course notes for osu me 5134 introduction to … harne me 5134, intro. vib. deform. solids the ohio...

RL Harne ME 5134, Intro. Vib. Deform. Solids The Ohio State University

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Course Notes for OSU ME 5134 Introduction to Vibration of Deformable Solids Prof. Ryan L. Harne*

Department of Mechanical and Aerospace Engineering, The Ohio State University, Columbus, OH 43210, USA

*Email: [email protected]

Last modified: 2018-01-01 11:43


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Table of contents

1 Course introduction ............................................................................................................................... 5

1.1 Course objectives .......................................................................................................................... 7

2 Lumped parameter system methods of analysis .................................................................................... 8

2.1 Lumped parameter elements in structural dynamics ..................................................................... 8

2.1.1 Free response of a spring-mass system ................................................................................. 9

2.1.2 Mass-spring-damper system ............................................................................................... 11

2.1.2.1 Underdamped mass-spring-damper system .................................................................... 12

2.1.2.2 Overdamped mass-spring-damper system ...................................................................... 13

2.1.2.3 Critically damped mass-spring-damper system .............................................................. 14

2.1.2.4 Response of mass-spring-damper systems to arbitrary force excitation ......................... 15

2.1.3 Harmonic force excitation of mass-spring-damper systems ............................................... 17

2.2 Multi-degree-of-freedom (DOF) system equations of motion .................................................... 19

2.3 Free response: natural frequencies and modes ............................................................................ 20

2.4 Modal analysis ............................................................................................................................ 24

2.4.1 Modal analysis of damped systems ..................................................................................... 27

2.4.2 Modal analysis of forced systems ....................................................................................... 28

3 Principles of continuum mechanics and variational methods ............................................................. 30

3.1 Review of variational methods ................................................................................................... 30

3.1.1 Stationary values of a function ............................................................................................ 30

3.1.2 Stationary value of a functional .......................................................................................... 31

3.1.2.1 Natural boundary conditions for one-dimensional problems .......................................... 34

3.1.3 Stationary value of a functional in multiple dimensions ..................................................... 37

3.1.3.1 Natural boundary conditions for two-dimensional problems .......................................... 39

3.1.4 Variational operator ............................................................................................................ 41

3.1.4.1 Stationary value of a functional with second-order derivatives ...................................... 43

3.1.5 Hamilton's principle ............................................................................................................ 44

3.1.5.1 Hamilton's principle for Lagrangians in multiple dimensions and higher-order derivatives ....................................................................................................................................... 47

3.2 Strain and kinetic energies .......................................................................................................... 47

3.2.1 Kinetic energy ..................................................................................................................... 47


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3.2.2 Strain energy ....................................................................................................................... 48

3.2.2.1 Axial strain energy .......................................................................................................... 49

3.2.2.2 Bending strain energy ..................................................................................................... 50

3.2.2.3 Shear strain energy .......................................................................................................... 52

3.2.2.4 Torsional strain energy .................................................................................................... 53

3.2.2.5 Summary of kinetic and strain energies .......................................................................... 54

3.3 The Ritz method in applied mechanics: a first look at approximate methods of analysis .......... 54

4 Distributed parameter system methods of analysis ............................................................................. 60

4.1 Newton's laws of motion: an equilibrium approach to governing equation derivation............... 60

4.1.1 Boundary conditions ........................................................................................................... 62

4.2 Free response .............................................................................................................................. 63

4.3 Free response by assumed modes method of analysis ................................................................ 67

4.4 Forced response .......................................................................................................................... 69

4.5 Longitudinal vibration of bars with continuously varying cross-section .................................... 72

4.6 Summary ..................................................................................................................................... 73

5 Transverse vibrations of a stretched string.......................................................................................... 75

5.1 Equations of motion and boundary conditions ............................................................................ 75

5.1.1 Equilibrium approach .......................................................................................................... 75

5.1.2 Variational approach ........................................................................................................... 77

5.2 Analysis of the free response ...................................................................................................... 79

5.3 Forced response .......................................................................................................................... 83

5.4 Vibration of discontinuous strings .............................................................................................. 88

6 Torsional vibrations of a bar or shaft .................................................................................................. 95

6.1 Equations of motion and boundary conditions ............................................................................ 95

6.1.1 Equilibrium approach .......................................................................................................... 95

6.1.2 Variational approach ........................................................................................................... 97

6.2 Analysis of the free response ...................................................................................................... 98

6.3 Forced response ........................................................................................................................ 100

6.4 Forced torsional vibration of a damped shaft ............................................................................ 106

7 Transverse vibrations of a beam ....................................................................................................... 113

7.1 Equations of motion and boundary conditions .......................................................................... 113


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7.1.1 Equilibrium approach ........................................................................................................ 113

7.1.1.1 Accounting for axial force, rotary inertia, and shear flexibility .................................... 116

7.1.2 Variational approach ......................................................................................................... 119

7.2 Analysis of the free response .................................................................................................... 122

7.3 Forced response ........................................................................................................................ 126

8 Approximate methods of analyzing vibrations of deformable solids ............................................... 132

8.1 The Ritz method ........................................................................................................................ 132

8.2 Finite element method (FEM) ................................................................................................... 142

References ................................................................................................................................................. 143


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1 Course introduction

Prior to this course, we have studied the mechanics and dynamics of systems composed of discrete or lumped parameter elements. Figure 1 presents examples of such systems. Discrete or lumped parameter elements hypothetically occupy no volume. In other words, a spring is simply the manifestation of a force generated as two mass particles displace relative to each other. In fact, a mass particle is a size-less body that possesses a means to attract other mass particles to it via Newton's theory of gravitation.

Thus, in one example, the see-saw in Figure 1(e) may be modeled as the two boys' masses at ends of a rigid, steel bar that is supported by two springs in the middle. After applying Newton's 2nd law, governing equations of motion for the angle of rotation of the rigid bar and translation of the mass center may be derived, studied, and analyzed to design the next generation of see-saws.

Figure 1. Examples of systems composed of nearly discrete or lumped parameter elements. (a) Triple pendulum model of human balance. (b) Multi-story buildings. (c) Isolation tables. (d) Engine mounts between chassis and engine. (e) See-saw. (f) DC motor. (g) Automotive suspension.

Yet, we may ask whether this discretization of engineering systems is always appropriate. We may also ask at what point does the approximation 'break down' and become clearly unable to reproduce observed behaviors of engineering systems. Of course, the steel bar of the see-saw will flex under the load of the boys, and especially under the load of two large adults. In fact, even the finite dimensions of the coil springs is important, since the diameter of the coil is important for the see-saw to maintain balance.


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Thus, at a certain point in the study of engineered structures, one must take into account the actual distribution of material and examine the dynamic response of the continuous system.

In fact, our world is composed of many structures that are continuously distributed along their dimensions in ways that may apparently prevent their discretization in modeling methods to lumped parameter analogs. Figure 2 presents several examples of engineered systems that contain continuous structural elements. From strings and cables, to strut bars and subframes, to cantilevered beams and wings, Figure 2 reveals that these structural elements are not discrete. Instead, these engineering systems are plainly distributed in space in ways that contribute significantly to their dynamic response.

Figure 2. Examples of common and emerging engineered systems containing continuous structures. (a) Guitar strings. (b) Strut bar to stiffen chassis. (c) Bridge with beams and cable suspensions. (d) Microcantilever for sensing gas diffusion. (e) Vibration energy harvesting beam for self-powering microelectronics. (f) Pipelines. (g) Automotive subframe. (h) Cantilevered airplane wing. (i) Eiffel Tower.

To complement our understanding of methods to study lumped parameter systems, this course will develop our knowledge of the approaches available to model, analyze, and predict the dynamic response of continuous engineering systems found throughout our world. In fact, there are only a few distinct structural elements that may be identified in the composition of a large variety of practical engineering systems. This course will examine those such structural elements that primary undergo oscillations around one dimension. These are the stretched string or cable, the rod as it deflects in longitudinal and torsional ways, and the beam that bends and flexes. This course will study these structural elements. While it is rare that these


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structural elements are exactly replicated in practice (excepting perhaps guitar strings), small variations of these elements may be identified throughout the examples of Figure 2. For instance, the strut bar in Figure 2(b) is readily modeled as a bar of varying cross-section and mass center along its length, while the cantilevered wing in Figure 2(h) is similar to a cantilevered beam that reduces in cross-sectional area from the clamped end at the aircraft fuselage to the wing tip.

Additional common structural elements not investigated in this course are the plate, shell, and membrane. On the other hand, these additional structural elements are recognized to be two- or three-dimensional analogs to the beam, curved beam, and the string, respectively. As such, the mathematics of the two-dimensional structures are handled in similar ways as for the one-dimensional counterparts [1] [2] [3]. The foundations developed in this course will prepare the emerging engineer to examine two- and three-dimensional structures with relative ease by compatible analytical approaches.

1.1 Course objectives

The objectives of this course are to

• Establish methods of deriving equations of motion for the dynamics of continuous, one-dimensional structural elements using equilibrium relations and variational approaches

• Acquire the understanding on how to formulate and employ linear modal analyses to examine the underlying dynamic behaviors of one-dimensional engineering structures, especially in terms of how such responses are influenced by boundary conditions

• Acquire skills needed to investigate the free and forced response of structural elements and account for various common damping mechanisms

• Establish technical understanding on the use of approximate analytical methods to predict the spatially distributed vibration of continuous systems

Following a review of mathematical preliminaries in the context of lumped parameter systems, the course will include sections that will be devoted to meet each objective in full.

The skills acquired through a successful undertaking of this course will significantly enrich the young engineer's toolset of analytical methods and fundamental knowledge on how to investigate the dynamics and vibrations of numerous engineering structures that are found throughout diverse engineering applications. The skills will likewise accelerate the engineer's ability to fully and effectively utilize common simulation methods of structural dynamics study, such as the finite element method, that are created out of the principles articulated in this course.


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2 Lumped parameter system methods of analysis

This chapter reviews mathematical principles employed to study the free and forced vibrations of dynamic lumped parameter systems. These principles are applied to the continuous structural elements studied in later chapters. In addition, concepts of linear modal analysis are presented which apply equally well to the continuous systems studied in later chapters.

2.1 Lumped parameter elements in structural dynamics

The spring-mass system is the simplest mechanical system that exemplifies harmonic oscillation, also termed vibration. The spring shown in Figure 3(a) is mass-less and originally rests with an undeformed length. The spring constant is k such that a force kF [N] applied at an end of the spring deflects the free

end by an amount y [m] results in an equal reaction force kF ky= . The units of k are [N/m]. When the

mass m is attached to the end of the spring, with the gravitational acceleration pointing down the page, an equilibrium is achieved between the spring force and gravitational force, so that k mg∆ = . The gravitational

acceleration is g in units [m/s2] while the mass m has units of [kg].

Figure 3. (a) Undeflected spring affixed on one side to a rigid base. (b) Mass applied to spring free end showing equilibrium. Different ways by which to denote dynamic motion of the mass about the original equilibrium, whether (c) about the spring undeflected position or (d) about the spring-mass equilibrium point.

If the mass then deflects in time around this equilibrium point as shown in Figure 3(c), then an application of Newton's 2nd law results in the equation of motion (EOM).

( ) ( )2

2

d x tm kx t mg

dt= − +

(2.1.1)

The EOM is an ordinary differential equation (ODE) due to dependence only on time t . The (2.1.1) is often rewritten as

mx kx mg+ =

(2.1.2)

where the functional dependence notation on ( )t is dropped for brevity and the over-dot operator indicates

a time derivative. The consequence of describing the mass displacement x as referenced from the spring


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undeflected position is that gravitational force remains in the EOM. A positive sign on the gravitational force is given on the right-hand side (RHS) of (2.1.2) due to the fact that the force works in the direction considered as positive in x .

On the other hand, as shown in Figure 3(d), if the mass displacement is referenced with respect to the

equilibrium of the mass-spring system, thus using ( )x t , the EOM becomes

( )mx k x mg= − + ∆ + (2.1.3)

which may be simplified to

0mx kx+ = (2.1.4)

due to the equilibrium condition k mg∆ = .

Consequently, expressing the response coordinate x of mass motion using the static equilibrium configuration enables one to neglect the body forces that are exerted upon the mass particles. This is true regardless of the body force involved, although gravity is the most common example since it is often uniform over the volume of a single body in many engineering and science applications.

2.1.1 Free response of a spring-mass system

Experience tells us that a mass suspended from a spring oscillates in harmonic motion. Thus, for the mass

originally displaced by ( ) 00x t x= = and given an initial velocity ( ) 00x t x= = , we may assume that the

mass oscillates in the way described by

( ) [ ]sin nx t A tω φ= + (2.1.1.1)

where A is an amplitude of displacement and φ is a phase angle. Substitution of (2.1.1.1) into (2.1.4) leads

to

[ ] [ ]2 sin sin 0n n nm A t kA tω ω φ ω φ− + + + = (2.1.1.2)

so that we find

2 ;n nk km m

ω ω= = (2.1.1.3)

We define the angular natural frequency as nω with units of [radians/s]. The frequency in [Hz=cycles/s]

is / 2n nf ω π= . The natural period of oscillation is 2 /n nT π ω= having units [s].

To determine the unknowns A and φ , we apply the initial conditions.

[ ]0 sin 0 sinnx A Aω φ φ= + = (2.1.1.4)

[ ]0 cos 0 cosn n nx A Aω ω φ ω φ= + = (2.1.1.5)


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Solving (2.1.1.4) and (2.1.1.5) simultaneously leads to

2 2 20 0n

n

x xA

ωω

+=

and 0

0

tan n xx

ωφ =

(2.1.1.6)

Putting these pieces back together, we find that the mass motion of the spring-mass system vibrates according to

( )2 2 2

0 0 1 0

0

sin tann nn

n

x x xx t tx

ω ωω

ω− +

= +

(2.1.1.7)

Note that by Euler's identity

cos sinje jθ θ θ± = ± (2.1.1.8)

where 1j = − is the imaginary number, the assumed solution made via (2.1.1.1) is just one way of

obtaining the final result (2.1.1.7). Indeed, we may also assume

( ) 1 2cos sinn nx t A t A tω ω= + (2.1.1.9)

By applying the initial conditions to (2.1.1.9), we obtain

( ) 00 cos sinn n

n

xx t x t tω ωω

= +

(2.1.1.10)

Trigonometric manipulations will show that (2.1.1.10) is equivalent to (2.1.1.7).

