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Linear Vibration analysis of Mechanical Systems

linear system theory in vibration engineering and linear vibration motor using resonance frequency linear or oscillating vibration machine linear modes of vibration
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Dr.NaveenBansal,India,Teacher
Published Date:25-10-2017
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Ch6Frame Page 171 Friday, June 2, 2000 6:43 PM Introduction to 6 Linear Vibrations 6.1 INTRODUCTION A question one may ask is “Why do bodies vibrate?” The explanation is rooted in the energy conservation principle. Consider as an example a mass m suspended on a spring having stiffness k (Figure 6.1) and assume that this mass is pushed up (or down) from its static equilibrium position by the amount y and then released. max By deforming the spring some energy is stored in it, which will be denoted by V . This energy is called the potential energy, and V represents the maximum max max energy transferred to the spring by deforming it by y . After releasing the mass, it max will start moving back to its original position. The motion of the mass means that it acquires some kinetic energy, which is equal to 1 2 T = - my ˙ (6.1) 2 where y ˙ is the time derivative of mass displacement, i.e., its velocity. At any intermediate position y the potential energy of the deformed spring is equal to 1 2 V = - ky (6.2) 2 If one assumes that the spring is ideal, i.e., its deformation does not lead to any energy losses, then according to the energy conservation principle the sum of the kinetic energy of the mass and the potential energy in the spring must be equal to the original energy introduced into this system, V . Thus, max 1 1 2 2 - my ˙ + - ky== V const. (6.3) max 2 2 The above equation shows that in a spring–mass system an energy transformation takes place, from potential to kinetic and back. More than that, one can see that this process is periodic. Indeed, when yy = – then VV = and it follows that at max max these extreme positions y˙0 = . Thus, there are two extreme positions of the mass and they are equal in magnitude, but at the opposite sides of the static equilibrium position. On the other hand, when y = 0 (or, more correctly, a static displacement), y ˙ = y ˙ . This process of a mass moving between two extreme positions in a periodic max fashion is called oscillation. It is characterized by two parameters: the amplitude of oscillation and the period of oscillation. The former is the maximum mass displace- 171 Ch6Frame Page 172 Friday, June 2, 2000 6:43 PM 172 Fundamentals of Kinematics and Dynamics of Machines and Mechanisms FIGURE 6.1 Spring–mass system. ment from the position of static equilibrium, y , whereas the latter is the time max between two consecutive maximum displacements. The relationship given by Equation 6.3 is a differential one. Since it is known that the motion of the mass is periodic with the amplitude y , one can assume it max to be described by a function: t yy = sin 2p - + a (6.4) max Łł T where t is the time, T is the period of oscillation, and a is some constant which is defined by the requirement that at t = 0, y = y . max In Equation 6.4 the maximum displacement y is known, and T is the only max unknown (besides a ) to be found in order to describe the motion of the mass in time. The expression for T follows from the requirement that the function (Equation 6.4) must satisfy Equation 6.3. Substitute yy and ˙ into Equation 6.3. After simple derivations, the differential Equation 6.3 is reduced to an algebraic relationship: 2 m2() p t 2 2 - - – k y cos 2p - + a = 0 (6.5) max 2 Łł T T For the above equality to be true at any time, the term in the brackets must be equal to zero, i.e., 2 m2() p - - – k = 0 (6.6) 2 Łł T The latter equation defines the period of oscillation as a function of mass and spring stiffness: m T = 2p - - (6.7) k Ch6Frame Page 173 Friday, June 2, 2000 6:43 PM Introduction to Linear Vibrations 173 The important point is that the period of oscillation is an intrinsic property of the mass–spring system and does not depend on the initial disturbance (or on the initial energy pumped into the system). The inverse of T gives the number of oscillations per unit time and it is called the frequency of oscillation, f. If the time is measured in seconds, then one oscillation per second is called a hertz. 