Bar Bending Schedle

# Bar Bending Schedule for Reinforcement Slab

## B.B.S for R.C.C Slab.

In This article will will discuss about How to prepare bar bending schedule for RCC Slab

To preparing bar bending schedule for a Slab. there are some criteria and the assumptions that require to be met. Let’s get started.. ### Forces on RCC Slab

You should know about the normal forces action on a Slab.

If u see the shear force and bending moment diagram, The slab has high moments at the center and less at ends. The shear force is maximum at the ends while at the center shear force is 0 or minimum.

• Use less cutting of the steel and provide desirable lapping.
• Lap another steels at the same time.
• Don’t Overlap at same position and place.
• Between the laps end 2 end spacing should be provided.

### Bar Bending Schedule Is Used by.

Bar bending schedule is used by different kind of the peoples and officials on the construction site which is given below.

1. Quantity Surveyor
2. Steel fixer
3. Contractor
4. Project Manager
5. Inspectors
6. Clerks of Construction

In the Construction Site all Estimation and Costing Operation is Responsibilities of ( QS ) Quantity Surveyor. The bar bending schedule gives the details to ( QS ) Quantity Surveyor for finding out the number of steel its type, size and shape. Contractor are using bar bending schedule for ordering the steel bar for proceeding the construction. Steel fixer using bar bending schedule  for doing their work or installing the bar in column, beam or slab according to the bar bending schedule that how much cut length or bending of bar will be needed and which size of steel should be use. The inspector and clerks of the construction refers to the B.B.S to make are that the reinforcement work in the field is in proper with the design of reinforcement as per drawing.

### Bar Bending Schedule for RCC Slab. ### Details Of Diagram

1. Thickness of the slab is150 mm
2. Development Length is 50d
3. Bottom bar of the slab will be provided 12 mm dia of the steel is at 200 mm center 2 center spacing.
4. Top bar of the slab will be provided 12 mm dia of the steel is at 200 mm center 2 center spacing.
5. Extra bar of the slab at x- axis will be provided 12 mm dia of the steel is at 125 mm center 2 center spacing.
6. Extra bar of the slab at y- axis will be provided 12 mm dia of the steel is at 125 mm center 2 center spacing.
7. Stirrup Spacing is 8mm @ 150 mm C to C.
8. Concrete Cover of the Slab is 25 mm.

### Steps for (B.B.S) of the Slab.

Generally, There are the five steps for preparing the bar bending schedule of the column these five is given below.

• Number of the steel.
• Cut length of the steel.
• Find out the lapping.
• Bar Bending Schedule.

### Numbers of the steel in Main bar.

Number of steels = (Total length of slab ÷ spacing of bar) +1

Number of stirrups = 5000 ÷ 200 + 1

Number of stirrups = 26 numbers

### Numbers of the steel in Distribution bar.

Number of steels = (Total length of slab ÷ spacing of bar) +1

Number of stirrups = 5000 ÷ 200 + 1

Number of stirrups = 26 numbers

### Numbers of the Extra bar at x-axis

Number of steels = (Total length of slab ÷ spacing of bar) +1

Number of stirrups = 5000 ÷ 125 + 1

Number of stirrups = 41 numbers

#### Numbers of the Extra bar at y-axis.

Number of steels = (Total length of slab ÷ spacing of bar) +1

Number of stirrups = 5000 ÷ 125 + 1

Number of stirrups = 41 numbers

### Cut length of the steel.

#### Cut Length of the Main Bar.

Length of Main bar = slab length – 2(concrete cover) + Development length (Ld) + Overlap Length

= 5000 – 2(25) + (12×12) + (50×12)

= 5694 mm or 5.69 m

### Now We Can find the length of bar.

The total length of Main bar = 5694 x number of steels in slab

The total length of bar = 5694 x 26

The total length of bar = 148044 mm or 148 meter

### Cut Length of the Distribution Bar.

Length of Distribution bar = slab length – 2(concrete cover) + Development length (Ld) + Overlap Length

= 5000 – 2(25) + (12×12) + (50×12)

= 5694 mm or 5.69 m

### Now we find the length of bar.

The total length of Distribution bar = 5694 x number of steels in slab

The total length of Distribution bar = 5694 x 26

The total length of bar = 148044 mm or 148 meter

### Cut Length of Extra Bar at x-axis.

Total Length of extra bar = Length of extra bar – concrete cover

= 1500 – 2(25)

= 1450 mm or 1.450 meter

### Now we find the length of bar.

The total length of extra bar at x- axis = 1450 x number of extra bar in slab

The total length of bar = 1450 x 41

The total length of bar = 59450 mm or 59.45 meter

### Cut Length of Extra Bar at y-axis.

Length of Extra Bar at y-axis = slab length – 2(concrete cover)

= 1500 – 2(25)

= 1450 mm or 1.40 m

### Now We Find the length Of The bar.

The total length of the Extra bar = 1450 x number of extra bar in slab

The total length of the bar = 1450 x 41

The total length of the bar = 59450 mm or 59.45 meter

### Main Bar.

length of lapping is 50d

Lapping length = 50 x Dia of steel

Length of lapping = 50×12 = 600 mm.

We know that the length of the typical steel is 12.25 m or 40 feet length

Total Length of the Main Bar = 148044 m that is more than 12.25 m or 40 feet so lapping should be provided.

Length of the lapping is 600 mm the result will be 148044 + 600 = 148644 m

It is mentioned already in this article, Consider the points while lapping the bar. There is no one right process we can use. Just remember the lines.

1. Don’t tie a column at the bottom & also at the top
2. Lap at the another bars every time
3. Overlap top bars with the bottom bars always

### Distribution Bar.

length of lapping is 50d

Lapping length = 50 x Dia of steel

Length of lapping = 50×12 = 600 mm.

We know that the length of typical steel is 12.25 m or 40 feet length

Total Length of the Main Bar = 148044 m that is more than 12.25 m or 40 feet so lapping should be provided.

Length of the lapping is 600 mm the result will be 148044 + 600 = 148644 m

It is the mentioned already in this article, consider the points while lapping the bar. There is no one right process we can use. Just remember the lines.

• Don’t tie a column at the bottom and also at the top.
• Lap at another bars every times.
• Overlap top bars with the bottom bars always.

### Numbers of The steel in Main bar

Number of the steels = (Total length of slab ÷ spacing of bar) +1

Number of the stirrups = 5000 ÷ 200 + 1

Number of the stirrups = 26 numbers

### Numbers of the steel in Distribution bar:

Number of the steels = (Total length of slab ÷ spacing of bar) +1

Number of the stirrups = 5000 ÷ 200 + 1

Number of the stirrups = 26 numbers

### Numbers of The Extra bar at x-axis.

Number of the steels = (Total length of slab ÷ spacing of bar) +1

Number of the stirrups = 5000 ÷ 125 + 1

Number of the stirrups = 41 numbers

### Numbers of The Extra bar at y-axis.

Number of the steels = (Total length of the slab ÷ Spacing of bar) +1

Number of the stirrups = 5000 ÷ 125 + 1

Number of the stirrups = 41 numbers

### Bar Bending Schedule ( BBS ).

The Bar bending schedule of  the column is shown below in table

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