Chapter 4: Spring and Damper Elements in Mechanical Systems PDF

4. 1 Spring Elements 169 Velocity-dependent forces such as ﬂuid drag are the subject of Section 4.4. Elements exerting a resisting force that is a function of velocity are called damping or damper elements. This section and Section 4.5 provide additional examples of how to model systems containing mass, spring, and damping elements. MATLAB can be used to perform some of the algebra required to obtain transfer functions of multimass systems, and can be used to ﬁnd the forced and free response. Section 4.7 shows how this is accomplished. ■ 4.1 SPRING ELEMENTS All physical objects deform somewhat under the action of externally applied forces. When the deformation is negligible for the purpose of the analysis, or when the corresponding forces are negligible, we can treat the object as a rigid body. Sometimes, however, an elastic element is intentionally included in the system, as with a spring in a vehicle suspension. Sometimes the element is not intended to be elastic, but deforms anyway because it is subjected to large forces or torques. This can be the case with the boom or cables of a large crane that lifts a heavy load. In such cases, we must include the deformation and corresponding forces in our analysis. The most familiar spring is probably the helical coil spring, such as those used in vehicle suspensions and those found in retractable pens. The purpose of the spring in both applications is to provide a restoring force. However, considerably more engineering analysis is required for the vehicle spring application because the spring can cause undesirable motion of the wheel and chassis, such as vibration. Because the pen motion is constrained and cannot vibrate, we do not need as sophisticated an analysis to see if the spring will work. Many engineering applications involving elastic elements, however, do not contain coil springs but rather involve the deformation of beams, cables, rods, and other mechanical members. In this section we develop the basic elastic properties of many of these common elements. 4.1.1 FORCE-DEFLECTION RELATIONS A coil spring has a free length, denoted by L in Figure 4.1.1. The free length is the length of the spring when no tensile or compressive forces are applied to it. When a spring is compressed or stretched, it exerts a restoring force that opposes the compression or extension. The general term for the spring’s compression or extension is deﬂection. The greater the deﬂection (compression or extension), the greater the restoring force. The simplest model of this behavior is the so-called linear force-deﬂection model, f = kx (4.1.1) where f is the restoring force, x is the compression or extension distance (the change in length from the free length), and k is the spring constant, or stiffness, which is deﬁned to be always positive. Typical units for k are lb/ft and N/m. Some references, particularly in the automotive industry, use the term spring rate instead of spring constant. Equation (4.1.1) is commonly known as Hooke’s Law, named after Englishman Robert Hooke (1635–1703). When x = 0, the spring assumes its free length. We must decide whether extension is represented by positive or negative values of x. This choice depends on the particular application. If x > 0 corresponds to extension, then a positive value of f represents the force of the spring pulling against whatever is causing the extension. Conversely, Figure 4.1.1 A spring element. x L f k 170 CHAPTER 4 Spring and Damper Elements in Mechanical Systems because of Newton’s law of action-reaction, the force causing the extension has the same magnitude as f but is in the opposite direction. The formula for a coil spring is derived in references on machine design. For convenience, we state it here without derivation, for a spring made from round wire. Gd 4 64n R 3 where d is the wire diameter, R is the radius of the coil, and n is the number of coils. The shear modulus of elasticity G is a property of the wire material. As we will see, other mechanical elements that have elasticity, such as beams, rods, and rubber mounts, can be modeled as springs, and are usually represented pictorially as a coil spring. k= 4.1.2 TENSILE TEST OF A ROD A plot of the data for a tension test on a rod is given in Figure 4.1.2. The elongation is the change in the rod’s length due to the tension force applied by the testing machine. As the tension force was increased the elongation followed the curve labeled “Increasing.” The behavior of the elongation under decreasing tension is shown by the curve labeled “Decreasing.” The rod was stretched beyond its elastic limit, so that a permanent elongation remained after the tension force was removed. For the smaller elongations the “Increasing” curve is close to a straight line with a slope of 3500 pounds per one thousandth of an inch, or 3.5 × 106 lb/in. This line is labeled “Linear” in the plot. If we let x represent the elongation in inches and f the tension force in pounds, then the model f = 3.5 × 106 x represents the elastic behavior of the rod. We thus see that the rod’s spring constant is 3.5 × 106 lb/in. This experiment could have been repeated using compressive instead of tensile force. For small compressive deformations x, we would ﬁnd that the deformations are related to the compressive force f by f = kx, where k would have the same value as before. This example indicates that mechanical elements can be described by the linear law f = kx, for both compression and extension, as long as the deformations are not 15000 Figure 4.1.2 Plot of tension test data. Increasing Tension Force (lbs) Linear 10000 Decreasing 5000 0 0 1 2 3 4 5 6 7 8 Elongation (Thousands of an Inch) 9 10 4. 1 Spring Elements 171 too large, that is, deformations not beyond the elastic limit. Note that the larger the deformation, the greater the error that results from using the linear model. 4.1.3 ANALYTICAL DETERMINATION OF THE SPRING CONSTANT In much engineering design work, we do not have the elements available for testing, and thus we must be able to calculate their spring constant from the geometry and material properties. To do this we can use results from the study of mechanics of materials. Examples 4.1.1 and 4.1.2 show how this is accomplished. Rod with Axial Loading E X A M P L E 4.1.1 ■ Problem Derive the spring constant expression for a cylindrical rod subjected to an axial force (either tensile or compressive). The rod length is L and its area is A. ■ Solution From mechanics of materials references, for example [Young, 2011], we obtain the forcedeﬂection relation of a cylindrical rod: L 4L x= f = f EA πED2 where x is the axial deformation of the rod, f is the applied axial force, A is the cross-sectional area, D is the diameter, and E is the modulus of elasticity of the rod material. Rewrite this equation as f = EA πED2 x= x L 4L Thus we see that the spring constant is k = E A/L = π ED2 /4L. The modulus of elasticity of steel is approximately 3 × 107 lb/in.2 . Thus a steel rod 20 in. long and 1.73 in. in diameter would have a spring constant of 3.5 × 106 