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• very good ppt sir, nice explanation and animations, can you please send me the same on my mail skumar.kumbhalkar@raisoni.net

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• Sinusoidal periodic
Show sinusoid in both time and frequency domains
Complex Periodic
Can be defined mathematically by time-varying function and repeats exactly at regular intervals. Show half circle, triangle, square, multi-component sine. With few exeptions, complex periodic data can be expanded into Fourier Series:
X(t) = a0/2 + sum [ an cos(2 pi n f1 t) + bn sin(2 pi n f1 t) ]
Or
X(t) = X0 + sum [ Xn cos(2 pi n f1 t - pn) ]
This means that complex periodic data consists of a static component X0 and an infinite number of sinusoidal components called harmonics. The harmonic frequencies are all integral multiples of f1. Example: Voice.
Almost Periodic.
When the ratio of frequencies in a signal comprising two or more sine waves do not form rational numbers, the fundamental period is infinitely long.
Transients
• Sinusoidal periodic
Show sinusoid in both time and frequency domains
Complex Periodic
Can be defined mathematically by time-varying function and repeats exactly at regular intervals. Show half circle, triangle, square, multi-component sine. With few exeptions, complex periodic data can be expanded into Fourier Series:
X(t) = a0/2 + sum [ an cos(2 pi n f1 t) + bn sin(2 pi n f1 t) ]
Or
X(t) = X0 + sum [ Xn cos(2 pi n f1 t - pn) ]
This means that complex periodic data consists of a static component X0 and an infinite number of sinusoidal components called harmonics. The harmonic frequencies are all integral multiples of f1. Example: Voice.
Almost Periodic.
When the ratio of frequencies in a signal comprising two or more sine waves do not form rational numbers, the fundamental period is infinitely long.
Transients
• Sinusoidal periodic
Show sinusoid in both time and frequency domains
Complex Periodic
Can be defined mathematically by time-varying function and repeats exactly at regular intervals. Show half circle, triangle, square, multi-component sine. With few exeptions, complex periodic data can be expanded into Fourier Series:
X(t) = a0/2 + sum [ an cos(2 pi n f1 t) + bn sin(2 pi n f1 t) ]
Or
X(t) = X0 + sum [ Xn cos(2 pi n f1 t - pn) ]
This means that complex periodic data consists of a static component X0 and an infinite number of sinusoidal components called harmonics. The harmonic frequencies are all integral multiples of f1. Example: Voice.
Almost Periodic.
When the ratio of frequencies in a signal comprising two or more sine waves do not form rational numbers, the fundamental period is infinitely long.
Transients
• Sinusoidal periodic
Show sinusoid in both time and frequency domains
Complex Periodic
Can be defined mathematically by time-varying function and repeats exactly at regular intervals. Show half circle, triangle, square, multi-component sine. With few exeptions, complex periodic data can be expanded into Fourier Series:
X(t) = a0/2 + sum [ an cos(2 pi n f1 t) + bn sin(2 pi n f1 t) ]
Or
X(t) = X0 + sum [ Xn cos(2 pi n f1 t - pn) ]
This means that complex periodic data consists of a static component X0 and an infinite number of sinusoidal components called harmonics. The harmonic frequencies are all integral multiples of f1. Example: Voice.
Almost Periodic.
When the ratio of frequencies in a signal comprising two or more sine waves do not form rational numbers, the fundamental period is infinitely long.
Transients
• ### Mechanical Vibrations all slides

1. 1. 1 Mechanical Vibrations - Introduction V. Rouillard © 2003 - 2013 Mechanical Vibrations 04:36:37 Some Figures Courtesy Addison Wesley
2. 2. 2 Mechanical Vibrations - Introduction V. Rouillard © 2003 - 2013 CONTENT • Fundamentals of vibrations • Single degree-of-freedom systems • Free vibrations • Harmonic forcing functions • General forcing functions • Two degree-of-freedom systems • Free vibrations • Forced vibrations • Multi degree-of-freedom systems • Free vibrations • Forced vibrations 04:36:37
3. 3. 3 Mechanical Vibrations - Introduction Mechanical vibrations • • • Defined as oscillatory motion of bodies in response to disturbance. Oscillations occur due to the presence of a restoring force Vibrations are everywhere: • • • • • Vehicles: residual imbalance of engines, locomotive wheels Rotating machinery: Turbines, pumps, fans, reciprocating machines Musical instruments Excessive vibrations can have detrimental effects: • • • • • • Human body: eardrums, vocal cords, walking and running Noise Loosening of fasteners Tool chatter Fatigue failure Discomfort When vibration frequency coincides with natural frequency, resonance occurs. 04:36:37 V. Rouillard © 2003 - 2013
4. 4. 4 Mechanical Vibrations - Introduction V. Rouillard © 2003 - 2013 Mechanical vibrations • Aeolian, wind-induced or vortex-induced vibration of the Tacoma Narrows bridge on 7 November 1940 caused it to resonate resulting in catastrophic failure. Tacoma Narrows Bridge Collapse Video 04:36:37
5. 5. 5 Mechanical Vibrations - Introduction V. Rouillard © 2003 - 2013 Mechanical vibrations • Millennium Bridge, London: Pedestrians, in reaction to lateral motion of the bridge, altered their gait and started behaving in concert to induce the structure to resonate further (forced periodic excitation): Video link 04:36:37
6. 6. 6 Mechanical Vibrations - Introduction V. Rouillard © 2003 - 2013 Fundamentals • In simple terms, a vibratory system involves the transfer of potential energy to kinetic energy and vice-versa in alternating fashion. • • When there is a mechanism for dissipating energy (damping) the oscillation gradually diminishes. In general, a vibratory system consists of three basic components: • • • A means of storing potential energy (spring, gravity) A means of storing kinetic energy (mass, inertial component) A means to dissipate vibrational energy (damper) 04:36:37
7. 7. 7 Mechanical Vibrations - Introduction V. Rouillard © 2003 - 2013 Fundamentals • • This can be observed with a pendulum: At position 1: the kinetic energy is zero and the potential energy is mgl(1 − cos θ ) • • • At position 2: the kinetic energy is at its maximum At position 3: the kinetic energy is again zero and the potential energy at its maximum. In this case the oscillation will eventually stop due to aerodynamic drag and pivot friction → HEAT 04:36:37
8. 8. 8 Mechanical Vibrations - Introduction V. Rouillard © 2003 - 2013 Degrees of Freedom • The number of degrees of freedom : number of independent coordinates required to completely determine the motion of all parts of the system at any time. • Examples of single degree of freedom systems: 04:36:37
9. 9. 9 Mechanical Vibrations - Introduction Degrees of Freedom • Examples of two degree of freedom systems: 04:36:37 V. Rouillard © 2003 - 2013
10. 10. 10 Mechanical Vibrations - Introduction Degrees of Freedom • Examples of three degree of freedom systems: 04:36:37 V. Rouillard © 2003 - 2013
11. 11. 11 Mechanical Vibrations - Introduction V. Rouillard © 2003 - 2013 Discrete and continuous systems • Many practical systems small and large or structures can be describe with a finite number of DoF. These are referred to as discrete or lumped parameter systems • Some large structures (especially with continuous elastic elements) have an infinite number of DoF These are referred to as continuous or distributed systems. • In most cases, for practical reasons, continuous systems are approximated as discrete systems with sufficiently large numbers lumped masses, springs and dampers. This equates to a large number of degrees of freedom which affords better accuracy. 04:36:37
12. 12. 12 Mechanical Vibrations - Introduction V. Rouillard © 2003 - 2013 Classification of Vibration • Free and Forced vibrations • • • Free vibration: Initial disturbance, system left to vibrate without influence of external forces. Forced vibration: Vibrating system is stimulated by external forces. If excitation frequency coincides with natural frequency, resonance occurs. Undamped and damped vibration • • • Undamped vibration: No dissipation of energy. In many cases, damping is (negligibly) small (steel 1 – 1.5%). However small, damping has critical importance when analysing systems at or near resonance. Damped vibration: Dissipation of energy occurs - vibration amplitude decays. Linear and nonlinear vibration • Linear vibration: Elements (mass, spring, damper) behave linearly. Superposition holds - double excitation level = double response level, mathematical solutions well defined. • Nonlinear vibration: One or more element behave in nonlinear fashion (examples). Superposition does not hold, and analysis technique not clearly defined. 04:36:37
13. 13. 13 Mechanical Vibrations - Introduction V. Rouillard © 2003 - 2013 Classification of Vibration • Deterministic and Random vibrations • • Deterministic vibration: Can be described by implicit mathematical function as a function of time. Random vibration: Cannot be predicted. Process can be described by statistical means. 04:36:37
14. 14. 14 Mechanical Vibrations - Introduction Vibration Analysis • • • • Input (excitation) and output (response) are wrt time Response depend on initial conditions and external forces Most practical systems very complex – (mathematical) modelling requires simplification Procedure: → → → → Mathematical modelling Derivation / statement of governing equations Solving of equations for specific boundary conditions and external forces Interpretation of solution(s) 04:36:37 V. Rouillard © 2003 - 2013
15. 15. 15 Mechanical Vibrations - Introduction Vibration Analysis 04:36:37 Example (1.3 Ed.3) V. Rouillard © 2003 - 2013
16. 16. 16 Mechanical Vibrations - Introduction V. Rouillard © 2003 - 2013 Spring Elements • • Pure spring element considered to have negligible mass and damping Force proportional to spring deflection (relative motion between ends): F = k ∆x • For linear springs, the potential energy stored is: U = 1 k ( ∆x ) 2 • Actual springs sometimes behave in nonlinear fashion • Important to recognize the presence and significance (magnitude) of nonlinearity • Desirable to generate linear estimate 04:36:37 2
17. 17. 17 Mechanical Vibrations - Introduction Spring Elements • Equivalent spring constant. • • Eg: cantilever beam: Mass of beam assumed negligible cf lumped mass Deflection at free end: mgl 3 δ= 3EI • Stiffness (Force/defln): k= • mg 3EI = 3 δ l This procedure can be applied for various geometries and boundary conditions. (see appendix) 04:36:37 V. Rouillard © 2003 - 2013
18. 18. Mechanical Vibrations - Introduction 18 Spring Elements • Equivalent spring constant. • Springs in parallel: w =mg=kδ +k 2 δ 1 w=mg=keqδ • where keq =k1 + k2 • In general, for n springs in parallel: i=n keq = ∑ ki i=1 04:36:37 V. Rouillard © 2003 - 2013
19. 19. Mechanical Vibrations - Introduction 19 Spring Elements • Equivalent spring constant. • Springs in series: δt =δ1 + δ2 • Both springs are subjected to the same force: mg = k1δ 1 = k2δ 2 mg=keqδ t • Combining the above equations: k1δ 1 = k2δ 2 = keqδ t δ 1= keqδ t k 04:36:37 1 and δ 2 = keqδ t k2 V. Rouillard © 2003 - 2013
20. 20. 20 Mechanical Vibrations - Introduction V. Rouillard © 2003 - 2013 Spring Elements • Springs in series (cont’d): • Substituting into first eqn: δt = • keqδ t k1 + keqδ t k2 Dividing by keqδt throughout: 1 1 1 = + keq k1 k2 • For n springs in series: 1 i=n  1  =∑   keq i=1  ki  04:36:37
21. 21. 21 Mechanical Vibrations - Introduction V. Rouillard © 2003 - 2013 Spring Elements • Equivalent spring constant. • • When springs are connected to rigid components such as pulleys and gears, the energy equivalence principle must be used. Example: Example (1.10 Ed.3) 04:36:37
22. 22. 22 Mechanical Vibrations - Introduction Mass / Inertia Elements • • • • • Mass or inertia element assumed rigid (lumped mass) • Modelling with lumped mass elements. Example: assume frame mass is negligible cf mass of floors. Its energy (kinetic) is proportional to velocity. Force ∝ mass * acceleration Work = force * displacement Work done on mass is stored as Kinetic Energy 04:36:37 V. Rouillard © 2003 - 2013
23. 23. 23 Mechanical Vibrations - Introduction V. Rouillard © 2003 - 2013 Mass / Inertia Elements • Equivalent mass - example: • The velocities of the mass elements can be written as: l2 x x &2 = &1 l1 • and l3 x x &3 = &1 l1 To determine the equivalent mass at position l1: x x & eq = & 1 04:36:37
24. 24. Mechanical Vibrations - Introduction 24 V. Rouillard © 2003 - 2013 Mass / Inertia Elements • Equivalent mass – example (cont’d) • Equating the kinetic energies: 1m & 2 x 2 1 1 • 1 + 2 m2& 2 + 2 m3& 3 = 2 meq& eq x2 1 x2 1 x2 Substituting for the velocity terms: meq 04:36:37 2 2 l  l  = m1 +  2 ÷ m2 +  3 ÷ m3  l1   l1 
25. 25. 25 Mechanical Vibrations - Introduction V. Rouillard © 2003 - 2013 Damping Elements • • • Absorbs energy from vibratory system → vibration amplitude decays. Damping element considered to have no mass or elasticity Real damping systems very complex, damping modelled as: • Viscous damping: • • • • Based on viscous fluid flowing through gap or orifice. Damping force ∝ relative velocity between ends Eg: film between sliding surfaces, flow b/w piston & cylinder, flow thru orifice, film around journal bearing. Coulomb (dry Friction) damping: • • 04:36:37 Based on friction between unlubricated surfaces Damping force is constant and opposite the direction of motion
26. 26. 26 Mechanical Vibrations - Introduction V. Rouillard © 2003 - 2013 Damping Elements • Hysteretic (material or solid) damping: • • 04:36:37 Based on plastic deformation of materials (energy loss due to slippage b/w grains) Energy lost due to hysteresis loop in force-deflection (stress-strain) curve of element when load is applied:
27. 27. 27 Mechanical Vibrations - Introduction V. Rouillard © 2003 - 2013 Damping Elements • Equivalent damping element: • 04:36:37 Combinations of damping elements can be replace by equivalent damper using same procedures as for spring and mass/inertia elements.
