Baixe Course Of Ships Stability and Trim e outras Notas de estudo em PDF para Engenharia Naval, somente na Docsity! MARITIME UNIVERSITY CONSTANTA FACULTY OF NAVIGATION COURSE OF SHIP’S STABILITY AND TRIM (For the use of students of Constanta Maritime University ) CONTENTS Section 1 Basic principles……………………………………………………………….. Section 2 Form Coefficients…………………………………………………………….. Section 3 Tonnes per Centimetre Imersion (TPC)……………………………………. Section 4 Load Lines……………………………………………………………………. Section 5 Centre of Gravity (G) and Centre of Buoyancy (B)……………………….. Section 6 Introduction to Transverse Statical Stability………………………………. Section 7 Conditions of Stability……………………………………………………….. Section 8 Initial Transverse Metacentre………………………………………………. Section 9 Free Surface Effect…………………………………………………………... Section 10 Curves of Statical Stability (GZ Curves)…………………………………. Section 11 List…………………………………………………………………………… Section 12 Introduction to Trim……………………………………………………….. Section 13 Suspended Weights…………………………………………………………. Section 14 Assessing Compliance of Ship’s Loaded Condition with IMO Criteria… Section 15 Curves of Statical Stability For Varying Conditions……………………... Section 16 Wall-Sided Formula………………………………………………………… Section 17 Factors Affecting the Shape of the Curve of Statical Stability…………... Section 18 The International Grain Code (IMO)……………………………………... Section 19 Inclining Experiment……………………………………………………….. Section 20 Trim Using Hydrostatic Data……………………………………………… Section 21 Wind Heeling and Ice Accretion…………………………………………… Section 22 Practical Ship Loading Problems………………………………………….. Bibliography …………………………………………………………………………….. Model of Stability Information and Tank Sounding Book………………………….. 1 7 12 20 29 36 45 49 56 70 75 87 106 115 137 145 154 168 198 206 235 245 263 265 1.1.2 Relative Density (RD) Quite often the Relative Density (RD) ot a substance is quoted instead of Density. This is simply a ratio of the density of the substance in question to that of Fresh Water. The density of fresh water is 1.000 tm'. In the previous example the density of the oil was 0.84 Ym?. The relative density of the oil was 0.84; in other words, the density of the oil is 0.84 times that of fresh water! 1.1.3 Density of water in which a ship typically floats A ship is presumed to always float in water that lies in the following density range: Fresh water (FW): 1.000tm? | (AD1.000) to Salt water (SW): 1.025tm” (RD 1.025) Water that lies between these two extremes is termed Dock Water (DW). lf a question states that a ship is floating in salt water (SW) then it can be always assumed that the water density is 1.025 tim”. Similarly, if in fresh water (FW) then a density of 1.000 tm? can be assumed. 1.2 THE LAWS OF FLOTATION 1.21 Archimedes' principle This states that when a body is wholly or partially immersed in a liquid, it experiences an upthrust (apparent loss of mass - termed Buoyancy force (Bf)), equal to the mass of liquid displaced. Consider a block of steel measuring 2m x2m x2m thathas a density of 7.84 tm”, Example 5 (a) if the block were to be suspended by a ship's crane that has a very accurate load gauge, what mass would register on the ga block were suspended over the ship's side in air? Solution (a) The block is suspended in air! Since: Mass = Volume x Density; uge if the Mass of the block = (em x 2m x 2m) x 7.84 tm" =62.72t The crane driver now lowers the block so that it becomes half submerged'in the dock water that has a density of 1.020 tim”. Example 5 (b) What mass will the gauge now indicate? Solution (b) The block is now displacing a volume of water where: Volume of water displaced = (2m x 2m x tm) = am? DOCK WATER DENSITY | 10H um 2m Fig 1.2 Mass of water displaced = Volume x Density of the dock water; =4m x 1.020tm? = 4.08 t which represents the upihrust due to buoyancy force (Bf) created by the displaced water. Mass of block 62.721 UpilrustduetoBf 4.08! Gauge reading 58.64 t Example 5 fc) What mass will the gauge indicate if the crane driver now lowers the block so that it is completely submerged in the dock water? Solution (c) The biock is now dispiacing a volume af water where: Volume of water displaced = (2m x 2m x 2m)= gm? EXIGYANCY FORCE AGTING AT CÊ DE UNDUFIWATER VOLUME 14 dm TOCA MATER DENSITY 1020 tm WEIGHT LORGE ACTIN GRAVITY OF THE 2: 0CK Fig. 1.3 DOCK WATER DENSITY. 4.020 bm? Fig. 1.4 Mass of water displaced = Volume x Density of the dock water; =8mº x 1.020 tm? =8.16t which represents the upthrust of the buoyancy force (Bt) created by the dispiaced water. UutrrAnis FORCE ATINO AT CENTROS DF UNDERWATER VOLUME [0,184 Mass of block 62.72t UpthrustduetoBt 8.161 Gauge reading 54.56 t E DENSITY 020 tie? À tus eoace acuno GRAVITV CE cup E ca Fig. 1.5 122 Law of flotation This states that every floating body displaces it's own mass of the liquid in which it floats. The displacemant ot a ship (or any floating object) is defined as the number of tonnes of water it displaces. It is usual to consider a ship displacing sait water of density 1.025 tmê, however, fresh water values of displacement (1.000 tim?) are often quoted in ship's hydrastatic data. The volume of displacement is the undenwater volume of a ship ailoat i.e. the volume belowthe waterline. To calculate the displacement (W) of a ship the following needs to be known: The volume of displacement (V) The density of the water in which it floats (p) Since: MASS = VOLUME x DENSITY the mass, or displacement, ot a ship is calculated by: DISPLACEMENT = VOLUME OF DISPLACEMENTx WATER DENSITY ie. W=Vxp 1.23 Draught and Freeboard Consider the ship shown. ap a PP as A 1 Fig. 1.6 Draught is the distance from the keel to the waterline (WL), as measured at the forward and aft ends of the ship. (More precisely the draught readings are taken as those read at the Fonward and Aft Perpendiculars - these terms are defined in Section 12). his expressed in metres. If the draughis forward and aft are the same then the ship is said to be on an even keel (as shown). Freebaard is the distance between the waterline (WL) and the top of the uppermost continuous deck. It is usually expressed in millimetres and is measured amidships. Hull Depth = Draught + Freeboard 1.24 Reserve Buoyancy This is the volume of the enclosed spaces above the wateriine. Because reserve buoyancy is a very important factor in determining a ship's seaworthiness minimum freeboards are assigned to a ship to ensure that there is adequate reserve buoyancy at all times. 21 COEFFICIENT OF FINENESS OF THE WATERPLANE AREA (Cu) Is defined as the ratio of the ship's water-piane area (WPA) to the area of a rectangle having the same length (L) and breadth (B) of the ship at the waterline in question. inceecadereseeaderereee 5 [us Wator-Plane Area (WPA) (mi?) es L Fig. 2.1 —D— Cw=WPA LxB Since the ship's WPA is less in area than the rectangle formed around it, the value of Cy must always be less than 1.00. Example 1 A ship has a length and breadth at the waterline of 40.1 m and 8.6 m respectively. If the water- plane area is 280 rrÊ calculate the coefficient of fineness of the water-plane area (Cw). Solution Cw=WPA = 280 = 0.812 LxB 40.1x82 Note that the answer has no units; it is simply a comparison of one area to another! 22 BLOCK COEFFICIENT (Co) The block coefticient (Co) of a ship is the ratio of the underwater volume of a ship to the volume of the circumscribing block. Fig. 2.2 Cp = Volume of displacement LxBxd Therefore: Displacementsup = (Lx Bx dx Ca)xp Since the ship's volume of displacement is less than the volume of displacement of the surrounding block, the value of Ca must always be less than 1.00. Example 2 A ship floats at a draught of 3.20 m and has a waterline length and breadth of 46.3 m and 15.5 mM respectively. Calculate the block coefficient (Ca) it its volume of displacement is 1800 mê, Solution Ca= Volumeofdisplacement = 1800 LxBxd 46.3x 15.5 x 32 Co =0.784 Example 3 A ship has length 200 m and breadth 18 m at the waterine. if the ship floats at an even keel draught of 7.56 m in water AD 1.012 and the block coefficient is 0.824 calculate the dispiacement. Solution Displacement = Volume of displacement x Density Displacement = (Length x Breadth x draught x Ca) x Density Displacement = (200x 18x 7.56x 0.824)x 1.012 = 22695t 10 2.3 —MIDSHIPS AREA COEFFICIENT (Cu) The midships coefficient (Cu) Of a ship à any draught is the ratio of the underwater transverse area of the midships section to the product of the breadth and draught (the surrounding rectangle). .— Wiiships —— ss Breadth Fig. 2.3 Cu = Underwater transverse area of midships section (Am) Breadth x Draught Cu= Bxd Similarly, the value of Cy must always be less than 1.00. This coefficient may be used to determine the prismatic coefficient (Cp). Example 4 A ship floats at a draught of 4.40 m and has a wateriine breadth of 12.70 m. Calculate the underwater transverse area of the midships section if Cy is 0.922. Solution Cu= Am Bxd 0922 = Am 12.70 x 4.40 0.922 = Am 55.88 0.922 x 55.88 = Am = 51.521 mº 13 3.1 —TONNES PER CENTIMETRE IMMERSION (TPC) The TPC for any given draught is the weight that must be loaded or discharged to change the ship's mean draught by one centimetre. Consider the ship shown floating in salt water (RD 1.025) with a water-plane area (WPA) at the waterline as shown. gu Wator-plane Ares (WPA) (m?) Fig. 3.1 A weight of 30 tonnes is loaded on deck so that the mean draught increases by 1 em. Ema 30 t Em Ca O (o) Fig. 3.2 Since the ship's displacement is equal to the mass of water displaced (Law of Flotation) it follows that the mass of the additional 'stice' of displaced water is equal to the added weight of 30 tonnes. In this instance, 30 tonnes represents the value of the Tonnes per Centimetre Immersion (TPC) for the ship at the initial draught before the weight was loaded. 