Vibrations Fundamentals and Practice ch02 Maintaining the outstanding features and practical approach that led the bestselling first edition to become a standard textbook in engineering classrooms worldwide, Clarence de Silva''s Vibration: Fundamentals and Practice, Second Edition remains a solid instructional tool for modeling, analyzing, simulating, measuring, monitoring, testing, controlling, and designing for vibration in engineering systems. It condenses the author''s distinguished and extensive experience into an easy-to-use, highly practical text that prepares students for real problems in a variety of engineering fields.
de Silva, Clarence W “Time Response” Vibration: Fundamentals and Practice Clarence W de Silva Boca Raton: CRC Press LLC, 2000 Time Response Vibrations are oscillatory responses of dynamic systems Natural vibrations occur in these systems due to the presence of two modes of energy storage Specifically, when the stored energy is converted from one form to the other, repeatedly back and forth, the resulting time response of the system is oscillatory in nature In a mechanical system, natural vibrations can occur because kinetic energy, which is manifested as velocities of mass (inertia) elements, can be converted into potential energy (which has two basic types: elastic potential energy due to the deformation in spring-like elements, and gravitational potential energy due to the elevation of mass elements against the Earth’s gravitational pull) and back to kinetic energy, repetitively, during motion Similarly, natural oscillations of electrical signals occur in circuits due to the presence of electrostatic energy (of the electric charge storage in capacitor-like elements) and electromagnetic energy (due to the magnetic fields in inductor-like elements) Fluid systems can also exhibit natural oscillatory responses as they possess two forms of energy But purely thermal systems not produce natural oscillations because they, as far as anyone knows, have only one type of energy These ideas are summarized in Appendix A Note, however, that an oscillatory forcing function is able to make a dynamic system respond with an oscillatory motion (usually at the same frequency as the forcing excitation) even in the absence of two forms of energy storage Such motions are forced responses rather than natural or free responses This book concerns vibrations in mechanical systems Nevertheless, clear analogies exist with electrical and fluid systems as well as mixed systems such as electromechanical systems Mechanical vibrations can occur as both free (natural) responses and forced responses in numerous practical situations Some of these vibrations are desirable and useful, and others are undesirable and should be avoided or suppressed The sound that is generated after a string of a guitar is plucked is a free vibration, while the sound of a violin is a mixture of both free and forced vibrations These sounds are generally pleasant and desirable The response of an automobile after it hits a road bump is an undesirable free vibration The vibrations felt while operating a concrete drill are desirable for the drilling process itself, but are undesirable forced vibrations for the human who operates the drill In the design and development of a mechanical system, regardless of whether it is intended for generating desirable vibrations or for operating without vibrations, an analytical model of the system can serve a very useful function The model will represent the dynamic system, and can be analyzed and modified more quickly and cost effectively than one could build and test a physical prototype Similarly, in the control or suppression of vibrations, it is possible to design, develop, and evaluate vibration isolators and control schemes through analytical means before they are physically implemented It follows that analytical models (see Appendix A) are useful in the analysis, control, and evaluation of vibrations in dynamic systems, and also in the design and development of dynamic systems for desired performance in vibration environments An analytical model of a mechanical system is a set of equations, and can be developed either by the Newtonian approach where Newton’s second law is explicitly applied to each inertia element, or by the Lagrangian or Hamiltonian approach, which is based on the concepts of energy (kinetic and potential energies) These approaches are summarized in Appendix B A time-domain analytical model is a set of differential equations, with respect to the independent variable time (t) A frequency-domain model is a set of input-output transfer functions with respect to the independent variable frequency (ω) The time response will describe how the system moves (responds) as a function of time Both free and forced responses