Size of brick with Mortar = 20 cm × 10 cm × 10 cm

No. of bricks required per 1 m^{3} brickwork = 1 / (0.2 × 0.1 × 0.1) = 500 nos.

No. of bricks for 10 m^{3} brickwork = 500 × 10 = 5000 nos.

Size of brick without Mortar = 19 cm × 9 cm × 9 cm

Actual volume of 5000 bricks = 5000 × (0.19 × 0.09 × 0.09) = 5000 × 0.001539 = 7.695 m^{3}

approx 7.7 m^{3}

Volume of mortar = 10 – 7.7 = 2.3 m^{3}

Adding 10% wastage = 2.3 + 2.3 × 10% = 2.53 m^{3}

Adding 25% for dry mortar weight (Dry mortar weight > wet mortar weight)

Volume of dry mortar = 2.53 + 2.53 × 25% = 3.16 m^{3}

approx 3.2 m^{3}

Now,

**For 1: 6 cement mortar,**

Volume of cement = 3.2 / 7 = **0.45 m**^{3}

Volume of sand = 3.2 – 0.45 = **2.7 m**^{3}

1 m^{3} OPC (Ordinary Portland Cement) = 1440 kg

Weight of 1 bag of cement = 50 kg

Volume of 1 bag of cement = 50 / 1440 = 0.0347 m^{3}

For 10 m^{3} of brickwork, no. of cement bags needed = 0.45 / 0.0347 = 12.97

approx **13 bags**

Similarly,

**For 1: 5 cement mortar,**

Volume of cement = 3.2 / 6 = **0.53 m**^{3}

Volume of sand = 3.2 – 0.53 = **2.67 m**^{3}

1 m^{3} OPC (Ordinary Portland Cement) = 1440 kg

Weight of 1 bag of cement = 50 kg

Volume of 1 bag of cement = 50 / 1440 = 0.0347 m^{3}

For 10 m^{3} of brickwork, no. of cement bags needed = 0.53 / 0.0347 = 15.27

approx **15 bags**

**For 1: 4 cement mortar,**

Volume of cement = 3.2 / 5 = **0.64 m**^{3}

Volume of sand = 3.2 – 0.64 = **2.56 m**^{3}

1 m^{3} OPC (Ordinary Portland Cement) = 1440 kg

Weight of 1 bag of cement = 50 kg

Volume of 1 bag of cement = 50 / 1440 = 0.0347 m^{3}

For 10 m^{3} of brickwork, no. of cement bags needed = 0.64 / 0.0347 = 18.44

approx **18 bags**

**For 1: 3 cement mortar,**

Volume of cement = 3.2 / 4 = **0.8 m**^{3}

Volume of sand = 3.2 – 0.8 = **2.4 m**^{3}

1 m^{3} OPC (Ordinary Portland Cement) = 1440 kg

Weight of 1 bag of cement = 50 kg

Volume of 1 bag of cement = 50 / 1440 = 0.0347 m^{3}

For 10 m^{3} of brickwork, no. of cement bags needed = 0.8 / 0.0347 = 23.05

approx **22.5 bags**

**For 1: 2 cement mortar,**

Volume of cement = 3.2 / 3 = **1.07 m**^{3}

Volume of sand = 3.2 – 1.07= **2.13 m**^{3}

1 m^{3} OPC (Ordinary Portland Cement) = 1440 kg

Weight of 1 bag of cement = 50 kg

Volume of 1 bag of cement = 50 / 1440 = 0.0347 m^{3}

For 10 m^{3} of brickwork, no. of cement bags needed = 1.07 / 0.0347 = 30.83

approx **30 bags**

**Ratio** |
**Cement** |
**Sand** |

**1:2** |
**1 m**^{3} (30 bags) |
**2 m**^{3} |

**1:3** |
**0.75 m**^{3} (22.5 bags) |
**2.25 m**^{3} |

**1:4** |
**0.6 m**^{3} (18 bags) |
**2.4 m**^{3} |

**1:5** |
**0.5 m**^{3} (15 bags) |
**2.5 m**^{3} |

**1:6** |
**0.45 m**^{3} (13 bags) |
**2.7 m**^{3} |

**As per standard values,**

### Kanwarjot Singh

Kanwarjot Singh is the founder of Civil Engineering Portal, a leading civil engineering website which has been awarded as the best online publication by CIDC. He did his BE civil from Thapar University, Patiala and has been working on this website with his team of Civil Engineers.
If you have a query, you can **ask a question here**.

Very good useful

I need more knowledge about the basic information in civil engineer work at field work site because I am fresh graduate

This formula assumes that all the faces of the brick are filled with mortar which isn’t true because the front and back faces of the brick are left for plastering

very simple way to explain everything needed…..thumbs up

Thanks, for the Very useful Information and Guide.

Actually, this is very important for the all Construction Workers.

Very Technical, and Useful. Thank You.

easily we can understand thanks

Thank you for guidance. Can you determine the motor requirements for all the sizes of the bricks. Will be very grateful.