Applied Structural and Mechanical Vibrations 2009 Part 8 pdf

Imposing the boundaryconditions 8.30 poses a serious limitation to the possible harmonic motionsbecause we get A=0 and the frequency equation 8.31which implies n integer and is satisfie

8.2.1 which are specified at a given time (usually t=0)—must be satisfied

at all times Let us suppose that our string is attached to a rigid support

at x=0 and extends indefinitely in the positive x-direction (semi-infinite

string) This is probably the simplest type of boundary condition and it

is not difficult to see that such a ‘fixed-end’ situation mathematicallytranslates into

(8.23)

for all values of t: a condition which must be imposed on the general solution

f(x–ct)+g(x+ct) The final result is that the incoming wave g(x+ct) is reflected

at the boundary and produces an outgoing wave –g(x–ct) which is an exact

replica of the original wave except for being upside down and travelling inthe opposite direction The fact that the original waveform has been reversed

is characteristic of the fixed boundary

Another simple boundary condition is the so-called free end which can beachieved, for example, when the end of the string is attached to a slip ring

of negligible mass m which, in turn, slides along a frictionless vertical post

(for a string this situation is quite artificial, but it is very important in manyother cases) In physical terms, we can write Newton’s second law stating

that the net transverse force F y (0, t) (due to the string) acting on the ring is

equal to Since and m is negligible, the

free-end condition is specified by

(8.24)

which asserts that the slope of string at the free end must be zero at all times

By enforcing the condition (8.24) on the general d’Alembert solution, it isnow easy to show that the only difference between the original and thereflected wave is that they travel in opposite directions: that is, the reflectedwave has not been inverted as in the fixed-end case Note that, as expected,

in both cases—fixed and free end—the incoming and outgoing waves carrythe same amount of energy because neither boundary conditions allow thestring to do any work on the support Other end conditions can be specified,for example, corresponding to an attached end mass, a spring or a dashpot

or a combination thereof

Mathematically, all these conditions can be analysed by equating thevertical component of the string tension to the forces on these elements For

instance, if the string has a non-negligible mass m attached at x=0, the

boundary condition reads

(8.25)

or, say, for a spring with elastic constant k0

(8.26)

Enforcing such boundary conditions on the general d’Alembert solutionmakes the problem somewhat more complicated However, on physicalgrounds, we can infer that the incident wave undergoes considerable distortionduring the reflection process More frequently, the reflection characteristics

of boundaries are analysed by considering the incident wave as pure harmonic,thus obtaining a frequency-dependent relationship for the amplitude andphase of the reflected wave

8.3 Free vibrations of a finite string: standing waves and

normal modes

Consider now a string of finite length that extends from x=0 to x=L, is fixed

at both ends and is subjected to an initial disturbance somewhere along itslength When the string is released, waves will propagate both toward theleft and toward the right end At the boundaries, these waves will be reflected

back into the domain [0, L] and this process, if no energy dissipation occurs,

will continually repeat itself In principle, a description of the motion of thestring in terms of travelling waves is still possible, but it is not the mosthelpful In this circumstance it is more convenient to study standing waves,whose physical meaning can be shown by considering, for example, twosinusoidal waves of equal amplitude travelling in opposite directions, i.e thewaveform

(8.27a)

which, by means of familiar trigonometric identities, can be written as

(8.27b)Two interesting characteristics of the waveform (8.27b) need to be pointedout:

1 All points x j of the string for which sin(kx j)=0 do not move at all times,

i.e y(x j , t)=0 for every t These points are called nodes of the standing

wave and in terms of the waveform (8.27a), we can say that wheneverthe crest of one travelling wave component arrives there, it is alwayscancelled out by a trough of the other travelling wave

2 At some specified instants of time that satisfy all points x of the

string for which reach simultaneously the zero position and theirvelocity has its greatest value At other instants of time, when

all the above points reach simultaneously their individual maximum

amplitude value A sin(kx), and precisely at these times their velocity is zero Among these points, the ones for which sin(kx)=1 are alternatively crests

and troughs of the standing waveform and are called antinodes.