32 TPPC FORMULA Consider the previous situation. Since: Mass = Volume x Density then: Mass of additional slice of water = Volume of the additional slice of water x Density. lí the WPA is assumed to not significantly change between the two waterlines, then: Volume of the slice = WFPA (m?) x 1 em; Water-plane Area (m?) Fig. 3.3 we cannot multiply m? by cms, therefore: volume af slice = WPA (m?) x 1 (m); 100 Added displacement (t) = WPA (mx 1 (m)x density (um?); 100 Therefore, the formula for TPC is: TPC=WPAxp 3 100 Example 1 Calculate the TPC for a ship with a water-plane area of 1500 m” when it is floating in: (a) fresh water; (b) dock water of RD 1.005; (c) sait water. Solution TPC=WPAxp 100 (a) — TPC=1500x 1.000= 15.000 100 (b) TPC = 1500x 1.005 = 15.075 100 (co) TPC=1500x1.025=15.375 100 14 15 33 —FACTORS AFFECTING TPC Consideration ofthe TPC formula: TPC=WPAxp 100 shows that: + TPC increases with WPA and for a normal ship-shape the WPA will increase with draught. TPC increases with density. * Two values of TPC are often quoted in ship's hydrostatic data, TPCsw and TPCrw. However, hydrostatic data for M. V. Almar is given for saltwater only. Consider the hydrostatic particulars for M.V. Almar on page 12ot the stability data book. lf the ship were floating at a draught of 5.00 m in sait water (RD 1.025) the displacement of the ship would be 15120 tones. To sink the ship by exactly 1 cm, 31.96 tonnes would have to be loaded. Consider the situation if the ship were to float at the same draught of 5.00 m but in fresh water (AD 1.000). Would the displacement and TPC values be the same as they were in salt water? Consider the fallowing diagrams showing the ship floating at the same craught but in different water densities. 4 Salt Water (RD 1.025 Um: y 9 Fresh Water (RD 1.000 tm? yr Fig. 34 DISPLACEMENT = VOLUME OF DISPLACEMENT x WATER DENSITY For both situations the volume of displacement is the same! It follows that the displacement of the ship when at a draught of 5.00 m in saft water must be greater than the displacement of the ship when at the same draught in fresh water (since salt water is denser than fresh water!). Consider now the TPC value for both situations. ) Satt Water (RD 1.025 tim” , , Fresh Water (RD 1.000 Ym* / Fig. 3.5 18 The answers differ slightly for two reasons: 1. In using the mean value of TPC it is assumed that the TPC value changes linear!y between the range of draughts concemed. This is not so, as the underwater form of a ship does not (usually) change uniformly with draught. 2. Displacement values taken from the hydrostatic data in using method 7 will be rounded to the nearest whole tone. lfthe change in draught is only small it is usual to use the TPC value for the initial waterline instead of the mean TPC value as shown in the examples. Obviously the greater the amount of cargo loaded or discharged; the greater will be the error! Example 4 M.V. Aimar has an initial mean draught of 5.80 m in sait water and loads 11300 t of cargo. Using the hydrostatic particulars calculate the fina! displacement and mean draught. Solution Method 1 Initial draught 5.80 m DISPLsw 17690t Cargo loaded 11300t FINAL DISPL. 28990t Enter data with final displacement gives a final mean draught of 9.200 m. Method 2 Initial draught 5.80 m DISPLsw 17690t | TPCow32.26t Cargo loaded 11900t FINAL DISPL. 28990! TPCow 3443t Mean TPCow = 32.26 + 34.43 = 33.345 t 2 Sinkage (cms) = w TP Cow MEAN Sinkage (ems) = 11300 =3389cms =3.389m 33.345 Initial draught 5.800m Sinkage 3.389 m FINAL DRAUGHT 2.189 m It should be evident from Example 4 that direct use of the Dispiacement and TPC values given in the hydrostatic data results in a more accurate answer (Method 1). Using the tormula method leads to unnecessary working and results in a less accurate answer. Example 5 M.V. Almar arrives in port with a mean draught of 5.30 m in dock water AD 1.076. How much cargo may be loaded to ensure that the maximum draught on completion is 5.70 m in the dock water? 19 Solution Method 1 Initial draught 5.30 m DISPLsw 16080 t DISPLow = 16080x 1.016 = 15939t 1.025 Final draught 5.70 m DISPLow 17370t DISPLom = 17370x 1.016 = 17217t 1.025 Initial DISPLow 16080 t Final DISPLow 172171 Cargo to load 1278t Method 2 initial draught 5.30 m TPCsw 32.07 t TPCom =32.07x 1.016 = 31.791 1.025 Fina! draught 5.70 m TPCow 32.22 t TPCow = 32.22x 1.016 =31.9Mt 1.025 Mean TPCow = 31.79 +31.94 =31.665t 2 Sinkage (cms) = 5.70 m- 5.30m =0.40m=40 cms Sinkage (cms) = w Mean TPCow Cargo to load (w) = Sinkage x Mean TPCow =40 x 31.865 = 1274.6t It should be evident from the previous example that: The displacement for the correct density must be used in the calculation. + The TPC for the density of water in which the ship is loading in should be used in the calculation. lt is usual to calculate the amount to load on the basis of the required salt water draught since seasonal load lines assigned to the ship apply to the ship at sea in salt water. 20 SECTION 4 - LOAD LINES INTRODUCTION Most ships will be assigned a minimum freeboard and a corresponding set of laad lines. These will be permanently marked on each side of the ship (certain classes of ship are exempt from these requirements). Load lines assigned to a ship correspond to ocean areas or 'zones', Oceans around the world are divided into these zones in terms of both geographical location and time of year (season). By ensuring that the appropriate seasonal load line mark is not submerged at sea in salt water (RD 1.025) the ship will always have the necessary reserve buoyancy to ensure seaworthiness. To ensure that the appropriate load line is never submerged at sea, il is essential that the learmer has a thorough knowledge of the load line markings, their spacing and dimensions. The ability to perform calculations to determine the maximum amount to load is also important, especially to the ship owner, as the absolute maximum cargo in terms of weight should be carried whenever possible. It is also essential that the ship is never 'overloaded”, as contravention of the conditions of load line assignment will arise, resulting in the ship being unseaworthy with respect to legislative requirements. Learning Objectives On completion of this section the leamer will achieve the following: E Know the dimensions of a set of load lines as would be assigned to a ship. 2. Understand the term Fresh Water Allowance (FWA) and derive the formula for FWA. 3. Understand the term Dock Water Allowance (DWA). 4. Perform calculations relating to the loading of a ship to the appropriate load line mark. 23 Rearranging this gives: WPA = TPC sw x 100 (ii) 1.025 where the WPA is that for the Summer load draught waterline. Substituting equation (ii) into equation (i): 0.025DISPLew = 100 x TPCsw x FWA 1.025 FWA (m) = 0.025 x DISPLew x 1.025 100 x TPCsw To express FWA in mm then: FWA= 0.025 x DISPLew x 1.025 1000 100 x TPCsw Rearranging gives: FWA (mm) = 0.025 x DISPLsw x 1000 100 x TPCsw Therefore: FWA (mm) = 0.25 x DISPLow TPCsw Thus: FWA (mm) = DISPL, Summer 4TPCsw Example 1 A ship floats in SW at the Summer displacement of 1680 tones. If the TPCsw is 5.18, how much will the draught change by if the ship is towed to a berth where the density of the water is 1.000 tim? Solution in moving from SW to FW the ship will experience sinkage by an amount equal to the FWA. FWA (mm) = DISPL. Summer 4TPCsw FWA= 1680 = 81./mm 4x5.18 The draught will increase by 81.1 mm! 43 DOCK WATER ALLOWANCE (DWA) The Dock Water Allowance (DWA) of a ship is the number of millimetres by which the mean draught changes when a ship passes from salt water to dock water, or vice-versa, when the ship is loaded to the Summer displacement. Consider the load line marks shown. The top of the Summer mark and the top of the Fresh mark both act as the limits af a scale af density, indicating the position of the sait water and fresh water waterlines respectively far a ship loaded to the Summer displacement. If such a ship was to be floating in water of an intermediate density, termed Dock Water, the change in draught when going from sait waterto dock water can be easily determined. Consider the scale marked on the section of load line shown. 