Prepared by Dr. Sudhir Sindagi Introduction to the Course Geometry Transverse Stability Seakeeping Maneuvering Naval Architecture Propulsion By Dr. Sudhir Sindagi Longitudinal Stability Strength Resistance 1 Prepared by Dr. Sudhir Sindagi Introduction to the Faculty Working in TMI since July 2011 as an Associate Professor Ph.D from IIT Madras M.Tech. from IIT Kharagpur Published more than 10 Research Papers in International WoS / Scopus indexed Journals Executed consultancy works for the industries across the Globe Received no. of awards including one from GOI 4 Yrs of work experience in Larsen and Toubro, Mumbai as an Assistant Manager – Naval Architect. https://www.linkedin.com/in/sudhir-sindagi-8aa47622/ 2 Prepared by Dr. Sudhir Sindagi 2019 End Sem Result 3 Prepared by Dr. Sudhir Sindagi Introduction to the Course Geometry Transverse Stability Seakeeping Maneuvering Naval Architecture Propulsion Longitudinal Stability Strength Resistance 4 Prepared by Dr. Sudhir Sindagi Introduction to the Course Naval Architecture -I – – – – – – – – – – – Geometry, Lines plan, Coefficients of Form Flotation, FWA, DWA & load line Numerical Methods/Integration Calculation of CG & Angle of List Stability concepts - Small Angle of Heel Stability at large angles of Heel Longitudinal Stability – Trim Damaged Stability Strength of Ships IMO Rules related to Stability Critical stability during docking / Grounding 5 Prepared by Dr. Sudhir Sindagi Introduction to the Course Naval Architecture -II – – – – Resistance of the Ship Propulsion of the Ship Manoeuvring of the Ship Motion of Ships on waves 6 Prepared by Dr. Sudhir Sindagi Lecture Plan Class Chapter No 1 2 3 4 5 6 7 8 9 Topic Detailed Introduction to the course Representation of ship geometry: Body plan, Profile Geometry, and Half breadth plan, Bonjean Curves. Definition of Lines plan, offset and table of offsets. Coefficinet of Areas - WPA & WSA, CG& CB, Coefficient of Forms Numericals on Coefficient of Forms Forms TPC, MCTC & Numericals Numericals- DWT, GT, NT, TPC, MCTC Hydrostatic Curves, Effect of Density, FB, Reserve Flotation, FWA, Buoyancy DWA & loadline FWA, DWA, Loadline FWA, DWA, Loadline- Numericals Marks 7-12 4-12 7 Prepared by Dr. Sudhir Sindagi Lecture Plan 10 Trapezoidal Rule, Simpson’s 1st & 2nd rules, 5-8 1 Rule. 11 Area of Water plane, LCF, TPC, MI - Numericals Marks 12 Numerical Use of Second Rule, 1/2 ordinates 13 Methods Use of Second Rule, 1/2 ordinates 14 Calculation of Displacement, VCB 15 Use of TPC to calculate the displacement and VCB 16 Use of 5-8-1 and 3-8-1 rule Find the change in CG when weight is added, removed or Calculation 17 shifted within the ship. of CG & 18 Numericals Angle of 19 Effect of suspended mass - Numericals List 20 Calculation of angle of list - Numericals 14 Marks 8 12 Prepared by Dr. Sudhir Sindagi Lecture Plan Explain stability, criterion for positive, negative and neutral 21 Marks stability, righting and heeling moment. Stability 22 concepts - BM= IT/V - Derivation 23 Small Numericals on GM= KB+BM-KG 24 Angle of Free Surface Effect 14-24 25 Heel Free Surface Effect - Numericals 26 Inclining Experiment 27 Inclining Experiment - Numericals 28 Derivation GZ= (GM+ BM/2 Tan^2θ)*sinθ - Wall Sided Formula 29 Stability at Angle of Loll 30 large Curve of Statical Stability - GZ Curve 7-12 31 angles of Curve of Statical Stability - GZ Curve - Numericals 32 Heel Curve of Statical Stability - KN tables & Curves- Numericals 9 33 Dynamical Stability Prepared by Dr. Sudhir Sindagi Lecture Plan 34 Trim, Concepts of IL, GML and MCTC. Find the drafts at the AP & FP and its trim. Change in mean 35 Longitudinal draft, trim and sinkage due to addition / removal of weight 36 Stability Numericals Calculate the change in mean draft, trim and sinkage due to 37 change in water density. Calculate the change in drafts & trim due to bilging by 38 Added weight and Lost Buoyancy methods. Consideration of the permeability of compartment and 39 Damaged stowage factors. Margin lines. Stability Calculate the damaged stability after flooding. Flooding 40 calculation. 41 Numericals Marks 14 14 10 Prepared by Dr. Sudhir Sindagi Lecture Plan 43 44 45 Strength 46 of ships 47 Ship as Girder, theory, Weight Curve, Buoyancy Curve, Load curve, SFD, BMD Properties of all curves with interlinking Problems Stress on a Section, Problems Determining MI of Section - Problems Determining MI of Section - Problems 48 Finding bending Stress in Decks 42 Marks 14-24 11 Prepared by Dr. Sudhir Sindagi Lecture Plan 49 50 51 IMO Rules related to Stability SOLAS Stability requirements of Merchant vessels. Stability of Vessels subjected to Wind Loading and grain carrier Floodable length curves, Factors of sub- division, permissible length, compartment standard Marks 4-10 12 Prepared by Dr. Sudhir Sindagi Text Book and Ref books Text Book: NIL Reference Books: – Buoyancy & Stability of Ships – by IR R F Scheltema De Heere & Bakker (George Haarp & Co. Ltd. London) – Ship Hydrostatics and Stability – by Adrian B Biran – Principles of Naval Architecture -Vol I – by Edward V Lewis (SNAME) – Ship Stability for Masters & Mates – by Derrett & Barrass – Naval Architecture for Marine Engineers – Reeds Volume - 4 – Introduction to Naval Architecture – Eric Tupper – Ship and Naval Architecture – R.Munro-Smith – Ship Construction – D.J.Eyers – Naval Architecture, Principles & Theory – B.Baxter 13 Prepared by Dr. Sudhir Sindagi Evaluation Program Internal Assessment as per guidelines issued by IMU Sr. No. Component Weightage Nature Date As per the Time Table declared by Exam 1 Class Test 20 Marks Written Teacher 2 Assessment 10 Marks Observation Continuous Term End 3 Examination 70 Written To be announced Note: There shall be a common minimum pass mark 50% in the External 14 Examinations and 50% overall. Prepared by Dr. Sudhir Sindagi 15 Prepared by Dr. Sudhir Sindagi Ship's Geometry and Flotation By Dr. Sudhir Sindagi 1 Prepared by Dr. Sudhir Sindagi Chapter Content Ship geometry, Definition of hull surface Lines plan drawing of ships Offset Table Bonjean Curves Archimedes principle, Displacements Coefficients of form Effect of Density and relative density of a liquid on drafts. Meaning of buoyancy, reserve buoyancy Center of Gravity and Center of Buoyancy of ship TPC, FWA, DWA Plimsoll line or Load line markings 2 Prepared by Dr. Sudhir Sindagi Ship’s Geometry 3 Prepared by Dr. Sudhir Sindagi Ship’s Geometry 4 Prepared by Dr. Sudhir Sindagi Ship’s Geometry 5 Prepared by Dr. Sudhir Sindagi Ship’s Geometry A ship’s hull form helps determine most of its main attributes; its stability characteristics; its resistance and therefore the power needed for a given speed; its seaworthiness; its manoeuvrability and its load carrying capacity. It is important, therefore, that the hull shape should be defined with some precision and unambiguously. The after perpendicular is the perpendicular drawn to the Summer Load Line - SLL (Design Water Line - DWL) at the after side of the rudder post, where fitted, or the line passing through the centreline of the rudder pintles. The fore perpendicular is the perpendicular drawn to the to the Summer Load Line - SLL (Design Water Line - DWL) at the intersection of the 6 forward side of the stem with the summer load line. Prepared by Dr. Sudhir Sindagi Ship’s Geometry 7 Prepared by Dr. Sudhir Sindagi Ship’s Geometry The length overall (LOA) is the horizontal distance between the extreme points at the forward and aft end of the ship, measured parallel to the SLL or DWL. The Length on the waterline (LWL) is the length on the waterline between the intersections of SLL or DWL at the bow and at the after end of the ship. The Length between perpendiculars (LBP) is the horizontal distance measured between the aft perpendicular and the forward perpendicular of the ship, measured parallel to the SLL or DWL. The mid-point between the two perpendiculars is called amidships or midship. The transverse section of the ship at the midship is called the midship 8 section. Prepared by Dr. Sudhir Sindagi Ship’s Geometry The draught of the ship at any point along its length is the vertical distance from the keel to the waterline. If a moulded draught is quoted it is measured from the inside of the keel plating. Freeboard is the difference between the depth at side and the draught, that is it is the height of the deck above the waterline. The freeboard is usually greater at the bow and stern than at amidships. This helps create a drier ship in waves. Air draught is the vertical distance from the summer waterline to the highest point in the ship, usually the top of a mast. This dimension is important for ships that need to go under bridges in navigating rivers or entering port. 9 Prepared by Dr. Sudhir Sindagi Ship’s Geometry The ship’s depth varies along the length but is usually quoted at midship. It is the vertical distance measured between the upper surface of the main deck at side and the base line of the ship. At midship, if the depth at the centreline of the ship is greater than at the side, then the ship is said to have the camber or the round of beam. Camber is generally provided only to weather decks to drain the water at side. Camber can be defined as the curvature given to the main deck in the transverse direction and is measured as the difference between the depth of the ship at the centreline and at the side. The curvature given to the main deck in the longitudinal direction is known as the Sheer. It is measured as the difference between the depth of the ship at the midship and/or at the forward perpendicular and/or at 10the aft perpendicular. Prepared by Dr. Sudhir Sindagi Ship’s Geometry 11 Prepared by Dr. Sudhir Sindagi Ship’s Geometry 12 Prepared by Dr. Sudhir Sindagi Ship’s Geometry 13 Prepared by Dr. Sudhir Sindagi Ship’s Geometry 14 Prepared by Dr. Sudhir Sindagi Ship’s Geometry The bottom of a ship, in the midships region, is usually flat but not necessarily horizontal. If the line of bottom is extended out to intersect the moulded breadth line, the height of this intersection above the keel is called the rise of floor or deadrise. Many ships have a flat keel and the extent to which this extends athwartships is termed the flat of keel or flat of bottom. In some ships the sides are not vertical at amidships. If the upper deck beam is less than that at the waterline it is said to have tumble home, the value being half the difference in beams. If the upper deck has a greater beam the ship is said to have flare. All ships have flare at a distance from amidships 15 Prepared by Dr. Sudhir Sindagi Ship’s Geometry Inward curvature given to the side shell above the SLL is known as tumble home and the outward curvature given to the side shell above the SLL is known as Flare. 16 Prepared by Dr. Sudhir Sindagi Ship’s Geometry Reserve buoyancy is the volume of the enclosed spaces above the waterline (SLL). It is a very important factor in which, minimum freeboards are assigned to a ship to ensure that there is adequate reserve buoyancy at all times especially when there is a case of flooding. Freeboard(F) = Depth(D) – Draft (T) 17 Prepared by Dr. Sudhir Sindagi Ship’s Geometry- Line Plan Drawing The hull shape is defined by its intersection with three sets of mutually orthogonal planes famously known as Lines Plan Drawing of the ship. Lines plan drawing basically consists of three different views created by the intersection of these planes viz. Body Plan, Profile View and Half Breadth Plan The horizontal planes (Longitudinal Horizontal Sections) are known as waterplanes and the lines of intersection are known as waterlines and are represented in Half Breadth Plan. The planes parallel to the middle line plane (Longitudinal vertical Sections) cut the hull in buttock (or bow and buttock) lines, the middle line plane itself defining the profile and are represented in Profile view. The intersections of the athwartships planes define the transverse 18 sections and are represented in Body Plan. Prepared by Dr. Sudhir Sindagi Ship’s Geometry- Line Plan Drawing 19 Prepared by Dr. Sudhir Sindagi Ship’s Geometry- Line Plan Drawing Transverse Sections Longitudinal Horizontal Sections Waterplanes Longitudinal vertical Sections- Buttocks 20 Prepared by Dr. Sudhir Sindagi Ship’s Geometry- Line Plan Drawing Transverse sections Waterlines Buttocks 21 Prepared by Dr. Sudhir Sindagi Ship’s Geometry- Line Plan Drawing 22 Prepared by Dr. Sudhir Sindagi Ship’s Geometry- Line Plan Drawing 23 Prepared by Dr. Sudhir Sindagi Ship’s Geometry- Line Plan Drawing 24 Prepared by Dr. Sudhir Sindagi Ship’s Geometry- Line Plan Drawing 25 Prepared by Dr. Sudhir Sindagi Ship’s Geometry- Line Plan Drawing 26 Prepared by Dr. Sudhir Sindagi Ship’s Geometry- Line Plan Drawing Stations: These are the longitudinal positions on the ship placed at an equal intervals, wherein, transverse sections are taken, to generate the Body Plan. Generally, the ship is divided into 10 equal parts longitudinally, resulting in 11 in number stations, with numbering 0,1,2,3 …8,9,10. The station with number 5 will be placed at the midship. To capture the abrupt changes in the shape of the hull at the aft and the forward end, the half stations are inserted. 27 Prepared by Dr. Sudhir Sindagi Ship’s Geometry- Line Plan Drawing Waterlines: These are the vertical positions on the ship placed at an equal intervals, wherein, Longitudinal Horizontal Sections (Waterplanes) are taken to generate the half breadth plan. Since the ship is symmetric about the centreline, hence in the half breadth plan, only half side of the ship is shown. 28 Prepared by Dr. Sudhir Sindagi Ship’s Geometry- Line Plan Drawing Offsets (ordinates): These are the transverse positions on the ship placed at an equal intervals, wherein, Longitudinal Vertical Sections (Buttocks) are taken to generate the Profile view. Offset of any point on any water line is the transverse distance from the center line 29 Prepared by Dr. Sudhir Sindagi Ship’s Geometry- Line Plan Drawing Body Plan: It is obtained by taking transverse sections of the ship at various stations. Generally, the left side of the body plan represents the sections of the ship taken at the aft end till midship. While, the right side of the body plan represents the sections of the ship taken at the forward end from midship 30 Prepared by Dr. Sudhir Sindagi Ship’s Geometry- Line Plan Drawing Half Breadth Plan: It is obtained by taking Longitudinal Horizontal Sections (Waterplanes) of the ship taken at various waterlines. Since the ship is symmetric about the centreline, hence in the half breadth plan, only half side of the ship is shown. 31 Prepared by Dr. Sudhir Sindagi Ship’s Geometry- Line Plan Drawing Profile View: It is obtained by taking Longitudinal Vertical Sections (Buttocks) of the ship taken at various offsets (ordinates). 32 Prepared by Dr. Sudhir Sindagi Ship’s Geometry- Line Plan Drawing 33 Prepared by Dr. Sudhir Sindagi Ship’s Geometry- Line Plan Drawing 34 Prepared by Dr. Sudhir Sindagi Ship’s Geometry- Line Plan Drawing 35 Prepared by Dr. Sudhir Sindagi Ship’s Geometry- Offset Table Offsets: Offset of any intersection point on any waterline is its transverse distance from the center line of the ship. A collection of such offsets in the form of table for different transverse sections taken at different stations is known as the offset table. The offset data or Table must be measured and prepared at every intersection points on each stations and waterlines including deck line, chines, knuckles and bulwarks (if any). Offset data also called as half breadth data, because it represents the half breadth of the ship at every station and waterlines. It is also a standard practice to indicate the data of height above for deck, chine, bulwark, and knuckles lines. The height above base of buttock lines may also be included whenever necessary. 36 Prepared by Dr. Sudhir Sindagi Ship’s Geometry- Offset Table 37 Prepared by Dr. Sudhir Sindagi Ship’s Geometry- Offset Table 38 Prepared by Dr. Sudhir Sindagi Ship’s Geometry- Offset Ta 39 Prepared by Dr. Sudhir Sindagi Ship’s Geometry- Bonjean Curves Bonjean Curve is a curve obtained by joining points at different waterlines, distance of which from the vertical line represents 40area of that transverse section till the particular waterline. Prepared by Dr. Sudhir Sindagi Ship’s Geometry- Bonjean Curves Collection of such curves prepared for different stations is known as the Bonjean Curves. Bonjean curves are a representation of the area of the transverse sections at successive waterlines. These are used in calculating the volume of displacement and the center of buoyancy at any waterline, or angle of trim. They are used in stability calculations, determining the capacity of the ship, or in launching calculations. 41 Prepared by Dr. Sudhir Sindagi Ship’s Geometry- Bonjean Curves Bonjean curves are presented in two ways – With common vertical axis – With separate vertical axes 42 Prepared by Dr. Sudhir Sindagi Ship’s Geometry- Bonjean Curves When plotted on a profile of the vessel, each Bonjean curve begins on the base of the station it represents. It rises forward on the profile, indicating the transverse area of the hull at that location on the hull. 43 Prepared by Dr. Sudhir Sindagi Coefficients of Form Archimedes' principle: It states that the upward buoyant force that is exerted on a body immersed in a fluid, whether fully or partially submerged, is equal to the weight of the fluid that the body displaces. Weight of the body = Weight of the water displaced Volume of the water displaced = underwater volume of the ship If Δ = Displacement or weight of the ship in tonnes= weight of the water displaced by the ship ๐= Volumetric displacement of the ship = Volume of water displaced, then โ= ๐ ∗ ๐ Here ๐ is the density of water displaced. 44 ρSW = 1.025 tonnes/m3 & ρFW = 1.0 tonnes/m3 Prepared by Dr. Sudhir Sindagi Coefficients of Form Coefficients of form or coefficients of fineness are dimensionless numbers that describe hull fineness and overall shape characteristics. The coefficients are ratios of areas or volumes for the actual hull form compared to prisms or rectangles defined by the ship’s length, breadth, and draft. Since length and breadth on the waterline as well as draft vary with displacement, coefficients of form also vary with displacement. Length between perpendiculars (LBP) is most often used, although some designers prefer length on the waterline. Coefficients of form can be used to simplify area and volume calculations for stability or strength analyses. As hull form approaches that of a rectangular barge, the coefficients 45 approach their maximum value of 1.0. Prepared by Dr. Sudhir Sindagi Coefficients of Form Coefficients of form: These are broadly classified into two categories Area Coefficients: – Midship Coefficient (CM) – Waterplane coefficient (CWP) Volume Coefficients: – Block Coefficient (CB) – Prismatic Coefficient (CP) • Longitudinal Prismatic Coefficient (CPL) • Vertical Prismatic Coefficient(CPV) – Volumetric Coefficient (CV) 46 Prepared by Dr. Sudhir Sindagi Coefficients of Form Midship Coefficient (CM) It is the ratio of the area of the Midship Section till SLL or DWL and area of a circumscribing rectangle whose sides are equal to the draught and the breadth of ship 47 Prepared by Dr. Sudhir Sindagi Coefficients of Form Midship Coefficient (CM) ๐๐ = area of the Midship Section till SLL or DWL area of a circumscribing rectangle with sides equal to B & T ๐๐ = ๐๐ ๐∗๐ Values of CM may range between 0.75 to 0.995 for normal ships. In some cases vessels have been built with bulges or blisters below the design waterline. If B is taken at the SLL, then CM may be greater than unity on such vessels 48 Prepared by Dr. Sudhir Sindagi Coefficients of Form Waterplane Coefficient (๐๐๐ ) or Coefficient of fineness: 49 Prepared by Dr. Sudhir Sindagi Coefficients of Form Waterplane Coefficient (๐๐๐ ) or Coefficient of fineness: It is the ratio of the area of the waterplane taken at SLL or DWL and area of a circumscribing rectangle whose sides are equal to the Length and the breadth. ๐๐๐ = area of the waterplane taken at SLL or DWL area of a circumscribing rectangle with sides equal to L & B ๐๐๐ = ๐๐๐ ๐ ∗๐ The values of CWP at the DWL range from about 0.65 to 0.95, depending upon type of ship, speed, and other factors. 50 Prepared by Dr. Sudhir Sindagi Coefficients of Form Block Coefficient (CB) 51 Prepared by Dr. Sudhir Sindagi Coefficients of Form โผ Block Coefficient: CB = ๏ณ / (L x B x T) A P Beam, B L (along DWL} F P Hull Volume, ๏ณ (under DWL) Block Volume, LxBxT Draft, T 52 Prepared by Dr. Sudhir Sindagi Coefficients of Form Block Coefficient (CB): It is the ratio of the volumetric displacement of the ship and the volume of the circumscribing rectangular prism with dimensions Length, Breadth and Draft of the ship. ๐๐ = volumetric displacement of the ship volume of the circumscribing rectangular prism with sides L,B & T ๐๐ = ๏ณ ๐ ∗๐ ∗๐ 53 Prepared by Dr. Sudhir Sindagi Coefficients of Form Block Coefficient (CB): The block coefficient indicates whether the form is full or fine and whether the waterlines will have large angles of inclination to the middle line plane at the ends. Values of CB at design displacement may vary between 0.36 for a fine high-speed vessel to about 0.92 for a slow and full Great Lakes bulk carrier. Large values signify large wave-making resistance at speed. A slow ship can afford a relatively high block coefficient as its resistance is predominately frictional. A high value is good for cargo carrying and is often obtained by using a length of parallel middle body, perhaps 15–20 per cent of the total 54 length. Prepared by Dr. Sudhir Sindagi Coefficients of Form Longitudinal Prismatic Coefficient (CP or CPL) 55 Prepared by Dr. Sudhir Sindagi Coefficients of Form Longitudinal Prismatic Coefficient (CP or CPL) It is the ratio of the volumetric displacement of the ship and the volume of the prism whose length equals the length of the ship and whose cross section equals the midship section area till SLL remaining constant throughout the length of the ship. ๐๐๐ = ๐๐๐ = Volumetric displacement of the ship Volume of the prism with ( ๐๐∗๐) ๏ณ ๐๐∗๐ Usual range of values is from about 0.50 to about 0.90. A vessel with a low value of ๐๐๐ (or CB) is said to have a fine hull form, while one with a high value of ๐๐๐ has a full hull form.. 56 Prepared by Dr. Sudhir Sindagi Coefficients of Form Longitudinal Prismatic Coefficient (CP or CPL) If two ships with equal length and displacement have different prismatic coefficients, the one with the smaller value of CPL will have the larger midship sectional area and hence a larger concentration of the volume of displacement amidships. The ship with the smaller CPL is also characterized by a protruding bulbous bow, which causes the swelling in the sectional area curve right at the bow, and its extension forward of Station 0. 57 Prepared by Dr. Sudhir Sindagi Coefficients of Form Vertical Prismatic Coefficient (CPV) 58 Prepared by Dr. Sudhir Sindagi Coefficients of Form Vertical Prismatic Coefficient (CPV) It is the ratio of the volumetric displacement of the ship and the volume of the prism whose depth equals the draft of the ship and whose vertical cross section equals the waterplane area taken at SLL remaining constant throughout the draft of the ship. ๐๐๐ = ๐๐๐ = Volumetric displacement of the ship Volume of the prism with ( ๐๐๐∗๐) ๏ณ ๐๐๐∗๐ Unless specifically mentioned, by default, it will be Longitudinal Prismatic Coefficient is used everywhere. 59 Prepared by Dr. Sudhir Sindagi Coefficients of Form Volumetric Coefficient (๐๐ ) or fatness ratio: This coefficient (or fatness ratio) is defined as the volume of displacement divided by the cube of one tenth of the vessel's length ๐๐ = ๐ ๐ (๐๐)๐ Ships with low volumetric coefficients might be said to be "thin", while those with a high coefficient are "fat." Values of the volumetric coefficient range from about 1.0 for light, long ships like destroyers, to 15 for short heavy ships like trawlers. 60 Prepared by Dr. Sudhir Sindagi Coefficients of Form Ranges of different coefficients of form for different types of vessels 61 Prepared by Dr. Sudhir Sindagi Coefficients of Form Importance of coefficients of form The angle at the bow is termed as the angle of entry and influences resistance. As speed increases a designer will reduce the length of parallel middle body to give a lower prismatic coefficient, keeping the same midship area coefficient. As speed increases still further the midship area coefficient will be reduced, usually by introducing a rise of floor. A low value of midship section coefficient indicates a high rise of floor with rounded bilges. A large value of vertical prismatic will indicate body sections of U-form; a low value will indicate V-sections. These features will affect the seakeeping performance. 62 Prepared by Dr. Sudhir Sindagi Sectional Area Curve It is curve obtained by joining points at each station, distance of which from the horizontal line represents the area of the transverse for that particular station section till SLL or DWL. 63 Prepared by Dr. Sudhir Sindagi Sectional Area Curve One of the fundamental hull form characteristics required to prepare the hydrostatic curves are the immersed sectional areas at ordinate stations. A curve whose ordinates are areas of cross sections up to a given waterline corresponding to each point in the length 64 Prepared by Dr. Sudhir Sindagi Sectional Area Curve Area under SAC= Volumetric displacement of the ship = ๏ณ Longitudinal location of centroid of SAC = LCB of the ship Length of the parallel middle body, Entrance and Run are obtained from the SAC 65 Prepared by Dr. Sudhir Sindagi Problems on Coefficients of Form Find the relationship between CB, CPL & CM We know that: ๐๐ = ๐Τ๐∗๐∗๐ From ๐๐ = ๐เต๐๐∗๐ ๐ฉ๐ฎ๐ญ๐ญ๐ข๐ง๐ ๐ = ๐๐๐ ∗ ๐ ๐ ∗ ๐ ๐ข๐ง ๐๐๐จ๐ฏ๐ ๐๐ช๐ฎ๐๐ญ๐ข๐จ๐ง ๐๐ = ๐๐๐ ∗๐๐ ∗๐ ๐๐ = ๐๐๐ ∗๐๐ เต๐∗๐∗๐ เต๐∗๐ ๐๐ฎ๐ญ ๐๐ = ๐๐เต๐∗๐ ๐๐ = ๐๐๐ ∗ ๐๐ 66 Prepared by Dr. Sudhir Sindagi Problems on Coefficients of Form Important Formulae: โ= ๐ ∗ ๐ ρSW = 1.025 tonnes/m3 & ρFW = 1.0 tonnes/m3 ๐ ๐ ๏ณ ๏ณ ๐๐ = ๐ โถ ๐๐๐ = ๐๐ : ๐๐ = : ๐๐๐ = ๐∗๐ ๐๐๐ = ๏ณ ๐๐๐∗๐ ๐ ∗๐ : ๐๐ = ๐ ∗๐ ∗๐ ๐๐∗๐ ๐ ๐ (๐๐)๐ ๐๐ = ๐๐๐ ∗ ๐๐ Form Merchant Vessels typical ranges are Ratio of ๐๐๐ง๐ ๐ญ๐ก โถ ๐๐๐๐ฆ = ๐๐๐ญ๐ฐ๐๐๐ง ๐. ๐ ~ ๐๐. Ratio of ๐๐๐ง๐ ๐ญ๐ก โถ ๐๐ซ๐๐๐ญ = ๐๐๐ญ๐ฐ๐๐๐ง ๐๐ ~ ๐๐. Ratio of ๐๐๐๐ฆ โถ ๐๐ซ๐๐๐ญ = ๐๐๐ญ๐ฐ๐๐๐ง ๐. ๐ ~ ๐. 67 Prepared by Dr. Sudhir Sindagi Problems on Coefficients of Form Following data of a ship are given L=200m, B=22m, T=7m, CPL=0.75, AWP=3500m2 Displacement in SW=23000 tonnes. Calculate CB, CWP & CM Given Data: L=200m, B=22m, T=7m, CPL=0.75, AWP=3500m2 Δ=23000 tonnes. CB=? CWP =? CM=? Using equation โ= ๐ ∗ ๐, ๐๐๐ฅ๐๐ฎ๐ฅ๐๐ญ๐ ๐ = 22439.02m3 ๏ณ Using equation: ๐๐ = Calculate: ๐๐ = 0.728 Using equation: ๐ ∗๐ ∗๐ ๐๐๐ ๐๐๐ = Calculate: ๐ ∗๐ Using equation : ๐๐๐ = ๏ณ ๐๐∗๐ ๐๐๐ = 0.795 Calculate ๐๐ = 149.59m2 68 Prepared by Dr. Sudhir Sindagi Problems on Coefficients of Form Following data of a ship are given L=200m, B=22m, T=7m, CPL=0.75, AWP=3500m2 Displacement in SW=23000 tonnes. Calculate CB, CWP & CM Given Data: L=200m, B=22m, T=7m, CPL=0.75, AWP=3500m2 Δ=23000 tonnes. CB=? CWP =? CM=? Using equation: ๐๐ = ๐๐ ๐∗๐ Calculate: ๐๐ = 0.971 69 Prepared by Dr. Sudhir Sindagi Problems on Coefficients of Form Following data of a ship are given L=122m, B=20m, T=14m, CB=0.82, CPL=0.93, CWP=0.75. Calculate Midship area and waterplane area of ship. Given Data: L=122m, B=20m, T=14m, CB=0.82, CPL=0.93, CWP=0.75 AM =? AWP =? ๏ณ Using equation: ๐๐ = Calculate: ๏ณ= 28011.2 m3 ๐ ∗๐ ∗๐ Using equation : ๐๐๐ = Using equation: ๐๐๐ = ๏ณ Calculate ๐๐ = 246.88m2 ๐๐∗๐ ๐๐๐ Calculate: ๐ ๐๐ = ๐ ∗๐ 1830m2 70 Prepared by Dr. Sudhir Sindagi Problems on Coefficients of Form A lifeboat, when fully laden, displaces 7.2 tonnes. Its dimensions are 7.5m, 2.5m, 1m, and its block coefficient 0.6. Find the percentage of its volume under water when floating in fresh water. Given Data: L=7.5m, B=2.5m, T= 1m, CB=0.6, โ= 7.2 tonnes %of underwater volume=? Using equation โ= ๐ ∗ ๐, ๐๐๐ฅ๐๐ฎ๐ฅ๐๐ญ๐ ๐๐๐ญ๐ฎ๐๐ฅ ๐ฎ๐ง๐๐๐ซ๐ฐ๐๐ญ๐๐ซ ๐ฏ๐จ๐ฅ๐ฎ๐ฆ๐ ๐๐ = 7.2m3 ๏ณ However, Using equation: ๐๐ = Calculate: ๐ ∗๐ ∗๐ ๐๐จ๐ญ๐๐ฅ ๐ฏ๐จ๐ฅ๐ฎ๐ฆ๐ ๏ณT= 11.25 m3 %of underwater volume= ๐๐๐ญ๐ฎ๐๐ฅ ๐ฎ๐ง๐๐๐ซ๐ฐ๐๐ญ๐๐ซ ๐ฏ๐จ๐ฅ๐ฎ๐ฆ๐ ๐๐จ๐ญ๐๐ฅ ๐ฏ๐จ๐ฅ๐ฎ๐ฆ๐ = ๐๐ ๏ณ =64% 71 Prepared by Dr. Sudhir Sindagi Problems on Coefficients of Form Two similar right circular cones are joined at their bases. Each one has a height equal to the diameter of its base. The composite body floats so that both apexes are in the water surfaces. Calculate the midship section coefficient, prismatic coefficient and the waterplane area coefficient. Given Data: Right circular cones floating with both the apexes in water surfaces H=D. CM=? CPL =? CWP=? Since, Right circular cones floating with both the apexes in water surfaces, hence T= D/2 72 Prepared by Dr. Sudhir Sindagi Problems on Coefficients of Form AM= (π/4 D2)/2 Using equation ๐๐ = ๐๐ ๐∗๐ Putting AM and B=D , T=D/2 and simplifying above equation, we get CM =0.7853. 