Engineering Structures 27 (2005) 1519–1527

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Experimental study on high-strength concrete beams failing in she

A. Claderaa,∗, A.R. Maríb,1

aUniversity of Balearic Islands, Department of Physics, Ctra. Valldemossa km 7.5, 07122 Palma de Mallorca, SpainbTechnical University of Catalonia, Department of Construction Engineering, Jordi Girona 1-3, 08034 Barcelona, Spain

Received 8 September 2004; received in revised form 16 March 2005; accepted 25 April 2005Available online 24 May 2005

Abstract

To better understand the response of high-strength concrete beams failing in shear with and without shear reinforcement,reinforced concrete beams were tested as a part of extensive research on shear design of reinforced high-strength concrete bconcrete compressive strength of the beams at the age of the tests ranged from 50 to 87 MPa. The primary design variables were tof shear and longitudinal reinforcement. A minimum amount of shear reinforcement related to the concrete tensile strength was also pThe details of the beam specimens, material properties, instrumentation and the testing procedure used are carefully described inThey will be useful for researchers to compare and analyze other design approaches. The test results are presented and discussinfluence of each design parameter is studied separately. Furthermore, the minimum amount of web reinforcement proposed in tis validated using experimental data from the literature. Test results are also compared with different shear design approaches anconclusions are drawn.© 2005 Elsevier Ltd. All rights reserved.

Keywords: High-strength concrete; Reinforced concrete; Beams; Tests; Shear failure; Building codes

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1. Introduction

The use of High-Strength Concrete (HSC) has increaconsiderably during the last decade, since it can be produreliably in the field using low water–cement ratios thankshigh-quality water-reducing admixtures. Furthermore, Hwill be more and more frequently used in columns, in precelements and in structures where durability is an importdesign parameter.

To give a simplified explanation, HSC is obtained bimproving the compactness of the concrete mix, whincreases the strength of both the paste and the interbetween the paste and the coarse aggregate. Howan increase in the strength of the concrete producesincrease in its brittleness and smoother shear failure surf(Fig. 1), leading to some concerns about the applicat

∗ Corresponding author. Tel.: +34 971 171378; fax: +34 971 173426.E-mail addresses: antoni.cladera@uib.es (A. Cladera),

antonio.mari@upc.es (A.R. Marí).1 Tel.: +34 93 4016508.

0141-0296/$ - see front matter © 2005 Elsevier Ltd. All rights reserved.doi:10.1016/j.engstruct.2005.04.010

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Fig. 1. Crack in high-strength concrete. The crack goes through taggregates.

of high-strength concrete. Since most of the current sheprocedures are based on tests carried out on beams wiconcrete compressive strength lower than 40 MPa, and oof the shear transfer mechanisms is shear-friction acrosscracks, the failure shear strength needs to be re-evaluate

1520 A. Cladera, A.R. Marí / Engineering Structures 27 (2005) 1519–1527

Table 1Details of the beam specimens and summary of the experimental results

Beam fc fsp bw d a/d ρw fya ρw fyb ρl Vfail Vcr(MPa) (MPa) (mm) (mm) (MPa) (MPa) (%) (kN) (kN)

H50/1 49.9 3.46 200 359 3.01 0 0 2.24 99.69 95H50/2 49.9 3.46 200 353 3.06 0.577 0.605 2.28 177.64 85H50/3 49.9 3.46 200 351 3.08 1.291 1.291 2.29 242.07 90H50/4 49.9 3.46 200 351 3.08 1.291 1.291 2.99 246.34 110H50/5 49.9 3.46 200 359 3.01 0 0 2.24 129.65 85H60/1 60.8 4.22 200 359 3.01 0 0 2.24 108.14 104H60/2 60.8 4.22 200 353 3.06 0.747 0.747 2.28 179.74 95H60/3 60.8 4.22 200 351 3.08 1.267 1.374 2.29 258.78 100H60/4 60.8 4.22 200 351 3.08 1.267 1.374 2.99 308.71 –H75/1 68.9 3.69 200 359 3.01 0 0 2.24 99.93 99H75/2 68.9 3.69 200 353 3.06 0.747 0.783 2.28 203.94 95H75/3 68.9 3.69 200 351 3.08 1.267 1.362 2.29 269.35 95H75/4 68.9 3.69 200 351 3.08 1.267 1.362 2.99 255.23 100H100/1 87.0 4.05 200 359 3.01 0 0 2.24 117.85 117H100/2 87.0 4.05 200 353 3.06 0.906 0.929 2.28 225.55 110H100/3 87.0 4.05 200 351 3.08 1.291 1.355 2.29 253.64 110H100/4 87.0 4.05 200 351 3.08 1.291 1.353 2.99 266.53 85H100/5 87.0 4.05 200 359 3.01 0 0 2.24 140.09 85

