Chapter 3 Harmonically Excited Vibration

Chapter Outline 3.1 Introduction 3.2 Equation of Motion 3.3 Response of an Undamped System Under Harmonic Force 3.4 Response of a Damped System Under Harmonic Force 3.5 Response of a Damped System Under

Chapter Outline 3.6 Response of a Damped System Under the Harmonic Motion of the Base 3.7 Response of a Damped System Under Rotating Unbalance 3.8 Forced Vibration with Coulomb Damping 3.9 Forced Vibration with Hysteresis Damping

Chapter Outline 3.10 Forced Motion with Other Types of Damping 3.11 Self-Excitation and Stability Analysis

3.1 Introduction Forced vibrations occurs when external energy is supplied to the system during vibration The external force can be supplied through either an applied force or an imposed displacement excitation, which may be harmonic, nonharmonic but periodic, nonperiodic, or random in nature. Harmonic response results when the system responses to a harmonic excitation Transient response is defined as the response of a dynamic system to suddenly applied nonperiodic excitations

3.2 Equation of Motion From Figure, the equation of Motion Using Newton’s Second Law of Motion: The homogeneous solution of the equation:

3.2 Equation of Motion The variations of homogeneous, particular, and general solutions with time for a typical case are shown in Figure below.

3.3 Response of an Undamped System Under Harmonic Force Consider an undamped system subjected to a harmonic force. If a force acts on the mass m of the system, The homogeneous solution is given by: where is the natural frequency. Because the exciting force and particular solution is harmonic and has same frequency, we can assume a solution in the form:

3.3 Response of an Undamped System Under Harmonic Force where X is the max amplitude of xp(t) where denotes the static deflection Thus,

3.3 Response of an Undamped System Under Harmonic Force Using initial conditions , Hence, The max amplitude can be expressed as:

3.3 Response of an Undamped System Under Harmonic Force where the quantity is called the magnification factor, amplification factor, or amplitude ratio. The variation of the amplitude ratio with the frequency ratio is shown in the Figure. The response of the system can be identified to be of three types from the figure.

3.3 Response of an Undamped System Under Harmonic Force Case 1. When 0 < < 1, the denominator in Eq.(3.10) is positive and the response is given by Eq.(3.5) without change. The harmonic response of the system is in phase with external force, shown in figure.

3.3 Response of an Undamped System Under Harmonic Force Case 2. When > 1, the denominator in Eq.(3.10) is negative and the steady-state solution can be expressed as where the amplitude, The variations are shown in figure.

3.3 Response of an Undamped System Under Harmonic Force Case 3. When = 1, the amplitude X given by Eq.(3.10) or (3.12) becomes infinite. This condition, for which the forcing frequency is equal to the natural frequency of the system, is called resonance. Hence, the total response if the system at resonance becomes

3.3 Response of an Undamped System Under Harmonic Force Total Response The total response of the system, Eq.(3.7) or Eq.(3.9), can also be expressed as

3.3 Response of an Undamped System Under Harmonic Force and

3.3 Response of an Undamped System Under Harmonic Force Beating Phenomenon If the forcing frequency is close to, but not exactly equal to, the natural frequency of the system, beating may occur. The phenomenon of beating can be expressed as:

3.3 Response of an Undamped System Under Harmonic Force The time between the points of zero amplitude or the points of maximum amplitude is called the period of beating and is given by with the frequency of beating defined as

Example 3.1 Plate Supporting A Pump A reciprocating pump, having a mass of 68kg, is mounted at the middle of a steel plate of thickness 1cm, width 50cm, and length 250cm, clamped along two edges as shown in Figure. During operation of the pump, the plate is subjected to a harmonic force, F(t) = 220 cos 62.832t N. Find the amplitude of vibration of the plate.

Example 3.1 Solution The plate can be modeled as a fixed-fixed beam having Young’s modulus (E) = 200GPa, length = 250cm, and area moment of inertia, The bending stiffness of the beam is given by: Thus,

3.4 Response of a Damped System Under Harmonic Force The equation of motion can be derived as Using trigonometric relations, we obtain The solution gives and

3.4 Response of a Damped System Under Harmonic Force The figure shows typical plots of the forcing function and steady-state) response. Substituting the following,

3.4 Response of a Damped System Under Harmonic Force We obtain and

3.4 Response of a Damped System Under Harmonic Force The following characteristics of the magnification factor (M) can be noted from figure and as follows:

3.4 Response of a Damped System Under Harmonic Force For an undamped system , Eq.(3.30) reduces to Eq.(3.10), and as . Any amount of damping reduces the magnification factor (M) for all values of the forcing frequency. For any specified value of r, a higher value of damping reduces the value of M. In the degenerate case of a constant force (when r = 0), the value of M = 1.

