EUROCODE 2 WORKED EXAMPLES - Federbeton

EC2 – worked examples 6-14
Taking d = 450 mm it results:
VRd,c = 0,12⋅1.63⋅ (30) 1/3 ⋅ 300 ⋅ 450 = 82.0 kN
The diagram is shown in Fig.6.5
The section, in the range of action effects defined by the interaction diagram, should have a
minimal reinforcement in accordance with [9.2.2 (5)-EC2] and [9.2.2 (6)-EC2]. Namely,
the minimal quantity of stirrups must be in accordance with [9.5N-EC2], which prescribes
for shear:
(Asw / s⋅bw) min = (0.08 ⋅ √fck)/fyk = (0.08 ⋅ √30)/450 = 0.010
with s not larger than 0.75d = 0,.75⋅450 = 337 mm.
Because of the torsion, stirrups must be closed and their pitch must not be larger than u/8,
i.e. 200 mm. For instance, stirrups of 6 mm diameter with 180 mm pitch can be placed. It
results : Asw/s.bw = 2⋅28/(180⋅300) = 0.0010
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Fig. 6.5 V-T interaction diagram for lightly stressed section