How to Calculate Column Load, Reinforcement, and Section Sizing? - Lceted

Column Load Calculation: Reinforcement and Section Sizing

The column is a vertical structural element meant to carry axial loads from slabs, beams, and walls. Proper column design refers to making sure that the building can hold loads without collapsing. In the paper, we will calculate the axial load a column needs to withstand, compute the reinforcement needed, and dimension the column section 

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1. Loads on Columns
Columns normally have two kinds of loads:

• Vertical Load Axial Loads: These are loads coming vertically from slabs, beams, and other structural members.

• Horizontal Forces Bending Moments: These are actually forces and normally induced by wind or seismic activities.

To make this article, primarily focus majorly on axial loads.

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2. Explain Axial Load Calculation
To determine the total load a column has to carry, we add loads transferred to it, dead load, live load, and beam loads.

(a) Dead Load (DL)
The dead load is the weight of permanent building components. In columns, the dead load of slabs, beams, and walls carried by the column above are calculated.

• Slab Load (W_slab):
We use to calculate the load from the slab:
W_slab = Slab area × Slab thickness × Unit weight of concrete

Example: For a 6m × 6m slab with a 150mm thickness and concrete density of 25 kN/m³, Then
W_slab = 6 × 6 × 0.15 × 25 = 135 kN

If the column supports one-quarter of the slab area,
W_slab_column = 135 / 4 = 33.75 kN

Beam Load (W_beam):
Given a beam weight 50 kN, half its weight transmitted down to the column.
W_beam_column = 50 / 2 = 25 kN

 

(b) Live Load (LL)
The live loads depend on the application of the building (residential, office, etc.). The live load is calculated as;

• Live Load (W_live)
W_live = Slab area x Live load per unit area

Example:
Area: 6m × 6m : Live load = 2 kN/m²
W_live = 6 × 6 × 2 = 72 kN

On the column on which quarter load is supported by the slab,
W_live_column = 72 / 4 = 18 kN

The live loads of 2 kN/m² are based on the general requirements that have been set in the building codes. For example, IS 875: Part 2 outlines the live loads of residential, commercial, and office buildings.

Live Load Standards:

• For dwelling, the live load is considered to be 2 kN/m² for rooms and hallways due to people and furniture.

• For office types, for example, the live load goes between 2.5 kN/m² to 4 kN/m², considering the function: e.g., office area, corridor, or conference.

In the above example, a standard value of 2 kN/m² is often used in residential buildings, as occupancy and utilization of space through furniture, people, and what not is moderate. Live loads would be a reasonable estimate and ensure that such everyday loads would not cause the column to cave in.

If it is an office building, you could enhance the live load to 3 or 4 kN/m² according to the use of floors or building usage as per IS 875.

Always refer to relevant building codes (IS, Eurocodes etc.) for the exact values according to usage.

 

(c) Wall Load
Compute the load from the walls that the column supports as follows:

• W_wall :
W_wall = Length of wall × Height of wall × Thickness of wall × Unit weight of brick

For example, for a 3m length, 3m high and 230mm thick wall and brick density is 20 kN/m³:
W_wall = 3 × 3 × 0.23 × 20 = 41.4 kN

(d) Total Axial Load (P)
Add all loads together to obtain total axial load in column
P = W_slab_column + W_beam_column + W_live_column + W_wall

P = 33.75 + 25 + 18 + 41.4 = 118.15 kN

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3. Column Section Sizing
The next step is to determine the required cross-sectional area of the column. The required area is calculated as:
A_g = P / f_c

Where:
• P = Total axial load on the column

• f_c = Permissible stress in concrete (N/mm²)

For P = 118.15 kN and f_c = 20 N/mm² (M20 concrete),
A_g = 118.15 × 1000 / 20 = 5,907.5 mm²

Therefore the column gross cross-sectional area should be at least 5,908mm².

Choosing Column Dimensions
We now choose column dimensions to satisfy this requirement. Usual column dimensions are limited by practical considerations (for example, architectural or structural). Let us assume we choose a column section of 250mm × 250mm:

A_actual = 250 × 250 = 62,500 mm²

Since A_actual > A_g, the column section is adequate to support the axial load.