are useful The frequency response will describe the way the system moves when excited by a harmonic (sinusoidal) forcing input, and is a function of the frequency of excitation This chapter introduces some basic concepts of vibration analysis ©2000 CRC Press FIGURE 2.1 A mechanical dynamic system using time-domain methods The frequency-domain analysis will be studied in subsequent chapters (Chapters and 4, in particular) 2.1 UNDAMPED OSCILLATOR Consider the mechanical system that is schematically shown in Figure 2.1 The inputs (or excitation) applied to the system are represented by the force f(t) The outputs (or response) of the system are represented by the displacement y The system boundary demarcates the region of interest in this analysis This boundary could be an imaginary one What is outside the system boundary is the environment in which the system operates An analytical model of the system can be given by one or more equations relating the outputs to the inputs If the rates of changes of the response (outputs) are not negligible, the system is a dynamic system In this case, the analytical model in the time domain becomes one or more differential equations rather than algebraic equations System parameters (e.g., mass, stiffness, damping constant) are represented in the model, and their values should be known in order to determine the response of the system to a particular excitation State variables are a minimum set of variables that completely represent the dynamic state of a system at any given time t These variables are not unique (more than one choice of a valid set of state variables is possible) The concepts of state variables and state models are introduced in Appendix A and also in this chapter For a simple oscillator (a single-degree-of-freedom mass-spring-damper system as in Figure 2.1), an appropriate set of state variables would be the displacement y and the velocity y˙ An alternative set would be y˙ and the spring force This chapter provides an introduction to the response analysis of mechanical vibrating systems in the time domain In this introductory chapter, single-degree-of-freedom systems that require only one coordinate (or one independent displacement variable) in their model, are considered almost exclusively Higher-degree-of-freedom systems will be analyzed elsewhere in the book (e.g., Chapter 5) Mass (inertia) and spring are the two basic energy storage elements in a mechanical vibrating system A mass can store gravitational potential energy as well when located against a gravitational force These elements are analyzed first In a practical system, mass and stiffness properties can be distributed (continuous) throughout the system But in this present analysis, lumped-parameter models are employed where inertia, flexibility, and damping effects are separately lumped into single parameters, with a single geometric coordinate used to represent the location of each lumped inertia This chapter section first shows that many types of oscillatory systems can be represented by the equation of an undamped simple oscillator In particular, mechanical, electrical, and fluid systems are considered Please refer to Appendix A for some foundation material on this topic The conservation of energy is a straightforward approach for deriving the equations of motion for undamped oscillatory systems (or conservative systems) The equations of motion for mechanical systems can ©2000 CRC Press FIGURE 2.2 A mass element be derived using the free-body diagram approach with the direct application of Newton’s second law An alternative and rather convenient approach is the use of Lagrange equations, as described in Appendix B The natural (free) response of an undamped simple oscillator is a simple harmonic motion This is a periodic, sinusoidal motion This simple time response is also discussed 2.1.1 ENERGY STORAGE ELEMENTS Mass (inertia) and spring are the two basic energy storage elements in mechanical systems The concept of state variables can be introduced as well through these elements (see Appendix A for details), and will be introduced along with associated energy and state variables Inertia (m) Consider an inertia element of lumped mass m, excited by force f, as shown in Figure 2.2 The resulting velocity is v Newton’s second law gives m dv = f dt (2.1) Kinetic energy stored in the mass element is equal to the work done by the force f on the mass Hence, ∫ fdx = ∫ f dt dt = ∫ fvdt = ∫ Energy E= Kinetic energy KE = dx ∫ mdv vdt = m vdv dt or mv (2.2) Note: v is an appropriate state variable for a mass element because it can completely represent the energy of the element – Integrate equation (2.1) from a time instant immediately before t = (i.e., t = ) ( ) v( t ) = v − + m t ∫ fdt (2.3) − Hence, with t = 0+, for a time instant immediately