Longitudinal Vibration Bars

4.Natural Frequencies and Modes of Continuous Systems Using the “Exact” Method 4.1 Equations of Motion Using Newton's Second Law We have seen that for discrete systems, the equations of motion may be obtained by applying Newton's second law to the masses along the directions corresponding to their degrees of freedom. The number of equations is equal to the number of degrees of freedom. The number of natural frequencies (and modes) is also the same. On the other hand continuous systems have distributed elastic properties as well as distributed mass. They have infinite degrees of freedom. For example, a simply supported beam can vibrate freely in any one of the infinite number of modes having any number of half-sine waves. So one may wonder whether there will be an infinite number of equations of motion. Fortunately this is not the case, and for most common engineering problems such as vibration of bars, shafts and thin beams, there is only one equation of motion having an infinite number of possible solutions. There are cases with more than one equation, for example two equations for arches and three equations for shallow shells, but in any case the number of equations of motion is finite. This leads to the question: "How does one obtain an equation of motion for a continuous system that has infinite number of possible solutions?". What is required is a general equation, which represents Newton's second law at any point in the continuous system. This is obtained by applying Newton's second law to an infinitesimal element at an arbitrary location in that system. This results in a partial differential equation. This is the equation of motion for the continuous system. Once an equation of motion for a given system is obtained then applying the boundary conditions to its general solution gives a set of equations. The solution of these simultaneous boundary condition equations results in a determinantal equation involving transcendental functions. This equation is the frequency equation. In some simple cases closed form solutions to the frequency equation is possible and one can get explicit expressions for the natural frequencies. For many practical problems however, the roots of the frequency equation have to be found by a trial and error computational procedure, and in this way the frequencies can be delimited to any desired accuracy. This method is referred to as the exact method because it does give exact results to several sets of standard problems but it must be noted that the derivation of the frequency equation can be very difficult or impossible for many engineering systems. Other approximate numerical techniques are more suitable for such problems but the approximate methods come with a price and that is the uncertainty in the error. For this reason, engineers, researchers and software developers who want to use or develop an approximate method should check the For educational use at Canterbury and TUHH © S. Ilanko 2005 1 accuracy of their numerical procedure by testing it on a related problem for which an exact solution is obtainable. For example one may need to calculate the natural frequencies of a mechanical arm with a complex geometry and support conditions (See Figure 4.1.1). Before using the results of an alternative numerical procedure or software package, one should check if the results obtained using the alternative method agree with results from exact method for a simpler case (Figure 4.1.2). While the actual problem involving a tapered beam with varying axial force is very difficult to solve exactly, any procedure to analyze this system could be tested on a uniform beam carrying an end mass subject to a constant axial end force which could be solved analytically. P Figure 4.1.1 Figure 4.1.2 Numerical procedures can only tell us information about specific systems and to study the effect of varying system parameters one needs to generate results for a number of sets of input data. The exact method gives us an opportunity to get a deep understanding of the general behavior of various systems. For these reasons, we will devote a considerable amount of space to discuss the exact method and the results of its application to some common continuous systems. We will start by describing the general procedure for deriving an equation of motion for a continuous system. Basic Steps for Deriving Equations of Motion Step 1: Step 2: An infinitesimal element of the vibrating body is isolated and a free body diagram is sketched showing all the forces acting on that element. The induced elastic actions (forces or moments) are obtained in terms of the coordinate corresponding to the vibratory motion using the following relationships: (a) force (or moment) - stress relationship; For educational use at Canterbury and TUHH © S. Ilanko 2005 2 Step 3: Step 4: Step 5: (b) constitutive (stress - strain) relationship; (c) strain - displacement relationship. The induced actions in the direction of motion are summed to obtain the net elastic action. If there is any imposed dynamic loading (force vibration), its component in the direction of motion is added to the net elastic action to obtain the net action (force or moment). The net action (force or moment) is related to the corresponding acceleration (linear or angular) using Newton's second law. The resulting equation is the equation of motion. This would be in the form of a partial differential equation. Free Vibration Analysis For free vibration analysis, step 4 is not necessary as there are no applied dynamic loads. Substituting the prescribed boundary conditions into the general solution of the equation of motion yields a frequency equation, often in a determinantal form. The roots of the frequency equation give the natural frequencies. The corresponding natural modes can also be determined using the general solution and the boundary conditions at each of the natural frequencies. This is best illustrated using the examples in the next section. We will first study the free vibrational behaviour of some simple skeletal elements such as bars, shafts and beams before proceeding to two-dimensional systems. In this chapter we will be concerned only with the calculation of natural frequencies and modes. The determination of the actual displacement of a system for any given set of initial conditions will be discussed in chapter 6. For educational use at Canterbury and TUHH © S. Ilanko 2005 3 4.2 Longitudinal Vibration of Bars Derivation of Equation of Motion: We will start with the free longitudinal vibration of bars which is one of the simplest cases of continuous systems. Figure 4.2.1 shows a finite segment of a bar which has a cross sectional area A (which may vary along the length) and is made of an elastic material having a density ρ and Young's modulus E. The positions of an infinitesimal element of the bar having length δx at rest, and at time t during vibration are also shown in the figure. Let u(x,t) be the longitudinal displacement of the bar at time t, and F(x,t) be the dynamic axial force induced at the same instance. F is taken as positive if it is tensile. u O δu+δx δx x Displaced Element Element in Equilibrium Figure 4.2.1 Let us now proceed to derive the equation of motion. Step 1: Sketch the free body diagram of an infinitesimal element: ρAδx F+δF F u Figure 4.2.2 Step 2: (a) Force - Stress relationship is: F = Aσ (b) Stress - Strain relationship is: σ = Eε (c) Strain - Displacement relationship: ε = For educational use at Canterbury and TUHH © S. Ilanko 2005 4 (4.2.1) (4.2.2) ∂u ∂x (4.2.3) 6) reduces to E ∂ 2u ∂ 2u − =0 ρ ∂x 2 ∂t 2 This may be solved using the method of separation of variables.5) into the above gives δF = (ρAδx ) ∂ ∂u ∂ 2u EA = ρA 2 ∂x ∂x ∂t (4.b) into equation (4.6) This is the partial differential equation governing the longitudinal motion of an elastic bar.∂u ∂x (4. EA is constant and equation (4.5) Using the last three equations. Thus g(t) = sin (ωt + φ ) (4. Let u ( x.7) gives the following ordinary differential equation: E d2 f +ω2 f = 0 ρ dx 2 (4. For a homogeneous.2. F = EA At any given time t. Ilanko 2005 5 . as usual. the general solution to equation (4.F = δF Since there are no external forces. t ) = f ( x) g (t ) (4. one can apply the simple harmonic time function.9) General Solution for Uniform Bars of Finite Length For a uniform bar of length L.8b) Substituting equations (4. δF = Step 3: The net elastic force is: Step 4: F + δF .2.2.2. this is the net force on the element and is given by equation (4.8a) where f and g are functions of x and t only respectively.2.2.7) (4.9) may be written (see Appendix A for the mathematical derivation) For educational use at Canterbury and TUHH © S.2.2.2.2.4) ∂ ∂u EA δx ∂x ∂x (4.2.8a. To determine the natural frequencies and modes of linear systems.5) Step 5: Using Newton's second law: ∂ 2u ∂t 2 Substituting equation (4.2. uniform bar.2. 10) into equation (4.11) E For convenience we will later switch to a non-dimensional axial coordinate ξ which is defined by x/L but until we complete the present case we will retain the axial coordinate x. Therefore a complete solution requires two boundary conditions and two initial conditions. u(0. Substituting the general solution into the boundary conditions results in frequency equations the roots of which are the natural frequencies.2.2. Ilanko 2005 6 .2. Case 1: Both Ends Longitudinally Restrained x=L x=0 Figure 4.7) is a second order partial differential equation.2. The order of differentiation is two with respect to both x and t. Equation (4. The following examples illustrate this procedure. However. for the calculation of natural frequencies and modes only the boundary conditions are required. u=0.e.11) may be obtained by substituting equation (4.2. In this book the general solution for various systems will be formulated to make the frequency parameters non-dimensional.3 Since both ends are restrained the displacement u is zero at the boundaries. The general solution may be written in other forms also.2. At x=0. The natural frequencies do not depend E on the form of general solution as explained later. Equation (4. For educational use at Canterbury and TUHH © S.f ( x ) = H 1 sin λ x x + H 2 cos λ L L (4. For example another valid form is f ( x ) = [H 1 sin (λx ) + H 2 cos(λx )] where λ = ω ρ .2.i.t) = 0.9).10) where H1 and H2 are undetermined coefficients and λ is a dimensionless frequency parameter given by λ = ωL ρ (4. The natural modes are obtained by substituting the calculated natural frequencies and boundary conditions into the general solution (or its derivative). Substituting this into equation (ii) gives: H1 sin λ = 0 Therefore either H1 = 0. For educational use at Canterbury and TUHH © S..∞). This gives u = H 1 sin The nth mode is sin nπx sin (ωt + φ ) L nπx L The first two modes are shown in Figure 4.2.2. The lowest (fundamental) natural frequency is given by ω1 = The nth natural frequency is ωn = π E L ρ nπ L E ρ To obtain the modes.0) = 0 (i) Similarly. f(0) = 0 Substituting equation (4.t) would be zero. Hence sin λ = 0 is the frequency equation. However if sin λ = 0.10) into this gives: H1 (0. H1= 0 is a trivial solution as H2 is also zero. ω= nπ L E ρ As there are infinite possible values of n (1. It can be seen that λ = 0 also results in a trivial solution as u(x.2. H1 is not necessarily zero.4.Since this is true for any t. f(L) = 0 gives: H1 sin λ + H2 cos λ =0 (ii) From equation (i). Using equation (4. or sin λ = 0. It should be noted that this figure is only a representation of the displacement that takes place longitudinally. there are infinite number of natural frequencies.2. where n is any integer.11).3.. Ilanko 2005 7 . This means vibration is possible if and only if sin λ = 0. the roots of which are λ = 0 or λ = nπ. H2 =0. H2 = 0 and λ = nπ may be substituted into the general solution for u.0) + H2 (1.. which means there is no motion. The above solution may be obtained in a general manner using a matrix approach. Generally. (iii) where [ B ] = 0 1 sin λ cos λ and {H } = H1 H2 This is an eigen value problem.0) cos λ . the frequency expression would still be ω n = nπ L E ρ . but in this case.0) sin λ = 0.2. could be difficult.n=2 n=1 x=L x=0 Figure 4. by calculating the determinant for various trial values of frequency and searching for a change in the sign of the determinant. giving λL = nπ. It gives (0. ie. Ilanko 2005 8 . Case 2: Both Ends Longitudinally Free Let us now consider free boundaries. (iv) This is the frequency equation in determinantal form.4 It is worth noting here that if f ( x ) = [H 1 sin (λx ) + H 2 cos(λx )] had been chosen as the general solution. The roots may have to be found by a trial and error procedure. This approach is suitable for programming and will be used in some of the examples for which interactive multi-media programs are available on the CD. sin λ = 0. the frequency equation would be sin (λL) = 0. since λ = ω ρ E .