The half-ordinates of a ship’s waterplane at equidistant intervals from forward are as follows: 0, 1.3, 5.2, 8.3, 9.7, 9.8, 8.3, 5.3 and 1.9m, respectively If the common interval is 15.9 metres, find the second moment of the waterplane area about the centreline and a transverse axis through the centre of flotation.

The half-ordinates of a ship’s waterplane at equidistant intervals from forward are as follows: 0, 1.3, 5.2, 8.3, 9.7, 9.8, 8.3, 5.3 and 1.9m, respectively If the common interval is 15.9 metres, find the second moment of the waterplane area about the centreline and a transverse axis through the centre of flotation.

Reference: Ship Stability for Masters & Mates (6^th edition), Exercise – 11.8

 

Solution:

1/2 ord (y)	SM (dx)	f(A) y*dx	y3	Y3*dx	Lever (x)	xydx	X2ydx
0	1	0	0	0	-4	0	0
1.3	4	5.2	2.19	8.76	-3	-15.6	46.8
5.2	2	10.4	140.6	281.2	-2	-20.8	41.6
8.3	4	33.2	571.78	2287.12	-1	-33.2	33.2
9.7	2	19.4	912.67	1825.34	0	0	0
9.8	4	39.2	941.19	3764.76	+1	39.2	39.2
8.3	2	16.6	571.78	1143.56	+2	33.2	66.4
5.3	4	21.2	148.87	595.48	+3	63.6	190.8
1.9	1	1.9	6.85	6.85	+4	7.6	30.4
		∑ = 147.1	∑ = 3295.34	∑ = 9913.07		∑ = 74	∑ = 448.4

 

Common interval, h = 15.9 m

I_CL = 2 * h/3 * 1/3 * ∑ Y³*dx

= 1 * 15.9/3 * 1/3 * 9913.07

= 35026.18 m⁴

 

Area = 2 * h/3 * ∑ ydx

= 2 * 15.9/3 * 147.1

= 1559.26 m²

 

I_midship = 2 * h³/3 * ∑ x²*y dx

= 2 * 15.9³/3 * 448.4

= 1201616.042 m⁴

 

Centre of flotation = ∑FM / ∑f(A) * h

        = +74 / 147.1 * 15.9

        = 7.99 m

I_LCF = I_midship – AX²

      = 1201616.042 - 1559.26 * 7.99²

      = 1101857 m⁴