Applied Structural and Mechanical Vibrations 2009 Part 9 pot

Different boundary conditions lead to more complicated calculations: for example, if our plate is simply supported at r=R the boundary conditions to be imposed on the solution 8.150 are,

The eigenfuncions are written

(8.154)

where the constant A—which, a priori, can depend on both n and m—can

be fixed by means of normalization

Different boundary conditions lead to more complicated calculations: for

example, if our plate is simply supported at r=R the boundary conditions to

be imposed on the solution (8.150) are, from eq (8.147a)

at r=R, and in polar coordinates the bending moment M r is written explicitly

Things are even worse for a completely free plate; in fact, in this case theboundary conditions read (eq (8.147c))

where M r is as above and the transverse shearing force Q r and the twisting

moment M rθ are given by

and we can express its solution as Obviously, it is now convenient

to adopt a system of rectangular coordinates so that the Laplacian andbiharmonic operators are written explicitly as

The function u1 satisfies the equation by separating the spacevariables and looking for a solution in the form we arrive

at the two equations

write the complete solution u(x, y) as

(8.157)

where the values of the constants A j and parameters α and ß depend on theboundary conditions The simplest case is when all edges are simply supportedand we must enforce the boundary conditions

(8.158a)

(8.158b)

where the conditions on the second derivative are obtained (eq (8.147a)) by

noting that, in rectangular coordinates, the bending moments M x and M y

are given by

(8.159)

By inserting the conditions of eqs (8.158a and b) into the solution (8.157)

we obtain that only A1 is different from zero so that

requirement yields the following mass-orthonormaleigenfunctions:

Let us investigate this point a bit further If we take a step back and writethe solution in the form substitution into the eigenvalue

(8.164)and we can separate it into two independent equations if

(8.165a)

Fig 8.8 A few lower-order modes for a rectangular plate simply supported on all

edges.

(8.165b)

or both Let us suppose that eq (8.165a) holds, this implies and

(8.166)

If now we consider the boundary conditions of simply supported (SS), clamped

(C) and free (F), along x=0 we have

which come from the expression of eqs (8.147a–c) in rectangular coordinates

by noting that Mx is given in eq (8.159) and that the Kirchhoff condition reads

(8.167)

A set of similar conditions apply at x=a Now, it is not difficult to show that

only the SS conditions can be satisfied by a function of the form (8.166)and, more specifically, we need a sine function which satisfies i.e

If also eq (8.165b) holds, all sides are simply supported and ananalogous line of reasoning yields Moreover, substitution of eqs(8.165a and b) and of into eq (8.164) yields

(8.168)which can be solved for the frequency to give eq (8.162)

When the edges at x=0 and x=a are simply supported and we exclude the

case of the other two edges simply supported, we are left with five possibilitiesfor which we must solve the equation

(8.169)whose solution depends on whether or However,even if separation of variables is possible in these latter cases, the information

on natural frequencies and mode shapes is not easily obtained and theinterested reader is urged to refer to the wide body of specific literature onthe subject

A final comment of general nature can be made on the orthogonality ofthe eigenfunctions From our preceding discussion, we know that mass andstiffness orthogonality are guaranteed by the symmetry of the eigenvalueproblem; however, it may be of interest to approach the problem from a

different point of view Let us consider two different eigenfunctions, say u nm and u lk : the equations

(8.170)

are identically satisfied Now, since from static classical plate theory thedifferential equation of static deflection is written we caninterpret the first of eqs (8.170) as the equation of the static deflection of theplate under the action of the load and, by the same token, we

can say that our plate assumes the deflected shape u lk when the load

is acting In other words, the loads q1 and q2 represent two

systems of generalized forces while u nm and u lk are the displacements caused

by such forces

We now invoke Betti’s theorem which states that:

For a linearly elastic structure the work done by a system q1 of forces

under a distortion caused by a system q2 of forces equals the work

done by the system q2 under a distortion caused by the system q1

In our case, this translates mathematically into

For our purposes, we can finish here our treatment on the free vibration

of continuous systems referring the interested reader to the specific literature

on the subject In particular, an interesting discussion on one-dimensional

eigenvalue problems in which boundary conditions contain the eigenvaluecan be found in Humar [19] and Meirovitch [6, 9].

