stability problems 6.pps

• 1. A ship has a displacement of 2,400 1. A ship has a displacement of 2,400 tons and KG 10.8 meters. Find the new tons and KG 10.8 meters. Find the new KG if a weight of 50 tons mass already KG if a weight of 50 tons mass already on board is raised 12 meters verticallyon board is raised 12 meters vertically..

• Answer: 11.05 meters

K

M

G

B

• THE POS. OF THE POS. OF KK IS FIXED IS FIXED• THE POS OF THE POS OF MM VARIES VARIES

W/ THE DFT OF VSL.W/ THE DFT OF VSL.G

G

• GG MOVES MOVES TOWARDSTOWARDS WT. LOADEDWT. LOADED

G

G

• GG MOVES MOVES AWAY FROMAWAY FROMWT. DISCHARGEWT. DISCHARGE

• GG MOVES MOVES PARALLELPARALLEL TO THE DIRTO THE DIR. OF THE WT. OF THE WT

BEING SHIFTEDBEING SHIFTED

• B MOVES TOWARDSB MOVES TOWARDSTHE THE LOW SIDE OFLOW SIDE OFINCLINED V/LINCLINED V/L

G

K

GM

• 1. Solution:

• GG’ = Wt. x Dist.

Δ

• = 50 tons x 12m

2,400 tons

• GG’ = 0.25 m

• Old KG = 10.80 m ( + )

• New KG = 11.05 m

• 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. 10.41 m

K

GM

• 2. Solution:

• GG’ = Wt x Dist

Δ

• = 40 tons x 4.5 m

2,000 tons

• GG’ = 0.09 m

• Old KG = 10.50 m ( - )

• New KG = 10.41 m

3. A ship of 2,000 tons displacement has 3. A ship of 2,000 tons displacement has KG 4.5 meters. A heavy lift of 20 tons KG 4.5 meters. A heavy lift of 20 tons mass is in the lower hold and has KG 2 mass is in the lower hold and has KG 2 meters. The weight is then raised 0.5 meters. The weight is then raised 0.5 meter clear of the tank top by a derrick meter clear of the tank top by a derrick whose head is 14 meters above the keel. whose head is 14 meters above the keel. Find the new KG of the ship?Find the new KG of the ship?

Ans. 4.62 m

K

GM

• 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’ = 0.12 m• Old KG = 4.50 m ( + )• New KG = 4.62 m•

4. A ship has a displacement of 7,000 4. A ship has a displacement of 7,000 tons and KG 6 meters. A heavy lift in tons and KG 6 meters. A heavy lift in the lower hold has KG 3 meters and the lower hold has KG 3 meters and mass 40 tons. Find the new KG when mass 40 tons. Find the new KG when this weight is raised through 4.5 meters this weight is raised through 4.5 meters vertically and is suspended by a derrick vertically and is suspended by a derrick whose head is 17 meters above the whose head is 17 meters above the keel.keel.

Ans. 6.08 m

• 4. Solution:

• Ht of Derrick = 17.0 m

• KG of Cargo = 3.0 m ( - )

• Distance = 14.0 m

• GG’ = Wt. x Dist.

Δ

• = 40 tons x 14.0 m

7,000 tons

• GG’ = 0.08 m

• Old KG = 6.00 m ( + )

• New KG = 6.08 m

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. 0.25 m

• 5. Solution:

• GG’ = Wt. x Dist.

Δ

• = 25 tons x 15 m

1,500 tons

• GG’ = 0.25 m

6. A tank holds 120 tons when full of 6. A tank holds 120 tons when full of fresh water. Find how many tons of oil of fresh water. Find how many tons of oil of relative density 0.84 it will hold, allowing relative density 0.84 it will hold, allowing 2% of the volume of the tank for 2% of the volume of the tank for expansion in the oil.expansion in the oil.

Ans. 98.78 tons

• 6. Solution:

• Capacity = Wt. of FW x Expansion

• = 120 tons x 2 % or ( .02 )

• Capacity = 2.4

• (-) 120.0

• Capacity = 117.6 tons

• Wt. of Oil = Capacity x R.D. of Oil

• 117.6 tons x 0.84

• Wt. of oil = 98.78 tons

7. A tank when full will hold 130 tons of 7. A tank when full will hold 130 tons of salt water. Find how many tons of oil salt water. Find how many tons of oil relative density 0.909 it will hold relative density 0.909 it will hold allowing 1% of the volume of the tank allowing 1% of the volume of the tank for expansion.for expansion.

