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Page 1 Lecture 2 Free Vibration of Single Degree of Freedom Systems ERT 452 VIBRATION MUNIRA MOHAMED NAZARI SCHOOL OF BIOPROCESS UNIVERSITI MALAYSIA PERLIS.

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Presentation on theme: "Page 1 Lecture 2 Free Vibration of Single Degree of Freedom Systems ERT 452 VIBRATION MUNIRA MOHAMED NAZARI SCHOOL OF BIOPROCESS UNIVERSITI MALAYSIA PERLIS."— Presentation transcript:

1 Page 1 Lecture 2 Free Vibration of Single Degree of Freedom Systems ERT 452 VIBRATION MUNIRA MOHAMED NAZARI SCHOOL OF BIOPROCESS UNIVERSITI MALAYSIA PERLIS

2 Page 2 COURSE OUTLINE  Free Vibration with Viscous Damping  Free Vibration with Coulomb Damping

3 Page 3 Free Vibration with Viscous Damping

4 Page 4 where c = damping constant Free Vibration with Viscous Damping Equation of Motion: From the figure, Newton’s law yields the equation of motion: or

5 Page 5 Free Vibration with Viscous Damping Assume a solution in the form: Hence, the characteristic equation is the roots of which are These roots give two solutions to Eq.(2.59)

6 Page 6 Free Vibration with Viscous Damping Thus the general solution is: where C 1 and C 2 are arbitrary constants to be determined from the initial conditions of the system.

7 Page 7 Free Vibration with Viscous Damping Critical Damping Constant and Damping Ratio: The critical damping c c is defined as the value of the damping constant c for which the radical in Eq.(2.62) becomes zero: or The damping ratio ζ is defined as:

8 Page 8 Case1. Underdamped system Free Vibration with Viscous Damping Thus the general solution for Eq.(2.64) is: Assuming that ζ ≠ 0, consider the following 3 cases: For this condition, (ζ 2 -1) is negative and the roots are:

9 Page 9 Free Vibration with Viscous Damping where (C’ 1,C’ 2 ), (X,Φ), and (X 0, Φ 0 ) are arbitrary constants to be determined from initial conditions. and the solution can be written in different forms:

10 Page 10 Free Vibration with Viscous Damping and hence the solution becomes For the initial conditions at t = 0, Eq.(2.72) describes a damped harmonic motion. Its amplitude decreases exponentially with time, as shown in the figure below.

11 Page 11 Free Vibration with Viscous Damping Underdamped Solution The frequency of damped vibration is:

12 Page 12 Due to repeated roots, the solution of Eq.(2.59) is given by Free Vibration with Viscous Damping In this case, the two roots are: Case2. Critically damped system Application of initial conditions gives: Thus the solution becomes:

13 Page 13 It can be seen that the motion represented by Eq.(2.80) is aperiodic (i.e., nonperiodic). Since, the motion will eventually diminish to zero, as indicated in the figure below. Free Vibration with Viscous Damping Comparison of motions with different types of damping

14 Page 14 In this case, the solution Eq.(2.69) is given by: The roots are real and distinct and are given by: Free Vibration with Viscous Damping For the initial conditions at t = 0, Case3. Overdamped system

15 Page 15 Free Vibration with Viscous Damping Logarithmic Decrement: Using Eq.(2.70), The logarithmic decrement can be obtained from Eq.(2.84):

16 Page 16 Free Vibration with Viscous Damping For small damping, Hence, or Thus, where m is an integer.

17 Page 17 The energy dissipated in a complete cycle is: Free Vibration with Viscous Damping Energy dissipated in Viscous Damping: In a viscously damped system, the rate of change of energy with time is given by:

18 Page 18 The energy dissipated in a complete cycle will be Thus, Eq.(2.95) becomes Consider the system shown in the figure below. The total force resisting the motion is: Free Vibration with Viscous Damping If we assume simple harmonic motion:

19 Page 19 where W is either the max potential energy or the max kinetic energy. Computing the fraction of the total energy of the vibrating system that is dissipated in each cycle of motion, Free Vibration with Viscous Damping The loss coefficient, defined as the ratio of the energy dissipated per radian and the total strain energy:

20 Page 20 The equation of motion can be derived as: Torsional systems with Viscous Damping: Free Vibration with Viscous Damping Consider a single degree of freedom torsional system with a viscous damper, as shown in figure (a). The viscous damping torque is given by: where J 0 = mass moment of inertia of disc k t = spring constant of system θ = angular displacement of disc

21 Page 21 In the underdamped case, the frequency of damped vibration is given by: Free Vibration with Viscous Damping where and where c tc is the critical torsional damping constant

22 Page 22 Example 2.11 Shock Absorber for a Motorcycle An underdamped shock absorber is to be designed for a motorcycle of mass 200kg (shown in Fig.(a)). When the shock absorber is subjected to an initial vertical velocity due to a road bump, the resulting displacement-time curve is to be as indicated in Fig.(b). Find the necessary stiffness and damping constants of the shock absorber if the damped period of vibration is to be 2 s and the amplitude x 1 is to be reduced to one-fourth in one half cycle (i.e., x 1.5 = x 1 /4). Also find the minimum initial velocity that leads to a maximum displacement of 250 mm.

