## Calculate the Quantity of Cement, Sand & Aggregate in M20 Concrete

First of all, I want to assure you that it is very easy to determine the quantity of sand, **cement**, and aggregate in M20, M15, and M10 Concrete. So take a look carefully and will easily be understood.

Let’s take some examples,

**Example: 1**

Let’s assume that the volume of total concrete(wet concrete) required for a particular work is 2 m^{3}. Now find out the quantity of cement, sand, and aggregate.

**Solution:**

**In the **case of M 20 concrete, the ratio of cement, sand, and aggregate is 1: 1.5: 3.

Dry volume of concrete = wet volume of concrete × 1.54 [ You can take this factor from 1.54 to 1.57 ]

Hence, the dry volume of concrete is = 2 × 1.54 = 3.08 m^{3}

➤ **The required quantity of cement** is = (Cement ratio **÷** Sum of all ratio) × Total volume of concrete = [latex]\frac{1}{(1 + 1.5 + 3)} \times 3.08[/latex] = 0.56 m^{3}

### How do you calculate cement in bags?

First of all, we need to calculate it in kg. We know that the density of cement is 1440 kg/ m^{3}. Therefore, 0.56 m^{3} cement means 0.56 × 1440 = 806.4 kg cement.

In general, each bag of cement weight is 50 kg.

So, the required number of cement bags are = 806.4 **÷** 50 = 16.12 ≈ 17 No.s.

➤ **Quantity of sand required** = ( Sand ratio **÷** Sum of all ratio) × Total volume of concrete = [latex]\frac{1.5}{(1 + 1.5 + 3)} \times 3.08[/latex] = 0.84 m^{3}

### How do you calculate sand in CFT?

Our total quantity of sand is 0.84 in m^{3}, but in CFT(cubic feet) it will be ( 0.84 × 35.31) = 29.66 ≈ 30 CFT( cubic feet) [** Note**: 1 m^{3} = 35.31 CFT or cubic feet ]

➤ **Quantity of aggregate** is = ( Aggregate ratio **÷** Sum of all ratio) × Total volume of concrete = [latex]\frac{3}{(1 + 1.5 + 3)} \times 3.08[/latex] = 1.68 m^{3}.

### How do you calculate aggregates in CFT?

We know, 1 m^{3} is equal to 35.31 CFT. Therefore, 1.68 m^{3} aggregate means 1.68 × 35.31 = 59.32 ≈ 60 CFT.

**Example: 2**

A section of footing is given below. Find out the quantity of cement sand and aggregates required for this footing.

**Solution:**

At first, we need to calculate the total volume of concrete required for this footing. The total volume of concrete is equal to the total volume of footing.

So, the total volume of concrete is = 1 × 7 × 6 = 42 cubic feet.

Dry volume of concrete will be = 42 × 1.54 = 64.68 cubic feet.

In the case of M20 concrete,

➤ **The Quantity of cement **required for this footing is = [latex]\frac{1}{(1 + 1.5 + 3)} \times 64.68 [/latex] = 11.76 cubic feet. Or, 0.33 in m^{3} [ **Note**: 1 m^{3} = 35.31 CFT or cubic feet]

The weight of cement = 0.33 × 1440 = 475.2 kg. Or, 475.2/50 = 9.50 ≈ 10 No.s bags.

➤ **Quantity of sand** required = [latex]\frac{1.5}{(1 + 1.5 + 3)} \times 64.68 [/latex] = 17.64 cubic feet or CFT .

➤ **The Quantity of aggregate** required = [latex]\frac{3}{(1 + 1.5 + 3)} \times 64.68 [/latex] = 35.25 CFT.

## Calculate the Quantity of Cement, Sand & Aggregate in M15 Concrete.

Now, we will calculate the quantity of cement and aggregate in M15 Concrete. Let’s take an example,

**Example:**

Let’s assume that the volume of total concrete(wet concrete) required for a particular work is 1 m^{3}. Now find out the quantity of cement, sand, and aggregate.