Finally, a further extension of Euler's identity and grasp of solutions to ordinary differential equations [4] encourages one to assume that the harmonic mass motion response is the real part of

( ) j tt e ω=x A (2.1.1.11)

where the bold font indicates a complex number. Thus, a solution to (2.1.4) would be [ ]Re x . To examine

this approach, substitution of (2.1.1.11) into (2.1.4) leads to

2 0j tm k e ωω − + = A (2.1.1.12)

Non-trivial solutions to (2.1.1.12) are found when / nk mω ω= = , which agrees with the prior result. In

addition, the use of the initial conditions results in

[ ]0 Re Re cosjx Ae Aθ θ = = = A (2.1.1.13)

with A as the magnitude and θ as the phase of the complex amplitude A , and

0 Re sinjn nx j Ae Aθω ω θ = = − (2.1.1.14)


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Then solving (2.1.1.13) and (2.1.1.14) simultaneously results in

2 2 20 0n

n

x xA

ωω

+=

and 0

0

tann

xx

θω−

=

(2.1.1.15)

Because [ ]Re x is the solution to (2.1.4), we have that

[ ] ( ) [ ]Re cos nx t A tω θ= = +x (2.1.1.16)

Simple trigonometry shows the equivalence of (2.1.1.16) to (2.1.1.1) according to the results via (2.1.1.15) and (2.1.1.6). It is therefore clear that either complex exponentials or trigonometric functions in their various forms may be suitable assumed response solutions to the harmonic oscillation behaviors of mechanical systems governed by ODEs and subjected to initial conditions.

2.1.2 Mass-spring-damper system

The oscillations of many engineering systems diminish in displacement amplitude as time elapses. This is a damping phenomenon associated with a conversion of the kinetic energy into heat. A common lumped parameter element to model damping behaviors is the dashpot damper, Figure 4(a), that is realized in shock absorbers in vehicle systems. In dashpot dampers, fluid is forced through orifices, which results in turbulence. The turbulence induces energy dissipation according to transfer of the flow energy to heat energy.

The force exerted by a dashpot due to a relative end deformation rate y [m/s] is cF cy= [N]. The damper

constant c has units [N.s/m].

Figure 4. (a) Schematic of dashpot damper, forcing liquid or air through orifices according to the rate of relative

deformation y of dashpot ends. (b) Schematic of spring-mass-damper system.

Thus, when the dashpot is placed in parallel with the spring, a mass-spring-damper system, Figure 4(b). Using Newton's 2nd law, the EOM is

0mx cx kx+ + = (2.1.2.1)

A more general means of solving (2.1.2.1) is adopted from foundations of solving ODEs [4]. We assume that

( ) tx t aeλ= (2.1.2.2)


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and substitute (2.1.2.2) into (2.1.2.1).

2 0tm c k aeλλ λ + + = (2.1.2.3)

Non-trivial results for a in (2.1.2.3) require that the bracketed term is equal to zero. This reveals the characteristic equation

2 0m c kλ λ+ + = (2.1.2.4)

which is solved for

21,2

1 42 2c c kmm m

λ = − ± − (2.1.2.5)

For convenience, we define a critical damping coefficient as

2 2cr nc km mω= = (2.1.2.6)

This enables one to define the damping ratio

2 2r n

c c cc m km

ζω

= = = (2.1.2.7)

The damping ratio quantifies the relative proportion of damping forces in the system with respect to inertial forces. Using the damping ratio definition, (2.1.2.5) is given by

21,2 1n nλ ζω ω ζ= − ± − (2.1.2.8)

2.1.2.1 Underdamped mass-spring-damper system

When the system is underdamped such that 0 1ζ< < , we express (2.1.2.8) using

21,2 1n njλ ζω ω ζ= − ± − (2.1.2.1.1)

and define the damped natural frequency

21d nω ω ζ= − (2.1.2.1.2)

The damped natural frequency quantifies the deviation of the natural frequency from the undamped value

nω that is associated with the damping forces.

It is consequently found that, by way of the two roots in (2.1.2.1.1), the total solution (2.1.2.2) to the EOM (2.1.2.1) is

( ) 1 2n d dt j t j tx t e a e a eζω ω ω− − = + (2.1.2.1.3)

By Euler's identity, the result of (2.1.2.1.3) simplifies to


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( ) [ ]sinntdx t Be tζω ω β−= + (2.1.2.1.4)

where the amplitude B and phase β are determined by the initial conditions to be

( ) ( )2 20 0 0

2n d

d

x x xB

ζω ωω

+ +=

and 0

0 0

tan d

n

xx x

ωβ

ζω=

+ (2.1.2.1.5)

The type of dynamic behavior exemplified by (2.1.2.1.4) is transient decay of oscillation. For instance,

given an initial displacement, the mass will return to rest at the static equilibrium *x =0 [m] after a sufficient time has elapsed, Figure 5.

Figure 5. Initial conditions of 0x =1 [m] and 0x =0 [m/s] for underdamped mass-spring-damper oscillator.

2.1.2.2 Overdamped mass-spring-damper system

When the damping ratio ζ >1, the system is termed overdamped and the roots obtained from the

characteristic equation (2.1.2.4) are two real numbers. Then, the motion of the mass is

( ) 2 21 11 1

n nn t ttx t e a e a eω ζ ω ζζω − − −− = + (2.1.2.2.1)

with

20 0

1 2

1

2 1

n

n

x xa

ζ ζ ω

ω ζ

− + − + − =−

(2.1.2.2.2)

and

20 0

2 2

1

2 1

n

n

x xa

ζ ζ ω

ω ζ

+ + − =−

(2.1.2.2.3)

Exemplary free response time series for the overdamped mass-spring-damper system are shown in Figure 6.

0 10 20 30 40 50 60 70 80-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1Response to Initial Conditions

Time (seconds)

Am

plitu

de (m

)


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Figure 6. Initial conditions of (blue solid curve) 0x =1 [m] and 0x =0 [m/s], (green dash-dot curve) 0x =0 [m] and 0x =1

[m/s], and (red dashed curve) 0x =0 [m] and 0x =-1.5 [m/s] for overdamped mass-spring-damper oscillator.

2.1.2.3 Critically damped mass-spring-damper system

When the damping ratio is exactly ζ =1, the system is termed critically damped and a pair of repeated roots

is obtained by solving the characteristic equation (2.1.2.4). Then we find

( ) ( )0 0 0n nt t

nx t x e x x teω ωω− −= + + (2.1.2.3.1)

Figure 7 contrasts the under-, over-, and critically damped systems. The critically damped system returns the mass to rest in the shortest time, while the underdamped system undergoes oscillations about the equilibrium. The overdamped system is the slowest system to return to equilibrium and it does so without oscillations.

Figure 7. Initial conditions of 0x =1 [m] and 0x =0 [m/s] for the mass-spring-damper oscillation that is (blue solid)

underdamped with ζ =0.5, (green dash-dot curve) critically damped with ζ =1, and (red dashed curve) overdamped with

ζ =2.

0 5 10 15 20 25-0.4

-0.2

0

0.2

0.4

0.6

0.8

1Response to Initial Conditions

Time (seconds)

Am

plitu

de (m

)

0 5 10 15-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

1.4Response to Initial Conditions

Time (seconds)

Am

plitu

de (m

)


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2.1.2.4 Response of mass-spring-damper systems to arbitrary force excitation

A forced mass-spring-damper system is shown in Figure 8. Due to the wide-ranging relevance of harmonic force excitation in structural dynamics applications, a significant proportion of this course will give attention to harmonic (e.g. sinusoidal) force excitation ( ) 0 cosF t F tω= . Prior to adopting such particular

focus, this Sec. 2.1.2.4 reviews a primary method by which the response of linear mass-spring-damper systems is determined given knowledge about an arbitrary force excitation.

Figure 8. Arbitrary force excitation on mass-spring-damper system.

The equation of motion for the system shown in Figure 8 is

( )mx cx kx F t+ + = (2.1.2.4.1)

The Laplace transform method is the primary method by which the response of the system to an arbitrary force ( )F t may be determined. A review of the use of Laplace transforms as a means to solve ordinary

differential equations is given in Refs. [5] [6]. The Laplace transform of (2.1.2.4.1) is (2.1.2.4.2).

( ) ( ) ( ) ( ) ( ) ( ) ( )2 0 0 0m s X s sx x c sX s x kX s F s − − + − + = (2.1.2.4.2)

The Laplace transform operator is denoted by ( )L . Thus, the Laplace transform of the mass displacement

( )x t is ( ) ( )L x t X s= while the Laplace transform of the force excitation ( )F t is ( ) ( )L F t F s= . The

initial conditions appear in (2.1.2.4.2) in the general case.

Collecting together the terms of (2.1.2.4.2), we have

( ) ( ) ( ) ( ) ( )2 0 0ms cs k X s ms c x mx F s + + = + + + (2.1.2.4.3)

The Laplace transform of the mass displacement is therefore found from

( ) ( ) ( ) ( )2 2 2

10 0ms c mX s x x F sms cs k ms cs k ms cs k

+= + +

+ + + + + + (2.1.2.4.4)

which is alternatively written as

( ) ( ) ( ) ( )1 2 3X s X s X s X s= + + (2.1.2.4.5)


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The linearity of the system under consideration permits one to take the inverse Laplace transform of each component in (2.1.2.4.5) individually and superpose the three components to find the total mass displacement response in time via

( ) ( ) ( ) ( )( ) ( ) ( ) ( )

1 1 1 11 2 3

1 2 3

L X s L X s L X s L X s

x t x t x t x t

− − − −= + + = + +

(2.1.2.4.6a,b)

Numerous examples of this procedure are given in Refs. [5] [6]. We consider an example here where the initial conditions are zero-valued. As such, we aim to find the inverse Laplace transform of

( ) ( )2

1X s F sms cs k

=+ +

(2.1.2.4.7)

To this end, knowledge of the force ( )F t is required in order to find the Laplace transform ( )F s such as

from available tables or by direct integral computation. As one example, consider a step input form of the force excitation in time

( )0

0; 0; 0t

F tF t

<= >

(2.1.2.4.8)

The Laplace transform of (2.1.2.4.8) is (2.1.2.4.9), as found from the direct integration or by assessing available tabulated transforms.

( ) 0FL F t

s= (2.1.2.4.9)

As a result, (2.1.2.4.7) becomes

( ) ( )0

2

FX s

s ms cs k=

+ + (2.1.2.4.10)

In order to solve for the inverse Laplace transform of (2.1.2.4.10) in order to determine the time response

of the mass displacement, ( ) ( )1x t L X s−= , it is wise to put (2.1.2.4.10) into a form that may be identified

from tables.

( ) 0

2

1FX s

c km s s sm m

= + +

(2.1.2.4.11)

The quadratic polynomial in the denominator of (2.1.2.4.11) must be simplified. In general, a quadratic polynomial will have two roots. These roots may both be real or may be a complex conjugate pair. Thus, a general way of writing the polynomial in the denominator of (2.1.2.4.11) is to expand it in terms of its roots, or poles. The (2.1.2.4.12) adopts this notation.

( ) ( )( )0

1 2

1FX s

m s s p s p=

+ + (2.1.2.4.12)


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Partial-fraction expansion is then used on (2.1.2.4.12) to decompose the higher-order denominator into low order components that are readily identified from tabulated Laplace transforms.

( ) 0 31 2

1 2

F cc cX sm s s p s p

= + + + + (2.1.2.4.13)

Once the constants 1c , 2c , and 3c are identified using the standard partial-fraction expansion approach [5]

[6], we take the inverse Laplace transform of (2.1.2.4.13) to find the time response of the mass displacement

( )x t .

( ) ( ) 1 21 01 2 3

p t p tFL X s x t c c e c e

m− −− = = + + (2.1.2.4.14)

When the two roots 1p and 2p are real, then (2.1.2.4.14) does not require simplification. When the two

roots are a complex conjugate pair, the (2.1.2.4.14) may be simplified using Euler's identity cos sinje jθ θ θ± = ± .

Refs. [5] [6] provide numerous examples of the complete Laplace transform method procedure, along with partial-fraction expansion. Interested readers are referred to those references for further examples. Much of the remainder of these notes pertain exclusively to harmonic force excitation of linear mass-spring-damper or continuous structural systems, for which the Laplace transform method is not the preferred solution technique.

2.1.3 Harmonic force excitation of mass-spring-damper systems

Consider that a harmonic force ( ) 0 cosF t F tω= acts on the mass-spring-damper system, Figure 9. The

EOM is therefore

0 cosmx cx kx F tω+ + = (2.1.3.1)

Dividing the equation by the mass leads to the result that

202 cosn nx x x f tζω ω ω+ + = (2.1.3.2)

where / 2 nc mζ ω= , /n k mω = , and 0 0 /f F m= . In Sec. 2.1.2, the homogeneous form of Eq. (2.1.3.2)

is solved for the case of different damping ratios ζ . Of course for an equation like (2.1.3.2) having

harmonic force excitation, the total solution for the displacement response ( )x t consists of the

complementary solution to the homogeneous equation part associated with the initial conditions and the particular solution associated with the harmonic force acting in the steady-state. The steady-state refers to response trends that occur at a time sufficiently great such that responses associated with the initial conditions are negligible. This sub-section, and many sections throughout the latter part of this course, consider only the steady-state response behaviors of mechanical and structural systems. In fact, this attention to steady-state response is conventional in many engineering and science practices of structural


18

dynamics. This is due to the principle of linear superposition and Fourier's theorem that together dictate the response of a structural system to an arbitrary input is equal to the individual responses of the system to an infinite number of single-frequency component inputs found from a Fourier series expansion of the arbitrary input.

Figure 9. Harmonic force excitation of the mass-spring-damper system.

To solve for the steady-state response, we assume that the oscillator mass moves with the same period of

motion as the harmonic force. As a result, the particular response ( )px t of the mass is assumed to be

( ) [ ]cos cos sinp p px t X t A t B tω θ ω ω= − = + (2.1.3.3)

where 2 2p pX A B= + and tan /p pB Aθ = .

By substituting (2.1.3.3) into (2.1.3.2), we have

2 2 2 202 cos 2 sin 0p n p n p p n p n pA B A f t B A B tω ζω ω ω ω ω ζω ω ω ω − + + − + − − + = (2.1.3.4)

The equation (2.1.3.4) must be valid for all time t . Thus, the coefficients to the sinusoidal terms must be individually zero.

