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VIBRATION OF CONTINUOUS SYSTEMSIntroductionModels of vibratory systems can be divided into two broad classes, lumped andcontinuous, depending on the nature of the parameters. In the case of lumped systems, thecomponents are discrete, with the mass assumed to be rigid and concentrated atindividual points, and with the stiffness taking the form of massless springs connectingthe rigid masses. The masses and springs represent the system parameters, and we refer tosuch models as discrete or lumped-parameter models. The motion of discrete systems isgoverned by ordinary differential equations. Continuous systems, on the other hand,differ from discrete systems in that the mass and elasticity are continuously distributed.Such systems are also known as distributed-parameter systems, and examples includestrings, rods, beams, plates and shells. While discrete systems possess a finite number ofdegrees of freedom, continuous systems have an infinite number of degrees of freedombecause we need an infinite number of coordinates to specify the displacement of everypoint in an elastic body. The displacement in this case depends on two independentvariables, namely x and t. As a result, the motion of continuous systems is governed bypartial differential equations to be satisfied over the entire domain of the system, subjectto boundary conditions and initial conditions.Although discrete systems and continuous system may appear entirely different in nature,the difference is more in form than concept. As a matter of fact, a certain physical systemcan be modeled either as discrete or as distributed, depending on the objectives of theanalysis. It turns out that discrete and continuous systems are indeed closely connected,and thus it comes as no surprise that both systems possess natural frequencies and normalmodes of vibration.In this topic we will study the free and forced vibration of continuous systems. Emphasiswill be placed on studying the vibration of taught strings, rods and beams. This covers abroad class of engineering applications, as many practical systems can be modeled by oneor more of such elements in order to study the dynamic behavior1

Vibration of StringsThe figure shows a fixed-fixed string of length L. The string is initially under tension Tand the aim is to study the transverse vibrations denoted by the displacement y(x,t),measured from the equilibrium position. It is assumed that both displacement and slopeare small.yy(x,t)xLIt is also assumed that the tension force in the string remains constant during vibration,which follows from the previous assumptions of small displacements. As in allcontinuous systems, the displacement variable depends on both the spatial (x) andtemporal (t) coordinates. A free body diagram of a string element is shown below.Neglecting gravity effects, we can apply Newton’s second law on the string element toobtain the governing equation of motion.Ty dy ydx x dx xydxTx dxxApplying Fy m a y gives: y T sin dx T sin dx 2 x t 2where is the mass per unit length of the string. For small displacements, sin (1) ,hence we obtain:T 2 y 2 x t2(2)

But yhence: x 2 y 2 y x 2 t 2(3) 2 y 1 2 y x 2 c 2 t 2(4)Twhich can be written as:which is known as the one-dimensional wave equation and c Tis the velocity of wave propagation along the string. The wave equation is a partial differential equation,and the same form will be encountered in similar problems involving the dynamics ofdistributed-parameter models. The equation must be satisfied over the entire domain andis subject to boundary conditions as well as initial conditions. Accordingly, theproblem posed is both a boundary value problem (BVP) and an initial value problem(IVP) from a mathematical point pf view.We now seek the solution of the wave equation, which represents the variation of thetransverse displacement at any point along the string and at any time for an arbitrarystring that is set in motion by certain initial conditions and left to vibrate freely. Thissolution is emulated by the using the principle of separation of variables. In this way, thetransverse displacement can be expressed as:y( x, t ) Y ( x) G(t )(5) 2 y d 2Y G x 2 dx 2(6) 2 yd 2G Y t 2dt 2(7)It follows that:andSubstitution into the equation of motion (4) yields:d 2Y1 d 2GG 2Y 2dx 2cdt(8)1 d 2Y 1 1 d 2G Y dx 2 c 2 G dt 2(9)which can also be written as:It is noted that the left-hand-side (LHS) of the above equation depends only on the spatialvariable x, whereas the RHS depends only on the temporal variable, t. In order to satisfy3

the equation, both sides of equation (9) must be equal to a constant. Let this constant be c . A negative constant was conveniently selected because this choice leads to an2oscillatory motion. The choice of a zero or positive constant does not yield a vibratorymotion, and therefore must be excluded. For example, if a zero constant was chosen, thisleads to:1 1 d 2G 0c 2 G dt 2ord 2G 0dt 2whose solution is given by:G c1t c2which is rejected because it indicates a solution that increases linearly with time. It can beshown that the choice of a positive constant gives rise to two terms; one exponentiallyincreasing and the other exponentially decreasing.Adopting the negative constant choice, and substituting into the equation of motion gives:1 d 2Y2 c 2Y dx(10)d 2Y2 c Y 02dx(11)1 1 d 2G2 c 22c G dt(12)which can be written as:Furthermore,which can similarly be expressed as:d 2G 2G 02dt(13)Y ( x) A sin c x B cos c x(14)G(t ) C sin t D cos t(15)These have the general solutions:And4

