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 of
composite beams
taking into account the time eﬀects
Quang-Huy NGUYEN
INSA de Rennes - Structural Engineering Research Group
University of Wollongong - Faculty of Engineering
13 July 2009
Q-H. Nguyen PhD Thesis Defense

Outline
1 Introduction
2 Elastic analysis of composite beams

Outline
1 Introduction
3 Time-Dependent Behaviour

Outline
1 Introduction
4 Nonlinear Behaviour of Materials

Outline
1 Introduction
5 Finite Element Formulations

Outline
1 Introduction
6 Time-Dependent Behaviour In the Plastic Range

Outline
1 Introduction
7 Conclusions and Futur works

Outline
1 Introduction
General
Background: Analysis of composite beams
Research questions
Objectives

General
Introduction to steel-concrete composite beam
Steel-concrete composite structure are widely used in the construction
industry
Economic
Reduced live load deflections
Reduced weight
Fast erection process
Increased span lengths are possible
Stiffer floors
Composite beam system (Ricker 1989)

General
Introduction to steel-concrete composite beam
Composite beams consist of steel beam and concrete slab joint
together as a unit by shear studs
steel beam
shear stud
concrete slab
profile sheeting
reinforcement

Bond models
A
A
B
B
section A-A section B-B

Bond models
A
A
B
B
Discrete bond model
Aribert (1982, France)
Schanzenback (1988, Germany)

Bond models
A
A
B
B
Discrete bond model
Aribert (1982, France)
Schanzenback (1988, Germany)
Distributed bond model
Newmark (1951, US)
Adekola (1968, Nigeria)

Analysis Type
Elastic Analysis Inelastic Analysis
Newmark, 1951
Adekola, 1968
N
N
tM
x
scd
X
Y
2
2
12
d ( )
( ) ( )
d
t
N x
N x C M x
x
μ− = ⇒ Analytical solution

Inelastic Analysis
Displacement-based Force-based Mixed
1v
2v
2u
1u
θ1
θ2
x
( )v x
( )u x
X
Y
θ
θ
⎡ ⎤
⎢ ⎥
⎢ ⎥
⎡ ⎤ ⎢ ⎥
=⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥⎣ ⎦
⎢ ⎥
⎢ ⎥
⎢ ⎥⎣ ⎦
1
1
1
2
2
2
( )
( )
( )
u
v
u x
x uv x
v
a
Assumed displacement field

Inelastic Analysis
Arizumi et al., 1981
Schanzenbach, 1988
Daniels, 1989
Boerave, 1990

Inelastic Analysis
⎡ ⎤
⎡ ⎤ ⎢ ⎥
=⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥⎣ ⎦
⎢ ⎥⎣ ⎦
1
2
2
( )
( )
( )
M
N x
x M
M 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
1
1
2
2
22
2
( ) ( )
( ) & ( )
( ) ( )
u
v
M
u x N x
x x Muv x M x
Nv
a b
X
Y
1M
1M
2N
( )N x
( )M x
x
1v
2v
2u
1u
θ1
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 i
iA
n
y i i i
iA
N A A
M z A A z
Fiber 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 Surface
Axial Force
Moment
Loading Surface
Compression 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 the
behaviour of composite beams?
4 How to take into account the time eﬀects in inelastic
analysis?

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 to
study their performances for both bond models
4 To combine time eﬀects and cracking of concrete

Outline
Basic assumptions
Governing Equations of Composite Steel-Concrete Beams
Exact Stiﬀness Matrix - Elastic behaviour
Comparison 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 through
bond

1 Equilibrium
Distributed bond
Discrete bond
2 Compatibility
3 Constitutive relations
2
2
d ( )
( ) 0
d
d ( )
( ) 0
d
d ( ) d ( )
0
dd
c
sc
s
sc
sc
z
N x
D x
x
N x
D x
x
M x D x
H p
xx
+ =
− =
+ + =
=sc sc eD∂ − ∂ −D P 0
Matrix form
zp
cH
cM dc cM M+
dc cN N+
dc cT T+
cN
cT
scV
scD
dx
ds sM M+
ds sN N+
ds sT T+
sM
sN
sT
scD
sH
x
z
y

1 Equilibrium
Distributed bond
Discrete bond
2 Compatibility
unconnected element
cN +
cM +
sN +
sM +
sN −
sM −
cN −
cM −
cN
cM
sN
sM
stQ
stQ
0xΔ =
connector element
=e∂ −D P 0
Unconnected beam segment Single connector
1
1
s
c st
N
N Q
M H
⎡ ⎤ ⎡ ⎤
⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥
= −⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥−
⎣ ⎦ ⎣ ⎦

1 Equilibrium
Distributed bond
Discrete bond
2 Compatibility
H
scd
su
cu θ
θ
v
x
z
y
2
2
d ( )
( )
d
d ( )
( )
d
d
d ( )
( ) (
( )
) ( )
d
( )
d
c
c
s
sc s c
s
u x
x
x
u x
x
x
v x
x
v x
d x u x u x H
x
x
ε
ε
κ
=
=
=
+
−
= −
Matrix form
T
sc scd = ∂
= ∂e d
d

