Flexural design span AB

Span AB - Flexure

where where bf = (0.2b + 0.1/0) ^ 0.2 /0 < b, where b1 = distance between webs / 2

Assuming beams at 7000 mm cc = (7000 - 350) / 2 = 3325 mm /0 = 0.85 x /1 = 0.85 x 9000 = 7650 mm+'

Figure 4.10

Effective flange width bf

Effective Flange Width

Figure 4.10

Effective flange width bf

Figure 4.11

Elevation showing definition of /o for calculation of flange width

Figure 4.11

Elevation showing definition of /o for calculation of flange width

beff1 = 0.2 x 3325 + 0.1 x 7650 < 0.2 x 7650 < 3325 = 1430 < 1530 < 3325 = 1430 mm bw = 350 mm bf2 = (0.2b2 + 0.1/0) < 0.2 ¡0 < b2

+++ The distance /0 is described as the distance between points of zero shear, 'which may <5.3.2.1(2)> be obtained from Figure 5.2'. From the analysis, /0 could have been taken as 7200 mm. <Figure 5.2>

where b2 = 0 mm, b = 1430 + 350 + 0 = 1780 mm d = 750 - 35 - 10 - 32 / 2 = 689 mm assuming 10 mm link and H32 in span fck = 30 MPa

K < K' section under-reinforced and no compression reinforcement required z = (d / 2) [1 + (1 - 3.53K)05] < 0.95d = (689 / 2) (1 + 0.917) < 0.95 x 689 = 661 < 654 ' z = 654 mm But z = d - 0.4x

' by inspection, neutral axis is in flange and as x < 1.25 hf design as rectangular section.

Check spacing of bars

Spacing of bars = [350 - 2 x (35 + 10) - 32] / (5 - 1) = 57

Clear spacing = 57 - 32 mm = 25 mm between bars Minimum clear distance between bars

= max[bar diameter; aggregate size + 5 mm] = max[32; 20 + 5] = 32 mm i.e. > 25 mm

' 5 no. H32 B no good For 4 bars in one layer, distance between bars = 44 mm so

<Appendix A1>

<Appendix A1> <Appendix A1>

/

-

35 cover

ft.______«

[32 bar

32 spacers 32 bar 10 |jnk

35 cover

350

Span AB bottom reinforcement d = 750 - 35 - 10 - 32 / 2 - 0.333 x 2 x 32 = 668 mm

K < K' .'. section under-reinforced and no compression reinforcement required z = (d / 2) [1 + (1 - 3.53K)05] < 0.95d = (668 / 2) (1 + 0.911) < 0.95 x 668 = 639 < 635 .'. z = 635 mm

.'. by inspection, neutral axis in flange so 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)

= 4158 / [350 x 668 + (1780 - 350) x 300] = 4158 / 662800 = 0.63%

N = 11 + 1.5 fck05 P0 / (p- p) + fck05 (p' / p0)05 / 12

= 11 + 1.5 (3005 x 0.055 / (0.063 - 0) + 3005 (0 / 0.55)15 = 11 + 7.2 + 0 = 18.2

<Appendix A1> <Appendix A1>

<Appendix A1>

F1 = (beff / bw = 1780 / 350 = 5.1) = 0.80

<Table 7.4N & NA>

<7.4.2(2), Concise EC2 10.5.2>

F2 = 7.0 / f (span > 7.0 m)

<7.4.2(2)>

where

f= 9000 mm

<5.3.2.2(1)>

F2 = 7.0 / 9.0 = 0.77

F3 = 310 / as

where as in simple situations = (fyk / y) (A^, / Aeprov) (SLS loads / ULS loads) (1 / ¿). However in this case separate analysis at SLS would be required to determine crs. Therefore as a simplification use the conservative assumption:

= (500 / 500) x (4824 / 4158) = 1.16 . Permissible l / d = 18.2 x 1.3 x 0.80 x 0.77 x 1.16 = 16.9 Actual l / d = 9000 / 668 = 13.5

As permissible less than actual .•. OK

*** 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.

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