# BBS of Beam Bar Bending Schedule Tutorial

Now we are going to learn how to create bbs of a beam, we have previously learnt the bbs of column and footing. 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 beam. One is long  U shaped Main bars, and the other one is a Stirrups that keep the Main bars together, in beam we have stirrups but in column we don’t call them stirrup we call them column ties.

Length of rebar

Beam has a width of 2.5m  (Our rebars extend the entirety of this width)

Our bars extends to the columns on both sides which has a width of 250mm minus 25mm cover on both sides = (250mm – 25mm) + (250mm – 25mm) (We have 2 columns)

= 450mm = 0.45m

Now our rebars bends 90 degree and extends a distance equal to 25 times diameter

= 25*12 + 25*12 ( 12mm diameter and two bends one on each side)

=600mm

= 0.6m

Total Length of Main rebars = Beam Width + Part that’s extend to column + L bends extensions

= 2.5 + 0.45  0.6

= 3.55m

Deduction

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

Here we have two 90 degree bend so we must apply a deduction of 2d + 2d to the total length of the bar

Deduction = 2*12 +2*12=48mm = 0.048m

Effective Length =  Total Length – Deduction

= 3.55 – 0.048

= 3.502m

Number of bars

In this case we can see that number of bars are 4

In other much complicated cases we use the formulae

No of bars = Distance/spacing + 1

But we won’t do this here because we can simply count the number of bars here.

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

= 3.502 * 4* 0.89

= 12.467 Kg

Beam Stirrups

Length if rebar

As we can see form the drawing, beam has a width of 0.25m

But there is a cover of 25mm on both sides

Total Cover = 25mm+25mm=50mm= 0.05m

Effective width= 0.25-0.05=0.2m

Similarly Effective length = 0.2m

Our stirrup has four sides and 2 hooks

Hooks have a length of 5 times diameter = 5*8 (8mm diameter of column ties)

=40mm

2 Hooks  = 2*40mm = 80mm

= 0.08m

Length of stirrup = Length of 4 sides + Length of 2 hooks

= 4*0.2 + 0.08

=0.88m

Deduction

As we know for every 45 degree bend we apply a deduction equal to d(diameter of bar)

Here we have hook of 135 degree = 3*45 degree

So we will apply deduction = 3*d

Because we have 2 hooks So total Deduction = 3d+3d=6d

= 6*8 (8mm diameter of ties)

= 48mm

= 0.048m

Effective Length = Length – deduction

= 0.88 – 0.048

= 0.832m

Number of bars

We know that the formulae to calculate the number of bar

Number of bars = Distance/ spacing +1

Stirrups are placed on the entire length of the beam

So distance is equal to length of beam = 2.5m= 2500mm

Spacing is given in the drawing equal to 120mm centre to centre

No of bars = 2500/120 +1

= 21 + 1 = 22

Unit weight

Unit weight of any rebar = d2/162

= 8*8/162

= 0.395 Kg/m

Total Weight

Total weight =Length*number*Unit Weight

= 0.832 * 22* 0.395

= 7.23 Kg

So here is the full bbs of beam

In the next article we are going to learn how to make bar bending schedule of a slab

If you want a free software to make bar bending schedule I highly recommend you download the official BBS app