In a later example we will show how to modify the least-squares method to obtain a function that passes through the origin. If we can perform an experiment to measure the force f as a function of velocity v, we can estimate the value of the damping coefficient c as follows.
Thus an exponential function can describe the data. However, in general the least- squares method will give a nonzero value for b because of the scatter or measurement error that is usually present in the data. In this case we know that the relation is a power function with an exponent of 2, and we need to estimate the value of the coefficient c.
The function that gives the smallest J value gives the best fit. Thus, the closer r 2 is to 1, the better the fit. The largest r 2 can be is 1. The value of S is an indication of how much the data are spread around the mean, and the value of J indicates how much of the data spread is left unaccounted for by the model. It is possible for J to be larger than S, and thus it is possible for r 2 to be negative.
Such cases, however, are indicative of a very poor model that should not be used. When the least-squares method is applied to fit quadratic and higher-order polynomials, the resulting equations for the coefficients are linear algebraic equations that are easily solved. The function fits a polynomial of degree n to data described by the arrays x and y, where x is the independent variable.
The first element p1 of the array p will be m, and the second element p2 will be b. The first element p1 of the array p will be m, and the second element p2 will be log b. The first element p1 of the array p will be m, and the second element p2 will be ln b. So we cannot use the polyfit function without modification to compute k1 and k2.
The following example shows how this is done. In this chapter we established the basic principles for developing mathematical models of these elements and applied these principles to several commonly found examples.
To be useful, numerical values eventually must be assigned to the parameters of a mathematical model of a spring or a damping element in order to make predictions about the behavior of the physical device. Therefore, this chapter included two sections that show how to obtain numerical parameter values from data by using the least-squares method.
However, if you are such a reader, you are strongly encouraged to consider learning MATLAB, because it will be very useful to you in the future.
Now that you have finished this chapter, you should be able to do the following: 1. Develop models of spring elements from the basic principles of mechanics of materials.
Develop models of damping elements from the basic principles of fluid mechanics. Apply the least-squares method to obtain numerical parameter values for a spring or damping element model when given the appropriate data. Compute the values of A and t. Obtain the expesssions for its velocity and acceleration as functions of time. What is the amplitude of the velocity and of the acceleration? Obtain the expressions for its velocity and acceleration as functions of time.
What is the amplitude of the displacement, of the velocity, and of the acceleration? Compute the value of the phase angle f. Compute the values of A and f. Determine the speed of the motor.
The following table gives the spring length y that was produced in a particular spring by the given applied force f. How should you connect the springs, end-to-end or side-by-side? Assume that the resulting motion is small and thus essentially horizontal, and determine the expression for the equivalent spring constant that relates the applied force f to the resulting displacement x. The coil has six turns. The coil diameter is 0. Assuming the system is in static equilibrium, sketch the plot of f versus x.
We will model the cable as a rod. Denote the translational spring constant of the beam by kb and the translational spring constant of the cable by kc. The displacement x is caused by the applied force f. Determine the equivalent translational spring constant for this arrangement. Hint: Are the two springs in series or in parallel? The legs are 1 m in length and 0. Compute the equivalent spring constant due to the legs, assuming the table top is rigid.
The steel beam is 0. The mass m is kg. All the springs have the same spring constant k. Investigate the accuracy of this approximation by creating three plots. For the first plot, plot sin x and x versus x for 0 x 1. Spring 2 is at its free length in the configuration shown. Figure P1. The mass m in the middle moves a distance x horizontally. Derive the relation between the horizontal restoring force f and the displacement x, assuming that x is small.
Neglect friction. The contact area is 0. Compute the uncertainty in the value of the damping coefficient. The damper resists with a force T as a function of the piston velocity v.
A thin disk of diameter D, suspended by a wire, is in contact with the surface of the liquid, whose thickness is h. In later chapters we will derive an expression for the damping coefficient c in terms of the observed oscillation frequencies of the disk when in and out of contact with the liquid. To compute the viscosity m, we need an expression for m as a function of c, D, and h.
