18.12 FORCED-DAMPED VIBRATIONS

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Emil Wells
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1 8. ORCED-DAMPED VIBRATIONS Vibrations A mass m is attached to a helical spring and is suspended from a fixed support as before. Damping is also provided in the system ith a dashpot (ig. 8.). Before the mass is set in motion, let B- B be the static equilibrium position under the eight of the mass. No, if the mass is subjected to an oscillating force sin t, the forces acting on the mass at any instant ill be ig. 8. l Impressed oscillating force sin t (donards) l Inertia force mẋ (upards) l Damping force cẋ (upards) l Spring force (restoring force) sx (upards) Thus, the equation of motion ill be mẋ + cẋ + sx sin t or mẋ + cẋ + sx sin t (8.36) Complete solution of this equation consists of to parts, the complementary function (C) and the particular integral (PI). C Xe -z n t sin ( d t + j ) [refer to Eq. (8.3)] To obtain the PI, let c m a, s m b, and m d. Then, using the operator D, the equation becomes (D + ad + b) x d sin t PI d sin t D + ad + b d sin t - + ad + b ( b- ) -ad d sin t ( b- ) + ad ( b- )- ad d d È sin tb ( - ) - adsin t Î ( b - ) - a D È Î tb a t ( b- ) + ( a) sin ( - ) - cos

2 Theory of Machines Take (b ) R cos j and a R sin j Constants R and j are given by R ( b - ) + ( a) and j tan a b - PI dr (sin t cos j - cos t sin j) ( b- ) + ( a) d ( b- ) + ( a) ( b- ) + ( a) sin(t j) d ( b - ) + ( a) sin(t j) / m Ê s c - + Ê Ëm Ëm sin(t j) x C + PI -z Xe dt - n t sin( j ) + sin(t j) sin(t j) (8.37) The damped-free vibrations represented by the first part (C) becomes negligible ith time as e. The steady-state response of the system is then given by the second part PI. The amplitude of the steady-state response is given by A (8.38) Ê m Êc Á - + Ë s Ës È Ê Ê - Á z + Á n Ë Ë n Î (8.39)

3 Vibrations 3 The equation is in the dimensionless form and is more convenient for analysis. It may be noted that the numerator /s is the static deflection of the spring of stiffness s under a force. The frequency of the steady-state forced vibration is the same as that of the impressed vibrations. j is the phase lag for the displacement relative to the velocity vector. c z a n tan j m c (8.4) s b- - s- m Ê m - Á Ë n The particular solution of the equation of motion can also be obtained graphically as follos: Assuming that the displacement of the vibrating mass under the action of the applied simple harmonic force o sin t is also simple harmonic and lags by an amount j. Then x A sin (t j) and ẋ Èp A cos (t j) A sin ( t j) + - Î ẋ A sin (t j) here A is the amplitude of vibrations. Substituting these values in the equation mẋ + cẋ + sx sin t m Èp A sin(t j) + c Asin ( t j) + - Î + sa sin(t j) sin t sin t + m Èp A sin( t j) c A sin ( t j) + - Î sa sin(t j) The forces and the vector sum of the same have been shon in ig. 8.. In triangle abc. ig. 8.

4 4 Theory of Machines ( sa - m A) + ( ca) or A ( s- m or A and tan j c s- m The vectors as shon in the diagram are fixed relative to one another and rotate ith angular velocity. 8.3 MAGNIICATION ACTOR The ratio of the amplitude of the steady-state response to the static deflection under the action of force is knon as magnification factor (M). M / ( s - s Ê m c - + Ê Ë s Ës È Ê Ê - Á z + Á n Ë Ë n Î (8.4) Thus, the magnification factor depends upon: (a) the ratio of frequencies,, and n (b) the damping factor. The plot of magnification factor against the ratio of frequencies (/ n ) for different values of z is shon in ig. 8.(a). The curves sho that as the damping increases or z increases, the maximum value of the magnification factor decreases and vice-versa. When there is no damping (z ), it reaches infinity at / n, i.e. hen the frequency of the forced vibrations is equal to the frequency of the free vibration. This condition is knon as resonance.

