REEDS NAVAL ARCHITECTURE PDF
Reeds Vol 4: Naval Architecture for Marine Engineers Format: PDF eBook ( Watermarked) Series: Reeds Marine Engineering and Technology Series. Naval Architecture Reed's Vol 4 Textbook. Download as PDF, TXT or read online from Scribd. Flag for inappropriate Reeds Naval Architecture for Engineers. Reed's Naval Architecture Centre of gravity - Free download as PDF File .pdf), Text File .txt) or view presentation slides online. Naval Architecture textbook.
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pdf. REED'S NAVAL ARCHITECTURE FOR MARINE ENGINEERS .. 4 REED'S NAVAL ARCHITECTURE FOR ENGINEERS (a) Pressure on top == Qgh q interval from 3 m to 7 m waterline ;; h 44 REED'S NAVAL ARCHITECTURE FOR. First published as Naval Architecture for Marine Engineers, Reprinted , Second edition published as Muckle's Naval Architecture, Reeds Vol 4: Naval Architecture for Marine Engineers (Reeds Marine Engineering and Technology Series) Richard Pemberton and E A Stokoe pdf download.
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Thus the relative density rd of fresh water is 1. It is weful to know that the density of a substance expressed in tim is numerically the same as the relative density. If a substanet has a relative density of x, then one cubic metre of the substance will have a mass of x tonnes. V cubit metres will have a mass of Vx tannes or lVx kilogrammes. If the relative density of lead is Density of lead. A plank 6m Calculate the density of the wood.
Jtn - Distance h is known as the hetld. The pressure at the supply tank depends upon the height of the header tank. Container iii could represent a double bottom tank hAving a vertical overflow pipe. The pressure inside the tank depends upon the height to which the liquid rises in tbe pipe. The total load exerted by a liquid on a horizontal plane is the product of the pressure and the area of the plane.
A rectangular double bottom tank is 20 m long 12 m wide and 1. It , 1 , should be noted that a very smaIl volume of liquid in a vertical pipe may cause a considerable increase in load. The total load on such a plane may be determined as follows.
Consider an irregular plane of area A. Let one such strip. Since tbe strip is thin. If H is the distance of the centroid of the p1aDe from the liquid surface. A rectanJU1ar bulkhead is 10 m wide and 8 m deep. It is loaded on one side only with oil of relative density 0. Calculate the load on tbe bulkhead if the oil is: COtre of pressure on an immersed plane is the point at which the whole liquid load.
Let the strip be distance r from the axis If the plane is vertical. O represents the surface of tbt: BJ l I, FAcr Centre of pressure from. Eumple- A peak bulkhead is in the form of a triangle. The tank is fiDed with sea water. Calcula1e the load on tm bulkhead and the position of the centre of pressure relative to tbe top of the bulkbead if the water is; a at the top of tbe bulkhead f b 4 m up the sounding pipe.
TICS 9 plane is obtained. If this is repeated at a number of points, the resultant values may be plotted to form the load diagram for the plane. The arm of this load diagram represents the load on the plane. For a rectangular plane with its edge in the surface, the load diagram is in the fonn of a triangle. For a rectangular plane with its edge below the surface, the load diagram is in the form of a trapezoid. The load diagrams for triaDCular plaDcs are parabolic.
LoadIm at top of bulkhead.. KHEAD "" Reactions are set up by the end connections at the lOp RT and at the bottom R. Taking moments about the lOp, R. This is also the position of the maximum bending moment. A bulkhead. The bulkhead is flooded to the top edge with sea water on one side only. P A piece of aluminium has a mass of I and itS volume is 42 an'.
A rectaDlu1ar double bottom tank 12 m long and 10 m wide is full of sea water. Calculate the head of water above the tank top if the load due to water pressure on the tank top is 9. A double bottom tank is J. The tank is filled with oil rd 0. The double bottom noon are spaced mID apart aDd are connected to the tank top by riveted a. A ballast tank is IS m long, 12 m wide and 1. Calculate the load on tbe top and shan side.
