design of steel deck for concentrated and non-uniform loading

Presentedby:

Copyright©2018SteelJoistInstitute.AllRightsReserved.

M A R C H 2 1 , 2 0 1 8

DesignofSteelDeckforConcentratedandNon-UniformLoading

MichaelMartignetti,CANAMMikeAntici,NUCOR

PollingQuestion

2

•  NewrequirementtoearnPDHcredits

•  Twoquestionswillbeaskedduringthedurationoftoday’spresentation

•  ThequestionwillappearwithinthepollingsectionofyourGoToWebinarControlPaneltorespond

Disclaimer

3

Theinformationpresentedhereinisdesignedtobeusedbylicensedprofessionalengineersandarchitectswhoarecompetenttomakeaprofessionalassessmentofitsaccuracy,suitabilityandapplicability.TheinformationpresentedhereinhasbeendevelopedbytheSteelJoistInstituteandisproducedinaccordancewithrecognizedengineeringprinciples.TheSJIanditscommitteeshavemadeaconcertedefforttopresentaccurate,reliable,andusefulinformationonthedesignofsteeljoistsandJoistGirders.ThepresentationofthematerialcontainedhereinisnotintendedasarepresentationorwarrantyonthepartoftheSteelJoistInstitute.Anypersonmakinguseofthisinformationdoessoatone’sownriskandassumesallliabilityarisingfromsuchuse.

4

LearningObjectives• Recognizeloadcasesthatrequireadditionalanalysisbeyonddistributionasauniformload• Understandthelimitstatesfordesignunderconcentratedloads• Examinedifferentloadpathsforvaryingconcentratedloadconditions• ReviewcurrentSDIdesignapproachforconcentratedloads• Demonstratepotentialshortcutstoconcentratedloaddesign• Presentexampleproblemsfordesignwithconcentratedloads

PresentationOutline

5

ü IdentifyTypicalDeckTypes

ü IntroductiontoConcentratedLoadsTypes

ü RoofDeckLimitStatesandDesignExample

ü FloorDeckLimitStatesandCurrentDesignMethodology

ü CompositeDeckDesignExamples–ShortcutsforMultipleLoads

ü FormDeckandSteelFibers

6

DeckTypes

RoofDeck• PermanentStructuralMember• NoConcreteTopping

CompositeDeck• DeckandConcreteWorkTogether• Embossments–CompositeAction

FormDeck• DeckisPermanentForm• DeckOftenCarriesSlabWeight

7

ConcentratedLoadsonRoofDeck

SafetyAnchors

SuspendedLoads SolarPanels

RoofDrains

8

ConcentratedLoadsonRoofDeckConstructionLoads

• People• Dollies

• Pallets• ToolChests

• RoofingMachinery

ConcentratedLoadsonFloorDeck

9

StorageRacks

ConcentratedLoadsonFloorDeckEquipmentLoads

10

11

ConcentratedLoadsonFloorDeckWallLoads

Parallel Transverse

12

RoofDeckDesignStandard/Manual

Availableatwww.sdi.org

13

RoofDeckDesignLimitStates

Deflection

WebCrippling

Shear

Stress

Bending/ShearInteraction

14

RoofDeck–TransverseDistribution

L=SpanX=%ofSpan

Basedon1½”Deck…

15

RoofDeckDesignExampleExample7FromRDDM…

16

RoofDeckDesignExample

L=SpanX=%ofSpan

17

RoofDeckDesignExampleP=210lb

W=30lb/ft

18

RoofDeckDesignExample

19

RoofDeckDesignExample

20

FloorDeckDesignStandards/Manual

Availableatwww.sdi.org

21

FloorDeckDesignLimitStates

My Bending(+ifsimplespan,+/-ifmultiplespan)Vn OneWayBeamShearVpr PunchingShearΔ DeflectionMwTransverse(Weakaxis)BendingMn,Mr ProprietaryDeck-SlabBending(nostuds)

22

wbe

LoadDistribution

SDICDDM/FDDM/C-2017

23

CurrentSDIDesignMethod

24

LimitStates

25

PollingQuestion#1

WhichLimitStateisNOTApplicableforDesigningConcentratedLoadsonConcreteSlabsonFLOORDeck?

a) WeakAxisBending

b) WebCrippling

c)  PunchingSheard)  PositiveBendinge) NegativeBending

P

a b

L

Thiswebinarmakesoneassumption....thewebinee(that’syou)cansolvethissimplebeamforshearandbending.Additionallimitstates(deflection,punching)aredefinedinthestandards,butunlikelytocontrol.Shearandbendingwillbediscussedindetail.Problemsolutionsareshown,butintendedasexamplesandguidesforfuturereference.Pleasefocusonthediagramsandtechniquesforloaddistribution,notthemathematicalsolution.

