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TALAT Lecture 2301: Design of Members Example 5.3: Resistance of cross section with radiating outstands
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TALAT Lecture 2301: Design of Members Example 5.3: Resistance of cross section with radiating outstands

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

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

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  • 1. TALAT Lecture 2301 Design of Members Axial Force Example 5.3 : Resistance of cross section with radiating outstands 6 pages Advanced Level prepared by Torsten Höglund, Royal Institute of Technology, Stockholm Date of Issue: 1999  EAA - European Aluminium Association TALAT 2301 – Example 5.3 1
  • 2. Example 5.3. Resistance of cross section with radiating outstands Flange width b 80 . mm Length L 1200 . mm kN 1000 . newton Inner circle c 18 . mm fo 300 . MPa γ M1 1.0 MPa 10 . Pa 6 Inner thickness ti 10 . mm E 70000 . MPa G 27000 . MPa Outer thickness to 5 . mm π .k 1 π k 1 , 4 .. 16 φk 1 φ k φ k 1 4 3 4 π .k 2 π φ k 1 φ 18 φ 17 4 3 4 Nodes no., i 0 .. 16 yk 1 b . cos φ k 1 zk 1 b . sin φ k 1 co-ordinates, thickness yk c . cos φ k zk c . sin φ k yk 1 c . cos φ k 1 zk 1 c . sin φ k 1 y18 y12 z18 z0 tlk 1 o th k 1 o tlk to th k ti tlk 1 ti . 0.8 th k 1 tlk 1 tl18 ti th 18 to yi zi tli th i = = = = i = φi = mm mm mm mm 1 -0.785 12.728 -12.728 5 10 2 0 18 0 8 8 3 0 80 0 0 0 4 0 18 0 5 10 5 0.785 12.728 12.728 8 8 6 0.785 56.569 56.569 0 0 7 0.785 12.728 12.728 5 10 90 8 1.571 0 18 8 8 9 1.571 0 80 0 0 45 10 1.571 0 18 5 10 11 2.356 -12.728 12.728 8 8 0 12 2.356 -56.569 56.569 0 0 13 2.356 -12.728 12.728 5 10 14 3.142 -18 0 8 8 45 15 3.142 -80 0 0 0 16 3.142 -18 0 5 10 90 17 3.927 -12.728 -12.728 8 8 90 45 0 45 90 18 3.927 -56.569 -56.569 10 5 Nodes i 1 .. rows ( y ) 1 TALAT 2301 – Example 5.3 2
  • 3. Cross-section constants, varying thickness 0 . mm 2 Sectorial ω 0 co-ordinates ω 0 yi .z yi . zi 1 ωi ω ω 0 i 1 i i 1 i 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 rows ( y ) 1 tli th i Area A .l A = 3.916 . 10 mm 3 2 i 2 i =1 rows ( y ) 1 yi 1 . t1 i yi 1 . t2 i . li S z = 1.186 . 10 11 3 First moments Sz yi mm of area i =1 rows ( y ) 1 zi 1 . t1 i zi 1 . t2 i . li S y = 2.457 . 10 mm 4 3 Sy zi i =1 rows ( y ) 1 Sω ω . t1 ω ω . t2 . l S ω = 2.692 . 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 I z = 4.551 . 10 mm 6 4 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 I y = 3.402 . 10 mm 6 4 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 i =1 Mixed I ω = 2.638 . 10 mm 9 6 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 I yz = 1.059 . 10 10 4 mm 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 I yω = 5.018 . 10 mm 7 5 TALAT 2301 – Example 5.3 3
  • 4. 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 I zω = 1.689 . 10 mm 7 5 l th i . tli . i 2 2 Iai tli th i 48 rows ( y ) 1 Iai . yi yi 1 . zi zi 1 I yz = 1.061 . 10 10 4 Influence of I yz I yz mm thickness 2 li i =1 rows ( y ) 1 Iai . zi 2 zi 1 I z = 4.56 . 10 mm 6 4 Iz Iz 2 li i =1 rows ( y ) 1 Iai . yi 2 yi 1 I y = 3.413 . 10 mm 6 4 Iy Iy 2 li i =1 rows ( y ) 1 l th i . tli . i . 1.05 I t = 8.602 . 10 mm 2 2 4 4 It tli th i 12 i =1 Sy A . z gc I y = 3.259 . 10 mm 2 6 4 z gc z gc = 6.274 mm Iy Iy A Sz y gc = 3.028 . 10 A . y gc I z = 4.56 . 10 mm 15 2 6 4 y gc mm Iz Iz A S y .S z S z.S ω I yz = 1.805 . 10 I yω = 5.018 . 10 mm 10 4 7 5 I yz I yz mm I yω I yω A A S y .S ω 2 Sω I ω = 7.875 . 10 mm I zω = 3.309 . 10 8 6 9 5 Iω Iω I zω I zω mm A A 1 2 . I yz I z . 10 , 0 , . atan 9 Principal α if I z Iy < Iy 180 α. = 7.944 . 10 15 axis 2 Iz Iy π 1. 4 . I yz I ξ = 4.56 . 10 mm 2 2 6 4 Iξ Iy Iz Iz Iy 2 1. 4 . I yz I η = 3.259 . 10 mm 2 2 6 4 Iη Iy Iz Iz Iy 2 TALAT 2301 – Example 5.3 4
  • 5. I zω . I z I yω . I yz I yω . I y I zω . I yz y sc = 1.625 . 10 15 Shear y sc z sc mm I y .I z I y .I z centre 2 2 I yz I yz z sc = 11.005 mm Warping Iw Iω z sc . I yω y sc . I zω I w = 2.353 . 10 mm constant 8 6 Polar radius Iy Iz 2 2 gyration ip y sc y gc z sc z gc i p = 44.932 mm A 0 50 Dashed line = gravity centre 0 0 Point = shear centre 50 50 0 50 Axial force resistance, flexural-torsional buckling Buckling length l L l = 1200 mm π .E .I y π .E .I z π .E .I ω Reference 2 2 2 buckling N Ey N Ez NT G .I . 1 2 2 t 2 2 loads l l l ip N Ey = 1.56 . 10 kN N Ez = 2.19 . 10 kN N T = 1.34 . 10 kN 0.2 . N Ez 3 3 3 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 N cr = 1.32 . 10 kN 3 Solution N cr minerr( N ) TALAT 2301 – Example 5.3 5
  • 6. Cross-section entirely of radiating outstands, no welding, unsymmetical section Table 5.5 α 0.2 λ 1 0.6 k2 1 and 5.8 f o .A λ c λ bar λ c λ c 0.945 = N cr 1 0.5 . 1 α . λ c 2 (5.37) φ λ 1 λ c χ φ = 0.981 χ = 0.804 2 2 φ φ λ c min z z gc max z z gc Table 5.5 ψ min z z gc = 62.8 mm ψ = 0.08 min z z gc max z z gc 2 max z z gc = 73.7 mm λ c 2.4 . ψ 2. Table 5.5 k1 1 k 1 = 0.998 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 = 942.32 kN γ M1 TALAT 2301 – Example 5.3 6