oe4680 2015 - lecture 15 - discussion exercise & exam questions

MScOffshore&DredgingEngineeringFacultyCEG,DepartmentHydraulicEngineeringFaculty3mE,DepartmentMaritime&TransportTechnology

116June,2015

DiscussionExercise&ExamQuestionsOE4680ArcticEngineeringMScOffshore&DredgingEngineering

BonusExerciseSummaryAim:

Determinetheglobaliceloadsfor2differentsubstructuredesignsofaGBSintheKaraSea

Stepstobetaken: Reviewoficeconditionsandproperties; Reviewoflimitingmechanisms; Crushingversusbending; Reviewofthestructuralconfiguration;Forthisexercise,youneed: Theexercisehandout; Thematerialgiventoyouduringthelectures; ExcerptfromISO19906(onBlackboard)

16June,2015 2

Figure1:Mapof1)Southwestand2)NortheastKaraSea

Rusanovskoye


BonusExerciseSummaryNoteherethat: Rusanovskoye islocatedinBaidaratskaya Bay

inthesouthwesternpartoftheKaraSea ThelocationoftheGBSisdefinitelyoffshore

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Figure1:Mapof1)Southwestand2)NortheastKaraSea

Rusanovskoye

Parameter AverageAnnualValuesOccurrence,firstice October

Occurrence,lastice July

FYlevellandfastice,thickness 1,6m

FYlevelicefloes,thickness 1,4 1,8m

FYlevelicefloes,eq.diameter 4,5km

MYlevelicefloes,thickness

FYridges,keeldepth 6,5 7,5m

MYridges,keeldepth

Icemovement,nearshore 0,4m/s

Icemovement,offshore 0,3m/s


Structuralconfiguration

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Figure2:SketchofPlatformDesignOptionsfortheKaraSeaGBS


Givenparameters

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Parameters Symbol Value UnitAirDensity(at19C) a 1,37 kg/m3

Windvelocity Va 21 m/sDensityseawater w 1027 kg/m3

Currentvelocity Vw 3,4 m/sSalinityseawater S 31 Densityseaice ice 910 kg/m3

Youngsmodulusseaice E 5,0 GPaPoisson'sratio 0,3 Rubbleheight hr 5 mIcetoicefrictioncoefficient i 0,05 Porosityicerubble e 0,35 Rubbleangleofrepose A10 Cohesionoficerubble c 1,7 kPaInternalfrictionangleicerubble 40

VariableParameters Symbol Unit Value Value ValueExtremeicefloethickness H m H1=1,6 H2=1,9 H3=2,2Icestructurefrictioncoefficient C C1=0,01 C2=0,02 C3=0,03

A1=40 A4=49 A7=58Coneangle A A2=43 A5=52 A8=61

A3=46 A6=55 A9=64


OverviewExerciseScoring Answeringallproblemscorrectly,yieldsa0,75 bonustoyourexamgrade. Thisbonusisvalidfor:

TheexaminQ4,onWednesday24June2015,09:0012:00,and TheretakeinQ5,onThursday13August2015,09:0012:00. Ergo:doingtheexamnextyear,meansredoingthebonusassignment!

Intotal,therewere32pointstobeearned,dividedamongthe4problemsas:1a. [2] 2a. [2] 3a. [1] 4a. [1]1b. [2] 2b. [2] 3b. [4] 4b. [8]

2c. [3] 3c. [1] 4c. [3]2d. [1]

1. [4] 2. [8] 3. [6] 4. [12]

Thus,everysinglepointequalsa(0,75/30=)0,025bonustoyourexam.16June,2015 6


ProblemStatement1[4pts]Inthewinter20142015,themeandailyairtemperatureattheconsideredlocationintheKaraSeawasbelowtheseawaterfreezingpointfrom16September2014untilandincluding18April2015.Theaveragemeandailyairtemperatureduringthisperiodwas17,1C.

Forthecalculationoficethickness,theKaraandChukchiSeashavethesamesitespecificconstants:intheChukchiSea4096freezingdegreedaysyieldanicethicknessof2,24m.

1. Forthegivenweatherconditionsinthewinter20142015,a. Calculatethenumberofaccumulatedfreezingdegreedaysin

theKaraSea.b. DeterminethemaximumundisturbedicethicknessintheKaraSea

assuminglinearheatconduction.

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Problem1aForthegivenconditionsinthewinter20142015:a. calculatethenumberofaccumulatedfreezingdegreedaysintheKaraSea.

Thenumberofaccumulateddegreedaysisfoundas:

meandailyairtemperatureattheconsideredlocationintheKaraSeawasbelowtheseawaterfreezingpointfrom16September2014untilandincluding18April2015.Theperiodfrom16September2014until18April2015yieldsatotalof:

15+31+30+31+31+28+31+18=215days.

Theaveragemeandailyairtemperatureduringthisperiodwas17,1C.Ta isgivenas17,1C.Theseawatersalinityis31,andthus(fromthelectureslideswefindthat)thefreezingpointoftheseawaterTbis1,705C.

Andthus:

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a b a bFDD daysC T T avg T T n

17,1 1,705 215 3310FDDC


Problem1bForthegivenweatherconditionsinthewinter20112012:b. DeterminethemaximumundisturbedicethicknessintheChukchiSea

assuminglinearheatconduction.

