Mechanical vibration william j palm iii pdf free download

Mechanical vibration william j palm iii pdf free download

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Section 1. The Least-Squares Method A systematic and objective way of obtaining a functional description of a set of data is the least-squares method. Suppose we want to find the coefficients of the straight line y ¼ mx þ b that best fits the following data: x 0 5 10 y 2 6 11 According to the least-squares criterion, the line that gives the best fit is the one that minimizes the sum of the squares of the vertical differences between the line and the data points Figure 1. We denote this sum by J.

If we evaluate this equation at the data values x ¼ 0, 5, and 10, we obtain the values y ¼ , 6. These values are different from the given data values y ¼ 2, 6, and 11 because the line is not a perfect fit to the data. No other straight line will give a lower value of J for this data. The General Linear Case We can generalize the above results to obtain formulas for the coefficients m and b in the linear equation y ¼ mx þ b. Use the least-squares method to Fitting the Linear estimate the spring constant k for the linear model f ¼ kx. Spring Function Solution Here our model is f ¼ mx þ b, so f plays the role of y in Equations 1. The required terms for these equations are n ¼ 9 and Xn Xn x2i ¼ xi ¼ i¼1 i¼1 X n X n yi xi ¼ fi xi ¼ i¼1 i¼1 X n X n yi ¼ fi ¼ i¼1 i¼1 Thus Equations 1. The plot is shown in Figure 1. Note that it does not pass through the origin.

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 a power function f ¼ cvm can describe the data. This gives c ¼ 10B ¼ Thus the desired function is f ¼ v , and the the estimate of the damping constant is c ¼ Thus an exponential function can describe the data. Thus gives b ¼ eB ¼ Note that this value of vð0Þ does not equal the first data value v ¼ If we know the mass m, we can compute c from c ¼ m.

For example, the force-deflection model of a spring must pass through the origin ð0; 0Þ because the spring exerts no force when it is not stretched. Thus a linear model y ¼ mx þ b sometimes must have a zero value for b. 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. To obtain a zero-intercept model of the form y ¼ mx, we must derive the equation for m from basic principles. Linear Spring Model Solution The solution is given by Equation 1. For example, the orifice damping relation states that f ¼ cv2.

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. In such cases we can modify the least-squares method to find the best-fit function of the form f ¼ cv2. The error ei is the difference between the data value yi and the value of y obtained from the function; that is, f ðxi Þ. The least-squares criterion used to fit a function f ðxÞ is the sum of the squares of the residuals, J. 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. If you are unfamiliar with MATLAB, you should study Appendix A before proceeding. In this section we show how to use MATLAB to fit functions to data. 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. Their solution forms the basis for the algorithm contained in the MATLAB polyfit function.

The function fits a polynomial of degree n to data described by the arrays x and y, where x is the independent variable. The result p is the row array of length n þ 1 that contains the polynomial coefficients in order of descending powers. With a suitable transformation, the power and exponential functions can be transformed into a linear function, but the polynomial w ¼ p1 z þ p2 has a different interpretation in each of the three cases: The Linear Function: y ¼ mx þ b. The first element p1 of the array p will be m, and the second element p2 will be b. The Power Function: y ¼ bxm. In this case log y ¼ m log x þ log b, which has the form w ¼ p1 z þ p2 , where the polynomial variables w and z are related to the original data variables x and y by w ¼ log y and z ¼ log x. The first element p1 of the array p will be m, and the second element p2 will be log b.

We can find b from b ¼ 10 p2. The Exponential Function: y ¼ bemx. In this case ln y ¼ mx þ ln b, which has the form w ¼ p1 z þ p2 , where the polynomial variables w and z are related to the original data variables x and y by w ¼ ln y and z ¼ x. The first element p1 of the array p will be m, and the second element p2 will be ln b. We can find b from b ¼ ep2. Use MATLAB to fit Fitting a Linear a linear model to the data. Use MATLAB to fit a power Fitting a Power function to the data. So we transform the data with the transformations V ¼ log v and F ¼ log f , then apply the polyfit function. Thus the fitted model is f ¼ v Use MATLAB to fit an Fitting an exponential function to the data. However, using the MATLAB polyfit function polyfit x,f,3 will give a function of the form f ¼ a0 þ a1 x þ a2 x2 þ a3 x3 , in which a0 and a2 are nonzero in general.

So we cannot use the polyfit function without modification to compute k1 and k2. The following example shows how this is done. The MATLAB program to implement Equation 1. Quality of Fit with MATLAB Solution Continue the script file given in Example 1. 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.

