Kingdom of Saudi Arabia Ministry of Higher Education Umm Al-Qura University College of Engineering & Islamic Architecture Mechanical Engineering Department Mechanical Vibrations 804 420 – 3 Lecture No. 9 Dr. Mohammad S. Alsoufi BSc, MSc PhD Room No.: 1080 Tel.: 00966 (012) 5270000 Ext.: 1163 E-mail: [email protected] 4. Harmonic Motions Kingdom of Saudi Arabia Ministry of Higher Education Umm Al-Qura University College of Engineering & Islamic Architecture Mechanical Engineering Department 9. Two Degree of Freedom Systems 9.1 Introduction Systems that require two independent coordinates to describe their motion are called “twodegree-of-freedom systems”. Some examples of systems having two degrees of freedom were shown in Figure. Two-degree-of-freedom systems (2DoF) Consider the automobile shown in Figure (a). For the vibration of the automobile in the vertical plane, a two-degree-of-freedom model shown in Figure (b) can be used. The body is idealized as a bar of mass m and mass moment of inertia supported on the rear and front wheels (suspensions) of stiffness and . The displacement of the automobile at any time can be specified by the linear coordinate x(t) denoting the vertical displacement of the C.G. of the body and the angular coordinate ( ) indicating the rotation (pitching) of the body about its C.G. Alternately, the motion of the automobile can be specified using the independent coordinates, ( ) ( ) of points A and B. Multistory building subjected to an earthquake Automobile Consider the motion of a multistory building under an earthquake. For simplicity, a twodegree-of-freedom model can be used as shown in Figure. The building is modeled as a rigid bar having a mass m and mass moment of inertia . The resistance offered to the motion of the building by the foundation and surrounding soil is approximated by a linear spring on stiffness k and a torsional spring of stiffness . The displacement of the building at any time can be specified by the horizontal motion of the base x(t) and the angular motion ( ) about the point O. Kingdom of Saudi Arabia Ministry of Higher Education Umm Al-Qura University College of Engineering & Islamic Architecture Mechanical Engineering Department Consider the system shown in Figure (a), which illustrates the packaging of an instrument of mass m. Assuming that the motion of the instrument is confined to the xy-plane, the system can be modeled as a mass m supported by springs in the x and y directions, as indicated in Figure (b). Thus the system has one point mass m and two degrees of freedom, because the mass has two possible types of motion (translations along the x and y directions). Packaging of an instrument. The general rule for the computation of the number of degrees of freedom can be stated as follows: (Number of masses in the system) Number × = of degrees of freedom of the system (Number of possible types of motion of each mass) 9.2 Equations of Motion for Forced Vibration Consider a viscously damped two-degree-of-freedom spring-mass system, shown in Figure (a). ( ), which The motion of the system is completely described by the coordinates ( ) define the positions of the masses m1 and m2 at any time t from the respective equilibrium positions. ( ), act on the masses m1 and m2 respectively. The external forces ( ) The free-body diagrams of the masses m1 and m2 are shown in Figure (b). Kingdom of Saudi Arabia Ministry of Higher Education Umm Al-Qura University College of Engineering & Islamic Architecture Mechanical Engineering Department The application of Newton’s second law of motion to each of the masses gives the equations of motion: ( ) ̇ ( ) ̇ ̈ ̈ ̇ ( ,⏟- ⃗̇ ( ) ,⏟- ) [ ] ⃗( ) ⃗( ) Where , - ( It can be seen that 1st equation contains terms involving x2 (namely, ̇ and ). nd It can be seen that 2 equation contains terms involving x1 (namely, ̇ and ). Hence, they represent a system of two coupled “second-order differential equations”. The motion of the mass m1 will influence the motion of the mass m2 and vice versa. Both equations can be written in “matrix form” as: ,⏟- ⃗̈ ( ) ) ̇ , - 0 1 , - [ ] ⃗( ) and ⃗( ) are called the “displacement and force vectors”, and are given by ⃗( ) { ( ) ( ) ⃗( ) { ( ) ( ) It can be seen that [m], [c], and [k] are all 2×2 matrices whose elements are the known masses, damping coefficients, and stiffnesses of