2 202p n p n pA B A fω ζω ω ω− + + = (2.1.3.5)

2 22 0p n p n pB A Bω ζω ω ω− − + = (2.1.3.6)

Simultaneously solving (2.1.3.5) and (2.1.3.6) leads to

( )2 20 /p nA fω ω σ= − and 02 /p nB fζω ω σ= where ( ) ( )2 22 2 2n nσ ω ω ζω ω= − + (2.1.3.7)

With the steady-state response solved, the total response of the underdamped mass-spring-damper system due to the initial conditions and harmonic force excitation is

( ) [ ] [ ]sin cosntdx t Be t X tζω ω β ω θ−= + + − (2.1.3.8)

As observed via (2.1.3.7), when the harmonic excitation frequency ω equals the natural frequency nω , the

denominator term σ is minimized, while the cosine response coefficient pA is eliminated. This results in

the displacement of the mass oscillating with 90 degrees of lag respecting the harmonic force. Although the term pA vanishes in this condition, it is found through a transform of the coefficients pA and pB to the


19

magnitude X that the response amplitude of the mass is approximately maximized in the case of nω ω=

which is termed resonance. At resonance, the velocity of the mass is perfectly in phase with the driving force and the response amplitude of the mass is most magnified.

2.2 Multi-degree-of-freedom (DOF) system equations of motion

The single degree-of-freedom (DOF) system studied in Sec. 2.1 is a simplification of the built-up nature of many engineering systems that are composed of numerous elements that may be considered as possessing unique degrees-of-freedom of motion coupled through elastic and dissipative elements. Here, we extend the basic concepts of EOM formulation and solution to the case of multi-DOF systems.

Figure 10. Multi-DOF system composed of mass-spring-damper sub-systems.

Consider Figure 10 that shows a series attachment of mass-spring-damper sub-systems, totaling a number N of such sub-systems. Using Newton's 2nd law to derive the EOM for the first mass shows

( ) ( )1 1 1 1 1 2 1 2 2 1 2m x c x k x c x x k x x+ + = − − − − (2.2.1)

The same routine for the second mass shows

( ) ( ) ( ) ( )2 2 2 2 1 2 2 1 3 2 3 3 2 3m x c x x k x x c x x k x x= − − − − − − − − (2.2.2)

Repeating the routine for the thn mass gives an EOM of

( ) ( ) ( ) ( )1 1 1 1 1 1n n n n n n n n n n n n n nm x c x x k x x c x x k x x− − + + + += − − − − − − − − (2.2.3)

Finally, the thN mass is governed by the EOM

( ) ( )1 1N N N N N N N Nm x c x x k x x− −= − − − − (2.2.4)

It is convenient to express equations such as (2.2.1) to (2.2.4) in a matrix form


20

1 1

2 2

3 3

1 1

1 2 2 1

2 2 3 3 2

3 3 4

1

0 0 0 00 0 0 00 0 0 0

0 0 0 00 0 0 0

0 0 00 0

0 0 0

0 0 00 0 0

N N

N N

N N N

N N

m xm x

m x

m xm x

c c c xc c c c x

c c c

c c cc c

− −

−

+ − − + − − +

+ + −

−

3

1

1 2 2 1

2 2 3 3 2

3 3 4 3

1 1

0 0 0 00 0 0

0 0 0 0

0 0 0 00 0 0 0

N

N

N N N N

N N N

x

xx

k k k xk k k k x

k k k x

k k k xk k x

−

− −

+ − − + − − +

+ = + −

−

(2.2.5)

For convenience, (2.2.5) is expressed via

M C K 0x x x+ + = (2.2.6)

The upright capitals denote matrices. Here, the matrices are square. The over-bar symbols denote vectors. The association of the matrices M , C , and K in (2.2.6) with the components in (2.2.5) is made by observation.

The matrices in (2.2.6) are termed the mass matrix M , damping matrix C , and stiffness matrix K . If

external, time-dependent forces are applied to each mass, [ ]1 2 3 1, , ,..., , TN Nf f f f f f−= , where ( )T is the

transpose operation, then the RHS of (2.2.6) may be replaced with f .

2.3 Free response: natural frequencies and modes

To solve for the free response of (2.2.6) due to a set of initial conditions

1,0

2,0

3,00

1,0

,0

N

N

xxx

x

xx

−

=

and

1,0

2,0

3,00

1,0

,0

N

N

xxx

x

xx

−

=

(2.3.1)


21

the undamped [ ]C=diag 0 and unforced 0f = problem is first solved. The equation of interest is therefore

M K 0x x+ = (2.3.2)

Using a general approach to solve the ODE system in (2.3.2), we assume

j te ω=x a (2.3.3)

The term [ ]1 2, ,..., TN=a a a a contains complex-valued displacement amplitudes, with the understanding that

the real part of (2.3.3) is the actual (or physical) solution to (2.3.2).

By substitution of (2.3.3) into (2.3.2), we have

2M K 0j te ωω − + = a (2.3.4)

Non-trivial solutions for the complex amplitudes a are obtained when the determinant of the matrix in brackets in (2.3.4) is zero. In other words, the non-trivial solutions exist so long as the bracketed matrix is non-invertible. A non-invertible matrix has a determinant that is equal to zero [4].

An example clarifies the concept and outcomes. Consider a 2-DOF system, so that the mass and stiffness matrices are

1

2

0M

0m

m

=

and 1 2 2

2 2

Kk k k

k k+ −

= − (2.3.5)

Then a solution to (2.3.4) is such that

21 1 2 2

22 2 2

det 0m k k k

k m kω

ω − + + −

= − − + (2.3.6)

Consequently, taking the determinant operation yields

( )4 21 2 1 2 2 1 2 2 1 2 0m m m k m k m k k kω ω− + + + = (2.3.7)

In a similar way as obtained for the single DOF system via (2.1.2.4), the (2.3.7) is the characteristic

equation for the 2-DOF system. Using the values of ( ) ( )1 2 1 2, , , 9,1,48,6m m k k = results in

1 4ω = ± [rad/s] and 2 8ω = ± [rad/s] (2.3.8)

The negative frequencies resulting from solution of the characteristic equation (2.3.7) are the mathematical outcome of using the complex exponential formulation of assumed solution. By substitution of the values in (2.3.8) into (2.3.3) and using Euler's identity, only the positive frequencies are retained.

While the results of (2.3.8) determine the frequencies at which the free oscillations occur, the amplitudes a shown in (2.3.4) must be determined for each frequency from (2.3.8). To compute the amplitudes, we substitute one of the frequencies of (2.3.8) into (2.3.4) and solve the simultaneous equations. In the case of the 2-DOF example studied here, this becomes


22

211 1 2 2

222 2 2

00

m k k kk m k

ωω

− + + − = − − +

aa

(2.3.9)

Considering only the positive frequencies found from (2.3.8), the equations of (2.3.9) may be written in terms of only the real parts of the complex amplitudes, so that

211 1 2 2

222 2 2

00

am k k kak m k

ωω

− + + − = − − +

(2.3.9)

Then, to determine each pair ( )1 2,a a that corresponds to each frequency in (2.3.8), the individual

substitutions are made

2111 1 1 2 2

2212 1 2 2

00

am k k kak m k

ωω

− + + − = − − +

and 2

122 1 1 2 22

222 2 2 2

00

am k k kak m k

ωω

− + + − = − − +

(2.3.10)

Using the values given above, we find that

( )( )

1111 21

21

4 9 48 6 6 06 2 0

6 4 1 6 0a

a aa

− + + − = → − + = − − +

(2.3.11)

( )( )

1212 22

22

8 9 48 6 6 06 2 0

6 8 1 6 0a

a aa

− + + − = → − − = − − +

(2.3.12)

It is found that the two equations each (2.3.11) are not unique, so that only a ratio of 11a and 12a may be

determined. Here it is computed that 21 113a a= . A similar situation results in (2.3.12), so that 22 123a a= − .

Thus, two vectors may be expressed using

1

13

a =

and 2

13

a = −

(2.3.13)

Therefore, returning to the assumed solution (2.3.3), by substitution of the results obtained above, we find

( ) 1 1 2 21 2

j t j t j t j tt ae be a ce de aω ω ω ω− − = + + + x (2.3.14)

Application of Euler's identity shows that only the real part of (2.3.14) is retained so that

( ) [ ] [ ]1 1 1 1 2 2 2 2sin sinx t A t a A t aω φ ω φ= + + + (2.3.15)

Whether considering (2.3.14) or (2.3.15), the unknown constants ( ), , ,a b c d or ( )1 2 1 2, , ,A A φ φ are

determined by an application of the initial conditions to the final expressions for the system response.

Thus, when given initial conditions, the 2-DOF system responds according to (2.3.15) which are harmonic oscillations at frequencies iω and with proportions of displacement given by the vectors ia .


23

The frequencies iω are called natural frequencies. The corresponding proportions of displacement response

ia are called mode shapes. By ordering the frequencies in increasing values, we see that the natural

frequency 1ω =2 [rad/s] is associated with the first mode shape where the proportions of displacement are

21 113a a= . The second mode shape occurs at frequency 2 8ω = [rad/s] in the proportions 22 123a a= − .

Using initial conditions in an example helps to understand how these motions correspond to the oscillations

of the 2-DOF system. Consider ( ) ( )10 20, 10 20, , 1,0,0,0x x x x = where the units are [m] for the displacements

and [m/s] for the initial velocities. Using (2.3.15), we have

( )( ) [ ] [ ]

[ ]1 1 1 1

1 22 2 2 2

sinsin

x t A ta a

x t A tω φω φ

+= +

(2.3.16)

( )( )

[ ][ ]

1 1 1 1

2 2 2 2

sin1 1sin3 3

x t A tx t A t

ω φω φ

+ = +−

(2.3.17)

Applying the initial conditions leads to 4 equations

1 1 2 2

1 1 2 2

1 1 2 2

1 1 2 2

1 sin sin0 3 sin 3 sin

0 2 cos 8 cos

0 6 cos 3 8 cos

A AA A

A A

A A

φ φφ φ

φ φ

φ φ

= += −

= +

= −

(2.3.18)

Solving leads to 112

A = [m], 212

A = [m], 1 2πφ = [rad], and 2 2

πφ = [rad]. Consequently, by (2.3.16) or

(2.3.17), we find that the response of the 2-DOF system due to the initial conditions is a combination of the modes with the amplitudes and phases

( )11 1cos 2 cos 82 2

x t t t= + [m] and ( )23 3cos 2 cos 82 2

x t t t= − [m] (2.3.19)

Figure 11 shows a portion of this time series of displacement resulting from the initial conditions on the system. Two frequencies may be identified in the time series of Figure 11: a higher frequency term and a slower modulation frequency. A spectrum analysis of the transient dynamics would confirm that these

frequencies are 1ω =2 [rad/s] and 2 8ω = [rad/s].


24

Figure 11. Time series of displacement for 2-DOF system.

Were the initial conditions selected such that 2A =0 and 1A ≠ 0, this would result in oscillation of the system

only in the first mode in which case the masses oscillate in-phase with the proportion of displacement amplitude given as 1/3 between the mass 1m and mass 2m . Conversely, were the initial conditions chosen

such that 1A =0 and 2A ≠ 0, then the system vibration would occur only in the second mode.

2.4 Modal analysis

A systematic method termed modal analysis is available to determine the modes and natural frequencies of a multi-DOF system and to assess its free and forced response. The overview of this approach is provided here. The presentation here is based on the description of the method given in Ref. [7]. Greater mathematical background on this technique is provided in Ref. [7].

First, we return to the undamped and unforced system equation (2.3.2) repeated here for convenience

M K 0x x+ = (2.4.1)

Along with this system, the initial conditions are known

( ) 00x x= and ( ) 00x x= (2.4.2)

A coordinate transformation is defined, using

1/2Mx q−= (2.4.3)

The matrix square-root 1/2M− operation is defined such that 1/2 1/2M M M= , so that 1/2 1/2M MM I− − = where

I is the identity matrix. Then, (2.4.3) is substituted into (2.4.1) and is then pre-multiplied by 1/2M− .

1/2 1/2

1/2 1/2 1/2 1/2

1/2 1/2

MM KM 0M MM M KM 0I M KM 0

q qq q

q q

− −

− − − −

− −

+ =

+ =

+ =

(2.4.4a,b,c)

0 2 4 6 8 10 12 14 16 18 20-3

-2

-1

0

1

2

3

time [s]

disp

lace

men

t [m

]red dash, x1. blue solid, x2


25

We now define the symmetric matrix 1/2 1/2K M KM− −= . A symmetric matrix is such that it equals its own

transpose K=KT .

I K 0q q+ = (2.4.5)

The eigenvalue problem stemming from (2.4.5) is then solved assuming j te ω=q v . We substitute this

assumed solution into (2.4.5) to find

2

2

- +K 0K

j t j te eω ωω

ω

=

=

v vv v

(2.4.6a,b)

For an N -DOF system, there are N natural frequencies ω and mode shapes v resulting from solution of

(2.4.6b). If we replace 2λ ω= in (2.4.6b), we have

K λ=v v (2.4.7)

The (2.4.7) is called the symmetric eigenvalue problem. The solutions to (2.4.7) are the eigenvalues λ and eigenvectors v . Thus, hereafter we will interchangeably refer to the v as either mode shapes or eigenvectors. The term eigenvalues λ will likewise be understood to denote the squared natural frequencies

2iω (specifically, the angular natural frequencies iω in units [rad/s]).

The next step is to normalize the eigenvectors. The normalization sought is to make the norm of the eigenvectors equal to 1. In other words, we find a constant α for any given eigenvector so that

( ) ( ) 1Ti iu uα α = (2.4.8)

Thus, the normalized eigenvector is v uα= .

Such normalization is achieved by the operation

1T

v uu u

= (2.4.9)

This normalization procedure results in a set of orthonormal eigenvectors so that

Ti j ijv v δ= (2.4.10)

where ijδ is the Kronecker delta

1,0,ij

i ji j

δ=

= ≠ (2.4.11)

As of MATLAB 2017a, if one computes (2.4.7) using the command [eig_vec,eig_val]=eig(K_tilde), where

K_tilde = K , then the resulting eigenvectors given by the columns of eig_vec are already normalized to be orthonormal.


26

Of course, it is wise for one to check of the orthonormalization. This can be completed by the MATLAB command v_iii=1./(eig_vec(:,iii).'*eig_vec(:,iii)).^(1/2)*eig_vec(:,iii), where iii is an index of the eigenvector columns. If v_iii.'*v_jjj = the Kronecker delta according to (2.4.11), then the desired orthonormalization is achieved.

Then, a matrix is defined from the N orthonormal eigenvectors

[ ]1 2 3 1P N Nv v v v v−= (2.4.12)

The corresponding eigenvalues of K are denoted by [ ]1 2 3 1T

N Nλ λ λ λ λ λ−= .

