Compression Members

Compression Members

• Compression members are susceptible to BUCKLING

• BUCKLING – Loss of stability– Axial loads cause lateral deformations (bending-like deformations)

P is applied slowlyP increasesMember becomes unstable - buckles

Column Theory

Axial force that causes Buckling is called Critical Load and is associated to the column strength

Pcr depends on

• Length of member• Material Properties• Section Properties

Column Theory - Euler Elastic Buckling

02

2

yEIP

dzyd

Governing Equation

EIPk),kzcos(B)kzsin(Ay 2 Solution

(kL)sinAyyLz@Byz@

0 0 0 0Apply Boundary

Conditions

nLEIPnkL

A

Ignore Solution Trivial - deflection No0

22

2

2

,1rLEA

LEIPn cr

P

P

y

z

Column Theory - Euler Buckling

2

22

LEInPcr

Column Theory - Euler Buckling

22

2

2

,1rLEA

LEIPn cr

gyration of radiusAIr

Elastic Buckling ya fAPf

Assumptions

• Column is perfectly straight

• The load is axial, with no eccentricity

• The column is pinned at both ends

No Moments

Need to account for other boundary conditions

Other Boundary Conditions

22

2rLEAPcr

22

5.0rLEAPcr

22

7.0rLEAPcr

Fixed on bottom

Free to rotate and translate

Fixed on bottom

Fixed on top

Fixed on bottom

Free to rotate

Other Boundary Conditions

In generalIn general

22

rKLEAPcr

K: Effective Length FactorK: Effective Length Factor

LRFD Commentary Table C-C2.2 p 16.1-240

Effective Length Factor

Column Theory - Column Strength Curve

Example I

A W12x50 is used as a column to support a compressive load of 145 kips. The length is 20 ft and the ends are pinned. Without regard to LRFD or ASD investigate the stability of the column

For a W12x50

in.96.1Minimum yrr

4.12296.1

)12(20Maximum yrL

rL

kips 9.278

4.122)6.14)(000,29(

2

2

2

2

rLEAPcr

kips 145 kips 9.278 crP OK – Column is Stable

Example I

A W12x50 is used as a column to support a compressive load of 145 kips. The length is 20 ft the bottom is fixed and the top is free. Without regard to LRFD or ASD investigate the stability of the column

For a W12x50

in.96.1Minimum yrr

4.12296.1

)12(20Maximum yrL

rL

kips 248.63

)4.122)(1.2()6.14)(000,29(

2

2

2

2

rKLEAPcr

kips 145 kips 248.63 crP NG – Column is Unstable

AISC Requirements

CHAPTER E pp 16.1-32

Nominal Compressive Strength

gcrn AFP

AISC Eqtn E3-1

AISC Requirements

LRFD

ncu PP

loads factored of Sum uP

strength ecompressiv design ncP

0.90 ncompressiofor factor resistance c

AISC Requirements

ASD

c

na

PP

loads service of Sum aP

strength ecompressiv allowable cnP

1.67 ncompressiofor factor safety c

AISC Requirements

ASD – Allowable Stress

aa Ff

gaa APf stress ecompressiv axial computed

crcr

c

cr

a

FFFF

6.067.1

stress ecompressiv axial allowable

To compute Fcr – ELASTIC BUCKLING

22

rKLEAPe

4)- E3 Eq.(AISC 2

2

rKLE

APF e

e

Recall Assumptions

Assumptions

• Column is perfectly straight

• The load is axial, with no eccentricity

• The column is pinned at both ends

To compute Fcr – ELASTIC BUCKLING

4)- E3 Eq.(AISC 2

2

rKLE

APF e

e

877.0 ecr FF Accounts for Imperfections

To compute Fcr – INELASTIC BUCKLING

22

rKL

AEP te

yF

F

cr FF ey

658.0

Design Strength

Alternatively

e

e

FE

rKL

rKLEF

2

2

2

yFE

rKL 71.4

ye FE

FE 71.4

2

ye FF 44.0Inelastic Buckling

In Summary

877.0

44.0or

71.4 658.0

otherwiseF

FF

FE

rKLifF

F

e

ye

yy

F

F

cr

ey

200rKL

Example

A W14x74 of A992 steel has a length of 20 feet and pinned ends. Compute the design strength for LRFD and the allowable compressive strength for ASD

Slenderness Ratio

20077.9648.2

)12)(20)(1(Maximum yrKL

rKL

11350000,2971.471.4

yFE

BucklingInelasticFE

rKL

y

71.4

Example

ksi 56.3077.96

)000,29(2

2

2

2

rKLEFe

ksi 21.25658.0658.0 56.3050

e

y

FF

crF

kips 6.549)8.21(21.25 gcrn AFP

Example

kips 6.549)8.21(21.25 gcrn AFP

LRFD

kips 4956.549)9.0(9.0 gcrnc AFPASD

ksi 31.15)21.25)(6.0(6.0 cra FF

kips 330)8.21(31.15 gaAF

Homework

4.3-14.3-44.3-6