Svetlitsky ~ Eng Vib Analysis,Vol.1

;;;;;;;;:: r--J 1 (S)

C

-

J

-

bM ~

Fig. 171.

Fig. 172.

78

2 Vibrations of Systems with Several Degrees of Freedom

0, stiffnesses C and c i of the springs and wheel tires, and trailer speed v. It is assumed that vertical displacements of the point 0 are negligible and the road profile is given by h = ho[l - cos(27tX11 1)), where x = vt. Take into account that the trailer vibration causes the friction force F f = a(dyldt - dy/dt) proportional to the relative vertical velocity of the trailer and its wheel. Ignoring the friction in the springs, determine the eigenfrequencies of the system and the critical velocities v., i.e., the velocities at which the amplitude of forced vibrations of the trailer attains the maximum.

230 A car moves on a rough road whose profile is given by h =ho[ I - cos(2ruilo))' where x = vt (Fig. 174). Determine the amplitudes of vertical and angular accelerations of the car under sustained forced vibrations. Use the following numerical values: car mass m = 6000 m, moment of inertia of the car about its central transverse axis J = 1200 kg m2, spring rates c i =200 kN/m and c 2 =250 kN/m, distances of the springs from the car center of mass II = I m and 12 = 2 m, car velocity v =25 kmlh, 10 =5 m, and ho =0.1 m.

Fig. 173

y

2ho

Fig. 174.

2.2 Forced Vibrations

79

231 Determine the critical velocity of the car described in Problem 230. 232 Derive the differential equations of forced vibrations of the car described in Problem 198 in the case when the car moves at speed v on a rough road whose profile is given by the equation h = ho[1 - cos(2nx1lo)]. 233 A centrifuge for separating liquids in a mixture is shown schematically in Fig. 175. When the centrifuge vessel rotates at a constant angular velocity OJ, heavier fractions of the mixture accumulate near the vessel walls, while lighter fractions are displaced into a region around the axis of rotation. The lighter fractions flow out the vessel through the central massless pipe having bending stiffness c. The time dependence of the liquid mass in the vessel is given by the equation mHq =mJl - tlt1), where m 1 is the liquid mass at t =0 and t1 is the total time of outflow. It is known that because of production errors, the center of mass of the centrifuge vessel filled with liquid is at a distance e (eccentricity of mass) from the axis of rotation. Set up the equations of small vibrations of the centrifuge, treating its vessel filled with liquid as a point mass mo + mliq • Changes in the vertical coordinate of the center of mass, vibrations of the liquid, the pipe mass, and effects dependent on the outflow rate may be ignored. 234 A unit impulse of force 1 = mdu(O)/dt acts on a body of mass m fixed at the end of an elastic rod with bending rigidity Elx (Fig. 176). The moment of inertia of the body about its central axis perpendicular to the figure plane is 1 = mI2/4. At t =0, the horizontal displacement, the angle of rotation, and the angular velocity of the body are equal to zero: u(O) = qJ(0) = dqJ(O)/dt = O. Determine the law of mo

I

Fig. 175.

Fig. 176.

80

2 Vibrations of Systems with Several Degrees of Freedom

tion of the body and the maximum bending moment in the rod. The gravity force may be ignored. 235 A torsional damper consists of elastic element 4 with torsional rigidity c I ' contact ring 1 at the free end of the element, and flywheel 2 pressed to the ring by spring 3 with torsional rigidity c2 (Fig. 177). A periodic torque M =Mo sin(mt) is applied to the flywheel. The moment of friction forces between the flywheel and ring is proportional to their relative angular velocity, with the proportionality factor equal to a l • The total moment of inertia of the flywheel and shaft is 1. Determine the amplitude of sustained forced vibrations of the ring. The masses of elastic element 4, ring 1, and spring 3 may be neglected. 236 A massive container 2 with mass m is suspended by spring 1 with stiffness C I (Fig. 178). In order to limit lateral displacements of the container and to damp lateral shocks, the system is equipped with light vertical guides 3 pressed to the container by springs 4 each with bending stiffness c2 • The coefficient of viscous friction between the container and guides is a. Set up the differential equation of vertical vibrations of the container subjected to a periodic force F = Fo sin( mt). Find the time dependence of the vertical displacement under sustained forced vibrations. The guide masses may be neglected. 237 When moving on a rough road, a car experiences vibrations. The vibration isolation of the car seat is shown schematically in Fig. 179, where the point 0 corresponds to the car floor, c I and C are the seat spring rates, and a is the coefficient of viscous friction. The springs and damper are assumed as massless. The influence of the total mass m of driver and seat on the car motion may be ignored because the mass m is much less than the car mass. The correlation function for vertical displacements of the point 0 is given by the formula Ky(') = Dy exp( - all ~), where Dy is the variance and a l is the damping parameter. (Such a dependence was found by processing the oscillogram of a stationary random process with zero

Fig. 177.

2.3 Critical States and Vibration Stability

81

4

c1 F Fig. 178.

Jbt

~

Fig. 179.

mathematical expectation, which describes the car motion at a constant speed v.) Determine the variance of vertical acceleration of the seat.

2.3 Critical States and Vibration Stability 238 A disk with mass m is fixed on a balanced elastic massless shaft of length 2/, which rotates at a constant angular velocity w (Fig. 180). The eccentricity of the disk mass is zero. Derive the differential equations of small vibrations of the disk as seen from the coordinate system rotating at the angular velocity w about the zaxis. Determine the frequencies of natural vibrations of the disk in the rotating coordinate system. Because of the symmetry of the system, angular deflections of the disk in the figure plane may be neglected, i.e., the disk is assumed to experience only vertical displacements. The gravity force may be ignored. 239 Consider the system shown in Fig. 78. Derive the equations of small vibrations of the disk with allowance for its eccentricity e. The equations should be written out in the rotating coordinate system.

240 Set up the differential equations of small vibrations of the system describe din Problem 238 (Fig. 180) in the case ofthe axial moments of the shaft cross section being different, i.e., Jx *" J, (for example, in the case of the shaft of a rectangular cross section). Determine the critical angular velocity w•.

241 Set up the differential equations of small vibrations of the system described in Problem 238 with allowance for the disk weight (Fig. 180). Determine the amplitudes of steady-state vibrations. The shaft mass may be neglected.

82

2 Vibrations of Systems with Several Degrees of Freedom

m

d .............-

d

I

m, J

I

Fig. ISO.

I

(J)

Fig. lSI.

242 Set up the differential equations of small free vibrations of the system considered in Problem 240 with allowance for the disk weight (Fig. 180). The shaft has a rectangular cross section bxh, with b < h. Determine the amplitudes of steady-state vibrations of the disk and the critical angular velocity OJ,. 243 A steel cantilever shaft of diameter d and length 1 rotates with a constant angular velocity OJ (Fig. 181). A disk with mass m and moment of inertia J about the rotation axis is fixed at the shaft end. Assuming the system as precessing regularly, determine the critical angular velocities of the shaft with (J::f. 0) and without (J = 0) allowance for the gyroscopic effect. Use the following numerical values: 1= 1 m, Jx =rtd"164 =490 cm', J =4.05 kg m2, E =200 GPa, and m = 175 kg. 244 A rotating steel shaft of diameter d is supported by a long bearing (Fig. 182). A flywheel of radius R and thickness h is fixed at the shaft end. The shaft may be treated as a rotating massless cantilever. Evaluate the critical angular velocities of the shaft with and without allowance for the gyroscopic effect in the case of R =0.3 m, h =80 mm, d =0.1 m, and 1= 1 m. 245 Evaluate the critical angular velocity of a steel shaft supported in two short bearings, which may be treated as hinges (Fig. 183). Use the following numerical values: D =0.6 m, h =80 mm, d =0.12 m, a = 1 m, 1=3 m, and E =200 GPa. 246 A disk with mass m is fixed on an elastic shaft of length 21 with bending rigidity EJx (Fig. 184). The system is in the field cPof a permanent magnet. The disk mass eccentricity is e. Determine the critical angular velocity OJ, of the shaft.

2.3 Critical States and Vibration Stability

83

d

a

I

I

Fig. 183.

Fig. 182.

Instruction. If a body is at the midpoint of an interpole gap, the attractive magnetic forces acting on it are counterbalanced. As soon as the body is displaced from the midpoint, these forces become different, F, =kf/Jo'/(a, - x)' and F, =kf/Jo'/(a, + x)', so that the net force L1F = F, - F, is nonzero (see Problem 107).

247 A thin rectangular perfectly rigid steel plate with mass m is supported by four springs (Fig. 185). The system is placed into an air flow. The lifting force F, is related to the flow rate v and the rotation angle B by the equation

= dC z hlpv 2()

F 1

d()

2

Here, Cz is the lift coefficient, the derivative dC/dB is assumed as given, and p is the density of air. The point 0 being at the distance 114 from the plate leading edge

Fig. 184.

84

2 Vibrations of Systems with Several Degrees of Freedom

is the center of the lift F,. Analyze the stability of small free vibrations of the plate. 248 Evaluate the critical flow rate v. at which the vibrations of the plate considered in Problem 247 become unstable. Use the following numerical values: dCjd() = 3, c, = 0.1 kNlm, c2 = 0.05 kNlcm, p= 1.25 kg/m\ h = 20 cm, and ly =mt112. 249 The level flight of an aircraft is always accompanied with various perturbances (Fig. 186). Assume that all the perturbances are in the vertical plane, i.e., the aircraft can be considered as a system with two degrees of freedoms: the small vertical displacement y and the rotation angle rp. Set up the differential equations of perturbed motion of the aircraft and find the stability conditions of steady-state level flight. It is assumed that the tail unit is subjected to only aerodynamic forces F 20 and LJF2, the velocity vector v is essentially horizontal, and the thrust vector R is constant in magnitude and directed along the aircraft axis 00,. Instruction. When an aircraft is in steady-state level motion, the aerodynamic forces FlO and F2o ' gravity force mg, thrust force R, and drag force Xo' as well as the corresponding moments, are in balance. In the case of perturbed motion of the aircraft, there appear additional aerodynamic forces LiF, and LiF2 and moment LIM" which are linear in the increments Lla, and Lla2 of the angles of attack, i.e., LiF, =C,Lla" LiF2 = C2L1a2, and LIM, =C,Lla" where Cj are constants (j = l, 2, 3). To derive the equations of motion, it is necessary to express LI~ in terms of y and qJ.

250 Analyze the stability of small free vibrations of the transmission pulleys considered in Problem 200 (Fig. 152). Assume that vibrations of the system cause the two perturbing torques LiM, = - p,drp/dt and LiM2 = - P2drp/dt. Here, drp/dt and drp/dt are the angular velocities of the pulleys, and p, and P2 = 10 are constants. Determine the values of p, such that the vibrations be stable. 251 A system consists of shaft 1 supported in bearings, flywheel 2 fixed on the shaft, and elastic massless coupling 3 (Fig. 187). Flywheel 2 is subjected to an ex

Fig. ISS.

2.3 Critical States and Vibration Stability

85

mg Fig. 186.

temal torque M =adqJ/dt proportional to its angular velocity. The force of viscous friction between the flywheel and coupling is proportional to their relative angular velocity, with the proportionality coefficient a1 =0.005 N m s. The total moment of inertia of the flywheel and shaft is J = 10-4 kg m\ and the torsional rigidity of the coupling is c 1 = 1 N m. Find the values of a 1 such that the motion of the flywheel be stable.

z

A~~--""X,

Fig. 187.

Fig. 188.

86

2 Vibrations of Systems with Several Degrees of Freedom

z

I

1

Fig. 189.

Fig. 190.

252 A top rotates about its axis at an angular velocity q (Fig. 188). The top center of gravity is at the distance I from the point 0, and the gravity force is mg. Determine the frequencies of vibrations of the top axis (z-axis) with respect to the vertical axis. Find the critical angular velocity Q o• of the top.

253 The axis of a gyro can rotates freely about the point 0 (Fig. 189). The gyro axis is held by four mutually orthogonal springs each with stiffness c, which are fastened to the axis at the point 0,. Determine the angular moments of the gyro such that its motion be stable. 254 Derive the equations of motion of the gyro considered in Problem 253 when the stiffnesses of the springs are different and equal to c, and C2 (in the x and y directions, respectively). Find the condition of small periodic motion of the gyro axis and determine the critical angular velocity q. of the gyro.

255 A gyro in gimbal mount serves as a pendulum mass (Fig. 190). Inner ring 1 with a balance of mass m) is supported in two bearings fixed on outer ring 2 which is fastened to massless rod 3. The gyro rotor is in the inner ring. The weights of the rotor and the inner ring are Q, and Q2' respectively; the moments of inertia of the rotor are ix" iyl' and i,l' and those of the inner ring are i x2 ' 1", and i,,; and the angular velocity of the rotor is Q o' Determine the frequencies of small vibrations of the system.

2.3 Critical States and Vibration Stability

87

Fig. 191.

256 A monorail car is stabilized by a vertical gyro (Fig. 191). The gyro axis is hinged at the point 0 1 so that it can rotate only in the XPZI plane of the rectangular coordinate system (XIYIZI) of the car. The z-axis of the rectangular coordinate system (xyz) is the symmetry axis of the gyro, with the y-axis coinciding with the YIaxis. The weights of the car and gyro are Qo and Q, the points CI and C2 are their centers of mass. The moment of inertia of the car about the xI-axis is JxO • The moments of inertia of the gyro are Jx ' Jy' and J" with Jy =J,. Assuming that the angles B and rp are small, set up the differential equations of small free vibrations of the gyro with respect to the vertical axis. Derive the characteristic equation and find the critical angular velocity .q. of the gyro.

z ,

Fig. 192.

88

2 Vibrations of Systems with Several Degrees of Freedom

Fig. 193.

257 A gyro in gimbal mount (rate gyro) is used for measuring angular velocity (Fig. 192). Outer ring 1 is fixed on a base rotating about the YI-axis at an angular velocity Q to be measured. The angular velocity of gyro rotor 2 is q. While the outer ring is rotating, the gyro axis turns by an angle 0 about the x-axis. Vibration damper 4 having the coefficient of viscous friction a is fastened to inner ring 3. The total moments of inertia of the rotor and inner ring about the x- and y-axes of the inner ring are lx and l y• The moment of inertia of the rotor about its symmetry axis (z-axis) is 10 • Set up the differential equations of motion in the angle 0. Find the angular velocity Q assumed as essentially constant, i.e., dadt ~ o. 258 A disk of radius r with mass m rolls on a horizontal rough surface without slipping (Fig. 193). The angular velocity of the disk is .Q. Set up differential equations of small free vibrations of the disk in the xOy plane and analyze the vibration stability for various values of Q. The disk thickness is taken as zero.

2.4 Approximate Methods of Determining the Lowest Frequency 259 Using the Rayleigh method, determine the lowest vibration frequency for the system with point masses m l and m 2 = 2ml shown in Fig. 194. The beam mass may be neglected. 260 Solve Problem 259 for the system shown in Fig. 195. 261 Solve Problem 259 for the system shown in Fig. 196. 262 Solve Problem 259 for the system shown in Fig. 129 (see Problem 171).

2.4 Approximate Methods of Detennining the Lowest Frequency

-f

89

1? 12 ;: EJ,

I

I

I

m2

m,

Fig. 194.

I

I

m

Fig. 195.

Fig. 196.

263 Detennine the lowest vibration frequency for the system shown in Fig. 196 with allowance for the beam mass m =2mi, where mo is the mass per unit length. 264 Determine the lowest vibration frequency for the system shown in Fig. 197. The stiffness of each spring is c =EJj2t. The beam mass may be neglected. 265 Using the Rayleigh method, determine the lowest frequency of longitudinal vibrations of three identical masses fixed on a massless rod (Fig. 198). The cross section area of the rod is F, and the Young modulus is E. 266 Solve Problem 289 for the system shown in Fig. 199. 267 Solve Problem 289 for the system shown in Fig. 200.

2 Vibrations of Systems with Several Degrees of Freedom

90

Fig. 197.

268 Using the Rayleigh method, determine the lowest frequency of torsional vibrations of the system shown in Fig. 20 I 269 Solve Problem 268 for the system shown in Fig. 202. The moment of inertia of each flywheel is J, and the torsional rigidity of the massless rod is c. 270 Using the Dunkerley method, determine the lowest vibration frequencies for the systems considered in Problems 259 and 261. 271 Using the Dunkerley method, determine the lowest frequency of longitudinal vibrations of the system considered in Problems 264.

272 Solve Problem 270 for the systems considered in Problems 266 and 267.

I

m 2

1m 1m 2EF

EF

~

~

Fig. 198.

m

I

I

I

I

Fig. 199.

Fig. 200.

2.5 Random Vibrations

---- - ---C2

C1

..... J1

91

r~

-

J2

Fig. 202.

Fig. 201.

2.5 Random Vibrations 273 Solve Problem 75 in the case of the road roughness being random and its spectral density given by Sh(OJ) =avl[21t(bv2+ OJ2)], where v is the trailer velocity. The trailer vibration may be considered as a stationary random process. Determine the standard deviation a,j.y) of angle rp as a function of v. Plot the dependence a,j.,v) in the case of 10 =5000 kg m\ c =2 105 N/m, 1= 2.5 m, a = 105 N s/m, a 1000 mm2/m, and b 2.5 105 m2 • The corresponding integrals are presented in Appendix 2.

=

=

274 Plot the standard deviation a,j.,c) of angle rp as a function of the spring rate c of the trailer considered in Problem 273 for the same numerical values. The trailer velocity is v = 10 mls. 275 Evaluate the standard deviation of angular velocity dqtdt of the trailer considered in Problem 273 for the same numerical values. 276 A point mass m fixed on a massless elastic beam of length 21 (Fig. 203) is subjected to a random perturbing force f characterized by the spectral density SjOJ) = P/[21t(J322+OJ2)]. The mass is connected to a vibration damper with the coefficient of viscous friction a. Evaluate the standard deviation of acceleration

Fig. 203.

92

2 Vibrations of Systems with Several Degrees of Freedom

d2yIM of the mass for the following values of the parameters: m = 10 kg, 1= 0.5 m, E = 200 GPa, a = 2000 N slm, p. = 105 N 2/s, fJ2 = 100 1/s, b = 20 mm, andh =50 mm.

277 Evaluate the standard deviation of reaction force and the maximum normal stress in the beam considered in Problem 276 for the same numerical values.

278 Under steady-state conditions, the torque Me developed by an engine is balanced by a moment of resistance M, depending on the angular velocity Q of the engine shaft (Fig. 204). The shaft is subjected to a stationary random torque AM characterized by the correlation function Kt.M( 1') =D exp(-a 11' I) and the mathematical expectation mt.M' The reduced moment of inertia of the system is J. Determine the variance of angular velocity Q. The changes of moment Me which are caused by small variations in the angular velocity may be ignored.

279 A magnetoelectric oscilloscope with loop vibrators was considered in Problem 69 (see Fig. 60). In practice, there is a random current L1i in the oscilloscope electric circuit, i.e., i =io + L1i, where io is a mean electric current. The current L1i can be due to, for example, thermal motion of electrons in the electric circuit. After all transients (occurring upon switching on the current) have died out, the mirror vibration induced by the random current can be considered as a stationary random process. Supposing that the spectral density of L1i has the form St.;Cm) = N ("white noise"), determine the standard deviation 0'", of oscilloscope reading rp. 280 An electric motor with point mass m is fixed on a cantilever flange massless beam (Fig. 205). The beam end is connected to a vibration damper having the coefficient of viscous friction a. Evaluate the maximum normal stresses arising in the beam when its restraint point (i.e., point 0) undergoes random vertical displacements Yo' The displacements can be considered as a stationary random process with the spectral density SyO(m) =20/[21t(100 + ( 2 )]. Use the following numerical values: E = 200 GPa, m = 300 kg, a = 5 104 N slm, 1= 2 m, Jx = 1.98 106 m4, and Wx = 3.97 10-5 m3 •

281 The system shown in Fig. 206 is subjected to a stationary random perturbing force pet) with the spectral density Sp(m) = al[21t(b2 + ( 2 )]. Determine the standard

1-~HJ J

Engine

a

Fig. 204.

b

n

2.5 Random Vibrations

=0===:::;t:!:Jm

93

-I~

I

Fig. lOS.

deviation ~ of vertical coordinate of the point mass m and the variance DM of bending moment M at the restraint point.

282 A motor with mass m is fixed on a cantilever beam having rigidity EJx (Fig. 205). The mass m is connected to a vibration damper with the coefficient of viscous friction a. The vertical motion of the restraint point is given by the equation Yo =YOlsin(mt)+LlYt(t), where YOt is the amplitude of regular displacements and Lly/t) is random displacements with the spectral density St(m) = aI[2rc(P + ( 2 )]. determine the standard deviation of difference &= Lly - Llyt' where Lly are random displacements of the mass. 283 The device shown schematically in Fig. 5 is used for measuring fluid or vapor pressure. The pressure LIp applied to the piston with cross section area F has both a regular component Llpo and a random component Llpt with the spectral density St = aI[2rc(jJ + ( 2)]. The friction force acting on the piston is Ff = a dMdt. Treating random displacements Llz t of the piston as a stationary random process, determine the standard error of readings of the device. 284 In general, the piston considered in Problem 5 can be under the action of two perturbations, namely, random pressure Llpt (see Problem 283) and random vertical displacement Llzo of the cylinder. The latter is equivalent to a kinematic pertur

P(t)

Fig. 206.

94

2 Vibrations of Systems with Several Degrees of Freedom

bation of the system. The corresponding spectral densities are given by the equations

The resistance force acting upon the piston is equal to adL1zldt. Assuming that the mutual correlation of the perturbations is absent, determine the standard deviation of random vertical displacement LIz\ of the piston. 285 The piston considered in Problem 5 is subjected to random pressure L1p\ characterized by the spectral density St,p =N ("white noise"). The cylinder is fixed on a vibrating base, and the spectral density of random displacement Llzo of the base is given by St" =txz/[21t(P22 + ol)]. Vertical displacements LIz\ of the rod which are caused by the perturbations L1p\ and Llzo lead to an instrumental error. The mutual correlation of the two perturbations is assumed to be absent. Determine the optimum value of spring stiffness c for which the standard error of readings of the device becomes smallest. 286 A body with mass m is suspended by a spring with stiffness c and connected to a vibration damper with the coefficient of viscous friction a (Fig. 207). The correlation function of random vertical displacements L1yo of the point 0 is given by the equation Ky =Dyexp(-a!.!). Determine the standard deviation of acceleration of the body. 287 A random perturbing force P(t) is applied to a body with mass m = 100 kg suspended by a spring with stiffness c 400 N/m (Fig. 208a). This force varies with time arbitrarily within a tolerance range (Fig. 208b). Find the time dependence P(t) such that the displacement of the body at the moment of time t\ =2 s be greatest. Compare the displacement found to that caused by a step force equal to the greatest possible value of P(t). Suppose that at t = 0 both the displacement with respect to the eqUilibrium position and the velocity of the mass are zero.

=

P(t)

+1

ami

1a

Fig.207.

t

0 -1 b Fig. 208.

2.5 Random Vibrations

95

288 The body considered in Problem 287 is subjected to the force P(I) such that its displacement at II = 2 s is equal to the greatest possible value. Determine the velocity at this moment of time. 289 A body with mass m = 100 kg is kept on a horizontal plane by a spring with stiffness c = 400 N/m (Fig. 209a). The force of friction on the plane is Ff = adxldl, with a = 400 N s/m. The tolerance range of a perturbing force acting on the body is shown in Fig. 209b. Find the time dependence P(t) such that at II =2 s the displacement be greatest (under zero initial conditions x(O)=dx(O)/dt=O). 290 For determining the thrust of a jet engine, it is installed on a test bed consisting of a truck and a spring balance (Fig. 2IOa). Because of an irregularity of charge combustion, the trust R varies arbitrarily within a range: R =Ro + L1R (Fig. 2IOb). Determine the greatest standard deviation of R at II =0.5 s (under zero initial conditions). Use the following numerical values: total mass of the systemm = 1000 kg, spring stiffness c =900 kN/m, coefficient of viscous friction a =4 104 N s/m, nominal thrust Ro = 2 105 N, and thrust spread L1R = ±O.OS Ro. The mass decrease due to the bum-out may be ignored. 291 A random perturbing force P(t) is applied to the mass considered in Problem 287. The force varies with time arbitrarily within a tolerance range (Fig. 208). The instrumental error LI of a device measuring the x-coordinate of the mass depends linearly on both the displacement x and the speed dx/dl caused by the force: LI =alx + a 2dx/dl, where a l and a 2 are constants and md 2xldl + ex =P(I). Determine the greatest possible error LIm" at II =0.5 s (under zero initial conditions) in the case of m = I kg, c = 0.16 kN/m, a l =0.173, and a2 = 0.008 s.

P(t)

AR

R b Fig. 209.

Fig. 210.

292 A dynamically stable motion of a rocket can be accompanied with small vibrations induced, primarily, by thrust spread, both linear and angular eccentricities

96

2 Vibrations of Systems with Several Degrees of Freedom

of the thrust, etc. The differential equation of small variations in the angle of pitch (Fig. 211a) has the form 10d2 rp1M + arp= M,. Here, 10 is the moment of inertia with respect to the axis through the center of gravity 0, arp is the restoring moment, and values of the random perturbing torque M, are assumed to be within the tolerance range ± b (Fig. 211b). Determine the range of variables rp and drpldt on the phase plane at the moment of time t, =5 s (under zero initial conditions) in the case of 10 = 104 kg m\ a =1.6 104 N m/rad, and b =2 MN/m. 293 The instrumental error of an oscilloscope (namely, the standard deviation of deflection angle rp of its mirror) was found in Problem 270 under the assumption that variations in the angle rp would be considered as a random process. This assumption is valid only if the transient period is far less than the observing time and if the current Lii can be treated as a stationary random process. For low damping (when nlpo « 1), the transient period can be of the same order of magnitude as the observing time. Moreover, in this case, the theory of stationary random processes is inapplicable. In order to analyze small vibrations of the oscilloscope mirror by means of the theory of nonstationary random processes, additional detailed data on the current Lii are needed. Namely, it is necessary to determine both its autocorrelation function and its mathematical expectation as functions of time. Determine the greatest possible instrumental error of the oscilloscope in the case of a nonstationary random current Lii being within a tolerance range ± a, where

a a

a

+b

0 ·b

t1

+a

t

0 ·a b

b Fig. 211. a

Fig. 212.

= ILii I =fJio(for example, ILii I =0.05 io). max

P(t)

max

2.5 Random Vibrations

97

294 A body of mass m is fixed on a perfectly rigid beam having moment of inertia 10 with respect to the left hinged support (Fig. 2l2a). The right support is a spring with stiffness c and a vibration damper with coefficient of viscous friction a. The body is subjected to a random perturbing force P(t) whose values are within a tolerance range (Fig. 212b). Determine the greatest possible dynamic reaction force N rm.x acting upon the body and find its dependence on the coefficient a for small nlpo. The initial conditions are assumed to be zero. 295 Solve Problem 294 in the case of a simple resonance, i.e., when the perturbing force acting upon the body is given by P(t) =a sin(wt) with r.o =Po. 296 Consider the system described in Problem 294. Find the time dependence P(t) such that the angular velocity rAt) at the moment of time tl =31tlpo be largest possible (Po is the frequency of free vibrations). The initial conditions at to =0 are

rAO)

=0 and drAO)/dt =ip o. The damping is absent (a =0).

297 A point mass m is fixed on a massless elastic rod of length 21, which is subjected to a Gaussian random torque M (Fig. 213). Its mathematical expectation mM=0 and correlation function KM =DM exp(-aJ~) are given. The mass is connected to a vibration damper with coefficient of viscous friction a o• Determine (1) the vertical displacement y of the mass m, (2) the maximum normal stress O"max and its variance in the cross section k, and (3) the system reliability, i.e., the probability of survival P[( 0;. - 0"....) > 0], where 0;. is the tensile yield stress. The quantities 0;. and

O"max

are assumed as independent Gaussian random variables.

298 Solve Problem 297 in the case of the rod fixed at one end (Fig. 214).

299 A car moves uniformly on a rough road with Gaussian random irregularities. If the car moves at the speed v, the spectral density of vertical displacement h of the wheels is given by Sh = av/[21t(r.o2 + bv2)]. The car suspension is shown schematically in Fig. 215. The contraction LI of the suspension is a Gaussian random variable, with its mathematical expectation mLl and variance O"Ll given. Determine the probability of survival of the car, i.e., the probability P[(LI- y) > 0],

I

Fig.213.

I

M

z

98

2 Vibrations of Systems with Several Degrees of Freedom

v

y

M

I

~====::$-_Z

Fig. 213.

Fig. 214.

assuming that Li and y are independent random variables. 300 A system consists of a shaft and two disks with moments of inertia J 1 and J2 about the z-axis (Fig. 216). Disk 1 is subjected to a stationary Gaussian random torque M. with its mathematical expectation (m M=0) and spectral density SM(m) given. Determine the maximum torque at the restraint (at z =0) and the probability P[( 1", - 'rna,} > 0]. Here, 1", is the shear yield stress assumed as a Gaussian variable and 'max is the maximum tangential stress at the restraint point. 301 A top-loaded vertical antenna with a radio-transparent spherical enclosure is shown in Fig. 217 a. The total mass of the antenna and enclosure is m. The total moment of inertia with respect to the axis perpendicular to the figure plane and passing through the center of mass 0 is J. At the initial moment of time t = 0, the antenna is subjected to a short-time random aerodynamic force F due to a gust of wind (Fig. 217b). Other aerodynamic forces acting upon the top may be ignored. The duration Lit of the force F is given, being much less than the free vibration period corresponding to the first eigenfrequency of the system. The impulse of the force Fis Lit

JF

= fF (t )dt ~Fojjt , o

where Fo is a Gaussian random quantity, with its mathematical expectation mF and

2.5 Random Vibrations

99

F

,•

Mo I

I

J1 Fig. 216.

I I

I

z t---

t

At

J2 Fig. 217.

standard deviation (J'F given. Therefore, IF is also a Gaussian random variable, with mJ = m~t and CJ'.J = (J'~t. The force F is directed along a line passing through the center 0 1 of the enclosure; therefore, the antenna is subjected to the moment of forces M =FlI and the angular momentum 1M =IFl I. The top can be considered as a straight rod of a constant cross section. Determine the maximum deflection angle rp of the antenna beam.

Answers and Solutions

1 Vibrations of Systems with a Single Degree of Freedom

1.1 Free Vibrations 1

= ~3EJx(a+b)1 ma 2b2;

p

(a)

p

(b)

=~/l2EJx Ima 2(4a+3b); p = ~(cl +3EJx I mI 3 ).

(c)

Ins t r u c t ion. In accordance with the force method, the canonical form of the differential equation of motion for a system with the single degree of freedom is

y Here,

811

= 811 (-my).

is the displacement of the load under the action of a unit force in the direction

coinciding with this displacement. Therefore, the frequency of free vibrations is

p =11 Jm811 • 2

p

(a)

p

(b)

p

(c) (d)

p

=~EJx Im1 3 ;

=~48EJx I 7ml 3 ;

=~6EJx I mh2(h+b);

= ~12EJx Ima 2 (4a+3b).

As an example of the determination of the compliance (coefficients for statically indeterminate systems), we consider Problem 2d. Breaking the constraint (under the assumption that the constraint between the load of mass m and the base surface is bilateral), we apply the force N (the constraint reaction

104

1 Vibrations of Systems with a Single Degree of Freedom

, tl

N Fig. 218.

force) to the load. As a result, we arrive at the frame loaded by both the force N and the inertial force J (Fig. 218). Using the force method, we obtain two equations in terms of the displacement of the load along the axes y and z:

= d}J + d22 N;

(1)

0= d 2 }J +d22 N;

(2)

y

-my.

where J = Excluding the force N, we arrive at the expression

-o}\my,

y= where Thus, the vibration frequency is

£5;\ = (£5;}b;2 - £5;2b;\) / b;2·

p = 1/ ~mo;l. The approach used makes it possible the determination of the system compliance

01

11

for statically indeterminate problems. This approach does not require

traditional calculations (as a rule, rather cumbersome) and preliminarily uncovering the static indeterminacy with subsequently determining the displacement of the unit force in the direction of the load shift under vibrations. If several constraints are imposed on the system, (e.g., the system is trice statically indeterminate), then the same number of algebraic equations can be obtained for finding constraint reaction forces. Furthermore, these forces are excluded from the differential equation.

3 (a)

(b)

1.1 Free Vibrations

105

4 Similarly to the solution to Problem 2, the differential equation for small torsional vibrations in all the systems has the form

.. 1 rp+-rp=O. J81l

Therefore, the eigenfrequencies of the systems 4a-4d (see Fig. 4) are correspondingly equal to (a) (b) (c) (d) Ins t r u c t ion: For variants (c) and (d), the systems are statically indeterminate. Thus, in order to find q, it is necessary initially to exclude the static indeterminacy.

5

~ ml~"'2· 6

7

P =~iC 9 m· 8 The rotation of the gear wheel through a certain small angle rp causes a displacement of the gear-wheel center by the distance rpD / 2 . In this case, the point A (the point of fixation of the springs to the carrier) shifts by the distance 8 = rpl - rpD / 2. With allowance for the moments of inertia of the gear wheel

J 1 = mlD 2 /8 and of the carrier J 2 = m2D 2 112, we can find the kinetic energy of the system

T _ (rfIJ+"'2)D2¢ + (rflJ D2 -

8

+

mlP) ¢2 _ miD¢2 8324·

We express the variation of the system potential energy in terms of the spring deformation and of the carrier helix angle:

II = 2

cf + m2g -t (1- cos rp) =[c (I - ~ ) + ~gl 2

]

rp2 .

Furthermore, using the Lagrange equation of the second kind, we arrive at the expression

106

1 Vibrations of Systems with a Single Degree of Freedom

"'2g1+4c (l-DI2)2 =0 (9mJ+ 6"'2)d2+8m2z2-12"'2lD rp .

.. + 12

rp

Hence, it follows that the eigenfrequency of the system is

p=

12m.gl-48c(l-DI2)2 ..." (9mJ+ 6"'2)D2+8m212-12"'2lD .

9 In the process of small free vibrations of the cylinder, its kinetic energy is the sum of the kinetic energies of the rotational and translational motions

T = (1 12)J¢/ + (11 2)m(rrp)2. With allowance for the moment of inertia J = Qr2 12g of the cylinder, the equation for the total kinetic energy can be rewritten as

T=(3/4)(QI g)r2rp2. The variation of the system potential energy is caused by stretching (compressing) the springs and lifting the cylinder when it moves over the concave surface. In the case of rotation of the cylinder through angle rp, its upper point A shifts by the distance 8 = 2rrp . The potential energy of the deformed springs is

II J = 2c8 212 = 4cr2 rp2 . While deviating the cylinder from its equilibrium position through angle rp, the angular coordinate If/ of the cylinder center of mass is

Iff = rpr I(R - r).

In this case, the variation of the potential energy of the cylinder depends on its position

II2

Q rrp2 (R-r) .

= Q(R - r)(1- cos '1') = 2

Substituting expression for kinetic and potential energies into the Lagrange equation, we arrive at the equation for small free vibrations

rp+( t (R~r) + 136 d)rp = 0,

whence it follows that the system eigenfrequency is _ 2 q + 16 cq

P-

"3 (R-r)

3Q·

As follows from the formula obtained, p ~ 00 as R ~ r , i.e., the eigenfrequency increases with a decrease in the radius of curvature of the concave surface.

10

p = ~c I (2m) - g I (2/). 11

p

= .Jc I Jm = 224 S·I.

1.1 Free Vibrations

107

12 To solve the problem, it is convenient to employ the Lagrange equation of the second kind. In the case of deviation of rod AB in the course of the vibration process through a certain angle rp from the equilibrium position, the kinetic energy of the disk is

T

= mv 2 /2+ Jmqi;

(1)

where rp is the torsion angle of the rod. The linear velocity v of the disk displacement and the angular velocity related to the rod deviation velocity

v = lip;

ip

iff

are

by the dependences

iff = (l / R)ip.

(2)

The potential energy of the system is

II = cc/ / 2 + mgy,

where y is the variation of the vertical coordinate of the disk center of mass. It is evident that

y

= 1(1- cos rp) sin a

~ (I rp2 sin a) / 2.

(3)

Substituting these relationships into the Lagrange equation of the second kind, we arrive at the differential equation

.. + c(lIR)2+mglsina

rp

(3/2)mI2

-

0

rp - .

Hence, the system eigenfrequency is

r---=-----

p=

c(ll R)2+mglsina (3/2)mI2

It follows from the latter relationship that the frequency p vanishes (i.e., the system becomes unstable) when

sin a

= -cl(mgR2).

13 In the case of shifting the float by the distance L1x in the vertical direction, e.g., downwards, a complementary buoyancy force

(mJ2 / 4)rL\x arises.

Thus, the

differential equation of motion can be represented in the form

[~L\X+(mJ2 /4)rL\x]/1 +m2L\x(l; /11)+cL\x(l;/I)

= 0,

Therefore, the system eigenfrequency is

p

= ~r-(mJ-2-r-/-4-+-c-I;-:-/-/-:-12-)(-~-+-m-2-1;-I-/-/12-).

14 We set up the differential equation of motion using the Lagrange equation of the second kind. The kinetic energy of the disk is

T

= mv2/2 + J (P12 / 2,

where v is the velocity of the disk motion and (PI is its angular velocity.

108

1 Vibrations of Systems with a Single Degree of Freedom

The potential energy of the disk is

II = mg( R - r )(1- cos rp).

The angles rp and

rpl

(see Fig. 219a) and also the velocity v are related to

each other by the relationships

rpl

= R;:r rp;

V= (R - r)ifJ.

(1)

With allowance for the latter expression, the disk kinetic energy is

T

= [(R -

r) / r2](mr2 + J)(ifJ2 /2).

We substitute the expressions for the kinetic and potential energies into the Lagrange equation

(mr2 + J)rp + mg[r2 / (R - r)]sin rp

= O.

(2)

In the case of small vibrations ( sin rp ~ rp), the vibration frequency is

p

= ~mgr2 / [(R -

r)(mr2 + J)].

We now derive the differential equation of motion using the d' Alembert method. According to Fig. 219b, we may write out

mX my

= - N cos rp + F cos rp;

= N cos rp + F sin rp -

(3)

mg;

JijJ = -Fr,

(4) (5)

Here, N is the normal pressing force and F is the friction force between the hub and the guide, which hampers slipping the disk.

r a

b

Fig. 219.

Multiplying Eqs. (3) and (4) by costp and sintp, respectively, we obtain after summation

1.1 Free Vibrations

m(x cos cP + ji sin cp) = F - mg sin cpo

109

(6)

Furthermore, using Eqs. (1) and (5), we write out the expression for the friction force

(7) We now express the coordinates of the disk center of mass in terms of the angle rp:

x

= (R - r)sin cP;

y

= (R -

r)(1- cos cp).

(8)

Substituting relationships (7) and (8) into Eq. (6), we arrive at Eq. (2) derived above by the Lagrange method. 15 The solution of this problem is similar to that of Problem 14. The differential equation for small vibrations has the form

(mR2 + 2J)ijJ + CR2 cp = 0,

where

({J is

the wheel rotation angle.

16 ..

cP + p

= 31\:~~2g;

CPOmin

3(R-a)g [2+4a2

-

0'

cP - ,

= f.l(l2 + 4a 2 ) / (12 + a 2 + 3Ra).

17

18 Analyzing motion of the load, we consider an arbitrary moment of time when the contact point is shifted to the position B (Fig. 220a). We decompose the total contact force into two components N x and Ny. Then, the differential equation

of motion can be represented in the form (1)

where Ax and ~y are the displacements of the load center of mass in the course of the vibration process. Using the d' Alembert principle, we consider the sum of the moments of all forces with respect to the center

-

Jc~

cP =

Ct

of gravity of the load:

Ny(ABcos(~ cP- ~ CPt)- ~x)cos~

Taking into account the fact that

AB = Rt ~ CPt

cP- Nx(a + ~y).

= R2 (~cpt -

(2)

~ cp), we find

1 Vibrations of Systems with a Single Degree of Freedom

110

y

a Fig. 220.

Am -

U r1 -

The quantity

R

2

i1({J.

M'

~y = AB· ~rp = R1R2

't:

~x

= a~rp.

is of a higher order of smallness.

Therefore, substituting Eq. (1) into Eq. (2), we obtain the differential equation of motion

(J c + a 2m)11 rp + mg(R1R2 / I1R - a)11 rp = O.

(3)

Thus, we find the eigenfrequency

p = ~r-[m-g-/-(-J-c-+-a-2-m-)]-(R-1-R-2-/-11-R---a-).

