Ship Stability for Masters and Mates 5E英文原版机械工程与材料科学教材

ShipShip StabilityStability forfor MastersMasters andand MatesMates 5E5E英文原版机械工程与材料科学教材英文原版机械工程与材料科学教材 Ship Stabilityfor Mastersand MatesShip StabilityforMasters andMatesFifth editionCaptainD R DerrettRevised byDr C B BarrassOXFORD AUCKLANDBOSTON JOHANNESBURGMELBOURNE NEWDELHI 9 ContentsPreface viiIntroductionixShip typesand generalcharacteristics xi1Forces and moments12Centroids and the centre of gravity93Density andspecific gravity194Laws offlotation225Effect of density ondraft and displacement336Transverse statical stability437Effect offree surface of liquidson stability508TPC and displacement curves559Form coefficients6110Simpson s Rulesfor areas and centroids6811Final KG9412Calculating KB BM andmetacentric diagrams9913List11414Moments ofstaticalstability12415Trim13316Stability andhydrostatic curves16217Increase indraft due to list17918Water pressure18419Combined listand trim18820Calculating the effect offree surface of liquids FSE 19221Bilging andpermeability20422Dynamical stability21823Effect ofbeam andfreeboard onstability22424Angle ofloll22725True meandraft23326The incliningexperiment23827Effect oftrim ontank soundings24328Drydocking and grounding24629Second momentsof areas25630Liquid pressureand thrust Centres ofpressure26631Ship squat27832Heel due to turning28733Unresisted rollingin stillwater29034List dueto bilgingside partments29635The DeadweightScale30236Interaction30537Effect ofchange of density ondraft andtrim31538List withzero metacentricheight31939The Trim and Stabilitybook32240Bending ofbeams32541Bending of ships34042Strength curvesfor ships34643Bending andshear stresses35644Simplified stabilityinformation372Appendix IStandard abbreviationsand symbols378Appendix IISummary ofstability formulae380Appendix IIIConversion tables387Appendix IVExtracts from the M S Load Lines Rules 1968388Appendix VDepartment ofTransport Syllabuses Revised April1995 395Appendix VISpecimen examinationpapers401Appendix VIIRevision one liners429Appendix VIIIHow topass examsin MaritimeStudies432Appendix IXDraft Surveys434Answers toexercises437Index443vi Contents closely followedbyher LBPand C bvalues Type ofship TypicalDWT LBPBR MLD TypicalCbService speedorname tonnes orm3 m m fully loaded knots ULCC VLCC565000440to25070to400 85to0 8213to153 4and supertankersto100000Medium sized100000250to17540to250 82to0 8015to1534oil tankersto50000OBO carriersup to200to300up to450 78to0 8015to16173000Ore carriersup to200to320up to580 79to0 831412to1512323000General cargo3000to100to15015to250 70014to16ships15000Liquefied natural130000m3up to28046to250 66to0 682034to16gas LNG and to75000m3liquefied petroleum LPG shipsPassenger liners5000to200to30020to400 60to0 6424to30 2examples below 20000QE2 built 19706002812Oriana built1994 727022432 20 6252 4Container ships10000to200to30030to450 56to0 6020to2872000Roll on roll off2000to100to18021to280 55to0 5718to24car andpassenger5000ferries 1998Dr C B BarrassChapter1Forces and momentsThe solutionof manyof theproblems concernedwith shipstabilityinvolves anunderstanding of the resolution of forces and moments Forthis reason a briefexamination of the basicprinciples will be advisable ForcesA forcecan bedefined as any pushor pullexerted ona body The S I unit offorce is theNewton one Newtonbeing the force requiredto producein amass of onekilogram anaeleration ofone metreper second per second When consideringa forcethe followingpoints regardingthe forcemust beknown a The magnitude of the force b The directionin which theforce is applied and c The point at which theforce is applied The resultant force When two or moreforces are acting ata point theirbined effect can berepresented byone forcewhich willhave the sameeffect as the ponent forces Such a force isreferred to as the resultantforce and theprocess of finding it is calledthe resolutionof theponent forces The resolutionof forces When resolvingforces it will beappreciated thataforce acting towards a point willhave the same effectasan equal forceactingaway from the point so longas both forces act in the same directionandin the same straight line Thus a force of10Newtons N pushing tothe right ona certainpoint can be substitutedfor aforce of10Newtons N pulling to the rightfrom the same point a Resolving two forces which act in thesame straight lineIfbothforcesact in thesamestraightlineandin thesame direction theresultant is theirsum but if the forcesact inopposite directionstheresultantis the difference of the two