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

1. TALAT Lecture 2301
Design of Members
Axial Force
Example 5.4 : Axial force resistance of channel cross
section
7 pages
Advanced Level
prepared by Torsten Höglund, Royal Institute of Technology, Stockholm
Date of Issue: 1999
 EAA - European Aluminium Association
TALAT 2301 – Example 5.4 1

2. Example 5.4. Axial force resistance of channel cross section
Width/depth b 80 . mm h 100 . mm fo 300 . MPa kN 1000 . newton
E
8 . mm 25 . mm 70000 . MPa MPa 10 . Pa
6
Inner thickness/outstand ti c E G
2.6
Outer/bottom thickness to 3.5 . mm tu 4 . mm γ M1 1.0
Length L 1200 . mm
i 0 .. 8 k 1 .. 3
Nodes no., tli to th i to yk 0.5 . h zi b
co-ordinates,
thickness th 1 ti tl8 ti yk 4 0.5 . h zk 2 0 . mm
y4 0 . mm z2 0.5 . z1
th 4 tu tl4 tu
y0 0.5 . h c z6 0.5 . z1
th 5 th 4 tl5 tl4
y8 y0 z1 z1 0.5 . ( ti to )
yi zi tli th i z7 z1
= = = =
i = mm mm mm mm
0 25 80 3.5 3.5
1 50 77.75 3.5 8
2 50 40 3.5 3.5
3 50 0 3.5 3.5
4 0 0 4 4
5 -50 0 4 4
6 -50 40 3.5 3.5
7 -50 77.75 3.5 3.5
90
8 -25 80 8 3.5
65
40
Nodes i 1 .. rows ( y ) 1 15
Cross-section constants, varying thickness 10
60 30 0 30 60
0 . mm
2
Sectorial ω0
co-ordinates
ω 0 yi .z yi . zi 1 ωi ω ω 0
i 1 i i 1 i
Thicknesses Dti tli Dti tli Dti
to simplify Dti th i tli t1 i tli t2 i t3 i
expressions 2 2 3 3 4
Length of 2 2
elements li yi yi 1 zi zi 1
rows ( y ) 1
tli th i
Area A .l A = 1.233 . 10 mm
3 2
i
2
i =1
rows ( y ) 1
yi 1 . t1 i yi 1 . t2 i . li S z = 8.47 . 10
13 3
First moments Sz yi mm
of area
i =1
TALAT 2301 – Example 5.4 2

3. TALAT 2301 – Example 5.4 3

4. rows ( y ) 1
zi 1 . t1 i zi 1 . t2 i . li S y = 4.388 . 10 mm
4 3
Sy zi
i =1
rows ( y ) 1
Sω ω . t1 ω ω . t2 . l S ω = 7.328 . 10 mm
6 4
i 1 i i i 1 i i
i =1
rows ( y ) 1
yi 1 . t1 i 2 . yi 1 . yi yi 1 . t2 i yi 1 . t3 i . li
2 2
Second Iz yi
moments
of area i =1
rows ( y ) 1
zi 1 . t1 i 2 . zi 1 . zi zi 1 . t2 i zi 1 . t3 i . li
2 2
Iy zi
i =1
rows ( y ) 1
2.
2 . ω i 1. ω i . t2 2.
Iω ω i 1
t1 i ω i 1 i
ω i
ω i 1
t3 i . li
Mixed i =1
rows ( y ) 1
I yz yi 1 . zi 1 . t1 i yi 1 . zi zi 1 zi 1 . yi yi 1 . t2 i yi yi 1 . zi zi 1 . t3 i . li
i =1
rows ( y ) 1
I yω yi 1 . ω i 1 . t1 i yi 1 . ω i ω i 1
ω i 1
. y
i
yi 1 . t2 i yi yi 1 . ω i ω i 1
. t3 . l
i i
i =1
rows ( y ) 1
I zω zi 1 . ω i 1 . t1 i zi 1 . ω i ω i 1
ω i 1
. z
i
zi 1 . t2 i zi zi 1 . ω i ω i 1
. t3 . l
i i
i =1
rows ( y ) 1
l Iai . yi yi 1 . zi zi 1
th i . tli . i
2 2
Iai tli th i I yz I yz
Influence of 48 2
thickness li
i =1
rows ( y ) 1 rows ( y ) 1
Iai . zi Iai . yi
2 2
zi 1 yi 1
Iz Iz Iy Iy
2 2
li li
i =1 i =1
rows ( y ) 1
l
th i . tli . i . 1.05 I t = 8.425 . 10 mm
2 2 3 4
It tli th i
12
i =1
Sy
A . z gc I y = 1.325 . 10 mm
2 6 4
z gc z gc = 35.593 mm Iy Iy
A
Sz
A . y gc I z = 2.151 . 10 mm
2 6 4
y gc y gc = 0 mm Iz Iz
A
S y .S z S z.S ω
I yz = 1.816 . 10 I yω = 1.057 . 10 mm
10 4 8 5
I yz I yz mm I yω I yω
A A y
S y .S ω
2
Sω
I ω = 9.438 . 10 mm I zω = 5.294 . 10
9 6 8 5
Iω Iω I zω I zω mm
A A
TALAT 2301 – Example 5.4 4

