# How to Create Bar Bending Schedule Of Slab

TweetBar Bending Schedule is a list of reinforcement bars depicted in a tabular form giving the detailed information of Bars, the shape of the bending with sketches, length of each bar, total weight, and the total length of bars.

The Bar bending is a process of cutting and bending reinforcement steel bar into Desire shape as per structural drawing was given by structural engineer for various structural elements like footing, column, beam, slab, etc.

While working on reinforced cement concrete, a schedule of bars is usually maintained. With the help of the Bar Bending Schedule, the details and requirements of different sizes of bars and the lengths of bars may be known and may be arranged and bent up during the time of construction. Thus, by creating a bar bending schedule it keeps us informed about

- The shape of bending along with its sketch
- Number of sets or number of bars of each set
- Length of each and the total length of a running meter
- The total weight of bars
- Size and Type of Bars

In this article, we will majorly focus on the process of bar bending schedule, How to create it, and the major advantages of creating it.

## Types of slabs used in Bar Bending schedule

Slabs can be classified into two types

- One way Slab
- Two-way slab

### One way Slab

In one way slabs the main bars are provided in shorter direction whereas distribution bars are provided in the longer direction

### Two-way Slab

In two-way slabs, the main bar are provided in both directions.

## Estimating Bar bending schedule of one way slab

For a perfect bar bending schedule in one way slab, we should consider the following example

Length of the slab = 5000 mm

Width of slab = 2000 mm

Main Bar = 12mm @ 150 mm/cc

Ly/Lx = Longer span /Shorter span

=5000/2000

=2.5 > 2

Distribution Bar = 8mm @ 150 mm/cc

Clear Cover (Top and Bottom) = 25 mm

Thickness of slab =150 mm

Development Length = 40 d

Where d is the diameter of Bar

### To calculate the bar bending schedule of the one-way slab.

### Step 1 :

Calculate no of bars

Calculate no of bars required for main bars and distribution bars

No of bars = Length of Slab/ spacing + 1

No of main bars = Lx/spacing +1

=5000/150

= 34

Then we have to calculate no of distribution bars

= Ly/Spacing +1

=2000/150 +1

=14

### Step 2:

Calculate cutting length of main bars and distribution bars

Cutting length of main bar

= clear span of slab + (2 x development length) + inclined length - (bend length)

Clear Span of Slab = 2000 mm

Development Length = 40 d

Inclined length = 0.42 d

1 d is for every 45º bend

Where d = diameter of Bar

Now let us calculate D

D = Thickness of slab - Both side clear cover (Top and Bottom) - diameter of bar

= 150 - (25 + 25) - 12

= 88 mm

Length of main bar

=Ly +2Ld + (1 x 0.42d) - (1d x 4)

= 2000 + (2 x 40 x 12) + (2 x 0.42 x 88) - (1 x 12 x 4)

=2838 mm

=2.838 m

Weight of main bars = d2 x L/162 X 34

=122 x 2.838/162

=86 kg

### Length of Distribution Bar :

Clean span of slab + 2 x development length

= Lx + 2 Ld

= 5000 +2 x 40 x 8

=5640 mm

=5.64 m

### Weight of distribution bars

= d2 x L/162 x14

=82 x 5.64/162 x 14

= 31 kg

### Step 3:

Calculate the top bar which is provided at the top of critical length (L/4) area

No of Top Bars = {(Ly/4)/spacing+1}x 2

= {(2000/4)/150+1} x 2 = 9

Length of extra bar = Ly - 2x Ly/4 +2 x 100

{Here 2 x 100 is for both side Lapping of 100 mm for extra bar}

= 2000 - (2 x 2000/4) +200

=1200 mm

=1.2 m

Weight of Extra Bar = d2 x L/162 x 9

= 82 x 1.2/162 x 9

=4.266 kg

## Bar Bending Schedule Of Two Way Slab

So we know, Longer span/Shorter span = Ly/Lx

=5000/3000

=1.66 < 2

It is a two-way slab

Given,

Length of Longer Span = 5000 mm

Length of Shorter Span = 3000 mm

Main Bar = 12 mm @150 mm/cc

Clear Cover (Top and Bottom) = 25 mm

Thickness of slab = 20 mm

Development Length = 40 d

Where d is the diameter of the bar

## Step By Step Method To calculate the bar bending schedule of the two way slab

### Step 1: For section A-A

Calculate number of bars required for main bars and distribution bar

No of Bars = Length of slab/spacing +1

Total no of bars = Ly/150 +1

=4000/150+1

=27

No of main bars =14

No of distribution bars =13

### Step 2:

#### Calculate cutting length of main bars and distribution bars

Cutting length of Main Bar = Clear span of slab +(2 x development length) + (inclined length) - (Bend length)

