Miscelleneous

# Stability – I : Chapter 10

###### Solution:

W = 5000t, RD = 0.88, I = 242m4

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

###### Solution:

RD = 1.025

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

###### 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

###### 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.

###### 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.

#### Vikrant_sharma

• yes, question number 20, solid gm is 0.9, subtract fsc and get fluid gm and then apply RM=WGMSinx

• makarand says:

the breath of the load water plan of a ship 90m long measured at equal intervals from forward are as follow 0,3.96,8.53,11.58,12.19,12.5,11.58,5.18,3.44, and 0.3m respectively.if the load draft is 5m and the block coefficient is 0.6. find the freshwater allowance and position of the centre of floatation

• Smd says: