Shear strength of RC beams with stirrups using an improved Eurocode 2 truss model with two variable-inclination compression struts

Engineering Structures 198 (2019) 109359

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Shear strength of RC beams with stirrups using an improved Eurocode 2 truss model with two variable-inclination compression struts

T

Dario De Domenico , Giuseppe Ricciardi ⁎

Department of Engineering, University of Messina, Contrada Di Dio, 98166 Sant’Agata, Messina, Italy

ARTICLE INFO

ABSTRACT

Keywords: Reinforced concrete beam Shear strength Strut-and-tie model Truss model Concrete compression struts Transverse reinforcement Eurocode 2 Stirrups Variable strut inclination method

The shear strength of reinforced concrete beams with stirrups can be assessed with strut-and-tie models based on the truss analogy. The Eurocode 2 formulation exploits such a possibility with a variable-angle inclination of the compression struts that is determined through the lower-bound theorem of plasticity. This approach tends to underestimate the shear capacity of lightly shear-reinforced beams where the concrete contribution, which is neglected in this formulation, is more significant. Based on this consideration but adopting an identical theoretical framework without adding a concrete contribution explicitly, this paper presents an upgrade of the EC-2 truss model incorporating two variable-inclination compression struts, whose inclination angles are determined through equilibrium conditions and the theory of plasticity. In this model, the upper compression strut may have lower inclination than the lower compression strut, which is mechanically motivated by the increase of shear stresses observed in the upper portion of the beam, in the neighborhood of the crack tip, following the trend of the principal compressive stress direction. This improved truss model enables the construction of a wider class of statically admissible solutions compared to the EC-2 single-strut approach. Closed-form expressions are derived for a variant of this model in which the two compression struts extend equally for half of the inner lever arm, which allows a simple and compact formulation for practical design purposes. Based on comparison with the well-established ACI-DafStb databases including more than 200 tests, the proposed model leads to values of the shear strength that are in very good agreement with experimental results and more accurate than the values obtained from the EC-2 procedure, despite the relatively comparable simplicity.

1. Introduction Although in the majority of cases shear and bending in reinforced concrete (RC) beams take place concurrently, shear failure is associated with a bending moment that is lower than that corresponding to pure flexural failure [1,2]. Shear collapse becomes manifest with inclined cracks and is generally sudden and brittle, in contrast to more ductile flexural failure with pseudo-vertical cracks. The presence of transverse reinforcement, typically in the form of stirrups, increases the shear strength and mitigates these brittle phenomena [3,4]. The shear strength of RC beams with stirrups can be assessed with basic equilibrium principles applied to strut-and-tie models, based on the truss analogy. Introduced by Ritter and Mörsch [5,6] at the beginning of the past century, the simplest strut-and-tie model is formed by two parallel chords, representing top concrete stress blocks and bottom longitudinal reinforcement, and web members, representing inclined compression struts corresponding to concrete stress fields between adjacent cracks and tensile ties simulating the distribution of

⁎

transverse reinforcement along the beam (e.g., steel stirrups). In the original Ritter-Mörsch (RM) truss model the compression struts are inclined at an angle = 45 with respect to the beam longitudinal axis, which also corresponds to the first shear cracking angle. However, experimental investigations [7] revealed that the RM truss model underestimated the shear strength of RC beams. This is due to a twofold reason: (1) the RM model neglects certain contributions in the shear transfer mechanism such as aggregate interlock along inclined cracks, shear contribution of the uncracked concrete in the compression zone, residual tensile stresses across cracks and dowel action exerted by longitudinal reinforcement [8]; (2) the actual angle of the shear cracks at failure may be less than 45°, depending upon various factors such as amount of transverse reinforcement, concrete strength and beam crosssection characteristics. In order to obtain more realistic estimates of the shear strength accounting for the two above-mentioned limitations of the RM model, many rational models for predicting the shear capacity of RC beams have been proposed in the literature over the last decades, some of which also inspired design formulations introduced in building

Corresponding author. E-mail address: [email protected] (D. De Domenico).

https://doi.org/10.1016/j.engstruct.2019.109359 Received 14 March 2019; Received in revised form 30 May 2019; Accepted 28 June 2019 Available online 29 August 2019 0141-0296/ © 2019 Elsevier Ltd. All rights reserved.

Engineering Structures 198 (2019) 109359

D. De Domenico and G. Ricciardi

Nomenclature

w

inclination angle of the transverse reinforcement with respect to the beam longitudinal axis dimensionless parameter identifying the transition depth from one strut inclination to the other ratio of the two cotangent function of the strut inclinations in the proposed model, = cot 2/cot 1 ( w) function of the shear reinforcement, equal to ( w ) = 1 + 4 w 8 w2 ( w) inclination of the compression strut with respect to the beam longitudinal axis in the EC-2 model lower and upper inclination angle of the compression strut 1, 2 with respect to the beam longitudinal axis mean inclination angle of the compression strut with rem spect to the beam longitudinal axis ( w) function of the shear reinforcement, equal to ( w ) = 1 + 8 w 16 w2 128 w3 (x , w ) function of the shear reinforcement 2 (x , w ) = x 2 + 4 w (1 + x 2)(1 ) w w , with x = cot 1 efficiency factor for concrete cracked in shear 1 fck /250) 1 = 0.6(1 longitudinal reinforcement ratio equal to = As /(bd ) transverse reinforcement ratio equal to w w = Asw /(sbw sin ) principal tensile and compressive stresses I , II cw1, cw2 web concrete stress in the lower and upper branch of the compression strut having angles 1 and 2 sw1, sw2 stirrups stress in the lower and upper branch of the compression strut having angles 1 and 2 shear ratio of transverse reinforcement equal to c c = w f ywd /fcd

a b bw d fck , fcd fcm fywd fy r s v vexp vRd vRd, s vRd, c

z As Asw C T VRd, c VRd, s VRd

codes for structural concrete. Some of these models are briefly summarized below. The European pre-standard ENV 1992-1-1 [9], and the ACI 318 Building Code [10] adopted a truss model in which = 45 (as in the RM model), but an additional “concrete contribution” (whose expressions were derived on a mostly empirical basis) was explicitly included as a correction term of the original RM shear strength (additive approach). Another class of truss models involves an inclination angle of the compression struts that may be different from 45° (variable strut inclination method). This angle actually varies as the shear force increases. Indeed, after the first shear cracks develop at 45°, the compression struts rotate and may cross two adjacent cracks due to aggregate interlock phenomena (enhanced by the presence of transverse reinforcement), thus unveiling inclination angles at failure less than 45° [11]. However, it is rather difficult to determine an appropriate value of and various options were studied. Well-established and accurate models incorporated compatibility conditions (in addition to equilibrium conditions) to determine the shear capacity of RC beams. Among these compatibility-aided theories, in the Modified Compression Field Theory (MCFT) [12–15] the inclination of the diagonal compressive stresses is determined though compatibility of deformations in the web of the beam (assuming principal stress direction and principal strain direction coincident), and the shear resistance is provided by a steel contribution (stirrups) plus a concrete contribution (representing the shear stress transferred vertically across the crack). Simplified expressions derived from the complete MCFT were incorporated into the AASHTO [16] as well as in the general method of the Canadian building code [17]. Other variants of these theories are the rotating angle softened truss model (RA-STM) [18], the

mechanical ratio of transverse reinforcement w = w f ywd /( 1 fcd ) shear span width of the beam cross section minimum web width of the beam cross section effective depth of the beam characteristic and design value of cylinder concrete compressive strength mean value of cylinder concrete compressive strength design yield strength of shear reinforcement design yield strength of longitudinal reinforcement normalization parameter equal to r = b w z 1 fcd spacing of stirrups in the longitudinal direction normalized shear equal to v = V / r normalized experimental shear strength equal to vexp = Vexp/r normalized actual shear strength of the beam vRd = VRd/r normalized shear strength due to yielding of stirrups, equal to vRd, s = VRd, s / r normalized shear strength to crushing of web concrete, equal to vRd, c = VRd, c /r inner lever arm of the beam (assumed equal to 0.9d for RC members) area of longitudinal reinforcement area of transverse reinforcement resultant force of compressive stress block in concrete resultant force of tensile stress in longitudinal reinforcement shear failure load due to crushing of web concrete shear failure load due to yielding of steel stirrups actual shear failure load of the beam, determined via VRd = min(VRd, c , VRd, s )

fixed angle softened truss model (FA-STM) [19], and the Disturbed Stress Field Model (DSFM) [20]. In the Critical Shear Crack Theory (CSCT) [21–23] the shear stress transferred across the crack is an explicit function of the crack width in the critical shear region. The results of this theory inspired the Swiss code for structural concrete [24]. Based on similar theoretical concepts but adopting specific presliding failure criteria, a mechanics-based segmental analysis procedure was developed and applied to RC beams with any kind of concrete and reinforcement with the assumption of a linearly varying crack width [25–28], showing a good accuracy of predictions. On the other hand, the general design philosophy of Model Code 2010 [29] presents different levels of approximation, where the most precise method belongs to the additive approach (add a concrete contribution calculated through compatibility conditions), while the most basic approach ignores the concrete contribution and adopts a fixed inclination angle less than 45° [30]. Alternatively, another possibility to estimate the shear capacity of a RC beam is to exploit the lower-bound theorem of plasticity [31,32]. The final angle maximizes the shear strength within the class of statically and plastically admissible solutions: a lower bound of the true failure load can be expediently determined through simple equilibrium principles (without any consideration to compatibility) provided the stress fields nowhere violate the yield condition. A plasticity-based variable strut inclination method was first incorporated in the Model Code 90 [33] and then adopted in the final version of the Eurocode 2 (EC2) [34], as well as in various National building codes of European countries, for instance NTC 18 in Italy [35] and DIN 1045 in Germany [36], although with different limitations on the variability of the strut inclination angle. The reason to decide for this model lies in its simplicity, transparency and freedom in design (through the choice of the angle). 2

