members in compression - iv

7/29/2019 Members in Compression - IV

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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

g cr n A F P

AISC Eqtn E3-1



AISC Requirements

LRFD

ncu P P

loadsfactoredof Sumu P

strengthecompressivdesignnc P

0.90ncompressiofor factor resistance c



Design Strength



In Summary

877.0

44.0or

71.4658.0

otherwise F

F F

F

E

r

KLif F

F

e

ye

y

y F

F

cr

e

y

200

r

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: If elements of cross section are thin LOCAL buckling occurs

The strength corresponding to any buckling mode cannot be developed




• Stiffened Elements of Cross-Section

• Unstiffened Elements of Cross-Section




• 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)




If local buckling occurs cross section is not fully effective Avoid whenever possible

Measure of susceptibility to local buckling

Width-Thickness ratio of each cross sectional element: l

If cross section has slender elements - l lr

Reduce Axial Strength (E7 pp 16.1-39 )



Slenderness Parameter - Limiting Values

AISC B5 Table B4.1

pp 16.1-16




AISC B5 Table B4.1

pp 16.1-17




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

71.4658.0

otherwise F

QF F

QF

E

r

KLif QF

F

e

ye

y y

F

QF

cr

e

y



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

71.4658.0

otherwise F

QF F

QF

E

r

KLif F

F

e

ye

y

y F

QF

cr

e

y

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 member Only 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, structuraltees, double-angle, equal

length single angles

No Axis of Symmetry: unequal length singleangles



Torsional Buckling

y x z

we

I I GJ L K

EC F

12

2 Eq. 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

24

112 ez ey

ez eyez ey

e F F

H F F

H

F F F AISC Eq. E4-5

22

y y

eyr L K

E F

22

21

o g z

wez

r AGJ

L K

EC F

2

22

1o

oo

r

y x H g

y xooo

A I I y xr

222

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



Lateral Torsional Buckling No Axis of Symmetry

0

2

2

2

2

o

oexee

o

oeyee

ez eeyeexe

r

y F F F

r x F F F

F F F F F F

AISC Eq. E4-6

Fe is the lowest root of the

Cubic equation



In Summary - Definition of F e

Elastic Buckling Stress corresponding to the controlling mode of

failure (flexural, torsional or flexural torsional)F

e :

Theory of Elastic Stability (Timoshenko & Gere 1961)

Flexural Buckling Torsional Buckling

2-axis of symmetry

Flexural Torsional

Buckling

1 axis of symmetry

Flexural Torsional

Buckling

No axis of symmetry

22

/ r KL E F e AISC Eqtn

E4-4AISC EqtnE4-5

AISC EqtnE4-6



Column Strength

877.0

44.0658.0

otherwise F

F F if F

F

e

ye y F

F

cr

e

y

g cr n A F P



EXAMPLE

Compute the compressive strength of a WT12x81 of A992 steel.

Assume (K xL) = 25.5 ft, (K yL) = 20 ft, and (K z L) = 20 ft

20043.8750.3

125.25

x

x

r

L K

r

KL

OK

43.8711350

000,2971.471.4

y F

E

ksi44.3743.87

000,292

2

2

2

r KL

E

F e

ksi59.28)50(658.0658.0 44.37

50

y

F

F

cr F F e

y

Inelastic Buckling

FLEXURAL Buckling – X axis

WT 12X81

Ag=23.9 in2

r x=3.50 in

r y=3.05 in

kips3.683)9.23(59.28 g cr n

A F P



EXAMPLE

20069.7805.3

1220

y

y

r

L K OK

ksi22.46

69.78

000,292

2

2

2

y y

eyr L K

E F

FLEXURAL TORSIONAL Buckling – Y axis (axis of symmetry)

WT 12X81

Ag=23.9 in2

r x=3.50 in

r y=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

20

f t y y

87.259.23

221293)09.2(02

222

g

y x

ooo A

I I

y xr

Shear Center



EXAMPLE


WT 12X81

Ag=23.9 in2

r x=3.50 in

r y=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

r AGJ

L K EC F

o g z

wez

4.16787.259.23

1)22.9(200,11

1220

)8.43)(000,29(

1

22

2

22

2



EXAMPLE


WT 12X81

Ag=23.9 in2

r x=3.50 in

r y=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

F F

H F F

H

F F F

ez ey

ez eyez ey

e

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

y x H



EXAMPLE


WT 12X81

Ag=23.9 in2

r x=3.50 in

r y=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

e y F ksi F

877.0

44.0658.0

otherwise F

F F if F

F

e

ye y F

F

cr

e

y



EXAMPLE


WT 12X81

Ag=23.9 in2

r x=3.50 in

r y=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

F F y F

F

cr

e

y

59.2850658.0

658.0

63.43

50

kips7.739)70.2(95.30 g cr n A F P

Compare to FLEXURAL Buckling – X axis

kips3.683)9.23(82.21 g cr n A F P



Column Design Tables

Assumption : Strength Governed by Flexural Buckling

Check Local Buckling

Column Design Tables

Design strength of selected shapes for effective length KL

Table 4-1 to 4-2, (pp 4-10 to 4-316)

Critical Stress for Slenderness KL/r

table 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

ksi P cr 67.22

ksi A P P g cr n 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

f t KL 20)20)(1(Maximum

Tabular values based on minimum radius of gyration

Fy=50 ksi

kips P nc 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

L K

20055.54

28.5

)12)(24(1

x

x

r

L K

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

28.5 xr

51.2 yr ksi P cr 24.36

kips

A P P g cr n

616

)17(24.36

17 g A



Example II




20025.3851.2

)12)(8(1

y

y

r

L K

20055.54

28.5

)12)(24(1

x

x

r

L K

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

28.5 xr

51.2 yr

ksi F

c

cr 09.24

kips A

F P

g c

cr

c

n

410

17 g A



Example II




20025.3851.2

)12)(8(1

y

y

r

L K

20055.5428.5

)12)(24(1

x

x

r

L K CAN I USE Column Load Tables?

y x

x

r r

L K KL

Not Directly because they are

based on min r (y axis buckling)

If x-axis buckling enter table with



Example II




20025.3851.2

)12)(8(1

y

y

r

L K

20055.5428.5

)12)(24(1

x

x

r

L K X-axis buckling enter table with

ft r r

L K KL

y x

x 43.111.2

)24)(1(

kips P n 616

members in compression - iv

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