Concrete flexural members reinforced with fiber reinforced polymer ...

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Concrete flexural members reinforced with fiber reinforced polymer: design for cracking and deformability John Newhook, Amin Ghali, and Gamil Tadros

Abstract: Fiber reinforced polymer (FRP) bars have lower modulus of elasticity than steel bars. For this reason when FRP bars are used as flexural nonprestressed reinforcement in concrete sections, the stress in the FRP is limited to a relatively small fraction of its tensile strength. This limit, necessary to control width of cracks at service, governs design of the required cross-sectional area of the FRP. Parametric studies on rectangular and T-sections are presented to show that the design based on allowable strain in the FRP results in sections that exhibit large deformation before failure. The concept of deformability, given in the Canadian Highway Bridge Design Code, as a requirement in the design of sections is discussed and modifications suggested. Using the new definition, it is shown that when, in addition to the crack control requirement, an upper limit is imposed on the cross-sectional area of the FRP, no calculations will be necessary to check the deformability. Key words: fibre reinforced polymer, reinforcement, concrete, design, deformability. Résumé : Les barres en polymères renforcés de fibres (FRP : « fiber reinforced polymer ») ont un module d’élasticité plus bas que celui de barres d’acier. Pour cette raison, lorsque des barres en FRP sont employées en tant que renforcement non précontraint en flexion dans des sections en béton, il est nécessaire de limiter la contrainte dans le FRP à une fraction relativement petite de sa résistance en tension. Le respect de cette limite, nécessaire au contrôle de la largeur des fissures durant le service, détermine quelle est l’aire de la section transversale du renforcement en FRP qui est requise. Des études paramétriques sur des sections rectangulaires et en T sont présentées afin de montrer que la conception basée sur le respect de la tension limite permise dans les FRP résulte en des sections qui présentent de larges déformations avant la rupture. Le concept de l’état de déformation, donné dans le Code canadien sur le calcul des ponts routiers en tant qu’exigence dans la conception de sections, est discuté et des modifications sont suggérées. En utilisant la nouvelle définition, il est montré que, si une limite supérieure est imposée sur l’aire de la section transversale de FRP en plus de l’exigence du contrôle des fissures, alors il n’est pas nécessaire de procéder à des calculs afin de vérifier l’état de déformation. Mots clés : polymères renforcés de fibres, renforcement, béton, conception, état de déformation. [Traduit par la Rédaction]

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Introduction The design of concrete sections in flexure that are reinforced with fibre reinforced polymers (FRP) is different from that of sections reinforced with steel because of the difference in mechanical properties of FRP and steel. Generally, the FRP bars used as reinforcement in concrete have Received 1 May 2001. Revised manuscript accepted 22 November 2001. Published on the NRC Research Press Web site at http://cjce.nrc.ca on 10 February 2002. J. Newhook. Department of Civil Engineering, Dalhousie University, 1360 Barrington Street, Halifax, NS B3H 3J5, Canada. A. Ghali.1 Department of Civil Engineering, The University of Calgary, 2500 University Drive N.W., Calgary, AB T2N 1N4, Canada. G. Tadros. ISIS Canada, 43 Schiller Crescent N.W., Calgary, AB T3L 1W7, Canada. Written discussion of this article is welcomed and will be received by the Editor until 30 June 2002. 1

Corresponding author (e-mail: [email protected]).

Can. J. Civ. Eng. 29: 125–134 (2002)

I:\cjce\cjce_29\cjce-01\L01-085.vp Tuesday, February 05, 2002 11:24:38 AM

tensile strength varying between 500 and 2200 MPa and modulus of elasticity varying between 40 and 150 GPa. The stress–strain relationship for FRP is linear up to rupture when the ultimate strength is reached. Unlike steel reinforcing bars, FRP bars do not undergo yield deformation or strain hardening before rupture. For this reason, design of sections in flexure has been based upon consideration of ultimate strength, serviceability, and deformability. The purpose of design for deformability is to ensure that failure of a section in flexure occurs only after development of sufficiently large curvature. Because of the relatively low modulus of elasticity of FRP bars compared with that of steel reinforcing bars, it is necessary to limit the stress in FRP bars in service to a relatively small fraction of their strength in order to control crack width. This paper demonstrates that this serviceability requirement controls the design, without the need in practice for special calculations to verify the deformability as specified in Section 16 of the Canadian Highway Bridge Design Code (CHBDC) (CSA 2000). The importance that sufficient deformation occurs before failure of nonprestressed members reinforced with FRP has

