Vibration and Shock Handbook 04

Vibration and Shock Handbook 04 Every so often, a reference book appears that stands apart from all others, destined to become the definitive work in its field. The Vibration and Shock Handbook is just such a reference. From its ambitious scope to its impressive list of contributors, this handbook delivers all of the techniques, tools, instrumentation, and data needed to model, analyze, monitor, modify, and control vibration, shock, noise, and acoustics. Providing convenient, thorough, up-to-date, and authoritative coverage, the editor summarizes important and complex concepts and results into “snapshot” windows to make quick access to this critical information even easier. The Handbook’s nine sections encompass: fundamentals and analytical techniques; computer techniques, tools, and signal analysis; shock and vibration methodologies; instrumentation and testing; vibration suppression, damping, and control; monitoring and diagnosis; seismic vibration and related regulatory issues; system design, application, and control implementation; and acoustics and noise suppression. The book also features an extensive glossary and convenient cross-referencing, plus references at the end of each chapter. Brimming with illustrations, equations, examples, and case studies, the Vibration and Shock Handbook is the most extensive, practical, and comprehensive reference in the field. It is a must-have for anyone, beginner or expert, who is serious about investigating and controlling vibration and acoustics.

4 Distributed-Parameter

Systems

Clarence W de Silva

The University of British Columbia

4.1 Introduction 4-14.2 Transverse Vibration of Cables 4-2

Wave Equation † General (Modal) Solution † Cable with Fixed Ends † Orthogonality of Natural Modes † Application

of Initial Conditions

4.3 Longitudinal Vibrations of Rods 4-13

Equation of Motion † Boundary Conditions

4.4 Torsional Vibration of Shafts 4-19

Shaft with Circular Cross Section † Torsional Vibration of Noncircular Shafts

4.5 Flexural Vibration of Beams 4-26

Governing Equation for Thin Beams † Modal Analysis †

Boundary Conditions † Free Vibration of a Simply Supported Beam † Orthogonality of Mode Shapes † Forced Bending Vibration † Bending Vibration of Beams with Axial Loads † Bending Vibration of Thick Beams † Use of the Energy Approach † Orthogonality with Inertial Boundary Conditions

4.6 Damped Continuous Systems 4-50

Modal Analysis of Damped Beams

4.7 Vibration of Membranes and Plates 4-52

Transverse Vibration of Membranes † Rectangular Membrane with Fixed Edges † Transverse Vibration of Thin Plates †

Rectangular Plate with Simply Supported Edges

Summary

This chapter presents the analysis of continuous (or distributed-parameter) mechanical vibrating systems In thesesystems, inertial, elastic, and dissipative effects are found continuously distributed in one, two, or three dimensions.Examples such as strings, rods, shafts, beams, membranes, and plates are studied Modal analysis is carried outusing the separation of time and space The orthogonality property of mode shapes is established Boundaryconditions are derived Free vibration and forced vibration are analyzed

4.1 Introduction

Often in vibration analysis, it is assumed that inertial (mass), flexibility (spring), and dissipative (damping)characteristics can be “lumped” as a finite number of “discrete” elements Such models are termed lumped-parameter or discrete-parameter systems Generally, in practical vibrating systems, inertial, elastic, anddissipative effects are found continuously distributed in one, two, or three dimensions Correspondingly,

4-1

we have line structures, surface/planar structures, or spatial structures They will possess an infinitenumber of mass elements, continuously distributed in the structure, and integrated with some connectingflexibility (elasticity) and energy dissipation In view of the connecting flexibility, each small element ofmass will be able to move out of phase (or somewhat independently) with the remaining mass elements Itfollows that a continuous system (or a distributed-parameter system) will have an infinite number of degrees

of freedom (DoFs) and will require an infinite number of coordinates to represent its motion In otherwords, extending the concept of a finite-degree-of-freedom system as analyzed previously, an infinite-dimensional vector is needed to represent the general motion of a continuous system Equivalently, a one-dimensional continuous system (a line structure) will need one independent spatial variable, in addition totime, to represent its response In view of the need for two independent variables in this case, one for timeand the other for space, the representation of system dynamics will require partial differential equations(PDEs) rather than ordinary differential equations (ODEs) Furthermore, the system will depend on theboundary conditions as well as the initial conditions

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 Ageneral structure may consist of more than one such member, and furthermore, boundary conditions(BCs) could be various, individual members may be nonuniform, and the material characteristics may beinhomogeneous and anisotropic Closed-form analytical solutions would not be generally possible in suchcases Nevertheless, the insight gained by analyzing the vibration of standard members will be quitebeneficial in studying the vibration behavior of more complex structures