Clear Span of slab= 5000 mm

Development Length = Ld°°

Ld = 40d

Inclined length = 0.42d

1d is for every 45° bend

Where d is the diameter of the bar

Now Calculate D

D = Thickness of the slab - both side clear cover (top and bottom) - diameter of bar

200 - (25 +25) - 12 = 138 mm

Length of Main bar = Lx +(2 x Ld) + (2 x 0.42d) - (1d x 4)

= 5000 +(2 x 40 x 12) + (2 x 0.42 x 138) - (1 x 12 x4)

=2838 mm

=6.02 m

For 14 bars total length

=6.02 x 14

= 84.28 m

Weight of main bars= d2 x L/162

= 122 x 84.28/162

=75 kg

Cutting Length of Distribution Bar

= clear span of slab + 2 x development length

= Lx +2 Ld

= 5000 +2 x 40 x 8

=5640 mm

=5.64 m

Total Length of Distribution Bar

=5.64 x 13

= 73.32

Weight of Distribution Bars

=d2 x L/162

= 82 x 73.32/162

=29 kgs

### Step 3:

#### Calculate Top Bar which is provided at the top of critical length(L/4) Area

No. Of Extra Bars ={(Ly/4)/spacing +1}x 2

={(4000/4)/150 +1} x 2 =16

Length of Extra Bar = Lx - 2 x Lx/4+2 x100 [ here 2 x100 is for both side lapping of 100 mm of extra bar]

=5000 - (2x 5000/4) + 200

=2800 mm

=2.7 mm

Total Length Of Extra Bar

= 2.7 x 16

= 43.8

Weight of Extra Bar = d2 x L/162

= 82 x 43.2/162

=17 Kg

### Step 1 For Section B-B

#### Calculate number of bars required for main bars and distribution bar

Total No.of Bars = Length of Slab/Spacing +1

Total No.of Bars = Lx/150 +1

=5000/150 +1

=35

No.Of Main Bars = 18

No. Of Distribution Bars = 17

#### Calculate The Cutting Length of Main Bars

Cutting length of Main Bar = Clear span of slab +(2 x development length) + (inclined length) - (Bend length)

Clear Span of Slab = 4000mm

Development length = Ld = 40d

Inclined Length = 0.42d

1d is for every 45° bend

Where d = Diameter of The bar

Now let us claculate d

d = Thickness of Slab - Both Side Clear Cover - Diameter of Bar

= 200 - (25+25) - 12

=138 mm

Cutting Length Of Main Bar

= Ly + 2 x Ld

= 4000 + 2 x 40 x 8

=4640 mm

= 4.64 m

Total Length Of Distribution Bar

=4.64 x 17

= 78.88

Weight of Distribution Bars

= d2 x L/162

= 82 x 78.88/162

=31 kg

### Step 3 :

Calculate No of Extra Bars

No.Of Extra Bars = {( Lx /4)/spacing +1 } x 2

={(5000/4)150 + 1} x 2 = 9

Length of Extra Bar = Ly -(2 x -Ly/4) + (2 x 100) [here 2 x 100 is for both side lapping of 100 mm for extra bar]

= 4000 - (2 x 4000/4) + 200

=2200 mm

=2.2 m

Total Length of Extra Bar

=2.2 x 9

=19.8 m

Weight of Extra Bar

= d2 x L/162

=82 x 19.8/162

=8kg

Thus in these easy step-by-step methods, one can create bar bending schedule.

## Advantages of Bar Bending Schedule

The following are the main advantages of Bar Bending schedule

- It helps in stock management at site
- It helps in auditing of reinforcement and provides a check on theft.

## 3) One can estimate the steel quantity through Bar Bending Schedule

## Conclusion:

Hope this article was useful and now one can easily estimate the Bar Bending Schedule if they follow these easy steps.