The Design of RCC Columns Design may be of different types, according to its behavior under the action of the loading. Primarily we can differentiate as Short Column Design and Long Column Design . In this article, I will deal with Long Column Design Only, Design of Axially Loaded Short Column has already been discussed by me step by step which is Topping the Google since last Two Years.
Okay, now before going into the actual design procedure I would like to ask you, Do You Know What is Column?
If not then here it is in simple way of understanding, A Column is a structural member which is vertical and loaded through its Longitudinal Axis (in case of Axially Loaded Column) or through parallel to its Longitudinal Axis (in case of eccentricity of Load) and also there is another condition, that is the effective height of the column must be greater than three times its least lateral dimension.
Suppose a rectangular column is having a section of 250mmx350mm then its least lateral dimension will be 250mm, and also considering that the column is fixed at its both ends rigidly and having its length as 1500mm, then the calculation will be as follows to know that at first if it is a column or not.
As the actual length/height (l) is 1500mm and the column is having its both ends fixed then as per I.S.: 4562000, effective length will be 0.65xl i.e. 0.65×1500 = 975mm
Now, (Effective Length/Least Lateral Dimension) = (975/250) = 3.90 >Three, hence it is a column, it was less than three, then it would have been called as Pedestal.
Now, let us understand the behavior of the Short Columns and Long Columns under the action of loads.
Short Column :
A Short column is a column whose effective length/height does not exceed 12 times its Least Lateral Dimension. A short column being relatively not very long as compared to its least lateral dimension, hence when a load acts through its axis, it mainly tends to be compressed, and when its compression bearing capacity exceeds the ultimate value, then it crushes and fails. Therefore, in case of short column failure mainly occurs by Compression Failure.
Long Column :
A Long column is a column whose effective length/height does exceeds 12 times its Least Lateral Dimension. A Long column being relatively long (and sometimes very long) as compared to its least lateral dimension hence it tends to bend sideways deviating from its axis under the action of load. It is similar to like why a stick of jute or other similar long stick if placed vertically on a firm hard surface and pressed vertically downward from its top, then it will be observed that the stick will try to bend itself deviating from the central axis and eventually will fail mainly due to its bending than direct compression.
So, what is the end result then between Long and Short Column?
Now, as the Long Column tends to bend under the action of load than compressing vertically, and as the Concrete is stronger in compression but is very week to withstand the bending stresses (especially tensile stresses) that develop hence it is found that A Short Column will be able to take much load than a Long Column considering every other property like reinforcement, Concrete Grade, section etc. constant.
In much more convenient way it can explained as like suppose there are two columns both having same cross sectional dimensions say 250mmx350mm, having same end conditions say Both Ends Fixed (which is most common in RCC Structures), Same Grade of Steel Fe415, Same amount of Steel and Ties, Same Grade of Concrete, everything be same, but only one difference that if Column A is Short Column, then another Column, i.e. Say Column B is a Long Column. Then by practical experiment it can be seen that Column A will be able to take up much load before failure than Column B. say Column A if can take 1500KN then Column B will be able to take 1300KN. Actually, more the column will be slender that means more the column will be longer with respect to its least lateral dimension, more the load taking capacity will be reduced as the tendency of bending will be increased.
Now let us go through the actual designing procedure of a Long Column. If you have already go through my article of Designing Short Axially Loaded Column then you can skip first Three Steps, as it is exactly same for both Design of Long Column and Design of Short Column and you can directly read from Step 4.
STEP 1 : Calculation of the Influence Area of the Columns :
The first step is to find out the Influence Area of the Column to be Designed. The Influence Area of a column is the area of which load is being transferred to the column to be designed for. For this purpose in a framed structure small and medium building the design of column is done for the column whose Influence Area is the largest hence the load coming on the column will be so the greater of the any other column in that building hence all the other column having lesser Influence Area hence lesser Loads if provided with the same Designed parameters that required for the column having largest Influence Area, then the whole Structure will automatically become safe against the Loads.
STEP 2: Calculation of the Loads Coming on Columns from the Influence Area :
In this step the Load Calculation is being done. This is done by calculating all the loads acting within the influence area.