Themaximumundisturbedicethicknessisfoundas:

Forthecalculationoficethickness,theKaraandChukchiSeashavethesamesitespecificconstants; intheChukchiSea4096freezingdegreedaysyieldanicethicknessof2,24m.Assuminglinearheatconductionitfollowsthatb=0,5,fromthedatafortheBeaufortSea,wethusfind:

WepreviouslyfoundthatCFDD =3310,thuswefindtheicethicknessfortheChukchiSeaas:

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bFDDh aC

;;

2,240,035

4096b Chukchi

Chukchi FDD Chukchi bFDD Chukchi

hh aC aC

0,035 3310 2,014bFDDh aC m


Insteadoftheundisturbedicethickness,extremeicefloethicknessesshouldbeusedforthedesignloads.Inotherwords,fromthispointonwardsusetheextremeicefloethicknessvaluespecificallygiventoyourgroup.

AssumethatforextremeicefloethicknessesintheKaraSea,theicetemperatureatthefloesurfaceis19C.

2. ForsubstructureA,thusforthesubstructurethatiscylindrical atthewaterline,a. Determinetheiceactionforanaveragesizedisolatedicefloeforlimitforce;b. CalculatethedesignactionforicecrushingfailureaccordingtoISO19906;

ProblemStatement2[8pts]

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SubstructureA

50m24m

18m


ForsubstructureA,thusforthesubstructurethatiscylindrical atthewaterline,a. Determinetheiceactionforanaveragesizedisolatedicefloeforlimitforce;Foranisolatedicefloe,wheretheicefloeisdescribedasanequivalentcircularfloewithadiameterDeq,wehavenoiceiceinteractionandthethermalexpansiondoesnotleadtoadditionalforces.Thus,thelimitforceactioncanbedescribedas:

Fromthelectureonicemechanics,wefindthatCd,a =0,025andCd,w =0,002. Andfromtable1:a =1,37kg/m3,Va =21m/s,w =1027kg/m3,Vw =3,4m/s. Additionally,table1givesanaverageannualvalueforthediameterofaFYlevel

icefloeasDeq =4,5km.

Thus: 2 22 3 2 3, 8 8,

0,025 1,37 21 4,5 10 0,002 1027 3,4 4,5 10

120,11 188,82 308,9LF floe

LF floe

F

F MN

Problem2a

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2 2 2 2 2, , ,4 8 8floe eq LF floe d a a a eq d w w w eqA D F C V D C V D

Isthisconservativeornot?


Problem2bb. CalculatethedesignactionforicecrushingfailureaccordingtoISO19906;

AccordingtoISO19906,wefindtheicecrushingloadthroughtheglobalicepressureduetocrushing(eqs.A.820andA.821)as:

TheicestrengthcoefficientforArcticareasisequalto:CR =2,8MPa. mandnareempiricalcoefficientsthatdependontheicethickness,butforthe

possiblethicknessesarealwaysfoundas:m=0,16,n=0,3. Thewidthofthestructureis:w=24mandh1isaunitvariable:h1=1m.

Withh=H2=1,9m,wefind:

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1 1

, n nm m

G G G R G Rh w h wF p hw p C F C hwh h h h

0,160,31,9 242,8 1,9 24 70,2 1,539

1 1,9G GF MN p MPa


Insteadoftheundisturbedicethickness,extremeicefloethicknessesshouldbeusedforthedesignloads.Inotherwords,fromthispointonwardsusetheextremeicefloethicknessvaluespecificallygiventoyourgroup.

AssumethatforextremeicefloethicknessesintheKaraSea,theicetemperatureatthefloesurfaceis19C.

2. ForsubstructureA,thusforthesubstructurethatiscylindrical atthewaterline,c. Determinethepenetrationofthestructureintoanaveragesizedisolatedicefloefor

alimitenergyeventanddeterminethecorrespondinglimitenergyiceaction;d. Concludewhichlimitingmechanismgovernstheiceactionandexplainwhy.

ProblemStatement2[8pts]

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SubstructureA

50m24m

18m


Problem2c(1)c. Determinethepenetrationofthestructureintoanaveragesizedisolatedice

floeforalimitenergyeventanddeterminethecorrespondinglimitenergyiceaction;

Theiceactionforlimitenergyisdeterminedfromtheworkenergyprincipal:

Anaveragesizedicefloehasadiameterof4,5km.Thus,withice=910kg/m3 andh=H2=1,9m,themassoftheicefloebecomes:

Thestructureisoffshore,sofromtable1:vbeg =0,3m/s.Obviously,vend =0m/s.

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2 21 12 2beg endF x dx p x w x h x dx mv mv

22 34 4910 4,5 10 1,9 27,5ice eqm D h Gkg Somehavechosenvbeg =0,4m/sandclaimedthistobeconservative.Isthiscorrect?


IceFloe

Recap RoadmapforLimitEnergyIceAction

Fromtheworkenergyprinciplewefound:

Astheanalyticalintegrationoverxisnotalwayseasytoapply,theiceactionforlimitenergycanbeapproximatedusingthefollowingroadmap:1. Increasethepenetrationx usingsmallincrementsx,2. Ateachpenetrationxi determineicepressure,interactionwidthand

icethickness,aswellasthecorrespondingiceaction.3. Perincrementassumethattheicepressure,interactionwidthandice

thicknessareconstantsothatthevelocityattheendofeachincrementcanbedeterminedusing:.

