Vibration and Shock Handbook 06

Vibration and Shock Handbook 06 Every so often, a reference book appears that stands apart from all others, destined to become the definitive work in its field. The Vibration and Shock Handbook is just such a reference. From its ambitious scope to its impressive list of contributors, this handbook delivers all of the techniques, tools, instrumentation, and data needed to model, analyze, monitor, modify, and control vibration, shock, noise, and acoustics. Providing convenient, thorough, up-to-date, and authoritative coverage, the editor summarizes important and complex concepts and results into “snapshot” windows to make quick access to this critical information even easier. The Handbook’s nine sections encompass: fundamentals and analytical techniques; computer techniques, tools, and signal analysis; shock and vibration methodologies; instrumentation and testing; vibration suppression, damping, and control; monitoring and diagnosis; seismic vibration and related regulatory issues; system design, application, and control implementation; and acoustics and noise suppression. The book also features an extensive glossary and convenient cross-referencing, plus references at the end of each chapter. Brimming with illustrations, equations, examples, and case studies, the Vibration and Shock Handbook is the most extensive, practical, and comprehensive reference in the field. It is a must-have for anyone, beginner or expert, who is serious about investigating and controlling vibration and acoustics.

Computer Techniques II II-1 © 2005 by Taylor & Francis Group, LLC Numerical Techniques 6.1 6.2 6.3 Introduction Single-Degree-of-Freedom System Forced Single-Degree-of-Freedom System Single-Degree-of-Freedom System † Summary of Systems with Two or More Degrees of Freedom Example † Summary of Two-Degree-of-Freedom System 6-1 6-2 6-8 6.4 Finite Difference Method for a Continuous System 6-11 6.5 Matrix Methods 6-14 6.6 Approximation Methods for the Fundamental Frequency 6-18 Bar † Beam † Summary of Finite Difference Methods for a Continuous System Example: Three-Degree-of-Freedom System † Bisection Method † Directly Calculating the Eigenvalues and Eigenvectors from the Matrix Equation † Summary of Matrix Methods Rayleigh Method † Dunkerley’s Formula † Summary of Approximations for the Fundamental Frequency Marie D Dahleh Harvard University 6.7 Finite Element Method 6-20 Bar Element † Beam † Summary of Finite Element Method Appendix 6A Introduction to MATLABw 6-24 Summary This chapter gives an overview of numerical techniques for vibration analysis The centered difference approximation for the first, second, and fourth derivative are given These form the basis for the finite difference approximation of both spring – mass systems and the continuous problem The fourth-order Runge – Kutta method is presented Both of these approaches are used to solve the single-degree-of-freedom (single-DoF) system In order to demonstrate these techniques for the multiple-degree-of-freedom (multi-DoF) system a two-degree-of-freedom (two-DoF) system is explored Finite element and finite difference methods are presented as solution techniques for the continuous problem (also see Chapter 9) The bar and beam are used for examples The Rayleigh method and Dunkerley’s formula are presented These are methods for computing the fundamental frequency 6.1 Introduction This chapter presents numerical techniques for three classes of vibration problems; the single spring mass, the multiple spring mass, and the continuous model The simplest oscillatory problem results when the problem can be modeled by the single spring– mass system This system is modeled by a second-order constant coefficient differential equation, which can be solved analytically when either there is no forcing, or the forcing is harmonic When the forcing is not harmonic, the finite difference method can be used to 6-1 © 2005 by Taylor & Francis Group, LLC 6-2 Vibration and Shock Handbook solve this second-order equation or to solve the system of first-order equations, which is equivalent to the second-order equation A larger computational problem arises when there are multiple springs and masses subjected to nonharmonic forcing Again, the finite difference method can be used to compute the full solution It is not always necessary to compute the full solution Often, all that is needed is the fundamental frequency This can be computed from the eigenvalues of a matrix problem Both the numerical solution to the full problem and the matrix problem will be discussed Finally, one may want to look at the vibration of a continuous element like a beam Here, both the finite difference method and the finite element method will be discussed 6.2 Single-Degree-of-Freedom System The simple undamped spring–mass system oscillates with a frequency known as the natural frequency The natural frequency is determined by the spring constant pffiffiffiffiffiand the mass in the following way, v ¼ k=m: This relationship is derived by applying Newton’s second law to the basic spring–mass system The resulting equation is given by m€x ¼ 2kx: The solution to this equation is harmonic with the frequency specified pffiffiffiffiffi FIGURE 6.1 Single spring – mass system by the expression v ¼ k=m: A more realistic vibration model of a simple oscillatory system includes a mass, a spring and a damper (see Figure 6.1) For simplicity, the mass is concentrated at the center of mass of the object, the spring is assumed to be of negligible mass, and, for the purposes of this discussion, the damping will be modeled by viscous damping This is described by a force proportional to the velocity, denoted by x_ : In the case of no external forcing, the system can be described by the following equation: mx ỵ c_x ỵ kx ẳ where m; c; and k are constants and m is the mass, c is the damping coefficient, and k is the spring constant This equation can be solved analytically by assuming a solution of the form x ¼ est where s is a constant Substitution into the differential equation yields the following quadratic equation: ms2 ỵ cs ỵ kịest ẳ This equation is satisfied for all values of t when the following quadratic equation, known as the characteristic equation, is satised: s2 ỵ The characteristic equation has two roots c k sỵ ẳ0 m m s1 ẳ q c ỵ c=2mị2 k=mị 2m s1 ẳ q c ðc=2mÞ2 ðk=mÞ 2m and The general solution is x ẳ A e s1 t ỵ B e s2 t The constants A and B are determined by the initial conditions xð0Þ and x_ ð0Þ: The single spring– mass system exhibits three types of behavior, overdamped, underdamped and critically damped © 2005 by Taylor & Francis Group, LLC Numerical Techniques 6-3 The overdamped state exists when k=m is larger than ðc=2mÞ2 : No oscillations exist in this state The underdamped case is oscillatory and results when ðc=2mÞ2 is larger than k=m: The limiting case between oscillatory and nonoscillatory motion occurs when c=2mị2 ẳ k=m: When this condition, is met, the system is said to be critically damped The next level of complexity results when the system is forced harmonically The following equation serves as a model for this system: mx ỵ c_x ỵ kx ẳ F sin vt The solution of this equation is found by first computing the complementary function, which is the solution of the homogeneous equation discussed above, and then the particular solution The details for computing a particular solution can be found in books on differential equations or mechanical vibrations (Thomson and Dahleh, 1998) 6.2.1 Forced Single-Degree-of-Freedom System In general, when the oscillatory system is forced in a nonharmonic way, the resulting differential equation cannot be solved in closed form, and numerical methods must be employed to predict the behavior of the system In this section, we consider two finite difference methods chosen for their simplicity For a more detailed discussion of numerical methods for ordinary differential equations see Atkinson (1978) and Isaacson (1966) The spring–mass system that is subjected to general forcing, Fðx; x_ ; tÞ; can be modeled by the following differential equation: mx ỵ c_x ỵ kx ẳ f x; x_ ; tị x0 ẳ x0ị x_ ẳ x_ 0ị where m; c; and k are constants and F is an arbitrary function The two initial conditions, x0 and x_ , are known In the first method, the second-order equation is integrated without change in form; in the second method, the second-order equation is rewritten as a system of two first-order equations and then the system of equations is integrated Both methods approximate the first and second derivatives with the centered difference approximation for the derivatives Finite difference methods are based on the Taylor expansion The centered difference method for the first and second derivative results from a combination of the forward and backward Taylor expansion about the point xi : To get the forward expansion, one writes the Taylor expansion for xiỵ1 : Similarly, the backward expression is obtained from the Taylor expansion about xi21 : These are given by xiỵ1 ẳ xi ỵ h_xi ỵ h2 h3 x i ỵ x ỵ ããã xi21 ẳ xi h_xi ỵ h2 h3 x i x ỵ ããã where h ẳ Dt: A second-order approximation is one which matches the Taylor expansion exactly up to and including terms of order h2 ; that is, to determine a second-order expansion one neglects terms of order h3 and higher A second-order approximation for the first derivative is obtained by subtracting the backward difference from the forward difference The resulting centered difference approximation is given by h2 xiỵ1 xi21 ị x þ ··· x_ i ¼ 2h Errors result when this expression is truncated after the first term These errors depend on h2 and the third derivative of xi : If the error in the computed derivative is larger than order h2 it may well arise from the neglected third derivative term © 2005 by Taylor & Francis Group, LLC 6-4 Vibration and Shock Handbook The centered difference approximation for the second derivative is found by adding the forward and backward difference expansions and ignoring terms of order h4 and higher The resulting approximation is x€ i ¼ h2 iv x ỵ ããã xi21 2xi ỵ xiỵ1 ị þ h 12 The truncation error that occurs for this expression depends on h2 and the fourth derivative of xi : Both the first and second derivative centered difference approximations are order h2 : They can be used together to create a second-order approximation to a second-order ordinary differential equation such as that describing the spring–mass system 6.2.1.1 Centered Difference Approximation After substituting these two centered difference approximations into the differential equation and rearranging terms, one gets the following recurrence relation for the single springmass system: xiỵ1 ẳ h2 f ðxi ; ti Þ ð fh2 2mÞxi m ch=2ịxi21 with the initial conditions x0 ẳ x0ị x_ ẳ x_ 0ị The recurrence relation should be used to compute all values of x from the initial condition By letting i ¼ in the recurrence relation, we get x2 ¼ h2 f ðx1 ; t1 Þ ðkh2 2mÞx1 ðm ch=2Þx0 In order to compute x2 ; both x0 and x1 are needed The initial conditions provide x0 : However, to start the calculation, we need an additional equation for x1 : This equation is derived by substituting i ¼ into the Taylor expansion for xiỵ1 ; using the initial conditions and ignoring terms of order h2 and higher Since the centered difference approximation is of order h2 ; it is consistent to compute x1 with an error of order h2 : For the point i ¼ 0; the forward difference is given by x1 ẳ x0 ỵ h_x0 This equation allows one to find x1 in terms of the two initial conditions, after which the recurrence relation can be used to find all subsequent discrete values of x: 6.2.1.2 Pseudocode for Centered Difference Approximation The following is an example of the MATLABw routine for the solution to the centered difference approximation of the general single-DoF spring–mass equations The function f needs to be specified in a separate function file: %initial conditions (index has to start at 1) x1ị ẳ x0 xd ¼ x00 %specify a time step h¼H %specify the constants m; k; c: mẳM kẳK cẳC %compute x2ị from the Taylor expansion x2ị ẳ x1ị ỵ h p xd %specify the total number of steps TEND ẳ tnal â 2005 by Taylor & Francis Group, LLC Numerical Techniques 6-5 %compute the solution for all remaining times using the recurrence relation t1ị ẳ h t2ị ẳ p h for I ¼ 3:TEND Xi ¼ h2 p f ðxi21 ; ti21 Þ ðk p h2 2 p mÞ p xi21 ðm c p h=2Þ p xi22 Tiị ẳ tI 1ị ỵ h end The method just presented has ignored terms of order h2 and higher This is known as the truncation error The calculation will contain other errors such as round-off error due to the loss of significant figures This loss is related to both the machine and the language used for the calculations The round-off errors are also related to the time increment h ¼ Dt in a complicated way that is beyond the scope of this chapter Better accuracy can be obtained by choosing a smaller Dt: However, the smaller the Dt; the larger the number of computations needed to reach the solution in a fixed time T: The increased number of computations affects both the total time of the calculation and the overall accuracy 6.2.1.3 Runge–Kutta Methods The centered difference approximation is not a self-starting method In other words, the calculation that determines x1 from the initial conditions does not come from the discretized equation, but rather from the Taylor expansion directly An alternate way to solve this equation is to use what is known as a Runge – Kutta method All of these methods approximate the differential equation with Taylor series expansions The advantage of this class of methods is that they are self-starting The disadvantage is that they only work on first-order equations (or systems of first-order equations) Before the Runge –Kutta method is discussed in detail, the second-order spring–mass equation has to be written as a system of two firstorder equations