Vibrations Fundamentals and Practice ch06

Vibrations Fundamentals and Practice ch06 Maintaining the outstanding features and practical approach that led the bestselling first edition to become a standard textbook in engineering classrooms worldwide, Clarence de Silva''s Vibration: Fundamentals and Practice, Second Edition remains a solid instructional tool for modeling, analyzing, simulating, measuring, monitoring, testing, controlling, and designing for vibration in engineering systems. It condenses the author''s distinguished and extensive experience into an easy-to-use, highly practical text that prepares students for real problems in a variety of engineering fields.

de Silva, Clarence W “Distributed Parameter Systems”

Vibration: Fundamentals and Practice

Clarence W de Silva

Boca Raton: CRC Press LLC, 2000

discrete-Generally, in practical vibrating systems, inertial, elastic, and dissipative effects are foundcontinuously distributed in one, two, or three dimensions Correspondingly, there are line structures,surface/planar structures, or spatial structures They will possess an infinite number of masselements, continuously distributed in the structure, and integrated with some connecting flexibility(elasticity) and energy dissipation In view of the connecting flexibility, each small element of masswill be able to move out of phase (or somewhat independently) with the remaining mass elements.

It follows that a continuous system (or a distributed-parameter system) will have an infinite number

of degrees of freedom and will require an infinite number of coordinates to represent its motion

In other words, when extending the concept of a finite-degree-of-freedom system as analyzedpreviously, an infinite-dimensional vector is needed to represent the general motion of a continuoussystem Equivalently, a one-dimensional continuous system (a line structure) will need one inde-pendent spatial variable, in addition to time, to represent its response In view of the need for twoindependent variables in this case — one for time and the other for space — the representation ofsystem dynamics will require partial differential equations (PDEs) rather than ordinary differentialequations (ODEs) Furthermore, the system will depend on the boundary conditions as well as theinitial conditions

The present chapter concerns vibration analysis of continuous systems Strings, cables, rods,shafts, beams, membranes, plates, and shells are example of continuous members In special cases,closed-form analytical solutions can be obtained for the vibration of these members A generalstructure may consist of more than one such member and, furthermore, boundary conditions may

be various, individual members may be nonuniform, and the material characteristics may beinhomogeneous and anistropic Closed-form analytical solutions would not be generally possible

in such cases Nevertheless, the insight gained by analyzing the vibration of standard members will

be quite beneficial in studying the vibration behavior of more complex structures The vibrationanalysis of a few representative continuous members is discussed in this chapter

The concepts of modal analysis can be extended from lumped-parameter systems to continuoussystems In particular, because the number of principal modes is equal to the number of degrees

of freedom of the system, a distributed-parameter system will have an infinite number of naturalmodes of vibration A particular mode can be excited by deflecting the member so that its elastic curve assumes the shape of the particular mode, and then releasing from this initial condition

When damping is significant and nonproportional, however, there is no guarantee that such aninitial condition could accurately excite the required mode A general excitation consisting of aforce or an initial condition will excite more than one mode of motion But, as in the case ofdiscrete-parameter systems, the general motion can be analyzed and expressed in terms of modalmotions, through modal analysis As discussed in Chapter 5, in a modal motion, the mass elementswill move at a specific frequency (the natural frequency), and bearing a constant proportion indisplacement (i.e., maintaining the mode shape), and passing the static equilibrium of the systemsimultaneously In view of this behavior, it is possible to separate the time response and spatialresponse of a vibrating system in a modal motion This separability is fundamental to modal analysis

of a continuous system Furthermore, in practice, all infinite number of natural frequencies andmode shapes are not significant and typically the very high modes can be neglected Such a modal-truncation procedure, although carried out by continuous-system analysis, is equivalent to approx-imating the original infinite-degree-of-freedom system by a finite-degree-of-freedom system Vibra-tion analysis of continuous systems can be applied in modeling, analysis, design, and evaluation

of such practical systems as cables; musical instruments; transmission belts and chains; containers

of fluid; animals; structures including buildings, bridges, guideways, and space stations; and transitvehicles, including automobiles, ships, aircraft, and spacecraft