3.4 Response of a Damped System Under Harmonic Force The reduction in M in the presence of damping is very significant at or near resonance. The amplitude of forced vibration becomes smaller with increasing values of the forcing frequency (that is, as ). For , the maximum value of M occurs when which can be seen to be lower than the undamped natural frequency and the damped frequency .

3.4 Response of a Damped System Under Harmonic Force The maximum value of X (when ) is given by: and the value of X at by For when r = 0. For , the graph of M monotonically decreases with increasing values of r.

3.4 Response of a Damped System Under Harmonic Force The following characteristics of the phase angle can be observed from figure and Eq.(3.31) as follows:

3.4 Response of a Damped System Under Harmonic Force For an undamped system , Eq.(3.31) shows that the phase angle is 0 for 0 < r < 1 and 180° for r > 1. This implies that the excitation and response are in phase for 0 < r < 1 and out of phase for r > 1 when . For and 0 < r < 1, the phase angle is given by 0 < Φ < 90°, implying that the response lags the excitation. For and r > 1, the phase angle is given by 90° < Φ < 180°, implying that the response leads the excitation.

3.4 Response of a Damped System Under Harmonic Force For and r = 1, the phase angle is given by Φ = 90°, implying that the phase difference between the excitation and the response is 90°. For and large values of r, the phase angle approaches 180°, implying that the response and the excitation are out of phase.

3.4 Response of a Damped System Under Harmonic Force Total response For an underdamped system, where For the initial conditions, Eq.(3.35) yields

Example 3.2 Total Response of a System Find the total response of a single degree of freedom system with m = 10kg, c = 20 N-s/m, k = 4000 N/m, x0 = 0.01 m, under the following conditions: An external force acts on the system with and . Free vibration with F(t) = 0.

Example 3.2 Solution a. From the given data,

Example 3.2 Solution Using initial conditions , Substituting Eq.(E.3) into (E.4),

Example 3.2 Solution Hence, and or a. From the given data,

Example 3.2 Solution b. For free vibration, the total response is Using the initial conditions,

3.4 Response of a Damped System Under Harmonic Force Quality factor and bandwidth: For values of damping , we can take Power absorbed by damper: From figure, R1 and R2 is the bandwidth of the system. Set , hence,

3.4 Response of a Damped System Under Harmonic Force Solving the equation for small values of , Using the relation, Thus, we obtained

3.5 Response of a Damped System Under The equation of motion becomes Assuming the particular solution Substituting,

3.5 Response of a Damped System Under Using the relation , Hence, the steady-state solution becomes,

3.5 Response of a Damped System Under Frequency Response The complex frequency response is given by: The absolute value becomes, where Thus, the steady-state solution becomes,

3.5 Response of a Damped System Under If , the corresponding steady-state solution is given by the real part of Eq.(3.53) If , the corresponding steady-state solution is given by the imaginary part of Eq.(3.53)

3.5 Response of a Damped System Under Complex Vector Representation of Harmonic Motion. Differentiating Eq.(3.58) with respect to time, The various terms of the equation of motion can be represented in the figure,

3.6 Response of a Damped System Under the Harmonic Motion of the Base From the figure, the equation of motion is If , where

3.6 Response of a Damped System Under the Harmonic Motion of the Base The steady-state response of the mass can be expressed as where or where and

3.6 Response of a Damped System Under the Harmonic Motion of the Base The variations of displacement transmissibility is shown in the figure below.

3.6 Response of a Damped System Under the Harmonic Motion of the Base The following aspects of displacement transmissibility can be noted from the figure: The value of Td is unity at r = 0 and close to unity for small values of r. For an undamped system (ζ = 0), Td →∞ at resonance (r = 1). The value of Td is less than unity (Td < 1) for values of r >√2 (for any amount of damping ζ ). The value of Td = 1 for all values of ζ at r =√2.

