# BBS of a Slab Bar Bending Schedule Tutorial

Now we are going to learn how to create bbs of a slab, we have previously learnt the bbs of column, footing and beam. We are going to use all the formulae we learnt in the basics of bbs.

As you can see from the drawing above there are two kinds of bars in this slab. One bar along the main x direction and another one along the y direction. It might seem that we have 4 bars but if you look carefully the black and red bars are actually the same bar but in the opposite direction.

Length of rebar

Slab has a width of 4 m

There is cover of 25mm on each side = 25mm+25mm= 50mm = 0.05m

Effective Slab length = slab width – cover = 4m – 0.05m = 3.95m

Bar Length = Slab length + inclined length

D = Slab Height – cover = 150 – (25 +25) = 100mm

Inclined length = 0.42*D

= 0.42*100= 42mm

Total Length of Main X bar = Effective Slab length  + inclined length

= 3.95 + 0.042= 3.992m

Bend Deduction

As we already have learned in the basic of bar bending schedule that for every 45 degree bend we apply a deduction equal to d( 1 times diameter)

Here we have two 45  degree bend so we must apply a deduction of d+d to the total length of the bar

Deduction = 12 +12=24mm = 0.024m

Effective Length =  Total Length – Deduction

= 3.992 – 0.024

= 3.968m

Number of bars

No of bars = Distance/spacing + 1

Distance = Slab length – cover = 3.95m

Spacing = 120mm

No of Bars = 3950/120 +1

= 33+1

= 34

Unit Weight

We know that Unit weight of any bar = d2/162

Here out diameter= 12mm

Unit weicht = 12*12/162 = 0.89 kg/m

Total Weight

Total weight = Length*number*Unit Weight

= 3.968 * 33* 0.89

= 116.54 Kg

Main x rebar

Main Y rebar

Length of rebar

Slab has a width of 5 m

There is cover of 25mm on each side = 25mm+25mm= 50mm = 0.05m

Effective Slab length = slab width – cover = 5m – 0.05m = 4.95m

Bar Length = Slab length + inclined length

D = Slab Height – cover = 150 – (25 +25) = 100mm

Inclined length = 0.42*D

= 0.42*100= 42mm

Total Length of Main X bar = Effective Slab length  + inclined length

= 3.95 + 0.042= 4.992m

Bend Deduction

As we already have learned in the basic of bar bending schedule that for every 45 degree bend we apply a deduction equal to d( 1 times diameter)

Here we have two 45  degree bend so we must apply a deduction of d+d to the total length of the bar

Deduction = 8 +8=16mm = 0.016m

Effective Length =  Total Length – Deduction

= 4.992 – 0.016

= 4.976m

Number of bars

No of bars = Distance/spacing + 1

Distance = Slab length – cover = 4.95m

Spacing = 120mm

No of Bars = 4950/120 +1

= 41+1

= 42

Unit weight

Unit weight of any rebar = d2/162

= 8*8/162

= 0.395 Kg/m

Total Weight

Total weight =Length*number*Unit Weight

= 4.976 * 42* 0.395

= 82.55 Kg