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Mechanical vibration

1. 1 Ing . Zina G. (M.Sc. in Mechanical System Design) Mekelle University Ethiopian Institute of Technology- Mekelle School of Mechanical and Industrial Engineering Mechanical Vibrations MEng 3054

2. Contents 1.1 Introduction 1.2 Periodic functions and Fourier Approximations 1.3 Modeling of Dynamic Systems for vibration 1.4 Characteristics of Discrete System Components 1.5 Differential Equation of Motion for 1 st Order and 2 nd Order system Chapter One Introduction to Mechanical Vibrations

3. 3 Importance of the Study of Vibration Why study vibration ? Vibrations can lead to excessive deflections and failure on the machines and structures To reduce vibration through proper design of machines and their mountings To utilize profitably in several consumer and industrial applications To improve the efficiency of certain machining, casting, forging & welding processes To stimulate earthquakes for geological research and conduct studies in design of nuclear reactors

4. 1.1 . Basic Concepts of Vibration What is vibration? Vibrations are oscillations of a system about an equilbrium position. Vibration is a dynamic behavior of physical system in which it oscillates about a certain equilibrium position. Mechanical vibration is the motion of a particle or body which oscillates about a position of equilibrium.

5. 5 Basic Concepts of Vibration All bodies having mass and elasticity are capable of producing vibration . The mass is inherent of the body and elasticity causes relative motion among its parts. When body particles are displaced by the application of external force, the internal forces in the form of elastic energy are present in the body. These forces try to bring the body to its original position. Involves transfer of potential energy to kinetic energy and vice versa In this way, vibratory motion is repeated indefinitely and exchange of energy takes place.

6. 6 Degree of Freedom ( d.o.f .) = min . no. of independent coordinates required to determine completely the positions of all parts of a system at any instant of time Examples of single degree-of-freedom systems: Basic Concepts of Vibration

7. 7 Examples of single degree-of-freedom systems: Basic Concepts of Vibration

8. 8 Examples of Two degree-of-freedom systems: Basic Concepts of Vibration

9. 9 Examples of Three degree of freedom systems: Basic Concepts of Vibration

10. 10 Example of Infinite- number-of-degrees-of-freedom system: Infinite number of degrees of freedom system are termed continuous or distributed systems Finite number of degrees of freedom are termed discrete or lumped parameter systems More accurate results obtained by increasing number of degrees of freedom Basic Concepts of Vibration

11. 11 Free Vibration: A system is left to vibrate on its own after an initial disturbance and no external force acts on the system. E.g. simple pendulum Forced Vibration: A system that is subjected to a repeating external force. E.g. oscillation arises from diesel engines Resonance occurs when the frequency of the external force coincides with one of the natural frequencies of the system Classification of Vibration

12. 12 Classification of Vibration Undamped Vibration: When no energy is lost or dissipated in friction or other resistance during oscillations Damped Vibration: When any energy is lost or dissipated in friction or other resistance during oscillations Linear Vibration: When all basic components of a vibratory system, i.e. the spring, the mass and the damper behave linearly Nonlinear Vibration: If any of the components behave nonlinearly

13. 13 Deterministic Vibration: If the value or magnitude of the excitation (force or motion) acting on a vibratory system is known at any given time Nondeterministic or random Vibration: When the value of the excitation at a given time cannot be predicted Examples of deterministic and random excitation: Classification of Vibration

14. Vibration… It is also an everyday phenomenon we meet on everyday life

15. Introduction … Useful Vibration Harmful vibration Noise Destruction Compressor Ultrasonic cleaning Testing Wear Fatigue Most vibrations in machines and structures are undesirable due to increased stresses and energy losses .

16. 16 Definition: Periodic function : is any function that repeats itself in time, i.e. any function for which there exists a fixed time T called the period such that f(t) = f( t+T ) for all values of t . Example: forcing function that is the sum of two sinusoids of different frequencies T f(t) is a general periodic function of period T f(t) t 1.2. Periodic functions and Fourier Approximations Periodic input (excitation) results in a periodic response It could be sinusoidal driving function at a single or multiple frequency

17. 17 Harmonic Motion This is the fundamental and simplest type of periodic motion expressed by sinusoidal functions. It has a significant importance in the study of mechanical vibrations. It represents mostly encountered types of motion in many dynamic systems. It is also used as a fundamental model to describe complex motions.