5. I zω . I z I yω . I yz I yω . I y I zω . I yz
y sc = 4.668 . 10
14
Shear y sc z sc mm
I y .I z I y .I z
centre 2 2
I yz I yz
z sc = 49.159 mm
z sc . I yω y sc . I zω I w = 4.24 . 10 mm
9 6
Warping Iw Iω
constant
Polar radius
gyration
0
50
Iy Iz 2 2
ip y sc y gc z sc z gc
A
i p = 100 mm 0
0
50
50 0 50
Buckling of stiffened flanges
5.4.5 Local buckling, internal elements i 3 , 5 .. 7
250 . MPa 2
zi 2 .
2 2
ε ε = 0.913 βi yi yi 2 zi
fo tli 1 th i
β 32 220
5.4.5 (3) c) ρ c if
i
22 , 1.0 , t efl .
ρ ctli t efh .
ρ cth i
heat-treated i ε β i β
2 i i i i
i
unwelded t efl .
ρ ctli t efh .
ρ cth i
ε ε i 1 i i 1 i t efl t efh
i i
ρ c= = =
βi = i mm mm
22.214 0.943 3.302 3.302
25 0.875 3.501 3.501
22.214 0.943 3.302 3.302
5.4.5 Local buckling, outstand elements i 1 , 8 .. 8
1 1
4 4 2 1
tli . th i th i . tli zi 1 .
3 3 2
t fic if tli > th i , , βi yi yi 1 zi
i t fic
i
TALAT 2301 – Example 5.4 5

6. β 10 24
5.4.5 (3) c) ρ c if
i
6 , 1.0 , t efl .
ρ ctli t efh .
ρ cth i
heat-treated i ε β i β
2 i i i i t fic t efl t efh
i i i i
unwelded
ε ρ c= = = =
ε βi = i mm mm mm
3.858 1 6.506 3.5 8
3.858 1 6.506 8 3.5
Reference: Pekoz, T., Flanges of nonuniform thickness.
Aluminium Design Workshop, Cornell University Oct. 1997
Buckling of edge stiffener i 1 .. 2 z2 = 40 mm
t efh t efl t efl t efh t efl
Effective i i i i i
t1 i t efl t2 i
area i 2 2 3
Length of 2 2
elements li yi yi 1 zi zi 1
2 t efl t efh
i i. 2
Area Ar li A r = 269 mm
2
i =1
2
Centre of 1
gravity zr zi 1 . t1 i zi zi 1 . t2 i . li . z r = 69.528 mm
Ar
i =1
Second moment
i 1 .. 2 z2 z1 15 . tl2 th 2 . 0.5 z6 z2 z2 = 25.25 mm
of area
Dti tli Dti tli Dti
Dti th i tli t1 i tli t2 i t3 i
2 2 3 3 4
Length of 2 2
elements li yi yi 1 zi zi 1
2
tli th i
Area A r1 .l A r1 = 328.1 mm
2
i
2
i =1
2
Centre of 1
gravity yr yi 1 . t1 i yi yi 1 . t2 i . li . y r = 45.218 mm
A r1
i =1
TALAT 2301 – Example 5.4 6