8.9 Forced vibrations and response analysis: the modal

approach

The action of external time-varying loads on a continuous system leads to anonhomogeneous partial differential equation In general, in the light of thepreceding sections, it is not difficult to obtain the governing differentialequation also because—once the mass and stiffness operators have beenintroduced—its formal structure is similar to the matrix equation governingthe forced vibrations of MDOF systems However, now the boundaryconditions come into play and we must pay due attention to them.Two general methods of solutions can be identified:

1 integral transform methods, which are particularly well suited to systemswith infinite or semi-infinite extension in space (note, in fact, that inthese cases the concept of normal mode loses its meaning) and forproblems with time-dependent boundary conditions;

2 the mode superposition method

Here, we concentrate our attention on method 2, which we can call ‘themodal approach’ The relevant equation of motion is written as

(8.172)

where and, for brevity of notation, we indicate with x the set of space variables Note that now f is a forcing function representing

an external action (it is the counterpart of vector f of eq (7.1)) and must not

be confused with the f functions of the preceding sections in this chapter,

where this symbol was often adopted to identify a general function to beused for the specific needs of that part only Equation (8.172) must be

supplemented by the set of p boundary conditions

(8.173)

Now, by virtue of the results given in Section 8.7.1, the general response w

can be expressed in terms of a superposition of eigenfunctions φi multiplied

by a set of time-dependent generalized coordinates y i (t), i.e.

(8.174)

which is the infinite-dimensional counterpart of eq (7.2) If we substitute thesolution (8.174) into eq (8.172) and then take the inner product of theresulting equation by we get

which, owing to the orthogonality relationships (8.128), reduces to the infiniteset of 1-DOF uncoupled equations

(8.175)

whose solution is given by (Chapter 5 and eq (7.7))

(8.176)

where y j(0) and j (0) are the initial jth modal displacement and velocity,

respectively Moreover, it is not difficult to show that we can obtain thesequantities from the inner products

(8.177)

where and are the initial conditions in physicalcoordinates The analogy with eq (7.5b) is evident

A few remarks can be made at this point

1 First, it is apparent that the first step of the whole approach is the solution

of the differential (symmetrical) eigenvalue problem satisfyingthe appropriate boundary conditions The resulting set of eigenvalues andorthonormal eigenfunctions make it possible to express the general solution

to the free vibrations problem (i.e eq (8.172) with f=0) as in eq (6.50),

which reads

(8.178)

For example, suppose that we want to consider the longitudinal motion of

a uniform clamped-free rod of length L and µ mass per unit length Let the

initial conditions be such that and Since the massorthonormal functions are given by

we can calculate the rod free motion by means of eq (8.178) To this end wemust first evaluate the inner product

(the calculation is not difficult and is left to the reader) and then obtain thefree motion as

where we know from eq (8.49) that

2 When expressed in normal coordinates, the kinetic and potential energy

of free vibration assume the particularly simple forms

(8.179)

so that the Lagrangian has no coupling terms between the coordinates and

is simply the Lagrangian function of an infinite number of independentharmonic oscillators

3 The final remark has to do with the apparent discrepancy according towhich our system may be equally well described by a continuous system of

coordinates w(x, t) or by a discrete one, i.e y j (t) The general feeling is that

this second set of coordinates cannot describe the same number of degrees

of freedom as the first one, although this number is infinite in both cases.However, it must be noted that, broadly speaking, the definition of normalcoordinates itself somehow incorporates the boundary conditions from the

outset, and this is not the case for w(x, t) In fact, the coordinates w(x, t) could be a priori chosen as completely independent of each other and this

would allow us—in principle—to describe also discontinuous motions ofour system By contrast, the expansion in normal modes requires that thefunctions representing the point-by-point displacement of our system bereasonably well-behaved

