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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.

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

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• 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