Numericals on Floatation and metacentric Heights
Dr. J. Badshah
Dairy Engineering Department Sanjay Gandhi Institute of Dairy
Technology
Numerical on ships
1. When a load of 20 tonnes is shifted by a distance of 6 m, the ship tilts through an angle of 4°. The ship has a displacement of 5000 tonnes of sea water ( specific weight of sea water = 1025 kg/m
3. If the moment of inertia of the area at the water line is 8000 m
4, calculate (a) The metacentric height, and
(b) the height of the centre of gravity of the vessel above the centre of Buoyancy.
Solution: Volume of the displaced water V = 5000 x10
3/ 1025 = 4870 cu. Meter
BM = I/V = 8000/4870 = 1.64 m
MG = w x/W tanθ = 20x 6 / 5000 (π/180) x4 = 0.344 m (Taking tan θ = θ)
BG = BM – MG = 1.64 – 0.344 = 1.296 m
Numerical on angle of heel
When a weight of 588.5 KN is moved through 5 m on the deck of a ship, it causes a tilt of 0.07 radians in the ship. If the ship has a
displacement of 98.1 MN , determine its metacentric height. Hence, find the angle of heel when the ship is going ahead and 4780 kN-s is being transmitted to a single propeller shaft which is rotating at 10.5 rad/s.
Solution: Metacentric Height MG = wx / W tan θ =
MG = 588.5 x 5 x 1000/98.1x106 xtan (0.07 radian) = 0.428 m
In the second case, if T is the torque, then then work transmitted is
wT = 4780 x 103
Therefore, Torque T = 4780 x 1000/10.5 = 4.55x 105 N-m.
Let θ be the angle of heel in second case. Tan θ = T/W x MG
Tan θ = 4.55 x 105 / (98.1 x 10 6 x 0.428) = 0.0108
Angle of heel θ = 0.621°
Numerical on metacentric Heights of Pontoon
A rectangular pontoon weighing 240 tonnes has a length of 20 m. The centre of gravity is 30 cm above the cetre of cross section. The metacentric height is 1.33 mm, when the angle of heel is 10°. The free board is not to be more than 0.67 m when the pontoon is vertcal. Find the breadth and height of the pontoon if it is floating in fresh water. Solution: Volume of water displaced = 240 x 10 3/ 103 = 240 m3
Let b be the breadth and D be the depth of pontoon. Therefore volume of water displaced by pontoon, 240 = b (D- 0.67) x20
Because BM = I/V = 20x b3 x1/12/240 = (1/144) b3
Height of centre of Buoyancy above base A AB= ( D- 0.67) /2
Height of centre of gravity above base A, AG = D/2 + 0.30
Therefore BG = (D/2 +0.30) – (D-0.67)/2 = 0.635 m
MG = BM – BG, 1.33 = 1/144 b3 – 0.635
B = 6.58 m and 6.58 (D -0.67) 12 or D = 2.49 m
