1994.pdf · 2010-08-24

EUROCODESBackground and Applications

“Dissemination of information for training” workshop 18-20 February 2008 Brussels

EN 1994 Eurocode 4: Design of composite steel and concrete structures Organised by European Commission: DG Enterprise and Industry, Joint Research Centre with the support of CEN/TC250, CEN Management Centre and Member States

Wednesday, February 20 – Palais des Académies EN 1994 - Eurocode 4: Design of composite steel and concrete structures Prigogine room

9:00-9:20 General presentation of EN1994 J. Raoul SETRA

9:20-10:40 Structural analysis and ultimate limit state

U. Kuhlmann Universität Stuttgart

10:40-11:00 Coffee

11:00-12:00 Serviceability limit state G. Hanswille Bergische Universität Wuppertal

12:00-13:30 Lunch

13:30-14:30 Composite columns G. Hanswille Bergische Universität Wuppertal

14:30-15:40 Composite slabs S. Hicks Steel Construction Institute

15:40-16:00 Coffee

16:00-17:30 Composite bridges L. Davaine & J. Raoul SETRA

All workshop material will be available at http://eurocodes.jrc.ec.europa.eu

http://eurocodes.jrc.ec.europa.eu/

GENERAL PRESENTATION OF EN1994

J. Raoul SETRA

Brussels, 18-20 February 2008 – Dissemination of information workshop 1


General presentation of Eurocode 4

Joël RAOUL



EUROCODE 4 : Design of composite steel and concrete structures

EN 1994-1-1 : general rules and rules for buildingsEN 1994-1-2 : structural fire designEN 1994-2 : general rules and rules for bridges

The general rules valid for bridges from part 1-1 are repeated in part 2 to get a self sufficient document.

EUROCODE 4


EUROCODESBackground and Applications EN 1994-1-1

ForwardSection 1 GeneralSection 2 Basis of designSection 3 MaterialSection 4 DurabilitySection 5 Structural analysisSection 6 ULSSection 7 SLSSection 8 Composite joints in frames for buildingsSection 9 Composite slabs for buildingsAnnex A (informative) Stiffness of joint in buildingsAnnex B (informative) Standard testsAnnex C (informative) Shrinkage of concrete for buildings

Common to all EC

Layoutcommon to all EC



ForwardSection 1 GeneralSection 2 Basis of designSection 3 MaterialSection 4 DurabilitySection 5 Structural analysisSection 6 ULSSection 7 SLSSection 8 Precast concrete slabs in bridgesSection 9 Composite plates in bridgesAnnex C Headed studs that cause splitting in the slab

thickness

EN 1994-2


EUROCODESBackground and Applications Rules for drafting

The paragraphs specific to buildings are put at the end to be easily modified.

EN 1994-1-1



The paragraphs specific to bridges are added at the end of the clauses.

EN 1994-2



Avoid cascades of referencesBrussels, 18-20 February 2008 – Dissemination of information workshop 8

EUROCODESBackground and Applications Scope of EN 1994-1-1

Composite members

Composite beamsComposite columns

Composite slabs

Composite joints


EUROCODESBackground and Applications Composite beams

Solid slab

Composite slab

Partially encased


EUROCODESBackground and Applications Composite columns

Partially encasedConcrete encased

Concrete filled


EUROCODESBackground and Applications Composite slabs


EUROCODESBackground and Applications Composite joints


EUROCODESBackground and Applications Scope of EN 1994-2

Composite bridgesI girdersBox sectionsCable stayed bridges not fully covered

Composite members

Filler beam decks

Tension members

Composite plates


EUROCODESBackground and Applications Composite bridges


EUROCODESBackground and Applications Composite members



Filler beam decks

transversal

longitudinal


EUROCODESBackground and Applications Tension members


EUROCODESBackground and Applications Composite plates


EUROCODESBackground and Applications Coordination EC4-EC3 : materials

12 10-610 10-6

equal for steel and concrete

Coefficient of expansion

S 235 – S 460+ EN 1993-1-12 (S 690)

S 235 – S 460Grade of steel

EC3EC4


EUROCODESBackground and Applications Coordination EC4-EC2 : materials

As in EC2 or annex C(3,25x10-4 in dry environment)

shrinkage

200 000210 000 (as in EC3) equal for steel and reinforcement

Modulus of elasticity

C12 – C90C20 – C60Concrete strength

EC2EC4


EUROCODESBackground and Applications Coordination EC4-EC3 : design rules

EN 1993-1-5(SLS ≠ ULS)

Slab : EC4 (same at SLS/ULS) steel flange : EN 1993-1-5

Effective width

EC3EC4


EUROCODESBackground and Applications Coordination EC4-EC2 : design rules

fcd = αcc fck / γCfcd = fck / γC0.85 is a calibration factor of Mpl,Rd

Design value

Vertical shear resistance of the cracked slab in EC2 has been modifiedShear

Effective width

EC2EC4

STRUCTURAL ANALYSIS AND ULTIMATE LIMIT STATE

U. Kuhlmann

Universität Stuttgart

Institute of Structural DesignUniversität Stuttgart Prof. Dr.-Ing. U. Kuhlmann

Prof. Dr.-Ing. Ulrike Kuhlmann

Institute of Structural DesignUniversität Stuttgart

Germany

Design of composite beams

according to Eurocode 4-1-1

Lecture:

Ultimate Limit States


A short introduction

Prof. Dr.-Ing. Ulrike Kuhlmann

Universität StuttgartInstitute of Structural DesignMain Fields: Steel, Timber and CompositePfaffenwaldring 770569 StuttgartGermany

Phone +49 711 685 66245fax +49 711 685 66236Email [email protected]

http://www.uni-stuttgart.de/ke/


Contents

1 - SCOPE

2 - SPECIFIC CHARACTERISTICS OF STRUCTURAL ANALYSIS

3 - METHODS OF GLOBAL ANALYSIS

4 - VERIFICATION FOR BENDING AND SHEAR FOR ULS

5 - SHEAR CONNECTION

Design of composite beams according to Eurocode 4-1-1


Part 1:

SCOPE

Car park, Messe Stuttgart


Definitions according EN 1994-1-1 [§1.5.2]

COMPOSITE MEMBERa structural member with components of concrete and of structural or cold-formed steel, interconnected by shear connection so as to limit the longitudinal slip between concrete and steel and the separation of one component from the other

SHEAR CONNECTIONan interconnection between the concrete and steel components of a composite member that has sufficient strength and stiffness to enable the two components to be designed as parts of a single structural member

COMPOSITE BEAMa composite member subjected mainly to bending

1 Scope


composite behaviour no composite behaviour

acting as one section

composite beam steel beam with concrete slab

COMPOSITE BEHAVIOUR

acting as two individual sections

1 Scope


TYPICAL COMPOSITE BEAMS

Seite 4 von Hanswille einfügen

1 Scope

[Source: Hanswille]


Materials according EN 1994-1-1 [§ 3]

CONCRETE

REINFORCEMENT

STRUCTURAL STEEL

> C 20/25; LC 20/25

< C 60/75; LC 60/75

Acc. EN 1992-1-1 § 3.2

strength: 400 N/mm2 fy,k 600 N/mm2

ductility: 1,05 (ft/fy)k 1,35

fy 460 N/mm2

CONNECTING DEVICESHeaded stud shear connector acc. EN 13918

structural steel

connecting devices

reinforcement

concrete

1 Scope


Part 2:

SPECIFIC CHARACTERISTICS

OF STRUCTURAL ANALYSIS

source:[ESDEP]


2 Specific characteristics of structural analysis

• Non-linear material behaviour

• Influence of erection and load history

• Influence of creep and shrinkage

• Influence of composite interaction

Characteristics

source:[ESDEP]


Non-linear material behaviour

q1

w1 w2 w3 w4

q2

q3

q4

q

w

w w


M-pl,Rd

M+pl,Rd

Cross-section

at supportin span



q1

w1 w2 w3 w4

q2

q3

q4

q

w

w w


q1 – first cracking (concrete slab) at support

M-pl,Rd

M+pl,Rd

Cross-section

at supportin span



q1

w1 w2 w3 w4

q2

q3

q4

q

w

w w


q2 – first yielding (steel section) at support

M-pl,Rd

M+pl,Rd

Cross-section

at supportin span



q1

w1 w2 w3 w4

q2

q3

q4

q

w

w w


q3 – first plastic hinge M-pl.Rd at support

M-pl,Rd

M+pl,Rd

Cross-section

at supportin span



q1

w1 w2 w3 w4

q2

q3

q4

q

w

w w

Cross-section

at supportin span


q4 – last plastic hinge M+pl.Rd in span

M-pl,Rd

M+pl,Rd


M+pl,Rd

M+pl,Rd

M-pl,Rd

M-pl,Rd

Cross-section in span

Cross-section at support



+

-cfcd

fyd

fyd

fyd

fsd

++

-

q1

q2

q3

q4

q MM


High efficiency of plastic hinge theory due to difference of plastic bending moment in span and at support - requires rotation capacity of section with first plastic hinge (at support)


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

2,0

4,0

6,0

8,0

10,0

12,0

q4 = qpl

q3

q

ql2q =

Mpl,F

Mpl,F

Mpl,Stpl =

load level q3

load level q4= qpl

Mpl,St

Mpl,St

Mpl,F

l l

qNon-linear material behaviour




[Source: Hanswille]


[Source: Hanswille]




[Source: Hanswille]

Classes 1 and 2






[Source: Hanswille]

Class 3


[Source: Hanswille]

Classification with partial concrete encasement




Reinforcement in tension flanges



[Source: Hanswille]


Influence of erection and load history

Example:Bridge Arminiusstraße in Dortmund

- erection steel structure

3 spansR = 900 m



Raising at inner supportsExample:Bridge Arminiusstraße in Dortmund

- raising at inner supports- scaffolding hanging at steel structure- concreting and hardening of concrete- lowering at inner supports- finalizing (pavement etc.)- traffic opening




[Source: Hanswille]




A

B

C

unpropped construction

propped construction

propped construction + jacking of props


[Source: Hanswille]



The bending capacity Mpl,Rd isindependent of the loading history in case of Class 1 or Class 2 cross sections

Using Class 3 or Class 4 cross sections the elastic behaviour of the loading history has to be taken into account in ULS

[Source: Hanswille]




Influence of creep and shrinkage


The effects of shrinkage and creep of concrete result in internal forces in cross sections, and curvatures and longitudinal strains in members

[Source: Hanswille]



creep and shrinkage has to be consideredFor Class 3 and 4 sections

bending capacity independent of creep and shrinkageFor Class 1 and 2 sections

only external deformationsFor statically determinate structures:

Due to creep and shrinkage:

[Source: Hanswille]



[Source: Hanswille]


In statically indeterminate structures the primary effects of shrinkage and creep are associated with additional action effects, such that the total effects are compatible;

These shall be classified as secondary effects and shall be considered as indirect actions in any case



Influence of composite interaction

[Source: Hanswille]


MRd MEd

MR

Mpl,Rd

Mpl,a,Rd

hi 1,0 Ncf

Nch=

A

B

CA

B

C

Ncf normal force in the concrete slab

due to Mpl,Rd

e

Nc=0

...degree of shear connections



MRd MEd

MR

Mpl,Rd

Mpl,a,Rd

hi 1,0 Ncf

Nch=

A

B

CA

B

C

Ncf normal force in the concrete slab

due to Mpl,Rd

e

Nc=0

= 0

0 < < 1

= 1

no shear connection acting as 2 independent sections

full shear connection acting as one section without slip full plastic resistance Mpl,Rd

partial shear connection acting as one section with slip at interfacebending resistance depending on shear connection

Influence of composite interaction


Part 3:

METHODS OF GLOBAL ANALYSIS

Bridge crossing Mosel at Bernkastel-Kues


• Structural stability

• Calculation of action effects

based on elastic theory

• Rigid plastic analysis

• Stresses based on elastic theory

3 Methods of global analysis

Bridge crossing Mosel at Bernkastel-Kues


General case

• Portal frames w/ shallow roofslopes

• Beam-and-column type plane frames

undeformedgeometry

10Ed

crcr F

F

deformed geometry

10Ed,HEd

Edcr

hVH

n

5.2.1(3)

y

n

3 alternatives of verification

5.2.1(4)B

y

Structural stability




Global second-order analysis + individual stability check of

membersEquivalent column method

EN 1994-1-1 5.2.2 (3)EN 1994-1-1 6.7.3.6 / 7

EN 1994-1-1 5.2.2 (6) b) and 5.2.2 (6) c)

Only for steel columns:EN 1993-1-1 5.2.2 (3) c)

5.2.2 (8)

Second-order analysisof whole system





accounting forglobal and localimperfections

0

w0

w0





including global imperfections

Individual stability check of members acc. to EN 1994-1-1

6.7.3.4 or 6.7.3.5

Buckling length = system length


members





Equivalent column method

neither global nor localimperfections

Equivalent columnmethod for member acc. EN 1993-1-1 6.3.1/2/3

Buckling length by global eigenvalue determination






accounting forglobal and localimperfections

including global imperfections

Individual stability check of members acc. to EN1994-1-1

6.7.3.4 or 6.7.3.5

Buckling length = member length


membersEquivalent column method

neither global nor localimperfections

Buckling length by global eigenvalue determination


Equivalent columnmethod for member acc. EN 1993-1-1 6.3.1/2/3




Calculation of action effects based on elastic theory


[Source: Hanswille]



Calculation of action effects based on elastic theory - General method

[Source: Hanswille]


[Source: Hanswille]






[Source: Hanswille]


Relation Classification - method of global analysis - resistance


[Source: Hanswille]


Rigid plastic analysis

[Source: Hanswille]


Rigid plastic global analysis may be used for ultimate limit state verifications other than fatigue, where second-order effects do not have to be considered and provided that all the members and joints of the frame are steel or composite, the steel material satisfies ductility requirements EN 1993-1-1, the cross-sections of steel members have sufficientrotation capacity and the joints are able to sustain their plastic resistance moments for a sufficient rotation capacity.



pl,Rd

+

-zpl

h

L

Fd

qd

> 0,5zpl

h0,15 if

Fd

Fd + qd L

Le Li

Lmax Lmin

Limitation of span ratio:

exterior span: Le < 1,15 Li

interior span: Lmax/Lmin 1,50

Beam with single load and rotation requirements at span:



Where rigid-plastic global analysis is used, at each plastic hinge location:

a) the cross-section of the structural steel section shall be symmetrical about a plane parallel to the plane of the web or webs,

b) the proportions and restraints of steel components shall be such that lateral-torsional buckling does not occur,

c) lateral restraint to the compression flange shall be provided a tall hinge locations at which plastic rotation may occur under any load case,

d) the rotation capacity shall be sufficient, when account is taken of any axial compression in the member or joint, to enable the required hinge rotation to develop and

e) where rotation requirements are not calculated, all members containing plastic hinges shall have effective cross-sections of Class 1 at plastic hinge locations.




For composite beams in buildings, the rotation capacity may be assumed to be sufficient where:

a) the grade of structural steel does not exceed S355,

b) the contribution of any reinforced concrete encasement in compression is neglected when calculating the design resistance moment,

c) all effective cross-sections at plastic hinge locations are in Class1; and all other effective cross-sections are in Class1 or Class2,

d) each beam-to-column joint has been shown to have sufficient design rotation capacity, or to have a design resistance moment at least 1,2 times the design plastic resistance moment of the connected beam,

e) adjacent spans do not differ in length by more than 50% of the shorter span,

f) end spans do not exceed 115% of the length of the adjacent span,

g) in any span in which more than half of the total design load for that span is concentrated within a length of one-fifth of the span, then at any hinge location where the concrete slab is in compression, not more than 15% of the overall depth of the member should be in compression; this does not apply where it can be shown that the hinge will be the last to form in that span,

h) the steel compression flange at a plastic hinge location is laterally restrained.





[Source: Hanswille]





[Source: Hanswille]


Stresses based on elastic theory


[Source: Hanswille]



Modular ratios taking into account effects of creep


[Source: Hanswille]



Elastic cross section properties taking into account creep


[Source: Hanswille]




[Source: Hanswille]



Primary effects due to shrinkage


[Source: Hanswille]



Primary effects due to shrinkage


[Source: Hanswille]


Part 4:

VERIFICATION FOR BENDING AND SHEAR

FOR ULTIMATE LIMITE STATE


4 Verification for bending and shear for ULS

• General

• Resistance of class 1 and 2 sections

• Resistance of class 3 and 4 sections

• Lateral torsional buckling


General - Basis of design


Rd=Mpl,Rd

Ed Rd

Ultimate limitstate:

Ed Cd

Serviceabilitliylimit state:

[Source: Hanswille]



Partial safety factor for concrete C according to EN 1992-1-1 e.g. C = 1.5

Partial safety factor for reinforcement steelS according to EN 1992-1-1 e.g. S= 1.15

Partial safety factor for structural steela according to EN 1993-1-1 e.g. M0 = 1.0


[Source: Hanswille]






General - Required verifications for composite beams

[Source: Hanswille]



General - Required verifications for composite beams

[Source: Hanswille]



General – Critical cross section

[Source: Hanswille]



General – Effective width

[Source: Hanswille]



General – Effective width of concrete flanges

[Source: Hanswille]



General – Non-linear bending resistance

[Source: Hanswille]



Classification girders

[Source: Hanswille]


Resistance of class 1 and 2 sections - classification


[Source: Hanswille]


Reduction of plastic bending resistance


[Source: Hanswille]



Reduction of plastic bending resistance

[Source: Hanswille]



Resistance of class 1 and 2 sections

[Source: Hanswille]

0,5 1,0


Resistance of class 1 and 2 sections - Full and partial shear connection


[Source: Hanswille]


Resistance of class 1 and 2 sections - Partial shear connection - general


[Source: Hanswille]

design resistance of studs


Resistance of class 1 and 2 sections Partial shear connection – determination of moment resistance


[Source: Hanswille]



Resistance of class 1 and 2 sections Partial shear connection – determination of moment resistance

[Source: Hanswille]


Resistance of class 3 and 4 sections - class 3


[Source: Hanswille]



Resistance of class 3 and 4 sections - class 4

[Source: Hanswille]


Resistance of class 3 and 4 sections Class 4 – Determination of stresses


[Source: Hanswille]


Resistance of class 3 and 4 sections Cross section: class 4 – bending resistance (method I)


[Source: Hanswille]


Resistance of class 3 and 4 sections Resistance to vertical shear


[Source: Hanswille]



Resistance of class 3 and 4 sections Resistance to vertical shear

[Source: Hanswille]


Resistance of class 3 and 4 sections Method I – Interaction of bending and shear


[Source: Hanswille]


Lateral torsional buckling


[Source: Hanswille]


Lateral torsional buckling – reduction factor

[Source: Hanswille]


Lateral torsional buckling – elastic critical bending moment


[Source: Hanswille]


Lateral torsional buckling – simplified verification

[Source: Hanswille]



Lateral torsional buckling – stabilizing forces on lateral frames


[Source: Hanswille]


Lateral torsional buckling – without direct calculation


[Source: Hanswille]


Part 5:

SHEAR CONNECTION


• Longitudinal shear forces• Determination of longitudinal shear forces• Full and partial shear connection• Requirements for shear connectors

• Headed studs• Head studs as shear connector• Horizontally lying studs• Headed studs used with profiled steel sheeting

• Longitudinal shear forces in concrete slab Part 5:

SHEAR CONNECTION

5 Shear connection


Part 5:

SHEAR CONNECTION



• Longitudinal shear forces in concrete slab

5 Shear connection


Longitudinal shear forces

5 Shear connection

[Source: Hanswille]


Determination of longitudinal shear forces - general

5 Shear connection

[Source: Hanswille]


Determination of longitudinal shear forces - by simplified method for Nc

5 Shear connection

[Source: Hanswille]


Partial shear connection – determination of longitudinal shear forces

5 Shear connection

[Source: Hanswille]


M

Requirements for shear connection – uniformly distribution

qd

5 Shear connection

[Source: Hanswille]


Requirements for shear connection – minimum degree

5 Shear connection

[Source: Hanswille]


Requirements for shear connection – ductility

5 Shear connection

[Source: Hanswille]


Part 5:

SHEAR CONNECTION




5 Shear connection


Headed studs

5 Shear connection

[Source: Hanswille]


Headed studs – typical load-slip behaviour

Pw … flashPZ … stud inclinationPB … stud bendingPR … frictionflash

5 Shear connection


Headed studs – typical load-slip behaviour

5 Shear connection

[Source: Hanswille]


Headed studs – design shear resistance

[Source:EC4-1

& Hanswille]

d diameter of stud shank 16 d 25mmfu specified ultimate tensile strength of the stud

material fu 500 N/mm²fck cylinder strength of concreteEcm secant modulus of elasticity of concretea =0.2 [(h/d)+1] for 3 h/d 4

=1.0 for h/d > 4

V =1.5 partial safety factor concrete failure=1.25 partial safety factor steel failure

5 Shear connection


Headed studs – detailing

5 Shear connection

[Source: Hanswille]


Headed studs – uplift forces

5 Shear connection

[Source: Hanswille]


Horizontally lying studs – examples

cast-in-place concrete

prefabricatedconcrete slab

5 Shear connection


Horizontally lying studs – examples

5 Shear connection

[Source: Hanswille]


Longitudinal sheardue to beam bending

Vertical sheardue to vertical beam support

Edge position Middle position

Horizontally lying studs – failure modes and position

Concrete edgefailure

Splittingfailure

Vertical shear

Longitudinalshear

5 Shear connection


[Source: EN1994-2]

Horizontally lying studs – load resistance for longitudinal shear

middle position edge positionsection A-A

v

..'rckv

L,Rds/aadfk.P

304041

a‘r effective edge distancea‘r = ar – cv - s/2 50 mm

kv factor for position of shear connectionkv = 1 edge positionkv = 1.4 middle position

v partial factor 1.25

d … diameter of the stud shank 19 d 25 mmh … overall height of the stud h/d 4s … spacing of stirrups a/2 s a

s/a‘r 3

s… diameter of stirrups s 8 mm

l… diameter of longitudinal reinforment l 10 mm

L,Rdd P.T 30stirrups

5 Shear connection


[Source: EN1994-2]

Horizontally lying studs – load resistance for vertical shear

middle position edge positionsection A-A

v

v.'

o,r...

ckV,Rd

kas/adf.P

703040500120 a … spacing of studs110 a 440 mm

h … overall height of the studh 100 mm

s… diameter of stirrups

s 12 mm

l… diameter of longitudinal reinforment

l 16 mm

l s

12121 .

