stability and second-order analysis of frames

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tosolvetheeigenvalueproblem.However,theanalyticalpro-cedureandthedesignstepsdescribedhereinarethesame.Example1

UsingtheAISCLRFDapproach(1999),determinethecol-umnsizesoftheframeloadedasshowninFig.3.Assumethat(1)thecolumnshavecontinuouslateralsupportperpen-diculartotheplaneofthestructure;(2)?b=0.20atthebaseofcolumnAB;(3)columnCDisconnectedatthebasewitha?xityfactor?d=0.90;and(4)the20-ft(6.096-m)girderACconsistsofaW16?31shapeconnectedtocolumnsABandCDbothwith?xityfactors?a=?c=http://wendang.chazidian.comeFy=36-ksi(248.2-MPa)steel.Solution:

1.TryW14?120forbothcolumns:A=35.3in.2(22,774mm2),Ix=1,380in.4(5.744?108mm4),andrx=6.24in.(158mm).

2.CalculatePu/A:(Pu/A)AB=700/35.3=19.83ksi(136.7MPa)and(Pu/A)CD=500/35.3=14.16ksi(97.6MPa).3.DeterminePcrfromthecharacteristicequation?[K]?P[K?]?=0andfrom(9)and(10):Takingintoaccountthat?=P2/P1=5/7,h1=12ft(3.658m),h2=18ft(5.586108m),L=20ft(6.096m),andIg=375in.4?(1.561mm4),withthesevaluesacriticalloadofPcr=2,015.1kip(8,964kN)isobtained.Then

?Pcr=(5/7)?2,015.1kip=1,439.4kip(6,403kN)

Therefore

2

K2

AB=

?

P=crh2

?

2,015.1?(12?12)2

=3.075

KCD=

?

21,439.4?(18?12)2

=2.425

4.Calculate?c

(?Khc)AB=

??r

?FyE

?

=0.7942<1.5

AB

therefore,useFcrinelastic

(?c)?KhCD=

?r

?FyE

?

=0.9396<1.5

CD

therefore,useFcrinelastic.5.Calculate?cFcr

(?F2

ccr)AB=0.85?(0.658)?cFy=0.85?0.768?36=23.50ksi(162MPa)>(Pu/A)=19.83ksi(136.7MPa)(?2

cFcr)CD=0.85?(0.658)?cFy=0.85?0.691?36=21.15ksi(145.8MPa)>(Pu/A)=14.16ksi(97.6MPa)

Becausebothcolumnsareapparentlyoverdesigned,anad-ditionaltrialwascarriedoutasfollows:

1.TryW14?109(A=32in.2,Ix=1,240in.4,andrx=6.22in.)forcolumnABandW14?82(A=24.1in.2,Ix=882in.4,andrx=6.05in.)forcolumnCD.

2.CalculatePu/A:(Pu/A)AB=700/32=21.88ksi(150.8MPa)and(Pu/A)CD=500/24.1=20.75ksi(143MPa).3.DeterminePcrfromthecharacteristicequation?[K]?P[K?]?=0:Takingintoaccountthat?=P2/P1=5/7,h1=12ft(3.658m),h2=18ft(5.586m),L=20ft(6.096m),andIg=375in.4(1.561?108mm4)

FIG.3.Example1:DesignofSingle-StoryFramewithoutLateral

Bracing

Pcr=1,580.9kip(7,032kN)and

?Pcr=(5/7)?1,580.9kip=1,129.2kip(5,023kN)

Therefore

KAB=

?21,580.9?(12?12)2

=3.29

KCD=

?

1,129.2?(18?12)2

=2.189

4.Calculate?c

(?KhFyc)AB=

??r?E?

=0.854<1.5

AB

therefore,useFcrinelastic

(?KhFyc)CD=

??r

?E

?

=0.877<1.5

CD

therefore,useFcrinelastic.5.Calculate?cFcr:

(?2

cFcr)AB=0.85?(0.658)?cFy=0.85?0.759?36=22.56ksi(155.5MPa)>(Pu/A)=21.88ksi(150.8MPa)

(?2

cFcr)CD=0.85?(0.658)?cFy=0.85?0.725?36=22.18ksi(152.9MPa)>(Pu/A)=20.75ksi(143MPa)

Therefore,this?naldesignisacceptable.

