Most of the engineer prefer ETABS for the analysis of structural members of building but for this they have to do design of slab either manually or using other software like SAFE. After detailed design of slab, for the initiation of analysis, there is necessities of preliminary value of other structural members. As we have previously discussed about the preliminary design of beam, now here we discuss about the preliminary design of column.
There is no different in preliminary design either it is axially, uniaxially or bi-axially; long or short column; braced or unbraced column. Since preliminary design is done on the basis of load only so for eccentric column, the size should be increase a little bit more.
Note: All the design procedure are based on Indian Standard codes but if you use other codes than this, that your country accept, then also it would not be wrong for the preliminary case.
Preliminary design of column:
Our first step will be calculation of all dead and live loads.
Load calculation:
1. Dead load
(i) Floor finish
(ii) Weight of slab and beam
(iii) Partition load.
2. Live load
Note: The intensity of live load decreases by 10% to the floor above up to 40%.
For self-weight of column, increase the total load by 20% and calculate factored load. [ There is not exact rule of taking 20%, it is totally based on the experiences result]
Assumes: percentage of steel = 2%
Using formula,
Pu = 0.4fckAc + 0.67fy As
Where,
Pu = factored load calculated
fck = characteristics strength of concrete cube at 28 days
fy = characteristics strength of steel reinforcement
Ac = Area of concrete
As = Area of longitudinal steel reinforcement
Ac = (Ag – As )
From this equation, we get Ag (gross area of section).
Depending upon the eccentricity, select square or rectangular shaped column
And dimension of column b and D is derived.
Ex 1. Find the preliminary dimension of D-6 column of G+3 stories building which is shown in figure below.
Solution,
Our first task is to find out the area of slab whose load is transmitted to beam and finally to that targeted column.
For D-6 column, let’s calculate slab area.
Length in x-axis = half the span of beam to the right and half the span of beam to the left
= 6/2+6/2
= 6 m
Length in y-axis = half the span of beam to the upward and half the span of beam to the downward
= 6/2+6/2
= 6 m
1. Dead load
Load from floor slab:
Thickness of slab = 160 mm [This is computed from detailed design of two-way slab but also can get idea from preliminary design of slab]
Weight of slab = unit weight * thickness of slab
= 25 * 0.16
= 4 KN/mm2
Assume, floor finish and partition = 1.5 KN/mm2
So, total load from slab = (weight of slab + floor finish and partition) * area of slab
= (4 + 1.5) * 6 * 6
= 198 KN
Note: The thickness of slab is taken as 160 mm but it was explained before that there should be provision of secondary beam for slab thickness greater than 150 mm. This is done just for simplicity in the design and will result greater dead load.
Load from beam:
Size of beam = 230 mm * 400 mm [size can be calculated as explained in preliminary design of beam]
Total weight of beam = weight of x-axis beam + weight of y-axis beam
= unit weight * area of beam * length of x and y-axis beam
= 25 * 0.23 * 0.4 * (6+6)
= 30 KN
2. Live load
Assume live load = 2.5 KN/m2
And roof live load = 1.5 KN/m2
| Load calculation with reduction | Total load | |
| Basement | 2.5 * 6 * 6 * 1 | 90 KN |
| First floor | 2.5 * 6 * 6 * 0.9 | 81 KN |
| Second floor | 2.5 * 6 * 6 * 0.8 | 72 KN |
| Third floor | 2.5 * 6 * 6 * 0.7 | 63 KN |
| Top floor | 1.5 * 6 * 6 * 0.6 | 54 KN |
| Total | 360 KN |
Total live and dead load = (198 + 30) * 5 + 360
= 1500 KN
Add 20% for self-weight = 1.2 * 1500 = 1800 KN
Factored load = 1.5 * 1800
= 2700 KN
Assumption:
percentage of steel = 2%
fck= 25 N/mm2
fy = 415 N/mm2
Using formula,
Pu = 0.4fckAc + 0.67fy As
where, As = 0.02 Ag
Ac = ( Ag – As )
= 0.98 Ag
Now,
Pu = 0.4fckAc + 0.67fy As
2700 * 1000 = 0.4 * 25 * 0.98 * Ag + 0.67 * 415 * 0.02 * Ag
Ag= 175769.8 mm2
Since the span of beam on each side is same, so it may be axially loaded. Hence, it would be more suitable to choose square or circular column. Due to ease in construction and availability of form work, square column is more suitable.
So b=D = √ Ag
= √ 175769.8
= 419.25 mm
Adopt size of column = 500 mm * 500 mm
Ex 2. Find the preliminary design of column D-7 of G+3 stories building which is shown in previous example.
Solution,
As previous, firstly, we will find out the area of slab whose load is transmitted to beam and finally to that targeted column.
For D-7 column, let’s calculate slab area.
Length in x-axis = half the span of beam to the right and half the span of beam to the left
= 6/2+6/2
= 6 m
Length in y-axis = half the span of beam to the upward and half the span of beam to the downward
= 6/2+6/2
= 6 m
Since there is no slab in 1/4th portion so area of slab = ¾ * 6 * 6 = 27 m2
1. Dead load
Load from floor slab:
Thickness of slab = 160 mm[This is computed from detailed design of two-way slab but also can get idea from preliminary design of slab]
Weight of slab = unit weight * thickness of slab
= 25 * 0.16
= 4 KN/m2
Assume, floor finish and partition = 1.5 KN/mm2
So, total load from slab = (weight of slab + floor finish and partition) * area of slab
= (4 + 1.5) * 27
= 148.5 KN
Load from beam:
Size of beam = 230 mm * 400 mm [size can be calculated as explained in preliminary design of beam]
Total weight of beam = weight of x-axis beam + weight of y-axis beam
= unit weight * area of beam * length of x and y-axis beam
= 25 * 0.25 * 0.4 * (6+6)
= 30 KN
2. Live load
Assume live load = 2.5 KN/m2
And roof live load = 1.5 KN/m2
| Load calculation with reduction | Total load | |
| Basement | 2.5 * 27 * 1 | 67.5 KN |
| First floor | 2.5 * 27 * 0.9 | 60.75 KN |
| Second floor | 2.5 * 27 * 0.8 | 54 KN |
| Third floor | 2.5 * 27 * 0.7 | 47.25 KN |
| Top floor | 1.5 * 27 * 0.6 | 24.3 KN |
| Total | 253.8 KN |
Total live and dead load = (148.5 + 30) * 5 + 253.8
= 1146.3 KN
Add 20% for self-weight = 1.2 * 1146.3 = 1375.56 KN
Factored load = 1.5 * 1375.56
= 2063.34 KN
Assumption:
percentage of steel = 2%
fck = 25 N/mm2
fy = 415 N/mm2
Using formula,
Pu = 0.4fckAc + 0.67fy As
where, As = 0.02 Ag
Ac = ( Ag – As )
= 0.98 Ag
Now,
Pu = 0.4fckAc + 0.67fy As
2064 * 1000 = 0.4 * 25 * 0.98 * Ag + 0.67 * 415 * 0.02 * Ag
Ag = 13436.25 mm2
Since the span of beam on each side is same, but there is not slab in one portion so distribution of load is different which seems like bi-axially loaded. Hence, it would be more suitable to choose square column of greater dimension for moment consideration in detailed design.
So b=D = √ Ag
= √ 134366.25
= 366.56 mm
Adopt size of column = 450 mm * 450 mm
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