68402 Slide # 1

Design of Compression Members

Monther DwaikatAssistant Professor

Department of Building EngineeringAn-Najah National University

68402: Structural Design of Buildings II61420: Design of Steel Structures62323: Architectural Structures II

68402 Slide # 2

Short and long columns

Buckling load and buckling failure modes

Elastic and Inelastic buckling

Local buckling

Design of Compression Members

Effective Length for Rigid Frames

Torsional and Flexural-Torsional Buckling

Design of Singly Symmetric Cross Sections

Design of Compression Members

68402 Slide # 3

Axially Loaded Compression Members

Columns Struts Top chords of trusses Diagonal members of trusses

Column and Compression member are often used interchangeably

68402 Slide # 4

Axially Loaded Compression Members

Commonly Used Sections:

• W/H shapes

• Square and Rectangular or round HSS

• Tees and Double Tees

• Angles and double angles

• Channel sections

68402 Slide # 5

Columns Failure modes (limit states):

• Crushing (for short column)

• Flexural or Euler Buckling (unstable under bending)

• Local Buckling (thin local cross section)

68402 Slide # 6

Short Columns Compression Members: Structural elements that are

subjected to axial compressive forces only are called columns. Columns are subjected to axial loads thru the centroid.

Stress: The stress in the column cross-section can be calculated as

f - assumed to be uniform over the entire cross-section.

Short columns - crushing

A

Pf

68402 Slide # 7

Long Columns This ideal state is never reached. The stress-state will be

non-uniform due to:• Accidental eccentricity of loading with respect to the centroid

• Member out-of –straightness (crookedness), or

• Residual stresses in the member cross-section due to fabrication processes.

Accidental eccentricity and member out-of-straightness can cause bending moments in the member. However, these are secondary and are usually ignored.

Bending moments cannot be neglected if they are acting on the member. Members with axial compression and bending moment are called beam-columns.

“Long” columns

68402 Slide # 8

Long Columns The larger the slenderness ratio (L/r), the greater the

tendency to buckle under smaller load Factors affecting tendency to buckle:

• end conditions

• unknown eccentricity (concentric & eccentric loads)

• imperfections in material

• initial crookedness

• out of plumbness

• residual stress

• buckling can be on one or both axes (major or minor axis)

68402 Slide # 9

Column Buckling Consider a long slender

compression member. If an axial load P is applied and increased slowly, it will ultimately reach a value Pcr that will cause buckling of the column. Pcr is called the critical buckling load of the column.

Pcr

Pcr

P

P

(a) (b)Pcr

Pcr

P

P

P

P

(a) (b)

Figure 1. Buckling of axially loaded compression members

68402 Slide # 10

Buckling Load

PM

• Now assume we have a pin connected column. If we apply a similar concept to that before here we find

Pcr

P

• The internal resisting moment M in the column is

P

Pcr

M

• We can write the relationship between the deflected shape and the Moment M

EI

P

EI

M

dx

d cr

2

2

02

2

EI

P

dx

d cr

x

x

crP - The load at which bucking starts to happen (Critical buckling load)

2

2

L

EIPcr

68402 Slide # 11

Column Buckling What is buckling?

Buckling occurs when a straight column subjected to axial compression suddenly undergoes bending as shown in the Figure 1(b). Buckling is identified as a failure limit-state for columns.

• The critical buckling load Pcr for columns is theoretically given by Equation (3.1):

• [3.1]

I - moment of inertia about axis of buckling.

K - effective length factor based on end boundary conditions.

22

LK

IEPcr

68402 Slide # 12

Effective Length

KL- Distance between inflection points in column.