In order to progress further along this line of reasoning, we must investigatethe possibility of motions satisfying the wave equation in which all parts ofthe string oscillate in phase with simple harmonic motion of the samefrequency From the discussions of previous chapters, we recognize thisstatement as the definition of normal modes

The mathematical form of eq (8.27b) suggests that the widely adoptedapproach of separation of variables can be used in order to find standing-wave, or normal-mode, solutions of the one-dimensional wave equation

So, let us assume that a solution exists in the form y(x, t)=u(x)z(t), where

u is a function of x alone and z is a function of t alone On substituting this

solution in the wave equation we arrive at

which requires that a function of x be equal to a function of t for all x and

t This is possible only if both sides of the equation are equal to the same

constant (the separation constant), which we call –ω2 Thus

(8.28)

The resulting solution for y(x, t) is then

(8.29)where and it is easy to verify that the product (8.29) results in aseries of terms of the form (8.27b) The time dependent part of the solutionrepresents a simple harmonic motion at the frequency ω, whereas for the

space dependent part we must require that

(8.30)

because we assumed the string fixed at both ends Imposing the boundaryconditions (8.30) poses a serious limitation to the possible harmonic motions

because we get A=0 and the frequency equation

(8.31)which implies (n integer) and is satisfied only by those values of

frequency ωn for which

(8.32)

These are the natural frequencies or eigenvalues of our system (the flexible string

of length L with fixed ends) and, as for the MDOF case, represent the frequencies

at which the system is capable of undergoing harmonic motion Qualitatively,

an educated guess about the effect of boundary conditions could have led us toargue that, when both ends of the string are fixed, only those wavelengths forwhich the ‘matching condition’ (where n is an integer) applies can satisfy the requirements of no motion at x=0 and x=L This is indeed the case

The first four patterns of motion (eigenfunctions) are shown in Fig 8.2:

the motion for n=1, 3, 5,…result in symmetrical (with respect to the point

x=L/2) modes, while antisymmetrical modes are obtained for n=2, 4, 6,…

So, for a given value of n, we can write the solution as

(8.33)where, for convenience, the constant of the space part has been absorbed in

the constants A n and B n Then, given the linearity of the wave equation, thegeneral solution is obtained by the superposition of modes:

(8.34)

where the (infinite) sets of constants A n and B n represent the amplitudes of thestanding waves of frequency ωn The latter quantities, in turn, are related to theallowed wavenumbers by the equation On physical grounds, since weobserved that there is no motion at the nodes and hence no energy flow betweenneighbouring parts of the string, one could ask at this point how a standingwave gets established and how it is maintained To answer this question wemust remember that a standing wave represents a steady-state situation; duringthe previous transient state (which, broadly speaking, we may call the ‘travellingwave era’) the nodes move and allow the transmission of energy along the string.Moreover, it should also be noted that nodes are not perfect in real strings

Fig 8.2 Vibration of a string with fixed ends: (c) third and (d) fourth modes.

which we recognize as Fourier series with coefficients A n and ωnB n ,

respectively Following the standard methods of Fourier analysis, we multiply

both sides of eqs (8.35) by sin k m x and integrate over the interval [0, L] in

order to obtain, by virtue of

(8.36)

the expressions

(8.37)

which establish the motion of our system Note that eq (8.34) emphasizesthe fact that the string is a system with an infinite number of degrees offreedom, where, in the normal mode representation, each mode represents asingle degree of freedom; furthermore, from the discussion above it is clearthat the boundary conditions determine the mode shapes and the naturalfrequencies, while the initial conditions determine the contribution of eachmode to the total response (or, in other words, the contribution of eachmode to the total response depends on how the system has been started intomotion) If, for example, we set the string into motion by pulling it aside atits centre and then letting it go, the ensuing free motion will comprise onlythe odd (symmetrical) modes; even modes, which have a node at the centre,will not contribute to the motion

A final important result must be pointed out: when the motion is written

as the summation of modes (8.34), the total energy E of the string—i.e the

integral of the energy density

over the length of the string—is given by

(8.38)

where it is evident that each mode contributes independently to the totalenergy, without any interaction with other modes (recall Parseval’s theoremstated in Chapter 2) The explicit calculation of (8.38), which exploits therelation (8.36) together with its cosine counterpart