A new coordinate transformation is next expressed by

Pq r= (2.4.13)

Due to the composition of the matrix P , we find that P P IT = . By substituting (2.4.13) into (2.4.5) and pre-

multiplying by PT , we find

P KP 0P P P KP 0I P KP 0

T T

T

r rr r

r r

+ =

+ =

+ =

(2.4.14a,b,c)

It is found that

2P KP=Tidiag ω Λ =

(2.4.15)

In other words, the spectral matrix Λ is a diagonal matrix composed of the eigenvalues λ (also the squared natural frequencies 2ω ) along the diagonal.

What is the significance of the steps to this stage? Consider (2.4.16)

I 0r r+ Λ = (2.4.16)

which is the alternative expression of (2.4.14c) using the spectral matrix notation (2.4.15). The identity and spectral matrices are both diagonal. In other words, (2.4.16) expresses that

21 11

22 22

2

21 1 1

22 2 2

2

1 0 0 00 00 1 0 00 0

0 000 0 0 1 00 0 0

00

0

N NN

N N N

r rr r

r r

r rr r

r r

ωω

ω

ωω

ω

+ =

+ = + =→ + =

(2.4.17a,b)


27

The (2.4.17b) shows that the transformed coordinate ir is independent of coordinates jr for i j≠ . In other

words, the response of each coordinate may be examined by solution to its independent ordinary differential equation.

Solutions to (2.4.17b) are readily obtained from principles described in Sec. 2.1. The transform of the initial conditions proceeds via

( ) 1/200 P MTr x= and ( ) 1/2

00 P MTr x= (2.4.18)

Once solutions to (2.4.17b) are obtained with the initial conditions (2.4.18), the physical responses are reconstructed by undertaking the reverse transforms

( ) ( )1/2M Px t r t−= (2.4.19)

The method outlined above is termed modal analysis. Modal analysis is central to the investigation of many engineering structures, including the reverse identification of system parameters for sake of simplified modeling [7]. Additional information may be found in the text [8] and review [9].

2.4.1 Modal analysis of damped systems

The extension of linear modal analysis to accommodate the inherent dissipation observed in many engineering contexts is straightforward when a certain and common form of damping is manifest. Specifically, when the damping matrix C may be composed from a linear combination of the mass and stiffness matrices by

C M Kα β= + (2.4.1.1)

a form of damped modal analysis may be undertaken. The damping form given in (2.4.1.1) is termed Rayleigh damping and also termed proportional damping. The constants α and β are respectively the

mass- and stiffness-proportional damping constants.

When (2.4.1.1) is introduced to (2.4.1), we have

( )M M K K 0x x xα β+ + + = (2.4.1.2)

Then, undertaking the procedures of linear modal analysis, we first obtain

( )I K K =0q q qα β+ + + (2.4.1.3)

The undamped eigenvalue problem is then solved to find Λ and P as exemplified in Sec. 2.4. Then the substitution Pq r= is made to (2.4.1.3) to yield

( )I+ 0r r rα β+ Λ + Λ = (2.4.1.4)

The matrix I+α βΛ is observed to be diagonal and is of a form

[ ]I+ 2 i idiagα β ζ ωΛ = (2.4.1.5)


28

In other words, similar to the decoupling of equations observed in (2.4.17), the decoupled modal equation

for the thi DOF that is proportionally damped is

22 0i i i i i ir r rζ ω ω+ + = (2.4.1.6)

The modal damping ratio for the thi mode is then computed to be

2 2i

ii

βωαζω

= + (2.4.1.7)

By the similarity of solving (2.4.1.6) to that shown in Sec. 2.1.2, the modal coordinates ir are computed

using the transformed initial conditions (2.4.18), after which the physical response vector is computed from the inverse transformation (2.4.19).

As this Sec. 2.4.1 reveals, when the damping of the system is proportional to mass and/or stiffness, a direct extension of linear modal analysis may be undertaken. Fortunately, a large number of problems in engineering demonstrate this form of proportional damping [9].

2.4.2 Modal analysis of forced systems

In the case of the single-DOF system, Sec. 0 shows that the steady-state response due to harmonic force excitation is solved by an assumption that the system responds in the same period of oscillation as the force. Then, superposition of complementary and particular solutions is found to yield the total solution of the displacement response. The use of linear modal analysis in Sec. 2.4.1 is extended in this Sec. 2.4.2 to account for harmonic force excitation acting on each of the N DOF. These forces are expressed by

( ) ( ) ( ) ( )1 2T

NF t F t F t F t= (2.4.2.1)

Then, the governing EOM of the mass-spring-damper system with N DOF is

( )M C Kx x x F t+ + = (2.4.2.2)

Undertaking the steps of modal analysis outlined in Secs. 2.4 and 2.4.1, one arrives at the transformed equation in modal coordinates

[ ] 1/22 P MTi ir diag r r Fζ ω −+ + Λ = (2.4.2.3)

In general, 1/2P MT F− is a linear combination of the N forces acting on the thi modal coordinate ir . In

other words, 1/2P MT F− contains a contribution from each of the N forces. Although each of the N forces physically acts only on the associated thi DOF, in the modal domain (2.4.2.3), the influence of the thi force is seen to distribute to all of the N modes.

By linearity of the system under consideration, the total steady-state response of (2.4.2.3) for the thi DOF

is the superposition of all steady-state responses given in the thi row of 1/2P MT F− .


29

Finally, the total response of the N DOF is due to the summation of complementary and particular solutions to (2.4.2.2).

Once the solutions are obtained according to the modal vector r , the physical response is reconstructed using (2.4.19). These steps emulate the efforts of Sec. 0 in relation to the single-DOF system since the equation decoupling resulting from linear modal analysis facilitates such straightforward mathematics.


30

3 Principles of continuum mechanics and variational methods

The prior Sec. 2 determined governing equations for lumped parameter systems using Newton's 2nd law. That approach may be termed an equilibrium approach to deriving the governing equations of motion. It is also possible to use these equilibrium conditions, built upon Newton's 2nd law, to derive the equations of motion that govern the vibrations of continuous systems, which is explored in Sec. 4.1.

On the other hand, for continuously distributed structures, there are times when it is unclear how to compose the necessary free-body diagrams that help one to yield such governing equations by the equilibrium approach. This concern is especially manifest when the properties or dimensions of the structure change in time and/or space.

This Sec. 3 provides an alternative approach to derive the governing equations of motion for continuously distributed systems using principles from continuum mechanics. Continuum mechanics refers to the branch of mechanics that analyzes materials as continuous media rather than as collections of discrete springs and masses so far examined. Through energy methods of analysis, the EOMs of continuous systems may be readily obtained. In addition, these methods are useful to also obtain the EOMs for lumped parameter systems, although such application is not the focus of this course.

The following sections introduce the necessary principles of continuum mechanics that lead into the analytical approaches available to obtain EOMs for continuous systems.

3.1 Review of variational methods

Variational calculus was motivated by Johann Bernoulli who postulated the brachistochrone problem that seeks to find the shortest time path between two points. Interested readers are invited to learn about the historic brachistochrone problem and other history of variational calculus that emerged from the problem's scrutinization by the Bernoulli brothers, l'Hopital, Euler, and other famous scientists†.

3.1.1 Stationary values of a function

A function ( )f x of a single variable x is shown in Figure 12. In many engineering problems, we need to

identify the extrema, or stationary points, of such functions. The stationary points *x of the function are computed from

( ) 0df x

dx= (3.1.1.1)

The stationary points *x may be local maxima, local minima, or points of inflection. Such properties are determined by the second derivative of the function evaluated at the stationary points of interest:

† https://en.wikipedia.org/wiki/Calculus_of_variations


31

local minima ( )*

2

2 0x

d f xdx

> ; local maxima ( )*

2

2 0x

d f xdx

< ; inflection points ( )*

2

2 0x

d f xdx

= (3.1.1.2)

Figure 12. Extrema of a function.

When the function is dependent upon multiple variables, such as ( )1 2, ,..., nf x x x , the stationary points

* * * *1 2, ,...,

T

nx x x x = are determined by simultaneously solving the set of equations

( ) ( )1 2, ,...,0;1n

i i

f x x x f xi n

x x∂ ∂

= = ≤ ≤∂ ∂

(3.1.1.3)

where [ ]1 2, ,..., Tnx x x x= . The stationary point (set) * * * *

1 2, ,...,T

nx x x x = are local minima when the

eigenvalues of the Hessian matrix H evaluated at the stationary point are positive. The Hessian matrix is computed from

2

2 2 2

21 1 2 1

2 2 2

22 1 2 2

2 2 2

21 2

H

H

iji j

n

n

n n n

fx x

f f fx x x x xf f f

x x x x x

f f fx x x x x

∂=∂ ∂

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

= ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

(3.1.1.4a,b)

When the eigenvalues of H are all negative when evaluated at *x , then the stationary point *x is a local

maximum. If the H has both positive and negative eigenvalues when evaluated at *x , the values *x denotes a saddle point which is a general multi-dimensional inflection point.

3.1.2 Stationary value of a functional

A functional is a function of a function. For example, consider that the function F is dependent upon a variable y which is itself dependent upon another variable x . Then, we may express a functional I as


32

( )2

1

, ,x

xI F x y y dx′= ∫ (3.1.2.1)

where the prime indicator means that the ( )y x and /y y x′ = ∂ ∂ .

Determining the stationary points of a functional results in equations whose solutions are the functions

( )y x that make the functional I to be stationary.

Consider that the values of the function ( )y x are known at two points 1x and 2x as shown in Figure 13.

We then express the essential boundary conditions as

1 1

2 2

,,

y x xy

y x x=

= = (3.1.2.2)

The essential boundary conditions are also termed geometric, kinematic, or Dirichlet boundary conditions.

Figure 13. Function and its variation.

We seek to then answer the question: what is the ( )y x such that the functional I exhibits a stationary value

and the essential boundary conditions are satisfied? To this end, we introduce a new function shown in Figure 13:

( ) ( ) ( )y x y x xεη= + (3.1.2.3)

The parameter ε is a small (ε <<1) arbitrary parameter independent of ( )xη . This variation function ( )xη

can also be arbitrary. The variation function ( )xη must also satisfy the essential boundary conditions

( ) ( )1 2 0x xη η= = (3.1.2.4)

We can realize any function in the neighborhood of ( )y x by selecting ε and ( )xη .

The function ( )xεη can be considered to be a weak variation to the function ( )y x . In other words, when

ε → 0 the original function ( )y x is recovered by y y→ . Consequently, we consider the replacement of

( )y x into the functional


33

( ) ( )2

1

, ,x

xI F x y y dxε εη εη′ ′= + +∫ (3.1.2.5)

The new functional ( )I ε may be expanded in a Taylor series so that

( ) ( )2

22

0 0

10 ...2

dI d II Id d

ε ε

ε ε ε εε ε

= =

= = + + +

(3.1.2.6)

For the stationary value of I to likewise be the stationary value of I , a necessary condition is that

0

0dId

εε

=

=

(3.1.2.7)

Because higher-order derivatives in the expansion (3.1.2.6) are taken from (3.1.2.7), only (3.1.2.7) needs to be met.

By the chain rule, we have that

2

1

x

x

dI F dy F dy dxd y d y dε ε ε

′ ∂ ∂= + ′∂ ∂ ∫

(3.1.2.8)

It is apparent that

( )dy xd

ηε=

and ( )dy xd

ηε′

′=

(3.1.2.9)

Then, evaluating (3.1.2.8) at ε =0 according to the necessary condition of (3.1.2.7), we have

( ) ( ) ( ) ( )2 2

1 10

x x

x x

dI F F F Fx x dx x x dxd y y y y

ε

η η η ηε

=

∂ ∂ ∂ ∂′ ′= + = + ′ ′∂ ∂ ∂ ∂ ∫ ∫

(3.1.2.10)

Integration by parts is taken on the second integration in (3.1.2.10). The result is that

( ) ( ) ( )

( )

22 2

1 11

2

1

xx x

x xx

x

x

F F d Fx dx x x dxy y dx y

d F x dxdx y

η η η

η

∂ ∂ ∂′ = − ′ ′ ′∂ ∂ ∂

∂= − ′∂

∫ ∫

∫ (3.1.2.11a,b)

Meeting the essential boundary conditions by (3.1.2.4) indicates that the boundary-related terms in the (3.1.2.11a) are zero. Therefore, we find that

( )2

10

x

x

dI F d F x dxd y dx y

ε

ηε

=

∂ ∂= − ′∂ ∂ ∫

(3.1.2.12)

For I to have the extremum value, the right hand side of (3.1.2.12) must be zero.


34

( )2

1

0x

x

F d F x dxy dx y

η ∂ ∂

− = ′∂ ∂ ∫ (3.1.2.13)

For this integral to hold for for an arbitrary ( )xη , the integrand must itself be equal to zero.

0F d Fy dx y

∂ ∂− = ′∂ ∂

(3.1.2.14)

The (3.1.2.14) is the Euler-Lagrange equation. The solution ( )y x to this equation, along with the boundary

conditions ( )1 1y x y= and ( )2 2y x y= , yields the stationary value of the functional I .

3.1.2.1 Natural boundary conditions for one-dimensional problems

Not all boundary conditions are essential. Occasionally the displacements y (using terms of continuum

mechanics) are not known at ends of the domain. These boundary conditions are termed natural, force, or Neumann boundary conditions.

Consider the prior example where the value of the function ( )2y x can take on any value. In the terms of

mechanics, this is a free end.

Repeating the steps above, now considering ( )1 0xη = and ( )2xη is arbitrary, we find that the integration

by parts in (3.1.2.11) does not reduce to so great extent:

( ) ( ) ( )

( ) ( )

22 2

1 11

2

12

xx x

x xx

x

xx

F F d Fx dx x x dxy y dx y

F d Fx x dxy dx y

η η η

η η

∂ ∂ ∂′ = − ′ ′ ′∂ ∂ ∂

∂ ∂= − ′ ′∂ ∂

∫ ∫

∫ (3.1.2.1.1)

As a result, the stationary value of the functional I is found from

( ) ( )2

120

0x

xx

dI F F d Fx x dxd y y dx y

ε

η ηε

=

∂ ∂ ∂= + − = ′ ′∂ ∂ ∂

∫

(3.1.2.1.2)

so that

( ) ( )2

12

2 0x

xx

F F d Fx x dxy y dx y

η η ∂ ∂ ∂

+ − = ′ ′∂ ∂ ∂ ∫ (3.1.2.1.3)

For an arbitrary ( )xη and ( )2xη , both terms must independently be zero in order for the sum of (3.1.2.1.3)

to be zero.

