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Compression Members

Compression Members. COLUMN STABILITY A. Flexural Buckling Elastic Buckling Inelastic Buckling Yielding B. Local Buckling – Section E7 pp 16.1-39 and

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Compression Members

COLUMN STABILITY

A. Flexural Buckling• Elastic Buckling• Inelastic Buckling• Yielding

B. Local Buckling – Section E7 pp 16.1-39 and B4 pp 16.1-14

C. Lateral Torsional Buckling

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

Design Strength

In Summary

877.0

44.0or

71.4 658.0

otherwiseF

FF

F

E

r

KLifF

F

e

ye

yy

F

F

cr

ey

200r

KL

Local Stability - Section B4 pp 16.1-14

Local Stability: If elements of cross section are thin LOCAL buckling occurs

The strength corresponding to any buckling mode cannot be developed

Local Stability - Section B4 pp 16.1-14

Local Stability: If elements of cross section are thin LOCAL buckling occurs

The strength corresponding to any buckling mode cannot be developed

Local Stability - Section B4 pp 16.1-14

• Stiffened Elements of Cross-Section

• Unstiffened Elements of Cross-Section

Local Stability - Section B4 pp 16.1-14

• Compact– Section Develops its full plastic stress before buckling

(failure is due to yielding only)

• Noncompact– Yield stress is reached in some but not all of its compression elements

before buckling takes place

(failure is due to partial buckling partial yielding)

• Slender– Yield stress is never reached in any of the compression elements

(failure is due to local buckling only)

Local Stability - Section B4 pp 16.1-14

If local buckling occurs cross section is not fully effectiveIf local buckling occurs cross section is not fully effectiveAvoid whenever possible

Measure of susceptibility to local bucklingMeasure of susceptibility to local bucklingWidth-Thickness ratio of each cross sectional element:

If cross section has slender elements - If cross section has slender elements - rr

Reduce Axial Strength (E7 pp 16.1-39 )

Slenderness Parameter - Limiting Values

AISC B5 Table B4.1 pp 16.1-16

Slenderness Parameter - Limiting Values

AISC B5 Table B4.1 pp 16.1-17

Slenderness Parameter - Limiting Values

AISC B5 Table B4.1 pp 16.1-18

Slender Cross Sectional Element:Strength Reduction E7 pp 16.1-39

Reduction Factor Q:

Q: B4.1 – B4.2 pp 16.1-40 to 16.1-43

877.0

44.0or

71.4 658.0

otherwiseF

QFF

QF

E

r

KLifQF

F

e

ye

yy

F

QF

cr

ey

Slender Cross Sectional Element:Strength Reduction E7 pp 16.1-39

Reduction Factor Q:

Qs, Qa: B4.1 – B4.2 pp 16.1-40 to 16.1-43

877.0

44.0or

71.4 658.0

otherwiseF

QFF

QF

E

r

KLifF

F

e

ye

yy

F

QF

cr

ey

Q=QsQa

COLUMN STABILITY

A. Flexural Buckling• Elastic Buckling• Inelastic Buckling• Yielding

B. Local Buckling – Section E7 pp 16.1-39 and B4 pp 16.1-14

C. Torsional, Lateral/Torsional Buckling

Torsional & Flexural Torsional Buckling

When an axially loaded member becomes unstable overall(no local buckling) it buckles one of the three ways

• Flexural Buckling• Torsional Buckling• Flexural-Torsional

Buckling

Torsional Buckling

Twisting about longitudinal axis of memberOnly with doubly symmetrical cross sections with slender cross-

sectional elements

Standard Hot-Rolled Shapes are NOT susceptible

Built-Up Members should be investigated

Cruciform shape particularly vulnerable

Flexural Torsional Buckling

Combination of Flexural and Torsional BucklingOnly with unsymmetrical cross sections