lb/in., the same as the rod whose curve is plotted in Figure 4.1.2. Beams used to support objects can act like springs when subjected to large forces. Beams can have a variety of shapes and can be supported in a number of ways. The beam geometry, beam material, and the method of support determine its spring constant. Spring Constant of a Fixed-End Beam E X A M P L E 4.1.2 ■ Problem Derive the spring constant expression of a ﬁxed-end beam of length L, thickness h, and width w, assuming that the force f and deﬂection x are at the center of the beam (Figure 4.1.3). Figure 4.1.3 A ﬁxed-end beam. f x w L h 172 CHAPTER 4 Spring and Damper Elements in Mechanical Systems ■ Solution The force-deﬂection relation of a ﬁxed-end beam is found in mechanics of materials references. It is L3 x= f 192EI A where x is the deﬂection in the middle of the beam, f is the force applied at the middle of the beam, and I A is the area moment of inertia about the beam’s longitudinal axis [Roark, 2001]. The area moment I A is computed with an integral of an area element d A. IA = r2 d A Formulas for the area moments are available in most engineering mechanics texts. For a beam having a rectangular cross section with a width w and thickness h, the area moment is IA = w h3 12 Thus the force-deﬂection relation reduces to L3 12L 3 f = f x= 192Ew h 3 16Ew h 3 The spring constant is the ratio of the applied force f to the resulting deﬂection x, or k= Figure 4.1.4 A leaf spring. 16Ew h 3 f = x L3 Table 4.1.1 lists the expressions for the spring constants of several common elements. Note that two beams of identical shape and material, one a cantilever and one ﬁxed-end, have spring constants that differ by a factor of 64. The ﬁxed-end beam is thus 64 times “stiffer” than the cantilever beam! This illustrates the effect of the support arrangement on the spring constant. A single leaf spring is shown in Table 4.1.1. Springs for vehicle suspensions are often constructed by strapping together several layers of such springs, as shown in Figure 4.1.4. The value of the total spring constant depends not only on the spring constants of the individual layers, but also on the how they are strapped together, the method of attachment to the axle and chassis, and whether any material to reduce friction has been placed between the layers. There is no simple formula for k that accounts for all these variables. 4.1.4 TORSIONAL SPRING ELEMENTS Figure 4.1.5 Symbol for a torsional spring element. Table 4.1.2 shows a hollow cylinder subjected to a twisting torque. The resulting twist in the cylinder is called torsion. This cylinder is an example of a torsional spring, which resists with an opposing torque when twisted. For a torsional spring element we will use the “curly” symbol shown in Figure 4.1.5. The spring relation for a torsional spring is usually written as T = kT θ kT T (4.1.2) where θ is the net angular twist in the element, T is the torque causing the twist, and k T is the torsional spring constant. We assign θ = 0 at the spring position where there is no torque in the spring. This is analogous to the free length position of a translational spring. Note that the units of the torsional and translational spring constants are not the same. FPS units for k T are lb-ft/rad; in SI the units are N · m/rad. 4. 1 Spring Elements Table 4.1.1 Spring constants of common elements. Coil spring 2R Gd 4 64n R 3 d = wire diameter n = number of coils k= Solid rod A k= EA L k= 4Ew h 3 L3 L Simply supported beam L w h Cantilever beam w Ew h 3 4L 3 w = beam width h = beam thickness k= h L Fixed-end beam w h k= 16Ew h 3 L3 k= 2Ew h 3 L3 L Parabolic leaf spring L w h Torsional spring constants of two elements are given in Table 4.1.2. They depend on the geometry of the element and its material properties, namely, E and G, the shear modulus of elasticity. If the cylinder is solid, the formula for k T given in Table 4.1.2 becomes kT = π GD4 32L 173 174 CHAPTER 4 Spring and Damper Elements in Mechanical Systems Table 4.1.2 Torsional spring constants of common elements. Coil spring Ed 4 64n D d = wire diameter n = number of coils kT = D Hollow shaft kT = π G(D 4 − d 4 ) 32L D d L Figure 4.1.6 A torsion-bar suspension. Side view Torsion bar Fixed end Frame Suspension arm Wheel Note that a rod can be designed for axial or torsional loading, such as with a torsionbar vehicle suspension. Thus, there are two spring constants for rods, a translational constant k = πED2 /4L, given previously, and a torsional constant k T . Figure 4.1.6 shows an example of a torsion-bar suspension, which was invented by Dr. Ferdinand Porsche in the 1930s. As the ground motion pushes the wheel up, the torsion bar twists and resists the motion. A coil spring can also be designed for axial or torsional loading. Thus there are two spring constants for coil springs, a translational constant k, given in Table 4.1.1, and the torsional constant k T , given in Table 4.1.2. 4.1.5 SERIES AND PARALLEL SPRING ELEMENTS In many applications, multiple spring elements are used, and in such cases we must obtain the equivalent spring constant of the combined elements. When two springs are connected side-by-side, as in Figure 4.1.7a, we can determine their equivalent spring constant as follows. Assuming that the force f is applied so that both springs have the same deﬂection x but different forces f 1 and f 2 , then f2 f1 = x= k1 k2 4. 1 Figure 4.1.7 Parallel spring elements. Figure 4.1.8 Series spring elements. f f f k1 x ke k2 f x x x Spring Elements k1 ke (a) k2 (b) (a) (b) If the system is in static equilibrium, then f = f 1 + f 2 = k1 x + k2 x = (k1 + k2 )x For the equivalent system shown in part (b) of the ﬁgure, f = ke x, and thus we see that its equivalent spring constant is given by ke = k1 + k2 . This formula can be extended to the case of n springs connected side-by-side as follows: ke = n ki (4.1.3) i=1 When two springs are connected end-to-end, as in Figure 4.1.8a, we can determine their equivalent spring constant as follows. Assuming both springs are in static equilibrium, then both springs are subjected to the same force f , but their deﬂections f /k1 and f /k2 will not be the same unless their spring constants are equal. The total deﬂection x of the system is obtained from 1 f 1 f + = + f x= k1 k2 k1 k2 For the equivalent system shown in part (b) of the ﬁgure, f = ke x, and thus we see that its equivalent spring constant is given by 1 1 1 = + ke k1 k2 This formula can be extended to the case of n springs connected end-to-end as follows: 1 1 = ke k i=1 i n (4.1.4) The derivations of (4.1.3) and (4.1.4) assumed that the product of the spring mass times its acceleration is zero, which means that either the system is in static equilibrium or the spring masses are very small compared to the other masses in the system. The symbols for springs connected end-to-end look like the symbols for electrical resistors similarly connected. Such resistors are said to be in series and therefore, springs connected end-to-end are sometimes said to be in series. However, the equivalent electrical resistance is the sum of the individual resistances, whereas series springs obey the reciprocal rule (4.1.4). This similarity in appearance of the symbols often leads people to mistakenly add the spring constants of springs connected end-to-end, just as series resistances are added. Springs connected side-by-side are sometimes called parallel springs, and their spring constants should be added. According to this usage then, parallel springs have the same deﬂection; series springs experience the same force or torque. 