73 Prepared by Dr. Sudhir Sindagi Problems on Coefficients of Form ๐ ๐ Using ๏ณ = ∗ ๐๐๐ ๐ก ๐ ∗ ๐ and putting in the equation : ๐๐๐ = Putting B=D , T=D/2 and simplifying above equation, we get ๐๐๐ =0.3333 Using equation: ๐๐๐ = ๏ณ ๐๐∗๐ ๐๐๐ ๐ ∗๐ Here, ๐ ๐๐ =2*1/2*D*D, L=2D Putting above and simplifying equation, we get ๐๐๐ =0.5 74 Prepared by Dr. Sudhir Sindagi Problems on Coefficients of Form A ship displaces 9450 tonnes and has a block coefficient of 0.7. The area of immersed midship section is 106 sq. meter. Assume sea water density =1.025 t/m3. If the beam =0.13 Length=2.1 draught, Calculate the length of the ship and its prismatic coefficient. Given Data: AM= 106 m2, B=0.13L=2.1T, CB=0.7, โ= 9450 tonnes L= ? CPL=? Using equation โ= ๐ ∗ ๐ , Calculate: ๏ณ= 9219.51 m3 ๏ณ Using equation: ๐๐ = and putting the terms, we get ๐ ∗๐ ∗๐ 0.7= 9219.51 ๐.๐๐ ๐ ∗๐.๐๐๐ ∗ ๐.๐ ๐ 75 Prepared by Dr. Sudhir Sindagi Problems on Coefficients of Form A ship displaces 9450 tonnes and has a block coefficient of 0.7. The area of immersed midship section is 106 sq. meter. Assume sea water density =1.025 t/m3. If the beam =0.13 Length=2.1 draught, Calculate the length of the ship and its prismatic coefficient. Given Data: Simplifying and solving above equation for Length, we get L= 117.84m ๏ณ Using equation : ๐๐๐ = Calculate: ๐๐๐ =0.738 ๐๐∗๐ 76 Prepared by Dr. Sudhir Sindagi Displacements A ship’s displacement significantly influences its behaviour at sea. A ship is said to be fully loaded, when it is floating at its minimum statutory freeboard. Freeboard is the vertical distance between the SLL or DWL or the load line corresponding to the particular zone and the deckline. Deckline is obtained by the intersection of the upper surface of the main deck and the outer surface of the side shell. It is a line having length of 300mm and thickness of 25 mm marked amidships on both sides of the ship, which is a part of Plimsol line (Load line). The main deck of a ship is the uppermost continuous weatherproof deck extending from bow to stern (running for the entire length). 77 Prepared by Dr. Sudhir Sindagi Displacements 78 Prepared by Dr. Sudhir Sindagi Displacements Lightweight of the ship is the weight of the newly built ship. It includes the weights of – – – – – Hull and Superstructure Machinery and Equipment Piping and ducting Cables and furnishings Accommodation Lightship or lightweight measures the actual weight of the ship with no fuel, passengers, cargo, water. Deadweight = fully loaded displacement – lightweight. Deadweight tonnage (often abbreviated as DWT) is the displacement at any loaded condition minus the lightship weight. It includes the crew, 79 passengers, cargo, fuel, water, provisions, and stores. Prepared by Dr. Sudhir Sindagi Effect of density on draft For Box Barge, as the weight in both SW and FW remains same โ๐๐ = โ๐ ๐ โ๐๐ = ๐๐๐ ∗ ๐๐๐ = โ๐ ๐ = ๐๐ ๐ ∗ ๐๐ ๐ L*B∗ ๐๐๐ * ๐๐๐ = L*B∗ ๐๐ ๐ ∗ ๐๐ ๐ ๐๐๐ * ๐๐๐ = ๐๐ ๐ ∗ ๐๐ ๐ Note: This equation is applicable only for the box barge Since, ๐๐ ๐ < ๐๐๐ < ๐๐๐ , hence ๐๐ ๐ > ๐๐๐ > ๐๐๐ Fresh Water Allowance (FWA) = ๐๐ ๐ − ๐๐๐ Dock Water Allowance (DWA) = ๐๐๐ − ๐๐๐ 80 Prepared by Dr. Sudhir Sindagi Effect of density on draft 81 Prepared by Dr. Sudhir Sindagi Problems on Displacements A ship is 150m long, has 20m beam, load draft 8m, light draft 3m. The block coefficient at the load draft is 0.766, and at the light draft is 0.668. Find the ship's deadweight. Given Data: L=150m, B= 20m, TSW=8m, TL=3m, CBFL=0.766, CBL=0.668, Deadweight=? ๏ณ Using equation: ๐๐ = , we calculate ๐ ∗๐ ∗๐ ๏ณFL= 18384 m3 and ๏ณL= 6012 m3 putting the terms, we get Deadweight = fully loaded displacement – lightweight Using equation โ= ๐ ∗ ๐ 82 Prepared by Dr. Sudhir Sindagi Problems on Displacements A ship is 150m long, has 20m beam, load draft 8m, light draft 3m. The block coefficient at the load draft is 0.766, and at the light draft is 0.668. Find the ship's deadweight. Deadweight = fully loaded displacement – lightweight. Deadweight = (๏ณFL– ๏ณL)* ρSW Deadweight=12681.3 tonnes 83 Prepared by Dr. Sudhir Sindagi Problems on Displacements A ship 120m long 15m beam has a block coefficient of 0.700 and is floating at the load draft of 7m in fresh water. Find how much more cargo can be loaded if the ship is to float at the same draft in salt water. Given Data: L=120m, B= 15m, TFW=7m, CB=0.7, TSW=7m To maintain same draft in SW how much more cargo to be added? Since, ๐๐ ๐ < ๐๐๐ < ๐๐๐ , hence ๐๐ ๐ > ๐๐๐ > ๐๐๐ Hence, to maintain same draft ins SW, need to add weight. Here, to maintain same draft, it will have similar underwater volume both in SW and FW. ๏ณSW = ๏ณFW 84 Prepared by Dr. Sudhir Sindagi Problems on Displacements A ship 120m long 15m beam has a block coefficient of 0.700 and is floating at the load draft of 7m in fresh water. Find how much more cargo can be loaded if the ship is to float at the same draft in salt water. ๏ณ Using equation: ๐๐ = , we calculate ๐ ∗๐ ∗๐ ๏ณFW= 8820 m3 = ๏ณSW ΔFW= ๏ณFW* ρFW = 8820 tonnes ΔSW=๏ณSW* ρSW = 9040.5 tonnes Weight to be added = ΔSW- ΔFW Weight to be added = 220.5 tonnes 85 Prepared by Dr. Sudhir Sindagi Problems on Displacements A homogeneous log of rectangular cross-section is 5m long, 60 cm wide, 40 cm deep, and floats in fresh water at a draft of 30 cm. Find the mass of the log and its relative density. Given Data: L=5m, B=0.6m, D=0.4 TFW=0.3m Δ =? Relative density=? Since it is the homogeneous rectangular log, hence its block coefficient will be ๐๐ = 1.0 Mass of the log = ๏ณ * ρFW Mass of the log =๐๐ * LB TFW * ρFW 86 Mass of the log =0.9 tonnes Prepared by Dr. Sudhir Sindagi Problems on Displacements A homogeneous log of rectangular cross-section is 5m long, 60 cm wide, 40 cm deep, and floats in fresh water at a draft of 30 cm. Find the mass of the log and its relative density. Density of the body = ๐๐๐ฌ๐ฌ ๐จ๐ ๐ญ๐ก๐ ๐๐จ๐๐ฒ ๐๐จ๐ญ๐๐ฅ ๐ฏ๐จ๐ฅ๐ฎ๐ฆ๐ ๐จ๐ ๐ญ๐ก๐ ๐๐จ๐๐ฒ Density of the body = ๐.๐ ๐๐๐ Putting the value ,we get Density of the body = 0.75 t/m3 Relative density is the ratio of the density of a substance to the density of a standard, usually water for a liquid or solid, and air for a gas. Relative Density of the body = 0.75 87 Prepared by Dr. Sudhir Sindagi CG and CB of the ship B - Center of Buoyancy: The geometric center of the ship's underwater hull body. Centroid of the underwater volume is the CB of the ship. It is the point at which all the forces of buoyancy may be considered to act in a vertically upward direction. The Center of Buoyancy will move as the shape of the underwater portion of the hull body changes. The centre of gravity of the ship is commonly denoted as point G. It is the point on which whole mass of the ship is supposed to be acting. When a ship is at equilibrium, the centre of buoyancy is vertically in line with the centre of gravity of the ship 88 Prepared by Dr. Sudhir Sindagi CG and CB of the ship 89 Prepared by Dr. Sudhir Sindagi CG and CB of the ship Longitudinal Center of Gravity (LCG): It is the Longitudinal distance of the CG from the AP. Vertical Center of Gravity (VCG of KG): It is the vertical distance of the CG from the keel. Transverse Center of Gravity (TCG): It is the transverse distance of the CG from the centreline of the ship. Longitudinal Center of Buoyancy (LCB): Vertical Center of Buoyancy (VCB or KB): Transverse Center of Buoyancy (TCB): For the upright condition of the ship TCG=TCB=0 as both CG and CB of the ship lie on centreline. 90 For Box barge, always KB=T/2 Prepared by Dr. Sudhir Sindagi CG and CB of the ship If TCG≠ ๐๐๐, ๐ญ๐ก๐๐ง ๐ญ๐ก๐ ๐ฌ๐ก๐ข๐ฉ ๐ฐ๐ข๐ฅ๐ฅ ๐๐ข๐ฌ๐ญ If LCG≠ ๐๐๐, ๐ญ๐ก๐๐ง ๐ญ๐ก๐ ๐ฌ๐ก๐ข๐ฉ ๐ฐ๐ข๐ฅ๐ฅ ๐๐ซ๐ข๐ฆ 91 Prepared by Dr. Sudhir Sindagi Problems on Displacements The KB of a rectangular block which is floating in fresh water is 50 cm. Find the KB in salt water. Given Data: KBFW=0.5m KBSW=? We know that, for Box Barge, as the weight in both SW and FW remains same ๐๐๐ * ๐๐๐ = ๐๐ ๐ ∗ ๐๐ ๐ ๐๐๐ =0.9756m Moreover, it is also known to that, for Box barge, always KB=T/2 ๐๐๐๐ = ๐๐๐ /2 92 ๐๐๐๐ =0.4878m Prepared by Dr. Sudhir Sindagi Problems on Displacements A box-shaped barge 55mX10mX6m is floating in fresh water on an even keel at 1.5m draft. If 1800 tonnes of cargo is now loaded, find the difference in the height of the centre of buoyancy above the keel. Given Data: L=55m, B= 10m, D= 6m, T OLD =TFW=1.5m, w= 1800 tonnes added. ๐๐๐๐๐ - ๐๐๐๐๐ = ? ΔOLD= ๏ณOLD* ρFW = LBT OLD * ρFW ΔOLD= 825 tonnes ΔNEW= ΔOLD +weight added = 2625 tonnes ΔNEW= ๏ณNEW* ρFW = LBT NEW * ρFW Solving it for T NEW we get T NEW = 4.77m 93 KB NEW = T NEW/2 = 2.38m Prepared by Dr. Sudhir Sindagi Problems on Displacements A box-shaped barge 55mX10mX6m is floating in fresh water on an even keel at 1.5m draft. If 1800 tonnes of cargo is now loaded, find the difference in the height of the centre of buoyancy above the keel. KB OLD = T OLD/2 = 0.75m ๐๐๐๐๐ - ๐๐๐๐๐ = 1.63m 94 Prepared by Dr. Sudhir Sindagi Problems on Displacements A homogeneous log is 3m long, 60 cm wide, 60 cm deep, and has relative density 0.9. Find the distance between the centres of buoyancy and gravity when the log is floating in fresh water. Given Data: L=3m, B=0.6m, D=0.6, RD=0.9 KG-KB= ? In case of solids density of the body=relative density=0.9 t/m3 Weight of the water displaced by the body = weight of the body Weight of the body = Total volume of the body * density of the body Δ = LBD *density of the body Δ = 0.972 tonnes 95 Prepared by Dr. Sudhir Sindagi Problems on Displacements A homogeneous log is 3m long, 60 cm wide, 60 cm deep, and has relative density 0.9. Find the distance between the centres of buoyancy and gravity when the log is floating in fresh water. Weight of the body=underwater volume of the body*density of the water Δ= ๏ณFW* ρFW = LBT FW * ρFW Solving it for T FW TFW = 0.54m. KB = T/2 = 0.27m Assuming KG is at the centroid of the body KG= D/2= 0.3m KG-KB= 0.03m 96 Prepared by Dr. Sudhir Sindagi TPC, FWA and DWA Tonnes per centimeter immersion (TPC): It is the weight required to be added or to be removed from the ship to change the draft of a vessel by 1 cm. Let AWP is the waterplane area of the vessel at any draft and is presumed to remain constant for small change of draft. 97 Prepared by Dr. Sudhir Sindagi TPC, FWA and DWA TPC =AWP (m2) x 1 (cm) x ρSW TPC = AWP (m2) x 1/100 (m) x ρSW (note 1 is divided by 100 to convert centimeter to meter). ๐๐๐ = ๐๐๐ ∗๐๐๐ ๐๐๐ The value of TPC will be different at different water density at the same draft. TPC is not constant for all the draft of a vessel. The water plane area of a merchant vessel changes with the draft. TPC is also not constant when the vessel is floating at same draft but in water of different density ๐๐๐๐๐ ๐๐๐๐๐ = ๐๐๐ ๐๐๐ 98 Prepared by Dr. Sudhir Sindagi TPC, FWA and DWA Derivation for Fresh Water Allowance (FWA = ๐๐ ๐ − ๐๐๐ ) Let us presume that the displacement of a vessel is Δ tonnes. The vessel is floating in seawater, density ρSW (1.025 t/m3). It is presumed that a giant crane lifts the vessel and put it in freshwater ρFW, (1.000 t/m3). Volumetric change in displacement due to change in water density d๏ = (Δ/ρFW)– (Δ/ρSW) d๏ = Δ x (ρSW – ρFW) / (ρSW * ρFW) If AWP is the area of the waterplane is assumed to remain constant for small change of draft dT, then d๏ = AWP x dT = Δ x (ρSW – ρFW) / (ρSW * ρFW) dT = FWA= Δ/ AWP x (ρSW – ρFW) / (ρSW * ρFW) 99 Prepared by Dr. Sudhir Sindagi TPC, FWA and DWA Derivation for Fresh Water Allowance (FWA = ๐๐ ๐ − ๐๐๐ ) FWA = Δ / AWP x (ρSW – ρFW) / (ρSW * ρFW) Putting the values of ρSW=1.025 t/m3 and ρFW=1.000 t/m3 and multiplying and dividing by 100 FWA = Δ / AWP x (1.025 – 1.0) / (ρSW *1.0*100/100 Rearranging the terms FWA = Δ x 0.025 / (ρSW*AWP /100) *100 FWA = Δ x 0.025 / TPCsw *100 FWA = Δ/(4000*TPCsw) m FWA = Δ/(40*TPCsw) cm FWA =๐๐ ๐ − ๐๐๐ = Δ/(4*TPCsw) mm 100 Prepared by Dr. Sudhir Sindagi TPC, FWA and DWA Derivation for Dock Water Allowance (DWA = ๐๐๐ − ๐๐๐ ) ๐๐๐ = ๐๐๐ −๐๐๐ = ๐ ๐๐ ∗ ๐๐๐ ๐๐๐ −๐๐๐ ๐๐๐ −๐๐ ๐ 101 Prepared by Dr. Sudhir Sindagi Problems on TPC, FWA and DWA The waterplane area of the ship is 4525.3m2. If a mass of 100 tonnes is added what will be increase in the draft (SW). Given Data: AWP=4525.3m2 w=100 tonnes added. ΔTSW= ? Using equation ๐๐๐๐๐ = ๐๐๐ ∗๐๐๐ , ๐๐๐ we get ๐๐๐๐๐ = 46.38 tonnes/cm Change in draft due to the addition or the removal of weight ๐๐๐ข๐ ๐ก๐ญ ๐๐๐๐๐ ๐จ๐ซ ๐ซ๐๐ฆ๐จ๐ฏ๐๐ ๐๐๐ ๐๐๐ ΔT= = 2.15 cm ๐๐.๐๐ ΔT= ๐๐ฆ 102 Prepared by Dr. Sudhir Sindagi Problems on TPC, FWA and DWA A ship with draft of 8.2m in dock water of density 1.01t/m3. TPC in SW =40. How much cargo the vessel will be able to load to bring her draft to 8.4m in the DW Given Data: TDW= 8.2m,, ๐๐๐๐๐ = 40 t/cm, To have TDW= 8.4m, w=? Using ๐๐๐๐๐ equation, ๐๐๐๐๐ = ๐๐๐ , ๐ฐ๐ ๐๐๐ ๐ ๐๐ญ ๐๐๐๐๐ =39.41 t/cm Change in draft expected = 0.2m Weight to be added = (ΔT) in cm* TPCDW Weight to be added = 788.29 tonnes. 103 Prepared by Dr. Sudhir Sindagi Problems on TPC, FWA and DWA A ship if floating in fresh water at a draft of 6.8m. Its maximum allowable draft in the fresh water is 7m. Its TPC in the SW at the corresponding draft of 6.8m is 40. Find the available dead weight tonnage. Given Data: TFW= 6.8m (TFW)max=7.0m ๐๐๐๐๐ = 40 t/cm Available deadweight=? Available deadweight = (ΔT) in cm* TPCFW Change in draft expected = 0.2m Using equation, ๐๐๐๐ ๐ ๐๐๐๐๐ = ๐๐ ๐ , ๐ฐ๐ ๐๐๐ ๐ ๐๐ญ ๐๐๐๐ ๐ =39.0243 t/cm Available deadweight = 780.48 tonnes 104 Prepared by Dr. Sudhir Sindagi Problems on TPC, FWA and DWA A ship of 10000tonnes displacement is floating in salt water. The ship has to proceed to a berth where density of water is 1008kg/m3. Find how much cargo must be discharged, if same salt water draft is to be maintained. FWA= 200mm. Given Data: ΔSW=10000 tonnes ๐๐๐ 1.008 t/m3 ,TSW= TDW , FWA= 200mm Weight to be removed = ? Change in draft expected by moving the ship from DW to SW = DWA Using equation ๐๐๐ = ๐๐๐ −๐๐๐ = ๐ ๐๐ ∗ ๐๐๐ DWA = 134.92mm Weight to be removed = DWA in cm * TPCDW ๐๐๐ −๐๐๐ ๐๐๐ −๐๐ ๐ 105 Prepared by Dr. Sudhir Sindagi Problems on TPC, FWA and DWA A ship of 10000tonnes displacement is floating in salt water. The ship has to proceed to a berth where density of water is 1008kg/m3. Find how much cargo must be discharged, if same salt water draft is to be maintained. FWA= 200mm. We know that, FWA =๐๐ ๐ − ๐๐๐ = Δ/(4*TPCSW) mm , we get TPCSW = 12.5 t/cm Using equation, ๐๐๐๐๐ ๐๐๐๐๐ = ๐๐๐ , ๐ฐ๐ ๐๐๐ ๐ ๐๐ญ ๐๐๐๐๐ =12.2926 t/cm Weight to be removed = DWA in cm* TPCDW Weight to be removed =165.85 tonnes. 106 Prepared by Dr. Sudhir Sindagi Problems on TPC, FWA and DWA A vessel of FWA 200mm goes from water of RD 1.018 to a water of RD 1.006. Find the change in draft and state whether there will be sinkage or rise? Given Data: FWA= 200mm, ๐๐๐๐ =1.018 t/m3 , ๐๐๐๐ = 1.006 t/m3 ΔT= ? When moved from ๐๐๐๐ to ๐๐๐๐ Since, ๐๐๐๐ < ๐๐๐๐ < ๐๐๐ , hence ๐๐๐๐ > ๐๐๐๐ > ๐๐๐ ΔT When moved from ๐๐๐๐ to ๐๐๐๐ = DWA2- DWA1 107 Prepared by Dr. Sudhir Sindagi Problems on TPC, FWA and DWA A vessel of FWA 200mm goes from water of RD 1.018 to a water of RD 1.006. Find the change in draft and state whether there will be sinkage or rise? Using equation ๐๐๐ = ๐๐๐ −๐๐๐ = ๐๐๐๐ = ๐๐๐๐ −๐๐๐ = ๐ ๐๐ ๐๐๐ ๐ ๐๐๐๐ = ๐๐๐๐ −๐๐๐ = ๐ ๐๐ ๐๐๐ ๐ ๐ ๐๐ ∗ ๐๐๐ ∗ ๐๐๐ −๐๐๐๐ ๐๐๐ −๐๐ ๐ ∗ ๐๐๐ −๐๐๐๐ ๐๐๐ −๐๐ ๐ ๐๐๐ −๐๐๐ ๐๐๐ −๐๐ ๐ ๐๐๐๐ = 55.009mm ๐๐๐๐ = 151.093mm ΔT When moved from ๐๐๐๐ to ๐๐๐๐ = DWA2- DWA1 ΔT When moved from ๐๐๐๐ to ๐๐๐๐ =96.084mm 108 Prepared by Dr. Sudhir Sindagi Problems on TPC, FWA and DWA A vessel arrives at a port X at the mouth of a river. Her displacement at that time was 12000tonnes and arrival draft is 5.77m in water with RD of 1.02. She has to cross a bar upriver before entering port Y. The depth of water over the bar is 6m and RD 1.005. If her TPC in SW is 25, find the minimum quantity of cargo to be offloaded at port X so that she can cross the bar with an under-keel clearance of 0.5m Given Data: Δ= 12000 tonnes, ๐๐๐๐ = 5.77m ๐๐๐๐ =1.02 t/m3 , ๐๐๐๐ = 1.005 t/m3 Depth of water at Y port = 6m, ๐๐๐๐๐ =25, UKC at port Y =0.5m Weight to be removed=? 109 Prepared by Dr. Sudhir Sindagi Problems on TPC, FWA and DWA We know that, ΔT When moved from ๐๐๐๐ to ๐๐๐๐ = DWA2- DWA1 Draft to be maintained in river Y = Depth of water in river Y – UKC Draft to be maintained in river Y =6-0.5 =5.5m Again, final draft in river Y = draft in river X + ΔT When moved from ๐๐๐๐ to ๐๐๐๐ Final draft in river Y = draft in river X + (DWA2- DWA1) 110 Prepared by Dr. Sudhir Sindagi Problems on TPC, FWA and DWA Using equation ๐๐๐ = ๐๐๐ −๐๐๐ = ๐๐๐๐ = ๐๐๐๐ −๐๐๐ = ๐ ๐๐ ๐๐๐ ๐ ๐๐๐๐ = ๐๐๐๐ −๐๐๐ = ๐ ๐๐ ๐๐๐ ๐ ๐ ๐๐ ∗ ๐๐๐ ∗ ๐๐๐ −๐๐๐๐ ๐๐๐ −๐๐ ๐ ∗ ๐๐๐ −๐๐๐๐ ๐๐๐ −๐๐ ๐ ๐๐๐ −๐๐๐ ๐๐๐ −๐๐ ๐ ๐๐๐๐ = 23.5mm ๐๐๐๐ = 95.5mm ΔT When moved from ๐๐๐๐ to ๐๐๐๐ = DWA2- DWA1 ΔT When moved from ๐๐๐๐ to ๐๐๐๐ =72mm Final draft in river Y = draft in river X + (DWA2- DWA1) Final draft in river Y = 5.77+0.072 = 5.842m Expected draft in River Y = 5.5m 111 Prepared by Dr. Sudhir Sindagi Problems on TPC, FWA and DWA Expected change in draft = 5.842- 5.5 =0.342m Weight to be removed = Expected change in draft in cm* TPCDW2 Using equation, ๐๐๐๐๐๐ ๐๐๐๐๐ = ๐๐๐๐ , ๐ฐ๐ ๐๐๐ ๐ ๐๐ญ ๐๐๐๐๐ =24.512 t/cm Weight to be removed = Expected change in draft in cm* TPCDW2 Weight to be removed =838.31 tonnes 112 Prepared by Dr. Sudhir Sindagi Load Line or Plimsol Line Load line is a special marking positioned amidships which depicts the draft of the vessel to which the ship can be loaded. The fundamental purpose of a Load Line is to allot a maximum legal limit up to which a ship can be loaded by cargo. By prescribing such limits, the risk of having the vessel sailing with inadequate freeboard and buoyancy can be limited. A vessel should be having sufficient freeboard at all times, any exceptions made will result in insufficient stability and excessive stress on the ship’s hull. However, since the buoyancy and immersion of the vessel largely depend on the type of water and its density, it is not practical to define a standard freeboard limit for the ship at all times. 113 Prepared by Dr. Sudhir Sindagi Load Line or Plimsol Line For this reason, the load line convention has put regulations which divides the world into different geographical zones each having different prescribed load line. For example, A vessel sailing in Winter on North Atlantic Ocean will have a greater freeboard than on a voyage in Tropical Zones and Fresh waters. All vessels of 24 meters and more are required to have this Load line marking at the centre position of the length of summer load water line. There are two types of Load line markings:– Standard Load Line marking – This is applicable to all types of vessels. – Timber Load Line Markings – This is applicable to vessels carrying timber cargo 114 Prepared by Dr. Sudhir Sindagi Load Line or Plimsol Line Load lines are horizontal lines extending forward and aft from a vertical line placed at a distance of 540mm from the centre of the disc. They measure 230mm by 25mm. The upper surfaces of the load lines indicate the maximum depths to which the ships maybe submerged in different seasons and circumstances. 115 Prepared by Dr. Sudhir Sindagi Load Line or Plimsol Line S – Summer :- It is the basic freeboard line at the same level as the Plimsol Line. Other load lines are marked based on this Summer freeboard line. T – Tropical :- It is 1/48th of summer draft marked above the Summer load line. W – Winter :- It is 1/48th of summer draft marked below the SLL. WNA – Winter North Atlantic :- It is marked 50mm below the Winter load line. It applies to voyages in North Atlantic (above 36 degrees of latitude) during winter months. F – Fresh Water :- It is the summer fresh water load line. The distance between S and F is the Fresh Water Allowance (FWA). TF – Tropical Fresh Water :- It is the fresh water load line in Tropical. It is 116 marked above the T at an amount equal to 1/48th of summer draft. Prepared by Dr. Sudhir Sindagi Load Line or Plimsol Line 117 Prepared by Dr. Sudhir Sindagi Load Line or Plimsol Line 118 Prepared by Dr. Sudhir Sindagi Problems on Load line Calculation of Freeboard in various zones FSW = Minimum Statutory Freeboard in SW ( Summer Freeboard) ๐๐๐ mm ๐๐ ๐๐๐ mm ๐๐ Freeboard in Tropical Zone = FT = ๐ ๐๐ − Freeboard in Winter Zone = FW = ๐ ๐๐ + Freeboard in Winter North Atlantic Zone = FWNA = ๐ ๐ + ๐๐ ๐ฆ๐ฆ Freeboard in Fresh water Zone = FFW = ๐ ๐๐ − ๐ ๐๐ mm Freeboard in Tropical Fresh water Zone = FTF = ๐ ๐๐ − ๐ ๐๐ − ๐๐๐ ๐๐ 119 mm Prepared by Dr. Sudhir Sindagi Problems on Load line Calculation of Drafts in various zones TSW = Draft in SW ( Summer Freeboard) ๐๐๐ mm ๐๐ ๐ ๐๐๐ − ๐๐ mm ๐๐ Draft in Tropical Zone = TT = ๐๐๐ + Draft in Winter Zone = TW = Draft in Winter North Atlantic Zone = TWNA = ๐๐๐ − ๐๐ ๐ฆ๐ฆ Draft in Fresh water Zone = TFW = ๐๐๐ + ๐ ๐๐ mm Draft in Tropical Fresh water Zone = TTF = ๐๐๐ + ๐ ๐๐ + ๐๐๐ ๐๐ mm 120 Prepared by Dr. Sudhir Sindagi Problems on Load line A ship of 4477 tonnes displacement is less than 100m long. It floats at a summer load line draft of 7m. Find the drafts of it in Winter, WNA, Tropical, FW, Tropical FW load line markings. Area of waterplane is 622m2 Given Data: Δ= 4477 tonnes, , T๐๐ =7m, ๐จ๐๐ =622m2 ๐๐๐๐๐ = ๐๐๐ ∗๐๐๐ ๐๐๐ = 637.55 t/cm FWA =๐๐ ๐ − ๐๐๐ = Δ/(4*TPCsw) mm FWA =175mm 121 Prepared by Dr. Sudhir Sindagi Problems on Load line Calculation of Drafts in various zones TSW = Draft in SW ( Summer Freeboard) =7m ๐๐๐ mm = 7.1458m ๐๐ ๐ ๐๐๐ − ๐๐ mm =6.8541m ๐๐ Draft in Tropical Zone = TT = ๐๐๐ + Draft in Winter Zone = TW = Draft in Winter North Atlantic Zone = TWNA = ๐๐๐ − ๐๐ ๐ฆ๐ฆ =6.8041m Draft in Fresh water Zone = TFW = ๐๐๐ + ๐ ๐๐ mm= 7.175m Draft in Tropical Fresh water Zone = TTF = ๐๐๐ + ๐ ๐๐ + ๐๐๐ ๐๐ mm TTF = 7.32m 122 Prepared by Dr. Sudhir Sindagi Problems on Load line A vessel is lying in a river berth of RD 1.01 with her summer load line mark 20mm above the water on the starboard side and 50mm above on the port side. Find how much cargo she can load to bring her to her “S” line markings in SW, if her summer displacement is 15000 tonnes and TPC 25. Given Data: ๐๐๐ = 1.01 t/m3 Δ= 15000 tonnes, ๐๐๐๐๐ = 25 Weight to be added = ? When draft in SW will matching with S line. If we assume mean WL in DW, then it will be 35mm below the S line. Moreover, when ship moves from DW to SW, then draft will reduce by the amount DWA. Total expected change in draft in SW= DWA+35mm 123 Prepared by Dr. Sudhir Sindagi Problems on Load line Using equation ๐๐๐ = ๐๐๐ −๐๐๐ = ๐ ๐๐ ∗ ๐๐๐ ๐๐๐ −๐๐๐ ๐๐๐ −๐๐ ๐ ๐๐๐ = 89.1mm Total expected change in draft in SW= DWA+35mm = 124.1 124 Prepared by Dr. Sudhir Sindagi Problems on Load line FWA =๐๐ ๐ − ๐๐๐ = Δ/(4*TPCsw) mm FWA =150mm Weight to be added = ΔT in cm* TPCSW Weight to be added =310.25 tonnes 125 Prepared by Dr. Sudhir Sindagi Summary Ship geometry, Definition of hull surface Lines plan drawing of ships Offset Table Bonjean Curves Archimedes principle, Displacements Coefficients of form Effect of Density and relative density of a liquid on drafts. Meaning of buoyancy, reserve buoyancy Center of Gravity and Center of Buoyancy of ship TPC, FWA, DWA Plimsoll line or Load line markings 126 Prepared by Dr. Sudhir Sindagi Numerical Integration in Naval Architecture By Dr. Sudhir Sindagi 1 Prepared by Dr. Sudhir Sindagi Chapter Content Introduction Use of integration in finding various properties Rules of Integration – Trapezoidal Rule – Simpson’s (1/3)rd rule – Simpson’s (3/8)th rule Numerical Simpson’s 5-8-(-1) rule Simpson’s 3-10-(-1) rule 2 Prepared by Dr. Sudhir Sindagi Introduction As previously stated, the shape of a ship's hull cannot usually be described by mathematical equations. In order to calculate fundamental geometric properties of the hull, naval architects use numerical methods. This chapter describes two methods, the trapezoidal and Simpson’s rules. The latter yields better approximations, but imposes a condition on the input. The following different properties of a hull form are estimated – Area of Waterplane, centroid of waterplane: Center of Flotation, LCF & TPC – Moment of Inertia of the Waterplane about Longitudinal Axis: Transverse MI (IT) – Moment of Inertia of the Waterplane about Transverse Axis passing through centroid: Longitudinal MI (IL) 3 – Underwater volume and its centroid (Center of Buoyancy), KB & LCB Prepared by Dr. Sudhir Sindagi Use of Integration In order to find the various properties of an area under the curve y=F(x) Consider a small elemental area dxdy Let ๐ฑเดค is the distance of the centroid of the small elemental area from the Y axis and ๐ฒเดค is the distance of the centroid of the small elemental area from the X axis Let, x and y is the distance of any point A on the curve from the x & y axis as shown dA= area of small elemental area = ydx Total Area under the curve will be (A) = ๐ โซ๐ฑ๐๐ฒ ๐ืฌโฌ 4 Prepared by Dr. Sudhir Sindagi Use of Integration Position of Centroid from Y axis เดฅ = ๐ ๐ข๐ซ๐ฌ๐ญ ๐๐จ๐ฆ๐๐ง๐ญ ๐จ๐ ๐๐ซ๐๐ ๐๐๐จ๐ฎ๐ญ ๐๐ฑ๐ข๐ฌ ๐๐ ๐ ๐๐ซ๐๐ ๐ เดฅ= ๐ โซ๐ฑ๐๐ฒ๐ฑ ๐ืฌโฌ ๐ โซ๐ฑ๐๐ฒ ๐ืฌโฌ Position of Centroid from X axis เดค = ๐ ๐ข๐ซ๐ฌ๐ญ ๐๐จ๐ฆ๐๐ง๐ญ ๐จ๐ ๐๐ซ๐๐ ๐๐๐จ๐ฎ๐ญ ๐๐ฑ๐ข๐ฌ ๐๐ ๐ ๐๐ซ๐๐ เดค= ๐ ๐ ๐ ๐ ∗โซ๐ฑ๐ ๐ฒ ืฌโฌ ๐ ๐ ๐ โซ๐ฑ๐๐ฒ ๐ืฌโฌ 5 Prepared by Dr. Sudhir Sindagi Use of Integration Moment of Inertia of the area about Longitudinal Axis: Transverse MI (IT) IT= Second moment of area about OX ๐ IT= ๐ ∗ ๐ ๐ โซ๐ฑ๐ ๐ฒ ๐ืฌโฌ Moment of Inertia of the area about Transverse Axis : Longitudinal MI (IL) IL= Second moment of area about OY ๐ ๐ IL= โซ๐ฑ๐๐ฒ ๐ฑ ๐ืฌโฌ 6 Prepared by Dr. Sudhir Sindagi Numerical Integration What does an integral represent? ๏ฒ b a d b c a ๏ฒ๏ฒ f ( x )dx = area f ( x)dxdy = volume Basic definition of an integral: ๏ฒ b a f(x) n f ( x )dx = lim๏ฅ f ( xk )๏x where n →๏ฅ ๏x = k =1 b−a n ๏x 7 Prepared by Dr. Sudhir Sindagi Trapezoidal Rule f fp 8 Prepared by Dr. Sudhir Sindagi Trapezoidal Rule Function f approximately by function fp. Then, ๏ฒ fdx ๏ ๏ฒ f p dx where fp is a linear polynomial interpolation, that is fp = ๏ฒ fdx ๏ ๏ฒ where (x − b ) f (a) + (x − a ) f (b) (a − b ) (b − a ) h f p dx = ๏ป f (a ) + f (b)๏ฝ 2 h =b−a 9 Prepared by Dr. Sudhir Sindagi Trapezoidal Rule f fp 10 Prepared by Dr. Sudhir Sindagi Trapezoidal Rule For two interval, we can use summation operation to derive the formula of two interval trapezoidal that is h h ๏ฒ fdx ๏ 2 ๏ป f ( x0 ) + f ( x1 )๏ฝ+ 2 ๏ป f ( x1 ) + f ( x2 )๏ฝ h = ๏ป f ( x0 ) + 2 f ( x1 ) + f ( x2 )๏ฝ 2 where h = x2 − x0 2 11 Prepared by Dr. Sudhir Sindagi Trapezoidal Rule f fp 12 Prepared by Dr. Sudhir Sindagi Trapezoidal Rule Similar to two interval trapezoidal, we can derive three interval trapezoidal formula that is h ๏ฒ fdx ๏ 2 ๏ป f ( x0 ) + 2 f ( x1 ) + 2 f ( x2 ) + f ( x3 )๏ฝ x3 − x0 where h= 3 Thus, for n interval we have ๏ฒ where n −1 h๏ฌ ๏ผ fdx ๏ ๏ญ f ( x0 ) + 2๏ฅ f ( xi ) + f ( xn )๏ฝ 2๏ฎ i =1 ๏พ xn − x0 h= n Prepared by Dr. Sudhir Sindagi Simpson’s 1/3rd Rule f fp 14 Prepared by Dr. Sudhir Sindagi Simpson’s 1/3rd Rule Function f approximately by function fp. Then, ๏ฒ fdx ๏ ๏ฒ f p dx where fp is a quadratic polynomial interpolation, that is ( ( ( x − x0 )(x − x2 ) x − x0 )(x − x1 ) x − x1 )(x − x2 ) fp = f ( x0 ) + f ( x1 ) + f ( x2 ) (x0 − x1 )(x0 − x2 ) (x1 − x0 )(x1 − x2 ) (x2 − x0 )(x2 − x1 ) ๏ฒ fdx ๏ ๏ฒ where h f p dx = ๏ป f ( x0 ) + 4 f ( x1 ) + f ( x2 )๏ฝ 3 x2 − x0 h= 2 15 Prepared by Dr. Sudhir Sindagi Simpson’s 1/3rd Rule f fp 16 Prepared by Dr. Sudhir Sindagi Simpson’s 1/3rd Rule For 4 subinterval we have h ๏ฒ fdx ๏ 3 ๏ป f ( x0 ) + 4 f ( x1 ) + 2 f ( x2 ) + 4 f ( x3 ) + f ( x4 )๏ฝ x4 − x0 h= where 4 Thus, for n subinterval we have ๏ฒ n −1 n−2 ๏ผ h๏ฌ fdx ๏ ๏ญ f ( x0 ) + 4 ๏ฅ f ( xi ) + 2 ๏ฅ f ( x j ) + f ( xn )๏ฝ 3๏ฎ i =1, 3, 5 j = 2, 4, 6 ๏พ where xn − x0 h= n 17 Prepared by Dr. Sudhir Sindagi Simpson’s 3/8th Rule fp f 18 Prepared by Dr. Sudhir Sindagi Simpson’s 3/8th Rule Similar to 1/3 Simpson’s method, f approximately by function fp where fp is a cubic polynomial interpolation, that is ( ( x − x1 )(x − x2 )(x − x3 ) x − x0 )( x − x2 )( x − x3 ) fp = f ( x0 ) + f ( x1 ) (x0 − x1 )(x0 − x2 )(x0 − x3 ) (x1 − x0 )(x1 − x2 )(x1 − x3 ) (x − x0 )(x − x1 )(x − x3 ) f ( x ) + (x − x0 )(x − x1 )(x − x2 ) f ( x ) + (x2 − x0 )(x2 − x1 )(x2 − x3 ) 2 (x3 − x0 )(x3 − x1 )(x3 − x2 ) 3 ๏ฒ fdx ๏ ๏ฒ where 3h f p dx = ๏ป f ( x0 ) + 3 f ( x1 ) + 3 f ( x2 ) + f ( x3 )๏ฝ 8 x3 − x0 h= 3 Prepared by Dr. Sudhir Sindagi Simpson’s 3/8th Rule Thus, for n subinterval we have 3โ เถฑ๐๐๐ฅ ≅ ๐(๐ฅ0 ) + 3 8 where ๐−1 เท ๐=1,2,4,5,7 ๐−2 ๐(๐ฅ๐ ) + 2 เท ๐(๐ฅ๐ ) + ๐(๐ฅ๐ ) ๐=3,6,9.. xn − x0 h= n 20 Prepared by Dr. Sudhir Sindagi Formulae Trapezoidal Rule n −1 h๏ฌ ๏ผ f ( xi ) + f ( xn )๏ฝ ๏ฒ fdx ๏ 2 ๏ญ๏ฎ f ( x0 ) + 2๏ฅ i =1 ๏พ Simpson’s 1/3rd Rule ๏ฒ xn − x0 h= n n −1 n−2 ๏ผ h๏ฌ fdx ๏ ๏ญ f ( x0 ) + 4 ๏ฅ f ( xi ) + 2 ๏ฅ f ( x j ) + f ( xn )๏ฝ 3๏ฎ i =1, 3, 5 j = 2, 4, 6 ๏พ Simpson’s 3/8th Rule 3โ เถฑ๐๐๐ฅ ≅ ๐(๐ฅ0 ) + 3 8 ๐−1 เท ๐=1,2,4,5,7 ๐−2 ๐(๐ฅ๐ ) + 2 เท ๐(๐ฅ๐ ) + ๐(๐ฅ๐ ) ๐=3,6,9.. 