after t = 0, one obtains ( ) = v( ) v0 + − + m 0+ ∫ fdt ©2000 CRC Press − (2.4) FfIGURE 2.3 A spring element Since the integral of a finite quantity over an almost zero time interval is zero, these results imply that a finite force will not cause an instantaneous change in velocity in an inertia element In particular, for a mass element subjected to finite force, since the integral on the RHS of equation (2.4) is zero, one obtains ( ) ( ) v 0+ = v 0− (2.5) Spring (k) Consider a massless spring element of lumped stiffness k, as shown in Figure 2.3 One end of the spring is fixed and the other end is free A force f is applied at the free end, which results in a displacement (extension) x in the spring Hooke’s law gives f = kx df = kv dt or (2.6) Elastic potential energy stored in the spring is equal to the work done by the force on the spring Hence, Energy : E= = ∫ fdx = ∫ kxdx = kx ∫ f dt dt = ∫ fvdt = ∫ f k dt dt = k ∫ fdf = 2k f df dx 1 or Elastic potential energy PE = f2 kx = 2 k (2.7) Note: f is an appropriate state variable for a spring, and so is x, because they can completely represent the energy in the spring Integrate equation (2.6) ( ) f (t ) = f − + k t ∫ vdt (2.8) − Set t = 0+ Then, ( ) = f (0 ) f ©2000 CRC Press + − + k 0+ ∫ vdt 0− (2.9) FIGURE 2.4 A mass element subjected to gravity From these results, it follows that at finite velocities, there cannot be an instantaneous change in the force of a spring In particular, from equation (2.9) one sees that at finite velocities of a spring ( ) ( ) (2.10) ( ) ( ) (2.11) f 0+ = f 0− Also, it follows that x 0+ = x 0− Gravitational Potential Energy The work done in raising an object against the gravitational pull is stored as gravitational potential energy of the object Consider a lumped mass m, as shown in Figure 2.4, that is raised to a height y from some reference level The work done gives Energy : E = ∫ fdy = ∫ mgdy Hence, Gravitational potential energy : PE = mgy 2.1.2 CONSERVATION OF (2.12) ENERGY There is no energy dissipation in undamped systems, which contain energy storage elements only In other words, energy is conserved in these systems, which are known as conservative systems For mechanical systems, conservation of energy gives KE + PE = const (2.13) These systems tend to be oscillatory in their natural motion, as noted before Also, as discussed in Appendix A, analogies exist with other types of systems (e.g., fluid and electrical systems) Consider the six systems sketched in Figure 2.5 System (Translatory) Figure 2.5 (a) shows a translatory mechanical system (an undamped oscillator) that has just one degree of freedom x This can represent a simplified model of a rail car that is impacting against a snubber The conservation of energy (equation (2.13)) gives ©2000 CRC Press 1 mx˙ + kx = const 2 (2.14) Here, m is the mass and k is the spring stiffness Differentiate equation (2.14) with respect to time t to obtain ˙˙˙ + kxx˙ = mxx Since x˙ ≠ at all t, in general, one can cancel it out Hence, by the method of conservation of energy, one obtains the equation of motion ˙˙ x+ k x=0 m (2.15) System (Rotatory) Figure 2.5(b) shows a rotational system with the single degree of freedom θ It may represent a simplified model of a motor drive system As before, the conservation energy gives ˙2 Jθ + Kθ = const 2 (2.16) In this equation, J is the moment of inertia of the rotational element and K is the torsional stiffness of the shaft Then, by differentiating equation (2.16) with respect to t and canceling θ˙ , one obtains the equation of motion K ˙˙ θ+ θ=0 J (2.17) System (Flexural) Figure 2.5(c) is a lateral bending (flexural) system, which is a simplified model of a building structure Again, a single degree of freedom x is assumed Conservation of energy gives 1 mx˙ + kx = const 2 (2.18) Here, m is the lumped mass at the free end of the support and k is the lateral bending stiffness of the support structure Then, as before, the equation of motion becomes ˙˙ x+ k x=0 m (2.19) System (Swinging) Figure 2.5(d) shows a simple pendulum It may represent a swinging-type building demolisher or a skilift and has a single-degree-of-freedom θ Thus, KE = Gravitational ©2000 CRC Press ( ) m lθ˙ 2 PE = Eref − mgl cos θ FIGURE 2.5 Six examples of single D.O.F oscillatory systems: (a) translatory, (b) rotatory, (c) flexural, (d) swinging, (e) liquid slosh, and (f) electrical Here, m is the pendulum mass, l is the pendulum length, g is the acceleration due to gravity, and Eref is the PE at the reference point, which is a constant Hence, conservation of energy gives 2˙2 ml θ − mgl cos θ = const (2.20) Differentiate with respect to t after canceling the common ml: ˙˙˙ + g sin θ θ˙ = lθθ Since θ˙ ≠ at all t, the equation of motion becomes g ˙˙ θ + sin θ = l (2.21) This system is nonlinear, in view of the