.12
= 0.75m

Tanthi  =   LM / W x GM
= 450 / 10000 x 0.7
= 3.4p
= 3d26.