K- Effective length factor

L- Column unsupported length

KL=L KL=0.5L LKL=0.7L L

K = 1.0 K = 0.5 K = 0.7

68402 Slide # 13

Column Buckling

Boundary conditions

Table C-C2.2Approximate Values of Effective Length Factor, K

68402 Slide # 14

Ex. 3.1- Buckling Loads Determine the buckling strength

of a W 12 x 50 column. Its length is 6 m. For minor axis buckling, it is pinned at both ends. For major buckling, is it pinned at one end and fixed at the other end.

x

y

68402 Slide # 15

Ex. 3.1- Buckling LoadsStep I. Visualize the problem

• For the W12 x 50 (or any wide flange section), x is the major axis and y is the minor axis. Major axis means axis about which it has greater moment of inertia (Ix > Iy).

Step II. Determine the effective lengths

• According to Table C-C2.2:• For pin-pin end conditions about the minor axis

• Ky = 1.0 (theoretical value); and Ky = 1.0 (recommended design value)

• For pin-fix end conditions about the major axis

• Kx = 0.7 (theoretical value); and Kx = 0.8 (recommended design value).

• According to the problem statement, the unsupported length for buckling about the major (x) axis = Lx = 6 m.

68402 Slide # 16

Ex. 3.1- Buckling Loads• The unsupported length for buckling about the minor (y) axis = Ly =

6 m.

• Effective length for major (x) axis buckling = Kx Lx = 0.8 x 6 = 4.8 m.

• Effective length for minor (y) axis buckling = Ky Ly = 1.0 x 6 = 6 m.

Step III. Determine the relevant section properties

• Elastic modulus of elasticity = E = 200 GPa (constant for all steels)

• For W12 x 50: Ix = 163x106 mm4. Iy = 23x106 mm4

68402 Slide # 17

Ex. 3.1- Buckling LoadsStep IV. Calculate the buckling strength

• Critical load for buckling about x - axis = Pcr-x =

Pcr-x = = 13965 kN.

Critical load for buckling about y-axis = Pcr-y =

Pcr-y = = 1261 kN.

Buckling strength of the column = smaller (Pcr-x, Pcr-y) = Pcr = 1261 kN.

Minor (y) axis buckling governs.

2 6

2

200 163 10

4800

22

yy

y

LK

IE

2 6

2

200 23 10

6000

22

xx

x

LK

IE

68402 Slide # 18

Ex. 3.1- Buckling Loads

Notes:• Minor axis buckling usually governs for all doubly

symmetric cross-sections. However, for some cases, major (x) axis buckling can govern.

• Note that the steel yield stress was irrelevant for calculating this buckling strength.

68402 Slide # 19

Let us consider the previous example. According to our calculations Pcr = 1261 kN. This Pcr will cause a uniform stress f = Pcr/A in the cross-section.

For W12 x 50, A = 9420 mm2. Therefore, for Pcr = 1261 kN; f = 133.9 MPa.

The calculated value of f is within the elastic range for a 344 MPa yield stress material.

However, if the unsupported length was only 3 m, Pcr =would be calculated as 5044 kN, and f = 535.5 MPa.

This value of f is ridiculous because the material will yield at 344 MPa and never develop f = 535.5 kN. The member would yield before buckling.

Inelastic Column Buckling

68402 Slide # 20

Inelastic Column Buckling Eq. (3.1) is valid only when the material everywhere in the cross-

section is in the elastic region. If the material goes inelastic then Eq. (3.1) becomes useless and cannot be used.

What happens in the inelastic range? Several other problems appear in the inelastic range.

• The member out-of-straightness has a significant influence on the buckling strength in the inelastic region. It must be accounted for.

• The residual stresses in the member due to the fabrication process causes yielding in the cross-section much before the uniform stress f reaches the

yield stress Fy.

• The shape of the cross-section (W, C, etc.) also influences the buckling strength.

• In the inelastic range, the steel material can undergo strain hardening. All of these are very advanced concepts and beyond the scope of this

course.

68402 Slide # 21

AISC Specifications for Column Strength The AISC specifications for column design are based on

several years of research. These specifications account for the elastic and inelastic

buckling of columns including all issues (member crookedness, residual stresses, accidental eccentricity etc.) mentioned above.

The specification presented here will work for all doubly symmetric cross-sections.

