NAVAL ARCHITECTURE 1 Class Notes d G G 1 W tonnes d G 1 G w tonnes d G 1 G w tonnes Omar bin Yaakob
Chapter 1 Introduction Naval Architecture Notes Introduction To carry out various activities at sea, rivers and lakes, man uses various types of marine structures, fixed and floating. The structures must be designed and built in various sizes, shapes and sophistication. Some of them are small and simple such as a canoe or a raft while others are large and complicated such as an aircraft carrier or a semi-submersible oil drilling platform. Naval architecture is an engineering field covering the technology in design of ships and floating structures. The persons having this expertise are called naval architects. To build these structures, shipbuilders requires design plans and guidelines prepared by naval architects. Knowledge in naval architecture is used to carry out design calculation and to produce plans which can be used by the shipyards. Although man has been using marine transport for a long time, not all these vehicles are designed and constructed using naval architecture knowledge. In fact the discipline of knowledge on ship design and naval architecture only appeared in the seventeenth century. Prior to that, shipbuilding is not based on science and technology but rather on the skills of the master craftsmen. This dependence on master craftsmen for shipbuilding can be traced back to the earliest civilization of Egypt, Greek and China. Similarly the war ships and exploration vessels built by the Romans, Muslims as well as the European colonial powers were not built using scientific methods. By the seventeenth century a number of scientists and engineers tried to apply science and mathematical methods in ship design. Among the earliest was sir Anthony Deane who wrote Doctrine of Naval Architecture in 1670. Among others, he put forward a method to determine the draught of the ship before it was built. Since then, a number of scientists and engineers continued to study and document various fields of naval architecture. In 1860, a professional body comprising of naval architects was formed under the name Institution of Naval Architects. A hundred years later the name was changed to Royal Institution of Naval Architects. A naval architects works to determine the size and shape of a ship tailored to its intended use. In addition, he estimates its stability, propulsive power as well as calculates the size and strength of its structure and the impact of waves on the vessel. The types of machinery and equipment to be installed, materials to be used and layout of ship are also determined based on naval architectural knowledge. Ship hydrostatics and stability is one of the most important subject in Naval Architecture. The safety of ships, crew, passengers and cargo will be jeopardised if ships are not stable. In this book, readers will be able to appreciate the basic terminologies, carry out simple hydrostatics calculations and will be equipped with basic tools to assess stability of vessels. Omar bin Yaakob, July 008
Chapter Ship Types, Basic Terms, Terminologies and Symbols Naval Architecture Notes 1. Types of Ships Ship types can be classed according to: 1. No of Hull a) Monohull/Single hull b) Multi-hull Catamaran Trimaran Quadramaran Pentamaran. Shape of hull form a) Roundbilge b) Chine Single Chine Multiple Chine 3. How the body is supported in water a) Hydrostatic b) Hydrodynamic c) Aeropowered Lift 4. Its function/mission a) Transport Tanker Bulk Carrier Containership Passenger ship General Cargo LNG Carrier b) Navy Aircrft Carrier Submarine Frigate Destroyer Patrol Craft Minesweeper Omar bin Yaakob, July 008 3
Naval Architecture Notes c) Work/Service Vessels Tugs Supply boat Crew Boats Heavy Lift Crane ships Fuel Supply Ships Fishing Boat Fire Fighting Boats Rescue Boats d) Leisure Vessels Cruise ships Tourist Boats Water Taxi Boat Houses Omar bin Yaakob, July 008 4
. Basic Terms, Terminologies and Symbols Naval Architecture Notes After perpendicular(ap) This is represented by a line which is perpendicular to the intersection of the after edge of the rudder-post with the designed load water-line. This is the case for both single- and twin-screw merchant ships. For some classes of warships, and for merchant ships having no rudder-post, the after perpendicular is taken as the centre-line of the rudder stock. Amidships ( ) This is the point midway between the forward and after perpendiculars. Breadth (B) Bilge BM L This is the maximum beam, or breadth, of the ship measured at amidships. This is the rounded plating at the lower corners between the vertical shell plating and the outer bottom plating. Longitudinal metacentric radius measured from centre of buoyancy BM T Block coefficient (C B) Camber or round of beam Coefficients of form Centre of flotation (F) Centre of buoyancy (B) Centre of gravity (G) Transverse metacentric radius measured from centre of buoyancy This is a measure of the fullness of the form of the ship and is the ratio of the volume of displacement to a given water-line, and the volume of the circumscribing solid of constant rectangular cross-section having the same length, breadth and draught as the ship. ie: C B = (L x B x T) The L PP is normally used in calculating the value of C B which varies with the type of ship. Fast ships 0.50-0.65 (fine form) Ordinary ships 0.65-0.75 (moderate form) Slow ships 0.75-0.85 (full form) This is the transverse curvature given to the decks, and is measured by the difference between the heights of the deck at side and centre. The amount of camber amidships is often one-fiftieth of the beam of the ship. Form is used as a general term to describe the shape of the ship's hull; and when comparing one ship's form with another, the naval architect makes use of a number of coefficients. These coefficients are of great use in power, stability, strength and design calculations. Examples are Cb, Cp, Cw etc. This is the centre of the area, or centroid, of the water-plane of a ship. For small angles of trim consecutive water-lines pass through F. The location is normally on the centerline and longitudinally the distance from AP or amidships is referred to as LCF This is the centroid of the underwater form of a ship, and is the point through which the total force of buoyancy may be assumed to act. Its position is defined by: (a) KB the vertical distance above the base, sometimes referred to as VCB (b) LCB the longitudinal distance measured either from amidships or AP or FP. This is the point through which the total weight of the ship may be assumed to act. It also is defined by: (a) KG the vertical distance above the base (b) LCG the longitudinal distance measured either from amidships or AP or FP Omar bin Yaakob, July 008 5
