TALAT Lecture 2301: Design of Members Example 10.1: Transverse bending of unsymmetrical flange
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This document is a technical lecture on the design of members, focusing on the transverse bending of an unsymmetrical flange with specific geometrical parameters and loading conditions. It includes calculations for shear force gradients, moment and stresses on the web and flanges, and provides a detailed example to illustrate the principles involved. The document references relevant regulations and design rules within the context of aluminium structures.
![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](https://image.slidesharecdn.com/talat-lecture-2301-design-of-members-example-101-transverse-bending-of-unsymmetrical-flange1579/85/TALAT-Lecture-2301-Design-of-Members-Example-10-1-Transverse-bending-of-unsymmetrical-flange-2-320.jpg)
![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](https://image.slidesharecdn.com/talat-lecture-2301-design-of-members-example-101-transverse-bending-of-unsymmetrical-flange1579/85/TALAT-Lecture-2301-Design-of-Members-Example-10-1-Transverse-bending-of-unsymmetrical-flange-3-320.jpg)
TALAT Lecture 2301: Design of Members Example 10.1: Transverse bending of unsymmetrical flange
- 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