Mathematical Modeling and response analysis of mechanical systems are the subjects of this chapter.

Chapter 3 Mechanical Systems A. Bazoune 3.1 INRODUCION Mathematical Modeling and response analysis of mechanical systems are the subjects of this chapter. 3. MECHANICAL ELEMENS Any mechanical system consists of mechanical elements. here are three types of basic elements in mechanical systems: Inertia elements Spring elements Dampers elements Inertia Elements. Mass and moment of inertia. Inertia may be defined as the change in force (torque) required to mae a unit change in acceleration (angular acceleration). hat is, change in force N inertia (mass) = or g change in acceleration m/s change in torque N-m inertia (moment of inertia) = or g change in ang. accel. rad/s Spring Elements. A linear spring is a mechanical element that can be deformed by eternal force or torque such that the deformation is directly proportional to the force or torque applied to the element. ranslational Springs or translational motion, (ig 3-1(a)), the force that arises in the spring is proportional to and is given by = (3-1) where is the elongation of the spring and is a proportionality constant called the spring constant and has units of [force/displacement]=[n/m] in SI units. 1/31

At point P, the spring force acts opposite to the direction of the force applied at point P. X X+ P f f Q X+ 1 X+ 1 P f igure 3-1 (a) One end of the spring is deflected; (b) both ends of the spring are deflected. ( X is the natural length of the spring) igure 3-1(b) shows the case where both ends P and Q of the spring are deflected due to the forces f applied at each end. he net elongation of the spring is 1. he force acting in the spring is then = (3-) 1 Notice that the displacement X + 1 and of the ends of the spring are measured relative to the same reference frame. Practical Eamples. Pictures of various types of real-world springs are found below. /31

orsional Springs Consider the torsional spring shown in igure 3- (a), where one end is fied and a torque τ is applied to the other end. he angular displacement of the free end is θ. he torque in the torsional spring is = θ (3-3) where θ is the angular displacement and is the spring constant for torsional spring and has units of [orque/angular displacement]=[n-m/rad] in SI units. θ τ τ θ θ 1 τ igure 3- (a) A torque τ is applied at one end of torsional spring, and the other end is fied; (b) a torque τ is applied at one end, and a torque τ, in the opposite direction, is applied at the other end. At the free end, this torque acts in the torsional spring in the direction opposite to that of the applied torque τ. or the torsional spring shown in igure 3-(b), torques equal in magnitude but opposite in direction, are applied to the ends of the spring. In this case, the torque acting in the torsional spring is = θ θ (3-4) 1 At each end, the spring acts in the direction opposite to that of the applied torque at that end. or linear springs, the spring constant may be defined as follows spring constant for translational spring spring constant for torsional spring change in force N = change in displacement of spring m change in torque = change in angular displacement of spring N-m rad Spring constants indicate stiffness; a large value of or corresponds to a hard spring, a small value of or to a soft spring. he reciprocal of the spring 3/31

=. constant is called compliance or mechanical capacitance C. hus C 1 Compliance or mechanical capacitance indicates the softness of the spring. Practical Eamples. Pictures of various types of real-world springs are found below. Practical spring versus ideal spring. igure 3-3 shows the force displacement characteristic curves for linear and nonlinear springs. All practical springs have inertia and damping. An ideal spring has neither mass nor damping (internal friction) and will obey the linear force displacement law. o igure 3-3 orce-displacement characteristic curves for linear and nonlinear springs. Damper Elements. A damper is a mechanical element that dissipates energy in the form of heat instead of storing it. igure 3-4(a) shows a schematic diagram of a translational damper, or a dashpot that consists of a piston and an-oilfilled cylinder. Any relative motion between the piston rod and the cylinder is resisted by oil because oil must flow around the piston (or through orifices provided in the piston) from one side to the other. 4/31

