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

1. TALAT Lecture 2301
Design of Members
Axial Force
Example 5.7 : Axial force resistance of orthotropic
double-skin plate
9 pages
Advanced Level
prepared by Torsten Höglund, Royal Institute of Technology, Stockholm
Date of Issue: 1999
 EAA - European Aluminium Association
TALAT 2301 – Example 5.7 1

2. Example 5.7. Axial force resistance of orthotropic double-skin plate
N newton
Input (highlighted)
kN 1000 . N
MPa 10 . Pa
6
Plate thickness t 5 . mm t1 t fo 240 . MPa
Plate width b 300000 . mm fu 260 . MPa
Plate length L 5000 . mm E 70000 . MPa
w
"Pitch" (2a or d) w 160 . mm a γ M1 1.1
2
If heat-treated alloy, then = 1 else ht = 0
ht cf 0 ht 1
a) Profiles with groove and tongue
Half bottom flange a2 40 . mm
Thickness of bottom flange t2 5 . mm
Profile depth h 70 . mm
Web thickness t3 5 . mm
Half width of trapezoidal a1 80 . mm
stiffener at the top
Number of webs nw 4
2 2
Width of web a3 a1 a2 h a 3 = 80.6 mm
TALAT 2301 – Example 5.7 2

3. Local buckling 0
a1 2 .a 2 a3 16
Internal elements β i β i β i β i=
1 t1 2 t2 3 t3 16
16.125
β max β i β = 16.1
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 50
0
50
100
100 50 0 50 100
Overall buckling, uniform compression
nw
Cross sectional area A 2 .t 1 .a 2 .t 2 .a 2 2 .t 3 .a 3 .
A = 2.812 . 10 mm
3 2
2
h nw
2 .t 2 .a 2 .h 2 .t 3 .a 3 . .
Gravity centre 2 2
e e = 30.022 mm
A
2
h .n w
2 .t 2 .a 2 .h 2 .t 3 .a 3 . A .e I L = 2.059 . 10 mm
2 2 6 4
Second moment of area IL
3 2
4. h. a 1 a 2
2
I T = 3.517 . 10 mm
6 4
Approx. torsional constant IT
2 .a 1 2 .a 2 a3
2.
t1 t2 t3
Rigidities of orthotropic plate
E .I L N . mm
2
B x = 9.007 . 10
8
Table 5.10 Bx ν 0.3
2 .a mm
N . mm
2
E
0.001 . N . mm B y = 1 . 10
3
Table 5.10 By G
2 .( 1 ν) mm
G .I T N . mm
2
H = 5.918 . 10
8
Table 5.10 H
2 .a mm
TALAT 2301 – Example 5.7 3

4. Elastic buckling load
2 2 4
π . Bx L L Bx
2 .H B y. N cr = 1.067 . 10 kN
5
(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 A
A ef = 2.812 . 10 mm
3 2
A ef . f o
(5.69) λ c
N cr λ c 0.08
=
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 = 0.501
1
(5.33) χ c χ c = 1.004
2 2
φ φ λ c
fo
(5.68) N cRd A ef . χ .c for one stiffener N cRd = 616.164 kN
γ M1
TALAT 2301 – Example 5.7 4

5. b) Truss cross section
a a
Half bottom flange a2 a1 a = 80 mm
2 2
Stiffener depth h 70 . mm a 1 = 40 mm
Thickness of bottom flange t2 5 . mm a 2 = 40 mm
Web thickness t3 5 . mm
2 2
Width of web a3 a1 h a 3 = 80.623 mm
Local buckling 0
2 .a 1 2 .a 2 a3 16
Internal elements β i β i β i β i=
1 t1 2 t2 3 t3 16
16.125
β max β i β = 16.1
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
50
0
50
100
100 50 0 50 100
TALAT 2301 – Example 5.7 5

