Nominal Mix Concrete
In this content we learn how to prepare 1 cubic meter of concrete. As we know that concrete is in wet condition during concreting when we place wet concrete, after place it gets harden. When concrete gets harden, concrete will shrink due to water evaporation, and air voids get fill by cement and sand. So, the total volume of ingredients (such as cement, sand and aggregate) require will be more than 1 m^{3}. To counter that shrinkage, take the factor of safety ranging from 1.52 to 1.57.
i.e. volume of dry Concrete = 1.52 to 1.57 times Volume of wet concrete
For, M25, M20, M15 – take 1.57
M10 – 1.55 and for M7.5 & M5 – 1.52
Let’s calculate material required to prepare one cubic meter of M15 grade to concrete
The usual mix proportions of cement concrete and where to use are given in Table –
Grade of Concrete | Mix Proportion | Maximum Size of Aggregate | Nature of Work |
M5 | 1:5:10 | 60 mm | Mass concrete work for heavy walls, foundation footings, etc. |
M7.5 | 1:4:8 | 60 mm | Mass concrete work for heavy walls, foundation footings, etc. |
M10 | 1:3:6 | 50 mm | Mass concrete work in culverts, retaining walls, etc. |
M15 | 1:2:4 | 40 mm | For all general R.C.C. works in building such as stair, beam, column, weather shed, slab, lintel, etc., machine foundations subjected to vibrations. |
M20 | 1:1.5:3 | 20 mm | Water retaining structures, pile, pre-cast products and for all general R.C.C. works in building. |
M25 | 1:1:2 | 12 to 20 mm | Heavy loaded R.C.C. columns and R.C.C. arches for long span R.C.C. column footings. |
Method – 1: Empirical Method / Volume Batching of M15 Grade Concrete Using Nominal Mix Method
As per IS456:2000, M15 Grade concrete proportion is = 1: 2: 4
The cement concrete is specified by proportions of different ingredients, 1 (cement) : 2 (fine aggregate) : 4 (coarse aggregate).
Total volume = 1 + 2 + 4 = 7
For M15, here we are assuming 1.57 as factor of safety
So, Total volume of ingredients required per cubic meter of concrete= 1.57
A. Calculate the Volumes of Material Required In 1 cum Concrete
- Cement
- Fine Aggregate / Sand
- Coarse Aggregate
B. Calculate the weights of materials required in 1 cum concrete
As above we calculate the quantity of contents but these quantities not in use in practically. Cement is often sold in 50kg bags by the suppliers and sand and aggregates are generally sold in cubic ft. so we must convert our calculated quantity to bags and ft^{3}. Use following conversions –
- 1 cum of cement is approximately 28.8 bags
- One cubic meter is 35.32 Cubic feet
- Cement
Method – 1:Density of Cement (loose) = 1440 kg/m^{3}
So, weight of cement required = 1440 × 0.22 = 322.56 kg
For 1m^{3} of Concrete = 322.56 kgs of cement is required.
1 Bag of Cement = 50 kg
So, number of Cement Bags for 1m^{3} of Concrete = 323/50 = 6.45 bags
Method – 2: Generally, 1 cum of cement is approximately 28.8 bags
Weight of cement needed = volume of cement × 28.8
= 0.224 × 28.8 = 6.45
Hence, we require 323 Kg or 6.5 bags of cement for 1m^{3 }of concrete
- Sand
Usually sand consists of moisture content that will increases the volume of sand due to bulking of sand effect. That is density of sand varies = 1600 – 1800 kg/m^{3.} here we use 1600kg/m^{3}.
Weight of sand required = 1600 × 0.448 = 716.8 kg
Volume of sand needed = 0.448 × 35.32 = 15.82 ft^{3} ………………………… (1m^{3} = 35.32 ft^{3})
Hence, we require 717 Kg or 16 ft^{3} of sand for 1m^{3 }of concrete
- Aggregate
As per IS 456-2000, usually quantity of aggregate use two time (lower limit 1.5 and upper limit 2.5) of sand by mass. So,
Weight of aggregate required = 2 × 716.8 = 1433.60 kg
OR
Use density = 1500 – 1800 kg/m^{3}, use 1600
Weight of aggregate required = 1600 × 0.896 = 1433.60 kg
Volume of aggregate needed = 0.896 × 35.32 = 31.54 ft^{3} ……………………….(1m^{3} = 35.32 ft^{3})
Hence, we require 1434 Kg or 31.5 ft^{3} of aggregate for 1m^{3 }of concrete.
C. Water Required for 1m^{3}^{ }of Concrete-
Water content should be used as recommended water-cement ratio. if the quantity of water in a mix has to be increased to overcome the difficulties of placement and compaction, also add cement to maintain the specified water-cement ratio. Water quantity is depending upon the climatic factors and workability required. We generally maintain 0.4-0.60 of water cement ratio. As per IS 456-2000, table 5, we use w/c 0.6 for M15.
Cement Required for 1m^{3 }of
Concrete = 323Kg
W/C ratio = 0.60 (60% of cement)
Water required for 1m^{3 }of Concrete.
= 323 x 0.60
Amount of Water Required for 1m^{3}^{ }of Concrete = 193.8kg = 195 litres
Hence, “195 Litres“ of Water is required for 1m^{3 }of M20 Concrete
Final Result of Calculation: –
- Quantities required for 1 m^{3 }of M15 grade concrete
- Cement – 323kg / 6.5bags
- Sand – 717 kg / 16 ft^{3}
- Aggregate – 1434 kg / 31.5 ft^{3}
- Water – 193 litres
METHOD – 2: Calculate The Quantities of Concrete Ingredient Using IS 456-2000
Proportions for Nominal Mix Concrete As per IS 456-2000, Table 9 –
NOTE: 1- General ratio of fine aggregate to coarse aggregate is 1:2, but it can be adjusted between 1:1.5 to 1:2.5 based on the grading of fine aggregate and size of coarse aggregate.