Naval Architecture Notes C P Prismatic coefficient, C P = A M x L Depth (D) This is the vertical distance between the base line and the top of the uppermost continuous deck measured at the side amidships. Draught (T) This is the depth of immersion from the keel to any waterline. Displacement This equals the volume ( ) or weight ( ) of water displaced by the hull. Displacement as a volume ( ) Displacement as a weight ( ) Deadweight Displacement tonnage Entrance and run Forward perpendicular (FP) Freeboard GM L GM T Gross tonnage (GRT) Heel () I L I LCF I T KM L KM T Length between perpendiculars (L PP) Length on the designed load water-line (L WL) This is volume of water displaced by the ship. It can be imagined as the volume of the hole in the water occupied by the ship measured in cubic metres. This is the weight of water displaced by the ship. It equals the volume displaced multiplied by a constant representing the density of water, ie: In fresh water = x 1000 kg/m³ In sea water = x 105 kg/m³ Weight (or mass) displacement equals the total weight of the ship when the ship is at rest in equilibrium in still water. This is the difference between the weight displacement and the lightship weight. This is the measure of a ship's capacity to carry cargo, fuel, passengers, stores, etc, expressed in tonnes. The size of tankers is often given in terms of deadweight tonnage, which is the design deadweight. Ships are usually chartered on the deadweight tonnage. This represents the designed total weight of the ship. It is the sum of lightship weight and deadweight. The size of warshipsand government ships is always given in terms of displacement tonnage. These are the shaped underwater portions of the ship forward and aft of the parallel middle body. This is represented by a line which is perpendicular to the intersection of the designed load water-line with the forward side of the stem. This may be considered to be the height amidships, of the freeboard deck at side above the normal summer load water-line. Longitudinal metacentric height measured from centre of gravity Transverse metacentric height measured from centre of gravity This is a measure of the total volume of enclosed spaces in a ship including the under-deck, 'tween-deck spaces and enclosed spaces above the upper deck. The size of most ordinary merchant ships is quoted in terms of gross tonnage. Although it unit is tons, it must be remembered that it is a measure of volume, not weight. 1 ton = 100 ft3. This is the amount of inclination of the ship in the transverse direction, and is usually measured in degrees. Longitudinal moment of inertia of waterplane about amidship Longitudinal moment of inertia of waterplane about F Transverse moment of inertia of waterplane about centreline Height of longitudinal metacentre above keel line Height of transverse metacentre above keel line This is the horizontal distance between the forward and after perpendiculars. This is the length, as measured on the water-line of the ship when floating in still water in the loaded, or designed, condition. Length overall (L OA) This is the length measured from the extreme point forward to the Omar bin Yaakob, July 008 6
Lightship weight extreme point aft. Naval Architecture Notes This equals the weight of an empty ship i.e. a ship without load. It is fully equipped and ready to proceed to sea, but with no crew, passengers, stores, fuel, water, or cargo on board. The boiler or boilers, however, are filled with water to their working level. MCT 1CM Moment to change trim 1 cm, MCT 1CM = GM L BM L = pi L 100 L 100L 100L Midship section This is the transverse section of the ship amidships. Base line Midship section area coefficient(c M) Net or register tonnage Parallel middle body (L P) Prismatic coefficient (C P) Rise of floor Sheer Trim Tumble-home Tonnes per centimetre (TPC) Water-plane area coefficient(c WP) This represents the lowest extremity of the ship. At the point where this line cuts the midship section a horizontal line is drawn, and it is this line which acts as the datum, or base line, for all hydrostatic calculations. Normally, this is the underside of keel. This is the ratio of the immersed area of the midship section to the area of the circumscribing rectangle having a breadth equal to the breadth of the ship and a depth equal to the draught. ie: C M = A M (B x T) C M values range from about 0.85 for fast ships to 0.99 for slow ships. This represents the tonnage of a ship after certain approved deductions, ie nonfreight earning spaces, have been made from gross tonnage. A register ton represents 100 cubic feet of volume. This is the length over which the midship section remains unchanged. This is the ratio of the volume of displacement of the ship to the volume of the circumscribing solid having a constant section equal to the immersed midship section area A M, and a length equal to the L PP i.e. C P = (A M x L) The Cp is a measure of the longitudinal distribution of displacement of the ship, and its value ranges from about 0.55 for fine ships to 0.85 for full ships. This is the amount by which the line of the outer bottom plating amidships rises above the base line, when continued to the moulded breadth lines at each side. This is the curvature given to the decks in the longitudinal direction, and is measured at any point by the difference between the height at side at that point and the height at side amidships. This is the difference between the draughts forward and aft. If the draught forward is greater than the draught aft it is called trim by the head, or bow. If the draught aft is greater, it is called trim by the stern. This is the amount by which the midship section falls in from the halfbreadth line at any particular depth. This is the mass which must be added to, or deducted from, a ship in order to change its mean draught by 1 cm. This is the ratio of the area of the water-plane to the area of the circumscribing rectangle having a length equal to the L PP and a breadth equal to B. ie: C WP = A W (L x B) The range of values is from about 0.70 for a fine ship to 0.90 for a full ship. Omar bin Yaakob, July 008 7