21 Prepared by Dr. Sudhir Sindagi Formulae Trapezoidal Rule โ เถฑ๐๐๐ฅ ≅ ๐ ๐ฅ0 + ๐ ∗ R + ๐(๐ฅ๐ ) 2 Simpson’s 1/3rd Rule xn − x0 h= n Trapezoidal multipliers โ เถฑ๐๐๐ฅ ≅ ๐ ๐ฅ0 + ๐ ∗ Odd + ๐ ∗ Even + ๐(๐ฅ๐ ) 3 Simpson’s 3/8th Rule Simpson’s multipliers 3โ เถฑ๐๐๐ฅ ≅ ๐ ๐ฅ0 + ๐ ∗ Triplet + 3 ∗ R + ๐(๐ฅ๐ ) 8 22 Prepared by Dr. Sudhir Sindagi Applicability of Rules Following conditions are necessary in using different rules Sr Condition for the Method Remark No applicability Trapezoidal Can be used for any number Least Accurate 1 Rule of ordinates method of all Must be used for any odd More accurate rd Simpson’s 1/3 2 number of ordinates method than Rule N=1,3,5,7…. trapezoidal method Most accurate Must be used when number Simpson’s 3/8th method, if in clash of ordinates = 3*n+1, where 3 Rule with any method, n=1,2,3… then this method23to No of ordinates = 4,7,10,13 be used. Prepared by Dr. Sudhir Sindagi Problems A ship of 180m long has ½ widths of waterplane as follows 1, 7.5, 12, 13.5, 14, 14, 13.5, 12, 7 and 0m respectively. Calculate waterplane area, LCF, TPC & waterplane area coefficient. Given Data: No of ordinates = 10 Hence using Simpson’s 3/8th rule. ๐๐ฉ๐๐๐ข๐ง๐ ๐๐๐ญ๐ฐ๐๐๐ง ๐ญ๐ฐ๐จ ๐ฌ๐ญ๐๐ญ๐ข๐จ๐ง๐ฌ = ๐ก = ๐ก= ๐๐๐ ๐๐−๐ ๐๐๐ง๐ ๐ญ๐ก ๐จ๐ ๐ญ๐ก๐ ๐ฌ๐ก๐ข๐ฉ ๐ง๐ฎ๐ฆ๐๐๐ซ ๐จ๐ ๐จ๐ซ๐๐ข๐ง๐๐ญ๐๐ฌ −๐ = 20m Use tabular method to solve such problems, as it will be very easy to solve problems. 24 Prepare table as shown in the next slide Prepared by Dr. Sudhir Sindagi Problems As shown, the first station will be AP and the last will be FP. Since, it is half breadth plan, the values obtained through calculations shall be multiplied by the factor of 2. In between station 4 and 5, the midship lies with the biggest section. 25 Prepared by Dr. Sudhir Sindagi Problems Center of Flotation (F or CF): It is centroid of the water plane area taken at the SLL. It is the point about which the ship trims. Longitudinal distance of the F from the AP is known as the LCF 26 Prepared by Dr. Sudhir Sindagi Problems Col(1) Col (2) Station ½ Widths 0(AP) 1 1 7.5 2 12 3 13.5 4 14 5 14 6 13.5 7 12 8 7 9 (FP) 0 Col (3) Col(4)= (2)X(3) Col(5) Col(6)=(4)X(5) Simpson’s Multiplier For Area Lever from AP For Moment @ AP 27 ΣA=00m2 ΣM=00m3 Prepared by Dr. Sudhir Sindagi Problems Col(1) Col (2) Station ½ Widths Col (3) Col(4)= (2)X(3) Col(5) Col(6)=(4)X(5) Simpson’s Multiplier For Area Lever from AP For Moment @ AP 0(AP) 1 1 1 0 0 1 7.5 3 22.5 1 22.5 2 12 3 36 2 72 3 13.5 2 27 3 81 4 14 3 42 4 168 5 14 3 42 5 210 6 13.5 2 27 6 162 7 12 3 36 7 252 8 7 3 21 8 168 9 (FP) 0 1 0 ΣA=254.5m2 9 028 ΣM=1135.5m3 Prepared by Dr. Sudhir Sindagi Problems As per the Simpson’s 3/8th rule Total Area of the ๐๐ก waterplane=A=2* ๐ * ΣA A = 3817.5 m2 We know that, Position of Centroid from Y axis เดฅ = ๐ ๐ข๐ซ๐ฌ๐ญ ๐๐จ๐ฆ๐๐ง๐ญ ๐จ๐ ๐๐ซ๐๐ ๐๐๐จ๐ฎ๐ญ ๐๐ฑ๐ข๐ฌ ๐๐ ๐ ๐๐ซ๐๐ ๐๐ก ๐ 1st Moment of the area of the waterplane about the AP= 2*h* * ΣM 1st Moment of the area about the AP = 340650 m3 LCF = ๐ ๐ข๐ซ๐ฌ๐ญ ๐๐จ๐ฆ๐๐ง๐ญ ๐จ๐ ๐๐ซ๐๐ ๐๐๐จ๐ฎ๐ญ ๐๐ ๐๐ซ๐๐ LCF =89.23m from AP 29 Prepared by Dr. Sudhir Sindagi Problems One more way to get location of centroid LCF = ๐ ๐ข๐ซ๐ฌ๐ญ ๐๐จ๐ฆ๐๐ง๐ญ ๐จ๐ ๐๐ซ๐๐ ๐๐๐จ๐ฎ๐ญ ๐๐ ๐๐ซ๐๐ 2∗h∗๐๐ก ∗ ΣM ΣM ๐ = =h∗ ๐๐ก ΣA 2∗ ๐ ∗ ΣA ΣM ΣA LCF =89.23m from AP LCF = h∗ ๐๐ฌ๐ข๐ง๐ , ๐๐๐ = ๐๐๐ ∗๐๐๐ ๐๐๐ TPC= 39.1293 t/cm Using equation: ๐๐๐ = Here, B = 14*2 = 28m ๐๐๐ = 0.7475 ๐๐๐ ๐ ∗๐ 30 Prepared by Dr. Sudhir Sindagi Problems A ship is floating on an even keel at 6m draft. The areas of the waterplane are as follows. Find the ship’s KB at this draft. Draft(m) 0 1 2 3 4 5 6 AWP(m2) 5000 5600 6020 6025 6025 6025 6025 Given Data: No of ordinates = 7 Using Simpson’s 1/3rd rule. ๐ก= 1m When waterplanes are provided, then we will get underwater Volume and KB 31 Prepared by Dr. Sudhir Sindagi Problems Col(1) Col (2) Col (3) Col(4)= (2)X(3) Col(5) Col(6)=(4)X(5) Draft AWP Simpson’s Multiplier For Volume Lever from Keel For Moment @ Keel 0 5000 1 5600 2 6020 3 6025 4 6025 5 6025 6 6025 ΣV=00m3 ΣM=00m4 32 Prepared by Dr. Sudhir Sindagi Problems Col(1) Col (2) Col (3) Col(4)= (2)X(3) Col(5) Col(6)=(4)X(5) Draft AWP Simpson’s Multiplier For Volume Lever from Keel For Moment @ Keel 0 5000 1 5000 0 0 1 5600 4 22400 1 22400 2 6020 2 12040 2 24080 3 6025 4 24100 3 72300 4 6025 2 12050 4 48200 5 6025 4 24100 5 120500 6 6025 1 6025 6 36150 ΣV= 105715 ΣM= 323630 33 Prepared by Dr. Sudhir Sindagi Problems As per the Simpson’s 1/3rd rule Total ๐ก Volume=V= ๐ * ΣV V = 35238.33 m3 We know that, Position of Centroid from Keel เดฅ = ๐ ๐ข๐ซ๐ฌ๐ญ ๐๐จ๐ฆ๐๐ง๐ญ ๐จ๐ ๐ญ๐ก๐ ๐๐จ๐ฅ๐ฎ๐ฆ๐ ๐๐๐จ๐ฎ๐ญ ๐ญ๐ก๐ ๐๐๐๐ฅ ๐ ๐๐จ๐ฅ๐ฎ๐ฆ๐ ๐ก ๐ 1st Moment of the volume about the keel= h* * ΣM 1st Moment of the volume about the keel = ?? KB = ๐ ๐ข๐ซ๐ฌ๐ญ ๐๐จ๐ฆ๐๐ง๐ญ ๐จ๐ ๐ญ๐ก๐ ๐ฏ๐จ๐ฅ๐ฎ๐ฆ๐ ๐๐๐จ๐ฎ๐ญ ๐ญ๐ก๐ ๐ค๐๐๐ฅ ๐๐จ๐ฅ๐ฎ๐ฆ๐ KB =3.06m from Keel ΣM = =h∗ ΣV 34 Prepared by Dr. Sudhir Sindagi Problems A 200m long vessel has half ordinates of a waterplane as below commencing from AP. Calculate – – – – Waterplane area & LCF from AP TPC in SW MI of the waterplane area @ transverse axis through LCF MI of the waterplane area @ longitudinal axis (CL) Station 0(AP) ½ Ord 0 1 2 3 4 5 6 10 13 14 14.2 14.2 14.1 7 8 9 10 14 11.5 6.2 0 Given Data: No of ordinates = 11, hence using using Simpson’s 1/3rd rule. ๐๐ฉ๐๐๐ข๐ง๐ ๐๐๐ญ๐ฐ๐๐๐ง ๐ญ๐ฐ๐จ ๐ฌ๐ญ๐๐ญ๐ข๐จ๐ง๐ฌ = ๐ก = ๐ก = 20m ๐๐๐ง๐ ๐ญ๐ก ๐จ๐ ๐ญ๐ก๐ ๐ฌ๐ก๐ข๐ฉ ๐ง๐ฎ๐ฆ๐๐๐ซ ๐จ๐ ๐จ๐ซ๐๐ข๐ง๐๐ญ๐๐ฌ −๐ 35 Prepared by Dr. Sudhir Sindagi Problems (1) (2) (3) (4)= (2)X(3) (5) (6)=(4)X(5) (7)=(5)*(6) Station ½ Ord SM For Area Lever For 1st M@AP For IL 0(AP) 0 1 10 2 13 3 14 4 14.2 5 14.2 6 14.1 7 14 8 11.5 9 6.2 10 (FP) 0 (8)=(2)3 (9)=(3)*(8) For y3 For IT 36 ΣA=00m2 ΣM1=00m3 ΣM2=00m4 ΣIT=00m4 Prepared by Dr. Sudhir Sindagi Problems (1) (2) (3) (4)= (2)X(3) (5) (6)=(4)X(5) (7)=(5)*(6) Station ½ Ord SM For Area Lever For 1st M@AP For IL (8)=(2)3 (9)=(3)*(8) For IT 0(AP) 0 1 0 0 0 0 0 0 1 10 4 40 1 40 40 1000 4000 2 13 2 26 2 52 104 2197 4394 3 14 4 56 3 168 504 2744 10976 4 14.2 2 28.4 4 113.6 454.4 2863.288 5726.576 5 14.2 4 56.8 5 284 1420 2863.288 11453.15 6 14.1 2 28.2 6 169.2 1015.2 2803.221 5606.442 7 14 4 56 7 392 2744 8 11.5 2 23 8 184 1472 9 6.2 4 24.8 9 223.2 2008.8 238.328 10 (FP) 0 1 0 10 0 0 0 ΣM =1626 ΣM =9762.4 ΣA=339.2 2744 10976 1520.875 3041.75 953.312 37 0 ΣI =57127 Prepared by Dr. Sudhir Sindagi Problems As per the Simpson’s 1/3rd rule ๐ก Total Area=A=2* ๐ * ΣA A = 4522.66m2 LCF= ๐ ๐ข๐ซ๐ฌ๐ญ ๐๐จ๐ฆ๐๐ง๐ญ ๐จ๐ ๐ญ๐ก๐ ๐๐ซ๐๐ ๐๐๐จ๐ฎ๐ญ ๐๐ ๐๐ซ๐๐ = =h∗ ΣM1 ΣA LCF=95.87m from AP ๐๐ฌ๐ข๐ง๐ , ๐๐๐ = ๐๐๐ ∗๐๐๐ ๐๐๐ TPC= 46.357 t/cm Moment of Inertia of the area about Transverse Axis : Longitudinal MI (IL) =IL= Second moment of area about AP ๐ IL= โซ๐ฑ๐๐ฒ ๐ ๐ฑ ๐ืฌโฌ 38 Prepared by Dr. Sudhir Sindagi Problems ๐ก ๐ IL@ AP = 2*h2* * ΣM2 IL@ AP =52066133.33m4 However, we need to estimate the MI about an axis passing through center of flotation. Hence, using parallel axis theorem MI about any axis = MI about F +AWP*d2 MI about F = MI about AP - AWP*d2 Here d= LCF= 95.87m IL@ F =10498107.93m4 39 Prepared by Dr. Sudhir Sindagi Problems Moment of Inertia of the area about Longitudinal Axis: Transverse MI (IT) IT= Second moment of area about centreline ๐ ๐ ๐ IT= ∗ โซ๐ฑ๐ ๐ ๐ฒ ๐ืฌโฌ IT = 2* ๐ ๐ก ** * ๐ ๐ ΣIT IT = 253897.77m4 40 Prepared by Dr. Sudhir Sindagi Problems The TPC values for a ship at 1.2m intervals of draught commencing at the keel are 8.2, 16.5, 18.7, 19.4, 20.0, 20.5 and 21.1 respectively. Calculate displacement at 7.2m draught Given Data: No of ordinates = 7 Hence using Simpson’s 3/8th rule. ๐ก =1.2m ๐ ∗๐ When TPC values are provided, u๐ฌ๐ข๐ง๐ ๐๐๐ = ๐๐ ๐๐, one needs to ๐๐๐ calculate AWP at different drafts. And then using those values of , one calculate the displacement and KB, if required, as per the procedure mentioned in problem 2. 41 Prepared by Dr. Sudhir Sindagi Problems Col(1) Col (2) Col (3) Col (4) Col(5)= (3)X(4) Col(6) Col(7)=(5)X(6) Draft TPC AWP Simpson’s Multiplier For Volume Lever from Keel For Moment @ Keel 0 8.2 1 16.5 2 18.7 3 19.4 4 20.0 5 20.5 6 21.1 ΣV=00m3 ΣM=00m4 42 Prepared by Dr. Sudhir Sindagi Problems Col(1) Col (2) Col (3) Col (4) Col(5)= (3)X(4) Col(6) Col(7)=(5)X(6) Draft TPC AWP Simpson’s Multiplier For Volume Lever from Keel For Moment @ Keel 0 8.2 800 1 800 0 0 1 16.5 1609.75 3 4829.25 1 4829.25 2 18.7 1824.39 3 5473.17 2 10946.34 3 19.4 1892.68 2 3785.36 3 11356.08 4 20.0 1951.21 3 5853.63 4 23414.52 5 20.5 2000 3 6000 5 30000 6 21.1 2058.53 1 2058.53 ΣV=28799.94m3 6 12351.18 ΣM=92897.37m4 43 Prepared by Dr. Sudhir Sindagi Problems As per the Simpson’s 3/8th rule Total ๐๐ก Volume=V= ๐ * ΣV V = 12959.97 m3 Δ=ρSW *V Δ=13283.97 tonnes KB = ๐ ๐ข๐ซ๐ฌ๐ญ ๐๐จ๐ฆ๐๐ง๐ญ ๐จ๐ ๐ญ๐ก๐ ๐ฏ๐จ๐ฅ๐ฎ๐ฆ๐ ๐๐๐จ๐ฎ๐ญ ๐ญ๐ก๐ ๐ค๐๐๐ฅ ๐๐จ๐ฅ๐ฎ๐ฆ๐ ΣM = =h∗ ΣV KB =3.87m from Keel 44 Prepared by Dr. Sudhir Sindagi Problems A ship is floating upright in SW on an even keel at 7m draft. The TPCs are as follows. Draft(m) 1 2 3 4 5 6 7 TPC in Tonnes 60 60.3 60.5 60.5 60.5 60.5 60.5 The volume between the outer bottom and 1m draft is 3044m3 and centroid is 0.5m above the keel. Find the Ship’s KB Given Data: No of ordinates = 7 Hence using Simpson’s 3/8th rule. ๐ก =1m 45 Initially, follow the similar procedure as mentioned in previous problem. Prepared by Dr. Sudhir Sindagi Problems 46 Prepared by Dr. Sudhir Sindagi Problems Col(1) Col (2) Col (3) Col (4) Col(5)= (3)X(4) Col(6) Col(7)=(5)X(6) Draft TPC AWP Simpson’s Multiplier For Volume Lever from WL at 1m For Moment @ WL at 1m 1 60 2 60.3 3 60.5 4 60.5 5 60.5 6 60.5 7 60.5 ΣV=00m3 ΣM=00m4 47 Prepared by Dr. Sudhir Sindagi Problems Col(1) Col (2) Col (3) Col (4) Col(5)= (3)X(4) Col(6) Col(7)=(5)X(6) Draft TPC AWP Simpson’s Multiplier For Volume Lever from WL at 1m For Moment @ WL at 1m 0 60 5853.659 1 5853.659 0 0 1 60.3 5882.927 3 17648.78 1 17648.78 2 60.5 5902.439 3 17707.32 2 35414.63 3 60.5 5902.439 2 11804.88 3 35414.63 4 60.5 5902.439 3 17707.32 4 70829.27 5 60.5 5902.439 3 17707.32 5 88536.59 6 60.5 5902.439 1 5902.439 ΣV=94331.71m3 6 35414.63 ΣM=283258.54m4 48 Prepared by Dr. Sudhir Sindagi Problems As per the Simpson’s 3/8th rule Total ๐๐ก Volume=V= ๐ * ΣV V = 35374.39 m3 Total underwater volume = VT= V+ Additional volume of appendages Total underwater volume = VT= 35374.39 +3044 = 38418.39 m3 ๐ ๐ข๐ซ๐ฌ๐ญ ๐๐จ๐ฆ๐๐ง๐ญ ๐จ๐ ๐ญ๐ก๐ ๐ฏ๐จ๐ฅ๐ฎ๐ฆ๐ ๐๐๐จ๐ฎ๐ญ ๐ญ๐ก๐ ๐ค๐๐๐ฅ ΣM KB = = h∗ ๐๐จ๐ฅ๐ฎ๐ฆ๐ ΣV KB =3.002m from WL at 1m But actual KB =3.002 +1 = 4.002m from Keel KB for total volume = ๐∗๐๐+๐๐๐๐∗๐.๐ VT KB for total volume = 3.724m from Keel 49 Prepared by Dr. Sudhir Sindagi Problems A 300m long vessel has a waterplane with half ordinates commencing from AP are as below. Station 0 ½ 1 2 3 4 5 5½ 6 ½ ord 0.1 7.5 10 12 12.3 11.4 8 5.2 1.0 Find the area of the waterplane if there are appendages forward and aft with a total area of 2.8m2 to be added to the main area Given Data: No of stations = 7. ๐ก= ๐๐๐ง๐ ๐ญ๐ก ๐จ๐ ๐ญ๐ก๐ ๐ฌ๐ก๐ข๐ฉ ๐ง๐ฎ๐ฆ๐๐๐ซ ๐จ๐ ๐จ๐ซ๐๐ข๐ง๐๐ญ๐๐ฌ −๐ ๐ก = 50m 50 Prepared by Dr. Sudhir Sindagi Problems It is the problem with half stations, which are included at the forward and the aft end. In such cases, needed to divide stations and ordinates as per the length between stations. The 1st group will contain 0, ½ and 1 stations, the 2nd group will contain 1,2,3,4&5 stations, the 3rd group will contain 5,5 ½ and 6 stations. After divisions of stations in the subgroups, need to check how many stations are included in the groups and then accordingly one can use particular method to estimate. Here, in this particular problem, all 3 groups have odd number of ordinates, hence we can use Simpson’s 1/3rd rule for all of them. For the group 1 and 3, the Simpson's multiplier will be multiplied with ½ and for the common stations, SMs will be added 51 Levers will be similar to that of stations. Prepared by Dr. Sudhir Sindagi Problems Col(1) Col (2) Station ½ Widths 0 0.1 ½ 7.5 1 10 2 12 3 12.3 4 11.4 5 8 5½ 5.2 6 1.0 Col (3) Col(4)= (2)X(3) Col(5) Col(6)=(4)X(5) Simpson’s Multiplier For Area Lever from AP For Moment @ AP Group 1 Group 2 Group 3 ΣA=00m2 ΣM=00m3 52 Prepared by Dr. Sudhir Sindagi Problems Col(1) Col (2) Station ½ Widths Col (3) Col(4)= (2)X(3) Col(5) Col(6)=(4)X(5) Simpson’s Multiplier For Area Lever from AP For Moment @ AP 0 0.1 1*½ =½ ½ 7.5 4*½ =2 1 10 2 12 4 3 12.3 2 4 11.4 4 5 8 5½ 5.2 4*½ =2 6 1.0 1*½ =½ 1*½ +1=1 ½ 1*½ +1= 1 ½ ΣA=0m2 ΣM=0m3 53 Prepared by Dr. Sudhir Sindagi Problems Col(1) Col (2) Station ½ Widths Col (3) Col(4)= (2)X(3) Col(5) Col(6)=(4)X(5) Simpson’s Multiplier For Area Lever from AP For Moment @ AP 0 0.1 1*½ =½ 0.05 0 0 ½ 7.5 4*½ =2 15 0.5 7.5 1 10 15 1 15 2 12 4 48 2 96 3 12.3 2 24.6 3 73.8 4 11.4 4 45.6 4 182.4 5 8 12 5 60 5½ 5.2 4*½ =2 10.4 5.5 57.2 6 1.0 1*½ =½ 1*½ +1=1 ½ 1*½ +1= 1 ½ 0.5 ΣA=171.15m2 6 3 ΣM=494.9m543 Prepared by Dr. Sudhir Sindagi Problems As per the Simpson’s 1/3rd rule Total Area of the ๐ก waterplane=A=2* ๐ * ΣA +2.8 A = 5707.8 m2 ΣM ( Here, the effect of additional area is neglected, as ΣA distance of its centroid from AP is not provided). LCF =144.58m from AP LCF from AP =h∗ 55 Prepared by Dr. Sudhir Sindagi Problems The Water plane of a ship at a particular draught has the ordinates, 20 meters apart, of magnitude 1.6m, 13.6m, 26.5m, 25.1m, 10.6m, and 2.1m. There is also an intermediate ordinate, midway between the first two ordinates, of magnitude 9.4m. Find the area of the water plane Given Data: This problem is exactly similar to the previous problem, except there is only one half station inserted between stations 1 and 2. h=20m. Here there will be only two groups: The 1st group will contain 0, ½ and 1 stations, the 2nd group will contain 1,2,3,4&5 stations. As both the groups have the odd number of ordinates, hence we can use Simpson’s 1/3rd rule for them. 56 Levers will be similar to that of stations. Prepared by Dr. Sudhir Sindagi Problems Col(1) Col (2) Station ½ Widths 0 1.6 ½ 9.4 1 13.6 2 26.5 3 25.1 4 10.6 5 2.1 Col (3) Col(4)= (2)X(3) Col(5) Col(6)=(4)X(5) Simpson’s Multiplier For Area Lever from AP For Moment @ AP Group 1 Group 2 ΣA=00m2 ΣM=00m3 57 Prepared by Dr. Sudhir Sindagi Problems Col(1) Col (2) Station ½ Widths Col (3) Col(4)= (2)X(3) Col(5) Col(6)=(4)X(5) Simpson’s Multiplier For Area Lever from AP For Moment @ AP 0 1.6 1*½ =½ ½ 9.4 4*½ =2 1 13.6 2 26.5 4 3 25.1 2 4 10.6 4 5 2.1 1 1*½ +1=1 ½ ΣA=0m2 ΣM=0m3 58 Prepared by Dr. Sudhir Sindagi Problems Col(1) Col (2) Station ½ Widths Col (3) Col(4)= (2)X(3) Col(5) Col(6)=(4)X(5) Simpson’s Multiplier For Area Lever from AP For Moment @ AP 0 1.6 1*½ =½ 0.8 0 0 ½ 9.4 4*½ =2 18.8 0.5 9.4 1 13.6 20.4 1 20.4 2 26.5 4 106 2 212 3 25.1 2 50.2 3 150.6 4 10.6 4 42.4 4 169.6 5 2.1 1 5 10.5 ΣM=572.5m3 1*½ +1=1 ½ 2.1 ΣA=240.7m2 59 Prepared by Dr. Sudhir Sindagi Problems As per the Simpson’s 1/3rd rule Total Area of the ๐ก waterplane=A=2* ๐ * ΣA A = 3209.33 m2 ΣM ΣV LCF =47.569m from AP LCF from AP =h∗ 60 Prepared by Dr. Sudhir Sindagi Problems The areas of water plane, 2.5 metres apart, of a Tanker is given below:Calculate the volume of displacement and the position of VCB. Water plane AWP(m2) 1 2 3 4 5 5½ 6 4010 4000 3800 3100 1700 700 200 Given Data: Answers are Underwater volume= 36716.667m3 Δ= 37634.58 tonnes. KB= 7.65m from keel. 61 Prepared by Dr. Sudhir Sindagi Simpson’s 5-8-(-1) Rule Simpson's 3rd rule Also known as the 5–8–1 rule, Simpson's third rule is used to find the area between two consecutive ordinates when three consecutive ordinates are known. The following equation estimates the area in the left half of the figure between ordinates y0 and y1. Area= ๐ก ๐๐ ∗ (๐๐ฒ๐ + ๐๐ฒ๐ − ๐ฒ๐) 62 Prepared by Dr. Sudhir Sindagi Simpson’s 3-10-(-1) Rule Simpson’s 3-10-(-1) Rule It is used to find moment of the area between two consecutive ordinates about an axis passing through the first ordinate, when three consecutive ordinates are known. The following equation estimates moment of the area between ordinates y0 and y1 about the first ordinate Moment of Area @ y0= ๐ก๐ ๐๐ ∗ (๐๐ฒ๐ + ๐๐๐ฒ๐ − ๐ฒ๐) 63 Prepared by Dr. Sudhir Sindagi Problems A vessel has the following waterplane areas at the given drafts. Draft (m) AWP(m2) 1 2 3 4 5 6 350 2500 3450 3960 4000 4030 When in light weight condition, the vessel floats at draft of 2m and when in fully loaded condition, it floats at the draft of 6m. Find its lightweight and total displacement in tonnes. Also calculate its load carrying capacity Given Data: No of ordinates are 6, hence cannot be solved by earlier Simpson's rules. For the first two WLs 3rd rule can be applied to get volumetric displacement in lightship condition and for next 5 WLs 1st rule can 64 applied to get volumetric displacement corresponding to deadweight. Prepared by Dr. Sudhir Sindagi Simpson’s 5-8-(-1) Rule Applying Simpson's 3rd rule for 1st two WLs Volume between WL1 and WL2 = ๐ก ๐๐ ∗ (๐๐ฒ๐ + ๐๐ฒ๐ − ๐ฒ๐) Volume between WL1 and WL2 = 1525m3. Applying Simpson’s 1st rule for next five WLs ๐ก Volume between WL2 and WL6 = ∗ [๐๐๐๐ + ๐๐๐๐ + ๐ ∗ ๐๐๐๐ + ๐ ∗ ๐ ๐๐๐๐ + ๐๐๐๐ ] Volume between WL2 and WL6 =14750m3. Total displacement = Total volume* ρSW Total displacement= 16681.875 tonnes Load carrying capacity = deadweight= 14750* ρSW =16681.875 -1525* ρSW 65 Load carrying capacity = deadweight=15118.75tonnes Prepared by Dr. Sudhir Sindagi Problems The half ordinates of a midship section of a ship at 5 waterlines at 4m intervals are. WL 1 2 3 4 5 ½ Breadth (m) 6.4 10.9 12.8 13.6 14.2 Find the area between waterlines 4&5 and height of the centroid of this layer above the base Given Data: Here both rules 5-8-1 and 3-10-1 rule can be used 66 Prepared by Dr. Sudhir Sindagi Simpson’s 5-8-(-1) Rule Applying Simpson's 3rd rule between WL4 and WL5 Area between WL4 and WL5 =2* ๐ก ๐๐ ∗ (๐๐ฒ๐ + ๐๐ฒ๐ − ๐ฒ๐) Area between WL4 and WL5 = 111.33m2. Applying Simpson’s 3-10-1 rule Moment of Area @ WL5 = ๐ ∗ ๐ก๐ ๐๐ ∗ (๐๐ฒ๐ + ๐๐๐ฒ๐ − ๐ฒ๐) Moment of Area @ WL5 = 221.06m3. We know that, Position of Centroid from any axis ๐ ๐ข๐ซ๐ฌ๐ญ ๐๐จ๐ฆ๐๐ง๐ญ ๐จ๐ ๐๐ซ๐๐ ๐๐๐จ๐ฎ๐ญ ๐ญ๐ก๐ ๐๐ฑ๐ข๐ฌ ๐๐๐.๐๐ เดฅ ๐= = =1.98m from WL5 ๐๐ซ๐๐ ๐๐๐.๐๐ Distance of he centroid from the keel = 4*4-1.98 Distance of he centroid from the keel = 14.02m 67 Prepared by Dr. Sudhir Sindagi Problems The ½ ordinates of a waterplane 120 m long are as follows. Calculate – waterplane area – Distance of centroid from midship. – Second moment of area of waterplane about centroid Draft( AP m) ½ Ord 1.2 (m) 0.5 1 3.5 5.3 1.5 2 3 4 5 6 7 8 8.5 9 9.5 FP 6.8 8 8.3 8.5 8.5 8.5 8.4 8.2 7.9 6.2 3.5 0 68 Prepared by Dr. Sudhir Sindagi Summary Introduction Use of integration in finding various properties Rules of Integration – Trapezoidal Rule – Simpson’s (1/3)rd rule – Simpson’s (3/8)th rule Numerical Simpson’s 5-8-(-1) rule Simpson’s 3-10-(-1) rule 69 Prepared by Dr. Sudhir Sindagi CG and angle of list of the ship By Dr. Sudhir Sindagi 1 Prepared by Dr. Sudhir Sindagi Chapter Content Introduction CG and CB of the ship: List and Trim Estimation of the CG of the ship Shift in CG of the ship – – – – Due to shifting in existing weight Addition of weight Removal of existing of weight Effect of suspended weight Angle of List 2 Prepared by Dr. Sudhir Sindagi Introduction The location of a ship's vertical center of gravity (LCG, TCG and VCG) is an important measurement needed to estimate the initial stability of a vessel. Centre of gravity is the point of a body at which all the mass of the body may be assumed to be concentrated. The force of gravity acts vertically downwards from this point with a force equal to the weight of the body. B - Center of Buoyancy: The geometric center of the ship's underwater hull body. Centroid of the underwater volume is the CB of the ship. It is the point at which all the forces of buoyancy may be considered to act in a vertically upward direction. The Center of Buoyancy will move as the shape of the underwater 3 portion of the hull body changes. Prepared by Dr. Sudhir Sindagi CG and CB of the ship When a ship is equilibrium, the centre buoyancy is vertically line with the centre gravity of the ship 4 at of in of Prepared by Dr. Sudhir Sindagi CG and CB of the ship Longitudinal Center of Gravity (LCG): It is the Longitudinal distance of the CG from the AP. Vertical Center of Gravity (VCG of KG): It is the vertical distance of the CG from the keel. Transverse Center of Gravity (TCG): It is the transverse distance of the CG from the centreline of the ship. 5 Prepared by Dr. Sudhir Sindagi CG and CB of the ship For the upright condition of the ship TCG=TCB=0 as both CG and CB of the ship lie on centreline. If TCG≠ ๐๐๐, ๐ญ๐ก๐๐ง ๐ญ๐ก๐ ๐ฌ๐ก๐ข๐ฉ ๐ฐ๐ข๐ฅ๐ฅ ๐๐ข๐ฌ๐ญ 6 Prepared by Dr. Sudhir Sindagi CG and CB of the ship For the upright condition of the ship TCG=TCB=0 as both CG and CB of the ship lie on centreline. If TCG≠ ๐๐๐, ๐ญ๐ก๐๐ง ๐ญ๐ก๐ ๐ฌ๐ก๐ข๐ฉ ๐ฐ๐ข๐ฅ๐ฅ ๐๐ข๐ฌ๐ญ Heel. A ship is said to be heeled when she is inclined by an external force. For example, when the ship is inclined by the action of the waves or wind. List. A ship is said to be listed when she is inclined by forces within the ship. For example, when the ship is inclined by shifting a weight transversely within the ship. This is a fixed angle of heel. 7 Prepared by Dr. Sudhir Sindagi CG and CB of the ship If LCG≠ ๐๐๐, ๐ญ๐ก๐๐ง ๐ญ๐ก๐ ๐ฌ๐ก๐ข๐ฉ ๐ฐ๐ข๐ฅ๐ฅ ๐๐ซ๐ข๐ฆ 8 Prepared by Dr. Sudhir Sindagi Estimation of the CG of the ship Consider a ship as shown in the figure, consisting of three major weights resulting in CG of the ship located at G. w1, w2 and w3 are the weights placed at a distances of d1, d2 and d3 distances from the AP. The total displacement of the ship = Δ = w1+w2+w3 Let, LCG = Longitudinal Center of Gravity from AP 9 Prepared by Dr. Sudhir Sindagi Estimation of the CG of the ship Taking moment about AP, we get LCG *( w1+w2+w3)= w1*d1+w2*d2+w3*d3 w ∗d +w ∗d +w ∗d w ∗d +w ∗d +w ∗d LCG= 1 1 2 2 3 3 = 1 1 2 2 3 3 w1+w2+w3 Δ To generalize it, LCG= ๐ ๐ข๐ซ๐ฌ๐ญ ๐๐จ๐ฆ๐๐ง๐ญ ๐จ๐ ๐ฐ๐๐ข๐ ๐ก๐ญ๐ฌ ๐๐๐จ๐ฎ๐ญ ๐๐ ๐๐จ๐ญ๐๐ฅ ๐ฐ๐๐ข๐ ๐ก๐ญ ๐จ๐ ๐ญ๐ก๐ ๐ฌ๐ก๐ข๐ฉ If d1, d2 and d3 distances from the Keel Taking moment about Keel, we get VCG * (w1+w2+w3)= w1*d1+w2*d2+w3*d3 w1∗d1+w2∗d2+w3∗d3 VCG=KG= Δ 10 Prepared by Dr. Sudhir Sindagi Estimation of the CG of the ship VCG=KG= ๐ ๐ข๐ซ๐ฌ๐ญ ๐๐จ๐ฆ๐๐ง๐ญ ๐จ๐ ๐ฐ๐๐ข๐ ๐ก๐ญ๐ฌ ๐๐๐จ๐ฎ๐ญ ๐๐๐๐ฅ ๐๐จ๐ญ๐๐ฅ ๐ฐ๐๐ข๐ ๐ก๐ญ ๐จ๐ ๐ญ๐ก๐ ๐ฌ๐ก๐ข๐ฉ If d1, d2 and d3 distances from the center of the ship, then Taking moment about center line, we get TCG *(w1+w2+w3)= w1*d1+w2*d2+w3*d3 w1∗d1+w2∗d2+w3∗d3 TCG= Δ TCG= ๐ ๐ข๐ซ๐ฌ๐ญ ๐๐จ๐ฆ๐๐ง๐ญ ๐จ๐ ๐ฐ๐๐ข๐ ๐ก๐ญ๐ฌ ๐๐๐จ๐ฎ๐ญ ๐๐ ๐๐จ๐ญ๐๐ฅ ๐ฐ๐๐ข๐ ๐ก๐ญ ๐จ๐ ๐ญ๐ก๐ ๐ฌ๐ก๐ข๐ฉ 11 Prepared by Dr. Sudhir Sindagi Problems A ship of 6000 tonnes displacement is composed of masses of 300, 1200 & 2000 tonnes at a distances of 60, 35 & 11m aft of amidships and masses of 1000, 1000 & 500 tonnes at distances of 15, 30 & 50 m forward of amidships. Calculate the LCG of the ship from midship. Solution: When number weight are provided with their locations, then it is recommended to use the tabular method to solve such problems 12 Prepared by Dr. Sudhir Sindagi Problems Weight ΣW=00 tonnes Distance from Midship Moment ΣM= 00 t-m 13 Prepared by Dr. Sudhir Sindagi Problems Weight Distance from Midship Moment 300 -60 -18000 1200 -35 -42000 2000 -11 -22000 1000 15 15000 1000 30 30000 500 50 25000 ΣW=6000 tonnes ΣM=-12000 t-m LCG from midship= ๐ ๐ข๐ซ๐ฌ๐ญ ๐๐จ๐ฆ๐๐ง๐ญ ๐จ๐ ๐ฐ๐๐ข๐ ๐ก๐ญ๐ฌ ๐๐๐จ๐ฎ๐ญ ๐๐ข๐๐ฌ๐ก๐ข๐ฉ ๐๐จ๐ญ๐๐ฅ ๐ฐ๐๐ข๐ ๐ก๐ญ ๐จ๐ ๐ญ๐ก๐ ๐ฌ๐ก๐ข๐ฉ LCG from midship= of the midship. −๐๐๐๐๐ =-2m ๐๐๐๐ from the midship. It means CG is142m aft Prepared by Dr. Sudhir Sindagi Problems A vessel of displacement 12500 tonnes has KG= 9.6m. On completion of loading, she is required to have a KG of 9.5m. Cargo of 1000 & 850 tonnes are loaded at KG 5.5m & 13.6m respectively. Find the KG at which to load further 1600 tonnes to produce the required final KG. Given Data Δold= 12500 tonnes KGold= 9.6m KGfinal=9.5 15 Prepared by Dr. Sudhir Sindagi Problems Weight Distance from Keel Moment Weight Distance from Keel 12500 9.6 12500 9.6 120000 1000 5.5 1000 5.5 5500 850 13.6 850 13.6 11560 1600 x 1600 x 1600x ΣW=000 tonnes Moment ΣM=-000 tm ΣW=15950 tonnes ΣM=137060+ 1600x ΣM Final KG= = 9.5m ΣW 137060+1600x 9.5 = ๐๐๐๐๐ X=9.046m from the keel 16 Prepared by Dr. Sudhir Sindagi Problems A vessel displacing 6200 tonnes KG=8m. Distribute 9108 tonnes of cargo between spaces KG 0.59m and 11.45m, so that the vessels completes loading with KG 7.57m. Given Data Δold= 6200 tonnes KGold= 8m KGfinal=7.57m 17 Prepared by Dr. Sudhir Sindagi Problems Weight Distance from Keel Moment Weight Distance from Keel 6200 8 6200 8 49600 w 0.59 w 0.59 0.59w 9108-w 11.45 9108-w 11.45 104286.6-11.45w ΣW=000 tonnes Moment ΣM=-000 tm ΣW=15308 tonnes ΣM=-153886.610.86w ΣM Final KG= = 7.57m ΣW 153886.6−10.86w 9.5 = ๐๐๐๐๐ W= 3499.54 tonnes and the other weight = 5608.46tonnes 18 Prepared by Dr. Sudhir Sindagi Shift in CG of the ship Shift in CG of the ship due to removal of existing weight: Consider a rectangular plank as shown: Now cut the length of plank of mass ‘w’ kg whose CG is ‘d’ meters away from CG of the plank. Note that a resultant moment of ‘w x d’ kg-m has been created in an anti-clockwise direction about ‘G’. The CG of the new plank shifts from ‘G’ to ‘G1’. 19 Prepared by Dr. Sudhir Sindagi Shift in CG of the ship Shift in CG of the ship due to removal of existing weight: The new mass (W-w) kg now creates a tilting moment of (Δ-w) x GG1 about G. Since both are referring to the same moment, (Δ-w) x GG1 = w x d Shift in CG due to the removal of weight = GG1 wxd wxd GG1 = = Δ−w Final mass w : the mass removed d : the distance between the CG of the mass removed and the original location of the CG of the body. When a weight is removed from a body, the CG shifts directly away from 20 the CG of the mass removed. Prepared by Dr. Sudhir Sindagi Shift in CG of the ship Shift in CG of the ship due to the addition or loading weight: Equating the tilting moments created due to the added weight, which must again be equal: (W + w) x GG1 = w x d wxd wxd GG1 = = Δ+w Final mass d : the distance between the CG of the mass added and the original CG of the body. To generalize, Shift in CG of the ship due to the addition or removal of weight= wxd wxd GG1 = = Δ±w Final mass 21 When a weight is added, the CG shifts towards the CG of the mass. Prepared by Dr. Sudhir Sindagi Shift in CG of the ship Shift in CG of the ship due to the addition or removal of weight wxd wxd GG1 = = Δ±w Final mass 22 Prepared by Dr. Sudhir Sindagi Shift in CG of the ship Shift in CG of the ship due to the shifting of existing weight: To calculate the height of the ship’s center of gravity after a vertical weight shift, the following equation is used: wxd wxd GG1 = = Δ Total weight w = weight shifted d = The distance by which the weight is shifted Shifting of weight, no matter where onboard it is, will always cause the ship’s center of gravity to move in the same direction as the weight shifted. 