1 f ẋ ẋ 1 f τ θ θ 1 τ igure 3-4 (a) ranslational damper; (b) torsional (or rotational) damper. ranslational Damper In ig 3-4(a), the forces applied at the ends of translation damper are on the same line and are of equal magnitude, but opposite in direction. he velocities of the ends of the damper are ẋ 1 and ẋ. Velocities ẋ 1 and ẋ are taen relative to the same frame of reference. In the damper, the damping force that arises in it is proportional to the velocity differences 1 where 1 of the ends, or = b = b (3-5) 1 = and the proportionality constant b relating the damping force to the velocity difference ẋ is called the viscous friction coefficient or viscous friction constant. he dimension of b is [force/velocity] = [N/m-s] in SI units. orsional Damper or the torsional damper shown in igure 3-4(b), the torques τ applied to the ends of the damper are of equal magnitude, but opposite in direction. he angular velocities of the ends of the damper are θ 1 and θ and they are taen relative to the same frame of reference. he damping torque that arises in the damper is proportional to the angular velocity differences 1 ( 1 ) θ θ of the ends, or = b θ θ = b θ (3-6) where, analogous to the translation case, 1 θ = θ θ and the proportionality constant b relating the damping torque to the angular velocity difference θ is called the viscous friction coefficient or viscous friction constant. he dimension of b is [torque/angular velocity] = [N-m/rad] in SI units. 5/31

A damper is an element that provides resistance in mechanical motion, and, as such, its effect on the dynamic behavior of a mechanical system is similar to that of an electrical resistor on the dynamic behavior of an electrical system. Consequently, a damper is often referred to as a mechanical resistance element and the viscous friction coefficient as the mechanical resistance. Practical Eamples. Pictures of various eamples of real-world dampers are found below. Practical damper versus ideal damper All practical dampers produce inertia and spring effects. An ideal damper is massless and springless, dissipates all energy, and obeys the linear force-velocity law (or linear torque-angular velocity law). Nonlinear friction. riction that obeys a linear law is called linear friction, whereas friction that does not is described as nonlinear. Eamples of nonlinear friction include static friction, sliding friction, and square-law friction. Square law-friction occurs when a solid body moves in a fluid medium. igure 3-5 shows a characteristic curve for square-law friction. igure 3-5 Characteristic curve for square-law friction. 6/31

3.3 MAHEMAI AHEMAICAL AL MODELING O SIMPLE MECHANICAL SYSEMS A mathematical model of any mechanical system can be developed by applying Newton s laws to the system. Rigid body. When any real body is accelerated, internal elastic deflections are always present. If these internal deflections are negligibly small relative to the gross motion of the entire body, the body is called rigid body. hus, a rigid body does not deform. Newton s laws. Newton s first law: (Conservation of Momentum) he total momentum of a mechanical system is constant in the absence of eternal forces. Momentum is the product of mass m and velocity v, or m v, for translational or linear motion. or rotational motion, momentum is the product of moment of inertia J and angular velocity ω, or Jω, and is called angular momentum. Newton s second law: ranslational motion: If a force is acting on rigid body through the center of mass in a given direction, the acceleration of the rigid body in the same direction is directly proportional to the force acting on it and is inversely proportional to the mass of the body. hat is, or force acceleration = mass ( mass)( acceleration ) = force Suppose that forces are acting on a body of mass m. If is the sum of all forces acting on a mass m through the center of mass in a given direction, then = m a (3-7) where a is the resulting absolute acceleration in that direction. he line of action of the force acting on a body must pass through the center of mass of the body. Otherwise, rotational motion will also be involved. Rotational motion. or a rigid body in pure rotation about a fied ais, Newton s second law states that ( moment of inertia)( angular acceleration ) = torque 7/31

