Concrete column plays an important role in the framed structure to transmit loads from slabs to the foundation. Hence due importance should be given while constructing the column.

To construct a strong column, one should know the optimum quantities of materials required so any mistakes like shortage of materials or excess of materials shall be avoided during the construction process.

In this article, let’s see how we can calculate the quantity of concrete required for constructing the columns.

There are two calculation methods to find the materials required. They are,

- Manual calculation
- Automatic excel calculation

Let’s go through one by one.

**1) Manual calculation**

There are different shapes of columns available. Some of the common shapes of columns generally used in the construction are,

- Round/Circular,
- Rectangular and
- Square.

Let’s discuss the materials required for each of them in detail.

**a) Materials required for a round/circular concrete column**

**Let’s assume you are planning to construct a column of diameter 500 mm and height 3m using M20 grade concrete.**

**Step 1:** Calculate the volume of the column which you plan to construct.

Where,

- r = radius of the column
- d = diameter of the column
- h = height of the column

Hence, volume of the given column = pi x (0.5/2)^{2} x 3 = 0.589 m3.

The volume we derived here is the wet volume of the concrete. But we want to measure materials in their dry state, hence to convert from wet volume to dry volume, add **54% extra** of the wet volume.

Dry volume of concrete required = 0.589 x 1.54 = 0.907 m3.

**Step 2: **Adopt the grade of concrete you plan to use to construct a column.

Here we use, the M20 grade of concrete. The ratio of M20 grade of concrete is **1: 1.5: 3.** (cement: Fine Aggregate(FA): Coarse aggregate(CA))

**Step 3: **Calculate the quantity of cement

Volume of cement required = 0.907 m3 x (1/5.5) = 0.1649 m3.

To convert into Kg, multiply the above by the **unit weight of cement which is 1440 kg/m3.**

Quantity of cement required = 0.1649 x 1440 = 238 kg.

Number of cement bags required = Total Quantity of cement required / weight of one cement bag

If the weight of one cement bag is 50kg then,

**Number of cement bags required** = 238/50 = 4.76 ≈ 5 bags.

**Step 4: **Calculate the quantity of Aggregates

Volume of FA (sand) req. = 0.907 x (1.5/5.5) = 0.2473 m3.

To convert m3 to cft, multiply it by **35.31**. (m3 = cubic meter, cft = cubic feet)

**Volume of FA (sand) req.** = 0.2473 m3 (or) 0.2473*35.31 = 8.73 cft.

**Volume of CA (gravel) req.** = 0.907 x (3/5.5) = 0.494 m3.

Therefore, the materials required to construct the column of the given dimensions are,

**Number of cement bags required = **5 bags.

**Volume of FA (sand) req.** **= **0.2473 m3 or 8.73 cft.

**Volume of CA (gravel) req.** **=** 0.494 m3.

**b) Materials required for a Rectangular concrete column**

**Let’s assume you are planning to construct a column of cross section 0.30m x 0.22m and height 3m using M20 grade of concrete.**

Volume of Rectangular column = L x B x H = 0.30 x 0.22 x 3 = 0.198 m3.

Dry volume of concrete required = 0.198 x 1.54 = 0.30492 m3

Volume of cement required = 0.30492 m3 x (1/5.5) = 0.05544 m3.

Quantity of cement required = 0.05544 x 1440 = 80 kg.

**Number of cement bags required** = 80/50 = 1.6 ≈ 2 bags.

**Volume of FA (sand) req.** = 0.30492 x (1.5/5.5) = 0.08316 m3 x 35.31 = 2.93 cft.

**Volume of CA (gravel) req.** = 0.30492 x (3/5.5) = 0.16632 m3.

**C) Materials required for a Square concrete column**

**Let’s assume you are planning to construct a column of cross section 0.30m x 0.30m and height 3m using M20 grade of concrete.**

Volume of column = L x B x H = 0.30 x 0.30 x 3 = 0.27 m3.