Engineering Structures 198 (2019) 109359

D. De Domenico and G. Ricciardi

However, as Regan pointed out [37], simpler models necessarily have to neglect some factors, considered secondary, although what is secondary in one case may be primary in another. In this regard, an evident limitation of the EC2 shear model is that the concrete contribution to the shear resistance is neglected. Therefore, despite the improvements over the original RM truss model, the Eurocode 2 truss model may produce conservative estimates of the shear strength in comparison with experimental results, especially for low amounts of transverse reinforcement where such concrete contribution is more significant [38–42]. Additionally, the distribution of shear stresses at imminent shear failure was found to increase in the upper portion of the beam (in the neighborhood of the crack tip) near the uncracked concrete chord [43,44], thus producing a marked rotation of the direction of principal compressive stresses compared to the bottom part. This variability of principal compressive stress direction cannot be incorporated in the EC-2 model because a single strut inclination angle is assumed throughout the beam height. These physical observations and mechanical considerations suggest introducing a second strut inclination in the upper portion of the beam with reduced angle compared to that of the lower compression strut. Building on these motivations, this paper presents an upgrade of the EC-2 truss model with two (rather than one) variable-inclination compression struts to better capture the trend of the isostatic lines of compression. Based on an identical theoretical framework to that of the Eurocode approach, the two strut inclination angles are determined through equilibrium conditions and the theory of plasticity. A wider class of statically admissible solutions is determined compared to the EC-2 single-strut approach. A broad comparison with the well-established ACI-DafStb databases [45,46] proves that the proposed truss model with two variable-inclination compression struts outperforms the EC-2 single strut approach in terms of both accuracy and precision, despite the relatively comparable simplicity.

2. Motivations for introducing the second inclination of the compressions strut The observation of typical crack pattern in RC beams failing in shear reveals that inclined shear cracks arise, which are directed from the beam supports and propagate up to the loading position, cf. Fig. 1(a). Closer inspection of the beam at failure also highlights a certain variability of the crack widths, decreasing from the bottom to the upper portion (compressive zone) of the beam. The first branch of the critical crack develops along the web of the beam with inclined flexural cracks, ideally at an angle 45° that corresponds to the initial principal compressive strain direction. This, however, does not correspond to the actual directions of principal compressive stresses at incipient failure shown in Fig. 1 because of a number of reasons. At low levels of load, the shear force is transferred through the inclined cracks by means of frictional stresses (aggregate interlock phenomena) and residual tensile stresses. In reality, aggregate interlock phenomena are more significant in the lower part of the crack, while the residual tensile stresses are more pronounced in the neighborhood of the crack tip and, according to ACI-ASCE Committee 445 [8], until crack widths of around 0.05–0.15 mm. Therefore, the variability of the crack width along the beam height governs the shear transfer mechanisms, which is also corroborated by refined experimental measurements from the literature [47]. Obviously, this variability is intimately related to concurrent bending effects taking place in the beam cross-section. Indeed, the lower part of the beam is subject to tensile stresses (from bending), thus experiencing larger crack widths, while the approximately upper-mid part of the beam is subjected to compressive stresses (also from bending) [48]. With further increase of the shear force, the crack widths increase again and a redistribution of shear stresses takes place so that the uncracked compression chord comes into play to satisfy the equilibrium, becoming a significant contribution [49]. Consequently, at incipient failure the shear stresses in the neighborhood of the crack tip and, in general, in the upper

Fig. 1. Motivation for the second inclination of the compression strut: (a) typical crack pattern at incipient shear failure of RC beam (modified from Regan [1]); (b) construction of Morh’s circles for two points along the beam height. 3

Engineering Structures 198 (2019) 109359

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portion of the beam are markedly higher than in the lower part of the beam, which in turn implies a variability of the strut inclination angle as described below. A preliminary stress analysis based on Mohr’s circle principles is effective for identifying the isostatic lines of compression in a graphical manner. Assuming plane stress conditions, element P, located in a portion of the beam where cracks widths are large, is characterized by a shear stress 1 that is relatively low, and a normal stress . The presence of the shear stress 1 0 gives rise to a lower inclination of the compression strut 1 < 45 . This physical condition is supported by experimental findings whereby several cracks associated with lower inclinations than 45° pass through the first cracks having inclinations at around 45°, especially for lightly shear-reinforced beams [11]. On the other hand, element Q, located in the upper-mid portion of the beam where the cracks have reduced widths, cf. Fig. 1(b), is characterized by a higher value of shear stress 2 > 1 . This higher shear stress 2 , in combination with the same normal stress as point P, produces a lower inclination angle of the compression strut than the previous case 2 < 1 , see again Fig. 1(b). Besides the two exemplificative points P and Q discussed above, stress analysis of any other point of the compression stress fields is expected to produce situations that vary gradually along the beam height, in a continuous manner. Several well-established models, already mentioned in the introduction [12–15,21–23,25–28], address the problem of explicitly modeling the variability of the crack width through compatibility conditions, by linking the inclination of the compression strut with the axial deformation in the web of the beam. Evidently, modeling the variability of the crack width with a plasticity-based model is not possible explicitly, since compatibility conditions are not used. Therefore, in the EC-2 truss model the determination of the inclination angle is carried out in the framework of the lower-bound theorem of plasticity [32,34], based on equilibrium conditions only. However, the single strut inclination angle of the EC-2 approach does not allow capturing the different shear stress states along the beam height observed at incipient failure and described above. The simplest way to incorporate this variability without much more analytical effort than the EC-2 truss model is to introduce two (rather than one) representative inclinations of the compression strut along the beam height. The second inclination 2 < 1 aims to describe, in a simplified way, how the shear stresses increase in the upper part of the beam and in the neighborhood of the crack tip in comparison to the lower part where the crack widths are higher. In analogy to the variable strut inclination method of the EC-2, these two inclination angles 1 and 2 will be determined through the lower-bound theorem of limit analysis in the sequel of the paper. It is worth noting that the above-described variability of the crack

width along the beam height is more pronounced for low amount of transverse reinforcement, where web crushing is preceded by stirrup yielding and the phenomenon of strut rotation is more evident [11,41]. As the amount of transverse reinforcement increases, the crack openings are controlled by stirrups, so that the shear stresses tend to be more uniformly distributed, without significant differences along the beam height and with minimum strut rotation [11]. Therefore, in this case we can imagine that 1 2 so that a single inclination of the compression strut (as in the Eurocode 2 truss model) may suffice for calculating a reasonable value of the shear strength. 3. Proposed truss model with two variable-inclination compression struts As said above, the proposed model is an upgrade of the Eurocode 2 truss model [34] and, as such, it complies with the same assumptions and theoretical principles that are briefly recalled here. At ultimate limit state the resistant mechanism of RC beams with shear reinforcement is represented by a truss model with two parallel chords (top concrete stress blocks with resultant C and bottom longitudinal reinforcement with resultant T ) and the web members (carrying the shear force) consisting of compression struts (describing the concrete stress fields) and transverse reinforcement (stirrups or web longitudinal reinforcement). It is assumed that transverse reinforcement undergoes a purely axial force (i.e. dowel action is neglected). Moreover, it is assumed that the spacing of stirrups is small enough (in comparison with the structural member size) so that their effects can be modelled with uniform stress fields – according to a “smeared truss model” [50–52]. Similarly, the concrete compression struts can be idealized as uniform compressive stress fields. In the Eurocode 2 truss model, the inclination angle of the transverse reinforcement with respect to the beam longitudinal axis is denoted as , while that of the concrete compression struts is indicated with . The inclination angle may be different from 0 = 45 that is the first shear cracking angle. In the Eurocode 2 truss model, the maximum shear capacity is determined through the static such that theorem of limit analysis by exploring values of 1 cot 2.5 [34], which corresponds to 21.8 45 . These limitations (cot ) min = 1 and (cot ) max = 2.5, though, may differ in alternative variants of the variable strut inclination method adopted from country to country, as overviewed in [53]. In order to evaluate the shear strength of a structural member with two inclinations of the concrete compression struts, it is sufficient to impose equilibrium conditions on different beam segments according to Ritter’s method (method of sections). Let us consider vertical equilibrium across the first beam segment obtained through a section parallel to the compression struts, see Fig. 2. For a general treatment, we

Fig. 2. Ritter’s method (method of sections) parallel to the compression struts to determine individual member forces. 4

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assume that the transition depth between the two inclination angles is governed by a parameter with 0 < < 1, whose physical role in the model will be further discussed later in this paper. The upper portion, having inclination 2 , has depth z , with z being the inner lever arm, while the lower portion, having inclination θ1, extends for the com) z . Normally, z can be taken as 0.9d , where d plementary depth (1 denotes the effective depth of the beam cross-section. The number of stirrups crossing the beam segment APB depends upon the spacing of transverse reinforcement. Assuming a constant spacing s , by simple geometrical considerations the number of stirrups is equal to

AQ PR ns = ns1 + ns2 = + s s (1 ) z (cot 1 + cot ) = + s

z (cot

2

+ cot )

Similarly, let us consider vertical equilibrium across another beam segment obtained through a section parallel to the transverse reinforcement, see Fig. 3. Let cw1 and cw2 denote the uniform stress field of the concrete lower and upper compressive struts, respectively, while Sc1 and Sc2 indicate the corresponding resultant compressive forces. The values of such resultants are easily calculated as follows

Sc1 = Sc 2 =

sw1 Asw ns1

=

Ss2 =

sw 2 Asw ns 2

=

bw z

sw1 f ywd

(1

)(cot

(2)