6. Overall buckling, uniform compression
2 .t 1 .a 1 2 .t 2 .a 2 2 .t 3 .a 3 A = 1.606 . 10 mm
3 2
Cross sectional area A
h
2 .t 2 .a 2 .h 2 .t 3 .a 3 .
2
Gravity centre e e = 35 mm
A
2
Second moment h
2 .t 2 .a 2 .h 2 .t 3 .a 3 . A .e I L = 1.309 . 10 mm
2 2 6 4
of area IL
3
4. h. a 1 a 2
2
I T = 1.952 . 10 mm
6 4
Torsion constant IT
2 .a 1 2 .a 2 a3
2.
t1 t2 t3
Orthotropic plate constant
E .I L N . mm
2
B x = 5.728 . 10
8
Table 5.10 Bx
2 .a mm
E .t 1 .t 2 .h
2
N . mm
2
B y = 8.575 . 10
8
Table 5.10 By
t1 t2 mm
G .I T N . mm
2
H = 3.285 . 10
8
Table 5.10 H
2 .a mm
Elastic buckling load
2 2 4
π . Bx L L Bx
2 .H B y. N cr = 6.786 . 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 = 1.606 . 10 mm
3 2
A ef A
A ef . f o λ c 0.08
=
(5.69) λ c
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 = 0.5
1
(5.33) χ c χ c = 1.005
2 2
φ φ λ c
fo
(5.68) N cRd 2 . A ef . χ .c for two pitches N cRd = 704.4 kN
γ M1
TALAT 2301 – Example 5.7 6

7. c) Frame cross section
Half bottom flange a 37.5 . mm a1 a
Stiffener depth h 70 . mm a2 a
Thickness of top flange t1 5 . mm a 1 = 37.5 mm
Thickness of bottom flange t2 5 . mm a 2 = 37.5 mm
Web thickness t3 5 . mm
Width of web a3 h a 3 = 70 mm
Local buckling 0
2 .a 1 2 .a 2 a3 15
Internal elements β i β i β i β i=
1 t1 2 t2 3 t3 15
14
β max β i β = 15
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
50
0
50
100
100 50 0 50 100
TALAT 2301 – Example 5.7 7

8. Overall buckling, uniform compression
2 .t 1 .a 2 .t 2 .a 2 t 3 .a 3 A = 1.1 . 10 mm
3 2
Cross sectional area A
h
2 .t 2 .a 2 .h t 3 .a 3 .
Gravity centre 2
e e = 35 mm
A
2
Second moment h
2 .t 2 .a 2 .h t 3 .a 3 . A .e I L = 1.062 . 10 mm
2 2 6 4
of area IL
3
4. h. a 1 a 2
2
I T = 1.901 . 10 mm
6 4
Torsion constant IT
2 .a 1 2 .a 2 a3
2.
t1 t2 t3
Orthotropic plate constant
E .I L N . mm
2
B x = 9.909 . 10
8
(5.80d) Bx
2 .a mm
a .t 2 t 3
3. 3
a .t 3 6 .h .t 2
3 3
E .t 1
3 3 2
. 2 t1 t
(5.80a By . 10 b . . 1
12 . 1 ν 32 . a 3 .h .t 1 .t 2 L
2 2 2 3 3 2
a .t 3 2 .h . t 1
3 3 3
t2
a .t 3
3
N . mm
2
B y = 1.118 . 10
7
3 3
2 .E .
t1 t2 mm
(5.80b) H
t3 6 .t 1 6 .t 2 N . mm
2
3. 1 H = 8.75 . 10
6
1 1
2 .a 2 .a t3 2 .a t3 mm
Elastic buckling load
2 2 4
π . Bx L L Bx
2 .H B y. N cr = 1.174 . 10 kN
5
(5.77) N cr if <
b 2 b b By
L
b
2 .π .
2
2 .π B x .B y
2
. H = 7.501 kN
(5.78) B x .B y H otherwise b
b
A ef = 1.1 . 10 mm
3 2
Buckling resistance A ef A
A ef . f o
(5.69) λ c λ c 0.047
=
N cr
α if ( ht > 0 , 0.2 , 0.32 ) λ 0 if ( ht > 0 , 0.1 , 0 ) α = 0.2
0.5 . 1 α . λ c
2
φ λ 0 λ c λ 0 0.1
=
φ = 0.496
1
(5.33) χ c
2 2 χ c = 1.011
φ φ λ c
fo
N cRd 2 . A ef . χ .c for two pitches N cRd = 485.112 kN
(5.68) γ M1
TALAT 2301 – Example 5.7 8

9. Summary
N Rd.a N
A a = 2.812 . 10 mm
3 2
a) N Rd.a = 616.2 kN = 219.1
Aa 2
mm
N Rd.b N
A b = 3.212 . 10 mm
3 2
b) N Rd.b = 704.4 kN = 219.3
Ab 2
mm
N Rd.c N
A c = 2.2 . 10 mm
3 2
c) N Rd.c = 485.1 kN = 220.5
Ac 2
mm
TALAT 2301 – Example 5.7 9