Bar bending schedule for floor columns. The part of the column which projected towards the sky on the superstructure is called Floor columns. And the part of the column which is inside of substructure is called Neck column. Finding out the steel quantity required for the neck column is already discussed in our previous article.
Bar bending schedule of Neck column
If you are new to Bar Bending Schedule, Refer the Basics of Bar Bending schedule
To make you clear in finding out the required steel quantities for the construction of the column, I am considering the below plan of different shapes of columns with reinforcement details.
Observations from above fig:
C1 is a Square Column
C2 is a T type column
C3 is a Circular column
C4 is a Rectangular column
C5 is a Y type Column
C6 is an L Type Column
True dimensions and shapes of the floor columns are decided and designed by the structural engineer based on soil history, type of construction, the total expected load of the structure. All the dimensions of the above columns considered only for explanation purpose.
Steps involved in calculating the bar bending schedule of a floor column:-
Steel required for construction is ordered in Kgs or Number of Bars. The standard size of each bar is 12m.
Calculating the Bar Bending schedule of the column is divided into two parts Main reinforcement calculation and ties calculation.
Ties are also called as Longitudinal reinforcement and Main bars are called as Main reinforcement.
Bar bending schedule of a floor column Main Reinforcement:
- Find the length of the single main bar
- Find the total length of the Main reinforcement
- Calculate the weight of steel required per 1m of Main reinforcement
- Find out the total weight of steel required for Main reinforcement
Look at the above image for a pictorial representation of Main reinforcement in a column. Observe the concrete cover and check here how to deduct concrete cover to find the length of each tie.
Look at the longitudinal reinforcement and main bars in the above image to understand finding the number of longitudinal bars.
Bar bending schedule of a floor column Longitudinal reinforcement:
- Deduct the concrete cover from all sides of ties and find out the length of stirrup using formulae.
- Obtain the total number of ties required using formula
- Find out the total length of ties required
- Calculate the weight of steel required per 1m of longitudinal reinforcement.
- Find out the total weight of steel required for longitudinal reinforcement
Important standards used in Bends & Hooks:
The below standards are most important in calculating the hook length and bend lengths at corners while finding the cutting length to estimate the bar bending schedule of a floor column.
- 1 Hook length = 9d or 75mm
- 45° Bend length = 1d
- 90° Bend length = 2d
- 135° Bend length = 3d
Remember, d = Diameter of Bar
Bar Bending schedule of Square floor column (C1):-
For the calculation of the total quantity of steel required for the square column, we are adopting these dimensions for bars.
Adopted:-
Dia of Main bars = 16mm =0.016m
Dia of Longitudinal bars (ties)= 8mm = 0.008m
Spacing between longitudinal bars = 100mm = 0.1m
BBS of Main reinforcement in Square column:-
Length of the Main bar runs parallel with the height of the floor. Length of the main bar is an addition of height of the column, height of the top slab and overlap length which is added to the top end of the column for the next floor purpose.
Length of each Main Bar = Height of the floor+Height of the slab+Overlap Length
Below bar dimensions are assumed for calculation purposes. Remember, the size of every bar is decided and adopted by considering the expected load of a building.
Length of Each Main Bar | =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 =1.37bars |
Weight of steel required for 1m of 16mm bar | =D^{2}/162 =16^{2}/162 =1.58kg/m |
Total weight of steel required for main bars | =Total length of main bars x 1.58 =15.8 x 1.58 =24.96Kgs |
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:
Formulas:
Total Length of each tie = Perimeter of tie+Total Hook length – Total Bend length
No. of ties = (Floor height/Spacing) + 1
Observations From figure:
No. of Hooks = 2;
No. of Bends bent @90^{0 } =3;
No. of Bends bent @135^{0 } =2;
Total Hook length | Hook Length = 9d No. of Hooks =2 Total hook length = 9d+9d = 18 x 0.008 =0.144m |
Total Bend length bent @135^{0} | No. of bends bent @135^{0} = 2; Bend length = 3d = No. of bends x 3d = 2 x 3 x 0.008 =0.048m |
Total Bend length bent @90^{0} | No. of bends bent @90^{0} = 3; Bend length = 2d = 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 =31ties |
Total length of ties | =31 x 1.248 =38.68m |
Total No. of ’12m’ bars | =38.68/12 =3.22bars |
Weight of steel required for 1m of 8mm bar | =D^{2}/162 =8^{2}/162 =0.39kg/m |
Total Wt of steel required for ties | =Total length of ties x 0.39 =38.68 x 0.39 =15.08kgs |
Bar Bending schedule of Rectangular floor column (C4):-
BBS of Main reinforcement in Rectangular column:-
Length of Main bar runs parallel with the height of floor. Length of main bar is an addition of height of the column, height of the top slab and overlap length which is added to the end of column for the next floor purpose.
