ME459/659 Notes# 1 A Review to the Dynamic Response of SDOF Second Order Mechanical System with Viscous Damping by Luis San Andrés Mast-Childs Chair Professor ME 459/659 Sound & Vibration Measurements Reproduced from material in ME 363 and ME617 http://rotorlab.tamu.edu ME459/659 Notes# 1a (pp. 1-31) Dynamic Response of SDOF Second Order Mechanical System: Viscous Damping d2 X d X MDKXF d t2 d t ext() t Free Response to F(t) = 0 + initial conditions and Underdamped, Critically Damped and Overdamped Systems Forced Response to a Step Loading F(t) = Fo MEEN 459/659 Notes 1a © Luis San Andrés (2019) 1-1 Second Order Mechanical Translational System: d2 X d X Fundamental equationMDKXF of motion about equilibrium position (X=0) d t2 d t ext() t dX2 FMFFF Xdt2 ext() t D K dX FD : Viscous Damping Force D dt FKXk : Elastic restoring Force ( M, D ,K ) represent the system equivalent mass, viscous damping coefficient, and stiffness coefficient, respectively. equation of motion + Initial Conditions in velocity and displacement; at t=0: X(0) Xoo and X (0) V MEEN 459/659 Notes 1a © Luis San Andrés (2019) 1-2 Second Order Mechanical Torsional System: Fundamental equation of motion about equilibrium position, θ=0 d Torques I Text() t T D T K ; dt Tθ D = D : Viscous dissipation torque T θ K = K : Elastic restoring torque ( I, D ,K ) are the system equivalent mass moment of inertia, rotational viscous damping coefficient, and rotational (torsional) stiffness coefficient, respectively. dd2 Equation of motion IDKT2 ext() t d t d t + Initial Conditions in angular velocity and displacement at t=0: (0) ooand (0) MEEN 459/659 Notes 1a © Luis San Andrés (2019) 1-3 (a) Free Response of Second Order SDOF Mechanical System Let the external force Fext=0 and the system has an initial displacement Xo and initial velocity Vo. EOM is d2 X d X MDKX 0 d t2 d t (1) Divide Eq. (1) by M and define: K n M : natural frequency of system D : viscous damping ratio, Dcr where DKMcr 2 is known as the critical damping magnitude. With these definitions, Eqn. (1) d2 X d X 20 2 X d t2 nn d t (2) The solution of the Homogeneous Second Order Ordinary Differential Equation with Constant Coefficients is of the form: st X() t Ae (3) where A is a constant found from the initial conditions. Substitute Eq. (3) into Eq. (2) and obtain: 22 s20 nn s A (4) MEEN 459/659 Notes 1a © Luis San Andrés (2019) 1-4 A is not zero for a non trivial solution. Thus, Eq. (4) leads to the CHARACTERISTIC EQUATION of the SDOF system: 22 ss20 nn (5) The roots of this 2nd order polynomial are: 2 1/ 2 s1,2 nn 1 (6) The nature of the roots (eigenvalues) depends on the damping ratio (>1 or < 1). Since there are two roots, the solution is s12 t s t X() t A12 e A e (7) A1, A2 determined from the initial conditions in displacement and velocity. From Eq. (6), differentiate three cases: Underdamped System: 0 < < 1, D < Dcr Critically Damped System: = 1, D = Dcr Overdamped System: > 1, D > Dcr Note that 1 n has units of time; and for practical purposes, it is akin to an equivalent time constant for the second order system. MEEN 459/659 Notes 1a © Luis San Andrés (2019) 1-5 Free Response of Undamped 2nd Order System For an undamped system, = 0, i.e., a conservative system without viscous dissipation, the roots of the characteristic equation are imaginary: s12 inn; s i (8) where i 1 Using the complex identity eiat = cos(at) + i sin(at), renders the undamped response as: X( t ) C12 cosnn t C sin t (9.a) K n M where is the natural frequency of the system. At time t = 0, apply the initial conditions to obtain V0 CXC1 0and 2 (9.b) n Eq. (9.a) can be written as: X( t ) XMn cos t (9.c) 2 2 V0 V0 where XXM 0 2 and tan n n X 0 XM is the maximum amplitude response. Notes: In a purely conservative system ( = 0), the motion never dies. Motion always oscillates about the equilibrium position X = 0 MEEN 459/659 Notes 1a © Luis San Andrés (2019) 1-6 Free Response of Underdamped 2nd Order System For an underdamped system, 0 < < 1, the roots are complex conjugate (real and imaginary