28. 28. Mechanical Vibrations - Introduction V. Rouillard © 2003 - 2013 Damping Elements FORD AU IRS 2 1 0 -2 Force [KN] 28 -1.5 -1 -0.5 0 -1 -2 -3 -4 -5 Velocity [m/s] 04:36:37 0.5 1 1.5 2
29. 29. 29 Mechanical Vibrations - Introduction Harmonic Motion • Harmonic motion: simplest form of periodic motion (deterministic). • • Pure sinusoidal (co-sinusoidal) motion • The motion of mass m is described by: Eg: Scotch-yoke mechanism rotating with angular velocity ω - simple harmonic motion: x = Asin( θ ) = A sin( ωt ) • Its velocity and acceleration are: dx = ω A cos( ωt ) dt and d 2x dt 2 04:36:37 = − ω 2 A sin( ωt ) = − ω 2 x V. Rouillard © 2003 - 2013
30. 30. 30 Mechanical Vibrations - Introduction V. Rouillard © 2003 - 2013 Harmonic Motion • • • 04:36:37 Sinusoidal motion emanates from cyclic motion The rotating vector generates a sinusoidal and a co-sinusoidal components along mutually perpendicular axes. Can be represented by a vector (OP) with a magnitude, angular velocity (frequency) and phase.
31. 31. Mechanical Vibrations - Introduction 31 Harmonic Motion • Often convenient to represent sinusoidal and co-sinusoidal components (mutually perpendicular) in complex number format • • Where a and b denote the sinusoidal (x) and co-sinusoidal (y) components a and b = real and imaginary parted of vector X 04:36:37 V. Rouillard © 2003 - 2013
32. 32. Mechanical Vibrations - Introduction 32 V. Rouillard © 2003 - 2013 Harmonic Motion Definition of terms: • Cycle: motion of body from equilibrium position → extreme position → equilibrium position → extreme position in other direction → equilibrium position . • • Amplitude: Maximum value of motion from equilibrium. (Peak – Peak = 2 x amplitude) Period: Time taken to complete one cycle τ= ω = circular frequency • Frequency: number of cycles per unit time. f = ω : radians/s 04:36:37 f Hertz (cycles /s) 1 ω = τ 2π 2π ω
33. 33. Mechanical Vibrations - Introduction 33 V. Rouillard © 2003 - 2013 Harmonic Motion • Phase angle: the difference in angle (lead or lag) by which two harmonic motions of the same frequency reach their corresponding value (maxima, minima, zero up-cross, zero down-cross) 04:36:37
34. 34. Mechanical Vibrations - Introduction 34 V. Rouillard © 2003 - 2013 Harmonic Motion • Phase angle: the difference in angle (lead or lag) by which two harmonic motions of the same frequency reach their corresponding value (maxima, minima, zero up-cross, zero down-cross) 04:36:37
35. 35. Mechanical Vibrations - Introduction 35 V. Rouillard © 2003 - 2013 Harmonic Motion • Natural frequency: the frequency at which a system vibrates without external forces after an initial disturbance. The number of natural frequencies always matches the number of DoF. • Beats: the effect produced by adding two harmonic motions with similar (close) frequencies. x1 = A sin( ωt ) x2 = A sin( ωt + δωt ) xt = x1 + x2 = A [sin( ωt ) + sin( ωt + δωt )] M +N M −N cos 2 2 δωt   δωt  xt = 2 A sin  ωt + cos   ÷ ÷ 2    2  Since sin M + sin N = 2 sin Eg: ω=40 Hz and δ= -0.075 • 04:36:37 In mechanical vibratory systems, beats occur when the (harmonic) excitation (forcing) frequency is close to the natural frequency.
36. 36. Mechanical Vibrations - Introduction 36 V. Rouillard © 2003 - 2013 Harmonic Motion • • • Octave: doubling of any quantity. Used mainly for frequency. Octave band (frequency): maximum is double of minimum. Eg: 64 – 128 Hz, 1000 – 2000 Hz. Decibel: defined as 10 x log(power ratio) P dB = 10Log  ÷  P0  In electrical systems (as in mechanical vibratory systems) power is proportional to the value squared hence:  X  dB = 20Log  ÷  X0  04:36:37
37. 37. Mechanical Vibrations - Introduction 37 V. Rouillard © 2003 - 2013 Harmonic (Fourier) Analysis • • Many vibratory systems not harmonic but often periodic Any periodic function can be represented by the Fourier series – infinite sum of sinusoids and co-sinusoids. ao + a1 cos( ωt ) + a2 cos( 2ωt ) + ........ 2 + b1 sin( ωt ) + b2 sin( 2ωt ) + ....... x( t ) = ao ∞ = + ∑ [an cos( nωt ) + bn sin( nωt )] 2 n =1 • To obtain an and bn the series is multiplied by cos(nωt) and sin(nωt) respectively and integrated over one period. 04:36:37
38. 38. Mechanical Vibrations - Introduction 38 Harmonic (Fourier) Analysis • Example: 04:36:37 V. Rouillard © 2003 - 2013
39. 39. Mechanical Vibrations - Introduction 39 Harmonic (Fourier) Analysis • Example: 04:36:37 V. Rouillard © 2003 - 2013
40. 40. Mechanical Vibrations - Introduction 40 Harmonic (Fourier) Analysis • As for simple harmonic motion, Fourier series can be expressed with complex numbers: eiωt = cos( ωt ) + i sin( ωt ) e −iωt = cos( ωt ) − i sin( ωt ) eiωt + e −iωt cos( ωt ) = 2 eiωt − e −iωt sin( ωt ) = 2i • The Fourier series: ao ∞ x( t ) = + ∑ [an cos( nω t ) + bn sin( nω t )] 2 n =1 Can be written as:  eiω t − e−iω t   ao ∞   eiω t + e −iω t    x( t ) = + ∑ an  ÷+ bn  ÷ ÷  ÷ 2 n =1   2 2i       04:36:37 V. Rouillard © 2003 - 2013
41. 41. Mechanical Vibrations - Introduction 41 V. Rouillard © 2003 - 2013 Harmonic (Fourier) Analysis • Defining the complex Fourier coefficients cn = • an − ibn 2 and cn −1 = The (complex) Fourier series is simplified to: x( t ) = ∞ ∑ n =−∞ 04:36:37 cn einωt an + ibn 2
42. 42. Mechanical Vibrations - Introduction 42 V. Rouillard © 2003 - 2013 Harmonic (Fourier) Analysis ao ∞ x( t ) = + ∑ [an cos( nω t ) + bn sin( nω t )] 2 n =1 • • The Fourier series is made-up of harmonics. Their amplitudes and phases are defined as: 2 2 An = ( an + bn ) b  φn = a tan  n ÷  an  04:36:37 harmonics
43. 43. Mechanical Vibrations - Introduction 43 V. Rouillard © 2003 - 2013 Harmonic (Fourier) Analysis • The amplitudes (magnitudes) and phases of the harmonics can be plotted as a function of frequency to form the frequency spectrum of spectral diagram: An 04:36:37
44. 44. 44 Mechanical Vibrations – Single Degree-of-Freedom systems V. Rouillard © 2003 - 2013 Free undamped vibration single DoF • • • Recall: Free vibrations → system given initial disturbance and oscillates free of external forces. Undamped: no decay of vibration amplitude Single DoF: • • • • mass treated as rigid, limped (particle) Elasticity idealised by single spring only one natural frequency. The equation of motion can be derived using • • • • Newton’s second law of motion D’Alembert’s Principle, The principle of virtual displacements and, The principle of conservation of energy. 04:36:37
45. 45. 45 Mechanical Vibrations – Single Degree-of-Freedom systems Free undamped vibration single DoF • Using Newton’s second law of motion to develop the equation of motion. 1. Select suitable coordinates 2. Establish (static) equilibrium position 3. Draw free-body-diagram of mass 4. Use FBD to apply Newton’s second law of motion: “Rate of change of momentum = applied force” F( t ) = As m is constant F( t ) = m For rotational motion d  dx( t )  m ÷ dt  dt  d 2 x( t ) dt 2 = mx && M ( t ) = J && θ For the free, undamped single DoF system 04:36:37 F( t ) = −kx = mx && or mx + kx = 0 && V. Rouillard © 2003 - 2013
46. 46. 46 Mechanical Vibrations – Single Degree-of-Freedom systems V. Rouillard © 2003 - 2013 Free undamped vibration single DoF Principle of virtual displacements: • “When a system in equilibrium under the influence of forces is given a virtual displacement. The total work done by the virtual forces = 0” • Displacement is imaginary, infinitesimal, instantaneous and compatible with the system • When a virtual displacement dx is applied, the sum of work done by the spring force and the inertia force are set to zero: −( kx )δ x − ( mx )δ x = 0 && • Since dx ≠ 0 the equation of motion is written as: 04:36:37 kx + mx = 0 &&
47. 47. 47 Mechanical Vibrations – Single Degree-of-Freedom systems Free undamped vibration single DoF Principle of conservation of energy: • • • No energy is lost due to friction or other energy-dissipating mechanisms. If no work is done by external forces, the system total energy = constant For mechanical vibratory systems: KE + PE = cons tan t or d ( KE + PE ) = 0 dt • Since 1 KE = mx 2 & 2 then and d 1 2 1 2 &  mx + kx ÷ = 0 dt  2 2  or mx + kx = 0 && 04:36:37 1 PE = kx 2 2 V. Rouillard © 2003 - 2013
48. 48. 48 Mechanical Vibrations – Single Degree-of-Freedom systems Free undamped vibration single DoF Vertical mass-spring system: 04:36:37 V. Rouillard © 2003 - 2013
49. 49. Mechanical Vibrations – Single Degree-of-Freedom systems 49 Free undamped vibration single DoF Vertical mass-spring system: mg • From the free body diagram:, using Newton’s second law of motion: mx = − k( x + δ st ) + mg && sin ce kδ st = mg mx + kx = 0 && • ∀ • Note that this is the same as the eqn. of motion for the horizontal mass-spring system ∴ if x is measured from the static equilibrium position, gravity (weight) can be ignored 04:36:37 be also derived by the other three alternative methods. This can V. Rouillard © 2003 - 2013
50. 50. 50 Mechanical Vibrations – Single Degree-of-Freedom systems V. Rouillard © 2003 - 2013 Free undamped vibration single DoF • • The solution to the differential eqn. of motion. As we anticipate oscillatory motion, we may propose a solution in the form: x( t ) = Acos( ωn t ) + B sin( ωn t ) or x( t ) = Aeiωnt + Be −iωnt alternatively,if we let s = ±iωn x( t ) = C e ± st • By substituting for x(t) in the eqn. of motion: C( ms 2 + k ) = 0 sin ce c ≠ 0, ms 2 + k = 0 and s = ±iωn = ± or 04:36:37 ωn = k m ¬ Characteristic equation k m ¬ roots = eigenvalues