So, although the normal mode description is, as a matter of fact,mathematically more restrictive, it really does not put any significantrestriction on the physical problem because the boundary conditions andthe good behaviour of the displacement functions are dictated by the physicalnature of our system and, ultimately, by the physics of the phenomena weare trying to describe Also, note that for MDOF systems there was no need

to make a similar remark because, whatever the coordinate system we choose

to adopt, the elements of the stiffness matrix automatically take care of theboundary conditions

We can now go back to the main discussion of this section and note that

gave special emphasis to the modal impulse response functions and to themodal frequency response functions, also showing their relationships withimpulse response functions and frequency response functions expressed inphysical coordinates We want now to extend those concepts to the case ofdistributed parameter systems

Let us consider an undamped system: from the preceding chapters we

know that the jth modal impulse response function h j (t) is the solution of the

equation

(8.180)where δ is the Dirac’s delta function Moreover, we also know that (eqs(5.7a) and (5.7b))

(8.181)

where the term representing the jth generalized mass does not appear

in the denominator of the right-hand side of eq (8.181) because we areconsidering mass-orthonormal eigenfunctions

Now, if for simplicity we assume eq (8.176) reduces to

(8.182)

If we further assume that the excitation is an impulse applied at position

x=x k at time we can t=0 write

so that

(8.183)

which must be substituted in eq (8.182) to obtain y j (t).

Then, from eq (8.174) we can obtain the response in physicalcoordinates as

(8.184)

More specifically, we can consider the response at the point x=x m , i.e.

(8.185a)

and note that this is just the displacement response of point x m to a unit

impulse applied at point x k at the instant t=0 In other words, eq (8.185a)

represents the impulse response function so that we can write

(8.185b)

which is the continuous systems counterpart of eq (7.37b) where h jk denoted

the physical coordinates (displacement) impulse response function at the jth DOF due to an unit force impulse applied at the kth DOF at t=0.

Alternatively, we can consider the frequency domain and note that the jth

modal frequency response function (receptance in this case) can be obtained

by assuming a forcing function in sinusoidal form This means that

so that eqs (8.175) become

(8.186)hence, assuming a response which is also in sinusoidal form, we have

(8.187a)

so that the jth receptance FRF is

(8.187b)

If we now consider a harmonic forcing function of unit amplitude applied

at the point x=x k , i.e.

we have the modal response is then

where is, by definition, the physical coordinates receptance

function corresponding to points x m and x k , we get from eq (8.189b)

of the second of eqs (8.191) to a general FRF function H(ω) other than

receptance is straightforward and reads

(8.192)Note that, for reasons outlined in Section 7.4.1, we only consider FRFfunctions in the forms of receptances, mobilities and accelerances; hence, for

our purposes and unless otherwise stated, the symbol H will always be tacitly

assumed to mean a FRF functions in one of these forms

For its importance in modal testing, we refer to eqs (8.189b) and (8.190)

to point out that a concentrated excitation force applied at a point x k where

for some index j results in no excitation of all these modes In

other words, for example, if we excite a beam by means of a concentratedload at mid-span, all even natural frequencies and modes

will not contribute to the measured response By the same token,

if we pluck at mid-length a guitar string tuned to give the note A at 440 Hz,the second (A at 880 Hz), fourth (A at 1760 Hz) etc harmonics will bemissing and the sound we hear will be the superposition of the fundamentalnote A (440 Hz) plus the third (E at 1320 Hz), the fifth (C# at 2200 Hz) etc.harmonics On the other hand, if we pluck the same string at one-third of itslength we will still hear the same fundamental note A at 440 Hz, but theharmonic content of the sound will be different

