SDOF linear oscillator Response to Harmonic Loading Giacomo Boffi - PowerPoint PPT Presentation

SDOF linear oscillator Giacomo Boffi SDOF linear oscillator Response to Harmonic Loading Giacomo Boffi Dipartimento di Ingegneria Civile e Ambientale, Politecnico di Milano March 19, 2014

Outline of parts 1 and 2 SDOF linear oscillator Giacomo Boffi Response of an Undamped Oscillator to Harmonic Load The Equation of Motion of an Undamped Oscillator The Particular Integral Dynamic Amplification Response from Rest Resonant Response Response of a Damped Oscillator to Harmonic Load The Equation of Motion for a Damped Oscillator The Particular Integral Stationary Response The Angle of Phase Dynamic Magnification Exponential Load Measuring Acceleration and Displacement The Accelerometre Measuring Displacements

Outline of parts 3 and 4 SDOF linear oscillator Giacomo Boffi Vibration Isolation Introduction Force Isolation Displacement Isolation Isolation Effectiveness Evaluation of damping Introduction Free vibration decay Resonant amplification Half Power Resonance Energy Loss

SDOF linear oscillator Giacomo Boffi Undamped Response Part I Response of an Undamped Oscillator to Harmonic Load

The Equation of Motion SDOF linear oscillator Giacomo Boffi Undamped For an undamped SDOF system subjected to an harmonic Response EOM Undamped excitation, characterized by the amplitude p 0 and frequency The Particular Integral Dynamic Amplification ω , we can write without loss of generality this equation of Response from Rest Resonant Response dynamic equilibrium: m ¨ x + k x = p 0 sin ω t . The solution to the above differential equation is the homogeneous solution plus a particular integral ξ ( t ) , x ( t ) = A sin ω n t + B cos ω n t + ξ ( t ) with m ¨ ξ ( t ) + k ξ ( t ) = p 0 sin ω t .

The Particular Integral SDOF linear oscillator Giacomo Boffi Undamped Response The particular integral can be written in terms of an EOM Undamped undetermined coefficient C times a sin: ξ ( t ) = C sin ω t , The Particular Integral Dynamic Amplification ξ ( t ) = − ω 2 C sin ω t . ¨ Response from Rest Resonant Response

The Particular Integral SDOF linear oscillator Giacomo Boffi Undamped Response The particular integral can be written in terms of an EOM Undamped undetermined coefficient C times a sin: ξ ( t ) = C sin ω t , The Particular Integral Dynamic Amplification ξ ( t ) = − ω 2 C sin ω t . ¨ Response from Rest Resonant Response 1. Substituting in m ¨ ξ ( t ) + k ξ ( t ) = p 0 sin ω t and symplyfing we get C ( k − ω 2 m ) = p 0 .

The Particular Integral SDOF linear oscillator Giacomo Boffi Undamped Response The particular integral can be written in terms of an EOM Undamped undetermined coefficient C times a sin: ξ ( t ) = C sin ω t , The Particular Integral Dynamic Amplification ξ ( t ) = − ω 2 C sin ω t . ¨ Response from Rest Resonant Response 1. Substituting in m ¨ ξ ( t ) + k ξ ( t ) = p 0 sin ω t and symplyfing we get C ( k − ω 2 m ) = p 0 . p 0 2. Solving for C we get C = k − ω 2 m ,

The Particular Integral SDOF linear oscillator Giacomo Boffi Undamped Response The particular integral can be written in terms of an EOM Undamped undetermined coefficient C times a sin: ξ ( t ) = C sin ω t , The Particular Integral Dynamic Amplification ξ ( t ) = − ω 2 C sin ω t . ¨ Response from Rest Resonant Response 1. Substituting in m ¨ ξ ( t ) + k ξ ( t ) = p 0 sin ω t and symplyfing we get C ( k − ω 2 m ) = p 0 . p 0 2. Solving for C we get C = k − ω 2 m , 3. collecting k in the right member divisor: C = p 0 1 1 − ω 2 m k k

The Particular Integral SDOF linear oscillator Giacomo Boffi Undamped Response The particular integral can be written in terms of an EOM Undamped undetermined coefficient C times a sin: ξ ( t ) = C sin ω t , The Particular Integral Dynamic Amplification ξ ( t ) = − ω 2 C sin ω t . ¨ Response from Rest Resonant Response 1. Substituting in m ¨ ξ ( t ) + k ξ ( t ) = p 0 sin ω t and symplyfing we get C ( k − ω 2 m ) = p 0 . p 0 2. Solving for C we get C = k − ω 2 m , 3. collecting k in the right member divisor: C = p 0 1 1 − ω 2 m k k 4. but k / m = ω 2 n hence C = p 0 1 1 − ω 2 /ω 2 k n

The Particular Integral SDOF linear oscillator Giacomo Boffi Undamped Response The particular integral can be written in terms of an EOM Undamped undetermined coefficient C times a sin: ξ ( t ) = C sin ω t , The Particular Integral Dynamic Amplification ξ ( t ) = − ω 2 C sin ω t . ¨ Response from Rest Resonant Response 1. Substituting in m ¨ ξ ( t ) + k ξ ( t ) = p 0 sin ω t and symplyfing we get C ( k − ω 2 m ) = p 0 . p 0 2. Solving for C we get C = k − ω 2 m , 3. collecting k in the right member divisor: C = p 0 1 1 − ω 2 m k k 4. but k / m = ω 2 n hence C = p 0 1 1 − ω 2 /ω 2 k n 5. with β = ω/ω n , we get: C = p 0 1 1 − β 2 . k

