Modelling of the non-linear behaviour of composite beams

Introduction Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Conclusions

Modelling of the non-linear behaviour ofcomposite beams

taking into account the time effects

Quang-Huy NGUYEN

INSA de Rennes - Structural Engineering Research GroupUniversity of Wollongong - Faculty of Engineering

13 July 2009

Q-H. Nguyen PhD Thesis Defense


Outline

1 Introduction

2 Elastic analysis of composite beams

3 Time-Dependent Behaviour

4 Nonlinear Behaviour of Materials

5 Finite Element Formulations

6 Time-Dependent Behaviour In the Plastic Range

7 Conclusions and Futur works



Outline

1 IntroductionGeneralBackground: Analysis of composite beamsResearch questionsObjectives



General

Introduction to steel-concrete composite beamSteel-concrete composite structure are widely used in the constructionindustry

EconomicReduced live load deflectionsReduced weightFast erection processIncreased span lengths are possibleStiffer floors

Composite beam system (Ricker 1989)



General

Introduction to steel-concrete composite beamComposite beams consist of steel beam and concrete slab jointtogether as a unit by shear studs

steel beam

shear stud

concrete slab

profile sheeting

reinforcement



Background: Analysis of composite beams

Bond models

A

A

B

B

section A-A section B-B

Discrete bond modelAribert (1982, France)Schanzenback (1988, Germany)

Distributed bond modelNewmark (1951, US)Adekola (1968, Nigeria)




Analysis Type

Elastic Analysis Inelastic Analysis

Newmark, 1951

Adekola, 1968

N

N

tM

x

scd

X

Y

22

12

d ( )( ) ( )

d t

N xN x C M x

xμ− = ⇒ Analytical solution




Inelastic Analysis

Displacement-based Force-based Mixed

1v

2v

2u

1uθ1

θ2 x

( )v x

( )u x

X

Y

θ

θ

⎡ ⎤⎢ ⎥⎢ ⎥

⎡ ⎤ ⎢ ⎥=⎢ ⎥ ⎢ ⎥

⎢ ⎥ ⎢ ⎥⎣ ⎦⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

1

1

1

2

2

2

( )( )

( )

uv

u xx uv xv

a

Assumed displacement field




Inelastic Analysis


Arizumi et al., 1981

Schanzenbach, 1988

Daniels, 1989

Boerave, 1990




Inelastic Analysis


⎡ ⎤⎡ ⎤ ⎢ ⎥

=⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦

⎢ ⎥⎣ ⎦

1

2

2

( )( )

( )

MN x

x MM x

N

b

X

Y

1M

1M

2N

( )N x( )M x

x

Assumed force field




Inelastic Analysis



Schanzenbach, 1988

Daniels, 1989

Boerave, 1990

Salari et al., 1998

Vieira, 2000

Alemdar, 2001




Inelastic Analysis


θ

θ

⎡ ⎤⎢ ⎥⎢ ⎥ ⎡ ⎤

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥= =⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥

⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎢ ⎥ ⎢ ⎥⎣ ⎦⎢ ⎥⎢ ⎥⎣ ⎦

1

1

11

22

22

2

( ) ( )( ) & ( )

( ) ( )

uv

Mu x N x

x x Muv x M xNv

a b

X

Y

1M

1M

2N

( )N x( )M x

x

1v

2v

2u

1uθ1

θ2 x

( )v x

( )u x

Both fields are assumed




Inelastic Analysis



Schanzenbach, 1988

Daniels, 1989

Boerave, 1990

Salari et al., 1998

Vieira, 2000

Alemdar, 2001

Salari et al., 1998

Ayoub, 1999

Alemdar, 2001




Inelastic Analysis

Concentrated Plasticity Distributed Plasticity

-endi -endj

Inelasticity is lumped at member ends

elastic member




Inelastic Analysis


The element behavior is monitored along its length

-endi -endj

Fiber element model

Fiber section




Inelastic Analysis


Fiber section model

εc

σc

σs

εs

Concrete fiber

Steel fiber

σ σ

σ σ

=

=

=

=

∑∫

∑∫1

1

d

d

n

i iiAn

y i i iiA

N A A

M z A A zFiber discretization

of cross-section

y

z


Fiber element model

Cross-section behavior




Inelastic Analysis


Fiber element model

Cross-section behavior

Fiber section model

Macro model

El-Tawil and Deierlein, 2001

Bounding SurfaceAxial Force

Moment

Loading SurfaceCompression Region

Tension Region

Stress-resultant Plasticity Models




Analysis Type

Elastic Analysis

Time effects

Inelastic Analysis

Time Effects

Gilbert, 1989

Boerave, 1991

Amadio and Fragiacomo, 1993

Dezi and Tarantino, 1993...



Research questions

1 Discrete bond model or distributed bond model?

2 Displacement-based, Force-based or Mixed formulation?

3 What is the influence of creep and shrinkage on thebehaviour of composite beams?

4 How to take into account the time effects in inelasticanalysis?



Objectives

The main objectives are:

1 Discrete versus distributed bond modelling

2 To study the time effects in composite beams(viscoelastic model)

3 To develop three non-linear F.E. formulations and tostudy their performances for both bond models

4 To combine time effects and cracking of concrete



Outline

2 Elastic analysis of composite beamsBasic assumptionsGoverning Equations of Composite Steel-Concrete BeamsExact Stiffness Matrix - Elastic behaviourComparison of the two bond models



Basic assumptions

1 Euler-Bernoulli’s assumption for both the slab and the profile

2 Slip can occur at the slab/profile interface but no uplift

3 Deformations and displacements remain small

4 Local buckling and torsional stress are not accounted for

5 Fiber discretization to describe section behaviour

6 Spring model to describe the force transfer mechanism throughbond



Governing Equations of Composite Steel-Concrete Beams

1 EquilibriumDistributed bondDiscrete bond

2 Compatibility

3 Constitutive relations

2

2

d ( ) ( ) 0dd ( ) ( ) 0dd ( ) d ( ) 0

dd

csc

ssc

scz

N x D xxN x D xxM x D xH p

xx

+ =

− =

+ + =

=sc sc eD∂ − ∂ −D P 0

Matrix form

zp

cH

cM dc cM M+

dc cN N+

dc cT T+

cN

cT

scVscD

dx

ds sM M+

ds sN N+

ds sT T+

sM

sN

sT

scD

sH

x

z

y





2 Compatibility3 Constitutive relations

unconnected elementcN+

cM+

sN+

sM+

sN−

sM−

cN−

cM−

cNcM

sNsM

stQstQ

0xΔ =

connector element

=e∂ −D P 0

Unconnected beam segment Single connector

1

1s

c st

N

N Q

M H

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥= −⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦





2 Compatibility


H

scd

su

cu θ

θ

v

x

z

y

2

2

d ( )( )dd ( )( )dd

d ( )( ) (

( )