Naval Architecture Notes Courtesy http://www.dynagen.co.za/eugene/hulls/terms.html Exercise: Visit these websites and get acquainted with more ship terms: 1. http://www.midwestconnection.com/glshpng/glossary.htm. http://www.scribd.com/doc/180086/ship-terms-glossary 3. http://cruises.about.com/od/cruiseglossary/cruise_ship_and_na utical_term_glossary.htm 4. http://www.islandregister.com/terms.html 5. http://phrontistery.info/nautical.html Omar bin Yaakob, July 008 8
Chapter 3 Hydrostatics and Floatation Naval Architecture Notes 3.1 Archimedes Law of Floatation Archimedes (born 87 B.C) Law states that An object immersed in a liquid experience a lift equivalent to the mass of liquid the object displaces. A man immersed in water for example will feel a weight reduction because part of the weight is supported by buoyancy. This buoyancy is equal to the weight of water displaced by his immersed body. 3. Reduction of Weight of Immersed Objects The maximum buoyancy is when the object is fully immersed and this equal the total outside volume of the object multiplied by the density of the fluid. When maximum available buoyancy is less than the weight of the object, the object will sink. That is why an anchor will sink to the bottom. However the object will still feel the weight reduction. Example 3.1: Consider a cuboid having dimensions 1m x 1m x m. If it weighs 3 tonnes in air, what is its apparent weight in water density 1000 kg/m 3? If the object is immersed in liquid, it will displace liquid around it equivalent to its external volume.? tonne In this case, displaced volume = 1 x 1 x = m 3 This is the volume of liquid pushed aside by the cuboid. Archimedes says that the weight of this object in liquid is reduced due to the support given by liquid on the object. The apparent weight equals the weight in air minus the reduction in weight of the object; or the buoyancy i.e. m 3 Buoyancy = volume diplacement x density of liquid = mass displacement = m 3 x 1000 kg/m 3 = 000 kg = tonnes = reduction in weight Omar bin Yaakob, July 008 9
Apparent weight = weight in air buoyancy Naval Architecture Notes Since the object weighs 3 tonne in air, it will apparently weigh only 1 tonne in water. Exercise 3.1 Do similar calculations to find out the apparent weight in oil (density 0.85 tonne/m 3 ) and muddy water (density 1.3 tonne/m 3 ) and mercury (density 13,000 kg/m 3 ) Fluid Oil Density ( ) Fluid Support ( ) Apparent Weight ( ) Fresh Water Muddy Water Mercury What can be concluded about relationships between buoyancy of objects and the densities of fluids in which they are immersed? 3.3 What make a Ship Floats? When the maximum available buoyancy is more than the weight of the object, the object will rise to the surface. It will rise to the surface until the weight of the object balances the buoyancy provided by its immersed portions. When the object is floating, its buoyancy is just enough to support its weight. At that point: Total weight W = Buoyancy = Displaced volume x liquid This principle explains why a steel or concrete ship can float. As long as the outer shell of the ship can provide enough volume to displace the surrounding water exceeding the actual weight of the ship, the ship will float. A floating ship is such that the total weight of its hull, machinery and deadweight equals to the weight of water displaced by its outer shell. If, while it is floating weights are added until the total weight exceeds the maximum buoyancy provided by the outer shell of the ship, the ship will sink. 3.4 Effect of Density An object experiences buoyancy force equivalent to the weight of fluid it displaces. For a particular object, the buoyancy force will depend on the density of the fluid, since its volume is constant. This explains for example why a bather will feel more buoyant while swimming at sea compared to in the river or lake. Also, a floating Omar bin Yaakob, July 008 10
Naval Architecture Notes object of constant weight will sink at a deeper draught in freshwater compared to in seawater. Total weight W = Buoyancy = Displaced volume x liquid Since weight does not change, the buoyancy is also constant. So displaced volume will be inversely proportional to the density of fluid. For floating object, this will determine its level of sinkage or draught. 3.5 Some Simple Problems The fact that a floating object displaces fluid equivalent to its weight can be used to solve a number of problems. Total weight W = Buoyancy = Displaced volume x water From this equation, we can obtain the weight of the object if we know the volume of water displaced. On the other hand, if we know its weight, we can work out its displaced volume. Just to understand the concept, consider a floating box of dimension L x B x D, floating at a draught T. CASE 1: We know its weight, we can find its draught In this case, we know the weight of the object, we can find the displaced volume: Displaced volume = W water i.e. for a box-shaped vessel: Displaced volume = L x B x T Hence draught T of the cuboid can be found. Omar bin Yaakob, July 008 11
Example 3. Naval Architecture Notes A cuboid shaped wooden block (L x B x D) 1.45m x 0.5m x 0.5m floats in water. If the block weighs 0.154 tonnes, find its draught if it floats in freshwater density 1.00 tonne/m 3. Solution: The weight of the block of 0.154 tonnes must be supported by displaced water i.e. the block must displace 0.154 tonnes of water: In fresh water, Volume of displaced water = L x B x T Weight of displaced water = x FW = 1.45 x 0.5 x T x FW This must equal 0.154 tonne 1.45 x 0.5 x T x fw = 0.154 tonnes T = 0.1 m Exercise 3. Do similar calculations for salt water (density 105 kg/m 3 and oil density 0.85 tonne/m 3 ) CASE : If we know its draught, we can know its volume displacement, we can find its weight If we know the draught of the cuboid, we can find its volume displacement and hence the weight of the object; Say if we know its draught T, volume displacement = L x B x T Weight = Buoyancy = Volume Displacement x water Weight = L x B x T x water Example 3.3 A box barge length 100m breadth 0m floats at a draught of 5m in sea water 1.05 tonne/m 3. Find its weight. Solution While floating in sea water density 1.05 tonne/m 3 : Volume Displacement = = L x B x T Weight of barge = Weight displacement, W = = x salt water = 100 x 0 x 5 x 1.05 = 1050 tonnes Omar bin Yaakob, July 008 1
