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# TALAT Lecture 2301: Design of Members Example 10.1: Transverse bending of unsymmetrical flange

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

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### TALAT Lecture 2301: Design of Members Example 10.1: Transverse bending of unsymmetrical flange

1. 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. 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. 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. 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