Lecture 4 (EC4 Version)(4)

7 ( )CV6107 (Part II)Lecture 4 – Composite ColumnLecture 4 – Composite Column

Chiew Sing-PingS h l f Ci il d E i t l E i iSchool of Civil and Environmental EngineeringNanyang Technological University

Scope of Lecture 4Scope of Lecture 4

GeneralLocal buckling of steel elementSimplified calculation methodAxial compressionAxial compressionResistance to compression and bending momentmomentWorked example

2

Composite column

Composite columns often offer significant economic advantages over either structural steel or reinforced concrete alternatives.

High load carrying capacities and high flexural rigidities with smaller sizes at reduced costs.

Excellent inherent fire resistances.

Can be very strong range of capacities for the same Can be very strong - range of capacities for the same external dimensions resulting in same column size externally in all floors of a building.externally in all floors of a building.

3

Application of composite column

4

5

Cross-section of composite column

Composite columns may be classified into two main types:

Concrete encased composite section

• Partially encased steel section

• Fully encased steel section

Concrete in-filled composite sectionConcrete in filled composite section

• Circular steel hollow section

• Rectangular steel hollow section

6

Concrete encased section

a) Full encased b) partially encased

7

Concrete-filled hollow section

a) concrete-filled rectangular hollow

section

b) concrete-filled circular hollow

section

c) concrete-filled circular hollow

ti ithsection section section with encased H-section

8

Local buckling

bc

cycy bFully encased steel sections

The effects of local b ckling ma be

cz

The effects of local buckling may be neglected if:

tw h hc

The concrete cover thickness is not less than the lager of the two following

tf cz

• ≥ 40mm• ≥ b/6

9

Local buckling- partially encased/concrete fill d tifilled section

Cross-section Max (d/t), max (h/t) and max (b/t)

Circular hollow dt

Circular hollowSteel section

d

z

y ( )y

23590/maxf

td =

Rectangular hollowSteel section h

zzz

t

( ) 23552/max th =Steel section hz

z

y( )

y52/max

fth

Partially encasedI-section

b

( )f23544/max tb =I section

tf

z

y( )

yf 44/max

ftb =

10

Calculation method

Two methods for calculating the resistance of composite columns:

General Method• Second order effects and imperfections taken into account in• Second-order effects and imperfections taken into account in

calculation• Can be used for asymmetric sections• Needs suitable numerical software

Simplified MethodSimplified Method• Full interaction between the steel and concrete sections until failure• Geometric imperfections and residual stresses taken into account p

in calculation, usually using European buckling curves• Plane sections remain plane

11

Simplified design method -Li it tiLimitations

Th i lifi d th d i bj t The simplified method is subject to the following limitations: bc

0.2 ≤ δ ≤ 0.9

Depth to width ratio: 0.2 < hc/bc <5.0

cycy b

p c c

The relative slenderness tw

cz0.2<λ

Steel reinforcement area:

0.3% ≤ As/Ac < 0.6%

tw

tf c

h hc

Concrete cover :• y-direction: 40 mm < cy < 0.4 b• z direction: 40 mm < c <0 3 h

tf cz

• z-direction: 40 mm < cz <0.3 h

12

Cross-sections resistance to axial icompression

Cross section resistance to axial compression is the sum of the plasticCross-section resistance to axial compression is the sum of the plastic compression resistances of each of its elements:

takes as 1 0 for concrete filled cross sections 0 85 for fully or

sdscdccydaRdpl, fAfAfAN ++= ααc

takes as 1.0 for concrete filled cross-sections ；0.85 for fully or partially concrete encased steel cross-sections

0.85 f cd f yd f sd0.85 f cd f yd sd

Npl,Rdpl,Rd

13

Cross-sections resistance to axial icompression

Concrete filled circ lar steel hollo sectionConcrete filled circular steel hollow section

ft ⎟⎞

⎜⎛

sdsck

yccdcydaaRdpl fA

ff

dtfAfAN +⎟⎟

⎠

⎞⎜⎜⎝

⎛++= ηη 1,

tUse in design if:

d

Use in design if:

• Relative slenderness 5.0≤λ

• e/d < 0.1, where e =MEd/NEd

14

Concrete filled circular steel hollow sectionConcrete filled circular steel hollow section

For 0 < e/d < 0.1:

( )( )d/101 ηηη ++t

( )de/10-1coc ηη =

( )( )de/101 aoaoa ηηη ++=

For e = 0:d

( )( ) 0.12325.0aoa ≤+== ληη

0175.18-9.42

coc ≥+== λληη

For e/d > 0.1:

coc

1a =η 0c =η

15

Relative slenderness

The relative slenderness λ is needed to check that the composite column is within the limits of applicability of the simplified method.

klN

cr

Rkpl,

NN

=λ

Npl,Rk is the Npl,Rd calculated using 0.1sac =γ=γ=γ

for fully or partially concrete encased steel section sksckcyaRkpl 85.0 fAfAfAN ++=for fully or partially concrete encased steel section

for concrete filled section

sksckcyaRkpl, 85.0 fffN

sksckcyaRkpl, fAfAfAN ++=

Ncr is the elastic critical force

( )2 EI( )2

eff2

cr LEIN π

=

16

Effective flexural stiffness

The effective flexural stiffness for short time loading:The effective flexural stiffness for short-time loading:

Steel section

( ) ff IEKIEIEEI ++=Ke = 0.6

E t d l

( ) ccmessaaeff IEKIEIEEI ++

Reinforcement Concrete

Ecm, secant modulusof concrete EC2

17

Effective flexural stiffness

The effective flexural stiffness for long-time loading:The effective flexural stiffness for long time loading:

( ) IEKIEIEEI( ) ceffc,essaaeff IEKIEIEEI ++=

Ke = 0.6

( ) tEdEdG,cmeffc, /1

1ϕ+

=NN

EE

tϕ creep coefficient EC2

NEd the total design load

NG,Ed permanent part of axial design load

18

Plastic Resistance in axial compression

A composite column has sufficient resistance to axial compression if:A composite column has sufficient resistance to axial compression if:

0.1Ed ≤NN

Rdpl,Nχ

the reduction factor (EC3)

0.1-

122≤

+=

λφφχ

[ ]( )[ ]22.0-15.0 λλαφ ++=

cr

Rkpl,

NN

=λ

19

Buckling curves for composite columns

α the imperfection factor which allows for different levels of imperfection in the columns (EC3)columns (EC3)

=0.21 for buckling curve a

= 0.34 for buckling curve b

0 49 f b kli=0.49 for buckling curve c

1 0Plastic resistance

ab

1.0

x Perfect critical load

λ0.0 1.0 2.0

20

Buckling curves for composite columnCross-section Limits Axis of buckling Buckling curve