Naval Architecture Notes Exercise 3.3 A block of wood length 5m, breadth 0.5m and depth 0.m is floating in seawater at a draught of 0.1m. Find the weight of the block. Exercise 3.4 Find the new draught of the box in example 3.3 when it goes into river, water density 1.000 tonne/m 3. Also find a new draught if it is in sea water with density 1.100 tonne/m 3. Exercise 3.5 1.0m A cylindrical container weighing 5 tonne floats with its axis vertical. If the diameter is 1.0m, find its draught in: i. sea water ii. oil of density 870 kg/ m 3. Exercise 3.6 A cylindrical tank diameter 0.6m and mass 00kg floats with its axis vertical. Find its present draught in oil ( = 0.95 tonne/m 3 ). Find the weight of cargo to be added to ensure it will float at a draught of 0.85m. 3.6 Hydrostatic Particulars A floating object will be at a certain draught depending on the total weight of the object, density of water and the shape of the object. For a ship, the shape of the object has strong influence on the draught of the ship; the shape and draught have to provide enough buoyancy to support the ship. Omar bin Yaakob, July 008 13
Naval Architecture Notes When a ship is floating at a certain draught, we can find the mass displacement and weight of the ship if we can find its displaced volume. Also we can know its waterplane area, calculate its TPC, KB, C b etc. These particulars which are properties of the immersed part of the ship are called hydrostatic particulars. Examples of hydrostatic particulars are:,, KB, LCB, A w, BM T, BM L, TPC, C B, C P, C M, C W, LCF, MCTC, WSA These particulars describe the characteristics of the underwater portion of ship at a particular draught. It is related to volumes, areas, centroids of volumes and areas and moments of volumes and areas of the immersed portion. If the ship is out of water, and draught becomes zero, the particulars ceased to exist. As long as draught and trim is maintained, the size and shape of the underwater immersed parts of the ship remains the same. The volumes, areas and moments of areas and volumes remain the same. Once draught or trim changes, the particulars will also change. This change in draught will normally occur due to changes in total weight of the ship, or if a force is applied to the ship to make it sink to a deeper draught. Example 3.4 A box m x 1m (LxB) in sea water is floating at a draught of 0.3m. Calculate its,, C B, C WP and TPC. i. = L x B x T = x 1 x 0.3 = 0.6m 3 ii. = L x B x T x = x = 0.6 x 1.05 = 0.615 tonnes iii. C B = = 0.6 = 1.00 L x B x T 0.6 iv. C WP = A wp = x 1 = 1.00 L x B x 1 v. TPC = A wp x = x 1 x 1.05 = 0.005 100 100 Exercise 3.7 Calculate the particulars at draught of 0.4, 0.5, 0.6 and 0.7m. Exercise 3.8 Find hydrostatic particulars in sea water (,,Awp,LCB, LCF,TPC) of a box barge with dimension L=100m, B=0m, at draughts of 1.0m, 3.0m, 5.0m, 7.0m, 9.0m. If the barge weighs 300 tonne, what is its draught? If the barge is floating at a draught of 4m, what is its C B? Omar bin Yaakob, July 008 14
Naval Architecture Notes It can be seen from Exercise 3.8 that for a box-shaped object at different draughts, the waterplane areas are constant. Hence, many hydrostatics particulars remain constant. Exercise 3.9: An empty cylindrical shaped tank is floating in sea water (density 1.05 t/m 3 ) at a draught of 8.0 m with its axis vertical. The external diameter of the tank is 1.0 m, internal diameter 11.0 m, thickness of base 1.0 m and the overall height is 16.0 meter. Its centre of gravity is 6 meter above its inner base. Calculate:. i. Find Hydrostatic particulars, Awp, LCB, Cb, Cp, TPC, WSA at T=1,, 4, 6, 8m. ii. Plot hydrostatic curves similar to page 19 showing all data. iii. Final draught of the tank after 500 m 3 diesel oil (density 850 kg/m 3 ) is poured into the tank. The second moment of area of a circle about its diameter is D 4 64. 3.7 Hydrostatic Particulars of a Ship Hydrostatic particulars of a real ship will be different. Consider the ship whose lines plan is shown below. At different draughts, the ship will have different waterplane areas, volumes and centroids. Hence, the hydrostatic particularly will vary as the draughts changes. Omar bin Yaakob, July 008 15
Naval Architecture Notes If areas, volumes, moments, centroids of the waterplanes and sections of the ships can be calculated, hydrostatic particulars of a ship can be obtained. These are calculated at the design stage, once the shape and size of the ship has been decided. Exercise 3.10 A ship with length 100m, breadth m has the following volumes and areas at different waterlines. Calculate its, C B, C W and TPC in saltwater density 1.05tonnes/m 3. Draught (m) Aw (m ) (m 3 ) 1800.0 3168.0 4 000.0 6547. 6 100.0 10137.6 8 10.0 1378.0 10 130.0 1744.0 (tonnes) x ro Cb Cw TPC LBT Aw (LB) Aw x ro 100 The particulars can be presented in two forms, either as a set of curves or in tabular format. Table 3.1 shows a typical table of hydrostatic particulars while an example of hydrostatic curves is shown on page 18. Draught (m) Table 3.1 Hydrostatic Particulars of Bunga Kintan LBP 100m Displacement (tones) Cb KB (m) BMT (m) BML (m) MCTC (tonne-m) LCB (m from ) LCF (m from ) 8.00 1480.00 0.7 4.07 3.66 180.00 190.00.50.00 7.50 13140.00 0.71 3.67 3.98 195.00 183.00.30 1.50 7.00 11480.00 0.70 3.6 4.46 19.00 180.00.00 0.70 6.50 9870.00 0.69.85 5.0 44.00 17.00 1.80-0.06 6.00 880.00 0.67.44 5.66 79.00 165.00 1.50-1.00 5.50 6730.00 0.66.04 6.67 37.00 157.00 1.10 -.00 5.00 50.00 0.64 1.63 8.06 39.00 146.00 0.00-3.00 Omar bin Yaakob, July 008 16