The concepts of modal analysis may be extended from lumped-parameter systems to continuoussystems In particular, since the number of principal modes is equal to the number of DoFs of thesystem, a distributed-parameter system will have an infinite number of natural modes of vibration Aparticular mode may 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 significantand nonproportional, however, there is no guarantee that such an initial condition could accuratelyexcite the required mode A general excitation consisting of a force or an initial condition will excitemore than one mode of motion However, as in the case of discrete-parameter systems, the generalmotion may be analyzed and expressed in terms of modal motions, through modal analysis In amodal motion, the mass elements will move at a specific frequency (the natural frequency), andbearing a constant proportion in displacement (i.e., maintaining the mode shape), and passing thestatic equilibrium of the system simultaneously In view of this behavior, it is possible to separate thetime response and spatial response of a vibrating system in a modal motion This separability isfundamental to modal analysis of a continuous system Furthermore, in practice an infinite number

of natural frequencies and mode shapes are not significant and typically the very high modes may beneglected Such a modal-truncation procedure, even though carried out by continuous-systemanalysis, is equivalent to approximating the original infinite-degree-of-freedom system by a finite-degree-of-freedom one Vibration analysis of continuous systems may be applied in the modeling,analysis, design, and evaluation of such practical systems as cables; musical instruments; transmissionbelts and chains; containers of fluid; animals; structures including buildings, bridges, guideways, andspace stations; and transit vehicles, including automobiles, ships, aircraft, and spacecraft

4.2 Transverse Vibration of Cables

The first continuous member which we will study is a string or cable in tension This is a line structurewhose geometric configuration can be completely defined by the position of its axial line with reference

to a fixed coordinate line We will study the transverse (lateral) vibration problem; that is, the vibration in

a direction perpendicular to its axis and in a single plane Applications will include stringed musicalinstruments, overhead transmission lines (of electric power or telephone signals), drive systems (beltdrives, chain drives, pulley ropes, etc.), suspension bridges, and structural cables carrying cars (e.g., skilifts, elevators, overhead sightseeing systems, and cable cars)

As usual, we will make some simplifying assumptions for analytical convenience However, 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 with thelongitudinal dimension (normally in the x direction)

2 The structure stays in a single plane and the motion of every element of the structure will be in afixed transverse direction ð yÞ:

3 The cable tension ðTÞ remains constant during motion In other words, the initial tension issufficiently large that the variations during motion are negligible

4 Variations in slope ðuÞ along the structure are small Hence, for example, u ø sin u ø tan u ¼ ›v›x:

A general configuration of a cable (or string) is shown in Figure 4.1(a) Consider a small element oflength dx of the cable at location x; as shown in Figure 4.1(b) The equation (Newton’s Second Law) ofmotion (transverse) of this element is given by

f ðx; tÞdx 2 T sinu þ T sinðu þ duÞ ¼ mðxÞdx›2vðx; tÞ›t2 ð4:1Þ

in which

vðx; tÞ ¼ transverse displacement of the cable

f ðx; tÞ ¼ lateral force per unit length of cable

mðxÞ ¼ mass per unit length of cable

T ¼ cable tension

u ¼ cable slope at location x:

Note that the dynamic loading f ðx; tÞ may arise due to such causes as aerodynamic forces, fluid drag,and electromagnetic forces, depending on the specific application

T

θ

FIGURE 4.1 (a) Transverse vibration of a cable in tension; (b) motion of a general element.

Using the small slope assumption we have sinu ø u and sinðu þ duÞ ø u þ du with u ¼ ›v=›x and

du ¼ ð›2v=›x2Þdx as dx ! 0: On substitution of these approximations into Equation 4.1 and cancelingout dx, we obtain

›z

›twith

v001¼ d2v1

dz2Clearly, then, v1satisfies Equation 4.3

Now, let us examine the nature of the solution v1ðx 2 ctÞ: It is clear that v1 will be constant when

x 2 ct ¼ constant: However, the equation x 2 ct ¼ constant corresponds to a point moving along the xaxis in the positive direction at speed c: What this means is that the shape of the cable at t ¼ 0 will

“appear” to travel along the cable at speed c: This is analogous to the waves we observe in a pond whenexcited by dropping a stone Note that the particles of the cable do not travel along x: it is the deformation

