# Concrete Material Calculation / Concrete Quantity

Important Point

## What is Material Calculation In Civil?

Material Calculation in civil Engineer, as like work Concrete, Bar Bending, Plumbing, Flooring, Electrical all work requirement minimum materials calculations.

### Material Use in Concrete?

Same as concrete is different parts of one product. All materials name Cement + Sand + Aggregate + Water + Addmixuer a part of concrete.

Concrete is defined as a different grade like M15, M20, M25, as per IS Code 456. All concrete Grade different mix designs. All mix design different materials, different positions.

All different material types of measurement units like Water measurement in liter and Cement in Bag, Aggregate in k.g. But Concrete is a measurement in Cubic Meter.

Also, read: Rate Analysis of Concrete

## How to Concrete Material Calculation?

Concrete material calculation first proportions of matrial such as 1 : 2 : 4 (M15), 1 : 1.5 : 3 (M20), 1 : 1 : 2 (M25), 1 : 0.5 : 1 (M30), Above M 25 Grade concrete need mix design.

Here 1 : 2 : 4 is Material how to define which material first number 1 is a Cement, Secon number material is a Fine Aggregate (Sand), here the third number is Coarse Aggregate, as per the sequence of proportions.

This method is based on a principle that the volume of fully compacted concrete is equal to an absolute volume of all the materials of concrete, i.e., cement, sand, coarse aggregates, and water.

This concrete structure may consist of footing, slabs, columns, beams, and foundations, etc. based on the types of structure.

A volume of concrete required for concrete structure may be calculated by summing up to the volumes of each structural member or each part of members.

The volume of a rectangular cross-sectional member may be calculated as Length x Width x Height/Depth/Thickness. The suitable formula shall be used for different cross-sectional shapes of members.

Also, read: Trapezoidal Footing Formula with Calculation

**The formula for calculation of Concrete Material** for the required volume of concrete is given by:

**Vc =Volume of Concrete**

**Vce = Volume of Cement **

**Vf = Volume of Fine Aggregate **

**Vca = Volume of Coarse Aggregate **

**Vw = Volume of Water**

**Sc = Specific Gravity of Concrete**

**Sce = Specific Gravity of Cement **

**Sf = Specific Gravity of Fine Aggregate **

**Sca = Specific Gravity of Coarse Aggregate**

**ce= Cement **

**f = Fine Aggregate **

**ca = Coarse Aggregate **

**dr = Dry Volume**

**we = Wet Volume**

**sa =Sum propositions Ratio**

**Above all Short Form Use in Equation**

**Lest Go Find Materials of Cement in Concrete**

First Grade of Concrete

So We Going M-15 Grade of Concrete.

M-15 Concrete Propositions ratio 1:2:4 **(ce:f:ca)**

Here, 1 is a Cement, 2 is a Fine Aggregate, 4 is a Course Aggregate.

Dry Volume of concrete = Wet Volume of Concrete X 55 % (Because of working time Dry Volume less as per using qty so, we need 30% and 25 % of wastage qty, Here 30% + 25% = 55% access of this qty)

Sum propositions Ratio **(sa)** = 1+2+4 = 8

**Vce = Volume of Cement / Cement Volume**

Vce = dr x (ce/sa) Cu.m.

Vce = 1.55 x (1/8) Cu.m.

Vce = 1.55 x 0.125 Cu.m.

Vce = 0.19375 Cu.m.

**How to Converter Cement Qty in Cu.m. to Bag**

The volume of Cement Cu.m. = 0.034722 x Cu.m. qty **(0.0347222 is Volume of qty for one bag)**

How to find **0.034722** = 50/1440 **(Each Cement Bag Weight = 50 kg, Density of Cement / Cement Volume = 1440 kg/ Cu.Meter., Volume of Cement Bag in Cubic Feet = 35.3147** )

**Vce in bag** = **0.19375 x (1/0.034722)**

**Vce in bag = 5.58 bag**

**or **

Vce in kg = **0.19375 x 1440** (1440 kg/Cu.m is a cement density)

Vce in kg **=279 kg Cement **

Vce in bag** = 279kg /50kg** (1 bag = 50 kg) **= 5.58 Bag**

**Vf = Volume of Fine Aggregate/Sand**

Vf = dr x (f/sa) Cu.m.

Vf = 1.55 x ( 2/8) Cu.m.

Vf = 1.55 x 0.25 Cu.m.

**Vf = 0.3875 Cu.m. This qty of fin aggregate in Cu.m. **

### How to Find Fine Aggregate in Cu.m. to Kg

**Sf = Specific Gravity of Fine Aggregate = 1700 kg/ Cu.m.**

Vf in kg = 0.3875 Cu.m. x 1700 kg/ cu.m.

**Vf in kg = 658.75 kg **

**Vca = Volume of Coarse Aggregate **

Vca = dr x (ca/sa) Cu.m.

Vca = 1.55 x (3/8) Cu.m.

Vca = 1.55 x 0.375 Cu.m.

**Vca = 0.58125 Cu.m. This qty of Coarse Aggregate in Cu.m.**

### How to Find Coarse Aggregate in Cu.m. to Kg

**Sca = Specific Gravity of Coarse Aggregate = 1650 kg/ Cu.m.**

Vca = 0.58125 Cu.m. x 1650 kg/ Cu.m.

**Vca in kg= 959.06 kg.**

**Vw = Volume of Water**

**w/c** = 0.45 as per IS Code 456:2000 Table-5

Required water for 1 cement bag= 0.45 × 0.034722 **=** **0.015625 cum**

1 Cu.m. water = 1000 litr.

Required water for 1 cement bag=0.015625 × 1000 = 15.63Litre.

Vw for 1 Cu.m. Concrete in Water = 15.63 litr x 5.58 bag = 87.22 litr

**Vw = 87.22 litr**

### Summary of Concrete Material Calculation

**Vce in bag = 5.58 bag (Cement in Bag)**

**Vf = 658.75 kg (Fine Aggregate in kg)**

**Vca = 959.06 kg. (Coarse Aggregate in kg)**

**Vw = 87.22 liters (Water in Liter)**

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