(cot

( ) (1

+ cot ) b w z

1 fcd

sin2

1 fcd

(1

where the parameter w , called “mechanical ratio of transverse reinforcement”, is defined as follows

VRd, s = r

vRd, c =

VRd, c = (1 r

w

=

Asw bw s sin

fywd 1 fcd

=

f ywd w

1 fcd

.

w

In Eqs. (3) and (4) bw indicates the minimum web width, = Asw /(sbw sin ) is the transverse reinforcement ratio, fywd is the design yield strength of transverse reinforcement, fcd is the design value of concrete compressive strength and 1 a strength reduction factor (or efficiency factor) for concrete cracked in shear whose value may be different from country to country (typically in the range 0.5 0.6). In this work, the recommended formula proposed in the Eurocode 2 [34] is adopted, namely 1 = 0.6(1 fck /250) , with fck being the characteristic concrete compressive strength. In the limit condition sw1 = sw 2 = f ywd Eq. (3) produces the shear strength due to yielding of steel transverse reinforcement 1 fcd

w [cot

+ cot

1

+ (cot

2

cot

2 1)]sin

(6)

+ Sc2 sin +

2=

( ) cw2

1 fcd

(cot 2 + cot ) 1 + (cot 2 )2

(7)

)

(cot x ) 2]

(cot 1 + cot ) (cot 2 + cot ) + 1 + (cot 1 ) 2 1 + (cot 2 ) 2

(8)

w [cot

)

+ cot

1

+ (cot

2

cot

2 1)]sin

(cot 1 + cot ) (cot 2 + cot ) + 1 + (cot 1 )2 1 + (cot 2 )2

(9) (10)

It is worth noting that vRd, s is a linear function of the variables cot 1 and cot 2 , whereas vRd, c has a nonlinear dependence upon the two cotangent functions. For = 0 the improved truss model reduces to the Eurocode 2 truss model with single compression strut inclination = 1. In the spirit of the lower-bound theorem of limit analysis [32], the actual shear strength is the smaller value of vRd, s and vRd, c , that is

(4)

w

VRd, s = bw z

1

(cot 1 + cot ) ) 1 + (cot 1 )2

1 fcd

vRd, s =

2

1]

In the remainder of the paper use will be made of dimensionless expressions to facilitate the application of the improved truss model for practical design and verification purposes. For this reason, the two expressions (5) and (8) are divided by a normalization parameter r = b w z 1 fcd (having dimension of force) in order to obtain the following dimensionless counterparts

+ Ss2 sin = sw 2 f ywd

) z (cot 1 + cot ) sin 2 + cot ) sin 2]

= 1/[1 + has been where the trigonometric identity used (with x = 1, 2 ). In the limit condition cw1 = cw2 = 1 fcd Eq. (7) produces the shear strength due to crushing of concrete compression struts

(1

+ cot ) +

1 fcd

(3)

1

z (cot

(sin x )2

The vertical equilibrium is a balance between the shear force V and the vertical component of the resultant force, which is obtained by projecting Ss = Ss1 + Ss2 along the vertical axis V = Ssv = Ssv1 + Ssv2 = Ss1 sin

cw1 bw [(1 cw 2 b w [

cw1

VRd, c = bw z

) z (cot 1 + cot ) s z (cot 2 + cot ) sw 2 Asw s

sw1 Asw

= =

V = Scv = Scv1 + Scv2 = Sc1 sin

where ns1 and ns2 denote the number of stirrups crossing the lower and upper compression struts, respectively. Let sw1 and sw2 denote the uniform stress fields of the steel transverse reinforcement in the two portions, and Asw the area of the corresponding cross-section. The resultant tensile forces of the two sets of transverse reinforcement are calculated as follows:

Ss1 =

cw 2 bw PK

The vertical equilibrium is a balance between the shear force V and the vertical component of the resultant force, which is obtained by projecting Sc = Sc1 + Sc 2 along the vertical axis

(1)

s

cw1 b w AH

vRd = vRd ( ,

w,

,

1,

2)

(11)

= min(vRd, s, vRd, c )

A rational design method leads to combinations ( 1, 2 ) that, for given , w , , correspond to the simultaneous attainment of the two types of failure modes

vRd, s ( 1,

2)

= vRd, c ( 1,

where the dependence upon the variables , compactness.

(5)

(12)

2) w,

has been omitted for

Fig. 3. Ritter’s method (method of sections) parallel to the transverse reinforcement to determine individual member forces. 5

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As said above, in the Eurocode 2 model the range of variability of the strut inclination angle is limited. In order to extend this concept to the proposed truss model, it is necessary to introduce here two sets of limitations for the range of variability of the two inclination angles of the compression struts, i.e.

(cot 1) min (cot 2) min

cot cot

1 2

(cot (cot

1)max

11.3 45 . 2 In the framework of the static theorem of limit analysis [32], a lower bound of the actual shear capacity can be found in the class of the statically and plastically admissible solutions (i.e. complying with the equilibrium conditions from which the above expressions are derived and nowhere violating the yielding conditions) corresponding to the combination of the cot 1 and cot 2 variables that, for given = ¯ , = ¯ , produce the largest shear strength value subject to the w = ¯ w, constraint conditions (13), that is

(13)

2 )max

With regard to the first angle 1, the same limitations as the Eurocode 2 can be assumed, namely (cot 1) min = 1 and (cot 1) max = 2.5, which 45 . With regard to the second angle 2 it is corresponds to 21.8 1 reasonable to allow a wider range of variability, considering that values of 2 < 1 can occur according to the sketch of Fig. 1 and the motivations outlined in Section 2. Indeed, the single strut inclination angle adopted in the Eurocode 2 truss model can be considered representative of an average compression strut inclination, which, in reality, is variable along the beam height. Therefore, the Eurocode 2 inclination angle can be seen as an intermediate value between the two angles of the improved truss model proposed in this work. These considerations, along with the observations of experimental crack pattern, justify values of 2 lower than 21.8 . In the remainder of the paper we propose (cot 2) min = 1 and (cot 2) max = 5.0 , which corresponds to

Fig. 4. Couples ( 1,

2)

max vRd ( ¯, ¯ w , ¯, 1, 2

s. t .

(cot (cot

1) min 2 ) min

cot cot

1 2

1,

2)

(cot 1) max (cot 2) max

(14)

The optimization problem (14) can be solved numerically for any transition depth and for any inclination of the transverse reinforcement . In the following Section this problem is particularized for the widely used case of members with vertical shear reinforcement ( = 90 ). In a variant of this model in which = 1/2 (corresponding to two compression struts that extend equally for half of the inner lever arm), the determination of the strut inclination angles 1, 2 and the evaluation of the shear strength is carried out with simple closed-form

that satisfy the rational design condition (12). 6

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(analytical) expressions. This assumption allows an expedient formulation for practical design purposes and is extensively discussed in the next section. The influence of this choice = 1/2 is also critically commented in the sequel of the paper.

or, at the limit, lying on the bisector line wherein the two inclination angles are equal to each other. In particular, for the different w values considered, the Eurocode 2 truss model with single strut inclination angle 1 = 2 leads to rational design solutions identified by the blue circles. These solutions are obtained as the intersection points between the curve described by Eq. (17) and the bisector line cot 2 = cot 1, which leads to the well-known value of the inclination angle

4. Shear strength evaluation for vertical shear reinforcement 4.1. Governing equations for model with

= 1/2

The two shear contributions (9) and (10) for reduce to the following expressions w

vRd, s = vRd, c =

2

[cot

1

+ cot

= 1/2 and

2]

cot

= 90

2

=

cot

1

±

(cot

1)

2

+4 2

w (1

+ (cot

EC 2 vRd =

(16)

w (1

+ (cot

2)(1

1) 1

w

w (cot 1 )

2)

(17) that is illustrated in Fig. 4 in a cot 1 cot 2 plane. Different curves are drawn for increasing values of w in the range 0.10–0.25. The condition cot 2 = cot 1 is also reported as a blue dotted line to discriminate between different relative magnitudes of the two inclinations 1 and 2 : points lying below this line are characterized by 2 > 1, whereas points lying above this line have 2 < 1. It emerges that the physically meaningful solutions must be sought in the area above the bisector line

2)

w

(18)

w (1

(19)

w)

However, it can be seen that for the improved truss model with two compression struts, within the admissible domain defined by the constraint conditions on the two strut inclination angles (shaded rectangular area) there exists a much wider range of rational design solutions that are also physically meaningful (i.e. above the bisector line). In other words, the presence of the second compression strut enables the construction of a wider class of statically admissible solutions ( 1, 2 ) , some of which are identified by the red squares in Fig. 4, that are all potential candidates for the evaluation of the shear capacity. According to the principles of the static theorem of limit analysis, the lower bound of the actual shear resistance is the largest possible value of the shear strength associated with such statically admissible couples ( 1, 2 ) , in accordance with Eq. (14). Indeed, while in the Eurocode 2 truss model the rational design method EC2 produces only one point = for a given w (that is, the intersection of the two lines vRd, s ( ) and vRd, c ( ) ), in the proposed truss model the shear contributions vRd, s ( 1, 2 ) and vRd, c ( 1, 2) are two surfaces (rather than two lines) because they are functions of two (rather than one)

)2)

Fig. 5. Optimal couple ( 1,

1

In these points the maximum shear capacity corresponds to the simultaneous attainment of crushing of compressive struts and yielding of steel transverse reinforcement (rational design method) and is equal to

The couples ( 1, 2 ) that, for given amount of shear reinforcement w , satisfy the rational design condition (12) of simultaneous attainment of maximum shear capacity due to yielding of steel transverse reinforcement and crushing of compressive struts can be found by equating (15) and (16), which leads to the following relationship

cot

=

w

(15)

