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2 CODE FORMULATIONS FOR SHEAR CAPACITY
The variable inclination method proposed in Eurocode 2 (EN 1992-1-1: 2004) is a simple equilibrium method which gives a transparent view of the flow of force in the structure. If the shear reinforcement yields the truss can, by rotation of the compression struts to a lower inclination, activate more reinforcement, Fig. 2; the shear resistance provided by stirrups is given in
Eq. (1): VR,s = Asws• z • fyw • cot θ (1)where Asw and s are the cross sectional area and the
spacing of the shear reinforcement, z is lever arm, fyw is the steel yield strength and bw is the width of the web.Due to strut rotation the stress in the concrete struts increases, consequently rotation can only continue until the crushing of the concrete occurs, Fig. 3; the corresponding shear force is given in Eq. (2):VR,c = bw • z • fc1 • αc cot θ + tan θ(2) where fc1 = νfc and αc is a coefficient that takes into account the prestressing effects; it increases the shear
resistance and depends on the compressive stress level and it is used as a multiplier of the compressive concrete strength. If the beam is not prestressed αc can betaken equal to 1.
According to ACI 318-08 model (ACI Committee,2005) the shear capacity VR is the sum of the concrete
contribute, the shear reinforcement contribute and aterm taking into account the effect of prestressing force N which gives a benefit in the ultimate shear capacity calculation, Eq. (3) (Joint ACI-ASCE, 1962):VR = 16 1 + N14 • A fc + ρ swfyw bwd (3)where A is the gross area of the section, ρsw is the shear reinforcement ratio and d is the depth of the beam. The CSA formulation, as mentioned above, is based on a variable inclination truss model and refers to the MCFT theory (Vecchio, Collins, 1986). According to this formulation the ultimate shear capacity is, as in
ACI Code, the sum of the concrete contribute and the steel contribute and it is given as:VR =
0.41 + 1500εx fcbwdv + Aswfywdv cot θs(4) where dv is taken as max{0.72h; 0.9d} and the strut
inclination, given by Eq. (5), depends on the longitudinal strain εx at mid-depth of the member, Eq. (6):θ = 29◦ + 7000εx (5)εx = V(1 + M/Vdv) + 0.5N − Apfp02(EsAs + EpAp)(6)where Ap, Ep, fp0 are the area, Young modulus and yielding stress of prestressing reinforcement; while Es and As are the Young modulus and area of longitudinal reinforcing bars on the flexural tension side of the member.Several authors during the years (Had dad in and Hong, 1971, Gupta, 1995, Yoshida, 2000) have studied the shear failure mechanism and compared the experimental results obtained with the international codes mentioned above. Because of the difficulties of the physical reality in many cases the semi-empirical formulations adopted in the international codes are too
conservative in the calculation of the shear capacity.In this paper the effects of the coefficients that influence the shear capacity, especially those factors regarding the crushing of concrete and the effect of the prestress, will be better analyzed. The coefficient
ν takes into account that after cracking the compressivestrength of the struts decreases because of the occurrence of transversal tension. In the years ν has become into several changes according to experimental results. This coefficient has been studied and modified by
several authors, such as Watanube and Kabeyasawa (Watanube and Kabeyasawa, 1998) whom theories
have been included in the design guidelines of the Architectural Institute of Japan (AIJ, 1988). In Table 1 different relationship for ν calculation are shown according to different codes.prestressed members with shear reinforcement it can be seen that there is apparently an increase of both safety and scatter. In Fig. 4(a) and 4(b) respectively the experimental results obtained for prestressed beams are compared with the shear resistance calculated according to the variable inclination method proposed by Eurocode 2 without considering and by considering