Design of compression members- Design of Column

In this article, we will discuss the Design of compression members step by step. So, to get a clear understanding read all carefully without skipping any lines.

What is compression members?

A compression member is defined as a structural member subjected t compressive force in a parallel directional parallel to its longitudinal axis.

The most common type of compression members are columns ( vertical elements in the building ). When the effective length of the structural member exceeds three times the least lateral dimension, it is known as a column.

If the effective length is less than or equal to three times the least lateral dimension, it is known as a pedestal.

Types of column

On the basis of slenderness ratio (l_ex/d or l_ey/b ), column are classified as:

Short  column

When the slenderness ratio l_ex/d and l_ey/b both are less than 12, the column is known as a short column. Where,

L_ex = Effective length of column in major axis

L_ey = Effective length of column in minor axis

d = depth of column in respect to major axis

b = breadth of column

Long  column

when either or both of slenderness ratio l­ex/d and l_ey/b are greater than or equal to 12, the column is known as long column. Where,

L_ex = Effective length of column in major axis

L_ey = Effective length of column in minor axis

d = Depth of column in respect to major axis

b = breadth of column

Design of compression members- Short Column

Partially no column is supposed to be subjected under perfect axial load. So the design of short column is based on the assumption of maximum eccentricity as 5 % of the lateral dimension.

Design steps of short column:

1) Calculate the factored load of the column.

2) Assume percentage of longitudinal steel.

( Note: Theoretically the percentage of steel lies between 0.8 % to 6 % in column. However, from practical consideration, the range of steel in column is 0.8 % to 4%)

3) From clause 39.3 of IS code 456:2000, the relation of axial load with load in the concrete and load in the steel is simplified to obtain gross area, area of concrete and area of steel.

Mathematically,

 Pu = 0.4fck.Ac + 0.67 fy.Asc

Where,

Pu =  Axial load on the member

fck =  characteristic compressive strength of the concrete

fy =  characteristic strength of the compression reinforcement

Asc = Area of longitudinal reinforcement of the column

       =  p % of gross area (Ag)

Ac = Area of concrete

     = Ag – p % of Ag

4) From the area steel ( Asc), the number of longitudinal reinforcement bars and the diameter of the reinforcement bars is fixed.

( note : Minimum 12 mm diameter bar should be used in column. Minimum number of bars in the cross section should be equal to the edges in cross section, i.e. four for rectangular or square, five for pentagon, six for hexagon, etc. Reinforcement bars should be distributed symmetrically. )

5) Lateral ties should be provided of diameter ɸ. Where, ɸ = (ɸ_{longitudinal bar} /4 ) or 6 mm ( whichever is greater ). Similarly, pitch of lateral ties should be least of

• Least lateral dimension
• 16 * diameter of the longitudinal bar
• 300 mm

6) Design summary needs to be provided with suitable diagram.

Example

Q)  Design a short column in section to carry an axial load of 100 KN using M20 mix and Fe 415 steel.

Solution;

Given, axial load = 100 KN

Grade of concrete = M20

Grade of steel  =  Fe415

We have,

Factored axial load (Pu) = 1.5 * 1000 KN

                                              = 1500 KN

Assume 1 % steel of total area of section and eccentricity of 5 % of lateral dimension.

 From clause 39.3 of IS cod 456:2000

   Pu = 0.4 fck.Asc + 0.67 fy Asc

Or, 1500 * 1000 = 0.4*20*(Ag – 1% of Ag) + 0.67*415*(1% of Ag)

Or, 1500* 1000 = 0.4*20*0.99Ag + 0.67*415*0.01 Ag

Or, 10.7005 * Ag = 1500 * 1000

Or, Ag = 140280.36

Since the column has square section,

Provide 380 mm * 380 mm square column

Here,

Area of steel (Asc) = 1 % of (380 *380) = 1444 mm²

If 4 nos of longitudinal bars of diameter ɸ­_L is provided, then,

Thus, provide 4 nos. of 22 mm diameter bars as longitudinal bar.

Design of  lateral ties

Pitch distance of lateral ties is least of the following;

1. Least lateral dimension = 380 mm
2.  16 * ɸ_L = 16 * 22 = 352 mm
3. 300 mm

Among 3 option no. 3 has least value. Thus provide 6 mm diameter of lateral ties at a spacing of 300 mm center to center.

Design summary

The design detail is illustrated in figure below:

Size of column = 380 mm * 380 mm

Longitudinal reinforcement = 4 nos. of 22 mm diameter bar

Traverse reinforcement = 6 mm diameter of lateral ties at a spacing of 300 mm c/c.

Design of compression members- Long Column

The design of a long column is similar to that of a short column except that additional moments need to be calculated as suggested by clause 39.7 of IS code 456:2000.

Example:

Q) Design of braced column 300 mm * 400 mm in size subjected to following conditions:

L_ex =L_ey = 6 m

Factored axial load = 1000 KN

Mu_x = 40 KNm at top and 30 KNm at bottom

Mu_y = 30 KNm at top and 25 KNm at bottom

Use M20 Concrete and Fe 415 steel

Solution:

Given,

Size of column = 300 mm * 400 mm

L_ex =L_ey = 6 m = 6000 mm

Factored axial load (Pu) = 1000 KN

Mu_x = 40 Knm at top and 30 KNm at bottom

Mu_y = 30 KNm at top and 25 KNm at bottom

Here,

Thus,

Since the design is greater of the imposed moment and moment due to eccentricity,

M’_ux = 25.33 KNm

M’_uy = 22 KNm

From clause 39.7.1 of IS code 456:2000,

Additional moments are:

Therefore, total design moments are:

Now, assume reinforcement to be distributed equally on four sides with clear cover (c.c.) = 40 mm and diameter of reinforcing bar (ɸ) = 20 mm

Assume percentage of steel (P) = 3%

Calculation of moment capacity along longer side

Using chart 45 of Sp 16, we get

Or, Mux1 = 0.165 *20*300*400²

                 = 158.4 KNm

Again.

Calculating moment capacity along shorter side,

Or, Muy₁ = 0.165*20*400*300²

                 = 118.8 KNm

Now,

From clause 39.6 of IS code 456:2000

Puz = 0.45 fck.Ac + 0.75 fy. Asc

       = 0.45 *20*0.97*300*400 + 0.75*415*0.03*300*400

      = 2168.1 KN

Now,

For the design to be safe

This is true

So, the design is safe with 3% steel

If n numbers of 20 mm ɸ bars are used then

Area of steel = 3% of b*D

Or. N*314.16 = 3600

Or, n = 11.46

To distribute equally in four sides, we use 12 numbers of 20 mm ɸ bars

Design of traverse reinforcement

Diameter of tie bar is greater of,

Thus a spacing of 300 mm c/c can be adopted for ties.

Design summary

Size of column = 300 mm*400 mm

Longitudinal  reinforcement = 12 nos. of 20 mm ɸ bar distributed equally on four sides

Traverse reinforcement = 6 mm diameter lateral ties at spacing of 300 mm c/c

Clear covert = 40 mm

The design detail is illustrated in figure below:

I hope you understood how to Design of compression members i.e short column and long column.

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