24 4.000 [EM If the ship were to go from SW to dock water of RD 1.010, the draught would change by the DWA. The amount of the DWA is simply a fraction of the FWA as shown, in this case 3/sths or 15/25ths of the FWA value. The DWA, as a fraction ot the FWA, is found by the formula: Fig. 4.5 DWA (mm) = FWA x (1025 - RD dock water) 25 NoteThe densities are multiplied by 1000 to simplify the working. Example 2 A ship has a FWA of 200 mm. Calculate the change in draught that will occur ff the ship proceeds from SW to a berth where the RD of the dock water is 1.078. Solution DWA (mm) = FWA x (1025 - RD dock water) 25 Therefore: DWA (mm) = 200 x [1025 - 1018) 25 DWA = 56 mm The draught will increase by 56 mm! The DWA formula is easily modified to calculate the change in draught that would occur if the ship were to proceed from dock water of one density to dock water of another. DWA (mm) = FWA x [RDoy — RDowa) 25 (- means difference between; take smaller value from greater value.) Example 3 A ship is loaded to its summer displacement and is to procesd down river from a berth where the dock water AD is 1.004 to another berth where the dock water RD is 1.016. If the FWA is 260 mm, calculate the change in draught that will occur and state whether it is an increase or a decrease. Solution DWA (mm) = FWA x [ADowr- ADowel Therefore: DWA (mm) = 260 x [1016 - 1004) 25 25 DWA = 124.8 mm The draught will decrease by 125 mm since the ship is moving into more dense water! Note Answers need onty be to the nearesi mm! 25 4.4 LOAD LINE CALCULATIONS 4.4.1 The need for appiying DWA/FWA to ensure maximum cargo Is foaded When loading a ship it is desirable to load as much cargo as possible. Kf the ship is loading in water that is /ess densethan salt water, such as dock water, then allowance should be made for the ship rising out of the water on reaching the sea, salt water density being greater than that of the dock water. Consider the following situation: A ship is loading in the Summer zone in dock water RD 1.012. lt can legally load so that the salt-water wateriine is leve! with the top edge of the Summer Load Line. Consider the situation where the officer in charge of loading, loads cargo until the dock water waterline becomes level with the top edge of the Summer load line! 1.012 (DW) Fig. 4.6 When the ship proceeds to sea, on reaching sait water (RD 1.025) the ship will rise and be light of the Summer marks as shown. MORE CARGO COULD HAVE BEEN LOADED SINCE THE SHIP IS LIGHT OF "00000005 seem Roqulnod (EM) THE SUMMER LOAD LINE MARK! To avoid this situation, but to also ensure ad ad that too much cargo is rever loaded, the Fig. 4.7 amount to safely load can be readily calculated. The aim of the problem is to ensure that on proceeding to sea the ship rises to the desired seasonai load line mark. This is achieved by considering the 'Fresh Water Allowance' or Dock Water Allowance' as appropriate in the calculatian. 442 Procedure for conducting Load Line catculations The following examples illustrate the method to be used to determine the maximum amount of cargo to load when the ship is floating in dock water. lt is important that the calculation procedure is foliowed exactly, particularly in step 2 of the next example. Example 4 A ship has a Summer load draught of 5.80 m, FWA 140 mm and TPC of 21.82. The ship is loading at a berth in dock water AD 1.007 and the present draught is 5.74 m. Calculate the maximum amount of cargo that can still be loaded tor the ship to be at the Summer load line mark on reaching the sea allowing for 26 tonnes of fuel still to be loaded prior to sailing. Solution 1. Calculate DWA (to the nearest mm). DWA (mm) = 140x (1025 - 1007) = 100.8 mm= 10t mm 25 2. Calcuiate the permitted sinkage” in dock water. Always start with the required load line draught and work as follows: Caiculate the maximum amount that can still be loaded in dock water. Permitted sinkage (ems) = w TP Therefore: w = Permitted sinkage (cms) x TPC w=202x 18.6x 1.006 = 368.8 tonnes 1.025 28 29 SECTION 5 - CENTRE OF GRAVITY (G) AND CENTRE OF BUOYANCY (B) INTRODUCTION Consider a ship hegled over by some external force, such as the wind. G represents the centre of gravity of the ship and B the centre of buayancy. These are the points of application of the weight force (Wf), acting vertically downwards, and the buoyancy force (Bf) acting verticaly upwards. Ship stability is concerned with the relative positions of G and B as the ship is heeled. Consider what will happen to the ship once the external heeling force is removed. If the lines of action of Wf and Bf are considered, they will act to return the ship to the upright condition. This section concerns the vertical position of the ship's centre of gravity (G) and how ils position changes when weights are shifted, loaded and discharged. The factors influencing the position of the centre of buoyancy will also be discussed. Learning Objectives On completion of this section the learner will achieve the following: 1. Understand the term Centre of Gravity (G). 2. Understand the effect on G of shifting a single weight vertically up/down. 3. Understand the effect of loading a single weight in a position vertically above/below G. 4. Understand the effect ot discharging a single weight from a position vertically above/Delow G. 5. Calculate the effect on G when shífting, loading or discharging multiple weights. 6. Understand the term Centre of Buoyancy (B) and identify the factors that influence it's position. 7. Perform calculations based on (2) to (5) above. 30 5.1 CENTRE OF GRAVITY (G) “Centre of gravity' (G) of a ship may be defined as being the point where the total weight force (Wit) ofthe ship is considered to act vertically downwards. Provided weighis within the ship are properly secured, the position of G is assumed to not move as the ship heels. (Qbviousty if the ship heels excessively lashings may give way causing cargo to shift!) Weight force (Wf) always acis vertically downwards! Fig. 5.3 When weights are shifted on board, loaded or discharged G will move. Whenever G is caused to move the 'shift of G'must be calculated. The position of the centre of gravity within the ship is the most influential factor in determining its stability. The officer in charge of loading the ship must be fully conversant with the way that G moves when shífting, loading and discharging weights. The vertical position of G is expressed in terms of 'metres above the keel" (KG). The vertical position of the centre of gravity of a weight on board is also expressed in terms of “metres above the keel (Kg). 33 523 Effect of discharging a weight Whenever a weight is discharged, G will move directly away from the centre of gravity of the discharged weight (g). Consider the ship shown where a weight is discharged from the upper deck. G moves to G. Fig.5.9 In this case the KG of the ship will decrease The vertical component of the shift of G is calculated by the formula:| GGy= wxd W-w where: “w'is the weight discharged. “e” is the vertical distance between G of the ship and g of the discharged weight. *Wis the ship's initial displacement. Example 4 A ship has a displacement of 13400 t and an initial KG of 4.22 m. 320 t of deck cargo is discharged from a position Kg 7.14 m. Calculate the final KG of the ship. Solution GGv=wxa = 320 x (7.14 - 4.22) = 0.071 m down W-w 73400 - 320 Initial KG 4.220 Mm Ggv oo7im FINAL KG 4.149m Example 8 A ship displaces 18000 t and has an initial KG of 5.30 m. Calculate the final KG if 10000 t of cargo is discharged from the lower hold (Kg 3.0 m). Solution GG,=wxd = 10000 x [5.3 - 3.0) = 2.875m W-w 18000 - 10000 initial KG 5.300 m GGv up 2.875 m FINAL KG 8.175m 34 5.3 MULTIPLE WEIGHT PROBLEMS it would be very tedious to do a calculation tor every single weight that is shifted, loaded on or discharged from the ship. In practice 'moments about the keer are taken to determine the final KG of the ship, where: MOMENTS (t-m) = WEIGHT (t)x DISTANCE (m) tfa ship is considered, then: MOMENTS (t-m) = DISPLACEMENT (t) x KG (m) Therefore: KG(m)= MOMENTS (t-m) DISPLACEMENT (t) When a number of weights are shifted, loaded or discharged, the moments for each weight are calculated. These are summed up and simply divided by the final displacement of the ship to give the final KG. A tabular approach is adopted and the method is easily illustrated by way of an example. One important point to note is that the first weight to be entered into the table is that of the ship's initial displacement along with the ship's initial KG. Example 6 A ship displaces 10000 t and has a KG of 4.5 m. The following cargo is worked: Load: 120tat Kg 60 m; 730 tat kg 3.2 m. Discharge: 68tfrom Kg 2.0m; 100 t from Kg 62 m. Shift: 86 tfrom Kg 2.2 mto Kg 6.0 m. Calculate the final KG. WEIGHT KG (m MOMENTS (t-m Consider the table shown. Each weight is multiplied by its KG to give a moments value. The sign of this valve (+ or -) depends on whether the weight is loaded or discharged. In the case of the weight that is shifted, this is simply treated as two separate weights: one that is discharged: and another of same weight that is loaded! The final KG (4.459 m) is simply found using the formula: KG(m)= MOMENTS (tm) DISPLACEMENT (t) 2 8. Discha a 138.0! ischarge (-i - -85 - “Load = Ste: FINAL 10682 47626,80 ie. 47626.80 = 4459m 106882.00 Note Answers should be given io 3 decimal! places. This method may be used for single weight problems also, with the advantage being that the directian of movement of G (either up or down) need not he considered. A final KG is 'automaticaliy' calculated! To prove this you should rework the previous single weight examples. A prereguisite for any KG calculation to be correct Is that the ship's Lighiweight Displacement and KG values be accurate. This is the subject of Section 19 - Inclining Experiment. 