A semi- -analytical mathematical formulation as a analytical - PowerPoint PPT Presentation

National Technical University of Athens School of Naval Architecture and Marine Engineering Division of Marine Structures A semi- -analytical mathematical formulation as a analytical mathematical formulation as a A semi tool for the dynamic analysis of catenary catenary shaped shaped tool for the dynamic analysis of slender structures slender structures by by Ioannis K. K. Chatjigeorgiou Chatjigeorgiou Ioannis chatzi@naval.ntua.gr chatzi@naval.ntua.gr

Dynamic equilibrium in normal direction Dynamic equilibrium in normal direction ∂ ∂ φ ∂ φ 3 v = − + − φ + M EI T w cos F ∂ ∂ 0 nd ∂ 3 t s s 2 2

Separation of variables Separation of variables ∂ ∂ φ ∂ φ 3 v = − + − φ + M EI T w cos F 0 nd ∂ ∂ ∂ 3 t s s = + φ = φ + φ ( , ) ( ) ( , ) ( s , t ) ( s ) ( s , t ) T s t T s T s t 0 1 0 1 ∂ ∂ φ ∂ ∂ ∂ ∂ ∂ 2 2 2 4 d q q q q q 1 q q = + + + φ − − ρ 0 M T T T w sin EI dC 0 1 1 0 0 Dn ∂ ∂ ∂ ∂ ∂ ∂ ∂ 2 2 2 4 d s s 2 t t t s s s Uncoupled- -undamped undamped system system – – free vibrations free vibrations Uncoupled ∂ ∂ ∂ ∂ 2 2 4 q q q q = + φ − M T ( s ) w sin ( s ) EI 0 0 0 ∂ ∂ ∂ ∂ 2 2 4 s t s s 3 3

Natural frequencies and mode shapes - - Free Free Natural frequencies and mode shapes vibrations vibrations Nondimensional form form Nondimensional ∂ ∂ ∂ ∂ 2 4 2 y y ( ) y ( ) x y = − + α + β K x x ∂ ∂ τ ∂ ∂ 2 4 2 x x Harmonic motions Harmonic motions 4 2 d y ( x ) ( ) d y ( x ) ( ) d y ( x ) − + α + β + ω = 2 0 0 0 K x x y ( x ) 0 0 4 2 d x d x d x 4 4

Simplified model – – untensioned untensioned vertical structure vertical structure Simplified model Triantafyllou, MS and , MS and Triantafyllou Triantafyllou, GS (1991), The paradox of a , GS (1991), The paradox of a Triantafyllou hanging string: an explanation using singular perturbations, Journal of rnal of hanging string: an explanation using singular perturbations, Jou Sound and Vibration, 148, 343- -351 351 Sound and Vibration, 148, 343 5 5

Simplified model – – vertical structure under tension: vertical structure under tension: Simplified model effective tension at the top and bending stiffness of the effective tension at the top and bending stiffness of the same order of magnitude same order of magnitude } { ( ) ( ) − ε = κ Ω + + κ Ω + − + ε + c T / x y a J 2 T x Y 2 T x 1 T / x e 2 1 0 2 0 { } { } [ ] ( ) ( ) ( ) ( ) − + − ε + − + + − ε + − κ + κ − Ω κ + κ T 1 ( 1 x ) / c 1 T 1 ( 1 x ) / e a J z Y z 1 / T J z Y z x 2 2 1 0 2 0 1 1 2 1 } [ ] { ( ) ( ) ( ) ( ) − κ χ + κ χ + Ω + κ χ + κ χ − a J Y 1 /( T 1 ) J Y ( 1 x ) 2 1 0 2 0 1 1 2 1 Chatjigeorgiou, IK (2006), Solution of the boundary layer problems for , IK (2006), Solution of the boundary layer problems for Chatjigeorgiou calculating the natural modes of riser type slender structures, Proc Proc calculating the natural modes of riser type slender structures, OMAE 2006 Conf, Hamburg, Germany, OMAE2006- -92390 92390 OMAE 2006 Conf, Hamburg, Germany, OMAE2006 6 6

Semi- -analytical formulation of the solution: Can be used in analytical formulation of the solution: Can be used in Semi both time and frequency domain – – Requires the calculation Requires the calculation both time and frequency domain of mode shapes. Application of the WKB* method Application of the WKB* method of mode shapes. 4 2 d y ( x ) ( ) d y ( x ) ( ) d y ( x ) − + α + β + ω = 0 0 0 2 K x x y ( x ) 0 0 4 2 d x d x d x Introduction of a perturbation coordinate z= ε Introduction of a perturbation coordinate z= ε x, x, ε ε =K<<1 =K<<1 4 2 d ( ) d ( ) d ( ) y z ( ) y z ( ) y z − ε 5 + ε 2 α + εβ + ω 2 = 0 0 0 z z y ( z ) 0 0 4 2 d z d z d z Two possible solutions Two possible solutions ( ) ε ( ) ε = = i u z u z ( ) e y ( z ) e y z 0 0 *Logan JD, Applied Mathematics Applied Mathematics , Wiley , Wiley Interscience Interscience, second edition , second edition *Logan JD, (1997) (1997) 7 7