the system, respectively. These matrices can be seen to be “symmetric”, so that , - , - , - , - , - , - Where the superscript T denotes the transpose of the matrix. Kingdom of Saudi Arabia Ministry of Higher Education Umm Al-Qura University College of Engineering & Islamic Architecture Mechanical Engineering Department 9.3 Free-Vibration Analysis of an Undamped System The solution of 1st and 2nd equations involves four constants of integration (two for each equation). We shall first consider the “free vibration solution”. For the free-vibration analysis of the system shown in Figure (a), we set ( ) If the damping is disregarded The1st and 2nd equations of motion reduce to ̈ ( ) ( ) ( ) ̈ ( ) ( ) ( ) ( ) ( ) We are interested in knowing whether m1 and m2 can oscillate harmonically with the same frequency and phase angle but with different amplitudes. Assuming that it is possible to have harmonic motion of m1 and m2 at the same frequency and the same phase angle . The solutions becomes as ( ) ( ) ( ) ( ) Where and are constants that denote the maximum amplitudes of ( ) and is the phase angle. ( ) ( ) and ( ) ( ) into Substituting ( ) ( ) ( ) ( ) ( ) ( ) and ̈ ( ) ,* , ( ) ( * )+ ( )+ ( - ( ( ) and ̈ ( ) ) ) Since the above equation must be satisfied for all values of the time t, the terms between brackets must be “zero”. This yields… * ( )+ * ( )+ Kingdom of Saudi Arabia Ministry of Higher Education Umm Al-Qura University College of Engineering & Islamic Architecture Mechanical Engineering Department Which represent two simultaneous homogenous algebraic equations in the unknowns . For “trivial solution” , this implies that there is no vibration. For “nontrivial solution” of , the determinant of the coefficients of be zero: * ( )+ [ ] * ( )+ Or ( ) *( ) ( ) + *( )( ) + ( ) ( ) 8 ( )( ( ) 9] ) The values of and remain to be determined. These values depend on the natural frequencies and . We shall denote the values of and ( ) Corresponding to as and ( ) Corresponding to as and Since * ( )+ homogenous, only the ratios 9 The roots are called “natural frequencies” of the system. This shows that it is possible for the system to have a nontrivial harmonic solution of the form of ( ) ( ) ( ) must This equation is called the “frequency” or “characteristic” equation because its solution yields the frequencies or the characteristic values of the system. The roots are given by ( ) ( ) 8 9 [8 and For ( ( ) ) and ( ) * and ( ) ( ) ( ( ) ( ) ( ) can be found. and ( ) ( ( ) ) ( ) ( ( ) ) ( ( ) ( ) ( ) ) )+ is Kingdom of Saudi Arabia Ministry of Higher Education Umm Al-Qura University College of Engineering & Islamic Architecture Mechanical Engineering Department The normal modes of vibration corresponding to as ) ⃗( ) { ( ) ) { ( ) ( ) ⃗ ( )( ) ( ) { ( ) } ( ) } { ( ) } { ( ) ( ) ( ) ( ) ( ) ( ) } { ( ( ) ( ( ) } { ) ) ( ( ) ) ( ) } } ⃗ ( )( ) ⃗ ( )( ) Where and are constants. Since ⃗ ( ) ( ) and ⃗ ( ) ( ) already involve the unknown constants choose with no loss of generality. The components of the vector ⃗( ) can be expressed as ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) Where the unknown constants , conditions: ( ) ( ) ( { ( ) The resulting motion can be obtained by a linear superposition of the two normal modes ( ) } ( ) ⃗( ) ( ) can be expressed, respectively, The vectors ⃗ ( ) and ⃗ ( ) which denote the normal modes of vibration are known as the “modal vectors” of the system. The free-vibration solution or the motion in time can be expressed as ⃗ ( )( ( ) ⃗( and ) Then, ( ̇ ( ( ̇ ( ) ) ) ) ( ) ( ) ( ) ( ) ( ) , and ( ) ( ) ( ) ( ) ( ̇ ( ) ̇ ( ) ( ) ( ) we can ) can be determined from the initial ̇ ( ) ( ) and ) ) ( ) ( ) ( ̇ ( ( ) ( ) ( ) ( ) Kingdom of Saudi Arabia Ministry of Higher Education Umm Al-Qura University College of Engineering & Islamic Architecture Mechanical Engineering Department The solution can be expressed as ( ) 8 ( ) 8 ( ) ( ) ̇ ( ) ( ̇ ( ) 9 ) ( ) 9 ( ) 8 ( ) 8 ̇ ( ) ( From which we obtain the desired solution: ( ) ( ) ) ( ) 3 ( ) 6* ( ) ( [2 [2 6* * ( )+ ( ) 3 ( ) ( )+ ( ) 2 2 ̇ ( ) ( ) * ( ) } 8 , ( ) { ( ) } 8 ̇ ( )+ , 7 3 ] ̇ ( ) ( ) { 3 ] ̇ ( ) ( ) ̇ ( )+ 7 ̇ ( ) 9 ( )- ̇ ( ) ̇ ( ) 9 ( ) ( )- Example No.1: Free-Vibration Response of a Two-Degree-of-Freedom System Find the free-vibration response of the system shown in Figure with: k1 = 30, k2 = 5 and k3 = 