6.1 TRANSVERSE VIBRATION OF CABLES

The first continuous member to be studied in this chapter is a string or cable in tension This is aline structure for which its geometric configuration can be completely defined by the position ofits axial line, with reference to a fixed coordinate line We will study the transverse (lateral) vibrationproblem; that is, the vibration in a direction perpendicular to its axis and in a single plane.Applications will include stringed musical instruments, overhead transmission lines (of electricpower or telephone signals), drive systems (belt drives, chain drives, pulley ropes, etc.), suspensionbridges, and structural cables carrying cars (e.g., ski lifts, elevators, overhead sightseeing systems,and cable cars)

As usual, some simplifying assumptions will be made for analytical convenience; but the resultsand insight obtained in this manner will be useful in understanding the behavior of more complexsystems containing cable-like structures The main assumptions are:

1 The system is a line structure The lateral dimensions are much smaller compared to thelongitudinal dimension (normally in the x direction)

2 The structure stays in a single plane, and the motion of every element of the structurewill be in a fixed transverse direction (y)

3 The cable tension (T) remains constant during motion In other words, the initial tension

is sufficiently large that the variations during motion are negligible

4 Variations in slope (θ) along the structure are small Hence, for example,

A general configuration of a cable (or string) is shown in Figure 6.1(a) Consider a smallelement of length dx of the cable at location x, as shown in Figure 6.1(b) The equation (Newton’ssecond law) of motion (transverse) of this element is given by

v(x,t) = transverse displacement of the cable

f(x,t) = lateral force per unit length of the cable

m(x) = mass per unit length of the cable

T = cable tension

θ = cable slope at location x.

Note that the dynamic loading f(x,t) may arise due to such causes as aerodynamic forces, fluiddrag, and electromagnetic forces, depending on the specific application

Using the small slope assumption, one obtains sinθ ≅ θ and sin(θ + dθ) ≅ θ + dθ with

On substitution of these approximations into equation (6.1) and canceling out dx, one obtains

(6.2)Now consider the case of free vibration, where f(x,t) = 0 Then,

2 2

v x t

v x t x

(6.4)Also assume that the cable is uniform so that m is constant

The solution to any equation of the form (6.3) will appear as a wave, traveling either in the forward(+ve x) or in the backward (–ve x) direction at speed c Hence, (6.3) is called the wave equationand c is the wave speed To prove this fact, first show that a solution to equation (6.3) can take theform

Clearly then v1 satisfies equation (6.3)

Now examine the nature of the solution v1(x – ct) It is clear that v1 will be constant when

x – ct = constant But, the equation x – ct = constant corresponds to a point moving along the

x-axis in the positive direction at speed c What this means is that the shape of the cable at t = 0will “appear” to travel along the cable at speed c This is analogous to the waves one observes in

a pond when excited by dropping a stone Note that the particles of the cable do not travel along x,and it is the deformation “shape” (the wave) that travels

Similarly, it can be shown that

dv dz

z x

v t

dv dz

z t

2 1 2 2 1

=

v x t( ), =v x2( +ct)

is also a solution to equation (6.3), and this corresponds to a wave that travels backward (–ve x

direction) at speed c The general solution, of course, will be of the form

(6.7)which represents two waves, one traveling forward and the other backward

As usual in modal analysis, one looks for a separable solution of the form

(6.8)for the cable/string vibration problem given by the wave equation (6.3) If a solution of the form of

equation (6.8) is obtained, it will be essentially a modal solution This should be clear from the

separability of the solution Specifically, at any given time t, the time function q(t) will be fixed, and

the structure will have a shape give by Y(x) Hence, at all times, the structure will maintain a particular

“shape” Y(x) and this will be a mode shape Also, at a given point x of the structure, the value of

the space function Y(x) will be fixed, and the structure will vibrate according to the time response

q(t) It will be shown that q (t) will obey the simple harmonic motion of a specific frequency This

is the natural frequency of vibration corresponding to that particular mode Note that, for a continuous

system, there will be an infinite number of solutions of the form (6.8), with different natural

frequencies The corresponding functions Y(x) will be “orthogonal” in some sense Hence, they are

called normal modes (normal meaning “perpendicular”) The systems will be able to move

indepen-dently in each mode, and this collection of solutions of the form (6.8) will be a complete set With

this qualitative understanding, one can now seek a solution of the form of equation (6.8) for the

system equation (6.3)

Substitute equation (6.8) in (6.3) to obtain

or

(6.9)

In equation (6.9), because the left-hand terms are a function of x only and the right-hand terms are

a function of t only, for the two sides to be equal in general, each function should be a constant

(that is, independent of both x and t) This constant is denoted by –λ2, which is called the separation

constant and is designated to be negative There are two good reasons for that If this common

constant were positive, the function q(t) would be nonoscillatory and transient, which is contrary

to the nature of undamped vibration Furthermore, it can be shown that a nontrivial solution for

Y(x) would not be possible if the common constant were positive.

