A213-STPinho-Dyn-lecture5

A213-STPinho-Dyn-lecture5 - Overview of last lectures...

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

View Full Document Right Arrow Icon
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 0000000000000000000 0000000000000000000000000000000000000 0000000000000000000 0000000000000000000 0000000000000000000000000000000000000 0000000000000000000 0000000000000000000 0000000000000000000000000000000000000 0000000000000000000 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 0000000000000000000000000000000000000 0000000000000000000 Structural Mechanics and Dynamics Structural Dynamics Lecture 5/7 a θ1 a m (a) 1 θ2 k Dr Silvestre Pinho m (b) 2 Overview of last lectures Vibration modes 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 Overview of last lectures g 3g and ω 2 = 2 2a 2a 00 00 00 00 00 00 00 00 00 2 ω1 = Dynamic forcing functions R(t) 1 1 -1 Mode 1 1 φ1 = 1 1 φ2 = − 1 Overview of last lectures Equation of motion: M && + K r = G sin(ωτ ) r Initial assumed response: r (τ) = g s sin(ωτ ) + g c cos(ωτ ) Acceleration: &&( t ) = − ω 2 g s sin(ωτ ) − ω 2 g c cos (ωτ ) r 5 6 STPinho 00 00 00 00 00 00 00 00 00 Mode 2 periodic forces 1 Repeat Period R(t) time transient forces 3 time 4 Overview of last lectures Solution: g s = (K − ω 2 M ) G −1 1 00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0000000000000000 0 0 0 0 0 0 0 0 0 0 0 0000000000000000000000000000000 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 0 0 0 000000000000000000000000000000000000000000000 0 0 0 0 0 0 0 0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0000000000000000000000000000000 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000 000000000000000000000000000000000000 00000000000000000000000 00000000000000 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 00 00 00 0 0 0 0 0 0000000000000000000000000000000000000000000000000000000000000000000000 00000000000000 0000000000 Force = Overview of last lectures Force Force Time Force Curve Σ Overview of last lectures Force R 00000 0000000000 00000 000000000000000 00000 00000 00000 0000000000 00 00 00 00 00 00000000000000 00000 0000 00000000 00000000000000 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 0 0 0 0 0 00 00 00 00 00 00 00 00 00 00 00 00 00 0 0 0 0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 0 0 0 0 0 0 0 0 00 00 00 00 00 00 00 00 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0000000000000000 0 0 0 0 0 0 0 0 0 0 0 0000000000000000 0 0 0 0 0 0 0 0 0 0 0 0 0000000000000000000000000000000 0 0 0 0 0 0 0 0000000000000000 0 0 0 0 0000000000000000000000000000000000000000000000000000000000000000000000 0000000000000000000000000000000 0 0 0 0000000000000000000000000000000 0 0 0 0 0 0 0 000000000000000000000 0 0 0 0 000000000000000000000000000000000000 0000000000000000000000000000000000000000000000000000000000 000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0000000000000000000 0 0 0 0 0 0 0 0 0 00000000000000000000000000000000000000000 000000000 00000000 000000000 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 000000000000000 0 0 0 0 00000000000000000000000 0 0 0 0000000000000000000000000000000000000000000000000000000000000000000000 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 0 0 0 0 0 00000000000000 00000000000000000000 STPinho Objectives of this lecture Overview of last lectures transient loading To learn how to obtain the response for Find response to each impulse THIS lecture