or = J α (3-8) where is the sum of all torques acting about a given ais, J is the moment of inertia of a body about that ais, and α is the angular acceleration of the body. Newton s third law. It is concerned with action and reaction and states that every action is always opposed by an equal reaction. orque or moment of force. orque, or moment of force, is defined as any cause that tends to produce a change in the rotational motion of a body in which it acts. orque is the product of a force and the perpendicular distance from a point of rotation to the line of action of the force. [ ] [ ] orque = force distance = N-m in SI units Moments of inertia. he moment of inertia J of a rigid body about an ais is defined by J = r dm (3-9) Where dm is an element of mass, r is the distance from ais to dm, and integration is performed over the body. In considering moments of inertia, we assume that the rotating body is perfectly rigid. Physically, the moment of inertia of a body is a measure of the resistance of the body to angular acceleration. igure 3-6 Moment of inertia Parallel ais theorem. Sometimes it is necessary to calculate the moment of inertia of a homogeneous rigid body about an ais other than its geometrical ais. 8/31

C X C R X igure 3-7 Homogeneous cylinder rolling on a flat surface As an eample to that, consider the system shown in igure 3-7, where a cylinder of mass m and a radius R rolls on a flat surface. he moment of inertia of the cylinder is about ais CC is J c 1 = m R he moment of inertia J of the cylinder about ais is 1 3 J = J c + m R = m R + m R = m R orced response and natural response. he behavior determined by a forcing function is called a forced response, and that due initial conditions is called natural response. he period between initiation of a response and the ending is referred to as the transient period. After the response has become negligibly small, conditions are said to have reached a steady state. igure 3-8 ransient and steady state response Parallel l and Series Springs Elements. In many applications, multiple spring elements are used, and in such cases we must obtain the equivalent spring constant of the combined elements. 9/31

Parallel Springs. or the springs in parallel, igure 3-9, the equivalent spring constant eq is obtained from the relation 1 e q igure 3-9 Parallel spring elements where = + = + = 1 1 eq eq = 1 + for parallel springs (3-9) his formula can be etended to n springs connected side-by-side as follows: n eq = i ( for parallel springs) (3-10) i= 1 Series Springs. or the springs in series, igure 3-10, the force in each spring is the same. hus y 1 e q igure 3-10 Series spring elements 1 = y, = y Eliminating from these two equations yields or = 1 1 + = = + = 1 1 1 10/31

or where + 1 1 = = = 1 1 1 1 + + 1 1 eq 1 1 1 = + for series springs eq 1 (3-11) which can be etended to the case of n springs connected end-to-end as follows n 1 1 = for series springs eq i= 1 i (3-1) ree Vibration without damping. Consider the mass spring system shown in igure 3-11. he equation of motion can be given by or where m + = 0 + = + = m 0 ωn 0 ω = n is the natural frequency of the system and is epressed in rad/s. aing L of both sides of the above equation where ( 0) = and ( 0) = gives rearrange to get m [ ] 0 0 + ω = 0 n s X s s X s + ωn L ( t) s + X ( s) =, Remember poles are s = ± jω s n m igure 3-11 Mass Spring System comple conjugates 11/31

and the response ωn s n s + n s + n X ( s) = + ω ω ω t is given by It is clear that the response t nt nt ω = sin( ω ) + cos( ω ) n t consists of a sine and cosine terms and depends on the values of the initial conditions and. Periodic motion such that described by the above equation is called simple harmonic motion. ( t) slope = Im s plane ω n t Re ω n π Period = = ω n igure 3-1 ree response of a simple harmonic motion and pole location on the s-plane 0 = = 0, if cos( ω ) = t t he period is the time required for a periodic motion to repeat itself. In the present case, n Period π π = = seconds ωn m he frequency f of a periodic motion is the number of cycles per second (cps), and the standard unit of frequency is the Hertz (Hz); that is 1 Hz is 1 cps. In the present case, 1/31

