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## BBS OF Circular Slab:

Circular slabs are not common in use like normal rectangular slabs. Circular slabs are mostly used in water tanks, swimming pools, manhole covers, towers, soak pit chambers, etc. In this article, I will prepare BBS of circular slab to show you how to calculate steel quantity for a circular slab.

Unlike other RCC members, this calculation is slightly difficult since the length of reinforcement bars will be changed according to the diameter of the circle.

Where L = Diameter of the Circular Slab Reinforcement

To find L1, we have to use Pythagoras theorem,

L1 = √(R^{2} – h_{1}^{2}) X 2

By using same formula

L2 = √(R2 – h_{2}^{2}) X 2

L3 = √(R^{2} – h_{3}^{2}) X 2

And so on..

Where L is the length of the reinforcement bar,

R = Radius of the circular slab reinforcement (Excluding clear cover on both sides)

h = Distance between two bars

### Circular Slab Reinforcement Calculation:

Assume,

Diameter of the circular slab = 3 metre = 3000 mm

Diameter of main bars = 12 mm

Diameter of distribution bars = 8 mm

### Calculation For Main Bars:

First, we need to calculate the number of rods required for the circular slab.

No. of rods = (Diameter of the circular slab – Clear Cover on Both Sides) / Centre to centre distance

= {(3000 – (2 X 25)}/200 = 15 no’s

Now we need to calculate the length of 8 no’s main rods separately.

As we know, L =( Diameter of the circular slab – Clear Cover on Both Sides) = 2.95 m

Now we need to calculate length of L1, L2, L3, L4, L5, L6, and L7.

Here we will use Pythagoras theorem,

The formula is L1 = √(R^{2} – h_{1}^{2}) X 2

R = 2.95/2 = 1.475

So,

L1 = √(R^{2} – h_{1}^{2}) X 2 = √(1.475)2 – (0.2)2 X 2 = 2.92 m

L2 = √(R^{2} – h_{2}^{2}) X 2 = √(1.475)2 – (0.4)2 X 2 = 2.84 m

L3 = √(R^{2} – h_{3}^{2}) X 2 = √(1.475)2 – (0.6)2 X 2 = 2.70 m

L4 = √(R^{2} – h_{4}^{2}) X 2 = √(1.475)2 – (0.8)2 X 2 = 2.48 m

L5 = √(R^{2} – h_{1}^{2}) X 2 = √(1.475)2 – (1)2 X 2 = 2.16 m

L6 = √(R^{2} – h_{1}^{2}) X 2 = √(1.475)2 – (1.2)2 X 2 = 1.72 m

L7 = √(R^{2} – h_{1}^{2}) X 2 = √(1.475)2 – (1.4)2 X 2 = 0.93 m

Total length of main bars Lx = L+2(L1)+2(L2)+2(L3)+2(L4)+2(L5)+2(L6)+2(L7)

= 2.95+(2×2.92)+(2×2.84)+(2×2.70)+(2×2.48)+(2×2.16)+(2×1.72)+(2×0.93)

= 34.45 m

Weight of Main Bar = 12^2 x 34.45/162 = 30.62 kg

### Calculation For Distribution Bars:

The calculation for distribution bar for circular slab reinforcement will be exact same as main reinforcement. There for I am skipping the calculation part.

Total length of distribution bars Ly = L+2(L1)+2(L2)+2(L3)+2(L4)+2(L5)+2(L6)+2(L7)

= 2.95+(2×2.92)+(2×2.84)+(2×2.70)+(2×2.48)+(2×2.16)+(2×1.72)+(2×0.93)

= 34.45 m

Weight of distribution bars = 8^2 x 34.45/162 = 13.60 kg

Total weight of steel reinforcement = 30.62 + 13.60 = 44.22 kg

### BBS OF Circular Slab:

#### For Main Reinforcement:

BAR | NUMBERS | CUTTING LENGTH | DIAMETER | WEIGHT |
---|---|---|---|---|

L | 1 | 2.95 m | 12 mm | 2.62 kg |

L1 | 2 | 2.92 m | 12 mm | 2.59 kg |

L2 | 2 | 2.84 m | 12 mm | 2.52 kg |

L3 | 2 | 2.70 m | 12 mm | 2.39 kg |

L4 | 2 | 2.48 m | 12 mm | 2.20 kg |

L5 | 2 | 2.16 m | 12 mm | 1.91 kg |

L6 | 2 | 1.72 m | 12 mm | 1.52 kg |

L7 | 2 | 0.93 m | 12 mm | 0.82 kg |

Total | 15 | 34.45 m | 30.62 kg |

#### For Distribution Bars:

BAR | NUMBERS | CUTTING LENGTH | DIAMETER | WEIGHT |
---|---|---|---|---|

L | 1 | 2.95 m | 8 mm | 1.16 kg |

L1 | 2 | 2.92 m | 8 mm | 1.15 kg |

L2 | 2 | 2.84 m | 8 mm | 2.52 kg |

L3 | 2 | 2.70 m | 8 mm | 1.12 kg |

L4 | 2 | 2.48 m | 8 mm | 2.20 kg |

L5 | 2 | 2.16 m | 8 mm | 1.06 kg |

L6 | 2 | 1.72 m | 8 mm | 0.67 kg |

L7 | 2 | 0.93 m | 8 mm | 0.36 kg |

Total | 15 | 34.45 m | 13.60 kg |

Also Read –

Difference Between Main Bars & Distribution Bars

Bar Bending Schedule Basics Rules

How To Calculate Of Bricks In Wall

Reinforcement Details In Column

Distribution Of Stress Between Steel & Concrete

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Thank you for bbs of circular slab.