0F d Fy dx y

∂ ∂− = ′∂ ∂

(3.1.2.1.4)


35

2

0x

Fy∂

=′∂

(3.1.2.1.5)

The boundary condition given by (3.1.2.1.5) is termed the natural boundary condition. In certain cases, the functional I may be expressed with additional forces or actions so that, for example,

( ) ( )2

12, ,

x

xI F x y y dx G y′= −∫ (3.1.2.1.6)

Repeating the derivation given above and assessing the limits in the integration by parts, we would find that the natural boundary condition becomes

2 22x x

F Gy y∂ ∂

=′∂ ∂

(3.1.2.1.7)

Example Consider the longitudinal deformations ( )u x of the bar shown in Figure 14. The total potential energy Π

of the bar is

U WΠ = −

( ) ( )2

2

0

1 12 2

L duU EA x dx ku Ldx

= + ∫

( ) ( )0

LW p x u x dx= ∫

where U is the strain energy of the bar, W is the potential energy of the applied loads associated with the

positive work done by the loads. We express the end deflection as ( ) Lu L u= .

Use the principle of minimum total potential energy to obtain the boundary value problem that governs the deflections of the bar.

Figure 14. Fixed-free bar with axially distributed load and end-spring boundary at free bar end.

Answer Using the problem statement information, the total potential energy is


36

( ) ( ) ( ) ( )2

2

0

1 12 2

L duEA x p x u x dx ku Ldx

Π = − +

∫

The total potential energy is therefore a functional. Consequently, we know that the minimization of a

functional results in a governing equation for the deformation ( )u x whose solution is the stationary point

of the functional. Thus, we can use the variational principles reviewed in Sec. 3.1.2 to find the boundary value problem for the deformations of the bar.

We first aim to find the correlations among the expression

( ) ( )2

12, ,

x

xI F x y y dx G y′= −∫

to the problem statement terms. Here, the functional I is the total potential energy Π . The remaining terms are

( ) ( ) ( ) ( )21, ,

2duF x u u EA x p x u xdx

′ = −

( ) 212L LG u ku= −

( )F p xu

∂= −

∂

( )F duEA xu dx∂ = ′∂

L L

Lx x

F G kuu u∂ ∂

= = −′∂ ∂

Then, we use the Euler-Lagrange equation to obtain the governing equation

( ) ( )

( ) ( )

0 0F d F d dup x EA xu dx u dx dx

d duEA x p xdx dx

∂ ∂ − = → − − = ′∂ ∂ = −

along with the associated boundary conditions

( )0 0u = and ( )L

Lx

duEA x kudx

= −


37

3.1.3 Stationary value of a functional in multiple dimensions

The approach described in Sec. 3.1.2 can be extended to the case when the functional is dependent upon multiple dimensions.

We start by introducing a two-dimensional functional

( ), , , ,x yI F x y u u u dΩ

= Ω∫ ∫ (3.1.3.1)

Here, the dependent variable is ( ),u x y and it is similar to a surface in two-dimensions. One could envision

a "height" value ( ),u x y prescribed for the variable at every combination of ( ),x y , like contour or surface

plots. The subscript x or y respectively denote partial derivatives of the function in those dimensions.

Thus /xu u x= ∂ ∂ . The problem domain extends over the ( ),x y space of Ω . Essential boundary conditions

are defined on Γ which is the boundary of the domain Ω . Thus, on Γ , we know ( ),u x y u= .

Like for the case of the one-dimensional functional, now we consider a variation of the functional. The varied surface is

( ) ( ) ( ), , ,u x y u x y x yεη= + (3.1.3.2)

where ( ),x yη is a weak variation applied to the function ( ),u x y , and ε is a small parameter independent

of the variation. The variation of the functional ( ),u x y can be thought of as a deviation of the surface that

meets the essential boundary conditions on the perimeter. This is visualized in Figure 15 using sinusoidal variations over the surface.

Figure 15. Two-dimensional function and its variation. The variation, shown as the colored surface, goes to zero on the

boundaries of the surface ( ),u x y which is the underlying mesh grid shown.

-1 -0.5 0 0.5 1-1

-0.5

0

0.5

1

-1

0

1

2

3

y

x

u(x,y) as mesh grid. u \tilde as varied function


38

Table 1. MATLAB code used to generate Figure 15.

x=linspace(-1,1,31); % set variable range y=linspace(-1,1,31); u=[]; vu=[]; for iii=1:length(y), u(:,iii)=2*x.^2+x.*y(iii); % define dependent function vu(:,iii)=u(:,iii)+.5*cos(3*pi*x/2)'*cos(3*pi*y(iii)/2); % define varied function end; figure(1); % plot clf; mesh(x,y,u','edgecolor','k','facecolor','none'); hold on; surf(x,y,vu','linestyle','none','facealpha',0.85); xlabel('x'); ylabel('y'); title('u(x,y) as mesh grid. u^\tilde as varied function'); view(5,35)

Because the weak variation ( ),x yη meets the essential boundary conditions, we have that

( ),x yη =0 on Γ (3.1.3.3)

Substitution of the varied function ( ),u x y into the functional gives a new functional

( ) ( ), , , ,x yI F x y u u u dεΩ

= Ω∫ ∫

(3.1.3.4)

( ) ( ), , , ,x x y yI F x y u u u dε εη εη εηΩ

= + + + Ω∫ ∫ (3.1.3.5)

For the functional ( )I ε to have a stationary value, following the principle of (3.1.2.6) regarding a Taylor's

series expansion, we must have that

0

0dId

εε

=

=

(3.1.3.6)

By the chain rule, we have

x yx y

dI F F F dd u u u

η η ηε Ω

∂ ∂ ∂= + + Ω

∂ ∂ ∂ ∫ ∫

(3.1.3.7)

Gauss identities are then used to remove the variation derivatives on the RHS of the (3.1.3.7). These identities generally read

x x xwv d n wvds w vdΩ Γ Ω

Ω = − Ω∫ ∫ ∫ ∫ ∫ (3.1.3.8)

y y ybv d n bvds b vdΩ Γ Ω

Ω = − Ω∫ ∫ ∫ ∫ ∫ (3.1.3.9)


39

where xn and yn are the direction cosines of the outer unit normal acting on the boundary Γ . Seeing the

parallels of the Gauss identities and (3.1.3.7), we have the analogs

x

Fwu∂

=∂

, ( ),x xv x yη= , y

Fbu∂

=∂

, ( ),y yv x yη= (3.1.3.10)

so that by substitution we have

( ) ( ), ,x yx y x y

dI F F F F Fx y d n n x y dsd u x u y u u u

η ηε Ω Γ

∂ ∂ ∂ ∂ ∂ ∂ ∂= − − Ω + + ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∫ ∫ ∫

(3.1.3.11)

For the surface ( ),u x y to make the functional I stationary, we must therefore have

( ) ( )0 , ,x yx y x y

F F F F Fx y d n n x y dsu x u y u u u

η ηΩ Γ

∂ ∂ ∂ ∂ ∂ ∂ ∂= − − Ω + + ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∫ ∫ ∫ (3.1.3.12)

Considering the essential boundary conditions on Γ and the fact that the (3.1.3.12) must be satisfied for all

( ),x yη , from (3.1.3.12) we obtain the Euler-Lagrange equation for the two-dimensional functional

(3.1.3.13) and the associated essential boundary conditions (3.1.3.14).

0x y

F F Fu x u y u

∂ ∂ ∂ ∂ ∂− − = ∂ ∂ ∂ ∂ ∂

in Ω (3.1.3.13)

( ),u x y u= on Γ (3.1.3.14)

3.1.3.1 Natural boundary conditions for two-dimensional problems

Extending the concept of natural, force, or Neumann boundary conditions to two-dimensional problems is straightforward. Recall that a natural boundary condition is one for which the displacement (or generalized coordinate more generically) is not specified. Here, we assume that the total boundary Γ of the domain Ω may be divided into a portion 1Γ where the essential boundaries are defined, and 2Γ where there are no

essential boundaries defined. It is important to recall that the variation ( ),x yη needs to exactly satisfy only

the essential boundaries 1Γ .

Considering (3.1.3.12), if the variation function ( ),x yη is not zero over the boundary, the second part of

(3.1.3.12) is not always zero. Consequently, the resulting equations obtained are

0x y

F F Fu x u y u

∂ ∂ ∂ ∂ ∂− − = ∂ ∂ ∂ ∂ ∂

in Ω (3.1.3.1.1)

( ),u x y u= on 1Γ (3.1.3.1.2)


40

0x yx y

F Fn nu u∂ ∂

+ =∂ ∂

on 2Γ (3.1.3.1.3)

In (3.1.3.1.3), the LHS term is referred to as flux.

In a similar way as for the one-dimensional problems, the flux may be specified rather than zero. Yet, in these cases, the functional will be of a form

( ) ( )2

, , , , , ,x yI F x y u u u d G x y u dsΩ Γ

= Ω +∫ ∫ ∫ (3.1.3.1.4)

where G is a differentiable function. Then, the associated natural boundary condition will be found

0x yx y

F F Gn nu u u∂ ∂ ∂

+ + =∂ ∂ ∂

on 2Γ (3.1.3.1.5)

Example Consider the functional

( )22

, , , , 2x yu uI x y u u u fu dx yΩ

∂ ∂ = + + Ω ∂ ∂ ∫ ∫

and assume that 0u = on the total boundary Γ .

Determine the boundary value problem whose solution is the stationary point of the functional I .

Answer From the developments of Sec. 3.1.3, the associated terms of the functional to be used in the Euler-Lagrange equation are

( )22

, , , , 2x yu uF x y u u u fux y

∂ ∂ = + + ∂ ∂

2F fu

∂=

∂

2 xx

F uu∂

=∂

2 yy

F uu∂

=∂

This leads to the Euler-Lagrange equation for the two-dimensional problem of

2 2 2 0 xx yyu uf u u f

x x y y ∂ ∂ ∂ ∂ − − = → + = ∂ ∂ ∂ ∂


41

with the essential boundary condition of 0u = on Γ . This is referred to as Poisson's equation. Solving Poisson's equation is therefore analogous to solving for the stationary point of the functional.

3.1.4 Variational operator

The method of finding the extrema of the one- and multi-dimensional functionals described above can be

generalized. First, we introduce our function ( )y x whose associated functional

( )2

1

, ,x

xxI F x y y dx= ∫ (3.1.4.1)

we wish to extremize. As before, the subscript x on the function ( )y x indicates /y x∂ ∂ . The varied

function ( )y x is introduced that satisfies essential boundary conditions.

( ) ( ) ( )y x y x xεη= + (3.1.4.2)

We define the variation operator δ according to the relationship

( ) ( ) ( )y x y x y xδ = − (3.1.4.3)

Equivalently, we have that ( ) ( )y x xδ εη= . The properties of the variation operator are:

1. The variation operator causes an infinitesimal change in the function ( )y x for fixed value of x so

that

0xδ = (3.1.4.4)

2. A variation operation is commutative with differentiation. Thus, the variation of the derivative is the derivative of the variation

( )dy d ydx dx

δ δ =

(3.1.4.5)

3. A variation operation is commutative with integration. Thus, the variation of an integral is the integration of the variation

( ) ( )2 2

1 1

x x

x xy x dx y x dxδ δ=∫ ∫ (3.1.4.6)

In a similar way as described in Sec. 3.1.2, for the extremization of a functional I , the first variation must be zero

0Iδ = (3.1.4.7)

Using the property (3.1.4.6), we have that the extremization is the same as


42

2 2

1 1

0x x

x xI Fdx Fdxδ δ δ= = =∫ ∫ (3.1.4.8)

Recalling (3.1.4.2) and (3.1.4.3), we have by substitution

( ) ( ), , , ,x x xF F x y y F x y yδ εη εη= + + − (3.1.4.9)

Using a first-order Taylor's series expansion on the first RHS term of (3.1.4.9), we have

( ) ( ), , , ,x x x xx

F FF x y y F x y y y yy y

εη εη δ δ∂ ∂+ + ≈ + +

∂ ∂ (3.1.4.10)

Then, we have that

2

1

0x

xxx

F FI y y dxy y

δ δ δ ∂ ∂

= + = ∂ ∂ ∫ (3.1.4.11)

Integration by parts on the second term of the RHS results in

22 2

1 11

xx x

xx xx x xx

F F d Fy dx y ydxy y dx yδ δ δ

∂ ∂ ∂= − ∂ ∂ ∂

∫ ∫ (3.1.4.12)

In this way, the extremization (3.1.4.11) is given by

22

11

0x

x

xx xx

F F d Fy ydxy y dx yδ δ

∂ ∂ ∂+ − = ∂ ∂ ∂ ∫ (3.1.4.13)

where ( )1y xδ and ( )2y xδ are arbitrary and independent. Since (3.1.4.13) must be true for all ( )1y xδ and

( )2y xδ , the terms must be individually equal to zero

2

1

0x

xx

F d F ydxy dx y

δ ∂ ∂

− = ∂ ∂ ∫ (3.1.4.14)

2

1

0x

x x

F yyδ∂

=∂

(3.1.4.15)

Since (3.1.4.14) must be true for any arbitrary variation yδ for slowly changing function ( )y x , we have

that

0x

F d Fy dx y

∂ ∂− = ∂ ∂

(3.1.4.16)

Thus, the variational approach results in the Euler-Lagrange equation (3.1.4.16) for the system. The solution to (3.1.4.16) minimizes the associated functional I . In addition, the boundary conditions shown in (3.1.4.15) indicate that


43

0x

Fy∂

=∂

or 0yδ = at 1x (3.1.4.17)

0x

Fy∂

=∂

or 0yδ = at 2x (3.1.4.18)

In (3.1.4.17) and (3.1.4.18), the prior expressions are natural boundary conditions and the latter expressions are essential boundary conditions.

It is evident that the variational approach yields the same result as that realized in Sec. 3.1.2. This may lead one to question the utility of the more general variational approach for extremizing the functionals. In fact, the concept and use of the variational operator is leveraged to translate this method to the study of dynamics and vibrations of continuous systems via Hamilton's principle.

3.1.4.1 Stationary value of a functional with second-order derivatives

The study of beam bending deformation involves potential energy expressions of higher-order spatial derivatives that those addressed in Sec. 3.1.2. In this case, the relevant functional takes on a general form of

( )2

1

, , ,x

x xxxI F x y y y dx= ∫ (3.1.4.1.1)

As before, we wish to identify the ( )y x that causes the functional I in (3.1.4.1.1) to be stationary.

Consequently, the first variation of the functional must be equal to zero.

0Iδ = (3.1.4.1.2)

The (3.1.4.1.2) becomes an expression of

2

1

0x

x xxxx xx

F F FI y y y dxy y y

δ δ δ δ ∂ ∂ ∂

= + + = ∂ ∂ ∂ ∫ (3.1.4.1.3)

The second term of (3.1.4.1.3) is then integrated by parts.

