# How can the axial load-carrying capacity of a steel column be determined?

Following on from our previous article on evaluating the Euler buckling load, we’re going to look at some other calculations we need when looking at buckling of columns.

## Eccentrically loaded steel column design.

The axial force can’t act centrically on the column, so, in addition, the column is always exposed to bending. As a result of the action of the bending moment, there is a further increase in the bending moment.

For the design of eccentrically loaded steel columns, the following conditions must be met:

NEdNcr,Nc,Rd +MEdMRd<1

Where:

- MEd=NEd x e
- e – eccentricity (m)
- M
_{Rd}– design resistance bending moment, MRd=W·fyM0 (kNm).

Euler critical buckling force is defined as: Ncr=2EILe2

The Euler critical buckling force is defined as:

Ncr=2EILe2

Take a 3-meter-long column with one end pinned and the other fixed. It has a cross-section area of A = 3.125 x 10-3 m2, moment of inertia of I = 4.7526 x 10-5 m4, and a width of 2.20 x 10-4 m3. Column is loaded with NEd = 150 kN at eccentricity of e = 0.02 m.

We’re going to calculate the axial force and bending moment interaction for steel grade S235 and E = 210 x 10^{9} Pa.

Nc,Rd=A x fyM0=0.003125 mm2 x 235 MPa1.1=667.6 kN

Le=K x L = 0.7 x 3.0m = 2.1m.

Ncr=2EILe2=9.86 x 210 x 109 Pa x 4.7526 x 10-52.12=223.14 kN

MEd=150 kN 0.02m=3.0 kNm

MRd=W x fyM0=2.2 x 10-4 m3 x 235 MPa1.1=47.0 kNm

NEdNcr,Nc,Rd +MEdMRd=150 kN223.14 kN+3.0 kNm47.0 kNm=0.67+0.064=0.734<1

**The axial load carrying capacity of real columns.**

Due to various imperfections of real steel columns, failure due to inelastic buckling occurs before the Euler buckling force. The procedure for determining the buckling load-bearing capacity of a real steel column consists of several steps:

- column plastic resistance calculation,
- column effective length determination,
- slenderness λ determination,
- slenderness λ
_{1}determination, - non-dimensional slenderness determination,
- buckling curve determination,
- reduction factor χ determination,
- column buckling load-bearing capacity calculation.

To consider geometric and material imperfections, dimensionless buckling curves were defined. The figure below shows dimensionless buckling curves.

Except for buckling curves, the reduction buckling factor (χ) can be calculated by the following expression:

χ=1ɸ+ɸ2-2<1.0

Where are:

ɸ=0.5·[1+α·̿-0.2+̿2]

Column buckling resistance (kN) is defined by the following expression:

Nb,Rd=χ x Nc,Rd.

Let’s check out another example. We’ll take another 3m long column with one end pinned and the other fixed. It has a cross-section area A=3.125×10^{-3} m^{2} and moment of inertia I=4.75026×10^{-5} m^{4}. Buckling curve a. Calculate the column buckling resistance for steel grade S235 and E=210 x 10^{9} Pa.

Nc,Rd=A x fyM0=0.003125 m2 x 235 MPa1.1=667.6 kN

Le=K x L=0.7 x 3.0m=2.1m

imin=IA=4.75026 x 10-50.003125=0.039 m

λ=Leimin=2.1m0.039m=53.8

1=π x Efy=3.14 x 210000 x 106 Pa235 x 106 Pa=29.9

=1=53.829.9=1.8

From the diagram above χ = 0.28.

Nb,Rd=χ x Nc,Rd=0.28 x 667.6 kN=186.9 kN

We’re going to finish off our series of buckling calculations in our next article, so be sure to check it out.

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