23 Page 23 Example 2.11 Solution Approach Approach: We use the equation for the logarithmic decrement in terms of the damping ratio, equation for the damped period of vibration, time corresponding to maximum displacement for an underdamped system, and envelope passing through the maximum points of an underdamped system.

24 Page 24 From which ζ can be found as 0.4037. The damped period of vibration given by 2 s. Hence, Hence the logarithmic decrement becomes Example 2.11 Solution Since,

25 Page 25 The displacement of the mass will attain its max value at time t 1, given by Thus the damping constant is given by: and the stiffness by: Example 2.11 Solution The critical damping constant can be obtained: or This gives:

26 Page 26 Since x = 250mm, Example 2.11 Solution The envelope passing through the max points is: The velocity of mass can be obtained by differentiating the displacement: as When t = 0,

27 Page 27 Free Vibration with Coulomb Damping

28 Page 28 Free Vibration with Coulomb Damping 28  Coulomb’s law of dry friction states that, when two bodies are in contact, the force required to produce sliding is proportional to the normal force acting in the plane of contact. Thus, the friction force F is given by: where N is normal force, μ is the coefficient of sliding or kinetic friction μ is usu 0.1 for lubricated metal, 0.3 for nonlubricated metal on metal, 1.0 for rubber on metal  Coulomb damping is sometimes called constant damping

29 Page 29 Equation of Motion: 29 Consider a single degree of freedom system with dry friction as shown in Fig.(a) below. Since friction force varies with the direction of velocity, we need to consider two cases as indicated in Fig.(b) and (c). Free Vibration with Coulomb Damping

30 Page 30 30 Case 1. When x is positive and dx/dt is positive or when x is negative and dx/dt is positive, the equation of motion can be obtained using Newton’s second law (Fig.b): Hence, where ω n = √k/m is the frequency of vibration A 1 & A 2 are constants Free Vibration with Coulomb Damping

31 Page 31 31 Case 2. When x is positive and dx/dt is negative or when x is negative and dx/dt is negative, the equation of motion can be derived from Fig. (c): The solution of the equation is given by: where A 3 & A 4 are constants Free Vibration with Coulomb Damping

32 Page 32 32 Fig.2.34 Motion of the mass with Coulomb damping Free Vibration with Coulomb Damping

33 Page 33 Free Vibration with Coulomb Damping Solution: 33 Eqs.(2.107) & (2.109) can be expressed as a single equation using N = mg: where sgn(y) is called the sigum function, whose value is defined as 1 for y > 0, -1 for y< 0, and 0 for y = 0. Assuming initial conditions as

34 Page 34 34 The solution is valid for half the cycle only, i.e., for 0 ≤ t ≤ π/ω n. Hence, the solution becomes the initial conditions for the next half cycle. The procedure continued until the motion stops, i.e., when x n ≤ μN/k. Thus the number of half cycles (r) that elapse before the motion ceases is: That is, Free Vibration with Coulomb Damping

35 Page 35 35 Note the following characteristics of a system with Coulomb damping: 1.The equation of motion is nonlinear with Coulomb damping, while it is linear with viscous damping 2.The natural frequency of the system is unaltered with the addition of Coulomb damping, while it is reduced with the addition of viscous damping. 3.The motion is periodic with Coulomb damping, while it can be nonperiodic in a viscously damped (overdamped) system. 4.The system comes to rest after some time with Coulomb damping, whereas the motion theoretically continues forever (perhaps with an infinitesimally small amplitude) with viscous damping. Free Vibration with Coulomb Damping

36 Page 36 36 Note the following characteristics of a system with Coulomb damping: 5.The amplitude reduces linearly with Coulomb damping, whereas it reduces exponentially with viscous damping. 6.In each successive cycle, the amplitude of motion is reduced by the amount 4μN/k, so the amplitudes at the end of any two consecutive cycles are related: As amplitude is reduced by an amount 4μN/k in one cycle, the slope of the enveloping straight lines (shown dotted) in Fig 2.34. Free Vibration with Coulomb Damping

37 Page 37 Free Vibration with Coulomb Damping 37 Torsional Systems with Coulomb Damping: The equation governing the angular oscillations of the system is and The frequency of vibration is given by

38 Page 38 Free Vibration with Coulomb Damping 38 and the amplitude of motion at the end of the rth half cycle (θ r ) is given by: The motion ceases when

39 Page 39 Example 2.14 Pulley Subjected to Coulomb Damping A steel shaft of length 1 m and diameter 50 mm is fixed at one end and carries a pulley of mass moment of inertia 25 kg-m 2 at the other end. A band brake exerts a constant frictional torque of 400 N-m around the circumference of the pulley. If the pulley is displaced by 6° and released, determine (1) the number of cycles before the pulley comes to rest and (2) the final settling position of the pulley. 39

40 Page 40 Example 2.14 Solution (1) The number of half cycles that elapse before the angular motion of the pullet ceases is: 40 where θ 0 = initial angular displacement = 6° = 0.10472 rad, k t = torsional spring constant of the shaft given by

41 Page 41 Example 2.14 Solution 41 and T = constant friction torque applied to the pulley = 400 N- m. Eq.(E.1) gives Thus the motion ceases after six half cycles. (2) The angular displacement after six half cycles:

42 Page 42


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