51. 51. Mechanical Vibrations – Single Degree-of-Freedom systems 51 V. Rouillard © 2003 - 2013 Free undamped vibration single DoF • • The solution to the differential eqn. of motion. Applying the initial conditions to the general solution: x( t =0 ) = A = x0 & ( t =0 ) = Bωn = & 0 x x • The solution becomes: x( t ) = Acos( ωn t ) + B sin( ωn t ) initial displacement initial velocity x( t ) = x0 cos( ωn t ) + &o x sin( ωn t ) ωn  x 2 &  if we let A0 =  x0 +  0 ÷    ωn     x( t ) = A0 sin( ωn t + φ ) 1 2 2 • x ω  and φ = a tan  0 n ÷ then x  &o  This describes motion of harmonic oscillator: • • • Symmetric about equilibrium position Thru equilibrium: velocity is maximum & acceleration is zero At peaks and valleys, velocity is zero and acceleration is maximum ∀ ωn = √(k/m) is the natural frequency 04:36:37
52. 52. 52 Mechanical Vibrations – Single Degree-of-Freedom systems Free undamped vibration single DoF • Note: for vertical systems, the natural frequency can be written as: ωn = k m sin ce k = ωn = 04:36:37 g δ st mg δ st or fn = 1 g 2π δ st V. Rouillard © 2003 - 2013
53. 53. 53 Mechanical Vibrations – Single Degree-of-Freedom systems Free undamped vibration single DoF • • Torsional vibration. Approach same as for translational system. Laboratory exercise. 04:36:37 V. Rouillard © 2003 - 2013
54. 54. 54 Mechanical Vibrations – Single Degree-of-Freedom systems Free undamped vibration single DoF • • Compound pendulum. • Assume rigid body → single DoF Given an initial angular displacement or velocity, system will oscillate due to gravitational acceleration. Restoring torque: mgd sin θ ∴ Equation of motion : J o&& + mgd sin θ = 0 θ ¬ nonlinear2nd order ODE Linearity is approximated if sin θ ≈ θ Therefore : J o&& + mgdθ = 0 θ Natural frequency : ωn = 04:36:37 mgd Jo V. Rouillard © 2003 - 2013
55. 55. 55 Mechanical Vibrations – Single Degree-of-Freedom systems Free undamped vibration single DoF Natural frequency : mgd ωn = Jo sin ce for a simple pendulum g ωn = l J 2 Then, l = o and since J o = mko then md 2 ko gd ωn = and l = 2 d ko 2 2 Applying the parallel axis theorem ko = kG + d 2 2 kG l= +d d Let l = GA + d = OA ωn = g 2 ko / d = g g = l OA 2  kG  04:36:37 The location A  GA = ÷ is the " centre of percussion ′′  d ÷   V. Rouillard © 2003 - 2013
56. 56. 56 Mechanical Vibrations – Single Degree-of-Freedom systems Free undamped vibration single DoF • • Stability. Some systems may have inherent instability 04:36:37 V. Rouillard © 2003 - 2013
57. 57. 57 Mechanical Vibrations – Single Degree-of-Freedom systems Free undamped vibration single DoF • • • Stability. Some systems may have inherent instability When the bar is deflected by θ, The spring force is : 2kl sin θ The gravitational force thru G is : mg The inertial moment about O due to the angular acceleration && is : θ 2 ml && J o&& = θ θ 3 The eqn. of motion is written as : ml 2 && l θ + ( 2kl sin θ ) l cos θ − mg sin θ = 0 3 2 04:36:37 V. Rouillard © 2003 - 2013
58. 58. 58 Mechanical Vibrations – Single Degree-of-Freedom systems Free undamped vibration single DoF For small oscillations, sinθ = θ and cos θ = 1 .Therefore ml 2 mgl θ + 2kl 2θ − θ =0 3 2 or  12kl 2 − 3mgl  && +  θ ÷θ = 0 2  ÷ 2ml   The solution to the eqn. of motion depends of the sign of ( ) (1) If ( ) >0, the resulting motion is oscillatory (simple harmonic) with a natural frequency  12kl 2 − 3mgl  ωn =  ÷  ÷ 2ml 2   04:36:37 V. Rouillard © 2003 - 2013
59. 59. 59 Mechanical Vibrations – Single Degree-of-Freedom systems V. Rouillard © 2003 - 2013 Free undamped vibration single DoF 2   && +  12kl − 3mgl ÷θ = 0 θ  ÷ 2ml 2   (2) If ( ) =0, the eqn. of motion reduces to: && θ =0 The solution is obtained by int egrating twice yielding : θ ( t ) = C1t + C2 & & Applying initial conditions θ ( t = 0 ) = θ0 and θ ( t = 0 ) = θ0 & θ ( t ) = θ0 t + θ0 Which shows a linear increase of angular displ. at cons tan t velocity. & And if θ = 0 the bar remains in static equilibrium at θ ( t ) = θ 0 04:36:37 0
60. 60. 60 Mechanical Vibrations – Single Degree-of-Freedom systems Free undamped vibration single DoF  12kl 2 − 3mgl  && θ +  ÷θ = 0 2  ÷ 2ml   (3) If ( ) < 0, we define:  12kl 2 − 3mgl   3mgl − 12kl 2  α = − ÷=  ÷ 2 2 2ml 2ml     The solution of the eq.of motion is : θ ( t ) = B1eα t + B2 e −α t & & Applying initial conditions θ ( t = 0 ) = θ0 and θ ( t = 0 ) = θ0 1  & & ( αθ0 + θ0 ) eα t + ( αθ0 − θ0 ) e−αt   2α  which shows that θ ( t ) increases exp onentially with time and is therefore unstable because the restoring moment ( springs ) θ( t ) = is less than the non − restoring moment due to gravity. 04:36:37 V. Rouillard © 2003 - 2013
61. 61. 61 Mechanical Vibrations – Single Degree-of-Freedom systems V. Rouillard © 2003 - 2013 Free undamped vibration single DoF • • Rayleigh’s Energy method to determine natural frequency Recall: Principle of conservation of energy: T1 + U1 = T2 + U 2 • Where T1 and U1 represent the energy components at the time when the kinetic energy is at its maximum (∴ U1=0) and T2 and U2 the energy components at the time when the potential energy is at its maximum (∴ T2=0) T1 + 0 = 0 + U 2 • For harmonic motion Tmax = U max 04:36:37
62. 62. Mechanical Vibrations – Single Degree-of-Freedom systems 62 Free undamped vibration single DoF • Rayleigh’s Energy method to determine natural frequency: Application example: • Find minimum length of mercury u-tube manometer tube so that f n of fluid column < 2 Hz. • • Determine Umax and Tmax: Umax = potential energy of raised fluid column + potential energy of depressed fluid column. U = mg x x + mg 2 raised 2 depressed = ( Axγ ) x x + ( Axγ ) 2 raised 2 depressed = Aγ x 2 A : cross sec tional area and γ : specific weight of mercury • Kinetic energy: 04:36:37 1 T = ( mass of mercury col ) vel 2 2 1  Alγ  2 =  & x 2 g ÷  V. Rouillard © 2003 - 2013
63. 63. 63 Mechanical Vibrations – Single Degree-of-Freedom systems Free undamped vibration single DoF • Rayleigh’s Energy method to determine natural frequency: Application example: • If we assume harmonic motion: x( t ) = X cos( 2π f n t ) where X is the max . displacement & t ) = 2π f n X sin( 2π f n t ) where 2π f n X is the max . velocity x( • Substituting for the maximum displacement and velocity: U max = Aγ X 2 U max = Tmax fn = • and ∴ 1  2g   ÷ 2π  l  Minimum length of column: 04:36:37 1  Alγ  Tmax =  ( 2π f n ) 2 X 2 2 g ÷  1  Alγ  Aγ X 2 =  ( 2π f n ) 2 X 2 2 g ÷  1  2g   ÷ ≤ 1.5 Hz 2π  l  l ≥ 0.221 m fn = V. Rouillard © 2003 - 2013
64. 64. 64 Mechanical Vibrations – Single Degree-of-Freedom systems Free single DoF vibration + viscous damping • Recall: viscous damping force ∝ velocity: F = −cx & c = damping cons tan t or coefficient [ Ns / m ] Applying Newton' s sec ond law of motion to obtain the eqn.of motion : mx = − cx − kx && & or mx + cx + kx = 0 && & If the solution is assumed to take the form : x( t ) = Ce st where s = ±iωn then : & t ) = sCe st and &x( t ) = s 2Ce st x( & Substituting for x, & and && in the eqn.of motion x x ms 2 + cs + k = 0 The root of the characteristic eqn. are : 2 −c ± c 2 − 4mk c c  k s1,2 = =− ±   ÷ − ÷ 2m 2m 2m   m   The two solutions are : x1 ( t ) = C1e s1t 04:36:37 and x 2 ( t ) = C 2 e s 2t V. Rouillard © 2003 - 2013
65. 65. 65 Mechanical Vibrations – Single Degree-of-Freedom systems Free single DoF vibration + viscous damping • The general solution to the Eqn. Of motion is: x( t ) = C1e s1t + C2 e s2t or x( t ) = C1   2  c  c   k  − +  ÷ − ÷t 2m  2m   m     e + C2 where C1 and C2 are arbitrary cons tan ts det er min ed from the initial conditions . 04:36:37   2  c  c   k  −  − ÷ − ÷t 2m  2m   m     e V. Rouillard © 2003 - 2013
66. 66. Mechanical Vibrations – Single Degree-of-Freedom systems 66 V. Rouillard © 2003 - 2013 Free single DoF vibration + viscous damping • Critical damping (cc): value of c for which the radical in the general solution is zero: 2  cc   k   ÷ −  ÷= 0  2m   m  • k = 2mωn = 2 km m cc = 2m or Damping ratio (ζ): damping coefficient : critical damping coefficient. ζ = c cc c c cc = = ζω n 2m cc 2m or The roots can be re − written : 2 ( ) c  c  k 2 s1,2 = − ±  ÷ −  ÷ = −ζ ± ζ − 1 ω n 2m  2m   m  And the solution becomes :  −ζ + ζ 2 −1 ω t  ÷ n   x( t ) = C1e • + C2  −ζ − ζ 2 −1 ω t  ÷ n   e The response x(t) depends on the roots s 1 and s2 → the behaviour of the system is dependent on the damping ratio ζ. 04:36:37
67. 67. Mechanical Vibrations – Single Degree-of-Freedom systems 67 V. Rouillard © 2003 - 2013 Free single DoF vibration + viscous damping  −ζ + ζ 2 −1 ω t  ÷ n  x( t ) = C1e • + C2 2    −ζ − ζ −1 ÷ωn t  e When ζ <1, the system is underdamped. (ζ2-1) is negative and the roots can be written as: ( ) s1 = −ζ + i 1 − ζ 2 ω n and ( ) s2 = −ζ − i 1 − ζ 2 ω n And the solution becomes : x( t ) = C1 x( t ) = e  −ζ +i 1−ζ 2 ω t  ÷ n  e −ζω nt x( t ) = e −ζω nt x( t ) = e −ζω nt  i   C1e   + C2 1−ζ 2 ω nt ÷   −ζ −i 1−ζ 2 ω t  ÷ n  e + C2  −i 1−ζ 2 ω t   ÷ n   e    ( ) ( { { C cos ( 1 − ζ ω t ) + C sin ( 1 − ζ ω t ) } sin ( 1 − ζ ω t + φ ) or x( t ) = X e ( C1 + C2 ) cos x( t ) = Xe −ζω nt ' 1 1 − ζ 2 ω nt + i ( C1 − C2 ) sin 2 2 n n ' 2 2 1 − ζ 2ωnt )} n 0 −ζωn t cos ( 1 − ζ 2 ω n t − φo 04:36:37 Where C’1, C’2; X, φ and Xo, φo are arbitrary constant determined from initial conditions. )
68. 68. Mechanical Vibrations – Single Degree-of-Freedom systems 68 V. Rouillard © 2003 - 2013 Free single DoF vibration + viscous damping { ' x( t ) = e −ζωnt C1 cos • ( ) 1 − ζ 2 ωn t + C'2 sin ( 1 − ζ 2 ωn t )} For the initial conditions: x( t = 0 ) = x0 and & t = 0 ) = & 0 x( x Then ' C1 = x0 and C'2 = & 0 + ζωn x0 x 1 − ζ 2ωn Therefore the solution becomes x( t ) = e • −ζωn t    x0 cos   ( 2 ) 1 − ζ ωnt + x & 0 + ζωn x0 2 1 − ζ ωn sin ( )   1 − ζ ωnt    2 This represents a decaying (damped) harmonic motion with angular frequency √(1-ζ2)ωn also known as the damped natural frequency. The factor e -( ) causes the exponential decay. 04:36:37
69. 69. 69 Mechanical Vibrations – Single Degree-of-Freedom systems V. Rouillard © 2003 - 2013 Free single DoF vibration + viscous damping τd = 2π ωd Xe −ζωnt Exponentially decaying harmonic – free SDoF vibration with viscous damping . 04:36:37 Underdamped oscillatory motion and has important engineering applications.