Derive this expression. Are the dampers in series or in parallel? The following table gives the measured deflection x that was produced in a particular beam by the given applied force f. Plot the data to see whether a linear relation exists between f and x, and determine the spring constant by drawing a straight line through the data by eye.
Discuss the significance of the intercept b. The subject consists of kinematics, the study of motion without regard to the forces causing the motion, and kinetics, the relationship between the forces and the resulting motion. The single most important step in analyzing a vibratory system is to derive the correct description of the dynamics of the system. If this description is incorrect, then any analysis based on that description will also be incorrect.
The description may be in the form of a differential equation of motion, but there are analysis methods that do not require the equation of motion, such as the work-energy methods of Section 2. These methods, however, are based on the principles of dynamics covered in this chapter. The chapter begins with a discussion of kinematics principles.
Vibrating systems are often categorized by their number of degrees of freedom, abbreviated DOF; this concept is developed in Section 2. This chapter focuses on the simplest systems: those having one degree of freedom. The kinetics of rigid bodies is covered in Section 2. Section 2. These concepts use kinetic energy equivalence to simplify the process of obtaining descriptions of systems containing both translating and rotating parts, systems containing distributed mass, and systems consisting of multiple masses whose motion is coupled.
Although the chapter focuses on motion with one degree of freedom, its principles can be easily extended to obtain models of systems having two or more degrees of freedom. This is done in Chapters 6 and 8. Kinematics is primarily concerned with the relationship among displacement, velocity, and acceleration, and thus it provides the basis for describing the motion of the systems we will be studying.
For the most part we will restrict our attention to motion in a plane. This means that the object can translate in two dimensions only and can rotate only about an axis that is perpendicular to the plane containing these two dimensions.
The completely general motion case involves transla- tion in three dimensions and rotation about three axes. This type of motion is considerably more complex to analyze. For such details, consult a reference on dynamics. We present a summary of the pertinent relations without derivation or extensive discussion. Categories of Plane Motion A particle is a mass of negligible dimensions. We may consider a body to be a particle if its dimensions are irrelevant for specifying its position and the forces acting on it.
We may represent the position of a particle by a position vector r measured relative to any convenient coordinate system. It is important to understand that the validity of the laws of physics does not depend on the coordinate system used to express them; the choice of a coordinate system should be made entirely for convenience.
If, however, the dimensions of the body are not negligible, we must consider the possibility of rotational motion.
There are four categories of plane motion. These are illustrated in Figure 2. In translation every line in the body remains parallel to its original position, and there is no rotation of any line in the body. Rectilinear translation occurs when all points in the body move in parallel straight lines, whereas curvilinear translation occurs when all points move on congruent curves.
In both cases, all points on the body have the same motion, and thus the motion of the body may be completely described by describing the motion of any point on the body. Rectilinear translation is usually easily recognized, but curvilinear translation is not always obvious. For example, the swinging plate in Figure 2.
The third category of plane motion is rotation about a fixed axis. In such motion all points in the body move in circular paths. The fourth category, general plane motion, consists of both translation and rotation.
The connecting rod in Figure 2. You may use any convenient measure of displacement as a coordinate. The proper choice of coordinates can simplify the resulting expressions for position, velocity, or acceleration, and even a simple problem can be made difficult if you select the coordinates unwisely. As we just saw, the motion of a simple pendulum see Figure 2. So the constant length L acts as a constraint on the motion of the mass at the end of the pendulum.
Not only do the rectangular coordinates result in more complicated expressions for velocity and acceleration, but the two coordinates are unnecessary because the pendulum rotating in a single plane has only one degree of freedom, abbreviated as DOF.
Thus only one of the two coordinates is independent. The presence of constraints that restrict the motion of an object usually means that fewer coordinates are needed to describe the motion of the object. Generalized coordinates are those that reflect the existence of constraints and yet still provide a complete description of the motion of the object.
Referring again to the pendulum, the angle u is a generalized coordinate, but the coordinates x and y are not. However, if a spring replaces the rigid rod, as in Figure 2. If the base of the pendulum is free to slide, as in Figure 2. Two pendula coupled as shown in Figure 2. Similarly, the system shown in Figure 2.