parts), i.e. 2 1/ 2 si1,2 nn 1 (10) Using the complex identity eiat = cos(at) + i sin(at), the response is: n t X( t ) e C12 cosdd t C sin t (11) 1/ 2 where 2 is the system damped natural frequency. dn 1 At time t = 0, applying initial conditions gives VX00n CXC1 0and 2 (11.b) d Eqn. (11) can be written as: n t X( t ) e XMd cos t (11.c) 22 C2 where XCCM 12 and tan C1 Note that as t , X(t) 0, i.e. the equilibrium position only if > 0; and XM is the largest amplitude of response only if =0 (no damping). MEEN 459/659 Notes 1a © Luis San Andrés (2019) 1-7 Free Response of Underdamped 2nd Order System: initial displacement only damping ratio varies Xo = 1, Vo = 0, ωn = 1.0 rad/s = 0, 0.1, 0.25 Motion decays exponentially for > 0 Faster system response as increases, i.e. faster decay towards equilibrium position X=0 Free response Xo=1, Vo=0, wn=1 rad/s damping ratio=0.0 1.5 damping ratio=0.1 damping ratio=0.25 1 0.5 0 X(t) 0 10 20 30 40 -0.5 -1 -1.5 time (sec) MEEN 459/659 Notes 1a © Luis San Andrés (2019) 1-8 Free Response of Underdamper 2nd Order System: Initial velocity only damping varies Xo = 0, Vo = 1.0 ωn = 1.0 rad/s; = 0, 0.1, 0.25 Motion decays exponentially for > 0 Faster system response as increases, i.e. faster decay towards equilibrium position X=0 Note the initial overshoot Free response Xo=0, Vo=1, wn=1 rad/s damping ratio=0.0 1.5 damping ratio=0.1 damping ratio=0.25 1 0.5 0 X(t) 0 10 20 30 40 -0.5 -1 -1.5 time (sec) MEEN 459/659 Notes 1a © Luis San Andrés (2019) 1-9 Free Response of Overdamped 2nd Order System For an overdamped system, > 1, the roots of the characteristic equation are real and negative, i.e., 1/ 2 1/ 2 ss 22 1 ; 1 (12) 12nn The free response of an overdamped system is: n t X( t ) e C1 cosh * t C 2 sinh * t (13) 1/ 2 where 2 has units of 1/time. Do not confuse this term * n 1 with a frequency since the response of motion is NOT oscillatory. At time t = 0, apply the initial conditions to get VX00n CXC1 0and 2 (14) * Note that as t , X(t) 0, i.e. the equilibrium position. Note: An overdamped system does to oscillate. The larger the damping ratio >1, the longer time it takes for the system to return to its equilibrium position. MEEN 459/659 Notes 1a © Luis San Andrés (2019) 1-10 Free Response of Critically Damped 2nd Order System For a critically damped system, = 1, the roots are real negative and identical, i.e. ss12 n (15) st The solution form X(t) = A e is no longer valid. For repeated roots, the theory of ODE’s dictates that the family of solutions satisfying the differential equation is n t X() t e C12 tC (16) At time t = 0, applying initial conditions gives Then CXCVX1 0and 2 0 n 0 (17) Note that as t , X(t) 0, i.e. the equilibrium position. A critically damped system does to oscillate, and it is the fastest to damp the response due to initial conditions. MEEN 459/659 Notes 1a © Luis San Andrés (2019) 1-11 Free Response of 2nd order system: Comparison between underdamped, critically damped and overdamped systems initial displacement only Xo = 1, Vo = 0 ωn = 1.0 rad/s = 0.1, 1.0, 2.0 Motion decays exponentially for > 0 Fastest response for = 1; i.e. fastest decay towards equilibrium position X = 0 Free response Xo=1, Vo=0, wn=1 rad/s damping ratio=0.1 1.2 damping ratio=1 1 damping ratio=2 0.8 0.6 0.4 0.2 X(t) 0 -0.2 0 5 10 15 20 25 30 35 40 -0.4 -0.6 -0.8 -1 time (sec) MEEN 459/659 Notes 1a © Luis San Andrés (2019) 1-12 Free Response of 2nd order System: Comparison between underdamped, critically damped and overdamped systems Initial velocity only Xo = 0, Vo = 1.0, ωn = 1.0 rad/s = 0.1, 1.0, 2.0 Motion decays exponentially for > 0 Fastest response for = 1.0, i.e. fastest decay towards equilibrium position X=0. note initial overshoot Free response Xo=0, Vo=1, wn=1 rad/s damping ratio=0.1 1 damping ratio=1 damping ratio=2 0.8 0.6 0.4 0.2 X(t) 0 0 5 10 15 20 25 30 35 40 -0.2 -0.4 -0.6 -0.8 time (sec) MEEN 459/659 Notes 1a © Luis San Andrés (2019) 1-13 E X A M P L E: A 45 gram steel ball (m) is dropped from rest through a vertical height of h=2 m.