V,Rd

V,d.

L,Rd

L,d

PF

PF

Interaction:

5 Shear connection


Headed studs used with profiled steel sheeting

5 Shear connection

[Source: Hanswille]


Headed studs used with profiled steel sheeting – load-slip behaviour

5 Shear connection

[Source: Hanswille]


Headed studs used with profiled steel sheeting – load resistance

5 Shear connection

[Source: Hanswille]


Headed studs used with profiled steel sheeting – load resistance

5 Shear connection

[Source: Hanswille]


Part 5:

SHEAR CONNECTION




5 Shear connection


Longitudinal shear forces in concrete slab - determination

Slab in compression

Slab in tension

5 Shear connection

[Source: Hanswille]


Longitudinal shear forces in concrete slab – strut-and-tie model

5 Shear connection

[Source: Hanswille]


Longitudinal shear forces in concrete slab – shear plane

section a-a: Acv= hc av

section b-b, c-c, d-d: Acv = Lv av

with Lv = Lb-b, Lc-c, Ld-d

section

5 Shear connection

[Source: Hanswille]


Acknowledgement

and many thanks to

My co-workersDipl.-Ing. Gunter HaufDipl.-Ing. Matthias KonradDipl.-Ing. Ana OžboltDipl.-Ing. Lars RölleDipl.-Ing. Markus Rybinskifor their support

Prof. Dr.-Ing. Gerhard Hanswille

for allowanceto base on his ppt - presentationprepared for lectures in Riga in 2006


Thank you very much

for your kind attention !

Universität StuttgartInstitute for Structural DesignSteel, Timber and Composite StructuresProf. Dr.-Ing. Ulrike KuhlmannPfaffenwaldring 770569 Stuttgart

Phone +49 711 685 66245Fax +49 711 685 66236Email [email protected]


Literature

Bode, H.: Euro-Verbundbau, Konstruktion und Berechnung, publisher Werner Verlag, Düsseldorf, 1998

Deutsches Institut für Bautechnik: Slim-Floor Träger mit UPE-Profilen, Allgemeine bauaufsichtliche ZulassungNr. Z-26.2-48, Technical Approval, 2005.

DIN 18800-5: Steel structures – Part 5: Composite structures of steel and concrete – Design and Construction, German Standard, 2006.

DIN EN 1994-1-1: Design of composite steel and concrete structures: General rules and rules for buildings, European Standard, 2002

Hanswille, G., Schäfer, M., Verbundtragwerke aus Stahl und Beton, Bemessung und Konstruktion - Kommentar zu DIN V 188000 Teil 5 Ausgabe November 2004, Stahlbaukalender 2005, editor Ulrike Kuhlmann, publisher Verlag Ernst & Sohn, Berlin

Hanswille G.: The new German design code for composite bridges,Engineering Foundation Conferences Composite Construction V, South Africa, Juli 2004

Hanswille G., Bergmann R.: New design methods for composite columns including high strength steel,Engineering Foundation Conferences Composite Construction V, South Africa, Juli 2004

Hanswille G., Piel W.: Composite shear head systems for improved punshing shear resistance of flat slabs,Engineering Foundation Conferences Composite Construction V, South Africa, Juli 2004

Hanswille G., Porsch M.: Load introduction in composite columns with concrete filled hollow sections,Engineering Foundation Conferences Composite Construction V, South Africa, Juli 2004

Roik, K., Bergmann, R., Haensel, J., Hanswille, G. Verbundkonstruktionen: Bemessung auf der Grundlage des Eurocode 4 Teil 1, Betonkalender 1993, publisher Verlag Ernst & Sohn, Berlin


Literature

Breuninger, U.; Kuhlmann, U.: Tragverhalten und Tragfähigkeit liegender Kopfbolzendübel unterLängsschubbeanspruchung, Stahlbau 70, p. 835-845, 2001.

Breuninger, U.: Zum Tragverhalten liegender Kopfbolzendübel unter Längsschubbeanspruchung, PhD-Thesis, Universität Stuttgart, Mitteilung Nr. 2000-1, 2000.

Kuhlmann, U.; Breuninger, U.: Behaviour of horizontally lying studs with longitudinal shear force, In: Hajjar, J.F., Hosain, M., Easterling, W.S. and Shahrooz, B.M. (eds), Composite Construction in Steel and Concrete IV, American Society of Civil Engineers, p.438-449, 2002.

Kuhlmann, U.; Kürschner, K.: Structural behaviour of horizontally lying shear studs, In: Leon, R.T. and Lange, J. (eds), Composite Construction in Steel and Concrete V, American Society of Civil Engineers, p.534-543, 2006.

Kuhlmann, U.; Rieg, A.; Hauf, G.; Effective Width Of Composite Girders With Reduced Height, Prof. Aribert - Symposium, July2006, Institut National des Sciences Appliquées (Rennes), France, 2006.

Kürschner, K.; Kuhlmann, U.: Trag- und Ermüdungsverhalten liegender Kopfbolzendübel unter Quer- und Längsschub, Stahlbau 73, p.505-516, 2004.

Kürschner, K.: Trag- und Ermüdungsverhalten liegender Kopfbolzendübel im Verbundbau, PhD-Thesis, Universität Stuttgart, Mitteilung Nr. 2003-4, 2003.

Raichle, J.: Fatigue behaviour and application of horizontally lying shear studs, In: 6th International PhD Symposium in Civil Engineering, Zurich, Switzerland, 2006.

Rybinski, M.: Structural behaviour of steel to concrete joints on basis of the component method, In: 6th International PhD Symposium in Civil Engineering, Zurich, Switzerland, 2006.

SERVICEABILITY LIMIT STATE

G. Hanswille Bergische Universität Wuppertal

1

1

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

Institute for Steel and Composite Structures

University of Wuppertal-Germany

Serviceability limit states of composite beams

Institute for Steel and Composite StructuresUniversity of Wuppertal

Germany

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

Eurocode 4

EurocodesBackground and Applications

Dissemination of information for training18-20 February 2008, Brussels

2




Contents

Part 1: Introduction

Part 2: Global analysis for serviceability limit states

Part 3: Crack width control

Part 4: Deformations

Part 5: Limitation of stresses

Part 6: Vibrations

3




Serviceability limit states


Limitation of stresses

Limitation of deflections

crack width control

vibrations

web breathing4





{ }∑ ∑ ψ+++= i,ki,01,kkj,kd QQPGEEcharacteristic combination:

frequent combination: { }∑ ∑ ψ+ψ++= i,ki,21,k1,1kj,kd QQPGEE

quasi-permanent combination: { }∑ ∑ ψ++= i,ki,2kj,kd QPGEE

serviceability limit states Ed ≤ Cd:

- deformation- crack width - excessive compressive stresses in concrete

Cd= - excessive slip in the interface between steel and concrete

- excessive creep deformation- web breathing- vibrations

5




Part 2:

Global analysis for serviceability limit states

6




Global analysis - General

Calculation of internal forces, deformations and stresses at serviceability limit state shall take into account the followingeffects:

shear lag;

creep and shrinkage of concrete;

cracking of concrete and tension stiffening of concrete;

sequence of construction;

increased flexibility resulting from significant incompleteinteraction due to slip of shear connection;

inelastic behaviour of steel and reinforcement, if any;

torsional and distorsional warping, if any.

2

7




Shear lag- effective width

σmax

σmax

b

be

The flexibility of steel or concrete flanges affected by shear in their plane (shear lag) shall be used either by rigorous analysis, or by using an effective width be

2,0bb

i

ei ≥

σmax

bei

bi

5 bei

y

bi

y

σmax

bei

σ(y)

σ(y)

2,0bb

i

ei <

σR

[ ]4

iRmaxR

maxi

eiR

by1)y(

2,0bb25,1

⎥⎦

⎤⎢⎣

⎡−σ−σ+σ=σ

σ⎥⎦

⎤⎢⎣

⎡−=σ

4

imax b

y1)y( ⎥⎦

⎤⎢⎣

⎡−σ=σ

shear lag

real stress distribution

stresses taking into account the effective width

8




midspan regions and internal supports:

beff = b0 + be,1+be,2

be,i= Le/8

Le – equivalent lengthend supports: beff = b0 + β1 be,1+β2 be,2

βi = (0,55+0,025 Le/bi) ≤ 1,0

Effective width of concrete flanges

Le=0,85 L1 for beff,1 Le=0,70 L2 for beff,1

Le=0,25 (L1 + L2) for beff,2 Le=2L3 for beff,2

L1 L2L3

beff,0

beff,1beff,1beff,2

beff,2

L1/4 L1/2 L1/4 L2/2L2/4 L2/4

bobe,1 be,2

bob1 b2

9




Initial sectional forces

redistribution of the sectional forces due to creep

ML

-Nc,o

Mc,o

Mst,o

Nst,o

Nc,r

-Mc,r

Mst,r

-Nst,r

zi,st

-zi,c ast

Effects of creep of concrete

primary effects

The effects of shrinkage and creep of concrete and non-uniform changes of temperature result in internal forces in cross sections, and curvatures and longitudinal strains in members; the effects that occur in statically determinate structures, and in statically indeterminate structures when compatibility of the deformations is not considered, shall be classified as primary effects.

10




Effects of creep and shrinkage of concrete

Types of loading and action effects:

In the following the different types of loading and action effects are distinguished by a subscript L :

L=P for permanent action effects not changing with timeL=PT time-dependent action effects developing affine to the creep coefficientL=S action effects caused by shrinkage of concreteL=D action effects due to prestressing by imposed deformations (e.g. jacking of

supports)

MPT(t)MPT (t=∞)

ϕ(t,to)ϕ(t∞,to)ϕ(ti,to)

time dependent action effects ML=MPT:

action effects caused by prestressing due to imposed

deformation ML=MD:

δ

ML=MD +

MD

MPT(ti)

11




Modular ratios taking into account effects of creep

[ ]cm

aooLoL E

En)t,t(1nn =ϕψ+=Modular ratios:

centroidal axis of the concrete section

centroidal axis of the transformed composite section

centroidal axis of the steel section (structural steel and reinforcement)

-zic,L

zist,Lzi,L

zczis,Last

zst

ΨPT=0,55time-dependent action effectsΨD=1,50prestressing by controlled imposed deformationsΨS=0,55shrinkageΨP=1,10permanent action not changing in time

Ψ=0short term loadingcreep multiplieraction

12




)t(EEn

ocm

sto =

Modular ratio taking into account creep effect:centroidal axis of the

concrete section

L,i2stL,cstL,cstL,i A/aAAJJJ ++=L,cStL,i AAA +=

L,iststL,ic A/aAz −=LcL,cLcL,c n/JJn/AA ==

-zic,L

zist,Lzi,L

zc-zis,L

ast

zst

Transformed cross-section properties of the concrete section:

Transformed cross-section area of the composite section:

Second moment of area of the composite section:

Distance between the centroidal axes of the concrete and the composite section:

))t,t(1(nn 0L0L ϕψ+=

Elastic cross-section properties of the composite section taking into account creep effects

centroidal axis of the composite section

centroidal axis of the steel section

3

13




Effects of cracking of concrete and tension stiffening of concrete between cracks

ε

εs(x)

εc(x)

Ns Ns

c

ctEf

s

2s2,s E

σ=ε

r,sεΔβ

εsr,1 εsr,2 εsm,y εsy

Ns

Nsy

Nsm

Ns,cr

B C

σs,2σs(x)

σc(x)

τv

xstage A: uncracked sectionstage B: initial crack formationstage C: stabilised crack formation

σc(x)

fully cracked section

A

σc(x)

σs(x)

mean strain εsm=εs,2- βΔεs,r

r,sεΔ

εsm

r,ss εΔβ=εΔ

ss

eff,cts E

fρ

β=εΔ

css A/A=ρ

4,0=β

14




-κ

εsm

MMs≈0

Ma

Na

εa

a

zs

Ns equilibrium:

aNMM sa −=

sa NN −=

εs,m

εs,2Δεs=β Δεs,r

εc

εs

compatibility:

aasm κ+ε=ε

aaaa

2s

aa

ssm JE

aMAEaN

AEN

=++ε

ss

eff,ct

ss

ssr2ssm E

fAE

Nρ

β−=εΔβ−ε=ε

mean strain in the concrete slab:

mean strain in the concrete slab:

za

Influence of tension stiffening of concrete on stresses in reinforcement

15




Ns,2

-Ms,2

ΔNts

-Ma,2

-Na,2 -ΔNts

ΔNts aa

Ns,2

-MEd

Ns

MEd

M

Ns

Nsε

zst,a

-zst,s

tsst

s,stsEdts2ss N

JzA

MNNN Δ+=Δ+=sts

seff,ctts

AfN

αρβ=Δ

tsst

a,staEdts2aa N

JzA

MNNN Δ−=Δ−=

aNJJMaNMM ts

st

aEdts2aa Δ+=Δ+=

Sectional forces:

st

sEds J

JMM =

aa

ststst JA

JA=α

Ns

-Ms

-Ma

-Na

fully cracked section tension stiffening

+ =z2=zst

ΔNts

Redistribution of sectional forces due to tension stiffening

2st JJ =

16




Stresses taking into account tension stiffening of concrete

Ns,2

-Ms,2

ΔNts

-Ma,2

-Na,2 -ΔNts

ΔNts azst,a

-zst,s

sts

sctmts

AfNαρ

β=Δ

aa

ststst JA

JA=α

Ns

-Ms

-Ma

-Na

fully cracked tension stiffening

+ =zst-MEd

sts

ctms,st

st

Eds

sts

ctm2,ss

fzJ

M

f

αρβ+=σ

αρβ+σ=σ

aa

ts

a

tsst

st

Eda

aa

ts

a

ts2,aa

zJ

aNANz

JM

zJ

aNAN

Δ+

Δ−=σ

Δ+

Δ−σ=σ

reinforcement: structural steel:

za

a

17




Influence of tension stiffening on flexural stiffness

EstJ1 uncracked sectionEstJ2 fully cracked sectionEstJ2,ts effective flexural

stiffness taking into account tension stiffening of concrete

κ

EstJ1

EstJ2EstJ2,ts

εsm

M

-M

κ

Ns

-Ms

-Ma

-Na

εa

azst

ast

s

ast

a

ts,2st JEaNM

JEM

IEM −

===κ

EJ

MR MRn

Est J1

Est J2,ts

EstJ2

Ma)NN(

1

JEJE,ss

aats,2st

ε−−

=

M

Curvature:

Effective flexural stiffness:

18




• Determination of internal forces by un-cracked analysis for the characteristic combination.

• Determination of the cracked regions with the extreme fibre concrete tensile stress σc,max= 2,0 fct,m.

• Reduction of flexural stiffness to EaJ2 in the cracked regions.

• New structural analysis for the new distribution of flexural stiffness.

L1 L2L1,cr L2,cr

EaJ2EaJ1 EaJ1

ΔM

un-cracked analysiscracked analysis

ΔM Redistribution of bending moments due to cracking of concrete

EaJ1 – un-cracked flexural stiffness

EaJ2 – cracked flexural stiffness

Effects of cracking of concrete - General method according to EN 1994-1-1

4

19




L1 L2

ΔMII

EaJ1

0,15 L1 0,15 L2

EaJ2

6,0L/L maxmin ≥

Effects of cracking of concrete –simplified method

For continuous composite beams with the concrete flanges above the steel section and not pre-stressed, including beams in frames that resist horizontal forces by bracing, a simplified method may be used. Where all the ratios of the length of adjacent continuous spans (shorter/longer) between supports are at least 0,6, the effect of cracking may be taken into account by using the flexural stiffness Ea J2 over 15% of the span on each side of each internal support, and as the un-cracked values Ea J1 elsewhere.

20




Part 3:

Limitation of crack width

21




Control of cracking

General considerations

If crack width control is required, a minimum amount of bonded reinforcement is required to control cracking in areas where tension due to restraint and or direct loading is expected. The amount may be estimated from equilibrium between the tensile force in concrete just before cracking and the tensile force in the reinforcement at yielding or at a lower stress if necessary to limit the crack width. According to Eurocode 4-1-1 the minimum reinforcement should be placed, where under the characteristic combination of actions, stresses in concrete are tensile.

minimum reinforcement

control of cracking due to direct loading

Where at least the minimum reinforcement is provided, the limitation of crack width for direct loading may generally be achieved by limiting bar spacing or bar diameters. Maximum bar spacing and maximum bar diameter depend on the stress σs in the reinforcement and the design crack width.

22




Recommended values for wmax

decompressionXD1,XD2,XS1,XS2,XS3

0,2 mm (2)0,3 mm

XC2, XC3,XC4

0,2 mm0,4 mm (1)XO, XC1

frequent load combinationquasi - permanentload combination

prestressed members with bonded tendons

reinforced members, prestressedmembers with unbonded tendons

and members prestressed by controlled imposed deformations

Exposure class

(1) For XO and XC1 exposure classes, crack width has no influence ondurability and this limit is set to guarantee acceptable appearance. In absence of appearance conditions this limit may be relaxed.

(2) For these exposure classes, in addition, decompression should bechecked under the quasi-permanent combination of loads.

23




Exposure classes according to EN 1992-1-1(risk of corrosion of reinforcement)

parts of marine structurespermanently submergedXS2

car park slabs, pavements, parts of bridges exposed to spray containingcyclic wet and dryXD3Corrosion induced by chlorides from sea water

structures near to or on the coastexposed to airborne saltXS1

parts of marine structurestidal, splash and spray zonesXS3

swimming pools, members exposed to industrial waters containing chlorides

wet, rarely dryXD2

concrete surfaces exposed to airborne chlorides moderate humidityXD1Corrosion induced by chlorides

concrete surfaces subject to water contact not within class XC2cyclic wet and dryXC4

external concrete sheltered from rainmoderate humidityXC3

concrete surfaces subjected to long term water contact, foundationswet, rarely dryXC2

concrete inside buildings with low air humiditydry or permanently wetXC1Corrosion induced by carbonation

concrete inside buildings with very low air humidityfor concrete without reinforcement, for concrete with reinforcement : very dry

XOno risk of corrosion or attack

ExamplesDescription of environmentClass

24




Cracking of concrete (initial crack formation)

ε εs

εc

LesLes

NsNs

w

c

ss A

A=ρ

As cross-section area of reinforcementρs reinforcement ratiofctm mean value of tensile strength of concrete

c

so E

En =

c1,cs1,sss AAA σ+σ=σ

Equilibrium in longitudinal direction:

Compatibility at the end of the introduction length:

c

1,c

s

1,s1,c1,s EE

σ=

σ⇒ε=ε

⎥⎦

⎤⎢⎣

⎡ρ+

ρσ=σ

osos

s1,s n1n

os

s1,sss n1 ρ+

σ=σ−σ=σΔ

Change of stresses in reinforcement due to cracking:

LesLes

σsσs,1

σc,1

Δσs

σc,1

σs,1

σs,2

Les

σ

( )oscctmr,s n1AfN ρ+=

5

25




Cracking of concrete – introduction length

ε εs

εc

LesLes

NsNs

w

c

ss A

A=ρ

Us -perimeter of the barAs -cross-section areaρs -reinforcement ratioτsm -mean bond strength

c

so E

En =

4ddL

AUL2s

ssmses

sssmsesπ

σΔ=τπ

σΔ=τ

oss

1,sss n1 ρ+σ

=σ−σ=σΔ

Change of stresses in reinforcement due to cracking:

Equilibrium in longitudinal direction

LesLes

σsσs,1

σc,1

Δσs

σc,1

σs,1

σs,2

Les

τsm

σ

sosm

sses n1

14

dLρ+τ

σ=

introduction length LEs

crack width

)(L2w cmsmes ε−ε=

26




Determination of the mean strains of reinforcement and concrete in the stage of initial crack formation

ctmL

os

esm,s f8,1dx)x(

L1 Es

≈τ=τ ∫

Mean bond strength:

s

smsssm,s σΔ

σΔ−σ=β⇒σΔβ−σ=σ

∫ τ=σΔx

0s

ss dx)x(

U4)x(∫ σΔ=σΔ

esL

0s

essm dx)x(

L1

εεs

εc(x)

LesLes

NsNs

w

LesLes

σs σs,1

σc,1

Δσs

σ

εcr

Δεs,cr

Mean strains in reinforcement and concrete:

crm,c εβ=ε

Mean stress in the reinforcement:εs,m

εc,m

βΔσs

x

σs,m

σs(x)

εs(x)

cr,s2,sm,s εΔβ−ε=ε

27




Determination of initial crack width

ε εs(x)

εc(x)

LesLes

NsNs

w

LesLes

σsσs,1

σc,1

Δσs

σs

εcr

Δεs,crεs,m

εc,m

βΔσs

x

σs,m

crack width

)(L2w cmsmes ε−ε=

sosm

sses n1

14

dLρ+τ

σ=

2,scmm,s )1( εβ−=ε−ε

εs,2

sossms

2s

n11

E2d)1(w

ρ+τσβ−

=

with β= 0,6 for short term loading und β= 0,4 for long term loading

ctmsm f8,1≈τ

28




Maximum bar diameters acc. to EC4

-56450

468400

5810360

61012320

81216280

121620240

162532200

253240160

wk= 0,2wk= 0,3wk= 0,4

maximum bar diameter forσs

[N/mm2]

∗sd

sm,ct

s2s

sossm

s2s

Ef6d

n11

E2d)1(w σ

≈ρ+τ

σβ−=

Crack width w:

Maximum bar diameter for a required crack width w:

)1()n1(E2wd 2

s

sossms

β−σ

ρ+τ=

2s

so,ctmk*s

2s

soso,ctmk

*s

Efw6d

)1(

)n1(Ef6,3wd

σ≈

β−σ

ρ+=

With τsm= 1,8 fct,mo and the reference value for the mean tensile strength of concrete fctm,o= 2,9 N/mm2 follows:

β= 0,4 for long term loading and repeated loading

29




Crack width for stabilised crack formation

ε

εs(x)

εc(x)

sr,max= 2 Les

Ns

w

c

ctEf

s

2s2,s E

σ=ε

εs(x)- εc(x)

sr,min= Les

)(sw cmsmmax,r ε−ε=

Crack width for high bond bars

cctm

cm

ssctm

2,sss

ctmc2,sm,s

s2,sm,s

Ef

Ef

AEfA

β=ε

ρβ−ε=β−ε=ε

εΔβ−ε=ε

Mean strain of reinforcement and concrete:

β= 0,6 for short term loading


)n1(Ef

E soss

ctm

s

scmsm ρ+

ρβ−

σ=ε−ε

30




Crack width for stabilised crack formation

ε

εs(x)

εc(x)

sr,max= 2 Les

Ns

w

c

ctEf

ss

2,s Eσ

=ε

εs(x)- εc(x)

sr,min= Les

sms

sctm

smscctm

es 4df

UAfL

τρ=

τ=

The maximum crack spacing sr,max in the stage of stabilised crack formation is twice the introduction length Les.


