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

1. TALAT Lecture 2301
Design of Members
Axial Force
Example 5.6 : Axial force resistance of orthotropic
plate. Open or closed stiffeners
5 pages
Advanced Level
prepared by Torsten Höglund, Royal Institute of Technology, Stockholm
Date of Issue: 1999
 EAA - European Aluminium Association
TALAT 2301 – Example 5.6 1

2. Example 5.6. Axial force resistance of orthotropicplate. Open
or closed stiffeners
Dimensions (highlighted)
Plate thickness t 16.5 . mm t1 t fo 240 . MPa N newton
Plate width b 1500 . mm fu 260 . MPa kN 1000 . newton
2200 . mm 70000 . MPa MPa 10 . Pa
6
Plate length L E
w
Stiffener pitch w 300 . mm a γ M1 1.1
2
If heat-treated alloy, then = 1 else ht = 0
ht ht 1
If cold formed (trapezoidal), then = 1 else cf. = 0
cf. cf 0
a) Open stiffeners
Half stiffener pitch a = 150 mm
Half bottom flange a2 50 . mm
Thickness of bottom flange t2 10 . mm
Stiffener depth h 160 . mm
Half web thickness t3 4.4 . mm
Half width of trapezoidal a1 0 . mm
stiffener at the top
Width of web a3 h a 3 = 160 mm
Local buckling
2 .a a3 18.182
Internal elements β i β i β = β max β i β = 18.182
2 .t 3
t1 i
1 2 18.182
250 . MPa
[1] Tab. 5.1 ε β 1 9 .ε β 1 9.186
=
fo
β 2 11 . ε β 2 11.227
=
β 3 18 . ε β 3 18.371
=
class i if β > β 1 , if β > β 2 , if β > β 3 , 4 , 3 , 2 , 1 class i = 3
a2
Outstand β β 1 2.5 . ε β 1 2.552
=
t2
β 2 4 .ε β 2 4.082
=
[1] Tab. 5.1 β =5
β 3 5 .ε β 3 5.103
=
class o if β > β 1 , if β > β 2 , if β > β 3 , 4 , 3 , 2 , 1 class o = 3
class if class o > class i , class o , class i class = 3
No reduction for local buckling
TALAT 2301 – Example 5.6 2

3. 5.11.6 Overall buckling, uniform compression
2 .t 1 .a 2 .t 2 .a 2 2 .t 3 .a 3 2 .t 1 .a 1 A = 7.358 . 10 mm
3 2
Cross sectional area A
h
2 .t 2 .a 2 .h 2 .t 3 .a 3 .
Gravity centre 2
e e = 37.054 mm
A
Second moment of area
2
h
2 .t 2 .a 2 .h 2 .t 3 .a 3 . A .e I L = 2.751 . 10 mm
2 2 7 4
IL
3
s if cf 1 , 2 . a 2 .a 3
. a
2 1 a2 , 2 .a s = 300 mm
Rigidities of orthotropic plate
E .I L ν 0.3
N . mm
2
B x = 6.42 . 10
9
Table 5.10 Bx
2 .a G
E mm
2 .( 1 ν )
2 .a . E .t N . mm
3 2
B y = 2.88 . 10
7
Table 5.10 By
s 12 . 1 ν
2 mm
2 .a . G .t N . mm
3 2
H = 2.016 . 10
7
Table 5.10 H
s 6 mm
Elastic buckling load
2 2 4
π . Bx L L Bx
2 .H B y. N cr = 2.031 . 10 kN
4
(5.77) N cr if <
b 2 b b By
L
b
2 .π
2
(5.78) . B x .B y H otherwise
b
Buckling resistance
A ef = 7.358 . 10 mm
3 2
A ef A
A ef . f o
(5.69) λ c λ c 0.295
=
N cr
α if ( ht > 0 , 0.2 , 0.32 ) λ 0 if ( ht > 0 , 0.1 , 0 ) α = 0.2
Table 5.6
λ 0 0.1
=
0.5 . 1 α . λ c
2
φ λ 0 λ c
φ = 0.563
1
(5.33) χ c
2 2 χ c = 0.959
φ φ λ c
fo
A ef . χ .c N cRd = 1.54 . 10 kN
3
(5.68) N cRd for one stiffener
γ M1
TALAT 2301 – Example 5.6 3

