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# TALAT Lecture 2301: Design of Members Example 5.6: Axial force resistance of orthotropic plate. Open or closed stiffeners

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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 5.6: Axial force resistance of orthotropic plate. Open or closed stiffeners

1. 1. TALAT Lecture 2301 Design of Members Axial Force Example 5.6 : Axial force resistance of orthotropic plate. Open or closed stiffeners 5 pages Advanced Level prepared by Torsten Höglund, Royal Institute of Technology, Stockholm Date of Issue: 1999  EAA - European Aluminium Association TALAT 2301 – Example 5.6 1
2. 2. Example 5.6. Axial force resistance of orthotropicplate. Open or closed stiffeners Dimensions (highlighted) Plate thickness t 16.5 . mm t1 t fo 240 . MPa N newton Plate width b 1500 . mm fu 260 . MPa kN 1000 . newton 2200 . mm 70000 . MPa MPa 10 . Pa 6 Plate length L E w Stiffener pitch w 300 . mm a γ M1 1.1 2 If heat-treated alloy, then = 1 else ht = 0 ht ht 1 If cold formed (trapezoidal), then = 1 else cf. = 0 cf. cf 0 a) Open stiffeners Half stiffener pitch a = 150 mm Half bottom flange a2 50 . mm Thickness of bottom flange t2 10 . mm Stiffener depth h 160 . mm Half web thickness t3 4.4 . mm Half width of trapezoidal a1 0 . mm stiffener at the top Width of web a3 h a 3 = 160 mm Local buckling 2 .a a3 18.182 Internal elements β i β i β = β max β i β = 18.182 2 .t 3 t1 i 1 2 18.182 250 . MPa [1] Tab. 5.1 ε β 1 9 .ε β 1 9.186 = fo β 2 11 . ε β 2 11.227 = β 3 18 . ε β 3 18.371 = class i if β > β 1 , if β > β 2 , if β > β 3 , 4 , 3 , 2 , 1 class i = 3 a2 Outstand β β 1 2.5 . ε β 1 2.552 = t2 β 2 4 .ε β 2 4.082 = [1] Tab. 5.1 β =5 β 3 5 .ε β 3 5.103 = class o if β > β 1 , if β > β 2 , if β > β 3 , 4 , 3 , 2 , 1 class o = 3 class if class o > class i , class o , class i class = 3 No reduction for local buckling TALAT 2301 – Example 5.6 2
3. 3. 5.11.6 Overall buckling, uniform compression 2 .t 1 .a 2 .t 2 .a 2 2 .t 3 .a 3 2 .t 1 .a 1 A = 7.358 . 10 mm 3 2 Cross sectional area A h 2 .t 2 .a 2 .h 2 .t 3 .a 3 . Gravity centre 2 e e = 37.054 mm A Second moment of area 2 h 2 .t 2 .a 2 .h 2 .t 3 .a 3 . A .e I L = 2.751 . 10 mm 2 2 7 4 IL 3 s if cf 1 , 2 . a 2 .a 3 . a 2 1 a2 , 2 .a s = 300 mm Rigidities of orthotropic plate E .I L ν 0.3 N . mm 2 B x = 6.42 . 10 9 Table 5.10 Bx 2 .a G E mm 2 .( 1 ν ) 2 .a . E .t N . mm 3 2 B y = 2.88 . 10 7 Table 5.10 By s 12 . 1 ν 2 mm 2 .a . G .t N . mm 3 2 H = 2.016 . 10 7 Table 5.10 H s 6 mm Elastic buckling load 2 2 4 π . Bx L L Bx 2 .H B y. N cr = 2.031 . 