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# Chapter 10 - Stability Calculation

### 1. On a ship of 5000 t displacement , a tank is partly full of DO of RD 0.88. If the moment of inertia of the tank about its centreline is 242m4 , find the FSC .

Solution:

W = 5000t, RD = 0.88, I = 242m4 FSC = (i di /W ) = (242 x 0.88 ) /5000 = 0.042m.

### 2. If the tank in Question 1 was partly full of SW instead of DO, find the FSC.

Solution:

RD = 1.025 FSC = (i di/W ) = (242 x1.025) /5000 = 0.049m.

### 3. On a ship of W 6000 t, KM 7.4m, KG 6.6m, a double bottom tank of i 1200m4 is partly full of FW . Find the GM fluid .

Solution:

W = 6000 t & KM = 7.4m, KG = 6.6m, i = 1200m4, RD = 1.00

GM (solid) = (KM – KG) = (7.4 – 6.6) = 0.8m

FSC = (i di /W ) = (1200 x 1) /6000 = 0.2m

Fluid GM = GM (solid) – FSC = (0.8 – 0.2) = 0.6m

4. Given the following particulars, Find the GM fluid : W= 8800 t, tank of i = 1166 m4 is partly full of HFO of RD 0.95, KM 10.1m, KG 9.4m .

Solution:

W = 8800 t & i = 1166m4, RD = 0.95, KM = 10.1 m & KG = 9.0m.

GM (solid) = (KM – KG) = (10.1 – 9.0) = 1.1mFSC = (i di /W ) = (1166 x 0.95)/ 8800 = 0.126m

GM fluid = GM (solid) – FSC = (1.1 – 0.126) = 0.974m.

### 5. On a vessel of W 16000 t, NO.4 port DB tank 20m long and 8m wide is partly full of DW ballast of RD 1.010. Find the FSC.

Solution:

W = 16000t, L = 20m & B = 8m,RD = 1.010

FSC = (i di / W) FSC =( LB3 X di )/( 12 x W) = (20 x 83 x 1.010)/(12 x 16000) = 0.054m.

### 6. A vessel has a deep tank on the starboard side 12m long 9m wide which is partly full of coconut oil of RD 0.72 . If W = 1200 t , KM = 9m and KG = 8.5m ,find the GM fluid .

Solution:

L = 12m, B = 9m, RD = 0.72 W = 12000 t, KM = 9m & KG = 8.5m

GM (solid) = (KM – KG) = (9 – 8.5) = 0.5m

FSC = (i di /W ) = (LB3 x di ) / (12 x W ) = (12 x 93 x 0.72) / (12 x 1200) = 0.044m

Fluid GM = GM (solid) -FSC = (0.5 – 0.044) = 0.456m.

### A vessel displacing 8000t , has a rectangular deep tank 10m long 8m wide and 9m deep full of SW . The KM is 7m and KG 6.2m. Find the GM when 1 / 3 of this tank as pumped out.

Solution:

W = 8000t. L = 10m, B = 8m & D = 9m RD = 1.025, KM = 7m, KG = 6.2m

Mass of the tank = (Volume x Density) = (L x B x D) x 1.025 = (10 x 8 x 9) x (1.025) = 738t

According to question, 1/3 of water pumped out = (1/3 x738) = 246t

After pumping out the water the KG of the tank = (6 + 1.5) = 7.5m Final W = 7754 t Final VM = 47755tm

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

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

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.