35 5.4 CENTRE OF BUOYANCY (B) “Centre of Buoyancy' of a ship is defined as being at the geometric centre of the underwater volume of the ship at a particular instant and is the point through which the total buoyancy force (Bi) is considered to act vertically upwards. lts position will constantly move as the ship moves at sea. er Br Fig. 5.10 Although the centre of gravity (G) is assumed to remain in the same place as the ship heels (provided weights do not shift within the ship), the centre of buoyancy constantly moves as the ship pitches, rofis and heaves. As the displacement (and draught) of the ship changes, so will the position of the centre of buoyancy when the ship is upright. The vertical position of the centre of buoyancy is termed the KB, being the vertical distance from the keel (K) to the centre of buoyancy (B). Fig. 5.11 For a box-shaped vessel on even keel KB is half the draught. 38 62 RIGHTING LEVER (GZ) Fighting lever (GZ) is defined as the horizontal distance, measured in metres, between the centre of gravity (G) and the vertical line of action of the buoyancy force (Bt) acting through the centre of buoyancy (B) when the ship is heeled. Righting lever (GZ) increases to some maximum value and then decreases as the ship progressively heels further. i Fig. 6.3 The righting levers tor specified angles of heel are represented on a Curve of Statical Stability, commonty known as a GZ Curve as shown. os 02 o2 Hael (dog Fig. 6.4 The procedure for producing such a curve is discussed in Section 10. 39 40 6.3 MOMENT OF STATICAL STABILITY (RIGHTING MOMENT) The moment of staticaí stability, commonhy referred to as the righting moment, at any given angle of heel is found by: RIGHTING MOMENT (t-m) = GZ (m) x DISPLACEMENT (t) which results from the buoyancy force (Bf) (being equal to the ship's displacement (Nf), acting on the end of the lever GZ, which pivots about G. The righting moment at any angle of heel represents the instantaneous 'value” of the ship's ability to retum to the upright, expressed in tonnes-melres, when G Z the ship is in “still water: conditions and is momentarily at rest i.e. acceleration forces as the ship ralis are ignored. Example 1 B Calculate the moment of statica! stability (righting moment) for a ship with a displacement of 12000 tonnes if the righting ever (GZ) is 0.46 m when heeled over. Fig. 6.5 Solution RIGHTING MOMENT = GZ x DISPLACEMENT RM = 0.46 x 12000 RM = 5520 tm 43 6.6 —CALCULATING THE MOMENT OF STATICAL STABILITY AT SMALL ANGLES OF HEEL In triangle GZM: Sing =OPP = GZ HYP GM Therefore: GZ=GMxSinb Having found GZ: RIGHTING MOMENT = GZ x DISPLACEMENT Note The above formula for GZ can only be used for small angles of heel. Example 2 Fig. 6.8 A ship has a dispiacement of 9420 tonnes and a KM of 9.22 m. In its present loaded condition the KG is 7.46 m. Calculate the moment of statica! stability available if the ship is heeled to (a) 2 deg. (b) 4 deg. (o) 8 deg. Solution KM 922m KG 746m GM 1.76m (a) GZ=GM x Sin8 GZ = 1.76 x Sin 2º = 0.08142m RM = GZ x DISPLACEMENT. AM = 0.06142 x 9420 = 578,6 tm (b) GZ=GMxSin6 GZ=1.76xSin4º=0,12277m AM = GZ x DISPLACEMENT RM = 0.12277 x 9420 = 1156.5 tm (c) GZ = GM x Sin 6 GZ = 1.76 x Sin 8º = 0.24494m RM = GZ x DISPLACEMENT RM = 0.24494 x 9420 = 2307.4 tem Example 3 A ship has a displacement of 8900 tonnas, a corresponding KM of 9.400 m and a KG of 7.620 m. (a) Caiculate the moment of statical stability when the ship is heeled to 5 degrees. (b) A weight of 200 tonnes is shifted from the lower hold (Kg 4.26 m) to the upper deck (Kg 12.60 m). Calculate the moment of statical stability that will! now exist if the ship is again heeled to 5 degrees. Solution (a) KM | 92.400m KG 7.620 Mm GM 1.780m GZ = GM x Sin 8; RM = GZ x DISPLACEMENT; (b) Take moments about the keel: GZ=1.780x8in5º=0.15514m AM =0.15514 x 8900 = 1380.7 tm WEIGHT () | KG(m) | MOMENTS (tm) | [KM 9,400 Initial dispt. seco 2.82 gr8180 FINALKG 7.807 Discharge -200) 4.26 -8520 FINAL GM 593 Load 200 12.60 25200 FINAL 8900.0 7.807 89486.0 GZ=GMxSing; GZ = 1.593 x Sin 5º = 0.13884 m RM = GZ x DISPLACEMENT; RM = 0.13884 x 8900 = 1295.7 tm Available righting moment has reduced as a result of the increased KG! In this section the transverse statical stability of a stable ship at smal! angles of heel has been discussed. A smail angle of heel is often considered to be any inclination of the ship up to approximately 10º. A more accurate definition of a smai! angle ot heel for a particular ship is the subject of discussion tor Section 16 - The Wall Sided Formula. 45 SECTION 7 - CONDITIONS OF STABILITY INTRODUCTION In the previous section, transverse statical stability was discussed in terms of a ship that was in a stable condition only. It is essential to discuss the behaviour of a ship when it may become unstable. A ship may become unstable if the centre ot gravity (G) is allowed to rise too high. There are a number of possible causes of this, principle ones being the loading of too much weight high up in the ship and the effect of free surfaces in slack tanks. This section simply discusses stability and instability in terms of the relative positions of G, B and M. Learning Objectives o Understand the term “stable condition: 2. Understand the term “neutral condition: 8. Understand the terms “unstable conditior! and 'angie of lol. 48 7,3 — UNSTABLE CONDITION AND ANGLE OF LOLL A ship is in an unstabfe condition if, when heeled by an external force in still water to a small angle, it continues to hee! Iurther when the external force is removed. E Consider a ship that is floating upright in still water with G and B M as shown. The initial transverse metacentre (M) is below G. E+ i.e. KM - KG = GM; which has a negative value. | The ship now heels to a smal) angle ot inclination. Consider the lines of action of Wf and Bf. They are acting in such a way to cause the ship to hee! further over. GZ is a capsizing lever! The question that now comes to mind is: Mil the ship capsize? Possibly! As the ship continues to heel, the centre of buoyancy (B) will move outward as the underwater volume of the ship changes shape. Provided that the centre of buoyancy can move sufficienily ouiboard to attain a new position vertically below G then the capsizing lever will disappear and the ship will come to rest at an angle of loft. !f the centre of gravity were very high then the ship would possibly capsize. If the ship is heeled beyond the angle of loll the centre of buoyancy (B) wil move outboard of the centre of gravity (G). This causes a positive righting lever (GZ) thai will act to return the ship back to the angie of lol. Note A ship Iying at an angle of loll is in a potentialy dangerous situation. If wind and/or waves were to cause the ship to roll through Fig 7.9 the vertical it would, in theory, come to rest at the same angle of loll on the other side. However, the momentum of the ship as it rolls over may be sufficient to cause itto capsize. In any event, cargo shift would be likely which would cause the situation to worsen further. The effects of “free liquid surfaces' in slack tanks are a principal cause of instability in ships. Section 9 considers free surface effect in detail. Whenever instability is suspected the procedures in Section 15 must be strictly folioweo. 49 SECTION 8 - INITIAL TRANSVERSE METACENTRE INTRODUCTION When designing a ship the factors that influence the height of the initial transverse metacentre (KIM) are of prime importance. It follows that the greater the KM value, then the greater will be the GMfor any given KG. It is important to appreciate that KG alone is not the influencing factor on the ship's initial condition of stability. lt wil be seen in this section that KM changes with draught/displacement; this means that a particular KG value may give adequate initial stability with respect to GM at one draught but not at another. In this section, the leamer will calculate KM values for box-shaped vessels whereby it will be seen at first hand the factors that influence KM. Metacentric diagrams are introduced as a means of graphical representation of the ship's initial stability. Leaming Objectives On completion of this section, the learner will achieve the following: 1. Understand more comprehensively the term initial Transverse Metacenire. 2. Calculate KM values tor a box-shaped vessel and produce a metacentric diagram. 3 Use a metacentric diagram to determine the condition ot stability of a ship at various draughts for a given assumed KG. 4. Use a metacentric diagram to determine the required final KG to ensure that a ship completes loading with a required GM. 5. Understand the factors affecting KM. 50 81 INITIAL TRANSVERSE METACENTRE EXPLAINED The initial transverse metacentre is the point of intersection of the lines of action of buoyancy force (Bi) when the ship is in the initial upright condition and subsequent!y heeled conditions, within small angles of heel. Consider the ship shown, heeled to some small angle of inclination. It can be seen that a wedge of buoyancy has been transferred from the high side to the heeled side (bb,). The resultant movement of B to B, at this instant is one that is parallel to and in the same direction as the shift of the centroid of the transferred volume of buoyancy. BB, could be calculated using the formula: BB,=yxbb, V Fig. 8.1 where: vis the volume of the transferred wedge; bb, is the distance through which it's centroid has moved, and; Vis the volume of displacement of the ship. (Note that this formula is simitar to the shift ot a single weight formula!) 