Concrete encased section y-y by y b

z-z c

P ti ll t d

y

zPartially concrete encasedSection

y-y b

z-z czy

Circular and rectangularhollow section

y-y a

z-z bzy

z%3s ≤ρ

%6%3 ≤< ρ z z b

Circular hollow section withadditional I-section

y-y b

z %6%3 s ≤< ρ

z-z b

Partially encased H sectionwith crossed H section

zy

with crossed H section any by

z

21

Interaction curve for compression and bending

Concrete filled rectangular hollow section

22

Interaction curve for compression and bending

Concrete filled rectangular hollow section—— point A and B

Point A fcd fyd fsd

Npl,Rd

- -

-

No moment-

f fPoint B ffcd fsdPoint B

Mpl,Rd-

fyd

- -hn

No axial force++

n

23

Interaction curve for compression and bending

Concrete filled rectangular hollow section —— point C and D

Point C fcdfyd fsd

Mc = Mpl Rd- c pl,Rd

N N

- -hn

hn

f d fyd f dPoint D

+ Nc= Npm,Rd++

fcd fyd fsd

- - -hn

MD = Mmax,Rd

ND= Npm,Rd / 2+ +

n

Npm,Rd = fcdAc

24

Resistance of composite section to bendingbending

Moment resistance of steel, concrete and reinforcements

f d fyd f dPoint D

idip,Rdi,max, fWM =

fcd fyd fsd

- - -Mmax,Rd

ND+ +

Hence, pssdpccdpaydRdi,max,Rdmax, 21 WfWfWfMM

n

++==∑Wpa,Wps, Wpc are the plastic section moduli for the steel section, the

i f t d th t f th it

1 2i=

reinforcement and the concrete of the composite cross-section respectively (for the calculation of Wpc, the concrete is assumed to be uncracked)

25

concrete is assumed to be uncracked).

f d fPoint B fydfcd fsdPoint B

Mpl,Rd

yd

-hn

- -

No axial force++

fcd fcd fcd

= _hn

M pl.c,RdMmax.c,RdM pl,n.c,RdM

= _hn

pl.a,RdMmax.a,RdM pl,n.a,RdM

26Similarly, for reinforcement Rds,n,pl,Rds,max,Rds,pl, - MMM =

Resistance of composite section to bending

Rdn,pl,Rdmax,Rdpl, - MMM =

where:

( ) ( ) ( ) cdcnpc,pcsdnps,psydnpa,paRdpl, -5.0-- fWWfWWfWWM α++=

where:αc = 1.0 for all concrete filled hollow sections

= 0.85 for fully or partially encased sections

Wpa,Wps, Wpc are the plastic section moduli for the steel section, the reinforcement and the concrete of the composite cross-section respectively (for the calculation of Wpc, thesection respectively (for the calculation of Wpc, the concrete is assumed to be uncracked).

Wpan, Wpsn, Wpcn are the plastic section moduli of the correspondingcomponents within the region of 2hn from the middle line of the composite cross-section.

h i th d th f th t l i f th iddl li fhn is the depth of the neutral axis from the middle line of the cross-section.

27

Concrete filled rectangular hollow sectionThe plastic section modulus of structural steel, reinforcement, concrete are:

Wpa = From Steel Section Tables

[ ]∑n

AW [ ]∑=

=i

isips eAW1

( )( ) ( ) WrthrrthtbW 422-2- 232

⎟⎞

⎜⎛π( )( ) ( ) pspc WrtrrW ---

2-4-

3-

4⎟⎠

⎜⎝

= π

( )cdsdsncdc ffAfAh -2-=

For z-axis, the dimensions h and b are to be exchanged to calculate the n

( )( )cdydcd

n fftbfh

-242 +=

grelated perimeters[ ]∑

=

=n

iizisnnps eAW

1,,,

( ) 2( ) psnnnp WhtbW -2- 2c =

WWbhW --2=

28

pcnpsnnnp WWbhW a

Concrete filled circular hollow section

The plastic section modulus of structural steel, reinforcement concrete are:reinforcement, concrete are:

pspcpa WWdW --6

3

=

[ ]∑=

=n

iisips eAW

1

6

( )pspc WtdW -

62- 3

=

( )ffAfA 2( )( )cdydcd

cdsdsncdcn fftdf

ffAfAh-242-2-

+=

[ ]∑=

=n

iizisnnps eAW

1,,, ( ) psnnnp WhtdW -2- 2

c =

pcnpsnnnp WWdhW --2a =

29

Interaction curve compression and b di tbending moment

Interaction curve for compression and bendingInteraction curve for compression and bending-Fully encased H section

30

Interaction curve for compression and b di tbending moment

0.85fcd fyd fsd

Points A :

-

- -

-Npl,Rd

Fully encased H-section --- Points A

NA= Npl,Rd MA=0

31

Interaction curve for compression and b di tbending moment

fsd0.85fcd

Points B :fyd

Mpl,Rdhn--

-

hn

2hn+

+No axial force

Fully encased H-section --- Points B

NB= 0 MB= Mpl,Rd

32

Interaction curve for compression and b di tbending moment

fsd0.85fcd fyd

Points C :

