EN1994_4_Hanswille

8/8/2019 EN1994_4_Hanswille

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1

G. Hanswille

Univ.-Prof. Dr.-Ing.

Institute for Steel and

Composite Structures

University of Wuppertal-Germany

Composite Columns

Institute for Steel and Composite Structures

University of WuppertalGermany

Univ. - Prof. Dr.-Ing. Gerhard Hanswille

Eurocode 4

EurocodesBackground and Applications

Dissemination of information for training

18-20 February 2008, Brussels


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G. Hanswille





Part 1: Introduction

Part 2: General method of design

Part 3: Plastic resistance of cross-sections and interaction curve

Part 4: Simplified design method

Part 5: Special aspects of columns with inner core profiles

Part 6: Load introduction and longitudinal shear

Contents


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Part 1: Introduction


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Composite columns

concrete

filled hollow

sections

partially concrete

encased sections

concrete encased

sections


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advantages:

high bearing resistance

high fire resistance

economical solution with regard to

material costs

disadvantages:

high costs for formwork

difficult solutions for connections

with beams

difficulties in case of later

strengthening of the column

in special case edge protection isnecessary

Concrete encased sections


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advantages:

high bearing resistance, especially in case

of welded steel sections

no formwork

simple solution for joints and loadintroduction

easy solution for later strengthening and

additional later joints

no edge protection

disadvantages:

lower fire resistance in comparison with

concrete encased sections.

Partially concrete encased sections


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advantages:

high resistance and slender columns

advantages in case of biaxial bending

no edge protection

disadvantages :

high material costs for profiles

difficult casting

additional reinforcement is needed for fire

resistance

Concrete fil led hollow sections


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advantages:

extreme high bearing resistance in

combination with slender columns

constant cross section for all stories is

possible in high rise buildings

high fire resistance and no additionalreinforcement

no edge protection

disadvantages:

high material costs

difficult casting

Concrete fil led hollow sections with

additional inner profiles


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Composite columns with hollow sections

and additional inner core-profi les

Commerzbank

Frankfurt


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Resistance of the member for

structural stability

Resistance to local Buckling

Introduction of loads

Longitudinal shear outside the areas of load

introduction

General method

Simplified method

Verifications for composite columns

Design of composite columns

according to EN 1994-1-1


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general method:

simpli fied method:

• double-symmetric cross-section

• uniform cross-section over the member length

• limited steel contribution factor δ

• related Slenderness smaller than 2,0

• limited reinforcement ratio

• limitation of b/t-values

• any type of cross-section and any

combination of materials

Methods of verification

Methods of verif ication in accordance

with EN 1994-1-1


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Verification is not necessary where

concrete encased cross-sections

partially encased I sections

concrete filled hollow section

b

t

t

dd

tε=⎟

⎠ ⎞⎜⎝ ⎛

52t

dmax

290t

dmax ε=⎟

⎠ ⎞⎜⎝ ⎛

ε=⎟ ⎠ ⎞⎜⎝ ⎛

44t

dmax

yk

o,yk

f

f =ε

f yk,o = 235 N/mm2

bc

hc

b

h

cy cy

cz

cz

y

z

⎩⎨⎧≥

6/b

mm40cz

Resistance to lokal buckling


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Part 2:

General design method

G H ill


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geometrical

imperfection

wo

1000Lwo =

L

σE+ -

residual

stresses due

to rolling or

welding

+

-

Design for structural stability shall take account of

second-order effects including residual stresses,

geometrical imperfections,

local instability,

cracking of concrete,

creep and shrinkage of concrete

yielding of structural steel and of reinforcement.

The design shall ensure that instability does not occur for themost unfavourable combination of actions at the ultimate limit

state and that the resistance of individual cross-sections

subjected to bending, longitudinal force and shear is not

exceeded. Second-order effects shall be considered in any

direction in which failure might occur, if they affect the structuralstability significantly. Internal forces shall be determined by

elasto-plastic analysis. Plane sections may be assumed to

remain plane. Full composite action up to failure may be

assumed between the steel and concrete components of the

member. The tensile strength of concrete shall be neglected.The influence of tension stiffening of concrete between cracks

on the flexural stiffness may be taken into account.

General method

G Hans ille


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General method of design

F

plastic zones in structural steel

e

cracked concrete

stresses in structural steel section

stresses in concrete and reinforcement

f y

f c

-

-

-

+

+

- - -

-

w

f s

εc

εs

εa

f cm

0,4 f cEcm

εc1uεc1f ct

σc

σs

f smf tm

Es

Ea

Ev

εv

σa

concrete

reinforcement

structural

steel

f y

f u

--

G Hanswille


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Typical load-deformation behaviour of

composite columns in tests

F [kN]

Deflection w [mm]

e

e=100mm

e=160mm

e=130mm

0 20 40 60 80 100

1600

1200

800

400

F

w A

B

C

concrete encased section and

bending about the strong axis:

Failure due to exceeding the

ultimate strain in concrete, buckling

of longitudinal reinforcement andspalling of concrete.

concrete encased section and

bending about the weak axis :Failure due to exceeding the

ultimate strain in concrete.

concrete filled hollow section:cross-section with high ductility

and rotation capacity. Fracture

of the steel profile in the tension

zone at high deformations and

local buckling in thecompression zone of the

structural steel section.

