Columns are structural elements that supports beams, slabs and other members through axial compression in vertical direction. Columns also resist moments in both directions with its flexure capacity.

Design of Columns as Compressive Members

Columns undergo compression due to loads and moments acting on it. Due to Axial loads and biaxial moments, they can fail in compression, buckling, shear and rupture. We design Columns against all these possibilities.

For Buckling we consider slenderness effects and for Crushing failure due to compression, we design as per axial load carrying capacity of a section. Shear is taken up partly by concrete and partly by shear link steel bars provided at section.

Minimum Eccentricity in Columns

All Compression members are to be designed for a minimum eccentricity of loads in two principal directions.

\color{blue}e_{min}=\frac{L}{500}+\frac{D}{30}\;\;(minimum\;20 \;mm)

L : Unsupported length of column (mm).

D : Lateral dimention in X direction or Y direction whichever to be considered. (mm)

Now, Moment due to minimum eccentricity is calculated as following :

\color{blue}M_{min}=P_{u}\times Minimum\; Eccentricity

Finally, We take maximum of the Analysis Moment (Given Moment) and Minimum Eccentricity Moment for Design.

Types of Columns

• Short Column (Axial)
• Long Column (Slender)
• Pedestal

Slenderness Ratio λ

\color{blue}\lambda=\frac{Effective \; length}{Least \; Lateral \; Dimention}

Short Column : λ < 12 and λ > 3

Long Column : λ ≥ 12

Pedestal : λ < 3

Slender Columns

Slender Columns have large tendency of buckling due to its length. We consider Additional moments as per IS Code for Slender Columns.

\color{blue}M_{ax}=\frac{P_{u}D}{2000} \left ( \frac{l_{ex}}{D} \right )^2

\color{blue}M_{ay}=\frac{P_{u}b}{2000} \left ( \frac{l_{ey}}{b} \right )^2 

We add these additional moments to the applied moments (Analysis Moments) to get Total moment for design. These Additional Moments also include the P-Δ effects.

So, Finally we Get our Design Moments after considering Additional moments due to slenderness and minimum eccentricity moments.

Design Moment = Maximum of (Analysis Moment and Minimum Eccentricity Moment) + Moment due Slenderness

Axial Load Capacity of Columns

\color{blue}P_{u}=0.4f_{ck}A_{c} +0.67f_{y}A_{sc}

A_g : Gross Area of section

A_sc : Area of steel in compression

A_c : Area of Concrete at section = A_g – A_sc

For Pedestal, Nominal steel of 0.15 % is provided longitudinally.

Axial Load Capacity for Helical Reinforcement

\color{blue}P_{u}=1.05[0.4f_{ck}A_{c} +0.67f_{y}A_{sc}]

In case of Helical reinforcement, Concrete undergoes triaxial compression in short column.

Interaction Diagram

Each Column size we specify, there is an interaction curve associated with it, which gives its capacity to resist Axial force and Moments in specified directions. Our Indian Code provides us with a set of normalized Interaction curves for two types of bar arrangements. We use these curves to determine the Capacity of the section and then compare it with given force and moments for design.

In order to find out required values of Moment capacity in X and Y direction, we find out the value of P_u/f_ckbd which along with the value of d’/D and f_y pin-points some value along horizontal axis. This is the value M_u1/f_ckbd² in which M_u1 is maximum uniaxial moment capacity along considered axis. If the given applied moment along the axis M_u is less than M_u1 , the section is safe.

Demand-Capacity Ratio (D/C ratio) is ratio of Applied moment by Maximum uniaxial moment carrying capacity. It must be less than 1, for safe sections.

Types of bar arrangements mentioned in code :

• Reinforcement Distributed on 2 sides
• Reinforcement distributed equally on 4 sides.

P_u : Factored Axial Force

M_ux : Moment applied in x direction

M_uy : Moment applied in y direction ; M_u : Combined moment

L : Effective length

Effective length is calculated by multiplying total length by factor given in Figure 26 and 27 (IS 456) where, Figure 26 gives effective length ratio for No Sway condition and Figure 27 provides effective length ratio for Sway condition.