1 1 k f== - - - - - (6.8) T 2p m In Equation 6.4 2p /T gives the angular displacement per unit time; i.e., it gives the angular velocity. This angular velocity is called the circular frequency of oscillation, w , and it is equal to 2p k w== - - - - (6.9) T m The mass–spring system shown in Figure 6.1 represents a single-degree-of-freedom (SDOF) system. The motion of the mass discussed above was caused by its initial displacement and Equation 6.3 describes this motion. One can simplify this equation by differentiating both sides of it with respect to time. The result is () my ˙˙+ ky y˙0 = (6.10) Since y˙0 „ at all times, Equation 6.10 can be satisfied only if the first term equals zero, i.e., my ˙˙+0 ky = (6.11) Now it can be seen that the obtained equation expresses Newton’s second law, where ky is the external spring force acting on the mass treated as a free body. In general, if a periodic force Pt() = P sin() w t acts upon the mass, then its motion 0 equation can be obtained by considering the equilibrium of forces, including inertial force, acting on this mass (Figure 6.2). Taking into account that forces ky and P(t) are opposite, the motion equation becomes my ˙˙+ ky = P sin() w t (6.12) 0 Equations 6.11 and 6.12 have been derived assuming that the system is not experi- encing any energy losses during oscillations. If this assumption cannot be made, then a force associated with the energy losses should be added to the resultant force acting on the mass in Figure 6.2. The causes of energy losses may be many, such as friction between the moving parts, oil or air resistance, internal losses in materials. These different causes may entail different mathematical models to describe them. In general, unless there is a dominating factor, all of them contribute to the total energy loss. The most convenient way to model energy losses is to assume that they are Ch6Frame Page 174 Friday, June 2, 2000 6:43 PM 174 Fundamentals of Kinematics and Dynamics of Machines and Mechanisms FIGURE 6.2 External force applied to a mass–spring system (a), and a free-body diagram of the mass (b). FIGURE 6.3 Mass–spring–damper system (a), and free-body diagram of mass (b). caused by forces proportional to the body velocity, similar to the resistance experi- enced by a body moving through a viscous liquid. Accordingly, these forces are called viscous forces and are taken in the form: F () t = cy ˙ (6.13) r where c is called the damping coefficient. In Figure 6.3a in addition to the spring a damping element is shown, which is conventionally used to identify the presence of viscous resistance force having damp- ing coefficient c. This additional force acts in the same direction as the spring resistance force (Figure 6.3b). The dynamic equilibrium of forces shown in Figure 6.3b gives the following differential equation: my ˙˙++ cy ˙ ky = P sin() w t (6.14) 0 Equations 6.11, 6.12, and 6.14 have one important property; they are linear differential equations. Their linearity is due to the assumptions that resistance forcesCh6Frame Page 175 Friday, June 2, 2000 6:43 PM Introduction to Linear Vibrations 175 are proportional to either displacement or velocity. If any of these assumptions is not adequate, then the equation becomes nonlinear. There are well-developed math- ematical techniques for solving linear equations, whereas nonlinear equations very often require an individual analytical approach or a numerical solution. This book will be limited to linear equations. Equation 6.14 is a second-order nonhomogeneous linear equation with constant coefficients. The next section considers solutions of Equation 6.14. 