0, m1 = 10 and m2 = 1 c1 = c2 = c3 = 0 For the initial conditions: ( ) ̇ ( ) ( ) ̇ ( ) ( ) 9 ̇ ( ) 9 ) Kingdom of Saudi Arabia Ministry of Higher Education Umm Al-Qura University College of Engineering & Islamic Architecture Mechanical Engineering Department Solution No.1: For the given data [ ]{ [ By setting ]{ the [ determinant ]{ } } of coefficient matrix in 2 3 to zero, we obtain the frequency equation from which the natural frequencies can be found as leads to , while thus, the normal modes (or eigenvectors) are given by ( ) ( ) ⃗( yields ( ) ⃗( ) { ( ) } 2 3 ( ) ) { ( ) } 2 3 ( ) ( ) , ( ) ( ) The free-vibration responses of the masses m1 and m2 are given by ( ) ( ) 2 3 2 3 the } ( ) ( ) ( ) ( ) Where the unknown constants conditions: ( ) ( ) ) ̇ ( ) ̇ ( ( ) , ( ) , ( ) ( ) and ( ) ( ) ( ) can be determined from the initial ( ) ( ) ( ) ( ) ( ) ( ) ( ) Kingdom of Saudi Arabia Ministry of Higher Education Umm Al-Qura University College of Engineering & Islamic Architecture Mechanical Engineering Department The solution ( ) of ( ( ) ) ( ) The solution of ̇ ( ( ) ( ) ( ) and ̇ ( yields ( ) ( ) So, ( ) ( ) Thus the free-vibration responses of masses m1 and m2 are given by ( ) ( ) ) ( ) ) ( ) ( and yields ( ) ( ) The graphical representation of the free-vibration responses is shown below ) Kingdom of Saudi Arabia Ministry of Higher Education Umm Al-Qura University College of Engineering & Islamic Architecture Mechanical Engineering Department 9.4 Torsional System Consider a torsional system consisting of two discs mounted on a shaft, as shown in Figure. The three segments of the shaft have rotational spring constants kt1, kt2 and kt3 as indicated in the figure. Also shown are the discs of mass moments of inertia J1 and J2, the applied torques Mt1 and Mt2 and the rotational degrees of freedom and The differential equations of rotational motion for the discs J1 and J2 can be derived as ̈ ( ̈ ( ) Which upon rearrangement become ̈ ( ) ̈ ) ( ) For the free-vibration analysis of the system, the above equation reduces to ̈ ( ̈ Torsional system with discs mounted on a shaft ) ( ) Example No.2: Natural Frequencies of a Torsional System Find the natural frequencies and mode shapes for the torsional system shown in Figure for: Torsional system Solution No.2: From the given data ̈ ̈ Kingdom of Saudi Arabia Ministry of Higher Education Umm Al-Qura University College of Engineering & Islamic Architecture Mechanical Engineering Department Rearranging and substituting the harmonic solution ( ) ( gives the frequency equation: The solution of gives the natural frequencies √ ) ( √ ) √ ( √ ) The amplitude ratios are given by ( ) ( ( ) ) √ ( ) ( √ ) ( ) 9.5 Forced-Vibration Analysis The equations of motion of a general two-degree-of-freedom system under external forces can be written as ̈ 1{ } ̈ 0 Equations ̈ ̈ ( ̇ ̇ 1{ } ̇ 0 ) ̇ ( ̇ ) ̇ [ ( ]2 3 ( ) { } ) Can be seen to be special cases of the equations of motion of a general two-degree-of-freedom system under external forces, with and and We shall consider the external forces to be harmonic ( ) Where is the forcing frequency. We can write the steady-state solutions as ( ) Where and are, in general, complex quantities that depend on and the system parameters. Substitution of ( ) and ( ) into equations of motion of a general two-degree-of-freedom system under external forces leads to Kingdom of Saudi Arabia Ministry of Higher Education Umm Al-Qura University College of Engineering & Islamic Architecture Mechanical Engineering Department [ ( ( ) ) ) ]{ ) ( we define the mechanical impedance ( ( ( )- [ ( ( , ( )- ⃗ ) ) ( ( ⃗ Equation , ( )- ⃗ ) ] ) { } { } , ( )- ⃗ Where the inverse of the impedance matrix is given by , ( ⃗ ⃗ can be solved to obtain ⃗ } ) as ⃗ { ) Where , ( } )- ( ) ( ) ( ) ( ) ( The solution ( ( ) ) ( ) ( ( ) ) ) ( ) ( ) [ ( ) ( ) ( ) ( ) ( ) ( ) Example No.3: Steady-State Response of Spring-Mass System Find the steady-state response of the system shown in Figure, when the mass m1 is excited by the force ( ) . Also, plot its frequency-response curve. A two-mass system subjected to harmonic force ( ) ( ) ( ) ] ( ) Kingdom of Saudi Arabia Ministry of Higher Education Umm Al-Qura University College of Engineering & Islamic Architecture Mechanical Engineering Department Solution No.3: The equations of motion of the system can be expressed as ̈ 