3.6 Response of a Damped System Under the Harmonic Motion of the Base For r <√2, smaller damping ratios lead to larger values of Td. On the other hand, for r >√2, smaller values of damping ratio lead to smaller values of Td. The displacement transmissibility, Td, attains a maximum for 0 < ζ < 1 at the frequency ratio r = rm < 1 given by:

3.6 Response of a Damped System Under the Harmonic Motion of the Base Force transmitted: The force transmissibility is given by:

3.6 Response of a Damped System Under the Harmonic Motion of the Base Relative Motion: The equation of motion can be written as The steady-state solution is given by: where, the amplitude, and

3.6 Response of a Damped System Under the Harmonic Motion of the Base

Example 3.3 Vehicle Moving on a Rough Road The figure below shows a simple model of a motor vehicle that can vibrate in the vertical direction while traveling over a rough road. The vehicle has a mass of 1200kg. The suspension system has a spring constant of 400 kN/m and a damping ratio of ζ = 0.5. If the vehicle speed is 20 km/hr, determine the displacement amplitude of the vehicle. The road surface varies sinusoidally with an amplitude of Y = 0.05m and a wavelength of 6m.

Example 3.3 Solution The frequency can be found by For v = 20 km/hr, ω = 5.81778 rad/s. The natural frequency is given by, Hence, the frequency ratio is

Example 3.3 Solution The amplitude ratio can be found from Eq.(3.68): Thus, the displacement amplitude of the vehicle is given by This indicates that a 5cm bump in the road is transmitted as a 7.3cm bump to the chassis and the passengers of the car.

3.7 Response of a Damped System Under Rotating Unbalance The equation of motion can be derived by the usual procedure: The solution can be expressed as The amplitude and phase angle is given by

3.7 Response of a Damped System Under Rotating Unbalance By defining , and

3.7 Response of a Damped System Under Rotating Unbalance The following observations can be made from Eq.(3.82) and the figure above: All the curves begin at zero amplitude. The amplitude near resonance is markedly affected by damping. Thus if the machine is to be run near resonance, damping should be introduced purposefully to avoid dangerous amplitudes. At very high speeds (ω large), MX/me is almost unity, and the effect of damping is negligible. For 0 < ζ < 1/√2 , the maximum of MX/me occurs when

3.7 Response of a Damped System Under Rotating Unbalance The solution gives: With corresponding maximum value Thus the peaks occur to the right of the resonance value of r = 1. For , does not attain a maximum. Its value grows from 0 at r = 0 to 1 at r → ∞ .

Example 3.5 Francis Water Turbine The schematic diagram of a Francis water turbine is shown in the figure in which water flows from A into the blades B and down into the tail race C. The rotor has a mass of 250 kg and an unbalance (me) of 5kg-mm. The radial clearance between the rotor and the stator is 5mm. The turbine operates in the speed range 600 to 6000rpm. The steel shaft carrying the rotor can be assumed to be clamped at the bearings. Determine the diameter of the shaft so that the rotor is always clear of the stator at all the operating speeds of the turbine. Assume damping to be negligible.

Example 3.5 Francis Water Turbine

Example 3.5 Solution The max amplitude can be obtained from Eq.(3.80) by setting c = 0 as The value of ω ranges from: to While the natural frequency is given by

Example 3.5 Solution For ω = 20π, Eq.(E.1) gives The stiffness of the cantilever beam is given by

Example 3.5 Solution and the diameter of the beam can be found: or

3.8 Forced Vibration with Coulomb Damping The equation of motion is given by

3.8 Forced Vibration with Coulomb Damping The energy dissipated by dry friction damping is If the equivalent viscous damping constant is denoted as ceq, Thus the steady-state response is:

3.8 Forced Vibration with Coulomb Damping where the amplitude can be found from Eq.(3.60): with Substituting Eq.(3.91) into (3.90) gives:

3.8 Forced Vibration with Coulomb Damping The solution of this equation gives: To avoid imaginary values of X, we need to have

3.8 Forced Vibration with Coulomb Damping The phase angle can be found: The energy directed into the system over one cycle when it is excited harmonically at resonance and that Φ = 90°,

3.8 Forced Vibration with Coulomb Damping For the nonresonant condition, the energy input is

Example 3.6 Spring-Mass System with Coulomb Damping A spring-mass system, having a mass of 10kg and a spring of stiffness of 4000 N/m, vibrates on a horizontal surface. The coefficient of friction is 0.12. When subjected to a harmonic force of frequency 2 Hz, the mass is found to vibrate with an amplitude of 40 mm. Find the amplitude of the harmonic force applied to the mass.