70. 70. 70 Mechanical Vibrations – Single Degree-of-Freedom systems V. Rouillard © 2003 - 2013 Free single DoF vibration + viscous damping x( t ) = Xe −ζωnt sin ( 1 − ζ 2 ωn t + φ ) or x( t ) = X 0 e −ζωnt cos ( 1 − ζ 2 ωnt − φo The cons tan ts ( X ,φ ) and ( X 0 ,φ0 ) representing the magnitude and phase become : X = X0 = ( ) ( ) ' C1 '  C1  φ = a tan  ' ÷  C2  04:36:37 2 + C'2 and 2  C'2  φ0 = a tan  − ' ÷  C1  )
71. 71. Mechanical Vibrations – Single Degree-of-Freedom systems 71 V. Rouillard © 2003 - 2013 Free single DoF vibration + viscous damping When ζ = 1, c=cc , system is critically damped and the two roots to the eqn. of motion become: • s1 = s2 = − cc = −ωn 2m and solution is x( t ) = ( C1 + C2t )e −ωnt Applying the initial conditions x( t = 0 ) = x0 and & t = 0 ) = & 0 yields x( x C1 = x0 C2 = & 0 + ωn x0 x The solution becomes : x( t ) = [ x0 + ( & 0 + ωn x0 ) t ] e −ωnt x • As t→∞ , the exponential term diminished toward zero and depicts aperiodic motion 04:36:37
72. 72. 72 Mechanical Vibrations – Single Degree-of-Freedom systems V. Rouillard © 2003 - 2013 Free single DoF vibration + viscous damping • When ζ > 1, c>cc , system is overdamped and the two roots to the eqn. of motion are real and negative: ( = ( −ζ − ) −1 ) ω < 0 s1 = −ζ + ζ 2 − 1 ωn < 0 s2 ζ2 n with s2 = s1 and the initial conditions x( t = 0 ) = x0 and & t = 0 ) = & 0 x( x the solution becomes :  −ζ + ζ 2 −1  x( t ) = C1e where C1 = C2 = ω t ÷ n  ( + C2  −ζ − ζ 2 −1  e ) x0ωn −ζ + ζ 2 − 1 + & 0 x 2ωn ζ 2 − 1 ( ) − x0ωn −ζ − ζ 2 − 1 − & 0 x 2ωn ζ 2 − 1 04:36:37 Which shows aperiodic motion which diminishes exponentially with time. ω t ÷ n 
73. 73. 73 Mechanical Vibrations – Single Degree-of-Freedom systems V. Rouillard © 2003 - 2013 Free single DoF vibration + viscous damping Underdamped ( ζ = 0 ) Overdamped ( ζ > 1 ) Critically damped ( ζ = 1 ) Underdamped ( ζ < 1 ) 2π ωn 2π ωd Critically damped systems have lowest required damping for aperiodic motion and mass returns to equilibrium position in shortest possible time. 04:36:37
74. 74. Mechanical Vibrations – Single Degree-of-Freedom systems V. Rouillard © 2003 - 2013 Free single DoF vibration + viscous damping Example 1.2 1 0.8 -] 0.6 0.4 0.2 0 [ t n m e c a l p s i D 74 -0.2 0 0.5 1 1.5 2 -0.4 -0.6 -0.8 -1 04:36:37 Elapsed Time [s] 2.5 3 3.5
75. 75. 75 Mechanical Vibrations – Single Degree-of-Freedom systems V. Rouillard © 2003 - 2013 Free single DoF vibration + viscous damping • Logarithmic decrement: Natural logarithm of ratio of two successive peaks (or troughs) in an exponentially decaying harmonic response. • • • Represents the rate of decay Used to determine damping constant from experimental data. Using the solution for underdamped systems: x1 X 0 e −ζωnt1 cos( ωd t1 − φ0 ) = x2 X 0 e −ζωnt2 cos( ωd t2 − φ0 ) Let t2 = t1 + τ d = t1 + x1 x2 2π then ωd cos( ωd t2 − φ0 ) = cos( 2π + ωd t1 − φ0 ) = cos( ωd t1 − φ0 ) and x1 e −ζωnt1 = = eζωnτ d x2 e −ζωn ( t1 +τ d ) Applying the natural ln on both sides, the log arithmic decrement δ is obtained : x  2π 2πζ 2πζ δ = ln  1 ÷ = ζωnτ d = ζωn = = 04:36:37  x2  1 − ζ 2 ωn 1 − ζ 2 ωd τd t1 t2
76. 76. 76 Mechanical Vibrations – Single Degree-of-Freedom systems V. Rouillard © 2003 - 2013 Free single DoF vibration + viscous damping • Logarithmic decrement: For low damping ( ζ << 1 ) 14 x  δ = ln  1 ÷ = 2πζ  x2  Valid for ζ < .3 12 10 δ 8 6 4 2 0 0 0.2 0.4 0.6 ζ 04:36:37 0.8 1
77. 77. 77 Mechanical Vibrations – Single Degree-of-Freedom systems V. Rouillard © 2003 - 2013 Free single DoF vibration + viscous damping • Logarithmic decrement after n cycles: x1 • Since the period of oscillation is constant: x x x x = 1 2 3 .... m xm +1 x2 x3 x4 xm +1 xj Since = eζω nτ d then x j +1 x1 x1 xm +1 ( ) ζω nτ d m = e = emζω nτ d The log arithmic decrement can therefore be obtained from a number m of successive decaying oscillations δ= 1  x1  ln  m  xm +1 ÷  04:36:37 x2 Xm+1
78. 78. 78 Mechanical Vibrations – Single Degree-of-Freedom systems V. Rouillard © 2003 - 2013 Free single DoF vibration + Coulomb damping • • • Coulomb or dry friction dampers are simple and convenient Occurs when components slide / rub Force proportional to normal force: F = µN F = µ mg for free − s tan ding systems where µ is the coefficient of friction. • • Force acts in opposite direction to velocity and is independent of displacement and velocity. Consider SDOF system with dry friction: 04:36:37 Case 1. Case 2.
79. 79. 79 Mechanical Vibrations – Single Degree-of-Freedom systems V. Rouillard © 2003 - 2013 Free single DoF vibration + Coulomb damping • • Case 1: Mass moves from left to right. x = positive and x’ is positive or x = negative and x’ is positive. The eqn. of motion is: ¬ 2nd order hom ogeneous DE mx = −kx − µ N or && mx + kx = − µ N && For which the general solution is : x( t ) = A1 cos( ωn t ) + A2 sin( ωn t ) − where the fre quency of vibration ωn is conditions of this portion of the cycle. • • µN k (1) k and A1 and A2 are constants dependent on the initial m Case 2: Mass moves from right to left. x = positive and x’ is negative or x = negative and x’ is negative. The eqn. of motion is: mx = − kx + µ N or && mx + kx = µ N && For which the general solution is : µN x( t ) = A3 cos( ωn t ) + A4 sin( ωn t ) + k (2) k and A3 and A4 are constants dependent m 04:36:37 on the initial conditions of this portion of the cycle. where the fre quency of vibration ωn is again
80. 80. 80 Mechanical Vibrations – Single Degree-of-Freedom systems V. Rouillard © 2003 - 2013 Free single DoF vibration + Coulomb damping • The term µN/k [m] is a constant representing the virtual displacement of the spring k under force µN. The equilibrium position oscillates between +µN/k and -µN/k 1 for each harmonic half cycle of motion. 04:36:37
81. 81. 81 Mechanical Vibrations – Single Degree-of-Freedom systems Free single DoF vibration + Coulomb damping • To find a more specific solution to the eqn. of motion we apply the simple initial conditions: x( t = 0 ) = x0 & t =0)=&0 x( x The motion starts from the extreme right ( ie. velocity is zero ) Substituting int o µN x( t ) = A3 cos( ωn t ) + A4 sin( ωn t ) + (2) k and & t ) = − A3ωn sin( ωn t ) + A4ωn cos( ωn t ) + 0 x( gives µN A3 = x0 − and A4 = 0 k Eqn.( 2 ) becomes µN  µN x( t ) =  x0 − cos( ωn t ) + ( 2a ) valid for 0 ≤ t ≤ π / ωn  ÷ k  k  04:36:37 and V. Rouillard © 2003 - 2013
82. 82. 82 Mechanical Vibrations – Single Degree-of-Freedom systems V. Rouillard © 2003 - 2013 Free single DoF vibration + Coulomb damping µN  µN x( t ) =  x0 −  ÷cos( ωn t ) + k  k  valid for 0 ≤ t ≤ π / ωn = Initial displacement for next half cycle 04:36:37
83. 83. 83 Mechanical Vibrations – Single Degree-of-Freedom systems V. Rouillard © 2003 - 2013 Free single DoF vibration + Coulomb damping • The displacement at π/ωn becomes the initial displacement for the next half cycle, x 1.  π   µN  µN 2µ N  − x1 = x  t = =  x0 − cos( π ) + = −  x0 − ÷  ÷ ÷ k  k k    ωn    π  and the initial velocity & ( t = 0 ) is = &  t = x x ÷ in eqn ( 2a )  ωn  Substituting these initial conditions int o eqn.( 1) µN x( t ) = A1 cos( ωn t ) + A2 sin( ωn t ) − ( 1) k and its derivative & t ) = −ωn A1 sin( ωn t ) + ωn A2 cos( ωn t ) x( gives 3µ N and k such that eqn.( 1 ) becomes : A1 = x0 − A2 = 0 3µ N  µN x( t ) =  x0 − cos( ωn t ) −  ÷ k  k  04:36:37 ( 1a ) valid for π / ωn ≤ t ≤ π 2 / ωn
84. 84. 84 Mechanical Vibrations – Single Degree-of-Freedom systems V. Rouillard © 2003 - 2013 Free single DoF vibration + Coulomb damping µN  µN x( t ) =  x0 − valid for 0 ≤ t ≤ π / ωn  ÷cos( ωn t ) + k  k  3µ N  µN x( t ) =  x0 − valid for π / ωn ≤ t ≤ π 2 / ωn  ÷cos( ωn t ) − k  k  = Initial displacement for next half cycle 04:36:37 This method can be applied to successive half cycles until the motion stops.
