Practical example of constructing a single-angle section

June 5, 2024 Roberto Magalhães

This article explains the design of individual angle profiles according to EN 1993-1-1 using a practical example. Angled sections subject to compressive forces are designed with reference to the appropriate section of the Code.

The procedure for sizing compression elements is the same as described in the article. Steel support structure in accordance with Eurocode 3.

Let's discuss the design with a practical example.

Construction data

Angular profile 60x60x6
Angular profile length 600mm
Applied load 100kN
Single row of screws
Steel grade S275

Division of sections

Let's find the material data

Profile thickness = 6 mm ≤ 40 mm and

Steel grade S275

From Table 3.1 of EN 1993-1-1,

fy = 275 N/mm ²

From Table 5.2 (Sheet 3 of 3)

ε = √(235/f _j ) = √(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 the section is class 3

Cross section compressive load capacity

N _Ed. /N _c,Rd ≤ 1.0

N _c,Rd = Af _j /γ _M0

γ _M0 = 1.0 from section 6.1

A = 695 mm ²

N _c,Rd = Af _j /γ _M0 = 695 x 275 / 1.0 = 191.1 kN

N _Ed. /N _c,Rd = 100/191.1 = 0.524 <1

The section meets the compression capacity.

torsion resistance

N _Ed. /N _{b, Road} ≤ 1.0

Calculate the buckling length L _cr

We need to check the buckling resistance at the critical axis. Therefore, we take into account the YY, ZZ and VV axes and determine the critical reduction factor (the lower the value, the greater the reduction in bending moment). This is mandatory when designing angular profiles.

Find the buckling length on each axis. The buckling length may be taken into account as specified in BS 5950.

YY axis; 0.85 x 600 = 510mm

ZZ axis; 1.0 x 600 = 600mm

VV axis; 0.85 x 600 = 510mm

Determine the radius of gyration (i) from the cross-section properties.

i – YY axis = 18.3 mm

i – ZZ axis = 18.3 mm

i-VV axis = 11.8 mm

λ1 = 93.9ε = 93.9 x 0.924 = 86.4

λ‾ = (Lcr / i)(1 / λ1)

Note: The symbol λ¯ must be as specified in the code. Compared to the code, there is a little difference between the symbol and the code.

Calculate for each direction to obtain the critical factor.

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

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

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

In construction with a simple angular profile, the connection is made by a single screw, λ¯ _{and f} must be taken into account. If there are two or more screws, eccentricities can be neglected.

In this example, consider the angle connected by a leg. Therefore, consider λ¯ _{and f}

YY axis; λ¯ _ef = 0.5 + 0.7λ‾ = 0.5 + 0.7 × 0.323 = 0.726

ZZ axis; λ¯ _ef = 0.5 + 0.7λ‾ = 0.5 + 0.7 × 0.379 = 0.765

VV axis; λ¯ _ef = 0.35 + 0.7λ‾ = 0.35 0.7×0.5 = 0.7

The critical reduction factor is the smallest value of “χ”. This value becomes smaller when Ø is larger. This value becomes larger when λ¯ _ef is higher. Therefore we choose

λ¯ _ef = 0.765

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

For angular profiles, the buckling curve is “b”.

α is derived from Table 6.1 of EN 1993-1-1.

α = 0.34

Now calculate Ø of Cl.
Ø = 0.5(1 + 0.34(λ¯0.2) + λ¯ ² ) = 0.5 (1 + 0.34 (0.765 – 0.2) + 0.765 ² ) = 0.889

Calculated reduction factor, χ

χ = 1 / (Ø + √(Ø ² –λ¯ ² ) ) = 1 / (0.889 + √(0.889 ² +0.765 ² )) = 0.742

Now calculate the buckling resistance

N _{b, Road} = χAf _j /γ _M1

N _{b, Road} = 0.742 x 695 x 275 / 1.0 = 141.8 kN

N _Ed. /N _{b, Road} <1.0