Fracture Mechanics of Concrete Structures Proceedings FRAMCOS-3 AEDIFICATIO Publishers, D-79104 Freiburg, Germany

LOAD DEFLECTION DIAGRAM OF OVER-REINFORCED CONCRETE BEAMS

A. Bascoul Laboratoire Materiaux et Durabilite des Constructions M. Duprat, M. Pinglot Laboratoire d'Etudes Thermiques et Mecaniques INSA - Genie Civil, Complexe scientifique de Rangueil 31077 Toulouse Cedex 4 - FRANCE

Abstract The load - deflection behaviour of reinforced concrete beams is calculated using a specifically designed Finite Element method. The beams are discretized into multi - layered short elements. The Moment - Curvature diagram of each element is calculated applying the assumptions that plane sections remain plane and that the strain in the reinforcement is the same as that in the surrounding concrete. Any stress - strain diagram can be used for the compressed concrete and for steel. It may be defined either according to experimental curves or using analytical laws.

The calculation method consists of defining successive equilibrium states, increasing step by step the maximum concrete strain. This process allows us to get the entire moment - curvature diagram, including the softening branch. Up to now, the effects of shear are not taken into account. The global deformations such as rotations and deflections are deduced from a numerical integration of curvatures. Key words: Numerical simulation, Navier's hypothesis, moment -curvature relationship, integration of curvature.

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1 Introduction

This contribution to the Round Robin analysis on modelling of Over­Reinforced Concrete Beams consists of presenting a numerical method of predicting the behaviour of any reinforced concrete beam in a first section and of giving the results of its application in a second section. The experimental data concerning the behaviour laws of the materials are shared by all the contributors so as it is not necessary to present them agam.

2 Numerical method

Beams are composed of n multi - layer elements, (Fig. la). The equilibrium of each element is calculated assuming that Navier's hypothesis apply.

2.1 Equilibrium and moment - curvature diagram for one element The elements are composed of m layers (50 layers in this application). Each one is considered to be subjected to uniaxial states of stress and strain, (Fig. I b ). For a given value of the curvature, the compressive concrete strain Ee of the upper layer is calculated by iteration in such a way that the axial resulting force Nnt tends toward zero CINintl < 10 daN).

Nnt is calculated through relation (1):

m m,

N int = L A ci ·CJ c ( Ei) + L A si ·CJ s ( Ei) (1) 1 1

Ei : strain of the layer i,

Ad: section of the concrete layer i,

CT' c ( E i) : stress of the concrete layer i, according to the strain - stress

relationship of concrete, Asi: section of the steel layer i,

as ( £i) : stress of the steel layer I, according to the strain - stress

relationship of steel.

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p p

1 il

.. ~. ~ ... ~ ';

i~t..l Fig. la

r--.....-~ ~ ~~--cr-s-(E-i)_~_+--+-----------~-+-----N-s~ section strain I streskes

"' Fig. lb

~ - u ii"'"'

i-i-

... ' I 1 I r

ii II ...

r

Fig. le

x x

Fig. ld

Fig.Id

Fig.1. Principle of the numerical method

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Then the bending moment can be deduced from relation (2): m m,

Mint= LAci.0-c(Ei).(YG -yi)+ LAsi·CYs(Ei).(YG -di) (2) i=l 1

Ya: distance of the upper concrete fibre to the neutral axis, Yi : distance of the concrete layer i to the neutral axis, di : distance of the steel layer i to the upper fibre.

Finally, the moment - curvature diagram of an element j is calculated by

incrementing step by step the curvature.!., (Fig. le). r

The tension stiffening effect is taken into account by modifying the curvature in relation to CEB-FIP 1990 recommendations.

2.2 Curvature diagram of the beam The loading of the beam is controlled by the curvature of the central element. The moment curvature diagram of this element is calculated as

described above. For a given step I, the curvature (~) / determines the

bending moment in the central element. Thus, both the corresponding load P1 and the moment diagram of the beam can be calculated (Fig. ld). Using the . moment - curvature diagram of each element, the curvature diagram of the beam can be drawn (Fig. le). However, beyond the peak load, some elements may be subjected to decreasing curvature. In this case, it is assumed that the moment - curvature relationship is linear down to zero.

2.3 Load - deflection diagram Rotations and deflections are obtained by numerical integration of the well known Bresse formulae:

eK =80 + t[:J.Axj (3) . J I

8K = ()OxK + f [_!_J .&j.(xK -x) j=I rj I

() K : rotation of the element K, 8K : deflection of the element K,

&.i : length of the elementj,

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(4)

( rl J : curvature of the element j at a given load level l .

.I I

The final position of the beam is obtained by a dichotomic approximation of the rotation 8 0 which induces a deflection 8

0 close to

zero Cl8n I< 5.10-2 mm ).

3 Application

3.1 Modelling of the beams

1655 1655

15 elements

I. J r ·i

3312 576 3312 . ~ -

20 elements 9 elements 20 elements

Fig.2b. Large beam 3.2 Results Results are presented on one page for one type of beam and concrete with both boundary conditions of concrete compression tests ( low fiction -high friction - HF) : - load - deflection diagrams (load = P, according to Fig. 1), - variations of the strains at the upper fibre and at the lower fibre.

In addition, load - support rotation diagrams are given in the four last figures.

