Damped Vibrations

SOLID MECHANICSDYNAMICS TUTORIAL – DAMPED VIBRATIONS This work covers elements of the syllabus for the Engineering Council Exam D225 – Dynamics of Mechanical Systems, C105 Mechanical and Structural Engineering and the Edexcel HNC/D module Mechanical Science. On completion of this tutorial you should be able to do the following.  Define a free damped oscillation.  Explain the purpose of damping.  Define damping coefficient.  Define damping ratio.  Derive formulae that describe damped vibrations.  Determine the natural frequency and periodic time for damped systems.  Define amplitude reduction factor.  Calculate damping coefficients from observations of amplitude. This tutorial covers the theory of natural vibrations with damping and continues the studies in the tutorial on free vibrations. To do the tutorial fully you must be familiar with the following concepts.  Simple harmonic motion.  Free vibrations. In reality. Figure 1 Friction dissipates the energy as heat.INTRODUCTION In the tutorial on free vibrations it was explained that once set vibrating. The diagram shows a displacement – time graph of a typical damped oscillation. The type of friction that is easiest to deal with mathematically is that created by a dashpot (also called a damper). The diagram shows two types of dashpot. the force opposing motion (the damping force) is directly proportional to the velocity of the piston. the system would carry on oscillating for ever because the energy put into the system by the initial disturbance cannot get out of the system. dx Fd  constant x velocity  c Figure 2 dt The constant of proportionality ‘c’ is called the damping coefficient and has units of N s/m. Because of the restriction. The equation for this force is as follows. air and oil. Often the piston is replaced with a simple diaphragm. It can be shown that for both cases. pressure is needed to force the fluid through the restrictor and this produces a force opposing motion.Dunn 2 . Again. © D. Such vibrations are called free and damped. Let us now examine two kinds of dashpots. pressure is needed to make the air flow through the orifice and this pressure will produce a force opposing the motion. OIL-The oil is contained in the cylinder and motion of the piston pushes the oil through restrictors in the piston to the other side. 1 DASHPOTS Here we will examine dashpots for linear motion.J. AIR-The piston moves inside the cylinder and pumps or sucks air through the orifice. the oscillation always dies away with time because some form of friction is present. The value of ‘c’depends on the size of the restriction. J. M d 2 x c dx 0  x k dt 2 k dt © D. First divide each term by k.Dunn 3 . the applied force must be zero (e.2. when you let go of it). These are: d2x The inertia force given by Fi  Mass x acceleration  M dt 2 dx The damping force given by Fd  constant x velocity  c dt The spring force given by Fs = k x The total force is then F  Fi  Fd  Fs d2x dx F M 2 c  kx dt dt Because the vibration is free. We make the following changes. it is opposed by three forces. DAMPED LINEAR VIBRATIONS Consider a mass suspended on a spring with the dashpot between the mass and the support. Figure 3 If a force F is applied to the mass as shown. d2x dx 0 M 2 c  kx and this is a linear second order differential equation and it is dt dt much discussed in most maths books.g. 1} b = {n(2-1) + 1} The resulting graph of displacement x with time t depends upon the damping ratio and hence the value of c. c c  4Mk This has the same units as the damping coefficient (N s/m). c c c c k c δ     ωn cc 4M k 2 M k 2k M 2k c 2δ  k ωn We can now replace the term c/k in our equation. The ration c/c c is called the damping ratio and this is defined as follows. 1 d 2 x 2δ dx 0   x and finally multiply each term by ω 2n 2 ω n dt 2 ω n dt d2x dx 0 2  2 δ ωn  ω 2n x dt dt This is now a standard equation and the solution may be found in standard text as x = Aeat + Be-bt A and B are constants of integration and a = {n(2-1) . c δ cc This may now be developed as follows.J.Dunn 4 . We may replace (k/M) with n2. We need to start with a definition. Three important cases should be considered. 