The 4 constants A, B, C and D are to be determined from the boundary conditions (BC’s)and initial conditions (IC’s). It also worthy to note that equation (14) defines thedeformation shape, whereas equation (15) defines the motion to be harmonic in time. Itbecomes appropriate now to define the unknown constant as the natural frequency ofthe system, and c as the wave number or spatial frequency. The general solutionmay then be expressed as:y ( x, t ) A sin( c) x B cos( c) x C sin t D cos t (16)Alternatively, and after some algebraic manipulation, the above solution may also bewritten as:y( x, t ) a1 sin ( c) x t a2 cos ( c) x t a3 sin ( c) x t a4 cos ( c) x t (17)Once again, the solution must contain 4 unknown constants.Example: Fixed-fixed stringLet us now consider the case of a string that is fixed at both ends, as shown.yy(x,t)xLThe imposed boundary conditions indicate that the string displacement at both ends mustbe equal to zero, or:y(0, t ) 0 and y(L, t ) 0 . The general solution is:y ( x, t ) Y ( x) G(t ) A sin( c) x B cos( c) x C sin t D cos t Substitution of the first BC into the general solution gives:0 B C sin t D cos t which implies B 0 . The general solution hence becomes:5

y ( x, t ) A sin( c) x C sin t D cos t Substitution of the second BC into the solution gives:0 A sin( L c) C sin t D cos t which implies:sin( L c) 0hence L c n , n 1, 2,3,The above equation is termed the frequency equation or characteristic equation of thesystem, as it gives values of the system natural frequencies. Clearly, the system possessesan infinite number of natural frequencies, as suggested earlier. Having obtained thenatural frequencies, the solution at any frequency or mode is expressed by:yn ( x, t ) An sin(n x L) Cn sin nt Dn cos nt Yn ( x) Gn (t )Therefore, at each natural frequency, there corresponds a certain mode shape or aneigenfunction defined byYn ( x) An sin n x L where each “n” represents a normal mode vibration with a natural frequency n n cLand mode shapeYn ( x) An sin n x L where An are arbitrary constants. The figure below shows the first few modes of thestring, as obtained from the above analysis.6

n 1n 2n 3n 4The general solution is given by:yn ( x, t ) An sin(n x L) Cn sin nt Dn cos nt where the terms in the first bracket define the displacement pattern at mode “n”, and theterms on the last bracket define a harmonic motion at the corresponding naturalfrequency. The free vibration solution is finally obtained as the sum of all modes ofvibration, or: y( x, t ) Cn sin nt Dn cos nt sin(n x L)n 1where Cn , Dn are constants to be determined from the IC’s.The eigenfunctions can also be shown to possess an orthogonality property (see nextsection) which is given by: 0Y(x)Y(x)dx nm 0 hnLn mn mFor a fixed-fixed string, this becomes:7

0L sin(n x L) sin(m x L)dx L 20n mn mIn order to study the complete free vibration problem, the initial conditions must also bedefined. Assume that the string is subjected to the following initial conditions:y( x,0) f ( x) , y( x,0) g( x)Substitution into the general solution gives: f ( x) Dn sin( n x L)n 1and g ( x) Cn n sin(n x L)n 1In order to obtain the constants Cn , Dn we multiply the above equations by sin(n x L)and integrate over the string length. By using the orthogonality codition, we get:LDn 2f ( x) sin(n x L)dxL 0andLCn 2g ( x) sin(n x L)dxL n 0It thus follows that the constants Cn , Dn determine the contribution of each mode to thegeneral solution. Now consider the special case where the initial conditions impose adisplacement pattern that coincides with one of the natural modes, say mode “k”, and thatthe initial velocity is zero. In this way, we have:f ( x) sin(k x L)andg ( x) 0It follows that: 0 k nDn 1 k nandCn 08