1 Equilibrium
Distributed bond
Discrete bond
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

Distributed bond
Equilibrium in term of the displacements
5 3
2
15 3
3 4 2
2 33 4 2
2
4 2
d d
d d
d d d
d d d
d d
dd
s s
s s
s
c s
u u
x x
v u u
x x x
u v
u u H
xx
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 ( ) ( ) 2
sinh 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 q
4 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
( )i
cu
( )i
su
( )i
θ
( )j
cu
( )j
su
( )j
θ
( )j
cu
( )i
v
( )i
cu
( )i
su
( )i
θ
( )j
cu
( )j
su
( )j
v
( )j
θ
( )j
su
( )j
θ
( )j
v
( )j
θ
+= +
( )i
v
( )i
cu
( )i
su
Connector
element
Unconnected
beam element
Connector
element

Discrete bond
with discrete bond
( )i
cu
( )i
su
( )i
θ
( )j
cu
( )j
su
( )j
θ
( )j
cu
( )i
v
( )i
cu
( )i
su
( )i
θ
( )j
cu
( )j
su
( )j
v
( )j
θ
( )j
su
( )j
θ
( )j
v
( )j
θ
+= +
( )i
v
( )i
cu
( )i
su
Connector
element
Unconnected
beam element
Connector
element
nc
eK
Exact sitffness
matrix
Analytical
solution

Discrete bond
with discrete bond
( )i
cu
( )i
su
( )i
θ
( )j
cu
( )j
su
( )j
θ
( )j
cu
( )i
v
( )i
cu
( )i
su
( )i
θ
( )j
cu
( )j
su
( )j
v
( )j
θ
( )j
su
( )j
θ
( )j
v
( )j
θ
+= +
( )i
v
( )i
cu
( )i
su
Connector
element
Unconnected
beam element
Connector
element
nc
eK
Exact sitffness
matrix
Analytical
solution
st
iK
Exact sitffness
matrix
Analytical
solution
st
jK
Exact sitffness
matrix
Analytical
solution

Discrete bond
with discrete bond
( )i
cu
( )i
su
( )i
θ
( )j
cu
( )j
su
( )j
θ
( )j
cu
( )i
v
( )i
cu
( )i
su
( )i
θ
( )j
cu
( )j
su
( )j
v
( )j
θ
( )j
su
( )j
θ
( )j
v
( )j
θ
+= +
( )i
v
( )i
cu
( )i
su
Connector
element
Unconnected
beam element
Connector
element
nc
eK
Exact sitffness
matrix
Analytical
solution
st
iK
Exact sitffness
matrix
Analytical
solution
st
jK
Exact sitffness
matrix
Analytical
solution
Exact sitffness
matrix
eK
assembly

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
300MPast
sc
k
k
s
= =
Discrete bond model: using 18 elements
Distributed 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]
Deflection[mm]
Discrete 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]
Discrete bond model
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
Curvature[1/m]
Discrete bond model
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
Axialforceintheconcreteslab[kN]
Discrete bond model
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 as
eighteen discrete bond elements
The discrete bond model represents the true connection and it is
simple to use but it requires a large number of elements

Outline
Time Eﬀects in Concrete
Time-discretized analytical solution for composite beams
Applications
Conclusions

Time Eﬀects in Concrete
1 Strain in concrete grow in
time
2 Shrinkage
3 Creep
4 Aging material
5 Play an important role in
serviceability
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 - free
Companion
Specimen
Loaded
(Creep)
Specimen
Specimen
Unloaded
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( )tε
1( )tε
t
t
t
t
t
t

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( )tε
1( )tε
t
t
t
t
t
t
Integral-type relation
εc(t) = σc(t1)J(t, t1) +
t
t1
J(t, τ)
dσc(τ)
dτ
dτ + εsh(t)

Time discrete approach
General method (step-by-step)
1
,
1
( ) ( )( ) ( ) ( )
n
c n sh n n i c i
i
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 i
n n n n
n i k i
n n n n
J t t J t t
J t t J t t
E t
J t t J t t
J t t J t t
J t t J t t
−
−
+ −
−
⎧ −
⎪ =
+⎪
⎪
= Ψ = ⎨
+ ⎪
−⎪ >
⎪ +⎩
[ ][ ]
1
1
1 1
1
d ( ) 1
( , ) d ( , ) ( , ) ( ) ( )
d 2
nt n
n n i n i i i
it
J t J t t J t t t t
σ τ
τ τ σ σ
τ
−
+ +
=
≅ + −∑∫
Trapezoidal rule
Time-discrete constitutive relation
where

Algebraic method (one step)
[ ]
1
1 1
d ( )
( , ) d ( ) ( )
d
( , ) with
nt
n n n
t
nJ t t tJt t tτ
σ τ
τ τ σ σ τ
τ
≅ − ≤ ≤∫
Effective modulus (EM) method (McMilan, 1916)
1
1
1
1 ( , )
(( , )) ,
( )
n
n
n
c
J
t t
J t t
E
t
t
ϕ
τ
+
= =
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
c
n
n
J
t t t t
E t
t
χ ϕ
τ
+
=
creep coefficient
aging coefficient

Rate-type (internal variables) method (Bažant, 1971)
0 1
1 1
( , ) 1 exp
( )
m
i ii
t
J t
E D
τ
τ
τ τ=
⎡ ⎤⎛ ⎞−
≅ + −⎢ ⎥⎜ ⎟
⎝ ⎠⎣ ⎦
∑
The creep function is approximated by the Dirichlet series
0 1 0
( ) 1
( ) ( ) with ( ) 1 exp
( )
tm
i i
i ii
t t
t t t d
E D
σ τ
ε 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

