Prepare a bar bending schedule of the fig Shown. The Roof slab is 3meters clear span and 6meters long. (In plan, top and bent up bars are as shown in dotted lines)

Assume Cover to be 40mm

Main straight bars 12mm Φ @ 12cm c/c

1) No. of bars $= \frac{3+(2×0.15)-(2×0.04)}{0.12}+1 \\ =27.83 =28no. (Say)$

∴ Straight bars = 14 bars and

Bent up bars = 14 bars

2) Length of bottom straight bars

L= Clear span +bearing – 2×end cover + (2×9×d)

L= 6 + (2×0.15) – (2×0.04) + (2×9×0.012)

=6.43m

3) Length of bent up bars

L= Clear span +bearing – 2×end cover + (2×9×d) + [No.of bent up×0.42D]

L= 6 + (2×0.15) – (2×0.04) + (2×9×0.012) + (2×0.42×0.100)

L=6.52m

4) Distribution bars

No. of bars at bottom $= \frac{6+(2×0.15)-(2×0.04)}{0.180} \\ =34.55 = 35no.(say)$ Length of bars

L= 3 + (2×0.15) – (2×0.04) + (2×9×0.006)

L=3.328m

5) Bent up R/F on top

Bent up =$\frac{L}{5}$ OR $\frac{L}{4}$ = $\frac{6}{5}$ = 1.2m

L=1.2 + (0.15×1) – (1×0.04)

L=1.39m

No.of bars $= \frac{1.39}{0.18}×2 \\ =15.44m = 16 no.s (say)$

Total no. of distribution bars

= 35 + 16 = 51 no.