8.9.1 Forced response of continuous systems: some examples

Example 8.1 Consider a vertical clamped-free beam of length L, mass per

unit length µ, and flexural rigidity EI subjected to an excitation in the form

of a lateral base displacement It is easy to realize that thissituation, for example, can be used as a first approximation to model theresponse of a tall, slender building to an earthquake excitation For our

purposes, we ignore the fact that the definition of an appropriate g(t) is very

difficult in this case and it is one of the most uncertain steps of the analysis

To solve this problem, we need to consider eq (8.55a); we write the beam

displacement y(x, t) as

(8.193)

and substitute it into eq (8.55a) Note that u(x, t) represents the displacement

of the beam relative to the rigid-body translation of the ground We get

(8.194)

where it is evident that the inertia forces depend on the total motion, whereas

the stiffness (and damping, if it were included in the analysis) forces dependonly on the relative motion The r.h.s of eq (8.194) is the effective earthquake

force and it is usually indicated with the symbol f eff In the light of the

discussion of preceding sections, it follows that—assuming the system initially

at rest—we have (eq (8.182))

and hence

where ωj are given by eq (8.66b) and the φj are the eigenfunctions (8.67), inwhich the constant C1 has been chosen so as to make them mutually mass-orthonormal

The relative response in physical coordinates is then obtained as thesuperposition

(8.195)

where, in general, for seismic excitation the minus sign in (8.195) is irrelevantand—owing to the complicated expressions of the eigenfunctions and of theground acceleration —the integrations must be performed numerically

In general, only a few terms of the series must be considered in order toobtain a satisfactory representation of the actual response so that, in the

end, we are brought back to the case of an n-DOF system, where only a finite number (n) of modal coordinates is needed to describe the response The total response y(x, t) is finally obtained according to eq (8.193).

A simpler case arises if we consider the longitudinal motion of our system;

the excitation g(t) is now a vertical displacement and the relevant equation

of motion is eq (8.42) The general form of the relative displacement is stillgiven by eq (8.195) but the eigenvalues ωj are given by the first of eq (8.49)and the eigenfunctions are given by eq (8.50), where In thiscircumstance, it is not difficult to show that the space integration in (8.195)results in

(8.196)

and only the time integration needs to be performed numerically

Example 8.2 Let us now consider a uniform clamped-free rod of length L

and mass per unit length µ excited by a tip load at the free end, i.e.

If the rod is at rest before the excitation occurs the first twoterms on the r.h.s of eq (8.176) are zero and

(8.197)

so that eq (8.176) reduces to

(8.198)

form of p(t) is a unit step (Heaviside) function θ(t), i.e.

we can substitute θ(t) into eq (8.198), perform the integration by noting that

since we have assumed a unit force, the dimensions of w(x, t) are displacement

per unit force (i.e m/N)

At this point, it is interesting to note that the system above can be analysedeither as:

1 an excitation-free system with a time-dependent boundary condition at

x=L, or

2 a forced vibration problem with homogeneous boundary conditions

As stated at the beginning of the preceding section, free-vibration problemswith nonhomogeneous boundary conditions are often tackled by an integraltransform (Laplace or Fourier) approach; however, the modal approach canalso be adopted in consideration of the fact that a boundary value problem

of type (1) can usually be transformed into a boundary value problem oftype (2) (e.g Courant and Hilbert [20] or Mathews and Walker [21])

In general, it appears that in these cases a disadvantage of the modalapproach—which is essentially a ‘standing waves solution’—is that theresultant series converges quite slowly and many terms must be included inorder to achieve a reasonable accuracy By contrast, depending on how theinverse transformation is carried out, the Laplace transform method allows

us the possibility to obtain a solution either in terms of standing waves or interms of travelling waves (waves being reflected back and forth within therod) This latter possibility—the travelling wave approach—leads to a solution

in the form of a rapidly converging series, thus making this strategy moreattractive However, on physical grounds, we may argue that the time scale

in which we are interested suggests the type of solution to adopt; in fact, thetravelling wave solution converges rapidly when we consider the short-termresponse of our system whereas, if the long-term response is desired, moreand more terms are needed The situation is reversed for the modal solution:

as time progresses, the terms corresponding to higher modes die out because

of damping and we are left with a series in which, say, only the first two orthree terms have a significant contribution