a Calculated using the real yielding stress of the stirrups.b Calculated using the stress of the stirrups in the test just before reaching the failure load.

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Moreover, shear failure in a beam without web reinforcement is sudden and brittle. Therefore, it is necessaryprovide a minimum amount of shear reinforcement, whicmust prevent sudden shear failure on the formation of firdiagonal tension cracking and, in addition, must adequatcontrol the diagonal tension cracks at service load leveThus, the minimum area of web reinforcement is intendeto ensure that the capacity of the member after crackiexceeds the load at which inclined cracking occurs. Duethe higher tensile strength of high-strength concrete, a highcracking shear is expected and hence, would require a laramount of minimum shear reinforcement [1].

The experimental campaign described in this paper wpart of extensive research that led to the proposal ofnew shear design procedure for normal and high-strenconcrete beams [2,3].

2. Objectives of the experimental campaign

The main objectives of the experimental campaigcarried out were:

– To study the influence of the concrete compressistrength on the shear strength in beams with and withoshear reinforcement. Some current procedures hold tthe failure shear strength does not increase when conccompressive strength is higher than 60 MPa for beamboth with and without web reinforcement.

– To propose and verify a minimum amount of wereinforcement for high-strength concrete beamsaccordance with the increase in concrete tensile strenfor high-strength concretes.

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– To evaluate the efficiency of the amount of shereinforcement as a function of the concrete compressstrength. Some authors believe that for high-strengconcrete beams, stirrups are more efficient thannormal-strength beam specimens [4,5].

– To evaluate the influence of the amount of longitudinreinforcement on the shear strength. The majoritycurrent codes’ limitation of the amount of longitudinareinforcement to 2% will be studied for high-strengtconcrete beams.

– To study the influence of web-distributed longitudinal reinforcement for high-strength members without stirrupas this variable has an important effect on the failure shstrength according to Ref. [6].

3. Design of the test specimens

In order to achieve the prior objectives, eighteen beaspecimens were designed and tested.Table 1 and Fig. 2show the details of the 200 mm wide× 400 mm deep beamspecimens that were tested with a shear span of 1080 mm

The test program [7] consisted of four series of beams(1) the H50 series, designed to have a concrete strengt50 MPa; (2) the H60 series, designed to have a concrstrength of 60 MPa; (3) the H75 series, designed to haa concrete strength of 75 MPa with silica fume; and (the H100 series, designed to have a concrete strength100 MPa. The actual concrete strength at the time of testis presented inTable 1.

Beam specimen number one in each series (H50H60/1, H75/1, and H100/1) did not have shear reinforcment. The longitudinal reinforcement consisted of tw

A. Cladera, A.R. Marí / Engineering Structures 27 (2005) 1519–1527 1521

Fig. 2. Test set-up and cross-section of the beam specimens.

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32 mm diameter bars(ρl = 2.24%), with a characteristicyielding stress of 500 MPa.

Beam number two in each series (H50/2, H60/2, H75/2and H100/2) contained the proposed minimum amountshear reinforcement. The longitudinal reinforcement waequal to that provided in series 1.