In the case of rotation of the flywheel, we can neglect the free-fall acceleration and represent the differential equation of free vibrations of the load in the form (see Fig. 220b)

= -Nx + m{j}2r~ '1'; ~ 'I' = (a / r)~ rp; Jc~ rp = Nxa - m{j}2r(R1~ rp1 - ~x),

m~x where

11 rp1 = 11 rpR2 / ~R and I1x = a~ rp. Excluding ~ '1'; 11 rp1; ~; and N from the equation, we arrive at

(J c + a2m)~ rp + m{j}2r[(R1R2 - a~R - a2~R / r)~R]~ rp = O. Hence, the eigenfrequency of load vibrations is

p = ~[mo/r / (J c + a 2m)](R1R2 / ~R - a - a 2 / r).

1.1 Free Vibrations

111

19

P=OJ.fRiT. 20 The position of the pillar at an arbitrary moment of time is shown in Fig. 221a. Taking the sum of the moments with respect to the hinge, we arrive at the relationship

- PI - mgy + ~a

= o.

(1)

Since ~ = 2c 2 (Yl - ~y) and Yl = ya / I, we obtain after certain transformations the differential equation for small free vibrations of the pendulum

my I + 2c2 (ya / 1- ~y)a - ngy = O.

(2)

The flexure ~y (Fig. 221b) is

~y

= P(l-a)2 a / (3EJJ = [a(l-a)2 / (3EJ x )](my).

(When determining Lly, we ignore an effect of the gravity force, therefore, ~

= PI/ a.)

After the transformations have been performed, we obtain from Eq. (2)

.. =

y

2c2a 2 /l-mg

a2u-a)2] y m 1+2c2 3EJ [

x

=0

.

In the case of small vibrations, the eigenfrequency of the load is

a Fig. 221.

b

v

112

1 Vibrations of Systems with a Single Degree of Freedom

p

=

2c2a2 !l-mg

_

0

2a23~:f]Y - .

m[I+2C As

EJx

~

00

and

a = 1/2 , we arrive at

p=~c/(2m)-g/21. 21

p

= w~mrl / J B •

22 In the case of small vibrations of point mass m, it is subjected to the forces shown in Fig. 222a (in the rotating coordinate system). They are the inertial force

J = -myo' the centrifugal inertial force F = mw 2 (R + 1), the Coriolis iner-

tial force Fe = 2mwyo (owing to the relative character of the velocity of motion), as well as the forces Q and N acting from the side of the elastic rod. Projecting the force F onto the axes y and z and assuming both vibrations and the angle pto be small, we may consider that the load is shifted only along the y axis. Therefore, using the d' Alembert principle, we arrive at the two following equations:

myo + Q- mw 2 yo = 0, 2mOJj;o + mw (R + 1) = N. 2

1r-c;:~-~

O~--~--~--~--~--~~~

b

Fig. 222.

(1)

1.1 Free Vibrations

113

In the case of small vibrations, an effect of the Coriolis force on the axial force can be ignored, i.e., we may assume

= mal(R + 1).

N

(2)

Analyzing the flexure of the rod (see Fig. 222b), we arrive at the equation

EJxY"

= -(Yo -

y)N + Q(l- z),

or (3)

The solution to Eq. (3) is of the form

= C1 coshaz+C2 sinhaz+ Yo -(QI N)(l-z). From the boundary conditions z = 0, y = 0, y' = 0, we find

(4)

y

C1 and C2 :

(5) Since y

= Yo

as z

= 1, from solution (4) with allowance for relationships (5),

we obtain

Q[(l I N)coshal-(l1 aN)sinhal]- Yo cosh a

= O.

(6)

Using Eq. (1), we exclude Q from Eq. (6). As a result, we arrive at the equation

. +[

Yo

cosha/a 3£Jx (alcoshal-sinhal)m

m =0

2J

0

m Yo = .

(7)

0), after the uncertainty has been unraveled, the bracket in Eq. (7) is equal to 3EJx I ml 3 . This corresponds We can show that in the case of

(as

a

~

to the eigenfrequency of the point mass. The vibration frequency of the rod is

P = ity

(al)3 coshal

3(alcoshal-sinhal)

_

(al)2 ( = 3(l+R/l) ' Po

3EJ I ( /2)) x

m

.

(8)

The product al that enters into expression (8) is related to the angular velocand to the load vibration eigenfrequency Po by the relationship

OJ

at

= ~3(l + R I /)(m I Po)'

23 If a force directed along the y axis is (statically) applied to the load of mass m (Fig. 223a) then the forces Q and N act on the rod. These forces can be expressed as

Q = P+ mw 2 yo; N

= Mm2(R + I).

(1)

(2)

1 Vibrations of Systems with a Single Degree of Freedom

114

Fig. 223.

We imply that the rod rigidity is described by the ratio

Yo

c = PI Yo and find

using the approximate formula for flexures in the case of the transverse-

longitudinal bending: (3)

where

y

= Ql3 1(3EJJ.

Substituting relationships (1) and (2) into expression (3), we obtain after certain transformations

Yo

[1- 3EJx(~2~/ Pc) ] 3~!x (1 + N 1Pc) = P.

Hence, the system rigidity (with due regard to the centrifugal inertial forces) is C

= Co [1 + (41 ;r2)(ali - moi,

where Co

=3EJx /13; a 2 = N I(EJx) = moi(R+l)1 EJx'

Based on the known system rigidity, we find the approximate expression for the eigenfrequency of load vibrations: r-----------------~

PI=PO

1+(4/;r2)(al)2-3(\~~~l)"

The numerical values of the ratios

pi Po

and

PI 1Po

(4)

for

R =I

Table 1.1. The curves for the ratios

pi Po

and

solution are presented in Fig. 223b.

PI 1Po

related to the exact

are given in

1.1 Free Vibrations

115

Table 1.1.

all p/ Po PI/ Po

0.5

1

2

3

4

5

6

7

0.701

1.111

1.382

1.732

2.108

1.443

1.091

0

1.031

1.112

1.391

1.761

2.176

2.614

3.066

3.527

[We imply formula (8) in the solution to Problem 22 and formula (4) in the solution to Problem 23, respectively.] As follows from the curves, only within the range

0 < al < 4, the quantities Po

and PI are little different from one another

(the difference is no more than 4%). In the case of use formula (4) for determining the frequency.

al > 4,

it is not possible to

24 The position of the load of mass m in the equilibrium state and with allowance for inertial forces (the load is shifted by the distance YIO) is shown in Fig. 224a. In this figure, the position of the load is indicated for the case that there is an additional time-dependent shift Y2k. The forces acting on the load are shown in Fig. 224b. Here,

F2y and F2x are the components of the force F2 (see

Fig. 224a); J is the inertial force, Q and N are the forces acting on the load from the side of the elastic rod. Apart from these forces, the load is subject to the Coriolis force (see the solution to Problem 22) that can be ignored in the case of small vibrations. The equation of motion for the load has the form

J +F2y -Q=O, or

mY2k - moi(ll + YIO + Y2k) + Q = O.

(1)

The forces Q and N can be represented as

Q = Qo + ~Q; NF2z = moll, where

Qo = mol (II +110)

(2)

is the force acting on the mass m in the deviated

equilibrium state; ~Q is an additional force caused by the vibrations. In the case of small vibrations, the force N (without allowance for the Coriolis force) is invariable in both static state and vibratory state. Equation (1) with due regard to relationship (2) takes the form

mY2k - molY2k + ~Q = O.

(3)

The differential equation for the arched rod axis can be written out as

EJxY" = -N(ylO + Y2k - y) + (1- z).

(4)

1 Vibrations of Systems with a Single Degree of Freedom

116

a :1

---

b

Fig. 224.

Since

Y = Yl + Y2

(see Fig. 224b), excluding from Eq. (4) tenns that characterize

the equilibrium condition, we arrive at the equation

Y2, -aY2

f1Q (/ ) = -a 2Y2k + EJ -z. x

(5)

The solution to Eq. (5) is of the fonn

Y2 As

=

C1 chaz + C2 shaz + Y2k -(I:1QI N)(l-z).

z = 0, the conditions Y2

=0

and

Y;

=0

(6)

must be valid, which is possi-

ble provided that (7) As z = / , taking into account expression (7), we obtain from relationship (6) the dependence

1.1 Free Vibrations

~Q

=

11 7

RJxCal)3coshal

Y2k t3(alcoshal-sinhal) .

Finally, the equation describing small vibrations of the load of mass m can be written out in the form .. Y2k

+

[

EJx (al)3 coshal mt3(alcoshal-sinhal)

0/ ]Y2k =0.

The vibration frequency of the load is _

P - Po

(al)3coshal ( OJ 3(alcoshal-sinhal) Po

)2 ' (2 _ 3EJ Po - ml3

x )

.

25 The equation describing small vibrations of the load of mass m (see Fig. 225) is of the form

mj\ -mO/Yk +Q=O.

(1)

We now write out the equation for the arched rod axis:

EJxY" = N(Yk - y) + Q(l- z), or

y"+a 2y=a 2Yk+(QIEJJ(l-z),

(a 2 = E~J·

(2)

After Eq. (2) has been solved, we have (3) = C] cosaZ + C2 sinaZ + h + (Q I N)(l- z). Y = Y' = 0 as z = 0, we obtain the following expressions for the con-

Y Since

stants C] and C2 :

C]

= -(Yk + Ql IN);

C2

= Q l(aN). 11

, R

Fig. 225.

As z

= I, we have Y = Yk

Thus, it follows from Eq. (3) the dependence

1 Vibrations of Systems with a Single Degree of Freedom

118

Q = [EJx a 2cosal I(sinal-al cosal)]Yk' Excluding Q from expression (1), we arrive at the equation for small vibrations of the load:

Yk + p~ {(al)3 cosal/[3(sinal-al cosal)] - (wi PO)2}Yk

= O.

Thus, the vibration frequency is

P = po~(al)3 cosal 1[3(sinal-al cosal)]-(w I PO)2. 26 In this case, the load deflected from the equilibrium position by the distance

Yo ' in addition to the inertial force my0 the centrifugal force

mw 2Yo'

and to the elastic force

cyo' is subject to

Therefore, the differential equation of motion has

the form

myo +(c-mw2)yo

= O.

Thus, the frequency of natural vibrations is

.J

P = elm - w2 , i.e., in the case that

elm - w2 P = 0 , the motion becomes unstable.

27

P = )21'0 I(ml). 28 We consider the state of mass m, which corresponds to a deviation from the equilibrium state (Fig. 226a). In accordance with the d' Alembert principle, we can write out or

y+ 1'oyl I[mxo(l-xo)] = O.

Thus, the load eigenfrequency is

P = ~1'o1 I[mxo(l-xo)]. The plot for this dependence is presented in Fig. 226b. 29 In the process of vertical vibrations, the load is subject to the following forces: the inertial force mv, the gravity force mg, and the restoring force N of .

=

The corresponding differential equation of motion has the form

mv+aF=mg, where

(1)

1.1 Free Vibrations

119

..

-fI1!J

11

JJ

,·x,

XI}

a 11

Fig. 226.

v=Rm+tu. The total stress

(j

in the rope can be considered as a sum of the static stress

and the stress caused by elongation

8.x of the vibrating rope:

aF=mg+EFAx/x,

(2)

where

x= Rmt =vt. With allowance for expression (2), we can rewrite Eq. (1) as

mD.i + 8.xEF /( vt) = O.

(3)

The differential equation (3) is the Bessel equation whose complete integral can be represented in the general form as

(4) where

CI and C2 are the integration constants, r = 2.JEFt / mv is the dimen-

sionless variable, and

II'

~ are the first-order Bessel functions of the first and

second kind, respectively. We now calculate the velocity of motion for the mass m:

1 Vibrations of Systems with a Single Degree of Freedom

120

(5)

Using the well known relation between the Bessel functions of the zero and first order, we arrive at

rloCr) =

iT [rIl(r)];

r1'o(r) =

iT [r~(r)].

We now represent Eq. (5) in the form

AX = [C/o(r)+ C2 1'o(r)][2EF /(mv)]. The integration constants are found from the boundary conditions and

L1x = L1Xo as t = to): CI =-A~(ro); C2 A=

(6)

(,ix = 0

= All (ro);

~04to1z-6

II(TO)YO(TO)-Io(z-o)l!(TO) .

We obtain the final solution to the problem in the form

,ix = Ar[II (ro)~ (r) -

~ (ro)II (r)].

(7)

We now find the sequential moments of time for which the moving mass passes by the equilibrium position, i.e., the moments of time corresponding to ,ix = O. As follows from solution (7), these moments of time are determined by the relationship

II (r)/ ~ (r) = II (ro)/ ~ (ro) = const.

(8)

In accordance with the setting of the problem,

ro

= 2~EFto / mv = 2~r--1O-1O-.-1O---5.-1-/(-10-0-.1-) = 63.2.

From the solution to Eq. (2), we obtain the moments of time Tn for which the rope elongation is equal to the static elongation. We now analyze a particular case of initial conditions. We assume that to = 0, therefore,

10 (0) = 1 and 1'0 (0) = 00 . By virtue of the boundedness of the veloc-

ity, it follows from Eq. (6) that

C2

= 0, CI = ,ixmv /(2EF) , and the solution

acquires the form

,ix = rII(r)AXomv/(2EF). Hence, it follows that

,ix =0 as II (rn) =O. For finding the roots rn, the

handbook (Janke E. et al., 1960) can be used. Thus, we determine r l

r 2 =7.01, r3

= 10.17, etc.

Moments of time

tn for which ,ix = 0 are tn

= r~mv /( 4EF).

=3.83,

1.1 Free Vibrations

121

30 In order to derive the differential equation for the motion of the load, we employ the Lagrange equation of the second kind. At any arbitrary moment of time, the position of the body is determined by two coordinates, namely, by the angle rp of the deviation from the static position of the load and by the length I of the unwound portion of the rope. Since I = OJRt is the known function of time, we choose the angle rp as the generalized coordinate. In a Cartesian coordinate system, the kinetic energy of the moving load is expressed as where

x = I sin qJ, y = I cos qJ. Therefore, the kinetic energy of the load is

T = O.5m[(l sin rp + lq,cosrp)2 + (l cosrp -lq,sin rp)2 + 0.5m(12 + 12rp2)].

(1)

The potential energy corresponding to the deviated state of the load is

II = mgl(l- cos qJ).

(2)

Substituting Eqs. (1) and (2) into the Lagrange equation

d(oT)+m_O t3ip &p - ,

dt

we arrive at the differential equation of motion in the form

iP + (2i / l)riJ + (g / I) sin qJ = O. Assuming the deviations to be small (sin qJ ~ qJ ), we obtain with allowance for the time dependence of the unwound portion of the rope (I = OJRt) that in the case of small free vibrations, the differential equation can be written out as

iP + 2ip / t + gqJ /(mRt) = O.

(3)

In the case of ascending the load, the free length of the rope at an arbitrary moment of time (under the assumption that 10 = OJRto) is

1 = 10 - OJRt = OJR(to - t) = OJRt" (tl

= to - t),

so that the differential equation of motion takes the form

iPt1

-

2qJt1 / tl + gqJ / (OJRt l ) = o.

(4)

The solution to Eq. (3) can be represented as

qJ = (CI / 1')11(1') + (C2 / 1')1';(1'), Here,

(5)

II ( 1'), 1'; ( 1') , are the first-order Bessel functions of the first and sec-

ond kind, respectively, and l'

= 2~ gtl / (OJR).

The integration constants are found from initial conditions of the problem (qJ = qJo and riJ = 0 as t = to).

122

1 Vibrations of Systems with a Single Degree of Freedom

Differentiating Eq. (5) with respect to time with due regard to the well-known relationships

1. [tII(T) ] = tI (T); 1. [t ~ (T) ] =- t 1'; (T), 2

(Janke E. et al., 1960), where 12 ( T) , 1'; ( T) are the second-order Bessel functions of the first and second kind, respectively, we arrive at

cP = -(Cl2 (T)/ T+ C2 1';(T)/ T).

From initial conditions, we have

CI = tpoTo1';(To) / [II ( To)1';( To) - I 2 ( To)~( To)]; C2 = -tpo TO 12 (To) / [II (To )1';( To) - I 2( To)~ (To)].

Substituting the expressions found for the integration constants

CI and C2

into equation of motion (5), we obtain _ TO II (')Y2('0)-l! (.)12('0) tp - tpo -; II('0)Y2 ('0)-I2 ('o)l!('0)·

We now represent the solution to Eq. (4) in terms of the Bessel functions: where 13 ( T), 1; ( T), are the third-order Bessel functions of the first and second kind, respectively. The integration constant are determined from initial conditions of the problem much as it was done in the case of solving Problem 29. In the final form, we have _ .3 13 (')Y2('O)-Y3(.)12 ('0)

tp - tpo

'5 13 ('0)Y2 ('0)-I2 ('O)}3 ('0) .

The case of descending the load should be analyzed independently.

31 The differential equation describing small free vibrations of a system with the single degree of freedom has the form

my+cy = 0,

(1)

where y is the vertical displacement of the rope and c is the bending rigidity. In order to determine the latter quantity, we apply the unit force P at the fixation point of the load. Then, we find that the bending rigidity is c = 1 / y I ' where Y1 is the flexure caused by the unit force. The differential equation for bending the beam lying on the elastic base is of the form

or (2)

where

1.1 Free Vibrations

123

We now represent the solution to Eq. (2) as

Y = y(O)VI + y'(0)V2 + y"(0)V3 + ym(0)V4·

(3)

y(O), y'(O), ... are the values of the beam flexure and of its derivatives at the origin (i.e., at X = 0) and V i(X) is the Krylov function:

Here,

VI V3

=coshfJxcosfJx; v2

= 2~(coshfJxsinfJx+sinhfJxcosfJx);

= 2~2 sinh fJx sin fJx; v4 = 4~3 (cosh fJx sin fJx -

sinh fJx cos fJx).

Assuming that the origin is chosen in the middle of the beam (at the load fixation point), we can write out the boundary conditions that allow determining y(O), y'(O), '" , etc.:

0, y'(O) = 0, y'''(O) = -1/ 2EJx' for X = I / 2, yeO) = 0, y"(O) = o. for

X

=

hVI' = Tak· mg·mto account t at tain

Y

(0) -

-

- YI -

-4fJ4, v 4' v2

=

Vi'V3" = V2 , v4 = v 3" we 0 b-

2fJ4 V2 (fJlI2)v3(fJlI2)-vl (fJlI2)v4(fJlI2) b v[(fJlI2)+4fJ4 v'f(fJlI2)

We now calculate the system rigidity

c = 1/ YI . To this aim, we initially find

the quantity /3:

fJ=~k/(4EJJ; fJ=Iml. In this case, fJI / 2 = 1· 2 / 2 = 1, c = 1/ YI . After the value of Y has been calculated, we find c = 1/ YI = 9.44 kN/m. Then, the frequency of natural vi1

brations is

= -Jc / m = -J9440 /100 == 9.7 c-

p

1

•

32 By analogy with the solution to Problem 31, the equation for bending the beam

has the form

Y

= CIVI + C2 V2 + C3V3 + C4V4·

Placing the origin at the left edge point of the beam, we find

CI = C2 = 0 .

For the cross section X = I, we have Y" = 0, EJxY'" = -1. Using these boundary conditions, we can find the flexure caused by the unit force:

YI

(I) -

1

v2(fJI)v3 (fJI)-vl (fJI)v4 (fJI)

- EJx v[(fJI)+4fJ4v2(fJI)v4(fJI)

With allowance for fJ =

0.01 cm- 1 and fJI = 1, we find the rigidity of the

system

c = 1 / YI' or and the natural frequency of vibrations

C

= 3.8kN/m,

1 Vibrations of Systems with a Single Degree of Freedom

124

p =,Jc / m = ,J3800 /20 = 13.8 c·'. 33 The rigidity of the system is determined in the same manner as in Problem 31. For the determination of initial parameters entering into Eq. (3) of the solution to Problem 31, we deal with the following conditions: y = 0, y' = 0 as X = 0

= EJx y = -1 / 2 as X = I. In this case, the flexure of the beam at the load fixation point (x = I) under the action of a unit force is

and Q

_

v'j (PI)-vz(PI)v4 (PI)

1

y\ (I) - 2EJx

v\ (Pl)vz (Pl)+4p4v3 (Pl)v4 (PI)

.

In the case under consideration

f31 = HJr--k -/E-'j-x'

or

f31 = 0.25,

and the load eigenfrequency is

p = ~ my~ (I)'

or

p

= 87 c·'.

34 The Rayleigh method is based on the equality of the maximal values of the kinetic and potential energies in the case of vibratory conservative systems:

Tmax

= IImax . The maximal kinetic energy of the system is

Tmax

= 0.5my; +0.5! mol dz.

Here, the second term represents the maximal value of the beam kinetic energy. The maximal potential energy (beam bending energy) is

IImax

= 0.5ey; ,

where

c = 3EJx /13. The beam flexure under vibration can be represented in the form y(t, z) =y/z)sinpt. The function y\ (z) must qualitatively correspond to the expected shape of vibrations and satisfy the boundary conditions of the problem. In the case being discussed, the equation for beam bending under the action of a certain force Q applied to the mass m (see Fig. 227) can be taken as y\ (z). In this case, _ y\(z)-

Q[3 EJ x

[327 ( )2 -21(z)3] 7 . Z

This function satisfies all boundary conditions of the problem (y\ = y{ = as Z = 0 and y" = 0 as system kinetic energy:

Z

0

= I). We now calculate the maximum value of the

1.1 Free Vibrations

125

~------~----~~Q

z

11 Fig. 227.

or Tmax

where

_ p2 ( QI3 ) 3EJx (m+ f.lmol),

-2

P=}l[1(f)' _t(f)3]dZ

=

0.236

is the coefficient used in the

determination of the system reduced mass. The maximal value of the potential energy is

TImax = (c / 2)[QI3 / (3EJJf and TI max , we find the natural frequency

Equating Tmax system according to the formula

for vibration of the

p = ~c / (m + JIfI'lol) = ~c / [m + (33 / 140)mol]' 35 The solution to the problem is similar to that presented above. We choose the function of beam flexure in the form

Yt(z)

= Yo sin1lZ / 2/.

In this case, the coefficient of the mass-reduction is r21

f.l = il.h sin 2 ~i dz, and the vibration eigenfrequency is p = ~,....~-/(-m-+-m-o/-), where

C

= 6EJx

/

13 •

After the calculations have been performed, we arrive at f.l

=1 .

126

1 Vibrations of Systems with a Single Degree of Freedom

If we take as the flexure function the flexures caused by a certain concentrated force applied to the mass m then the beam mass reduction coefficient becomes equal to. J1 = 34/ 33 ~ 1.03. 36 The formula for calculating the eigenfrequency of the system has the form

p

= ~ c / (m + f11'YlJ),

where c is the system rigidity, /J is the mass reduction coefficient, and m, is the distributed mass of the system. The parameters c and /J, which correspond to the schematic diagrams presented in Figs. 32a-32f, are, respectively, equal to

c = Gd 4 / (8D;), J1 ~ 0.333; c = EJ)/3,

J1 ~ 0.121;

c = 96EJx /7/3, J1 ~ 0.445; c = 24EJ) 13 , J1 ~ 0.371; c = 3EJ)3 / (a 3b 3), J1 ~ 0.522;

c = mJ 4G / (66a), where i is the number of turns in the spring. The coefficient /J is determined with allowance for particular boundary conditions for schematic diagrams shown in Fig. 32 (see the solution to Problem 34). In the case of the schematic diagram presented in Fig. 32e, it is necessary to replace the mass m by the moment of inertia J in the formula for the determination of eigenfrequency.

1.2 Free Vibrations of a System with Allowance for Resistance Forces 37 The differential equation for small free vibrations of the mixer shaft is of the

form

JijJ + alp + crp = 0,

or (1)

where

2n = a / J,

p2

= c / J.

n = 1.2 /(2 . 0.5) = 1.2 s-\ and p2 = GmJ4 / (32lJ) = 19.6 s-', or p = 4.43 s-'. Since p > n, the damping

According to the formulation of the problem,

periodic regime arises, and the solution to equation (1) describing free vibrations takes the form

1.2 Free Vibrations of a System with Allowance for Resistance Forces

rp = e-l.2t (C1 cos4.25t + C2 sin 4.25/).

127

(2)

The integration constants C1 and Cz can be found from the initial conditions of the problem: rp = rpo and cp = 0 as 1 = 0 (rp is the twist angle of the uniformly rotating shaft). the shaft In the case of the shaft uniformly rotating at an angular velocity twist angle is

n,

rpo

= -aQl c.

Using the initial conditions, we find C1 = -aQ I c and C2 = O. With allowance for this fact, equation of motion (2) acquires the form

rp = _( d~ )e1.2t cos 4.25/. The time 1* necessary for the vibration amplitude to decrease by a factor of 10 can be found from the condition e -1.2t. = 0.1. Hence, it follows that I. = 1.9 s.

38 The differential equation of motion has the form

iP + 2ncp + p~rp = 0, where 2n =

"y 4(D;~) pv; and J is the moment of inertia of the cylinder.

We now calculate

Po2 =

GJp Ll

= 8.10 "d5·32 ::::: 1 .02 5 L-32,,~4Hp

S-2

and 2n = 0 046v ..

Since '0=

~=6.5s, po-n

we finally arrive at

n = p~ - 4~2

=

(1.02)2

4'i!:~~2

= 0.295 s-' and

V

= 12.826 mls.

39 Based on the force method, we obtain the following differential equation for small free vibrations of mass m: (1)

Equation (1) contains yet another unknown quantity Yk because the resistance force is applied to not the mass m but to the inertialess beam at the point K. Therefore, we should analyze a further equation relating the displacement of the point K to the forces acting on the system. In accordance with the force method, we have (2)

Thus, the analysis of free vibrations of mass m is reduced to solving the set of equations (1) and (2). We seek the solution in the form

128

1 Vibrations of Systems with a Single Degree of Freedom

= AeAt ;

y

Yk

= BeAt.

As a result, we arrive at the characteristic equation of the third degree in the form a(022 _ 012 )..1,3 021 all

+_1_..1,2 + 021

a022 ..1,+_1_= 0 011 021m 011021m .

The problem being analyzed corresponds to problems with a fractional number of degrees of freedom (in the case under consideration, we deal with one and a half degrees of freedom).

40 We set up the differential equation of disk motion. The disk is subject to the inertial moment - J ijJ, to the moment - C rp of the spring resistance forces, and to the moment - a¢> of the damper viscous resistance. Therefore, we can write out

JijJ + a¢> + crp = 0, or

ijJ + 2n¢> + p2rp = 0,

where J

= Jl'D;~P = 1.57 ·10-3 kg.m\ 2n = a / J, and p2 = C /

J.

The spring stiffness is C --

EJx -

Edt

Jl'~i - 67 ~i,

0938

C =.

m.

N

In the absence of damping, the system eigenfrequency is

p The To

logarithmic

= 27'( / ~ p2 - n 2.

= ~c / J,

decrement

is

p

= 24.3s-

determined

l

•

as

<5 = n To '

where

Hence, it follows that the reduced coefficient of the vis-

cous-resistance force is

n = <5p / ~02 + 47'(2; and the coefficient of viscous resistance is a

n = 5.64,

= 2n Ii = 2·5.64/ 1.57.10-3 = 7.18.10 3 N m s.

41. We write out the differential equation describing free vibrations of the system in the form .. a ' cl+mg 0 rp + 2ml rp + --zmrrp = ,

where a is the coefficient of viscous resistance to the motion. The system eigenfrequency in the absence of damping is

= ~(cl + mg) / 2ml. Using the formula c = Gd 4 / (80 3 i), we find c = 333 kN/m, which yields p = 14.0 S-I. The damping period of vibrations is p

1.2 Free Vibrations of a System with Allowance for Resistance Forces

whence it follows

n = ~ p2 - 41r2 /

129

,g .

Upon substituting values of the variables and calculating, we obtain n == 12.5 s·!. The coefficient of resistance to the motion, we find from the condition 2n = a / (2ml), i.e., a = 4nml = 4 ·12.5 ·10·0.15 = 75 N m s.

42 In the case of motion of the piston in liquid, it is subject to the inertial force my , to the spring elastic force ey, and to the resistance force developed by the damper. The resistance force is proportional to the velocity of the fluid flow through the orifice in the piston, and, hence, to the velocity of piston motion

F=aj. Using the d'Alembert principle, we arrive at the differential equation of piston motion

Y+ 2ny + p2 Y = O. (y

(1)

Taking into account initial conditions of the problem under consideration = Yo and = 0 as t = 0 ), we represent the solution to Eq. (1) in the form

y

y = yoe-nt cos(~ p2 - n2 . t).

(2)

The reduced coefficient of viscous resistance can be calculated by the formula

n

= [41r.uH / (mz)](D / e)4.

After the calculations have been carried out, we arrive at

p2 The time

tl/2

= e / m = 3.10 2 /2.73 = 1098.9 S·2.

n

~

5.42

S·I

and

required to decrease the deviation of the piston from the equilib-

rium position by a factor of two, we find from the condition y = 0.5 Yo' Substituting the values obtained for nand pinto Eq. (2), we determine the desired time using the relationship -nt 05 . Yo =yoe ,

whence it follows

tl/2

= 0.14 s.

43 The moving load is subject to the inertial force

-mi, to the spring elastic re-

sistance force 2ex, and to the resistance force ax of liquid. With allowance for these forces, the differential equation of load motion can be represented in the form where

X + 2m + p2 X = 0, n = a / 2m = 500/ 2·50 = 5 S·I and p2 = 2e / m = 2.~~OO = 40

S·2.

130

1 Vibrations of Systems with a Single Degree of Freedom

Taking into account initial conditions, we obtain the solution to the equation of -nt motion of the load in the form X = xoe cos PIS, where

PI

= ~ p2 -

n 2 = 3.87

S-I.

The vibration damping time

t.

can be found from

the condition whence it follows

44 In accordance with the d' Alembert principle, we set up the differential equation of load motion in the form of the equilibrium equation under the deviation of the rod through a certain small angle f{J'.

mglrp = ml2ip - ca 2ip - a 2aip = O. Using the denotation 2n

(1)

= a 2a / (mI2), p2 = (ca 2 - mg/) / m1 3 ,

we

write out differential equation (1) as

ip + 2nip + p2rp = O. For n

(2)

= 0, the system under consideration has the eigenfrequency P = ~(ca2-mg/) / m1 3 •

In order to determine the logarithmic decrement of damped vibrations, we cal-

8D 3 i . Upon substituting numerical values, we obtain c = 10 kN/m and p2 = 1600 S-2, or P = 40 S-I.

culate the spring stiffness The

reduced

n = a 2a / (2mI2)

C

= Gd 4

/

coefficient

= (0.2

of

damper

resistance

is

= 24 S-I. The eigenfrequency of damped vibrations is PI = ~ p2 - n 2 = .J1600 - 576 = 32 S-I. The period of damped vibrations is To = 27r / P = 6.28 / 40 = 0.157 s. Thus, the logarithmic decrement is 8 = nTo = 24·0.157 = 3.768. 2

·3 .10

2) /

2 ·1· 0.5

2

45 In the case of deviation of the level of the liquid column from the equilibrium position by the height x, the differential equation of vibrations, which is obtained from the Lagrange equation of the second kind, acquires the form

.. 2 nx+px= 2 0, x+ where

2n = a / m = 4a sin 13 / [mi 2 ps(1 + sin 13)] ;

p2

= (2g / s)sinfJ,

1.2 Free Vibrations of a System with Allowance for Resistance Forces

(since the gravity force of an unbalanced column of liquid is 2xgpm1 2

= ~ p2 -

Hence, it follows that the vibration frequency is PI tion period is To

131

/

4 ),

n 2 and the vibra-

= 27r / PI.

46 While moving (from the right to the left) the load of mass m from the initial position, we can write out the differential equation of motion as mXI

Since x

= Xo

and X = 0 as t XI -_( Xo

where

+cXI -F, =0.

(1)

= 0, the solution to Eq. (1) has the form

F, ) COSp t +-2' F, --2 mp mp

(2)

P =.Jc / m.

We can find the time tl corresponding to the load motion from the right to the left from the condition XI (tl)

= O. Therefore, tl = 7r / p.

The maximal deviation of the load to the left from the neutral position is

xlmax = -xo + 2F, / mp2. In the case of the load motion from the left to the right, the differential equation of motion (the Coulomb friction force changes its sign) is represented in the form (3)

For initial conditions of the form X2

x2 = XI max , X2 = 0

= (Xo -

as t

~~ )cos pt -

= tl , the solution to Eq. (3) is

(:;2)'.

After each half-period, the vibration amplitude decreases by

2F, /(mp2).

Therefore, we may write out for the nth half-period

xn

= [xo -(2n-l)F, /(mp2) cos pt+ (-ly-I F, /(mp2).

Vibrations of the load continue so long as

F, < C /

x(m-I)max •

47 When twisting of a free spring through a certain angle rp, the spring length changes by A. In the structure under consideration, the axial displacement of the disk is absent, i.e., A. = O. Therefore, the reaction force N arises in the supports. In order to determine the structure rigidity, we write out the following equations for displacements:

132

1 Vibrations of Systems with a Single Degree of Freedom

(1)

811 , 812

Here,

8 21 , 822

are twist angles and

are axial elongations of the free

spring subject to a unit moment and a unit force, respectively. We now calculate the coefficients 8ij. To do this, we apply to the spring an external torque 8 = 1, which causes the internal torque internal bending moment

Mlb

= 1· cos a

MIt

= 1· sin a and the

(a is the helix angle of spring turns).

The application of a unit axial force also causes the appearance of the internal torque

M2t

=1· R cos a

and the internal bending moment

M 2b

=1· R sin a

(R is the radius of friction surfaces). Following the Mohr method, we have

8

II

=

r27rRi Mlb GJp

.l:J

ds +

r27rRi

.l:J

a

MI~ d s = 27r Ri (sin2 + cos2

EJx

GJp

a).

EJx

'

(2) (3)

(4) where i is the number of turns in the spring. The solution to the set of equations (1) with allowance for possible to obtain the following relationship:

A =0

makes it

rp=c/ M, where and CI

= 812 /(811 812 -81;).

In accordance with the solution obtained, the moment of the friction force is

Mf

= JlNR = JlclrpR = cOrp.

Here, C * = pRcl and J1. is the coefficient of the Coulomb friction, and R is the radius if the friction surfaces. In the last relationship, the minus sign is absent because ofthe existence of two friction surfaces, and M f is independent of the fact what is the contact surface. Thus, in the process of vibration, the following moments act on the disk: the moment of inertial forces J rp, the moment of the spring resistance force crp, and the moment of the friction force the velocity form

ip

c· rp, which is always directed oppositely to

of motion. The differential equation of motion, we represent in the

1.2 Free Vibrations of a System with Allowance for Resistance Forces

J¢ + cljJ + sign~· qJ = 0, where the term

sign qJ

133

(5)

implies the sign ofthe velocity of motion.

According to the formulation of the problem, at I = from the equilibrium position through angle qJo.

0, the disk is deviated

For the fIrst quarter of the vibration period (as qJo ~ qJ ~ qJOI ), the differential equation of motion has the form

ip+(c/l +c· / J)qJ =O.

The necessary condition for disk motion is the validity of the inequalities

cqJo > c· qJo or M > M f as is indicated in the formulation of the problem. The solution to the equation of motion for the fIrst quarter of the vibration period is of the form where

kl

= ~(c-c·)/ J o.

= 1l / 2kl . In the case of passing by the equilibrium position ( qJ = 0), the signs of both

The time of motion in the fIrst quarter of the period is II

the resistance force moment and inertial moment change. At the same time, the sign of the Coulomb friction force moment conserves because of the conservation of the sign of the velocity ljJ. Therefore, we can describe motion in the second quarter of the period by the differential equation

¢ + (c + c· )qJ / J = O. Taking into account initial conditions of motion (qJ = I

= II)' the solution to Eq. (6) can be represented in the form

(6)

0, ljJ = -ljJokl as

· k qJ -- -qJo rkl sm zI, 2

= ~(cI + c· ) / J .Hence, it follows that at the end of a half-period, as = 1l / 2k2 ,i.e., I = II + 12 = 0.51l(1 / kl + 1/ k 2), the deviation of the disk

where k2

12

from the equilibrium condition is (7)

the velocity of motion being equal to qJ~1 = O. In the third quarter of the period, the differential equation of motion has form (1) but with the initial angle qJOI being determined by expression (7). In the fourth quarter of the period, the motion is described by differential equation (6).

1 Vibrations of Systems with a Single Degree of Freedom

134

Solving the problem in a manner similar to that of preceding Problem 46, we arrive at the conclusion that the amplitude of the disk rotation angle decreases at the end of the first period down to

qJOI = qJo(kl / k2)2 = qJo(C - CO) / (c + c·). At the end of the nth period, the vibration amplitude is

qJOn = qJo (kl / k 2 ) 2n = qJo [( C - c" ) / (c + c" ) The vibration period is

T = 2(tl +t2 ) =

n-(11 +

r.

lJ+ l~. (~C7· +~C'/).

48 In the presence of Coulomb friction between the disk and the supports, the differential equation of free vibrations of the system (see Problem 47) can be written out as

JijJ + c(p + sign tpe. qJ = O.

(1)

We consider that, at an initial moment of time (t = 0 ) the disk has a certain deviation qJo being at the rest state «(Po = 0). In this case, for the starting phase of motion, as angle qJ varies within the limits form

.. qJo

0 < qJ ::::; qJo' equation (1) has the

+ k I2 qJ = 0,

(2)

where

k~=(c-c")/ J. When passing the disk by the equilibrium position, the sign of the inertial force changes. and Eq. (1) acquires the following form:

.. + k 22 qJ2

qJ2 where

= 0,

k; = (c + c· ) / J.

Thus, in the third and fourth quarters of the period, Eqs. (2) and (3) are valid, respectively, etc. The solution to Eq. (2), we represent in the form qJI = CI cos kit + C2 sin kit. With due regard to qJI

qJI

= qJo

and (PI

= qJo cos kit,

(PI

= 0 as t = 0, we arrive at

= -klqJo sin kit.

(4)

1.2 Free Vibrations of a System with Allowance for Resistance Forces

135

Fig. 228.

In the phase plane ( ffJ; rP), expressions (4) can be represented as ellipsis equations (Fig. 228):

ffJI2 / ffJ;

+ ffJI2 / (kl ffJo)2 = l.

The solution to Eq. (3), we obtain with allowance for the fact that

ffJ2

= ffJI = 0

and rP2

= rPl = -klffJo as I = II = 1r / 2kl . In this case,

ffJ2 =-(kl / k2)ffJosink21,

(P2 =-klffJocosk21.

(5)

Here, the time t in the second quarter of the motion varies from 0 to 12 , The end of the second quarter of motion is determined from the condition that rP2

=0

as 1 = 12, Hence, it follows that 12 = 1r / 2k2 . The set of equations (5) can also be represented in the phase plane as the ellipsis equation (see Fig. 228): 2

f/J2

·2

2

(klf/Jo/ k2)

+~-1 2 •

(klf/Jo)

We find the solutions for other segments of motion in a similar manner. Since

kl < k2' we deal with the damped motion of the system. 49 The differential equation for free mechanical vibrations of the systems has the form

m.x+cx = O.

(1)

The differential equations of electric-current oscillations and voltage oscillations in the circuit can be represented as

136

1 Vibrations of Systems with a Single Degree of Freedom

Li +.l/=O c '

(2)

cO +u / L =0.

(3)

In accordance with the structure of Eqs, (2) and (3), an analog electric circuit is a circuit containing a capacitor of capacity C, inductance coil L, and a switch. If we accumulate electric charge on the capacitor plates and close the circuit, then oscillations of the electric current I and of the voltage U arise there at a frequency

p

=1/ .JLC.

50 The differential equation of free damped piston vibrations is written out as

m.x+ax+cx=O. The differential equations for oscillations of the electric current and of the voltage, we represent in the form

Li + JU + / / C = 0,

cO +U / R+U / L =0.

(1) (2)

Equation (1) describes oscillations of the electric current in a circuit that contains an inductance coil L, a resistor R, and a capacitor C connected in parallel. Equation (2) describes voltage oscillations in a circuit consisting of the same components but connected in series.

1.3 Forced Vibrations 51 When the vibrograph base moves according to the harmonic law Y Yo sin mt , the magnet executes vibrations being described by the differential

=

equation

my = -C(YI - y).

(1)

Here, YI and Y are displacement of the lower and upper spring ends, respectively, and YI - Y is the spring stretching, or, which is the same, the displacement of the magnet with respect to the coil. We now make use of denotation Z=

YI-Y.

(2)

Then, ••

Yo

••

••

=Z+ Y =z- Yom

2

•

slllmt.

(3)

Substituting the value of Y from Eq. (1) with allowance for Eq. (2) into Eq. (3), we arrive at the differential equation of small forced vibrations

1.3 Forced Vibrations

Z

where p~

137

2 . = Poz = Yom 2 slnml,

= c 1m.

