Complete Bar Bending Schedule for Different Structure (Free e-Book)
1) What is BBS (Bar Bending Schedule)?
The process of cutting and bending of steel rod or reinforcement recommended by the structural engineer is known as a bar bending schedule (BBS).
2. Why Bar Bending Schedule is important?
The main purpose of the Bar Bending Schedule is to clarify the location and nature of steel rods in the structural member like a beam, slab, column, stairs, etc.
Another purpose of Bar Bending schedule
- To know the shape of the reinforcement
- To know the quantity of steel rod in the structure
- To know the diameter and length of a steel rod at different locations of the structure.
- To know the spacing and bending of a steel rod at different lengths.
- To know the weight of reinforcement required.
- To calculate the cutting length of steel rod-like, stirrups, ties, ring, hooks, etc.
- It helps to guide the workmen in a systematic way of cutting steel rods.
- It helps to make the project economical by minimizing the wastage of steel rods.
3. Standard Size of steel bars in RCC Work
6mm, 8mm, 10mm, 12mm, 16mm, 20mm, 25mm, 32mm, 36mm, 40mm
But, the constructors should order to the factory to construct the desired diameter of steel according to their design of the structure.
So it may be, 45mm, 50mm, 60mm, etc.
4. Grades of TMT steel?
TMT bars are typically graded as Fe 415, Fe 500, Fe 550, Fe 600, etc. where the numbers indicate the level of stress that must be applied to deform it.
For example, Fe 550 starts to deform when a pressure of 550 Newton/mm2 is applied to it.
5. General Application of TMT Steel?
Fe-415, Fe-500, Fe-550, and Fe-600 are the different grades of TMT Steel.
Fe – 415: Due to its higher uniform elongation, it can be used in RCC Constructions in corrosion and as well as an earthquake-prone zone.
Fe – 500: The high corrosion resistance, excellent bendability, and great resistance on dynamic loading of Fe – 500, it is used in RCC constructions in building, bridges, and other concrete structures.
Fe – 550: It has similar properties to Fe – 500, but differs in the yield and tensile strength. It is also used in RCC construction that is exposed to coastal, marine, or underground environments.
Fe – 600: Since it provides better toughness and has more yield and tensile strength when compared to other TMT Steel grades, it is used for large RCC construction purposes
6. Why steel used on concrete Structure?
Concrete is strong in compression but weak in tension. This is the reason we place the steel in concrete.
It allows the concrete to withstand tensile forces that would otherwise crack apart unreinforced concrete.
Concrete is said to be reinforced when it contains a steel bar.
The reinforcing bar must be totally encased in concrete to a minimum depth of 50mm or 2″ to prevent contact with water and prevent rusting of the steel.
7. What is a clear cover?
The clear cover is simply the small distance between the outside surface of concrete and the reinforcement. i.e. steel rod provided inside of that concrete structure.
This clear cover for different structures depends upon the design of a structure that how much is its length, width, and load on the structure.
8. Types of Clear cover
1) Nominal Cover:
It is the distance measured from the face of the member to the outermost face of the reinforcement including Stirrups or links. It is the dimension shown in drawings and detailing.
2) Effective Clear cover:
This is the distance measured from the face of the member to the center of the area of the main reinforcement, which is tension or compression reinforcement.
This is the dimension usually used for design calculations.
Effective Cover = Overall depth – effective depth.
9. Why is clear cover important?
If I say you in one line, then it will be “clear cover is provided to make the RCC structure long-lasting“.
But let us discuss some of the importance of clear cover,
It is provided to prevent steel rods inside the structure from corrosion due to weather and thermal effects.
For example, in the case of fire in the building, the metal begins to melt and lose its strength. So, the clear cover acts as an insulator and prevents melting.
- Also in case of weather, it acts as a cover that does not let steel come in contact with the atmosphere and prevent it from corrosion.
- It makes the bond between the steel and concrete which makes it capable of bearing compression as well as tensile force.
10. How much should be a clear cover of the structure?
It depends on types of structure but the minimum clear cover recommended for the different structure is given as,
- The minimum clear cover of the beam should be 25 mm and the column should be 40 mm.
- The minimum clear cover of the slab should be 15 mm. But, for safety rather than economical, we should provide 20 mm.
- The minimum clear cover for footing should be above 50 mm. because all the loads of superstructure are transferred to the ground through the footing. So, it should be more.
- The minimum clearance of the column should be 40 mm.
- The minimum clear cover for the raft foundation should be 75 mm.
- The minimum clear cover for the staircase should be 15 mm.
- The minimum clear cover for retaining structure or retaining wall should be 20 mm.
1. What is Lap length, Development length, Anchorage Length?
i) Lap length
Lap length is the length of the overlap of the bar required to safely transfer stress from one bar to another.
Lap length is different in the case of tension and compression zones and mainly depends on the grade of concrete and steel.
If two different diameter bars are to be lapped, lap length is based on a smaller diameter.