4. Repeatuntil,theiceactionforlimitenergyis:.16June,2015 15

RigidStructure 2 21 12 2beg endF x dx p x w x h x dx mv mv

2 21 1 12 2i i i i ipw h x mv mv 1 0iv ,maxLE iF F


Problem2c(2)c. Determinethepenetrationofthestructureintoanaveragesizedisolatedice



Anaveragesizedicefloehasadiameterof4,5km.Thus,withice=910kg/m3 andh=H2=1,9m,themassoftheicefloebecomes:

Thestructureisoffshore,sofromtable1:vbeg =0,3m/s.Obviously,vend =0m/s. Usingtheroadmap:

Forh=H1=1,6m,wefindtheexactpenetrationas:19,48m.Forh=H2=1,9m,wefindtheexactpenetrationas:19,90m.Forh=H3=2,2m,wefindtheexactpenetrationas:20,27m.

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22 34 4910 4,5 10 1,9 27,5ice eqm D h Gkg


Problem2c(3) FastApproximationc. Determinethepenetrationofthestructureintoanaveragesizedisolatedice



Anaveragesizedicefloehasadiameterof4,5km.Thus,withice=910kg/m3andh=H2=1,9m,themassoftheicefloewasfoundas27,5Gkg.

Againfromtable1wehave:vbeg =0,3m/s.Obviously,vend =0m/s. Now,asafirstestimateletusassumethatF(x)=FG isconstant.Thepenetrationmaythenbeapproximatedas:

Solvingtheproblemusingtheroadmap,yieldedapenetrationof19,90m.

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2 9 2212 627,5 10 0,3 17,632 2 70,2 10begG beg Gmv

F x dx F x mv x mF


Problems2c(4)and2dc. Determinethepenetrationofthestructureintoanaveragesizedisolatedicefloe

foralimitenergyeventanddeterminethecorresponding limitenergyiceaction; Usingtheroadmap,wefoundanexactpenetrationof19,90m,

whileafastapproximationyieldedapenetrationof17,63m. Withadiameterof24m,thestructureisfullyenveloped atapenetrationof12m. Thelimitenergyiceactionisthusequaltothelimitstressload:

d. Concludewhichlimitingmechanismgovernstheiceactionandexplainwhy. Fromquestionsa.andb.wefind:FLS


ProblemStatement33. UsingtheISO19906provisions:

a. Calculatetheaverageicesalinityfordesignconditions;b. Calculatethecorrespondingbrinevolumeandtotalporosity;c. Determinetheflexuralstrengthoftheiceforpreliminarydesign.

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Problem3aUsingtheISO19906provisions:a. Calculatetheapproximateicesalinityfordesignconditions;TheAVERAGEicesalinityofagrowingfirstyearlevelicesheetisfoundaccordingtotheISO19906provisionsbythefollowingequation:

Clearlyallgivenextremeicethicknessesare>0,34mandthus,substitutingthepossiblevaluesgivesasalinity(inppt)as:

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13,4 17,4 for 0,348,0 1,62 for 0,34

h h mS

h h m

1 1,6: 5,4088,0 1,62 2 1,9: 4,922

3 2,2: 4,436

H SS h H S

H S


Problem3b(1)UsingtheISO19906provisions:b. Calculatethecorrespondingbrinevolumeandtotalporosity;

ThecorrespondingbrinevolumefollowsfromISO19906as:

Forthecalculationofthebrinevolumethatcorrespondstotheaverage salinitySoveranicesheet,wemustalsousetheaverage temperatureovertheicesheet.

AssumethatforextremeicefloethicknessesintheKaraSea,theicetemperatureatthefloesurfaceis19C.Inthelectureonicemechanics,itwasexplainedthatthetemperatureinanicefloebyapproximationchangeslinearlyovertheheight: atthefloesurfacethetemperatureisgivenas19C. Lookingattheheatfluxthroughtheice,thetemperatureatthebottomofthe

icesheetmustbeequaltothefreezingpoint,i.e.1,705C.

Andthuswefindthat:

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49,18 0,53brineV S T

19 1,705 2 10,3525avgT T C


Problem3b(2)UsingtheISO19906provisions:b. Calculatethecorrespondingbrinevolumeandtotalporosity;SubstitutingtheaveragesalinitySandtheaveragetemperatureTyields:

Accordingtotheicemechanicslecture,theairvolume(inppt)maybeapproximated as:

Andthustheporosityisfound(inppt)as:

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1: 28,5649,18 0,53 5,28 2: 25,99

10,35253: 23,42

brine

brine brine

brine

H VV S S H V

H V

9101 1 1 8,76

918,05bulk sea ice

airparticles pure ice i

VT

1: 28,56 8,76 37,322: 25,99 8,76 34,753: 23,42 8,76 32,18

brine air

HV V H

H

916,7 0,13pure ice i iT T


Problem3cUsingtheISO19906provisions:c. Determinetheflexuralstrengthoftheiceforpreliminarydesign;

TheflexuralstrengthoftheiceisdefinedinISO19906as:

Here,thebrinevolumeshouldbesubstitutedasthebrinevolumefraction,thusabrinevolumeof34,75(inppt)correspondstoabrinevolumefraction0,03475.

Thus,theflexuralstrengthscorrespondingtothedifferenticethicknessesare:

NotethatthesevaluesarehigherthanwhatISO19906notesreasonable!