Therefore, the general solution reduces to:y( x, t ) sin(k x L) cos k tThat is, the string vibrates in its “kth” mode due to the fact that the initial conditions causeonly the “kth” mode to be excited. The continuous system in this special case behaves likea single DOF system. The same notion can be observed for a multi-DOF system. If, onthe other hand, an impulse is given to the system, it can be shown that a wide spectrum offrequencies or modes are excited, and hence the system response will contain asummation of a large number of modes. In practice, the contribution of the higher modesis usually smaller, and the system response will be almost predominantly be described bythe fundamental (i.e. first) mode, together with only a few higher modes.9

Orthogonality of EigenfunctionsLet us now prove the orthogonality condition for a fixed-fixed string. The same techniquecan be applied to other forms of boundary conditions. Recall Eq. (11) obtainedpreviously:d 2Y2 c Y 02dx(18)This equation must be satisfied at all natural frequencies or all normal modes ofvibration. Consequently, at an arbitrary mode “m”, we have m , Ym as the naturalfrequency and associated eigenfuction, respectively. Similarly, at mode “n”, wehave n , Yn . Let us first assume that these two modes are distinct. It follows from Eq. (18)that:d 2Ym2 m c Ym 02dx(19)d 2Yn2 2 n c Yn 0dx(20) andMultiplying Eq. (19) by Yn and integrating over the string length yields:LLd 2Ym2 Yndx m c YmYn dx2dx00(21)The LHS of the above equation can be integrated by parts: dYm L L dYm dYn L dYm dYnLHS Yn dx dx dx 0 0 dx dx 0 dx dx(22)and the first term vanishes due to the boundary conditions defined (both ends fixed). Itfollows that:LLdY dY2 0 dxm dxn dx 0 m c YmYn dx(23)Similarly, we can multiply equation (20) by Ym and integrate over the string length. Thisyields:LLdY dY2 0 dxn dxm dx 0 n c YmYn dx10(24)

It is noted that the LHS of Eqs. (23) and (24) are identical, irrespective of the order ofmultiplication. Subtracting Eq. (24) from Eq. (23) gives:0 m2 n2L Y Y dx(25)m nc20But m , n are two distinct modes, i.e. m n , therefore:L Y Y dx 0m n, m n(26)0which proves the orthogonality condition for eigenfunctions of a fixed-fixed string. Thiscondition also holds true for other types of BC’s. Moreover, since eigenfunctions can bearbitrarily scaled (or normalized), we can write:L Y Y dx hm mm, m n(27)0where hm is a constant.Elastic or Inertial AttachmentsFinally, let us now consider other forms of boundary conditions. Figure 6 shows a stringthat is fixed at one end and attached to a spring at the other. The boundary condition atthe fixed end x 0 is given by y(0, t ) 0 . This is called a geometric boundarycondition, because it describes a specified displacement. Such a condition is also knownas an essential or imposed boundary condition.Lky(x,t)xElastic attachments.11

The boundary condition at the other end is not so obvious at first sight. Indeed it becomesappropriate to draw a free-body diagram of the string in order to investigate the forceinteraction. Such a free-body diagram is shown in Fig.7. At x L we need to balanceforces in the vertical direction. Thus we have:T y( L, t ) ky ( L, t ) x(28)This is called a natural boundary condition (also known as dynamic or additionalboundary condition) as it describes forces and moments acting on the system. We canthen proceed with the solution in the same way described above in order to obtain thenatural frequencies, eigenfunctions and response to initial conditions.ykz ky(L, t )y(x,t) y( L, t ) xTFigure 7. Free-body diagram.12zx

Vibration of RodsIn this section, let us study the free longitudinal vibration of rods (bars). Consider afixed-free rod of length L undergoing longitudinal vibration, as shown below.LUndeformedxDeformedu(x,t)The nomenclature adopted in this case is listed below. Density (mass per unit volume)PAxial forceu( x, t )Longitudinal displacementACross-sectional areaEYoung’s modulus of elasticityThe longitudinal displacement, which is assumed to be small, depends on both the spatial(x) and temporal (t) variables. It is assumed that the displacement is small. In order tostudy the rod vibration, we need to write down its equation of motion. Co