DOI: 10.1139/L01-085

© 2002 NRC Canada


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Can. J. Civ. Eng. Vol. 29, 2002 Table 1. Typical properties of reinforcing bars. Reinforcement

Ef or Es (GPa)

ffu or fy (MPa)

Glass (GFRP) Aramid (AFRP) Carbon (CFRP) Steel

40 80 150 200

550 1200 2000 400

been discussed by Jaeger et al. (1997), Kakizawa et al. (1993), and Gangarao and Vijay (1997). The deformability of members prestressed with FRP has been discussed by Naaman and Jeong (1995), Abdelrahman et al. (1997), and Grace and Sayed (1997). The present paper is concerned only with internally reinforced, nonprestressed members.

Reinforcement properties Three types of FRP bars used as reinforcement for concrete are considered in this paper: glass, aramid, and carbon (GFRP, AFRP, and CFRP). A linear elastic stress–strain relationship will be assumed for the three types of FRP in tension, with rupture of the FRP occurring when ultimate strength is reached. FRP bars are weak in compression and thus the presence of FRP bars in the compression zone of a concrete section in flexure is ignored. The commonly used bilinear stress–strain relationship is assumed for steel reinforcement with a linear elastic portion up to yield followed by an extensive perfectly plastic portion. Table 1 lists the modulus of elasticity, Ef, and the tensile strength, ffu, for each of the three most common types of FRP together with the modulus of elasticity, Es, and nominal yield strength, fy, for steel bars. Flexural strength Failure of a reinforced concrete section in flexure can be caused by rupture of FRP or yielding of steel reinforcement in tension or by crushing of the concrete. Research (Gangaroa and Vijay 1997; Jaeger et al. 1997) has established that the ultimate flexural strengths for both failure types can be calculated using the same equations when the reinforcement is steel or FRP bars. This approach is adopted in design guidelines by both the American Concrete Institute (ACI 2001) and ISIS Canada (2001) and in the CHBDC design criteria (CSA 2000). These equations are reviewed here and used in the discussions in this paper. The actual deformation and flexural strength of sections is investigated in this paper; therefore, the material resistance and strength reduction factors required by codes are not included in the equations. Failure of tension reinforcement Figure 1a shows the stress and strain distributions in a section at failure by rupture of the FRP reinforcement. Such a section is said to be under-reinforced. The theory for sections under-reinforced with steel bars is well documented in textbooks. Initially the steel yields and the curvature increases rapidly until the strain in concrete, εc, at the extreme compressive fibre reaches an ultimate value εcu = 0.0035 and

thus failure occurs. The stress in concrete is idealized by the rectangular stress block shown in Fig. 1b. However, when a section is under-reinforced with FRP, no yielding occurs and the failure is caused by rupture of the FRP. The strain in the reinforcement will be εfu = ffu/Ef, where ffu and Ef are the tensile strength and the modulus of elasticity of the FRP. The corresponding strain εc at the extreme compressive fibre will be less than εcu. Thus, the distribution of compressive stress on the concrete cannot be idealized by the traditional rectangular block in Fig. 1b. One way to determine the distribution of stress on the concrete is to use a stress–strain relationship such as the one proposed by Todeschini et al. (1964) and adjusted by MacGregor (1997) (Fig. 2): [1]

fc = 1.8 fc′

ε /ε 0 1 + ( ε / ε 0) 2

where ε0 = 1.71 fc′ /Ec; fc′ is the specified compressive strength of concrete; Ec is the modulus of elasticity of concrete, which may be taken as 4750 fc′ MPa (Todeschini et al. 1964); fc is the magnitude of the concrete stress corresponding to a strain ε at any fibre. The assumption that plane cross section remains plane after deformation leads to the linear strain distribution shown in Fig. 1a, from which [2]

c εc = d ε c + ε fu

where c is the depth of the compression zone and d is the distance from the extreme compressive fibre to the tension reinforcement. The absolute value of the resultant tensile force, T, in the reinforcement and the resultant compression force in the concrete, C, are equal. Thus, [3]