) ( )d

( )d

cc

s

sc s c

s

u xxxu xxxv xx

v xd x u x u x Hx

x

ε

ε

κ

=

=

=

+

−

= −

Matrix form

Tsc scd = ∂

= ∂e d

d





2 Compatibility


[ ]nonlinear( ) ( )x f x=D e

( ) ( )x x=D k e

Section constitutive law

Section stiffness matrix

[ ]nonlinear( ) ( )sc scD x f d x=

( ) ( )sc sc scD x k d x=

Bond constitutive law

linear elastic behaviour

Bond stiffness

Fiber discretization of cross-section

y

z

linear elastic behaviour



Exact Stiffness Matrix - Elastic behaviour

Distributed bond

Equilibrium in term of the displacements5 3

215 3

3 4 2

2 33 4 2

2

4 2

d dd dd d dd d d

d ddd

s s

s s

sc s

u ux xv u ux x x

u vu u Hxx

μ ζ

ζ ζ

ζ

⎧− =⎪

⎪⎪ = +⎨⎪⎪

= + +⎪⎩

Analytical solution

compatibility relations

constitutive relations

Exact displacement fields

( ) ( )

( ) ( ) ( )

( ) ( ) ( )

( ) ( )

)

) (

(s s

c c

s

c

v v

u x Z x

u x x Z x

x

v x x Z x

x x Z xθ θθ

= +

= +

= +

= +

X C

X C

X C

X C

Exact force fields( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

s s

c c

s N N

c N N

M M

T T

N x x Z x

N x x Z x

M x x Z x

T x x Z x

= +

= +

= +

= +

X C

X C

X C

X C ( ) ( ) 2sinh cos( h 1 0 0) 0s x x xx xμ μ⎡ ⎤= ⎣ ⎦X




Distributed bond

0= +eK q Q Q

1 1,Q q

2 2,Q q

3 3,Q q4 4,Q q

5 5,Q q

6 6,Q q

7 7,Q q

8 8,Q q

L

1 8( 0) ... ( )

z

c

p

Q N x Q M x L= − = = =

↔ = +Q YC Q

Static boundary conditions

( )1 8

1

( 0) ... ( )

z

c

p

q u x q x Lθ−

= = = =

→ = −C X q q

Kinematic boundary conditions

Exact siffness matrix




Discrete bond

Composite beam element with discrete bond

( )icu

( )isu

( )iθ( )jcu

( )jsu

( )jθ( )jcu

( )iv

( )icu

( )isu

( )iθ

( )jcu

( )jsu

( )jv

( )jθ

( )jsu

( )jθ

( )jv( )jθ

+= +

( )iv

( )icu( )isu

Connector element

Unconnected beam element

Connector element




Discrete bond


( )icu

( )isu

( )iθ( )jcu

( )jsu

( )jθ( )jcu

( )iv

( )icu

( )isu

( )iθ

( )jcu

( )jsu

( )jv

( )jθ

( )jsu

( )jθ

( )jv( )jθ

+= +

( )iv

( )icu( )isu

Connector element


Connector element

nceK

Exact sitffness matrix

Analytical solution




Discrete bond


( )icu

( )isu

( )iθ( )jcu

( )jsu

( )jθ( )jcu

( )iv

( )icu

( )isu

( )iθ

( )jcu

( )jsu

( )jv

( )jθ

( )jsu

( )jθ

( )jv( )jθ

+= +

( )iv

( )icu( )isu

Connector element


Connector element

nceK


Analytical solution

stiK


Analytical solution

stjK


Analytical solution




Discrete bond


( )icu

( )isu

( )iθ( )jcu

( )jsu

( )jθ( )jcu

( )iv

( )icu

( )isu

( )iθ

( )jcu

( )jsu

( )jv

( )jθ

( )jsu

( )jθ

( )jv( )jθ

+= +

( )iv

( )icu( )isu

Connector element


Connector element

nceK


Analytical solution

stiK


Analytical solution

stjK


Analytical solution


eK assembly



Comparison of the two bond models

20kN/m

6m 12m

40kN/m

12mmφ

800mm

100mm

IPE 200 200mm

80mm

Nelson 75-16

34GPa

210GPa

300000kN/m

1m

c

s

st

E

E

k

s

=

=

=

=

:stiffness of a single row of shears studs

:connector spacing

:equivalent distributed bond stiffness

st

sc

k

s

k

300MPastsckks

= =

Discrete bond model: using 18 elementsDistributed bond model: using 2 elements




0 2 4 6 8 10 12 14 16 18

-20

0

20

40

60

80

100

120

140

160

180

200

Distance from left support [m]

Def

lect

ion

[mm

]

Discrete bond modelDistributed bond model

176 mm

180 mm

20kN/m40kN/m

Deflection distribution along the beam




0 2 4 6 8 10 12 14 16 18-1.5

-1

-0.5

0

0.5

1

1.5

2


Slip

[mm

]


20kN/m40kN/m

0.7−

1.1−

Slip distribution along the beam




0 2 4 6 8 10 12 14 16 18

-0.03

-0.02

-0.01

0

0.01

0.02


Cur

vatu

re [1

/m]


20kN/m40kN/m

Curvature distribution along the beam




0 2 4 6 8 10 12 14 16 18-1000

-800

-600

-400

-200

0

200

400

600

800

1000


Axi

al fo

rce

in th

e co

ncre

te s

lab

[kN

]


20kN/m40kN/m

Axial force distribution along the beam




Conclusions

Discrete bond model: Discontinuities of axial force and curvature

Distributed bond model: all fields are continuous

Two distributed bond elements gives nearly identical results aseighteen discrete bond elements

The discrete bond model represents the true connection and it issimple to use but it requires a large number of elements



Outline

3 Time-Dependent BehaviourTime Effects in ConcreteTime-discretized analytical solution for composite beamsApplicationsConclusions



Time Effects in Concrete

1 Strain in concrete grow intime

2 Shrinkage

3 Creep

4 Aging material

5 Play an important role inserviceability

Curves of shrinkage, creep and recovery after unloading

0t (start of drying)

loading

2t unloading1t

εsh= DRYING SHRINKAGE

ELASTIC RECOVERY

CREEP RECOVERY

σε ε ε= − sh

εsh

σ

εv= CREEP

εe= INITIAL ELASTIC STRAIN

t

t

t

εsh(t)σε ( )t ε( )t

Recovery

Load - freeCompanionSpecimen

Loaded(Creep)Specimen

SpecimenUnloaded

σ



Linear viscoelastic model for concrete

Linear creep assumption: εc(t) = σcJ (t, t1)

Principle of superposition in time (Boltzmann, 1874)

cε

1t

2t

2t

1σ

2σ

1 2σ σ+

2( )tε

1( )tε

cσ cε

cε

1t

cσ

cσ

1t

2t

2t1t

2( )tε1( )tε

t

t

t

t

t

t

Integral-type relation

εc(t) = σc(t1)J (t, t1) +∫ t

t1

J (t, τ)dσc(τ)

dτdτ + εsh(t)