⎟⎟⎠

⎞⎜⎜⎝

⎛ρ+

ρβ−

σρτ

= )n1(E

fE2

dfw soss

ctm

s

s

ssm

sctm

maximum crack width for sr= sr,max

β= 0,6 for short term loading


6

31




Crack width and crack spacing according Eurocode 2

)(sw cmsmmax,r ε−ε=Crack width

s

s21max,r

d425,0kkc4,3sρ

⋅⋅+= ds-diameter of the bar

c- concrete cover

In Eurocode 2 for the maximum crack spacing a semi-empirical equation based on test results is given

k1 coefficient taking into account bond properties of the reinforcement with k1=o,8 for high bond bars

k2 coefficient which takes into account the distribution of strains (1,0 for pur tension and 0,5 for bending)

ss

soss

ctmss

cmsm E6,0)n1(

Ef

Eσ≥ρ+

ρβ−σ=ε−ε

Crack spacing

β= 0,6 for short term loading β= 0,4 for long term loading and repeated loading

32




Determination of the cracking moment Mcr and the normal force of the concrete slab in the stage of initial cracking

cracking moment Mcr:

[ ]

[ ])z2/(h1(z

JnfM

2/hzJn

fM

oco,ic

ioo,ceff,ctcr

co

ioo,ceff,ctcr

+σ−=

+σ−=

ε

ε

( )ε+

ε

ε+

++

ρ+σ−=

++

=

,scoc

os,ceff,ctccr

,scio

issococrcr

N)z2/(h1

n1)f(AN

NJ

zAzAMNprimary effects due to shrinkage

cracking moment Mcr

hc

zio

zo

zi,st

Nc+s

Mc+s

Mc,ε

Mcr

Nc,ε

σc

σcε

ast

ctm1eff,ct,cc fkf ==σ+σ ε

sectional normal force of the concrete slab:

( )

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

ρ++

ρ+σ−

++

ρ+=

εε+

)n1(fA)z2/(h1n1A

N

)z2/(h11)n1(fAN

0seff,ctc

oc

os,cc,sc

oc0seff,ctccr

kc

kc,ε≈ 0,3

33




Simplified solution for the cracking moment and the normal force in the concrete slab

simplified solution for the normal force in the concrete slab:

primary effects due to shrinkage

cracking moment Mcr

hc zo

zi,st

Nc+s

Mc+s

Mc+s,ε

Mcr

Nc+s,ε

σc

σcε

csctmccr kkkfAN ⋅⋅≈

0,13,0

z2h1

1k

o

cc ≤+

+=

shrinkage

k = 0,8 coefficient taking into account the effect of non-uniform self-equilibrating stresses

ks= 0,9 coefficient taking into account the slip effects of shear connection

cracking moment

34




k = 0,8 Influence of non linear residual stresses due to shrinkage and temperature effects ks = 0,9 flexibility of shear connectionkc Influence of distribution of tensile stresses in concrete immediately prior to

crackingmaximum bar diameter

ds modified bar diameter for other concrete strength classes σs stress in reinforcement acc. to Table 1fct,eff effective concrete tensile strength

css

eff,ctcs kkk

fAA

σ≥ 0,13,0

zh11k

occ ≤+

+=

Mcr

Mc

NcNc,ε

Mc,ε

cracking moment

Na,ε

Ma,ε

shrinkage

hc zo

o,ct

eff,ctss f

fdd ∗=

∗sd

fcto= 2,9 N/mm2

zi,o

Determination of minimum reinforcement

35




stresses in reinforcement taking into account tension stiffening for the bending moment MEd of the quasi permanent combination:

c

ss A

A=ρ

sts

eff,cts,st

2

Eds

ts2,ss

fz

JM

αρβ+=σ

σΔ+σ=σ

aa

22st JA

JA=α4,0=β

Control of cracking due to direct loading –Verification by limiting bar spacing or bar diameter

Ns,2

-Ms,2

ΔNts

-Ma,2

-Na,2 -ΔNts

ΔNts azst,a

-zst,s

Ns

-Ms

-Ma

-Na

fully cracked tension stiffening

+ =zst

-MEd

za

a

The bar diameter or the bar spacing has to be limited

The calculation of stresses is based on the mean strain in the concrete slab. The factor βresults from the mean value of crack spacing. With srm≈ 2/3 sr,max results β ≈ 2/3 ·0,6 = 0,4

Ac

As

Aa

36




Maximum bar diameters and maximum bar spacing for high bond bars acc. to EC4

-56450

468400

5810360

61012320

81216280

121620240

162532200

253240160

wk= 0,2wk= 0,3wk= 0,4

maximum bar diameter forσs

[N/mm2]

-50100360

-100150320

50150200280

100200250240

150250300200

200300300160

wk= 0,2wk= 0,3wk= 0,4

maximum bar spacing in [mm] for

σs

[N/mm2]

∗sd

Table 1: Maximum bar diameter Table 2: Maximum bar spacing

7

37




Direct calculation of crack width w for composite sections based on EN 1992-2

zst

-zst,s

As σs,

MEd

sts

eff,cts,st

stEd

sf

zJ

Mαρ

β+=σ

aa

ststst JA

JA=α

c

ss A

A=ρ 4,0=β


Ns

-Ms

-Ma

-Na

ss

soss

ctmss

cmsm E6,0)n1(

Ef

Eσ≥ρ+

ρβ−σ=ε−ε

ss

max,rd34,0c4,3sρ

+=

crack width for high bond bars:

c - concrete cover of reinforcement38




Stresses in reinforcement in case of bonded tendons – initial crack formation

As, ds

Ap, dp

Les

Lep

τsmτpm

Δσp

σs=σs1+Δσs

Equilibrium at the crack:

)n1(AfNAA totoceff,ctppss ρ+==σΔ+σEquilibrium in longitudinal direction:

s,esmsss LdA τπ=σ

eppmppp LdA τπ=σΔ

Compatibility at the crack:

epp

1ppes

s

1ssps L

EL

EσΔ−σΔ

=σ−σ

⇒δ=δ

v

s

sm

pm1

p1s

1p

p1ss

dd

AAN

AAN

τ

τ=ξ

ξ+

ξ=σΔ

ξ+=σ

N

σs,1

Δσp1

Stresses:

With Es≈Ep and σs1=Δσp1=0 results:Δσs

σp

σp=σpo+σp1+Δσp

39




Stresses in reinforcement for final crack formation

Maximum crack spacing:

p

1p2p2,p

s

1s2s2sps E

)(E

)( σΔ−σΔβ−σΔ=

σ−σβ−σ=δ=δ

[ ]

)AA(2Afd

s

dndn2

sAf

p2

ssm

ceff,ctsmax,r

pppmsssmmax,r

cct

ξ+τ=

πτ+πτ=

Compatibility at the crack:

pmp

pmax,r1p2psm

s

smax,r1s2s A

U2

sAU

2s

τ=σ−στ=σ−σ

Equilibrium in longitudinal direction:

Equilibrium at the crack:

p2ps2so AAPN σΔ+σ=−

σs

As, ds

Ap, dp

σs2

Δσp2

Δσp2

Δσp1

σs1

sr,max

σcσc=fct,eff

x

mean crack spacing: sr,m≈2/3 sr,max

40




Determination of stresses in composite sections with bonded tendons

⎥⎥⎦

⎤

⎢⎢⎣

⎡

ρξ

−ρ

−σ=⎥⎥⎦

⎤

⎢⎢⎣

⎡

ξ+

ξ−

+−σ=σΔ

⎥⎦

⎤⎢⎣

⎡ρ

−ρ

+σ=⎥⎥⎦

⎤

⎢⎢⎣

⎡

+−

ξ++σ=σ

eff

21

toteff,ct

*s

p21s

c21

ps

ceff,ct

*sp

toteffeff,ct

*s

ps

c

p21s

ceff,ct

*ss

1f4,0AA

AAA

Af4,0

11f4,0AA

AAA

Af4,0

c

p21s

eff

c

pstot

AAA

AAA

ξ+=ρ

+=ρ

Stresses σ*s in reinforcement

at the crack location neglecting different bond behaviour of reinforcement and tendons:

sttot

ctms,st

st

Ed*s

fzJ

Mαρ

β+=σ

aa

ststst JA

JA=α 4,0=β

zst

-zst,s

AsAp σs,σp

MEd

Stresses in reinforcement taking into account the different bond behaviour:

41




Part 4:

Deformations

42




Deflections

Deflections due to loading applied to the composite member should be calculated using elastic analysis taking into account effects from

- cracking of concrete,

- creep and shrinkage,

- sequence of construction,

- influence of local yielding of structural steel at internal supports,

- influence of incomplete interaction.

L1 L2

ΔMEaJ1

0,15 L1 0,15 L2

EaJ2

Effects of cracking of concrete

Sequence of construction

F

F

steel member

composite member

gc

8

43




Deformations and pre-cambering

δ1δ2

δ3

δ4

δ1 – self weight of the structureδ2 – loads from finish and service workδ3 – creep and shrinkageδ4 – variable loads and temperature effects

δp

δmax

δw

quasi –permanent(better frequent)

risk of damage of adjacent parts of the structure (e.g. finish or service work)

quasi -permanent

generallimitationcombination

250/Lmax ≤δ

500/Lw ≤δ

δ1

δc

δ1 deflection of the steel girder

δc deflection of the composite girder

Pre-cambering of the steel girder:

δp = δ1+ δ2+ δ3 +ψ2 δ4

δmax maximum deflection

δw effective deflection for finish and service work

44




Effects of local yielding on deflections

For the calculation of deflection of un-propped beams, account may be taken of the influence of local yielding of structural steel over a support.

For beams with critical sections in Classes 1 and 2 the effect may be taken into account by multiplying the bending moment at the support with an additional reduction factor f2 and corresponding increases are made to the bending moments in adjacent spans.

f2 = 0,5 if fy is reached before the concrete slab has hardened;

f2 = 0,7 if fy is reached after concrete has hardened.

This applies for the determination of the maximum deflection but not for pre-camber.

45




More accurate method for the determination of the effects of local yielding on deflections

EaJ1

ΔM

EaJ2

z2 Mel,Rk

σa=fyk fyk

--

+

(EJ)eff

EaJ2

Mel,Rk

Mpl,Rk

Mpl,RkMEd

EaJeff

L1 L2lcr lcr

EaJeff

46




Effects of incomplete interaction on deformations

P

ss

PPcD

su

PRd

The effects of incomplete interaction may be ignored provided that:

The design of the shear connection is in accordance with clause 6.6 of Eurocode 4,

either not less shear connectors are used than half the number for full shear connection, or the forces resulting from an elastic behaviour and which act on the shear connectors in the serviceability limit state do not exceed PRd and

in case of a ribbed slab with ribs transverse to the beam, the height of the ribs does not exceed 80 mm.

47




Differential equations in case of incomplete interaction

Nc

Na

Ma

Mc

Va

Nc+dNc

Na+dNa

Ma+ dMa

Mc+ dMc

Va+dVadx

a

za (w)

zc

x

Ec, Ac, Jc

Ea, Aa, Ja

vL

vL

awuus cav ′+−=

VcVc+dVc

aa

ac

Slip:

ccc u,u ε=′

aaa u,u ε=′

q)awuu(acw)JEJE(0)awuu(cuAE0)awuu(cuAE

casaacc

cascaa

casccc

=′′+′−′−′′′′+

=′+−−′′=′+−+′′

cccc uAEN ′=

aaaa uAEN ′=

wJEM ccc ′′−=

wJEM aaa ′′−=

0wJEV ccc ≈′′′−=

wJEV aaa ′′′−=

48




⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

λ

−λ

λα−

λα+=

)2

cosh(

1)2

cosh(15

38415

481JELq

3845w 42o,ia

4

q

1JJ

J1

o,ca

o,i −+

=α

²LcAAAE

so,i

ao,ca=β

βαα+

=λ12

F

L

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

λ

λ

λα−

λα+=

)sinh(

)2

(sinh48121IE48

LFw2

32o,ia

3

L

Deflection in case of incomplete interaction for single span beams

Aio, Jio

composite sectionconcrete section

steel section Aa, Ja

Aco=Ac/no, Jco= Jc/no

no=Ea/Ec

w

9

49




Mean values of stiffness of headed studs

P

ss

PP

cD

su

PRd

eLnt=2

Rd

uD P

sC =spring constant per stud:

spring constant of the shearconnection: L

tDs e

nCc =

type of shear connection

1500headed stud ∅ 22mmwith Holorib-sheeting and one stud per rib

1250headed stud ∅ 19mmwith Holorib-sheeting and one stud per rib

3500headed studs ∅ 25mmin solid slab

3000headed stud ∅ 22mmin solid slabs

2500headed stud ∅ 19mmin solid slabs

]cm/kN[CD

cs

50




Simplified solution for the calculation of deflections in case of incomplete interaction

( ) πξ=ξ sinqq

L

Lx

=ξ

za

zc

Ec, Ac, Jc

Ea, Aa, Ja

a

Nc

Na

Ma

Mc 2

aeff,c

aeff,cao,ceff,io a

AAAA

JJJ+

++=

The influence of the flexibility of the shear connection is taken into account by a reduced value for the modular ratio.

eff,ioa4

4

2

ccmoaa

aaccmoaaccm

4

4

o JE1Lq

aAEAE

AEAEJEJE

1Lqwπ

=

β+

β++π

=

)1(nn soeff,o β+=s

2ccm

2

s cLAEπ

=β

eff,o

ceff,c n

AA =

effective modular ratio for the concrete slab

εa

εc

51




Comparison of the exact method with the simplified method

w/wc

L [m]1,1

1,0

1,2

1,3

1,4

1,5

5,0 10,0 15,0 20,0

w/wc

L [m]

1,1

1,0

1,15

1,2

1,25

5,0 10,0 15,0 20,0

cD = 2000 KN/cm

q

L

w

beff

5199

450 mm

Ecm = 3350 KN/cm²

exact solutionsimplified solution with no,eff

1,05 η=0,8

η=0,4

η=0,8

η=0,4

cD = 1000 KN/cm

wo- deflection in case of neglecting effects from slip of shear connection

η degree of shear connection

52




1875 1875

1875 18753750

7500

F

F

F/2 F/2

load case 2

load case 1

δ[mm]

F [kN]

20 40 60

50

100

150

200

0

Deflection at midspan

1500

445

270

175

50

IPE 270

load case 2

load case 1

Deflection in case of incomplete interaction-comparison with test results

53




780

50125Load case 2F=145 kN

Load case 1F= 60 kN

second moment of area

cm4

19,4 (97%)11,7 (106%)Jio,eff= 21.486,0Theoretical value, taking into account flexibility of shear connection

12,9 (65%)7,8 (71%)Jio= 32.387,0Theoretical value, neglecting flexibility of shear connection

20,0 (100 %)11,0 (100%)-Test

Deflection at midspan in mm

F F

s

s[mm]10 20 30 40

40

80

120

160 push-out test

Deflection in case of incomplete interaction-Comparison with test results

54




Part 5:

Limitation of stresses

10

55




Limitation of Stresses

σc MEd

σa

+

+

-

-

MEd

σa

σs

+

-

PEd ≤ ks PRd

σEd ≤ kc fck

σEd ≤ ks fsk

σEd ≤ ka fyk

stress limit

ks = 0,75characteristicheaded studs

kc= 0,60characteristicconcrete

ks = 0,80characteristicreinforcement

ka = 1,00characteristicstructural steel

recommended values ki

combination

Stress limitation is not required for beams if in the ultimate limit state,

- no verification of fatigue is required and

- no prestressing by tendons and /or

- no prestressing by controlled imposed deformations is provided.

56




composite section steel sectiongc

MEd(x)

VEd

MEd

VEd(x)

+

-+

bc

beff

y

z

ziox

Concentrated longitudinal shear force at sudden change of cross-section

Nc

vL,Ed,max

Lv=beff

effioca

ioeff,cEdmax,Ed,L bJE/E

zAM2v =

Ac,eff

longitudinal shear forces

+-

Local effects of concentrated longitudinal shear forces

57




z

y

bc=10 m

300

500x2014x2000

800x60

δ

P

CD = 3000 kN/cmper stud

L = 40 m

gc,d

cross-section

FE-Model

system

shear connectors

Local effects of concentrated longitudinal shear forces

58




Ultimate limit state - longitudinal shear forces

FE-Model:

FE-Model

L = 40 m

x

s

P

cD

s

P

cD

EN 1994-2

x [cm]

ULS

59




200 400 600 800 1000 1200 1400 1600 1800 2000

-4000

-3500

-3000

-2500

-2000

-1500

-1000

-500

0

500

Serviceability limit state - longitudinal shear forces

EN 1994-2

FE-Model:

FE-Model

L = 40 m

vL,Ed[kN/m]

x [cm]

x

s

P

cD

s

P

cD

SLS

60




Part 6:

Vibrations

11

61




Vibration- General

EN 1994-1-1: The dynamic properties of floor beams should satisfy the criteria in EN 1990,A.1.4.4

EN 1990, A1.4.4: To achieve satisfactory vibration behaviour of buildings and their structural members under serviceability conditions, the following aspects, among others, should be considered:

the comfort of the user

the functioning of the structure or its structural members

Other aspects should be considered for each project and agreed with the client

62




Vibration - General

EN 1990-A1.4.4:

For serviceability limit state of a structure or a structural member not to be exceeded when subjected to vibrations, the natural frequency of vibrations of the structure or structural member should be kept above appropriate values which depend upon the function of the building and the source of the vibration, and agreed with the client and/or the relevant authority.

Possible sources of vibration that should be considered include walking, synchronised movements of people, machinery, ground borne vibrations from traffic and wind actions. These, and other sources, should be specified for each project and agreed with the client.

Note in EN 1990-A.1.4.4: Further information is given in ISO 10137.

63




xk

ls

Span length

xls

time tF(x,t)

F(x,t)

tsxk

tk

Vibration – Example vertical vibration due to walking persons

1,755,5> 3,2fast running (sprint)

1,303,3∼2,5slow running (jog)

1,002,2∼2,3fast walk0,751,5∼2,0normal walk0,61,1∼1,7slow walk

stride length

ls[m]

forward speedvs = fs ls[m/s]

pacing rate

fs [Hz]

The pacing rate fs dominates the dynamic effects and the resulting dynamic loads. The speed of pedestrian propagation vs is a function of the pacing rate fs and the stride length ls.

64




Vibration –vertical vibrations due to walking of one person

time t

Fi(t)left foot

right foot

F(t)

both feet

1. step 2. step 3. step

ts=1/fs

( )⎥⎥⎦

⎤

⎢⎢⎣

⎡Φ−πα+= ∑

=

3

1nnsno tfn2sin1G)t(F

Go weight of the person (800 N)αn coefficient for the load component of n-th harmonicn number of the n-th harmonicfs pacing rateΦn phase angle oh the n-th harmonic

During walking, one of the feet is always in contact with the ground. The load-time function can be described by a Fourier series taking into account the 1st, 2nd and 3rd harmonic.

α1=0,4-0,5 Φ1=0

α2=0,1-0,25 Φ2=π/2

α3=0,1-0,15 Φ3=π/2

Fourier-coefficients and phase angles:

65




Vibration – vertical vibrations due to walking of persons

( )sE v/Lf

gen

namax e1

MFka δ−−

δπ

=

Mgen

F(t)

cδ

w(t)maximum acceleration a, vertical deflection w and maximum velocity v

acceleration

Emax

2E

max

f2av

)f2(aw

π=

π=

fE natural frequencyFn load component of n-th harmonicδ logarithmic damping decrementvs forward speed of the personka factor taking into account the different

positions xk during walking along the beamMgen generated mass of the system

(single span beam: Mgen=0,5 m L)

m

L

Fn(t)xk

w(xk,t)

L/2ka Fn(t)

w(t)

( )tfE

gen

na Ee1)tf2(sin

MFk)t(w δ−−π

δπ

=sv

Lt =

66




Logarithmic damping decrement

For the determination of the maximum acceleration the damping coefficient ζ or the logarithmic damping decrement δmust be determined. Values for composite beams are given in the literature. The logarithmic damping decrement is a function of the used materials, the damping of joints and bearings or support conditions and the natural frequency.