1 For one-step method
Faella et al. 2002
Ranzi and Bradford 2005
2 For rate-type (internal variables) method
Jurkiewiez et al. 2005
3 For general method
The 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
d5
u
(n)
s
dx5
− µ2
n
d3
u
(n)
s
dx3
= ζ
(n)
1 + ζ
(n)
2
n−1
i=1
Ψn,i
d2
M
(i)
co
dx2
+ ζ
(n)
3
n−1
i=1
Ψn,i
d2
N
(i)
co
dx2
d3
v(n)
dx3
= ζ
(n)
4
d4
u
(n)
s
dx4
+ ζ
(n)
5
d2
u
(n)
s
dx2
+ ζ
(n)
6
n−1
i=1
Ψn,i
dN
(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 n
Q q
Exact siffness matrix at the instant
( ) ( )
2 2,n n
Q q
( ) ( )
3 3,n n
Q q
( ) ( )
4 4,n n
Q q
( ) ( )
8 8,n n
Q q
( ) ( )
5 5,n n
Q q
( ) ( )
6 6,n n
Q q
( ) ( )
7 7,n n
Q 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
nt

Applications
2300mm
200mm
934 mm
64.56kN/m
25m 25m
cG
sG
2
20 300mm×
2
1550 15mm×
2
30 450mm×
2
3
8 4
2
3
8 4
100mm
460000mm
0m
15.3310 mm
934 mm
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
10
1
10
2
10
3
10
4
10
5
-2017
-2016
-2015
-2014
-2013
-2012
-2011
-2010
-2009
-2008
-2007
-2006
Time [days]
RedundantreactionR[kN]
Distributed bond - general method
Dezi and Tanrantino
64.56kN/m
25m 25m
= 0.4 kN/mm²sck
= 0.1kN/mm²sck
2007.16
2012.30
2013.51
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
Deflection[mm]
Discrete bond model (80 elements)
Distributed bond model (2 elements)
64.56kN/m
25550days
30days

Comparison of time-discrete approaches
10
1
10
2
10
3
10
4
10
5
17
18
19
20
21
22
23
24
Time [days]
Midspandeflection[mm]
Step-by-step method
"Rate-type" method
AAEM method
EM method
MS 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
Deflection[mm]
30 days
25550 days: creep only
t=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
Bendingmoment[kN.m]
30 days
25550 days: creep only
t=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 model
leads to a quite complexe solution but it reduces signiﬁcantly the
number of elements
General method gives precise results but it requires the storage of
the whole stress history
"Rate-type" method gives nearly identical results as general
method. This method avoids almost data storage but the
determination of model parameters is quite complexe
Among algebraic methods, AAEM method seems to perform very
well
A signiﬁcant impact of shrinkage

Outline
Constitutive Models of Steel
Constitutive Models of Shear Stud
Constitutive Models of Concrete

Constitutive models of steel
1 Models based on explicit stress-strain relationship
2 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 1
ξ ε ε−
( )2 0 r 2
ξ 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

( )e p
Eσ 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

More involved
Cyclic behaviour is included in the model
Serveral physical phenomena can be coupled: damage, time effects ...
yε
σ
ε
yσ
uσ
hε uε
yε−hε−uε−
yσ−
uσ−
O 1O2O 3O4O
Monotonic loading
Cyclic loading
Linear isotropic hardening model

Constitutive models of shear stud
1 Discrete bond: The following models are selected to describe the
behaviour of one shear stud
P
δ
δ δ δ
P P P
uPuP uP
1 u0.95P P=
2 fu1.05P P=
fuP
uδ 1δ 2δ
( ) 2
u 11 exp
c
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 and
stiﬀness are calculated by dividing the strength and stiﬀness of a
single row of shear studs by their distance along the beam.

Constitutive models of concrete
Explicit stress-strain relationship recommended by CEB-FIP Model
Code 1990
Good approach of stress-strain
monotonic curve in compression
No unloading information
No ascending branch of stress-strain
monotonic curve in tension 1cE
c
E 1c
ε ,limc
ε
cm
f−
0.5 cm
f−
0.9 ctm
f
ctm
f
15%
cε
Compression
Tension
cσ
Goals: To develop a model for concrete based on the elasto-plastic
damage theory
1 Reproduce exactly stress-strain monotonic curve in compression
of CEB-FIP Model Code 1990
2 Take into account the degradation of the elastic moduli
3 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)d
D
( ) ( )
1 2d 0 pdp p1
( , , ) ( )
2
DD Dp pεε ε ε
−
Ψ − = + − + Ψ
Elastic damage part plastic part
Thermodynamically associated variables
( )p
2
d
1ˆ; ;
2p
R Y
D
σ
ε ε
σ
∂Ψ ∂Ψ ∂Ψ
= = = = −
∂ − ∂ ∂
ε
σ
σ
p
ε
ε
0E
d
ε
e
ε
0EE
( )
1d d
D E
−
=
e
ε
( )
2d d d1
2
E εΨ =
( )
2e 0 e1
2
E εΨ =

