Let us calculate the total load over the columns as shown below.

__Given data:__

Height of column = 2.7m.

Sectional dimension of the column = 230mm x 450mm. = 0.23m. x 0.45m.

No. of columns = 4 nos.

**Calculation:**

Lx /Ly = 4500mm. / 3000mm.

= 1.5 < 2.

Therefore it is a two-way slab.

The load distribution of the **2-way slab** over the beams and columns is as shown below.

**Total load over the columns**

= [Self wt. of the column + superimposed load from the beams]

Here,

**a.) Self wt. of the column**

= [(area of cross-section) × density of RCC]

= [ (0.23m. x 0.45m.) × 25 KN/m³]

= **2.588 KN/m.**

**Factored self wt. of the column**

= [1.5 × 2.588KN/m.]

= **3.88 KN/m.**

**Total factored self-wt. of the column**

**=** [ factored wt./m × height of the column]

= [3.88KN/m × 2.7m.]

= **10.476 KN.**

**b.) Load transferred from beam to the columns.**

From the above drawing, you can observe that the load from beams **B1 & B2** are equally distributed to all 4 columns of the structure.

So, the load transferred over an individual column

= half of the [total load from beam B1 + from beam B2]

**Note:** The above values of the load from the beams are taken from the article **ðŸ‘‡**

**ðŸ‘€. ****How to calculate the total load over the RCC beam?**

= 1/2 × [116.445]

= **58.222KN.**

**Total factored load over the columns**

= [Self wt. of the column + load from the beam]

= [ 10.476 + 58.23]

= **68.706 KN.**

**= (68.706 × 101.97) kg.**

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**Thank you for going through these calculation steps❤. Have a good day ðŸ˜„.**

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