Time Impulse Representation 11 9 7 Outline Summary The Convolution or Duhamel Integral Example This lecture τ dτ Time Time Impulse Representation Typical Impulse Applied at time τ Rd τ 05/03/2009 12 10 8 2 05/03/2009 The Convolution or Duhamel Integral Two difficulties: The Convolution or Duhamel Integral Transient loading n DOFs Jean Marie Constant Duhamel 1797 - 1872 13 14 The Convolution or Duhamel Integral Eigenvectors turn multi-DOF set of The Convolution or Duhamel Integral M && + K r = R(τ ) r dynamic equations into a series of uncoupled, single degree of freedom set m 1 0 Φt M Φ = m = 0 . 0 15 0 m2 0 . 0 0 0 m3 . 0 . . . . . 0 0 0 . mp 16 The Convolution or Duhamel Integral M && + K r = R(τ ) r The Convolution or Duhamel Integral Equation of motion: M && + K r = R(τ ) r k 1 0 Φt K Φ = k = 0 . 0 0 k2 0 . 0 0 0 k3 . 0 . . . . . 0 0 0 . kp Multiply by Φt: Define q: Replace q: Diagonal matrices: Φt M && + Φt K r = Φt R(τ ) r r = Φq && Φt MΦ q + Φt KΦ q = Φt R(τ ) && m q + k q = Φt R (τ ) Solve independent equations 17 18 STPinho 3 05/03/2009 The Convolution or Duhamel Integral Eigenvectors: n-DOF n x 1-DOF The Convolution or Duhamel Integral Two time variations involved meaning of time becomes complicated Time associated with the forcing function τ Time associated with displacement t t constant τ variable between 0 and t 20 Solution process only has to be developed for a single DOF system: & m&& + cr + kr = R (τ) r 19 The Convolution or Duhamel Integral Force history series of continuous impulses of magnitude Rdτ lim(dτ→0) Solution: Find response to each impulse Superposition sum (integral) 21 The Convolution or Duhamel Integral & r Single DOF system: m&& + cr + kr = R (τ) continuous original force Can be written as: Impulse at time zero: && + 2ω u ξr + ω 2 r = & r u force 1 R (τ ) m The Convolution or Duhamel Integral Immediately before impulse Immediately after impulse Impulse change 00 00 00 00 00 00 000000000000000000 000000000000000000000000 000000000000000000000000 00 00 00 00 00 00 000000 000000000000000000 000000000000000000 force The Convolution or Duhamel Integral Impulse = change in momentum & r= R(τ ) dτ at time τ = 0 + m force τ = 0τ = 0+ force is zero for τ > 0+ Impulse R(τ) dτ in momentum t response time time 23 STPinho 000000 000000000000000000000000 00 00 00 00 00 00 000000000000000000 00 00 00 00 00 00 000000000000 00 00 00 00 00 00 000000000000000000 00 00 00 00 00 00 00 00 00 00 00 00 000000000000000000 000000000000000000000000 000000000000000000000000 00 00 00 00 00 00 000000 000000000000000000 000000000000000000 Impulse R(τ) dτ t response time time 22 Impulse R(τ) dτ t response time time 24 4 05/03/2009 The Convolution or Duhamel Integral Equilibrium equation: && + 2ωuξ r + ωu2 r = 0 for time τ ≥ 0 + & r The Convolution or Duhamel Integral Known solution: r = r01 e − αt sin β t + r02 e − αt cos βt Known solution: r = r01 e − αt Initial conditions give: − αt sin β t + r02 e cos βt 12 r01 = With α = ω u ξ and β = ω u (1 − ξ 2 ) 1 R (τ) dτ βm r02 = 0 So response to impulse at τ = 0 is: 25 r(t) = 1 − αt e sin βt R (τ) dτ βm 26 The Convolution or Duhamel Integral Response at time t to impulse at τ = 0: 1 − αt r(t) = e sin βt R (τ) dτ βm force t-τ 000000000000000000000000000000000000 00 00 00 00 00 00 000000000000000000000000000000 00 00 00 00 00 00 000000000000000000000000 00 00 00 00 00 00 000000000000000000000000000000 000000000000000000000000000000000000 Impulse R(τ) dτ The Convolution