1 d 2 x c dx 0 2 2  x ω n dt k dt Next we examine the term c/k.In the work on undamped vibrations it was shown that without a dashpot a natural oscillation occurs with angular frequency n = (k/M) rad/s. We will be using a term called the critical damping coefficient and it is defined as follows. © D. δω n t x1 Ce δω mT The ratio is  e n  AM PLITUDEREDUCTION FACTOR x 2 Ce . If we assume that t = 0 and x = C at the moment the mass is released we get a decaying cosinusoidal oscillation as shown.J.2).δω n (t  mT) 2π ω Now examine the term δω n mT. In this case C = -10.Dunn 5 .δ 2 and δω n mT  hence 1. UNDER DAMPED This occurs when  < 1 and c < cc. The damped frequency is  = n(1. The damped frequency is f = /2 and the periodic time of the damped angular oscillation is T = 1/f = 2/ AMPLITUDE REDUCTION FACTOR Consider two oscillations. .δω n t The amplitude of the first oscillatio n at time t is x 1  Ce The amplitude of the oscillatio n that occurs ' m' cycles later occcurs mT seconds later so .1. one occurring m cycles after the first.δω n (t  mT) x 2  Ce .δ2 © D.  x2  1. Put T  and this becomes 2π δ m n ω ω 2π δ m Now substituteω  ω n 1 . Figure 4 The natural undamped angular frequency is n = (k/M)½ .  δω t x  Ce n cosωt  The graph shows the result if the mass is pulled down 10 units and released. The displacement is described by the following equation.δ2 2π δ m x1  e 1 .2.δ and next we take the natural log to get rid of the exponential so 2 x2 x  2π δ m ln  1   and this is probably a more useful expression to use. If the damping is increased. the oscillations die away quicker and eventually a critical point is reached where the mass just returns to the rest position with no overshoot or oscillation.2. This occurs when  = 1 and c = cc. OVERDAMPED This occurs when  > 1 and c > cc. 2. CRITICALLY DAMPED. Figure 5 © D.3.Dunn 6 . The result is an exponential decay with no oscillations but it will take longer to reach the rest position than with critical damping.J. The result is an exponential decay as shown.2. SOLUTION For successive amplitudes m = 1 x   3  ln  1   ln    ln6  1.478δ 2 3.298δ 2 1.21  so 1 .5 mm respectively.274 13.075  0.792  square both sides 1.Dunn 7 .J.δ2 1 13.1 A vibrating system is analysed and it is found that two successive oscillations have amplitudes of 3 mm and 0.δ2 39.298 © D.δ 2  12. Calculate the amplitude reduction factor and the damping ratio.298δ 2  1 and δ 2   0.5  2π δ 1. WORKED EXAMPLE No.075 and δ  0.792  amplitude reduction factor  x2   0. 785 Hz simplify and 8   so k  3158 N/m 1 k 1 k k fn  so 4  2 2π M 2π 5 5 The critical damping coefficien t is given by c c  4M k  4 x 5 x 3158  251.3 Ns/m © D. The natural frequency. v.δ2 2π δ2 ln 20   2.δ2 17. iii. ii. The actual damping coefficient. It is observed that the amplitude reduces to 5% of its initial value after 2 oscillations.δ2 (4 π δ ) 2  (2.0. vi. It takes 0. i.J. The critical damping coefficient.3 x 0.Dunn 8 . The spring stiffness.δ 2 18.5     The damped frequency is given by f  f n 1 .2 A mass of 5 kg is suspended on a spring and set oscillating.232  58.976 1. The actual frequency.3 Ns/m The actual damping coefficien t is c  c c x δ  251.0538 δ  0. iv.δ 2  4 1 . Calculate the following. WORKED EXAMPLE No.232 The natural frequency is the number of oscillatio ns  time taken 2 fn   4 Hz 0.δ 2 157.9 δ 2  8. SOLUTION For 2 oscillations m = 2 and x1/x2 = 100/5 = 20  x  2π δ m ln  1    x2  1.0538  3.5 seconds to do them. The damping ratio.996) 2 1.59 δ 2  1 δ 2  0.996 and next square both sides 1.59δ 2  1 . 07 mm VELOCITY  50sin ωt ωe-δω n t  50cos ωt δωn e -δ ωn t dx v dt v  50sin 8. The mass is pulled down 50 mm and released.2 2  4.2 2  27.3135. Calculate the following.50 mm x  Ce -δω n t cos ωt   .5 1  0.313(0.628  22.50e -0.443)  4. The damped frequency. velocity and acceleration after 0.183 x (0.71t  x  50 x 0. The system is fitted with a damper with a damping ratio of 0.28x 0.28 rad/s M 5 ωn fn   4.313(27.71 rad/s The initial displacement is down 50 mm so C  .183) a  3112.2)(28.313(27.7 a  5554 mm/s 2 © D.2.313δωn e -δ ωn t v  227.J.Dunn 9 .712 )(0. i. k 4000 ωn    28.2  129.28 1  0.28)(0.28) 2 (0.71)(0.5 Hz 2π f  f n 1  δ 2  4.3cos 27. SOLUTION You will need to be able to do advanced differentiation in order to follow this solution.2) 2 (28.2 x 28.183)  100sin 8.078  50cos 8.183)  50cos 8.41 Hz (Answer i) ω  ω n 1  δ 2  28.7 mm/s ACCELERATION  50cos ωt ω 2e -δ ωn t  100sin t ωδωn e -δ ωn t  50cos t δ 2ω 2n e -δ ωn t dv a dt a  50cos 8. The displacement.93  204.3 A mass of 5 kg is suspended on a spring of stiffness 4000 N/m.3 seconds. ii.9  2571. WORKED EXAMPLE No. A mass of 5 kg is suspended from a spring of stiffness 46 kN/m. It is set oscillating and it is observed that two successive oscillations have amplitudes of 10 mm and 1 mm.J. The damped frequency. Calculate the following.523 Hz) ii.3.976) . A dashpot is fitted between the mass and the support with a damping ratio of 0. The amplitude reduction factor.1 1. (959 N s/m) 3. i.4. (6. The undamped frequency. .Dunn 10 . (2. (-4. 