The Loads acting are broadly classified as Dead Load (DL) and Live Load (LL). Dead Loads are the load of objects which cannot be moved from on place to another like the loads of Brick Work, Beams, Slabs etc. and the Live Loads are the loads coming from movable objects such as Humans, Chair, Table etc.
Thus We Need to Calculate the Dead Loads as well as Live Loads within the Influence Area, these are as follows in the general case of a Building :
A) Dead Loads :
 Due to weight of Slab [25000 N/m3 ]
 Due to weight of Floor Finish [500 N/m2]
 Due to weight of Brick Masonry [19200 N/m3]
 Due to weight of Beam Rib [25000 N/m3]
 Due to weight of Self Weight of Column [25000 N/m3]
B) Live Load :
It depends upon the Nature of the Structure, and it values for different structural nature are given in the concerned Code of Practice, like in India these are given in I.S.: 875Part II.
For Residential Buildings, it is generally considered @ 2KN/m2 = 2000 N/m2
Now after correct calculation of above loads the Total Load is Calculated by,
Total Load on each floor = Dead Load + Live Load
Now this the actual load which will be acting on column for each floor, now if the building say 5 storied, then just multiply the value with the nos. of floors, like for five storied building multiply the Total Load on each story with 5.
Now thus the Total load acting on column at Column Base is Obtained and it is denoted with ‘P’.
Hence P= Total Load on each Floor X Nos. of Stories = (Dead Load + Live Load) X Nos. of Stories.
Now we shall move to the actual Designing to determine suitable Column sections and its Reinforcements so that the above load is safely resisted by the column Designed.
It can be done by Three main Methods of Design : a) Working Stress Method b) Ultimate Load Method and c) Limit State Method.
The Modern Practice is to use Limit State Method for all types of Designing, Hence I’ll discuss here the Limit State Method Of Design Of Column.
STEP 3 : Finding The Gross CrossSectional Area Required For The Columns :
This is the one of the most important and main step of the Design of Columns.
First in the Limit State Method of Design we must increase the load acting on the column with a Load Factor so that if there will be any accidental increase of loads the column will be still safe to resist the load without a failure. The Factor of Safety for Dead Load + Live Load Combination is 1.5, hence we must multiply the load action on column (P) with the 1.5 to obtain the Ultimate Load that is the Factored Load of the Column that is Pu.
Hence Factored Load, Pu = 1.5 X P
For Design we will work with this value of load.
Now before going on I’m here to say that I will design according to the Code Of Practice of I.S.: 4562000
The Ultimate Load of a Column is given by,
Pu = 0.4.fck.Ac + 0.67.fy.Asc [Equation I]
Where, Pu = Ultimate Load of the Column in N/mm2
fck= Yield Strength of Concrete in N/mm2
Ac = Area of Concrete (CrossSectional Area) of Column in mm2
fy = Yield Strength Of Steel in N/mm2
Asc = Area of Steel (CrossSectional Area) in Column in mm2
Now the column consists of Concrete and as well as Steel in the form of Reinforcements hence the Total CrossSectional Area of Column is made of Area of Concrete and Area of Steel.
The Total CrossSectional area of Column can be also termed as Gross CrossSectional Area of Column and it’s denoted by Ag.
Hence, Gross CrossSectional Area of Column = C/S Area of Concrete + C/S Area of Steel
Therefore, Ag = Ac + Asc
And hence, Ac = Ag – Asc
Now putting the above obtained value in the original equation (Equation I) we get,
Pu = 0.4.fck.(AgAsc) + 0.67.fy.Asc [Equation II]
Now Assume the Percentage of Steel you want to use ranging anywhere from 0.8% to 6% with Respect to Gross CrossSectional Area of the Column (Ag). Say Assuming Steel as 1% of Ag it means Area of Steel Asc = 1% of Ag = 0.01Ag
The higher will be the percentage of steel used the lower will be Ag and thus lesser will be the crosssectional dimension of the column. But the as the Price of Steel is very high as compared to the Concrete hence it is desirable to use as less as steel possible to make the structure economical, again if the percentage of steel is lowered then the Ag will increase at higher rate, about 30% with decrease of just 1% of steel and so each lateral dimension of the column will increase and will cause a gigantic section to be provided to resist the load. Therefore both the factors are to be considered depending upon the amount of loading.
