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TALAT Lecture 2301: Design of Members Example 5.7: Axial force resistance of orthotropic double-skin plate

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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 5.7: Axial force resistance of orthotropic double-skin plate

1. 1. TALAT Lecture 2301 Design of Members Axial Force Example 5.7 : Axial force resistance of orthotropic double-skin plate 9 pages Advanced Level prepared by Torsten Höglund, Royal Institute of Technology, Stockholm Date of Issue: 1999  EAA - European Aluminium Association TALAT 2301 – Example 5.7 1
2. 2. Example 5.7. Axial force resistance of orthotropic double-skin plate N newton Input (highlighted) kN 1000 . N MPa 10 . Pa 6 Plate thickness t 5 . mm t1 t fo 240 . MPa Plate width b 300000 . mm fu 260 . MPa Plate length L 5000 . mm E 70000 . MPa w "Pitch" (2a or d) w 160 . mm a γ M1 1.1 2 If heat-treated alloy, then = 1 else ht = 0 ht cf 0 ht 1 a) Profiles with groove and tongue Half bottom flange a2 40 . mm Thickness of bottom flange t2 5 . mm Profile depth h 70 . mm Web thickness t3 5 . mm Half width of trapezoidal a1 80 . mm stiffener at the top Number of webs nw 4 2 2 Width of web a3 a1 a2 h a 3 = 80.6 mm TALAT 2301 – Example 5.7 2
3. 3. Local buckling 0 a1 2 .a 2 a3 16 Internal elements β i β i β i β i= 1 t1 2 t2 3 t3 16 16.125 β max β i β = 16.1 250 . 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 50 0 50 100 100 50 0 50 100 Overall buckling, uniform compression nw Cross sectional area A 2 .t 1 .a 2 .t 2 .a 2 2 .t 3 .a 3 . A = 2.812 . 10 mm 3 2 2 h nw 2 .t 2 .a 2 .h 2 .t 3 .a 3 . . Gravity centre 2 2 e e = 30.022 mm A 2 h .n w 2 .t 2 .a 2 .h 2 .t 3 .a 3 . A .e I L = 2.059 . 10 mm 2 2 6 4 Second moment of area IL 3 2 4. h. a 1 a 2 2 I T = 3.517 . 10 mm 6 4 Approx. torsional constant IT 2 .a 1 2 .a 2 a3 2. t1 t2 t3 Rigidities of orthotropic plate E .I L N . mm 2 B x = 9.007 . 10 8 Table 5.10 Bx ν 0.3 2 .a mm N . mm 2 E 0.001 . N . mm B y = 1 . 10 3 Table 5.10 By G 2 .( 1 ν) mm G .I T N . mm 2 H = 5.918 . 10 8 Table 5.10 H 2 .a mm TALAT 2301 – Example 5.7 3
4. 4. Elastic buckling load 2 2 4 π . Bx L L Bx 2 .H B y. N cr = 1.067 . 10 kN 5 (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 A A ef = 2.812 . 10 mm 3 2 A ef . f o (5.69) λ c N cr λ c 0.08 = 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 = 0.501 1 (5.33) χ c χ c = 1.004 2 2 φ φ λ c fo (5.68) N cRd A ef . χ .c for one stiffener N cRd = 616.164 kN γ M1 TALAT 2301 – Example 5.7 4
5. 5. b) Truss cross section a a Half bottom flange a2 a1 a = 80 mm 2 2 Stiffener depth h 70 . mm a 1 = 40 mm Thickness of bottom flange t2 5 . mm a 2 = 40 mm Web thickness t3 5 . mm 2 2 Width of web a3 a1 h a 3 = 80.623 mm Local buckling 0 2 .a 1 2 .a 2 a3 16 Internal elements β i β i β i β i= 1 t1 2 t2 3 t3 16 16.125 β max β i β = 16.1 250 . 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 50 0 50 100 100 50 0 50 100 TALAT 2301 – Example 5.7 5