For typical composite floor beams in buildings with natural frequencies between 3 and 6 Hz the following values for the logarithmic damping decrement can be assumed:

δ=0,10 floor beams without not load-bearing inner walls

δ=0,15 floor beams with not load-bearing inner walls

1

2

3

4

5

6

Dampingratioξ [%]

3 6 9 12fE [Hz]

ξπ=δ 2

results of measurements in buildings

with finishes

without finishes

12

67




Vibration –vertical vibrations due to walking of persons

People in office buildings sitting or standing many hours are very sensitive to building vibrations. Therefore the effects of the second and third harmonic of dynamic load-time function should be considered, especially for structure with small mass and damping. In case of walking the pacing rate is in the rage of 1.7 to 2.4 Hz. The verification can be performed by frequency tuning or by limiting the maximum acceleration.

In case of frequency tuning for composite structures in office buildings the natural frequency normally should exceed 7,5 Hz if the first, second and third harmonic of the dynamic load-time function can cause significant acceleration.

Otherwise the maximum acceleration or velocity should be determined and limited to acceptable values in accordance with ISO 10137

F(t)/Go

( )∑=

Φ−π+=3

1nnsno tfn2sinFG)t(F

0,4

0,2

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

0,1

fs=1,5-2,5 Hz

2fs=3,0-5,0 Hz

3fs=4,5-7,5 Hz

68




Limitation of acceleration-recommended values acc. to ISO 10137

1 5 10 50 100

0,01

0,05

0,1

acceleration [m/s2]

frequency [Hz]

0,005 basic curve ao

Multiplying factors Ka for the basic curve

Residential (flats, hospitals) Ka=1,0Quiet office Ka=2-4General office (e. g. schools) Ka=4

ao Kaa ≤

natural frequency of typical compositebeams

COMPOSITE COLUMNS

G. Hanswille Bergische Universität Wuppertal

1

1




Composite Columns

Institute for Steel and Composite StructuresUniversity of Wuppertal

Germany

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

Eurocode 4

EurocodesBackground and Applications

Dissemination of information for training18-20 February 2008, Brussels

2





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

3





4




Composite columns

concrete filled hollow

sections

partially concrete encased sections

concrete encased sections

5




Special Cross-Sections

hollow sections with additional inner profiles

partially concrete encased sections

6




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 is necessary

Concrete encased sections

2

7




advantages:

high bearing resistance, especially in case of welded steel sectionsno formwork simple solution for joints and load introductioneasy solution for later strengthening and additional later jointsno edge protection

disadvantages:lower fire resistance in comparison with concrete encased sections.

Partially concrete encased sections

8




Casting of partially concrete encased sections

reinforcing pocket 1

casting pocket 1

turning the steel profile

reinforcing pocket 2

casting pocket 2

9




advantages:

high resistance and slender columns advantages in case of biaxial bendingno edge protection

disadvantages :high material costs for profilesdifficult castingadditional reinforcement is needed for fire resistance

Concrete filled hollow sections

10




H ≈ 0,2L

hole for vent

Opening for casting

Casting from the top

Pumping in vertical

direction

pumping in inclined position

hole for vent

e ≤ 5mOutside compactor

L

Casting of concrete in case of concrete filled hollow sections

11




advantages:

extreme high bearing resistance in combination with slender columnsconstant cross section for all stories is possible in high rise buildingshigh fire resistance and no additional reinforcementno edge protection

disadvantages:high material costsdifficult casting

Concrete filled hollow sections with additional inner profiles

12




Composite columns with hollow sections and additional inner core-profiles

CommerzbankFrankfurt

3

13




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

14




general method:

simplified 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 verification in accordance with EN 1994-1-1

15




Verification is not necessary where

concrete encased cross-sections

partially encased I sections

concrete filled hollow section

bt

t

dd tε=⎟

⎠⎞

⎜⎝⎛ 52

tdmax

290tdmax ε=⎟

⎠⎞

⎜⎝⎛

ε=⎟⎠⎞

⎜⎝⎛ 44

tdmax

yk

o,yk

ff

=ε

fyk,o = 235 N/mm2

bc

hc

b

h

cy cy

cz

cz

y

z

⎩⎨⎧

≥6/b

mm40cz

Resistance to lokal buckling

16




Part 2:

General design method

17




geometrical imperfection

wo

1000Lwo =

L

σE

+ -residual stresses due to rolling or welding

+

-

General method

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 the most 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 structural stability 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.

18




General method of design

F

plastic zones in structural steel

e

cracked concrete

stresses in structural steel section

stresses in concrete and reinforcement

fy

fc

-

-

-

+

+

- - -

-

w

fs

εc

εs

εa

fcm

0,4 fcEcm

εc1uεc1fct

σc

σs

fsm

ftm

Es

Ea

Ev

εv

σa

concrete

reinforcement

structural steel

fyfu

--

4

19




elasto - plastic resistance of the cross-section

elasto-plastic full plasticfcm

fsm

fyfsm

fsm

- - -+

fy

MRu,N

NRuNRu

Mpl,u,,N

R,pl

RMM

1,0

R,pl

RNN

1,0

A

B

A B

R,pl

RuNN

III

R,pl

N,RuMM

F

w+wo

M=F (w+wo)

Case I: ultimate load of the system is reached due to elasto-plastic failure of the critical cross-section

Case II: stability failure before reaching the elasto-plastic resistance of the cross-section.

20




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

wA

B

C

concrete encased section and bending about the strong axis:Failure due to exceeding the ultimate strain in concrete, buckling of longitudinal reinforcement and spalling 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 the compression zone of the structural steel section.

A

F

B

C

21




M

N

NEd

MEd

Ed

Rpl,d

Rpl,mEd

wo=L/1000 wo

w

d,pl

m,plR R

R=γ

e

E

λu Ed

wu

Verification λu ≥ γR

λu : amplification factor for ultimate system capacity

εc

fcm

0,4 fc Ecm

εc1uεc1fct

σc

concrete

εs

σs

fsm

ftm

Es

reinforcementεa

Ea

εv

σa

structural steel

fyfu

+

+

-

-

+ -

geometrical Imperfection

Residual stresses

Ev

General Method – Safety concept based on DIN 18800-5 (2004) and German

national Annex for EN 1994-1-1

22




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

800550

1200

700

t=25mm

t=50mmS355

S235

Residual stresses

λ

23




Part IV-3:

Plastic resistance of cross-sections and interaction curve

24




Resistance of cross-sections

sdscdcydaRd,pl

Rd,plsRd,plcRd,plaRd,pl

fAfAfAN

NNNN

+ν+

++=

=

Design value of the plastic resistance to compressive forces:

ckcsksykaRk,pl fAfAfAN ν++=

c

ckcd

s

sksd

a

ykyd

fffff

fγ

=γ

=γ

=Design strength:

85,0=ν0,1=ν

fyd

Npla,Rd

ν fcd

Nplc,Rd

fsdNpls,Rd

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:

5

25




Confinement effects in case of concrete filled tubes

σc,r

σc,r

σaϕσaϕ

dt

concretestructural steel

ηa fyd

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

fck,c

ck

r,c

fσ

ck

c,ck

ff

2.0

1.5

1.0

0.5

0

1.25

0d-2t

σc,r

For concrete stresses σc>o,8 fck the Poisson‘s ratio of concrete is higher than the Poisson‘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 ultimate strains 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 the longitudinal shear resistance.

rcckcck ff ,21, σα+α=

α1=1,125α2= 2,5α1=1,00

α2= 5,0

26




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

e101de)1(10 ,ccao,aa

EdEd

NMe =

5,0≤λ

⎟⎟⎠

⎞⎜⎜⎝

⎛η++η=

ck

ykccdcaydaRd,pl f

fdt1fAAfN

Design value of the plastic resistance to compressive forces taking into account the confinement effect:

9,425,0 coao =η=ηBasic values η for stocky columns centrically loaded:

d

t

y

z

MEd NEd

fc

fye/d>0,1 : ηa=1,0 and ηc=0

27




Plastic resistance to combined bending and compression

NNpl,Rd

NEd

Mpl,Rd

Mpl,N,Rd= μ Mpl,Rd

fyd

(1-ρ) fyd

0,85 fcd fsd

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, if the 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 design steel strength (1 - ρ) fyd in the shear area Av.

interaction curve

28




MRd

fyd

fsd

0,85fcdfyd

+

zpl

Vc,EdVa,Ed

Nc+s

Na

Ma Mc,+s

+

-

fyd

-

-

NEd

VEdfsd

fsd

fsd -

Rd,cEd,cRd,plaEd,a VVVV ≤≤

Ed,aEdEd,c

Rd,pl

Rd,pla

Rd

aEdEd,a

VVVMM

MMVV

−=

≈=

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 reinforced concrete section by :

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

Mpl,Rd is the plastic resistance moment of the composite section.

29




Determination of the resistance to normal forces and bending (example)

tf

yzpl

zc

Nc

0,85fcd

-

fyd

zs

zs

Naf

Naf

Naw,c

zaw,c

zaw,t

Naw,t(1-ρ) fyd

VEd

NEd

Mpl,N,Rd

Position of the plastic neutral axis: Edi NN∑ =

Edydplwwydplwcdplw Nf)1()zh(tf)1(ztf85,0z)tb( =ρ−−−ρ−+−

fsd

b

hw

ydwcdw

ydwwEdpl f)1(t2f85,0)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)ftwh(afNt,awzt,awNc,awzc,awNczcNRd,N,plM +++++=

sdss

cdplwc

ydfaf

ydwplwt,aw

ydwplc,aw

fA2N

f85,0z)tb(N

ftbN

f)1(t)zh(N

f)1(tzN

=

−=

=

ρ−−=

ρ−=

Ns

Ns

tw-

+

Edt,awc,awc NNNN =−+

As

30




Simplified determination of the interaction curve

N

M

2N Rd,pm

Rd,pmN

Rd,plM Rdmax,M

Rd,plNA

B

D

C

Rd,plN

Rd,plRd,B MM =

Rd,pmN

Rd,pmN5,0

A

B

C

D

0,85fcd fsd

fsd

fsd

fsd

- -

-+

+

+

zpl 2hn

hn

-

--

-

-

-

+

-

Rd,plRd,C MM =

Rdmax,Rd,D MM =

fyd

fyd

fyd

fyd

fyd

fydfyd

zpl

zpl

As a simplification, the interaction curve may be replaced by a polygonal diagram given by the points A to D.

0,85fcd

0,85fcd

0,85fcd

6

31




Resistance at points A and D

fcdfsd

- --

0M

NNNN

Rd,A

Rd,plsRd,plcRd,plaRd,pl

=

++=

fyd

Point A

Mpla,RdMpls,Rd0,5 Mplc,Rd

0,85 fcdfsd

fyd-+

-

+

zsi

zsi

bc

hc

h

Rdmax,Rd,D

Rd,plcRd,D

MM

N5,0N

=

=

[ ] yssisisds,plRd,pls fzAfWM ∑==

cds,pla,pl

2cc

cdc,plRd,plc f85,0WW4

hbf85,0WM

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡−−==

Point D

Nplc,Rd Npls,Rd Npla,Rd

ydffw

2f

yda,plRd,pla f)th(tb4

t)t2h(fWM⎥⎥⎦

⎤

⎢⎢⎣

⎡−+

−==

b

h

tftw

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 32




Bending resistance at Point B (Mpl,Rd )

Mpl,Rd

Mpln,Rd

0,85fcd

fsd

fyd

-+

-

+ 2fyd

+

+

2 fsd

-

+

0,85fcd 0,85fcd

- -

fyd

hn

fsd

MD,Rd

ND,Rdzpl

+ =

+

+

+

+

ND,Rd

hn

hn

hn

+ =

At point B is no resistance to compression forces. Therefore the resistance to compression forces at point D results from the additional cross-section zones in compression. With ND,Rd the depth 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 the depth 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

33




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 fcd

fsd

fyd

-

+

-

+

2fyd

-

-

2 fsd

-

+

0,85fcd 0,85fcd

- -

fyd

2hn fsd+ =

+

+ +

+

+

NC,Rd

2hn

2hn

2hn

+ =

+

Nc,Rd

Mc,Rd2hn

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

hn

zpl

Point B Point C

34




Determination of Mn,Rd

hn

hn2hn

0,85 fcd2 fsd 2 fyd

---

-fsd fyd

-+

Mn,Rd

ND,Rd= Npm,Rd

hn

( )2

hth)f2(zA2)f2(zf)A2(2

hf85,0h)tb(M nwnydsssdscds

ncdnwRd,n ++⎥⎦

⎤⎢⎣⎡ −−=

( )2h2thfz2fA2zf)A2(

2hf85,0h)tb(M n

wnydssdsscdsn

cdnwRd,n ++⎥⎦⎤

⎢⎣⎡ −−=

A

B Mn,Rd

cdplccn,pl f85,0W21M = sds,plsn,pl fWM = yda,plan,pl fWM == + +

The stress blocks acc. to A and B give the same bending moment Mn,Rd

The depth hn results from the equilibrium condition ∑Ni= Npm,Rd

Rd,pm

ydnw

scdsd

cdnwN

f2)ht()A2()ff2(

f85,0h)tb(=

⎪⎭

⎪⎬

⎫

+−+

−

Wpl,c, Wpl,s und Wpl,a are the plastic section moduliof the cross-section within the depth 2hn.

Rd,nM

0,85 fcd

35




cdcRd,pm f85,0AN =

sds,plcdc,plyda,plRdmax, fWf85,0W21fWM ++=

zin

1snisn,pl eAW ∑=

=i

sdsn,plcdcn,plydan,plRd,n fWf85,0W21fWM ++=

Rd,nRdmax,Rd,pl MMM −=

d

ey

ez

y

z

b

2nan,pl ht2W =

t rari

s,pla2i

3i

2c,pl W)r

2d)(4(rr

32

4)t2d()t2b(W −−π−−−

−−=

s,plc,pla2a

3a

2a,pl WW)r

2d)(4(

32

4W −−−π−−−= rrdb

ziss,pl eAn

1iW i=

Σ=

)ff2(t4fb2)ff2(AN

hcdydcd

cdsdsnRd,pmn −+

−−=

sn,pl2ncn,pl Wh)t2b(W −−=

Depth hn and plastic section moduliwithin the depth hn:

Plastic section moduli of the composite section:

Determination of hn and Mpl,Rd for concrete filled rectangular hollow sections

36




zin

1snisn,pl eAW ∑=

=i

d

ey

ez

y

z

t

s,pl3

c,pl W6

)t2d(W −−

=

s,plc,pl3

a,pl WW6dW −−=

)ff2(t4fd2)ff2(AN

hcdydcdcdsdsnRd,pm

n −+

−−=

2nan,pl ht2W =

zisis,pl eAn

1iW

=Σ=

zin

1snisn,pl eAW ∑=

=i

Plastic section moduli of the composite section:

Determination of hn and Mpl,Rd for concrete filled circular hollow sections

Depth hn and plastic section moduliwithin the depth hn:

cdcRd,pm f85,0AN =




7

37




)(4

)2(W ffw

2f

a,pl thtbtth−+

−=

bc

hc

b

h

ey

ez

y

z

s,pla,pl

2cc

c,pl WW4

W −−=hb

zisis,pl eAn

1iW

=Σ=

Plastic neutral axis outside the steel section: h/2 ≤ hn < hc/2

cdc

cdsdsncdydaRd,pmn f85,0b2

)f85,0f2(A)f85,0f2(ANh

−−−−= a,plan,pl WW =

Plastic neutral axis in the flange of the steel section: h/2-tf < hn < h/2

)f85,0f2(b2f85,0b2)f85,0f2(A)f85,0f2()hbA(N

hcdydcdc

cdsdsncdydaRd,pmn −+

−−−−−= )h4h(

4bWW 2

n2

a,plan,pl −−=

Plastic neutral axis in the web of the steel section: 2 hn ≤ h/2-tf

)f85,0f2(t2f85,0b2)f85,0f2(AN

hcdydwcdc

cdsdsnRd,pmn −+

−−= 2

nwan,pl htW =

sn,plan,pl2nccn,pl WWhbW −−=

zin

1snisn,pl eAW ∑=

=i

twtf

Determination of hn and Mpl,Rd for concrete encased sections – strong axis

cdcRd,pm f85,0AN =




38




yisis,pl eAn

1iW

=Σ=

plastic neutral axis outside the steel section b/2 ≤ hn < bc/2

a,plan,pl WW =

plastic neutral axis in the web of the steel section : 2 hn ≤ tw/2

Plastic neutral axis in the flange of the steel section : tw/2 < hn < b/2

sn,plan,pl2nccn,pl WWhhW −−=

yin

1snisn,pl eAW ∑=

=i

24)2(W

2f

2wf

a,plbttth

+−

=

s,pla,pl2

ccc,pl WW

4W −−=

bh

cdc

cdsdsncdydaRd,pmn f85,0h2

)f85,0f2(A)f85,0f2(ANh

−−−−=

)f85,0f2(t4f85,0h2)f85,0f2(A)f85,0f2)(ht2A(N

hcdydfcdc

cdsdsncdydfaRd,pmn −+

−−−−−=

)h4b(2tWW 2

n2f

a,plan,pl −−=

)f85,0f2(h2f85,0h2)f85,0f2(AN

hcdydcdc

cdsdsnRd,pmn −+

−−= 2

nan,pl hhW =

bc

hc

b

h

ey

ez y

ztw tf

Determination of hn and Mpl,Rd for concrete encased sections – weak axis

cdcRd,pm f85,0AN =




39




Part 4:

Simplified design method

40




Methods of verification acc. to the simplified method

Design based on the European buckling curves

Design based on second order analysis 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

41



University of Wuppertal-GermanyScope of the simplified method

double symmetrical cross-sectionuniform cross-sections over the member length with rolled, cold-formed or welded steel sectionssteel 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

NfA

9,02,0 =δ≤δ≤

0,2N

N

cr

Rk,pl ≤=λ

c

sss A

A%0,6%3,0 =ρ≤ρ≤

42




LoadF [kN]

deflection w [mm]

2000

1500

1000

500

020 40 60 80 100

short term testFu = 2022 kN

Fu = 1697 kN

long term test

Fv = 534 kN

30 cm

30 c

m

e=3 cm

L =

800

cm

e

F The horizontal deflection and the second order bending moments increase under permanent loads due to creep of concrete. This leads to a reduction of the ultimate load.

wo

permanent load

The effects of creep of concrete are taken into account in design by a reduced flexural stiffness of the composite cross-section.

wt

Effects of creep of concrete

8

43




Effects of creep on the flexural stiffness

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 at loading to, the time t considered and the notional size of the cross-section

b b

h h

UA2h c

o =

)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, which results 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

44




b

b

b

a

c

b

buckling curve

cross-section

%3s ≤ρ

%6%3 s ≤ρ<

cr

k,plNN

=λ

Rd,pl

RdNN

=χ

0,2 1,0

1,0

0,8

0,6

0,4

0,2

0,6 1,4

a

b

c

0,1NN

Rd

Ed ≤

Rd,plRd NN χ=

cdcsdsydaRd,pl fAfAfAN ν++=

Verification:

Design value of resistance

0,2N

N

cr

Rk,pl ≤=λ

1,8

Verification for axial compression with the European buckling curves

buckling about strong axis

buckling about weak axis

85,0=ν

00,1=ν

85,0=ν

85,0=ν

00,1=ν

00,1=ν

45




relative slenderness

relative slenderness:

0,2N

N

cr

Rk,pl ≤=λ

sksckcykaRk,pl fAfAfAN +ν+=

2eff

2

cr )L()EJ(N

βπ

=

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

elastic critical normal force

85,0

ffc

ckcd

=ν

γ=

effective flexural stiffness

00,1

ffc

ckcd

=ν

γ=

Ke=0,6

characteristic value of the plastic resistance to compressive forces

β - buckling length factor

46




22II,eff

2

cr

cr

EdoEdEd

L)JE(

N

NN1

1wNMmax

β

π=

−=

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


Rd,plMRdEd MMMmax μα=≤

Effective flexural stiffness

Npl,Rd

NEd

N

wo

Mpl,RdMRd

αM μ Mpl,Rd

MMpl,N,Rd

fsd

Mpl,N RdNEd

VEd

fyd

(1-ρ) fyd

0,85fcd

-

+

-

Verification for combined compression and bending

wo equivalent geometrical bow imperfection

The factor αM takes into account the difference between the full plastic and the elasto-plastic resistance of the cross-section resulting from strain limitations for concrete.

47




Favourable effects of the compression force on the bending resistance of the cross-section

Rd,plNM

=μ

1,0

Npl,Rd

0,1=μ

max,dμ

Case A Case B

MEd,max MEd,max

Case A: Bending moment depends directly on the action of the normal force

Case B: Bending moment and normal force result from independent actions.

NEd,A NEd,B

MEd,R

MEd,R=NEd e

A,dμ

B,dμ

max,dμ

N

NEd,A

NEd,B

0,5 Npm,Rd

Values μd greater than 1,0 should only be used where the bending moment MEd depends directly on the action of the normal force NEd, for example where the moment MEd results from an eccentricity of the normal force NEd. Otherwise an additional verification is necessary, because an overestimation of the normal force leads to an increased bending resistance (see normal forces NEd,A and NEd,B). For composite compression members subjected to bending moments and normal forces resulting from independent actions, the partial factor γF for those internal forces that lead to an increase of resistance should be reduced by 20%.

48




Global analysis

First-order analysis may be used if the increase of the relevant internal forces or moments caused by the deformations given by first-order analysis is less than 10%. This condition may be assumed to be fulfilled if the following criterion is satisfied:

αcr ≥ 10

where αcr is the factor by which the design loading would have to be increased to cause elastic instability.