or Duhamel Integral Response at time t to impulse at τ = 0: r(t) = 1 − αt e sin βt R (τ) dτ βm Impulse at time τ: Just shift origin from 0 to τ. Response at time t to impulse at τ: r(t) = 1 −α (t − τ ) e sin β(t − τ) R (τ) dτ βm (t ≥ τ) 28 τ t-τ t response time time 27 The Convolution or Duhamel Integral force The Convolution or Duhamel Integral Linear problem r(t) = ∑ 0 t t-τ 0000000000000000000 000000000000000000000000000000000000000 00 00 00 00 00 00 00 00 00 00 0000000000 00000000000000000000000000000 00000000000000000000000000000 00 00 00 00 00 00 00 00 00 00 00000000000000000000000000000 00 00 00 00 00 00 00 00 00 00 00000000000000000000000000000 000000000000000000000000000000000000000 00 00 00 00 00 00 00 00 00 00 00000000000000000000000000000 superposition Impulse R(τ) dτ 1 −α (t − τ ) e sin β(t − τ ) R (τ ) dτ βm In the limit: r(t ) = ∫ 29 t τ t-τ t response time time 0 e − α (t − τ ) sin β(t − τ) R (τ) dτ βm 30 STPinho 5 05/03/2009 The Convolution or Duhamel Integral Neglect damping Include non-zero initial conditions r(t) = ∫ 0 t The Convolution or Duhamel Integral Solution: r(t ) = ∫ t 0 e − α (t − τ ) sin β(t − τ) R (τ) dτ βm t & sin ω u (t − τ) r R (τ) dτ + r0 cos ω u t + 0 sin ω u t ωu m ωu r ( t ) = ∫ W (t − τ) R (τ) dτ Impulse response function: W (t − τ ) = e − α ( t − τ ) sin β(t − τ) βm 32 0 Convolution or Duhamel integral 31 Example Example Undamped, 1 DOF Zero initial displacement and velocity Load: Force R(τ) R0 R(τ) = R0(1 - τ/td) R(τ) = 0 33 td time 34 Example Force R(τ) Example 0 ≤ t ≤ td R0 R(τ) = R0(1 - τ/td) Undamped, 1 DOF m&& + kr = R (τ) r R(τ) = 0 r (t ) = ∫ 0 t t sin ωu (t − τ ) sin ωu (t − τ ) τ 1 − R(τ ) dτ = ∫ t ωu m ωu m d 0 dτ td ≤ t ≤ ∞ time General solution: r(t) = ∫ 0 t td r (t ) = ∫ 0 td sin ω u (t − τ) R (τ) dτ ωu m 35 sin ωu (t − τ ) sin ωu (t − τ ) R(τ ) dτ + ∫ 0 dτ ωu m ωu m td t =∫ 0 td sin ωu (t − τ ) τ 1 − t ωu m d dτ 36 STPinho 6 05/03/2009 Example The integrals to solve might be more or less complex depending on the loading Example 00000000000000000000000000000000000 000000000000000000 000000000000000000 00000000000000000 0000000000000000000000000000000000 00000000000000000 R0 R(τ) = R0(1 - τ/td) R(τ) = 0 Back to the double pendulum… R(τ) td time 37 38 Example Equations of motion: g θ 1 0 &&1 a 0 1 && + g θ 2 − 2a 0000000000000 00 00 00 00 00 00 00 00 00 00 00 00 00 Example Generalised coordinates: θ 1 − 1 q1 θ = 1 = = Φq θ 2 1 1 q2 − g 2a θ1 = R (τ) g θ 2 0 a R0 R(τ) = R0(1 - τ/td) R(τ) = 0 R(τ) Example Equation of motion multiplied by Φt: && 1 1 1 0 1 − 1 q1 − 1 1 0 1 1 1 q + &&2 g g − 1 − 1 q 1 1 R(τ ) 1 1 a 1 2a + = g g 1 1 q2 − 1 1 0 − 1 1 − a 2a 41 STPinho 00 00 00 00 00 00 00 00 00 00 00 00 td time 39 40 Example Equation of motion multiplied by Φt: q 0 q 1 R (τ) 2 0 &&1 g a 0 2 && + 0 3g a q = − R (τ) q 2 2 Independent equations! 42 7 05/03/2009 Example Independent equations: g q 1 = R (τ ) a 3g q 2 = − R (τ) 2 && 2 + q a 2 &&1 + q Example Once solved for q, obtain θ: θ1 (t ) 1 − 1 q 1 (t ) θ (t ) = 1 1 q (t ) 2 2 Each can be (has been) solved 43 44 Transient loading Assimilating this will require independent study!!! Summary The Convolution or Duhamel Integral Example Any question? Next lecture: Continuous systems 45 46 STPinho 8 ...
View Full Document

This note was uploaded on 03/24/2010 for the course AE A.213 taught by Professor S.pinhoandm.aliabadi during the Winter '09 term at Imperial College.

Ask a homework question - tutors are online