Single Angle Section Design Worked Example

Single angle section design using EN 1993-1-1 is discussed in this article by a worked example. Angle section subject to compressive force is designed with referring relevant section in the code.

The design procedure of the compression members is the same as discussed in the article Steel Column Design using Eurocode 3.

Let’s discuss the design with a worked example.

Desing Data

• Angle section 60x60x6
• Lenght of the angle section 600mm
• Applied Load 100kN
• Single raw of bolts
• Steel grade S275

Classification of the Sections

Let’s find the material data

Thickness of section = 6 mm ≤ 40 mm and

Steel grade S275

From Table 3.1 of EN 1993-1-1,

fy = 275 N/mm²

From Table 5.2 (sheet 3 of 3)

ε = √(235/f_y) = √(235/275) = 0.924

From Table 5.2 (sheet 3 of 3)

h/t = 60 / 6 = 10 ≤ 15ε = 15 x 0.924 = 13.86

(b+h) / 2t = (60+60) / (2×6) = 10 ≤ 11.5ε = 11.5 x 0.924 = 10.626

Therefore, section is class 3

Compression Capacity of Cross Section

N_Ed / N_c,Rd ≤ 1.0

N_c,Rd = Af_y / γ_M0

γ_M0 = 1.0 from Clause 6.1

A = 695 mm²

N_c,Rd = Af_y / γ_M0 = 695 x 275 / 1.0 = 191.1 kN

N_Ed / N_c,Rd = 100 / 191.1 = 0.524 < 1 

Section satisfy the compression capacity.

Buckling Resistance

N_Ed / N_b,Rd ≤ 1.0

Calculate the buckling length, L_cr

We need to check the buckling resistance in the critical axis. Therefore, we consider axis Y-Y, Z-Z, and V-V and find the critical reduction factor(less the value higher the reduction in the bending moment. This is a must in angle section design.

Find the buckling length in each axis. Buckling length may be considered as specified in the BS 5950.

Y-Y axis; 0.85 x 600 = 510 mm

Z-Z axis; 1.0 x 600 = 600 mm

V-V axis; 0.85 x 600 = 510 mm

From section property data, obtain the radius of gyration (i)

i – Y-Y axis = 18.3 mm

i – Z-Z axis = 18.3 mm

i- V-V axis = 11.8 mm

λ1 = 93.9ε = 93.9 x 0.924 = 86.4

λ‾ = (Lcr / i)( 1 / λ1)

Note: the symbol λ¯ shall be as indicated in the code. There is a slight difference in the symbol when compared with the code.

Calculate for each direction to obtain the critical factor.

λ‾ Y-Y axis = (510 / 18.5)( 1 / 86.4) = 0.323

λ‾ Z-Z axis = (600 / 18.5)( 1 / 86.4) = 0.379

λ‾ V-V axis = (510 / 11.8)( 1 / 86.4) = 0.5

When the single angle section design the connection is made by a single bolt, λ¯_eff shall be considered. If there are two or more bolts, eccentricities may neglect.

In this example, consider the angle connected through one leg. Therefore, consider, λ¯_eff

Y-Y axis; λ¯_eff = 0.5 + 0.7λ‾ = 0.5 + 0.7×0.323 = 0.726

Z-Z axis; λ¯_eff = 0.5 + 0.7λ‾ = 0.5 + 0.7×0.379 = 0.765

V-V axis; λ¯_eff = 0.35 + 0.7λ‾ = 0.35 0.7×0.5 = 0.7

The critical reduction factor is the less value of “χ”. This value become lesser when Ø is larger. This value becomes larager when is λ¯_eff higher. Therefore, we select

λ¯_eff = 0.765

Now obtain Buckling curve from Table 6.2 of EN 1993-1-1.

For angle sections, the buckling curve is ” b “.

Obtain α from Table 6.1 of EN 1993-1-1.

α = 0.34

Now calculate Ø from Cl. 6.3.1.2
Ø = 0.5[ 1 + 0.34(λ¯0.2 ) + λ¯²] = 0.5 [ 1 + 0.34 (0.765 – 0.2) + 0.765²] = 0.889

Calcualte reduction factor, χ

χ = 1 / [ Ø + √(ز – λ¯²) ] = 1 / [ 0.889 + √( 0.889² + 0.765²)] = 0.742

Now calculate the buckling resitance

N_b,Rd = χAf_y / γ_M1

N_b,Rd = 0.742 x 695 x 275 / 1.0 = 141.8 kN

N_Ed / N_b,Rd < 1.0

N_Ed / N_b,Rd = 100 / 141.8 = 0.7 < 1

Section is adequate.

Further reading on structural steel, you may refer to the article Wikipedia article.