ASE 365 - Lecture 31

# ASE 365 - Lecture 31 - Free vibration of continuous systems...

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Unformatted text preview: Free vibration of continuous systems approach. Newtonian a by vibration, transverse in beams and vibration, torsional in shafts vibration, axial in rods vibration, transverse in strings for ICs) BCs, (PDEs, problems value boundary derived have We • • • • length. unit per torque or force nt; displaceme angular or axial , transverse ; or , ; or where : problems order Second = = = = = ′ ′- f u GJ EA T s J A m f u s u m ρ ρ ) ( f v EI v A = ′ ′ ′ ′ + ) ( ρ : problems order Fourth L x t x f t x v T t x v x A < < = = ′ ′- , ) , ( ) ) , ( ( ) , ( ) ( ρ : string a of vibration free for PDE visualize. to easiest is vibration transverse problems, order second the Of ) variables. of n (Separatio ? which in one I.e., solution? motion" s synchronou " a there Is : question Key ) ( ) ( ) , ( t F x V t x v = ) ! (separable : or : is which : becomes PDE so, If ) ( ) ( ) ) ( ( ) ( ) ( ) ( ) ) ( ( ) ( ) ( ) ( , ) ) ( ) ( ( ) ( ) ( ) ( x V x A x V T t F t F t F x V T t F x V x A L x t F x V T t F x V x A ρ ρ ρ ′ ′ = ′ ′ = < < = ′ ′- . : time in ODE an gives This =- CF F C x V x A x V T t F t F x x V x A x V T t F t F = ′ ′ = ′ ′ = ) ( ) ( ) ) ( ( ) ( ) ( ) ( ) ( ) ) ( ( ) ( ) ( ρ ρ : constant a to equal be must they Therefore, only. of function a is RHS the and only, time of function a is LHS the , Since system. ve conservati a being string the with consistent not is This lly. exponentia decay and grow solutions two s ' ) ( t F . is equation stic characteri the ODE, this For 2 =- C s negative. one and positive one : roots real two are there , If C ). if (linear solution harmonic a has which , becomes ODE time the Then . or , So 2 2 2 = = +- = ≤ ϖ ϖ ϖ F F C C ). ( ) ( ) ) ( ( ) ( ) ( ) ) ( ( ) ( ) ( 2 2 x V x A x V T x x V x A x V T t F t F ρ ϖ ϖ ρ = ′ ′-- = ′ ′ = : in ODE an obtain we which from , have we Then vibration. of modes and s frequencie natural obtain to solved are types EVP Both . EVP the to analogous (EVP), problem eigenvalue a is This u u M K 2 ϖ = algebraic al differenti problems. eigenvalue al differenti analogous at arrive to taken be can steps same the problems, vibration...
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ASE 365 - Lecture 31 - Free vibration of continuous systems...

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