The equation for the single-DoF system, subjected to arbitrary forcing f is mx ỵ c_x ỵ kx ẳ f ðx; x_ ; tÞ It can be rewritten as the following system of first-order equations x_ ¼ u u_ ẳ x ẳ ẵf x; u; tị cu þ kx m These equations are coupled and need to be solved together Any Runge–Kutta method for the system of equations requires a Taylor series expansion of x and u about xi and ui : The Taylor expansion is given by xiỵ1 ẳ xi ỵ x_ i h ỵ x h2 ỵ ããã uiỵ1 ẳ ui ỵ u_ i h ỵ u i h2 ỵ ããã As above, the time increment will be denoted by h ¼ Dt: The first-order Runge –Kutta method (also known as Euler’s method) is obtained by retaining terms of first-order and lower, i.e., the Euler approximation is given by xiỵ1 ẳ xi ỵ x_ i h ohị uiỵ1 ẳ ui ỵ u_ i h oðhÞ The truncation error for Euler’s method is order h: The error depends linearly on h: Matching more terms in the Taylor expansion generates higher order Runge–Kutta methods The most commonly used Runge –Kutta method is the fourth-order method that matches the Taylor expansion up to terms of order h4 : This is a significant reduction in the error For a derivation of this method see Cheney and Kincaid (1999) © 2005 by Taylor & Francis Group, LLC 6-6 Vibration and Shock Handbook TABLE 6.1 Fourth-Order Runge– Kutta Method for Spring – Mass Equation t T1 T2 T3 T4 X_ ¼ u x ¼ ti ¼ ti þ h=2 ¼ ti þ h=2 ¼ ti þ h X1 X2 X3 X4 ẳ xi ẳ xi ỵ U1 h=2 ẳ xi ỵ U2 h=2 ẳ x i ỵ U3 h U1 U2 U3 U4 ¼ ui ¼ ui þ G1 h=2 ¼ ui þ G2 h=2 ¼ ui þ G3 h X€ ¼ G G1 G2 G3 G4 ¼ GðT1 ; X1 ; U1 Þ ¼ GðT2 ; X2 ; U2 ị ẳ GT3 ; X3 ; U3 Þ ¼ GðT4; ; X4 ; U4 Þ Source: Thomson and Dahleh 1998 Theory of Vibration Applications, 5th ed With permission For the system of first-order equations given above, the fourth-order Runge–Kutta method requires four values of t; x; u, and, G where G ẳ 1=mẵf x; u; tị cu ỵ kx : It can be computed for each point i as follows in Table 6.1 Combining these quantities in the following method gives the fourth-order RungeKutta method: xiỵ1 ẳ xi ỵ h=6U1 ỵ 2U2 ỵ 2U3 ỵ U4 ị uiỵ1 ẳ ui ỵ h=6G1 ỵ 2G2 ỵ 2G3 þ G4 Þ where it is recognized that the four values of U divided by six represent the average slope dx=dt and the four values of G divided by six result in an average of du=dt: A way to check the accuracy of a Runge– Kutta method is to Taylor expand G1 ; G2 ; G3 ; and G4 and collect like terms One will find that the above combination produces an expansion which is exact up to order h4 : 6.2.1.4 Pseudocode for the Fourth-Order Runge –Kutta Method For simplicity, the code is given for the single first-order equation x_ ẳ Gt; xị with the initial data x0ị ẳ x0 : The function G should be specified in a function file %initial conditions (index has to start at 1) x1ị ẳ x0 ; %time step needs to be specied h ¼ H; h2 ¼ p H; %TEND is the total number of time steps TEND ¼ Tfinal; T ¼ h; for I ¼ 1:TEND G1 ¼ H p Gt; xị; G2 ẳ H p Gt ỵ h2; x ẳ 0:5 p G1ị; G3 ẳ H p Gt ỵ h2; x ẳ 0:5 p G2ị; G4 ẳ h p Gt ỵ h; x ỵ G3ị; X ẳ x ỵ G1 ỵ p G2 ỵ p G3 ỵ G4ị=6:0; t ẳ I ỵ 1ị p h; end 6.2.1.5 Example Solve numerically the differential equation 4x ỵ 2000x ẳ Ftị with the initial conditions x0 ẳ x_ ẳ The forcing is as shown in Figure 6.2 © 2005 by Taylor & Francis Group, LLC FIGURE 6.2 The forcing function for Example 6.2.1.5 (Source: Thomson and Dahleh 1998 Theory of Vibration Applications, 5th ed With permission.) Numerical Techniques 6-7 6.2.1.5.1 Centered Difference Using the centered difference approximation for the second derivative, one gets the following discrete equation xiỵ1 ẳ h2 =4Ftị 500h2 xi xi21 ỵ 2xi This equation is valid for i $ 1: From the initial conditions x0 ¼ x_ ¼ 0; x1 is computed by the following x1 ẳ x0 ỵ x_ ẳ ỵ 0: A time step of h ẳ 0.02 sec has been used in the calculation The numerical solution as compared with the exact solution is given in Table 6.2 Table 6.3 contains the absolute errors for the two discrete solutions 6.2.1.5.2 Runge–Kutta In order to use the Runge–Kutta method, the second-order equation needs to be written as a system of first-order equations Let u ¼ x_ ; then u_ ẳ Gx; tị ẳ :25 p Ftị 500x where x0ị ẳ and u0ị ẳ 0: This is now in a form which can be directly input into a fourth-order Runge –Kutta solver The Runge–Kutta method is more accurate then the centered difference approximation This can be seen in Table 6.3, which gives the absolute value of the difference between the exact solution and the computed solutions This is known as the absolute error Moreover, the Runge–Kutta method is selfstarting and can be used for a single variable or a system of variables as in the above example The price that is paid is that the fourth-order Runge –Kutta method requires four function evaluations for the first derivative for each time step This is offset by the fact that this method has higher accuracy and has a large stability region so one can take larger time steps TABLE 6.2 Time t 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0.18 0.20 0.22 0.24 0.26 0.28 0.30 0.32 0.34 0.36 0.38 0.40 0.42 0.44 0.46 0.48 Solution to Example 6.2.1.5 Exact Solution 0.00492 0.01870 0.03864 0.06082 0.08086 0.09451 0.09743 0.08710 0.06356 0.02949 20.01005 20.04761 20.07581 20.08910 20.08486 20.06393 20.03043 0.00906 0.04677 0.07528 0.08898 0.08518 0.06436 0.03136 © 2005 by Taylor & Francis Group, LLC Central Difference 0.00500 0.01900 0.03920 0.06159 0.08167 0.09541 0.09807 0.08712 0.06274 0.02782 20.01267 20.05063 20.07846 20.09059 20.08461 20.06171 20.02646 0.01407 0.05180 0.07916 0.09069 0.08409 0.06066 0.02511 Runge–Kutta 0.00492 0.01869 0.03862 0.06076 0.08083 0.09447 0.09741 0.08709 0.06359 0.02956 20.00955 20.04750 20.07571 20.08903 20.08485 20.06400 20.03056 0.00887 0.04656 0.07509 0.08886 0.08516 0.06423 0.03157 6-8 Vibration and Shock Handbook TABLE 6.3 Absolute Error for Centered Difference and for the Fourth-Order Runge– Kutta Time t 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0.18 0.20 0.22 0.24 0.26 0.28 0.30 0.32 0.34 0.36 0.38 0.40 0.42 0.44 0.46 0.48 Error for Central Difference Error for Runge–Kutta All Times 1.0 £ 1023 0.0001 0.0003 0.0006 0.0008 0.0008 0.0009 0.0006 0.0000 0.0008 0.0017 0.0026 0.0030 0.0026 0.0015 0.0003 0.0022 0.0040 0.0050 0.0050 0.0039 0.0017 0.0011 0.0037 0.0062 0.0 0.0100 0.0180 0.0600 0.0300 0.0400 0.0200 0.0100 0.0300 0.0700 0.0500 0.1100 0.1000 0.0700 0.0100 0.0700 0.1300 0.1900 0.2100 0.1900 0.1200 0.0200 0.1300 0.2100 Stability is a measure of how quickly errors in the computed solution grow or decay There are very few numerical methods that are stable for all choices of time step For most methods, there is a range of time steps which produce a stable method The fourth-order Runge –Kutta method is stable for larger values of the time step than are the lower-order Runge–Kutta methods A numerical method will not converge to a solution if the time step does not produce a stable method Often, the stability criterion places a stricter limitation on the time step than accuracy does For a more