**Solution:**

**In the case of M15 concrete,** the ratio of cement, sand, and aggregate is 1: 2: 4.

Dry volume of concrete = wet volume of concrete × 1.54

Hence, the dry volume of concrete is = 1 × 1.54 = 1.54 m^{3}

➤ **The required quantity of cement** is = (Cement ratio **÷** Sum of all ratio) × Total volume of concrete = [latex]\frac{1}{(1 + 2+ 4)} \times 1.54[/latex] = 0.22 m^{3}

0.22 m^{3} cement means 0.22 × 1440 = 316.8 kg cement.

In general, each bag of cement weight is 50 kg.

So, the required number of cement bags are = 316.8 **÷** 50 = 6.33 ≈ 7 No.s.

➤ **Quantity of sand required** = ( Sand ratio **÷** Sum of all ratio) × Total volume of concrete = [latex]\frac{2}{(1 + 2 + 4)} \times 1.54[/latex] = 0.44 m^{3}

Our total quantity of sand is 0.44 in m^{3}, but in CFT(cubic feet) it will be ( 0.44 × 35.31) = 15.53 ≈ 16 CFT( cubic feet) [** Note**: 1 m^{3} = 35.31 CFT or cubic feet ]

➤ **Quantity of aggregate** is = ( Aggregate ratio **÷** Sum of all ratio) × Total volume of concrete = [latex]\frac{4}{(1 + 2 + 4)} \times 1.54[/latex] = 0.88 m^{3}.

We know, 1 m^{3} is equal to 35.31 CFT. Therefore, 0.88 m^{3} aggregate means 0.88 × 35.31 = 31.07 ≈ 32 CFT.

## Calculate the Quantity of Cement, Sand & Aggregate in M10 Concrete.

Now, we will calculate the quantity of cement and aggregate in M10 Concrete. Let’s take an example,

**Example:**

Let’s assume that the volume of total concrete(wet concrete) required for a particular job is 3 m^{3}. Now find out the quantity of cement, sand, and aggregate.

**Solution:**

**In the case of M10 concrete,** the ratio of cement, sand, and aggregate is 1: 3: 6.

Dry volume of concrete = wet volume of concrete × 1.54

Hence, the dry volume of concrete is = 3 × 1.54 = 4.62 m^{3}

➤ **The required quantity of cement** is = (Cement ratio **÷** Sum of all ratio) × Total volume of concrete = [latex]\frac{1}{(1 + 3+ 6)} \times 4.62[/latex] = 0.462m^{3}

0.22 m^{3} cement means 0.462× 1440 = 665.28 kg cement.

In general, each bag of cement weight is 50 kg.

So, the required number of cement bags are = 665.28 **÷** 50 = 13.3 ≈ 14 No.s.

➤ **Quantity of sand required** = ( Sand ratio **÷** Sum of all ratio) × Total volume of concrete = [latex]\frac{3}{(1 + 3 + 6)} \times 4.62[/latex] = 1.386 m^{3}

Our total quantity of sand is 1.386 m^{3} , but in CFT(cubic feet) it will be ( 1.386 × 35.31) = 48.93 ≈ 49 CFT( cubic feet) [** Note**: 1 m^{3} = 35.31 CFT or cubic feet ]

➤ **Quantity of aggregate** is = ( Aggregate ratio **÷** Sum of all ratio) × Total volume of concrete = [latex]\frac{6}{(1 + 3 + 6)} \times 4.62[/latex] = 2.77 m^{3}.

We know, 1 m^{3} is equal to 35.31 CFT. Therefore, 2.77 m^{3} aggregate means 2.77 × 35.31 = 97.8 ≈ 98 CFT.

**Read More:**

Advantages and Disadvantages of Concrete

Types of Aggregates

Different Types of Cement Tests