(8.39)

is left to the reader

We close this section with a word of caution Traditionally, cable vibrationobservations of natural frequencies and mode shapes are compared to those

of the taut string model However, a more rigorous approach must take intoaccount the axial elasticity and the curvature of the cable (for example, power-line cables hang in a shape called ‘catenary’ and generally have a sag-to-span ratio between 0.02 and 0.05) and may show considerable discrepanciescompared to the string model In particular, the natural frequencies and modeshapes depend on a cable parameter (E=Young’s modulus,

A=cross-sectional area, ρg=cable weight per unit volume, L0=half-span length)and on the sag-to-span ratio The interested reader may refer, for example,

to Nariboli and McConnell [3] and Irvine [4]

8.4 Axial and torsional vibrations of rods

In the preceding sections we considered in some detail a simple case of continuoussystem—i.e the flexible string However, in the light of the fact that our interestlies mainly in natural frequencies and mode shapes, we note that we can explorethe existence of solutions in which the system executes synchronous motionsjust by assuming a simple harmonic motion in time and asking what kind ofshape the string has in this circumstance This amounts to setting

and substituting it into the wave equation to arrive directly atthe first of eqs (8.28) so that, by imposing the appropriate boundary conditions(fixed ends), we arrive at the eigenvalues (8.32) and the eigenfunctions

(8.40)

where C n are arbitrary constants which, a priori, may depend on the index n.

If now we consider the axial vibration of a slender rod of uniform density ρ

and cross-sectional area A in presence of a dynamically varying stress field σ(x, t),

we can isolate a rod element as in Fig 8.3 and write Newton’s second law as

(8.41)

where y(x, t) is the longitudinal displacement of the rod in the x-direction.

If we assume the rod to behave elastically, Hooke’s law requires that

where is the axial strain, and upon substitution in

eq (8.41) we get

(8.42)

and physical dimensions, the discussions of the preceding sections apply alsofor the cases above.

If now we separate the variables and assume a harmonic solution in time,

we arrive at the ordinary differential equations

(8.46)

where, for convenience, we called u(x) the spatial part of the solution in

both cases (8.42) and (8.45) and the parameter γ2 is equal to ω2ρ/E for the

first of eqs (8.46) and to ω2ρ/G for the second.

Again, in order to obtain the natural frequencies and modes of vibrations

we must enforce the boundary conditions on the spatial solution

(8.47)One of the most common cases of boundary conditions is the clamped-free(cantilever) rod where we have

(8.48)

so that substitution in (8.47) leads to

which, in turn, translates into meaning that the naturalfrequencies are given by

(8.49)

for the two cases of axial and torsional vibrations, respectively Like the

eigenvectors of a finite DOF system, the eigenfunctions are determined towithin a constant In our present situation

(8.50)

where u n (x) must be interpreted as an axial displacement or an angle,

depending on the case we are considering

If, on the other hand, the rod is free at both ends, the boundary conditions

the free-free rod the solution with n=0 is a perfectly acceptable root and

does not correspond to no motion at all (see the taut string for comparison)

In fact, for n=0 we get and, from eq (8.46), so that

where C1 and C2 are two constants whose value is irrelevantfor our present purposes Enforcing the boundary conditions (8.51) gives

which corresponds to a rigid body mode at zero frequency As for the discretecase, rigid-body modes are characteristic of unrestrained systems

For the time being, we do not consider other types of boundary conditionsand we turn to the analysis of a more complex one-dimensional system—thebeam in flexural vibration This will help us generalize the discussion oncontinuous systems by arriving at a systematic approach in which thesimilarities with discrete (MDOF) systems will be more evident

In fact, if now in eq (8.46) we define replace the differentialoperator with a stiffness matrix and the mass density ρ with a mass matrix,

we may note an evident formal similarity with a matrix (finite-dimensional)eigenvalue problem Moreover, it is not difficult to note that the same applies

to the case of the flexible string

8.5 Flexural (bending) vibrations of beams

Consider a slender beam of length L, bending stiffness EI(x) and mass per unit length µ(x) We suppose further that no external forces are acting.

