FRP Reinforcement, According to European Normatives, of Flexural R.C Elements under Shear Force

FRP Reinforcement, According to European Normatives, of Flexural R.C Elements under Shear Force

Igli Kondi

M.Sc. Civil Engineer Polytechnic University of Tirana, Faculty of Civil Engineering, Tirana, Albania

Elfrida Shehu

Ph.D. 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

In this article is presented the reinforcement with fiber polymer of flexural reinforced concrete elements under the action of shear force. The reinforcement from shear force action is needed in those cases when the shear force from external actions exceeds the internal ultimate shear strength of the reinforced concrete members. Ultimate shear strength of r.c elements is a product of the contribution from both concrete and transversal reinforcement steel of the elements. Along the article are presented some reinforcement details of a member both in length and in members cross section. Also, some design formulas of a fiber polymer reinforced members shear strength are presented. Closing this article, to have a better idea how this reinforced member react from shear force action, a solved example of a beam under flexural condition is given.

Keywords-reinforcement, FRP, shear, Eurocode 2

1. Introduction

   Design of a flexural element, for shear force, is an important aspect of the design procedure. For a certain case of loading, the shear strength of the element is product of concrete and transversal steel reinforcement strength. When the element is reinforced with FRP, which is the focus of this article, the shear strength is not only the product of concrete and steel reinforcement, but an important role plays also the FRP reinforcement. A great significance, to FRP shear strength, has the positioning of FRP in the elements length and in the cross section. Following the article, through the figures, some ways of positioning of FRP are presented. Each case of positioning is followed by

   the characteristic of implementation and with the difficulties that each case present.

2. FRP shear-strengthened of an element

   Shear strengthening of a flexural RC element consist on setting FRP strips (action only in one direction) or a FRP mesh (action in both directions). The ideal case will be if the FRP strip follow the principal tensile stress, but is more practical the positioning of FRP strips perpendicular with the elements axis. In this case the strips must be positioned close or distant to each other. Some ways of positioning are given in Figure 1a,b,c and 2.

   Fig. 1a.

   Fig. 1b.

   shear strength of the reinforcement. As is seen in Figure 3, the lateral FRP reinforcement is not preferred.

   Fig. 3.

   Always try to realize U-shape or circumferential reinforcement. Figure 4 shows the positioning of FRP in the element length.

   Fig. 4.

   If a strip FRP reinforcement is used, the distance between two consecutive strips must not be great than 0.8A (see Figure 2), where A is the height of the element cross section, in this case every diagonal crack will be traversed but at least one strip. During site implementation, the FRP could be damaged and in this case is advisable to reduce the FRP modulus of elasticity by divide it with 1.2 for one way FRP, and by

   1.5 for two ways FRP. Also, deformation values during design procedure must be 50% of failure deformation values. This depends from the reinforcement type (cutting, bending, compression or torsion)

3. Shear strength of a FRP reinforced element

   Ultimate shear strength of a FRP reinforced element is given by equation (1):

   VRd min{VRds VRdf , VRd max}

   (1)

   VRds – Design value of shear force sustained by the RC element reinforced with transversal reinforcement, given by equation (2):

   V A z f (ct g ct g) sin

   (2)

   Rds sw ywd s

   Fig. 2.

   The role of the adhesive material in the shear strength is negligible so, only the FRP fibers contribute in the

   VRdmax – design value of shear force which can be sustained by the element, limited by crushing of compression struts, given by equation (3):

   V a bz f

   • ct g ct g

     (3)

     1 led sin

     w

     w

     Rdmax cw 1 cd

     1 ct g2

     fed ffdd 1 3 min0.9d, h

     (9)

     1

     1

     Equation (1), (2) and (3) are according to Eurocode 2. In case of a RC element with rectangular or T cross section, reinforced with circumferential FRP shape,

     ffdd -FRP maximal strength without disconnecting from the element, is given by equation (10):

     VRd,f

     is given by equation (4):

     ffdd

     2 Ef Fd

     t

     (10)

     1

     1

     VRd,f

     0.9 d f

     fed 2 tf

     (ct g ct g) bf

     p

     (4)

     f ,d f

     fd – Coefficient accepted 1.2 – 1.5

     Rd f

     d Cross section effective height

     led – Optimal design anchorage length given by:

     Rd – Partial safety factor

     1

     2 E t

     f – FRP effective design strength

     led min

     f f Fd , 200mm

     (11)

     fed

     Rd fbd 2

     tf – FRP thick

     bf , pf -strips width and strips distance between each

     Ef – FRP modulus of elasticity

     other measured perpendicular with fiber direction. When the fiber are positioned close to each other, without distance between, or the FRP is in a two directional mesh, then bf/pf = 1. For more see Figure 5.

     tf – FRP thick

     Fd – Specific energy of disconnection

     fbd =2 Fd / su – Disconnection tangential stress between FRP and the RC element.

     su =0.25 mm

     Rd 1.25 – Correctional factor.

