5Vibration AnalysisAll real engineering structures exhibit some amount of energy dissipation. A helical spring in the suspension system of an automobile will dissipate some energy due to the internal interaction of grains of steel within the spring. A leaf spring adds friction between the leaf elements to the internal energy dissipation. While both of these do aid the damping effect in a suspension system, they are not significant because of the presence of the shock absorber.At other times these internal effects are the only energy dissipating means available. One example of this situation is the cable used in the tether satellite experiments conducted jointly by the United States and Italy in the late 1990’s. These experiments investigated the possibility of using a single cable attached to a controlled mass as an alternative to solar cells for generating electricity on future space flights.By dragging a conductor through the magnetic field of the earth, it is possible to create an electric current which will provide power to the
6Vibration Analysisspacecraft. One major point of concern related to the potential problem that drag on the approximately sixty miles of cable might create a skip rope effect that would lead to instability and/or breaking of the 0.5 mm diameter cable.This situation was compounded by the fact that the only available energy dissipation was the movement between the wire and the insulation. Fortunately, this small amount of damping proved to be sufficient enough to prevent any problems.As will be seen as the equations are developed, most engineering systems exhibit only a small amount of damping even thought they may be extremely large and complex. Once again, this amount of damping is a built-in characteristic that comes about from the materials and construction methods employed. It is the result of effects such as friction between connected elements, internal friction that occurs during deformation, and windage.
7Vibration AnalysisBecause the amount of damping is generally small, it is acceptable to ignore its presence when determining the vibrational characteristics of the system. However, the need to know the amplitude of vibration requires that the damping be included. This is especially true when the damping effect is large.Due the complex nature of most engineering systems, it is generally difficult to determine or estimate damping exactly especially due to the interaction of these damping effects. This situation can be dealt with by studying each type damping individually and then determining which type of damping dominates in the system of interest. This procedure generally results in the ability to do an adequate analysis.The most common types of damping are viscous, dry friction, and hysteretic. Hysteretic (or structural) damping arises in structural elements due to hysteresis losses in the material. The type and amount of damping has a large effect on the dynamic response level.
9Vibration AnalysisViscous damping is a very common form of damping which is found in many engineering systems such as instruments and automobiles. The viscous damping force is proportional to the difference in the velocity of the ends of the damping element and opposes the motion. This makes the viscous damping force a linear continuous function of the velocity.As the equations that result from analysis of viscous damping are thesimplest mathematical treatment, other forms of damping are expressed in the form of an equivalent viscous damper.For the damped SDOF system shown to the right, the damping force is equal to the damping coefficient times the velocity as shown in the free body diagram.
10Vibration Analysis The equation of motion is For the damped vibration case, the solution still has the form which repeats itself every second derivative with a negative coefficient. However, the complex exponential form is the easiest to work with and will be used here. This means the equation for the displacement, x(t) will have the formDifferentiating twice and substituting the results into the equation of motion yieldsAs X = 0 is a trivial solution and est g 0 for all time, the quadratic in s is solved for the roots.
11Vibration Analysis The roots are Hence where X1 and X2 are arbitrary constants which are found by using the initial conditions. The system response depends on both the algebraic sign of c and whether c2 is greater than,less than, or equal to 4km.If c is positive, x(t) will decrease in amplitude over time. If c is negative, x(t) will increase in amplitude over time meaning the system is unstable. If c2 is less than 4km, x(t) will oscillate. If c2 is equal to or greater than 4km, x(t) will not oscillate.
12Vibration AnalysisTo establish that these effects do occur, substitute the values for s1 and s2 into x(t) and manipulate the resulting expression as follows:This equation must be evaluated for each of the three possibilities. To accomplish this, the transitional value of c2 = 4km is used to define the critical damping coefficient, cc, which represents the value of damping where oscillation ceases. This yieldsTo eliminate the need to have a different equation for every possible combination of m, c, and k, the damping ratio, z, is defined as
13Vibration AnalysisThis allows the expression for x(t) to be rewritten as follows:Case 1: Underdamped (0 [ z < 1)For this case, the exponents involving z2 – 1 are complex and x(t) will oscillate for at least a short time, depending on the value of z. Euler’s identity allows the complex exponential terms to be rewritten in the following manner:Substitution of this equivalence into the equation for x(t) yields
14Vibration AnalysisSince x(t) is a real quantity (i.e. it can be felt and observed), the values for X1 and X2 must be defined in such a way as to produce a real value for x(t). This means that X1 and X2 must be complex conjugates orX1 = a - ib X2 = a + ibSubstituting this result into the equation for x(t) gives
15Vibration AnalysisIt is also convenient to define the damped natural frequency asSubstituting this definition into the equation for x(t) yieldsx(t) is a harmonic function that decays exponentially with time as shown in the figure to the right. The values for A and B depend on the initial conditions and the decay rate on z.
16Vibration AnalysisA more compact form for x(t) can be had by including a phase angle f as shown belowFrom the system response shown, it is easily seen that both the initial displacement and velocity were positive and that z was small enough to allow complete oscillations
17Vibration Analysisto occur before reaching a level that can be considered effectively zero.There are three things about any transient response that are important. These are 1) rise time which indicates how quickly the system response reaches a specified percentage of its maximum value, 2) maximum value, 3) settling time which indicates how quickly the system response decays to within a specified percentage above theequilibrium or steady-state value. These values have an impact on one another and it is important to know what range is acceptable for each when designing a system.