By invoking the Euler-Bernoulli theory of beams—namely that plane sections initially perpendicular to the axis of the beam remain plane andperpendicular to the neutral axis during bending—and by deliberately neglectingthe (generally minor) contribution of rotatory inertia to the kinetic energy, wecan refer back to Example 3.2 to arrive at the governing equation of motion,

cross-(8.54)

where the function y(x, t) represents the transverse displacement of the beam.

Equation (8.54) is a fourth-order differential equation to be satisfied at every

point of the domain (0, L) and it is not in the form of a wave equation If,

for simplicity, we also assume that the beam is homogeneous throughout itslength, eq (8.54a) becomes

(8.55a)

or, alternatively

(8.55b)

where ρ is the mass density

Note that a does not have the dimensions of velocity We do not enter

into the details of flexural wave propagation in beams, but is worth notingthat substitution of a harmonic waveform into eq (8.55) leads tothe dispersion relation and since the phase velocity of wavepropagation is given by it follows that

(8.56)

which shows that the phase velocity depends on wavelength and impliesthat, as opposed to the cases of the previous sections, a general nonharmonicflexural pulse will suffer distortion as it propagates along the beam Energy,

in this case, propagates along the beam at the group velocity

which can be shown (e.g Kolsky [5] and Meirovitch [6]) to be related to thephase velocity by

Furthermore, eq (8.56) predicts that waves of very short wavelength (veryhigh frequency) travel with almost infinite velocity This unphysical result isdue to our initial simplifying assumptions—i.e the fact that we neglectedrotatory inertia and shear deformation—and the price we pay is that theabove treatment breaks down when the wavelength is comparable with thelateral dimensions of the beam Such restrictions must be kept in mind alsowhen we investigate the natural frequencies and normal bending modes ofthe beam unless, as it often happens, our interest lies in the first lower modesand/ or the beam cross-sectional dimensions are small compared to its length.When this is the case, we can assume a harmonic time-dependent solution

substitute it into eq (8.55) and arrive at the fourth-orderordinary differential equation

(8.57)

where we define

We try a solution of the form and solve the characteristic equation

which gives and so that

(8.58)

where the arbitrary constants A j or C j (j=1, 2, 3, 4) are determined from

the boundary and initial conditions The calculation of natural frequenciesand eigenfunctions is just a matter of substituting the appropriateboundary conditions in eq (8.58); we consider now some simple andcommon cases

Case 1 Both ends simply supported (pinned-pinned configuration)

The boundary conditions for this case require that the displacement u(x)

and bending moment vanish at both ends, i.e

(8.59)

where, in the light of the considerations of Section 5.5, we recognize that thefirst of eqs (8.59) are boundary conditions of geometric nature and hencerepresent geometric or essential boundary conditions On the other hand,the second of eqs (8.59) results from a condition of force balance and hencerepresents natural or force boundary conditions

Substitution of the four boundary conditions in eq (8.58) leads to

and to the frequency equation

(8.60)which implies and hence

(8.61)

The eigenfunctions are then given by

(8.62)

and have the same shape as the eigenfunctions of a fixed-fixed string

Case 2 One end clamped and one end free (cantilever configuration)

Suppose that the end at x=0 is rigidly fixed (clamped) and the end at x=L is free; then the boundary conditions require that the displacement u(x) and slope du/dx both vanish at the clamped end, i.e.

(8.63a)

and that the bending moment and shear force both vanish atthe free end, i.e.

