A ship’s waterplane is 120 metres long. The half-ordinates at equidistant intervals from forward are as follows: 0, 3.7, 7.6, 7.6, 7.5, 4.6 and 0.1m, respectively 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 at equidistant intervals from forward are as follows:

0, 3.7, 7.6, 7.6, 7.5, 4.6 and 0.1m, respectively

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 – 11.6

Solution:

FOR	Half ord.(y)	SM	Y*SM	Y3	Y3*SM	Lever(x)	X*Y*SM	X2*Y*SM
	0	1	0	0	0	-3	0	0
	3.7	4	14.8	50.65	202.6	-2	-29.6	59.2
	7.6	2	15.2	438.97	877.94	-1	-15.2	15.2
	7.6	4	30.4	438.97	1755.88	0	0	0
	7.5	2	15	421.87	843.74	+1	+15	15
	4.6	4	18.4	97.33	389.32	+2	+36.8	73.6
	0.1	1	0.1	0.001	0.001	+3	+0.3	0.9
AFT			∑ = 93.9		∑ = 4069.481		∑ = +7.3	∑ = 163.9

Common Interval, h = 120/6 = 20 m

I_CL = 2*(h/3) * (1/3) * ∑Y³*SM

       = 2 * (20/3) * (1/3) * 4069.481

       = 18086.58 m⁴

 

I_midship = 2 * (h³/3) * ∑X²*Y*SM

                = 2 * (20³/3) * 163.9

                = 874133.33 m⁴

 

Area = 2*(h/3) * ∑Y*SM

        = 2 * (20/3) * 93.9

         = 1252 m²

Center of Flotation = (∑f(M))/(∑f(A)) * h

                                = (+7.3/93.9) * 20

                                = 1.55 m

I_LCF = I_midship - AX²

       = 874133.33 – 1252* 1.55²

       = 871106.56 m4