(1. the frequency equations are transcendental functions and sometimes they are only expressed in a determinantal form. and for non-trivial solution |B|= 0. For educational use at Canterbury and TUHH © S. as expanding large determinants analytically. Equations (i) and (ii) may be written [B]{H} ={0}. and as before λ = nπ. 0 ) − H 1 cos λ − λ L λ L (4.displacement relationship f′= λ L (H 1 cos(λξ ) + H 2 sin (λξ )) .10) Using the strain .2. and the natural frequencies are given by ω n = For educational use at Canterbury and TUHH © S.x=L x=0 Figure 4. f ′ = 0 ∂x Differentiating the expression for f in equation (4. [B] = λ/L 0 (λ / L) cos λ − (λ / L) sin λ and {H } = H1 H2 For non-trivial solution of equations (i) and (ii) |B|= 0 gives − λ L 2 sin λ = 0 This is possible if either λ = 0.12) H 2 (0. Therefore. The relative positions of all points in the bar will remain unchanged. or sin λ = 0. the bar remains free of any dynamic axial strain. unlike for case 1 the solution is not trivial. This corresponds to a rigid body motion of the bar with zero frequency. But ω is zero since λ is zero. From equation (4. If λ = 0. ∂u = 0 . where ξ = x/L λ f ′(0) = 0 gives: L λ f ′( L) = 0 gives: L H 1 (1. If λ ≠ 0. Since this is true for any time. As u′ is zero. u = H2 sin(ωt+φ). the axial strain is zero. Let us investigate the two possibilities. Such systems that have at least one zero natural frequency associated with one or more modes of rigid body motion are classified as semi-definite systems.8).0 ) = 0 (i) H 2 sin λ = 0 (ii) Once again.2.2. we may write this in the form [B]{H} ={0}. Ilanko 2005 9 nπ L E ρ as for case 1. . then λ = nπ. The displacement is independent of time. u = H2 sin φ.2.5 At a free boundary. the condition is the same as that for case 1 resulting in the following equation: H1 (0. This means u(x. For educational use at Canterbury and TUHH © S.7 At x=0.u = H 2 cos(nπξ )sin (ωt + φ ) From equation (i).2.0) + H2 (1.t) would be zero. From here onwards these steps will not be included except for some complicated systems). x=L x=0 Figure 4. for λ ≠ 0. (Note we have missed some obvious steps including the definition of the elements of [B] and the statement |B| = 0. the condition is the same as that for case 2.2. ie. Either λ = 0 or cos λ = 0. λ L [H 1 cos λ − H 2 sin λ ] = 0 (i) (ii) For non-trivial solution of equations (i) and (ii). which is a trivial solution.2. From equation (i). If λ = 0 then sin (λξ ) would also be zero. Other End Free Now let us consider a combination of fixed and free boundaries. H2 = 0. There are two possible solutions.6.0) = 0 At x=L. Ilanko 2005 10 . λ cos λ = 0.6 Case 3: One End Restrained. n = 0 (Rigid Body Motion) n=1 n=2 x=L x=0 Figure 4. H1 = 0 giving cos(nπξ ) The nth mode is: The natural modes for this system are shown in Figure 4. 2. m0 x=0 x=L Figure 4. n=1 n=2 x=L x=0 Figure 4.. f(x) = 0. ω= Using equation (4. H1 (0.2. such a rigid body displacement is possible. Ilanko 2005 11 (i) .Therefore for non-trivial solution.0) = 0 At x = L. the boundary condition we need is not so simple.2. In the case of a free-free bar.. as for Case 1.11). cos λ = 0 The roots are: λ = (n-½)π.0) + H2 (1. Case 4: One End Restrained. (2n − 1)π 2L for n = 1. ie.2. Other End Carrying A Concentrated Mass We have seen cases where either the displacement or the force is zero at a boundary.2.9 At x = 0. E ρ Substituting H2 = 0 and the roots for λ into the general solution.9. The nth mode is: sin ((n − 1 2 )πξ ) Figure 4.8 It may be noted that whether a system is semi-definite or not could be easily checked by considering the question whether the system can be displaced freely without inducing any internal strains. For educational use at Canterbury and TUHH © S.8 illustrates the modes for this case. but for Cases 1 and 3 it is not so.2.. There are many situations where neither of these is zero. One such example is a bar carrying a concentrated mass (let us say a particle possessing mass or a completely rigid body) at one end as shown in Figure 4. the modes may be determined.11) this may be reduced to λ tan λ .2. the motion is simple harmonic giving ü = -ω2 u. this relationship is obtained by applying Newton's second law to the mass.t) = 0 Since this is true for any t. t ) Figure 4.10) and (4.t) i. one has to look for a relationship between the two.2. F(L.2. m0 Once the frequencies are found.2. In the case of a connected mass.m0 ü(L.When dealing with a boundary where neither the displacement nor the force is zero. In the present case since H2 is zero.t) . In order to do this for a finite mass m0.t) m0 u ( L. it is best to sketch a free-body diagram (see Figure 4.F(L. This gives the frequency equation: EAλ cos λ − m0ω 2 sin λ = 0 L Using equation (4. where η is the ratio of the total ALρ mass of the bar to the tip mass m0 given by η = . Ilanko 2005 12 . Mode calculation means finding the relationship between the constants H1.t) = m0 ü(L.10).e. the mode is given by the term associated with H1. H2 by using all but one of the boundary conditions. Therefore -EA u′(L. For educational use at Canterbury and TUHH © S.t) + m0 ω2 u(L.10 .η = 0. -EA u′(L.12) into the above gives: EA λ L [H 1 cos λ − H 2 sin λ ] − m0ω 2 [H 1 sin λ + H 2 cos λ ] = 0 Rearranging we get H1 EAλ cos λ EAλ sin λ − m0ω 2 sin λ − H 2 + m0ω 2 cos λ = 0 L L (ii) For non-trivial solution of equations (i) and (ii) |B|= 0.2. That is cos (λx/L). -EA f ′(L) + m0 ω2 f(L) = 0 Substituting equations (4.t) = 0 But in determining natural modes. which is partially axially restrained by linear elastic springs of stiffnesses k1 and k 2.Special cases: If m0 → 0. If m0 → ∞. By changing the magnitude of the end mass to a very large value one can observe that the first few natural modes except for the lowest one. This frequency may be estimated by the formula ω = EA k where k = . For educational use at Canterbury and TUHH © S.2. η → 0. This agrees with the roots of the frequency equation for case 1. have very little translation at the end carrying the mass. The roots are given by λ = (n-½)π which agrees with the result for case 3 as to be expected. This means either λ or tan λ must be zero. it should be noted that there would be another mode corresponding to a low frequency vibration of the end mass in which the bar provides the elastic restraint only. in the case of an attached mass setting its magnitude to zero and a very large value) and see if the solution for the special cases agree with what is readily available. but it may be estimated using the simple formula given above. The lowest frequency will monotonically decrease with mass. This demonstrates the fact that an infinite mass corresponds to a fully restrained end. at x = 0 and at x = L respectively. Consider the vibration of the bar shown in Figure 4.exe. When dealing with non-standard boundary conditions. However.11. it is good to check the results by putting limiting values (for example. Therefore from the frequency equation tan λ → ∞. m0 L The solution to Case 4 is available as an interactive tutorial modes951. and approach zero as the mass approaches infinity. Ilanko 2005 13 . Case 5: Both Ends Partially Restrained Another end condition where one needs to obtain a relationship between the force and displacement is a bar that is connected to a support that is not fully rigid (immovable). η → ∞. It is easy to miss this frequency if one looks for the roots of the transcendental frequency equation. The flexibility of a support may be modelled by an elastic spring of finite stiffness. its frequencies and modes may be calculated by considering only one half of the structure and by using conditions of symmetry or anti-symmetry at the centre as explained below.8) and (4. This means the system can vibrate either in a symmetrical mode or anti-symmetrical mode.11 At x = 0. k1 u(0.0) + H 2 (1.0) − H 2 (0.t).k2 f(L) = EA f′(L) (The sign of spring force is negative because a positive f(L) causes compression in the spring) Substituting equations (4.2.k2 u(L.t) This is true for any t. λ L λ L [H 1 (1.0 )] = EA − EA ie. .t) = EA u′(L.10) into the above gives: k1 [H 1 (0.2. Substituting equations (4.t) = EA u′(0.2. Finding modes: For educational use at Canterbury and TUHH © S.k2 k1 x=L x=0 Figure 4. the force in the spring = the force induced in the bar (Newton's third law). Therefore . The above frequency equation reduces to: 2kLEAλ (iii) tan λ + 2 2 2 k L − (EAλ ) [ ] If a system is structurally symmetrical.0)] H 1 + k1 H 2 = 0 (i) Similarly at x = L. Ilanko 2005 14 .10) and (4.12) into the above equation yields: − k 2 sin λ + EAλ cos λ EAλ sin λ H 1 − k 2 cos λ − H2 = 0 L L (ii) For non-trivial solution of equations (i) and (ii) |B|= 0.2. ie.2. This results in the following frequency equation: (k1 + k2 )EAλL = 0 tan λ + 2 k1k2 L2 − (EAλ ) [ ] Special Case: If k1 = k2 then the system is symmetrical. t) = -u(L. 2 L L + ∆. Ilanko 2005 15 . If the term that is assumed to be unity were zero there would be arithmetic overflow as a division by zero is involved.t) = u(L-x. if k1 is zero the above expression for the mode would become infinity. the mode on one side of the centre line (x = L/2) is a mirror image of that on the other side.Using the first boundary condition. In such a case H1 should be set to zero and the mode would then be cos (λx/L) Case 6: Symmetrical and Anti-Symmetrical Vibration of the System in Case 5 In a symmetrical vibration. t 2 2 δu = 0 For educational use at Canterbury and TUHH © S.t). This means u(0. t = −u . This means at x = u L .t) As ∆ → 0. u(x.t) and for 0<x<L.t = 0 2 (a) Equation(a) forms a boundary condition for symmetrical vibration. At x = L L L . t = u − ∆.t) = -u(L-x. u . Therefore if any denominator term has a very small magnitude then a different term should be selected as unity. For example. the ratios of the constants is calculated by taking one of them as unity. in the current problem. In an anti-symmetrical vibration. u(x. H 2 = f(x) = H 1 sin Mode is sin λx L λx L λEA + + λEA Lk1 cos Lk1 H1 λx cos Lk1 λEA L λx L When calculating modes. the displacement of both sides of the centreline will be in phase. t 2 2 2 This is possible only if u L . EA λ L sin λ 2 − k cos For educational use at Canterbury and TUHH © S. To illustrate this procedure. As the system is symmetrical let us consider the left half of it.2.10) into equation (b) gives: H 1 cos λ 2 − H 2 sin λ =0 2 For non-trivial solution of equations (i) and (iv). This means the strain. stress and the force at the centre are zero. the spring force is equal to the force induced in the bar. For anti-symmetrical vibration. − EA ie. λ cos L λ tan 2 λ λ − k sin 2 + EA λ (kL ) 2 =0 =0 (iii) This is the frequency equation for the symmetrical vibration. It is safer to consider the anti-symmetrical boundary condition for longitudinal system in terms of force being zero as this would hold even in cases where there is a concentrated mass at the centre (see Case 9). Ilanko 2005 16 λ 2 =0 (iv) .u′ ie. To obtain the natural frequencies and modes corresponding to symmetrical vibration. This yields: − EA λ L H 1 + kH 2 = 0 (i) For symmetrical vibration. Similarly for analysing the vibration corresponding to anti-symmetrical modes. L . At x = 0. substituting equation(4.2.8) into equation (a) gives: λ H 1 sin 2 + H 2 cos λ =0 2 (ii) For non-trivial solution of equations (i) and (ii). equation (b) is used in conjunction with any one of the end conditions.t = 0 2 (b) Equation (b) is a boundary condition applicable for the anti-symmetrical vibration of a symmetrical system. substituting equation (4. equation (a) is used in addition to the boundary condition at any one of the ends. let us consider the special case of k1 = k2 = k in case 5. 