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

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1 Chapter 3 Mechanical Systems Dr. A.Aziz. Bazoune 3.1 INTRODUCTION Mathematical Modeling and response analysis of mechanical systems are the subjects of this chapter. 3. MECHANICAL ELEMENTS Any mechanical system consists of mechanical elements. There 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 make a unit change in acceleration (angular acceleration). That is, change in force N inertia (mass) = or kg change in acceleration m/s change in torque N-m inertia (moment of inertia) = or kg 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. Translational Springs For translational motion, (Fig 3-1(a)), the force that arises in the spring is proportional to and is given by F = k (3-1) where is the elongation of the spring and k is a proportionality constant called the spring constant and has units of [force/displacement]=[n/m] in SI units. 1/34

2 At point P, the spring force F acts opposite to the direction of the force F applied at point P. X X+ P f f Q X+ 1 X + 1 P f Figure 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) Figure 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. The net elongation of the spring is 1. The force acting in the spring is then F = k (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. /34

3 Torsional Springs Consider the torsional spring shown in Figure 3- (a), where one end is fied and a torque τ is applied to the other end. The angular displacement of the free end is θ. The torque T in the torsional spring is T k θ = T (3-3) where θ is the angular displacement and k T is the spring constant for torsional spring and has units of [Torque/angular displacement]=[n-m/rad] in SI units. θ τ τ θ θ 1 τ Figure 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. For the torsional spring shown in Figure 3-(b), torques equal in magnitude but opposite in direction, are applied to the ends of the spring. In this case, the torque T acting in the torsional spring is T T = k θ θ (3-4) 1 At each end, the spring acts in the direction opposite to that of the applied torque at that end. For linear springs, the spring constant k may be defined as follows spring constant k for translational spring spring constant k T 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 k or k T corresponds to a hard spring, a small value of k or k T to a soft spring. The reciprocal of the spring 3/34

4 constant k is called compliance or mechanical capacitance C. Thus C 1 k 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. Figure 3-3 shows the force displacement characteristic curves for linear and nonlinear springs. All practical springs have inertia and damping. F An ideal spring has neither mass nor damping (internal friction) and will obey the linear force displacement law. o Figure 3-3 Force-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. Figure 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 4/34

5 by oil because oil must flow around the piston (or through orifices provided in the piston) from one side to the other. 1 f & & 1 f τ θ & θ & 1 τ Figure 3-4 (a) Translational damper; (b) torsional (or rotational) damper. Translational Damper In Fig 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. The velocities of the ends of the damper are & 1 and &. Velocities & 1 and & are taken relative to the same frame of reference. In the damper, the damping force F that arises in it is proportional to the velocity differences & 1 & of the ends, or F = b & & = b& (3-5) 1 & = & & and the proportionality constant b relating the damping force F where 1 to the velocity difference & is called the viscous friction coefficient or viscous friction constant. The dimension of b is [force/velocity] = [N-s/m] in SI units. Torsional Damper For the torsional damper shown in Figure 3-4(b), the torques τ applied to the ends of the damper are of equal magnitude, but opposite in direction. The angular velocities of the ends of the damper are & θ 1 and & θ and they are taken relative to the same frame of reference. The damping torque T that arises in the damper is proportional to the angular velocity differences & θ1 & θ of the ends, or T ( 1 ) T = b & θ & θ = b & θ (3-6) & θ = & θ & θ and the proportionality b T relating the damping torque T to the angular velocity difference & θ is where, analogous to the translation case, 1 constant T 5/34

6 called the viscous friction coefficient or viscous friction constant. The dimension of b is [torque/angular velocity] = [N-m-s/rad] in SI units. 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. Friction 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. Figure 3-5 shows a characteristic curve for square-law friction. Figure 3-5 Characteristic curve for square-law friction. 6/34

7 3.3 MATHEMATICAL MODELING OF SIMPLE MECHANICAL SYSTEMS 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. Thus, a rigid body does not deform. Newton s laws. Newton s first law: (Conservation of Momentum) The 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 mv, for translational or linear motion. For 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: Translational 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. That is, or force acceleration = mass ( mass)( acceleration ) = force Suppose that forces are acting on a body of mass m. If F is the sum of all forces acting on a mass m through the center of mass in a given direction, then F = ma (3-7) where a is the resulting absolute acceleration in that direction. The 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. For a rigid body in pure rotation about a fied ais, Newton s second law states that 7/34

8 or ( moment of inertia)( angular acceleration ) = torque T = Jα (3-8) where T 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. Torque or moment of force. Torque, 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. Torque is the product of a force and the perpendicular distance from a point of rotation to the line of action of the force. [ ] [ ] Torque = force distance = N-m in SI units Moments of inertia. The moment of inertia J of a rigid body about an ais is defined by J = rdm (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. Figure 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. 8/34

