# BBS of Column Bar Bending Schedule Tutorial

We are going to learn how to create bbs of a column in this article, this is in continuation to our previous article where we learned how to create bbs of a footing.

As you can see from the drawing above there are two kinds of bars in this column. One is long L shaped main bars, and the other are ties 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

Column has a height of 2.4m

Footing has a height of 0.5m

Total height = height of column + height of footing = 2.4+0.5=2.9m

Now as we can see the Main column bars do not go all the way to the bottom of the footing

At the bottom there is a cover of 50mm on top of that we have main bottom bar of 10mm on top of that we have bottom distribution bar of 10mm and on top of that our main column bar sits

Main column bar is 50mm+10mm+10mm=70mm above the bottom of fitting

Height of column bar = 2.9m – 0.07m= 2.83m

There is also L bend in our main column bar of length 25* diameter of bar=25*10 (diameter = 10mm)

Total Length = Height of column + L bend = 2.83+ 0.25 = 3.08m

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 one 90 degree bend so we must apply a deduction of 2d to the total length of the bar

Deduction = 2*10=20mm = 0.02m

Effective Length =  Total Length – Deduction

= 3.08 – 0.02

= 3.06

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= 10mm

Unit weicht = 10*10/162 = 0.617

Total Weight

Total weight. = Length*number*Unit Weight

= 3.06 * 4* 0.617

= 7.55 Kg

Column Ties

Length

As we can see form the image column 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 column 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 column = 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

Ties are placed on the entire length of the bar

So distance is equal to hight of column main bar = 2.83m= 2830

Spacing is given in the drawing equal to 12mm centre to centre

No of bars = 283/12 +1

= 24 + 1 = 25

Unit weight

Unit weight of any rebar = d2/162

= 8*8/162

= 0.395

Total Weight

Total weight =Length*number*Unit Weight

= 0.832 * 25* 0.395

= 8.216 Kg

So here is the full bbs of column