The solution to the last equation is of the form _ Z -

Yo • 2/ 2 1 slnml.

Po (j) It is evident that the least distortion of vibrations takes place in the case of

p~

«

m 2 . In this case,

Z

~ yosinwt. The minus sign implies that the phase of

magnet vibrations is opposite to that of base vibrations, i.e., in this case, the magnet remains immobile in space.

52 While rotating the piece at an angular velocity

OJ, the law of motion for the

measuring rod is expressed by the equation

x

= I sin mI.

(1)

-mi, reaction force R, of the spring. (Here, cl' c2 and 80 are,

In this case, the rod is subject to the inertial force and the elastic force c)x + c2 (x + 80 )

respectively, the total stiffness of plane springs, the stiffness of the helical spring, and the preliminary elongation of the helical spring.) Thus, the differential equation of motion takes the form

mi+c)x+c2 (x+80 ) = R, or (2)

where

c2 80

= Ro .

The condition of continuous contact implies that the reaction force R is always directed towards the same side (upward). Then, the ultimate angular velocity m. can be found from the condition that the minimal value of R over a vibration period if only once becomes equal to zero. Substituting Eq. (1) into Eq. (2), we find

(p~ - ( where

Po

2)

= ~(c) + c2 ) I m

sin ml + Ro I( me) = R,

(3)

is the eigenfrequency of the measuring rod.

We now consider possible cases.

> ( 2 ), the minimal reaction tained at the moments of time corresponding to sin m/) = -1 . In this case, In the pre-resonance operating mode (P~

- p~ + m2 + Ro I(me) > 0, or

is at-

(4)

138

1 Vibrations of Systems with a Single Degree of Freedom

Ro /(me) > m~ -ai.

(5)

In order to have the pennanent contact, it is necessary for condition (5) to be valid within the entire range of possible values of OJ dition is valid for OJ

(0 < OJ < Po ), If this con-

= 0 , then it will be valid for all OJ *- 0 .

As OJ = 0, Ro > (CI + c2 )e .

If condition (5) is invalid, then violation of the contact is possible, i.e., R = 0 . From condition (4), we obtain the following expression for the ultimate angular velocity OJ:

One more case of contact violation can occur as OJ mode). In this case,

2

> Po2

(post-resonance

R = 0 , at the time moment corresponding to OJt2 = 1. Thus,

we determine another value of the ultimate velocity, using the fonnula

m;. = p~ + Ro /(me).

The dynamic reaction at OJ

= 0.lOJ2* can be found from Eq. (2):

R = Ro + [( cl + c2) - m( 0 .1OJ2.)2]e sin OJ;.tl .

(6)

Furthennore, we calculate the parameters entering into Eqs. (3) and (5). The stiffness of the helical spring is C

=

Gd4

=

1010·0.0014

8lYi 0.0053.20 The stiffness of two plane springs is 2

cl

=4

kN/m

.

x = 24·2·101I ·5·1O-2 U5·1O- 12 = 9 26 = 24EJ t3 27.10-6 .12 .

kNl

m.

The ultimate angular velocity is

m2•

-

cl+ c2

m

+ Ro _ ml -

+ 0.05.10-3 5 = 6004 s-' ..

.-~----~-------

4.103+9.26.103 0.05

Finally, the measuring force of the device detennined according to fonnula (6) is

R =5±1.3 N. 53 Eliminating the hinge (i.e., replacing it by the reaction force N) and applying an unknown moment M to the end cross section (Fig. 229), we obtain, in accordance with the force method, three equations

y

= 811 (-my) + 812 M + 813 N;

(). =821 (-my) + 822 M + 823 N; YI

= 813 (-my) + 832 M + 833 N.

1.3 Forced Vibrations

139

v

Fig. 229.

Here,

6ij

are the compliance coefficients;

8 = 80 cos OJ!; Y\

is the vertical dis-

y\ = 0 . U sing two last equations of the set obtained, we express M and N . After certain transformations, we arrive at the following equation with respect to y : y + ay = b80 cos OJ!, placement of the cross section associated with the hinge, and

where

a and b are coefficients depending on 6ij'

For stable vibrations, we have bf)o cos OJ!. Y = a-OJ 2

After y has been determined, we find M and N. The bending moment in the restraint is (see Fig. 229)

Mb = JI+2IN +M. The maximal normal stress in the restraint is O"max

where

= Mb /Wx '

w.. is the resistance moment in the cross section X .

54 The problem should be solved in a similar manner as Problem 53. In the cross section whose displacement is given, we apply an unknown force P. Furthermore, introducing again the second unknown force N , we eliminate the support. Then, using the force method, we arrive at the set of three equations:

= 6\\ (-my) + 612 N +613~'

(1)

0= 6 21 (-my) + 622 N + 623~'

(2)

= 631 (-my) + 632 N +633~'

(3)

y

YI

Here, ~ is an unknown force applied at the point k and

Yk

= Yo cos OJ! .

140

1 Vibrations of Systems with a Single Degree of Freedom Based on Eqs. (2) and (3), we now find the forces ~ and

N. The latter is

determined from the expression N (02\033-03\023)mji+023Yocoswt 022033 -023032

=

Excluding termine

N and

(4)

~ from Eq. (1), we obtain the following expression to de-

y: (5)

where

P and b are coefficients depending on 8ij.

Determining y from Eq. (5), we can find the force

N.

55 The displacement by a certain distance of rod 1 causes the rotation of the lever through a certain angle ({J = x/I. In this case, when the motion occurs upward, the force in spring 2 and spring 5, respectively, increases and decreases. Thus, the inertial force - mX , the spring resistance force 2c\ (x + X 0 ), the support reaction

force

R , and the force caused by lever 3 and equal to

[J({J + C2«({J - ({Jo)] /10

act on rod 1. In this case, the differential equation of motion of rod 1 has the form:

(m+J / 12)x+(c\ -c2 / 12)x+(c\xO +c2({Jo /1)= Ro. Using denotations m + J / 12 = M, c\ = c 2 / f = C, and c\X O+ C2({Jo / 1 = Po, we can represent the differential equation of motion of the measuring rod as

MX + CX + Po = R.

(1)

The condition of permanent contact consists in the fact that at the point B 1, the reaction force does not change its sign, i.e., R ~ 0 . Since the surface profile is described by the equation X = a sin 2Jrz/A , the equation of motion of the measuring rod has the form X

= a sin OJI,

(2)

where

mt = 2Jrz/A = (2Jr/A)vt. Substituting Eq. (2) into Eq. (1) and taking into account the condition we obtain

(p~ - m2)asin OJI + Po where

Po

= .JC / M

/ M ~ 0,

is the eigenfrequency of the device lever.

(3)

R

~

0,

(4)

1.3 Forced Vibrations

141

Two cases of violating the contact are possible (see the solution to Problem 52). For the pre-resonance regime (P~ the case of

> 0/), violating the contact is possible in

= P~ - Po / aM.

0)2

If the system parameters are chosen in such a way that the condition c P+c

D £0

_

2

~f

l 2 > -Z2a - Po 1V1a is fulfilled, then violating the contact will not occur.

For operations in the post-resonance region (0/ > P~), we can find the value of OJ for which violating contact is possible from the condition of vanishing relationship (4) (in the case of sin 0Jt1

= 1). Then,

= p~ + Fa / aM.

OJ2

With allowance for relationship (3), we find that the ultimate velocity termined by the formula

v.

v.

is de-

= (A / 2Jr)~ p~ + Fa / (aM).

56

v.

Mol = 27CA ~ Po2 +;;;-,

where

Po2 =c2 / J •

57 The differential equation for stimulated vibrations of the trimmer has the form

J qJ + [c + (ci + C2 )/ 2 ]tp = c21xO sin 0Jt. In the case of a resonance, OJ;

= [c + (ci

+ c2 )/ 2 ]/ J

= p~ + (ci + c2 )f / J,

whence it follows

58 The dynamic coefficient is expressed by the formula

kdyn = a

/ Xo

= 1/ 11

0)2 /

p21.

For schematic diagrams shown in Fig. 50a and 50b, we have, respectively,

kdyn = 1 / 11-

m0)2 /

cJ,

and

59 The dynamic coefficient is given by the formula

k dyn Since k dyn

= 1/15,

= 1/ 11- J0)2 /

cl.

and the motor operates in the post-resonance region

(p~ > 0)2), the stiffness of the system must be equal to

1 Vibrations of Systems with a Single Degree of Freedom

142

c = J oi

= 20 . 105 2 116 = 1.3 8 . 104

We assume the segment

1\ to be

N m.

absolutely rigid, and the fixation of supports

to the base to be a restraint. After uncovering the static uncertainty and calculating the angular displacement of the motor support, we find that the stiffness of one support is (1) Here,

the

of

moment

inertia

for

the

cross

sectional

area

is

= bh 3 112 = b4 I c, and the Young modulus for steel is E = 200 GPa. Since C = 2c\, with allowance for 0), we find b = 0.654 cm. We take b = 6.5 mm, hence h = 2b = 13 mm. In this case, after certain calculations, we obtain C = 2c\ = 1.34 . 104 N m, k dyn = 1115.1. Jx

As the calculations show, the maximal bending moment in the pillar is = O.3M. Therefore, the alternating moment

Mmax

Mmax

= O.3M = 0.3( Mo + kdyn M\ sin at).

acts in the dangerous cross section. The average value of the maximal stress in the cross section is

am The stress amplitude is aa

=0.3Mo In: =275

MPa.

= kdyn ·O.3M\ In: = 110 MPa.

Thus, the coefficient of the reserve of endurance is

n

(j

= ualu\+umlu 1 = 1.8. a

60 Since the perturbing force has a periodic character, the amplitude of stimulated vibrations of the system is

a = a st 111- oi I p~l. The amplitude of the support reaction force is proportional to the amplitude of the elastic-coupling deformation

Rdyn = ca = RsJll-m2 I p~l. Hence, it follows that the dynamic coefficient is

kdyn

= Rdyn I Rst = 1I1l-m2 I p~l.

Thus, for schematic diagrams presented in Figs. 52a--52c, we have, respectively,

1111- mm I cl; kdyn = 1111- mm I cl; kdyn

=

2

2

1.3 Forced Vibrations

kdyn

143

= 1/11- 9mai /20cl·

61 We denote the shifts of the upper spring end and of the load as XI and X 2 , respectively. Therefore, the force acting on the load in the case of motion of the upper end of the spring is F = C(XI - x 2 ). The differential equation of motion of the load has the form

mX 2 + aX 2 or

C(XI -

x2 )

= 0,

x2 + 2nX2 + P~X2 = (CXO / m)sin ax.

The amplitude of the load vibration with respect to the equilibrium position (assuming (j) = Po) is

x 20

= ~ -Jc / m = liV ~7J.~0 = 1.615

In the resonance regime, the phase shift equals elongation of the spring is

1[ /

cm.

2, therefore, the entire

= x 20 + x st = x 20 + Q/ C = 1.615 + 1.26 = 2.875 x st is the spring elongation under the static force Q = mg . xmax

cm,

where The maximal force stretching the spring is

Pmax

= cXmax 7200.2.875.10-2 = 207

N.

We now find the mean and amplitude stress in the spring

8.91.5.10-2 = 92 74 MPa 'm =8QD = 3.14-(5.10-2)3 . , ;rd3

.7200.5.10'a = 8x;rd3cD = 8.1.615.103.14.(5.10-2)3 2

20

2

= 1185 .

MP

a.

The safety factor with respect to the yield strength is

ny

= 'y / 'max = 'y / ('m + 'a) = 92.7~~~18.5 = 2.84.

Finally, the safety factor for the fatigue strength is

nf

= 1/ ('a

/ '_I + 'm / 'B)

= 1/ Cio14 + IJ~05) = 2.22.

62 We denote the vertical displacement of the rod and the load as XI and x 2 ' re-

F = a(xi - X 2 ) being transferred through the damper, to the inertial force mx 2 ' and to the resistance spectively. In this case, the load is subject to the force

force 2cx 2 of the springs. In accordance with the d'Alembert principle, the differential equation of motion can be represented in the form

mX2 + cX2

= a(xi -

x2 )·

Taking into account that the displacement of the rod depends on the angular velocity (J) of the crank

1 Vibrations of Systems with a Single Degree of Freedom

144

XI

we obtain

= rsinat,

mX2 + 00 2 + 2ex2 = aor cos at,

or where

2n = a / m , P~

= 2e / m.

The solution to this equation has the form

x 2 = 2nar sine m t+ ~) / ~r-(p-~-_-m-2)-+-4-n-2-m-2; tg ~ = 2nm / (p~ - ( 2). The maximal stress in the spring arises in the case of its maximal compression, i.e., in the case of sine m t+ ~) = 1. We note that x 2mzx = r since by the condition, m = Po. The force, corresponding to this compression is ~

=er.

The total force in the spring is

p

= ~ + mg / 2 = cr + mg / 2.

Thus, the maximal stress is Tmax

= 8(er + mg / 2)D(Jld3).

63

P = ~ p~ - n2 (where p~

= e / n, n = a / 2m);

ml

=0;

m2 = ~2(p~ - 2n 2). 64 The differential equation of motion for the system being analyzed has the form

X + 2ni + p~x

= Fo sinxt,

(1)

where X is the horizontal displacement of the system center of mass. This displacement is counted off from the position of static equilibrium;

2n=a/(~ +m2); p~ =e/(ml +~); Fo =mom2e/(~ +m2)· We now write out the solution to Eq. (1) for the case of stable motion: X

= Fo sin(wt- c)/ ~(p~ _( 2)2 + 4n 2ai ,

(2)

where C is the vibration phase. Furthermore, we calculate the work of friction forces (i.e., the dissipated energy) over a vibration period T = 27r / m :

Af = t

':2.7r /OJ. t':2.7r /OJ • axdx= axdt=

(2

aFo7rOJ

Po-OJ

2)24 2 2.

+ n OJ

(3)

1.3 Forced Vibrations

Using the condition

d Af

/

d a = 0, we now find the value

145

a. of the coeffi-

cient of viscous friction for which the dissipated energy becomes maximal. Differentiating Eq. (3) with respect to a and taking into account that

n = a I [2("", + m2 )] , we arrive at the expression (p~ - 0/)2 -

a.o/ 1("", + m2)2 = 0,

whence it follows

(4) Substituting relationship (4) into Eq. (3), we can calculate the maximum value of the energy dissipated over a vibration period:

~ax = 1#02 ("", +~) I [2(p~ - 0/)]1.

(5)

As follows from Eq. (3), in the resonance regime (Po = OJ), the work of the friction forces is inversely proportional to the coefficient a of viscous friction. 65 Using the solution to Problem 64, we find from Eq. (2) of this solution the amplitude of stable system vibrations: Xo

= mo0J 2e / [("'" + ~)~(p~ -

In this case, the force acting on the springs is

OJ2)2

+ 4n 2OJ2].

F; = ex .

We now find the value of the angular velocity for which the amplitude of the force acting on the springs equals the amplitude of the perturbing force: . cmom 2e 2

F =ex = 10

0

With allowance for p~

(ml+"'z)~(p6-m2)2+4n2m2

=mOJe 0

•

= e I ("'" + ~) , we obtain Po2

= VI( Po2 -

W

2)

+ n 2 WI2 ,

whence it follows OJ1 =

~2p~ -4n 2 = ~2(p~ -2n 2 ).

In Fig. 230, the plot demonstrating the amplitude ofthe force function of the perturbing-force frequency from the plot,

F;o

< mo0J 2[ and

OJ

F;o = exo

as a

is presented. As immediately follows

F;o > lrloOJ2[

in the case of OJ ~ OJ1 and

OJ ~ OJ1 , respectively.

66 The differential equation of motion for mass m has the form

my + a.Y + ey = Po sin at,

or

y + 2ny + Poy = (Po / m)sinat, where

(1)

146

1 Vibrations of Systems with a Single Degree of Freedom

Fig. 230.

2n = a / m ; p~

= c / m;

c = 3EJx / 12.

We now write out the solution to Eq. (1) in the case of the steady-state vibration regime

Y = Asin(ox + ffJ), where

A = Yst / ~(1- ai / p~)2 + (2n(1) / p~)2 ; Yst

tan ffJ = 2n(1) /(p~ -

= Po / c;

(1)2) .

The total flexure of the beam is Y + Yo . Here, Yo is the flexure caused by the gravity force mg of the motor. Thus, the flexure of the beam and therefore its inner stresses vary according to an asymmetric cycle. We now calculate the stresses in the restraint: O'm

= mgl/ u;: = 6mgl/ (bh 2 ) = (6·50 ·10 ·1) / (125 ·10--6) = 24 MPa.

The stresses are proportional to the flexures and thus can be calculated by the formula where 0'

st

3). = P.1I Wx = 125.10--6 2000-1-6 = 96 MPa'' p20 = C / m = 3EJ x / (mI ' 0

p~ = 6250 S-2. The safety factor of the fatigue strength, we determine by the formula

n=

1 aa/a_l+am /ab' Here, the coefficient n is only unknown quantity. Using formula (2), we obtain n 62.64

~

0"

a

= 2nm =

2·62.64·50

=

(N·s)/m.

S-I,

whence

it follows

1.3 Forced Vibrations

147

67 The dynamic coefficient is k dyn = 11~r-(1-_-(1)-2-I-p-~-)2-+-(-2-n-(1)-I-p-~)-2 . After numerical data have been substituted, we arrive at

kdyn

= 0.316 .

68 The differential equation of motion for the load of mass m with respect to the base has the form

y + 2ny + p~y = Yo (1) 2(sin ax + 200 sin lOax),

y is the displacement of the load with respect to thew base, 2n = aim; p~ = elm.

where

and We now write out the solution to this equation for the steady-state motion: _

yom 2 • ( 2 2 sm (1)1 v(Po -m ) +4n m

Y - I 2 22

200m Yo . (10 ) a, ) + v(Po I 2 22 2 2 sm (1)1 - a 2 , -100m) +400n m 2

=2n(1) 1«(1)2 - p~); tan a 2 = 20n(1) 1(1 00(1)2 - p~) . Taking into account that p~ = 0.01(1)2 and n = 0.02(1) , we finally obtain

where

tan a,

that the relative displacement of the load, which is recorded by the device, is determined according to the expression

y

= Yo sin(ax - a,) + 2Yo(lOax - a 2 )·

69 The differential equation of small free vibrations of the mirror has the form JijJ + alp + crp = M(i), or

ip+ 2nip+ p~rp = M(i)1 J, where M( i) = Fma is the perturbing moment. We now calculate the rigidity of the system. To this aim, we apply to the mirror the unit moment M, = F;a = 1 (Fig.231a).

i I

a

Fig. 231.

~

148

1 Vibrations of Systems with a Single Degree of Freedom

In this case, the mirror rotates through the angle axis. Assuming

h << II

rpl

with respect to the vertical

(Fig.231 b), we find

F;

= 2QP= 2QfAa / II·

Therefore,

MI

= F;a = 2Qrpl a 2 /

II

= Crpl·

Hence, the rigidity of the system is

c = 2Qa 2 / II. The moment of inertia of the mirror with respect to its rotation axis is

J

= ms b 2 /12

Thus, we find the frequency of free vibrations of the mirror:

P = ~24Qa2 / (m 3/ I b2) - (a / 2J)2. 70 The differential equation of motion for the vibrator frame is of the form

JiP + aip + crp = Mo.

(1)

We now represent Eq.(1) as

iP + 2nip + P~rp = M,

(2)

where M = Mo / J. Since the perturbing moment i.e., satisfies the condition

Mo

Mo

is applied suddenly,

0 as t < 0,

={ Mo as t > 0,

the initial condition being of the form out the solution to Eq. (2) as

rp = 0

and

ip = 0

as t

= 0 , we can write

rp = rpst [1- ~~ e -nt sin(pl t + P)]'

r

,"51

Z

~------~------~~----~.~t Fig. 232.

1.3 Forced Vibrations

Here, and

rpst

149

= Mo / c = yi / c; Po = Fcfj ; n = a / 2J ; PI = ~ P~ - n 2

;

tan fJ = PI / n = ~ (p~ - n2 ) / n .

After the calculations have been performed, we arrive at the relationships C

= GJ /1= 40·109 ·3.l4·16·10- 12 = 21 N/m' 32-3.10-2

p

Po

'

=.Jc / J = ~2.1 / (1.02 ·10-6 ·10) .

In the absence of damping, we have

= Po = 1320 s-'; fJ = 1l /2. In the presence of damping (n = OJ Po), we obtain PI

PI

= 1259

s-'; tan fJ

= 3.17 ; fJ == 72°30' .

Figure 232 shows the plots of the ratio rp / different coefficients n.

rpst

as a function of PI t for two

71 We denote the absolute angular velocity of the flywheel as

rif

rif. Then, the iner-

tial torque acting on the flywheel is J (J is the polar moment of inertia of the flywheel). The spring resistance force is determined by the difference between the rotation angles of the flywheel and of the pulley and is c( '1/ - rp), where rp is the pulley rotation angle. Thus, the differential equation of motion for the flywheel has the form

Ji;J+a(,p-ip) + ('I/-rp) = O.

'1/ - rp . Then, we have Jii+av = cv = -Jep. Taking into account that ip = W = Wo + WI cos kt , we arrive at

We now use the denotation V =

ii + 2nv + P~V = kWI sin kt, where

2n = a / J

and p~

=C / J .

The solution to the last equation is of the form V

where

= A sin(kt - fJ),

A = kWI / ~ (p~ - k 2 )2 + 4n 2k 2

The phase shift attains

•

fJ =arctg[2nk /(p~ -e)],

and the logarithmic decrement is

g = 21ln / ~ p~ - n2 •

1 Vibrations of Systems with a Single Degree of Freedom

150

We now calculate the flywheel moment of inertia

J

4 (1-~) by = 1CD 32 D4 g'

Upon substituting numerical values, we obtain J = 3.68.10-2 kg m2 • The frequency of natural vibrations (in the absence of damping) and the resistance coefficient are, respectively,

Po = ~r-14-6-/(-3.-68-.-1O--2-) = 62.92 s-', n = a 12J = 0.73/(2.3.68.10-2) = 9.92 Thus, we find

8 =1 and the phase shift fJ

~

53°30'.

72 The differential equation of motion of the vibrator case can be written out as

MX+ax+ex = pet),

where

M =m+2(mo +ml ). P(t) , we find by the projecting onto the vertical axis the

The perturbing force inertial forces for the unbalanced masses:

pet) = 2moaiesinlOt.

Thus, the amplitude of forced vibrations of the vibrator is

Xo where

= xst 1~(1-lO2 1p;)2 + (2nlOl p;)2 ,

xst =2molO2ele, n=aI2M,and Po =.JeIM.

73 In the case of pressure variations, the differential equation of motion of the indicator measuring rod has the form

mX + ax + ex = pet) = IIp(t)F fJ,

or

x+ 2m + k; x = IlpF fJ 1m = B sin lOt,

(1)

where

2n = aim;

kg = elm;

B = ApoFefJ I(mll).

(2)

The solution to Eq. (1) is

x = Xo sine lOt - ffJ), where

xO1sx t

(3)

= e01"I(e - lO 2)2 + 4n 2lO 2• 0

Substituting numerical data into relationships (2) and (4), we arrive at

2n=200, xst =1.5

mm.

(4)

1.3 Forced Vibrations

151

-~~--~----~~~~~

p.

Fig. 233.

The plot of the ratio Xo 1xst as a function of

(1) 1ko (i.e., amplitude-fre-

quency characteristic of the device) is shown in Fig. 233 (curve 1). 74 The differential equation of motion for cylinder 4 (see Fig. 65) can be written

out as or

x+ 2n.x+ P~X = B sin (1)t, where

2n = aim = 0.75 .10-2 /0.1.10-2 = 75 S-I; p~ = (c1 +c2 +c3 )1 m = (2000+50+ 1000)/0.1 = 3.5.104

S-2;

B = poeAF l(m!1) = (3.5 .104 ·10-4 ·1· 4 .10-4)/(0.1.10-2 ) = 14 m1s2 • The static deviation from the cylindrical shape is

xst =BI p~ = 14/(3.5.104 ) =4·10-4

m

=O.4mm.

The vibration frequency of the cylinder as a function of the vibration amplitude has the form

Xo 1xst = [(1- (1)2 1p~)2 In the case under consideration,

2n 1Po

+ (2n(1) 1p~ i r1l2.

= 75.J3.5 .104 = 0.4

r

Therefore,

Xo 1xst = [(1- (1)2 1p~)2 + (0.4(1) 1Po)2 l/2 This function is plotted in Fig. 233 (curve 2). In the same figure, the tolerance zo-

152

1 Vibrations of Systems with a Single Degree of Freedom

ne for the dynamic measurement error (shaded band) is shown. This zone can be described by the equation Xo

1X st

=1± 0.1.

The intersection point of the boundary of the tolerance zone with the amplitude--frequency characteristic yields the maximum admissible value for the angular velocity of the piece being tested: lUmax

:-;

P~ 13 = 62.5

S·I.

75 In order to solve the problem, we employ the Lagrange equation of the second kind. The kinetic and potential energies of the system, as well as the Rayleigh dissipative function are, respectively,

T=mv 2 12+J/l 12, TI = c(lcp - h)2 12, R =a(lip-h)2 12. We now introduce these expressions into the Lagrange equation d (aT) dt aqJ -

aT -_ - an aR . aqJ aqJ - arp

As a result, we have

Jip + al(lip - h) + cl(lcp - h) = 0, or (1)

where

27r1lboal )2 + Po4(Tho )2 tan ( ----g;-

b --

2n = al 2 1J o; p~

r -_ Po21) 1(4n7rV).,

= cl 2 1J o'

We can represent the solution to Eq. (1) in the form

cp = A sin(27rvt 1I) - r) + B cos(27rvt 1I) - r) + D. It is evident that

D = ho 1I , so that we have the following set of equations

determining constants A and B:

[p~ - (27rv 11))2]A - 2n(27rv 1I))B

= b;

2n(27rv 1I))A + [p~ - (27rv 11))2]B = O. Thus, we find

A_

b[p6-(27Z"vll))2] • - [p6-(27Z"vll))2f+4n2(27Z"VII))2 ,

B =_

2nb(27Z"VII)) [p6-(27Z"VI 1))2 ]2+4n2 (27Z"VII))2 •

The solution to equation of motion (1) can be represented as

1.3 Forced Vibrations

153

ffJ = ffJo sin(21Z"Vt / 10 + fi) + ho / I, where

=.JA 2+B2 =

ffJo

b

J[P5-(21lV/ll)2f+4n2(2trv/ll)2 .

The maximum value of the vibration amplitude as a function of the trailer velocity of motion, we find from the condition

dffJo /d V = 0, or

2(2n/ 1)2 h;[p~ -2p~oi +W4 +4n 2w2]+ 4[(2n/ 1)2 h;w 2+ p~(ho / 1)2](p~ _w 2+ 2n2) = 0,

(2)

where W = 21Z'v / II . After certain transformations of expression (2), we arrive at the relationship

2 4 4 2 6 0 n w + PoW -Po = . In the absence of damping (n = 0 ), we obtain from the last equation that the

maximum amplitude of forced vibrations is attained in the case of a resonance at W =

Po or v. = poll /21Z'.

In the presence of damping in the system,

W = ~r_-'-p-~-/-2-n-2-±-~rp=~=/4=n=4=+=p=~=/=n=2. Since only the real-valued root has the physical sense, we obtain as a final result

76 By analogy with Eq. (5) of the solution to Problem 47, we represent the differential equation of the disk motion in the form

lip + cip + signipc· ffJ = Mo sin(wt + r), where C=b22/(bl1b21-bl;),and c· =jJRc\. Assuming that at a certain moment of time t = 0 in the steady-state vibration mode the deviation of the system from the eqUilibrium position becomes maximal, we represent the differential equation in the first segment of motion in the form

fiJI + P;ffJI =jJsin(wt+r),

where

P; = (c-c*)/ J and jJ = Mo / J.

The solution to Eq. (1) can be written out as

ffJI

= CI cos Pit + C2 sin Pit +[jJ /(p: -(2)]sin(wt + r),

where ris the initial phase of the perturbing moment.

(1)

1 Vibrations of Systems with a Single Degree of Freedom

154

The equation of motion in the second segment of motion (see solution to Problem 47) is the following:

= ,u sin(wI + r),

ip2 + P; flJ2

where

P; = (c + c.) / J .

(2)

We now write out the solution to Eq. (2):

flJ2

= C3cos pi + C4 sin pi + [,u /(p; -

( 2)] sin(wl + r),

(3)

In order to determine integration constants, we make use of the following conditions (i) in the case of the system motion in the fIrst segment,

fIJI

=0

and fIJI

as 1= 0, and

=0

and

By the periodicity condition, we fInd at the end of the second segment flJ2

=0

(ii) in the case of the system motion in the second segment,

flJ2

= flJlmax

= flJlmax

=

as I

flJ2

= II .

= =

and flJ2 -flJlmax as I 12 ff / W . Thus, when solving the problem, we deal with the following set of six equations with six unknown quantities

Cp C2 , C3 , C4 , II

and

r:

PIC2 + [,uw/(p; - ( 2)]cosr = 0; flJlmax = CI cos Pili + C2 sin Pltl + [,u /(p; - ( 2)] sin(wtl + r); -PICI sinp/I + PIC2 cos Pili + [,uw/(p; -(2)]COS(W/I + r) =

=-P2 C3sinp2/, + P2C4cos P2t, + [,uw/(p; _(2)]COS(wt, + r); C3cos P2tl + C4 sin P2tl + [,uw /(p; - ( 2)] sin(W/l + r) = 0; CI + [,u/(p; -(2)]sinr = -{C3cos P2ff / w + C4 sinp2ff / w+ +[,u/(p; -(2)]sin(ff+ r)}; - P2C3 sin(p2ff / w) + P2C4 COS(P2ff / w) +[,uw /(p; _(2)]COS(ff +

+r)=O. 77 The exact solution to the problem on steady-state vibrations in the presence of Coulomb friction in the system turns out to be rather complicated (see Problem 76). The simplifIcation is attained under the assumption that even in the presence of Coulomb friction, the motion corresponding to the harmonic law

fIJ

= flJo sin( OX + r)

does occur in the system. In this case, the energy dissipated in one cycle can be determined by the formula

1.3 Forced Vibrations

155

v =4pNRfIlo /2. In this expression, we took into account that the friction force pN is directly proportional to the deviation angle 9' of the system since N = c,fIl (see the solution to Problem 47). In the presence of viscous friction in the system, the dissipated energy is

V,

= a1£OXP~

(see the solution to Problem 64). Equating the energies U and UJ ' we find the coefficient of equivalent viscous friction:

a

= 2JlC, / (1lm).

The amplitude of stimulated vibrations of the system possessing viscous resistance equivalent to Coulomb friction is

fIlo=

fils! / ~r-(1-_-m-2-/-p-~-)2:-+-(-2-n-m-/-p--:-~--:-)2 ,

where fils! = Mo / c is the static deviation of the system under the moment Mo and c is the angular rigidity of the system. Furthermore,

2n = a / J

= 2JlC, / (1lmJ);

p~

= c / J.

The phase shift angle yis determined from the relationship

tanr = 2nm/(p~ _m 2). 78 The differential equation of motion of the system can be written out as

JijJ + aifJ + cfIl + sign rix·ofll = Mo + ~ sin ai.

(1)

The equation of an equivalent linear system has the form

JijJ + a·ifJ + cfIl = Mo + M, sinai,

(2)

where a· is the coefficient of equivalent viscous resistance. This coefficient can be determined from the condition of the equality of the supplied and dissipated energies in one cycle. Assuming the steady-state motion to be harmonic, i.e.,

fIl = fIlo + tA sine ai + r),

where fIlo = Mo / C, and using Eq. (1), we can calculate the energy dissipated in the system. The energy dissipated over a period in the viscous-friction damper is 1£afll,m, whereas the energy dissipated in the Coulomb friction damper can be determined from the dependence of the moment of the Coulomb friction force M f = pNR = JlC,RfIl on the disk twist angle 9'. As can be seen from the expression for MJ' the dissipative work of the friction forces over a half-period in the case of an asymmetric cycle of deviations is

[(Mfmax

-

M fmin )/ 2]2tA

= 2JlC,RfIlofll,.

156

1 Vibrations of Systems with a Single Degree of Freedom

Equating the energies dissipated over a period in the given and equivalent systems, we arrive at whence it follows

a*

= a + (4J1CJR / "OJ )ffJo / fA.

The amplitude of forced vibrations in the presence of viscous friction can be calculated according to the formula ffJJ

where

= M J / c~r-(1-_-OJ-2-/-P-;)-2-+-(-2-n-OJ-/-p-;-)2,

2n = a * / J ;

P; = c / J .

The phase shift angle y, we determine in the same manner as in the case of viscous friction:

tanr = 2nOJ/(p; _OJ2). 79 The differential equation of motion of mass m is written out as

X + p;x = P(t) / m.

(1)

The perturbing force P(t) can be expanded into the Fourier series

pet) = P"o [.2._±(cos2wt + cos4wt + cos6wt + ... 1[ 1[ 1·3 3·5 5·7

)J.

(2)

The solution to Eq. (1) is a sum of the solutions for each of the terms in series (2), i.e., _ 2Po 4Po cos2nwt ( X -1iC - 1[m LoIn=l (2n-l)(2n+l)(P5-4n2m2)" 3

,,00

Po = 2nOJ, i.e., at OJ = Po /2; Po /4; Po /6;

)

The resonance takes place in the case of

The Fourier method is approximate since it fails to allow in the calculations for an infinite number of terms of series (2). In addition, this series is rather cumbersome, which hampers summing a large number of terms in expression (3) for attaining a higher accuracy. We now consider another (exact) method of solving Eq. (1) that describes small vibrations under the action of an arbitrary periodic force P( t) . In the case of steady-state vibrations, the periodicity condition must be fulfilled:

x(t) = x(t + T), x(t) = x(t + T).

(4)

The general solution to Eq. (1) at arbitrary initial conditions is of the form

x = xo cos pt+ ;~ sin Pot + m10 1P(T)sinpo(t-T)d T.

(5)

Conditions (4) must be valid at any arbitrary moment of time including the moment t = 0 . Therefore, we obtain two equations

1.3 Forced Vibrations

Xo Xo

= Xo cos PoT + ;~ sin PoT + m10 , P(t)sinpo(T -

r)d r,

= -xoPo sin PoT +xo cos PoT + ~ , P(t) cos poeT -r)d r,

157

(6)

The set of equations (6) makes it possible to determine the unknown values of Xo and xo: q(I-cos PoT)+bz (sin PoT)/ Po

= 2(I-cospoT) , . _ bz(l-cosPoTHVosinpoT Xo 2(I-cosPoT) ,

Xo

(7)

where

bl

= m10 'P(t)sinpo(T-r)dr,

b2 = ~, P(t)cosPo(T-r)dr,

(7)

Substituting the obtained values of Xo and Xo into general solution (5) makes it possible to determine in a close form (i.e., without expanding the force p(t) into the Fourier series) the law of motion for mass m within the interval 0:::;; t :::;; T, i.e., within the intervals

nT:::;; t :::;; (n + l)T.

This algorithm of solution was proposed by H. Duffing. As follows from expressions (7), they tend to infinity (resonance) as PoT = 2mr, or (because of

T = 7r /

OJ, see Fig. 67)

Po

= 2nOJ. In other words, we have obtained the same

resonance as in the case 0 the expansion of the force P( t) into the Fourier series. The Duffing method can be used in solving equations describing small vibrations, which take into account viscous-resistance forces, and in the case of systems with a finite number of degrees of freedom for which other methods (operational calculus, expanding into the Fourier series) are of minor usefulness.

80 The solution to the problem is similar to that of Problem 79. In the case being analyzed, the perturbing force can be represented in the form of the Fourier series

pet) = P.

o

[1. +.lsin OJt _1.( cos2mt + cos4mt + cos6mt + ... )J. 2 1·3 3·5 5·7 7f

7f

(1)

We can write out the solution to the equation ofload vibrations with respect to the equilibrium position in the form _ Po Posinmt 2Po oo cos2nmt (

x--+ -tfm 7fC 2m(p6-m2 )

I

n=1 (2n-l)(2n+l)(P5-4n2m2)"

The total displacement of the load is

Xo =x+mg/ c.

2 )

1 Vibrations of Systems with a Single Degree of Freedom

158

The algorithm of the solution, which is based on the Duffing method, completely coincides with that used in the solution to Problem 79. However, while and b2 , it is necessary to take into account features of the periodic determining

q

force P{t) (see Fig. 68b), which vanishes within the interval T / Therefore (see the solution to Problem 79),

bl

2~t

• 1 rr· I rr = mpo .lJ P(t)smpo(T-r)dr= mpo.lJ Posmpo(T-r)dr,

~

T.

/2

rr

rr

(3)

b2 = ~.lJ P(t)cosPo(T-r)dr=~.lJ 12 PocosPo(T-r)dr,

After Xo and Xo have been determined, the general solution to the equation describing vibrations of mass m coincides with the solution to Problem 79 [see Eq. (5)]: within the interval 0 ~ t ~ T /2,

! P(r)sinpo(t-r)d r, x(t) = -xoPo sin Pot+xo cos Pot + ~ !P(r) cos po(t-r)d r; x(t) = Xo cos Pot + ;~ sin Pot + m10

and within the interval T /

.

2~t~T ,

(4)

rr /2

x(t) = Xo cos Pot + ;~ sin Pot + m10.lJ P(r) sin Po(t - r)d r,

(5)

x(t) = -xoPo sin Pot + Xo cos Pot + ~.lJrr P(r) cos Po(t - r)d r; /2

81 After the action of the impulse of force and in accordance with the theorem of angular momentum, the velocity of mass m is varies (the displacement Yo of mass

m for the time 8t is zero): . _ Jp

Yo- m ·

(1)

After the impulse of force has acted, free vibrations are described by the equation

ji + P~Y = 0 (p~

= c / m; c = 3EJx /

/3).

The solution to Eq. (2) with allowance for initial conditions

Yo = -',:;

as t

(2)

Yo

= 0,

= 0 is the following: _ Jp

•

Y - mpo sm Pot,

(3)

. _ Jp

Y-mCosPot. 82 The answer to the problem is Po

.

Po·

Y = -mpo 2 (1- COS Pot), Y = mno sm Pot. r'

1.3 Forced Vibrations

159

As follows from the solution obtained, in the case of a sudden application of the constant force P, the flexure y is twice as large as in the case of a static force. As

tk

= :0 '

y(tk) = 2P / mp~ Here,

Yo

= 2yo.

is the static flexure.

83 The force p(t) can be represented in the form (see Fig. 71a)

p(t) = PoH(t) - PoH(t - tl ),

(1)

where H is the Heaviside function. For zero initial conditions, the solution to the equation of stimulated vibrations of mass m is of the form

y

1

= m10 P(t)sinpo(t-r)d r.

(2)

After integration, we arrive at

yet) = {

P02 (I-cos Pot) mpo

for

0 ~ t ~ tl (3)

p,

---\[cos Po(t -tl)] - cos Pot fort ~ t. mpO In the case that the force pet) varies with time as shown in Fig. 71b, p(t) can be represented as a sum

P(t) = Po [H(t) - H(t - tl)] + 2Po[H(t - tl ) - H(t - t2)]

(4)

For the given case of loading, we find after necessary transformations the following solution:

y(t)=

0 ~ t ~ tl

P02 (1- cos Pot), mpo

for

P0 [2-cosPot-cosPo(t-tl )], mpo2

fortl~t~t2

P02 [2cosPo(t-t2)-cosPo(t-tl)-cosPot], mpo 84 For initial conditions

Yo = Yo = 0

we obtain

y= m10 In the case under study with

y

= a(t -

for

(5)

t > t 2.

[see the solution to Problem 83, Eq. (2)],

l P(r)sinpo(t-r)dr.

pet) = at, we have

sin Pot / Po) / (mp~);

y = a(I- cos Pot) / (mp~).

(1)

(2)

160

1 Vibrations of Systems with a Single Degree of Freedom

85 We introduce the impulse of force J p into the equation describing vibrations of mass m: •• 2 _ 1 J S:(t) x+ Poxm pU ,

(1)

where 8(t) is the Dirac delta-function. The general solution to Eq. (1) is

1 +~ 1

x(t) = Xo cos Pot + ;~ sin Pot + m10 x(t) = -XoPo sin Pot +xo cos Pot

J p8(T)sinpo(t- T)d T,

(2)

J p8(T) cos po(t-T)d T;

where Xo and Xo are values of x(t) and x(t) atthe moment of time t = O. By virtue of properties of the delta-function, the integrals entering into solution (2) are

b1(t) = ~;o b2 (t) ="',:;

18 (T)sinpo(t-T)dT= ~;osinpoH(t),

1

8(T) cos po(t - T)d T ="',:; cos PotH(t),

(3)

where H is the Heaviside function. In the case of steady-state vibrations, the conditions

x(t + T) = x(t), x(t + t) = x(t),

must be fulfilled, or

x(T) = x(O) = xo' x(T) = x(O) = xo'

(4)

With allowance for expressions (3) and (4), we obtain from general solution (2) the set of two equations with respect to Xo and xo:

bl (T)

b2 (T)

= (1- cos PoT)xo - ;~ sin poT,

= xoPo sin PoT + xo(1- cos poT).