2. Ilanko 2005 17 . even if that centre is in the middle of a mass or spring. This is schematically illustrated in Figure 4. This illustrates the fact that the vibration modes of a symmetrical system consist of symmetrical modes and antisymmetrical modes.ie. For educational use at Canterbury and TUHH © S.2. Note: It can be shown that the left hand side of equations (iii) and (v) are the factors of the left hand side of equation (iii) for case 5. then the condition of antisymmetry is that the force at the centre of the system is zero.12. tan λ 2 − kL =0 EAλ (v) This is the frequency equation for the anti-symmetrical vibration. Anti-Symmetrical Mode Symmetrical Mode x=0 x=L Axis of Symmetry Figure 4.12 If a symmetrical system has a mass or spring element at the centre. 15) (4.x2 are the longitudinal coordinates measured from the left hand end of the segments.13) we get H2 = 0 For educational use at Canterbury and TUHH © S. A1 x1 = 0 A2 x1 = L1 x2 = 0 x2 = L2 Figure 4.Case 7: Stepped Bar The system shown in Figure 4.t) to represent the longitudinal displacement in the two segments.12 has a geometrical discontinuity.16) E It should be noted that the time function sin(ωt+φ) is the same for both segments. as the system contains a discontinuity.2. The second segment has length L2 and cross sectional area A2.2.2. 2 (4.t) and u2(x2.2. as the entire system vibrates together at the same frequency. u1 = 0.2. The first segment of the bar is of length L1 and cross sectional area A1. Boundary Conditions: At x1 = 0.13 The dynamic displacements and their derivatives may be expressed in the following form: u1 = H 1 sin λ1 x1 L1 where λ1 = ωL1 u 2 = H 3 sin ρ 1 λ1 x1 L1 sin (ωt + φ ) . x1. Using equation (4.13) (4. The bar is made of a material having density ρ and Young's modulus E. 2 L2 ρ + H 4 cos 1 (4.2. Ilanko 2005 18 (i) .14) E λ2 x2 where λ2 = ωL2 + H 2 cos λ 2 x2 L2 sin (ωt + φ ) . it is convenient to use two functions u1(x1.2. t).t) Differentiating equations (4.At x2 = L2.13) and (4.t) = u2(0.t) ie. cross sectional area A.2. u1(L1. and a concentrated mass m0 attached to the bar at distance a from the left hand end as shown in the Figure. Substituting equations (4. (iii) Since the force is continuous (or by Newton's third law). Continuity Conditions: Since the displacement is continuous.15) into the above gives: H1 sin λ1 + H2 cos λ1 .H4 = 0. These are obtained using continuity conditions at x1 = L1 (or x2 = 0).t) = EA2u2′(0.15) and substituting into this ∂x2 λ2 condition gives: L2 [H 3 cos λ 2 − H 4 sin λ 2 ] = 0 . It consists of a uniform bar of length L.2. density ρ and Young's modulus E. A1λ1 L2 Case 8: Restrained Bar Carrying A Concentrated Mass at A Mid-Point Consider the longitudinal vibration of the system shown in Figure 4.2. (ii) Two more equations are required as we have four undetermined coefficients H1-4. For educational use at Canterbury and TUHH © S.2.2.15) and substituting into the above gives: H1 EA1λ1 cos λ1 EA1λ1 sin λ1 λ − H2 − H 3 EA2 2 = 0 L1 L1 L2 Equations (i) to (iv) may be expressed in matrix form [B]{H}={0} For non-trivial solution |B|= 0. Ilanko 2005 19 .2. F1(L1. strain = ∂u2 = 0 . The frequency equation of this system is required.14.t) = F2(0. Both ends of the bar are fully axially restrained. This yields the following frequency equation: (iv) (v) A2 λ2 L1 sin λ1 sin λ2 − cos λ1 cos λ2 = 0 . Differentiating equation (4. EA1u1′(L1.13) and (4. 19) (4.t) = u2(0.18) E For 0<x2<L-a. Consider the freebody shown in Figure 4.m0 x1 = 0 x1 = a x2 = 0 x2 = L-a Figure 4.20) (i) (ii) (iii) . For educational use at Canterbury and TUHH © S. u1 = 0 gives At x2 = L-a. t ) = H 3 sin λ 2 x2 x2 + H 4 cos λ 2 (L − a ) (L − a ) where λ2 = ω ( L − a) Boundary Conditions: At x1 = 0.2.t). As in case 7. Ilanko 2005 20 (4.H4 = 0 Continuity of force at x1 = a is ensured by applying Newton's second law to the mass.2. This yields H1 sin λ1 + H2 cos λ1 .2.2.2. u2 = 0 gives sin (ωt + φ ) ρ E H2 = 0 H3 sin λ2 + H4 cos λ2 = 0 Continuity Conditions: Continuity of displacement at x1 =a is u1(a.17) (4. It is not possible to apply a single governing differential equation for the whole bar. let us use two functions to represent the dynamic axial displacements in the segments on either side of the mass.15. u 2 (x 2 . u1 ( x1 .2. t ) = H 1 sin λ1 x x1 + H 2 cos λ1 1 a a λ1 = ωa where sin (ωt + φ ) ρ (4. For 0<x1<a.14 The presence of m0 introduces a discontinuity. 2.F1(a. t ) Figure 4.t) F2(0. The general approach used in the treatment of non-standard boundary conditions in Cases 4.t) u1 ( a.2. That is. the origin of the coordinate for the second bar (x2) could have been set at the right end. This would have made the solution slightly simpler because H4 would have been zero.2.17) and (4. |B|= 0 This gives the following frequency equation: λ1 a cos λ1 + sin λ1 λ2 L−a cot λ2 − m0ω 2 =0 EA In this case.m0 F1(a.t) = EA[u2′(0. Ilanko 2005 21 .t) Differentiating equations (4.19) into the above yields: H1 λ1 a cos λ1 − H 2 λ1 a sin λ1 − H 3 λ2 L−a − H4 m 0ω 2 =0 EA (iv) For non-trivial solution of equations(i)-(iv).t)] = -m0ω2u2(0. finite masses would require the application of Newton’s second law of motion while elastic elements such as springs would require the use of constitutive equations (F = k e). 5 and 8 for the longitudinal vibration holds for other types of vibrations too.t)] The acceleration is ü1(a.t)-u1′(a.ω2u1(a.t) = .t) .t) Therefore EA[u2′(0.15 The net longitudinal force = F2(0.t)-u1′(a. Case 9: Symmetrical and Anti-Symmetrical Vibration of a Bar with a Mass at the Centre For educational use at Canterbury and TUHH © S. This means treating the problem as that of Case 4. where the displacement form u will not be the same as that for the bar. but with the magnitude of mass changed to m0/2 and the length changed to L/2. Since we assume that the mass is rigid. Using the results for Case 1. we get an expression for the nth natural frequency in symmetrical vibration as ω n = 2 nπ L E ρ The anti-symmetrical case is not so straightforward. Since we are not interested in the straining within the mass (which for the purpose of analysis has been taken as rigid) one has to resort to the condition associated with the force. t ) = 0 as for Case 6.2.16 This is a special case of Case 8. This means the system could be treated as a fixedfixed bar with a length of L/2. we would get the wrong answer. and then apply Newton’s second law of motion to half the mass to find a boundary condition for the bar. For educational use at Canterbury and TUHH © S. Ilanko 2005 22 . If one were to apply the condition that u ′( L / 2. At the centre the displacement is zero.m0 x= 0 L/2 x= L/2 L/2 Figure 4. That is we can say the force at the centre of the mass is zero. where the mass is located at the centre. the displacement at the right end of the left bar is also zero. Let us consider the symmetrical case first and take the left half of the structure. because this condition only holds at the centre of the mass. 3.3.3 Torsional Vibration of Shafts Derivation of Equation of Motion: The partial differential equation governing the torsional vibration of a circular shaft is of the same form as the equation of motion for longitudinal vibration of bars. It will be shown that for many simple boundary conditions the natural frequencies and modes of torsional vibration can be deduced from the solution for equivalent longitudinal vibration problems.t) x1 x θ x2 Figure 4. Let θ(x.3.t) be the dynamic torque induced at the same instance. Let us now derive the equation of motion following the basic steps. Ilanko 2005 23 .2 Step 2: From Mechanics of Materials.t) T(x1 .t) be the angle of twist at time t. Figure 4. and T(x. T(x2.1 Step 1: Sketch the free body diagram of an infinitesimal element: δx T+δT T θ Figure 4. Figure 4.3. Shear modulus G.4. the torque-twist relationship is: For educational use at Canterbury and TUHH © S.1 also shows the sign convention for T and θ.1 shows a segment of a shaft having circular cross section with the following properties: Polar second moment of area J. Density ρ. It may be noted that changing E and u in equation (4.2 into the above gives: δT = ρJ (δx) 2 ∂ ∂θ ∂θ GJ = ρJ 2 ∂x ∂x ∂t (4.1) δT = ∂ ∂θ GJ δx ∂x ∂x (4. General Solution for Uniform Shafts of Finite Length The general solution takes the form θ ( x.3.3) This is the partial differential equation governing the torsional vibration of an elastic shaft of circular cross section.6) .∂θ T = ∂x GJ At any time t.3.3.3.3) reduces to G ∂ 2θ ∂ 2 θ − = 0.7) to G and θ results in equation (4.3.2.3.2) Step 3: The net elastic action is T + δT .3. ρ ∂ x 2 ∂t 2 (4.3. Step 4: Since there are no external dynamic torques.2. Ilanko 2005 24 ρ G (4. t ) = [H 1 sin (λξ ) + H 2 cos(λξ )]sin (ωt + φ ) (4.4) Again a wave equation is obtained. uniform shaft GJ is constant and equation (4.5) where H1 and H2 are undetermined coefficients. ξ is x/L. this is the net torque on the element and is given by equation 4.3. Step 5: Using Newton's second law in the rotational sense: ∂ 2θ ∂t 2 Substituting equation 4.4).3. For a homogeneous. and the non-dimensional parameter λ is defined by λ = ωL For educational use at Canterbury and TUHH © S.T = δT. (4. The natural frequencies and modes of torsional vibration may be obtained by using the general solution and the end conditions as illustrated in the case of longitudinal vibration. sin λ = 0. by replacing the elastic modulus E with shear modulus G. Substituting equation(4. Using equation (4.3.3 At x=0.5) into this gives: H1(0.3.6) the nth natural frequency is obtained. Hence θ(0.t) = 0 gives: H1 sin λ + H2 cos λ = 0 For non-trivial solution of equations (i) and (ii). Substituting condition (i) and the expression for the natural frequency into the general solution results in the expression for the nth mode sin For educational use at Canterbury and TUHH © S. ωn = nπ L G ρ This result could have been deduced from the result for the natural frequency of longitudinal vibration of a bar having restrained ends. Ilanko 2005 25 nπ x . This is illustrated in the following example.0) = 0 Similarly θ(L.t) = 0.0) + H2(1. The mathematical similarity between the equations of motion for longitudinal and torsional vibration may be used to deduce the results for one case if the corresponding results for the other case are known.3. L . θ=0 for any t. (i) (ii) This is the frequency equation the roots of which are given by λ = nπ where n is any integer. Case 1: Both Ends Torsionally Restrained x=L x=0 Figure 4. The radii of gears I2 and I3 are r2 and r3 respectively.Similarly the results for other simple boundary conditions may be deduced from the corresponding results for the longitudinal vibration of a bar. For example. and G2. The dynamic rotations of the two shafts may be expressed in the following form: θ 1 = [H 1 sin (λ1ξ1 ) + H 2 cos(λ1ξ1 )]sin (ωt + φ ) . J1. where ξ1 = x1 / L1 and λ1 is defined by For educational use at Canterbury and TUHH © S.3. The same roots of the frequency parameter as that for Case 4 of the longitudinal system could be used to find the natural frequencies of a corresponding shaft this time using equation (4.7) . Case 2: A Geared System x1 = L 1 x1 = 0 I2 Shaft 1 I1 r2 θ 1 ( x1 .3. Therefore the torsional vibration systems for other common boundary conditions will not be discussed here. the mass terms in the frequency equations for longitudinal systems should be replaced by polar moment of inertia terms if corresponding results for torsional systems are required. I3 and I4.4 Consider the vibration of the geared system shown in Figure 4. provided the ratio η is defined as JLρ/I0. ρ1 and L1 for the first shaft. We will however consider the torsional oscillations of geared system since we have not dealt with a corresponding longitudinal vibration problem. The properties of the two shafts are: G1. J2.6).3.4. results from Case 4 of the longitudinal system could be used to find the natural frequencies of a shaft carrying a mass with polar moment of inertia I0. The polar moments of inertia of the rotors (which are assumed rigid) are I1. t ) I3 x2 = 0 x2 = L 2 Figure 4. t ) Shaft 2 r3 I4 θ 2 ( x2 .3. ρ2 and L2 for the second shaft. I2. It should be noted that for systems with rigid end masses. Ilanko 2005 26 (4. 