Chapter 12 vibration of two degree of freedom system

CHAPTER 12 TWO-DEGREE- OF-FREEDOM-SYSTEMS Introduction to two degree of freedom systems: • The vibrating systems, which require two coordinates to describe its motion, are called two-degrees-of –freedom systems. • These coordinates are called generalized coordinates when they are independent of each other and equal in number to the degrees of freedom of the system. • Unlike single degree of freedom system, where only one co-ordinate and hence one equation of motion is required to express the vibration of the system, in two- dof systems minimum two co-ordinates and hence two equations of motion are required to represent the motion of the system. For a conservative natural system, these equations can be written by using mass and stiffness matrices. • One may find a number of generalized co-ordinate systems to represent the motion of the same system. While using these co-ordinates the mass and stiffness matrices may be coupled or uncoupled. When the mass matrix is coupled, the system is said to be dynamically coupled and when the stiffness matrix is coupled, the system is known to be statically coupled. • The set of co-ordinates for which both the mass and stiffness matrix are uncoupled, are known as principal co-ordinates. In this case both the system equations are independent and individually they can be solved as that of a single- dof system. • A two-dof system differs from the single dof system in that it has two natural frequencies, and for each of the natural frequencies there corresponds a natural state of vibration with a displacement configuration known as the normal mode. Mathematical terms associated with these quantities are eigenvalues and eigenvectors. • Normal mode vibrations are free vibrations that depend only on the mass and stiffness of the system and how they are distributed. A normal mode oscillation is defined as one in which each mass of the system undergoes harmonic motion of same frequency and passes the equilibrium position simultaneously. • The study of two-dof- systems is important because one may extend the same concepts used in these cases to more than 2-dof- systems. Also in these cases one can easily obtain an analytical or closed-form solutions. But for more degrees of 209 freedom systems numerical analysis using computer is required to find natural frequencies (eigenvalues) and mode shapes (eigenvectors). The above points will be elaborated with the help of examples in this lecture. Few examples of two-degree-of-freedom systems Figure 1 shows two masses with three springs having spring stiffness free to move on the horizontal surface. Let 1 2 and mm 12 , and kk k 3 1 2 and x x be the displacement of mass respectively. 