term sin θ For small θ, sin θ is approximately equal to θ Hence, the linearized equation of motion is g ˙˙ θ+ θ=0 l ©2000 CRC Press (2.22) System (Liquid Slosh) Consider a liquid column system shown in Figure 2.5(e) It may represent two liquid tanks linked by a pipeline The system parameters are Area of cross section of each column = A Mass density of liquid = ρ Length of liquid mass = l Then, KE = Gravitational (ρlA) y˙ 2 PE = ρA(h + y) g g (h + y) + ρA(h − y) (h − y) 2 Note that the center of gravity of each column is used in expressing the gravitational PE Hence, conservation of energy gives 1 2 ρlAy˙ + ρAg(h + y) + ρAg(h − y) = const 2 (2.23) Differentiate: ˙˙˙ + g(h + y) y˙ − g(h − y) y˙ = lyy But, one has y˙ ≠ for all t Hence, ˙˙ y + g(h + y) − g(h − y) = or ˙˙ y+ 2g y=0 l (2.24) System (Electrical) Figure 2.5(f) shows an electrical circuit with a single capacitor and a single inductor Again, conservation of energy can be used to derive the equation of motion First, an alternative is given Voltage balance gives v L + vC = where vL and vC are the voltages across the inductor and the capacitor, respectively ©2000 CRC Press (2.25) The constitutive equation for the inductor is L di = vL dt (2.26) dvC =i dt (2.27) The constitutive equation for the capacitor is C Hence, by differentiating equation (2.26), substituting equation (2.25), and using equation (2.27), one obtains L dv d i dv L i = =− C =− dt dt dt C or LC d 2i +i = dt (2.28) Now consider the energy conservation approach for this electrical circuit, which will give the same result Note that power is given by the product vi Capacitor Electrostatic energy : E = ∫ vidt = ∫ vC dv Cv dt = C vdv = dt ∫ (2.29) Here, v denotes vC Also, v= C ∫ idt (2.30) 0+ Since the current i is finite for a practical circuit, then ∫ idt = 0− Hence, in general, the voltage across a capacitor cannot change instantaneously In particular, ( ) ( ) v 0+ = v 0− (2.31) Inductor Electromagnetic energy E = ©2000 CRC Press ∫ vidt = ∫ L di Li idt = L idi = dt ∫ FIGURE P2.17 A vibrating cam-follower mechanism Would the natural frequency of motion be different if the plane of motion is vertical with the carriage (M) moving a Horizontally? b Vertically? 2.17 Cam-follower mechanisms are commonly used to realize timed, periodic motions having some desired characteristics For example, they are used in synchronized opening and closing of valves in internal combustion (IC) engines A schematic representation of such an arrangement is shown in Figure P2.17 The rocker arm is supported on a smooth pivot One end of it carries a spring-loaded valve The other end (drive end) has the follower (roller-type) for which the input motion is determined by the shape profile and the rotatory speed of the cam that is in intimate contact with the follower The following parameters are given: J = moment of inertia of the rocker arm and follower combination about the supporting pivot m = mass of the valve and stem combination (not included in J) l = lever arm length of the valve weight from the pivot point k = stiffness of the valve spring a Determine expressions for the equivalent mass meq and the equivalent stiffness keq of the entire system as located at the valve b What is the undamped natural frequency of rocking motions? What is the significance of this frequency in the proper operation of the cam-follower system? c If the equivalent (linear, viscous) damping constant at the valve location is b, what is the damped natural frequency and the damping ratio of the system? 2.18 Consider a heavy coil of helical spring, as shown in Figure P2.18 The force (f) vs deflection (x) relationship of the spring is given by f = Gd x 8nD3 where d = diameter of the coil wire of the spring D = mean diameter of the spring (coil) n = number of (active) turns in the spring G = shear modulus of the coil wire ©2000 CRC Press Assuming that the mass density of the spring material is ρ, derive an expression for the first undamped natural frequency of oscillation of the spring in the fixed-free configuration of end conditions shown in the figure FIGURE P2.18 A heavy helical spring 2.19 2.20 An excavator boom/stick along with its bucket is modeled as a light rigid rod of length l with a lumped end mass m, as shown in Figure P2.19 The flexibility in the system is modeled as a torsional spring of stiffness k located at the base O Write an equation of motion for the boom for small rotations α from the static equilibrium configuration where the boom is inclined at an angle θ with the vertical What is the corresponding natural frequency of oscillation? Determine the condition of stability Neglect damping effects Consider a uniform elastic post of mass m, length l, and area of cross section A that is vertically mounted on a rigid concrete floor There is a mss M attached