8) θ 2 = [H 3 sin (λ 2ξ 2 ) + H 4 cos(λ 2ξ 2 )]sin (ωt + φ ) .3. T1 (0.3..3. Ilanko 2005 27 ∂θ into the above equation gives: ∂x ..11) ∂θ 2 λ = 2 [H 3 cos(λ 2ξ 2 ) + H 4 sin ((λ 2ξ 2 ))]sin (ωt + φ ) ∂x L2 (4. First consider the vibration of the first rotor (see Figure 4.t) View from positive x direction θ1 (0. t ) = I 1 Substituting the torque-twist relationship T = GJ For educational use at Canterbury and TUHH © S. t ) ∂t 2 =-ω2I1θ1(0.10) Differentiating yields: ∂θ 1 λ = 1 [H 1 cos(λ1ξ 1 ) + H 2 sin (λ1ξ1 )]sin (ωt + φ ) ∂x L1 (4.t) I1 θ1 (0.5 Applying Newton's second law to I1 gives: ∂ 2θ 1 (0. four conditions are required.3.11) Boundary Conditions: As there are four constants H1 .5). t ) T1 (0. (4.λ1 = ωL1 ρ1 G1 and (4. t ) Figure 4.3.3. H4.t) T1 (0.3.9) where ξ 2 = x 2 / L2 and λ2 is defined by λ2 = ωL2 ρ2 G2 and (4. G1J1θ1′(0,t) + ω2I1θ1(0,t) = 0 Substituting the general solution for θ1 and its derivative [equations (4.3.7) and (4.3.11)] into the above equation yields: G1 J 1λ1 H 1 + ω 2 I1H 2 = 0 L1 (i) At x1=L1 (or x2=0) the gear constraint equation gives: θ 2 (0, t ) = − r2 θ1 (L1 , t ) r3 Substituting equations (4.3.7) and (4.3.9) into the above gives: r2 r H 1 sin λ1 + 2 H 2 cos λ1 + H 4 = 0 r3 r3 (ii) The third condition is obtained by applying Newton's second law to the rotors 2 and 3, and eliminating the contact force between these rotors from the two equations. Figure 4.3.6 gives an exploded view of these rotors. For educational use at Canterbury and TUHH © S. Ilanko 2005 28 I2 T1(L1,t) θ 1 ( L1 , t ) θ 1 ( L1 , t ) T1(L1,t) (P)(r2 ) (P)(r2 ) P I3 P θ 2 (0, t ) T2(0,t) T2(0,t) θ 2 (0, t ) View from positive x direction Figure 4.3.6 Applying Newton's second law to I2 gives: − T1 (L1 , t ) + r2 P = I 2θ (L1 , t ) (iiia) Similarly for I3 we get T2(0,t) + P r3 = -ω2I3θ2(0,t) Eliminating P from equations (iiia) and (iiib) gives: T1 (L1 , t ) + (iiib) r2 r T2 (0, t ) − ω 2 I 2θ 1 (L1 , t ) − 2 I 3θ 2 (0, t ) = 0 r3 r3 Substituting the gear constraint condition and the torque-twist relationships gives G1 J 1θ 1′(L1 , t ) + r2 r G 2 J 2θ 2′ (0, t ) − ω 2 I 2 + 2 r3 r3 2 I 3 θ 1 (L1 , t ) = 0 Substituting equations (4.3.7),(4.3.11) and (4.3.12) into the above yields: B31 H1 + B32 H2 + B33 H3 + B34 H4= 0 For educational use at Canterbury and TUHH © S. Ilanko 2005 29 (iii) where B31 = G1 J 1 λ1 L1 B32 = − G1 J 1 B33 = G2 J 2 cos λ1 − ω λ1 L1 sin λ1 + ω λ2 L2 r I2 + 2 r3 2 2 r I2 + 2 r3 2 I 3 sin λ1 (iiic) 2 I3 cos λ1 (iiid) r2 r3 (iiie) B34 = 0 The last equation may be obtained by applying Newton's second law to I4. Consider the free body shown in Figure 4.3.7: I4 θ 2 ( L2 , t ) θ 2 ( L2 , t ) T2(L2,t) (iiif) T2(L2,t) Figure 4.3.7 From Newton's second law, − T2 (L2 , t ) = −ω 2 I 4θ 2 (L2 , t ) Substituting the torque-twist relationship and equations (4.3.9) and(4.3.12) into the above gives: B41 H1 + B42 H2 + B43 H3 + B44 H4 = 0 (iv) where B41 = 0 (iva) B42 = 0 (ivb) B43 = G2 J 2 λ2 L2 B44 = − G2 J 2 cos λ2 − ω 2 I 4 sin λ2 λ2 L2 sin λ2 + ω 2 I 4 cos λ2 For educational use at Canterbury and TUHH © S. Ilanko 2005 30 (ivc) (ivd) Equations (i),(ii),(iii) and (iv) may be written in matrix form [B]{H} = {0}. For the non-trivial solution of {H}, |B| = 0. This is the frequency equation. The coefficients of [B] in the third and fourth rows are explicitly given in equations (iiic) ...(ivd). Other coefficients can easily be obtained from equations (i) and (ii). For educational use at Canterbury and TUHH © S. Ilanko 2005 31 Again the natural frequencies and modes of strings may be deduced from the corresponding solution for the longitudinal vibration of a bar or the torsional vibration of a shaft. and v(x.t) be the lateral dynamic displacement.4. It has no flexural rigidity. For small amplitude vibration.t) x1 T0 x x2 Figure 4.4 Lateral Vibration of Strings Derivation of Equation of Motion: Another interesting category of vibration that is governed by the one dimensional wave equation is the small amplitude lateral vibration of strings. For these reasons some examples of vibration of strings is discussed following the derivation of the equation of motion. T0+ F2 v o T0+ F1 T0 v(x2. However. some physical boundary conditions are not common to all these three systems. Let m be the mass of the string per unit length. the fluctuation in the axial tension F may be neglected as it is small compared to the static tension T0.1 shows a segment of a vibrating string subject to a uniform static tension T0. There are some similarities between the lateral vibration of strings and the more complicated lateral vibration of beams (which is more commonly encountered in engineering). The elastic restoring action is provided primarily by the component of the tension in the direction of motion.4.t) v(x1. Figure 4.1 Step 1: Sketch the free body diagram of an infinitesimal element.4. Ilanko 2005 32 . Figure 4.2: For educational use at Canterbury and TUHH © S.4. A string is a structural element that can only sustain axial tensile straining. 4. T0 ∂v ∂ 2v = m(δx ) 2 ∂t ∂t δ (∂v / ∂x) ∂ 2v =m 2 δx ∂t as δx → 0. For educational use at Canterbury and TUHH © S.e. this step is not necessary.2) For a uniform string (if m is a constant) this may be solved using the same form of general solution as those used in the previous cases. (4.2 Step 2: As the change in tension during vibration is small (it would be of the order of the dynamic strain). this is the net force on the element Step 5: Using Newton's Second law in the v direction: T0δ i. Step 3: The net elastic action in the direction of motion is given be (T0) sin(θ+δθ) .T0 + F+δF θ+δθ v θ T 0+ F o v+δv v x x δx Figure 4. Ilanko 2005 33 .4. Therefore the net elastic action is T0 (δθ) + F (δθ) + (δF)θ which may be written as T0 δ(∂v/∂x). this becomes T0 m ∂ 2v ∂ 2v − =0 ∂x 2 ∂t 2 (4.4. sin θ→θ.1) since θ→∂v/∂x as θ→0 Step 4: Since there are no external dynamic forces.(T0) sin θ As the amplitude of vibration is small. 0 i.0) = 0. Substituting equation (i) into (ii) gives H1 sin λ = 0 For non-trivial solution sin λ = 0. which means λ = nπ For educational use at Canterbury and TUHH © S.4.0 (b) Similarly.3) where H1 and H2 are undetermined coefficients. t ) = [H 1 sin (λξ ) + H 2 cos(λξ )]sin (ωt + φ ) (4. H2 =0. Ilanko 2005 34 (i) (ii) .3 (a) at x=0. Case 1: String Fixed at Both Ends First Mode Second Mode x=0 x=L Figure 4.e. v=0.0) + H2 (1. x is given by ξ = x / L and λ is given by: λ = ωL m T0 (4.General Solution for Uniform Strings of Finite Length The general solution is v(x. Substituting this into the general solution gives: H1 (0. v=0 at x=L gives: H1 sin λ + H2 cos λ = 0.4.4) Substituting the above equations into the boundary conditions yield frequency equations and modes as illustrated in the following examples.4. 3a is also negligible.4 For educational use at Canterbury and TUHH © S. The results are valid even for the lateral vibration of the systems shown in Figure 4.4.4.4.) the nth natural frequency is obtained as ωn = The corresponding mode is: sin nπ nπ L T0 m x L It is worth noting that the above expressions for the natural frequency and mode was obtained using only the transverse boundary conditions. k1 k1 x=0 (a) x=L T0 m 2 T0 k x=0 m1 x=L (b) Figure 4. which have different longitudinal end conditions. The springs and/or masses in these cases can only cause a dynamic fluctuation in the axial tension which is small compared to the static tension T0.4.4. The effect of the end springs in Figure 4. The longitudinal motion of the end masses are secondary for small amplitude lateral vibration. Ilanko 2005 35 .From this and equation (4.4. 4.5 The discontinuity at the centre means two functions v1 and v2 should be used and four boundary conditions would be required.6. A sketch of a symmetrical mode is shown in Figure 4.t) = 0.4. The far ends of the strings are laterally restrained.5. Substituting the general solution in terms of non-dimensional coordinate ξ1 = x1 / L v1 = [H 1 sin (λ1ξ1 ) + H 2 cos(λ1ξ1 )]sin (ωt + φ ) (i) gives: (ii) H2 = 0 Now consider the symmetrical vibration. However. The tension in the strings is T0. the symmetrical and anti-symmetrical vibrations can be analysed separately as shown here. The sphere translates laterally without any rotation. the boundary condition at x1 = 0 (which is common to both symmetrical and anti-symmetrical vibrations) is: (a) v1(0.4. Ilanko 2005 36 . Taking the left side of the centre line. which consists of a uniform solid sphere of radius R and mass m0 connected to two strings symmetrically. m0 2R x=0 x=L Figure 4. For educational use at Canterbury and TUHH © S.Case 2: Two Identical Strings Connected to a Solid Sphere Consider the lateral vibration of the system shown in Figure 4. m0 ω2 v1(L.e. there is no net moment on the sphere from the string reactions.e. Ilanko 2005 37 (iv) .t) T0 θ(L.4.t) T0 Free Body Diagram for m0 m0 2R x=L x=0 Figure 4.t) . ↑ -2T0 v1′(L.t) Hence. The lateral force components from the strings are equal and their sum is given by ↓ 2 T0 sin ∂v1 ∂x1 at x = L As v1 is very small this may be written as 2T0 v1′(L.4. t ) 2T0 v1′(L.t) = m0 v1 ( L1 .4) into the above gives: m m0 λ cot 1 = 0 = λ2 2mL (mass of both strings) For educational use at Canterbury and TUHH © S.6 As can be seen from the free body diagram in the Figure.m0 ω2 sin λ1 = 0 i.θ(L. cot λ1 = m0ω 2 L 2T0 λ0 Substituting equation (4. using Newton's second law: i.t)= 0 (iiia) Substituting the general solution for v1 and equation (ii) into the above yields: 2(λ1/L)T0 cos λ1 . λ1 = nπ. cot λ1 → 0. i. This corresponds to the case of a fixed-fixed string of length L or the anti-symmetrical modes of a fixed-fixed string of length 2L.7 For educational use at Canterbury and TUHH © S. Special Cases: (1) (2) If m0 → 0.7. The fundamental modes corresponding to the two special cases of symmetrical vibration are shown in Figure 4.e. (2n − 1)π i. λ1 = 2 This corresponds to the case of a fixed-free string of length L (a difficult condition to achieve physically) or the symmetrical modes of a fixed-fixed string of length 2L.This is the frequency equation. If m0 → ∞.4. cot λ1 → ∞. m0→0 2R x=0 x=L (a) Negligible Mass m0→∞ x=0 x=L (b) Large Mass Figure 4. Ilanko 2005 38 2R .4.e. 4. cos Θ = cos γ → 1 Substituting these into equation (v) gives ΣM = 2T0R( Θ +γ) ΣM = I0 γ But from Newton's second law: 2T0R(γ+ Θ ) = -I0 γ Therefore Substituting the compatibility condition. the sphere undergoes a pure rotation as shown in Figure 4. In anti-symmetrical vibration.Let us now consider the anti-symmetrical modes.4. t ) R (v) . sin Θ → Θ .t) Free Body Diagram for m0 γ m0 2R x=0 x=L Figure 4. sin γ → γ. Rγ=v1(L.8. t )] = Using equations (i) and (ii) this becomes: For educational use at Canterbury and TUHH © S.t) = Θ The net moment on the sphere is 2[T0 sin Θ (R cos γ) + T0 cos Θ (R sin γ)] As Θ and γ are small. T0 θ(L. t ) + v1 (L.t) T0 θ(L.t) into the above equation gives: 2T0 [Rv (L.8 Let θ1(L. Ilanko 2005 39 ω 2 I 0 v1 (L. 9 For educational use at Canterbury and TUHH © S.4. this would correspond to the case of a fixed-fixed string of length L) (b) If I0 → ∞. then tan λ1 = 0 corresponding to the vibration of a fixedfixed string of length L. Ilanko 2005 40 .2T0 R λ1 L tan λ1 This gives: ω 2I0 cos λ1 + sin λ1 2T0 − λ1 =− R =0 R I ω 2L L− 0 2T0 R (vi) Special Cases: (a) If I0 → 0. (If R=0. The modes for these special cases are shown in Figure 4. then ΣM = 0 giving Θ = -γ tan λ1 R =− λ1 L Note that Θ = -γ means the cable forces on the mass would be acting radially.9 I0→0 γ 2R x=0 x=L (a) Negligible Inertia I0→∞ 2R x=0 x=L (b) Large Inertia Figure 4.4. This is to be expected since otherwise there will be a net moment. that is for a point mass. 1 For educational use at Canterbury and TUHH © S.t) o Sf(x1.5 Lateral Vibration of Beams Derivation of Equation of Motion All the systems considered so far had equations of motion in the form of a wave equation. the slope of the beam could be large and the rotary inertia may not be negligible. The effect of neglecting rotary inertia will be discussed later. 