complete discussion of stability, see Strang (1986) In Table 6.3, the absolute error for both methods grows but it does not grow exponentially Controlled error growth is the signature of a stable time step 6.2.2 * * * * * 6.3 Summary of Single-Degree-of-Freedom System Unforced equation of motion m€x þ c_x þ kx ¼ 0: Forced equation of motion mx ỵ c_x ỵ kx ẳ f x; x_ ; tÞ: Centered difference approximation for the first derivative x_ i ẳ x xi21 ị: 2h iỵ1 Centered difference approximation for the second derivative x€ i ¼ xi21 2xi ỵ xiỵ1 ị: h Fourth-order RungeKutta method Systems with Two or More Degrees of Freedom A system that requires more than one coordinate to describe its motion is a multi-DoF system These systems differ from single-DoF systems in that n DoF are described by n simultaneous differential equations and have n natural frequencies When these systems are written in matrix notation, they look © 2005 by Taylor & Francis Group, LLC Numerical Techniques 6-9 just like the single-DoF system The equations of motion of a viscously damped multi-DoF system can be written as follows: ẵM x ỵ ẵC x_ ỵ ẵK x ẳ f where ẵM ; ½C ; and ½K are the mass, damping, and stiffness matrices, respectively; x is the displacement vector; and f is the force vector Both the central difference method and Runge– Kutta can be applied to the matrix equation The method follows exactly that given above where the scalar quantities are replaced by the matrix quantities 6.3.1 Example As an example, consider the two-DoF system shown in Figure 6.3 In this example, k1 ¼ k2 ¼ 36 KN=m; m1 ¼ 100 kg; m2 ¼ 25 kg; and f ¼ 4000 N for t and for t # 0: The initial conditions are all zero, i.e., x1 ¼ x_ ¼ x2 ¼ x_ ¼ 0: The equation of motion for this system is 100x1 ỵ 36;000x1 36;000x2 x1 ị ẳ 25x2 ỵ 36;000x2 x1 ị ẳ f This can be written in matrix notation as " # " # 100 21 x ỵ 36;000 ẳf 25 21 where x ¼ ðx1 ; x2 Þt and f ¼ ð0; f Þt : 6.3.1.1 FIGURE 6.3 Two-DoF system (Source: Thomson and Dahleh 1998 Theory of Vibration Applications, 5th ed With permission.) Centered Difference Using the centered difference approximation for the second derivatives, one obtains the following two recurrence relations x1iỵ1 ẳ 2720x1i ỵ 360x2i ịDt ỵ 2x1i x1i21 x2iỵ1 ẳ 1440x1i x2i ị ỵ 160ịDt ỵ 2x2i x2i21 These two equations are only valid for i $ 3: From the single-DoF system, we know that this method is not self-starting In order to compute x12 and x22 ; one needs to use an additional equation, which one obtains from the Taylor expansion First one needs to compute the initial acceleration for the system This is obtained from the differential equation and the initial data For this problem, the initial acceleration is given by x€ ¼ and x€ ¼ 160: Next, values for x121 and x221 are computed from the Taylor expansion x121 ¼ x10 Dt x_ ỵ Dt x ẳ x221 ¼ 80Dt Now, the recurrence relations can be used to compute the rest of the terms An example of the MATLAB code for this calculation is the following © 2005 by Taylor & Francis Group, LLC 6-12 Vibration and Shock Handbook and the temporal equation d2 G=dt ỵ v2 G ẳ The displacement equation can be solved using a finite difference approximation The finite difference approximation requires dividing the FIGURE 6.5 Discretized bar bar of length L into n pieces each of length h ẳ L=n 1ị: The nite difference mesh consists of n points, 1; 2; 3; …; n: The first node (labeled 1) is the left-hand endpoint of the bar, and the last node, labeled n; is the right-hand endpoint of the bar (see Figure 6.5) The value of U at each point i is denoted by Ui : Using the centered difference approximation for the second derivative, the discretized version of the displacement equation becomes Uiỵ1 2Ui ỵ Ui21 ị=h2 ỵ aUi ẳ Collecting like terms, the equation becomes Uiỵ1 2 lịUi ỵ Ui21 ¼ where l ¼ h2 a2 : As is typical for the finite difference approximation, this equation holds at the interior mesh points ði ¼ 2; 3; …; n 1Þ and not at the endpoints To see that this equation does not hold at the endpoints, substitute i ¼ into the recurrence relationship The resulting equation is U2 2 lịU1 ỵ U0 ẳ The problem is that there is no value U0 : Recall that U1 represents the displacement of the lefthand endpoint Since U0 represents a point even further to the left, it is not on the bar Similarly, the recurrence relation does not work for i ¼ n: When the relationship is evaluated for i ẳ n; the value Unỵ1 is needed and it does not exist The displacement equation evaluated for i ẳ n is Unỵ1 2 lịUn þ Un21 ¼ When the relationship is evaluated for i ẳ 2; 3; ; n 1ị, the resulting matrix problem is tridiagonal This is characteristic of the centered difference approximation It is given by the following matrix: 32 3 21 … 21 2 l U1 76 7 7 21 2 l 21 … 76 U2 7 76 7 76 7 6 7 … 21 2 l 21 76 U3 7 76 7 76 ¼ 76 7 76 7 76 7 76 7 76 7 76 7 76 7 54 5 21 2 l 21 Un This matrix problem does not have a unique solution since there are n unknowns and there are only n 2 equations The boundary conditions for the bar provide the needed two additional constraints For a bar that is fixed at both ends, the deflection is zero at the ends (x ¼ and x ¼ L) Under these conditions, U1 ¼ Un ¼ 0: If one of the ends is free then the stress is zero at that end, which means that dU=dx ¼ 0: For the purposes of illustration, let us assume that the end at x ¼ is fixed and the other end is free This means that U1 ¼ and dU=dxln ¼ 0: If we use the centered difference approximation for the derivative, we get dU=dxln ẳ Unỵ1 Un21 ị=2h ẳ which means that Unỵ1 ẳ Un21 : â 2005 by Taylor & Francis Group, LLC Numerical Techniques 6-13 Unỵ1 is the displacement of a point off the bar We never compute Unỵ1 ; however, as one can see from a previous calculation, this value is used when the recurrence relationship is evaluated for i ẳ n; Unỵ1 2 lịUn ỵ Un21 ẳ 0: Under the assumption that the stress is zero at the right-hand endpoint, the remaining constraint becomes 2 lịUn 2Un21 ẳ For a bar that has one fixed left-hand endpoint and a free right-hand end, the matrix problem is the following: 32 3 21 0 … 22l U2 7 76 6 U3 7 21 2 l 21 … 7 76 7 76 6 U4 7 7 … 21 2 l 21 76 7 76 6 7¼6 7 76 6 7 76 7 76 7 76 7 6 7 l 21 21 2 607 56 U n21 5 Un 22 2 l This matrix problem has a unique solution because there are n linearly independent equations and there are n unknowns 6.4.1.1 Example Consider a bar of unit length that is divided into four equal pieces The left-hand end is fixed so that U1 ¼ 0; and the right-hand end is free The eigenvalue problem for this system is the following: 32 3 0 U2 2 l 21 76 7 21 2 l 21 7 76 U3 7 76 ¼ 76 7 6 7 21 2 l 21 54 U4 5 0 22 2 l U5 The eigenvalues, l; are l ẳ ẵ0:1522; 1:2346; 2:7654; 3:8478 : The natural frequencies can be recovered pffiffiffiffiffi from the eigenvalues since l ¼ h2 v=c and c ¼ E=r is determined by the material properties of the bar 6.4.2 Beam A second example of finite difference approximation is given by looking at a centered