It is proposed that the minimum amount of sheareinforcement be proportional to the tensile strength of thconcrete, fct,m . In the ‘Design guidance for high-strengthconcrete’ [8] it is suggested that traditional tensile strengthcan be unconservative for high-strength concrete, andproposes the following equation:

fct,m = 0.30 3√

f 2c MPa if fc ≤ 60 MPa (1)

fct,m = 0.58 2√

fc MPa if fc > 60 MPa. (2)

The proposed amount of minimum shear reinforcemis given in Eq. (3) and compared inFig. 3 with some othercode proposals.

Aw,min = fct,m

7.5

bws

fyMPa (3)

where Aw,min is the amount of minimum shear reinforcement,bw is the web width,s the stirrup spacing andfy is theyield strength of the reinforcing steel.

The above formulation was derived taking into accouthat the minimum shear reinforcement is necessary to ava sudden failure after the diagonal cracking and, thereforshould be proportional to the tensile strength of the concrTo totally derive the proposed equation, the minimuamount was fixed to around 0.35 MPa for a 25 MPa concr

1522 A. Cladera, A.R. Marí / Engineering Structures 27 (2005) 1519–1527

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Fig. 3. Comparison of the minimum amount of web reinforcementdifferent specifications and proposed equation proportional to the conctensile strength.

(Fig. 3), as the ACI 99 Code proposed for conventionconcretes.

The third specimen in each series (H50/3, H60/3, H75and H100/3) had the same amount of web reinforcementall beams, 8 mm diameter stirrups spaced by 210 mm. It wdesigned to have the highest amount of web reinforcemin order to produce a shear failure with the provideamount of longitudinal reinforcement. Additionally, thflexural tension reinforcement for all the beams consisted2φ32 bars.

The fourth beam in each series (H50/4, H60/4, H75and H100/4) had the same shear reinforcement asthird series but the flexural tension reinforcement consisof 2φ32 bars plus a 1φ25 bar. Hence, the amount olongitudinal reinforcement was equal to 2.99%.

The fifth beam specimen in each series (H50/5 aH100/5) did not have stirrups but instead contained fosmall longitudinal bars (8 mm diameter) distributed alonthe web plus two bars in the compression zone, resulta distance between them of 110 mm. As was mentionearlier, Collins et al. [6] postulated that the size effecis not only a function of the beam depth, but alsothe distance between distributed longitudinal reinforcem(sz). According to the AASHTO Specifications, each layeof this crack control reinforcement must have an area ofleast 0.004bwsz . The area provided in specimens H50/5 anH100/5 verified the previous expression.

4. Fabrication of the test specimens, material properties,instrumentation and testing procedure

The beams were cast at the Alvisa precast concrplant, located in Selgua (Huesca, Spain). The concrcomponents, reinforcement bars, moulds, and proceduwere those actually used at that plant. A maximum aggreg

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size of 12 mm was used throughout the series. Stand150 mm× 300 mm cylinders were cast with the specimeto obtain the compressive strength and splitting strenof each concrete mix (Table 1). The splitting strength ofthe concrete H60 was higher than the H75 and H1splitting strength. No reason was found for this behavioThese cylinders were kept under the same environmeconditions as the beam specimens until the time of testin

Spanish standard B 500 S reinforcing bars, withcharacteristic yielding stress of 500 MPa, were used.Table 2lists the actual yield stress,fy, and the ultimate stress,fu, forthe web reinforcing bars.

Table 2Properties of web reinforcing bars

Size-series Area fy fu(mm2) (MPa) (MPa)

φ6-H60 and H75 28.27 530 680φ8-H60 and H75 50.27 530 685φ6-H50 and H100 28.27 530 680φ8-H50 and H100 50.27 540 672

To monitor the behaviour of the tested beams, tapplied loads, strains at the reinforcement and atconcrete surface, and displacements were measured udifferent instruments such as load cells, strain gaugesmagnetostrictive transducers (LVDTs). All the variablwere monitored continuously by the data acquisition systePhotography and video equipment were also utilised.