Mpl,Rdhn

2h

-- -

Npm,Rdhn

2hn

++

Fully encased H-section --- Points C

NC= Npm,Rd = 0.85fcdAc

MC= M l RdMC= Mpl,Rd

33

Interaction curve for compression and b di tbending moment

fsd0.85fcd fyd

Points D :

sd

---

cd ydMmax,Rd

++ Npm,Rd/2

Fully encased H-section --- Points D

ND= NC

Rd d d d1 0.85DM M f W f W f W= = + × +max,Rd yd pa cd pc sd ps0.852DM M f W f W f W+ +

34

Concrete encased I-sectionMajor -axis

The plastic section modulus of structural steel, reinforcement, concrete are:

Wpa = From Steel Section Tables

p , ,

[ ]∑n

[ ]∑n

A[ ]∑=

=i

isips eAW1

[ ]∑=

=i

izisnnps eAW1

,,,

cc WWhbW2 WWhbW --2=

bFor neutral axis in the web, hn<h/2-tf

papscc

pc WWW --4

= panpsnncnp WWhbW --c =

bc

b( )( )cdydwcdcc

cdsdsncdccn fftfb

ffAfAhc

c

-222-2-αααα

+= 85.0=cα

h hc

ez

( )cdydwcdcc fff c

2a nwnp htW =

cy

p

35z

Concrete encased I-sectionMajor -axis

For neutral axis in the flange, h/2-tf<hn<h/2For neutral axis in the flange, h/2 tf hn h/2

( ) ( )( )( )( )

cdcydfwcdsdsncdccn ffbfb

ffthtbffAfAh c -22---2- ααα +

= ( )cdydcdccn ffbfb c-222 αα +

( )( )2-- 22 fw thtb

bhW

bFor neutal axis outside the steel h/2<hn<hc/2

( )( )4

-2a

fwnnp bhW =

bc

b( ) ( )cdcydacdsdsncdccn fb

ffAffAfAh

αααα

2-2--2- c=

h hc

ez

cdcc fb α2

aa pnp WW = cy

aa pnp

36z

Concrete encased I-sectionMinor -axis

The plastic section modulus of structural steel, reinforcement, concrete are:

Wpa = From Steel Section Tables

[ ]∑=n

eAW [ ]∑=n

eAW[ ]∑=

=i

isips eAW1

[ ]∑=

=i

iyisnnps eAW1

,,,

hb2panpsnncnp WWhhW --2

c =

bc

bFor h ≤ t /2

papscc

pc WWhbW --4

= panpsnncnpc

bFor hn≤ tw/2

( )( )

cdsdsncdcc ffAfAh c-2- αα=

h hcy

( )cdydcdccn ffhfh

hc-222 αα +

2hhW =

zey

a nnp hhW =

37

y

Concrete encased I-sectionMinor -axis

For tw/2 < hn < b/2w n

( ) ( )( )( )

cdcydfwcdsdsncdccn fffh

ffhttffAfAh c -2-2-2- ααα +

= ( )cdydfcdccn fftfh c-242 αα +

( )2-2

22 fw tht

htW =

bc

bFor b/2<hn<bc/2

4-2a nfnp htW =

b( ) ( )d

cdcydacdsdsncdccn fh

ffAffAfAh

αααα

2-2--2- c=

h hcy

cdcc fh α2

WW =

zey

aa pnp WW

38

y

Second-order amplification ofb di tbending moments

The influence of the second order effects on bending moments for compositeThe influence of the second order effects on bending moments for composite columns may be neglected if

N10

Ed

effcr,cr ≥=

NN

α

is the critical normal forceeffcr,N is the critical normal force,

( )2

IIeff,2

effcr, LEI

Nπ

=L

( ) ( )ccmIIe,ssaaoIIeff, IEKIEIEKEI ++=

Ke,II take as 0.5;Ko take as 0.9

39

Ko take as 0.9

Second-order amplification ofb di tbending moment

If the influence of the second order effects on bending moments for composite

The amplification factor k:

If the influence of the second order effects on bending moments for composite columns is included.

