Rcs1-chapter3-constitutive-law

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Reinforced Concrete
Structures 1 - Eurocodes
RCS 1
Professor Marwan SADEK
https://www.researchgate.net/profile/Marwan_Sadek
https://fr.slideshare.net/marwansadek00
Email : marwansadek00@gmail.com
If you detect any mistakes, please let me know at : marwansadek00@gmail.com

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RCS1
M. SADEK
Ch 1 : Generalities – Reinforced concrete in practice
Ch 2 : Evolution of the standards – Limit states
Ch 3 : Mechanical Characteristics of materials – Constitutive
relations
Ch 4 : Durability and Cover
Ch 5 : Beam under simple bending – Ultimate limit state ULS
Ch 6 : Beam under simple bending – serviceability limit state SLS
Ch 7 : Section subjected to pure tension

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Selected References
French BAEL Code (91, 99)
 Règles BAEL 91 modifiées 99, Règles techniques de conception et de calcul des
ouvrages et constructions en béton armé, Eyrolles, 2000.
 J. Perchat (2000), Maîtrise du BAEL 91 et des DTU associés, Eyrolles, 2000.
 J.P. Mougin (2000), BAEL 91 modifié 99 et DTU associés, Eyrolles, 2000.
 ….
EUROCODES
 H. Thonier (2013), Le projet de béton armé, 7ème édition, SEBTP, 2013.
 Jean-Armand Calgaro, Paolo Formichi ( 2013) Calcul des actions sur les
bâtiments selon l'Eurocode 1 , Le moniteur, 2013.
 J. M. Paillé (2009), Calcul des structures en béton, Eyrolles- AFNOR, 2009.
 Jean Perchat (2013), Traité de béton armé Selon l'Eurocode 2, Le moniteur,
2013 (2ème édition)
 Manual for the design of concrete building structures to Eurocode 2, The
Institution of Structural Engineers, BCA, 2006.
 A. J. Bond (2006), How to Design Concrete Structures using Eurocode 2, The
concrete centre, BCA, 2006.
https://usingeurocodes.com/
M. SADEK

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In addition to Eurocodes, the references that are mainly
used to prepare this course material are :
 Thonier 2013
 Perchat 2013
 Paillé 2009
Some figures and formulas are taken from
 Cours de S. Multon - BETON ARME Eurocode 2 (available on internet)
 Cours béton armé de Christian Albouy
M. SADEK

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CHAPTER III
Mechanical Characteristics of Materials
Behaviour Constitutive Law
M. SADEK
1. Concrete
2. Steel
Annexes

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Concrete compressive strength - Test
M. SADEK
 France : Cylindrical Concrete specimen, H/D = 2
(diameter16 cm, height 32 cm)
1. Concrete 2. Steel  Annexes

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M. SADEK
Concrete compressive strength - Test

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M. SADEK
 fcm : Mean value of concrete cylinder compressive strength
 fck : Characteristic compressive cylinder strength of concrete at 28 days
(fck  90 MPa)
Characteristic value & 5% Fractile
% specimen
fck(t)=fcm(t) – 8 (in MPa)
Concrete compressive strength - Standard

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Concrete compressive strength

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> 200 MPa
 Ultra high performance concrete
Fiber Reinforced Concrete

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fcm(t) = cc x fcm
 Normal Concrete
(t < 28 jours)
 fcm(t) is the mean concrete compressive strength at an age of t days

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M. SADEK
 fcd : Design value of concrete compressive strength
cc= 1 (ANF)
C = 1.5 (persistent situation) ; 1.2 (accidental)
 fck : Characteristic compressive cylinder strength of concrete at 28 days

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Another test for tensile strength of Concrete
 The tensile strength by splitting test
(Essai brésilien)
 Design of rigid pavements
Difficult
 Flexure Test - 4 pts

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fctk : Characteristic axial tensile strength of concrete
fctk 0,05  fctm  fctk 0,95

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 Behaviour of Concrete – Stress-Strain Diagram
1. Structural Analysis
2. Sections Design

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1. Structural Analysis/ Second order effect

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Modulus of Elasticity
Secant modulus Ecm (@ 0.4 fcm), for short term loading
Effective modulus of elasticity of concrete Ec,eff (Creep Effect)
(fcm en MPa, Ecm en GPa)
1. Structural Analysis/ Second order effect
The EC2 defines a design Elastic modulus :

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2. Section Design
a) Parabola-rectangle
diagram
b) Bi-linear stress-strain relation

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c) Rectangular stress distribution
Note : In the present lecture, the design of section at ULS is conducted using the
diagram c (simpler diagram)
The use of diagrams a and b are authorized by the EC2.