Length of each Main Bar = Height of the floor+Height of the slab+Overlap Length
Overlap length = 50d (Diameter of Bar)
Why 50d? Refer here
Adopted:-
Dia of Main bars = 18mm =0.018m
Dia of Longitudinal bars (ties)= 8mm = 0.008m
Spacing between longitudinal bars = 100mm = 0.1m
Length of Each Main Bar | =Height of the floor + Height of the slab + Overlap Length =3.0m+0.15+50×0.018 =4.05m |
No. of Main bars | =4 |
Total Length of Main bars | =4 x 4.05m =16.2m |
Total No. of bars | =16.2/12 =1.35bars |
Weight of steel required for 1m of 16mm bar | =D^{2}/162 =18^{2}/162 =2kg/m |
Total weight of steel required for main bars | =Total length of main bars x 1.58 =16.2 x 2 =32.40Kgs |
BBS of Longitudinal reinforcement in Rectangular column:
Concrete Cover Deduction:
As per concrete cover condition considering the below rectangular column, deduct the 0.02m from all sides of the column to find the dimensions of ties.
Finding the total length of Ties:
Formulas:
Total Length of each tie = Perimeter of tie+Total Hook length – Total Bend length
No. of ties = (Floor height/Spacing) + 1
Observations From figure:
No. of Hooks = 2;
No. of Bends bent @90^{0 } =3;
No. of Bends bent @135^{0 } =2;
Total Hook length | Hook Length = 9d No. of Hooks =2 Total hook length = 9d+9d = 18 x 0.008 =0.144m |
Total Bend length bent @135^{0} | No. of bends bent @135^{0} = 2; Bend length = 3d = No. of bends x 3d = 2 x 3 x 0.008 =0.048m |
Total Bend length bent @90^{0} | No. of bends bent @90^{0} = 3; Bend length = 2d = 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 = 0.41+0.19+0.41+0.19 +0.144-0.096 = 1.248m |
No. of ties | =(3.0/0.1) + 1 =31ties |
Total length of ties | =31 x 1.248 =38.68m |
Total No. of Bars required | =38.68/12 =3.22bars |
Weight of steel required for 1m of 8mm bar | =D^{2}/162 =8^{2}/162 =0.39kg/m |
Total Wt of steel required for ties | =Total length of ties x 0.39 =38.68 x 0.39 =15.08kgs |
Bar Bending schedule of Circular floor Column (C3):-
BBS of Main reinforcement in Circular column:-
Length of Main bar runs parallel with the height of floor. Length of main bar is an addition of height of the column, height of the top slab and overlap length which is added to the end of column for the next floor purpose.
Length of each Main Bar = Height of the floor + Height of the slab + Overlap Length
Overlap length = 50d (Diameter of Bar)
Why 50d? Refer here
Adopted:-
Dia of Main bars = 20mm =0.020m
Dia of Longitudinal bars (ties)= 8mm = 0.008m
Spacing between longitudinal bars = 100mm = 0.1m
Length of Each Main Bar | =Height of the floor + Height of the slab + Overlap Length =3.0m+0.15+50×0.020 =4.15m |
No. of Main bars | =8 |
Total Length of Main bars | =8x 4.15m =33.2m |
Total No. of bars | =33.2/12 =2.67bars |
Weight of steel required for 1m of 16mm bar | =D^{2}/162 =20^{2}/162 =2.47kg/m |
Total weight of steel required for main bars | =Total length of main bars x 2.47 =33.2 x 2.47 =82Kgs |
BBS of Longitudinal reinforcement in Circular column:
Concrete Cover Deduction:
As per concrete cover condition considering the below circular column, deduct the 0.025m from all sides of the column to find the dimensions of ties. Therefore dia of the circular stirrup is 950mm.