For taking out value from this chart, we need to calculate β₁ (Column top) and β₂ (Column bottom) values, which is ratio of flexural stiffness of column to the total flexural stiffness of column and beams spanning into the column.

Here, Effective length ratio is ratio of effective length and unsupported length. Unsupported length is the clear length of columns between two lateral supports.

b : Width

d’ = Top Cover + Diameter of bar/2

D : Overall depth of Column

Units to be used for ratio : N, mm

Methods of Columns Design

• Resultant Method
• Discrete Method

We Design Columns for combined axial force and bending.

Resultant Method

In this method we take combined effect of biaxial moments for design of section.

In order to take Resultant of Moment in X and Y direction, we do the vector addition of these Moments. We calculate Load Angle α .

\color{blue}\alpha=\tan^{-1}\frac{M_{uy}}{M_{ux}} 

\color{blue}M_{u}=  \sqrt{M_{ux}^2 + M_{uy}^2 }

For Our Manuel Way of Calculation, We assume a certain percentage of steel and then use interaction curves for safety check. Alternatively, We can directly find required percentage of steel from charts given in SP 16 (charts 27-62).

The values of P_u/f_ckbd and M_u/f_ckbd² along with d’/D , f_y and arrangement types pin points the value of p/f_ck

Now, Percentage of steel p = f_ck times the value.

But, In case of Softwares, They generate Resultant capacity of section – M_u.res.cap value through internal calculations and then find out Resultant capacity in x and y direction as follows:

M_ux.cap = M_u.res.cap Cos α

M_uy.cap = M_u.res.cap Sin α

Then, Softwares finally compare the values with design moments in both directions in order to check safety. This method is adopted widely as it is more precise and it provided more economic sections.

Discrete Method

In this method, we check given moments with respective uniaxial capacities in each directions. The given moment should be less than the uniaxial moment capacity for a given direction.

Furthermore, The section must also be safe against the combined action of biaxial moments, So Our Indian Codes gives us the following criteria that ensures safety if fulfilled.

\color{blue} \left ( \frac{M_{uy}}{M_{uy1}} \right )^{\alpha_{n}}   + \left (\frac{M_{ux}}{M_{ux1}}  \right )^{\alpha_{n}}\leq 1

M_ux1 and M_uy1 are maximum uniaxial capacities along x and y directions. M_ux and M_uy are given x and y moments applied.

α_n : Exponent factor which depends upon P_u/P_uz value as follows

\color{blue}P_{uz}=0.45f_{ck}A_{c} +0.75f_{y}A_{sc}

Design For Shear

For Shear Design, First we calculate Design Shear Force through Analysis. Afterwards, We find out Design Shear Stress at the section.

Shear stress along Depth D

\color{blue}\tau _{vy}=\frac{V_{u}}{B\times D_{eff.}}

Where, V_u is Design Shear force (N)

D_eff is effective depth (mm)

B : Width of section (mm)

For, Shear stress along Width B

\color{blue}\tau _{vx}=\frac{V_{u}}{D\times B_{eff.}}

Now, For values of percentage of main bars steel at section and Grade of concrete, We figure out value of Shear Strength τ_c , as per Table 19, IS 456.

But, Before using this value, we need to multiply it with shear strength enhancement factor (As per IS 456, clause 40.2.2, Shear strength of member under axial compression), This shear enhancement factor is given by following formula subjected to the maximum value of 1.5

\color{blue}Factor=1 +\frac{3P_{u}}{D\times f_{ck}\times B}

Finally, If the value of shear stress at section is less than the enhanced shear strength value, then shear links are not required. Otherwise, We will provide shear links to columns. In case of no shear steel requirement (τ_v < τ_c), we provide nominal ties with sufficient spacing.

Detailing Provisions

• Diameter should be greater than 12 mm.
• Minimum 4 bars for rectangular section and 6 bars for circular section.
• Minimum steel = 0.8 %
• Maximum Steel = 8 % (Practically 2.5 %).
• Diameter of lateral ties must be greater than One-forth of Maximum main bar diameter.
• For lateral ties, spacing should be minimum of 300 mm, 16 times diameter and width of section.
• If Spacing of Main bars is less than 75 mm, ties goes around corner bars only.