My suggestion is to use the following Percentage of steel for the Design, Which I’ve found to be effective and to produce economical and safe section of Column.
Loading (Pu) in N Percentage Of Steel for Satisfactory Design
Below 250000 ——————————————–0.8%
250,000 to 500,000 ————————————–1.0%
500,000 to 750,000 ————————————–1.5%
750,000 to 1000,000 ————————————2.0%
1000,000 to 1500,000 ———————————–2.5%
1500,000 to 2000,000 ———————————–3.0%
And so on, with increase of each 250,000 N increasing the Percentage of Steel as 0.5%.
Now input the value of the Asc in the form of Ag in the Equation I. For example suppose 1% Steel is used then the equation will be like the one below :
Pu = 0.4.fck.(Ag0.01Ag) + 0.67.fy.0.01Ag [Equation III]
Therefore, if we know the Grade of Concrete and Grade of Steel to be used and Factored Load coming on the Column and Assuming the Percentage of steel required appropriately then we can Very Easily Calculate the GrossSectional Area (Ag) of the Column required from the above form of the equation.
Now as the Ag is obtained thus the Lateral Dimensions of the Column that are the sides of the column can be easily determined.
The Ag or GrossSectional Area of the Column means that it is the product of the two lateral sides of a column [i.e. Breadth (b) X Depth (D)], hence reversely knowing the Ag we can determine the Lateral Dimensions.
For making a Square Section just Determine the Root Value of the Ag. Like if the Value of Ag is 62500 mm^2 Then considering square section of a column we can get each side
Also Rectangular Column Sections Can be made by using different proportion say b : D = 1 : 2 , Hence D=2b , Therefore, Ag = b X D = b X 2b = 2b2 or b=
Hence D can be also determined as D=2b after Calculating the b.
Most of the times after calculating the sides of a column it will give results such as 196.51mm or 323.62 etc. values, which practically cannot be provided at field, hence we must increase those values to the nearest greater multiple of 25mm (i.e. 1 inch). For examples a value of 196.51mm may be increased to 200mm or 225mm or 250 mm even, and a value of 323.62mm may be increased to 350mm. more it will be increased the more it will be safer, but it is uneconomical to increase by a very high amount, it should not be increased more than by 75mm to consider the economical factor.
STEP 4 : Check For Long/Short Columns:
Depending upon the ratio of Effective Length to the Least Lateral Dimension of a column, a column may be classified as Long Column and Short Column. If the value of this ratio is less than 12 then it’s called as a short column and if the value is more than 12 then it’s called as a Long Column.
Let us now, concentrate on Long Column. First of all, we need to find out the effective length of a column, which can be obtained by multiplying a factor with the actual unsupported length of the column. The factor depends upon the end condition of the column. In most general cases, we use a Both End Fixed Column for which The Factor is 0.65.
Therefore, Effective Length = Effective Length Factor (0.65) x Unsupported Length (l). suppose a column has an unsupported length of 5.5m = 5500mm, hence the effective length will be lef = 0.65×5500 = 3575mm. Least lateral dimension means the shorter of the two dimensions of column that is length and breadth. But in case of a circular column as there is only diameter, hence we will use the diameter.
Suppose a column is of 350mmx250mm section and has an unsupported length of 5500mm, then the Ratio of Effective length t the Least Lateral Dimension will be as follows:
(Effective Length/Least Lateral Dimension) = (lef/b) = (3575/250) = 14.3 which is greater than 12 and hence it is a Long Column.
STEP 5 : Calculation of Reduction Coefficient :
As we already know that load taking capacity of a Long Column is less than that of a Short Column provided everything else conditions are same. Therefore, if we provide the section as obtained in case of a short column design formula will be not sufficient enough for taking up the same amount of load.
Now the question is how much load carrying capacity of the column will be reduced in case of Long Column?