(5)

Xo and xo: X JpsinPoT . . __ Jp o - 2mPo(1-cosPoT) , Xo - 2m' (6) As a result, we arrive at the functions x(t) and x(t) in the case of steady-state vibrations (i.e., for PoT"* 2mr ) within the intervals nT::;; t ::;; (n + 1)T , where n = 0,1, 2, ... : From the set of equations (5), we find

(7)

1.3 Forced Vibrations

161

86 We write out the equation for vibrations of mass m within the time interval [0,

n

(1)

In the given setting the problem, the expressions for

bl and b2 (see the solu-

tion to Problem 85) are

f )H2 (I - f), b2(/) = ~ cos poIHI(/) - ~ cos Po(/-f)H2(1 -f)·

bl (t) = ~;o sin PolHI(I) - ~;o sin Po (I -

Thus, we can find

(2)

bl (T) and b2 (T) :

bl(T) = ~;o (sinpoT-sinpof), b2(T)

= ~ (cos PoT -cos Po f)· Xo X(/):

Then, we can find finally,

x(/) and

and

Xo

[see the solution to Problem 79, Eq. (7)], and,

Xo smpo/+ • Jp [ . elI • ( T)H2' ] ( ) --xoCOSPo/+ Po xl mpo SmpoII11 -smpo 1-"2

X(/) = -xo sinpo/+Xo cos Pol + ~ [COSPoIHI -cosPo(t-f)H2].

(3)

87 The equation describing small vibrations of mass m can be written out as ••

2

P(t)

x+ PoX =1i/.

(1)

In order to determine the steady-state periodic motion of mass m, we employ the Duffing method (see the solutions to Problems 79 and 80). The general solution to Eq. (1) containing an arbitrary right-hand side has the form

x=xoCOSPo/+ ;~ sin Pol + m10 !p(/)sinpo(/-r)dr, where

Xo

and

Xo

are unknown quantities taken at the initial moment of time

1=0. For the determination of

r r

Xo

and

xo' we should find

periodicity conditions:

bl(T) = m10 b2(T) = ~ x(/):

bl (T) and b2(T) using

P(r)sinpo(T-r)dr=

':;0 !lsinpo(T-r)dr,

P(r)cosPo(T-r)dr=

~

After the quantities tain

(2)

bl (I), b2(I), xo'

and

(3)

!lCOSPo(T-r)dr.

Xo

have been determined, we ob-

1 Vibrations of Systems with a Single Degree of Freedom

162

x(t) = Xo cos Pot + pXo sin Pot + o

(1- cos Pot),

P02

mpo

(0 ~ t ~ t,) x(t) = Xo cos Pot + pXo sin Pot + o

P02

[cos Po (t - t,) - cos Pot].

(4)

mpo (t, ~ t ~ T)

88 We write out the differential equation of motion of mass m as

X+ 2ny + p~x = (1/ m)P(t), where

(1)

2n = a o / m .

The solution to Eq. (1) has the form

x(t) =e- nt [Yo cosmt+ nyo+yo sinm0t+-l-sinm tJ Wo mwo 0' where

mo = ~ p~ - n2

•

(2)

In order to provide the periodic nature of the solution

with a period T, we need to choose initial values of

Yo and Yo using periodicity

conditions:

Yo = yeT);

Yo = yeT).

Furthermore, we differentiate expression (2) and find (3), we determine the desired values of _

Yo -

Yo and

yet) . From condition

yo:

e"T sinwoT . mwo(e 2nT -2enT coswoT+l) '

. _ e"T( coswoT -c%sinwoT )-1

Yo -

(3)

(4)

m(e2nT -2enT coswoT +1)

Substituting Eqs. (4) into formula (2), we arrive at the solution related to the interval 0 ~ t ~ T , which is also valid for consequent intervals.

1.4 Critical States and Vibration Stability 89 We consider an elementary volume of liquid bounded by the tube cross sections with the coordinates x and X + dx (Fig. 234). The mass of liquid inside this volume is liquid.)

dm, = pFodx . (Here, m, = pFo ·l

is the mass of the unit column of

1.4 Critical States and Vibration Stability

163

dm,g,

to the

The volume under consideration is subject to the gravity force tangential component of the inertial force

dm/px,

and to the Coriolis force

2dm, voifJ . The moment of the forces with respect to the point of suspension is dM or

M

=

= -(gcp + ipx + 2voifJ)x d m,

! dM = -(m/gcpI2+m,ipI3 13 +vom/ifJ).

We now write out the moments of inertial forces for the tube mass and of the gravity force:

M,

= -(moip/2 13 + mogcpl 12).

In accordance with the d' Alembert principle,

M+M2 =0, or .. 3m,vo· 3 g - 0 cp+ "'o+rflJ.l Cp+'2ycp - .

Thus, we find that the eigenfrequency of the tube filled with moving liquid is

P = ~3g 1(2/) - {3m,vo 1[2(mo + m,/)]}2. The motion is aperiodic if v>

v.

Fig. 234.

v., where

= [(mo + m,/) I m, ]~2g 1(3/).

164

1 Vibrations of Systems with a Single Degree of Freedom

90 The deformed state of the vibratory system is illustrated in Fig. 235. Bending the tube is accompanied by the appearance in liquid of centrifugal inertial forces. Their magnitude per unit tube length is

q=m,v2 / Pc' The differential equation of the arched tube axis is of the form dd:2

where

Jx

(EJxY") = q,

(1)

= Jrd 3£5 / 8 is the axial moment of inertia for the tube cross section.

Since in the chosen coordinate system zOy (see Fig. 235) the second derivative of the flexure is negative, we have

q = m,v2 / Pc = -m,v 2 d2 y /d 2 z. With allowance for this expression, Eq. (1) takes the form

EJxY" +m,viy" =0. We represent the solution to Eq. (2) as y = CFosk + C 2sink where

It

= ~r-m..-V2-/-E-'J-x •

(2)

+ C;z + C4 ,

We determine the integration constants from the boundary conditions: (i) y =0 as z =0;

MB =EJxY" =0 asz=O; (iii) y' = 0 as Z =I; (iv) Q = -EJxy'" = -myo /2 as z =I; (ii)

The last condition represents the expression for the cutting force Q acting in the rod cross section in the case of z = l.

11

Fig. 235.

1.4 Critical States and Vibration Stability

165

Based on these boundary conditions, we can write out the equation for the arched tube axis as

y = [(myo)(2A?EJJ](sink / cosk -

k).

It follows from this expression that for z = I, the equation of motion of the load has the form

Yo + 2EJxJ.} cosAl[m(sinAl + Al cosAl}] = o.

The vibration frequency of the load depends on the fluid flow velocity in the tube:

P = ~2EJxJ.} cosAl / [m(sinAl + Al cosAl)] . (3) If the fluid flow velocity v =0, then expression (3) yields the well-known formula for the determination of the eigenfrequency

P = ~6EJx /(mI3). The vibration frequency is zero for cosAl = 0, i.e., for Al = 7r / 2. Hence, the critical fluid flow velocity is

v. = J,.,~rEJ-x-/-~-=-(-1C-/2-1)'~EJ=x=/~=. 91 We consider an elementary volume of fluid flowing in a radial orifice at a velocity Vo (Fig. 236). In the process of small vibrations, the volume is subject to both the additional inertial force Coriolis force

dmlrip

of the translational motion and the

2dmlvoip . The moment of these forces with respect to the rotation

axis is

dM = -dmlr 2rp - 2dmlVo ipr , where ~ is the fluid mass per unit length of the orifice channel, tion of the disk angular velocity from the stationary value The total moment of inertia of fluid is

M = -n

e

ip

is the devia-

OJ.

(r2¢+ 2voipr)ml dr = -nml[¢(r; -r)/3 +voip(r22 -'12)].

I

In addition, as the disk vibrates, it is subject to both the moment of inertial forces J ip and the moment of elastic forces crp from the shaft. With allowance for all the forces acting on the disk, the differential equation describing its small vibrations is written out as

.. + nvoml (ri- r?)/3 • + c =0 rp J+nml(ri-r?)/3 rp J+nml(ri-r?)/3 rp .

This makes it possible to find the natural frequency of disk small vibrations with due regard to the fluid flow, as well as the velocity

PI =0):

v.

(from the condition

166

1 Vibrations of Systems with a Single Degree of Freedom

Fig. 236.

v• =

4c[J+nml(ri-r?)/3] n"'I(ri-r?)

92 The solution to this problem differs from that to Problem 91 only by determination of the cutting force. In the presence of the spring, we may consider

Q = -EJxY'" = -(myo +cYo)/ 2,

Therefore, the expression for the critical velocity of the fluid flow has the form

tan vI = (vlc-2EJxv2)/ c.

93 We consider the deviated state of the disk (Fig. 237a). In this position, the disk is subject to the elastic restoring force

F: = -c~, where c is the bending rigidity of the shaft ( c Since

= 6EJx / 13 ).

1; = r - e , the elastic restoring force is

F: = -c(r -

e).

In this case, the differential equation of motion can be written out in the vector form as

1.4 Critical States and Vibration Stability

167

y

a Fig. 237.

mr + c(r - e) = o.

(1)

where m is the disk mass. We now project Eq. (1) onto the immobile coordinate axes x and y. As a result, we obtain

mX + cx = cecosax,}

(2)

my+cy = cesinax.

In the case of the steady-state motion, the solution to the set of equations (2) is

= [ep~ / (p~ - (i)2)]cosax; y = [ep~ / (p~ - (i)2)]sinax,

X

(3)

.J m.

where Po = c / It follows from Eqs. (3) that the vibration amplitude unboundedly increases as (i).

-+ Po

=.Jc / m.

< (i). , the deviation of the disk center of mass from the rotation 1j + e. When (i) > (i). , the displacement of the disk center of mass occurs

In the case

(i)

axis is towards the rotation center. In other words, the disk center of mass resides at the distance 1j - e from the rotation axis (see Fig. 237b). As follows from expressions (3), in the case of rather high rotation velocities ( (i) -+ 00), self-centering of the disk occurs. (i.e., the displacement r tends to zero). In the case under consideration, the disk mass is

m = trf)2 hp / 4 = 0.625 kg. The shaft bending rigidity is

c = 6EJx / [3 The eigenfrequency is

Po

= 2.35

kN/m.

=.Jc / m= 61.4 s-'.

168

1 Vibrations of Systems with a Single Degree of Freedom

The deviation of the mass from the equilibrium position [see Eq. (3)] is

1) = r - e =

~X2 + l- e = eoi / (p~ -

oi),

or

1)

=1

-ll)

J1Po2

=1.6

rom.

We now determine the force that deflects the rod by 1.6 rom: P=Co=c1) =3.77 N. Using

the

formula

(j

= Ms / w"

w" = trd3 /32), we have as a final result,

(j

(where

Ms

= PI/ 2,

and

= 38.4 MPa.

94

95 In the deflected position (Fig. 238), the rotating shaft is subject to the inertial distributed load with the intensity

q = fYlooiy, where

fYlo is the mass of the shaft unit length,

OJ

is the shaft angular velocity, and

y is the shaft flexure. We set up the differential equation of the arched shaft axis in the coordinate system rotating with the shaft:

EJxY IV

= q = moOJ 2 y.

The corresponding characteristic equation is of the form

.t -e = (A? -e)(A? + e)= o.

where

k4

=

fYloOJ 2 / (EJx )'

t--.....,~..-

---.....

"-~J

• Fig. 238.

We now write out the solution to Eq. (1) as

y = C1 sin kz + C2 cos kz + C3 sh kz + C4 ch kz,

(1)

1.4 Critical States and Vibration Stability

169

In the case of the pivoted support of the shaft ends, the boundary conditions are y = y" = 0 as z =0,

y

= y" = 0 as z =l.

In order to determine the integration constants, we make use of the set of four uniform equations:

C; sinkl + C2 coskl + C3 shkl + C4 chkl = 0;

C2 +C4 =0;

- eC2 + k2 C4 = 0; - C1k 2sinkl- C2 cosk! + C3k2 shkl + C4 k 2chkl = O.

e

Equating to zero the determinant of this set, we find (sink!) x (sinhkl) = O. The least nonzero root of this equation is kl = ;r, i.e., the critical angular velocity of the shaft rotation is OJ.

= (;r2 / 12)~EJx /1rlo'

96 In the case of bending the shaft, the equation of the elastic line in the coordinate system rotating with the shaft has the form

y = C1 sinkz + C2 coskz + C3 shkz + C4 chkz,

where

k4

= mo0J 2 / EJx •

For the cantilever restraint (see Fig. 8la), we deal with the following boundary conditions y = y' = 0 as z = 0,

y'" = y" = 0 as z =I. This allows us to find the equation of the form (coskl) x (coshkl) = --1. (l ) which determines the critical angular velocity. The solution to Eq. (1) can be simply obtained using the plot. To do this, it is sufficient to find the coordinates (kl)i' (i = 1, 2, ... ) for the intersection of the

left-hand side of Eq. (1) with the straight line" = -1 (Fig. 239). The least root of this equation is (kl)i ~ ity is OJ.

1.875, and the critical angular veloc-

= (3.52/ 12 )~ EJx / 1rlo .

In the case that the shaft is fixed in a manner shown in Fig. 81b, we have, correspondingly,

1 Vibrations of Systems with a Single Degree of Freedom

170

1to-_... ..;-

-I

\ \ \

,'--"

/

I

I

Fig. 239.

y y

= y' = 0 as z = 0, = y' = 0 as z = t.

This allows us to obtain the following equation for the determination of the critical angular velocity:

cos kl . sh kl = 1. The least nonzero root of this equation is (kl)] = 4.9, and the critical value of the angular velocity is

97 In the state of the tube, which is deviated from the straightforward equilibrium position, each unit length of the tube is subject to two inertial forces. The first one

(mo + ~ )w 2Y is caused by the rotation of the tube containing liquid. The second force, ~ v2P = -~ v2 y" , is the centrifugal inertial force of moving liquid. This force is associated with curving the shaft (see Fig. 240). (The sign minus is determined by the sign of curvature.) The differential equation of the arched shaft axis can be represented in the form

E'Jxy IV

= ( mo + ~ )OJ2 Y -

mo v 2y" ,

or where

k]2

= ~ v2 /

(EJ x ), k;

= (mo + ~ )w 2 / (EJJ.

We can write out the characteristic equation corresponding to Eq. (1) as

(1)

1.4 Critical States and Vibration Stability

171

11

Fig. 240.

whence it follows

~,2 = ±i~(kJ2 + ~ kJ4 + 4ki) / 2; ~,4 = ±~(~k: +4ki -kJ2)/ 2. The solution to Eq. (1), we obtain in the form

y

= CJ sin~z+C2 COS~Z+C3 Sh~Z+C4 ChA4Z.

The integration constants are determined according to the conditions y = y" = 0 as z =0,

y = y" = 0 as Z =l.

The critical angular velocity can be found from the equation

(sinA) x (sinhA}) The left-hand side of Eq. (2) vanishes in the cases of ~

=

at OJ 0 . For velocity OJ*

k = 1, i.e.,

= ~ 1[4EJx /

~I

= be,

(2)

~

= 0, i.e.,

= 1[ , we obtain the value of the critical angular

[l4(mo + ~)]_1[2~ v2 / W(mo + ~)],

98 In accordance with the d' Alembert principle, the equation of motion of lever 2 can be obtained as a sum of the moments (with respect to the point A) of all forces acting on the lever:

Jip+ miplJ2 /2 + clJ2rp/2 -mJoil~rp =O.

1 Vibrations of Systems with a Single Degree of Freedom

172

Here, J ift is the moment of the inertial force of the vibrating lever; the moment of the rod inertial force;

m/#12

is

cll2CfJ2 is the moment of the spring compres-

sion force, ~oil;CfJ is the moment of an additional inertial centrifugal force arising as a result of the deviation of the vibrating lever through the angle rp, and

OJ=~RI/ R. Thus, the rod eigenfrequency is

,....----P = elll2-m1w2ti J A +mzIll2 .

In the case of

oi »OJ: = cll2 / (2~/;),

small vibrations of masses

ml

become unstable so that usual operations of the senSor of the velocity regulator are violated. Thereby, the critical value of the angular velocity (i). is the upper limit for current angular velocities of the regulator.

99 In the case of small vibrations of the loads

~,

their horizontal displacements

I!i. r are related to the vertical shift I!i. y of the sleeve of mass 111z by the expression

l!i.y = [/1/ (/1 + 12)](tga + tgfi)l!i.r = kl!i.r.

In order to derive the differential equation for the loads, we make use of the Lagrange equation of the second kind: -lLo~ = dq oq

oT +m =0 . oq oq

(1)

The kinetic and potential energies of the loads can be written out as

T-

ml~2 2

+

2 ml~y2

2k2 +

2 ml(r+~ylk)2aw2 . 2

'

II = 111zg(yo + l!i.y) + c(Yo + l!i.y)2 /2., where Yo is the spring compression corresponding to the steady-state operation mode of the regulator. Taking in Eq. (1) the displacement l!i.y as the generalized coordinate q, we arrive at the equation

(m2 + 2k; )l!i.y+(c- 2i~2 )l!i.y + m2g + CYo - 2nyw2 steady-state regime,

= O.Since in the

I!i. y = I!i.Y= 0, we have

111zg + cYo - 2~roi / k = O. Hence, we can reduce the differential equation for small vibrations of the regulator to the form

I!i.Y + p 2l!i.y = 0, where p is the system eigenfrequency,

1.4 Critical States and Vibration Stability

p=~(ec-2m.ol)/(k2~ +2m.).

173

(2)

It follows from expression (2) that there exists a certain critical angular velocity of the regulator, OJ.

= ec /(2m.),

for which the eigenfrequency vanishes.

100

I _ ~g/4c-D •-

2

+

(m2g)2 _ 8c

~gD

8c .

101 In the case of small vibrations of the rod with respect to the position of dynamic equilibrium, the rod is subject to the moment the gravity force

Jip

of the inertial force, to

Q= mg , and to the centrifugal inertial force directed along the

perpendicular to the rotation axis and equal to O.5m1OJ 2 sin(a + tp). (Here, the angle

a

determines the position of dynamic eqUilibrium, whereas tp is the small

angular deviation from this position.) We derive the equation of small rod vibrations, in the coordinate system rotating at angular velocity OJ. The rod can rotate with respect to the axis passing through the hinge. For the position of the rod, which is shown in Fig. 85, this axis is normal to the picture plane and passes through the point O. Equating to zero the sum of the moments of all the forces acting on the rod and of the rod moment of inertia with respect to the axis, we arrive at the equation

-Jip+ ~l OJ2 sin(a+ tp)fl cos(a +tp)-Qtsin(a+tp) = O. Taking into account that J o = ml 2 /3 and assuming the angle tp to be small, we obtain

iP+
(1)

Two last terms entering into this equation correspond to the position of dynamic equilibrium for which tp 0 , i.e., the sum of these terms is zero:

=iP =

1.5sina / I-O.5OJ 2 sin 2a = O. Hence, the eqUilibrium position turns out to be possible for two values of the angle a: (2)

As

a l = 0, it immediately follows from Eq. (1) that vibrations are stable if OJ2

< 1.5g II.

The critical angular velocity for which the equilibrium state is unstable corresponds to OJ.

= ~1.5 g / I.

174

1 Vibrations of Systems with a Single Degree of Freedom

Substituting

a2

into Eq. (1), we obtain that in the case of

oi > 3g /(2/). any arbitrary angle of deviation corresponds to the stable equilibrium state.

102

103 In the case of small vibrations, the kinetic energy of the load is

T = 0.5m.:e + 0.5ma/ x 2 , where x corresponds to the displacement of the load, which is measured from the rotation axis. The potential energy of the spring deformation is

II =O.5c[(Oo + X)2 + (00 - X)2] =c(O; + x 2). write out the Lagrangian function L = T - II and substitute it into

We now the Lagrange equation. Thus, we arrive at the differential equation

x+(2c-moi)x/ m =O.

Hence, the eigenfrequency and the critical angular velocity are, respectively,

p

=.J2c / m - oi

and

(j).

= .J2c / m.

104 The differential equation of relative motion of mass m has the form

x+(c-m{j)2)x/ m = 0,

where x is the deviation of the mass m from the position of dynamic equilibrium. The vibration frequency and critical angular velocity are, respectively,

p

= .Jc / m _{j)2

and

(j).

=.Jc / m.

105 In the process of small vibrations, the plate is subject to the lifting force Y, to the moment Me = ca produced by the elastic rod, and to the inertial force. Since the plate surface area is s

y where p is the air density and

[c =GJp /(2/) = ;rd4 G /(64/)]. The

J

= (bh

moment 2 PI

of

the

= b x h, we can write out the lifting force as = ;rpv2bha, C

plate

is the angular stiffness of the system inertial

forces

is

J Ii ,

where

/ 3)(h + 3d / 2) is the moment of inertia of the plate with respect to

the rotation axis (PI is the density of the plate material). Using the d' Alembert principle, we set up the differential equation of motion as a sum of moments of all the forces with respect to the rotation axis:

1.4 Critical States and Vibration Stability

175

Jii+ca-Y(3h/3+d 12) = 0, or

ii+[cl l-trpv2bhh. 1(2J)]a = 0, where

h. =3hI4+dI2. Hence, the natural frequency of plate vibrations is

p = ~c I J -trpv2bhh. 1(2J). It is easy to see that there exists a certain critical velocity

v. = ~2c I(trpbhh.), for which the vibration frequency is zero. In this case, the system loses its stability (divergence of the plate occurs). 106 The differential equation describing small free vibrations of the lever can be written out as

J ip + crp = PaRrp, or

ip+(c- PaR)rp = O.

The frequency of natural vibrations and the critical value of the force are, respectively,

p

= ~(c- PaR) I J.

and

P. =cl R.

107 The attraction force acting on mass m and developed by the magnet is

F =k
(x« al ) (see Fig. 241), a force arises equal to the difference between the at-

traction forces in each of the positions:

M' = F; - F;

= k
X)2 - k
where

F; =k
mX + 21'ox I 1 = 4k
x+[21'o I(ml) -4k
176

1 Vibrations of Systems with a Single Degree of Freedom

Fig. 241.

In the case of small vibrations, the natural frequency of mass m in the magnetic field is

p

= ~21'0 /(ml) -4k~ /(mai).

A peculiar feature of the obtained expression for the natural frequency is its dependence on the magnetic-flux intensity <1>0' This dependence is equivalent to an effect of the so-called negative elasticity. we can put in correspondence to the critical Each value of the tension

1'0,

value of the magnetomotive force

<1>. = ~1'oai /(2kl), for which the frequency p vanishes. In this case, the equilibrium state of mass m is unstable, i.e., its vibrations are impossible. 108

p=~bh3E/(4/3m)-4k~/(mai); <1>. =~bh3Eai /(16kI 3). 109 Based on the force method, we write out the differential equation of motion for point mass m as

y Here,

0"

=13 /48EJ

point of mass m;

0'2

= 0Il (-my) + 0'2Ro·

is the beam flexure caused by a unit force at the fixation is the beam flexure at the fixation point of mass m and

caused by a unit force applied to the beam at a distance

z from

the origin of the

1.5 Parametric Vibrations

177

coordinate system, or (which is the same in accordance with the reciprocity theorem for displacements) the beam flexure caused by a unit force applied at a distance z from the fixation point of mass m. We take the following approximate expression for 012 :

012

=0ll sin ~z

•

Since z = vt , we finally obtain the following differential equation describing vibrations of mass m: (1) y + p~y = (Ro / m)sin:rvt II. For uniform initial conditions, yeO) = yeO) =0, the solution to Eq. (1) is of

the form

~

y

[ 110 '] ( sin ")" - f sin P6-(~) Po

m

Pot).

(2)

Analysis of the denominator in expression (2) shows that a certain critical vecan exist, which is found from the condition locity of motion of the force

Ro

p~ - (:rv. /1)2 = 0, whence it follows

v.

= Ipo /:r = (l / :r)~48EJ /(l3 m).

However, for this value of the velocity v, there is an uncertainty in Eq. (2). After this uncertainty has been revealed (as :rv / 1 ~ pression

y

Po), we arrive at the ex-

= -[Ro /(2mpo) ](t cos Pot - sin Pot / Po)'

As far as the time of motion of the force Ro along the beam is finite, the displacement of mass m is also finite, i.e., there exists no critical velocity.

1.5 Parametric Vibrations 110 In order to derive the differential equation of motion, we make use of the d' Alembert method. To do this, we project all the forces acting on the load onto the immobile xqy axes (Fig. 242). Thus, we arrive at the equations

mX = -N sin cp,

(1)

178

1 Vibrations of Systems with a Single Degree of Freedom

y

Fig. 242.

my = N cos rp - mg.

(2)

Excluding from Eq. (1) tension force N, we obtain

m(x cosrp + ysin rp)

= -mgsin rp.

(3)

We now express the coordinates of the load position in terms of both the angle Yl of the point of suspension: rp of the rod deflection from the vertical and the displacement

Y

= I-I cos rp + yp

X

= I sin rp.

(4)

Thus, we have in this case,

y

= I(rpsinrp+i/ cosrp)- Y ow2 sin ax,

x = I(rpcosrp -

ip2 sin rp).

(5)

Substituting expressions (5) into Eq. (3) and assuming angle rp to be small, i.e.,

cos rp ~ 1 , sin rp ~ rp , and rp2 ~ 0, we obtain rp + [g /1- (YoW2 / I)sin ax]rp = O.

(6)

Equation (6) is usually represented in the form of the Mathieu equation

d 2 rp /d r2 + (a + 2q cos 2r)rp = 0, where a = 4 g / (0/ I); 2q = 4Yo / I ; and r = ax / 2 . Depending on particular system parameters, small vibrations of the pendulum can be either stable or unstable. Stable vibrations correspond to the position of the

1.5 Parametric Vibrations

179

representation point in the unshaded region of the Ince--Strutt diagram (see Appendix 1). If the parameters a and q are such that this point lies within the shaded region, then the vibrations tum out to be unstable. In the case under consideration, a = 0.784 and q = 0.04, which implies the position of the representation point in the stability region of the Ince--Strutt diagram (see Appendix 1).

111 The solution to the problem is similar to the case of Problem 110. However, in contrast to this case, the coefficient a in the Mathieu equation is negative. Thus, the pendulum motion is stable for OJ ~ 396 S-I (provided that the condition

lal < q2 /2

is valid).

112 lUmin ~

125

S-I.

113 The differential equation describing small free vibrations of the plate is similar to that obtained while solving Problem 110. We now ignore the drag resistance since it is rather weak in the case of small deflections of the plate. We also consider the flow velocity in the above solution to vary with time and neglect the terms containing v~. Thus, we arrive at the differential equation of motion in the form

d 2 cp /d r2 + (a + 2qcos2r)cp = 0, where

_.-£( cP _

a-

0)2

J

6) ,q 2 __ 4 3ll" bl 2pVov, J -1/3 bJ. - 0)2 2 J ' - 3 ijJ .

3ll" bl 2PV 4 J

For the parameters indicated while formulating the problem, and in the case of

a = 1.53, the coefficients of the Mathieu equation are

2q =--0.132; Iql =0.066 for VI =5 mls; 2q =--0.264; Iql =0.132 for VI = 10 mls. It follows from the Ince-Strutt diagram (see Appendix 1) that in the case of the flow velocity that varies within the indicated limits, the vibrations are stable.

114 In order to derive the differential equation of motion, we can employ the approximate formula for determining rod flexures in the case of transverselongitudinal bending:

y = Yo / (1- P /

Po),

where Yo = -my / c is the flexure of a rod under the action of an only transverse load. In the case under study, this is the inertial force of the load of mass m. (Here, c is the bending stiffness of the rod and

P

0=

1[2 EJx /4/2 is the critical

value of the compressing force.) Thus, the differential equation of motion can be represented in the form

y + (1- P / Po) ~ = 0, or

180

1 Vibrations of Systems with a Single Degree of Freedom

y + (a + 2qcos2r)y = 0, where

a

=....L( 6EJ

x

(j)2

mil

24Po ).

1[2Im'

2

q

=_....L 24P1o

(j)2 1[2 m '

On substituting numerical values, we find a = 6.63 and Iql = 0.195. In the Ince-Strott diagram (see Appendix 1), the representation point having these coordinates corresponds to stable motion.

115 In the case under investigation, the beam stiffness corresponding to transversely bending is c = 24EJx / 13 , and the critical force is ~ = ;r2 EJx / 12. The coefficients of the Mathieu equation are determined by the following expressions:

a=(I- Po)~. P mOJ2' e

2q=-~~ Pe m(j)2 •

On substituting numerical data, we find a =4.2 and Iql = 0.02. In the InceStrott diagram, the representation point having these coordinates lies in the unstable region (see Appendix 1).

116 The differential equation of motion for mass m can be written out in the form

y + 2(To + 1; sin at)y / (ml) = 0,

or

d2 y/d 2 r+(a+2qcos2r)y=0, where a -

2q --

4 2To -

(j)2

rnr -

4 2·20 -512' 25 2 0.25·0.2 - . ,

2·10 mr - 2542 0.25·0.2 -

4 2TI -

(j)2

2 56' .

,

r=at/2. The motion of mass m is stable because the representation point with the coordinates a =5.12 and Iql = 1.28 lies within the unshaded region of the Ince-Strott diagram (see Appendix 1).

117 The differential equation of motion for the load, we can write out as

y

= -t\Imy,

where t\I = 13 / (3EJx ) = (10 -II sin at)2 / (3EJ x )'

We restrict our analysis by only the linear part in the expansion of the function 1 / t\ I in series with respect to powers of II' Thus, we arrive at the differential equation

1.5 Parametric Vibrations

Y.. + 3EJ mlij

x

181

(1 + 3 10II'sm lOt ) Y = 0.

It can be represented in the form of the Mathieu equation

d 2 y/d 1"2 + (a+ 2q cos 21")y = 0, where a_...±...3EJx • - m2 mlij ,

2 _...±...9EJx i.

q- m2

mlij 10'

lOt =1'+21". On substituting numerical values of system parameters into the expressions for the coefficients a and 2q, we find a 2.5 and q 0.75. The solution to the Mathieu equation can be either stable or unstable depending on the relation between the coefficients a and q. In the case under discussion, the representation point lies within the unshaded region of the Ince-Strutt diagram (see Appendix 1). In other words, for the parameters chosen, the motion of the load is stable.

=

=

118 At an arbitrary moment of time, the inflection point of the line shifts from the middle point by the distance YI' and the line is deflected from the equilibrium position through the angle rp (Fig. 243). Using the theorem for variation of the angular momentum, we can write out the differential equation of motion in the form

Jt [m(lo + Yl)2 cp] = -mg(lo + Yl )rp,

(1)

whence it follows

(2) In the case of application of the d' Alembert method, we analyze the balance of forces acting on the load at an arbitrary moment of time. These forces are the line tension force N, the gravity force mg, and the inertial forces mX and my . Furthermore, we project the forces acting on the load onto the vertical and horizontal axes. As a result, we have

my = N cos rp - mg,

(3)

mX = -N sin rp.

(4)

The coordinates x and y of the load can be expressed in terms of the angle rp and the length 10:

X = (10 + YI) sin rp; Y = (10 + Yl )(1- cos rp). On substituting these relationships into Eqs. (3) and (4) and excluding the tension force, we also arrive at Eq. (2).

182

1 Vibrations of Systems with a Single Degree of Freedom

Fig. 243.

119 The differential equation of motion of the pendulum has the form

where

I

= 10 + II sin oX .

rn + KI""m = 0,

ih + 1. dl .." I dt""

120 The variation in the position of mass m causes a change in the moment of inertia of the system. In this case, for deriving the equation of motion, it is convenient to use the theorem for variation of angular momentum: (1)

where J o is the total moment of inertia of the system with respect to the point of suspension 0; rp is the angular deviation of the system from the eqUilibrium position;

MOi are the moments with respect to the point of suspension 0 for all exter-

nal forces acting on the system. The equation of motion for mass m with respect to the frame has the form

x = Xo sin OJI.

The moment of inertia of the system is

J o = J I + m(lo + Xo sin OJI)2.

(2)

In the case of the deflection of the system through angle rp, the total moment of gravity forces with respect to the point of suspension 0 is

,LMOi = -mlg/ICP-mg(lo +xo sinOJt)cp.

(3)

On substituting expressions (2) and (3) into Eq. (1), we arrive at the differential equation of motion of the system in the form

1.5 Parametric Vibrations

183

[J1 + m(lo +xo sin wt)2]qj + [2mxow(l0 +xo sinwt) cos wt]ifJ +

+g[ml/l + m(lo +xo sinwt)]qJ = O. 121 The solution is similar to that of Problem 120. The differential equation of the pendulum motion can be written out as .. 2XO coswt. g 0

qJ + lo+xosinwt WqJ + lo+xosinwt qJ =

,

where rp is the angle of deviation of the pendulum from the vertical.

122 In the case of deviation of the tub by the distance x from the equilibrium position and rotation of the pulley through angle rp, the variation in the rope length is (Fig. 244)

M = RqJ + ~Ig1 + x 2 -/01' or, in the case of small vibrations ( X

<< 10l ), III = RqJ.

Hence, the change in the rope tension is

IlT = MEF 110

= REFqJI 10,

Then, the differential equation of motion for the tub can be rewritten as

mX = -(1'0 + IlT) sin a,

where a is the angle of deviation of the rope from the vertical in the course of tub vibrations. For small vibrations, we may write out

sin a

=a =xl 10l'

(1)

With allowance for expression (1), we can represent the differential equation of motion in the form

x+(a+2qcos2r)x = o.

(2)

We now numerically calculate the parameters entering into expression (2):

a = 41'0 l(mw2/ol) = 4.10 2 /(500.10 2 ·0.7) = 0.114; 2q = 4REFqJo l(mw2Iolol) = 4·0.1· 200.109 .10 4 ·0.1/(500x X 102

·1·0.7) = 2.28; q = 1.14.

As follows from the Ince-Strott diagram, (see Appendix 1), the small vibration of the tub are unstable.

1 Vibrations of Systems with a Single Degree of Freedom

184

Fig. 244.

123 The differential equation of motion of mass m in a magnetic field has the same form as the equation in the solution to Problem 108:

m.X + (c - k2 / a;)x = 0, where

c = 3EJx I [3 and =

<1>0

(1)

+ <1» sin wI.

With due regard to small amplitude of the magnetomotive force <1», Eq. (1) takes the following form:

.. (c

kij x+ - - - - 2k0
m mar

mar

'

or

d 2 xl d r2 +(a+ 2qcos2r)x = 0, where

k2 a =4- ( .£. _ _ _ 0) 0)2

m mar

and

4 2k 0) 2q = __ 0)

Plotting the point with the coordinates

mar

.

a =1.03,

Iql =0.02

in the Ince-

Strutt diagram, we find that this point lies in the unshaded region, and therefore, the motion of mass m is stable. 124 In the course of motion of the bus, its distance a I from the conductor varies as a) =ao +x. The force acting on the bus in the process of vibration is

1.5 Parametric Vibrations

where

185

Po = 2pl/2ol / ao;

F2

= 2pIl(I20+I21 sinmt) ~ 2pIII (I 1 . ) (1- L) l1o+x a 20+ 21 smmt a· o

0

In this case, the differential equation of motion is of the form

mi+2cx=M, or .. 2c X+m

[1

pIli (1

cao

--2

1 smmt . )] X-_ 2pIII21/ . am smmt.

20+ 21

0

125 The differential equation that describes vibrations of the lever (see Fig. 90) has the form .. ( c RoR RIR. ) 0

rp+ J--:r-T smmt rp= .

Introducing the notation mt = 1r / 2 + 2, , we obtain after certain transformations the coefficients of the Mathieu equation _ 4 c-PoR 2 _ 4 llR a- m2-J-' q-- m2 T . On calculating numerical values of the coefficients a conclude that the motion is stable.

=6.0 and q =0.4, we may

126 At an arbitrary moment of time, the position of the load is determined by the coordinate

Xo = Xoo +vt. In this case, the differential equation describing the vibrations is of the form (see the solution to Problem 28) .. +To I =0

Y

m (l-xoo-vt)(Xoo+vt)

Y

.

127 Let at an arbitrary moment of time, the point of the pendulum suspension be shifted from the equilibrium position by a certain distance Yo, and the pendulum be deflected through angle rp from the vertical (Fig. 245). Then, the differential equation of motion for mass m has the form m· miI =-Nsin 'f"

(1)

myI =-Ncosrp+mg.

(2)

We now substitute into these equations the expressions for the coordinates

= SI + II sin rp; YI = Yst + Yo +/1 cosrp, XI

(3)

1 Vibrations of Systems with a Single Degree of Freedom

186

Fig. 245.

(l 1SIS2)To(yo + Yst) = mg - m[y -II (q;sintp+ qi costp)]' and exclude the force N. As a result, we find

11q; -

Yo sin tp = -g sin tp,

Assuming also the angle rp to be small, we arrive at the equation

(4) As follows from the equilibrium condition for the node of the pendulum point of suspension,

N costp = T(yo + Yst ) 1SI + To (Yo + Yst)1 S2'

y

Expressing N in terms of on the basis of relationship (2) with due regard to formula (3), we arrive at the equation Next, taking into account that

IToYst l(sls2) = mg , and ignoring the nonlin-

ear terms in the right-hand side of the equation (because of the smallness of angle rp), we can rewrite the last expression in the form

Yo = -ToYoll(msls2),

or (5)

where

p2

=Tol l(msls2 ).

Solving Eq. (5) with allowance for initial conditions as

(Yo = Yoo

and

Yo = 0

t = 0), we obtain the equation of motion for the pendulum point of suspension: Yo = Yoo cos pt.

1.5 Parametric Vibrations

187

We now substitute this expression into Eq. (4):

t

iP + I (g + Yop2 cos pt)rp = O.

Performing the replacement

pt = 2r, we arrive at the Mathieu equation

d 2 rp/d r2 + (a + 2q cos 2r)rp = 0, where

a =4q /(p2/1) and 2q =4yoo / II'

Furthermore, we make use of numerical values of related parameters and find

a = 0.0687 and q = 0.665, which correspond to stable motion.

In the second case under consideration, unboundedly growing angular vibrations, of course, cannot exist because the system is conservative. Here, we imply that the unstable motion corresponds to the instability of vertical vibrations (the tendency for vibrations to grow with angle rp). For the exact determination of angle rp, the vibrations should be analyzed with nonlinear terms taken into account.

where

p2

= 1'01/(ms l s2).

Solving Eq. (5) with allowance for initial conditions as

(Yo = Yoo and Yo = 0

t = 0), we obtain the equation of motion for the pendulum point of suspension: Yo = Yoo cos pt. We now substitute this expression into Eq. (4):

iP+t(g+ yop2 cospt)rp = O. I

pt = 2r, we arrive at the Mathieu equation d 2 rp/d 1'2 + (a+ 2qcos21')rp = 0,

Performing the replacement

where

a = 4q /(p2/1) and 2q = 4yoo / II'

Furthermore, we make use of numerical values of related parameters and find

a = 0.0687 and q = 0.665, which correspond to stable motion.

In the second case under consideration, unboundedly growing angular vibrations, of course, cannot exist because the system is conservative. Here, we imply that the unstable motion corresponds to the instability of vertical vibrations (the tendency for vibrations to grow with angle rp). For the exact determination of angle rp, the vibrations should be analyzed with nonlinear terms taken into account. 128 The solution is analogous to that obtained for Problem 127. In the first case ( a = 0.0981, and q = 0.5), the pendulum motion is unstable, whereas in the second case ( a = 0.0981 , and

q = 0.07 ), the motion is stable.

1 Vibrations of Systems with a Single Degree of Freedom

188

1.6 Nonlinear Vibrations 129 The kinetic energies of the upper and lower disks are given by the equations

T=2l[m 8D rp + m '2] j

j

2

j

' 2

jX j

(1)

,

2 ( ·2 . 2)] '

(2)

. T 2-- 1 [m2D22 8 rp+m 2 x 2+Y2

2

where XI is the horizontal coordinate of the upper disk, x 2 and Y2 are the horizontal and vertical coordinates of the lower disk, and ({J is the rotation angle of the system. The potential energy of the system is

II= m 2 gy 2 Taking the rotation angle X

j

({J as

(3)

•

a generalized coordinate, we have

=O.5rpDj , X 2 = 1 sinrp-0.5rpDj , Y 2 = 1(I-cosrp)

hence,

(4)

x

Xj =0.5ipD j , 2=(1 cosrp-0.5Dj)ip, Y2 =1 sinrpip .

