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# TALAT Lecture 2301: Design of Members Example 4.2: Hollow cross section (polygon)

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This example provides calculations on bending moment of members based on Eurocode 9.

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### TALAT Lecture 2301: Design of Members Example 4.2: Hollow cross section (polygon)

1. 1. TALAT Lecture 2301 Design of Members Bending Moment Example 4.2 : Hollow cross section (polygon) 5 pages Advanced Level prepared by Torsten Höglund, Royal Institute of Technology, Stockholm Date of Issue: 1999  EAA - European Aluminium Association TALAT 2301 – Example 4.2 1
2. 2. Example 4.2 Hollow cross section (polygon) b Half width b 50 . mm Half width of a flat parts Thickness s 1.2 . mm 1 2 O 0 . mm a = 20.711 mm Co-ordinates b a s and thickness a b s a b s Comment: 300 . MPa MPa 10 . Pa b a s 6 Nodes 4 and 9 fo are added to b O s γ M1 1.1 kN 1000 . newton indicate shift in y b z a t s neutral axis Nodes i 1 .. rows ( y ) 1 a b s a b s 50 b a s 0 b O s b a s z 0 0 mm Gross cross section Area of cross ti . section 2 2 dAi yi yi 1 zi zi 1 50 elements 50 0 50 Cross y rows ( y ) 1 section mm area A dAi 2 A = 397.6 mm i =1 First rows ( y ) 1 dAi Sy moment Sy zi zi 1 . z gc of area. 2 A Gravity centre i =1 z gc = 0 mm z gc.gr z gc Second rows ( y ) 1 dAi zi . zi 1 . A . z gc moment 2 2 2 Iy zi zi 1 Iy Iy of area 3 i =1 I y = 5.255 . 10 mm 5 4 rows ( y ) 1 Area within 0.5 . yi yi 1 . zi A v = 8.284 . 10 mm 3 2 mid-line Av zi 1 Check: i =1 ( 2 .b ) 2 .( b a ) = 8.284 . 10 mm 2 2 3 2 rows ( y ) 1 2 2 yi yi 1 zi zi 1 Sum of l/t Dn Dn = 276.142 ti i =1 Torsion 4 .A v 2 I v = 9.941 . 10 mm constant 5 4 Iv Dn 2 . A v . min ( t ) W v = 1.988 . 10 mm 4 3 Torsion Wv min ( t ) = 1.2 mm resistance TALAT 2301 – Example 4.2 2
3. 3. Effective cross section. First iteration Zero position zd 0 ρ c 0 zi 1 z gc zi z gc zd zi z gc 5.4.3 (1) ψ i if zi 1 z gc < zi z gc , , i zi z gc zi 1 z gc max z max z d 0.8 (5.7 or 5.8) gi if ψ i > 1 , 0.7 0.30 . ψ i , max z = 50 mm 1 ψ i 2 2 yi yi 1 zi zi 1 250 . MPa 5.4.3 (1) c) βi if t i > 0 , gi . ,1 ε o ε o 0.913 = ti fo max z max z 5.4.4 (5) εi if zi > z gc zi 1 > z gc , if zi 1 z gc < zi z gc , ε o . , ε o. , 99 zi z gc zi 1 z gc 2 5.4.5 (3) c) β ε ε Comment: εi is given a large value (99) > 22 , 32 . 220 . i i i heat-treated, ρ c if , 1.0 i ε i β i β i for elements entirely on the tension side unwelded Effective βi t eff t eff ρ .c t = i thickness ε ρ c= = ψi = gi = βi = εi = i i mm 0.414 0.824 28.452 0.913 31.167 0.8 0.96 1 1 34.518 0.913 37.812 0.692 0.831 0.414 0.824 28.452 0.913 31.167 0.8 0.96 0 0.7 12.081 1.418 8.518 1 1.2 Area of effective 0 0.7 12.081 99 0.122 1 1.2 thickness t eff . 2 2 0.414 0.824 28.452 99 0.287 1 1.2 elements dAi yi yi 1 zi zi 1 i 1 1 34.518 99 0.349 1 1.2 0.414 0.824 28.452 99 0.287 1 1.2 Cross rows ( y ) 1 0 0.7 12.081 99 0.122 1 1.2 section A eff dAi 0 0.7 12.081 1.418 8.518 1 1.2 area i =1 2 2 A eff = 362.498 mm A = 397.645 mm First rows ( y ) 1 moment dAi Sy of area. Sy zi zi 1 . z gc 2 A eff i =1 Gravity centre z gc = 4.046 mm 50 The nodes in the vertical webs are moved to the neutral axis from first iteration 0 z mm i 0 .. rows ( y ) 2 0 0 zi if zi > z gc . zi 1 < z gc , z gc , zi z GC1 mm i 1 .. rows ( y ) 1 zi if zi > z gc . zi 1 < z gc , z gc , zi 50 50 0 50 Check z4 = 4.046 mm z9 = 4.046 mm OK ! y y GC1 , mm mm TALAT 2301 – Example 4.2 3