A

F

B

C

G Hanswille

General Method Safety concept based


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M

NEd

MEd

Ed

Rpl,d

Rpl,mEd

wo=L/1000 wo

w

N

d,pl

m,plR

R

R=γ

e

E

λu Ed

wu

Verification λu ≥γ

R

λu : amplification factor for ultimatesystem capacity

εc

f cm

0,4 f c Ecm

εc1uεc1f ct

σc

concrete

εs

σs

f sm

f tm

Es

reinforcement

εa

Ea

εv

σa

structural

steel

f y

f u

+

+

-

-

+-

geometrical

Imperfection

Residual

stresses

Ev

General Method – Safety concept based

on DIN 18800-5 (2004) and German

national Annex for EN 1994-1-1

G Hanswille


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0,5 1,0 1,5 2,0

0,5

1,0

Composite columns for the

central station in Berlin

-

χbuckling curve a

buckling curve b

buckling curve c

buckling curve d

800

550

1 2 0 0

7 0 0

t=25mm

t=50mm

S355

S235

Residual stresses

λ

G. Hanswille


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Part IV-3:

Plastic resistance of cross-sections and

interaction curve

G. Hanswille


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Resistance of cross-sections

sdscdcydaRd,pl

Rd,plsRd,plcRd,plaRd,pl

f Af Af AN

NNNN

+ν+

++=

=

Design value of the plastic

resistance to compressive forces:

ckcsksykaRk,pl f Af Af AN ν++=

c

ckcd

s

sksd

a

ykyd

f f

f f

f f

γ

=

γ

=

γ

=Design strength:

85,0=ν0,1=ν

f yd

Npla,RdNplc,Rd

f sdNpls,Rd

ν f cd

y

z

Characteristic value of the plastic

resistance to compressive forces:

Increase of concrete

strength due to better

curing conditions in case

of concrete filled hollow

sections:

G. Hanswille

C fi ff i f


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Confinement effects in case of concrete

fil led tubes

σc,r

σc,r

σaϕσaϕ

dt

concretestructural steel

ηa f yd

ydaRd,a

2yd

2,aRd,a

2,a

2Rd,a

f

f

η=σ

=σσ−σ+σ ϕϕ

0.05 0.10 0.15 0.20 0.25 0.30 0.35

f ck,c

ck

r ,cf σ

ck

c,ck

f

f

2.0

1.5

1.0

0.5

0

1.25

0d-2t

σc,r

For concrete stresses σc>o,8 f ck the Poisson‘s ratio of concrete is higher than thePoisson‘s ratio of structural steel. The confinement of the circular tube causes radial

compressive stresses σc,r . This leads to an increased strength and higher ultimatestrains of the concrete. In addition the radial stresses cause friction in the interface

between the steel tube and the concrete and therefore to an increase of thelongitudinal shear resistance.

r cckcck f f ,21, σα+α=

α1=1,125α

2= 2,5

α1=1,00α2= 5,0

G. Hanswille

C fi t ff t t E d 4 1 1


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Confinement effect acc. to Eurocode 4-1-1

( ) 092,015,18

0,15,0

KKco,c

Kao,a

≥λ−λ−η=η

≤λ+η=η

λ

λinfluence of

slenderness for

influence of load

eccentricity : ⎟ ⎠ ⎞

⎜⎝ ⎛ −η=ηη−+η=η λλ

d

e101

d

e)1(10 ,ccao,aa

Ed

Ed

N

Me =

5,0≤λ

⎟⎟ ⎠ ⎞⎜⎜

⎝ ⎛ η++η=

ck

ykccdcaydaRd,plf f

dt1f A Af N

Design value of the plastic resistance to compressive forces

taking into account the confinement effect:

9,425,0 coao Basic values η for stocky columns

centrically loaded:

d

t

y

z

MEd NEd

f c

f ye/d>0,1 : ηa=1,0 and ηc=0

G. Hanswille

U i P f D IPl ti i t t bi d b di d


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Plastic resistance to combined bending and

compression

N

Npl,Rd

NEd

Mpl,Rd

Mpl,N,Rd= μ Mpl,Rd

f yd

(1-ρ) f yd

0,85 f cd f sd

Mpl,N Rd

NEd

VEd

2

Rd,pla

Ed,aRd,plaEd,a

Rd,plaEd,a

1V

V2V5,0V

0V5,0V

⎥⎥⎦⎤

⎢⎢⎣⎡ −=ρ⇒>

=ρ⇒≤

zply

z

M

--

+

The resistance of a cross-section to combined

compression and bending and the corresponding

interaction curve may be calculated assuming

rectangular stress blocks.

The tensile strength of the concrete should be

neglected.