Time discrete approach

General method (step-by-step)

1

,1

( ) ( )( ) ( ) ( )n

c n sh n n i c ii

c n c nE t t tt t εε σσ−

=≅ − + Ψ⎡ ⎤⎣ ⎦ ∑

2 1

1

,1

1 1

1

( , ) ( , ) if i 1( , ) ( , )2( )

( , ) ( , )( , ) ( , ) if i 1( , ) (

d

,

an

)

n n

n n n n

c n n in n n n

n i k i

n n n n

J t t J t tJ t t J t t

E tJ t t J t t

J t t J t tJ t t J t t

−

−+ −

−

⎧ −⎪ =+⎪⎪= Ψ = ⎨+ ⎪−⎪ >

⎪ +⎩

[ ][ ]1

1

1 11

d ( ) 1( , ) d ( , ) ( , ) ( ) ( )d 2

nt n

n n i n i i iit

J t J t t J t t t tσ ττ τ σ στ

−

+ +=

≅ + −∑∫

Trapezoidal rule

Time-discrete constitutive relation

where




Algebraic method (one step)

[ ]1

1 1d ( )( , ) d ( ) ( )d

( , ) withnt

n n nt

nJ t t tJt t tτσ ττ τ σ σ ττ

≅ − ≤ ≤∫

Effective modulus (EM) method (McMilan, 1916)

11

1

1 ( , )(( , )) ,( )nn

nc

J t tJ t tE

tt

ϕτ += =


1( ) ( ))( ) ( ( )c nc n c n sh n ct tE t t tσ ε ε σ≅ − + Ψ⎡ ⎤⎣ ⎦

[ ]1( , ) 1 ( , ) ( , )2 n n nn J t t JJ t tt τ = +

Mean stress (MS) method (Hansen, 1964)

Age adjusted effective modulus (AAEM) method (Bažant, 1972)

1 1

1

1 ( , ) ( , )( , )

( )n

cn

nJt t t t

E tt

χ ϕτ

+=

creep coefficient

aging coefficient




Rate-type (internal variables) method (Bažant, 1971)

0 1

1 1( , ) 1 exp( )

m

i ii

tJ tE D

τττ τ=

⎡ ⎤⎛ ⎞−≅ + −⎢ ⎥⎜ ⎟

⎝ ⎠⎣ ⎦∑

The creep function is approximated by the Dirichlet series

0 1 0

( ) 1( ) ( ) with ( ) 1 exp( )

tm

i ii ii

t tt t t dE Dσ τε ε ε τ

τ τ=

⎡ ⎤⎛ ⎞−= + = −⎢ ⎥⎜ ⎟

⎝ ⎠⎣ ⎦∑ ∫

The integral-type relation becomes


( )shE εσ ε ε′Δ ′− Δ − ΔΔ=

0E( ) ( ) ( )i i i iE t D t D tτ= −

( ) ( )i i it D tη τ= ( )m tη

( )mE t

1( )tη

1( )E t

( )tσ( )tσ

Aging Kelvin chain



Time-discretized analytical solution for composite beams

1 For one-step methodFaella et al. 2002Ranzi and Bradford 2005

2 For rate-type (internal variables) methodJurkiewiez et al. 2005

3 For general methodThe solution is presented in the following




Time-discrete force-deformation relations

N (n)s = (EA)s ε(n)

s

N (n)c = (EA)(n)

c ε(n)c − (EA)(n)

co ε(n)sh + (EB) (n)

c κ(n) +n−1∑i=1

Ψn,i N (i)co

M (n) = (EB) (n)c ε(n)

c − (EB)(n)co ε

(n)sh + (EI ) (n)κ(n) +

n−1∑i=1

Ψn,i M (i)co

where

N (i)co = α

(i)1 x2 + α

(i)2 x + α

(i)3 +

i∑j=1

(β

(i,j)1 sinh(µjx) + β

(i,j)2 cosh(µjx)

)

M (i)co = α

(i)4 x2 + α

(i)5 x + α

(i)6 +

i∑j=1

(β

(i,j)3 sinh(µjx) + β

(i,j)4 cosh(µjx)

)




Equilibrium in term of displacements

d5u(n)s

dx5 − µ2n

d3u(n)s

dx3 = ζ(n)1 + ζ

(n)2

n−1∑i=1

Ψn,id2M (i)

co

dx2 + ζ(n)3

n−1∑i=1

Ψn,id2N (i)

co

dx2

d3v(n)

dx3 = ζ(n)4

d4u(n)s

dx4 + ζ(n)5

d2u(n)s

dx2 + ζ(n)6

n−1∑i=1

Ψn,idN (i)

co

dx

Analytical solution

u(n)s = X(n)

s C(n) + Z (n)s (x) +

n−1∑i=1

(a(n,i)

s sinh(µix) + b(n,i)s cosh(µix)

)v(n) = X(n)

v C(n) + Z (n)v (x) +

n−1∑i=1

(a(n,i)

v sinh(µix) + b(n,i)v cosh(µix)

)




Time-discretized exact stiffness matrix

( )( )(0

) ( )n nn n= +e q Q QK

( ) ( )1 1,n nQ q

Exact siffness matrix at the instant

( ) ( )2 2,n nQ q

( ) ( )3 3,n nQ q

( ) ( )4 4,n nQ q

( ) ( )8 8,n nQ q( ) ( )

5 5,n nQ q

( ) ( )6 6,n nQ q

( ) ( )7 7,n nQ q

L

( ) ( )( ) ( )1 8( 0) ... ( )n nn n

cq u x q x Lθ= = = =Kinematic boundary conditions

( ) ( )( ) ( )1 8( 0) ... ( )n nn n

cQ N x Q M x L= − = = =Static boundary conditions

Composite beam element

nt



Applications

2300mm

200mm

934mm

64.56kN/m

25m 25m

cG

sG

220 300mm×

21550 15mm×

230 450mm×

2

3

8 4

2

3

8 4

100mm

460000mm

0m

15.3310 mm

934mm

42800mm

0m

159.4910 mm

c

c

c

c

s

s

c

s

H

A

S

I

H

A

S

I

=

=

=

=

=

=

=

=666mm

Creep and shrinkage functions are defined in CEB-FIP Model Code 1990

0 0 28 030days, 30MPa, 80%, 196mm, 0.25sh ct t f RH h s= = = = = =

Two-span composite beam analyzed by Dezi and Tarantino, 1993



Comparison with existing model

101

102

103

104

105

-2017

-2016

-2015

-2014

-2013

-2012

-2011

-2010

-2009

-2008

-2007

-2006

Time [days]

Red

unda

nt re

actio

n R

[kN

]