Finding the total length of Ties:
Formulas:
Total Length of each tie = Perimeter of tie+Total Hook length – Total Bend length
The perimeter of Circle or we usually called as a Circumference of a circle is = 2πR or πD
No. of ties = (Floor height/Spacing) + 1
Observations From figure:
Dia of a circular tie (D) = 950mm
No. of Hooks = 2;
No. of Bends bent @135^{0 } =2;
No. of bends bent @90^{0} = 6;
Total Hook length | Hook Length = 9d No. of Hooks =2 Total hook length = 9d+9d = 18 x 0.008 =0.144m |
Total Bend length bent @135^{0} | No. of bends bent @135^{0} = 2; Bend length = 3d = No. of bends x 3d = 2 x 3 x 0.008 =0.048m |
Total Bend length bent @90^{0} | No. of bends bent @90^{0} = 6; Bend length = 2d = No. of bends x 3d = 6 x 3 x 0.008 =0.048m |
Total Length of each tie | = Perimeter of tie+ Total Hook length – Total Bend length =πD+0.144m – 0.048 =3.14 x 0.95+0.144m – 0.048 =3.079m |
No. of ties | =(3.0/0.1)+1 =31ties |
Total length of ties | =31×3.079 =95.45m |
Total No. of Bars required | =95.45/12 =7.95bars |
Weight of steel required for 1m of 8mm bar | =D^{2}/162 =8^{2}/162 =0.39kg/m |
Total Wt of steel required for ties | =Total length of ties x 0.22 =95.45 x 0.39 =37.22kgs |
Bar Bending Schedule of T Shape Floor Column (C2):
BBS of Main reinforcement in T Shape column:-
Length of Main bar runs parallel with the height of floor. Length of main bar is an addition of height of the column, height of the top slab and overlap length which is added to the end of column for the next floor purpose.
Length of each Main Bar = Height of the floor+Height of the slab+Overlap Length
Overlap length = 50d (Diameter of Bar)
Why 50d? Refer here
Adopted:-
Dia of Main bars = 20mm =0.020m
Dia of Longitudinal bars (ties)= 8mm = 0.008m
Spacing between longitudinal bars = 100mm = 0.1m
Length of Each Main Bar | =Height of the floor + Height of the slab + Overlap Length =3+0.15+50×0.20 =4.15 |
No. of Main bars | =10 |
Total Length of Main bars | =10×4.15m =41.5m |
Total No. of bars | =41.5/12 =3.46bars |
Weight of steel required for 1m of 16mm bar | =D^{2}/162 =20^{2}/162 =2.47kg/m |
Total weight of steel required for main bars | =Total length of main bars x 2.47 =41.5 x 2.47 =102.5Kgs |
BBS of Longitudinal reinforcement in T shape column:
Concrete Cover Deduction:
As per concrete cover condition considering the below T Type column, deduct the 0.25m from all sides of the column to find the dimensions of ties.
Finding the total length of Ties:
Formulas:
Total Length of each tie = Perimeter of tie + Total Hook length – Total Bend length
No. of ties = (Floor height/Spacing) + 1
Observations From figure:
No. of Hooks = 4;
No. of Bends bent @90^{0 } =6;
No. of Bends bent @135^{0 } =4;
Total Hook length | Hook Length = 9d No. of Hooks =4 Total hook length = 4x9d = 36 x 0.008 =0.288m |
Total Bend length bent @135^{0} | No. of bends bent @135^{0} = 4; Bend length = 3d = No. of bends x 3d = 4 x 3 x 0.008 =0.096m |
Total Bend length bent @90^{0} | No. of bends bent @90^{0} = 6; Bend length = 3d = No. of bends x 3d = 6 x 3 x 0.008 =0.144m |
Total Length of each tie | = Perimeter of tie+ Total Hook length – Total Bend length =0.15+0.45+0.15+0.45+ +0.55+0.15+0.55+ 0.15+ 0.288-0.096-0.144 =2.64m |
No. of ties | =(3.0/0.1)+1 =31ties |
Total length of ties | =31×2.64 =81.84m |
Total No. of Bars required | =81.84/12 =6.82bars |
Weight of steel required for 1m of 8mm bar | =D^{2}/162 =8^{2}/162 =0.39kg/m |
Total Wt of steel required for ties | =Total length of ties x 0.22 =81.84 x 0.39 =31.91Kgs |
Bar Bending Schedule of L Shape floor column(C6):
BBS of Main reinforcement in L Shape column:-
Length of Main bar runs parallel with the height of floor. Length of main bar is an addition of height of the column, height of the top slab and overlap length which is added to the end of column for the next floor purpose.
