Hello Abigail,
M15 grade concrete has the proportion of 1:2:4 of cement, fine aggregate, and Course aggregate.
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)
As the volume of 1kg of cement is calculated, the volume of FA and CA is also calculated based on the given ratio 1:2:4.
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)
Convert Volume requirements to Weight:
So, FA required = 0.072*1600 = 115 kgs
and CA 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, 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)
Cement required for 1cum of concrete = 1/0.167 = 5.98 Bags ~ 6 Bags
