Bar Bending Schedule of a Floor Column | Steel calculation | BBS


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:

  1. Find the length of the single main bar
  2. Find the total length of the Main reinforcement
  3. Calculate the weight of steel required per 1m of Main reinforcement
  4. Find out the total weight of steel required for Main reinforcement

bar bending schedule of a floor column (Column top view)

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.

bar bending schedule of a floor column (Column Sectional view)

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:

  1. Deduct the concrete cover from all sides of ties and find out the length of stirrup using formulae.
  2. Obtain the total number of ties required using formula
  3. Find out the total length of ties required
  4. Calculate the weight of steel required per 1m of longitudinal reinforcement.
  5. 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.

 

Top view and sectional view of Square column main reinforcement

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

=D2/162

=162/162

=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.

 

Top View of Square Column for tie length calculation

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 =3;

No. of Bends bent @135 =2;

Total Hook lengthHook Length = 9d

No. of Hooks =2

Total hook length = 9d+9d

= 18 x 0.008

=0.144m

Total Bend length

bent @1350

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

= 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

=D2/162

=82/162

=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 

 

Top view and sectional view of Square column main reinforcement

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

=D2/162

=182/162

=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.

bar bending schedule of a floor column (Top View of Rectangular Column)

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 =3;

No. of Bends bent @135 =2;

Total Hook lengthHook Length = 9d

No. of Hooks =2

Total hook length = 9d+9d

= 18 x 0.008

=0.144m

Total Bend length

bent @1350

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

= 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

=D2/162

=82/162

=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.

Sectional view of a Circular Column

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

=D2/162

=202/162

=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.

Longitudinal reinforcement of a Circular column

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 =2;

No. of bends bent @900 = 6;

Total Hook lengthHook Length = 9d

No. of Hooks =2

Total hook length = 9d+9d

= 18 x 0.008

=0.144m

Total Bend length

bent @1350

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 = 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

=D2/162

=82/162

=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.

Topview and Sectional view of a T Shape column

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

=D2/162

=202/162

=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 =6;

No. of Bends bent @135 =4;

longitudinal reinforcement details of T shape column -I

longitudinal reinforcement details of T shape column -II

Total Hook lengthHook Length = 9d

No. of Hooks =4

Total hook length = 4x9d

= 36 x 0.008

=0.288m

Total Bend length

bent @1350

No. of bends bent @1350 = 4;

Bend length = 3d

= No. of bends x 3d

= 4 x 3 x 0.008

=0.096m

Total Bend length

bent @900

No. of bends bent @900 = 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

=D2/162

=82/162

=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 

Top view and sectional view of L Column

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

=D2/162

=202/162

=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

longitudinal reinforcement of L Shape column

longitudinal reinforcement details of T shape column -II

Total Hook lengthHook Length = 9d

No. of Hooks =4

Total hook length = 4x9d

= 36 x 0.008

=0.288m

Total Bend length

bent @1350

No. of bends bent @1350 = 4;

Bend length = 3d

= No. of bends x 3d

= 4 x 3 x 0.008

=0.096m

Total Bend length

bent @900

No. of bends bent @900 = 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

=D2/162

=82/162

=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:-

BBS of Y Type 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

Y Shape Column Main reinforcement details

As per Pythagorean theorem,

Hypotenuse2= Opposite2 + Adjacenet2

Hypotenuse2= 0.5752 + 0.352

Hypotenuse = 0.673m

From fig.  No. of BarsInclined 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

=D2/162

=122/162

=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

Longitudinal reinforcement of Y Shape Column

Part1: 

Total Hook lengthHook Length = 9d

No. of Hooks =2

Total hook length = 9d+9d

= 18 x 0.008

=0.144m

Total Bend length

bent @1350

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

= 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.

Longitudinal reinforcement of Y Shape Column

Hook length for each tieHook 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 @1350

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

= 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

=D2/162

=82/162

=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

Top Mesh of Y Column

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

=D2/162

=122/162

=0.88kg/m

Total weight of steel

required for top mesh

= 6.55×0.88

=5.764Kgs

Abstract for the steel calculation of floor column of the above plan:

Read More about Bar Bending schedule



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