Bar Bending Schedule of an R.C.C Beam

The preparation of the bar bending schedule (BBS) of an R.C.C beam requires a proper understanding of the design and the detailing of the structural member. The calculations involved in the preparation of the BBS of an R.C.C beam are explained step by step in this article.

Understanding Detailing of an R.C.C. Beam

The figure below shows a beam supported between two columns. Figure 1 shows the top view of the beam and column arrangement. Figure 3 and Figure -4 show the sectional view from A – A and B – B perspectives, respectively. We will analyze each view one by one.

1. Figure 1 shows the top view of the column and the beam. The dimension of the column used is 600 x 600 mm. Two sections are drawn, A – A and B -B.

Fig.1. Top View of Beam-Column Arrangement - Sections Taken

2. Figure-2 shows the view from A-A section. This sectional view will make us see the longitudinal arrangement of the bottom and the top bars that is used in the concrete. The clear span is given by ‘L’, where the value of L = 5m. The supports shown in the figure are the columns that were seen in figure-1.

Fig.2. Reinforcement Detailing of an R.C.C Beam

You can observe an extension of bars (which are marked in Red ) to both the columns. This is called the development length, Ld.

Note: The choice of development length is based on the standard codes of the region. This value is sometimes chosen based on the design firm and their practise. The development length provided here is Ld = 50d, IS 456:2000, Clause.26.2.1,where d is the diameter of the bar used.

The figure-2 also shows the division of the span into three sections of length L/3. This is the left, center and the right zone of the beam where the arrangement of stirrups and the reinforcement bars will vary. This is based on the ductility-based design of the beam. More details are explained in the step below.

3. Figure 3 shows the B-B section. This is the cross-sectional view of the beam. Table 1 below shows the details of bars that are obtained and provided based on the beam design.

Fig.3. Section B-B Crossection of R.C.C Beam © PRODYOGI

As mentioned in before step, the beam is divided into three zones. And from the figure-3 you can see that the bottom reinforcement for the beam is provided in two layers. The layer that is closer to the stirrup forms the first layer, and the next layer placed forms the second layer, which is represented by a ‘+ ‘(As in Table 1).

Similarly, considering the top reinforcement, the layer that is closer to the stirrup from the first layer. The layer that is placed below forms the second layer for the top reinforcement.

Bottom Reinforcement			Top Reinforcement			Stirrups
Left	Centre	Right	Left	Centre	Right	8mm Diameter bars @ 100, 150 and 100 c/c
2- 25ø	2- 25ø	2- 25ø	2- 20ø	2- 20ø	2- 20ø
+ 2- 20ø	+2- 20ø	+ 2- 20ø

Table.1: The Reinforcement Details of an R.C.C. Beam

For this beam under consideration, we have 2 layers of bottom reinforcement and 1 layer of top reinforcement. 

From the table, the stirrup spacing is provided at three values. This means that the 8mm diameter bars are to be placed at 100, 150, and 100mm spacing center to center for the left, center, and right zones of the beams, respectively. This is clearly seen in Figure figure-2.

Calculation For Preparing Bar Bending Schedule of an R.C.C Beam

Before understanding the calculations, you must know the different columns used in the bar bending schedule list.

Also Read: What is a Bar Bending Schedule in Construction?

Now consider the beam explained in figure-1,2 and 3 being sketched as per the shape codes in Indian Standard ( IS 2502 -1963).

Fig.4.A sketched Figure of a BBS details of an R.C.C Beam

Determination of Length of Reinforcement Bars

Now, we need to determine the length of each member that is used in the detailing. As shown in the figure, we need to find the length of 3 sets of bars. We will calculate the length one by one.

1. Bottom Reinforcement - Bar No: 1

We have two layers of 25mm and 20mm diameter bars ( d ) at the bottom layer. . The red portion forms the development length. Then the total length of a single bottom reinforcement bar for

First Layer: d = 25mm

Lb = Clear Span + Development length  both sides - Bending Reductions on both sides

                                                  Lb = 5000 + ( 50d ) * 2  - ( 2d ) * 2

Note : For Bar bending of 45 degrees the bending reduction is 1d, For 90 degrees bend it is 2d, for 135 degree bend it is 3d.

Therefore,

Lb = 5000 + 50 x 25 x 2 - 2 x 25 x2 = 7400 mm = 7.4m

We have 2 numbers of bars in first layer. Then total length  = 2 * 7.4m = 14.8m

Second layer: d = 20mm

Lb = Clear Span + Development length on both sides - Bending Reductions on both sides

            = 5000 +  ( 50d ) * 2  - ( 2d ) * 2

            = 5000 + 50 x 20 x 2 - 2 x 20 x2 = 6920 mm = 6.92 m

For 2 bars of second layer, total length = 2 * 6.92 = 13.84m

2. Top Reinforcement - Bar No: 2

This bar will have the same length as of the second layer of bottom reinforcement = 13.84m

3. Cutting length of stirrup Calculation - Bar No:3

From the figure -5 the cutting length of stirrup is

Ls = a + b + Hook Length - Bending lengths Deductions

Hook length = 2 x 2d ( There are two hooks); Stirrup diameter d = 8mm;

Fig.5: Beam Cross Section for Determining Cutting length of Stirrup

Note: Hook Length is taken as 10d; it must have a minimum length of 75mm, as per IS 2502-1963. The bending length reduction will be for 135 degrees (3d) at the junction of hook formation and for 90 degrees (2d) for other three bending as shown in figure-5.

From the figure-5,

a = Total Beam breadth - 2 * ( clear cover) - 2* (half the diameter of stirrup)

   = 300 - 2x 30 - 2 x 4

a = 232mm

b = Total Beam depth - 2 * ( clear cover) - 2* (half the diameter of stirrup)

   = 600 - 2 x 30 - 2 x 4

b = 532mm

Fig.6. Hook Angles 135 degrees and 90 degrees

Therefore, the cutting length of the stirrup is

Ls = a + b + Hook Length - Bending lengths Deductions

     = 232 + 532 + 2x10d - [ 2*3d ] - [ 3 * 2d]

     = 232 + 532 + 2 x 10 x 8 - [ 2 x 3 x 8 ] - [ 3 x 2 x 8]

     = 1592 mm = 1.592m

Determination of the Number of Stirrups for a Beam

The beam is divided into three zones, all have a length of L/3. The zone 1 and zone 3 use 8mm diameter stirrups at 100mm c /c, so both zones have an equal number of stirrups. The Central part is zone 2 which have stirrup placed at 150mm c/c. So,

Zone 1 : 8mm diameter bars @ 100mm c/c

No: of stirrups = (Length of zone/ spacing )+ 1

                                                                     = ( (L/3) /100 ) + 1

                                                                      =  (1666.67 / 100) +1 = 17.7 == 18Nos

Zone 2 : 8mm diameter bars @ 150mm c/c

No: of stirrups = (Length of zone/ spacing ) - 1

     = ( (L/3) /100 ) - 1

                          =  (1666.67 / 150) -1  == 10Nos

Zone 3: 8mm diameter bars @ 100mm c/c

Similar to Zone 1 = 18 Nos

Total Number of Stirrups =  18 + 10 + 18 = 46 No:s

Total Cutting Length = 46 x cutting stirrup length of single one = 46 * 1.592 = 73.23m

The details of the beam are scheduled in tabular form as shown below. This is just an example of how the above calculations are meant to be scheduled in the list. The column requirements may change from site to site. The shape code column in the list below is required for complex bending details.

Fig.6. Bar Bending Schedule of a R.C.C Beam

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