22 2

1 11

xx x

xx xx x xx

F F d Fy dx y ydxy y dx yδ δ δ

∂ ∂ ∂= − ∂ ∂ ∂

∫ ∫ (3.1.4.1.4)

The third term of (3.1.4.1.3) is integrated by parts twice.

22 2

1 11

222

11 1

2

2

xx x

xx x xx xxx xx xxx

xxx

x xxx xx xxx x

F F d Fy dx y y dxy y dx y

F d F d Fy y ydxy dx y ydx

δ δ δ

δ δ δ

∂ ∂ ∂= − ∂ ∂ ∂

∂ ∂ ∂= − + ∂ ∂ ∂

∫ ∫

∫ (3.1.4.1.5)

Consequently, the first variation of the functional is


44

2 22

111

2

2 0x x

x

xxx xx x xx xx xx

F d F d F F d F FI ydx y yy dx y dx y y dx y y

δ δ δ δ ∂ ∂ ∂ ∂ ∂ ∂

= − + + − + = ∂ ∂ ∂ ∂ ∂ ∂ ∫ (3.1.4.1.6)

Each part of (3.1.4.1.6) must individually be zero in order for the stationary value to exist for arbitrary selection of yδ . Thus, the Euler-Lagrange governing equation for the system whose solution is the

stationary value of the function is

2

2 0x xx

F d F d Fy dx y dx y

∂ ∂ ∂− + = ∂ ∂ ∂

(3.1.4.1.7)

The associated boundary conditions of the problem are

0xx

Fy∂

=∂

or 0xyδ = at 1x and 2x (3.1.4.1.8)

0x xx

F d Fy dx y

∂ ∂− = ∂ ∂

or 0yδ = at 1x and 2x (3.1.4.1.9)

3.1.5 Hamilton's principle

Hamilton's principle states that among all possible time histories of displacement that satisfy the boundary conditions and compatibility equations of the system from initial 1t to final 2t times, the time history

corresponding to the actual solution of response causes the Lagrangian functional to be minimized. This is expressed by

2

1

0t

tI Ldtδ δ= =∫ (3.1.5.1)

The Lagrangian L is the combination of total kinetic energy T , total potential energy U , and total applied loads and body force W where the latter is associated with positive work.

L T U W= − + (3.1.5.2)

An example helps us to consolidate the use of Hamilton's principle.

Example Consider again the longitudinal deformations of the bar in Figure 14. Now the deformations are assumed

to occur in time t and space x so that ( ),u x t . Such dynamic deformation are now referred to as vibrations.

The applied force is also time-varying, so that ( ),p x t . The total potential energy is

( ) ( )2

2

0

1 1 ,2 2

L uU EA x dx ku L tx∂ = + ∂ ∫

The total potential energy U contains the term associated with strain energy in the bar and the term associated with stretching of the end spring. The potential of the applied loads is


45

( ) ( )0

, ,L

W p x t u x t dx= ∫

For the bar, the total kinetic energy is

( )2

0

12

L uT A x dxt

ρ ∂ = ∂ ∫

Find the governing equation of motion for the vibrations of the bar and determine the associated boundary conditions.

Answer The Lagrangian for the system is

( ) ( ) ( ) ( )2 2

2

0

1 1 1, ,2 2 2

L

Lu uL T U W T A x EA x p x t u x t dx kut x

ρ ∂ ∂ = − + = = − + − ∂ ∂

∫

The associated functional considered by Hamilton's principle is

( ) ( ) ( ) ( )1

0

2 22

0

1 1 1, ,2 2 2

t L

Lt

u uI A x EA x p x t u x t dx ku dtt x

ρ ∂ ∂ = − + − ∂ ∂

∫ ∫

The statement for the minimization of the first variation of the functional is

( ) ( ) ( ) ( )1

0

2 22

0

1 1 1, , 02 2 2

t L

Lt

u uI A x EA x p x t u x t dx ku dtt x

δ δ ρ ∂ ∂ = − + − = ∂ ∂

∫ ∫

Commuting the variation and integration operations, we have

( ) ( ) ( ) ( )1

0

2 22

0

1 1 1, ,2 2 2

t L

Lt

u uA x EA x p x t u x t dx k u dtt x

ρ δ δ δ δ ∂ ∂ − + − ∂ ∂

∫ ∫

( ) ( ) ( ) ( )1

0 0, ,

t L

L Lt

u u u uA x EA x p x t u x t dx ku u dtt t x x

ρ δ δ δ δ ∂ ∂ ∂ ∂ − + − ∂ ∂ ∂ ∂

∫ ∫

We use the second property of the variation operator to commute the differentiations and variations

( ) ( ) ( ) ( ) ( )1

0 0,

t L

L Lt

u uA x u EA x u p x t u dx ku u dtt t x x

ρ δ δ δ δ ∂ ∂ ∂ ∂ − + − ∂ ∂ ∂ ∂

∫ ∫

Integration by parts is then used on the relevant terms of the expression above

( ) ( ) ( )

( )

21 2

0 11

2

1

2

2

2

2

tt t

t tt

t

t

u u uA x u dt A x u udtt t t t

uA x udtt

ρ δ ρ δ δ

ρ δ

∂ ∂ ∂ ∂ = − ∂ ∂ ∂ ∂ ∂

= −∂

∫ ∫

∫


46

( ) ( ) ( ) ( )0 0

0

LL Lu u uEA x u dx EA x u EA x udx

x x x x xδ δ δ∂ ∂ ∂ ∂ ∂ − = − + ∂ ∂ ∂ ∂ ∂ ∫ ∫

Subsequently, the minimization of the Lagrangian functional by its first variation is obtained by substituting and grouping the several components derived above.

( ) ( ) ( ) ( )1

0

2

200

0 ,L

t L

L Lt

u u uA x EA x p x t udx EA x u ku u dtt x x x

ρ δ δ δ ∂ ∂ ∂ ∂ = − + + − − ∂ ∂ ∂ ∂

∫ ∫

Since the time integration must hold for any arbitrary initial and final times, 1t and 2t , respectively, we

have that

( ) ( ) ( ) ( )2

200

, 0L

L

L Lu u uA x EA x p x t udx EA x u ku u

t x x xρ δ δ δ

∂ ∂ ∂ ∂ − + + − − = ∂ ∂ ∂ ∂ ∫

Then, because the variations at 0x = and x L= are also arbitrary, the integration term and the boundary relations must be independently equal to zero.

( ) ( ) ( )2

20, 0

L u uA x EA x p x t udxt x x

ρ δ ∂ ∂ ∂ − + + = ∂ ∂ ∂

∫

( ) ( ) ( ) 00

0 0, 0 0L

L L L L Lu u uEA x u ku u EA L u ku u EA ux x xδ δ δ δ δ∂ ∂ ∂

− − = → − − = − =∂ ∂ ∂

And finally, because the variation uδ must be arbitrary, we have that

( ) ( ) ( )2

2,u uEA x p x t A xx x t

ρ∂ ∂ ∂ + = ∂ ∂ ∂

In conclusion, we find the governing equation for the longitudinal vibrations of the rod

( ) ( ) ( )2

2,u uEA x p x t A xx x t

ρ∂ ∂ ∂ + = ∂ ∂ ∂

The associated boundary conditions are likewise found to be

( )0, 0u t = at 0x =

( ) LuEA L kux∂

− =∂

at x L=

Note that the boundary conditions given above are specific to the situation shown in Figure 14.


47

3.1.5.1 Hamilton's principle for Lagrangians in multiple dimensions and higher-order derivatives

Hamilton's principle is easily extended to the case of a structure extending in multiple dimensions or having multiple specific deformations, like translation and rotation. When the Lagrangian includes several such function dependencies including

( ), , , , ,x y tL x y t u u u (3.1.5.1.1)

then the first variation of the Lagrangian functional must account for these factors according to the chain rule:

2

1

0t

x y ttx y t

L L L LI u u u u dtu u u u

δ δ δ δ δ ∂ ∂ ∂ ∂

= + + + = ∂ ∂ ∂ ∂

∫ (3.1.5.1.2)

The procedures of simplifying (3.1.5.1.2) using integration by parts, in order to collect common terms of uδ , are then undertaken in the same manner as that performed in the Example above. The Euler-Lagrange

equation, which is the equation of motion for the system, results from such procedures as observed above.

Similarly, when the functional contains higher-order derivatives such as

( ), , , , ,x xx tL x y t u u u (3.1.5.1.3)

then Hamilton's principle becomes an expression of

2

1

0t

x xx ttx xx t

L L L LI u u u u dtu u u u

δ δ δ δ δ ∂ ∂ ∂ ∂

= + + + = ∂ ∂ ∂ ∂ ∫ (3.1.5.1.4)

In this case, integration by parts is utilized twice to reduce the derivative order of the variation from xxuδ

to uδ . We will encounter this scenario in deriving the governing equations of motion for the vibrations of beams. As a result, we will defer presentation of such an example until Sec. 7 when the topic is especially relevant.

3.2 Strain and kinetic energies

The methods described in Sec. 3.1 enable the derivation of governing equations of motion and associated boundary conditions for arbitrary continuous structures. To utilize these tools, one must identify the total kinetic and strain energies of the structure. This Sec. 3.2 provides a sufficient introduction to the techniques employed to determine these energies for different engineering structures (or solid media, in general) and concludes with a summary of relevant strain and kinetic energy expressions to be used in this course.

3.2.1 Kinetic energy

The total kinetic energy for a continuous structure having displacements [ ]T, ,u u v w= , where the

superscript T denotes a transpose operator, is determined from


48

T12 V

T u udVρ= ∫ ∫ ∫ (3.2.1.1)

When the motion of the structure is uni-axial, for instance only in the x axis that pertains to the displacement component u , then the total kinetic energy becomes

212 V

T u dVρ= ∫ ∫ ∫ (3.2.1.2)

Moreover, when the structure has cross-sectional area ( )A x that varies along the length from 0x = to

x L= , then (3.2.1.2) is simplified further to be

( ) ( )2

2

0 0

1 12 2

L L uT A x u dx A x dxt

ρ ρ ∂ = = ∂ ∫ ∫ (3.2.1.3)

The (3.2.1.3) was used for the total kinetic energy of longitudinal deformation ( ),u x t of the bar in the

Example of Sec. 3.1.5.

Of course, when the body is small such that we may refer to it as merely a particle with total mass m , the integration of the mass density ρ over the body volume V results in a kinetic energy of

212

T mu= (3.2.1.4)

We are familiar with (3.2.1.4) in the study of lumped parameter dynamic systems.

3.2.2 Strain energy

The total potential energy U is composed from potentials due to applied loads and body forces and the potential associated with the strain energy. The strain energy is energy that is stored in an elastic body undergoing deformation.

Figure 16 illustrates the scenario for an arbitrarily shaped three-dimensional system subjected to loads. Strain energy is developed in the system to maintain a state of equilibrium. The system is considered to be composed from an infinite number of infinitesimally small cubes under general states of strain due to applied forces that induce stresses on the cubes.


49

Figure 16. General state of strain for a differential cubic element in a solid.

The differential strain energy of the infinitesimal cubic element is expressed by

12 xx xx yy yy zz zz xy xy xz xz yz yzdU dVσ ε σ ε σ ε τ γ τ γ τ γ = + + + + + (3.2.2.1)

Integrating (3.2.2.1) over the system volume results in the total strain energy U .

12 xx xx yy yy zz zz xy xy xz xz yz yzV

U dVσ ε σ ε σ ε τ γ τ γ τ γ = + + + + + ∫ (3.2.2.2)

If we express (3.2.2.2) in a vector notation, we arrive at

T12 V

U dVσ ε= ∫ (3.2.2.3)

where

T, , , , ,xx yy zz xy xz yzσ σ σ σ τ τ τ = (3.2.2.4)

T, , , , ,xx yy zz xy xz yzε ε ε ε γ γ γ = (3.2.2.5)

In many cases, the total strain energy expression (3.2.2.2) may be simplified due to symmetries and assumptions regarding the state of strain of the structure.

3.2.2.1 Axial strain energy

When a not-too-slender beam is subjected to loads predominantly acting along the beam axis, we refer to the continuous structure as a "bar" or "rod". When a force difference occurs over the length extent of a differential length of a bar, a state of uniaxial stress occurs over the cross-section, as seen at right in Figure 17.


50

Figure 17. Axial strain due to force difference along differential element of a bar.

The uniaxial strain that occurs in the bar differential element is

xxdudx

ε = (3.2.2.1.1)

If the bar is composed of a linear elastic isotropic material, by Hooke's law the stress is related to the strain via

xx xxEσ ε= (3.2.2.1.2)

The constant of proportionality E is termed the Young's modulus. Therefore, using (3.2.2.1) the axial strain energy over the differential element is

221 1 1

2 2 2xx xx xxdudU dV E dV E dVdx

σ ε ε = = =

(3.2.2.1.3)

The remaining terms of (3.2.2.1) are not present in (3.2.2.1.3) due to the geometry considered and assumptions regarding uniformity of the applied force over the cross-section area.

The total strain energy due to the axial deformation is then found from a volume integration of (3.2.2.1.3).

Considering a cross-section of the bar differential element to have area ( )A x that varies in the x spatial

coordinate, we have

( )2

0

12

L duU EA x dxdx

= ∫ (3.2.2.1.4)

If the deformation of the bar changes in time, such as due to vibrations, then the derivatives in the derivation above are partial derivatives.

3.2.2.2 Bending strain energy

While bars permit longitudinal or axial stress, beams support both bending and shearing stresses. In practice, a distinction between "bars" and "beams" is artificial. This is because all such slender continuous structures are subjected to both axial loads, bending moments, and shear loads. It is simply a matter of context whether axial, bending, and/or shear effects are more prominent for the given structure and its


51

loading scenario. This note is intended to alleviate confusion potentially observed in studying other references that refer to all such structural members as "beams" according to their support of axial load, bending moment, and shear loads. Indeed, this reduction of terms is true but may cause confusion. Here, we strictly refer to "bars" in the context of uniaxial deformation while we refer to "beam" when the deformation is out-of-axis as induced by moments and/or shear.

When a moment difference occurs over the spatial extent of a prismatic beam differential element, a bending

stress is induced according to the transverse deflection ( )w x . The situation is schematically shown in

Figure 18. The bending stress at a distance y from the neutral axis is

( ) yxx

z

M yy

Iσ = (3.2.2.2.1)

where the moment of inertia is 2zI y dA= ∫ ∫ if the origin is at the neutral axis of the beam element.

Figure 18. Bending of a differential, prismatic beam element due to moment difference between the differential element ends.