(5)

Substituting Eqs. (1)-(5) into the Lagrange equation of the second kind, we arrive at the equation of motion

m2 g1 sinrp

..

rp+ (3m jD j2/8)+m2[ (D22 /8)+(D j2/4)+/2 -lD jcosrp]

0

.

130

J

2 12" cl 2cosrpcosa( --;:======= 1 -m a+----''---1 =0 , 3 sinrp ~1-sinasin2rp ~I +sinasin2rp p

=~3c/m sinrp .

131

a+ 3g sin a + 3c (.J 1+ sin a - ,.h - sin a) =0 , 41

p

4m

=~(3g /4)[c/(mg )+1/1].

132 There are four time intervals in a vibration period: two intervals in which one of the springs contacts the wall and two intervals of free motion of the load. When the load is being subjected to the elastic force of one spring, its motion is described by the equation

1.6 Nonlinear Vibrations

x =(V 0/Po)sinPot,

189

(1)

where Po2 = elm and Vo is the impact velocity of the body (spring). Therefore, the contact duration is t, = nlpo' the duration of free motion of the body is t2 = 2L11vo' and the vibration period is

T =2(t\ +t2)=2(1Z'/Po +2L1/v 0) .

(2)

We now express the period in terms of the amplitude Xo of free vibrations. The smallest length of the compressed spring is evidently equal to Xo - ,1. According to Eq. (1), Xo -,1 = v/Po or

Vo=Po(Xo-L1) .

(3)

Substituting Eq. (3) into Eq. (2), we find

T =(2/Po)[1Z'+2L1/(xo-L1)] . Finally, the frequency of free vibration is

P =21Z'/T=1Z'Po/[1Z'+2L1/(x o-L1)] . 133

134 When the load is being subjected to the elastic force of the left spring, its motion is described by the equation .. 2 0, x\+p\x\=

(1)

where p,2 = elm. Under the initial condition X1 (0) = --V 10 at the equilibrium point x,(O) =0, the solution to Eq. (1) has the form

X1=-(VIO/Pl)Sin(P1t) . The contact duration t, is equal to the half-period: t, =nip,. Similarly, the duration of contact with the right spring is t2 = nlp2' The period and frequency of steadystate vibrations are, respectively,

T =t\ +t2 = ~P\ + P2)/(P\P2) , P =21Z'/T =2~C\C2/ml(..fc: +.fc:). 135 The stiffness e of a system (spring) is defined as the ratio e =Pix, where P is a force applied to the spring and x is the spring contraction due to the force. The frequency of small free vibrations of a load with mass m is p =(elm)'l2. For the ratio elm to be constant, either the load P must be constant or the spring characteristic must be nonlinear. In the latter case, the spring stiffness is defined as e =dPldx. It follows from the constant frequency condition that

190

1 Vibrations of Systems with a Single Degree of Freedom

P 2 =(g Ip)( dP Idx )=const Therefore,

or

= exp[ (p 2/g )(X +C I)] The constant C} is found from the condition P =Po at x =xo: P

CI

=(g/p2)logPo -x o .

Finally, the contraction of the shock absorber as a function of the force applied to it takes the form

136 The kinetic and potential energies of the system are

T

=mi 2/2, II=c(00+ot/2.

Here, c~ = To and c8 = T} are an initial tension and a tension caused by vibrations, respectively. The line elongation 8 is related to the vertical displacement x by the equation 8 =2/(1 - cosa) ~ ill, where a ~ xII. Therefore, the strain energy of the system is

II=C(00+x 2/lf/2. Substituting these expressions into the Lagrange equation, we arrive at the equatioll of motion

i' +(2cOo/m )[ xl / +x 3/(001 2)] = 0 .

In order to find the ftrst integral of this differential equation, we take into account that

Then,

i2=AI_}4cOox (l+~Jdx = A ml 0 01 o AI -(2cOo/ml )[X 2 -Ag +(X 4 _A 04)/2001] . The solution satisfying the initial conditions has the form

i = ~(2cOo/ml)[ Xg-x 2 +(x ~ -X 4)/2001] . This expression can be presented as a curve on the phase plane. 137 The tension in the springs is (see Fig. 246)

1.6 Nonlinear Vibrations

191

Fig. 246.

(1)

where x is the horizontal displacement of the load, Finding the horizontal component of the force acting on the load, we arrive at the equation of motion:

mi" =-'lI'x /10 =-cX 3/1 02 , or X' +kx 3 =0 ,

(2)

where k =c/(m102), We introduce the notation dx/dt =y, then

.. dy dy dx , dy dy x=--=----=x--=y-dt dxdt dx dx' Substituting this expression into Eq, (2), we find

y dy =-kx 3dx ,

(3)

=

=

The integration of Eq, (3) under the initial conditions x(O) Xo and dx(O)/dt 0 yields l =k(X04 - X4)/2, Solving this equation with regard to the definition y =dx/dt, we find

where & =xlxo is a dimensionless displacement. Because of the symmetry of the problem, the vibration period is

192

1 Vibrations of Systems with a Single Degree of Freedom

T

=-±- {2J

=-±- {2K(vv'2)

de

xoV-ko~l-e4 xoVk ..fi

Here, K(k) is the complete elliptic integral of the first kind, which is a limiting case of the elliptic integral of the first kind: K(k) = F(7d2,k), where

F(q.>,k)=

. f ~(l-t2)( cosdta+sin at2) ,k =sma .

1

2

cosql

2

In the case under consideration, rp =n/2 and a =n14. According to the tables of the elliptic integrals, F(n12, 1I21n) = 1.8541 (Janke E.et al., 1960). Therefore, the period and the vibration frequency are, respectively,

T =1.8541(4/0/xo)~m/c , p =O.85(xo/lo)~c/m . 138 According to the Rayleigh method, the maximum kinetic energy of a conservative system must be equal to its maximum potential energy: Tmax = il.n.., where

T max = mv ; /2, IImax = Since T

Lll max

f2T d( ill) .

o

=cL11 (see Problem 137) and L11 ~ il(2Io)' then IImax =cx;/(4I

Therefore,

mv;/2=cx;/(4I

n,

n.

xo=~V2m/c

.

Taking into account the relation between the frequency of free vibrations and the initial displacement (see Problem 137), we find

p

=(0.85/lo)~V2c/m

139 The differential equation of motion is similar to that derived in Problem 137:

mi" +2Tox /10 +CX 3/1~ =0

Introducing the dimensionless variable form

&

=xlxo'

.

we rewrite this equation in the (1)

e

Substituting the relation "& = de / de into Eq. (1), we find the solution satisfying the initial conditions:

e2=[2cj{mlo)][{To/c)(I-e2)+(x;/4/0)(I-e4)]. Equation (2) describes the motion of the mass m on the phase plane (Fig. 247).

(2)

1.6 Nonlinear Vibrations

193

140 Consider the motion of the load with x > 0 and dx/dt < 0 (see Fig. 246). Since the Coulomb friction force is always directed oppositely to the velocity of motion, the differential equation of vibrations has the form (see Problem 139)

mi" +2Tox /lo+cX 3/{ml;)=F or (1)

=

where & x/xo is the dimensionless coordinate. With regard to the relation 'j; = &de / dE and the initial conditions &(0) = &0 and d&(O)/dt = 0, we find from Eq. (1)

. _+ t,; E-_",2

_

F(Eo-E) To{E~-E2) cxg{E~-E4) + + 2' mxo mlo 4mlo

Substituting the numerical values, we have

&=±Ji~-100(Eo-E)+200{E~-E2)+50{E~-E4)

.

(2)

=1 and &< O. Ac& =- 17.3 lis at the point & =O. At the moment of time t =T/2

We now consider the motion in the first half-period when

&0

cording to Eq. (2), when &=0, the displacement is &f':! - 0.63. Equation (2) with &0 f':! - 0.63 can also be used to consider the motion in the second half-period, but in this case &> 0 and the sign of the friction force should be changed. As a result, we find that &f':! 6.8 lis at the point &= 0 and &f':! 0.18 at t = T when = 0 (Fig. 248). The subsequent trajectory on the phase plane can be found in much the same way as described above. The body stops moving when the elastic force F at an end position becomes less than (or equal to) the friction force. =0). As follows from Eq. (1), this takes place already at the point & f':! 0.18

e

(e

ii,1ls

i, 11s

22.3

-1

Fig. 247.

Fig. 248.

194

1 Vibrations of Systems with a Single Degree of Freedom

141 The differential equation of motion has the form

or

mx•• +CX ++C1X 3 = 0

(1)

•• 2 3 0, X+pX+f.lX=

(2)

where 11 is a small parameter. We seek for the solution to Eq. (2) in the form 2

X =XO+f.lXl+P x 2+... ,

(3)

with

(4) where PI' a" and P2 are constants to be found. Substituting sums (3) and (4) into Eq. (2) and neglecting terms above the fIrst power of 11 , we find

2 0 + P(X.. 1 + PIX 2 1 + a1x 0 + X03) =0 X..0 + PIX

.

(5)

Equation (5) must be valid for arbitrary 11, hence, ••

2

XO+PIXO=

0

,

(6)

X"1+PI2X 1=-(a1Xo+xg) .

(7)

Under the initial conditions xo(O) =A and dxo(O)/dt =0, the solution to Eq. (6) has the form Xo =A cos(P,t). Substituting this solution into the right hand side of Eq. (7), we arrive at the equation X"I

+ PI2X 1 =-(aIA +3A 3 /4)cOS(Pl )-(A 3 /4)cos(3Pl) .

(8)

We seek for a bounded solution to Eq. (8), hence, a, =-3A2/4. Otherwise, the perturbation frequency would be equal to the eigenfrequency of the system so that x, ~ 00. Solution to Eq. (8) satisfying the initial conditions x,(O) =dx,(O)ldt = 0 has the form (9)

Therefore, in the first approximation, the solution to Eq. (1) can be written out as

X =A COSpl + P[A 3/(32pn](COS3P1t -cosp1t) .

=p2 + 3J.IA2/4 ~ 1.17 lis. 142 According to the method of least squares, the function F =cx + c/ can be approximated by a linear function F =coX provided that the coeffIcient Co is found

The eigenfrequency is found from Eqs. (4) and (9): p,2

from the minimum condition for the integral A

2

ncoX -(CX +C1X 3)] dx

o

1.6 Nonlinear Vibrations

195

considered as a function of co. Carrying out the integration and then differentiating the result with respect to CO' we find

Co =c +3c I A

2/5 .

Therefore, the frequency of free vibrations of the load is

P

=~clm =~(5c +3c I A 2)/5m

143 Assuming the motion to be harmonic, we seek for the solution to the equation

X- + P 2X +,ur 3 =0 in the form (see Problem 141) X0

=A COs(p1t +a)

(1)

.

(2)

According to the Galerkin method, the function xit) is an approximate solution to Eq. (1) provided that 211:/p]

(3)

J(X- o+p2xo +,urg)x odt=0.

o

Substituting solution (2) into condition (3), we find p]2 = l + 0.75pA2. This result coincides with that obtained by the Lyapunov-Poincare method in Problem 141.

144 In the case under consideration, the characteristic of the elastic elements can be presented in the form

=c (x - .,1), for .,1:::; x : :; A . To find a linear approximation F] =coX to this nonlinear dependence, we evaluF

= 0,

for 0:::; X

:::; .,1 ,

F

ate the root-mean-square deviation

J

LI

A

.,10

A-L1L1

=~ fcgx 2dx +-l-nc(x -L1)-c ox

r

dx .

Using the condition dJ/dco =0, we find

Co =c(2A 2 -2AL1-d)/(2A

2

+2AL1+5d) .

The vibration frequency is given by the expression

p

=~C(2A 2-2AL1-d)/m(2A 2+2AL1+5d)

Vibrations are possible if 2A 2 - 2AA- A2 = 0, i.e., A > 1.36 A.

145 The equation of free vibrations has the form (see Problem 137)

mi· +2Tx II =0 ,

where T = cAl + To is the spring tension, with AI;::: x2/(2/o). Hence, •• 2 3 0 x+px+,ur = ,

1 Vibrations of Systems with a Single Degree of Freedom

196

where l find

=2T/(mlo) and J-l =c/(m102). Solving this equation (see Problem 141), we

For the numerical values given, Xo

~

91.2 mm.

146 The force acting upon the load is

F

=klP~/(al-x)2 -klP~/(al +X)2 ~4(klP~/an(x /al +2x 3/an.

The differential equation of motion has the form

mi"+cx =F, ori"+p2x +J.lX 3 =O, where (1)

The fundamental frequency of free vibrations is found by the method of small parameter (see Problem 141):

/4 .

Pl2 = P 2 +3J.IX g

Substituting Eq. (l) into Eq. (2), we find the critical value xO' (PI

X g. =(c -4klP~/ an( 6klP~/ an or &2

=2(I-a)/(3a) ,

(2)

=0): (3)

(4)

where a = 4kr/Jo2/(ca I 3 ) and Ii= xoJa l • Dependence l( a) is shown in Fig. 249. The shaded domain with Ii < 1 is the stability region, with the curve corresponding to the critical values of Ii and a .

\

a Fig. 249.

1.6 Nonlinear Vibrations

197

For a ~ 0.4, & = 1. In this case, the mass strikes against the magnet poles, and the solution given above is inapplicable.

147 The differential equation of motion of bus B has the form (see Problem 124)

mi"+2cx -2pl/2[Z/aO-Z/(ao+x )]=0.

Expanding lI(ao + x) in powers of x and introducing the dimensionless variable = x/xo' we arrive at the equation

XI

.£1

+ P 2X I + pllx 12 - p;flx 13 =0 ,

(1)

where 111 =x/ao' P=2J1lJp(mao2) and l =2c/m - p. We seek for the periodic solution xl(t) to Eq. (1) and the vibration frequency P in the form 2

XI =X 10 + PIX II + PIX 12 , PI =P + 2

2

C

IPI

+

C

(2)

2 2PI'

where x lO' XII' X 12 and CI' C 2 are, respectively, functions and constants to be found. Substituting sums (2) into Eq. (1) and neglecting terms above the second power of 111, we arrive at the system of equations ••

2

0

XIO +PI XIO= , X•• 11 + P I2 X11 =C IX 10 - flx210'

(3)

(4) (5)

Under the initial conditions xlO(O) = 1 and dxlO(O)/dt =0, Eq. (3) has the solution =COS(Plt) (if l > 0). No secular terms can enter into the solution to Eq. (4); therefore, CI =O. Under the initial conditions xlI(O) = dxll(O)ldt = 0, the solution to Eq. (4) is

XIO

X 11

P PI

=- - 2 +

PCOspl pcos2pl psmplt 2 + 2 + 2' 3PI 6PI 3PI

Similarly, substituting xlO(t) and xlI(t) into the right hand side of Eq. (5), we solve Eq. (5). The factor mUltiplying COS(PIt) in the solution xlit) must be zero; hence,

C 2 =5p2/(4pn- 3P/4 . The fundamental frequency of free vibrations is found from the second equation of (2) with C l = 0:

PI2=P 2 +(X o/aot[5p2/(4pn- 3P/4] . Equation (6) should be solved for P1 2 • Taking into account that P1 we can take P1 2 =p2 in the right hand side of the equation. Then,

PI2= pg -P+(x o/aoY {5p2j[4(pg -P)]-3P/4} ,

(6) 2

=l

+ CzJ1\

198

1 Vibrations of Systems with a Single Degree of Freedom

where P02 =elm.

148 The solution to this problem differs in sign of lem 147.

P from

that found in Prob-

149 The equation of motion coincides with Eq. (1) in Problem 147, with p, =xrm./ao serving as a small parameter. The solution satisfying the initial conditions x(O) =0 and dx(O)/dt =Vo is given by Xo =(v,jp)sin(p,t). Therefore, xOmax =v,jp" i.e., p, =v/arP,. The fundamental frequency of free vibrations of the bus is found in much the same way as it is done in Problem 147: Pl2

= p2 +(vo/aop )2[5P2/{5p2)_3P/4] ,

where

150 We use the following approximation to the equation of motion:

iP+grp/l-grp3/(61)=O.

(1)

Taking p =g/(6l) as a small parameter, we obtain the fundamental frequency accurate to the first order in p: Pl2

=(g /1){1-rp~/8) .

(2)

Using the linearization method, we rewrite Eq. (1) in the form

iP+erp=O,

where the constant c is to be found from the minimum condition for the weighted root-mean-square deviation of the nonlinear characteristics from a linear one, i.e.,

d de

IDf( grp grp3 J]2 ll-l -61- erp rp drp=O.

Performing the calculations, we find

-

e

= Pl2 = p;{1-5rp~/42) ,

(3)

where Po2=gll. The difference of expressions (2) and (3) is

L1 = p;{ 1-5rp~/42)- p;{1-rp~/8) = O.006p;rp~

.

151 The differential equation of motion of the lever has the form (see Problem 106)

iP+ P2rp+ f1rp3 =0 ,

where p2 =(c - FR)/J and p =FR/(6l). The fundamental frequency found by the method of small parameter is given by the expression Pl2

=P2 +(3/4){mo/Pl)2(FR/6J) .

(1)

1.6 Nonlinear Vibrations

199

According to the Galerkin method, 2tr/ p,

f( ijJl + P 2tpl + ,utpi)tpldt =0 ,

o

where f/J, = f/Josin(Pl + a) and f/Jo = (j)JPI" Performing the integration, we find

PI2 =P 2 +(3/4)(01 0 / Plt(FR/6J) . Therefore, frequencies (1) and (2) found by the two methods are identical.

152 The differential equation of small free vibrations of the lever has the form

JijJ+(mglo+c )tp+sign(~)k~2 =0 . Here, the function

sign( ~)

(1)

implies that the sign of the moment

Mf =k~2

coincides with that of the angular velocity ~. Equation (1) can be written out as

ijJ+p2tp+asign(~)~2 =0, where l =(mglo + c)/J and a =kll. In order to find the vibration frequency as a function of the initial deflection rp(0), we solve this equation under the initial conditions rp(0) = a and dqidt = =0. In the first half-period,

sign( ~) =- 1 and

the equation of motion takes the

form

ijJ+ P 2tp_a~2 = 0 .

(2)

We seek for the solution to Eq. (2) in the form

22 2 2C tp=tpo+atpl+atpl+···'p =PI+ la+ C2a2 +....

(3)

Substituting (3) into Eq. (2) and neglecting the terms of higher powers of a than the second one, we obtain •• 2 0 tpO+PltpO= ,

(4)

C

= tpl - ItpO' ijJ2 + PI2tp2 = 2~1~0 -C Itpl -C 2tpO . ••

2

tpl + PI tpl

• 2

With regard to the initial conditions, we find from the first equation

tpo = a cos(Pl)

.

Substituting this solution into the second equation of (4) and setting C, secular terms to be zero), we find

tpl

=a2/2-2a2cos(Plt)j3-a2cos2(Pl)j6 .

(5)

=0 (for the (6)

In turn, the substitution of Eqs. (5) and (6) into the right hand side of the third equation of (4) yields

200

1 Vibrations of Systems with a Single Degree of Freedom

(P2 + PI2rp2 =-c 2aCOsPIt -(2/3)a 3PI2(sinpl +sin2Pl t )

.

This equation can be written out in the form

9'1, +PI'("

=( p~a3 -C,a)cOSPl +p~a3 (2cos2Plt-cos 3Pl) .

For the secular term to be zero, the factor multiplying COS(Plt) must be zero, i.e., C2 = p 12a2/3. The frequency offree vibrations is found from Eq. (3):

PI=p/~1+a2a2/3 . Taking (jJ = (jJo + a(jJp we find the angular amplitude at t = nlpl:

rp(Il/PI)=al =-(a-4a 2a/3) . Similarly, the equation of motion in the second half period has the form

(p+p 2rp+ai{i=0. Solving this equation, we find the frequency of free vibrations and the angular amplitude at the end of the second half-period:

p2=p/~1+a2a\2/3, a2=aJ-4a~/3. The motion in the subsequent half-periods is found in much the same way.

153 The equation of motion of the mass m has the form

mi' +CX +CIX 3 =Posinmt .

(1)

In accordance with the Galerkin method, we substitute

X ~XI =xosinmt into Eq. (1) and require that the integral 27rlm

I =

f( mi' l +CX I +C IX 13 -

o

Po sinmt ~ Jdt

be equal to zero. Carrying out the integration, we find

(3/4)c l x g+(c -m(2)x 0 -Po =0 .

(2)

It is convenient to solve Eq. (2) graphically. The point of intersection of the plots 171 =(3/4)C 1xo3 and 172 =Po - (c - moi)xo is Xo ~ 2.27 cm (Fig. 250).

154 The roots of the equation

(3/4)c\x g+(c -m(2)x 0 -Po =0

(1)

depend on the frequency of the perturbing force (see Eq. (2) in Problem 153). Two real roots of this equation will be equal to each other if D =0, where D = l + /, q = - 2Pr/3cp P = 4(c - mal)/(9c), Therefore,

1.6 Nonlinear Vibrations

t/

al. =c/m +{9cJ4m ){2Po/3c 1

3

201

•

For the numerical values given in Problem 153, we find m. = 16.6 1/s, with q = - 13.33 and p = - 5.63. In order to determine the amplitudes of the roots of Eq. (1), they should be written out in trigonometric form:

(XO)1 =-2rcos(cp/3), (X O)2,3 =2rcos[(Jl'±cp)/3] . 17, N

2

1 Fig. 250.

6

4

-2 ~

Fig. 251.

22.27 3

202

1 Vibrations of Systems with a Single Degree of Freedom

Since r =sign(q) IPltl2 =- 2.38 and cos(tp) =q/r3 = 1, then (xo)t =4.76 cm and (XO)2.3 = - 2.38 cm. The same result can be obtained graphically (see Fig. 251). As the frequency (j) increases, the straight line 112 =Po - (c - m{j)2)xo rotates counterclockwise, with at (j) = (j). it being tangent to the curve 11ixo)'

155 The differential equation of motion derived in Problem 153 can be rewritten in the form

mX' +maix =(mm 2-c)x -clx 3 +Posinmt .

(1)

In accordance with the Duffing method, we take the function XI

=xosinmt

(2)

as a fIrst approximation. Substituting xt(t) into the right hand side of Eq. (1), we arrive at the equation in the second approximation:

mi' 2 +mm 2 x 2 =[( mm 2 -C )x 0 -(3/4)c lx g+Po]sinmt +

(1/4)c xgsin3mt . l

For the secular term to be zero, it is necessary that

(mm2 -c )x 0 -(3/4)c lx g+ Po =0 . This equation coincides with the expression determining the forced-vibration amplitude found by the Galerkin method. If condition (2) is satisfIed, the second approximation is found from the differential equation

mi' 2 +mm 2 x 2 = (I/4)c 1xgsinmt

;

hence, X2

=A sinmt +B cosmt -[cx g/(32m 2m )]Sin3mt

The constants of integration, A and B, are determined from the initial conditions. If the friction is absent, the phase shift between the perturbing force and the displacement should be equal to zero. In this case, the displacement of the mass attains its maximum at the same time as the force does. Hence, the moment of time t =T/4 can be taken as a zero time when x 2 =Xo and dx/dt =O. It follows from these conditions, that A = Xo - Ctxo3/(32{j)2m) and B = O. Therefore, the approximate solution to the equation of forced vibrations has the form X

=xosinmt-[c1xg/(32m 2m )](sinmt-Sin3mt)

According to Eq. (2), Xo

=2.27 cm;

hence, x

= 2.27 sin(10t) -

. O.036[sin(10t)-

sin(30t)].

156 The differential equation of motion has the form

mi' +ai +cx +c1x 3 =Posin(mt+p) ,

(1)

1.6 Nonlinear Vibrations

203

17, N 4~-""'---o-

2

xocm 1.65 2 Fig. 252.

where f3 is the phase shift between the displacement x and the force F(t). Substituting the approximate solution X

=X 0 sin( mt )

(2)

into Eq. (1) and neglecting the terms with sin(3wt), we obtain

(-mm 2x 0 +CX 0 +0.75c IX g- Po cosjJ)sinmt +

(3)

(amx 0 - Po sinjJ)cosmt = O. Since the functions sine wt) and cos( wt) are linearly independent, the coefficients ofEq. (3) must be equal to zero:

-mm 2x 0 +CX 0 +0.75c l x g-PocosjJ= 0,

(4)

amx 0 - Po sin/3= O. The amplitude Xo and phase shift f3 are found from Eqs. (4). Eliminating fJ,

arrive at the equation

[(c -m(2)x 0 +O.75c l x

gf +(amx or =P0

2 •

we (5)

Figure 252 illustrates the graphical method of solving Eq. (5), with

llJ =[(c-mm 2 )x o+0.75c l xgf +(amx o)2, 172 =P02. The point of intersection is at Xo

~

1.65 cm, with f3 =arcsin( awxr!Po)

~

54°40'.

157 If the pendulum experiences vibrations about the equilibrium position inclined to the vertical, then the equation of its motion has the form

J(p+aip+mgl sin
(.0- ip)

in powers of

ip

in the vicinity of the point

204

1 Vibrations of Systems with a Single Degree of Freedom

and allowing for only the linear term, we arrive at the equation

JijJ+[a+M'(n)]¢J+mg1qJ=M(n) . ¢J = ijJ =0 , its deflection angle is qJo=M(n)/(mgl) .

Hence, if the pendulum is at rest, i.e.,

The equation of motion in the variable If! = cP - CPo takes the form

JV/+[a+M'(n)]~+mgllfl=O . For a + M'(il) > 0, the system is dissipative, i.e., its vibrations are damped. For a + M'(il) < 0, the system is self-exciting; in this case, vibrations increasing with time are possible. In order to analyze the actual behavior of the system, we should allow for nonlinear expansion terms in

M (n- ¢J) . As a result,

JV/+[a+M'(n)]~+mgllfl=M"(n)~2 /2-M"'(n)~3 For M"(il)

/6.

=0, this equation can be written out in the form JV/-bl~+b2~3 +mgllfl=O ,

where bl =- a - M'(il) and b2 =M' " (il)/6. Finally, introducing x = lfIPib/bYI2 and [) =Pl, with Po =(mgl/J)112 and J.i =b/(Jpo)' we arrive at the Rayleigh equation

£+x

=,u(1-x2)x, where x =dx/d8.

(1)

158 The nonlinear equation of angular vibrations of the brake block has the form

JijJ+cqJ-M(n-¢J)=o, where J is the moment of inertia of the brake block with respect to the shaft axis. Allowing for only the linear term in the expansion of M

(n- ¢J), we arrive at

the equation

JijJ+cqJ+a¢J=M(n) ,

(1)

where a =dM/dQ. Introducing CPI = cP - CPo' where CPo is the rotation angle in steady-state regime, with cCPo =M(il), we have

JijJl +CqJI +a¢JI =0 .

(2)

As follows from Eq. (2), angular vibrations of the brake block are stable if a> 0, i.e., if Inl ~ In.1 (Fig. 253).

1.6 Nonlinear Vibrations

205

MinI

n. n

n n.

•'ig.253.

159 As is shown in Problem 157, in the case of friction forces depending nonlinearly on the velocity, the equation of motion of the pendulum is reduced to the Rayleigh equation

2 = J.lXPo . (I -X.2/Po2) ' X. + PoX

(1)

where p= - [a+ M'(il)]/(Jpo). We seek for the solution in the form x =a cos(Pot + where a and r are indefinite functions slowly varying with time, i.e., their derivatives with respect to time is of the same order of magnitude as the small parameter p. The corresponding velocity dxldt is

n,

i

=acos(pot+r)-£¥7o sin(pot+r)-aysin(pot+r) .

There are two indefinite functions, a and y, in the solution; hence, the condition

acos(pot +r )-aysin(pot +r) = 0

(2)

should be satisfied. In this case,

i

=-apo sin(Pof +r) ,

.£ = -cipo sin(Pot +r )-apg cos(Pof +r )-aWocos(pot +r)

(3)

.

(4)

The substitution of expressions (3) and (4) into Eq. (1) yields

-cipo sin(Pot +r )-aWocos(pot +r) =

(5)

=-,uapg[ l-a 2 sin 2 (P ot +r )]sin(pof +r). The unknown functions daldt and d}1'dt are found from Eqs. (2) and (5):

y = ,LgJo[ l-a 2 sin 2 (Pot +r )]sin(Pof +r )cos(Pot +r), a = a,LgJo[ l-a 2 sin 2 (Pot + r) ]sin 2 (P ot + r).

(6)

206

1 Vibrations of Systems with a Single Degree of Freedom

According to the Van der Pol method, functions (6) can be substituted by their values averaged over a period, with y taken as a parameter rather than a function of time: 21f/Po

21f/Po

r =J!JL Jr( T)dT, a= Po Ja( T)dT . 27r

27r

0

(7)

0

It is convenient to introduce the new independent variable B = Po r: + 1. then, dB= Pod r: + (d}1dr:) dr:. Since, dtfdt is of the same order of magnitude as the parameter )1, we take dB:::: pod r: . In this case,

r= f.1Po J(1-a sin B)sinBcosBdB. 27r 21f

2

2

0

Performing the integration, we obtain

a= (f.1Poa/2)( 1- 3a /4) = f.1PolP( a) /2, 2

r=f.1Polf/(a)/27r =O.

(8)

Equation (8) allows us to determine the time dependence of the amplitude a. As is known from the general theory of nonlinear vibrations, the system has limiting cycles provided that the right hand side of Eq. (8) is equal to zero. The roots of the equation da/dt = 0 are a l = 0 and a 2 = 2/3 1(2. They correspond to the equilibrium position and the periodic motion of the pendulum, respectively. We now analyze the stability of these states of the pendulum. Let

a = a;

+ All ,

(9)

where a, are roots of the equation l1X..a) = 0 and L1a is a deflection of the pendulum from its eqUilibrium position or from its periodic motion. The substitution of (9) into Eq. (8) yields

We allow for only the linear expansion term in the right hand side of this equation:

dLla/dt =(f.1Po/ 2)(diP/da; )L1a Solving the equation, we find

Aa =c

exp(f.1PO diP t) 2 dai

.

(10)

.

Therefore, for )1 > 0, the periodic motion (or the eqUilibrium position) of the pendulum is stable if d O. In the case under consideration, )1 = O.oI5 (see Problem 157) and

1.6 Nonlinear Vibrations

207

tP(a) =a(1-3a /4) , dl/JI da=(1-9a /4) . 2

2

For a =a, =0, dlUda, = 1 > 0, i.e., the equilibrium position is unstable. For a =~ =2/3'12, dlUda 2 =-1 < O. Therefore, the limiting cycle of vibrations is stable. In order words, after the equilibrium of the pendulum has been disturbed, the amplitude of angular vibrations is slowly varying with time t and tends to the limiting value

'If =(az/PO)~bl/b2 :::::0.224 rad. 00

160 Since the initial deflection angle 1jI(0) =0.1 rad is less than the limiting value '1/00 =0.224, the vibration amplitude tends to this limit, increasing monotonically. Figure 254 shows the pendulum motion on the phase plane ('1/, drp/dt). The limiting cycle (the steady periodic motion) is shown by solid line. In the case of 1jI(0) > '1/00, the vibration amplitude decreases with increasing time (dashed line on the outside of the limiting cycle). 161 The solution found in Problem 159 has the form x =a cos(Pot + y), where drtdt =O. Therefore, the frequency of the stable limiting cycle is Po' 162 This problem is solved in much the same way as Problem 159. As a result,

tP(a)=(a/8)(4-a 2 ), 'If(a) =0 . The first root a, =0 of the equation f/X...a) =0 corresponds to the unstable equilibrium of the pendulum, the second root a2 =2 is the limiting-cycle amplitude. Differentiating the function f/X...a) with respect to a (see Problem 159), we find

" rad/s

Fig. 254.

208

1 Vibrations of Systems with a Single Degree of Freedom

(dlP/da)la=a2 =(4/S)-(3a;/S)=-1<0. Therefore, the limiting cycle is stable.

163 In steady-state vibration regime, the drag force balances the rope tension:

cx o=c xSpvg/2,

(1)

where Xo is the rope elongation. The differential equation of small longitudinal vibrations of the glider has the form

mX"+c(xo+x)=cxS,o(vo-xt/2,

(2)

where x is the coordinate relative to the equilibrium position. There is the minus sign in the right hand side of the equation since the glider speed decreases with increasing x. With regard to equality (1), Eq. (2) can be written out as

X" + pgx = -,u(2v oX

-x 2) =Fr

where P02 = elm and f.1 = cpS/(2m). It is more accurate to write the force F, as

Fr

=- ,u(2v oX - X Ix I)

because the drag force must be an odd function of dx/dt. We seek for the solution to Eq. (2) in the form x = a cos(Pot + rp) = a cos(B). The Van der Pol method yields the following equation for a(t) (see Problem 159):

d

d; = - :n-

2Jr

f(2v oa-a 2Polsin~)sin2 mB ,

or daldt = rJX...a), where rJX...a) = - (aj.ilpo)[2v o'IT - (8/3)ap o]· The roots a = 0 and a2 = 3vo'IT/(4po) of the equation rJX...a) = 0 correspond to the equilibrium state and the limiting cycle with amplitude a2 , respectively. The derivatives of rJX...a) at the points a and a2 are j

j

dlP(a )/da-2v on-IPo <0, dlP(a2)/da=~ on-/(4po»0. l

Therefore, the equilibrium state of the glider, i.e., its steady flight, is stable.

2 Vibrations of Systems with Several Degrees of Freedom

2.1 Free Vibrations 164. Using the d' Alembert principle, we arrive at the equations of motion

mli"1 +C(X I-x 2) = 0, m 2i"2 +C(X 2 -X 1)=0. We seek for the solution to these equations in the form XI

=A sinpt, x 2=B sinpt ,

where the coefficients A and B are to be found from the system of equations

(c -mIP2)A -cB =0, -cA +(c -m2p2)B =0 .

As A and B are not zero, the determinant of this system must be equal to zero:

c-mlP

2

-c Finally, we find

PI =0, P2

=~(ml +m 2)c /(m lm2) .

165 We take the rotation angle (jJ of the cylinder and the deflection angle .9 of the pendulum as generalized coordinates. The kinetic energy of the cylinder is

TI =(Jqi+Mr2qi)/2, where J =Mr212. The velocity of the mass m is the vector sum of the velocity ld.9ldt perpendicular to the rod and the horizontal velocity rdqidt of the cylinder. The kinetic energy of the mass m is

T2 = m(12a 2+r2qi +218r~cos8)1 2 . The total potential energy depends only on the vertical displacement of the mass m:

210

2 Vibrations of Systems with Several Degrees of Freedom

1I=mg/(1-cos8) . Substituting the Lagrange function L

=T, + T2 -

d(bt/ aj;)/dt-bt/

l1into the Lagrange equation

dii

=0 ,

we obtain

(3Mr2/ 2+mr 2)ijJ+mlr"./)cos8-mlr9 2sin 8 =0,

(1)

mlrijJcos 8 + ml 2"./) + mgl sin 8 = O. In the case of small vibrations, we can neglect the nonlinear terms in (1) and take sin f) ~ f) and cos f) ~ 1. Then, Eqs. (1) take the form

(3M / 2+m )r2ijJ+mlr"./) =0,

(2)

mlrijJ+ m1 2"./) + mgl8 = O.

To determine the eigenfrequencies, we set ({J =A sin(pt) and f) = B sin(pt). Substituting these expressions into Eqs. (2) and equating the determinant of the homogeneous system of the equations to zero, we find

(I.5M +m)g ~ 32·10 ) = ::::::4.6 105M +m I-ml 3.0·0.5

PI =0, P2 = (

1/s.

166. The differential equations of motion of the disk are derived by the force method (Timoshenko S. and Young D.H, 1945): X

= 0 11 (-mi·)+012 (-JijJ) ,

qJ

= t512 ( -mi· )+022(-JijJ).

(1)

Here, o,j are disk displacements caused by unit forces in the directions of virtual displacements. The quantities 0.2 were found in Problem 47:

(Sin 2a + cos 2aJ 0 = 1CI1D (cos2a + sin 2aJ 4 EJx GJ p '22 EJx GJ P '

o = 1CI1D 11

3

2

(1 1J

1CI1D • °12=021=--sm2a --+-- . 4 EJ x GJ p Substituting these expressions into Eqs. (1), we seek for the solution in the form 0. =Ai sin(pt). As a result, we find

PI2

,

2

=

8(EJx +GJp ) (_ 1+ nnD3m(1+3cos22a)

2.1 Free Vibrations

211

10

8 6 4 L -____I~--~I----~I ~

15

30

b

45° 2a

Fig. 255.

Fig. 256.

For Glp = 0.8Elx and P02 = El/(nrcD3m), the frequencies PI and P2 as functions of a are shown in Fig. 255. 167 Let us consider the position of the bodies at an arbitrary moment of time (Fig. 256a). The bodies are subjected to the forces shown in Fig. 256b. Using the d' Alembert principle, we arrive at the differential equations

mlY'1 +Toa , +T Oa 2 =0,

m 2Y'2 +T Oa 2+T Oa 3 =0,

(1)

where a l =y/l, a2 = (Y2 - YI)1l and a3 =yfl. Therefore,

mlY'1 +[ToY ,-To(Y 2- Y 1)]/1 =0, m2Y'2+[ToY2+TO(Y2-YI)]j1 =0. The eigenfrequencies are found from this equation:

p 1,2

= ~T;( ~IT(m--+-mI 2 +-~--;m=12=+=m=2=2=-=m=l=m=2=\)}'(-m-Im2)

168. According to the force method, the differential equations of motion of the mass m have the form

Y = b'll( -mY' )+b'12{ -J01ifJ) , tp= b'12{ -mY' )+b'22{ -J01ifJ)

We seek for the solution to these equation in the form

Y =A sinpt, tp=B sinpt . Performing usual transformations, we arrive at the characteristic equation of the system:

p4_[(b'2JOI+b'llm )p2+1]/(L1mJOl)=0, where L1 = J, I~2 - J,AI' Therefore, the eigenfrequencies squared are

212

2 Vibrations of Systems with Several Degrees of Freedom

As 101 ~ 0, the first eigenfrequency PI tends to (~lmtl\ i.e., to the eigenfrequency of a point mass. To find a correction to this quantity for small, but nonzero, values of 101' we rewrite the above equation in the form

PI2 =(022J OI +Ol1 m )/(2L1mJo1 )

•

Expanding the root in powers of 101' we find

PI2 = 1/(011m )-(0;2J0\ ) / (4L1m 2011 ) . The second term in this expression is the desired error associated with the assumption of point mass.

169 In order to determine eigenfrequencies for statically indeterminate many-body systems, it is convenient to use the force method to formulate the equations of motion. In this case, the static indefinability of the system remains unsolved, but the superfluous constraints should be substituted by unknown forces Xi. We choose the basic system as it is shown in Fig. 257. The beam is subjected to the inertial forces -md2xldt2 and -md2y/dl and an unknown reaction X. The displacements of points of application of these forces obey the equations

Y 1 = -m l Y"1 0 11- m 2Y"2 0 12 +X 0\3'

Y 2=-m l Y"1 0 21- m 2Y"2 0 22 +X 0 23 , Y 3 =-m l Y"1 0 31- m 2Y"2 0 32 +X0 33 .

Assuming Yl =Cl sin(pt), Y2 =C2 sin(pt), and Xl =Xo sin(pt) and taking into account that Y3 =0, we obtain the system of three homogeneous equations in the unknowns Cl' C2, and Xo and the corresponding characteristic equation

(l-mIP2011) -m 1P 2021 -miP 2031

-m 2p 2012

0\3

°

(l-m 2p 2022) 23 =0 -m 2P 2032 0 33

or

Oll m 1m 2 0 21 0 31

012 0 22 0 32

013 0'23 P 4 -[ m 1(Oll033 -0;3 )+m2(0 22 0 33 -O~ 0 33

)]p

2

+033 = o.

(1)

Using the Mohr method, we determine the coefficients o.} by mUltiplying bending moment diagrams (1), (2), and (3) in accordance with the Vereshchagin's rule (Fig. 257) (Timoshenko S. and Young D.H, 1945). As a result, we find

2.1 Free Vibrations

213

b p.a..U¥-..........~&..,&..&.&..&,..I,;~

a(/1a)//

1

.. 2

....-=~~~~~~

l.:r..-h-rrrrTrr'l'"'i-rrr'l':P'l3

/ Fig. 258.

Fig. 257.

0 12 =021 = ab - -(2 / -a 2-b 2) , 6lEJx

b (b3-4/2J '

0 23 =032 = 4EJx

2

where 1= 2(a + b). Substituting the numerical values and solving Eq. (1), we find PI = 101 lis and P2 = 150 lis.