difference approximation to compute the transverse vibration of a uniform beam We will consider a special case of the Euler equation for the beam The following fourth-order equation results when the flexural rigidity, ðEIÞ; is constant: d4 W=dx4 b4 W ¼ where b contains all of the material properties of the beam (i.e., b4 ¼ rAv2 =EI) The centered difference formula for the fourth derivative is given by f iv ẳ 1=h4 fiỵ2 4fiỵ1 ỵ 6fi 4fi21 ỵ fi22 ị In order to see that this expansion is in fact second-order, combine the Taylor expansions for fiỵ2 ; fiỵ1 ; fi ; fi21 ; and fi22 ; and notice that you obtain an approximation for the fourth derivative which exactly matches the combined Taylor approximations up to order h2 : Using this approximation, the transverse beam equation at the interior mesh point becomes Wiỵ2 4WIỵ1 ỵ lịWi 4Wi21 ỵ Wi22 ẳ © 2005 by Taylor & Francis Group, LLC 6-14 Vibration and Shock Handbook where l ¼ h4 b4 : This equation is valid for i ¼ 3; n 2: As before, the boundary conditions have to be used in order to have sufficent conditions to determine all n values for the deflection, W: There will be two additional constraints placed at each end In this case, all of the types of boundary condition require fictitious points In other words, they require that the beam be extended beyond the physical boundaries to include W21 ; W0 ; Wnỵ1 ; and Wnỵ2 : Let us consider three types of boundary conditions: fixed end, free end, and simply-supported end For simplicity, we will derive all of the conditions for the right-hand endpoint x ¼ L: Fixed end: When the end of the beam is fixed, both the deflection W ¼ and dW=dx ¼ 0: As above, the centered difference approximation for the first derivative requires the introduction of a fictitious point These conditions translate into Wn ¼ and dW=dxln ẳ 1\h2 Wnỵ1 Wn21 ị ẳ These constraints reduce to the following: Wnỵ1 ẳ Wn21 Free end: The bending moment and the shear force are zero at the free end These are given by a second and third derivative, respectively Assuming the free end is located at N; the two ctitious points introduced are at n ỵ and n ỵ 2: The discrete version of the two constraints is: d2 W=dx2 ln ¼ 1=h2 ðWn21 2Wn ỵ Wnỵ1 ị ẳ d3 W=dx3 ln ẳ 1=2h3 ịWnỵ2 2Wnỵ1 ỵ 2Wn21 Wn22 ị ẳ Simply-supported end: The boundary conditions of this type require that the deflection and the moment are zero When these are applied to the right-hand endpoint Wn ¼ d2 W=dx2 ẳ 1=h2 Wnỵ1 2Wn ỵ Wn21 ị ẳ Combining these two equations, one sees that the value of W at the fictitious point is equal to the value on the beam, i.e., Wnỵ1 ẳ Wn21 : All three of these boundary condition types allow one to replace the fictitious values in the recurrence relations with values on the beam The procedures followed are very similar to those outlined for the longitudinal vibration of a beam 6.4.3 Summary of Finite Difference Methods for a Continuous System Equation of motion for the bar: ›2 u=›x2 ¼ 1=c2 ›2 u=›t : Displacement equation for the bar: d2 U=dx2 ỵ a2 U ẳ 0: Finite difference approximation: Uiỵ1 2 lịUi ỵ Ui21 ẳ 0: Equation of motion for the beam: d4 W=dx4 b4 W ¼ 0: Centered difference approximation for the fourth derivative: f iv ¼ 1=h4 ð fiỵ2 4fiỵ1 ỵ 6fi 4fi21 ỵ fi22 ị: Finite difference approximation for the beam equation: Wiỵ2 4Wiỵ1 ỵ lịWi 4Wi21 ỵ Wi22 ẳ 0: 6.5 Matrix Methods In the preceding section, the equations of motion are solved for the system Often, one is not interested in the complete solution Rather, one is interested in the natural frequencies and the normal modes When the number of DoF is very high, only the lowest natural frequency needs to be computed Several methods to find some, but not all, of the eigenvalues of the system will be presented © 2005 by Taylor & Francis Group, LLC Numerical Techniques 6.5.1 6-15 Example: Three-Degree-of-Freedom System A three-DoF system will be used as an example The larger DoF systems work exactly in the same manner but produce larger calculations Let us consider the following spring – mass system in Figure 6.6 The equation of motion for this system is 32 32 0 x€ 21 x1 76 76 76 76 M 76 x€ ỵ K 21 21 76 x2 54 54 0 7 ¼ 607 x€ 21 x3 FIGURE 6.6 Three spring – mass system (Source: Thomson and Dahleh 1998 Theory of Vibration Applications, 5th ed With permission.) Assuming a harmonic solution of the form x ẳ A sin vtị gives the following eigenvalue problem: 312 3 2 0 21 x1 7C6 7 B B2l6 ỵ k6 21 21 7C6 x2 ¼ 5A4 5 @ 0 21 x3 where l ¼ v2 m=k: The eigenvalues are found by setting the characteristic equation of the determinant equal to zero This gives the following polynomial: l3 4:5l2 ỵ 5l ¼ It is a simple matter to graph this polynomial and see that there are three real roots to this equation 6.5.2 Bisection Method One method for calculating the roots of a nonlinear function is called the bisection method This method finds the zeros of nonlinear functions by bracketing the zero in an interval ½a; b : The interval is chosen so that f ðaÞ and f ðbÞ are of opposite sign If f is a continuous function and it is positive at one endpoint, say f ðaÞ 0; and it is negative at the other endpoint, f ðbÞ , 0; then it has had to go through zero at some point in the interval It is possible that it has gone through zero more then once in the interval From Figure 6.7, one can see that one root is between [0, 0.5]; another is between [1, 1.5]; and a third root is in [2.5, 3] In order to find all three roots, the bisection method has to be used three times; one for each interval As an example, the MATLAB code for the first root is given below 6.5.2.1 MATLAB Code for the Bisection Method clear %set an acceptable tolerance for the root tol ¼ 10e %endpoints of the interval a¼0 b ẳ 0:5 for i ẳ 1:20 c ẳ 0:5pa ỵ bị if abs( f(c)) , tol, break, end â 2005 by Taylor & Francis Group, LLC 6-16 Vibration and Shock Handbook −2 −4 −6 −8 −10 −12 −1 −0.5 0.5 1.5 2.5 3.5 FIGURE 6.7 Graph of the cubic polynomial l3 4:5l2 ỵ 5l ẳ 0: (Source: Thomson and Dahleh 1998 Theory of Vibration Applications, 5th ed With permission.) if f aịpf cị , bẳc else aẳc end end The root 0.2554 is found after ten iterations As long as the initial interval is chosen so that the function has different signs at the endpoints, this method will always converge There are nonlinear root finding methods, such as Newton’s method, which when they converge so more quickly However, they may or may not converge 6.5.2.2 MATLAB Function for Finding the Roots of a Polynomial MATLAB has a built in root-finding method which requires the creation of a vector c that contains the coefficients of the polynomial in descending order The MATLAB command roots (c), produces the roots of the polynomial For our example, c ẳ ẵ1; 24:5; 5; 21 : The roots of this equation are given by l ¼ 2.8892, 1.3554, 0.2554 6.5.2.3 Mode Shapes The mode shapes are determined by substituting each eigenvalue into the matrix problem and computing the corresponding eigenvector As an example of