1 andm 2 m 1 m 3 k 1 k 2 m 2 k 2 x 1 x Figure 1 As described in the previous lectures one may easily derive the equation of motion by using d’Alembert principle or the energy principle (Lagrange principle or Hamilton’s principle) 11 mx  11 kx 1 m 21 2 ()kx x − Using d’Alembert principle for mass , from the 1 m free body diagram shown in figure 1(b) 11 1 2 1 2 2 ()mx k k x kx++ − =  0 0 (1) and similarly for mass m 2 22 11 2 3 2 ()mx kx k k x−++ =  (2) 22 mx  22 1 ()kx x − 32 kx 2 m Important points to remember • Inertia force acts opposite to the direction of acceleration, so in both the free body diagrams inertia forces are shown Figure 1 (b), Free body diagram towards left. • For spring , assuming 2 k 21 x x> , 210 The spring will pull mass towards right by 1 m 22 1 (kx x) − and it is stretched by 21 x x − (towards right) it will exert a force of 22 1 (kx x) − towards left on mass . Similarly assuming 2 m 12 x x> , the spring get compressed by an amount 21 x x − and exert tensile force of . One may note that in both cases, free body diagram remain unchanged. 21 2 (kx x− ) Now if one uses Lagrange principle, The Kinetic energy = 2 11 2 2 11 22 Tmx mx=+  2  and (3) Potential energy = 22 11 2 1 2 32 11 1 () 22 2 Ukx kxx k=+ −+ 2 x (4) So, the Lagrangian 22 2 2 11 2 2 11 2 1 2 3 2 11 11 1 () 22 22 2 2 L TU mx mx kx kx x kx ⎛⎞⎛ =−= + − + − + ⎜⎟⎜ ⎝⎠⎝  ⎞ ⎟ ⎠ (5) The equation of motion for this free vibration case can be found from the Lagrange principle 0 kk dL L dt q q ⎛⎞ ∂∂ −= ⎜⎟ ∂∂ ⎝⎠  , (6) and noting that the generalized co-ordinate 11 2 and qx qx 2 = = which yields 11 1 2 1 2 2 ()mx k k x kx++ − =  0 0 = (7) 22 11 2 3 2 ()mx kx k k x−++ =  (8) Same as obtained before using d’Alembert principle. Now writing the equation of motion in matrix form 12 2 11 1 223 22 2 0 0 0 0 kk k mx x kkk mx x +− ⎡⎤ ⎡⎤⎛⎞ ⎛⎞ ⎛⎞ + ⎜⎟ ⎜⎟ ⎜⎟ ⎢⎥ ⎢⎥ −+ ⎝⎠ ⎣⎦⎝⎠ ⎝⎠⎣⎦   . (9) Here it may be noted that for the present two degree-of-freedom system, the system is dynamically uncoupled but statically coupled. 211 Example 2. Consider a lathe machine, which can be modeled as a rigid bar with its center of mass not coinciding with its geometric center and supported by two springs, . 12 ,kk In this example, it will be shown, how the use of different coordinate systems lead to static and or dynamic coupled or uncoupled equations of motion. Clearly this is a two- degree-of freedom system and one may express the co-ordinate system in many different ways. Figure 3 shows the free body diagram of the system where point G is the center of mass. Point C represents a point on the bar at which we want to define the co-ordinates of this system. This point is at a distance from the left end and from right end. Distance between points C and G is . Assuming 1 l 2 l e c x is the linear displacement of point C and c θ the rotation about point C, the equation of motion of this system can be obtained by using d’Alember’s principle. Now summation of all the forces, viz. the spring forces and the inertia forces must be equal to zero leads to the following equation. 