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5,881,76 bVf MP ea

5,88

1: 0,6521,76 2: 0,682

3: 0,716

b

fV

f f

f

H MPae H MPa

H MPa


Withtheaimtoreducetheiceaction,thestructureisredesignedandequippedwithaconeatthewaterline.

4. ForsubstructureB,thusforthesubstructurethatisconical atthewaterline,andforthespecificcombinationofparametersspecifiedforyourgroup,a. Determinethediameteroftheconeatthestillwaterlevel;b. Calculatethetotalhorizontal andverticaldesignloadforbendingfailure

accordingtoISO19906(wehereactuallyassumethatthecodeprovisionsforslopedsurfacesalsoholdforconicalcollars);

c. Determinethereductionofthetotalhorizontaldesignloadbyapplyingadownwardconewiththesamewaterlinediameterastheupwardcone.

ProblemStatement4.[12pts.]

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SubstructureB

50m

A

18m6m24m


Problem4aForsubstructureB,thusforthesubstructurethatisconical atthewaterline,andforthespecificcombinationofparametersspecifiedforyourgroup,a. Determinethediameteroftheconeatthestillwaterlevel;Thetopoftheconeislocated6mabovethewaterline(MSL)asshowninFigure2,andthediameterofthetopoftheconeis24m.Thediameteroftheconeatthewaterlineisthusfoundas:

Consequently,thediameteroftheconeatthestillwaterlevelisfoundas:

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,

1: 7,15 4 : 5,22 7: 3,756

2: 6,43 5: 4,69 8: 3,33tan tan

3: 5,79 6: 4,20 9: 2,93

add add addtop cone

add add add add

add add add

A r A r A rh

r A r A r A rA A

A r A r A r

bot,

1: 38,30 4 : 34,43 7: 31,502 2: 36,87 5: 33,38 8: 30,65

3: 35,59 6: 32,40 9: 29,85c c add

A w A w A ww b r A w A w A w

A w A w A w


Problem4b(1)b. Calculatethetotalhorizontal andverticaldesignloadforbendingfailure

accordingtoISO19906;

ThetotalhorizontaldesignloadforbendingfailureisfoundaccordingtoISO19906as:

IngeneralthebreakingcomponentHB isthemaincomponent,whichisfoundas:

Here,theflexuralstrengthf istheresultofquestion3c.16June,2015 26

1

B P R L TH

B

f c

H H H H HF Hh

: Loadrequiredtobreaktheiceblocksagainsttheslope: Loadrequiredtopushtheiceblocksuptheslope: Loadrequiredtoturntheiceblockatthetopoftheslope: Loadrequiredtopushthe

B

R

T

P

HHHH sheeticethroughtherubble

: LoadrequiredtolifttheicerubblewiththeunbrokenicefloeLH

0,25

2 3

0,255 2with:

4 12 10,68sin coscos sin

C C Cw w

B f C

Ehw L Lgh g vHE




Wefind:

Where:

Thisyields:

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0,25

2 3

0,255 2with:

4 12 10,68sin coscos sin

C C Cw w

B f C

Ehw L Lgh g vHE

sin cos sin 5 2cos 5: 1,334cos sin cos 5 2sin 5

A C AA C A

C2 = 0.02,A5 = 52

0,25

9 3

2

1: 20,77 89,555 10

2: 23,63 91,6812 1027 9,81 1 0,3

3: 26,38 94,93

C C

C C C

C C

H L m mHL H L m m

H L m m

0,2551, 1, 1: 2,30

0,68 2, 2, 5: 4,773, 3, 9: 10,34

Bw

B f C B

B

H C A H MNghH H C A H MNE

H C A H MN





Theremainingloadcomponentscanbewrittenas:

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1

B P R L TH

B

f c

H H H H HFH

h


B

R

T

P



22 tan 11 1

tan 2tan

sin cos tan sin cos0,5 1 cos 1cos sin tan tan tan sin

tan 1 1 tan1 0,5 1 tan 1tan tan tan t

P r i i

i rR i r i

L r r i

H wh g e

w ghH e h h

H wh h g e

2

ancos1,5 0

sin cosT i

c

H wh g



accordingtoISO19906;ThehorizontalbreakingloadHB ,aswellastheothercomponentsarenowfoundbysimplysubstitutingthecalculatedvaluesintotheISO19906equations.

ForexampleusingH2,C2&A5,wefind:(i.e.h=1,9m; =0,02; =52)

Andthetotalhorizontalandverticalforcesbecomes:

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4,77 0,01 4,83 0,67 1,28 12,04 9,034,771 10,682 91,68 1,9

B P R L T HH V

B

f c

H H H H H FF MN F MNHh

4,7711,84,830,671,28

B

P

R

L

T

H MNH kNH MNH MNH MN


UpwardBendingHorizontalActionFH(angle)

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6,0

10,0

14,0

18,0

22,0

20 22 24 26 28 30 32 34 36 38 40 42 44 46 48 50 52 54 56 58 60 62 64

H1.6C0.01H1.6C0.02H1.6C0.03H1.9C0.01H1.9C0.02H1.9C0.03H2.2C0.01H2.2C0.02H2.2C0.03

angle

MN


Problem4c(1)c. Determinethereductionofthetotalhorizontaldesignloadbyapplyinga

downwardconewiththesamewaterlinediameterastheupwardcone.

Upwardconespushtheiceup,whereasdownwardconespushtheicedown.