requency f 1 = = m Hz π he undamped natural frequency ω n is the frequency in the free vibration of a system having no damping. If the natural frequency is measured in Hz or in cps, it is denoted by f n. If it is measured in rad/sec, it is denoted by ω n. In the present system, ωn = π fn = rad/sec m Rotational System. Rotor mounted in bearings is shown in figure 3-13 below. he moment of inertia of the rotor about the ais of rotation is J. riction in the bearings is viscous friction and that no eternal torque is applied to the rotor. J ω b J ω bω igure 3-13 Rotor mounted in bearings and its BD. form Apply Newton s second law for a system in rotation or M = J θ = J ω J ω + bω = 0 ω + b / J ω = 0 1 ω + ω = 0 ( J / b) Define the time constant τ = ( J / b), the previous equation can be written in the 1 ω + ω = 0, ω ( 0) = ω τ which represents the equation of motion as well as the mathematical model of the system shown. It represents a first order system. o find the response ω ( t ) L of both sides of the previous equation., tae 13/31

1 sω( s) ω ( 0) + Ω ( s) = 0 τ L[ ω] L[ ω ] 1 s + Ω ( s) = ω τ + 1 τ ( s) ω Ω = s + ( 1 τ ) s is nown as the characteristic polynomial and the where the denominator equation s + ( 1 τ ) = 0 is called the characteristic equation. aing inverse L of the above equation will give the epression of ω ( t ) b / J t 1/τ t αt ω t = ω e = ω e = ω e It is clear that the angular velocity decreases eponentially as shown in the figure ( t / τ ) below. Since lim e = 0; then for such decaying system, it is convenient to t depict the response in terms of a time constant. 1.5 ree response of a rotor bearing system α=0 ω o ω(t) α=0. α=0.5 α=0.7 0.5 α=1 α= α=5 α=10 0 0 5 10 15 0 time (t) igure 3-14 Graph of ω α t e for ranges of α. 14/31

A time constant is that value of time that maes the eponent equal to -1. or this system, time constant τ = J / b. When t = τ, the eponent factor is ( t / τ ) ( τ / τ ) 1 e = e = e = 0.368 = 36.8 % his means that when time constant of its final value. We also have ω ω( t) = τ, the time response is reduced to 36.8 % τ = J / b = time constant ω τ = 0.37 ω ω 4τ = 0.0 ω ωo e b J t 0.37ω 0.0ω τ τ 3τ 4τ 5τ t igure 3-15 Curve of angular velocity ω versus time t for the rotor shown in igure 3-13. http://www.sciences.univ-nantes.fr/physique/perso/gtulloue/equadiff/equadiff.html Spring-Mass Mass-Damper System. Consider the simple mechanical system shown involving viscous damping. Obtain the mathematical model of the system shown. b b m m + ( t) igure 3-16 Mass -Spring Damper System and the BD. 15/31

i) he BD is shown in the figure 3-16. ii) Apply Newton s second law of motion to a system in translation: or = m b = m m + b + = 0 ree Vibration of a second order system If in SI units m = 0. 1 g, b = 0. 4 N/m-s, and = 4 N/m, the above differential equation becomes 0.1 + 0.4 + 4 = 0 + 4 + 40 = 0 o obtain the free response ( t ), assume ( 0) = and Laplace transform of ( t ) as L ( t) = X ( s) both sides of the given equation 0 = 0. ẋ By writing the, we obtain Laplace transform of s X s s 0 0 + 4 sx s 0 + 40 X s = 0 L t L t L t Substitute in the transformed equation ( 0) obtain or solving for = and ( 0) = 0, [ ] s + 4s + 40 X s = s + 4 X s yields X s Which can be written as ( s + 4 ) ( s + 4) = = s + + 4 4 G s 4s + 40 s s + 0 Characteristic polynomial ( s + 4) X s = = s + 4s + 40 Characteristic polynomial ẋ and rearrange, we where G( s ) is referred to as the transfer function that gives the relationship and the output X ( s ). G( s ) can be shown graphically as: between the input 16/31