85. 85. Mechanical Vibrations – Single Degree-of-Freedom systems 85 Free single DoF vibration + Coulomb damping During each half period π/ωn the reduction in magnitude (peak height) is 2µN/k • • Any two succesive peaks are related by: 4µ N  xm = xm −1 −   ÷  k  • • The motion will stop when xn < µN/k The total number of half vibration cycles, r, is obtained from: 2µ N   µ N  x0 − r   ÷≤  ÷  k   k  or  µN  x0 −    k  r≥   2µ N    ÷  k    04:36:37 V. Rouillard © 2003 - 2013
86. 86. 86 Mechanical Vibrations – Single Degree-of-Freedom systems V. Rouillard © 2003 - 2013 Free single DoF vibration + Coulomb damping µN  µN x( t ) =  x0 − valid for 0 ≤ t ≤ π / ωn  ÷cos( ωn t ) + k  k  3µ N  µN x( t ) =  x0 − valid for π / ωn ≤ t ≤ π 2 / ωn  ÷cos( ωn t ) − k  k  Final position = Initial displacement for next half cycle 04:36:37
87. 87. 87 Mechanical Vibrations – Single Degree-of-Freedom systems V. Rouillard © 2003 - 2013 Free single DoF vibration + Coulomb damping • Important features of Coulomb damping: 1. The equation of motion is nonlinear (cf. linear for viscous damping) 2. Coulomb damping does not alter the system’s natural frequency (cf. damped natural frequency for viscous damping). 3. The motion is always periodic (cf. overdamped for viscous systems) 4. Amplitude reduces linearly (cf. exponential decay for viscous systems) 5. System eventually comes to rest – number of vibration cycles finite (cf. sustained vibration with viscous damping) 6. The final position is the permanent displacement (not equilibrium) equivalent to the friction force (cf. approaches zero for viscous systems) 04:36:37
88. 88. Mechanical Vibrations – Single Degree-of-Freedom systems 88 V. Rouillard © 2003 - 2013 Forced (harmonically excited) single DoF vibration • • External energy supplied to system as applied force or imposed motion (displacement, velocity or acceleration) • Harmonic forcing function takes the form: This section deals only with harmonic excitation which results in harmonic response (cf. steady-state or transient response from non-harmonic excitation). F( t ) = F0 ei ( ω t +φ ) • • • • or F( t ) = F0 cos( ωt + φ ) or F( t ) = F0 sin( ωt + φ ) Where F0 is the amplitude, ω the frequency and φ the phase angle. The response of a linear system subjected to harmonic excitation is also harmonic. The response amplitude depends on the ratio of the excitation frequency to the natural frequency. Some “common” harmonic forcing functions are: • • Rotating machine / element with (large) residual imbalance • • Vehicle travelling on pavement corrugations or sinusoidal surfaces Regular shedding of vortices caused by laminar flow across slender structures (VIV) – ie: chimneys, bridges, overhead cables, mooring cables, tethers, pylons… Structures excited by regular (very narrow banded) ocean / water waves 04:36:37
89. 89. Mechanical Vibrations – Single Degree-of-Freedom systems 89 V. Rouillard © 2003 - 2013 Forced (harmonically excited) single DoF vibration • Equation of motion when a force is applied to a viscously damped SDOF system is: mx + cx + kx = F ( t ) && & • ¬ non hom ogeneous differential eqn. The general solution to a nonhomogeneous DE is the sum if the homogeneous solution x h(t) and the particular solution xp(t). • The homogeneous solution represents the solution to the free SDOF which is known to decay over time for all conditions (underdamped, critically damped and overdamped). • The general solution therefore reduces to the particular solution x p(t) which represents the steady-state vibration which exists as long as the forcing function is applied. 04:36:37
90. 90. Mechanical Vibrations – Single Degree-of-Freedom systems 90 V. Rouillard © 2003 - 2013 Forced (harmonically excited) damped single DoF vibration • Example of solution to harmonically excited damped SDOF system: Homogenous solution: decaying vibration @ natural frequency Particular solution: steady-state vibration @ excitation frequency Complete solution 04:36:37
91. 91. Mechanical Vibrations – Single Degree-of-Freedom systems 91 V. Rouillard © 2003 - 2013 Forced (harmonically excited) single DoF vibration – undamped. • Let the forcing function acting on the mass of an undamped SDOF system be: F( t ) = F0 cos( ωt ) • The eqn. of motion reduces to: mx + kx = F0 cos( ωt ) && • Where the homogeneous solution is: xh ( t ) = C1 cos( ωn t ) + C2 sin( ωn t ) where ωn = k / m • As the excitation is harmonic, the particular solution is also harmonic with the same frequency: x p ( t ) = X cos( ωt ) • Substituting xp(t) in the eqn. of motion and solving for X gives: X= • F0 k − mω 2 The complete solution becomes 04:36:37 x( t ) = xh ( t ) + x p ( t ) = C1 cos( ωnt ) + C 2 sin( ωnt ) + F0 k − mω 2 cos( ωt )
92. 92. Mechanical Vibrations – Single Degree-of-Freedom systems 92 V. Rouillard © 2003 - 2013 Forced (harmonically excited) single DoF vibration – undamped. • Applying the initial conditions x( t = 0 ) = x0 C1 = x0 − • and F0 & t =0)=&0 x( x and k − mω 2 C2 = gives: &0 x ωn The complete solution becomes: &  F0  x F0  x( t ) =  x0 − cos( ωnt ) +  0 ÷sin( ωn t ) + cos( ωt ) ÷ 2 2 ωn   k − mω  k − mω  • The maximum amplitude of the steady-state solution can be written as: X = δ st • 1 2 ω  1−  ÷  ωn  F where δ st = 0 k X/δst is the ratio of the dynamic to the static amplitude and is known as the amplification factor or amplification ratio and is dependent on the frequency ratio r = ω/ωn. 04:36:37
93. 93. Mechanical Vibrations – Single Degree-of-Freedom systems 93 V. Rouillard © 2003 - 2013 Forced (harmonically excited) single DoF vibration – undamped. • When ω/ωn < 1 the denominator of the steady- X / δ st state amplitude is positive and the amplification factor increases as ω approaches the natural frequency ωn. The response is in-phase with the excitation. • When ω/ωn > 1 the denominator of the steadystate amplitude is negative an the amplification factor is redefined as: X 1 = δ st  ω  2 ω ÷ −1  n and the steady − state response becomes : x p ( t ) = − X cos( ωt ) which shows that the response is out-of-phase with the excitation and decreases (→ zero ) as ω increases (→ ∞) 04:36:37 r= ω ωn
94. 94. Mechanical Vibrations – Single Degree-of-Freedom systems 94 Forced (harmonically excited) single DoF vibration – undamped. • When ω/ωn < 1 the denominator of the steadystate amplitude is positive and the amplification factor increases as ω approaches the natural frequency ωn. The response is in-phase with the excitation. • When ω/ωn > 1 the denominator of the steadystate amplitude is negative an the amplification factor is redefined as: X 1 = δ st  ω  2 ω ÷ −1  n and the steady − state response becomes : x p ( t ) = − X cos( ωt ) which shows that the response is out-of-phase with the excitation and decreases (→ zero ) as ω increases (→ ∞) 04:36:37 X / δ st V. Rouillard © 2003 - 2013
95. 95. Mechanical Vibrations – Single Degree-of-Freedom systems 95 Forced (harmonically excited) single DoF vibration – undamped. • When ω/ωn = 1 the denominator of the steadystate amplitude is zero an the response becomes infinitely large. This condition when ω=ωn is known as resonance. X / δ st 04:36:37 V. Rouillard © 2003 - 2013
96. 96. Mechanical Vibrations – Single Degree-of-Freedom systems 96 V. Rouillard © 2003 - 2013 Forced (harmonically excited) single DoF vibration – undamped. • The complete solution &  F0  x F0  x( t ) =  x0 − cos( ωn t ) +  0 ÷sin( ωn t ) + cos( ωt ) 2÷ 2  k − mω  k − mω  ωn  can be written as: x( t ) = Acos( ωn t + φ ) + x( t ) = Acos( ωn t + φ ) − δ st 2 cos( ωt ) for ω / ωn < 1 2 cos( ωt ) for ω / ωn > 1 ω  1−  ÷  ωn  δ st ω  1−  ÷  ωn  where A and φ are functions of x0 and & 0 as before. x • The complete solution is a sum of two cosines with frequencies corresponding to the natural and forcing (excitation) frequencies. 04:36:37
97. 97. 97 Mechanical Vibrations – Single Degree-of-Freedom systems Forced (harmonically excited) single DoF vibration – undamped. ω /ωn < 1 ω /ωn > 1 04:36:37 V. Rouillard © 2003 - 2013
98. 98. Mechanical Vibrations – Single Degree-of-Freedom systems 98 V. Rouillard © 2003 - 2013 Forced (harmonically excited) single DoF vibration – undamped. • When the excitation frequency ω is close but not exactly equal to the natural frequency ωn beating may occur. • Letting the initial conditions x0= x’0 =0 , the complete solution: &  F0  x F0  x( t ) =  x0 − cos( ωn t ) +  0 ÷sin( ωnt ) + cos( ωt ) 2÷ 2 ωn   k − mω  k − mω  reduces to : x( t ) = ( F0 / m ) ( ωn2 − ω 2 ) [ c os( ωnt ) − cos( ωt )] = ( F0 / m )    ω + ωn     ω − ωn    2 sin  ÷t  ×sin   ÷t   2 2  ( ωn − ω )   2   If we let the excitation frequency be slightly less than the natural frequency: ωn − ω = 2ε where ε is a small positive number. Then ωn ≈ ω and ωn + ω = 2ω therefore : 2 ( ω n − ω ) ( ωn + ω ) = ω n − ω 2 = 4 ε ω 2 Substituting for ωn − ω , ωn + ω and ωn − ω 2 in the complete solution yields : 04:36:37 x( t ) = ( F0 / m ) sin ( ε t ) ×sin ( ωt ) ( 2εω )  2  
99. 99. Mechanical Vibrations – Single Degree-of-Freedom systems 99 V. Rouillard © 2003 - 2013 Forced (harmonically excited) single DoF vibration – undamped. x( t ) = • ( F0 / m ) sin ε t ×sin ωt ( ) ( ) ( 2εω ) Since ε is small, sin(ε t) has a long period. The solution can then be considered as harmonic motion with a principal frequency ω an a variable amplitude equal to X(t ) = 04:36:37 ( F0 / m ) sin ε t ( ) ( 2εω )
100. 100. 100 Mechanical Vibrations – Single Degree-of-Freedom systems V. Rouillard © 2003 - 2013 Forced (harmonically excited) single DoF vibration – Damped. • • Steady-state Solution • The equation of motion of a SDOF system with viscous damping is: If the forcing function is harmonic: F( t ) = F0 cos( ωt ) mx + cx + kx = F0 cos( ωt ) && & • The steady-state response is given by the particular solution which is also expected to be harmonic: x p ( t ) = X cos( ωt − φ ) where the amplitude X and the phase angle φ are to be det er min ed 04:36:37
101. 101. 101 Mechanical Vibrations – Single Degree-of-Freedom systems Forced (harmonically excited) single DoF vibration – Damped. • Substituting xp into the steady-state eqn. of motion yields: ( ) X  k − mω 2 cos( ωt − φ ) − cω sin( ωt − φ ) = F0 cos( ωt )   applying the trigonometric relationships : cos( ωt − φ ) = cos( ωt )cos( φ ) + sin( ωt ) sin( φ ) sin( ωt − φ ) = sin( ωt )cos( φ ) − cos( ωt ) sin( φ ) we obtain : ( ) X ( k − mω 2 ) sin( φ ) − cω cos( φ ) = 0   X  k − mω 2 cos( φ ) + cω sin( φ ) = F0   which gives : X= F0 ( )  2 k − mω − ( cω )     for the particular solution 2 2 x p ( t ) = X cos( ωt − φ ) 04:36:37 1 and 2  cω  φ = a tan  ÷  k − mω 2  V. Rouillard © 2003 - 2013
102. 102. 102 Mechanical Vibrations – Single Degree-of-Freedom systems V. Rouillard © 2003 - 2013 Forced (harmonically excited) single DoF vibration – Damped. • Alternatively, the amplitude and phase can be written in terms of the frequency ratio r = ω/ωn and the damping coefficient ζ: X = δ st 1  2 ω   ω    1 −  ÷  +  2ζ   ωn    ωn        2 2  ω   2ζ ÷ ωn ÷ φ = a tan  = 2÷  1−  ω  ÷  ω ÷ ÷   n  04:36:37 1 = 2  2ζ r  a tan  ÷  1 − r2  1  2  1 − r  + [ 2ζ r ]    2 2 1 2
103. 103. 103 Mechanical Vibrations – Single Degree-of-Freedom systems Forced (harmonically excited) single DoF vibration – Damped. X = δ st 04:36:37 1 1  2 2 2 2  1 − r  + [ 2ζ r ]      2ζ r  φ = a tan  ÷  1 − r2  V. Rouillard © 2003 - 2013
104. 104. 104 Mechanical Vibrations – Single Degree-of-Freedom systems V. Rouillard © 2003 - 2013 Forced (harmonically excited) single DoF vibration – Damped. • The magnification ratio at all frequencies is reduced with increased damping. • The effect of damping on the magnification ratio is greatest at or near resonance. • The magnification ratio approaches 1 as the frequency ratio approaches 0 (DC) • The magnification ratio approaches 0 as the frequency ratio approaches ∞ • For 0 < ζ < 1/ √2 the magnification ratio maximum occurs at r = √(1 - 2ζ2) or ω = ωn √(1 - 2ζ2) which is lower than both the undamped natural frequency ωn and the damped natural frequency ωd = ω n √(1 - ζ2) • When r = √(1 - 2ζ2) Mmax= 1/[2ζ √(1 - ζ2)] → if Mmax can be measured, the damping ratio can be determined. • • 04:36:37 When ζ = 1/√2 dM/dr = 0 at r = 0. When ζ > 1/√2 M decreases monotonically with increasing frequency.
105. 105. 105 Mechanical Vibrations – Single Degree-of-Freedom systems V. Rouillard © 2003 - 2013 Forced (harmonically excited) single DoF vibration – Damped. • • For damped systems (ζ > 0) when r < 1 the phase angle is less than 90o and response lags the excitation and when r >1 the phase angle is greater than 90 o and the response leads the excitation (approaches 180 o for large frequency ratios.. • 04:36:37 For undamped systems (ζ = 0) the phase angle is 0o (response in phase with excitation) for r<1 and 180 o (response out of phase with excitation) for r>1. For damped systems (ζ > 0) when r =1 the phase lag is always 90o.
