TALAT Lecture 2301: Design of Members Example 9.2: Beam-column with rectangular hollow section

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

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TALAT Lecture 2301: Design of Members Example 9.2: Beam-column with rectangular hollow section

  1. 1. TALAT Lecture 2301 Design of Members Axial force and bending moment Example 9.2 : Beam-column with rectangular hollow section 4 pages Advanced Level prepared by Torsten Höglund, Royal Institute of Technology, Stockholm Date of Issue: 1999  EAA - European Aluminium Association TALAT 2301 – Example 9.2 1
  2. 2. Example 9.2 Beam-column with rectangular hollow section Dimensions and material properties MPa 10 . Pa 6 Length lc 3800 . mm kN 1000 . newton Thickness tw 6 . mm tf 6 . mm Width hy 180 . mm hi hy 2 .t f h i = 168 mm by 120 . mm bi by 2 .t w b i = 108 mm [1] Table 3.2b Alloy: EN AW-6060 T6 EP t < 15 mm f 0.2 140 . MPa fu 170 . MPa [1] (5.4), (5.5) fo f 0.2 fa fu E 70000 . MPa γ M1 1.1 γ M2 1.25 Forces and moment Axial force N Ed 110 . kN Transverse force F Ed 8 . kN F Ed . l c Bending moment M y.Ed M y.Ed = 7.6 kN . m 4 Classification of the cross section in axial compression hy 2 .t f 250 . MPa Web β w β w 28 = ε ε = 1.336 tw fo [1] Tab. 5.1 β 1w 11 . ε β 2w 16 . ε β 3w 22 . ε class c if β w β 1w , if β w > β 2w , if β w > β 3w , 4 , 3 , 2 , 1 > class c = 3 [1] 5.8.4.1 Ae A η 1.0 TALAT 2301 – Example 9.2 2
  3. 3. Classification of the cross section in y-y axis bending hy 2 .t f Web β w 0.4 . β w 11.2 = tw [1] Tab. 5.1 β 1w 11 . ε β 2w 16 . ε β 3w 22 . ε [1] 5.4.5 class w if β w β 1w , if β w > β 2w , if β w > β 3w , 4 , 3 , 2 , 1 > class w = 1 by 2 .t w Flange β f β = 18 f tf [1] Tab. 5.1 β 1f 11 . ε β 2f 16 . ε β 3f 22 . ε [1] 5.4.5 class f if β > β 1w , if β f > β 2f , if β f > β 3f , 4 , 3 , 2 , 1 f class f = 2 Classification of the total cross-section iny-y axis bending: class y if class f > class w , class f , class w class y = 2 Classification of the cross section in z-z axis bending by 2 .t w Web β w 0.4 . β w 7.2 = tf [1] Tab. 5.1 β 1w 11 . ε β 2w 16 . ε β 3w 22 . ε [1] 5.4.5 class w if β w β 1w , if β w > β 2w , if β w > β 3w , 4 , 3 , 2 , 1 > class w = 1 hy 2 .t f Flange β f β = 28 f tw [1] Tab. 5.1 β 1f 11 . ε β 2f 16 . ε β 3f 22 . ε [1] 5.4.5 class f if β > β 1w , if β f > β 2f , if β f > β 3f , 4 , 3 , 2 , 1 f class f = 3 Classification of the total cross-section inz-z axis bending: class z if class f > class w , class f , class w class z = 3 Cross section constants b y .h y b i.h i A = 3.456 . 10 mm 3 2 A b y .h y b i.h i h y .b y h i.b i 3 3 3 3 I y = 1.565 . 10 mm I z = 8.28 . 10 mm 7 4 6 4 Iy Iz 12 12 12 12 I y .2 I z.2 W el.y = 1.738 . 10 mm W el.z = 1.4 . 10 mm 5 3 5 3 W el.y W el.z hy by b y .h y b i.h i 2 2 W pl.y = 2.1 . 10 mm 5 3 W pl.y 4 4 W pl.y [1] 5.6.2.1 class y = 2 α y α y = 1.208 class z = 3 α z 1 W el.y Iy Iz iy i y = 67.3 mm iz i z = 49 mm TALAT 2301 – A Example 9.2 3 A
  4. 4. Flexural buckling lc η .f o lc η .f o TALAT (5.6) λ y . λ y 0.804 = λ z . λ z 1.105 = π .i y E π .i z E φ y 0.5 . 1 0.20 . λ y φ z 0.5 . 1 0.20 . λ z 2 2 [1] 5.8.4.1 0.1 λ y 0.1 λ z 1 1 χ y χ y = 0.779 χ z χ z = 0.586 2 2 2 2 φ y φ y λ y φ z φ z λ z [1] Table 5.5 k1 1 k2 1 A .f o N Rd N Rd = 439.9 kN γ M1 Exponents in interaction formula [1] 5.9.4.2 (4) ψ α z.α y ψ if ( ψ > 2 , 2 , ψ ) ψ = 1.208 ψ c χ z.ψ ψ c if ψ c < 0.8 , 0.8 , ψ c ψ c 0.8 = Cross weld in mid section [1] Tab. 5.2 HAZ softening factor ρ haz 0.65 ρ haz . f u . γ M1 [1] 5.9.4.5 ω o ω o 0.695 = ω x ω o f o . γ M2 fo M y.Rd α y . W el.y. M y.Rd = 26.721 kN . m M y.Ed = 7.6 kN . m γ M1 fo M z.Rd α .z el.z. W M z.Rd = 17.572 kN . m M z.Ed 0 . kN . m γ M1 χ min χ y χ min = 0.779 Flexural buckling check ψc 1.7 1.7 0.6 N Ed 1 . M y.Ed M z.Ed [1] 5.9.4.2 (4) = 0.939 < 1,0 OK ! χ min x . N Rd .ω ω o M y.Rd M z.Rd TALAT 2301 – Example 9.2 4

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