For the total load calculation on Columns, Beams, and Slab we must know about various loads coming on the column. Generally, the Column, Beam, and Slab arrangement are seen in a frame type of structure.

In the frame structure, the load is transferred slab to beam, beam to column and ultimately it reached the foundation of the building. For column design calculation, a loads on the following elements are to be calculated,

**What Is Column**

Column length is generally 3 times their least lateral cross-sectional dimension. The Strength of any column mainly depends on its shape and size of cross-section, length, location, and position of the column.

A Column is a vertical component in a building structure, which is mainly designed to carry the compressive and buckling load. The column is one of the important structural members of the building structure. As per the Load coming on the column, size is increased or decreased.

**Load Calculation on Column**

## What Is Beam?

The Beam is a horizontal structural member in building construction, which is designed to carry shear force, and bending moment, and transfer the load to columns on both ends of it.

Beam’s bottom portion experiences tension force and upper portion compression force. Therefore, More steel reinforcement is provided at the bottom compared to the top of the beam.

## What is Slab?

The slab is a level structural element of the building that is provided to create a flat hard surface. These flat surfaces of slabs are utilized for making floors, roofs, and ceilings.

It is a horizontal structural member whose size may vary depending upon the structure size and area and its thickness** **also may vary.

But the slab minimum thickness is specified for normal construction around 125 mm. Generally, every slab is supported by a beam, column, and wall around** **it.

**Load On Column, Beam & Slab**

**1) Column Self Weight X Number of floors**

**2) Beams Self Weight per running meter**

**3) A load of walls per running meter**

**4) The total load on Slab (Dead load + Live load + Self-weight)**

Besides this above loading, the columns are also subjected to bending moments that have to be considered in the final design.

The most effective method for designing structure is to use advanced structural design software like ETABS or STAAD Pro.

These tools are reduced laborious and consuming methods of manual calculations for structural design, this is highly recommended nowadays in the field.

for professional structural design practice, there are some basic assumptions we use for structural loading calculations.

Read More: Steel Quantity Calculation Excel Sheet

**Column Design Calculation** [pdf]

** 1. Load Calculation on Column **

we know that the Self-weight of Concrete is around 2400 kg/m3, which is equivalent to 240 kN and the Self-weight of Steel is around 8000 kg/m3.

So, if we assume a column size of 230 mm x 600 mm with 1% steel and 3 meters standard height, the self-weight of the column is around 1000 kg per floor, which is equal to 10 kN.** **

- Volume of Concrete = 0.23 x 0.60 x 3 =0.414mÂ³
- Weight of Concrete = 0.414 x 2400 = 993.6 kg
- Weight of Steel (1%) in Concrete = 0.414x 0.01 x 8000 = 33 kg
- Total Weight of Column = 994 + 33 = 1026 kg = 10KN

While doing column design calculations, we assume the self-weight of columns is between **10 to 15 kN per floor.**

**2. Beam Load Calculation**

We adopt the same method of calculations for beams** **also.

we assume each meter of the beam has dimensions of 230 mm x 450 mm excluding slab thickness.

Assume each (1m) meter of the beam has a dimension

230 mm x 450 mm excluding slab.

Volume of Concrete = 0.23 x 0.60 x 1 =0.138mÂ³

Weight of Concrete = 0.138 x 2400 = 333 kg

Weight of Steel (2%) in Concrete = 0.138 x 0.02 x 8000 = 22 kg

Total Weight of Column = 333 + 22

So, the self-weight will be around **3.5 kN** per running meter.

**3. Wall Load Calculation**

we know that the Density of bricks varies between 1500 to 2000 kg per cubic meter.** **

For a 6-inch thick Brick wall of 3-meter height and a length of 1 meter,

The** **load / running meter to be equal to 0.150 x 1 x 3 x 2000 = 900 kg,

which is equivalent to **9 kN/meter. **

This method can be adopted for load calculations of Brick per running meter for any brick type using this technique.

For aerated concrete blocks and autoclaved concrete blocks, like Aerocon or Siporex, the weight per cubic meter is between 550 to 700 kg** **per cubic meter.

if you are using these blocks for construction, the wall loads per running meter can be as low as **4 kN/meter**, the use of this block can significantly reduce the cost of the project.

### 4. **Slab Load Calculation**

Let, Assume the slab has a thickness of 125 mm.