1 Axis of Symmetry: channels, structural tees, double-angle, equal length single angles

No Axis of Symmetry: unequal length single angles

Torsional Buckling

yxz

we IIGJLK

ECF

1

2

2Eq. E4-4

Cw = Warping Constant (in6)Kz = Effective Length Factor for Torsional Buckling

(based on end restraints against twisting)G = Shear Modulus (11,200 ksi for structural steel)J = Torsional Constant

Lateral Torsional Buckling 1-Axis of Symmetry

2411

2 ezey

ezeyezeye FF

HFF

H

FFF AISC Eq. E4-5

2

2

yy

eyrLK

EF

22

2 1

ogz

wez

rAGJ

LK

ECF

2

22

1o

oo

r

yxH

g

yxooo

A

IIyxr

222

oo yx , Coordinates of shear center w.r.t centroid of section

Lateral Torsional Buckling No Axis of Symmetry

02

2

22

o

oexee

o

oeyee

ezeeyeexe

r

yFFF

r

xFFF

FFFFFF

AISC Eq. E4-6

Fe is the lowest root of theCubic equation

In Summary - Definition of Fe

Elastic Buckling Stress corresponding to the controlling mode of failure (flexural, torsional or flexural torsional)

Fe:

Theory of Elastic Stability (Timoshenko & Gere 1961)

Flexural Buckling Torsional Buckling2-axis of symmetry

Flexural Torsional Buckling1 axis of symmetry

Flexural Torsional BucklingNo axis of symmetry

2

2

/ rKL

EFe

AISC EqtnE4-4

AISC EqtnE4-5

AISC EqtnE4-6

Column Strength

877.0

44.0 658.0

otherwiseF

FFifF

F

e

yeyF

F

cr

ey

gcrn AFP

EXAMPLE

Compute the compressive strength of a WT12x81 of A992 steel.

Assume (KxL) = 25.5 ft, (KyL) = 20 ft, and (Kz L) = 20 ft

20043.8750.3

125.25

x

x

r

LK

r

KLOK

43.8711350

000,2971.471.4

yF

E

ksi 44.3743.87

000,292

2

2

2

rKL

EFe

ksi 59.28)50(658.0658.0 44.37

50

yF

F

cr FF e

y

Inelastic Buckling

FLEXURAL Buckling – X axisWT 12X81

Ag=23.9 in2

rx=3.50 in

ry=3.05 in

kips 3.683)9.23(59.28 gcrn AFP

EXAMPLE

20069.7805.3

1220

y

y

r

LKOK

ksi 22.46

69.78

000,292

2

2

2

yy

eyrLK

EF

FLEXURAL TORSIONAL Buckling – Y axis (axis of symmetry)

WT 12X81

Ag=23.9 in2

rx=3.50 in

ry=3.05 in

y=2.70 in

tf=1.22 in

Ix=293 in4

Iy=221 in4

J=9.22 in4

Cw=43.8 in6

00 x

20ft

yy

87.259.23

221293)09.2(0 2

222

g

yxooo

A

IIyxr

Shear Center

EXAMPLE

FLEXURAL TORSIONAL Buckling – Y axis (axis of symmetry)

WT 12X81

Ag=23.9 in2

rx=3.50 in

ry=3.05 in

y=2.70 in

tf=1.22 in

Ix=293 in4

Iy=221 in4

J=9.22 in4

Cw=43.8 in6

ksi

rAGJ

LK

ECF

ogz

wez

4.167

87.259.23

1)22.9(200,11

1220

)8.43)(000,29(

1

22

2

22

2

EXAMPLE

FLEXURAL TORSIONAL Buckling – Y axis (axis of symmetry)

WT 12X81

Ag=23.9 in2

rx=3.50 in

ry=3.05 in

y=2.70 in

tf=1.22 in

Ix=293 in4

Iy=221 in4

J=9.22 in4

Cw=43.8 in6

ksi

FF

HFF

H

FFF

ezey

ezeyezeye

63.53

4.16722.46

8312.04.16722.46411

8312.02

4.16722.46

411

2

2

8312.0

87.25

090.2011

2

2

22

o

oo

r

yxH

EXAMPLE

FLEXURAL TORSIONAL Buckling – Y axis (axis of symmetry)

WT 12X81

Ag=23.9 in2

rx=3.50 in

ry=3.05 in

y=2.70 in

tf=1.22 in

Ix=293 in4

Iy=221 in4

J=9.22 in4

Cw=43.8 in6

Elastic or Inelastic LTB?