9 Figure 3-7 Homogeneous cylinder rolling on a flat surface As an eample to that, consider the system shown in Figure 3-7, where a cylinder of mass m and a radius R rolls on a flat surface. The moment of inertia of the cylinder is about ais CC is Jc 1 = mr The moment of inertia J of the cylinder about ais is 1 3 J = Jc + mr = mr + mr = mr Forced response and natural response. The behavior determined by a forcing function is called a forced response, and that due initial conditions is called natural response. The 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. Figure 3-8 Transient and steady state response Parallel 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. 9/34

10 Parallel Springs. For the springs in parallel, Figure 3-9, the equivalent spring constant k eq is obtained from the relation k 1 k e q k F F Figure 3-9 Parallel spring elements where F = k + k = k + k = k 1 1 eq keq = k1+ k for parallel springs (3-9) This formula can be etended to n springs connected side-by-side as follows: n k eq i= 1 i ( for parallel springs) = k (3-10) Series Springs. For the springs in series, Figure 3-10, the force in each spring is the same. Thus y k k1 F k eq F Figure 3-10 Series spring elements 1 F = k y, F = k y Eliminating from these two equations yields or F F = k k1 k k k1+ k F = k F k= F+ F = F k1 k1 k1 10/34

11 or where k + k k k 1 1 = F F = = kk k1+ k + k1 k k eq = + for series springs k k k 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 k k eq i= 1 i (3-1) Free Vibration without damping. Consider the mass spring system shown in Figure The equation of motion can or where be given by 1 m && + k= 0 k && + = && + = m 0 ωn 0 ω = n k m K is the natural frequency of the system and is epressed in rad/s. Taking LT of both sides of the above equation where ( 0) = o and & ( 0) = & o gives & [&&] sx s s ω X s = n L () t M Figure 3-11 Mass Spring System rearrange to get 1 The weight W = mg of the mass m is equal to the static deflection of the spring. Therefore, they will cancel each other, (See Appendi A). 11/34

12 so + & o X( s) =, Remember poles are s=± jω s + ω 1443 n n & o ωn s X() s = + o ω s + ω s + ω and the response t is given by It is clear that the response n n n & t nt o nt ω o () = sin( ω ) + cos( ω ) n comple conjugates t consists of a sine and cosine terms and depends on the values of the initial conditions o and & o. Periodic motion such that described by the above equation is called simple harmonic motion. ( t) slope = & o Im s plane o ω n t Re ω n π Period = T = ω n Figure 3-1 Free response of a simple harmonic motion and pole location on the s-plane & 0 = & = 0, if o cos( ω ) t t = o n The period T is the time required for a periodic motion to repeat itself. In the present case, Period π T = seconds ω The 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, n 1/34

13 Frequency f 1 = = T ω π Hz The 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, k ωn = π fn = rad/sec m Rotational System. Rotor mounted in bearings is shown in figure 3-13 below. The moment of inertia of the rotor about the ais of rotation is J. Friction in the bearings is viscous friction and that no eternal torque is applied to the rotor. Tt () J ω b J ω bω Figure 3-13 Rotor mounted in bearings and its FBD. Apply Newton s second law for a system in rotation form 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 o τ which represents the equation of motion as well as the mathematical model of the system shown. It represents a first order system. To find the response ω ( t) LT of both sides of the previous equation., take 13/34

14 1 sω( s) ω ( 0) + Ω ( s) = τ { L[ ω] L[ ω & ] 1 s+ Ω ( s) = ω τ + 1 τ + 1τ = 0 o ( s) where the denominator equation above equation will give the epression of ω ( t) ωo Ω = s + ( 1 τ ) s is known as the characteristic polynomial and the s is called the characteristic equation. Taking inverse LT of the b/ J t 1/τ t αt o o o 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 Free response of a rotor bearing system α=0 ω o ω(t) α=0. α=0.5 α= α=1 α= α=5 α= time (t) Figure 3-14 Graph of ω α t o e for ranges of α. 14/34

15 A time constant is that value of time that makes the eponent equal to -1. For this system, time constant τ = J/ b. When t = τ, the eponent factor is ( t / τ) ( τ / τ) 1 This means that when time constant=, the time τ response is reduced to 36.8 % of its final value. We also have τ = J / b = time constant ωτ = 0.37 ωo ω 4τ = 0.0ω e = e = e = = 36.8 % o Figure 3-15 Curve of angular velocity ω versus time t for the rotor shown in Figure Spring-Mass-Damper System. Consider the simple mechanical system shown involving viscous damping. Obtain the mathematical model of the system shown. b k Fb = b& Fk = k m m ( t) FreeBody Diagram(FBD) Figure 3-16 Mass -Spring Damper System and the FBD. i) The FBD is shown in the figure ii) Apply Newton s second law of motion to a system in translation: Rearranging F = m&& b& k = m&& 15/34

