When carrying out the manual design of reinforced concrete structures, column loads are usually assessed by considering the support reactions from beams they are supporting, or by the tributary area method. The latter is more popular due to its simplicity and speed, but usually fails to capture all the loads that are imposed on the columns (such as wall loads), while the former is more complex and time-consuming, it usually very representative of all the possible loads that are imposed on the column.

**Steps on how to load a column from beam support reaction**

The following steps can be adopted when using beam support reactions to obtain the axial load on columns.

(1) Load the floor slab adequately, and factor the loads at ultimate limit state, using the appropriate load combination.

(2) Load all the beams that are connected to the column.

(3) Transfer the loads from slab to the beam using the appropriate relationship. Based on yield lines, the loads are usually triangular or trapezoidal, but this is cumbersome to analyse manually. An equivalent UDL (reasonably accurate) can be used to transfer slab loads to beams by employing the formulas below.

*One way slabs*

Long span q = nl_{x}/2

Short slab q = nl_{x}/5

*Two way slabs*

Long span q = nl_{x}/2(1 – 1/3k^{2})

Short span q = nl_{x}/3

Where;

q = Load transferred from slab to the beam

n = load at ultimate limit state

1.4g_{k} + 1.6q_{k }(BS 8110)

1.35g_{k} + 1.5q_{k} (Eurocode 2)

k = L_{y}/L_{x}

L_{y }= Length of long span of slab

L_{x} = Length of short span of the slab

(4) Analyse the floor beams completely using any suitable method of your choice, while also considering any additional load that may be on the beam such as wall load and finishes.

(5) Obtain the support reactions of the beam, which represents the load that is transferred from the floor to the column.

**Design Example**

**Design Example**

In this post, the floor plan shown below is for a shopping complex, and it desired to obtain the column axial loads at ultimate limit state.

**Design Data**

Size of all columns = 230 x 230mm

Size of all beams = 450 x 230mm

Thickness of slab = 200mm

Unit weight of concrete = 25 kN/m^{3}

Unit weight of sandcrete block = 3.47 kN/m^{2}

f_{ck }= 25 N/mm^{2}

f_{yk} = 500 N/mm^{2}

Load combination = 1.35g_{k }+ 1.5q_{k}

Design variable load (q_{k}) = 4 kN/m^{2}

k = L_{y}/L_{x} = 6/5 = 1.2 (in all cases)

Only external beams are carrying block work load.

**Load Analysis**

*Roof beam*

Permanent load on roof beams g_{k} = 6 kN/m

Variable load on roof beam q_{k} = 1.5 kN/m

(These values are assumed)

At ultimate limit state, n = 1.35(6) + 1.5(1.5) = 10.35 kN/m

*Slab Load Analysis*

Concrete own weight = 25 kN/m^{3 }× 0.2m = 5.0 kN/m^{2}

Screeding and Finishes (say) = 1.35 kN/m^{2}

Partition allowance = 1.5 kN/m^{2}

Total (g_{k}) = 7.85 kN/m^{2}

Variable action (q_{k}) = 4 kN/m^{2}

n = 1.35gk + 1.5qk = 1.35(7.85) + 1.5(4) = 16.6 KN/m^{2}

*Wall load on beam*

Unit weight of sandcrete block = 3.47 kN/m^{2}

Height of wall = 3.5m

Wall load on beam = 3.47 kN/m^{2 }× 3.5m = 12.145 kN/m

**Load on Beams***External Longitudinal beams (Axis 1:A-D and Axis 3:A-D)*

Self weight of beam (factored) = 1.35 × 0.23m × 0.25m × 25 = 1.94 kN/m

Load from slab = nl_{x}/2(1 – 1/3k^{2}) = [(16.6 × 5)/2] × (1 – 1/(3 ×1.2^{2})) = 31.893 kN/m

Load from block work = 12.145 kN/m

Total load = 45.978 kN/m

*External Transverse beams (Axis A:1-3 and Axis D:1-3)*

Self weight of beam (factored) = 1.35 × 0.23m × 0.25m × 25 = 1.94 kN/m

Load from slab = nl_{x}/3 = (16.6 × 5)/3 = 27.667 kN/m

Load from block work = 12.145 kN/m

Total load = 41.752 kN/m

*Internal Longitudinal Beam (Axis 2:A-D)*

Self weight of beam (factored) = 1.35 × 0.23m × 0.25m × 25 = 1.94 kN/m

Load from slab = nl_{x}/2(1 – 1/3k^{2}) = [(16.6 × 5)/2] × (1 – 1/(3 ×1.2^{2})) = 2 × 31.893 kN/m = 63.786 kN/m (we multiplied by two because this beam is receiving slab load from two directions.)