Example 3.6 Solution The vertical force (weight) of the mass is N = mg = 10 X 9.81 = 98.1N. The natural frequency is And the frequency ratio is The amplitude of vibration X is given by:

Example 3.6 Solution Thus, the solution gives .

3.9 Forced Vibration with Hysteresis Damping From figure, the equation of motion can be derived The steady-state solution can be assumed as Substituting, we obtained

3.9 Forced Vibration with Hysteresis Damping and And thus the amplitude ratio attains its maximum value at the resonant frequency in the case of hysteresis damping, while it occurs at a frequency below resonance in the case of viscous damping.

3.9 Forced Vibration with Hysteresis Damping The phase angle has a value of tan-1(β) at ω = 0 in the case of hysteresis damping, while it has a value of zero at ω = 0 in the case of viscous damping. This indicates that the response can never be in phase with the forcing function in the case of hysteresis damping. Thus, the equation of motion becomes where the quantity is called the complex stiffness or complex damping.

3.9 Forced Vibration with Hysteresis Damping The steady-state solution is given by the real part of

3.10 Forced Motion with Other Types of Damping Example 3.7 Quadratic Damping Find the equivalent viscous damping coefficient corresponding to quadratic or velocity squared damping that is present when a body moves in a turbulent fluid flow.

Example 3.7 Solution The damping force is assumed to be where a is a constant. The energy dissipated per cycle is We obtain the equivalent viscous damping coefficient

Example 3.7 Solution The amplitude of the steady-state response is where r = ω/ωn, and Hence,

3.11 Self-Excitation and Stability Analysis Dynamic Stability Analysis Consider the equation of motion of a single degree of freedom system: This leads to a characteristic equation The roots of the equation are:

3.11 Self-Excitation and Stability Analysis Let the roots be expressed as where p and q are real numbers so that Hence,

Example 3.8 Instability of Spring-Supported Mass on Moving Belt Consider a spring-supported mass on a moving belt, as shown in figure (a). The kinetic coefficient of friction between the mass and the velt varies with the relative (rubbing) velocity, as shown in figure (b). As rubbing velocity increases, the coefficient of friction first decreases from its static value linearly and them starts to increase. Assuming that the rubbing velocity, v, is less than the transition value, vQ, the coefficient of friction can be expressed as where a is a constant and W=mg is the weight of the mass. Determine the nature of free vibration about the equilibrium position of the mass.

Example 3.8 Solution Let the equilibrium position of mass m correspond to an extension of x0 of the spring. Then, where V is the velocity of the belt. Hence, the rubbing velocity v is given by:

Example 3.8 Solution The equation of motion for free vibration is i.e., The solution is given by where C1 and C2 are constants and,

3.11 Self-Excitation and Stability Analysis Dynamic Instability Caused by Fluid Flow The figure illustrates the phenomenon of galloping of wires:

3.11 Self-Excitation and Stability Analysis The figure illustrates the phenomenon of singing of wires: Experimental data show that regular shedding occurs strongly in the range of Reynolds number (Re) from about 60 to 5000. In this case,

3.11 Self-Excitation and Stability Analysis For Re > 1000, the dimensionless frequency of vortex shedding, expressed as Strouhal number (St), is approximately equal to 0.21. where f is the frequency of vortex shedding. The harmonically varying lift force (F) is given by where c is a constant (c = 1 for a cylinder), A is the projected area of the cylinder perpendicular to the direction of V, ω is curcular frequency and t is time.

Example 3.10 Flow-Induced Vibration of a Chimney A steel chimney has a height of 2m, an inner diameter 0.75m, and an outer diameter 0.80m. Find the velocity of the wind flowing around the chimney which will induce transverse vibration of the chimney in the direction of airflow.

Example 3.10 Solution Approach: Model the chimney as a cantilever beam and equate the natural frequency of the transverse vibration of the chimnet to the frequency of vortex shedding. The natural frequency of transverse vibration of a cantilever beam is where For the chimney, E = 207X109 Pa, ρg = 76.5X103 N/m3, l = 20m, d = 0.75m, D = 0.80m,

Example 3.10 Solution and Thus,

Example 3.10 Solution The frequency of vortex shedding is given by Strouhal number: The velocity wind (V) which causes resonance can be determined as