Input ( s + 4 ) s + 4 s + 4 0 Characteristic polynomial ransfer function X s Output igure 3-17 ransfer function between input and output. iii) It is clear that the characteristic equation of the system is has comple conjugates roots. s + 4s + 40 = 0 and ( s+ ) s + 4s + 40 = s + 4s + 4 + 36 = s + + 6 = 0 he roots of the above equation are therefore comple conjugate poles given by s = + j6 and s = j 6 1 iv) he epression of X ( s ) can be written now as: s + 4s + 40 ( s + ) 1 6 ( s + ) + 3 ( s + ) + s + 4 s + + s + X ( s) = = = + s + + 6 s + + 6 s + + 6 = + 6 6 t = L X s yields 1 v) Solving for t 1 t ( t) = e cos6t + e sin 6t 3 17/31

( t ) Im t e t 1 t ( t) = e cos6t + e sin 6t 3 s 1 = + j6 j6 s plane t j6 Re Damped period, d π = ω d s 1 = j6 igure 3-18 ree Vibration of the mass-spring-damper system described by + 4 + 40 = 0 0 ẋ 0 = 0,. with initial conditions = and 10 Pole-Zero Map 5 Imaginary Ais 0-5 -10-4 -3.5-3 -.5 - -1.5-1 -0.5 0 Real Ais 1.5 ree vibration of mass-spring-damper system o (t) 0.5 0-0.5 0 0.5 1 1.5.5 3 3.5 4 4.5 5 t (sec) igure 3-18 ree Vibration of the mass-spring-damper system described by + 4 + 40 = 0 0 ẋ 0 = 0,. with initial conditions = and 18/31

Remar: he epression of cos( ω ) + sin( ω ) cos( ωt ) or sin( ωt ), that is a t b t can be written in terms of cos sin a b a t b t a b cos t cos t a + b a + b ( ω ) + ( ω ) = + ( ω ) + ( ω ) Define ψ such that herefore a b 1 b cosψ = and sinψ = ψ = tan a + b a + b a ( ω ) + ( ω ) = + ψ ( ω ) + ψ ( ω ) acos t bsin t a b cos cos t sin sin t Using the identity herfore, or Where ( t ) = ( t) + ( t ) cos ω ψ cos ω cosψ sin ω sinψ ( ω ) + ( ω ) = + ( ω ψ ) acos t bsin t a b cos t ( ω ) + ( ω ) = + ( ω + ϕ ) acos t bsin t a b sin t a b 1 a sinϕ = and sinϕ = ϕ = tan a + b a + b b herefore, 10 t e t 3 t = ( sin 6 + 71.56 ) 3.4 WORK ENERGY, AND POWER Wor. he wor done in a mathematical system is the product of a force and a distance (or a torque and the angular displacement) through which the force is eerted with both force and distance measured in the same direction. d d θ W = d W = d cosθ igure 3-19 Wor done by a force 19/31

he units of wor in SI units are : [ wor] = [ force distance] = [ N-m] = [ Joule] = [ J] he wor done by a spring is given by: = = = =. 1 W = d = 0 l l + igure 3-19 Wor done by a spring. Energy. Energy can be defined as ability to do wor. Energy can be found in many different forms and can be converted from one form to another. or instance, an electric motor converts electrical energy into mechanical energy, a battery converts chemical energy into electrical energy, and so forth. According to the law of conservation of energy, energy can be neither created nor destroyed. his means that the increase in the total energy within a system is equal to the net energy input to the system. So if there is no energy input, there is no change in the total energy of the system. Potential Energy. he Energy that a body possesses because of its position is called potential energy. In mechanical systems, only mass and spring can store potential energy. he change in the potential energy stored in a system equals the wor required to change the system s configuration. Potential energy is always measured with reference to some chosen level and is relative to that level. h m m g igure 3-0 Potential energy 0/31

