Unstiffened Element with torsional Restraint - an Analytical Approach for Postbuckling Behavior using GBT

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1 Unstiffened Element with torsional Restraint - an Analytical Approach for Postuckling Behavior using GBT Institute for material and mechanics in civil engineering Technical University of Darmstadt Lissaon Septemer 2006

2 Prismatic Structures Richard Schardt and Christof Schardt

3 System s, f s + 2 W 1 W x, u l

4 Modes - FSM vs. GBT (1) Before Orthogonalization deformation k=3 After Orthogonalization deformation k=3 deformation k=4 deformation k=4 deformation k=5 deformation k=5 deformation k=6 deformation k=6 deformation k=7 deformation k=7 deformation k=8 deformation k=8

5 Resulting deformation, overlay and contained modes SingleModes Comined -0,80-0,60-0,40-0,20 0,00 1,00 2,00 3,00 4,00 5,00 6,00 7,00 8,00 9,00 10,00 Reihe1 0,00 Reihe2 0,20 Reihe3 Reihe4 Reihe5 0,40 0,60 0,80 1, Reihe1 Reihe2 Reihe3 Reihe4 Reihe5

6 Numerical GBT-Solution (10 stripes) Deformation Transverse ending-moment x x

7 One single function for the unit-displacement. ending moment s = 0 deflection 1 τ k m m m 1 1 s = s = 0 s = f (η) = m 1 2 2K η2 ḟ (η) = m 1 K η ( η + 17 ( η + 17 m(s) = m 1 ( η η2 30 η2 ) ) + (1 τ)η (1) ) + (1 τ) (2) 60 η2 (3)

8 Section-values 3 C = K E 3 D = t µ G f (s) 2 ds = K3 ( τ τ 3E s ( 2 f s η ) ds+ 2K m(s) f (s) ds = s G (1 µ)(1 0.03τ τ 3 B = 1 K s m(s)2 ds + (1 τ)m 1 = 5.2 K τ(1 0.24τ) 133 κ σ = u f (s) 2 t ds = 2 ( C A τ τ κ σ = u f (s) 2 t ds = ( τ 0.01τ 2) C A 3333 ν = f (s) 4 ds = 5 t ( τ τ 5 s

9 Section-values 3 C = K E 3 D = t µ G f (s) 2 ds = K3 ( τ τ 3E s ( 2 f s η ) ds+ 2K m(s) f (s) ds = s G (1 µ)(1 0.03τ τ 3 B = 1 K s m(s)2 ds + (1 τ)m 1 = 5.2 K τ(1 0.24τ) 133 κ σ = u f (s) 2 t ds = 2 ( C A τ τ κ σ = u f (s) 2 t ds = ( τ 0.01τ 2) C A 3333 ν = f (s) 4 ds = 5 t ( τ τ 5 s

10 Section-values 3 C = K E 3 D = t µ G f (s) 2 ds = K3 ( τ τ 3E s ( 2 f s η ) ds+ 2K m(s) f (s) ds = s G (1 µ)(1 0.03τ τ 3 B = 1 K s m(s)2 ds + (1 τ)m 1 = 5.2 K τ(1 0.24τ) 133 κ σ = u f (s) 2 t ds = 2 ( C A τ τ κ σ = u f (s) 2 t ds = ( τ 0.01τ 2) C A 3333 ν = f (s) 4 ds = 5 t ( τ τ 5 s

11 Section-values 3 C = K E 3 D = t µ G f (s) 2 ds = K3 ( τ τ 3E s ( 2 f s η ) ds+ 2K m(s) f (s) ds = s G (1 µ)(1 0.03τ τ 3 B = 1 K s m(s)2 ds + (1 τ)m 1 = 5.2 K τ(1 0.24τ) 133 κ σ = u f (s) 2 t ds = 2 ( C A τ τ κ σ = u f (s) 2 t ds = ( τ 0.01τ 2) C A 3333 ν = f (s) 4 ds = 5 t ( τ τ 5 s

12 Section-values 3 C = K E 3 D = t µ G f (s) 2 ds = K3 ( τ τ 3E s ( 2 f s η ) ds+ 2K m(s) f (s) ds = s G (1 µ)(1 0.03τ τ 3 B = 1 K s m(s)2 ds + (1 τ)m 1 = 5.2 K τ(1 0.24τ) 133 κ σ = u f (s) 2 t ds = 2 ( C A τ τ κ σ = u f (s) 2 t ds = ( τ 0.01τ 2) C A 3333 ν = f (s) 4 ds = 5 t ( τ τ 5 s