to the top end of the post, as shown in Figure P2.20(a) In studying longitudinal vibrations of the post, we wish to obtain an equivalent model as shown in Figure P2.20(b), where me is the equivalent mass of the post as concentrated at the top end and ke is the equivalent stiffness for longitudinal motions Assuming a linear variation of longitudinal deflection and velocity along the post, determine keq and meq What is the natural frequency of undamped longitudinal vibrations? The Young’s modulus of the post material is E FIGURE P2.19 An excavator stick (boom) with the bucket ©2000 CRC Press Hint: FIGURE P2.20 (a) A vertical post with an end mass.; (b) an equivalent model for longitudinal vibrations 2.21 2.22 This problem deals with resolving the direction of a spring force in vibration studies Consider the spring-loaded carriage mechanism sketched in Figure P2.21 The carriage unit of mass m is supported by two springs (with restraining guide plates), one vertical having stiffness k and the other inclined at θ to the horizontal, with stiffness ka Show that the equivalent stiffness keq on the carriage M in the vertical direction is k + ka /sin2θ What is the undamped natural frequency of vibration of the carriage system? A centrifugal water pump is to be located at the free end of an overhung beam, as shown in Figure P2.22 It is required that the operating speed of the pump does not correspond to a natural frequency (strictly, a resonant frequency) of the structural system Given that: M m l I mass of the pump mass of the beam length of the beam 2nd moment of area of the beam cross section about the horizontal neutral axis of bending E = Young’s modulus of the beam material ©2000 CRC Press = = = = FIGURE P2.21 A carriage with bi-directional spring restraints Hint: FIGURE P2.22 A water pump mounted on an overhung beam 2.23 determine an expression for the equivalent linear stiffness keq and mass meq of the beam, as located at the free end What is the corresponding natural frequency of vibration of the pump-beam system? A gear transmission has a meshed pair of gear wheels of moments of inertia Jd and Jl about their axes of rotation, and a step-down gear ratio r The load gear wheel is connected to a purely torsional load of stiffness kl as shown in Figure P2.23 Suppose that the angle of rotation of the drive gear wheel is θ and that of the load gear wheel is α in the opposite direction, so that r = θ/α Determine the equivalent moment of inertia and the equivalent torsional stiffness of the system, both with respect to the drive side and the load side of the transmission For each case, what is the natural frequency of torsional vibration? Justify the results ©2000 CRC Press Hint: FIGURE P2.22 A water pump mounted on an overhung beam FIGURE P2.23 A gear transmission with a torsional load 2.24 The handle of a hoist is modeled as a rigid light rod pivoted at the bottom and restrained by a torsional spring of stiffness k, along with a uniform circular disk of mass m and radius r attached to the top end, as shown in Figure P2.24 The distance from the bottom pivot to the center of the disk is l Initially, the handle is in its vertical configuration, where the spring is in its relaxed position Obtain an equation for angular motion θ of ©2000 CRC Press 2.25 the handle with respect to this configuration What is the natural frequency of small vibrations of the handle about its upright position? Under what conditions would such vibrations be not possible? Neglect energy dissipation A simplified model of an elevator is shown in Figure P2.25 Note that: J r k m = = = = moment of inertia of the cable pulley radius of the pulley stiffness of the cable mass of the car and occupants a Which system parameters are variable? Explain b Suppose that the damping torque Td (ω) at the bearings of the pulley is a nonlinear function of the angular speed ω of the pulley Taking the state vector x as x = [ω f v] T in which f = tension force in the cable v = velocity of the car (taken positive upwards) the input vector u as [ ] u = Tm in which Tm = torque applied by the motor to the pulley (positive in the direction indicated in figure) and, the output vector y as y = [v] obtain a complete, nonlinear state-space model for the system c With Tm as the input and v as the output, convert the state-space model into the nonlinear input-output differential equation model What is the order of the system? d Give an equation for which the solution provides the steady-state operating speed v of the elevator car e Linearize the nonlinear input/output differential-equation model obtained in part (c), for small changes Tˆ m of the input and vˆ of the output, about an operating point Note: Tm = steady-state operating-point torque of the motor (assumed known) Hint: Denote d T (ω ) as b(ω ) dω