4 compression members - 2011 new
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Compression Members
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Introduction
Resistance of Cross-SectionsSections not prone to local buckling
Sections prone to local buckling
Buckling Resistance of MembersSections not prone to local buckling
Sections prone to local buckling
Reduction Factor for Buckling ResistanceElastic Critical Force & Buckling LengthNon-Dimensional Slenderness for Flexural BucklingBuckling Curve of Perfect ColumnBuckling Curves of Imperfect ColumnsSelection of Buckling Curve and Imperfection Factor
Design Procedure
ExamplesExample CM-1 (UC with intermediate restraint under compression)Example CM-2 (CHS under compression)
Outline
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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 thepresence of transverse forces. They are referred to as beam-columnsand 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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Class 1, 2 and 3 cross-sections areunaffected by local buckling.
Design resistance of cross-sectionNc,Rdequals the plastic resistanceNpl,Rd.
Resistance of Cross-Sections
Sections NOT PRONE to local buckling
Class 4 sections suffers from local bucklingwhich prevents the attainment of squash load.
Design resistance of cross-sectionNc,Rdlimited to local buckling resistance.
Sections PRONE to local buckling
EN 1993-1-1 Clause 6.2.4 (1)
The design value of the compression forceNEdat each cross-sectionshall satisfy:
RdcEd NN ,
EN 1993-1-1 Clause 6.2.4 (2)
0M
y
RdcAfN
=,0M
yeff
RdcfAN
=,
If Class 4 section is unsymmetrical, it has to bedesigned as beam-columndue to the additionalmomentarising from eccentricity of the
centroidal axis.
M0= 1.00
42tc /
14tc /
Internal element
Outstand elementNon slender
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Buckling Resistance of Member
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Elastic Buckling of Columns
2
2cr
EI
L
Euler Buckling Load
Ncr =L
2
crcr 2
cr
N Ef
A (L / i)
= =
Buckling stress
I = i2A
i = radius of gyration
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=21
2
E
fy
f
fy
Failure byCross section yielding
Failure by elastic buckling
Euler elastic buckling
Buckling Curve of Perfect Column
1
where is the column slenderness.i
Lcr=
2
2
2
22
2
2
E
L
Ei
AL
EI
A
Nf
crcr
crcr ====
Whenfcr=fy, =1
yf
E
=1
is the radius of gyration.I
Ai =
i
Lcr=
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0.0
0.2
0.4
0.6
0.8
1.0
0 1 2 3 4
f/fy
Factors Influence the Buckling of Columns
/1
=0
Euler buckling curve
1. Effective length of Column
2. Residual Stresses
3. Member initial out-of-straightness4. Types of cross section
5. Local buckling of component plate
Practical Region
Inelastic buckling
Elastic buckling
Imperfectcolumns
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= L / 1000
Initial out of straightness
+260 N/mm
-125 N/mm
+55 N/mm
Rolled Section
C
T
C C
C C
T
2
2
2
CC
T
CC
T
C
T
TWeb Distribution
Welded section
Factors that affect overall buckling of columns
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Design buckling resistanceNb,Rdshould betaken as:
Buckling Resistance of Members
Sections NOT PRONE to local buckling
Design buckling resistanceNb,Rdshould betaken as:
Sections PRONE to local buckling
EN 1993-1-1 Clause 6.3.1.1 (1)
The design value of the compression forceNEdshall be checked against the design
buckling resistance:
RdbEd NN ,
EN 1993-1-1 Clause 6.3.1.1 (2)
1M
y
Rdb
AfN
=,
1M
yeff
Rdb
fAN
=,
If Class 4 section is unsymmetrical, it has to bedesigned as beam-column.
Holes for fasteners at the column ends need NOT
to be taken into account in determiningAandAeff.
M1= 1.00
/ /
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Reduction Factor for Buckling Resistance
EN 1993-1-1 Clause 6.3.1.2
1.0but +=
22
1
]0.2)(0.5[1 2 ++=where
imperfection factor (refer to Table 6.1 & 6.2)
non-dimensional slenderness
sections-cross4Classfor
sections-cross3and21,Classforcr
y
N
Af=
cr
yeff
N
fA=
For , or for , the buckling resistance check
can be ignored and only cross sectional checks apply.