The Particular Integral SDOF linear oscillator Giacomo Boffi Undamped Response EOM Undamped The Particular Integral We can now write the particular solution, with the Dynamic Amplification Response from Rest dependencies on β singled out in the second term: Resonant Response ξ ( t ) = p 0 1 1 − β 2 sin ω t . k

The Particular Integral SDOF linear oscillator Giacomo Boffi Undamped Response EOM Undamped The Particular Integral We can now write the particular solution, with the Dynamic Amplification Response from Rest dependencies on β singled out in the second term: Resonant Response ξ ( t ) = p 0 1 1 − β 2 sin ω t . k The general integral for p ( t ) = p 0 sin ω t is hence x ( t ) = A sin ω n t + B cos ω n t + p 0 1 1 − β 2 sin ω t . k

Response Ratio and Dynamic Amplification SDOF linear oscillator Factor Giacomo Boffi Undamped Response EOM Undamped The Particular Integral ∆ st = p 0 / k being the static deformation , defining the Dynamic Amplification Response from Rest 1 Response Ratio , R ( t ; β ) = 1 − β 2 sin ω t , we can write Resonant Response ξ ( t ) = ∆ st R ( t ; β ) . 1 Introducing the dynamic amplification factor D ( β ) = 1 − β 2 ξ ( t ) = ∆ st D ( β ) sin ω t .

Response Ratio and Dynamic Amplification SDOF linear oscillator Factor Giacomo Boffi Undamped Response EOM Undamped The Particular Integral ∆ st = p 0 / k being the static deformation , defining the Dynamic Amplification 1 Response from Rest Response Ratio , R ( t ; β ) = 1 − β 2 sin ω t , we can write Resonant Response ξ ( t ) = ∆ st R ( t ; β ) . 1 Introducing the dynamic amplification factor D ( β ) = 1 − β 2 ξ ( t ) = ∆ st D ( β ) sin ω t . 4 D( β ) D ( β ) is stationary and almost equal to 1 when 3 2 ω ≪ ω n (this is a quasi- static behaviour), it 1 grows out of bound when β ⇒ 1 (resonance), it 0 -1 is negative for β > 1 and goes to 0 when ω ≫ ω n -2 (high-frequency loading). -3 0 0.5 1 1.5 2 2.5 3 β = ω / ω n

Dynamic Amplification Factor, the plot SDOF linear oscillator Giacomo Boffi Undamped Response 4 EOM Undamped D( β ) The Particular Integral Dynamic Amplification 3 Response from Rest Resonant Response 2 1 0 -1 -2 -3 0 0.5 1 1.5 2 2.5 3 β = ω / ω n

Response from Rest Conditions SDOF linear oscillator Giacomo Boffi Starting from rest conditions means that x ( 0 ) = ˙ x ( 0 ) = 0. Undamped Response EOM Undamped The Particular Integral Dynamic Amplification Response from Rest Resonant Response

Response from Rest Conditions SDOF linear oscillator Giacomo Boffi Starting from rest conditions means that x ( 0 ) = ˙ x ( 0 ) = 0. Undamped Let’s start with x ( t ) , then evaluate x ( 0 ) and finally equate Response this last expression to 0: EOM Undamped The Particular Integral Dynamic Amplification x ( t ) = A sin ω n t + B cos ω n t + ∆ st D ( β ) sin ω t , Response from Rest Resonant Response x ( 0 ) = B = 0 .

Response from Rest Conditions SDOF linear oscillator Giacomo Boffi Starting from rest conditions means that x ( 0 ) = ˙ x ( 0 ) = 0. Undamped Let’s start with x ( t ) , then evaluate x ( 0 ) and finally equate Response this last expression to 0: EOM Undamped The Particular Integral Dynamic Amplification x ( t ) = A sin ω n t + B cos ω n t + ∆ st D ( β ) sin ω t , Response from Rest Resonant Response x ( 0 ) = B = 0 . We do as above for the velocity: x ( t ) = ω n ( A cos ω n t − B sin ω n t ) + ∆ st D ( β ) ω cos ω t , ˙ x ( 0 ) = ω n A + ω ∆ st D ( β ) = 0 ⇒ ˙ ω ⇒ A = − ∆ st D ( β ) = − ∆ st β D ( β ) ω n

Response from Rest Conditions SDOF linear oscillator Giacomo Boffi Starting from rest conditions means that x ( 0 ) = ˙ x ( 0 ) = 0. Undamped Let’s start with x ( t ) , then evaluate x ( 0 ) and finally equate Response this last expression to 0: EOM Undamped The Particular Integral Dynamic Amplification x ( t ) = A sin ω n t + B cos ω n t + ∆ st D ( β ) sin ω t , Response from Rest Resonant Response x ( 0 ) = B = 0 . We do as above for the velocity: x ( t ) = ω n ( A cos ω n t − B sin ω n t ) + ∆ st D ( β ) ω cos ω t , ˙ x ( 0 ) = ω n A + ω ∆ st D ( β ) = 0 ⇒ ˙ ω ⇒ A = − ∆ st D ( β ) = − ∆ st β D ( β ) ω n Substituting, A and B in x ( t ) above, collecting ∆ st and D ( β ) we have, for p ( t ) = p 0 sin ω t , the response from rest: x ( t ) = ∆ st D ( β ) ( sin ω t − β sin ω n t ) .