The amplification factor k:

01≥=k β 0.1-1 effcr,Ed

≥=NN

k

is an equivalent moment factor, shown in the following table

β

40

Second-order amplification ofb di tbending moment

Factor β for the determination of moment

Moment distribution Moment factors β

Factor β for the determination of moment

First –order bending moment from

member imperfectionMEd member imperfection or lateral load:

β=1.0

End moments:

MEd

M

-1≤ r ≤1

End moments:

β=0.66+0.44r

but β≥0 44

MEd rMEd

-1≤ r ≤1 but β≥0.44

41

Member Imperfection of bending moment

For the initial imperfection e caused by the designNEdFor the initial imperfection e0 caused by the design axial load NEd on a composite column, there will be a bending moment of NEde0.d 0

The design bending moment for the composite columnff fe0 length considered both second-order effects of end

moment and imperfection is given by:

0Ed2Ed1Ed.max eNkMkM +=

NEd k1, k2 are the factors of second order effects

42

A i f B kli M b

Member imperfections for composite column

Cross-section Axis of buckling

Buckling curve

Member imperfection

Concrete encased section y-y b L/200y y b L/200

z-z c L/150Partially concrete encased

y

zPartially concrete encasedSection

y-y b L/200

z-z c L/150zy

Circular and rectangularhollow section

y-y a L/300

z-z b L/200zy

z

z z b L/200Circular hollow section withadditional I-section

y-y b L/200

z

z-z b L/200

Partially encased H sectionwith crossed H section b L/200

zy

with crossed H section any b L/200yz

43

Second-order amplification ofb di tbending moment

0 M1,Ed M1,Ed

Lk M1

M NEde0 βM

k1M1,Ed k2NEde0

M2,EdNEde0 βM1,Ed

(a) First order (b) First order (c) end moment (d) Imperfection(a) First order bending moment single curve

(b) First order bending moment double curve

(c) end momentincreased by second order effect

(d) Imperfectionmoment increasedby second ordereffecteffect

44

Resistance of columns to compression and i i l b di tuniaxial bending moment

N/Npl,Rd

1 0 Resistance locus of1.0the cross-section

χd=NEd/Npl,Rd

M/Mpl,RdM /M

1.00μd=Mpl,N,Rd/Mpl,Rd

45

Resistance of columns to compression and i i l b di tuniaxial bending moment

MMM

Rdpl,d

Ed

RdN,pl,

Ed αμ

≤=MM

MM

MEd the greatest of the end moments and the maximumbending momentbending moment

Mpl,N,Rd the plastic bending resistance =μdMpl,Rd

αM a reduction factor=0.9 for S235 and S335

0 8 f S420 d S460=0.8 for S420 and S460

μd if >1.0 should not be used in practice

46

Influence of transverse shear force

• It can be assumed that the transverse shear Va,Ed is carried by the steela, dsection only.

• The effect of shear only needs to be taken into account if the shearforce is more than 50% of the shear resistance Vpl a Rd of the steelforce is more than 50% of the shear resistance Vpl.a.Rd of the steelsection.

• the shear area is reduced over the sheared zone (usually the web of the steel section) A reduced design steel strength in shear area is:steel section). A reduced design steel strength in shear area is:

with( ) yd-1 fρ2

Eda, 1-2

⎟⎟⎞

⎜⎜⎛

=V

Vρwith( ) ydfρ

Rda,pl,⎟⎠

⎜⎝V

ρ

47

Influence of transverse shear force

The shear force VEd may be distributed into Va,Ed acting on the structural t l ti d V ti th i f d t tisteel section and Vc,Ed acting on the reinforced concrete section:

Rda,pl,MVV

Rdpl,

Rda,pl,EdEda, M

VV =

Where:

Eda,EdEdc, -VVV =

Where:

Mpl,a,Rd is the plastic resistance moment of the steel section

M i th l ti i t t f th it tiMpl,Rd is the plastic resistance moment of the composite section.