(8.63b)

We recognize eqs (8.63a) as geometric boundary conditions and eqs.(8.63b)

as natural boundary conditions Substitution of eqs (8.63a and b) into (8.58)gives

Note that for the approximation is generally good Theeigenfunctions can be obtained from the first three of eqs (8.64a) which give

and, upon substitution into eq (8.58) lead to

(8.67)

where C1 is arbitrary One word of caution: because of the presence ofhyperbolic functions, the frequency equation soon becomes rapidly divergentand oscillatory with zero crossings that are nearly perpendicular to the γL-

axis For this reason it may be very hard to obtain the higher eigenvaluesnumerically with an unsophisticated root-finding algorithm

Case 3 Both ends clamped (clamped-clamped configuration)

All the boundary conditions are geometrical and read

(8.68)

We can follow a procedure similar to the previous case to arrive at

and to the frequency equation

(8.69)The first six roots of eq (8.69) are

(8.70)

which, for can be approximated by Note that the root

of eq (8.69) implies no motion at all, as the reader can verify by solving

eq (8.57) with and enforcing the boundary condition on the resultingsolution

As in the previous case, the eigenfunctions can be obtained from the

relationships among the constants C j and are given by

(8.71)where now

(8.72)

Case 4 Both ends free (free-free configuration)

The boundary conditions are now all of the force type, requiring that bending

moment and shear force both vanish at x=0 and x=L, i.e.

(8.73)

which, upon substitution into eq (8.58) give

Equating the determinant of the 4×4 matrix to zero yields the frequencyequation

(8.74)which is the same as for the clamped-clamped case (eq (8.69)), so that theroots given by eq (8.70) are the values which lead to the first lower frequenciescorresponding to the first lower elastic modes of the free-free beam Theelastic eigenfunctions can be obtained by following a similar procedure as inthe previous cases They are

(8.75)

where is the same as in eq (8.72) In this case, however, the system isunrestrained and we expect rigid-body modes occurring at zero frequency,i.e when On physical grounds, we are considering only lateraldeflections and hence we expect two such modes: a rigid translationperpendicular to the beam’s axis and a rotation about its centre of mass.This is, in fact, the case Substitution of in eq (8.57) leads to

(8.76a)

where A, B, C are constants Imposing the boundary conditions (8.73) to

the solution (8.76a) yields

(8.76b)which is a linear combination of the two functions

(8.77)

where we omitted the constants because they are irrelevant for our purposes

It is not difficult to interpret the functions (8.77) on a mathematical and on

a physical basis: mathematically they are two eigenfunctions belonging tothe eigenvalue zero and, physically, they represent the two rigid-body modesconsidered above

We leave to the reader the case of a beam which is clamped at one end andsimply supported at the other end The frequency equation for this case is

(8.78)and its first roots are

(8.79)

Also, note that we can approximate

Finally, one more point is worthy of notice For almost all of theconfigurations above, the first frequencies are irregularly spaced; however,

as the mode number increases, the difference between the two frequencyparameters and γnL approaches the value π for all cases This result

is general and indicates, for higher frequencies, an insensitivity to theboundary conditions

8.5.1 Axial force effects on bending vibrations

Let us consider now a beam which is subjected to a constant tensile force T

parallel to its axis This model can represent, for example, either a stiff string

or a prestressed beam

On physical grounds we may expect that the model of the beam with noaxial force should be recovered when the beam stiffness is the dominantrestoring force and the string model should be recovered when tension is byfar the dominant restoring force This is, in fact, the case The governingequation of motion for the free motion of the system that we are consideringnow is

(8.80)

Equation (8.80) can be obtained, for example, by writing the two equilibriumequations (vertical forces and moments) in the free-body diagram of Fig 8.4and noting that, from elementary beam theory,

Alternatively, we can write the Lagrangian density

and arrive at eq (8.80) by performing the appropriate derivatives prescribed

in eq (3.109) The usual procedure of separation of variables leads to asolution with a harmonically varying temporal part and to the ordinarydifferential equation

(8.81)

Fig 8.4 Beam element with tensile axial force (schematic free-body diagram).

where, as in the previous cases, we called u(x) the spatial part of the solution.