“shape” (the wave) that travels

Similarly, it can be shown that

is also a solution to Equation 4.3 and this corresponds to a wave that travels backward (negative xdirection) at speed c: The general solution, of course, will be of the form

vðx; tÞ ¼ v1ðx 2 ctÞ þ v2ðx þ ctÞ ð4:7Þwhich represents two waves, one traveling forward and the other backward

4.2.2 General (Modal) Solution

As usual, we look for a separable solution of the form

for the cable/string vibration problem given by the wave equation 4.3 If a solution in the form ofEquation 4.8 is obtained, it will be essentially a modal solution This should be clear from the separabilityitself of the solution Specifically, at any given time t; the time function qðtÞ will be fixed and the structurewill have a shape given by YðxÞ: Hence, at all times the structure will maintain a particular shape YðxÞ andthis will be a mode shape Also, at a given point x of the structure, the space function YðxÞ will be fixed andthe structure will vibrate according to the time response qðtÞ: It will be shown that qðtÞ will obey thesimple 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 of Equation 4.8 with different natural frequencies The corresponding functions YðxÞ will be

“orthogonal” in some sense Hence, they are called normal modes (normal meaning perpendicular) Thesystems will be able to move independently in each mode and this collection of solutions in the form ofEquation 4.8 will be a complete set With this qualitative understanding, let us now seek a solution of theform of Equation 4.8 for the system Equation 4.3

Substitute Equation 4.8 in Equation 4.3 We obtain

YðxÞd2qðtÞ

dt2 ¼ c2d2YðaÞ

dx2 qðtÞor

1YðxÞ

d2YðxÞ

dx2 ¼ c2qðtÞ1 d2dtqðtÞ2 ¼ 2l2 ð4:9Þ

In Equation 4.9, since 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 2l2, which is called the separation constant and isdesignated to be negative There are two reasons for this If this common constant were positive, thefunction qðtÞ would be nonoscillatory and transient, which is contrary to the nature of undampedvibration Furthermore, it can be shown that a nontrivial solution for YðxÞ would not be possible if thecommon constant were positive

The unknown constant l is determined by solving the space equation (mode shape equation) ofEquation 4.9; specifically

The general solution is

YðxÞ ¼ A1ejl þ A2e2jl ¼ C1coslx þ C2sinlx ð4:13ÞNote that, since YðxÞ is a real function representing a geometric shape, the constants A1and A2have

to be complex conjugates and C1 and C2 have to be real Specifically, in view of the fact thatcoslx ¼ ðej l þ e2j l = Þ2 and sinlx ¼ ðej l 2 e2j l =2j Þ, we can show that

A1¼ 1

2ðC12 jC2Þ and A2¼

1

2ðC1þ jC2ÞFor analytical convenience, we will use the real-parameter form of Equation 4.13

Note that we cannot determine both constants C1and C2using BCs Only their ratio is determinedand the constant multiplier is absorbed into qðtÞ in Equation 4.8 and then determined using theappropriate initial conditions (at t ¼ 0) It follows that the ratio of C1 and C2and the value ofl aredetermined using the BCs Two BCs will be needed Some useful situations and appropriate relations aregiven inTable 4.1

4.2.3 Cable with Fixed Ends

Let us obtain the complete solution for the free vibration of a taut cable that is fixed at both ends Theapplicable BCs are

where l is the length of the cable Substitution into Equation 4.13 gives

C1£ 1 þ C2£ 0 ¼ 0

C1cosll þ C2sinll ¼ 0Hence, we have

A possible solution for Equation 4.15 is C2¼ 0: However, this is the trivial solution, which corresponds toYðxÞ ¼ 0 (i.e., a stationary cable with no vibration) It follows that the applicable, nontrivial solution is

sinll ¼ 0which produces an infinite number of solutions forl given by

Note: If we had used a positive constantl2instead of 2l2in Equation 4.9, only a trivial solution (with

C1¼ 0 and C2¼ 0) would be possible for YðxÞ: This further justifies our decision to use 2l2: SubstituteEquation 4.16 into Equation 4.9 to determine the corresponding time response (generalized coordinates)