forces and acts inthe direction of thelarger of the two forces Example1Whilst moving an objectone manpulls on it with aforce of200Newtons andanother pushesinthesamedirectionwith aforce of300Newtons Find theresultantforce propellingthe object Component forces300N A200NThe resultantforce isobviously500Newtons the sum of the twoforces andacts inthe directionof each of the ponent forces Resultant force500N Aor A500NExample2A force of5Newtons is applied towards a pointwhilst aforce of2Newtons isappliedat thesame pointbut inthe oppositedirection Find the resultantforce Component forces5N A2NSince theforces areapplied inopposite directions the magnitudeof theresultantis the differenceof the twoforces and actsinthedirectionof the5Nforce Resultant force3N Aor A3N b Resolving twoforces whichdo not act inthesamestraight lineWhen thetwoforces donotact inthesamestraightline their resultantcanbe foundby pletinga parallelogramof forces Example1A force of3Newtons and aforce of5N acttowardsa pointatan angle of120degrees to each other Find thedirection andmagnitudeof the resultant Ans Resultant4 36N at36 34120to the5N force Note Notice thateachof the ponentforcesand the resultantall acttowards the pointA 2Ship Stabilityfor Mastersand MatesFig 1 1 E EEEEE Example2A shipsteams dueeast foran hourat9knots through a currentwhich sets120degrees T at3knots Find thecourse and distance madegood The ship s forcewould propelher fromA toB inone hourand thecurrentwould propelher fromA toC inone hour The resultantis AD 0 9712 11 6miles andthis willrepresent thecourse anddistance madegood inone hour Note In theabove exampleboth of theponentforcesand the resultantforceall actaway from the pointA Example3A force of3N actsdownwards towardsa pointwhilst anotherforce of5N actsaway from the point to therightas shown in Figure1 3 Find the resultant In thisexample one forceisactingtowards the pointand thesecond forceisacting away from the point Before pletingthe parallelogram substituteeither aforce of3N actingaway from the pointfortheforce of3N towardsthe point as shown in Figure1 4 oraforce of5N towards the pointfor theForces andmoments3Fig 1 2Fig 1 3Fig 1 4forceof5N away from the point asshown in Figure1 5 In thisway bothofthe forcesact eithertowards orawayfrom the point The magnitudeanddirection of the resultantis thesame whicheversubstitution ismade i e 5 83Nat anangle of59 to thevertical c Resolving twoforces whichact inparallel directionsWhen twoforcesactinparallel directions their bined effectcan berepresented byoneforcewhose magnitudeis equalto thealgebraic sum ofthe twoponentforces and whichwill actthrough a point aboutwhichtheir momentsare equal The followingtwo examplesmay helpto makethis clear Example1In Figure1 6the parallelforces Wand Pareactingupwards throughA andBrespectively Let Wbe greater than P Their resultant W P acts upwardsthroughthe pointC such that P y W x Since W is greaterthan P thepoint Cwill benearer toB thanto A Example2In Figure1 7the parallelforces Wand Pactinopposite directionsthrough AandB respectively If Wis againgreaterthanP their resultant W P actsthrough pointC onAB producedsuch thatP y W x 4Ship Stabilityfor Mastersand MatesFig 1 5Fig 1 6Fig 1 7Moments ofForcesThe moment of aforceisa measure of theturning effect of theforce aboutapoint The turningeffect willdepend upon the following a The magnitudeof theforce and b The length of the lever uponwhich theforce acts the leverbeing theperpendiculardistance betweenthe lineof actionof theforce and thepoint aboutwhich the moment is being taken The magnitudeof themoment is the productof theforce and the lengthof thelever Thus iftheforceis measured in Newtons and the length of theleverin metres themomentfound will be expressed inNewton metres Nm Resultant moment Whentwoor moreforces areacting abouta pointtheirbinedeffectcanberepresented byone imaginarymoment calledthe Resultant Moment The processoffindingthe resultant moment isreferredto as the Resolution of the ComponentMoments Resolution ofmoments To calculatethe resultant moment abouta point find thesumof the momentsto producerotation ina clockwise directionabout the point and thesumof the momentsto producerotation in ananti clockwise direction Take thelesser of these twomoments from thegreater and thedifferencewillbe the magnitudeof the resultant Thedirection inwhich itacts willbe thatof thegreater of thetwoponentmoments Example1A capstanconsists of a drum2metres indiameter aroundwhicharope iswound and fourlevers atright anglestoeachother