The unknown constant λ is determined by solving the space equation (mode shape equation)

2 2 2

α

2 2 2

Y x

d Y x

d q t dt

( ) ( ) = ( ) ( )= −λ

Note that, since Y(x) is a real function representing a geometric shape, the constants A1 and A2 have

to be complex conjugates and C1 and C2 have to be real Specifically, in view of the fact that

For analytical convenience, use the real-parameter form of equation (6.13)

Note that one cannot determine both constants C1 and C2 using boundary conditions Only their

ratio is determined, and the constant multiplier is absorbed into q(t) in equation (6.8) and then determined using the appropriate initial conditions (at t = 0) It follows that the ratio of C1 and C2

and the value of λ are determined using the boundary conditions Two boundary conditions will

be needed Some useful situations and appropriate relations are given in Table 6.1

One can obtain the complete solution for free vibration of a taut cable that is fixed at both ends.The applicable boundary conditions are

C1 = 0 and

(6.15)

A possible solution for equation (6.15) is C2 = 0 But this is the trivial solution, which corresponds

to Y(x) = 0; that is, a stationary cable with no vibration It follows that the applicable, nontrivial

o

, 0

Y x i( )o = 0

dY x dx

which produces an infinite number of solutions for λ given by

(6.16)

As mentioned earlier, the corresponding infinite number of mode shapes is given by

(6.17)

for i = 1, 2, , ∞

Note: If one had used a positive constant λ2 instead of –λ2 in equation (6.9), only a trivial solution

(with C1 = 0 and C2 = 0) would be possible for Y(x) This further justifies the decision to use –λ2.Substitute equation (6.16) into (6.9) to determine the corresponding time response (generalized

coordinates) q i (t); thus,

(6.18)

in which

(6.19)

Equation (6.18) represents a simple harmonic motion with the modal natural frequencies ωi given

by equation (6.19) It follows that there are an infinite number of natural frequencies, as mentionedearlier The general solution of equation (6.18) is given by

(6.20)

where the amplitude parameter c i and the phase parameter φi are determined using two initial

conditions of the system It should be clear that it is redundant to use a separate constant C i for

Y i (x) in equation (6.17) as it can be absorbed into the amplitude constant in equation (6.20), to

express the general free response of the cable as

(6.21)

In this manner, the complete solution has been expressed as a summation of the modal solutions

This is known as the modal series expansion Such a solution is quite justified because of the fact that the mode shapes are orthogonal in some sense, and what was obtained above is a complete set of normal modes (normal in the sense of perpendicular or orthogonal) The system is able to

move independently in each mode, with a unique spatial shape, at the corresponding natural

frequency, because each modal solution is separable into a space function Y i (x) and a time function (generalized coordinate) q i (t) Of course, the system will be able to simultaneously move in a linear combination of two modes (say, C1Y1(x)q1(t) + C2Y2(x)q2(t)) because this combination satisfies the

original system equation (6.3), in view of its linearity, and because each modal component satisfiesthe equation But, clearly, this solution (with two modes) is not separable into a product of a space

function and a time function Hence, it is not a modal solution In this manner, it can be arguedthat the infinite sum of modal solutions is the most general solution to system (6.3).Orthogonality of mode shapes plays a key role in this argument and, furthermore, it is useful in

the analysis of the system In particular, in equation (6.21), the unknown constants c i and φi aredetermined using the system initial conditions, and the orthogonality property of modes is useful

in that procedure This will be addressed next

A cable can vibrate at frequency ωi called natural frequency while maintaining a unique natural

shape Y i (x) called mode shape of the cable It has been shown that for the fixed-ended cable, the

natural mode shapes are given by , with the corresponding natural frequencies ωi given by

equation (6.19) It can be easily verified that

(6.22)