TABLE 4.1 Some Useful Boundary Conditions for the Cable Vibration Problem

Type of End Condition Nature of End x ¼ x 0 Boundary Condition Modal Boundary Condition Fixed

k M

Equation 4.18 represents a simple harmonic motion with the modal natural frequencies vi given byEquation 4.19 It follows that there are an infinite number of natural frequencies, as mentioned earlier.The general solution of Equation 4.19 is given by

qiðtÞ ¼ cisinðvit þfiÞ ð4:20Þwhere the amplitude parameter ciand the phase parameterfiare determined using two of the initialconditions of the system It should be clear that it is redundant to use a separate constant Cifor YiðxÞ inEquation 4.17, and that this may be absorbed into the amplitude constant in Equation 4.20 to express thegeneral free response of the cable as

vðx; tÞ ¼Xcisin ipx

In this manner, the complete solution has been expressed as a summation of the modal solutions This isknown as the modal series expansion Such a solution is quite justified because of the fact that the modeshapes are orthogonal in some sense, and what we obtained above were a complete set of normal modes(normal in the sense of perpendicular or orthogonal) The system is able to move in each modeindependently, with a unique spatial shape, at the corresponding natural frequency, because each modalsolution is separable into a space function, YiðxÞ, and a time function (generalized coordinate), qiðtÞ: Ofcourse, the system will be able to simultaneously move in a linear combination of two modes (say,

C1Y1ðxÞq1ðtÞ þ C2Y2ðxÞq2ðtÞ), since this combination satisfies the original system Equation 4.3 because ofits linearity and because each modal component satisfies the equation However, clearly, this solution,with two modes, is not separable into a product of a space function and a time function Hence, it is not amodal solution In this manner, it can be argued that the infinite sum of modal solutionsPciYiðxÞqiðtÞ isthe most general solution to the system (Equation 4.3) The orthogonality of mode shapes plays a key role

in this argument and, furthermore, it is useful in the analysis of the system, as we shall see In particular,

in Equation 4.21, the unknown constants ciandfiare determined using the system initial conditions,and the orthogonality property of modes is useful in that procedure

4.2.4 Orthogonality of Natural Modes

A cable can vibrate at frequencyviwhile maintaining a unique natural shape YiðxÞ; called the mode shape

of the cable We have shown that, for the fixed-ended cable, the natural mode shapes are given bysinðipx=lÞ with the corresponding natural frequencies,vi: It can be easily verified that

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 responsecan be formed as a linear combination of these normal mode solutions

Orthogonality holds for other types of BCs as well To show this, we observe from Equation 4.9 that

Multiply Equation 4.23 by YjðxÞ; Equation 4.24 by YiðxÞ; subtract that second result from the first,and integrate with respect to x along the cable length from x ¼ 0 to l: We obtain

2ðl0

dYidx

2ðl0

dYidx

dYj

dx dxHence, the first term of Equation 4.25 becomes

YjdYi

dx 2 Yi

dYjdx

l 0which will vanish for common BCs Then, sinceli–lj for i – j, we have

ðl

0YiðxÞYjðxÞdx ¼

0 for i – jl

If the cable tension varies along the length x; what is the corresponding equation of free lateral vibration?

A hoist mechanism has a rope of freely hanging length l in a particular equilibrium configuration andcarrying a load of mass M; as shown inFigure 4.2(a).Determine the equation of lateral vibration and theapplicable BCs for the rope segment

Solution

With reference toFigure 4.1(b),Equation 4.1 may be modified for the case of variable T as

2T sinu þ ðT þ dTÞ sinðu þ duÞ ¼ m dx››t2v2 ð4:27Þ

where f ðx; tÞ ¼ 0 for free vibration Now, with the

assumption of small u; and by neglecting the

second-order product term dT du; we obtain

T du þ u dT ¼ m››x2v2dx

Next, using

u ¼ ›v›x; du ¼ ››x2v2; and dT ¼ ›T

›xdxand canceling dx; we obtain the equation of lateral

negligible for the case of a stationary hoist Then,

longitudinal equilibrium (in the x direction) of the

small element of rope shown in Figure 4.2(b) gives

ðT þ dTÞ cosðu þ duÞ 2 T cos u 2 mg dx ¼ 0

For small u; we have cos u ø 1 and cosðu þ

duÞ ø 1 up to the first-order term in the Taylor

series expansion Hence,

dT ¼ mg dx ð4:29ÞIntegration gives

T ¼ T0þ mgx ð4:30Þwith the end condition

T ¼ Mg at x ¼ 0Hence,

mgdx

T θ+dθ

0

Mg T+dT

θ

l

FIGURE 4.2 (a) Free segment of a stationary hoist; (b) a small element of the rope.