106. 106. 106 Mechanical Vibrations – Single Degree-of-Freedom systems V. Rouillard © 2003 - 2013 Forced (harmonically excited) single DoF vibration – Damped. • Complete Solution • The complete solution is the sum of the homogeneous solution xh(t) and the particular solution xp(t): x( t ) = X 0 e −ζωnt cos( ωd t − φ0 ) + X cos( ωt − φ ) where ωd = ωn 1 − ζ 2 , X and φ are given as before, and X 0 and φ0 are det er min ed from the initial conditions 04:36:37
107. 107. Mechanical Vibrations – Single Degree-of-Freedom systems 107 V. Rouillard © 2003 - 2013 Forced (harmonically excited) single DoF vibration – Damped. • Quality Factor & Bandwidth • • When damping is small (ζ < 0.05) the peak magnification ratio corresponds with resonance ( ω =ωn). The value of the magnification ratio (Quality factor or Q factor) becomes:  X  1 1 Q= = = ÷ 1 2ζ  δ st ω =ωn  2  2 2 2  ω    ω   1 −  ÷  +  2ζ     ωn    ωn       • The points where the magnification ratio falls below Q/√2, are called the half power points R 1 and R2. (Power is proportional to amplitude squared: Power = Fv = cv2 = c(dx/dt)2 • The difference between the half power frequencies is called the bandwidth. 1 2ζ Q 2 The Quality factor Q can be used to estimate the equivalent viscous damping of systems. • Q= 04:36:37 Bandwidth R1 1 R2
108. 108. 108 Mechanical Vibrations – Single Degree-of-Freedom systems V. Rouillard © 2003 - 2013 Forced (harmonically excited) single DoF vibration – Damped. • The values of the half power frequencies are determined as follows:  X  1 Q 1 = = = δ ÷ 2 2 2 2ζ  st  2   ω 2   ω 1 −  ÷  +  2ζ    ωn    ωn    In terms of the frequency ratio r : r 4 − r 2 ( 2 − 4ζ 2 ) + ( 1 − 8ζ 2 ) = 0 Which, when solved gives : r12 = 1 − 2ζ 2 − 2ζ 1 + ζ 2 and 2 r2 = 1 − 2ζ 2 + 2ζ 1 + ζ 2 When ζ is small , ζ 2 is negligible and the solutions can be reduced to : 2 2 ω  r12 = R1 =  1 ÷ ; 1 − 2ζ  ωn  ∴ ( ) 2 and 2 2 2 2 2 2 ω2 − ω1 = R2 − R1 ωn ; 4ζωn 04:36:37 2 2 ω  r2 = R2 =  2 ÷ ; 1 + 2ζ  ωn 
109. 109. 109 Mechanical Vibrations – Single Degree-of-Freedom systems V. Rouillard © 2003 - 2013 Forced (harmonically excited) single DoF vibration – Damped. ω2 + ω1 2 2 = ωn and ω2 − ω1 = ( ω2 + ω1 ) ( ω2 − ω1 ) , 2 the bandwidth ∆ω = ω2 − ω1 can be written as : Since 2 2 2 ω2 − ω1 4ζωn ∆ω = ; ; 2ζωn ω2 + ω1 2ωn The qualily factor Q can then be exp ressed in terms of the natural frequency and bandwidth : Q; ω 1 ; n 2ζ ∆ω 04:36:37
110. 110. 110 Mechanical Vibrations – Single Degree-of-Freedom systems V. Rouillard © 2003 - 2013 Forced (harmonically excited) single DoF vibration – Damped. • • Complex notation. Recall that a harmonic function may expressed as follows: F( t ) = F0 cos( ωt + φ ) • • • = F0 sin( ωt + φ ) F0 ei( ωt +φ ) If the harmonic forcing function is expressed in complex form: F = F0 eiωt The equation of motion for a damped SDOF system becomes: mx + cx + kx = F0 eiωt && & The actual excitation function is real and is represented by the real part of the complex function. Consequently, the steady-state response is also real and is represented by the real part of the complex particular solution which takes the form: x p ( t ) = Xeiωt Therefore : & p ( t ) = iω Xeiωt x • = and && p ( t ) = −ω Xeiωt x Substituting in the eqn. of motion gives: − mω 2 Xeiωt + icω Xeiωt + kXeiωt = F0 eiωt 04:36:37
111. 111. 111 Mechanical Vibrations – Single Degree-of-Freedom systems V. Rouillard © 2003 - 2013 Forced (harmonically excited) single DoF vibration – Damped. • The response amplitude becomes: X= F0 ¬ X / F0 is called the RECEPTANCE ( Dynamic compliance )  k − mω 2 + icω    multiplying the numerator & deno min ator on the RHS by k − mω 2 − icω and separating real and imaginary components : ( ) ( )   k − mω 2 cω  X = F0  −i 2 2   k − mω 2 + c 2ω 2 k − mω 2 + c 2ω 2     ( ) ( ) e −iφ where y applying the complex relationships : x + iy = Aeiφ where A = x 2 + y 2 and φ = a tan    ÷ x The magnitude of the response can be written as : X= F0 ( 1 )   2 k − mω + c 2ω 2     And the steady − state solution becomes : xp( t ) = 04:36:37 2 2 F0 (   k − mω  ) 2 2  + c 2ω 2   1 ei( ωt −φ ) 2  cω  φ = a tan  ÷  k − mω 2 
112. 112. 112 Mechanical Vibrations – Single Degree-of-Freedom systems V. Rouillard © 2003 - 2013 Forced (harmonically excited) single DoF vibration – Damped. • As before the response amplitude: X= ( F0 )  k − mω 2 + icω    can be written in terms of the frequency ratio r and the damping ratio ζ : kX 1 = ≡ H( iω ) ¬ Complex Frequency Re sponse Function ( FRF ) F0 1 − r 2 + i2ζ r The magnitude of H( iω ) is given by : H( iω ) = kX = F0 1 ( 1− r ) 2 2 which is the same as the magnification ratio M : + ( 2ζ r ) 2 It can be shown that the complex FRF and its magnitude are related by : H( iω ) = H( iω ) e −iφ where e −iφ = cos φ + i sin φ and  2ζ r  φ = a tan  ÷  1 − r2  The steady − state response can therefore be exp ressed as : F x p ( t ) = 0 H( iω ) ei( ωt −φ ) k • Measurements of the magnitude FRF can be used to experimentally determine the values of m, c and k. 04:36:37
113. 113. 113 Mechanical Vibrations – Single Degree-of-Freedom systems V. Rouillard © 2003 - 2013 Forced (harmonically excited) single DoF vibration – Damped. • • When the excitation function is described by: F( t ) = F0 cos( ωt ) The steady-state response is given by the real part of the solution: xp( t ) = F0 ( ) 1 ei( ωt −φ ) = 2   2 k − mω 2 + c 2ω 2     F = Re  0 H( iω )eiωt  k    F = Re  0 H( iω ) ei( ωt −φ )  k    • • F0 (  k − mω 2   Conversely, when the excitation function is described by: ) 2  + c 2ω 2   1 cos( ωt − φ ) 2 F( t ) = F0 sin( ωt ) The steady-state response is given by the imaginary part of the solution: xp( t ) = F0 ( ) 1 2   2 k − mω 2 + c 2ω 2     F = Im  0 H( iω )eiωt  k     F0 H( iω ) ei( ωt −φ )  = 04:36:37 Im k    ei( ωt −φ ) = F0 (  k − mω 2   ) 2  + c 2ω 2   1 sin( ωt − φ ) 2
114. 114. 114 Mechanical Vibrations – Single Degree-of-Freedom systems Forced (harmonically excited) single DoF vibration – Damped. • • Complex Vector Notation of Harmonic Motion: Harmonic excitation and response can be represented in the complex plane Steady − state displacement : F x p ( t ) = 0 H( iω ) ei( ω t −φ ) k Steady − state velocity : F & p ( t ) = iω 0 H( iω ) ei( ω t −φ ) = iω x p ( t ) x k Steady − state acceleration : F0 H( iω ) ei( ω t −φ ) = −ω 2 x p ( t ) k Since i and − 1 respectively can be written as : 2 x && p ( t ) = ( iω ) π π  π  i2 i = cos  ÷+ i sin  ÷ = e 2 2 • and − 1 = cos ( π ) + i sin ( π ) = eiπ It can be seen that: • • The velocity leads the displacement by 90 o and is multiplied by ω. The acceleration leads the displacement by 180 o and is multiplied by ω2. 04:36:37 V. Rouillard © 2003 - 2013
115. 115. 115 Mechanical Vibrations – Single Degree-of-Freedom systems V. Rouillard © 2003 - 2013 Forced (harmonically excited) single DoF vibration – Damped. • Complex Vector Notation of Harmonic Motion: xp( t ) = 04:36:37 F0 H( iω ) ei( ω t −φ ) k x & p ( t ) = iω x p ( t ) x && p ( t ) = −ω 2 x p ( t )
116. 116. Mechanical Vibrations – Single Degree-of-Freedom systems 116 V. Rouillard © 2003 - 2013 Forced (harmonically excited) single DoF vibration – Damped. • • • Response due to base motion (harmonic) • At any time, the length of the spring is x – y and the relative velocity between the two ends of the damper is x’ – y’. • The equation of motion is: In this case, the excitation is provided by the imposed harmonic motion of the supporting base. The displacement of the base about a neutral position is denoted by y(t) and the response of the mass from its static equilibrium position by x(t). mx + c( & − & ) + k( x − y ) = 0 && x y If y( t ) = Y sin( ωt ) the eqn.of motion becomes : mx + cx + kx = cy + ky && & & k( x − y ) = cωY cos( ωt ) + kY sin( ωt ) = A sin( ωt − α ) cω where A = Y k 2 + ( cω )2 and α = a tan  −   ÷  k  • The applied displacement has the same effect of applying a harmonic force of magnitude A to the mass. 04:36:37 c( & − &y ) x y( t ) = Y sin( ω t )
117. 117. 117 Mechanical Vibrations – Single Degree-of-Freedom systems V. Rouillard © 2003 - 2013 Forced (harmonically excited) single DoF vibration – Damped. • The steady-state response of the mass is given by the particular solution x p(t): xp( t ) = Y k 2 + ( cω )2 ( 1 ) sin( ωt − φ1 − α ) 2   2 k − mω 2 + c 2ω 2     cω  cω  where α = a tan  −  and φ1 = a tan   ÷ ÷  k   k − mω 2  The solution can be simplified to : x p ( t ) = X sin( ωt − φ ) where   2 2 X  k + ( cω )  = Y 2 2 2 2 +c ω   k − mω   ( ) 1 2   2 1 + ( 2ζ r )  =  2 2 2 + ( 2ζ r )   1− r   ( ) 1 2 ¬ Displacement Transmissibility and     mcω 3 2ζ r 3  ÷ = a tan  φ = a tan ÷  k k − mω 2 + ( cω )2 ÷ 1 + ( 4ζ 2 − 1)r 2     04:36:37 ( )
118. 118. 118 Mechanical Vibrations – Single Degree-of-Freedom systems Forced (harmonically excited) single DoF vibration – Damped.   2 X  1 + ( 2ζ r )  = Y 2 2 2 + ( 2ζ r )   1− r   ( 04:36:37 ) 1 2 and   2ζ r 3 φ = a tan  ÷ 1 + ( 4ζ 2 − 1 )r 2   V. Rouillard © 2003 - 2013
119. 119. 119 Mechanical Vibrations – Single Degree-of-Freedom systems Forced (harmonically excited) single DoF vibration – Damped. • Characteristics of the displacement transmissibility: • The transmissibility is 1 when r = 0 (DC) and close to 1 when r is small. • For undamped systems (ζ = 0), Td → ∞ at resonance (r = 1) • For all damping values Td<1 for r >√2 and Td = 1 for r = √2 • For r <√2 Td is inversely proportional to ζ • For r >√2 Td is proportional to ζ 04:36:37 V. Rouillard © 2003 - 2013
120. 120. Mechanical Vibrations – Single Degree-of-Freedom systems 120 V. Rouillard © 2003 - 2013 Forced (harmonically excited) single DoF vibration – Damped. • • Transmitted Force The force transmitted to the base/support is caused by the reaction of the spring and damper: F = k( x − y ) + c( & − & ) = − mx x y && Since the steady − state ( particular ) solution is x p ( t ) = X sin( ωt − φ ) ,F can be written as : F = mω 2 X sin( ω t − φ ) = FT sin( ωt − φ ) • Where FT is the amplitude of the transmitted force and is given by: 1 2  2 FT 1 + ( 2ζ r ) 2 =r  ¬ Force Transmissibility 2 2 2 kY  ( 1 − r ) + ( 2ζ r )  • Note that the transmitted force is always in–phase with the motion of the mass x(t): k( x − y ) c( & − &y ) x y( t ) = Y sin( ω t ) 04:36:37
121. 121. 121 Mechanical Vibrations – Single Degree-of-Freedom systems V. Rouillard © 2003 - 2013 Forced (harmonically excited) single DoF vibration – Damped.  1 + ( 2ζ r )  FT = r2   kY ( 1 − r 2 ) + ( 2ζ r )2   2 04:36:37 1 2
122. 122. 122 Mechanical Vibrations – Single Degree-of-Freedom systems V. Rouillard © 2003 - 2013 Forced (harmonically excited) single DoF vibration – Damped. • • Relative Motion If z = x – y represents the motion of the mass relative to the base, the eqn. of motion: k( x − y ) mx + c( & − & ) + k( x − y ) = 0 && x y can be written as : mz + cz + kz = − my = mω 2Y sin( ωt ) && && & The ( steady − state ) solution of which is : z( t ) = mω 2Y sin( ωt − φ1 ) ( )  2 2 2 k − mω + ( cω )     where the amplitude Z is given by : Z= mω 2Y ( )  2 k − mω + ( cω )     and the phase φ1 is given by : 2 2 1 1 2 = Z sin( ωt − φ1 ) =Y 2 c( & − &y ) x r2 (   1− r   cω   2ζ r  φ1 = a tan  = a tan  ÷ ÷  1 − r2   k − mω 2  04:36:37 ) 2 2 2 + ( 2ζ r )   1 2
123. 123. 123 Mechanical Vibrations – Single Degree-of-Freedom systems Forced (harmonically excited) single DoF vibration – Damped. • Relative Motion Z = Y r2 (   1− r  ) 2 + ( 2ζ r )    2ζ r  φ1 = a tan  ÷  1 − r2  04:36:37 2 2 1 2 V. Rouillard © 2003 - 2013