In Figure 2. Either x or u may be used as a generalized coordinate. A vibrating string or cable, such as shown in Figure 2. There are an infinite number of possible values of x; therefore there are an infinite number of possible values of y.
Finally, when interpreting the illustrations in this text, you should assume that the number of degrees of freedom is the smallest number that is consistent with the figure; that is, assume that the object has the simplest motion consistent with the figure, unless otherwise explicitly stated.
For example, an object attached to a spring, as in Figure 2. Thus it has six degrees of freedom. However, with the representation shown in Figure 2. In our Pulley-Cable examples we will assume that the cords, ropes, chains, and cables that drive the pulleys do Kinematics so without slipping and are inextensible; if not, then they must be modeled as springs.
Figure 2. Suppose the pulley is free to rotate and to translate vertically, but not horizontally. Let L be the length of the cable, which is attached to a support. Solution Suppose the system is initially in the configuration shown in Figure 2. Let x be the displacement of the endpoint and y be the displacement of the pulley center, as shown in Figure 2. This says that the speed of the cable end A is twice the vertical translational speed of the pulley center.
A Pulley System a How many degrees of freedom does it have? Solution a If the cable does not stretch, the system has one degree of freedom, because the motion of mass m2 is not independent of the motion of mass m1.
To see why this is true, write the expression for the cable length. As another example, consider Figure 2. Suppose we need to determine the relation between the velocities of mass mA and mass mB. Define x and y as shown from a common reference line attached to a fixed part of the system. So the speed of point A is three times the speed of point B, and in the opposite direction. Such motion can involve both translation and rotation.
A particle is a mass of negligible dimensions whose motion can consist of translation only. For example, we normally need not know the size of an earth satellite in order to describe is orbital path.
The second law implies that the vector sum of external forces acting on a body of mass m must equal the time rate of change of momentum. Note that the acceleration vector and the force vector lie on the same line. The free-body diagram provides a reliable way of accounting for all these forces.
To draw the diagram, you mentally isolate the particle from all contacting and influencing bodies and replace the bodies so removed by the forces they exert on the particle. For any unknown forces, when drawing the free-body diagram, you should assume a direction for the force and denote its magnitude with a variable. If the object can rotate, then the translational equations must be supplemented with the rotational equations of motion, which are treated in this section. The simplest case in this category of motion occurs when the body is constrained to rotate about a fixed axis.
Slightly more complicated cases occur when the object is free to rotate about an axis passing through the center of mass, which may be accelerating, or when the object is constrained to rotate about an axis passing through an arbitrary but accelerating point. Thus, Equation 2. The term torque and the symbol T are often used instead of moment and M.
The expressions for I for some common shapes are given in Table 2. More extensive tables can be found in handbooks and engineering mechanics texts for example, see [Meriam, ]. The pendulum Equation of swings about the fixed pivot at point O.
Assume that the rod mass is negligible compared Motion of a to m. Pendulum Solution We can use Equation 2. In the free-body diagram in Figure 2. From Table 2. From Equation 2. The pulley inertia will be negligible if either its mass or its radius is small.
Thus, when we neglect the inertia of a pulley, the tension forces in the cable may be taken to be the same on both sides of the pulley.
General Plane Motion We now consider the case of an object undergoing general plane motion, that is, translation and rotation about an axis through an accelerating point.
Assume that the object in question is a rigid body that moves in a plane passing through its mass center, and assume also that the object is symmetrical with respect to that plane. Thus it can be thought of as a slab with its motion confined to the plane of the slab. We assume that the mass center and all forces acting on the mass are in the plane of the slab.
We can describe the motion of such an object by its translational motion in the plane and by its rotational motion about an axis perpendicular to the plane. Two force equations describe the translational motion, and a moment equation is needed to describe the rotational motion.
Consider the slab shown in Figure 2. Define an x-y coordinate system as shown with its origin located at a nonaccelerating point. The mass center is located at point G. The quantities aGx and aGy are the accelerations of the mass center in the x and y directions, relative to the fixed x-y coordinate system. The terms IG and a are the mass moment of inertia and angular acceleration of the body about this axis.
