207.NA November 2022 Q.8(b)

  The ½ ordinates of a water plane at 15m intervals, commencing from aft, are 1, 7, 10.5, 11, 11, 10.5, 8, 4 and 0 m. Calculate:

i. TPC

ii. Distance of the centre of flotation from midships

iii. Second moment of area of the water plane about a transverse axis through the centre of flotation.

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Solution

½ ordinate	SM	Product of area	lever	Product for 1st moment	lever	Product for 2nd moment
1	1	1	+4	+4	+4	+16
7	4	28	+3	+84	+3	+252
10.5	2	21	+2	+42	+2	+84
11	4	44	+1	+44	+1	+44
11	2	22	0	0	0	0
10.5	4	42	-1	-42	-1	+42
8	2	16	-2	-32	-2	+64
4	4	16	-3	-48	-3	+144
0	1	0	-4	0	-4	0
		∑m = 190		∑m1 = +52		∑m2 = +646

 

i) Common interval h = 15m

Area of water plane Aw = 2 x ((h/3) x product of area ) --------- 2 given because 1/2 ordinates are given

                                       = 2 x (15/3) x 190

                                       = 1900 m3

TPC =  (Aw x density of water ) / 100

        = (1900 x 1.025) / 100

        = 19.475

ii)Distance of the centre of flotation from midships = (h x product of first moment) / product of area

                                                                               X  = (15 x +52) / 190

                                                                                X = 4.11 m towards aft (As value is +ve)

iii) Second moment about midship Im = (2/3) x h3 x Product of 2nd moment

                                                              = (2/3) x 153 x 646

                                                              = 1453500 m4

 

Second moment of area about centroid = Im - (Aw x X2)

                                                               = 1453500 - (1900 x 4.112)

                                                               = 1421405 m4