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Oct 30, 2023

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NE 224 Spring 2023 Homework 02 (Due 3/6, Monday, by 11:59PM, Canvas upload) 1. A ship has the following characteristics when floating in seawater: ∆ = 44 , 500 tons C B = 0 . 689 KM = 59 . 35 ft L = 720 ft C W = 0 . 770 KB = 19 . 00 ft B = 105 . 8 ft C M = 0 . 968 Calculate the following and indicate units: a) Draft in seawater; b) Waterplane area; c) Midship section area; d) Prismatic coefficient; e) Transverse moment of inertia of water plane; f) Tons per inch immersion. Solution: a) specific weight of seawater: ρ s g = 1 / 35 tons/ft 3 , the volume of displacement: ∇ = ∆ ρ s g = 44500 ton 1 / 35 ton / ft 3 = 1557500 ft 3 block coefficient C B = ∇ LBT = 0 . 689 Therefore, the draft: T = ∇ C B LB = 1557500 ft 3 0 . 689 × 720 ft × 105 . 8 ft = 29 . 7 ft b) The waterplane corfficient: C W = A w BL 1

Waterplane area: A w = C W BL = 0 . 770 × 720 ft × 105 . 8 ft = 58655 . 5 ft 2 c) Midship section area: A M = C M BT = 0 . 968 × 105 . 8 ft × 29 . 7 ft = 3041 . 7 ft d) Prismatic coefficient: C P = C B C M = 0 . 689 0 . 968 = 0 . 712 e) Metacentric ratius: BM = KM − KB = 59 . 35 ft − 19 ft = 40 . 35 ft Transverse moment of inertia of waterplane: I T = BM ∇ = 40 . 35 ft × 1557500 ft 3 = 6284525 ft 4 f) Tons per inch immersion: TPI = A w 420 = 58655 . 5 420 = 139 . 66 2

2. A cylinder is placed in water with its axis vertical. Show that, if the center of gravity is in the waterplane, the cylinder will just float upright ( GM = 0) if the radius divided by the draft is greater than √ 2. Hint: Moment of inertia of a circle about its diameter is 0 . 25 AR 2 where A is the area of circle and R is the radius of the circle. Solution: Metacentric radius BM = I T ∇ = 0 . 25 AR 2 ∇ = 0 . 25( πR 2 ) R 2 ∇ = 0 . 25 πR 4 ∇ Volume of displacement ∇ = AT = πR 2 T , where T is the draft. Metacentric radius BM = 0 . 25 πR 4 ∇ = 0 . 25 πR 4 πR 2 T = 0 . 25 R 2 T Center of buoyancy height: KB = T 2 vertical distance from metacenter to baseline: KM = KB + BM = 0 . 25 R 2 T + T 2 G is located at the water surface, so the height of center of gravity: KG = T The metacentric height: GM = KM − KG = 0 . 25 R 2 T + T 2 − T = 0 . 25 R 2 T − T 2 If GM ≥ 0: GM = 0 . 25 R 2 T − T 2 ≥ 0 Therefore, R T ≥ √ 2 3

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3. A ship is floating with angle of heel ϕ = 2 . 5 ◦ toward the starboard. In order to return the ship to upright condition, a weight equal to 3% of the total displacement will be moved from startboard to port. The metacentric height GM = 1 . 3 m. Determine the distance the weight needs to be moved. Solution: The tangent of the heeling angle after moving weight w transversely from y 0 to y 1 tan ϕ = w ( y 1 − y 0 ) ∆ GM Therefore, the distance: y 1 − y 0 = ∆ GM tan ϕ w = ∆ GM tan ϕ 0 . 03∆ = GM tan 2 . 5 ◦ 0 . 03 = 1 . 3 m × tan 2 . 5 ◦ 0 . 03 = 1 . 892 m 4

4. The only bridge across the river within a hundred kilometers has collapsed. A local entrepreneur has a rectangular barge 22 meters long, 4.4 meters wide and 2.4 meters deep (keel to deck), which floats at a draft of 1.0 meter and has KG = 1 . 2 meters when nothing is aboard. The barge owner offers (for a fee) to carry a large truck across the river on the deck of the barge. The truck mass is 35 metric tons and its center of gravity is 2.0 meters above the ground. Density of water: ρ = 0 . 998 MT/m 3 . a) Calculate the initial stability GM of barge before the truck is loaded; b) Calculate the initial stability G 1 M 1 of the truck-barge combination. Would you advise the trucker to accept the offer? Note: After the truck is loaded to the barge, both the center of gravity and metacenter will shift vertically. Solution: Dimension of the barge: L = 22 m, B = 4 . 4 m and D = 2 . 4 m. Before loading, the draft of the barge: T = 1 . 0 m and KG = 1 . 2 m. The waterplane is a rectangle with dimension L × B . The waterplane area A w = LB = 22 m × 4 . 4 m = 96 . 8 m 2 The transverse moment of inertia of the water plane: I T = 1 12 LB 3 = 1 12 × (22 m) × (4 . 4 m) 3 = 156 . 1707 m 4 Volume of displacement: ∇ = LBT = 22 m × 4 . 4 m × 1 . 0 m = 96 . 8 m 3 The metacentric radius: BM = I T ∇ = 156 . 1707 m 4 96 . 8 m 3 = 1 . 6133 m Height of center of buoyancy: KB = T 2 = 0 . 5 m 5