8.1.1 Metacentric Radius If B is plotted for several small angles o! heel, it may be assumed that it follows the arc of a circle centred at M. BM is termed the metacentric radius. Fig. 82 81.2 Calculating KM for box-shaped vesseis it is convenient to consider the KM for a box-shaped vessel because the maths is simple, however, the same principles will apply for ship shapes. KM is calculated by the formula: KM = KB + BM For a box-shaped vessel on an even keel: KB = draught 2 53 ao Eua 2 mM 1 KG Em) ao “5 10 a E o ol. pio 12324 8 6 78 9 40 42 15 44 15 S2m | Dravght jm) t28m | () For the vessel to be unstable the KG must be greater than the KM at the draught concerned. This occurs between the draughts 5.2 m and 12.8 m. (ii) Ata draught of 3.00 m KM was calculated to be 12.61m. KM 12.61 m KG 9200m GM 3.61m DISPLACEMENTa0x = L x B x d x density DISPLACEMENT:0x = 100 x 20 x 3.00 x 1.025 DISPLACEMENT 50x = 6150 lonnes GZ=GMx Sino GZ=3.61 xSinsº GZ = 0.91463....metres Therefore: Righting moment = GZ x Displacement Righting Moment = 0.31463..... x 6150 Righting moment = 1835 tm 822 To determine the final KG required to complete loading with a required GM The box-shaped vessel for which the metacentric diagram was drawn had a length of 100 m and a breadth of 20 m. Consider the following example using the same metacentric diagram: Example 2 It is intended to load the vessel to a maximum permissible draught in sait water of 4.5 m. (a) What is the maximum displacement of the vessel? Solution (a) DISPLACEMENT50x = (L xBxd)xp DISPLACEMENTa0x = (100 x 20 x 4.5) x 1.025 DISPLACEMENT ox = 9225 t 54 (b) The required GM on completion of loading is 1.20 m. What is the maximum permissible KG? Solution (b) From the graph the KM for a draught of 4.5 m is approximately 9.5 m. KM 95m Required: GM 12m Maximum: KG 83m (0) The vessel is currentiy loaded to displacement of 8465 t and has a KG of 8.40 m. What is the maximum Kg at which to load the final 760 t of cargo to ensure thai the final GM requirement of 1.20 m is achieved? Solution fc; in (b) it was determined that the maximum KG required was 8.3 m. Take moments about the keel in the normal way but let 'x' equal the Kg at which to load the final 250 t. in the formula: KG (m) = MOMENTS (tm) DISPLACEMENT tt) the final KG is already known as 8.3 m! 3 a Ke mi zo Draught tm BD AD 1 2 13 1415 WEIGHT () | KG(m) | MOMENTS (tm) KG (m)= MOMENTS (tm) InitiaLdisol 8465 84 z1106 DISPLACEMENT (t) Load 760 x 760x FINAL 9225 71106 + 760x 8.3=([71106 + 760x) 9225 Solving x will give the answer! 8.3 x 9225 = 71106 + 760x 76567.5 = 71106 + 760x 76567.5 - 71106 = 760x 5461.5 = 760x Theretore: 5461.5 = x = 7.186m 760 The maximum Kg at which to load the fina! 760 t weight is 7.186 m to ensure that the final KG does not exceed 8.3 m, thus ensuring that the final GM is at least 1.2 m. Had the value of KM been calculated using the formula instead of taking it from the graph a more accurate answer would have resulted. in practice the metacentric diagram for a ship (if available) will have to be used as presented in the stability data book, since the KM for a ship shape is not readily determined. Note A question might ask for the maximum weight that can be loaded at a specified Kg to ensure that a final KG value is not exceeded. In this case the same method as in Answer (c) would be used except that x" equals the amount of cargo to load at the specified Kg instead. 83 FACTORS AFFECTING KM 8.31 Beam 55 Consider two ships of different beam each heeled to the same angje of inclination as showm. In the narrow ship a smal! wedge ol buoyancy is transterred from the high side ta the low side (bb,) causing B to move to B,. Initial transverse metacentre is at M:. In the broader ship, a larger wedge of buoyancy is transferred from the high side to the low side (bb;) causing B to move further out to B;. Initial transverse metacentre is higher at Mo. lf the formula: BB,=vx bb, v Fig. 8.6 is considered the larger the lume of the transferred wedge of buoyancy and the greater the distance through which the centroid af the wedge is caused to shift, the greater will be the outward movement of B as the ship is heeled. Thus: KM increases as beam increases resulting in broader ships being more stable. 8.32 Draught Consider the formula: BB,=vx bb, v At the Joad draught (displacement) the volume oi the transterred wedge of buoyancy (v) represents a smaller part ot the total volume of displacemnent of the ship (V) than it would at the light draught (displacement). Thus: KM decreases as draught increases for the normai range of operational draughts of a ship. 58 9.2 FREE SURFACE DATA For a rectangular shaped tank, the calculation of the effects of free surface is straightlorward. However, not all tanks are this convenient regular shape and data relating to tanks of all shapes on board are included in the ship's Tank Sounding Data tables. This data can be provided in a number of ways, the person on board conducting the ships stability calculations must be familiar with the data supplied. 92.1 Calculating the effect of free surface In a rectangular shaped tank For a tank that has a rectangular free surface the virtual rise of Gin metres can be calculated by: GGy= mb? x dt 124 ds where: GGvis the virtual rise of G in metres; fis the tank length; bis the tank breadth; dtis the density of the liquid in the tank; dsis the density of the water in which the ship floats (1.025 Um); and; Vis the volume of displacement of the ship. Since: DISPLACEMENT = VOLUME OF DISPLACEMENT x DENSITY Le. W=Vxds it follows that: GG,= b? x dt 12W Example 1 A ship has an initial displacement of 10500 t and KG 7.60 m. À rectangular cargo oil tank of length 30 m and breadth 20 m is partialy filled with 9600 t of oil (RD 0.86). ! the Kg of the oil is 8.00 m, calculate the effective GM if the KM for the final displacement is 8.80 m. Solution 1. Taking moments about the keel, calculate the new solid KG. WEIGHT (t) | KG (m) MOMENTS (t-m) Initial disp. 10500 7.60 79800 Cargo oil 9600 8.00 76800 FINAL 20100 TIM 156600 2. Calculate the effect of free surface. GGyv=ib' x dt = 30x20 x 086=0856m 12W 12x 20100 (The final dispiacement must be used!) 3 Calculate the solid GM and then apply the free surface correction to obtain the fluid GM. KM 8.800 SOLID KG ZIMN SOLID GM 1.009 FSE (GGv) 0.856 FLUID GM 0.153 59 9.22 Free surface moments The moment of inertia (), aten termed the second moment of area, of the free liquid surface ofa rectangular tank may be determined by: = 16 (mº) 12 lf the value of | is multiplied by the liquid density then a value of Free Surface Moments" (FSM's) (t-m) is obtained. ESM's (tm)= bx dt 12 Consider the previous free surface effectformula: GGy= lb? x dt 12W Therefore: G& = FSMs Displacement Since: Final KG = Sum of moments Displacement it is evident from the above that the greater the value of the free surface moments, the greater the value of the effective KG and the greater the loss of GM (GGy). In calculating the effective GM it is usual to make allowances for free surfaces by incorporating the FSM's in the KG table where they must always be ADDED. Consider the previous example. Example 2 A ship has an initial displacement of 10500 t and KG 7.60 m. A rectangular cargo oil tank of length 30 m and breadth 20 m is partiaily filled with 9600 t of oil (RD 0.86). f the Kg of the oil is 8.00 m calculate the effective GM if the KM for the final displacement is 8.80 m. Solution 1. Calculate the FSM's using: FSM's (tm) =? x dt 12 FSM's (tm) = 30 x 20 x 0.86 = 17200 tm 12 2. Taking moments about the keel, also adding the FSM's in the 'moments' column”, calculate the fluid KG. WEIGHT (t) KG (m) | MOMENTS (tm) Initial displ. 10509 7.80 79800 Cargo oil 8600 8.00 76800 FSM's 17200 FINAL 20100 8.647 173800 3. Apply the fluid KG value to the final KM to obtain the final fluid GM. KM 8.800 FLUID KG 8.647 FLUID GM 0.153 60 923 Representation of free surface data in tank sounding/ullage tables In tank sounding or ullage tables free surtace data for use in calculating the ship's effective KG and GM can be represented using alternative methods. 9.2.3.1 Method 1 - Free surface moments for an assumed density value Consider the extract from a tank sounding table shown paying particular attention to the column headings. [Tank: 2C0.Stbd Cargo Ol Density: [ ozão ] Sounding | Weight LcG TCG vca FSM's ems; tonnes, (m foa m m tm The table is for a cargo oil tank in a tanker - No. 2 Cargo Oil tank Starboard. Free Surface Moments (FSM's) in tonnes-metres are tabulated for an assumed liquid density of 0.740 tm Weight of liquid in the tank is tabulated against sounding for an assumed liquid density of 0.740 Ym?. VCG (Vertical Centre of Gravity or Kg) indicates the vertical position of the oil within the ship in terms ot metres above the keel for the appropriate sounding. Example 3 A ship displaces 5400 t and has a KG of 7.860 m. No. 2 Cargo Oil tank Stbd. is filled to a sounding of 150 cms with cargo oil RD 0.740. Calculate the fina! effective KG and GM if the KM for the final condition is 8.000 m. (Use the sounding table extract given.) [Tank: 2C0.Stbd Carga OIl Density: [oz | Sounding Welght TCG vca FSM's (ul) tonnes) m m [t-m 3 Moments of mertia (1) in metres? (m') are tabulated. Volume of liquid in the tank in cubic metres (mº) is tabulated. Density of the liquid is not considered. To obtain the weight (mass) of the liquid and the Free Surface Moments which are to be incorporated into the KG moments table both volume and ! values as tabulated must be multipiied by the density of the liquid in the tank. Example 5 A ship displaces 5400 t and has a KG o! 7.860 m. No. 2 Cargo Oil tank Sibd. is filled to a sounding of 150 cms with cargo oil RD 0.740. Caiculate the final effective KG and GM ff the KM for the final condition is 8.000 m. (Use the sounding table extract given.) Solution Obtain cargo data from table. [Tank: 2C0.Sthd [ ] Sounding Val Lca cms; GU. m [m foa) Caiculate the mass of oil in the tank. Mass = Volume x Density;Mass = 145.22 x 0.740 = 107.46 t Caiculate the FSM's for the oil. FSM's =! x Density; FSM's = 505.0 x 0.740 = 373.7 tm Calculate the final effective KG and hence the final effective GM as before. WEIGHT (t) | KG(m) | MOMENTS (tm) Initial disol 5409,00. 