We will not consider an integral transform strategy of solution here (theinterested reader is referred to Meirovitch [22]) but, using the rod exampleabove, we will show how a problem of type (1) can be transformed in aproblem of type (2)

Our rod problem can be formulated as a type (1) problem in the followingform:

(8.201a)

(8.201b)

where eq (8.201a) is the homogeneous equation of motion and eq (8.201b)are the boundary conditions Since one boundary condition (the second) isnonhomogeneous we assume the solution of our problem in the form

(8.202)

where the term —which we can define in compact notation as

—is a so-called ‘pseudostatic’ displacement brought about by the

boundary motion and v(x, t) is the displacement relative to the support displacement Mathematically, the function u st is chosen in such a way as to

make the boundary conditions for v(x, t) homogeneous On physical grounds, the usual assumption made for the choice of u st is that no inertia forces (i.e.,

no accelerations) are produced by the application of the support motion;

hence the name ‘pseudostatic’ For our case, this assumption implies that u st

obeys the equation

(8.203)

from which follows (provided that

(8.204)Moreover, given the expression (8.202), the boundary conditions (8.201b)become

from which—if we want homogeneous boundary conditions for v(x, t)—it

follows

(8.205)

Enforcing the boundary conditions (8.205) on the solution (8.204) leads to

(8.206)

The transformation of the problem (8.201) into a type (2) problem iscomplete when we determine the nonhomogeneous equation of motion for

the relative displacement v(x, t): this is simply accomplished by substituting

eq (8.202) into eq (8.201a) and results in

(8.207a)

where the r.h.s of eq (8.207) has clearly the dimensions of N/m and, for

short, can be indicated with the symbol f eff (effective force)

Equations (8.207a), (8.206) plus the homogeneous boundary conditions

(8.207b)

constitute our type (2) boundary value problem which fits into the scheme

of problems that can be more effectively tackled by the modal approach In

this light, we expand v(x, t) in a series of eigenfunctions and calculate the

normal coordinates as prescribed in eq (8.182) (note that, from eq (8.206),

Now, in the problem we are considering, we assumed that p(t) is the

Heaviside function θ(t); since (eq (2.67a) or (2.84)) the timeintegral of eq (8.209) becomes

where we take into account the properties of the Dirac’s delta function (eq

(2.69)) and the explicit expression of h j The final steps consist of substituting

this result in eq (8.209), writing explicitly the series expansion of v(x, t), i.e.

and putting it all back together into the solution (8.202) which becomes

(8.210)

This result must be compared with eq (8.200) and it is not difficult toshow that they are equal This is due to the fact that the function (π2x/8L)

can be expanded in a Fourier series as (the proof is left to the reader)

so that—after performing the product in eq (8.200)—the first term is exactly

(x/EA), i.e the function ␺(x) of the pseudostatic displacement The advantage

of including explicitly the pseudostatic displacement from the outset lies inthe more rapid convergence of the series (8.210) as compared to the series(8.200), the pseudostatic displacement representing the average position aboutwhich the vibration takes place

Example 8.3 In modal testing, we are often concerned with the response of

a given system to an impulse loading So, consider the rod of Example 8.2

subjected to a unit impulse applied at x=L at t=0 The response in physical coordinates at x=L is given by eqs (8.185) and reads

(8.211)

This result should be hardly surprising because we know from Chapter 5

(eq (5.42)) that the impulse response function is the time derivative of theHeaviside response function So, in this circumstance we could have ignored

eq (8.185) by simply noting that the result (8.211) can be obtained by

calculating the time derivative of eq (8.200) and by substituting x=L in it.