These functions were presented in a separate section that may be skipped by readers who will not be using MATLAB. 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. Use MATLAB to implement the least-squares method. PROBLEMS SECTION 1. SECTION 1. 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. Force f ðNÞ Spring length y ðmÞ 0 1. 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. Determine the equivalent spring constant ke for this arrangement, where f ¼ ke x. k1 k2 f k1 x FIGURE P1. 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. a Are the two springs in series or in parallel? b What is the equivalent spring constant for this arrangement? Determine the equivalent translational spring constant for this arrangement. Hint: Are the two springs in series or in parallel? D1 D2 x f L1 L2 FIGURE P1. 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. m k FIGURE P1. All the springs have the same spring constant k. f FIGURE P1. 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. pffiffiffi 1. Assuming the system is in static equilibrium and that the angle u is small, determine the equivalent spring constant ke for this arrangement, where f ¼ ke x. k2 x θ f k1 L2 L1 L3 FIGURE P1. When x ¼ 0, spring 1 is at its free length. Spring 2 is at its free length in the configuration shown. D1 x D2 k1 k2 m FIGURE P1. Figure P1. L q m g FIGURE P1. D x m r 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.

L m x L FIGURE P1. Neglect friction. The contact area is 0. a Assuming that the oil film is 1 mm thick, compute the damping coefficient c. b How much force is required to keep the plate moving at 0. Compute the uncertainty in the value of the damping coefficient. The damper resists with a torque T as a function ω h ε D ε d FIGURE P1. The damper resists with a force T as a function of the piston velocity v. f v Cylinder Piston L ε D ε FIGURE P1. A thin disk of diameter D, suspended by a wire, is in contact with the surface of the liquid, whose thickness is h. The wire acts like a torsional spring and the liquid acts like a torsional ω h D FIGURE P1. 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. The shaft diameter is D ¼ 2 in.

Are the dampers in series or in parallel? b Obtain the expression for the equivalent damping coefficient ce for Figure P1. 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. Use the least-squares method to estimate the spring constant k for the linear model f ¼ kx þ b. 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. a Rectilinear translation. b Curvilinear translation. c Fixed-axis rotation. d General plane motion. a The plate shows curvilinear translation if the links are identical and parallel; each link rotates about a fixed axis. b The connecting rod has general plane motion, while the piston has rectilinear translation and the crank has fixed-axis rotation.

EXAMPLE 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.

The coordinates x and y are constrained by the fact that x2 þ y2 ¼ L2. 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. The coordinates ðx; uÞ may be used as generalized coordinates. Two pendula coupled as shown in Figure 2. The coordinates ðu1 ; u2 Þ may be used as generalized coordinates. Similarly, the system shown in Figure 2. The coordinates ðx1 ; x2 Þ may be used as generalized coordinates. In Figure 2. The object has one degree of freedom because Dx ¼ RDu. 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 The mass attached to a spring can move with six degrees of freedom in general translation in three directions and rotation about three axes. b Representation of a mass constrained to have one degree of freedom, vertical translation.

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. Noting that the cable lengths wrapped around the pulleys are constant, we can write x þ 3y ¼ constant. 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. speed, will remain that way as long as it is not acted upon by an unbalanced external force. 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. Thus dðmvÞ ¼f 2. For an object treated as a particle of mass m, if the mass is constant, the second law can be expressed as dv m ¼ ma ¼ f 2. Note that the acceleration vector and the force vector lie on the same line. If the mass is constrained to move in only one direction, say along the direction of the coordinate x, then the equation of motion is the scalar equation dv ma ¼ m ¼f 2.

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.

The magnitude of the angular momentum can be expressed as H ¼ Iv 2. 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, ]. Axis I u v M FIGURE 2. b Free-body diagram showing the reaction force in the rod. 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. The force on the support at the pulley center is F3 ¼ F1 þ F2. 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. Then the two force equations can be written as fx ¼ maGx 2.

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. For this case we can apply the following moment equation: MO ¼ IO a 2. The following moment equation applies regardless of whether the axis of rotation is constrained or not: MG ¼ IG a 2. 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 ,. and any couples applied to the body. 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 following moment equation applies for an accelerating point P, which need not be fixed to the body: MP ¼ IG a þ maG d 2. 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, in general, MP does not equal MG and IP does not equal IG. If point P is fixed at some point O, then aPx ¼ aPy ¼ 0, and the moment equation, Equation 2. 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.

where fn and ft are the net external forces in the n and t directions, and an and at are the acceleration components in those directions. Because the coordinates are perpendicular, the total acceleration magnitude a can be found from qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a ¼ a2n þ a2t 2. In this case the following relations are also true: v ¼ Rv 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. b Free-body diagram. c Forces acting at point G. 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. a b b Free-body diagram. 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. In a later chapter we will see how to solve this equation for uðtÞ. 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. In this case, v ¼ Rv.

b Free-body diagram of the equivalent pendulum model. Pure sliding motion. This occurs when the wheel is prevented from rotating such as when a brake is applied. In this case v ¼ 0 and v 6¼ Rv. Sliding and rolling motion. In this case v 6¼ 0. Because slipping occurs in this case, v 6¼ Rv. 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. If slipping occurs, the tangential force is computed from the dynamic friction equation: ft ¼ md N. 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.

