TALAT Lecture 2301: Design of Members Example 5.5: Axial force resistance of laced column
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This 3-page document provides an example calculation for determining the axial force resistance of a laced column. It includes dimensions, material properties, and calculations of various parameters needed for the analysis. Key steps and results are shown, such as determining the effective length, flexural buckling resistance, and checking that the lacing can resist the required shear force. In the end, it is determined that the lacing can adequately resist the applied axial load of 270 kN.
TALAT Lecture 2301: Design of Members Example 5.5: Axial force resistance of laced column
- 1. TALAT Lecture 2301 Design of Members Axial Force Example 5.5 : Axial force resistance of laced column 3 pages Advanced Level prepared by Torsten Höglund, Royal Institute of Technology, Stockholm Date of Issue: 1999 EAA - European Aluminium Association TALAT 2301 – Example 5.5 1
- 2. Example 5.5. Axial force resistance of laced column Width b 400 . mm fo 200 . MPa kN 1000 . newton 30 . mm 20 . mm E 70000 . MPa MPa 10 . Pa 6 Longitudinal members d o di γ M1 1.0 Lacing bd 40 . mm td 10 . mm 500 . mm 2 2 a d a b d = 640.3 mm lc Effective length lc 5 .m = 10 a Axial force N Ed 270 . kN 5.4.3 Slenderness parameters 2 2 do di π. 2 Single main Al A l = 392.699 mm member 4 4 4 do di π. I l = 3.191 . 10 mm 4 4 Il 64 Il il i l = 9.014 mm Al Effective 2 .a . 1 . f o λ l λ = 1.888 length 2 . a il π E l 4 .A l A = 1.571 . 10 mm 3 2 Four main A member b 2 4 .I l 4 .A l. I = 6.296 . 10 mm 7 4 I 2 I i i = 200.2 mm A Effective lc 1 fo λ o . . λ o 0.425 = length l c i π E b d .t d 2 Lacing Ad A d = 400 mm b d .t d 3 I d = 3.333 . 10 mm 3 4 Id 12 Effective Id d . 1 . fo id i d = 2.9 mm λ d λ d 3.774 = length d Ad id π E f o .A .d 3 Composite 2 λ c λ o λ c 0.466 = member E .A d .a .b 2 TALAT 2301 – Example 5.5 2
- 3. 5.8.4 Flexural buckling, buttoned strut Table 5.5 α 0.2 λ o 0.1 k1 1 k2 1 λ λ c λ c = 0.466 and 5.6 1 0.5 . 1 α . λ 2 (5.33) φ λ o λ χ φ = 0.645 χ = 0.916 2 2 φ φ λ fo 5.8.3 (1) N b.Rd χ .k 1 .k 2 . .A N b.Rd = 287.8 kN γ M1 5.8.4 Flexural buckling, single longitudinal member Table 5.5 α 0.2 λ o 0.1 k1 1 k2 1 λ λ l and 5.6 1 0.5 . 1 α . λ 2 (5.33) φ λ o λ χ φ = 2.46 χ = 0.248 2 2 fo φ φ λ 5.8.3 (1) N l.Rd χ .k 1 .k 2 . .A l N l.Rd = 19.4 kN γ M1 5.8.4 Flexural buckling, lacing 0.015 . N Ed q χ .l c q χ = 1.022 kN . m 1 TALAT qχ Vχ V χ = 2.555 kN (5.22) N Ed .λ 2 2 c l c. 1 d 0.9 . A . f o Nl V χ. 0.02 . N b.Rd = 5.757 kN N l = 4.091 kN b λ λ d 1 0.5 . 1 α . λ 2 (5.33) φ λ o λ χ φ = 7.989 χ = 0.067 2 2 φ φ λ fo 5.8.3 (1) N l.Rd χ .k 1 .k 2 . .A d N l.Rd = 5.3 kN < N l OK ! γ M1 TALAT 2301 – Example 5.5 3