The influence of transverse shear forces on the

resistance to bending and normal force should be

considered when determining the interaction curve, ifthe shear force Va,Ed on the steel section exceeds 50%

of the design shear resistance Vpl,a,Rd of the steel

section. The influence of the transverse shear on the

resistance in combined bending and compression

should be taken into account by a reduced designsteel strength (1 - ρ ) f yd in the shear area Av.

interaction curve

G. Hanswille

Univ Prof Dr IngInfluence of vertical shear


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MRd

f sd

0,85f cdf yd

+

f yd

zpl

Vc,EdVa,Ed

Nc+s

Na

Ma Mc,+s

+

--

f sd

f sd -

f yd-

NEd

VEdf sd

Rd,cEd,cRd,plaEd,a VVVV ≤≤

Ed,aEdEd,c

Rd,pl

Rd,pla

Rd

aEdEd,a

VVV

M

M

M

MVV

−=

≈=

Verification for vertical

shear:

-

MRd= Ma + Mc+s NEd = Na +Nc+s

zpl

Influence of vertical shear

The shear force Va,Ed should not exceed

the resistance to shear of the steel section.

The resistance to shear Vc,Ed of the

reinforced concrete part should be verified

in accordance with EN 1992-1-1, 6.2.

Unless a more accurate analysis is

used, VEd may be distributed into

Va,Ed acting on the structural steel

and Vc,Ed acting on the reinforcedconcrete section by :

Mpl,a,Rd is the plastic resistance

moment of the steel section.

Mpl,Rd is the plastic resistance momentof the composite section.

G. Hanswille

Univ Prof Dr IngDetermination of the resistance to normal


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Determination of the resistance to normal

forces and bending (example)

tf

y

zplz

c

Nc

0,85f cd

-

f yd

zs

zs

Naf

Naf

Naw,c

zaw,c

zaw,t

Naw,t(1-ρ) f yd

VEd

NEd

Mpl,N,Rd

Position of the plastic neutral axis: Edi NN∑ =

Edydplwwydplwcdplw Nf )1()zh(tf )1(ztf 85,0z)tb( =ρ−−−ρ−+−

f sd

b

hw

ydwcdw

ydwwEdpl

f )1(t2f 85,0)tb(

f )1(thNz

ρ−+−

ρ−+=

Plastic resistance to bending Mpl,N,Rd in case of the

simultaneously acting compression force NEd and the

vertical shear VEd:

szsN2)f twh(af Nt,awzt,awNc,awzc,awNczcNRd,N,plM +++++=

sdss

cdplwc

ydf af

ydwplwt,aw

ydwplc,aw

f A2N

f 85,0z)tb(N

f tbN

f )1(t)zh(N

f )1(tzN

=

−=

=

ρ−−=

ρ−=

Ns

Ns

tw-

+

Edt,awc,awc NNNN =−+

As

G. Hanswille

Univ -Prof Dr -IngSimplif ied determination of the interaction curve


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Simplif ied determination of the interaction curve

N

M

2

N Rd,pm

Rd,pmN

Rd,plM Rdmax,M

Rd,plN A

B

D

C

Rd,plN

Rd,plRd,B MM =

Rd,pmN

Rd,pmN5,0

A

B

C

D

0,85f cd f sd

f sd

f sd

f sd

- -

-+

+

+

zpl 2hn

hn

-

--

-

-

-

+

-

Rd,plRd,C MM =

Rdmax,Rd,D MM =

f yd

f yd

f yd

f yd

f yd

f ydf yd

zpl

zpl

0,85f cd

0,85f cd

0,85f cd As a simplification, the interactioncurve may be replaced by a polygonal

diagram given by the points A to D.

G. Hanswille

Univ -Prof Dr -IngResistance at points A and D


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Univ. Prof. Dr. Ing.




Resistance at points A and D

f cdf sd

---

0M

NNNN

Rd, A

Rd,plsRd,plcRd,plaRd,pl

=

++=

f yd

Point A

Mpla,RdM

pls,Rd0,5 Mplc,Rd

0,85 f cdf sd

f yd-

+

-

+

zsi

zsi

bc

hc

h

Rdmax,Rd,D

Rd,plcRd,D

MM

N5,0N

=

=

[ ] yssisisds,plRd,pls f z Af WM ∑==

cds,pla,pl

2cc

cdc,plRd,plc f 85,0WW4

hbf 85,0WM

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡−−==

Point D

Npla,RdNplc,Rd Npls,Rd

ydf f w

2f

yda,plRd,pla f )th(tb4

t)t2h(f WM

⎥⎥⎦

⎤

⎢⎢⎣

⎡−+

−==

b

h

tf tw

Rd,plcRd,plsRd,plaRdmax, M21MMM ++=

Wpl,a plastic section modulus of the

structural steel section

Wpl,s plastic section modulus of the cross-section of reinforcement

Wpl,c plastic section modulus of the concrete

section

G. Hanswille

Univ.-Prof. Dr.-Ing.Bending resistance at Point B (M )


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Bending resistance at Point B (Mpl,Rd )

Mpl,Rd

Mpln,Rd

0,85f cd

f sd

f yd

-

+

-

+2f

yd

+

+

2 f sd

-

+

0,85f cd 0,85

f cd

- -

f yd

hn

f sd

MD,Rd

ND,Rdzpl

+ =

+

+

+

+

ND,Rd

hn

hn

hn

+ =

At point B is no resistance tocompression forces. Therefore

the resistance to compression

forces at point D results from the

additional cross-section zones in

compression. With ND,Rd thedepth hn and the position of the

plastic neutral axis at point B can

be determined. With the plastic

bending moment Mn,Rd resulting

from the stress blocks within thedepth hn the plastic resistance

moment Mpl,Rd at point B can be

calculated by:

Rdln,pRd,DRd,pl MMM −=

+

+

zpl

hn

Point D

Point B

G. Hanswille

Univ.-Prof. Dr.-Ing.Plastic resistance moment at Point C


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The bending resistance at point C

is the same as the bending

resistance at point B.

