Principles of Naval Architecture and Ship Construction, Assignments of Marine Engineering

Download Principles of Naval Architecture and Ship Construction and more Assignments Marine Engineering in PDF only on Docsity! Task 1 - Ship Construction Q1 Materials used in the construction of ship vessels are steel materials, aluminum alloys, wooden materials and composite materials can be given as examples. These materials are preferred according to the area of use on the ship. Materials to be used in shipbuilding must satisfy the compression, tensile and shear stresses, hardness, brittleness and strength values. 1. Steel Materials: Because steel blends impressive strength with malleability, almost every ship that is built today has a hull made of steel. High-tensile steel hulls allow ships to carry enormous weight without capsizing. DISADVANTAGES: The biggest disadvantage of using steel for shipbuilding is its weight. Steel is much heavier than other materials like aluminum or wood, meaning that ships built with steel will have more drag in the water and will be slower than those made with lighter materials. 2. ALUMINUM ALLOYS : Naval architects and builders are turning to aluminium because of its durability, light weight, versatility and recyclability. These characteristics can ensure higher speeds, fuel efficiency, bigger payloads, better controllability and superior design. DISADVANTAGES: I am seeking formal statements on what are some possible disadvantages of aluminum alloy as a replacement of steels in ship building.  It includes following factors  Expensive raw material.  Expensive fabrication.  Requires special techniques to weld.  Flammable.  Corrodes rapidly in salt water.  Softer than steel. 3. Fiberglass / composite Materials: The objective in the fiberglass boatbuilding is to achieve lightness in weight, vibration damping, corrosion resistance, and impact resistance, low construction cost and ease of construction. Fiber glass is much stronger and more flexible than aluminum. This allows us to create boats that maintain superior strength throughout decades of use. They can stand up to the abuse of crashing waves and high speeds while giving you the confidence to reach even the most remote fishing holes. DISADVANTAGES: The inherently brittle nature of fiberglass tends to crake easily, meaning you might be more prone to spending money on minor repair costs. In conjunction with the above point, fiberglass tends to be on the more expensive side of the spectrum, both upfront and over the long term. 4. Wooden Materials: Wood is the traditional boat building material used for hull and spar construction. It is buoyant, widely available and easily worked. It is a popular material for small boats (of e.g. 6-metre (20 ft) length; such as dinghies and sailboats). DISADVANTAGES:  It can be very expensive.  It requires much maintenance.  It is susceptible to marine organisms such as wood borer. Figure 1 shows a summary of the manufacturing cycle. MECHANICAL SCORIA SHEET METAL OR CUTTING He REMOVAL PLATE FORMING _ NAVAL WORKSHOP ACETYLENE TORCH PROFILE CUTTING FORMING SMALL PREFABRICATION ASSEMBLY fe ssony A wnc i LARGE _PREFABRICATION | ASSEMBLY WELDING inno a HULL ASSEMBLY = =— Question 3 UMM Tt MELE LMM al LECT 3000 WH 6000 Lay-Up Methods for Fiberglass (GRP) Composites This article provides an overview of layup methods for fiberglass composites, linings and coatings. Fiberglass Laminates Fiberglass laminates have revolutionized a huge range of industries and sectors, providing enhanced performance and structural strength without adding additional weight. Easy to use, they can be molded and applied to a variety of surfaces, creating seamless, all-in-one shapes and sizes (difficult to achieve with metals and wood) that are incredibly strong and lightweight. And those traditional materials are also susceptible to rusting and have either limited strength or excessive weight in the case of steel. Question 4 In 1969, IACS was given consultative status with the International Maritime Organization (IMO). It remains the only non-governmental organization with Observer status which is able to develop and apply Rules. Compliance with the IACS Quality System Certification Scheme (QSCS) is mandatory for IACS Membership. IACS is a not-for-profit membership organization of classification societies