 ## How to Calculate Rcc Column Size

The RCC column size calculation depends on the loads on the column. For the calculating size of the column, you need to know the following:

• Factored load on the column
• The grade of steel, and
• The grade of concrete. After which you can put the value of parameters in the factored load formula. This formula is used for calculating the size of the square, rectangular and circular column. The minimum size of the column should not be less than 230 mm x 230 mm.

Moreover, With this formula, you can also calculate the bearing capacity of the column and area of steel on the column, which further helps you in calculating steel ratio in the reinforced concrete column. If you have not read our post of how to calculate RCC beam size, then skim it quickly as it will help to understand this post more promptly.

## WHAT IS COLUMN?

It is a vertical element or pillar of the structure which transmit loads of slab and beam to the footing of the structure for proper distribution or balancing the loads through compression on the weight of the structures. They are different other kinds of column design. Example, a small support wooden or metal is typically called a post and support with a rectangular, or other non-round section are usually called piers.

Size of Column:

Column sizes for different stories –

NOTE: These are the standard RCC column sizes, but it will increase or decrease with respect to the load.

We can calculate the column size according to the load on the column:

This formula used for column design calculations, with the help of this formula we will find out the size of the column, and You can also calculate the bearing capacity of the column to put the value of column size.

According to IS 456:2000 clause 39.3, page no.71

Pu = 0.4 fck Ac + 0.67 fy Asc

Assume percentage of steel = 1%

Ac = Ag – Asc

Asc = 0.01 Ag

Then, Ac = Ag – 0.01 Ag

Ac = 0.99 Ag

Where,

Pu = factored load on column (Pu = 1.5 x P)

Ac = area of concrete

Asc = area of steel

Fy = yield strength of steel

Fck = compressive strength of concrete

Ag = area of section (size of column)

There are some examples of RCC column design calculation.

Example-

If Pu = 1420 KN, fck = 25 N/mm2, fy = 415 N/mm2

Assume percentage of steel = 1%

Pu = 0.4 x 25 x (0.99Ag) + 0.67 x 415 x 0.01Ag

Pu = 9.9 Ag + 2.78 Ag

1420 x 10^3= 12.68 Ag

Ag = 1420 x 10^3/12.68

You can choose rectangular square or circular column as per requirement

For rectangular column

Area of rectangle = l x b

111987.38 = l x 230 (Assume B = 230 mm)

l = 486.90 mm ≈ 490 mm

For circular column, you can use the area of the circle

Area of circle = π r^2

There is some column design example mentioned in steps below which can help you to check the compression, to determine whether the column is short or long and a check on eccentricity. For this, you need some design codes like IS 456:2000 and SP 16 for column design chart

After you determine the size of the column, a grade of steel and grade of concrete, you can use many different structural design software available in the market which can help you to design the building like STAAD Pro, ETABS, TEKLA Structural Designer and others for building design.

Compression and Buckling check:

Compression:

σ allowable = P/ A

Where,

σ allowable = allowable stress on column

A = cross- sectional area of column

If  σ allowable > σ applied  then column will be SAFE

Here, σ applied = applied yield strength of steel like Fe- 250, Fe – 415, Fe – 500

Example-

Here,

P = Pu /1.5

P = 1420/1.5

P = 946.67 KN

as we know, σ = P / A

σ allowable = 8.4 N/mm2

here,

σ applied or f= 415 N/mm2

So,

σ applied > σ allowable, SAFE

Buckling:

Where,

E = young modulus of concrete or steel

For steel (E= 2 x 105 N/mm2)

I = moment of inertia

Moment of Inertia for rectangular shape

If Per > P then column will be safe from buckling

Example –

We are taking grade of concrete M- 25

So, fck = 25 N/mm2

E = 5000(25)1/2

E = 25000 N/mm2

For rectangular column

I = bd3/12

I = 230 x 4903 /12

I = 2254.93 x 106 mm4

Therefore,

Per = 92691657.86N

Per = 92691.65 KN

Here, P = 946.67 KN

So,

Per > P, SAFE

IF Per < P the size of the column is not suitable to bear the load.

Then we will change the size of the column and repeat all the steps.

K value for column buckling:

Therefore,

Effective length = L x K

Example

Assume length of column or height = 3.5 m

And fixed from both ends

Then,

Effective length = 3.5 x 0.7

Slenderness ratio(λ):

If λ > 12 then the column is long column

If λ < 12 then short column

You can identify the column is short column or long column

Example-

Assume effective length = 2450 mm and least lateral dimension = 230

Then, this is a short column

Eccentricity:

For minimum eccentricity

For maximum eccentricity

emax = 0.05 B or D

If emax > emin then column will be safe

Example –

For minimum eccentricity

L = 3500 mm and B = 230 mm, D = 490 mm

emin = L /500 + B or D /30

emin = 3500/ 500 + 490 /30

emin = 23.33 > 20

then,

emin  = 23.33

For maximum eccentricity

emax = 0.05 x 490

emax = 24.5

here,

max  > emin, SAFE

Span between two columns: it  is more important to know about the maximum and minimum distance between two columns.

Hope this information on calculating the RCC column size will help you in preparation and study of structure analysis. If you have any doubts in solving problems of calculating the RCC column size or any fundamental topics related to structural analysis, then feel free to write to me!

And if you like my articles and efforts, please tell us through commenting below. We will be pleased to get your valuable feedbacks. Like the column concrete calculation, Eye on structures will try to bring you concepts of structural engineering, structural design through software and real-life structure issues and possible solutions for them. So get ready to build up your strong structure knowledge with us. You can also join us through our social media handles on Instagram and Facebook for getting daily doses of information related to impressive structures and their concepts. Happy reading!

### This Post Has 10 Comments

1. Barebente Pascal

It’s been excellent and educative to visit you here. I would like to know and get more from you. The vast knowledge is enriching.
Thank you
Pascal Barebente – Uganda Kampala

2. Thank you for the appreciation. 🙂

3. Minu Sharma

This is very helpful and easy to understand. We are very thankful to you for giving your valuable time and attention towards this. Keep it up👍 Pls also share some designing tools link or name if possible..

1. Thank you for such kind words 🙂

4. Abhinav Awchitte

Thank you so much for brief and easy explaination of problem.i would love to understand more concepts and more knowledge from you.
But i have one question
Assume percentage of steel = 1%

Pu = 0.4 x 25 x (0.99Ag) + 0.67 x 415 x 0.01Ag

Pu = 9.9 Ag + 2.78 Ag

1420 x 103= 12.68 Ag

Ag = 1420 x 103/12.68
At this example what is 103 value and where it comes from??
1. 