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# TALAT Lecture 2301: Design of Members Example 5.2: Axial force resistance of symmetric hollow extrusion

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This example provides calculations on axial force of members based on Eurocode 9.

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### TALAT Lecture 2301: Design of Members Example 5.2: Axial force resistance of symmetric hollow extrusion

1. 1. TALAT Lecture 2301 Design of Members Axial Force Example 5.2 : Axial force resistance of symmetric hollow extrusion 4 pages Advanced Level prepared by Torsten Höglund, Royal Institute of Technology, Stockholm Date of Issue: 1999  EAA - European Aluminium Association TALAT 2301 – Example 5.2 1
2. 2. Example 5.2 Axial force resistance of symmetric hollow extrusion Half width b 50 . mm fo 300 . MPa Indent c 10 . mm E 70000 . MPa Half width of a 18 . mm γ M1 1.0 flat parts Thickness s 1.2 . mm Length L 1200 . mm Nodes no., co-ordinates, 0 b a s thickness 1 b c a s 2 a b c s 3 a b s MPa 10 . Pa 6 4 a b s 5 a b c s kN 1000 . newton 6 b c a s 7 b a s i 8 y b z a t s 9 b c a s 10 a b c s 11 a b s 12 a b s 13 a b c s 50 14 b c a s 0 15 b a s z 0 16 b a s 0 mm 50 50 0 50 Nodes i 1 .. rows ( y ) 1 y mm 5.4.5 Local buckling 250 . MPa ε fo 2 2 yi yi 1 zi zi 1 5.4.3 (1) c) βi if t i > 0 , ,1 ti 5.4.5 (3) c) β ε ε 2 > 22 , 32 . 220 . i heat-treated, ρ c if , 1.0 i ε β i β i unwelded TALAT 2301 – Example 5.2 2
3. 3. Effective thickness t eff ρ .c t βi t eff i = ρ c= = Effective area i = βi = ε t eff . 2 2 i mm of elements dAi yi yi 1 zi zi 1 i 1 8.333 9.129 1 1.2 rows ( y ) 1 2 25.927 28.402 0.854 1.025 Area of effective 2 3 8.333 9.129 1 1.2 cross section A eff dAi A eff = 356.591 mm 4 30 32.863 0.77 0.924 i =1 5 8.333 9.129 1 1.2 6 25.927 28.402 0.854 1.025 rows ( y ) 1 First dAi Sy 7 8.333 9.129 1 1.2 moment Sy zi zi 1 . z gc 2 A eff 8 30 32.863 0.77 0.924 of area. i =1 9 8.333 9.129 1 1.2 Gravity centre 10 25.927 28.402 0.854 1.025 z gc = 0 m 11 8.333 9.129 1 1.2 12 30 32.863 0.77 0.924 13 8.333 9.129 1 1.2 14 25.927 28.402 0.854 1.025 Axial force resistance, 15 8.333 9.129 1 1.2 no flexural buckling 16 30 32.863 0.77 0.924 fo N Rd A eff . N Rd = 107 kN γ M1 50 0 z mm 0 0 z GC mm 50 50 0 50 y y GC 5.8.4 Flexural buckling , mm mm Second rows ( y ) 1 ti . 2 2 moment of yi yi 1 zi zi 1 zi . zi 1 . 2 2 area of effective I zi zi 1 3 cross section i =1 I = 4.701 . 10 mm 5 4 Radius of I gyration r A eff K .L l = 1.2 . 10 mm 3 Table 5.7 K 1 l Table 5.5 α 0.2 λ o 0.1 k1 1 k2 1 and 5.6 See to the l. 1 . f o right λ λ = 0.689 r π E TALAT 2301 – Example 5.2 3
4. 4. 1 0.5 . 1 α . λ 2 (5.33) φ λ o λ χ φ = 0.796 χ = 0.837 2 2 φ φ λ fo 5.8.3 (1) N b.Rd χ .k 1 .k 2 . .A eff N b.Rd = 89.5 kN γ M1 TALAT 2301 – Example 5.2 4