Af ffu = ∫ fc da

where da is an elemental area of the compressive zone and Af is the cross-sectional area of the tension reinforcement. Equations [1]–[3] can be used to determine the strain εc at the extreme compressive fibre, the stress distribution, and hence the location of the resultant compressive force and yCT (using numerical integration). The nominal moment resistance of the under-reinforced section and the corresponding curvature are [4]

M u = Af ffu y CT

[5]

ψ u = (ε c + ε fu )/d

where yCT is the distance between the resultant concrete compressive force and the tension reinforcement. Failure by crushing of concrete When the flexural failure is induced by crushing of concrete, without rupture of FRP or yielding of steel, the section is said to be over-reinforced. For a T-section to be overreinforced, it must have a large amount of reinforcement, which is considered impractical. Figure 1b shows the strain and stress distribution at ultimate for an over-reinforced rectangular section. The nonlinear distribution of concrete in the compression zone is replaced by an equivalent uniform stress over a part of the © 2002 NRC Canada



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Fig. 1. Stress–strain distributions in flexure: (a) failure by rupture of FRP; (b) failure by crushing of concrete; and (c) service condition with strain in FRP equal to ε fs.

compression zone such as is defined in Fig. 1b in accordance with Standard A23.3-94 (CSA 1994). The nominal moment resistance of over-reinforced sections is given by [6]

M u = Af ff y CT

[7]

y CT = d − β1c / 2

Fig. 2. Stress–strain relationship adopted in analysis of underreinforced sections (MacGregor 1997).

The stress in the reinforcement at failure, ff, which has a value smaller than ffu (or smaller than the yield stress when steel is used), is given by [8]

1/ 2   β f′  ff = 0.5 Ef ε cu  1 + 4α1 1 c  − 1   ρf Ef ε cu   

where εcu is the strain in concrete at the extreme compression fibre and ρf = Af/bd is the reinforcement ratio. Equation [8] can be derived by equating the resultants of tensile and compressive stresses and using eqs. [6] and [7]. According to A23.3-94 (CSA 1994), failure by concrete crushing is considered to have occurred when εcu = 0.0035 and the values of α1 and β1 are empirically taken equal to

[9]

α1 = 0.85 − 0.0015 fc′ ≥ 0.67 β1 = 0.97 − 0.0025 fc′ ≥ 0.67

The curvature at ultimate is given by © 2002 NRC Canada



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Can. J. Civ. Eng. Vol. 29, 2002 Table 2. Balanced and minimum reinforcement ratios for FRP reinforced rectangular sections. Balanced reinforcement ratio (%)

[10]

ψu =

Reinforcement

fc′ = 30 MPa

fc′ = 60 MPa

fc′ = 30 MPa

fc′ = 60 MPa

Glass (GFRP) Aramid (AFRP) Carbon (CFRP)

0.80 0.34 0.23

1.38 0.59 0.39

0.42 0.19 0.11

0.59 0.27 0.16

ε c + ( ff / Ef ) d

with εc = 0.0035. For T-sections having c > hf (where hf is the depth to flange), eqs. [7] and [8] need to be adjusted (McGregor 1997). Balanced condition The reinforcement ratio, ρfb, for which failure of reinforcement and crushing of concrete occur simultaneously is referred to as the balanced reinforcement ratio. For a rectangular section, ρfb can be determined by the equation [11]

c   f′  ρfb = α1β1  b   c   d   ffu 

with cb Ef ε cu = d Ef ε cu + ffu The values of ρfb for various reinforcement types are given in Table 2.

Minimum reinforcement ratio for Mu to be greater than the cracking moment Cracking occurs when the stress in concrete at the extreme tension fibre reaches the tensile strength of concrete in tension, fr; the moment value, Mcr, at which cracking occurs is [12]

M cr = fr

I yt

where I is the second moment of area of the uncracked transformed section about its centroid axis. The transformed section is composed of the areas of concrete plus (Ef/Ec)Af. yt is the distance between the centroid of the transformed section and the extreme tension fibre. After cracking, the tension is assumed to be resisted entirely by the reinforcement. To avoid sudden failure by rupture or yielding of the reinforcement, its amount must be sufficient to have a moment of resistance, Mu (calculated by eq. [4]), greater than Mcr by an appropriate “safety factor”. Thus, [13]

ρ=

Minimum reinforcement ratio (%)

Af f I  1  > r   (safety factor) bd ffu y CTy t  bd 

where yCT is given by eq. [14] or eq. [18].