4. b) Closed stiffeners
Half stiffener pitch a = 150 mm
Plate thickness t 1 = 16.5 mm
Half bottom flange a 2 = 50 mm
Thickness of bottom flange t 2 = 10 mm
Stiffener depth h = 160 mm
Web thickness t3 9 . mm
Half width of trapezoidal a1 80 . mm
stiffener at the top
2 2
width of web a3 a1 a2 h a 3 = 162.8 mm
a4 a a1 a 4 = 70 mm
Local buckling 9.697
2 .a 1 2 .a 2 a3 2 .a 4 10
Internal elements β i β i β i β i β =
i
1 t1 2 t2 3 t3 4 t1 18.088
β max β i β = 18.1 8.485
250 . newton
[1] Tab. 5.1 ε β 1 9 .ε β 1 9.186
=
fo 2
mm
β 2 13 . ε β 2 13.268
=
β 3 18 . ε β 3 18.371
=
class i if β > β 1 , if β > β 2 , if β > β 3 , 4 , 3 , 2 , 1 class i = 3
No reduction for local buckling
2 .t 1 .a 2 .t 2 .a 2 2 .t 3 .a 3 A = 8.88 . 10 mm
3 2
Cross sectional area A
h
2 .t 2 .a 2 .h 2 .t 3 .a 3 .
Gravity centre 2
e e = 44.415 mm
A
2
h
2 .t 2 .a 2 .h 2 .t 3 .a 3 . A .e I L = 3.309 . 10 mm
2 2 7 4
Second moment IL
of area 3
4. h. a 1 a 2
2
I T = 3.097 . 10 mm
7 4
Torsion constant IT
2 .a 1 2 .a 2 a3
2.
t1 t2 t3
t2
4. 1 ν . a2 a 3 .a 1 .a 4 .h .
2 2 2 2
(5.79c) C1
C 1 = 3.076 . 10 mm
9 4
3 .a .t 1
3
E .t
3
B = 2.88 . 10 N . mm
7
B
(5.79d)
12 . 1
2
ν
TALAT 2301 – Example 5.6 4

5. a1 a2 2
a
4. a 1 a 2 .a 1 .a 4 . 1
2
a 1 .a 3 t 2
3
a2 a1
(5.79e) C2 . C 2 = 4.851
a 2 . 3 .a 3 4 .a 2
3 t1
Rigidities of orthotropic plate
E .I L E N . mm
2
B x = 7.72 . 10
9
Table 5.10 Bx ν 0.3 G
2 .a 2 .( 1 ν ) mm
2 . B. a
(5.79a) By
2 .a 1 .a 3 .t 1 . 4 .a 2 .t 3 a 3 .t 2
3 3 3
2 .a 4
a 3 .t 1 . 4 .a 2 .t 3 a 3 .t 2 a 1 . t 3 . 12 . a 2 . t 3 4 .a 3 .t 2
3 3 3 3 3 3
G .I T N . mm
2
B y = 3.929 . 10
7
6 .a mm
(5.79b) H 2 .B
1.6 . G . I T . a 4
2
N . mm
1 2
1 . 1 H = 7.305 . 10
8
L .a .B
2 C1 mm
π .
4
C2
4
L
Elastic buckling load
2 2 4
π . Bx L L Bx
2 .H B y. N cr = 3.378 . 10 kN
4
(5.77) N cr if <
b 2 b b By
L
b
2 .π
2
(5.78) . B x .B y H otherwise
b
Buckling resistance
A ef = 8.88 . 10 mm
3 2
A ef A
A ef . f o
(5.69) λ c λ c 0.251
=
N cr
α if ( ht > 0 , 0.2 , 0.32 ) λ 0 if ( ht > 0 , 0.1 , 0 ) α = 0.2
λ 0 0.1
=
0.5 . 1 α . λ c
2
φ λ 0 λ c
φ = 0.547
1
(5.33) χ c χ c = 0.969
2 2
φ φ λ c
fo
A ef . χ .c N cRd = 1.877 . 10 kN
3
(5.68) N cRd for one stiffener
γ M1
TALAT 2301 – Example 5.6 5