23 Prepared by Dr. Sudhir Sindagi Shift in CG of the ship wxd wxd = Δ Total weight Shift in CG of the ship will be in the transverse direction GG1 = 24 Prepared by Dr. Sudhir Sindagi Shift in CG of the ship Effect of suspended weights As we are aware that, the CG of a body is the point at which the force of gravity may be considered to act vertically downwards. For a suspended weight, whether the vessel is upright or inclined, the point through which the force of gravity may be considered to act vertically downwards is g, the POINT OF SUSPENSION. 25 Prepared by Dr. Sudhir Sindagi Shift in CG of the ship Important points: The CG of a body will move directly TOWARDS the CG of any weight ADDED. The CG of a body will move directly AWAY from the CG of any weight DISCHARGED. The CG of a body will move PARALLEL to the shift of the CG of any weight MOVED within the body. When a weight is SUSPENDED, CG of the weight is considered to be at the POINT OF SUSPENSION. wxd wxd = Δ±w Final mass d : the distance between the CG of the mass added and the original CG of the body. GG1 = wxd wxd = Δ Total weight w = weight shifted and d = The distance by which the weight is shifted GG1 = 26 Prepared by Dr. Sudhir Sindagi Problems A heavy lift derrick is used to discharge a 100t package from a ship of displacement 8000 tonnes with KG at 8.2m. If the KG of the weight while on board is 3m and if the derrick head is 25m above the keel, find the KG of ship – While discharging – After discharge Given Data: 27 Prepared by Dr. Sudhir Sindagi Problems While Discharging: To shift the weight, the moment weight is lifted by the derrick, the weight will act at the point of suspension, which is derrick head. Hence, it will be typical case of shift in weight vertically upwards wherein, d= 25-3= 22m Shift in CG of the ship, due to vertical shift in weight wxd wxd GG1 = = Δ Total weight 28 Prepared by Dr. Sudhir Sindagi Problems While Discharging: GG1 =0.275m upwards New KG =Old KG + GG1 = 8.475m After Discharging: This can be solved by two ways – A. Considering the weight is removed from the derrick head – B. Considering the weight is removed from the deck at Kg= 3m A. Considering the weight is removed from the derrick head wxd GG1 = Δ− w d : the distance between the CG of the mass removed and the original CG of the body 29 d = 25-8.475 = 16.525m Prepared by Dr. Sudhir Sindagi Problems After Discharging: wxd GG1 = = 0.2091m downwards Δ− w New KG =Old KG - GG1 = 8.475 – 0.2091 = 8.265m B. Considering the weight is removed from the deck at Kg= 3m Here, the CG of the ship will be at 8.2 and the Kg of the weight is at 3m d : the distance between the CG of the mass removed and the original CG of the body d = 8.2-3 = 5.2m wxd GG1 = = 0.065m upwards Δ− w New KG =Old KG + GG1 = 8.2 + 0.065 = 8.265m 30 Prepared by Dr. Sudhir Sindagi Problems A vessel of 8000 tonnes displacement has 75 tonnes of cargo on the deck. It is lifted by a derrick whose head is 10.5m above the CG of the cargo and be placed in the lower hold 9m below the deck and 14m forward of its original position. Calculate the shift of vessel’s CG from its original position when the cargo is – Just clear of the deck – In its final position Given Data 31 Prepared by Dr. Sudhir Sindagi Problems Just Clear of the deck: To shift the weight, the moment weight is lifted by the derrick i.e. when the weight is Just Clear of the deck, the weight will act at the point of suspension, which is derrick head. Hence, it will be typical case of shift in weight vertically upwards wherein, d= 10.5m Shift in CG of the ship wxd GG1 = Δ GG1 = 0.098m upwards 32 Prepared by Dr. Sudhir Sindagi Problems In its final position: Here the weight is not only moved downwards, it also being moved forward For the downward movement This can be solved by two ways – A. Considering the weight is shifted from the derrick head – B. Considering the weight is shifted from the main deck A. Considering the weight is shifted from the derrick head In this case, d= 19.5m wxd G1G 2 = =0.1828m downwards Δ 33 Prepared by Dr. Sudhir Sindagi Problems In its final position: B. Considering the weight is shifted from the main deck In this case, d= 9m wxd GG2 = =0.084m downwards Δ This can also be obtained by subtracting the shift in the CG of the ship as follows GG2 = G1G2 - GG1 GG2 = 0.1828-0.098 GG2 = 0.1828-0.098 GG2 = 0.084m downwards 34 Prepared by Dr. Sudhir Sindagi Problems In its final position: For the forward movement In this case, d= 14m wxd G2G 3 = Δ G2G3 = 0.131m forward 35 Prepared by Dr. Sudhir Sindagi Problems A ship of 4000 tonne displacement has its centre of gravity 1.5m aft of midships and 4m above keel. 200 tonne of cargo are now added 45m forward of midships and 12m above the keel. Calculate the new position of the centre of gravity and the angle in which the centre of gravity moves relative to the horizontal Given Data 36 Prepared by Dr. Sudhir Sindagi Problems It is the typical case of weight addition, wherein, weight added =200 tonnes wxd GG1 = Δ+ w d : the distance between the CG of the mass addedd and the original CG of the body For the horizontal movement of CG of the ship d = 45+1.5 = 46.5m GG1 =2.21m forward For the vertical movement of CG of the ship d = 12-4= 8m 37 G1G2 =0.3809m upwards Prepared by Dr. Sudhir Sindagi Problems For the new position of the ship KG2 = KG+G1G2 KG2 =4.3809m above the keel LCG=2.21-1.5 =0.71m forward of midship For the angle of in which the centre of gravity moves relative to the horizontal G1G2 GG1 ฦ= 9.76 degrees tan θ = 38 Prepared by Dr. Sudhir Sindagi Angle of List The angle of list is the degree to which a vessel heels (leans or tilts) to either port or starboard at equilibrium—with no external forces acting upon it. Listing is caused by the off-centerline distribution of weight onboard due to uneven loading or due to flooding. By contrast, roll is the dynamic movement from side to side caused by waves 39 Prepared by Dr. Sudhir Sindagi Angle of List Let an existing weight on board a ship is shifted to starboard side transversely as shown in the figure. This causes a shift in the CG of the ship from G to G1 40 Prepared by Dr. Sudhir Sindagi Angle of List In this new position, if one considers no shift in the CB of the ship, then buoyancy force and gravity force acting at CB and CG of the ship respectively would create a clockwise moment and the ship will not be in equilibrium. To achieve the equilibrium condition with zero moment acting on the ship, slowly CB of the ship will start moving towards starboard side in such a way that, it will set itself exactly below the CG of the ship. This is typically a equilibrium condition and the ship will remain in that state unless any weight is being shifted/removed/added. This condition/state of the ship is known as the listed condition. In the listed condition, the new line of action of buoyancy will intersect the center line of the ship at M, which is known as the Transverse 41 Metacentre of the ship. Prepared by Dr. Sudhir Sindagi Angle of List Transverse Metacentre is the point about which the ship heels/lists/rolls. The vertical distance between the original CG of the ship and the metacentre is known as the metacentric height (GM). The angle of list can found by GG1 TCG ๐ญ๐๐ง ๐ = = GM GM Here, GG1 is known as the TCG of the ship, which is transverse distance of the CG of the ship from the centreline and can be found using wxd TCG= GG1= if the existing weight is being shifted Δ wxd 42 TCG= GG1= if the weight is added/removed Δ±๐ฐ Prepared by Dr. Sudhir Sindagi Problems A vessel of 6500 tonnes displacement has a KM 7.2m and KG 6.8m. A weight of 100 tonnes is shifted 2.3m to starboard and 3.9m upwards. If the vessel is initially upright, calculate the resulting list. Given data: Δ= 6500 tonnes w=100 tonnes dh= 2.3m dv= 3.9m Here, since there are two movements of the weight, one in transverse direction and the other being vertically upwards, hence accordingly, the CG of the ship will also move in the similar direction. 43 Prepared by Dr. Sudhir Sindagi Problems In the first case, dh= 2.3m w x dh GGH = Δ GGH = 0.03538m In the second case, dv= 3.9m w x dv G HG V = Δ GHGV = 0.06m vertically upwards The angle of list can found by GGH TCG ๐ญ๐๐ง ๐ = = New GM Old GM−GHGV Φ= 5.9407 degrees When there is a vertical shift in CG of ship, GM value changes. 44 Prepared by Dr. Sudhir Sindagi Problems A ship 8000 Tonnes displacement has KM =8.7 meters KG= 7.6 meters. The following weights are either loaded or discharged as mentioned. – Load 250 T cargo, VCG 6.1 m, TCG 7.6 m to Stbd side – Load 300 T Fuel Oil, VCG 0.60 m, TCG 6.1 m to Port side – Discharge 50 T of Ballast, VCG 1.2 m, TCG 4.6 m to Port side. Find the Final List of the ship after completion of these activities. Here, it is recommended to use tabular method 45 Prepared by Dr. Sudhir Sindagi Problems Weight KG 8000 7.6 8000 250 6.1 250 300 0.6 300 -50 1.2 -50 ΣW=000 tonnes Moment ΣM=-000 tm Weight ΣW=000 tonnes TCG Moment ΣM=-000 tm 46 Prepared by Dr. Sudhir Sindagi Problems Weight KG Moment Weight TCG Moment 8000 7.6 60800 8000 0 0 250 6.1 1525 250 7.6 1900 300 0.6 180 300 -6.1 -1830 -50 1.2 -60 -50 -4.6 +230 ΣW=8500 tonnes ΣM=62445 t-m ΣW=8500 tonnes ΣM=300 tm ΣM 62445 Final KG= = ΣW 8500 Final KG=7.346m 47 Prepared by Dr. Sudhir Sindagi Problems ΣM 300 = ΣW 8500 Final TCG=0.035m on starboard side. The angle of list can found by GGH TCG ๐ญ๐๐ง ๐ = = New GM ๐๐ −๐ง๐๐ฐ ๐๐ 0.035 ๐ญ๐๐ง ๐ = Final TCG= ๐.๐−๐.๐๐๐ Φ = 1.49 degrees on starboard side. 48 Prepared by Dr. Sudhir Sindagi Problems A ship of 12500 tonne displacement and 15m beam has a metacentric height of 1.10m. A mass of 80tonne is lifted from its position in the centre of lower hold by one of ship’s derricks and placed on the quay 2m away from the ship’s side. The ship heels to a maximum angle of 3.5 deg when the mass is being moved. – a) Does the GM alter during operation? – b) Calculate the height of the derrick head above the original centre of gravity of the mass 49 Prepared by Dr. Sudhir Sindagi Problems 50 Prepared by Dr. Sudhir Sindagi Problems In this case, as shown in the second figure, the ship will list with angle 3.5 degree, wherein, using derrick, the weight will be shifted in both horizontal and vertical distances causing both the horizontal and vertical shifts in the CG of the ship, respectively. Since there is vertical shift in the CG of the ship, hence GM of the ship will alter. GGH due to dh and GHGV due to dV Here, dh is known but dV is unknown. w x dh GGH = Δ here dh = 9.5m GGH = 0.0608m 51 Prepared by Dr. Sudhir Sindagi Problems GGH GGH = New GM Old GM−GHGV Putting the values, one can obtain the value of GHGV 0.0608 ๐ญ๐๐ง ๐. ๐ = 1.1−GHGV GHGV =0.10593m Putting it in the below equation, we can calculate the value of dv w x dv G HG V = Δ dV = 16.5515m ๐ญ๐๐ง ๐ = 52 Prepared by Dr. Sudhir Sindagi Problems A ship of 13750 tonnes displacement, GM = 0.75 m, is listed 2.5 degrees to starboard and has yet to load 250 tonnes of cargo. There is space available in each side of No. 3 between deck (centre of gravity, 6.1 m out from the centreline). Find how much cargo to load on each side if the ship is to be upright on completion of loading. Assume, no vertical shift in CG of ship due to addition of weight. 53 Prepared by Dr. Sudhir Sindagi Problems Since there is no vertical shift in CG of the ship, hence GM remains same. Using below equation, one can find the value of GGH GGH ๐ญ๐๐ง ๐ = New GM GGH = 0.032m Since, it is the case of weight addition, hence w1 x d๐−w2 x d2 GGH = Δ+๐ฐ๐+๐ฐ๐ w x 6.1−(250−w1)∗ 6.1 0.032 = 1 13750+250 Solving this equation, we get 54 w1 =88.04 tonnes and w2 =161.97tonnes Prepared by Dr. Sudhir Sindagi Summary Introduction CG and CB of the ship: List and Trim Estimation of the CG of the ship Shift in CG of the ship – – – – Due to shifting in existing weight Addition of weight Removal of existing of weight Effect of suspended weight Angle of List 55 Prepared by Dr. Sudhir Sindagi Transverse stability of the ship By Dr. Sudhir Sindagi 1 Prepared by Dr. Sudhir Sindagi Chapter Content Introduction – Types of Equilibrium Conditions for the stability of the ship and submarine Stiff and Tender Ship Stability for the small angles of heel – Derivation for the Metacentric radius – Free Surface Effect – Inclining Experiment Stability at the large angles of heel – – – – – Derivation for the wall sided formula Angle of Loll GZ Curve, Cross curves of Stability (KN Curves) Dynamical Stability of the ship Requirements of the IMO for the GZ Curve 2 Prepared by Dr. Sudhir Sindagi Introduction The body is said to be in equilibrium when the summation of forces and moments acting on it is zero. Normally two forces act on a floating body, one is gravitational force and the other one is the buoyancy force. The gravitational force acts through the center of gravity. Similarly the buoyancy force also acts through the center of buoyancy. The position of the center of gravity depends on the distribution and magnitude of the weights. But the center of buoyancy depends only on the shape of the immersed portion of the vessel. It is the centroid of the under water portion of the floating body. 3 Prepared by Dr. Sudhir Sindagi Introduction The conditions of equilibrium for floating bodies are: The buoyancy force is equal to the gravitational force. The center of buoyancy and the center of gravity lie in the same vertical line. The common line of action of these forces will be perpendicular to the new water line. 4 Prepared by Dr. Sudhir Sindagi Introduction MT Wind Water Resistance When a ship is affected by outside forces (Upsetting Forces), it will alter its state of equilibrium. The forces of wind- and the opposing force of the water below the waterline- will cause an external 5 moment couple about the ship’s center of flotation. Prepared by Dr. Sudhir Sindagi Introduction The ship reacts to this external moment couple by pivoting about F, causing a shift in the center of buoyancy. The center of buoyancy will shift because the submerged volume shifts MT Wind B Water Resistance Note that there is no change in weight or it’s distribution so there is 6 NO change in the location of G! Prepared by Dr. Sudhir Sindagi Introduction Because the location of B changes, the location of where the FB is applied also changes. Because G does not move, the location of the Δs force does not change. The weight force and the buoyancy are no longer aligned. The heeling over causes the creation of an internal moment couple known as the Righting Moment (RM) ๏s F MT B FB As a result, the ship is now back into equilibrium, even as it heels over due to the 7 wind force. Prepared by Dr. Sudhir Sindagi Introduction Upsetting forces of stability: Beam winds, with or without of rolling, Lifting of heavy weights over the side Free swinging of weights, High speed of turns, Grounding and docking, Strain on mooring lines, Towline pull of tugs, Entrapped water on deck. 8 Prepared by Dr. Sudhir Sindagi Types of Equilibrium Stable equilibrium A ship is said to be in stable equilibrium if, when inclined, she tends to return to the initial position. For this to occur the centre of gravity must be below the metacentre, that is, the ship must have positive metacentric height (M above G & GM>0) If moments are taken about G there is a moment to return the ship to the upright (GM>0). This moment is referred to as the Moment of Statical Stability or Righting Moment and is equal to the product of the force 'Δ' and the lever GZ. Moment of Statical Stability = Δ x GZ (tonnesmetres). 9 Prepared by Dr. Sudhir Sindagi Types of Equilibrium 10 Prepared by Dr. Sudhir Sindagi Types of Equilibrium Stable equilibrium The lever GZ is referred to as the righting lever and is the perpendicular distance between the centre of gravity and the vertical through the new centre of buoyancy. At a small angle of heel (less than 100) GZ = GM sin ฯ Moment of Statical Stability = RM= Δ x GM sin ฯ 11 Prepared by Dr. Sudhir Sindagi Types of Equilibrium Neutral Equilibrium When G coincides with M i.e. GM=0 as shown in Figure, the ship is said to be in neutral equilibrium, and if inclined to a small angle she will tend to remain at that angle of heel until another external force is applied. The ship has zero GM. Note that KG = KM. Moment of Statical Stability = Δ x GZ, but in this case GZ = 0; Moment of Statical Stability = 0. Therefore there is no moment to bring the ship back to the upright or to heel her over still further 12 Prepared by Dr. Sudhir Sindagi Types of Equilibrium 13 Prepared by Dr. Sudhir Sindagi Types of Equilibrium Unstable Equilibrium When a ship which is inclined to a small angle tends to heel over still further, she is said to be in unstable equilibrium. For this to occur the ship must have a negative GM. G is above M. The moment of statical stability=Δ*GZ, is clearly a capsizing moment which will tend to heel the ship still further. 14 Prepared by Dr. Sudhir Sindagi Types of Equilibrium 15 Prepared by Dr. Sudhir Sindagi Types of Equilibrium When a ship in unstable or neutral equilibrium, which can be made stable by lowering the effective centre of gravity of the ship. To do this one or more of the following methods may be employed: – – – – Weights already in the ship may be lowered, Weights may be loaded below the centre of gravity of the ship, Weights may be discharged from positions above the centre of gravity, Free surfaces within the ship may be removed 16 Prepared by Dr. Sudhir Sindagi Stability of Submarine For the submarine which is fully submerged in water, as shape of the submerged volume doesn't alter, the position of center buoyancy remains unaltered and it remains on the center line of the ship. Hence, it is interesting to know the condition to generate the righting moment, in turn, the submarine to remain stable. The condition for the stability of the submarine is G lies below B. Unstable submarine Stable submarine 17 Prepared by Dr. Sudhir Sindagi Metacentric Height 18 Prepared by Dr. Sudhir Sindagi Metacentric Height 19 Prepared by Dr. Sudhir Sindagi Metacentric Height The centre of gravity of a body `G' is the point through which the force of gravity is considered to act vertically downwards with a force equal to the weight of the body. KG is VCG of the ship. The centre of buoyancy `B' is the point through which the force of buoyancy is considered to act vertically upwards with a force equal to the weight of water displaced. It is the centroid of the underwater volume. KB is VCB of the ship. To float at rest in still water, a vessel must displace her own weight of water, and the centre of gravity must be in the same vertical line as the centre of buoyancy. KM = KB + BM Also KM = KG + GM GM= KB+ BM – KG 20 BM= Metacentric Radius and GM = Metacentric Height Prepared by Dr. Sudhir Sindagi Metacentric radius The metacentric radius of a ship is the vertical distance between its center of buoyancy and metacenter. 21 Prepared by Dr. Sudhir Sindagi Metacentric Height GM is crucial to ship stability. The table below shows typical working values for GM for several shiptypes all at fully-loaded drafts. Ship type General cargo ships Oil tankers Double-hull supertankers GM at fully-loaded condition 0.30–0.50m 0.50–2.00m 2.00–5.00m Container ships 1.50–2.50m Ro-Ro vessels Bulk ore carriers 1.50 m approximately 2–3m 22 Prepared by Dr. Sudhir Sindagi Stiff Ship The Rolling time period of a ship is the time taken by the ship to roll from one side to the other and back again to the initial position. When a ship has a comparatively large GM, for example 2 m to 3 m, the righting moments at small angles of heel will also be comparatively large. It will thus require larger moments to incline the ship. When inclined she will tend to return more quickly to the initial position. The result is that the ship will have a comparatively short time period, and will roll quickly – and perhaps violently – from side to side. A ship in this condition is said to be ‘stiff’, and such a condition is not desirable. The time period could be as low as 8 seconds. The effective centre of 23 gravity of the ship should be raised within that ship. Prepared by Dr. Sudhir Sindagi Tender Ship When the GM is comparatively small, for example 0.16 m to 0.20 m the righting moments at small angles of heel will also be small. The ship will thus be much easier to incline and will not tend to return so quickly to the initial position. The time period will be comparatively long and a ship, for example 25 to 35 seconds, in this condition is said to be ‘tender’. As before, this condition is not desirable and steps should be taken to increase the GM by lowering the effective centre of gravity of the ship. The officer responsible for loading a ship should aim at a happy medium between these two conditions whereby the ship is neither too stiff nor too tender. A time period of 15 to 25 seconds would generally be acceptable for 24 those on board a ship at sea. Prepared by Dr. Sudhir Sindagi Derivation for BM The metacentric radius of a ship is the vertical distance between its center of buoyancy and metacenter. This parameter can be visualized as the length of the string of a swinging pendulum of the center of gravity of the pendulum coincides the center of buoyancy of the ship. In other words, the ship behaves as a pendulum swinging about its metacenter. It is a different fact that, the metacenter of the ship changes itself, every moment. Because with every angle of heel, the transverse shift in center of buoyancy will vary, therefore creating a new metacenter. For small angle of heel, this vertical shift in M is neglected. The importance of this parameter can be realised when the 25 mathematical expression of metacentric radius is investigated. Prepared by Dr. Sudhir Sindagi Derivation for BM 26 Prepared by Dr. Sudhir Sindagi Derivation for BM Consider a ship as shown in the figure, is initially floating upright at waterline WL. Due to the external force or the heeling moment by wind or wave, let the ship heels by an angle ฯ and floats at new waterline W1L1. In this case, a triangular wedge on port side will come out of the water, known as emerging wedge and a wedge with similar volume on starboard side immerses into the water, known as immerses wedge. This is simply a case of shifting of volume from port to the starboard, causing a shift in the center of buoyancy of the ship. Consider a small elementary length of the wedge dx along the length of the ship. Volume of the elementary wedge (v) = area *length of the wedge 27 Prepared by Dr. Sudhir Sindagi Derivation for BM Volume of the elementary wedge (v) = Volume of the elementary wedge (v) = ๐ ๐ฒ ∗ ๐ฒ tan φ ๐๐ฑ ๐ ๐ ๐ ๐ฒ tan φ ๐๐ฑ ๐ Here, the distance by which the volume of wedge is shifted = distance by which the centroid is shifted from g to g1 d= ๐ ๐ ๐ฒ+ ๐ฒ ๐ ๐ = ๐ ๐ฒ ๐ As mentioned earlier, this is case of the shift in volume, hence there is a shift in the center buoyancy of the ship. Small shift in the center buoyancy of the ship due to the shift in Volume of the elementary wedge = ๐๐จ๐ฅ๐ฎ๐ฆ๐ ๐จ๐ ๐ฐ๐๐๐ ๐ ∗ ๐๐ข๐ฌ๐ญ๐๐ง๐๐ ๐๐ฒ ๐ฐ๐ก๐ข๐๐ก ๐ญ๐ก๐๐ข๐ซ ๐๐๐ง๐ญ๐ซ๐จ๐ข๐ ๐ฌ๐ก๐ข๐๐ญ๐ฌ ๐ฏ∗๐ = ๐ฏ๐จ๐ฅ๐ฆ๐๐ญ๐ซ๐ข๐ ๐๐ข๐ฌ๐ฉ๐ฅ๐๐๐๐ฆ๐๐ง๐ญ ๐จ๐ ๐ญ๐ก๐ ๐ฌ๐ก๐ข๐ฉ ๐ 28 Prepared by Dr. Sudhir Sindagi Derivation for BM Small shift in the center buoyancy of the ship = Small shift in the center buoyancy of the ship = ๐ ๐ ๐ ๐ฒ tan φ ๐๐ฑ∗ ๐ฒ ๐ ๐ ๐ ๐ ๐ ๐ฒ ๐๐ฑ ๐ ๐ ∗ tan φ Total shift in the center of buoyancy of the ship = BB1 which will be obtained by integrating above equation for the entire length BB1 = ๐ ๐ ๐ ๐ โซ๐ฑ๐ ๐ฒ ๐ืฌโฌ tan φ ๐ ๐ ๐ ๐ โซ๐ฑ๐ ๐ฒ ืฌโฌ ๐ ๐ ∗ Here, the term = IT = Transverse Moment of inertia of the ship’s waterplane area taken at SLL about the longitudinal axis passing through the centerline. 29 Prepared by Dr. Sudhir Sindagi Derivation for BM BB1 = IT tan φ ๐ From the triangle MBB1, we can write tan φ = BB1 BM Putting this in above equation and after simplification, we get BM = IT ๐ 30 Prepared by Dr. Sudhir Sindagi Derivation for BM Assumptions in the derivation: The angle of heel is assumed to be lesser than equal to 7 or 10 degrees. Since, the angle of heel is assumed to be smaller, hence the vertical shift in the center of buoyancy of the ship due to the vertical shift in the centroid of the wedge is neglected. BB1 is perpendicular to the center line of the ship. The side walls (shells) of the ship are assumed to be vertical. 31 Prepared by Dr. Sudhir Sindagi Derivation for BM Derive the expression for the Transverse Metacentric Radius (BMT)and Longitudinal Metacentric Radius (BML) of the rectangular barge floating at the draft T. 32 Prepared by Dr. Sudhir Sindagi Derivation for BM IT = Transverse Moment of inertia of the ship’s waterplane area taken at SLL about the longitudinal axis passing through the centerline. IT = ๐๐ ๐ ๐๐ IL = Longitudinal Moment of inertia of the ship’s waterplane area taken at SLL about the Transverse axis passing through the centroid of the waterplane. IL = ๐๐๐ ๐๐ Volumetric displacement = ๐ = ๐๐๐ Transverse Metacentric Radius = BMT = IT ๐ = ๐๐ ๐ ๐๐ ๐๐๐ 33 Prepared by Dr. Sudhir Sindagi Derivation for BM Transverse Metacentric Radius BMT = ๐๐ ๐๐๐ Longitudinal Metacentric Radius = BML = Longitudinal Metacentric Radius = BML = IL ๐ = ๐๐๐ ๐๐ ๐๐๐ ๐๐ ๐๐๐ 34 Prepared by Dr. Sudhir Sindagi Problems The second moment of area of the water plane about the centre line is 20000 m4. The displacement of the ship is 7000 tonnes, while floating in a water of density 1.008 tonnes/m3. KB=1.9m. KG=3.2m. Calculate the initial metacentric height of the ship. Given Data Δ=7000 tonne, GM=?, KB=1.9m, IT= 20000 m4. KG=3.2m We know that, GM=KB + BM – KG But BM = IT ๐ Using equation โ= ๐ ∗ ๐, ๐๐๐ฅ๐๐ฎ๐ฅ๐๐ญ๐ ๐ = 6944.44m3 Putting in the equation of BM, we get BM= 2.88m GM=1.58m 35 Prepared by Dr. Sudhir Sindagi Problems A ship of 12000 tonne displacement has a metacentric height of 0.6m and a centre of buoyancy of 4.5 m above the keel. The second moment of area of the water plane about the centre line is 42.5x1000 m4. Calculate the height of centre of gravity above the keel. Given Data Δ=12000 tonne, GM=0.6m, KB=4.5m, IT= 42.5x1000 m4. KG=? We know that, GM=KB + BM – KG But BM = IT ๐ Using equation โ= ๐ ∗ ๐, ๐๐๐ฅ๐๐ฎ๐ฅ๐๐ญ๐ ๐ = 11707.317m3 Putting in the equation of BM, we get BM= 3.63m KG=7.53m 36 Prepared by Dr. Sudhir Sindagi Problems A rectangular barge 100mX16mX6m depth is floating in SW at draft of 4m. Calculate its KB, BM & GM of the vessel if it’s KG=5.9m. Also comment on KG of ship at which the ship will become unstable. Given Data T= 4m, KG=5.9m KB=?, BM=? GM= ?& KG=?, when ship becomes unstable โ= ๐๐๐ ∗ ๐๐ ∗ ๐ ∗ ๐. ๐๐๐= 6560 tonnes. T KB = =2m ๐ BMT = IT ๐ = ๐๐ ๐๐๐ BMT = 5.33m 37 Prepared by Dr. Sudhir Sindagi Problems KM= KB+BM = 7.33m GM= KB+BM-KG = KM-KG GM=1.433m Ship becomes unstable when GM becomes negative and is possible when KG>KM KG>7.33m 38 Prepared by Dr. Sudhir Sindagi Problems A vessel of constant triangular cross section has a depth of 12m and breadth at the main deck of 15m. Calculate the draft at which the vessel becomes unstable if the KG is equal to 6.675m Given Data D=12m, B=15m, KG= 6.675m, T=? When vessel becomes unstable. Vessel become unstable when GM<0, GM=KB + BM – KG <0 KB + BM < KG In this case, KB + BM < 6.675m 39 Prepared by Dr. Sudhir Sindagi Problems For ๐๐ , take the water plane at the SLL and obtain its second moment of area of the waterplane about longitudinal axis. Assuming length of the prism =L ๐๐ = ๐๐๐ ๐๐ ๐= Area of the section*Length of the section ๐ ๐ ๐= ๐ ๐*L 40 Prepared by Dr. Sudhir Sindagi Problems But BM = BM = ๐ ๐ IT ๐ ๐๐๐ ๐๐ ๐ ๐∗L Simplifying above equation, we get BM=0.2604T KB= 2T/3 Putting all these values in the equation, KB + BM < 6.675m 2T/3 + 0.2604T < 6.675m T< 7.2m. Ship becomes unstable when the draft < 7.2m 41 Prepared by Dr. Sudhir Sindagi Problems A Conical buoy has a displacement of 0.73 tonnes when floating with its vertex down in sea water. Vertex angle is 60 degrees and second ๐๐๐ . ๐๐ moment of inertia of the circular waterplane is The CG of the ship is 1.22 from the vertex. Find the metacentric height and the depth of water from the vertex. Given Data Δ = 0.73 tonnes, ๐๐ = ๐๐๐ , ๐๐ KG= 1.22m GM= ? T=? 