## Flexural design span BC

Span BC - Flexure

= 50.3 x 106 / 900 x 2572 x 35 = 0.02 i.e. < K' (as before K' = 0.168) By inspection, z = 0.95d = 0.95 x 257 = 244 mm By inspection, neutral axis is in flange.

Span BC - Deflection

p = 474 / (159 (x 257 + (900 - 159) x 100) = 474 / 114963 = 0.41% p0 = 0.59% (for fck = 30)

N = 11 + 1.5 fck05 po / p + 3.2fck05 (po /p - 1)15 <Exp. (7.16a)>

= 11 + 1.5 x 3505 x 0.055 / 0.041 + 32 x 3505 (0.055 / 0041 - 1)15 = 11 + 11.9 + 3.8 = 26.7 K = (internal span) 1.5 F1 = (beff / bw = 6.0) 0.8

F2 = 7.0 / leff = 7.0 / 9.0 = (span > 7.0 m) 0.77 <7.4.2(2)>

O = (fyk / fc) (As,e, / A,pmv) (SLS loads / ULS loads) (1 / = 4348 x (474/628) [(430 + 0.3 x 50) /1338](61.1 / 55.9) = 434.8 x 0.75 x 0.43 x 1.09 = 153 MPa

Permissible l / d = 26.8 x 1.5 x 0.8 x 0.77 x 1.50 = 37.1

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