Design of a column./ How to design the diameter & no. of reinforcement bars required in a column?/ How to calculate the dia. & no. of bars in a column?

 Eg:

Calculate the diameter & no. of steel bars required in a column of dimension 225mm. x 300mm. The column has an axial load of 570KN. M25 grade concrete & Fe-415 steel are used in the column.

Given data:

Axial load over column = 570KN = 570 x 10³N

Column size = 225mm. x 300mm.

Characteristic compressive strength of concrete=fck=25N/mm²

Yield strength of steel = fy= 415 N/mm².

Calculation:

Ⅰ. Longitudinal reinforcement:

As per IS code

Pu = [0.4 x fck x Ac + 0.67 x fy x Asc] -----------①

Where,

Pu = Ultimate factored load 

     = [factor of safety x axial load over column]

     = [ 1.5 x 570 x 10³N]

 Pu = 855 x 10³N.

Ac = Sectional area of concrete

      = [Ag - Asc]

Here,

 Ag = Gross sectional area of the column

       = [225mm. x 300mm.]

       = 67,500mm²

Asc = Sectional area of steel =?

Putting all the values in formula ①

    Pu = [0.4 x fck x (Ag- Asc) + 0.67 x fy x Asc]

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

855 x 10³ =[6,75,000 - 10Asc + 278Asc]

855 x 10³ = 6,75,000 +268Asc

268Asc = [8,55,000 -6,75,000]

268Asc = 1,80,000

Asc = [1,80,000 ➗ 268]

    Asc = 671.6 mm²

Let us provide a minimum size diameter bar as per IS code, i.e. 12mm.

The sectional area of 1 bar 

=[ π x d² ➗ 4]

= [  3.142 x 12² ➗ 4]

= [452.448 ➗ 4]

= 113.11 mm².

No. of rebars required

= [Asc ➗ area of 1 bar]

= [671.6 ➗ 113.11]

= 5.94 nos. 

By rounding off, the no. of bars required = 6nos.

Now, 

 The actual sectional area of steel

   = [6nos. x 113.11]

   = 678.66 mm².

Ⅱ. Transverse reinforcement:

1. Stirrup ( lateral tie) diameter:

As per IS code,

The diameter of lateral tie bars should be at least 1/4th the dia. of the main bars.

i.e. = [1/4 x 12mm.]

      = 3mm.

So, we can provide a 6mm. or 8mm. dia. lateral tie which is > 3mm., & hence safe.

2. Spacing of stirrup:

The c/c spacing of lateral ties shall not be more than the least of the following three criteria.

a. The least sectional dimension of the column 

      = breadth of the column 

      = 225mm.

b. 16 times the smallest dia. bar of the longitudinal reinforcement

  = [16 x 12mm. ]

  = 192mm.

   &

c. 300mm.

The least of all the 3 conditions is 192mm.

Hence, we can provide c/c spacing of lateral tie as 192mm.

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Thank you for going through these calculation steps❤. Have a good day 😄.

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