M15 concrete ratio in 1 m3
Material requirement for producing 1 Cum of Nominal Concrete Mix
| Grade of Concrete | Nominal Proportion | Cement (bag)/cum |
|---|---|---|
| M15 | 1:3:6 | 4.5 |
| M20 | 1:2:4 | 6 |
| M25 | 1:1.75:3.5 | 6.75 |
| M30 | 1:1.5:3 | 7.5 |
Material Requirement For Producing 1 Cum Of Nominal Concrete Mix
Method-1: DLBD Method To Determine Material Requirement For Nominal Concrete Mix
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
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 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 k
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.
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