175 176 CHAPTER 4 Spring and Damper Elements in Mechanical Systems E X A M P L E 4.1.3 Determining Equivalent Stiffness ■ Problem Determine the equivalent stiffness ke of the arrangement shown in Figure 4.1.9(a). Figure 4.1.9 Determination of an equivalent spring contact. k k k k f x k (a) k1 = k/3 f k2 = 2k x (b) ke = 2k/7 f x (c) ■ Solution The three springs connected end-to-end have an equivalent stiffness k1 that is found from 1 3 1 1 1 = + + = k1 k k k k or k1 = k/3. The two springs connected side-by-side have an equivalent stiffness k2 that is found from k2 = k + k = 2k The equivalent arrangement is shown in Figure 4.1.9(b). From this we see that the equivalent stiffness ke can be found from 1 1 7 1 1 3 = = + = + ke k1 k2 k 2k 2k which gives ke = 2k/7. The simplest equivalent arrangement is shown in Figure 4.1.9(c). Note that the arrangement in Figure 4.1.9(a) is less stiff than a single spring of stiffness k. Note that springs connected end-to-end result in an arrangement that is less stiff than the springs taken individually. Consider two springs with stiffnesses k1 and k2 connected end-to-end. Their equivalent stiffness is found from 1 1 1 = + ke k1 k2 Thus ke = k1 k2 k1 + k2 4. 1 Spring Elements 177 Now if we write k1 = k and k2 = αk, then ke = α k 1+α It can be shown that the ratio α/(1 + α) < 1, and therefore ke < k. Similarly, springs connected side-by-side result in an arrangement that is more stiff than the springs taken individually. Consider two springs with stiffnesses k and αk connected side-by-side. Their equivalent stiffness is found from ke = k + αk = (1 + α)k Because α > 0, then 1 + α > 1, and therefore ke > k. Spring Constant of a Stepped Shaft E X A M P L E 4.1.4 ■ Problem Determine the expression for the equivalent torsional spring constant for the stepped shaft shown in Figure 4.1.10. L1 L2 Figure 4.1.10 A stepped shaft. D1 D2 ■ Solution Each shaft sustains the same torque but has a different twist angle θ. To see why this is true, imagine that shaft 1 is steel and shaft 2 is a soft licorice stick. Which twists more? Therefore, they are in series so that T = k T1 θ1 = k T2 θ2 , and the equivalent spring constant is given by 1 1 1 = + k Te k T1 k T2 where k T1 and k T2 are given in Table 4.1.2 as k Ti = Gπ Di4 32L i i = 1, 2 Thus k Te = k T1 k T2 k T1 + k T2 Although it is frequently useful to recognize when two or more spring elements are connected side-by-side or end-to-end, it is often necessary to rely on the basic deﬁnitions of parallel and series elements. For example, are the springs in Figure 4.1.11 in series or in parallel? Note that although these two springs appear to be connected side-byside, they are not in parallel, because they do not have the same deﬂection. Thus their equivalent spring constant is not given by the sum, 2k. To determine the equivalent spring constant in any situation, we can always go back to the basic principles of statics. 178 CHAPTER 4 Spring and Damper Elements in Mechanical Systems x Figure 4.1.11 A lever-spring system. k x f f L 2 kx x 2 kx 2 k L 2 O (a) E X A M P L E 4.1.5 O (b) Rx O Ry (c) Spring Constant of a Lever System ■ Problem Figure 4.1.11 shows a horizontal force f acting on a lever that is attached to two springs. Assume that the resulting motion is small enough to be only horizontal and determine the expression for the equivalent spring constant that relates the applied force f to the resulting displacement x. ■ Solution From the triangle shown in part (b) of the ﬁgure, for small angles θ, the upper spring deﬂection is x and the deﬂection of the lower spring is x/2. Thus the free body diagram is as shown in part (c) of the ﬁgure. For static equilibrium, the net moment about point O must be zero. This gives f L − kx L − k Therefore, x f =k x+ 4 xL =0 22 = 5 kx 4 and the equivalent spring constant is ke = 5k/4. Note that although these two springs appear to be connected side-by-side, they are not in parallel, because they do not have the same deﬂection. Thus their equivalent spring constant is not given by the sum, 2k. 4.1.6 NONLINEAR SPRING ELEMENTS Up to now we have used the linear spring model f = kx. Even though this model is sometimes only an approximation, nevertheless it leads to differential equation models that are linear and therefore relatively easy to solve. Sometimes, however, the use of a nonlinear model is unavoidable. This is the case when a system is designed to utilize two or more spring elements to achieve a spring constant that varies with the applied load. Even if each spring element is linear, the combined system will be nonlinear. An example of such a system is shown in Figure 4.1.11a. This is a representation of systems used for packaging and in vehicle suspensions, for example. The two side springs provide additional stiffness when the weight W is too heavy for the center spring. 4. 1 Spring Elements 179 Deﬂection of a Nonlinear System E X A M P L E 4.1.6 ■ Problem Obtain the deﬂection of the system model shown in Figure 4.1.12a, as a function of the weight W . Assume that each spring exerts a force that is proportional to its compression. W Slope = k1 + 2k2 W Figure 4.1.12 A nonlinear spring arrangement. Platform x d Slope = k1 k1 k2 k2 d (a) x (b) ■ Solution When the weight W is gently placed, it moves through a distance x before coming to rest. From statics, we know that the weight force must balance the spring forces at this new position. Thus, W = k1 x if x < d W = k1 x + 2k2 (x − d) if x ≥ d We can solve these relations for x as follows: W x = if x < d k1 W + 2k2 d x = if x ≥ d k1 + 2k2 These relations can be used to generate the plot of W versus x, shown in part (b) of the ﬁgure. The system in Example 4.1.5 consists of linear spring elements but it has a nonlinear force-deﬂection relation because of the way the springs are arranged. However, most spring elements display nonlinear behavior if the deﬂection is large enough. Figure 4.1.13 is a plot of the force-deﬂection relations for three types of spring elements: the linear spring element, a hardening spring element, and a softening spring element. The stiffness k is the slope of the force-deﬂection curve and is constant for the linear spring element. A nonlinear spring element does not have a single stiffness value because its slope is variable. For the hardening element, sometimes called a hard spring, f f f 5 kx Figure 4.1.13 Forcedeﬂection curves for (a) a hardening spring and (b) a softening spring. f 5 kx x (a) x (b) 180 CHAPTER 4 Spring and Damper Elements in Mechanical Systems its slope and thus its stiffness increases with deﬂection. The stiffness of a softening element, also called a soft spring, decreases with deﬂection. From the plot in Figure 4.1.12(b), we can see that the spring arrangement in part (a) of the ﬁgure creates a hardening spring. 