Compression New v0 Unprotected

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COMPRESSION

MEMBER/COLUMN:

Structural member

subjected to axial load

P

P

2Compression Module


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3Compression Module


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Compression Module 4


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8/92Compression Module 8


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Strength design requirements:

PuP

n (P

aP

n/)ASD

Where = 0.9 for compression(= 1.67)ASD

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Axial Strength

Strength Limit States:

Squash Load

Global Buckling

Local Buckling

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Global

Buckling

Local

Flange

Buckling

Local

Web

Buckling

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14

Over-all buckling Flexural

Torsional Torsional-flexural


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Flexural Buckling


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Torsional buckling


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Flexural-Torsional buckling


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Flexural-Torsional buckling


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INDIVIDUAL COLUMN

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Squash Load

Fully Yielded Cross Section

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When a short, stocky column is loaded the strength is limited by

the yielding of the entire cross section.

Absence of residual stress, all fibers of cross-section yield

simultaneously at P/A=Fy.

P=FyA

yL0P

P

L0

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Results in a reduction in the effective

stiffness of the cross section, but the

ultimate squash load is unchanged.

Reduction in effective stiffness caninfluence onset of buckling.

24Compression Theory

RESIDUAL STRESSES


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RESIDUAL STRESSES


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P=FyA

yL0

No Residual Stress

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With residual stresses, flange tips yield first at

P/A + residual stress = FyGradually get yield of entire cross section.

Stiffness is reduced after 1styield.

RESIDUAL STRESSES

= Yielded


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P=FyA

yL0

RESIDUAL STRESSES

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P=(Fy-Fres)A

1

No Residual Stress

= Yielded

Steel

1

= Yielded


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P=FyA

yL0

RESIDUAL STRESSES


P=(Fy-Fres)A

1

Yielded

Steel

2

No Residual Stress

1

2

= Yielded


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P=FyA

yL0

RESIDUAL STRESSES

29

P=(Fy-Fres)A

1

Yielded

Steel

1

2

2

3

3

No Residual Stress

= Yielded


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P=FyA

yL0

RESIDUAL STRESSES

P=(Fy-Fres)A

1

Yielded

Steel

1

2

2

3

3

Effects of Residual

Stress

4

304

No Residual Stress


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Euler Buckling

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Assumptions:

Column is pin-ended.

Column is initially perfectly straight.

Load is at centroid.

Material is linearly elastic (no yielding).

Member bends about principal axis (no twisting).

Plane sections remain Plane.

Small Deflection Theory.

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Euler Buckling


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PE =

divide byA, PE/A = , then with r2 =I/A,

PE/A = FE =

FE = Euler (elastic) buckling stress

L/r= slenderness ratio

2

2

L

EI

2

2

AL

EI

( )22

rLE

Re-write in terms of stress:

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Euler Buckling


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E

P

2

2

L

EIPE=

Stable Equilibrium

Bifurcation Point

Euler Buckling

P

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Dependant onIminandL2.

Independent of Fy.

L

PE 2

2

L

EIx

2

2

L

EIy

Minor axis buckling

For similar unbraced length in each direction,minor axis (Iyin a W-shape) will control strength.

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Major axis buckling

Euler Buckling


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Buckling controlled by largest value ofL/r.

Most slender section buckles first.

L/r

FE

( )22

rL

EFy

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Euler Buckling


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( )

2

2

e

EIP

KL

=

( )

2

2

e

EI

FKL

r

=

2

2

2

2

)2/1(

4

L

EI

L

EIPE

==

Similar to pin-pin,

withL =L/2.

Load Strength =

4 times as large.

EXAMPLE

KL

Set up equilibrium and solve

similarly to Euler buckling

derivation.

Determine a K-factor.

End Restraint(Fixed)

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Length of equivalent pin ended

column with similar elastic

buckling load,

Effective Length = KL

End Restraint(Fixed)

Distance between points ofinflection in the buckled shape.

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Compression Theory 45


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EULER ASSUMPTIONS

(ACTUAL BEHAVIOR)

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Fy

KL/r

2

2

=

r

KL

EFE

Experimental Data

Overall Column Strength

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0 = initial mid-span deflection of column

Initial Crookedness/Out of Straight

P

P

M = Po

o

50

o


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P

2

2

L

EIPE =

o= 0

o

Elastic theory

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P

2

2

L

EIPE =

o

= 0

o

Elastic theory

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Actual Behavior



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Buckling is not instantaneous.

ASTM limits of 0 = L/1000 or 0.25 in 20 feetTypical values are 0 = L/1500 or 0.15 in 20 feet

Additional stresses due to bending of the column,

P/AMc/I.