26

P

CanWeSolveThisLoadDiagram?

NEWforthisPresentation

27

ShortcutTheory

Influencezonesmay(andusuallydo)overlapasillustrated.Thissuggeststhestressintheseareasisgreaterthanthestressinnon-lappedzones.Theeffectivewidthsoftheseinfluencezones(be1andbe2)changeasloadsP1andP2movealongthespan.Insituationswhereloadlocationsarefixed(storageracks,scaffolds),asimplebeamdiagramforshearandbendingcaneasilybedefined.

28

2Loads“In-Line”

P1be1

P2be2

a b

L

ForanalysispurposesofMyandVn,twoloadsareonthebeamandequationsforshearandbendingarecumbersome,butsimplistic.Forcalculationpurposes,P1andP2aretypicallyequalloads,butdistributionwidthsbe1andbe2maydiffer;hence,loadsareillustratedasbeingdifferent.Variables“L”,“a”and“b”areconsistentwithtraditionalengineeringloaddiagrams.Nothingnewsofar,exceptbeamsaretobeanalyzedusingdistributedconcentratedloads,P/be,inlieuofuniformloadssuggestedintheliterature.

29

2Loads“In-Line”,MyandVn

P1/be1 P2/be2

Mn

30

ThisgraphillustratesbendingmomentsforP1/be1,P2/be2andanyuniformloadalongthebeam.Noticethatthemomentsarecumulativeandmustnotexceedtheallowable.

2Loads“In-Line”,My

P1/be1

P2/be2

Vn

ΣV

VP1

VP2

31

Asimilargraphforshear.Again,P1/be1,P2/be2andanyuniformloadalongthebeamarecumulativeandmustnotexceedallowable

2Loads“In-Line”,Vn

32

Weakaxisbendingfor“in-line”loadswilltakealittlemoreexplanation.Thebasicpremiseis“Loadsareuniformlydistributedalongthelength“w”.”Ifinfluencezonesoverlap(andtheyusuallydo),thegenericweakaxisbendingequationprovidedbySDIneedsaslightmodification.

2Loads“In-Line”,Mw

1

2

3

4

5

6

7

0 w

P/w

Overlap

Thenewequationformultiple“in-line”loadsforweakaxisbendingissimplyalinearinterpolationbetweenasingleloadanalysisandtwoloadscombined.Thegreatadvantagetothisequationis“ ITWORKSEVERYWHERE” regardlessoftheoverlap.

33

2Loads“In-Line”,Mw

• 2x12x20gacompositedeck• 8-0span• 5”NWslab(t=3”)• W6xW6-W2.1xW2.1(d=1.5”)• Scaffoldpost,b=4”• WL=0• Wd=(1.2)52psfFDDM2C• φMy=4140ft-lbs/ftFDDM4C• φVn=5116lb/ftFDDM8BφMw=2757in-lb/ft

Todemonstratethemechanicsfor“in-line”loads,considerscaffoldingduringconstruction.Thesubcontractorhasaskedtousescaffoldingforthebrickfascia.Howshouldyourespond?PunchingshearanddeflectionareunlikelytolimitPandwillnotbeshowninthisexample.

34

2Loads“In-Line”,ScaffoldExample

φP13.30

φP24.78

1.5’ 4.5’

8.0’

2.0’

Shear:FromFDDM8B,φVn=5116lbs.DistributeloadsP1andP2overtheireffectivewidths,be1andbe2,assumeP1=P2andsolveforP.Don’tforgettoadddeadandapplicableliveloads.

• be1=3.30ft• be2=4.78ft• w=4.33ft• Lap=2.33ft(use2.0)• Wd=62psf

35

2Loads“In-Line”,ScaffoldExample,Vn

φP13.30

φP24.78

1.5’ 4.5’

8.0’

2.0’

Bending:FromFDDM4C,φMy=4140ft-lbs.Again,distributeloadsP1andP2overtheireffectivewidthsandsolveforP.

36

• be1=3.30ft• be2=4.78ft• w=4.33ft• Lap=2.33ft(use2.0)• Wd=62psf

2Loads“In-Line”,ScaffoldExample,My

withNEWMwequationφP4.33

Weak:Thiswilltakemoreexplanation.1. NoticethattheloadPisdistributedoveraneffectivewidth“w”,not“be”.2. Theweakaxisbeamlength=beandwilldifferforP1andP2.3. bemaxwillcontrol.4. Withmultiple“in-line”loads,usethenewφMwtocorrectforinfluencezoneoverlap.5. Useφ=0.75andΩ=2.0,notACIfactors.