Many design codes such as CSA A23.3-94 (1994) and ACI 318-99 (1999) transform eq. [13] to empirical expressions for the minimum steel reinforcement, Afmin , permitted in rectangular and in T-sections, in terms of fc′ divided by the specified yield strength. For simplicity, the moment of inertia of gross concrete section about its centroidal axis is used in lieu of I in eq. [13]. The minimum ratio of FRP, ρfmin, has to be greater than the value given by eq. [13], with a safety factor relevant for this material. The appropriate value to be used is not within the scope of this paper; this should be given in codes on design of concrete structures reinforced with FRP. Table 2 gives ρfmin for a rectangular section for the three types of reinforcement in Table 1 and two concrete strengths. The values in the table are equal to ρfmin = 5 fc′ /(12ffu), in accordance with the guidelines of American Concrete Committee 440 (ACI 2001). These are approximately equal to the values that can be calculated by eq. [13], assuming fr = 0.6 fc′ , d = 0.9 (height of section), and yCT 0.9d, with a factor of safety = 3.

Crack width control FRP bars have higher strength, ffu, than the specified yield strength fy of steel reinforcing bars commonly used in North America. However, because the modulus of elasticity of FRP bars, Ef, is lower than that for steel, Es, the higher strength cannot be effectively exploited. This is because of the need to control the width of flexural cracks in the tension zone. The main parameters that influence crack width include the crack spacing, the quality of bond between the concrete and reinforcing bars, and, above all, the strain in the reinforcement. Through the use of a crack control parameter, CSA A23.3-M94 (1994) implicitly limits the maximum crack width for flexural members with steel reinforcement to 0.40 and 0.33 mm respectively for exterior and interior exposure. ACI 318-99 (1999) also limits crack widths to 0.40 mm for all exposure conditions. In design, CSA A23.3-M94 requires that the nominal moment multiplied by a material resistance factor, 0.85 for steel reinforcement, be equal to or greater than the applied service moment multiplied by an appropriate combination of dead load and live load factors, 1.25 and 1.5 respectively. Taking an average load factor of 1.38, the service stress in the steel is approximately (0.85/1.38) times the yield stress, or 0.6fy. CSA A23.3-M94 permits the use of this 60% value directly in the calculation of crack control parameters. Similarly, ACI 318-99 allows the use of 0.6fy as the permissible reinforcement stress in calculations related to crack control. When fy = 400 MPa, the corresponding reinforcement strain is 1200 × 10–6. An advantage of FRP bars over steel reinforcement is that there is no risk of corrosion. Thus, CHBDC (CSA 2000) rec© 2002 NRC Canada



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Table 3. Reported crack widths (Kobayashi et al. 1997).

Amount of reinforcement required to control cracking

Specimen

Crack width (mm)

ρf (%)

C C C C C C

0.50 0.50 0.60 0.55 0.60 0.70

0.40 0.28 0.23 0.23 0.19 0.16

– – – – – –

120 160 190 190 225 260

ommends limiting the crack width to 0.71 and 0.50 mm for interior and exterior exposure, respectively, while JSCE (1997) recommends maximum crack width of 0.50 mm for both interior and exterior exposure. The limits adopted by ACI 440 (ACI 2001) are similar to those of CHBDC. From this it is seen that the width of cracks allowed for FRP reinforced members is 1.8 or 1.5 times the value allowed for steel-reinforced members. In the following discussion, it is assumed that this ratio between allowable crack widths is 5/3. Furthermore, it is assumed that the width of cracks is approximately proportional to the strain in the reinforcement. Thus, the corresponding permissible strain in the FRP reinforcement at service is 5/3 × 1200 × 10–6, or 2000 × 10–6. For a specified strain in the reinforcement, the width of cracks can vary substantially from member to member, depending on parameters such as duration or repetition of loading, shape and dimensions of the cross section and cover. Even identical members may exhibit different crack widths for the same load. In experiments (Kobayashi et al. 1997) on beams in flexure of cross section 400 mm wide and depths 120–260 mm, reinforced with carbon FRP, the crack width listed in Table 3 have been observed when the strain in the reinforcement reached 2000 × 10–6. The table also includes the reinforcement ratio ρf. The values listed above indicate that for this test series, the crack width is 0.5–0.7 mm. This test series shows crack widths not substantially different from 0.7 mm, which is anticipated when εf = 2000 × 10–6. However, in other tests more difference can occur between the observed and the anticipated crack width. The parametric study described below, therefore, examines the effect of choosing either a higher or a lower value of permissible strain associated with the acceptable crack width. For a permissible strain of 2000 × 10–6, the corresponding stresses at service, ffs, are 80, 160, and 300 MPa for GFRP, AFRP, and CFRP, respectively. With these relatively low stresses, it will be shown that the deformability requirements will be satisfied. In other words, when the design does not allow the stress in service in FRP bars to exceed these values, no special calculations are necessary to verify that the requirements for deformability are satisfied. It is noted that certain types of FRP, particularly glass FRP, require that service strains in the FRP be limited to a fraction of ultimate strain to avoid problems such as creep rupture, static fatigue, and alkali deterioration (ACI 2001; Benmokrane and Rahman 1998). The allowable limits suggested above are more stringent than those service strain limits.