4.2 MODELING MASS-SPRING SYSTEMS If we assume that an object is a rigid body and if we neglect the force distribution within the object, we can treat the object as if its mass were concentrated at its mass center. This is the point-mass assumption, which makes it easier to obtain the translational equations of motion, because the object’s dimensions can be ignored and all external forces can be treated as if they acted through the mass center. If the object can rotate, then the translational equations must be supplemented with the rotational equations of motion, which were treated in Sections 3.2 and 3.4. If the system cannot be modeled as a rigid body then we must develop a distributedparameter model that consists of a partial differential equation, which is more difﬁcult to solve. 4.2.1 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 one or more rigid bodies. If the system consists of an object attached to a spring, the simplest way to do this 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. 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 homework problems and some of the examples to follow, the numerical values are not given. In such cases, unless otherwise explicitly stated, you should assume that the spring mass can be neglected. In Section 4.3 we will develop a method to account for the spring mass without the need to develop a partial differential equation model. This method will be useful for applications where the spring mass is neither negligible nor the dominant mass in the system. An ideal spring element is massless. A real spring element can be represented by an ideal element either by neglecting its mass or by including it in another mass in the system. 4.2.2 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 4.2.1a. 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 and the mass m is in equilibrium when the spring is at its free length. The equilibrium location of G is the point marked E. Part (b) of the ﬁgure shows the mass displaced a distance x from the equilibrium position. In this position, the spring has been stretched a 4. 2 E L a E L k Modeling Mass-Spring Systems Figure 4.2.1 Models of mass-spring systems. x E a k m G m (a) 181 G (b) k m kx x m (d) (c) distance x from its free length, and thus its force is kx. The free body diagram, displaying only the horizontal force, is shown in part (c) of the ﬁgure. From this diagram we can obtain the following equation of motion: m ẍ = −kx (4.2.1) Note that neither the free length L nor the cube dimension a appears in the equation of motion. 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 4.2.1d. 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. You should also assume that the location of the equilibrium position is known. The shaded-line symbol shown in Figure 4.2.1a is used to indicate a rigid support, such as the horizontal surface and the vertical wall, and also to indicate the location of a ﬁxed coordinate origin, such as the origin of x. 4.2.3 EFFECT OF GRAVITY Now suppose the object slides on an inclined frictionless surface. In Figure 4.2.2a the spring is at its free length. If we let the object slide until it reaches equilibrium (Figure 4.2.2b), the spring stretches a distance δst , which is called the static spring deﬂection. Because the mass is in equilibrium, the sum of the forces acting on it must be zero. Thus, for the forces parallel to the inclined surface, mg sin φ − kδst = 0 Figure 4.2.2c shows the object displaced a distance x from the equilibrium position. In this position the spring has been stretched a distance x + δst from its free length, and thus its force is k(x + δst ). The free-body diagram displaying only the forces parallel to the plane is shown in part (d) of the ﬁgure. From this diagram we can obtain the Figure 4.2.2 Effect of inclination on a mass-spring model. L L L a k m G st aE k g st m m G m G (b) a k(x 1 st) x k (a) E (c) mg sin kst 5 mg sin (d) 182 CHAPTER 4 Spring and Damper Elements in Mechanical Systems following equation of motion: m ẍ = −k(x + δst ) + mg sin φ = −kx + (mg sin φ − kδst ) Because mg sin φ = kδst the term within parentheses is zero and the equation of motion reduces to m ẍ = −kx, the same as for the system shown in Figure 4.2.1. 4.2.4 CHOOSING THE EQUILIBRIUM POSITION AS COORDINATE REFERENCE The example in Figure 4.2.2 shows that for a mass connected to a linear spring element, the force due to gravity is canceled out of the equation of motion by the force in the spring due to its static deﬂection, as long the displacement of the mass is measured from the equilibrium position. We will refer to the spring force caused by its static deﬂection as the static spring force and the spring force caused by the variable displacement x as the dynamic spring force. We need not choose the equilibrium location as the coordinate reference. If we choose another coordinate, however, the corresponding equation of motion will contain additional terms that correspond to the static forces in the system. For example, in Figure 4.2.3a if we choose the coordinate y, the corresponding free body diagram is shown in part (b) of the ﬁgure. The resulting equation of motion is m ÿ = −k(y − L) + mg sin φ = −ky + k L + mg sin φ Note that k L + mg sin φ = 0 so the static force terms do not cancel out of the equation. The advantages of choosing the equilibrium position as the coordinate origin are (1) that we need not specify the geometric dimensions of the mass and (2) that this choice simpliﬁes the equation of motion by eliminating the static forces. Now suppose we place the mass m on a spring as shown in Figure 4.2.4a. Assume that the mass is constrained to move in only the vertical direction. If we let the mass settle down to its equilibrium position at E, the spring compresses an amount δst and thus mg = kδst (Figure 4.2.4b). If the mass is displaced a distance x down from equilibrium (Figure 4.2.4c), the resulting spring force is k(x + δst ) and the resulting free body diagram is shown in part (d) of the ﬁgure. Thus the equation of motion is m ẍ = −k(x + δst ) + mg = −kx + (mg − kδst ) = −kx So the equation of motion reduces to m ẍ = −kx. Figure 4.2.3 Choice of coordinate origin for a mass-spring model. Figure 4.2.4 Static deﬂection in a mass-spring system. G y mg a m st E L st G E m x x k(y 2 L) k L x G m g k(x + st) m m mg sin (a) (b) m (a) (b) (c) (d) 4. 