6.2 SOLUTION OF SECOND-ORDER NONHOMOGENEOUS EQUATIONS WITH CONSTANT COEFFICIENTS Linear equations have a very important property; they allow linear combination of solutions to form new solutions. Namely, if f () t and f () t are two solutions, then 1 2 their combination with some constant coefficients, ft() = a f () t + a f () t , is also 1 1 2 2 a solution of this equation. This is called the superposition principle. The following will use this principle in various situations, but its first application is to split the solution of Equation 6.14 into two parts: one describing the behavior of the system without the external forces (the corresponding differential equation is called homo- geneous and its solution the general solution) and the other describing the behavior of the system subjected to external forces (the corresponding differential equation is called nonhomogeneous and its solution the particular solution). The solution of the homogeneous equation will first be considered. 6.2.1 SOLUTION OF THE HOMOGENEOUS EQUATION The homogeneous equation to be solved is my ˙˙++ cy ˙ ky =0 (6.15) It is more convenient for the following to transform this equation into one with nondimensional constants by dividing each term by m. The transformed equation is as follows: 2 y ˙˙++ 2xw y ˙ w y = 0 (6.16) n n where it is denoted c 2 k - - - - x== and w (6.17) n 2mw m n In Equation 6.17 x is called the nondimensional damping coefficient, and w is called n the natural frequency (see Section 6.3). It is known that an exponential function l t yt() = e (6.18)Ch6Frame Page 176 Friday, June 2, 2000 6:43 PM 176 Fundamentals of Kinematics and Dynamics of Machines and Mechanisms is a possible solution of this equation, where l is some constant. If this function is substituted into Equation 6.16, the following algebraic equation is obtained: 2 2 l t () l + 2xw lw + e = 0 (6.19) n n l t Since e in the above equation cannot be equal to zero, the nontrivial solution of Equation 6.19 exists only if the quadratic polynomial equals zero: 2 2 l++ 2xw lw =0 (6.20) n n The above equation is called the characteristic equation, since it defines two unique roots and thus two possible solutions of the differential equation. The two roots are 2 l =() – x – x – 1 w (6.21) 12 , n If l and l are two distinct roots, then it means that there are two possible solutions 1 2 of Equation 6.16. The general solution is a combination (again, the principle of superposition is used) of these two solutions with some constants: l t l t 1 2 yt() = c e + c e (6.22) 1 2 The specific form of the solution given by Equation 6.22 depends on the type of roots in Equation 6.21: real and distinct, complex, or real and equal. Each of these cases will be considered separately since, as will be seen later, the type of roots reflects the type of system behavior. • Roots are real and distinct This is possible if x 1 in Equation 6.21; then the solution given by Equation 6.22 is an exponentially decreasing in time function since both roots are negative. • Roots are complex This is possible if x 1 in Equation 6.21; then they can be written in the form 2 l = a – ib , where ax = – w and b = 1 – x w are real numbers and i is the 12 , n n imaginary unit (square root of –1). Now the solution Equation 6.22 takes the following form: () a + ib t () a – ib t yt() = c e + c e (6.23) 1 2 By using the Euler formula – ib t e = cosbti – sinb t (6.24)Ch6Frame Page 177 Friday, June 2, 2000 6:43 PM Introduction to Linear Vibrations 177 the solution Equation 6.23 can be transformed to a t a t yt() = c e() cosb t + isinb t + c e() cosb t – isinb t 1 2 a t (6.25) = e () c + c cosbti +() c – c sinb t 1 2 1 2 a t = e a cosbta + sinb t 1 2 where a = c + c and a = i(c – c ) are two new arbitrary constants. 