Naval Architecture Notes 3.8 Using Hydrostatic Curves and Tables Hydrostatic curves and tables can be used to obtain all hydrostatic particulars of a ship once the draught or any one of the particulars is known. Example 3.5 From MV Bulker hydrostatic Curves (pg18) at a draught of 7m, we can obtain displacement = 31,000 tonnes, LCF =.0m forward of amidships and MCTC = 465 tonne-m etc. Also if we know the ship weighs 40,000 tonnes, its draught, TPC, MCTC, LCF and LCB can be obtained. Exercise 3.11 Using MV Bulker Hydrostatic Curves, find displacement, LCB, LCF, TPC at draught of 9.5m. If 1500 tonnes is added to the ship, what is its new draught? Hydrostatic tables can be used in a similar manner to obtain hydrostatic particulars once draught is known or to obtain draught and other particulars once the displacement or another particular is known. There is however a need to interpolate the table to obtain intermediate values. HOMEWORK 1: By using the hydrostatic particulars of Bunga Kintan shown in Table 3.1: i. Draw full hydrostatic curves of the ship ii. Find values of displacement, KB, LCB, BM T, BM L, MCTC, C B, LCF of the ship if it is floating at a draught of 6.75m. iii. Find values of T, KB, LCB, BM T, BM L, MCTC, C B, LCF of ship if the ship weighs 11,480 tonnes. iv. When the ship is floating at a draught of 5.5m, 3000 tonne cargo was added. What is its new draught? Submission Date: Exercise 3.11: Calculate,, KB, LCB, A w, TPC, C B, C P, C M, C W, LCF of a cylinder radius 1m floating with axis vertical at draughts of 1.0, 1.5,.0 and.5m. Omar bin Yaakob, July 008 17
Naval Architecture Notes MV Bulker Hyrostatic Curves Omar bin Yaakob, July 008 18
4.1 Introduction Chapter 4 Basic Stability Consideration Naval Architecture Notes One of the factor threatening the safety of the ship, cargo and crew is the lost or lack of stability of the vessel. Stability calculation is an important step in the design of the ship and during its operation. While designing the ship, the designers must be able to estimate or calculate to check whether the ship will be stable when constructed and ready to operate. For the ship's master, he must be able to load and stow cargo and handle the ship while ensuring that the ship will be stable and safe. 4. What is stability? Stability is the tendency or ability of the ship to return to upright when displaced from the upright position. A ship with a strong tendency to return to upright is regarded as a stable vessel. On the other hand, a vessel is said to be not stable when it has little or no ability to return to the upright condition. In fact, an unstable ship may require just a small external force or moment to cause it to capsize. Figure 4.1 (a) (b) (c) An analog y for stability is often given of the marble. In Figure 1 (a), the marble in the bowl will return to its original position at the bottom of the bowl is it is moved to the left or the right. This marble is in a condition called positively stable. A slight push on the marble which is put on an upside down bowl as in Figure 1 (b) will cause it to roll off, a condition equivalent to instability. A neutrally stable ship is analogous to a marble put on a flat surface, it will neither return nor roll any further. 4.3 Longitudinal and Transverse Stability Ship initial stability can be seen from two aspects, longitudinally and transversely. From longitudinal viewpoint, the effect of internal and external moments on ship's trim is considered. Important parameters to be calculated are trim and the final draughts at the perpendiculars of the ship. In any state, there is a definite relationship between trim, draughts and the respective locations of the centres of buoyancy and centre of gravity. The trim angle is rarely taken into consideration. Transverse stability calculation considers the ship stability in the port and starboard direction. We are interested in the behaviour of the ship when external statical moment is applied such as due to wind, waves or a fishing net hanging from the side. The effect of internally generated moment such as movement of masses on-board transversely is also studied. An important relationship considered is that Omar bin Yaakob, July 008 19
Naval Architecture Notes between heeling and righting moments and the resulting angle of heel and its consequence on the safety of the boat. This Chapter will focus on basic transverse stability particularly the relationships between the metacentre and the centre of gravity. 4.4 Basic Initial Stability: The role of GM M T w M T w G W L G w 1 L 1 W B K W K B W B 1 L (a) (b) Figure 4. Consider the ship floats upright in equilibrium as in the above figure 4. (a). The weight of the ship equals its displacement and the centre of buoyancy is directly below the centre of gravity. When the ship is slightly disturbed from upright, the centre of buoyancy being centre of underwater volume moves to the right. The line of action of buoyancy vertically upward crosses the original centreline at the metacentre, M. Since G does not move, a moment is generated to turn the ship back to its original position. This moment is called the returning moment. In this case, M was originally above G and we can see that the returning moment is positive. If M was below G i.e. GM negative, the returning moment will be negative hence the ship is unstable. If M is at G, then the ship is neutrally stable. Righting moment is the real indication of stability i.e. the ability of the ship to return to oppose any capsizing moment and return the ship to upright position. The larger the righting moment, the better stability is. Omar bin Yaakob, July 008 0
Naval Architecture Notes Consider the triangle shown below: M T Righting moment = x GZ and GZ = GM T sin G Z For any displacement, righting moment depends on GZ. And GZ depends on GM. The bigger GZ, the bigger Righting Moment. M T Relationships between K, B, G and M T are important. G B KM T = KB + BM T K KM T = KG + GM T For any particular draught or displacement at low angle of heel, keel K and metacentre M are fixed. Therefore the values of KB, BM and hence KM are fixed, as can be obtained from hydrostatic particulars. Therefore the distance GM T will only depend on the height of centre of gravity. In other words, to ensure a large GM T, we can only control KG. 4.5 Determining Centre of Gravity, Areas or Volumes of Composite Bodies The above section has shown that the relative position of M and G are important in determining ship stability. Since M is constant for any particular draught, only G will finally determine the value of GM. Before we go into the details of stability calculations, we have to consider how to determine the location of G. Consider a composite body consisting of two portions shown in Figure 4.3. Area A ca C cb Area B Figure 4.3 xa X xb Omar bin Yaakob, July 008 1