170 Differential equations of small free vibrations, which are found by the force method, are similar to those obtained in Problem 169. In this case, the eigenfrequencies of the systems shown in Figs. 128a-d are, respectively, (a) PI = 3.68 Po, P2 = 10.62 Po; (b) PI = 2.34 Po, P2 = 3.05 Po; (c) PI = 2.54 Po, P2 = 4.37 Po; (d) PI = 1.05 Po, Pz = 6.06 Po· where P02 =EJj(ml\ 171 Using the force method, we arrive at the differential equations of motion

214

2 Vibrations of Systems with Several Degrees of Freedom (1)

Y I = 011 (-m lil)+OI2( -m 2i2)' Y 2 = 021( -m lil)+022( -m 2i2).

To determine the coefficients ~i' we take unit forces to be applied to each load, construct bending moment diagrams (1), (2), and (3) shown in Fig. 258, and em ploy Vereshchagin's rule. As a result,

01l=022=41 3/(9EJ x )' 012=021=713/(18EJx).

(2)

(0I1mIP2-I)AI+0I2m2p2A2 =0,

(3)

Substituting the solutions YI =Al sin(pt) and Y2 =A2 sin(pt) into Eq. (1), we obtain the system of algebraic equations

021mlP 2AI

+(022 m2P

2

-1)A2 = O.

Therefore, the characteristic equation is

P 4 -P 2

mloll +m 20 22 1 0 +----,-----~ m lm 2(01l022 -0;1) m lm 2(01l022 -0;1)

Substituting expressions (2) into Eqs. (4), we arrive at

r

P 4_P 272EJ x/ (5ml 3) +54/5(EJ x / ml 3 The roots of this equation are

3), P2 PI =0.89~'---EJ-'x /(-:--ml---:-

(4)

=0 .

=3.7~EJx /( m1 3) .

To find the eigenmodes, we set All = 1 and P = PI in Eqs. (3). In this case,

A21 =(1-0I1mlpn/(0I2m2pn~1.052. Similarly, setting AI2 = 1 and P =P2' we find A22 ~ - 0.475 (see curves (3) and (4) in Fig. 258). The numerical values of the eigenfrequencies are PI = 336 1/s and P 2 = 1400 1/s.

172 The equations of motion found by the force method have the form

yJm =-0I1il- 0 l2i2- 0 I3i3' Y21m =-02Iil-022i2- 0 23i3' Y 31m =-03Iil-032i2 -033i3'

where D:I =25A, D:2 =39A, D:3 = In, ~2 =8U, ~3 =39A, ~3 =25A, andA =t/(3888EJJ We seek for the solution in the form Yi =Ai sin(pt). In much the same way as in Problem 171, we obtain the characteristic equation

2880,tl-1344,tl + 131,u-1 = 0 ,

(1)

where 11 =mtl/(3888EJ). The left hand side of Eq. (1) as a function of 11 is shown in Fig. 259. The roots found from this plot are III = 0.00825, J4 = 0.126, and J.L; =0.334. Using these values, we find

2.1 Free Vibrations

215

173 The center of gravity of the vibrating plate is defined by its horizontal x and vertical y displacements and the rotation angle cp. In this case, the differential equations of small free vibrations have the form

= -ollmX" -012mji -013Jip, Y = -02lmX" -022 mji -023J ip, (jJ = -03l mX" -032 mji -033 J ip· X

(1)

The coefficients o,j are determined by multiplying, according to the Vereshchagin's rule, bending moment diagrams (1), (2), and (3), which correspond to unit forces and unit torque applied to the plate center of gravity in x-, y-, and ~ directions, respectively (Fig. 260 a-c): £\1 = (1/3)<\, £\2 = b;1 = b;2 = (1/4) <\, £\3 = ~I = b;3 = ~2 = (1/2) o,jl, and ~3 = 0/12, where <\ = t/(El). In the case of small vibrations, y = Z({Y2 and Eqs. (1) are reduced to X

= -Ollmi· -(mOI2 +2JOI3 /I)ji,

y

= -021 mi· -( m022 +2J0 23 / I )ji .

(2)

Assuming x = AI sin(pt) and y = A2 sin(pt) and taking the equality 1= mZ2/12 into account, we find the eigenfrequencies

Pl=1.185po' P2=4.75po'

Po=~EJx/(mI3).

(3)

To find the eigenmodes, we set x = All sin(Plt) and y = A21 sin(pJ) in Eqs. (2). For P = PI and P = P2' respectively, we have

f(P)

4

Fig. 259.

216

2 Vibrations of Systems with Several Degrees of Freedom

1

e

d Fig. 260.

A2l =A ll (1-mpI20ll)/[pNmOI2 +JOI3 /1)] =1.15A lP An =A12 (l-mp;Oll)/[p;(mo I2 +Jo13 /1 )]=-0.87A I2 . The orthogonality condition

mlA llA12 +m2A21A22 =0 for the eigenmodes can be used to verify the solutions obtained. In the case under consideration, m, =mz =m, and this condition is evidently satisfied. The eigenmodes found are shown in Fig. 260d,e.

174 The eigenfrequencies are p, = 0.97 Po and Pz = 3.2 Po' with Po = [EJjCmt)]"2. The eigenmodes and displacements are shown in Figs. 261a,b for the masses m, and IDz, respectively.

.......-

-

\

11

a Fig. 261.

b

,I

2.1 Free Vibrations

217

175 The frame has three degrees of freedom, namely, the horizontal, vertical, and angular displacements of the plate. We choose a basic statically determinate system as it is shown in Fig. 262. The differential equations of motion found by the force method have the form

X =-ollm:i" -o12mj" -o13J ip+R014 ,

(1)

y =-021mi' -o22m)/ -o23J ip+Ro24 , (jJ

=-031mi' -032 m)/ -033J ip+ R034 ,

where R is an unknown reaction force. System (1) should be completed by a kinematic constraint, namely, the linear displacement in the direction of the unknown reaction R must be equal to zero:

mi'041 +m)/042 +Jip043 -Ro44 =0.

(2)

Here, x, y, and qJ are the horizontal, vertical, and angular displacements of the plate, respectively. To determine the coefficients of Eqs. (1) and (2), we take unit forces applied to the basic system and construct the bending-moment curves just as in Problem 173. To simplify Eqs. (1), we take into account that the vertical displacement of the left end of the plate is zero, i.e., the vertical y and angular qJ displacements of the plate center of gravity are related by the equality y = lrp/2. Taking the relation J = m12/12 into account, we arrive at the system of equations

1

a

Fig. 262.

1

b

c

e

f

218

2 Vibrations of Systems with Several Degrees of Freedom

x =-Ollmi· -(m012 +mlo13/6)ji +Roi4' y

(3)

=-021mi· -( m022 + mlo23 /6)ji + R024 ,

O=-041mi· -(m042 +mlo43 /6)ji +Ro44 • We then find the coefficients b;j' using the bending moment diagrams shown in ~I = (1/3)~, ~2 = ~I = ~2 = 824 = ~2 = (1/4)8 0' ~3 = b;1 Fig. 262a-d: = ~3 = b;2 =(112)8/1, b;3 = 8/1\ ~I = ~4 =(1/6)~, b;4 =~3 = 8/1, and 844 =(2/3)8/1, where ~ =t/(EJ). Assuming x =Al sin(pt), y =A2 sin(pt), and R =Ro sin(pt), we determine the eigenfrequencies of the system: PI = 1.45 Po andp2 =5.83 Po, with Po =[EJ/(mt)tll • To find the eigenmodes, we set x = XI = All sin(PI t) and y = YI = A21 sin(Pl). As follows from Eqs. (3), A21 = 1.05 All. Similarly, setting x =X 2 = AI2 sin(P2 t) and Y = Y2 = A22 sin(Pl), we find A22 =-0.95 A 12 · The eigenmodes found for All =AI2 =1 are shown in Fig. 262e,f by dashed lines.

=71.2 1/s and P =466 1/s. 177 PI =0.685 Po and P = 1.4 Po, with Po = [EJj(mt)]II2. The corresponding eigen176 PI

2

2

modes are shown in Fig. 263.

178 The bending-strain energy of the system is II =cll2, where c =3EJjt is the bending rigidity of the cantilever rod (see Fig. 264a). The kinetic energy of the load is

T

= mug /2+Jip~/2,

In the case of small vibrations, Y = method, we arrive at the equations

where Uo =u +ipl /4 .

(qJ-

u) cosqJ""

qJ-

mit· +mlip/4+c(u -qi)=O, mit· +5mlip/4-4c(u -qi)= O.

0.46

a Fig. 263.

4.38

b

u. Using the Lagrangian

2.1 Free Vibrations

a

219

b

Fig. 264.

To derive the characteristic equation, we substitute u = AI cos(pt) and

ffJ

=A2 cos(pt) into these equations. As a result,

(c/m-p2) -(lp2/4 +cl / m )_0 -(p 2+4c/m) (4cl/m-5lp2/4)or

Hence,

PI =0, P2

=~87EJ/(4mI 3 )

The eigenfrequencies PI and P2 correspond to the motion of the system as a rigid body and to the flexural vibration, respectively (Fig. 264b,c). The orthogonality condition muOl u02 + iffJllffJ22 =0 provides a relation between the mode amplitudes. Setting UOI =u02 = 1, we have ffJlI =4/(5/) and ffJ I2 =- 511. 179 The total potential energy of the system with allowance for the gravity force is

cy 2 5 cy 2 5 2 II =---mgl (l-coscp) ~ -+-mglcp . 2 4 2 8 The kinetic energy is the same as found in Problem 178. The equations of motion are

mii+mlip/4+c(u -cpt)=O, mit" +5mlip/4-4c(u -cpt )+5mgcp= O. The corresponding characteristic equation

220

2 Vibrations of Systems with Several Degrees of Freedom

p4 -29cp2 j(4m )+5cg j(ml )=0 has the roots

P\,2

=~29c/(8m )±~(29c/8m t -5cg j(ml)

,

where c = 3EJ/t

180 We consider the system at an arbitrary moment of time (Fig. 265a). The kinetic energy is given by the equation

T =m\u\2 /2+m 2u;

/2 .

When determining the bending-strain energy, we can consider the rod as a beam supported at its ends (Fig. 265b). Since the bending moments at the end points are zero, the strain energy is II= cu II 2/2, where C = 6EJjt In the case of small vibrations when coso. Ri 1, u ll Ri al - u t and al Ri u/2, the equations of motion have the form

mil '\-cu\-cu 2/2=0, mii2 +cu 2/4-cuJ2=0. The corresponding characteristic equation

p2[m\m2p2 -(cmJ4+cm 2)]=0 has the roots

a

a Fig. 265.

b

c

m112m 2

2.1 Free Vibrations

221

The eigenfrequencies PI and P2 correspond, respectively, to the rotation of the rod as a rigid body and to the vibrational mode shown in Fig. 265c.

181 According to the d' Alembert principle, a system will be at rest if loaded by inertial forces (Fig. 266a). Consequently, the supporting force is

R =-miit -2mu·2-mii3 .

(1)

The moment of the inertial forces with respect to the hinge is zero, i.e.,

mait +4mhi·2+ 3mhi·3 = 0 .

(2)

There are four unknowns (R, up u2' and u3 ) in Eqs. (1) and (2). To set up additional equations, we consider the bending caused by the inertial forces:

EJxY "=-Rz -mu·t(z -/)-2mu·2(z -2/) . Integrating this equation two times and taking Eq. (1) into account, we arrive at the equation

EJxY

=(u· t +2u·2+u·3)mz 3/6_

(3)

r

mu·t(z _/)3 /6-2mu·2(z - 2/ /6+Cz +D . The constants of integration C and D are found from the boundary conditions:

y =0 for z =0 and z =31. It follows from these conditions that D =0 and

C

= -(19u·t + 52u·2+27u·3)mI 2/18 .

Finally, we arrive at the equation for the curved rod axis:

EJ Y x

2

. ((z _/)3

mU t

J

z= m (u..t + 2··u 2 +u..3 )(z-63- 3/ -26

(4)

.. ((z _2/)3

4/ 2ZJ

- - -2mu 9 2

6

As is seen from Fig. 266a,

Yt =-Y (z =/)=u)3- u l ' Y2 =-Y (z =21)=2u 3/3-u 2 ·

(5)

Using Eqs. (4) and (5), we obtain the two additional equations

8 .. 23 .. 4 .. 3EJx EJ x 0 "3mut +3 mu2 + mU3+-1-3-Ut-~U3= ,

17

..

28

..

5

..

EJ x

2EJ x

18mut+9mu2+"3mu3+~ut-~u3=

0

(6)

.

Equations (2) and (6) are a complete system. The corresponding characteristic equation

2 Vibrations of Systems with Several Degrees of Freedom

222

c

b

a

d

Fig. 266.

x 2+ 108 (EJ x )2]= 0 2[ 4_16EJ rnl 3 P 5 rnl3

P P has the roots

PI

=0, P2 =1.23~EJx f(rnl 3) , P3 =3.8~EJx f(rnl 3)

The zero frequency corresponds to the rotation of the rod as a perfectly rigid body (Fig. 266b). To find relations between the eigenmode amplitudes at p = P2' we substitute ui = Ai2 sin(N) into Eqs. (2) and (6). As a result,

4A 22 +3A 32 =-A I2 , 11.5A 22 +7A32 =-A 12 For A 23

AI2

= 1, we have

=- 0.485 and

Figs. 266c,d.

A33

A22

= 0.61 and

A32

•

= - 1.15. Similarly, for P = P3.' we have

=0.31. The second and third eigenmodes found are shown in

182 The eigenmodes can be found from the following considerations. The system has two zero eigenfrequencies corresponding to the translatory and angular motions of it as a rigid body (Fig. 267a). These eigenmodes are shown in Figs. 267b,c. Two bending eigenmodes, symmetric and antisymmetric, can be found from the orthogonality condition. We introduce the following notation: ulJ' U 12 ' U 13' U I4 and rtJlJ' rtJ12 , rtJ13 , rtJI4 (u 2J' U 22 ' U 23 ' U 24 and rtJ2J' rtJ22 , rtJ23 , rtJ24) are linear and angular displacements corresponding to the four eigenmodes for the left (right) body, respectively. The third (symmetric bending) mode must be orthogonal to the translatory mode, i.e.,

rnu IIU 13 + rnu 21u 23 +JCfJIICfJI3 +JCfJ2ICfJ23 = 0 .

2.1 Free Vibrations

223

Since flJlI = flJlZ = 0, we have u l ) = U ZJ = 0, assuming that U II = U ZI = 1 (for the first mode, u lJ =u lJ 0). The third mode must be orthogonal to the second mode:

*

mu 12U 13 + mu 22 U23 +J'P12'PI3 +J'P22'P23 = 0 .

Setting U IZ = - 1 and uzz = 1 and taking into account that u l ) = un = 0 and flJlZ = flJ2) = 3/(41), we find flJl3 = - flJ2)" Similarly, from the conditions that the fourth mode must be orthogonal to the first, second, and third modes,

mU l4 +mu 24 +J'P14 ·0+J'P24 ·0=0,

3

3

m( -1)( -1)+m +J'P14-+J'P24-=0, 41 41 m .0+m(-I).0+J'PI3'P14 +J'P23'P24 =0, we obtain U I4 = - U Z4 = - 1 and flJI4 = flJZ4 = - 16/(3/). The third and fourth modes are shown in Figs. 267 d, e. Using the eigenmodes found, we now determine their frequencies. Since the third mode is symmetric, we consider only the right half of the system (Fig. 267f). To determine the bending rigidity of the system, we find its angle of deflection caused by the unit moment of forces applied to the body: f/J;) = 8,) =1/(2£1). Corresponding eigenfrequencies are

PI

a

=0,P3 =~"""""-;I/(-J8---'-11) =2.84~EJx j{mI3)

£i1

3112

cil=J

_ _ _""U21=1 b ~=r-1 ....... _ _ _ _---A.

r_

c

f

u -1

Uet!.:81 fI1Jf=3I(41)

Fig. 267.

Similarly, the fourth, antisymmetric mode corresponds to Fig. 267g. In this case, the kinetic and potential energies are

224

2 Vibrations of Systems with Several Degrees of Freedom

T =Ji/ /2+mu~ /2, Il=cy 2/2 , where Uo =U + lrpl2, c =24EJjl\ and y =lrpl2 - u. The equations of motion and characteristic equation have the form

(J + mZ 2/16)ip+mlit'/4+cl 2 rp/4-clu /2 =0, mlip/4+mu' -clrp/2+cu =0, P 2 [Jmp 2 -9C(Z2m +J

)/16] =0.

The frequencies corresponding to the antisymmetric mode are

P2=0,

p4=8.2~EJx/(mI3)

.

183 The system has four degrees of freedom and, therefore, four eigenmodes: two anti symmetric and two symmetric modes (Figs. 268a-d). The frequencies of the anti symmetric modes can be found just as in Problem 182:

PI

=0, P3 =4.7~EJx /(mI 3)

5/14

a

c

e

Fig. 268.

To determine the frequencies of the symmetric modes, we consider the right half of the system as a cantilever. The differential equations of motion found by the force method have the form

2.1 Free Vibrations

225

The coefficients b.j are determined by the Mohr method with the use of the Vereshchagin's rule:

8 11 =31l 3/(48EJ x ) ,8'2 = 8 2, =3/ 2 /(4EJ x ) ,822 = 1/(EJx ) . Finally, the frequencies of the symmetric modes are

P 2 = 1.1~EJ x/ (m1 3) , P 4 = 6.5~'--EJ----;x / (-=--ml--:-3) . Assuming U22 = u24 = 1, we find the angles f/J22 = 0.93/1 and f/J24 = - 3.3/1.

184 A position of the system is defined by the four coordinates y" Y2' (Fig. 269a). The kinetic energy is

f/J"

and f/J2

T = m,y ,2 /2+m 2 y; /2+J/p; /2+Jip;/2 . It is convenient to express the strain energy in terms of the relative linear Yk and angular f/Jk displacements of the bodies. Assuming the left body to be fixed, we take a force Q and a moment M of forces as applied to the end of the elastic rod (Fig. 269b). In this case, the deflection of the rod and its rotation angle are, respectively,

Y k =8 11Q +8'2 M , ffJk

=8 2,Q +822 M ,

where ~, = tl(3EJ), ~2 =~, = [2/(2EJ), and~, = l/(EJ). Therefore,

Q = Y k 8 22 / L1-ffJk 8 22 / L1, M

= -Y k 8 2,/ L1+ffJk 8 11 /

L1 = 8 11 8 22 -8;2 = (1/6)[/ 2/( EJ x

)f

The strain energy of the rod is

b Fig. 269.

L1,

2 Vibrations of Systems with Several Degrees of Freedom

226

2

2

allfPk +a12 fPkY k +a22 y k . Here,

(2

2)

I 12 EJ x all = Lf £5'1l -/£5'12£5'1l +-£5'12 = - - , 2 EJ x 3 21 [( 2 ) I2 ] 3EJ x a12 = LfI I £5'1l£5'22 +£5'12 -2£5'12£5'1l--£5'12£5'22 =--2-' 2 EJ x 3 21

a22 =

(2 / I 2 2) 3EJ x £5'21- £5'12£5'22+-£5'22 =--3-' 2 EJ x 3 21

LfI

As is seen from Fig. 269a,

Y k =Y 2 - Y I -/2fP2/2-/lfP2/2-/lfPI/2-lfPI

(2)

=Y2-YI-/ 2fP2/ 2 - / I'fPJ2, fPk =fP2-fPI'

where II'

=1+ 1/2. Substituting expressions (2) into Eq. (1), we have

11 =bllfP; +b 22 fP; +b 33 y 12 +b 44 y; +b 12 fPlfP2

+b 13 fPIY I +b 14 fPIY 2 +b 23 fP2Y I +b 24 fP2Y 2 +b 34 y IY I , where

bll =all +(/'IY Clz2/4 - I '1 a12 /2, b22

= all +/ 22Clzz/4-/ 2a12 /2, b33 =b44 =Clz2'

bl2 =-2all + 1'1 a12 /2+/ 2a12 /2-1 'I 12a22 /2 ,

b13 = al2 -1'1 a22 , bl4 = -a12 + I' I Clz2 ' b23 = -a12 + 12Clz2 , b24 = a12 -/ 2Clz2' b34 = -2Clz2 . Using the Lagrange equations, we arrive at the differential equations of motion:

J/PI + 2b ll fPI +b 12 fP2 +b 13y I +b l4 y

2

=0,

J iP2 + b12 fPI + 2b 22 fP2 + b23 y I + b24 y 2 = 0 ,

mJ; I +b 13 fPI +b23fP2 + 2b33 y I +b34 y 2 = 0, m2Ji 2 +b l4 fPI +b24 fP2 +b34 y 1+ 2b44 y

2

=O.

185 Substituting Yi = Ai sin(pt) and rp, = Bi sin(pt) into the equations of motion considered in Problem 184, we arrive at the characteristic equation

2.1 Free Vibrations

2b ll

-JIP Z

b12

-JZp 2

b l3

b l4

b24 b34 2b44 -J4P Z

b l3

b Z3

b23 2b33 -J3P 2

b l4

b Z4

b34

biZ

2b zz

227

=0

or

JJ 8 6 4 Z 0 . I Zm lm 2P +alP +a2P +a3P +a4 = Since two degrees of freedom of the system correspond to its motion as a perfectly rigid body, four roots of the characteristic equation must be zero, i.e., the coefficients a 3 and a4 must vanish. Substituting the expressions for bij into the reduced equation

we find

a l =-2[JI J Zm lb44 +JIJ2m2b33 +Jlmlmzb2Z +JZm lm 2bll ]

=-(13/16)m 3lEJx

'

a2=J/ ZB34 +J lm 2B 23 +Jlm lB z4 -J2m 1B 14 -J2m 2B\3 -m lm ZB 12 =(21/32)(mEJx /lt, whereBg =b gZ-4bii bjj

•

The corresponding eigenfrequencies are found from Eq. (1): PI,2

= 0, P3 = 0.447 ~EJx /( mI 3), P 4 = 3.52~EJx /( m1 3)

186 The system has five degrees of freedom: the vertical displacements and the rotation angles of the two loads and the vertical displacement of the central point mass. The zero eigenfrequencies P, = P2 = 0 correspond to the motion of the system as a perfectly rigid body (Fig. 270a,b). The third mode must be orthogonal to these zero modes:

mu lJu \3 + mu 2l U 23 + mu 31U33 +JCfJI\CfJ\3 +JCfJ31CfJ33 =0,

(1)

mu 12U 13 + mu Z2UZ3 + mu 32U33 +JCfJ12CfJ\3 +JCfJ32CfJ33 =O. Since rpll = rp31 = 0 and u ll = U 21 = u31 = 1, Eqs. (1) give U 31 + U 23 + U 33 =0. Taking u ,3 = U 33 = - 1, we find U 23 = 2. Similarly, substituting U 32 = 1, U ,2 = U ,3 = U 33 = - 1, and rp12 = rp32 = lit into Eq. (1), we find U 22 = O. Therefore, the third eigenmode is symmetric (Fig. 270c). The orthogonality conditions for the fourth eigenmode have the form

Ul4 +U Z4 +U 34 =0 ,

(2)

228

2 Vibrations of Systems with Several Degrees of Freedom

J

-U 14 +U 34 +-(qJ14 +qJ34)=O ,

(3)

ml

J

-U 14 +2U 24 -U 34 -(qJ13qJ14 +qJ33qJ34)=O .

(4)

m

Since the fourth mode must be anti symmetric, i.e., U l4 = - u34 = - 1, Eq. (2) gives U24 = O. In addition, ({J14 = ({J34' Setting U34 = 1, we find from Eq. (3) ({J14 = ({J34 = - mill = - 411 (Fig. 270d). The fifth, symmetric mode must be orthogonal to the four modes considered above. Similar consideration yields ({J15 = - ({J35 = 3mll(l({J13) = 121(l({J13) and U25 =-2 (with U I5 = U35 = 1). This eigenmode is shown in Fig. 270e. The eigenfrequencies can be evaluated by using the modes found above. To find the frequencies of symmetric modes, we treat the system as a hinged beam loaded by the moments M = l / ({J at its ends and by the inertial forces PI' P2'

a

Fig. 270.

2.1 Free Vibrations

229

and P 3 (Fig. 270f). As is follows from the equilibrium condition, PI + P 3 =P 2, i.e., u2 =2u l • The differential equation describing the beam profile has the form

EJxY "=M +mu,p2z - mu2p 2(Z -I) . The solution to this equation has the form

Since dy(O)/dz =- ffJ, it follows from (5) that

EJxrp=MI +mu,/2p2

/2.

(7)

Taking the equality M =Jl ffJ into account, we reduced Eqs. (6) and (7) to the form

(-mep 3 /3+3EJ x )u, -Jp 2/ 2rp/2=0,

(8)

mI2p 2uJ2+(Jlp2_EJx )rp=O. The roots of the corresponding characteristic equation

p 4 -52EJxp 2/(mI 3)+ 144[EJx/(m/ 3 )f =0 are

To find the angles ffJl3 and ffJ15' we use one of equations (8) and the eigenfrequencies found: ffJl3 =- 5.5/1 for U l3 =- 1, and ffJI5 =- 2.211 for U l5 = 1. The frequencies of antisymmetric modes are found in much the same way as in Problem 182:

187 To set up the differential equations of motion, we use the d' Alembert principle. In the process of vibrations, the flywheels are SUbjected to the moments of inertial and elastic forces. Assuming that ffJI < ffJ2 < ffJ3 (ffJj are the rotation angles), we have

-J/p,+M, =0, -J2ijJ2+(M2 -M,)=0, J3ijJ3-M3=0, or

230

2 Vibrations of Systems with Several Degrees of Freedom

-JjiPj +C j(tp2 -tpj)=O, -J2iP2 +C 2(tp3 -tp2)-C j(tp2 -tpl)=O, -JjiPj -C 2( tp3 -tp2) = 0. Assuming f/J; =Ai sin(pt), we arrive at the characteristic equation

J

°.

j Cj c2 +C jC2J 1+J2+J3P 2 = P 6-p 4(C +C2 +-+J2

J j J3

J 1J 2J 3

With regard to c t = GJ/l t and c2 = GJ/Z2, we find the eigenfrequencies

p j = 0, P 2 = 92.411 S ,p 3 = 151.511 S

.

188 We denote by fPt' fP2' fPt', and fP/ the rotation angles of the left disk, right disks, wheel (1), and wheel (2), respectively. The kinetic and potential energies are

T =J1ipi /2+J 2ip;j2,

II=c1(tpj-tpj't /2+C 2(tp2 -tp2't

/2.

(1)

The rotation angles are interrelated by the equation fPt' =- fP2' u. Eliminating from Eq. (1) the angle fPj and using the Lagrange equation, we obtain the equations of motion

JjiPj +Cj(tpj +Utp2')=0,

(2)

J 2iP2 -c 2(tp2 '-tp2) = 0, uc j(tpj +Utp2 ,)+C 2(tp2 '-tp2) = 0. The last equation implies that the moments of elastic forces acting on the gear wheels are balanced at all moments of time. Eliminating the angle fP2" we arrive at the equations

.. CjC2 (tpj +Utp2) Jjtpj + 2 =0, cju +c 2 .. c jc 2u (tpj +Utp2) J 2tp2 + 2 =0. cju +c 2 The corresponding characteristic equation

p 2 [J/ 2P 2 -(J1u 2 +J2) has the roots

C~C2

]= b1u +C 2

°

2.1 Free Vibrations

231

ClC2 (Jlu 2 +J2) (C IU 2+C 2) Jl 2

Pl=0,P2=

189 The equations of angular vibrations of the disks are set up just as in Problems 187 and 188. The eigenfrequencies for systems shown in Figs. 145a-e are, respectively,

(a)pl =0, P2 =~3c/(2J);

(b)Pl =~c/(2J), P2 =~2ii/J ;

(C)Pl =0.8~c/J ,P2 =1.97~c/J ;

(e)pl=0.651~c/J ,p2=1.256~c/J. 190 The equations of motion are derived by the Lagrangian method. The kinetic and potential energies are

T =Jip;

/2 + 2Jip;/3+3Jip;/2 ,

Il=c(rpl-rp2t /2+C(rp3-rp2t /2. The differential equations of motion have the form

J(P1 +C(rpl-rp2)=0,

(1)

2J(P2 -C(rpl-rp2)+C(rp2 -rp3)=0, 3J(P3 -C(rp2 -rp3)=0. We seek for a solution to Eqs. (1) in the form stants Ai are determined from the equations

(c -Jp 2)A

I

(fJi

=Ai sin(pt),

-cA 2 =0,

(2)

-CAl +(2c -2Jp 2)A2 -CA3 =0, -cA 2+(c -3Jp2)A3 =0. The characteristic equation

p2[p 4-7cp 2/(3J )+(c/J )2]= has the roots

where the con-

°

232

2 Vibrations of Systems with Several Degrees of Freedom

1J'......

O.!],!

-~0.77

P3

Fig. 271.

We now find relationships between the amplitudes of these eigenmodes. For zero frequency (first mode), we find from Eq. (2) A" =A 2, =A 3,. For the second mode, A22 =(c - Jp2 2)AJc =0.435A'2 and A32 =cA 2i(c - 3Jp22) =- O.615A 22 . For the third mode, A 23 = - O.77A'3 and A33 = O.I79A'3" The eigenmodes are shown in Fig. 271 for A" =A'2 =A13 = 1. To verify the eigenmodes found above, we use the orthogonality condition. For the first and second modes, this condition has the form

JAJlA 12 +2JA z1 A zz +3JA 3IA 3Z =0

(3)

Substituting the amplitudes Aij found above, we have

(1.1-2.1.0.435-3.1.0.615)J =0.025J ~O . Similarly, the orthogonality conditions for the first (or second) and third modes yield, respectively,

(1.1-2.1.0.77+3.0.179)J =0.003J ~O, (1.1-2.0.77 .0.435-3.0.615+0.179)J ~ O. 191 The differential equations of motion of the disks can be written out as

Jipl +C(
2.1 Free Vibrations

233

1

Fig. 273.

Fig. 272.

where c =Gi/l. Taking

P

2

[P

6

(jJi

=Ai sin(pt), we arrive at the characteristic equation c ( )2P -"25 ()3] J =0 ,

lIe 4 c - 2J P +8 J

2

which has the roots

PI=0,p2=0.656~c/J ,P3=1.335~c/J ,p4=1.82~c/J . The eigenmodes found are shown in Fig. 272. It is easy the verify that the orthogonality conditions are satisfied.

192 The characteristic equation

p'[p'- ~~ p4+18(~)' p'-tr]=o

has the roots

PI =0, pi =0.613c/J ,p; =2.26c/J ,p; =5.8c/J These eigenmodes are characterized by the vibration amplitudes, respectively,

234

2 Vibrations of Systems with Several Degrees of Freedom

tJJll = tJJZI = tJJ31 = tJJ41 = 1, tJJ12 =1,tJJzz =0.387, tJJ3Z =-0.0375, tJJ4Z =-0.455, tJJ13 = 1, tJJZ3 = -1.26, tJJ33 = -0.966, tJJ 43 = 0.412,

tJJI4 = 1, tJJZ4 = -4.8, tJJ34 = 6.22, tJJ 44 = -0.8. 193 The expression for the strain energy is the same as in Problem 188, while the expression for the kinetic energy contains two additional terms:

T =J/pi /2+J zip;/2+J I'(ipl ,)Z /2+J z'(ipz')Z /2 . Eliminating the angle fIJI' (see Problem 188) and using the Lagrange equations, we obtain

J/PI +CltJJl +C IUtJJ2'=0, (J z'+J I'u z)~z '+CIUtJJl +(c IU 2 +C 2)tJJ2 '-C 2tJJ2 =0, J2~2 -C 2tJJ2 '+C 2tJJ2 Taking f/J;

=O.

=sin(pt), we arrive at the characteristic equation P 2 {lIJ 2J 3P 4 -[J2(C IJ 3+C 3J I)+J/3C2]P 2

+(CP2 -C I2U2)J2+C lc 2J 3+(C 2C3-C;YI }=O, where 13 = 11 'u 2 + 12 ' and c3= C I U 2 + c2•

194 The kinetic and potential energies of the system are given by the equations

T =J1ip; /2+J 2ip;j2+J3ip;/2, II =C1(tJJl _tJJI,)2 /2+C I(tJJ2 -tJJ2 ,)2 /2+C 2(tJJ3 -tJJ3,)2 /2. Here, fIJI" fIJ/, and fIJ/ are the rotation angles for the gear wheels in the gear reduction (see Fig. 273), with

tJJl ,= -tJJ3 'Z / 2 Z I = -UtJJ3 ', " tJJl = tJJ2 .

Eliminating the angles fIJI' and the equations of motion

flJ2'

and using the Lagrange equations, we arrive at

JI~I +C1(tJJl +UtJJ3')=0, J2~2 +C 1(tJJ2 +UtJJ3')=0,

J3~3 +C 2(tJJ3 -tJJ3')=0, CIU (tJJl +UtJJ3 ,)+CIU (tJJ2 +UtJJ3 ')+C 2(tJJ3 '-tJJ3) = O. The last equation gives

tJJ3'=[C 2tJJ3 -clu (tJJl +tJJ2)]/(2c 1u 2 +C 2) .

2.1 Free Vibrations

Eliminating

(jJ3"

235

we obtain the system of equations

J/p, +allqJ, -a'2qJ2 +a13 qJ3 =0, JiP2 -a'2qJ, +all qJ2 +a13 qJ3 =0, J 3(P3 + a13 qJ, + a13 qJ2 + 2ua13 qJ3 = 0 , where 2

2

2

2

c, U +c,c 2 c, U c,c 2u all = 2 ,a'2 = 2 ' a13 = 2 2c,u +c 2 2c U +c 2 2c,u +c 2 The characteristic equation is

J/ 2J 3P6 + [2ua13J,J 2-a,,(J,J3+J2J 3)]P 4

+[2ua13all (J, +J2)-a:3(J, +J2)-a,22J 3]P 2 -2a13 [U(a:, -a,22 )-a13 (a'2 +a,,)] = O. It is easy to prove that the free term of this equation must vanish. Indeed, it is evident that the uniform rotation of all the gear wheels is possible; therefore, the characteristic equation must have zero roots.

195 Considering the dynamic equilibrium of each flywheel, we arrive at the equations

J,(p, +c,(qJ, -qJ2)+a(~'-~3)=O,

(1)

J 2(P2 -c,(qJ,-qJ2)-C 2(qJ3 -qJ2)=O, where (jJI' (jJ2' and (jJ3 are the rotation angles for the left and right flywheels and for the contact ring of the damper, respectively. The third equation is found from the equilibrium condition for the elastic shell:

a(~, -~3)-C2(qJ3 -qJ2)= Substituting the solutions acteristic equation

(jJi

(/!?J, +2a+c,)

°.

(2)

=Ai exp(AJ) into Eqs. (1) and (2), we obtain the char-pa

or

In this case, there is a system with a fractional number of degrees of freedom. Equation (3) has two zero roots (11.1•2 ~ 0), which correspond to the rotation of the system as a rigid body. The remaining three roots are found from the equation

236

2 Vibrations of Systems with Several Degrees of Freedom

13

/I,

C21 2+ J 1+J2 ( C +C ) /1,+ 1 J 1+J2 C 1C 2 -- 0 . +-/1, 1 2

a

J/ 2

a

J 1J 2

Substituting the numerical values, we have

1 3 +6012 +6301+12600=0 ; therefore, A3 =- 53.1 and A4,5 =- 3.45 + i 16.5.

196 We choose the vertical displacement y of the center of gravity of the case and its rotation angle rp as generalized coordinates (Fig. 274). The potential and kinetic energies are given by the equations

II=cl(y +rp/lt /2+c 2(y -rp/2)2/2 , T =my 2 /2+Jil

/2.

Substituting these expressions into the Lagrange equations, we find

mji +C1(Y +rp/I)+C 2(Y -rp/2)=0, JijJ+clll(y +rp/I)+c 212(y -rp/2)=0, or

ji +ally +a12 CP=O, ijJ+llzIY +a22 CP =0, ' where

all =(C I +c 2 )/m , al2 =(cil l -c 2 12 )/m

,

n/J .

a21 =(c 1l l -c 212)/J , llz2 =(c I I J2 +c 21

Fig.274.

We seek for the solution in the form y = A sin(pt) and rp = B sin(pt). The corresponding characteristic equation has the roots

2.1 Free Vibrations

P,,2 = i.e., PI

{all +a22)/2+~{all -a22

= 16.5 1/s and P =50 1/s.

237

t /4 +a'2a2' '

2

197 For the eigenfrequencies to be equal to each other, the following conditions must be satisfied:

c,l, =c 212, (c,I,2 +cin/J ={c, +c 2)/m . These conditions are reduced to JIm = 11/z198 We take the vertical y and angular rp displacements of the case center of gravity and the vertical displacements YI and Y2 of the front and back axles as generalized coordinates (Fig. 151). The kinetic and potential energies and the Rayleigh dissipative function are given by the equations

T =Jil/2+my 2 /2+m,y ,2 /2+m 2y /

a {. R =2 Y

/2,

a {.. .)2 +ipb -y,.)2 +2 Y -rpa-Y2 .

Substituting T, II, and R into the Lagrange equations, we arrive at the equations of motion

J oq,+a(a 2+b2)~+(c2a2 +c,b 2)tp+a{b -a)y +{c,b -c 2a)y -abj, +~2-c,by, +c 2Y 2a =0,

mJi +2aj +(c 1 +c 2)Y +a{b -a)~+{clb -c 2a)tp-a{y, + y 2) -c,Y I - C 2Y 2 =0, m,Ji, +ay, +{c, +c 2)Y ,-ab~-c,btp-ay -cltp=O, m2i2+ay2+{2c+c2)Y2+aa~+c2atp-aj -c 2y =0. 199 We take the displacements YI and Y2 of the front and back axles as generalized coordinates. The viscous forces are ignored (a= 0). We introduce new unknowns Z, =Y + rpa and Z2 =Y - rpb. Then,

Y =(z ,b +z 2a)/{a+b); tp={z, -z 2)/{a+b) . Just as in Problem 198, we obtain the two independent systems of differential equations

238

2 Vibrations of Systems with Several Degrees of Freedom

Fig. 275.

mbil +cl(a+b)z I-c(a+b)y 1=0,

(1)

mlil +(C +CI)Y I-CIZ 1=0; mai2+cl(a+b )Z2 -c(a+b)Y 2 =0,

(2)

mli2 +(C +CI)Y 2 -CIZ 2 =0.

The special case Jo =mab is equivalent to the design diagram presented in Fig. 275. Equations (1) [(2)] describe vibrations of the front (back) axle and a mass m2 =mbl(a+b) [m 3 =mal(a+b)], with the mass m3 (m 2 ) fixed. We find from Eqs. (1) and (2), respectively, PI =8.27 lis, P3 =48.5 lis and P2 =9.9 lis, P4 =48.5 lis. 200 The elongations of the upper and lower transmission sides,

..111 =R I(jJI- R2(jJ2' ..112 =-RI(jJ1 +R 2(jJ2'

(1)

can be expressed in terms of the additional stresses defined in the formulation of the problem: (2)

When vibrating, each pulley is subjected to both the moment of inertial forces and the moment of tensions. The differential equations of free vibrations of the pulleys are

JliPl +RIF (..10"1-..10"2)=0,

(3)

J 2iP2 - R2F (..10"1 - ..10"2) = o. With regard to Eqs. (1) and (2), Eqs. (3) can be written out as

=0, J l iP2 - RIR2Fa3(jJ1 +R; Fa3(jJ2 = 0, JliPl + RI2Fa3(jJ1 - RIR2Fa3(jJ2

(4)

2.1 Free Vibrations

Fig. 276.

239

Fig. 277.

where a3 =(a, + a2)la,a2• The eigenfrequencies are found from the characteristic equation corresponding to Eq. (4):

PI =0, P 2 =~r-R-IF.-a---:3(-R-I/-JI+-R-2-/J-2-:-) ~ 146.3 1/ S

.

201 The kinetic and potential energies of the system shown in Fig. 276 are given by the equations

T =J/p~ /2+J/p;/2+J I' fi/2

/2

(1)

n =c,,1/ 12/2 +c,,1/ 22/2+ mg [hi Sirlll' +h2(COSIj/-1)],

where c = EFIl. The elongations of the elastic elements are

or, with regard to x

~

,,1/ 1 =R l rpl-R 2rp2 -x

COSrplO'

Lll2 =R 2rp2 -Rlrpl-x

COSrplO'

(2)

h,ljI,

,,1/ 1 =Rlrpl - R 2rp2 - hi Ij/COSrplO , Ll/2 =R 2rp2 -Rlrpl-hllj/cosrplO'

Substituting Eqs. (1) and (2) into the Lagrange equations, we obtain

J/PI + 2cRI2rpi - 2cR IR 2rp2 = 0, JiP2 +2cR IR 2rpl-2cR/rp2 =0, J I' ifr +2chl2 II' cosrp 10 + mgh l = 0 .