The design strength of columns for the flexural buckling limit state is equal to cPn

Where, c = 0.9 (Resistance factor for compression members)

68402 Slide # 22

Inelastic Buckling of Columns

PFF

• Elastic buckling assumes the material to follow Hooke’s law and thus assumes stresses below elastic (proportional) limit.

• If the stress in the column reaches the proportional limit then Euler’s assumptions are violated.

L/rL/r

Stress “F”Stress “F”

Euler assumptionsEuler assumptions

2

2

)/( rL

EFcr

Elastic Buckling (Long Columns)

Inelastic Buckling (Short columns)

Proportional Proportional limitlimit

68402 Slide # 23

AISC Specifications for Column Strength

Pu Pn

Pn = Ag Fcr [E3-1]

[E3-4]

Fe- Elastic critical Euler buckling load

Ag - gross member area

K - effective length factorL - unbraced length of the member r - governing radius of gyration

The 0.877 factor in Eq (E3-3) tries to account for initial crookedness.

22

rKL

EFe

ye

yyeFyF

crFErKLF

FErKLFF

71.4877.0

71.4658.0

Inelastic [E3-2]

Elastic [E3-3]

68402 Slide # 24

AISC Specifications For Column Strength

c = E

F

r

LK y

Fcr/Fy

1.0

1.5

0.39 Fcr = Fy

2c

877.0

Fcr = Fy 2c658.0

c = E

F

r

LK y

c =

E

F

r

LK y

Fcr/Fy

1.0

1.5

0.39 Fcr = Fy

2c

877.0Fcr = Fy

2c

877.0

Fcr = Fy 2c658.0 Fcr = Fy 2c658.0

r

KLyF

E71.4

yeFyF

cr FF

658.0

ecr FF 877.0

Inelastic Buckling (Short columns)

Elastic Buckling (Long columns)

68402 Slide # 25

AISC Specifications For Column Strength

For a given column section:

• Calculate I, Ag, r

• Determine effective length K L based on end boundary conditions.

• Calculate KL/r

• If KL/r is greater than , elastic buckling occurs and use Equation (E3.4)

• If KL/r is less than or equal to , inelastic buckling occurs and use Eq. (E3.3)

Note that the column can develop its yield strength Fy as KL/r approaches zero.

yF

E71.4

yF

E71.4

68402 Slide # 26

Ex. 3.2 - Column Strength Calculate the design strength of W14 x 74 with length of 6

m and pinned ends. A36 steel is used.

• Step I. Calculate the effective length and slenderness ratio for the problem

Kx = Ky = 1.0

Lx = Ly = 6 m

Major axis slenderness ratio = KxLx/rx = 6000/153.4 = 39.1

Minor axis slenderness ratio = KyLy/ry = 6000/63 = 95.2

• Step II. Calculate the buckling strength for governing slenderness ratio

The governing slenderness ratio is the larger of (KxLx/rx, KyLy/ry)

68402 Slide # 27

Ex. 3.2 - Column Strength• KyLy/ry is larger and the governing slenderness ratio;

• MPa

• Therefore, MPa

• Design column strength = cPn = 0.9 (Ag Fcr) = 0.9 (14060x154)/1000= 1948.7 kN.

• Design strength of column = 1948.7 kN.

200000/ 4.71 4.71 133.7

248y y yy

EK L r

F

0.658 154y eF F

cr yF F

2 2

2 2

3.1416 200000217.8

95.2/e

y y y

EF

K L r

68402 Slide # 28

Local Buckling Limit State

Figure 4. Local buckling of columns

The AISC specifications for column strength assume that column buckling is the governing limit state. However, if the column section is made of thin (slender) plate elements, then failure can occur due to local buckling of the flanges or the webs.

68402 Slide # 29

Local Buckling Limit State

• Local buckling is another limitation that represents the instability of the cross section itself.

• If local buckling occurs, the full strength of the cross section can not be developed.

68402 Slide # 30

If local buckling of the individual plate elements occurs, then the column may not be able to develop its buckling strength.