The web strain,υxy , was measured by means of a roseof two LVDTs mounted in one side of the beams of seriH50 and H100 (seeFig. 4). Using the Mohr circle it is easyto deduce how the web strain can be obtained fromreading of the transducers [7]. However, the direction of thestrains in the web may not be found with this set-up.

The load was applied at midspan of the beam specimby a 150 mm wide and 28 mm thick neoprene pad un

Fig. 4. Rosette of two magnetostrictive transducers used to obtain the sstrain.

A. Cladera, A.R. Marí / Engineering Structures 27 (2005) 1519–1527 1523

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Fig. 5. Typical crack patterns at failure in the tested beams. (a) Bespecimen without web reinforcement. (b) Beam specimen with longitudinreinforcement distributed along the web. (c) Beam specimen with stirru(d) Beam specimen H60/3 – shear cracks did not reach the load applicazone.

a spherical bearing. The beam specimen was supporteda sliding pin bearing, on the instrumented side, and a fixpin bearing on the opposite side, both of 40 mm diametThe tests were carried out under displacement control usa closed-loop hydraulic compression machine with a loadicapacity of 1000 KN.

5. Failure modes

All beam specimens failed in shear. Nevertheless, bespecimen H60/3 collapsed due to a combination of sheand high longitudinal strain, and shear cracks did not crothe compression zone of the beam.Fig. 5 shows the typicalcracking patterns at failure observed in the experimencampaign.

The mode of failure for beams without stirrups idifferent from that of beams with shear reinforcemenBeams H50/1, H60/1, H75/1 and H100/1 failed suddenwith the appearance of a single shear crack. In general,higher the beam’s concrete compressive strength, the brisits failure.

On the other hand, beams containing stirrups presena more ductile response. After the formation of the firshear crack, stirrups started to work and further shear cradeveloped. At failure, the compressed top part of the beawas crushed due to the combination of compressive ashear stresses. In the photographs ofFig. 6 the spalling ofthe concrete next to the crack prior to the failure can be seThis spalling was best observed in beams with the highconcrete compressive strength.

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Beam specimens containing distributed longitudinreinforcement developed more than one shear cra(seeFig. 5) and failure shear strength was higher thanother similar beams without web reinforcement. Howevefailure was also sudden.

6. Discussion of the test results

6.1. Beam specimens without web reinforcement

Specimens H50/1, H60/1, H75/1, and H100/1 did ncontain shear reinforcement. The only parameter whivaried for all beams was the concrete mix. Longitudinreinforcement,ρl , was constant and equal to 2.24%Their failure shear strengths were 99.69 KN, 108.14 KN99.93 KN, and 117.85 KN respectively. Therefore, there wa slight increase in failure shear strength as the concrcompressive strength increased, except for beam specimH75/1, whose splitting concrete strength,fsp, was lower thanthe H60 splitting strength (Table 1).

6.2. Beam specimens with stirrups

Beam specimens H50/2, H60/2, H75/2, and H100/2 weprovided with the minimum amount of web reinforcemenas proposed in this paper in Eq. (3). The value of the shearforces as measured at stirrup yielding,Vy in Table 3, wastaken to be the shear strength when the two stirrups crossthe crack yielded.Fig. 7 and Table 3demonstrate that theamount of web reinforcement provided was appropriabecause the beams showed a significant reserve of strenafter cracking.

For elements with stirrups the influence of the concrecompressive strength can be studied from the beaspecimens in Series 3 and 4. The failure shear strengththese test specimens is shown inFig. 8. As a general trend,it increases as the concrete compressive strength increaexcept for beam H100/3. Beam specimen H60/4 collapsunder a very high force after the longitudinal reinforcemehad yielded. No apparent reason for this behaviour wdiscovered.

This general increment can be explained as an increasthe web reinforcement effectiveness, as for members withoweb reinforcement the increment was not so remarkable.

The addition of web reinforcement improves the sheresponse of the specimen by increasing the failure shstrength and a higher ductile response. The cracking pattalso changed. For example, in beam H50/1, a single shcrack was reported, while two shear cracks were noticedbeam H50/2 and three to four shear cracks in beam specimH50/3.

