STABILITY PROBLEM 4

1. A box-shaped vessel 24m x 6m x 3m displaces 150 tonnes of water 1. A box-shaped vessel 24m x 6m x 3m displaces 150 tonnes of water. Find the draft when the vessel is floating in salt water. Ans. 1.016 m

1. Solution: ∆ = L x B x Draft x Rel. Density 150 tons = 24m x 6m x Draft x 1.025 Draft = 150 24m x 6m x 1.025 Draft = 150 147.6 Draft = 1.016 m

2. A box-ship 150m x 20m x 12m on an even keel at 5m draft 2. A box-ship 150m x 20m x 12m on an even keel at 5m draft. A compartment amidships is 15m long and contains timber of relative density 0.8 and a stowage factor 1.5 cubic meters per ton. Calculate the new draft if this compartment is now bilged. Ans. 5.085 m

Permeability = Broken Stowage Stowage Factor = 0.25 / 1.50 x 100 2. Solution: Space occupied = 1 By timber 0.8 = 1.25 m³ Stowage Factor = 1.50 m³ Space Occupied = 1.25 m³ ( - ) Broken Stowage = 0.25 m³ Permeability = Broken Stowage Stowage Factor = 0.25 / 1.50 x 100 Permeability = 16.67% or 0.1667 x 100

Inc. in Draft = Vol. of Lost Buoyancy Area of Intact WP = 0.1667 x 15 x 20 x 5 150 x 20 – 01667 x 15 x 20 250 / 2950 Increase in Draft = 0.085 m Old Draft = 5.000 m ( + ) New Draft = 5.085 m

3. A box-ship draws 7. 5 m in dock water of density 1. 006 ton per cu 3. A box-ship draws 7.5 m in dock water of density 1.006 ton per cu.m. Find the draft when she is floating in sea water. Ans. 7.361 m

3. Solution: New Draft Old Density Old Draft New Density New Draft 1.006 ton/m³ 7.5 m 1.025 ton/m³ New Draft = 7.5 m x 1006 ton/m³ 1.025 ton/m³ New Draft = 7.361 m

4. A box-ship 40 meters long, 6 meters beam, is floating at a draft of 2m. F and A. She has an amidships compartment 10m long which is empty. If the original GM is 0.6m, find the new GM if this compartment is bilged. Ans. 0.558 meter

4. Solution: A. Find the KG KB = ½ x Draft KB = ½ x 2m KB = 1.00 m BM = I or LB³ V 12 x V = 40m x 6³m 12 x 40m x 6m x 2m = 8,640 / 5,760 BM = 1.50m KB = 1.00m ( + ) KM = 2.50m GM = 0.6m ( - ) KG = 1.90m

B. Find the New Draft Inc. in Draft = Vol. of Lost Buoyancy Area of Intact Waterplane = 10m x 6m x 2m 40m x 6m – 10m x 6m = 120 / 240 – 60 = 120 / 180 Inc. in Draft = 0.67m Old Draft = 2.00m ( + ) New Draft = 2.67m

C. Find the New GM KB = ½ x New Draft = ½ x 2.67m KB = 1.335m BM = I or LB³ V 12 x V = 30m x 6³m 12 x 40m x 6m x 2m = 6480 /5760 BM = 1.125m New BM = 1.125m New KB = 1.335m ( + ) New KM = 2.460m New KG = 1.900m ( - ) New GM = 0.56m

5. A box-ship 80m x 10m x 6m is floating upright in salt water on an even keel at 4 meters draft. She has an amidships compartment 15m long which is filled with timber (SF=1.5 cu.m. of ton). One ton of solid timber would occupy 1.25 cu.m. of space. What would be the increase in draft if this compartment is now bilged. Ans. 0.129 meter

