WATERTIGHT INTEGRITY. Ship is divided into watertight compartments by means of transverse and longitudinal bulkheads bulkheads.

Transcription

1 Damage Stability

2 WATERTIGHT INTEGRITY Ship is divided into watertight compartments by means of transverse and longitudinal bulkheads bulkheads. When a watertight compartment (or a group of compartments) is damaged depending on the damage length according to the rules, the ship is expected to sustain its watertight integrity.

3 WATERTIGHT INTEGRITY

4 WATERTIGHTNESS? Watertightmeans;resistance of the structure against static water pressure without any leakage. It prevents water or any other liquid to flow other compartments (or in the opposite direction). The main deck and main bulkheads of a ship must be watertight. Watertight bulkheads must extend to the main deck. Any opening in the watertight bulkheads must be equipped with watertight equipments (doors, windows, valves etc.).

5 WATERTIGHT DOORS

6 WATERTIGHT DOORS

7 WATERTIGHT DOORS

8 Damage If the shell of a ship is damaged, leakage will take place between the sea and the damaged spaces until stable equilibrium is established or until the ship capsizes or sinks. It is impractical to design a ship to withstand any possible damage due to collision, grounding or military action

9 Damage

10 Watertight Bulkheads Watertight bulkheads : the hull is subdivided into watertight compartments by means of watertight bulkheads and decks.

11 Bulkhead Decks Bulkhead decks : the deck up to which these bulkheads extends.

12 Margin Line Margin line : is a line defining the highest permessible location on the side of the vessel of any damaged waterplane in the final condition of sinkage,trim and heel. After flooding of a prescribed number of compartments the ship shall not submergebeyondalinesituatedatleast76mm.belowthedeckatside.

13 Margin Line

14 Margin Line

15 Permeability, μ Permability is the ratio of amount of water that can a enter or a compartment or tank to the total volume of compartment or tank. PERMEABILI TY= AVAILABLE VOLUME TOTAL VOLUME

16 Surface Permeability, μ a Surface permability is the percentage of a waterplane that can be occupied by water.

17 Typical Permeability Watertight compartment (warship) 97% Watertight compartment (merchant) 95% Accommodation spaces 95% Machinery spaces 85% Dry cargo spaces 70% Bunkers, stores, cargo holds 60%

18 Example 1 A cargo hold with dimensions of l = 15.24m., w = 9.14 m., h=6.1 m. is completely flooded. Determine the volume of the space and the corrected flooded volume: Volume = m. x 9.14 m. x 6.1 m. = m 3 Flooded volume = Volume x permeability Flooded volume = m 3 x 60% Flooded volume = m 3

19 Floodable Length Floodable length: at a given point of the ship length is the maximum length with the center at that point that can be flooded without submerging the ship beyond margin line. Yaralı Bölme Boyu: Gemi boyunca herhangi bir noktada, bu nokta merkez olmak üzere gemiyi sınır hattına teğet hale getirecek en büyük bölme boyuna o noktadaki yaralı bölme boyu denir.

20 Floodable Length Curve Floodable Length Curve: It is obtained by the connection of floodable lengths at different points along the ship length. Yaralı Bölme Boyu Eğrisi: Gemi boyunca değişik noktalarda hesaplanan yaralı bölme boyu değerlerinin birleştirilmesi ile elde edilir.

21 Factor of Subdivision Factor of Subdivision: The factor of subdivision is a factor prescribed by the applicable regulations and by international convention that depends on ship length and criterion of service.

22 Permissible Length Permissible Length is obtained by multiplying the floodable length at that point by the factor of subdivision.

23 Determination of the location of the volume centre of damaged compartment Δ: Initial displacement Δ 1 : Displacement after damaged B: Initial centre of buoyancy G: Initial centre of gravity x 0 : Initial centre of longitudinal buoyancy, LCB v w : The volume of water entering the vessel x w : Distance of volume centre of damaged compartment from maestory

24 Determination of the location of the volume centre of damaged compartment

25 Determination of the volume of water entering the vessel Initial displacement and longitudinal centre of buoyancy (LCB) can be read from hydrostatcic curves. However, displacement and LCB are unknown after damaged. Therefore, if trim, forward and aft draught are known, trimmed waterline can be drawn over Bonjean area curves.

26 Determination of damaged length and location of mid of compartment v w : The volume of water entering the vessel x w : Distance of volume centre of damaged compartment from maestory µ: Permability of the compartment v 0 : Volume of the compartment x m : Distance of the volume centre of the compartment to the mid of damaged length x l : Distance of the mid of damaged length of compartment to maestory x c : Distance of the computed volume centre of compartment to maestory (x c =x l -x m )

27 Determination of damaged length and location of mid of compartment Damaged length and location of compartment can be determined by iterative procedure.