2 Figure 4.2.5 Choice of coordinate direction for a mass-spring model. E x m m x k E E k k x m Modeling Mass-Spring Systems Figure 4.2.6 Modeling an external force on a mass-spring system. E x k kx m m f f (a) (b) kx x m kx (a) x x m m kx (b) (c) In Figure 4.2.4c, we imagined the mass to be displaced downward from equilibrium and thus we chose the coordinate x to be positive downward. However, we are free to imagine the mass being displaced upward; in this case, we would choose x to be positive upward and would obtain the same equation of motion: m ẍ = −kx. You should draw the free body diagram for this case to make sure that you understand the principles. Figure 4.2.5 shows three situations, and the corresponding free body diagrams, that have the same equation of motion, m ẍ = −kx. It is important to understand that any forces acting on the mass, other than gravity and the spring force, are not to be included when determining the location of the equilibrium position. For example, in Figure 4.2.6a a force f acts on the mass. The equilibrium position E is the location of the mass at which kδst = mg sin φ when f = 0. From the free body diagram shown in part (b), the equation of motion is m ẍ = f − kx. The previous analysis is based on a system model that contains a linear spring and a constant gravity force. For nonlinear spring elements, the gravity forces may or may not appear in the equation of motion. The gravity force acts like a spring in some applications and thus it might appear in the equation of motion. For example, the equation of motion for a pendulum, derived in Chapter 3 for small angles, is m L 2 θ̈ = −mgLθ . The gravity term is not canceled out in this equation because the effect of gravity here is not a constant torque but rather is a torque mgLθ that is a function of the coordinate θ . 4.2.5 SOLVING THE EQUATION OF MOTION We have seen that the equation of motion for many mass-spring systems has the form m ẍ + kx = f , where f is an applied force other than gravity and the spring force. Suppose that the force f is zero and that we set the mass in motion at time t = 0 by pulling it to a position x(0) and releasing it with an initial velocity ẋ(0). The solution form of the equation can be obtained from Table 2.3.2 and is x(t) = C1 sin ωn t + C2 cos ωn t, where we have deﬁned k (4.2.2) ωn = m 183 184 CHAPTER 4 Spring and Damper Elements in Mechanical Systems Using the initial conditions we ﬁnd that the constants are C1 = ẋ(0)/ωn and C2 = x(0). Thus the solution is ẋ(0) x(t) = sin ωn t + x(0) cos ωn t (4.2.3) ωn This solution shows that the mass oscillates about the rest position x = 0 with a √ frequency of ωn = k/m radians per unit time. The period of the oscillation is 2π/ωn . The frequency of oscillation ωn is called the natural frequency. The natural frequency is greater for stiffer springs (larger k values). The amplitude of the oscillation depends on the initial conditions x(0) and ẋ(0). As shown in Example 2.8.3, the solution (4.2.3) can be put into the following form: x(t) = A sin(ωn t + φ) (4.2.4) where sin φ = x(0) A A= [x(0)]2 cos φ = ẋ(0) + ωn ẋ(0) Aωn (4.2.5) 2 (4.2.6) If f is a unit-step function, and if the initial displacement x(0) and initial velocity ẋ(0) are zero, then you should be able to show that the unit-step response is given by 1 π 1 1 + sin ωn t − (4.2.7) x(t) = (1 − cos ωn t) = k k 2 The displacement oscillates about x = 1/k with an amplitude of 1/k and a radian frequency ωn . E X A M P L E 4.2.1 Beam Vibration ■ Problem The vertical motion of the mass m attached to the beam in Figure 4.2.7a can be modeled as a mass supported by a spring, as shown in part (b) of the ﬁgure. Assume that the beam mass is negligible compared to m so that the beam can be modeled as an ideal spring. Determine the system’s natural frequency of oscillation. ■ Solution The spring constant k is that of the ﬁxed-end beam, and is found from Table 4.1.1 to be k = 16Ew h 3 /L 3 . The mass m has the same equation of motion as (4.2.1), where x is measured from the equilibrium position of the mass. Thus, if the mass m on the beam is initially displaced Figure 4.2.7 Model of a mass supported by a ﬁxed-end beam. m x x m k (a) (b) 4. 2 Modeling Mass-Spring Systems 185 vertically, it will oscillate about its rest position with a frequency of k = m ωn = 16Ew h 3 m L3 The source of disturbing forces that initiate such motion will be examined in later chapters. If the beam mass is appreciable, then we must modify the equation of motion. We will see how to do this in Section 4.3. A Torsional Spring System E X A M P L E 4.2.2 ■ Problem Consider a torsional system like that shown in Figure 4.2.8a. A cylinder having inertia I is attached to a rod, whose torsional spring constant is k T . The angle of twist is θ . Assume that the inertia of the rod is negligible compared to the inertia I so that the rod can be modeled as an ideal torsional spring. Obtain the equation of motion in terms of θ and determine the natural frequency. ■ Solution Because the rod is modeled as an ideal torsional spring, this system is conceptually identical to that shown in Figure 4.2.8b. The free body diagram is shown in part (c) of the ﬁgure. From this diagram we obtain the following equation of motion. I θ̈ = −k T θ This has the same form as (4.2.1), and thus we can see immediately that the natural frequency is ωn = kT I If the cylinder is twisted and then released, it will oscillate about the equilibrium θ = 0 with a √ frequency of k T /I radians per unit time. This result assumes that the inertia of the rod is very small compared to the inertia I of the attached cylinder. If the rod inertia is appreciable, then we must modify the equation of motion, as will be discussed in Section 4.3. Figure 4.2.8 A torsional spring system. kT kT kT I I I (a) (b) (c) Cylinder on an Incline ■ Problem Determine the equation of motion of the cylinder shown in Figure 4.2.9(a) in terms of the coordinate x. The mass moment of inertia about the cylinder’s center of mass is I and its mass is m. Assume the cylinder rolls without slipping. E X A M P L E 4.2.3 186 CHAPTER 4 Figure 4.2.9 (a) A cylinder supported by a spring on an incline (b) Free body diagram. Spring and Damper Elements in Mechanical Systems mg cos x x k m k(x + D) R mg sin x ω ft α N (a) (b) ■ Solution Taking x = 0 to be the equilibrium position, f t to be the tangential force acting on the cylinder, and to be the static deﬂection, we see from the free body diagram in part (b) of the ﬁgure that m ẍ = mg sin α − k(x + ) − f t (1) I ω̇ = R f t (2) From the geometry of a circle, we know that if the cylinder does not slip, then Rθ = x, where θ is the cylinder rotation angle, so that ω = θ̇ . Thus Rω = ẋ and R ω̇ = ẍ. From (2) we ﬁnd that f t = I ω̇/R = I ẍ/R 2 Substituting this into (1) and using the fact from statics that mg sin α = k we obtain from (1) m ẍ = −kx − or I m+ 2 R I ẍ R2 ẍ + kx = 0 The natural frequency is ωn = k m + I /R 2 The cylinder will roll back and forth with this radian frequency. Note that because the acceleration due of gravity g does not appear in the two previous equations, this system would have the same oscillation frequency in the moon, say, as it does on Earth. Only the static deﬂection and thus the location of the equilibrium point are affected by g. Note that if the cylinder is solid, then I = m R 2 /2, and the previous relations reduce to 3 m ẍ + kx = 0 2 ωn = 2k 3m 4. 