42 Prepared by Dr. Sudhir Sindagi Problems Using equation โ= ๐ ∗ ๐, ๐ ๐ 0.73 = ๐๐๐ ๐ ∗ ๐ From the geometry, we get relationship between R & T tan ๐๐ = ๐ ๐ Putting R= tan ๐๐ ∗ ๐ in above equation, we get ๐ ๐ 0.73 = ๐(tan ๐๐ ∗ ๐)๐ ๐ ∗ ๐. ๐๐๐ Solving this equation for the T, we get T= 1.27m D= 2R=2 * tan ๐๐ ∗ ๐ D= 1.4664m 43 Prepared by Dr. Sudhir Sindagi Problems Metacentric height is calculated using GM= KB + BM – KG Here, KB= 2T/3 But BM = IT ๐ , ๐๐ = ๐๐๐ ๐๐ , ๐ = Δ/ ρ = 0.73/1.025 Putting all these values in the above equation, we get GM= 2T/3 + GM= 2T/3 + IT ๐ – 1.22 ๐๐๐ ๐๐ ๐.๐๐/ρ – 1.22 Putting values of T and D, We get, GM GM= -0.0583m and the buoy is unstable. 44 Prepared by Dr. Sudhir Sindagi Problems A pontoon raft 10m long is formed by two cylinders 0.75m radius spaced 2m apart between centers and is planked over wood forming a platform 10m x 3m. When laden, the raft floats with cylinders half immersed in river water and its centre of gravity when laden is 1m above the waterline. Calculate the transverse and longitudinal metacentric heights. Given Data L=10m, D=0.75m, T=D/2 45 Prepared by Dr. Sudhir Sindagi Problems KG= 1m above the waterline KG=1+0.75=1.75m Distance of center of buoyancy from the center for semicircle= KB= R - ๐๐ ๐๐ ๐๐ ๐๐ KB= 0.4316m BM = IT ๐ ๐=volume of two half cylinders ๐= ๐๐๐ ๐ ๐= 17.6714 m3 46 Prepared by Dr. Sudhir Sindagi Problems IT = MI of outer rectangle – MI of inner rectangle ๐๐∗๐.๐๐ IT = ๐๐ − ๐๐∗๐.๐๐ ๐๐ IT =35.625 m4 BM = IT ๐ BM = 2.0159m GM=KB + BM – KG GM=0.6975m 47 Prepared by Dr. Sudhir Sindagi Free Surface Effect When any tank is not fully filled and the ship heels, the liquid in the tank moves across the tank in the direction of heel, causing CG of the ship to move away from the centerline reducing the righting lever, GM and in turn, the stability of the ship reduces. This effect is known as the free surface effect. The free surface effect can become a problem in tankers, wherein, cargo holds are partially full, fuel tanks, or water tanks (especially if they span the full breadth of the ship), or from accidental flooding. If a compartment or tank is either empty or full, there is no change in the craft's center of mass as it rolls from side to side. However, if the compartment is only partially full (Filled less than 90% or filled more than 10%), the liquid in the compartment will respond to the vessel's 48 heave, pitch, roll, surge, sway or yaw. Prepared by Dr. Sudhir Sindagi Free Surface Effect 49 Prepared by Dr. Sudhir Sindagi Free Surface Effect Let there be a tank which contains a liquid of density ρl and iT is the transverse moment of inertia of the tank about its longitudinal axis. Due to shift of the liquid in the tank, a horizontal shift in the CG from G to G1 occurs. Here it is assumed that, for a small angle of heel up to 70, the vertical shift in the CG of the ship is neglected. The line of action of gravity force intersects the centerline of the ship at GV, resulting in a vertical shift in the CG from G to GV. As it has been observed in the past, FSE causes a vertical shift in the CG of the ship reducing stability, hence, it was concluded that, this vertical shift from G to GV is the FSE GGV= FSE= FSC = Free Surface Correction GVM= Fluid GM which is considering FSE 50 GM= Solid GM without considering FSE Prepared by Dr. Sudhir Sindagi Free Surface Effect GVM= Fluid GM = Solid GM - FSE GM= Solid GM without considering FSE If v is the volume of the liquid inside the tank shifted GG1 = w x d (vρl) d ρl(vd) = = Δ Δ Δ (vd) = ๐ ๐ ๐ฒ tan φ ๐ ๐๐ฑ ๐ ๐ฒ ๐ For the entire length, it will be (vd) = ๐ ( ๐ Here, iT = ๐ ๐ โซ) ๐ฑ๐ ๐ฒ ๐ืฌโฌtan ๐ ๐ ๐ ( โซ)๐ฑ๐ ๐ ๐ฒ ๐ืฌโฌ ๐ ρl(vd) ρl(iT tan ๐) GG1 = = Δ Δ 51 Prepared by Dr. Sudhir Sindagi Free Surface Effect But tan φ ๐๐๐ = ๐๐๐ ๐๐ ρl(iT๐๐ ) GG1= Δ ๐ ๐ iT∗ρl Δ FSE does not depend upon FSE=FSC=GGV= – Weight of the liquid inside the tank – Location of the tank 52 Prepared by Dr. Sudhir Sindagi Free Surface Effect Reducing FSE by inserting transverse bulkhead at the center of length Since density and displacement of the ship remains same for the particular loading, FSE depends on iT of the tank Here, iT = iT1+ iT2 New iT = New iT = (๐/๐)๐๐ ๐๐ ๐๐๐ ๐๐ + (๐/๐)๐๐ ๐๐ = Old iT No change in the iT is observed. Hence no change in FSE with this modified configuration 53 Prepared by Dr. Sudhir Sindagi Free Surface Effect Reducing FSE by inserting longitudinal bulkhead at the center of Breadth Since density and displacement of the ship remains same for the particular loading, FSE depends on iT of the tank Here, iT = iT1+ iT2 New iT = New iT = New iT = New iT = New i = ๐(๐/๐)๐ ๐๐ + ๐(๐/๐)๐ ๐๐ ๐๐๐ ๐ ๐๐๐ ๐ = = Old iT ๐๐ ๐ ๐๐ ๐ ๐ Old iT for one BHD ๐๐ ๐ Old iT for two BHDs ๐๐ ๐ Old i for three BHDs 54 Prepared by Dr. Sudhir Sindagi Free Surface Effect Reducing FSE by inserting longitudinal bulkhead at the center of length New iT = ๐ (๐+๐)๐ Old iT for n in number longitudinal bulkheads inserted FSE = ๐ iT∗ρl (๐+๐)๐ Δ FSE = ๐ iT∗ρl (๐)๐ Δ where n is number longitudinal bulkheads inserted where n is number longitudinal compartments created 55 Prepared by Dr. Sudhir Sindagi Problems A vessel displacing 10000 tonnes, KG=8.9m, KM=9.4. The vessel loads Ballast water of RD 1.01 into a rectangular tank 30mX20mX2m up to 1m depth of the tank. The tank has a single centerline division. KG of BW is 0.5. Find the fluid GM of the vessel. Assume KM remains constant. Given Data Δ = 10000 tonnes, KG=8.9m, KM=9.4 ρl= 1.01 tonnes/m3 KG of BW=0.5m Fluid GM=? Weight added into the tank= volume of the tank* ρl= w= 30*20*1*1.01= 606 tonnes Shift in CG of the ship due to the addition of weight wxd GG1 = 56 Δ±w Prepared by Dr. Sudhir Sindagi Problems Here d : the distance between the CG of the mass added and the original location of the CG of the ship w x d 606 x (๐.๐ −๐.๐) GG1 = = Δ±w 10606 GG1 =0.4799m (downwards) FSE can be calculated in two ways FSE = ๐ iT∗ρl (๐)๐ Δ where n is number longitudinal compartments created Since there is one centerline division creating two compartments, hence n=2. ๐ 30∗203∗1.01 FSE = ๐ (๐) 10606∗12 57 FSE =0.4761m upwards Prepared by Dr. Sudhir Sindagi Problems FSE also can be calculated for individual compartments Since both compartments are identical, hence Total FSE will be twice of the FSE for the individual tank iT∗ρl FSE = *2 Δ 30∗103∗1.01 FSE = *2 10606 ∗12 FSE =0.4761m (upwards) Fluid GM= KM-KG + downward shift in CG of ship-FSE Fluid GM=9.4-8.9+0.4799-0.4761 Fluid GM=0.503m 58 Prepared by Dr. Sudhir Sindagi Problems A vessel of 10000 tonnes displacement, KM=9.3, KG=7.3 has two rectangular identical deep tanks, port and starboard, each 15m long, 10m wide and 8m deep. The starboard tank is full of SW while the port deep tank is empty. Calculate the fluid GM of the vessel when one quarter of the water in the starboard deep tank is transferred to the port deep tank. Given Data Δ = 10000 tonnes, KG=7.3m, KM=9.3 ρl= 1.025 tonnes/m3 Fluid GM=? This is the typical problem of shifting of weight with FSE in both tanks Weight shifted= volume of the liquid* ρl 59 w= 15*10*2*1.025= 307.5 tonnes Prepared by Dr. Sudhir Sindagi Problems Shift in CG of the ship due to shifting of weight wxd GG1 = Δ 307.5x 6 GG1 = 10000 GG1 =0.1845m (downwards) Since both compartments are identical Total FSE will be twice FSE of each tank iT∗ρl FSE = *2 Δ 15∗103∗1.025 FSE = *2 10000∗12 FSE =0.2562m (upwards) 60 Prepared by Dr. Sudhir Sindagi Problems Fluid GM= Solid GM-FSE Fluid GM= KM-KG + downward shift in CG of ship-FSE Fluid GM=9.3-7.3+0.1845-0.2562 Fluid GM=1.92825m 61 Prepared by Dr. Sudhir Sindagi Problems A ship of 8000 tonnes displacement has KM=7.5m, KG=7.0m. A double bottom tank is 12m long, 15m wide and 1m deep. The tank is divided longitudinally at the center line and both sides are full of salt water. Calculate the list if one side is pumped out until it is half empty. Given Data Δ = 8000 tonnes, KG=7.0m, KM=7.5 ρl= 1.025 tonnes/m3 Angle of List=? This is the typical problem of removal of weight with FSE in only one tank Weight removed= volume of the liquid* ρl w= 12*7.5*0.5*1.025= 46.125 tonnes 62 Prepared by Dr. Sudhir Sindagi Problems Shift in CG of the ship due to the removal of weight Here d : the distance between the CG of the mass added and the original location of the CG of the ship For horizontal shift d= 3.75m w x d 46.125∗3.75 GGH = = Δ−w 8000−46.125 GGH = 0.0217m (Port side) For vertical shift d= 6.25m w x d 46.125∗6.25 GGV = = Δ−w 8000−46.125 GGV = 0.03624m (upwards) 63 Prepared by Dr. Sudhir Sindagi Problems Since there is only one tank which is half empty, hence Total FSE will be for only one tank iT∗ρl FSE = Δ 12∗7.53∗1.025 FSE = (8000−46.125)∗12 FSE =0.0543m (upwards) Fluid GM= Solid GM-FSE Fluid GM=0.50-0.03624-0.0543 Fluid GM=0.40912m 64 Prepared by Dr. Sudhir Sindagi Problems ๐ญ๐๐ง ๐ = ๐ญ๐๐ง ๐ = GGH Fluid GM 0.02174 ๐.๐๐๐๐๐ Φ = 3.05 degrees on port side. 65 Prepared by Dr. Sudhir Sindagi Inclining Experiment Purpose: To Find KG or GM of a vessel when it is as near to the completion as possible, that is as near to the light condition as possible. The purpose of this procedure is to achieve a satisfactory accuracy in the determination of the lightship weight and of the coordinates of its CG. Reason: The above are initial conditions which must be known before the stability of a ship in any particular condition of loading can be determined. For example, when dealing with height of CG above the keel, the initial position of the CG must be known to decide the final one. Also since the KG is very great comparing to the GM, it must be accessed very accurately if GM is to be found with reasonable 66 accuracy. Prepared by Dr. Sudhir Sindagi Inclining Experiment l= Length of the pendulum y= reading on the batton sin ๐ = ๐ ๐ wxd But GG1 = Δ GG1 Also, tan ๐ = ๐๐ From here, we get GM Now, GM=KM- KG KG = KB + BM - GM 67 Prepared by Dr. Sudhir Sindagi Inclining Experiment Procedure: During the experiment, the ship is forcibly inclined by shifting weights at fixed distance across the deck under controlled condition and finding the resultant angle of heel. The weights are usually concrete blocks and inclination is measured by using long pendulum one forward and one aft of the midship section. If third pendulum to be used, it is usually placed amidships. The movement of the pendulum is made across battens which lie perfectly horizontal when the ship is upright. The pendulums are fixed at a height of about 10m above the battens at the centerline of the ship. 68 their The pendulum bob may be immersed in water or oil to damp motion. Prepared by Dr. Sudhir Sindagi Inclining Experiment Preparations before execution of an Inclining Experiment The experiment should be carried out in calm water & nice weather. No wind, no heavy rain, no tides. It is essential that the ship be free to incline (mooring ropes should be as slack as possible, but be careful.) The vessel should be upright prior to the inclining. However, an initial list of the ship not exceeding 0.5° is permissible. Prior to the inclining test, lists of all items which are to be added, removed, or relocated should be prepared. These weights and their locations should be accurately recorded. All objects should be secured in their regular positions. All weights which may swing or shift must be secured in their known position.69 Prepared by Dr. Sudhir Sindagi Inclining Experiment Preparations before execution of an Inclining Experiment Normally, the total value of missing weights should not exceed 2 percent and surplus weights, excluding liquid ballast, not exceed 4 percent of the light ship displacement State of all tanks to be noted carefully. Drafts to be accurately read amidships and on both sides of the ship. The ship should be cleared of residues of cargo, tools, debris, scaffolding and snow. Icing of the inner and outer surfaces, the underwater hull included, is not permitted Density of water is measured at a number of positions and depths around the ship. 70 All cross connections between tanks are to be closed Prepared by Dr. Sudhir Sindagi Inclining Experiment Preparations before execution of an Inclining Experiment Preferably, all tanks should be either full or empty. The number of tanks containing liquids should be kept to a minimum. Soundings and density of liquids in tanks should be taken. Shapes of tanks which are partly filled are to be known in order to determine the free liquid surface effect Minimum number of people on board and should remain on a specified positions during each reading. All service tanks and machinery plant pipings are to be filled as for the working condition. The angle of inclination should be small enough with the range of validity of the theory. 71 The ship in experiment should not have a large trim. Prepared by Dr. Sudhir Sindagi Inclining Experiment Procedure: During the experiment, the ship is forcibly inclined by shifting weights at fixed distance across the deck under controlled condition and finding the resultant angle of heel. The weights are usually concrete blocks and inclination is measured by using long pendulum one forward and one aft of the midship section. If third pendulum to be used, it is usually placed amidships. The movement of the pendulum is made across battens which lie perfectly horizontal when the ship is upright. The pendulums are fixed at a height of about 10m above the battens at the centerline of the ship. 72 their The pendulum bob may be immersed in water or oil to damp motion. Prepared by Dr. Sudhir Sindagi Problems The inclining experiment is carried out on a vessel. The following data is noted – – – – – – – Displacement when inclined: 9550 tonnes Mass of inclining weight: 10 tonnes Distance of Transverse shift of weight: 18m Length of Pendulum line: 9.5m Mean deflection of pendulum line: 100mm KG of inclining weight: 12.5m KM=8.35. Calculate the GM of the vessel. A tank containing 150tonnes of FW is full at the time of experiment. KG of water is 1m. Calculate the lightship KG of the ship. 73 Prepared by Dr. Sudhir Sindagi Problems Given Data Δ = 9550 tonnes, w=10 tonnes, d= 18m, l= 9.5m, y=0.1m, KG of weight=12.5m, KM=8.35m GM=? FW weight = 150 tonnes and its KG=1m, KG light=? sin ๐ = ๐ ๐ Φ = 0.603 degrees. wxd But GG1 = Δ GG1 =0.01884m GG1 tan ๐ = ๐๐ From here, we get GM = 1.787m 74 Prepared by Dr. Sudhir Sindagi Problems Now, GM=KM- KG KGold= KM - GM KGold= 6.5596m To calculate the KG in lightship condition, the inclining weights and FW are to be removed from the ship as they are not part of the lightweight condition. Hence using the tabular method Weight Distance from Keel Moment 9550 6.5596 62644.18 -10 12.5 -125 -150 1 -150 ΣW=9390 tonnes ΣM=62369.18 t-m ΣM Final KG= ΣW Lightship KG= 6.642m 75 Prepared by Dr. Sudhir Sindagi Problems While loading a cargo of timber on a deck at KG=12m, it is noted that a sling of timber weighing 8 tonnes moved from one side of the ship to the other side by 16m and inclined the ship by 1 deg. KM at the draft was 10.5m. Calculate the approx. KG. Displacement =13000 tonnes. How much more cargo would be safe to load on the deck, if the GM was not to be less than 0.5m. Given Data Δ = 13000 tonnes, w=8 tonnes, d= 16m, KG of weight=12.5m, KM=10.5m, Φ = 01 degrees If GM=0.5m How much more cargo can be loaded=? wxd But GG1 = Δ 76 GG1 =0.009846m Prepared by Dr. Sudhir Sindagi Problems tan ๐ = GG1 ๐๐ From here, we get GM = 0.564m Now, GM=KM- KG KGold= KM - GM KGold= 9.935m Now to have GM>0.5 KM- KG >0.5 KG <10m GGV= KGNEW-Kgold wxd GGV=0.0641m = = Δ+๐ฐ W=416.65 tonnes 77 Prepared by Dr. Sudhir Sindagi Problems In an inclining experiment, a mass of 12.5 tonnes was moved 10 metres across the deck and caused a plumb line, 12 metres long to move out 320 mm. A double bottom tank in the ship was full of water, during the experiment. Mass of water in the tank is 450 tonnes and had its centre of gravity 0.9 metres above the keel, without which the ship would have been in the light condition. If the ship’s displacement at the time of experiment was 3750 tonnes and her KM was 9.0 metres, find: a) The KG at the time of experiment. b) The light KG Given Data Δ = 3750 tonnes, w=12.5 tonnes, d=10m, l=12m, y=0.32m, KG of weight=12.5m, KM=9m, BW=450 tonnes with KG=0.9m 78 KG= ? At the experiment and at the lightship condition Prepared by Dr. Sudhir Sindagi Problems sin ๐ = ๐ ๐ Φ = 1.528 degrees. wxd But GG1 = Δ GG1 =0.033m GG1 tan ๐ = ๐๐ From here, we get GM = 1.237m Now, GM=KM- KG KGold= KM - GM KGold= 7.762m 79 Prepared by Dr. Sudhir Sindagi Problems To calculate the KG in lightship condition, the inclining weights and BW are to be removed from the ship as they are not part of the lightweight condition. However, KG of inclining weights are not provided here, hence assuming they are part of the lightship condition. Hence using the tabular method Weight Distance from Keel Moment 3750 7.762 29107.5 -450 0.9 -405 ΣW=3300 tonnes ΣM=-28702.5 t-m ΣM Final KG= ΣW Lightship KG= 8.697m 80 Prepared by Dr. Sudhir Sindagi Empirical expressions for KB 81 Prepared by Dr. Sudhir Sindagi Stability at the Large Angles of Heel At the large angles of heel (greater than 7 degrees) the centre of buoyancy will move further out and the low side and the force of buoyancy can no longer be considered to act vertically upwards though fixed position of M, the initial metacentre. M will no longer remain in one position and it is no longer of direct relevance to the calculation. By definition, M remains the point at which successive lines of buoyancy intersect as the ship heels. The ship will, in general, trim as well as heel. This means that the CB moves fore and aft as well as transversely. The deck edge and the turn of bilge will become immersed or exposed. The metacentre moves as per: when B moves up, M moves down; and when B moves down, M moves up. 82 Prepared by Dr. Sudhir Sindagi Stability at the Large Angles of Heel 83 Prepared by Dr. Sudhir Sindagi Stability at the Large Angles of Heel 84 Prepared by Dr. Sudhir Sindagi Stability at the Large Angles of Heel Derivation for the Wall sided formula (๐๐ = (๐๐ + ๐๐ tan๐ ๐) sin φ)) ๐ 85 Prepared by Dr. Sudhir Sindagi Stability at the Large Angles of Heel Derivation for the Wall sided formula (๐๐ = (๐๐ + ๐๐ tan๐ ๐) sin φ)) ๐ In the case of large angles of heel, the vertical shift in the CB of the ship due to the vertical shift in the centroid of the wedge cannot be neglected. This vertical shift in the CB causes vertical shift in the metacenter from M to M’, causing an increase in the metacentric height from GM to GM’ resulting in an increase in the righting lever from GZ to GZ’. New Righting Lever = GZ’ = GZ+ ZZ’ New Righting Lever = GM sin ฯ+ NB2 New Righting Lever = GM sin ฯ+ B1B2 sinฯ New Righting Lever = (GM + B1B2)sinฯ 86 Prepared by Dr. Sudhir Sindagi Stability at the Large Angles of Heel Volume of the elementary wedge (v) = Volume of the elementary wedge (v) = ๐ ๐ฒ ∗ ๐ฒ tan φ ๐๐ฑ ๐ ๐ ๐ ๐ฒ tan φ ๐๐ฑ ๐ Here, the distance by which the volume of wedge is shifted vertically upwards = distance by which the centroid is shifted from g to g1 d= ๐ ๐ฒ ๐ tanฯ+ ๐ ๐ฒ ๐ tanฯ = ๐ ๐ฒ ๐ tanฯ As mentioned earlier, this is case of the shift in volume, hence there is a shift in the center buoyancy of the ship. Small shift in the center buoyancy of the ship due to the shift in Volume of the elementary wedge = ๐๐จ๐ฅ๐ฎ๐ฆ๐ ๐จ๐ ๐ฐ๐๐๐ ๐ ∗ ๐๐ข๐ฌ๐ญ๐๐ง๐๐ ๐๐ฒ ๐ฐ๐ก๐ข๐๐ก ๐ญ๐ก๐๐ข๐ซ ๐๐๐ง๐ญ๐ซ๐จ๐ข๐ ๐ฌ๐ก๐ข๐๐ญ๐ฌ ๐ฏ∗๐ = ๐ฏ๐จ๐ฅ๐ฆ๐๐ญ๐ซ๐ข๐ ๐๐ข๐ฌ๐ฉ๐ฅ๐๐๐๐ฆ๐๐ง๐ญ ๐จ๐ ๐ญ๐ก๐ ๐ฌ๐ก๐ข๐ฉ ๐ 87 Prepared by Dr. Sudhir Sindagi Stability at the Large Angles of Heel Small shift in the center buoyancy of the ship = Small shift in the center buoyancy of the ship = ๐ ๐ ๐ ๐ฒ tan φ ๐๐ฑ∗ ๐ฒ ๐ ๐ tanฯ ๐ ๐ ๐ ๐ฒ ๐๐ฑ ๐ ๐ ∗ tan๐ ๐ Total shift in the center of buoyancy of the ship = BB1 which will be obtained by integrating above equation for the entire length B1B2 = ๐ ๐ ๐ ๐ โซ๐ฑ๐ ๐ฒ ๐ืฌโฌ ๐ ๐ ∗ tan๐ ๐ ๐ ๐ ๐ โซ๐ฑ๐ ๐ฒ ๐ืฌโฌ IT Here, the term = =Transverse Moment of inertia of the ๐ ship’s waterplane area taken at SLL about the longitudinal axis passing through the centerline. 88 IT tan๐ ๐ BM ๐ B1B2 = ∗ = ∗ tan ๐ Prepared by Dr. Sudhir Sindagi Stability at the Large Angles of Heel New Righting Lever = (GM + B1B2)sinฯ Righting Lever (GZ) = (GM + BM ๐ ∗ tan๐ ๐)sinฯ This equation is applicable when angle of heel is greater than 70. For small angles of heel below 70 Righting Lever (GZ) = GMsinฯ Increase in the Righting lever = ZZ’ = BM ๐ ∗ tan๐ ๐sinฯ Vertical shift in the Metacenter is MM’ = B1B2 = BM ๐ ∗ tan๐ ๐ 89 Prepared by Dr. Sudhir Sindagi Stability at the Large Angles of Heel Angle of Loll: When a vessel has negative metacentric height (GM) i.e., is in unstable equilibrium, any external force applied to the vessel will cause it to start heeling. As it heels, the moment of inertia of the vessel's waterplane increases, which increases the vessel's BM. Since there is relatively little change in KB of the vessel, the KM of the vessel increases. At some angle of heel (say 10°), KM will increase sufficiently equal to KG, thus making GM of vessel equal to zero. When this occurs, the vessel goes to neutral equilibrium, and the angle of heel at which it happens is called angle of loll. In other words, when an unstable vessel heels over towards a progressively increasing angle of heel, at a certain angle of heel, the 90 centre of buoyancy (B) may fall vertically below the centre of gravity (G). Prepared by Dr. Sudhir Sindagi Stability at the Large Angles of Heel 91 Prepared by Dr. Sudhir Sindagi Stability at the Large Angles of Heel Angle of Loll BB1 = tan φ IT ๐ From the triangle MBB1, we can write tan φ = BB1 BM Putting this in above equation and after simplification, we get BM = IT ๐ 92 Prepared by Dr. Sudhir Sindagi Stability at the Large Angles of Heel Angle of Loll The angle at which a ship with a negative initial metacentric height will achieve the equilibrium condition is the angle of Loll. In a seaway, such a ship will oscillate between the angle of loll on the starboard and the on Port side. Depending upon external forces such as wind and waves a ship may suddenly flop over from PS to SB and then back again to PS. Such abrupt oscillation, different from a continuous roll, is characteristic for negative metacentric heights. An angle of loll can be corrected only by lowering the centre of gravity, not by moving loads transversely. This can be done by moving weight downwards, adding water ballast in double bottom tanks or removing 93 weight above the ship vertical centre of gravity. Prepared by Dr. Sudhir Sindagi Stability at the Large Angles of Heel Derivation for the Angle of Loll: We know the Righting Lever (GZ) = (GM + BM ๐ ∗ tan๐ ๐)sinฯ However, for the Angle of Loll, the Value of GZ=0 as it is in the equilibrium condition and more so GM= -ve. BM 0= (-GM + ∗ tan๐ ๐)sinฯ ๐ This will yield either sinฯ = 0 giving ฯ=0 for the equilibrium condition or BM -GM + ∗ tan๐ ๐ =0 ๐ BM ๐ ∗ tan๐ ๐ = GM 94 Prepared by Dr. Sudhir Sindagi Stability at the Large Angles of Heel Derivation for the Angle of Loll: BM ∗ tan๐ ๐ = GM ๐ tan๐ ๐ ๐= = 2GM ๐๐ ± tan−๐ 2GM ๐๐ This is the expression by which one can estimate the angle of Loll for the initial unstable ship. 95 Prepared by Dr. Sudhir Sindagi Problems A ship 12000 t displacement has second moment of area about the centerline 72x103 m4. If the metacentric height is -0.05. Calculate the Angle of Loll. At one point during Voyage, the above vessel is found to have angle of Loll 130. Calculate the initial Metacentric height. Given Data Δ=12000 tonne, GM=-0.05m, IT= 72x103m3, ๐ = ? If Angle of Loll (๐) = 130. then GM=? Angle of Loll ๐ = But BM = IT ๐ BM = 6.15m ± tan−๐ 2GM ๐๐ , ๐๐ =72x103m3 , ๐ = Δ/ ρ = 12000/1.025 96 Prepared by Dr. Sudhir Sindagi Problems Angle of Loll ๐ = ± tan−๐ ๐ = ± tan −๐ 2GM ๐๐ 2∗๐.๐๐ ๐.๐๐ ๐ = ±7.268 degrees Now in the second case ๐) = 130 ๐๐ = ± tan−๐ 2∗๐๐ ๐.๐๐ GM= -0.1638m 97 Prepared by Dr. Sudhir Sindagi GZ Curve The easiest and handiest tool for analysing a surface ship’s stability, is by graphs or curves. Since the stability of a ship can be directly commented on by the nature and value of its metacentric height (GM), a direct method to track the stability of a ship for a range of heel angles would be, to generate a curve that relates this parameter to the angle of heel. Since metacentric height is directly related to the righting lever (GZ) and angle of heel, the curve of statical stability is a plot between the righting lever and angle of heel. 98 Prepared by Dr. Sudhir Sindagi GZ Curve 99 Prepared by Dr. Sudhir Sindagi GZ Curve 100 Prepared by Dr. Sudhir Sindagi GZ Curve MAIN FEATURES OF THE GZ CURVE Slope at the origin. For small angles of heel, the righting lever is proportional to the angle of inclination, the metacentre being effectively a fixed point. Initial GM: The tangent to the GZ curve at the origin represents the metacentric height at the angle of heel equal to 1 rad = 57.3 degrees. Maximum GZ. This is proportional to the largest steady heeling moment that the ship can sustain without capsizing, and its value and the angle at which it occurs are both important. 101 Prepared by Dr. Sudhir Sindagi GZ Curve MAIN FEATURES OF THE GZ CURVE Range of stability. At some angle, often greater than 90 degrees, the GZ value reduces to zero and becomes negative for larger inclinations. This angle is known as the angle of vanishing stability and the range of angle for which GZ is positive is known as the range of stability. For angles less than this, a ship will return to the upright state when the heeling moment is removed. Angle of deck edge immersion. For most ship forms, there is a point of inflexion in the curve corresponding roughly to the angle at which the deck edge becomes immersed. In general, of course, the angle at which the deck edge is immersed varies along the length, but is within a fairly narrow band for the larger sections amidships which exert most 102 influence upon the stability. Prepared by Dr. Sudhir Sindagi GZ Curve MAIN FEATURES OF THE GZ CURVE Area under the curve. The area under the curve represents the ability of the ship to absorb energy imparted to it by winds, waves or any other external agency. The dynamical stability of a ship at any inclination is defined as the work done in heeling the vessel to that inclination. Dynamical stability = Δ* Area under the stability curve The dynamic stability of a ship is the area enclosed within its static stability curve. It gives us the magnitude of external heeling energy that the ship can absorb before capsizing. 103 Prepared by Dr. Sudhir Sindagi Problems A ship displacing 5000t has GZ values as follows when KG= 6m. Draw the Statical Stability Curve for this condition and for KG= 5m. Discuss the effect on Stability of lowering the CG of ship. Angle of Heel (θ) GZ(m) 0 15 30 45 60 75 90 0 0.14 0.36 0.73 0.67 0 -0.65 Given Data Δ=5000 tonne, KG1=6m, KG2=5m, To draw the statical stability curve(GZ Curve), need to select suitable scale to fit on the graph paper. Suggested scales on X and y axis of the graph papers are 104 Prepared by Dr. Sudhir Sindagi Problems On X axis, which is longer side of the graph paper 1cm = 5 degree (always) On Y axis, which is shorter side of the graph paper 1cm ≈ (Difference of maxima and minima of GZ values)/15 In this case, it will be 1cm≈(0.73-(-0.65)/18≈0.08m Write these scales on the graph paper(Top right location) Position location of the X axis in such way that, you can locate the negative value of GZ if given on the graph paper. Plot the graph by locating the points with values in the table. Join these points with smooth curves and from graphs obtain following Angle of Vanishing Stability, Range of Stability, GZ max and Initial GM. 105 Prepared by Dr. Sudhir Sindagi Problems To draw GZ curve at different value of KG Need to calculate values of GZ for new KG values as per the following formula. G1Z1 = GZ ± GG1* sinฯ When G goes down new values of GZ increases and Vice versa. With these new values of GZ values for different values of angles of heel, one can plot similar GZ curve as per the procedure mentioned earlier. 