Permeability = Broken Stowage Stowage Factor = 0.25 / 1.50 x 100 5. Solution: Stowage Factor = 1.50 m³ Space Occupied = 1.25 m³ ( - ) Broken Stowage = 0.25 m³ Permeability = Broken Stowage Stowage Factor = 0.25 / 1.50 x 100 Permeability = 16.67% or 0.1667 x 100

Inc. in Draft = Vol. of Lost Buoyancy Area of Intact WP = 0.1667 x 15 x 10 x 4 80 x 10 – 0.1667 x 15 x 10 100.02 / 774.95 Increase in Draft = 0.129 m

6. A box-ship 75m long x 10m wide x 6m deep is floating in salt water on an even keel at a draft of 4.5 meters. Find the new mean draft if a forward compartment 5 meters long is bilged. Ans. 4.821 m

6. Solution: W = X x B x d x R. Density W = Trimming Moment X = Length of the Bilged Compartment W = 5m x 10m x 4.5m x 1.025 ton/m³ W = 230.625 tons TPC = WPA / 97.56 TPC = 70 x 10 97.56 TPC = 7.175

6. Solution: Inc. in Draft = W / TPC = 230.625 tons / 7.175 Inc. in Draft = 0.3214 m or 32.14 cm Old Draft = 4.5 m ( + ) New Mean Draft= 4.821 m

7. A ship displaces 7500 cu. m. of water density 1,000 kgs per cu. m 7. A ship displaces 7500 cu.m. of water density 1,000 kgs per cu.m. Find the displacement in tons when the ship is floating at the same draft in water density 1,015 kgs per cu.m. Ans. 7,612.5 tons

7. Solution: New ∆ New Density Old ∆ Old Density New ∆ 1,015 kgs./m³ 7,500t 1,000 kgs./m³ New ∆ = 7,500t x 1,015 kgs./m³ 1,000 kgs/m³ New ∆ = 7,612.5 tons

8. A vessel of 10,000 tonnes displacement has a KG of 8 8. A vessel of 10,000 tonnes displacement has a KG of 8.00 meters and KM of 9.78. Find GM if weight of 20 tonnes is removed 10 meters vertically above the base line. Ans. 1.784 m

9. A vessel of 10,000 tonnes displacement has a KG of 8 9. A vessel of 10,000 tonnes displacement has a KG of 8.29 meters and KM of 9.78. Find GM if weight of 20 tons is added 10 meters vertically above the base line. Ans. 1.487 m

10. A vessel of 8,000 tonnes displacement has a KG of 7 10. A vessel of 8,000 tonnes displacement has a KG of 7.00 meters and KM of 9.78. Find GM if weight of 20 tons is added 10 meters vertically above the base line. Ans. 2.773 m

11. A vessel of 8,000 tons displacement has a KG of 7 11. A vessel of 8,000 tons displacement has a KG of 7.00 meters and KM of 9.78. Find GM if weight of 20 tons is removed 10 meters vertically above the base line. Ans. 2.805 m

12. A ship is floating in salt water on an even keel at 6 meters draft 12. A ship is floating in salt water on an even keel at 6 meters draft. TPC is 20 tonnes. A rectangular-shaped compartment amidships is 20 meters long, 10 meters wide, and 4 meters deep. The compartment contains cargo with permeability 25 percent. Find the new draft if this compartment is bilged. Ans. 6.1025 meters

13. A box-ship of 75m long x 10m wide x 6m deep is floating in salt water on an even keel at a draft of 4.5 meters. Find the new drafts (final drafts) if a forward compartment 5 meters long is now bilged. Ans. A= 3.788 meters; F=6.002 meters

14. A box-ship of 75m long x 10m wide x 6m deep is floating in salt water on an even keel at a draft of 4.5 meters. Find the Change of Trim if a forward compartment 5 meters long is now bilged. Ans. 221.4 cm by the head

15. A box-ship floats upright on an even keel in a fresh water and the center of buoyancy is 0.50 meter above the keel. Find the KB when she floating in salt water. Ans. 0.488 m