We can easily calculate the BBS of Slab From the drawings of the Structure.
We know Main reinforment bars are provided in the shorter span of the slabs and Distribution bars along the longer span.
For better understanding let I take an example;
Suppose we have a one way slab having a length of 5m and a width of 2m (clear span)
Suppose we provide main bars of diameter d 12mm @100 mm spacing. And the distribution bars of dia d 8mm @125mm c/c.
Let the thickness of slab be 150mm.and the clear cover of 25mm
Now we have to:
- Calculate the quantity of steel.
2.Calculate the weight of steel.
So we have given data:
Length L= 5m Or 5000mm
Width W = 2m Or 2000mm
Clear cover cc= 25mm
Main bar dia d= 12mm @100mm c/c
Distribution bar dia d= 8mm @125mmc/c
Thickness of slab = 150mm.
Let I solve it now:
First I will calculate the no. Of bars.
Step 1. No. Of bars main and distribution
Formula= (Total length – 2 cc )/100 +1
Main bars =( 5000 – 2×25) /100 +1
Main bars= (5000–50)/100 +1
Main bars= 4950/100 +1
Main bars= 49.5 +1
Main bars= 50.5 say 51 bars
Distribution bars=(2000–2×25) /100 +1
Distribution bars= 17bars
Step2. Calculating Cutting length:
Main bars: Formula =L+2Ld+ (1×0.42D) – 2×1d
Where L= clear span of slab
Ld= development length (which is 40 d)
d is dia of bar
O. 42D is inclined length (bend length)
1d = 45° bends
Now calculating first D:
Putting the values in above formula we have,
Cutting length= 2000+(2×40×12) +(1×.42×88) -2×1×12
Distribution bar cutting length=
Formula=clear span +2Ld
Main bar=51 numbers
= 12×12/162 ×151.62
Distribution bars= 17numbers
So total weight of steel =173kgs!