To ensure that the stress in the FRP does not exceed the allowable limit for crack control, the cross-sectional area of the bars should be determined by [14]

Af = Ms/(ffs yCT)

where yCT = d – c/3, with c being the depth of the compression zone (Fig. 1c). In service, the stress in concrete is commonly sufficiently low to justify the use of linear stress– strain relationship for concrete. Thus the stress in concrete varies linearly over the depth of the compression zone (Fig. 1c). With these assumptions, the neutral axis passes through the centroid of area of the transformed section composed of the area of the concrete in compression and nAf, where n = Ef/Ec. The distance, c, between the extreme compression fibre and the neutral axis is given by the solution of the quadratic equation (Ghali et al. 2002): [15]

c2 + a1c + a2 = 0

[16]

c=

1 冸 −a1 + a12 − 4a 2 冹 2

where a1 =

2hf 2nAf (b − bw) + bw bw

a2 =

hf2 2nAf d (b − bw) − bw bw

[17]

The dimensions b, bw, d, and df are defined in Fig. 1c. When c < hf, set bw = b in eq. [17]. In this case, the resultant compressive force is located at a distance c/3 from the extreme compressive fibre and yCT is given by eq. [14]. The same equation may also be used when c > hf to give an approximate value of yCT. More accurately, when c > hf, the distance between the resultant compressive and tensile forces is given by [18]

y CT = d −

c  bc2 − (b − bw) (c − hf ) 2 (c + 2hf )/ c    3 bc2 − (b − bw) (c − hf ) 2 

For a rectangular or T-section with a given value of the reinforcement ratio ρf (= Af/bd) and the strain εfs, the corresponding values of the moment at service Ms and curvature ψs are given by [19]

Ms = ρf bdεfsEf yCT

[20]

ψs =

ε fs d−c

Deformability requirements Failure in flexure of concrete sections reinforced by steel bars is accompanied by large curvature. The relative amount of curvature is largest when the failure is by yielding of the reinforcement in tension (under-reinforced sections) and reduces markedly when the failure is by crushing of the concrete (over-reinforced sections). For this reason, design © 2002 NRC Canada



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Fig. 3. Deformability factors, DF, for steel-reinforced sections.

codes have generally recommended that flexural members be designed as under-reinforced, or that the reinforcement ratio be significantly less than ρb. However, CHBDC (CSA 2000) recommends, and other code committees are considering recommending, that the reinforcement ratio of FRP for sections in flexure be greater than ρfb to ensure that at ultimate the failure will be caused by crushing of concrete without rupture of the FRP bars. For T-sections with wide flange, the reinforcement areas Af > ρfbbd is too large to be practical. Furthermore, it will be shown below that both Tand rectangular under-reinforced sections, with FRP bars, will also exhibit sufficiently large curvature at failure. Thus, it is proposed to allow reinforcement ratio ρf to be less than ρfb provided that the product (moment × curvature) in service be an acceptable fraction of the same product at ultimate. This is referred to as the deformability requirement and is measured by the following dimensionless factors: ψu ψs

[21]

Curvature factor =

[22]

Moment factor =

[23]

Deformability factor, DF =

Mu Ms ψu Mu ψ sM s

CHBDC (CSA 2000) and Jaeger et al. (1997) check the deformability by a “performance factor” defined by an equation similar in form to eq. [23], but with ψsMs replaced by the product of the curvature and the moment corresponding to a maximum concrete compressive strain of 0.001. CHBDC requires that the performance factor be greater than 4 or 6 for rectangular and T-sections, respectively. In the present paper the deformability factor, DF, is defined as the ratio of ψuMu to ψsMs, with the subscripts u and s referring to actual ultimate and service states, respectively. For a member in flexure, the strain energy per unit length is equal to the integral of the area below the M–ψ graph.

