bar bending schedule

Bar Bending Schedule (BBS) | Preparation of Bar Bending Schedule | Numerical Example

 

Bar Bending Schedule is also commonly referred to as the BBS. It is the comprehensive list that details the type, size, location, mark, length, number and the bending of each bar or fabrication in the reinforcement drawing.

In other words, bar bending schedule is the list of all the reinforcement bars for the reinforced concrete work of any construction. Such process of enlisting the type, size, location, mark, length, number and the bending details of each bar in the reinforced concrete work is known as Scheduling.

The primary objective of preparing the bar bending schedule is to organize all the rebars efficiently in the structural unit as well as to give the detailed descriptions of the reinforcements. It is mostly expressed in tabular format.

 

Importance of Preparing Bar Bending Schedule

The main importance of preparing the bar bending schedule can be listed as follows:

a. It is essential for determining the quantities of materials particularly steel.

b. It enlists all the details of the rebars beforehand thereby the wastage is eliminated.

c. It is essential for the calculation of the cutting length of various types of rebars.

d. It assists the workmen and labours in completing the reinforced concrete works efficiently.

e. It helps to prepare the bills that can be claimed to the clients.

 

Personnel Using The Bar Bending Schedule

The following personnel make use of the prepared bar bending schedule:

a. The contractor who is responsible for ordering the required reinforcements.

b. Fabricator who is responsible for fabricating the reinforcements.

c. Quantity Surveyor

d. Site Incharge

e. Detailer, etc.

 

Information Given By The Bar Bending Schedule

The bar bending schedule, in general, delivers the following information:

a. Identification of Member

b. Mark of each Bar

c. Diameter of each Bar

d. Length of each Bar

e. Bar Mark

f. Number of Members

g. Number of Members in each Bar

h. Total Number of Bars

i. Total Length

j. Shape Code

k. Bending Dimensions

 

Framework For The Preparation of Bar Bending Schedule

a. The bars must always be listed in the order of floors i.e. floor by floor for building structures.

b. Each structural unit must consist of bars grouped together. For example; beam, column etc.

c. The fabrication schedule and the form of bars must be in compliance with BS 8666.

d. The schedules for the cutting and bending purpose must be provided in a separate A4 size paper and should not be included in the detailed reinforcement drawings.

e. The bar mark reference on the label attached to a particular bundle of bars must significantly represent the unique properties of that particular group only. For example; length, size, type of bars etc.

f. In the schedule, as far as possible, the bars must be ordered in numerical order.

 

Preparation Of Bar Bending Schedule

The necessary column for the bar bending schedule can be listed as follows:

a. The shape of Bars

b. The diameter of Bars

c. Bar Number or the Bar Reference

d. Spacing

e. Length of Bars

f. Cutting Length

g. Number of Bars

The detailed reinforcement drawing provides various information required for the bar bending schedule. Such information includes the shape of bars, length of bars, the diameter of bar and spacing of bars.

The formula for calculating the various parameters are as follows:


a. Number of Bars

Number of Bars = (Length/ Spacing) + 1

 

b. Cutting Length

Cutting Length = True Length of Bar – Deductions

The things that must be considered during the preparation of the bar bending schedule can be listed as follows:

a. The bars greater than 36mm in size should not be bundled.

b. Welding should not be done at the bends.

 

Steps Involved In The Preparation Of Bar Bending Schedule

a. Each bar is first labelled using an appropriate reference mark. With reference to the structural drawing, the shapes of all the bars are recorded. Then, a table is prepared and the diameter of each bar is labelled in the table.

For example; The number of bars may be expressed as #1, #2 etc. If the diameter of the bar is 12mm then it is expressed as #[email protected]”.

 

b. Then, the number of bars is calculated using the above-listed formula.

For example; if the length along which the stirrups are placed is 7000 mm and the c/c spacing is 150 mm then, the number of bars is calculated as:

Number of Bars = (length/ Spacing) + 1

= (7000/ 150) + 1

= 47.67  i.e. 48 number of bars

 

c. The length of the reinforcement bars is then measured.

 

d. Then, the unit weight of the reinforcement bar is computed as,

Unit Weight of Bar = Volume of Bar * Density of Bar

 

Benefits Of Preparing The Bar Bending Schedule

The major benefits of the bar bending schedule can be summarized as follows:

a. When the bar bending schedule is made, the cutting and bending of the reinforcement can be done at the manufacturing industry itself which can be easily transported to the site later. This results in the saving of time for bending and cutting of reinforcement at the site.

b. The bar bending schedule also helps to prevent unnecessary wastages. The wastage of steel is reduced by 5 to 10% thereby reducing the overall expense of the project.

c. It ensures efficient reinforcement works at the site and improves the quality of the work achieved.

d. It assists in the auditing of reinforcements.

e. It can also be used as a means of checks in case of theft and pilferage.

f. It aids in the preparation of the estimate of the quantity of steel required. Thus, the shortage of reinforcement as well as the over buying or stocking of the reinforcements can be duly reduced.

 

Points To Remember For Bar Bending Schedule

 

a. The commonly used diameter of Reinforcement Bars (in mm)

6mm, 8mm, 10mm, 12mm, 16mm, 20mm, 25mm, 28mm, 32mm, 36mm, 40mm

 

b. Standard Length of Reinforcement Bars (in m)

The standard length of reinforcement bars is taken 12m.