A vertical bulkhead 9 m wide and 8 m deep bas sea water on one side Only to a depth of 6 m. Find the load on the bulkhead if it bas oil rd 0. The end bulkhead of an oil fuel bunker is in the form of a rectangle 10 m wide and 12 m high. A dock gate 6 m wide and S m deep has fresh water on one lioicie to a depth of 3 m and sea waler on the other side to a d. Calculate the resultant load and position of the centre of pressure:. A triangular bulkhead is S m wide at the top and 7 m deep.
It is loaded to a depth D with sea water, when it is found that the load. A trianauIar bulkhead is 7 m wide at the top and has a vertical depth of 8 m. Calculate the load on the bulkhead and the position of the centre of pressure if the bulkhead is flooded with sea water on only one side: II, A watertij;ht bulkhead is 8 m hiah and is supported by vertical stiffeners mm apart.
The bulkhead is flooded to its top edge with sea water. A bulkhead is supported by vertiea1 stiffeners. The distance between the stiffeners is one ninth of the dcpt. When the bulkhead is flooded to the top with sea water on ooe side only, the muimum shearing force in the stitreoers is tN. This loss in weight is the upthrust exerted by the liquid on the body and is equal to the weight of the volume of liquid which the body displaces.
If a solid body is suspended in fresh water, completely im- mersed, the upthrust is the weight of fresh water having the same volume as the body. This is acceptable since tbe value of g used to ob- tain the weight in air is the same as tbe value of g used to obtain the upthrust in water. A solid block of cast iron bas a mass of kg. When it is completely immersed in fresb water the mass appears to be reduced to kg. Calculate the relative density of cast iron. A piece of brass rd 8. Calculate the apparent mas.
The body will then rise: A body of fd oflcss than Example, A block of wood 4 m loog, 0. Calcuhue thr: Calculate the mass of the barge. Thus if the volwnc of tbe underwater portion of the ship is k. It is usual to assume that a ship floats in sea water of density Since the volume of water displaced depends upon the draught, it is useful to calculate values of displacement for a range of draughts.
These values may then be plotted to forril a displacement C'UJ'lIe. The following symbols will be used tbrougbout the text: This confUSion may be reduced if the. A ship displaces mJ of sea water at a par- ticular draught.
The force of buoyancy acts at the centre ofbuoyancy, which is the centre of gravity of the underwater volume of the ship.
The venical position of the centre of buoyancy VCB is usuaUy liven as a distance above the ked. The distance from the waterline to the VeB may be found by two other methods: OJ Aw At different drauahts.
Reed's Naval Architecture Centre of gravity
It may be usu. The watef1 lane area of a ship is mJ. TPC z 0. JolT OF m. XBXd c. A ship m loo,a.: Dt of 14 toDAe. IlUp'lJ bull which is in contact with tbt: This area may bt: To this area should bt: Calculate the wetted surface area of the ship: This principle may be seen in a projector where a small image is projected from a slide onto a screen.
The height of the image depends upon the distance of the screen from the light source, but the proportions and shape of the image remain the same as the image on the slide. Thus the image on the screen is a scaled-up version of the image on the slide. This may be shown by comparing two circles having diameters D and d respectively.
NT, T. NTS OF roa. S the area of the latter. This may be shown by c: Volume of small sphere "" i tP Since i is constant: D' ratio of volumes -Qf Thus it diameccr D is rwiu diameter, d, tbe voJume of the former is tight limes the volume of the latter.
Ai and S 0: A ship m lona Qis;pUtca tonne and has a wetted. Ca1culate the displacement and wetted surface area of a 6 m model of the ship. If the values of upthrust at different positions along the leDlth of the ship are plotted on a base representing the ship's lenath, a buoyancy curw is formed Fia;. The area of this curve represents the tota. I upthrust exerted by the water on the ship. The total weiaht of a ship con.