how to compute the eigenvector for a given eigenvalue, we will compute the eigenvector associated with l ¼ 2:8892: The first step is to substitute this value of l into the eigenvalue equation This gives the following problem: 32 3 2:489 21 x1 76 7 21 1:745 21 76 x2 ¼ 54 5 21 0:744 x3 The solution vector found by Gaussian elimination is given by x ¼ ðx1 ; x2 ; x3 ịt ẳ 0:2992; 0:7446; 1:0ịt â 2005 by Taylor & Francis Group, LLC Numerical Techniques 6-17 6.5.3 Directly Calculating the Eigenvalues and Eigenvectors from the Matrix Equation Ideally one would like to use a computer program such as MATLAB to compute both the eigenvalues and the eigenvectors directly from the matrix equation In order to this, the matrix equation needs to be rewritten slightly In general, the matrix equation for the normal mode vibration is given by ẵ2lM ỵ K x ẳ where M and K are the mass and stiffness matrices, respectively, both are square symmetric matrices, and l is the eigenvalue related to the natural frequency by l ¼ nv2 : Premultiplying the preceding equation by M21 , we have another form of the equation: ẵ2lI ỵ A x ẳ where A ẳ M21 K and A is called the dynamic matrix In general, A is not symmetric In order to use MATLAB to compute the eigenvectors and eigenvalues, the matrix must be symmetric There is a standard transformation of coordinates that converts the matrix problem into a standard eigenvalue problem which can be solved by a computer Assume that we have a transformation of coordinates which has the following form: x ¼ U 21 When this transformation is substituted into the equation, we get ẵ2lMU 21 ỵ KU 21 y ẳ Premultiplying the equation by U2t gives ½2lU 2t MU 21 þ U 2t KU 21 y ¼ From this equation, one can see that if either M or K equals Ut U; then the preceding equation is reduced to standard form, at which time it is possible to compute both the eigenvalues and eigenvectors directly To see how this works, let us assume that M ¼ Ut U: The equation becomes ẵ2lI ỵ U2t KU21 y ẳ where l ¼ v2 : This equation is in standard form since U2t KU21 is symmetric 6.5.3.1 Example To illustrate the use of the dynamic matrix and the standard computer form, we can use MATLAB to calculate the eigenvalues and eigenvectors for Example 6.5.1 The first step is to convert the problem into standard form For this example, the dynamic matrix is given by 1:5 20:5 7 A¼6 21:0 2:0 21:0 21:0 1:0 We can compute both the eigenvalues and the eigenvectors in MATLAB using the command ẵU; D ẳ eigAị: The result of this command is two matrices, U and D: Matrix U contains the eigenvectors as column vectors and D is a diagonal matrix which has the eigenvalues on the diagonal Continuing with our example, we get the following two matrices: 0:7569 0:3031 0:2333 7 U¼6 0:2189 20:8422 0:5808 20:6158 © 2005 by Taylor & Francis Group, LLC 0:4458 0:7799 6-18 Vibration and Shock Handbook and D¼6 1:3554 0 2:8892 0 0:2544 7 In order to use this transformation then, one has to be able to write M or K as Ut U: This is known as the Cholesky decomposition In MATLAB, if one has a positive definite matrix M; the matrix can be Cholesky-decomposed with the built in function chol The command U ¼ cholðMÞ produces an upper triangular matrix U such that Ut U ¼ M: Cholesky decomposition is a special case of LU factorization where L is a lower triangular matrix and U is an upper triangular matrix The computational procedure begins by writing the algebraic equations that result from the following calculation: 32 3 u11 0 u11 u12 u13 m11 m12 m13 76 7 u12 u22 76 u22 u23 ¼ m21 m22 m23 54 5 u13 u23 u33 0 u33 m31 m32 m33 If the matrix M is positive definite, then the above matrix multiplication results in six linearly independent equations 6.5.4 Summary of Matrix Methods Bisection method for the roots of a polynomial MATLAB command roots for finding the roots of a polynomial Cholesky decomposition 6.6 Approximation Methods for the Fundamental Frequency The smallest natural frequency, known as the fundamental frequency, of a multi-DoF system is often of greater interest than the high natural frequencies because its forced response in many cases is the largest One approach to this problem is to extend the Rayleigh method to matrix problems We will see that the Rayleigh frequency approaches the fundamental frequency from the high side 6.6.1 Rayleigh Method Let M and K be the mass and stiffness matrices, respectively, and x is the assumed displacement vector for the amplitude of vibration For harmonic motion, the maximum kinetic energy is Tmax ¼ 1=2vxt Mx and the maximum potential energy is Umax ¼ 1=2xt Kx Since the maximum kinetic energy equals the maximum potential energy, these two quantities are equal Equating these two and solving for v2 gives the Rayleigh quotient: v2 ¼ xt Kx xt Mx It can be shown (Thomson and Dahleh, 1998) that this quotient approaches the lowest natural frequency from above and it is somewhat insensitive to the choice of amplitudes © 2005 by Taylor & Francis Group, LLC Numerical Techniques 6.6.2 6-19 Dunkerley’s Formula Dunkerley’s formula produces a lower bound for the fundamental frequency and can be used in conjunction with the Rayleigh method to get a good approximation for the fundamental frequency Dunkerley’s formula is based on the characteristic equation for the flexibility coefficients The flexibility influence coefficient, aii , is defined as the displacement at i due to a unit force applied at j with all other forces equal to zero This concept is most easily understood through an example 6.6.2.1 Computation of the Flexibility Matrix The procedure for computing the flexibility matrix and in particular, the computation of the flexibility matrix for the three spring –mass matrix systems shown in Figure 6.6, are discussed Example First, one applies a unit force to mass with no other forces present, i.e., f1 ¼ 1; f2 ¼ f3 ¼ 0: The displacements are located in the first column of the flexibility matrix This gives 32 x1 1=k1 0 76 x2 ¼ 1=k1 0 76 54 x3 1=k1 0 In this case, springs k2 and k3 are not stretched and are displaced equally with mass Now, a unit force is applied to mass and there are no other forces This allows us to write the second column of the matrix to get 32 1=k1 0 x1 76 x2 ¼ 1=k1 þ 1=k2 76 54 1=k1 ỵ 1=k2 x3 0 This time, the unit force is transmitted through k1 and k2 : The spring k3 is not stretched Finally, the force is applied to mass and there are no other forces present This gives the third column of the matrix: 32 3 0 1=k1 x1 76 7 76 x2 ẳ 0 1=k1 ỵ 1=k2 54 5 x3 1=k1 ỵ 1=k2 ỵ 1=k3 Since the flexibility matrix is the sum of the three previous matrices, it is given by x1 1=k1 6 x2 ¼ 1=k1 x3 1=k1 32 f1 76 76 f 54 1=k1 1=k1 1=k1 þ 1=k2 1=k1 þ 1=k2 1=k1 þ 1=k2 1=k1 þ 1=k2 ỵ 1=k3 f3 An interesting feature of the exibility matrix is that it is symmetric about the diagonal For simplicity of notation, let the ij element of the flexibility matrix be given by aij mj : Dunkerley’s formula is