Naval Architecture Notes Distance of Centre of Composite to the reference axis: i.e., X = A x xa + B x xb A + B X = Total moment of area about the reference axis Total area If the composite consists of volumes, Centre of Volume X = Total moment of volume about the reference axis Total volume If the composite consists of weights, Centre of Gravity X = Total moment of weight about the reference axis Total weight Example 4.1 Find centre of area (from AP) for an object consisting of four components shown in the figure below. Component Area (m ) Distance from AP (m) 1 -.5 3 4 TOTAL Moment of Area about AP (m 3 ) Centroid from AP = Total moment of area about AP Total area = m Omar bin Yaakob, July 008
Example 4. Naval Architecture Notes A trimaran has three hulls and the respective volume displacements, LCB and KB are shown below. Find the total displacement, LCB and KB. Hull Volume Displacemen t (m 3 ) Lcb (m aft of amidships) Side 1 158.7 13.0.5 Main 1045.8.0.0 Side 158.7 13.0.5 Kb (m above keel) Hull Volume Displace ment (m 3 ) lcb (m aft of amidships) Moment about amidships (m 4 ) Side 1 158.7 13.0.5 Main 1045.8.0.0 Side 158.7 13.0.5 TOTAL Kb (m above keel) Moment about keel (m 4 ) LCB= Total moment about amidships = 4.56 m aft of amidships Total Volume KB= Total moment about keel =.1 m Total Volume Example 4.3 A stack of weights consists of one 3kg weight and two kg weights. Find centre of gravity of the stack above the floor: Omar bin Yaakob, July 008 3
Item Weight (kg) CG above floor (cm) Wt A Wt B Wt C JUMLAH Naval Architecture Notes Moment about Floor (kgcm) Final CG = = cm Example 4.4 A ship has three parts and the respective weights and Kg are as follows. Find the total weight and KG. Part Weight (tonnes) Kg (m above keel) Lightship 000 5.5 Cargo 1 300 7.6 Cargo 500.5 Part Weight (tonnes) Lightshi 000 5.5 p Cargo 1 300 7.6 Cargo 500.5 TOTAL Kg (m above keel) Moment about keel (tonne-m) KG = Total moment about Keel = Total weight m Example 4.5 A ship of 6,000 tonnes displacement has KG = 6 m and KM = 7.33 m. The following cargo is loaded: 1000 tonnes, Kg.5 m 500 tonnes, Kg 3.5 m 750 tonnes, Kg 9.0 m The following cargo is then discharged: 450 tonnes of cargo Kg 0.6 m And 800 tonnes of cargo Kg 3.0 m Find the final GM. Omar bin Yaakob, July 008 4
Naval Architecture Notes Item Weight (tonne) Kg Moment about keel (tonne-m) Ship Loaded Cargo1 Cargo Cargo3 Unloaded Cargo 6000 1000 500 750-450 -800 6.0.5 3.5 9.0 0.6 3.0 Final moment Final KG = Final displacement = Final KG = m Final KM = m Final KG = m Ans. Final GM = m Homework A box-shaped barge is floating in sea water at a draught of 5m. The extreme dimensions of the barge (L x B x D) are 1m x 11m x 10m. The wall and floor are 0.5m thick. Its centre of gravity is 4m above keel. Calculate: i. The displacement and GM T of the empty barge. ii. The barge is to be used to carry mud (density1500 kg/m 3 ). If the draught of the barge cannot exceed 7.5m, find the maximum volume of mud that can be loaded into the barge. iii. For the barge loaded as in (ii), find its GM T. Submission Date: Omar bin Yaakob, July 008 5
4.6 Movement of Centre of Areas, Volumes and Weights Naval Architecture Notes When a portion is added or removed from an object, its centre moves. Consider a homogenous object as shown below: i. If weight is moved a distance y: Centre of gravity moved x = GG = m x y M i.e. total moment divided by total weight ii. If weight m is removed: The remaining weight M-m Movement of centre of gravity x = GG = m x d M-m i.e. total moment divided by remaining weight. Example 4.6 A ship weighing 7000 tonnes is floating at the wharf. At that time, KM = 6.5 m and GM 0.5m. Find new GM when a 30 tonnes box is loaded at Kg 10.0m. Assume no change in KM. Method 1: Find rise in KG Original KG = KM - GM = m Distance 30 tonnes box from original G = GG = 30 x 4.0 = 0.017m 7030 KG = KG+ GG = KM does not change, therefore, GM = = m m m M G G B K 10 Omar bin Yaakob, July 008 6
Naval Architecture Notes Method : Find final KG using table of moment about keel Portion Mass (m) Kg (m) Moment about keel (tonne-m) Ori. Ship 7000 6.0 Box 30 10.0 Total 7030 KG = Sum of moment Sum of weight KG = m GM = KM - KG KM - KG = m 4.7 Hanging Weights, The Use Of Derricks And Cranes The use of cranes and derricks will make the weights suspended. Suspended weights acts at the point of suspension. Therefore a weight that was initially located on the lower deck for example will instantly be transferred to the point of suspension at the instant the weight is lifted by the derrick. The centre of gravity KG will suddenly increase and since KM is constant, GM will reduce suddenly. If the rise in KG is more than the original GM, the net GM will be negative, leading to instability. Example 4.7 A ship of 7,500 tonnes displacement is upright and has GM 0.0m and KM 6.5 m. A heavy cargo of 100 tonnes already on the lower deck (kg=m) is to be unloaded using the ship s crane. When lifting the cargo crane head is 15 m above keel. What is GM during lifting. Comment of the safety of the operation. Treat as if the weight is suddenly transferred from lower deck to the point of suspension, a distance of 15 meters. The KG will rise, and since KM constant, GM will be reduced. Original KG = KM-GM= 6.5 0. = 6.3m Rise in KG = 100 x 13 7,500 =0.173m KG during lifting = KG + Rise = 6.473m GM during lifting = KM- Kgnew = 6.5-6.473 = 0.07m Omar bin Yaakob, July 008 7
Naval Architecture Notes 4.8 Free Surface Correction When free surface exists on board the ship, stability of ship is affected. The free surface gives rise to free surface moment which in effect reduce GM. The reduction is called Free Surface Correction (F.S.C). FSC is calculated from the second moment of area of the surface of the fluid; FSC = Free surface moment Ship displacement Free Surface Moment (FSM) = i x ρ fluid Where i the second moment of area of the surface of the fluid and ρ fluid density of the fluid being considered. is the Once the FSC is known, the new reduced GM called GM fluid is obtained GM fluid = GM solid - FSC It is important that free surface be avoided or at least minimised. Note also that KG in ships having free surface is called KG fluid and regarded increased by FSC. KG fluid = KG solid + FSC For tanks with a rectangular surface: Free surface moment = 1 x tank length x tank breadth 3 x density of fluid 1 Free surface correction = 1 x tank length x tank breadth 3 x density of fluid 1 ship displacement EXERCISE 4 1. Bunga Kintan (Hydrostatic data given on page 1) is floating at draught of 6.5m. If its KG is 6.8m, what is its GM?. A ship has a displacement of 1,800 tonnes and KG = 3m. She loads 3,400 tonnes of cargo (KG =.5 m) and 400 tonnes of bunkers (KG = 5.0m). Find the final KG..84m 3. A ship sails with displacement 3,40 tonnes and KG = 3.75 m. During the voyage bunkers were consumed as follows: 66 tonnes (KG = 0.45 m) and 64 tonnes (KG =1 m). Find the KG at the end of the voyage. Omar bin Yaakob, July 008 8