28 Determination of damaged length and location of mid of compartment The volume of water entering the vessel (v w ) Distance of volume centre of damaged compartment from amidships Θ (x w ) : Volume of the compartment (v 0 =v w /μ) First trial 1) Assume x m =0 (x l =x w ) 2) Damaged Length, l 1 = (Volume of the compartment, v 0 ) / (mean sec. Area, A mean ) where A mean is determined from trimmed sectional area curve

29 Determination of damaged length and location of mid of compartment 3) In order to determine v m and x m s = l 1 / 4 v m = (s /3) 3 x m = s 4 / 3 x c = x l x m v m -v 0 Tolarence for volume x c -x w Tolarence for location of volume centre l=l 1, x l =x w

30 Determination of damaged length and location of mid of compartment Second Trial: Damaged length is changed proportionally by the volume. l 2 = l 1 (v 0 /v m ) v m : Computed volume at first trial If the value of x m compartment slides to the bow. computed at first trial is negative, then the mid of the x l = x w +x m where x m is the distance of the volume centre of compartment to amidship computed at first trial

31 Determination of damaged length and location of mid of compartment In order to determine v m and x m s = l 2 / 4 v m = (s /3) 3 x m = s 4 / 3 x c = x l x m v m -v 0 Tolarence for volume x c -x w Tolarence for location of volume centre If tolarances for both volume and location are not satisfied, third, fourth,..trials are made similar to second trial. l=l 2

32 Determination of damaged length and location of mid of compartment If either tolarence for volume or tolarance for location are satisfied at a trial, satisfied one is fixed, trials are continued to make for the other one.

33 Example 2 A ship is 116 meter long and displaces 7842 ton in salt water (ρ 1 =1.025 t/m 3 ) when her mean draught is 6.4m. Longitudinal Centre of Buoyancy (LCB) is m aft from amidships. After one of her stern compartment is damaged, she floats trimmed to the aft. Her sectional areas after damaged are given in the following table. a) Determine the volume of water entering the vessel (v w ) and the location of volume centre of damaged compartment (x w ) b) Assume permability of compartment, µ is 1. Determine damaged length and Distance of the mid of damaged length of compartment to amidship

34 Example / /

35 Sec. Sec. No: Area SM M. Arm [1] [2] [1 2] [3] [1 2 3] 0 13,85 0,5 6, ,625 0,5 37,3 2 74,6-4,5-335,7 1 70,15 1,5 105, , , , , , , , , , , , , ,2 7 94, , ,2 8 60, , ,4 9 24,3 1,5 36, ,8 9,5 10, ,7 4,5 93, , = 2506,5 4 = -1196,88 Example 2-SOLUTION s=l s /10= 11,6m New displacemetvolume = (s/3)* 3 = 9691,8m 3 Initial displacement volume= 7651m 3 v w = 2040,8m 3 LCB 1 =s 3 / 4 = -5,5391m Both initial LCB and LCB 1 after damaged are in the same side Therefore, x w = -24,000 m.

36 Example 2-SOLUTION Predicted x w is between section 2 and section 3. Sec. No. Sec. Area Dist. Amiship 2 101,35-34, ,8-23,2 Sec. Area at -24m from amidship= 119,5m 2 l 1 =v 0 /A mean =2040.8/119.5= m. s=l 1 /4=4.27 m.

37 Example 2-SOLUTION y i Loc. A mean SM M. Arm [1] [2] [1] [2] [3] [1] [2] [3] = = v m = (s/3) 3 = m 3 x m =s 4 / 3 = m v 0 -v m = =42.2 m3 %Error= /2040.8=%2.07

38 Methods of Calculation There two available method calculating the final stage of the damaged vessel : Lost Buoyancy Method Added Weight Method

39 Comparision of Lost Buoyancy and Added Weight Methods

40 Added Weight Method Consider a vessel that has been damaged such that a portion of the bottom is now open to the sea... The vessel s draft will increase because an amount of the buoyancy was lost...

41 Added Weight Method In this method, flood water is treated as a fixed weight (for example, the weight of water of volume ABEF). This added weight is balanced by the additional buoyant force of the layer WL 1 WL. The added weight can be determined by either the volume ABEF or waterline WL 1. Therefore, WL 1 waterline is not known can be just guessed. For the predicted waterline, added weight and additional buoyant force caculation are done iteratively until both of them are equal. This method is known as Added weight method.