2 Modeling Mass-Spring Systems 187 4.2.6 DISPLACEMENT INPUTS AND SPRING ELEMENTS Consider the mass-spring system and its free body diagram shown in Figure 4.2.10a. This gives the equation of motion m ẍ + kx = f . To solve this equation for x(t), we must know the force f (t). Now consider the system shown in Figure 4.2.10b, where we are given the displacement y(t) of the left-hand end of the spring. This represents a practical application in which a rotating cam causes the follower to move the left-hand end of the spring, as in part (c) of the ﬁgure. If we know the cam proﬁle and its rotational speed then we can determine y(t). Suppose that when x = y = 0, both springs are at their free lengths. To draw the free body diagram we must make an assumption about the relative displacements of the endpoints of the spring element. The free body diagram has been drawn with the assumption that y > x. Here we are not given an applied force as an input, but nevertheless we must draw the free body diagram showing the forces acting on the mass. The force produced by the given displacement y(t) is the resulting spring force k(y − x). The equation of motion is m ẍ = k1 (y − x) − k2 x. Note that we must be given y(t) to solve this equation for x(t). If we need to obtain the force acting on the follower as a result of the motion, we must ﬁrst solve for x(t) and then compute the follower force from k1 (y − x). This force is of interest to designers because it indicates how much wear will occur on the follower surface. When displacement inputs are given, it is important to realize that ultimately the displacement is generated by a force (or torque) and that this force must be great enough to generate the speciﬁed displacement in the presence of any resisting forces or system inertia. For example, the motor driving the cam must be able to supply enough torque to generate the motion y(t). x Figure 4.2.10 Force and displacement inputs. k m f f m kx (a) x y k2 k1 m k1( y 2 x) (b) x y k1 k2 m (c) m k 2x 188 Figure 4.2.11 Plots of displacement, velocity, and acceleration for simple harmonic motion. CHAPTER 4 Spring and Damper Elements in Mechanical Systems 4 d 2x/dt 2 3 dx/dt 2 1 x 0 –1 –2 –3 –4 0 1 2 3 t 4 5 6 7 4.2.7 SIMPLE HARMONIC MOTION From the equation of motion m ẍ = −kx we can see that the acceleration is ẍ = −kx/m = −ωn2 x. This type of motion, where the acceleration is proportional to the displacement but opposite in sign, is called simple harmonic motion. It occurs when the restoring force—here, the spring force—is proportional to the displacement. It is helpful to understand the relation between the displacement, velocity, and acceleration in simple harmonic motion. Expressions for the velocity and acceleration are obtained by differentiating x(t), whose expression is given by (4.2.4): π ẋ(t) = Aωn cos(ωn t + φ) = Aωn sin ωn t + φ + 2 ẍ(t) = −Aωn2 sin(ωn t + φ) = Aωn2 sin(ωn t + φ + π) The displacement, velocity, and acceleration all oscillate with the same frequency ωn but they have different amplitudes and are shifted in time relative to one another. The velocity is zero when the displacement and acceleration reach their extreme values. The sign of the acceleration is the opposite of that of the displacement, and the magnitude of the acceleration is ωn2 times the magnitude of the displacement. These functions are plotted in Figure 4.2.11 for the case where x(0) = 1, ẋ(0) = 0, and ωn = 2. 4.2.8 SYSTEMS WITH TWO OR MORE MASSES Up to now our examples have included only a single mass or inertia. If two or more masses or inertias are connected by spring elements, the basic principles remain the same, but we must ﬁrst make some assumptions regarding the relative motion of the mass or inertia elements. This allows us to assign directions to the force or moment vectors on the free body diagrams. If you are consistent with your assumptions, the resulting equations of motion will correctly describe the dynamics of the system even when the relative motion is different from what you assumed. This is often a difﬁcult concept for beginners, and a common mistake is to make one assumption for one mass and make the opposite assumption for the other mass. 4. 2 Modeling Mass-Spring Systems 189 We usually choose the coordinates as the displacements of the masses from their equilibrium positions because in equilibrium the static deﬂection forces in the spring elements cancel the weights of the masses. Thus the free body diagrams show only the dynamic forces, and not the static forces. Thus, for linear spring elements, the equations of motion expressed in such coordinates will not contain the weight forces or the static deﬂection forces. These concepts are illustrated in the following example. Equation of Motion of a Two-Mass System E X A M P L E 4.2.4 ■ Problem Derive the equations of motion of the two-mass system shown in Figure 4.2.12a. ■ Solution Choose the coordinates x1 and x2 as the displacements of the masses from their equilibrium positions. In equilibrium the static forces in the springs cancel the weights of the masses. Thus the free body diagrams showing the dynamic forces, and not the static forces, are as shown in Figure 4.2.12b. These diagrams have been drawn with the assumption that the displacement x1 of mass m 1 from its equilibrium position is greater than the displacement of m 2 . From these diagrams we obtain the equations of motion: m 1 ẍ 1 = f − k1 (x1 − x2 ) m 2 ẍ 2 = k1 (x1 − x2 ) − k2 x2 If we move all terms to the left side of the equal sign except for the external force f , we obtain m 1 ẍ 1 + k1 (x1 − x2 ) = f (1) m 2 ẍ 2 − k1 (x1 − x2 ) + k2 x2 = 0 (2) In drawing the free body diagrams of multimass systems, you must make an assumption about the relative motions of each mass. For example, we could have assumed that the displacement x1 of mass m 1 from its equilibrium position is less than the displacement of m 2 . Figure 4.2.12c shows the free body diagrams drawn for this assumption. If your assumptions are correct, the forces shown on the diagram must be positive. Note that the directions of the forces associated with spring k1 are the opposite of those in part (b) of the ﬁgure. You should k2 x 2 k2 x 2 m2 m2 k1(x1 2 x2) k1(x2 2 x1) m1 m1 k2 x2 m2 k1 x1 m1 f f f (a) (b) (c) Figure 4.2.12 A system with two masses. 190 CHAPTER 4 Spring and Damper Elements in Mechanical Systems conﬁrm that the diagram in part (c) results in equations of motion that are identical to equations (1) and (2). You must be consistent in your assumptions made about the relative motion when drawing the free-body diagrams. A common mistake is to use one assumption to obtain the free-body diagram for mass m 1 but another assumption for mass m 2 . 