The net moment MG is caused by the action of the external forces f1 , f2 , f3 ,. The positive direction of MG is determined by the right-hand rule counterclockwise if the x-y axes are chosen as shown. Note that point G in the preceding equations must be the mass center of the object; no other point may be used. However, in many problems the acceleration of some point P is known, and sometimes it is more convenient to use this point rather than the mass center or a fixed point.
The terms rx and ry are the x and y components of the location of G relative to P. IP is the mass moment of inertia of the body about an axis through P. Note that the angular acceleration a is the same regardless of whether point O, G, or P is used. CurvilinearTranslation If the mass center moves along a curved path curvilinear translation , the following forms of the force equations may be more useful.
In terms of the mutually perpendicular normal and tangential coordinates n and t, the force equations are written as Figure 2. In summary, the equations of motion for a rigid body in plane motion are given by two force equations expressed in an appropriate coordinate system, and the moment equation—either Equation 2. All the equations need not be used for given problem; the appropriate choice depends on the nature of the problem.
The Pendulum Revisited Many engineering devices rotate about a fixed point and are examples of a pendulum, although they might not appear to be so at first glance. In the pendulum analysis of Example 2. This is not always the case, however, and the following example examines this situation. Its mass moment of inertia about its mass center G Pendulum is IG.
Its mass center is a distance L from point O. Obtain its equation of motion in terms of the angle u and determine the reaction forces on the support at point O, in terms of u and u. Solution Choosing the moment equation, Equation 2. The moment MO is caused by the weight mg of the object acting through the mass center at G Figure 2. Because the mass center G does not translate in a straight line, it is more convenient to use normal-tangential coordinates to find the reaction forces.
The reaction forces acting on the body at point O are f1 and f2. For the normal direction, Equations 2. The distance h of M from the mass center G is called the metacentric height.
Solution Use the moment equation about the mass center G and the force diagram in Figure 2. This has the same form as the pendulum equation, Equation 1 of Example 2.
We can use this solution to determine the roll frequency of the ship. Its hook and suspended load act A Crane like a pendulum whose base is moving. The hook pulley and load have a combined weight Application mg. The trolley is moving to the left with an acceleration a.
Treat the pulley cable as a rigid rod of length L and negligible mass, and obtain the equation of motion in terms of u. Solution This is an application for the moment equation, Equation 2. From Figure 2. Equation 2. The wheel shown in Figure 2. Pure rolling motion. This occurs when there is no slipping between the wheel and the surface.
Pure sliding motion. This occurs when the wheel is prevented from rotating such as when a brake is applied. Sliding and rolling motion. The wheel will roll without slipping pure rolling if the tangential force ft is smaller than the static friction force ms N, where N is the force of the wheel normal to the surface Figure 2.
In this case the tangential force does no work because it does not act through a distance. If the static friction force is smaller than ft , the wheel will slip. Work-energy methods are based on an integration of the equation of motion over displacement, whereas impulse-momentum methods are based on an integration over time. The term on the left is called the kinetic energy KE , often denoted by the symbol T. This equation expresses the principle of conservation of mechanical energy, which states that the change in kinetic energy plus the change in potential energy is zero.
This principle is not valid if nonconservative forces do work on the object. Note that the potential energy has a relative value only. The choice of reference point for measuring x determines only the value of the constant C, which Equation 2. Elastic Potential Energy An elastic element has a potential energy whenever it is stretched or compressed.
Note that this work is the same regardless of whether the spring is compressed or stretched. In many of our applications the potential energy consists of gravitational potential energy and the potential energy due to any spring elements. To distinguish them, we will denote the gravitational potential energy by Vg and the potential energy due to a spring element by Vs.
Following the development leading to Equation 2. The kinetic energy of a rigid body is the sum of the kinetic energy of translation and the kinetic energy of rotation. The Work-Energy Principle Let us look more closely at the concept of work.
Thus the work may be interpreted either as the component of F, F cos u, along the dr direction times the displacement ds or as the force F times the component of dr, ds cos u, along the F direction.
The work is positive if F cos u is in the direction of the displacement. Note that the component of F normal to dr, F sin u, does no work. Let T1 be the kinetic energy of the particle at point 1 and T2 the kinetic energy at point 2. Similarly, let V1 and V2 be the potential energy of the particle at point 1 and point 2.
