Dimensioning for the ULS in shear

To preclude shear failure, each large wall is dimensioned for a shear force, VFd. obtained by multiplying the shear force from the analysis for the design seismic action, VEd, by a magnification factor e:

For the usual value of q = 3 applying to systems of large lightly reinforced walls, the value of e is equal to 2, and exceeds that given by equation (D5.19) for ductile walls of the same ductility class (M). Moreover, as

• the rules for dimensioning the vertical reinforcement explicitly request minimization of the flexural overstrength, MRd/MEd

• the period of the fundamental mode in the direction of the length of the wall, Tx, is normally not (much) longer than the corner period of the spectrum, Tc, the value of e from equation (D5.52) is of the order of that given by equation (D5.18) for slender ductile walls of DCH, and exceeds those given by equation (D5.17) for squat ductile walls of DCH.

As the magnification factor e provides a large margin between the design shear force, VEd = eVEd, and the value from the analysis, VEd, and, moreover, the vertical reinforcement is dimensioned for minimum flexural overstrength, it is allowed not to place in large lightly reinforced walls the minimum amount of smeared horizontal reinforcement, if the design shear force, VEd = eVEd, is less than the design shear resistance of concrete members without shear reinforcement, FRd c, according to Eurocode 2, given by equation (D5.49). The requirement for horizontal reinforcement is more relaxed than for non-seismic actions because, if inclined cracks form despite fulfilment of the verification VEd < VRd c, their width will not grow uncontrolled as in walls without horizontal reinforcement under force-controlled actions (e.g. wind), but will soon close due to the transient and deformation-controlled nature of the seismic action. Moreover, due to the large horizontal dimension of the wall, /w, any inclined cracks will intersect a floor and mobilize the horizontal ties required to be placed at its intersection with the wall, as well as part of the slab reinforcement in the immediate vicinity of the wall that runs parallel to /w.

Strut And Tie Model
Fig. 5.11. Design of a large wall with openings, using the strut-and-tie model

If ^Ed > ^Rd c horizontal reinforcement should be calculated according to Eurocode 2, on Clauses the basis of either a variable strut inclination model for shear resistance, or a strut-and-tie 5.4.3.5.2(2), model, depending on the geometry of the wall. The first type of model is appropriate for 5.4.3.5.2(3) walls without openings. Eurocode 2 provides for a strut inclination with respect to the vertical, 9, between 22 and 45° and allows calculating the required horizontal reinforcement on the basis of the minimum value of the shear force within lengths of z cot 9, where z is the internal lever arm, normally taken equal to 0.8/w. Experimental and field evidence suggests that in large walls under lateral loading the struts follow a fan pattern up to a distance z from the base of the wall; from then up, they are at an angle 9 of 45°, intersecting the floors and mobilizing them as ties. The implication for design is that wall horizontal reinforcement should be calculated for 9 = 45°, starting with the value of the shear force atz = 0.8/w from the base and taking into account as part of the shear reinforcement the cross-section of the ties placed at the intersection of the wall with the floors. The floors should be included as ties in any strut-and-tie model due to be used in the presence of significant openings in the wall (see Fig. 5.11). If the geometry of the wall and its openings is not symmetric with respect to the centreline, a different strut-and-tie model should be constructed for each sense of the seismic action parallel to the plane of the wall (positive or negative). Struts should avoid intersecting the openings, and their width should not be chosen to be more than 0.25/w or 4bvo, whichever is smaller.

If VEd > Vm c, and horizontal reinforcement needs to be calculated according to Eurocode 2, then a minimum amount of smear horizontal reinforcement should be placed. For large lightly reinforced walls this minimum amount is a Nationally Determined Parameter with a recommended value equal to the minimum horizontal reinforcement required by Eurocode 2 in walls subjected to non-seismic actions. It should be recalled that, according to Eurocode 2, wall horizontal reinforcement should be placed at a maximum bar spacing of 0.4 m and at a minimum ratio which is a Nationally Determined Parameter, with a recommended value of 0.1% or of the ratio of web vertical reinforcement, whichever is greater. Clause The shear force Vf d computed at construction joints at floor levels from equation (D5.52)

5.4.3.5.2(4) should be verified against the design resistance of the interface in sliding, VRdi, taken according to Eurocode 2. This latter is equal to the shear stress given by equation (D5.51), multiplied by ¿>woz. The values of the coefficients 0.35 and 0.6 for cohesion and friction, respectively, apply for a naturally rough free concrete surface without treatment. If the surface is artificially roughened through raking and exposure of aggregates to an average of 3 mm of roughness about every 40 mm, these values may be increased to 0.45 and 0.7, respectively. An additional requirement with respect to Eurocode 2 is that the anchorage length of the clamping bars included in pv should be increased by 50% over the normal value required in Eurocode 2. This does not mean that all vertical bars crossing the interface need to have their anchorage length increased: the requirement applies only to those bars that need to be included in pv so that VEd < VRdj.

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