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

1. TALAT Lecture 2301
Design of Members
Shear Force
Example 6.7 : Shear force resistance of orthotropic
plate. Open or closed stiffeners
6 pages
Advanced Level
prepared by Torsten Höglund, Royal Institute of Technology, Stockholm
Date of Issue: 1999
 EAA - European Aluminium Association
TALAT 2301 – Example 6.7 1

2. Example 6.7. Shear 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
L
L = 2.2 . 10 mm
3
Length of web panel a = 150 mm = 7.333
2 .a
2 .a 2 .a
2 2
L
(5.97) kτ if > 1.00 , 5.34 4.00 . , 4.00 5.34 . k τ = 5.414
2 .a L L
0.81 . 2 . a . f o
(5.96) λ w
t E λ w 0.371
=
kτ
0.48
Table 5.12 ρ v
λ w ρ v = 1.295
fu
Table 5.12 η 0.4 0.2 . η = 0.617
fo
Table 5.12 ρ v if ρ v > η , η , ρ v
ρ v= 0.617
fo
ρ .b .t 1 . V w.Rd = 3.33 . 10 kN
3
(5.95) V w.Rd v
γ M1
TALAT 2301 – Example 6.7 2

3. 5.11.6 Overall buckling, shear force
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 )
E .t N . mm
3 2
B y = 2.88 . 10
7
Table 5.10 By
12 . 1
2 mm
ν
G .t N . mm
3 2
H = 2.016 . 10
7
Table 5.10 H
6 mm
Elastic buckling load
1
4 H
L. B y η
φ φ = 0.38
(5.83) (5.84) b Bx B x .B y
η = 0.047
0.567 . φ 1.92 . φ 0.1 . φ 2.75 . φ .η
2 2
(5.82) kτ 3.25 1.95 k τ = 3.423
1
k τ .π
2
4
(5.81) V o.cr . B .B 3 V o.cr = 2.506 . 10 kN
3
b x y
Buckling resistance
b .t 1 .f o
(5.120) λ ow λ ow 1.54
=
V o.cr
0.6
(5.119) χ o χ o if χ o > 0.6 , 0.6 , χ o χ o 0.189
=
2
0.8 λ ow
fo
χ .o . t 1 . V o.Rd = 1.022 . 10 kN
3
(5.118) V o.Rd b
γ M1
V Rd = 1.022 . 10 kN
3
V Rd if V w.Rd < V o.Rd , V w.Rd , V o.Rd
TALAT 2301 – Example 6.7 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
L
L = 2.2 . 10 mm 2 .a 2 .a 1
3
Length of web panel am a m = 140 mm = 15.714
am
2 2
L am am
(5.97) kτ if > 1.00 , 5.34 4.00 . , 4.00 5.34 .
am L L k τ = 5.356
0.81 . a m . f o
(5.96) λ w
t E λ w 0.174
=
kτ
0.48
Table 5.12 ρ v
λ w ρ v = 2.76
fu
Table 5.12 η 0.4 0.2 . η = 0.617
fo
Table 5.12 ρ v if ρ v > η , η , ρ v
ρ v= 0.617
fo
ρ .b .t 1 . V w.Rd = 3.33 . 10 kN
3
(5.95) V w.Rd v
γ M1
TALAT 2301 – Example 6.7 4

5. 5.11.6 Overall buckling, shear force
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
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
1
4 H
L. B y η
φ φ = 0.392
(5.83) (5.84) b Bx B x .B y
η = 1.326
0.567 . φ 1.92 . φ 0.1 . φ 2.75 . φ .η
2 2
(5.82) kτ 3.25 1.95 k τ = 6.52
1
k τ .π
2
4
(5.81) V o.cr . B .B 3 V o.cr = 6.311 . 10 kN
3
b x y
TALAT 2301 – Example 6.7 5

6. TALAT 2301 – Example 6.7 6

7. Buckling resistance
b .t 1 .f o
(5.120) λ ow λ ow 0.97
=
V o.cr
0.6
(5.119) χ o χ o if χ o > 0.6 , 0.6 , χ o χ o 0.345
=
2
0.8 λ ow
fo
χ .o . t 1 . V o.Rd = 1.861 . 10 kN
3
(5.118) V o.Rd b
γ M1
TALAT 2301 – Example 6.7 7