Series 4 beam specimens had 2.99% of longitudinreinforcement compared with 2.24% in the Series 3 beamThe failure shear strength increased slightly as the amoof longitudinal reinforcement increased except for in beaH75/4 (Fig. 8). The average increase was approximately 5%

1524 A. Cladera, A.R. Marí / Engineering Structures 27 (2005) 1519–1527

Fig. 6. Cracking prior to failure and at failure in a beam with web reinforcement. Concrete spalling near the diagonal crack.

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Table 3Minimum amount of web reinforcement, observed failure, yielding and cracking shear for each specimen provided with the proposed minimum amo

Beam fc ρwa Vfail Vy Vcr

b Vy/Vcr Vfail/Vcr(MPa) (MPa) (KN) (KN) (KN)

H50/2 49.9 0.577 177.64 158 85 1.86 2.09H60/2 60.8 0.747 179.74 140 95 1.47 1.89H75/2 68.9 0.747 203.94 144 95 1.52 2.15H100/2 87.0 0.906 225.55 194 110 1.76 2.32

a Calculated using the real yielding stress of the stirrups.b Approximated cracking shear force.

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Fig. 7. Shear deformation in beams without shear reinforcement and wthe proposed minimum.

6.3. Beam specimens with distributed longitudinal rein-forcement

Beams H50/5 and H100/5 were based on the tecarried out by [6]. They demonstrated that the size effecdisappears when beams without stirrups contain wedistributed longitudinal reinforcement.

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Fig. 8. Beam specimens with web reinforcement. Failure shear strenversus concrete compressive strength for series 3 and 4.

The failure mechanism was considerably different fbeams with distributed longitudinal reinforcement whecompared with similar beams without any kind of wereinforcement. Beams H50/1 and H100/1 failed suddeafter the formation of the first shear crack. Failuwas especially brisk for the beam with the higheconcrete compressive strength. Beam specimens contaidistributed longitudinal reinforcement developed more th

A. Cladera, A.R. Marí / Engineering Structures 27 (2005) 1519–1527 1525

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Fig. 9. Beam specimens with distributed longitudinal reinforcement. Shestrain in beams H50/5 and H100/5.

one shear crack and failure shear strength was 30.5% higin beam H50/5 and 18.9% in beam specimen H100/5 thin other, similar beams without web reinforcement (Fig. 9).However, failure was also sudden.

7. Comparison of test results with different designapproaches

Table 4summarises the shear procedures included in tACI 318-02 Code [9], AASHTO LRFD 2001 [10], the 2002Final Draft of Eurocode 2 [11] and the method proposedby Cladera and Marí [2,3]. To calculate the predictions(Table 5) all the safety factors were taken equal to 1.0The amount of web reinforcement was obtained by consering the stress state of the stirrups during the test just befreaching the failure load (Table 1). For the beams with webreinforcement, the EC predicted failure loads depend onangle of the compression struts,θ . It has been taken, to ob-tain the highest prediction, as cotgθ equal to 2.5 (Table 5).For the ACI procedure the angle is fixed to 45◦(cotgθ = 1).For the AASHTO LRFD shear procedure and the methoproposed by Cladera and Marí the angle of the compressstrut is obtained by compatibility (Table 5). Both methodsare based on the Modified Compression Field Theory.

For the eighteen beam specimens, the averageVtest/

Vpredictedratio is 1.34 for the ACI 318-99 formulation, 1.26for the EC-2, 1.18 for the AASHTO LRFD, and 1.14 for theRefs. [2,3] predictions. The coefficient of variation (standardeviation over the average) is 11.21% for the ACI Cod6.66% for the AASHTO LRFD, 20.2% for the EC-2, and6.93% for the Cladera and Marí shear procedure.

Both the AASHTO LRFD and Cladera and Marí predictions prove very satisfactory when compared with the ECpredictions.

The ACI procedure correlates better with membewithout web reinforcement, than with members with she

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Fig. 10. Validation of the proposed amount of minimum reinforcement wexperimental data of the literature.

reinforcement. For members with stirrups, the procedis more conservative, especially for high-strength concrspecimens.