Thus,onanupwardcone:1. Therewillberubblepileup,asaconsequence

a. operationsarenegativelyinfluencedbyrubblegettingintheway,andb. iceactionsincreaseastheicerubbleblockstheslope.

2. Lowericeactionsduetoalowerflexuralstrengthoftheice3. Highericeactionsaswehavetotakeintoaccounttheweightofice,

insteadofitsbuoyancyonadownwardslope.4. Highericeactionsasthefrictioncoefficientishigheronanupwardslope;

downwardslopesarelubricatedduetowaterinbetweentheslopeandtheice.

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Problem4c(2)Onanupwardcone:1. Therewillberubblepileup,asaconsequence

a. operationsarenegativelyinfluencedbyrubblegettingintheway,andb. iceactionsincreaseastheicerubbleblockstheslope.

2. Lowericeactionsduetoalowerflexuralstrengthoftheice3. Highericeactionsaswehavetotakeintoaccounttheweightofice,

insteadofitsbuoyancyonadownwardslope.4. Highericeactionsasthefrictioncoefficientishigheronanupwardslope;

downwardslopesarelubricatedduetowaterinbetweentheslopeandtheice.

InISO19906,theflexuralstrengthiscalculatedasanaverageflexuralstrengthandthereforecannotdistinguishbetweenup/downwardbending!

Furthermore,ISO19906considersonlyoneicestructurefrictioncoefficient,anddoesnotdistinguishbetweenfrictionaboveandunderwater!

16June,2015 32


Problem4c(3)UpwardversusDownwardbending:

Theremainingdifferencesbetweenupwardanddownwardbending: Theweightoftheiceonanupwardslopeshouldbereplacedbyitsbuoyancy

foradownwardslope,i.e.theicedensityshouldbereplacedbythesubmergeddensity.(i wi ;alsoseethelecturesonIceActions)

Differenceiceactionsduetodryandwetrubblepileup

So: Forupwardanddownwardbendingthehorizontalbreakingload HB isthesame. Forupwardbending,wecalculateHP,HR,andHT withtheicedensityi. Fordownwardbending,HT =0,butallothercomponentsthatareafunctionof

theicedensity,i.e.HP,HR,HLarecalculatedwiththesubmergeddensitywi.

16June,2015 33



downwardconewiththesamewaterlinediameterastheupwardcone.


Theremainingloadcomponentscanbewrittenas:

16June,2015 34

1

B P R L TH

B

f c

H H H H HFH

h


B

R

T

P



22 tan 11 1

tan 2tan

sin cos tan sin cos0,5 1 cos 1cos sin tan tan tan sin

tan 1 1 tan1 0,5 1 tan 1tan tan tan t

P r i i

i rR i r i

L r r i

H wh g e

w ghH e h h

H wh h g e

2

ancos1,5 0

sin cosT i

c

H wh g

3with 117i w i kg m

NotethatthecomponentHT maybeneglectedcompletely,asasubmergediceblockwillturnbeforetheendoftheslope.



downwardconewiththesamewaterlinediameterastheupwardcone.ThehorizontalbreakingloadHB remainsthesameandtheothercomponentsarefoundfromtheISO19906equations.

ForexampleusingH2,C2&A5,wefind:(i.e.h=1,9m; =0,02; =52)

Andthetotalhorizontalandverticalforcesbecomes:

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,4,77 0,00 0,62 0,18 1,28 5,81 6,244,771 1

0,682 91,68 1,9

B P R L TH H red

B

f c

H H H H HF MN F MNHh

4,771,520,620,180,00

B

P

R

L

T


4,7711,84,830,671,28

B

P

R

L

T


12,04HF MN


DownwardBendingHor.ActionFH(angle)

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2,0

3,0

4,0

5,0

6,0

7,0

8,0

9,0

10,0

11,0

12,0

10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48 50 52 54 56 58 60 62 64

H1.6C0.01H1.6C0.02H1.6C0.03H1.9C0.01H1.9C0.02H1.9C0.03H2.2C0.01H2.2C0.02H2.2C0.03

MN

angle


Commonerrors Insertingthewrongunitsintheequations,forexample:

cohesionc=1,7kPa,sointheequations,thevalueofc=1700,not1,7.

Forexample,inexcel,trigonometricfunctionsarecalculatedusinganglesinradians,notdegrees.

Calculatetheaveragetemperatureas:

Usetherightequation,usingtherightvaluesforallofthevariablesandthensomehowmessupthecalculationandendupwiththewronganswer.

Usethenearshoreicevelocityof0,4m/s,insteadoftheoffshoreicefloevelocityof0,3m/sforastructurethatisapproximately200kmoffshore.

Forgettingpartsofequations

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21 1,595 9,72avg

T C 21 1,595 11,3

2avgT T C


Generalremarks 1groupmanagedtogettheirgroupnumberwrong,andthereforeused

differentparametersthangiven.

2groupsmanagedtonotregisteratallandusedselfchosenparameters.(butdidnotdoatallbad!)

Multiplegroupsforgottoincludeasinglepageoverviewoftheanswers Quiteafewmanagedtopresent,forexample,aresultingiceforceof

22658254,86NPleasewritethisas:22,66MN(!)

Pleaseproperlyincludethecorrectunits,sometimesIcantevendistinguishwhetherIamlookingatN,kN orMN!