366 mm/s and -3475 mm/s) © D. A mass of 50 kg is suspended from a spring of stiffness 10 kN/m.188) . The displacement.5 N s/m) .56 Hz) . (15. SELF ASSESSMENT EXERCISE No.1 s. (14. (254.812 mm. The system is damped and the damping ratio is 0. The damping ratio. velocity and acceleration after 0. (1. The damping coefficient. The critical damping coefficient. The damped frequency. (0. Determine the following. A mass of 30 kg is supported on a spring of stiffness 60 000 N/m. . The mass is raised 5 mm and then released. Calculate the following.21 Hz) 2. The true frequency.26 Hz) . J.δ2 displacement. The torsional spring stiffness is torque per unit angle so the units of k are Nm/radian.3.2)   2π δ m The amplitude reduction factor is ln  1   where  is the angular 2  1. The critical damping coefficient is cc= (4 I k) The natural frequency is n= (k/I) Remember that for a shaft the torsional stiffness kt = GJ/L The polar second moment of area for a circular section is D4/32 The actual angular frequency is  = n(1. © D. These are Mass m 2nd moment of inertia I Force F Torque T Distance x angle  Velocity v angular velocity  Acceleration a angular acceleration  Figure 6 A dashpot for a torsional system would consist of a vane which rotates inside a pot of oil. The damping torque is directly proportional to the angular velocity such that Damping Torque = c  c is the torsional damping coefficient with units of N m s/radian.Dunn 11 . the linear quantities are replaced with angular quantities in the formula. TORSIONAL OSCILLATIONS The theory is the same for torsional oscillations. 9163 and next square both sides 1.9 kg m2.5 x 0.014 doubled.δ 2 48 δ 2  1 δ 2  0.5   2π δ m ln  1  2  1.9     The damped frequency is given by f  f n 1 . The system has proportional damping and it is observed that the amplitude reduces by 60% after one oscillation. The shaft material has a modulus of rigidity of 90 GPa. The actual damping coefficient. The natural frequency.54 N m s/radian © D.J.98 x 10 -9 m 4 32 32 2 x 90 x 10 9 x 0.4 A horizontal shaft is fixed at both ends and carries a flywheel at the middle.δ 2 39.1443 The torsional stiffness is k t  GJ/L and there is one each side of the flywheel so it is πD 4 π x 0.0208 δ  0. The spring stiffness. iii.09 1 .04 Hz The critical damping coefficien t is given by c c  4 I k t  4 x 1.98 x 10 -9 k t  2 GJ/L   1944 N m s/radian 1 1 kt 1 1944 The natural frequency is f n    5.9163) 2 1. SOLUTION For 1 oscillations m = 1 and 1 = 100% 2 = 40% 1/2= 100/40 = 2.9 x 1944  121.0208  5. The damping ratio.1443  17.478 δ 2  0. v. ii. i.δ 2  5.δ2 47δ 2  1 .8396 1. Calculate the following. The actual frequency.5 Ns/m The actual damping coefficien t is c  c c x δ  121.5   0.09 Hz 2π I 2π 1. The flywheel has a moment of inertia of 1. WORKED EXAMPLE No. J    0. The critical damping coefficient. The shaft is 1 m long either side of the flywheel and is 10 mm diameter.0.δ2 2π δ ln 2.Dunn 12 . vi. iv.δ2 (2 π δ ) 2  (0. 4. The shaft has a torsional stiffness of 5000 N m s/rad. (10053 N m s/radian) The critical damping coefficient for the system. velocity and acceleration after 0. The flywheel has a moment of inertia of 30 kg m 2. The damping ratio is 0. (1 418 Nms/rad) The frequency of damped oscillations? (2.6%) 2. A torsional damper is fitted with a damping ratio of 0.24 Hz).2 1. i. The damped frequency.02 radian and released.J.8 x 10-4 rad/s and -0.125. The flywheel is rotated 0. The amplitude reduction factor. (1. It is suspended on the end of a vertical shaft 2 m long and 40 mm diameter.226 rad/s2) © D. Calculate the following.5 seconds after being released. (45.Dunn 13 . SELF ASSESSMENT EXERCISE No.833 Hz) ii. Calculate the following. -8.00141 rad. (0. The angular displacement. A shaft with negligible inertia has a flywheel suspended from the end and a damper to damp the vibrations. The torsional stiffness of the shaft. A flywheel has a moment of inertia of 50 kg m 2. The modulus of rigidity of the shaft is 80 GPa.