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

Contents of the Article

### 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 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.

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 🙂

Bala is a Planning Engineer & he is the author and editor of Civil Planets.

1. Hi Bala, I find your blog very informative. Please review the following step, if i am wrong please excuse.

Step 2 – Find the Number of (L) Bars required
Number of L Bars = (Diameter of the slab – concrete cover on both sides)/Spacing = 2950/200 = 14.75 or 15 Nos

Apart from the Centre Rod (L), we need 7 more bars on both sides such as L1, L2, L3, L4, L5, L6 & L7

• APPLY THE SAME FORMULA AND FOR L4 H4 =( 4*200) FOR L5 H5 =( 5*200)

2. Kowshik

Hi bro, all of your Explanations is very informative.
But in this Circular Slab i’m not getting the same lengths for L1,L2,L3… as you explained when drawn in AutoCAD