10 kN 4 (5.77) N cr if < b 2 b b By L b 2 .π 2 (5.78) . B x .B y H otherwise b Buckling resistance A ef = 7.358 . 10 mm 3 2 A ef A A ef . f o (5.69) λ c λ c 0.295 = N cr α if ( ht > 0 , 0.2 , 0.32 ) λ 0 if ( ht > 0 , 0.1 , 0 ) α = 0.2 Table 5.6 λ 0 0.1 = 0.5 . 1 α . λ c 2 φ λ 0 λ c φ = 0.563 1 (5.33) χ c 2 2 χ c = 0.959 φ φ λ c fo A ef . χ .c N cRd = 1.54 . 10 kN 3 (5.68) N cRd for one stiffener γ M1 TALAT 2301 – Example 5.6 3
4. 4. b) Closed stiffeners Half stiffener pitch a = 150 mm Plate thickness t 1 = 16.5 mm Half bottom flange a 2 = 50 mm Thickness of bottom flange t 2 = 10 mm Stiffener depth h = 160 mm Web thickness t3 9 . mm Half width of trapezoidal a1 80 . mm stiffener at the top 2 2 width of web a3 a1 a2 h a 3 = 162.8 mm a4 a a1 a 4 = 70 mm Local buckling 9.697 2 .a 1 2 .a 2 a3 2 .a 4 10 Internal elements β i β i β i β i β = i 1 t1 2 t2 3 t3 4 t1 18.088 β max β i β = 18.1 8.485 250 . newton [1] Tab. 5.1 ε β 1 9 .ε β 1 9.186 = fo 2 mm β 2 13 . ε β 2 13.268 = β 3 18 . ε β 3 18.371 = class i if β > β 1 , if β > β 2 , if β > β 3 , 4 , 3 , 2 , 1 class i = 3 No reduction for local buckling 2 .t 1 .a 2 .t 2 .a 2 2 .t 3 .a 3 A = 8.88 . 10 mm 3 2 Cross sectional area A h 2 .t 2 .a 2 .h 2 .t 3 .a 3 . Gravity centre 2 e e = 44.415 mm A 2 h 2 .t 2 .a 2 .h 2 .t 3 .a 3 . A .e I L = 3.309 . 10 mm 2 2 7 4 Second moment IL of area 3 4. h. a 1 a 2 2 I T = 3.097 . 10 mm 7 4 Torsion constant IT 2 .a 1 2 .a 2 a3 2. t1 t2 t3 t2 4. 1 ν . a2 a 3 .a 1 .a 4 .h . 2 2 2 2 (5.79c) C1 C 1 = 3.076 . 10 mm 9 4 3 .a .t 1 3 E .t 3 B = 2.88 . 10 N . mm 7 B (5.79d) 12 . 1 2 ν TALAT 2301 – Example 5.6 4
5. 5. a1 a2 2 a 4. a 1 a 2 .a 1 .a 4 . 1 2 a 1 .a 3 t 2 3 a2 a1 (5.79e) C2 . C 2 = 4.851 a 2 . 3 .a 3 4 .a 2 3 t1 Rigidities of orthotropic plate E .I L E N . mm 2 B x = 7.72 . 10 9 Table 5.10 Bx ν 0.3 G 2 .a 2 .( 1 ν ) mm 2 . B. a (5.79a) By 2 .a 1 .a 3 .t 1 . 4 .a 2 .t 3 a 3 .t 2 3 3 3 2 .a 4 a 3 .t 1 . 4 .a 2 .t 3 a 3 .t 2 a 1 . t 3 . 12 . a 2 . t 3 4 .a 3 .t 2 3 3 3 3 3 3 G .I T N . mm 2 B y = 3.929 . 10 7 6 .a mm (5.79b) H 2 .B 1.6 . G . I T . a 4 2 N . mm 1 2 1 . 1 H = 7.305 . 10 8 L .a .B 2 C1 mm π . 4 C2 4 L Elastic buckling load 2 2 4 π . Bx L L Bx 2 .H B y. N cr = 3.378 . 10 kN 4 (5.77) N cr if < b 2 b b By L b 2 .π 2 (5.78) . B x .B y H otherwise b Buckling resistance A ef = 8.88 . 10 mm 3 2 A ef A A ef . f o (5.69) λ c λ c 0.251 = N cr α if ( ht > 0 , 0.2 , 0.32 ) λ 0 if ( ht > 0 , 0.1 , 0 ) α = 0.2 λ 0 0.1 = 0.5 . 1 α . λ c 2 φ λ 0 λ c φ = 0.547 1 (5.33) χ c χ c = 0.969 2 2 φ φ λ c fo A ef . χ .c N cRd = 1.877 . 10 kN 3 (5.68) N cRd for one stiffener γ M1 TALAT 2301 – Example 5.6 5