Question { L&T, 188392 }

Method of findingthe dry ingredient quantity of 1 m3
concrete.

Answer

I heard DLBD method is more accurate that using the factor of 1.54 to convert wet volume to dry volume. In DLBD method, To calculate dry ingredients volume in 1:2:4 concrete

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.

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

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