each being2metres long If aman on the end of eachlever pusheswith aforceof500Newtons whatstrain isput on the rope See Figure1 8 a Moments aretaken aboutO the centre of thedrum Total momentin ananti clockwisedirection 4 2 500 NmThe resultant moment 4000Nm Anti clockwise Let thestrain on the rope P NewtonsThemoment aboutO P 1 Nm P 1 4000or P 4000NAns The strainis4000N Note For a body toremain atrest theresultantforce actingon the body mustbe zero and theresultant moment aboutits centre of gravitymust also bezero ifthe centre of gravity beconsidered afixed point Forcesandmoments5MassIn theS I system ofunits it is mostimportant todistinguish betweenthemass of a bodyand itsweight Mass is the fundamentalmeasureof thequantity ofmatter ina bodyand isexpressed in terms of the kilogramandthe tonne whilst the weight of a body is theforce exertedonitby theEarth s gravitationalforce and ismeasuredinterms of theNewton N andkilo Newton kN Weight andmass areconnected bythe formula Weight Mass AelerationExample2Find the weight of a bodyof mass50kilograms ata placewhere theaelerationdueto gravity is9 81metres persecondpersecond Weight Mass Aeleration 50 9 81Ans Weight 490 5NMoments ofMassIf theforceof gravity is considered constantthen the weight ofbodies isproportionalto their mass and theresultantmoment oftwoormoreweights abouta pointcanbeexpressedintermsof theirmassmoments Example3A uniformplank is3metres longand issupported ata pointunder itsmid length A loadhaving amass of10kilograms is placed ata distance of0 56Ship Stabilityfor MastersandMates P NFig 1 8 a metres from one end and asecond loadof mass30kilograms is placed atadistance ofone metre from the other end Find theresultantmomentabout themiddleof the plank Moments aretaken aboutO the middleof the plank Clockwise moment 30 0 5 15kgmAnti clockwise moment 10 1 10kgmResultant moment 15 10Ans Resultant moment 5kgm clockwiseForcesandmoments7Fig 1 8 b 8Ship Stabilityfor Mastersand MatesExercise11A capstanbar is3metres long Two menare pushingon the bar each withaforceof400Newtons If oneman is placed half way alongthebarand theotherat theextreme endof thebar find theresultantmomentabout thecentre of thecapstan 2A uniformplank is6metres longand issupported ata pointunder itsmid length A10kg mass is placedon the plank ata distance of0 5metres fromone endanda20kg mass is placedonthe plank2metres from the otherend Find theresultantmomentabout the centre of theplank 3A uniformplank is5metres longand issupported atapointunder itsmid length A15kg massis placed1metrefrom oneendanda10kg massisplaced1 2metres from theother end Find wherea13kg massmust beplacedontheplank so that theplank willnot tilt 4A weightlessbar2metres longis suspended fromtheceiling ata pointwhichis0 5metres infrom oneend Suspended fromthesameend isamass of110kg Find the mass whichmust besuspended fromapoint0 3metres infromtheotherendof thebar sothat thebar willremainhorizontal 5Three weightsare placedonaplank One of15kg massisplaced0 6metres infromoneend the nextof12kg massisplaced1 5metres infromthe sameend and thelast of18kg massisplaced3metres fromthis end Ifthe mass of theplank beignored find theresultantmomentabout theendof theplank Chapter2Centroids andthecentre of gravityThe centroidof anarea issituated atits geometricalcentre In eachof thefollowingfigures G represents thecentroid and ifeach areawassuspended fromthis pointit wouldbalance The centre of gravity of a body is thepointat whichall the mass of thebody may be assumedto beconcentrated and is thepoint through whichthe forceof gravity isconsidered to act vertically downwards with aforceequal to the weight of the body It isalso thepoint aboutwhich thebodywould balance Fig 2 1The centre of gravity of a homogeneous bodyis atits geometricalcentre Thus the centre of gravity of ahomogeneousrectangular blockis half wayalong its length half way acrossits breadthand athalf itsdepth Let usnow consider theeffectonthe centre of gravity of a bodywhenthe distributionof mass within thebodyischanged Effect ofremoving ordischarging massConsidera rectangularplank ofhomogeneous wood Its centre of gravitywillbe atits geometricalcentre that is half way alongitslength half wayacross itsbreadth and athalf depth Let the mass of theplankbe Wkgandlet itbe supportedby means of awedge placedunder the centre of gravityas shown in Figure2 2 The plankwill balance Now leta shortlength of theplank of mass wkg be cutfromoneendsuch thatits centre of gravityis d metres fromthe centre of gravity of