Height of metacenter from baseline: KM = KB + BM = 0 . 5 m + 1 . 6133 m = 2 . 1133 m KG = 1 . 2 m is given, therefore, the metacentric height before loading: GM = KM − KG = 2 . 1133 m − 1 . 2 m = 0 . 9133 m b) The original displacement mass: ∆ m = ρ ∇ = 0 . 998 MT / m 3 × 96 . 8 m 3 = 96 . 6064 MT The mass of the truck m = 35 MT. After the truck is loaded to the barge, the increase of the draft: δT = w ρgA w = mg ρgA w = m ρA w = 35 MT 0 . 998 MT / m 3 × 96 . 8 m 2 = 0 . 3623 m The new draft after loading: T 1 = T + δT = 1 . 0 m + 0 . 3623 m = 1 . 3623 m < D = 2 . 4 m The new draft is till below the depth of the barge. Now check the stability after loading. The center of gravity of the truck is 2.0 m above ground. Therefore, after the truck is loaded to the top of the barge, the center of gravity of the truck will be located at z = D + 2 . 0 m = 2 . 4 m + 2 . 0 m = 4 . 4 m The new metacentric height: G 1 M 1 = GM + w w + ∆ T + δT 2 − z − GM = GM + m m + ∆ m T + δT 2 − z − GM = 0 . 9133 m + 35 MT 35 MT + 96 . 6064 MT 1 . 0 m + 0 . 3623 m 2 − 4 . 4 m − 0 . 9133 m = 0 . 9133 m + 0 . 2659 × ( − 4 . 1322 m) = − 0 . 1855 m The new metacentric height G 1 M 1 is negative. Therefore, the barge will no longer be stable after loading the truck. The offer should not be accepted. 6

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5. A rectangular barge 120 ft long, 30 ft wide and 13 ft deep floats in seawater at a draft of 7 ft when the only weights aboard are those for the inclining experiment. It is then inclined and a transverse shift of 5 tons through 20 ft heels the barge 2 ◦ . After the barge is returned to upright (the inclining equipment remains aboard), 480 tons of fuel oil having a specific gravity of 0.90 are loaded equally into three double-bottom tanks, each 10 ft wide and 120 ft long, formed by two equally spaced longitudinal bulkheads. Free surface is created in each of the tanks. Calculate the virtual metacentric height (metacentric height after the free surface correction) in this condition. Specific weight of seawater: ρ s g = 1 35 ton / ft 3 . Solution: When the barge is unloaded, the displacement ∆ = ρ s gLBT = 1 35 ton / ft 3 × 120 ft × 30 ft × 7 ft = 720 ton The equation for inclining experiments gives the metacentric height of the unloaded barge: GM = wd ∆ tan ϕ = 5 ton × 20 ft 720 ton × tan 2 ◦ = 3 . 9773 ft After the fuel is loaded, the vertical distance from baseline to the fuel free surface: T f = w f 0 . 9 ρ s g 1 LB = 480 ton 0 . 9 × 1 35 ton / ft 3 × 120 ft × 30 ft = 5 . 1852 ft The vertical distance from baseline to the center of gravity of the loaded fuel: z f = T f 2 = 5 . 1852 ft 2 = 2 . 5926 ft After loading the fuel, the increase in the barge’s draft: δT = w f ρ s gA w = w f ρ s gLB = 480 ton 1 35 ton / ft 3 × 120 ft × 30 ft = 4 . 6667 ft The new metacentric height after loading without considering the free surface correc- tion: G 1 M 1 = GM + w f w f + ∆ T + δT 2 − z f − GM = 3 . 9773 ft + 480 ton 480 ton + 720 ton 7 ft + 4 . 6667 ft 2 − 2 . 5926 ft − 3 . 9773 ft = 3 . 9773 ft + 0 . 4 × (2 . 7634 ft) = 5 . 0827 ft 7

The transverse moment of inertia of each fuel tank: i t = 1 12 L B 3 3 = 120 ft × (30 ft) 3 12 × 3 3 = 10000 ft 4 Now add the free surface correction to the metacentric height. The new metacentric height after free surface correction: G 1 M 1 ′ = G 1 M 1 − X ρ f gi t ∆ + w f = G 1 M 1 − X 0 . 9 ρ s gi t ∆ + w f = G 1 M 1 − 3 × 0 . 9 ρ s gi t ∆ + w f = 5 . 0827 ft − 3 × 0 . 9 × 1 35 ton / ft 3 × 10000 ft 4 480 ton + 720 ton = 5 . 0827 ft − 0 . 6429 ft = 4 . 4398 ft 8