48

Resistance of columns to compression and bi i l b di tbiaxial bending moments

N/Npl,Rd N/Npl,Rd

1.0 1.0

NEd/Npl,Rd NEd/Npl,RdαMμdy αMμdy

1.00 μdyM /M 1.00 μdzMy/Mpl,y,Rd

0 μdzMz/Mpl,z,Rd

b) Axis(z-z)a) Axis of anticipated failure (y-y)

49

Resistance of columns to compression and bi i l b di tbiaxial bending moments

0M Ed/M l Rd

αMμdy

μdy

My,Ed/Mpl,y,Rd

αMμdz

μ

Mz,Ed/Mpl,z,Rd

μdz

c) Biaxial bending resistance

50

Resistance of columns to compression and bi i l b di tbiaxial bending moments

For combined compression and biaxial bending the following conditionsFor combined compression and biaxial bending the following conditionsshould be satisfied for the stability check with the column length and for the check at the end.

yM,Rdypldy

Edy, αμ

≤M

M

ddy μμ =

Rdy,pl,dyμ

MEdz, α≤

M

01Edz,Edy, ≤+MM

αα =

ddz μμ =where

zM,Rdz,pl,dz

αμ

≤M

0.1Rdz,pl,dzRdy,pl,dy

≤+MM μμ

yM,M αα =

zM,M αα =

51

Design procedure

Find NEd and M Ed at both ends of the column

Calculate Npl Rd and Ncrpl,Rd cr

Calculate , and thenλ χCalculate , and then

Is N ≤ N ? C l t t hNo

λ χ

χIs NEd ≤ Npl,Rd ? Column not strong enough

YYes

χ

Is the column in axial compression only? YesColumn verified

N

See next page

No

52

Find Vpl, a, Rd. Is VEd > 0.5Vpl, a, Rd?Yes No

Determine Mpl, a, Rd and Mpl, Rd, and hence Va, Ed and Vc, Ed. Is Va, Ed > 0.5Vpl, a, Rd?

Determine the interaction curve for the cross-section.

No

Calculate ρ and hence reduced fyd. Find member imperfection, e0.Yes

Can first-order member analysis be used?Yes N No

Determine MEd , the maximum firstorder

Find MEd, max by secondorderanalysis of the pin ended

Calculate Ncr, eff = π2(EI)eff, II/L2

find β for end moments MEd t and

No No

maximum firstorder bending moment within the column length.If M M it i

analysis of the pin-ended column length with force NEd

and end moments MEd, 1 and

find β for end moments MEd, top and MEd, bot and hence k (= k1); find k2 for β = 1; fi d th d i t f th lIf MEd, 1 = MEd, 2 it is

MEd, max = MEd, 1 + NEde0

MEd, 2.find the design moment for the column,MEd, max = k1MEd + k2NEde0

From NEd and the interaction diagrams, find μdy and μdz. Check that the cross-section can resist M Ed and M Ed

53

can resist My, Ed, max and Mz, Ed, max.

Worked Example

d = 406.4mm

t = 12.5mm

y 160.

0mm

138.