When the M–ψ graph is linear, the area is equal to (1/2)Mψ and the ratio of Mψ at any two load levels is equal to the ratio of the strain energy values. By the use of the deformability factor as defined in eq. [23], the ratio of Mψ at ultimate and at service is used as a practical means to indicate approximately the ratio of the strain energy values at the two load levels. Using the definition of deformability provided in eq. [23], the deformability factors for a rectangular section with steel reinforcement are shown in Fig. 3. The strain in the steel at service is taken as 1200 × 10–6 with fy = 400 MPa, Es = 200 GPa, and fc′ = 30 and 60 MPa. Each curve is extended to the respective balanced reinforcement ratio, ρbal, as calculated by eq. [11] with fy replacing ffu. Due to the ductility of steel, large curvature factors (eq. [21]) and hence high values of DF are achieved at low reinforcement ratios. However, the values drop rapidly as the reinforcement ratio increases reaching a value of 4 at approximately 85% of ρbal. An allowable deformability factor, DF ≥ 4, is adopted here for all concrete sections in flexure, because as will be shown, there is no substantial difference between T- and rectangular sections in terms of their deformability. It will also be shown that DF is greater than 4 for FRP-reinforced sections except when the reinforcement ratio is impractically high, that is, when ρf > ρfmax. An empirical equation will be given for the maximum allowable reinforcement ratio, ρfmax. It is noted, however, that even the value of ρfmax given here may be impractically high.

Parametric study The magnitude of the deformability factor, DF, is determined below for rectangular and T-sections in flexure reinforced with GFRP, AFRP, or CFRP. The allowable strain at service in the FRP, εfs = 2000 × 10–6, is adopted. The parametric study shows that DF decreases with the increase in ρf and the decrease of fc′ . The value ρf is varied, starting by a minimum value, ρfmin, given in Table 2. The maximum value is taken as the smaller of 0.04 and the value that corresponds to DF = 4. A relatively low value, fc′ = 30 MPa, and a relatively high value, fc′ = 60 MPa, are adopted. The range of ρf is extended to 4%, although this reinforcement ratio may be considered impractical; this is done to cover the range where DF approaches the allowable limit, DF = 4.0. Rectangular sections Table 4 gives the ratios Mu/Ms, ψu/ψs, and DF for rectangular sections in flexure; b is the width of the section; d is the distance from the extreme compression fibre to the centroid of tension reinforcement. GFRP, AFRP, and CFRP are used as reinforcement for the sections analyzed in Table 4. Results are presented for fc′ = 30 MPa and fc′ = 60 MPa for each FRP type. The lower limits of the reinforcement ratio, ρf, adopted in the tables are 0.40%, 0.19%, and 0.11% for GFRP, AFRP, and CFRP, respectively, with fc′ = 30 MPa. For the high strength concrete, with fc′ = 60 MPa, the lower limits adopted for ρf are 0.60%, 0.28%, and 0.16%, respectively, for GFRP, AFRP, and CFRP. These lower limits approximate the minimum reinforcement ratios given in Table 2. The values of Mu and ψu are calculated by eqs. [4] © 2002 NRC Canada



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131 Table 4. Representative values of deformability factors for FRP reinforced rectangular sections. ρf (%)