0.2 0.04cr
Ed
N
N
14/8/2013
2
2
cr
crL
EIN
=
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Buckling length
Lcr= effective length in BS5950
No guidance given in EC3
NCCI
Use same factors as BS5950 Applied to system length, L
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Elastic Critical Force & Buckling Length
I Second moment of area which is determined based on thegross cross sectional propertiesfor all classes of cross-sections.
Lcr
Buckling length in thebuckling plane considered.2
2
cr
crL
EIN
=
Nominal buckl ing lengthsLcrfor compression members
Non-sway mode
Sway mode
where
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Non-Dimensional Slenderness for Flexural Buckling
2 2
1
1
( / )
/
y y y
cr
cr cr
ycr
Af Af fAL
N EI L I E
fL
i E
= = =
= =
2 2
1
( / )
eff y eff y eff y
cr
cr cr
eff ycr
A f A f A fAL
N EI L A I E
A fL
i A E
= = =
=
For Class 1, 2 and 3cross-sections,
For Class 4cross-sections,
= Lcr/i 1= (E/fy)0.5
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0.0
0.2
0.4
0.6
0.8
1.0
0 1 2 3 4
f/fy
Buckling Curves of Imperfect Columns
/1
=0.13 =0.21 =0.34 =0.49 =0.76
=0
Euler buckling curve
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Buckling curve a0 a b c d
Imperfection factor 0.13 0.21 0.34 0.49 0.76
Table 6.2: Selection of buckling curve for a cross-section Table 6.1: Imperfection factors fo r buck ling curves
Selection of Buckling Curve and Imperfection Factor
Imperfections can be attributed to the following:
initial out-of-straightness
eccentricity of applied loads
material variations
residual stresses
Typical residual stress p rofil e in a hot-rolled I-section
Residual compressive stressResidual tensile stress
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Design Procedure
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For each axis of buckling, determine
buckling lengthLcr
Limiting slenderness1
non-dimensional slenderness
appropriate strut curve(a0, a, b, c or d)from Table 6.2
imperfection factorfrom Table 6.1
buckling reduction factor
Design Procedure
Check ifNb,Rd>NEd. Else, repeat steps.
Use the smaller value ofto determine buckling resistanceNb,Rd.
Determine design axial forceNEd.
Select a trial section such thatNEd
/A
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Examples
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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.N
N
2.5m
2.5m
AA
Section AA
z z
y
y
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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
= (235/fy)0.5= 0.92
Classification of flange
Flange is Class 1 (Plastic).
Classification of web
Web is Class 1 (Plastic).
8.320.92996.20 === */ ff
tc
30.40.92*333317.1 === ww tc /
Section is Class 1 (PLASTIC).
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Resistance of CrossSection
2101kN1.0
10*)(275)10*(76.4 32
0
====
M
y
RdplRdc
AfNN
,,
Flexural Buckling abouty-yaxis
Use buckling curveb= 0.34
86.8210000/275 === yfE/1
0.643
86.8
1
8.96
5001=
==
1
,
y
ycr
y
i
L
&1.21.02205.8
209.6
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Flexural Buckling aboutz-zaxis
Use buckling curvec= 0.49
0.55486.81
5.202501 =
==
1
,
z
zcrz
iL
100mm&1.2
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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
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Resistance of CrossSection
2400kN2616kN1.0
10*)(355)10*(73.7 32
0
>===
M
y
Rdc
AfN
,
Buckling Resistance of Member
Use buckling curvea= 0.21
76.4210000/355 === yfE/1
0.631
76.4
1
8.30
4001=
==
1
i
Lcr
2400kNkN210*1.0
355*)10*(73.7*0.854 3
2
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Effective length Lcrof
compression members
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Effective length of
column in frame
Lcr=KL
K
L
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Projects to Illustrate theConcept of Steel Design
Prof. Richard Liew
Dept of Civil & EnvironmentalNational University of Singapore
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Buckling of compression members Axial compressioncauses failure by buckling(out-of-plane
deflection) in slender members.
Buckling about major(x-x) axis. Buckling about minor(y-y) axis.
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Members with intermediate lateral restraints
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Pin connected space frame
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Pin-connected space frame
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Adequate Bracing during
Construction
1
1
2
3
6
4
5
Sequence of erection
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Cantilever Structure
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Cantilever trusses
1
1
2
3
6
4
5
49
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Lateral bracing for
cantilever trusses
1
1 2
3
6
4
5
Purlin
Cantilever trussFly bracing
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Question
y
y
z z
Lcr,y= ?
Lcr,z= ?