1 1 2 2 1 2 • Roots are real and equal This is possible if x = 1 in Equation 6.21 so that l ==lx– w . In this case, 12 , n in addition to the solution given by Equation 6.18, there is another one given by the l t function te , and thus, the general solution becomes l t l t yt() = c e + c te (6.26) 1 2 6.2.2 PARTICULAR SOLUTION OF THE NONHOMOGENEOUS EQUATION The nonhomogeneous equation with nondimensional coefficients is 2 y ˙˙++ 2xw y ˙ w y = p sinw t (6.27) n n o where p = P /m. The particular solution caused by the harmonic forcing function o o is also harmonic and can be taken in either of the following two forms: y () t== d cosw td + sinw t Dsin() w t – f (6.28) p 1 2 where d , d , D, and f are constants. 1 2 Taking the second form of the solution in Equation 6.28 and substituting it into Equation 6.27 obtains 2 2 D () w – w sin() w t – f + 2xw ww cos() t – f = p sinw t (6.29) n n o By using the trigonometric relations cos() w t – f = cosw t cosfw + sin t sinf (6.30) and sin() w t – f = sinw t cosfw –cos t sinf (6.31)Ch6Frame Page 178 Friday, June 2, 2000 6:43 PM 178 Fundamentals of Kinematics and Dynamics of Machines and Mechanisms Equation 6.29 is transformed into the following: 2 2 D –() w – w sinf + 2xw wf cos cosw t n n (6.32) 2 2 + D () w – w cosf + 2xw wf sin sinwtp = sinw t n n o The above equation can be satisfied at any moment in time if the harmonic functions on the left and on the right have equal amplitudes. This requirement leads to two equations: 2 2 D –() w – w sinf + 2xw wf cos = 0 (6.33) n n and 2 2 D () w – w cosf + 2xw wf sin = p (6.34) n n o To solve the above system for the unknowns D and f , transform it into a simpler system. This is done by first multiplying Equation 6.33 by –sin f and Equation 6.34 by cos f and adding them. The result is 2 2 D() w – w = p cosf (6.35) n o Now, multiply Equation 6.33 by cos f and Equation 6.34 by sin f and add them. The result is D2xw w = p sinf (6.36) n o The new system, Equations 6.35 and 6.36, can be easily solved. Square both sides in Equations 6.35 and 6.36 and add the equations. This gives the expression for the unknown D p o D = - - (6.37) 2 2 2 2 () w – w +() 2xw w n n Now if Equation 6.36 is divided by Equation 6.35, the expression for the angle f is obtained: 2xw w n f = arctan (6.38) 2 2 Łł w – w n Note that the constants d , d in Equation 6.28 are equal to 1 2 d = –Dsinf and d = Dcosf (6.39) 1 2Ch6Frame Page 179 Friday, June 2, 2000 6:43 PM Introduction to Linear Vibrations 179 Using Equations 6.35 and 6.36 to express sin f and cos f through D gives the following expressions for these coefficients: 2 2D xw w n - - d = – (6.40) 1 p o and 2 2 2 D() w – w n d = - - (6.41) 2 p o where D is given by Equation 6.37. It is of interest to consider the case when, instead of forcing function p sin w t, 0 a function p cos w t is used in Equation 6.27. It can be checked that the solutions 0 for the amplitude, Equation 6.37, and for the phase angle, Equation 6.38, remain the same. 6.2.3 COMPLETE SOLUTION OF THE NONHOMOGENEOUS EQUATION The complete solution is a superposition of the general solution of the homogeneous equation (Equation 6.22) and the particular solution of the nonhomogeneous equation (Equation 6.28). Thus, l t l t 1 2 y () t = y () t + y () t = c e++ c e d cosw td+ sinw t (6.42) c g p 1 2 1 2 where c and c are unknown constants, and d and d are given by Equations 6.40 1 2 1 2 and 6.41. As was discussed in Section 6.1, the vibration of the mass on a spring is caused by the initial displacement of this mass. By displacing and releasing the mass, one introduces into the mass–spring system some initial potential energy. It was tacitly assumed that this initial displacement was slow enough so that the corresponding kinetic energy of motion could be ignored. However, if this is not the case, then the total energy transferred to the system is a sum of both potential and kinetic energies. The former is associated with the initial displacement of the mass, while the second is associated with its initial velocity. Thus, in general, the motion of a body from the undisturbed position starts with some initial displacement and with some initial velocity. These are called initial conditions. For a single body in a uniaxial motion, there are two initial conditions, which are stated as follows dy t c - - y() 0 = Y and = V (6.43) c 0 