1groupmanagedtoproperlycopyallthecorrectequations,butwithoutgivingtheresultinganswers

Noneedtorepresentthewholeexercise.Especially,ifyoudonotintendtogivepartialanswers

16June,2015 38


Overviewexerciseresults

Therewere55groups,ofthese2groupsfailedtodelivertheiranswers Theaveragescoreofall106 participantswas 22,0 points;

16June,2015 39

0

1

2

3

4

5

6

7

8

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32


BonusExerciseResults

16June,2015 40

Grp Score Bonus Grp Score Bonus Grp Score Bonus

1 23 0,575 20 29 0,725 39 14 0,350

2 25 0,625 21 30 0,750 40 13 0,325

3 29 0,725 22 26 0,650 41 20 0,500

4 25 0,625 23 42 22 0,550

5 26 0,650 24 43 28 0,700

6 29 0,725 25 10 0,250 44 21 0,525

7 12 0,300 26 18 0,450 45 17 0,425

8 26 0,650 27 27 0,675 46 17 0,425

9 19 0,475 28 28 0,700 47 28 0,700

10 24 0,600 29 30 0,750 48 14 0,350

11 27 0,675 30 10 0,250 49 25 0,625

12 18 0,450 31 9 0,225 50 13 0,325

13 30 0,750 32 25 0,625 51 16 0,400

14 26 0,650 33 28 0,700 52 14 0,350

15 14 0,350 34 24 0,600 53 14 0,350

16 21 0,525 35 28 0,700 54 18 0,450

17 23 0,575 36 28 0,700 55 24 0,600

18 29 0,725 37 27 0,675

19 28 0,700 38 15 0,375

MScOffshore&DredgingEngineeringFacultyCEG,DepartmentHydraulicEngineeringFaculty3mE,DepartmentMaritime&TransportTechnology

4116June,2015


16June,2015 42

Examdates&ConfigurationoftopicsExamdates

Exam: Wednesday24June, 09:00 12:00, 2Czaal2Reexam: Thursday13August, 09:00 12:00, t.b.a.

Theexamroughlyfollowsthesetupofthecourseschedule:

GeneralArcticEngineering: 7080% Arcticregions,Arcticstructuresandicefeatures; Icephysicsand/oricemechanics; Iceactionsandicestructureinteraction; Icemanagementand/orISO19906; ScalemodellingandArcticOceanography.

Dynamicsoficestructureinteraction: 2030% Frequencylockinandiceinducedvibrations; Physicsbasedandnumericalmodelling; Industryexperience;


1

2

16June,2015 43

Arcticregionsandicefeatures

Q: Nametheseas/areasoftheselocationsanddescribetheicefeaturesthatyouexpecttoencounterhere.

1. BaffinBay Firstyearlevelicefloes, Firstyeariceridges, Multiyearlevelice floes,and (Many)Icebergs.

2. (Southwestern)KaraSea Firstyearlevelicefloes, Firstyeariceridges,and (Rarely)Multiyeariceridges.

Atthelocationsgiveninthefigurebelow,offshorehydrocarbonfieldsarebeingdevelopedorwillbedevelopedinthenextfewyears.


16June,2015 44

ArcticconceptsQ: Discusstheadvantages(pros)andlimitations(cons)ofthefollowing

hydrocarbonproductionconceptsforuseintheArctic: GravityBasedStructure ShipshapedFloatingProduction,StorageandOffloadingunit(FPSO) ArcticTLP

LetsdiscusstheShipshapedFPSOunit(pros &cons): FPSOscommonlyusedfordeeporverydeepwaters. Goodrubbleclearing Canbedisconnectedforpossiblecollisionswithseverefeaturessuchasicebergs. Canbereused. Largedeckspaceavailable. Requiresnooffshoretopsideinstallation. Reliesforabigpartonicevaning capability Canonlyresistfirstyearicefeaturesandoftenrequiresicemanagement. SmalloperationalweatherwindowcomparedtoaGBSoranArtificialisland. Relativelyexpensivetomake. Theriserandmooringsystemsmaybeexposedtobrokenice.


2

16June,2015 45

Arcticregions/conceptsandicefeaturesQ: Whichofthegivenconceptswouldyouchoose

foreachlocationandexplainwhy.

1. BaffinBay GBS:Toodeep(200300m) FPSO:Goodopenwaterbehaviourandassumingitsdisconnectablemakesthisaviableoption.

ArcticTLP:perfectforopenwaterbehaviour,butitcannotbedisconnectedforicebergs.

Bestoption:FPSO(disconnection,openwater)

2. (Southwestern)KaraSea GBS:Viableoptionespeciallyintheshallowparts FPSO:Notaviableoptionastheoccurringiceloadsaretoohigh,evenwithicemanagement.

ArcticTLP:Probablytooshallow.Bestoption:GBS(ifwaterdepthallows)

GBS, Ship-shaped FPSO or Arctic TLP?

1


16June,2015 46

Arcticregions/structuresandicefeatures

Q: Name2(3)areaswhereoffshoreplatformsmustbedesignedforbothiceandseismicloads?

SeaofOkhotsk(Sakhalin), CookInlet(Alaska),or Bohai Bay(China)

Notehere:theCaspianSeaisawronganswer: ThereareearthquakesinthesouthernpartoftheCaspianSea,but

thereisnoseaicethere. InthenorthoftheCaspianSeathereisseaiceduringthewinters,

buttherearenoearthquakes.