5mm

80.0

mm Design Data

C30/37 concreteS355 steel gradey S355 steel grade

Effective length =3.0 m

z

12 Φ12 bar Es=210 kN/m2

fsk= 460 N/mm2

zNEd=4000 kNMEd=50 kNm

54

Design strength:2

yd N/mm3550.1/355 ==f

2cd

2sd

N/mm2015.1/0.30

N/mm40015.1/460

==

==

f

f

Cross sectional areas:( ) 2

22

a mm4.154684

4.3814.406=

−π=A

( )

( ) 22

22

s

4381

mm2.1357412x12

4

π

=π

=A

( ) 2c mm5.1128912.1357

44.381

=−π

=A

Ratio of reinforcement:

Note: if has to be limited to 6%for calculation

6.0%1.2%x100%5.112891/2.1357/ cs <===ρ AA

%6>ρ ρNote: if , has to be limited to 6%for calculation.%6>ρ ρ

55

Check for local buckling:

( ) 6.59355/235905.32125/4.406/ 2 =<==td

Plastic resistance to compression:

( )kN

fAfAfAN

082921000/.01357.2x400x20.05.112891355x4.15468

sdscdcydaRdpl,

++=

++=

( ith t fi t ff t)kN0.8292= (without confinement effect)

Steel contribution ratio:

( )9.02.066.00.8292/1000/355x4.15468/ Rdpl,yda ≤≤=== δNfAδ

56

M t f i tiMoment of inertia:( ) 4

44

a cm7.3003064

14.3864.40=

−π=I

( ) 44

4222s

cm21021344173614.38

cm4.17360.8x131.1x485.13x131.1x40x1.131x16.264

=−π

=

=++=

I

I

Effective stiffness:

c cm2.1021344.173664

==I

( )( ) 2

ccmessaaeff

6.86078100/2.102134x32x6.04.1736x2107.30030x210 kNm

IEKIEIEEI

=++=

++=

Slenderness: assume short-time loading

fAfAfAN ++=

( ) kN

fAfAfAN

3.95021000/460x2.135730x5.112891355x4.15468sksckcyaRkpl,

=++=

++=

( ) kNlEIN 7.943953/6.86078x/ 222eff

2cr =π=π=

57

Reduction factor for column buckling:

317.07.943953.9502crRkpl, ===λ NN

Reduction factor for column buckling:

( )[ ] ( )[ ] 563.0317.02.0317.021.015.02.015.0 22 =+−+=λ+−λα+=φ

Confinement effect of concrete:

[ ] [ ] 973.0317.0563.0563.011 2222 =−+=λ−φ+φ=χ

Confinement effect of concrete:

5.0<λConfinement effect is considered since and e/d<0.1 (taking e=0 cm)

( ) 0.1909.02325.0744.0175.189.4

aoa

2coc

<=λ+=η=η

=λ+λ−=η=η

kNfAff

dtfAfAN 7.84031 sds

ck

yccdcydaaRdpl, =+⎟⎟

⎠

⎞⎜⎜⎝

⎛η++η=

58

Check axial compression

0 973 8403 7 8176 8kN > NNχ ×

Note: the increase in axial compression due to confinement effect

, 0.973 8403.7 8176.8kN > Npl Rd EdNχ = × =

014.10.8292/7.8403 ==

Therefore a 1 4% increaseTherefore a 1.4% increase.

Interaction Curve

Plastic section moduli:W 1940cm3Wpa = 1940cm3

[ ]12

3

1135cmps si iW A e= =∑

1

59

( )( )2 32 2 2 ( 2 )b t h t h d t⎛ ⎞( )( ) ( )3 2

33

2 2 2 ( 2 )44 3 2 6

(406.4 25) 135 = 9112 cm

pc ps ps

b t h t h d tW r r t r W Wπ− − −⎛ ⎞= − − − − − − = −⎜ ⎟

⎝ ⎠− -135 = 9112 cm

6=

Neutral axis positionAssume 2 reinforcements lies within the region 2hAssume 2 reinforcements lies within the region 2hn

( ) ( )( )

2 112891.5 20 113 2 460 2042.5mm

2 4 (2 ) 2 406.4 20 4 12.5 2 355 20c cd sn sd cd

ncd yd cd

A f A f fh

df t f f− − × − × × −

= = =+ − × × + × × × −( )( )cd yd cdf f f

Hence, assumption for Asn is verified

Plastic section moduli in the region2h :