1{ } ̈ 0 ( ) ( ) 2 3 ( ) ( ( ) Since 8 ( ) ( ) ( ) ( )( ) ( ) ( )( ) ) [8 12 3 ( ) ( ) 0 ( ( ) ) 9 ( ) 9 8 ( )( ) 9] Therefore, { ( ) . / } [. / . / ][ . / ] [. / . / ][ . / ] ( ) Frequency-response curves Kingdom of Saudi Arabia Ministry of Higher Education Umm Al-Qura University College of Engineering & Islamic Architecture Mechanical Engineering Department 9.6 Vibration Absorbers The vibration absorber, also called dynamic vibration absorber, is a mechanical device used to reduce or eliminate unwanted vibration. It consists of another mass and stiffness attached to the main (or original) mass that needs to be protected from vibration. Thus the main mass and the attached absorber mass constitute a two-degree-of-freedom system; hence the vibration absorber will have two natural frequencies. The vibration absorber is commonly used in machinery that operates at constant speed, because the vibration absorber is tuned to one particular frequency and is effective only over a narrow band of frequencies. Common applications of the vibration absorber include reciprocating tools, such as sanders, saws, and compactors, and large reciprocating internal combustion engines which run at constant speed (for minimum fuel consumption). In these systems, the vibration absorber helps balance the reciprocating forces. Without a vibration absorber, the unbalanced reciprocating forces might make the device impossible to hold or control. Vibration absorbers are also used on high-voltage transmission lines. In this case, the dynamic vibration absorbers, in the form of dumbbellshaped devices (see Figure), are hung from transmission lines to mitigate the fatigue effects of wind induced vibration. A machine or system may experience excessive vibration if it is acted upon by a force whose excitation frequency nearly coincides with a natural frequency of the machine or system. In such cases, the vibration of the machine or system can be reduced by using a vibration neutralizer or dynamic vibration absorber, which is simply another spring-mass system. The dynamic vibration absorber is designed such that the natural frequencies of the resulting system are away from the excitation frequency. We shall consider the analysis of a dynamic vibration absorber by idealizing the machine as a single-degree-of-freedom system. 9.6.1 Undamped Dynamic Vibration Absorber When we attach an auxiliary mass m2 to a machine of mass m1 through a spring of stiffness k2, the resulting two-degree-of-freedom system will look as shown in Figure. The equations of motion of the masses m1 and m2 are ̈ ̈ ( ) ( ) Undamped dynamic vibration absorber Kingdom of Saudi Arabia Ministry of Higher Education Umm Al-Qura University College of Engineering & Islamic Architecture Mechanical Engineering Department By assuming harmonic solution ( ) we can obtain the steady-state amplitudes of the masses ( ) ( )( ) ( )( ) We are primarily interested in reducing the amplitude of the machine In order to make the amplitude of zero, the ( ( ) )( ) numerator of should be set equal to zero. This gives If the machine, before the addition of the dynamic vibration absorber, operates near it resonance, . Thus if the absorber is designed such that The amplitude of vibration of the machine, while operating at its original resonant frequency, will be zero. By defining ( as the natural frequency of the machine or main system, and ( ) ) As the natural frequency of the absorber or auxiliary system, equations rewritten as . [ . / ][ / . / ] and can be Kingdom of Saudi Arabia Ministry of Higher Education Umm Al-Qura University College of Engineering & Islamic Architecture Mechanical Engineering Department [ . / ][ . / ] Figure shows the variation of the amplitude of vibration of the machine with the machine speed The two peaks correspond to the two natural frequencies of the composite system. As seen before, at At this frequency, equation gives This shows that the force exerted by the auxiliary spring is opposite to the impressed ) and neutralizes it, thus force ( reducing to zero. The size of the dynamic vibration absorber can be found Effect of undamped vibration absorber on the response of machine The values of and depend on the allowable value of It can be seen from Figure that the dynamic vibration absorber, while eliminating vibration at the known impressed frequency