3.t) v(x2. The displacement due to shear strain is small. M(x2.t) M(x1. elastic properties. The derivations presented in this section are based on the following assumptions: Assumptions: 1. 2. Ilanko 2005 41 .4. 4. The transverse cross sections of the beam remain plane during bending. A beam satisfying the above assumptions is called an "Euler-Bernoulli Beam".t) x x1 x2 Figure 4.t) v(x1. and (b).t) v Sf(x2. 6. It should be noted here that even if the beam is slender (cross sectional dimensions are small compared to its span) and the amplitude of displacements remain small. The stresses induced are within the limit of proportionality. The cross section of the beam has at least one axis of symmetry. The cross sectional dimensions of the beam are small compared to the span so that: (a). The lateral vibration of beams is governed by a fourth order partial differential equation. The beam is made of a material having linear. and is often referred to as a thin beam. The rotary inertia of the beam is negligible. for higher modes of vibration. The displacements are small.5. 5. Mass per unit length (density × area) m.1 shows a segment of a laterally vibrating beam.2 It is convenient to replace the transverse shearing force Sf(x.5.t)] are shown acting in their positive directions.t) in the longitudinal direction as shown in Figure 4.t) be the lateral dynamic displacement of the beam. S(x.t) in the lateral direction and F(x. Let v(x. Step 1: Sketch the free body diagram of an infinitesimal element: v(x.t) M+δM M F+δF vF o S x Step 2: Neutral Surface v+δv x δx Figure 4. S+δS v(x. The induced actions [dynamic shearing force Sf(x.t) and bending moment M(x. Ilanko 2005 42 .5.5.3. The properties of the beam are: Young's modulus E.Figure 4. Sagging moments and anticlockwise shear forces are positive.t) M v v x δΜ M+ Neutral Surface Sf+δSf v+δv Sf o x δx x Figure 4. Second moment of area about the neutral axis I.5.t) with its components. Let us now follow the basic steps in deriving the equation of motion.3 From Mechanics if Materials the moment curvature relationship for a thin beam is For educational use at Canterbury and TUHH © S. 3) this becomes: − ∂2 ∂ 2v ∂ 2v EI δ x = (m δ x) ∂t 2 ∂x 2 ∂x 2 This gives: 2 2 2 ∂ ∂ v ∂ v EI +m 2 = 0 2 2 ∂x ∂x ∂t (4.1) Neglecting the rotary inertia of the beam.F (δv) = 0 But F (δv) being a product of two small dynamic terms is negligible As δx is infinitesimal.2) The net elastic action in the positive lateral direction is -S + (S + δS) = δS But from equation (4.2) δS = δ − =- ∂M ∂x ∂2M δx ∂x 2 =− ∂2 ∂ 2v EI δx ∂x 2 ∂x 2 Step 4: Since there are no external dynamic forces this is the net lateral force. δM + S (δx) . Ilanko 2005 43 .5. summing the moments on the infinitesimal element and equating to zero gives: (M + δM) .5.3) ∂ 2v ∂t 2 Using equation (4. this gives S = Step 3: ∂M ∂x (4.e. Step 5: Applying Newton's second law in the lateral direction we get δS = (m δx) (4.F (δv) = 0 i.5.5.5. For a homogeneous.5.M = EI ∂ 2v ∂x 2 (4. uniform beam this reduces to the following form: For educational use at Canterbury and TUHH © S.M + S (δx) .4) This is the partial differential equation governing the lateral vibration of an Euler-Bernoulli beam. 5. Ilanko 2005 44 (4.5) .EI ∂ 4v ∂ 2v + m =0 ∂x 4 ∂t 2 For educational use at Canterbury and TUHH © S. (4.5) may be written in the following form: v(x. M=0. ∂x 2 Substituting equation (4. (4. Case 1: Simply Supported at Both Ends n=2 n=1 x=L x=0 Figure 4.5.5.0) + H4 (0. where.5.5.6) into equation (4.5. in terms of the non-dimensional axial coordinate ξ = x/L. ∂ 2v = 0.6) into this yields: i. Substituting the general solution into this equation gives: H1 (1.5.0) + H2 (0.4 (a) (b) At x=0.6b) Substituting equations (4.t) = f(x) g(t). v=0. The natural frequencies and modes can be determined as illustrated in the following examples. at x = 0.General Solution for Uniform Beams of Finite Length The general solution to equation (4.5.7) The above equations are valid for any boundary conditions. Ilanko 2005 45 (i) .5) gives λ = L4 mω 2 EI (4.5.6) f ( x ) = H 1 cosh (λξ ) + H 2 sinh (λξ ) + H 3 cos(λξ ) + H 4 sin (λξ ) (4.5.0) + H3 (1.6a) g(t) = sin(ωt+φ) and. EI For educational use at Canterbury and TUHH © S.e.0) = 0 At x=0. or H1 = H3 =0 Let us consider the possibility that λ = 0.0) . Substituting this into equation (iii) and (iv) gives: H2 sinh λ + H4 sin λ = 0 H2 λ L 2 sinh (λ ) − H 4 (ia) λ L 2 sin (λ ) = 0 (iia) (iiia) (iva) For non-trivial solution of equations (iiia) and (iva) sin λ = 0 For educational use at Canterbury and TUHH © S.t) = [H1 + 0 + H3 + 0] sin(ωt+φ) = 0 [using equation (ia)] This is a trivial solution. for non-trivial solution H1 = H3 = 0. at x = L.0) .0) + H2 (0. M= 0 results in the following equation: λ L 2 [H 1 cosh(λ ) + H 2 sinh (λ ) − H 3 cos(λ ) − H 4 sin (λ )] = 0 (iv) Equations (i) to (iv) may be written in matrix form as [B] {H} = {0}.t) gives v(x. v = 0 leads to: H1 cosh λ + H2 sinh λ + H3 cos λ + H4 sin λ = 0 (iii) Also at x = L. Substituting this into the general solution for v(x.H3 (1.0)] = 0 (ii) Similarly. Ilanko 2005 46 .(c) (d) (λ/L)2 [H1 (1. Therefore.H4 (0. 1 where [B ] = λ 0 1 0 −λ sinh λ cos λ 2 cosh λ λ cosh λ λ sinh λ − λ2 cos λ 2 H1 H2 0 and {H } = H3 sin λ 2 H4 − λ sin λ 0 2 2 The characteristic equation (frequency equation) is |B| = 0 which gives λ4 sinh λ sin λ = 0 For this particular case the frequency equation may be obtained as follows: From equation (i) H3 = -H1 (λ / L )2 (H 1 − H 3 ) = 0 From equation (ii) Substituting equation (ia) into the above gives: either λ = 0. this gives: H1 (1. ∂v/∂x=0. It can be shown that this determinantal equation reduces to the following frequency equation: λ2(1-cosh λ cos λ) = 0. v=0. v=0 yields: H1 cosh λ + H2 sinh λ + H3 cos λ + H4 sin λ = 0 (iii) L (c) (d) (i) At x=L.0) − H 3 (0.0) + H 2 (1.5.nπ Hence λ = nπ and ω = L 2 EI m The nth mode is: sin (nπx/L) Case 2: Both Ends Clamped n=1 n=2 x=L x=0 Figure 4.0) + H 4 (1.0) + H4 (0. ∂v/∂x=0 gives: λ L [H 1 sinh (λ ) + H 2 cosh(λ ) − H 3 sin (λ ) + H 4 cos(λ )] = 0 (iv) This may be written in matrix form as [B]{H}={0} For non-trivial solution of this equation |B| = 0.0) = 0 At x=0. Substituting the general solution for v into this equation gives λ [H 1 (0. As for case 1. For educational use at Canterbury and TUHH © S. Ilanko 2005 47 .0) + H3 (1.5 Boundary Conditions: (a) (b) At x=0.0) + H2 (0.0)] = 0 (ii) At x=L. 73 L 2 EI rad/s m The expression for mode is: cosh λ where r = x x x x .0)] = 0 For educational use at Canterbury and TUHH © S.5.. ∂x 2 Using the general solution we get: i.6 This is an example of a semi-definite system.Free Beam n=1 Rigid Body Translation (n=0) n=2 x=L x=0 Figure 4. ∂ 2v = 0. Ilanko 2005 48 (i) . which gives the fundamental frequency ω1 = 4.0) + H 2 (0. The first non-trivial root of this equation is λ = 4. sinh λ − sin λ Case 3: Free .0) + H 4 (0. This means λ=0 is a trivial solution.73. The boundary conditions are: (a) At x=0..M=0.e EI λ L 2 [H 1 (1. substituting λ=0 into the general solution gives v = (H1+H3) g(t).0) − H 3 (1. − (r ) sinh λ − cos λ + (r ) sin λ L L L L cosh λ − cos λ . But H1 + H3 = 0 from equation (i).While λ=0 satisfies the above equation. Ilanko 2005 49 . This means the non-zero natural frequencies of free-free beams and clamped-clamped beams are the same.] The non-zero roots of equation (v) are identical to the roots for the clamped .0)] = 0 (iii) Similarly at x=L. (c) At x=0.0) + H 3 (0. This corresponds to a rigid body translation.clamped beam. (v) For this case λ=0 is not a trivial solution as illustrated below: Substituting λ=0 into the general solution for v gives: v = [H1 (1) + H2 (0) + H3 (1) + H4 (0)] sin(0t+φ) = (H1+H3) sin(φ) = constant. However.5. the modes are different. M=0 gives λ L 2 ∂ 2v = 0 resulting in the following equation: ∂x 2 [H 1 cosh(λ ) + H 2 sinh (λ ) − H 3 cos(λ ) − H 4 sin (λ )] = 0 (i) These are the same as equations (ii) and (iv) for case 1.(b) Similarly at x =L.0) + H 2 (1.6)] into the above yields: For a uniform beam this means λ 3 L (d) [H 1 (0. [There is another semi-definite mode corresponding to a rigid body rotation. The remaining boundary conditions are obtained by setting the shearing force to zero at both ends. S=0 gives: − ∂M =0 ∂x ∂ 3v = 0.0) − H 4 (1. For educational use at Canterbury and TUHH © S. ∂x 3 Again substituting the general solution [equation (4. S=0 gives: λ L 3 [H 1 sinh (λ ) + H 2 cosh (λ ) + H 3 sin (λ ) − H 4 cos(λ )] = 0 (iii) For non-trivial solution of equations (i)-(iv) the determinantal equation can be shown to be λ10(1-cosh λ cos λ) = 0. sinh λ − sin λ For educational use at Canterbury and TUHH © S.For free-free beams the expression for the modes is: cosh λ where r = x x x x . − (r )sinh λ + cos λ − (r ) sin λ L L L L cosh λ − cos λ . Ilanko 2005 50 . EI1.t) = f2(x2).t) where v1(x1. in terms of the non-dimensional axial coordinates ξ1 (=x1/L1) and ξ2 (=x2/L2). which may be treated as simple supports. Both ends are supported on short bearings.g(t) (i) in which.7. Ilanko 2005 51 .5. Let the centroid G be located at distances e1 and e2 from the ends of the two shaft segments to which it is connected.t) and v2(x2.7) it is often necessary to consider the flexural vibration of shafts carrying masses.g(t). Let the flexural rigidity length and mass per unit length of the first shaft segment be EI1. m2 G L2 L1 e2 e1 Figure 4.Case 4: Beams Carrying Masses In finding the whirling speed of shafts (see section 4.5. Let the mass of the rotor be m0 and its moment of inertia about its centroidal axis parallel to the neutral axis of the cross section be I0. f 1 ( x ) = H 1 cosh (λ1ξ 1 ) + H 2 sinh (λ1ξ1 ) + H 3 cos(λ1ξ1 ) + H 4 sin (λ1ξ1 ) (ii) g(t) = sin(ωt+φ) and λ1 = L1 4 (iii) m1ω 2 EI 1 (iv) Also f 2 (x ) = H 5 cosh (λ 2ξ 2 ) + H 6 sinh (λ 2ξ 2 ) + H 7 cos(λ 2ξ 2 ) + H 8 sin (λ 2ξ 2 ) in which λ2 = L2 4 m2ω 2 EI 2 (v) (vi) For educational use at Canterbury and TUHH © S.7 Let the displacements of the two shaft/beam segments be v1(x1. m1 EI2. and let the corresponding parameters for the second shaft be EI2.t) = f1(x1). Let us consider the general case of a shaft system carrying a rotor at x1=L1 and x2 =-L2 as shown in Figure 4. and v2(x2. L1 and m1 respectively. L2 and m2. t)+S2(-L2. In the program all parameters are input in non-dimensional form as follows: For the mass m0 use m0/m1L1 where m1 is the mass per unit length of the first beam/shaft segment.S1(L1.t 2 = −ω m0 (v1(L1. v2=0 and v ′2′ = 0 will yield: H5=H7=0 Substituting these into equations (ii) and (v) results in the following simplified expressions for the functions f1 and f2. S2(-L2. For non-trivial solution |B|=0. t ) L1 .t).t) e1 = I0 And for the translation. An interactive multimedia program modes951. t ) + e1 1 2 ∂x1 ∂t L1 . Similarly at the right hand end x2=0. v1=0 and v1′′ = 0 give: H1+H3=0 and (λ1)2 (H1-H3)=0 giving H1=H3=0 as in case 1.a) f 2 (x ) = H 6 sinh (λ 2ξ 2 ) + H 8 sin (λ 2ξ 2 ) (v.a) The remaining conditions are found by applying the following conditions at x1 = 0 and x2 = -L2.t)+ e1 v1′ ( L1 .t ∂v ∂2 v1 ( L1 . For educational use at Canterbury and TUHH © S. At x1=0. Continuity of slope gives: f 1′( L1 ) − f 2′(− L 2 ) =0 Continuity of displacement gives: f 1 ( L1 ) + (e1 + e2 ) f 1′( L1 ) − f 2 (− L2 ) = 0 Applying Newton’s second law in the