›2vð0; tÞ

›t2 ¼ Yið0Þd2qiðtÞ

dt2 ¼ 2v2

iYið0ÞqiðtÞwhich holds for all t and whereviis the ith natural frequency of vibration Hence, the modal BC at x ¼ 0is

gdYdxið0Þ þv2

iYið0Þ ¼ 0 for i ¼ 1; 2; … ð4:33ÞThe BC at x ¼ l is

which holds for all t: Hence, the corresponding modal BC is

4.2.5 Application of Initial Conditions

The general solution to the cable vibration problem is given by

vðx; tÞ ¼XciYiðxÞ sinðvit þfiÞ ð4:36Þwhere YiðxÞ are the normalized mode shapes which satisfy the orthogonality property (Equation 4.26).The unknown constants ciandfiare determined using the initial conditions

of the orthogonality condition (Equation 4.26) We obtain

Oncefjis determined in this manner, we can obtain cj by using

Consider a taut horizontal cable of length l and

mass m per unit length, as shown in Figure 4.3,

excited by a transverse point force f0sinvt at

location x ¼ a; where v is the frequency of

(harmonic) excitation and f0 is the forcing

amplitude Determine the resulting response of

the cable under general end conditions and initial

conditions For the special case of fixed ends, what

is the steady-state response of the cable?

where the Dirac delta function (unit impulse function)dðxÞ is such that

ða 2

a 1

for an arbitrary function gðxÞ; provided that the point a is within the interval of integration ½a1; a2 : Weseek a “modal superposition” solution of the form

FIGURE 4.3 A cable excited by a point harmonic force.

Now sincevj ¼lj

ffiffiffiffiffiffiT=m

p(see Equation 4.19), we obtain

€qjðtÞ þv2

jqjðtÞ ¼ 2flm0YjðaÞ sinvt for j ¼ 1; 2; 3; … ð4:46ÞThis has the familiar form of a simple oscillator excited by a harmonic force and its solution is wellknown The initial conditions qjð0Þ and _qjð0Þ are needed Suppose that the initial transverse displacementand the speed of the cable are

vðx; 0Þ ¼ dðxÞ and _vðx; 0Þ ¼ sðxÞThen, in view of Equation 4.45, we can write

ðl

_qjð0Þ ¼ 2l

Some important results for transverse vibration of strings and cables are summarized in Box 4.1

4.3 Longitudinal Vibrations of Rods

It can be shown that the governing equation of the longitudinal vibration of line structures such as rodsand bars is identical to that of the transverse vibration of cables and strings Hence, it is not necessary torepeat the complete analysis here We will first develop the equation of motion, then consider BCs, nextidentify the similarity with the cable vibration problem, and will conclude with an illustrative example

vðx; tÞ ¼XYiðxÞqiðtÞwith

Traveling-wave solution (long cable, independent of end conditions):

vðx; tÞ ¼ v1ðx 2 ctÞ þ v2ðx þ ctÞOrthogonality:

ðl

0YiðxÞYjðxÞdx ¼

0 for i – jl

ðl

0dðxÞYiðxÞdx_qið0Þ ¼ 2

l

ðl

0sðxÞYiðxÞdxVariable-tension problem:

4.3.1 Equation of Motion

Consider a rod that is mounted horizontally (so that the gravitational effects can be neglected) as shown

in Figure 4.4(a) A small element of length dx (the limiting case of dx) at position x is shown inFigure 4.4(b) The longitudinal strain at x is given by

1 ¼ ›u

where uðx; tÞ ¼ longitudinal displacement of the rod at x from a fixed reference

Note that the fixed reference may be chosen arbitrarily but, if the assumption of small u is needed, therelaxed (unstrained) position of the element must be chosen as the reference The longitudinal stress atthe cross section at x iss ¼ E1 and, hence, the longitudinal force is

P ¼ EA›u

where

E ¼ Young’s modulus of the rod

A ¼ 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 4.4(b) is

rA dx›2uðx; tÞ›t2 ¼ P þ dP 2 P þ f ðx; tÞdxor

rA››t2u2dx ¼ dP þ f ðx; tÞdx ð4:55ÞNow, from Equation 4.54, we have

x+dx

Mass density = r Force per unit length = f(x,t)

0

u(x,t)

P+dP P

Area of cross section = A(x)

FIGURE 4.4 (a) A rod with distributed loading and in longitudinal vibration; (b) a small element of the rod.