D
( ) ( )
1 2d 0 pdp p1
( , , ) ( )
2
DD Dp pεε ε ε
−
Ψ − = + − + Ψ
( )p
2
d
1ˆ; ;
2p
R Y
D
σ
ε ε
σ
∂Ψ ∂Ψ ∂Ψ
= = = = −
∂ − ∂ ∂
( ) ( )ˆ, , ( ) 0yf R Y pRσ σ σ= − − ≤
Yield/damage condition
ε
σ
σ
p
ε
ε
0E
d
ε
e
ε
0EE
( )
1d d
D E
−
=
e
ε
( )
2d d d1
2
E εΨ =
( )
2e 0 e1
2
E εΨ =

D
( ) ( )
1 2d 0 pdp p1
( , , ) ( )
2
DD Dp pεε ε ε
−
Ψ − = + − + Ψ
( )p
2
d
1ˆ; ;
2p
R Y
D
σ
ε ε
σ
∂Ψ ∂Ψ ∂Ψ
= = = = −
∂ − ∂ ∂
Flow rule, damage and hardning/softening laws
( ) ( )ˆ, , ( ) 0yf R Y pRσ σ σ= − − ≤
Yield/damage condition
( ) ( )p d 1
1 1 sign( ) ; sign( ) ;
ˆ
f f f
D p
RY
ε λ λ σ λ λ σ λ λ
σ
β β β β
σ
∂ ∂ ∂
= − = − − = = − − = = −
∂ ∂∂
β : scalar paramater, proposed by Meschke et al, 1997
ε
σ
σ
p
ε
ε
0E
d
ε
e
ε
0EE
( )
1d d
D E
−
=
e
ε
( )
2d d d1
2
E εΨ =
( )
2e 0 e1
2
E εΨ =

Determination of the hardening/softening functions
Introducing a scalar parameter ζ
( )
0
1 p
E
ζ
σ
ε − = +
ε
σ
σ
0E 0EE
e
ε
p
ε d
ε
e
ε
d
E
( )p d
1 pζε ε+ = +
Assumption

Monitonic loading condition
( ) ( )ˆ, , ( ) 0yf R Y R pσ σ σ= − + =
( )
0
1 p
E
ζ
σ
ε − = +
ε
σ
σ
0E 0EE
e
ε
p
ε d
ε
e
ε
d
E
( )p d
1 pζε ε+ = +
Assumption

( ) ( )ˆ, , ( ) 0yf R Y R pσ σ σ= − + =
( )σ σ ε=
( )
0
1 p
E
ζ
σ
ε − = +
Explicit stress-strain relationship ε
σ
σ
0E 0EE
e
ε
p
ε d
ε
e
ε
d
E
( )p d
1 pζε ε+ = +
Assumption

( ) ( )ˆ, , ( ) 0yf R Y R pσ σ σ= − + =
( )σ σ ε=
( )
0
1 p
E
ζ
σ
ε − = +
( )R R p=
Hardening/softening functions may be explicitly obtained
Explicit stress-strain relationship ε
σ
σ
0E 0EE
e
ε
p
ε d
ε
e
ε
d
E
( )p d
1 pζε ε+ = +
Assumption

In compression
In tension
( )
1
2 2 2
1 2 2 3 4
1 3
2 2 32 2
1 2
0 0 0
ˆ( )
1
ˆ cos arccos
3
c
i i i
i 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
ct
t
u
f
R p
p
p
=
⎛ ⎞
+⎜ ⎟
⎝ ⎠ 0E
ctf
σ
εctε
p
tR
relation σ ε−
( )tR p
0
d
01
E
E D+
tension
( )1
t
u
ct c
G
p
f 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, εp
n, pn, Dd
n}
2 Give at the time tn+1: ∆ε ⇒ εn+1 = εn + ∆ε
3 Predictor: compute elastic trial stress and test for inelastic
loading
σtrial
n+1 =
1
Dn
(εn+1 − εp
n)
Rtrial
= R(pn)
f trial
n+1 = σtrial
n+1 − Rtrial
IF f trial
n+1 ≤ 0 THEN
εp
n+1 = εp
n
pn+1 = pn
Dd
n+1 = Dd
n
σn+1 = σtrial
n+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
εp
n+1 = εp
n + (1 − β) ∆λsign(σtrial
n+1 )
σn+1 =
1
Dn
(εn+1 − εp
n) −
∆λ
Dn
sign σtrial
n+1
Dd
n+1 = Dd
n + ∆Dd
= Dd
n + β∆λ
sign(σtrial
n+1 )
σn+1
Compute the tangent modulus
Etg
n+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 model
Karsan and Jirsa, 1969
Simulation of cyclic compression test
0 0.1 0.2 0.3 0.4 0.5 0.6
0
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 model
Gopalaratnam and Shah, 1987
Simulation of cyclic tension test
Good agreement of the calculated curve with the experiments is
observed

Outline
Displacement-Based Formulation
Force-Based Formulation
Two-ﬁeld Mixed Formulation
State Determination Algorithm
Applications

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
8q
9q
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 i
sc sc sc scD D k d− − − −
= + Δ = + ΔD D k e
( )T
d d 0 dsc sc e
L
D xδ δ∂ − ∂ − = ∀∫ D P
8 DOF: Xu and Aribert, 1995
10 DOF: Daniels and Crisinel, 1989
16 DOF: Dall'Asta and Zona, 2002