Service condition ε f = 2000 × 10–6 Ms/bd2

ψ s*d

Ultimate condition Mu/bd2

Factors

ψ u*d

Mu/Ms

ψ u/ψ s

DF

0.01564 0.01725 0.01561 0.01163 0.00885

6.8 6.6 5.8 3.8 2.4

7.0 7.3 6.5 4.5 3.1

47.6 48.3 37.6 17.1 7.6

0.01542 0.01725 0.01553 0.01466 0.01096

6.8 6.7 5.9 5.4 3.6

6.9 7.2 6.4 5.9 4.1

46.9 48.4 37.2 32.1 14.5

0.01705 0.01850 0.01561 0.01163 0.00734

7.4 7.2 5.8 3.8 1.7

7.6 8.0 6.5 4.5 2.3

56.4 57.2 37.5 17.1 4.0

0.01689 0.01850 0.01662 0.01003 0.00734

7.4 7.3 6.4 3.1 1.8

7.5 7.8 6.9 3.6 2.2

56.0 57.4 44.1 11.2 4.0

0.01556 0.01691 0.01483 0.01194 0.00733

6.8 6.4 5.4 4.0 1.7

6.9 7.2 6.1 4.7 2.3

47.3 45.7 33.0 18.5 4.0

0.01542 0.01683 0.01508 0.00957 0.00734

6.9 6.5 5.6 2.9 1.8

6.9 7.0 6.1 3.3 2.2

47.0 45.6 34.5 9.7 4.0

(a) GFRP reinforcement, f c′ = 30 MPa 0.40 0.80 1.00 2.00 4.00

0.3086 0.6065 0.7563 1.4808 2.8796

0.00224 0.00235 0.00239 0.00258 0.00286

2.1059 3.9873 4.3592 5.6285 7.0381

(b) GFRP reinforcement, f c′ = 60 MPa 0.60 1.38 1.75 2.00 4.00

0.4624 1.0440 1.3158 1.4978 2.9230

0.00225 0.00239 0.00244 0.00248 0.00271

3.1671 7.0018 7.7032 8.1176 10.4904

(c) AFRP reinforcement, f c′ = 30 MPa 0.19 0.34 0.50 1.00 3.35

0.2937 0.5208 0.7570 1.4822 4.7046

0.00224 0.00232 0.00240 0.00258 0.00318

2.1746 3.7430 4.3592 5.6285 8.1042

(d) AFRP reinforcement, f c′ = 60 MPa 0.28 0.59 0.75 2.50 5.80

0.4325 0.8967 1.1338 3.6229 8.0535

0.00224 0.00236 0.00241 0.00281 0.003334

3.2174 6.5656 7.2434 11.3213 14.5965

(e) CFRP reinforcement, f c′ = 30 MPa 0.11 0.23 0.30 0.50 1.80

0.3180 0.6546 0.8479 1.3909 4.7319

0.00225 0.00236 0.00242 0.00256 0.00318

2.1691 4.1820 4.5615 5.5029 8.1193

( f ) CFRP reinforcement, f c′ = 60 MPa 0.16 0.39 0.50 1.50 3.10

0.4623 1.1013 1.4065 4.0492 8.0598

0.00225 0.00240 0.00246 0.00286 0.00333

3.1669 7.1603 7.9154 11.7697 14.6049

and [5] when ρf < ρfb and by eqs. [6] and [10] when ρf ≥ ρfb. The values of Ms and ψs are determined by eqs. [19] and [20] when the strain in the FRP is taken as εfs = 2000 × 10–6. Figures 4–6 show the variation of DF with the reinforcement ratio, ρf. T-sections The equations used in preparing Table 4 for rectangular sections apply also to T-sections when the distance c from the extreme compression fibre to the neutral axis is less than or equal to the thickness of the flange, hf. When c > hf, the distance yCT between the extreme compression fibre and the

tension reinforcement will be different from that for a rectangular section. However, this has only a small influence on the value of DF. Figure 7 shows the variation of DF with hf/d for a T-section with CFRP. The T-section used for this figure has the ratio b/bw = 9. The highest reinforcement ratio used is ρf = Af/(bd) = 0.44%; the cross-sectional area Af is equal to 4.0% of bwd. Any larger reinforcement ratio is considered impractical. Effect of varying the allowable strain, εfs The allowable strain in service in the FRP, εfs = 2000 × 10–6, adopted in preparation of Figs. 4–6 corresponds to al© 2002 NRC Canada



132 Fig. 4. Deformability factors, DF, for GFRP-reinforced sections.

Can. J. Civ. Eng. Vol. 29, 2002 Fig. 7. Deformability factors, DF, for T-sections with variable hf/d and b/bw (CFRP, fc′ = 30 MPa, ε fs = 2000 × 10–6).

Fig. 5. Deformability factors, DF, for AFRP-reinforced sections. Fig. 8. Effect of varying ε fs on the deformability factors, DF (CFRP, rectangular section fc′ = 30 MPa).