obtained from the characteristic equation of the flexibility matrix, which is obtained by computing the determinant of the following matrix: a11 m1 1=v2 a21 a31 © 2005 by Taylor & Francis Group, LLC a12 a22 m2 1=v a32 a13 ¼0 a23 a33 m3 1=v 6-20 Vibration and Shock Handbook A third-degree equation in 1=v2 is obtained by expanding the determinant One obtains the following cubic equation: 1=v2 ị3 a11 m1 ỵ a22 m2 ỵ a33 m3 ị1=v2 ị2 ỵ ã ã ã ẳ A cubic equation has three roots that are denoted by 1=v2i ị for i ẳ 1; 2; 3: This allows the cubic equation to be factored: ð1=v2 1=v21 Þð1=v2 1=v22 ị1=v2 1=v23 ị ẳ The highest two powers of this equation are given by ð1=v2 Þ3 1=v21 ỵ 1=v22 ỵ 1=v23 ị1=v2 ị2 ỵ ã · · ¼ The coefficient of the second highest power is equal to the sum of the roots of the characteristic equation, which is also equal to the sum of the diagonal elements of the matrix A21 : This relationship is not just true for n ¼ but is more generally true for n greater than or equal to For the general n-DoF system 1=v21 ỵ 1=v22 þ · · ·1=v2n ¼ a11 m1 þ a22 m2 þ · · · þ ann mn The fundamental frequency is the smallest natural frequency Since v2 ; v3 ; … are larger than v1 ; the reciprocal of these frequencies is smaller then the reciprocal of the fundamental frequency An estimate for the fundamental frequency is obtained by neglecting all of the higher modes in the left-hand side of the above equation This estimate gives a value for v1 that is smaller then the true value of the fundamental frequency Dunkerley’s formula is a lower bound for the fundamental frequency and it is given by 1=v21 , a11 m1 þ a22 m2 þ · · · þ ann mn 6.6.3 Summary of Approximations for the Fundamental Frequency Rayleigh method v2 ¼ xt Kx xt Mx Dunkerley’s formula 1=v21 , a11 m1 ỵ a22 m2 ỵ ã ã ã ỵ ann mn : 6.7 Finite Element Method In the finite element method, complex structures are replaced by assemblages of simple structural elements known as finite elements The elements are connected by joints or nodes The force and moments at the ends of the elements are known from structural theory, the joints between the elements are matched for compatibility of displacement, and the force and moment at the joints are established by imposing the condition of equilibrium The accuracy obtainable from the finite element method depends on being able to duplicate the vibration mode shapes Using only one finite element between structure joints or corners gives good results for the first lowest mode, because the static deflection curve is a good approximation to the lowest dynamic mode shape For higher modes, several elements are necessary between structural joints This leads to large matrices The eigenvalues and eigenvectors need to be computed numerically This section introduces the basic idea of the finite element method as it applies to the simple vibration problem The basic idea behind the finite element method is to break up the structure into simple component structures The structural elements for the bar and the beam are discussed here © 2005 by Taylor & Francis Group, LLC Numerical Techniques 6.7.1 6-21 Bar Element The force–displacement relationship for a uniform rod is F ẳ EA=LịU where E is the young’s modulus, A is the cross sectional area, L is the length of the element and U is the displacement Figure 6.8 shows this one-dimensional element The two endpoints are the nodes For simplicity, assume the axial displacement at any point ¼ x=L is linear: FIGURE 6.8 One dimensional element Ux; tị ẳ atị ỵ btịx and U1 tị ẳ U0; tị; U2 tị ẳ UL; tị: These two conditions uniquely determine the coefficients aðtÞ and bðtÞ: They are given by the following: atị ẳ U1 tị and FIGURE 6.9 btị ẳ U2 tị U1 tịị=L Linear mode shapes Using the linear element, the displacement anywhere along the beam is given by Ux; tị ẳ x=LịU1 tị ỵ x=LU2 tị ẳ w1 U1 tị ỵ w2 U2 tị where w1 ẳ x=L and w2 ¼ x=L FIGURE 6.10 Superposition of the linear mode shapes w1 and w2 are known as the mode shape, which can be seen in Figure 6.9 These two mode shapes can be superimposed to create a linear function An example of such a function is given in Figure 6.10 The kinetic energy of the bar is given by T ¼ :5 6.7.1.1 ðl u_ mdx ¼ :5m l 6ị_u1 tị ỵ 6u_ tịị2 ld6 ẳ :5ml1=3_u21 ỵ 1=3_u1 u_ ỵ 1=3_u22 ị Mass Matrix The generalized mass matrix is derived from the Lagrange equations using the following: D ›T Dt ›u_ Given the kinetic energy for the bar, the Lagrange equations become D T ẳ mL1=3u1 ỵ 1=6u2 ị Dt u_ D T ẳ mL1=3u1 ỵ 1=6u2 ị Dt u_ © 2005 by Taylor & Francis Group, LLC 6-22 Vibration and Shock Handbook The mass matrix for an axial element with a uniform mass distribution per unit length is given by " # mL 6.7.1.2 Stiffness Matrix The force–displacement relationship for a uniform bar is given by " # " #" # F1 u1 EA 21 ¼ L 21 F2 u1 6.7.1.3 Variable Properties A simple approach to problems with variable properties is to use a large number of elements of short length The variation of mass or stiffness over each element is small and can be neglected In this model, the mass and stiffness for each element is constant and can be placed outside the integral If a large number of elements are needed to capture the behavior, this will lead to a large matrix problem Example A tapered rod is modeled as two uniform sections where EA1 ¼ 2EA2 and m1 ¼ 2m2 ; k2 ¼ 2EA2 =L: The displacement equation for this system is given by " # " #" # " # u1 2v2 m2 22 ỵ 1ị 2EA2 ỵ 21 ẳ ỵ L u2 21 This can be written as a standard eigenvalue problem where l ¼ v2 m2 L=12EA2 The eigenvalue problem is the following 6l 21 ỵ lị 21 ỵ lị 2lị ẳ0 The solution of the determinant requires computing the roots of a quadratic equation The two roots are l ¼ ½0:6140; 1:1088 : The two natural frequencies can be computed from l They are pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi v1 ¼ 1:4029 ððEA2 Þ=ðM2 LÞÞ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi v2 ¼ 3:6477 ððEA2 Þ=ðM2 LÞÞ The modes shapes are calculated by solving the eigenvalue problem The two eigenvectors are x1 ẳ ẵ0:5773; t x2 ẳ ½25258; t and 6.7.2 Beam When the ends of the element are rigidly connected to the adjoining structure, the elements act like a beam with the moments and lateral forces acting at the joints Generally, the axial displacement u2 u1 is small compared to the lateral displacement V of the beam The local coordinates of the beam element are lateral displacements, V; and rotation, u, at the two ends This results in four coordinates V1 ; u1; V2; u2 : Four constraints uniquely determine a cubic polynomial Therefore, the lateral displacement of a beam is assumed to be described by a cubic © 2005 by Taylor & Francis Group, LLC Numerical Techniques 6-23 polynomial Vxị ẳ p1 ỵ p2 ỵ p3 62 ỵ p4 64 where ¼ x=L and pi ; the coefficients of the polynomials, are related to the lateral displacement and the rotation The lateral displacement at the left-hand side determines V1 : Vð0; tị ẳ V1 tị ẳ p1 The rotation