13 Section-values 3 C = K E 3 D = t µ G f (s) 2 ds = K3 ( τ τ 3E s ( 2 f s η ) ds+ 2K m(s) f (s) ds = s G (1 µ)(1 0.03τ τ 3 B = 1 K s m(s)2 ds + (1 τ)m 1 = 5.2 K τ(1 0.24τ) 133 κ σ = u f (s) 2 t ds = 2 ( C A τ τ κ σ = u f (s) 2 t ds = ( τ 0.01τ 2) C A 3333 ν = f (s) 4 ds = 5 t ( τ τ 5 s

14 Differentialequations E 1 C 1 V E1 C 133 κ(1 + 2ϕ) 3 V 2 = 1 W (4) E 2 C 2 V E2 C 233 κ(1 + 2ϕ) 3 V 2 = 2 W (5) E 3 C 3 V G 3 D 3 V + 3 B 3 V +( 133 κ 1 W u κ 2 W u )(1 + ϕ) 3 V E 3333 ν(1 + ϕ)(1 + 2ϕ)( 3 V 3 ) = 0 (6) 2

15 Differentialequations E 1 C 1 V E1 C 133 κ(1 + 2ϕ) 3 V 2 = 1 W (4) E 2 C 2 V E2 C 233 κ(1 + 2ϕ) 3 V 2 = 2 W (5) E 3 C 3 V G 3 D 3 V + 3 B 3 V +( 133 κ 1 W u κ 2 W u )(1 + ϕ) 3 V E 3333 ν(1 + ϕ)(1 + 2ϕ)( 3 V 3 ) = 0 (6) 2

16 Case: uckling Condition - reduced from (3): E 3 C 3 V G 3 D 3 V + 3 B 3 V + ( 133 κ 1 W κ 2 W) 3 V = 0 Sinus-solution and areviations: w(x, s) = 3 V m 3f ( πx ) sin (7) l ( π ) 2 ( l ) 2 3 P := E 3 C + G 3 D + 3 B (8) l π R L := 133 κ 1 W κ 2 W 0 (9) Buckling condition 3 P γ R L = 0

17 Accuracy of single-function-approach compared to full solution k 2 k, l char l char 1 exact solution (numerical) analytical approach (eq. ***) ψ 2 1 ψ 0 1 1

18 Buckling Values 2 k τ = 1 τ = 0.8 τ = 0.6 τ = 0.4 τ = 0.2 τ = τ = ψ 2 1 ψ 0 1 1

19 DGL for postuckling [ 1 3 P 1 + ϕ (1 W u 133 κ + 2 W u 233 κ) + Eπ2 4 ] ( V ) 2 m ν(1 + 2ϕ) 3 V m sin πξ = 0 l

20 Stress-development (1): statically defined load

21 Stress-development (2): geometrically defined load

22 Stress-components σ σ u σ ω 2 σ u 2 σ σ ω σ u, 1 σ u σ x σ 1 σ u 1 σ 1 σ σ 2 σ u = 0 0 s 0 s statically defined geometrically defined

23 Increasing stress σ How stress σ x increases for oth cases: -20 statically defined -30 geometrically defined σ x s w/t= w/t= s

24 The two load-cases: statically defined: (SD) geometrically defined: (GD) i W = γ s iw 0 i W u = γ g iw 0 1 W u = 1 W 1 W ω 1 W = 1 W u + 1 W ω 2 W u = 2 W 2 W ω 2 W = 2 W u + 2 W ω

25 Model for geometrically defined load (a) () F γ F F 0 1 W, 2 W E A = (c) δu (d) 1 W 0 γ F 2 W 0 γ F 1 W, 2 W

26 Load-deformation curves -70 W W ψ 1 = 0 ψ 1,2 = 1 1 W 1 W exact computation approxiation reaching ultimate load ψ 2 = 0 1 W w0 + w 10 2 W

27 Ultimate loads over slenderness k k 1 1 τ = 0 τ = N λ 0 N λ 1 τ = 0 1 τ = w0 = Euler λ 0 w0 = λ

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