Steel Column Design EC3 Worked Example

Steel column design according to Eurocode 3, EN 1993-1-1, is discussed in this article by a worked example. Clarifications on design are done by referring to the relevant clauses of the code.

A detailed explanation of the steel column design is discussed in the article steel column design using Eurocode 3.

The design process given in the above article is as follows.

Let’s start the design process with a worked example.

Design Data

• Size of the column 150x150x 31.1 kg/m
• Steel Grade S275
• Height of the column 3m
• Axial load 500 kN
• Consider pin ended column for simplicity in this example

Classification of Section

Let’s find the strength parameters.

t < 40mm

From Table 3.1 of EN 1993-1-1,

fy = 275 N/mm2

ε = √(235/f_y) = √(235/275) = 0.92

From Clause 6.1,

γ_m0 = 1.0

γ_m1 = 1.0

A = 39.65 cm²

We can calculate the slenderness limits.

Flange

C_f/t_f = 63.5 / 10 = 6.35  < 9ε = 9 x 0.92 = 8.82; Section Class 1

Web

C_w/t_w = 114 / 7 = 16.3 < 33ε = 33 x 0.92 = 30.36; Section Class 1

• Plastic Resistance or Resistance of Cross Section, N_{c, Rd}
  • Non-Slender Sections: Class 1

N_c,Rd = Af_y / γ_m0 = 3965 x 275 / 1 = 1090.4 kN

After calculating the resistance, the following capacity check shall be done.

N_Ed = 500 kN

N_Ed / N_c,Rd = 500 / 1090.4 = 0.46 < 1 

Cross section satisfy the plastic resistance.

• Buckling resistance, N_b,Rd

Firstly, calculate the buckling resistance depending on the class of the section.

let’s follow the procedure given in the design process.

λ1 = 93.9ε = 93.9 x 0.92 = 86.4

In column designs, we have to consider the resistance around each axis[ Y-Y and Z-Z].

i = radius of gyration

I_yy = 6.4 cm

i_zz = 3.77 cm

L_cr = Buckling Length

Sine the both ends of the column is pined,

L_cr = 1.0 x L = 3 m

λ¯ = L_cr / i λ₁

Y-Y  axis; λ¯ = L_cr / i λ₁ = 3000 / ( 64 x 86.4 ) = 0.54

Z-Z  axis; λ¯ = L_cr / i λ₁ = 3000 / ( 37.7 x 86.4 ) = 0.92

Let’s find the α – imperfection factor from Table 6.1 of EN 1993-1-1

Find the Buckling Curve

h / b = 150 / 150 = 1 < 1.2

t_f = 10mm < 100 mm

Now let’s refer to the part of Table 6.2 of EN 1993-1-1,

Thus, the bucking curve are as follows

Y-Y axis – b

Z-Z axis – c

From Table 6.1

Buckling Curve	a0	a	b	c	D
Imperfection factor α	0.13	0.21	0.34	0.49	0.76

α at Y-Y axis = 0.34

α at Z-Z axis = 0.49

Now calculate, Ø

Ø = 0.5 [ 1 + α( λ¯ – 0.2) + λ¯²]

About Y-Y axis

Ø = 0.5 [ 1 + 0.34(0.54 – 0.2) + 0.54²] = 0.7

About Z-Z axis

Ø = 0.5 [ 1 + 0.49(0.92 – 0.2) + 0.92²] = 1.1

Calculate Reduction factor, χ

χ =1 /[ Ø + √(ز – λ¯² )] ≤ 1.0

About Y-Y axis

χ =1 /[ 0.7 + √(0.7² – 0.54² )] = 0.873 < 1.0

About Z-Z axis

χ =1 /[ 1.1 + √(1.1² – 0.92² )] = 0.587 < 1.0

The lowest χ is the critical axis against buckling and less buckling resistance is available.

Thus,

χ = 0.587

Now calculate Buckling Resistance N_b,Rd,

N_b,Rd = χ A f_y / γ_M1 – Non slender – Class 1

N_b,Rd = 0.587 x 3965 x 275 / 1 = 640 kN

• Capacity Check

N_Ed / N_b,Rd ≤ 1.0

N_Ed / N_b,Rd = 500 / 640 = 0.781 < 1

Section is Ok.

As additional information, you may refer to the article published in Wikipedia Eurocode 3: Design of Steel Structures.