Therefore, the local buckling limit state must be prevented from controlling the column strength.

Local buckling depends on the slenderness (width-to-thickness b/t ratio) of the plate element and the yield stress (Fy) of the material.

Each plate element must be stocky enough, i.e., have a b/t ratio that prevents local buckling from governing the column strength.

The AISC specification provides the slenderness (b/t) limits that the individual plate elements must satisfy so that local buckling does not control.

Local Buckling Limit State

68402 Slide # 31

Local Buckling Limit State

• Local buckling can be prevented by limiting the width to thickness ratio known as “” to an upper limit r

yf F

E

t

b56.0

yw F

E

t

h49.1

b

hwt

ft

b

ft

68402 Slide # 32

The AISC specification provides two slenderness limits (p and r) for the local buckling of plate elements.

Local Buckling Limit State

Compact

Non-Compact

Slender

Compact

Non-Compact

Slender

b

t

F

Axial shortening,

Axi

al F

orce

, F

Fy

Compact

Non-Compact

Slender

Compact

Non-Compact

Slender

b

t

F

Axial shortening,

Axi

al F

orce

, F

Fy

68402 Slide # 33

• If the slenderness ratio (b/t) of the plate element is greater than r then it is slender. It will locally buckle in the elastic range before reaching Fy

• If the slenderness ratio (b/t) of the plate element is less than r but greater than p, then it is non-compact. It will locally buckle immediately after reaching Fy

• If the slenderness ratio (b/t) of the plate element is less than p, then the element is compact. It will locally buckle much after reaching Fy

If all the plate elements of a cross-section are compact, then the section is compact.• If any one plate element is non-compact, then the cross-section is

non-compact

• If any one plate element is slender, then the cross-section is slender.

Local Buckling Limit State

68402 Slide # 34

Local Buckling Limit State• Cross section can be classified as “compact”, “non compact” or

“slender” sections based on their width to thickness ratios

• If the cross-section does not satisfy local buckling requirements its critical buckling stress Fcr shall be reduced

• If then the section is slender, a reduction factor for capacity shall be computed from

• It is not recommended to use slender sections for columns.

r

AISC Manual for Steel Design

68402 Slide # 35

The slenderness limits p and r for various plate elements with different boundary conditions are given in the AISC Manual.

Note that the slenderness limits (p and r) and the definition of plate slenderness (b/t) ratio depend upon the boundary conditions for the plate.

• If the plate is supported along two edges parallel to the direction of compression force, then it is a stiffened element. For example, the webs of W shapes

• If the plate is supported along only one edge parallel to the direction of the compression force, then it is an unstiffened element. Ex., the flanges of W shapes.

The local buckling limit state can be prevented from controlling the column strength by using sections that are compact and non-compact.

Avoid slender sections

Local Buckling Limit State

68402 Slide # 36

Local Buckling Limit State

68402 Slide # 37

Ex. 3.3 – Local Buckling

2000001.49 1.49 42.3

248ry

E

F

Determine the local buckling slenderness limits and evaluate the W14 x 74 section used in Example 3.2. Does

local buckling limit the column strength? • Step I. Calculate the slenderness limits

See Tables in previous slide.

For the flanges of I-shape sections in pure compression

For the webs of I-shapes section in pure compression

Use E = 200000 MPa

2000000.56 0.56 15.9

248ry

E

F

68402 Slide # 38

Ex. 3.3 – Local Buckling• Step II. Calculate the slenderness ratios for the flanges and webs

of W14 x 74

• For the flanges of I-shape member, b = bf/2 = flange width / 2

Therefore, b/t = bf/2tf.

For W 14 x 74, bf/2tf = 6.41 (See Section Property Table)

• For the webs of I shaped member, b = hh is the clear distance between flanges less the fillet / corner radius of each flange

For W14 x 74, h/tw = 25.4 (See Section Property Table).

