We have discussed Bar Bending Schedule for different structural members such as

Here we are going to discuss the Bar Bending Schedule for Circular Slab

Circular Slab

Circular slabs are not usually common like flat rectangular slabs that we use almost every floor. Circular slabs are used in the following areas

  • Water tank cover
  • Manhole covers
  • A room specifically designed for circular slab
  • Towers
  • Pump houses constructed above tube wells
  • Soak Pit Design

Unlike one way slabs, circular slabs are subjected to bending moments in both directions.

Let’s discuss with an example calculation. Note here we are dealing with a simple manhole cover for calculation purposes; we have taken greater value as the diameter of the slab. 

In practice, the circular slab of 3m diameter may have detailed reinforcement.

Circular Slab Reinforcement Detailing

Circular Slab Reinforcement Detailing

Circular Slab Reinforcement Calculation

Assume you have to prepare a bar bending schedule for a circular slab of 3 metre diameter with 12mm main & distribution bars with 200mm spacing; concrete cover of 25mm.

Circular Slab Reinforcement Calculation

Splitting the slab into two parts for easy calculation.

Step 1 – Find the Value of L

The only formula you need to remember for the entire calculation is

L = √(R2 – h2) x 2 where

L = Length of the reinforcement

R – Radius of the slab

h – Distance between the reinforcement

From the above diagram, We can calculate the value of L

L = Diameter of Circular Slab – Concrete Cover on both sides = 3000-(2*25) = 2950 mm or 2.95 m

Radius of the slab = (Diameter of the slab – concrete cover on both sides ) / 2 = 2950/2 = 1475 mm or 1.475m

Step 2 – Find the Number of (L) Bars required

Number of L Bars = (Diameter of the slab – concrete cover on both sides)/Spacing = 1475/200 = 7.375 or 7 Nos

Apart from the Centre Rod (L), we need 3 more bars on both sides such as L1, L2, L3

Step 3 – Find the Consecutive Bar Length

For L = √(R2 – h12) x 2 where h1 = distance between L & L1 or spacing of bars – 200 mm

Therefore, L1 = √(R2 – h12) x 2 =  √((1475)2 – (200)2)x2 = 2066.7 mm

L2 = √(R2 – h22) x 2= √((1475)2 – (400)2) x 2 = 2007.8 mm

L3 = √(R2 – h32) x 2= √((1475)2 – (600)2) x 2 = 1905.58 mm

Note – The circular slab has reinforcement on both sides, so the value will be multiplied by 2 to get the reinforcement length on the perpendicular side.

Step 4 – Bar Bending Schedule Format – Circular Slab

Description Dia of bar (mm) Number of Bars Cutting Length (m) Total Cutting Length (m) Weight Per Metre (Kg/m) Total Weight (Kg)
L 12 2 2.95 5.9 0.89 5.251
L1 12 4 2.07 8.28 0.89 7.369
L2 12 4 2.01 8.04 0.89 7.155
L3 12 4 1.91 7.64 0.89 6.799
TOTAL 26.57 Kg

Happy Learning 🙂

Author

Bala is a Planning & QS Engineer working at Megha Engineering & Infrastructure Limited who has more than 9+ Years of experience in Building & Infrastructure. He is the author, editor of Civil Planets.

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