0 dt t = 0 These two initial conditions define the constants c and c in Equation 6.42. Satisfying 1 2 the initial displacement condition givesCh6Frame Page 180 Friday, June 2, 2000 6:43 PM 180 Fundamentals of Kinematics and Dynamics of Machines and Mechanisms c++ c d = Y (6.44) 1 2 1 0 Then taking the derivative of y (t) and then satisfying the initial velocity requirement c gives c l++ c l d w = V (6.45) 1 1 2 2 2 0 Solving Equations 6.44 and 6.45 for the unknowns c and c gives 1 2 () Y – d l – V + d w 0 1 2 0 2 c = - - (6.46) 1 l – l 2 1 and () Y – d l – V + d w 0 1 1 0 2 c = - - (6.47) 2 l – l 1 2 where l and l are given by Equation 6.21. 1 2 The type of roots (real or complex) in Equation 6.21 leads to different forms of complete solution. Each case will be looked into separately. • Roots are real and distinct () x 1 In this case the complete solution is – l t – l t 1 2 y () t = c e++ c e d cosw td + sinw t (6.48) c 1 2 1 2 One can see that the first two terms in Equation 6.48, associated with the initial disturbance, tend to zero exponentially and so after some time, practically, only the periodic terms caused by the external load will remain. A system in which the nondimensional damping coefficient x 1 is said to be overdamped. • Roots are complex () x 1 In this case the complete solution is (see Equation 6.25 for the general part of the solution) –xw t n y () t = e() a cosbta + sinb t + d cosw td + sinw t (6.49) c 1 2 1 2 One can see that in this case the amplitude of the initial disturbance is also decreasing exponentially while oscillating. A system in which the nondimensional damping coefficient x 1 is said to be underdamped. However, in this case, like in the previous one, the effect of initial disturbance disappears after some time and only the periodic oscillation remains.Ch6Frame Page 181 Friday, June 2, 2000 6:43 PM Introduction to Linear Vibrations 181 • Roots are real and equal (x = 1) In this case the complete solution is –xw t n y () t = e () c + c t + d cosw td + sinw t (6.50) c 1 2 1 2 One can see that the initial disturbance tends to zero and after some time only the oscillation caused by the periodic external force remain. This is the boundary case between the oscillating and nonoscillating initial disturbance, and the corresponding damping is called critical damping. The next section considers various applications of the above solution to an SDOF system. 6.3 FREE VIBRATIONS OF AN SDOF SYSTEM WITH NO DAMPING (x== 0, p 0) o The characteristic equation, Equation 6.20, in this case has imaginary roots (see Equation 6.21) l = – i() w . This is a particular case of the complete solution 12 , n obtained in Section 6.2.3 and the corresponding solution can be obtained from Equation 6.49 by taking, x=== 0, d 0, d 0, and b= w . As a result, one 1 2 n obtains y () t = a cosw ta + sinw t (6.51) g 1 n 2 n where a and a are constants defined by the initial conditions: 1 2 y() 0== a Y (6.52) g 1 0 and y ˙() 0== a w V (6.53) g 2 n 0 Thus, the final form of the solution for an SDOF system without damping is V 0 y () t = Y cosw t + - - sinw t (6.54) g 0 n n w n This equation once more shows that any disturbance causes the system to oscillate with the circular frequency w , which is defined by the system properties. n This frequency is called the natural frequency of the system. One can also see that the effects of initial displacement and initial velocity are uncoupled, which is another manifestation of the principle of superposition in linear systems. In other words, one can solve first for initial displacement, second for initial velocity, and then combine the results.Ch6Frame Page 182 Friday, June 2, 2000 6:43 PM 182 Fundamentals of Kinematics and Dynamics of Machines and Mechanisms y (t) g y st 7.5 5 2.5 Time 1 2 3 4 5 6 -2.5 -5 -7.5 FIGURE 6.4 Normalized amplitude of free vibrations of an SDOF system without damping (y is the static displacement). Dashed line = initial velocity; solid thin line = initial displace- st ment; solid thick line = both. In Figure 6.4 oscillations of an SDOF system are shown for the case of initial displacement only, initial velocity only, and for both, displacement and velocity, conditions. The time between two consecutive peaks is the period of oscillation. One can see that the period is the same for any type of initial disturbance. Recall that this period is equal to T = 2p /w and w does not depend on the type of n n disturbance. 