For linear elastic isotropic materials, we again have (3.2.2.1.2). In this case, the bending strain ( )xx yε

becomes

( )2

2xxd wy y ydx

ε κ= ≈ (3.2.2.2.2)

The bending strain energy associated with the differential element is therefore

22 2

2 2

1 1 12 2 2

zxx xx

z

M y d w d wdU dV y dV E y dVI dx dx

σ ε

= = =

(3.2.2.2.3)

Then, the total strain energy due to the bending of the beam is then

2 22 22

2 20 0

1 12 2

L L

zA

d w d wU E y dA dx EI dxdx dx

= =

∫ ∫ ∫ ∫ (3.2.2.2.4)


52

If the transverse deflection v changes in time, the derivatives above are partial derivatives on ( ),w x t .

3.2.2.3 Shear strain energy

Consider the differential beam element cross-section shown at left in Figure 19. The total transverse displacement of the beam centerline is

s bw w w= + (3.2.2.3.1)

where the subscripts s and b respectively refer to shear and bending contributions. The total slope of the center at the differential element location x is therefore

s bdw dwdwdx dx dx

= + (3.2.2.3.2)

Figure 19. Shearing of a differential beam element.

The beam cross-section rotates only due to bending. As a result, the small-angle rotation of the cross-section is

bdw dwdx dx

φ β= = − (3.2.2.3.3)

where /sdw dxβ = is the shear deformation or shear angle. Note that the cross-section rotation φ is a

function of the location along the beam length ( )xφ . The associated strain-displacement relation is then

xydwdx

ε φ= − + (3.2.2.3.4)

The shear stress xyτ is then related to shear strain through

xy xykGτ ε= (3.2.2.3.5)

where the G is the shear modulus and k is a shear correction factor, whose values are oftentimes tabulated [10]. Then, the strain energy associated with shear of the differential beam element is

21 12 2xy xy

dwdU kG dVdx

τ ε φ = = −

(3.2.2.3.6)

The total strain energy due to shear is therefore


53

0 0

1 12 2

L L

A

dw dwU kG dAdx kAG dxdx dx

φ φ = − = − ∫ ∫ ∫ ∫ (3.2.2.3.7)

If the transverse deflection w changes in time, the derivatives above are partial derivatives on ( ),w x t and

( ),x tφ .

3.2.2.4 Torsional strain energy

A differential shaft element is twisted by an angle dθ due to a torque difference between the ends of the element, as seen in Figure 20.

The shear strain that develops in the shaft at a radial distance r from the centroid is

( ) ( ) /x xdr r G rdxθ θθγ τ= = (3.2.2.4.1)

Figure 20. Twisting of differential shaft element due to end torque difference between the element extents.

As a consequence, the strain energy of the differential shaft element is

21 12 2x x

ddU G r dVdxθ θθτ γ = =

(3.2.2.4.2)

Integrating this differential strain over the volume of the shaft leads to the total strain energy in torsion

22

0

12

L

A

dU G r dAdxdxθ =

∫ ∫ ∫ (3.2.2.4.3)

where the polar moment of inertia J is identified.

2

AJ r dA= ∫ ∫ (3.2.2.4.4)

In this way, the total torsional strain energy of the shaft is

2

0

12

L dU GJ dxdxθ =

∫ (3.2.2.4.5)

As before, if the angular rotations θ change in time, the derivatives above are partial derivatives on ( ),x tθ

.


54

3.2.2.5 Summary of kinetic and strain energies

Table 2 summarizes the strain and kinetic energy expressions to be used in this course pertaining to common engineering structures and their vibrations. Note that beams have one kinetic energy while, depending on how loads act on them, they may possess one or two strain energies: bending and shear. Thus, there is only a single kinetic energy expression associated with beams and it is arbitrarily collected with "beam in bending deformation" in Table 2. Note that partial differentials are used throughout Table 2 as required considering both strain and kinetic energies. In each case, the material or geometric properties are verified to be most generally functions along the continuous structure length coordinate x .

Table 2. Summary of strain and kinetic energies of common continuous structures.

strain energy kinetic energy

bar in axial deformation ( )

2

0

12

L uU EA x dxx∂ = ∂ ∫ ( )

2

0

12

L uT A x dxt

ρ ∂ = ∂ ∫

shaft in torsion ( )

2

0

12

LU GJ x dx

xθ∂ = ∂ ∫ ( )

2

0

12

LT J x dx

tθρ ∂ = ∂ ∫

beam in bending deformation ( )

22

20

12

L

zwU EI x dx

x ∂

= ∂ ∫ ( )

2

0

12

L wT A x dxt

ρ ∂ = ∂ ∫

beam in shear deformation ( )

0

12

L dwU kAG x dxdx

φ = − ∫

see above for the beam T

3.3 The Ritz method in applied mechanics: a first look at approximate methods of analysis

The Ritz method is an approximate method of analysis that is foundational in the formulation of the finite element method for problems in mechanics and dynamics. Because the Ritz method is first encountered as a suitable vehicle to approximate problems in applied mechanics, we introduce the Ritz method here as it relates to static equilibria of loaded continuous structures. Sec. 8.1 will introduce the Ritz method in the context of studying the dynamics of continuous structures.

A key strategy used in approximate methods of analysis is to convert partial or ordinary differential equations into algebraic equations. The algebraic equations are often in a matrix form amenable to solution by efficient computing algorithms. The equations are converted by way of an approximation of the response via a linear combination of unknown constants and trial functions.

( ) ( ) ( )01

N

n nn

u x x a xφ φ=

= +∑ (3.3.1)


55

In (3.3.1), the unknown constants are na and the trial functions are ( )xφ . The trial functions must be (1)

linearly independent, (2) must satisfy all essential boundary conditions of the problem, and (3) must form a complete set. The trial function ( )0 xφ is introduced in order to satisfy natural boundary conditions, if they

are relevant for the problem at hand.

The Ritz method is applied to problems of mechanics according to the principle of minimum total potential energy. The (3.3.1) is substituted into the expression for the total potential energy (3.3.2)

U WΠ = − (3.3.2)

which is a combination of total potential of strain energies U and total potential of applied loads W . Minimization of the total potential energy is therefore expressed according to the stationary values obtained from the solution to the system of equations

0; 1n

n Na∂Π

= ≤ ≤∂

(3.3.3)

An example helps to illuminate how the Ritz method may be used in problems of applied mechanics.

Consider the fixed-free beam in Figure 21. The potential associated with the strain energy is

( )22

220

1 12 2

L d wU EI dx kw Ldx

= +

∫ (3.3.4)

The potential of the applied loads is

( ) ( )0

LW p x w x dx= ∫ (3.3.5)

We then must select our trial functions according to the form of assumed solution (3.3.1) and knowledge of boundary conditions. For the fixed-free beam, we recall from principles of mechanics that the essential boundary conditions that must be met are the displacement and slope must be zero at x =0, ( )0 0w = and

( )0 0w′ = . We do not need to select the trial functions based on the spring at x L= .

Therefore, as a first example, we may choose an assumed solution of the form

( )1 1 2 1 12 2

1 21 1

n

n n nn n

x x xw x a a a aL L L

φ+ + +

= =

= + = =

∑ ∑ (3.3.6)


56

Figure 21. Fixed-free beam with end spring and distributed force per unit length.

Then, we substitute (3.3.6) into (3.3.2).

( ) ( )2 22 2 2

0 01 1 1

1 12 2

L L

n n n n n nn n n

EI a dx k a L p x a dxφ φ φ= = =

′′Π = + − ∑ ∑ ∑∫ ∫ (3.3.7)

Then, the stationary value of the total potential energy is obtained by (3.3.3). For (3.3.7), the system of equations obtained by use of (3.3.3) is

( ) ( ) ( )2 2

0 01 1

0L L

m m n m m n nn nn

EI a dx k a L L p x dxa

φ φ φ φ φ= =

∂Π ′′ ′′= + − = ∂ ∑ ∑∫ ∫ (3.3.8)

Simplifying, we have

( ) ( ) ( )2

0 01

L L

n m n m m nn

EI dx k L L a p x dxφ φ φ φ φ=

′′ ′′ + =∑ ∫ ∫ (3.3.9)

2

1nm m n

nK a P

=

=∑ (3.3.10)

In (3.3.10), we identify the stiffness matrix nmK and generalized forces nP as

( ) ( )0

L

nm n m n mK EI dx k L Lφ φ φ φ′′ ′′= +∫ (3.3.11)

( )0

L

n nP p x dxφ= ∫ (3.3.12)

To take the problem further, we simplify by assuming that the spring stiffness 3/k EI Lα= and that the applied load is distributed in a way such that ( ) ( )0 /p x P x L= . Using our assumed solutions (3.3.6), we

substitute these results into(3.3.11) and (3.3.12) to find the respective matrix components.

( ) ( )

( )( )

( )( )

1 2 1 2

0

3 3

3

1 1

1 11

1 11

n m n mL

nmx x L LK EI n n m m dx EALL L L L

mn m nEI EIm nL L

mn m nEIm nL

α

α

α

+ − + − = + + +

+ += +

+ −+ +

= + + −

∫

(3.3.13)


57

1

00

01

3

nL

nx xP P dxL L

P Ln

+ =

=+

∫ (3.3.14)

For N =2, we have the set of linear algebraic equations given by

103

2

4 6 1/ 46 12 1/ 5

aEI P LaL

α αα α

+ + = + +

(3.3.15)

As a result, the unknown constants na are found to be

( )4

1 0

2

361420 12 4

a P La EI

ααα

+ = − −+

(3.3.16)

The Figure 22 presents results of the beam bending displacement, with respective values used for computing shown in the caption. In this example, the α =1 and the beam deflects upwards at the beam end. We note that the boundary conditions are met according to the shape of the bending deflection in Figure 22.

Figure 22. Beam bending displacement of fixed-free beam with end spring and distributed load. 0P =1 [N/m], EI =5 [N.m4],

α =1, and L =1 [m].

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.002

0.004

0.006

0.008

0.01

0.012

0.014

normalized length along beam

beam

ben

ding

dis

plac

emen

t [m

]

number of trial functions= 2. alpha=1.


58

We may repeat the process for a larger α and increasing number of trial functions N . The results in Figure 23 are for the case of α =100. In this case, the end spring is considerably stiff with respect to the bending stiffness EI of the beam. As a result, we would intuitively anticipate that the spring will restrain the beam of deflecting greatly at the end. The results in Figure 23 confirm this intuition. The results also show that increasing the number of trial functions N from 2, to 3, to 4, and then to 5 shows a convergent trend in the computed result. Namely, a large difference in the result is obtained from N =2 to N =3. In contrast, from N =3 to N =4 or 5, a minor change in result occurs. In fact, the N =4 and N =5 curves (blue and cyan, respectively) are on top of each other in the Figure 23.

Figure 23. Increasing number of trial functions used for larger α to constrain beam motion.

This first look exemplifies that the Ritz method may be used to approximate the mechanical response of continuous structures due to applied loads. The Ritz method is an approximate method, which means that increasing the computational effort and by the appropriate selection of semi-arbitrary trial functions ones is able to accurately predict the behavior of a loaded structure without resort to a more rigorous exact solution approach.

The Ritz method is founded upon variational calculus and energy methods of analysis. These are key strategies on which our subsequent investigations of the vibrations of continuous structures will be established.

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1x 10

-3

normalized length along beam

beam

ben

ding

dis

plac

emen

t [m

]

number of trial functions=, red=2; green=3, blue=4, cyan=5. alpha=100.


59

Table 3. MATLAB code used to generate Figure 22 and Figure 23.

P_0=1; % [N/m] distributed force per length EI=5; % [N.m^4] bending stiffness of beam alpha=1; % [dim] ratio of end spring stiffness to beam bending stiffness L=1; % [m] length of beam num_trial_functions=2; % number of trial functions used K=zeros(num_trial_functions); % preallocate the stiffness matrix P=zeros(num_trial_functions,1); % preallocate the generalized force vector for iii=1:num_trial_functions for jjj=1:num_trial_functions K(iii,jjj)=EI/L^3*(iii*jjj*(iii+1)*(jjj+1)/(iii+jjj-1)+alpha); % stiffness matrix entry end P(iii)=P_0*L*1/(iii+3); % generalized force vector entry end a_const=K\P; % compute the unknown constants x=linspace(0,L,41); % [m] create spatial range of response beam_bending_disp=zeros(length(x),1); for iii=1:num_trial_functions beam_bending_disp=beam_bending_disp+a_const(iii)*(x.'/L).^(iii+1); end figure(1) clf hold on plot(x/L,beam_bending_disp,'r'); xlabel('normalized length along beam'); ylabel('beam bending displacement [m]'); title(['number of trial functions= ' num2str(num_trial_functions) '. alpha=' num2str(alpha) '.']); box on


60

4 Distributed parameter system methods of analysis

The vibration of continuous systems may be analyzed once equations of motion are obtained. Sec. 3 surveyed fundamentals of variational methods and continuum mechanics required to determine the governing equations of motion for continuous structures. While the equilibrium approach built upon Newton's second law of motion was demonstrated to reveal the governing equations for lumped parameter systems in Sec. 2, here in Sec. 4.1 a review is provided for the means to use Newton's laws of motion to derive the governing equations of motion for continuous structures. This serves as a complement to the variational approach. Based on the specific structural geometry and loading conditions observed in a particular application, one decides whether to employ variational or equilibrium methods to derive the governing equations for further investigation.

Following the review of the equilibrium approach for deriving EOMs for continuous systems, this Sec. 4 also describes the means to examine the free and forced response of continuous structures which is used throughout the remainder of the course. In these examples, unless otherwise denoted, we use the example of the longitudinal vibrations of a bar for consistency.

4.1 Newton's laws of motion: an equilibrium approach to governing equation derivation

In the same way as undertaken for systems of lumped parameter masses, springs, and dampers, the force balance represented by Newton's 2nd law of motion may be applied for a differential element of a

continuous structure. Consider the one-dimensional longitudinal deformations ( ),u x t of the bar in Figure

24(a). The bar has varying cross-sectional area ( )A x and Young's modulus ( )E x along its length L , while

the density ρ is presumed to remain independent of the length coordinate x . In Figure 24(a), a bar element

of differential length dx is seen at initial and later times. At the later time t , the cross-sectional area on one

side of the differential element has deformed forward by an amount ( ),u x t . The cross-sectional area on

the other side of the differential element has deformed forward by an amount ( ),u x dx t+ .

An application of Newton's 2nd law on a free-body diagram of this differential element, Figure 24(b), indicates that the inertial force, computed from the time-rate of change of momentum, is equal to the difference in forces at the faces of the differential element.

( ) ( ) ( ) ( ),, ,

u x tF x t F x dx t A x dx

t tρ ∂ ∂

− + + = ∂ ∂ (4.1.1)

The force at the LHS face of the differential element is related to the stress ( ),x tσ acting on the face of

the bar.