Perturbation expansion and equation of the perturbation Perturbation expansion and equation of the perturbation terms terms u ′ υ = ( ) ( ) ( ) ( ) ′ ′ ′ ′ ′ ′ ′ ′ ′ − ε 4 υ + ε 3 υ υ + ε 3 υ υ + ε 2 υ υ 2 + ευ 4 + α ε υ + υ 2 + β υ + ω 2 = 4 3 6 z z 0 ( ) ( ) ( ) ( ) 2 υ = υ + ευ + ε z z z O 0 1 ( ) ( ) ( ) ( ) ( ) ( ) ′ − ευ 4 + εα υ + α υ 2 + εα υ υ + β υ + εβ υ + ω 2 + ε 2 = z z 2 z z z O 0 0 0 0 0 1 0 1 ( ) ( ) ( ) − β ± β 2 − ω 2 α ( ) ( ) z z 4 z α υ + β υ + ω = 2 2 υ = z z 0 ( ) 0 0 0 α 2 z ( ) ( ) ( ) ′ − υ + α υ + α υ υ + β υ = ( ) 4 z 2 z z 0 ′ υ 4 − α υ z 0 0 0 1 1 υ = 0 0 ( ) ( ) 1 α υ + β 2 z z 0 8 8

Final solution for the mode shapes Final solution for the mode shapes ⎛ ⎞ ⎛ ⎞ ( ) ( ) ( ) ( ) ( ) ( ) x x − β ξ + β ξ − ω α ξ − β ξ − β ξ − ω α ξ 2 2 2 2 ⎜ ⎟ ⎜ ⎟ ∫ ∫ 4 4 = ξ + ξ y ( x ) C exp d C exp d ⎜ ( ) ⎟ ⎜ ( ) ⎟ 0 1 2 α ξ α ξ ⎜ ⎟ ⎜ ⎟ 2 2 ⎝ ⎠ ⎝ ⎠ 0 0 ⎛ ⎞ ⎛ ⎞ ( ) ( ) ( ) ( ) ( ) ( ) x x − β ξ + β ξ − ω α ξ − β ξ − β ξ − ω α ξ 2 2 2 2 ⎜ ⎟ ⎜ ⎟ ∫ 4 ∫ 4 + ξ + ξ exp i d exp i d C C ⎜ ( ) ⎟ ⎜ ( ) ⎟ 3 4 α ξ α ξ ⎜ ⎟ ⎜ ⎟ 2 2 ⎝ ⎠ ⎝ ⎠ 0 0 Natural frequencies and unknown constants are Natural frequencies and unknown constants are determined by the specified boundary conditions determined by the specified boundary conditions 9 9

Mode shapes for normal motions and curvature Mode shapes for normal motions and curvature 2 1 1.5 0.9 1 0.8 0.7 0.5 yo(x) 0.6 Static term 0 0.5 −0.5 0.4 vertical coordinate tension 0.3 angle −1 curvature/10 0.2 −1.5 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0.1 x 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1000 x (s/L) 500 E. Passano and C.M. Larsen, 0 Efficient analysis of a catenary riser, d 2 yo(x)/dx 2 Proc OMAE 2006 Conf, OMAE paper 92308, (2006), −500 Hamburg, Germany. −1000 −1500 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 x 10 10

Solution in the time domain – – Fundamental assumption: Fundamental assumption: Solution in the time domain linear variation of tangential motions* linear variation of tangential motions* ∂ ∂ ∂ ∂ φ ∂ ∂ ∂ 2 4 2 2 ( ) ( ) ( ) d ( ) y y y y y y y = − + α + β + ν ⋅ τ + ν ⋅ τ − 0 K x x r r b 0 ∂ a a ∂ τ ∂ τ ∂ τ ∂ ∂ ∂ 2 4 2 2 d x x x x x Using orthogonality orthogonality of mode shapes of mode shapes Using 1 N N N ( ) ∑ ∑ ∑ ∫ b ( ) ( ) η = Ω τ Λ + Ν + Ω τ Μ η − η ϕ η ϕ ϕ 0 & & & & r cos r cos d x k a 0 k kj a 0 kj j j j j j k ( 0 ) C = = = j 1 k j 1 j 1 0 Solution in the time domain suing Runge Runge- -Kutta Kutta Solution in the time domain suing *V.J. Papazoglou, S.A. Mavrakos and M.S. Triantafyllou, Nonlinear cable response and model testing in water, Journal of Sound and Vibration, 140(1990), 103-115. 11 11

Steady state response for bending vibrations for constant Steady state response for bending vibrations for constant amplitude 20cm and variable frequency 1.0, 0.5 rad/s rad/s amplitude 20cm and variable frequency 1.0, 0.5 −3 −3 x 10 x 10 4 2 3 1.5 2 1 1 0.5 y(x,t) y(x,t) 0 0 −1 −0.5 −2 −1 −3 −1.5 −4 0.8 1.5 0.6 1 1 1 0.8 0.8 0.4 0.6 0.6 t t 0.5 0.4 0.4 0.2 x x 0.2 0.2 0 0 0 0 12 12

Steady state response for bending moments for constant Steady state response for bending moments for constant amplitude 20cm and variable frequency 1.0, 0.5 rad/s rad/s amplitude 20cm and variable frequency 1.0, 0.5 0.6 1.5 0.4 1 0.2 0.5 0 y’’(x,t) y’’(x,t) 0 −0.2 −0.5 −0.4 −1 −0.6 −0.8 −1.5 0.8 1.5 0.6 1 1 1 0.8 0.8 0.4 0.6 t 0.6 t 0.5 0.4 0.4 0.2 x x 0.2 0.2 0 0 0 0 13 13

Parametric excitation of the structure Parametric excitation of the structure r a0 =20cm, ω ω =0.5rad/s =0.5rad/s r a0 =20cm, 14 14