Compression Members

Compression Members

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Outline Introduction Resistance of Cross-Sections

Sections not prone to local buckling Sections prone to local buckling

Buckling Resistance of Members

Sections not prone to local buckling Sections prone to local buckling

Reduction Factor for Buckling Resistance Elastic Critical Force & Buckling Length Non-Dimensional Slenderness for Flexural Buckling Buckling Curve of Perfect Column Buckling Curves of Imperfect Columns Selection of Buckling Curve and Imperfection Factor

Design Procedure Examples

Example CM-1 (UC with intermediate restraint under compression) Example CM-2 (CHS under compression) 2

Introduction  Compression members are structural components that are subject to axial compression loads only.  These generally refer to compressed pin-ended struts found in trusses, lattice girders or bracing members.  Most real columns are subjected to significant bending moments in addition to the axial loads, due to the eccentricities of axial load and the presence of transverse forces. They are referred to as beam-columns and are covered in a separate chapter.  Compression members must be checked for → resistance of cross-sections → buckling resistance of members

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Resistance of Cross-Section

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Resistance of Cross-Sections EN 1993-1-1 Clause 6.2.4 (1)

The design value of the compression force NEd at each cross-section shall satisfy:

N Ed ≤ N c , Rd

Internal element

c / t ≤ 42ε

EN 1993-1-1 Clause 6.2.4 (2)

Outstand element

Non slender

c / t ≤ 14ε

Sections NOT PRONE to local buckling

Sections PRONE to local buckling

Class 1, 2 and 3 cross-sections are unaffected by local buckling.

Class 4 sections suffers from local buckling which prevents the attainment of squash load.

Design resistance of cross-section Nc,Rd equals the plastic resistance Npl,Rd.

Design resistance of cross-section Nc,Rd limited to local buckling resistance.

N c , Rd =

Af y

γM0

γM0 = 1.00

N c , Rd =

Aeff f y

γM0

If Class 4 section is unsymmetrical, it has to be designed as beam-column due to the additional moment arising from eccentricity of the centroidal axis. 5

Buckling Resistance of Member under Axial Force NEd

NEd

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Buckling of compression members •

Axial compression causes failure by buckling (out-of-plane deflection) in slender members.

Buckling about major (y-y) axis.

Buckling about minor (z-z) axis.

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Elastic Buckling of Columns Euler Buckling Load Ncr = π EI Lcr 2 2

L

I = i2 A i = radius of gyration

Buckling stress N cr π2 E = f cr = A (Lcr / i) 2

Buckling Curve of Perfect Column

Failure by elastic buckling

Ncr =

f/fy 1.0 Failure by Cross section yielding

Euler elastic buckling

λ=1.0

λ=

Af y N cr

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Factors Influence the Buckling of Columns f /fy

α =0

1.0

Euler buckling curve

0.8

1. 2. 3. 4. 5.

Imperfect columns

0.6

Effective length of Column Residual Stresses Member initial out-of-straightness Types of cross section Local buckling of component plate

0.4 0.2

Elastic buckling Practical Region Inelastic buckling

0.0 0

1

2

3

λ=

Af y N cr

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Factors that affect overall buckling of columns 2 T -125 N/mm C +260 N/mm

2

C

+55 N/mm

2

C C

C

δ =L / 1000

T

Rolled Section T T

C

C

Initial out of straightness

C C

C T

T Web Distribution

Welded section 11

Buckling Resistance of Members EN 1993-1-1 Clause 6.3.1.1 (1)

The design value of the compression force NEd shall be checked against the design buckling resistance:

N Ed ≤ N b , Rd EN 1993-1-1 Clause 6.3.1.1 (2)

Sections NOT PRONE to local buckling

Sections PRONE to local buckling

Design buckling resistance Nb,Rd should be taken as:

Design buckling resistance Nb,Rd should be taken as:

N b , Rd

χAf y = γ M1

γM1 = 1.00

N b , Rd

χAeff f y = γ M1

If Class 4 section is unsymmetrical, it has to be designed as beam-column.