which when substituted into Equation 4.55 gives

s

ð4:59Þ

which should be compared with Equation 4.4 The analysis of the present problem may be carried outexactly as for the cable vibration In particular, the traveling wave solution will hold Mode shapeorthogonality will hold also Even the BCs are similar to those of the cable vibration problem.4.3.2 Boundary Conditions

As for the cable vibration problem, two BCs will be needed along with two initial conditions in order toobtain the complete solution to the longitudinal vibration of a rod Both free and forced vibration may beanalyzed as before For a fixed end at x ¼ x0; we will have no deflection Hence,

with the corresponding modal end condition

Xiðx0Þ ¼ 0 for i ¼ 1; 2; 3; … ð4:61ÞFor a free end at x ¼ x0; there will not be an end force Hence, in view of Equation 4.54, the applicable BCwill be

on the entire column due to a seismic jolt.

Determine the subsequent vibration motion of

the column from its initial equilibrium

›2uðx; tÞ

›t2 ¼ c2›2uðx; tÞ

›x2 þ Mg

rAlSince M ¼rAl, we have

›2uðx; tÞ

›t2 ¼ c2›2uðx; tÞ

›x2 þ g ð4:66ÞBoundary conditions are

Substitute Equation 4.74 into Equation 4.73 We have C2¼ 0: Next, use Equation 4.75 We obtain

EAliC1coslil þ kC1sinlil ¼ 0Since, C1– 0 for a nontrivial solution, the required condition is

EAlicoslil þ k sinlil ¼ 0

0

x

l k

FIGURE 4.5 A column suspended from a fixed form and supported on a flexible base.

plat-which may be expressed as

This transcendental equation has an infinite number of solutionsli; which correspond to the modes ofvibration The solution may be made computationally and the corresponding natural frequencies areobtained using

vi¼lic ¼li

ffiffiffiEr

s

¼li

ffiffiffiffiffiffiEAlM

ð4:81Þand

ðl

0sinljx dx ¼ 1

lj½1 2 cosljlAccordingly, Equation 4.79 becomes

€qjðtÞ þv2

jqjðtÞ ¼ g

ljlj½1 2 cosljl ð4:82Þwhere the right-hand side is a constant and is completely known from Equation 4.81 and Equation 4.76,and vj is given by Equation 4.77 Now Equation 4.82, which corresponds to a simple oscillator with

a constant force input, may be solved using any convenient approach For example, the particularsolution is

4.4 Torsional Vibration of Shafts

Torsional vibrations are oscillating angular motions of a device about some axis of rotation Examples arevibration in shafts, rotors, vanes, and propellers The governing PDE of the torsional vibration of a shaft

is quite similar to that we previously encountered of the transverse vibration of a cable in tension and thelongitudinal vibration of a rod However, in the present case, the vibrations are rotating (angular)motions with resulting shear strains, shear stresses, and torques in the torsional member Furthermore,the parameters of the equation of motion will take different meanings When bending and torsionalmotions occur simultaneously, there can be some interaction, thereby making the analysis more difficult.Here, we neglect such interactions by assuming that only the torsional effects are present or that themotions are quite small

Since the form of the torsional vibration equation is similar to forms we have studied before, the sameprocedures of analysis may be employed and, in particular, the concepts of modal analysis will be similar.However, the torsional parameters will be rather complex for members with noncircular cross sections.Nevertheless, vast majority of torsional devices have circular cross sections

4.4.1 Shaft with Circular Cross Section

Here, we will formulate the problem of the torsional vibration of a shaft having a circular cross section.The general case of a nonuniform cross section along the shaft is considered, but the usual assumptionssuch as homogeneous, isotropic, and elastic material are made

First, we will obtain a relationship between torque ðTÞ and angular deformation or twist ðuÞ for acircular shaft Consider a small element of length dx along the shaft axis and the cylindrical surface at ageneral radius r (in the interior of the shaft segment), as shown in Figure 4.6(a) During vibration, thiselement will deform (twist) through a small angle du:

A point on the circumference will deform through r du as a result, and a longitudinal line on thecylindrical surface will deform through angleg; as shown in Figure 4.6(a) From the strength of materialsand elasticity theory of solid mechanics, we know thatg is the shear strain Hence,

However, allowing for the fact that the angular shiftu is a function of t as well as x in the general case ofdynamics, we use partial derivatives and write

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

where G ¼ shear modulus

This shear stress acts tangentially Consider a small annular cross section of width dr at radius r ofthe shaft, as shown inFigure 4.6(b).By symmetry, the shear stress will be the same throughout thisregion and will form a torque of rt £ 2pr dr ¼ 2pr2t dr: Hence, the overall torque at the shaft crosssection is

