# Bar Bending Schedule of Column

## Bar Bending Schedule of Column

The calculation of bar bending schedule of column given below:

The height of the column is 4 m and having a cross-sectional area is 300 x 400 mm and having a 40 mm of clear cover.Six bars are going to use has a dia of 16 mm.The 8 mm dia of stirrups is going to use @ 150mm and @200 mm at L/3.So calculate the cutting length bars and stirrups?

### Given Data:

Height = 4 m

Cross section = 300 x 400 mm

Clear Cover = 40 mm

No of vertical bar = 6 no’s

Diameter of vertical bar = 16 mm

Diameter of Stirrup = 8 mm

Stirrups center to center spacing = @150 or @ 200 mm

BBS of Column = ? ### Solution:

The Calculation was to proceed into two steps.
1.Vertical bar calculation

2.Cutting length of stirrups

### Step 1: Vertical Bar Calculation

Length of 1 bar = H + Ld

Where

Ld = development length

H = Height of column

### Length of 1 bar

= 4000 mm + 40d (d is dia of the bar)

= 4000 + 40 x 16

= 4000 + 640

= 4640 mm or 4.640 m

So the length of 1 vertical bar is 4.640 m and we have a total 6 number of bars,

### Total length

= 6 x 4.640

= 27.84 m long vertical bar is required.

### Step 2: Cutting the length of stirrups

Cross-sectional area column is 300 mm* 400mm

A is vertical cross-section area of the stirrup

B is a horizontal cross-section area of the stirrup ### Calculation of A

= 300 – 2 x clear cover

= 300 – 2 x 40

= 300 – 80

= 220 mm

### Calculation of B

= 400 – 2 x clear cover

= 400 – 2x 40

= 400 – 80

= 320 mm

No of stirrups

= 4000/3

= 1333.3mm or 1.33m Formula = L/3 / spacing + 1

(No of stirrups in end zone)

= 1333.3 / 150

= 8.8 nos say 9 nos

There are total two zones of 150mm spacing and one zone of 200mm spacing

= 2 x 9

= 18 nos (at end zones)

At Mid Zones

= 1333.3 / 200

= 6.6 nos say 7 nos

Total no of stirrups

= 18 + 7

= 25 nos

Cutting Length of one stirrup

### Formula:

= (2 x A) + (2 x B) + hook – bend

Cutting length

= (2 x A) + (2 x B) + 2 x 10d – 5 x 2d

Where

Hook = 10d

Bend = 5x 2d (because we have 5 bends in stirrup)

d = dia of bar

= (2 x 220) + (2 x 320) + 2 x 10 x 8 – 2 x 5 x 8

= 440 + 640 +160 – 80

= 1160 mm or 1.16 m

So we have a total 25 nos of stirrups are going to use,

Total length

= 25 x 1.16

= 29m long 8 mm bar