We design columns based on Ordinary and Ductile Conditions as per requirements.

Numerical Illustration for Design of a Slender Column by Biaxial Discrete Method

Given Data : 
Grade of Concrete = M40
Grade of Steel = Fe500
Width of Column B (Along x)= 600 mm
Depth of Column D (Along y) = 600 mm (Square Column)
Clear floor Height h = 900 mm  (both in respect of D and B Dimensions)
d' = 30 + 16/2 = 38 mm
Sway Condition : Non Sway
Effective length factor for both axis = 0.8
Effective length = 0.8 x 900 = 720 mm
Pu = 7321.31 kN
Mux = -201.24 kNm
Muy = -116.1 kNm
Vux = -116.14 kN
Vuy = -155.6 kN

FLEXURE AND AXIAL DESIGN 

Minimum Eccentricity along B (governing) = 900/500 + 600/30 = 21.8 mm > 20 mm
Eccentricity Moment = 7321.3 x 21.8/1000 = 159.6 kNm
Along D : Slenderness Ratio = Effective length / D = 720/600 = 1.2 < 12 (Not slender)
Along B : Slenderness Ratio = Effective length / D = 720/600 = 1.2 < 12 (Not slender)
Mux(Design Final) = 201.24 kNm
Muy(Design Final) = 159.6 kNm (Since eccentricity moment is greater)
Pu = 7321.31 kN
Uniaxial capacity along x-x (along D)
Pu/fckbd = (7321.31 x 1000)/(40x600x600) = 0.5
Let us assume percentage of steel provided to be 1.5 %
Now, p/fck = 2/40 = 0.03
d'/D = 38/600 = 0.06 (about 0.05)
fy = 500 N/mm2
Our section has reinforcement uniformly distributed on two sides.

So, Using Code SP 16, Chart 35 (d'/D = 0.05,fy = 500)
For Pu/fckbd = 0.5 & p/fck = 0.03, We get corresponding value 
Mux1/fckbd2 = 0.03
Now, Mux1 = 40 x 600 x 6002 x 0.03 / 106 = 259.2 kNm
Similarly, Along y-y (along B)
We have same values of d' and D (As B = D), which gives same d'/B
So, we get Muy1/fckdb2 = 0.03 and hence, Muy1 = 259.2 kNm
Here, we have 
Mux =  201.24 kNm < Mux1 (SAFE)
Muy =  159.6 kNm  < Muy1 (SAFE)
Now, Check for Combined Action : 
Puz = 0.45 x 40 x (600x600)(1-0.015) + 0.75 x 500 x 0.015 x 600 x 600 = 8407.8 kN
Pu/Puz = 7321.31/8407.8 = 0.87 
SO, αn = 2 (Pu/Puz > 0.8)
Now, Biaxial moment action check -
(Muy/Muy1)αn + (Mux/Mux1)αn = (159.6/259.2)2 + (201.24/259.2)2 = 0.98 < 1  (SAFE)

Hence, Area of Main bars Provided = 1.5 x 600 x 600/100 = 5400 mm2
Let us Provide 12 nos. of 25 mm bars equally distributed along both sides.

SHEAR DESIGN

Governing shear force Vuy = -155.6 kN    (we generally design along both directions and take maximum of two)
Deff = 600 - 38 =562 mm
Shear Stress (Along D or y axis) = τv = Vuy/(B x Deff) = 155.6 x 1000/(600 x 562) = 0.46 N/mm2
For M40 and pt= 1.5 %, We get τc = 0.79 N/mm2 
Enhanced shear factor = 1 + (3 x 7321.31 x 1000)/(40 x 600 x 600) = 2.52, which becomes 1.5 due to maximum limit criteria.
Enhanced shear stress = 1.5 x 0.79 = 1.18 N/mm2 
Here, we can see that enhanced shear strength is greater than design shear stress at section, Hence, there is no requirement for shear bars. But We need to provide nominal ties for binding the main bars and create some confinement. Let us provide 8 mm ties at spacing of 300 mm c/c.