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Dennis Holt
Entry level engineer
Asked a question last year

How many cement bags and sand is required for M20 grade of concrete?

How many cement bags and sand is required for M20 grade of concrete?

Where am I?

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Actually, the answer to your question is simple.

  • Since the bulk density of cement is 1440kg/cu.m.
  • Mix proportion of M 20 grade of concrete = 1:1.5:3
  • Weight of one cement bag = 50kg.

Required total dry volume of Material = 1.57 cu.m

  • Required Volume of cement = Ratio of cement x 1.57/(1+1.5+3)
  • Required Volume of cement = 1 x 1.57/5 .5
  • Required Volume of cement = 0.285 cu.m
  • Required Weight of cement = volume of cement x bulk density of cement
  • Required Weight of cement = 0.285 cu.m x 1440 kg/cu.m
  • Required Weight of cement = 410 kg
  • No of cement bags required = 410/50
  • No of cement bags required = 8.2 bags i.e. approximately 8 Nos.

Hence, 8 bags of cement are required to prepare one cubic meter of M20 grade concrete.

Thank You.

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.

Seth Morgan
Construction Superintendent

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

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

Jayden Adams
Construction Assistant

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