Flexural design span BC and CD similar

Flexure where where beff, = (0.2b| + 0.1/0) < 0.2 /0 < b1 where b1 referring to Figure 3.9

= (7500 - 1000 - 550) / 2 = 2975 mm /0 = 0.70 x /2 = 0.7 x 7500 = 5250 mm beff1 = (0.2 x 2975 + 0.1 x 5250 < 0.2 x 5250 < 2975 = 1120 < 1050 < 2975 = 1050 mm bw = 2000 mm bf2 = (0.2b2 + 0.1/0) < 0.2 /0 < b2 where

Referring to Figure 3.9

5250 mm as before beff2 = 0.2 x 3725 + 0.1 x 5250 < 0.2 x 5250 < 3725 = 1270 < 1050 < 3725 = 1270 mm b = 1050 + 2000 + 1270 = 4320 mm

MEd / bd2fCk

b2 = distance between webs / 2

d = 252 mm as before assuming 10 mm link and H25 in span fck = 30

By inspection, K < K' .'. section under-reinforced and no compression reinforcement required z = (d / 2) [1 + (1 - 3.53K)05] < 0.95d = (252 / 2) (1 + 0.924) < 0.95 x 252 = 242 > 239 . z = 239 mm

By inspection, x < 1.25 hf design as rectangular section

<Appendix A1> <Appendix A1>

<Appendix A1>

Deflection

By inspection, compared to span AB but for the purposes of illustration:

N = Basic l / d: Check whether to use Exp. (7.16a) or Exp. (7.16b) p0 = 0.59% (for fck = 35)

where bw = 2000 mm p = 3783 / (2000 x 252 + (4320 - 2000) x 100) = 3783 / 736000 = 0.51%

 p < p0 .'. use Exp. (7.16a) N = 11 + 1.5 fck05 Po / p + 3.2fck05 (p0 / p - 1)15 = 11 + 1.5 x 3505 x 0.059 / 0.051 + 3.2 x 3505 (0.059 / 0.051 - 1)15 = 11 + 10.2 + 23.5 = 44.7 K = (internal span) = 1.5 F1 = (beff / bw = 4320 / 2000 = 2.16) = 0.88 <7.4.2(2), Concise EC2 10.5.2> F2 = 7.0 / leff = 7.0 / 7.4 = (span > 7.0 m) = 0.95 <7.4.2(2)> F3 = 310 / G;, where o; = (fk / r6) (Aare, / ApJ (SLS loads / ULS loads) (1 / S) = 434.8 x (3783 / 3828) [(47.8 + 0.3 x 45.8) /

(1.25 x 47.8 + 1.5 x 45.8)] x (1 / 0.908) = 434.8 x 0.99 x 0.48 x 1.10 = 227 MPa

. Permissible l / d = 44.7 x 1.37 x 0.88 x 0.95 x 1.37 = 70.1

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

Hogging

Assuming curtailment of top reinforcement at 0.301 + al <How to: Detailing

From analysis MEd at 0.3l from BC (& DC) = 216.9 kNm at 0.3l from CB (& CD) = 185.6 kNm

K = 216.9 x 106 / (2000 x 2262 x 35) = 0.061 By inspection, K < K' z = (226/2)[1 + (1 - 3.53 K')05] = (226/2)[1 + (1 - 3.53 x 0.061)05] = (226/2) (1 + 0.89) < 0.95d = 214 mm < 215 mm

Use 12 no. H20 T (3748 mm2) (to suit links and bottom steel)

Top steel at supports may be curtailed down to 12 no. H20 T

at 0.3l + al = 0.3 x 7500 + 1.25 x 214 = 2518 say 2600 mm <9.2.1.3(2)>

from centreline of support

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