Flexural design support A

= 195 kNm in hogging

MEd,min = 1148 x 0.25 in hogging and in sagging = 287 kNm

MEd / bd2fck where where bf = (0.2bi + 0.1/0) < 0.2 l0 < bi where b1 = distance between webs / 2

l0 = nominal: assume 01

:. b = bw = 350 mm d = 750 - 35 - 10 - 32 / 2 = 689 mm assuming 10 mm link and H32 in support fck = 30 MPa

= 287 x 106 / (350 x 6892 x 30) = 0.058 Restricting x / d to 0.45 K = 0.168

K < K' :. section under-reinforced and no compression reinforcement required z = (d / 2) [1 + (1 - 3.53K)05] < 0.95d = (689 / 2) (1 + 0.89) < 0.95 x 689 = 652 < 654 : z = 652 mm

where fyd = 500 / 1.15 = 434.8 MPa = 287 x 106 / (434.8 x 652) = 1012 mm2

<Appendix A1>

<Appendix A1>

' The distance l0 is described as the distance between points of zero moment, 'which may <5.3.2.1(2)>

be obtained from Figure 5.2'. In this case l0 = 0. (see figure 4.11)

Check anchorage of H32 U-bars

Bars need to be anchored distance 'A' into column Figure 4.9 Distance 'A'

Assuming column uses 35 mm cover 10 mm links and 32 mm bars Distance 'A' = 2 [350 - 2 (35 + 10) ] - 32 / 2 - 32 / 2 + 750 - [2 (35 + 10)] - 2 x 32 / 2 - (4 - n) (3.5 + 0.5) x 32 = 488 + 628 - 110 = 1006 mm Anchorage length,

'bd = a'b.rqd — 'b.min where a = conservatively 1.0

b.req

where

asd = design stress in the bar at the ULS

= 434.8 x 1012 / 1608 = 274 MPa fbd = ultimate bond stress

= 2.25 n n fct,d where n = 1.0 for good bond conditions n2 = 1.0 for bar diameter < 32 mm fct,d = ®ct fctk / ?c

= 1.0 x 2.0 / 1.5 = 1.33 MPa fbd = 2.25 x 1.33 = 3.0 MPa /b,rqd = (32 / 4) (274 / 3.0) = 731 mm"" /bmin =max[10y 100 mm] = 250 mm /bd = 731 mm i.e. < 1006 mm .'. OK

Use 2 no. H32 U-bars

*** Anchorage lengths may be obtained from published tables. In this instance, a figure of 900 mm may be obtained from Table 13 of the How to on Detailing. <How to: Detailing > Greener Homes for You

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