 Open Access
 Total Downloads : 398
 Authors : Igli Kondi, Julian Kasharaj, Elvis Capo
 Paper ID : IJERTV2IS90710
 Volume & Issue : Volume 02, Issue 09 (September 2013)
 Published (First Online): 23092013
 ISSN (Online) : 22780181
 Publisher Name : IJERT
 License: This work is licensed under a Creative Commons Attribution 4.0 International License
Design of FRP (Fiber Reinforced Polymer) for Bending Moment, According to European Normatives, for Reinforcement of Flexural Reinforced Concrete Elements
Igli Kondi
M.Sc. Civil Engineer Polytechnic University of Tirana, Faculty of Civil Engineering, Tirana, Albania
Julian Kasharaj M.Sc. Civil Engineer
Polytechnic University of Tirana, Faculty of Civil Engineering, Tirana, Albania
Elvis Capo
Civil Engineer Polytechnic University of Tirana, Faculty of Civil Engineering, Tirana, Albania
Abstract
Reinforcement is the process of strengthening existing structural members in order to make them, or all the structure, resistant to a new system of action that was not taken in consideration during the design period. This article is focused on FRP method of reinforcement. As we know, nearly all the members on an engineering object are subject of flexural condition. Taking in consideration the importance of bending moment in the structure, this article bring a design procedure of FRP for bending moment, as a reinforcement method for a RC member. This design procedure is according the European Normatives and all the approaches are made using the ultimate limit state design.
Keywordsreinforcement, FRP, bending moment, Eurocode 2

Introduction
In order to design FRP for bending moment, at first some assumption about how the elements react under flexural condition needs to be made. To fully understand the behaviour of a RC member reinforced with FRP the ways of failure of this member must be determined. Along this article, a stressstrain relationship of RC member under the bending moment action is given. With the use of equilibrium equations written for the element cross section the design formula for FRP design are founded. At the end of the article, to create a better idea of FRP design for bending moment, a solved case of a beam under flexural condition is provided.

Basic Assumption
Some of the basic assumptions taken in consideration during design procedure are:

concrete works only in compression zone

plain section before deformation remain plain even after deformation

Hooks law take place

FRP are placed in tensile zone

FRP are perfectly attached to concrete

The material used for attaching the FRP to concrete doesnt affect the material behaviour
– FRP behaviour is the same as reinforcing bar

Existing tensile steel bars area could or could not be taken in consideration

Deformation from shear may be not taken in consideration.

The effect from previous loading before reinforcement should be taken in consideration.


Types of failure
Types of failure of a RC member reinforced with FRP under bending moment can be categorized in two groups

Failure because of destruction of one of the components
This type of failure comes from the crush of concrete in compressed zone (compressive stresses exceed the compressive strength), rebar in tensile zone breaks (tensile stresses on rebar exceed the tensile strength) or FRP breaks (tensile stresses on FRP exceed the tensile strength). Those types are called the classic type of failure.

Premature types of failure
Premature failure is also called peelingof and concreteripping. In this case there is no components destruction but the failure comes from the separation of FRP from concrete surface. Failure because the separation of FRP from concrete or separation of a piece of concrete together with FRP from the element must be studied carefully because a perfect connection between FRP and concrete could make possible common work of FRP and concrete. The disconnection might not happen in the entire length of element, but only in a certain area. In this case the beam may not fail, but is certain that its strength is decreased. When the disconnection happens in entire length of the element, we could say that the reinforcement has failed. FRP doesnt support any action and the element without reinforcement fails. This type of failure is called peelingof and in this case the failure is sudden, without visible signs, so the failure is not ductile but fragile. FRP disconnection from concrete may happen because the contact surface isnt ready for the implementation of FRP, poor ability of adhesive used and especially low adhesive resistance in high temperatures. A premature failure that happens because of the loss of links that leads to disconnection of concrete form FRP and longitudinal reinforcing is called concrete ripping. This happens because the tensile and shear strength of adhesive used is higher than concrete. In this case concrete cracking begins from FRP and progresses with 45Â° angle inclination till the longitudinal tensile rebar. Another type of failure happens when FRP components fiber and resin split from each other. This type of failure is particular for high strength concrete. In conclusion we can say that those types of failures leads to the failure of the flexural element (beam) for loads much smaller compared with the classical types of failure.