18Vibration AnalysisShown below are plots of the damped transient response of the same system with increasingly amounts of damping. Notice how the maximum value and settling time both decrease as z increases. But the amount of energy needed to begin moving the system also increases with damping.
20Vibration Analysis Case 2: Critically damped (z = 1) For this case, the roots are real and equal which means the exponents involving z2 – 1 are zero and x(t) will simply decay exponentially back to the equilibrium or steady-state value.
22Vibration Analysis Case 3: Overdamped (z > 1) For this case, the roots are real and distinct which means the exponents involving z2 – 1 are positive, real numbers and x(t) will simply decay exponentially back to the equilibrium or steady-state value but at a rate that is slower than the critical damping case.
23Vibration AnalysisTo understand what happens to x(t) as z increases, each exponent needs to be examined.
26Vibration AnalysisThe transient response can be used to determine the amount of damping present in a single degree-of-freedom model of a system by imparting an initial displacement and/or velocity to the system and monitoring the amplitude of the response.As shown below, a single degree-of-freedom was disturbed from equilibrium and the amplitude of the motion plotted versus time. The motion is given byThe maximum value will occur whenAnd the period, td, is
27Vibration Analysis Therefore, But, Substituting yields, Dividing XI by XII yields
28Vibration AnalysisTaking the natural logarithm of each side gives the logarithmic decrement, d, asSolving for z yields,If z is small, it is possible to use
29Vibration AnalysisHowever, if z is small enough that 1 – z2 is approximately 1, the difference between XI and XII is likely to be small enough that it cannot be measured accurately. In this case instead of using two successive peaks, use two peaks that are p cycles apart. This changes the definition for tII toAnd the equation for zbecomesWhich for small z can be written as
30Vibration AnalysisEven if z is large enough that 1 – z2 cannot be set equal to 1, good accuracy for the estimate of z can be obtained using the approximate equation as shown below.As seen in the figure, the error in the estimate of z is small up to a value of z of about 0.45 or 0.5.
32Vibration AnalysisGiven - The machine shown below weighs 3600 lb and rests on a set of vibration isolators. As the machine is lowered onto the vibration isolators, the four springs (one on each corner) each deflect 3 inches. Determine the undamped natural frequency of vibration, w, and the stiffness of each spring. What should the damping coefficient for this set of isolators be if a damper is located with each spring and the damped natural frequency of vibration, wd, is to be 10% less than w?Solution – Since all four springs deflect the same amount, the springs are in parallel and k in the figurerepresents the sum of the individual stiffnesses. Therefore, the value of k can be found using
33Vibration Analysis Each spring has a stiffness of and the undamped natural frequency isThe damped natural frequency, wd, is to be 90% of w so it has a value of s-1.The damping coefficient for the set must be found using the definitions for the critical damping coefficient and wd.
34Vibration AnalysisThe critical damping coefficient for the set is found usingThe damping ratio for the set is found using
35Vibration Analysis The damping coefficient for the set is found using Each damper needs a damping coefficient of lb-s/in.
37Vibration AnalysisGiven – A light rigid rod of length L is pinned at O and has a body of mass m attached at the other end as shown. A spring and viscous damper are connected in parallel and attached to the rod at a distance a from the pivot. The system is set up in a horizontal plane.Assuming that the damper is adjusted to provide critical damping, obtain the motion of the rod as a function of time if it is rotated through a small angle q0 and then released. Given that q0 = 2o and that the undamped natural frequency of the system is 3 rad/s, calculate the displacement 1 s after release.If the damping is reduced to 80% of its critical value, calculate the logarithmic decrement for the system.
38Vibration AnalysisSolution – If the rod is rotated slightly in clockwise direction, the spring and damper will exert a restoring force which is vertically upward and they will create a counter clockwise moment about O equal toFor the critical damping case, the solution has the form ofTo determine A and B we need
39Vibration Analysis Therefore, For q0 = 2o, w = 3 rad/s, and t = 1 s, The logarithmic decrement is given bySo if z is 0.8, d is
41Vibration AnalysisGiven – Shown below is the schematic representation of a drop hammer forge. This system is composed of an anvil (which weighs 5,000 N and is mounted on a foundation that has a stiffness of 5 x 106 N/m and a viscous damping coefficient of 104 N-s/m) and a drop hammer called the tub which weighs 103 N. During a particular operation the tub falls 2 m before striking the anvil. If the anvil is at rest prior to impact by the tub, determine the response of the anvil afterimpact. Assume the coefficient of restitution between the anvil and the tub is 0.4.Solution – The first task is to determine the velocity of the tub just prior to impact and the velocities of both the tub and anvil just after impact.
42Vibration Analysis M(va2 - va1 ) = m(vt2 – vt1 ) To determine these velocities, we use the principle of conservation of momentum with the tub velocity before and after impact represented by vt1 and vt2, respectively. The velocity of the anvil before and after impact are given by va1 and va2, respectively. The principle of conservation of momentum statesM(va2 - va1 ) = m(vt2 – vt1 )Both va1 and vt1 have values which are either known or can be easily found. As the anvil is initially at rest, va1 is zero. The velocity of the tub just prior to impact is found using½ mv2t1 = mgh
43Vibration Analysis Or the value for vt1 is Substituting these values for vt1 and va1 into the momentum equation yields- OR -The definition of the coefficient of restitution is
44Vibration Analysis - OR - Solving these two equations simultaneously yieldsThus the initial conditions for the anvil areThe damping coefficient is equal to
45Vibration AnalysisThe undamped and damped natural frequencies are found usingThe displacement of the anvil is given by