This example provides calculations on axial force of members based on Eurocode 9.

1. TALAT Lecture 2301
Design of Members
Axial Force
Example 5.2 : Axial force resistance of symmetric
hollow extrusion
4 pages
Advanced Level
prepared by Torsten Höglund, Royal Institute of Technology, Stockholm
Date of Issue: 1999
 EAA - European Aluminium Association
TALAT 2301 – Example 5.2 1

2. Example 5.2 Axial force resistance of symmetric hollow extrusion
Half width b 50 . mm fo 300 . MPa
Indent c 10 . mm E 70000 . MPa
Half width of a 18 . mm γ M1 1.0
flat parts
Thickness s 1.2 . mm
Length L 1200 . mm
Nodes no.,
co-ordinates, 0 b a s
thickness
1 b c a s
2 a b c s
3 a b s
MPa 10 . Pa
6
4 a b s
5 a b c s kN 1000 . newton
6 b c a s
7 b a s
i 8 y b z a t s
9 b c a s
10 a b c s
11 a b s
12 a b s
13 a b c s 50
14 b c a s 0
15 b a s
z 0
16 b a s 0
mm
50
50 0 50
Nodes i 1 .. rows ( y ) 1 y
mm
5.4.5 Local buckling
250 . MPa
ε
fo
2 2
yi yi 1 zi zi 1
5.4.3 (1) c) βi if t i > 0 , ,1
ti
5.4.5 (3) c) β ε ε
2
> 22 , 32 . 220 .
i
heat-treated, ρ c if , 1.0
i ε β i
β i
unwelded
TALAT 2301 – Example 5.2 2

3. Effective
thickness t eff ρ .c
t
βi t eff
i
= ρ c= =
Effective area i = βi = ε
t eff .
2 2 i mm
of elements dAi yi yi 1 zi zi 1
i 1 8.333 9.129 1 1.2
rows ( y ) 1 2 25.927 28.402 0.854 1.025
Area of effective 2 3 8.333 9.129 1 1.2
cross section A eff dAi A eff = 356.591 mm
4 30 32.863 0.77 0.924
i =1 5 8.333 9.129 1 1.2
6 25.927 28.402 0.854 1.025
rows ( y ) 1
First dAi Sy 7 8.333 9.129 1 1.2
moment Sy zi zi 1 . z gc
2 A eff 8 30 32.863 0.77 0.924
of area. i =1 9 8.333 9.129 1 1.2
Gravity centre
10 25.927 28.402 0.854 1.025
z gc = 0 m 11 8.333 9.129 1 1.2
12 30 32.863 0.77 0.924
13 8.333 9.129 1 1.2
14 25.927 28.402 0.854 1.025
Axial force resistance,
15 8.333 9.129 1 1.2
no flexural buckling
16 30 32.863 0.77 0.924
fo
N Rd A eff . N Rd = 107 kN
γ M1
50
0
z
mm
0
0
z GC
mm
50
50 0 50
y y GC
5.8.4 Flexural buckling ,
mm mm
Second rows ( y ) 1
ti .
2 2
moment of yi yi 1 zi zi 1
zi . zi 1 .
2 2
area of effective I zi zi 1
3
cross section i =1
I = 4.701 . 10 mm
5 4
Radius of I
gyration r
A eff
K .L l = 1.2 . 10 mm
3
Table 5.7 K 1 l
Table 5.5 α 0.2 λ o 0.1 k1 1 k2 1
and 5.6
See to the l. 1 . f o
right λ λ = 0.689
r π E
TALAT 2301 – Example 5.2 3

4. 1
0.5 . 1 α . λ
2
(5.33) φ λ o λ χ φ = 0.796 χ = 0.837
2 2
φ φ λ
fo
5.8.3 (1) N b.Rd χ .k 1 .k 2 . .A
eff N b.Rd = 89.5 kN
γ M1
TALAT 2301 – Example 5.2 4