Presentation on theme: "1 MEEG 5113 Set 2. 2 3 4 5 Vibration Analysis All real engineering structures exhibit some amount of energy dissipation. A helical spring in the suspension."— Presentation transcript:
1 MEEG 5113 Set 2
5 Vibration Analysis All 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
6 Vibration Analysis spacecraft. 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.
7 Vibration Analysis Because 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.
9 Vibration Analysis Viscous 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 the simplest 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.
10 Vibration Analysis The equation of motion is Differentiating twice and substituting the results into the equation of motion yields 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 form As X = 0 is a trivial solution and e st g 0 for all time, the quadratic in s is solved for the roots.
11 Vibration Analysis The roots are Hence where X 1 and X 2 are arbitrary constants which are found by using the initial conditions. The system response depends on both the algebraic sign of c and whether c 2 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 c 2 is less than 4km, x(t) will oscillate. If c 2 is equal to or greater than 4km, x(t) will not oscillate.
12 Vibration Analysis To establish that these effects do occur, substitute the values for s 1 and s 2 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 c 2 = 4km is used to define the critical damping coefficient, c c, which represents the value of damping where oscillation ceases. This yields To eliminate the need to have a different equation for every possible combination of m, c, and k, the damping ratio, , is defined as
13 Vibration Analysis This allows the expression for x(t) to be rewritten as follows: Case 1: Underdamped (0 [ < 1) For this case, the exponents involving 2 – 1 are complex and x(t) will oscillate for at least a short time, depending on the value of . 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
14 Vibration Analysis Since x(t) is a real quantity (i.e. it can be felt and observed), the values for X 1 and X 2 must be defined in such a way as to produce a real value for x(t). This means that X 1 and X 2 must be complex conjugates or X 1 = a - ibX 2 = a + ib Substituting this result into the equation for x(t) gives
15 Vibration Analysis It is also convenient to define the damped natural frequency as Substituting this definition into the equation for x(t) yields x(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 .
16 Vibration Analysis A more compact form for x(t) can be had by including a phase angle as shown below From the system response shown, it is easily seen that both the initial displacement and velocity were positive and that was small enough to allow complete oscillations
17 Vibration Analysis to 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 the equilibrium 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.
18 Vibration Analysis Shown 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 increases. But the amount of energy needed to begin moving the system also increases with damping.
20 Vibration Analysis Case 2: Critically damped ( = 1) For this case, the roots are real and equal which means the exponents involving 2 – 1 are zero and x(t) will simply decay exponentially back to the equilibrium or steady-state value.
21 Vibration Analysis
22 Vibration Analysis Case 3: Overdamped ( > 1) For this case, the roots are real and distinct which means the exponents involving 2 – 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.
23 Vibration Analysis To understand what happens to x(t) as increases, each exponent needs to be examined.
24 Vibration Analysis
26 Vibration Analysis The 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 by The maximum value will occur when And the period, d, is
27 Vibration Analysis Therefore, But, Substituting yields, Dividing X I by X II yields
28 Vibration Analysis Taking the natural logarithm of each side gives the logarithmic decrement, , as Solving for yields, If is small, it is possible to use
29 Vibration Analysis However, if is small enough that 1 – 2 is approximately 1, the difference between X I and X II 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 t II to And the equation for becomes Which for small can be written as
30 Vibration Analysis Even if is large enough that 1 – 2 cannot be set equal to 1, good accuracy for the estimate of can be obtained using the approximate equation as shown below. As seen in the figure, the error in the estimate of is small up to a value of of about 0.45 or 0.5.
32 Vibration Analysis Given - 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, , 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, d, is to be 10% less than ? Solution – Since all four springs deflect the same amount, the springs are in parallel and k in the figure represents the sum of the individual stiffnesses. Therefore, the value of k can be found using
33 Vibration Analysis Each spring has a stiffness of and the undamped natural frequency is The damped natural frequency, d, is to be 90% of 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 d.
34 Vibration Analysis The critical damping coefficient for the set is found using The damping ratio for the set is found using
35 Vibration Analysis The damping coefficient for the set is found using Each damper needs a damping coefficient of lb-s/in.
37 Vibration Analysis Given – 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 0 and then released. Given that 0 = 2 o 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.
38 Vibration Analysis Solution – 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 to For the critical damping case, the solution has the form of To determine A and B we need
39 Vibration Analysis Therefore, For 0 = 2 o, = 3 rad/s, and t = 1 s, The logarithmic decrement is given by So if is 0.8, is
41 Vibration Analysis Given – 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 10 6 N/m and a viscous damping coefficient of 10 4 N-s/m) and a drop hammer called the tub which weighs 10 3 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 after impact. 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.
42 Vibration Analysis To determine these velocities, we use the principle of conservation of momentum with the tub velocity before and after impact represented by v t1 and v t2, respectively. The velocity of the anvil before and after impact are given by v a1 and v a2, respectively. The principle of conservation of momentum states M(v a2 - v a1 ) = m(v t2 – v t1 ) Both v a1 and v t1 have values which are either known or can be easily found. As the anvil is initially at rest, v a1 is zero. The velocity of the tub just prior to impact is found using ½ mv 2 t1 = mgh
43 Vibration Analysis Substituting these values for v t1 and v a1 into the momentum equation yields Or the value for v t1 is - OR - The definition of the coefficient of restitution is
44 Vibration Analysis Solving these two equations simultaneously yields - OR - Thus the initial conditions for the anvil are The damping coefficient is equal to
45 Vibration Analysis The undamped and damped natural frequencies are found using The displacement of the anvil is given by