MEEN 617 Notes: Handout 2a © Luis San Andrés (2008) 2-1Handout #2a (pp. 1-39) Dynamic Response of Second Order Mechanical Systems with Viscous Dissipation forces 2()2ext tdX dXMDKXFdt dt++= Free Response to initial conditions and F(t) = 0, Underdamped, Critically Damped and Overdamped Systems Free Response for system with Coulomb (Dry) friction Forced Response for Step Loading F(t) = FoMEEN 617 Notes: Handout 2a © Luis San Andrés (2008) 2-2Second Order Mechanical Translational System: Fundamental equation of motion (about equilibrium position, X=0) ()X ext t D KdVFMFFFdt==−−∑ DdXFDVDdt== : Viscous Damping Force kFKX= : Elastic restoring Force 22IdXFMaMdt== : Inertia Force where ( M, D ,K ) represent the equivalent mass, viscous damping coefficient, and stiffness coefficient, respectively. Since dXVdt=write the equation of motion as: 2()2ext tdX dXMDKXFdt dt++= + Initial Conditions in velocity and displacement; at t=0: (0) and (0)ooVV X X==MEEN 617 Notes: Handout 2a © Luis San Andrés (2008) 2-3Second Order Mechanical Torsional System: Fundamental equation of motion (about equilibrium position, θ=0) ()ext t D KdTorques I T T Tdtθθω==−−∑ Tθ D = Dθ ω : Viscous dissipation torque T θ K = Kθ θ : Elastic restoring torque T θI = I dω /dt : Inertia torque where ( I, Dθ ,Kθ ) are equivalent mass moment of inertia, rotational viscous damping coefficient, and rotational (torsional) stiffness coefficient, respectively. Since ω = dθ /dt , then write equation of motion as: 2()2ext tddIDKTdt dtθθθθθ++= + Initial Conditions in angular velocity and displacement at t=0: (0) and (0)ooωωθθ== MEEN 617 Notes: Handout 2a © Luis San Andrés (2008) 2-4(a) Free Response of Second Order Mechanical System Pure Viscous Damping Forces Let the external force be null (Fext=0) and consider the system to have an initial displacement Xo and initial velocity Vo. The equation of motion for a 2nd order system with viscous dissipation is: 220dX dXMDKXdt dt++= (1) with initial conditions (0) and (0)ooVV X X== Divide Eq. (1) by M and define: nKMω=: undamped natural frequency of system crDDζ= : viscous damping ratio, where 2crDKM= is known as the critical damping value With these definitions, Eqn. (1) becomes: 22220nndX dXXdt dtζω ω++= (2) The solution of the Homogeneous Second Order Ordinary Differential Equation with Constant Coefficients is of the form: ()stXt Ae= (3) Where A is a constant yet to be found from the initial conditions. Substitute Eq. (3) into Eq. (2) and obtain:MEEN 617 Notes: Handout 2a © Luis San Andrés (2008) 2-5()2220nnssAζω ω++= (4) Note that A must be different from zero for a non trivial solution. Thus, Eq. (4) leads to the CHARACTERISTIC EQUATION of the system given as: ()2220nnssζω ω++= (5) The roots of this 2nd order polynomial are: ()1/221,21nnsζω ω ζ=− −∓ (6) The nature of the roots (eigenvalues) clearly depends on the value of the damping ratio ζ . Since there are two roots, the solution to the differential equation of motion is now rewritten as: 1212()ststXt Ae Ae=+ (7) where A1, A2 are constants determined from the initial conditions in displacement and velocity. From Eq. (6), differentiate three cases: Underdamped System: 0 < ζ < 1, → D < Dcr Critically Damped System: ζ = 1, → D = Dcr Overdamped System: ζ > 1, → D > Dcr Note that ()1nτζω= has units of time; and for practical purposes, it is regarded as an equivalent time constant for the second order system.MEEN 617 Notes: Handout 2a © Luis San Andrés (2008) 2-6Free Response of Undamped 2nd Order System For an undamped system, ζ = 0, i.e. a conservative system without viscous dissipation, the roots of the characteristic equation are imaginary: 12;nnsisiωω=− = (8) where 1i=− is the imaginary unit. Using the complex identity eiat = cos(at) + i sin(at), renders the undamped response of the conservative system as: ()()12() cos sinnnXtC tC tωω=+ (9.a) where nKMω= is the natural frequency of the system. At time t = 0, the initial conditions are (0) and (0)ooVV X X== hence 010 2andnVCX Cω== (9.b) and equation (9.a) can be written as: ()() cosMnXt X tωϕ=− (9.c) Where 22002MnVXXω=+and ()00tannVXϕω= XM is the maximum amplitude response. Notes: In a purely conservative system, the motion never dies out, it is harmonic and periodic. Motion always oscillates about the equilibrium position X = 0MEEN 617 Notes: Handout 2a © Luis San Andrés (2008) 2-7Free Response of Underdamped 2nd Order System For an underdamped system, 0 < ζ< 1, the roots are complex conjugate (real and imaginary parts), i.e. ()1/221,21nnsiζω ω ζ=− −∓ (10) where 1i =− is the imaginary unit. Using the complex identity eiat = cos(at) + i sin(at), write the solution for underdamped response of the system as: ()()()12() cos sinntddXt e C t C tζωωω−=+ (11) where ()1/221dnωω ζ=− is the system damped natural frequency. At time t = 0, the initial conditions are (0) and (0)ooVV X X== Then 0010 2andndVXCX Cζωω+== (11.b) Equation (11) representing the system response can also be written as: ()() cosntMdXt e X tζωωϕ−=− (11.c) where 2212MXCC=+and ()21tanCCϕ= Note that as t→ ∞, X(t) → 0, i.e. the equilibrium position only if ζ > 0; and XM is the largest amplitude of response only if ζ=0 (no damping).MEEN 617 Notes: Handout 2a © Luis San Andrés (2008) 2-8 Free Response of Underdamped 2nd Order System: initial displacement only damping ratio varies Xo = 1, Vo = 0, ωn = 1.0 rad/s ζ = 0, 0.1, 0.25 Motion decays exponentially for ζ > 0 Faster system response as ζ increases, i.e. faster decay towards equilibrium position X=0 Free response Xo=1, Vo=0, wn=1 rad/s-1.5-1-0.500.511.50 10203040time (sec)X(t)damping ratio=0.0damping ratio=0.1damping ratio=0.25MEEN 617 Notes: Handout 2a © Luis San Andrés (2008) 2-9Free Response of Underdamper 2nd Order System: Initial velocity only damping varies Xo = 0, Vo = 1.0 ωn = 1.0 rad/s; ζ = 0, 0.1, 0.25 Motion decays exponentially for ζ > 0 Faster system response as ζ increases, i.e. faster decay towards equilibrium position X=0 Note