Appropriate allowances shall be incorporated in the structural analysis to cover the effects of imperfections, including residual stresses and geometrical imperfections such as lack of verticality, lack of straightness, and unavoidable minor eccentricities in joints of the unloaded structures.

The assumed shape of imperfections shall take account of the elastic buckling mode of the structure or member in the plane of buckling considered, in the most unfavourable direction and form.

9

49




Equivalent initial bow imperfections

Buckling curve

a b c

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

%3s ≤ρ

%6%3 s ≤ρ<

Member imperfection

50




Imperfections for global analysis of frames

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

m115,0m

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

51




Frames sensitive against second order effects

1,Ed

Rk,plNN

5,0≤λw0

φ2

φ1

NEd,1 NEd,2

equivalent forces

NEd,1 φ1

2,Ed22

o NLw8q=

2,Ed2

o NLw4

2,Ed2

o NLw4

L2

L1 2,Ed

Rk,plNN

5,0>λ

NEd,1NEd,2

NEd,1 φ1

NEd,2 φ2

NEd,2 φ2

i,Ed

Rk,plNN

5,0≤λ

2i

eff2

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 neglected within the global analysis if:

cr

Rk,plN

N=λ

imperfections

52




Second order analysis

⎟⎠

⎞⎜⎝

⎛−

εξ−ε

+⎟⎠

⎞⎜⎝

⎛ε

ξε+ξ−ε=ξ 1

)2/(cos)5,0(cosM

sinsin)1(sinrM)(M oR

⎟⎠

⎞⎜⎝

⎛−

εξ−ε

+⎟⎠

⎞⎜⎝

⎛ε

ξε+ξ−εε=ξ 1

)2/(cos)5,0(sinM

sincos)1(cosr

LM)(V o

Rz

Bending moments including second order effects:

Maximum bending moment at the point ξM:

[ ] 0

2

omax M)5,0cos(

c1M)r1(M5,0M −ε

+++=

)5,0(tan1

M2)r1(M)1r(Mc

o ε++−

=ε

+=ξcarctan5,0M

2o2

o1)wN8Lq(Mε

+=II,eff

Ed)JE(

NL=ε

⎟⎠

⎞⎜⎝

⎛ =ξ

0ddM

maxM

ζM

ζ

Lwoq

MR

r MRN

EJ

MR

r MR

53




Simplified calculation of second order effects

4,0

3,0

2,0

1,0

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

exact Solution

simplified solution

r=1,0

r=0,5

r= - 0,5

r=0

k

crNN

MR

r MR N

L

EJ

ζM

Exact solution:

)5,0cos(c1)r1(M5,0M

2

Rmax ε+

+=

)5,0(tan1

r11rc

ε+−

=

ε+=ξ

carctan5,0MII,eff

Ed)JE(

NL=ε

simplified solution:

cr

EdR

max

NN1M

Mk−

β== r44,066,0 +=β

maxM

ζ

44,0≥β

MR

r MR

54




N

NRd wo

Mpl,RdMRd

μ Mpl,Rd

NEd=NRd

⎥⎦

⎤⎢⎣

⎡ −ε

= 1)2/cos(

1L

)EJ(w8M 2

II,effo

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

Background of the member imperfections

The initial bow imperfections were recalculated from the resistance to compression calculated with the European buckling curves.

Bending resistance:

10

55




500

400

300

200

j

λ1,0 2,0

C20/S235

C40/S355

C60/S355

owLj =

λ

1

2

3

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

1,0

1,1

1,2

0,9

0,8

1

12

2

3

3

δ )w(N)(NoRd

Rd κ=δ

Geometrical bow imperfections –comparison with European buckling 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 on the European buckling curves.

56




Rd,plNN

Rd,plMM

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,plN

N=λ

Plastic cross-section resistance

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

57




Resistance to combined compression and biaxial bending

InteractionMy, Mz, NEd

Interaction Mz, N Interaction

My, N

InteractionMy, Mz

Rd,y,pl

y

MM

Rd,z,pl

zM

M

Rd,plNN

The 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

EdNN

Rd,y,pldyEdRd,y M)N(M μ=

Rd,y,pldzEdRd,z M)N(M μ=Ed,zμ Ed,yμ

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

MM

M

Rd,z,pldz

Ed,z

Rd,y,pldy

Ed,y ≤μ

+μ

0,1dz

Ed,z

dy

Ed,y ≤μ

μ+

μ

μ

58




Verification in case of compression an biaxial bending

MRd,y,pldz

Ed,zM

Rd,y,pldy

Ed,yM

MM

Mα≤

μα≤

μ

0,1M

MM

M

Rd,y,pldz

Ed,z

Rd,y,pldy

Ed,y ≤μ

+μ

Rd,plNN

Rd,y,pl

Rd,yMM

dyμ

Rd,plNN

Rd,z,pl

Rd,zMM

dzμ

For both axis a separate verification is necessary.

Verification for the interaction of biaxial bending.

Imperfections should be considered only 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 both planes.

Rd,pl

EdNN

Rd,pl

EdNN



59




Part 5:

Special aspects of columns with inner core profiles

60




Composite columns – General Method

Commerzbank Frankfurt

Highlight CenterMunich

New railway station in Berlin (Lehrter Bahnhof)

Millennium Tower Vienna

11

61




Composite columns with concrete filled tubes and steel cores – special effects

Resistance based on stress blocks (plastic resistance)

Non linear resistance with strain limitation for concrete

tube core concretefy fy fc

fy fy fc

N

M

M

N

strains ε

Rd,pl

RdM M

Mμ

=α

Cross-sections with massive inner cores have a very high plastic shape factor and the cores can have very high residual stresses. Therefore these columns can not be design with the simplified method according to EN 1944-1-1.

σED

r62




distribution of yield strenghtσED [N/mm2]

U/A [1/m]

dK [mm]10 20 30 40 50

80100200400 130

50

100

150

200

250

300

⎥⎥⎦

⎤

⎢⎢⎣

⎡−σ=σ 2

K

2EDE

rr21)r(

fyk fy(r)

2

kyk

y

rr1,095,0

f)r(f

⎟⎟⎠

⎞⎜⎜⎝

⎛+=

fyk – characteristic value of the yield strenght

σED

r, rk

residual stresses:

rk

dk

kdU π=

4/dA 2kπ=

Residual stresses and distribution of the yield strength

ddK

63




General method – Finite Element Model

initial bow imperfectionstresses in the tube

stresses in concrete

load introduction

cross-section

64




dK

σE=fy

tube 406x6,3 - S235

concrete C30core dK=300 mm fy=265 N/mm2

buckling curve d(considering residual stresses)

χ

1,0

0,5

0,5 1,0 1,5Kλ

NRd= χ Npl,Rd

buckling curve a(no residual stresses)

Influence of residual stresses - comparison of resistance with the European buckling curves

65




Part 6:

Load introduction and longitudinal shear

66




Load introduction and shear connection

Provision shall be made in regions of load introduction for internal forces and moments applied from members connected to the ends and for loads applied within the length to be distributed between the steel and concrete components, considering the shear resistance at the interface between steel and concrete. A clearly defined load path shall beprovided that does not involve an amount of slip at this interface that would invalidate the assumptions made in design.

Where composite columns and compression members are subjected tosignificant transverse shear, as for example by local transverse loads and by end moments, provision shall be made for the transfer of the corresponding longitudinal shear stress at the interface between steel and concrete. For axially loaded columns and compression members, longitudinal shear outside the areas of load introduction need not to be considered

Basic requirements

12

67




Load introduction over the steel section

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/Ld2

LE

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

NN

NNNN

NN ===

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

Ns,Ed

⎥⎥⎦

⎤

⎢⎢⎣

⎡−=+=

Rd,pl

a,plEdEd,sEd,cEd,L N

N1NNNV RdRd,L PnV =

PRd – design resistance of studs68




Load introduction for combined comression and bending

NEdMEd

Ma,EdNa,Ed

Mc,Ed +Ms,Ed

Nc,Ed +Ns,Ed

Rd,pl

EdNN

sectional forces due to NEdund MEd

sectional forces based on plastic theory

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

2

Rd,pl

Ed2

Rd,pl

Edd N

NMMR ⎟

⎟⎠

⎞⎜⎜⎝

⎛+⎟

⎟⎠

⎞⎜⎜⎝

⎛=

2

Rd,pl

Rd2

Rd,pl

Rdd N

NMME ⎟

⎟⎠

⎞⎜⎜⎝

⎛+⎟

⎟⎠

⎞⎜⎜⎝

⎛=

Rd,sc

Ed,sc

Rd,sc

Ed,sc

Rd,a

Ed,a

Rd,a

Ed,a

dd

MM

NN

MM

NN

RE

+

+

+

+ ====

Rd,plNN

Rd,pl

RdNN

Rd,pl

EdMM

Rd,pl

RdMM

Rd,plMM

1,0

1,0

dE

dR

fyd

zpl

Nc+s,RdNa,Rd

Ma;Rd Mc,+s,Rd

+

--

fsd

fsd0,85fcd MRd

NRd

+ =

69




Load introduction – Example

fyd

zpl

-Nc+s,RdNa,Rd

Ma;Rd Mc,+s,Rd

+

- -

fsd

fsdfcd

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)

ri ehPed,v

Ped,h

n5,0eM

nN

Pmaxh

Ed,scEd,scEd

++ +=

xizi

2

i2i

Ed,sc2

i2i

Ed,scEd,scEd z

r

Mx

r

Mn

NPmax

⎥⎥⎦

⎤

⎢⎢⎣

⎡

∑+

⎥⎥⎦

⎤

⎢⎢⎣

⎡

∑+= +++

n – number of studs within the load introduction length

sectional forces based on stress blocks:

Mc+s,Ed

70




Load introduction by end plates

bcx bcy

bcx

bcy

I I

Section I-I Ac1

NEdNEd, c+sNEd,a

Aa1

Ed,aEdsEdc

Rd,pl

a,plEdEd,a

NNNNN

NN

−=

=

+

sectional forces of the steel and concrete section of the column:

sectional forces in Section I-I:

Ed,1aEdEd,1c

1a,pl1c,pl

1a,plEdEd,1a

NNN

NNN

NN

−=

⎥⎥⎦

⎤

⎢⎢⎣

⎡

+=

distribution with a ratio 1:2,5

Longitudinal shear force:

Rd,LEd,1cEd,scEd,L V)NN(V ≤−= +

1ccdc1c

cd1c1c,pl

yd1a1a,pl

Af3AAfAN

fAN

≤=

=

0,1NN

0,1NN

1c,pl

Ed,1c

1a,pl

Ed,1a ≤≤

verification in Section I-I

71




Shear resistance of stud connectors welded to the web of partially encased I-Sections

µ PRd / 2 µ PRd / 2

PRd

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 may be 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 μis 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.

Dc

Rd,LRRdRd,L VPV +=

RdRd,LR PV μ=72




PRd

< 300 < 400 < 600

PRd PRdPRd PRd PRd

VLR,Rd/2

RdRd,LRRd,LRRdRd,L PVVPnV μ=+=

Shear resistance of stud connectors welded to the web of partially encased I-Sections

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

vcmck

21,Rd

1Efd29,0Pγ

α=

v

2

u2,Rd1

4df8,0P

γ⎟⎟⎠

⎞⎜⎜⎝

⎛ π⋅=

PRd= min

13

73




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

Shear resistance of stud connectors welded to the web of partially encased I-sections

74




Load introduction – longitudinal shear forces in concrete

NEd

Dc Dc

Zs Zs

ZsDc

θθ

I I

Longitudinal shear force in section I-I:

not directly connected concrete area As1

sdscdc

sd1scd1c

Rd,pl

a,plEdEd,L fAf85,0A

fAf85,0ANN

1NV++

⎥⎥⎦

⎤

⎢⎢⎣

⎡−=

NEd

bccy cy

LE

I I

As- cross-section area of the stirrups

Longitudinal shear resistance of concrete struts:

Ecdy

max,Rd,L Ltancot

f85,0c4V

θ+θ

ν=

longitudinal shear resistance of the stirrups:

θ =45o

Eydw

ss,Rd,L Lcotf

sA4V θ=

sw- spacing of stirrups

2ckck mm/Ninfwith))250/f(1(6,0 −=ν

75




test I/1

FF

w [mm]

F [kN]

Fu = 1608 kN

2000

1500

1000

500

00 2 4 6 8 10 12 14

Load introduction – longitudinal shear forces in concrete – test results

w

76




Load introduction – Examples (Airport Hannover)

Load introduction with gusset plates

77




Load introduction with partially loaded end plates

Load introduction with partially loaded end plates

78




Load introduction with distance plates for columns with inner steel cores

distance plates

Post Tower Bonn

14

79




Stiffener

Distance plate

σc σc σc

Composite columns with hollow sections –Load introduction

gusset plate stiffeners and end plates distance plates

80




σc,r

σa,tσa,y

σc

σa

The confinement by the tubes leads to a high resistance in partially loaded areas.

Confinement effects in partiallly loaded areas

81




Typical load-deformation curves

0.5

1.5

2.5

2.0

1.0

0 5 10 15 20 25 30

Pu

35

δ [mm]

P [MN]

Pu,stat

series SXIII

P

δ

P [MN]

5 10 15

Pu

1.0

3.0

5.0

4.0

2.0

0

δ [mm]

20

series SV

P

δ

Pu,stat

82




σc,r

σa,tσa,t

σcσa

Mechanical model

⎥⎦

⎤⎢⎣

⎡η+=

c

ycL

1

c1cm,cR f

fdt1

AAAfP

A1

Ac

Effect of partially loaded area

Effect of confinement by the

tube

ηc,L = 3,5 ηc,L = 4,9

83




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

2,0

4,0

re [MN]

41 testsVr= 0.14

rt [MN]

Pc,Rm

Pc,Rk = 0.78 Pc,Rm

Pc,Rd = 0.66 Pc,Rm

6,0

σc,r

σa,y σa,y

σcσa,x

8,0

10,0

Test evaluation according to EN 1990

1

c

c

ycL1cm,cR A

Aff

dt1AfP

⎥⎥⎦

⎤

⎢⎢⎣

⎡η+=

A1

Ac

84




Fu = 6047 kNFu,stat = 4750 kNδu = 7.5 mm

F [kN]

δ [mm]

6000

4000

2000

5,0 10,0 15,0

Fu

ts

bc

σc

tp

psc t5tb +=

Load distribution by end plates

d1~

15

85




yd1

cdc

1

c

ck

ykcLcdRd,c f

AfA

AA

ff

dt1f ≤≤⎥

⎦

⎤⎢⎣

⎡η+=σ

fck concrete cylinder strength t wall thickness of the tubed diameter of the tubefyk yield strength of structural steelA1 loaded areaAc cross section area of the concreteηc,L confinement factor

ηc,L = 4,9 (tube) ηc,L = 3,5 (square hollow sections)

20AA

1

c ≤

A1

Rd,cσ

Load distribution 1:2,5

bc bc

psc t5tb +=

tp

ts

Design rules according to EN 1994-1-1

86




Contribution of reinforcement

ssdss fAN η=

2/AAwith/A A0.5 - 1.5η

mm30e

1c

1cs≤

=≤

eg

A1

σc

4.0 8.0 12.0

δ [mm]

P [kN]

concrete C40/50

Pu= 8258 kN

Pu= 5917 kN ΔP

2000

4000

6000

8000

tube 323,9 x 5,6

reinforced- 8∅25

unreinforced

e = 20 mm

00.0

δ

For concrete filled circular hollow sections, longitudinal reinforcement may be taken into account for the resistance of the column, even where the reinforcement is not welded to the end plates or in direct contact with the endplates, provided that verification for fatigue is not required and the gap eg between the reinforcement and the end plate does not exceed 30 mm.

Reinforcement outside the load introduction area A1is not effective.

87




Verification outside the areas of load introduction

Outside the area of load introduction, longitudinal shear at the interface between concrete and steel should be verified where it is caused by transverse loads and / or end moments. Shear connectors should be provided, based on the distribution of the design value of longitudinal shear, where this exceeds the design shear strength τRd.

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

δA

B

C

pure bond (adhesion)

mechanical interlock

friction

σr

88




Design shear strength τRd

concrete encased sections

bc

hc

b

h

cy cy

cz

cz

y

z

5,2c

c1c02,01

z

min,zzc

co,RdRd

≤⎥⎦

⎤⎢⎣

⎡−+=β

βτ=τ

τRd,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)

COMPOSITE SLABS

S. Hicks Steel Construction Institute


Background and ApplicationsEUROCODES

EN 1994 - Eurocode 4: Design of composite steel and concrete structures

Composite Slabs

Stephen Hicks


EUROCODESBackground and Applications Composite slabs

Concrete cast in situWelded mesh reinforcement for crack control, transverse load distribution and fire resistance

Headed stud connectors for shear connection to the composite beam and, when required, end anchorage to the slab



Through-deck welding of headed stud shear connectors



Conventional composite construction


EUROCODESBackground and Applications Benefits of composite beams

• Bending resistance increased by a factor of 1.5 to 2.5

• Stiffness increased by a factor of 3 to 4.5

• Steel weight reduced by typically 30 to 50%

• Reduction in beam depth (span:depth ≈ 25)

• Lightweight construction


EUROCODESBackground and Applications Benefits of composite slabs

• Profiled steel sheeting acts as a safe working platform and permanent formwork.

• Unpropped construction may be achieved.

• Sheeting can stabilise beams during construction.

• Sheeting can provide all, or part, of the main tension reinforcement to the slab.


EUROCODESBackground and Applications Examples of composite construction in UK

Commercial sector Residential sector


EUROCODESBackground and Applications Examples of composite construction in UK

Health sector



Types of profiled steel sheeting defined in EN 1994-1-1

Re-entrant profiled steel sheet

Open trough profiled steel sheet



Practical examples of open trough and re-entrant profiled steel sheets used for composite slabs

Cover width: 1000

Multideck 60

323 mm

9 mm

60 mm

15 mm

Cover width: 600300 mm

60 mm ComFlor 60

207 mm

58 mm Confraplus 60

Cover width: 1035

183 mm

73 mm Cofrastra 70

Cover width: 732

Cover width: 900

Multideck 80

300 mm

9 mm

80.5 mm

15 mm

80 mm

180 mm 120 mm

145 mm

70 mm ComFlor 80

Cover width: 600

Cover width: 750

Cofrastra 40

150 mm

40 mm

152.5 mm

51 mm

Cover width : 610

Super Holorib 51



Composite construction with services passed under structural zone


EUROCODESBackground and Applications Examples of fixings for ceilings and services

Wedge attachment Clip attachment

Alternative wedge attachment


EUROCODESBackground and Applications EN 1994-1-1 detailing requirements

Scope limited to sheets with narrowly spaced ribs : br / bs ≤ 0,6

Slab thicknessWhen slab is acting compositely with beam or is used as a diaphragm:h ≥ 90 mm & hc ≥ 90 mm

When slab is not acting compositely with beam or has no stabilizing function:h ≥ 80 mm & hc ≥ 40 mm

Reinforcement ≥ 80 mm²/m in both directions

Spacing of reinforcement barss ≤ 2h & 350 mm

Maximum aggregate sizedg ≤ 0,4 hc, b0 / 3 and 31,5 mm

o

o

b b r

r

b

b

s

s

Re-entrant trough profile

b b

b b

b b

p

p

p

h

h

h

h

h

h

c

c

h1/2

Open trough profile


EUROCODESBackground and Applications EN 1994-1-1 composite slab bearing requirements

The bearing length shall be such that damage to the slab and the bearing is avoided; that fastening of the sheet to the bearing can be achieved without damage to the bearing and that collapse cannot occur as a result of accidental displacement during erection.

•For bearing on steel or concrete: lbc = 75 mm and lbs = 50 mm•For bearing on other materials: lbc = 100 mm and lbs = 70 mm

bs bs bs

bc

bc

(a) (b)bs bs

(c)


EUROCODESBackground and Applications Actions and action effects on profiled steel sheeting

a) Imposed load on a 3 m × 3 m working area (or the length of the span if less), with an intensity of 10% of the self-weight of the concrete but ≤ 1,5kN/m² and ≥ 0,75 kN/m

b) Imposed load of 0,75 kN/m²

c) Self weight load corresponding to the design thickness of the slab plus ponding effects if δ > h / 10

b b b b

3000 3000

a ac c



Analysis for internal forces and moments - set-up for double span tests on profiled steel sheeting

Combined bending and crushing at internal support



Typical forms of shear connection in composite slabs

(a) (c)

(b) (d)

a) Mechanical interlock through the provision of indentations or embossments rolled into the profile.

b) Frictional interlock for re-entrant profiles. c) End anchorage from through-deck welded stud connectors or other

local connection.d) End anchorage from deformation of the ends of the ribs at the end of the

sheeting.


EUROCODESBackground and Applications Longitudinal shear resistance

Test set-up from EN 1994-1-1, Annex B


EUROCODESBackground and Applications Classification of ductile or brittle behaviour

1. Brittle behaviouro m-k method

2. Ductile behaviour - failure load exceeds the load causing a recorded end slip of 0,1 mm by more than 10%o Partial connection methodo m-k method

LoadP(kN)

50

40

30

20

10

Slip atfirst end

Slip at second end

10 20 30 40 50

P/2 P/2

Deflection (mm)

δ

δ

1

2Load

F(kN)

F/2 F/2



Mean value for the ultimate shear stress with additional longitudinal shear resistance caused by the support reaction:

Determination of longitudinal shear resistance without end anchorage for the partial connection method

-

-

-

-

+

+

+

f

f

f

f

f

yp

yp

yp

cm

cm

N

N

cf

cf

c

AB

C1.0

p,Rm

p,Rm

MM

test

test test

MM

ηη = cN

N

F2

F2

L L o s

MM

Mean value for the ultimate shear stress:

( )os

cftestu LLb

N+

=η

τ ( )os

tcftestu LLb

VN+−

=µη

τ


EUROCODESBackground and Applications Determination of design value for τu,Rd from tests

For each variable investigated:

• 3 test specimens with the shear span Ls as long as possible, whilst still providing failure in longitudinal shear.

