Bar Bending Schedule of an R.C.C Beam
The preparation of bar bending schedule of an R.C.C beam requires proper understanding of the design and the detailing of the structural member. The calculations involved in the preparation of an R.C.C beam is 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. The figure1 shows the top view of the beam and column arrangement. The Figure3 and Figure 4 shows 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.
2.The figure2 shows the view from AA 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 figure1.

Fig.1. Top View of Beam Column Arrangement  Sections Taken 
2.The figure2 shows the view from AA 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 figure1.

Fig.2. Reinforcement Detailing of an R.C.C Beam 
You can observe an extension of bars (That are marked in Red ) to both the column. This is called as 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 figure2 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.

Fig.3. Section BB Crossection of R.C.C Beam 
As mentioned in before step, the beam is defined into three zones. And from the figure3 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 Table1).
Similarly, considering the top reinforcement, the layer that is closer to the stirrup form 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 the right zones of the beams respectively. This is clearly seen in the figure2.
Before understanding the calculations, you must know the different columns used in the bar bending schedule list.
Now Consider the beam explained in figure1,2 and 3 being sketched as per the shape codes in Indian Standard ( IS 2502 1963).
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 bar. We will calculate the length one by one.
Similar to Zone 1 = 18 Nos
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.
You can read What is Bar Bending Schedule in Construction?
Now Consider the beam explained in figure1,2 and 3 being sketched as per the shape codes in Indian Standard ( IS 2502 1963).

Fig.4.A sketched Figure of Bar and bending details of an R.C.C Beam 
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 bar. 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;
Note: Hook Length is taken as 10d, it must have a minimum length of 75mm, as per IS 25021963. The bending length reduction will be for 135 degrees (3d) at the junction of hook formation and for 90 degrees (2d) for other three bendings as shown in figure5.
From the figure5,
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
Therefore, the cutting length of 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 number of stirrups
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 the 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 how the above calculations are meant to be scheduled in list. The column requirements may change from site to site. The shape code column in the below list is required for complex bending details.
The details of the beam are scheduled in tabular form as shown below. This is just an example how the above calculations are meant to be scheduled in list. The column requirements may change from site to site. The shape code column in the below list is required for complex bending details.