# How do you calculate the slab BBS?

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:

D=thickness -2cc-d

D= 150–2×25–12

D=88mm

Putting the values in above formula we have,

Cutting length= 2000+(2×40×12) +(1×.42×88) -2×1×12

CL=2000+960+36.96–24

CL=2973mm

Cutting length=2.97m

Distribution bar cutting length=

Formula=clear span +2Ld

=5000+2×40×8

=5.64m

Conclusion:

Main bar=51 numbers

Total length=51×2.97

=151.62m

Weight:

=D^2/162×L

= 12×12/162 ×151.62

Weight=134.77kg

Distribution bars= 17numbers

Total length=17×5.64=96m

Weight=8×8/162 ×96

Weight= 38kg

So total weight of steel =173kgs!