1 cot 1 cot 2 + 2 1 + (cot 1 )2 1 + (cot 2 )2

EC2

identified within the family of rational design solutions for 7

w

= 0.15

Engineering Structures 198 (2019) 109359

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The corresponding shear strength of the improved truss model with two compression struts is obtained by substituting the two optimal angles (21) into one of the two relations (15) or (16), which yields opt vRd =

Fig. 6. Optimal inclination angles ( EC2

opt 1 ,

opt 2 )

and EC2 rational design angle

inclination angles. Therefore, the rational design method produces an intersection line (rather than an intersection point) between these two surfaces, within which the actual shear capacity is to be evaluated. These concepts are better illustrated in the surface and contour plots reported in Fig. 5. The surface vRd defined in (11) (for given w = 0.15) is plotted in the symmetric range 1 cot i 5 (i = 1,2), although only the domain defined by constraint conditions (13) is meshed. The bisector line cot 2 = cot 1 is superimposed (dashed blue line). The Eurocode 2 solution (single strut inclination method) lies on such bisector line and is represented by the blue circle located on the ridge of the surface: this is the only point satisfying the rational design condition vRd, s = vRd, c along such bisector line. In particular, from (18) and (19) EC2 = 22.79 and evaluated for w = 0.15, this point is characterized by EC2 vRd = 0.357 . However, there exists a family of rational design solutions of the improved truss model determined by the condition (12), which is shown in thick black line (representing the intersection line between vRd, s and vRd, c ). Among these rational design solutions, the optimal solution is represented by the combination of ( 1, 2 ) angles associated with the highest value of shear capacity: this point (marked with a red square) is located on the ridge of the surface like the EC-2 point but is associated with a higher shear capacity than the Eurocode 2 solution. The coordinates of this optimal design point can be found analytically. First, we substitute the relationship (17) (ensuring a rational design solution) into (11) so that vRd becomes function of only one cotangent function, i.e. vRd = vRd (cot 1) . In order to identify the maximum of the resulting function vRd , we calculate the first derivative with respect to x = cot 1

wx

(1 + 4

w) x

3

+ [1 + 2 w (1 4 4(1 + x 2) 2 (x , w )

w )x

2

+2

wx

4]

(x ,

w)

where (x , w ) = x 2 + 4 w (1 + x 2)(1 The maximum w point can be determined by setting (20) equal to zero, which produces eight solutions in terms of x = cot 1. Only two of them have a physical meaning and are discussed here. The first solution is valid in the range 0< w w,lim , where w,lim is a limit value that will be defined below. This optimal solution is called cot 1opt , while the corresponding cot 2opt value is found by substituting cot 1opt into (17). The final result (optimal couple of angles 1opt , 2opt ) is reported below

=

1+8

w

16

(

w)

=

1+4

w

8

2 w

128

2 w

cot

opt 1

=

( w) 2 2 w

cot

opt 2

=

4 w 1 + 8 w + 2 ( w) 1+4 w ( w)

(

w ))

+ 16 ( w ))

2 w

1+8

w

.

(22)

EC2

(20)

w)

+8 w ( 8(1 + 4 w

Fig. 6 along with the EC2 angle given by (18). For the moment, we purposely ignore any constraint condition, as those introduced in (13), but we will incorporate these limitations in a following discussion in Section 4.2. 0 the compressions struts would have ideal In the limit as w inclinations 1opt = 45 and 2opt = 0 , which corresponds to the first shear crack inclination for a beam without shear reinforcement. With increasing amounts of transverse reinforcement the inclination of the lower compression strut ( 1opt ) decreases, while that of the upper compression strut ( 2opt ) increases. For w = w,lim = 0.25 the two compression struts reach the same inclination, equal to 30°, and from this point onwards their inclinations are variable with increasing w and EC2 = cot 1 (1 equal to each other, namely 1opt = 2opt = w )/ w . Based on this analysis, it emerges that the proposed truss model modifies the Eurocode 2 (single strut inclination) truss model only in the range w < 0.25, that is, for beams with low amounts of transverse reinforcement. Coming back to the discussion of Fig. 5, for the assumed w = 0.15 Eq. (21) give 1opt = 41.29 , 2opt = 14.72 and the associated strength opt = 0.371. As expected, the value of the EC2 angle from (22) is vRd

2 w) .

(

w )(1

As said above, the optimal solution (21) is valid only for a certain range of mechanical ratios of transverse reinforcement, within which the resulting expressions are real-valued. Indeed, it can be verified that the function ( w ) becomes complex-valued for w > w,lim = 0.25. Interestingly, beyond this amount of transverse reinforcement, the only physically meaningful (real-valued) solution obtained by setting (20) equal to zero is the same as that of the single strut inclination method EC2 EC2 (Eurocode 2 solution), i.e. cot 1opt cot with cot given by EC2 opt cot (18). Substituting this value into (17) gives cot 2 and, obviously, the associated shear resistance is given by (19), i.e. opt EC2 . In other words, for w > w,lim = 0.25 the optimal solution vRd vRd of the improved truss model with two variable-inclination compression struts coincides exactly with the shear strength given by the Eurocode 2 truss model with single compression strut, because the two inclination angles are equal to each other. This circumstance, obtained with mere mathematical considerations, has evident physical justifications in view of the motivations outlined at the end of Section 2: higher amounts of transverse reinforcement contribute to reducing the crack widths and to mitigating the consequent variability of the shear stresses along the beam height. Therefore, the second compression strut, introduced in the proposed truss model to capture these variability effects, turns out to be useless for beams with high amount of transverse reinforcement. Graphs of the two optimal angles ( 1opt , 2opt ) given by (21) (for w w,lim = 0.25) and by (18) (for w > w,lim = 0.25) are depicted in

(obtained without any constraint condition) versus mechanical ratio of transverse reinforcement w .

dvRd (x ) dx x 4 =

2 (

EC2

= 22.79 (single strut inclination method) is intermediate between the two angles found in the improved truss model. Moreover, the shear strength obtained with the improved truss model is slightly higher than the EC2 value. However, it seems that the introduction of the second compression strut leads to a modest strength increase in comparison with the Eurocode 2 single strut solution. Recalling again the motivations of the proposed truss model stated in Section 2, more significant strength increases are expected for lower amounts of transverse reinforcement whereby the shear stresses are expect to vary more markedly along the beam height, increasing from bottom to top. This is demonstrated in Fig. 7 that essentially reports similar graphs to the previous Fig. 5 but corresponding to a lower amount of transverse reinforcement, namely w = 0.115. In this case, EC2 = 19.82 , the rational design solution lying on the bisector line is which is beyond the range of admissible values because the

3 w

w)

(21) 8

Engineering Structures 198 (2019) 109359

D. De Domenico and G. Ricciardi EC2

> (cot )max (cyan circle). This implies that, accorresponding cot cording to the limitations of the Eurocode 2, the actual design point would be located at the boundary of the admissible domain and associated with cot = (cot ) max (blue circle). This point has coordinates EC2 EC2 = 21.80 = 0.2875. Nevertheless, the introduction of the and vRd second compression strut in the improved truss model allows the construction of a wider class of statically admissible solutions. Within such wider class of solutions, the optimal design point given by (21) is also an admissible solution complying with the limitations (13) (red square). Substituting the given w = 0.115 into Eqs. (21) and (22) gives the coordinates of this point: 1opt = 42.74 , 2opt = 11.44 and the shear opt strength vRd = 0.346 , respectively. As a result, around 20% shear strength increase is obtained in comparison with the Eurocode 2 solution. In the authors’ opinion, it is exactly in this range of beams having low amount of transverse reinforcement that the single strut inclination method produces conservative results, as also documented in the relevant literature [38–44]. Therefore, the proposed truss model with two strut inclinations may lead to better estimates of the actual shear strength in this specific range, which is verified in the next Section by comparison with experimental results. From another perspective, the trend of the shear strength in terms of the two inclination angles 1, 2 can be assessed by drawing different sections along the vRd surface for fixed = cot 2/cot 1 . Three values of are considered, namely = 1 (corresponding to the Eurocode 2 single strut solution 1 = 2 ), = 2 and = opt that is obtained from (21) as cot 2opt /cot 1opt . Relevant results are shown in Fig. 8 for the two amounts of shear reinforcement discussed above (the sections are illustrated with the same colors adopted in the 3D plots, for consistency and easier interpretation). It is clearly seen that when the rational design point of the EC-2 single strut method is beyond the admissible domain (which occurs for low amounts of shear reinforcement as for the case w = 0.115), the shear strength produced by the improved truss model is much higher.

Fig. 7. Optimal couple ( 1,

2)

Fig. 8. Sections along the vRd surface for fixed = cot 2 /cot 1 and for two different mechanical ratios of transverse reinforcement: (a) w = 0.15 ; (b) w = 0.115 .

4.2. Design regions based on amount of transverse reinforcement for = 1/2 Combining the above considerations on the search for the optimal design point given in (21) (and in (18) for w > w,lim ) and the constraint conditions given in (13) makes it possible to identify different design regions based on the amount of transverse reinforcement w . Different design expressions for design and verification purposes can then be obtained in each design region.

identified within the family of rational design solutions for 9

w

= 0.115 .