that establish minimum technical standards and requirements that address maritime safety and environmental protection and ensures their consistent application. IACS members includes ABS (US of A), BV (France), CCS (China), CRS (Croatia), DNV GL (Germany), IRS (India), KR (South Korea), LR (UK), PRS (Poland), RINA (Italy), RS (Russia) and ClassNK (Japan). TASK 2 Question 1 Simpson's Rule is a numerical integration method that can be used to calculate the displacement of a vessel. The formula for Simpson's Rule is: Displacement = (L/3) * (A0 + 4A1 + 2A2 + 4A3 + 2A4 + 4A5 + 2A6 + 4A7 + 2A8 + 4A9 + A10) where L is the waterline length and A0 to A10 are the half areas of each section. Using the given values, we can calculate the displacement as: Displacement = (29.5/3) * (0 + 4(3.04) + 2(4.9) + 4(5.02) + 2(4.62) + 4(4.21) + 2(3.8) + 4(3.38) + 2(2.93) + 4(1.79) + 0) Displacement = 441.15 cubic meters To calculate the position of the longitudinal center of buoyancy, we need to find the moment of each half area about a reference point and then sum them up. The reference point is usually the aft perpendicular. Using the given values, we can calculate the moment of each half area as: M1 = 1.7 * 3.04 / 2 = 2.584 M2 = (2.6 + 1.7) * 4.9 / 2 = 9.695 M3 = (3.4 + 2.6) * 5.02 / 2 = 16.964 M4 = (3.4 + 3.4) * 4.62 / 2 = 15.708 M5 = (3.4 + 3.4) * 4.21 / 2 = 14.374 M6 = (3.4 + 3.4) * 3.8 / 2 = 12.88 M7 = (3.4 + 3.4) * 3.38 / 2 = 11.386 M8 = (3.4 + 3.4) * 2.93 / 2 = 9.893 M9 = (2.9 + 3.4) * 1.79 / 2 = 4.6345 The moment of the last section (A10) is zero because the half area is zero. Now we can sum up the moments and divide by the displacement to get the position of the longitudinal center of buoyancy: LCB = (2.584 + 9.695 + 16.964 + 15.708 + 14.374 + 12.88 + 11.386 + 9.893 + 4.6345) / 441.15 LCB = 6.989 meters from the aft perpendicular Therefore, the displacement of the vessel is 441.15 cubic meters and the position of the longitudinal center of buoyancy is 6.989 meters from the aft perpendicular. Question 2 GZ Curve is a curve indicating the transverse distance between the center of gravity (G) and center of buoyancy (B) in a condition when the vessel is heeled to a certain angle. The shipyard provides the information to calculate GZ for various angles of heel and for various displacements. The vessel's center of gravity (G) has a distinct effect on the righting lever (GZ) and consequently the ability of a vessel to return to the upright position. The lower the center of gravity (G), the bigger is the righting lever (GZ). GZ calculation: GZ=GM sin φ and is called the righting lever. GM is known as the metacentric height. For a given position of G, as M can be taken as fixed for small inclinations, GM will be constant for any particular waterline. What is "M": The Metacenter (M) “M” is a point, generally on the centerline, at which the initial upward forces acting through B cut the upward forces acting through B¹ after the vessel has been inclined by an external force. For a vessel to be stable G must be below M. Therefore, KG = 14.75 - 0 = 14.75 meters. Now we can calculate BM as: BM = I / (V * KG) Using the previously calculated displacement of 441.15 cubic meters, we can calculate V as: V = 441.15 * 1.025 V = 451.84 metric tons Therefore, BM = 1203.3 / (451.84 * 14.75) BM = 0.580 meters So, the BM of the vessel is 0.580 meters. Question 4 Missing Question 5 Missing Question 6 G2 Curve According to KN ratio table at a given displacement of 350 tonne. Heel Angle o KN value (Lever) (Meters) 100 0.51 200 0.98 300 1.15 400 1.8 500 2.09 600 2.26 700 2.33 800 2.31 900 2.20 Question 7 Tonnes per centimeter immersion is a measure of a vessel's buoyancy or displacement. It represents the weight in tonnes required to immerse the vessel by one centimeter. This measurement is often used in naval architecture and marine engineering to determine a vessel's stability and loading conditions. The value of tonnes per centimeter immersion is affected by a vessel's shape, size, and weight distribution. It can be used to calculate the vessel's draft and the amount of cargo or ballast required to achieve a desired draft. Question 8 L = 100 M Displacement = 6300 tons The mean draught = 6 M KB = 3 M Longitudinal BM = 104.7 M KG = 54 M Center of F = midship Weight moved forward to aft = 60 tons MCT = 1 cm Task 3 Q1. In the event of ship flooding, it is important to follow emergency protocols and procedures to ensure the safety of everyone onboard. Here is a general emergency checklist during ship flooding: 1. Master should be informed. 