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
8q
9q
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 i
sc sc sc scD D k d− − − −
= + Δ = + ΔD D k e
( )T
d d 0 dsc sc e
L
D xδ δ∂ − ∂ − = ∀∫ D P
0
1i
R
−
Δ = + −q Q Q QK
T 1 T 1
d di i
sc sc sc
L L
x k x− −
= +∫ ∫B k B BK BElement stiffness matrix
Element resisting forces T 1 T 11
d di i
s sc
L L
i
R 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

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 2scQ 3scQ
/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
0
d
d
0
d
d d
0
dd
c
sc
s
sc
sc
N
D
x
N
D
x
M D
H p
xx
+ =
− =
+ + =
Distributed bond
Element is internally determinate
Equilibrium Exact force fields
Element is internally indeterminate
A 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 i
sc sc sc scd d f D− − − −
= + Δ = + Δe e f D
( ) ( )T T
d d 0 ,sc
L L
x D d x Dδ δ δ δ∂ − + ∂ − = ∀∫ ∫D d e d Dsc sc sc
1Q
2Q
3Q
4Q
5Q
1scQ 2scQ 3scQ
/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 i
sc sc sc scd d f D− − − −
= + Δ = + Δe e f D
( ) ( )T T
d d 0 ,sc
L L
x D d x Dδ δ δ δ∂ − + ∂ − = ∀∫ ∫D d e d Dsc sc sc
1Q
2Q
3Q
4Q
5Q
1scQ 2scQ 3scQ
/2L
( )scD x
( )scD x
/2L
zp
1
0
i
r
−
Δ = − − ΔQ qF q q
Element flexibility matrix

0( ) ( ) ( )x x x= + DD b Q
( ) ( )x x=d a q
Ayoub and Filippou 2000
2Q
zp
1Q
3Q 4Q
5Q
6Q
6 force DOF
1q
2q
3q4q
5q
6q
7q
8q9q
10q
10 displacement DOF

Slip compatibility is satisfied in strict sense
0( ) ( ) ( )x x x= + DD b Q
( ) ( )x x=d a q
( ) ( )scd x x= scB q
1 1 11
; i i i i
sc sc sc s
i i i
cD D k d−− − −
= + Δ= + Δe e f D Ayoub and Filippou 2000
2Q
zp
1Q
3Q 4Q
5Q
6Q
6 force DOF
1q
2q
3q4q
5q
6q
7q
8q9q
10q
10 displacement DOF

0( ) ( ) ( )x x x= + DD b Q
( ) ( )T T
d d d ,sc sc e
L L
D x xδ δ δ δ∂ − ∂ − + ∂ − ∀∫ ∫D P D d e d D
( ) ( )x x=d a q
1 1 11
; i i i i
sc sc sc s
i 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
1Q
3Q 4Q
5Q
6Q
6 force DOF
1q
2q
3q4q
5q
6q
7q
8q9q
10q
10 displacement DOF

0( ) ( ) ( )x x x= + DD b Q
( ) ( )T T
d d d ,sc sc e
L L
D x xδ δ δ δ∂ − ∂ − + ∂ − ∀∫ ∫D P D d e d D
( ) ( )x x=d a q
1 1 11
; i i i i
sc sc sc s
i 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 element
1 1
0
T 1
i i
sc e sc
i
r
− −
−
⎡ ⎤ ⎡ ⎤Δ + − −⎡ ⎤
⎢ ⎥ ⎢ ⎥=⎢ ⎥
⎢ ⎥ ⎢ ⎥− ⎢ ⎥⎣ ⎦⎣ ⎦ ⎣
Δ
⎦
K G q Q Q GQ Q
G F qQ 0
1
e
i
R
−
Δ = + −q Q Q QK
condense out
ΔQ
2Q
zp
1Q
3Q 4Q
5Q
6Q
6 force DOF
1q
2q
3q4q
5q
6q
7q
8q9q
10q
10 displacement DOF

State determination algorithm: Displacement vs. Force models
Structure equilibrium
-1
Solve i i i
g gUΔ = ΔK q P
Displacement-based element
Equilibrium?
Element resisting forces
T T
d di i
sc sc
L L
i
R 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
yes
Exit
no
1i i= +
General purpose finite element
program
Given displacements at the
structural nodes
Determinate resisting forces
and stiffness matrix
1i
g
−
q
1i
gR
−
P
1i
gU
−
ΔP
A
B
D
i
gΔq
gq
gP
1i
g
−
K
1

-1
Solve i i i
g gUΔ = ΔK q P
Displacement-based element
Equilibrium?
T T
d di i
sc sc
L L
i
R x D x= +∫ ∫BQ B D
Constitutive laws
( ) ; ( )i i
( ) ; ( )i i
scx d xe
yes
Exit
no
1i i= + Element resisting forces
i
RQ
Constitutive laws
( ) ; ( )i i
( ) ; ( )i i
scx d xe
Compute element forces
;i i
scQ Q
Force-based element
Compute internal forces
( ) ; ( )i i
scx D xD

i
RQ
-1
Solve i i i
g gUΔ = ΔK q P
Force-based element
Introduce an iteration scheme at the
element level
Consider element distributed loading
For regular beams: Spacone, 1994; Spacone
et al., 1996
For composite beams: Salari 1999; Alemdar 2001
Iteration sheme at the element level
No element internal loading
-1
Solve i i i
g gUΔ = ΔK q P
Force-based element
Itearative element
state determination
Nodal displacements
i
q

i
RQ
-1
Solve i i i
g gUΔ = ΔK q P
Force-based element
Introduce an iteration scheme at the
element level
Consider element distributed loading
For regular beams: Spacone, 1994; Spacone
et al., 1996
For composite beams: Salari 1999; Alemdar 2001
Iteration sheme at the element level
No element internal loading
Propose a new state determination for composite
beam with element distributed loading
-1
Solve i i i
g gUΔ = ΔK q P
Force-based element
Itearative element
state determination
Nodal displacements
i
q

