BBS : Bar Bending Schedule for Beam | BBS for Beam | LCETED | PART - 2 -lceted LCETED INSTITUTE FOR CIVIL ENGINEERS

## Jun 13, 2021

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

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

Lets start,

CONSIDER GIVEN DATA AS

Bar Bending Schedule for Beam

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 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 more detailed and technical implementation of design aspects. So click to read this article

GIVEN DATA

Clear Span of Beam = 5500 mm

Development Length Ld = 41d (assumption)

Clear Cover on any ends = 25 mm

Bottom – 2 numbers of 16 (DIA) - 2 numbers of 20 (DIA)

Top – 2 numbers of 12 (DIA)

Stirrups zone A,C = 8 (DIA) @ 150mm clear cover zone B = 8 (DIA) @ 200mm clear cover

CUTTING LENGTH OF TOP BAR

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

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 = 5.5 + (41d x 2) – (2 x 25) = 3.5 + (41(12) x 2) – (2 x 25)

= 5500+(2 x 41d) – 2 x 25 = 5500 + (41(12) X2) – 50

Length of 1 Rod Cutting length of top bar = 6434

The length of one Rod Cutting length of top bar is 6.434. we have a total bar three bars

= 3 x 6.434 = 19.30m

The total length of top bar = 19.30m bar – 2bar of 12 mm needed

FIND CUTTING LENGTH OF BOTTOM BAR

2 numbers of 16mm dia

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

= 5500+(2 x 41d) – 2 x 25 = 5500 + (41(16) X2) – 50

Length of 1 Rod Cutting length of top bar = 6762

2 numbers of 20mm dia

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

= 5500+(2 x 41d) – 2 x 25 = 5500 + (41(20) X2) – 50

Length of 1 Rod Cutting length of top bar = 7090

Total length of the bottom bar

2 numbers of 16mm dia = 2 x 6.762 = 13.52mm

2 numbers of 20mm dia = 2 x 7.090 = 14.18mm

CUTTING THE LENGTH OF STIRRUPS

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

Take it has,

500mm is side A

300mm is side B

Minimum clear cover to the reinforcement15 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.

 SIDE A SIDE B A =500 – 2 Side clear cover A = 500  – 2 x clear cover A = 500  – 2 x 40 A = 500  – 80 A = 420 mm B =300 –  2 x Top & Bottom cover B = 300 – 2 x clear cover B = 300 – 2 x 40 B = 300 – 80 B = 220 mm

NO OF STIRRUPS REQUIRED

As mentioned Stirrups spacing 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

Formula = L/3 +1

NO OF STIRRUPS NEEDED IN ZONE A & C

=1800 / 150 + 1

= 13 nos

NO OF STIRRUPS NEEDED IN ZONE B

=1900 / 200 + 1

= 10.5 nos say 11 nos

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(500+300) + 2(9x8) – 3(2x8) – 2(3x8)

= 2(800) + 2(72) – 3(16) – 2(24)

= 1600 + 144 – 48 – 48

1648mm say as 1.64m

We have a total of (11 +13) 24 nos of stirrups, which are going to use,

Total length

= 25 x 1.64 = 39.36m

CUTTING THE LENGTH OF RECTANGULAR STIRRUPS = 40 m

RESULT:

 Diameter of Bar Numbers Cutting Length Total Length No of bar Top Bar 12mm 3 6.434m 19.30 m 2 Bottom Bar 16mm 2 6.762 m 13.52 m 2 Bottom Bar 25mm 2 7.090 m 14.18 m 2 Stirrups 8mm 24 1.44 m 40 m 1

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