• be1=3.30ft• be2=4.78ft• w=4.33ft• Lap=2.33ft(use2.0)

4.78’

37

2Loads“In-Line”,ScaffoldExample,Mw

38

Influencezonesfor“adjacent”loadswilloverlap,buttheoverlapdoesnotmeantwicethestress.Intuitively,weknowstressesaregreatestdirectlyundertheloadanddissipatealongtheedges.Effectivewidthformulasfor“be”and“w”compensateforthisstressgradient.Forshearandbending,adjustbesoconcreteisnotusedtwice.be’=be/2+loadspacing/2.Forweakaxisbending,ΣMwwillrequireamoredetaileddiscussion.

2Loads“Adjacent”

Pbe’

a b

L

ForanalysispurposesofMyandVn,loadPisdistributedoverbeorbe’.Simple.

39

2Loads“Adjacent”,MyandVn

Pw

be+Loadspacing

Overlappinginfluencezonesmayresultincumulativeweakaxisbendingmoments,andtraditionalengineeringmechanicsarenotappropriateforatwo-wayslabproblemwithsinusoidalstressdistribution.

Sinusoidalstressdistribution?Twowayslabdesign?Thissoundscomplicated,butthenextfewgraphsandexampleproblemmakesunderstandingandanalysisrelativelyeasy.

Pw

40

2Loads“Adjacent”,Mw

beoverlap<loadspacing

MWn

ΣM

P/w P/w

41

2Loads“Adjacent”,Mw

MWn

ΣM

P/w P/w

42

beoverlap>loadspacing

2Loads“Adjacent”,Mw

MWnΣM

P/w P/w

43

beoverlap>>>loadspacing

2Loads“Adjacent”,Mw

Samedeckas“in-line”example• WL=0• Wd=(1.2)52psfFDDM2C• φMy=4140ft-lbs/ftFDDM4C• φVn=5116lb/ftFDDM8B• φMw=2757in-lb/ft

Todemonstratethemechanicsfor“adjacent”loads,let’srotatethescaffoldfromourpreviousexample.Atx=3-6,thedistributionwithbe=4.78ft,andadjacentinfluencezonesoverlap.ThemechanicsforMyandVnaresimilartothepreviousexampleusingamodifiedbe.Again,punchingshearanddeflectionareunlikelytolimitPandwillnotbeshowninthisexample.

44

2Loads“Adjacent”,ScaffoldExample

φP3.14

3.5’ 4.5’

8.0’

• be=4.78ft• be’=3.14ft• W=4.33ft• Wd=62psf

45

2Loads“Adjacent”,ScaffoldExample,MyandVn

4.33’

Aloaddevelopsasinusoidalmomentenvelopeoverabeamlength=beandisresistedbytheavailableweakaxisbendingmoment=φMw

46

2Loads“Adjacent”,ScaffoldExample,Mw

4.33’

2.16’

x=0.65’

Andwecancalculatethemomentatanypointxalongthiscurve.Inthisexample,weareinterestedinthemomentatx=0.65’.

47

2Loads“Adjacent”,ScaffoldExample,Mw

Mx=0.65

ΣMx<2757

Focusonthe

picture,nottheequation.

48

18”

2Loads“Adjacent”,ScaffoldExample,Mw

Youguessedit....4loads....“In-line”and“adjacent”.Iftheseloadsarestatic,thecalculationsaretedious,butnotdifficult.Ifloadsaremoving,hireaninternforthesummer.ForMyandVn,useP1/be1’andP2/be2’withsimpleshearandmomentenvelopes.ForMw,usenewMwlapequationandnewsinusoidalmomentenvelope.

49

4Loads“In-Line”and“Adjacent”

50

“Whatsizeliftcanthisfloorsupport?”Slab(FDDMExample4)• 2x12compositedeck• 20gage• 4½”totaldepth• 3ksiNWconcrete• 9-0clearspan• 25psfconcurrentLL• 6x6–W2.1xW2.1WWR• d=1.25”AssumedLift• 52“length• 30“width• 12”x4.5“tires• 2.5mph

ExampleProblem

Asageneralruleforscissorliftshear,locateonetirenearthesupportandtheshortaxle“adjacent”createsmaximumshear.Ifso,be1=1.12ft,be2=4.94ft,andw=4.88ft.Forshear,P2adjacentinfluencezonesoverlapandbe2’shouldbeused.P1influencezonesdonotoverlap,sodistributionwidthbe1needsnocorrection.

be2’=4.94/2+2.66/2=3.80ft.

