Solution Naval Numerical 25

Numerical 25: A ship of length 120 m displaces 11750 tonne when floating in sea water of density 1025 kg/m3-cube. The center of gravity is 2m above the center of buoyancy and the water plane is defined by the following equidistant half-ordinates given in Table station AP 1, 2, 3, 4, 5, 6, 7 .FP½ breadth (m) 3.3, 6.8, 7.6, 8.1, 8.1, 8.0, 6.6, 2.8, 0. Calculate EACH of the following: (a) The area of the waterplane; (b) The position of the centroid of the waterplane from midships; (c) The second moment of area of the waterplane about a transverse axis through the centroid; (d) The moment to change trim one centimeter (MCT1cm).The center of gravity is 2m above the center of buoyancy.

Solution: Given that Length (L) = 120m
Displacement (Δ) =11750 tonne
water density (ρ) = 1025 kg/cu-m
The center of gravity is 2m above the center of buoyancy

½ breadth	SM	Product for area	Lever for first moment from mid-ship	Product for first moment	Lever for 2nd moment	Product for 2nd moment
3.3	1	3.3	-4	-13.2	-4	52.8
6.8	4	27.2	-3	-81.6	-3	244.8
7.6	2	15.2	-2	-30.4	-2	60.8
8.1	4	32.4	-1	-32.4	-1	32.4
8.1	2	16.2	0	0	0	0
8.0	4	32	1	32	1	32
6.6	2	13.2	2	26.4	2	42.8
2.8	4	11.2	3	33.6	3	100.8
0	1	0	4	0	4	0
		∑a = 150.7		∑m = -65.6		∑I= 576.4

Common interval (h) = 120/8 =15
(a) Water plane area (Aw) = 2 x (h/3) x ∑a 
Aw = 2x (15/3)×150.7
= 1507 m-sq
(b) Position of centroid of water plane from mid ship = h x ∑m/∑a 
=15 x (-65.6/150.7)
= -  6.52
i.e  6.25m aft of the mid-ship
(c)The second moment of area of the waterplane about a transverse axis through the centroid .
We know that the second moment of area of waterplane about the centroid is given bywhere x is the distance of centroid from midships.
Im is the second moment of area of water plane about midships and is given by
thus,
  
=1296900- 64063.173
= 1232836.82

(d) The moment to change trim one centimetre (MCT1cm).
 where,

and

 

thus,


MCT1cm = 103.3464tonne-m,