In other words, the natural modes are orthogonal Equation (6.22) represents the principle oforthogonality of natural modes in this case

Orthogonality makes the modal solutions independent and the corresponding mode shapes

“normal.” It also makes the infinite set of modal solutions a complete set or a basis so that any

arbitrary response can be formed as a linear combination of these normal mode solutions.The orthogonality holds for other types of boundary conditions as well To show this, one seesfrom equation (6.9) that

0( )+λ ( )=

i j i j

l

2 2 2 2 0

dY dx

dY

dx dx

j i l

j i l

i j

l

2 2

∫ = −∫

Hence, the first term of equation (6.25) becomes

which will vanish for typical boundary conditions Then, since λi≠λj for i ≠ j, one has

One can pick the value of the multiplication constant in the general solution for Y(x), given by equation (6.13), so as to normalize the mode shapes such that

which is consistent with the result (6.22) Hence, the general condition of orthogonality of naturalmodes, may be expressed as

dY dx

dY

dx dx

j j i

2 2

j i i j l

for

where f(x,t) = 0 for free vibration Now, with the assumption of small θ and by neglecting the

second-order product term dT dθ, one obtains

T x

v x

2 +

Longitudinal (axial) dynamics of the rope are negligible for the case of a stationary hoist Then,

longitudinal equilibrium (in the x direction) of the small element of rope shown in Figure 6.2(b)

gives

For small θ, cosθ≅ 1 and cos(θ + dθ) ≅ 1 up to the first-order term in the Taylor series expansion.Hence,

(6.29)Integration gives

(6.30)with the end condition

∂

∂ =( + ) ∂∂ + ∂

∂

2 2

2 2

∂

∂ = +  ∂∂ + ∂∂

2 2

2 2

v t

The general solution to the cable vibration problem is given by

∂ ( )

∂ = ( ) ( )= − ( ) ( )

2 2

2 2

20

Multiply equations (6.39) and (6.40) by Y j (x) and integrate with respect to x from 0 to l, making

use of the orthogonality condition (6.26) Thus,

Solving these two equations, one obtains

(6.41)

Once φj is determined in this manner, one can obtain c j using

(6.42)

E XAMPLE 6.2

Consider a taut horizontal cable of length l and mass m per unit length, as shown in Figure 6.3,

excited by a transverse point force f0sinωt at location x = a, where ω is the frequency of (harmonic)

excitation and f0 is the forcing amplitude Determine the resulting response of the cable undergeneral end conditions and initial conditions For the special case of fixed ends, what is the steady-state response of the cable?

S OLUTION

It has been shown that the forced transverse response of a cable is given by equation (6.2):

(6.2)

where v(x,t) is the transverse displacement and f(x,t) is the external force per unit length of the cable.

FIGURE 6.3 A cable excited by a point harmonic force.

j l

j l

j l

v x t

v x t x

f x t m

For the point force F at x = a, the analytical representation of the equivalent distributed force

per unit length is

(6.43)where the Dirac delta function (unit impulse function) δ(x) is such that

This has the familiar form of a simple oscillator excited by a harmonic force and its solution is

well known The initial conditions q j(0) and j(0) are needed Suppose that the initial transversedisplacement and the speed of the cable are

Then, in view of equation (6.44), one can write

2( )= −λ ( )

(6.48)

Multiply equations (6.47) and (6.48) by Y j (x) and integrate from x = 0 to l using the orthogonality

property (6.26) The necessary initial conditions are obtained:

BOX 6.1 Transverse Vibration of Strings and Cables

T x

v x

6.2 LONGITUDINAL VIBRATION OF RODS

It can be shown that the governing equation of longitudinal vibration of line structures such as rodsand bars is identical to that of the transverse vibration of cables and strings Hence, it is notnecessary to repeat the complete analysis here This section first develops the equation of motion,then considers boundary conditions, next identifies the similarity with the cable vibration problem,and concludes with an illustrative example

u(x,t) = longitudinal displacement of the rod at distance x from a fixed reference.