4.3 ENERGY METHODS The force exerted by a spring is a conservative force. If the spring is linear, then its resisting force is given by f = −kx and thus the potential energy of a linear spring is given by 1 2 kx (4.3.1) 2 where x is the deﬂection from the free length of the spring. A torsional spring exerts a moment M if it is twisted. If the spring is linear the moment is given by M = k T θ , where θ is the twist angle. The work done by this moment and stored as potential energy in the spring is θ θ 1 M dθ = k T θ dθ = k T θ 2 (4.3.2) V (θ ) = 2 0 0 V (x) = So the potential energy stored in a torsional spring is V (θ ) = k T θ 2 /2. The conservation of energy principle states that T + V = T0 + V0 = constant, where T and V are the system’s kinetic and potential energies. For the system shown in Figure 4.3.1a, for a frictionless surface, the principle gives 1 2 1 2 1 2 1 2 m ẋ + kx = m ẋ 0 + kx 0 = constant 2 2 2 2 This relation can be rearranged as follows: m 2 k ẋ − ẋ 20 + x 2 − x 20 = 0 2 2 which states that T + V = 0. For the system shown in Figure 4.3.1b, we must include the effect of gravity, and thus the potential energy is the sum of the spring’s potential energy Vs and the gravitational potential energy Vg , which we may choose to be zero at y = 0. The conservation of energy principle gives T + Vg + Vs = constant or T + V = T + Vg + Vs = 0 Figure 4.3.1 (a) A system having kinetic and elastic potential energy. (b) A system having kinetic, elastic potential, and gravitational potential energy. x L k m (a) L k y g m (b) 4. 3 Energy Methods 191 The spring is at its free length when y = 0, so we can write 1 2 1 m ẏ − mgy + ky 2 = constant 2 2 Note that the gravitational potential energy has a negative sign because we have selected y to be positive downward. The numerical value of the gravitational potential energy depends on the location of the datum and it may be negative. We are free to select the location because only the change in gravitational potential energy is signiﬁcant. Note, however, that a spring’s potential energy is always nonnegative and that the potential energy is positive whenever the deﬂection from the free length is nonzero. A Force Isolation System E X A M P L E 4.3.1 ■ Problem Figure 4.3.2 shows a representation of a spring system to isolate the foundation from the force of a falling object. Suppose the weight W is dropped from a height h above the platform attached to the center spring. Determine the maximum spring compression and the maximum force transmitted to the foundation. The given values are k1 = 104 N/m, k2 = 1.5 × 104 N/m, d = 0.1 m, and h = 0.5 m. Consider two cases: (a) W = 64 N and (b) W = 256 N. ■ Solution The velocity of the weight is zero initially and also when the maximum compression is attained. Therefore T = 0 and we have T + V = T + Vs + Vg = 0 or Vs + Vg = 0 That is, if the weight is dropped from a height h above the platform and if we choose the gravitational potential energy to be zero at that height, then the maximum spring compression x can be found by adding the change in the weight’s gravitational potential energy 0 − W (h + x) = −W (h + x) to the change in potential energy stored in the springs. Thus 1 k1 (x 2 − 0) + [0 − W (h + x)] = 0 2 which gives the following quadratic equation to solve for x: 1 k1 x 2 − W x − W h = 0 2 If x ≥ d, Vs + Vg = 0 gives if x < d if x < d (1) 1 1 k1 (x 2 − 0) + (2k2 )[(x − d)2 − 0] + [0 − W (h + x)] = 0 2 2 which gives the following quadratic equation to solve for x: (k1 + 2k2 )x 2 − (2W + 4k2 d)x + 2k2 d 2 − 2W h = 0 if x ≥ d if x ≥ d (2) For the given values, equation (1) becomes 104 x 2 − 2W x − W = 0 if x < 0.1 (3) and from equation (2), 4 × 104 x 2 − (2W + 6000)x + 300 − W = 0 if x ≥ 0.1 (4) Figure 4.3.2 A forceisolation system. W h Platform d x k2 k1 k2 192 CHAPTER 4 Spring and Damper Elements in Mechanical Systems For case (a), the positive root of equation (3) gives x = 0.0867, which is less than 0.1. So only the middle spring is compressed, and it is compressed 0.0867 m. The resulting maximum force on the foundation is the spring force k1 x = 104 (0.0867) = 867 N. For case (b), the positive root of equation (3) gives x = 0.188, which is greater than 0.1. So all three springs will be compressed. From equation (4), 4 × 104 x 2 − (512 + 6000)x + 300 − 256 = 0 which has the solutions x = 0.156 and x = 0.007. We discard the second solution because it is less than 0.1 and thus corresponds to compression in the middle spring only. So the outer springs will be compressed 0.156 − 0.1 = 0.056 m and the middle spring will be compressed 0.156 m. The resulting maximum force on the foundation is the total spring force k1 x + 2k2 (x − 0.1) = 104 (0.156) + 2(1.5 × 104 )(0.156 − 0.1) = 3240 N. 4.3.1 OBTAINING THE EQUATION OF MOTION In mass-spring systems with negligible friction and damping, we can often use the principle of conservation of energy to obtain the equation of motion and, for simple harmonic motion, to determine the frequency of vibration without obtaining the equation of motion. E X A M P L E 4.3.2 Figure 4.3.3 A mass-spring system. L st k x m Equation of Motion of a Mass-Spring System ■ Problem Use the energy method to derive the equation of motion of the mass m attached to a spring and moving in the vertical direction, as shown in Figure 4.3.3. ■ Solution With the displacement x measured from the equilibrium position, and taking the gravitational potential energy to be zero at x = 0, the total potential energy of the system is 1 1 1 V = Vs + Vg = k(x + δst )2 − mgx = kx 2 + kδst x + kδst2 − mgx 2 2 2 Because kδst = mg the expression for V becomes V = 1 2 1 2 kx + kδst 2 2 The total energy of the system is T +V = 1 2 1 2 1 2 m ẋ + kx + kδst 2 2 2 From conservation of mechanical energy, T + V is constant and thus its time derivative is zero. Therefore, d d (T + V ) = dt dt 1 2 m ẋ 2 + d dt 1 2 kx 2 Evaluating the derivatives gives m ẋ ẍ + kx ẋ = 0 Canceling ẋ gives the equation of motion m ẍ + kx = 0. + d dt 1 2 kδ 2 st =0 4. 3 Energy Methods Example 4.3.2 shows that if we can obtain the expression for the sum of the kinetic and potential energies, T + V , the equation of motion can be found by differentiating T + V with respect to time. Although this was a simple example, it illustrates that with this method we need not draw the free body diagrams of every member of a multibody system whose motion can be described by a single coordinate. 4.3.2 RAYLEIGH'S METHOD We can use the principle of conservation of energy to obtain the natural frequency of a mass-spring system if the spring is linear. This approach is sometimes useful because it does not require that we ﬁrst obtain the equation of motion. The method was developed by Lord Rayleigh (John William Strutt) and was presented in his Theory of Sound in 1847. A modern reprint is [Rayleigh, 1945]. Rayleigh is considered one of the founders of the study of acoustics and vibration. We illustrate Rayleigh’s method here for a second-order system, but it is especially useful for estimating the lowest natural frequency of higher-order systems with several degrees of freedom and distributed parameter systems with an inﬁnite number of natural frequencies. In simple harmonic motion, the kinetic energy is maximum and the potential energy is minimum at the equilibrium position x = 0. When the displacement is maximum, the potential energy is maximum but the kinetic energy is zero. From conservation of energy, Tmax + Vmin = Tmin + Vmax Thus Tmax + Vmin = 0 + Vmax or Tmax = Vmax − Vmin (4.3.3) For example, for the mass-spring system oscillating vertically as shown in Figure 4.3.3, T = m ẋ 2 /2 and V = k(x + δst )2 /2 − mgx, and from (4.3.3) we have, 1 1 1 Tmax = m(ẋ max )2 = Vmax − Vmin = k(xmax + δst )2 − mgxmax − kδst2 2 2 2 or 1 1 m(ẋ max )2 = k(xmax )2 2 2 where we have used the fact that kδst = mg. In simple harmonic motion |ẋ max | = ωn |xmax |, and thus, 1 1 m(ωn |xmax |)2 = k|xmax |2 2 2 √ 2 Cancel |xmax | and solve for ωn to obtain ωn = k/m. In this simple example, we merely obtained the expression for ωn that we already knew. However, in other applications the expressions for T and V may be different, but if the motion is simple harmonic, we can directly determine the natural frequency by using the fact that |ẋ max | = ωn |xmax | to express Tmax as a function of |xmax | and then equating Tmax to Vmax − Vmin . This approach is called Rayleigh’s method. A common mistake when applying Rayleigh’s method is to assume that Vmin = 0, but this is not always true, as shown by the following example. 