Let us calculate the total load over beams B1 & B2 as shown below.
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Given data:
A span of beam B1= 5500mm. = 5.5m.
A span of beam B2 = 2500mm. = 2.5m.
Sectional dimension of all the beams = 230mm. x 450mm. = 0.23m. x 0.45m.
Calculation:
Lx /Ly = 5500mm. / 2500mm.
= 2.2 > 2.
Therefore it is a one-way slab.
The load distribution of the one-way slab over the beams is as shown below.
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1. Beam - B1:
Total load over the beam B1
= [Self wt. of the beam + superimposed load from the slab]
Here,
a.) Self-wt. of the beam /m.
= [(area of cross-section) × density of RCC]
= [ (0.23m. x 0.45m.) × 25 KN/m³]
= 2.588 KN/m.
Total self-wt. of the beam
= [ beam wt./m × span of the beam]
= [2.588KN/m × 5.5m.]
= 14.234 KN.
Factored self-wt. of the beam
= [1.5 × 14.234]
= 21.351KN.
b.) Load transferred from slab to the beam B1
= [1/2 x (area of slab) x W]
Before proceeding further, Go through the article 👇
👀. Understanding the concept of the load distribution from slab to beam.
Where all the load distribution formula is derived.
= [ 1/2 x (5.5 x 2.5) x 12.94]
The value of W is taken from the article
How to calculate the total load over the RCC slab?
= 88.96KN.
Total factored load over the beam B1
= [Factored self wt. of the beam + factored load from the slab]
= [21.351 + 88.96]
= 110.311KN.
2. Beam - B2:
Total load over the beam B2
= [Self wt. of the beam]
Here,
a.) Self wt. of the beam
= [(area of cross-section) × density of RCC]
= [ (0.23m. x 0.45m.) × 25 KN/m³]
= 2.588 KN/m.
Factored self wt. of the beam
= [1.5 × 2.588KN/m.]
= 3.88 KN/m.
Total factored self-wt. of the beam
= [ factored wt./m × span of the beam]
= [3.88KN/m × 2.5m.]
= 9.70 KN.
Total factored load over the beam B2
= 9.70KN.
Note: Beam-B2 does not carry any load from the slab.
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Thank you for going through these calculation steps❤. Have a good day 😄.
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