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**show

**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 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 = √(R^{2}– h^{2}) 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 = √(R^{2} – h1^{2}) x 2 where h1 = distance between L & L1 or spacing of bars – 200 mm

Therefore, L1 = √(R^{2} – h1^{2}) x 2 = √((1475)^{2} – (200)^{2})x2 = 2066.7 mm

L2 = √(R^{2} – h2^{2}) x 2= √((1475)^{2} – (400)^{2}) x 2 = 2007.8 mm

L3 = √(R^{2} – h3^{2}) 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 🙂

## 3 Comments

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)

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