CIVIL ENGINEERING

(BBS) Bar Bending Schedule for Column | Sample calculation | Estimation of Steel

(BBS) Bar Bending Schedule for Column | Sample calculation | Estimation of Steel

Finding difficulties in bar-bending-schedule? You're in the right place. Learn bar-bending-schedule of the column in an easy step by step process with an Example. Learn bar bending schedule format for the column.

BAR BENDING SCHEDULE


What is Bar-bending schedule?

Bar bending schedule is a type of schedule in which we can calculate how much steel is required for any construction work.
The quantity of steel, cutting-length of reinforcement bar everything can be determined with the help of bar bending format.

Bar bending schedule is very important terms for civil engineers as it helps to calculate the number of steel which is required to construct a column, beam, foundation, slab etc.

By using the bar bending schedule we can simply calculate the total amount of steel required for a project.

 Now, I will clarify every detail about the bar bending schedule calculation of COLUMN.


Let's take the example of a column for which we will apply Bar-bending-schedule format.

➤ EXAMPLE-Schedule of column


Column of (250 mm*250 mm),  4-16Φ+2-12Φ,2 L8Φ @150 mm c/c.

Given data:
column height 3 meters, clear cover 40 mm.

Given-



  • Column length = 3 meters
  • clear cover = 40 mm
  • 4 nos of 16 mm dia bar are used
  • 2 nos of 12 mm dia bars used in column
  • The diameter of stirrup bar= 8 mm
  • The spacing between each stirrup = 150 mm




∎ STEP I


 Construct a table as follows. It will be very easy to understand through table format because it simplifies all our calculation.


BAR BENDING SCHEDULE FOR COLUMN

  1. SL. NO
  2. TYPE
  3. LENGTH OF BAR (in meter)
  4. No.OF BAR
  5. UNIT WEIGHT OF BAR( in KG/meter)
  6. TOTAL LENGTH OF BAR( in meter)
  7. TOTAL WEIGHT( in kg)

∎ STEP II- (calculation of stirrup)


In the case of the column, we have two types of reinforcement steel bars-


  1. Main bars (Vertical bars)
  2. Stirrups (Horizontal bars)

bar-bending-schedule for column



Here you have to understand that while calculating the length of stirrup you need to calculate the stirrup legs also.


(*See the above image to know about stirrup leg)


As mentioned in the schedule of the column, stirrup is 2 legged.

Now, assuming the length of each leg is 12*d, where d is the diameter of stirrup bars.
We take this as 12*d (d is diameter of stirrup)

Total stirrup leg length=2*12*d =24 d
(Multiplying 12*d by 2 because of the 2leg)

Here, dis of stirrup bar=d= 8mm,

Thus the length of stirrup leg=24*8=192mm

bar-bending-schedule for column


Now we have to find out the remaining length of the stirrup.

To calculate the length you need to look at the above image.

From the above figure, we can say that the remaining length of the stirrup is 2*(b+l)

Dimension of column= (B*L) mm^2
b=(250-clear.cover)= 170 mm
l=(250-clear.cover)= 170 mm

Thus,Total length of each stirrup =(2*(b+l)+192)mm=872 mm.
[ where 192 mm is the leg length of each stirrup]

So we have calculated the stirrup length for 1 stirrup. Now we need to find out the no. of stirrup required for the given column.

No.of stirrup = (column height/spacing of stirrup)+1 =(3000/150)+1=24 nos.

Total length of stirrup =no.of stirrup*total length of each stirrup =(24*872)mm= 20.93 m




STEP III (calculation of the main bar)





You need to notice that total no. of the main bar is 6, where
  1. 4 no of bars of 16 mm diameters
  2. 2 no. of bars of 12 mm diameters

 Main bar length=column height-clear cover for top and bottom=3000-2*40=2.92 m


Total length of 4 no. of 16 mm dia bars=4*2.92=11.68 m (4 nos) [2.92 m= length of 1 bar]


Total length of 2 no.of 12 mm dia bars
=2*2.92=5.84 m(2 nos)


Unit weight of bar=(d^2/162)kg/m [ MAIN FORMULA]

Keep in mind that, in the case of unit weight we will use diameter(d) in millimeter.

(d=dia of the main bar in mm)

Unit weight of stirrup bar=(8*8)/162=0.39 kg/m


Unit weight of 12 mm bar=(12*12)/162=0.89 kg/m


Unit weight of 16 mm bar=(16*16)/162=1.58 kg/m


[Total weight =unit weight*total length ]

Hence,
Total weight of stirrup=0.39*20.93=8.2 kg


Total weight of 16 mm bar=1.58*11.68=18.45 kg


Total weight of 12 mm bar
=0.89*5.84=5.20 kg


Now,Total weight of steel =(8.20+18.45+5.20)=31.85 kg


bar-bending-schedule for column



Conclusion


I hope you liked this post about bar bending schedule for the column. If you still have any unanswered question please comment below

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Here we have recommended a must buy the book to practice Bar-Bending Schedule.

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(BBS) Bar Bending Schedule for Column | Sample calculation | Estimation of Steel (BBS) Bar Bending Schedule for Column | Sample calculation | Estimation of Steel Reviewed by Civil049seminar on July 11, 2018 Rating: 5

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