Holes for fasteners at the column ends need NOT to be taken into account in determining A and Aeff. 12

12/8/2016

Reduction Factor χ for Buckling Resistance EN 1993-1-1 Clause 6.3.1.2

χ=

1 Φ+ Φ −λ 2

2

but χ ≤ 1.0

where Φ = 0.5[1 + α (λ − 0.2) + λ 2 ]

α – imperfection factor (refer to Table 6.1 & 6.2) λ – non-dimensional slenderness

N cr =

π 2 EI 2 cr

L

λ=

Af y

λ=

Aeff f y

N cr N cr

for Class 1, 2 and 3 cross - sections for Class 4 cross - sections

For λ ≤ 0.2 , or for N Ed ≤ 0.04 , the buckling resistance check N cr

can be ignored and only cross sectional checks apply. 13

Elastic Critical Load N cr =

π 2 EI L2cr

where

I – Second moment of area which is determined based on the gross cross sectional properties for all classes of cross-sections. Lcr – Effective buckling length in the buckling plane considered.

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Effective buckling lengths Lcr for compression members

Non-sway mode

Sway mode

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Non-Dimensional Slenderness for Flexural Buckling For Class 1, 2 and 3 cross-sections,

λ = λ

Af y = N cr

 A  1 Af y Lcr  =   2 2 I (π EI / Lcr )   π

Lcr f y = λ / λ1 πi E

λ = Lcr/i

fy   E 

λ1 = π(E/fy)0.5

For Class 4 cross-sections,

= λ

Aeff f y = N cr

⇒ λ = Lcr πi

 Aeff Aeff f y = Lcr  2 2  A (π EI / Lcr ) 

Aeff

fy

A

E

A  1  I   π

fy   E 

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Buckling Curves of Imperfect Columns f /fy

α =0

1.0

Euler buckling curve

0.8

α =0.13 α =0.21 0.6 α =0.34 α =0.49 α =0.76

0.4 0.2 0.0 0

1

2

3

λ=

Af y N cr

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Selection of Buckling Curve and Imperfection Factor Table 6.2: Selection of buckling curve for a cross-section

Table 6.1: Imperfection factors for buckling curves Buckling curve Imperfection factor α

a0

a

0.13 0.21

b

c

d

0.34 0.49 0.76

Imperfections can be attributed to the following: 〄 initial out-of-straightness 〄 eccentricity of applied loads 〄 material variations 〄 residual stresses Typical residual stress profile in a hot-rolled I-section Residual compressive stress Residual tensile stress

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Design Procedure Determine design axial force NEd. Select a trial section such that NEd / A < fy for non class 4 cross-sections NEd /Aeff < fy for Class 4 cross-sections. Perform section classification.

N b , Rd = χAf y

For each axis of buckling, determine ➥ buckling length Lcr χ= ➥ Limiting slenderness λ1 Φ+ ➥ non-dimensional slenderness λ ➥ appropriate strut curve (a0, a, b, c or d) from Table 6.2 ➥ imperfection factor α from Table 6.1 ➥ buckling reduction factor χ

1 Φ −λ 2

2

but χ ≤ 1.0

Use the smaller value of χ to determine buckling resistance Nb,Rd. Check if Nb,Rd > NEd. Else, repeat steps –. 19

Example CM-1: Universal column with intermediate restraint under compression Determine the maximum compression load that can be taken by a 5m column using 203x203x60UC in S275 steel. Both ends of the column are pin supported about both y-y and z-z axes. A lateral restraint, that is aligned to the y-y axis, is provided at mid-height. y

N

2.5m

z A

z

A

y Section A–A 2.5m

N

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Yield Strength tw = 9.4mm, tf = 14.2mm. Maximum thickness = 14.2mm < 16mm (EN 10025-2) For S275 steel, fy = 275N/mm2

Section Classification ε = (235/fy)0.5 = 0.92 Classification of flange c f / t f = 6.20 ≤ 9ε = 9 * 0.92 = 8.32 ⇒ Flange is Class 1 (Plastic). Classification of web cw / t w = 17.1 ≤ 33ε = 33 * 0.92 = 30.4 ⇒ Web is Class 1 (Plastic).

Section is Class 1 (PLASTIC).