• 1 test specimen with the shear span Ls as short as possible (but not less than 3 × overall slab thickness), whilst still providing failure in longitudinal shear to classify the behaviour

Characteristic value of the longitudinal shear strength τu,Rkcalculated from the test values as the 5% fractile from EN1990, Annex D

τu,Rk is divided by the partial safety factor γVS to obtain a design value τu,Rd



Neutral axis above the sheeting and full shear connection (η = 1)

Design compressive normal force in the concrete flange:Nc,f = Np = Ape fyp,d

Depth of the concrete in compression xpl = Nc,f / (0,85 fcd b) ≤ hc

Design moment resistance of the composite slab in sagging bending MRd = Nc,f (dp - 0,5 xpl)

d p

xpl

z

c,f

p

N

N

M pl,Rd

f

f

yp,d

cd0.85

+

-

Centroidal axis of the profiled steel sheeting



Neutral axis within the sheeting and full shear connection (η = 1)

Design compressive normal force in the concrete flange: Nc,f = 0,85 fcd b hc

Reduced plastic moment resistance of the sheeting:

Lever arm:

Design moment resistance of the composite slab in sagging bending

MRd = Nc,f z + Mpr

p

c,fN

M

f

f

f

yp,d

yp,d

cd0.85

+

+

+

- - -

e e

Centroidal axis of the profiled steel sheetingPlastic neutral axis of the profiled steel sheeting

hcprz

+=

−=

dyp,pe

cfpapr 125,1

fANMM

( )dyp,pe

cfppc5,0

fANeeehhz −+−−=



Partial shear connection (0 < η < 1)

Design compressive normal force in the concrete flange: Nc = τu,Rd b Lx ≤ Nc,f

Reduced plastic moment resistance of the sheeting:

Lever arm:

Design moment resistance of the composite slab in sagging bending

MRd = Nc z + Mpr

−=

dyp,pe

cpapr 125,1

fANMM

( )dyp,pe

cppc5,0

fANeeehhz −+−−=

p

N

M

f

f

f

yp,d

yp,d

cd0.85

+

+

+

- -

--

e e

Centroidal axis of the profiled steel sheetingPlastic neutral axis of the profiled steel sheeting

hcpr

+=

c

z


EUROCODESBackground and Applications End anchorage

According to EN 1994-1-1, design resistance of a headed stud welded through the steel sheet used for end anchorage should be taken as the lesser of:

PRd kt

or

Ppb,Rd = kφ ddo t fyp,d

where PRd is the design resistance of a headed stud embedded in concrete, kt is a reduction factor for deck shape, ddo is the diameter of the weld collar (which may be taken as 1,1 times the shank diameter), t is the sheet thickness and kφ = 1 + a / ddo ≤ 6,0

d

f

f

yp

yp

/2

/2

d0

≥ 1.5 d0a d

Stud



Variation of bending resistance along a span: uniform distributed load

M

M

pl,Rd

pl,p,RdM

Ve,Rdb u,Rdτ

L LLsf x

M Ed

L

q

L x

MRd with end anchorage

MRd without end anchorage

MRd

Mpa

MEd

Rdu,

Rdpb,

bPτ b

kP

Rdu,

tRd

τor whichever is the lesser



Variation of bending resistance along a span: Point load

M

M

pl,Rd

pl,p,Rd

L LLsf x

M Ed

L

L x

M

FMRd without end anchorage

MEd

Mpa

MRd


EUROCODESBackground and Applications Classification of ductile or brittle behaviour

1. Brittle behaviouro m-k method

2. Ductile behaviour - failure load exceeds the load causing a recorded end slip of 0,1 mm by more than 10%o Partial connection methodo m-k method

LoadP(kN)

50

40

30

20

10

Slip atfirst end

Slip at second end

10 20 30 40 50

P/2 P/2

Deflection (mm)

δ

δ

1

2Load

F(kN)

F/2 F/2


EUROCODESBackground and Applications Determination of m-k values from tests

For each variable investigated:

• 3 test specimens with the shear span Ls as long as possible, whilst still providing failure in longitudinal shear.

• 3 test specimens with the shear span Ls as short as possible (but not less than 3 × overall slab thickness), whilst still providing failure in longitudinal shear to classify the behaviour

If behaviour brittle, Vt = 0,8 (F / 2)

m

Vertical shear

Longitudinal shear1

2

Vb.d

tp

k

Flexural

A b L

L

ps

s s

sLong Short

L L

F/2 F/2



Characteristic regression line calculated from the test values as the 5% fractile

Determination of m-k values from tests

Design shear resistance

m

Vertical shear

Longitudinal shear1

2

Vb.d

tp

k

Flexural

A b L

L

ps

s s

sLong Short

L L

Mean value

+= k

bLmAbd

Vs

p

VS

pRdl, γ

F/2 F/2


EUROCODESBackground and Applications Disadvantages of m-k method

• The results contain all the influencing parameters, but are impossible to separate from one another.

• Methodology is not based on a mechanical model and is therefore less flexible than the partial connection approach (contribution from end anchorage and reinforcement need to be evaluated from additional tests).

• Other loading arrangements that differ from the test loading can be problematical.


EUROCODESBackground and Applications Effective width for slabs with concentrated loads

For hp / h ≤ 0,6

For bending and longitudinal shear:i) for simple spans and exterior spans of continuous slabs

ii) for interior spans of continuous slabs

For vertical shear

Width of slab over which load is distributed

bm = bp + 2 (hc + hf)

Case – Concentrated loads applied parallel to the spanCase – Concentrated loads applied perpendicular to the span

b p

Finishes Reinforcement

b

b

hc

m

cm

h

h f

p

L

bbp

Lp

b bp

1

2

bLL

Lbb p ≤

−+= 12 pmem

bLL

Lbb p ≤

−+= 133,1 pmem

bLL

Lbb p ≤

−+= 1pmev


EUROCODESBackground and Applications Transverse reinforcement for concentrated loads

If the characteristic imposed loads do not exceed the values given below, a nominal transverse reinforcement of not less than 0,2% of the area of concrete above the ribs of the sheet (which extends ≥ the minimum anchorage length beyond bem), may be provided without any further calculation:

• concentrated load: 7,5 kN;• distributed load: 5,0 kN/m².

For characteristic imposed loads greater than these values, the distribution of bending moments and the appropriate amount of transverse reinforcement should be evaluated according to EN 1992-1-1.

b p

Finishes Reinforcement

b

b

hc

m

cm

h

h f

p


EUROCODESBackground and Applications Vertical shear resistance of composite slabs

Vv,Rd should be determined using EN 1992-1-1, 6.2.2 which gives the following:

Vv,Rd = [CRd,c k(100ρl fck)1/3 + k1 σcp] bsd (6.2a)

with a minimum of

Vv,Rd = (vmin + k1 σcp) bsd (6.2b)

where ρl = Asl / bs d, Asl is the area of the tensile reinforcement which extends ≥ (lbd + d) beyond the section considered and other symbols are defined in EN1992-1-1.

For normal loading conditions, and the fact that the sheeting is unlikely to be fully anchored, the vertical shear resistance will commonly be based on Eq (6.2b).

For heavily loaded slabs, additional reinforcement bars may be required at the support and the vertical shear resistance based on Eq (6.2a). According to the ENV version of EN 1994-1-1, it is permitted to assume that the sheeting contributes to Asl provided that it is fully anchored beyond the section considered.


EUROCODESBackground and Applications Punching shear resistance

c

c

c c

p

p

p

p p

Section A - A

d

d

AA

p

p

p

d

Loaded area ofdimensions a x b

h f

h

hh

h

Criticalperimeter

b + 2h p f

f

b

a +2h a

The punching shear resistance Vp,Rdshould be calculated according to EN 1992-1-1. For a loaded area ap × bp, which is applied to a screed with a thickness hf, the critical perimeter is given by:

cp = 2πhc+ 2(bp+ 2hf) + 2(ap+ 2hf+ 2dp –2hc)


EUROCODESBackground and Applications Serviceability limit states for composite slabs

Crack widthsFor continuous slabs that are designed as simply-supported, the minimum cross-sectional area of the anti-crack reinforcement within the depth hc should be:

• 0,2% of the cross-sectional area of the concrete above the ribs for unpropped construction• 0,4% of the cross-sectional area of the concrete above the ribs for propped construction.

The above amounts do not automatically ensure that wmax ≤ 0,3 mm as given in EN1992-1-1 for certain exposure classes.

If cracking needs to be controlled, the slab should be designed as continuous, and the crack widths in hogging moment regions evaluated according to EN 1992-1-1, 7.3.

Deflection

Deflections due to loading applied to the composite member should be calculated using elastic analysis, neglecting the effects of shrinkage.

For an internal span of a continuous slab, the deflection may be estimated using the following approximation:• the average value of the cracked and uncracked second moment of area may be taken.• for the concrete, an average value of the modular ratio for long-term and short-term effects may

be used.

For external, or simply supported spans, calculations of the deflection of the composite slab may be omitted if:

• the span/depth ratio of the slab does not exceed 20 for a simply-supported span and 26 for an external span of a continuous slab (corresponding to the lightly stressed concrete limits given in EN 1992-1-1; and

• the load causing an end slip of 0,5 mm in the tests on composite slabs exceeds 1,2 times the design service load.


EUROCODESBackground and Applications Standard push test

150

250

250

150 150260

Cover 15P

PRkLo

ad p

er s

tud

P (k

N)

Slip (mm)δδu

6 mm



Position of studs in open trough sheeting and reduction factor formula according to EN 1994-1-1

b Edge ofbeam

p,g p,nh h h

e

sc

o

Compressionin slab

Force from stud

(a) Central (b) Favourable (c) Unfavourable

kt = 0.85 / √nr (b0 / hp) {(hsc / hp) – 1} ≤ kt,max

0,600,60

0,700,80

≤ 1,0> 1,0

nr = 2

0,750,75

0,851,00

≤ 1,0> 1,0

nr = 1

Profiled steel sheeting with holes and studs 19 mm or 22 mm in diameter

Studs not exceeding 20 mm in diameter and welded through profiled steel sheeting

Thickness t of sheet(mm)

Number of stud connectors per rib



Stud ductility demonstrated in full-scale composite beam tests with studs through-deck welded in open trough sheeting

0

20

40

60

80

100

120

140

160

180

0 5 10 15 20 25 30

Slip (mm)

Axi

al fo

rce

(kN

)

7th pair 6th pair

Point at which maximummoment was appliedin Cycle 5

Point at which deck delamination wasobserved

-40

-20

0

20

40

60

80

100

120

140

160

0 5 10 15 20 25 30

Slip (mm)

Axi

al fo

rce

(kN

)

Strong Central Weak


EUROCODESBackground and Applications Load-slip curves for push tests cf. beam tests

0

10

20

30

40

50

60

70

80

90

0 5 10 15 20 25 30

Slip (mm)

Load

per

stu

d (k

N)

Push test Beam test

-40

-20

0

20

40

60

80

100

120

140

0 5 10 15 20 25 30

Slip (mm)

Load

per

stu

d (k

N)

Push test Beam test

nr = 1

nr = 2



Recommended detailing to push test with open trough profiled steel sheeting

Back-breaking failure

150 150260

Bedded in mortar or gypsum

4d minimum 750

250250250

P

Steel section:254 x 254 89 UC or HE 260 B

30 recessoptional

A

T C


EUROCODESBackground and Applications Where can I get further information?

[email protected]

http://www.access-steel.com/

COMPOSITE BRIDGES

L. Davaine & J. Raoul SETRA

1

Eurocodes - Background and applications

Dissemination of information workshop

Brussels, 18-20 February 2008

EN 1994 Part 2EN 1994 Part 2

Composite bridgesComposite bridges

Joël RAOUL

Laurence DAVAINE

Ministry for Ecology, Sustainable development and Spatial Planning (Paris)

Technical Centre for Highways & Motorways

46, avenue Aristide BriandBP 100F - 92 225 Bagneux Cedex


EUROCODESBackground and Applications Contents : 7 parts

1. Introduction to composite bridges in Eurocode 4

2. Global analysis of composite bridges

3. ULS verifications

4. SLS verifications

5. Connection at the steel–concrete interface

6. Fatigue (connection and reinforcement)

7. Lateral Torsional Buckling of members in compression

All points are illustrated with numerical applications to a twin-girder bridge with upper reinforced concrete slab.



Box-girder bridges

Composite bridges with steel girders under the slab

Introduction to composite bridges in EN1994

© Sétra

© Sétra

Half through composite bridges



NOTE : The cable stayed bridges with composite deck are not completely covered by EN 1994-2.

Filler beam decks

in transverse direction (National Annex)in longitudinal direction

Bowstring arches

© Sétra

© Sétra

Introduction to composite bridges in EN1994


EUROCODESBackground and Applications The main others EN called when using EN1994-2

EN 1992 – 1-1 : General rules for concrete

EN 1993 – 1-1 :General rules for steel

EN 1993 – 1-5 :Stiffeners ; Plate buckling EN 1993 – 1-8 :

JointsEN 1993 – 1-9 :Fatigue

EN 1993 – 1-10 :Brittle fracture

EN 1993 – 1-12 :S690 EN 1993 – 1-11 :

Cables

EN 1990 :Basis of designCombinationsAnnex A2 : application to bridges

EN 1991 :1-1 Permanent loads1-3 Snow1-4 Wind1-5 Temperature1-6 Loads during execution1-7 Accidental loads2 Traffic

EN 1090 :Execution

EN 1993 – 2 : Steel bridges

EN 1994 – 2 : Composite bridges

EN 1992 – 2 : Concrete bridges










2


EUROCODESBackground and Applications Global analysis for composite bridges

• Elastic global analysis without bending redistribution

• Second order effect to be considered for structures where

,

10crcr

Ed ULS

FF

α = ≤

In this elastic global analysis, the following points should be taken into account :• effects of creep and shrinkage of concrete,

• effective width of flanges for shear lag,

• stages and sequence of construction,

• effects of cracking of concrete,

• temperature effects of heat of hydration of cement (only for construction stages).

• Non-linear global analysis may be used (no application rules)



CLASS 2 sections which can develop M pl,Rd withlimited rotation capacity

CLASS 3 sections which can develop M el,Rd

CLASS 1 sections which can form a plastic hingewith the rotation capacity required for a global plasticanalysis

COMPOSITE BRIDGESIn general, non-uniform section(except for small spans)CL. 1 CL.3/4

BUILDINGS



P

M

θ

Mel,Rd

Mpl,Rd

θ

M at mid-span with increase of P

Class 1

Cracking of concrete

Static structure

Deformed structure Yielding

1

2

Actual behaviour of a continuous composite girder

When performing the elastic global analysis, two aspects of the non-linear behaviour are directly or indirectly considered.



• Determination of the stresses σc in the extreme fibre of the concrete slab under SLS characteristic combination according to a non-cracked global analysis

• In sections where σc < - 2 fctm, the concrete is assumed to be cracked and its resistance is neglected

Cracked global analysis

An additional iteration is not required.

1

!

EI1EI2

EI1

EI1 = un-cracked composite inertia (structural steel + concrete in compression)

EI2 = cracked composite inertia (structural steel + reinforcement)



Ac = 0

As

EI2

EI1

L1 L2

0.15 (L1+ L2)

Simplified method usable if :

- no pre-stressing by imposed deformation

- Lmin/Lmax>0.6

In the cracked zones EI2 :

• the resistance of the concrete in tension is neglected

• the resistance of the reinforcement is taken into account

1Cracked global analysisBrussels, 18-20 February 2008 – Dissemination of information workshop 12


Yielding at mid-span is taken into account if :– Class 1 or 2 cross-section at mid-span (and MEd > Mel,Rd )– Class 3 or 4 near intermediate support– Lmin/Lmax < 0.6

• Elastic linear analysis with an additional verification for the cross-sections in sagging bending zone (M>0) :

MEd < 0.9 Mpl,Rd

or

• Non linear analysis

Class 1 or 2 Class 3 or 4

Lmax Lmin

Yielding 2

3



• To calculate the internal forces and moments for the ULS combination of actions

– elastic global analysis (except for accidental loads)» linear» non linear (behaviour law for materials in EC2 and EC3)

– cracking of the concrete slab– shear lag (in the concrete slab : Le/8 constant value

for each span and calculated from the outside longitudinal rows of connectors)

– neglecting plate buckling (except for an effectivep area of an element ≤ 0.5 * gross area)

Global analysis of composite bridges - SynthesisBrussels, 18-20 February 2008 – Dissemination of information workshop 14


• To calculate the internal forces and moments for the SLS combinations of actions

– as for ULS (mainly used for verifying the concrete slab)

• To calculate the longitudinal shear per unit length (SLS and ULS) at the steel-concrete interface

– Cracked global analysis, elastic and linear– Always uncracked section analysis– Specific rules for shear connectors design in the elasto-

plastic zones for ULS (Mel,Rd < MEd < Mpl,Rd)

Global analysis of composite bridges - Synthesis


EUROCODESBackground and Applications Shear lag in composite bridges

• Concrete slab ⇒ EN 1994-2– Same effectives width beff for

SLS and ULS combinations of actions

• Steel flange ⇒ EN 1993-1-5– Used for bottom flange of a

box-girder bridge– Different effectives width for

SLS and ULS combinations of actions

– 3 options at ULS (choice to be performed in the National Annex)

eff ,flangeb

flangeb

eff ,slabbslabb

xσ



• Global analysis : constant for each span for simplification (with a value calculated to that at mid-span)

• Section analysis : variable on both sides of the vertical supports over a length Li /4

Effectives width of the concrete slab – EN1994-2



Application to a steel-concrete composite twin girder bridge

Global longitudinal bending

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bri

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EUROCODESBackground and Applications Example : Composite twin-girder road bridge

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60 m 80 m 60 m

C0 P1 C3P2

Note:

IPE600 every 7.5m in side spans and every 8.0m in central span

7 m 2.5 m2.5 m

2.8 m

34 cm

IPE 600

fib 1200mm=

fsb 1000mm=

4



2618 18 26 18

P1C0 P2 C360 m 60 m80 m

35 m 5 10 18 8 10 28 10 8 18 10 5 35

40 mm 55 80 120 80 55 40 55 80 120 80 55 40

h = 2800 mm

bfi = 1200 mm

bfs = 1000 mmNote : Bridge dimensions verified according to Eurocodes (cross-section resistance at ULS, SLS stresses and fatigue)

Longitudinal structural steel distribution of each main girder

Example : Structural steel distribution

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EUROCODESBackground and Applications Example : Used materials

Structural steel (EN1993 + EN10025) :

S355 N for t ≤ 80 mm (or S355 K2 for t ≤ 30 mm)

S355 NL for 80 < t ≤ 150 mm

Cross bracing and stiffeners : S355Shear connectors : headed studs with fu = 450 MPaReinforcement : high bond bars with fsk = 500 MpaConcrete C35/45 defined in EN1992 : fck,cyl (at 28 days) = 35 MPa

fck,cube (at 28 days) = 45 MPa fctm = -3.2 MPa

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295315S 355 NL

325335345355S 355 N

100 < t ≤ 15080 < t ≤ 10063 < t ≤ 8040 < t ≤ 6316 < t ≤ 40t ≤ 16

thickness t (mm)Yield strength fy (MPa)

Note : the requirements of EN 1993-1-10 (brittle fracture and through-thickness properties) should also be fulfilled.



( )L 0 L tn n . 1= + ψ φ

a0

cm

EnE

=

( )t 0t tφ = φ − creep function defined in EN1992-1-1 with : t = concrete age at the considered instantt0 = mean value of the concrete age when a long-term

loading is applied (for instance, permanent loads)t0 = 1 day for shrinkage action

{Lψ correction factor for taking account of the slight variations in the long-term load

intensity that could occur during the bridge life

Permanent loads

Shrinkage

Pre-stress by imposed deformations (for instance, jacking on supports)

1.1

0.55

1.5

Creep - Modular ratios for bridges

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for short term loading (ψL = 0)


EUROCODESBackground and Applications Example : construction phasing

1. Concreting order of the slab 12.5-m-long segments

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3 x 12.5 m 4 x 12.5 m

6 x 12.5 m3 x 12.5 m

1 2 3 16 15 14 7 13 12 11 10 9 8654

A B

CD

2. Construction timing

Steel structure put in place

Time (in days)

t = 016 concreting phases in a selected order assuming :

• 3 working days per segment

• only 1 mobile formwork (2 kN/m²)

t = 66

End of slab concreting

t = 80

Note : 14 days are required in EN1994-2 before introducing pre-stressing by imposed deformations.

t = 110

Non-structural equipments (pavement, safety barriers,…) put in place

assembling bridge equipments

......1st 16th

...Pre-stressing



t = 0

......1st 16th

Time (in days)t = 66 t = 80 t = 110

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3…6366Phase 16

…………

58Phase 2

3Phase 1

Mean value of the ages of concrete segments :

used for all concreting phases (simplification of EN1994-2).

066 63 ... 3t 35.25 days

16 phases+ + +

= =

( )1 0t , tφ = φ = ∞

( )L,1 0 1n n 1 1.1.= + φ

+ 14 days

0t 49.25 days=

( )2 0, tφ = φ ∞

( )L,2 0 2n n 1 1.5.= + φ

+ 30 days

0t 79.25 days=

( )3 0, tφ = φ ∞

( )L,3 0 3n n 1 1.1.= + φ

Example : age of concrete

Note : t0 = 1 day when shrinkage is applied to a concrete segment.

