Bar Bending Schedule(BBS) for Circular Column | Excel Sheet

byCivil Engineering Information -July 28, 2022

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Bar bending schedule Of Circular Column

B.B.S. for Circular R.C.C. Column

Hello there, Civil Engineers. How to Prepare B.B.S. for Circular Column is the title of this article.

To Prepare B.B.S. ready for a column. There are some criteria and assumptions.

It must be fulfilled. Let's get this started.

Forces on Column

You should be familiar with the regular forces acting on a column.

Just take a look at the diagram below.

The Column has fewer moments in the centre and more at the bottom and top, according to the shear force and bending moment diagram.

When the bending moment is greatest, never tie a column at the top or bottom; always tie the Column at the centre.

Tie the bars at the site height of 4 to 5 feet. The maximum size of the roof should be 11 feet.

Use less cutting of steel and provide desirable lapping.

Lap another steel at the same time.

Overlap the top steels with the bottom steels.

Between the laps, end to end, spacing should be provided.

Bar Bending Schedule (B.B.S.) Used by

On construction sites, the following types of people and officials use B.B.S.:

Quantity Surveyor

Steel fixer

Contractor

Project manager

Inspectors

Clerks of construction

All Estimation and Costing operations on the construction site are the responsibility of

a Quantity Surveyor The v gives the quantity surveyor the information he needs to find

Determine the number of steels, their type, size, and shape. A bar is being used by contractors.

Bending schedule (B.B.S.) for ordering steel bar to proceed with the

Construction. Steel fixers complete their work using a bar bending schedule (B.B.S.).

or installing the Bar according to the B.B.S. in the Column, beam, or slab (B.B.S.)

That how much cut length or bar bending will be required, and what size of

Steel must be used. The B.B.S. is referred to by construction inspectors and clerks.

Ensure that the reinforcement work in the field is in accordance with the design

reinforcement in accordance with the drawing

B.B.S. for Column

Details from the diagram

The thickness of the slab is 150 mm

Floor height is 3 m or 3000 mm

Footing depth is 400 mm

The development Length is 50d

The Column has 8 numbers 16 mm dia bars

Stirrup spacing is 8mm @ 150 mm C/C

The concrete Cover of the Footing is 40 mm

The concrete Cover of the slab is 25 mm

Steps for B.B.S. of Column

In general, there are five steps to preparing a column bar bending schedule, which are as follows:

Find the length of the Vertical Bar

Find out lapping

Cutting Length of Stirrups

Number of Stirrups

B.B.S.

Find the length of the Vertical Bar.

Length of Vertical bar = Development length (Ld) + footing depth + Floor Heights + Slab Thicknesses + Overlap Length

= (12×16) +400+ (3000) +(150) +(50×16)

= 4734 mm or 4.73 m

Now we find the length of one vertical Bar.

The total length of the vertical bar = 4732 x the number of steels in the Column.

The total length of the vertical bar = 4732 x 8

The total length of the vertical bar = 37856 mm

Find out lapping

length of lapping is 50d

Lapping length = 50 x Dia of steel.

Length of lapping = 50×20 = 1000 mm.

We know that the length of typical steel is 12.25 m or 40 feet.

Total Length of Vertical Bar = 37.856 m, more than 12.25 m or 40 feet, so lapping should be provided.

The length of lapping is 1000 mm the result will be 37856 + 1 = 38.856 m

It is mentioned already in this article, consider the points while lapping the Bar. There is no one right process we can use. Just remember the lines.

Don't tie a column at the bottom and also at the top

Lap at another bar every time

Overlap top bars with bottom bars always

Cutting Length of Stirrups

The cut length of the Stirrup was already discussed in the B.B.S. of beam article.

Length of One Hook = 9d

Cutting length of Stirrup = (Perimeter of Stirrup – concrete cover) + Number of Hooks

Cutting length of stirrup = (πd-cc) + (2 hooks)

Cutting length of stirrup = (3.14×450-25) +(2x9d)

= (1413-25) + (2×9×16)

= 1388 + 288

Cutting length of Stirrup = 1676 mm

Number of Stirrups

Number of stirrups = (Total length of Column ÷ spacing of stirrups) +1

Number of stirrups = 4773 ÷ 150 + 1

Number of stirrups = 32.82, say 33 numbers