16June,2015 47

IceactionsQ: Whatarethemechanismsthatlimittheiceloadonastructureduringice

structureinteraction? LimitEnergy LimitStress LimitForce

Q: Explainbrieflywhatismeantbyeachoftheselimitingmechanisms.

LimitEnergy:Themechanismthatoccurswhentheactionislimitedbythe(relative)kineticenergyormomentumoftheicefeature.ThismechanismisalsoreferredtoasLimitMomentum.

LimitStress:Themechanismthatoccurswhenthedrivingforcesworkingontheicefeaturearesufficientfortheicetofailasitinteractswiththestructure.

LimitForce:Themechanismthatoccurswhenanicefeatureisdrivenagainstthestructureandthedrivingforcesareinsufficientfortheicetofailandenvelopthestructure.


16June,2015 48

IceactionsQ: Theiceloadsexertedonaslopingstructurearedifferentfromtheiceloads

exertedonaverticalstructure.Explainonwhichofthestructurestheicefloeexertsthelowestloads,explainwhy,andnamethecorrespondingfailuremodes.Againstslopingstructurestheicefailsthrough(ice) bending,whileonaverticalstructuretheicefailsthrough(ice) crushing.

Theiceactionbybendingdependsmainlyontheflexuralstrengthofice,whilecrushingactionmainlydependsonthecompressivestrengthofice.Theflexuralstrengthoficeismuchlowerthenthecompressivestrengthoficeandthereforebendingexertslowerloadsonastructurethancrushing.

Ergo,theloadswillbelowestonslopingstructures.Notehoweverthatrubblepilingupand/oradfreeze maydiminishtheadvantagesofslopingstructures

Tocalculatestaticiceactions,weprincipallyapplytheISO19906,butwehaveextensivelydiscussedthisduringthefirstpartofthislecture.


Adfreeze theformationoficebustles

Tidescausewatervariationsandwhenthewaterlowers:thewaterfilmleftonthepilewillfreezeandaccumulate

16June,2015 49


IceactionsQ: IntheArctic,offshorestructuresarevulnerabletomanydifferent

environmentalloads.Someoftheseenvironmentalloadswilloccuratthesametime.Forthegivencombinationsofenvironmentalloadsexplainwhetherornotyouwouldchoosetoconsiderthemsimultaneouslyandexplainwhy:

MaximumwaveandiceloadsWaveandIceloadswillnormallynotoccurtogether,sincewavesareattenuatedbythepresenceofseaice.Therefore,thecombinationofthesetwoloadsIsnotaviableloadcombination.

MaximumwaveandmaximumwindloadsUsuallywavesareatitshighestwhenthewindloadsarehighest,thereforetheloadcombinationofmaxwaveandmaxwindisveryviable.

LeveliceandicebergcollisionloadsLeveliceandicebergcollisionloadswillcausefailureagainstastructureusingacompletelydifferentlimitingmechanism.Leveliceisusuallylimitstress,whileicebergcollisionsarelimitenergy.Althoughtheoccurrenceofbothphenomenonatthesametimeispossible,designingforthecombinationofthetwotogetherisirrelevant,duetotheirdifferenceinloadingmechanism.

16June,2015 50


Crystallography:BernalFowler/icerulesQ: TheBernalFowlerrulesdescribethearrangementofwatermoleculesand

hydrogenatomsintheidealcrystallinestructureofice.Givethe4BernalFowlerrules.

1. Thewatermoleculeispreservedintheicelattice.Ergo,1Oatomwith2Hatoms.

2. Eachwatermoleculeistetrahedrically bondedto4adjacentwatermolecules.

3. Thereisonly1hydrogenatomperoxygenoxygenbond.4. Thehydrogenatomsaremobilesorules13maybesatisfiedinany

configuration.

16June,2015 51


IcegrowthQ: Whenseaiceformsandgrowsitobtainsdifferentformsandshapeswhile

goingthroughthedifferentstagesofitslife.Withrespecttothis,explainthefollowingterms:

FrazilIceFinespiculesorplatesofice,suspendedinwater.

CongelationiceCongelationiceisalsoknownassecondaryiceandthisisthepartofanicelayerthatisgrowninadditiontotheprimaryice.Congelationiceconsistsofthetransitionzone,thecolumnarzoneandtheskeletonlayer.

BrineBrineiswaterthatissupersaturatedwithsaltthatisenclosedinseaice;Assaltisexpelledfromthefirsticeplateletsthatform,thesalinityofthesurroundingwaterincreases.Duringgrowth,theiceplateletstakeinwaterfromthesurroundingseawater,increasingthesalinityofsurroundingwaterfurther.Duringfurthericegrowththehighsalinitywaterisincludedalongtheplateletboundariesintheformofliquidorsolidinclusions.Thenowisolatedbrineinclusionsarecalledbrinepockets.

16June,2015 52


IceMechanicsQ: Whatarethe4physicalicepropertiesthatinfluencethestrengthofice? Temperature Porosity Salinity Crystallography

Q: Howdoestherelativevelocitybetweenastructureandanicefloeinfluencetheloadsonthatoffshorestructureduringicestructureinteraction? Lowvelocity ductilefailure High(er)velocity brittlefailure

16June,2015 53


DynamicsoficestructureinteractionQ: Whatarethe3maintypesofmodelsthatareavailabletomodeldynamic

interactionbetweenseaiceandoffshorestructures?Giveashortexplanationofeachtypeofmodel. Physicsbased models

Thistypeofmodelling triestoapproachrealityasmuchaspossiblebytakingintoaccountthefundamentalphysical(micro)propertiesofthephenomenontobemodelled.