51

“Whatsizeliftcanthisfloorsupport?”

ExampleProblem

φP1.12

4.5’

9.0’

4.33’

φP3.8

52

“Whatsizeliftcanthisfloorsupport?”

ExampleProblem

Asageneralruleforscissorliftbending,locateonetireatmidspanandtheshortaxle“in-line”createsmaximumpositivebending.Ifso,be1=3.9ft.be2=4.94ftandw=4.88ft.Forpositivebending,P2adjacentinfluencezonesoverlapandbe2’shouldbeused.P1influencezonesdonotoverlap,sodistributionwidthbe1needsnocorrection.

be2’=4.94/2+4.33/2=4.64ft.

53

“Whatsizeliftcanthisfloorsupport?”

ExampleProblem

φP3.90

2.0’ 4.5’

9.0’

2.5’

φP4.64

54

“Whatsizeliftcanthisfloorsupport?”

ExampleProblem

Thelimitingliftlocationforweakaxisbendingandpositivebendingaresimilar...Locateonewheelatmidspanwiththeshortaxlein-line.Noticethatin-lineloadsP1andP2overlapandlap=4.88’–2.5’=2.38’;therefore,in-linecorrectionsarerequired.AdjacentloadsP2andP2overlap,buttheoverlap<wheelspacing,sonoadjacentcorrectionsarerequired.

55

“Whatsizeliftcanthisfloorsupport?”

ExampleProblem

Whencomparingbe2andthewheelspacing,influencelinesoverlap,buttheoverlapislessthan52”.Thisisgoodnews;ΣMwcalculationsarenotrequired.Weonlyneedtocorrectforin-lineloadswiththenewMwequation.

φP2w

52”

φP2w

be2=4.94’

be2=4.94’

56

“Whatsizeliftcanthisfloorsupport?”

ExampleProblem

57

5.0 6.0 7.0 8.0 9.0 10.02740 2648 2422 2284 2190 2272 2043

22 3730 2881 2876 2867 2854 2657 20435470 2881 2876 2867 2854 2657 20432740 2648 2422 2284 2190 2272 22933730 3122 3120 3112 2985 3097 27375470 3122 3120 3112 3102 3466 27372740 2648 2422 2284 2190 2272 22933730 3570 3302 3113 2985 3097 31265470 3570 3571 3568 3560 4535 38443300 3117 2861 2704 2598 2539 2657

22 4520 3442 3438 3430 3417 3377 26576640 3442 3438 3430 3417 3377 26573300 3117 2861 2704 2598 2539 28174520 3704 3703 3694 3549 3468 34966640 3704 3703 3697 3686 4353 34963300 3117 2861 2704 2598 2539 28174520 4198 3908 3694 3549 3468 38486640 4198 4202 4200 4193 5100 4885

Pleaseconsultwithappropriateprofessionalforφ,impactorunbalancedloadfactors.

30"x52"(52"x30")loadfootprintconcurrentwith25psfconstructionliveload. φMw

4.5"wheel φVn

WWRd=t/2 φMy

6x6-W2.9xW2.94x4-W2.9xW2.9

206x6-W2.1xW2.16x6-W2.9xW2.94x4-W2.9xW2.9

186x6-W2.1xW2.1

6x6-W2.9xW2.94x4-W2.9xW2.9

6x6-W2.1xW2.16x6-W2.9xW2.94x4-W2.9xW2.9

6x6-W2.1xW2.16x6-W2.9xW2.94x4-W2.9xW2.9

6x6-W2.1xW2.16x6-W2.9xW2.94x4-W2.9xW2.9

6x6-W2.1xW2.1

φP/SpanSlab Gage WWR φMw

20

18

4.5"(t=2.5")

5.0"(t=3.0")

FDDMScissorLiftTables?

58

Slab• 1.5x6x18gacompositedeck• 5.0”TotalDepth• 3ksiNWConcrete• 7-0ClearSpan• 40psfConcurrentLL• 6x6–W2.9xW2.9WWR• d=1.0”DataRack• 42“deep• 28“overallwidth• 21”casterspacing• 3“casters• 3000#staticcapacityFirstthought–3000#/(28”x42”)+40psf=407psf

FDDMTable6A=400psfNoGood!

“Canmyfloorsupportthisdatarack(s)?”

ExampleProblem

59

750# 750# 750# 750# 750# 750# 750# 750# 750# 750#

“Canmyfloorsupportthisdatarack(s)?”

ExampleProblem

60

The“stacked”datarackorientationmayvary.Ifstackedadjacent,castersmayonlybe14”apart,soloadswouldcombine(1500lbs)withamodifieddistributedwidthof=2.33’.Ifstackedin-line,multiple750lbloadsoccuralongthespanwithamodifieddistributionwidth=3.57’width.