(3)

240

2 Vibrations of Systems with Several Degrees of Freedom

Here, '1/= 'I/o + '1/" where the angle 'I/o is determined by the static load mg. If the system is at rest, i.e., d2 1f11dt2 =0, then

2ch l21f/ 0 COSfPlO + mghl = 0 . In this case, the third equation in system (3) takes the form

J I ' /ifl + 2chl21f/1 COSfPlO = 0 . The eigenfrequencies are found just as in Problem 169:

PI =0, P2 =18711s, P3=2701ls . 202 The kinetic and potential energies are given by the equations (Fig. 277)

T =J/pU2+Jip;/2+m lx 2/2,

(1)

II =cL11 12/2+cL11; /2+m l g (rl +r2Xl-coSIf/).

Here, c =EFIl is the stiffness of the elastic rods, AI) and Al2 are their elongations, with

L111 =R l fPl- R2fP2 -x

COSfPlO'

L112 =-RlfPl +R2fP2 -x The rotation angles equation

({JI

COSfPlO·

(driving pulley 4)and '1/ (link 3) are interrelated by the (2)

where If/= xI(r) + r 2). Hence, x = r) ({J" and Eqs. (1) take the form

T = Jip;!2+(J I +mlrn~; /2

r

II = ~ [(RI - r l COSfPlO )fPI - R2fP2 + ~ [-(RI + r l COSfPlO )fPI + R2fP2 +m l g (rl +r2)(I-COS

r.

f l fP l ) . f l +r2

Using the Lagrange equations, we arrive at the equations of motion

.. [2EF(RI2+rI2COS2fPlO) ] 2EFRIR2 _ fPI+ ( +mlgrl fPI-( fP2- 0 , J I +mlrl I J I +mlrl I

2)

.. 2EF R R 2EF R 2 fP2 - - - I 2fPl +-- 2 fP2 J 21 J 21

2)

=0 .

The eigenfrequencies are found from the corresponding characteristic equation: PI = 34.6 lis and P2 = 179 lis.

2.1 Free Vibrations

241

a..-_ I ,+b/2

Fig. 278.

203. The body at a moment of time is shown in Fig. 278. Since the tensions To are assumed to be constant, horizontal displacements of the body may be neglected. The equations of motion have the form

my" =-To(sina l +sinaz) , J;P=-To(h l +h2)· Taking into account that

sinal =(Y +brp/ 2)//1' sina 2 =(Y -brp/ 2)/11'

hI =(b /2 )sin( a l +a), h2 =(b /2 )sin(rp-a 2),

The corresponding characteristic equation has the form

2 ToUI +Iz) -p + mll/2 T obU2 -II) or

T obU2 -II)

2 Vibrations of Systems with Several Degrees of Freedom

242

In the special case of II =12 =I, these equations of motion become independent and the eigenfrequencies are

PI =~Zfo/(ml), P2 =~(2+b/I)(Tob /2J) . 204. A position of the cable car is specified by three coordinates, namely, x, y + h, and qJ (Fig. 279). The horizontal coordinate is an independent function of the car speed (x =Xo + vt), while the two other coordinates are related by the equations

mjl =To(sina l +sina 2)-mg , JiP = -T Oh2 +T ohl . Taking into account that

sinal ~(Y +bcp/2)/sp sina 2 ~(Y -bcp/2)/S2'

hi ~ h + Y - x sin a I , h2 ~ h + Y - (I - x ) sin a 2, we write out the equations in the form

jI +(Zfo/mb )AI(t)y +(To/m )A2(t )cp= -g ,

iP+(To/J )A2(t)y +(Tob /2J )AI(t )cp= 0, where

AJ(t) =b(l -b )j[2(x 0 +vt -b /2)(1 -x 0 -vt -b /2)], A2(t) =b[1 -2(x 0 +vt )]/[ 2(x 0 +vt -b /2)(1 -x 0 -vt -b/2)].

x

Fig. 279.

mg

2.1 Free Vibrations

243

a

c

b Fig. 280.

205 We consider an arbitrary deflection from the equilibrium state (Fig. 280a). According to the d' Alembert principle, the differential equations of motion can be written out in the form

mi' =-T Icosa l +T2 cosa 2 ,my" =-(TI sinal +T2 sina 2)+mg.

(1)

As is seen from Fig. 280a,

cosa l = (x 0 +X )/(s 10 + Lis I)' cosa 2= [II -(x 0 +X )]/(s 20 + Lis Z), sinal =(Yo+Y )/(SIO+LisI)' sina2=(yo+Y )/(S20+ Lis 2)'

TI =TIO

+ L1TI ,

T2 =T20

+ L1T2 ·

Keeping only the terms linear in displacements, we write out the equations as

..

T lOX

T 20X

L1T IX 0

SIO

S20

SIO

mx = - - - - - - -

T 20 (zI-X 0)Lis 2 2

S20

+

L1T 2 (z I - X0) T lOX oLis I + 2 S20

SIO

(2)

244

2 Vibrations of Systems with Several Degrees of Freedom

.. TlOy T 20 y LiTtY 0 my = - - - - - - SIO S20 SIO

LiT2y 0 TIOYoAs t T 20 y oAs 2 + 2 + 2. S20 SIO S20

The elongations of the line branches are related to the tension increments by the equations

As t = LiTtSIO/(EF)=LiTtSIO/(clt) , ,As 2= LiT 2s 20 /(cl t ) .

(3)

There are six unknowns in Eqs. (2) and (3), namely, x, y, ..::::IT" ..::::IT2 , ..::::Is" and ..1s2 • The identities ·2 . 2 a + cos 2 a =1 sm at + cos2 at =1 ,sm 2 2

can serve as additional geometric equations which allow us to obtain the following two relations (with regard to the expressions written above):

As t =(YO/SIO)Y +(XO/SIO)X ,

(4)

As 2 =(/t -X o)X / S20 +(Y 0/ S20)Y The static tensions entering into Eqs. (2) are interrelated by the equilibrium equations

which give

_ (It-xo)StO _ XOS20 TIO-mg ,T20 -mg--. ltY 0 ltY 0

(5)

Substituting Eqs. (3)-(5) into Eqs. (2), we have

i" +allx +at2 y =0, )i +a2l x +a22 y =0 ,

Setting x = A sin(pt) and y = B sin(pt), we arrive at the characteristic equation

all-p 2

(6)

2.1 Free Vibrations

245

Hence,

PI,2 = (all +a22)/2+~(aIl-~2t /4 + al22

If the body is at the middle point (i.e., Xo =1/2, SIO =S20 = so), the initial tensions are TIO = T20 = To. In this case, system (6) is reduced to the two independent equations

mi·+2[To{y;/sg)+EF{x;/sg)]x =0, mji +2[To{x;/sg)+EF{y;/sg)]y =0.

(7)

The eigenfrequencies corresponding to vertical and horizontal vibrations are, respectively,

PI =~2{ToY; +EFx;)/{msg) ,

(8)

P21 =~2{T oX; +EFy ;)/{msg). For a series of special cases, the eigenfrequencies can be found from Eqs. (8). In the case of Xo =0, So =Yo ' To = mg/2, and II =0, i.e., vertical vibrations of a body suspended by a tensile line (see Fig. 280b),

PI =~g Iso, P2 =~r--2E-J/"""""(m-s-'-o) . In the case of Yo =0, i.e., vibrations of a body at the middle point of a horizontal elastic line (see Fig. 28c),

PI =~"---4E-F/"-:--(m---:-ll) ' P2 =~2To,/(mll) , where To' is the tension for which Yo l':::: O.

206 The differential equations of motion are similar to Eqs. (6) of Problem 205, but the quantities x O' Yo' TlfV and T20 are now functions of velocity v and time t. According to the initial conditions, the body moves along arc of an ellipse (Fig. 281) given by the following equation in polar coordinates:

SIO =soo/(l-e cosal)=soo/(1-exO/s IO ) .

Taking into account that SIO =Soo + vt, we find Xo = vt/e, where e =1/1. The vertical coordinate of the body is

YO=~SI20-X; .

(1)

Finding x- and y-components of the force applied to the body, we arrive at the equations

246

2 Vibrations of Systems with Several Degrees of Freedom

mg

y

Fig.2SI.

mi' o=T 20(1, -X O)/S 20 -T,er 0/S10

(2)

,

mjio =TlOy O/S,O +T 20 y 0/S20 -mg.

Taking into account that d\/dt2 =0, we find from Eqs. (2)

_ m (Y" 0 + g )(1, - X o)s 10

T IO -

Y 0/ ,

_

,T20 -

m (ji 0 + g )x oS 20

yoi,

,

where d\/M are determined by differentiating Eq. (1). 207. The voltage drop in the armature winding is equal to the applied voltage:

U L +U R +U'P =U 0 sinmt .

(1)

Substituting UL =Ldi/dt, UR =Ri' and U'P = a)dq)"dt, we have

L di /dt + R; +a,dcp/dt =U 0 sinmt

(2)

The differential equation of the armature rotation is

Jd 2cp/dt 2+ccp=a 2i,

(3)

where f4i is the electromagnetic perturbing moment. Substituting i=A)exp(at) and tp=A 2exp(at) into Eqs. (1) and (2), we arrive at the characteristic equation

A? +(R/L )A2 +(a,a 2+cL )A/(JL ) + Rc/(JL )=0

.

In the special case of negligible ohmic resistance (R ~ 0), 1 /l"

1 • C a,a 2 = 0, /l,2 3 = ±l - + - . J JL

208 In the equilibrium state, the capacitor charge is qo and the springs are contracted by Xo (by the force between the capacitor plates). The charge and displace-

2.1 Free Vibrations

247.

ment are q = qo + L\q and x =Xo +LIx, respectively. The capacitance is C =a/(l- x), where a is a constant and I is the electrode spacing in the case when no charge is present. To derive the equations of motion, we use the Lagrange equations. The total potential energy (including both electric and elastic energies) of the system is

II = -UL1q +cx 2/2 . Expanding this expression in powers of !J.q and !J.x up to the second order, we have

II=

] (/-x o)L1q 2 qoL1qL1x cL1x 2 [qo(t-x o) +--+ U L1q + 2a a 2 a

+x o-;i}h 17 +

0.

As follows from the equilibrium condition, the bracketed expressions must be equal to zero. The total kinetic and magnetic energy and the energy lost in the circuit are, respectively,

T = mLii 212 + LL14 2/2, D = RL14 2/2 . Therefore, the differential equations of free vibrations of the system take the form

=0,

mL1X· +cL1x -!l.!LL1q a

LL1I/ + RL14 +

(I -x )L1q 0

a

209. The kinetic and poteritial energies of the system are, respectively,

T =J/p~/2+J2ip;/2, II=c(rp] +rp2f

/2 ,

where (jJ1 and (jJ2 are the rotation angles of the disks. Using the Lagrange equations, we arrive at the differential equations of motions

J/p] +C(rp]-rp2)=0,

(1)

J/P2 +C(rp2 -rp2)=0. Substituting (jJ1 = A1 sin(pt) and teristic equation

(jJ2

= A2 sin(pt) into Eqs. (1), we obtain the charac-

which has the roots

p] =0, P2

=~c(J] +J2)/(J\J2) .

With allowance for the multiple zero roots, the rotation of the disks is described by the equations

248

2 Vibrations of Systems with Several Degrees of Freedom

2.2 Forced Vibrations 210 The differential equations of motion derived by the force method have the form

y 1 = 8 11 (-m lil +P )-812 m 2i2'Y 2 =821 ( -mlil +P )-822 m 2i2' Substituting YI =AI sin( mt) and Y2 =A2 sin( mt) into these equations, we find the amplitudes AI and A 2:

_

811(1-m2822w2)+8i2m2ul

A I - Po (1-8 m w 2 )( 1-8 m w 2) -82 m m W4 l 22 2 12 l 2 11

8 21 A 2 =Po (1-811 m w 2 )( 1-822 m w 2) -8212 m m w 4 1 1 2 2

'

'

o

Since 01 = 1I(m2 n ), we have

AI = -8Pol 3 /(9EJ x ), A21 = - 64Pol 3 / ( 63EJx ) , The loads m l and m 2 are subjected to the periodic forces FI =- P - m l d2y/dl and F2 =- P - m2d2y/dt2with amplitudes FlO =0 and F 20 =2PA.!o..2' As is seen from the bending moment diagram shown in Fig, 282, the cross section passing through the load m2 is the most critical section where the bending moment is equal to Mmax =46Po1l21. Therefore, the maximum stress in the beam is CY max

=Mmax/Wx =4Pol822/(381}Vx) ,

211 The differential equations of motion of the system have the form

XI =(Posinwt-ml.i'I)811 +-m 2.i'2 8 12' X2 =-mi'1 8 21- m2:i'2 8 22'

(1)

2.2 Forced Vibrations

249

Fig. 282.

where XI and x2 are the displacements of the loads m l and m2, respectively. For the case shown in Fig. 161a, the coefficients of Eqs. (1) are

8" = 1/(2c )+ / 3/(6EJ

x )

=

2/( 3c ) ,

8 12 =8 21 =8 22 = /3/( 6EJx ) =1/(6c ).

= X =

Substituting XI

10

X

XIO

sin(wt) and x 2 =x20 sin(wt) into Eq. (1), we find the amplitudes

P0 8,,(1-m2822ai)

(1-8"m1ai)(1-822m2w2)-8~2mlm2w4' POmlw2821811 (1-811m1w2)(1-822m2w2)-8~2m1m2w4'

=~------~~~~~~~---------

20

Hence, The load m2 is subjected to the force P IO =2c(x lO - x20) =2.31 Po. The coefficient of dynamic force transfer is kdyn =P/P2 =2.31. For the case shown in Fig. 161b, kdyn = 4.5.

212 We resolve the inertial force Fo into the components Fy =Fo sin(wt) and

F, =Fo cos(wt), perpendicular and parallel to the beam axis, respectively. A posi-

2 Vibrations of Systems with Several Degrees of Freedom

250

tion of the motor center of gravity is determined by the vertical displacement y and the rotation angle rp. The force method yields the equations

)8 cp=(Fy -my" )8

y =(Fy -my"

11

21

+(FzR -JijJ)812 ,

(1)

+(FzR -JijJ)822 ,

where J =Jo + mR2 is the moment of inertia of the motor with respect to its base. The coefficients of Eqs. (1) are found by the Vereshchagin's rule:

8 11 = 4/ 3/(9EJ x )' 8 12 = 8 21 = 2/ 3/(9EJ x ) , 8 22 = 13/( 3EJx ) . Setting Fy =F, =0 and substituting y =c I sin(pt) and rp =c2 sin(pt) into Eq. (1), we find the frequencies of free vibrations:

PI =1.43~EJx/(mI3) ,P2 =3.96~EJx/(mI3) . The solution to Eqs. (1) is sought in the form

y =AIsinwt+A 2coswt, cp=BIsinwt+B 2coswt .

(2)

Substituting expressions (2) into Eqs. (1) and taking the relation between the centrifugal force components into account, we arrive at the following equations in the constants AI' A2, R I, and R2:

(l-mw2811)AI-Iw2812BI =F o8 1p (l-mw 28 11 )A 2-Iw2812B2 =FoR 8 12 , (1-Iw2822)BI-mw2812AI = Fo8 12 , (1-Iw 28 22 )B 2-mw 28 12A2 =FoR822 • Solving this system, we find

Al = (Fo/ A)[ (1-Jw 28 22 )811 +Jw 28;2 ] = -104A o/227 , A2 = (Fo/ A)[ (1-mw 2 8 11 )812 +mw2812822] = -72A o/227 , BI = (FoR / A)[ (1-Jw 28 22

)8

12

+Jw 28 12 8 22 ]= -18Bo/227,

B2 = (FoR / A)[ (1-mw 28 11 )822 +mw 28;2] =5B o/227 , where

A=(1-mw2811)(1-Jw2822)-mJw48i2 . As follows from expressions (2), the amplitudes of forced linear and angular vibrations are, respectively,

2.2 Forced Vibrations

y

0

251

=~A,2 +A; =0.467Fo/ 3/(EJ x )'

f/Jo =~B,2 +B22

=0.319Fol2 /(EJ x ).

213 Using the force method, we arrive at the equations

x, = -tJl1 m,x', +tJ'2( -m 2i"2 + p),

(1)

X2' = -tJ2,m,i", +tJ22 (-m 2i"2 + p).

Substituting x,

=A sine 0Jt) and x =B sine 0Jt) into Eqs. (1), we have 2

(l-tJ l1 m,ai)A -tJ12 m 2oiB =POtJ'2' -tJ2,mlo 2A +(1-tJ22 m2(02)B =POtJ22 · Hence, the vibration amplitude for the mass m2 is

_ tJ22 ( I-tJ l1 m,(02)+m,tJi2(02 B - Po ( I-tJ l1 m,(O 2)( I-tJ 22 m2(O 2) -m,m 2tJ12 tJ2,(O4 ' where

The amplitude B is equal to zero if

tJ22(1-tJl1m,(02)+m,tJi2(02 =0. Therefore,

(02=tJ22/[m,(tJl1tJ22-tJi2)] .

(2)

The left hand side of Eq. (2) is the partial frequency for the mass m" i.e., the vibration frequency for the mass m, provided that the mass m2 is fixed.

214 The vibration amplitude for the mass m2 vanishes if

(02 = tJ22 /[ m,( tJl1 tJ22 -tJi2)] ' where

215 The equations of motion derived by the force method have the form

y, =-tJl1 m,Y", -tJ12 m2i2 +tJ13 M ,

(1)

y 2 =-tJ2,m,i,-tJ22 m2i2 +tJ23 M . Since y, must be equal to zero, Eqs. (1) are reduced to

tJ'2 m2Y"2 =tJ13 M , Y 2 =-tJ22 m 2i2 +tJ23 M Substituting Y2

=A sine 0Jt) into Eqs. (2), we arrive at the equations

(2)

252

2 Vibrations of Systems with Several Degrees of Freedom

-alol2 m2A =013M , (1-022 m2m2 )A =023M , which give the condition of zero vibration amplitude for the mass m l :

m2=013/[m2(022013-023012)] . 216 A position of the system is shown in Fig. 283, with the mast substituted by a rod. The coordinates of the mass m l can be expressed in terms of the angle rp and the displacement of the mass m2 :

Xl =x 2 +1 I sincp, Yl =ll(l-coscp) . To set up the equations of small vibrations of the masses m l and m2, we use the Lagrangian method. Writing out the kinetic and potential energies of the system,

T=m~~~+m~~~+m~:~, II =mlg (l-coscp)+cx:

/2,

where c =3EJjt is the bending stiffness of the rod, we arrive at the equations of motion

(ml +m 2).i"2 +cx 2+mll/p=Fosinmt , m/ lx· 2+mII121p+mlglICP=O.

(1)

Substituting x 2 =A sin( mt) and rp =B sin( mt) into Eqs. (1), we find the vibration amplitude A for the mass m2 :

A

=(1/ L1)PO(gll-112m2)ml '

n-------+-~~X X

Fig. 283.

2.2 Forced Vibrations

253

where LI is the determinant of the algebraic system corresponding to Eqs. (1). The amplitude A goes to zero as 0/ ~ gill' where (gll/12 is the vibration frequency of the pendulum in the case of the mass m2 fixed. 217 To set up the differential equations of small vibrations of the system, we use the Lagrangian method. The angular velocity of the disk is =q + LIn, where q is the steady-state angular velocity and Lln= rp is an additional angular velocity due to vibrations. Both the vibrating masses m are displaced by the interval x from their dynamic equilibrium positions at the distances 1from the rotation axis. The kinetic and potential energies of the system are

n

T

=J(no+L1n)2 /2+2m [(no +L1nt(t +X)2 +X 2]/2,

t

TI=cqi /2+2c J(xo+x /2, where 1=lo + Xo and Xo is the spring elongation under steady rotation. In this case,

the torque M serves as a generalized force. The differential equations of motion are derived from the Lagrange equations of the second kind:

..

ap

4mnol· M .

rp+ J +2m1 2 + J +2m1 2X = :i" x

oSIDWf

2nolip+(c J /m -tio~ =0.

We seek for the solution to these equations in the form rp =A sin( liJt) and =B sin(liJt). As a result, we arrive at a system of algebraic equations, which has

the solution

Fig. 284.

254

2 Vibrations of Systems with Several Degrees of Freedom

where L1 is the determinant of this system. The damper parameters for which the vibration amplitude vanishes are determined by the relation 0/ =(c/m) - il,/ . Therefore, the damping takes place if the frequency of the disturbing torque is equal to the eigenfrequency of the mass m supported by a spring on the disk rotating at a constant angular velocity .q.

218 A modified design of the Pringle damper is shown in Fig. 284. In this case, the springs have preloads (cIXo) so that the masses m will be motionless relative to the disk until the centrifugal m1o.q.2 is less than the preload, i.e., m1o.q.2 < c l xo'

219 For forced vibrations with a frequency (j) to be damped, the eigenfrequency p of the pendulum on the disk must be equal to (j). In the case under consideration, p = .q(bll)ll2. Therefore, the pendulum parameters must satisfy the condition (j)= .q(bll)ll2. 220 The differential equations of forced vibrations of the system have the form

YI=-81ImlY'I+812(-m2Y'2+P)'

(1)

Y 21 =-82I m l Y'1 +822 (-m 2i2 +p), where ~I = ~2 = ~I = tl(3EJ), ~2 = 23t/(l2EJ), m l = 2m, m2 = m, and P = Po sin(OJt). We assume that GJk = 0.8EJ,. Substituting YI = AI sin(OJt) and Y2 =A2 sin(OJt) into Eqs. (1), we find the forced-vibration amplitudes

Al = l.085Po13 /(EJx

)'

A2 = l.085 Pol 3 /(EJ x )

•

The forces at the mass points are

FI =-mlY'1 =2mA l co 2 sinwt = O.8Posinwt , F2 =P -m 2Y'2 ={Po+A 2co 2m )sinwt =2.94Posinwt . The cross section passing through the clamped end is the most critical section at which the magnitudes of bending and torsional moments are, respectively,

Mb=(F\O+F2o)1

=3. 74Pol ,Mt =F2ol =2.94Pol .

According to the maximum shearing-stress theory, the equivalent stress in the critical cross section is O'eq

=(l/Wx )~M;+Mt2 =4.75Po/Wx '

where W, =m/132 is the modulus of bending resistance for a rod of a circular cross section. As follows from the strength condition O'eq < O'/n" the allowed disturbing-force amplitude is equal to Po ~ 106 N.

221 This problem is solved in much the same way as Problem 210. Since

~I

=~2

and ~2 =2~1 =2l/(GJp ), the steady-state vibration amplitudes for the disks are equal to AI =- ~IMo and A2 = ~#o' The maximum torques acting on the shaft are

=l/(GJ

p)

2.2 Forced Vibrations

255

Mo

4Mo

Mo

Fig. 285.

M\O=Mo+AlalJ l =-3M o' M20=A20iJ2=-Mo' The corresponding torsional moment diagram is shown in Fig. 285. The maximum shearing stress and the safety factor are, respectively, 'max

=M /(O.2d 3) =4Mo/(0.U 3)= 100MPa, max

222 The equations of forced vibrations are similar to the equations of free vibrations considered in Problem 215 except for the right hand side of the fourth equation which now contains the disturbing torque M. Substituting (fJi =Bi sin( wt) into these equations, we obtain

(c -J( 2)B I -cB 2 =0, -CBl +(2c -J( 2)B2 -CB3 =0, -cB 2+(2c -J( 2)B3 -CB4 =0, -CB 32 +(c -2J( 2)B4 =Mo' Solving these equations, we find the following angles of torsion: B, = 0.373 Mrlc, B2 =9.274 Mrlc, B3 =- 0.841 Mrlc, and B4 =- 0.352Mrlc. Hence, the amplitudes of

torsional moments are

Ml2 =c(B l -B 2 )=O.l02Mo , M 23 =c(B 2-B 3)= 1.112Mo ' M34 =C(B3 - B 4)=-1.193M o , with Mmax = IM341.

223 This problem is solved just as Problem 222. The amplitudes of forced vibrations and of torsional moments are, respectively, B, =3.08 Mrlc, B2 = 1.135 Mrlc, B3=- 0.1 Mrlc, B4 = - 0.02 Mrlc, and M'2 = 1.945 Mo' M23 = 1.235 Mo'

2 Vibrations of Systems with Several Degrees of Freedom

256

M34 = - 0.19 Mo' The maximum shearing stress in the shaft is =1.945 MJWp'

'max

= Mm.lWp

224 A deflection of the system from its equilibrium position is shown in Fig. 286a. The elongations of the upper and lower links are, respectively,

,,1l] =R]rp]-R 2rp2' ,,1l2 =-R]rp]-R 2rp2 . Additional stresses in the links can be found from the relations Litl =Liu/fE and Lit2 =Lia;lIE (see Fig. 286b). As follows from the equilibrium condition for the pulleys,

J/p] + R]F (,,10'] - ,,10'2) = 0, J 2q,2 + R2F (,,10'2 - ,,10']) = M , or

where

all =2R/EF 1(J]l) , a]2 =-2R]R2EF /(J]l) , a2]=-2R]R2EF /(J2 l ) , a22 =2R 22EF /(Ji). Assuming

({Jl

=A exp(ipt) and ({J2 =B exp(ipt), we find the eigenfrequencies

J

2EF R2 p] =0, P2 = -(R2 __ I + _ 2 ~187 11 s . 1 J] J 2 The zero root corresponds to small rotational displacements of the pulleys in the same direction, with no additional stresses appearing in the links. The amplitudes of small forced angular vibrations of the pulleys are given by the equations

rp20 = - J 2 [( all -

OJ

2)(a22 - 2)

_ Mo(all rplO- J 2[(all 2)(a22 -OJ

i.e.,

({JIO

= - 0.595 10-4 Mo and

({J2

OJ

-

a]2a2]

],

-OJ 2 ) -OJ

2) -a]2a2] ] ,

= 0.51 10-4 Mo' The stresses in the links are

0'] =0'0+,,10'] =0'0+E(R]rp]-R 2rp2)/l , 0'2 =

0'0 +,,10'2 = 0'0 +E (R 2rp2

-R]rpl)/l ,

or 0; = 0'0 - 0.0269 Mo sin(mt) and u2 = 0'0 + 0.0269 Mo sin(mt). The stress in one of the links is equal to zero if Mo =73.5 N m.

225 The differential equations of small forced vibrations have the form (see Problem 224)

2.2 Forced Vibrations

257

a Fig. 286.

iP1 + R12 Fa3(jJJJ 1-R 1R 2Fa3(jJ2 / J 1= 0, iP2 -R1R2Fa/P1/J2+R/Fa3(jJ)J2 =Mosinwt.

Substituting CfJ1 = CfJiO sine rot) and CfJ2 = CfJ20 sine 0Jt) into these equations, we find

MoR1R2Fa3

(jJ1O = J 1J 2OJ 2[ OJ 2- (R12/J 1+ R22/)] J 2 Fa3 , _

Mo(0J2-R12Fa)J1)

(jJ 20 - J 20J 2[OJ 2- (R12/ J 1+ R22/)] J 2 Fa3

.

The stress variations are related to the link elongations by the equations A _ Lll1 _ R (jJ1O - R 2(jJ20 . LJO"] ---- 1 smOJt, a1 a1

A _Ll/2 _R (jJ20- R 1(jJ1O . LJ0"2 ---- 2 smOJt. a2 ~

Therefore, the stresses in the links depend on time as

0"1 = 0"10 + LlO"IO sinOJt , 0"2 = 0"20 + LlO"20 sinOJt 226 The differential equations of motion of flywheels 2 and 3 have the form

J 1iP1 +c 1(jJ1 - M f

=M 0 sinOJt ,

(1)

J 2iP2 +C 2(jJ2 +M f =0. Here, CfJ1 and CfJ2 are the rotation angles; C1CfJI and C2 CfJ2 are the elastic forces; and Mf is the moment of friction forces between the flywheels, which is related to the pressure Pn by the equation

258

2 Vibrations of Systems with Several Degrees of Freedom

= J,UPnfXlF

Mf

.

F

Here, F is the contact surface area, Pn = N/F, P is the radial coordinate of a surface element dF, N =No + &/ is the total pressing force, No = t5r1 ~ I is the preload of the spring, and &/ is a dynamic contribution to the pressing force. The shaft rigidity is c I = GJJII = mlI4G/(32IJ The pressure P. varies periodically with f/Jr We now find a relation between f/J2 and &/. Since the spring length determined by the preliminary contraction Do is independent of the angle f/J2' the displacement equations can be written out as

8 1P + LiN8" = 0, 8 2P + LiN 8 21 = ffJ2

.

(2)

Here, ~p and ~ I are the spring contractions due to the external torque M and the unit axial force, respectively; and ~p and ~I are the corresponding angular twists in the case of free spring. According to the Mohr method,

8. = 9

M.(I)M. (I)

J

s

I

GJ p

j

ds +

M.(b)M. (b)

J

I

s

EJ x

j

ds.

Taking into account that Mp(l) = M sin a, Mp(b) = M cos a, M 2(t) = sin a, M 2(b) = cos a, MI(I) = (D/2) cosa, M2(b) = (D/2) sina and that the spring wire length is s =TtDi/cos a, where i is the number of coils, we find

8= IP

trD 2 Mi

4

(_1_+_1_) sin2a GJ p

EJ x

cosa

,

8" = trD 3 i (cos 2 a + sin 2 a)_l_, 4 GJ p EJ x cosa s:

__n

U 2P -IUJ

8= 21

.(Sin

MI

trD 2i

4

a + cos 2 a)_I_ , GJ p EJ x cosa 2

(_1_+_1_) sin2a GJp

EJ x

cosa·

Therefore,

8 1P

M(EJ x + GJ p )sin 2a

I1N = - 8 11 = - D(EJ x cos 2 a + GJ p sin 2 a )'

2.2 Forced Vibrations

6 1P 611

s:

rp2 = U 2P

s:

- - U 21

=

259

JrDMi [ . 2 2 EJ sm a + GJ p cos aEJxGJ p X

] 1 (EJ x + GJpy sin 2 a 4(EJx cos 2 a + GJ p sin 2 a) cos a' In the case of small a(sina~ a, cosa~ 1, and sin2a~ 0),

&I _ M (EJx +GJ p)2a

-

DEJ

x

_ JrDMi ,rp2 - EJ x

The frictional torque is D)/2

Mf = LIPn,udF =PnJi f2Jrp2dP=JrJiPn(DI3 -dn/12 , d)/2

or

Mf

= Ji(N 0 +&1 )(D I3 -dn/[ 3(D12 -dn]

Taking into account that the torque Mj is always opposite in direction to the relative angular velocity of the flywheels, we arrive at the equations of motion

i(

)

.

.. . ( . .) D 13 - d J1rpl +C1rpl-Slgn rpl-rp2 2 2 N O+n 1rp2 =Mosmmt, DI -d 2

iz (N O+n 1rp2 )=0 , .. . (. . )JiD/-d J 2rpZ+C z 2rp2- S1 gnrpl-rp2 3 DI -d 2 where

227 The differential equations of motion of the flywheels have the form

J1iPi +C Irpl + a 2 ( tPl - tP2) = M 0sinmt ,

(1)

J 2 iP2 +C zrpz - a 1(tPl -tPz) =0, where c) = nd)4G/(321) and c2= d)4E/(64Di). From the second equation of system (1), we find

(2) Differentiating the first equation of system (1) with respect to time and substituting Eq. (2) into the result, we arrive at the equation

JIJ2~z/ a l +(J 1+JZ )iP2 +(J 1C2+JzC})iPz/a} +(c} +C 2)tP2 + +C}C2rp2/a} =Momcosmt. The solution to Eq. (3) is sought in the form

(3)

260

2 Vibrations of Systems with Several Degrees of Freedom

(4) The substitution of solution (4) into Eq. (3) yields the system of equations in A2 andB2 :

[J/ 2m4 -VIC2 +J2c l )al +C IC2]A 2/a l -

-[VI +J2)m3 +(C I+c 2)W]B2 =0, "[J/ 2W4 -(J IC2+J2c )m 2+C IC2]B 2/a -[VI +J2)W 3-(C I+c 2)w]A2 =Mow. l

l -

The maximum angular deflection of flywheel 3 from its equilibrium position is

lP20=~A22+B; . Using the solution found in Problem 226, we determine the internal forces in the spring:

M=

EJx N = 8 0 + M(EJ x +KJp )2a = lP20 nDi ' 8 11 DEJx

=8 4GJp + (EJ x +GJp )2a o nD 3i lP20 nD 2i Hence, the largest stresses in the spring are O"max

=M jwx

T

=8ND=8~+

max

n:d 3

=

ElP20/(2;rr) , 0

nD 2i

lP20

(E+2G)da nDi .

As follows from Eq. (2),

lPI =_1_ fV2~2 +a l
al

where

C2)B2 A 1= (J 2W+-+ A 2' BI =B 2- ( J 2C w+-2)A2 - . W al W al The amplitude of angular vibrations of flywheel 2 and the largest stress in the shaft are, respectively,

lPlO = ~r-A-12-+-B-I-2 , Tmax = 32c I lPlO / ( n:d

n.

228 Using the d' Alembert principle, we arrive at the following differential equations of motion for the disk and ring, respectively:

2.2 Forced Vibrations

We seek the solution to these equations in the form

rpl = A sin OJt + B cosOJ! , rp21 = C sin OJ! + D cosOJ! , where the constants A, B, C, and D are found from the system of equations

(c -J IOJ 2 )A -aOJ(B -D )=Mo' aOJ(A -C )+(c -J IOJ 2 )B =0, aOJ(B +D )+J2OJ 2C =0, aOJ(A +C )+J2OJ 2D =0. Hence,

A =Mo[ (c -J l ol)(J;0J 4+a2OJ2)-J2a2OJ4]/Lt, B

=M oaf; OJ5 / Lt ,

C =MO[(C-JIOJ2)a2OJ2_J2a2OJ4]/Lt,

D =MOJ2aOJ3(c-JIOJ2)/Lt, where

c -JI OJ2 Lt=

Fig. 287.

aOJ 0 aOJ

-aOJ c -JI OJ2 aOJ 0

aOJ -aOJ 0 J 2OJ2 aOJ aOJ J 2OJ2 0

261

262

2 Vibrations of Systems with Several Degrees of Freedom

The amplitudes of stable vibrations of the disk and ring are determined by the equations, respectively, rplO

=~A 2 +B 2 ,rp20 =~C 2 +D 2

•

229 The system under consideration is shown schematically in Fig. 287. The kinetic and potential energies and the Rayleigh dissipative function are given by the equations 2 J' 2 ·2 T =mv +~+ mlYI

2

2

2'

II= c(lrp-YIY + cI(YI-hY 2 2

R = a(lip- YIY

2

Substituting these expressions into the Lagrange equations, we have

Jolp+al (lip- YI)+cl (lrp- Y 1)= 0,

mlil-a(lip- YI)-c(lrp- Y I)+CI(Y I-h )=0, or

.. 2' 2 2n. pg 0 rp+ nrp+ Porp--l-Y I--l-Y I = , il

+2n ly I +PI20Y I +P;oY 1- 2n l1ip- P1201rp= p;oh,

where 2n = atIJ2, 2n, = aim" P02 = ctIJo' pIQ2 = elm" and P202 = c/m,.In the case of a= 0, the eigenfrequencies and the critical towing velocities are, respectively, 2 2 2 P 1,2 = PO+PIO+P20+ 2 -

2 2 J2 PO+PIO+P20 _p2p2

(2

2

0

20

,

230 The strain energy of the car springs is

II=i(y +rp/I-hIY + C;(y -rp/2- h2)2 Here, h, and h2 are the vertical displacements of the front and back wheels, respectively:

hI = ho[1- cos( 2mt / I0)]' h2 = ho {1- cos[ 21Z"(vt +L)/1 0]} , where L = I, + 12, The kinetic energy of the system is

T =my 2 /2+Jip2 /2+mv

2/2 .

The differential equations of small vibrations follow from these expressions:

2.2 Forced Vibrations

263

mji +(C I +C 2)Y +(CI/I-C2/2)lP=(clhl +c 2h2), JiP+(cl/ l +c 2/ 2)y

+(c

2 I/ 1

+c 2/;)lP=(c l/ lhl +c 2/ 2h2).

It is convenient to rewrite these equations in the form

1" +ally +aI2 lP=b l -b2COS( mt )+b3 sin( cot), iP+a2l y +a22 lP=b l'-b 2'COS( mt )+b 3 'sin( mt),

where

OJ

(1)

=21CVllo and

bl =(C I+c 2)h o/m, b2 =[C I+c 2cos(2ni.,jlo)]ho/m,

b3=sin(27rL/l o)hoc 2/m, bl '=(cl/ l +c 2/ 2)ho/J, b2' =[c III +C 2/2 cos( 2JZ"L/ I 0) ]ho / J ,b3' =c 2/2 hO sin( 2JZ"L/ 10 ) /J. We seek for the solution to Eq. (1) in the form

=Y 0 + A cos( mt ) + B sin( mt ) , lP = lPo + A I cos( cot) + B I sin( mt ) ,

Y where

Yo = (b la22 -b l'a12 B

(2)

)j~pA =[aI2bl'-b2(~2 _(02)]/~,

=[b3(~2 -co2)-aI2b3']I~,

lPo =(allbl'-~Ibl)/~'

Al =[~lb2 -b2'(all -m 2)]1~,BI

=[b3'(all-co2)-~lb3]1~p

~I =all~2 -aI2~1'~ =(all-m2)(~2 -(2)-aI2~1. Rewriting Eq. (2) in the form

y

= y 0 +~A 2 + B 2 sin( cot + 13), lP =lPo +~AI2 + B/ sin( cot + 13) ,

we find the amplitudes of vertical and angular accelerations: ..

2

tA2 + B 2 , lPmax ·· = co 2 "VtA2I + B2I

Y max = CO 'V

Substituting the numerical values, we find

1" max =7.5 mls

2

and

.

iP max

=0.2 rad/s 2 •

231 Since the vibration damping in the system under consideration is disregarded (see Problem 230), the critical velocities of travel correspond to resonances, i.e., V*I = IrPl2n and V*2 = IrPf2n, where PI and P2are the eigenfrequencies. 232 If the car considered in Problem 198 moves on a rough road, its tires have additional contractions hI and hr Consequently, the expression for the potential energy of the car coincides with that found in Problem 198 except for the two last terms which must be substituted by C(YI - h)2/2 + 2C(y2 - h2)2/2, where

264

2 Vibrations of Systems with Several Degrees of Freedom

hi

~ hO( I-cos 2~t). h, ~ hO(1- cos 2n(vt ~ a +h )J

The differential equations of motion coincide with those derived in Problem 198 except for the two last equations whose right hand sides should now take the form ch1 and 2ch2• respectively.

233 The differential equations of motion for the total mass m =mo + mJl - tit) of the vessel and liquid are

mi" +cx =ce coswt , mji +cy =ce sinwt . where e is the deflection of the mass m from its equilibrium position. Denoting the total mass at t =0 by min =mo + m1 and introducing the dimensionless time ,,= t/t1• we rewrite these equations in the form

d 2x + ct l2x _ ectI2coS(wtl ,) d,2 min -ml' min -ml'

234 The differential equations of motion were derived in Problem 178:

Fig. 288.

Fig. 289.