( )4 0, tφ = φ ∞ ( )L,4 0 4n n 1 0.55.= + φ



EN1992-1-1, Annex B :

( ) ( )0.3

0 0 c 0 00

0H

0t

tt

tt

t, t . t . t →+∞

−φ = φ β − = φ → φ β + −

( ) ( )0 RH cm 1 2 0.230 c

00m

RH1 16.8 1100. f . 1 . . . .0.

tt10.10. h f

− φ = φ β β = + α α +

• RH = 80 % (relative humidity)

• h0 = notional size of the concrete slab = 2Ac/u where u is the part of the slab perimeter which is directly in contact with the atmosphere.

• C35/45 : as fcm = 35+8 > 35 MPa, α1 = (35/fcm)0.7, α2 = (35/fcm)0.2

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Example : creep function and modular ratio values

Bridge equipments

Pre-stressing

Shrinkage

Concrete self-weight

Long term loading

nL,3 = 14.15

nL,2 = 18.09

nL,4 = 15.23

nL,1 = 15.49

Short term loading

a0

cm

En 6.16

E= =

5


EUROCODESBackground and Applications Example: shear lag in the concrete slab

60 m 80 m 60 m

C0 P1 C3P2

on support

in span 0.85x60 = 51m0.7x80 = 56m0.85x60 = 51m

0.25 x (60+80) = 35m 0.25 x (60+80) = 35m

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Equivalent spans Le :

where:eff 0 1 e1 2 e2b b .b .b= + β + β eei i

Lb min ;b

8

=

i 1.0β = except at both end supports where:e

iei

L0.55 0.025 1.0

bβ = + ≤

•

•



5.83 < 6.01.129 < 1.00.9482.23.251End supports C0 and C46.0/ /2.23.235Internal supports P1 and P26.0//2.23.256Span 26.0//2.23.251Spans 1 and 3

beff (m)β2β1be2be1Le (m)

Example: shear lag in the concrete slab

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b1 b1=3.5 m b2=2.5 m

be1 be2

beff

b0=0.6 m

b2

=> No reduction for shear lag in the global analysis

=> Reduction for shear lag in the section analysis :

beff linearly varies from 5.83m at end supports to 6.0 m at a distance L1/4.


EUROCODESBackground and Applications Applied loads on the road bridge example

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EN1991 part 2Fatigue load model (for instance, the equivalent lorry FLM3)FLM3

EN1991 part 2Road traffic (for instance, load model LM1 with uniform design loads UDL and tandem systems TS)

UDL, TS

EN1991 part 1-5Thermal gradientTk

Variable loads

Possibly, pre-stressing by imposed deformations (for instance, jacking on internal supports)

P

Creep (taken into account through modular ratios)

EN1992 part 1-1EN1994 part 2

Shrinkage (drying, autogenous and thermal shrinkage strains)

S

EN1991 part 1-1Self weight:• structural steel• concrete (by segments in a selected order)• non structural equipments (safety barriers, pavement,…)

Gmax , Gmin

Permanent loads


EUROCODESBackground and Applications Effects of shrinkage in a composite bridge

-

-

+ ze.n.a.

bc,eff

( ).= +σ cs cscs

steel

N z zNA I

( ),

.1. .

= − + +

σ ε cs csc cs

concrete c csc eff

N z zb NE

b n A I

1- Auto-equilibrated stress diagram in every section and an imposed rotation due to the bending moment Miso = Ncszcs :

csε

Free shrinkage strain applied on concrete slab only (no steel – concrete interaction)

cs N−

csze.n.a.

+

Shrinkage strain applied on the composite section (after steel – concrete interaction)

cs c cs c cN E .b h− = − εch


EUROCODESBackground and Applications Effects of shrinkage in a composite bridge

2- Curvature in an isostatic bridge due to the imposed deformations :

3- Compatibility of deformations to be considered in an hyperstatic bridge :

isoMisoM

L

( )v xP1 P2

P1 P2P31L 2L

hyperM

( )v P3 0=

1+2 = isostatic (or primary) effectsEffects of shrinkage

3 = hyperstatic (or secondary) effects


EUROCODESBackground and Applications Shrinkage and cracked global analysis

Concrete in tension

Cracked zone

isoM isoMisoM

Isostatic effects neglected in cracked zones for calculating hyperstatic effects

isoM

SLS combinations iso + hyper effects

hyper effects

hyper

hyper hyper

iso + hyper

ULS combinations

hyperM

hyperM

-

-

+-

-

+

6



0.6h

400

16 °C

4 °C

-5 °C

-8 °C

+15 °C -18 °C

2- Linear gradients :1- Non linear gradients :

• could be neglected if all cross-sections are in Class 1 or 2

3- Difference +/- 10 °C : +/- 10 °C

Thermal gradient from EN 1991 part 1-5Brussels, 18-20 February 2008 – Dissemination of information workshop 32

EUROCODESBackground and Applications Traffic load LM1 from EN 1991 part 2

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girder no. 2girder no. 1

3 m3 m 3 m 2 m

Lane no. 1 Lane no. 2

3.5 m 3.5 m

0.5 m1 m

Longitudinal axis of the most loaded girder Bridge axis

Safety barrier

Safety barrier

Lane no. 3Residual

area

9 kN/m²2.5 kN/m²

UDL (Uniform Design Load)

TS (Tandem System)

2 m

300 kN / axle 200 kN / axle 100 kN / axle

Characteristic values

of traffic loads from LM1


EUROCODESBackground and Applications Combinations of actions

For every permanent design situation, two limit states of the bridge should be considered :

Serviceability Limit States (SLS)• Quasi permanent SLS

Gmax + Gmin + S + P + 0.5 Tk

• Frequent SLSGmax + Gmin + S + P + 0.75 TS + 0.4 UDL + 0.5 TkGmax + Gmin + S + P + 0.6 Tk

• Characteristic SLSGmax + Gmin + S + P + (TS+UDL) + 0.6 TkGmax + Gmin + S + P + Qlk + 0.75 TS + 0.4 UDL + 0.6 TkGmax + Gmin + S + P + Tk + 0.75 TS + 0.4 UDL

Ultime Limite State (ULS) other than fatigue1.35 Gmax + Gmin + S + P + 1.35 (TS + UDL) + 1.5 (0.6 Tk)1.35 Gmax + Gmin + S + P + 1.35 Qlk + 1.35 (0.75 TS + 0.4 UDL) + 1.5 (0.6 Tk)1.35 Gmax + Gmin + S + P + 1.5 Tk + 1.35 (0.75 TS + 0.4 UDL)

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EUROCODESBackground and Applications Un-cracked global analysis

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Cracked zone on

P1

-12

-10

-8

-6

-4

-2

0

2

4

6

8

0 20 40 60 80 100 120 140 160 180 200

x = 49.7 m x = 72.5 m x = 121.6 m x = 150.6 m

17 %.L1 15.6 %.L2 23 %.L2 17.7 %.L3

2. 6.4 MPa− = −ctmf

x (m)

σ (MPa) : Stresses in the extreme fibre of the concrete slab, under Characteristic SLS combination when considering concrete resistance in every cross-section

L1 = 60 m L2 = 80 m L3 = 60 m

Note : Dissymmetry in the cracked lengths due to sequence of slab concreting.

EI2 EI2EI1 EI1EI1

Cracked zone on P2


EUROCODESBackground and Applications Cracked global analysis: bending moments

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37.5937.06 41.33

-80.69 -77.66

50.8456.0750.16

-103.54-107.25-120

-100

-80

-60

-40

-20

0

20

40

60

80

0 20 40 60 80 100 120 140 160 180 200

ELS caractéristiqueELU fondamental

Ben

din

g m

omen

t (M

N.m

)

Fundamental ULSCharacteristic SLS

x (m)



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Cracked global analysis: shear forces

5.54 5.49

3.24

-5.49-5.54

-3.26

1.09

7.47 7.39

4.38

-3.09 -2.92

-7.46 -7.41

-4.40

3.09

-10

-8

-6

-4

-2

0

2

4

6

8

10

0 20 40 60 80 100 120 140 160 180 200

ELS caractéristiqueELU fondamentalFundamental ULS

Characteristic SLS

x (m)

Shea

r fo

rce

(MN

)

7











EUROCODESBackground and Applications ULS verifications of a composite bridge

• resistance of the composite cross-sections

- for bending moment M (EN 1994-2, 6.2.1)

- for shear force V (EN 1994-2, 6.2.2.1 to 6.2.2.3)

- for interaction M+V (EN 1994-2, 6.2.2.4)

• shear resistance in the concrete slab (EN 1994-2, 6.2.2.5(3) )

• concrete slab (EN 1992)

• shear connection (see below, point 5)

• fatigue ULS (see below, point 6)

• LTB around intermediate supports (see below, point 7)



p.n.a

e.n.a

Elastic resistance(for classes 1 to 4)

Plastic resistance(for classes 1 and 2)

0.85 fck/γc

fy/γM

(+)

(−)

fck/γc

(+)

fy/γM

(−)

compression

tension

ULS section resistance under M > 0

e.n.a. = elastic neutral axis

p.n.a. = plastic neutral axis



p.n.a

e.n.a

Elastic resistance(for classes 1 to 4)

Plastic resistance(for classes 1 and 2)

compression

tension fsk/γs

(−)

fy/γM

(+)

fy/γM

0.85 fck/γc

(−)

(+)

fy/γM

fsk/γs

ULS section resistance under M < 0


EUROCODESBackground and Applications Class 4 composite section with construction phases

• Use of the final ULS stress distribution to look for the effective cross-section

• If web and flange are Class 4 elements, the flange gross area is first reduced. The corresponding first effective cross-section is used to re-calculate the stress distribution which is then used for reducing the web gross area.

a,EdM

+

c,EdM

=

Ed a,Ed c,EdM M M= +

Recalculation of the stress distributionrespecting the sequence of construction

1- Flange

2- Web

eff eff effA ,I ,G

Justification of the recalculated stress distribution



• For Class 1 or 2 sections :

– If VEd< 0.5.VRd, no interaction occurs.

– If not, the criterion MEd < Mpl,Rd is verified using a reduced Mpl,Rd value

• For Class 3 or 4 sections : See Eurocode 3 part 1-5.

Plastic resistance : ensured by the steel web Vpl,a,Rd is calculated by using Eurocode 3 part 1-1.

Shear buckling resistance :

See Eurocode 3 part 1-5.

Interaction between M and V :

ULS resistance under V and interaction M + V

yw w wRd b,Rd bw,Rd bf ,Rd

M1

f h tV V V V

3η

= = + ≤γ

yRd pl,a ,Rd V

M0

fV V A .

3= =

γ

2

Ed

Rd

V2 1V

η = −

8



• For the solid slab of a composite bridge:

Ed Rd,cV V≤ => Shear reinforcement (Ast for b = 1 m) is not necessary (nor the minimum shear reinforcement area according to EN1992-2,9.2.2)

1 3Rd,c Rd,c l ck 1 cp c min 1 cp cV C k(100 f ) k bh (v k )bh = ρ + σ ≥ + σ

Rd,cC

0.15C 0.12= =

γ1k 0.12=

c

200k 1

h= +

stl

c

Abh

ρ =

Edcp,0 cp

c

N1.85 MPa

bhσ = − ≤ σ =

1.5min ckv 0.035.k f=

ULS shear resistance in the concrete slab

• If the concrete flange is in tension :

-10

-8

-6

-4

-2

0

2

4

6

8

10

0 20 40 60 80 100 120 140 160 180 200

x (m)

Stre

sses

in th

e sl

ab a

t ULS

(MPa

)

Lower fibreUpper fibresigma_cp,0



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60 m 80 m 60 m

AΣ BΣ

Concrete in tension

M<0

Class 3 (elastic section analysis)

MULS = -107.25 MN.m

VULS = 7.47 MN

Section AΣConcrete in compression

M>0

Class 1 (plastic section analysis)

MULS = +56.07 MN.m

VULS = 1.04 MN

Section BΣ

Example: Analysis of 2 different cross-sections



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Example: Cross-section ΣA under bending

-171.2 MPa-149.2 MPa

-275.8 MPa

261.3 MPa

2.5 m 3.5 m

Stress diagram under bending

ysteel,inf

M0

f295 MPaσ ≤ =

γ

ysteel,sup

M0

f295 MPa− = − ≤ σ

γ

skre inf .

S

f434.8 MPa− = − ≤ σ

γ

1000 x 120 mm²

1200 x 120 mm²2560 x 26 mm²

Elastic section analysis :


EUROCODESBackground and Applications Example: Cross-section ΣA under shear force

2

whk 5.34 4 5.75

aτ = + =

cr Ek 19.58 MPaττ = σ =

hw = 2560 mm

a = 8000 mm

First cross-bracing in central spanP1

VEd = 7.47 MN

VEd = 6.00 MN

tw = 26 mm

w

w

h 31k

t τ

ε≥

η

Shear buckling to be considered:

yw w wRd b,Rd bw,Rd bf ,Rd

M1

f h tV V V V

3

η= = + ≤

γ

Contribution of the flange Vbf,RdContribution of the web Vbw,Rd

yww

cr

f1.33 1.08

3λ = = ≥

τ

ww

1.370.675

0.7χ = =

+ λ

ywbw,Rd w w w

M1

fV h t 8.14 MN

3= χ =

γ

bf ,RdV 0.245 MN= can be neglected.

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EUROCODESBackground and Applications Example: Cross-section ΣA under M+V interaction

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Ed

Rd

V0.5

V≥ so the M+V interaction should be checked, and as the section is in

Class 3, the following criterion should be applied (EN1993-1-5) :

2f ,Rd1 3

pl,Rd

M1 2 1 1.0

M

η + − η − ≤

at a distance hw/2 from internal support P1.

f ,RdM 117.3 MN.m= : design plastic resistance to bending of the effective composite section excluding the steel web (EN 1994-2, 6.2.2.5(2)).

f ,RdEd1

pl,Rd pl,Rd

MM0.73 0.86

M Mη = = ≤ =

Ed3

bw,Rd

V0.89

Vη = =

pl,RdM 135.6 MN.m= : design plastic resistance to bending of the effective composite section.

As MEd < Mf,Rd, the flanges alone can be used to resist M whereas the steel web resists V.

=> No interaction !



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Example: Cross-section ΣB (Class 1)

9.2 MPa

202.0 MPa

-305.2 MPa

2.5 m 3.5 m

p.n.a.+

-

ck

C

f0.85

γ

yf

M0

fγ

yw

M0

f−

γ

1000 x 40 mm²

1200 x 40 mm²

2720 x 18 mm²

Plastic section analysis under bending : Ed pl,RdM 56.07 M 79.59 MN.m= ≤ =2

whk 5.34 4 5.80

aτ = + =

w

w

h 31k

t τ

ε≥

η, so the shear buckling has to be considered:

yw w wEd Rd b,Rd bw,Rd bf ,Rd bw,Rd

M1

f h tV 2.21 MN V V V V V 4.44 MN 10.64 MN

3

η= ≤ = = + ≈ = ≤ =

γ

and

Ed

Rd

V0.5

V≤ => No M+V interaction !

9











EUROCODESBackground and Applications SLS verifications in a composite bridge

• Limitation of stresses in cross-sections at characteristic SLS

• Crack width control

• Limitations of deflections (national regulations)

• Web breathing (fatigue phenomenon, see EN1993-2)

Note : for shear connectors, see section 5 below

M>0

c ck0.6.fσ ≤ (concrete in compression)

a yk1.0.fσ ≤

M<0

s sk0.8.fσ ≤

a yk1.0.fσ ≤

(reinforcement in tension)



1. Minimum reinforcement required

- in cross-sections where tension exists in the concrete slab for characteristic SLS combinations of actions

- estimated from equilibrium between tensile force in concrete just before cracking and tensile force in the reinforcement (at yielding or at a lower stress level if necessary to limit the crack width)

2. Control of cracking due to direct loadingThe design crack width wk should be limited to a maximum crack width wmaxby limiting :

- bar spacing s ≤ smax

- or bar diameter Φ ≤ Φmax

wmax depends on the exposure class of the considered concrete face

smax and Φmax depend on the calculated stress level σs = σs,0 + ∆σs in the reinforcement and on the design crack width wk

Crack width control

3. Control of cracking due to indirect loadingFor instance, concrete shrinkage.



* : for bridges near sea water

** : for bridges subjected to (very) frequent salting

*** : for the bottom surface of a bridge deck located within 6 m above a road with (very) frequent salting

XC3, XS1*

XC4, XS1*, XD3***

waterproofing layer

XC4, XS1*, XD3**,

XF1 or XF2**

XC4, XS1*, XD3**, XF3 or XF4**

XC4, XS1*, XD3***,

XF1 or XF2**

Mechanical abrasionXM

Chemical attackXA1 to XA3

Freeze/thaw attackXF1 to XF4Attack to concrete

Corrosion induced by chlorides from sea waterXS1 to XS3

Corrosion induced by chloridesXD1 to XD3

Corrosion induced by carbonationXC1 to XC4Risk of corrosion of reinforcement

No risk of corrosion or attack of concreteXO

Description of the environmentClass

Exposure classes for composite bridges (durability)


EUROCODESBackground and Applications Exposure classes for composite bridges (durability)

XC3

XC4

waterproofing layer

XC4, XF1

XC4, XF3

XC4, XF1

Hypothesis : Bridge in a low-level frost areaThe choice of exposure classes leads to define :

• a minimum resistance for concrete (according to EN1992 and EN206), for instance C30/37

• a concrete makeup (maximum E/C ratio, minimum cement content) according to EN206

• a structural class (S1 to S6) for every face of the slab, chosen according to Table 4.3 in EN1992 and to the retained concrete

• a minimum concrete cover for every face of the slab according to the exposure class and the structural class

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hc

z0

e.n.a.calculated with n0

fct,eff

change in the location of the

neutral axis

before cracking

after cracking

σsσc

s c ct,eff ct s sk k. k f A A = σ

cc

0

1k 0.3 1.0

h1

2z

= + ≤+

stress distribution within the tensile concrete height hc before cracking (including indirect loading) + change in the location of the neutral axis at cracking time

reduction of the normal force in the concrete slab due to initial cracking and local slip of the shear connection

ks = 0.9

k = 0.8 effect of non-uniform shape in the self-equilibrating stresses within hc

Minimum reinforcement

fct,eff = fctm σs = fskand give the minimum reinforcement section As,min.

10


EUROCODESBackground and Applications Example : minimum reinforcement

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The elastic neutral axis is located in the steel web for every section of the bridge, so Act is the slab section : Act = 6 x 0.34 = 2.04 m²

hc = 0.34 m

fct,eff = fctm = -3.2 Mpa

z0 = 0.52 m

cc

0

1k min 0.3;1.0 1.0

h1

2z

= + = +

fsk = 500 MPa

As,min = 94 cm² which means a minimum reinforcement ratio s,min 0.46%ρ =

For the design, the following reinforcement ratios have been considered :

• Top layer : high bonded bars with φ = 16 mm and s = 130 mm, so

• Bottom layer : high bonded bars with φ = 16 mm and s = 130 mm, so

s,top 0.46%ρ =

s,bottom 0.46%ρ =

We verify : s,top s,bottom s,min0.92%ρ + ρ = ≥ ρ



Recommended values defined in EN1992-2 (concrete bridges) :

Maximum crack width wmax

The stress level σs,0 in the reinforcement is calculated for the quasi-permanent SLScombination of actions (in case of reinforced concrete slab).

The tension stiffening effect ∆σs should be taken into account.



Ast is put in place through n high bonded bars of diameter φ per meter.

or

Diameter φ∗

(Table 7.1)

Spacing s = 1/n

(Table 7.2)

Crack width control

ct ,eff* f2.9 MPa

Φ = Φ



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The stress level σs due to direct loading at quasi-permanent SLS combinations of actions can be calculated :

• Top and bottom layers : Ast with φ = 16 mm and s = 130 mm, so

• σs,0 = 106 Mpa (maximum tension) at quasi-permanent SLS in the top layer

-150

-100

-50

-

50

100

0 20 40 60 80 100 120 140 160 180 200

x (m)

Stre

sses

at Q

P SL

S (M

Pa)

Stresses in the upper layer of reinforcement, calculated by neglecting concrete resistance (in tension).

Example : crack width control for direct loading

s,top s,bottom 0.46%ρ = ρ =



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Example : crack width control for direct loading

• Tension stiffening effect :

• in the considered cross-section (where σs,0 is maximum) :

ctms

s st

f0.4∆σ =

ρ α

sta a

AI1.31

A Iα = = s 0.92%ρ = (Reinforcement ratio)

ctms

s st

f0.4 106.2 MPa∆σ = =

ρ α•

•

•

•

•

•

s s,0 s 212.2 MPaσ = σ + ∆σ =

*max 22.3 mmΦ = (interpolation in Table 7.1 of EN 1994-2)

*max max16 mm 3.2 /2.9 24.6 mmΦ = ≤ Φ = Φ =

maxs 235 mm= (interpolation in Table 7.2 of EN 1994-2)

or

maxs 130 mm s 235 mm= ≤ =


EUROCODESBackground and Applications Example : crack width control for indirect loading

The stress level σs due to indirect loading (for instance, concrete shrinkage) can not be calculated in the reinforcement.

In the sections where the concrete slab is in tension for characteristic SLS combinations of actions, σs is estimated using :

cts s c ct,eff

s

A 2.04k kk f 0.9 0.8 1.0 3.2 250.4 MPa

A 0.92% 2.04σ = = =

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The reinforcement layers are designed using high bonded bars with φ = 16 mm.

