How many cement bags and sand is required for M20 grade of concrete?
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RiyaJames
For an M20 grade of concrete, the ratio is (1:1.5:3) ie, (1 part of cement: 1.5 part of sand: 3 part of Aggregate)
Let us assume for 1 Cum. Of M20 concrete,
So, Wet volume of concrete = 1 Cum
Dry volume of concrete = 1.54 times of the wet volume of concrete (Here 1.54 stands
as “Factor of safety” to counter the shrinkage)
= 1.54 Cum
Total of ratio = 1 + 1.5 + 3 =5.5
Therefore, Volume of Cement = (1/5.5) x 1.54 = 0.28 m3
No. of Cement Bags = 0.28 x (1440 Kg/cum / 50 Kg/Cement Bag)
= 0.28 x 28.8 = 8.064 Bags
Say, 8 Bags of Cement
Volume of Sand = (1.5/5.5) x 1.54 = 0.42 m3
Say, 0.42 Cum of Sand
Volume of Aggregate = (3/5.5) x 1.54 = 0.86 m3
Say, 0.86 Cum of Aggregate
NOTE: Density of Cement = 1440 Kg/Cum for one bag of 50 Kg cement
Vivek Patel
for 1 cubic meter, we calculate as follow
let’s take dry mixture of aggregate sand and cement required is 1.52 cubic meter.
M20 have an approximate proportion of cement sand and aggregate is 1:1.5:3
cement in bag= (1/5.5.)x1.52/0.035 =7.89 say=8 bags
sand in kg = (1.5/5.5)x1.52 x 1650 =684 kg
aggregate in kg =(3/5.5) x 1.52 x 1700 =1410 kg
Komal Bhandakkar
Thank You.
Jitendra45
For calculating the quantity of cement, sand, aggregate, and water follow these steps:
For one cubic meter of M20 (1:1.5:3)
Wet volume of Concrete = 1 cum
Dry volume of Concrete = 1.54 cum (increased by 54% for dry volume)
Cement = 1.54/(1+1.5+3) × 28.8 = 8.064 say 8 bags. (1 cum = 28.8 bags)
Sand = 1.54/(1+1.5+3) × 1.5 = 0.42 cum or 14.83 cft.
Aggregate = 1.54/(1+1.5+3) × 3 = 0.84 cum or 29.66 cft.
Water Quantity
Assume water-cement ratio = 0.5
Water required = water-cement ratio × Cement quantity (kg)
= 0.5 × 400
= 200 litres.
nikeetasharma
Mostly the ratio of 1:1.5:3 is used for M20 grade of concrete
So, Volume of dry concrete = 1.54 to 1.57 times volume of wet concrete
Assume, 1cum of concrete work ratio sum = 1+1.5+3 = 5.5
shrinkage or safety factor = 1.57
Total volume of wet concrete required is = 1.57cum
Volume of broken stone require = (3/5.5) * 1.57 = 0.856 m3
Volume of sand require = (1.5/5.5) * 1.57 = 0.471 m3
Volume of cement = (1/5.5) * 1.57 = 0.285 m3
=0.285*1440 =411 kg
For 1 m3 of M20 (1:1.5:3)
Broken stone = 0.856 m3
Sand = 0.472 m3
Cement = 8.22 bag
Therefore,
8 bag of cement is required for 1cum of concrete work in M20