106 Prepared by Dr. Sudhir Sindagi Problems A ship of 5000 tonnes displacement has righting levers as follows. Angle of Heel (θ) GZ(m) 10 0.21 20 0.33 30 0.40 40 0.43 Calculate the dynamical stability to 40 degrees heel We know that, Dynamical stability = Δ* Area under the stability curve To calculate the area under the GZ curve, one can make use Simpson’s rule. Here, n=5 (Including value of GZ=0 at 0 degree angle of heel) Hence using Simpson’s 1st rule h= 10 degrees=0.1745 rad (Always use angles in degrees 107during multiplication or division) Prepared by Dr. Sudhir Sindagi Problems Col(1) Col (2) Col (3) Col(4)= (2)X(3) Angle GZ Simpson’s Multiplier For Area 0 0 1 0 10 0.21 4 0.84 20 0.33 2 0.66 30 0.40 4 1.6 40 0.43 1 0.43 ΣA=3.53 rad-m Total Area under the GZ ๐ก curve=A= ๐ * ΣA=0.2053 rad-m Dynamical stability = 1026.5 tonnes-m- rad = 1026.5* 9.81 kJ Dynamical stability = 10069.965 kJ 108 Prepared by Dr. Sudhir Sindagi IMO Criteria for GZ Curve The area under the righting lever curve (GZ curve) shall not be less than – 0.055 metre-radians up to 30° angle of heel and – not less than 0.09 metre-radians up to 40° or the angle of downflooding, whichever is smaller. – Additionally, the area under the righting lever curve (GZ curve) between the angles of heel of 30° and 40° or between 30° and angle of down-flooding, whichever is smaller, shall not be less than 0.03 metre-radians. The righting lever GZ shall be at least 0.2 m at an angle of heel equal to or greater than 30°. The maximum righting lever shall occur at an angle of heel not less than 25°. 109 The initial metacentric height GM shall not be less than 0.15 m. Prepared by Dr. Sudhir Sindagi IMO Criteria for GZ Curve 110 Prepared by Dr. Sudhir Sindagi IMO Criteria for GZ Curve Severe wind and rolling criterion (weather criterion) The ability of a ship to withstand the combined effects of beam wind and rolling shall be demonstrated, 111 Prepared by Dr. Sudhir Sindagi IMO Criteria for GZ Curve Severe wind and rolling criterion (weather criterion) The ship is subjected to a steady wind pressure acting perpendicular to the ship’s centreline which results in a steady wind heeling lever (lwl1) ๐ฅ๐ฐ๐ฅ๐ = ๐∗๐∗๐ ๐๐๐๐∗๐ ∗โ & ๐ฅ๐ฐ๐ฅ๐ = 1.5 ๐ฅ๐ฐ๐ฅ๐ P = 504 N/m2 (wind speed = 29 m/s). A = projected lateral area of the portion of the ship and deck cargo above the waterline in m2. Z = vertical distance from the centre of A to the centre of the underwater lateral area or approximately to a point at one half the draught in m. ๐ฅ๐ฐ๐ฅ๐ = a gust wind heeling lever 112 Prepared by Dr. Sudhir Sindagi IMO Criteria for GZ Curve Severe wind and rolling criterion (weather criterion) From the resultant angle of equilibrium (ฯ0), the ship is assumed to roll owing to wave action to an angle of roll (ฯ1) to windward. The angle of heel under action of steady wind (ฯ0) should be limited to a certain angle to the satisfaction of the Classification Society. As a guide, 16° or 80% of the angle of deck edge immersion, whichever is less, is suggested. The ship is then subjected to a gust wind pressure which results in a gust wind heeling lever lwl2 Under these circumstances, area “b” should be equal to or greater than area “a” 113 Prepared by Dr. Sudhir Sindagi IMO Criteria for GZ Curve Severe wind and rolling criterion (weather criterion) ฯ0 = angle of heel under action of steady wind ฯ1 =angle of roll to windward due to wave action ฯ2 =angle of down-flooding (ฯf) or 500 or ฯc, whichever is less, where: ฯf = angle of heel at which openings in the hull, superstructures or deckhouses which cannot be closed weathertight immerse. ฯc = angle of second intercept between wind heeling lever ๐ฅ๐ฐ๐ฅ๐ and GZ curves. 114 Prepared by Dr. Sudhir Sindagi IMO Criteria for GZ Curve International Code for the Safe Carriage of Grain in Bulk The angle of heel due to the shift of grain shall not be greater than 12° or in the case of ships constructed on or after 1 January 1994 the angle at which the deck edge is immersed, whichever is the lesser; In the statical stability diagram, the net or residual area between the heeling arm curve and the righting arm curve up to the angle of heel of maximum difference between the ordinates of the two curves, or 40° or the angle of down flooding whichever is the least, shall in all conditions of loading be not less than 0.075 metre-radians; and The initial metacentric height, after correction for the free surface effects of liquids in tanks, shall be not less than 0.30 m 115 Prepared by Dr. Sudhir Sindagi IMO Criteria for GZ Curve International Code for the Safe Carriage of Grain in Bulk 116 Prepared by Dr. Sudhir Sindagi IMO Criteria for GZ Curve International Code for the Safe Carriage of Grain in Bulk 117 Prepared by Dr. Sudhir Sindagi IMO Criteria for GZ Curve Additional criteria are recommended for passenger ships 1. The angle of heel on account of crowding of passengers to one side should not exceed 10°. 2. The angle of heel on account of turning should not exceed 10° when calculated using the following formula: V2 T MR = 0.02 Δ KG − L 2 MR : heeling moment in metre-tons V : service speed in m/sec., L : length of ship at waterline in m., ๏ : displacement in metric tons, T : mean draught in m. , KG : height of centre of gravity above keel in m. 118 Prepared by Dr. Sudhir Sindagi Cross Curves of Stability – KN Curves The vertical position of centre of gravity (G) of the ship is not always fixed. It changes with every voyage, depending on the loading conditions and the amount of ballast. The CG of the ship also changes when the ship is in transit. This varying nature of CG makes it difficult for the designer to assume the loading conditions at which GZ curves should be obtained, because different values of KG would result in different metacentric heights (GM), and righting lever (GZ), the stability curves for each of the loading condition would be different. The cross curves of stability were developed, so that, for any loading condition (where KG is already known), values of righting lever (GZ) can be obtained for all angles of heel. 119 Prepared by Dr. Sudhir Sindagi Cross Curves of Stability – KN Curves The point ‘K’ represents the keel of the ship. If at any angle of heel (ฯ), a line parallel to that of GZ is drawn from ‘K’, then the point of intersection of this line with the line of action of buoyancy, is represented as ‘N’. GZ = KN- KG* sinฯ The only unknown in the above expression is the value of KN which are obtained from KN curves. 120 Prepared by Dr. Sudhir Sindagi Cross Curves of Stability – KN Curves Once the KN curves are obtained, it is now possible for the designer to obtain the stability curve / GZ curve for any loading condition 121 Prepared by Dr. Sudhir Sindagi Hydrostatic Curves A series of graphs drawn to a vertical scale of draught and a base of length, which gives values such as, Vertical, Longitudinal and Transverse Center of Buoyancy, Mass Displacement (Δ), Volume Displacement (∇), Longitudinal and Transverse Centre of Floatation, Metacentric Radius, Moment to Change Trim 1 cm (MCT), Tonnes per cm Immersion (TPC). All the hydrostatic parameters are calculated by a stability analysis and are plotted on a graph against different drafts. This graph is collectively called hydrostatic curves. In practice tables with hydrostatic parameters calculated for different draughts are used. 122 Prepared by Dr. Sudhir Sindagi Hydrostatic Curves 123 Prepared by Dr. Sudhir Sindagi Hydrostatic Curves 124 Prepared by Dr. Sudhir Sindagi Summary Introduction – Types of Equilibrium Conditions for the stability of the ship and submarine Stiff and Tender Ship Stability for the small angles of heel – Derivation for the Metacentric radius – Free Surface Effect – Inclining Experiment Stability at the large angles of heel – – – – – Derivation for the wall sided formula Angle of Loll GZ Curve Dynamical Stability of the ship Requirements of the IMO for the GZ Curve 125 Prepared by Dr. Sudhir Sindagi Longitudinal and Damaged stability By Dr. Sudhir Sindagi 1 Prepared by Dr. Sudhir Sindagi Chapter Content Introduction- Trim Trimming Moment, MCT 1cm, IL, BML, GML Trim due to – Shifting of existing weight – Addition or removal of weight Damaged Stability – Deterministic damage stability • Added Weight Method • Lost Buoyancy Method – Permeability – Probabilistic damage stability – Floodable length, Margin Line, Permissible Length, Factor of Subdivision, Floodable Length curve 2 Prepared by Dr. Sudhir Sindagi Introduction Trim may be considered as the longitudinal equivalent of list. Trim is also known as ‘longitudinal stability’. It is in effect transverse stability turned through 90°. Instead of trim being measured in degrees it is measured as the difference between the drafts forward and aft. If difference is zero then the ship is said to be on even keel. Trim =0 , if TA=TF 3 Prepared by Dr. Sudhir Sindagi Introduction If forward draft is greater than aft draft, the vessel is said to be trimming by the bow (forward). Trim by forward (t) = TF-TA If aft draft is greater than the forward draft, the vessel is said to be trimming by the stern (aft). Trim by aft (t) = TA-TF 4 Prepared by Dr. Sudhir Sindagi Introduction In case of Trim by aft, if TM is the mean draft at F, then x is the increase in the draft at AP above F (mean draft) x= increase in the draft at AP above F (mean draft) x=LCF* tan θ = ๐๐ −๐๐ LCF∗ ๐ = ๐๐๐ ∗ ๐ญ ๐ y= reduction in the draft at FP below F= t-x 5 Prepared by Dr. Sudhir Sindagi Introduction ๐๐ +๐๐ Mean draft at F = TM= Applicable when F lies at Midship ๐ ๐ญ TM= ๐๐ -x = ๐๐ - ๐๐๐ ∗ Applicable when F lies anywhere. ๐ TM= ๐๐ +y = ๐๐ + (๐ญ − ๐ฑ) 6 Prepared by Dr. Sudhir Sindagi Introduction In case of Trim by forward, if TM is the mean draft at F, then x is the reduction in the draft at AP below F (mean draft) x= reduction in the draft at AP below F (mean draft) x=LCF* tan θ = ๐๐ −๐๐ LCF∗ ๐ = ๐๐๐ ∗ ๐ญ ๐ y= increase in the draft at FP above F= t-x 7 Prepared by Dr. Sudhir Sindagi Introduction ๐๐ +๐๐ Mean draft at F = TM= ๐ ๐ญ TM= ๐๐ +x = ๐๐ + ๐๐๐ ∗ ๐ TM= ๐๐ −y = ๐๐ - (๐ญ − ๐ฑ) 8 Prepared by Dr. Sudhir Sindagi Trimming Moment, MCTC, IL, BML, GML 9 Prepared by Dr. Sudhir Sindagi Trimming Moment, MCTC, IL, BML, GML 10 Prepared by Dr. Sudhir Sindagi Trimming Moment, MCTC, IL, BML, GML Consider a ship to be floating at rest in still water and on an even keel as shown in Figure 1 Now let a weight ‘w’, already on board, be shifted aft through a distance ‘d’, . This causes the centre of gravity of the ship to shift from G to G1, parallel to the shift of the centre of gravity of the weight shifted, so that: wxd GG1 = Δ Δ*GG1 = w x d Trimming Moment =Δ*GG1 = w x d is generated The ship will now trim until the centres of gravity and buoyancy are again in the same vertical line as shown in the Figure 2. 11 Prepared by Dr. Sudhir Sindagi Trimming Moment, MCTC, IL, BML, GML When trimmed, the wedge of buoyancy LFL1 emerges and the wedge WFW1 is immersed. Since the ship, when trimmed, must displace the same weight of water as when on an even keel, the volume of the immersed wedge must be equal to the volume of the emerged wedge and F, the point about which the ship trims, is the centre of gravity of the water-plane area. The point F is called the ‘centre of flotation’ or ‘tipping centre’. A vessel with a rectangular water-plane has its centre of flotation on the centre line amidships but, on a ship, it may be a little forward or abaft amidships, depending on the shape of the water-plane. In trim problems, unless stated otherwise, it is to be assumed that the centre of flotation is situated amidships. 12 Prepared by Dr. Sudhir Sindagi Trimming Moment, MCTC, IL, BML, GML Trimming moments are taken about the centre of flotation since this is the point about which rotation takes place. The longitudinal metacentre (ML) is the point of intersection between the verticals through the longitudinal positions of the centres of buoyancy. The vertical distance between the centre of gravity and the longitudinal metacentre (GML) is called the longitudinal metacentric height. BML is the height of the longitudinal metacentre above the CB IL = Longitudinal Moment of inertia of the ship’s waterplane about the Transverse axis passing through the centroid of the waterplane. ๐๐๐ IL IL = , BML = ๐๐ ๐ Longitudinal Metacentric Radius = BML = ๐๐ ๐๐๐ for rectangular barge13 Prepared by Dr. Sudhir Sindagi Trimming Moment, MCTC, IL, BML, GML 14 Prepared by Dr. Sudhir Sindagi Trimming Moment, MCTC, IL, BML, GML Righting Moment in the longitudinal direction = Δ * GZL sin θ = ๐๐๐ ๐๐๐ Since the angle of trim is very small sin θ = tan θ tan θ = ๐๐๐ ๐๐๐ ๐๐๐ = ๐๐๐ tan θ Righting Moment in the longitudinal direction = Δ * GZL Righting Moment in the longitudinal direction = Δ * ๐๐๐ tan θ t Righting Moment in the longitudinal direction = Δ * ๐๐๐ ๐ This is the moment which acts on the ship, to float it on the even keel. 15 Prepared by Dr. Sudhir Sindagi Trimming Moment, MCTC, IL, BML, GML If Trim (t) = 1cm Moment required to change the trim by 1cm= Δ * ๐๐๐ MCT1cm= 1 cm ๐ Δ ∗ ๐๐๐ ๐๐๐๐ Since the angle of trim is very small, hence the distance GML is very high as compared to the distance BG. In the view of above, in case no data is available, it is assumed that GML≈BML Δ ∗ ๐๐๐ MCT1cm≈ ๐๐๐๐ 16 Prepared by Dr. Sudhir Sindagi Steps to solve Problems of Trim 1. Calculate the Trimming moment using 1. 2. w*d- when weight is shifted by a distance d Tabular method – when weights are added or removed from and to the ship 2. For the trimming moment all distances are to be taken from F. 3. If MCT 1cm is not given then calculate it using Δ ∗ ๐๐๐ 4. MCT1cm≈ ๐๐๐๐ 5. Calculate the trim using Trimming moment and MCT1cm using Trimming Moment 6. Trim in cm= MCT1cm 7. Calculate change in draft at aft using t 8. x=LCF* tan θ = x=LCF* ๐ 17 Prepared by Dr. Sudhir Sindagi Steps to solve Problems of Trim 9. Calculate change in draft at forward using 10. y= t – x 11. In the case, weight is added or removed from or to the ship, then calculate the change in draft using TPC as follows weight added or removed 12. Change in draft in cm (ΔT)= ๐๐๐ 13. Calculate final drafts at AP and FP using tabular method. Drafts Original Change in draft due to Trimming Moment Change in draft due to weight addition or removal Final Drafts AP (m) FP (m) ±x ±ΔT ±y ±ΔT 18 Prepared by Dr. Sudhir Sindagi Problems A vessel displacing 30000 tonnes is floating at drafts of 8.3m at FP and 9.6m at AP. MCT1 is 300. Center of flotation is 109m forward of AP, Length of the ship is 210m. Find the drafts at FP and AP if 1000 tonnes of ballast is moved from a tank CG of 175m forward of AP to a tank with CG 205m forward of AP. Given Data Δ = 30000 tonnes, ๐๐ = 8.3m, ๐๐ = 9.6m, MCT1=300, LCF=109m forward of AP, L=210m, w=1000 tonnes, d=(205-175)=30m ๐๐ = ?, ๐๐ = ? Trimming moment= w*d = 30000 t-m Trimming Moment 30000 Trim in cm= = 300 MCT1cm 19 Trim in cm (t)= 100 cm =1m Prepared by Dr. Sudhir Sindagi Problems Since the weight is being shifted from aft to forward, hence draft at AP will reduce and will increase at FP. Reduction in draft at AP = x= t 100 x=LCF* = 109* ๐ ๐๐๐ Reduction in draft at AP =x =51.904cm=0.51904m Increase in draft at FP = y = t - x Increase in draft at FP = y = 0.4809m Since, weight is not added and not removed hence there is no need to calculate the change in draft using TPC. 20 Prepared by Dr. Sudhir Sindagi Problems Making table for Final draft calculations. Drafts AP (m) FP (m) Original Change in draft due to Trimming Moment Change in draft due to weight addition or removal Final Drafts 21 Prepared by Dr. Sudhir Sindagi Problems Making table for Final draft calculations. Drafts Original Change in draft due to Trimming Moment Change in draft due to weight addition or removal Final Drafts AP (m) FP (m) 9.6 8.3 ( - ) 0.51903 ( + ) 0.4809 0 0 9.0810 8.7809 22 Prepared by Dr. Sudhir Sindagi Problems A ship 100 m in length floats at draughts forward 7.00 m and aft 6.80 m. Calculate the final draughts if 150 tonnes is loaded 20 m forward of aft perpendicular given that TPC is 15 and MCTC is 150 tm and LCF is 45 m forward of aft perpendicular. Given Data L=100m, ๐๐ = 7.0m, ๐๐ = 6.8m, MCT1=150, LCF=45m forward of AP, w=150 tonnes, TPC=15 ๐๐ = ?, ๐๐ = ? Here, since weight is added, hence to calculate the trimming moment, it will be calculated from F d= (45-20) = 25m 23 Trimming moment= w*d = 3750 t-m Prepared by Dr. Sudhir Sindagi Problems Trimming Moment 3๐๐๐ Trim in cm= = 150 MCT1cm Trim in cm= 25 cm =0.25m Since the weight is added aft of CF, hence draft at AP will increase and will decrease at FP. Increase in draft at AP = x= t 0.25 x=LCF* = 45* ๐ ๐๐๐ Increase in draft at AP =x =0.1125m Reduction in draft at FP = y = t - x Reduction in draft at FP = y = 0.1375m 24 Prepared by Dr. Sudhir Sindagi Problems Here, weight is added to the ship, hence the change in draft using TPC will be weight added or removed Change in draft in cm = ๐๐๐ Change in draft in cm = 150 ๐๐ Change in draft in cm = 10 cm = 0.1m Drafts Original Change in draft due to Trimming Moment Change in draft due to weight addition or removal Final Drafts AP (m) 6.8 FP (m) 7.0 + 0.1125 0.1 7.0125 - 0.1375 0.1 25 6.9625 Prepared by Dr. Sudhir Sindagi Problems A ship 100 m long arrives in port with draughts of 3m at FP and 4.3 m at AP. The hydrostatic particulars are TPC=10, MCTC= 120 tm, LCF= 3m aft of amidships. 80 tonnes of cargo is now loaded at a position of 24 m ford of amidships and 40 tonnes of cargo is discharged from 12m aft of amidships. Find out the new draughts. Given Data L=100m, ๐๐ = 3.0m, ๐๐ = 4.3m, MCT1=120, LCF=3m aft of Midship, w1=80 tonnes, and w2=40 tonnes, TPC=10 ๐๐ = ?, ๐๐ = ? Trimming Moment = w1*d1+w2*d2 (forward trim) Trimming Moment = 80*27+40*9 26 Trimming Moment = 2520 t-m (forward trim) Prepared by Dr. Sudhir Sindagi Problems Trimming Moment 2520 Trim in cm= = 120 MCT1cm Trim in cm= 21 cm =0.21m Since it is creating forward trim, hence draft at AP will decrease and will increase at FP. Reduction in draft at AP = x= t 0.21 x=LCF* = 47* ๐ ๐๐๐ Reduction in draft at AP =x =0.09827m Increase in draft at FP = y = t - x Increase in draft at FP = y = 0.1113m 27 Prepared by Dr. Sudhir Sindagi Problems Here, one weight is added to the ship and the other is removed, hence the change in draft using TPC will be weight added or removed Change in draft in cm = ๐๐๐ Change in draft in cm = (80−40) ๐๐ Change in draft in cm = 4 cm = 0.04m Drafts Original AP (m) 4.3 FP (m) 3.0 Change in draft due to Trimming Moment Change in draft due to weight addition or removal Final Drafts - 0.0987 0.04 4.2413 + 0.1113 0.04 28 3.1513 Prepared by Dr. Sudhir Sindagi Problems A vessel floating at drafts forward 9.84m and aft 10.62m. She loads following weights. TPC=26, MCT1=148 tonnes-m/cm, LCF=64m forward of AP, Length=120m. Find final drafts. Is it trimmed by aft or forward? . Weight (Tonnes) LCG from AP (m) Loads 450 25 Loads 320 100 Discharges 140 110 Given Data L=120m, ๐๐ = 9.84m, ๐๐ = 10.62m, MCT1=148, LCF=64m forward of AP, TPC=26 ๐๐ = ?, ๐๐ = ? 29 Prepared by Dr. Sudhir Sindagi Problems Loads Loads Discharges Weight (Tonnes) 450 320 140 Lever from F (m) 39 36 46 Trimming Moment = w1*d1+w2*d2 +w3*d3(Aft trim) Trimming Moment = 450*39-320*36+140*46 Trimming Moment = 12470 t-m (aft trim) Trimming Moment 12470 Trim in cm= = 148 MCT1cm Trim in cm= 0.8425m Since it is creating aft trim, hence draft at AP will increase and will 30 reduce at FP. Prepared by Dr. Sudhir Sindagi Problems Increase in draft at AP = x= t 0.8425 x=LCF* = 64* ๐ ๐๐๐ Increase in draft at AP =x =0.4493m Reduction in draft at FP = y = t - x Reduction in draft at FP = y = 0.3931m weight added or removed Change in draft in cm = ๐๐๐ Change in draft in cm = (450+320−140) ๐๐ Change in draft in cm = 0.2423m 31 Prepared by Dr. Sudhir Sindagi Problems Drafts AP (m) FP (m) Original Change in draft due to Trimming Moment Change in draft due to weight addition or removal 10.62 + 0.4493 0.2423 9.84 - 0.3931 0.2423 Final Drafts 11.3116 9.6896 32 Prepared by Dr. Sudhir Sindagi Problems A vessel has following details TF=10.8m, TA=11.4m, TPC=28 tonnes/cm, L=138m, LCF=65m forward of AP. It has to complete loading with final drafts as TF=10.9m, TA=11.4m. Find the amount of cargo load to achieve above. Given Data L=138m, ๐๐ = 10.8m, ๐๐ = 11.4m, LCF=65m forward of AP, TPC=28 To achieve ๐๐ = 10.9, ๐๐ = 11.4 what load to be added? Here, no MCT1 is provided and as no other required data is provided to calculate it, hence calculating required values from mean draft, initial & final trim and TPC. Initial Trim = ti = ๐๐ - ๐๐ = 0.6m 33 Final Trim = tf = ๐๐ - ๐๐ = 0.5m Prepared by Dr. Sudhir Sindagi Problems TMi= ๐๐๐ข - ๐๐๐ ∗ ๐ญ๐ข ๐ TMi= 11.1173m TMf= ๐๐๐ข - ๐๐๐ ∗ ๐ญ๐ ๐ TMf= 11.1644m Change in mean draft expected = ΔTmean= TMf- TMi= 0.0471m Weight to be added = ΔTmean* TPC Weight to be added =131.88 tonnes 34 Prepared by Dr. Sudhir Sindagi Problems A ship 150 m long floats at draughts of 8.20 m Forward and 8.9 m Aft. MCT 1 cm 260 tonne-m TPC 28 and LCF 1.5m Aft of midship. It is necessary to bring the ship to an even keel and a double bottom tank 60 m forward of midships is available. Calculate the mass of water required and the final draughts. Given Data L=150m, ๐๐ = 8.2m, ๐๐ = 8.9m, MCT 1 cm= 260, LCF= 1.5m Aft of midship, TPC=28 To bring the ship to an even keel calculate mass of water required to be added? ๐๐ = ?, ๐๐ = ? 35 Prepared by Dr. Sudhir Sindagi Problems Here, the initial trim and the trimming moment needs to be nullified. Initial Trim = t = ๐๐ - ๐๐ = 0.7m Trimming Moment = trim * MCT1 cm Trimming Moment = 18200 t-m Trimming Moment = weight to added * distance from F 18200= w* (1.5+60) w= 295.93 tonnes. Since weight is added at the forward end, hence it will cause increase in the draft at FP and the reduction in draft at AP. Reduction in draft at AP = x= t 0.7 36 x=LCF* = 73.5* ๐ ๐๐๐ Prepared by Dr. Sudhir Sindagi Problems Reduction in draft at AP =x =0.343m Increase in draft at FP = y = t - x Increase in draft at FP = y = 0.357m weight added Change in draft in cm = ๐๐๐ Change in draft in cm = (295.93) ๐๐ Change in draft in cm = 0.1056m 37 Prepared by Dr. Sudhir Sindagi Problems Drafts AP (m) FP (m) Original Change in draft due to Trimming Moment Change in draft due to weight addition or removal 8.9 - 0.343 0.1056 8.2 + 0.357 0.1056 Final Drafts 8.6626 8.6626 38 Prepared by Dr. Sudhir Sindagi Change in trim due to change in density When a ship passes from sea water into river water, or vice versa without change in displacement, there is a change in trim in addition to the change in mean draught. This change in trim is always very smaIl. Consider a ship of displacement Δ floats in sea water wherein G and B are in the same vertical line. If the vessel moves into river water of lesser density there is a bodily sinkage resulting in increase in draught. 39 Prepared by Dr. Sudhir Sindagi Change in trim due to change in density The volume of displacement has been increased by a layer of volume v, whose centroid is assumed to be at the centre of flotation F, resulting in the total shift as FB. This causes the centre of buoyancy to move from B to B1 and the centre of gravity remaining at G. Let ๐๐ & ๐๐ be the volumetric displacements in SW and RW, wherein, change in underwater volume= v= ๐๐ - ๐๐ 40 Prepared by Dr. Sudhir Sindagi Change in trim due to change in density Shift in Center of Buoyancy of the ship due to additional sinkage in the ๐ฏ∗๐ ๐ฏ∗๐ ๐ ship is BB1= = ๐ ๐๐ (๐๐ −๐๐ )∗๐ ๐ BB1= ๐๐ BB1= (๐๐ −๐๐ )∗๐ ๐ ๐๐ Since B1 is no longer in line with G, a trimming moment of Δ* BB1 acts on the ship causing a change in trim by the bow Trimming Moment Δ∗ BB1 Trim in cm= = MCT1cm MCT1cm Δ∗FB (๐๐ −๐๐ ) Trim in cm= MCT1cm ๐๐ 41 If the ship moves from the RW into SW, It will change trim by the stern. Prepared by Dr. Sudhir Sindagi Change in trim due to change in density Four important cases: If F lies aft of B Ship moves from SW to FW or RW (moves from higher density to lower density), the Ship will trim by the Bow. Ship moves from RW or FW to SW (moves from lower density to higher density), the Ship will trim by the stern. If F lies forward of B Ship moves from SW to FW or RW (moves from higher density to lower density), the Ship will trim by the stern. Ship moves from RW or FW to SW (moves from lower density to higher density), the Ship will trim by the bow. 42 Prepared by Dr. Sudhir Sindagi Change in trim due to change in density A ship 120 m long and 9100 tonne displacement floats at a level keel draught of 6.50 m in fresh water of 1.000 t/m3 MCT1=130 tonnes-m, TPC in sea water 16.5. LCB 2.30 m forward of midships. LCF 0.6 m aft of midships. Calculate the new draughts if the vessel moves into sea water of 1.024 t/m3 without change in displacement Given Data L=120m, Δ=9100 tonnes ๐๐ = 6.5m, ๐๐ = 6.5m, LCF=0.6m aft of midship, TPCSW=16.5, LCB 2.30 m forward of midships New drafts=? When it moves from FW to SW of ρ= 1.024 t/m3. When the ship moves from FW to SW, the draft reduces, which can be calculated using FWA as Δ ρSW – ρRW FWA = TPCsw ρRW 43 Prepared by Dr. Sudhir Sindagi Change in trim due to change in density FWA =13.2cm =0.132m Change in trim to change in density of water is Δ∗FB (๐๐ −๐๐ ) Trim in cm= MCT1cm ๐๐ Since the ship moves from the RW into SW & LCF is aft of LCB, It will change trim by the stern. Trim in cm (t)=4.87cm by stern = 0.0487 by stern t ๐.๐๐๐๐ Increase in draft at aft =x= LCF* = 59.4* ๐ ๐๐๐ Increase in draft at aft =x = 0.0241m (+ve) Reduction in draft at forward = y = t-x = 0.0246m (-ve) 44 Prepared by Dr. Sudhir Sindagi Problems Drafts Original Change in draft due to Trimming Moment Change in draft due to change in density Final Drafts AP (m) FP (m) 6.5 + 0.0241 -0.132 6.5 - 0.0246 -0.132 6.3921 6.3434 45 Prepared by Dr. Sudhir Sindagi Change in trim due to change in density A vessel with forward draft 8.72m and aft draft =9m is floating in SW. She enters in DW of RD=1.004. Find the initial and the mean drafts and her drafts in DW. Given that, MCTC=162, LCF=82m forward of AP, LCB=90m forward of AP, L=170m, TPC=29.8 and Δ=2700 tonnes Given Data Follow the similar procedure to solve the question. 46 Prepared by Dr. Sudhir Sindagi Determination of trim, due to the addition of large masses When a large mass is added to a ship, the resultant increase in draught is sufficient to cause changes in all the hydrostatic details. It then becomes necessary to calculate the final draughts from first principles. Such a problem exists every time a ship loads or discharges the major part of its deadweight. The principle is that after loading or discharging, the vessel is in equilibrium and the final G is in the same vertical line as the final B. For any given condition of loading, it is possible to calculate the displacement Δ and the longitudinal position of the centre of gravity G relative to midships. From the hydrostatic curves or data, the mean draught may be obtained at this displacement, and hence the value of MCT1 and the distance of 47 the LCB and LCF from midship. Prepared by Dr. Sudhir Sindagi Determination of trim, due to the addition of large masses These values are calculated for the level keel condition and it is unlikely that the LCB will be in the same vertical line as G. Thus a trimming moment acts on the ship which is the product of the displacement and the longitudinal distance between B and G, known as the trimming lever. The trimming moment, divided by the MCT1 cm gives the change in trim from the level keel condition. The vessel changes trim about the LCF and hence it is possible to calculate the end draughts. When the vessel has changed trim in this manner, the new centre of buoyancy B1 lies in the same vertical line as G. 48 Prepared by Dr. Sudhir Sindagi Problems A ship 125 m long has a light displacement of 4000 tonne with LCG 1.60 m aft of midships. The following items are now added: – – – – Cargo 8500 tonne LCG 3.9 m forward of midships Fuel 1200 tonne LCG 3.1 m aft of midships Water 200 tonne LCG 7.6 m aft of midships Stores 100 tonne LCG 30.5 m forward of midships. At 14000 tonne displacement the mean draught t is 7.80 m, MCT1cm 160 tonne m, LCB 2.00 m forward of midships and LCF 1.5m aft of midships. Calculate the final draughts. Given Data L=125m, Δ=4000 tonnes,๐๐ฆ๐๐๐ง = 7.8m at 14000 tonnes, LCF=1.5m aft of midship, LCB 2m forward of midships, LCG 1.60 m aft of midships 49 New drafts=? When large weights are added Prepared by Dr. Sudhir Sindagi Problems Weight Lever from (Tonnes) Midship Cargo 8500 3.9 Fuel 1200 -3.1 Water 200 -7.6 Stores 100 30.5 Lightweight 4000 -1.6 ΣW=14000 LCG= ΣM/ ΣW =1.754m forward of midship LCB =2.0 forward of midship Trimming Lever = 2.0- 1.745 =0.246m Trimming Moment = 14000*0.246 Trimming Moment =3444 tm ---It will be trim by aft Moment 33150 -3720 -1520 3050 -6400 ΣM=24560 50 Prepared by Dr. Sudhir Sindagi Change in trim due to change in density Trimming Moment MCT1cm 3444 Trim in cm=(t)= = 21.525cm 160 t ๐.