So, the Self-weight of each square meter of the slab would be

= 0.125 x 1 x 2400 = 300 kg which is equivalent to 3 kN.

Now, If we consider the Finishing load to be 1 kN per meter and the superimposed live load to be 2 kN per meter.

So, from the above data, we can estimate the slab load to be around **6 to 7 kN per square meter.**

**5. The Factor of Safety**

In the end, after calculating the entire load on a column, do not forget to add in the factor of safety, which is most important for any building design for the safe and convenient performance of the building during its design life duration.

This is important when Load Calculation on Column is done.

As Per IS 456:2000, the factor of safety is 1.5.

how to calculate the load of a building pdf download

## How to Calculate Column Size For Building

A** **column is one of the important elements of any building structure. The column size for the building is calculated as per load coming on the column from the superstructure.

For buildings with heavy loading conditions, the column size is increased. The column size is an important factor while designing any building structure.

**Difference column sizes used in building design**,

- 9″ x 9″
- 9″ x 12″
- 12″ x 12″
- 12″ x 15″
- 15″ x 18″
- 18″ x 18″
- 20″ x 24″
- As per Structural load, more sizes

For Column size calculation we required the following data,

- Grade of Steel
- Grade of Concrete
- Factored Load on Column

(Note: Minimum size of the column should not be less than 9″ x 9″ ( 230 mm x 230 mm)

The following are column design calculation steps to decide the size of the column for the building.

**Pu = 0.4 f _{ck} A_{c }+ 0.67 f_{y} A_{sc} ( Clause No: 39.3 Page No: 71 IS 456:2000)**

**Pu = Axial Load on Column****f**_{ck}= Characteristics compressive strength of concrete**A**_{c}= Area of Concrete**f**_{y}= Characteristics Tensile strength of concrete**A**_{sc}= Area of Steel Reinforcement**A**_{c}= A_{g}– A_{sc}**A**_{sc}= 0.01 A_{g}**A**_{c}= 0.99 A_{g}

**Where A _{g} = Gross Area of Column**

**Consider 1% of Steel in Column,**

**A _{c} = A_{g â€“} A_{sc}**

**Example:** Design an RCC square short column subjected to an axial compressive load of 600 KN. The grade of concrete is M -20 and the Grade of steel is Fe -500. Take Steel 1% and Factor of safety = 1.5.

**Pu = 600 KN, f _{ck} = 20 N/mm^{2}, f_{y} = 500 N/mm^{2}, Steel = 1%, Factor of Safety = 1.5**

**Pu = Axial Compressive Load on Column = 600 KN**

**Factored load on column = Pu = 600 x 1.5 = 900 KN**

**P _{u }= 0.4 f_{ck} A_{c }+ 0.67 f_{y} A_{sc}**

**900 x 10 ^{3} = 0.4 x 20 x (0.99 A_{g}) + 0.67 x 500 x (0.01 A_{g})**

**900 x 10 ^{3} = 7.92 A_{g} + 3.35 A_{g} **

**900 x 10 ^{3} = 11.27 A_{g} **

** A _{g} = 79858 mm^{2}**

**For Square Column**,

**Size of Column** = âˆš79858

**Size of Column **= 282.59 mm

**Provide square column size 285 mm x 285 mm**

A_{g} = Provided = 81225 mm^{2}

A_{sc} = 0.01 A_{g} = 0.01 x 81225

** A _{sc} = 812.25 mm^{2}**

Provide** **8 Nos of 12 mm Dia steel with an area of steel = 905 mm^{2}

The size of the column for 600 KN load is 285 mm x 285 mm (12″ x12″)

**Watch Video:** **Load Calculation on Column**

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Md Khaibul AlamVery useful tips its help every engineer in their field.please send that type of tips for civil engineering frequently.

Gopal Behera#Useful

Habibur rahmanIndeed these tips are very useful & helpful in the working field

Abdul Ajeesif we have 4 columns in a floor , should we multiple 10kn x 4 columns to find load? or we can put 10 kN for one floor randomly?

DheerajInterestingly this article has the same errors as seen on another post from 2011

Manjunath hukkeriSeriously I just loved the explanation

U explained each and every bit of it very cleanly

Thanks for this article sir

bodrick ikumbukothanks for the help i really needed

V Ranjith Kumargood

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