4. 4. Second iteration zi 1 z gc zi z gc z gc = 4.046 mm 5.4.3 (1) ψi if zi 1 z gc < zi z gc , , zi z gc zi 1 z gc zd zi z gc 0.8 i (5.7 or 5.8) gi if ψ i > 1 , 0.7 0.30 . ψ i , 1 ψ i max z max z d 2 2 max z = 54.046 mm yi yi 1 zi zi 1 5.4.3 (1) c) βi if t i > 0 , gi . ,1 ti max z max z 5.4.4 (5) εi if zi > z gc zi 1 > z gc , if zi 1 z gc < zi z gc , ε o . , ε o. , 99 zi z gc zi 1 z gc 2 5.4.5 (3) c) β ε ε > 22 , 32 . 220 . i i i heat-treated, ρ c if , 1.0 unwelded i ε i β i β i Effective t eff ρ .c t βi t eff thickness = i ε ρ c= = β ψi = gi = βi = εi = i i mm Max slender- i ness β e β max max β e 0.458 0.837 28.906 0.913 31.665 0.791 0.949 i ε i - Internal 1 1 34.518 0.913 37.812 0.692 0.831 elements β max = 37.812 0.458 0.837 28.906 0.913 31.665 0.791 0.949 Cross section 0 0.7 14.441 1.349 10.707 1 1.2 class class if β max 22 , 3 , 4 class = 4 0 0.7 9.721 99 0.098 1 1.2 Area of 0.363 0.809 27.918 99 0.282 1 1.2 effective 1 1 34.518 99 0.349 1 1.2 t eff . thickness 2 2 dAi yi yi 1 zi zi 1 0.363 0.809 27.918 99 0.282 1 1.2 elements i 0 0.7 9.721 99 0.098 1 1.2 Cross rows ( y ) 1 0 0.7 14.441 1.349 10.707 1 1.2 section A eff dAi area i =1 2 A eff = 361.596 mm First rows ( y ) 1 dAi moment Sy zi zi 1 . Sy of area. 2 z gc i =1 A eff Gravity centre z gc = 4.144 mm 50 0 Gross cross z section I y.gr Iy mm I y.gr = 5.255 . 10 mm 5 4 z GC1 0 0 mm z GC2 mm 50 50 0 50 y y GC1 , mm mm TALAT 2301 – Example 4.2 4
5. 5. A . z gc 2 Second rows ( y ) 1 I eff Iy moment of dAi zi . zi 1 . 2 2 area of effective I y zi zi 1 I eff = 5.187 . 10 mm 3 5 4 cross section i =1 I eff W eff = 9.579 . 10 mm 3 3 Section zd zi z gc max z.eff max z d W eff modulus i max z.eff Compare I y.gr W gr = 1.051 . 10 mm 4 3 gross zd zi z gc.gr max z.gr max z d W gr i max z.gr section Moment fo resistance M Rd W eff . M Rd = 2.613 kN . m γ M1 Axial force Move node close zi if zi z GC1 < 1 . mm , zi 1 , zi to GC1 to nearest node 2 2 yi yi 1 zi zi 1 5.4.3 (1) c) βi if t i > 0 , ,1 ti 2 5.4.5 (3) c) β ε o ε o > 22 , 32 . 220 . i heat-treated, ρ c if , 1.0 ε o 0.913 = i ε o β i β i unwelded βi t eff Effective = ρ .c i t eff t ε o ρ c= = thickness i = βi = i mm 1 34.518 37.812 0.692 0.831 2 34.518 37.812 0.692 0.831 Area of effective 3 34.518 37.812 0.692 0.831 thickness 4 0 0 1 1.2 t eff . elements 2 2 dAi yi yi 1 zi zi 1 5 34.518 37.812 0.692 0.831 i 6 34.518 37.812 0.692 0.831 Cross rows ( y ) 1 7 34.518 37.812 0.692 0.831 2 section A eff dAi A eff = 275.335 mm 8 34.518 37.812 0.692 0.831 area 9 0 0 1 1.2 i =1 10 34.518 37.812 0.692 0.831 First rows ( y ) 1 dAi Sy moment Sy zi zi 1 . z gc of area. 2 A eff Gravity centre i =1 z gc = 0 mm 50 0 z mm 0 0 fo z GC Axial force N Rd A eff . resistance γ M1 mm N Rd = 75.091 kN 50 50 0 50 y y GC , mm mm TALAT 2301 – Example 4.2 5