TALAT Lecture 2301
Design of Members
Shear Force
Example 6.8 : Shear 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 6.8 1

Example 6.8. Shear 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 6.8 2

Local buckling of top flange
Length and width
L
L = 5 . 10 mm
3
of web panel am a1 a m = 80 mm = 62.5
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 = 5.341
0.81 . a m . f o
(5.96) λ w
t1 E λ w 0.328
=
kτ
0.48 fu
Table 5.12 ρ v η 0.4 0.2 .
λ w fo ρ v= 1.462
η = 0.617
Table 5.12 ρ v if ρ v > η , η , ρ v
ρ v= 0.617
fo
ρ .b . t 1 t3 . V w.Rd = 4.036 . 10 kN
5
(5.95), two flanges V w.Rd v
γ M1
No reduction for local buckling 50
0
50
100
100 50 0 50 100
Overall buckling, shear
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
TALAT 2301 – Example 6.8 3

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
Elastic buckling load
1
4
L. B y H
φ = 1.711 . 10
5
(5.83) (5.84) φ η
B x .B y
b Bx
η = 6.236 . 10
5
0.567 . φ 1.92 . φ 0.1 . φ 2.75 . φ .η .
2 2 6
(5.82) kτ 3.25 1.95 k τ = 1.216 10
1
k τ .π
2
4
(5.81) V o.cr . B .B 3 V o.cr = 0.039 kN
b x y
Buckling resistance
b .t 1 .f o
λ ow 3.039 . 10
3
(5.120) λ ow =
V o.cr
0.6
if χ o > 0.6 , 0.6 , χ o χ o 6.495 . 10
8
(5.119) χ o χ o =
2
0.8 λ ow
fo
(5.118) V o.Rd χ .o . t 1 .
b V o.Rd = 0.021 kN
γ M1
V Rd if V w.Rd < V o.Rd , V w.Rd , V o.Rd V Rd = 0.021 kN
TALAT 2301 – Example 6.8 4

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 of top flange
Length and width
L
L = 5 . 10 mm 2 .a 1
3
of web panel am a m = 80 mm = 62.5
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 = 5.341
0.81 . a m . f o
(5.96) λ w
t1 E λ w 0.328
=
kτ
0.48 fu
Table 5.12 ρ v η 0.4 0.2 .
λ w fo ρ v= 1.462
η = 0.617
Table 5.12 ρ v if ρ v > η , η , ρ v
ρ v= 0.617
fo
ρ .b . t 1 t3 . V w.Rd = 6.055 . 10 kN
5
(5.95), three flanges V w.Rd v t2
(web) γ M1
50
0
50
100
100 50 0 50 100
TALAT 2301 – Example 6.8 5

Overall buckling, shear
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
1
4
L. B y H
(5.83) (5.84) φ η φ = 0.018
B x .B y
b Bx
η = 0.469
0.567 . φ 1.92 . φ 0.1 . φ 2.75 . φ .η
2 2
(5.82) kτ 3.25 1.95 k τ = 4.156
1
k τ .π
2
. B .B 3 4
(5.81) V o.cr x y V o.cr = 105.983 kN
b
Buckling resistance
b .t 1 .f o
(5.120) λ ow λ ow 58.282
=
V o.cr
0.6
if χ o > 0.6 , 0.6 , χ o χ o 1.766 . 10
4
(5.119) χ o χ o =
2
0.8 λ ow
fo
(5.118) V o.Rd χ .o . t 1 .
b V o.Rd = 57.795 kN
γ M1
V Rd if V w.Rd < V o.Rd , V w.Rd , V o.Rd V Rd = 57.795 kN
TALAT 2301 – Example 6.8 6

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 of top flange
Thickness tm if t 1 < t 2 , t 1 , t 2 t m = 5 mm
Length and width L
L = 5 . 10 mm 2 .a 1
3
of web panel am a m = 75 mm = 66.667
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 = 5.341
0.81 . a m . f o
(5.96) λ w
tm E λ w 0.308
=
kτ
0.48 fu
Table 5.12 ρ v η 0.4 0.2 . ρ v= 1.559
λ w fo
η = 0.617
Table 5.12 ρ v if ρ v > η , η , ρ v
ρ v= 0.617
fo
ρ .b . t 1 t2 . V w.Rd = 4.036 . 10 kN
5
(5.95), two flanges V w.Rd v
γ M1
No reduction for local buckling 50
0
50
100
100 50 0 50 100
TALAT 2301 – Example 6.8 7

Overall buckling, shear
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
1
4
L. B y H
φ = 5.432 . 10
3
(5.83) (5.84) φ η
B x .B y
b Bx
η = 0.083
0.567 . φ 1.92 . φ 0.1 . φ 2.75 . φ .η
2 2
(5.82) kτ 3.25 1.95 k τ = 3.409
1
k τ .π
2
4
(5.81) V o.cr . B .B 3 V o.cr = 3.848 kN
x y
b
Buckling resistance
b .t 1 .f o
(5.120) λ ow λ ow 305.887
=
V o.cr
0.6
χ o 6.412 . 10
6
(5.119) χ o χ o if χ o > 0.6 , 0.6 , χ o =
2
0.8 λ ow
fo
(5.118) V o.Rd χ .o . t 1 .
b V o.Rd = 2.099 kN
γ M1
V Rd if V w.Rd < V o.Rd , V w.Rd , V o.Rd V Rd = 2.099 kN
TALAT 2301 – Example 6.8 8

Summary
V Rd.a N
A a = 2.812 . 10 mm = 7.6 . 10
3 2 3
a) V Rd.a = 0 kN
Aa 2
mm
V Rd.b N
A b = 3.212 . 10 mm
3 2
b) V Rd.b = 57.8 kN = 18
Ab 2
mm
V Rd.c N
A c = 2.2 . 10 mm
3 2
c) V Rd.c = 2.1 kN =1
Ac 2
mm
TALAT 2301 – Example 6.8 9