Load from block work = 0 (there are no block works inside the building)

Total load = 65.726 kN/m

*Internal Transverse beams (Axis B:1-3 and Axis C:1-3)*

Self weight of beam (factored) = 1.35 × 0.23m × 0.25m × 25 = 1.94 kN/m

Load from slab = nlx/3 = (16.6 × 5)/3 = 2 × 27.667 kN/m = 55.334 kN/m (slab load is coming from two directions)

Load from block work = 0

Total load = 57.274 kN/m

*Self weight of Columns*

Self weight of column (factored) = 1.35 × 0.23m × 0.23m × 3.75m × 25 kN/m^{3} = 6.695 kN

**Structural Analysis**

A little consideration will show that the support reactions for the floor beams (equal spans) can be obtained by considering the shear force coefficients given below;

**Fig 4**: Shear force coefficient (3 span beam)

**Load Analysis of Column A1***1st Floor*

Load from longitudinal roof beam (1:A-D) = 0.4 × 10.35 kN/m × 6m = 24.84 kN

Load from transverse roof beam (A:1-3) = 0.375 × 10.35 kN/m × 5m = 19.41 kN

Self weight of column = 6.695 kN

Total = 50.945 kN

*Ground Floor*

Load from above = 50.945 KN

Load from longitudinal floor beam (1:A-D) = 0.4 × 45.978 kN/m × 6m = 110.347 kN

Load from transverse floor beam (A:1-3) = 0.375 × 41.752 kN/m × 5m = 78.285 kN

Self weight of column = 6.695 kN

Total = 246.272 kN**Load Analysis of Column A2***1st Floor*

Load from longitudinal roof beam (2:A-D) = 0.4 × 10.35 kN/m × 6m = 24.84 kN

Load from transverse roof beam (A:1-3) = 2(0.625 × 10.35 kN/m × 5m) = 64.687 kN

Self weight of column = 6.695 kN

Total = 96.222 kN

*Ground Floor*

Load from above = 96.222 KN

Load from longitudinal floor beam (2:A-D) = 0.4 × 65.726 kN/m × 6m = 157.742 kN

Load from transverse floor beam (A:1-3) = 2(0.375 × 41.752 kN/m × 5m) = 156.57 kN

Self weight of column = 6.695 kN

Total = 417.229 kN

**Load Analysis of Column B1***1st Floor*

Load from longitudinal roof beam (1:A-D) = (0.5 × 10.35 kN/m × 6m) + (0.6 × 10.35 kN/m × 6m) = 68.31 kN

Load from transverse roof beam (B:1-3) = 0.375 × 10.35 kN/m × 5m = 19.41 kN

Self weight of column = 6.695 kN

Total = 94.415 kN

*Ground Floor*

Load from above = 94.415 KN

Load from longitudinal floor beam (1:A-D) = (0.5 × 45.978 kN/m × 6m) + (0.6 × 45.978 kN/m × 6m) = 303.455 kN

Load from transverse floor beam (B:1-3) = 0.375 × 57.274 kN/m × 5m = 107.388 kN

Self weight of column = 6.695 kN

Total = 511.953 kN

**Load Analysis of Column B2***1st Floor*

Load from longitudinal roof beam (2:A-D) = (0.5 × 10.35 kN/m × 6m) + (0.6 × 10.35 kN/m × 6m)= 68.31 kN

Load from transverse roof beam (A:1-3) = 2(0.625 × 10.35 kN/m × 5m) = 64.687 kN

Self weight of column = 6.695 kN

Total = 139.692 kN

*Ground Floor*

Load from above = 139.692 kN

Load from longitudinal floor beam (2:A-D) = (0.5 × 65.726 kN/m × 6m) + (0.6 × 65.726 kN/m × 6m) = 433.7916 kN

Load from transverse floor beam (A:1-3) = 2(0.625 × 57.274 kN/m × 5m) = 357.9625 kN

Self weight of column = 6.695 kN

Total = 938.141 kN

As you can see, that was very straightforward with the use of shear force coefficients, given the fact that the beams were of equal span. When the beams are not equal, you have to carry out the analysis, and transfer the shear forces accurately. Now, let us compare load analysis of column B2, by the tributary area method.