Note that the fixed reference can be chosen arbitrarily However, if the assumption of small u is

needed, the reference may be chosen as the relaxed (unstrained) position of the element The

longitudinal stress at the cross section at x is σ = Eε and, hence, the longitudinal force is

FIGURE 6.4 (a) A rod with distributed loading and in longitudinal vibration; and (b) a small element of the rod.

ε = ∂

∂

u x

E = Young’s modulus of the rod

A = area of cross section.

It is not necessary at this point to assume a uniform rod Hence, A may depend on x.

The equation of motion for the small element shown in Figure 6.4(b) is

or

(6.55)From equation (6.54), one has

(6.56)which when substituted into (6.55) gives

u

u x t x

,

ρ

and forced vibration can be analyzed as before For a fixed end at x = x0, there will be no deflection.Hence,

(6.60)with the corresponding modal end condition

00,

and equation (6.57) becomes

But M = ρAl Hence,

(6.66)Boundary conditions are

(6.67)

(6.68)Initial conditions are

(6.74)

l

,( )=

u x t

u x t x

Mg Al

0( )+λ ( )=

X x i( )=C1sinλi x+C2cosλi x

X i( )0 =0

Substitute equation (6.74) into (6.73) We have C2 = 0 Next, use equation (6.75) and obtain

Since C1≠ 0 for a nontrivial solution, the required condition is

which can be expressed as

(6.76)

This transcendental equation has an infinite number of solutions λi that correspond to the modes

of vibration The solution can be made computationally and the corresponding natural frequenciesare obtained using

and the overall solution is

(6.84)

The constants A j and B j are determined using the initial conditions q j(0) and j(0) These are obtained

by substituting equation (6.71) into (6.69) and (6.70), multiplying by X j (x), and integrating from

x = 0 to l, making use of the orthogonality property (6.64) Specifically, one obtains

j j l

j j

1

12

and

(6.86)



6.3 TORSIONAL VIBRATION OF SHAFTS

Torsional vibrations are oscillating angular motions of a device about some axis of rotation.Examples are vibration in shafts, rotors, vanes, and propellers The governing partial differentialequation of torsional vibration of a shaft is quite similar to that encountered previously for transversevibration of a cable in tension and longitudinal vibration of a rod But, in the present case, thevibrations are rotating (angular) motions with resulting shear strains, shear stresses, and torques inthe torsional member Furthermore, the parameters of the equation of motion will take differentmeanings When bending and torsional motions occur simultaneously, there can be some interaction,thereby making the analysis more difficult Here, one neglects such interactions by assuming thatonly the torsional effects are present or the motions are quite small

Because the form of the torsional vibration equation is similar to what was studied before, thesame procedures of analysis can be employed and, in particular, the concepts of modal analysiswill be similar But, the torsional parameters will be rather complex for members with noncircularcross sections For this reason, and also because a vast majority of torsional devices have circularcross sections, this case will be considered first

Here, one can formulate the problem of torsional vibration of a shaft having circular cross section.The general case of non-uniform cross section, along the shaft, is considered However, the usualassumptions such as homogeneous, isotropic, and elastic material are made

First, obtain a relationship between torque (T) and angular deformation or twist (θ) for a circular

shaft Consider a small element of length dx along the shaft axis and observe the cylindrical surface

at a general radius r (in the interior of the shaft segment), as shown in Figure 6.6(a) During

vibration, this element will deform (twist) through a small angle dθ

FIGURE 6.6 (a) Small element of a circular shaft in torsion, and (b) shear stresses in a small annular cross

section carrying torque.

j l

A point on the circumference will deform through rdθ as a result, and a longitudinal line onthe cylindrical surface will deform through angle γ as shown in Figure 6.6(a) From solid mechanics(strength of materials or theory of elasticity), one knows that γ is the shear strain Hence,

But, allowing for the fact that the angular shift θ is a function of t as well as x in the general case

of dynamics, one uses partial derivatives, and writes

(6.87)

The corresponding shear stress at the deformed point at radius r is

(6.88)where

G = shear modulus.