• Step III. Make the comparisons and comment

For the flanges, b/t < r. Therefore, the flange is non-compact

For the webs, h/tw < r. Therefore the web is non-compact

Therefore, the section is non-compactTherefore, local buckling will not limit the column strength.

68402 Slide # 39

Design of Compression Members • Steps for design of compression members

• Calculate the factored loads Pu

• Assume (a cross section) or (KL/r ratio between 50 to 90)

• Calculate the slenderness ratio KL/r and the ratio Fe

• Calculate c Fcr based on value of Fe

• Calculate the Area required Ag

• Choose a cross section and get (KxL/rx )and (KyL/ry) (KL/r) max

• Recalculate c Fcr and thus check

• Check local buckling requirements

crc

urequired F

PA

ucrcgnc PFAP

22

rKL

EFe

68402 Slide # 40

Ex. 3.4 – Design Strength Determine the design strength of an ASTM A992 W14 x

132 that is part of a braced frame. Assume that the physical length L = 9 m, the ends are pinned and the column is braced at the ends only for the X-X axis and braced at the ends and mid-height for the Y-Y axis.

• Step I. Calculate the effective lengths.

From Section Property Table

For W14 x 132: rx = 159.5 mm; ry = 95.5 mm; Ag =25030 mm2

Kx = 1.0and Ky = 1.0

Lx = 9 m and Ly = 4.5 m

KxLx = 9 m and KyLy = 4.5 m

68402 Slide # 41

Ex. 3.4 – Design Strength

344 620.50.658 344 272.8crF

• Step II. Determine the governing slenderness ratio

KxLx/rx = 9000/159.5 = 56.4

KyLy/ry = 4500/95.5 = 47.1

The larger slenderness ratio, therefore, buckling about the major

axis will govern the column strength.

• Step III. Calculate the column strength

2 2

2 2

200000620.5

56.4e

x x x

EF

K L r

MPa

200000/ 4.71 4.71 113.6

344x x xy

EK L r

F

MPa

68402 Slide # 42

Ex. 3.4 – Design Strength

5.1356.0 y

r F

E

• Step IV. Check the local buckling limits

For the flanges, bf/2tf = 7.15 <

For the web, h/tw = 17.7<

Therefore, the section is non-compact. OK.

5.1356.0 y

r F

E

0.9 25030 272.8 1000 6145nP kN

68402 Slide # 43

Ex. 3.5 – Column Design A compression member is subjected to service loads of 700 kN DL and 2400

kN of LL. The member is 7.8 m long & pinned at each end. Use A992 steel and select a W shape.

• Step I. Calculate the factored design load Pu

Pu = 1.2 PD + 1.6 PL = 1.2 x 700 + 1.6 x 2400 = 4680 kN.

• Step II. Calculate Fcr by assuming KL/r = 80

2 2

2 2

200000308.4

80e

EF

KL r

MPaMPa

344 308.40.658 344 215.3crF

200000/ 4.71 4.71 113.6

344y

EKL r

F

MPaMPa

68402 Slide # 44

Ex. 3.5 – Column Design• Step III. Calculate the required area of steel

A = 4680*1000/(0.9*215.3) = 24156 mm2

• Step IV. Select a W shape from the Section Property TablesSelect W14 x 132. It has A = 25030 mm2 OR W12 x 136 A = 25740 mm2

Select W14 x 132 because it has lower weight.

KyLy/ry =7800/95.5 = 81.7

Fe = 295.7 Fcr = 211.4 Pn = 4897.3 kN OKW14 x 145 is the lightest.

Section is non-compact but students have to check for that Note that column sections are usually W12 or W14. Generally sections bigger

than W14 are not used as columns.