The differmee between the weiaht and buoyancy at any point is the lotId at tbat point. A load diagram is formed. Because of this unequal loadinI, however, shearioa forces and bencliDJ moments are set up in the ship. A box. TIle "';lIltu uniformly distrib. Sketch the sMaciDg force and bending moment diqrams aDd. It is therefore only necessary to consider the added calIO IUd the additional buoyancy required. S , kN excess weiiht-- Companmmu 2, 3 and 4 load.
Cf I DUa. M Fig. A piece of metal cm in volume is attached to the bot- tom of a block of wood 3. The system floats in fresh water with cm projecting above the water. Calculate the relative density of the metal. A raft 3 m long and 2 m wide is constructed of timber 0. Calculate the minimum mass which must be placed on top of the raft to sink it.
A box barge 65 m long and 12 m wjde floats at a draught of 5. A ship has a constant cross-section in the form of a triangle which floats apex down in sea water. The ship is 85 m long, 12 m wide at the deck and has a depth from keel to deck of 9 m.
Draw the displacement curve using 1. From this curve obtain the displaCement in fresh water at a draught of 6. A cylinder 15 m long and 4 m outside diameter floats in sea water with its axis in the waterline.
A vessel 40 m long has a constant cross-section in the form of a trapezoid 10 m wide at the top, 6 m wide at the bottom and 5 m deep, It floats in sea water at a draught of 4 m. Calculate its displacement. The waterplane areas of a ship at 1. A ship ISO m 10Dl and The TPC is A ship displaces The area of immersed midship section is mI.
Ith of the ship aDd the prismatic ooefflCient. The length of a ship is 18 times the drauJb. At the load waterplane. Deter- mine the length of the ship and the TPC in fresh water. The wetted surface area of the appendages is 30 mI and -iOh is to 'be added for longitudinal curvature. A ship of 14 lXJO toone displacement, m S. A box barge is 75 m lona, 9 m beam and 6 m deep.
A similar barge having a volume of m] is to be constructed. Calculate the length, breadth and depth of the new barp. The wetted surface area of a ship is twice that of a similar ship. The displacement of Ute latter is tOODe less lhan the former. Determine the displaeement of the latter. The dist: Wben calculating the area of a waterplane it is usual to divide the lenJtb of the ship into about 10 equal pans, sMnI J 1 sections. These sec: It is convenient to measure distances from the centreline to the ship side, Jiving half ordinates.
These half ordinates are used in conjunction with Simpson's Rule and the answer multiplied by 2. The equally-spaced half ordinates of a watertight flat 27 m long au J.
If the immersed cross-sectional areaS of a ship at a number of positions along the length of the ship are plotted on a base representing the ship's length Fig. Hence tbe displacement of the ship at any given draught may be calculated.
The longitudinal centroid of this figure represents the longitudinal centre of buoyancy of the ship. The vertical centroidl of these two curves represent tbe vertical centre of buoyancy of the ship. The immersed cross-sectional areas throuah a ship m Calculate tbe displacement of the ship in sea water of 1.
Displacement - vol of displacement x density 41 x 1. The TPC values for a ship at 1. Calculate the displacement at 7. Intermediate ordinates are introduced to reduce the spacing to half or quane..
While it is possible to calculate the area of such a waterp1aDe by dividing it into separate sections. The foUawing method may be used. The 1 ordinates at sections AP, i. Thus Area from 0 to J'2 -;- 4y: The following example shows how these multipliers may be determined. Wl keel 0.
Let the common interval.. Such a curve may be drawn on a base equal to Be, with ordinates of xy, and the area under this curve may be found by putting the values of xy through Simpson's Rule.