5 Vibrations 5 ig. 8. In practice, the magnification factor cannot reach infinity oing to friction hich tends to dampen the vibration. Hoever, the amplitude can reach very high values. ig. 8.(b) shos the plots of phase angle vs. frequency ratio (/ n ) for different values of z. Observe that, l Irrespective of the amount of damping, the maximum amplitude of vibration occurs before the ratio / n reaches unity or hen the frequency of the forced vibration is less than that of the undamped vibrations. l Phase angle varies from zero at lo frequencies to 8 at very high frequencies. It changes very rapidly near the resonance and is 9 at resonance irrespective of damping. l In the absence of any damping, phase angle suddenly changes from zero to 8 at resonance. Example 8. A machine part having a mass of.5 kg vibrates in a viscous medium. A harmonic exciting force of 3 N acts on the part and causes a resonant amplitude of 4 mm ith a period of. second. ind the damping coefficient. If the frequency of the exciting force is changed to 4 Hz, determine the increase in the amplitude of the forced vibrations upon the removal of the damper. Solution m.5 kg, 3 N, A 4 mm, T. s, p p 8.56 rad/s T. (i) At resonance, n or n or s m No, A s m 8.56 rad/s 8.56, s 39 N/m or.39 N/mm È Ê Êz - Á + Á n Ë Ë n Î

6 6 Theory of Machines or A z or.4 3/39 z or z.56 c m n z N/m/s.75 4 N/mm/s (ii) f n p 4 p 5.3 rad/s With damper: 3/39 A È Ê5.3 È Ë 8.56 Î 8.56 Î 3/ ( ) ( ) Ê Á Ë.55 m Without damper: z A 3/ m \ Increase in magnitude m or 49.7 mm Example 8. A single-cylinder vertical diesel engine has a mass of 4 kg and is mounted on a steel chassis frame. The static deflection oing to the eight of the chassis is.4 mm. The reciprocating masses of the engine amounts to 8 kg and the stroke of the engine is 6 mm. A dashpot ith a damping coefficient of N/mm/s is also used to dampen the vibrations. In the steady-state of the vibrations, determine: (i) the amplitude of the vibrations if the driving shaft rotates at 5 rpm (ii) the speed of the driving shaft hen the resonance occurs. Solution m 4 kg N 5 rpm c N/m/s r 8 mm D.4 mm.4 m p rad/s 6 No s D mg \ s , s N/m Centrifugal force due to reciprocating parts (or the static force), mr 8.8 (5.36) 3948 N (i) A [Eq. (8.38)] [.635-4(5.36) ] + ( 5.36).7 m or 7. mm n

7 Vibrations 7 (ii) Resonant speed: 6 s.635 n rad/s m 4 p N or 63.93, N 6.5 rpm 6 Example 8.3 A body having a mass of 5 kg is suspended from a spring hich deflects mm under eight of the mass. Determine the frequency of the free vibrations. What is the viscous damping force needed to make the motion aperiodic at a speed of mm/s? If, hen damped to this extent, a disturbing force having a maximum value of N and vibrating at 6 Hz is made to act on the body, determine the amplitude of the ultimate motion. Solution m 5 kg, D mm, N, f 6 Hz g 9.8 f n 4.55 Hz p D p. The motion becomes aperiodic hen the damped frequency is zero or hen it is critically damped (z ) and g 9.8 n 8.59 rad/s D. c c c m n N/m/s.857 N/mm/s Thus, the force needed is.857 N at a speed of mm/s. A But, p f p rad/s and s can be found from f n sm / p or 4.55 s/5 p or s 6 N/m \ A È 6 5 (37.7) Î - + ( ). 98 m.98 mm

This equation of motion may be solved either by differential equation method or by graphical method as discussed below:

This equation of motion may be solved either by differential equation method or by graphical method as discussed below: 2.15. Frequency of Under Damped Forced Vibrations Consider a system consisting of spring, mass and damper as shown in Fig. 22. Let the system is acted upon by an external periodic (i.e. simple harmonic)