63.430.22)50(44.044.0 ey FksiF

877.0

44.0 658.0

otherwiseF

FFifF

F

e

yeyF

F

cr

ey

EXAMPLE

FLEXURAL TORSIONAL Buckling – Y axis (axis of symmetry)

WT 12X81

Ag=23.9 in2

rx=3.50 in

ry=3.05 in

y=2.70 in

tf=1.22 in

Ix=293 in4

Iy=221 in4

J=9.22 in4

Cw=43.8 in6

ksi

FF yF

F

cr

ey

59.2850658.0

658.0

63.43

50

kips7.739)70.2(95.30 gcrn AFP

Compare to FLEXURAL Buckling – X axis

kips 3.683)9.23(82.21 gcrn AFP

Column Design Tables

Assumption : Strength Governed by Flexural BucklingCheck Local Buckling

Column Design Tables

Design strength of selected shapes for effective length KLTable 4-1 to 4-2, (pp 4-10 to 4-316)

Critical Stress for Slenderness KL/rtable 4.22 pp (4-318 to 4-322)

EXAMPLE

Compute the available compressive strength of a W14x74 A992 steel compression member. Assume pinned ends and L=20 ft. Use (a) Table 4-22 and (b) column load tables

(a) LRFD - Table 4-22 – pp 4-318

20077.9648.2

)12)(20)(1(Maximum

yr

KL

r

KL

Table has integer values of (KL/r) Round up or interpolate

Fy=50 ksi

ksiPcr 67.22

ksiAPP gcrn 494)8.21(67.22

EXAMPLE

Compute the available compressive strength of a W14x74 A992 steel compression member. Assume pinned ends and L=20 ft. Use (a) Table 4-22 and (b) column load tables

(b) LRFD Column Load Tables

ftKL 20)20)(1(Maximum Tabular values based on minimum radius of gyration

Fy=50 ksi

kipsPnc 494

Example II

A W12x58, 24 feet long in pinned at both ends and braced in the weak direction at the third points. A992 steel is used. Determine available compressive strength

20025.3851.2

)12)(8(1

y

y

r

LK

20055.5428.5

)12)(24(1

x

x

r

LK

Enter table 4.22 with KL/r=54.55 (LRFD)

28.5xr

51.2yrksiPcr 24.36

kips

APP gcrn

616

)17(24.36

17gA

Example II

A W12x58, 24 feet long in pinned at both ends and braced in the weak direction at the third points. A992 steel is used. Determine available compressive strength

20025.3851.2

)12)(8(1

y

y

r

LK

20055.5428.5

)12)(24(1

x

x

r

LK

Enter table 4.22 with KL/r=54.55 (ASD)

28.5xr

51.2yr

ksiF

c

cr 09.24

kipsAFP

gc

cr

c

n 410

17gA

Example II

A W12x58, 24 feet long in pinned at both ends and braced in the weak direction at the third points. A992 steel is used. Determine available compressive strength

20025.3851.2

)12)(8(1

y

y

r

LK

20055.5428.5

)12)(24(1

x

x

r

LKCAN I USE Column Load Tables?

yx

x

rr

LKKL

Not Directly because they are based on min r (y axis buckling)

If x-axis buckling enter table with

Example II

A W12x58, 24 feet long in pinned at both ends and braced in the weak direction at the third points. A992 steel is used. Determine available compressive strength

20025.3851.2

)12)(8(1

y

y

r

LK

20055.5428.5

)12)(24(1

x

x

r

LK X-axis buckling enter table with

ftrr

LKKL

yx

x 43.111.2

)24)(1(

kipsPn 616