42 Determination of Added Weight v c : Volume of damaged compartment (DCEF) a c : The surface area (AB) of damaged compartment on the waterline. µ: The volume permability of damaged compartment µ a : The area permability of the surface of damaged compartment : The initial displacement 1 : The displacement at final stage v w : The volume of water entering the vessel. (ABEH) w: The weight of water entering the vessel. p: The value of parallel sinkage

43 If we assume that the paralel sinkage cause a minor change of water plane area of a ship (A wp A wp1 ), we can determine the paralel sinkage by equating the volume of the layer WL 1 WL to the volume of ABEF. If the damaged compartment is bounded on top.

44 Vertical centre of gravity at final stage kg: vertical centre of water entering the ship Vertical centre of buoyancy at final stage Transverse Metacentre Radius at final stage

45 Longitudinal Metacentre Radius at final stage Free Surface Effect Damaged compartment can be considered as a half filled tank. Free Contact Effect Damaged compartment can be considered as a half filled tank.

46 Transverse Metacentre Height The angle of heel

47

48 Longitudinal Metacentre Height

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50 Example : A barge L = 20 m. B = 5 m. T = 1.5 m. KG = 1.5 m.

51 Intact Condition The displacement volume: I = L B TI = 20x5x1.5 = 150m The mass displacement: I = ρ I = 1.025x150 = tons The moment of inertia of the waterplane area about CL: 3 3 B L 5 x20 I I = = = m

52 Metacentric radius: Intact Condition BM I = = = m I 150 I Or, it could be calculated directly as: B L /12 B 5 BM = = = = 1. m I LBTI 12TI 12x Height of the center of buoyancy: T KB = I I 75m 2 = 0.

53 Intact Condition The metacentric height: GMI = KBI + BM I KGI = = m For small angles of heel, the righting moment: M RI = I GM I sin φ = x0.693x sin φ = sin φt. m.

54 LOST BUOYANCY METHOD According to the Lost Buoyancy Method, flooded compartment (DCEF) is assumed tobeapartofseaanymore(itdoesnotbelongtheship). Asthefloodedwaterdoesnotbelongtotheship,itcausesnofreesurfaceeffects. Neither before damaged nor after damaged ABCD region does not yield buoyant force. Because after damaged it does not belong the ship. Paralel Sinkage: Volume (DCEF)=Volume (W 1 ADW+BL 1 LC)

55 LOST BUOYANCY METHOD Paralel Sinkage: Volume (DCEF)=Volume (W 1 ADW+BL 1 LC) Damaged compartment is bounded on top.

56 LOST BUOYANCY METHOD It is assumed that displacement and location of vertical centre of gravity does not change after damaged.

57 LOST BUOYANCY METHOD

58 LOST BUOYANCY METHOD The heel angle of damaged ship is determined by where d is the distance between the volume centre of damaged compartment and the new longitudinal centre of floatation. w c =v c μ ρ

59 LOST BUOYANCY METHOD

60 LOST BUOYANCY METHOD Longitudinal Metacentre Height of a Damaged Ship

61 LOST BUOYANCY METHOD

62 Forward and aft draughts after damaged Table 1: Final Draughts for a starboard side damaged ship Port Aft Port Fore Starboard Aft Starboard Forward Initial Draught T T T T Paralel Sinkage +p +p +p +P Heel -m i -m i +m s +m s Trim -t k +t b -t k +t b Final Draught T+p-m i -t k T+p-m i +t b T+p+m s -t k T+p+m s +t b

63 Example A cargo hold with dimensions of l = 15.24m., w = 9.14 m., h=6.1 m. is completely flooded. Determine the volume of the space and the corrected flooded volume: Volume = m. x 9.14 m. x 6.1 m. = m 3 Flooded volume = Volume x permeability Flooded volume = m 3 x 60% Flooded volume = m 3 Once the volume is corrected, the determination of the new G and changes and list and trim are computed as previously discussed

64 Lost Buoyancy Method

65 Lost Buoyancy Method After loosing the central compartment, the waterplane area is: A L = ( L l) B= (20 4) x5= 80m To compensate for the loss of buoyancy of the central compartment, the draft increases to : 2 I 150 TL = = = A 80 L m

66 Lost Buoyancy Method The height of the center of buoyancy increases to : TL KBL = = = m 2 2 The moment of inertia of the waterplane area: 3 3 B ( L l) 5 (20 4) I L = = = The metacentric 12 radius: m 4 BM IL L = = = I m

67 Lost Buoyancy Method The metacentric height: GML = KBL + BM L KGI = = m Righting moment for small heel angles M RL = I GM L sin φ = x0.549sin φ = sin φt. m