How to Design RCC Column .

How to Design Column An RCC Column may be designed in the following step: 

1. Based on the concrete grade to be used, determine the allowable stresses in the concrete, main bars, and ties.

What is Column? - Types of Column, Reinforcement, Design Procedure  

2.Calculate the superimposed load the column is expected to carry. For this, add the column’s dead weight(assumed) to get the total load the column has to take at its base. 

3. Consider some suitable value of reinforcement Asc says between 0.8 to 2% of gross area A of the column.

 Determine area A from the following expression. 

P = σccAc + σscAsc 

P = σcc( Ag – p.Ag) + σsc .pA

 Ag = P/ σcc(1- p)+p.σsc  

Where A = Column’s gross area 

And p = ratio of steel to gross area = Asc /Ag. 

4. After calculating the area Ag, now determine the column’s dimension. If it is square of side b, then b =√Ag. If a circular column of diameter D is to be used, D (=b) will be equal to√4Ag/π.

5. For the known end condition, calculate the column’s effective length lef and calculate the lef /b ratio to decide whether it is a short column or a long column.

6. If lef /b<12, then the column will be designed as a short column for which dimensions have been determined in steps 3 and 4. Calculate the steel area Asc and distribute the bar suitably around the column’s periphery, keeping suitable cover. 

7. If lef /b > 12, then the column will be designed as a long column for which the reduction factor Cr is calculated from the expression, 

Cr = 1.25 – lef /48b

For an equivalent short column, calculate the design load p’. 

P’ = P/Cr 

Using the received value of P’, re-calculate the area Ag and determine the column’s size. Also, determine the steel Area Asc = pAg. 

8. Determine the diameter of bars used as ties and calculate its pitch as per known rules. 

Example – Find the safe load that column can carry for the data given below. 

Effective Length = 4 meter 

Diameter = 400 mm 

Main Bar = 8 nos of 20 mm diameter

Concrete Grade = M 20  

Steel = Fe 415 

Solution –  First we will found out given column is a short or long column. 

Effective length = 4 meter = 4000 mm

Hence l /b = 4000/400 = 10<12 

Hence the column is short. 

The formula for load calculation is 

P = P = σ A +  σ A Area of steel (Asc) =  8π/4 (20) = 2513.3 mm2 

Area of column Ag = π/4(400)2 =  125664 mm2

 Area of Concrete Ac = Ag – Asc = 125664 – 2513.3 = 123151 mm2 

For M 20 Concrete = 5 n/mm2

 For Fe 415 steel = 190 n/mm2 

P = (5 × 123151) + (190 × 2513.3) = 1093281 N or 1093.281KN 

If the effective length is increased to 8 m then what will be the safe load that column can carry?  

For effective length = 8 meter = 8000 mm 

and

D = 400 mm l /b  = 8000/400 = 20 >12 

Thus the column is a long column for which we have to calculate the reduction factor.

C = 1.25 –  l /48b     = 1.25 – 8000/(48 × 400) = 0.8333 

P = 0.8333 × 1093281 = 911067 = 911.067 KN