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# TALAT Lecture 2301: Design of Members Example 8.1: Torsion constants for open cross section

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This example provides calculations on torsion of members based on Eurocode 9.

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### TALAT Lecture 2301: Design of Members Example 8.1: Torsion constants for open cross section

1. 1. TALAT Lecture 2301 Design of Members Torsion Example 8.1 : Torsion constants for open cross section 6 pages Advanced Level prepared by Torsten Höglund, Royal Institute of Technology, Stockholm Date of Issue: 1999  EAA - European Aluminium Association TALAT 2301 – Example 8.1 1
2. 2. Example 8.1. Torsion constants for open cross section Comment: The following expressions are applicable to open cross sections. When there are branches, go back the same way but with thickness = 0. Example: The co-ordinates for node 7 and 8 are the same as for 5 and 4 and t = 0 for element 7 and 8. Nodes, co-ordinates and thickness 0 34 0 6 1 10 0 6 2 8 9 6 3 25 9 6 4 25 3 6 5 59 0 6 6 59 13 15 i y . mm z . mm t . mm 7 59 0 0 300 8 25 3 0 9 25 34 6 0 250 10 0 37 6 11 0 245 6 200 12 62 252 6 13 62 289 10 150 z Nodes i 1 .. rows ( y ) 1 mm 100 Area of cross ti . section 2 2 dAi yi yi 1 zi zi 1 elements 50 Cross rows ( y ) 1 0 section 0 A = 3.199 . 10 mm 3 2 area A dAi i =1 50 50 0 50 100 First moment rows ( y ) 1 y dAi Sy of area, Sy zi zi 1 . z gc mm gravity centre 2 A i =1 z gc = 117.14 mm Second rows ( y ) 1 dAi zi . zi 1 . A . z gc moment of 2 2 2 Iy zi zi 1 Iy Iy area of effective 3 i =1 cross section I y = 3.596 . 10 mm 7 4 rows ( y ) 1 First moment dAi Sz of area, Sz yi yi 1 . y gc 2 A gravity centre i =1 y gc = 19.073 mm TALAT 2301 – Example 8.1 2
3. 3. Second rows ( y ) 1 dAi yi . yi 1 . A . y gc moment 2 2 2 Iz yi yi 1 Iz Iz of area 3 i =1 I z = 2.104 . 10 mm 6 4 rows ( y ) 1 Second dAi S y .S z moment I yz 2 . yi 1 . zi 1 2 . yi . zi yi 1 . zi yi . zi 1 . I yz I yz of area 6 A i =1 I yz = 2.026 . 10 mm 6 4 1 2 . I yz 0 . mm , 0 , . atan 4 Principal α if I z Iy 180 axis 2 Iz Iy α. = 3.414 π 1. 4 . I yz I ξ = 3.608 . 10 mm 2 2 7 4 Iξ Iy Iz Iz Iy 2 1. 4 . I yz I η = 1.983 . 10 mm 2 2 6 4 Iη Iy Iz Iz Iy 2 0 . mm 2 Sectorial ω0 co-ordinates ω 0 yi .z yi . zi 1 i 1 i ωi ω i 1 ω 0 i ωmean = rows ( y ) 1 dAi Iω mean of Iω ω ω . ω mean i 1 i sectorial 2 A i =1 co-ordinates ω mean = 799.492 mm rows ( y ) 1 Sectorial dAi S z.I ω constant I yω 2 . yi 1 . ω i 1 2 . yi . ω i yi 1 . ω i yi . ω i 1 . I yω I yω 6 A i =1 I yω = 2.921 . 10 mm 8 5 rows ( y ) 1 Sectorial dA S y .I ω I zω 2 . ω i 1 . zi 1 2 . ω i . zi ω .z ω i . zi . Ii I zω constant i 1 i 1 6 zω A i =1 I zω = 9.181 . 10 mm 8 5 rows ( y ) 1 2 Sectorial dAi Iω ω i .ω i 1 . 2 2 constant I ωω ω i ω i 1 I ωω I ωω 3 A i =1 = 7.377 . 10 10 6 I ωω mm I zω . I z I yω . I yz I yω . I y I zω . I yz Shear y sc z sc y sc = 18.728 mm I y .I z I y .I z centre 2 2 I yz I yz z sc = 120.773 mm Warping Iw I ωω z sc . I yω y sc . I zω if t i > 0 . mm , t i , 100 . mm I w = 2.129 . 10 constant 10 6 to mm i rows ( y ) 1 2 TALAT 2301 – Example 8.1 3