7.860 42444.0 SM 8.00 Cargo oil 107.46 2.302 247.4 ELUID KG. BIS À ESTE a FLUID GM 0.181 FINAL 5507.46, 7819 43065.1 9.2.3.3 Summary You will not have a choice as to which of the two methods to use, it simply depends on the format of the tank sounding tables that are supplied to the ship. Consider the significant errors in the calculation of GM that will occur if: Tabulated FSM's for an assumed liquid density are not corrected for the actual density of the liquid in the tank! Volume is not converted to mass! Tabulated | values are not multiplied by the density of the liquid in the tank! ALWAYS CHECK! Tank sounding data for M.V. Almar are tabulated using assumed density values (Method 1). [E 9.3 FACTORS INFLUENCING FREE SURFACE EFFECT Consider the free surface formula for loss of GM (GG,): | GGy (m)= Ib? x dt 12W tis clear that the breadth ot the tank is the most important factor. 9.3.1 Tank breadth Ha tank is subdivided, the loss of GM can be greatly reduced. This can be demonstrated by way of the following three worked examples. Example 6 A ship has a displacement of 12000 t and initial KG of 7.84 m. A rectangular double bottom tank has the following dimensions; length 20 m, breadth 15 m and is filled with sait water ballast (RD 1.025) to a sounding of 2.00 m. If the KM for the final condition is 8.00 m, calculate the final effective GM. Solution 1. Calculate the mass and Kg of the ballast water. Mass = Volume x Densily; Mass = (| x b x sounding) x density; Mass = (20 x 15 x 2) x 1.025 = 615.01 Since it is a double bottom tank the Kg of the ballast water will be half the sounding: Kg=05x2.0=1.00m 2 Calculate the FSM's for the rectangular free liquid surface: FSM's (tm) =? x dt 12 FSM'S (tm) = 20x 15º x 1.025 = 5765.6tm 12 Taking moments about the keel calculate the final KG and hence the final KM: WEIGHT (t) | KG(m) | MOMENTS (t-m) | |Km 8.000 Initial disp 12900,00 840 94080,0 FLUI K: 7,904. [SW ballast 615.00 1,000 615.0 FLUID GM 0.036 ESM's 5765.6 FINAL 12615,00 7.964 100460.6 To satisfy the IMO intact stability requirements the minimum GM requirement for a shipis 0.15m. This ship clearly does not satisfy that requirement! Consider the same example but this time the tank will be equally subdivided into two tanks. Example 7 A ship has a displacement of 12000 t and initial KG of 7.84 m. A rectangular double bottom tank is equally subdivided has the following dimensions; is 75 m— ja 7,8 m——— length 20 m and breadth 15 m and is filled with sait water ballast (RD 1.025) to a sounding of eg 2.00m. + sm——— Fig. 9.4 65 if the KM for the final condition is 8.00 m calculate the final effective GM. lt can be seen that there are now two tanks each having a breadth of 7.5 m. Solution 1. Calculate the mass and Kg of the ballast water. Mass = Volume x Density; Mass = (x b x sounding) x density; Mass = (20 x 15 x 2) x 1.025 = 615.0t Alternatively: Mass per tank = Volume x Density: Mass per tank = (| x b x sounding) x censity; Mass = (20x 7.5 x 2) x 1.025= 307.51 Totaí mass = 307.5 x 2 tanks = 615.0t Since it is a double bottom tank the Kg of the ballast water will be half the sounding: Kg=05x2.0=1.00m 2 Caiculate the FSM's per tank for the rectangular free liquid suríace: FSM's (tm) =P xdt 12 FSM's (t-m) = 20 x 7.5º x 1.025 = 720.7 tm 12 Total FSM's = 720.7 x 2 tanks = 1441.4 tm 3. Taking moments about the keel, calculate the final KG and hence the fina! GM: WEIGHT (t) | KG(m) | MOMENTS (tm) | |km 2.000 ELUID KG 7.821 FLUID GM 0.379 FINAL 12815.00 7.821 96136,4 Subdividing the tank has resulted in the fina! GM being much improved. This is a direct result of the reduced free surface moments. For the undivided tank the total FSM's where 5765.6 tm. For the subdivided tank the total FSM's where 1441.4 tm. FREE SURFACE MOMENTS HAVE BEEN REDUCED TO ONE QUARTER OF THEIR ORIGINAL VALUE! Le. 5765.6 = 14414 tm 4 Consider the same example but this time the tank will be equally subdlvided into three tanks. Example 8 A ship has a displacement of 12000 t and initial KG of 7.84 m. A rectangular double bottom tank, which is equally subdivided into three compartments, has length 20 m and overall breadth 15 m and is filled with sait water ballast (RD 1.025) to a sounding of 2.00 m. if the KM for the final condition is 8.00 m, calculate the final effective GM. lt can be seen that there are now three tanks each having a breadth of 5.0 m. to subdivide tanks into three compariments at most. The benefit of any further subdivision which would improve efiective GM by a decreasing amount each time would be greatiy offset by the additional steel weight and piping arrangements required. An exception might be in the case of a product or chemical carrier where such ships are designed to carry a wide range of cargoes at any one time in relatively smaller quantities. 9.3.2 Tank length Free surface moments (and loss of GM) are directly proportional to the length of the tank i.e. ifthe tank 68 NE, Typical oil tanker tank arrangement Fig. 9.9 length is doubled so will be the value af the free surface moments (and loss of GM). 923.3 Density Free surface moments (and loss of GM) are directly proportional to the density of the liquid in the tank as discussed in 9.2.3.7, the greater the density af the liquid in the tank, the greater the FSM's and subsequent loss oi GM. 934 Ship displacement Free surface moments (and loss of GM) are inversely proportional to the displacement of the ship. For a given tank, the loss of GM will be smaller as the displacement increases and vice-versa. should be noted that the actual free surface moments for any tank are not affected by the ship's displacement (since ship displacement is not included in the formula for their calculation anyway). 9.4 IMPORTANT POINTS TO NOTE REGARDING FREE SURFACE MOMENTS These are summarised as follows: 1. For a tank to be considered subdivided, it must be fitted with an oiltight o watertight longitudinal bulkhead ensuring that there is no possibility of liquid transfer. This means that any valves connecting the subdivided tanks must be capable of being fully closed. 2. A wash plate is fitted to prevent damage to internal tank plating that may be caused by wave action within the tank. it does not reduce free surface elfect! (b) Tank fitted with Jongitudiral bulkhesd Fig. 9.10 69 two similar rectangular tanks are filed to different levels, the free surface moments for each will be the same. (Consider the formula for FSM's if you are unsure!) prende | lfa tank is empty or pressed-up, Vo) free surface moments will not exist in that tank. Fig. 9.11 Wen calculating the effective KG/GM, the free surtace moments of ai! slack tanks must be incorporated into the KG moments table. (Loss of GM due to free surface will be that which results from the cumulative effects of all the slack tanks on board.) 70 SECTION 10 - CURVES OF STATICAL STABILITY (GZ CURVES) INTRODUCTION The curve of statica! stability, or GZ curve as it is most commonty referred to, is a graphicai representation of the ship's transverse statical stability. Transverse statical stability is the term used to describe the ability of a ship to return to the upright, when it has been forcibly heeled by an external force and is momentarily at rest when floating in still water. RIGHTING MOMENT (tm) = GZ (m) x DISPLACEMENT (t) At any angle of heel, it is the horizontal disposition of G and B that determines the GZ value. As a ship progressively heels over the righting lever, GZ, increases to some maximum value and then decreases until at some angle of heel it becomes negative i.e. it becomes a capsizing lever. Calculating the value of GZ, at specified angles of heel for a ship's particular condition of loading, will allow a curve of statica! stability, or GZ curve, to be produced. es [) 01 az tm) o 6 o) Do mm o io im m o 4 s2 Hool (dog) Fig. 10.1 The greater the values of GZ, the greater will be the area under the curve. Minimum standards with respect to the area under the curve (and other criteria) are specified in the 'Code on intact stability (MO) and these are incorporated in the government legislation of most countries that adopt the IMO conventions. Assessing compliance of a ship's loaded condition is considered in Section 14. tis the aim of this section to review the method of actually producing a curve of statical stability and to be able to extract basic information from it. Learning Objectives On completion of this section, the leamner will achieve the following: 1. Understand the term KN and how KN values may be used to obtain GZ values for specified angles of heel. 2 Know the procedure for producing a curve of statical stability. 