Distributed bond - general methodDezi and Tanrantino

64.56kN/m

25m 25m

= 0.4 kN/mm²sck

= 0.1kN/mm²sck

2007.16

2012.302013.51

2015.60

2007.72

2013.08

2016.30

2013.87

R

Time evolution of the redundant reaction at intermediate support




0 10 20 30 40 50

0

5

10

15

20

25

30


Def

lect

ion

[mm

]

Discrete bond model (80 elements)Distributed bond model (2 elements)

64.56kN/m

25550days

30days




Comparison of time-discrete approaches

101

102

103

104

105

17

18

19

20

21

22

23

24

Time [days]

Mid

span

def

lect

ion

[mm

]

Step-by-step method"Rate-type" methodAAEM methodEM methodMS method

64.56kN/m

25m 25m

Creep effect only

Time evolution of midspan deflection



Effect of Creep and Shrinkage on Deflection

0 10 20 30 40 50

0

5

10

15

20

25

30

35

40


Def

lect

ion

[mm

]

30 days25550 days: creep onlyt=25550 days: creep + shrinkage

64.56kN/m

27%

39%




Effect of Creep and Shrinkage on Bending Moment

0 10 20 30 40 50

-10000

-8000

-6000

-4000

-2000

0

2000

4000


Ben

ding

mom

ent [

kN.m

]30 days25550 days: creep onlyt=25550 days: creep + shrinkage

70%

Bending moment distribution along the beam



Conclusions

An original time-discretized analytical solution has been derived

Compare to discrete bond model, the distributed bond modelleads to a quite complexe solution but it reduces significantly thenumber of elements

General method gives precise results but it requires the storage ofthe whole stress history

"Rate-type" method gives nearly identical results as generalmethod. This method avoids almost data storage but thedetermination of model parameters is quite complexe

Among algebraic methods, AAEM method seems to perform verywell

A significant impact of shrinkage



Outline

4 Nonlinear Behaviour of MaterialsConstitutive Models of SteelConstitutive Models of Shear StudConstitutive Models of Concrete



Constitutive models of steel1 Models based on explicit stress-strain relationship2 Models based on plasticity framework

One-dimensional problem

Easy to implement for monotonic loading

Cyclic models not easy to formulate

Menegotto-Pinto model

( )1 0 r 1ξ ε ε−

( )2 0 r 2ξ ε ε−

0E 0E0E

hE

hE

( )0,r rε σ( )0 0 1,ε σ

( )0 0 0,ε σ

( )0 0 2,ε σ

( )2,r rε σ

( )1,r rε σ

σ

ε



Constitutive models of steel

1 Models based on explicit stress-strain relationship2 Models based on plasticity framework

( )e pEσ ε ε= −Elastic stress-strain relationship

Yield condition and closure of the elastic range

Flow rule, isotropic and kineatic hardening laws

Kunh-Tucker complementarity conditions

Consistency condition

( ) ( )0 , , , 0 , , , 0f f XR RXλ σ λ σ≥ ≤ =

( ) ( ), , 0 if , , 0f fRX X Rλ σ σ= =

p sign( )Xε λ σ= −p λ=

sign( )Xα λ σ= −

Kinematic hardening stress-like variable

Kinematic hardening strain-like variable

( ) ( )(, 0), ( )yR R pf X X ασ σ σ= − − + ≤Isotropic hardening strain-like variable

Isotropic hardening stress-like variable



Constitutive models of steel

1 Models based on explicit stress-strain relationship2 Models based on plasticity framework

More involved

Cyclic behaviour is included in the model

Serveral physical phenomena can be coupled: damage, time effects ...

yε

σ

ε

yσ

uσ

hε uεyε−hε−uε−

yσ−

uσ−

O 1O2O 3O4O

Monotonic loadingCyclic loading

Linear isotropic hardening model



Constitutive models of shear stud

1 Discrete bond: The following models are selected to describe thebehaviour of one shear stud

P

δ

δ δ δ

P P P

uPuP uP1 u0.95P P=

2 fu1.05P P=fuP

uδ 1δ 2δ

( ) 2

u 11 expc

P P c δ= −⎡ ⎤⎣ ⎦0E0E

0E

0E

Elastic-perfectly plastic model Ollgaard et al., 1971 Salari, 1999

2 Distributed bond: The equivalent distributed bond strength andstiffness are calculated by dividing the strength and stiffness of asingle row of shear studs by their distance along the beam.



Constitutive models of concreteExplicit stress-strain relationship recommended by CEB-FIP ModelCode 1990

