# Bar Bending Schedule for Column [BBS]

## How do you prepare Bar Bending Schedule for column?

okay Let's go, we have a column design 2d drawing with Column data below,

Height = 4 meter

Cross section = 300 x 400 mm

Clear cover = 40 mm

No of vertical bars = 6 no's

Diameter of vertical bar = 16 mm

Diameter of stirrup = 8 mm

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

## Bar Bending Schedule for One Way Slab

### Step 1 : Vertical Reinforcement Bar calculation

Formula:-

Length of 1 bar = H + Ld

Where

• Ld = development length.
• h   = Height of column Length of 1 bar.
• d   = diameter of reinforcement bar.

= 4000 + 40d

4000+ 40 ×16

= 4640mm

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

Total length = 6 x 4.640m

= 27.84 m

long vertical reinforcement bar required.

## Bar Bending Schedule For Two Way Slab

### Step 2: Cutting length of stirrups

Cross sectional area of column is 300mm*400mm

#### Calculation of 'A' is vertical cross section area of stirrup

= 300 — 2 x clear cover

= 300 - 2 x 40

= 300 - 80

= 220mm

#### Calculation of 'B' is horizontal cross section area of stirrup

= 400 - (2 x clear cover)

= 400- (2 x 40)

= 400 - 80

= 320mm

No of stirrups = 4000/3

=1333.3mm.

Formula = L/3 / spacing + 1

(No of stirrups in end zone)

= 1333.3 / 150

= 8.8 nos ~ 9 nos

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

=2x9

= 18 nos (at end zones)

At Mid Zone

=1333.3 / 200

= 6.6 nos ~ 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 × 10 d 5×2d

where

Hook = I0 d

Bend = 5 x 2d (because we have 5 bends in stirrup) d = is diameter of bar

= (2 × 220) + (2 × 320) +2 ×10×8 -2×5×8

= 440 + 640 + 160 - 80

= 1160 mm.

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

Total length = 25 x 1.16 = 29m long 8mm reinforcement bar.

No Comment