Work is force times displacement, so the integral on the right represents the total work done on the mass by the force f ðxÞ. The term on the left is called the kinetic energy KE , often denoted by the symbol T. VðxÞ is called the potential energy function PE. If v and x have the values v0 and x0 at the time t0 , then mv20 þ Vðx0 Þ ¼ C 2 Comparing this with Equation 2. 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. For some problems, the following forms of the principle are more convenient to use: mv20 mv2 þ Vðx0 Þ ¼ þ VðxÞ 2. 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. The equation of motion of a body of inertia I not translating but in pure rotation about an axis through a fixed point or through the center of mass is dv I ¼M dt where v is the angular velocity and M is the applied moment. 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. For general plane motion, the kinetic energy is 1 1 T ¼ mv2G þ IG v2 2 2 where vG is the velocity of the center of mass and IG is the moment of inertia about the mass center. 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. a What is the Conservation of static deflection in the spring when the mass m is stationary? b Suppose the spring is Energy initially stretched 0. 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. b Free-body diagram before impact. 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. b Let d be the maximum spring deflection. The desired answer is the positive solution, d ¼ m. This can be expressed as G where the linear momentum of the particle is G ¼ mv. 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. If the linear impulse is zero, then G1 ¼ G2 , which states that the linear momentum is conserved. 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. Selecting the positive solution and adding it to the initial static deflection of m, we obtain the total deflection: þ ¼ m.

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. If the angular impulse is zero, then HO1 ¼ HO2 which states that the angular momentum is conserved. Impact When one particle impacts another, the coefficient of restitution e is defined as jrelative velocity of separationj e¼ 2.

This means that the particles rebound from each other and no energy is lost. The impact is perfectly plastic or inelastic if e ¼ 0. This means that the particles stick to one another; however, all the energy is not necessarily lost. where G ¼ mvG. For plane motion, the angular momentum and angular impulse vectors are normal to the plane of motion, and so vector notation is not needed. The angular impulse-momentum principle for this case states that Z t2 HG1 þ MG dt ¼ HG2 t1 where HG ¼ IG v. For any point O, fixed or moving, on or off the body, HO ¼ IG v þ mvG d 2. If the body is constrained to rotate about an axis through a fixed point O, which may be on or off the body, then H O ¼ IO v 2. Motion Solution a The free-body diagram is shown in Figure 2. b a b Free-body diagram. Assume that the pendulum is a slender rod of mass m2. a Determine the angular velocity v of the pendulum immediately after the impact.

b Determine the maximum angular displacement of the bar following impact. 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. b Equivalent translating sytem. section, it is much easier to obtain the equation of motion for this equivalent system because we need not deal with two separate masses and the reaction forces between them.

Instead of a block, suppose we have a wheel of mass mw , radius R, and moment of inertia Iw Figure 2. b Motion of a wheel and axle. If the wheel does not slip, its rotational speed is related to the translational speed by v ¼ Rv. One way to solve this problem is to find the equivalent mass of the wheel and use Equation 2. The equivalent mass is the fictitious mass value that makes the wheel have the same kinetic energy as the block if they are both moving at the same speed v. The equivalent mass can be interpreted as the mass value that a blindfolded person would feel while pushing on the axle, thinking he or she was pushing a block assuming that the wheel and block are moving horizontally. Thus the wheel is equivalent to a block having a mass Iw m ¼ mw þ 2.

Another way to compute the equivalent mass is to equate the force f required to give the block and the axle the same acceleration v. This force produces a torque Rft on the wheel. This approach gives the same equivalent-mass value obtained with energy equivalence, but the energy-equivalence method is easier because it does not involve the tangential force ft. Equivalent Mass of a Cart For the cart shown in Figure 2. Note that because the tangential wheel force does no work, we must treat the suface in Figure 2. The equivalent mass expresses only the equivalent inertial resistance to changes in motion. It expresses only the kinetic energy, not the potential energy.