Good approach of stress-strainmonotonic curve in compression

No unloading information

No ascending branch of stress-strainmonotonic curve in tension 1cE

cE 1cε ,limcε

cmf−

0.5 cmf−

0.9 ctmfctmf

15%cε

Compression

Tension

cσ

Goals: To develop a model for concrete based on the elasto-plasticdamage theory

1 Reproduce exactly stress-strain monotonic curve in compressionof CEB-FIP Model Code 1990

2 Take into account the degradation of the elastic moduli3 Take into account the tension softening response



Governing equation of the elasto-plastic damage model

Assumption of a Helmholtz-free energy

Change of the compliance as internal variable (Govindjee at al., 1995)dD

( ) ( )1 2d 0 pdp p1( , , ) ( )2

DD Dp pεε ε ε−

Ψ − = + − + Ψ

Elastic damage part plastic part

Thermodynamically associated variables

( )p2

d

1ˆ; ;2p

R YD

σε ε

σ ∂Ψ ∂Ψ ∂Ψ= = = = −∂ − ∂ ∂

ε

σ

σ

pεε

0E

dε eε

0EE

( ) 1d dD E−=

eε

( )2d d d12E εΨ =

( )2e 0 e12E εΨ =






( ) ( )1 2d 0 pdp p1( , , ) ( )2


Ψ − = + − + Ψ



( )p2

d

1ˆ; ;2p

R YD

σε ε

σ ∂Ψ ∂Ψ ∂Ψ= = = = −∂ − ∂ ∂

( ) ( )ˆ, , ( ) 0yf R Y pRσ σ σ= − − ≤

Yield/damage condition

ε

σ

σ

pεε

0E

dε eε

0EE

( ) 1d dD E−=

eε

( )2d d d12E εΨ =

( )2e 0 e12E εΨ =






( ) ( )1 2d 0 pdp p1( , , ) ( )2


Ψ − = + − + Ψ



( )p2

d

1ˆ; ;2p

R YD

σε ε

σ ∂Ψ ∂Ψ ∂Ψ= = = = −∂ − ∂ ∂

Flow rule, damage and hardning/softening laws

( ) ( )ˆ, , ( ) 0yf R Y pRσ σ σ= − − ≤

Yield/damage condition

( ) ( )p d 11 1 sign( ) ; sign( ) ;ˆf f fD p

RYε λ λ σ λ λ σ λ λ

σβ β β β

σ∂ ∂ ∂

= − = − − = = − − = = −∂ ∂∂

β : scalar paramater, proposed by Meschke et al, 1997

ε

σ

σ

pεε

0E

dε eε

0EE

( ) 1d dD E−=

eε

( )2d d d12E εΨ =

( )2e 0 e12E εΨ =



Determination of the hardening/softening functions

Introducing a scalar parameter ζ

( )0

1 pE

ζσε − = +

ε

σ

σ

0E 0EE

eε

pε dε eε

dE

( )p d 1 pζε ε+ = +Assumption





Monitonic loading condition

( ) ( )ˆ, , ( ) 0yf R Y R pσ σ σ= − + =

( )0

1 pE

ζσε − = +

ε

σ

σ

0E 0EE

eε

pε dε eε

dE







( ) ( )ˆ, , ( ) 0yf R Y R pσ σ σ= − + =

( )σ σ ε=

( )0

1 pE

ζσε − = +

Explicit stress-strain relationship ε

σ

σ

0E 0EE

eε

pε dε eε

dE







( ) ( )ˆ, , ( ) 0yf R Y R pσ σ σ= − + =

( )σ σ ε=

( )0

1 pE

ζσε − = +

( )R R p=

Hardening/softening functions may be explicitly obtained

Explicit stress-strain relationship ε

σ

σ

0E 0EE

eε

pε dε eε

dE





In compression

In tension

( )1

2 2 21 2 2 3 4

1 32 2 32 2

1 20 0 0

ˆ( )

1ˆ cos arccos3

c

i i ii i i

i i i

R p p p p p p

p p p p p p

ζ ζ ζ ζ ζ

β β η η μ

+

−

= = =

⎡ ⎤= − + + − − +⎢ ⎥

⎢ ⎥⎣ ⎦⎧ ⎫⎡ ⎤⎛ ⎞

⎛ ⎞ ⎛ ⎞⎪ ⎪⎢ ⎥⎜ ⎟+ − + +⎨ ⎬⎜ ⎟ ⎜ ⎟⎢ ⎥⎜ ⎟⎝ ⎠ ⎝ ⎠⎪ ⎪⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦⎩ ⎭

∑ ∑ ∑

By using the stress-strain relationship of the CEB-FIP Model Code 1990, we obtain

Hyperbolic softening law (Meschke et al., 1997)

2( )

1

ctt

u

fR ppp

=⎛ ⎞

+⎜ ⎟⎝ ⎠ 0E

ctf

σ

εctεp

tR

relation σ ε−

( )tR p

0d

01EE D+

tension

( )1t

uct c

Gpf l ζ

=+

fracture energy

characteristic length

:tG:cl



Integration algorithm for the elasto-plastic damage model I

1 Data at the time tn: {σn, εn, εpn, pn, Dd

n}2 Give at the time tn+1: ∆ε⇒ εn+1 = εn + ∆ε

3 Predictor: compute elastic trial stress and test for inelasticloading

σtrialn+1 =

1Dn

(εn+1 − εpn)

Rtrial = R(pn)

f trialn+1 =

∣∣σtrialn+1

∣∣− Rtrial

IF f trialn+1 ≤ 0 THEN

∣∣∣∣∣∣∣∣εp

n+1 = εpn

pn+1 = pnDd

n+1 = Ddn

σn+1 = σtrialn+1

END → EXIT

ELSE proceed to step 4



Integration algorithm for the elasto-plastic damage model II

4 Corrector: by using the Kuhn-Tucker’s conditions, compute∆λ and then update the other variables

∆λ = ∆p = pn+1 − pn

εpn+1 = εp

n + (1− β) ∆λsign(σtrialn+1 )

σn+1 =1

Dn(εn+1 − εp

n)− ∆λ

Dnsign

(σtrial

n+1)

Ddn+1 = Dd

n + ∆Dd = Ddn + β∆λ

sign(σtrialn+1 )

σn+1

Compute the tangent modulus

E tgn+1 =

∂σ

∂ε

∣∣∣∣n+1

=1

Dn+1− 1

Dn+1 − (Dn+1)2 ∂R

∂p

∣∣∣∣n+1

END → EXIT



Numerical comparisons

-7-6-5-4-3-2-10

-30

-25

-20

-15

-10

-5

0

ζ

β

=

=

= −

=

0

Material parameters

27.9[MPa]

30000[MPa]

0.647

0.4

cmf

E

[mm/m]ε

[MPa]σ

Proposed modelKarsan and Jirsa, 1969

Simulation of cyclic compression test

0 0.1 0.2 0.3 0.4 0.5 0.60

1

2

3

4[MPa]σ

[mm/m]ε

ζ

=

=

=

=

= −

0

Material parameters

3.5[MPa]

31000[MPa]

65[N/m]

68[mm]

0.2

ct

t

c

f

E

G

l

Proposed modelGopalaratnam and Shah, 1987

Simulation of cyclic tension test

Good agreement of the calculated curve with the experiments isobserved



Outline

5 Finite Element FormulationsDisplacement-Based FormulationForce-Based FormulationTwo-field Mixed FormulationState Determination AlgorithmApplications



Displacement-Based Formulation

Assuming continuous displacement fields

Compatibility is satisfied in strict sense

Linearization of the constitutive equations

Equilibrium is satisfied in weak form

1q

2q

3q

4q

5q

6q

7q

8q9q

10q

Element 10 DOF

( ) ( )x x=d a q

( ) ( ) ; ( ) ( )scx x d x x= = sce B q B q

1 1 1 1;i i i i i i isc sc sc scD D k d− − − −= + Δ = + ΔD D k e

( )Td d 0 dsc sc eL

D xδ δ∂ − ∂ − = ∀∫ D P

8 DOF: Xu and Aribert, 1995

10 DOF: Daniels and Crisinel, 1989

16 DOF: Dall'Asta and Zona, 2002



Displacement-Based Formulation

Assuming continuous displacement fields

Compatibility is satisfied in strict sense


Equilibrium is satisfied in weak form

Governing equation of the finite element

1q

2q

3q

4q

5q

6q

7q

8q9q

10q

Element 10 DOF

( ) ( )x x=d a q

( ) ( ) ; ( ) ( )scx x d x x= = sce B q B q

1 1 1 1;i i i i i i isc sc sc scD D k d− − − −= + Δ = + ΔD D k e

( )Td d 0 dsc sc eL

D xδ δ∂ − ∂ − = ∀∫ D P

01iR−Δ = + −q Q Q QK

T 1 T 1d di isc sc sc

L L

x k x− −= +∫ ∫B k B BK BElement stiffness matrix

Element resisting forces T 1 T 11 d di is sc

L L

iR cx D x−− −= +∫ ∫B D BQ

8 DOF: Xu and Aribert, 1995

10 DOF: Daniels and Crisinel, 1989

16 DOF: Dall'Asta and Zona, 2002



Force-Based Formulation

Assuming continuous force fields

Equilibrium is satisfied in strict sense

0

( ) ( ) ( )