theplank The otherend now beingof greatermass will tilt downwards Figure2 3 a shows thatby removingthe shortlengthof plank aresultantmoment of w dkgm has been createdinananti clockwise directionabout G Now considerthe newlengthofplank asshown inFigure2 3 b Thecentre of gravity willhave movedto the new half length indicatedby thedistanceG to G1 The new mass W w kg now producesa tiltingmomentof W w GG1kgm aboutG 10Ship Stabilityfor Mastersand MatesFig 2 2Fig 2 3 a Fig 2 3 b Since theseare simplytwo differentways ofshowing thesame effect themoments mustbethesame i e W w GG1 w dorGG1 w dW wmetresFrom thisit may be concluded that when massisremoved fromabody the centreof gravity of thebody will move directly awayfromthe centreof gravity of the massremoved andthe distance it moves willbe given bythe formula GG1 w dFinal massmetreswhereGG1is the shift of the centreof gravity of thebody w is the massremoved anddis the distance betweenthe centreof gravity ofthe massremoved andthe centreof gravity ofthebody Application toshipsIn eachoftheabove figures G represents the centreof gravityofthe shipwith amass of w tonneson boardata distanceofdmetres from G G to G1represents theshift ofthe ship s centreof gravityduetodischarging themass In Figure2 4 a it willbe noticedthat the massis vertically belowG andthat whendischarged G will move vertically upwards to G1 Centroids andthe centreof gravity11Fig 2 4 Discharging amass w In Figure2 4 b themassisverticallyabove G andthe ship s centre ofgravity will move directlydownwards to G1 In Figure2 4 c themassis directlyto starboardof Gandthe ship scentre of gravity will move directlyto portfrom G to G1 In Figure2 4 d themassis belowand toport ofG andthe ship s centreof gravity will move upwardsand tostarboard In eachcase GG1 w dFinal displacementmetresEffect of addingor loadingmassOnce againconsidertheplank ofhomogeneous woodshowninFigure2 2 Now adda pieceofplankof masswkg ata distanceofdmetres from G asshowninFigure2 5 a The heavierendoftheplankwill againtiltdownwards By addinga massofwkg ata distanceofdmetresfrom Ga tilting momentofw dkgm aboutGhas beencreated Now considerthenewplank asshowninFigure2 5 b Its centre ofgravity willbe atits newhalf length G1 andthenewmass W w kg will produceatiltingmomentof W w GG1kgm aboutG These tiltingmoments mustagain beequal i e W w GG1 w dorGG1 w dW wmetresFrom theabove itmaybeconcludedthatwhenmassis addedto abody the centreof gravityofthebody willmove directlytowardsthe centreof12Ship Stabilityfor Mastersand MatesFig 2 5 a Fig 2 5 b gravityofthemassadded andthe distance whichitmoveswillbegiven bytheformula GG1 w dFinal massmetreswhereGG1is theshift ofthe centreof gravityofthebody w is the massadded anddisthe distance betweenthe centresof gravity Application toshipsIn eachoftheabove figures G representsthe position ofthe centre ofgravityofthe ship beforethemass ofwtonnes hasbeen loaded After themasshasbeenloaded Gwillmove directlytowardsthe centreof gravity oftheadded mass i e from Gto G1 Also in eachcase GG1 w dFinal displacementmetresEffectof shiftingweightsIn Figure2 7 G representsthe originalposition ofthecentreof gravityof ashipwitha weight of w tonnes inthe starboard side ofthe lower holdhaving its centreof gravity inposition g1 If this weight isnow dischargedthe ship s centreof gravity willmove from Gto G1directly awayfrom g1 When thesame weight is reloadedon deckwith its centreof gravity at g2the ship s centreof gravitywillmove from G1to G2 Centroids andthecentreof gravity13Fig 2 6 Adding amassw From thisit canbe seenthat ifthe weighthad beenshifted from g1to g2the ship s centreof gravitywould havemoved from Gto G2 It canalsobeshown thatGG2is parallel to g1g2and thatGG2 w dWmetreswhere wisthemassofthe weightshifted disthedistancethrough whichitis shifted and Wisthe ship s displacement The centreof gravityofthebody willalways move parallel to theshift ofthe centreof gravityof any weight moved within thebody Effectofsuspended weightsThe centreof gravityofabodyisthepoint throughwhich theforce ofgravitymaybeconsidered toact verticallydownwards Consider the centreof gravityofa weight suspendedfromthehead ofa derrickasshowninFigure2 8 It canbe seenfrom Figure2 8that whether the shipis uprightor inclinedineither direction thepointinthe ship throughwhichtheforceof gravitymay beconsidered toactverticallydownwards isg1 thepoint ofsuspension Thus thecentreof gravityofa suspendedweight isconsideredto be at thepoint of suspension Conclusions1 The centreof gravityofabod