3. Identity the basic features of a curve of statical stability. 4. Understand the terms Stiff and Tender with respect ta the curve of statical stability. 73 10.3 BASIC INFORMATION AVAILABLE FROM THE CURVE OF STATICAL STABILITY Consider the curve in the previous example. The following information can be extracted from it: (a) The GZ value tor any angle of heel. This can be used to calculate the moment of statical stability for the ship at that particular angle of heel if the formula: RIGHTING MOMENT (tm) = GZ (m) x DISPLACEMENT (t) is applied. (b) The maximum GZ and the angle of heel at which it occurs. (c) The range of positive stability and the angle of vanishing stability (AVS). (d) The approximate angie of deck edge immersion (Bog). Figure 10.4 shows the ship heeled to the point where deck edge immersion takes place. The angle at which this occurs is identified on the curve as the point where the curve trend changes from increasing steepness to decreasing steepness. This is known as the point of inflexion of the curve. It is often dfficult to estimate its position but helps to identify the point of inflexion if a series of vertical lines are drawn on the curve. If each siice is taken in turn it may be considered if the trend is one of the following: increasing steepness; decreasing steepness; or; neither. ht is only an approximaton and open to interpretation! Consider the curve constructed in the previous example, the aforementioned information is illustrated. 2 E oz if cz tm qu = Inflexion o» a nm elim ms E Rana ot pesttive simbiliy E Rael fog) Fig. 10.6 Fig. 10.4 /. Fig. 10.5 From the curve: The maximum GZ value is 0.57 m and occurs at an approximate angle of hee! of ag. The range of stability is from o to 64º (the angle of vanishing stability being 649). The angle at which deck edge immersion takes place is approximately 23º. 74 10.4 CURVES OF STATICAL STABILITY FOR STIFF AND TENDER SHIPS 10.4,1 Stiff ships A strf ship is one with a very large GM caused by KG being too small. This occurs if too much weight is placed low down within the ship. The ship will be excessively stable, righting moments will be so large as to cause the ship to return to the upright very quickly when heeled. Roll period will be short. A very large GM should be avoided for the following reasons: e The ship will return to the upright very quickly whereby the motion will be jerky causing excessive strain on cargo Jashings and possible cargo shift. e Loose gear wil be thrown about. e tis uncomfortable for crew and injury may result from the Fig. 10.7 ship's quick motion. e Structural damage to the ship may occur due to racking. 10.4.2 Tender ships A tender ship is one with a very small GM caused by KG being too large. This occurs if too much weight is placed high up within the ship. The ship wil have insufficient stability, righting moments will be very small when heeled causing the ship to be sluggish and slow to return to the upright. Roll period will be long. (A tender ship is still a stabfe ship i.e. M is above G) A very small GM should be avoided for the following reasons: Because of the small righting moments the ship will oniy offer limited resistance to being rolled, causing the ship to be rolled to farger angles of heel. This will increase the risk of water being shipped on deck. The ship will be slow to return to the upright and will tend to remain at the extent of the roll tor a comparatively long time. This 7 will create greater and more prolonged strain on cargo lashings Pa and increase the risk of cargo shift. Eur Fig. 10.8 Rolling to excessive angies of heel is also uncomfortable for the crew and injury may result. As a guide, a GM of between 48% of the ships breadth is desirable. GR survos for std trocar ahipa: Container ships that have containers : stowed on deck may probably be more Ep suited to a GM value on the tender side e of these limits to minimise the stresses 2 on deck container lashings. az 18 La Typical curves of statical stability for não — both a stiff and tender ship are shown. om ag dA am mentos — 80 ---tmder 75 SECTION 11 - LIST INTRODUCTION So tar stability has oniy been considered for a ship that is upright, whereby G is on the centre line and the ship floats upright in still water. It is necessary to consider the position of G in the transverse sense as well as the vertical. There is a distinction to be made between the terms fistand heel, this often being overlooked or ignored completely. List is the term used to describe a ship that is inclined due to the distribution of weights within it. Heel is the term used to describe a ship that has been torcibly inclined by externa! forces (wind, waves eic.). Learning Objectives On completion of this section, the learner will achieve the following: 1. 2. Ds nam Calculate the list caused by a transverse shift of a single weight using the basic !list triangle” for a ship that is initially upright. Calculate the list caused by a transverse and vertical shift of a single weight for a ship that is initially upright. Calculate the list caused by a single weight being loaded ar discharged. Calculate the weight to shift to bring a listed ship upright. Calculate the final list when loading and/or discharging multiple weights for a ship that is initially upright. Calculate the final list when loading and/or discharging multiple weights for a ship that is initially listed. Calculate the weights to load each side of the centre line to ensure that the ship completes upright. Understands the effect of free surface on list. 78 11.3 CALCULATING THE LIST DUE TO A SINGLE WEIGHT BEING LOADED OR DISCHARGED l a weight is loaded or discharged then both the vertical and horizontal components of the shift of G must be considered and the final GM must be LoL used to calculate the final list. : À SyiyiG, Remember the rules: HH a weight is loaded G will move directty towards the centre of gravity of the loaded weight. : Hf a weight is discharged G will move À. i +) directiy away from the centre of fa. 115 gravity of ihe discharged weight. The procedure for single weight load/discharge problems is as follows: 1. Caleulate GGy using: GGy=wxd WEw “d' being the vertical distance between G of the ship and g of the loaded/discharged weight. 2. Apply GG, to the ship's initial KG to find the final KG. 3. Calculate the final GM using: GM = KM - KG 4. Calculate GGy using: GG, = w xd WEw 'd” being the horizontal distance between G of the ship and g of the loaded/discharged weight. 5. Using the formula: Tan Ousr= GG; calculate the list. GM Follow Examples 3 and 4, one for a weight being loaded, the other for a weight being discharged. may help your understanding of the working if you do a sketch for each case. Example 3 A ship initially upright displaces 6400 t and has KG 4.6 m and KM 6.5 m. À weight of 80 t is loaded on deck at Kg 10.2 m, 6.2 m off the centre line to starboard. Calculate the final list. Assume KM remains constant. Solution GGy=wxd GGv=80x(10.2- 4.6) =0.069m W+w 6400 + 80 Initial KG 4.600 m KM 6.500 m GGv (up) 0.089 m Final KG 4.669m Final KG 4.669m Final GM 1.83!m GGu=wxd GG,=80x62 =0.077m Wew 6400 + 80 Tan B.isr = GGy = 0.077 =0.04205 List=2.4º Stbd. GMena 1.831 79 Example 4 A ship inilially upright displaces 14480 t and has a KG 8.82 m and KM 10.96 m. A weight of 240 tis discharged from a position in the lower hold Kg 3.6 m, 2.8 m off the centre line to port. Calculate the final list. Assume KM remains constant. Solution GGy=wxd GGy = 240 x (8.82 - 3.6) = 0.088 m wW-w 14480 - 240 Initial KG 8.820 m KM 70.960 m Gar fu 0.088 m Final KG 8.908m Final KG 8.908 m Final GM 2.052 m GGu=wxd GG,=240x28 =0.047m Ww-w 14480 - 240 Tan Bisr = Ga = 0.047 =0,02290 List=1.3º Stbd. GMeinaL 2.052 so 11.4 SHIFTING A WEIGHT ALREADY ON BOARD TO BRING A LISTED SHIP UPRIGHT For a ship to be upright. PORT MOMENTS = STARBOARD MOMENTS where G must be on the centre line. Fig. 11.6 Example 5 A ship that is listed will have G off the centre line by a distance GGu as shown. Consider the formula for a shift of weight. GGh=wxd W Rearranging this gives: GGuxW=wxd (GGu x W) represents the listing moments that the ship initially has. (w x d) represents the moments required to equal (GG» x W) if the ship is to complete upright. Consider the following example A ship has a displacement of 12000 t and is initially listed 2º to starboara. If the KG of the ship is 11.60 m and the KM is 12.00 m, how much ballast water must be transferred from a starboard side ballast tank to a port side ballast tank through a distance of 16.00 m? Solution To complete upright: Port Moments = Starboard moments Ship is initially listed to starboard. KM - 1200m KG 11.60m SM 0.40m Tan Ousr= GG GM Tan2º = GG, 0.40 GG, = Tan 2ºx 0.40 = 0.074 Mm G is off the centre line ta starboard by 0.014 m. Required port moments to counteract list (w x d) must equal initial starboard listing moments (GGu xW). GG xW=wxd 0.014 x 12000 = w x 16.00 168 = 16w w= 10.5 tonnes to transfer 83 11.7 LOADING WEIGHTS ABOUT THE CENTRE LINE TO COMPLETE UPRIGHT A common question arises where the ship is near completion of loading and the remaining cargo has to be distributed between two compartments that are either side of the centre line in such a way that the ship completes upright. To complete upright: Port moments must equal starboard moments! There are two methods of approach to this type of problem as shown in the next example. Example 8 (Method 1) From the following details calculate the final GM and the amount of cargo to load in each space so that the ship will complete loading upright: Initial displacement 18000 t, KG 8.80 m, KM 9.40 m and listed 3º to starboard. 