Flexural design support B

Support B

At centreline of support M = 1394 kNm

From analysis, at face of support <5.3.2.2(3)>

MEdBA = 1209 kNm MEdBC = 1315 kNm

K = MEd / bf where bw = 350 mm d = 750 - 35 - 12 - 32 / 2 = 687 mm assuming 10 mm link and H32 in support but allowing for H12 T in slab fck = 30 MPa

for £= 0.85, K' = 0.168: to restrict x / d to 0.45, K' = 0.167

Compression steel required z = (d / 2) [1 + (1 - 3.53 K')05]

= (687 / 2) [1 + (1 - 3.53 x 0.167)05] = (687 / 2) (1 + 0.64) < 0.95d = 563 mm

As2 = (K - K')fckbd2 / fsc(d - d2) <Fig. 3.5, Appendix A1

where How to: Beams>.

d2 = 35 + 10 + 32 / 2 = 61 mm f6C = 700(x - d2) / x < fyd where x = 2.5 (d - z) = 2.5 (687 - 563) = 310 mm fsc = 700 x (310 - 61) / 310 < 500 / 1.15 = 562 MPa but limited to < 434.8 MPa As2 = (0.265 - 0.167) x 30 x 350 x 68T2 / [434.8(687- 61) ] = 1784 mm2

Try 4 no. H25 B (1964 mm2) A6 = M / fyAz + As2 fsC / fyd <Appendix A1>

= 0.167x 30 x 350 x 6S72/ (4348 x 563) +1570 x 4348 /4348 = 3380 + 1784 = 5164 mm2

Try 4 no. H32 T + 4 no. H25 T (5180 mm2) This reinforcement should be spread over beff <9.2.1.2(2), Fig. 9.1>

where b1 = distance between webs / 2.

Assuming beams at 7000 mm cc

= 0.15 x (9000 + 8000) = 2550 mm . beff1 = 0.2 x 3325 + 0.1 x 2550 < 0.2 x 2550 < 3325 = 920 < 510 < 3325 = 510 mm bw = 350 mm f = (0.2bz + 0.1/0) < 0.2 l0 < b2

Use 4 no. H32 T + 4 no. H25 T (5180 mm2) @ approx 100 mm cc

Eurocode Spanningsverdeling

Figure 4.13

Support B reinforcement

Figure 4.13

Support B reinforcement

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