68402 Slide # 45

Effective Length

Specific Values of K shall be known

For compression elements connected as rigid frames the effective length is a function of the relative stiffness of the element compared to the overall stiffness of the joint. This will be discussed later in this chapter

Values for K for different end conditions range from 0.5 for theoretically fixed ends to 1.0 for pinned ends and are given by:

Table C-C2.2 AISC Manual

End conditions K

Pin-Pin 1.0

Pin-Fixed 0.8

Fixed-Fixed 0.65

Fixed-Free 2.1

Recommended design values (not theoretical values)

68402 Slide # 46

• If we assume all connections are pinned then: Kx L = 3 m and Ky L = 6 m

• However the rigidity of the beams affect the rotation of the columns. Thus in rigid frames the K factor can be determined from the relative rigidity of the columns

• Determine a G factor

3 m

3 m

'10LK x

'20LK y

ggg

ccc

LIE

LIEG

/

/

• Where “c” represents column and “g” represents girder

• The G value is computed at each end of the member and K is computed factor from the monograms in

gg

cc

LI

LIG

/

/

AISC Manual – Figure C-C2.2

K Factor for Rigid Frames

3 m3 m

6 m6 m

68402 Slide # 47

Effective Length of Columns in Frames So far, we have looked at the buckling strength of

individual columns. These columns had various boundary conditions at the ends, but they were not connected to other members with moment (fix) connections.

The effective length factor K for the buckling of an individual column can be obtained for the appropriate end conditions from Table C-C2.2 of the AISC Manual .

However, when these individual columns are part of a frame, their ends are connected to other members (beams etc.). • Their effective length factor K will depend on the restraint offered

by the other members connected at the ends.

• Therefore, the effective length factor K will depend on the relative rigidity (stiffness) of the members connected at the ends.

68402 Slide # 48

Effective Length of Columns in Frames The effective length factor for columns in frames must be

calculated as follows:• First, you have to determine whether the column is part of a braced

frame or an unbraced (moment resisting) frame.

• If the column is part of a braced frame then its effective length factor 0 < K ≤ 1

• If the column is part of an unbraced frame then 1 < K ≤ ∞

• Then, you have to determine the relative rigidity factor G for both ends of the column

• G is defined as the ratio of the summation of the rigidity (EI/L) of all columns coming together at an end to the summation of the rigidity (EI/L) of all beams coming together at the same end.

It must be calculated for both ends of the column

b

b

c

c

L

IEL

IE

G

c: for columns

b: for beams

68402 Slide # 49

Effective Length of Columns in Frames

• Then, you can determine the effective length factor K for the column using the calculated value of G at both ends, i.e., GA and GB and the appropriate alignment chart

• There are two alignment charts provided by the AISC manual,

• One is for columns in braced (sidesway inhibited) frames. 0 < K ≤ 1

• The second is for columns in unbraced (sidesway uninhibited) frames. 1 < K ≤ ∞

• The procedure for calculating G is the same for both cases.

68402 Slide # 50

Effective LengthMonograph or

Jackson and Moreland

Alignment Chart

for Unbraced Frame

68402 Slide # 51

Effective LengthMonograph or

Jackson and Moreland

Alignment Chart

for braced Frame

68402 Slide # 52

Ex. 3.6 – Effective Length Factor Calculate the effective

length factor for the W12 x 53 column AB of the frame shown. Assume that the column is oriented in such a way that major axis bending occurs in the plane of the frame. Assume that the columns are braced at each story level for out-of-plane buckling. Assume that the same column section is used for the stories above and below.

10 ft.

10 ft.

12 ft.

15 ft.

20 ft.18 ft.18 ft.

W14 x 68

W14 x 68

W14 x 68

B

A

W12

x 7

9

W12

x 7

9

W12

x 7

9

10 ft.

10 ft.

12 ft.

15 ft.

20 ft.18 ft.18 ft.

W14 x 68

W14 x 68

W14 x 68

B

A

W12

x 7

9

W12

x 7

9

W12

x 7

9

5.4 m 5.4 m 6 m

4.5 m

3.6 m

3.0 m

3.0 m

68402 Slide # 53

Ex. 3.6 – Effective Length Factor• Step I. Identify the frame type and calculate Lx, Ly, Kx, and Ky if

possible.It is an unbraced (sidesway uninhibited) frame.