4. 4. i i i rows ( y ) 1 2 ti It dAi . I t = 5.856 . 10 mm 4 4 Torsion It Wt constant 3 min t o i =1 W t = 9.76 . 10 mm 3 3 Sectorial co-ordinate i 0 .. rows ( y ) 1 with respect ω s ω i ω mean z sc . yi y gc y sc . zi z gc ω mi min ω s ω ma max ω s to shear centre i Iw ω max 7.804 . 10 mm 3 2 ω max if ω mi > ω ma , ω mi , ω ma = Ww ω max i 1 .. rows ( y ) 1 A = 3.199 . 10 mm 3 2 y gc = 19.073 mm 300 z gc = 117.14 mm y sc = 18.728 mm 250 z sc = 120.773 mm I y = 3.596 . 10 mm 7 4 200 I z = 2.104 . 10 mm 6 4 I yz = 2.026 . 10 mm 6 4 150 I t = 5.856 . 10 mm 4 4 Torsion constants W t = 9.76 . 10 mm 3 3 100 Warping I w = 2.129 . 10 10 6 constants mm 50 W w = 2.729 . 10 mm 6 4 ω max 7.804 . 10 mm 3 2 = 0 α = 3.414 deg I ξ = 3.608 . 10 mm 7 4 50 50 0 50 100 I η = 1.983 . 10 mm 6 4 Comment: If the load is acting below the shear centre (the point) there is no torsional moment acting on the beam. TALAT 2301 – Example 8.1 4
5. 5. Sectorial co-ordinate 300 with respect to shear centre 250 i 0 .. rows ( y ) 1 ω s 200 i = mm . 1000 2 i= 0 -7.804 150 1 -4.906 2 -4.743 3 -0.46 100 4 0.065 5 3.938 6 2.927 50 7 3.938 8 0.065 9 1.42 0 10 -0.618 11 3.278 12 -4.293 13 -1.307 50 100 100 50 0 50 100 150 sectorial co-ordinate cross section kN 1000 . N 5 . kN . m 4.8 . m MPa 10 . Pa 1 6 Biaxial bending q L Moment 2 My 14.4 L My q. Mz 0 . kN . m = kN . m 8 Mz 0 cos ( α ) sin( α ) Mξ My Mξ 14.374 Principal axis R R. = kN . m bending sin( α ) cos ( α ) Mη Mz Mη 0.857 Torsion mt 0 . kN m t = 0 kN m t .L 2 B 3 = 0 kN . m 2 Bi-moment B3 8 St Venants m t .L torsion Tw T w = 0 kN . m 2 Warping B3 σ w .ω stress si i Iw TALAT 2301 – Example 8.1 5
6. 6. yi yi y gc y gc Rotation of co-ordinate i 0 .. rows ( y ) 1 R. R. zi zi z gc z gc system M ξ . yi y gc M η . zi z gc Bending stresses σ ξ σ η i Iξ i Iη Sum of stresses σ σ ξ σ η σ w σ mi min ( σ ) σ ma max ( σ ) Max stresses σ max if σ mi > σ ma , σ mi , σ ma σ max= 74.1 MPa Principal axis 300 250 σ w σ ξ σ η σi ti i i i = = = = = 200 i= MPa MPa MPa MPa mm 1 0 -8.78 51.3 42.52 6 2 0 -7.78 55.133 47.36 6 150 3 0 5.35 54.283 59.63 6 4 0 5.07 49.105 54.17 6 5 0 18.66 49.524 68.18 6 6 0 18.97 55.134 74.1 15 100 7 0 18.66 49.524 0 0 8 0 5.07 49.105 0 0 9 0 4.33 35.727 40.06 6 50 10 0 -5.68 35.076 29.39 6 11 0 -10.62 -54.687 -65.31 6 0 12 0 13.87 -59.304 -45.43 6 0 13 0 13 -75.271 -62.27 10 50 100 50 0 50 100 Cross section Axial stress TALAT 2301 – Example 8.1 6