Refer to igure 3-0, the potential energy, U of a mass m is given by: U = mg d = mgh 0 or a translational spring, the potential energy U (sometimes called strain energy which is potential energy that is due to elastic deformations) is: 1 U = d = d = 0 0 If the initial and final values of are 1 and, respectively, then 1 1 Change in potential energy U = d = d = 1 Similarly, for a torsional spring 1 1 θ 1 1 Change in potential energy U = dθ = θ d = θ θ θ θ θ 1 1 1 Kinetic Energy. mechanical systems. Only inertia elements can store inetic energy in 1 mv = Kinetic energy= 1 Jθ (ranslation) (Rotation) he change in inetic energy of the mass is equal to the wor done on it by an applied force as the mass accelerates or decelerates. hus, the change in inetic energy of a mass m moving in a straight line is Change in inetic energy d = W = d = dt dt 1 1 = v dt = mv v dt = m v dv 1 1 1 = 1 mv 1 mv t t t t v t t v 1 he change in inetic energy of a moment of inertia in pure rotation at angular velocity θ is 1/31

1 1 Change in inetic energy = J θ J θ 1 Dissipated Energy. Consider the damper shown in igure 3-1 in which one end is fied and the other end is moved from 1 to. he dissipated energy in the damper is equal to the net wor done on it: W t t d W = d = b d = b dt = b dt dt t t 1 1 1 1 1 b igure 3-0 Damper. Power. Power is the time rate of doing wor. hat is, Power = P = dw dt where dw denotes wor done during time interval dt. In SI units, the wor done is measured in Newton-meters and the time in seconds. he unit of power is : N-m Joule Power = = = Watt = W s s [ ] [ ]. Passsive Elements. Non-energy producing element. hey can only store energy, not generate it such as springs and masses. Active Elements. forces and torques. Energy producing elements such as eternal Energy Method for f Deriving Equations of Motion. Equations of motion are derive from the fact that the total energy of a system remains the same if no energy enters or leaves the system. /31

Conservative Systems. (damping) are called conservative systems. Systems that do not involve friction ( U) W + = Change in the total energy Net wor done on the system by eternal forces If no eternal energy enters the system ( 0 forces) then W =, no wor done by eternal or ( U) 0 + = ( + U) = constant Conservation of energy only for conservative systems (No friction or damping) An Energy Method for Determining Natural requencies. he natural frequency of a conservative system can be obtained from a consideration of the inetic energy and the potential energy of the system. Let us assume that we choose the datum line so that the potential energy at the equilibrium state is zero. hen in such a conservative system, the maimum inetic energy equals the maimum potential energy, or ma = U ma Solved Problems. Eample 3-5 Page 80 (etboo) Consider the system shown in the igure shown. he displacement is measured from the equilibrium position. he Kinetic energy is: he potential energy is: he total energy of the system is he change in the total energy is 1 = m 1 U = 1 1 + U = m + m 3/31

d dt U d 1 1 0 d m + = + = t 1 1 = m + = 0 = m + = 0 Since ẋ is not zero then we should have or where ( ) m + = 0 + = + = m 0 ωn 0 ω = is the natural frequency of the system and is epressed in rad/s. Another way of finding the natural frequency of the system is to assume a displacement of the form n = Asinωnt m Where Ais the amplitude of vibration. Consequently, 1 1 1 1 ( cosω ) = m = m A ω nt ( sinω ) U = = A nt Hence the maimum values of and U are given by Since ma rom which = Uma, we have 1 1 ma = m A ω, U = A n ma 1 1 m A ω = A n ω = n m 4/31

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ABLE 1. Summary of elements involved in linear mechanical systems Element ranslation Rotation Inertia 4 3 m = m a 1 J = J α θ Spring 1 θ 1 θ = ) = ( 1 = θ θ ) = θ ( 1 Damper ẋ ẋ 1 b θ 1 b θ = b ) = b ( 1 = b θ θ ) = b θ ( 1 PROCEDURE he motion of mechanical elements can be described in various dimensions as translational, rotational, or combination of both. he equations governing the motion of mechanical systems are often formulated from Newton s law of motion. 1. Construct a model for the system containing interconnecting elements.. Draw the free-body diagram. 3. Write equations of motion of all forces acting on the free body diagram. or translational motion, the equation of motion is Equation (1), and for rotational motion, Equation () is used. 31/31