This example provides calculations on deviation of linear stress distribution of members based on Eurocode 9.

1. TALAT Lecture 2301
Design of Members
Deviation of linear stress distribution
Example 10.1 : Transverse bending of unsymmetrical
flange
4 pages
Advanced Level
prepared by Torsten Höglund, Royal Institute of Technology, Stockholm
Date of Issue: 1999
 EAA - European Aluminium Association
TALAT 2301 – Example 10.1 1

2. Example 10.1. Transverse bending of unsymmetical flange
Check transverse bending of the web due to shear
force gradient in the flanges
Section height: hw 400 . mm
Flange widths: bo 300 . mm bu 120 . mm
Flange thickness: to 16 . mm tu 16 . mm
Web thickness: tw 6 . mm
Overall length: L 6 .m O 0 . mm
10 . kN . m
1
Load q
Cross section, half profile
Co-ordinates
bu tu
O
O O tu
y O z t
hw tw
bo
hw to
2
400
300
200
kN 1000 . newton
100
MPa 10 . Pa
6
0 E 70000 . MPa
100 0 100
[1] ENV 1999-1-1. Eurocode 9 - Design of aluminium structures - Part 1-1: General rules. 1997
[2] StBK-N5Regulations for cold-formed steel and aluminium structures. Svensk Byggtjänst
Stockholm 1979
[3] Hetenyi, M. Beams on elastic foundation. University of Michigan, 1958
TALAT 2301 – Example 10.1 2

3. Nodes i 1 .. rows ( y ) 1
rows ( y ) 1
Area of half
ti .
2 2
cross section dAi yi yi 1 zi zi 1 A dAi
i =1
First moment rows ( y ) 1
dAi Sy
of area, y axis, Sy zi zi 1 . z gc z gc = 214.3 mm
gravity centre 2 A
i =1
Second moment rows ( y ) 1
dAi
zi . zi 1 . A . z gc
2 2 2
of area, y axis, Iy zi zi 1 Iy Iy
section modulus 3
i =1 Iy
rows ( y ) 1 Wy
First moment dAi z gc
Sz yi yi 1 .
of area, z axis
W y = 9.493 . 10 mm
2 5 3
i =1
Second moment rows ( y ) 1
of area with dAi S y .S z
respect to I yz 2 . yi 1 . zi 1 2 . yi . zi yi 1 . zi yi . zi 1 . I yz I yz
6 A
y and z axes i =1
I yz = 5.811 . 10 mm
7 4
Lateral force k h .q on bottom flange
Lateral force I yz
kh k h = 0.143
constant [2] 2 .I y
Second moment 2
t u .b u 0.5 . b u . t u 0.5 . b u . t u
3 2 2
of area of bottom
flange + 0.27 . h w Au b u .t u 0.27 . h w w
.t I u.z yu
3 Au Au
[2]
E .t w
3
Modulus of
k = 59.062 kN . m
2
foundation [3] k (Transverse bending of top flange neglected)
4 .h w
3
1
4
k
λ . L = 2.867
1
Parameter λ [3] λ λ = 0.478 m
4 .E .I u.z
L . L
Lateral deflection of
k h .q 2 . cosh λ . cos λ .
the bottom flange 2 2
vc . 1 v c = 22.3 mm
at the middle of . L ) cos ( λ . L )
k cosh ( λ
the span [3]
Transverse bending m z.6
moment and stresses m z k .v c .h w m z = 0.527 kN σ transv σ transv= 87.9 MPa
2
in the web tw
L L
sinh λ . . sin λ .
Lateral moment k h .q 2 2 M c .y u
Mc . M c = 1.56 kN . m σ c
in bottom flange [3] 2 cosh ( λ . L ) cos ( λ . L ) I u.z
λ
σ c= 17.3 MPa
TALAT 2301 – Example 10.1 3

4. Comment : If the web resistk h . q by itself then
k h .q .h w
3
vc v c = 24.184 mm
k h .q .h w
3
mz m z = 0.571 kN tw
3 .E .
m z.6 12
σ transv σ transv= 95.2 MPa
2
tw
2 My
L
Main axis bending My q. σ x σ x = 47.4 MPa σ x σ c = 64.7 MPa
8 Wy
TALAT 2301 – Example 10.1 4