## Flexural design

Effective depth

Assuming 10 mm links: d = 450 - 35 - 10 - 25 / 2 = 392 mm Flexure in span

K = MEd / bd2fck = 137.2 x 106 / (300 x 3922 x 30) = 0.099 z / d = 0.90

z = 0.90 x 392 = 353 mm As = MEd / fydz = 137.2 x 106 / (434.8 x 353) = 894 mm2

Check spacing

Spacing = 300 - 2 x 35 - 2 x 10 - 25 = 185 mm Assuming 10 mm diameter link Steel stress under quasi-permanent loading O = (fyk / Ys) (As,e, / As,prov) (SLS loads / ULS loads) (1 / = fyd x (As.req / As.prJ x (3k + ¥2 Hk) / ^k + YqHJ (1 /

= (500/1.15) x (894/982) x [(30.2 + 03 x 11.5) /508] (1 / 1.03) = 434.8 x 0.91 x 0.66 x 0.97 = 253 MPa As exposure is XC3, max. crack width wmax = 0.3 mm .'. Maximum bar size = 14 mm or max. spacing = 185 mm - say OK

Deflection

Check span: effective depth ratio

Basic span: effective depth ratio for p = 0.76% = 27.4 Max. span = 27.4 x 392 = 10740 mm . OK

Flexure: support

<Fig. 3.5> <Appendix A.1> <Concise EC2 Table 15.5>

<7.3.1(5) & NA> <Table 7.2N & NA>

K = MEd / bd2fck where d = 450 - 35 - 10 - 25 / 2 = 392 mm K = 193.8 x 106 / (300 x 3922 x 30) = 0.142 By inspection, K < K' (0.142 x 0.168*) .'. no compression reinforcement required.

<Appendix A.1> <Concise EC2 Table 15.5>

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