“Canmyfloorsupportthisdatarack(s)?”

ExampleProblem

61

1.75’

7.0’

1.75’3.5’

750lbs0.83’

0’

7.0’

3.5’

1500lbs2.33’

3.5’

7.0’

643plf3.57’

1500lbs2.33’

1500lbs2.33’

DataRack–Vn,My

62

7.0’

643plf3.57’

φ=1.6or1.2?

DataRack–Mw

63

1.75’

L

1500w

1.75’3.5’

1500w

12”

be=4.33’

750w

12”21”

Adjacentloadspacing=7”and21”<be/2,soweakaxisbendingmomentswillbecumulative

14”14”

750w

750w

750w

DataRack–Mw

64

21”

be=4.33’

14” 14”26”

14” 12” 14” 12”

26”

DataRack–Mw-ShortAxleAdjacent

65

“Canmyfloorsupportthisdatarack(s)?”

Regardlessofdatarackorientation,shearandbendingcapacitiesweremorethanadequate.Ifthedatarackisconsideredaliveloadandφ=1.6,weakaxisbendingfails.Ifφ=1.2,weakaxisbendingcapacityisadequate.Mysuggestion.....dropWWRto1.25”.

ExampleProblem

66

AllCases Influencezonesfordataracks,lift,scaffoldswilloverlap. Deflectionandpunchingareunlikelytogovernwithtraditionalframing. Loadfactorsmaybesubjective(φ=1.2,1.4,1.6) FDDMtabulatesφMyandφVn. Ifslabisnotrestrained(nostuds),consultwithsupplierforφMn.

BeamShear Locateoneloadatmidspanandtheshortaxleadjacent

Usebe’soconcreteisnotusedtwice. Don’tforgetuniformloads.

PositiveBending Locateoneloadatmidspanandtheshortaxlein-line.

Usebe’soconcreteisnotusedtwice. Don’tforgetuniformloads.

WeakAxisBending Locateoneloadatmidspanandtheshortaxlein-line.

Usebeincalculations,notbe’ Uniformdeadandliveloadsaresupportedinpositivebending,sonota componentofweakaxisbending. Ifadjacentloadspacing>be/2,momentsarenotcumulative. Equationscompensatefor“w”overlap.Noothercorrectionsarerequired. Ifadjacentloadspacing<be/2,ΣMwusingsinusoidalequationisrequired.

SummaryPageforMultipleLoads

Priorexampleswerecompositedecksandsimplespans.Formdecksaretypicallymulti-spanwithnegativebendingandinteractionoverthesupports.Deadload(slab)issupportedbytheformdeck,sonotavariableforshearorbending;otherwise,thedesignapproachissimilar.DistributeP,compareVmaxtoVn,+Mmaxto+Myand-Mmaxto-My.

67

φP1be1

φP2be2

+M1-2

-M2

-V2

+V1

+V2

+M2-3 +M3-4

+V3

-V3 -V4

-M3

FormDeck

Intheory,fibersarenotareplacementforWWRasatensilecomponent,soAs=0.Ifso,Mw=0,whichsuggestsP=0.Thissimplycannotbetrue.Loaddistributionwithsteelfibersisun-known,butoldtestingshowedpositiveresults.Canwerationallyestimateloadcapacitywithsteelfibers?• Oneoptionisignoringthecontributionoftheconcreteandusingdeckonlyfortransversedistribution.ThisoptionreducesdistributionwidthbeandφPabout70%.

• Asecondoptionusesbe=1’.ThisoptionreducesφPabout75%.Areductioninloadcapacitywouldbeanticipated,but70-75%maybeconservative.AdditionaltestinganddesignproceduresusingsteelfibersisrequiredbeforeSDIcouldconfidentlyprovideguidance.

68

SteelFibers

69

PollingQuestion#2

TrueorFalse…Theuseofshearstudsonthebeamswillincreasetheallowablemagnitudeofconcentratedloadsonaslabmostofthetime.

a)True

b)False

PollingQuestionAnswers

WhichLimitStateisNOTApplicableforDesigningConcentratedLoadsonConcreteSlabsonFLOORDeck?

B)WebCrippling

TrueorFalse…Theuseofshearstudsonthebeamswillincreasetheallowablemagnitudeofconcentratedloadsonaslabmostofthetime.

B)False

70

Presentedby:

Copyright©2018SteelJoistInstitute.AllRightsReserved.

THANKYOU

MichaelMartignetti,CANAMMikeAntici,NUCOR