## Flexural design span AB

Span AB - Flexure

: 300 - 25 - 8 - 20 / 2 = 257 assuming 8 mm link at H20 in span

K < K :. section under-reinforced and no compression reinforcement required.

z = (d / 2) [1 + (1 - 3.53K)05] < 0.95d = (257 / 2) (1 + 0.951) < 0.95 x 257 = 251 < 244 : z = 244 mm But z = d - 0.4x

: x = 2.5(d - z) = 2.5(257 - 244) = 33 mm : By inspection, neutral axis is in flange

<Appendix A1>

<Appendix A1>

<Appendix A1>

Span AB - Deflection

N = Basic l / d: check whether p > p0 and whether to use Exp. (7.16a) or Exp. (7.16b) p0 = fck05 / 1000 = 3505 / 1000 = 0.59%

where bw = min. width between tension and compression chords. At bottom assuming 1 / 10 slope to rib: = 150 + 2 x (25 + 8 + 20 / 2) / 10 = 159 mm

ftt Section 2.18 of PD 6687 [5] suggests that p in T-beams should be based on the area of concrete above the centroid of the tension steel.

p = 523 / (159 ( 257 + (900 - 159) x 100) = 523 / 114963 = 0.45%

N = 11 + 1.5 fCk05 P0 / P+ 3.2fCk°5 (p0 / P"i)15] <Exp. (7.16a)>

= 11 + 1.5 x 3505 x 0.055 / 0.045 + 3.2 x 3505 (0.055 /

0.045 - 1)15 = [11 + 10.8 + 2.0] = 22.8 K = (end span) 1.3 F1 = (bef / bw = 5.66) 0.8

F2 = 7.0 / /eff = 7.0 / 7.5 = (span > 7.0 m) 0.93 <7.4.2(2)>

F3 = 310 / where a = (fk / Y) (A™, / ApJ (SLS loads / ULS loads) (1 / 8

= 434.8(523/628) [ (430 + 0.3 x 5.0) / 13.38] (653/61.7***) = 434.8 x 0.83 x 0.43 x 1.06 = 164 MPa

. Permissible / / d = 22.8 x 1.3 x 0.8 x 0.93 x 1.50 = 33.0

Support A (and D): flexure (sagging) at solid/rib interface

Reinforcement at solid/rib interface needs to be designed for <9.2.1.3.(2)>

both moment and for additional tensile force due to shear (shift rule)

At solid/rib interface

As = MEd / fydZ + AFtd / fyd <9.2.1.3.(2), Fig. 9.2>

where z = (d / 2) [1 + (1 - 3.53K)05] < 0.95d where

K = MEd / bdf where b = 900 mm d = 300 - 25 - 8 - 25 - 20 / 2 = 232

assuming 8 mm links and H25 B in edge beam fck = 30

*** In analysis, 15% redistribution of support moments led to redistribution of span moments: S= 61.7 / 65.3 = 0.94

888 Both Asprov / Asreq and any adjustment to N obtained from Exp. (7.16a) or Exp. (7.16b) is restricted to 1.5 by Note 5 to Table NA.5 in the UK NA. Therefore, 310 / ue is restricted to 1.5.

Figure 3.13

Section at solid/rib intersection

Figure 3.13

Section at solid/rib intersection z = (232 / 2) (1 + 0.980) < 0.95 x 232 = 230 < 220 z = 220 mm

<Appendix A1>

where

0 = angle between the concrete compression strut and the beam axis. Assume cot 0= 2.5 (as a maximum) a = angle between shear reinforcement and the beam axis. For vertical links, cot a = 0 AFtd = 1.25VE = 1.25 x 29.3 = 36.6 kN

<Appendix A2, Concise Table

: 18.3 x 106 / (434.8 x 220) + 36.6 x 103 / 434.8 : 191 + 84 mm2 = 275 mm2

.Try 1 no. H20 B in end supports"

Support B (and C) (at centreline of support)

MEd / bd2fCk

where d = 300 - 25 cover - 12 fabric - 8 link - 20 / 2 = 245

K = 69.4 x 106 / (900 x 2452 x 35) = 0.037 By inspection, K < K'

z = (245 / 2) [1 + (1 - 3.53 K)05] < 095d = (245 / 2) (1 + 0.932) < 0.95d = 237 mm

Support B (and C): flexure (hogging) at solid/rib interface

Reinforcement at solid/rib interface needs to be designed for both moment and for additional tensile force due to shear (shift rule)

An alternative method would have been to calculate the reinforcement required to resist MEd at the shift distance, al, from the interface

Ae = MEd / fydz + AFtd / fyd where z = (245 / 2) [1 + (1 - 3.53 K)05] < 0^95d where

<Concise Table 15.4, Appendix A>

A Section under-reinforced: no compression reinforcement required z = (245 / 2) (1 + 0.723) < 232 = 211 mm

0 = angle between the concrete compression strut and the beam axis. Assume cot 0= 2.5 (as a maximum) a = angle between shear reinforcement and the beam axis. For vertical links, cot a = 0 AFtd = 1.25VEd = 1.25 x 40.9 = 51.1 kN

As = 42.4 x 106 / (434.8 x 211) + 51.1 x 103 / 434.8 = 462 + 117 mm2 = 579 mm2 / rib To be spread over beff where by inspection, beff = 900. .'.Centre of support more critical (679 mm2 / rib required). Top steel may be spread across beff where bw + beff1 + beff2 < b bw + 2 x 0.1 x 0.15 x (/1 + y 150 + 0.03 x (7500 + 9000) < 900 645 mm

.'. Use 2 no.H16 above rib and 3 no.H12 between (741 mm2 / rib) where 2 no.H16 and 2 no.H12 are within beff

<Appendix A2, Concise Table

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