The design of a reinforced concrete (RC) column plays a vital role in ensuring the strength and stability of any structure. Columns carry loads from beams and slabs down to the foundation. In this article, we will discuss how to calculate the diameter and number of reinforcement bars required for a column based on given parameters using IS code design principles.

Given Design Data

Let’s consider a rectangular column with the following specifications:

• Axial Load (P) = 570 kN = 570 × 10³ N
• Column Size = 225 mm × 300 mm
• Concrete Grade (fck) = M25 = 25 N/mm²
• Steel Grade (fy) = Fe-415 = 415 N/mm²

Our goal is to determine the required area of steel (Asc), the number of bars, and the diameter of transverse reinforcement.

Step 1: Calculate Ultimate Load

The ultimate factored load (Pu) is given by:
Pu = 1.5 × Axial Load
Pu = 1.5 × 570 × 10³ = 855 × 10³ N

This factor of safety ensures the column can safely resist unexpected overloads.

Step 2: Formula for Axial Load Capacity

According to IS 456:2000, the axial load on a short column with lateral ties is:

Pu = 0.4 × fck × Ac + 0.67 × fy × Asc …(1)

Where,

• Pu = Ultimate load
• Ac = Area of concrete = Ag – Asc
• Ag = Gross sectional area of column(which is the total geometric area of the column’s shape, including the area of the concrete and the area occupied by the steel reinforcement)

• Asc = Area of steel reinforcement (to be found)

Step 3: Substitute Known Values

Calculate the gross area of the column:
Ag = 225 × 300 = 67,500 mm²

Substitute all known values into Equation (1):

855 × 10³ = [0.4 × 25 × (67,500 – Asc)] + [0.67 × 415 × Asc]

Simplify the equation:

855 × 10³ = 675,000 – 10Asc + 278Asc
855 × 10³ = 675,000 + 268Asc

Now,
268Asc = 855,000 – 675,000
268Asc = 180,000

Asc = 180,000 ÷ 268 = 671.6 mm²

Step 4: Select a Suitable Bar Diameter

As per IS 456, the minimum bar diameter for a column is 12 mm.
The area of one 12 mm bar is:

Area = (π × d²) ÷ 4
= (3.142 × 12²) ÷ 4
= 113.11 mm²

Step 5: Determine Number of Bars

Number of bars = Asc ÷ Area of one bar
= 671.6 ÷ 113.11 = 5.94

By rounding off, the required number of bars = 6-No.

Now, the actual steel area provided is:
6 × 113.11 = 678.66 mm², which is slightly greater than required and therefore safe.

Step 6: Design of Transverse Reinforcement

1. Diameter of Lateral Ties

As per the IS code,
Diameter of tie ≥ ¼ × diameter of main bar
= ¼ × 12 mm = 3 mm

Hence, we can provide 6 mm or 8 mm ties, both satisfying the minimum requirement.

2. Spacing of Lateral Ties

The spacing of ties should not exceed the least of the following three limits:

a) Least lateral dimension = 225 mm,  (Column Size = 225 mm × 300 mm)

b) 16 × diameter of smallest longitudinal bar = 16 × 12 = 192 mm
c) 300 mm

The least value is 192 mm, so we provide lateral ties @ 192 mm c/c.

Final Column Design Summary

For this example, a 225 mm × 300 mm column carrying 570 kN of axial load is safely supported by 6 bars of 12 mm diameter and 6 mm ties at a 192 mm spacing.

Click Here To See How Many Bars In A Column? Essential Guide For Reinforced Concrete Columns