MC,Rd= Mpl,Rd

The normal force results from the

stress blocks in the zone 2hn.

Plastic resistance moment at Point C

Mpl,Rd

0,85 f cd

f sd

f yd

-

+

-

+

2f yd

-

-

2 f sd

-

+

0,85f cd 0,85f cd

- -

f yd

2hn f sd+ =

+

+ +

+

+

NC,Rd

2hn

2hn

2hn

+ =

+

Nc,Rd

Mc,Rd2h

n

NC,Rd = 2 ND,Rd = Ncpl,Rd = Npm,Rd

hn

zpl

Point B Point C

G. Hanswille



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Part 4:

Simplified design method

G. Hanswille

Univ.-Prof. Dr.-Ing.Simplified Method


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Methods of verification acc. to the simplified method

Design based on the

European buckling curves

Design based on second orderanalysis with equivalent geometrical

bow imperfections

Kλ

κ

wo

wo

Axial

compression

Resistance

of member

in combined

compression

and bending

Simplified Method

Design based on second order

analysis with equivalent geometrical

bow imperfections

G. Hanswille


I tit t f St l d


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Scope of the simplified method

double symmetrical cross-section

uniform cross-sections over the member length with rolled,

cold-formed or welded steel sections

steel contribution ratio

relative slenderness

longitudinal reinforcement ratio

the ratio of the depth to the width of the composite cross-section should be within the limits 0,2 and 5,0

Rd,pl

yda

N

f A9,02,0 =δ≤δ≤

0,2N

N

cr

Rk,pl ≤=λ

c

sss

A

A%0,6%3,0 =ρ≤ρ≤

G. Hanswille


Institute for Steel andEffects of creep of concrete


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Load

F [kN]

2000

1500

1000

500

020 40 60 80 100

short term test

deflection w [mm]

Fu = 2022 kN

Fu = 1697 kN

long term test

Fv = 534 kN

30 cm

3 0

c m

e=3 cm

L = 8 0 0 c m

e

F

wo

The horizontal deflection and

the second order bending

moments increase underpermanent loads due to creep of

concrete. This leads to a

reduction of the ultimate load.

permanent

load

The effects of creep of

concrete are taken intoaccount in design by a

reduced flexural stiffness of

the composite cross-section.

wt

p

G. Hanswille



Effects of creep on the flexural stiffness


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The effects of creep of concrete are

taken into account by an effective

modulus of elasticity of concrete

)t,t(N

N1

EE

oEd

Ed,G

cmeff ,c

ϕ+

=

Ecm Secant modulus of concrete

NEd total design normal force

NG,Ed part of the total normal force that is

permanent

ϕ(t,to) creep coefficient as a function of the time atloading to, the time t considered and the

notional size of the cross-section

b b

h h

U

A2h co =

)hb(2U += b5,0h2U +≈

notional size of the cross-section for

the determination of the creep

coefficient ϕ(t,to)In case of concrete filled hollow section the drying of the

concrete is significantly reduced by the steel section. A

good estimation of the creep coefficient can be

achieved, if 25% of that creep coefficient is used, whichresults from a cross-section, where the notional size hois determined neglecting the steel hollow section.

ϕt,eff = 0,25 ϕ(t,to)

effective perimeter U of the cross-

section

G. Hanswille


Institute for Steel andVerification for axial compression with the


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cross-section bucklingcurve

b

c

a

b

b

b

%3s ≤ρ

%6%3 s ≤ρ<

cr

k,pl

N

N

=λ

Rd,pl

Rd

NN=χ

0,2 1,0

1,0

0,8

0,6

0,4

0,2

0,6 1,4

a

b

c

0,1

N

N

Rd

Ed ≤

Rd,plRd NN χ=

cdcsdsydaRd,pl f Af Af AN ν++=

Design value of

resistance

Verification:

0,2N

Ncr

Rk,pl ≤=λ

1,8

European buckling curves

buckling about

strong axis

buckling about

weak axis

85,0=ν

85,0=ν

00,1=ν

85,0=ν

00,1=ν

00,1=ν

G. Hanswille



Relative slenderness


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relative slenderness:

0,2N

N

cr

Rk,pl ≤=λ

sksckcykaRk,pl f Af Af AN +ν+=

2

eff 2

cr )L(

)EJ(N

β

π=

)JEJEKJE()EJ( ssceff ,ceaaeff ++=

elastic critical normal force

85,0

f f

c

ckcd

=ν

γ=

effective flexural stiffness

00,1

f f

c

ckcd

=ν

γ=

Ke=0,6

characteristic value of the plastic

resistance to compressive forces

β - buckling length factor

G. Hanswille


Institute for Steel andVerification for combined


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22II,eff

2

cr

cr

EdoEdEd

L

)JE(N

N

N1

1wNMmax

βπ=

−=

bending moments taking into

account second order effects:

9,0K5,0Kwith

)JEJEKJE(K)EI(

oII,e

ssceff ,ceaaoII,eff

==

++=

Verification

αM= 0,9 for S235 and S355

αM= 0,8 for S420 and S460

Rd,plMRdEd MMMmax μα=≤

Npl,Rd

NEd

N

wo

Mpl,RdMRd

αM μ Mpl,Rd

M

Effective flexural stiffness

Mpl,N,Rd

f sd

Mpl,N Rd

NEd

VEd

f yd

(1-ρ) f yd

0,85f cd

-

+

-

compression and bending

wo equivalent

geometrical bow

imperfection

The factor αM takes into account thedifference between the full plastic and the

elasto-plastic resistance of the cross-section

resulting from strain limitations for concrete.

G. Hanswille


Institute for Steel andEquivalent initial bow imperfections


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q p

Buckling curve

a b c

wo= L/300 wo= L/200wo= L/150

%3s ≤ρ

%6%3 s ≤ρ<

Member imperfection

G. Hanswille


Institute for Steel andImperfections for global analysis of frames


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hmo ααφ=φ

Global initial sway imperfection acc. to EN 1993-1-1:

φ φ Φo basic value with Φo = 1/200

αh reduction factor for the height h in [m]

αm reduction factor for the number of columns in a row

0,132but

h2 hh ≤α≤=α

⎥⎦⎤

⎢⎣⎡ +=α

m

115,0m

m is the number of columns in a row including only

those columns which carry a vertical load NEd

not

less than 50% of the average value of the column

in a vertical plane considered.

h

sway imperfection

equivalent forces

NEd,1 NEd,2

Φ NEd,1 NEd,1NEd,2

Φ NEd,1

Φ NEd,2

Φ NEd,2

G. Hanswille


Institute for Steel andFrames sensitive against second order effects


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1,Ed

Rk,pl

N

N

5,0≤λ

w0

φ2φ1

NEd,1 NEd,2

equivalent

forces

NEd,1 φ1

2,Ed22

o NL

w8q=

2,Ed

2

o N

L

w4

2,Ed2

o NL

w4

L2

L1 2,Ed

Rk,pl

N

N

5,0>λ

NEd,1NEd,2

NEd,1 φ1

NEd,2 φ2

NEd,2 φ2

i,Ed

Rk,pl

N

N

5,0≤λ

2i

eff 2

cr L

)EJ(N π=

)JEJE6,0JE()EJ( ssceff ,caaeff ++=

Within a global analysis, member imperfections in

composite compression members may be

neglected where first-order analysis may be

used. Where second-order analysis should be

used, member imperfections may be neglectedwithin the global analysis if:

cr

Rk,pl

N

N=λ

imperfections

G. Hanswille



C it St t

Second order analysis


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41



⎟

⎠

⎞⎜

⎝

⎛ −

εξ−ε

+⎟

⎠

⎞⎜

⎝

⎛ ε

ξε+ξ−ε=ξ 1

)2/(cos

)5,0(cosM

sin

sin)1(sinr M)(M oR

⎟ ⎞

⎜⎝

⎛ −

εξ−ε

+⎟ ⎠

⎞⎜⎝

⎛ ε

ξε+ξ−εε=ξ 1

)2/(cos

)5,0(sinM

sin

cos)1(cosr

L

M)(V o

Rz

[ ]

Bending moments including second order effects:

Maximum bending moment at the point ξM:

0

2

omax M)5,0cos(

c1M)r 1(M5,0M −

ε+++=

)5,0(tan

1

M2)r 1(M

)1r (Mc

o ε++

−=

ε+=ξ

carctan5,0

M

2o2o 1)wN8Lq(M

ε+=

II,eff Ed)JE(

NL=ε

⎟ ⎠

⎞⎜⎝

⎛ =ξ

0d

dM

maxM

ζMζ

Lwo

MR

r MRN

EJ

MR

r MR

q

G. Hanswille




Simplif ied calculation of second order effects


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42



4,0

3,0

2,0

1,0

0,25 0,50 0,75 1,00

exact method

simplified method

r=1,0

r=0,5

r= - 0,5

r=0

k

cr NN

MR

r MR N

L

EJ

ζM

Exact solution:

)5,0cos(

c1)r 1(M5,0M

2

Rmax ε+

+=

)5,0(tan

1

r 1

1r c

ε+−

=

ε+=ξcarctan

5,0MII,eff

Ed

)JE(

N

L=ε

simplified solution:

cr

EdR

max

N

N1

M

Mk

−

β== r 44,066,0 +=β

maxM

ζ

44,0≥β

MR

r MR

G. Hanswille




Background of the member imperfections


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43



N

NRd wo

Mpl,RdMRd

μ Mpl,Rd

NEd=NRd

⎥⎦

⎤⎢⎣

⎡−

ε

= 1

)2/cos(

1

L

)EJ(w8M

2

II,eff o

Bending moment based on second

order analysis:II,eff

Rd

)EJ(

NL=ε

Resistance to axial compression

based on the European buckling

curves:

Rd,plRd NN χ=

Determination of the equivalent bow

imperfection:

Rd,plMRd MM μα=M

⎥⎦⎤⎢

⎣⎡ −

ε−

μα=

1)2/(cos1

1)EJ(8

LMw

II,eff

2Rd,pldM

o

The initial bow imperfections were

recalculated from the resistance to

compression calculated with the

European buckling curves.