Fig. 6. Deformability factors, DF, for CFRP-reinforced sections.

lowable crack width of 5/3 times the width implicitly given by CSA A23.3-M94 or ACI 318-99 for steel-reinforced members. It may be desirable in design to tolerate wider or to require narrower cracks by the selection of the value for εfs greater or less than 2000 × 10–6. As well, other factors, including bond characteristics of individual bars, bar spacing, and bar size, may result in a variation in the value of εfs. It can be verified that the deformability factor is approximately proportional to (εfs)–2. This can be seen from the graphs in Fig. 8, which are prepared for CFRP with fc′ = 30 MPa and εfs = 2400 × 10–6, 2000 × 10–6, and 1600 × 10–6. Discussion of results of parametric study Examining the graphs in Figs. 4–6, it can be seen that the deformability factor, DF, has an almost constant very high value when ρf ≤ ρfb. The value of DF decreases with the increase of ρf and the decrease of fc′ . With the allowable value of strain in service in the FRP, ffs = 2000 × 10–6, used to pre© 2002 NRC Canada



Newhook et al.

133

pare Figs. 4–6, it is seen that DF ≥ 4 except in the following unusual situations: • For AFRP, when fc′ = 30 MPa and ρf > 3.4%. • For CFRP, when fc′ = 30 MPa and ρf > 1.75% or when fc′ = 60 MPa and ρf > 3.0%. The parametric study in which DF is calculated for variable ρf, fc′ , and ffs enabled to determine the value of ρf for which DF is equal to 4.0. These calculations gave the empirical inequality: [24]

ρf

ffs < 0.19 α1 fc′

This can be used to set a maximum FRP reinforcement ratio to be used in design: [25]

ρfmax = 0.2

α1 fc′ ffs

As ρf approaches ρfmax, the amount of FRP reinforcement becomes difficult to accommodate. Thus, for economy and ease of accommodating the FRP bars, ρf 4.0. Thus, there is no need to set a lower limit for ρf to achieve deformability. The condition that ρf be greater than ρfb is not necessary. Figures 4–6 indicate that rectangular sections with GFRP, AFRP, or CFRP bars have DF > 4 except when ρf is extremely high. Comparison of the three figures indicates that with CFRP bars the values of DF are smaller than with the other two types of bars. Figure 7 shows that a T-section with CFRP bars, having a flange width = 9bw also has DF > 4 even when the cross-sectional area of the FRP, Af, is as high as 4% of bwd. As the ratio b/bw approaches 1, the area properties of the section approach those of a rectangle. Thus, the parametric study performed for rectangular sections and for T-sections with b/bw = 9 covers a wide range of the T- or rectangular sections that can occur in practice, with the three kinds of FRP bars.

Design of sections in flexure Based on the above discussion and the parametric study, the amount of FRP reinforcement for a section in flexure, for which the concrete dimensions have been selected, should be determined as follows: (1) Due to service loads, the stress in the reinforcement should not exceed the allowable value, ffs. The reinforcement ratio, ρf = Af/(bd), should be between the limits ρfmin ≤ ρf < ρfmax, where ρfmax = 0.2α1 fc′ / ffs (eq. [25]). The minimum reinforcement should be selected to ensure that the calculated nominal moment exceeds the moment causing flexural cracking by an acceptable margin. The maximum reinforcement ratio is introduced to ensure that the section exhibits sufficient deformability before failure; however, it may be necessary not to approach the maximum to avoid congestion of the reinforcement. (2) The calculated moment resistance should be greater than the required demand due to the factored external loads. The strength reduction and load factors required

by codes or technical committees dealing with FRP reinforcement should be used in calculating the two moments. The moment resistance is given by eq. [4] or eq. [6] for under-reinforced or over-reinforced sections, respectively. Comments: The concrete dimensions of the section should be selected such that the deflection not be excessive. Deflection of flexural members can be more critical for members with FRP than with steel reinforcement. Selection of minimum thickness to control deflection is beyond the scope of the present paper; however, guidance is provided by ACI 440 (ACI 2001), the Japan Society of Civil Engineers (JSCE 1997), and Ghali et al. (2002). The purpose of the permissible stress ffs is to control the width of cracks. The permissible stress in the FRP in service may be taken as Efεfs with εfs selected according to the tolerable crack width. A value of εfs = 2000 × 10–6 is proposed here. This value permits a strain in FRP reinforcement of 5/3 times the strain allowed for steel reinforcement by various codes. In practice, ρf