11 2 2 ()() cccc cc mx me k x l k x l θθ θ ++−++=   0 0 (10) Again taking moment of all the forces about point C 1112 22 ()()() Gc c c c c c c Jmxmeekxllkxll θθ θ θ ++ −− + + =    (11) Noting 2 cG J Jme=+ , the above two equations in matrix form can be written as 12 2211 22 2 2 11 11 2 2 0 0 c cc c mmex x kk klkl me J kl kl kl kl θ +− ⎡ ⎤⎛⎞ ⎛⎞ ⎡⎤ ⎛⎞ + ⎜⎟ ⎜⎟ ⎜⎟ ⎢⎥ ⎢⎥ −+ ⎝⎠ ⎣⎦⎣ ⎦⎝⎠ ⎝⎠   c θ = (12) 1 k k 2 k c x c θ 11 () cc kx l θ − 22 () cc kx l θ + G 2 l e cc J θ  C () cc mx e θ +   1 l Figure 3: Free body diagram of the system Figure 2 212 Now depending on the position of point C, few cases can are studied below. Case 1 : Considering , i.e., point C and G coincides, the equation of motion can be written as 0e = x θ 11 ()kxl θ − 22 ()kxl θ + 12 2211 22 22 11 11 22 0 0 0 0 G m kk klkl xx J kl kl kl kl θθ +− ⎡⎤ ⎡⎤ ⎛⎞ ⎛⎞⎛⎞ + ⎜⎟ ⎜⎟⎜⎟ ⎢⎥ ⎢⎥ −+ ⎝⎠ ⎝⎠⎝⎠ ⎣⎦⎣⎦   = (13) So in this case the system is statically coupled and if 11 2 2 kl kl = , this coupling disappears, and we obtained uncoupled and x θ vibrations. Case 2 : If, , the equation of motion becomes 22 11 kl kl= 12 22 11 2 2 0 0 0 0 c cc c mmex x kk me J kl kl θθ + ⎡⎤⎛⎞ ⎛⎞ ⎡⎤ ⎛⎞ + ⎜⎟ ⎜⎟ ⎜⎟ ⎢⎥ ⎢⎥ + ⎝⎠ ⎣⎦⎣⎦⎝⎠ ⎝⎠   c = . (14) Hence in this case the system is dynamically coupled but statically uncoupled. Case 3: If we choose , i.e. point C coincide with the left end, the equation of motion will become 1 0l = . (15) 12 22 2 22 22 0 0 c cc c mmex x kk kl me J kl kl θ + ⎡ ⎤⎛⎞ ⎛⎞ ⎡⎤ ⎛⎞ + ⎜⎟ ⎜⎟ ⎜⎟ ⎢⎥ ⎢⎥ ⎝⎠ ⎣⎦⎣ ⎦⎝⎠ ⎝⎠   c θ = 0 Here the system is both statically and dynamically coupled. Normal Mode Vibration Again considering the problem of the spring-mass system in figure 1 with , , , the equation of motion (9) can be written as 1 mm= 2 2mm= 123 kkkk=== 1121 2122 () 0 2() mx k x x kx mx k x x kx +−+= −−+=   (16) 213 We define a normal mode oscillation as one in which each mass undergoes harmonic motion of the same frequency, passing simultaneously through the equilibrium position. For such motion, we let 11 2 2 , it it x Ae x Ae ω ω == (17) Hence, 2 12 2 12 (2 ) 0 (2 2 ) 0 km AkA kA k m A ω ω −−= −+ − = (18) or, in matrix form 2 1 2 2 0 2 0 22 A km k A kkm ω ω ⎡⎤ −− ⎛⎞ ⎛⎞ = ⎢⎥ ⎜⎟ ⎜⎟ −− ⎝⎠ ⎝⎠ ⎣⎦ (19) Hence for nonzero values of (i.e., for non-trivial response) 1 and A 2 A 2 2 2 0 22 km k kkm ω ω −− = −− . (20) Now substituting 2 ω λ = , equation 6.1. yields 2 3 (3 ) ( ) 0 2 kk mm λλ −+ 2 = (21) Hence, 1 31 (3)0.634 22 kk mm λ =− = and 2 31 ( 3) 2.366 22 kk mm λ =+ = So, the natural frequencies of the system are 11 2 0.634 and 2.366 kk mm ωλ ω == = Now from equation (1)., it may be observed that for these frequencies, as both the equations are not independent, one can not get unique value of . So one should find a normalized value. One may normalize the response by finding the ratio of . 1 and A 2 A 12 to AA From the first equation (19) the normalized value can be given by 1 2 2 22 Ak k A km km ω λ == −− (22) and from the second equation of (19), the normalized value can be given by 2 1 2 22 22 A km km Ak k ω λ −− == (23) Now, substituting 2 11 0.634 k m ωλ == in equation (22) and (23) yields the same values, as both these equations are linearly dependent. Here, 214 1 1 2 0.732 1 A A λλ = ⎛⎞ = ⎜⎟ ⎝⎠ (24) and similarly for 2 22 2.366 k m ωλ == 2 1 2 2.73 1 A A λλ = ⎛⎞ − = ⎜⎟ ⎝⎠ (25) It may be noted • Equation (19) gives only the ratio of the amplitudes and not their absolute values, which are arbitrary. • If one of the amplitudes is chosen to be 1 or any number, we say that amplitudes ratio