H' 6x "" YI' -I: It is usually neces. Since the ce: The roUowinC calcula- tion shows the met. YIO spaced It m apart commencing from aft. Tbe positive sign indicates an ordinate: It may be omitted rom the table as will be seen rom the worked examples. The half ordinates of a waterplane ISO m long are as follows: S The balf ordinaltS of a waterplane m long are as follows: FP iord 0 5.
O 19B. There arc many oecasions. The widths of the Wlk surface. It is necessary in tbis calculation to determine the area. FlJtST A. Em C6'1troD fiml. Q Second moment of area about centreline.
A double bouom tank 21 m long has a wateniiht centre girder. The 'Nidths of the tank top measured from the centreline to the ship's side are Calculate the second moment of area of the tank surface about a IonaitudinaJ axis through its centroid, for one side of the ship only.
Let the widths of a bulkhead at intervals of h. SM I "''''''''' for Area 1y, , """""w A fore peak bulkhead is 4. At regular intervals of 1. A ship m long has t widths of waterptane of 1, 7.
Reeds Vol 4 Calculation of area, volume, moments text book
Calculate the displacement at 9 m drauabt. A ship m long and 18 m beam floats at a draught of 9 m. The immersed cross-sectional areas at equal intervals are S, ScaionAPt 1 I! The i breadths of the load waterplane of a ship m loni. The displacement of a ship at draughts of 0, I, 2, 3 and 4 m are 0, ,, and tonne. Calculate the distance of the centre of buoyancy above the keel when floating at a draught of 4 m.
The widths of a deep tank bulkhead at equal intervals of 1. Calculate the load On the bulkhead and the position of the centre of pressure, if the bulkhead is flooded to the top edge with sea water on one side only. A forward deep tank 12 m long extends from a longitudinal bulkhead to the ship's side.
The widths of the tank surface measured from the longitudinal bulkhead at regular inten'als are 10,9,7,4 and I m. Calculate the second moment of area of the tank surface about a longitudinal axis passing tbrough its centroid. A ship m long has! Calculate tbe second moment of area of the waterplane about the centreline. The immersed cross-sectional areas of a ship m long, commencing from aft, are 2, 40, 79, , , , If the object is suspended from this point, then it will remain balanced and will not tilt.
Hence the weight may be represented by mass. Since both the thickness and the cknsity are constant, moments of aretl may be used. This system may also be applied to detnmine the centre of gravity. The actual calculation of the centre of gravity of a ship is a very lengthy process. Such a calculation is usually carried out for a passenger ship in the initial design stages, but the results are confirmed by an alternative method when the ship is completed.
It is usual to measure the 'Vertical position of the centre of gravity VCG of the ship above the keel and this distance is denoted by KG. The height of the centre of gravity of an item on the ship above the keel is denoted by Kg. The longitudinal position of the centre of gravity LCG is usually given as a distance forward or aft of midships.
Reeds Vol 4 Calculation of area, volume, moments text book
A ship of tonne displacement is composed of masses of Determine the height of the centre of gravity of the ship above the keel. A ship of tonne displacement is composed of masses of , and tonne at distances 60, 35 and 11 m aft of midships, and masses of , and tonne at distances IS, 30 and SO m forward of midships. Calculate the distance of the centre of gravity of the ship from midships.
The distance moved by the ship's centre of gravity depends upon the magnitude of the added mass, the distance of the mass from the ship's centre of gravity and the displacement of the ship. If a mass is placed on the port side of the ship in the forecastle, the centre of gravity moves forward, upwards and to port. The actual distance and diredion of this movement is selclom required but the separate components are most important, i.
When an item 00 a ship is removed. A ship of toone displacement has its centre of gravity I. S m aft of midships and 4 m above the keel. Calculate the new position of the centre of. The same answer may be obtained by taking moments abow the original oenttt of gravity thus: Cenue of gravity from t9tal m: Shift in cenue of aravity or, mass moved x distance moved Shift in centre of gnvity Ell TOTAL mass This expression is most useful in ship calculations and is applied throughout stability and trim work.