2. Emergency Stations signal should be sounded. 3. All watertight doors should be closed. 4. Main Engine(s) should be slowed down/stopped. 5. Crew should muster to damage control stations. 6. Survival crafts and other lifesaving appliances should be prepared for use, if needed. 7. Stakeholders and relevant authorities should be informed. Emergency Essentials Following are the types of Emergencies on ship: 1. Fire. 2. Collision. 3. Grounding. 4. Cargo hose burst. 5. Major leakage or spillage of oil cargo. 6. Any other event which calls for emergency action Emergency Alarms The general emergency alarm on the ship is recognized by 7 short ringing of the bell followed by a long ring or using the ship horn signal of 7 short blasts followed by 1 long blast. Question 2 Watertight subdivisions: Subdivision of a ship's hull into watertight compartments is called compartmentation. Vertically designed watertight divisions/walls within the ship's structure to avoid ingress of water in the compartment if the adjacent compartment is flooded due to damage in ship's hull. Broadly, the factor of subdivision ensures that one, two or three compartments can be flooded before the margin line is immersed leading to what are called one-, two- or three-compartment ships. That is, compartment standard is the inverse of the factor of subdivision. Design involves parameters such as maximum safe operating pressure and temperature, safety factor, corrosion allowance and minimum design temperature (for brittle fracture). Construction is tested using nondestructive testing, such as ultrasonic testing, radiography, and pressure tests. 1. The IMO instruments governing safe ship designs. 2. Design, construction, subdivision and stability, buoyancy, seakeeping and arrangements, including evacuation matters, of all types of ships, vessels, craft and mobile units covered. 3. IMO instruments testing and approval of construction and materials. 4. Load line matters. How to calculate Sinkage: Calculate Parallel Rise or Parallel Sinkage (PR/PS). Divide the net weight addition/removal by Tons per Inch Immersion (TPI, found using Draft Diagram and Functions of Form and ORIGINAL DISPLACEMENT.) Calculate Bodily Sinkage of ship: Your TPI can be found on the ship's deadweight scale, and changes depending on the draft of the vessel. Step 1: Calculate amount of weight to be added or removed. Step 2: Find TPI from your deadweight scale. Step 3: Divide the total weight added by TPI. Step 4: Add sinkage or subtract rise from your initial drafts. How do you calculate the change of trim? The change of trim is given by the trimming moment divided by MCT. The new draughts become 5.751 m forward and 5.97 m aft Question 5 Permeability Permeability is the quality or state of being permeable—able to be penetrated or passed through, especially by a liquid or gas. The verb permeate means to penetrate, pass through, and often become widespread throughout something. Where the compartment Permeability is the amount of water that can enter a compartment or tank after it has been bilged. When an empty compartment is bilged, the whole of the buoyancy provided by that compartment is lost. It changes with the compartment type because, the ratio of the floodable volume to the total volume of the compartment gives the permeability of the compartment. Task 4 Question 1 Force diagram showing a typical global load on cargo hull and briefly explain what it represents. Include at least hull weight, superstructure, machinery, cargo and fuel tonnage. 1. Superstructure. Question 3 Panting As waves pass along the ship, they cause fluctuation in water pressure which tends to create in & out movements of the shell plating. This is particularly the case at fore end. The rules of classification society require extra stiffening at the end of the ship in the forms of beams, brackets, stringers & plates in order to reduce the possibility of a damage. This in & out movements is called panting. Steps taken in respect of stress involved. 1. Since the ship is capable of bending in a longitudinal in vertical plane it follows that there must be material in the ship structure to resist this bending. 