EXPERIMENT (2) BUOYANCY & FLOTATION (METACENTRIC HEIGHT) 1 By: Eng. Motasem M. Abushaban. Eng. Fedaa M. Fayyad.
ARCHIMEDES PRINCIPLE Archimedes Principle states that the buoyant force has a magnitude equal to the weight of the fluid displaced by the body and is directed vertically upward. Buoyant force is a force that results from a floating or submerged body in a fluid. The force results from different pressures on the top and bottom of the object. W is the weight of the shaded area F 1 and F 2 are the forces on the plane surfaces F B is the buoyant force the body exerts on the fluid 2
ARCHIMEDES PRINCIPLE The force of the fluid on the body is opposite, or vertically upward and is known as the Buoyant Force. The force is equal to the weight of the fluid it displaces. The buoyant forces acts through the centroid of the displaced volume 3 The location is known as the center of buoyancy.
STABILITY: SUBMERGED OBJECT Stable Equilibrium: if when displaced returns to equilibrium position. Unstable Equilibrium: if when displaced it returns to a new equilibrium position. Stable Equilibrium: Unstable Equilibrium: C > CG, Higher C < CG, Lower 4
STABILITY: SUBMERGED OBJECT If the Centre of Gravity is below the centre of buoyancy this will be a righting moment and the body will tend to return to its equilibrium position (Stable). If the Centre of Gravity is above the centre of buoyancy,an overturning moment is produced and the body is (unstable). Note that, As the body is totally submerged, the shape of displaced fluid is not altered when the body is tilted and so the centre of buoyancy unchanged relative to the body. 5
BUOYANCY AND STABILITY: FLOATING OBJECT Slightly more complicated as the location of the center buoyancy can change: 6
METACENTRE AND METACENTRIC HEIGHT Metacentre point (M): This point, about which the body starts oscillating. Metacentric Height: Is the distance between the centre of gravity of floating body and the metacentre. 7
STABILITY OF FLOATING OBJECT If M lies above G a righting moment is produced, equilibrium is stable and GM is regarded as positive. If M lies below G an overturning moment is produced, equilibrium is unstable and GM is regarded as negative. If M coincides with G, the body is in neutral equilibrium. 8
DETERMINATION OF METACENTRIC HEIGHT 1- Practically : 2- Theoretically: MG = BM + OB OG...(2) In Water OB = 0.5 V b. d 9
h 10
PURPOSE: To determine the metacentric height of a flat bottomed vessel in two parts: PART (1) : for unloaded and for loaded pontoon. PART (2) : when changing the center of gravity of the pontoon. 11
EXPERIMENTAL SET-UP: The set up consists of a small water tank having transparent side walls in which a small ship model is floated, the weight of the model can be changed by adding or removing weights. Adjustable mass is used for tilting the ship, plump line is attached to the mast to measure the tilting angle. 12
PART (1) Determination of floatation characteristic for unloaded and for loaded pontoon. 13
PROCEDURE 1. Assemble the pontoon by positioning the bridge piece and mast. 2. Weigh the pontoon and determine the height of its center of gravity up the line of the mast. 3. Fill the hydraulic bench measuring tank with water and float the pontoon in it, then ensure that the plumb line on the zero mark. 4. Apply a weight of 50 g on the bridge piece loading pin then measure and record the angle of tilting and the value of applied weight 14
PROCEDURE 5. Repeat step 4 for different weights; 100, 150, & 200 g, and take the corresponding angle of tilting. 6. Repeat the above procedure with increasing the bottom loading by 2000 gm and 4000 gm. 7. Record the results in the table. 8. Calculate GM practically where, W has three cases. 9. Draw a relationship between θ (x-axis) and GM (y-axis), then obtain GM when θ equals zero. 10. Calculate GM theoretically. 15
Pontoon measurement: - Pontoon dimension : Depth (D) = 170 mm Length (L) = 380 mm, Width (W) = 250 mm. -The height of the center of gravity of the pontoon is OG vm = 125 mm from outer surface of vessel base. - The balance weight is placed at x = 123 mm from pontoon center line. - The weight of the pontoon and the mast W vm = 3000 gm Bilge Weight Wb (gm) Off balance wt. P (gm) Mean Def. θ (degree) Exp. GM (mm) GM at θ =0 from graph BM OB Theo. GM (mm) (mm) (mm) 0.00 50 100 150 200 2000.00 50 x1 = 30 100 150 200 4000.00 100 x1 = 37.5 150 200 16 250
PURPOSE: To determine the metacentric height of a flat bottomed vessel in two parts: PART (1) : for unloaded and for loaded pontoon. PART (2) : when changing the center of gravity of the pontoon. 17
Remember: - Pontoon dimension : Depth (D) = 170 mm Length (L) = 380 mm, Width (W) = 250 mm. - The height of the center of gravity of the pontoon is OG vm = 125 mm from outer surface of vessel base. - The balance weight is placed at x = 123 mm from pontoon center line. - The weight of the pontoon and the mast W vm = 3000 gm 18
PROCEDURE PART (2) : when changing the center of gravity of the pontoon. 1. Replace the bilge weights by 4x 50 gm weights. 2. Apply a weight of 300gm on a height of 190 mm from the pontoon surface. 3. Apply weights of 40, 80 &120 gms on the bridge piece loading pin, then record the corresponding tilting angle. 4. Calculate GM practically where GM = P(123) 3500.θ 5. Draw a relationship between θ in degrees (x-axis) and GM Practical (y-axis), then obtain GM when θ equals zero. 19
PROCEDURE 6. Move 50 gm bilge weight to the mast ahead, then repeat steps 3,4&5. 7. Repeat step 6 moving 100, 150 & 200 gm bilge weight to the mast. 8. Determine the height of the center of gravity for each loading condition according to equation OG = Wvm(125) + Wb(35) + Wb1(190) + Wm(790 + W L ) 2 20
3000(125) + 300(190) + Wb(35) + Wm(790 + OG = 3500 L ) 2 21
8. Calculate GM theoretically according to equation GM (Th.) = BM + OB OG Notice: BM & OB are constants for all loading conditions, since the dimensions & the weight of pontoon do not alter. 22
Table (2) \ Part (2) Off balance wt. Mean Def. Exp. GM BM OG Theo. GM P (gm) θ (degree) (mm) (mm) (mm) (mm) Mast Weight = 0.0 40 2.40 80 4.88 120 7.50 Mast Weight = 50.0 40 3.45 80 7.23 120 10.50 Mast weight = 100.0 20 3.28 40 6.35 80 12.00 Mast Weight = 150.0 10 3.70 20 10.23 40 14.78 Mast weight = 200.0 23 Unstable
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QUESTIONS 25