A213-STPinho-Dyn-lecture7

# A213-STPinho-Dyn-lecture7 - Overview of last lectures...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 05/03/2009 Overview of last lectures Structural Mechanics and Dynamics Structural Dynamics Lecture 7/7 Free vibrations Discrete systems (n DOFs) Continuous systems Periodic loading Transient loading Free vibrations Dr Silvestre Pinho 1 2 Objectives of this lecture This lecture To practice developing the theory and solving tutorial problems 3 4 Outline Developing the theory – Duhamel integral Tutorial problem – periodic vibrations Summary Developing the theory – Duhamel integral 5 6 STPinho 1 05/03/2009 Duhamel integral – section a. Q1 R(τ) Duhamel integral r (t ) = = t R0 t R0 t ∫ sin ω( t − τ) dτ = ωm ∫ sin ω t cos ωτ dτ − ∫ cos ω t sin ωτ dτ ωm 0 0 0 t m&& + kr = R (τ) r Ro Time τ Usual approximation: σdynamic = 2σstatic Is this reasonable? r (t ) = ∫ sin ω( t − τ ) R (τ ) dτ ωm 0 t R 0 sin ω t sin ωτ cos ω t cos ωτ + ωm ω ω 0 R0 sin 2 ωt + cos 2 ωt − cos ωt =2 ωm R = 0 (1 − cos ωt ) = δ static (1 − cos ωt ) k ( ) 7 8 Periodic vibrations – section b. Q4 Steady state response? Ro sinΩτ m1 Tutorial problem – periodic vibrations (a) 9 k1 Periodic vibrations – section b. Q4 Steady state response? Equation of motion: m&& + kr = R 0 sin Ωt r Ro sinΩτ Periodic vibrations – section b. Q4 1 r0 = = δ static 2 2 k−Ω m Ω 1 − ω m1 ( R0 ) General steady state solution: r = r0 sin Ωt k1 So what if Ω = ω ? ( ) (a) 11 STPinho 000000000000000000000000000000000000000000000000000000000000000000000000000000000 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 000000000000000000000000000000000000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000000000000000000000000000000000000 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00000000000000000000000000000000000000000000000000000000000000000000000000000000000 Replacing gives: R0 1 r0 = = δ static 2 k − Ω2m Ω 1− ω 000000000000000000000000000000000000000000000000000000000000000000000000000000000 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 000000000000000000000000000000000000000000000000000000000000000000000000000000000 10 Ro sinΩτ m1 k1 (a) 12 2 05/03/2009 Periodic vibrations – section b. Q4 Ro sinΩτ m2 Periodic vibrations – section b. Q4 Steady state response? m2 m1 k2 k2 Ro sinΩτ k1 m1 k 2 m 2 = k 1 m1 Ro sinΩτ m1 k 2 m 2 = k 1 m1 13 (b) Periodic vibrations – section b. Q4 Eqs. of motion: m1 0 0 &&1 k 1 + k 2 r + m 2 &&2 − k 2 r − k 2 r1 R 0 sin Ωt = k 2 r2 0 Ro sinΩτ m1 m2 Periodic vibrations – section b. Q4 Simplifying: ω 2 (1 + α ) − Ω 2 − ω2 − αω 2 r10 R 0 m 1 = ω 2 − Ω 2 r20 0 R 0 m 1 αω 2 ω (1 + α ) − Ω 2 0 2 k2 General s.s. solution: r1 r10 = r sin Ωt r2 20 r &&1 2 r10 && = − Ω r sin Ωt r2 20 ω 2 − Ω 2 r10 1 r = 2 2 2 2 2 4 20 ω (1 + α ) − Ω ω − Ω − αω ω ( )( ) k1 Define: k k ω2 = 1 = 2 m1 m 2 000000000000000000000000 000000000000000000000000000000000000000000000000000000000000000000000000 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 k m α= 2= 2 k1 m1 = 15 m 1 ω (1 + α ) − Ω 2 [( R0 2 )(ω 2 −Ω 2 )− αω ] 4 (b) Periodic vibrations – section b. Q4 ω 2 − Ω2 r10 R0 = r m ω 2 (1 + α ) − Ω 2 ω 2 − Ω 2 − αω 4 2 20 1 ω Periodic vibrations – section b. Q4 [( )( ) ( ) So what if Ω = ω ? Does it make any sense? Imagine free vibration of mass m2 What would reaction at ‘support m1’ be? Ro sinΩτ m1 17 STPinho 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 000000000000000000000000 00000000000000000000000000000000000000000000000 000000000000000000000000 00000000000000000000000000000000000000000000000 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00000000000000000000000000000000000000000000000000000000000000000000000 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 000000000000000000000000 000000000000000000000000000000000000000000000000 000000000000000000000000 000000000000000000000000000 000000000000000000000000000 000000000000000000000000000 000000000000000000000000000000000000000000000000000000 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 (a) k1 k1 14 (b) ω2 − Ω 2 2 ω 16 ( ) m2 k2 k 2 m 2 = k 1 m1 k1 18 (b) 3 05/03/2009 Summary Developing the theory – Duhamel integral Tutorial problem – periodic vibrations Any question? 19 STPinho 4 ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online