rotational sense gives: M2(-L2. Ilanko 2005 52 .a) and the expressions for the bending moments and shear forces into the above equations yields four equations of the form: [B]{H}={0}. simplifying the problem to that of a four by four matrix equation.a) and (v.There are eight constants H1…8 but in this case by setting the origins of the co-ordinates x1 and x2 at the supports. t ) Substituting the reduced general solutions in equations (ii. we can eliminate four of the constants.t)=m0 ∂ 2 ∂v1 ∂t 2 ∂x1 =−ω2I0 v1′ ( L1 .t) e2 + S1(L1.t)-M1(L1.exe (windows95 version) can be used to calculate the natural frequency parameter λ for the above problem. f 1 ( x ) = H 2 sinh (λ1ξ 1 ) + H 4 sin (λ1ξ ) (ii. This is because the program finds the roots of the frequency equation by searching for a change in the sign of the determinant. The trial frequency parameters λmin. Applying special conditions for each of these cases. All lengths are non-dimensionalised with respect to the length of the first beam segment L1. By using slightly unsymmetrical properties. but in some cases some roots may be missed. one can calculate the natural frequencies of a simply supported beam of length 2L1. L2/L1 and EI2/EI1 to unity. i.e. L2/L1 and EI2/EI1.For the moment of inertia I0 use I0/m1L13. The results may be tested against simple known cases. m2/m1 and EI2/EI1 are required as inputs. symmetrical vibration modes will normally have a zero slope at the centre (exceptions to this are described later) for the following reason: For educational use at Canterbury and TUHH © S. i. by setting m0 and I0 to zero. the coincident roots may be separated and delimited. It can also be used to study symmetrical and antisymmetrical modes. Symmetrical modes In the case of beams. Actual frequencies may be calculated using equation (iv). Antisymmetrical modes are symmetrical but opposite in sign. and analysing only half the structure often results in considerable savings in time. The results displayed give the roots λ1 for the first beam-segment. λmax. Symmetrical modes are the ones whose actual displacement form on one side of the centre-line coincides with the image (formed by a mirror placed along the axis of symmetry) of its displacement on the other side. Ilanko 2005 f((L-s)/2) L/2 f((L+s)/2) s/2 s/2 Figure 4. The properties of the second beam-segment are also required as ratios. and m2/m1 . and putting values other than unity for m2/m1.e. the eccentricities are e1/L1 and e2/L1. For example.8 53 L/2 . Putting large values for the mass m0 or inertia I0 would give results corresponding to constrained cases (with the exception of some low frequency modes where the system behaves like a discrete system). the program can also be used to calculate the natural frequencies of stepped beams by setting m0 and I0 to zero.5. dλ are also for the first segment. Symmetrical and Anti-symmetrical Vibration of Beams The vibration modes of any symmetrical structure will be either symmetrical or antisymmetrical. and for some symmetrical cases the determinant takes a stationary value at zero. The program may be used to study the behaviour of the system under various conditions. This may be used as a boundary condition for half of the structure.e. This would be true. if the displacement is continuous over the centre. However. δx →0. df = 0. located at equidistant from the centre.e. they must be equal. This means. or the mass is rigidly connected.e. For symmetry. a notable exception must be mentioned here. even if the beam carries a mass at the centre. a discontinuity may occur. continuity condition for slope. dx Hence the slope of a continuous curve representing the lateral displacement should vanish at the centre. L − δx L + δx =0 − f 2 2 as δx →0. f L−s L+s =0 −f 2 2 If we take two points that are very close to the centre (i. if two beam segments were hinged together at the centre. if only symmetrical modes are considered. provided the beam is continuous. will be replaced by the conditions that the bending moment will be zero in both beams at the hinge.Consider the displacement of two corresponding points on either side of the centre line. δf =0 δx i. This should not be confused with a continuous beam on which a mass is hinged/pinned at the centre. Ilanko 2005 54 free to rotate relatively Beams hinged to each other Figure 4. thus allowing a discontinuity in the slope to take place. let s = δx). For educational use at Canterbury and TUHH © S.5.9 . f i. and moment angular acceleration equation. The amplitude of displacement of these two points are f L−s 2 and f L+s 2 respectively. In such a case. In the case of a connected structure. S2(0. another symmetrical boundary condition may be obtained. the corresponding points on either side of the centre must have f = 0 at the centre. However.t) S1((L-b)/2).5.5.10 Anti-symmetrical modes Similar arguments may be presented for the antisymmetrical modes. A sudden jump in the shearing force would occur within the boundaries of the mass. The symmetrical boundary condition of a beam without any attached masses or springs may be conveniently modeled using a sliding boundary condition as shown here. we may obtain one of the free-bodies shown in the following diagram: By applying Newton's second law in the lateral direction.t) v1 (( L − b) / 2. if the beam carries a mass. or is attached to a spring.In the case of a symmetrical vibration. then the shear force in the beam at the centre will also be zero. If the beam has no attached mass. t = 0 v1 . and the end of a beam. equal and opposite displacement. Taking the origin of co-ordinate axes as the left most point on the beam segment. let us consider a beam carrying a mass at the centre.11 For educational use at Canterbury and TUHH © S. then the shear force in the beam will not be zero. for the lefthalf beam we get: L m L−b L−b − S1 . t ) = 0 v2 (0. For anti-symmetrical modes. t ) v 2 (0. Ilanko 2005 55 . Making an imaginary cut through the centre of the mass. For example. and is not restrained by any other structural elements including springs. t ) 2 m0/2 m0/2 If neither a mass nor a spring is attached to the beam. For any finite displacement at Figure 4. t ) zero slope & zero shear force Figure 4.t 2 2 2 b Alternatively considering the right-half beam m0 gives: m S 2 (0. even if it is close to the centre-line of the structure where the shear force is zero. which is connected to the mass. then the shear force at the centre of the beam would be zero. the shear force at a section through the centre line would be zero. the centre would therefore cause a contradiction. including partial elastic linear and rotational restraints. and will not be zero in the beam where it is rigidly connected to the mass. The solution manual contains the derivation of the boundary conditions in terms of displacement functions and its derivatives. application of Newton's second law of motion in the rotational I sense for half the mass (i. cable restraints. It may also be shown that the bending moment distribution will also be ant-symmetrical. For educational use at Canterbury and TUHH © S. which leads to the conclusion that there can be no displacement at the centre. The problems in Section 4. pin connections and rigid connections. Ilanko 2005 56 . Moment = 0 θ ) will yield a boundary condition for anti2 symmetry. for a beam carrying a mass with a moment of inertia. the bending moment may change abruptly within the boundaries of the mass. In such a case. becoming zero exactly at the centre of the ant-symmetric system.e.8 include a number of practical non-standard boundary and continuity conditions. Once again. t) o x x1 x2 Figure 4.5. Step 1: Sketch the free body diagram of an infinitesimal element: P v(x.6 Lateral Vibration of Statically Axially Loaded Beams Derivation of Equation of Motion v P M(x2..t) v(x1. it should be noted that the axial force can vary in some common applications such as in the case of rotating beams (the steady state centripetal force varies with distance). The solution procedure described in this section is not applicable for such cases.2 For educational use at Canterbury and TUHH © S.1 Figure 4.t) be the lateral dynamic displacement of the beam.t) M(x1.t) P v(x2.6. Let v(x.1.t) Sf(x2.t) M v v Neutral Surface M+δΜ Sf+δSf P v+δv Sf o x x δx Figure 4.t) Sf(x1. In many practical problems the induced axial force would remain constant. Ilanko 2005 57 .4. However.6.1 shows a segment of an axially loaded beam that is vibrating laterally. Comparing this with figure 4. it can be seen that the only change is the addition of the axial force P. Let us now follow the basic steps in deriving the equation of motion.6.. 3.t) M P+F+δF v+δv v P+F o S x δx x Figure 4.F (δv) –P (δv) = 0 i. δM + S (δx) . this gives S = − +P ∂x ∂x Step 3: (4.t) in the longitudinal direction as shown in Figure 4.1) Neglecting the rotary inertia of the beam. Neutral Surface S+δS M+δM v(x.6.M + S (δx) .t) in the lateral direction and F(x.e.6. S(x.6.5. let us replace the transverse shearing force Sf(x.F (δv) –P (δv) = 0 But F (δv) being a product of two small dynamic terms is negligible ∂M ∂v As δx is infinitesimal.2) The net elastic action in the positive lateral direction is -S + (S + δS) = δS But from equation (4.As in section 4.3) .3 Step 2: From Mechanics if Materials[ ] the moment curvature relationship for a thin beam is M = EI ∂ 2v ∂x 2 (4.6. summing the moments on the infinitesimal element and equating to zero gives: (M + δM) .t) with its components.6. Ilanko 2005 58 (4.6.2) δS = δ − = ∂M ∂v +P ∂x ∂x ∂ ∂M ∂v δx = − +P ∂x ∂x ∂x = − − ∂2M ∂ 2v + P δx ∂x 2 ∂x 2 ∂2 ∂ 2v ∂ 2v EI + P δx ∂x 2 ∂x 2 ∂x 2 For educational use at Canterbury and TUHH © S. in which (4.6.6. (4. For a homogeneous.6.6.6.5) may be written in the following form: v(x.6b) Substituting equations (4. Applying Newton's second law in the lateral direction we get δS = (m δx) ∂ 2v ∂t 2 Using equation (4.6.6.6.6a) (4.6) into equation (4.Step 4: Step 5: Since there are no external dynamic forces this is the net lateral force.6.d) For educational use at Canterbury and TUHH © S. Ilanko 2005 59 .5) gives β= γ = π 2γ 2 + λ4 + π 4γ 2 4 and α= − π 2γ 2 PL2 mω 2 4 .5) General Solution for Uniform Beams of Finite Length The general solution to equation (4.6. uniform beam this reduces to the following form: EI 4 2 2 ∂ v ∂ v ∂ v −P 2 +m 2 = 0 4 ∂x ∂x ∂t (4. f ( x ) = H 1 cosh β x x x x + H 2 sinh β + H 3 cos α + H 4 sin α L L L L g(t) = sin(ωt+φ) and.t) = f(x) g(t).3) this becomes: − ∂ 2v ∂2 ∂ 2v ∂ 2v = ( m δ x ) EI + P δ x ∂t 2 ∂x 2 ∂x 2 ∂x 2 This gives: 2 2 2 2 ∂ ∂ v ∂ v ∂ v EI -P 2 +m 2 = 0 2 2 ∂x ∂x ∂x ∂t (4.7c.7a.4) This is the partial differential equation governing the lateral vibration of an Euler-Bernoulli beam.6. and λ = L EI π 2 EI + λ4 + π 4γ 2 4 .6) where.b) (4. (4. The natural frequencies and modes can be determined as illustrated in the following examples.The above equations are valid for any boundary conditions. For educational use at Canterbury and TUHH © S. Ilanko 2005 60 . where B11 = 1.0) + H4 (0.0)] = 0 Similarly.4 (a) (b) At x=0.. For this particular case the frequency equation may be obtained as follows: For educational use at Canterbury and TUHH © S. v=0. . The characteristic equation (frequency equation) is |B| = 0.0)] − α L 2 [H 3 (1..0) + H2 (0.e.0) + H 2 (0. Ilanko 2005 61 (iv) . B44 = -(α)2 sin α.0.0) = 0 (i) At x=0.6. Substituting the general solution into this equation gives: H1 (1.6) into this yields: i. M=0 results in the following equation: β L 2 [H 1 cosh β + H 2 sinh β ] − α L 2 [H 3 cos α + H 4 sin α ] = 0 Equations (i) to (iv) may be written in matrix form as [B] {H} = {0}. v=0 leads to: H1 cosh β + H2 sinh β + H3 cos α + H4 sin α = 0 (d) (ii) (iii) Also at x=L.0) + H3 (1. EI β 2 L (c) [H 1 (1. ∂x 2 Substituting equation (4..0) + H 4 (0. at x=L.6. ∂ 2v = 0 at x = 0. M=0.Case 1: Simply Supported at Both Ends P n=2 n=1 P x=L x=0 Figure 4. 