This shear stress acts tangentially Consider a small annular cross section of width dr at radius r

of the shaft, as shown in Figure 6.6(b) By symmetry, the shear stress will be the same throughout

this region, and will form a torque of r ×τ× 2πrdr = 2πr2τdr Hence, the overall torque at the

shaft cross section is

which, in view of equation (6.88), is written as

Note: For a shaft with noncircular cross section, replace J by J t in this equation It follows that the

larger the torsional rigidity GJ, the higher the torsional stiffness K, as expected Furthermore, longer

members have a lower torsional stiffness (and smaller natural frequencies)

Now apply Newton’s second law for rotatory motion of the small element dx shown in Figure

where J is the polar moment of area, as discussed before Also, suppose that a distributed external

torque of τ(x,t) per unit length is applied along the shaft Hence, the equation of motion is

Substitute equation (6.93) and cancel dx to get the equation of torsional vibration of a circular shaft

as

(6.95)

For the case of a uniform shaft (constant J) in free vibration (τ(x,t) = 0),

(6.96)with

x t x

ρ

(6.99)Orthogonality property of mode shapes Θi (x):

(6.100)

Unlike for longitudinal and transverse vibrations of rods and beams, in the case of torsional vibration

of shafts, the same equation of motion for circular shafts (equations (6.95) and (6.96)) cannot beused for shafts with noncircular cross sections The reason is that the shear stress distributions inthe two cases can be quite different, and equation (6.88) does not hold for noncircular sections

Hence, the parameter J in the torque-deflection relations (e.g., equations (6.93) and (6.94)) is not

the polar moment of area in the case of noncircular sections In this case, one writes

(6.101)where

J t = torsional parameter

The Saint Venant theory of torsion and related membrane analogy, developed by Prandtl, have

provided equations for J t in special cases For example, for a thin hollow section,

(6.102)where

A s = enclosed (contained) area of the hollow section

p = perimeter of the section

t = wall thickness of the section

For a thin, solid section, one has

(6.103)where

a = length of the narrow section

t = thickness of the narrow section

Torsional parameters for some useful sections are given in Table 6.2

p

t s

= 4

2

J t= t a3

3

E XAMPLE 6.4

Consider a thin rectangular hollow section of thickness t, height a, and width a/2, as shown in

Figure 6.7(a) Suppose that the section is opened by making a small slit as in Figure 6.7(b) Study

the effect on the torsional parameter J t and torsional stiffness K of the member due to the opening.

TABLE 6.2

Torsional Parameters for Several Sections

s

t a3

3

0 1406 (a2−a1)

S OLUTION

a Closed section:

The contained area of the section

The perimeter of the section p = 3a

Using equation (6.102), the torsional parameter

b Open section:

Solid length of the section = 3a

Using equation (6.103), the torsional parameter

Ratio of the torsional parameters

FIGURE 6.7 (a) A thin closed section, and (b) a thin open section.

FIGURE 6.8 A torsional guideway transit system.

For members of equal length, their torsional stiffness satisfy the same ratio as given by this

expression Since t is small compared to a, there will be a significant drop in torsional stiffness

due to the opening (cutout)



E XAMPLE 6.5

An innovative automated transit system uses an elevated guideway with cars whose suspensionsare attached to (and slide on) the side of the guideway Due to this eccentric loading on the guideway,there is a significant component of torsional dynamics in addition to bending Assume that the

torque T j acting on the guideway due to the jth suspension of the vehicle is constant, acting at a point x j as measured from one support pier and moving at speed v j A schematic representation isgiven in Figure 6.8 The guideway span shown has a length l and a cross section that is a thin- walled rectangular box of height a, width b, and thickness t The ends of the guideway span are

restrained for angular motion (i.e., fixed)

a Formulate and analyze the torsional (angular) motion of the guideway

b For a single point vehicle entering a guideway that is at rest, what is the resulting dynamicresponse of the guideway? What is the critical speed that should be avoided?

c Given the parameter values:

and vehicle speed

v = 60 mph (26.8 m s–1)

compute the crossing frequency ratio given by

and discuss its implications

S OLUTION

a For a uniform guideway with distributed torque load τ(x,t), and noncircular cross section having torsional parameter J t, the governing equation is

J J

t a

to tC

Fundamental natural frequency of guideway

where A i (i = 1, 2) are the constants of integration The torsional boundary conditions

corresponding to the fixed ends (no twist) are

where l = guideway span length For a nontrivial solution, one needs

which is still unknown and is determined through initial conditions Here, the normalizedeigenfunctions are used:

(6.108)The orthogonality condition given by

2 2

,

,

d dx

2 2 20

where T j = torque exerted on guideway by the jth suspension

v j = speed of the jth suspension

x 0j = initial position of the jth suspension (at t = 0) along the guideway

δ(·) = Dirac delta function

The forced motion can be represented in terms of the normalized eigenfunctions as

(6.112)

where q i (t) = generalized coordinate for forced motion in the ith mode On substituting

relations (6.111) and (6.112) into (6.104) and integrating the result over the span length,after multiplying by a general eigenfunction while making use of the orthogonalityrelation (6.109), one obtains

(6.113)

b For a single suspension entering the guideway at t = 0, with the guideway initially at

rest , one has n =1 and x01 = 0 Then the complete solution of equation(6.113) is

j i j j j

n

2 2 2

0 1

Note from equation (6.114) that the critical speed corresponds to v C = 1 and should be

avoided In typical transit systems, v C is considerably less than 1

c For the given numerical values, by straightforward computation, it can be shown that:

Note: The expression for thin hollow section given in Table 6.2 was used to compute J t Finally,

the crossing frequency ratio is computed to be v C = 0.017, which is much less than 1.0, as expected



6.4 FLEXURAL VIBRATION OF BEAMS

This section will discuss yet another continuous member in vibration Specifically, a beam (or rod

or shaft) in flexural vibration is considered The vibration is in the “transverse” or “lateral” direction,which is accompanied by bending (or flexure) of the member Hence, the vibrations are perpen-dicular to the main axis of the member, as in the case of the cable or string that was studied inSection 6.1 But a beam — unlike a string — can support shear forces and bending moments atits cross section In the initial analysis of bending vibration, assume that there is no axial force atthe ends of the beam Further simplifying assumptions will be made, which will be clear in thedevelopment of the governing equation of motion The analysis procedure will be quite similar tothat followed in the previous sections for other continuous members

The study of bending vibration (or lateral or transverse vibration) of beams is very important

in a variety of practical situations Noteworthy, here, are the vibration analysis of structures likebridges, vehicle guideways, tall buildings, and space stations; ride quality and structural integrityanalysis of buses, trains, ships, aircraft and spacecraft; dynamics and control of rockets, missiles,machine tools, and robots; and vibration testing, evaluation, and qualification of products withcontinuous members

To develop the Bernoulli-Euler equation, which governs transverse vibration of thin beams, consider

a beam in bending, in the x-y plane, with x as the longitudinal axis and y as the transverse axis of

bending deflection, as shown in Figure 6.9 The required equation is developed by considering thebending moment-deflection relation, rotational equilibrium, and transverse dynamics of a beamelement

rad s Hz

ω

section A of the beam element, at a distance w (measure parallel to y) from the neutral axis of

where E = Young’s modulus (of elasticity) Then, the bending moment is

where I = second moment of area of the beam cross section, about the neutral axis Thus,

, where, v = lateral deflection of the beam at

element δx Hence, change in slope = , where δθ is the arc angle of bending for thebeam element δx, as shown in Figure 6.9

Also, δx = Rδθ Hence, Cancel δθ, and obtain

(6.119)Substitute equation (6.119) in (6.118) and obtain

(6.120)

Rotatory Dynamics (Equilibrium)

Again consider the beam element δx, as shown in Figure 6.10, where forces and moments acting

on the element are indicated Here, f(x,t) = excitation force per unit length acting on the beam, in the transverse direction, at location x Neglect rotatory inertia of the beam element.

The equation of angular motion is given by the equilibrium condition of moments:

(6.121)

where the previously obtained result (6.120) for M has been substituted Note that a uniform beam

is not assumed and, hence, I = I(x) will be variable along the beam length.

Transverse Dynamics

The equation of transverse motion (Newton’s second law) for element δx is

FIGURE 6.10 Dynamics of a beam element in bending.

