Tiêu đề HPGE 16 Solutions.pdf ✅

16 Hydraulics: Stability of Floating Bodies Solutions SITUATION 1. If the center of gravity of a ship in the upright position is 10 m above the center of gravity of the portion under water, the displacement being 10 MN, and the ship is tipped 30° causing the center of buoyancy to the shift sidewise by 8 m. ▣ 1. Find the metacentric height in tilted position. [SOLUTION] GB = 10 m, Wdisp = 10 MN, θ = 30°, Shift = 8 m MG = MB ± GB Compute for MB: MB = I V MB = shift sin θ MB = 8 m sin 30° MB = 16 m Compute for MG: MG = 16 m − 10 m MG = 6 m ▣ 2. What type of moment is generated? [SOLUTION] Since MG is positive, then it is a RIGHTING MOMENT. ▣ 3. What is the magnitude of the moment? [SOLUTION] RM = W(MG sin θ) RM = (10 MN)(6 m × sin 30°) RM = 30 000 kNm

(0.65)(7.5 cm) 2 (30 cm) = ( y 4 ) 2 y y = 25.99 cm ▣ 5. Determine the distance of the metacenter from the center of buoyancy. [SOLUTION] MB = I V From ratio and proportion, r = y 4 = 26 cm 4 = 6.5 cm MB = π 4 r 4 π 3 r 2h MB = 3r 2 4h MB = 3(6.5 cm) 2 4(26 cm) MB = 1.21875 cm ▣ 6. Determine the distance from the metacenter to the center of gravity. [SOLUTION] Compute for GB: G = h 4 from the top G = 30 cm 4 = 7.5 cm B is y 4 from the surface of water. B = 4 cm + 26 cm 4 B = 10.5 cm GB = 10.5 cm − 7.5 cm GB = 3 cm MG = 1.218 cm − 3 cm MG = −1.791 cm
SITUATION 3. A rectangular scow 9 m wide 15 m long and 3.6 m high has a draft in sea water of 2.4 m. Its center of gravity is 2.7 m above the bottom of the scow. ▣ 7. Determine the initial metacentric height. [SOLUTION] GB = 2.7 m − 1.2 m GB = 1.5 m Compute for MB MB = I V MB = 1 12 (15 m)(9 m) 3 (9 m × 2.4 m)(15 m) MB = 2.8125 m Compute for the metacentric height MG = MB − GB MG = 2.8125 m − 1.5 m MG = 1.3125 m ▣ 8. Determine the final metacentric height if the body is tilted until one end is just submerged in water. [SOLUTION] Formula for MB: MB = B 2 12D (1 + tan2 θ 2 )