2. The girder, beam, stringer, bracket & plates are sufficient thickness according or respect of stresses involved in ship. 3. Stresses loads distribute all the members not to be acted in single point. It may cause damage the structure. 4. Hence the plating must be stiffened so as be able to withstand against compressive load. 5. Sides of the ship are stiffened transversely while the bottom deck are stiffened longitudinally. The above-mentioned actions are measured in structure of ship. While cargo handling and weight distributing are also important. Question 4 Strain Gauge It was invented in 1938 by Edward-E-Simmons and Arthur C. Ruge. It is one of the significant sensors used in the geotechnical field to measure the amount of strain on any structure. Whenever an external force is applied to an object it tends to change its shape and size. Therefore, altering its resistance. The change in shape and size is called strain and it is measured by strain gauge.  Working principle A straining gauge works on the principle of electrical conductance and its dependence on the conductor geometry. Whenever a conductor is stretched within the limit of its elasticity it does not break but gets narrower and longer. Similarly, when I is compressed its get shorts and broader, ultimately changing its resistance. We know resistance is directly proportional on the length and cross-sectional area of the conduction. R = L/A Where. R= resistance L = length A = cross sectional area  Use of strain gauge in ship hull stresses Strain gauge are commonly electro resistive type. Although a new technology is at hand these days sometimes providing better accuracy. 1. Short baseline for 1” long sample. 2. Long baseline for 2M long sample. 3. Hull girder bending moment. Task 5 Question 1 Resistance of ship: When the ship moves through the water at any speed a force or resistance is exerted by the water on the ship. There are mainly two main sections of resistance. 1. Fractional resistance 2. Residuary resistance The main factor is of ship resistance are as follows.  The speed of a ship  The wetted surface area  The length of ship  The roughness of the hull  The density of water  Froude’s formula  it was William Froude who put forward the idea of dividing the total resistance of the ship.  Froude’s law of fractional resistance of the ship/vessel are as follows: Formula: Rf = fSVn Where. i. “f” is a coefficient which depend upon the length of the ship ii. “S” is wetted surface area. iii. “V” is the ship speed in knots. iv. “n” is an index of about 1.825. P is the engine power in kilowatts (kW). Rf is the frictional resistance in newtons (N) which we calculated in question 2 as 8189.9 N. V is the speed of the ship in meters per second (m/s) which is 8 knots * 0.5144 = 4.1152 m/s. eta is the propulsive efficiency which we will assume to be 0.65. AC is the Admiralty coefficient which is given as 383.45. Substituting the given values, we get: P = 8189.9 * 4.1152 / (0.65 * 383.45) P = 145.5 kW Therefore, the estimated engine power requirement for the ship is approximately 145.5 kW. Question 4  Propeller diameter “D” the diameter of the propeller is the diameter of a circle or disc cut out by the blade tips.  Pitch If the propeller is assumed to work in an un-yielding fluid, then in one revolution of the shaft of propeller will move forward a distance which is known as the pitch.  Slip The distance advanced by a propeller during one revolution when delivering no thrust is termed the analysis pitch. This is greater than the geometrical pitch of the propeller. When developing thrust the propeller advance per revolution is less than the analysis pitch. This difference is termed as the ship.  Cavitation The trust of propeller varies approximately as the square of the revolutions. Thus, the speed of rotation is increased there is considerable increase in thrust. The net pressure at any point on the back of the blade is algebraic sum of the atmospheric pressure, water pressure and suction caused by the thrust. When the suction is high at any point the net pressure may fall below the vapor pressure of the water. Causing a cavity or bubble to form on the blade. The cavity is filled with water and air. The forming and collapsing of these cavities are known as “Cavitation”.