It should be noted that it is not necessary to know either the position of the centre of gravity of the ship, or the position of the mass relative to the centre of aravity of the ship.
The rise in the centre of gravity is the same whether the mw is moved from the tank: The centre of gravity of the ship moves in the same direction as the centre of gravity of the mass. Thus if a mass is moved forward and down, the centre of gravity of the ship also moves forward and down.
A ship of tonne displacement has a mass of tonne on the fore deck SS m forward of midships. Calculate the shift in the centre of gravity of the ship if the mass is moved to a position 8 m forward of midships.
If the vessel now heels, the mass moves in the direction of the heel until it again lies vertically below the point of suspension, and no matter in which direction tbe vessel heels, the centre of gravity of the mass is always below this point. Thus it may be seen that the position of the centre of gravity of a hanging mass, relative to the ship, is at the point of suspension.
This principle proves to be very important when loading a ship by means of the ship's derricks. If, for example, a mass lying on the tank top is being discharged, then as soon as the mass is clear of the tank top its centre of gravity is virtually raised to the derrick head, causing a corresponding rise in the centre of gravity of the ship. If the mass is now raised to the derrick bead there is no further change in the centre of gravity of the ship.
Ships which are equipped to load heavy cargoes by means of heavy lift derricks must have a standard of stability which will prevent excessive beel when the cargo is suspended from the derrick. A similar principle is involved in the design of ships which carry hanging cargo such as chilled meat. The meat is suspended by hangers from the underside of the deck and there- fore the centre of gravity of the meat must be taken as the deck from which it hangs.
A ship of 10 tonne displacement has a mass of 60 tonne lying on the deck. A derrick, whose head is 7. Calculate the shift in the vessel's centre of gravity when the mass is: A ship of tonne displacement has its centre of gravity 6 m above the k. Find the new displacement and position of the cenlre of gravity when masses of Calculate the new position of the centre of e: A ship has tonne of cargo in the hold. The displacement of the vessel is tonne aDd its centre of gravity is 1.
Find tbe new position of the centre of gravity if this cargo is moved to an after bold, 40 m from midships. An oil tanker of 17 tonne displacement has its centre of gravity I m aft of midships and has 2: This fuel is transferred to the after oil fuel bunker whose centre is SO rn from midships.
Calculate the new position of the centre of gravity: A ship of tonne displacement has tonne of careo on board. Determine the alteration in position of tbe centre of gravity.
A ship of 10 tOIlDe displacement has its centre of ,ravity m above the keel. Masses of Find the new displacement and position of the centre of gravity. A vessel of tonne displacement has 75 tonne of eario on the deck.
It is lifted by a derrick whose bead is In theory it is possible to balance a pencil on its point on a fiat surface. The pend! In practice this is found to be impos- sible to achieve. It Thus it is euential to consider practical conditions and to assume that a ship is always movinl.
In the upriaht position Fia. Since the weiaht is equal to the upthrust, and tbe centre of gravity and the centre of buoyancy are in the same vatical line. It may be seen from Fig. GM is said to be positive when G lies below M and the vessel is stable.
A ship with a la. The ship is then said to be sriff. A stiff ship will be very uncomfortable, having a very small roll.. If the centre of aravity lies above the transverse metacaltte Fli.
S , there is DO moment acting on the ship which will tbcrefore remain iDc: The vessel is then said 10 be in MUtraJ equilibrium. S Since any reduction in the heilbt of G will make the ship stable.
KB is tbe distance of he centre of buoyancy above tbe keel and may be found by one of the methods shown previously.
