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

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

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 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)

Design of Members in Compression

• Selection of an economical shape: Find lightest shape

• Usually category is defined beforehand, e.g. W, WT etc

• Usually overall nominal dimensions defined in advance

because of architectural and other requirements.

USE OF COLUMN LOAD TABLES

IF NOT APPLICABLE - TRIAL AND ERROR

EXAMPLE I – COLUMN LOAD TABLES

A compression member is subjected to service loads pf 165 dead and 535 kips live. The member is 26 feet long and pinned at each end

LRFD

Calculate factored load

kips 054,1)535(6.1)165(2.16.12.1 LDPu

Required Design Strength

kips 054,1ncP

Enter Column Tables with KL=(1)(26)=26 ft

uPXW kips 230,1 :strengthdesign 14514 OK

EXAMPLE I – COLUMN LOAD TABLES

A compression member is subjected to service loads pf 165 dead and 535 kips live. The member is 26 feet long and pinned at each end

ASD

Calculate factored load

kips 700)535()165( LDPa

Required Allowable Strength

kips 700c

nP

Enter Column Tables with KL=(1)(26)=26 ft

aPXW kips 702 :strengthdesign 13214 OK

EXAMPLE Ii – COLUMN LOAD TABLES

Select the lightest W-shape that can resist a service dead load of 62.5 kips and a service live load of 125 kips. The effective length is 24 feet. Use ASTM A992 steel

LRFD

Calculate factored load and required strength

nu PLDP ckips 275)125(6.1)5.62(2.16.12.1

Enter Column Tables with KL=(1)(24)=24 ft

No Footnote: No need to check for local buckling

kips 275 with W8No:8 ncPφW

kips 282 ,5410 :10 c nPXWW

kips 293 ,5810 :12 c nPXWW

kips 293 ,6114 :14 c nPXWW

IF COLUMNS NOT APPLICABLE

1. Assume a value for Fcr

ycr FF

2. Determine required areaLRFD

crc

ugugcrc F

PAPAF

ASD

cr

ag

g

acr F

PA

A

PF

6.06.0

IF COLUMNS NOT APPLICABLE

3 Select a shape that satisfies area requirement

4 Compute Fcr for the trial shape

5 Revise if necessary• If available strength too close to required value try next tabulated value

• Else repeat 1-4 using Fcr of trial shape

6 Check local stability and revise if necessary

Example

Select a W18 shape of A992 steel that can resist a service dead load of 100 kips amd a service live load of 300 kips. Effective length KL=26 ft

Calculate factored load and required strength

kips 600)300(6.1)100(2.16.12.1 LDPu

Try ksi 333

2 ycr FF

Required Area 2in 2.20

339.0

600

crc

ug F

PA

Example

Select a W18 shape of A992 steel that can resist a service dead load of 100 kips amd a service live load of 300 kips. Effective length KL=26 ft

Try W 18x71

2.20in 8.20 2 gA

Slenderness 2005.18370.1

1226

min

r

KLOK

OKAvailable Area

Euler’s Stress

ksi 5.85.183

)000,29(

/ 2

2

2

2

rKL

EFe

11350

000,2971.471.4

yF

EElastic BucklingSlenderness Limit

5.183min

r

KL

ELASTIC BUCKLING

Example

Select a W18 shape of A992 steel that can resist a service dead load of 100 kips amd a service live load of 300 kips. Effective length KL=26 ft

Critical Stress ksi 455.75.8877.0877.0 ecr FF

NG

Design Strength

kips 140)8.20)(455.7(9.0 gcrcnc AFP kips 600

Example

Select a W18 shape of A992 steel that can resist a service dead load of 100 kips amd a service live load of 300 kips. Effective length KL=26 ft

Required Area 2in 3.33

209.0

600

crc

ug F

PA

Assume NEW Critical Stress

2

455.733 ksi 20crF

Example

Select a W18 shape of A992 steel that can resist a service dead load of 100 kips amd a service live load of 300 kips. Effective length KL=26 ft

Try W 18x119

3.33in 1.35 2 gA

Slenderness 2000.11669.2

1226

min

r

KLOK

OKAvailable Area

Euler’s Stress

ksi 27.210.116

)000,29(

/ 2

2

2

2

rKL

EFe

11350

000,2971.471.4

yF

EElastic BucklingSlenderness Limit

116min

r

KL

ELASTIC BUCKLING

Example

Select a W18 shape of A992 steel that can resist a service dead load of 100 kips amd a service live load of 300 kips. Effective length KL=26 ft

Critical Stress ksi 65.1827.21877.0877.0 ecr FF

Design Strength

kips 589)1.35)(65.18(9.0 gcrcnc AFP kips 600 NG

This is very close, try next larger size

Example

Select a W18 shape of A992 steel that can resist a service dead load of 100 kips amd a service live load of 300 kips. Effective length KL=26 ft

Try W 18x130

2in 2.38gA

Slenderness 2006.11570.2

1226

min

r

KLOK

Available Area

Euler’s Stress

ksi 42.216.115

)000,29(

/ 2

2

2

2

rKL

EFe

11350

000,2971.471.4

yF

EElastic BucklingSlenderness Limit

6.115min

r

KL

ELASTIC BUCKLING

Example

Select a W18 shape of A992 steel that can resist a service dead load of 100 kips amd a service live load of 300 kips. Effective length KL=26 ft

Critical Stress ksi 79.1842.21877.0877.0 ecr FF

OK

Design Strength

kips 646)2.38)(79.18(9.0 gcrcnc AFP kips 600

More on Effective Length Factor

Effective Length Factor-Alingnment Charts

Use alignment charts (Structural Stability Research Council SSRC)

AISC Commentary Figure C-C2.3 nad C-C2.4 p 16-.1-241

Connections to foundationsConnections to foundations(a) Hinge

G is infinite - Use G=10(b) Fixed

G=0 - Use G=1.0

Assumption of Elastic Behavior is violated whenInelastic Flexural Buckling min

71.4r

KL

F

E

y

Example

gg

cc

LI

LIG

Joint A

94.018/183020/1350

12/107012/833

AG

A

B

W12x120

C

W12x120

W24x68

W24x68W24x55

W24x55

12’

15’

12’W12x96

20’ 18’

Joint B

95.018/183020/1350

15/107012/1070

BG

Joint C

0.10CGPinned EndSway Uninhibited

Example

AISC Commentary Figure C-C2.3 nad C-C2.4 p 16-.1-241

COLUMN AB

94.0AG

95.0BG

3.1xK

COLUMN BC

0.10cG

95.0BG

85.1xK

More on Effective Length

Violated

Alingnment Charts & Inelastic Behavior

22

rKL

EFcr

elasticinelastic GE

E

LEI

LIEG t

gg

cct

elasticinelastic GG a

SRF: Table 4-21 AISC Manual pp 4-317

E

E

F

Ft

cr

cra

)elastic(

)inelastic(Stiffness Reduction Factor

22

rKL

EF t

cr

Elastic

Inelastic

Example

Compute Stiffness Reduction Factor per LRFD for an axial compressive stress of 25 ksi and Fy=50 ksi

ksi 25g

u

A

P

y

FF

gc

uinelasticcr F.

A

PF ey6580ksi 27.78

9.0

25)(

ksi 61.35506580ksi 27.78 50 eF F. e

23.3161.35877.08770)( eelasticcr F.F

890023.31

28.27

)(

)( .F

F

elasticcr

inelasticcra