Slope at A v Slope at

v x

or

Here, ρ = mass density of the beam material Thus,

or, in view of equation (6.121), one obtains the governing equation of forced transverse motion ofthe beam as

(6.122)

This is the well-known Bernoulli-Euler beam equation

The solution to the flexural vibration problem given by equation (6.122) can be obtained exactly

as followed previously for other continuous members Specifically, first obtain the natural cies and mode shapes, and express the general solution as a summation of the modal responses.The approach is similar for both free and forced problems, but the associated generalized coordinateswill be different This approach is followed here

frequen-For modal (natural) vibration, consider the free motion described by

2 4 4

v x t

v x t t

A

=ρ

Observe from equation (6.124) that it is fourth order in x and second order in t, whereas the

governing equations for transverse vibration of a cable, longitudinal vibration of a road, and

torsional vibration of a shaft are all identical in form and second order in x So, the behavior of

transverse vibration of beams will not be exactly identical to that of the other three types ofcontinuous system In particular, the traveling wave solution (6.7) will not be satisfied However,there are many similarities

In each mode, the system will vibrate in a fixed shape ratio Hence, the time and space functionswill be separable for a modal motion; seek a solution of the form

(6.126)This separable solution for a modal response is justified as usual Note that even in the lumpedparameter case (Chapter 5), the same assumption was made — except in that case, one had a modalvector:

instead of a mode shape function Y(x) For a given mode of a lumped parameter system, Y i values

denote the “relative” displacements of various inertia elements m i, as shown in Figure 6.11 Hence,

the vector Y corresponds to the mode shape Note that Y i can be either positive or negative Also,

q(t) is the harmonic function corresponding to the natural frequency.

It should be clear that Y and q(t) are separable in this lumped-parameter case of modal motion.

Then, in the limit, Y(x) and q(t) also should be separable for the distributed-parameter case Substitute equation (6.126) in (6.123), and bring terms containing x to the LHS and terms containing t to the RHS Then,

(6.127)

Because a function of x cannot be equal to a function of t in general (unless each function is equal

to the same constant), ω2 is defined as a constant

It has not been shown that this separation constant (ω2) should be positive This requirement

can be verified due to the nature of the particular vibration problem; that is, q(t) should have an

oscillatory solution in general It is also clear that the physical interpretation of ω is a naturalfrequency of the system Equation (6.127) corresponds to the two ordinary differential equations

— one in t and the other in x — as

(6.128)

(6.129)

The solution of these two equations will provide the natural frequencies ω and the corresponding

mode shapes Y(x) of the beam.

Y n

1 2

M

2 2 2

2 2 2

0( )−ω ρ ( )=

For further analysis of the modal behavior, assume a uniform beam Then, EI will be constant

and equation (6.129) can be expressed as

(6.130)where

(6.131)

The positive parameter λ is yet to be determined, and will come from the mode shape analysis.The characteristic equation corresponding to equation (6.130) is

(6.132)The roots are

(6.133)Hence, the general solution for a mode shape (eigenfunction) is given by

(6.134)

There are five unknowns (C1, C2, C3, C4, and λ) here The mode shapes can be normalized and one

of the first four unknowns can be incorporated into q(t) as usual The remaining four unknowns

are determined by the end conditions of the beam Thus, four boundary conditions will be needed

Note:

The four modal boundary conditions that are needed can be derived in the usual manner, depending

on the conditions at the two ends of the beam The procedure is to apply the separable (modal)solution (6.126) to the end relation with the understanding that this relation has to be true for all

possible values of q(t) The relation (6.128) can be substituted as well, if needed.

For example, consider an end x = x0 that is completely free Then, both bending moment andshear force have to be zero at this end From equations (6.120) and (6.121), one obtains

d Y x

4 4 40( )−λ ( )=

(6.136)

Substitute equation (6.126) into (6.135) and (6.136)

which are true for all q(t) Hence, the following modal boundary conditions result for a free end:

(6.137)

(6.138)For a uniform beam, equation (6.138) becomes

(6.138b)

Some common conditions and the corresponding modal boundary condition equations for bendingvibration of a beam are listed in Box 6.2

To illustrate the approach, consider a uniform beam of length l that is pinned (simply supported)

at both ends In this case, both displacement and the bending moment will be zero at each end.Accordingly, one has the following modal boundary conditions (BCs):

2 0

2 0

2 0

d Y x dx

3 0

Y( )0 = =0 Y l( )

d Y dx

d Y l dx

2 2

2 2

00( )= = ( )

C1+C3=0