Upcoming SlideShare
×

# TALAT Lecture 2301: Design of Members Example 6.8: Shear force resistance of orthotropic double-skin plate

657 views

Published on

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

Published in: Education
0 Likes
Statistics
Notes
• Full Name
Comment goes here.

Are you sure you want to Yes No
• Be the first to comment

• Be the first to like this

Views
Total views
657
On SlideShare
0
From Embeds
0
Number of Embeds
5
Actions
Shares
0
31
0
Likes
0
Embeds 0
No embeds

No notes for slide

### TALAT Lecture 2301: Design of Members Example 6.8: Shear force resistance of orthotropic double-skin plate

1. 1. TALAT Lecture 2301 Design of Members Shear Force Example 6.8 : Shear 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 6.8 1
2. 2. Example 6.8. Shear 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 6.8 2
3. 3. Local buckling of top flange Length and width L L = 5 . 10 mm 3 of web panel am a1 a m = 80 mm = 62.5 am 2 2 L am am (5.97) kτ if > 1.00 , 5.34 4.00 . , 4.00 5.34 . am L L k τ = 5.341 0.81 . a m . f o (5.96) λ w t1 E λ w 0.328 = kτ 0.48 fu Table 5.12 ρ v η 0.4 0.2 . λ w fo ρ v= 1.462 η = 0.617 Table 5.12 ρ v if ρ v > η , η , ρ v ρ v= 0.617 fo ρ .b . t 1 t3 . V w.Rd = 4.036 . 10 kN 5 (5.95), two flanges V w.Rd v γ M1 No reduction for local buckling 50 0 50 100 100 50 0 50 100 Overall buckling, shear 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 TALAT 2301 – Example 6.8 3
4. 4. 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 Elastic buckling load 1 4 L. B y H φ = 1.711 . 10 5 (5.83) (5.84) φ η B x .B y b Bx η = 6.236 . 10 5 0.567 . φ 1.92 . φ 0.1 . φ 2.75 . φ .η . 2 2 6 (5.82) kτ 3.25 1.95 k τ = 1.216 10 1 k τ .π 2 4 (5.81) V o.cr . B .B 3 V o.cr = 0.039 kN b x y Buckling resistance b .t 1 .f o λ ow 3.039 . 10 3 (5.120) λ ow = V o.cr 0.6 if χ o > 0.6 , 0.6 , χ o χ o 6.495 . 10 8 (5.119) χ o χ o = 2 0.8 λ ow fo (5.118) V o.Rd χ .o . t 1 . b V o.Rd = 0.021 kN γ M1 V Rd if V w.Rd < V o.Rd , V w.Rd , V o.Rd V Rd = 0.021 kN TALAT 2301 – Example 6.8 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 of top flange Length and width L L = 5 . 10 mm 2 .a 1 3 of web panel am a m = 80 mm = 62.5 am 2 2 L am am (5.97) kτ if > 1.00 , 5.34 4.00 . , 4.00 5.34 . am L L k τ = 5.341 0.81 . a m . f o (5.96) λ w t1 E λ w 0.328 = kτ 0.48 fu Table 5.12 ρ v η 0.4 0.2 . λ w fo ρ v= 1.462 η = 0.617 Table 5.12 ρ v if ρ v > η , η , ρ v ρ v= 0.617 fo ρ .b . t 1 t3 . V w.Rd = 6.055 . 10 kN 5 (5.95), three flanges V w.Rd v t2 (web) γ M1 50 0 50 100 100 50 0 50 100 TALAT 2301 – Example 6.8 5
6. 6. Overall buckling, shear 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 1 4 L. B y H (5.83) (5.84) φ η φ = 0.018 B x .B y b Bx η = 0.469 0.567 . φ 1.92 . φ 0.1 . φ 2.75 . φ .η 2 2 (5.82) kτ 3.25 1.95 k τ = 4.156 1 k τ .π 2 . B .B 3 4 (5.81) V o.cr x y V o.cr = 105.983 kN b Buckling resistance b .t 1 .f o (5.120) λ ow λ ow 58.282 = V o.cr 0.6 if χ o > 0.6 , 0.6 , χ o χ o 1.766 . 10 4 (5.119) χ o χ o = 2 0.8 λ ow fo (5.118) V o.Rd χ .o . t 1 . b V o.Rd = 57.795 kN γ M1 V Rd if V w.Rd < V o.Rd , V w.Rd , V o.Rd V Rd = 57.795 kN TALAT 2301 – Example 6.8 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 of top flange Thickness tm if t 1 < t 2 , t 1 , t 2 t m = 5 mm Length and width L L = 5 . 10 mm 2 .a 1 3 of web panel am a m = 75 mm = 66.667 am 2 2 L am am (5.97) kτ if > 1.00 , 5.34 4.00 . , 4.00 5.34 . am L L k τ = 5.341 0.81 . a m . f o (5.96) λ w tm E λ w 0.308 = kτ 0.48 fu Table 5.12 ρ v η 0.4 0.2 . ρ v= 1.559 λ w fo η = 0.617 Table 5.12 ρ v if ρ v > η , η , ρ v ρ v= 0.617 fo ρ .b . t 1 t2 . V w.Rd = 4.036 . 10 kN 5 (5.95), two flanges V w.Rd v γ M1 No reduction for local buckling 50 0 50 100 100 50 0 50 100 TALAT 2301 – Example 6.8 7
8. 8. Overall buckling, shear 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 1 4 L. B y H φ = 5.432 . 10 3 (5.83) (5.84) φ η B x .B y b Bx η = 0.083 0.567 . φ 1.92 . φ 0.1 . φ 2.75 . φ .η 2 2 (5.82) kτ 3.25 1.95 k τ = 3.409 1 k τ .π 2 4 (5.81) V o.cr . B .B 3 V o.cr = 3.848 kN x y b Buckling resistance b .t 1 .f o (5.120) λ ow λ ow 305.887 = V o.cr 0.6 χ o 6.412 . 10 6 (5.119) χ o χ o if χ o > 0.6 , 0.6 , χ o = 2 0.8 λ ow fo (5.118) V o.Rd χ .o . t 1 . b V o.Rd = 2.099 kN γ M1 V Rd if V w.Rd < V o.Rd , V w.Rd , V o.Rd V Rd = 2.099 kN TALAT 2301 – Example 6.8 8
9. 9. Summary V Rd.a N A a = 2.812 . 10 mm = 7.6 . 10 3 2 3 a) V Rd.a = 0 kN Aa 2 mm V Rd.b N A b = 3.212 . 10 mm 3 2 b) V Rd.b = 57.8 kN = 18 Ab 2 mm V Rd.c N A c = 2.2 . 10 mm 3 2 c) V Rd.c = 2.1 kN =1 Ac 2 mm TALAT 2301 – Example 6.8 9