Dry volume of concrete required = 0.27 x 1.54 = 0.4158 m3

Volume of cement required = 0.4158 m3 x (1/5.5) = 0.0756 m3.

Quantity of cement required = 0.0756 x 1440 = 109 kg.

**Number of cement bags required** = 109/50 = 2.18 bags.

**Volume of FA (sand) req.** = 0.4158 x (1.5/5.5) = 0.1134 m3 x 35.31 = 4 cft.

**Volume of CA (gravel) req.** = 0.4158 x (3/5.5) = 0.2268 m3.

**2) Automatic Excel calculation**

Even though the above calculation method is pretty simple and easy, it becomes tedious when you have too many columns of various cross sections, heights, and grades. In that situation, an automatic Excel calculator can be incredibly helpful.

Just enter the input and get the results within seconds.

Download the Excel file here.

**Frequently Asked Questions** **(FAQ’s)**

**1) The above calculations are made for a single column, what if I had to calculate materials for several columns?**

Just multiply the number of columns by the wet volume of concrete required and follow the same process as mentioned above.

For example,

**Let’s assume you are planning to construct 8 columns of cross section 0.30m x 0.30m and height 3m using M20 grade of concrete.**

Volume of single column = L x B x H = 0.30 x 0.30 x 3 = 0.27 m3.

Volume of all columns = Volume of single column x Number of columns = 0.27 x 8 = 2.16 m3.

Dry volume of concrete required = 2.16 x 1.54 = 3.3264 m3

Volume of cement required = 3.3264 m3 x (1/5.5) = 0.6048 m3.

Quantity of cement required = 0.6048 x 1440 = 870 kg.

**Number of cement bags required** = 870/50 = 17.4 bags.

**Volume of FA (sand) req.** = 3.3264 x (1.5/5.5) = 0.9072 m3 x 35.31 = 32 cft.

**Volume of CA (gravel) req.** = 3.3264 x (3/5.5) = 1.8144 m3 x 35.31 = 64 cft.

**2) The above calculations are made for columns with the same height or cross-section, what if I had columns of varying cross-section and height?**

**Let’s assume you are planning to construct 8 columns of cross section 0.30m x 0.30m with a height of 3m and 2 columns of cross section 0.20m x 0.20m with a height of 2m using M20 grade of concrete.**

Let’s calculate,

Volume of 0.30m cross-section columns = 0.3 x 0.3 x 3 x 8 = 2.16 m3.

Volume of 0.20m cross-section columns = 0.2 x 0.2 x 2 x 2 = 0.16 m3.

The total volume of all columns = 2.16 + 0.16 = 2.32 m3.

Dry volume of concrete required = 2.32 x 1.54 = 3.5728 m3

Volume of cement required = 3.5728 m3 x (1/5.5) = 0.6496 m3.

Quantity of cement required = 0.6496 x 1440 = 936 kg.

**Number of cement bags required** = 936/50 = 18.72 bags.

**Volume of FA (sand) req.** = 3.5728 x (1.5/5.5) = 0.9744 m3 x 35.31 = 34.4 cft.

**Volume of CA (gravel) req.** = 3.5728 x (3/5.5) = 1.9488 m3 x 35.31 = 68.8 cft.

Hope you understand everything you need to know about the **column concrete estimation**. If you have any queries (or) if you find this article helpful, let us know in the comment section.

**Read more:**

15 Types of Cement and Their Uses in the Construction Industry

How to Check the Quality of Cement at the Construction Site?

How Many Yards Are in a Ton of Gravel? A Comprehensive Guide

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Wanyama InnocentWow

This is the best

Thanks so much for coming up with this amazing and helpful site.

Do u have an app?

AdminThank you so much for your appreciation. At this time, we don’t have an official app, but it’s on our radar for the future. Stay tuned!

Eze EmmanuelLove this, thanks❤