How to find quantity of cement bags, sand, coarse aggregate, from 1m3 of concret

How to find quantity of cement bags, sand, coarse aggregate, from 1m3 of concrete of m20 design?

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How to find quantity of cement bags, sand, coarse aggregate, from 1m3 of concrete of m20 design?..

Answer / s k majumder

Meter Cubic wet volume = 1*1.52 =1.52 Meter Cubic dry vol.
= 1.52*35.33=53.70 cft

Hence for M-20 where nominal portion 1.1.5.3
Total volume comes to (1+1.5+3)= 5.5

Therefore Dry volume for one cubic meter.
Cement= 53.70/5.5=cft= 9.76cft= bag= 7.81 bag=8 bags
sand = 9.76*1.5=14.64 cft
stone = 9.76*3=29.28 ct

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How to find quantity of cement bags, sand, coarse aggregate, from 1m3 of concrete of m20 design?..

Answer / banti kumar

The dry volume of concrete is 15.2 for 10m³ of concrete.
The ratio of M20 is 1:1½:3
Therefore 1+1½+3 =5.5
Now 15.2÷5.5=2.76 say 2.8cum

Sand =2.8×1½=4.8 cum
Coarse aggregate=2.8×3=8.4cum
No of cement bags =2.8÷0.035=80 bags(10m3)
Therefore for 1m3 80÷10= 8 bags.

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How to find quantity of cement bags, sand, coarse aggregate, from 1m3 of concrete of m20 design?..

Answer / z.m mughal

M20 = 1:1½:3
Wet concrete = 1 cum
To convert wet to dry material of concrete, we multiply 1.54. (Note 1.54 is for concrete but if we calculate dry material of mortar, we will multiply 1.27)
1*1.54=1.54 cum (dry material)
Formula of material is
Ratio/sum of ratio * total material quantity
Sum of ratio = 1 1½ 3 = 5.5
Total material quantity = 1.54 cum

Cement, 1/5.5 * 1.54 = 0.28 cum
0.28/0.0354 = 7.909 bags , say 8 bags
(Note: 0.0354 is volume of 50kg cement bag in cum and 1.25 in cft).

Sand or fine aggregate, 1.5/5.5 * 1.54 = 0.42 cum

Coarse aggregate, 3/5.5 * 1.54 = 0.84 cum.