T ¼ð2pr2t drwhich, in view of Equation 4.88, is written as

Now, we apply Newton’s Second Law for rotatory motion of the small element dx; shown inFigure 4.6(a) The polar moment of inertia of the element isÐr2dm ¼Ðr2r dx dA ¼ r dxÐr2dA ¼rJdx; where J is the polar moment of area, as discussed before Also, suppose that a distributed externaltorque oftðx; tÞ per unit length is applied along the shaft Hence, the equation of motion is

rJ dx››t2u2 ¼ T þ dT 2 T þtðx; tÞdx ¼ ›T›xdx þtðx; tÞdx

Substitute Equation 4.93 and cancel dx to obtain the equation of torsional vibration of a circularshaft as

s

ð4:97Þ

Note that Equation 4.96 is quite similar to that for transverse vibration of a cable in tension and thelongitudinal vibration of a rod Hence, the same concepts and procedures of analysis may be used Inparticular, two boundaries conditions will be needed in the solution; for example

Fixed end at x ¼ x0:uðx0; tÞ ¼ 0 ð4:98ÞFree end at x ¼ x0: ›uðx0; tÞ

4.4.2 Torsional Vibration of Noncircular Shafts

Unlike the longitudinal and transverse vibrations of rods and beams, when considering the torsionalvibration of shafts, the equation of motion for circular shafts (Equation 4.95 and Equation 4.96) cannot

be used for shafts with noncircular cross sections The reason is that the shear stress distributions in thetwo cases can be quite different, and Equation 4.88 does not hold for noncircular sections Hence, theparameter J in the torque–deflection relations (e.g., Equation 4.93 and Equation 4.94) is not the polarmoment of area in the case of noncircular sections In the noncircular case, we write

T ¼ GJt›

where Jt¼ torsional parameter

The Saint-Venant theory of torsion and the related membrane analogy, developed by Prandtl, haveprovided equations for Jt in special cases For example, for a thin hollow section

Jt¼ 4tA2s

where

As¼ enclosed (contained) area of the hollow section

p ¼ perimeter of the section

t ¼ wall thickness of the section

For a thin, solid section, we have

a ¼ length of the narrow section

t ¼ thickness of the narrow section

Torsional parameters for some useful sections are given in Table 4.2

TABLE 4.2 Torsional Parameters for Several Sections

a

t 3 a 3

Solid square

a a

Example 4.4

Consider a thin, rectangular hollow section of thickness t; height a; and width a=2; as shown inFigure 4.7(a) Suppose that the section is opened by making a small slit as in Figure 4.7(b) Study theaffect on the torsional parameter Jt and torsional stiffness K of the member due to the opening.Solution

(a) Closed section:

The contained area of the section is As¼ a2=2

The perimeter of the section p ¼ 3a

Using Equation 4.102, the torsional parameter is

Jtc¼ ta33(b) Open section:

The solid length of the section ¼ 3a

Using Equation 4.103, the torsional parameter is

Jt0¼ at3The ratio of the torsional parameters is

Jt0

Jtc ¼

3t2

a2For members of equal length, torsional stiffness will also be in the same ratio as is given by thisexpression Since t is small compared with a; there will be a significant drop in torsional stiffness due tothe opening (cutout)

there is a significant component of torsional dynamics in addition to bending Assume that the torque Tjacting on the guideway due to the jth suspension of the vehicle to be constant, acting at a point xj asmeasured from one support pier, and moving at speed vj: A schematic representation is given inFigure 4.8 The guideway span shown has a length L and a cross section which is a thin-walled rectangularbox 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

l ¼ 60 ft ð18:3 mÞabt ¼ 5 ft £ 2:2 ft £ 12 ft ð1:52 m £ 0:67 m £ 0:15 mÞ

r ¼ 4:66 slugs=ft3ð2:4 £ 103kg=m3Þ

G ¼ 1:55 £ 106lb=in:2ð1:07 £ 1010N=m2Þand vehicle speed

v ¼ 60 mi=h ð26:8 m=secÞCompute the crossing frequency ratio given by

vc¼ Rate of span crossingFundamental natural frequency of guidewayand discuss its implications

SupportPier

Guideway

Car

FIGURE 4.8 A torsional-guideway transit system.