is normalized to that number. • The normalized amplitude ratios are called the normal modes and designated by ( ) n x φ . From equation (24) and (25), the two normal modes of this problem are: 12 0.731 2.73 ( ) ( ) 1.00 1.00 xx φφ − ⎧⎫ ⎧ == ⎨⎬ ⎨ ⎩⎭ ⎩ ⎫ ⎬ ⎭ 2 In the 1 st normal mode, the two masses move in the same direction and are said to be in phase and in the 2 nd mode the two masses move in the opposite direction and are said to be out of phase. Also in the first mode when the second mass moves unit distance, the first mass moves 0.731 units in the same direction and in the second mode, when the second mass moves unit distance; the first mass moves 2.73 units in opposite direction. Free vibration using normal modes When the system is disturbed from its initial position, the resulting free-vibration of the system will be a combination of the different normal modes. The participation of different modes will depend on the initial conditions of displacements and velocities. So for a system the free vibration can be given by 1112 2 sin( ) sin( )xA t B t φ ωψ φ ωψ =++ + (27) 215 Here A and B are part of participation of first and second modes respectively in the resulting free vibration and 1 and 2 ψ ψ are the phase difference. They depend on the initial conditions. This is explained with the help of the following example. Example: Let us consider the same spring-mass problem (figure 4) for which the natural frequencies and normal modes are determined. We have to determine the resulting free vibration when the system is given an initial displacement 12 (0) 5, (0) 1xx = = and initial velocity . 12 (0) (0) 0xx==  m 2m k k k 1 x 2 x Figure 4 Solution: Any free vibration can be considered to be the superposition of its normal modes. For each of these modes the time solution can be expressed as: 1 1 2 1 1 2 2 2 0.731 sin 1 2.731 sin 1.00 x t x x t x ω ω ⎧⎫ ⎧⎫ = ⎨⎬ ⎨ ⎬ ⎩⎭ ⎩⎭ − ⎧⎫ ⎧⎫ = ⎨⎬ ⎨ ⎬ ⎩⎭ ⎩⎭ The general solution for the free vibration can then be written as: 1 11 22 2 0.731 2.73 sin( ) sin( ) 1.00 1 x AtB t x ω ψω − ⎧⎫ ⎧⎫ ⎧⎫ =++ ⎨⎬ ⎨ ⎬ ⎨ ⎬ ⎩⎭ ⎩⎭ ⎩⎭ ψ + 2 where A and B allow different amounts of each mode and 1 and ψ ψ allows the two modes’ different phases or starting values. Substituting: 216 1 12 2 1 112 2 (0) 5 0.731 2.731 sin sin (0) 1 1 1 (0) 0 0.731 2.731 cos cos (0) 0 1 1 x AB x x AB x ψψ 2 ωψ ω ψ − ⎧⎫ ⎧⎫ ⎧ ⎫ ⎧ ⎫ == + ⎨⎬⎨⎬⎨⎬ ⎨ ⎬ ⎩⎭ ⎩ ⎭ ⎩ ⎭ ⎩⎭ − ⎧⎫ ⎧⎫ ⎧ ⎫ ⎧ ⎫ == + ⎨⎬⎨⎬ ⎨⎬ ⎨ ⎬ ⎩⎭ ⎩ ⎭ ⎩ ⎭ ⎩⎭ 0 12 12 cos cos 0 => 90 ψψ ψψ == == Substituting in 1 st set: 5 0.731 2.731 11 1 AB − ⎧⎫ ⎧ ⎫ ⎧ ⎫ =+ ⎨⎬ ⎨ ⎬ ⎨ ⎬ ⎩⎭ ⎩ ⎭ ⎩ ⎭ 0.731A-2.731B= 5 A= 2.233 A+B = 1 B=-1.233 Hence the resulting free vibration is 1 12 2 0.731 2.731 2.233 cos 1.233 cos 1.00 1.000 x tt x ω ω − ⎧⎫ ⎧⎫ ⎧ ⎫ =− ⎨⎬ ⎨ ⎬ ⎨ ⎬ ⎩⎭ ⎩ ⎭ ⎩⎭ Normal modes from eigenvalues The equation of motion for a two-degree-of freedom system can be written in matrix form as 0 M xKx+=  (28) where and M K are the mass and stiffness matrix respectively; is the vector of generalized co-ordinates. Now pre-multiplying x 1 M − in both side of equation 6.2. one may get 1 0 I xMKx − +  = (29) or, 0 I xAx+=  (30) Here is