42.NA April 2019 Q.6(b)

The immersed cross-sectional area of a ship 120m long, commencing from aft are 2,40,79,100,103,104,104,103,97,58 and 0 m2 calculate - (i) Displacement; (ii) Longitudinal position of the centre of buoyancy

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Given

L = 120 m

To find


i) Displacement 
ii) Longitudinal position of the centre of buoyancy

Solution

Area	SM	Product for volume	Lever	Product for 1stmoment
2	1	2	+5	+10
40	4	160	+4	+640
79	2	158	+3	+474
100	4	400	+2	+800
103	2	206	+1	+206
104	4	416	0	Σma = +2130
104	2	208	-1	-208
103	4	412	-2	-824
97	2	194	-3	-582
58	4	232	-4	-928
0	1	0	-5	0
		Σv = 2388		Σmf = -2542

It is given that L = 120 m and there are 10 equidistant spaces. So common interval

h = 120 / 10 

   = 12 m

i) We know that displacement = Density x Under water volume ----------------(1)

   But Under water volume = (h/3) x Σv

                                          = (12/3) x 2388

                                          = 9552 m3

   (1) -------------------------->

    

   Displacement = 1.025 x 9552

                          = 9790.8 tonne

ii) Centre of buoyancy from midship = h x (Σma + Σmf) / Σv

                                                         = 12 x (2130 - 2542 ) / 2388

                                                         = 2.070 m forward 

   NOTE : If Σma is greater than Σmf , the buoyancy will be forward of midship.