# Design of Column

By Deepanshu Rastogi|Updated : December 2nd, 2021

Through Champion Study Plan for GATE Civil Engineering (CE) 2022, we are providing Design of Column study notes and other important materials on every topic of each subject.

These topic-wise study notes are useful for the preparation of various upcoming exams like GATE CivilIESBARCISROSSC-JEState Engineering Services examinations and other important upcoming competitive exams.

The article contains fundamental notes on the "Design of Column"  topic of the "Design of Concrete Structures" subject.

## Design of Column

#### Working Stress Method(Not in the GATE Syllabus)

Slenderness ratio (λ)

If λ > 12, then the column is long.

Load carrying capacity for short column

where AC = Area of concrete,

σSC Stress in compression steel

σCC Stress in concrete

Ag Total gross cross-sectional area

ASC Area of compression steel

Load carrying capacity for long column

where,

Cr = Reduction factor

where leff = Effective length of the column

B = Least lateral dimension

imin = Least radius of gyration and

where, l = Moment of inertia and A = Cross-sectional area

The effective length of a column

The effective length of Compression Members

A column with helical reinforcement

The strength of the column is increased by 5%

for short column

for long column

Longitudinal reinforcement

(a) Minimum area of steel = 0.8% of the gross area of the column

(b) Maximum area of steel

(i) When bars are not lapped Amax = 6% of the gross area of the column

(ii) When bars are lapped Amax = 4% of the gross area of the column

Minimum number of bars for reinforcement

For rectangular column  4

For circular column  6

Minimum diameter of bar = 12 mm

Maximum distance between longitudinal bar = 300 mm

Pedestal: It is a short length whose effective length is not more than 3 times of least lateral dimension.

Transverse reinforcement (Ties)

where Φmin = dia of the main longitudinal bar

φ = dia of the bar for transverse reinforcement

Pitch (p)

where φmin = minimum dia of the main longitudinal bar

Helical reinforcement

(i) Diameters of helical reinforcement are selected such that

(ii) Pitch of helical reinforcement: (p)

where,

dC = Core diameter = dg – 2 × clear cover to helical reinforcement

AG = Gross area = Π(dg)2/4

dg = Gross diameter

Vh = Volume of helical reinforcement in a unit length of the column

φh = Diameter of steel bar forming the helix

dh = centre to centre dia of the helix

= dg – 2 clear cover - φh

φh = diameter of the steel bar forming the helix

Some other IS recommendations

(a) Slenderness limit

(i) Unsupported length between end restrains < 60 times least lateral dimension.

(ii) If in any given plane, one end of the column is unrestrained, then its unsupported length<100 B2/D

(b) All columns should be designed for a minimum eccentricity of

### Limit state method

1. Slenderness ratio (λ)
if
λ<12 Short column
1. Eccentricity

If emin≤0.05 D, it is a short axially loaded column.
where Pu = axial load on the column
2. A short axially loaded column with helical reinforcement
3. Some other IS code Recommendations

(a) Slenderness limit

(i) Unsupported length between end restrains <60 times least lateral dimension.

(ii) If in any given plane, one end of the column is unrestrained, then its unsupported length <100 B2/D

(b) All columns should be designed for a minimum eccentricity of

Where e = 0, i.e., the column is truly axially loaded.

This formula is also used for members subjected to combined axial load and bi-axial bending and is also used when e > 0.05 D.

For more information about the design of column, you can refer to the following video on the Byju Exam Prep's official youtube channel.

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