Flexural design span AB

K = Mf^ / bd2^ where where beffl = (0.2b, + 0.1 lo) < 0.2 lo < b, where b, = distance between webs / 2 Referring to Figure 3.9

= (7500 - 1000 - 550) / 2 = 2975 mm l0 = 0.85 x l, = 0.85 x 7500 = 6375 mm beff1 = 0.2 x 2975 + 0.1 x 6375 < 0.2 x 6375 < 2975 = 1232 < 1275 < 2975 = 1232 mm bw = 2000 mm bf2 = (0.2b2 + 0.1 y < 0.2 l0 < b2

where b2 = distance between webs / 2. Referring to Figure 3.9

= (9000 - 1000 - 550) / 2 = 3725 mm l0 = 6375 mm as before beff2 = 0.2 x 3725 + 0.1 x 6375 < 0.2 x 6375 < 3725 = 1382 < 1275 < 3725 = 1275 mm b = 1232 + 2000 + 1275 = 4507 mm d = 300 - 25 - 10 - 25 / 2 = 252 mm assuming 10 mm link and H25 in span fck = 35 MPa

<Appendix A1>

K < K section under-reinforced and no compression reinforcement required z = (d / 2) [1 + (1 - 3.53K)05] < 0.95d = (252 / 2) (1 + 0.886) < 0.95 x 252 = 238 < 239 ' z =238 mm But z = d - 0.4 x

' neutral axis in flange as x < 1.25 hf design as rectangular section

Span AB - Deflection

N = Basic l / d: check whether p > p0 and whether to use Exp. (7.16a) or Exp. (7.16b)

= 5835 / [2000 x 252 + (4507 - 2000) x 100] = 5835/754700 = 0.77%

p > p0 ' use Exp. (7.16b) N = 11 + 1.5 fCk05 p0 / (p- p) + fCk05 (p / P0)05 / 12

<Appendix A1>

<Appendix A1>

= 11 + 1.5 x 3505 x 0.059 / (0.077 - 0) + 3505 (0 / 059)15 = 11 + 6.8 + 0 =17.8

where os = (fk / r:) (A6ie, / 4fJ (SLS loads / ULS loads) (1 / 8) = 434.8 x (5835 /5892) [(47.8 + 03 x 45.8) / (1.25 x

47.8 + 1.5 x 45.8)] x (1 / 0.945) = 434.8 x 0.99 x 0.48 x 1.06 = 219 MPa

F1 = (beff / bw = 4057 / 2000 = 2.03) = 0.90

<Table 7.4N & NA>

<7.4.2(2), Concise EC2 10.5.2>

F2 = 7.0 / leff (span > 7.0 m)

<7.4.2(2)>

where

<5.3.2.2(1)>

leff = 7100 + 2 x 300 / 2 = 7400 mm

F2 = 7.0 / 7.4 = 0.95

F3 = 310 / 0

' Permissible l / d = 17.8 x 1.3 x 0.90 x 0.95 x 1.41 = 27.9

tttt 2.18 of PD 6687[5] suggests that pin T sections should be based on the area of concrete above the centroid of the tension steel.

— 310 / 219 x 13 / 12 — 1.53"** — say 1.50 . Permissible / / d — 17.8 x 1.3 x 0.90 x 0.95 x 1.50 — 29.7

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