Mix design is a process of determining the right quality materials and their relative proportions to prepare concrete of desired properties like workability, strength, setting time and durability.

While following a mix design is advised to optimise the material consumption, it is not possible at site to always come up with Mix design. Nominal mix concrete is prepared by approximate proportioning of cement, sand and aggregate to obtain target compressive strength.

Mix ratio of concrete defines the ratio of cement sand and aggregate by volume in that order. So a mix ratio of 1:2:4 represents Cement: Sand: Aggregate – 1:2:4 (by Volume)

## Material Requirement for Producing 1 Cum of Nominal Mix Concrete

The following are the materials required to produce 1 Cum of Concrete of a given Nominal Mix Proportion

### Cement, Sand and aggregate requirement for Different grade concrete

Grade of Concrete | Nominal Mix Proportion | Cement Required (bag)/cum | Sand Required (in cft) | Aggregate Required (in cft) |
---|---|---|---|---|

M15 | 1:3:6 | 4.5 | 16.04 | 32.07 |

M20 | 1:2:4 | 6 | 14.98 | 29.96 |

M25 | 1:1.75:3.5 | 6.75 | 14.75 | 29.14 |

M30 | 1:1.5:3 | 7.5 | 14.06 | 28.11 |

Incase you want to convert the requirement of Sand and Aggregate in Cum; **1 Cum = 35.31 Cft**

**Eyeopener** :Many popular blogs claim M20 nominal mix as 1:1.5:3 ,however we strongly differ by same.Through this blog,we are also trying to address the same myth which is being carried forward since last 4 decades.

**The reason being:** With constant research and development in the field of cement technology and its manufacturing process ,a M20 mix of “1:1.5:3″(by volume) would be too rich, over engineered and uneconomical (~7.5 bags of cement per cum) and will ultimately result into a M30 concrete and above (IS:456 too have the minimum cement/cementitious content of 06 bags for M20). As the latest generation of 53 grade OPC cement is ultimately giving a strength of 65 to 70 MPa at 28days, 1:2:4 will give a strength of M20.

A detail procedure to calculate the cement bags required for 1: 2 :4 mix (~6 bags of cement per cum) is shown below.

## Method-1: DLBD method to determine Requirement For Nominal Concrete Mix (M20 - 1:2:4)

The DLBD (Dry Loose Bulk Densities) method is an accurate method to calculate cement, sand and aggregate for a given nominal mix concrete. This gives accurate results as it takes into account the Dry Loose Bulk Densities of materials like Sand and Aggregate which varies based on the local source of the material

For calculation, We consider a nominal concrete mix proportion of 1:2:4 (~M20).

### Step-1: Calculate Volume of Materials required

Density of Cement = 1440 kg/cum (Approx)

Volume of 1 Kg of Cement = 1/1440 = 0.000694 cum

Volume of 01 bag (50 kg) of cement = 50 X 0.000694 = 0.035 cubic meter (cum)

Since we know the ratio of cement to sand (1:2) and cement to aggregate (1:4)

Volume of Sand required would be = 0.035*2 = 0.07 cubic meter (cum)

Volume of Aggregate required would be = 0.035*4 = 0.14 cubic meter (cum)

### Step-2: Convert Volume requirements to Weight

To convert Sand volume into weight we assume, we need the dry loose bulk density (DLBD). This density for practical purposes has to be determined at site for arriving at the exact quantities. We can also assume the following dry loose bulk densities for calculation.

DLBD of Sand = 1600 kgs/cum

DLBD of Aggregate = 1450 Kgs/Cum

So, **Sand** required = 0.072*1600 = **115 kgs**

and **Aggregate** required = 0.144*1450 = **209 kgs**

Considering water/cement (W/C) ratio of 0.55

We can also arrive at the Water required = 50*0.55 = 27.5 kg

**So, One bag of cement (50 Kgs) has to be mixed with 115 kgs of Sand, 209 Kgs of aggregate and 27.5 kgs of water to produce M20 grade concrete.**

Cement | Sand | Aggregate | Water |

1 bag (50kg) | 115 Kgs | 209 Kgs | 27.5 Kgs |

### Step-3: Calculate Material Requirement for producing 1 Cum Concrete

From the above calculation, we have already got the weights of individual ingredients in concrete.

So, the weight of concrete produced with 1 Bag of cement (50 Kgs) =50 kg + 115 kg + 209 kg + 27.5 kg = 401.5 kg ~ 400 kgs

Considering concrete density = 2400 kg/cum,

One bag of cement and other ingredients can produce = 400/2400 = 0.167 Cum of concrete (1:2:4)

01 bag cement yield = 0.167 cum concrete with a proportion of 1:2:4

*01 cum of concrete will require*

Cement required = 1/0.167 = 5.98 Bags ~ 6 Bags

Sand required = 115/0.167 = 688 Kgs or 14.98 cft

Aggregate required = 209/0.167 = 1251 kgs or 29.96 cft

## Method-2: Emperical method to Determine Material Requirement for Nominal Concrete Mix

Although empirical method is easy to use in determining the materials requirement for Nominal Concrete mix, it sometimes doesn’t give accurate results as it doesn’t take into factor the local variations in the materials.

Let’s design M20 grade concrete. Ratio for M20 concrete is **1 : 2 : 4**

### Step-1: Calculate the Volumes of Material Required in 1 Cum Concrete

The dry volume of concrete mixture is always greater than the wet volume. The ratio of dry volume to the wet volume of concrete is 1.54.

So 1.54 Cum of dry materials (cement, sand and aggregate) is required to produce 1 Cum of concrete

Volume of Cement required = 1/(1+2+4) X 1.54 = 1/7 X 1.54 = 0.22 Cum

Volume of Sand required = 2/7 X 1.54 = 0.44 Cum or 15.53 cft

Volume of Aggregate required = 4/7 X 1.54 = 0.88 Cum or 31.05 cft

Note: 1 cubic meter = 35.29 cubic feet

### Step-2: Calculate the weights of materials Required in 1 Cum Concrete

Density of Cement (loose) = 1440 kgs/cum

So weight of cement required = 1440 X 0.22 = 316.2 Kgs or 6.32 bags

Density of Sand = 1600 Kgs/cum

Weight of Sand required = 1600 X 0.44 = 704 kgs

Density of Aggregate = 1450 kgs/cum

Weight of aggregate required = 1450 X 0.88 = 1,276 Kgs