The ½ breadths of the load waterplane of a ship 150 m long, commencing from aft, are 0.3, 3.8, 6.0, 7.7, 8.3, 9.0, 8.4, 7.8, 6.9, 4.7 and 0 m respectively. Calculate: (a) area of waterplane (b) distance of centroid from midships (c) second moment of area about a transverse axis through the centroid

 The ½ breadths of the load waterplane of a ship 150 m long, commencing from aft, are 0.3, 3.8, 6.0, 7.7, 8.3, 9.0, 8.4, 7.8, 6.9, 4.7 and 0 m respectively. Calculate:

(a) area of waterplane

(b) distance of centroid from midships

(c) second moment of area about a transverse axis through the centroid

Reference: REED'S NAVAL ARCHITECTURE FOR MARINE ENGINEERS (4^th edition), Exercise – 3.6

 

Solution:

a)

	½ breadth	SM	f(A)	Levers	FM	Levers	SM
AFT	0.3	1	0.3	-5	-1.5	-5	7.5
	3.8	4	15.2	-4	-60.8	-4	243.2
	6.0	2	12	-3	-36	-3	108
	7.7	4	30.8	-2	-61.6	-2	123.2
	8.3	2	16.6	-1	-16.6	-1	16.6
	9.0	4	36	0	0	0	0
	8.4	2	16.8	+1	16.8	+1	16.8
	7.8	4	31.2	+2	62.4	+2	124.8
	6.9	2	13.8	+3	41.4	+3	124.2
	4.7	4	18.8	+4	75.2	+4	300.8
FOR	0	1	0	+5	0	+5	0
			∑f(A) = 191.5		∑f(FM) = 19.3		∑f(SM) = 1065.1

 

Common interval, h = 150/10 = 15 m

Waterplane area, Aw = 2* h/3 * ∑f(A)

                                    = 2 * 15/2 * 191.5

                                    = 1915 m²

b)

Centroid from midship = ∑f(FM)/ ∑f(A) * h

                                      = 19.3/191.5 * 15

                                      = 1.512 m FOR

c)

I_midship = 2* h³/3 * ∑f(SM)

             = 2 * 15³/3 * 1065.1

             = 2396475 m⁴

I_centroid = I_midship – AX²

             = 2396475 – 1915*1.512²

             = 2392097.034 m⁴