6. 6. Overall buckling, uniform compression 2 .t 1 .a 1 2 .t 2 .a 2 2 .t 3 .a 3 A = 1.606 . 10 mm 3 2 Cross sectional area A h 2 .t 2 .a 2 .h 2 .t 3 .a 3 . 2 Gravity centre e e = 35 mm A 2 Second moment h 2 .t 2 .a 2 .h 2 .t 3 .a 3 . A .e I L = 1.309 . 10 mm 2 2 6 4 of area IL 3 4. h. a 1 a 2 2 I T = 1.952 . 10 mm 6 4 Torsion constant IT 2 .a 1 2 .a 2 a3 2. t1 t2 t3 Orthotropic plate constant E .I L N . mm 2 B x = 5.728 . 10 8 Table 5.10 Bx 2 .a mm E .t 1 .t 2 .h 2 N . mm 2 B y = 8.575 . 10 8 Table 5.10 By t1 t2 mm G .I T N . mm 2 H = 3.285 . 10 8 Table 5.10 H 2 .a mm Elastic buckling load 2 2 4 π . Bx L L Bx 2 .H B y. N cr = 6.786 . 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 = 1.606 . 10 mm 3 2 A ef A A ef . f o λ c 0.08 = (5.69) λ c 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 = 0.5 1 (5.33) χ c χ c = 1.005 2 2 φ φ λ c fo (5.68) N cRd 2 . A ef . χ .c for two pitches N cRd = 704.4 kN γ M1 TALAT 2301 – Example 5.7 6
7. 7. c) Frame cross section Half bottom flange a 37.5 . mm a1 a Stiffener depth h 70 . mm a2 a Thickness of top flange t1 5 . mm a 1 = 37.5 mm Thickness of bottom flange t2 5 . mm a 2 = 37.5 mm Web thickness t3 5 . mm Width of web a3 h a 3 = 70 mm Local buckling 0 2 .a 1 2 .a 2 a3 15 Internal elements β i β i β i β i= 1 t1 2 t2 3 t3 15 14 β max β i β = 15 250 . 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 50 0 50 100 100 50 0 50 100 TALAT 2301 – Example 5.7 7
8. 8. Overall buckling, uniform compression 2 .t 1 .a 2 .t 2 .a 2 t 3 .a 3 A = 1.1 . 10 mm 3 2 Cross sectional area A h 2 .t 2 .a 2 .h t 3 .a 3 . Gravity centre 2 e e = 35 mm A 2 Second moment h 2 .t 2 .a 2 .h t 3 .a 3 . A .e I L = 1.062 . 10 mm 2 2 6 4 of area IL 3 4. h. a 1 a 2 2 I T = 1.901 . 10 mm 6 4 Torsion constant IT 2 .a 1 2 .a 2 a3 2. t1 t2 t3 Orthotropic plate constant E .I L N . mm 2 B x = 9.909 . 10 8 (5.80d) Bx 2 .a mm a .t 2 t 3 3. 3 a .t 3 6 .h .t 2 3 3 E .t 1 3 3 2 . 2 t1 t (5.80a By . 10 b . . 1 12 . 1 ν 32 . a 3 .h .t 1 .t 2 L 2 2 2 3 3 2 a .t 3 2 .h . t 1 3 3 3 t2 a .t 3 3 N . mm 2 B y = 1.118 . 10 7 3 3 2 .E . t1 t2 mm (5.80b) H t3 6 .t 1 6 .t 2 N . mm 2 3. 1 H = 8.75 . 10 6 1 1 2 .a 2 .a t3 2 .a t3 mm Elastic buckling load 2 2 4 π . Bx L L Bx 2 .H B y. N cr = 1.174 . 10 kN 5 (5.77) N cr if < b 2 b b By L b 2 .π . 2 2 .π B x .B y 2 . H = 7.501 kN (5.78) B x .B y H otherwise b b A ef = 1.1 . 10 mm 3 2 Buckling resistance A ef A A ef . f o (5.69) λ c λ c 0.047 = N cr α if ( ht > 0 , 0.2 , 0.32 ) λ 0 if ( ht > 0 , 0.1 , 0 ) α = 0.2 0.5 . 1 α . λ c 2 φ λ 0 λ c λ 0 0.1 = φ = 0.496 1 (5.33) χ c 2 2 χ c = 1.011 φ φ λ c fo N cRd 2 . A ef . χ .c for two pitches N cRd = 485.112 kN (5.68) γ M1 TALAT 2301 – Example 5.7 8
9. 9. Summary N Rd.a N A a = 2.812 . 10 mm 3 2 a) N Rd.a = 616.2 kN = 219.1 Aa 2 mm N Rd.b N A b = 3.212 . 10 mm 3 2 b) N Rd.b = 704.4 kN = 219.3 Ab 2 mm N Rd.c N A c = 2.2 . 10 mm 3 2 c) N Rd.c = 485.1 kN = 220.5 Ac 2 mm TALAT 2301 – Example 5.7 9