# Bar Bending Schedule – (BBS) Basic Formulas

## Bar Bending Schedule – (BBS) Basic Formulas

Bar bending Schedule is the process of cutting & Bending of reinforcement bar in to the required shape.

## Importance of Bar Bending Schedule?

a). Location & Marking of Bar.

b). Type of Bar.

c). Size of Bar.

d). Cutting Length of the bar.

e). The number of Bar.

f). Bending details of Bar

g). Total quantity.

## Advantages of Bar Bending Schedule?

1). Cutting length & bending of reinforcement can be done by Bar Bending Schedule.

2). B.B.S avoids wastage of reinforcement& thus saves the project cost.

3). It provides a better estimation of reinforcement steel requirements for each & every structure member.

4). B.B.S is very much useful during auditing of reinforcement & provides check or theft & pilferage.

5). It enables easy & fast preparation of bills of construction work for clients & contractors.

### Why Steel is used As Reinforcement Bar?

i). Coefficient of thermal expansion characteristics similar to concrete.

ii). Good strength for the economic design of R.C.C structure.

iii). Good gripping or good bonding with concrete.

iv). Good resistance against corrosion for higher durability.

v). Good bed ability for providing required shape.

Standard Length of Bar -12meter or 40 feet.

### Size & Unit Weight of Bars

 Dia. Of Bar Unit Weight of Bar (D2/162) 8mm (8×8)/162 = 0.395 kg/m. 10mm (10×10)/162 = 0.617 kg/m. 12mm (12×12)/162 = 0.889 kg/m. 16mm (16 x16)/162 = 1.580 kg/m. 20mm (20×20)/162 = 2.470 kg/m. 25mm (25×25)/162 = 3.858 kg/m. 32mm (32×32)/162 = 6.320 kg/m.

### Bend Deduction of bars & values

Bend increases the length of bars. So we need a lesser length than we see in the drawing. So, Cutting length is taken lesser than the required length.

For 45 degree Bend

Always take – 1d

For 90 Degree Bend

Always take – 2d

For 135 Degree Bend

Always take – 3d

Always take – 4d

### Clear Cover & Effective Cover

The cover is the thickness of concrete measured from the nearest outer surface of reinforcement. The cover is mainly provided to protect the reinforcement bars. In RCC structure reinforcement provided in concrete up to a specific distance from the face of the member.

Clear Cover

The clear cover is the distance between the outer surface of the concrete to the nearest surface of the reinforcing bar.

Effective Cover

The effective cover is the distance measured from the face of the member to the center of the area(centroid) of the main reinforcement. This is the dimension mostly used for design calculation.

Effective cover = Clear cover + Dia of stirrups/ links + 0.5 x Dia of main reinforcement bar.

### Values of Clear Cover

 FOOTING 50mm COLUMN 40mm BEAM 25mm SLAB 20mm STAIRCASE 15mm CHAJJA 15mm PILE 60mm PILE CAP 60mm

### Standard Codes Used For BBS

IS Code: 456-2000 Plain & Reinforced Concrete

1S Code 2502-1963 Code of Practice for bending & fixing.

IS: 5525-1969 Detailing of reinforcement in RCC work.

SP 34-1987 Concrete Reinforcement Detailing.]

### Symbols & Representation in BBS

 Symbols Representation ϴ Plain Round Bar □ Plain Square Bar # Deformed Bar @ Center to center spacing

### Cutting Length of Bars

 Shape of Bar Cutting Length of Bar CL – a + L + b – bend deduction Bend deduction – 2 x 2d (90*). CL – a + L – bend deduction Bend deduction – 1 x 2d (90*). CL – L + (2 x Hook Length)  Hook Length = 10D

### Cutting Length of Crank Bar

Crank length – 0.42d.

Cutting Length – Ld + Span Length (L) + { 0.42H x 2} + Ld – Bend Deduction (2 x 90* + 4 x 45*).

### Hook

Length of hook = 9d or 10d, (where d is the diameter of the bar)

Length of bar = L + 9d + 9d meter

### Cutting Length Of Rectangular Stirrups

Cutting Length Of Rectangular Stirrups – (a x 2) + (b x 2) + Hook Length

Where, a  =  A – (2 x clear cover) – d

a =  B – (2 x clear cover) – d

Hook Length = 2 x 10D

### Cutting Length Of Triangular Stirrups

Cutting Length Of triangular Stirrups – (H x 2) + a + Hook Length

Where, H  =  (x2 + y2)⅟2  Where,  X = a/2 & Y = b

a =  A – (2 x clear cover) – d

b =  B – (2 x clear cover) – d

### Cutting Length Of Circular Stirrups

Cutting Length of Circular Stirrups = (Pi x d) + Hook length

Where d = D – (2 x clear cover) – Dia of the bar.

### Cutting Length of Diamond Stirrups

Cutting length of diamond stirrups = (H x 4) + Hook Length

Where, H = (x2 + y2)

Where, x = a/2 & y = b/2.

### Cutting Length of Spiral or Helix Stirrup

Length of Helix/Spiral Bar = n√ c2 + p2

So, first, we should find out the value of ‘n’ & ‘c’.

No. of Pitches (n) = H/p+1

Where, Length of pile cage  = H

Pitch or spacing of spiral = P

Circumstances (C) = x d

Where,

Dia of spiral Ring (d) = D – (2 x cc) – Dia of the bar.

### Cutting Length of Chair bar

Cutting length of chair bar = 2A + 2D + B

Where, A = (2 x spacing) + 50

B = 50d

D = Footing height – [{ upper + lower cc} + ( Upper side main + dist bar dia + lower side main bar dia)]

I hope this information can be useful for you guys,

Thanks For the Great Attention!

Good Bye & Take Care

Happy Learning

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