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D. De Domenico and G. Ricciardi

Let us refer to the graphs shown in Fig. 9: similarly to Fig. 4, on the left hand-side in Fig. 9a) we report different rational design solutions expressed by Eq. (17) in the cot 1 cot 2 plane for five selected values of the mechanical ratio of transverse reinforcement w , called w1, w2 , w2.5 , w3 w,lim and w,max . In particular, w1 corresponds to a rational design solution intersecting the admissible domain in the point P1 having coordinates ((cot 1) max , (cot 2) max ) = (2.5, 5) . Any beam having 0 w w1 (design region 1) can be designed adopting cot 1(1) = (cot 1) max = 2.5 and cot 2(1) = (cot 2)max = 5. The value of w1 is found by substituting these limit values in Eq. (17) and solving for w , which leads to

cot

2|cot 1= 2.5

=5

w1

cot

cot

opt 2 | w = w2

= (cot

2) max

=5

w2

= 0.1136

25 + 104 52

w

2704

2 w

(26)

w

(27)

= 1.0801

5 + 260

w

+

25 + 104

w

2704

2 w

(28)

104

The path of the solutions in the design region 2 is the straight line from P1 to P2 in Fig. 9a). The design region 3 is characterized by w2 w w3 , where w3 is the value of w associated with the condition cot 1opt = cot 2opt from Eq. (21), that is

Therefore, beams with very low amounts of transverse reinforcement w w1 fall in the design region 1 and the shear strength is computed through Eq. (24), while the two inclination angles are set equal to their minimum allowed values, as illustrated in Fig. 9c). The design region 2 is characterized by w1 w w 2 where w2 is found by imposing that the value of cot 2opt given in (21)2 (unconstrained optimization) is equal to (cot 2) max , which yields

cot

(2) 1 | w = w2

opt (2) vRd =

(24)

w

5+

The cot 1(2) value in (26) decreases with increasing w , from (cot 1) max = 2.5 to the value 1.0801 in (27), which is illustrated in Fig. 9b), while the corresponding angles are reported in Fig. 9c). Introducing the two values of the inclination angles into one of the two expressions (15) or (16) produces the following expression of the shear strength in terms of w for the design region 2:

Introducing the two limit values of the inclination angles into one of the two expressions (15) or (16) produces a linear expression of the shear strength in terms of w for the design region 1 opt (1) vRd = 3.75

=

Expression (26) for w = w2 gives the abscissa of the limit point P2 in Fig. 9a), which is a bit higher than 1:

(23)

= 0.0716

(2) 1

cot

opt 1

= cot

opt 2

w3

(29)

= 0.25

It can be verified that w3 w,lim , where w,lim was identified before when discussing Fig. 6. In point P3 the two compression struts reach the same inclination, equal to 30°, thus the corresponding coordinates of point P3 in Fig. 9a) are (cot 30 , cot 30 ) = (1.7321, 1.7321) . The abscissa of point P3b is a bit smaller than cot 1 = (cot 1)max = 2.5 and can be found as the intersection of the rational design curve (17) evaluated for w = w3 with the straight line cot 2 = (cot 2 )min = 1, which produces cot 1 |P3b = 2.4142 . Any beam having w2 w w3 (design region 3)

(25)

Any beam having w1 w w 2 (design region 2) can be designed adopting cot 2(2) = (cot 2)max = 5, while the associated cot 1(2) is found as the intersection of the rational design curve with the straight line cot 2 = (cot 2)max = 5, which produces

Fig. 9. Design regions based on the amount of transverse reinforcement: (a) identification on the cot 10

1

cot

2

plane; (b) optimal cot

i

values; (c) optimal

i

values.

Engineering Structures 198 (2019) 109359

D. De Domenico and G. Ricciardi

can be designed adopting cot 1(3) = cot 1opt and cot 2(3) = cot 2opt (where cot 1opt and cot 2opt are given in (21)), while the corresponding shear opt strength is equal to the vRd value given in (22), that is opt (3) vRd

opt vRd =

2 (

w )(1

+8 w 8(1 + 4

(

w ))

w

+ 16 ( w ))

2 w

1+8

w

(30)

The path of the solutions in the design region 3 is represented by the curved line from P2 to P3 in Fig. 9a), whereby the values of cot 2 decrease significantly from (cot 2) max = 5 to cot 30 = 1.7321 and the values of cot 1 increase moderately from 1.0801 to cot 30 = 1.7321 (cf. also Fig. 9b) and the corresponding angles shown in Fig. 9c)). Finally, the design region 4 is characterized by w3 w w,max , where w,max = 0.5 represents the maximum effective transverse reinforcement because further increase of shear reinforcement w > w,max does not contribute to increase of the shear strength [34]. Any beam having w3 w w 4 (design region 4) can be designed adopting for the two inclination angles the same angle of the EC-2 EC2 single strut inclination method, namely cot 1(4) = cot 2(4) = cot , EC2 (where cot is given in (18)), while the associated shear strength is that of the EC-2 model opt (4) vRd

EC 2 vRd =

w (1

Fig. 10. Shear strength vRd versus mechanical ratio of transverse reinforcement w of the improved truss model with two compression struts compared to the EC2 truss model with single compression strut, plasticity theory and RitterMörsch truss model.

Strictly speaking, there is also a design region 5 characterized by > 0.5, in which the failure is sudden due to over-reinforcement. This additional amount of shear reinforcement is useless, because web crushing of concrete governs the shear failure in this range, with a (5) = 0.5 and angles of inclination strut equal to 45° (this constant vRd applies to the improved as well as to the classical EC-2 model).

(31)

w)

w

The path of the solutions in the design region 4 is represented by the straight line from P3 to P4 in Fig. 9a), whereby both the values of cot 1 and cot 2 decrease in equal proportion (lying on the bisector line) from cot 30 = 1.7321 to cot 45 = 1 (cf. also Fig. 9b)). On the other hand, the Eurocode 2 single strut truss model has only two design regions (I and II), separated by the value w2.5. This particular mechanical ratio of transverse reinforcement is found by setting the two expressions (15) or (16) equal to each other for cot 1 = cot 2 = 2.5 (maximum limit values of the EC-2 approach [34]), which produces w2.5

4.3. Influence and meaning of the parameter In the previous subsection the transition depth from one strut inclination to the other has been set equal to = 1/2 . This assumption has facilitated the derivation of analytical design expressions, because solutions of the general optimization problem (14) have been obtained in closed-form as reported in Table 1. In reality, the inclination angle of the compression strut does not change sharply at a specific depth of the beam cross-section, as assumed in the model, but a gradual variation is envisaged along the beam height as discussed in Section 2. It should be taken for granted that three, four or, in general, multiple piecewise inclinations could provide a more consistent representation of the compressive stress fields in web concrete due to the bending-shear interaction, but with a resulting detrimental effect on the simplicity of the formulation. Therefore, the two values of the inclination strut 1 and 2 should be interpreted as mean, representative values along two finite depths of the beam, whose length is governed by the parameter. In some papers from the literature, the crack inclination was analyzed and two branches of the critical crack were considered to write the equilibrium of a part over the shear critical crack. The first branch is usually considered up to the neighborhood of the neutral axis, and the second branch, inside the concrete chord, connects the first branch and the loading point. As an example, Marí et al. [43,44,49] proposed a simple expression to determine the first branch of the critical crack

(32)

= 0.1379

The corresponding values of the inclination angles of the two compression struts of the proposed model are the coordinates of the point P2.5b in Fig. 9a) and are calculated by substituting w2.5 value into the two expressions (21), which leads to

cot

(2.5b) 1

= cot

cot

(2.5b) 2

= cot

opt 1 | w = w2.5 opt 2 | w = w2.5

= 1.1170 (33)

= 4.1404

whose mean is indeed very close to cot = 2.5 of the EC-2 approach. A summary of the design regions of the improved truss model and the corresponding expressions of the shear strength are listed in Table 1. The implications of the improved truss model on the shear strength are depicted in Fig. 10: as said before, the improved truss model coincides with the Eurocode 2 single strut model for w > 0.25. The differences between the two models are more evident for beams with low amounts of transverse reinforcement, say for w < 0.2 , which corresponds to the design regions 1 and 2 and the first portion of the design region 3.

Table 1 Design regions of the proposed truss model with two variable-inclination compression struts. Design region

w

1 2

0

limitations w

cot (cot

w1

cot

1 1)max

w

w2

3

5 + 25 + 104 w 52 w

w2

w

w3

cot

4

w3

w

w,max

cot

5 w1

w,max

= 0.0716 ;

w2

= 0.1136 ;

w3

<

w

= 0.25 ;

cot w,max

= 0.5; (

opt 1

=

EC2

EC2

w)

(cot (cot

= 2.5

w1

2 2704 w

( w) 2 2 w 1

=

cot w

cot

w

=1

=

1+8

cot w

16

2 w

128 11

vRd

2

2) max 2) max opt 2

EC2

3 w;

1

=

opt vRd =

EC2 vRd

w)

=

1+4

w

8

2 w

(

w) .