( ) ( ) ( ) ( )sc sc sc sc

sc

D

x

x x x

x x x

= +

= + +

b Q c Q

D Db Q c Q

1Q

2Q

3Q

4Q

5Q

1scQ 2sc

Q3sc

Q

/2L

( )scD x

( )scD x

/2L

Parabolic approximation of scD

zp

Discrete bond: exact force fields

Distributed bond: parabolic bond force distribution

(cubic approximation:

Salari 1999; Alemdar 2001)

2

2 0

d 0dd 0dd d 0

dd

csc

ssc

sc

N DxN DxM DH p

xx

+ =

− =

+ + =

Distributed bond

Element is internally determinateEquilibrium Exact force fields

Element is internally indeterminateA bond force distribution is assumed

Discrete bond ( )0scD =

Is a particular solution of equilibrium equations0( )xDEquilibrium equations

→





Compatibility is enforced in a integral form



0

( ) ( ) ( )

( ) ( ) ( ) ( )sc sc sc sc

sc

D

x

x x x

x x x

= +

= + +

b Q c Q

D Db Q c Q

1 1 1 1;i i i i i i isc sc sc scd d f D− − − −= + Δ = + Δe e f D

( ) ( )T Td d 0 ,scL L

x D d x Dδ δ δ δ∂ − + ∂ − = ∀∫ ∫D d e d Dsc sc sc

1Q

2Q

3Q

4Q

5Q

1scQ 2sc

Q3sc

Q

/2L

( )scD x

( )scD x

/2L


zp









Compatibility is enforced in a integral form



Governing equation of the force-based element

0

( ) ( ) ( )

( ) ( ) ( ) ( )sc sc sc sc

sc

D

x

x x x

x x x

= +

= + +

b Q c Q

D Db Q c Q

1 1 1 1;i i i i i i isc sc sc scd d f D− − − −= + Δ = + Δe e f D

( ) ( )T Td d 0 ,scL L

x D d x Dδ δ δ δ∂ − + ∂ − = ∀∫ ∫D d e d Dsc sc sc

1Q

2Q

3Q

4Q

5Q

1scQ 2sc

Q3sc

Q

/2L

( )scD x

( )scD x

/2L


zp



10

ir−Δ = − − ΔQ qF q q

Element flexibility matrix





Two-field Mixed Formulation


0( ) ( ) ( )x x x= + DD b Q

( ) ( )x x=d a qAssuming continuous displacement fields

Ayoub and Filippou 2000

2Q

zp

1Q3Q

4Q

5Q 6Q

6 force DOF

1q

2q3q4q

5q

6q7q

8q9q

10q

10 displacement DOF





Slip compatibility is satisfied in strict sense0( ) ( ) ( )x x x= + DD b Q


( ) ( )scd x x= scB q

Linearization of the constitutive equations1 1 11 ; i i i i

sc sc sc si i i

cD D k d−− − −= + Δ= + Δe e f D Ayoub and Filippou 2000

2Q

zp

1Q3Q

4Q

5Q 6Q

6 force DOF

1q

2q3q4q

5q

6q7q

8q9q

10q

10 displacement DOF






( ) ( )T Td d d ,sc sc eL L

D x xδ δ δ δ∂ − ∂ − + ∂ − ∀∫ ∫D P D d e d D




sc sc sc si i i

cD D k d−− − −= + Δ= + Δe e f D

Equilibrium and section strain compatiblity are enforced in a integral form (Hellinger Reissner variational principle)


2Q

zp

1Q3Q

4Q

5Q 6Q

6 force DOF

1q

2q3q4q

5q

6q7q

8q9q

10q

10 displacement DOF






( ) ( )T Td d d ,sc sc eL L

D x xδ δ δ δ∂ − ∂ − + ∂ − ∀∫ ∫D P D d e d D




sc sc sc si i i

cD D k d−− − −= + Δ= + Δe e f D

Equilibrium and section strain compatiblity are enforced in a integral form (Hellinger Reissner variational principle)

Governing equation of the two-field mixed element1 1

0

T 1

i isc e sc

ir

− −

−

⎡ ⎤ ⎡ ⎤Δ + − −⎡ ⎤⎢ ⎥ ⎢ ⎥=⎢ ⎥⎢ ⎥ ⎢ ⎥− ⎢ ⎥⎣ ⎦⎣ ⎦ ⎣

Δ⎦

K G q Q Q GQ Q

G F qQ 01

eiR−Δ = + −q Q Q QKcondense out

ΔQ


2Q

zp

1Q3Q

4Q

5Q 6Q

6 force DOF

1q

2q3q4q

5q

6q7q

8q9q

10q

10 displacement DOF


State determination algorithm: Displacement vs. Force models

Structure equilibrium-1Solve i i ig gUΔ = ΔK q P

Displacement-based element

Equilibrium?

Element resisting forcesT Td di i

sc scL L

iR x D x= +∫ ∫BQ B D

Constitutive laws( ) ; ( )i i

R scR sc scD D d= =D D e

Compute deformations( ) ; ( )i i

scx d xe

yesExit

no1i i= +

General purpose finite element program

Given displacements at the structural nodesDeterminate resisting forces and stiffness matrix

1ig−q

1igR−P

1igU−ΔP

A

B

D

igΔq

gq

gP1ig−K

1



Displacement-based element

Equilibrium?

Element resisting forcesT Td di i

sc scL L

iR x D x= +∫ ∫BQ B D




scx d xe

yesExit

no1i i= + Element resisting forces

iRQ




scx d xe

Compute element forces;i iscQ Q

Force-based element

Compute internal forces( ) ; ( )i i

scx D xD


Element resisting forcesiRQ


Force-based element

Introduce an iteration scheme at the element levelConsider element distributed loading

For regular beams: Spacone, 1994; Spacone et al., 1996

For composite beams: Salari 1999; Alemdar 2001 Iteration sheme at the element levelNo element internal loading


Force-based element

Itearative element state determination

Nodal displacementsiq


Element resisting forcesiRQ


Force-based element

Introduce an iteration scheme at the element levelConsider element distributed loading

For regular beams: Spacone, 1994; Spacone et al., 1996

For composite beams: Salari 1999; Alemdar 2001 Iteration sheme at the element levelNo element internal loading

Propose a new state determination for composite beam with element distributed loading