State determination algorithm for force-based element
-1
Solve i i i
g gUΔ = ΔK q P
Element displacements
1i i i−
= + Δq q q
n
gP
1n
g
+
P
101in
gg
−=+
Δ=ΔPP
1 0i n
g g
− =
=q q 1i
g
−
q i
gq 1n
g
+
q
1i
gR
−
P
1i
gU
−
ΔP
A
B
D
i
gΔq
Convergence
gq
gP
1i
g
−
K
1
Structure level

-1
Solve i i i
g gUΔ = ΔK q P
1j i=
Δ = Δq q
Imposed displacements
Element forces
;j j
scQ Q
1i i i−
= + Δq q q
A
B
C
D
1i−
q i
q q
1j =
Δq
2j =
Δq
3j =
Δq
2j =
ΔQ
3j =
ΔQ
1i−
Q
i
Q
1j =
ΔQ
Q
1 1
1j =
K
1
1i−
K 2j =
K
j=3 convergence
1i−
q i
q q
1j =
Δq
2j =
Δq
3j =
Δq
scQ
1
12 2
sc sc
j j
sc
−= =
⎡ ⎤− Δ⎣ ⎦Q QF q
1
sc
i−
QK
1
sc
j =
QK
2
sc
j =
QK
1
1
13 3
sc sc
j j
sc
−= =
⎡ ⎤− Δ⎣ ⎦Q QF q
1j
sc
=
ΔQ
2j
sc
=
ΔQ
3j
sc
=
ΔQ
1i
sc
−
Q
i
scQ
j=3 convergence
A
B
C
D
Elementlevel
1
11 1
sc sc sc
j j j
j j j j j
sc sc
−
−− −
Δ = Δ
⎡ ⎤Δ = Δ − Δ⎣ ⎦Q Q Q
Q K q
Q K q F q

-1
Solve i i i
g gUΔ = ΔK q P
1j i=
Δ = Δq q
Element forces
;j j
scQ Q
Internal forces
;j j
scDD
1i i i−
= + Δq q q
1
0
j j j
sc sc sc sc
j j i jj
sc
D
= =
+
+ + Δ
=
=
b Q c Q
D bQ cQ D
Particular solution due to the
element distributed loads

-1
Solve i i i
g gUΔ = ΔK q P
1j i=
Δ = Δq q
Element forces
;j j
scQ Q
Internal forces
;j j
scDD
Deformations
;j j
scde
Constitutive laws
; ; ;j j j j
scR scRD fD f ;j j
scrr
Residual deformations
1i i i−
= + Δq q q
( )xD
1
( )xe
1j =
ΔD
1i−
D
i
D
2j =
ΔD
3j =
ΔD
1i−
e
1j =
Δe
i
e
2j =
Δe
3j =
Δe
1j
R
=
D 2j
R
=
D
1j =
r
2j =
r
1j =
f
2j =
f
1
1
1i−
f
A
B
C
D
1 1j j j j j j− −
Δ = Δ → = + Δe f D e e e
( )j j j j
R= −r f D D
Gauss-Labatto integration points

-1
Solve i i i
g gUΔ = ΔK q P
1j i=
Δ = Δq q
Element forces
;j j
scQ Q
Internal forces
;j j
scDD
Deformations
;j j
scde
Constitutive laws
; ; ;j j j j
scR scRD fD f ;j j
scrr
Residual deformations 1j j −
⎡ ⎤= ⎣ ⎦K F
Element stiffness
Convergence?
,j j j j
scR scRD D tol− − ≤D D
1i i i−
= + Δq q q

-1
Solve i i i
g gUΔ = ΔK q P
1j i=
Δ = Δq q
Element forces
;j j
scQ Q
Internal forces
;j j
scDD
Deformations
;j j
scde
Constitutive laws
; ; ;j j j j
scR scRD fD f ;j j
scrr
⎡ ⎤= ⎣ ⎦K F
Element stiffness
Element residual
1j+
Δq
displacements
Convergence?
,j j j j
1i i i−
= + Δq q q
no
next iteration j
( )
( )
T T
T T
1
1
d
dsc sc sc
j j
sc sc
L
j j j j
sc s
j
c
L
r x
r x
−
+
= − +
⎡ ⎤− ⎣ ⎦
Δ
+
∫
∫QQ Q Q
b r b
F F c r c
q
A
B
C
D
1i−
q i
q q
1j =
Δq
2j =
Δq
3j =
Δq
2j =
ΔQ
3j =
ΔQ
1i−
Q
i
Q
1j =
ΔQ
Q
1 1
1j =
K
1
1i−
K 2j =
K
j=3 convergence
Elementlevel