     Specific energy of disconnection Fd

     equation (11):

     is given by

     kb kG f f

     (12)

     Fd Fs

     cm ctm

     Fig. 5.

     According to Figure 5, in equation (4), pf

     can be

     fcm – Average concrete compressive strength

     fctm – Average concrete tensile strength

     Fs – Safety factor

     replaced by pf sin . In this case pf

     is the distance

     kb – Correctional factor given by equation (13):

     between two strips measured parallel to element axis. Distances between strips must be limited by:

     50mm bf 250mm (5)

     kb

     2 bf / b 1 1 bf / b

     (13)

     bf pf min{0.5d, 3bf, bf + 200mm} (6)

     If min {0.5d, 3bf, bf + 200mm} < bf then a different type of FRP, with other geometry or physical parameters, should be used. In case of circular cross

     To clarify bf and b see Figure 6.

     section RC elements, VRd,f

     is given by equation (7):

     Vrd,f

     1

     Rd

     D f

     t

     ed 2 f

     ctg

     (7)

     In this case, FRP is positioned perpendicular to element axis (=90°) and all the element is cover by FRP positioned close to each other.

     D – Cross section diameter.

     Getting started from equation (1) we can determine the value of maximum shear force supported by FRP:

     VRdf,max = VRdmax VRds (8)

     FRP effective strength on a rectangular cross section reinforced with U-shape FRP is given by equation (9):

     Fig. 6.

     Equation (12) is valid for bf/b0.25. If bf/b<0.25 than kb=1.18 corresponding to bf/b=0.25.

     kG – Correctional coefficient determined experimentally. For precast FRP kG=0.023mm and for FRP prepared in site kG=0.037mm. If an element s strengthened with the help of many FRP strips, each strips with a width bf, kb can be calculated with

     equation (12) accepting b values as the distance b Cross section width

     between to nearby strips axes.

     Asl

     4.62

     0.00331

     (21)

     – Angle between FRP strips and elements axes. See Figure 5.

     hw – See Figure 6.

     For a rectangular section strengthened with circumferential FRP:

     1 b d 30 46.5

     Asl Longitudinal reinforcement area Asl=4.62 cm2

     A

     A

     cp

     cp

     NEd (22)

     c

     1 l sin

     ffed ffdd 1 e

     NEd Axial force acting on the element, compression

     or tensile, in this case NEd=0 so cp = 0 and k1 = 0.15,

     6 min0.9d, hw

     (14)

     so:

     VRd,c = [0.12·1.655·(100·0.00331·25)1/3 +

     1 f f 1

     2 R fd fdd

     le sin

     min0.9d, hw

     0.15·0]·300·465 = 56000N = 5600daN

     Because VEd = 12000daN > VRd,c = 5600daN than the

     R = 0.2 + 1.6rc/b (15)

     0 rc/b 0.5 (16)

     rc – Rounding radius of FRP at elements edges. b Cross section width

     steel transversal reinforcement is needed so 8mm diameter stirrups every 10cm are used. Using equation

     (2) lets calculate VRds:

     V A z f (ct g ct g) sin (2)

     Rds sw ywd s

     On equation (14), the second part is taken in

     consideration only if its positive. In case of circular

     Asw

     lateral reinforcement area A

     sw=2·0.5=1cm2

     cross section with diameter D and FRP covers all the element and =90:

     f E (17)

     s Stirrups distance, s=10cm z=0.9d=0.9·46.5=41.85cm

     fywd Design yield strength of shear reinforcement

     fed f f max

     Ef – FRP modulus of elasticity

     f max – FRP relative deformation accepted 5·10-3.

4. Solved Case

Let study a RC element (beam) under flexural conditions. Shear force acting on the element VEd =

fywd= fyd=4348 daN/cm2. Is recommended that 1 cot

2.5, cot = 1 is accepted. For = 90°, cotg = 0 and sin = 1, so:

VRd,s = 1·41.85·4348·(1+0)·1/10 = 18196daN VEd = 12000daN < VRd,s = 18196daN

Also, we need to control if VEd is less than VRdmax the design value of shear force which can be sustained by the element, limited by crushing of compression struts, given by:

12000daN; rectangular cross section with width

V a b z f

• ct g ct g

(3)

b=30cm and height h=50cm; C25/30 concrete with fcd = 141.7daN/cm2, fctd =12daN/cm2; S-500 steel with fyd =

Rdmax cw 1 cd

1 ct g2

4348daN/cm2; FRP modulus of elasticity Ef = 2350000daN/cm2; As = 4.62cm2; 8mm stirrups every 10cm. Calculate FRP area needed to reinforce the element if the acting shear force is doubled. As on the compression zone is accepted equal to 0 and there is no

inclined steel reinforcement on the element. In the beginning lets calculate VRd,c which is the ultimate design shear force sustained by the element without transversal reinforcement:

Rd,c Rd,c l ck 1 cp

Rd,c Rd,c l ck 1 cp

V C k 100 f 1/3 k b d (18) fck cylindrical strength of concrete fck=25N/mm2 and CRd,c = 0.18/c = 0.18/1.5 = 0.12

When cp = 0 the recommended value for cw is cw=1.

