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2.4.5 Shear reinforcement 43 2

2.4.5.1 Minimum links

Here, for comparison with BS 8110 design, grade 250 reinforcement will be 5.4.2.2 used.

Interpolation from EC2 Table 5.5 gives Minimum

AJs = 0.0022 x 300 = 0.66 mm2/mm lf ^sd - ("if) ^fw - refer to Section 2.4.5.3 for VRd2

2.4.5.2 Capacity of section without shear reinforcement 4.3.2.3

Assume 2T25 effective

P; = 982/(300 x 440) = 0.00743 k = 1.6 - d = 1.6 - 0.44 = 1.16

^Rdi = 300 x 440 x 035 x 1-16 x (1.2 + 40 x 0.00743) x 10"3 = 80.2 kN

2.4.5.3 Shear reinforcement by standard method 4.3.2.4.3

Maximum capacity of section v = 0.7 - y200 = 0.7 - 32/200 = 0.54 < 0.5 Eqn 4.21

VRd2 = 0.5 x 0.54 x (32/1.5) x 300 x 0.9 x 440 x 10"3 = 684 kN Eqn 4.25

Design shear force is shear at a distance d from the face of the support. This 4.3.2.2(10) is 590 mm from the support centreline.

Design of shear reinforcement is summarized in Table 2.3.

Table 2.3 Design of shear reinforcement
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