18. 18 where: x o is the maximum value of the displacement, which is also called the amplitude, measured in meters . is the circular frequency of oscillation, measured in radians per second ( rad /sec ). is also another form of frequency measured in cycles-per-second (cps) or Hertz (Hz). is the phase shift given in radians . Harmonic Motion Representation:

19. 19 Fourier Theory : Any periodic function f(t) with period T, may be represented by an infinite series of the form: Fourier Coefficients It is twice the average of the function f(t) over one cycle

20. 20 Simplification: Orthogonality: Integral of the products of two orthogonal functions is zero.

21. 21 Symmetry If f(t)=f(-t) : Symmetry about the origin (y axis) Called even function The Fourier Series contains only cos components, i.e. b n =0 If f(t)=-f(-t) : Called an odd function The Fourier Series contains only sin components, i.e. a n =0

22. 22 1.3. Modeling of Dynamic Systems for Vibration Machines could be simple or complex Algorithm for vibration analysis: Vibration Problem Produce physical modeling Produce Mathematical Modeling Solve the Mathematical Model Check the solution with reality Modify the physical model in light of the solution Stop Yes No Vibration Analysis Procedure Step 1: Mathematical Modeling Step 2: Derivation of Governing Equations Step 3: Solution of the Governing Equations Step 4: Interpretation of the Results

23. 23 Model : Simplified but sufficiently sophisticated to represent the real system Depends on an individual perspective Depends on the accuracy of analysis needed Elements of Modeling: Mass Spring Damper Rod Beam Plate Building blocks in the development of physical models of dynamic systems

24. 24 Physically, all elements of a mechanical system are of continuous type and possess non-linear characteristic . To make the analysis simple and not to deal with the non-linear characteristics of all the components, we need to: take simplified assumptions and identify the basic parts which have relevance to the problem model by lumped or discretized equivalent: Mass Spring and Damper After Physical Modeling follows Mathematical Modeling Define physical quantities Establish equation of motion Then solution of equation of motion gives the motion of the dynamic system as function of time. With linear Characteristics PDE for Continuous Model ODE for Discretized Models

25. 25 Vibration Analysis Procedure Example of the modeling of a forging hammer :

26. 26 Discrete Truck Model for Vibration Analysis This model consists of seven masses with ten spring-damper-systems. The discretization has made the whole truck to have only ten DoF , consisted of six vertical displacements ( y 1 , y 2 , y 3 , y 4 , y 5 , y 6 ) and four rotations ( φ 3, φ 4 ,φ 5 ,φ 7 ). Note: The lumped elements should posses the same properties like the continuous element, which is satisfied through the equivalence in kinetic, potential & dissipation energies between the real system and the mechanical model.

27. Lumped (Rigid) Modelling Numerical Modelling Element-based methods (FEM, BEM) Statistical and Energy-based methods (SEA, EFA, etc.) Modelling of vibrating systems

28. 28 Essential properties of a machine or part to experience vibration are: Inertia: associated with mass moving in translation or rotation ( m or J ) The mass or inertia element is assumed to be rigid body It can gain or lose kinetic energy whenever the velocity of the body changes. Elasticity: Capability to deform and return to original form (deformation before plastic limit) Symbolized by spring Accumulation of elastic potential energy Damping: Represents energy dissipation in system Symbolized by a dashpot (damper) 1.4. Characteristics of Discrete System Components

29. 29 In real mechanical system we can model Inertia & elasticity are distributed Examples Distributed mass & tortional stiffness of crank shaft Distributed mass and bending stiffness of a bridge Actual Mechanical Model J 1 J 2 J 3 J 4 J 5 k 1 k 2 k 3 k 4 k 6 k 5 c 1 c 2 c 3 c 4 c 5 c 6 Actual Mechanical Model m 1 m 2 m n …… m 3 m n-1

30. 30 Note : Distributed properties are represented by equivalent lumped masses interconnected to one another through lumped damping and stiffness elements. This gives a discrete vibration model or discrete system . The lumped elements should posses the same properties like the continuous element, which is satisfied through the equivalence in kinetic, potential & dissipation energies .