Let W 1—2 denote the work done on the particle during displacement from point 1 to point 2 by all external forces other than gravity or spring forces. Calculate the velocity of the mass m after it has dropped 0. Solution a From the equilibrium free-body diagram shown in Figure 2. Application of The coefficient of dynamic friction between the mass and the surface is m.
Solution a The free-body diagram is shown in Figure 2. The force doing work is the friction force mN. Note that the work done by the friction force on the mass m is negative because the friction force is in the opposite direction of the displacement. Note also that the normal force N does not do work and thus does not appear in the work-energy equation. This equation says that the time rate of change of the linear momentum of a particle equals the resultant of all forces acting on the particle.
This equation expresses the linear impulse-momentum principle, which states that the linear impulse equals the change in linear momentum. Note that these results apply also to a system of particles. For example, in a system composed of two particles, if the two particles have equal but opposite forces applied to them, the vector sum of forces acting on the system is zero, and thus the momentum of the system is conserved.
Assume that the impact is inelastic so that m1 sticks Linear Momentum to m2. Calculate the maximum spring deflection caused by the impact. Solution We may think of this process as consisting of three phases: 1 the drop, 2 the impact, and 3 the motion immediately after the impact.
Let v1 be the velocity of m1 just before impact. Let v2 be the velocity of the combined mass after impact. This is the lowest point reached by the combined mass after the impact. Of course, the mass will not remain at rest in this position but will oscillate.
This relation can be extended to a system of particles. This equation expresses the principle of angular impulse and momentum, which states that the change in angular momentum of m about a fixed point O equals the total angular impulse acting on m about O. This means that the particles rebound from each other and no energy is lost. This means that the particles stick to one another; however, all the energy is not necessarily lost.
For plane motion, the angular momentum and angular impulse vectors are normal to the plane of motion, and so vector notation is not needed. Motion Solution a The free-body diagram is shown in Figure 2. Assume that the pendulum is a slender rod of mass m2. Solution a The angular momentum of the entire system consisting of the mass m1 and the pendulum is conserved because the force of impact acting on the pendulum is equal but opposite to that acting on the mass m1.
Thus we will take the mass center still to be located a distance L from the pivot point. The system in Figure 2. The cart body has a mass mb , and each wheel has a mass mw. The next section treats systems with springs, but for now let us concentrate on the cart dynamics.
One way to analyze the cart dynamics is to draw two free-body diagrams, one for the cart body and one for the wheel-axle system. To do this you must include the reaction forces between the axle and the cart body.
These diagrams will produce three equations of motion for the wheel-axle system one rotational and two translational equations and two translational equations for the cart body. These five equations must be combined to eliminate the unknowns to obtain a single equation in terms of the variable x. An easier way to solve this problem is to find the equivalent mass of the cart-wheel system and model the system as shown in Figure 2. Instead of a block, suppose we have a wheel of mass mw , radius R, and moment of inertia Iw Figure 2.
Roger Burton. Anonymous i6iMRM1C2. Daniel CB. Mihir Ashar. Wacko Asahan. Khalil Raza. Marcos Felipe. Mechanical Vibrations S. Yau Sai Lung. Yair Zarza. Majid Khan. More From Rogerio. Fainer Cerpa Olivera. Automatic Control for Mechanical Engineers by M. Bui van Trinh. Mahdi Daly. Andrei Ciungan. Amit Nirmal. With a clear explanation of worked examples, applications and modern computer tools, William Palm's Mechanical vibration system provides a solid foundation.
You will learn how to use math and science knowledge modeling and analysis system, from a single degree of freedom system, there are two or more degrees of freedom. Independent MATLAB at the end of most chapters show how to use this standard engineering tool, to solve vibration problems in the context of the function. The text introduced Simulink in MATLAB solutions may be difficult for the program, such as modeling Coulomb friction effects and simulation systems, including nonlinear.
Adequate throughout the text provides the opportunity to practice identification, formulation, to solve the vibration problem. Public Pastes.
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