Force-based element

Itearative element state determination

Nodal displacementsiq

State determination algorithm for force-based element


Element displacements1i i i−= + Δq q q

ngP

1ng+P

10

1i

ng

g−=

+Δ

=Δ

PP

1 0i ng g− = =q q 1i

g−q i

gq1n

g+q

1igR−P

1igU−ΔP

A

B

D

igΔq

Convergence

gq

gP1ig−K

1

Structure level



1j i=Δ = Δq qImposed displacements

Element forces;j jscQ Q


A

B

CD

1i−q iq q

1j =Δq

2j =Δq3j =Δq

2j =ΔQ3j =ΔQ

1i−Q

iQ1j =ΔQ

Q

1 11j =K

11i−K 2j =K

j=3 convergence

1i−q iq q

1j =Δq

2j =Δq3j =Δq

scQ

1

12 2sc sc

j jsc

−= =⎡ ⎤− Δ⎣ ⎦Q QF q

1sc

i−QK

1sc

j =QK

2sc

j =QK

1

1

13 3sc sc

j jsc

−= =⎡ ⎤− Δ⎣ ⎦Q QF q

1jsc=ΔQ

2jsc=ΔQ

3jsc=ΔQ

1isc−Q

iscQ

j=3 convergenceA

B

C

D

Element level

1

11 1sc sc sc

j j j

j j j j jsc sc

−

−− −

Δ = Δ

⎡ ⎤Δ = Δ − Δ⎣ ⎦Q Q Q

Q K q

Q K q F q





Internal forces;j jscDD


10

j j jsc sc sc sc

j j i jjsc

D= =

+

+ + Δ

=

=

b Q c Q

D bQ cQ D

Particular solution due to the element distributed loads






Deformations;j jscde

Constitutive laws; ; ;j j j j

scR scRD fD f ;j jscrr

Residual deformations


( )xD1

( )xe

1j =ΔD

1i−D

iD

2j =ΔD3j =ΔD

1i−e

1j =Δeie

2j =Δe3j =Δe

1jR=D 2j

R=D

1j =r

2j =r

1j =f2j =f1

11i−f

A

B

C

D

1 1j j j j j j− −Δ = Δ → = + Δe f D e e e

( )j j j jR= −r f D D

Gauss-Labatto integration points









Residual deformations 1j j −⎡ ⎤= ⎣ ⎦K F

Element stiffness

Convergence?,j j j jscR scRD D tol− − ≤D D











Element stiffness

Element residual

1j+Δqdisplacements



no

next iteration j( )

( )

T T

T T

1

1

d

dsc sc sc

j jsc sc

L

j j j jsc s

j

cL

r x

r x−

+ = − +

⎡ ⎤− ⎣ ⎦

Δ

+

∫

∫QQ Q Q

b r b

F F c r c

q

A

B

CD

1i−q iq q

1j=Δq

2j =Δq3j=Δq

2j =ΔQ3j =ΔQ

1i−Q

iQ1j=ΔQ

Q

1 11j =K

11i−K 2j =K

j=3 convergence

Element level










Element stiffness

Element residual

1j+Δqdisplacements



Element resisting forcesji

R Q=Q

yes

no

next iteration j










Element stiffness

Element residual

1j+Δqdisplacements




R Q=Q

Structure resisting forcesassemble( )i

RiR=P Q

1igU tol+Δ ≤P

Convergence?

1i igU ext R+Δ = −P P P

Structure unbalanced forces

yes

no

next iteration j










Element stiffness

Element residual

1j+Δqdisplacements




R Q=Q


RiR=P Q

1igU tol+Δ ≤P

Convergence?



Exit yes

yes

no

next iteration j










Element stiffness

Element residual

1j+Δqdisplacements




R Q=Q


RiR=P Q

1igU tol+Δ ≤P

Convergence?



Exit yes

yes

no

no

next iteration j

next NR iteration i


Comparison with experimental data

8φ

100

800

IPE 400

5 10φ

5 10φ

section AA

length unit: mm

650 650 650 650650 650 650 650

P

200 2500 2500 200

A

A

Poutre PI4

Load-deflection diagrams

Simply-supported composite beam (Aribert et al., 1983)

0 50 100 1500

100

200

300

400

500

Midspan displacement [mm]

Forc

e P

[kN

]

18 Displacement-based elements4 Force-based elements4 Mixed elementExperiment (Ariber al al., 1983)

0 50 100 1500

100

200

300

400

500


18 Displacement-based elements12 Force-based elements12 Mixed elementExperiment (Ariber al al., 1983)

33.3 mm30.7 mm

Distributed bond Discrete bond




Slip distribution

0

0.5

1

1.5

Glis

sem

ent [

mm

]

12 E.F. mixteRésultat exprérimental

650 mm 650 mm 650 mm 650 mm650 mm 650 mm 650 mm 650 mm

P=257 kN

P=334 kNP=366 kN

P

-1.5

-1

-0.5

0G

lisse

men

t [m

m]

6 E.F. mixteRésultat exprérimental

P=297 kN

P=257 kNP=297 kN

P=334 kN

P=366 kN

Discrete bond9 connector element

Distributed bond




P P

10 350× 10 350×

7 300×

100 2250 2250 2250 2250 100

B

B

C

CPoutre PH3

10φ

100

800

HEA200

7.67 cm²

section CC

8.04 cm²

10φ

100

800

HEA200

1.6 cm²

section BB

1.6 cm²

Two-span composite beam (Ansourian 1981)

0 10 20 30 40 50 600

50

100

150

200

250

300


Forc

e P

[kN

]

24 Displacement-based elements6 Force-based elements6 Mixed elementExperiment

Load-deflection diagrams

Distributed bond




-0.15 -0.1 -0.05 0 0.05 0.1 0.150

50

100

150

200

250

300

Courbures [1/m]

Forc

e P

[kN

]

24 E.F. déplacement6 E.F. équilibre6 E.F. mixteRésultat exprérimental

P P

100 2250 2250 2250 2250 100

B

B

Poutre PH3

200

A

A

[ ]L : mm

Distributed bond

Section B-B(negative bending)

Section A-A(Positive bending)

Curvature [1/m]

Forc

e P

[kN

]



Comparison of the three finite element formulations

0p

2000mm

50m

m50

mm

20mm

yσ

sE1

sE1

σ

ε

yD

scE1

scE1

scD

scd5300MPa 2 10 MPa

200N/mm 1000MPay s

y sc

E

D E

σ = = ×

= =

A

A

section AA

Cantilever composite beam




2000mm

0p

δ

0 100 2000

2

4

6

8

10

12

14

Deflection [mm]

1 element2 elementsConverged solution

δ0 100 200

0

2

4

6

8

10

12

14

Deflection [mm]

1 element2 elementsConverged solution

δ0 100 200

0

2

4

6

8

10

12

14

Deflection [mm]

Dis

tribu

ted

load

[kN

/m]

1 element2 elements4 elements64 elementsConverged solution

0p

δ

Displacement-based element Force-based element Mixed element

Load-deflection diagramsQ-H. Nguyen PhD Thesis Defense



0 500 1000 1500 2000

-20

-15

-10

-5

0

5

x [mm]

Ben

ding

mom

ent [

kN.m

] 1 Displacement-based element1 Force-based element1 mixed elementConverged solution

0p 7 kN/m=

0 500 1000 1500 2000-50

0

50

100

150

200

250

X [mm]

Axi

al fo

rce

Nc

[kN

/m]