at the left-hand end determines the second coefficient, u1 : ›V ð0; tị ẳ u1 ẳ p2 x The remaining two coefcients are combinations of the variables and they are given by the following two constraints: VL; tị ẳ V2 tị and V L; tị ẳ u2 x Applying these two conditions, one obtains P3 ẳ 1=L2 23v1 tị 2u1 tịL ỵ 3v2 tị u2 tịLị P4 ẳ 1=L3 2v1 tị ỵ u1 tịL 2v2 tị ỵ u2 tịLị These coefficients can be represented by the following matrix equation: p1 6p 6 27 6 7¼6 6 p3 23 p4 0 22 22 32 v1 76 07 76 Lu2 7 76 76 17 v2 Lu The shape functions for the beam elements are determined by equating a single displacement to one, and all other displacement to zero The first shape function is derived by letting V1 ¼ and the remaining three variables zero; i.e., V2 ¼ u1 ¼ u2 ¼ 0: This gives p1 ¼ p2 ¼ p3 ¼ 23 p4 ¼ The first shape function becomes w ẳ 36 ỵ 26 A similar calculation for u1 ¼ and V2 ¼ V1 ¼ u2 ¼ yields p1 ¼ and the shape function becomes p2 ¼ l p3 ¼ 22l p4 ẳ l w2 ẳ l6 2l62 ỵ l63 The remaining two shape functions are determined similarly The four shape functions for the beam are w ¼ 36 ỵ 26 w2 ẳ l6 2l62 þ l63 w ¼ 36 2 26 w4 ẳ 2l62 ỵ l63 â 2005 by Taylor & Francis Group, LLC 6-24 6.7.2.1 Vibration and Shock Handbook Mass Matrix Just as in the case of the bar, the generalized mass mij is given by ðl Mij ¼ wi wj mdx Substitution of the above shape functions and integration yields the following mass matrix for the uniform beam element in terms of the end displacements: 156 22L 54 213L 22L 2L2 13L 23L2 7 mL 6 7 420 54 13L 156 222L 213L 23L2 6.7.3 4L2 222L Summary of Finite Element Method Linear finite element for a bar w1 ¼ x=L and w2 ¼ x=L Mass matrix for the bar " # ML Stiffness matrix for the bar " F1 F2 Cubic finite elements for the beam # EA ¼ L " 21 21 #" u1 # u2 w1 ¼ 362 ỵ 263 w2 ẳ l6 2l62 ỵ l63 w3 ẳ 362 263 w4 ẳ 2l62 ỵ l6 Appendix 6A Introduction to MATLABw MATLAB is a software package for numerical computation, visualization, and symbolic manipulation (also see Appendix 32A) It is an interactive environment with hundreds of built-in functions, which are in essence subroutines These functions range from plotting commands, to those for finding the eigenvalues and eigenvectors of a matrix, to those for solving an ordinary differential equation, and much more In addition to the built-in functions, MATLAB contains a programming language, which allows the user to write their own functions The name MATLAB is an abbreviation of matrix laboratory The original versions of MATLAB concentrated on numerical analysis of linear systems of equation MATLAB is available from the Mathworks (www.mathworks.com) The best way to learn MATLAB is by playing around with the different functions (Pratap, 2001) The first thing to note when you launch MATLAB is that it is a window-based environment There are three windows: the command window, the graphics window, and the edit window The command © 2005 by Taylor & Francis Group, LLC Numerical Techniques 6-25 window is the main window and it is the one in which you run all functions (built-in or user created) This is the window which appears when the program is launched, and it has the symbol q as a prompt The graphics window is where all of the graphics are displayed, and the edit window is where users create and save all of their own programs, known as m files MATLAB provides routines for all of the basic areas of numerical analysis (numerical linear algebra, data analysis, Fourier transform, and interpolation), curve fitting, root-finding, numerical solution of ordinary differential equations, integration, and graphics There are specialized tool boxes for signal processes and control systems to name two With these hundreds of functions, it is imperative to have a good help facility In the command window, the command help functionname provides online help For example, type help help to get more information about help There are three other commands for information: lookfor, helpwin, and helpdesk Lookfor gives a list of functions with the keyword in their description Helpwin gives a help window Helpdesk is a web browser-based help Given the early history of MATLAB as a matrix laboratory, it should not be surprising that one of its strengths is its ability to manipulate vectors and matrices very well A row vector is created by typing in the command window the following: qx ẳ ẵ2; 3; 6; This command will produce x¼ A common vector is created by entering the following qx ẳ ẵ2; 1; 3; This produces x ¼ The elements of a vector or matrix are separated by commas or by spaces, and the rows are separated by semicolons Printing is suppressed by ending the line with a semicolon The following command will produce a £ matrix but it will not print it: " # A¼ Providing that the operations make sense, it is easy to operations on vectors and matrices in MATLAB For example, if A and B are two matrices of the same size, the command A ỵ B adds the matrices There are several MATLAB commands given throughout the text of this chapter Several books are available to get you started with MATLAB; for example, Hanselman and Littlefield (2001), Palm (2001), Pratap (2001), and Recktenwald (2000) The best way to learn more about these commands is to type help and the command name This way, you get the most up-to-date information concerning the function References Atkinson, K 1978 An Introduction to Numerical Analysis, 2nd ed., Wiley, New York Cheney, E and Kincaid, D 1999 Numerical Mathematics and Computing, 4th ed., Brooks Cole, Monterey, CA © 2005 by Taylor & Francis Group, LLC 6-26 Vibration and Shock Handbook Hanselman, D and Littlefield, B 2001 Mastering MATLAB 6: A Comprehensive Tutorial and Reference, Prentice Hall, Upper Saddle River, NJ Isaacson, E and Keller, H 1966 Analysis of Numerical Methods, Wiley, New York Palm, W 2001 Introduction to MATLAB for Engineers, McGraw Hill, Boston Pratap, R 2001 Getting Started with MATLAB 6: A Quick Introduction for Scientists and Engineers, Oxford University Press, New York Recktenwald, G 2000 Numerical Methods with MATLAB: Implementation and Application, Prentice Hall, Upper Saddle River, NJ Strang, G 1986 Introduction to Applied Mathematics, Wellesley-Cambridge Press, Wellesley, MA Thomson, W and Dahleh, M 1998 Theory of Vibration with Applications, 5th ed., Prentice Hall, Upper Saddle River, NJ © 2005 by Taylor & Francis Group, LLC ... Vibration and Shock Handbook The centered difference approximation for the second derivative is found by adding the forward and backward difference expansions and ignoring terms of order h4 and. .. Cheney, E and Kincaid, D 1999 Numerical Mathematics and Computing, 4th ed., Brooks Cole, Monterey, CA © 2005 by Taylor & Francis Group, LLC 6-26 Vibration and Shock Handbook Hanselman, D and Littlefield,... LLC 6-16 Vibration and Shock Handbook −2 −4 −6 −8 −10 −12 −1 −0.5 0.5 1.5 2.5 3.5 FIGURE 6.7 Graph of the cubic polynomial l3 4:5l2 ỵ 5l ¼ 0: (Source: Thomson and Dahleh 1998 Theory of Vibration