Example :- Design RCC column which carries 1200kN load with 3.5 m length. Assume M20 grade concrete and Fe415 grade steel

Example :- Design RCC column which carries 1200kN load with 3.5 m length. Assume M20 grade concrete and Fe415 grade steel

Since given load is 1200kN we need to make that to factored load

Factored load = 1.5X1200 =1800 KN

Step 1:  Calculation of Ac by assuming Asc

By assuming the percentage of steel as 1% of the gross area

Asc =(1/100)Ag

So Asc = 0.01Ag

And also we have Ag = Ac + Asc

By substituting the Asc value we can easily get Asc

So Ac = Ag-0.01Ag

So finally Ac = 0.99Ag

Step 2: Calculation of dimensions of the column by using Ag

Since we have

Pu=0.45fckAc+0.67fyAsc

By substituting the values

1800X10³=0.45X20X0.99Ag+0.67X415X0.01Ag

By calculation we can get Ag = 153971.173 mm²

By taking column as square section which is having a side area of the rectangle is a².

So a² = 153971.173 mm²

And Finally a = 392.39 mm

So, Take a as 400 mm.

Now the next step is checking for the e_min

Since

e_min = (L/500) + (D/30)

= (3500/500)  + (400/30)

= 7+13.333

= 20.33 mm

Hence it is safe as per emin

Since the emin value is less than 20 mm so we can proceed to the design section of 400 mm X 400 mm column

Step 3: Calculation of reinforcement values of main steel Ac

Again we will use same Pu formulae for calculating the reinforcement values

Since

Pu=0.45fckAc+0.67fyAsc

By substituting the values

1800X10³=0.45X20X(153971.173-Asc)+0.67X415XAsc

By calculating we can get Asc = 1539.711 mm²

So let us consider 1540 mm² area

By assuming the 16mm diameter bars

The area of 1 single bar is (Π/4)16² = 201 mm²

Number of bars required is given by is given by (Asc/Area of 16mm bar)

N = 1540/201 = 7.661 no’s

Take approximately 8 no’s

So take 8 no’s of 16mm diameter bars as main reinforcement

Step 4: Calculation of transverse reinforcement

In transverse reinforcement initially we need to determine the pitch

Pitch is the least for following 3 values

1. 16d =16X16 = 256mm
2. Minimum lateral dimensions =400mm
3. 300mm

The least of above three values is 256mm  so we will take 256mm is the pitch for the column section design.

Diameter of bars calculation

The diameter of the transverse reinforcement is calculated by using two expressions shown below

1. 0.25d = 0.25 X 16 = 4 mm
2. 5 mm

So the final reinforcement use of 8 no’s of 16mm diameter main reinforcement and 6mm diameter bars of 256 mm center to center distance is used in the 400mmX400mm size with 1200kN point load.

The complete reinforcement details are shown in the below figure