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TALAT Lecture 2301: Design of Members Example 5.4: Axial force resistance of channel cross section

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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 5.4: Axial force resistance of channel cross section

1. 1. TALAT Lecture 2301 Design of Members Axial Force Example 5.4 : Axial force resistance of channel cross section 7 pages Advanced Level prepared by Torsten Höglund, Royal Institute of Technology, Stockholm Date of Issue: 1999  EAA - European Aluminium Association TALAT 2301 – Example 5.4 1
2. 2. Example 5.4. Axial force resistance of channel cross section Width/depth b 80 . mm h 100 . mm fo 300 . MPa kN 1000 . newton E 8 . mm 25 . mm 70000 . MPa MPa 10 . Pa 6 Inner thickness/outstand ti c E G 2.6 Outer/bottom thickness to 3.5 . mm tu 4 . mm γ M1 1.0 Length L 1200 . mm i 0 .. 8 k 1 .. 3 Nodes no., tli to th i to yk 0.5 . h zi b co-ordinates, thickness th 1 ti tl8 ti yk 4 0.5 . h zk 2 0 . mm y4 0 . mm z2 0.5 . z1 th 4 tu tl4 tu y0 0.5 . h c z6 0.5 . z1 th 5 th 4 tl5 tl4 y8 y0 z1 z1 0.5 . ( ti to ) yi zi tli th i z7 z1 = = = = i = mm mm mm mm 0 25 80 3.5 3.5 1 50 77.75 3.5 8 2 50 40 3.5 3.5 3 50 0 3.5 3.5 4 0 0 4 4 5 -50 0 4 4 6 -50 40 3.5 3.5 7 -50 77.75 3.5 3.5 90 8 -25 80 8 3.5 65 40 Nodes i 1 .. rows ( y ) 1 15 Cross-section constants, varying thickness 10 60 30 0 30 60 0 . mm 2 Sectorial ω0 co-ordinates ω 0 yi .z yi . zi 1 ωi ω ω 0 i 1 i i 1 i Thicknesses Dti tli Dti tli Dti to simplify Dti th i tli t1 i tli t2 i t3 i expressions 2 2 3 3 4 Length of 2 2 elements li yi yi 1 zi zi 1 rows ( y ) 1 tli th i Area A .l A = 1.233 . 10 mm 3 2 i 2 i =1 rows ( y ) 1 yi 1 . t1 i yi 1 . t2 i . li S z = 8.47 . 10 13 3 First moments Sz yi mm of area i =1 TALAT 2301 – Example 5.4 2
3. 3. TALAT 2301 – Example 5.4 3
4. 4. rows ( y ) 1 zi 1 . t1 i zi 1 . t2 i . li S y = 4.388 . 10 mm 4 3 Sy zi i =1 rows ( y ) 1 Sω ω . t1 ω ω . t2 . l S ω = 7.328 . 10 mm 6 4 i 1 i i i 1 i i i =1 rows ( y ) 1 yi 1 . t1 i 2 . yi 1 . yi yi 1 . t2 i yi 1 . t3 i . li 2 2 Second Iz yi moments of area i =1 rows ( y ) 1 zi 1 . t1 i 2 . zi 1 . zi zi 1 . t2 i zi 1 . t3 i . li 2 2 Iy zi i =1 rows ( y ) 1 2. 2 . ω i 1. ω i . t2 2. Iω ω i 1 t1 i ω i 1 i ω i ω i 1 t3 i . li Mixed i =1 rows ( y ) 1 I yz yi 1 . zi 1 . t1 i yi 1 . zi zi 1 zi 1 . yi yi 1 . t2 i yi yi 1 . zi zi 1 . t3 i . li i =1 rows ( y ) 1 I yω yi 1 . ω i 1 . t1 i yi 1 . ω i ω i 1 ω i 1 . y i yi 1 . t2 i yi yi 1 . ω i ω i 1 . t3 . l i i i =1 rows ( y ) 1 I zω zi 1 . ω i 1 . t1 i zi 1 . ω i ω i 1 ω i 1 . z i zi 1 . t2 i zi zi 1 . ω i ω i 1 . t3 . l i i i =1 rows ( y ) 1 l Iai . yi yi 1 . zi zi 1 th i . tli . i 2 2 Iai tli th i I yz I yz Influence of 48 2 thickness li i =1 rows ( y ) 1 rows ( y ) 1 Iai . zi Iai . yi 2 2 zi 1 yi 1 Iz Iz Iy Iy 2 2 li li i =1 i =1 rows ( y ) 1 l th i . tli . i . 1.05 I t = 8.425 . 10 mm 2 2 3 4 It tli th i 12 i =1 Sy A . z gc I y = 1.325 . 10 mm 2 6 4 z gc z gc = 35.593 mm Iy Iy A Sz A . y gc I z = 2.151 . 10 mm 2 6 4 y gc y gc = 0 mm Iz Iz A S y .S z S z.S ω I yz = 1.816 . 10 I yω = 1.057 . 10 mm 10 4 8 5 I yz I yz mm I yω I yω A A y S y .S ω 2 Sω I ω = 9.438 . 10 mm I zω = 5.294 . 10 9 6 8 5 Iω Iω I zω I zω mm A A TALAT 2301 – Example 5.4 4