193 194 CHAPTER 4 Spring and Damper Elements in Mechanical Systems E X A M P L E 4.3.3 A Cylinder and Spring ■ Problem Figure 4.3.4 A spring connected to a rolling cylinder. Apply Rayleigh’s principle to determine the natural frequency of the cylinder shown in Figure 4.3.4. The mass moment of inertia about the cylinder’s center of mass is I and its mass is m. Assume the cylinder rolls without slipping. ■ Solution x k m ω α R Take x = 0 to be the equilibrium position and to be the static deﬂection. From the geometry of a circle, we know that if the cylinder does not slip, then Rθ = x where θ is the cylinder rotation angle and ω = θ̇ . Take the gravitational potential energy to be zero at the equilibrium position. The elastic potential energy at equilibrium is k2 /2 and the total potential energy is 1 k(x + )2 − mgh 2 where h = x sin α. Because mg sin α = k at equilibrium, the expression for V becomes V = V = 1 1 1 k(x 2 + 2x + 2 ) − mgx sin α = kx2 + k2 2 2 2 Thus Vmin = k2 /2. The kinetic energy is 1 1 1 1 ẋ 2 1 T = m ẋ 2 + I θ̇ 2 = m ẋ 2 + I 2 = 2 2 2 2 R 2 I m+ 2 R ẋ 2 because θ̇ = ẋ/R. Assuming simple harmonic motion occurs, we obtain x(t) = A sin(ωn t + ϕ) and ẋ = ωn A cos(ωn t + ϕ). Thus xmax = A, ẋmax = ωn A, and Tmax 1 = 2 I m+ 2 R (ωn A)2 1 2 1 2 k A + k 2 2 = Vmax − Vmin , and we have Vmax = From Rayleigh’s method, Tmax 1 2 This gives I m+ 2 R (ωn A)2 = m+ 1 2 1 2 1 2 1 k A + k − k = k A2 2 2 2 2 I R2 ωn2 = k or ωn = k m + I /R 2 Of course, this is the same answer found in Example 4.2.3, as it should be. The reader must decide which method is easier, the free body diagram method or an energy method such as Rayleigh’s. Rayleigh’s method is especially useful when it is difﬁcult to derive the equation of motion or when only the natural frequency must be found. The following example illustrates this point. 4. 3 Energy Methods 195 Natural Frequency of a Suspension System E X A M P L E 4.3.4 ■ Problem Figure 4.3.5 shows the suspension of one front wheel of a car in which L 1 = 0.4 m and L 2 = 0.6 m. The coil spring has a spring constant of k = 3.6 × 104 N/m and the car weight associated with that wheel is 3500 N. Determine the suspension’s natural frequency for vertical motion. ■ Solution Imagine that the frame moves down by a distance A f , while the wheel remains stationary. Then from similar triangles the amplitude As of the spring deﬂection is related to the amplitude A f of the frame motion by As = L 1 A f /L 2 = 0.4A f /0.6 = 2A f /3. Using the fact that kδst = mg, the change in potential energy can be written as Vmax − Vmin = 1 1 1 1 k(As + δst )2 − mg As − kδst2 = k A2s = k 2 2 2 2 2 Af 3 2 The amplitude of the velocity of the mass in simple harmonic motion is ωn A f , and thus the maximum kinetic energy is 1 m(ωn A f )2 2 = Vmax − Vmin , we obtain Tmax = From Rayleigh’s method, Tmax 1 1 m(ωn A f )2 = k 2 2 Solving this for ωn we obtain 2 ωn = 3 k 2 = m 3 2 Af 3 2 3.6 × 104 = 6.69 rad/s 3500/9.8 Figure 4.3.5 A vehicle suspension. Upper control arm Frame Wheel L1 L2 Lower control arm 4.3.3 EQUIVALENT MASS OF ELASTIC ELEMENTS If an elastic element is represented as in Figure 4.3.6a, we assume that the mass of the element either is negligible compared to the rest of the system’s mass or has been included in the mass attached to the element. This included mass is called the equivalent mass of the element. We do this so that we can obtain a lumped-parameter model of the system. As we did with rigid-body systems in Chapter 3, we compute the equivalent mass by using kinetic energy equivalence, because mass is associated with kinetic energy. Figure 4.3.6 An example of a spring element with distributed mass. k y m dy L mc (a) (b) x 196 CHAPTER 4 Spring and Damper Elements in Mechanical Systems E X A M P L E 4.3.5 Equivalent Mass of a Rod ■ Problem The rod shown in Figure 4.3.6b acts like a spring when an axially applied force stretches or compresses the rod. Determine the equivalent mass of the rod. ■ Solution In Figure 4.3.6b, the mass of an inﬁnitesimal element of thickness dy is dm r = ρ dy, where ρ is the mass density per unit length of the material. Thus the kinetic energy of the element is (dm r ) ẏ 2 /2, and the kinetic energy of the entire rod is 1 KE = 2 L 1 ẏ dm r = 2 0 L ẏ 2 ρ dy 2 0 If we assume that the velocity ẏ of the element is linearly proportional to its distance from the support, then ẏ = ẋ y L where ẋ is the velocity of the end of the rod. Thus KE = 1 2 L ẋ 0 y L 2 ρ dy = 1 ρ ẋ 2 2 L2 L y 2 dy = 0 1 ρ ẋ 2 y 3 2 L2 3 L 0 or 1 1 ρ ẋ 2 L 3 = KE = 2 L2 3 2 ρL 3 ẋ 2 = 1 mr 2 ẋ 2 3 because ρ L = m r , the mass of the rod. For an equivalent mass m e concentrated at the end of the rod, its kinetic energy is m e ẋ 2 /2. Thus the equivalent mass of the rod is m e = m r /3. So the mass m in Figure 4.3.6a is m = m c + m e = m c + m r /3. The approach of this example can be applied to a coil spring of mass m s ; its equivalent mass is m s /3. A similar approach using the expression for the kinetic energy of rotation, Ir θ̇ 2 /2, will show that the equivalent inertia of a rod in torsion is Ir /3, where Ir is the rod inertia. This type of analysis can also be used to ﬁnd the equivalent mass of a beam by using the appropriate formula for the beam’s static load-deﬂection curve to obtain an expression for the velocity of a beam element as a function of its distance from a support. The derivations of such formulas are given in basic references on the mechanics of materials and are beyond the scope of this text. The expressions for the equivalent beam masses given in Table 4.3.1 were derived in this manner. Because the static loaddeﬂection curve describes the static deﬂection, it does not account for inertia effects, and therefore the expressions given in Table 4.3.1 are approximations. However, they are accurate enough for many applications. 4. 3 Energy Methods Table 4.3.1 Equivalent masses and inertias of common elements. Translational systems Nomenclature: m c = concentrated mass m d = distributed mass m e = equivalent lumped mass System model: m e ẍ + kx = 0 Equivalent system Helical spring, or rod in tension/compression Cantilever beam Massless spring k Rest x position me mc md md md m e = m c + 0.23m d mc mc m e = m c + m d /3 Simply supported beam mc Ly2 Fixed-end beam L兾2 md mc Ly2 m e = m c + 0.38m d m e = m c + 0.50m d Rotational systems Nomenclature: Ic

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