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J

l

140-r-~~---~~~~~~~~~~~~

120

z 100 -

~ 80 I.-'

"'O 60 ~ 0 ~ 40 -

20 -

0

0 20

g -------- ---F

40 60 80 100

Deflection [mm]

Fig. 3a. Load - deflection diagrams for large beams and normal concrete

---------2-0---Hf-------------1-~~-----------

-0.01 0 0.01 0.02 0.03

Strain

Fig. 3b. Strains at upper and lower fibres for large beams and normal concrete

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so.-~~,-~----;~~-,~~-;=====~~---,

45 40

r--"' 35 ~ 30 -1--J 25 '"O g 20

,..:i 15

10 5

----+-------1 --LF __ -+--"""·-· ___ "+------1---------1 --HF -

0 _L-J._L_J_L...J.....l,_J._L-l-..1.-J......l-~..J-.J.....l-J-l......_._....___.._.._"'--'--'--'"--'-'

0 10 20 30 40 50 60

Deflection [mm]

Fig. 4a. Load - deflection diagrams for small beams and normal concrete

__ " ______ -LF ·--···----·-······-··---·· --- --40 -·" -----

-HF , ____ __,

-0.01

- HF _________ ___,

--l--·-··--11---¥~---.--- ___.,,-1---...------t--- -- LF 1-------1

-0.005 0

+------+-------!----"---!

0.005 Strain

0.01 0.015 0.02

Fig. 4b. Strains at upper and lower fibres for small beams and normal concrete

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150 140 ----· --------

130 120 110

,-._ 100 z 90 ~ 80 1--1

"'d 70 ~ 0 60 ~ 50

40 30 20 10 0

0 10 20 30 40 50

Deflection [mm]

Fig. 5a. Load deflection diagrams for small beams and high strength concrete

-0.01 -0.005 0

----+-----+--~- -------------

0.005

Strain

0.01 0.015 0.02

Fig. Sb. Strains at upper and lower fibres for small beams and high strength concrete

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60

160---~---r~~~~~...--~-.-~~-.-~--.

140

120 -- ---- ------l------+-·--·---------t- ~----!---·---~

:z100 ~ \-I 80 re ~

j 60

40 -·····-····-

20 -

-LF

-HF

0 ~'--4--4-+-+--!-~-!-~-+-+-+---+-'r-+--+-f-+--+-+-+-+-r-+--+-~

0 10 20 30 40 50 60 Deflection [mm]

Fig. 6a. Load - deflection diagrams for small beams and fibre concrete

-0.01 -0.005 0

1------H --LF -HF

-~---------H-LF

0.005

Strain

0.01

-HF

0.015 0.02

Fig. 6b. Strains at upper and lower fibres for small beams and fibre concrete

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140-.-~--,~~~~~~~~~~~~~

120

z 100 ~ 80+-------__,_~--~---+-<----+----+---~• "--'

~ 60 0

~ 40

20 0 ....+-+-+-+--+-+--+-+-+--+-+--+-+-+--+-+---+-1-+-+-+---+-+-+-+-+~f--+-l

0 0.005 0.01 0.015 0.02 0.025 0.03

Rotation [rd]

Fig. 7. Load - support rotation diagrams for large beams and normal concrete

so.....--~~~~...,..--~~~~..,--~~~___,,--~~~--.

4 5 --!·····------· ----·---+---

40 -1------·---------1------------>

35 Z30-~---------------·~,~---~ "--'25 ~

~ 20 -------·-------,f/L,,-+----i

~15-1---------1----4---~-10

5

_ ___,_LF ____ ___, __ ____, - HF

-LF

-HF

0 _.__._.....__._-'--...__...__..__..__..__..___..__..___.__.__..___.--'---'---1..___..l

0 0.01 0.02

Rotation [rd]

0.03 0.04

Fig. 8. Load - support rotation diagrams for small beams and normal concrete

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140 -

120

r-i100 z ~ 80 "e ~ 0 60 ~

40

20

0 0 0.005 om 0.015 0.02 0.025 o.o3 o.035 o.o4

Rotation [rd]

Fig. 9. Load - support rotation diagrams fors mall beams and high strength concrete

160~~~-,--~~---~~..--~~,--~--.

140

120

z 100 ~ '-J 80 "e ~

j 60

40

20

--1 ~: 1-

0 -f-t-t--+-+-t-+-+-+-+-+--+-+-1--+-_,__...,r-+-+-~r-+-+-+--l-l 0 0.01 0.02 0.03

Rotation [rd] 0.04 0.05

Fig. 10. Load - support rotation diagrams for small beams and fibre concrete

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Table 1. Peak load characterisation

Beam Concrete Friction Fmax Deflection Comp. Tens. Sup. KN mm Strain Strain Rot. (rd)

Large Normal LF 110 40 0.00236 0.00204 0.016 Large Normal HF 136 51.6 0.00391 0.00289 0.02 Small Normal LF 28.7 15.9 0.00227 0.00153 0.012 Small Normal HF 35 21.6 0.00393 0.00227 0.015 Small High. Str LF 99.7 36.5 0.00271 0.00349 0.029 Small High. Str HF 123 45.7 0.00345 0.00435 0.036 Small Fibre LF 108 41.1 0.00352 0.00408 0.032 Small Fibre HF 147 59.3 0.00595 0.00605 0.045

4 Conclusion

Looking forward to the experimental results of the tests on beams, we just give a summary of our numerical simulations by reporting above, in table 1, the peak load and the corresponding parameters for each type of beam. These values will be easy to compare at first glance with the tests carried out at Aalborg University.

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