16 m&& + b& + k= 0 The above equation represents the mathematical model as well as the free vibration for a second order model. If m= 01. kg, b= 04. N/m-s, and k = 4N/m, the above differential equation becomes 0.1 && + 0.4& + 4 = 0 && + 4& + 40 = 0 To obtain the free response ( t ), assume ( 0) = o and transform of both sides of the given equation & () () & Take Laplace 0 = 0. sx s s sx s X s = L && t L & t L t Substitute in the transformed equation ( 0) obtain or solving for which can be written as () = o and ( 0) = 0, [ ] s + 4s+ 40 X s = so + 4 o X s yields ( so + 4o) ( s + 4) X( s) = = s + 4s s + 4s+ 40 G s Characteristic polynomial ( s + 4) X s = = o 1443 s + 4s + 40 Characteristic polynomial & and rearrange, we where G( s ) is referred to as the transfer function that gives the relationship between the input and the output X( s ). G( s ) can be shown graphically as: o o o ( s + 4) s s + 40 Characteristic polynomial Transfer function Xs Figure 3-17 Transfer function between input and output. iii) It is clear that the characteristic equation of the system is s + 4s + 40= 0 and has comple conjugates roots. 16/34

17 ( s+ ) s + 4s+ 40= s s = s = 0 The roots of the above equation are therefore comple conjugate poles given by s1 = + j6 and s = j6 iv) The epression of X( s ) can be written now as: s + 4s+ 40 ( s + ) 1 6 o ( s+ ) + 3 ( s+ ) + s+ 4 s+ + s+ X( s) = = = + s+ + 6 s+ + 6 s+ + 6 o o o o = v) Solving for ( t 1 ) = L X ( s ) yields () t = cos + Or 10 t o t () = e o ( sin 6t ) t e t T d = π ω d 3 o 1 t t o e 6t e sin 6t 3 s1 = + j6 t 1 t t = o e cost 6 e sint t Im j6 j6 s = j6 s plane Re Figure 3-18 Free Vibration of the mass-spring-damper system described by && + 4& + 40 = 0 with initial conditions ( 0) = o and & 0 = 0,. (See Appendi B). 17/34

18 10 Pole-Zero Map 5 Imaginary Ais Real Ais 1.5 Free vibration of mass-spring-damper system o (t) t (sec) Figure 3-18 Free Vibration of the mass-spring-damper system described by && + 4& + 40 = 0 with initial conditions ( 0) = o and & 0 = 0,. 3.4 WORK ENERGY, AND POWER Work. The work 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 F d θ F W = F d W = F dcosθ Figure 3-19 Work done by a force The units of work in SI units are : [ work] = [ force distance] = [ N-m] = [ Joule] = [ J]. The work done by a spring is given by: 18/34

19 F 1 W = k d= k { 0 F k l l+ F Figure 3-19 Work done by a spring. Energy. Energy can be defined as ability to do work. Energy can be found in many different forms and can be converted from one form to another. For 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. This 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. The Energy that a body possesses because of its position is called potential energy. In mechanical systems, only mass and spring can store potential energy. The change in the potential energy stored in a system equals the work 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 mg F Figure 3-0 Potential energy Refer to Figure 3-0, the potential energy, U of a mass m is given by: U = mg d= mgh 0 19/34

20 For a translational spring, the potential energy U (sometimes called strain energy which is potential energy that is due to elastic deformations) is: 1 U = Fd= kd= k 0 0 If the initial and final values of are 1 and, respectively, then 1 1 Change in potential energy Δ U= Fd= kd= k k1 Similarly, for a torsional spring 1 1 θ 1 1 Change in potential energy Δ U = T dθ = k θ d= k θ k θ θ θ θ 1 1 T T T 1 Kinetic Energy. Only inertia elements can store kinetic energy in mechanical systems. T 1 mv = Kinetic energy= 1 Jθ& (Translation) (Rotation) The change in kinetic energy of the mass is equal to the work done on it by an applied force as the mass accelerates or decelerates. Thus, the change in kinetic energy T of a mass m moving in a straight line is Change in kinetic energy d Δ T =Δ W = Fd= F dt dt = Fvdt= mvv & dt= mvdv = 1 mv 1 mv t 1 t1 t t v t t v 1 The change in kinetic energy of a moment of inertia in pure rotation at angular velocity θ & is 1 1 Change in kinetic energy Δ T = J & θ J & θ 1 0/34