400 tonnes of cargo remains to be loaded where space is available in a tween deck Kg 10.5 m, 7.0 mto port of CL and 10.0 mto starboard of CL. (Assume KM remains constant). Solution Calculate initial GM. KM 9.400 Initial KG 8.800 Initial GM 0.600 Caiculate GGu using: GGu = Tan Busr x GM GG» = Tan 3º x 0.600 = 0.031 m. Take moments about the keel to determine the final KG and GM (note hat all 400 t of cargo is loaded at Kg 10.5 m so treat as a single weight!). WEIGHT KG (m) [| MOMENTS (tm KM 9,490 Load 400 10.50 4200.0 KG 8.837 FINAL 18400 162600.0 om [05 Taking moments about the centre line load al 400 tin the port side. Pont moments Stbd moments Distoft CL 7.000 2800.0 o 8.0 22420 HF all 400 t were loaded into the port side space lhe ship would complete with an excess of 2242 t- m moments to port. Therefore some of this 400 t must now be shifted to the space on the starbaard side (a distance of 17.0 m). Zedo=wxd 2242 =Wx (7.0 + 10.0) 2242 = 17w w= 2242 = 131.9tto shift from port to starboard 17 as To complete upright: Load 400.0-— 131.9 268.1 tport Load 131.9 t Starboard Solution (Method 2) Calculate initial GM. KM 9.400 initial KG 8.800 Initial GM 0.600 Caiculate GGu using: GGu= Tan Busr x GM GGu = Tan 3º x 0.600 = 0.031 m. Take moments about the keel to determine the final KG and GM (note that all 400 t of cargo is loaded at Kg 10.5 m so treat as a single weightt). WEIGHT (t) KG (mi MOMENTS (t-m; tia! di; KM 2.490 roses] FINAL 18400 152600.0 GM . Taking moments about the centre line: Let x = cargo to load to port; (400 — x) = cargo ta load to starboard. Dist off Port Sthd Weight CL moments (t-m]) moments (tm 400 - x; 10.000 (4000 - 10x, To complete upright: Port moments must equal starboard moments. Therefore: 7x = 558 + (4000 — 10x) 7x = 558 + 4000 — 10x 7x + 10x = 558 + 4000 17x = 4558 x =4558 7 x=268.1 tto port 400- 268.1 = 131.9tto starboard 85 11.8 LIST AND FREE SURFACE EFFECT Consider figure 11.8. The basic list triangle is GG;M. GM is the solid metacentric height, the GM that would exist if the ship had no slack tanks. GGu is the distance that G is off the centre line. GGy is the virtual rise of G due to tank free surfaces. Since GM is reduced to GM (the Fluid GM) it can be seen that the angte of list has increased for the same distance that G is off the centre line (GGu). The greater the free surface mamentsifree surface effect; the greafer will be the list for the same listing moments. Example 9 A ship displaces 13200 t KG 10.2 m and is initially upright. Baltast water AD 1.025 is run into a rectangular DB tank length 24 m, breadih 10 m to a sounding of 4.00 m. If the Kg of the ballast water is 2.00 m and it's transverse centre of gravity (TCG) is 5.00 mto starboard of the centre line calculate the final angle of list: (a) assuming no free surface moments; (b) accounting for free surface moments. Assume KM for the final displacement is 11.64 m. Solution mass of ballast water loaded = 24 x 10x 4 x 1.025 = 984 t (a) (assuming no FSM's) WEIGHT (8 KG (m MOMENTS (t-mi KM 11,849 Load O FINAL 14184 Ê 136608.0 GM Taking moments about ihe keel to determine finaí KG and GM. Taking moments about the centre line calculate GG. Dist off Port Stbd Weight EL moments (tm) moments (tm gas 5.0 9.0 4920.0 GGy = Net listing moments = 4920 = 0.347 m Final displacement 14184 Calculate the final list. Tan Ousr = Gu =0.347 =0.17272 GMena 2.009 Final list = 9.8º Stbd. (b) (including FSM's) FSM's = 1 x dt =24x10x1025 =2050tm 12 12 88 12.1 TERMS RELATING TO SHIP LENGTH The following terms relating to ship length should be understood. 12.1.1 Forward Perpendicutar (FP) This is the vertical line of reference that intersects the Summer Load waterline at the forward edge of the stem when the ship is on an even keel. 12.1.2 After Perpendicular (AP) This is the vertical line of reference that coincides with the after edge of the stem post, or, if no stern post, then the turning axis of the rudder. 12.1.3 Length between perpendiculars (LBP) Is the horizontal distance between the forward and after perpendiculars. It is this length that is considered when conducting trim calculations. 12.1.4 Length overall (LOA) Is the horizontal distance between the after most part and forward most part of the ship. 12.1,5 Amidships Is the mid point between the forward and aft perpendiculars. It is not ithe mid point in the length of the ship. Consider Figure 12.1. LOA LBR Amkinhipe Fig. 12.1 89 12.2 DRAUGHT MARKS AND READING THE DRAUGHT The draught marks on a ship should be marked at the Forward and Aff o 2 Perpendiculars on both sides. The numerals 2 fem are 10 centimetres in height and are spaced fo ema 10 centimetres apart as shown. 5.0 j 5.0 em O as 8 em “70 Bom The draught is read as shown in Figure 12.3 using the fower edge of the numerals. Intermediate values have to be estimated. If the water is quite choppy then great accuracy will not be possible. E 2 Rm 430 4.0 40 cs 8 Fig. 122 Fig. 12.3 Example 1 What is the draught reading for each of the waterlines shown in Figure 12.4? Solution 2 (a) 280m 3.0 (b) 2.50m “8 ta) (c) Approximately 2.37 m (d) — Approximately 1.82 Mm raise madie— (2) (o) 2.0 Ideally the draughts should be read on both sides of the ship and the mean draught forward and the mean draught aft determined. — Boa For obvious reasons this is rarely done so before the draughts are read the ship should be brought to the upright condition to liosuzasisarnssao| eliminate errors. E Fig. 12.4 At the outset it was stated that the draught marks should be in line with the forward and after perpendiculars but this will never be so. At the after end the curvature of the stern may make the draught marks difficult to ses. At the forward perpendicular there is nothing to mark them on! Therefore it is usual to set them a suitable distance forward and aft of the respective perpendiculars whereby the readings obtained will have to be corrected to the perpendicular. 90 123 TRIM Trim is the difference in centimetres or metres between the forward and aft draughts, as measured at the forward and aft perpendiculars. Consider the ship shown in Figure 12.5 with draughts Fwd. 2.20 m and Aft 2.68 m. ú Fig. 12.5 The trim of the ship is: 2.68 - 220 0.48 m by the stern, or; 48 cms by the stern. The same ship is now floating with draughts Fwd 2.70 m and Aft 2.32 m. Fig. 126 The trim ofthe ship is: 2.70- 2.92 0.38 m by the head, or; 38 ems by the head. 93 12.5 MOMENT TO CHANGE TRIM BY QNE CENTIMETRE (MCTC) This is the trimming moment required to change the ships trim by exactly one centimetre. ft is tabulated in the ships hydrostatic particulars and is used to determine the change of trim that takes place when weights are shifted, loaded or discharged. Consider the ship previously shown Figures 12.7-12.9 where a weight was shifted aft along the deck. The change of trim can be calculated by the formula: COT (cms) = Trimming moment MCTC where the trimming moment is: wxd 'w"being the weight shifted, and *d” being the distance through which the weight is shifted longitudinally. Thus: COT (cms)= wxd MCTC Example 4 A weight of 150 tonnes is moved aít by distance of 20 m. ff the MCTC for the current draught is 250 tm determine the final trim of the ship if the initial trim was 0.20 m by the ste. Solution COT(ems)=wxd =150x20=12cms MCTC 250 CcOT=0.120m initial trim: 0.200 m by the stem cor: 0.120 m further by the stern Final trim 0.320 m by STERN 94 12.6 FORMULA FOR CALCULATING MCTC In practice the MCTC value will always be found for the draught in question in the ship's hydrostatic particulars. However, in examinations it may have to be calculated and the formula for calculating MCTC is: MCTC = WxGM, 100LBP where: *Wis the ship's displacement; 'GMV'is the longitudinal metacentric height, &; “LBP"is the length between perpendiculars. The derivation of this formula is as follows. The shipin Figure 12.10 is on an even keel with a weight on deck. não The weight is shifted aft along the deck through [o db f distance “d' metres. In accordance with the Do SÍ . formula: GGi=wxd à wW Fig. 12.10 G will move aft to G; (parallel to and in the same direction as the shift of the weight). Fig. 12.11 G and B become horizontally separated creating a trimming lever. This causes the ship to trim by | the stern until B attains a new position vertically below the new longitudinal centre of gravity, G (Figure 12.12). Fig. 12.12 GG; Mh is a right angled triangle where: Tan 8 = OPP ADJ Since: aa then: Also, in Figure 12. therefore: Tan6=6GG, GM 1=wxd and; TanB=GG, w GM Tan6=wxd Wx GM 1& Tan 6 = TRIM (m) LBP (m) Fig. 12.13 lfthe change of trim due to the weight shifted is exactly 1 cm, then: Tan8=0.01 (m) LBP (m) Since: Tanô=wxd Wx GM which equals] Tan8= 1 100LBP equals: Tan6 = MCTC Wx GM (1) (2 (because (w x d) is the moment to change the trim by exactly 1 cm.) Bringing formulae (1) and (2) together gives: 1 = MCTC 100LBP Wx GM Rearranging this gives: MCTC=WxGM, 100LBP 95