How the load is transferred from the slab to the beams?/Understanding the concept of the load distribution from slab to beam.

 The load transfer from the slab to the beam can be determined using yield line theory.

But, the commonly adopted practice is

1. For a one-way slab 👉  The load is split equally between the beams that support the longer span of the slab.

When Ly/Lx ≥ 2, the slab is considered a one-way slab.

Suppose if the total load of the slab is Wu, the load over the individual beam B1 = Wu/2

If the total load per unit area of the slab is W, 

The load transferred to the individual beam B1

= [1/2 x area of slab  x W]

= [1/2 x Ly x Lx x W]

2. For a two-way slab 👉  The load is transferred to all 4 sides of the slab. Usually, the load of a triangular area is taken by the beam at the shorter side (Lx) & a load of the trapezoidal area is distributed to the longer side (Ly) of the slab. 

When Ly/Lx < 2, the slab is considered a two-way slab.

Now, let us understand the load distribution concept in a two-way slab from the following drawing.

1. Load transferred to the beam B1

= [ trapezoidal area x total load per unit area of the slab (W)]

Here,

Area of trapezoid AEFD

= [average length x perpendicular width]

= [{(side AD + side EF) / 2 } x (length of DG)]

= [{(Ly + (Ly - Lx) /2}  x (Lx / 2)]

= [{ 2Ly/2 - Lx/2} x (Lx / 2)]

= [Lx Ly /2 - Lx²/4]

= [1/2 ( Lx Ly - Lx²/2)]

Therefore,

The load transferred to the individual beam B1

= [1/2 ( Lx Ly - Lx²/2) x W]

2. Load transferred to the beam B2

= [ triangular area x total load per unit area of the slab (W)]

Here,

Area of triangle DFC

= [1/2 x base x height]

= [1/2 ✖ Lx ✖ Lx/2]

= Lx²/4

Therefore,

The load transferred to the individual beam B2

= [ Lx²/4 x W]

Note: This method will be accurate only when the load over the slab will be uniformly distributed.

Now, my question to you is, "what will be the load distribution pattern in a square-shaped slab?"

In a square-shaped slab Lx = Ly 

Ly /Lx = 1 < 2, the slab is considered a two-way slab.

Here, the load will be transferred equally to all 4 sides. 

The load distributed over the beam-B1 = Load over the beam-B2

The load over the individual beam ( B1 or B2)

 = [ triangular area x total load per unit area of the slab (W)]

=  [ Lx²/4 x W] or [ Ly²/4 x W]

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

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