124. 124. 124 Mechanical Vibrations – Single Degree-of-Freedom systems V. Rouillard © 2003 - 2013 Forced (harmonically excited) single DoF vibration – Damped. • • Rotating Imbalance Excitation With the horizontal components cancelled the vertical component of the excitation is: 2 F( t ) = meω sin( ω t ) The eqn. of motion is : Mx + cx + kx = meω 2 sin( ω t ) && & and the steady − state solution becomes :  me  ω 2  i( ω t −φ )  x p ( t ) = X sin( ω t − φ ) = Im   ÷ H( iω ) e  M  ωn     The response amplitude and phase are given by : X = 2 meω 2 ( ) 1 me  ω  MX =  ω ÷ H( iω ) or me = M  n  2 k − Mω + ( cω )      cω   2ζ r  φ = a tan  = a tan  ÷ ÷  1 − r2   k − Mω 2  04:36:37 2 2 2 r2 (   1− r  ) 2 2 2 + ( 2ζ r )   1 = r 2 H( iω ) 2
125. 125. 125 Mechanical Vibrations – Single Degree-of-Freedom systems Forced (harmonically excited) single DoF vibration – Damped. • Rotating Imbalance Excitation MX = me r2 (   1− r  ) 2 2 = r 2 H( iω )  2ζ r  φ = a tan  ÷  1 − r2  04:36:37 1  2 + ( 2ζ r ) 2   V. Rouillard © 2003 - 2013
126. 126. Mechanical Vibrations – Single Degree-of-Freedom systems 126 V. Rouillard © 2003 - 2013 Forced (harmonically excited) single DoF vibration – Damped. • • Forced Vibration with Coulomb Damping The equation of motion for a SDOF with Coulomb damping subjected to a harmonic force is: Mx + kx ± µ N = F0 sin( ω t ) && • • Solution complicated. • If µN << F0 motion of mass m will approximate harmonic motion • When µN << F0 an approximate solution to eqn. of motion may be used to determine equivalent viscous If µN is large cf F0, motion of mass m is discontinuous damping ratio. • • This is achieved by equating dissipated energy for both cases. For Coulomb damping, the energy dissipated during a cycle of amplitude X is: ∆W = 4 ( µ NX ) − 4 quarter cycles • For viscous damping, the energy dissipated during a cycle of amplitude X is: ∆W = 2π / ω ∫ Fv dt = t =0 = π ceqω X 2 04:36:37 2π / ω ∫ t =0 2 2π dx ceq   dt = ∫ ceq X 2ω cos 2 ( ωt ) d( ωt )  ÷  dt  t =0
127. 127. Mechanical Vibrations – Single Degree-of-Freedom systems 127 Forced (harmonically excited) single DoF vibration – Damped. • Equating the dissipated energies: ceq = • πω X 2 And the equivalent damping ratio is defined as: ζ eq = • 4µ N ceq cc = ceq 2mωn = 4µ N 2µ N = 2mωnπω X π mωnω X The amplitude X and the phase φ of the response becomes: 12    2   4µ N   1−   ÷  F0   π F0   X= k  2 2   1 −  ω       ωn ÷          • These approximations are only valid for 04:36:37    4µ N ±1 −   π F0  φ = a tan  1 2  2   4µ N      1−     π F0 ÷         µ N << F0 V. Rouillard © 2003 - 2013
128. 128. 128 Mechanical Vibrations – Single Degree-of-Freedom systems V. Rouillard © 2003 - 2013 SDoF systems – General forcing functions • • • • Methods to solve response due to general (nonharmonic) forcing functions. • When forcing function is periodic (not harmonic), it can be described with a series (sum) of harmonic or Fourier components. General forcing function may be periodic (nonharmonic) or aperiodic. Aperiodic forcing functions may be finite or infinite When the duration of a transient forcing function << natural period of system, forcing function called SHOCK. 04:36:37
129. 129. 129 Mechanical Vibrations – Single Degree-of-Freedom systems V. Rouillard © 2003 - 2013 Types of deterministic forcing functions. Deterministic Deterministic Periodic Periodic Sinusoidal Sinusoidal 04:36:37 Complex Periodic Complex Periodic Non-periodic Non-periodic Almost Periodic Almost Periodic Transient Transient
130. 130. 130 Mechanical Vibrations – Single Degree-of-Freedom systems V. Rouillard © 2003 - 2013 Types of deterministic forcing functions. Deterministic Deterministic Periodic Periodic Sinusoidal Sinusoidal Non-periodic Non-periodic Complex Complex Periodic Periodic Almost Periodic Almost Periodic Transient Transient Can be defined mathematically. Waveform contains harmonics which are multiples if the 04:36:37 fundamental frequency (show spectrum) Signal factory.vee
131. 131. 131 Mechanical Vibrations – Single Degree-of-Freedom systems V. Rouillard © 2003 - 2013 Types of deterministic forcing functions. Deterministic Deterministic Periodic Periodic Sinusoidal Sinusoidal Complex Periodic Complex Periodic Non-periodic Non-periodic Almost Periodic Almost Periodic Transient Transient Contains sine wave of arbitrary frequencies which frequency ratios are not rational numbers (show 04:36:37 spectrum) Signal factory.vee
132. 132. 132 Mechanical Vibrations – Single Degree-of-Freedom systems V. Rouillard © 2003 - 2013 Types of deterministic forcing functions. Deterministic Deterministic Periodic Periodic Sinusoidal Sinusoidal Complex Periodic Complex Periodic Half-sine pulse 04:36:37 Non-periodic Non-periodic Transient Transient Almost Periodic Almost Periodic Sin(x)/x Exp(T-x) All other deterministic data that can be described by a suitable function
133. 133. 133 Mechanical Vibrations – Single Degree-of-Freedom systems V. Rouillard © 2003 - 2013 SDoF systems – General forcing functions - Periodic • • For periodic forcing functions, the response of system is obtained by using the principle of superposition: • The periodic forcing function (period τ = 2π/ω) can be expressed as a Fourier series: The total response consists of sum of response functions due to individual harmonic functions in forcing function. a F( t ) = o + 2 ∞ ∑ a j cos( jω t ) j =1 + ∞ ∑ b j sin( jωt ) j =1 where τ 2 a j = ∫ F( t )cos( jω t ) dt τ0 for τ 2 b j = ∫ F( t ) sin( jω t ) dt, τ0 • j = 1, 2, 3..... for The eqn. of motion can be written as: a mx + cx + kx = o + && & 2 • j = 0, 1, 2..... ∞ ∑ a j cos( jω t ) j =1 + ∞ ∑ b j sin( jω t ) j =1 The RHS 04:36:37 is a constant + a sum of harmonic functions.
134. 134. 134 Mechanical Vibrations – Single Degree-of-Freedom systems V. Rouillard © 2003 - 2013 SDoF systems – General forcing functions - Periodic • Using the principle of superposition, the steady-state solution is the sum of the steady-state solution for the following equations: mx + cx + kx = && & mx + cx + kx = && & ao 2 (1) ∞ ∑ a j cos( jω t ) (2) j =1 mx + cx + kx = && & ∞ ∑ b j sin( jω t ) (3) j =1 • The steady-state solutions of (1), (2) and (3) are xp( t ) = xp( t ) = xp( t ) = 04:36:37 ao 2k aj k ( 1− j r ) 2 2 2 + ( 2ζ jr )2 bj k ( 1− j r ) 2 2 2 + ( 2ζ jr )2 cos( jω t − φ j ) sin( jω t − φ j )
135. 135. 135 Mechanical Vibrations – Single Degree-of-Freedom systems V. Rouillard © 2003 - 2013 SDoF systems – General forcing functions - Periodic • The entire steady-state solution is given by: a xp( t ) = o + 2k ∞ ∑ j =1 aj k ( 1− j r ) 2 2 2 + ( 2ζ jr )2 cos( jω t − φ j ) + ∞ ∑ j =1 bj k ( 1− j r ) 2 2 2 + ( 2ζ jr )2 sin( jω t − φ j ) where  2ζ jr  ω φ j = a tan  and r = ÷ ωn 1 − j2r2   • • • The response amplitude and phase for each harmonic (j th term) depend on j. • • Complete Solution • This requires setting the complete solution and its derivative to the specified initial displacement and velocity which produces a complicated expression for the transient part of the solution. When r = 1 the response amplitude is relatively high for any value j (more so when both j and ζ are small) As j becomes larger (higher harmonics) the amplitude response becomes smaller → the first few terms are usually needed to generate a reasonably accurate response. The complete solution is obtained by including the transient part of the solution which is dependent on the initial conditions. 04:36:37 Example: Triangular forcing function. Vee & Excel
136. 136. 136 Mechanical Vibrations – Single Degree-of-Freedom systems V. Rouillard © 2003 - 2013 SDoF systems – General forcing functions - Periodic • Situation sometimes arises when the periodic forcing function is given (obtained) experimentally (eg: wave, wind , seismic, topography..) and represented by discrete measurement data. • • When the (measured) data cannot be readily described by a mathematical function The discrete measurement data can be integrated numerically to obtain the Fourier coefficients. 2 N a0 = ∑ Fi N i =1 • 2 N  2 jπ ti  a j = ∑ Fi cos  ÷ and N i =1  τ  2 N  2 jπ ti  b j = ∑ Fi sin  ÷ N i =1  τ  for j = 1, 2..... The Fourier coefficients can then be used to find the solution with the excitation frequency taken as the lowest frequency component of the data. 04:36:37 2π ω= τ
137. 137. 137 Mechanical Vibrations – Single Degree-of-Freedom systems V. Rouillard © 2003 - 2013 SDoF systems – General forcing functions – Nonperiodic • When the forcing function is arbitrary and nonperiodic (aperiodic) it cannot be represented with a Fourier series • Alternative methods for determining the response must be used: • • • • Representation of the excitation function with a Convolution integral Using Laplace Transformations Approximating F(t) with a suitable interpolation method then using a numerical procedure Numerical integration of the equations of motion. 04:36:37
138. 138. 138 Mechanical Vibrations – Single Degree-of-Freedom systems V. Rouillard © 2003 - 2013 SDoF systems – General forcing functions – Nonperiodic • When the forcing function is arbitrary and nonperiodic (aperiodic) it cannot be represented with a Fourier series • Alternative methods for determining the response must be used: • • • • Representation of the excitation function with a Convolution integral Using Laplace Transformations Approximating F(t) with a suitable interpolation method then using a numerical procedure Numerical integration of the equations of motion. 04:36:37
139. 139. 139 Mechanical Vibrations – Single Degree-of-Freedom systems V. Rouillard © 2003 - 2013 SDoF systems – General forcing functions – Nonperiodic • • Convolution integral • An impulse can be measured by the resulting change in momentum: Consider one of the simplest nonperiodic exciting force: Impulsive force: which has a large magnitude F which acts for a very short time ∆t. Im pulse = F ∆t = m& 2 − mx1 x & where & 1 and & 2 represent the velocity of the lumped mass before and after the impulse . x x • The magnitude of the impulse F∆t is represented by F= % t + ∆t ∫ F dt t and a unit impulse is defined as f = lim % ∆t →0 • t + ∆t ∫ F dt = Fdt = 1 t For Fdt to have a finite value, F approaches infinity as ∆t nears zero. 04:36:37
140. 140. 140 Mechanical Vibrations – Single Degree-of-Freedom systems V. Rouillard © 2003 - 2013 SDoF systems – General forcing functions – Nonperiodic • • • Convolution integral – Impulse response Consider a (viscously) damped SDoF (mass-spring-damper system) subjected to an impulse at t=0. For an underdamped system, the eqn. of motion is: mx + cx + kx = 0 && & • And its solution: −ζωn t x( t ) = e   & 0 + ζωn x0 x   sin ( ωd t )   x0 cos ( ωd t ) + 1 − ζ 2 ωn     where c ζ = 2mωn • ωd = ωn 1 − ζ • 2 k  c  = − ÷ m  2m  ωn = If, prior to the impulse load being applied, the mass is at rest, then: − x( t < 0 ) = 0 and & t < 0 ) = 0 x( • 2 or x( t = 0 ) = 0 and & t = 0 − ) = 0 x( The impulse-momentum equation gives: − f = 1 = mx( t = 0 ) − mx( t = 0 ) = mx 0 & & & % And the initial conditions are given by: x( t 04:36:37 = 0 ) = x0 = 0 and k m & t =0)=&0 = x( x 1 m
141. 141. 141 Mechanical Vibrations – Single Degree-of-Freedom systems V. Rouillard © 2003 - 2013 SDoF systems – General forcing functions – Nonperiodic • • Convolution integral – Impulse response The solution reduces to: e −ζωnt x( t ) = g( t ) = sin ( ωd t ) mωd • g(t) is the impulse response function an represents the response of a viscously damped single degree of freedom system subjected to a unit impulse. 04:36:37
142. 142. 142 Mechanical Vibrations – Single Degree-of-Freedom systems SDoF systems – General forcing functions – Nonperiodic • • Convolution integral – Impulse response If the magnitude of the impulse is F instead of unity, the initial velocity x’0 = F/m and the response becomes: Fe −ζωnt x( t ) = % sin ( ωd t ) = F g( t ) % mωd • If the impulse is applied to a stationary system at an arbitrary time t = τ the response is x( t ) = F g( t −τ ) % 04:36:37 V. Rouillard © 2003 - 2013
