Step 1: Mathematical Modeling
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1 083 Mechanical Vibrations Lesson Vibration Analysis Procedure The analysis of a vibrating system usually involves four steps: mathematical modeling derivation of the governing uations solution of the uations and interpretation of the results. Step : Mathematical Modeling Mathematical model represents all the important features of the system for the purpose of deriving the mathematical uations governing the system s behavior. The mathematical model should include enough details to be able to describe the system in terms of uations without maing it too complex. The mathematical model may be linear or nonlinear depending on the behavior of the system s components. Linear models permit quic solutions and are simple to handle; however nonlinear models sometimes reveal certain characteristics of the system that cannot be predicted using linear models. First we can use a very crude or elementary model to get a quic insight into the overall behavior of the system. Subsuently the model is refined by including more components and details so that the behavior of the system can be observed more closely. Example : [] Consider a forging hammer shown in Figure. The forging hammer consists of a frame a falling weight nown as the tup an anvil and a foundation bloc. The anvil is a massive steel bloc on which material is forged into desired shape by the repeated blows of the tup. The anvil is usually mounted on an elastic pad to reduce the transmission of vibration to the foundation bloc and the frame. Come up with two versions of mathematical model of the forging hammer. Figure : A forging hammer. Copyright 007 by Withit Chatlatanagulchai
2 083 Mechanical Vibrations Lesson Figure : A motorcycle with a rider. Example : [] Consider a motorcycle with a rider in Figure. Develop a suence of four mathematical models of the system. Consider the elasticity of the tires elasticity and damping of the struts (in the vertical direction) masses of the wheels and elasticity damping and mass of the rider. Copyright 007 by Withit Chatlatanagulchai
3 083 Mechanical Vibrations Lesson Virtual wor method (d Alembert s Method) Lagrange s method We will study each method in details in the following lessons. Step 3: of the Governing Equations The uations of motion must be solved to find the response of the vibrating system. We can use the following techniques for finding the solution: Standard methods of solving differential uations Laplace transform methods Numerical methods Matrix methods In this course we will focus on the first three methods; each method will be discussed in the following lessons. The last method namely matrix methods are mostly used in higher degree of freedom and can be studied once the students are familiar with linear algebra. Step : Derivation of Governing Equations Once the mathematical model is available we use the principles of dynamics and derive the uations that describe the vibration of the system. The uations of motion can be derived by drawing the free-body diagrams of all the masses involved. The uations of motion of a vibrating system are in the form of a set of ordinary differential uations for a discrete system and partial differential uations for a continuous system. The uations may be linear or nonlinear depending on the behavior of the components of the system. Several approaches are commonly used to derive the governing uations. They are Newton s second law of motion Energy method Equivalent system method (Rayleigh s method) Step 4: Interpretation of the Results The solution of the governing uations gives the displacements velocities and accelerations of the various masses of the system. These results must be interpreted with a clear view of the purpose of the analysis and the possible design implications of the results. Vibration Model: Mechanical Elements As discussed before a vibratory system consists of three elements: mass spring and damper.. Spring Elements A spring is generally assumed to have negligible mass and damping. A force developed in the spring is given by F = x where F is the spring force x is the displacement of one end with respect to the other and is the spring stiffness or spring constant. 3 Copyright 007 by Withit Chatlatanagulchai
