Design for beam shear support A

At d from face of support

VEd = 646 - (350 / 2 + 0.689) x (1.35 x 46.0 + 1.5 x 63.3)

= 646 - 0.864 x 157.1 Check maximum shear resistance bw zvfCd / (cot 0 + tan 0)

510.3 kN

where

bw = 350 mm as before z = 0.9d v = 0.6 (1 - fck / 250) = 0.6 (1 - 30 / 250) = 0.528 fd = 30 / 1.5 = 20.0 MPa 0 = angle of inclination of strut.

= 0.5 sin-1 {vEd2 / [0.20 fCk (1 - fCk / 250) ] } > cot-12.5 where

^ = VEd / bz = VEd / (b x 0.9d) = 510.3 x 103 / (350 x 0.9 x 689) = 2.35 MPa 0 = 0.5 sin-1 {2.35 / [0.20 x 30 (1 - 30 / 250) ] } > cot-12.5 = 0.5 sin-1 (0.445) > cot-12.5 = 0.5 x 26.4° > 21.8° = 21.8°

: 1.0 x 350 x 0.90 x 689 x 0.528 x 20.0 / (2.5 + 0.4) = 790 kN

<6.2.1(8), BS EN 1990 A1.2.2,

NA & Exp. (6.10a)>

<Exp. (6.9) & NA>

<6.2.3 & NA> <6.2.3(1)>

<2.4.2.4(1) & NA> <Exp. (6.9), Appendix A2>

<2.4.2.4(1) & NA> <Exp. (6.9), Appendix A2>

Shear reinforcement

Shear links: shear resistance with links Vs = (Asw / s) z fywd cot 6

Asw / s = area of legs of links/link spacing z = 0.9d as before fywd = 500 / 1.15 = 434.8 cot 6 = 2.5 as before Asw / s > 510.3 x 103 / (0.9 x 689 x 434.8 x 2.5) = 0.76

Minimum A^ / s = p^Xsin a where pw min = 0.08 x fck05 / fyk = 0.08 x 3005 / 500 = 0.00088 bw = 350 mm as before a = angle between shear reinforcement and the longitudinal axis. For vertical reinforcement sin a = 1.0

Maximum spacing of links longitudinally = 0.75d = 516 mm

<9.2.2(5), Exp. (9.4) > <Exp. (9.5N) & NA>

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