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CHS PROBLEM SOLVING 1 - 13
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1. A ship has a displacement of 2,400 tons and KG 10. 8 meters
1. A ship has a displacement of 2,400 tons and KG 10.8 meters. Find the new KG if a weight of 50 tons mass already on board is raised 12 meters vertically. Answer: meters
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M G G G G G G B K THE POS. OF K IS FIXED THE POS OF M VARIES
W/ THE DFT OF VSL. G G G G MOVES TOWARDS G WT. LOADED G G G MOVES AWAY FROM WT. DISCHARGE B G MOVES PARALLEL TO THE DIR. OF THE WT BEING SHIFTED K B MOVES TOWARDS THE LOW SIDE OF INCLINED V/L
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M G K
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1. Solution: GG’ = Wt. x Dist. Δ = 50 tons x 12m 2,400 tons GG’ = m Old KG = m ( + ) New KG = m
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2. A ship has displacement of 2,000 tons and KG 10. 5 meters
2. A ship has displacement of 2,000 tons and KG 10.5 meters. Find the new KG if a weight of 40 tons mass already on board is shifted from the ‘tween deck to lower hold trough a distance of 4.5 meters vertically. Ans m
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M G K
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2. Solution: GG’ = Wt x Dist Δ = 40 tons x 4.5 m 2,000 tons GG’ = m Old KG = m ( - ) New KG = m
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3. A ship of 2,000 tons displacement has KG 4. 5 meters
3. A ship of 2,000 tons displacement has KG 4.5 meters. A heavy lift of 20 tons mass is in the lower hold and has KG 2 meters. The weight is then raised 0.5 meter clear of the tank top by a derrick whose head is 14 meters above the keel. Find the new KG of the ship? Ans m
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M G K
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3. Solution: Height of Derrick fr. Keel = 14 m KG. of Cargo = 2.0 m ( - ) Distance = 12.0 m GG’ = Wt x Dist. Δ = 20 tons x 12.0 m 2,000 tons GG’ = m Old KG = m ( + ) New KG = 4.62 m
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4. A ship has a displacement of 7,000 tons and KG 6 meters
4. A ship has a displacement of 7,000 tons and KG 6 meters. A heavy lift in the lower hold has KG 3 meters and mass 40 tons. Find the new KG when this weight is raised through 4.5 meters vertically and is suspended by a derrick whose head is 17 meters above the keel. Ans m
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4. Solution: Ht of Derrick = m KG of Cargo = m ( - ) Distance = m GG’ = Wt. x Dist. Δ = tons x 14.0 m 7,000 tons GG’ = m Old KG = m ( + ) New KG = m
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5. Find the ship in the center of gravity of a ship of 1,500 tons displacement when a weight of 25 tons mass is shifted from the starboard side of the lower hold to the port side on deck through a distance of 15 meters. Ans m
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5. Solution: GG’ = Wt. x Dist. Δ = 25 tons x 15 m 1,500 tons GG’ = m
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6. A tank holds 120 tons when full of fresh water
6. A tank holds 120 tons when full of fresh water. Find how many tons of oil of relative density 0.84 it will hold, allowing 2% of the volume of the tank for expansion in the oil. Ans tons
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6. Solution: Capacity = Wt. of FW x Expansion = tons x 2 % or ( .02 ) Capacity = (-) Capacity = tons Wt. of Oil = Capacity x R.D. of Oil 117.6 tons x 0.84 Wt. of oil = tons
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7. A tank when full will hold 130 tons of salt water
7. A tank when full will hold 130 tons of salt water. Find how many tons of oil relative density it will hold allowing 1% of the volume of the tank for expansion. Ans tons