4 The wor done in deforming a spring is stored as strain or potential energy in the spring and is given by U = x. Actual springs are linear only up to a certain deformation. Beyond a certain value of deformation the stress exceeds the yield point of the material and the force-deformation relation becomes nonlinear as in Figure 3. ( ) y x = ( 3 ) Px a x 6EI ( 3 ) Pa x a 6EI E is Young s modulus I is moment 083 Mechanical Vibrations Lesson 0 x a a x l where of inertia of the cross section of the beam and P is external force. Find the spring constant of a cantilever beam with an end mass m shown in Figure 5. () x l m y Figure 5: A cantilever beam with end mass. Figure 3: Nonlinearity beyond the yield point. Example 3: [] Deflection of the cantilever beam in Figure 4 is given as a P x l y Figure 4: A cantilever beam. 4 Copyright 007 by Withit Chatlatanagulchai
5 083 Mechanical Vibrations Lesson From () we have that the static deflection of the beam at the free end is given by ( 3 ) mgl l l mgl δ st = = 6EI 3EI Therefore the spring constant is 3 mg 3 = = EI. 3 δ l st. can be called an uivalent stiffness of the cantilever beam.... Combination of Springs Several springs are used in combination. They can be connected in parallel or in series or both. These springs can be combined into a single uivalent spring as follows. Springs in Parallel Consider two springs connected in parallel from the free-body diagram we have Figure 6: Springs in parallel. 5 Copyright 007 by Withit Chatlatanagulchai
6 Therefore the uivalent spring is W = δ + δ st = δ. In general if we have n springs with spring constants n in parallel then the uivalent spring constant can be obtained as Springs in Series Next we consider two springs connected in series in Figure 7. st = +. st = n. Therefore W = δ W = δ W = δ. st δ δ st = and δ = Substitute the uations above in to () we get therefore δ 083 Mechanical Vibrations Lesson δ st δ + = δ st st st = +. In general if we have n springs with spring constants n in series then the uivalent spring constant can be obtained as = n The static deflection of the system Figure 7: Springs in series. δ st is given by δ = δ + δ () st. Since both springs are subjected to the same force W we have Example 4: [] Figure 8 shows the suspension system of a freight truc with a parallel-spring arrangement. Find the uivalent spring constant of the suspension if each of the three helical springs is made of steel with a 9 shear modulus G = 80 0 N/ m and has five effective turns mean coil diameter D = 0 cm and wire diameter d = cm. 6 Copyright 007 by Withit Chatlatanagulchai
7 083 Mechanical Vibrations Lesson The spring constant of each helical spring is given by 9 ( 80 0 )( 0.0) 3 ( )( ) 4 Gd = = = N/ m. 3 8nD Since there are three springs connected in parallel we have = 3 = 0000 N/ m. Figure 8: Parallel springs in a freight truc. The uivalent spring constant of a helical spring under axial load is given by Example 5: [] Determine the torsional spring constant of the steel propeller shaft shown in Figure 9. A hollow shaft under torsion has uivalent spring constant 4 Gd = 8nD 3 πg = D d 3l ( ) 4 4 where d = wire diameter D = mean coil diameter n = number of active turns. l = length D = outer diameter d = inner diameter. where Figure 9: Propeller shaft. 7 Copyright 007 by Withit Chatlatanagulchai
8 083 Mechanical Vibrations Lesson There are two sections of torsional springs section - and section -3. Since δ + δ3 = δ st we treat the two sections as springs connected in series. The spring constant of each section is given by 3 9 ( 80 0 ) 4 4 ( ) π = 3( ) = Nm rad / 9 ( 80 0 ) 4 4 ( ) π = 3( ) 6 = Nm / rad. Since the springs are connected in series we have = + 3 = = Nm/ rad. 6 6 Example 6: [] Consider a crane in Figure 0(a). The boom AB has a cross-sectional area of 500 mm. The cable is made of steel with a cross-sectional area of 00 mm. Neglecting the effect of the cable CDEB find the uivalent spring constant of the system in the vertical direction. 8 Copyright 007 by Withit Chatlatanagulchai
9 083 Mechanical Vibrations Lesson Figure 0: Crane lifting a load. 9 Copyright 007 by Withit Chatlatanagulchai
