STABILITY PROBLEM 3

1. STABILITY PROBLEM 3

2. 1. A ship displaces 8,800 tons at a summer draught of 7.28m. Its Fresh Water Allowance is 315 mms. Find its change of draught when proceeding to a port of UNIT DENSITY. Answer = 31.5 cms

3. 1. Solution: • A.I. = FWA x ( Change in Density ) • 25 • = 315 mms. x ( 1,025 – 1,000 ) • 25 • = 315 mms. x 25 • 25 • A. I. = 315 mms. or 31.5 cms.

4. 2. A box has a dimension of 2ft x 2.5ft x 3ft and weighs 95 lbs. What is its stowage factor? • Answer = 353.7

5. 2. Solution: • Volume = 2 ft. x 2.5 ft. x 3 ft. • Volume = 15 ft³ • Stowage Factor = Volume • Weight / 2240 lbs. • = 15 ft³ • 95 lbs/ 2240 lbs • = 15 ft³ / 0.0424 lbs • Stowage Factor = 353.7

6. 3. You are going to load bales of abaca with SF 65 and lead with SF 18. The Remaining space is 257,000 cu ft. and the total weight to be loaded is 5,400 tons. How much of each cargo should be loaded to make the vessel FULL AND DOWN? • Answer = 2,000 t lead / 3,400 t abaca

7. 3. Solution: • WLF = Weight of the cargo having the large SF • WLF = Volume – ( Wt. of Cargo x Small SF ) • Difference in SF • = 257,000 ft³ - ( 5,400t x 18 ) • 65 – 18 • = 257,000 – 97,200 • 47 • = 159,800 / 47 • WLF = 3,400 t ( Abaca ) • Wt. of Lead = 5,400 t – 2,400 t = 2,000 t

8. 4. A vessel with a light displacement of 6,500t loaded cargoes of 1,300t, 1,400t and 1,200t with a KG of 6.9m, 6.4m and 8.4m respectively. After loading, the new KG was then found to be 7.07 m. What was the KG prior to loading? • Answer = 7.00 m

9. WTDISTMOMENT • LOADED = 1,300 x 6.9 = 8,970 • LOADED = 1,400 x 6.4 = 8,960 • LOADED = 1,200 x 8.4 = 10,080 • INIT.DISPL = (+) 6,500 • TOTAL DISPL= 10,400 • (-) 8,970 • (-) 8,960 • (-)10,080 • 45,518 x 7.07 =73,528 KG = MOMENT WEIGHT = 45,518 6,500 OLD KG = 7.00 m

10. 5. A bulk carrier has a displacement of 40,000 tons and the KG is 9 m .The KN at 10 degrees heel is 1.87 m. What is the righting arm at this heel? • Answer = 0.307 m

11. 5. Solution: • GZ = KN – ( KG x Sin θ ) • = 1.87 m – ( 9 m x Sin 10º ) • = 1.87 – 1.5628 • GZ = 0.307 m

12. 6. A bulk carrier has a displacement of 25,000 tons. KG is 8.6 m. At an angle of heel of 15 degrees, KN is 2.98. Solve for the righting moment. • Answer = 18,854 mtr-tons

13. 6. Solution: • GZ = KN – ( KG x Sin θ ) • 2.98 – ( 8.6 x Sin 15º ) • 2.98 – 2.2258 • GZ = 0.754 m • MSS = W x GZ • = 25,000 x 0.754 • MSS = 18,854 ton - meter

14. 7. While inspecting the log pond, you noticed that the average length of the logs to be loaded is 15 m and the average diameter is 70 cms. While floating in pond of density 1,012 kgs./m³, 90 percent of their volume is submerged. What is the average weight of each log? • Answer = 5.26 tons

15. 7.Solution; • Vol. of Cylinder = π x r² x h • = 3.1416 x 0.35²m x 15m • = 5.77m³ • x 0.9 (90%) • Volume = 5.2m³ • Weight = Volume x Density • = 5.2m³ x 1.012ton/m³ • Weight = 5.26 tons

16. 8. A box–shaped vessel with 12m beam is floating upright at a draught of 6.7m even keel. Find the draft if the vessel is now listed 18 degrees. • Answer = 8.226 mtrs

17. 8. Solution: New Draft = ½ x B x Sin List + ODr. x Cos List = ½ x 12m x Sin 18º + 6.7m x Cos 18º = 1.854 + 6.372 New Draft = 8.226 m

18. 9. On completion of loading, a vessel of 8,800 tons displacement and a GM of 0.45 m was listing 3 degrees to stbd. How many tons of ballast should be pumped into the port wing tank, 6 m from the centerline to make the vessel upright? • Answer = 34.6 tons

19. 9. Solution: • Tan List = Wt x Dist ∆ x GM • Weight = Tan List x ∆ x GM Distance • Weight = Tan 3º x 8,800 tons x 0.45m 6m • Weight = 34.6 tons

20. 10. The speed made good was 16 knots and the engine speed 15.4 knots. What is the percentage of slip? • Answer = - 3.9%

21. X 100 • 10. Solution: • % Slip = Eng. Spd – Obs. Spd • Eng. Spd • = 15.4 – 16.0 • 15.4 • - 0.6 • 15.4 • % Slip = - 3.9 % X 100 X 100

22. 11. A cylindrical drum, 1.4m high has a diameter of 80 cms. It contains oil of density 0.78 ton/m³ Find the weight of oil if filled to capacity. • Answer = 549 kgs

23. 11. Solution: • Volume = π x r² x l • = 3.1416 x 0.40² x 1.4m • Volume = 0.70 m³ • Weight = Volume x Density • = 0.70 m³ x 0.78 ton/m³ • Weight = 0.5489 ton x 1,000 kgs. • Weight = 549 kgs.

24. 12. If the circumference of a mooring rope is 250 mms, what is its diameter? • Answer = 80 mms

25. 12. Solution: • Circumference = 3.1416 x Diameter • Diameter = C / 3.1416 • = 250 mms / 3.1416 • Diameter = 79.58 mms

26. 13. A log is floating in a pond of relative density 1.008. Its length is 12 m and its diameter is 50 cms. If 95% of the log is submerged, what is its weight? • Answer = 2.26 tons

27. 13. Solution: • Vol. of Cylinder = π x r² x h • = 3.1416 x 0.25²m x 12m • Volume = 2.36m³ • x 0.95 (95%) • Volume = 2.24m³ • Weight = Volume x Density • = 2.24m³ x 1.008m³/ton • Weight = 2.26 tons

28. 14. On arrival at the discharging port, the displacement was 7,800 t. After Discharging 3,200 t of cargo with an average KG of 5.8 m the new KG was found to be 6.14 m. What was the vessel’s KG prior to discharge? • Answer = 6.00 m

29. WTDISTMOMENT • 7800 – 3,200 = 4600 • DISCH = 3,200 x 5.8 = 18,560 • FINAL DISPL= 4,600 x 6.14 = 28,244 (+) • INITIAL DISPL = 7,800 46,804 KG = MOMENT WEIGHT = 46,804 7,800 OLD KG = 6.00 m

30. 15. A box-shaped vessel with 14 m beam is floating upright at a draught of 5.4 m even keel. Find the draught if the vessel is now listed 15º. • Answer = 7.028 m

31. 15. Solution: New Draft = ½ x B x Sin List + ODr. x Cos List = ½ x 14m x Sin 15º + 5.4m x Cos 15º = 1.8133 + 5.22 New Draft = 7.028 m