21 Dissipated Energy. Consider the damper shown in Figure 3-1 in which one end is fied and the other end is moved from 1 to. The dissipated energy Δ W in the damper is equal to the net work done on it: t t d Δ W = Fd= b & d= b dt = b dt { & dt & F t t b Figure 3-0 Damper. Power. Power is the time rate of doing work. That is, Power = P = dw dt where dw denotes work done during time interval dt. In SI units, the work done is measured in Newton-meters and the time in seconds. The unit of power is : N-m Joule Power = = = Watt = W s s [ ] [ ]. Passsive Elements. Non-energy producing element. They 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 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. Conservative Systems. Systems that do not involve friction (damping) are called conservative systems. 1/34

22 ( T U) { W Δ + = Δ 1443 Change in the total energy Net work done on the system by eternal forces If no eternal energy enters the system ( W 0 forces) then Δ =, no work done by eternal or ( T U) 0 Δ + = ( T + U) = constant Conservation of energy only for conservative systems (No friction or damping) An Energy Method for Determining Natural Frequencies. The natural frequency of a conservative system can be obtained from a consideration of the kinetic 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. Then in such a conservative system, the maimum kinetic energy equals the maimum potential energy, or Solved Problems. Eample 3-5 Page 80 (Tetbook) T ma = U ma Consider the system shown in the Figure shown. The displacement is measured from the equilibrium position. The Kinetic energy is: The potential energy is: The total energy of the system is The change in the total energy is d dt 1 T = m& 1 U = k Since & is not zero then we should have 1 1 T + U = m& + k ( && k) T U d d m k + = & + = t 1 1 = m &&& + k& = 0 = & m+ = 0 m&& + k = 0 m k /34

23 or where k && + = && + = m 0 ωn 0 ω = n k m 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 = Asinωnt Where A is the amplitude of vibration. Consequently, 1 1 & 1 1 ( cosω ) T = m = m A ω nt ( sinω ) U = k = k A nt Hence the maimum values of T and U are given by Since Tma = Uma, we have From which 1 1 Tma = m A ω, U = k A n ma 1 1 maω = ka n ω = n k m 3/34

24 Problem 4/34

25 Problem 5/34

26 6/34

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30 TABLE 1. Summary of elements involved in linear mechanical systems Element Translation Rotation Inertia F 4 F 3 m F = m a F F 1 T J T = J α θ Spring F 1 k F T θ 1 k θ T F = k ) = k ( 1 T = k θ θ ) = kθ ( 1 Damper F & & 1 b F T & θ 1 b θ & T F = b & & ) = b& ( 1 T = b & θ & θ ) = b & θ ( 1 PROCEDURE The motion of mechanical elements can be described in various dimensions as translational, rotational, or combination of both. The 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. For translational motion, the equation of motion is Equation (1), and for rotational motion, Equation () is used. 30/34

31 APPENDIX A: Static Equilibrium k L o k k δ st m m kδ st m m k ( + δ st ) mg mg At static equilibrium, we have: mg = Kδ st Let the mass be displaced a distance downward from its equilibrium, the spring force becomes f = K( + δ st ). Apply Newton s second law for the mass-spring system, one can write: or or Therefore ( δ ) F = m && K + + mg = m && st K Kδ + mg = m && st 1443 = 0 Static Equilibrium m && + K = 0 31/34

32 The above equation represents the mathematical model for the mass-spring system. It is a second order ordinary differential equation with constant coefficients. It is clear that the weight mg cancels with the static deflection Kδ of the system. Therefore, it does not appear in the equation of motion. st 3/34

33 APPENDIX B: The epression of acos( ωt) + bsin( ωt ) can be written in terms of sin( ωt ), that is cos ωt or a b a cos ( ωt ) + b sin ( ωt ) = a + b cos ( ωt ) + sin ( ωt ) a + b a + b Define ψ such that cosψ = sin ψ = a a + b 1 b ψ tan b = a a + b Therefore a cos ( ωt ) + b sin ( ωt ) = a + b cosψ cos ( ωt ) + sin ψ sin ( ωt ) Using the identity ( ) = + cos ωt ψ cos ωt cosψ sin ωt sinψ Therfore, or Where ( ω ) + ( ω ) = + ( ω ψ) acos t bsin t a b cos t ( ω ) + ( ω ) = + ( ω + ϕ) a cos t b sin t a b sin t sin ϕ = cos ϕ = a + ϕ tan b = b a + b a b 1 a 33/34

34 Useful Sites Torsion Mass Spring System Response of First and second Order Systems 34/34

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