Thus we must be careful using the equivalent mass when gravity affects the motion. Incline Solution The equivalent mass is, from Equation 2. Thus the gravity force is mw g. In such cases the concept of equivalent inertia is useful. One finds Equivalent Inertia such a device on lathes, for example, where it is used to translate the cutting tool. It consists of a Lead Screw of a rotating threaded rod and a translating mass. A motor supplies the torque T required to rotate the rod. Obtain the expression for its equivalent inertia and obtain its equation of motion. Solution The expression for the equivalent inertia of the lead screw is found by equating the kinetic energy of the real system translation plus rotation with the kinetic energy of rotation of the fictitious, purely rotational system. The mass m rides on a support having ball bearings aligned with the screw threads.

When the screw rotates one revolution 2p radians , the mass translates the distance L, which is the distance between the threads. A system should be viewed as an equivalent mass if an applied force is specified and as an equivalent inertia if an applied torque is specified. Equivalent Mass of Spring Elements Kinetic energy equivalence is often used to determine the equivalent mass of the spring element, because kinetic energy is associated with the mass parameter. The following example shows how to include the mass of a spring. This fraction is obtained by assuming that the velocity of a particle along the length of the rod is proportional to its distance from the support. At the support, the velocity is of course zero. At a point halfway from the support, the velocity is one-half the velocity at the end of the rod, and so on.

The equivalent inertia of rotational spring elements can be found using kinetic energy equivalence. The rod in torsion, shown in Table 2. They have dynamics of their own, due to their mass, elasticity, and damping properties. For example, a lever can flex if the forces it transmits are great enough. For example, we treat a lever as a rigid rod whenever possible. This gives a simpler model whose coordinates are those of the reference location. Here we show how to use kinetic energy equivalence to facilitate this process. In a later chapter we will show how to use elastic potential energy and energy dissipation expressions to obtain equivalent spring and damping models.

Geared Systems Several types are gears are used in mechanical systems. These include helical-spur gears, worm gears, bevel gears, and planetary gears. We now use the spur gear pair to demonstrate how to use kinetic energy equivalence to obtain a simpler model. This approach can be used to analyze other gear types and other drive types, such as belt or chain drives. The gear inertias are IC and ID. The shaft inertias are IB and IE. The end view is shown in Figure 2. Represent the system by the equivalent single-shaft, gearless system shown in Figure 2. The torque TF acting on IF is an externally applied torque from the motor. Assume that the shaft torsional elasticity, gear tooth elasticity, and gear tooth friction are negligible.

Solution We can represent the geared system shown in Figure 2. The inertia of the pendulum about its pivot point is IA ¼ mL2. Equating the energy expression of the original system to that of the equivalent system, we obtain I2 1 Ie ¼ I 1 þ 2 ¼ mL2 þ IB þ IC þ 2 ðID þ IE þ IF Þ N N Remember that when an inertia is referenced to a slower shaft, its equivalent value must be greater than its true value so that its kinetic energy will remain the same. When referenced to a faster shaft, its equivalent value must be less than its true value. b End view. c Representation of the pendu- TF lum as an equivalent inertia.

d Single ω1 N inertia equivalent to the system shown d in part a. We can now obtain the equation of motion for the equivalent system, by noting that the torque TA is the torque about shaft AB due to the weight of the pendulum. It is TA ¼ mgL sin u1. For this a more detailed model is required. We will develop such models in later chapters. We now consider a number of examples in which the restoring force is due to spring elements. For now we restrict ourselves to systems with one degree of freedom. We will consider two or more degrees of freedom starting with Chapter 6. Real versus Ideal Spring Elements By their very nature, all real spring elements have mass and are not rigid bodies. Thus, because it is much easier to derive an equation of motion for a rigid body than for a distributed-mass, flexible system, the basic challenge in modeling mass-spring systems is to first decide whether and how the system can be modeled as a single rigid body.

If the system consists of an object attached to a spring, the simplest way to model the systems as a single rigid body is to neglect the spring mass relative to the mass of the object and take the mass center of the system to be located at the mass center of the object. With this assumption, we treat the spring as an ideal spring element, which is massless. This assumption is accurate in many practical applications, but to be comfortable with this assumption, you should know the numerical values of the masses of the object and the spring element. In some of the chapter problems and some of the examples in this text, the numerical values are not given. In such cases, unless otherwise explicitly stated, you should assume that the spring mass can be neglected. A real spring element can be represented as an ideal element either by neglecting its mass or by including it in another mass in the system, using the equivalent-mass methods developed in Section 2.