Consider a ship whose volume of displacement is v. If the ship is now inclined to a small angle 8. Since 8 is small it may be assumed that S is on the centreline. A box barae of lensth L and breadth B Ooat5 at a level keel draupt d. Calculate the height of the transverse meta- centre above the keel. A vuseI of conSWll trian. Calculate the dr. L Fig. The heiaht of the' transverse metacentre above the keel may then be found at any intermediate drauJht. A vessel of c: Oraw the mc: If the height of the cenue of gravity of the empty ship is blown, it is possible to calcu1ale its position for any given c: It is therefore neceuary to carry out the inc: The experiment is commenced with the ship upright.
A small mass m is moved across the ship tbrouih a distance d. This causes tbe centre of gravity to move from its oripnal position G on tbe ceut. But tbe vertical through B I intersects the centreline at M, the transverse: The deflection a of the pendulum may be measured when the mass is moved across the deck. Thus if I;: A mass of 6 tonne is moved transversely through a distance of 14 m on a ship of tonne displacement, when the deflection of an 11 m pendulum is found to be nun.
The transverse metacentre is 7. Determine the height of the centre of gravity above the keel. At leut rwo penduJums are used, one forward and ooe aft. The pendulum bobs are immersed in water or liaht oil to dampen tbe swing. Four musts A. C and D are placed on the deck. The mooriq ropes are slackened and Ibe sJUp-to-sbore gang.
The draughts and density of water are read as accurately as possible. The inclinina: The deflections of the penduJums are recorded for each movement of mass. An avcraae of these deflections is used to determine tbe metae: Tbw if there: Only those men required for the experimeD: Any movement of liquid affects the results and therefore all tanks should be empty or pressed up tight.
The magnitude and position of any mass which is not included in the lightweight of the ship should be noted and it is therefore necessary to sound all tanks and inspect the whole ship.
Corrections are made to the cmue of gravity for any such masses. An instrument for recording inclination is in use by many shipyards. Jt consists of a heavy metal pendulum balanced on knife edges, aeared to a pen arm which records the angJe of heel on a rotatiDJ drum.
The advantages of using this instrument, known as a StabiJogrQph, are tbat a penna. If, for jnstance, the mooring TOpeS are restricting tbe beet the irregular movement will be seen aD the drum. The centre of gravity of the ship moves away from the centreline, reducing tbe riJbring lever and increasinc the angle of heel. The movement of the cmtre of gravity from G to 01 has been caused by the uansfer of a wedge of liquid across the tank.
Thus 1. But the righting lever is the perpendicular distance between the verticals through the centre of buoyancy and the centre of gravity, and this distance may be measured at any point. A ship of SOOO lonne displacement bas a rectangular tank 6 m long and 10 m wide. Calculate the virtual reduaion in metaeentric height if this tank. It would, however. If a tank is sub-divided by n longitudinal divisions forming equal tanks, then 1. In such a ship, tanks which are re- , quired to carry liquid should be pressed up tight.
If the ship is c initially unstable and heeling to port, then any attempt to intro- duce water ballast will reduce the stability.
Before ballasting, therefore, an attempt should be made to lower the centre of gravity of the ship by pressing up existing tanks and lowering masses in the ship. If water is introduced ,into a double bottom tank on the starboard side the vessel will flop to starboard and may possibly capsize. A small tank on the port side should there- fore be filled completely before filling on the starboard side.
The angle of heel will increase due to free surface and the effect of the added mass but there will be no sudden movement of the ship. A panicularly dangerous condition may occur when a nre breaks out in the upper 'tween decks of a ship or in the accommodation of a passenger ship. If water is pumped into the space, the stability of the ship will be reduced both by the added mass of water and by the free surface effect.
Any accumulation of water should be avoided. It may be possible to discharge the water using a ponable pump. In calm weather or in pon a hole may be drilled in the side of the ship. A hole in the deck would allow the water to work its way into the bilges from where it may be pumped overboard, but it is doubtful if such a method would be possible except in rare circumstances.
It may, however, be possible to remove the 'tween deck hatch covers thus restricting the height of water to about mm. The cargo in the lower hold would be damaged by such a method but this would be preferable to losing the ship.