2 2 3( 2 ) 38.14 4.25 0 689 cmpcn n psnW d t h W= − − = × − =

Plastic section moduli in the region2hn:Wpsn = 0

p p

2 2 340.64 4.25 689 0 45 cmpan n pcn psnW dh W W= − − = × − − =

max, 0.5 833.8kNmRd yd pa cd pc sd psM f W f W f W= + + =

60

, max, 0.5 811kNmpl Rd Rd yd pan cd pcn sd psnM M f W f W f W= − − − =

The resistance force

, 2258kNpm Rd c cdN A f= =

Point

ABending Moment M=0 MA= 0 kNm

ACompression force N=Npl,Rd NA= 8403.7 kN

BBending Moment M=Mpl,Rd MB= 811 kNm

Compression force N=0 NB= 0 kN

CBending Moment M=Mpl,Rd MC= 811 kNm

C i f C 22 8 kCompression force N=Npm,Rd NC= 2258 kN

DBending Moment M=Mpm,Rd MD= 833.8 kNm

Compression force N=0.5Npm,Rd ND= 1129 kN

61

Interaction curve

8000

9000

Interaction curve

AA

5000

6000

7000

N

3000

4000

5000

C

0

1000

2000C

D

B00 200 400 600 800 1000

B

M

62

To check whether second order effects can be neglected, a reduced value of Ncr is required. The effective flexural stiffness is

( ) 6 2eff II o a a e II c eff c s s( ) = 63493 10 kNmmEI K E I K E I E I= + + ×( )eff,II o a a e,II c,eff c s s( )

Hence, the elastic critical force is:

2 2 6eff, II

cr,eff 2 2 6

π ( ) 63493 10 = 69628 kN3 10

EIN

Lπ × ×

= =×

The result is less than 10NEd for both major axis and minor axis, so thesecond order effects must be allowed for.seco d o de e ects ust be a o ed o

The bending moment, MEd,top = 150 kNm, MEd,bot = 0 kNm, so r = 0, then, β = 0.66. β

0.66 = 0.71 1 4000 69628

kN N

β= =

Ed cr,eff1- 1 - 4000 69628N N

63

The bending moment MEd should be modified with the effect of memberimperfection. The member imperfection is:

0 = /300 = 10 mme L

The mid-length bending moments due to NEd are

Ed 0 = 4000 0.01 = 40 kNm N e ×Ed 0

For the bending moment from the member imperfection, β = 1.0. Then,

impEd cr,eff

1.0 = 1.061- 1 - 4000 69628

kN N

β= =

Hence, the bending moment after modification is

y,Ed y y,Ed,top imp,y Ed 0,z= + = 0.7 50+1.06 40 = 77 kNM k M k N e × ×

64

Bending resistance

,/ 4000 / 8403.7 0.48d Ed pl RdN Nχ = = =

Bending resistance

1 1 0.48 0.711 1 0 27

dd

χμ

χ− −

= = =

, ,/ 2258 / 8403.7 0.27pm pm Rd pl RdN Nχ = = =

1 1 0.27pmχ− −

77 0.14 0.90 71 811

EdMMμ

= = <×, 0.71 811d pl RdMμ ×

Hence, the composite column is adequate

65

Conclusions

Design of composite column is rational and comprehensive for both concrete encased sections and concrete in-filled hollow sections.Both the strength and the deformation assessments should be carried out carefully. The flexural rigidities are essential in determining the column buckling behavior of composite column.The plastic stress block method is applicable to both steel and composite columns. However, although the design principles are the same, the calculation procedure for composite column is much more involved and complicated.

66

Conclusions

The design methodology for composite columns areThe design methodology for composite columns are as follows:

Compression resistanceCompression resistanceMoment resistanceI t ti b t i d b diInteraction between compression and bending moment Euler buckling formulation, column buckling curves, and reduction factorsSecond order effects and imperfection

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