design of steel deck for concentrated and non-uniform loading · has been developed by the steel...
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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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ConcentratedLoadsonRoofDeckConstructionLoads
• People• Dollies
• Pallets• ToolChests
• RoofingMachinery
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FloorDeckDesignLimitStates
My Bending(+ifsimplespan,+/-ifmultiplespan)Vn OneWayBeamShearVpr PunchingShearΔ DeflectionMwTransverse(Weakaxis)BendingMn,Mr ProprietaryDeck-SlabBending(nostuds)
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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?
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
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
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
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
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
20
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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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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“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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