Ultimate state design analyse of classic types of failure.
Which are the classic ways of failure?

Concrete fail in compression zone. Reinforcing steel has not reached the yield strength.

Reinforcing steel fail in tensile zone. In this case the reinforcing area is not enough to resist the tensile stresses from bending moment. This is the most typically case of usage of FRP for reinforcement of the flexural RC members from bending moment.

Simultaneous fail of both concrete and reinforcing steel.
To avoid the destruction of concrete compressive zone, the FRP area must be smaller than a certain value Afmax. EC 2 recommends show that maximum height (x) of the compression zone must be less than 0.45d for concrete with fck35N/mm2 and less than 0.35d for
concrete with fck>35N/mm2. This restriction makes the
cross section to show a ductile behaviour and reinforcement steel has reached the yield strength. So the section could rotate significantly. With increasing of concrete strength its ductility decreases. That why for concrete strength fck>35N/mm2 the maximum height of compressive zone is less than 0.35d. Figure 1 show stresses distribution of a cross section for a flexural element with rectangular section, maximal height of compression zone is accepted x=0.45d.
Fig 1.
As rebar area in tensile zone
As rebar area in compressive zone Af FRP cross section area
fcd concrete design compressive strength M bending moment from external load 's = EsÂ·s stress on As
Es modulus of elasticity of As (Es = Es)
Es modulus of elasticity of As s relative deformation of As
Ac = 0.8Â·xÂ·b concrete compressive zone x compression zone height
fyd design tensile strength of reinforcement Ef FRP modulus of elasticity
f relative deformation of FRP b cross section width
h cross section height
d effective depth of cross section d concrete cover
Figure 2 shows cross section strains condition (see line nr. 3) corresponding to Figure 1 stress condition.
h x 0.35% 50 20.92 0.35% 0.486%
f x 20.92
Fig.2.
Writing the equilibrium equation for forces on x direction:
AsÂ·s – AcÂ·fcd + AsÂ·fyd + AfÂ·EfÂ·f = 0 (1)
On equation (1) are known As, As, fcd, fyd = fyd, b, h, s = Es, d, d and also we know the FRP type used so Ef is known.
s = EsÂ·s (2)
From triangle identities of deformed shape of the cross section:
Finally: Afmax = (AsÂ·s + AcÂ·fcd AsÂ·fyd) / (EfÂ·f) = (6.03Â·4348 + 502.08Â·141.7 12.56Â·4348) / (2350000Â·0.00486) = 3.74cm2
Failure type with the breaking of steel reinforcement is more acceptable because is more ductile. If FRP area is smaller from what is needed the element will fail. According to Eurocodes, the ultimate deformation for reinforcing steel is accepted 1%, but this value is very conventional because, in reality, reinforcing steel deformations are grate than 1%. Most FRP have deformation less than 1% and according to this, we can suppose, if the element fail from tensile stress, the failure comes from FRP destruction (breaking). To design FRP from bending moment we use figure 3. Solving the equilibrium equation of forces on horizontal axes:
0.8Â·xÂ·bÂ·fcd AsÂ·fyd + AsÂ·fyd + AfÂ·EfÂ·f = 0 (6)
This equation is true if s yd and s yd. Only if this condition is fulfilled we can write say that s = s
s
s
x x d ' '
x d ' 0.35%
(3)
= fyd = fyd. So:
0.35%
's x
'
x d'
f'yd
'
(7)
On equation (3) x = 0.45d and c = 0.35% is the ultimate concrete deformation. Solving equation (3) for s we can calculate s with the help of equation (2). If
And:
s c x s
yd yd
s yd than s = fyd = fyd (yd = fyd / Es). Using again the triangle identites for the deformed shape of cross section:
f c
h x x 0
(8)
f
f
x h x h x 0.35% (4)
0.35% f x
In this case we can write: Af=(AsÂ·fyd+AcÂ·fcdAsÂ·fyd)/(EfÂ·f)=Afmax (5)
With the help of equation (5) we can calculate the FRP
area for maximal height of compressed area accepted 0.45d or 0.35d according to EC 2 recommendations. If FRP area is greater than the area calculated with equation (5), is the risk of element failure because the crush of compressed zone. Let suppose having a beam with b = 30cm and h = 50cm, d = 3.5cm, d = 46.5cm so x = 0.45d = 0.45Â·46.5 = 20.92cm. Concrete C25/30 with fcd = 141.7daN/cm2 and S500 steel with fyd =
4348daN/cm2, Es = 2100000daN/cm2. As = 316 =
3Â·2.01 = 6.03cm2. As = 420 = 4Â·3.14 = 12.56cm2. Ef =
2350000daN/cm2.
Fig. 3.
The design should take in consideration that during the reinforcement procedure the elements extreme tensile fiber have an initially deformation 0. FRP deformation is caused from the additional loading. This is the reason because on equation (7) from f is subtracted 0. For a better understanding, see figure 4.
' x d ' 0.35% 20.92 3.5 0.35% 0.291%
s x 20.92
And yd = fyd/Es = 4348/2100000 = 0.207%. Because s = 0.291% > yd = 0.207% than s = fyd = fyd = 4348daN/cm2 and also Ac = 0.8Â·xÂ·b = 0.8Â·20.92Â·30 = 502.08cm2 so:
Fig. 4.
The ultimate design bending moment is calculated with the help of the second equilibrium equation which is a
sum of moments about the axes passing on centre of gravity of concrete compressed zone:
MRd Asfyd(d 0.4x) Af Ef f (h 0.4x) A's Es's(0.4x d')
(9)
equilibrated. Graphically this is described on figure 5(the pink line):
Supposing that s yd:
MRd Asfyd(d 0.4x) Af Ef f (h 0.4x) A's fyd (0.4x d')
(10)
And also:
h x (11)
f c x
0 f , lim
f,lim must not exceed 50% of initially FRP deformation (fd) and also must not exceed 5 time the yield deformation of reinforcing steel (yd).


