STEP 1 :- Calculation of the Influence Area for Design of 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.

 

DETERMINATION OF INFLUENCE AREAS FOR LOAD DISTRIBUTION ON COLUMNS

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 :-

        I.            Due to weight of Slab                                    [25000 N/m³ ]

      II.            Due to weight of Floor Finish                        [500 N/m²]

    III.            Due to weight of Brick Masonry                   [19200 N/m³]

    IV.            Due to weight of Beam                                 [25000 N/m³]

      V.            Due to weight of Self Weight of Column        [25000 N/m³]

 

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/m² = 2000 N/m²

 

COLUMN FORM WORK OF BUILDING

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 Columns.

STEP 3 :-Finding The Gross Cross-Sectional Area Required For The Design of 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 of Columns 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/mm²

              f_ck= Yield Strength of Concrete in N/mm²

              A_c = Area of Concrete (Cross-Sectional Area) of Column in mm²

               f_y = Yield Strength Of Steel in N/mm²

            A_sc= Area of Steel (Cross-Sectional Area) in Column in mm²

 

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 loading.

 

My suggestion is to use the following Percentage of steel for the Design of Column, 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 mm²Then considering square section of a column we can get each side

 

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. A short column mainly fails by direct compression and has a lesser chance of failure by buckling. And in the case of a long column the failure mainly occurs due to the buckling alone. Long column being slender, that is being thin like stick as compared with its length it grows a tendency to get bend by deviating from its vertical axis under the action of loads. Due to this tendency of long column to get buckled (bend) a long column of all same properties and dimensions that of a short column will be able to carry much lesser load safely than that of the short column.

 

Suppose a 400mmx400mm short column can take a load of 1000KN , then a long column of 400mmx400mm having same grade of concrete, same amount of reinforcement and same workmanship will be able to carry a lesser load like say about 800KN only, hence we get a loss of 200KN which is 20% loss of load carrying capacity. So the above formula used in Step 3 holds good only for the Short Column. For using it in long column a little modification is needed, which I will update it later when I will get hands on this article again. For now let us concentrate on Short 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 a unsupported length of 2.7m = 2700mm, hence the effective length will be lef = 0.65×2700 = 1755mm. 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 400mmx200mm section and has an unsupported length of 2700mm, then the Ration of Effective length t the Least Lateral Dimension will be as follows :-

(Effective Length/Least Lateral Dimension) = (lef/b) = (1755/200) = 8.775 which is less than 12 and hence is a Short Column.

 

STEP 5  Check For Eccentricity for Design of Columns :

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. : 456-2000 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.

STEP 6 :-Calculating The Area Of Steel Required for Design of Columns :

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 mm²and 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 mm²

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,

 

STEP 7 :-Determining The Diameter and Spacing Of The Lateral Ties for Design of Columns:

 

The Diameter of the Ties shall not be lesser than the Greatest of the following two values 

1.  6mm 
2.  1/4^th of the Diameter of the Largest Diameter Bar

 

 

The Spacing of Ties shall not exceed the least of the followings three values 

1.  Least Lateral Dimension 
2. 16 Times of the Diameter of the Smallest Diameter Longitudinal Bar 
3. 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 6mm 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 6 = 288mm

Hence Provide 5mm Diameter Ties @ 250mm 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]

 

R.C.C. is actually consisting of two main different parts firstly Concrete and secondly Reinforcement in the form of Steel. As we all know that when ever loads come on a structure it tries to take the shape of bending, and the nature of bending depends upon it’s supports and their respective positions. When the shape of bending takes a shape convexity downward then it’s called as Sagging and the Bending the an corresponding Bending Moment is Treated as +ve , and when the shape of the bending takes a shape of convexity upwards then it’s called as Hogging and the Bending and corresponding Bending Moment is treated as –ve.

Why Reinforcement is Needed?

Due to the bending of a structure one side of the structure about Neutral Axis Compresses and the other side is tensed, due to the above fact Compressive Stress and Tensile Stress develops in the structure from the Neutral Axis towards the Extreme Fiber that is the exterior face of the structure in a increasing manner in terms of magnitude of the stresses. So the Structure is having both Compressive and Tensile Stresses in it.

In the case of a Concrete, it has enough resistance against Compressive Stresses but it is very very Weak in Tensile Stresses and it can only resist a very little Tensile Stress which is as good as negligible. Hence to make a structure safe and in working condition to take up the Design Loads the structural material must e able to withstand both Tensile as well as Compressive Stresses, now as the Concrete is not strong to withstand the Tensile stress, so some measures is to be taken and it is to be provided with something which will take up the Tensile Stresses developed in Concrete. Due to the above fact A Structural Member such as Column, Beam, Slab etc are not made of Plain Concrete, instead it’s Reinforced with Steel embedded in it, which take ups the Tensile Stresses developed.

Why Steel is Used As Reinforcement?

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