# How to Load Columns in RC Building Design

(Using Beam Support Reactions)

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.

(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 spanÂ  Â q = nlx/2
Short slab q = nlx/5

Two way slabs
Long spanÂ  Â  q = nlx/2(1 â€“ 1/3k2)
Short spanÂ  Â  Â q = nlx/3

Where;
q =Â  Load transferred from slab to the beam
n = load at ultimate limit state
1.4gkÂ + 1.6qk (BS 8110)
1.35gkÂ + 1.5qk (Eurocode 2)
k = Ly/Lx
Ly = Length of long span of slab
Lx = 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

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/m3
Unit weight of sandcrete block = 3.47 kN/m2
fck = 25 N/mm2
fyk = 500 N/mm2

Load combination = 1.35gkÂ + 1.5qk
Design variable load (qk) = 4 kN/m2

k = Ly/Lx = 6/5 = 1.2 (in all cases)
Only external beams are carrying block work load.

Roof beam
Permanent load on roof beams gk = 6 kN/m
Variable load on roof beamÂ  qk = 1.5 kN/m
(These values are assumed)
At ultimate limit state, n = 1.35(6)Â + 1.5(1.5) = 10.35 kN/m

Concrete own weight = 25 kN/m3Â Â Ã— 0.2mÂ  = 5.0Â  kN/m2
Screeding and Finishes (say)Â  Â  = 1.35Â  kN/m2
Partition allowanceÂ  Â = 1.5Â  kN/m2
Total (gk)Â  Â  =Â  7.85 kN/m2
Variable action (qk)Â  Â  = 4 kN/m2

n = 1.35gk + 1.5qk = 1.35(7.85) + 1.5(4) = 16.6 KN/m2

Unit weight of sandcrete block = 3.47 kN/m2
Height of wall = 3.5m
Wall load on beam =Â  3.47 kN/m2Â Â Ã— 3.5mÂ  = 12.145Â  kN/m

External Longitudinal beams (Axis 1:A-D and Axis 3:A-D)
Self weight of beam (factored) = 1.35 Ã— 0.23m Ã— 0.25m Ã— 25 = 1.94 kN/m
Load from slab = nlx/2(1 â€“ 1/3k2) = [(16.6 Ã— 5)/2] Ã— (1 â€“ 1/(3 Ã—1.22)) = 31.893 kN/m
Load from block work = 12.145Â  kN/m

External Transverse beams (Axis A:1-3 and Axis D:1-3)
Self weight of beam (factored) = 1.35 Ã— 0.23m Ã— 0.25m Ã— 25 = 1.94 kN/m
Load from slab = nlx/3Â = (16.6 Ã— 5)/3Â = 27.667 kN/m
Load from block work = 12.145Â  kN/m

Internal Longitudinal Beam (Axis 2:A-D)
Self weight of beam (factored) = 1.35 Ã— 0.23m Ã— 0.25m Ã— 25 = 1.94 kN/m
Load from slab = nlx/2(1 â€“ 1/3k2) = [(16.6 Ã— 5)/2] Ã— (1 â€“ 1/(3 Ã—1.22)) = 2 Ã— 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)

Internal Transverse beams (Axis B:1-3 and Axis C:1-3)
Self weight of beam (factored) = 1.35 Ã— 0.23m Ã— 0.25m Ã— 25 = 1.94 kN/m
Load from slab = nlx/3Â = (16.6 Ã— 5)/3Â = 2 Ã— 27.667 kN/m = 55.334 kN/m (slab load is coming from two directions)
Load from block work = 0

Self weight of Columns
Self weight of column (factored) = 1.35 Ã— 0.23m Ã— 0.23m Ã— 3.75m Ã— 25 kN/m3 = 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)

1st Floor
Load from longitudinal roof beam (1:A-D) = 0.4 Ã— 10.35 kN/m Ã— 6m = 24.84 kN
Load from transverse roof beam (A:1-3) = 0.375 Ã— 10.35 kN/m Ã— 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 Ã— 45.978 kN/m Ã— 6m = 110.347 kN
Load from transverse floor beam (A:1-3) = 0.375 Ã— 41.752Â kN/m Ã— 5m = 78.285 kN
Self weight of column = 6.695 kN
Total = 246.272 kN

1st Floor
Load from longitudinal roof beam (2:A-D) = 0.4 Ã— 10.35 kN/m Ã— 6m = 24.84 kN
Load from transverse roof beam (A:1-3) = 2(0.625 Ã— 10.35 kN/m Ã— 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 Ã— 65.726Â kN/m Ã— 6m = 157.742 kN
Load from transverse floor beam (A:1-3) = 2(0.375 Ã— 41.752Â kN/m Ã— 5m) = 156.57 kN
Self weight of column = 6.695 kN
Total = 417.229 kN

1st Floor
Load from longitudinal roof beam (1:A-D) = (0.5 Ã— 10.35 kN/m Ã— 6m)Â + (0.6 Ã— 10.35 kN/m Ã— 6m) = 68.31 kN
Load from transverse roof beam (B:1-3) = 0.375 Ã— 10.35 kN/m Ã— 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 Ã— 45.978 kN/m Ã— 6m)Â + (0.6 Ã— 45.978 kN/m Ã— 6m)Â Â  = 303.455 kN
Load from transverse floor beam (B:1-3) = 0.375 Ã—Â 57.274Â kN/m Ã— 5m = 107.388 kN
Self weight of column = 6.695 kN
Total = 511.953 kN

1st Floor
Load from longitudinal roof beam (2:A-D) = (0.5 Ã— 10.35 kN/m Ã— 6m)Â + (0.6 Ã— 10.35 kN/m Ã— 6m)= 68.31 kN
Load from transverse roof beam (A:1-3) = 2(0.625 Ã— 10.35 kN/m Ã— 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 Ã— 65.726Â kN/m Ã— 6m)Â + (0.6 Ã— 65.726Â kN/m Ã— 6m)Â  = 433.7916 kN
Load from transverse floor beam (A:1-3) = 2(0.625 Ã— 57.274Â kN/m Ã— 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.

Thank you for visiting, join the Structville Interaction on Facebook by visiting;

A dynamic civil engineer with vast experience in research, design, and construction of civil engineering infrastructures. He is a member of the Nigerian Society of Engineers. Reach him at ubani@structville.com

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

2. 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.
1st Floor
Load from longitudinal roof beam (1:A-D) = 0.4 Ã— 10.35 kN/m Ã— 6/2m = 12.42 kN
Load from transverse roof beam (A:1-3) = 0.375 Ã— 10.35 kN/m Ã— 5/2m = 9.703 kN
Self weight of column = 6.695 kN
Total = 28.818 kN

Likewise other columms analysis.

3. 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?

4. 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

5. 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

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

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

8. 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

9. 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) 25Ã—0.23Ã—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

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

11. 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

12. 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

13. Greetings Sir, Thank you for the post. Iâ€™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?