Presentedby:

Copyright©2018SteelJoistInstitute.AllRightsReserved.

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

DesignofSteelDeckforConcentratedandNon-UniformLoading

MichaelMartignetti,CANAMMikeAntici,NUCOR

PollingQuestion

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•  NewrequirementtoearnPDHcredits

•  Twoquestionswillbeaskedduringthedurationoftoday’spresentation

•  ThequestionwillappearwithinthepollingsectionofyourGoToWebinarControlPaneltorespond

Disclaimer

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

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

PresentationOutline

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ü IdentifyTypicalDeckTypes

ü IntroductiontoConcentratedLoadsTypes

ü RoofDeckLimitStatesandDesignExample

ü FloorDeckLimitStatesandCurrentDesignMethodology

ü CompositeDeckDesignExamples–ShortcutsforMultipleLoads

ü FormDeckandSteelFibers

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DeckTypes

RoofDeck• PermanentStructuralMember• NoConcreteTopping

CompositeDeck• DeckandConcreteWorkTogether• Embossments–CompositeAction

FormDeck• DeckisPermanentForm• DeckOftenCarriesSlabWeight

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ConcentratedLoadsonRoofDeck

SafetyAnchors

SuspendedLoads SolarPanels

RoofDrains

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ConcentratedLoadsonRoofDeckConstructionLoads

• People• Dollies

• Pallets• ToolChests

• RoofingMachinery

ConcentratedLoadsonFloorDeck

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StorageRacks

ConcentratedLoadsonFloorDeckEquipmentLoads

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ConcentratedLoadsonFloorDeckWallLoads

Parallel Transverse

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RoofDeckDesignStandard/Manual

Availableatwww.sdi.org

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RoofDeckDesignLimitStates

Deflection

WebCrippling

Shear

Stress

Bending/ShearInteraction

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RoofDeck–TransverseDistribution

L=SpanX=%ofSpan

Basedon1½”Deck…

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RoofDeckDesignExampleExample7FromRDDM…

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RoofDeckDesignExample

L=SpanX=%ofSpan

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RoofDeckDesignExampleP=210lb

W=30lb/ft

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RoofDeckDesignExample

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RoofDeckDesignExample

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FloorDeckDesignStandards/Manual

Availableatwww.sdi.org

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FloorDeckDesignLimitStates

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

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wbe

LoadDistribution

SDICDDM/FDDM/C-2017

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CurrentSDIDesignMethod

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LimitStates

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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.

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P

CanWeSolveThisLoadDiagram?

NEWforthisPresentation

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ShortcutTheory

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

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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.

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2Loads“In-Line”,MyandVn

P1/be1 P2/be2

Mn

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ThisgraphillustratesbendingmomentsforP1/be1,P2/be2andanyuniformloadalongthebeam.Noticethatthemomentsarecumulativeandmustnotexceedtheallowable.

2Loads“In-Line”,My

P1/be1

P2/be2

Vn

ΣV

VP1

VP2

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Asimilargraphforshear.Again,P1/be1,P2/be2andanyuniformloadalongthebeamarecumulativeandmustnotexceedallowable

2Loads“In-Line”,Vn

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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.

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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.

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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

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2Loads“In-Line”,ScaffoldExample,Vn

φP13.30

φP24.78

1.5’ 4.5’

8.0’

2.0’

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

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• 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’

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2Loads“In-Line”,ScaffoldExample,Mw

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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.

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2Loads“Adjacent”,MyandVn

Pw

be+Loadspacing

Overlappinginfluencezonesmayresultincumulativeweakaxisbendingmoments,andtraditionalengineeringmechanicsarenotappropriateforatwo-wayslabproblemwithsinusoidalstressdistribution.

Sinusoidalstressdistribution?Twowayslabdesign?Thissoundscomplicated,butthenextfewgraphsandexampleproblemmakesunderstandingandanalysisrelativelyeasy.

Pw

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2Loads“Adjacent”,Mw

beoverlap<loadspacing

MWn

ΣM

P/w P/w

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2Loads“Adjacent”,Mw

MWn

ΣM

P/w P/w

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beoverlap>loadspacing

2Loads“Adjacent”,Mw

MWnΣM

P/w P/w

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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.

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2Loads“Adjacent”,ScaffoldExample

φP3.14

3.5’ 4.5’

8.0’

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

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2Loads“Adjacent”,ScaffoldExample,MyandVn

4.33’

Aloaddevelopsasinusoidalmomentenvelopeoverabeamlength=beandisresistedbytheavailableweakaxisbendingmoment=φMw

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2Loads“Adjacent”,ScaffoldExample,Mw

4.33’

2.16’

x=0.65’

Andwecancalculatethemomentatanypointxalongthiscurve.Inthisexample,weareinterestedinthemomentatx=0.65’.

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2Loads“Adjacent”,ScaffoldExample,Mw

Mx=0.65

ΣMx<2757

Focusonthe

picture,nottheequation.

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18”

2Loads“Adjacent”,ScaffoldExample,Mw

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

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4Loads“In-Line”and“Adjacent”

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“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.

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“Whatsizeliftcanthisfloorsupport?”

ExampleProblem

φP1.12

4.5’

9.0’

4.33’

φP3.8

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“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.

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“Whatsizeliftcanthisfloorsupport?”

ExampleProblem

φP3.90

2.0’ 4.5’

9.0’

2.5’

φP4.64

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“Whatsizeliftcanthisfloorsupport?”

ExampleProblem

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

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“Whatsizeliftcanthisfloorsupport?”

ExampleProblem

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

φP2w

52”

φP2w

be2=4.94’

be2=4.94’

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“Whatsizeliftcanthisfloorsupport?”

ExampleProblem

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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

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18

4.5"(t=2.5")

5.0"(t=3.0")

FDDMScissorLiftTables?

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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

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

“Canmyfloorsupportthisdatarack(s)?”

ExampleProblem

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

“Canmyfloorsupportthisdatarack(s)?”

ExampleProblem

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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

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7.0’

643plf3.57’

φ=1.6or1.2?

DataRack–Mw

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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

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21”

be=4.33’

14” 14”26”

14” 12” 14” 12”

26”

DataRack–Mw-ShortAxleAdjacent

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“Canmyfloorsupportthisdatarack(s)?”

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

ExampleProblem

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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.

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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.

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SteelFibers

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PollingQuestion#2

TrueorFalse…Theuseofshearstudsonthebeamswillincreasetheallowablemagnitudeofconcentratedloadsonaslabmostofthetime.

a)True

b)False

PollingQuestionAnswers

WhichLimitStateisNOTApplicableforDesigningConcentratedLoadsonConcreteSlabsonFLOORDeck?

B)WebCrippling

TrueorFalse…Theuseofshearstudsonthebeamswillincreasetheallowablemagnitudeofconcentratedloadsonaslabmostofthetime.

B)False

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Presentedby:

Copyright©2018SteelJoistInstitute.AllRightsReserved.

THANKYOU

MichaelMartignetti,CANAMMikeAntici,NUCOR