The EC-2 equations are also excessively conservafor members with shear reinforcement, in particular fmembers with the minimum amount of web reinforcemewhile for beam specimens without stirrups it is slightunconservative.

8. Validation of the proposed minimum amount of webreinforcement

It has already been seen that, for the beam specimdescribed in this paper, the minimum amount of wreinforcement proportioned adequate reserve shear strenHowever, as significant amount of data exist in the curreliterature, a broader analysis was carried out using tresults by [1,7,12–17].

Fig. 10 represents the amount of web reinforcemeprovided in tested beams over the minimum amouproposed versus the reserve shear strength index(Vfail/Vcr).Ozcebe et al. [12] suggested that to insure an adequamargin of safety the value of the index should be greater th1.30. This condition is satisfied (Fig. 10) by the minimumamount of web reinforcement proposed by Eq. (3).

9. Conclusions

Based on the test results of the eighteen beam specimthe following conclusions can be drawn:

– Beams without web reinforcement presented a vebrittle behaviour. The higher their concrete compressstrength, the brisker their failure.

1526 A. Cladera, A.R. Marí / Engineering Structures 27 (2005) 1519–1527

Table 4Summary of different shear procedures

Equations Comments

EC-2 Final Draft 2002

Beams without web reinforcement:VRd,c = [0.18k(100ρl fck )1/3 + 0.15σ ′

cd ]bwd

VRd,c min = [0.035k3/2 f 1/2ck + σ ′

cd ]bwd

Beams with web reinforcement(VRd,c = 0)

VRd,s = Asw

sz f ywd cotθ

fck ≤ 90 MPa

k = 1 +√

200

d≤ 2.0

ρl = Al

bwd> 0.02

1 ≤ cotθ ≤ 2.5

AASHTO LRFD 2000Vc = β

√fc bw z β and θ are listed in a table as a function

of the longitudinal strain in the web and thenon-dimensional shear.

Vs = Av fys z cotθ Vmax = 0.25 fcbwd

ACI 318-02Vc =

(0.16

√fc + 17ρl

V · d

M

)bwd

Vs = Asw

sz f ywd ≤ 0.67

√fcbwd

fc < 70 MPaV d/M ≤ 1

Prop. Cladera and Marí [2,3]

Beams without web reinforcement:Vc = [0.225ξ(100ρl )

1/2 f 0.2c ]bwd

θ is expressed as a equation which dependson the longitudinal strain in the web and thenon-dimensional shear.

Beams with web reinforcement:Vc = [0.17ξ(100ρl )

1/2 f 0.2c τ1/3]bwd

Vs = dvAw

sfywd cotθ

See Refs. [2,3] for furtherdetails.

edures

Table 5Summary of the predictions made by the ACI Code 318-99, the 2001 AASHTO LRFD Specifications, April 2002 Final Draft of the Eurocode 2 procand method proposed in Refs. [2,3]

Beam Vfail θ Vpredicted Vtest/Vpredicted

(KN) ACI LRFD EC [2,3] ACI LRFD EC [2,3] ACI LRFD EC [2,3]

H50/1 100 – – – – 86 85 105 94 1.16 1.18 0.95 1.06H50/2 178 45 33 22 34 127 149 96 153 1.39 1.19 1.85 1.16H50/3 242 45 36 22 38 175 198 204 206 1.38 1.22 1.19 1.17H50/4 246 45 34 22 36 179 210 204 228 1.37 1.17 1.21 1.08H50/5 130 – – – – 86 100 105 124 1.51 1.30 1.24 1.05

H60/1 108 – – – – 94 92 112 98 1.14 1.17 0.96 1.10H60/2 180 45 34 22 35 145 167 119 171 1.24 1.08 1.51 1.05H60/3 259 45 36 22 38 188 211 217 216 1.38 1.23 1.19 1.20H60/4 309 45 34 22 36 192 223 217 240 1.61 1.39 1.42 1.29