Length of each Main Bar = Height of the floor+Height of the slab+Overlap Length
Overlap length = 50d (Diameter of Bar)
Why 50d? Refer here
Adopted:-
Dia of Main bars = 20mm =0.020m
Dia of Longitudinal bars (ties)= 8mm = 0.008m
Spacing between longitudinal bars = 100mm = 0.1m
Length of Each Main Bar | =Height of the floor + Height of the slab + Overlap Length =3+0.15+50×0.20 =4.15 |
No. of Main bars | =8 |
Total Length of Main bars | =8×4.15m =33.20m |
Total No. of bars | =33.2/12 =2.77bars |
Weight of steel required for 1m of 16mm bar | =D^{2}/162 =20^{2}/162 =2.47kg/m |
Total weight of steel required for main bars | =Total length of main bars x 2.47 =33.2 x 2.47 =82.00Kgs |
BBS of Longitudinal reinforcement in L shape column:
Concrete Cover Deduction:
As per concrete cover condition considering the below L Type column, deduct the 0.25m from all sides of the column to find the dimensions of ties.
Finding the total length of Ties:
Formulas:
Total Length of each tie = Length of tie+Total Hook length – Total bend Length
No. of Hooks = (Floor height/Spacing)+1
Each tie has two hooks (both should be added in the length of a tie)
Hook length = 9D
Why 9D? refer here
Total Hook length | Hook Length = 9d No. of Hooks =4 Total hook length = 4x9d = 36 x 0.008 =0.288m |
Total Bend length bent @135^{0} | No. of bends bent @135^{0} = 4; Bend length = 3d = No. of bends x 3d = 4 x 3 x 0.008 =0.096m |
Total Bend length bent @90^{0} | No. of bends bent @90^{0} = 4; Bend length = 3d = No. of bends x 3d = 4 x 3 x 0.008 =0.096m |
Total Length of each tie | = Perimeter of tie+ Total Hook length – Total Bend length =[0.15+0.55+0.15+0.55] +[0.15+0.55+0.15+0.55] + 0.096 -0.096 =2.8m |
No. of ties | =(3.0/0.1)+1 =31ties |
The total length of ties | =31×2.8 =86.8m |
Total No. of Bars required | =86.8/12 =7.23bars |
Weight of steel required for 1m of 8mm bar | =D^{2}/162 =8^{2}/162 =0.39kg/m |
Total Wt of steel required for ties | =Total length of ties x 0.22 =86.8 x 0.39 =33.85Kgs |
Bar Bending Schedule of Y Shape Floor Column (C5):
Y type column is generally provided outside of the building for transformer purpose and it has no extension to construct further columns. So, hook length is not added to the top of the main bars.
Overlap length = 50d (Diameter of Bar)
Why 50d? Refer here
Adopted:-
Dia of Main bars = 12mm =0.012m
Dia of Longitudinal bars (ties)= 8mm = 0.008m
Spacing between longitudinal bars = 100mm = 0.1m
BBS of Main reinforcement in Y Shape column:-
Length of each Main Bar = Adopted Height of the column
To find the length of the main bar required for Y type column, the column is divided into two parts
Inclined bar and straight bar. To keep it clear, we calculate each bar separately.
As there is no further construction on this column, the concrete cover is deducted on all the sides of the column.
After deduction of the concrete cover from all the sides, below are the dimensions of main reinforcement in Y Shape Column
As per Pythagorean theorem,
Hypotenuse^{2}= Opposite^{2} + Adjacenet^{2}
Hypotenuse^{2}= 0.575^{2} + 0.35^{2}
Hypotenuse = 0.673m
From fig. No. of Bars | Inclined Bars = 6 Straight bars = 2 |
Length of each straight bar | =1.90+0.575 =2.475m |
Length of each inclined bar | =1.90+0.673 =2.573m |
The total length of the Inclined bar | =No. of Inclined bars x Length of each inclined bar = 6 x 2.475m = 14.85m |
The total length of the Straight bar | =No. of Straight bars x Length of each straight bar = 2 x 2.573 =5.146m |
Total Length of Main reinforcement | =Total length of Inclined bars + the total length of straight bars = 14.85+5.146 = 20m |
Total No. of bars | =20/12 =1.67 bars |
Weight of steel required for 1m of 12mm bar | =D^{2}/162 =12^{2}/162 =0.88kg/m |
Total weight of steel required for main bars | =Total length of main bars x 0.88 =20 x0.88 =17.6Kgs |
BBS of Longitudinal reinforcement in Y shape column:
The vertical reinforcement calculation of Y type column is also divided into two parts.