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 Other aspects – Creep (EC2, 3.1.4)

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At t =  , under a constant compression stress
 Ec : Tangent modulus of elasticity (may be taken as 1,05 Ecm)
 t0 : the age of concrete at the time of loading (in days)
(Linear creep)
(Non-Linear creep)
 Other aspects – Creep (EC2, 3.1.4)

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Creep Coefficient
where Ac is the concrete cross-sectional area and u is
the perimeter of that part which is exposed to drying

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Creep Coefficient

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M. SADEK
Poisson ratio
 = 0 - Calculation of internal forces
 = 0.2 - strain Calculation

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M. SADEK
Steel

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M. SADEK
 High bond (twisted):
 bars
 welded wire mesh
 Round or plain bars : rarely used in Europe (only if folding out needed)
used in Lebanon especially in villages
because of the ease bending of stirrups
Types of Steel

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M. SADEK
Steel used in R.C

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M. SADEK


Tension Test

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M. SADEK
The ductility of Steel of the reinforced concrete is characterized by :
εuk Characteristic strain of reinforcement steel at maximum load
 the Characteristic value of
3 Classes of ductility
Stress/strain

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M. SADEK
 A : Normal ductility B500A (welded wire mesh in general, and bars with
low diameter)
 B : high ductility B500B (in general the HB bars with a diameter > 12)
 C : very high ductility C450 (generally used in seismic areas,
especially in USA)
 L’EN 10080 defines 3 classes of ductility :

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M. SADEK
The application rules for design and detailing in this Eurocode are valid for
a specified yield strength range (400 fyk600 Mpa)
 Note : for bridges and for construction in seismic zones, steel B and C are
authorized (The French N.A allows the use of steel A outside the critical zones)
 L’EN 10080 defines 3 classes of ductility :

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M. SADEK
Idealised / Design stress-strain diagrams
for reinforcing steel (ULS)
s = 1.15 (persistent)
1.0 (accidentel)

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M. SADEK
 Design stress-strain diagrams:
 a horizontal top branch without the need to check the strain limit
 an inclined top branch with a strain limit (s0  s  ud = 0.9uk )

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M. SADEK
 Note : the calculation of the slope using the diagram A gives different value from that of
diagram B, due to an error of the presentation of the diagram in EC2. It explains the difference
in the expression of the stress obtained by different authors (Perchat, Paillé, Thonier, Ricotier)
 In this document, we will calculate the slope on the basis of the diagram B.

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M. SADEK
Elastic modulus Es = 200 000 MPa
Density = 7.85 T/m3
 Thermal expansion coefficient= 1.10-5
 Diameters:
6, 8, 10, 12, 14, 16, 20, 25, 32, 40, (50 , 56)
 Welded wire mesh: with lower diameter (see Annex)
Other characteristics

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M. SADEK
Annexes

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M. SADEK

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 6 8 10 12 14 16 20 25 32 40
1 0.28 0.50 0.79 1.13 1.54 2.01 3.14 4.91 8.04 12.57
2 0.57 1.01 1.57 2.26 3.08 4.02 6.28 9.82 16.08 25.13
3 0.85 1.51 2.36 3.39 4.62 6.03 9.42 14.73 24.13 37.70
4 1.13 2.01 3.14 4.52 6.16 8.04 12.57 19.63 32.17 50.27
5 1.41 2.51 3.93 5.65 7.70 10.05 15.71 24.54 40.21 62.83
6 1.70 3.02 4.71 6.79 9.24 12.06 18.85 29.45 48.25 75.40
7 1.98 3.52 5.50 7.92 10.78 14.07 21.99 34.36 56.30 87.96
8 2.26 4.02 6.28 9.05 12.32 16.08 25.13 39.27 64.34 100.53
9 2.54 4.52 7.07 10.18 13.85 18.10 28.27 44.18 72.38 113.10
10 2.83 5.03 7.85 11.31 15.39 20.11 31.42 49.09 80.42 125.66
11 3.11 5.53 8.64 12.44 16.93 22.12 34.56 54.00 88.47 138.23
12 3.39 6.03 9.42 13.57 18.47 24.13 37.70 58.90 96.51 150.80
13 3.68 6.53 10.21 14.70 20.01 26.14 40.84 63.81 104.55 163.36
14 3.96 7.04 11.00 15.83 21.55 28.15 43.98 68.72 112.59 175.93
15 4.24 7.54 11.78 16.96 23.09 30.16 47.12 73.63 120.64 188.50
16 4.52 8.04 12.57 18.10 24.63 32.17 50.27 78.54 128.68 201.06
17 4.81 8.55 13.35 19.23 26.17 34.18 53.41 83.45 136.72 213.63
18 5.09 9.05 14.14 20.36 27.71 36.19 56.55 88.36 144.76 226.19
19 5.37 9.55 14.92 21.49 29.25 38.20 59.69 93.27 152.81 238.76
20 5.65 10.05 15.71 22.62 30.79 40.21 62.83 98.17 160.85 251.33
Wt kg / ml 0.222 0.395 0.617 0.888 1.208 1.578 2.466 3.853 6.313 9.865

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Welded wire mesh (anti cracking)

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Welded wire mesh (France)

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 Determination of the slope of the diagram (/) for steel A & B
 Deduce the equation of s for both Steel classes.
 Practice : determination of the value of (s) for different (s)
 Difference between diagrams with horizontal top branch and inclined
top branch
 (persistent situation / accidental)
 Determine the creep coefficient for a column
Exercices

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fck(t)=fcm(t) – 8
fcm(t) = cc x fcm
cc= 1 (FNA)
C = 1.5 (persistent) ; 1.2 (accidental)
Reminder of main formulas

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s = 1.2 (persistent)
1.0 (accidental)
or
Reminder of main formulas

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