6.4 FORCED VIBRATIONS OF AN SDOF SYSTEM WITH NO DAMPING (x = 0 ) It is assumed that forced vibrations start at time zero from the undisturbed state of the system. Thus, the initial conditions are Y = 0 and V = 0 at t = 0. The forcing 0 0 function is p sin(w t). The complete solution is given by Equation 6.49 in which the 0 first two terms on the right describing the vibrations caused by the initial conditions contain two unknown (a and a ) constants, whereas the constants in the last two 1 2 terms (d and d ) are given by Equations 6.40 and 6.41 and Equation 6.37. Taking 1 2 into account that, for the case of x = 0 , bw = and d = 0, the complete solution is n 1 y () t = a cosw ta + sinw td + sinw t (6.55) c 1 n 2 n 2 where the constants a and a are found by satisfying the initial conditions 1 2 y() 0== a 0 (6.56) c 1 and dy () t c == a w+0 d w (6.57) - - 2 n 2 dt t = 0Ch6Frame Page 183 Friday, June 2, 2000 6:43 PM Introduction to Linear Vibrations 183 It follows from the latter equation that w a = –d - - (6.58) 2 2 w n Substituting the above constants into Equation 6.55 yields w y () t = d sinw t – - - sinw t (6.59) c 2 n Łł w n where (see Equation 6.41) 2 2 2 D() w – w n d = - - (6.60) 2 p 0 and (see Equation 6.37) p 0 D = (6.61) 2 2 w – w n Thus, the final form of the solution of forced vibrations starting from zero initial conditions is p w 0 - - y () t = sinw t – sinw t (6.62) c n 2 2 Łł w n w – w n The obtained result shows that the forced vibration in this case is a superposition of two motions: one with the frequency of the external force w and the other with the natural frequency w . The resultant amplitude of vibrations is a function of the n frequency w , and at w = w it becomes undetermined since both numerator and n denominator equal zero at this frequency. One can use L’Hopital’s rule to resolve the uncertainty by taking the ratio of derivatives of the numerator and denominator with respect to w and then setting w = w . The result is n –sinw t + w t cosw t n n n y() ww= = - (6.63) c n –2w n One can see that the amplitude of vibrations in this case is proportional to the time t and thus grows to infinity. This phenomenon of unlimited amplitude growth is called resonance, and the corresponding frequency is called the resonance frequency. As is clear, the resonance frequency is the natural frequency of the system. It will be seen later that in real systems with damping these frequencies are close but not equal.Ch6Frame Page 184 Friday, June 2, 2000 6:43 PM 184 Fundamentals of Kinematics and Dynamics of Machines and Mechanisms y (t) c y st 40 20 Time 2 4 6 8 10 -20 -40 FIGURE 6.5 Dynamic response in time of an SDOF system without damping (y is the st static displacement). Dashed line = w /w = 0.5; solid thin line = w /w = 0.75; solid thick line n n = w /w = 1. n In Figure 6.5 the displacement of the mass in a system without damping is shown at different frequencies of external force. One can see that at the resonance frequency the amplitude of displacement grows linearly with time. 6.5 STEADY-STATE FORCED VIBRATIONS OF AN SDOF SYSTEM WITH NO DAMPING ( x = 0) In this case the effect of the initial conditions is neglected and thus only the particular solution is considered. The latter is given by Equation 6.55 in which both a and a 1 2 equal zero. Thus, Equation 6.62 in which the second term in the parentheses is absent gives the solution p 0 y () t = sinw t (6.64) c 2 2 w – w n This solution shows that the frequency of vibrations in this case is equal to the frequency of the forcing function, and the amplitude is a function of this frequency. One can see now that when w = w the amplitude