d f Linearize the state-space model obtained in part (b) and give the model matrices A, B, C, and D in the usual notation Obtain the linear input/output differential equation from this state-space model and verify that it is identical to what was obtained in part (e) ©2000 CRC Press FIGURE P2.24 A handle of a mechanical hoist FIGURE P2.25 A simplified model of an elevator 2.26 An automated wood cutting system contains a cutting unit that consists of a DC motor and a cutting blade, linked by a flexible shaft and coupling The purpose of the flexible shaft is to locate the blade unit at any desirable configuration, away from the motor itself A simplified, lumped-parameter, dynamic model of the cutting unit is shown in Figure P2.26 ©2000 CRC Press FIGURE P2.26 A wood cutting machine The following parameters and variables are shown in the figure: Jm = bm = k = Jc = bc = Tm = ωm = Tk = ωc = TL = axial moment of inertia of the motor rotor equivalent viscous damping constant of the motor bearings torsional stiffness of the flexible shaft axial moment of inertia of the cutter blade equivalent viscous damping constant of the cutter bearings magnetic torque of the motor motor speed torque transmitted through the flexible shaft cutter speed load torque on the cutter from the workpiece (wood) In comparison with the flexible shaft, the coupling unit is assumed rigid, and is also assumed light The cutting load is given by TL = c ω c ω c The parameter c, which depends on factors such as the depth of cut and the material properties of the workpiece, is assumed to be constant in the present analysis [ ] T a Using Tm as the input, TL as the output, and x = ωˆ m, Tˆk , ωˆ c as the state vector, develop a complete (nonlinear) state model for the system shown in Figure P2.26 What is the order of the system? b Using the state model derived in part (a), obtain a single input-output differential equation for the system, with Tm as the input and ωc as the output c Consider the steady operating conditions, where Tm = Tm, ωm = ωm, Tk = Tk, ωc = ωc, and TL = TL are all constants Express the operating point values ωm, Tk, ωc , and TL in terms of Tm and the model parameters You must consider both cases: Tm > and Tm < d Now consider an incremental change Tˆ in the motor torque and the corresponding m changes ωˆ m, Tˆ k, ωˆ c, and Tˆ L in the system variables Determine a linear state model (A, B, C, D) for the incremental dynamics of the system in this case, using [ x = ωˆ m , Tˆk , ωˆ c ©2000 CRC Press ] T [ ] [ ] as the state vector, u = Tˆm as the input and y = TˆL as the output e In the nonlinear model (see part (a)), if the twist angle of the flexible shaft (i.e., θm – θc) is used as the output what would be a suitable state model? What is the system order then? f In the nonlinear model, if the angular position θc of the cutter blade is used as the output variable, explain how the state model obtained in part (a) should be modified What is the system order in this case? Hint for Part (b): ( ) d ω ω = ω c ω« c dt c c ( ) d2 ««c + 2ω« c2 sgn(ω c ) ωc ωc = ωc ω dt 2.27 It is required to study the dynamic behavior of an automobile during the very brief period of a sudden start from rest Specifically, the vehicle acceleration a in the direction of primary motion, as shown in Figure P2.27(a), is of interest and should be considered as the system output The equivalent force f(t) of the engine, applied in the direction of primary motion, is considered as the system input A simple dynamic model that can be used for the study is shown in Figure P2.27(b) Note that k is the equivalent stiffness, primarily due to tire flexibility, and b is the equivalent viscous damping constant, primarily due to energy dissipations at the tires and other moving parts of the vehicle, taken in the direction of a Also, m is the mass of the vehicle a Discuss advantages and limitations of the proposed model for the specific purpose b Using force fk of the spring (stiffness k) and velocity v of the vehicle as the state variables, engine force f(t) as the input, and the vehicle acceleration a as the output, develop a complete state-space model for the system (Note: You must derive the matrices A, B, C, and D for the model.) c Obtain the input/output differential equation of the system d Discuss the characteristics of this model by observing the nature of matrix D, and the input and output orders of the input-output differential equation 2.28 a Briefly explain why a purely thermal system typically does not have a free oscillatory response, whereas a fluid system can b Figure P2.28 shows a pressure-regulated system that can provide a high-speed jet of liquid The system consists of a pump, a spring-loaded accumulator, and a fairly long section of piping that ends with a nozzle The pump is considered