Yes, the answer is application of Reduction Coefficient. Reduction coefficient is a fractioned value less than 1, which depends upon the effective length of the column and Least Lateral Dimension of the column under consideration. Applying this Reduction Coefficient to the load carrying capacity of a short column we can find the reduced load carrying capacity of the same column if it is long, and other things remaining unchanged by as follows:
Reduced Load Carrying Capacity of Long Column
= Factored Load Carrying Capacity of Short Column x Reduction Coefficient
As per I.S.: 4562000 Clause B3.3 Page No. 81 we get that,
Cr = 1.25 – (lef/48b)
Where, Cr = Reduction Coefficient
lef = Effective Length of the Column
b = Least Lateral Dimension of the column
This formula holds good for lef/b values not exceeding 40. But in cases where lef/b value exceeds 40 then the following formula is to be adopted as mentioned below,
Cr = 1.25 – (lef/160x i min)
Where, Cr = Reduction Coefficient
lef = Effective Length of the Column
i min = Least Radius of Gyration
Suppose a column is of 350mmx250mm section and has an unsupported length of 5500mm, which will be Long Column having Lef/b ratio of 14.90 as already discussed, then Reduction Coefficient will be as follows:
Reduction Coefficient, Cr = 1.25 – (lef/48b) = 1.25 – {3575/(48×250)} = 0.9521
Therefore, if the Short Column could carry a load of 1500KN then Long Column with same properties and condition will carry
Reduced Load Carrying Capacity of Long Column
= Factored Load Carrying Capacity of Short Column x Reduction Coefficient
= 1500 x 0.9521 = 1428.15KN
STEP 6 : Redesigning the section for Long Column Design :
Section can be Redesigned by two methods as mentioned below,

By designing for increased load :
In this case, the design will be done for increased amount of load which is equal to the amount of load carrying capacity which will be reduced as an effect of being a Long Column, hence the load reduction effect of being Long Column will be compensate in this way.
Like as we found previous step that load carrying capacity of a column is being reduced from 1500KN to 1428.15KN, hence reduction in the load carrying capacity is 1500KN1428.15KN = 71.85KN.
Therefore, we have to design the column for such a load that which will after applying Reduction Coefficient on it will come to the Factored Load for which actual design was required considering Short Column. Then after that the design procedure will be same to find out the Sectional Dimension and Reinforcement of the Column as mentioned in the Step 3 above, and thereafter will continue through Steps 7,8 and 9.
We can easily directly found the load to be designed for by just dividing the Factored Load for Short Column with the Reduction Coefficient, as follows:
Load to be Designed for Long Column
= Factored Load on the Short Column/ Reduction Coefficient = 1500/0.9521 = 1575.46KN
Cross Checking, 1575.46×0.9521 = 1499.995KN say 1500KN, hence column which will be designed for the above calculated higher load will due to the effect of being a Long Column its load carrying capacity will be reduced and come to 1500KN, that is actually what was required.

By reducing the permissible stress of the materials :
To be honest this is the process as mentioned in the I.S. Code, but in this case the formula of the Short Column Design will need to be a bit modified. Actually, I first discussed another method as it will be much easier to remember and you will find it easy as it is exactly same as that of Short Column Design.
I.S. Code advised to reduce the permissible stress used in the calculation to find out Gross CrossSectional Area and Area of Reinforcement by multiplying the Permissible Stress of the Steel and Concrete with the Reduction Coefficient. This method is also same just a little modification is to be done in the formula of the short column as mentioned below,
Pu = 0.4.fck.Cr.Ac + 0.67.fy.Cr.Asc [Equation I]
Where, Pu = Ultimate Load of the Column in N/mm2
fck= Yield Strength of Concrete in N/mm2
Ac = Area of Concrete (CrossSectional Area) of Column in mm2
fy = Yield Strength Of Steel in N/mm2
Asc = Area of Steel (CrossSectional Area) in Column in mm2
Cr = Reduction Coefficient
Here, Pu is not the increased load as obtained by dividing with Reduction Coefficient, but only the simple Factored Load as used in the design of short column.