๐๐๐๐๐ Increase in draft at aft =x= LCF* = 61* Trim in cm= ๐ ๐๐๐ Increase in draft at aft =x = 0.1050m (+ve) Reduction in draft at forward = y = t-x = 0.1102m (-ve) ๐๐ = ๐๐ฆ๐๐๐ง +x =7.905m ๐๐ญ = ๐๐ฆ๐๐๐ง − ๐ฒ=7.6898m 51 Prepared by Dr. Sudhir Sindagi Stability of Ship during Docking The process of docking and undocking of ships might not seem like an important operation. However, it’s a process that is carried out more than once, not only by shipbuilding yards during the construction of a ship, but also as regular part of the ship’s lifetime. The understanding of the process of docking is specialised, and hence, not many naval architects or engineers are thorough with the inner details of docking. From time to time, it becomes important to carry out repairs in the underwater portion of the hull, such as renewal of the sacrificial anodes, refit of the propellers, overhauling of the propulsion shafts, repair of rudders, underwater hull blasting to remove fouling, etc. 52 Prepared by Dr. Sudhir Sindagi Stability of Ship during Docking In order to carry out these repairs, the underwater portion of the hull needs to be made accessible, which is the purpose served by a dry dock. It has also become a common practice in large shipyards to build their ships on dry docks, and float it out when ready for trials. For such procedures, the docking plans need to be prepared taking into consideration the increase in weight of the ship structure along the building time. Once the ship has been built, the dry dock is flooded and the ship is undocked. The calculations for undocking also play a major role in the process because it is during undocking that the ship is at a risk of capsizing. 53 Prepared by Dr. Sudhir Sindagi Stability of Ship during Docking When the ship enters a dry dock, it must have a positive metacentric height; and is usually trimmed by stern. The floor of the dry dock is lined with keel blocks, which are so arranged such that they can bear the weight of the ship. The dock gates are then closed and the water is pumped out of the dock in stages. Since the ship has a trim by stern, the stern of the ship will first sit on the keel blocks. The rate of pumping out water is reduced as the stern is almost about to touch the keel blocks 54 Prepared by Dr. Sudhir Sindagi Stability of Ship during Docking When the ship’s stern just touches the keel blocks, part of the ship’s weight is being borne by the keel blocks. The contact between the stern and the keel block creates a normal reaction or upthrust. The magnitude of this upward normal reaction increases as the water level in the dry dock reduces. 55 Prepared by Dr. Sudhir Sindagi Stability of Ship during Docking It is this upthrust that creates a virtual reduction in the metacentric height of the ship. Hence it is very crucial to maintain sufficient positive metacentric height before docking, lacking which, the ship may heel over to either side, or even slip off the keel blocks and capsize. Since the location of the stern is a known point, its distance from the center of floatation (l) can be calculated instantly 56 Prepared by Dr. Sudhir Sindagi Stability of Ship during Docking Figure shows the longitudinal section of a ship during the critical period. ‘P’ is the upthrust at the stern and ‘l’ is the distance of the centre of flotation from aft. The trimming moment is given by P*l. But the trimming moment is also equal to MCTC * change of trim (t) P*l = MCTC * change of trim (t) 57 Prepared by Dr. Sudhir Sindagi Stability of Ship during Docking 58 Prepared by Dr. Sudhir Sindagi Stability of Ship during Docking Now consider Figure which shows a transverse section of the ship during the critical period after she has been inclined to a small angle (ฦ degrees) by a force external to the ship. For the sake of clarity the angle of heel has been magnified. The weight of the ship (W) acts downwards through the centre of gravity (G). The force P acts upwards through the keel (K) and is equal to the weight being borne by the blocks. For equilibrium the force of buoyancy must now be (W - P) and will act upwards through the initial metacentre (M). There are, thus, three parallel forces to consider when calculating the effect of the force P on the ship’s stability. Two of these forces may be replaced by their resultant in order 59to find the effective metacentric height and the moment of statical stability Prepared by Dr. Sudhir Sindagi Stability of Ship during Docking Righting Moment = (W - P) * GZ – P * GZ’ Righting Moment = (W - P) * GM sin ฦ – P * KG sin ฦ Righting Moment = W * GM sin ฦ - P * GM sin ฦ – P * KG sin ฦ Righting Moment = W * GM sin ฦ – (GM + KG) * P sin ฦ Righting Moment = W * GM sin ฦ – KM * P sin ฦ Righting Moment = W * (GM – ๐ โ ๐๐) sin ฦ The above equation is equivalent, Righting Moment = โ ∗ ๐๐ sin θ New GM during docking = GM Loss of GM during docking = If ๐ โ ๐ โ ๐ โ ๐๐ ๐๐ ๐๐ > GM, then the ship becomes unstable. 60 Prepared by Dr. Sudhir Sindagi Problems Just before entering drydock, a ship of 5000 tonnes floats at draft of 2.7m forward and 4.2m aft. The length between perpendiculars is 150m. Assuming blocks are horizontal and based on following hydrostatic data, find the force on the heel of the stern frame, which is at the AP, when the ship is just about to settle on the dock blocks and the GM at that instant. KG=8.5m, KM=9.3m, MCT1=107 t-m/cm, LCF=2.7m aft of amidships. Given Data L=150m, Δ=5000 tonnes,๐๐ = 4.2m, ๐๐ = 2.7m, LCF=2.7m aft of midship, KG=8.5m, KM=9.3m, MCT1=107 t-m/cm P=? GM during Docking=? 61 Prepared by Dr. Sudhir Sindagi Problems Initial Trim = ๐๐ -๐๐ = t=1.5m by aft Distance of CF from the point of resting the ship= l = 72.3m Using below equation, the upthrust(P) can be calculated P*l = MCTC * change of trim (t), we get P= 222 tonnes Loss of GM during docking = ๐ โ ๐๐, we get Loss of GM= 0.41m Final GM during Docking = Initial GM- Loss of GM Final GM during Docking = KM- KG – loss of GM Final GM during Docking = 0.39m 62 Prepared by Dr. Sudhir Sindagi Problems As ship of 3000 tonnes displacement and 100m length has KM=6m, KG=5.5m, LCF= 2m aft of amidships. MCT1=40 t-m/cm. Find the maximum trim for the ship to enter a drydock of the GM at the critical instant before the ship takes blocks forward and aft is not less than 0.3m. Given Data L=100m, Δ=3000 tonnes,๐๐ = 4.2m, ๐๐ = 2.7m, LCF=2m aft of midship, KG=5.5m, KM=6.0m, MCT1=40 t-m/cm t=? When GM during Docking>0.3m Loss of GM permitted = Initial GM- Final GM Loss of GM permitted = 6.0 – 5.5 – 0.3 = 0.2m Loss of GM during docking = ๐ โ ๐๐ = 0.2 63 Prepared by Dr. Sudhir Sindagi Problems We get, P= 100 tonnes Now we know that P*l = MCTC * change of trim (t), we get Here, l=48m Permitted trim < 1.2m 64 Prepared by Dr. Sudhir Sindagi Damaged Stability The term damage stability deals with the ability of a ship to float in water and regain its upright equilibrium position when some sort of structural damage has occurred. Generally, following an accident, the damage is hull fracture leading to flooding of ship compartments. If so many compartments are flooded that there is not enough buoyancy available to keep the vessel afloat, the ship may sink. Another critical scenario due to hull breach is ship capsizing due to loss of transverse stability as it can happen very quickly. To assess the behaviour of a ship after some damage two methods are considered: – Deterministic damage stability – Probabilistic damage stability 65 Prepared by Dr. Sudhir Sindagi Damaged Stability Deterministic Damage Stability This is a traditional method of assessment of the stability of a ship when it is flooded. In this process, the ship is divided into several subdivisions along its length with the help of transverse watertight bulkheads. Now the stability of the ship is calculated when one or more compartments get flooded due to a breach of hull. The changes in draft and stability when a compartment becomes flooded due to damage can be investigated by either of two methods: • Lost Buoyancy method • Added weight method 66 Prepared by Dr. Sudhir Sindagi Damaged Stability Deterministic Damage Stability: Lost Buoyancy method The damaged compartment(s) is considered open to the sea and therefore, does not contribute to the buoyancy of the ship. So, the lost buoyancy must be compensated by sinkage of the vessel and the moment due to change in LCB of the vessel is manifested through the heel or trim of the vessel. The assumptions considered in this method are that the flooded compartment does not provide buoyancy anymore and hence, there is no change in displacement or KG of the vessel and no free surface effect is observed. 67 Prepared by Dr. Sudhir Sindagi Damaged Stability Deterministic Damage Stability: Added Weight method This method considers that water ingresses in the damaged compartments up to the new water level and the weight of the ingressed water augments the displacement of the vessel that is compensated by the sinkage of the vessel. Consequently, the KG of the vessel changes due to the weight of ingressed water and Free Surface Effects has to be taken into account, if the compartment is partially filled with water. The weight added shifts the CG of the vessel that might lead to list or trim of the vessel. Both methods will give identical answers for final draughts, trim and RM, despite different values for GM. However, IMO/SOLAS recommends 68 the use of Lost buoyancy method for all calculations Prepared by Dr. Sudhir Sindagi Lost Buoyancy Method Bilging amidships compartments When a vessel floats in still water it displaces its own weight of water. Figure shows a box-shaped vessel floating at the waterline, WL. Now let an empty compartment amidships be holed below the waterline to such an extent that the water may flow freely into and out of the compartment. A vessel holed in this way is said to be ‘bilged’. 69 Prepared by Dr. Sudhir Sindagi Lost Buoyancy Method Bilging amidships compartments Figure shows the vessel in the bilged condition. The buoyancy provided by the bilged compartment is lost. The draft has increased and the vessel now floats at the waterline W1L1, where it is again displacing its own weight of water. ‘X’ represents the increase in draft due to bilging. 70 Prepared by Dr. Sudhir Sindagi Lost Buoyancy Method Bilging amidships compartments The volume of lost buoyancy (v) is made good by the volumes ‘y’ and ‘z’. v= y + z Let ‘A’ be the area of the water-plane before bilging, and let ‘a’ be the area of the bilged compartment. Then y + z = A*x –a*x v = x* (A - a) ๐ฏ Increase in draft = x = ๐ −๐ ๐๐จ๐ฅ๐ฎ๐ฆ๐ ๐จ๐ ๐ฅ๐จ๐ฌ๐ญ ๐๐ฎ๐จ๐ฒ๐๐ง๐๐ฒ x= ๐๐ซ๐๐ ๐จ๐ ๐ข๐ง๐ญ๐๐๐ญ ๐ฐ๐๐ญ๐๐ซ๐ฉ๐ฅ๐๐ง๐ 71 Prepared by Dr. Sudhir Sindagi Lost Buoyancy Method - Problems A box-shaped vessel is 50 metres long and is floating on an even keel at 4 metres draft. A amidships compartment is 10 metres long and is empty. Find the increase in draft if this compartment is bilged. Given Data: L=50m, T=4m, l=10m Volume of the compartment =l*B*T =10*B*4 =40B Increase in draft = x = Increase in draft = x = ๐ฏ ๐ −๐ 40B ๐๐∗๐ −๐๐๐ Increase in draft = x =1m 72 Prepared by Dr. Sudhir Sindagi Lost Buoyancy Method - Problems A box-shaped vessel is 150 metres long x 24 metres wide x 12 metres deep, and is floating on an even keel at 5 metres draft. GM = 0.9 metres. A compartment amidships is 20 metres long and is empty. Find the new GM if this compartment is bilged. Given Data: L=150m, B= 24m, TOLD=5m, GMOLD= 0.9m, l=20m, GMnew=? Volume of the compartment =l*b*T =20*24*5 =2400m3 Increase in draft = x ๐ฏ = ๐ −๐ Increase in draft = x = 2400 ๐๐๐∗๐๐−๐๐∗๐๐ Increase in draft = x =0.7692m≈0.77m New Draft≈ 5.77m 73 Prepared by Dr. Sudhir Sindagi Lost Buoyancy Method - Problems Since, in this problem, KG value is not provided, one need to estimate its KG value based on the old GM value, using GMOLD= KBOLD+BMOLD-KG Here KG remains same, as in the lost buoyancy method, it is assumed that KG does not changes. KBOLD= TOLD/2 =2.5m ๐๐ BMOLD- = =9.6m ๐๐๐๐๐๐ Putting these values in the equation for GMOLD= , we get KG= 11.2m To calculate, GMNEW,will use following equation GMNEW= KBNEW+BMNEW-KG 74 Prepared by Dr. Sudhir Sindagi Lost Buoyancy Method - Problems KBNEW= TNEW/2 =2.885m ๐๐ BMNEW- = ๐๐๐ =8.3188m NEW Putting these values in the equation for GMNEW= , we get GMNEW= KBNEW+BMNEW-KG GMNEW= 0.003m 75 Prepared by Dr. Sudhir Sindagi Lost Buoyancy Method Permeability (µ): 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. Typical values for permeability, are as follows: – – – – – Empty compartment - 100% Engine room - 80% to 85% Grain-filled cargo hold - 60% to 65% Coal-filled compartment - 36% approximately Filled water ballast tank (when ship is in salt water) - 0% Consequently, the higher the value of the permeability for a bilged compartment, the greater will be a ship’s loss of buoyancy when the 76 ship is bilged. Prepared by Dr. Sudhir Sindagi Lost Buoyancy Method Permeability (µ): When a bilged compartment contains cargo, the formula for finding the increase in draft must be amended to allow for the permeability. If ‘µ’ represents the permeability, expressed as a fraction, then the volume of lost buoyancy will be ‘µ*v’ and the area of the intact waterplane will be ‘A - µ*a’ square metres. The formula then reads. Increase in draft = x = ๐๐ฏ ๐ −๐๐ 77 Prepared by Dr. Sudhir Sindagi Permeability (µ) - Problems A box-shaped vessel is 64 metres long and is floating on an even keel at 3 metres draft. A compartment amidships is 12 m long and contains cargo having a permeability of 25 per cent. Calculate the increase in the draft if this compartment be bilged. Given Data: L=64m, T=3m, l=12m, µ=0.25, x=? Volume of water entering the compartment =µ*l*b*T =0.25*12*B*3 =9B µ๐ฏ Increase in draft = x = ๐ −µ๐ Increase in draft = x = 9B ๐๐๐−๐.๐๐∗๐๐∗๐ Increase in draft = x =0.1475m≈0.15m 78 Prepared by Dr. Sudhir Sindagi Permeability (µ) - Problems A box barge 60 m long and 10 m wide floats at an even keel draught of 4 m. It has a compartment amidships 12 m long. Calculate the new draughts if this compartment is laid open to the sea when permeability is (i) 100% (ii) 85% (iii) 60%. Given Data: L=60m, B= 10m, T=4m, l=12m, x=? When µ=1.0, 0.85 and 0.6. X= 1m when µ=1.0 X= 0.8192m when µ=0.85 X= 0.5454m when µ=0.6 79 Prepared by Dr. Sudhir Sindagi Permeability (µ) - Problems A box barge of 58m long and 8m wide floats a draught of 5m and has a mid- length compartment 9m long containing coal (relative density 1.28) which stows at 1.22 m³/t. Calculate the new draught if this compartment is bilged. Given Data: L=58m, B= 8m, T=5m, l=9m, x=? When coal (relative density 1.28) which stows at 1.22 m³/t. Stowage factor is space occupied by the cargo in m³ per tonnes of it. Total weight of the cargo = volume of the compartment / stowage factor. Total weight of the cargo = 9*8*5 / 1.22=295.08 tonnes Volume of the cargo occupied by the cargo = weight of the cargo / 80 Density of the cargo Prepared by Dr. Sudhir Sindagi Permeability (µ) - Problems Volume of the cargo occupied by the cargo = weight of the cargo / Density of the cargo Volume of the cargo occupied by the cargo = 295.08 / 1.28 Volume of the cargo occupied by the cargo =230.53 m3 Volume available for the water to enter in it once is flooded = Total volume of the compartment - Volume of the cargo in the compartment Volume available for the water to enter in it once is flooded = 360 230.53 = 129.467 m3 Permeability of the compartment = Volume available for the water to enter in it once is flooded / Total volume of the compartment µ = 0.3596 81 Prepared by Dr. Sudhir Sindagi Permeability (µ) - Problems Increase in draft = x µ๐ฏ = ๐ −µ๐ Increase in draft = x = ๐.๐๐๐๐∗๐๐๐ ๐๐∗๐ −๐.๐๐๐๐∗๐∗๐ Increase in draft = x = 0.2954m New Draft = 5.2954m 82 Prepared by Dr. Sudhir Sindagi Permeability (µ) - Problems A box-shaped vessel 75m long, 10m wide and 6m deep is floating in salt water on an even keel at a draft of 4.5m. Find the new drafts if a forward compartment 5m long is bilged. Given Data: L=75m, B= 10m, T=4.5m, l=5m, ๐๐ = ?, ๐๐ = ? When the bilged compartment is situated in a position away from amidships, the vessel’s mean draft will increase to make good the lost buoyancy but the trim will also change. There will be a horizontal component of the shift of the centre of buoyancy (B1B2) equal to half the length of the compartment flooded. A trimming moment of W * B1B2 by the head is produced and the vessel will trim about the centre of flotation (F). 83 Prepared by Dr. Sudhir Sindagi Permeability (µ) - Problems Trimming moment = W * B1B2 = w * d It can therefore be seen that the effect on trim is similar to that which would be produced if a mass equal to the lost buoyancy were loaded in the bilged compartment. Such questions shall be solved in a similar method of Trim problems. 84 Prepared by Dr. Sudhir Sindagi Permeability (µ) - Problems Trimming moment = w * d Weight of the water = 5*10*4.5*1.025= 230.625 tonnes d = Lever from new LCF. Here, due to flooding, the CF will shift aft wards from midship by a distance of half the length of the compartment and the weight acts at the centroid of the compartment. d = 37.5m Trimming moment = 230.625 * 37.5 = 8648.4375 t-m. Trimming Moment Trim in cm= MCT1cm Here, MCT1 cm is not given and hence to be calculated for the box barge. 85 Prepared by Dr. Sudhir Sindagi Permeability (µ) - Problems MCT1cm≈ Δ ∗ ๐๐๐ ๐๐๐๐ Δ = Volume * density = 75 * 10 * 4.5 *1.025 Δ = 3459.375 tonnes. ๐๐๐ = ๐๐ ๐ ๐๐๐๐๐ ๐ ๐๐ IL = , here the new length of the ship = 70m and ๐ will remain same. ๐๐๐ = 84.6914m ๐๐๐๐.๐๐๐∗๐๐.๐๐๐๐ MCT1cm≈ , here in the denominator L= 75m as drafts at ๐๐๐ ∗๐๐ AP and FP are to be calculated which are separated by a distance of 86 75m Prepared by Dr. Sudhir Sindagi Permeability (µ) - Problems MCT1cm≈ ๐๐. ๐๐๐๐ Trimming Moment 8648.4375 Trim in cm= = 39.0639 MCT1cm Trim in cm=221.39cm=2.21m by forward. Since the forward compartment is flooded, hence the ship will trim by forward. t 221.39 Reduction in draft at AP = x= x=LCF* = 35* , here LCF= 35 at ๐ ๐๐ new location of F from AP Reduction in draft at AP =x =103.32cm=1.033m Increase in draft at FP = y = t - x Increase in draft at FP = y = 1.1807m 87 Prepared by Dr. Sudhir Sindagi Permeability (µ) - Problems Change in draft in cm = TPC = ๐ ∗๐๐๐ ๐๐๐ = weight added or removed ๐๐๐ ๐.๐๐๐ ∗๐๐∗๐๐ ๐๐๐ Change in draft in cm = = 7.175, here for Area, new length is taken 230.625 ๐.๐๐๐ Change in draft in cm = 32.143 cm = 0.3214m Drafts Original AP (m) 4.5 FP (m) 4.5 Change in draft due to Trimming Moment Change in draft due to weight addition or removal Final Drafts - 1.033 0.3214 3.7884 + 1.1807 0.3214 88 6.0021 Prepared by Dr. Sudhir Sindagi Permeability (µ) - Problems A box shaped vessel 75m long, 10m wide and 6m deep is floating in SW on an even keel at a draft of 4.5. Find the new drafts if a forward compartment 7m long is bilged. Given Data: L=75m, B= 10m, T=4.5m, l=7m, ๐๐ = ?, ๐๐ = ? Trimming moment = w * d Weight of the water = 322.875 tonnes d = 37.5m Trimming moment = 12107.8125 t-m. Trimming Moment Trim in cm= MCT1cm 89 Prepared by Dr. Sudhir Sindagi Permeability (µ) - Problems MCT1cm≈ Δ ∗ ๐๐๐ ๐๐๐๐ Δ = 3459.375 tonnes. ๐๐๐ = ๐๐ ๐ ๐๐๐๐๐ ๐ ๐๐ IL = , here the new length of the ship = 68m and ๐ will remain same. ๐๐๐ = 77.64m MCT1cm≈ 35.81 Trimming Moment Trim in cm= MCT1cm Trim in cm=3.381 m by forward. 90 Prepared by Dr. Sudhir Sindagi Permeability (µ) - Problems Reduction in draft at AP = x= x=LCF* t ๐ Reduction in draft at AP =x =1.5329m Increase in draft at FP = y = t - x Increase in draft at FP = y = 1.8484m weight added or removed Change in draft in cm = ๐๐๐ TPC = ๐ ∗๐๐๐ ๐๐๐ = 6.97, here for Area, new length is taken Change in draft in cm = 0.4632m 91 Prepared by Dr. Sudhir Sindagi Permeability (µ) - Problems Drafts AP (m) FP (m) Original Change in draft due to Trimming Moment Change in draft due to weight addition or removal 4.5 - 1.5329 0.4632 4.5 + 1.8484 0.4632 Final Drafts 3.4303 6.8116 92 Prepared by Dr. Sudhir Sindagi Permeability (µ) - Problems A box-shaped vessel 100 metres long, 20 metres wide and 12 metres deep is floating in salt water on an even keel at 6 metres draft. A forward compartment is 10 metres long, 12 metres wide and extends from the outer bottom to a watertight flat, 4 metres above the keel. The compartment contains cargo of permeability 25 per cent. Find the new drafts if this compartment is bilged. Given Data: L=100m, B= 20m, T=6m, l=10m, b=12m, ๐๐ = ?, ๐๐ = ? When µ=0.25. 93 Prepared by Dr. Sudhir Sindagi Permeability (µ) - Problems In this question, as the compartment depth is limited to the height of 4m and hence is not touching the WL, hence there wont be changes in length of WL after flooding. Trimming moment = w * d Weight of the water = 123 tonnes d = Lever from LCF. Here, the CF will not shift aft wards. d = 45m Trimming moment = 5535 t-m. Trimming Moment Trim in cm= MCT1cm Here, MCT1 cm is not given and hence to be calculated for the box 94 barge. Prepared by Dr. Sudhir Sindagi Permeability (µ) - Problems MCT1cm≈ Δ ∗ ๐๐๐ ๐๐๐๐ Δ = Volume * density Δ = 12300 tonnes. ๐๐๐ = ๐๐ ๐ ๐๐๐๐๐ ๐ ๐๐ IL = , here the new length of the ship = old length and ๐ will remain same as usual. ๐๐๐ = 138.889m MCT1cm≈ 170.467 95 Prepared by Dr. Sudhir Sindagi Permeability (µ) - Problems Trimming Moment Trim in cm= MCT1cm Trim in cm=0.324m by forward. Since the forward compartment is flooded, hence the ship will trim by forward. t Reduction in draft at AP = x= x=LCF* ๐ Reduction in draft at AP =x =0.162m Increase in draft at FP = y = t - x Increase in draft at FP = y = 0.162m 96 Prepared by Dr. Sudhir Sindagi Permeability (µ) - Problems Volume of the compartment =µ*l*b*T =0.25 * 10 * 12 * 4 =120 m3 Increase in draft = x = Increase in draft = x = µ๐ฏ ๐ −µ๐ 120 ๐๐๐∗๐๐−๐.๐๐∗๐๐∗๐๐ Increase in draft = x =0.0609m Drafts Original AP (m) 6 FP (m) 6 Change in draft due to Trimming Moment Change in draft due to weight addition or removal Final Drafts -0.162 0.0609 5.8989 + 0.162 0.0609 97 6.2229 Prepared by Dr. Sudhir Sindagi Damaged Stability Probabilistic Damage Stability Damage stability calculations by probabilistic damage assessment is required by SOLAS Chapter II-1, part B. This is required for cargo ships 80 m in length and upwards and to all passenger ships regardless of length. This approach is based on probability of survival after certain compartment or group of compartments are damaged. Attained index is a measure of ship's safety after damage or Collision. Two ship's with same index are assumed to have same level of safety for damage or collision, irrespective of location of damage. Probability of survival is calculated as sum of probability of damage of a space or group spaces multiplied by probabilities of survival98 after corresponding space damage. Prepared by Dr. Sudhir Sindagi Damaged Stability Probabilistic Damage Stability This approach uses the concept of probability to ensure that ships can survive damage to its compartment(s). There are two probability factors that are used in this approach. Probability that a particular compartment(s) will damage in an incident (factor “p”) the probability that ship will survive if that compartment(s) is flooded (Factor “s”) Used as the requirement for the cargo ships and passenger ships. Multiplying these two factors (p x s) will give the probability of surviving that damage case. 99 Prepared by Dr. Sudhir Sindagi Damaged Stability Probabilistic Damage Stability The value of S in all these will either be 0 or 1. This is because when we have considered a damage, the ship will either survive (probability 1) or not survive (probability 0). So if this ship is three compartment ship, there is no need to consider the probability of survival for four and more compartments because it will be zero. But there is still one thing to consider. At what drafts we need to consider all these damages? SOLAS requires that these should be considered at three drafts. 100 Prepared by Dr. Sudhir Sindagi Damaged Stability Probabilistic Damage Stability Deepest subdivision draught (ds): Which corresponds to the Summer Load Line draught of the ship. Light service draught (dl): Service draught corresponding to the lightest anticipated loading and associated tankage, including, however, such ballast as may be necessary for stability and/or immersion. Partial subdivision draught (dp): light service draught plus 60% of the difference between the light service draught and the deepest subdivision draught.. 101 Prepared by Dr. Sudhir Sindagi Damaged Stability Probabilistic Damage Stability: Deepest subdivision draft (ds) will have 102 Prepared by Dr. Sudhir Sindagi Damaged Stability Probabilistic Damage Stability: Light service draught (dl) will have 103 Prepared by Dr. Sudhir Sindagi Damaged Stability Probabilistic Damage Stability: Partial subdivision draught (dp) will have 104 Prepared by Dr. Sudhir Sindagi Damaged Stability Probabilistic Damage Stability: Attained Subdivision Index Finally the bottom line. How would a ship comply with the damage stability requirements? As per SOLAS Chapter II-1, part B-1, Regulation 6, the ship complies with damage stability when Attained Subdivision Index > Required subdivision index As per SOLAS, attained subdivision index is calculated by the formula. 105 Prepared by Dr. Sudhir Sindagi Damaged Stability Probabilistic Damage Stability: Required subdivision index SOLAS chapter II-1, Reg 7 gives the formula to calculate the required subdivision index for a ship. These formulas are different for different type and size of the ship. This would be the minimum required value of subdivision index. If the actual value of subdivision index (Attained value) is less than the required, the subdivisions need to be re-arranged or increased to have attained subdivision index to be more than required subdivision index. 106 Prepared by Dr. Sudhir Sindagi Damaged Stability 107 Prepared by Dr. Sudhir Sindagi Damaged Stability Floodable length and factor of subdivision This is an old approach in which, the number of subdivisions required is calculated by knowing the floodable length along the ship. Floodable length is the length of the compartment which if flooded will cause the ship to sink up to the margin line. 108 Prepared by Dr. Sudhir Sindagi Damaged Stability Floodable length and factor of subdivision Margin line is the imaginary line drawn at least 76mm below the upper surface of the bulkhead deck at side, such that, in any condition of damage/flooding of the compartment, new waterline does not intersect the margin line. For a ship one needs to put subdivisions (bulkheads) to divide the ship into compartments. One such compartment is AB. The length of this compartment (Length AB) need to such that if this compartment is flooded, the ship will sink to a point where margin line is just submerged. 109 Prepared by Dr. Sudhir Sindagi Damaged Stability Floodable length and factor of subdivision Now we want to place another bulkhead aft of midship. Again this bulkhead needs to be at a location (C) such that if compartment AC is flooded, the ship will sink to a point where margin line is just submerged. And with this same approach, we can decide the location of other bulkheads along the ship’s length. 110 Prepared by Dr. Sudhir Sindagi Damaged Stability Permissible length and factor of subdivision The maximum permissible length of a compartment having its centre at any point in the ship’s length is obtained from the floodable length by multiplying the latter by an appropriate factor called factor of subdivision. The factor of subdivision shall be: FOR NEW CLASS B, C AND D SHIPS AND EXISTING CLASS B RO-RO PASSENGER SHIPS: 1.0 when the number of persons the ship is certified to carry is less than 400, and 0.5 when the number of persons the ship is certified to carry is 400 or more. 111 Prepared by Dr. Sudhir Sindagi Damaged Stability Floodable length Curve Floodable length curve represents the maximum floodable length of the ship along the ship’s length. This curve is obtained by vertically plotting the floodable length along the ship’s length. 