Effect of Spring Free Length and Object Geometry Suppose we attach a cube of mass m and side length 2a to a linear spring of negligible mass, and we fix the other end of the spring to a rigid support, as shown in Figure 2. We assume that the horizontal surface is frictionless. If the mass is homogeneous, its center of mass is at the geometric center G of the cube. The free length of the spring is L. The force f is the net horizontal force acting on the mass due to sources other than the spring. If the applied force f is zero, when we position the mass m so that the spring is at its free length, the mass will be in equilibrium. The equilibrium location of G is the point marked E. In this position the spring has been stretched a distance x from its free length, and thus its force is kx. The free-body diagram of this situation is shown in Figure 2. These two parameters need to be known only to locate the equilibrium position E of the mass center. Therefore we can represent the object as a point mass, as shown in Figure 2.

Unless otherwise specified, you should assume that the objects in our diagrams can be treated as point masses and therefore their geometric dimensions need not be known to obtain the equation of motion. a The block of side length 2a is in equilibrium at point E when the applied force f is zero. b The displacement x is measured from the equilibrium position of the mass center G. c Free-body diagram. The hatched-line symbol shown in Figure 2. The Equilibrium Position as Coordinate Reference Figure 2. and most chapters have a separate section devoted to the Simulink features relevant to the chapter. Read with Our Free App Audiobook Free with your Audible trial, Read book FormatPDFEBook, Ebooks Download PDF KINDLE, Download [PDF] and Readonline, Read book Format PDF EBook, Download [PDF]and Read OnlineLINK READ OR DOWNLOAD, CLICK NEXT PAGE.

Extended embed settings. You have already flagged this document. Thank you, for helping us keep this platform clean. The editors will have a look at it as soon as possible. EN English Deutsch Français Español Português Italiano Român Nederlands Latina Dansk Svenska Norsk Magyar Bahasa Indonesia Türkçe Suomi Latvian Lithuanian český русский български العربية Unknown. Self publishing. Login to YUMPU News Login to YUMPU Publishing. TRY ADFREE Self publishing Discover products News Publishing. Share Embed Flag. SHOW LESS. ePAPER READ DOWNLOAD ePAPER. TAGS download vibration mechanical ebook engineering chapters simulink audiobook ebooks register. Create successful ePaper yourself Turn your PDF publications into a flip-book with our unique Google optimized e-Paper software. START NOW. Read with Our Free App Audiobook Free with your Audible trial, Read book FormatPDF EBook, Ebooks Download PDF KINDLE, Download [PDF] and Readonline, Read book Format PDF EBook, Download [PDF] and Read Online LINK READ OR DOWNLOAD, CLICK NEXT PAGE Mechanical Vibration BOOK DETAIL Amazon Business : For business-only pricing, quantity discounts and FREE Shipping.

Register a free business account Hardcover: pages Publisher: Wiley; 1 edition March 3, Language: English ISBN ISBN Product Dimensions: 8. Click Button "DOWNLOAD" Or "READ ONLINE" 2. Sign Up To Acces "Mechanical Vibration" 3. Choose the book you like when you register 4. You can also cancel your membership if you are bored 5. Common applications of the subject include vehicle suspension systems and vibration isolators and absorbers designed to minimize the effects of vibration on people and machines. Prerequisite subject is dynamics, and Mechanical Vibration provides many examples of deriving vibration models from dynamics principles.

The first seven chapters form the core text material. One or more of the remaining four chapters may be covered, depending on the time available and on the emphasis desired by the instructor. Because MATLAB is now the standard applications language used in engineering curricula, this language is emphasized in the text, via separate sections at the end of chapter. In addition, Simulink, which provides a graphical user interface that is often much easier to use than traditional command-line programs, is covered in the text. This is the only vibration text to cover Simulink, and most chapters have a separate section devoted to the Simulink features relevant to the chapter. More documents Similar magazines Info. Read with Our Free App Audiobook Free with your Audible trial, Read book FormatPDFEBook, Ebooks Download PDF KINDLE, Download [PDF] and Readonline, Read book Format PDF EBook, Download [PDF]and Read OnlineLINK READ OR DOWNLOAD, CLICK NEXT PAGE Page 2 and 3: Mechanical VibrationBOOK DETAILAmaz.

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We can be fairly confident of this prediction because we have data below and above lb and we have seen that our model describes this data very well. This is the lowest point reached by the combined mass after the impact. Download [PDF] Cohen s Pathways of the Pulp Expert Consult, 11e By - Kenneth M. Download [PDF] The Bungalow Mystery Nancy Drew Mysteries By - C. Fernand Porsche in the s. This can be expressed as G where the linear momentum of the particle is G ¼ mv. A particle is a mass of negligible dimensions whose motion can consist of translation only.

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