Naval Architecture Notes 4. A ship has displacement,000 tonnes and KG = 4m. She loads 1,500 tonnes of cargo (KG = 6m), 3,500 tonnes of cargo (KG = 5m), and 1,50 tonnes of bunkers (KG = 1m). She then discharges,000 tonnes of cargo (KG =.5 m) and consumes 900 tonnes of oil fuel (KG = 0.5 m.) during the voyage. If KM= 5.5m, find the final GM on arrival at the port of destination. 5. A ship arrives in port with displacement 6,000 tonnes and KG 6 m. She then discharges and loads the following quantities: Discharge 150 tonnes of cargo KG 4.5 metres 675 tonnes of cargo KG 3.5 metres 40 tonnes of cargo KG 9.0 metres Load 980 tonnes of cargo KG 4.5 metres 550 tonnes of cargo KG 6.0 metres 700 tonnes of bunkers KG 1.0 metre 70 tonnes of FW KG 1.0 metres During the stay in port 30 tonnes of oil (KG 1 m.) are consumed. If the final KM is 6.8 m., find the GM on departure. 6. A ship of 9,500 tonnes displacement has KM 9.5 m and KG 9.3 m. A load 300 tonnes on the lower deck (Kg 0.6 m) is lifted to the upper deck (Kg 11 m). Find the final GM. 7. A ship of 4,515 tonnes displacement is upright and has KG 5.4 m and KM 5.5 m. It is required to increase GM to 0.5m. A weight of 50 tonnes is to be shifted vertically for this purpose. Find the height through which it must be shifted. 8. A ship of 7,500 tonnes displacement has KG 5.8 m. and GM 0.5 m. A weight of 50 tonnes is added to the ship, location Kg = 11m and 7m from centreline to the starboard side. Find final location of G above keel and from the centreline. What is its new GM? 9. A ship has a displacement of 3,00 tonnes (KG = 3 m. and KM = 5.5 m.). She then loads 5,00 tonnes of cargo (KG = 5. m.). Find how much deck cargo having a KG = 10 m. may now be loaded if the ship is to complete loading with a positive GM of 0.3 metres. 10. A ship of 4,500 tonnes displacement is upright and has KG 5.4 m and KM 5.5 m. It is required to move a weight of 50 tonnes already on the deck (kg=6m) using the ship s derrick. The derrick head is 13 m above keel. Is it safe to do so? 11. A ship of 9,500 tonnes displacement and has KM 9.5 m and KG 9.3 m. The ship has two fuel tanks in double bottoms, rectangular shape each 0 x 5m containing bunker density 900 kg/m 3. Find GM fluid when free surface exists in the tank. 1. Find Gm fluid for the ship in question 11 but with one tank only, length 0m breadth 10m. 13. What happens to i when there are three tanks with b = 3.33m in question 11. Omar bin Yaakob, July 008 9
Appendix A Naval Architecture Notes Second Moments of Areas The second moment of an element of an area about an axis is equal to the product of the area and the square of its distance from the axis. Let da in Figure A.1 represent an element of an area and let y be its distance from the axis AB da y A Fig. A.1 B The second moment of the element about AB is equal to da x y. To find the second moment of a rectangle about an axis parallel to one of its sides and passing through the centroid l dx x A G b B Fig. A. In Figure A., l represents the length of the rectangle and b represents the breadth. Let G be the centroid and let AB, an axis parallel to one of the sides, pass through the centroid. Omar bin Yaakob, July 008 30
Naval Architecture Notes Consider the elementary strip which is shown shaded in the figure. The second moment (i) of the strip about the axis AB is given by the equation:- i= l dx x x Let I AB be the second moment of the whole rectangle about the axis AB then:- b/ 1 l. x. dx AB -b/ b/ 1 l x. dx AB -b/ 3 x l 3 b / b / 1 AB 3 lb 1 3. To find the second moment of a rectangle about one of its sides. l dx b x A Fig. A.3 B Consider the second moment (i) of the elementary strip shown in Figure A.3 about the axis AB. i= l dx x x Omar bin Yaakob, July 008 31
Let I AB be the second moment of the rectangle about the axis AB. Then :- b 1 x AB l.. dx O Naval Architecture Notes x 3 l 3 b O or 1 AB 3 lb 3 4. The Theorem of Parallel Axes The second moment of an area about an axis through the centroid is equal to the second moment about any other axis parallel to the first reduced by the product of the area and the square of the perpendicular distance between the two axes. Thus, in Figure A.4, if G represents the centroid of the area (A) and the axis OZ is parallel to AB then:- IOZ IAB - Ay A B y O Z G Fig. A.4 Omar bin Yaakob, July 008 3
Naval Architecture Notes 5. Second moment of area of a circle X A D y B X Fig. A.5 For circle, the second moment of area about an axis AB. I AB D 64 4 What is I XX? 6. Applications. Second moment of areas are used in calculations of BM L and BM T : BM L I F and BM T I T Where I F is longitudinal second moment of area of the waterplane about the centre of floatation, I T is transverse second moment of area about the centreline and is volume displacement. Omar bin Yaakob, July 008 33
Naval Architecture Notes EXERCISES 1. Find BM L and BM T of a box shaped barge 10m x 0m x 10m floating at a draught of 7m.. A cylinder of radius r = 10m is floating upright at draught of 6m in fresh water. Find its KM L and KM T. 3. A fish cage consists of a wooden platform placed on used oil drums with the following dimensions. 6m 4m If the total weight of the structure is 3 tonnes, floating in sea water calculate: i) draught ii) KM T iii) KM L Diameter 1m. Homework 3: A catamaran consists of two box-shaped hulls spaced 5m apart, centreline to centreline. Each hull measures (L x B x D) 10m x 0.5m x 1m. If its draught is 0.3m, find its : i) and ii) KB iii) BM T iv) Maximum allowable KG if GM minimum is 0.m Submission Date: Omar bin Yaakob, July 008 34