Solved case
A flexural RC beam is studied. Bending moment form external loading MEd = 15000daNm; rectangular cross section with width b=30cm and height h=50 cm, concrete cover d=3.5cm, effective depth of cross section d=hd=503.5=46.5cm; C25/30 concrete with
fcd=141.7daN/cm2; S500 reinforcing steel with fyd=4348daN/cm2, Es = 2100000daN/cm2, yd = fyd/Es = 4348/2100000 = 0.207%; FRP modulus of elasticity Ef
= 2350000daN/cm2, ultimate design deformation fd =
1.8%. Calculate the FRP area for reinforcing the element for a new external loads moment MEd = 20000daNm. Initially let calculate the reinforcing steel area needed to resist a bending moment of MEd = 15000daNm:
= MEd/(bÂ·d2Â·fcd) = 1500000/(30Â·46.52Â·141.7) = 0.163
On equation:
Fig. 5.
The compressed zone height: x=0.259d=0.259Â·46.5 = 12.04cm. Since x = 10.36cm < 0.259d = 12.04cm then we can say that we are in the second case (see figure 5). In a more detailed way, adapted to our case, the deformed shape and stresses distribution are given on figure 6 and figure 7.
Fig. 6.
0.8 x (1 0.4 x )
d d
(12)
if x/d=0.45 is replaces than max is calculated: max = 0.8Â·0.45Â·(10.4Â·0.45) = 0.295 (13)
Because = 0.163 < max = 0.295 there is no need for reinforcing on compressed zone, so As = 0. However 2 diameter 16 mm bars are used as reinforcing on the compressed zone.
= 1.251.25Â·(12)1/2 = 1.251.25Â·(12Â·0.163)1/2 =
Fig. 7.
Some of the values given on figure 6 and figure 7 can be found as following. So, to calculate the concrete deformation c according to the deformed shape of the
So:
0.223 (14)
section we have:
x d x
As = 0.8Â·Â·dÂ·bÂ·fcd/fyd = 0.8Â·0.223Â·46.5Â·30Â·141.7/4348 =
(14)
8.11cm2