143. 143. 143 Mechanical Vibrations – Single Degree-of-Freedom systems V. Rouillard © 2003 - 2013 SDoF systems – General forcing functions – Nonperiodic • • Convolution integral – Arbitrary exciting force • • • The impulse acting at t = τ is given by F(τ )∆τ. If we consider the arbitrary force to comprise of a series of impulses of varying magnitudes such that at time τ, the force F(τ) acts on the system for a short period ∆ τ. At any time t the elapsed time is t - τ The system response at t due to the impulse is x( t ) = F g( t −τ ) = F( τ )∆τ g( t −τ ) % • The total response at time t is determined by summing the responses caused by the impulses acting al all times τ : x( t ) = ∑ F( τ ) g( t − τ ) ∆τ Making ∆τ → 0 the response can be exp ressed as : t x( t ) = ∫ F( τ ) g( t − τ ) dτ 0 Substituting the impulse response function g( t − τ ) : t • • 1 −ζω ( t −τ ) x( t ) = sin [ ωd ( t − τ ) ] dτ ¬ Convolution or Duhamel int egral ∫ F( τ )e n mωd 0 04:36:37 This solution does not account for initial conditions. Can be integrated explicitly or numerically depending on F(t)
144. 144. 144 Mechanical Vibrations – Single Degree-of-Freedom systems V. Rouillard © 2003 - 2013 SDoF systems – General forcing functions – Nonperiodic • • Convolution integral – Arbitrary exciting force In the case where the excitation is provided by an arbitrary imposed motion of the base, y(t), the relative displacement is given by: t 1 z( t ) = y( −ζω ( t −τ ) sin [ ωd ( t − τ ) ] dτ ∫&& τ )e n ωd 0 04:36:37 Example: Step load
145. 145. 145 Mechanical Vibrations – Single Degree-of-Freedom systems V. Rouillard © 2003 - 2013 SDoF systems – General forcing functions – Nonperiodic • When the forcing function is arbitrary and nonperiodic (aperiodic) it cannot be represented with a Fourier series • Alternative methods for determining the response must be used: • • • • Representation of the excitation function with a Convolution integral Using Laplace Transformations Approximating F(t) with a suitable interpolation method then using a numerical procedure Numerical integration of the equations of motion. 04:36:37
146. 146. 146 Mechanical Vibrations – Single Degree-of-Freedom systems SDoF systems – General forcing functions – Nonperiodic • • • • • • • Laplace Transformation Efficient method to generate solution of linear differential equations Converts differential equations into algebraic equations to facilitate solving Can be applied to discontinuous functions Can be used for any type of excitation including periodic & harmonic Automatically accounts for initial conditions The Laplace transform of x(t) is given by: ∞ x( s ) = L x( t ) = ∫ e − st x( t ) dt 0 • Where s the subsidiary variable and is usually complex. • To use Laplace Transform: 1. Write the equation of motion 2. Compute or look-up the Laplace transform of each term using known initial conditions 3. Solve the transformed (algebraic ) equation of motion 4. Use the inverse Laplace transform to obtain the response (solution) 04:36:37 V. Rouillard © 2003 - 2013
147. 147. 147 Mechanical Vibrations – Single Degree-of-Freedom systems V. Rouillard © 2003 - 2013 SDoF systems – General forcing functions – Nonperiodic • When the forcing function is arbitrary and nonperiodic (aperiodic) it cannot be represented with a Fourier series • Alternative methods for determining the response must be used: • • • • Representation of the excitation function with a Convolution integral Using Laplace Transformations Approximating F(t) with a suitable interpolation method then using a numerical procedure Numerical integration of the equations of motion. 04:36:37
148. 148. 148 Mechanical Vibrations – Single Degree-of-Freedom systems V. Rouillard © 2003 - 2013 SDoF systems – General forcing functions – Nonperiodic • • • Numerical Methods (interpolation) • Often more practical to represent the digitised data with a series of incremental functions: • • Step functions Used when the nonperiodic forcing function cannot be described mathematically It may be possible to “fit” a mathematical approximation (say polynomial) to data then use the convolution integral The arbitrary function is represented by a series of step functions of varying magnitudes ∆F1, ∆F2, ∆F3… and start times t1, t2, t3…. • Note that the polarity of ∆F changes with the slope of the function • Smaller intervals yield better accuracy. • The approximation is also improved by choosing the subsequent start times so that 04:36:37 F(t) intersects the step at midheight of the step.
149. 149. 149 Mechanical Vibrations – Single Degree-of-Freedom systems V. Rouillard © 2003 - 2013 SDoF systems – General forcing functions – Nonperiodic • Numerical Methods (interpolation) - Step functions • The system response due to a step excitation ∆Fi for any time interval ti - 1 < t < ti (i = 1, 2, 3 …..j-1) can be determined from the previous example:   1 j −1 x( t ) = ∆ Fi 1 − e −ζωn ( t −ti )  cos k i =1   ∑ • ( ωd ( t − ti ) ) +  ζωn sin ( ωd ( t − ti ) )   ωd  When t = tj the response is:   −ζω ( t − t )  ζω 1 j −1 x( t ) = ∑ ∆ Fi 1 − e n j i  cos ωd ( t j − ti ) + n sin ωd ( t j − ti )   k i =1 ωd    ( 04:36:37 ) ( )
150. 150. 150 Mechanical Vibrations – Single Degree-of-Freedom systems V. Rouillard © 2003 - 2013 SDoF systems – General forcing functions – Nonperiodic • • Numerical Methods (interpolation) - Rectangular impulses • The response of the system in any time interval ti - 1 < t < ti is obtained by adding the response caused by Fj The arbitrary function is represented by a series of rectangular impulses Fi the polarity of which depends on the polarity of F(t) at that instant. (applied over ∆tj to the response at t = tj which represent the initial condition: 04:36:37
151. 151. 151 Mechanical Vibrations – Single Degree-of-Freedom systems V. Rouillard © 2003 - 2013 SDoF systems – General forcing functions – Nonperiodic • • Numerical Methods (interpolation) – Ramps (linear) approximation The arbitrary function is represented by a series of linear functions and the response of the system in any time interval ti - 1 < t < ti is obtained by adding the response caused by the linear (ramp) during a specified interval to the response due to the previous ramp (initial condition) 04:36:37
152. 152. 152 Mechanical Vibrations – Single Degree-of-Freedom systems V. Rouillard © 2003 - 2013 SDoF systems – General forcing functions – Nonperiodic • • (Shock) Response Spectrum • The Shock Response Spectrum (SRS) is plotted for a range of natural frequencies usually at fractional octave intervals. • The SRS is used to determine the effect of a particular (shock) excitation function on damped SDoF systems. • Given the nature of real shocks, the SRS is usually computed using numerical means. Shows the variation in maximum response of a damped SDOF due to a particular transient (shock) excitation. 04:36:37
153. 153. 153 • Mechanical Vibrations – Two Degree-of-Freedom systems Two degree of freedom systems: 04:36:37 V. Rouillard © 2003 - 2013
154. 154. 154 • Mechanical Vibrations – Two Degree-of-Freedom systems Two degree of freedom systems: 04:36:37 V. Rouillard © 2003 - 2013
155. 155. 155 Mechanical Vibrations – Two Degree-of-Freedom systems V. Rouillard © 2003 - 2013 • • • • No. of DoF of system = No. of mass elements x number of motion types for each mass • Under an arbitrary initial disturbance, the system will vibrate freely such that the two normal modes are superimposed. • Under sustained harmonic excitation, the system will vibrate at the excitation frequency. Resonance occurs if the excitation frequency corresponds to one of the natural frequencies of the system For each degree of freedom there exists an equation of motion – usually coupled differential equations. Coupled means that the motion in one coordinate system depends on the other If harmonic solution is assumed, the equations produce two natural frequencies and the amplitudes of the two degrees of freedom are related by the natural, principal or normal mode of vibration. 04:36:37
156. 156. 156 Mechanical Vibrations – Two Degree-of-Freedom systems • • Equations of motion • • V. Rouillard © 2003 - 2013 Motion of system described by position x1(t) and x2(t) of masses m1 and m2 Consider a viscously damped system: The free-body diagram is used to develop the equations of motion using Newton’s second law 04:36:37
157. 157. 157 • Mechanical Vibrations – Two Degree-of-Freedom systems Equations of motion m1&& 1 + c1& 1 + k1 x1 − c2 ( & 2 − & 1 ) − k 2 ( x2 − x1 ) = F1 x x x x m2&& 2 + c2 ( & 2 − & 1 ) + k2 ( x2 − x1 ) + c3& 2 + k3 x2 = F2 x x x x or m1&& 1 + ( c1 + c2 )x1 − c2& 2 + ( k1 + k 2 )x1 − k 2 x2 = F1 x & x m2&& 2 − c2& 1 + ( c2 + c3 )x2 − k2 x1 + ( k2 + k3 )x2 = F2 x x & • • The differential equations of motion for mass m 1 and mass m2 are coupled. The motion of each mass is influenced by the motion of the other. 04:36:37 V. Rouillard © 2003 - 2013
158. 158. 158 • Mechanical Vibrations – Two Degree-of-Freedom systems V. Rouillard © 2003 - 2013 Equations of motion m1&& 1 + ( c1 + c2 )x1 − c2& 2 + ( k1 + k 2 )x1 − k 2 x2 = F1 x & x m2&& 2 − c2& 1 + ( c2 + c3 )x2 − k 2 x1 + ( k 2 + k3 )x2 = F2 x x & • The coupled differential eqns. of motion can be written in matrix form: r r r && t ) + [ c ] &r t ) + [ k ] x( t ) = F( t ) x( [ m] x( where [ m] , [ c ] and [ k ] are the mass, damping and stiffness matrices respectively and are given by:  m1 0  [ m] =    0 m2  − c2   c1 + c2 [ c] =  c2 + c3   − c2   k1 + k2 [ k] =   − k2 r r r r x(t), & && and F(t) are the displacement, velocity, acceleration and force vectors x(t), x(t) − k2  k 2 + k3   respectively and are given by : x x  x1( t )  r  & 1( t )  r && 1( t )  r x( t ) =  x( x(  & t)=   && t ) =   and x2 ( t )  & 2 ( t ) x && 2 ( t )  x    • r  F1( t )  F( t ) =   F2 ( t )   Note: the mass, damping and stiffness matrices are all square and symmetric [m] = [m] T and consist of the mass, damping and stiffness constants. 04:36:37
159. 159. 159 • • Mechanical Vibrations – Two Degree-of-Freedom systems V. Rouillard © 2003 - 2013 Free vibrations of undamped systems The eqns. of motion for a free and undamped TDoF system become: m1&& 1 + ( k1 + k2 )x1 − k2 x2 = 0 x m2&& 2 − k 2 x1 + ( k 2 + k3 )x2 = 0 x • Let us assume that the resulting motion of each mass is harmonic: For simplicity, we will also assume that the response frequencies and phase will be the same: x1( t ) = X 1 cos( ωt + φ ) • and x2 ( t ) = X 2 cos( ωt + φ ) Substituting the assumed solutions into the eqns. of motion: { }  − m1ω 2 + ( k1 + k 2 ) X 1 − k 2 X 2  cos( ω t + φ ) = 0   { }  − k2 X 1 + − m2ω 2 + ( k2 + k3 ) X 2  cos( ωt + φ ) = 0   As these equations must be zero for all values of t, the cosine terms cannot be zero. Therefore: { −m1ω 2 + ( k1 + k2 ) } X 1 − k2 X 2 = 0 − k2 X 1 + { − m2ω 2 + ( k 2 + k3 ) } X 2 = 0 • Represent two simultaneous algebraic equations with a trivial solution when X1 and X2 are both zero – no vibration. 04:36:37
160. 160. 160 • • Mechanical Vibrations – Two Degree-of-Freedom systems V. Rouillard © 2003 - 2013 Free vibrations of undamped systems Written in matrix form it can be seen that the solution exists when the determinant of the mass / stiffness matrix is zero:  − m1ω 2 + ( k1 + k 2 )  − k2    X1  = 0    − k2 − m2ω 2 + ( k2 + k2 )   X 2      { } { } or 2 m1m2ω 4 − { ( k1 + k 2 ) m2 + ( k 2 + k3 ) m1 } ω 2 + ( k1 + k2 ) ( k 2 + k2 ) − k 2 = 0 • • The solution to the characteristic equation yields the natural frequencies of the system. The roots of the characteristic equation are: 2 2 1  ( k + k ) m + ( k 2 + k3 ) m1  ω1 , ω 2 =  1 2 2  2 m1m2  1 2  2  ( k1 + k 2 ) ( k 2 + k 3 ) − k 2   1   ( k1 + k2 ) m2 + ( k 2 + k3 ) m1  ±   −4 2  m1m2     2 • m1m2 This shows that the homogenous solution is harmonic with natural frequencies 04:36:37    ω1 and ω2