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


Bar Bending Schedule


Always take – 1d


For 90 Degree Bend


Bar Bending Schedule


Always take – 2d


For 135 Degree Bend

Bar Bending Schedule


Always take – 3d


For 180 Degree Bend


Bar Bending Schedule

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


BEAM 25mm
SLAB 20mm
PILE 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*).







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


Bar Bending Schedule
Bar Bending Schedule


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



Bar Bending Schedule


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


Bar Bending Schedule



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


Bar Bending Schedule



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


Bar Bending Schedule



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




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


Cutting Length of Chair bar


Bar Bending Schedule
Bar Bending Schedule



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,


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Thanks For the Great Attention!


Good Bye & Take Care


Happy Learning 



Also, Read


Calculate Footing Bar Bending Schedule (BBS) & Steel Quantity in Footing

Bar Bending Schedule of Beam and BBS of Beam

Bar Bending Schedule of Column BBS





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