# BBS for Square Floor Columns - Bar Bending Schedule Hello people, today, we will examine the bar bending schedule (BBS) for floor columns. The aspect of the column which extended towards the sky on the superstructure is called Floor columns. Furthermore, the aspect of the column which is within base is called Neck column. Discovering the steel amount required for the neck column is as of now talked about in our past article.

In Bar bending schedule, the bars are sorted out for each auxiliary units (Beams or columns or sections or footings and so forth) and itemized list is readied which determines the Bar area (Bar in footings, pieces, pillars or columns), Bar Marking (to recognize the bar as per the drawing), Bar Size (length of the bar utilized), Quantity (No. of Bars utilized), Cutting length, Type of Bend and Shape of the bar in fortification drawings.

Genuine measurements and states of the floor columns are chosen and planned by the basic specialist dependent on soil history, sort of development, the complete anticipated heap of the structure. All the elements of the above columns considered uniquely for clarification reason.

Steel required for development is requested in Kgs or Number of Bars. The standard size of each bar is 12m. Computing the Bar Bending schedule of the column is separated into two sections Main fortification estimation and ties count. Ties are likewise called as Longitudinal support and Main bars are called as Main fortification.

Bar bending schedule of a floor column longitudinal reinforcement:

1. Discover the length of the single fundamental bar
2. Locate the total length of the Main support
3. Ascertain the weight of steel required per 1m of Main support
4. Discover the total weight of steel required for Main support.

Bar bending schedule of a floor column Longitudinal fortification:

1. Deduct the solid spread from all sides of ties and discover the length of stirrup utilizing formulae.
2. Acquire the complete number of ties required utilizing equation
3. Discover the total length of ties required
4. Compute the weight of steel required per 1m of longitudinal fortification.
5. Discover the absolute weight of steel required for longitudinal fortification

Significant principles utilized in Bends and Hooks

The beneath norms are generally significant in figuring the snare length and curve lengths at corners while finding the slicing length to gauge the bar bending schedule of a floor column.

a. 1 Hook length = 9d or 75mm
b. 45° Bend length = 1d
c. 90° Bend length = 2d
d. 135° Bend length = 3d

where d = Diameter of Bar

Bar Bending schedule of Square floor column

For the estimation of the total amount of steel required for the square column, we are embracing these measurements for bars.

a. Dia of Main bars = 16mm =0.016m
b. Dia of Longitudinal bars (ties)= 8mm = 0.008m
c. Dispersing between longitudinal bars = 100mm = 0.1m

BBS of Main fortification in Square columns

Length of the Main bar runs corresponding with the height of the floor. Length of the principle bar is an option of height of the column, height of the top section and cover length which is added to the top finish of the column for the following floor reason.

Length of every Main Bar = Height of the floor + Height of the slab + Overlap Length
= Height of the floor + Height of the slab + Overlap Length = 3.0m + 0.15 + 50 × 0.016 = 3.95m

No. of Main bars = 4
Total Length of Main bars = 4 × 3.95m =15.8m
Total No. of ’12m’ bars = 15.8 / 12 = 1.37 bars
Weight of steel required for 1m of 16mm bar = D2 / 162 = 162 / 162 = 1.58 kg/m
Total weight of steel required for main bars = Total length of main bars x 1.58 = 15.8 x 1.58 = 24.96 Kgs

BBS of Longitudinal reinforcement in Square column

Concrete Cover Deduction: As per concrete cover condition considering the below square column, deduct the 0.05m from all sides of the column to find the dimensions of ties.

Finding the total length of Ties:

Total Length of each tie = Perimeter of tie +Total Hook length – Total Bend length
No. of ties = (Floor height / Spacing) + 1

Observations:

a. No. of Hooks = 2;
b. No. of Bends bent @ 900 = 3;
c. No. of Bends bent @1350 = 2;
d. Total Hook length = 9d
e. No. of Hooks = 2

Total hook length = 9d + 9d = 18 x 0.008 = 0.144m
Total Bend length (bent @135)
No. of bends bent @1350 = 2;
Bend length = 3d
= No. of bends x 3d = 2 x 3 x 0.008 = 0.048m

Total Bend length
bent @900
No. of bends bent @900 = 3;
Bend length = 2d
So, No. of bends x 3d = 3 x 2 x 0.008 = 0.048m

Total bend length of each tie = 0.048 + 0.048 =0.096m
Total Length of each tie = Perimeter of tie + Total Hook length – Total Bend length
= 0.3+ 0.3 + 0.3 + 0.3 + 0.144 - 0.096 = 1.248m

No. of ties = (3.0/0.1) + 1 = 31 ties
Total length of ties = 31 x 1.24 = 38.68m
Total No. of ’12m’ bars =38.68/12 = 3.22bars
Weight of steel required for 1m of 8mm bar = D2/162 = 82/162 = 0.39 kg/m
Total weight of steel required for ties =Total length of ties x 0.39 =38.68 x 0.39 =15.08kgs