diameter 16mm bars are used with an area of As
c s
Solving equation (14) for c:
=4Â·2.01 = 8.04cm2 (0.86%).
x s
10.36 0.01
From = x/d = 0.223 we can find the compressed zone height: x = Â·d = 0.223Â·46.5 = 10.36cm
According to EC 2, if a 0.35% deformation of the compressed zone and 0.35% deformation of reinforcing steel are accepted than the cross section is balanced or
c d x 46.5 10.36 0.00286 0.286% Because c = 0.286% > 0.2% than c = fcd. Relative deformation of bottom extreme fiber, which is in the same time the initially deformation 0 is:
d x h x
(15)
design area of FRP in order that the reinforced element
s 0
(h x)
(h x)
Solving equation (15) for 0:
s (50 10.36)*0.01 0.0109 1.09% 0 d x 46.5 10.36
Before the application of FRP, the bottom part of the cross section will be relived so a part of the deformation will disappear, this deformation decrease is accepted 30% of the initially deformation. In this case the bottom extreme fiber of the cross section, also 0, the effective deformation is calculated:
0 (effective) = 0.7Â·0 = 0.7Â·1.09% = 0.763%
Let suppose that for different reason the bending moment increase to a new value of MEd = 20000daNm. For this new condition we need to calculate the FRP are in order that the element must not fail. To solve this problem we refer to figure 3. Equation (6) and (9) or
(10) are used. On equation (6) x, Af, and f are unknown. With the help of equation (11) we can calculate FRP deformation (f) function of x. So, now, there are to unknowns, x and Af. On equation (10), if MEd = MRd, there are also two unknowns, x and Af. Solving those two equations with two unknown variable we found:
x = 13.97cm dhe Af = 3.83cm2
In reality, we use another method. A certain value of FRP area is accepted and the values of x, f and MRd are calculated. If MRd> MEd the problem is solved, but if MRd< MEd than a bigger value of FRP area is accepted and the calculations are repeated. Equations (6) and (9) are true only if s yd. To calculate s equation (15) is used.
s = 0.0035Â·(46.513.97)/13.97 = 0.814% yd = 0.207%
Checking if FRP ultimate deformation passes the equation (11) condition:
flim = 0.5Â·fd = 0.5Â·1.8% = 0.9% flim = 5Â·yd = 5Â·0.207% = 1035%
flim = 0.9% is accepted. From equation (11) we calculate f = 0.139% < flim = 0.9%, the condition is fulfilled.


Conclusion
This article provided a design approach of FRP as a reinforcing material for flexural RC elements. All the design process was done by following some steps given along the material. The focus of this material was to give a better understanding of FRP behavior as a reinforcing material and to give e clearly stressstrain relationship of the reinforced element. All the design formulas are used from Eurocode 2 to calculate the
must resist a new load case. At the end of the article, a solved numerical case was given. This solved case gives a practical way of calculating the necessary FRP area for reinforcing.

References

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CNR (COMMISSIONE NAZIONALE DELLE RICERCHE) Istruzioni per la progettazione, lesecuzione ed il controllo di interventi di consolidamento statico mediante lutilizzo di compositi fibrorinforzati. Marzo 2012.

TRIANTAFILLOU T.C Guidelines for the dimensioning of reinforced concrete elements strengthened with CFRP (carbon fibre reinforced polymers). 1999.

RUBINO C., PINI D., IANELLI P. Rinforzo di strutture in cemento armato. 2005.

AICAP Guida alluso dellEurocodice 2. 2006.

BARBATO S. Adeguamento sismico di edifici in c.a. mediante l'impiego di materiali compositi fibrorinforzati (FRP). 2003.