It is imponant to note that the free surface effect depends upon the displacement of the ship and the shape and dimensions of the free surface. It is independent of the total mass of liquid in the tank and of the position of the tank in the ship.
The ship with the greatest free surface effect is, of course, the oil tanker, since space must be left in the tanks for expansion of oil. Originally tankers were built with centreline bulkhead and expansion trunlcs.
It is not possible to design dry cargo vessels in the same way, since the position of the centre of gravity of the ship varies considerably with the nature and disposition of the cargo. Thus, while the free surface effect in a tanker is greater than in a dry cargo ship, it is of more impor- tance in the latter.
The effect of a suspended mass on the stability of a ship may be treated in the same way as a free surface. It may be shown, as stated in Chapter 4, that the centre of gravity of the mass may be taken as acting at the point of suspension. The proof of the formula for BM was based on the assumption that the two waterplanes intersect at the centreline and that the wedges are right angled triangles. The righting lever is the perpendicular distance from a vertical axis through the centre of gravity G to the centre of buoyancy Bt.
This distance may be found by dividing the moment of buoyancy about this axis by the buoyancy. The method used is as follows. I" Fig. Sections through the ship are drawn at intervals along the ship's length. These sections are inclined to an angle of, say ISO, The integrator is set with its axis in the ,I, , , , , ': This is repealed for different waterlines and for angles of 30".
The displacement, height of centre of pavity and metacentric heigbt of a vessel may be cakulated for any loaded condition. At this displacement tbe righting levers may be obtained at the respective angles for the assumed position of the cCDtre of Vavity. These values must be amended to suit the actual beiaht of the centre of ,ravity. Let G - assumed position of cenue of gravity 0 1 "" actual position of centre of cravity , ,I I' I.
The initial slope of the curve lies along a line drawQ from the origin to OM plot: Draw the amended statical stability curve. Hi Fig. A tremendous change takes place in this curve when tbe weather deck edse becomes immersed. Thus a shj.
Vessel B is a raised quarter deck ship. It is essential for a vessel with small freeboard. If a vessel is initially unstable it will Dot remain upright but will either heel to the Angle of Loll or will capsize depending upon the degree of instability and the shape of the stability curve Fig. Vessel A will heel to an angle of loll of about 8 0 but still remains a fairly stable ship, and while this heel would be very inconvenient.
If vessel B is unstable it will capsize since at alI angles the righting lever is negative. B A Fig. The righting leve: Calculate the dynamical stability of the ship at 60 0 heel. The common interval must be expressed in rrzdians. If a vessel is initially unstable it wUl either capsize or beel to tbe angle of IoU. Hence jf the vessel is assumed to be wall- sided: Thus the angle of loll may be determined for any given unstable condition.
A more practical application of this expression may be found when the vessel is listing at sea. It is necessary first to bring the ship upright and then provide sufficient stability for the remainder of the voyage. Thus it is essential to estimate the negative metacentric height causing the angle of IoU in order to ensure tbat these conditions are realised.
A ship displaces 12 tonne. The following are the disposition of the masses on board the ship. A vessel of constant rectaIlau1a. Calculate tbe draught. A raft is made from two cylinders each 1. The distance between the centres of the cylinders i5 3 m. If the drauaht is 0. A vessel of constant trian.
Draw the metacenuic diagram usiOl O. S m intervals of draught up to the 3. An inclinina experiment was carried out on a ship of SOCK tonne displacement. S m lana to deflect mm. Calculate the metaeeDtric height and the beight of the centre of ITavity above lhe ked. An inclining experiment was carried out on a ship of tonne displacemeot, when masses of 6 tonne were moved traos- versely through U.
The deflections of a 7. S m pendulum were 81, 78, 8S, 83, 19, 82, 84 and 80 mm respectively. Calculate the metaeentric beight. A ship of tonne displacement has a rectangular double: Calculate the virtual reduction in metaeeDtric height due to free surface.