Empirical modelsModelsbasedondata.

Phenomenological modelsModelsthattrytomimicthebehaviour ofacertainphenomenonratherthanlookingintothesourceofthisbehaviour.

16June,2015 54


IcestructureinteractionQ: Theinteractionbetweenanicefloeandaslopingstructureisdescribedbya

loadingcyclein2alternatingphases.Describethisloadingcycleandidentifyits2phases.

1. Uponinitialcontactoftheicefloewiththehulloftheslopingstructure,thetipoftheicefloeispusheddownwardsandtheicefloestartsbendingdownwardsuptothepointwheretheicefloe,heremodelledasabeam,breaksinbendingatacertaindistancefromtheinteractionpointatthetipoftheicefloe.Thisisthefirstphasecommonlydescribedas:Bendinguptofailure.

2. Onceapieceoftheicefloe(beam)breaksofffromtheicefloe,thispiece(orpiecesofrubble)ispusheddowntheslopebytheremainingicefloe,untilthetipoftheremainingicefloehitsthehulloftheslopingstructure.Thisisknownasthesecondphase.Oncethetipoftheremainingicefloehitsthestructuretheicefloeisonceagainapplyingadirectloadtothestructure,andwethuscommonlydescribethisphaseas:Pushingrubble(down)untilreloading.

16June,2015 55


IcestructureinteractionQ: Whichbeamtheorywouldyouusetomodeltheinteractionbetweentheice

andthedownwardslopingstructureand why? EulerBernoullibeamtheory,alsocommonlyknownastheclassicalbeam

theory.

Whenmodellingtheiceasabeam,thebeamrepresentingtheicecanalwaysbeconsideredtobelongorslender;Whenbeamsarelong/slender,sheardeformationsandrotationalinertiamaybedisregardedasisassumedfortheEulerBernoullibeamtheory.

AdditionalNote: Whenconsideringshortbeams,sheardeformationsandrotationalinertia

shouldbetakenintoaccountaccordingtoTimoshenkoRayleighbeamtheory.

16June,2015 56


DynamicicestructureinteractionBecausetheicesheetismoving,thedisplacementoftheicesheetdoesnotonlydependontimebutalsodependsonitspositionintime.Themovingicefloe,modelledasabeamonanelasticfoundationisthereforeknownasaconvectivesystem.Consequently,itsverticalaccelerationisfoundasafunctionoficefloevelocityandaccelerationas:

Q: Assumingthatthevelocityoftheicesheetisconstant,givetheequationofmotionforthebendingofamovingicesheetonanelasticfoundationusingEulerBernoullibeamtheory.Here,theaxialcompressionalongtheicesheetandthecorrespondingdampingmaybeneglected.

Assumingthattheicefloevelocityisconstant,wecanwritetheverticalaccelerationoftheicesheetas:

Consequently,wefindtheequationofmotionas:

16June,2015 57

2 2 2 222 2 22 z z z z zD u u u u uv t v t a tDt t x t x x

2 2 22

2 2

2

2 2

z z z zu u u uv vt xD tt x

D

2 2 2 4

22 2 42 0 z z z z

zfoundation

bendingconvective inertia

u u u uA v v EI kut x t x x


Dynamicicestructureinteraction

16June,2015 58


DynamicsoficestructureinteractionDuringastructuresfirstoperationsintheChukchiSea,theicefloevelocityintheChukchiSeavariesisobservedtobebetween0,04and0,08m/sandthebreakinglengthoftheicethatfailsagainstthestructureinbending rangesfrom10to14m.Thenaturalfrequencyofthestructureis0,195rad/s.Q: Determinethefrequencyrangeoficefailureagainsttheslopingstructureandexplain

whetherfrequencylockinmayoccurwhileoperatingthisstructureintheChukchiSea.Thelowesticefailurefrequencyisfoundforacombinationofthebiggestbreakinglengthandthelowesticefloevelocity:

Accordingly,thehighesticefailurefrequencyisfoundforacombinationofthesmallestbreakinglengthandthehighesticefloevelocity:

Previously,wefoundanaturalfrequencyof0,195rad/s,whichis4timesbiggerthanthehighesticefailurefrequency.Duetothebigdifference,wemayexpectthattheicefailurefrequencyandthestructuresnaturalrollingfrequencywillnotsynchronize.Thus,wedonotexpectfrequencylockintooccurhere.16June,2015 59

min

max

2 2 0,04 0,01814

lowv rad s

max

min

2 2 0,08 0,05010

highv rad s


BeforeIgo

Doyouhaveanyquestions?

or

Isanythingunclearthatyouwouldliketoseefurtherexplained?

IfyouthinkofsomethinglatercometomyofficethisThursdayorthisFriday

16June,2015 60


16June,2015 61

Good luck with the examon Wednesday 24 June,

and enjoy the summer holiday!

oe4680 2015 - lecture 15 - discussion exercise & exam questions

Documents

thickness fyridges

kgm3youngsmodulusseaice

porosityicerubble e

keeldepth icemovement

kgm3currentvelocity

mssalinityseawater s

kgm3windvelocity va

densityseaice ice