Miscelleneous

Stability – I : Chapter 10

  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

We know that :
FSC = (i di /W )
= (242 x 0.88 ) /5000
= 0.042m.

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

RD = 1.025

We know that:
FSC = (i di/W )
= (242 x1.025) /5000
= 0.049m.

  1. 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

We know that:

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

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

  1. 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)
= 1.1m

We know that:

FSC = (i di /W )
= (1166 x 0.95)/ 8800
= 0.126m

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

  1. 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

We know that:

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

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