31. 31 The discrete mechanical system elements are: Spring relates force to displacement (or torque to rotation) Possess the property of elasticity Generally assumed to be mass less A spring operated in the linear range is characterized by its stiffness, k [N/m]. F s Linear Range, Slope of k, Spring Constant or stiffness ∆x=x 2 -x 1 Softening Spring ( ∆ F<< ∆ x ) stiffening Spring ( ∆ F>> ∆ x ) F s F s x 1 x 2 F s =k(x 2 -x 1 ) φ k t M s M s =k t φ Linear Spring Torsional Spring Discrete System Components

32. 32 Damper relates force to velocity consists of a piston fitted loosely in a cylinder filled with viscous fluid represented by viscous damper or dash pot assumed to be mass-less F d is the damping force and resists an increase in relative velocity. C [Ns/m] is the coefficient of viscous damping F d F d Slope = C F d Rotational Damper Linear Damper

33. Common Uses of Dashpots Door Stoppers Vehicle Suspension Bridge Suspension Flyover Suspension

34. 34 Mass Relates force to acceleration Slope = m F m m F m Translational Vibration Torsional Vibration Spring, k [N/m] Torsional Spring, k t [Nm] Viscous damper, C [Ns/m] Torsional Viscous Damper, C t [ Nms ] Mass, m [kg] Mass moment of inertia, J [ kg.m 2 ] F s , F d M s , M d F m = m Units and symbols for translational and rotational vibrations is summarized as shown below.

35. 35 1.5. Differential equation of motion for 1 st Order and 2 nd Order Linear systems Case i ) System with m=0 This is a 1 st order linear ODE with constant coefficients . Case ii) Spring-Damper-Mass system c F(t) k F(t) F s = kx x(t) FBD c F(t) k F(t) F s = kx x(t) FBD m m This is a 2 nd order linear ODE with constant coefficients .

36. 36 Case iii) Suspended Spring-Mass-Damper m x(t) y(t) Equilibrium position Undeformed position F(t) k C FBD 1 Ky(t) F(t) y(t) mg FBD 2 K(x-x st ) F(t) x(t) mg

37. 37 Simple Vibration Simulations Too hard suspension system of a car results in throwing of the car on uneven road, while too soft suspension system will swing the car, which results in lost of the contact between the wheels and the road.

38. Model the following simplified structural elements as mass-spring systems and determine the equivalent spring stiffness, k of the spring. a) A rod experiencing axial (longitudinal) vibration The mass of the rod is negligible as compared to that of the block attached at the end. The work has been done by the force F which will be accumulated in the deformed bar as strain energy and when F is released this energy will be converted to kinetic energy, so that the system will experience longitudinal vibrations. Assumptions: Example 1:

39. From strength of materials, the uniform normal strain is given by: From this equation we can get the applied force to be: Hence, from Hook’s law, the equivalent stiffness of the rod modeled as a spring will be: Where: x is the longitudinal deformation, L is the unstretched length, A is the crossectional area and E is the elastic modulus of the rod.

40. b) A beam experiencing Transverse Vibration Assumptions: The mass of the beam is negligible as compared to that of the block placed at a distance of ‘a’ from the left. The work has been done by the force F which will be accumulated in the bent beam as strain energy and when F is released this energy will be converted to kinetic energy, so that the system will experience lateral vibrations.

