** 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.**

**To go through all types of structural design articles,** **click here.**

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

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