φ∗ = φ fct,eff/fct,0 = 2.9/3.2 = 14.5 mm

The interpolation in Table 7.1 from EN 1994-2 gives : σs,max = 255 Mpa

We verify :

σs = 250.4 Mpa < σs,max = 255 Mpa

11











EUROCODESBackground and Applications Steel-concrete connection

• Full interaction required for bridges

• Elastic resistance design of the shear connectors at SLS and at ULS

• Plastic resistance design of the shear connectors at ULS in Class 1 or 2 cross sections where Mel,Rd ≤ MEd ≤ Mpl,Rd

• Shear connectors locally added due to concentrated longitudinal shear force (for instance, shrinkage and thermal action at both bridge deck ends or cable anchorage)

• ULS design of transverse reinforcement to prevent longitudinal shear failure or splitting in the concrete slab

Objective :

Transmit the longitudinal shear force vL,Ed per unit length of the steel-concrete interface

Performed by the use of shear connectors (only studs in EN1994) and transverse reinforcement


EUROCODESBackground and Applications Resistance of the headed stud shear connector

• Shank shear resistance :

• Concrete crushing :

16 d 25mm≤ ≤

1.5d≥

0.4d≥

h 3d≥

t

d

2(1)

Rk udP 0.8f .4

π=

(2) 2

Rk ck cmP 0.29 d f E= α

s Rdk .P

RkRd

V

PP =γ

National AnnexDesign resistanceLimit State

U.L.S.

S.L.S.

V 1.25γ =

sk 0.75=

(1) (2)Rk Rk RkP min P ;P =

h0.2 1d

α = +

ifh3 4d

≤ ≤ , then

else 1α =


EUROCODESBackground and Applications Elastic design of the shear connection

• SLS and ULS elastic design using the shear flow vL,Ed at the steel-concrete interface, which is calculated with an uncracked behaviour of the cross sections.

SLS ULS

( ) { }, .≤SLS iL Ed s Rd

i

Nv x k Pl

For a given length li of the girder (to be chosen by the designer), the Nishear connectors are uniformly distributed and satisfy :

For a given length li of the girder (to be chosen by the designer), the Ni

* shear connectors are uniformly distributed and satisfy :

( )*

, 1.1 .≤ULS iL Ed Rd

i

Nv x Pl

( )0 ≤ ≤ ix l ( ) *,

0

.≤∫il

ULSL Ed i Rdv x dx N P

( ), ( ). += Ed

c sL

c sEd

A zv x AV zxI

Shear force from cracked global

analysisUncracked

mechanical properties

2.5 m 3.5 m

e.n.a.sz

cz


EUROCODESBackground and Applications Example : SLS elastic design of connectors

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0

0.2

0.4

0.6

0.8

1

1.2

1.4

0 20 40 60 80 100 120 140

Shear flow at SLS (MPa/m)Shear resistance of the studs (MPa/m)

L1 = 29 m L2 = 41 m L3 = 41 m L4 = 29 m

Studs with :

d = 22 mm

h = 150 mm

in S235

L,Edv SLS

in MPa/m



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Example : ULS elastic design of connectors

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

0 20 40 60 80 100 120 140

Shear flow at ULS (MPa/m)Shear resistance of the studs (MPa/m)

• Using the same segment lengths li as in SLS calculation and the same connector type

L,Edv ULS

in MPa/m

12



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Example : longitudinal spacing of studs rows

0

100

200

300

400

500

600

700

800

0 20 40 60 80 100 120 140

spacing at SLS (mm)spacing at ULS (mm)

x (m)

e (mm)

=> Elastic design governed by ULS.



• Eventually adding shear connectors in the elasto-plastic zoneswhere Mpl,Rd > MEd > Mel,Rd

Elasto-plastic design (ULS) of the shear connection

• NB (or NB*) is determined by

using an interaction M-N diagram in the section B.

•

connectors to put between sections A and B.

P1 B

Mpl,Rd

MEd

A

First yielding in at least one fibre of the cross-section

where MEd = Mel,Rd

Elasticdesign

Elasto-plastic design

NA

P2C

NB

Elasto-plastic zone

( )*B B A

Rk V

N or N Nn

P

−=

γ


EUROCODESBackground and Applications Interaction diagram in the cross-section B

• Two options : simplified diagram (straight line GH) / more precise diagram (broken line GJH)

NB (N)

MB (N.m)

Mpl,RdMEdMel,Rd

Ma,Ed

Nel,B NB Npl,BNB*0

GG

JJ

HH

ckpl,B eff c

C

0.85.fN .b .h=γ

• Plastic resistance of the concrete slab (within the effective width) to compressive normal force :



+ k . =

Step 1 : stress diagram for load cases applied to the structure beforeconcreting Section B

Step 2 : stress diagram for load cases applied to the structure after concreting Section B

Step 3 : ULS stress diagram in Section B (if yielding is reached in the extreme bottom fibre)

σai(1) σai

(2) σai

σas(1) σas

(2) σas

σc(2) σc

fyk

fyk

fcd=fck/γc

k (< 1) is the maximum value for keeping step 3 within its elastic strength limits. =>

Elastic resistance moment in the section B

MMa,a,EdEd ++ MMc,c,EdEd == MMEdEd

(For instance, σai(1) + k.σai

(2) = fyk )Mel,Rd = Ma,Ed + k. Mc,Ed



-50

-40

-30

-20

-10

0

10

20

30

40

50

0 20 40 60 80 100 120 140

M_Ed+

M_Ed-

M_pl,Rd+

M_pl,Rd -

Example : Bending moment in section B

Ma,Ed(B) = 2.7 MN.m -----> MEd(B) = 22.3 MN.m < Mpl,Rd (B) = 25.7 MN.m

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(Class 1)

x (m)

1 2 3 4 5 6 7 8 910111213141516

Concreting phases

M (M

N.m

)



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Example : Normal stresses in section B

Mc,Ed(B) = 22.3 – 2.7 = 19.6 MN.m

σai(2) = (-360.3) – (-63.0) = -297.3 Mpa

k is defined by ( )− −

= = ≤σ

y(2)

ai

f 63.0k 0.95 1.0

Mel,Rd is then defined by Mel,Rd = Ma,Ed + k. Mc,Ed = 21.3 MN.m

-63.0 MPa σai(2)

σas(2)

σc11.9 MPa

151.7 MPa

-360.3 MPa

Ma,Ed(B) = 2.7 MN.m MEd(B) = 22.3 MN.mMc,Ed(B)

88.2 MPafy = -345 MPa

13


EUROCODESBackground and Applications Example : Interaction diagram in section B

26.9 cm

3.6 cm

beff = 5.6 m

0.65 m

0.95*11.9 MPa

0.95*3.0 MPa

Nel = 11.4 MN

k * ULS stresses

=γck

C

f0.85 19.8 MPa

= =γck

pl c,effC

fN 0.85 .A 30.3 MPa

NB (MN)

MB (MN.m)

Mpl,Rd = 25.7MEd = 22.3

Mel,Rd = 21.3

MaEd = 2.7

Nel = 11.4 NB = 25.8Npl = 30.3

NB* = 15.7

0

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-400

-300

-200

-100

0

100

200

300

400

0 20 40 60 80 100 120 140

ULS Stresses (MPa) in the bottom steel flange

fy

Section CSection A

Section B

(σmax = -360.3 Mpa)

fy = -345 MPa

3.3 m 2.8 m

Example : Limits of the elasto-plastic zone

26.9 cm

3.6 cm

beff = 5.6 m

0.65 m

11.8 MPa

3.1 MPa

11.3 MPa

2.9 MPa

Nel(C) = 11.5 MNNel(A) = 12.1 MN

Section A Section C


EUROCODESBackground and Applications Adding shear connectors by elasto-plastic design

• 9 rows with 4 studs and a longitudinal spacing equal to 678 mm (designed at ULS)

(15.7-11.5)/(4x0.1095) = 10 rows

spacing = 2800/10 = 280 mm

(15.7-12.1)/(4x0.1095) = 9 rows

spacing = 3300/9 = 367 mm

More precise interaction

diagram

(25.8-11.5)/(4x0.1095) = 33 rows

spacing = 2800/33 = 84 mm(which is even lower than 5d=110 mm !)

(25.8-12.1)/(4x0.1095) = 28 rows

spacing = 3300/28 = 118 mm

Simplified interaction

diagram

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Section B Section CSection A

3300 mm 2800 mm

e = 678 mm



•

• for a structural steel flange in tension, subjected to fatigue

•

•

•

Detailing for shear connectors

25 mm ≤ De• to allow a correct welding of the connector :

2 5≤ fd . .t1 5≤ fd . .td h

tf

• for solid slabs :

3≥h d1 5≥head . dΦ0 4≥headh . d

• and if the used shear connectors are studs :



Longitudinal spacing between shear connectors rows

– to insure the composite behaviour in all cross-sections :emax = min (800 mm; 4 h )where h is the concrete slab thickness

– if the structural steel flange in compression which is connected to the concrete slab, is a class 3 or 4 element :

• to avoid buckling of the flange between two studs rows :

• to avoid buckling of the cantilever eD-long part of the flange :

– and if the used shear connectors are studs :

Transversal spacing between adjacent studs

max fy

235e 22tf

≤

Detailing for shear connectors

min5.d e≤

2 5≥trans ,mine . .d

2359≤D fy

e tf

4≥trans ,mine .d

for solid slabs

in other cases


EUROCODESBackground and Applications Transverse reinforcement for solid slabs

Truss model for transverse reinforcement which supplements the shear strength of the concrete on potential surface of failure (a-a for instance)

1aa L,Ed

eff

bv v ULS .

b=

compression

tension

com

pres

sion

fθ

L,Edv

cracksa

ab1

beff

1m Ab+At

a

ab1

beff

Transverse reinforcement

Ab

At

hc

14


EUROCODESBackground and Applications Transverse reinforcement for solid slabs

• tension in reinforcement :

• compression in concrete struts :

• for slab in tension at ULS : (or )

• for slab in compression at ULS : (or )

• Other potential surfaces of shear failure defined in EN1994-2 :

( )aa c f b t sdv .h .(1m).tan A A .fθ ≤ +

ckaa cd f f

fv 0.6 1 f .sin cos

250 ≤ − θ θ

f1.0 co tan 1.25≤ θ ≤f38.6 45° ≤ θ ≤ °

f1.0 co tan 2.0≤ θ ≤ f26.5 45° ≤ θ ≤ °











EUROCODESBackground and Applications Fatigue ULS in a composite bridge

In a composite bridge, fatigue verifications shall be performed for :

• the structural steel details of the main girder (see EN1993-2 and EN1993-1-9)

• the slab concrete (see EN1992-2)

• the slab reinforcement (see EN1994-2)

• the shear connection (see EN1994-2)

Two assessment methods in the Eurocodes which differ in the partial factor γMf for fatigue strength in the structural steel :

Safe lifeNo requirement for regular in-service inspection for fatigue damage

Damage tolerantRequired regular inspections and maintenance for detecting and repairing fatigue damage during the bridge life

High consequenceLow consequence

Consequence of detail failure for the bridgeAssessment method(National Choice)

Mf 1.0γ = Mf 1.15γ =

Mf 1.15γ = Mf 1.35γ =


EUROCODESBackground and Applications Damage equivalent stress range ∆σE

In a given structural detail of the bridge which is subjected to repeated fluctuations of stresses due to traffic loads, a fatigue crack could initiate and propagate. The detail fails when the damage D in it reaches 1.0 :

crack

Longitudinal stiffener on a web

weld

∆σ

In term of D, the actual traffic (ni, ∆σi)i is equivalent to nE = Σ ni cycles of the unique equivalent stress range ∆σE.

i

i

nD

N= ∑Total damage in the detail :

damage Log N (cycles)

Log ∆σ (stress range)

ni Niii

i

nd

N=

Fatigue S-N curve of the studied detail (EN1993-2)

with 2 slopes (m=3 and m=5)

NC = 2.106

∆σC

ND=5.106 NL=100.106

1m

mN C. −= ∆σ


EUROCODESBackground and Applications Fatigue Load Model 3 « equivalent lorry » (FLM3)

axle = 120 kN

E p.∆σ = λΦ ∆σ

• 2.106 FLM3 lorries are assumed to cross the bridge per year and per slow lane defined in the project

• every crossing induces a stress range ∆σp = |σmax,f - σmin,f | in a given structural detail

• the equivalent stress range ∆σE in this detail is obtained as follows :

where :

• λ is the damage equivalence factor

• Φ is the damage equivalent impact factor (= 1.0 as the dynamic effect is already included in the characteristic value of the axle load)


EUROCODESBackground and Applications Damage equivalence factor λ

In a structural steel detail (in EN 1993-2):λ=λ1 λ2 λ3 λ4 < λmaxwhich represents the following parameters :

λ1 : influence of the loaded lengths, defined in function of the bridges spans (< 80 m) and the shape of the influence line for the internal forces and moments

λ2 : influence of the traffic volume

λ3 : life time of the bridge ( λ3=1 for 100 years)

λ4 : influence of the number of loaded lanes

λmax : influence of the constant amplitude fatigue limit ∆σD at 5.106 cycles

For shear connection (in EN1994-2):

For reinforcement (in EN1992-2):

For concrete in compression (in EN1992-2 and only defined for railway bridges):

v v,1 v,2 v,3 v,4. . .λ = λ λ λ λ

s fat s,1 s,2 s,3 s,4. . . .λ = ϕ λ λ λ λ

c c,0 c,1 c,2,3 c,4. . .λ = λ λ λ λ

15


EUROCODESBackground and Applications Example : Damage equivalence factor λv

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• for road bridges (with L< 100 m) : v,1 1.55λ =

• hypothesis for the traffic volume in the example (based for instance on the existing traffic description in EN 1991 part 2):

Mean value of lorries weight :

(1 8)

obsmlv,2 6

NQ 4070.848

480 0.5.10 480 λ = = =

1 55i i

mli

nQQ 407 kN

n

= =

∑∑

• bridge life time = 100 years, so v,3 1.0λ =

6obsN 0.5.10= lorries per slow lane and per year with the following distribution

1Q 200 kN= 2Q 310 kN= 3Q 490 kN= 4Q 390 kN= 5Q 450 kN=

40% 10% 30% 15% 5%

• only 1 slow lane on the bridge, so v,4 1.0λ = v 1.314λ =



Stress range ∆σp = | σmax,f – σmin,f | in the structural steel

FLM3+

In every section :

Fatigue loadsBasic combination of non-cyclic actions

max min kG (or G ) 1.0 (or 0.0)S 0.6T+ +

max min a,Ed c,EdM (or M ) M M= + FLM3,max FLM3,minM and M

Ed,max,f a,Ed c,Ed FLM3,maxM M M M= + +Ed,min,f a,Ed c,Ed FLM3,minM M M M= + +

L 0

1 1c,Ed,max,f c,Ed FLM3,max

1 1n n

v vM M

I I

σ = + L 0

1 1c,Ed,min,f c ,Ed FLM3,min

1 1n n

v vM M

I I

σ = +

• Bending moment in the section where the structural steel detail is located :

• Corresponding stresses in the concrete slab (participating concrete) :

σc,Ed,max,f > 0σc,Ed,min,f < 0

Case 3

σc,Ed,max,f < 0σc,Ed,min,f < 0

Case 2

σc,Ed,max,f > 0σc,Ed,min,f > 0

Case 1 a a1 1 1 1

a,Ed c,Ed FLM3,max a,Ed c,Ed FLM3,mina 1 1

1p FLM3

a 1 1 1

v vv v v vM M M M M M

I I I I I Iv

MI

= + + − + +

∆σ = ∆

2p FLM3

2

vM

I∆σ = ∆

1 2 1 2p c,Ed FLM3,max FLM3,min

1 2 1 2

v v v vM M M

I I I I

∆σ = − + +



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0

5

10

15

20

25

30

0 20 40 60 80 100 120 140 160 180 200x (m)

Stre

ss ra

nge

(MP

a)

Stress range from M_min Stress range from M_maxalways without concrete participation always with concrete participation

Stress range ∆σp for the upper face of the upper steel flange

1 2 3 16 15 14 7 13 12 11 10 9 8654Sequence of concreting



Stress range ∆σs,p = | σs,max,f – σs,min,f | in the reinforcement

σc,Ed,max,f > 0σc,Ed,min,f < 0

Case 3

σc,Ed,max,f < 0σc,Ed,min,f < 0

Case 2

σc,Ed,max,f > 0σc,Ed,min,f > 0

Case 1

1s,p FLM3

1

vM

I∆σ = ∆

c,Ed FLM3,max2s,p c,Ed FLM3,min s,f

2 c,Ed FLM3,min

M MvM M 1

I M M

+ ∆σ = + + ∆σ − +

( ) 1 2s,p c,Ed FLM3,max c,Ed FLM3,min s,f

1 2

v vM M M M

I I

∆σ = + − + + ∆σ

• influence of the tension stiffening effect

ctms,f

st s

f0.2∆σ =

α ρ !Fatigue : 0.2 SLS verifications : 0.4

• in case 3, Mc,Ed is a sum of elementary bending moments corresponding to different load cases with different values of v1/I1 (following nL).

sta a

AIA I

α = s,effs

c,eff

A.100

Aρ =



Slope v2/I2 (fully cracked behaviour)

Tension stiffening effect

σs Stresses in the reinforcement (>0 in compression)

Bending moment in the composite section

M

case 1

s,p,1∆σ

case 3

s,p,3∆σ

c,Ed FLM3,minM M+

c,Ed FLM3,maxM M+

case 2

s,p,2∆σ

s,f∆σ

Slope v1/I1

Tension stiffening effectBrussels, 18-20 February 2008 – Dissemination of information workshop 90

EUROCODESBackground and Applications Fatigue verifications

cFf E

Mf

∆τγ ∆τ ≤

γc

Ff EMf

∆σγ ∆σ ≤

γ• In a structural steel detail :

RskF,fat E

S,fat

∆σγ ∆σ ≤

γ• In the reinforcement :

3 5

Ff E Ff E

C Mf C Mf

1.0 γ ∆σ γ ∆τ

+ ≤ ∆σ γ ∆τ γ

S,fat 1.15γ =

1k1

2k1

* 6N 1.10= logN

Rsklog ∆σ

Rsk 162.5 MPa∆σ =

skf 1k 5=

2k 9=

16


EUROCODESBackground and Applications Classification of typical structural details



( ) ( )m m

R R C CN N∆τ = ∆τ

m=8∆τc= 90 MPa

NR (log)Nc =

2.106 cycles

∆τR (log)

m=5∆σc=80 MPa

m=3

NR (log)

∆σR (log)

Nc = 2.106 cycles

E∆τ

E∆σ

Fatigue verifications for shear connectors

1. For a steel flange in compression at fatigue ULS :

cFf E

Mf ,s

∆τγ ∆τ ≤

γFf 1.0γ =

Mf ,s 1.0γ =with the recommended values :

2. For a steel flange in tension at fatigue ULS :

cFf E

Mf

∆σγ ∆σ ≤

γc

Ff EMf ,s

∆τγ ∆τ ≤

γFf E Ff E

C Mf C Mf ,s

1.3γ ∆σ γ ∆τ

+ ≤∆σ γ ∆τ γ












1. Bridge with uniform cross-sections in Class 1,2 or 3 and an un-stiffened web (except on supports) : U-frame model

2. Bridge with non-uniform cross-sections : general method from EN1993-2, 6.3.4

• 6.3.4.1 : General method

• 6.3.4.2 : Simplified method (Engesser’s formula for σcr)

To verify the LTB in the lower bottom flange (which is in compression around internal supports), two approaches are available :

LTB around internal supports of a composite girder

ultLT

cr

αλ =

α

( )LTLT fχ = λ

withy

ulta

fα =

σcr

cra

σα =

σand

LT ult

M1

1.0 ?χ α

≥γ


EUROCODESBackground and Applications Example : lateral restraints

7000

2800

1100

600

1100IPE 600

Cross section with transverse bracing frame in span

Lateral restraints are provided on each vertical support (piles) and in cross-sections where cross bracing frames are provided:

• Transverse bracing frames every 7.5 m in end spans and every 8.0 m in central span

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• A frame rigidity evaluated to Cd = 20.3 MN/m (spring rate)



Dead loads (construction phases, cracked elastic analysis, shrinkage)

Traffic loads (with unfavourable transverse distribution for the girder n°1)

TS = 409.3 kN/axleudl = 26.7 kN/m

+

MEd = -102 MN.m

NEd = MEd / h

= 38 MN

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Maximum bending at support P1 under traffic

17



• EN 1993-2, 6.3.4.2 : ENGESSER

•• EN 1993EN 1993--2, 6.3.4.1: 2, 6.3.4.1: General methodGeneral method

t bI3 3

f f 120.120012 12

= =

N EIccr 2 192 MN= =

N Ncr cr Edα = =

• I and NEd are variable

• discrete elastic lateral support, with rigidity Cd

a = 8 ma = 7,5 m a = 7,5 m

c = Cd/a

x

uy

L = 80 m

Lcr = 20 m

(I)

(II)

(III)

5.1 < 10

N Ncr cr Edα = =

==

(Mode I at P1)(Mode II at P2)(Mode III at P1)

8.910.317.5

•NEd = constant = Nmax

• I = constant = Imax

Elastic critical load for lateral flange buckling

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0 20 40 60 80 100 120 140 160 180 200

Stre

sses

in th

e m

id-p

lane

of t

he lo

wer

flan

ge[M

Pa]

First order stresses in the mid plane of the lower flange (compression at support P1)

EN1993-2, 6.3.4.1 (general method)

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.fyf

ult,kf

295 1 18249

minασ

= =

=

= =

= ≥

ult,kop

cr,op

1.188.9

0.37 0.2

αλ

α

Using buckling curve d: op 0.875 1.0χ = ≤

ult,kop

M1

1.036 0.94 1.01.1

αχ

γ= = > NO !



More information about the numerical design example by downloading the PDF guidance book :

“Eurocodes 3 and 4 – Application to steel-concrete composite road bridges”

on the Sétra website :

http://www.setra.equipement.gouv.fr/In-English.html

Thank you for your kind attention

1994.pdf · 2010-08-24

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