2 2704 w

2 1+8 2 ( w )(1 + 8 w ( w )) + 16 w w 8(1 + 4 w ( w ))

EC 2 = vRd

w w

=1

(

w

5 + 260 w + 25 + 104 w 104

4 w 1 + 8 w + 2 ( w) 1+4 w ( w)

=

EC2

3.75

=5 =5

= 0.5

w (1

w)

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D. De Domenico and G. Ricciardi

assuming a constant horizontal projection equal to 0.85d , so that the resulting inclination angle is directly related to the neutral axis depth and to the cross-section characteristics [42]. The second branch of the critical crack was also considered, among others, by Zararis and Papadakis [54] to determine size effect assuming a splitting failure mode. On the contrary, in the formulation proposed in this paper these concepts are not applicable because a continuous truss model is considered, without any specific reference to the shear critical crack. The inclination angles 1 and 2 considered in the proposed model represent the principal compressive stress directions between two adjacent cracks that, according to sketch in Fig. 11, may be different from the crack inclinations. In the spirit of a strut-and-tie model with parallel compression chord and tensile tie (resultants C and T ) that do not contribute to the vertical equilibrium, the contribution to the shear resistance is calculated on a length equal to the inner lever arm z . Therefore, in contrast to the two branches of the critical crack recalled above, the transition depth from one strut inclination to the other has no direct relationship to the neutral axis depth. Indeed, according to the sketch in Fig. 11, the second compression strut penetrates into the neutral axis to a certain extent. This extent depends upon the assumed distribution of normal stresses for concrete (e.g., linear, parabolic [54–56], etc.) that influence the position of the resultant compressive force C with respect to the neutral axis and, in turn, the inner lever arm z. On the other hand, given a reasonable curvilinear profile for the principal compressive stress directions as that shown in Fig. 11, the choice of affects the meaning (and values) of the two inclination angles assumed in the model. As an example, taking = 0.3 implies that the two inclination angles 1(b) , 2 (b) would be flatter than those obtained with = 0.5, namely 1(a) , 2 (a) , because they would be average angles representative of an upper portion of the web (the lower branch of depth 0.7z , and the upper branch of depth 0.3z ). In this case, the second compression strut would be confined to a smaller depth than z/2 , and a lower-bound limitation for the second inclination angle 2 smaller than cot 1 5 = 11.3 might be reasonable.

Fig. 11. Influence of the parameter

Based on the above remarks, the most logical choice to capture the progressive variation of principal compressive stress directions throughout the web of the beam is to select the transition depth at the middle of the inner lever arm. Additionally, with any value of different from 1/2 , a computer program would be doubtlessly needed to find a numerical solution of the optimization problem stated in Eq. (14). Consequently, all the straightforward derivations discussed in Sections 4.1 and 4.2 would be no longer valid. We believe that the advantage of the proposed model lies in the possibility of improving the Eurocode 2 truss model without significant additional computational effort. Considering that any other arbitrary value of that is different from 1/2 still represents a simplification of the real behavior, the choice of = 1/2 is justified for inherent computational convenience and for practical design purposes. 4.4. Design and verification with the proposed truss model The closed-form expressions presented in the previous subsection and summarized in Table 1 along with the dimensionless shear strength graph of Fig. 10 for = 1/2 facilitate the application of the improved truss model for practical design and verification purposes. In particular, for verification purposes the mechanical strength of the materials (concrete and steel), the geometrical dimensions of the beam and the reinforcement details are known. The design value of the applied shear force VEd is also known. The normalization parameter r = b w z 1 fcd can be computed, where z can be assumed equal to 0.9d [50]. Then, the dimensionless value vEd = VEd/r is calculated. Moreover, the mechanical ratio of transverse reinforcement w is calculated through Eq. (4), from which the design region in which the beam falls is directly identified. The verification is carried out by comparing the value of vEd with the dimensionless shear strength vRd for the specific design region in which the beam falls (with application of appropriate safety factors). For design purposes, the amount of transverse reinforcement w (related to the spacing of the steel stirrups) is unknown, but the other

on the representative inclination angles of the two compression struts. 12

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D. De Domenico and G. Ricciardi

material and geometrical parameters are supposed to be known. As a first step, we verify whether vEd exceeds 0.5, in which case the dimensions of the beam (bw , d ) or the mechanical strength of concrete ( fcd ) have to be modified (increased). As a second step, we identify the design region in which vEd is located by drawing a horizontal line at vRd = vEd in Fig. 10 and intercepting the vRd curve. Once the design region is established, we refer to just one of the specific vRd w relationships reported in the last column of Table 1, say v¯Rd ( w ) . Thus, the minimum amount of transverse reinforcement w0 can be evaluated by imposing, at the limit, v¯Rd ( w 0 ) = vEd (with application of appropriate safety factors). The beam is designed with a higher amount of transverse reinforcement than the minimum value w > w0 (e.g. increasing the area Asw or reducing the spacing s in (4)). When designing the transfer reinforcement, a minimum transverse reinforcement ratio w, min should be provided even when no shear reinforcement is required. Since the proposed procedure modifies the EC-2 formulation but is based on an identical theoretical framework, the expression in §9.2.2 of the EC2 [34] can be adopted, depending on the class of steel and concrete. A flow-chart is presented in Fig. 12 to summarize the main steps of the shear strength calculation procedure for the proposed model in the closed-form ( = 1/2 ) and numerical (free ) formulations, while two worked examples are reported in Appendix A.

5. Comparison with experimental results 5.1. Verification of compression strut angles with experimental data The compression strut angles 1 and 2 have been determined via the lower-bound theorem of plasticity, solving the optimization problem (14). As said earlier in Section 4.3, they should be interpreted as mean, representative values of the strut inclination along a finite portion of the beam height (depending on the chosen ). In general, the angle of compression strut is that of the post-cracking principal compressive stress direction, which may be flatter than the actual angle of inclined crack observed in the test. It seems interesting to check how the two compression strut angles, determined via optimization, fit experimental observations. Walraven et al. [11] carried out experimental measurements on a set of I-shaped beams for different shear reinforcement ratios and for different classes of concrete. A system of LVDTs arranged in triangles were placed at the web of the beams to measure the web deformation along different directions. The principal compressive strain directions were then calculated in accordance with Mohr’s circle, and the strut rotation angle was reported under the assumption of principal stress directions coincident with principal strain directions (also used in the MCFT [12]). The experimental strut rotations for 12 beams having w = w fyw /( 1 fc ) < 0.5 are reported in Fig. 13 together with the two inclination angles 1 and 2

Fig. 12. Flow-chart of the shear strength calculation procedure of the proposed model. 13

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D. De Domenico and G. Ricciardi

of the proposed truss model, already illustrated in Fig. 9c. The mean inclination angle of the proposed model is computed through = 1/2 ). m = ( 1 + 2 )/2 (which is reasonable under the assumption There is a good agreement between predicted mean inclination m and strut rotation measured from the experiments, with an average ratio exp / m = 0.92 and a coefficient of variation (CoV) of 6.76%, as reported in Table 2. Other 6 beams tested in [11] having w > 0.5 are purposely excluded in this comparison because were associated with minimal strut rotations ( exp 40 45 ), which is of limited interest since in this range the two strut inclinations are both equal to 45° as in the EC-2 model (cf. Table 1). The predictions with the analytical formula proposed by He et al. [41] for a modified variable-angle truss model combining the theory of plasticity with the MCFT are also superimposed in Fig. 13 for comparison. These latter predictions are also reasonably consistent with the experimental results as well as with the mean inclination angles obtained with the proposed truss model via optimization.

normalization parameter r = b w z 1 fc . This allows a clear graphical comparison in the w v plane of the two truss models, which is reported in Fig. 14. It is clearly seen that the proposed truss model is better able to describe the mean trend of the experimental data in the range of lightly shear-reinforced beams (say w < 0.1) where the EC-2 formulation turns out to be conservative. From a mechanical point of view, this improvement is related to the introduction of the second compression strut that enables the construction of a wider class of statically admissible solutions than the EC-2 truss model. Many rational approaches to predicting shear capacity of RC beams have been developed, many of which can be solved analytically. Besides comparing with the EC-2 formulation, other simplified approaches are considered, namely the additive approach of the ACI 318-11 [10], the modified variable-angle truss model developed by He et al. [41], which is also a modification of the EC-2 method based on plasticity theory in conjunction with MCFT, the shear strength design formula proposed by Russo et al. [60,61] based on a modified version of the ACI 318-11 formulation, and the Model Code 2010 level III approximation [29], which is based on a simplified version of the MCFT. The aim of this comparison is neither to assess the best method nor to replace the existing code-based formulations. Indeed, only few simplified methods are selected, based on reasons of implementation simplicity or declared relationship with existing code formulations – the comparison with all codes and research-based methods is obviously beyond the scope of this paper. Rather, the purpose of this comparison is to demonstrate that the proposed improved EC-2 truss model provides accurate predictions, despite the relatively comparable simplicity to other methods proposed in codes or alternative analytical formulations based on different underlying mechanisms being included in the model. The expressions are applied without any partial safety factors, and the mean values of the material strength parameters are used for the calculations. Consequently, the results compared are not exactly the real predictions of the various models because fck or fc should be used depending on the formulation employed. The trend of predicted versus experimental shear strength can be assessed through the plots in the vpred vexp plane shown in Figs. 15 and in 16 for the two considered ACI DafStb databases, respectively. It can be seen that in the improved EC-2 model the majority of the couples (vexp, vpred ) fall close to the 45° line and are approximately equally distributed above and below the 45° line, meaning small bias on average. In the other considered code-based and analytical approaches the predictions tend to underestimate, on average, the actual shear capacity observed experimentally. The performance of the considered models in terms of vexp/ vpred ratios is summarized in Tables 4 and 5 for the two considered ACI DafStb databases, respectively. The improved EC-2 model with closed-form expressions ( = 1/2 ) produces a mean vexp/ vpred ratio equal to 1.05 and 1.10, and a CoV equal to 18.49% and 23.06% for the two databases, respectively. These simple statistics