Overlap Length for compression members like columns should be= 50d
The Overlap Length for tension members like beams should be = 40d
[Where d is the Diameter of the bar]
ii) Development length
Development length is the length of the bar required to transfer stress from steel to concrete.
If the figure below (Ld) is Development length.
iii) Anchorage Length
Anchorage Length is provided if sufficient development length cannot be able to be provided inside the support/fixed end.
- L value is generally considered as 8 times for a 90-degree bend
- 6 times the diameter for a 135-degree bend and
- 4 times the diameter of the bar for a 180-degree bend.
- In almost all cases, we use a 90-degree bend.
2. Different Type of mixes
i) Nominal Concrete Mix Ratios
ii) Standard Mixes Ratio
iii) Designed concrete ratio
i) Nominal Concrete Mix Ratios
The nominal concrete mix ratio is used for small construction where no actual strength calculation is required. Literary we can say this is the approx. value of the ratio of cement, sand, and aggregates for preparing concrete.
The mix proportions like 1:1.5:3, 1:2:4, 1:3:6, etc. are adopted in a nominal mix of concrete without any scientific base, only on the basis of past empirical studies.
ii) Standard Mixes Ratio (IS-456)
The nominal mixes of fixed cement-aggregate ratio (by volume) vary widely in strength and may result in under or over-rich mix.
For this reason, the minimum compressive strength has been included in many specifications. These mixes are termed standard mixes.
In this designation, the letter M refers to the mix and the number to the specified 28-day cube strength of mix in N/mm2.
iii) Designed concrete ratio:-
The designed concrete ratio is prepared by an expert in the construction. This is actual and calculated value by structural engineers.
There are no guidelines for any grade of concrete. It depends on the structure to be constructed which should bear the load.
3. Standard Length of Bars in meter
Standard Length of Bar is 12 meters.
4. Standard Length of Bars in feet
Standard length of bar = 40 feet
5. Diameter of reinforcement bars in inches and Millimeter:
#3(3/8″) (9.525 mm)
#4 (1/2″) (12.70 mm)
#5 (5/8″) (15.875 mm)
#6 (3/4″) (19.05 mm)
#7 (7/8″) (22.225 mm)
#8 (1) (26 mm)
#9 (9/8) (28.575 mm)
#10 (10/8) (31.75 mm)
6. How to calculate Weight of steel per bar (Kg/Bar):
Weight per bar = (D2/ 162.25) x standard length of bar
For example, Diameter of Bar = D = 8 mm
Weight per bar = (D2/ 162.25) x 12 = 82/162.25 x 12 = 4.74 Kg/ bar
7 .Weight of bar per ft (Kg/ft):
Weight per feet = where ‘D’ is in mm
For example, Diameter of bar = (D2/ 533) = 12.70 mm or (#4)
Weight per feet = = 12.702/533 = 0.302 kg/ft
If the diameter of bar is in inches then,
Weight per feet = where ‘D’ is in No. (#4)
For example, Diameter of bar = D = #4
Weight per feet = = 42/52.986 = 0.302 Kg/ft
8. Weight of steel per bar (Kg/Bar):
Weight per Bar = (D2/ 533) x standard length of bar
For example, diameter of bar = D = 12 mm
Weight per Bar = (D2/ 533) x 40 = 10.89 Kg/Bar
(Since length of 1 bar is 40 feet)
9. Overlapping of steel in footing:
Neck column to footing overlapping = 40d to 50d
Column to column or beam overlapping = 50d
Development length for dowel bars = 16d
General overlapping when 12 m bar finish = 50d
10. What is Spacing?
The Distance between two reinforcement bars in the structure like a beam, column, or slab is known as Spacing.
11. Formula to calculate Number of bars of any length
No. of bars = (Opposite Length/Spacing) + 1
12. Formula to calculate Number of stirrups
No. of stirrups = (Actual Length of Column or Beam/Spacing) + 1
Outer to Outer length of the beam is 5000mm and 8mm diameter stirrups are placed at 200mm Center to Center
Assume that you are providing a clear cover of 25mm on either side.
Available length = 5000-(2×25) =4950mm
The spacing of stirrups is 200mm.
=24.75 = Approx. 25
Hence, 26 stirrups. (No. of stirrups are spacing + one.)
13. Deduction of the length of bars due to bending at a different angle:
45 Degree = 1d
90 Degree = 2d
135 Degree = 3d
Why deduct the length of bars due to bending at a different angle?
At the time of bending the reinforced steel bar at some angles (degrees), owing to elongation in steel, the length of steel will be raised to some extent.
Prior to arranging the bars in structural components, the steel bar should be cut, for this purpose, it is necessary to work out the perfect length of the bar (cutting length). With the purpose of working out the cutting length, it is required to subtract the elongated length of the bar because of the bend.
14. Crank length & extra bars:
Crank = 0.42d ( for 45 Degree)
0.27d ( for 30 Degree)
0.58d (for 60 Degree)
Extra bars length = L/4
I hope this article on “Bar Bending Schedule” remains helpful to you.
Happy Learning – Civil Concept
Civil Engineer – Ranjeet Sahani