Stability – I : Chapter 10

Ship’s weight	KG	VM (t)
8000t	6.2m	49600tm
(-) 246t (discharge)	7.5m	(-) 1845tm

Final W = 7754 t                                                                  Final VM = 47755tm

We know that :

Final KG =   (Final VM)/ (Final W)
Final KG = (47755)/(7754)
= 6.159m

GM (solid) = (KM – KG)
= (7.0 – 6.159)
= 0.841m

We know that:

FSC = (i di / W)
= (LB³ x di)/ (12 x W)
= (10 x 8³ x 1.025)/(12 x 7754)
= 0.05m

So, GM fluid = GM (solid) – FSC
= (0.841 – 0.055)
= 0.785m.

8. A ship of W 5000t has a tank 16m long, 10m wide and 4m deep which is empty. KM is 7.2m and KG 7.0m. Find the GM fluid if 400t of oil of RD 0.95 are received in it.

Solution:

W = 5000 t
Volume of tank = (L x B x D)
= (16 x 10 x 4) m³

KM =7.2m & KG = 7.0m
W = 400 t, RD = 0.95

We know that, dimension of tank are given change of KG of tank due to change of sounding has to be considered

Mass = (Volume x Density)
400t = (Volume) x 0.95
Volume   = (400/0.95)
= 421.05m³

Now, Depth of the oil = (volume / area)
(421.052/ (16 x 10)
= 2.631m

Again, we know that:
KG   = (depth/2)
= (2.631/2)
= 1.315m