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.
Material requirement for producing 1 Cum of Nominal Concrete Mix
The following are the materials required to produce 1 Cum of Concrete of a given Nominal Mix Proportion
|Grade of Concrete||Nominal Proportion||Cement (bag)/cum||Sand (in cft)||Aggregate (in cft)|
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 material 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 = 0.00694 cum
- Volume of 01 bag (50 kg) of cement = 50 X 0.00694 = 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 requirement to weights
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.
|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: Empirical 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