1 Displacement-based element1 Force-based element1 mixed elementConverged solution

0p 7 kN/m=

Poor representation of internal forcesDisplacement-based & mixed models




0 500 1000 1500 2000

-0.3

-0.25

-0.2

-0.15

-0.1

-0.05

0

0.05

x [mm]

Cur

vatu

re [1

/m] 2 Displacement-based element

2 Force-based element2 mixed elementConverged solution

0p 7 kN/m=

0 500 1000 1500 2000

0

0.05

0.1

0.15

0.2

0.25

x [mm]

Slip

[mm

]

2 Displacement-based element2 Force-based element2 mixed elementConverged solution

Force-based modelsInter-element slip discontinuity

Inter-element curvature discontinuity

Displacement-based & mixed models



Conclusions

Three finite element formulations have been developed for twobond models

A new state determination algorithm for force-based elementincluded element distributed loads was presented

The numerical-experimental comparison shown validates themodels reliability and the capacity to determine the experimentalbehaviour of composite beams

Force-based element and mixed element are both computationallymore efficient than the displacement-based element

For the same number of elements, force-based element yieldsbetter results than mixed element



Outline

6 Time-Dependent Behaviour In the Plastic RangeIntroductionViscoelastic/plastic Model for ConcreteApplications



Introduction

Viscoelastic modelsSuitable to linear analysis onlyUnable to account for thecracking due to shrinkage

Viscoplastic modelsComplicate to implement

→ propose a viscoelastic/plastic model

101

102

103

104

-10

-8

-6

-4

-2

0

2

4

6

Time [days]

Stre

ss [M

Pa]

Tensile strength: 2.9 MPa

200mm

100mm

( ) 0.03mmtδ ∀ = −

100mm( )tσ

Concrete specimen C30, CEB-FIP model 1990

with shrinkagewithout shrinkage

Stress relaxation according to viscoelastic model



Viscoelastic/plastic Model for Concrete

Combination of linear visco-elasticity and continuum plasticity (Van Zijl et al., 2001)

Decomposition of total strain

ve ve( ) ( ) ( )t E t tσ ε σ= +

ve p sh( ) ( ) ( ) ( )t t t tε ε ε ε= + +

viscoelastic strain

plastic strain

shrinkage strain

Viscoelastic model Plastic modelYield condition

( ) ( ), , ( ) 0yf R R pσ σ σ= − − ≤

p fε λσ∂

=∂

Flow rule

( )ve p sh( ) ( ) ( ) ( ) ( )t E t t t tσ ε ε ε σ= − − +

0E1( )E t 2 ( )E t

H( )mE t

1( )tη 2 ( )tη ( )m tη

( )tσ ( )tσ

yσ

( )ve tε ( )p tε ( )sh tε

Rheological viscoelastic/plastic model



Integration algorithm for Viscoelastic/plastic model I

1 Data at the time tn: {σn, εn, εpn, pn, γ

(n)i }

2 Give at the time tn+1: {εn+1, εshn+1}

3 Compute the parameters of the viscoelastic model: Eevn+1, σ̃n+1

4 Viscoelastic Predictor: compute viscoelastic trial stress and testfor plastic loading

σ trialn+1 = Eve

n+1(εn+1 − εp

n − εshn+1

)+ σ̃n+1

f trialn+1 = f

(σtrial

n+1 , R(pn))

IF f trialn+1 ≤ 0

THEN viscoelastic step :

∣∣∣∣∣∣εp

n+1 = εpn

pn+1 = pnσn+1 = σtrial

n+1

END → EXIT

ELSE plastic step: proceed to step 5



Integration algorithm for Viscoelastic/plastic model II

5 Corrector: by using the Kuhn-Tucker’s conditions, compute∆λ and then update the other variables

pn+1 = pn + ∆λ

εn+1 = εn + ∆λsign(σtrialn+1 )

σn+1 = σtrialn+1 + Eve

n+1∆λsign(σtrialn+1 )

Compute the tangent modulus

E tgn+1 =

∂σ

∂ε

∣∣∣∣n+1

= Even+1

1−Eve

n+1

Even+1 +

dRdp

END → EXIT



Simulation of relaxation test

101

102

103

104

-10

-8

-6

-4

-2

0

2

4

6

Time [days]

Stre

ss [M

Pa]

Linear viscoelastic modelViscoelastic/plastic model

Tensile strength: 2.9 MPa

101

102

103

104

-0.08

-0.06

-0.04

-0.02

0

0.02

0.04

0.06

0.08

Time [days]

Stra

in [%

]

Total strainViscoelastic strainPlastic strainShrinkage strain

Time evolution of stress

Time evolution of strain

200mm

100mm

( ) 0.03mmtδ ∀ = −

100mm( )tσ

Concrete specimen C30, CEB-FIP model 1990

The proposed model is able to represent the cracking phenomena due to shrinkage



Simply-supported composite beam

0 100 200 300 400 5000

10

20

30

40

50

60

Force P [kN]

Flèc

he à

mi-t

ravé

e [m

m]

A

=

= →

t 30 days

P 0 400kN

LoadP [kN]

Mid

span

def

lect

ion

[mm

]

101

102

103

104

0

10

20

30

40

50

60

Temps [jours]

Flèc

he à

mi-t

ravé

e [m

m]

avec retraitsans retrait

A=

= →

P 400kN

t 30 days 50 years

Time [days]

Mid

span

def

lect

ion

[mm

]

without shrinkage effect

with shrinkage effect

8φ

100

800

IPE400

5 10φ

5 10φ

section AA

length unit: mm

650 650 650 650650 650 650 650

P

200 2500 2500 200

A

A

Poutre PI4

Simply-supported composite beam (Aribert et al., 1983)

Evolution of mid-span deflection



Summary

1 A finite element model with exact stiffness matrix based on theanalytical solution was developed (for linear elastic andviscoelastic behaviours)

2 A elasto-plastic damage model was proposed for concrete

3 Three finite element formulations was developed for compositebeams with partial interaction

4 A new state determination algorithm was developed for theforce-based element including element distributed load

5 A viscoelastic/plastic model was proposed for concrete in order tosimulate the interaction between the time effects and the cracking



Conclusions

1 The discrete bond model represents the true connection and it issimple to use but it requires a large number of elements

2 Compare to discrete bond model, distributed bond model is lesscomputationally expensive because it reduces significantlynumber of elements

3 Among three finite element formulations, force-based formulationperforms better

4 Significant influence of creep and especially of shrinkage on theglobal response of composite beams in serviceability

5 Time effects play an important role in the inelastic response ofcomposite beam



Future works

1 Realize a parametric study of time effects in nonlinear behaviourof composite beam

2 Modelling of the behaviour of composite beams usingTimoshenko beam theory (in progress)

3 Take into account the nonlinearity geometry using corotationalformulation

4 Take into account the uplift

5 Extend the F.E. tools to composite frame



Thanks for your attention !


Modelling of the non-linear behaviour of composite beams

Engineering

Transcript of Modelling of the non-linear behaviour of composite beams