The Design of R.C.C. Column is very important, as in the Framed Structured Building that is in the modern day construction the Frames are formed by networks of Columns and Beams. A Column is a Structural Member which is vertical and subjected to Compressive forces, and having it’s Length (Height) greater than the least cross-sectional dimension of it self. To Satisfactorily Complete the Design of Column The Following Steps must followed.

__STEP 1 :-__

__Calculation of the Influence Area of the Column :__
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 Column 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 :__

I.
Due to weight of Slab [25000 N/m

^{3 }]
II.
Due to weight of Floor Finish [500 N/m

^{2}]
III.
Due to weight of Brick Masonry [19200 N/m

^{3}]
IV.
Due to weight of Beam [25000 N/m

^{3}]
V.
Due to weight of Self Weight of Column [25000 N/m

^{3}]__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.: 875-Part II.

For Residential Buildings it is generally considered @
2KN/m

^{2}= 2000 N/m^{2}
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 Cross-Sectional Area Required For The Column :__
This is the one of the most important and main step of the
Design of Column.

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 P

_{u.}
Hence Factored Load, P

_{u}= 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.: 456-2000

The Ultimate Load of a Column is given by,

P

_{u}= 0.4.f_{ck}.A_{c}+ 0.67.f_{y}.A_{sc}[Equation I]
Where, P

_{u}= Ultimate Load of the Column in N/mm^{2}
f

_{ck}= Yield Strength of Concrete in N/mm^{2}
A

_{c }= Area of Concrete (Cross-Sectional Area) of Column in mm^{2}
f

_{y }= Yield Strength Of Steel in N/mm^{2}
A

_{sc}= Area of Steel (Cross-Sectional Area) in Column in mm^{2}
Now the column consists of Concrete and as well as Steel in
the form of Reinforcements hence the Total Cross-Sectional Area of Column is
made of Area of Concrete and Area of Steel.

The Total Cross-Sectional area of Column can be also termed
as Gross Cross-Sectional Area of Column and it’s denoted by A

_{g}.
Hence, Gross Cross-Sectional Area of Column = C/S Area of Concrete + C/S Area of Steel

Therefore, A

_{g}= A_{c}+ A_{sc}
And hence, A

_{c}= A_{g}- A_{sc}
Now putting the above obtained value in the original
equation (Equation I) we get,

P

_{u}= 0.4.f_{ck}.(A_{g}-A_{sc}) + 0.67.f_{y}.A_{sc }[Equation II]
Now Assume the Percentage of Steel you want to use ranging
anywhere from 0.8% to 6% with Respect to Gross Cross-Sectional Area of the
Column (A

_{g}). Say Assuming Steel as 1% of A_{g}it means Area of Steel A_{sc}= 1% of A_{g}= 0.01A_{g}
The higher will be the percentage of steel used the lower
will be A

_{g}and thus lesser will be the cross-sectional 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 A_{g}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 loadings.
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 (P___{u}) 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 A

_{sc}in the form of A_{g }in the Equation I. For example suppose 1% Steel is used then the equation will be like the one below :-
P

_{u}= 0.4.f_{ck}.(A_{g}-0.01A_{g}) + 0.67.f_{y}.0.01A_{g}
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 Gross-Sectional Area (A

_{g}) 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 A

_{g}or Gross-Sectional 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 A_{g}we can determine the Lateral Dimensions.
For making a Square Section just Determine the Root Value of
the A

_{g}. Like if the Value of A_{g}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,
A

_{g}= b X D = b X 2b = 2b^{2}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 :-__

__Calculating The Area Of Steel Required :__Now the Area of Steel Required A

_{sc }is to be calculated from the A

_{g}as the predetermined percentage of A

_{g}. For example if the Gross-Sectional Area of the Column is 78600 mm

^{2}and at the starting of calculation of Ag it was assumed that 1% Steel is used then we get,

A

_{sc}= 1% of A_{g}= 0.01A_{g}= 0.01 X 78600 = 786 mm^{2}
Now we shall provide such amount of Reinforcements that the
Cross-Sectional Area of the Reinforcement provided is Equal to or Greater than
the Cross-Sectional 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 Column and minimum diameter of Bars that to be used is 12mm Diameter. Hence 4 Nos. of 12mm Diameter Bars are must in any Column irrespective of their necessities.__

__STEP 5 :-__

__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
- 1/4
^{th}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 1/4

^{th}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]

[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]