1 = 0.6·(1-fck/250) = 0.6·(1-25/250) = 0.54. So 1 =

0.54 for fck = 25Mpa. Also cotg = tan = 1 is accepted. For = 90°, ctg = 0. So:

VRd,max = 1·30·41.85·0.54·141.7·(1+0)/(1+1) = 48033daN

VEd = 12000daN < VRd,max = 48033daN

The element is safe against shear force. Let suppose that the shear force value is double, so VEd = 24000daN. In this case:

VEd = 24000daN < VRd,s = 18196daN

So the element must be reinforced with FRP because there is a risk of failure. Let suppose that a striped reinforcement, like in Figure (4), is used. To calculate

k 1

200 2

(19)

the shear force sustained by FRP equation (4) is used:

d

d Effective depth of the cross section in mm

VRd,f

0.9 d f

fed 2 tf

(ct g ct g) bf

1

1

p

(4)

k 1

200 1.655 2

465

(20)

Rd f

Rd partial safety factor accepted 1.2 d = 46.5cm

tf = 1.2mm = 0.12cm

ctg = 1

f 240.8

1 200 sin 90 202.4 N

= 90°, ctg = 0. bf = 5cm

Ed 1

3 0.9 465

mm2

pf = 10cm

V 1

0.9 465 202.4 2 1.2 (1 0) 50

To calculate ffed, equations (9) and (10) are used:

Rd,f

1.2 100

f f

1

led sin

(9)

84704N 8470.4daN

w

w

ed fdd

1 3 min0.9d, h

Finally:

VEd = 24000daN < VRd,s + VRd,f = 18196 + 8470 =

1

1

ffdd

2 Ef Fd

t

(10)

26666daN

VEd = 24000daN < VRd,max = 48033daN

f ,d f

f,d coefficient, with value 1.2 1.5. Is accepted 1.35.

kb kG f f (12)

VRdf,max = VRdmax VRds = 48033 18196 = 29837daN

> VRd,f = 8470daN

All conditions are fulfilled and the element is safe

Fd Fs

cm ctm

against shear force.

fcm – Average concrete compressive strength

fcm =33N/mm2

5. Conclusion

fctm – Average concrete tensile strength N/mm2

Fs – Safety factor Fs =1.5

fctm =2.6

The use of FRP is a safe method for reinforcing flexural elements against the shear force. Till now, the basis of FRP positioning in element length and in its

kb – Correctional factor given by equation (13):

2 bf / b

cross section together with some design formulas given above, are well known. As was seen by the solved case,

kb

1

1 bf / b

(13)

FRP reinforcement plays an important role on reinforcing the element against shear force, increasing

Because bf/b = 5/30 = 0.166 < 0.25 than bf/b = 0.25 is accepted to calculate kb, so:

k 2 0.25 1.183 1

b 1 0.25

kG – Correctional coefficient determined experimentally. For precast FRP kG=0.023mm and for FRP prepared in site kG=0.037mm.

significantly the element strength. With all that, FRP is a relatively new material in material construction field and for these reason there is a great need for studies and experiments to fully understand this material in order that the structural engineer make use of them, not only in separate members but in all structure.

1. References

   1.183 0.037

   33 2.6 0.270 N

   Fd 1.5 mm

   1 2 235000 0.270 N

   1. Eurocode 2. November 2005.

   2. CNR (COMMISSIONE NAZIONALE DELLE

      f fdd 1.35 1.2

      240.8

      mm2

      RICERCHE) Istruzioni per la progettazione, lesecuzione ed il controllo di interventi di

      1

      led min

      2 E t

      f f Fd , 200mm

      consolidamento statico mediante lutilizzo di compositi fibrorinforzati. Marzo 2012.

      Rd fbd 2

      Rd 1.25 – Correctional factor.

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

      fbd =2 Fd / su – Disconnection tangential stress between FRP and the RC element.

      su =0.25 mm.

      fbd = 2·0.270/0.25 = 2.16N/mm2

      1 3.1412 235000 1.2 0.270

      polymers). 1999.

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

   5. AICAP Guida alluso dellEurocodice 2. 2006.

   6. BARBATO S. Adeguamento sismico di edifici in c.a. mediante l'impiego di materiali compositi fibrorinforzati

(FRP). 003.

led min 1.25 2.16 2

min{227mm, 200mm}

led=200 mm is accepted.

, 200mm