Lx = Ly = 3.6 m

Ky = 1.0

Kx depends on boundary conditions, which involve restraints due to beams and columns connected to the ends of column AB.

Need to calculate Kx using alignment charts.

• Step II. Calculate Kx

Ixx of W 12 x 53 = 425 in4 Ixx of W14x68 = 753 in4

021.1360.6

493.6

1220

723

1218

7231212

425

1210

425

L

IL

I

G

b

b

c

c

A

68402 Slide # 54

Ex. 3.6 – Effective Length Factor

Using GA and GB : Kx = 1.3 - from Alignment Chart on Page

16.1-242

• Step III. Design strength of the column KyLy = 1.0 x 12 = 12 ft.Kx Lx = 1.3 x 12 = 15.6 ft.rx / ry for W12x53 = 2.11(KL)eq = 15.6 / 2.11 = 7.4 ft.KyLy > (KL)eq

Therefore, y-axis buckling governs. Therefore cPn = 547 kips

835.0360.6

3125.5

1220

723

1218

7231215

425

1212

425

L

IL

I

G

b

b

c

c

B

68402 Slide # 55

Ex. 3.8 – Column Design Design Column AB of the frame shown below for a design

load of 2300 kN. Assume that the column is oriented in such a way that

major axis bending occurs in the plane of the frame. Assume that the columns are braced at each story level

for out-of-plane buckling. Assume that the same column section is used for the

stories above and below. Use A992 steel.

68402 Slide # 56

Ex. 3.8 – Column Design

10 ft.

10 ft.

12 ft.

15 ft.

20 ft.18 ft.18 ft.

W14 x 68

W14 x 68

W14 x 68

B

AW

12 x

79

W12

x 7

9

W12

x 7

9

10 ft.

10 ft.

12 ft.

15 ft.

20 ft.18 ft.18 ft.

W14 x 68

W14 x 68

W14 x 68

B

AW

12 x

79

W12

x 7

9

W12

x 7

95.4 m 5.4 m 6 m

4.5 m

3.6 m

3.0 m

3.0 m

68402 Slide # 57

Ex. 3.8 – Column Design• Step I - Determine the design load and assume the steel material.

Design Load = Pu = 2300 kN.

Steel yield stress = 344 MPa (A992 material).

• Step II. Identify the frame type and calculate Lx, Ly, Kx, and Ky if possible.

It is an unbraced (sidesway uninhibited) frame.

Lx = Ly = 3.6 m

Ky = 1.0

Kx depends on boundary conditions, which involve restraints due to beams and columns connected to the ends of column AB.

Need to calculate Kx using alignment charts.

Need to select a section to calculate Kx

68402 Slide # 58

Ex. 3.8 – Column Design

• Step III - Select a column section

Assume minor axis buckling governs.

Ky Ly = 3.6 m

Select section W12x53

KyLy/ry = 57.2 Fe = 604.4 Fcr = 271.1

cPn for y-axis buckling = 2455.4 kN

• Step IV - Calculate Kx

Ixx of W 12 x 53 = 177x106 mm4

Ixx of W14x68 = 301x106 mm4

68402 Slide # 59

Ex. 3.8 – Column Design

Using GA and GB : Kx = 1.3 - from Alignment Chart

177 1773 3.6

1.02301 3015.4 6

c

cA

b

b

I

LG

I

L

177 1773.6 4.5

0.836301 3015.4 6

c

cB

b

b

I

LG

I

L

68402 Slide # 60

Ex. 3.8 – Column Design Step V - Check the selected section for X-axis buckling

Kx Lx = 1.3 x 3.6 = 4.68 m

Kx Lx/rx = 35.2 Fe = 1590.4 Fcr = 314.2

For this column, cPn for X-axis buckling = 2846.3

Step VI - Check the local buckling limits

For the flanges, bf/2tf = 8.69 <

For the web, h/tw = 28.1 <

Therefore, the section is non-compact. OK, local buckling is not a problem

5.1356.0 y

r F

E

9.3549.1 y

r F

E