28.NA January 2019 Q.6(B)

A ship of 15000 tonne displacement has righting levers of 0, 0.38, 1.0, 1.41 and 1.2 m at angles of hell of 0°, 15°, 30°, 45° and 60° respectively and an assumed KG of 7.0 m. The vessel is loaded to this displacement but the KG is found to be 6.80m and GM 1.5m –
(i) Draw the amended stability curve; (ii) Estimate the dynamic stability at 60°

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Given 

△ = 15000 t
Kg = 7 m
New Kg = 6.8 m
GM = 1.5 m
GZ = 0, 0.38, 1.0, 1.41, 1.2
θ    = 0, 15, 30, 45, 60

To find

1. Stability curve
2. Dynamic stability at 60 deg

Solution 

GG1 = Kg - New Kg
         = 7 - 6.8
         = 0.2


New Kg is lies below Kg. Therefore 

G1Z = GZ + GG1 Sinθ

θ	Sinθ	GG1 Sinθ	GZ	G1Z	SM	Product
0	0	0	0	0	1	0
15	0.259	0.0518	0.38	0.43	4	1.72
30	0.500	0.100	1.0	1.1	2	2.2
45	0.707	0.1414	1.41	1.55	4	6.2
60	0.866	0.1732	1.20	1.37	1	1.37
Total						11.49

1. To draw stability curve we need angle of heel value for X axis and GZ value for Y axis.          Both we can get from the table.

    Draw X and Y axis & Plot the value as per the table.

    

    GM is given as 1.5 m , So mark a point GM to 57.3 deg angle of heel. And make a                straight line

    Draw a tangent from GM and 57.3 deg meeting point to the 0.

2. Dynamic stability at 60 deg = △ x Area of GZ curve up to 60 deg heel 

            Area under the curve   = 1/3 x h x product

                                             h  = 15/57.3

                

             Area under the curve  = 1/3 x 15/57.3 x 11.49
                                                 = 1.002

     Dynamic stability at 60 deg = 15000 x 9.81 x 1.002

     Dynamic stability at 60 deg = 147.5 MJ