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 the beam to the columns.
From the above drawing, you can observe that the load from beams B1 & B2 is 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 👇
👀. 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.
Thank you for going through these calculation steps❤. Have a good day 😄.
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