-1
Solve i i i
g gUΔ = ΔK q P
1j i=
Δ = Δq q
Element forces
;j j
scQ Q
Internal forces
;j j
scDD
Deformations
;j j
scde
Constitutive laws
; ; ;j j j j
scR scRD fD f ;j j
scrr
⎡ ⎤= ⎣ ⎦K F
Element stiffness
Element residual
1j+
Δq
displacements
Convergence?
,j j j j
1i i i−
= + Δq q q
ji
R Q=Q
yes
no
next iteration j

-1
Solve i i i
g gUΔ = ΔK q P
1j i=
Δ = Δq q
Element forces
;j j
scQ Q
Internal forces
;j j
scDD
Deformations
;j j
scde
Constitutive laws
; ; ;j j j j
scR scRD fD f ;j j
scrr
⎡ ⎤= ⎣ ⎦K F
Element stiffness
Element residual
1j+
Δq
displacements
Convergence?
,j j j j
1i i i−
= + Δq q q
ji
R Q=Q
Structure resisting forces
assemble( )i
R
i
R=P Q
1i
gU tol+
Δ ≤P
Convergence?
1i i
gU ext R
+
Δ = −P P P
Structure unbalanced forces
yes
no
next iteration j

-1
Solve i i i
g gUΔ = ΔK q P
1j i=
Δ = Δq q
Element forces
;j j
scQ Q
Internal forces
;j j
scDD
Deformations
;j j
scde
Constitutive laws
; ; ;j j j j
scR scRD fD f ;j j
scrr
⎡ ⎤= ⎣ ⎦K F
Element stiffness
Element residual
1j+
Δq
displacements
Convergence?
,j j j j
1i i i−
= + Δq q q
ji
R Q=Q
assemble( )i
R
i
R=P Q
1i
gU tol+
Δ ≤P
Convergence?
1i i
gU ext R
+
Δ = −P P P
Exit yes
yes
no
next iteration j

-1
Solve i i i
g gUΔ = ΔK q P
1j i=
Δ = Δq q
Element forces
;j j
scQ Q
Internal forces
;j j
scDD
Deformations
;j j
scde
Constitutive laws
; ; ;j j j j
scR scRD fD f ;j j
scrr
⎡ ⎤= ⎣ ⎦K F
Element stiffness
Element residual
1j+
Δq
displacements
Convergence?
,j j j j
1i i i−
= + Δq q q
ji
R Q=Q
assemble( )i
R
i
R=P Q
1i
gU tol+
Δ ≤P
Convergence?
1i i
gU ext R
+
Δ = −P P P
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 A A
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 50 100 150
0
100
200
300
400
500
Midspan displacement [mm]
ForceP[kN]
18 Displacement-based elements
4 Force-based elements
4 Mixed element
Experiment (Ariber al al., 1983)
0 50 100 150
0
100
200
300
400
500
12 Mixed element
Experiment (Ariber al al., 1983)
33.3 mm30.7 mm
Distributed bond Discrete bond

Slip distribution
0
0.5
1
1.5
Glissement[mm]
12 E.F. mixte
Résultat 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
0
Glissement[mm]
6 E.F. mixte
P=297 kN
P=257 kN
P=297 kN
P=334 kN
P=366 kN
Discrete bond
9 connector element
Distributed bond

P P
10 350× 10 350×
7 300×
100 2250 2250 2250 2250 100
B
B
C
C
Poutre PH3
10φ
100
800
HEA200
7.67 cm²
section CC
8.04 cm²
10φ
100
800
HEA200
1.6cm²
section BB
1.6cm²
Two-span composite beam (Ansourian 1981)
0 10 20 30 40 50 60
0
50
100
150
200
250
300
ForceP[kN]
6 Mixed element
Experiment
Distributed bond

-0.15 -0.1 -0.05 0 0.05 0.1 0.15
0
50
100
150
200
250
300
Courbures [1/m]
ForceP[kN]
24 E.F. déplacement
6 E.F. équilibre
6 E.F. mixte
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]
ForceP[kN]

Comparison of the three ﬁnite element formulations
0p
2000mm
50mm50mm
20mm
yσ
sE
1
sE
1
σ
ε
yD
scE
1
scE
1
scD
scd
5
300MPa 2 10 MPa
200N/mm 1000MPa
y s
y sc
E
D E
σ = = ×
= =
A
A
section AA
Cantilever composite beam

2000mm
0p
δ
0 100 200
0
2
4
6
8
10
12
14
Deflection [mm]
1 element
2 elements
Converged solution
δ
0 100 200
0
2
4
6
8
10
12
14
Deflection [mm]
1 element
2 elements
Converged solution
δ
0 100 200
0
2
4
6
8
10
12
14
Deflection [mm]
Distributedload[kN/m]
1 element
2 elements
4 elements
64 elements
Converged solution
0p
δ
Displacement-based element Force-based element Mixed element

0 500 1000 1500 2000
-20
-15
-10
-5
0
5
x [mm]
Bendingmoment[kN.m]
1 Displacement-based element
1 Force-based element
1 mixed element
Converged solution
0p 7 kN/m=
0 500 1000 1500 2000
-50
0
50
100
150
200
250
X [mm]
AxialforceNc[kN/m]
1 mixed element
Converged solution
0p 7 kN/m=
Poor representation of internal forces
Displacement-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]
Curvature[1/m]
2 mixed element
Converged 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 mixed element
Converged solution
Force-based models
Inter-element slip discontinuity
Inter-element curvature discontinuity
Displacement-based & mixed models