5.2. Verification of shear strength with experimental data The shear strength predictions of the improved truss model proposed in this paper are here compared to experimental results of RC beams reported in the literature. The selection of a reliable and homogeneous database is of utmost importance to provide a fair and representative validation of the proposed model. Many databases of shear test results were reported in the literature, see e.g. [27,57,58]. In this context, selected test results were reported by Reineck et al. [45,46] in the ACI-DafStb databases, which are adopted here to reduce the influence of database heterogeneity. These shear strength results were extracted from an extensive database of 886 collected tests of RC beams with vertical stirrups ( = 90 ) subjected to point loads, through a series of clear and transparent control criteria (e.g., flexural and anchorage checks, minimum compressive strength, maximum spacing of stirrups, lack of essential information in the source test reports, etc.). In particular, two databases are here separately considered: 1) the small evaluation database presented in [45], which comprises 87 tests; 2) a larger database presented in [46], which includes 170 tests. We do not consider the large evaluation database of 157 tests reported in [45], because some of these tests were included in a later and updated version of DafStb H. 617 database with 170 tests reported in [46], which is the second database adopted in this paper (with 42 different references). To make a consistent comparison with the proposed model, the shear reinforcement ratio c = w f yw / fc is calculated for each beam (where the subscript c in c refers to the use of the compressive strength of concrete at the denominator, to avoid confusion with w in which the reduced web compressive strength fcw = 1 fc appears). Based on the study of Lee and Hwang [59], when c > 0.2 over-reinforced shear failure takes place, which implies that the crushing of web concrete occurs before yielding of shear reinforcement. However, this is in contrast to the main assumptions of the proposed plasticity-based model, cf. Eqs. (3) and (5), as well as of the original EC-2 truss model, which was also noted by other authors [41]. In particular, 4 of the 87 beams in the ACI-DafStb small evaluation database [45] and 5 of the 170 beams in the ACI-DafStb H. 617 database [46] have c > 0.2 . It is, therefore, reasonable to exclude these beams in the comparison, and to consider the remaining 83 tests and 165 tests in the two databases, respectively. The range of significant input parameters for the two resulting databases are listed in Table 3. On the other hand, all the considered beams fulfilled the ACI 318-11 criterion on the minimum reinforcement ratio [10]. As first step, since the proposed truss model modifies the Eurocode 2 formulation in the range w < w lim = 0.25, it is interesting to compare the predictions of the EC-2 truss model against those of the improved truss model in such range. From the experimental shear strength Vexp the dimensionless shear strength vexp is calculated by dividing Vexp with the

Fig. 13. Experimental versus predicted strut inclination angles for I-shaped beams tested by Walraven et al. [11]. 14

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Table 2 Verification of compression strut angles for beams tested by Walraven et al. [11]. Beam designation AE30L AE30M GD30L LG30L LG30M LR30L LR30M LG60L LG60M GD90L GD90M GD90H

†

w (%)

0.43 0.887 0.43 0.43 0.887 0.43 0.887 0.66 1.25 0.68 1.52 2.65

f yw (MPa)

fc (MPa)

fcw =

560 608 560 560 608 560 608 608 608 540 567 567

21.90 19.80 17.20 16.80 22.70 29.00 28.10 41.10 41.20 83.00 83.90 84.80

11.99 10.94 9.61 9.40 12.38 15.38 14.96 20.61 20.65 33.27 33.45 33.62

The efficiency factor is computed through

1

= 0.6(1

1 fc (MPa)

†

[°]

w

exp

0.20 0.49 0.25 0.26 0.44 0.16 0.36 0.19 0.37 0.11 0.26 0.45

26.00 38.00 28.00 29.00 37.00 28.00 38.00 28.00 32.00 22.00 29.00 35.00

m

[°]

29.10 44.43 30.00 30.66 41.55 28.24 36.87 28.90 37.47 25.43 30.66 42.13 Mean std CoV (%)

exp/ m

0.89 0.86 0.93 0.95 0.89 0.99 1.03 0.97 0.85 0.87 0.95 0.83 0.92 0.06 6.76

fc /250) as recommended by the Eurocode 2 [34].

Table 3 Range of input parameters in the considered databases. ACI-DafStb small evaluation database (83 tests) [45]

ACI-DafStb H. 617 database (165 tests) [46]

Input parameter

min

max

mean

min

max

mean

b w (mm) d (mm) a/d ( ) (%)

50.0 198.0 2.448 0.473

457.2 1200.0 7.102 4.725

198.0 470.2 3.327 1.880

75.0 161.0 2.444 0.138

457.2 1200.0 7.102 5.204

196.8 381.1 3.257 2.319

0.079

1.678

0.435

0.070

2.646

0.348

15.7

125.3

50.4

13.4

125.3

53.7

fy w (%)

f yw (MPa)

fc (MPa) w

Vexp (kN)

271.0 270.0 0.018 94.0

990.0 820.0

0.334 1330.0

481.5 451.6

0.095 362.6

271.0 229.0 0.017 81.0

990.0 820.0

0.484 1330.0

501.0 459.0

0.071 277.3

highlight the improved performance of the proposed model over the EC-2 truss model, which is instead associated with a mean vexp/ vpred ratio equal to 1.39 and 1.56, and a CoV equal to 24.02% and 31.43%. Building on this observation, the proposed truss model, viewed as an improved version of the EC2 single strut model, provides better estimates of the shear strength that are more in line with the actual experimental results (smaller bias and higher precision). Moreover, the proposed model is associated with the average vexp/ vpred ratio closest to the unity (1.05 and 1.10) and with the lowest standard deviation value (0.194 and 0.253) among the considered formulations for both the databases (although not the lowest CoV for the small evaluation database). The comparison of simplified (closed-form) versus numerical formulations is shown in Fig. 17. The predictions of the numerical formulation are obtained with = 0.3, solving the optimization problem (14) numerically, while the simplified formulation is based on the closed-form expressions in Table 1. It is clearly seen that the two formulations are in very good agreement with each other. Moreover, based on the observations outlined in Section 4.3 on the meaning of the parameter, two different lower-bound limitations of the second inclination angles are investigated, namely (cot 2) max = 5 (as in the simplified formulation, corresponding to 11.31°) and (cot 2) max = 7 (corresponding to 8.13°). It can be observed that reducing the lowerbound limitation of the second angle, as motivated in subsection 4.3, leads to results that are almost identical to those of the simplified formulation with = 1/2 , because the same average inclination would be obtained considering the two different depths (mean vnumerical/ vsimplified = 0.97 and CoV of around 1%). Considering the

Fig. 14. Comparison of experimental data against shear strength predictions from EC-2 truss model and improved EC-2 truss model for the ACI-DafStb databases [45,46] in the w v plane for w < 0.25.

computational simplicity of the closed-form expressions reported in Table 1, the use of the simplified formulation is certainly preferable for use in daily engineering practice. Finally, in Fig. 18 the trend of the vexp/ vpred ratio is reported as a function of the shear reinforcement ratio c = w f yw / fc for the proposed approach compared to EC-2 [34], ACI 318–11 [10] and He et al. [41] formulations. It is clearly noticed that the underestimation of shear capacity for low amounts of shear reinforcement, observed in alternative formulations, is significantly mitigated in the proposed approach, with vexp/ vpred ratios being much closer to the unity. Obviously, this in turn implies that a slightly higher number of unconservative predictions (vexp/ vpred < 1) is obtained, so that appropriate safety factors for the model should be carefully determined in order to account for the inherent material strength and model uncertainties. The predictions of the proposed improved EC-2 model coincide with those of the Eurocode 15

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D. De Domenico and G. Ricciardi

Fig. 15. Comparison of predicted versus experimental shear strength using various methods for ACI-DafStb small evaluation database [45] (corresponding mean and CoV of the vexp/ vpred ratio are reported in Table 4).

Fig. 16. Comparison of predicted versus experimental shear strength using various methods for ACI-DafStb H. 617 database [46] (corresponding mean and CoV values of the vexp/ vpred ratio are reported in Table 5).

16

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D. De Domenico and G. Ricciardi

Table 4 Comparison of predicted versus experimental shear strength data for ACI-DafStb small evaluation database (83 tests) [45]. vexp/vpred ratio Mean Standard deviation CoV (%) Minimum Maximum

ACI 318-11

EC-2

Improved EC-2 (proposed)

Russo et al. [60,61] (shear strength formula)

He et al. [41] (plasticity theory + MCFT)

MC-2010 [29] (level III)

1.48 0.335 22.58 0.69 2.33

1.39 0.333 24.02 0.82 2.28

1.05 0.194 18.49 0.57 1.52

1.46 0.345 23.58 0.65 2.43

1.39 0.236 16.99 0.84 2.03

1.40 0.323 23.11 0.66 2.45

Table 5 Comparison of predicted versus experimental shear strength data for ACI-DafStb H. 617 database (165 tests) [46]. vexp/vpred Mean Standard deviation CoV (%) Minimum Maximum

ACI 318-11

EC-2

Improved EC-2 (proposed)

Russo et al. [60,61] (shear strength formula)

He et al. [41] (plasticity theory + MCFT)

MC-2010 [29] (level III)

1.50 0.358 23.91 0.69 3.03

1.56 0.489 31.43 0.52 3.11

1.10 0.253 23.06 0.39 2.07

1.26 0.412 32.77 0.65 2.63

1.46 0.349 23.99 0.62 2.54

1.45 0.363 25.09 0.66 2.89

2 for high shear reinforcement ratios c (which is clearly seen in the comparisons shown in the top part of Fig. 18). This justifies the relatively higher values of CoV reported in Table 5 for the ACI-DafStb H. 617 database (around 23%) in comparison with alternative formulations proposed in the literature, e.g. the mechanical model proposed by Marí et al. [49], which reported CoV values of around 17% for the same database.

motivated by the variability of principal compressive stress direction, due to the increase of shear stresses in the upper portion of the beam. We have pointed out that these effects are more pronounced for beams with low amount of transverse reinforcement, in which the crack widths are more evident in the bottom part and reduce moving upwards due to concurrent bending effects. These phenomena modify the isostatic lines of compression and, consequently, the inclinations of the compression strut, which justifies values of 2 lower than 1. In the authors’ opinion, it is exactly in this range of beams with low shear reinforcement that the proposed model with two strut inclinations may lead to better estimates of the actual shear strength than the EC-2 formulation. Conversely, for beams with high amount of transverse reinforcement, the crack openings are controlled by stirrups, so that shear stresses tend to be more uniformly distributed, without significant