112 Prepared by Dr. Sudhir Sindagi Topics to study from the Internet – Factor of Subdivision – Floodable length Curve – Permissible Length 113 Prepared by Dr. Sudhir Sindagi Summary Introduction- Trim Trimming Moment, MCTC, IL, BML, GML Trim due to – Shifting of existing weight – Addition or removal of weight Damaged Stability – Deterministic damage stability • Added Weight Method • Lost Buoyancy Method – Permeability – Probabilistic damage stability – Floodable length, Margin Line, Permissible Length, Factor of Subdivision, Floodable Length curve 114 Prepared by Dr. Sudhir Sindagi Strength of Bulkhead and Ship By Dr. Sudhir Sindagi 1 Prepared by Dr. Sudhir Sindagi Chapter Content Introduction Force and Center of Pressure for – Regular shaped bulkhead – Irregular shaped bulkhead Strength of Ship – – – – – Sagging Hogging Ship as a girder SFD & BMD for the ship Calculation of stresses in Main deck and Keel due to Bending Moment 2 Prepared by Dr. Sudhir Sindagi Introduction The problem of calculating the necessary strength of ships is made difficult by the many and varied forces to which the ship structure is subjected during its lifetime. These forces may be divided into two groups, namely statical forces and dynamical forces. The statical forces are due to: – The weight of the structure which varies throughout the length of the ship. – Buoyancy forces, which vary over each unit length of the ship and are constantly varying in a seaway. – Direct hydrostatic pressure. – Concentrated local weights such as machinery, masts, derricks, winches, etc. 3 Prepared by Dr. Sudhir Sindagi Introduction The dynamical forces are due to: – Pitching, heaving and rolling. – Wind and waves. These forces cause bending in several planes and local strains are set up due to concentrated loads. The effects are aggravated by structural discontinuities. A stress is the mutual actual between the parts of a material to preserve their relative positions when external loads are applied to the material. Thus, whenever external loads are applied to a material stresses are created within the material. 4 Prepared by Dr. Sudhir Sindagi Introduction When an external load is applied to a material in such a way as to cause an extension of the material it is called a ‘tensile’ load, whilst an external load tending to cause compression of the material is a ‘compressive’ load and corresponding stresses are called as ‘tensile’ and Compressive stresses, respectively. A shearing stress is a stress within a material which tends to break or shear the material across tending to cause deformation of a material by slippage along a plane 5 Prepared by Dr. Sudhir Sindagi Force and CP of Bulkhead The water pressure at any depth is due to the weight of water above the point in question and increases uniformly with depth below the surface. From this, it can be seen that the pressure at any depth will vary with the depth below the surface and the density of the water. Force acting on the bulkhead is given by the Area under the pressure diagram* width of the bulkhead. Lets consider a rectangular shaped bulkhead with dimension B X D, immersed completely into the water of density ρ in such a way that, WL is coinciding with top of the bulkhead. Pressure at any point on the bulkhead is proportional to the depth of water, hence it will be 0 at the top and maximum at the bottom of the bulkhead, as shown in the figure. 6 Prepared by Dr. Sudhir Sindagi Force and CP of Bulkhead Force = Area under the pressure diagram* width of the bulkhead ๐ ๐ ๐ ๐๐ ๐๐๐ ๐ Force = ๐๐ ๐ ∗ ๐ ∗ ๐ Force = This can be rearranged as follows Force = ๐๐ ∗ (๐ ∗ ๐) ∗ ๐ ๐ Force = ๐๐ * Area of the Bulkhead * distance of the centroid from WL Force = ๐๐ * First moment of Area about WL 7 Prepared by Dr. Sudhir Sindagi Force and CP of Bulkhead Center of Pressure is the point on the bulkhead where resultant force is assumed to be acting. Its distance from the WL is mentioned as CP and is calculated using ๐๐ = ๐ ๐ข๐ซ๐ฌ๐ญ ๐๐จ๐ฆ๐๐ง๐ญ ๐จ๐ ๐ ๐จ๐ซ๐๐ ๐๐๐จ๐ฎ๐ญ ๐๐ ๐๐จ๐ญ๐๐ฅ ๐ ๐จ๐ซ๐๐ ๐๐ ∗ ๐๐๐๐จ๐ง๐ ๐๐จ๐ฆ๐๐ง๐ญ ๐จ๐ ๐๐ซ๐๐ ๐๐๐จ๐ฎ๐ญ ๐๐ ๐๐ ∗๐ ๐ข๐ซ๐ฌ๐ญ ๐๐จ๐ฆ๐๐ง๐ญ ๐จ๐ ๐๐ซ๐๐ ๐๐๐จ๐ฎ๐ญ ๐๐ ๐๐ = ๐๐๐๐จ๐ง๐ ๐๐จ๐ฆ๐๐ง๐ญ ๐จ๐ ๐๐ซ๐๐ ๐๐๐จ๐ฎ๐ญ ๐๐ ๐ ๐ข๐ซ๐ฌ๐ญ ๐๐จ๐ฆ๐๐ง๐ญ ๐จ๐ ๐๐ซ๐๐ ๐๐๐จ๐ฎ๐ญ ๐๐ ๐๐ = 8 Prepared by Dr. Sudhir Sindagi Force and CP of Bulkhead Consider a arbitrary shape of the bulkhead, immersed completely in the water. Consider a small elemental area of depth dh with width b at the depth h. Force = ๐๐ * Area of the Bulkhead * distance of the centroid from WL Force on the elemental area = ๐๐ ∗ ๐๐๐ก ∗ (๐ก ๐๐ก + ) ๐ Since dh is very small, hence dh/2 can be neglected. Force on small area= ๐๐ ∗ ๐๐ก ๐๐ก ๐ Total Force = ๐๐ โซ๐ก๐ ๐ก๐ ๐ืฌโฌ ๐๐ = ๐๐๐๐จ๐ง๐ ๐๐จ๐ฆ๐๐ง๐ญ ๐จ๐ ๐๐ซ๐๐ ๐๐๐จ๐ฎ๐ญ ๐๐ ๐ ๐ข๐ซ๐ฌ๐ญ ๐๐จ๐ฆ๐๐ง๐ญ ๐จ๐ ๐๐ซ๐๐ ๐๐๐จ๐ฎ๐ญ ๐๐ ๐ ๐๐ = โซ๐ก๐ ๐๐ก๐ ๐ืฌโฌ ๐ โซ๐ก๐ ๐ก๐ ๐ืฌโฌ 9 Prepared by Dr. Sudhir Sindagi Force and CP of Bulkhead Lets verify above expressions for the rectangular shaped bulkhead ๐ Total Force = ๐๐ โซ๐ก๐ ๐ก๐ ๐ืฌโฌ Here b=B is constant, hence Total Force = ๐๐ ๐ Force = ๐ โซ๐ก ๐ืฌโฌ ๐๐ก ๐ ๐๐ ๐๐๐ ๐ ๐๐ = ๐ โซ๐ก๐ ๐๐ก๐ ๐ืฌโฌ ๐ โซ๐ก๐ ๐ก๐ ๐ืฌโฌ ๐๐ = ๐๐ ๐ = ๐ ๐ โซ๐ก๐ ๐๐ก ๐ืฌโฌ ๐ ๐ โซ๐ก๐ ๐ก ๐ืฌโฌ = ๐ โซ๐ก๐ ๐๐ก ๐ืฌโฌ ๐ โซ๐ก๐ ๐ก ๐ืฌโฌ = ๐๐ ๐ ๐๐ ๐ 10 Prepared by Dr. Sudhir Sindagi Force and CP of Bulkhead Derive the expression for the force and CP for the triangular shaped bulkhead shown in the figure ๐ Total Force = ๐๐ โซ๐ก๐ ๐ก๐ ๐ืฌโฌ ๐ ๐ ๐ก ๐ ๐ก ๐ Here b≠B is not constant, but = , b= B Total Force = ๐ ๐ ๐ ๐๐ โซ๐ก๐ ๐ก ๐ืฌโฌ ๐ ๐ ๐ Force = ๐๐ ๐๐๐ ๐ ๐๐ = ๐๐ = โซ๐ก๐ ๐๐ก๐ ๐ืฌโฌ ๐ โซ๐ก๐ ๐ก๐ ๐ืฌโฌ ๐๐ ๐ 11 Prepared by Dr. Sudhir Sindagi Force and CP of Bulkhead Derive the expression for the force and CP for the triangular shaped bulkhead shown in the figure ๐ Total Force = ๐๐ โซ๐ก๐ ๐ก๐ ๐ืฌโฌ ๐ ๐ Here b≠B is not constant, but = Total Force = ๐๐ ๐ ๐−๐ก โซ ๐ืฌโฌB ๐ ๐−๐ก ๐ ๐−๐ก ๐ , b= B ๐ก ๐๐ก ๐ ๐ Force = ๐๐ ๐๐๐ ๐ ๐๐ = ๐๐ = โซ๐ก๐ ๐๐ก๐ ๐ืฌโฌ ๐ โซ๐ก๐ ๐ก๐ ๐ืฌโฌ ๐ ๐ 12 Prepared by Dr. Sudhir Sindagi Force and CP of Bulkhead Calculate the force acting and its CP for a rectangular shaped bulkhead of dimensions 12m X 10m (B X D), if SW is completely filled on one side of the bulkhead and on the other side, the cargo of density 0.88 t/m3 is half filled. Force = ๐๐ * Area of the Bulkhead * distance of the centroid from WL FSW = ๐๐ ∗ ๐๐ ∗ ๐ ๐ FSW = ๐. ๐๐๐ ∗ ๐. ๐๐ ∗ ๐๐ ∗ ๐๐ ∗ ๐๐ ๐ FSW = 6033.15 kN FC = ๐. ๐๐ ∗ ๐. ๐๐ ∗ ๐๐ ∗ ๐ ∗ ๐ ๐ FC = 1294.92 kN FResultant = FSW - FC = 4738.23 kN 13 Prepared by Dr. Sudhir Sindagi Force and CP of Bulkhead To calculate the CP value for the resultant force, one needs to estimate either reaction at top or the bottom of the bulkhead. Taking moment about Top FSW * ๐๐ = ๐ FC * (5 + ๐๐ ) ๐ + FB * 10 After putting values of FSW & FC we get FB =2943.63 kN Again, Taking moment about Top FR *CP= FB * 10 After putting values of FR we get CP= 6.213m from top. 14 Prepared by Dr. Sudhir Sindagi Strength of Ship In a seaway, the ship is subjected to both static and dynamic forces which cause it to bend in a longitudinal vertical plane. For the purposes of structural design and for comparison, ship and the ship problem is considered as a static one, so that it resolves into the ship being poised statically on a wave and the resulting forces and moments acting on the ship are calculated. In order to determine the forces acting on a ship, it is treated as a girder. It is necessary to determine the distribution of weight and the buoyancy. The total weight must equal to the total buoyancy and the fore and aft position of the centre of gravity must be in the same athwartship section as the centre of buoyancy. 15 Prepared by Dr. Sudhir Sindagi Strength of Ship For the purpose of investigating the longitudinal bending of a ship certain assumptions are made. The calculations are, in general, carried out for two standard conditions, Hogging and Sagging. The assumptions are as follows: – a) the ship is head on to the waves and is poised statically on a wave; – b) the wave has a trochoidal profile of length equal to the length of the ship and a height as described below. – c) the wave crest is at amidships for the hogging condition; – d) the wave crest is at the ends for the sagging condition. A long accepted practice has been to take the height of the wave (h) as 1/20th of the length in strength calculations. However, observation of sea waves has shown that longer waves tend 16 to be less steep than shorter waves. Prepared by Dr. Sudhir Sindagi Strength of Ship Lloyd's Register suggested that a better approximation to the height would be h = 0.607 ๐ณ meters Later Murray of Lloyd's Register suggested that wave height should vary as L0.3 instead of L0.5 and Muckle has derived from data that the wave height can be written h = 1·632*L0.3 where L is in metres. 17 Prepared by Dr. Sudhir Sindagi Strength of Ship Consider the case of a homogeneous log of rectangular section floating freely at rest in still water. The total weight of the log is balanced by the total force of buoyancy and the weight (W) of any section of the log is balanced by the force of buoyancy (B) provided by that section. There is therefore no bending moment longitudinally which would cause stresses to be set up in the log. 18 Prepared by Dr. Sudhir Sindagi Strength of Ship Now consider the case of a ship floating at rest in still water, on an even keel, at the light draft. Although the total weight of the ship is balanced by the total force of buoyancy, neither is uniformly distributed throughout the ship’s length 19 Prepared by Dr. Sudhir Sindagi Strength of Ship Although the sections in the ship are not free to move in this way, bending moments, and consequently longitudinal stresses, are created by the variation in the longitudinal distribution of weight and buoyancy, and these must be allowed for in the construction of the ship. When a ship encounters waves at sea the stresses created differ greatly from those created in still water. The maximum stresses are considered to exist when the wavelength is equal to the ship’s length and either a wave crest or trough is situated amidships 20 Prepared by Dr. Sudhir Sindagi Strength of Ship In this case, although once more the total weight of the ship is balanced by the total buoyancy, there is an excess of buoyancy over the weight amidships and an excess of weight over buoyancy at the bow and the stern. This situation creates a tendency for the ends of the ship to move downwards and the section amidships to move upwards. Under these conditions the ship is said to be subjected to a ‘hogging’ stress. 21 Prepared by Dr. Sudhir Sindagi Strength of Ship Consider the effect after the wave crest has moved onwards and the ship is now supported by wave crests at the bow and the stern and a trough amidships. There is now an excess of buoyancy over weight at the ends and an excess of weight over buoyancy amidships. The situation creates a tendency for the bow and the stern to move upwards and the section amidships to move downwards 22 Prepared by Dr. Sudhir Sindagi Strength of Ship Under these conditions a ship is said to be subjected to a sagging stress. Sagging and hogging of ship occurs even when the ship in calm water. If the weight distribution is such that, more weights are distributed at the midship and lesser weights are distributed at the ends, then ship will Sag. However, the ship will hog, when the weights are distributed more at the ends rather than at the midship. 23 Prepared by Dr. Sudhir Sindagi Strength of Ship 24 Prepared by Dr. Sudhir Sindagi Strength of Ship As shown in the figure, under the static condition, weight of structure, machinery, cargo etc. acts downwards on the ship, while pressure from the surrounding fluid acts upwards or sideways depending on the position. This situation becomes more complex when the ship is moving in waves. Hence it becomes paramount to design ship’s structure so that it will withstand such loading conditions. In order to find scantlings(thicknesses of every structural member) of the ship, it is very important to calculate Bending Moment acting on the ship, which is estimated as per the procedure 25 Prepared by Dr. Sudhir Sindagi Strength of Ship 1. Distribution of weights per unit length known as the weight curve is plotted against the length of the ship. 2. Distribution of buoyancy force per unit length known as the Buoyancy curve is plotted against the length of the ship, which varies as per the sectional area of the ship. 3. Hence each and every point on the ship is subjected to a resultant load which can be calculated as 4. Load / length = Buoyancy Force / Length (b) – Weight Force / Length (w) 26 Prepared by Dr. Sudhir Sindagi Strength of Ship 5. These values of load / length are plotted against to get the Load Curve. 6. Due to the variation in the values of load at each and every point, these points are subjected to a shear force ๐ 7. Shear Force (SFx) = โซ ๐ ๐ืฌโฌ− ๐ฑ ๐๐ฑ 8. Variation in the values of shear forces are plotted against length to obtain Shear Force Diagram (SFD) 9. Similar to the values of SFx, Bending moment values are calculated as 10. Bending Moment ๐ (BMx)=โซ๐ฑ๐ ๐ฑ๐ ๐ ๐ืฌโฌ 27 Prepared by Dr. Sudhir Sindagi Strength of Ship 11. Variation in the values of BMx are plotted to generate Bending Moment Diagram (BMD). 12. Maximum value of Bending moment is then used to calculate the minimum values of section modulus of the ship required to withstand it using ๐๐ฆ๐๐ฑ 13. ๐ = ๐ ๐ฒ 14. Minimum section Modulus ๐ ๐ฒ 15. (Z) min =( )min= ๐๐ฆ๐๐ฑ ๐ 28 Prepared by Dr. Sudhir Sindagi Strength of Ship 29 Prepared by Dr. Sudhir Sindagi Problems A rectangular barge of 250m length 35m beam and 20m depth floats in SW at a draft of 2m when it is empty. The lightship weight may be assumed to be uniformly distributed over the barge length. It has five holds each 50m long. The barge floats in SW loaded with cargo as shown below. Cargo weights within each hold are to be assumed as uniformly distributed over the length of the hold. Calculate and plot diagrams of distributions of weight, buoyancy, load, shear and bending. Determine the values of the curves at each bulkhead and at their maximum points. Hold 1 Hold 2 14000t 17000t 50m length 50m length Hold 3 28000t 50m length Hold 4 17000t 50m length Hold 5 14000t 50m length 30 Prepared by Dr. Sudhir Sindagi Strength of Ship Given Data: L=250m, B=35m, T=2m floats in SW. Weight in lightship condition = Volume * Density Lightweight = 250*35*2*1.025 =17937.5 tonnes Weight of Cargo = 14000+17000+28000+17000+14000 Weight of Cargo = 90000 tonnes Displacement of ship = 17937.5 + 90000 Displacement of ship = 107937.5 Since it is given that, the lightship weight may be assumed to be uniformly distributed over the barge length ๐๐ข๐ ๐ก๐ญ ๐ฐ๐๐ข๐ ๐ก๐ญ 17937.5 31 = = 71.75 t/m ๐๐๐ง๐ ๐ญ๐ก ๐๐๐ Prepared by Dr. Sudhir Sindagi Strength of Ship Now, cargo weight may be assumed to be uniformly distributed over the length of the cargo hold. ๐๐๐ซ๐ ๐จ ๐ฐ๐๐ข๐ ๐ก๐ญ 14000 for the hold 1 = = 280 t/m ๐๐๐ง๐ ๐ญ๐ก ๐๐ ๐๐๐ซ๐ ๐จ ๐ฐ๐๐ข๐ ๐ก๐ญ ๐๐๐ง๐ ๐ญ๐ก for the hold 2 = ๐๐๐ซ๐ ๐จ ๐ฐ๐๐ข๐ ๐ก๐ญ ๐๐๐ง๐ ๐ญ๐ก for the hold 3 = ๐๐๐ซ๐ ๐จ ๐ฐ๐๐ข๐ ๐ก๐ญ ๐๐๐ง๐ ๐ญ๐ก ๐๐๐ซ๐ ๐จ ๐ฐ๐๐ข๐ ๐ก๐ญ ๐๐๐ง๐ ๐ญ๐ก for the hold 4 = for the hold 5 = 17000 ๐๐ 28000 ๐๐ 17000 ๐๐ 14000 ๐๐ = 340 t/m = 510 t/m = 340 t/m = 280 t/m 32 Prepared by Dr. Sudhir Sindagi Strength of Ship Now, since it is a rectangular barge with rectangular sectional area remaining constant throughout the length, hence buoyancy force may be assumed to be uniformly distributed over the length of the ship. ๐๐ฎ๐จ๐ฒ๐๐ง๐๐ฒ ๐ ๐จ๐ซ๐๐ Δ 107937.5 = b= = = 431.75 t/m ๐๐๐ง๐ ๐ญ๐ก ๐๐จ๐ญ๐๐ฅ ๐ฐ๐๐ข๐ ๐ก๐ญ ๐๐๐ง๐ ๐ญ๐ก ๐ at any point = ๐๐๐ ๐๐ข๐ ๐ก๐ญ ๐ฐ๐๐ข๐ ๐ก๐ญ ๐๐๐ง๐ ๐ญ๐ก + ๐๐ข๐ ๐ก๐ญ ๐ฐ๐๐ข๐ ๐ก๐ญ ๐๐๐ง๐ ๐ญ๐ก ๐๐๐ซ๐ ๐จ ๐ฐ๐๐ข๐ ๐ก๐ญ ๐๐๐ง๐ ๐ญ๐ก for the hold ๐๐จ๐ญ๐๐ฅ ๐ฐ๐๐ข๐ ๐ก๐ญ ๐๐๐ง๐ ๐ญ๐ก for the Hold 1 = ๐๐จ๐ญ๐๐ฅ ๐ฐ๐๐ข๐ ๐ก๐ญ ๐๐๐ง๐ ๐ญ๐ก for the Hold 1 (w1) = 71.75 t/m + 280 t/m = 351. 75 t/m ๐๐จ๐ญ๐๐ฅ ๐ฐ๐๐ข๐ ๐ก๐ญ ๐๐๐ง๐ ๐ญ๐ก for the Hold 2 (w2) = 71.75 t/m + 340 t/m = 411. 75 t/m + ๐๐๐ซ๐ ๐จ ๐ฐ๐๐ข๐ ๐ก๐ญ ๐๐๐ง๐ ๐ญ๐ก for the hold 1 33 Prepared by Dr. Sudhir Sindagi Strength of Ship ๐๐จ๐ญ๐๐ฅ ๐ฐ๐๐ข๐ ๐ก๐ญ ๐๐๐ง๐ ๐ญ๐ก for the Hold 3 (w3) = 71.75 t/m + 510 t/m = 631. 75 t/m ๐๐จ๐ญ๐๐ฅ ๐ฐ๐๐ข๐ ๐ก๐ญ ๐๐๐ง๐ ๐ญ๐ก for the Hold 4 (w4) = 71.75 t/m + 340 t/m = 411. 75 t/m ๐๐จ๐ญ๐๐ฅ ๐ฐ๐๐ข๐ ๐ก๐ญ ๐๐๐ง๐ ๐ญ๐ก for the Hold 5 (w5) = 71.75 t/m + 280 t/m = 351. 75 t/m Hence each and every point on the ship is subjected to a resultant load which can be calculated as Load / length = Buoyancy Force / Length (b) – Weight Force / Length (w) Buoyancy Force Weight Force for the hold 1 ๐๐จ๐๐ ๐๐๐ง๐ ๐ญ๐ก for the Hold 1 = ๐๐จ๐๐ ๐๐๐ง๐ ๐ญ๐ก for the Hold 1 = 431.75 – 351.75 = 80 t/m ๐๐๐ง๐ ๐ญ๐ก ๐๐๐ง๐ ๐ญ๐ก 34 Prepared by Dr. Sudhir Sindagi Strength of Ship ๐๐จ๐๐ ๐๐๐ง๐ ๐ญ๐ก for the Hold 2 = 431.75 – 411.75 = 20 t/m ๐๐จ๐๐ ๐๐๐ง๐ ๐ญ๐ก for the Hold 3 = 431.75 – 631.75 = -200 t/m ๐๐จ๐๐ ๐๐๐ง๐ ๐ญ๐ก for the Hold 4 = 431.75 – 411.75 = 20 t/m ๐๐จ๐๐ ๐๐๐ง๐ ๐ญ๐ก for the Hold 5 = 431.75 – 351.75 = 80 t/m With the available data we can get the weight curve, buoyancy curve and then the load curve. 35 Prepared by Dr. Sudhir Sindagi Strength of Ship 36 Prepared by Dr. Sudhir Sindagi Strength of Ship Shear Force at any point of consideration is the area under the load curve till that point. SFA= 0 SFB= 80 * 50 = 4000 t SFC= 4000+ 20*50 = 5000 t SFD= 5000 – 200*50 = -5000 t SFE= -5000 + 20*50 = -4000 t SFF= -4000 + 50*40 = 0 SFO= 0 37 Prepared by Dr. Sudhir Sindagi Strength of Ship Bending Moment at any point of consideration is the area under the SFD till that point. BMA= 0 BMB= ½ * 50 * 4000 = 100000 tm BMC= 100000 + ½ (5000+4000)* 50 = 325000 tm BMO= 325000 + ½ *25 *5000 = 387500 tm BMD= 387500 – ½ *25*5000 = 325000 tm BME= 325000 - ½ (5000+4000)* 50 = 100000 t-m BMF= 100000 - ½ * 50 * 4000 = 0 38 Prepared by Dr. Sudhir Sindagi Strength of Ship 39 Prepared by Dr. Sudhir Sindagi Strength of Ship Summary of Curves ( Relationship between strength Curves): The area under the weight curve and that under the buoyancy curve are equal. The centroids of the areas of weight and buoyancy are in the same athwartship section. For the load curve the areas above and below the base line are equal. The maximum values of the shearing force occur where the load curve crosses the base line. The maximum bending moment occurs where the shearing force curve crosses the base line. The shearing force and bending moment curves must close. The value 40 at the ends is zero. Prepared by Dr. Sudhir Sindagi Problems A box shaped barge of uniform construction is 32m long and displaces 352tonnes when empty. It is divided by transverse bulkheads into four equal compartments. Cargo is loaded into each compartments as below No 1 Hold -192 tonnes No 2 Hold- 224tonnes No 3 hold- 272 tonnes No 4 hold- 176 tonnes Given Data: L=32m, Lightweight = 352 tonnes. Weight of Cargo = 192+224+272+176 Weight of Cargo = 864 tonnes Displacement of ship = 352 + 864 Displacement of ship = 1216 t 41 Prepared by Dr. Sudhir Sindagi Problems Since it is given that, the lightship weight may be assumed to be uniformly distributed over the barge length ๐๐ข๐ ๐ก๐ญ ๐ฐ๐๐ข๐ ๐ก๐ญ 352 = = 11 t/m ๐๐๐ง๐ ๐ญ๐ก ๐๐ Now, cargo weight may be assumed to be uniformly distributed over the length of the cargo hold. ๐๐๐ซ๐ ๐จ ๐ฐ๐๐ข๐ ๐ก๐ญ 192 for the hold 1 = = 24 t/m ๐๐๐ง๐ ๐ญ๐ก ๐๐๐ซ๐ ๐จ ๐ฐ๐๐ข๐ ๐ก๐ญ ๐๐๐ง๐ ๐ญ๐ก ๐๐๐ซ๐ ๐จ ๐ฐ๐๐ข๐ ๐ก๐ญ ๐๐๐ง๐ ๐ญ๐ก ๐๐๐ซ๐ ๐จ ๐ฐ๐๐ข๐ ๐ก๐ญ ๐๐๐ง๐ ๐ญ๐ก ๐ for the hold 2 = 224 for the hold 3 = 272 for the hold 4 = ๐ ๐ 176 ๐ = 28 t/m = 34 t/m = 22 t/m 42 Prepared by Dr. Sudhir Sindagi Strength of Ship Now, since it is a rectangular barge with rectangular sectional area remaining constant throughout the length, hence buoyancy force may be assumed to be uniformly distributed over the length of the ship. ๐๐ฎ๐จ๐ฒ๐๐ง๐๐ฒ ๐ ๐จ๐ซ๐๐ Δ 1216 = b= = = 38 t/m ๐๐๐ง๐ ๐ญ๐ก ๐ ๐๐ ๐๐จ๐ญ๐๐ฅ ๐ฐ๐๐ข๐ ๐ก๐ญ ๐๐๐ง๐ ๐ญ๐ก ๐๐ข๐ ๐ก๐ญ ๐ฐ๐๐ข๐ ๐ก๐ญ ๐๐๐ง๐ ๐ญ๐ก at any point = ๐๐จ๐ญ๐๐ฅ ๐ฐ๐๐ข๐ ๐ก๐ญ ๐๐๐ง๐ ๐ญ๐ก for the Hold 1 = ๐๐จ๐ญ๐๐ฅ ๐ฐ๐๐ข๐ ๐ก๐ญ ๐๐๐ง๐ ๐ญ๐ก for the Hold 1 (w1) = 24 t/m + 11 t/m = 35 t/m ๐๐จ๐ญ๐๐ฅ ๐ฐ๐๐ข๐ ๐ก๐ญ ๐๐๐ง๐ ๐ญ๐ก for the Hold 2 (w2) = 39 t/m + ๐๐ข๐ ๐ก๐ญ ๐ฐ๐๐ข๐ ๐ก๐ญ ๐๐๐ง๐ ๐ญ๐ก ๐๐๐ซ๐ ๐จ ๐ฐ๐๐ข๐ ๐ก๐ญ ๐๐๐ง๐ ๐ญ๐ก + for the hold ๐๐๐ซ๐ ๐จ ๐ฐ๐๐ข๐ ๐ก๐ญ ๐๐๐ง๐ ๐ญ๐ก for the hold 1 43 Prepared by Dr. Sudhir Sindagi Strength of Ship ๐๐จ๐ญ๐๐ฅ ๐ฐ๐๐ข๐ ๐ก๐ญ ๐๐๐ง๐ ๐ญ๐ก for the Hold 3 (w3) = 45 t/m ๐๐จ๐ญ๐๐ฅ ๐ฐ๐๐ข๐ ๐ก๐ญ ๐๐๐ง๐ ๐ญ๐ก for the Hold 4 (w4) = 33 t/m Hence each and every point on the ship is subjected to a resultant load which can be calculated as Load / length = Buoyancy Force / Length (b) – Weight Force / Length (w) Buoyancy Force Weight Force for the hold 1 ๐๐จ๐๐ ๐๐๐ง๐ ๐ญ๐ก for the Hold 1 = ๐๐จ๐๐ ๐๐๐ง๐ ๐ญ๐ก for the Hold 1 = 38 - 35= 3 t/m ๐๐๐ง๐ ๐ญ๐ก ๐๐๐ง๐ ๐ญ๐ก 44 Prepared by Dr. Sudhir Sindagi Strength of Ship ๐๐จ๐๐ ๐๐๐ง๐ ๐ญ๐ก for the Hold 2 = 38 – 39 = -1 t/m ๐๐จ๐๐ ๐๐๐ง๐ ๐ญ๐ก for the Hold 3 = 38 - 45= -7 t/m ๐๐จ๐๐ ๐๐๐ง๐ ๐ญ๐ก for the Hold 4 = 38 - 33= 5 t/m With the available data we can get the weight curve, buoyancy curve and then the load curve. 45 Prepared by Dr. Sudhir Sindagi Strength of Ship 46 Prepared by Dr. Sudhir Sindagi Strength of Ship Shear Force at any point of consideration is the area under the load curve till that point. SFA= 0 SFB= 3 * 8 = 24 t SFC= 24 – 1*8= 16 t SFD= 16 – 7*8= -40 t SFE= -40 + 5*8= 0 t To find the location of 0 ๐ ๐๐ = ๐−๐ ๐๐ x = 2.285m from C 47 Prepared by Dr. Sudhir Sindagi Strength of Ship Bending Moment at any point of consideration is the area under the SFD till that point. BMA= 0 BMB= ½ * 8 * 24 = 96 tm BMC= 96 + ½ (24+16)* 8 = 256 tm BMO= 256 + ½ *2.285 *16 = 274.28 tm BMD= 274.28 – ½ *(8-2.285)*40 = 160 tm BME= 160 - ½*40*8= 0 t-m 48 Prepared by Dr. Sudhir Sindagi Problems 49 Prepared by Dr. Sudhir Sindagi Problems A deck beam is in the form of an H-girder as shown in the accompanying Figure. If the bending moment at the middle of its length is 15 tonnes metres, find the maximum stress in the steel. To find Maximum stress, we will use the equation ๐ ๐ = ๐ ๐ฒ 50 Prepared by Dr. Sudhir Sindagi Problems Since the section is has uniform thicknesses at the top and the bottom, hence, the neutral axis lies at the mid point of the section. Moment of Inertia of the area about the neutral axis will be ๐๐๐ = ๐.๐∗๐.๐๐ ๐๐ ๐.๐๐∗๐.๐๐๐ − *2 ๐๐ ๐๐๐ =3.1042 * 10-4 m4 ๐ ๐ = ๐ ๐ฒ ๐๐ ∗๐๐๐๐∗๐.๐๐ ๐ ๐.๐๐ = 3.1042 ∗ 10−4 ๐ = 71.108 MN/m2 51 Prepared by Dr. Sudhir Sindagi Problems A Midship Section Drawing is shown here. (See Fig.1). (a) Calculate the Moment of Inertia of the Ship Section. (b) Calculate the Section Modulus at Deck & Section Modulus at Keel Section Modulus = Z= ๐๐๐ ๐ In this case, need to find location of NA MI about NA Need to use tabular method for NA 52 Prepared by Dr. Sudhir Sindagi Problems (1) (2) (3) (4) =(2) * (3) (5) =(3) * (4) (6) Item Area of the section Lever from Keel 1st Moment about Keel 2nd Moment about Keel INA about their neutral axis ΣM1=00 m3 ΣM2=00 m4 ΣINA =00 m4 Deck 02 Side shells Bottom Shell Total ΣA=00 m2 53 Prepared by Dr. Sudhir Sindagi Problems (1) (2) (3) (4) =(2) * (3) (5) =(3) * (4) (6) Item Area of the section Lever from Keel 1st Moment about Keel 2nd Moment about Keel INA about their neutral axis 84.5 20∗0.0253 ≈0 12 Deck 20*0.025 = 0.5 02 Side 13*0.014*2 = 0.364 shells Bottom Shell 20*0.02 =0.4 Total ΣA=1.264 m2 13 6.5 0.01 6.5 2.366 15.379 2∗0.014∗133 = 12 5.1263 0.004 0.00004 20∗0.023 ≈0 12 ΣM1=8.87 m3 ΣM2=99.879 m4 ΣINA =5.1263 m4 54 Prepared by Dr. Sudhir Sindagi Problems Distance of NA from the keel Distance of NA from the keel = ๐ ๐ข๐ซ๐ฌ๐ญ ๐๐จ๐ฆ๐๐ง๐ญ ๐จ๐ ๐ญ๐ก๐ ๐๐ซ๐๐ ๐๐๐จ๐ฎ๐ญ ๐ค๐๐๐ฅ ๐๐ซ๐๐ Distance of NA from the keel =7.017 m from keel We need to estimate the MI about NA. Hence, using parallel axis theorem MI about Keel= Σ(INA)self +Σ(A*d2) MI about Keel= 5.1263 + 99.879 MI about Keel= 105.005 m4 MI about Keel= INA+A*d2 105.005 = INA+ 1.264*7.0172 INA=42.767m4 == 55 ΣM1 ΣA Prepared by Dr. Sudhir Sindagi Problems Section Modulus (Z) is given by Z= ๐๐๐ ๐ฒ (Z)Main Deck = (Z)Main Deck = ๐๐๐ ๐ฒ๐๐ 42.767 (๐๐−๐.๐๐๐) (Z) Main Deck =7.148 m3 (Z) Keel = (Z) Keel = ๐๐๐ ๐ฒKeel 42.767 ๐.๐๐๐ (Z) Keel = 6.094 m3 56 Prepared by Dr. Sudhir Sindagi Safe Stress Shear Stress at any point is given by เดฅ ๐ ๐๐ ๐ช= ๐๐ F= The shearing Force at the section under consideration เดฅ= Moment of area about the Neutral axis above or below the surface under ๐๐ consideration. เดฅ=Distance of the centroid of the area under consideration from NA ๐ I= Total Moment of Inertia about the NA b= Total thickness of material resisting shear The maximum allowable stress in terms of the length of a ship using P=77+0.25*L ๐ = ๐๐ ๐๐ก๐๐ซ๐ ๐ ๐ข๐ฌ ๐ข๐ง ๐๐ ๐๐ง๐ ๐ฆ๐ ๐ ๐ข๐ฌ ๐ข๐ง ๐๐๐ญ๐๐ซ๐ฌ. ๐ ∗ ๐๐ 57 Prepared by Dr. Sudhir Sindagi Problems เดค about the NA is In a ship the maximum shearing force is 44.76MN. ๐๐ 151000mcm2. I about NA is 2758000m2cm2 and the thickness of the shell plating at the neutral axis is 2.14cm. Determine the shear stress induced at the neutral axis. Given Data: เดค = 151000mcm2, I about NA is 2758000m2cm2, b= 2.14cm F= 44.7 MN, ๐๐ q=? ๐ช= ๐ช= ๐ช= เดค ๐ ๐๐ ๐๐ ๐๐∗m∗cm2 ๐๐ฆ ∗m2cm2 ๐๐ m2 *100 58 Prepared by Dr. Sudhir Sindagi Problems ๐ช= ๐ช= เดค ๐ ๐๐ ๐๐ ๐๐.๐๐∗๐๐๐๐๐๐ ๐.๐๐∗๐๐๐๐๐๐๐ ๐๐ ๐ช = ๐. ๐๐ *100 ๐๐ m2 m2 59 Prepared by Dr. Sudhir Sindagi Problems The midship section of a ship of breadth 16.5m and depth 11m can be assumed as shown in the figure. All the material has a thickness of 1.25cm. Determine the MI of the section about NA. Moments are taken about the base. What are the section modulus at the upper deck and at the keel? Determine bending stress (stating whether tensile or compressive) for a sagging BM of 44500 t-m. If the ships length is 200m, is it safe in bending stress? 60 Prepared by Dr. Sudhir Sindagi Problems (1) (2) (3) (4) =(2) * (3) (5) =(3) * (4) (6) Item Area of the section Lever from Keel 1st Moment about Keel 2nd Moment about Keel INA about their neutral axis UD 2nd Deck Tank Top Keel 2 Side Shells Central Girder Total 61 ΣA=00 m2 ΣM =00 m3 ΣM =00 m4 ΣI =00 m4 Prepared by Dr. Sudhir Sindagi (4) =(2) * (3) (5) =(3) * (4) Problems (1) (2) (3) Item Area of the section Lever from Keel UD 16.5*0.0125 =0.20625 2nd Deck Tank Top Keel 2 Side Shells 0.20625 0.20625 0.20625 2*11*0.0125=0.275 Central Girder 1.2*0.0125=0.015 Total ΣA=1.115 m2 11 8.4 1.2 0 5.5 0.6 1st Moment about Keel 2.2687 1.7329 0.2475 0 1.5125 0.009 (6) 2nd Moment about Keel INA about their neutral axis 24.9557 16∗0.01253 ≈0 12 14.5521 16∗0.01253 ≈0 12 0.297 16∗0.01253 ≈0 12 0 16∗0.01253 ≈0 12 8.3187 2∗0.0125∗113 = 12 0.0054 0.0125∗1.23 = 12 ΣM1=5.7702 ΣM2=48.0134 m3 m4 2.7729 0.0018 62 ΣINA =2.7747 m4 Prepared by Dr. Sudhir Sindagi Problems Distance of NA from the keel Distance of NA from the keel = ๐ ๐ข๐ซ๐ฌ๐ญ ๐๐จ๐ฆ๐๐ง๐ญ ๐จ๐ ๐ญ๐ก๐ ๐๐ซ๐๐ ๐๐๐จ๐ฎ๐ญ ๐ค๐๐๐ฅ ๐๐ซ๐๐ Distance of NA from the keel=5.1751 m from keel We need to estimate the MI about NA. Hence, using parallel axis theorem MI about Keel= Σ(INA)self +Σ(A*d2) MI about Keel= 50.9051 m4 MI about Keel= INA+A*d2 50.9051 = INA+ 1.115*5.17512 INA=21.053m4 == 63 ΣM1 ΣA Prepared by Dr. Sudhir Sindagi Problems Section Modulus (Z) is given by Z= ๐๐๐ ๐ฒ (Z)Main Deck = ๐๐๐ ๐ฒ๐๐ (Z)Main Deck = ๐๐.๐๐๐ (๐๐−๐.๐๐๐๐) (Z) Main Deck =3.6125 m3 (Z) Keel = ๐๐๐ ๐ฒKeel (Z) Keel = ๐๐.๐๐๐ ๐.๐๐๐๐ (Z) Keel = 4.066m3 64 Prepared by Dr. Sudhir Sindagi Problems Stresses at Main Deck and the keel are calculated as follows σ= ๐ ๐ (σ)Main Deck = (σ)Main Deck = ๐ ๐๐๐ ๐๐๐๐๐∗๐๐๐๐∗๐.๐๐ ๐.๐๐๐๐ (σ)Main Deck = =120.84 MN/m2 (Compressive) (σ) Keel = (σ) Keel = ๐ ๐Keel ๐๐๐๐๐∗๐๐๐๐∗๐.๐๐ ๐.๐๐๐ (σ) Keel = 107.36 MN/m2 (Tensile) 65 Prepared by Dr. Sudhir Sindagi Problems We know that, Safe stress is calculated using following P=77+0.25*L P=127 MN/m2 and ๐ = ๐๐ ∗ ๐ ๐๐ ๐ = ๐๐๐. ๐ MN/m2 Since induced stresses are lesser than the allowable stresses hence, the structure of the ship is safe. 66 Prepared by Dr. Sudhir Sindagi Summary Introduction Force and Center of Pressure for – Regular shaped bulkhead – Irregular shaped bulkhead Strength of Ship – – – – – Sagging Hogging Ship as a girder SFD & BMD for the ship Calculation of stresses in Main deck and Keel due to Bending Moment 67