becomes infinite. Now look more n closely at the amplitude as a function of forcing frequency w . p 0 D = (6.65) 2 2 w – w n From the latter equation it follows that, when w = 0, then p p m p 0 0 0 D== - - - -= -= D st 2 mk k w nCh6Frame Page 185 Friday, June 2, 2000 6:43 PM Introduction to Linear Vibrations 185 y c y st 20 10 w 2 4 6 8 10 w n -10 -20 FIGURE 6.6 Amplitude–frequency diagram of an SDOF system without damping (y is st the static displacement). where D is the static displacement of the mass caused by the force P . Furthermore, st 0 when w approaches w from the left (w w ), then the amplitude tends to infinity n n while remaining positive, whereas when w approaches w from the right (w w ), n n then the amplitude tends to infinity while remaining negative. Also, when w tends to infinity, then the amplitude tends to zero while being negative. This dependence of the amplitude on the frequency is shown in the diagram of Figure 6.6 in normalized coordinates. This diagram is called the amplitude–frequency diagram. The function in the diagram depicted in Figure 6.6 comprises two continuous functions. Such a function is called a piecewise continuous function and can be described as follows: p 0 sin() w t ifww n 2 2 w – w n y () t = (6.66) c p 0 sin() w t – p ifww n 2 2 w – w Ł n In Equation 6.66 the amplitude is now always positive, whereas the sign change after the resonance is controlled by the shift in the angle by p . This angle p is called the phase angle. Thus, up to the resonance frequency the phase angle is equal to 0, and after the resonance frequency it becomes –p . Below it will be seen that for systems with damping this change in the phase angle is continuous. 6.6 FREE VIBRATIONS OF AN SDOF SYSTEM WITH DAMPING (x „ 0, p = 0 ) 0 As was discussed in Section 6.2.1, the form of the general part of the solution depends on the degree of damping. Consider each case separately.Ch6Frame Page 186 Friday, June 2, 2000 6:43 PM 186 Fundamentals of Kinematics and Dynamics of Machines and Mechanisms • Overdamped system () x 1 In this case Equation 6.42, in which d and d equal zero, gives the complete 1 2 solution: l t l t 1 2 y () t = c e + c e (6.67) c 1 2 and c and c are determined by the initial conditions: 1 2 y() 0 = c + c = Y (6.68) c 1 2 0 and y ˙() 0 = c l + c l = V (6.69) c 1 1 2 2 0 2 2 Taking into account that l =() – x + x – 1 w , l =() – x – x – 1 w (see Equa- 1 n 2 n tion 6.21), the expressions for the constants c and c become 1 2 Y l – V 0 2 0 c = – - - (6.70) 1 2 2 x – 1w n and Y l – V 0 1 0 c = - - (6.71) 2 2 2 x – 1w n One can see that the two initial conditions are independent of each other. In Figure 6.7 the response of the overdamped system to initial conditions is shown for the case of , x = 1.2 w = 10 rad/s, Y = 1cm, and V = 5 cm/s. n 0 0 • Underdamped system () x 1 In this case the complete solution is given by Equation 6.49 in which d and d 1 2 equal zero. –xw t n y () t = e() a cosbta + sinb t (6.72) c 1 2 where a and a are obtained from the initial conditions requirements 1 2 y() 0 = a = Y (6.73) c 1 0Ch6Frame Page 187 Friday, June 2, 2000 6:43 PM Introduction to Linear Vibrations 187 y c y st 6 5 4 3 2 1 Time 2 4 6 8 10 FIGURE 6.7 Response of the overdamped (x = 1.2) SDOF system to initial conditions. and dy () t c == –x w a + a b V (6.74) - - n 1 2 0 dt t = 0 Taking into account Equation 6.73, the second constant is V + xw Y 0 n 0 a = (6.75) 2 b 2 where b = 1 – x w (see Section 6.2.1). n Recall that the undamped system oscillates with natural frequency w when n subjected to initial disturbance. In the case of a damped system, the frequency of oscillation becomes 2 bw== 1 – x w (6.76) d n One can see that this frequency of free damped vibrations is always smaller than the natural frequency, w w . d n The displacement of the mass in time is shown in Figure 6.8. As is seen from Equation 6.72 the amplitude of the oscillating motion is decreasing exponentially. This property of the declining amplitude is used to determine the damping coefficient experimentally. This