6. This would be true only if both α and β are zero. The variation of frequency with load is shown graphically in figure 4.6. Ilanko 2005 62 . Substituting this into equation (iii) and (iv) gives: H2 sinh β + H4 sin α = 0 (iiia) From equation (i) 2 H2 β L 2 sinh β − H 4 2 α sin α = 0 L (iva) For non-trivial solution of equations (iiia) and (iva) sin α = 0 Hence α = nπ Substituting equation (4. into the general solution for v(x.8) where ωn is the nth natural frequency of the axially loaded beam.H3 = -H1 (ia) 2 From equation (ii) β H1 . Therefore.d) results in the following equation: ωn Ωn 2 =1+ P Pcn (4.t) = [H1 + 0 + H3 + 0] sin(ωt+φ) = 0 [using equation (ia)] This is a trivial solution.5 sin nπ x L For educational use at Canterbury and TUHH © S.6.7c. or H1 = H3 =0 2 2 Let us consider the possibility that α +β = 0.α H3 = 0 Substituting equation (ia) into the above gives: (iia) either α2+β2 = 0. for non-trivial solution H1 = H3 = 0.t) gives v(x. Ωn is the corresponding natural frequency of the unloaded beam and Pcn is the nth critical load of the same beam which is given by EI nπ L The nth mode is: 2 .6.7b) into the above yields: nπ= − π 2γ 2 λ4 + + π 4γ 2 4 This. Substituting α=β=0. together with equations (4. 5 For educational use at Canterbury and TUHH © S.ω2 Ωn2 P -Pcn Figure 4.6. Ilanko 2005 63 . ...... acceleration due to vibration v and the acceleration due to centripetal acceleration Ω 2v.....(4.. δ S = ( m .....δ x = m ( v − v 2 )δ x For principal vibration....7.3.2) ......Ω2v v(x...4..... a= v . since its position will change due to the rotation. This is similar to the free-body in Figure 4....(4.......1) Using Newton’s 2nd law of motion....1) and (4.7...e. Consider the motion of the free-body in Figure 4.t) S+dS Neutral Surface M+dM a M F+dF vF o x S a Ω v+dv v x δx Figure 4......7. The lateral motion of an infinitesimal element may be regarded as a radial motion of that element...5. v = −ω 2 v giving δ S = − mv (ω 2 + 2 )δ x Equating the rights hand sides of equations (4.δ x ) a = m....1 Following the derivations for a non-rotating beam..7 Whirling Speed of Shafts Let us now consider the lateral vibration of a shaft that is rotating at a speed of Ω rad/sec...1.. but this time the free-body is vibrating as well as rotating about the original axis of the beam at a speed of Ω rad/sec.a .... For educational use at Canterbury and TUHH © S..7.. Hence its radial acceleration a consists of two components... i. Ilanko 2005 64 .7. the net lateral force in the element is given by: δS = − ∂2 ∂ 2v EI δx ∂x 2 ∂x 2 .7.2) we get. ..5) If Ω → ωS then...... and long-bearings that restrain lateral rotations may be taken as ‘clamped’ supports.......7...(4........4) For a non-rotating shaft.7.(4.... uniform shaft this reduces to the following form: EI ∂ v = m(ω 2 + 4 ∂x 4 2 )v . In modelling a shaft for calculating whirling speeds...∂2 ∂ 2v EI = mω2 + 2 2 ∂x ∂x ( 2 )v ... that the form of the solution will be the same whether or not the speed of rotation is zero. referred to as whirling. That is to say that (ω2+Ω2) and ω S2 may be interchanged..........7.. the natural frequencies ωS may be found by solving the above equation for Ω =0..(4... If the speed of rotation (Ω) approaches the natural frequency of the non-rotating shaft (ωS)...... The whirling speeds are therefore equal to the flexural natural frequencies of the non-rotating shafts. by using the equation (ω2+Ω2) = ω S2 ... For educational use at Canterbury and TUHH © S... then the natural frequency of the rotating shaft (ω) would approach zero! That is a state of instability... short bearings that permit lateral rotations may be taken as ‘simple’ supports. Ilanko 2005 65 .. The speeds at which this occurs are called whirling speeds..... One can deduce the flexural natural frequencies of a rotating shaft from the natural frequencies of nonrotating shafts..3) For a homogeneous. ω → 0... It may be noted here....... 12(1 − ν 2 ) …(4. w y=b x=a y x x=0 y=0 Figure 4. without deriving it.1) in which E and ν are the elastic modulus and Poisson’s ratio of the plate material.8 Transverse Vibration of Plates (I just commenced this section potential for error is high at this stage) So far we have considered one-dimensional structural elements. σy. Traditionally. y and t. Figure 4. which are given by For educational use at Canterbury and TUHH © S. Exact natural frequencies and modes are available in the literature for the out-of-plane vibration of rectangular plates. Equations of motion for two and three-dimensional structures can be derived using the same general approach. Ilanko 2005 66 .8.y. We will present the equation of motion for a plate subject to uniform static in-plane loading.1 shows the coordinate system and the notation for the plate properties. the in-plane loading is expressed as load per unit length Nx. then their components in the out-of-plane direction will affect the equation of motion in the same way the axial force affects the transverse vibration of beams. This is given by D= Eh 3 .t) is a function of x. and Nxy. and h is the plate thickness.8.1 Equation of Motion The net elastic flexural resisting force in an infinitesimal plate element is expressed in terms of a parameter D called the plate rigidity. even for systems with simple geometrical shapes such as rectangular plates. and then proceed to solve it for some specific boundary conditions.4. If the plate is subject to in-plane stresses σx.8. although the availability of exact solutions is limited. but only for the case where two opposite edges are simply supported. The out-ofplane displacement coordinate is w(x. and τxy. Ny. 0.0.8.5a.4a).8.6) and (4.3) that g ′′(0) f ( x) + g (0) f ′′( x) = 0 We have from equation (4.5a) Similarly we can show that g ′′(b) = 0 …(4. t ) = − D ∂ 2 w( x. t ) ∂ 2 w( x. Such a form of solution exits for a plate with two opposite edges simply supported in the absence of any in-plane shear loading (Nxy = 0).5b) The sinusoidal function g ( y ) = sin( nπy / b) …(4.8.2) gives: For educational use at Canterbury and TUHH © S.3).8.7) into equation (4. t ) = f ( x) g ( y ) sin(ωt + α ) …(4.b).9.3) This assumes that the mode consists of a product of functions in x and y directions.0.2) …(4.Nx Ny N xy σx =h σy σ xy The equation of motion is then: D∇ 4 w + N x ∂2w ∂2w ∂2w ∂2w + 2 N + N + ρ h =0 xy y ∂x∂y ∂x 2 ∂y 2 ∂t 2 Let us consider a general solution of the form w( x.4b) The bending moment intensity (moment per unit length) along the edge y = 0 is given [Ref] by: M y ( x. t ) = f ( x) sin( nπy / b) sin(ωt + α ) Let f ( x ) = G1 sinh(ϕx / a ) + G 2 cosh(ϕx / a ) + G3 sin(ϕx / a ) + G4 cos(ϕx / a ) …(4. g (0) = 0 Therefore g ′′(0) = 0 …(4.8.8.8.3) now becomes w( x.8. Ilanko 2005 67 . By substituting this into the governing differential equation we can show that it also satisfies the equation of motion as follows. This means the boundary conditions are w( x.8. it follows from equation (4. Equation (4. t ) = w( x.3) that: g ( 0) = 0 …(4. y. (4.8.4a) and g (b) = 0 …(4. Let us assume that the plate we consider is simply supported at the edges y = 0 and y = b. t ) = 0 It follows from equation (4.b) and (4.4a. t ) + ν =0 ∂y 2 ∂x 2 Since this is true for all t.8.8.8.7) Substituting equations (4.8.0.6) satisfies boundary conditions (4.8.8.8. y. b. 8) where f 1 ( x) = G1 sinh(ϕx / a ) + G2 cosh(ϕx / a ) and f 2 ( x) = G3 sin(ϕx / a ) + G4 cos(ϕx / a ) Since equation (4. For educational use at Canterbury and TUHH © S. i.8.. (b) two opposite edges supported and the other two clamped. the terms associated with f1(x) and f2(x) must vanish.8) is true for any x.f1 ( x) + f 2 ( x) 2 ϕ a ψ a nπ − b 2 nπ + b 2 2 + 2 ϕ 2 a 2 − ψ a Nx nπ − D b 2 2 Nx nπ − D b Ny D 2 − Ny D ρhω 2 D − ρhω 2 D =0 f ( x ) = G1 sinh(ϕx / a ) + G 2 cosh(ϕx / a ) + G3 sin(ϕx / a ) + G4 cos(ϕx / a ) …(4. ----this section has yet to be completed---.8. Ilanko 2005 68 .e……. and a brief warning on the limitation of the results for large in-plane loadings ( N x ) etc for which there may be a need to consider the effect of initial imperfections.solution will be given for two cases only: (a) all edges simply supported. Another 8 pages or so would tidy up this chapter. y. φkare the jth and kth modes respectively.4) From d’Alembert’s principle.t x. if this element is subject to a static force dFj then the resulting displacement would be φj(x. Ilanko 2005 69 . Let the net internal induced restoring action q dFjsin(ωt+α) dm th corresponding to the j mode be dFj sin(ωjt+α).9. the following equation was derived showing that the natural modes are orthogonal with respect to the mass.9 Orthogonality of Principal Modes In Chapter 3. equation (4..9..4. The integral would be a double integral for 2-D systems (for example.3) becomes: dFj = -ωj2 φj dm ………. and the integral would be a single integral.) sin(ωjt+α).9.(4. If the coordinate of dynamic displacement is q(x.1. where φj(x. with respect to x and y in a Cartesian coordinate system).. the dynamic displacement would be of the form: q = φj(x.1 dFj sin(ωjt+α) = dm q .) is the jth mode of vibration.y and z. n i =1 mi qi . For one-dimensional systems the modes would be functions of one spatial coordinate. and triple for 3-D systems (w..9.. Figure 4.k = 0 for j ≠ k (4.).) Proof of Orthogonality: Consider the vibration of an infinitesimal element of a continuous system having a mass dm as shown in Figure 4.r.9.9.9..2) m where φj. ……….3) th If the vibration corresponds to the j mode. For educational use at Canterbury and TUHH © S.t) then from Newton's second law of motion.y.1) For continuous systems we will derive an orthogonality equation of the form: φ jφ k dm for j ≠ k (4. j qi .y.(4.y. φ k + dFk .h.9.Similarly it may be shown that the net internal action corresponding to the kth mode and the resulting displacement are dFk and φk(x.9.6) where Wbjk is the work done by any attached masses at the boundaries as a result of forces corresponding to the jth mode and kth mode. Ilanko 2005 70 .9.2 illustrates the force-displacement relationship for the infinitesimal element. where the force dFk is given by: dFk = -ωk2 φk dm ………. Net work done = dF j .) respectively. If the order of application of the forces is reversed then the total work done is: dF j . and φk is caused by dFk and that during displacement φk.7) For linear elastic structures the total work done on the system is independent of the path. Noting that φj is caused by dFk dFj. Hence equating the r. φ j + + Wbkj 2 2 ………. as a pre-existing force. consider the work done by the internal actions if forces dFj and dFk are applied on the linear elastic system.9. in which the two triangular hatched areas represent the work done by the infinitesimal inertial forces while causing corresponding infinitesimal displacements.(4.s of equations (4.y. φ j 2 + dF j . and the rectangular shaded area represents the work done by dFj as it moves by φk which is caused by dFk.. φ j dFk . φ k Figure 4. φ k + dFk .(4.6) and (4. in Force the order stated.e. As for the discrete systems.2 + dF j . φ k + 2 2 Figure 4.. The total work done includes the integral of the work done on the element and any work done on the boundaries by inertia forces corresponding to rigid masses.y.9. φ j dFk .). applied in the order stated.(4.9. the net Displacement work done on the element is obtained as: φj φk dF j .9. φ k + Wbjk 2 ………. i.7) gives: For educational use at Canterbury and TUHH © S. dFj dFj would do work to its full potential.5) Here again we can say that a static force dFk would cause a displacement of φk(x. φ j dFk . Using such normal modes.9.9) m It is possible to include such terms as a part of the integral using the delta functions within the integral of equation (4.e.φk Substituting equations (4.. an extra term mo φk(a). 10) The above orthogonality relations hold only between two different modes (for j ≠ k) and if j=k then the integrals will give non-zero values. The above integral is used in dynamic response calculations. φ k + dFk . ………. φ j + + Wbkj 2 2 Canceling the common terms we get.7) and (4.8) into the above equation gives -ωk2 φk.φj + Wbkj ……….