The 1/2 ordinates of a waterplane at 15m intervals, commencing from aft, are 1, 7, 10.5, 11, 11, 10.5, 8, 4 and Om.
Calculate:
(a) TPC
(b) Distance of the centre of flotation from midships
(c) Second moment of area of the waterplane about a transverse axis through the centre of flotation.
Structured for DG Shipping MEO Class II examination scoring criteria.
1/2 Ord | SM | Product | Lever | Product | Lever | Product |
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 | ΣMA = 174 | 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 |
ΣA = 190 | ΣMF = -122 | ΣI = +646 |
$$Common\:interval\:\left(h\right)=15$$
$$Waterplane\:area\:\left(A_{w}\right)=2\times\frac{h}{3}\sum A$$
$$=2\times\frac{15}{3}\times190$$
$$=1900m^2$$
$$Longitudinal\:centre\:of\:flotation\:LCF=h\times\frac{\sum M_{A}+\sum M_{F}}{\sum A}$$
$$=15\times\frac{174-122}{190}$$
$$LCF=4.105m$$
$$A_{w}=\frac{100\:\times\:TPC}{\rho}$$
$$TPC=\frac{1.025\times1900}{100}$$
$$TPC=19.475m^2$$
$$Second\:moment\:about\:midships\:\left(I_{m}\right)=2\times\frac{h^3}{3}\sum I$$
$$=2\times\frac{15^3}{3}\times646$$
$$I_{m}=1453500m^4$$
$$Second\:moment\:of\:area\:about\:centroid\:\left(I_{F}\right)=I_{m}-A_{w}\times TPC^2$$
$$1453500-\left(1900\times41.05^2\right)$$
$$I_{F}=1421483m^4$$