And
Pu = 0.4.fck.Cr.(AgAsc) + 0.67.fy.Cr.Asc [Equation II]
And
Pu = 0.4.fck.Cr.(Ag0.01Ag) + 0.67.fy.Cr. 0.01Ag [Equation III] [Considering 1% Steel]
Notations being as usual. Rest of the thing in the Step 3 will be as usual to follow, and then Steps 7,8 and 9 is to be followed for completion of the Design of Axially Loaded Long Column.
STEP 7 : Check For Eccentricity :
Eccentricity means deviating from the true axis. Thus an Eccentric Load refers to a load which is not acting through the line of the axis of the column in case of column design. The eccentric load cause the column to bend towards the eccentricity of the loading and hence generates a bending moment in the column. In case of eccentric loading we have to design the column for both the Direct Compression and also for the bending moment also. Practically all columns are eccentric to some extent which may vary from few millimetres to few centimetres. In practical field it is almost impossible to make a perfectly axially loaded column, as a reason we have to consider a certain value of eccentricity for safety even though if we are designing for a axially loaded column. The conditions of considering eccentricity and its value may differ from code to code according to the country. Here I will tell you what I.S. : 462000 says. According to it the eccentricity which we have to consider for design must be taken as the greater of the followings :
i) 20mm.
ii) (lef/500) + (b/30)
Where,
lef = Effective Length of the Column
b = Lateral Dimension of the Column (We have to calculate two separate values for two sides in case of rectangular column)
Permissible Eccentricity : 0.05b where b is the dimension of a side of a column, we have to check for two sides separately in case of rectangular column.
The Permissible eccentricity must be greater than or equal to the actual eccentricity of the column. Or else we have to design it for bending also.
STEP 8 : Calculating The Area Of Steel Required :
Now the Area of Steel Required Asc is to be calculated from the Ag as the predetermined percentage of Ag. For example, if the GrossSectional Area of the Column is 78600 mm2 and at the starting of calculation of Ag it was assumed that 1% Steel is used then we get,
Asc = 1% of Ag = 0.01Ag = 0.01 X 78600 = 786 mm2
Now we shall provide such amount of Reinforcements that the CrossSectional Area of the Reinforcement provided is Equal to or Greater than the CrossSectional Area of Steel required above.
Hence in the above case we shall Provide 4 Nos. of 16mm Diameter Bars
Hence, The Actual Area of Steel Provided,
Hence the Area of Steel Provided is Greater than Area of Steel Required, Hence the Structure will be Safe.
NOTE : The minimum of 4 Nos. of Bars to be provided at the four corners of a rectangular or Square Columns and minimum diameter of Bars that to be used is 12mm Diameter. Hence 4 Nos. of 12mm Diameter Bars are must in any Columns irrespective of their necessities.
STEP 9 : Determining The Diameter and Spacing Of The Lateral Ties:
In this step, we will Determine the Diameter and the Spacing of the Lateral Ties or Transverse Links or Binders.
The Diameter of the Ties shall not be lesser than the Greatest of the following two values
 5mm
 1/4th of the Diameter of the Largest Diameter Bar
For an example if a Column has 16mm and 20mm both types of bar as Longitudinal Bars or main Reinforcement then 1/4th of 20mm = 5mm Hence we shall provide 5mm diameter Ties.
The Spacing of Ties shall not exceed the least of the followings three values
 Least Lateral Dimension
 16 Times of the Diameter of the Smallest Diameter Longitudinal Bar
 48 Times of the Diameter of Ties
For an example A Column of 250mm X 350mm Dimension having 20mm and 16mm Diameter Longitudinal Bars and 5mm Diameter Ties we get,
 Least Lateral Dimension = 250mm
 16 Times of the Diameter of the Smallest Diameter Longitudinal Bar = 16 X 16 = 256mm
 48 Times of the Diameter of Ties = 48 X 5 = 220mm
Hence Provide 5mm Diameter Ties @ 200mm C/C
[In this case our objective is to minimize the value to reduce the spacing and to make the structure more stable, hence we shall take least value and suitably in a multiple of 25mm]
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