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Resistance of Cross–Section N c , Rd = N pl , Rd =

Af y

γM0

(76.4 * 102 )(275) * 10 −3 = = 2101kN 1.0

Flexural Buckling about y-y axis Lcr , y = 500cm

λ1 = π E / f y = π 210000/275 = 86.8 λy =

Lcr , y 1  500  1  =  = 0.643  i y λ1  8.96  86.8 

h 209.6 = = 1.02 < 1.2 & t f = 14.2mm < 100mm b 205.8 Use buckling curve b ⇒ α = 0.34 Φ y = 0.5[1 + α (λ y − 0.2) + λ y2 ] = 0.782 1 χy = = 0.815 2 2 Φ y + Φ y − λy N b , y , Rd = χ y

Af y

γ M1

Buckling curve Imperfection factor α

(76.4 * 102 ) * 275 = 0.815 * * 10 −3 = 1712kN 1.0

a0

a

b

c

d

0.13

0.21

0.34

0.49

0.76

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Flexural Buckling about z-z axis Lcr , z = 250cm

λ1 = 86.8

Lcr , z 1  250  1  λz = =  = 0.554  iz λ1  5.20  86.8  h < 1.2 & t f < 100mm b Use buckling curve c ⇒ α = 0.49 Φ z = 0.5[1 + α (λz − 0.2) + λz2 ] = 0.740 1 χz = = 0.812 2 2 Buckling curve Φ z + Φ z − λz Imperfection factor α 2 Af y (76.4 * 10 ) * 275 = 0.812 * * 10 −3 = 1707kN N b , z , Rd = χ z 1.0 γ M1

a0

a

b

c

d

0.13

0.21

0.34

0.49

0.76

Buckling Resistance of Member N b , Rd = min( N b , y , Rd , N b , z , Rd ) = 1707kN

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Example CM-2: Circular hollow section under compression A circular hollow section (CHS) member is to be used in a 4m long column which is pinned at both ends. The design axial compression, NEd, is 2400kN. Assess the suitability of a hot-rolled 244.5x10 CHS in grade 355 steel for this application.

Yield Strength t = 10.0mm < 16mm (EN 10025-2) For S355 steel, fy = 355N/mm2

Section Classification ε = (235/fy)0.5 = 0.814

d/t = 244.5/10.0 = 24.5 ≤ 50ε2 = 33.1

Section is Class 1. 24

Resistance of Cross–Section N c , Rd =

Af y

γM0

(73.7 * 102 )(355) * 10 −3 = = 2616kN > 2400kN 1.0

Buckling Resistance of Member Lcr = 400cm

λ1 = π E / f y = π 210000/355 = 76.4 λ=

Lcr 1  400  1  =  = 0.631  i λ1  8.30  76.4 

Use buckling curve a ⇒ α = 0.21 Φ = 0.5[1 + α (λ − 0.2) + λ 2 ] = 0.744

χ=

1 Φ+ Φ −λ 2

2

= 0.854

(73.7 * 102 ) * 355 = 0.854 * * 10−3 = 2230kN < 2400kN N b,Rd = χ 1.0 γ M1 ∴The chosen cross-section, 244.5x10 CHS, in grade S355 steel is not acceptable. Af y

Try 244.5x12.5 CHS, in grade S355

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Effective buckling length Lcr of compression members

K

Effective length of column in frame Lcr =KL L

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Effective length factor for continuous columns based on stability analysis

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Projects to Illustrate the Concept of Steel Design

Prof. Richard Liew Dept of Civil & Environmental National University of Singapore

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Members with intermediate lateral restraints

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Pin-connected space frame

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There are conics, bolts, nuts and pins at both ends of pipes.

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Collapse of Hartford Civil Centre Stadium Connecticut, USA (1971)

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Top chord buckling length was assumed to be 4.57m 4.57m

9.14m

The top chord was free to deflect in the out of plane horizontally.

Collapse under heavy snow load

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Adequate Bracing during Construction 6 5 4 Sequence of erection

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2 1 1

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Cantilever Structure

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Cantilever trusses 6 5 4 3 2 1 1

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Lateral bracing for cantilever trusses 6 5 4 3 2 1 1 Purlin

Fly bracing

Cantilever truss

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