Introduction
1.
Bar
Bending Schedule (BBS) is commonly used in RCC construction such as footings,
foundations, columns, beams, and slabs.
Before
preparing a BBS, some important factors must be considered, including cutting
length of bars, lap length, development length, crank length, and hook length.
In
this article, we will explain the complete process of preparing a Bar Bending
Schedule in a very simple way.
BBS helps engineers
and contractors estimate steel quantity accurately and reduce reinforcement
wastage at the construction site.
BBS format in excel free download
What is Bar Bending Schedule (BBS)
Bar Bending Schedule (BBS) is used to calculate the cutting length and
shape of steel bars used in RCC work. Steel bars are bent into shapes like
straight, L-shape, U-shape, square, rectangular, or crank bars as per the
structural drawing. BBS helps to know the exact steel quantity and weight,
makes bar placing easier at site, and ensures accurate billing. Since steel is
about 20% of the total project cost, preparing a proper BBS is very important.
BBS format in excel free download
Important For Bar Bending Schedule
Important Indian Standard codes used in BBS preparation
Adding these references improves engineering accuracy and credibility.
IS Code | Description |
IS 2502 | Code of practice for bending and fixing of bars |
IS 456:2000 | Plain and reinforced concrete code |
SP 34 | Handbook on reinforcement detailing |
Adding these references improves engineering
accuracy and credibility.
BBS drawing in Pdf free download
Steel weight Calculation Formula
1.
Steel Weight Per Meter in KG
8 mm to 25 mm bar ka cross-section area
Dia of Bar
Weight (kg/m)
Cross Section Area of Bar
roundoff
8 mm
0.395 Kg
50.27 mm²
10 mm
0.617 Kg
78.54 mm²
12 mm
0.888 Kg
113.10 mm²
16 mm
1.58 Kg
201.06 mm²
20 mm
2.47 Kg
314.16 mm²
25 mm
3.858 Kg
490.87 mm²
Example cross section Area of Bar (8 mm bar):
Area = π × R²
= 3.14× 4 x 4 = 50.27 mm²
8 mm to 25 mm bar ka cross-section area
Dia of Bar
Weight (kg/m)
Cross Section Area of Bar
roundoff
8 mm
0.395 Kg
50.27 mm²
10 mm
0.617 Kg
78.54 mm²
12 mm
0.888 Kg
113.10 mm²
16 mm
1.58 Kg
201.06 mm²
20 mm
2.47 Kg
314.16 mm²
25 mm
3.858 Kg
490.87 mm²
Area = π × R²
= 3.14× 4 x 4 = 50.27 mm²
Bending and Cutting Tolerances
SP 34 Clause 13.3.3
Where an overall or an internal dimension of the bent bar is specified, the
tolerance, unless otherwise stated, shall be as follows:
|
Dimension (MM) |
Tolerance (mm) |
|
≤ 750 MM |
+3 / −5 |
|
> 750 ≤ 1500 MM |
+5 / −10 |
|
> 1500 ≤ 2500 MM |
+6 / −15 |
|
> 2500 MM |
+7 / −25 |
Tolerance for Straight Bars
|
Bar Type |
Tolerance (mm) |
|
All lengths |
+25 / −25 |
Cutting Length Calculation
as per is 2502 clause 3.1 page 12
|
Ref
No. |
Bar
Shape / Type |
Approx
Total Length Formula (L) |
Example
For 500 X 230 MM Beam 10mm Stirrups |
|
A |
|
L =
2(A + E) + 24d |
If |
|
B |
|
L =
2(A + E) + 20d |
If |
Cutting Length Calculation
Most Commonly use at site
|
Ref
No. |
Bar
Shape / Type |
Approx Total Length Formula (L) |
Example
For 500 X 230 MM Beam 10mm Stirrups |
|
A |
Stirrup
with 135° hooks |
L = 2(A + E) + 20d - 12D |
If |
|
B |
Stirrup
with 90° hooks |
L = 2(A + E) + 20d -10d |
If |
Cover block
Cover block and
chair shall be provided to keep the bar in position.
Cover blocks shall be of the same grade of concrete as that of concrete member
in which they are placed.
Clear cover to all reinforcement shall be as follows:
Slabs = 20 mm or dia of bar whichever is greater
Beams
– Stirrups/Link = 25 mm
R.C.C.
Walls (General)
i) Vertical bar (inner face) = 25 mm
ii) Vertical bar (earth face) = 40 mm
Footings
= 50 mm
Raft
i) Top = 50 mm
ii) Bottom = 60 mm
Columns
/ Boundary Elements:
Face
up to 230 mm → 40 mm to main bar
Face
more than 230 mm → 40 mm to ties/links
Water
Tank / STP Walls
i) Water face / earth face = 40 mm
ii) Dry face = 25 mm
Stitch
Slab
i) Top = 25 mm or dia of bar whichever is greater
ii) Bottom = 40 mm
OVER LAPS
No.
of laps shall be kept to a minimum and all lapping, juggling, bend etc. shall
conform to IS:456 and IS:2502.
Do
not lap more than 50% of the bar at any one section.
Embedment
length of beam bars within end column shall be Ld + 10 × dia of reinforcement
a)
For Foundation Beam:-
Avoid
laps in top bars near mid span.
Avoid
laps in bottom bars near supports (columns).
b)
For Super Structure Beams
Avoid
laps in bottom bars near mid span.
Avoid
laps in top bars near supports.
COLUMNS:-
Splicing
shall be done as per figure.
Bending
of bars shall be carried as per IS–2502.
REINFORCEMENT:
All
reinforcement used in this work shall conform to IS:1786 having Fe500 grade and
will be in the form of deformed bars, TMT bars. This shall be indicated in the
reinforcement details as “T”.
Reinforcing
bars shall conform accurately to dimensions shown on relevant drawings and
IS:2502.
Chairs
to support the top layer of reinforcement of slabs, raft, etc. shall be placed
at not more than the maximum horizontal distance between the legs of the chair.
The diameter of the bars used in preparing chairs shall be minimum 12 mm.
Dowel
length, anchorage embedment, lap/splice length shall as per Table–2. In case of
lap/splice between bars of different diameters, the smaller diameter will be
considered for the calculation of the lap/splice length.
Reinforcing bars
shall be tested whenever a new lot of steel arrives at site. Such tests shall
be conducted for each diameter of the reinforcing bars in accordance with
IS:1786, in designated laboratories
Conclusion
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