Ans. 114.13 tons

• 7. Solution:• Capacity = Capacity of FW x Expansion• = 130 x 1% or .01• Capacity = 1.3• (-) 130.0 tons• Capacity = 128.7 tons• R.D. Oil = 0.909 ( x )• Wt. of Oil = 116.98 tons• Actual Wt. = Weight of Oil / Density of SW• = 116.98 tons / 1.025• Actual Wt. of Oil = 114.13 tons

8. A tank measuring 8m x 6m x 7m is 8. A tank measuring 8m x 6m x 7m is being filled with oil of relative density being filled with oil of relative density 0.9. Find how many tons of oil in the tank 0.9. Find how many tons of oil in the tank when the ullage is 3 meters.when the ullage is 3 meters.

Ans. 172.8 tons

• 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 = 172.8 tons

9. Oil of relative density 0.75 is run into 9. Oil of relative density 0.75 is run into a tank measuring 6m x 4m x 8m until the a tank measuring 6m x 4m x 8m until the ullage is 2 meters. Calculate the number ullage is 2 meters. Calculate the number of tons of oil the tank then containsof tons of oil the tank then contains.

Ans. 108 tons

• 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

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. 83.3 tons

• 10. Solution:

• Capacity = Wt. of FW x Expansion

• = 100 t x 2% or .02

• = 2

• (-)100

• Capacity = 98 tons

• Weight of Oil = Capacity x R.D. of Oil

• = 98 tons x 0.85

• Weight of Oil = 83.3 tons

•

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. 1094.04 tons / 0.75 m

• 11. Solution:

• Δ = L x B x Draft x Cb x R. Density

• = 20m x 6m x 1.5m x 1 x 1.013 tons/m³

• Δ = 1,094.04 tons

• KB = ½ x Draft

• = ½ x 1.5 m

• KB = 0.75 m

• 12. A box-shaped barge 55m x 10m x 6m 12. A box-shaped barge 55m x 10m x 6m is floating in fresh water on an even keel is floating in fresh water on an even keel at 1.5 m draft. If 1,800 tons of cargo is at 1.5 m draft. If 1,800 tons of cargo is now loaded, find the difference in the now loaded, find the difference in the height of the centre buoyancy above the height of the centre buoyancy above the keel?keel?

• Ans. 1.636 m

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

• If 1,800 is now loadedIf 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 = 1800 tons• 55m x 10m x 1 x 1.000 ton/m³• Draft = 1800• 550• Draft = 3.27m• KB = ½ x Draft• = ½ x 3.27m• KB = 1.635m

• 12. Solution:

• New Draft New Δ

• Old Draft Old Δ

• New Draft 1,800 tons

1.5m 825 tons

• New Draft = 1.5m x 1,800 tons

825 tons

• New Draft = 3.27mNew Draft = 3.27m

• KB = ½ x Draft

• KB = ½ x 3.27m

• KB = 1.635m

13. A box-shaped barge 75m x 6m x 4m 13. A box-shaped barge 75m x 6m x 4m displaces 180 tons when light. If 360 tons displaces 180 tons when light. If 360 tons of iron are loaded while the barge is of iron are loaded while the barge is floating in fresh water, find her final draft floating in fresh water, find her final draft reserve buoyancy.reserve buoyancy.

Ans. 1.2m / 70%

• 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 = 180 tons

75m x 6m x 1 x 1.000

Light Draft = 180 tons

450

Light Draft = 0.4m

• New Draft New Δ

Old Draft Old Δ

• New Draft 360 tons

0.4m 180 tons

• New Draft = 0.4m x 360 tons

180 tons

• New Draft = 0.8m

• Light Draft = 0.4m ( + )

• Final Draft = 1.2m

•

•

•

• Reserve Buoyancy = Final Draft/ Depth

• = 1.2m / 4m

• Intact Buoyancy = 0.3 or 30 %

• ( - ) 100 %

• Reserve Buoyancy = 70 %

4m

Depth 1.2m Draft

RESERVE BUOYANCY