Rebars, the reinforcing steel bars used in reinforced concrete construction, are described in detail in a bar bending schedule, or Bar Bending Schedule. It is a kind of construction document that describes the number of rebars, as well as their dimensions, lengths, forms, and bending requirements for various structural elements, such as beams and columns. Contractors, engineers, and construction workers all rely on the bar bending schedule to make sure the reinforcing bars are properly inserted in the concrete structure.
The following details are often included in the Bar Bending Schedule:
- Bar markings and numbers: The BBS provides the distinctive mark or number used to distinguish each bar from others in the structure.
- Bar dimensions: The Bar Bending Schedule specifies the dimensions of each bar needed for the construction.
- Details on how each bar should be bent to match the contour of the building are shown in the BBS’s comprehensive drawings.
- The BBS indicates the number of bars needed for each component of the building.
- Weight of bars: To determine the overall weight of the reinforcing steel utilised in the construction, the BBS additionally supplies the weight of each bar.
The BBS is a crucial document in the building of reinforced concrete because it assures that the reinforcing steel is fitted properly and in accordance with the necessary requirements. The BBS assists with minimising mistakes and ensuring that the construction satisfies the necessary strength and durability criteria by giving specific information about the size, shape, and bending characteristics of each bar.
Making a bar bending schedule (BBS) in reinforced concrete construction has a number of benefits, such as:
- A precise list of the number of reinforcing bars, their dimensions, lengths, and bending information needed for each structural member is provided by the BBS. This lowers the possibility of material scarcity or waste by allowing engineers and contractors to calculate the precise amount of materials required for the project.
- Improved quality control: The BBS makes sure that the reinforcing bars are put properly and in accordance with the necessary requirements. By doing so, the possibility of mistakes is decreased, the completed construction is of higher quality, and the necessary criteria of strength and durability are met.
- Better project scheduling: By giving a thorough schedule of the reinforcing steel required, the BBS aids in better project scheduling. This enables contractors to efficiently organise their resources and operations, which can assist to reduce delays and guarantee that the project is finished on schedule.
- Simple communication: The BBS offers a standardised method of informing all parties engaged in the construction project about the need for reinforcing steel. This can lessen the possibility of arguments about adjustments or additional work and ensure that all stakeholders are aware of what is necessary.
- Reduced cost and time: By reducing the need for onsite alterations or revisions, the BBS can aid in reducing construction costs and time. The BBS helps to reduce mistakes and makes sure that the reinforcing steel is fitted correctly the first time by giving specific information on the size, shape, and bending features of each bar.
How To Calculate the Weight Of Bar:-
Bar’s weight (in kilogrammes) is equal to (D2 / 162) * L.
Where:
The bar’s diameter, D, is measured in millimetres.
L is the bar’s length in metres.
Example:
Consider a steel bar that is 6 metres long and has a diameter of 12 millimetres. The bar would weigh the following according to the formula:
Bar Weight = (122 / 162) * 6
Bar weighs 1.139 kg.
As a result, the bar would be around 1.139 kg in weight.
How To Calculate Length Of Bar.
The length of a bar can be calculated using the following formula:
Length of Bar = (Total Length of Bars) – [(Number of Bends) x (Bend Deduction)]
Example:-
Consider a structural drawing that calls for a 6-meter-long “U”-shaped bend in a bar with a 12 mm diameter and 90-degree bends on both ends. Indian Standard Code IS 2502 states that a 90-degree bend requires a bend reduction of 1d, or 12mm in this instance.
By applying the formula, the bar’s length would be:
Bar length = 6 m – [(2) x (0.012 m)]
Bar length is 5.976 metres.
As a result, the bar needed for this specific structural component would be around 5.976 metres long.
How To Calculate Spiral Ring Cut Length In Pile Foundation:-
Here’s an example:
Suppose a pile has a diameter of 1 meter, and the required spacing of the spiral reinforcement bars is 150 millimetres. The pitch of the spiral is 100 millimetres, and the number of turns required is 4. The Spiral Angle is 45 degrees, and the Extra Length for Lap is 300 millimetres.
Diameter of the pile = 1 meter
Circumference = π x Diameter = 3.14 x 1 = 3.14 meters
Spacing between the spiral reinforcement bars = 150 millimetres
The pitch of the spiral = 100 millimetres
Number of turns required for the spiral = 4
Spiral Angle = 45 degrees
Extra Length for Lap = 300 millimetres
Using the formula above, we get:
Length of spiral reinforcement bar = [(4) x (0.1)] / sin(45) + 0.3
Length of spiral reinforcement bar = 1.13 meters
Therefore, for this pile, the length of each spiral reinforcement bar would be approximately 1.13 meters.
How To Calculate Cut Length Of Bar For Column:-
Suppose a column has a height of 3 meters, a width of 0.3 meters, and a length of 0.3 meters. The reinforcement bar required is a 20-millimetre diameter bar, and the clear cover is specified as 40 millimetres. The required lap length is specified as 40 times the diameter of the bar.
Height of column = 3 meters
Width of column = 0.3 meters
Length of column = 0.3 meters
Diameter of the reinforcement bar = 20 millimetres
Clear cover = 40 millimeters
Lap length = 40 x Diameter of bar = 40 x 20 = 800 millimeters
Using the formula above, we get:
Cut Length = 3 – (2 x 0.04) – 0.8
Cut Length = 2.12 meters
Therefore, for this column, the cut length of the 20 millimeter diameter reinforcement bar would be approximately 2.12 meters.
How To Calculate Cut Length Of Square Stirrups :-
Suppose a column has a height of 3 meters, a width of 0.3 meters, and a length of 0.3 meters. The required size of the square stirrup is 8 millimeters, and the clear cover is specified as 40 millimeters. The spacing of the stirrups is specified as 100 millimeters. The Extra Length for Lap is 300 millimeters.
Height of column = 3 meters
Width of column = 0.3 meters
Length of column = 0.3 meters
Size of square stirrup = 8 millimeters
Length of one side of stirrup = Width of column – (2 x Clear cover)
Length of one side of stirrup = 0.3 – (2 x 0.04)
Length of one side of stirrup = 0.22 meters
Spacing of stirrups = 100 millimeters
Number of stirrups required = Height of column / Spacing of stirrups
Number of stirrups required = 3 / 0.1
Number of stirrups required = 30
Cut Length = 4 x (0.22) x (30) + 0.3
Cut Length = 26.7 meters.
Conclusion
A BBS’s conclusion entails a careful examination of the document to make sure that all the material is correct and comprehensive. The structural element’s size and shape, the placement and spacing of the reinforcing bars, the lap lengths, and the anchoring details should all be examined by the engineer or detailer.
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