known as the dynamic matrix. Now to find the normal modes, 1 AMK − = 11 2 2 , it it x Xe x Xe ω ω == , the above equation will reduce to [ ] 0AIX λ −= (31) 217 where { } 2 12 and = T Xxx λ ω = . From equation (31) it is apparent that the free vibration problem in this case is reduced to that of finding the eigenvalues and eigenvectors of the matrix A. Example: Determine the normal modes of a double pendulum. Solution Kinetic energy of the system = 22 22 22 11 1 211 22 1212 2 1 11 (2cos()) 22 l ml l llTm θ θθ θθθ =+ ++  θ − Potential energy of the system = { } {} 112112 121 1 22 2 (1 cos ) (1 cos ) (1 cos ) ( ) (1 cos ) (1 cos ) Umg mgl l gm ml ml 2 θ θθ θθ =− + − +− =+ −+− So Lagrangian of the system = {} 22 11 1 2 1 1 2 2 12 2 1 1 2 1 2 2 11 ( 2 cos( )) ( )(1 cos ) (1 cos ) 22 LTU ml m l l ll g m m m θ θθ θθ θ θ =− =+++ −−+−+−  1 θ Figure 5 2 θ 11 l θ 21 θ θ − 11 l θ 22 l θ So using Lagrange principle, and assuming small angle of rotation, the equation of motion can be written in matrix form as 2 121 1 121 212 1 2 22 2 212 22 2 ()0 0 () 0 0 mmlg mml mll mlg mll ml θ θ θ θ ⎛⎞ + ⎡⎤ + ⎡⎤⎛ ⎛⎞ += ⎜⎟ ⎢⎥ ⎜⎟ ⎜⎟ ⎢⎥ ⎝⎠ ⎣⎦⎝ ⎣⎦ ⎝⎠   ⎞ ⎠ l Now considering a special case when 12 12 andmmm ll = === 0 ⎛⎞ ⎝⎠ mlg θ θ θ θ ⎛⎞ ⎛⎞ ⎡⎤ ⎡ ⎤ ⎛⎞ += ⎜⎟ ⎜⎟ ⎜⎟ ⎢⎥ ⎢ ⎥ ⎣⎦ ⎣ ⎦ ⎝⎠ ⎝⎠ ⎝⎠   , the above equation becomes 22 1 1 22 2 2 20 2 00 mlg ml ml mlg ml ml θ θ θ θ ⎛⎞ ⎡⎤ ⎛⎞ ⎡⎤ += ⎜⎟ ⎢⎥ ⎜⎟ ⎜⎟ ⎢⎥ ⎣⎦ ⎝⎠ ⎣⎦ ⎝⎠   or, ml 1 2 1 2 2 21 20 0 11 01 0 Now 2 11 20 21 1 12 01 22 g Amlg ml l −− ⎛⎞⎡⎤⎛ == ⎜⎟ ⎜ ⎢⎥ −− ⎝⎠⎣⎦⎝ ⎞ ⎟ ⎠ To find eigenvalues of A, 218 [...]... the normal modes of vibration of a double pendulum with same length and mass of the pendulum 3 Develop a matlab code for determination of free -vibration of a general two- degree of freedom system 4 Derive the equation of motion for the double pendulum shown in figure p1 in terms of θ1 and θ2 using Lagrange principle Determine the natural frequencies and mode shapes of the systems If the system is started... s3,4 = − ± ⎜ ⎟ − 3 m m m ⎝m⎠ So the system has a pair of complex conjugate SEMI-DEFINITE SYSTEMS The systems with have one of their natural frequencies equal to zero are known as semidefinite or degenerate systems One can show that the following two systems are degenerate systems x1 m2 m1 θ2 θ1 x2 k I1 I2 Figure 10 Figure 11 From figure 10 the equation of motion of the system is ⎡ m1 0 ⎤ ⎛ x1 ⎞ ⎡ k ⎢ 0... proportional to the mass ratio the system Similarly one may show for the two- rotor system, θ1 I =− 2 θ2 I1 (47) (48) the ratio of angle of rotation inversely proportional to the moment of inertia of the rotors Forced harmonic vibration, Vibration Absorber Consider a system excited by a harmonic force F1 sin ωt expressed by the matrix equation ⎡ m11 m12 ⎤ ⎛ x1 ⎞ ⎡ k11 k12 ⎤ ⎛ x1 ⎞ ⎛ F ⎞ ⎢m ⎥⎜ ⎟+ ⎢ ⎥ ⎜... Consider a vibrating system of mass m1 , stiffness k1 , subjected to a force F sin ωt As studied in case of forced vibration of single -degree of freedom system, the system will have a steady state response given by F sin ωt x= , where ωn = k1 / m1 2 m(ωn − ω 2 ) (55) which will be maximum when ω = ωn Now to absorb this k1 