On the other hand, the receptance FRF can be obtained from eq (8.190):

at x=L this is

(8.212)

In the light of preceding chapters, we expect that h(L, L, t) and H(L, L,

ω) form a Fourier transform pair However, the Fourier transform of eq(8.211) does not exist, but we may note that the Laplace transform of eq(8.211) does exist and is given by

where s is the (complex) Laplace operator and can be expressed as

Hence, leaving aside mathematical rigour for a moment, we see that we canarrive at eq (8.212) by first taking the Laplace transform of eq (8.211) andthen letting This mathematical trick is just for purposes of illustrationand it would not be needed if the system had some amount of positive damping;

as a matter of fact, this is always the case for real systems whose time responseand FRFs (eq (8.211) and (8.212)) do not go to infinity when

Example 8.4 Consider now the case of a constant force P moving at a

constant velocity V along an Euler-Bernoulli beam simply supported at bothends The engineering importance of this case is evident because this examplecan be used to model a number of common situations, the simplest onebeing a heavy vehicle travelling across a bridge deck We also make thereasonable assumption that the mass of the vehicle is small in comparisonwith the beam mass (the bridge deck) and it does not alter appreciably itseigenvalues and eigenfunctions

Mathematically, the moving load can be represented as

(8.213)

and, with reference to eq (8.182), we obtain

so that, assuming the beam at rest at t=0, we get

(8.214)

where now

are the eigenfrequencies of our pinned-pinned Euler-Bernoulli beam

A double integration by parts in eq (8.214) leads to

(8.217a)

and the time of passage t j at these values of speed is given by

(8.217b)

so that, calling the fundamental period of vibration of the beam,

evaluate the response (8.216) at the critical speeds—i.e when —

we run into an indeterminate 0/0 situation However, we can use L’Hospital’srule and obtain

(8.218)

which, owing to the finiteness of the time of passage, is a bounded quantityand does not grow indefinitely with time

We will not consider an example of a two-dimensional system (say, amembrane or a plate) but it is evident that, besides the added mathematicalcomplexities, the extension of the modal approach to these systems followsthe same line of reasoning Obviously, now the expansion (8.174) must includeall modes and hence we must sum on all indexes, i.e the mode indexes andthe degeneracy indexes For example, recalling the eigenfunctions of a circularmembrane clamped at its outer edge (eqs (8.120) and (8.121)) the expansion

in series of eigenmodes reads

where the first series involves the Bessel functions of the first kind of orderzero (no degeneracy), while the second series involves the Bessel functions oforder and the twofold degeneracy (expressed by eq (8.121)) which,

in the expansion above, is taken into account by means of the summation

8.10 Final remarks: alternative forms of FRFs and the

introduction of damping

For the sake of completeness, two final remarks are needed before closing thischapter The first remark has to do with an alternative approach for finding aclosed-form solution of the frequency response function of a continuous system

In essence, if we assume a harmonic excitation in the form

measured at x=x m and the excitation is applied at the point x=x k—i.e

then we have

(8.221)

By this method, the solution is not obtained in the form of a series ofeigenfunctions (for example, like eq (8.200)) and one of the advantages isthat it can be profitably used in the case of support motion where, in general,

a set of orthonormal functions cannot be obtained

As an example of this method we can consider the longitudinal vibrations

of a vertical rod (see end of Example 8.1) subjected to a support harmonic

motion of unit amplitude If, to be consistent with eq (8.220), we call w(x,

t) the longitudinal rod displacement, the relevant equation of motion is

i.e in correspondence of the eigenvalues of a clamped-free rod.The same line of reasoning applies if we reconsider the system of example8.2 and we assume a harmonic excitation of unit amplitude at the free end

x=L Equations (8.222a) and (8.223a) still apply, but the boundary conditions

for W are now

of the eigenfunctions φj of the clamped—free rod and calculate the innerproduct

From eq (8.226) the FRF at x=L is obtained as

(8.228)

which must be compared with the series solution (8.212)

The second remark has to do with the fact that in none of the precedingsections have we taken into account the effect of energy dissipation However,the inclusion of damping—both in free and forced vibration conditions—leads to results that parallel closely the MDOF case

As stated on a number of occasions, damping is difficult to define and thegeneral assumption of viscous damping is mostly a matter of mathematicalconvenience rather than an effective explanation of the physical phenomenon

In this light, if we call w(x, t) the function that represents the displacement

of our continuous system, the general equation of motion (8.172) can bewritten as

(8.229)