41. From strength of materials, we know that the displacement due to static load at a distance of ‘a’ from the left (at the point where the mass is attached) is given by: where I is the cross sectional moment of inertia about the neutral axis. From this equation we can determine the static force as: But, from Hook’s law, since F = ky , the equivalent stiffness of the spring will be:

42. c) A shaft experiencing Torsional Oscillation The mass moment of the shaft is negligible as compared to that of the disk placed at the end of the shaft. The work has been done by the Torque T which will be accumulated in the twisted shaft as strain energy and when T is released this energy will be converted to kinetic energy, so that the system will experience Torsional Vibrations. Assumptions:

43. From strength of materials, we know that the angular deformation of the shaft due to static torque T is given by: where J is the polar moment of inertia of the shaft of length L and G is the shear modulus of the shaft. From this equation we can determine the static Torque as: But, from Hook’s law, since T = k t θ , the equivalent Torsional stiffness of the Torsional spring is:

44. Replace the combination of springs with a single spring of an equivalent spring constant with out altering the behavior of the system. parallel combination of springs Here, the block is considered not to rotate but only translate by x and hence it is true to say that all ends of the n springs are stretched by x, equal to the displacement of the block due to the static force F. From this we can say that: From which the equivalent stiffness, k eq , will be given by: Example 2:

45. b ) series combination of springs As shown the same force will be developed in each spring and is equal to the force, F acting on the block. However, the change in length of each spring is dependent upon the spring stiffness.

46. That is, the displacement of the block will be equal to the summation of all deformations of each spring as given by: From which the equivalent spring is given by:

47. c) Model the following combination of springs by a single spring of an equivalent stiffness attached to the given mass. These steps, step ( i ) and (ii), are reduction steps of the model. Step (ii), shows that the two springs attached at the left and right of the block are exerting force to the left and are deformed by same displacement, x. Hence, the spring forces add up to give the total applied force and the springs behave as if they are in parallel.

48. Here, these two springs are exerting force to the left and displaced by same displacement x. Hence, the spring forces add-up and behave as if they are in parallel, and hence the equivalent spring and the model (step ( iii)) are given by:

49. d) Model the flexibility of the simply supported beam and the spring as a single spring of equivalent stiffness, k eq , of the system shown hereunder. The deflection of the simply supported beam due to a unit load at x=2m is calculated by:

50. For a unit static load applied at 2 L/3, the equivalent stiffness of the simply supported beam will be given by: Considering a series connection of the two springs as shown hereunder, the equivalent stiffness will be given by: Note that, since springs are arranged in series, the equivalent spring stiffness is less than the stiffness of each springs, i.e. flexibility is increased by arranging springs in series.

51. Governing equation (Equation of motion) of SDOF systems By Newton’s Second Law By equivalent energy method

52. Equation of motion (Governing equation) a ) By Newton’s Second Law Governing equation (Equation of motion) of SDOF systems Translation Rotation

53. System FBD Free-Body Diagram (FBD)

54. Example 3 : Find equation of motion of the following system by using Newton’s Second Law. c c FBD

55. b ) By equivalent energy method Energy balance approach of the original system and the model Equation of motion (Governing equation)

56. Model the following mass-pulley and spring system to a single block of displacement, x and spring. Determine the equivalent mass and equivalent stiffness using the energy balance approach of the original system and the model, and make the equation of motion. Assuming: - the disk rolls with out slip, - no slip occurs at the pulley, and - the pulley is friction less, we have: Example 4:

57. Note : If the mass of the spring is much smaller than the mass of the block, but not negligible, a reasonable one-degree-of freedom approximation can be made by approximating the spring’s inertial effect. The inertial effects of a linear spring with one end fixed and the other end connected to a moving body can be approximated by placing a particle whose mass is 1/3 of the mass of the spring at the point where the spring is connected to the body. This is valid for helical coil springs, bars that are modeled as springs for longitudinal vibrations, and shafts acting as torsional springs.

58. Example: Find equation of motion of the following system by using Equivalent energy method.

59. Ex: Find governing equations by a) Equivalent energy b) Newton’s Second Law by defining equivalent mass, stiffness and damping constant Do it yourself !

60. Ex: Show that the system’s equation of motion is By equivalent energy method By Newton method Do it yourself !