introduces two resonant frequencies and at which the amplitude of the machine is infinite. The operating frequency must therefore be kept away from the frequencies and The values of and can be found by equating the denominator of ( ) to zero. Noting that ( . and setting the denominator of ( ) ( ) [ ( ) 6 The two roots of this equation are given by . ( / ][ ) / . / ] )( to zero leads to ) 7 Kingdom of Saudi Arabia Ministry of Higher Education Umm Al-Qura University College of Engineering & Islamic Architecture Mechanical Engineering Department ( ) ( ) {[ . /. / ] 8[ . . } /. / ] . / 9 } / Example No.4: Vibration Absorber for Diesel Engine A diesel engine, weighing 3,000 N, is supported on a pedestal mount. It has been observed that the engine induces vibration into the surrounding area through its pedestal mount at an operating speed of 6,000 rpm. Determine the parameters of the vibration absorber that will reduce the vibration when mounted on the pedestal. The magnitude of the exciting force is 250 N, and the amplitude of motion of the auxiliary mass is to be limited to 2 mm. Solution No.4: The frequency of vibration of the machine is Since the motion of the pedestal is to be made equal to zero, the amplitude of motion of the auxiliary mass should be equal and opposite to that of the exciting force | | ( ( ) ( ) ( Example No.5: Absorber for Motor-Generator Set A motor-generator set, shown in Figure, is designed to operate in the speed range of 2000 to 4000 rpm. However, the set is found to vibrate violently at a speed of 3000 rpm due to a slight unbalance in the rotor. It is proposed to attach a cantilever mounted lumped-mass absorber system to eliminate the problem. When a cantilever carrying a trial mass of 2 kg tuned to 3000 rpm is attached to the set, the resulting natural frequencies of the system are found to be 2500 rpm and 3500 rpm. Design the absorber to be attached (by specifying its mass and stiffness) so that the natural frequencies of the total system fall outside the operating-speed range of the motor-generator set. ) ) Motor-generator set Kingdom of Saudi Arabia Ministry of Higher Education Umm Al-Qura University College of Engineering & Islamic Architecture Mechanical Engineering Department Solution No.5: The natural frequencies of the motor-generator set and √ of the absorber are given by √ . / The resonant frequencies and of the combined system are given by equation }. . / Since the absorber (m = 2 kg) is tuned, rpm). Using the notation . (corresponding to 3,000 √. / / Since and are known to be 261.80 rad/s (or 2500 rpm) and 366.52 rad/s (or 3500 rpm), respectively, we find that Hence, . / √. Or 4 Since lower limit of / , then 5 and . The specified is 2000 rpm or 209.44 rad/s, and so With this value of , then and ( ) . With these values, the second resonant frequency can be found from . / √. / Which gives larger than the specified upper limit of 4,000 rpm. The spring stiffness of the absorber is given by ( ) ( ) Kingdom of Saudi Arabia Ministry of Higher Education Umm Al-Qura University College of Engineering & Islamic Architecture Mechanical Engineering Department Problem No.1: Find the natural frequencies of the system shown in Figure, with m1 = m, m2 = 2m, k1 = k and k2 = 2k. Determine the response of the system when k = 1,000 N/m, m = 20 kg, and the initial values of the displacements of the masses m1 and m2 are 1 and -1, respectively. Solution No.1: Kingdom of Saudi Arabia Ministry of Higher Education Umm Al-Qura University College of Engineering & Islamic Architecture Mechanical Engineering Department Problem No.2: One of the wheels and leaf springs of an automobile, traveling over a rough road, is shown in Figure (a). For simplicity, all the wheels can be assumed to be identical and the system can be idealized as shown in Figure (b). The automobile has a mass of m1 = 1,000 kg and the leaf springs have a total stiffness of k1 = 400 kN/m. The wheels and axles have a mass of m2 = 300 kg and the tires have a stiffness of k2 = 500 kN/m. If the road surface varies sinusoidally with amplitude of Y = 0.1 m and a period of l = 6 m, find the critical velocities of the automobile. (a) Solution No.2: (b) Kingdom of Saudi Arabia Ministry of Higher Education Umm Al-Qura University College of Engineering & Islamic Architecture Mechanical Engineering Department
© Copyright 2024