Bending resistance:

G. Hanswille




Geometrical bow imperfections –

comparison with European buckling


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44



500

400

300

200

j

λ1,0 2,0

C20/S235

C40/S355

C60/S355

ow

L j =

λ

1

2

3

0,4 0,8 1,2 1,6 2,0

1,0

1,1

1,2

0,9

0,8

11

2

2

3

3

δ )w(N)(N

oRd

Rd κ=δ

curves for axial compression

The initial bow imperfection is a

function of the related slenderness

and the resistance of cross-sections.

In Eurocode 4 constant values for w0are used.

wo= l/300

The use of constant values for wo leads to

maximum differences of 5% in

comparison with the calculation based onthe European buckling curves.

G. Hanswille




Comparison of the simplified method with non-linear calculations for combined compression

d b di


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45



Rd,plN

N

Rd,plM

M

0,2 0,4 0,6 0,8 1,0

0,2

0,4

0,6

0,8

1,050,0k =λ

00,1k =λ

50,1k =λ

00,2k =λ

Resistance as a function of the

related slenderness

general method

simplified method

cr

Rk,pl

N

N=λ

Plastic cross-sectionresistance

and bending

G. Hanswille




Resistance to combined compression

and biaxial bending


46/73

46

p

University of Wuppertal-Germanyand biaxial bending

Interaction

My, Mz, NEd

Interaction

Mz, N Interaction

My, N

Interaction

My, Mz

Rd,y,pl

y

M

M

Rd,z,pl

z

M

M

Rd,plN

NThe resistance is given by a three-dimensional interaction relation. For

simplification a linear interaction

between the points A and B is used.

Approximation: A

Bdyμ

dzμ

Rd,pl

Ed

N

N

Rd,y,pldyEdRd,y M)N(M μ=

Rd,y,pldzEdRd,z M)N(M μ=Ed,zμ Ed,y

μ

Rd,yEd,yEd,y MM μ=

Rd,yEd,zEd,z MM μ=

approximation for the

interaction curve:

0,1dz

Ed,z

dy

Ed,y ≤μ

μ+

μ

μ

0,1M

M

M

M

Rd,z,pldz

Ed,z

Rd,y,pldy

Ed,y

≤μ+μ

0,1dz

Ed,z

dy

Ed,y ≤μ

μ+

μ

μ

G. Hanswille




Verif ication in case of compression an biaxial

bending


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47

University of Wuppertal-Germanybending

MRd,y,pldz

Ed,zM

Rd,y,pldy

Ed,y

M

M

M

M

α≤μα≤μ

0,1M

M

M

M

Rd,y,pldz

Ed,z

Rd,y,pldy

Ed,y ≤μ+

μ

Rd,plNN

Rd,y,pl

Rd,y

M

M

dyμ

Rd,plNN

Rd,z,pl

Rd,z

M

M

dzμ

For both axis a separate verification

is necessary.

Verification for the interaction of biaxial

bending.

Imperfections should be consideredonly in the plane in which failure is

expected to occur. If it is not evident

which plane is the most critical,

checks should be made for bothplanes.

Rd,pl

Ed

N

N

Rd,pl

Ed

N

N

αM= 0,9 for S235 and S355αM= 0,8 for S420 and S460

G. Hanswille





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Part 5:

Special aspects of columns with inner core profiles

G. Hanswille




U i i f W l G

Composite columns – General Method


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49


Commerzbank Frankfurt

Highlight Center

Munich

New railway station in Berlin(Lehrter Bahnhof)

Millennium Tower Vienna

G. Hanswille




U i it f W t l G

Composite columns with concrete fil led

tubes and steel cores – special effects


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50

University of Wuppertal-Germanyp

Resistance based on stress blocks (plastic

resistance)

Non linear resistance with strain limitation for concrete

tube core concrete

f y f y f c

N

M

f y f y

f c

M

N

strains ε

Rd,pl

RdM

M

M

μ=α

Cross-sections withmassive inner cores have

a very high plastic shape

factor and the cores can

have very high residualstresses. Therefore these

columns can not be

design with the simplified

method according to EN

1944-1-1.σED

r

G. Hanswille




University of Wuppertal Germany

Residual stresses and distribution of the

yield strength


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51


distribution of yield strenghtσED [N/mm2]

U/A [1/m]

dK [mm]

10 20 30 40 50

80100200400 130

50

100

150

200250

300

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡−σ=σ

2

K

2

EDEr

r 21)r (

f yk f y(r)

2

kyk

y

r

r 1,095,0

f

)r (f ⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +=

f yk – characteristic value

of the yield strenght

σED

r, r k

residual stresses:

r k

dk

kdU π=

4/d A 2k

π=

ddK

G. Hanswille





General method – Finite Element Model


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52


initial bow imperfectionstresses in the tube

stresses in concrete

load introduction

cross-section

G. Hanswille






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53


Part 6:

Load introduction andlongitudinal shear

G. Hanswille





Load introduction over the steel section


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54


NEd

Nc,Ed

Na,Ed

LE < 2,0 d

d

Aa As

Ac

PD

load introduction by headed studs within the

load introduction length LE

⎩⎨⎧≤ 3/L

d2LE

d minimum transverse dimension of

the cross-section

L member length of the column

sectional forces of the cross-section :

Rd,pl

c,plEdEd,c

Rd,pl

s,plEdEd,s

Rd,pl

a,plEdEd,a

N

NNN

N

NNN

N

NNN ===

required number of studs n resulting from thesectional forces NEd,c+ NEd,s:

Ns,Ed

⎥⎥⎦

⎤

⎢⎢⎣

⎡−=+=

Rd,pl

a,plEdEd,sEd,cEd,L

N

N1NNNV RdRd,L PnV =

PRd – design resistance of studs

G. Hanswille





Load introduction for combined comression and

bending


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55

y pp y

NEdMEd

Ma,EdNa,Ed

Mc,Ed +Ms,Ed

Nc,Ed +Ns,Ed

Rd,pl

Ed

N

N

sectional forces due to NEdund MEd

sectional forces based on plastic theory

Rd,scRd,aRdRd,scRd,aRd NNNMMM ++ +=+=

2

Rd,pl

Ed

2

Rd,pl

Edd

N

N

M

MR ⎟

⎟ ⎠

⎞⎜⎜⎝

⎛ +⎟

⎟ ⎠

⎞⎜⎜⎝

⎛ =

2

Rd,pl

Rd

2

Rd,pl

RddNN

MME ⎟⎟

⎠ ⎞⎜⎜

⎝ ⎛ +⎟⎟

⎠ ⎞⎜⎜

⎝ ⎛ =

Rd,sc

Ed,sc

Rd,sc

Ed,sc

Rd,a

Ed,a

Rd,a

Ed,a

d

d

M

M

N

N

M

M

N

N

R

E

+

+

+

+ ====

Rd,plN

N

Rd,pl

Rd

N

N

Rd,pl

Ed

M

M

Rd,pl

Rd

M

M

Rd,plM

M

1,0

1,0

dE

dR

f yd

zpl

Nc+s,RdNa,Rd

Ma;Rd Mc,+s,Rd

+

--

f sd

f sd0,85f cd

MRd

NRd

+ =

G. Hanswille





Load introduction – Example


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56

f yd

zpl

-Nc+s,RdNa,Rd

Ma;Rd Mc,+s,Rd

+

- -

f sd

f sdf cd

Zs

zc Zs

shear forces of studs based on elastic theory shear forces of studs based on plastic theory

Nc+s,EdNc+s,Ed Mc+s,Ed

-Ns,i

Ns,i

-Nc∑+=

∑+=

+

+

sisiccRd,sc

sicRd,sc

zNzNM

NNN

PEd(N)

PEd(M)

r iPed,v

Ped,h

eh

n5,0e

M

n

NPmax

h

Ed,scEd,scEd

++ +=

xizi

2

i2i

Ed,sc

2

i2i

Ed,scEd,scEd z

r

Mx

r

M

n

NPmax

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

∑+

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

∑+= +++

n – number of

studs within the

load introduction

length

sectional forces based on stress blocks:

Mc+s,Ed

G. Hanswille





Shear resistance of stud connectors welded

to the web of partially encased I-Sections


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57

µ PRd / 2µ PRd / 2

PRd

Dc

Rd,LRRdRd,L VPV +=

RdRd,LR PV μ=

Where stud connectors are attached to the web of a fully or

partially concrete encased steel I-section or a similar

section, account may be taken of the frictional forces that

develop from the prevention of lateral expansion of the

concrete by the adjacent steel flanges. This resistance maybe added to the calculated resistance of the shear

connectors. The additional resistance may be assumed to

be on each flange and each horizontal row of studs, where μ

is the relevant coefficient of friction that may be assumed.

For steel sections without painting, μ may be taken as 0,5.

PRd is the resistance of a single stud.

G. Hanswille






to the web of partially encased I-Sections


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58

PRd

< 300 < 400 < 600

PRd PRdPRd PRd PRd

VLR,Rd/2

RdRd,LRRd,LRRdRd,L PVVPnV μ=+=

In absence of better information from tests, the clear distance betweenthe flanges should not exceed the values given above.

v

cmck2

1,Rd1

Ef d29,0P

γ

α=

v

2

u2,Rd1

4

df 8,0P

γ⎟⎟ ⎠

⎞⎜⎜⎝

⎛ π⋅=

PRd= min

G. Hanswille






to the web of partially encased I-sections


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test series S1 test series S2

F [kN]

w [mm]

