A ship’s waterplane is 120 metres long. The half-ordinates commencing from aft are as follows: 0, 1.3, 3.7, 7.6, 7.6, 7.5, 4.6, 1.8 and 0.1m, respectively. The spacing between the first three and the last three half-ordinates is half of that between the other half-ordinates. Calculate the second moment of the waterplane area about the centreline and about a transverse axis through the centre of flotation.

A ship’s waterplane is 120 metres long. The half-ordinates commencing from aft are as follows: 0, 1.3, 3.7, 7.6, 7.6, 7.5, 4.6, 1.8 and 0.1m, respectively. The spacing between the first three and the last three half-ordinates is half of that between the other half-ordinates. Calculate the second moment of the waterplane area about the centreline and about a transverse axis through the centre of flotation.

Reference: Ship Stability for Masters & Mates (6^th edition), Exercise – 11.9

 

Solution:

½ ord(y)	SM(dx)	f(A) ydx	Levers(x)	xydx	X2ydx	Y3	Y3dx
0	0.5	0	-3	0	0	0	0
1.3	2	2.6	-2.5	-6.5	16.25	2.19	4.38
3.7	1.5	5.55	-2	-11.1	22.2	50.65	75.97
7.6	4	30.4	-1	-30.4	30.4	438.97	1755.88
7.6	2	15.2	0	0	0	438.97	877.94
7.5	4	30	+1	+30	30	421.87	1687.48
4.6	1.5	6.9	+2	+13.8	27.6	97.33	145.99
1.8	2	3.6	+2.5	+9	22.5	5.83	11.66
0.1	0.5	0.05	+3	+0.15	0.45	0.001	0.0005
		∑ = 94.3		∑ = 4.95	∑ = 149.4		∑ = 4559.3

 

Common interval, h = 120/6 = 20 m

ICL = 2 * h/3 * 1/3 * ∑ Y³dx

       = 2 * 20/3 * 1/3 * 4559.3

       = 20263.55 m⁴

Center of flotation = ∑FM/∑f(A) * h

                                = 4.95/94.3 * 220

                                = 1.04 m

Area = 2 * h/3 * ∑ydx

          = 2 * 20/3 * 94.3

          = 1257.33 m²

I_midship = 2 * h³/3 * ∑ X²ydx

            = 2 * 20³/3 * 149.4

            = 796800 m⁴

I_LCF = I_midship – AX²

       = 796800 – 1257.33*1.04²

       = 795440.07 m⁴