 

c. Weight of Bar Per Length (in kg/m)

The weight per length of the reinforcement bars is expressed as,

Weight/ m = D²/ 160

For example; if the diameter of the reinforcement bar is 6mm, then,

Weight/ m = 6²/ 160

= 0.225 kg/m

(Note: you should keep the  value of D in mm and answer comes in m.)

 

d. Weight Per Bar (in kg/bar)

The weight per bar can be expressed as,

Weight/ bar = Weight of Bar per m x Standard Length of Bar

= 0.225 x 12

= 2.7 kg/bar

 

e. Diameter of Reinforcement Bars (in inches)

1. #3 Rebar = 3/ 8 in = 9.525 mm

2. #4 Rebar = 1/ 2 in = 12.70 mm

3. #5 Rebar = 5/ 8 in = 15.875 mm

4. #6 Rebar =  3/ 4 in = 19.05 mm

5. #7 Rebar = 7/ 8 in = 22.225 mm

6. #8 Rebar = 1 in = 25.4 mm

7. #9 Rebar = 9/8 = 28.575 mm

 

f. Standard Length of Reinforcement Bar (in feet)

The standard length of reinforcement bar is taken as 40 ft.

 

g. Weight of Reinforcement Bar per Length (in kg/ ft)

The weight per bar can be expressed as,

Weight/ Length = D² / 533

For example; if the diameter of the reinforcement bar is #4 i.e. ½ in or 12.70mm, then,

Weight/ Length= 12.70²/ 533 = 0.302 kg/ft (since, D must be in mm)

Also,

Weight/ Length = D²/ 52.896 

For example; if the diameter of the reinforcement bar is #4, then,

Weight/ Length = 4²/ 52.896 = 0.302 kg/ ft (since, D must be in #No.)

 

h. Weight of Reinforcement Bar per Bar (kg/ bar)

The weight of reinforcement bar per bar can be expressed as,

 Weight/ Bar = (D²/ 533 )* standard length of bar

For example; if the diameter of the reinforcement bar is 12.70mm (#4), then,

Weight/ bar = (12.70²/ 533) * 40

= 12.10 kg/ bar

 

i. Cutting Length of Main Reinforcement Bar with Hook:

Length of Hook = 9d, where d is the diameter

Length of Main Bar = L + 9d + 9d

 

j. Cutting Length of Main Bar with Bend 

Length of bend = 10d – 6d

Length of Main Bar = L + 12d + 12d

 

k. Overlapping

Neck Column to Footing Overlapping = 40d – 50d

Column to Column or Beam Overlapping = 50d

Development Length for Dowel Bars = 16d

General overlapping on Completion of 12m bar = 50d

 

l. Concrete Cover

Cover basically includes a clear cover and effective cover.

Clear cover for footing is taken as 75mm, for column 25mm to 50mm, for beam 25 to 50mm and for slab 25mm.

 

m. Bend Deduction 

For 45 deg bend = 1d

For 90 deg bend = 2d

For 135 deg bend = 3d

 

n. Crank Length & Extra Bar Length 

For 45 deg bend, crank length = 0.42d

For 30 deg bend, crank length = 0.27d

For 60 deg bend, crank length = 0.58d

Length of extra bar = L/ 4

 

Format for Bar Bending Schedule in Microsoft Excel

The standard format for the preparation of bar bending schedule in excel is as follows:

bbs

 

 

Numerical Example of Bar Bending Schedule

a. An RCC beam 350 mm wide and 500 mm deep with a length 5000 mm is reinforced with four numbers of 12mm bars that are placed in one single row. Out of the four bars, two of the bars are straight and two of the bars are bent up. Also, two additional anchor bars are provided on the top with 10 mm diameter. Stirrups of diameter 6 mm are provided at a c/c spacing of 150 mm. Determine the total quantity of steel required and the bar bending schedule.

Solution,

Length (L) = 5000 mm = 5 m

Depth (d) = 450 mm

Breadth (b) = 350 mm

Let the cover be 25mm.

Now,

Length of Straight Bar = (L – 2 x cover) + 18 x dia of bar

= (5000- 2 x 25) + 18 * 12 (as the dia of bars is 12mm)

= 5166 mm = 5.16 m

( Note: Length of main bar is L + 9d +9d but we have to reduce cover from both side hence it becomes L + 9d + 9d – 2 * cover and finally written as ((L-2* cover)+ 18d)

 

Length of Bent Up Bar = (L – 2 x cover) + 18 x dia of bar + 0.42 x d

= (5000 – 2 x 25) + 18 x 12 + 0.42 x 450

= 5355 mm = 5.35 m

( Note: Crank lenth is taken as 0.42 d for 45° bend.)

 

Length of Anchor Bar = (L – 2 x cover) + 18 x dia of anchor bar

= (5000 – 2x 25) + 18 x 10

= 5130 mm = 5.13 m

 

Length of Stirrup = 2 ( d’ + b’) + 24 * dia of stirrups

= 2 (300 + 400) + 24 * 6

= 1544 mm = 1. 54 m

Where, d’ and b’ are the depth and width without considering the cover of 25mm on either side.

 

Number of Stirrups = {(Total Length – 2 x total cover) / Spacing} + 1

= {(5000 – 2 x 50)/ 150} + 1

= 34

 

The bar bending schedule is thus as follows:

bar bending schedule


Extra Knowledge:

bar bending schedule

 

 

Read More: Grades of Concrete

Read More: Preliminary Estimate

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