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 a shorter span are called main bars and in a 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

Cutting Length:

As bar is placed in a 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 a 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

Cutting Length:

= 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…

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