STABILITY PROBLEMS NO. 1

1. You are loading a cargo of canned goods with a stowage factor of 65 1. You are loading a cargo of canned goods with a stowage factor of 65. If you allow 15% for broken stowage, how many tons can be loaded in a space of 55,000 cubic feet? Ans. 719 Tons

1. Solution; Weight = Volume x Allowance Stowage Factor 65 Weight = 719.23 tons

2. What is the stowage factor of cargo of canned goods packed in cartons weighing 340 pounds and 36 cubic feet each carton? Ans. 237 ft³/ton

2. Solution: SF = Volume Weight / 2240 lbs. SF = 36 ft³ 340 tons / 2240 lbs SF = 237 ft³/ton

3. If the stowage factor of cotton is 56, allowing a stowage loss of 15%, it will stow at a factor of ___. Ans. 65.9

3. Solution: S. Factor = Stowage Factor of Cotton Allowance S. Factor = 56 0.85 S. Factor = 65.9

4. A ship of 11,000 tons displacement has a moment of statical stability of 500 tons meters when heeled 5°. Find the initial metacentric height? Ans. 0.522 m

4. Solution; MSS = ∆ x GM x Sin θ ∆ x GM x Sin θ = MSS GM = MSS ∆ x Sin θ = 500 tons meter 11,000 tons x Sin 5º GM = 0.52 m

5. A vessel displacing 18,000 tons has a KG of 50 feet 5. A vessel displacing 18,000 tons has a KG of 50 feet. A crane is used to lift cargo weighing 20 tons from a supply vessel. When lifting, the head of the crane boom is 150 feet above the keel. What is the change in KG? Ans. 0.11 foot

5. Solution: GG’ = W x D ∆ GG’ = 20 t x ( 150 ft – 50 ft ) 18,000 t GG’ = 20 t x 100 ft. GG’ = 0.1111 foot

6. A weight of 500 tons is loaded into a ship so that its centre of gravity is 10 meters from that of the ship. Find the shift of G if the ship's original displacement was 3,000 tons. Ans. 1.428 m

6. Solution: GG’ = W x D ∆ GG’ = 500 t x 10 m 3,500 tons GG’ = 1.428 m

7. What will be the reduction of GM on a 5,000 tons displacement ship with 6.3 KG in sea water after half of the sea water ballast is discharged from the double bottom ballast tank 22 m x 10 m x 2 m with 1.3 m KG? Ans. 0.236 m.

7. Solution: Wt. of Ballast = L x B x D x R. Den. = 22 x 10 x 1 x 1.025 Wt. of Ballast = 225.5 tons Old ∆ = 5,000 t - 225.5 t New ∆ = 4774.5 tons Dist = KG of Vsl.– KG of Tank = 6.3 m – 1.3 m Dist = 5.0 m GG’ = W x D ∆ = 225.5 t x 5 m 4774.5 t GG’ = 0.236 m

8. The original displacement of a ship was 4,285 tons and her KG was 6 8. The original displacement of a ship was 4,285 tons and her KG was 6.00 meters. Find her new KG after she has loaded the following weights; 800 tons at 3.6 meters above the keel, 440 tons at 7.0 meters above the keel. 110 tons at 5.8 meters above the keel, 630 tons at 3.0 meters above the keel. Ans. 5.46 m

8. Solution: Weight KG Moments 4,285 6.0 m 25,710 800 3.6 m 2,880 440 7.0 m 3,080 110 5.8 m 638 630 3.0 m 1,890 6,265 34,198 New KG = Total Moments / Total Weights = 34,198 / 6265 New KG = 5.46 m

9. A vessel displacing 18,000 short tons and has a VCG of 70 ft 9. A vessel displacing 18,000 short tons and has a VCG of 70 ft. You dump 100 short tons of water, which had a VCG of 88 ft., what is the new KG? Ans. 69.9 feet

9. Solution: Weight VCG Moments 18,000 70 ft 1,260,000 ( - ) 100 88 ft 8,800 ( - ) 17,900 1,251,200 New KG = Total Moments Total Weights New KG = 1,251,000 / 17,900 New KG = 69.9 ft

10. A rectangular shaped vessel 200 m in length and 32 m in breadth, floats at an even keel draft of 9.0 m. The KG is 10.0 m, KB is 4.737, BM is 8.989 m. What is her GM? Ans. 3.726 m

10. Solution: KB = 4.737 m BM = 8.989 m ( + ) KM = 13.726 m KG = 10.00 m ( - ) GM = 3.726 m

11. At a given draft, a ship of 120 meters length and 15 meters beam has a coefficient of fineness of the waterplane of 0.770. What is her TPC at this draft? Ans. 14.20 tons

11. Solution: TPC = L x B x Cw x Rel. Density 100 TPC = 120 m x 15 m x 0.770 x 1.025 TPC = 14.20 tons

12. What is the TPC of a box-shaped vessel 120 m x 15 m x 10 m floating at even keel draft of 4.5 m? Ans. 18.45 tons

12. Solution: TPC = 1.025 x Area of Waterplane 100 TPC = 1.025 x 120 m x 15 m TPC = 1845 TPC = 18.45 tons

13. A rectangular shaped vessel 100 m in length, 20 m breadth is floating in salt water at an even keel draft of 6.0 m. A forward hold 10.0 m long was bilged. What is the TPC? Ans. 18.45 tons

13. Solution: TPC = 1.025 x Area 100 TPC = 1.025 x ( 100m – 10m ) x 20 m TPC = 1.025 x 90m x 20m TPC = 18.45 tons

14. A drum which is 1. 5 m long and 60 cm 14. A drum which is 1.5 m long and 60 cm. in diameter has mass of 20 kgs when empty. Find what will be her draft in water of density 1,024 kg/m³ If it contains 200 liters of paraffin of relative density 0.6. Ans. 0.484 m

14. Solution: Den. of Paraffin = RD of Paraffin x FW Density = 0.6 x 1,000 kgs./m³ Den. Of Paraffin = 600 kgs./m³ Mass of the Paraffin = Volume x Density = 0.2 m³ x 600 kgs./m³ Mass of the Paraffin = 120 kgs. Mass of the Drum = 20 kgs. ( + ) Total Mass = 140 kgs.

14. Solution: Vol. of water displaced = Mass / Density =140 kgs. / 1,024 kgs./m³ Vol. of water displaced = 0.137 m³ Vol. of water displaced (V) = π²d d = V / π² d = 0.137 m³ 3.1416 x .09 Draft = 0.484 m

15. A box-shaped barge 16m x 6m x 5 m is floating alongside a ship in fresh water at a mean draft of 3.5 m. The barge is to be lifted out of the water and loaded on to the ship with a heavy lift derrick. Find the load in tons borne by the purchase when the draft of the barge is reduced to 2 m? Ans. 144 tons

15. Solution: Mass of the Barge = L x W x Dr x Rel. Den. = 16 x 6 x 3.5 x 1.0 Mass of the Barge = 336 tons Mass of water = 16 x 6 x 2 x 1.0 Displaced at 2m = 192 tons Mass of the Barge = 336 tons Mass of Barge at 2 m = 192 tons ( - ) Load Borne by the = 144 tons Purchase