FFFF

test S1/3

test S3/3

0 2 4 6 8 10 12 14 160

500

1000

1500

2000

2500

3000

3500

G. Hanswille





Load introduction – longitudinal shear

forces in concrete


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NEd

Dc Dc

Zs Zs

ZsDc

θθ

I I

Longitudinal shear force in section I-I:

not directly connected concretearea As1

sdscdc

sd1scd1c

Rd,pl

a,plEdEd,L

f Af 85,0 A

f Af 85,0 A

N

N1NV

++

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡−=

NEd

bccy cy

LE

I I

As- cross-section area of the stirrups

Longitudinal shear resistance of concrete struts:

Ecdy

max,Rd,L Ltancot

f 85,0c4V

θ+θ

ν=

longitudinal shear resistance of the stirrups:

θ =45o

Eydw

ss,Rd,L Lcotf

s

A4V θ=

sw

- spacing of stirrups

2ckck mm/Ninf with))250/f (1(6,0 −=ν

G. Hanswille





Load introduction – longitudinal shear

forces in concrete – test results


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61

test I/1

FF

w [mm]

F [kN]

Fu = 1608 kN

2000

1500

1000

500

00 2 4 6 8 10 12 14

w

G. Hanswille





Load introduction – Examples

(Airport Hannover)


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Load introduction with

gusset plates

G. Hanswille





Load introduction with partially loaded

end plates


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63

Load introduction with

partially loaded end plates

G. Hanswille





Load introduction with distance plates for

columns with inner steel cores


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64

distance plates

Post Tower Bonn

G. Hanswille





Composite columns with hollow sections –

Load introduction


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65

Stiffener

Distance

plate

σc

σc σc

gusset platestiffeners and

end platesdistance plates

G. Hanswille





Mechanical model


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σc,r

σa,t σa,t

σcσa

⎥⎦

⎤⎢⎣

⎡η+=

c

ycL

1

c1cm,cR

f

f

d

t1

A

A Af P

A1

Ac

Effect of partially

loaded area

Effect of

confinement by thetube

ηc,L = 3,5 ηc,L = 4,9

G. Hanswille





Typical load-deformation curves


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0.5

1.5

2.5

2.0

1.0

0 5 10 15 20 25 30

Pu

35

[mm]

P [MN]

Pu,stat

series SXIII

P

P [MN]

5 10 15

Pu

1.0

3.0

5.0

4.0

2.0

0

[mm]

20

series SV

P

Pu,stat

G. Hanswille





Test evaluation according to EN 1990


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2,0 4,0 6,0 8,0 10,0

r e [MN]

2,0

4,0

41 testsVr = 0.14

r t [MN]

Pc,Rm

Pc,Rk = 0.78 Pc,Rm

Pc,Rd = 0.66 Pc,Rm

6,0

σc,r σa,y σa,y

σc

σa,x

8,0

10,0

1

c

c

y

cL1cm,cR A

A

f

f

d

t1 Af P

⎥⎥⎦

⎤

⎢⎢⎣

⎡η+=

A1

Ac

G. Hanswille





Load distribution by end plates


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Fu = 6047 kN

Fu,stat = 4750 kN

u = 7.5 mm

F [kN]

δ [mm]

6000

4000

2000

5,0 10,0 15,0

Fu

ts

bc

σc

tp

psc t5tb +=

d1~

G. HanswilleUniv.-Prof. Dr.-Ing.




Design rules according to EN 1994-1-1


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yd1

cdc

1

c

ck

ykcLcdRd,c f

A

f A

A

A

f

f

d

t1f ≤≤⎥

⎦

⎤⎢⎣

⎡η+=σ

f ck concrete cylinder strength

t wall thickness of the tube

d diameter of the tube

f yk yield strength of structural steel

A1 loaded area

Ac cross section area of the concrete

ηc,L confinement factorηc,L = 4,9 (tube)

ηc,L = 3,5 (square hollow sections)

20 A

A

1

c ≤

A1

Rd,cσ

Load distribution 1:2,5

bc bcpsc t5tb +=

tp

ts





Verif ication outside the areas of load introduction

F


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Outside the area of load introduction,

longitudinal shear at the interface

between concrete and steel should be

verified where it is caused bytransverse loads and / or end

moments. Shear connectors should

be provided, based on the distribution

of the design value of longitudinalshear, where this exceeds the design

shear strength τ Rd.

In absence of a more accurate

method, elastic analysis, considering

long term effects and cracking of

concrete may be used to determine

the longitudinal shear at the interface.

F

δ A

B

C

pure bond

(adhesion)

mechanical

interlock

friction

σr





Design shear strength τRd

bc


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72

concrete

encased

sections

bc

hc

b

h

cy cy

cz

czy

z

5,2c

c1c02,01

z

min,zzc

co,RdRd

≤⎥

⎦

⎤⎢

⎣

⎡−+=β

βτ=ττRd,o= 0,30 N/mm2

concrete filled tubes

τRd= 0,55 N/mm2

concrete filled

rectangular hollow

sections

τRd= 0,40 N/mm2

flanges of partially

encased I-sections

τRd= 0,20 N/mm2

webs of partially

encased I-sections

τRd= 0,0 N/mm2

cz- nominal concrete cover [mm]

cz,min=40mm (minimum value)






73/73

73

Thank you very

much for your kind

attention

EN1994_4_Hanswille

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Transcript of EN1994_4_Hanswille