How to Calculate Cement, Sand and Coarse Aggregate Quantity in Concrete?/mix design

The Dimension Of This Slab

Length – 6m or 20 feet

Breadth – 6m or 20 feet

Depth – 0.150m or 0.5 feet

Grade of concrete – M20

M20 Mix Ratio= 1 : 1.5 : 3

The Mix Ratio Denotes The Following

1– Cement

1.5– Sand (Fine aggregate – 1.5 Times of Cement Quantity)

3– Blue metal (Coarse aggregate – 3 Times of Cement Quantity)

Volume of concrete we need = length x breadth x depth =

L x b x h = 6 x 6 x 0.15 = 5.4 m^{3 } ^{
}

OK

First, we can find the volume of concrete for 1m3 so we can use on different volume

The volume of 1m^{3} concrete

We find out the volume of concrete in wet condition 1m3 (After the consolidation of Cement + sand + water). So in dry conditions, we have to add 30 to 35% sand bulkage and add 20% for wastage.

Accounting bulkage of sand - 34%

Accounting wastage of material -20%

Hence 1 cum becomes 1.54 (1+0.34+0.2) cum

Total Part of the Concrete = 1+1.5+3 = 5.5 Parts

Calculation of Volume of Cement in 1m3 of Concrete

Cement Quantity= (Cement Part / Concrete Parts) x Concrete Volume

Cement part = 1 (1:1.5:3)

Concrete parts = 1+1.5+3 = 5.5 Parts

Density of cement - 1440 kg/cum

Cement = (1/5.5) x 1.54 = 0.2798

Density of Cement = 1440 kg/ m3 = 0.2798 X 1440 = 403.19 kg

1 Cement BagWeight = 50 Kg

Hence 403.29/50 = 8.06 bags

LCETED TRIVIA

To be simple

Volume of,

One cement bag = 0.0347 cum

One cement bag = 1.225 cft

1 cft box = 1*1*1.25 ft

Hence Required cement = ((1/5.5) x (1.54))/(0.0347) = 8.06 bags

Calculation of The volume of sand in 1m3 of Concrete

Sand Quantity= (sand Part / Concrete Parts) x Concrete Volume

Sand part = 1.5 (1:1.5:3)

Concrete parts = 1+1.5+3 = 5.5 Parts

Density of sand = 1500–1550 kg/cum

Sand = (1.5/5.5) x 1.54 = 0.42

Density of Cement = 1500 kg/ m3 = 0.42 X 1500 = 630 kg

But we all are buying sand in cft

Therefore,

1 Cum = 35.31 Cft

Sand Quantity = 0.42

= 0.42 x 35.31= 14.83 Cft

100 cft = 1 Unit

2.83 cum = 1 unit

1 load - 4 unit

Generally on lifting for sand, 1 unit sand equals 35 cement bags of sand

1 cft box = 1 x 1 x 1.25 ft

Calculation of Volume of aggregate in 1m3 of Concrete

Aggregate Quantity = (sand Part / Concrete Parts) x Concrete Volume

Aggregate part = 3 (1:1.5:3)

Concrete parts = 1+1.5+3 = 5.5 Parts

Density of aggregate = 1520 - 1680 kg/cum

Aggregate = (3/5.5) x 1.54 = 0.84

But we all are buying aggregate in cft

Therefore,

1 Cum = 35.31 Cft

Aggregate Quantity = 0.82

= 0.84 x 35.31 = 29.66 Cft

Total Materials Required for 1m^{3}

^{
}

Cement - 8 bags

Sand - 14.83 Cft

Aggregate - 29.66 Cft

Therefore,

We need the volume of concrete for 5.4m^{3}

^{
}

Cement = 5.4 x 8 ( 8 bag of cement for 1m^{3}) = 43.2 – 43 bags

Sand = 5.4 x 14.83 (14.83 cft of sand for 1m^{3}) = 80.08 cft

Aggregate = 5.4 x 29.66 (29.66 cft of aggregate for 1m^{3}) =160.164 cft

As LCETED civil engineer, you should know how to use the Excel sheet to find this quantity. so use this sheet for your uses You Can Change This Value As Per Your Specification

Depending on the kind of building, a concrete structure may consist of beams, slabs, columns, and foundations, among other things. By adding up the volumes of all the structural components or individual portions of members, one may determine the volume of concrete needed for a concrete construction. The formula for calculating the volume of a rectangular cross sectional member is length x width x height (or depth x thickness). Suitable formulas must be utilized for various cross-sectional member shapes.

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