F sin ωt vibration, one may add a secondary spring and mass system as shown in... 1 ⎠ ⎝0⎠ ⎝ 1 ⎠ ⎝ For the second set of equations to be satisfied, cosψ 1 = cosψ 2 = 0 , so that ψ 1 = ψ 2 = 900 Hence A = 0.6035 and B = −0.1036 So the equation for free vibration can be given by ⎛ x1 ⎞ ⎛ 0.4142 ⎞ ⎛ −2.4142 ⎞ ⎜ ⎟ = 0.6035 ⎜ ⎟ cos ω1t − 0.1036 ⎜ ⎟ cos ω2t 1 ⎠ ⎝ 1 ⎠ ⎝ ⎝ x2 ⎠ Damped-free vibration of two- dof systems Consider a two degrees of freedom system with damping as shown in figure... m21ω k12 − m12ω 2 ⎤ ⎛ F ⎞ ⎥ ⎜ ⎟ k22 − m22ω 2 ⎦ ⎝ 0 ⎠ ⎡ k22 − m22ω 2 − k12 + m12ω 2 ⎤ ⎛ F ⎞ ⎢ ⎥⎜ ⎟ 2 k11 − m11ω 2 ⎦ ⎝ 0 ⎠ ⎣ −k21 + m21ω = k11 − m11ω 2 k12 − m12ω 2 k21 − m21ω 2 k22 − m22ω 2 (52) Hence X1 (k = 22 ) − m22ω 2 F Z (ω ) , (53) ⎡ k11 − m11ω 2 where [ Z (ω ) ] = ⎢ 2 ⎣ k21 − m21ω X2 (k = 21 k12 − m12ω 2 ⎤ ⎥ k22 − m22ω 2 ⎦ ) − m21ω 2 F (54) Z (ω ) Example Consider the system shown in figure 12. .. 3 ) = m 2 (ω 2 − 12 )(ω 2 − ω2 ) m m k k 2 where, 12 = and ω2 = 3 are normal mode frequencies of this system m m Hence, 2 2 2 2 4 2 2 2 4 ( 2k − mω ) F 2 X1 = X2 = 2 m 2 (ω 2 − 12 )(ω 2 − ω2 ) kF 2 m (ω − 12 )(ω 2 − ω2 ) 2 2 So it may be observed that the system will have maximum vibration when ω = ω1 or , ω = ω2 Also it may be observed that X 1 = 0, when ω 2 = 2k / m Tuned Vibration Absorber... m2 the stiffness and mass of the secondary spring and mass system, vibration can be k completely eliminated from the primary system For ω 2 = 2 , m2 Z (ω ) = k1k2 − m1k2 k2 k k k k − k1m2 2 − k2 m2 2 + m1m2 2 2 m2 m2 m2 m2 m2 k2 k2 2 2 = k1k2 − m1 2 − k1k2 − k2 + m1 2 = − k2 m2 m2 and X 2 = − k2 F F = 2 − k2 k2 (61) (62) 228 Centrifugal Pendulum Vibration Absorber The tuned vibration absorber is only... − m21ω 2 F (54) Z (ω ) Example Consider the system shown in figure 12 where the mass m1 is subjected to a force F sin ωt Find the response of the system when m1 = m2 and k1 = k2 = k3 k1 m1 x1 F sin ωt k2 x2 m2 Figure 12 Solution: The equation of motion of this system can be written as ⎡ m1 ⎢0 ⎣ 0 ⎤ ⎛ x1 ⎞ ⎡ k1 + k2 ⎜ ⎟+ m2 ⎥ ⎝ x2 ⎠ ⎢ −k2 ⎦ ⎣ ⎡ m 0 ⎤ ⎛ x1 ⎞ ⎡ 2k ⎢ 0 m ⎥ ⎜ x ⎟ + ⎢ −k ⎣ ⎦⎝ 2 ⎠ ⎣ −k2... the equation of motion If the lower mass is given an impulse F0 δ (t), determine the response in terms of normal modes θ1 L1 x1 θ2 m1 L2 m2 x2 Figure P1 5 A centrifugal pump rotating at 500 rpm is driven by an electric motor at 120 0 rpm through a single stage reduction gearing The moments of inertia of the pump impeller and the motor are 1600 kg.m2 and 500 kg.m2 respectively The lengths of the pump . CHAPTER 12 TWO- DEGREE- OF- FREEDOM- SYSTEMS Introduction to two degree of freedom systems: • The vibrating systems, which require two coordinates to describe its motion, are called two- degrees -of. Unlike single degree of freedom system, where only one co-ordinate and hence one equation of motion is required to express the vibration of the system, in two- dof systems minimum two co-ordinates. both the system equations are independent and individually they can be solved as that of a single- dof system. • A two- dof system differs from the single dof system in that it has two natural