Flexural design span BC

Span BC - Flexure

where bef = (0.2b1 + 0.110) < 0.2 I0 < b1 where b1 = distance between webs / 2.

Assuming beams at 7000 mm cc = (7000 - 350) / 2 = 3325 mm l0 = 0.85 x l1 = 0.85 x 8000 = 6800 mm beff1 = 0.2 x 3325 + 0.1 x 6800 < 0.2 x 6800 < 3325 = 1345 < 1360 < 3325 = 1360 mm bw = 350 mm befl2 = (0.2b2 + 0.110) < 0.2 I0 < b2 where

b = 1360 + 350 + 0 = 1710 mm b2 = 0 mm befl2 = 0 mm b = beff = beff1 + bw + beff2

b2 = 0 mm beff2 = 0 mm d = 750 - 35 - 10 - 32 / 2 = 689 mm assuming 10 mm link and H32 in span fck = 30 MPa

By inspection, K < K' .'. section under-reinforced and no <Appendix A1>

compression reinforcement required z = (d / 2) [1 + (1 - 3.53K)05] < 0.95d <Appendix A1>

= (689 / 2) (1 + 0.95) < 0.95 x 689 = 672 > 655 .'. z = 655 mm By inspection, x < 1.25 hf design as rectangular section <Appendix A1>

Span BC - Deflection

By inspection, compared with span AB OK

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