What is Bar Bending Schedule

Bar bending schedule or schedule of bars is a tabular representation of reinforcement bar. It is generally represented for each type of R.C.C work.

With the help of bar bending schedule the requirement of different length and sizes of bars may be known and can be arranged and bent-up during the time of construction.

Bar bending schedule generally describes the particulars of bars, shape of bending with sketches and total length and weight of the bars along with their numbers.

Bar bending schedule is generally prepared while estimating a R.C.C work or structure.

Advantages of BBS

• Quantities of steel reinforcement of different diameter and different grades are calculated easily.
• Ideas of different sizes of bars, bend and length of bars can be easily acquired through schedule of bars.
• During the auditing of reinforcement on construction site, bar bending schedule becomes very much helpful.
• Moreover it helps to avoid confusion on the construction site.
• It provides the exact quantity of steel required for work due to which optimization of reinforcement can be done in case of cost overrun.
• Bar bending schedule makes it easy for site engineers to check and verify the cutting length and bar bending while inspection on the site.
• At the end of the entire work the construction bills can be easily created with the help of these schedules of bars.

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How to Prepare Bar Bending Schedule

Generally reinforced cement concrete works may be calculated under 2 items :

1. Concrete work including centering and shuttering.
2. Steel reinforcement along with its bending, cutting, laying etc in Quintal or Tonnes.

The quantity of steel is very small in volume hence no deduction is made for the steel from the volume of concrete.

Steel reinforcement is calculate as per actual requirement including overlap, hooks, cranks etc and is determined from detailed drawing.

Generally the percentage of steel reinforcement depends on the design of structure. nu7

For lintel, slab etc	0.7 % to 1 %
Beam	1 % to 2 %
Column	1 % to 5 %
Footing	0.5 % to 0.8 %

#Extra Length of Bar :

1. Standard Hook (180° Bend) :

• Extra length for 1 hook = 9Φ
• Extra length for 2 hooks = 2 × 9Φ = 18Φ

2. For 90° Bend :

• 90° bend is generally provided for HYSD (High Yielding Strength Deformed) Bars.
• Extra Length for one 90° bend = 6Φ
• Extra length for two 90° bend = 2 × 6Φ = 12Φ

3. Bent-up Bars :

• Extra length for one bent-up =
• Extra length for two bent-up bars = 2 × 0.42 d = 0.84 d
• d = D – (top cover + bottom cover)

4. For Two Legged Stirrups :

• Extra length of hook = 24Φ
• A = b – 2 (side cover)
• B = D – (top cover + bottom cover)
• Total length of stirrups = 2 (A + B) + 24Φ         …… (Φ = dia of steel reinforcement)

#How to Calculate Weight of Bars in Bar Bending Schedule :

Weight of bars is generally calculated in Kilograms and it is calculated for every one meter length.

Weight of Bars in Kg/m =

Here, Φ = diameter of bars used.

#Calculation of Number of Bars :

Number of Bars =

Preparation of Bar Bending Schedule With Simple Example

Problem :  R.C.C simply supported beam of side300 mm × 650 mm is reinforced with 4 nos of 20 mm diameter bars. The main bars are provided in the one row and bent-up bars are provided on the second. Two anchor bars of 12 mm diameter are provided to top and 6 mm diameter stirrups are provided at 140 c/c. The span of beam is 5.6 m and end bearing is 30 cm. Calculate the total quantity of mild steel reinforcement and prepare bar bending schedule of the same.

Solution :  Now let us check the given data in the problem.

First of all we should assume the clear cover on all sides of the beam = 25 mm.

Width (b) = 300 mm
Overall Depth (D) = 650 mm
Depth (d) = = 650 – (2 × 25) = 600 mm
TL = 5600 + (2 × 300) = 6200 mm

Step 1 : Length of Main Bars :

4 bars are provided i.e 2 main  bars and 2 bent-up bars.

a. Straight bars (2, 20Φ)

Length of straight bar = [ TL – (2 × side cover ) + (2 × 9Φ) ]
= 6200 – (2 × 25) + (2 × 9 × 20)
=  6510 mm / 6.51 m

b. Bent-up Bars (2, 20Φ)

Length of Bent-up bar = [ TL – (2 × side cover ) + (2 × 0.42 d) + (2 × 9Φ) ]
= 6200 – (2 × 25) + (2 × 0.42 × 600) + (2 × 9 × 20)
= 7014 mm/ 7.014 m

Step 2 : Length of Anchor Bars (2, 12Φ) :

L = [ TL – (2 × side cover ) + (2 × 9Φ) ]
= 6200 – (2 × 25) + (2 × 9 × 12)
= 6366 mm/ 6.366m

Step 3 : Length of Stirrups (6 mm Φ) :

A = 300 – (2 × clear cover)
= 300 – (2 × 25) = 250 mm

B = 650 – (2 × clear cover)
= 650 – (2 × 25) = 600 mm

L = 2 (A + B) + 24Φ
= 2 (250 + 600) + (24 × 6)
= 1844 mm/ 1.84 m

 

Number of Stirrups =
=

= 45 Nos.

Bar Bending Schedule :

Sr No	Description of Bar	Shape of Bar	No	Փ (mm)	L(m)	Total Length of Bar (m)	Wt in Kg/m	Total Weight (Kg)
1	Main Straight Bar		2	20	6.51	13.02	2.47	32.15
2	Main Bent-up Bar		2	20	7.014	14.028	2.47	34.65
3	Anchor Bar		2	12	6.366	12.732	0.89	11.32
4	Stirrups		45	6	1.844	82.98	0.22	18.25
Total Qty =								96.37 Kg