Chapter 4 Calculations of Ship Hydrostatic Particulars Naval Architecture Notes 4.1 The Importance of Hydrostatic Particulars In the previous chapter, we have seen the importance of knowing the hydrostatic particulars of a vessel. If we have the hydrostatic particulars in the form of tables, curves, or our own direct calculation, we can obtain details about the ship in any particular condition. We can also determine or estimate what would happen when ship condition changes such as due to addition or removal of weights. To draw hydrostatic curves or to make the table, we need to calculate the particulars. The hydrostatic particulars can be obtained only if we carry out calculations of area, volumes and moment at various draughts or water plane area. Using some known relationships, the particulars can be derived from areas, volumes and moments, If the body has a uniform shape, such as cuboids, cones, spheres or prisms, calculation of areas, volumes and moments are easy. For example water plane areas, block coefficients, TPC, MCTC, KB and LCB of such objects can be found using simple formulae. We can easily obtain the particulars at any draught and if necessary plot the curves. However not all ships have simple and uniform shapes as above. In fact, most ships have hull shapes which are varying in three directions. This makes it difficult to calculate hydrostatic particulars. 4. Methods to Calculate Areas Figure 4.1 Typical Ship Half-Breadth Plan Consider the shape of the ship whose body plan is shown in Figure 4.1. If we want to find the area of the section or water plane for example, we do not have simple methods. Similarly to find volume displacement or LCF will not be easy. If we want calculate the water plane area of the ship in Figure 4.1 at a particular draught, we may use a few methods. Omar bin Yaakob, July 008 35
Naval Architecture Notes First is to plot the curve on a graph paper from where the area under the curve can be obtained by counting the squares. To improve accuracy, smaller boxes or triangles can be used. The method is tedious and it s accuracy depends on the size of the smallest grid. To use this method, we need to plot the curve first; a disadvantage when sometimes we are only provided with offset data, i.e. halfbreadth at various stations. The second method is to use an equipment called the planimeter. This equipment can be used to measure the area of a shape drawn on paper. Again, this equipment can only be used only when hard copy of the waterline drawing is available. Moreover, similar to graphical method, planimeter requires a lot of man power. Figure 3. A Planimeter The third method is to use mathematical approximation. In this method, an attempt is made to represent the curve or shape by a mathematical expression. By using calculus, area and moments of the area bounded by the curve can be found by integration. Mathematical methods are normally preferred for a number of reasons. First there is no need for a hard copy of the curves. Offset tables are normally available and the data can be used directly in the calculations. A very important feature of mathematical methods is the ability to make use of the technology offered by computers. The use of mathematical methods also enable us to obtain not only areas but all hydrostatic particulars. As we have seen in chapter 3, we need to calculate not only areas but also volumes, positions of centroids of waterplanes (LCF) and centroids of volumes (KB and LCB). In addition we require second moments of areas for calculations of MCTC and metacentric heights. Unlike graphical or planimeter methods, mathematical methods can easily be used to calculate these particulars. A very important caution should be noted when using mathematical methods. The accuracy of the calculations will mainly depend on the degree of fit of the actual curve to the mathematical expression representing it. 4.3 Mathematical Methods Omar bin Yaakob, July 008 36 ST1 ST ST3 ST4 ST5 h offset
Naval Architecture Notes Figure 4.3 Waterline or Sectional Area Curve Figure 4.3 shows a curve which may represent a half-waterplane area or a curve of sectional areas. A waterplane curve is represented by offsets made up of halfbreadth at various stations. Stations are positions along the length of the ship and normally separated by a common-interval, h. To cater for the fast changing slopes of the curve at the stern and bow regions, half stations may be used. To calculate the area, centroid and moment under such curve, its offsets and h are required. By assuming that the curve can be represented by a certain mathematical formulae, calculations can be made. A number of methods have been developed for these purpose such as Newton-Cottes, Tchebycheff, Trapezoidal and Simpson methods. In this course, we will concentrate on the two most popular methods; Trapezoidal and Simpson methods. 4.4 Trapezoidal Method When a curve can be assumed to be represented by a set of trapezoids, the area under the curve can be calculated. C D E B y1 y y3 y4 A h F Figure 4.4 Waterline or Sectional Area Curve In Figure 4.4, the area under the curve is the are area of trapezoid ABCDEF. Area = = 1 y y h y y h y y 1 1 h y 1 1 y y 3 3 y 4 1 3 4 Exercise 1. Find the Trapezoidal formulae for curves made up of Omar bin Yaakob, July 008 37
i) 6 offsets ii) 9 offsets iii) n offsets Naval Architecture Notes. The midship section of a chine vessel has the following offsets: Draught(m) 0 0.5 0.5 0.75 1.0 Half-breadth (m) 0 0.6 1.0 1.5 1.9 Calculate its midship section coefficient at draught of 1.00m. 3. Find the water plane area of a ship LBP = 10m made up of the following offsets: Station 0 1 3 4 Half-breadth 0 0.3 1.0 1. 1.1 (m) Find its area, TPC and waterplane coefficient 4.5 Simpson Rules for Areas. Simpson rule is the most popular method being used in ship calculations to calculate volumes, second moments of areas and centroid. This is because it is flexible, easy to use and its mathematical basis is easily understood. Basically, the rule states that the ship waterlines or sectional area curves can be represented by polynomials. By using calculus, the areas, volumes, centroids and moments can be calculated. Since the separation between stations are constant, the calculus has been simplified by using multiplying factors or multipliers. There are three Simpson rules, depending on the number and locations of the offsets. 4.5.1 Simpson First Rule Simpson s First Rule C D B y1 y y3 A -h O h E Figure 4.5 Waterline or Sectional Area Curve with Three Offsets Omar bin Yaakob, July 008 38