 HOW TO CALCULATE THE QUANTITY OF STEEL IN TWO WAY SLAB BBS

# HOW TO CALCULATE THE QUANTITY OF STEEL IN TWO WAY SLAB BBS EXAMPLE:
Preparing a bar bending schedule of two way slab. The slab size is 3 by 4 reinforcement is 16 mm@150 mm and 10 mm@200 mm c/c shorter and longer span direction respectively. Assume the clear cover is 20 mm?
SOLUTION:
Bars that are placed in shorter span is called main bars and in longer span it is called a distribution bar.
Main bars:
First, we find out the total no of bars in the shorter span
No of bars
= Total span /spacing
= 3960 mm/150 mm
26.4 say 27 No’s of bars

As bar is placed in shorter span which is 3 m.
= Total length of span – 2 x clear cover
= 3000 – 2 x 20
2960 mm or 2.960 m Ans…
So we have total 27 No’s of main bars,
Total length:
= 27 x 2.960 m
79.92 m Ans…
Weight:
= d2/162 x length
= 16 x 16/162 x 79.92

126.2 kg Ans…
Distribution bar:
No of bars
= total span /spacing
= 2960 mm/200 mm
=15.8 say 16 No’s
As bar is placed in longer span which is 4 m

= total length – 2 x clear cover
= 4000 – 2 x 20
3960 mm or 3.960 m
So we have total 16 No’s of distribution bars so the total length is
Total length:
= 3.960 x 16
63.36 m long bar
Weight:
= d2/162 x length
=10 x 10/162 x 63.36
39.1 kg
Total weight of bars:
= weight of main bar + distribution bar
= 126.2 + 39.1

165.3 kg Ans…