Conclusions
Three ﬁnite element formulations have been developed for two
bond models
A new state determination algorithm for force-based element
included element distributed loads was presented
The numerical-experimental comparison shown validates the
models reliability and the capacity to determine the experimental
behaviour of composite beams
Force-based element and mixed element are both computationally
more eﬃcient than the displacement-based element
For the same number of elements, force-based element yields
better results than mixed element

Outline
Introduction
Viscoelastic/plastic Model for Concrete
Applications

Introduction
Viscoelastic models
Suitable to linear analysis only
Unable to account for the
cracking due to shrinkage
Viscoplastic models
Complicate to implement
10
1
10
2
10
3
10
4
-10
-8
-6
-4
-2
0
2
4
6
Time [days]
Stress[MPa]
Tensile strength: 2.9 MPa
200mm
100mm
( ) 0.03mmtδ ∀ = −
100mm
( )tσ
Concrete specimen C30, CEB-FIP model 1990
with shrinkage
without shrinkage
Stress relaxation according to viscoelastic model

Introduction
Viscoelastic models
Suitable to linear analysis only
Unable to account for the
cracking due to shrinkage
Viscoplastic models
Complicate to implement
→ propose a viscoelastic/plastic model
10
1
10
2
10
3
10
4
-10
-8
-6
-4
-2
0
2
4
6
Time [days]
Stress[MPa]
200mm
100mm
( ) 0.03mmtδ ∀ = −
100mm
( )tσ
with shrinkage
without 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 model
Yield condition
( ) ( ), , ( ) 0yf R R pσ σ σ= − − ≤
p f
ε λ
σ
∂
=
∂
Flow rule
( )ve p sh
( ) ( ) ( ) ( ) ( )t E t t t tσ ε ε ε σ= − − +
0E
1( )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, εp
n, pn, γ
(n)
i }
2 Give at the time tn+1: {εn+1, εsh
n+1}
3 Compute the parameters of the viscoelastic model: Eev
n+1, σn+1
4 Viscoelastic Predictor: compute viscoelastic trial stress and test
for plastic loading
σ trial
n+1 = Eve
n+1 εn+1 − εp
n − εsh
n+1 + σn+1
f trial
n+1 = f σtrial
n+1 , R(pn)
IF f trial
n+1 ≤ 0
THEN viscoelastic step :
εp
n+1 = εp
n
pn+1 = pn
σn+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(σtrial
n+1 )
σn+1 = σtrial
n+1 + Eve
n+1∆λsign(σtrial
n+1 )
Compute the tangent modulus
Etg
n+1 =
∂σ
∂ε n+1
= Eve
n+1



1 −
Eve
n+1
Eve
n+1 +
dR
dp




END → EXIT

Simulation of relaxation test
10
1
10
2
10
3
10
4
-10
-8
-6
-4
-2
0
2
4
6
Time [days]
Stress[MPa]
Linear viscoelastic model
Viscoelastic/plastic model
10
1
10
2
10
3
10
4
-0.08
-0.06
-0.04
-0.02
0
0.02
0.04
0.06
0.08
Time [days]
Strain[%]
Total strain
Viscoelastic strain
Plastic strain
Shrinkage strain
Time evolution of stress
Time evolution of strain
200mm
100mm
( ) 0.03mmtδ ∀ = −
100mm
( )tσ
The proposed model is able to
represent the cracking
phenomena due to shrinkage

Simply-supported composite beam
0 100 200 300 400 500
0
10
20
30
40
50
60
Force P [kN]
Flècheàmi-travée[mm]
A
=
= →
t 30 days
P 0 400kN
LoadP [kN]
10
1
10
2
10
3
10
4
0
10
20
30
40
50
60
Temps [jours]
Flècheàmi-travée[mm]
avec retrait
sans retrait
A
=
= →
P 400kN
t 30 days 50 years
Time [days]
without shrinkage effect
with shrinkage effect
8φ
100
800
IPE 400
5 10φ
5 10φ
section A A
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 the
analytical solution was developed (for linear elastic and
viscoelastic behaviours)
2 A elasto-plastic damage model was proposed for concrete
3 Three finite element formulations was developed for composite
beams with partial interaction
4 A new state determination algorithm was developed for the
force-based element including element distributed load
5 A viscoelastic/plastic model was proposed for concrete in order to
simulate the interaction between the time effects and the cracking

Conclusions
1 The discrete bond model represents the true connection and it is
simple to use but it requires a large number of elements
2 Compare to discrete bond model, distributed bond model is less
computationally expensive because it reduces significantly
number of elements
3 Among three finite element formulations, force-based formulation
performs better
4 Significant influence of creep and especially of shrinkage on the
global response of composite beams in serviceability
5 Time effects play an important role in the inelastic response of
composite beam

Future works
1 Realize a parametric study of time eﬀects in nonlinear behaviour
of composite beam
2 Modelling of the behaviour of composite beams using
Timoshenko beam theory (in progress)
3 Take into account the nonlinearity geometry using corotational
formulation
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

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