H75/1 100 – – – – 99 96 117 98 1.01 1.04 0.85 1.02H75/2 204 45 34 22 35 152 175 124 177 1.34 1.17 1.64 1.15H75/3 269 45 36 22 38 192 214 215 218 1.40 1.26 1.25 1.24H75/4 255 45 34 22 36 196 228 215 242 1.30 1.12 1.19 1.05

H100/1 118 – – – – 100 100 126 98 1.18 1.18 1.07 1.20H100/2 226 45 35 22 36 150 196 147 182 1.50 1.15 1.53 1.24H100/3 254 45 37 22 38 179 222 214 210 1.42 1.14 1.18 1.21H100/4 267 45 35 22 36 183 237 214 232 1.46 1.13 1.25 1.15H100/5 140 – – – – 100 120 126 129 1.26 1.17 1.11 1.09

Average 1.34 1.18 1.26 1.14Stand. deviation 0.15 0.08 0.25 0.08

COV (%) 11.21 6.66 20.18 6.93Minimum 1.01 1.04 0.85 1.02Maximum 1.61 1.39 1.85 1.29

A. Cladera, A.R. Marí / Engineering Structures 27 (2005) 1519–1527 1527

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– For beams without web reinforcement, the failure sheastrength generally increased as the concrete compressstrength increased, except for beam H75/3.

– High-strength concrete beams with stirrups presentedless fragile response than similar beams without wereinforcement.

– The minimum amount of web reinforcement proposed ithis paper was sufficient in terms of the demand of reservof strength after shear cracking.

– For beams with the same geometric amount of transverreinforcement, the higher their concrete compressivstrength, the more effective stirrups are.

– For high-strength concrete beams with stirrups, thlimitation of the amount of longitudinal reinforcement to2% is not experimentally justified.

– Beam specimens with longitudinally distributed web reinforcement along the web showed a better behaviouthan similar beams without any kind of shear reinforcement. Although their failure was also fragile, severashear cracks, rather than a single one, were reported, athe failure shear strength increased about 25%.

– Methods based on the modified compression field theoshowed a close correlation to the empirical results icomparison to the other codes’ correlations.

Acknowledgments

The research described in this paper comprises partthe Spanish Ministry of Science and Technology’s projectTRA99/0974 and MAT2002-00615. The beam specimenwere cast at the Alvisa precast concrete plant locatedSelgua (Huesca, Spain), the authors extend their sincethanks to them. The beams were tested at the StructuTechnology Laboratory of the Construction EngineeringDepartment, with the assistance of the former undergraduastudents Josep Capell and Jorge Suárez.

Appendix. Notation

Aw,min amount of minimum shear reinforcement;a/d shear span to depth ratio;bw web width;d effective depth;fc concrete compressive strength;fct,m average concrete tensile strength;fsp average concrete splitting strength;fu ultimate tensile strength of the reinforcing steel;fy yield strength of the reinforcing steel;s stirrup spacing;sz vertical distance between distributed longitudina

reinforcement;Vcr cracking shear strength;Vfail failure shear strength;

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Vy shear strength at yielding of the stirrups;φ diameter of the rebar;ρl amount of longitudinal reinforcement (%);ρw fy amount of shear reinforcement (MPa).

References

[1] Yoon YS, Cook WD, Mitchell D. Minimum shear reinforcement innormal, medium and high-strength concrete beams. ACI Struct J 1993(5):576–84.

[2] Cladera A, Marí AR. Shear design procedure for reinforced concrbeams using artificial neural networks. Part I: Beams without stirruEng Struct 2004;26(7):917–26.

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Dr. A. Cladera is an assistant professor at the Department of Physat the University of Balearic Islands. He obtained his Ph.D. in 2003the Technical University of Catalonia researching on shear strengthreinforced high-strength concrete beams.

Dr. A.R. Marí is a Professor of Construction Engineering at the TechnicUniversity of Catalonia, Barcelona, Spain. He is a member of the fib tagroup 4.1 “Serviceability models” and general reporter of the Committeecharge of updating the Spanish Concrete Code.