Bar Bending Schedule : How to Prepare BBS

 

What is Bar Bending Schedule.

Bar bending schedule or agenda of bars is a tabular illustration of reinforcement bar. It is generally represented for every form of R.C.C paintings

With the help of bar bending agenda the requirement of various length and sizes of bars can be acknowledged and can be arranged and bent-up during the time of production.

Bar bending schedule usually describes the particulars of bars, shape of bending with sketches and general length and weight of the bars along side their numbers.

Bar bending time table is normally organized while estimating a R.C.C work or structure.

Advantages of BBS

·         Quantities of metallic reinforcement of various diameter and one of a kind grades are calculated without problems.

·         Ideas of various sizes of bars, bend and length of bars can be effortlessly obtained thru schedule of bars.

·         During the auditing of reinforcement on creation web page, bar bending time table turns into very a good deal beneficial.

·         Moreover it facilitates to keep away from confusion on the construction website.

·         It presents the precise quantity of metallic required for work because of which optimization of reinforcement can be achieved in case of price overrun.

·         Bar bending time table makes it clean for web site engineers to test and affirm the cutting duration and bar bending even as inspection on the site.

·         At the quit of the entire work the construction bills can be effortlessly created with the help of these schedules of bars.

How to Prepare Bar Bending Schedule

Generally bolstered cement concrete works can be calculated beneath 2 items :

1. Concrete paintings such as centering and shuttering.

2. Steel reinforcement together with its bending, cutting, laying and so forth. In Quintal or Tones.

The quantity of metallic may be very small in quantity hence no deduction is made for the metal from the quantity of concrete.

Steel reinforcement is calculate as in keeping with actual requirement which include overlap, hooks, cranks and so forth. And is determined from distinctive drawing.

Generally the share of metal reinforcement relies upon on the design of shape.

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.