7. Second 2
yi 1 . t1 i 2 . yi 1 . yi yi 1 . t2 i yi 1 . t3 i . li y r . A r1
moments 2 2 2
Ir yi
of area
i =1
z2 z1 . 0.5
I r = 1.668 . 10 mm
4 4
k 6 .. 8 i 0 .. 2 j 0 .. 8
z6 z2
yr zr
mm mm
Left: Cross section for
second moment of areaIr
based on t
Right: Effective cross section areaAr
based on t ef
s1 y3 y5 b1 zr
b 1 = 69.528 mm
I r . th 3
3
s 1 = 100 mm
N r.cr 1.05 . E .
1.5 . s 1 . th 4
3
N r.cr = 52.195 kN
b1 . 1
3
b 1 . th 3
3
f o .A r
Table 5.5 α 0.2 λ 1 0.6 λ c λ c 1.243
=
and 5.8 N r.cr
1
0.5 . 1 α . λ c
2
(5.37) φ λ 1 λ c χ r χ r if χ r > 1 , 1 , χ r
2 2
φ φ λ c φ = 1.337
χ r 0.547
=
i 1 .. 2 t efl χ .r efl
t t efh χ .r efh
t
i i i i
i 7 .. 8 t efl χ .r efl
t t efh χ .r efh
t
i i i i
i 1 .. 8
t efl t efh li
Length of 2 2 i i
elements li yi yi 1 zi zi 1 = = =
mm mm mm
8 t efl t efh 1.913 4.372 25.101
i i. 2
Area A eff li A eff = 904.9 mm 1.805 1.805 38.875
2 3.302 3.302 38.875
i =1
3.501 3.501 50
3.501 3.501 50
Axial force resistance, flexural-torsional buckling 3.302 3.302 38.875
1.805 1.805 38.875
Buckling length l L l = 1200 mm 4.372 1.913 25.101
TALAT 2301 – Example 5.4 7

8. π .E .I y π .E .I z π .E .I ω
Reference 2 2 2
buckling N Ey N Ez NT G .I t . 1
loads 2 2 2 2
l l l ip
N Ez = 1.03 . 10 kN 0.2 . N Ez
3
N Ey = 635.91 kN N T = 475.37 kN Guess: N
N . N Ez N . NT N .i p z gc . N . N Ey y gc . N . N Ez
2 2 2 2 2
Given N Ey z sc N y sc N 0
Solution N cr minerr( N ) N cr = 348.11 kN
General cross-section
Table 5.5 α 0.35 λ 1 0.4 k2 1
and 5.8
f o . A eff
λ c λ bar λ c λ c 0.883
=
N cr
1
0.5 . 1 α . λ c
2
(5.37) φ λ 1 λ c χ φ = 0.974 χ = 0.721
2 2
φ φ λ c
min z z gc max z z gc
Table 5.5 ψ min z z gc = 35.6 mm ψ = 0.11
min z z gc max z z gc
max z z gc = 44.4 mm
2
λ c
2.4 . ψ .
2
Table 5.5 k1 1 k 1 = 0.996
λ c . 1
2 2
1 λ c
fo
5.8.3 (1) N b.Rd χ .k 1 .k 2 . .A N b.Rd = 265.8 kN
γ M1
TALAT 2301 – Example 5.4 8