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

Join TheConstructor to ask questions, answer questions, write articles, and connect with other people. When you join you get additional benefits.

Log in to TheConstructor to ask questions, answer people’s questions, write articles & connect with other people. When you join you get additional benefits.

Lost your password? Please enter your email address. You will receive a link and will create a new password via email.

## 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 bagssand in kg = (1.5/5.5)x1.52 x 1650=684 kgaggregate in kg=(3/5.5) x 1.52 x 1700 =1410 kg## 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.

## 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 m

^{3}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^{3}Say,

0.42 Cum of Sand^{3}Say,

0.86 Cum of AggregateNOTE:Density of Cement = 1440 Kg/Cum for one bag of 50 Kg cement## Komal Bhandakkar

Thank You.

## 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