2.2 Forced Vibrations

265

u· + liP/4+c(u -lrp)/m =0, u· +SliP/4-4c(u -lrp)/m =0. Under the initial conditions given, the solution to this system has the form

U =20Jt/29m +9Jsin(P2t)/29mp2' rp=20Jt/29m 1-20J Sin(P2t )/29m/P2' where P2 = (29c/5myl2. The motion of the center of gravity is described by the equation

Uo=U +rpl

/4 =2SJt/29m +4J Sin(P2t )/29mp2'

The inertial forces acting upon the load are

F =-mu'O=4JP2sin(P2t)/29, M =-JiP=-SJ/P2 Sin(P2t )/29. The bending moment is maximal in cross section A of the rod (Fig. 288):

M max =M +Fl /4=4J/P2/29=0.643J ~EJx /(ml). 235 The differential equations of motion of the flywheel and ring are

JiPl +a 1(tPl-tP2)+C 2rpl =Mosinmt, a 1(tP2 -tPl)+C 1rp2 =0, where qJl and qJ2 are the rotation angles. Solutions to these equations can be presented in the form

rpl =Aj sinmt + B j cosmt , rp2 =A2 sinmt + B2 cosmt Substituting these expressions into the equations, we find

A2 =(C) +c 2-J(2)MojLI ,B 2=-M oJox l/(Lla)) , where L1 = (c l + c2 - lai)2 + (lax/af The amplitude of stable vibrations is found from the equation qJ202 =(A 22 + B22)112. 236 The differential equations of small vertical vibrations of the container and guide have the form

mX')+2a(x)-x 2)+C)X 1 = Fosinmt , -a(x j -x 2)+C 2X2 =0, where XI and x 2 are the vertical displacements of the container and the guide. These equations are solved in much the same way as those in Problem 235. Under steady-state vibrations,

Xl(t) =A1sinlUt + B)cosmt , X2(t) =A 2 sinmt + B 2cosmt Substituting these expressions into the equations, we find

266

2 Vibrations of Systems with Several Degrees of Freedom

237 The driver's seat position at a moment of time is shown in Fig. 289. The differential equation of motion of the mass m is

mY"2+ a (Y2-Yt)+(Y2-Yt)=0. One more equation follows from the equilibrium condition for the wheel suspension:

These equations can be written out in the form

Y"2 +2nj 2 +P;Y 2 -2nj t - P;Y t =0, 2

·

22'

2

2

(1) 2

ny2+POY2- nyl-POYI-PIYt=-PtYo'

where 2n = alm, p02 =elm, andp,2 =elm. In order to find the transfer functions, we take the Laplace transform of Eqs. (1) under zero initial conditions. As a result,

(P2 +2np + P;'2(P )-(2np + p;)Y1(p )=0, (2np +P;'2(P )-(2np +P;Pt2't(p )=-Pt2yo(p). Consequently, the transfer function relating a perturbation (input) to the mass displacement (output) is

W P _

( )-

PI2(2np+p;+pn

2np3+(p~+pnp2+2npI2p+p~+P~PI2

The spectral density of the acceleration is given by the equation

s Y2. =ois, Y2 S Y2 (01) =

fv (i01t s Yo (01)

is the spectral density of the vertical displacement, which is related to the correlation function given: 00

Syo= JKYo(r)exp(-iwt)dr= -00

=DyO

00

J{exp[-r(a t -i01)]+exp[-r(a 1 +i01)]}dr=

o The expression for S,. can be written out in the form

2D a 2 Yo ;. at +01

2.3 Critical States and Vibration Stability

where

267

t

L1=12n(imf +(2na l + P; + pn(imr +[ al(p; + pn+ 2npn( im + +[PI2(p; +2na l)+ p;]im+al(p; + p;pnr The variance of vertical acceleration

1 D . =Y2 21r

JS . dm

00

Y2

-00

where

2 a4 =a I (4 2 I2) ,ao=2n, al = 2 na I + P02+ PI' Po + PoP a2=al(p; + pn+ 2npl2 ,a3 = PI2(p; +2na l)+ P;

.

2.3 Critical States and Vibration Stability 238 The center of gravity of the disk at an arbitrary instant t is specified by the vector r (Fig. 290). To describe the disk motion, we use the Lagrange equations of the second kind. The kinetic energy of the disk is

T =m(v;+v:)j2+Jooi/2,

(1)

where Vx and Vy are the x- and y-components of the velocity of the disk center and 10 is the moment of inertia with respect to the axis perpendicular to the disk plane and passing through the center (point OJ The velocity of the center of gravity is specified by the vector v:

v =i i

+j y +[ mx r ], [m x r ] =

i

j

k

0

0

m

x

y

0

Here, i, j, and k are the unit coordinate vectors. Hence, the velocity components can be presented as the scalar products:

268

2 Vibrations of Systems with Several Degrees of Freedom

vx

=(v .j )=i -U?)!

,v y

=(v -j)= Y +M

The bending-strain energy of the shaft is

II =cr 2/2 =c (X 2+ Y

(2)

2)/2 ,

(3)

where c =6EJIt is the stiffness of the shaft. Substituting Eqs. (1)-(3) into the Lagrange equations

dOL OL dOL OL ----=0 ----=0 dtOi at 'dtry iY

'

with L = T - 17, we arrive at the equations of small vibrations of the disk relative to the rotating coordinate system:

i"+(p;-m2)x -2i1?Y =0, , () Y + Po - ()) y + 2M = 0 , ••

2

2

.

where P02 = elm. Taking x = A exp(ipt) and y = B exp(ipt), we obtain the characteristic equation

(Po2-m 2) -P 2 2mip

-2mip (2 2 2)2 2 2 (Po2-m 2) -P 2= Po -()) -P -4(0 P =0.

Hence, the two eigenfrequencies are PI = Po -

OJ

and P2 = Po +

OJ.

239 In the rotating coordinate system yO 'x, the disk center of gravity (point 0) is specified by the vector r (Fig. 291): r =r l + e, where r l is the displacement of the shaft axis. The relative position of points 0' and 0 1 ' shown in Fig. 291 corresponds to the case of OJ < OJ, (see Problem 93), when the steady-state equilibrium condition is satisfied: (1)

y

x

x

Fig. 290.

2.3 Critical States and Vibration Stability

y

269

x

x

Fig. 291.

In the case of m> m., Eq. (1) takes the form mm\ = c(xo+e). If the motion of the disk is stationary, the vectors r and r I are collinear since the points 0", 0, and 0 1 are on the straight line O"x. The angle rp determines small deflections of the vector e from the 0" x axis, while the coordinates x and y of the disk center of gravity determine its small displacements relative to the rotating coordinate system. The kinetic energy of the system and the bending-strain energy of the shaft are

T

=m(v; +v n/2+Jo(m+ipt

II =cr 2/2=C[(X where V x

/2 ,

-e COSqJ)2 +(Y -e SinqJt]/2

=i - C0l , v y = y + cux , (see Problem 238) and 10 is the moment of

inertia with respect to the point o. In the moving coordinate system xO'y, the disk center of gravity (point 0) is specified by the three independent variables x, y, and rp. Consequently, substituting these expressions into the Lagrange equations, we arrive at the three equations

m(X- -co/ )-mm(y +cux )+c(x -e COSqJ)=O,

(2)

m(Y- +aIi )+mm(i -C0l )+C(Y -e sinqJ) =0, JofiJ+n(x -e cosqJ)e sinqJ-C(Y -e sinqJ)e cosqJ= O. Since the vibrations under consideration are assumed as small, we can take x =Xo + XI' y =Yl' cosrp ~ 1, sinrp ~ rp, and XI rp ~ YI rp ~ O. In this case, we suppose that the X axis is in the direction of the straight line passing through the points 0", 0, and 0 1• Eliminating from Eqs. (2) the equations of equilibrium in the steadystate regime, we obtain

270

2 Vibrations of Systems with Several Degrees of Freedom

i"1

+(p; -al)x 1- 2mi 1 =0,

ji 1 +(p; -al)y 1 +2Cl?Y 1 - p;erp=O,

°

•• 2 2 rp+PIXOrp-PIYI= ,

where P02 =elm and PI 2 =cello. It should be noted that these equations are valid both for 0) < 0). and for 0) > 0),. 240 The bending stiffnesses of the shaft in the directions of the x- and y-axes (principal axes of inertia) are denoted by Cx and cy" The kinetic energy of the system .has the same form as that in Problem 238, and the potential energy is

II=c xx 2/2+c y y 2/2. In the rotating coordinate system, the differential equations of small vibrations have the form

i" +(k} -ai)x -2Cl?Y =0,

ji +(k: _m2)y +2mi =0, where kx2 =clm and ky2 =clm. Taking x =A exp(ipt) and y =B exp(ipt), we obtain the equation

P 4 _ P 2 (kx2 +k: +2m2)+(kx2 -m2)(k: _m2)=

°,

whence,

k 2+k2 +2m 2 (k2 _k2)2 +,1 x y +2m2(k2+k2) P 1,2 =,1V x y2 V 4 x y The motion of the disk is stable provided that the frequencies PI and P2 are realvalued. If PI •2 are complex conjugate numbers, i.e., PI ,2 =± iA, the solution is proportional to the function exp(A,t) which increases infinitely with time. In this case, the vibration amplitude for a certain angular velocity v. should grows infinitely, and, therefore, small vibrations become unstable. Such an angular velocity is referred to as a critical velocity. Equation (2) have complex conjugate roots if

r--------------------

kx2+k2y +2m2 < (k2x _k2)2 +8m2(k2x +ky2) y Whence, kx ~ 0). ~ kyo It follows from these inequalities that there is a range of critical angular velocities. 241 As in the case of Problem 238, it is convenient to solve the problem in the rotating coordinate system. However, the extra term ~ = mgY I should be added to the expression for the strain energy considered above, with YI being the coordinate of the center of gravity in the fixed coordinate system XPIYI (Fig. 292). In the rotating coordinate system xOly,

2.3 Critical States and Vibration Stability

271

Fig. 292.

III = mg (x sinwt + y coswt) . The total potential energy is

Il=c(x2+y 2)/2+mg(x sinwt+y coswt) , where c = 2End'/4t The kinetic energy is given by the same expression as in Problem 238. The differential equations of motion have the form

.i"+(k2-m2~ -2ll?Y +gsinwt=O,

(1)

ji +(k2 _m2)y +2mi + g coswt =0, where k = elm. The solution to Eqs. (1) is sought in the form x = A sine lOt) and y = B cos( lOt). Substituting these expressions into Eqs. (1), we arrive at an inhomogeneous system of equations in A and B, which has the solution A =B =-glf. The critical angular velocity is found from the solvability condition for the homogeneous system: m. = kIl. 2

242 The differential equations of motion relative to the rotating coordinate system (see Problem 241) are

.i" +( kx 2 -al)x -2£0/ = -g sinwt ,

ji

+( ky

2

_m2)y +2mi = g coswt ,

where kx2 =4Ebh3/(mP) and k/ =4Ebh3/(mP). Substituting x =A sine lOt) and y =B cos( lOt) into these equations, we find the amplitudes of steady-state vibrations:

The critical angular velocity of the roller is

272

2 Vibrations of Systems with Several Degrees of Freedom

01=

k}k:/[2(k +k:)]. 2

x

In addition to this value, there is the range of critical angular velocities found in Problem 240. 243 When the bent roller rotates about the AB axis, the inertial forces acting upon the disk are perpendicular to its axis (Fig. 293). In this case, the total force of inertia acting upon an element dm is dF dmolr,. Because r, r + y, the force dF can be presented as a sum: dF =dF, + dF2, where dF, =dmo/y and dF2 = dmolr. The force dF, is directed along the y axis. The moment Mx of the force dF2 with respect to the x axis impedes the disk's rotation. Since the rotation angle about the x axis is small,

=

=

m

where J is the moment of inertia about the rotation axis. Therefore, the roller is subjected to the force F, and moment Mx' The displacement equations for the attachment point of the disk are found by the force method:

(I-Ollm01 2)y +012 012Jrp=O, 021m012y -(I+02201 2J)rp=O,

A

Fig. 293.

(1)

2.3 Critical States and Vibration Stability

273

where ~! = tl(3EJ), ~2 = ~! = - tl(2EJ), ~2 = l/(EJ), and Jx = ntfl64 is the moment of inertia of the roller cross section. The equation for the critical angular velocity follows from the solvability condition for system (1):

co4+ 12EJx (m12 Jml 3 3

_JJC02_~(EJx)2 =0. Jm

(2)

12

Whence,

+~(EJx )2

2 co 2 = 6EJ x (J _ m1 3J+ [ 6EJ x (J _ m1 J]2 Jm13 3 Jm Jm13 3

(3)

12

The second root of Eq. (2) corresponds to imaginary values of the angular velocity. If the moment of inertia of the disk can be disregarded, i.e., the disk is treated as a point mass, the critical angular velocity is given by Eq. (2) with J ~ 0:

co 2=~1/(m811) =~3EJx /(mI 3) .

(4)

The numerical values of lO. evaluated by formulas (3) and (4) are equal to 165 and 130 lis, respectively. Thus, the gyroscopic effect results in increasing the critical angular velocity.

244 The critical angular velocity lO. evaluated with and without allowance for the gyroscopic effect is approximately equal to 138 and 127.8 lis, respectively. 245 The critical angular velocity lO. evaluated with allowance for the gyroscopic effect is approximately equal to 173 lis. If this effect is disregarded, lO. ~ 158.6 1/s. 246 The forces acting on the disk are shown in Fig. 294, where r! is the roller deflection, e is the eccentricity of the disk center of mass, Fe! =-crt is the elastic force acting on the disk, and M m is the total attractive force due to the magnets. According to the d' Alembert principle, the differential equations of motion in the case of constant angular velocity of the disk have the form

:i" + p 2X =ep 2 cosOJ{ + L1Fm m Y" + P 2Y =ep 2 sin OJ{ ,

,

(1)

where l = elm = 6EJ/mt The total attractive force is

L1Fm =k(/J~/(a-x

t -k(/J~/(a+x t·

Allowing for only the linear term in the expansion of this expression in powers of

x,

274

2 Vibrations of Systems with Several Degrees of Freedom

a

y

x

a

x

Fig. 294.

we write out the fIrst equation of system (1) in the form

X'+[p2-4ktPV(ma 3)]x =p 2eCOSOJI

(2)

It follows from Eqs. (1) and (2) that in the magnetic fIeld, the system has the two critical angular velocities lU. l

=~r-p-2-_-4-k-tP-~-/-'(-m-a-3-), lU. 2 =P .

247 The plate has three degrees of freedom: the vertical displacement z and the two angular displacements ({J and (). If the angle () of attack is nonzero, an aerodynamic lifting force appears. In this case, vertical displacements of the plate cause elastic forces of the springs, with the net force

F

=-Cl(Z -(1 /2 +rph /2)-c l(z -(1 /2-rph /2)

-c 2(z +(1 /2 + rph /2 )-c 2(z +(1 /2-rph /2). The moments of the elastic forces with respect to the x and y axes are

Mx

=-(C l+C 2)rph2 ,

My =2c l(z -(1 /2)/-2c 2(z +(1/2)1 . The differential equations of motion of the plate have the form

mi" +2(c l +c 2)Z +(C I -C 2 )(1- dc z pv 2 hIO=O, dO 2 2 J y B+.!.(c l +C 2)/ 20+(C 2-c/z _ dc z hl pv 20 = 0, 2 dO 8 )

1

(1) (2)

2.3 Critical States and Vibration Stability

.. h 2 ( c +c ) (jJ= 0. Jx(jJ+1 2

275

(3)

4

Equation (3) is independent of Eqs. (1) and (2) and allows us to find the frequency of angular vibrations about the x axis:

p

=h~(Cl +c 2 )/(4Jx )

•

In order to find the remaining two frequencies, we substitute z =A exp(ipt) and B exp(ipt) into Eqs. (1) and (2). As a result, we arrive at a homogeneous system of algebraic equations with the determinant ()=

2(c 1 +C 2 ) +P 2

(C 1 -c 2

1--'--'-----'-'-

D-

)l - 2FIO

- -(Cm1-c )l 2

where FlO =(dc/d9)hlplI4. The characteristic equation follows from the condition

D=O:

P 4 +~p 2 +a4 =0 where

(4)

_ 2(c 1 +C 2) + 2(c 1 +CJl2 -FlOl

~-

_[2(c

a4 -

1

'

4J y

m

+c 2)l-FIO ](C 1 +c 2)l 2mJ y

[(CI-C2)l-2FIO](CI-C2)l

+~----~----~~--~-

mJ y

For the roots of characteristic equation (4) to have negative real parts, its coefficients must be positive (for a biquadratic equation, this condition is necessary and sufficient):

l

-FIO --+

4J y

2(c +c 2) (C 1

m

+

1

+c 2)l2 2J y

>0,

Under these conditions, vibrations of the plate are stable.

248 Vibrations of the plate are stable provided that the real parts of the roots of characteristic equation (4) derived in Problem 246 are negative. For this condition to be met, the coefficients a, and a2 must be negative. The values of the aerodynamic force F for which the coefficients a, and a2 are equal to zero are critical; whence, the critical velocities v are determined by the

276

2 Vibrations of Systems with Several Degrees of Freedom

conditions a, = 0 and a2 = O. Substituting the numerical values into these conditions, we find v., =32.6 mls and v'2 =73 mls.

249 If the level flight of the aircraft is perturbed weakly, the equations of motion derived by the d' Alembert method take the following form (Fig. 295):

F j +L1FI +L1F2 +Rrp=O,

(1)

M j +AM I -L1F21 =0,

(2)

where Fj and Mj are, respectively, the inertial force and the moment of inertia with respect to the axis perpendicular to the plane of the figure and passing through the point 0'. We now find the increment of the angles of attack, Lla, and LI~: Lla, = rpdylvdt, where dylvdt is the angle of downwash. For small rp and y, the vertical displacement of the tail unit is y, =y - rpl. Therefore, the increment Lla2 can be assumed as dependent on only the angle of downwash:

Lia 2=C 2(ipl- Y)jv

Thus, the increments LIF" LlF2' and LIM, are

y

1=.,0 x v mg

Fig. 295.

L1FI =CI(rp- y

Iv ), L1F2 =C 2(ipl- y )/v ,

AMI =C 3 (rp-y

Iv).

The substitution of expressions (3) into Eqs. (1)-(2) yields

(3)

2.3 Critical States and Vibration Stability

~v· + C 1 +C2 ~V Y

mv

_ Y

277

C21 ¢J_(CI+R)qJ=O mv m '

.. (C 3 -C 2/) A C 2/2. C 3 0 qJ+ LlV +--qJ--qJ= , Jv Y Jv J where LtVy = dy/dt. Taking LtVy = A exp(AJ) and rp = B exp(At), we arrive at the characteristic equation

A? + alA? +a21+a3 =0, al =[C 21 2m +(C I +C 2)J ]/(Jmv ),

=[

a2 -C 3mv 2+C 2/2(C I+C 2)]/(Jmv

2),

a3=[(C I +R )(C 3-C 2/)-C 3(C I+C 2)]/(Jmv). The necessary condition of stability of the unperturbed motion is aj > 0, while the necessary and sufficient condition (Hurwitz criterion) is a l a2 - a3 > O.

250 The differential equations of motion of the flywheels have the form

(PI +allqJI -a12 qJ2 = PI¢JI , (P2 -a2l qJI +~2qJ2 =-P2¢J2· The characteristic equation for this system is

13+(P2 - PI)12+( all +a22 - PIP2)1+( all P2 -~2Pl) =0 , or

13 +(10- PI)12+(21420-lOPI)1+(51000-16320PI)= 0 . For the motion to be stable, the coefficients of the characteristic equation must be positive, i.e., A < 10, PI < 2142, and PI < 3.1. Therefore, PI < 3.l. The Hurwitz criterion is the sufficient condition of the stability of a motion. For the cubic equation, Lt2 > 0, i.e., (10 - A)(21420 - lOPI) - (51000 - 16320Lt) > 0; therefore, - 00 < PI < 34 and 480 < PI < 00. Comparing these inequalities to the necessary condition, we find the interval of allowed values of PI: - 00 < PI < 3.1.

251 When the flywheel rotates, it slips relative to the slip-ring of the coupling. Denoting the rotation angles of the flywheel and slip-ring by rpl nd rp2' respectively, we write out the differential equation of motion of the flywheel:

J(PI +a l(¢JI-¢J2)=a¢JI ' where

a I ( ¢J I - ¢J2 ) is the frictional torque.

The differential equation of motion of the slip-ring is

a l(¢J2 -¢JI)+C lqJ2 = 0 .

278

2 Vibrations of Systems with Several Degrees of Freedom

Taking 'PI

=A exp(AJ) and 'P2 =B exp(AJ), we arrive at the equation

A+A2(Jcl-aal)/(Jal)+ACI(al-a)/(Jal)=O , 3

(1)

which have the three roots

The motion of the flywheel is stable if (a) A,2 and ~ are negative real numbers, or (b) ~ and A,3 are complex conjugate numbers with negative real parts. In these cases, the perturbed motion ('PI and 'Pz) is either aperiodic or damped vi-

bration, respectively.

z'; z

mg /,

I

x'

Fig. 296.

The positiveness condition for the coefficients of Eq. (1) is the necessary and sufficient condition for the roots to be negative: Je l - aal > 0 and a l - a > O. Substituting the second condition into the first one, we obtain Je l > aal > a2, i.e., a < (C/)Ifl = 10-2 N m s. Because the condition a < a =0.005 N m s should also be met, the motion is stable if a < 0.005 N m s.

2.3 Critical States and Vibration Stability

279

252 The differential equation of rotation about a fixed point has the form

dKldt =M, where K is the principal angular momentum of the top about the point (Fig. 296). In general, K 1m, where 1 is the matrix of the moments of inertia with respect to the body coordinate axes and m is the angular velocity of the top. The matrix J is of the form

=

o

Jx J= -Jyx -Jzx

-Jxy

-Jxz

Jy -Jzy

-JyZ Jz

It is known that the total derivative of a vector K is related to the local derivative taken in a moving coordinate system by the equation

dK/dt=dK'/dt+[{l}'xK] ,

(1)

where the angular velocity m' of the coordinate system differs, in general, from

m* m'.

OJ:

If the local derivative is taken in the body coordinate system, then m =m', and we arrive at the Euler equation d'K/dt+ [()}Xl(] =M. The components of this equation relative to the moving coordinate system (x', y', z'), which differs from the body coordinate system, are (the prime in the local derivative is omitted, see Fig. 296)

dKx,jdt +Kz,{l}y' -K y ,{l}z' = M x"

(2)

dK y' jdt +Kx,{l}z' -K z ,{l}x' =M y"

dKz.fdt +K y ,{l}x' -Kx,{l}y' =Mz" The coordinate system (x', y', z') moves in conjunction with small vibrations of the top but does not rotate with it. The Oz' axis always coincides with the Oz axis of symmetry. Because of this, the angular velocities OJx " OJy " and OJ,. can be assumed as small. If the angular displacements of the symmetry axis of the gyroscope are small, the total angular velocity of the top takes the form m = ~ + m', with (in the smalldeviation approximation)

,,,' - r' ,LU ,,,' X''-- f)' ,LUY'-""· ,,,' - in

LU Z' -

The angular-moment components are

K x' =Jx{l}x' K y' =Jy{l}y, K z =Jz{l}z, where

{l}x :::!O, {l}y :::!ip, {l}z :::!.Qo+y,

principal axes of inertia. For small angles, when x' components

R:

x and y'

M z' :::! 0, My'

:::!

R:

and the Ox', Oy', and Oz' axes are y, the moment of gravity force has the

mgx , , M x'

:::!

mgy , ,

280

2 Vibrations of Systems with Several Degrees of Freedom

where x' =lrp and y' =lB. Because of the axial symmetry, Jx =Jy , and Eqs. (2) are reduced to the linear differential equations

J x B+Jziloip- mglB= 0,

(3)

J x ip-Jz iliJ- mglrp = 0,

mz =0. As follows from the third equation, two equations yields

OJ,

=const =,q. The superposition of the first

y-i rJzilo/Jx -ymgl /J x =0 ,

where tion

r= ()+ irp. Substituting y = C exp(At), we arrive at the characteristic equa-

which has the roots A12

=+,.ao/(2JJ±~[J,.ao/(2J,)]' -mgl /J, ] .

A perturbed motion of the top is stable if (J,Qj2J/ > mgl/Jx ' Therefore, the

Fig. 297.

critical angular velocity is ,q. = (2/J,)(mgUyn. 253 In the case of small deflections of the gyroscope axis from its vertical position, the differential equations of motion have the form (see Problem 252)

2.3 Critical States and Vibration Stability

281

B+Jzilip/Jx =Mx·/Jx , iP-Jzil/}/Jx =My·/Jx . In the case under consideration, there are elastic forces acting on the gyroscope together with the gravity force. A position of a point 0 1 of the gyroscope axis is shown in Fig. 297. Substituting the expressions for the moment components

Mx' ={mg /2-2cl)IB, My. ={mg /2-2cl)lrp, into these equations, we obtain

B+ Jzilip (mg -4cl)IB 0 2Jx ' Jx .. Jzil/J (mg - 4cl )lrp rpO. 2Jx Jx The roots of the corresponding characteristic equation (see Problem 252) are -<1,2

=i

[J,ilo/(2J, )±~[JA/(2J,)]' -I (mg -4cI)/(2J,)] ,

Therefore, the motion of the gyroscope is stable if

Jzilo>~2J){mg -4cl) . 254 The equations of motion of the gyroscope have the form (see Problem 253)

.. J ilo;/) (mg -4c 2 1)W B+ z 'f' =0, Jx 2Jx (mg -4c 2 1)lrp =0. 2Jx Taking B= Al exp(At) and tp = A2 exp(At), we arrive at the characteristic equation

+~( mg -4c l)( mg -4c 1) =..1,4 +a 4Jx

l

2

2 ..1,2

+a4 = O.

The roots of this equation are

..1,1,2 = ~'-Clz-/2-±-~-;:~=i=:=/4=-=a=4

.

If a2 > 0, a4 > 0, and a/ - a4 > 0, these roots can be written as

282

2 Vibrations of Systems with Several Degrees of Freedom

..1. 1,2 = i

~a2/2+~a; /4-a

4

'

i,e., in this case, the motion of the gyroscope axis is periodic. The critical velocity is found from the condition a22 - a4 > 0:

da.= J x ; [~(mg -4cJ)(mg -4c 1)+(mg -2c l- 2c i)] . 2

Jz

l

255 To derive the differential equations of small vibrations we use the Langrange equations of the second kind. The position of the system at an instant is shown in Fig. 298. The Ox, Oy, and Oz axes are the body axes for internal ring 1. The straight line 00' is always in the yz-plane. The kinetic and potential energies of the rotor, internal ring, and countertweight are given by the expressions, respectively,

m I (.]:2 + /2 .) +-2Loi J J T =_ +---.l::.L{02 I 2 <;'1 <;, I 2 Xl 2 Yl III =mlgl(l-cosr).

z Fig. 298.

J 2

+~(O2 Zl'

2.3 Critical States and Vibration Stability

m (.;:2 +/,2 .) +~W2 J J W2 T =_2 +-21.... 2 2 '='2 '='2 2 X2 2 Y2

J

283

+~W2

2

Z2 '

ll2 = m2gi (I-cosr) , T3 = ~3 (~;

+17; +?;),

ll3=m3g[(i +a)-(i +acosB)cosr].

r

Assuming the angles () and as small, we find the angular velocity components of the rotor and the internal ring:

wx1=B, wy1=y, WZl=Qo+yB, WX2 =B, wY2 =y, WZ2 =YBr::!O. The corresponding coordinates of the centers of mass are

;1 =ir , ;2 =ir, ;3 =(i +a)r,

SI=i, s2=i, s3=i+a, rh=aB. Substituting these expressions into the Lagrange equations

d or dt IJB -

ff..r -ll) IJB

~ or _~T -ll)

°, or dt

Or

0,

we arrive at the equations of motion

J/}-JZ1QoY + m3gaB= 0,

(1)

J/j+Jz1QoB+Ar=0, where

JJ =JX1 +JX2 +m3a2 , J 2=JY1 +JY2 +(mJ +m 2)i 2+m 3(i +ay , A =[(mJ +m 2)1 +(1 +a)m3]g . Taking ()= C1 cos(pt) and

r= C

2

°.

sin(pt), we obtain the characteristic equation

p4 _ p2(J2m3ga+J 1A +Jz2/io)/(J/2)+A m3ga/(JJJ2) = The eigenfrequencies PI and P2 are easily found from this equation.

256 The kinetic and potential energies of the gyro rotor and the monorail car are given by the expressions, respectively,

T = Q (j;2 +;2 + i 2)+ J x W2 + J y W2 + Jz W2 2g \'=' '=' J 2 x 2 Y 2 Z,

2 Vibrations of Systems with Several Degrees of Freedom

284

n = Qat(l-cosB)-Q(at +a2Xl-coscp1 J

·2

T.=~ o 2'

n = - Ql(l 0

0 0

-

cos cp )

Here, the coordinates of the rotor center of gravity are

~=(~ +a2)coscp+at cosB, t; =(a\ +a2)sincp,x t =at sinB . The rotor acquires additional angular velocities

e

and

ip

related to angular

ip is directed along the XI OJ z =no+ipsinB.

displacements of both the rotor and the car. The vector axis. The components of mare OJ x

=ipcosB

and

The angular displacements () result in the angular velocity OJ y

=

e

directed

along the y axis. The differential equations of motion follow from the Lagrange equations of the second kind:

(Jy +manB-Jznoip-Q~B=O, [J x

+m(~ +~t]tP+Jznoe-[Q(at +~)+Qolo]cp=O.

Taking the solution to these equation as () =Alcos(pt) and IP =A 2sin(pt), we arrive at the characteristic equation 4

p -

+

Jzn~-(mat2+Jy )[Q(a\+a2)+Qolo]- m(at +a2)2 +Jx {Jy +man[Jx +m(a\ +~Y]

Qat[Q(at +a2)+Qolo] _ 4 2 _ [ ]-P -a 2 P +a 4 -O. (J y + man J x +m(a\ +~)2

The motion of the gyroscope rotor is stable if the roots of this equation are realvalued, i.e., the necessary condition of stability is CXz2 - 4a4 > O. Therefore, the critical angular velocity of the rotor is determined by the condition CXz2 =4a4• 257 To derive the differential equations of motion, we use the Lagrange equation of the second kind. The kinetic energy of the system is

T =Jx OJ~ /2+J y OJ~ j2+JoOJ; /2. The position of the gyroscope at a moment of time is shown in Fig. 298. The quantities {i}x' my, and {i}, are the angular-velocity components. The axes Ox, Oy, and Oz move in conjunction with the ring but do not rotate with the rotor. Therefore,

2.3 Critical States and Vibration Stability

285

x

Fig. 299.

OJ=Qo+Q+8,

OJ x

=8,

OJ y

=Qcos8,

OJ z

=Qo -Qsin8.

For small 0, the kinetic and potential energies and the Rayleigh function of the system are given by the equations

T =J>S2 /2+J y d /2+J o(Q o_Q8)2 II =cl 282

/2,

R = a5 2

/2 .

/2,

Substituting these expressions into the Lagrange equation

d or dt 05 -

o(T -II) OR 08 + 05=0,

we arrive at the differential equation of motion of the gyroscope:

J+2n8+ p~5=-JoQQo/Jx ' where 2n = allx and Po' =cl'/lx· If the eigenfrequency Po is large, the vibrations under nonzero initial conditions are overdamped. In this case, the angle J is determined only by the right hand side of this equation, i.e.,

Hence, the angular velocity is

Q=-&:12/(JoQo) . 258 We consider the fixed coordinate system Oxyz with its origin at the point of contact of the disk to the plane (Fig. 299).

286

2 Vibrations of Systems with Several Degrees of Freedom

At a moment of time, the disk's plane makes a small angle B with the vertical, while the tangent to the disk at the point of contact makes an angle f/J with the OIY axis, i.e., the angles Band f/J between the disk axes and the Ox, Oy, and Oz axes are small. The angular velocity of the disk is

Q=Qlel +Q2e2+Q3e3' where e p e2 , and e3 are the unit vectors and

Q I = -(/JcosO, Q 2 = 0, Q 3 ;::::

l.oj

are the angular-velocity components. The angular moment about the point 0 is

K =Kle l +K 2e 2+K3e3 =-A (/Jcos(:kl +Afk2 +C.!£3 ' where A = mi/4 and C = mr212 are the equatorial and axial moments of inertia of the disk, respectively. Using the equation for angular moment, we arrive at the differential equations

A

!

(-cPcosO)+A OcPsinO+CQO=O,

AB+CQcPcosO+A cP2cosOsinO=N zr, CQ=-Nyr, where Ny and N, are the reaction force components. In the case of small vibrations, with .Q = const, we have ... ...

A O+CQ(/J=N zr, A (/J-CQO=O .

(1)

We now consider the motion of the disk center of mass. The velocity of the point 0 is

The components of the angular acceleration can be expressed in terms of the linear and angular velocities:

2Q3+v 3Q 2' W 3 =V· 3-v IQ 2+V 2Q 2. WI =v· I -v

W 2 =V· 2-v

3Q I +V IQ 3'

Finally, the differential equation of motion of the center of mass takes the form

mw I =m(v· I -v 2Q3+V 3Q 2)=N x -mg cosO, mw 2 =m(v· 2-V 3Q I +V IQ 3)=N y , mw 3 = m(v· 3-V IQ2+V 2Q I)=N z -mg sinO. Omitting the terms nonlinear in B, we rewrite the third equation in the form

-mr(B+QcP)=N z -mgO . After the elimination of N, and d(['¥dt from Eqs. (1) and (2), we have

(2)

2.4 Approximate Methods of Evaluating the Lowest Eigenfrequency

. (C +mr 2)Cd/A -mgr

0+

A +mr

2

287

_

O-C 1

'

where C1 is an arbitrary constant detennined from the initial conditions. The eigenfrequency of small vibrations of the disk is

P=

t

(e + mr led /A - mgr = 2

A +mr2

(3d _

5

g) . r

The small vibrations are stable if Q> (g/3r)ll2.

2.4 Approximate Methods of Evaluating the Lowest Eigenfrequency 259 Let a static load be applied to the beam. We take the corresponding deflection curve as its normal mode (Fig. 3(0). In this case, the maximal strain energy is equal to the work of the applied forces, i.e., llmax

=(1/2)I p iY i

'

where Yi are the deflections. The maximal kinetic energy is

Tmax=(1/2)Ip2Yi2mi. Equating II,.ax to Tmax' we obtain the formula for the lowest frequency:

p2=I PiYi/Im i Yi2 . It is convenient to take the weights as static loads, i.e., Pi =mig. In this case,

p2=g'L m iYi/'L m iYi2 .

(1)

To find the vertical deflections, we plot the bending moment diagrams corresponding to the applied loads and to unit forces (curves 1 and 2 in Fig. 300). Multiplying these diagrams according to the Vereshchagin's rule, we obtain

Y I = llP/ 3/(9EJ x), Y 2 = 23P1l 3/( 18EJx)

Substituting these expressions into Eq. (2), we find p

.

=0.891(EJjmtyl2.

260 p = 0.71 (EJjmt) 112.

261 In the case under consideration, the deflection profile can be taken approximately as Y =Yosin(nz/3l). The maximal strain energy of the beam and the maximal kinetic energy of the point masses are, respectively,

288

2 Vibrations of Systems with Several Degrees of Freedom

2P' 1

2

Fig. 300.

Tmax In the case of m, = m2 = m, the lowest eigenfrequency is p = 1.1(EJjmp)'I2. 262 The displacements of the masses are equal to y, =Y3 =81mgtl(3888EJ) and Y2 = 159mgtl(3888EJ), The lowest eigenfrequency is = 5.71(EJjmn'l2. 263 The maximal kinetic energy of the system with allowance for the distributed mass and the maximal strain energy are, respectively (see Problem 261),

T_

~ p2my;[(3/4) + (l/2}')Sin TImax

2

(1IZ/3/}1z ],

= (31EJx Y; /4 X,,/31t,

2.4 Approximate Methods of Evaluating the Lowest Eigenfrequency

289

The lowest eigenfrequency is

p

=~7rEJx j[324(m +mol)13] =O.91~EJx j(me)

.

264 The maximal strain and kinetic energies and the lowest eigenfrequency are, respectively,

II max

f

3t . 2 -21C) . 2 11Z dz cy ; ( . 2 1C = EJx Y ; ( -1C) 4 sm +-- sm -+sm

2

31

o

2

31

3

3'

T max =3p 2my ;/4, p = 1.35~EJx j{mI3) . 265 To find the lowest eigenfrequency by the Rayleigh method, we use the formula written out in the solution to Problem 259: 3

p2=gt;miYi

/

3

t;m i y i2

(1) •

The profiles of axial loads 1,2, and 3 are shown in Fig. 301. Calculating the deflections Yi,

Y 1 =3lmg / (EF ), Y 2 =5lmg / (EF ), Y 3 =6lmg / (EF) ,

and substituting these expressions into Eq. (1), we find p

266 p =0.775(EFlml)ll2. 267 p = 0.65(EFlml)ll2.

p

1

2

3

1

N Fig. 301.

=0.45(EFlml)ll2.

290

2 Vibrations of Systems with Several Degrees of Freedom

•

I I I I I I I I I I I I I I I r1

1

2

Fig. 302.

268 The lowest eigenfrequency of torsional vibrations can be presented in the form (1)

where Mi are static torques applied to the system and rpi are the angles of torsion of the flywheels. The torques Mi can be taken as proportional to the moments of inertia: Mi = ali' The moment diagrams corresponding to M, and M2 and to the unit moments are shown in Fig. 302. The angular twists are

rpl =(MI + Mz)/c)

Taking into account that M, eigenfrequency squared 2 _

P -

,

rpz = Mz/c z+(MI + Mz)/c)

.

=all and M2 =al2, we find from Eq. (1) the lowest

J1(J, +J )jc, +J [J /C +(J, +Jz)/C,] J1(JI +Jzt jc z+Jz[Jz/c z+(JI +J )/C 1 2

2

l

2

2

2

r

269 P =0.35(c/J)'n.. 270 According to the Dunkerley method, the lowest eigenfrequency is determined by the formula lIp2 = IIp/ + lIP22... + lIPn2, where Pi is the partial eigenfrequency of the mass mi' For the system considered in Problem 259, p,2 = lI(m,~) and P22= 1I(mA2)' where ~, =~2 = 12tl(27EJJ Hence, in the case of m, =m 2=m, P =0.865(EJjmt)'n.. If m 2 =2m, =m (see Problem 261), P = 1.06(EJjmt)'n..

2.5 Random Vibrations

291

271 The lowest eigenfrequency is found from 1// = 1/P 12+ 1/P22+ 1/P32, where p/ = 1/(m~I)' P22= 1/(m~2)' and P32= 1/(m~3)' with Oil = 3l/(EF), 022 =211(EF), and ~3 = l/(EF). Hence, P = 0.41(EFlml)ll2. 272 For the systems considered in Problems 266 and 267, P = 0.7(EFlml)112 and P =0.58(EFlml)ll2, respectively.

2.5 Random Vibrations 273 The differential equation of small angular vibrations of the trailer was obtained in Problem 75:

lp+a/ 2 ipjJ o+pgrp=(c+a)lhjJo '

(1)

where Po2=d/Jo. As is known, the spectral density Shew) of an input quantity h is related to the spectral density Slw) of the corresponding output quantity rp by the equation

S rp ( OJ ) =

fv (iOJt S

h(

OJ) ,

where I W(iw) I is the modulus of the frequency response Wr(w), which is related to the transfer function W(P): Wr(w) = W(P = iw) (Svetlitsky V., 2002). The transfer function is the ratio of the Laplace transform of the output quantity to that of the corresponding input quantity: W(P) =Xou,(P)/Xin(P). The transfer function for Eq. (1) has the form

4 3 2

1 O'---......- .......- .......---L~ 5 10 15 V, mls

Fig. 303.

292

2 Vibrations of Systems with Several Degrees of Freedom

Fig. 304.

W{p)=

{c+ap)1

J o(po2+p2)+aI 2p

Hence, the frequency response is

W (im)(c + iam) I - J o(pg-m 2)+iam/ 2 In the case under consideration, the spectral density Sh( liJ) can be written out as Sh( liJ) = lOvl[21t(25v 2+ liJ2)]. Then, the variance ofthe angle IP is given by the integral

or

D = rp

1

r

(101 ~ /21Z-)(C 2 +a 2 m2

)dm

v

vP;(

-"'IJo{im +(im )2 (5J o +aI 2)+ im(Jop; +j,aI 2)+5Jo

This expression coincides with the integral 13 given in Appendix 2. Hence,

D =10; -azbo+aob]-aoa]b2/a3 rp 2aO{aOa3-a]az) ,

2.5 Random Vibrations

293

bo= 0, bl = _a 2[2 , b2= C 2[2 , ao= J 0 , a l =5;Jo+at 2 , llz. = p;Jo+ 5; at 2 , a3=5p;Jo . Finally,

D _ 'I' -

The variance

5vJ oP; a 2 + (at 2 + 5vJ 0 ~ 2 arp - ap;Jo(p;Jo+5vat2 +25v 2J o) 2 _

urp is shown in Fig. 303 as a function of the velocity v.

274 The standard deviation of the angle rp is determined by the expression (see Problem 273)

a 'I' =

5vJop;a 2+( at 2+5vJo~2 ap;Jo(p;Jo+5vat 2+25v 2JO)

The deviation urp is shown in Fig. 304 as a function of the spring rate c. The limiting value of urp as c ~ 00 is

2 0)/( at 4) .

arp. = ~(a[ + 5vJ

275 The spectral density of a function is related to that of its time derivative by the equation

Hence, the variance of the angular velocity dqtdt is given by the integral 00

Dip = fal S '1'( m)dm . Since

S 'I' ( m)=~ (im )1 2 S h ( m)

,

the variance of the angular velocity is

Dip=

00

(1/2Jl)101~ (c 2 +a 2oi)oidm

f

-00~o(im)3 +(imnSJ