If now we let eq (8.81) yields

where is positive and is negative It follows that we have the fourroots ±η and where we defined

(8.82)

The solution of eq (8.81) can then be written as

(8.83)which is formally similar to eq (8.58) but it must be noted that the hyperbolicfunctions and the trigonometric functions have different arguments We cannow consider different types of boundary conditions in order to determinehow the axial force affects the natural frequencies The simplest case is whenboth ends are simply supported; enforcing the boundary conditions (8.59)leads to and to the frequency equation

(8.84)which results in because for any nonzero value of η.The allowed frequencies are obtained from this means

(8.85a)

which can be rewritten as

(8.85b)

where it is more evident that for small values of the nondimensional ratio

(i.e when ) the tension is the most important restoring

force and the beam behaves like a string At the other extreme—when R is

large—the stiffness is the most important restoring force and in the limit of

we return to the case of the beam with no axial force

In addition to the observations above, note also that:

• In the intermediate range of values of R, higher modes are controlled by stiffness because of the n2 factor under the square root in eq (8.85b)

• The eigenfunctions are given by (enforcement of the boundary conditions

leads also to C2=0)

(8.86)which have the same sinusoidal shape of the eigenfunctions of the beamwith no tension (although here the sine function has a different argument).The conclusion is that an axial force has little effect on the mode shapes butcan significantly affect the natural frequencies of a beam by increasing theirvalue in the case of a tensile force or by decreasing their value in the case of

a compressive force In fact, the effect of a compressive force is obtained by

just reversing the sign of T In this circumstance the natural frequencies can

be conveniently written as

(8.87)

where we recognize as the critical Euler buckling load When

the lowest frequency goes to zero and we obtain transverse buckling

In the case of other types of boundary conditions the calculations are, ingeneral, more involved For example, we can consider the clamped-clamped

configuration and observe that placing the origin x=0 halfway between the

supports divides the eigenfunctions into even functions, which come fromthe combination

and odd functions, which come from the combination

In either case, if we fit the boundary conditions at x=L/2, they will also fit at x=–L/2 For the even functions the boundary conditions

lead to the equation

are obtained from eq (8.88b)

8.5.2 The effects of shear deformation and rotatory inertia (Timoshenko beam)

It was stated in Section 8.5 that the Euler-Bernoulli theory of beams providessatisfactory results as long as the wavelength is large compared to thelateral dimensions of the beam which, in turn, may be identified by the radius

of gyration r of the beam section Two circumstances may arise when the

above condition is no longer valid:

1 The beam is sufficiently slender (say, for example, ) but we areinterested in higher modes

2 The beam is short and deep

In both cases the kinematics of motion must take into account the effects ofshear deformation and rotatory inertia which—from an energy point ofview—result in the appearance of a supplementary term (due to sheardeformation) in the potential energy expression and in a supplementary term(due to rotatory inertia) in the kinetic energy expression

Let us consider the effect of shear deflection first Shear forces t result in an

angular deflection θ which must be added to the deflection ␺ due to bendingalone Hence, the slope of the elastic axis ∂y/∂x is now written as

(8.89)

and the relationship between bending moment and bending deflection (fromelementary beam theory) now reads

(8.90)

Moreover, the shear force Q is related to the shear deformation θ by

(8.91)

where G is the shear modulus, A is the cross-sectional area and is an

adjustment coefficient (sometimes called the Timoshenko shear coefficient)whose value must generally be determined by stress analysis considerationsand depends on the shape of the cross-section In essence, this coefficient isintroduced in order to satisfy the equivalence

and accounts for the fact that shear is not distributed uniformly across thesection For example, for a rectangular cross-section and othervalues can be found in Cowper [7]

In the light of these considerations, the potential energy density consists

of two terms and is written as

(8.92)

where in the last term we take into account eq (8.89)

On the other hand, the kinetic energy density must now incorporate aterm that accounts for the fact that the beam rotates as well as bends If we

call J the beam mass moment of inertia density, the expression for the kinetic

energy density is written as

(8.93)

Moreover, J is related to the cross-section moment of inertia I by

(8.94)

where is the beam mass density and is the radius of gyration

of the cross-section Taking eqs (8.92) and (8.93) into account, we are now

in the position to write the explicit expression of the Lagrangian density

(8.95)

and perform the appropriate derivatives prescribed in eq (3.109) to arrive at

the equation of motion In this case, however, both y and ␺ are independent variables and hence we obtain two equations of motion As a function of y,

the Lagrangian density is a function of the type where, followingthe notation of Chapter 3, the overdot indicates the derivative with respect

to time and the prime indicates the derivative with respect to x.

So, we calculate the two terms