The general solution of Equation 4.105 is

Qđxỡ Ử A1sinlx ợ A2coslxwhere Aiđi Ử 1; 2ỡ are the constants of integration The torsional BCs corresponding to the fixedends (no twist) are

Qđ0ỡ Ử Qđlỡ Ử 0where l Ử guideway span length For a nontrivial solution, we need

A2Ử 0and

The solution corresponds to an infinite set of eigenfunctionsQiđxỡ satisfying Equation 4.105 Each

of these represents a natural mode in which the beam can undergo free torsional vibrations Theactual motion consists of a linear combination of the normal modes depending on beam initialconditions and the forcing term tđx; tỡ: The integration constant A1 may be incorporated(partially) into the generalized coordinate q; which is still unknown and is determined throughinitial conditions Here, we use the normalized eigenfunctions

Qiđxỡ Ửpffiffi2sinipx

The orthogonality condition given by

1l

ffiffiffiffiffiffiffiffiGJ=rJt

TjỬ torque exerted on guideway by the jth suspension

vjỬ speed of the jth suspension

x0jỬ initial position (at t Ử 0) along the guideway of the jth suspension

dđởỡ Ử Dirac delta function

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

orthogonality relation (Equation 4.109), one obtains

d2qi

dt2 þv2

iqi¼ rJ1tl

Xn j¼1

TjQiðx0jþ vjtÞ i ¼ 1; 2; … ð4:113Þ

(b) For a single suspension entering the guideway at t ¼ 0; with the guideway initially atrest ðqið0Þ ¼ _qið0Þ ¼ 0Þ; we have n ¼ 1 and x01¼ 0: Then, the complete solution ofEquation 4.113 is

qiðtÞ ¼

ffiffi2

4.5 Flexural Vibration of Beams

In this section, we will study a beam (or rod or shaft) in flexural vibration The vibration is in thetransverse or lateral direction, which is accompanied by bending (or flexure) of the member Hence, thevibrations are perpendicular to the main axis of the member, as in the case of a cable or string, which westudied in Section 4.2 However, a beam, unlike a string, can support shear forces and bending moments

at its cross section In the initial analysis of bending vibration, we will assume that there is no axial force

at the ends of the beam We will make further simplifying assumptions that will be clear in thedevelopment of the governing equation of motion The analysis procedure will be quite similar to that wehave followed in the previous sections

The study of the bending vibration (or lateral or transverse vibration) of beams is very important in avariety of practical situations Noteworthy are the vibration analyses of structures like bridges, vehicleguideways, tall buildings, and space stations; the ride quality and structural integrity analysis of buses,trains, ships, aircraft and spacecraft; the dynamics and control of rockets, missiles, machine tools androbots; and the vibration testing, evaluation, and qualification of products with continuous members.4.5.1 Governing Equation for Thin Beams

Now, we will develop the Bernoulli–Euler equation, which governs the transverse vibration of thin beams.Consider a beam bending in the x–y plane, with x as the longitudinal axis and y as the transverse axis ofbending deflection, as shown inFigure 4.9 We will develop the required equation by considering themoment–deflection relation, rotational equilibrium, and transverse dynamics of a beam element

(1) Moment–deflection relation

A small beam element of length dx subjected to bending moment M is shown Neglect any transversedeflections due to shear stresses Consider a lateral area element dA in the cross section A of the beamelement, at a distance w (measure parallel to y) from the neutral axis of bending

Normal strain (at dA)

1 ¼ ðR þ wÞdu 2 R du

R duNote that the neutral axis joins the points along the beam when the normal strain and stress are zero.Hence,

ð

w2dA ¼ EI

Rwhere I ¼ second moment of area of the beam cross section about the neutral axis So, we have

M ¼ EI

Slope at A ¼›v=›x; slope at B ¼ ð›v=›xÞ þ ð›2v=›x2Þdx; where v ¼ lateral deflection of the beam atelement dx: Hence, the change in slope ¼ ð›2v=›x2Þdx ¼ du; where du is the arc angle of bending for thebeam element dx; as shown in Figure 4.9

y

x

δx

w R

Also, we have dx ¼ R du: Hence, ð›2v=›x2ÞR du ¼ du: Cancel du: We obtain

(2) Rotatory dynamics (equilibrium)

Again, consider the beam element dx; as shown in Figure 4.10, where forces and moments acting onthe element are indicated Here, f ðx; tÞ ¼ excitation force per unit length acting on the beam, in thetransverse direction, at location x: Disregard the rotatory inertia of the beam element

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

M þ Q dx 2 M þ ›M

›x ›x ¼ 0or