φj dx = 0 ………. this results in equation (4.11) For educational use at Canterbury and TUHH © S. In such cases the above equation reduces to: dFk.e. For a one-dimensional structure (bars. That is: φ iφ j dm + mo φk(a). dFj. φ k + Wbjk = 2 dF j .9. φk.9. and it is therefore convenient to normalize the modes such that this integral is unity. i.φj(a) = 0 for j ≠ k (4.e. (φj)2 dm =1.…. φj. beams and shafts) having a constant mass per unit length m. (ωj2-ωk2) φk.φj = dFj.2). It should also be noted that mass should be replaced with an appropriate moment of inertia term. φ j 2 + dF j .(4.φk dm i. Ilanko 2005 71 .2).(4.9.9.9.φk + Wbjk = dFk.φj dm = (φj)2 dm ≠ 0..φj dm = 0 If ωk ≠ωj. φ k + dFk .9. the orthogonality relationship reduces to: φk.φj dm = 0 For a beam carrying a concentrated mass m0 at x = a.φj dm = -ωj2 φj.…. if the displacement coordinate q is a rotational coordinate.9.(4.φj(a) should be included..dF j . i.8) For beams without any attached masses the terms Wbjk and Wbkj would be zero. This is the orthogonality condition.. It should be noted here that in response calculations. Ilanko 2005 72 . the actual displacements would be expressed in terms of the modes (whether normalised or not) and weighting coefficients. It is also worth noting that the orthogonality relations may be used to check the accuracy of modes. In the case of approximate methods. calculated values of the integral in equation (4.9. which are dependent on initial conditions and/or any applied dynamic force.2) (which would be zero if the modes were exact) would give an estimate of any error in the mode calculations.Similar orthogonal relationship may also be obtained in terms of stiffness of a structure. This will be discussed in Chapter 5. For educational use at Canterbury and TUHH © S. Symmetrical Vibration of a Partially Restrained Cable. Derive the boundary condition equations for the torsional vibration of the geared system shown in Figure P.1. Review Problems in Natural Frequencies and Modes of Continuous Systems P. The cantilevers may be modelled by elastic springs of lateral stiffness k each.4.4.4. Show that the frequency equation for symmetrical lateral vibration of the cable is given by: λ tan λ 2 = kL m .10.4.2.4. Its ends are connected to two identical short cantilever beams of negligible mass (see figure P. A uniform cable of length L and mass per unit length m is under a static tensile force T0. where λ = ωL . Ilanko 2005 73 . For educational use at Canterbury and TUHH © S.4. in terms of the angles of twist θ1. T0 T0 T0 T0 k k Figure P. P.θ2 and their derivatives ∂θ1/∂x1 etc.2.1).1. and is made of a material having shear modulus G.4. u2’ etc.ρ1 I r x2 = L2 Shaft 2: G. and a concentrated mass m0 at the other end as shown in Figure P. Proceed to obtain a frequency equation.3.4. u1’. in determinantal form.x1 = L1 Shaft 1: G.4.ρ2 2r x1 = 0 16I 8I x2 = 0 Figure P. Geared System P.2. and a tube are rigidly connected to a wall.4. Obtain the frequency equation for the torsional vibration of this system. A solid shaft. Write down the necessary boundary to determine the longitudinal natural modes and frequencies in terms of u1.4.4. deduce the frequency equations for the torsional vibration of a shaft subject to the following conditions: (a) (b) Both ends torsionally restrained. From the above result. One end restrained and the other end free. The cutting edge has a polar moment of inertia I0.J2. for the longitudinal vibration of this system in terms of the parameters shown in FigureP.J1.4. The drill pipe of an oil well has a length L and polar second moment of area J.4. P. For educational use at Canterbury and TUHH © S. Ilanko 2005 74 . and density ρ. u2. and if not write down an approximate formula for it. Ilanko 2005 75 . The shaft is of circular cross section.ρ2 mass m0 L Figure P. radius R. density ρ and shear modulus G. E2.5. length L.Tube. 4I I I L/2 L/2 For educational use at Canterbury and TUHH © S. b) Derive the frequency equation for the anti-symmetrical case only.4. The end rotors have a polar moment of inertia I while the central rotor has an inertia of 4I.A2. P.4.ρ1 Shaft. E1. a) Write down the boundary condition equations for the symmetrical and antisymmetrical torsional vibration of this system and . d) Can the expression you deduced in part (c) be applied to determine the first nonzero natural frequency.A1.4. c) Deduce an expression for the natural frequencies corresponding the antisymmetrical vibration if the polar moment of inertia of the rotors are very large compared to the polar moment of inertia of the shaft. A shaft carrying three rotors is shown in Figure P.4.5. ρ.4.L String.6. For educational use at Canterbury and TUHH © S.Figure P.4.5 P. Write down the boundary conditions in terms of constants H1. for the vibratory system shown in the following diagram: E. Tension T0. length .4. vibrates laterally Figure P.6.A.. H3 etc. Ilanko 2005 76 . H2. mass per unit length m. 8. find the first flexural natural frequency of the rod in Hz. in terms of the deflection function f1 and or its derivatives and the parameters shown in the diagram.4. Second moment of area of a circle about its diameter is given by: I = (π r4)/4 where r is the radius of the circle. Ilanko 2005 77 .4.8) For educational use at Canterbury and TUHH © S.7 P. 1.P. The material properties are: E = 207 GPa and ρ = 7800 kg/m3.m mass m0 Spring stiffness k L/2 L/2 Figure (P.4. EI.4 m Figure P. Show that the frequency equation is of the form cosh λ cos λ + 1=0. If 1.4.4.875 is the first root of the frequency equation. write down the boundary conditions for obtaining the frequency equation for symmetrical flexural modes.7 A 20 mm diameter circular steel rod is clamped at one end and free at the other as shown in the diagram below.8 For the beam shown in Figure P. I. The mass m0 is rigidly connected to the beam. The joint also provides partial restraint against relative rotation (stiffness Kr) of the two beams. The cable may be assumed to be massless. Do not substitute the general solution to the equation of motion. The centroid of the mass is at distance e to the right end of the beam. One of the beams is fully restrained against rotation and partially restrained (stiffness k) against translation at one end.P.m L d System (b) EI . This mass has a moment of inertia IG about the neutral axis at its centroid.m E.m m0 G x 1= 0 x =L1 x2= 0 System (c) For educational use at Canterbury and TUHH © S. Ilanko 2005 78 x 2= L2 e . in terms of the lateral dynamic displacement v and/or its derivatives.4. System (c): This consists of a uniform beam supported on knife-edges at two points and carrying a mass m0 at the tip.9 (a)-1(j).I. Write down the boundary conditions that are necessary for obtaining the frequency equations corresponding to the small amplitude lateral vibration of the beams shown in Figure P. System (a): This system consists of two beams. Hinge E.I. that is fully restrained at one end and is connected to a cable of length d at the other end The tension in the cable and beam is T. 0 T E. Elastic Cable System (b): This consists of a uniform beam of length L and flexural rigidity EI. The other end of this beam is hinged to the second beam.9.m Kr k L1 L2 System (a) Light.4. The moments of inertia of these masses about the neutral axis of the L beam are I1 and I2.I1 E. EI. and segmental length of each beam segment are shown in the Figure. mass per unit length m and length L carrying end masses m1. is simply supported at two intermediate points. Ilanko 2005 79 m2. Also write down the two boundary conditions for the anti-symmetrical L/4 L/4 L/4 L/4 flexural vibration that are different from System (f) the symmetrical case. The mass m0 is pinned to the beam at the neutral axis. System (g): This consists of a beam having flexural m1.I1 . length L and mass per unit length m. a uniform beam of flexural rigidity EI. The beam is not supported at any point. I2 . mass per unit length. having a moment of inertia I0 about the neutral axis at mid-span. m2.L m0 k System (d) EI1 .m symmetrical flexural vibration of this system..System (d): In this system. and partially restrained against translation at the right end. Write down the m 0. a uniform beam of flexural rigidity EI and mass per unit length m. I1 . System (e): This consists of a stepped cantilever beam carrying a mass m0 at the tip. System (g) For educational use at Canterbury and TUHH © S.I.m2 x1= L1 x2 = 0 . m0 x2 = L2 System (e) System (f): In this system.m1 x1 = 0 EI2 . The flexural rigidity. It also carries a mass m0.I 0 eight boundary conditions for the EI.m rigidity EI.m. is restrained against rotation at the left end. It also carries a mass m0 which is pinned to the right end. Ilanko 2005 80 .For educational use at Canterbury and TUHH © S. Light.m System (i): This consists of two beams connected by a smooth hinge which is partially restrained by a spring of stiffness k. Elastic Cables E.I. The right end of the second beam is simply supported. All cables and the beam are under static tension T0 . Neglect any friction in the pin.I. A disc of mass m0 is pinned to the beam at the right end.I. For educational use at Canterbury and TUHH © S.m L System (j) m0 . System (j): The left end of this beam is simply k supported and partially restrained against rotation by a coil spring that has a stiffness k [moment/unit rotation].m Hinge E. The two lateral cables have length h and the longitudinal cable is of length l.I. E. the beam is clamped at one end and elastically restrained by three light cables having Young's modulus E and cross-sectional area A.m k L1 L2 System (i) .A 0 E.System (h): In this system. Ilanko 2005 81 T0 T0 h T0 h d L System (h) E. The other end of the first beam is permitted to slide transversely but restrained against rotation. 11 . find the first critical speed of the shaft in r.8. in mω 2 PL2 4 and = L λ π 2 EI EI (a).p.4. where β and α + G2 sinh β + G3 cos α + G4 sin α L L L L are given by: β = which γ = π 2γ 2 + λ4 + π 4γ 2 4 and α = − π 2γ 2 + λ4 + π 4γ 2 4 . It carries two rotors very close to the bearings which are located at a distance of 1. If 4. length L and mass per unit length m. ∂x ∂x ∂t The general solution of this equation is of the form: v(x.m.4. Using the above.cosh(λ)cos(λ)) = 0.t) = f(x) sin (ωt+φ).5 m Figure P.11. State any assumptions made. The properties of steel are: elastic modulus E = 207 Gpa and density ρ = 7800 kg/m3. derive an expression for the natural frequencies of this system.4.73 is the first root (λ1) of the frequency equation.3. Show that the frequency equation for the flexural vibration of this beam is of the form (1 . subject to a compressive force of 0.5 m apart as shown in Figure P. (b). List all assumptions that were made in the theory and in your model.10. The Material properties are: E = 207 GPa and ρ = 7800 kg/m3. Comment on whether or not any research is needed to fully understand the vibrational behaviour of the above or For educational use at Canterbury and TUHH © S.11 A 20 mm diameter circular steel shaft is supported on long bearings at both ends.2 Pc1 where Pc1 is the first critical load of the beam. The partial differential equation governing the small amplitude flexural vibration of a thin simply supported beam of flexural rigidity EI. under static tensile axial loading P is given by 4 2 2 v v v EI ∂ 4 − P ∂ 2 + m ∂ 2 = 0 . and explain any practical limitations. P. Calculate the fundamental natural frequency of a 1. Ilanko 2005 82 1.2 m long simply supported steel beam of 50mm x 50 mm square section. Function f(x) is expressed in the following form: f ( x ) = G1 cosh β x x x x . 4. Kr EI.m.12 For educational use at Canterbury and TUHH © S.12 An axially loaded simply supported beam is fully restrained at one end. and is restrained against translation and partially restrained against rotation at the other end as shown in Figure P.4. P. Substitute the general solution into the boundary conditions and set up the elements of this determinant.similar practical systems.12. Ilanko 2005 83 P . Write down the boundary conditions to form the frequency equation in determinantal form.4.L Figure P.