will be discussed in Section 6.8. It is important to point out that the coefficient of damping is not the property of the material only, since it depends also on the system mass and stiffness. Note that the material properties are not present explicitly in motion equations. It was only assumed in Section 6.2 that the material possesses some damping properties repre- sented by the damping coefficient.Ch6Frame Page 188 Friday, June 2, 2000 6:43 PM 188 Fundamentals of Kinematics and Dynamics of Machines and Mechanisms y c y st 6 4 2 Time 2 4 6 8 10 -2 -4 FIGURE 6.8 Response of the underdamped (x = 0.2) SDOF system to initial conditions. • Critical damping () x = 1 The solution is given by Equation 6.50 in which d and d equal zero. 1 2 –w t n y () t = e() c + c t (6.77) c 1 2 Satisfying the initial conditions yields y() 0== c Y (6.78) c 1 0 and dy () t c == –w c + c V (6.79) - - n 1 2 0 dt t = 0 From Equations 6.78 and 6.79 the coefficient c follows 2 c = V + w Y (6.80) 2 0 n 0 In Figure 6.9 the system response to initial conditions in the case of critical damping is shown. In Figure 6.10 the graphs from Figures 6.7 through 6.9 are shown together for the sake of comparison. 6.7 FORCED VIBRATIONS OF A DAMPED (x 1) SDOF SYSTEM WITH INITIAL CONDITIONS Equation 6.49, repeated here for the sake of convenience, gives the complete solution: –xw t n y () t = e() a cosb t + a sinb t + d cosw td + sinw t (6.81) c 1 2 1 2Ch6Frame Page 189 Friday, June 2, 2000 6:43 PM Introduction to Linear Vibrations 189 y c y st 6 5 4 3 2 1 Time 2 4 6 8 10 FIGURE 6.9 Response of the critically damped (x = 1) SDOF system to initial conditions. y c y st 8 6 4 2 Time 2 4 6 8 10 -2 -4 FIGURE 6.10 Comparison of an SDOF system response to initial conditions. Dashed line = overdamping (x = 1.2); solid thin line = underdamping (x = 0.2); solid thick line = critical damping (x = 1). Take the same initial conditions as in the case of no damping in Section 6.4, namely, that Y = 0 and V = 0 at t = 0, and the forcing function is also the same, 0 0 p sin(w t). The equations defining constants are given by Equations 6.46 and 6.47: 0 – d l + d w 1 2 2 c = - (6.82) 1 l – l 2 1 and – d l + d w 1 1 2 c = - (6.83) 2 l – l 1 2Ch6Frame Page 190 Friday, June 2, 2000 6:43 PM 190 Fundamentals of Kinematics and Dynamics of Machines and Mechanisms 2 where l terms are given by Equation 6.21, l =() – x – x – 1 w , and d and d 12 , n 1 2 2 2 2 2 by Equations 6.40 and 6.41, d = –2D xw w ⁄ p and d = D() w – w ⁄ p . One can 1 n 0 2 n 0 take into account that the roots of the characteristic equation in this case are complex and use simplified expressions for the l , l = a ± ib , where ax =w– and 1,2 n 2 b = 1 – x w . Then, the constants c and c become complex conjugate numbers: n 1 2 a w c = –0.5 d –i – d - + d - - (6.84) 1 1 1 2 Łł b b and a w c = –0.5 d +id – - + d - - (6.85) 2 1 1 2 Łł b b Thus, the constants a and a in Equation 6.81 become recall that a = c + c and 1 2 1 1 2 a = i(c – c ) 2 1 2 a== c + c –d (6.86) 1 1 2 1 and a w - - - a== ic() – c d – d (6.87) 2 1 2 1 2 b b In Figures 6.11 and 6.12 the vibrations of an SDOF initially undisturbed system are shown for two forcing frequencies and x = 0.4, p = 1. It is seen that for a 0 frequency close enough to the resonance frequency, w = 0.75w , the effect of initial n conditions is visible only for the first cycle of motion (Figure 6.11), while for a frequency far enough from the resonance frequency, w = 0.25w , the effect of initial n conditions is not visible at all (Figure 6.12). In general, the effect of initial conditions diminishes very quickly. This allows one to neglect it all together in applications and thus to consider forced vibrations of systems with damping as a steady-state process. 6.8 FORCED VIBRATIONS OF AN SDOF SYSTEM WITH DAMPING (x 1) AS A STEADY-STATE PROCESS In this case the coefficients a and a in Equation 6.81 are set to zero. Thus, the 1 2 motion is described by the particular solution of the nonhomogeneous equation, Equation 6.28: y () t = Dsin() w t – f (6.88) p