BBS : Bar Bending Schedule for Beam | BBS for Beam | LCETED | PART - 1

 In this article, we are going to look at how to make a bar bending schedule for beam or BBS for Beam.

 

Some helpful posts related to this article,

1. How Many Numbers Of Rods Per Bundle? How To Calculate It

2. How To Calculate Cutting Length For Square Stirrups

 

Now we are going to make BBS for a beam of two example data.

 

Let's start,

 

CONSIDER GIVEN DATA AS

Bar Bending Schedule for Beam

EXAMPLE -1

At the bottom of the figure, the beam has a clear span of 3.5 m, which has 2 numbers and 12 mm dia bars on top and bottom with 8 mm dia stirrups on the 150 mm and clear cover with 25 mm clear cover on both ends and sides of the beam

 

GIVEN DATA

Clear Span of Beam = 3500 mm

Development Length Ld = 41d (assumption)

Clear Cover on any ends = 25 m

Bottom – 2 numbers of 12⌀ (DIA)

Top – 2 numbers of 12⌀ (DIA)

Stirrups spacing – 8⌀ (DIA) @ 150mm clear cover

 

CUTTING LENGTH OF TOP BAR

 

Length of 1 Rod Cutting length of top bar = l + (Ld x 2) – (2 x c.c) Clear Cover on 2 ends

l = Clear Span of Beam

Ld = Development length (Anchorage) Ld on 2 sides (41d)

d = dia of bar

c.c = clear cover (2 ends)

 

Length of 1 Rod Cutting length of top bar = 3.5 + (41d x 2) – (2 x 25) = 3.5 + (41(12) x 2) – (2 x 25)

 

= 3500+(2 x 41d) – 2 x 25 = 3500 + (2 x 41 x 12) – 50

 

Length of 1 Rod Cutting length of top bar = 4434

Length of one Rod Cutting length of top bar is 4.43. we have total bar two bars

= 2 x 4.43 = 8.86m

Total length of top bar = 8.86m bar  – 1bar  of 12 mm needed

 

 

FIND CUTTING LENGTH OF BOTTOM BAR

 

Length of 1 Rod Cutting length of bottom bar = l + (Ld x 2) – (2 x c.c) Clear Cover on 2 ends

 

Follow above procedure

 

Total length of bottom bar = 8.86m bar – 1bar of 12 mm needed

 

 

CUTTING THE LENGTH OF STIRRUPS

 

The cross-sectional area of the beam is 300 mm x 400 mm

Take it has,

300mm is side A

400mm is side B

Minimum clear cover to the reinforcement: 15 mm to the bars in slabs, 25 mm to bars in beams and columns. In large columns, say 450 mm in thickness, the cover should be 40 mm.

 

Also Read: know about cover blocks used in construction

 

SIDE A	SIDE B
A =300 – 2 Side clear cover   A = 300  – 2 x clear cover   A = 300  – 2 x 40   A = 300  – 80   A = 220 mm	B =400 –  2 x Top & Bottom cover   B = 400 – 2 x clear cover   B = 400 – 2 x 40   B = 400 – 80   B = 320 mm

 

 

 

NO OF STIRRUPS REQUIRED

As mentioned Stirrups spacing – 8⌀ (DIA) @ 150mm clear cover

Formula = L/150 +1

L/150 + 1 = 3500/150 + 1 = 23.33 + 1 = 24.33 = 25nos

 

Also Read: How to calculate the quantity of steel in column (BBS)

CUTTING LENGTH OF ONE STIRRUP

 

Formula:

 

Cutting Length of Stirrups = Perimeter of Shape + Total hook length – Total Bend Length

 

= 2(a+b) + 2(9d) – 3(2d) – 2 (3d)

= 2(300+400) + 2(9x8) – 3(2x8) – 2(3x8)

= 2(700) + 2(72) – 3(16) – 2(24)

= 1400 + 144 – 48 – 48

= 1448mm say as 1.44m

 

 

We have a total of 25 nos of stirrups, which are going to use,

 

Total length

= 25 x 1.84

CUTTING THE LENGTH OF RECTANGULAR STIRRUPS = 46 m

 

RESULT:

 

	Diameter of Bar	Numbers	Cutting Length	Total Length	No of bar
Top Bar	12mm	2	4.3m	8.6 m	1
Bottom Bar	12mm	2	4.3m	9.3 m	1
Stirrups	8mm	25	1.44m	46m	2

 

EXAMPLE – 2

At the bottom of the figure, the beam has a clear span of 5.5 m, which has 3 numbers and 12 mm dia bars on top and the bottom consists of two layers of the bottom layer (2 numbers of 16mm dia) and one layer of the top (2 numbers of 20mm dia). with 8 mm dia stirrups on 3 Zones, where Zone A, C has stirrups of 150 mm spacing & Zone B, has stirrups of 200 mm spacing. and clear cover with 25 mm clear cover on both ends and sides of the beam

 

As you can see that this data has a more detailed and technical implementation of design aspects. So click to read this article

 

ALSO READ: BBS: Bar Bending Schedule for Beam | BBS for Beam | LCETED | PART - 2

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