as a flow source of value Qs The following parameters are important: A = k = L = Ap = Ao = Cd = ρ = area of cross section (uniform) of the accumulator cylinder spring stiffness of the accumulator piston wall length of the section of piping from the accumulator to the nozzle area of cross section (uniform, circular) of the piping exit area of the nozzle discharge coefficient of the nozzle mass density of the liquid Assume that the liquid is incompressible The following variables are important: P1r = P1 – Pr = pressure at the inlet of the accumulator with respect to the ambient reference Pr ©2000 CRC Press FIGURE P2.27 (a) A vehicle suddenly accelerating from rest; (b) a simplified model FIGURE P2.28 Pressure regulated liquid jet system ©2000 CRC Press Q = volume flow rate through the nozzle h = height of the liquid column in the accumulator Note that the piston (wall) of the accumulator can move against the spring, thereby varying h i Considering the effects of the movement of the spring-loaded wall and the gravity head of the liquid, obtain an expression for the equivalent fluid capacitance Ca of the accumulator in terms of k, A, ρ, and g Are the two capacitances that contribute to Ca (i.e., wall stretching and gravity) connected in parallel or in series? Note: Neglect the effect of bulk modulus of the liquid ii Considering the capacitance Ca, the inertance I of the fluid volume in the piping (length L and cross-sectional area Ap), and the resistance of the nozzle only, develop a nonlinear state-space model for the system The state vector x = [ P1r , Q] and the T [ ] input u = Qs For flow in the (circular) pipe with a parabolic velocity profile, the inertance is given by I= 2ρL Ap and the discharge through the nozzle is given by Q = A0 Cd 2.29 P2 r ρ in which P2r is the pressure inside the nozzle with respect to the outside reference pressure Pr Give reasons for the common experience that in the flushing tank of a household toilet, some effort is needed to move the handle for the flushing action but virtually no effort is needed to release the handle at the end of the flush A simplified model for the valve movement mechanism of a household flushing tank is shown in Figure P2.29 The overflow tube on which the handle lever is hinged is assumed rigid Also, the handle rocker is assumed light, and the rocker hinge is assumed frictionless The following parameters are indicated in the figure: lv = the lever arm ratio of the handle rocker lh m = equivalent lumped mass of the valve flapper and the lift rod k = stiffness of the spring action on the valve flapper r= The damping force fNLD on the valve is assumed quadratic and is given by f NLD = a v NLD v NLD where the positive parameter a = au for upward motion of the flapper (vNLD ≥ 0) = ad for downward motion of the flapper (vNLD ≤ 0) ©2000 CRC Press FIGURE P2.29 Simplified model of a toilet flushing mechanism with au >> ad The force applied at the handle is f(t), as shown We are interested in studying the dynamic response of the flapper valve Specifically, the valve displacement x and the valve speed v are considered outputs, as shown in Figure P2.29 Note that x is measured from the static equilibrium point of the spring where the weight mg is balanced by the spring force a Using valve speed (v) and the spring force (fk) as the state variables, develop a (nonlinear) state-space model for the system b Linearize the state-space model about an operating point where the valve speed is v For the linearized model, obtain the model matrices A, B, C, and D in the usual notation Note that the incremental variables xˆ and vˆ are the outputs in the linear model, and the incremental variable fˆ (t) is the input c From the linearized state-space model, derive the input-output model relating fˆ (t) and xˆ d Give expressions for the undamped natural frequency and the damping ratio of the linear model in terms of the parameters a, v, m, and k Show that the damping ratio increases with the operating speed 2.30 The electrical circuit shown in Figure P2.30 has two resistors R1 and R2, an inductor L, a capacitor C, and a voltage source u(t) The voltage across the capacitor is considered the output y of the circuit a What is the order of the system and why? b Show that the input-output equation of the circuit is given by ©2000 CRC Press FIGURE P2.30 An RLC circuit driven by a voltage source a2 d2y dy du + a1 + a0 y = b1 + b0 u dt dt dt Express the coefficients a0, a1, b0, and b1 in terms of the circuit parameters R1, R2, L, and C What is the undamped natural frequency? What is the damping ratio? c Starting with the auxiliary differential equation a2 ˙˙ x + a1 x˙ + a0 x = u T and using x = [ x x˙ ] as the state vector, obtain a complete state-space model for the system in Figure P2.30 Note that this is the “superposition