A ship of tonne displacement bas its centre of gravity 4. S m above the keel and transverse metacentre S. O m above the keel when a rectangular tank 7. S m lona: A mass of 10 tonne is moved 12 m across the deck..
A ship of 6Ol: X tonne displacement has its cenue of gravity S. A rectaoauIar double bottom tank Calculate the metaeentric height.
Reeds: Naval Architecture for Marine Engineers
An oil tanker 24 m wide displaces 2S tonne when loaded in Dine equal tanks, each 10 m long, with oil rd 0. Calculate the toW free surface effect with: A ship of A mass of 80 toDDe is lifted from its position in the centre of tbe lower hold by one of the ship's derriclcs, and placed on the quay 2 m from the ship's side.
The rightina levers of a ship, for an assumed KG of 3. When the ship is loaded to the same cl. A ship of 18 tonne displacement has KB 5. A ship of tonne displacement has KG 5. Thus if. If the draught forward is greater than the draught aft the vessel is said to trim. If a small mass m is added to a ship at the centre of flotation, there is an increase in mean drauiht but no change in trim. A large mass e. Hence the mean draught may be taken as the draught at the LCF. The mean of the end draughts may be compared with the actual draught amidships to determine whether the vessel is hogging or sagging, but is of little relevance in hydrostatic calculations.
Thus the effect of an added mass on the draughts may be divided into: The change in trim due to any longitudinal movement of mass may be found by considering its effect on the centre of gravity of the ship.
The centre of gravity G and the centre of buoyancy B lie in the same vertical line. If the: It is useful to mow the momC1lt which will cause a chanae in trim of one CD1. Chanae in trim. Since the vessel cb. By similar triangles. A ship of tonne displacement, '96 m long, floats.. GMI, lOS ttl and centre of flotation 2. A slUp ISO m has dtaug. It tb: Such a problem exists every time a ship loads or disebarges the major part of its deadweight.
The underlyioa: 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. This trimming moment is the displacement multiplied by the longituc1inal distance between B and G.
The vessel chanses trim about the LCF and hence it is possible to calculate the end draughts. When the vessel has changed trim in this manner, tbe new centre of buoyancy B l lies in the same vertical line as G. A ship 12S m long has a light displacement of tonne with LCG 1.
The following items are now added: Cargo tonne Leg 3. S m forward of midships. At 14 tonne displacement the mean clraUJht is 7. S m aft of midships. IA il Water 7. M IS20 Stores la. SF 10SO Liahtweight 1. S em Drauaht forward. For any aival displacement. Consider a ship of displacement A tonne. Since the increase: S Example. Fmd the c: OIOx 1. If the vessel moves from sea wue: Let b.
This change in uim is always smaIL. The volume of displacement has been increased by a layer of volume I wbose cenue of pvity is at tbe centre of flotation F. This causes the centre of buoyancy to move from B to B" the centre of cravity remaining at G. If the ship moves from the river water into sea water. A ship m 10Da: LCB 2. LCF 0. OR '" x 2. If the loss in buoyancy exceeds the reserve buoyancy the vessel will sink. CIt- pressed II a ratio or percentage of the total volume of the compartment.
The permeability of a macbiDay spaoe is about 85 per cent and accommodation about 9S per ceDt. The permeability of a carlO hold varies considerably with the type. The effects of bilgina; a mid-Ieftllh compartment may be shown most simply by considering a box buge of length L. S If this compartment is billed, buoyancy is lost and must be replac: The yolume of buoyancy lost is the volume of the compartment up to waterline WL. L - L-,. J iii may be regarded as the effective length of the bilged compartment.
Calculate the new draught if this compartment is bilged: S Consider a box barge of Ien,t. Search this site. A Beautiful Day for a Wedding: A Ransom Christmas: Aching Joy: Advanced French Vocabulary: Qui vivra verra!
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