10 083 Mechanical Vibrations Lesson The uivalent system is given in Figure 0(c). First we need to se trigonometry to find the length the angle θ Using law of u cosine we have l l l and. ( )( ) = cos35 =.3 m. The angle θ can be found using the law of cosine and the nowledge of l ( )( ) 0 = l + 3 l 3 cos θ θ = A vertical displacement x of point B will cause the spring to deform by an amount of x = xcos 45 and the spring to deform by an amount of x = x ( θ ) is cos 90. The potential energy stored in the spring of the uivalent system U = x. The potential energy stored in the springs of the actual system is where U = ( ) ( ) xcos 45 + xcos 90 θ 0 Copyright 007 by Withit Chatlatanagulchai
11 AE 6 9 ( 00 0 )( 07 0 ) 6 = = =.68 0 / l ( )( 07 0 ) AE l 0 N m 7 = = = N/ m. 083 Mechanical Vibrations Lesson Since the total potential energy stored in the springs of the uivalent system and the actual system must be ual we have U = U = N/ m. Figure : Translational masses connected by a rigid bar.. Mass or Inertia Elements In many practical applications several masses appear in combination. For a simple analysis we can replace these masses by a single uivalent mass. If the system has translational masses we can replace them with a single uivalent mass. If the system has rotational masses we can replace them with a single uivalent inertia. And if the system has both translational and rotational masses we can choose to replace them with a single uivalent mass or a single uivalent inertia. We can do this using the fact that the inetic energy of the actual system and its uivalent system must be ual. Example 7: [] Figure (a) shows a system whose uivalent system is given in Figure (b). Find the uivalent mass m. Copyright 007 by Withit Chatlatanagulchai
12 Let T be the inetic energy of the actual system and let T be the inetic energy of the uivalent system. We have T T = m x + m x + m x 3 3 lx lx 3 = mx + m + m3 l l = m x. Figure : Translational and rotational masses in a rac and pinion arrangement. 083 Mechanical Vibrations Lesson Therefore T = T lx lx 3 mx + m + m3 = mx l l l l 3 m = m+ m + m3. l l Example 8: [] Consider translational and rotational masses coupled in Figure. a) Find ui valent translational mass b) Find uivalent rotational inertia Copyright 007 by Withit Chatlatanagulchai
13 a) Let T be the inetic energy of the actual system and let T be the inetic energy of the uivalent system. We have T T = mx + J θ 0 x = mx + J0 = mx. R T = T m R J J J mr ( ) θ + 0θ = θ 083 Mechanical Vibrations Lesson = + Example 9: [] Find the uivalent mass of the system in Figure 3. J 0. Therefore x + 0 = mx J m x R J0 m = m+. R T = T b) Let T be the inetic energy of the actual system and let T be the inetic energy of the uivalent system. We have T T = mx + J θ 0 ( θ ) = m R + J 0θ = J θ. Figure 3: An example system. Therefore 3 Copyright 007 by Withit Chatlatanagulchai
14 083 Mechanical Vibrations Lesson Let T be the inetic energy of the actual system and let T be the inetic energy of the uivalent system. Neglecting the rotational inetic energy of lin we have T T = mx + J θ + J θ + m x + mx + J cθ c + mx c Therefore p p x ml x x l = mx + J p + + m r p r p rp x mr c c xl + m l + + m r p rpr c = m x. c x l rp x ml x x l mx = mx + Jp + + m r p r p rp + m l + + m l J ml ml ml ml ml m = m r r r r x mr c c xl x c r p rpr c r p p c c + +. p p 4 p p rp rp 4 Copyright 007 by Withit Chatlatanagulchai
15 .3 Damping Elements Damping is a mechanism by which the vibrating energy is gradually converted into other energy such as heat or sound. There are three common types of damping: viscous damping Coulomb or dry friction damping and hysteretic damping. Viscous damping dissipates energy by using the resistance offered by the fluid such as air gas water or oil. Coulomb or dry friction damping is caused by friction between dry rubbing surfaces. It has constant magnitude but opposite in direction to the motion of the vibrating body. Hysteretic damping happens when some materials are deformed and the vibrating energy are absorbed and dissipated by the material. Among the three types of damping viscous damping is the most commonly used and can be modeled as Figure Mechanical Vibrations Lesson Since in viscous damping the damping force is proportional to the velocity of the vibrating body we have where F = cv (4) c is called damping constant. Comparing (3) with (4) we have c μ A. h = (5) Combination of dampers is done similar to that of springs. The inetic energy due to damping is given by U = cv. Example 0: [] A bearing which can be approximated as two flat plates separated by a thin film of lubricant offers a resistance of 400 N when the relative velocity between the plates is 0 m/s. The area of the plates is 0. m. The absolute viscosity of the oil in-between is Pa- e the clearance between the s. Determin plates. Figure 4: Parallel plates with a viscous fluid in between. From Newton s law of viscous flow we have where μ is the absolute viscosity of the fluid. F du μv τ = = μ = (3) A dy h 5 Copyright 007 by Withit Chatlatanagulchai