6. Conclusions This paper has presented a shear strength model obtained by an upgrade of the EC2 truss model with two (rather than one) representative inclinations of the compression strut along the beam height, called 1 (lower inclination) and 2 (upper inclination). This is

Fig. 17. Comparison of shear strength predictions between simplified (closed-form) formulation (with vnumerical . 17

= 1/2 ) vsimplified versus numerical formulation (with

= 0.3)

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D. De Domenico and G. Ricciardi

Fig. 18. Comparative trend of vexp/ vpred ratio versus shear reinforcement ratio

differences along the beam height and with minimum strut rotation [11]. Consequently, in this range the introduction of the second inclination angle does not produce significant variations of the results compared to the EC-2 model. Similarly to the EC2 variable strut inclination method, the two inclinations, variable within a certain range, have been identified in the framework of the lower-bound theorem of plasticity, by solving an optimization problem. In particular, the improved EC-2 truss model enables the construction of a wider class of statically admissible solutions (combinations of 1, 2 ) that were excluded a priori by the EC-2 single strut approach. In a general variant of the model, the depth of the two compression struts has been set equal to z (upper strut) and (1 ) z (lower strut), with 0 < < 1 an arbitrary parameter and z the inner lever arm of the beam cross-section. Dimensionless analytical expressions have been derived, and numerical solutions of the shear strength evaluation problem can be determined for any value of transition depth by maximizing the shear strength within the statically admissible domain. The influence and meaning of the parameter has been analyzed. The most logical choice to capture the progressive variation of principal compressive stress directions throughout the web of the beam is to select the transition depth at the middle of the inner lever arm ( = 1/2 ). Moreover, this assumption facilitates the derivation of analytical solutions resulting from the general optimization problem, which is convenient for practical design purposes in daily engineering practice. Specific design regions have been identified depending upon the mechanical ratio of transverse reinforcement w , and

c

for different methods.

the evaluation of the shear strength can be easily performed through either simple closed-form design expressions (listed in Table 1) or a convenient dimensionless design graph (shown in Fig. 10). Guidelines on how to proceed with these two design tools have been provided in Section 4.4, with the aid of a flow-chart summarizing the main steps of the shear strength calculation procedure. Two worked examples have also been presented in Appendix A for the sake of clarity. The new model has been validated against experimental results from the literature. In particular, the two strut inclination angles determined via optimization seem to be well correlated to experimental strut rotation angles for I-shaped beams reported by Walraven et al. [11] for a range of shear reinforcement ratios. Moreover, the shear strength predictions from the improved EC-2 model have been compared to more than 200 experimental shear strength results reported in two well-established ACI-DafStb databases [45,46]. The proposed plasticity-based model with closed-form expressions, viewed as an improvement of the EC2 single strut approach, is better able to capture the experimental shear strength results than the EC-2 approach (with smaller bias and higher precision), despite the relatively comparable simplicity. In particular, the mean vexp/ vpred ratio is equal to 1.05 and 1.10, and the standard deviation is equal to 0.194 and 0.253 for the two databases, respectively, while the EC-2 model has mean equal to 1.39 and 1.56, and standard deviation equal to 0.333 and 0.489. The underestimation of the shear capacity for low amounts of shear reinforcement, observed in the EC-2 formulation as well as in other approaches, is mitigated in the proposed approach, with vexp/ vpred ratios 18

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D. De Domenico and G. Ricciardi

being much closer to the unity. Since the limitations on the upper inclination angle 2 are additional variables of the problem (while those on the lower angle 1 can be assumed the same as in the EC2 model), the practical implications of adopting two different variability ranges of 2 in combination two difparameters have been critically analyzed. The choice ferent (cot 2) max = 5, extensively discussed in Section 4, seems to provide a good prediction of the selected experimental results. However, different design choices can certainly be developed based on the general theoretical concepts elaborated in this paper.

Acknowledgements The corresponding author would like to express his personal gratitude to Professor Karl-Heinz Reineck for assistance in selecting the experimental data and for providing clarifications regarding the ACIDafStb databases used in this paper. The constructive comments of the three anonymous Reviewers have contributed to the improvement of the manuscript in comparison with an earlier version.

Appendix A. Examples of shear strength calculation The proposed analytical expressions reported in Table 1 for the closed-form formulation as well as the numerical formulation summarized in the flow-chart of Fig. 12 are here applied to two different beams whose mechanical and geometrical data were reported in the literature. One of the two selected beams has w1 < w < w2 and falls into design region 2, while the other beam has w2 < w < w3 and falls into design region 3. In this way, the two proposed examples cover most of the possible design situations for lightly shear-reinforced beams. The closed-form expressions can be implemented in any spreadsheet, while the optimization problem for values other than = 1/2 can be solved through any nonlinear programming solver. To this aim, the built-in MATLAB fmincon subroutine, which adopts an interior-point algorithm [62], has been used. A.1. Example 1 – Design region 2 The first example concerns beam T-2-B tested by Soerensen [63], which is also included in the ACI-DafStb large evaluation database presented by = 1.054%, w = 0.305%, fyw = 397 MPa , Reineck et al. [45,46]. The beam presents the following characteristics: bw = 110 mm , d = 298 mm , fc = 24.9 MPa . It failed under a shear force Vexp = 130.95 kN . Based on the above data, as a first step the efficiency factor (assuming fcd = fc ) and the mechanical ratio of transverse reinforcement w (assuming fywd = fyw ) are calculated 1

(

= 0.6 1

w

=

f ywd w f cd

fc 250

=

) = 0.6(1

0.305 100

×

24.9 250

) = 0.54024

397 0.54024 × 24.9

= 0.90013

(A1)

Since 0.0716 < w < 0.1136 this beam falls into design region 2, and the corresponding expressions of the inclination angles and normalized shear strength are given by Eqs. (26), (25) and (28), respectively, which lead to 5 + 25 + 104 × 0.90013

cot

1

=

cot

2

= cot

vRd =

2704(0.90013)2

52 × (0.90013) 2,max

=5

2

= 1.8221

1

= 28.76

= 11.30

5 + 260 × 0.90013 + 25 + 104 × 0.90013 104

2704(0.90013)2

= 0.30704

(A2)

The normalization parameter for this beam is r = 110 × (0.9 × 298) × 0.54024 × 24.9 = 3.9698 × 105 (in [N] units). Multiplying the normalized shear strength vRd from (A2) with the normalization parameter r gives VRd = 0.30704 × (3.9698 × 105)/103 = 121.89 kN , which compared to the failure shear force gives a ratio Vexp/ VRd = 1.075. On the other hand, the Eurocode 2 truss model for this beam provides a maximum shear force equal to VEC2 = 89.33 kN , which corresponds to a ratio Vexp/ VEC 2 = 1.466. This example confirms the conservative nature of the EC-2 formulation for lightly shear-reinforced beams and the better performance of the proposed formulation. The numerical formulation with not fixed a priori is also applied for comparative purposes, as summarized in Fig. 12. Assuming = 0.3 and (cot 2) max = 7 the optimization problem (14) gives vRd = 0.31707 , which, multiplied with the normalization parameter VRd = 0.31707 × (3.9698 × 105)/103 = 125.87 kN . Compared to the failure shear force, this gives a ratio Vexp/ VRd = 1.040 . A.2. Example 2 – Design region 3 The second example concerns beam ET3 tested by Leonhardt and Walther [64], which is also included in the ACI-DafStb small evaluation = 1.396%, w = 0.514%, database presented by Reineck et al. [45]. The beam presents the following data: bw = 100 mm , d = 300 mm , fyw = 314 MPa , fc = 23.75 MPa . It failed under a shear force Vexp = 126.25 kN . Repeating the same calculations for 1 and w gives 1

(

= 0.6 1

w

=

f ywd w f cd

fc 250

=

) = 0.6(1

0.514 100

×

23.75 250

314 0.543 × 23.75

) = 0.543

= 0.12515

(A3)

Since 0.1136 < w < 0.25 this beam falls into design region 3, and the corresponding expressions of the inclination angles and normalized shear strength are given by Eqs. (21) and (30), respectively, which lead to

19

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D. De Domenico and G. Ricciardi

( (

w) w)

cot cot vRd =

= =

1 2

1 + 8 × 0.12515 1 + 4 × 0.12515

16 × (0.12515) 2 8×

(0.12515) 2

=

0.38817 2 2 × 0.12515

=

4 × 0.12515 1 + 8 × 0.12515 + 2 × 0.38817 1 + 4 × 0.12515 1.2246

= 1.0966

1

128 × (0.12515)3 = 1.2246 1.2246 = 0.38817

= 42.36 = 4.5552

2

= 12.38

2 × 0.38817(1 + 8 × 0.12515 1.2246) + 16 × (0.12515)2 1 + 8 × 0.12515 8 × (1 + 4 × 0.12515 1.2246)

= 0.35366

(A4)

The normalization parameter for this beam is r = 100 × (0.9 × 300) × 0.543 × 23.75 = 3.482 × 105 (in [N] units). Multiplying the normalized shear strength vRd from (A4) with the normalization parameter r gives VRd = 0.35366 × (3.482 × 105)/103 = 123.14 kN , which compared to the failure shear force gives a ratio Vexp/ VRd = 1.024 . On the other hand, the Eurocode 2 truss model for this beam provides a maximum shear force equal to VEC2 = 108.94 kN , which corresponds to a ratio Vexp/ VEC 2 = 1.159. The numerical formulation with = 0.3 provides vRd = 0.3777 , which corresponds to VRd = 131.52 kN and a ratio Vexp/ VRd = 0.9599.

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