method” of developing a state model d Clearly explain why, for the system in Figure P2.30, neither the current ic through the capacitor, nor the time derivative of the output (i.e., y˙ ) can be chosen as a state variable 2.31 Consider two water tanks joined by a horizontal pipe with an on/off valve With the valve closed, the water levels in the two tanks were initially maintained unequal When the valve is suddenly opened, some oscillations were observed in the water levels of the tanks Suppose that the system is modeled as two gravity-type capacitors linked by a fluid resistor Would this model exhibit oscillations in the water levels when subjected to an initial-condition excitation? Clearly explain your answer A centrifugal pump is used to pump water from a well into an overhead tank This fluid system is schematically shown in Figure P2.31(a) The pump is considered as a pressure source Ps(t) and the water level h in the overhead tank is the system output The ambient pressure is denoted by Pa The following parameters are given: Lv, dv = length and the internal diameter of the vertical segment of the pipe ©2000 CRC Press FIGURE P2.31 (a) A system for pumping water from a well into an overhead tank; (b) lumped parameter model of the fluid system Lh, dh = length and the internal diameter of the horizontal segment of the pipe At = area of cross section of the overhead tank (uniform) ρ = mass density of water µ = dynamic viscosity of water g = acceleration due to gravity Suppose that this fluid system may be approximated by the lumped parameter model shown in Figure P2.31(b) a Give expressions for the equivalent linear fluid resistance of the overall pipe (i.e., combined vertical and horizontal segments) Req, the equivalent fluid inertance within ©2000 CRC Press the overall pipe Ieq, and the gravitational fluid capacitance of the overhead tank Cgrv in terms of the system parameters defined above [ ] b Treating x = P3a Q where T as the state vector, P3a = pressure head of the overhead tank Q = volume flow rate through the pipe c d 2.32 a b develop a complete state-space model for the system Specifically, obtain the matrices A, B, C, and D Obtain the input-output differential equation of the system What is the characteristic equation of this system? Using the following numerical values for the system parameters: Lv = 10.0 m, Lh = 4.0 m, dv = 0.025 m, dh = 0.02 m ρ = 1000.0 kg m3, = 1.0 ì 103 N s m2, and tank diameter = 0.5 m, compute the undamped natural frequency ωn and the damping ratio ζ of the system Will this system provide an oscillatory natural response? If so, what is the corresponding frequency? If not, explain why Define the following terms with reference to the response of a dynamic system: i Homogeneous solution ii Particular solution (or particular integral) iii Zero-input (or free) response iv Zero-state (or forced) response v Steady-state response Consider the first-order system τ 2.33 dy + y = u(t ) dt in which u is the input, y is the output, and τ is a system constant i Suppose that the system is initially at rest with u = and y = 0, and suddenly a unit step input is applied Obtain an expression for the ensuing response of the system Into which of the above five categories does this response fall? What is the corresponding steady-state response? ii If the step input in part (i) above is of magnitude A, what is the corresponding response? iii If the input in part (i) above was an impulse of magnitude P, what would be the response? Consider a mechanical system that is modeled by a simple mass-spring-damper unit with a forcing excitation Its equation of motion is given by the normalized form ˙˙ y + 2ζω n y˙ + ω 2n y = ω 2n u(t ) in which u(t) is the forcing excitation and y is the resulting displacement response The system is assumed to be underdamped (ζ < 1) Suppose that a unit step excitation is applied to the system At a subsequent time (which can be assumed t = 0, without loss of generality) when the displacement is y0 and the velocity is v0, a unit impulse is applied to the mass of the system Obtain an expression for y that describes the subsequent response of the system, with proper initial conditions ©2000 CRC Press ... both free (natural) responses and forced responses in numerous practical situations Some of these vibrations are desirable and useful, and others are undesirable and should be avoided or suppressed... of both free and forced vibrations These sounds are generally pleasant and desirable The response of an automobile after it hits a road bump is an undesirable free vibration The vibrations felt... operating without vibrations, an analytical model of the system can serve a very useful function The model will represent the dynamic system, and can be analyzed and modified more quickly and cost effectively