Design for high beam shear support B

As uniformly distributed load predominates consider at d from face of support

VEd = 1098 - (350 / 2 + 0.689) x (1.35 x 46.0 + 1.5 x 633)

= 1098 - 0.864 x 157.1 = 962.3 kN By inspection, shear reinforcement required and cot 0 < 2.5. Check VRd max (to determine 9) Check maximum shear resistance As before

6 = 0.5 sin-1 {^ / [0.20 fck (1 - fck / 250) ] } > cot-12.5 where

= 962.3 x 103 / (350 x 0.9 x 687) = 4.45 MPa 6 = 0.5 sin-1 {4.45 / [0.20 x 30 (1 - 30 / 250) ] } > cot-12.5 = 0.5 sin-1 (0.843) > cot-12.5 = 0.5 x 57.5° > 21.8° = 28.7°

. • ^Rd.max= 1.0 x 350 x 090 x 687 x 0528 x 20.0 / (1.824 + 0.548) = 963.4 kN OK

<Exp. (6.9) & NA> <6.2.3 & NA> <6.2.3(1)>

<Exp. (6.9), Appendix A2> <6.2.3(2) & NA>

Shear reinforcement

Shear links: shear resistance with links

Aew / s > VEd / z fywd cot 0 Asw / s > 962.3 x 103 / (0.9 x 687 x 434.8 x 1.824) = 1.96

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