Look at the below picture for more details
Part1:
Total Hook length | Hook Length = 9d No. of Hooks =2 Total hook length = 9d+9d = 18 x 0.008 =0.144m |
Total Bend length bent @135^{0} | No. of bends bent @135^{0} = 2; Bend length = 3d = No. of bends x 3d = 2 x 3 x 0.008 =0.048m |
Total Bend length bent @90^{0} | No. of bends bent @90^{0} = 3; Bend length = 2d = 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 = 0.25+0.25+0.25+0.25 +0.144-0.096 = 1.048m |
No. of ties | =(1.9/0.1) + 1 =20ties |
The total length of the 1st part longitudinal reinforcement | =20 x 1.048 =20.96m |
Part 2:
In Part-2 each tie length is varied, so the length of each tie is calculated separately for accurate dimensions. Use Pythagorean theorem to find the unknown length. However D is the same for all the ties, so the hook length and bend length is the same for all the ties.
Hook length for each tie | Hook Length = 9d No. of Hooks =2 Total hook length = 2x9d = 2x9x0.008 = 0.144m |
No. of ties | =(0.575/0.1)+1 =6ties |
Total Bend length bent @135^{0} | No. of bends bent @135^{0} = 2; Bend length = 3d = No. of bends x 3d = 2 x 3 x 0.008 =0.048m |
Total Bend length bent @90^{0} | No. of bends bent @90^{0} = 3; Bend length = 2d = 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 |
1st tie length from top | = Perimeter of tie+ Total Hook length – Total Bend length =0.341+0.25+0.25+0.341 +0.144-0.096 =1.23m |
2nd tie length from top | = Perimeter of tie+ Total Hook length – Total Bend length =0.463+0.25+0.463+0.25 +0.144-0.096 =1.47m |
3rd tie length from top | = Perimeter of tie+ Total Hook length – Total Bend length =0.584+0.25+0.584+0.25 +0.144-0.096 =1.71m |
4th tie length from top | = Perimeter of tie+ Total Hook length – Total Bend length =0.706+0.25+0.706+0.25 +0.144-0.096 =1.96m |
5th tie length from top | = Perimeter of tie+ Total Hook length – Total Bend length =0.828+0.25+0.828+0.25 +0.144-0.096 =2.204m |
The total length of the 2nd part longitudinal reinforcement | =2.204+1.96+1.71 +1.47+1.23 =8.57m |
Total Calculation of tie calculation used for Y Column:
Total length of ties | = 1st tie length + 2nd tie length = 20.96+8.57 = 29.53m |
No. of 12m length bars required | =29.53/12 =2.46m |
Weight of steel required for 1m of 8mm bar | =D^{2}/162 =8^{2}/162 =0.39kg/m |
Total Wt of steel required for ties | =Total length of longitudinal reinforcement x 0.39 =29.53 x0.39 =11.51Kgs |
Apart from this, there is also a top mesh above the Y column to hold the whole reinforcement intact
Look at the below image for more details about the top mesh
Provide spacing between bars is 0.1m
Length of each X Bar | =0.95m |
Length of each Y Bar | =0.25m |
No. of X Bars | =(0.25/0.1)+1 =4bars |
No. of Y Bars | =(0.95/0.1)+1 =11bars |
Total length of X bars | =4xlength of each X Bar =4×0.95 =3.8m |
Total length of Y bars | =11x length of each Y Bar =11 x 0.25 =2.75m |
Total length of Bar | =3.8+2.75 =6.55m |
Weight of steel required for 1m of 12mm bar | =D^{2}/162 =12^{2}/162 =0.88kg/m |
Total weight of steel required for top mesh | = 6.55×0.88 =5.764Kgs |