16 Area (A) From (4) 083 Mechanical Vibrations Lesson h v Figure 5: Flat plates separated by thin film of lubricant. From (5) F = cv ( ) 400 = c 0 c= 40 Ns/ m. μ A c = h ( 0.) 40 = h h= 0.86 mm. Example : [] A milling machine is supported on four shoc mounts as shown in Figure 6(a). The elasticity and damping of each shoc mount can be modeled as a spring and a viscous damper as shown in Figure 6(b). Find the uivalent spring constant and the uivalent damping constant c of the milling machine support. 6 Copyright 007 by Withit Chatlatanagulchai
17 083 Mechanical Vibrations Lesson Figure 6: A milling machine. 7 Copyright 007 by Withit Chatlatanagulchai
18 083 Mechanical Vibrations Lesson One Degree of Freedom Free Vibrations 3 Systems with Mass and Spring Consider a simple mass-and-spring system in Figure 7. Initially the mass stretches the spring to an amount of Δ below the unstretched position. From the Newton s second law we have Δ = mg. (6) From the free-body diagram of the actual system in Figure 6(c) we have F = F + F + F + F s s s s3 s4 = x + x + x 3 + x 4 Fd = Fd+ Fd + Fd3 + Fd4 = cx + cx + cx + cx. 3 4 From the free-body diagram of the uivalent system in Figure 6(d) we have Unstretched position m Δ m Δ mg x x ( Δ + x) m mg Static uilibrium position x F F s d = x = c x. Since the forces of the two systems are ual we have = c = c + c + c + c. 3 4 Figure 7: A system with mass and spring. With x chosen to point downward from the static uilibrium position and using (6) we have ( ) mg Δ+ x = mx x = mx. 8 Copyright 007 by Withit Chatlatanagulchai
19 Let the natural fruency be ω n and let we have ω n = (7) m x+ ω x = 0. n The solution of the second-order differential uation above is x = Asinω t+ bcos ω t. The constants A and are evaluated from initial conditions x ( 0) to be n B ( 0) ( ) x 0 x = sin ωnt+ x( 0 ) cos ωnt. ω n n x and One remar is that when x is chosen to be from the static uilibrium position the weight mg is cancelled with Δ and disappears from the differential uation. From (7) we have From (6) we have ω n = = = rad / s. m 0.5 ( ) mg Δ= = = 6 mm Mechanical Vibrations Lesson Example 3: [] Determine the natural fruency of the mass m on the end of a cantilever beam of negligible mass shown in Figure 8. l Figure 8: A cantilever beam with end mass. m x 4 Natural Fruency Example : [] A 0.5 g mass is suspended by a spring having a stiffness of N/mm. Determine its natural fruency in cycles per second. Determine its static deflection. 9 Copyright 007 by Withit Chatlatanagulchai
20 A cantilever beam with end load as an uivalent spring constant 083 Mechanical Vibrations Lesson 3EI =. 3 l Therefore the natural fruency is ω n = = m 3EI /. 3 rad s ml Example 4: [] A rod in Figure 9 has 0.5 cm in diameter and m long. When the dis is given an angular displacement and released it maes 0 oscillations in 30. seconds. Determine the polar moment of inertia of the dis. From the Newton s second law we have l J θ Figure 9: A rotating dis. J θ = θ. For a shaft under torsion the uivalent spring constant is From (7) we have 9 ( )( ) 4 πgd π = = =.455 Nm/ rad. 3l 3 ( ) 4 ω = n π = 30. J J = J g m Copyright 007 by Withit Chatlatanagulchai
21 Example 5: [] Consider a pivoted bar in Figure 0. The bar is horizontal in the uilibrium position with spring forces and Determine its natural fruency. P P. 083 Mechanical Vibrations Lesson a O CG.. b θ c P P Figure 0: A pivoted bar. From Newton s second law we have a + b θ + θ = 0 J 0 0 ( ) ( ) J θ = aθ a bθ b ω = n a + b J 0. Copyright 007 by Withit Chatlatanagulchai
22 Lesson Homewor Problems Mechanical Vibrations Lesson Homewor problems are from the ruired textboo (Mechanical Vibrations by Singiresu S. Rao Prentice Hall 004) References [] Mechanical Vibrations by Singiresu S. Rao Prentice Hall 004 [] Theory of Vibration with Applications by William T. Thomson and Marie Dillon Dahleh Prentice Hall 998 Copyright 007 by Withit Chatlatanagulchai
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