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4. Reinforcement Design
Steel reinforcement is provided to resist both axial and bending stresses. The following expression can be made use of to get the quantity of steel reinforcement:
A_s = (P − A_g × f_c) / f_y

Where,
• P = Axial load on the column

• A_g = Gross area of the column

• f_c = Concrete strength (N/mm²)

• f_y = Yield strength of steel, say for Fe 415 = 415 N/mm²

For P = 118.15 kN, A_g = 5,908 mm², f_c = 20 N/mm², and f_y = 415 N/mm²:
A_s = (118.15 × 1000 − 5,908 × 20) / 415

A_s = (118,150 − 118,160) / 415 = −10 / 415 = 0 mm²

This expression indicates that the concrete itself could be sufficient for axial load but conventional reinforcements are given to resist bending moment and also for safety purposes. Now let's take minimum reinforcement for IS Code.

 

Minimum Reinforcement
According to IS 456: 2000, minimum reinforcement percentage lies in between 0.8% and 6% of the cross sectional area of column:
A_s_min = 0.8% of A_g = 0.8 / 100 × 5,908 = 47.26 mm²

This is the minimum reinforcement area for column, and this should be distributed as bars.

Bar Size and Spacing
Let's use 4 bars of 12mm diameter:
Area of 1 bar = π / 4 × d²
Area = π / 4 × (12)² = 113 mm² per bar

For 4 bars:
A_s_total = 4 × 113 = 452 mm²

This area is adequate since it surpasses the minimum required area of 47.26 mm².

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5. More Considerations
Lateral Reinforcement (Ties/Stirrups)
Stirrups restrain the main reinforcement bars against buckling. The diameter and spacing of the stirrup are based on the shear forces and lateral loads.

 

Bending Moments and Seismic Design
Columns in seismic areas are designed to resist lateral forces. Further detailing will be needed in order to have ductility with closely spaced stirrups or increased reinforcement ratios.

 

Factors of Safety
The safety factors both in the strength of steel and concrete account for uncertainties involved in material properties, construction practices, and load estimations in actual design.

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Conclusion
Column design: axial loads, cross-sectional area for reinforcement. Calculations of the dead load, live load, and other factors ensure that columns are appropriately sized as well as suitably reinforced to maintain stability in the building. Reinforcement calculations include bending moments, lateral forces, and stresses in the column. IS codes ensure safety and economy in design to meet standard requirements.

 

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Example Summary:
• Total Load on Column: 118.15 kN

• Minimum Gross Area: 5,908 mm²

• Actual Area: 250mm × 250mm = 62,500 mm²

• Minimum Reinforcement: 47.26 mm²

• Provided Reinforcement: 452 mm² (4 bars of 12mm)

 

For Knowledge we added below the design of Bending Moment calculation in column.

Design for Bending Moments

The column must be designed not only to carry axial loads but also to resist the bending moments. The design process involves:

1. Calculating the Required Area of Steel (A_s): The area of steel required to resist bending can be calculated using the formula:

A_s = M / (0.87 × f_y × z)

Where:

• M = Bending moment (N·mm)
• f_y = Yield strength of steel (N/mm²)
• z = Lever arm (approximately 0.87d, where d is the effective depth of the column)

Example Calculation: Assuming f_y = 415 N/mm² and an effective depth d = 400 mm:

A_s = (30 × 10^6) / (0.87 × 415 × 0.87 × 400) ≈ 119.54 mm²

2. Minimum Reinforcement Requirements: According to IS 456: 2000, the minimum area of tensile reinforcement should not be less than:

A_s_min = 0.12% × A_g

3. Distribution of Reinforcement: If significant bending moments are expected, additional reinforcement bars may be added to the tensile side of the column to enhance its resistance to these forces. The arrangement should be balanced with adequate ties/stirrups to prevent buckling.

2. Additional Considerations for Bending Moments

• Lateral Reinforcement (Stirrups/Ties): Stirrups provide lateral support for the main reinforcement and help prevent buckling under compression. They also help resist shear forces that may occur due to bending moments. The spacing and size of stirrups can be calculated based on the shear force acting on the column.
• Ductility in Seismic Design: In regions prone to seismic activity, columns must be designed with ductility in mind. This includes using higher reinforcement ratios, closely spaced stirrups, and possibly larger cross-sections to accommodate greater lateral forces without failure.
• Use of Concrete Grades: The grade of concrete used also influences the moment capacity of the column. Higher-grade concrete can provide better resistance to bending moments due to its higher compressive strength.