Assuming elastic material theory (never yields),

Papproaches PE.

Actually, some strength loss

small 0=> small loss in strengthslarge 0 => strength loss can be substantial

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P

e

L

Load Eccentricity

55

P

2

2

L

EIPE =

o= 0

Elastic theory


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If moment is significant section must be designed

as a member subjected to combined loads.

Buckling is not instantaneous.

Additional stresses due to bending of the column,

P/AMc/I.

Assuming elastic material theory (never yields),

Papproaches PE.

Actually, some strength loss

small e => small loss in strengthslarge e => strength loss can be substantial

57

Load Eccentricity


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Fy

ET= Tangent Modulus

E

(Fy-Fres)

Test Results from an Axially Loaded Stub Column

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Inelastic Material Effects


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KL/r

2

2

=

r

KL

EFe


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Elastic Behavior


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KL/r

2

2

=

r

KL

EFe

60

Fy-Fres

Fy

2

2

=

r

KL

EF Tc

Inelastic

Elastic



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KL/r

2

2

=

r

KL

EFe

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Fy-Fres

Fy

2

2

=

r

KL

EF Tc

Inelastic

Elastic



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Elastic Buckling:ET =E

No yielding prior to bucklingFe Fy-Fres(max)Fe = predicts buckling (EULER BUCKLING)

Two classes of buckling:

Inelastic Buckling:Some yielding/loss of stiffness prior to buckling

Fe > Fy-Fres(max)

Fc - predicts buckling (INELASTIC BUCKLING)

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O ll C l St th


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Fy

KL/r

2

2

=

r

KL

EFE

Experimental Data


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O ll C l St th


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Major factors determining strength:

1) Slenderness (L/r).

2) End restraint (Kfactors).

3) Initial crookedness or load eccentricity.

4) Prior yielding or residual stresses.


The latter 2 items are highly variable between specimens.

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LOCAL BUCKLING

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Local Buckling is related to Plate Buckling

Flange is restrained by the web at one edge.

Failure is localized at areas of high stress

(maximum moment) or imperfections.

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Failure is localized at

areas of high stress

(maximum moment) or

imperfections.

Web is restrained by the flanges.

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(maximum moment) or

imperfections.


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(maximum moment) or

imperfections.


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Chapter E:

Compression Strength

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c= 0.90 (c= 1.67)

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Specification considers the following conditions:

Flexural Buckling

Torsional BucklingFlexural-Torsional Buckling

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Compressive Strength

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The following slides assume:

Non-slender flange and web sections

Doubly symmetric members

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Since members are non-slender and doubly symmetric,

flexural (global) buckling is the most likely potential failure

mode prior to reaching the squash load.

Buckling strength depends on the slenderness of the section,

defined as KL/r.

The strength is defined asPn= FcrAg Equation E3-1

81


EKLFy


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Fe = elastic (Euler) buckling stress, Equation E3-4

If , then Fcr = 0.877Fe. Equation E3-3

This defines the elastic buckling limitwith a reduction factor, 0.877, times the theoretical limit.

If , then . Equation E3-2

This defines the inelastic buckling limit.

yF

E.

r

KL714 y

F

cr F.F e

= 6580

yF

E.

r

KL714>

2

2

=

r

KL

E

Fe

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I l i M i l Eff


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KL/r

2

2

=

r

KL

EFe


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Elastic Behavior

I l ti M t i l Eff t


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KL/r

2

2

=

r

KL

EFe

84

Fy-Fres

Fy

2

2

=

r

KL

EF Tc

Inelastic

Elastic


I l ti M t i l Eff t


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KL/r

2

2

=

r

KL

EFe

85

Fy-Fres

Fy

2

2

=

r

KL

EF Tc

Inelastic

Elastic



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Design Aids

Table 4-22

cFcras a function of KL/r

Tables 4-1 to 4-20

cPnas a function of KLy

Useful for all shapes.Larger KL/rvalue controls.

Can be applied to KLxby

dividing KLyby rx/ry.

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L l B kli C it i


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Local Buckling Criteria

Slenderness of the flange and web, , are used as criteria todetermine whether local buckling might control in the elastic or

inelastic range, otherwise the global buckling criteria controls.

Criteria r are based on plate buckling theory.

For W-Shapes

FLB, = bf

/2tf

rf

=

WLB, = h/tw rw =

yF

E.560

yF

E.491

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Local Buckling


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> r slender element

Failure by local buckling occurs.

Covered in Section E7

Many rolled W-shape sections are dimensioned such

that the full global criteria controls.

90

g


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Slenderness Criteria

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Per Section E.2

Recommended to provide

KL/rless than 200

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