CE432_L12_EffectiveLength

7/29/2019 CE432_L12_EffectiveLength

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AnnouncementsAnnouncements10/8/0910/8/09

Lecture #12Lecture #12Effective LengthEffective Length

CE A432CE A432 Steel DesignSteel Design

T. Bart Quimby, P.E., Ph.D.T. Bart Quimby, P.E., Ph.D.

Fall 2009Fall 2009

General Buckling TermsGeneral Buckling Terms

Will occurWill occurABOUTABOUT a principlea principleaxis andaxis and IN A PLANEIN A PLANE normalnormalto the buckling axisto the buckling axis

about both axesabout both axes

Effective LengthEffective Length

Euler bucklingEuler bucklingequation is aequation is afunction of thefunction of the

Braced or Sidesway Inhibited

EFFECTIVEEFFECTIVELENGTHLENGTH in ain agivengiven PLANEPLANEOFOFBUCKLINGBUCKLING..

Braced Frame K always < 1.0

Effective LengthEffective Length

Euler bucklingEuler bucklingequation is aequation is a

function of thefunction of the

UnBraced or Sidesway Uninhibited

EFFECTIVEEFFECTIVELENGTHLENGTH in ain agivengiven PLANEPLANEOFOFBUCKLINGBUCKLING..

UnBraced Frame K always > 1.0

Lateral Support TermsLateral Support Terms

Columns are eitherColumns are either BRACED (BRACED (SideswaySideswayInhibited)Inhibited) oror UNBRACED (UNBRACED (SideswaySideswayUninhibited)Uninhibited) in the plane of bucklingin the plane of buckling


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Lateral Support by LevelLateral Support by Level

All columns on a level and in the sameAll columns on a level and in the sameplane generally have same lateral supportplane generally have same lateral supportconditionsconditions

The columns areall UNBRACED inthe plane shown

The columns areall BRACED in the

plane shown

Lateral Support by LevelLateral Support by Level

Columns in a level may be orientedColumns in a level may be orienteddifferently.differently.

l i l

Web in plane offrame

l i lframe

Note that frame jointsprovide only partial fixity

Lateral Support AND EndLateral Support AND EndConditionsConditions MAY BE DIFFERENTMAY BE DIFFERENT

in Each Principle Planein Each Principle Plane

ALWAYS DRAW TWO COLUMNPROFILES

with support conditions when doingcolumn design/analysis

Effective Length CoefficientEffective Length Coefficient

Theoretical Values: SCM pg 16.1Theoretical Values: SCM pg 16.1--240.240.

The theoretical values assume thatThe theoretical values assume thatjoints are completely fixed againstjoints are completely fixed against

..

Reality is usually somewhere inReality is usually somewhere inbetween. This affects the value of K.between. This affects the value of K.

A method accounting for rotationalA method accounting for rotationalstiffness of joints is given on SCM pgstiffness of joints is given on SCM pg

16.116.1--239 through 243.239 through 243.

AISC Method LimitationsAISC Method Limitations Alignment Chart: Braced FrameAlignment Chart: Braced Frame

gg

cc

LIG

= .


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Alignment Chart:Alignment Chart: UnbracedUnbraced FrameFrame Typical Framed JointTypical Framed Joint

Rotational Stiffness at JointRotational Stiffness at Joint

Look atLook atstiffnessesstiffnessesthat resistthat resist (I/L)c2

rotation inrotation inthe plane.the plane.

The I used is the onein the plane ofbending for eachmember that adds

resistance to rotationof the joint

(I/L)g1

(I/L)g2

(I/L)c1

22112211

gggg

cccc

LILI

LILIG


Look atLook atstiffnessesstiffnessesthat resistthat resist

(I/L)c2


The I used is the onein the plane ofbending for eachmember that adds

resistance to rotationof the joint

222211

gg

cccc

LI

LILIG

(I/L)g1

g2

(I/L)c1


Look atLook atstiffnessesstiffnesses

that resistthat resist

(I/L)c2


The I used is the onein the plane ofbending for eachmember that addsresistance to rotationof the joint

0

2211 cccc LILIG

(I/L)g1

(I/L)g2

(I/L)c1

This is apure PIN

Rotational StiffnessRotational Stiffness

Pinned: G = (I/L)c/ 0 = infiniteUse G = 10

Fixed: G = (I/L)c/ infinite = 0Use G = 1


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I to use for each planeI to use for each plane

Ic1 = Ix Ic1 = Iy

g1 = x g2 = x g1 = x g2 = x

Ic2 = Ix Ic2 = Iy

SampleSampleProblemProblem

DetermineDetermine

CapacityCapacityof theof thecolumncolumnshownshown

Strong Axis Buckling, KStrong Axis Buckling, Kxx Compute GCompute Gtoptop & G& Gbtmbtm..

Determine KDetermine Kxx from thefrom the

UNBRACEDUNBRACED frame nomographframe nomograph..

65.0181330151170

1054112541

49.01880015712

btm

top

G

G

FindFind KKxx fromfromNomographNomograph

GB = 0.65

GA= 0.49

Kx ~ 1.19

Weak Axis Buckling, KWeak Axis Buckling, Kyy ComputeCompute GGtoptop && GGbtmbtm..

DetermineDetermine KKyy from thefrom the BRACEDBRACED

frameframe nomographnomograph..

113.01251010510

107.57127.57

094.010510

.

btm

top

G

G

FindFind KKyy fromfrom NomographNomograph

GB = 0.094

GA= 0.113

Ky ~ 0.55


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Determine ColumnDetermine ColumnSlendernessSlenderness

1.29

in89.5

in14419.1x

r

LK

25.41.92in1

in14455.

y

y

r

LK

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CE432_L12_EffectiveLength