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7. Solution: Capacity = Capacity of FW x Expansion = x 1% or .01 Capacity = (-) tons Capacity = tons R.D. Oil = ( x ) Wt. of Oil = tons Actual Wt. = Weight of Oil / Density of SW = tons / 1.025 Actual Wt. of Oil = tons
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8. A tank measuring 8m x 6m x 7m is being filled with oil of relative density 0.9. Find how many tons of oil in the tank when the ullage is 3 meters. Ans tons
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8. Solution: Wt. of Oil = L x W x (H – Ullage) x R.D of Oil = 8m x 6m x ( 7m – 3m ) x 0.9 = 8m x 6m x 4m x 0.9 Wt. of Oil = tons
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9. Oil of relative density 0
9. Oil of relative density 0.75 is run into a tank measuring 6m x 4m x 8m until the ullage is 2 meters. Calculate the number of tons of oil the tank then contains. Ans tons
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9. Solution: Weight of Oil = L x W x ( H – Ullage ) x R.D. Oil = 6m x 4m x (8m – 2m) x 0.75 = 6m x 4m x 6m x 0.75 Weight of Oil = 108 tons
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10. A tank will hold 100 tons when full of fresh water
10. A tank will hold 100 tons when full of fresh water. Find how many tons of oil of relative density 0.85 may be loaded if 2% of the volume of the oil loaded is to be allowed for expansion. Ans tons
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10. Solution: Capacity = Wt. of FW x Expansion = 100 t x 2% or .02 = (-)100 Capacity = 98 tons Weight of Oil = Capacity x R.D. of Oil = 98 tons x 0.85 Weight of Oil = tons
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11. A box-shaped vessel 120m x 6m x 2. 5 m floats at a draft of 1
11. A box-shaped vessel 120m x 6m x 2.5 m floats at a draft of 1.5 m in water of density 1,013 kgs/m³. Find the displacement in tons, and the height of the centre of buoyancy above the keel. Ans tons / 0.75 m
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11. Solution: Δ = L x B x Draft x Cb x R. Density = 20m x 6m x 1.5m x 1 x tons/m³ Δ = 1, tons KB = ½ x Draft = ½ x 1.5 m KB = m
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12. A box-shaped barge 55m x 10m x 6m is floating in fresh water on an even keel at 1.5 m draft. If 1,800 tons of cargo is now loaded, find the difference in the height of the centre buoyancy above the keel? Ans m
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12. Solution: Δ = L x W x Draft x Cb x R. Density = 55m x 10m x 1.5m x 1 x 1.000ton/m³ Δ = 825 tons KB = ½ x Draft KB = ½ x 1.5m KB = 0.75m
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If 1,800 is now loaded Δ = L x W x Draft x Cb x R. Density 1,800 tons = 55m x 10m x Draft x 1 x 1.000 Draft = tons 55m x 10m x 1 x ton/m³ Draft = 550 Draft = 3.27m KB = ½ x Draft = ½ x 3.27m KB = m
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12. Solution: New Draft New Δ Old Draft Old Δ New Draft 1,800 tons 1.5m tons New Draft = 1.5m x 1,800 tons 825 tons New Draft = m KB = ½ x Draft KB = ½ x 3.27m KB = m
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13. A box-shaped barge 75m x 6m x 4m displaces 180 tons when light
13. A box-shaped barge 75m x 6m x 4m displaces 180 tons when light. If 360 tons of iron are loaded while the barge is floating in fresh water, find her final draft reserve buoyancy. Ans m / 70%
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13. Solution: Light Δ = L x W x Draft x Cb x R. Density 180 tons = 75m x 6 m x Draft x 1 x 1.0 Light Draft = tons 75m x 6m x 1 x 1.000 Light Draft = tons 450 Light Draft = 0.4m
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New Draft New Δ Old Draft Old Δ New Draft tons 0.4m tons New Draft = 0.4m x 360 tons 180 tons New Draft = 0.8m Light Draft = 0.4m ( + ) Final Draft = 1.2m
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Reserve Buoyancy = Final Draft/ Depth = 1.2m / 4m
Intact Buoyancy = or 30 % ( - ) % Reserve Buoyancy = % RESERVE BUOYANCY 4m Depth 1.2m Draft
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