5. 5. I zω . I z I yω . I yz I yω . I y I zω . I yz y sc = 4.668 . 10 14 Shear y sc z sc mm I y .I z I y .I z centre 2 2 I yz I yz z sc = 49.159 mm z sc . I yω y sc . I zω I w = 4.24 . 10 mm 9 6 Warping Iw Iω constant Polar radius gyration 0 50 Iy Iz 2 2 ip y sc y gc z sc z gc A i p = 100 mm 0 0 50 50 0 50 Buckling of stiffened flanges 5.4.5 Local buckling, internal elements i 3 , 5 .. 7 250 . MPa 2 zi 2 . 2 2 ε ε = 0.913 βi yi yi 2 zi fo tli 1 th i β 32 220 5.4.5 (3) c) ρ c if i 22 , 1.0 , t efl . ρ ctli t efh . ρ cth i heat-treated i ε β i β 2 i i i i i unwelded t efl . ρ ctli t efh . ρ cth i ε ε i 1 i i 1 i t efl t efh i i ρ c= = = βi = i mm mm 22.214 0.943 3.302 3.302 25 0.875 3.501 3.501 22.214 0.943 3.302 3.302 5.4.5 Local buckling, outstand elements i 1 , 8 .. 8 1 1 4 4 2 1 tli . th i th i . tli zi 1 . 3 3 2 t fic if tli > th i , , βi yi yi 1 zi i t fic i TALAT 2301 – Example 5.4 5
6. 6. β 10 24 5.4.5 (3) c) ρ c if i 6 , 1.0 , t efl . ρ ctli t efh . ρ cth i heat-treated i ε β i β 2 i i i i t fic t efl t efh i i i i unwelded ε ρ c= = = = ε βi = i mm mm mm 3.858 1 6.506 3.5 8 3.858 1 6.506 8 3.5 Reference: Pekoz, T., Flanges of nonuniform thickness. Aluminium Design Workshop, Cornell University Oct. 1997 Buckling of edge stiffener i 1 .. 2 z2 = 40 mm t efh t efl t efl t efh t efl Effective i i i i i t1 i t efl t2 i area i 2 2 3 Length of 2 2 elements li yi yi 1 zi zi 1 2 t efl t efh i i. 2 Area Ar li A r = 269 mm 2 i =1 2 Centre of 1 gravity zr zi 1 . t1 i zi zi 1 . t2 i . li . z r = 69.528 mm Ar i =1 Second moment i 1 .. 2 z2 z1 15 . tl2 th 2 . 0.5 z6 z2 z2 = 25.25 mm of area Dti tli Dti tli Dti Dti th i tli t1 i tli t2 i t3 i 2 2 3 3 4 Length of 2 2 elements li yi yi 1 zi zi 1 2 tli th i Area A r1 .l A r1 = 328.1 mm 2 i 2 i =1 2 Centre of 1 gravity yr yi 1 . t1 i yi yi 1 . t2 i . li . y r = 45.218 mm A r1 i =1 TALAT 2301 – Example 5.4 6
7. 7. Second 2 yi 1 . t1 i 2 . yi 1 . yi yi 1 . t2 i yi 1 . t3 i . li y r . A r1 moments 2 2 2 Ir yi of area i =1 z2 z1 . 0.5 I r = 1.668 . 10 mm 4 4 k 6 .. 8 i 0 .. 2 j 0 .. 8 z6 z2 yr zr mm mm Left: Cross section for second moment of areaIr based on t Right: Effective cross section areaAr based on t ef s1 y3 y5 b1 zr b 1 = 69.528 mm I r . th 3 3 s 1 = 100 mm N r.cr 1.05 . E . 1.5 . s 1 . th 4 3 N r.cr = 52.195 kN b1 . 1 3 b 1 . th 3 3 f o .A r Table 5.5 α 0.2 λ 1 0.6 λ c λ c 1.243 = and 5.8 N r.cr 1 0.5 . 1 α . λ c 2 (5.37) φ λ 1 λ c χ r χ r if χ r > 1 , 1 , χ r 2 2 φ φ λ c φ = 1.337 χ r 0.547 = i 1 .. 2 t efl χ .r efl t t efh χ .r efh t i i i i i 7 .. 8 t efl χ .r efl t t efh χ .r efh t i i i i i 1 .. 8 t efl t efh li Length of 2 2 i i elements li yi yi 1 zi zi 1 = = = mm mm mm 8 t efl t efh 1.913 4.372 25.101 i i. 2 Area A eff li A eff = 904.9 mm 1.805 1.805 38.875 2 3.302 3.302 38.875 i =1 3.501 3.501 50 3.501 3.501 50 Axial force resistance, flexural-torsional buckling 3.302 3.302 38.875 1.805 1.805 38.875 Buckling length l L l = 1200 mm 4.372 1.913 25.101 TALAT 2301 – Example 5.4 7
8. 8. π .E .I y π .E .I z π .E .I ω Reference 2 2 2 buckling N Ey N Ez NT G .I t . 1 loads 2 2 2 2 l l l ip N Ez = 1.03 . 10 kN 0.2 . N Ez 3 N Ey = 635.91 kN N T = 475.37 kN Guess: N N . N Ez N . NT N .i p z gc . N . N Ey y gc . N . N Ez 2 2 2 2 2 Given N Ey z sc N y sc N 0 Solution N cr minerr( N ) N cr = 348.11 kN General cross-section Table 5.5 α 0.35 λ 1 0.4 k2 1 and 5.8 f o . A eff λ c λ bar λ c λ c 0.883 = N cr 1 0.5 . 1 α . λ c 2 (5.37) φ λ 1 λ c χ φ = 0.974 χ = 0.721 2 2 φ φ λ c min z z gc max z z gc Table 5.5 ψ min z z gc = 35.6 mm ψ = 0.11 min z z gc max z z gc max z z gc = 44.4 mm 2 λ c 2.4 . ψ . 2 Table 5.5 k1 1 k 1 = 0.996 λ c . 1 2 2 1 λ c fo 5.8.3 (1) N b.Rd χ .k 1 .k 2 . .A N b.Rd = 265.8 kN γ M1 TALAT 2301 – Example 5.4 8