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This made what i have being working on easy for me. But please can this be used for steel structures too?. Thanks

I am happy that you found this helpful. The answer to your question is yes, shear force from beams are transferred to columns as axial load in steel structures.

Boss in fact you are too much! I always like your approach each subject. Thank you for your educational materials

A little observation here bro.

The loads from each beam to the columns should be divided by 2 since adjacent columns bear half the load of the beam they support not all.

You multiplied by 6m instead of 6/2 in the analysis of Col A1 and so on.

Pls review them coz I think they are wrong.

Eg.

Load Analysis of Column A1

1st Floor

Load from longitudinal roof beam (1:A-D) = 0.4 × 10.35 kN/m × 6/2m = 12.42 kN

Load from transverse roof beam (A:1-3) = 0.375 × 10.35 kN/m × 5/2m = 9.703 kN

Self weight of column = 6.695 kN

Total = 28.818 kN

Likewise other columms analysis.

may you please explain why we used value of 0.4 and 0.375? :'(

Thanks for this. You are doing a great job for us that wanna know more. Thumbs up!

No Sir. The values are the support reactions being directly transmitted to the columns. All load distribution has been done during the analysis of the beams.

So 0.4 x q x 6m = support reaction at A. Why share it again?

I see bro. Can you explain why or how the 0.4, 0.375,0.625,x q x l

. I thought its half loads transfered from beams. ie 0.5

Thanks man

OK. I got your work now. I just learnt a new thing.

How do we calculate the Shear force coefficients? In case of other beams of unequal span?

Pls thanjst

If the spans are unequal, it is still the same thing… You analyse the beam using any method of your choice (force method, 3 moment equation, slope deflection method etc). No rocket science involved

Thanks. I get it now.

Good job as usual. Pls can you come up with an example of rotating or vibrating equipment support foundation base design. Thanks.

halo i see your articles as very helpful. I have checked the calculations above and checked with Staad.pro i have realized that the loads are higher in staad.pro.

Can you give an explaination please. Thank you very much

Once in a while, they may incorporate on their depiction the tasks, which they have provided assembling materials. http://www.goldapple-alu.com；

Please my engr. Convert this to PDF, so that we young ENGR can able to download it directly.

I believe this is one of the so much vital information for me.

And i am satisfied studying your article. But want to

Observation on some basic things, the site taste is perfect, the articles is in reality

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Very awesome and impressive analysis I really appreciateappreciate you entire contributions to young potential engineers.

Thank you may God(ALLAH) bless you.

Pleas where do yo got 0.25 from load on beams

Axis1 a-d and axis 3 a-d

Self weight of beam (factored)

= 1.35*0.23*0.25*25??

Thanks sir

His roof beam depth is 250mm i.e (0.25m), so the beam self weight is unit weight of concrete x the width of the beam x it depth (i.e) 25×0.23×0.25 = 1.4375kn/m then he factor it by 1.35 to change to ultimate limit state

n = 1.35x 1.4375 = 1.94Kn/m

Thank you Sir. You cleared all of my doubts especially on the trapezonal and triangular load transfers from Slabs to beams.

Can you do an an example of a concrete column of more than 12 floors. Design a foundation, beam ,column hypothetically using moment distribution to bs 8110 for educational purposes only.can take any dimension and assumption

I really enjoy every subject you have treated so far .

Please kindly throw more lights on the design on Strip Foundations cast monothically with ground inverted T- beam usually used for wall fencing. Thanks

Okay, we will look into that. kindly check if the link below will answer your questions. https://structville.com/2016/04/example-on-the-analysis-and-design-of-continuous-slab-and-beam-footing-per-bs-8110-11997.html

Greetings Sir, Thank you for the post. I’m having a confusion when you are calculating the Load analysis on Column A2 FOR Ground floor Transverse beam(A:1-3): Questions I have are

1. Why are we multiplying by 2

2. Is the factor not supposed to be 0,625*ql? instead of 0,375*ql?

3. Another question is on the given data the beam sizes given as 450*230? yet we used 250mm? is this for the roof beams?

Thanks Engr. I really gained a lot from your article.

But what will you do when you have a beam that has more than 3 spans?