Noticethat(1)theobtainedequivalentlengthK-factorsinbothcolumnsarequitedifferentfromthosecalculatedusingthealignmentcharts[whicharestillincludedintheAISCLRFD(1999)speci?cations];and(2)thecurrentdesignresultsinlightercolumnsthanthoseobtainedusingthealignmentchartK-factors.

Duetomanyuncertaintiesinthestructuraldesignofframeswithsemirigidconnections,thestiffnessreductionfactor?=Et/E[suggestedbyAISCLRFD(1999)]shouldalwaysbetakenas1.0becausethisalwaysgivesconservativeresults.Example2

Using(9)and(10),determinethevariationsoftheeffectivelengthK-factorofbothcolumnsintheframeofExample1[Fig.4(a)]with(1)diagonalbracingSbaseofcolumnAB??;(2)degreeof?xityattheb;and(3)degreesof?xityattheendsofbeamAC?aand?c.AssumeW14?109[A=32in.2(20,645mm2),Ix=1,240in.4(5.161?108mm4),andrx=

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FIG.4.Example2:ParametricStudyonPlaneFrame:(a)StructuralModel;(b)EffectsofDiagonalBracingS?;(c)EffectsofColumnABFixityFactoratB?b;(d)EffectsofGirderACFixityFactors?aand?c

6.22in.(158mm)]forcolumnABandW14?82[A=24.1in.2(15,548mm2),Ix=882in.4(3.6712?108mm4),andrx=6.05in.(154mm)]forcolumnCDwith?d=0.90.Solution:Eachofthethreeeffectswasstudiedindependentlyasfollows:

?Effectsofdiagonalbracing—Thelateralstiffnessrestrain-ingtheframeagainstsideswayS?providedbyadiagonalbracingofhorizontallengthL,heighth,andcrossareaAbisgivenbySalmonandJohnston(1996,Chapter14,Eq.14.5.16)asfollows:

AbEL2/h3

S?=

[1?(L/h)2]3/2

AsensitivitystudywascarriedoutontheeffectsofS?onthestabilityoftheframeofExample1withthe?xityfactorsindicatedinFig.3.TheresultsareshowninFig.4(b).Forinstance,S?=100kip/in.(175.12kN/m)cor-respondstoaneffectivediagonalbracewithcrossarea=2.015in.2(1,300mm2)thatconnectsjointsAandD.Thebene?tsofdiagonalbracingonthestabilityofthisframe

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areapparent[Fig.4(b)],particularlyfromS?=0to46kip/in.(80.55kN/m).Withmerely0.93in.2(604.5mm2)ofdiagonalbracing(2.91%ofthecrossareaofcolumnAB)orS?=46kip/in.(80.55kN/m),theK-factorisre-ducedinhalfinbothcolumns.BothcurvesinFig.4(b)aresteepupto46kip/in.(80.55kN/m);additionalincre-mentsinS?donotshowsigni?cantreductionsintheK-factors.

?EffectsofcolumnABbase?xity?b—The?xityofthecolumnbasesalsohassigni?cantbene?tsonthestabilityoftheframe.Fig.4(c)indicatesthat,byincreasingthe?xityofcolumnABfromzero(perfectlyhinged)to0.20,theK-factorisreducedby25%ineachcolumn,repre-sentinga78%increaseinthecriticalloadofthispartic-ularframe.NoticethatbothcurvesinFig.4(c)aresteepfrom?b=0uptoapproximately?b=0.60.Thisindicatesthatanyadditional?xityofthebaseofcolumnABbeyond0.60showslittleadditionaleffectiveness.

?Effectsofgirderend?xities?aand?c—The?xityofthegirderconnectionshassomebene?tsonthestabilityoftheframebuttheyarenotassigni?cantasthoseatthecolumnbases[solidcurvesinFig.4(d)].NoticethatthevariationsoftheeffectivelengthK-factorswiththedi-