( ) ( ) ( ), ,F x t A x x tσ= (4.1.2)


61

The constitutive relation for linear elastic uniaxial deformation then relates the stress to the state of strain

( ),x tε at the same face of the bar.

( ) ( ) ( ), ,x t E x x tσ ε= (4.1.3)

Finally, the strain-displacement relation is

( ) ( ),,

u x tx t

tε

∂=

∂ (4.1.4)

Figure 24. (a) Bar element in one-dimensional longitudinal deformation. (b) Free-body diagram of bar differential element.

All together, the force at the LHS face of the differential element is

( ) ( ) ( ),,

u x tF x t EA x

x∂

=∂

(4.1.5)

A linearized Taylor's series expansion is then used to determine the force acting on the RHS face of the differential element.

( ) ( ) ( )

( ) ( ) ( ) ( )

,,

, ,

u x dx tF x dx t EA x dx

xu x t u x t

EA x EA x dxx x x

∂ ++ = +

∂∂ ∂ ∂

≈ + ∂ ∂ ∂

(4.1.6)

Finally, by Newton's 2nd law in (4.1.1), we have

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ), , , ,u x t u x t u x t u x tEA x EA x EA x dx A x dx

x x x x t tρ

∂ ∂ ∂ ∂ ∂ ∂− + + = ∂ ∂ ∂ ∂ ∂ ∂

(4.1.7)

Simplifying, we have the governing equation of motion for the uniaxial longitudinal deformations of the bar.


62

( ) ( ) ( ) ( ), ,u x t u x tEA x A x

x x t tρ

∂ ∂ ∂ ∂= ∂ ∂ ∂ ∂

(4.1.8)

( ) ( ) ( ) ( )2

2

, ,u x t u x tEA x A x

x x tρ

∂ ∂∂=

∂ ∂ ∂ (4.1.9)

When the bar cross-sectional area and Young's modulus are constant over the bar length, the EOM reduces to

( ) ( )2 2

2 2

, ,u x t u x tE

x tρ

∂ ∂=

∂ ∂ (4.1.10)

When the bar has a distributed axial load ( ),p x t , similar to the situation shown in Figure 14, a modified

free-body diagram will include the load. If the assumed direction of load action is in the positive direction of x , then the force is added to the total net forces working on the bar differential element. Then, the force balance results in the governing equation

( ) ( ) ( ) ( ) ( )2

2

, ,,

u x t u x tEA x p x t A x

x x tρ

∂ ∂∂+ =

∂ ∂ ∂ (4.1.11)

Note that the resulting governing equation of motion (4.1.11) is the same as that derived by the variational method in the Example of Sec. 3.1.5.

4.1.1 Boundary conditions

In the variational method of obtaining the EOMs of continuous structures, the boundary conditions are inherently derived from the undertaking (natural boundary conditions where displacements are unknown) and from knowledge of the system (essential boundary conditions where displacements are known).

When obtaining the EOMs by the equilibrium approach, the boundary conditions may be determined using expressions employed in Sec. 4.1 for sake of deriving the EOMs. Whether essential or natural, the boundary conditions are identified with help from knowledge of the system. Essential boundary conditions are recognized according to known constraints on the structure. For instance, with a bar fixed at x L= , the

displacements must be zero at this location for all time, ( ), 0u L t = . Essential boundary conditions are

therefore straightforward to identify and mathematically express.

Natural boundary conditions must be derived according to the force balance at the respective end.

For example, a free end to the bar at 0x = indicates that the force vanishes at the end since no reaction force occurs. Thus, using (4.1.5), a free end to the bar at 0x = indicates the boundary condition is

( ) ( )0,0 0

u tEA

x∂

=∂

.


63

Alternatively, if the bar terminates with a point mass m positioned at x L= , then application of a force balance at the end indicates that

( ) ( ) ( )2

2

, ,u L t u L tEA L m

x t∂ ∂

− =∂ ∂

(4.1.1.1)

Whereas, for a spring with stiffness k positioned at x L= , a similar force balance yields

( ) ( ) ( ),,

u L tEA L ku L t

x∂

− =∂

(4.1.1.2)

In other words, the general approach to determine the expressions for the boundary conditions via the equilibrium approach is to apply the force balance relations at the respective structure ends. When the boundary conditions are not essential, the resulting force balances give the expression for the natural boundary conditions.

4.2 Free response

To analyze the vibration response of continuous systems, the EOMs must be obtained whether by the variational or equilibrium approach. We repeat the EOM for the bar here for convenience.

( ) ( ) ( ) ( )2

2


x x tρ

∂ ∂∂=

∂ ∂ ∂ (4.2.1)

Then, we assume that EA and Aρ do not vary in x along the bar length, and then define a new parameter

/c E ρ= so that (4.2.1) is written

2 22

2 2

u ucx t∂ ∂

=∂ ∂

(4.2.2)

Formally, (4.2.2) is the linear wave equation. In the study of waves, the physical meaning of c is the speed of constant phase of a wave that travels along the bar. Here, we give attention to the study of vibrations due to the finite extent of the bar from 0x = to x L= . Thus, in such context the c is useful for brevity of notation.

Considering (4.2.2), we apply the mathematical method of separation of variables. In this approach, we assume that the one-dimensional longitudinal deformation may be decomposed into parts strictly dependent upon either space or time.

( ) ( ) ( ),u x t X x T t= (4.2.3)

By substitution of (4.2.3) into (4.2.2), we obtain

( ) ( ) ( ) ( )2c X x T t X x T t′′ = (4.2.4)


64

where ( )′ is used to denote / x∂ ∂ and the overdot is used to denote / t∂ ∂ . The terms of (4.2.4) may be

reorganized to collect together space-dependent terms on the LHS and time-dependent terms on the RHS.

2 X TcX T′′=

(4.2.5)

Since the LHS is independent of the RHS in (4.2.5), they must each be equal to the same constant. Here,

we denote the constant 2ω− .

22 2

2 0Xc X XX c

ωω′′

′′= − → + = (4.2.6)

2 2 0T T TT

ω ω= − → + =

(4.2.7)

In both cases, (4.2.6) and (4.2.7) are second-order ODEs whose general solutions are combinations of sinusoids given by

( ) cos sinx xX x A Bc cω ω

= + (4.2.8)

( ) cos sinT t C t D tω ω= + (4.2.9)

Application of the boundary and initial conditions enables one to identify the unknown amplitude constants

A , B , C , and D along with relations among these constants and the unknown ω .

Example

Consider Figure 25 that shows a bar fixed at both ends. With the initial conditions given by ( ),0 sin xu xLπ

=

and ( ),0 0u x = determine the free response for the bar ( ),u x t for all time.

Figure 25. Bar with mass at one end.

Answer The boundary conditions are that

( )0, 0u t = and ( ), 0u L t =

According to the method of separation of variables (4.2.3), the total assumed solution is


65

( ) ( ), cos sin cos sinx xu x t A B C t D tc cω ω ω ω = + +

Applying the boundary condition at 0x = , we have

( ) ( )( )0 0, cos sinu t A C t D tω ω= = +

Since the deformation of the bar at the boundary must be a zero displacement for all time, we determine

that 0A = . Then we remain with

( ) ( ), sin cos sinxu x t B C t D tcω ω ω= +

Applying the boundary condition at x L= , we have

( ) ( )0 , sin cos sinLu L t B C t D tcω ω ω= = +

Since the relation must be true for all time, we have that

sin 0 ; 1,2,...nL n c n

c Lω πω= → = =

where the index 0n = is omitted since it corresponds to the static problem. This indicates that an infinite

number of possible frequencies nω satisfy the governing equation. These are the natural frequencies of

vibration of the bar nω . As a result, we identify an infinite number of mode shapes where the spatial

variation of the bar longitudinal deformation for the thn mode shape is given by

( ) sin nn n

xX x Bc

ω=

Each mode shape occurs at a specific natural frequency. Therefore, the total deformation response of the bar is

( )1

, sin cos sin sinn nn n n n

n

x xu x t C t D tc c

ω ωω ω

∞

=

= +∑

where the constant B is absorbed into the constants C and D .

It is observed that the modes are orthogonal. Thus, we find that the general vibration response for the bar may be found from a superposition of an infinite number of the normal modes multiplied by time-dependent terms.

The initial conditions are applied to determine the remaining constants C and D .

( )1

sin ,0 sin cos 0 sin sin 0n nn n n n

n

x xx u x C DL c c

ω ωπ ω ω∞

=

= = +∑


66

1sin sin n

nn

xx CL c

ωπ ∞

=

=∑

We express the final relation in terms of the frequency values

1sin sinn

n

x n xCL Lπ π∞

=

=∑

Taking advantage of the orthogonality of the mode shapes, we multiply both sides of the equation by

sin m xLπ and integrate over the bar length.

0 01

sin sin sin sinL L

nn

x m x n x m xdx C dxL L L Lπ π π π∞

=

= ∑∫ ∫

The orthogonality of modes results in the finding that 1 1; 0, 1mC C m= = ≥ . This shows that only the first

mode shape contributes to the vibration due to the initial conditions.

To solve for the constants D , we apply the second initial condition

( )1

0 ,0 sin sin 0 sin cos 0n nn n n n n n

n

x xu x C Dc c

ω ωω ω ω ω

∞

=

= = − +∑

1

1

0 sin

sin

nn n

n

nn

xDc

n c n xDL L

ωω

π π

∞

=

∞

=

=

=

∑

∑

Because the relation must be satisfied for all time, we conclude that nD =0 for all n .

In conclusion, we find that the free vibration response of the bar due to the initial conditions, and meeting the constraints of the boundary conditions, is a summation of all of the vibrations modes as

( )1

, sin cosnn

n x n ctu x t CL Lπ π∞

=

=∑

Since only the first mode contributes due to the form of the initial conditions, we have that

( ), sin cosx ctu x tL Lπ π

=

which shows the spatial and time-varying nature of the vibration response.

The frequency nω related to the vibration of the thn mode shape is referred to as the natural frequency.

Thus, each mode shape occurs in a specific spatial distribution of displacement amplitude with oscillations at a rate associated with the natural frequency.


67

4.3 Free response by assumed modes method of analysis

Through the Example in Sec. 4.2, we found that the general response of the longitudinal vibrations of the bar is associated with a superposition of the orthogonal modes weighted by time-dependent terms. As a result, we may improve our assumed solution by assuming the expansion

( ) ( ) ( )1

, n nn

u x t X x tξ∞

=

=∑ (4.3.1)

where ( )nX x is the thn normal mode and ( )n tξ is the time-dependent weighting term. This strategy is

referred to as the assumed modes method of analysis.

By substitution of (4.3.1) into the EOM (4.2.2), we have

( ) ( ) ( ) ( )2

10n n n n

nc X x t X x tξ ξ

∞

=

′′− + =∑ (4.3.2)

The normal modes are orthogonal. Thus, we take advantage of this property and multiply (4.3.2) by ( )mX x

where m n≠ and integrate over the bar length.

( ) ( ) ( ) ( ) ( ) ( )2

0 01

0L L

n n m n n mn

c t X x X x dx t X x X x dxξ ξ∞

=

′′− + =∑ ∫ ∫ (4.3.3)

We then take advantage of the properties of normal modes, i.e. orthogonal functions. Namely, two

orthogonal, normal modes ( )nX x and ( )mX x have the following properties.

( ) ( )0

0;L

n mX x X x dx m n= ≠∫ (4.3.4)

( ) ( )0

0;L

n mX x X x dx m n′ ′ = ≠∫ (4.3.5)

( ) ( )0

0;L

n mX x X x dx m n′′ = ≠∫ (4.3.6)

In light of (4.3.4) and (4.3.6), we have from (4.3.3) a system of equations given by

( ) ( ) 0; 1,2,...,n n n nM t K t nξ ξ+ = = ∞ (4.3.7)

where

( ) ( )0

L

n n nM X x X x dx= ∫ (4.3.8)


68

( ) ( )2

0

L

n n nK c X x X x dx′′= − ∫ (4.3.9)

Note that the equations (4.3.7), albeit infinite in number, are decoupled and are each similar to the equation of motion pertaining to an individual mass-spring oscillator as detailed in Sec. 2.1.1. Thus, to obtain the associated free response for each decoupled mass-spring oscillator-type equation, we need to identify the associated initial conditions. Considering the initial conditions

( ) ( ) ( )1

,0 0n nn

u x X x ξ∞

=

=∑ (4.3.10)

( ) ( ) ( )1

,0 0n nn

u x X x ξ∞

=

=∑

(4.3.11)

We multiply both of the equations (4.3.10) and (4.3.11) by ( )mX x and integrate over the bar length.

( ) ( ) ( ) ( ) ( )0 0

1,0 0

L L

m n n mn

u x X x dx X x X x dxξ∞

=

=∑∫ ∫ (4.3.12)

( ) ( ) ( ) ( ) ( )0 0

1,0 0

L L

m n n mn

u x X x dx X x X x dxξ∞

=

=∑∫ ∫

(4.3.13)

As a result, we relate the initial to its contribution to each normal mode via

( ) ( ) ( )0

20 ,0L

m mu x X x dxL

ξ = ∫ (4.3.14)

( ) ( ) ( )0

20 ,0L

m mu x X x dxL

ξ = =∫

(4.3.15)

In the event that the modes have been normalized via the technique described in Sec. 2.4, which leads to

orthonormal modes, we find that ( )2

01

L

n nM X x dx= =∫ . In this case, we drop the 2 / L factor on the RHS

of both (4.3.14) and (4.3.15). The more general case wherein the modes are just orthogonal and not orthonormal uses the results (4.3.14) and (4.3.15).

Recalling our findings of Sec. 2.1.1 for the free response of the mass-spring system, the solution to an equation of the form (4.3.7) is

( ) ( ) ( ) ( )0, 0 cos sinn

n n n n nn

u x t X x t tξ

ξ ω ωω

= +

(4.3.16)

so that the total solution, and thus total free vibration response of the bar, is

( ) ( ) ( ) ( )1

0, 0 cos sinn

n n n nn n

u x t X x t tξ

ξ ω ωω

∞

=

= +

∑

(4.3.17)


69

4.4 Forced response

While initial conditions determine the free response vibration of a continuous system, the excitation conditions determine the forced response. In practical applications, a finite amount of damping exists that prevents free oscillations from persisting for all time. Therefore, often the steady-state structural dynamics associated with the forced response are the primary behavior evaluated in investigations of continuous systems. This Sec. 4.4 describes and exemplifies the strategy used to compute the forced response of continuous structures.

Consider the one-dimensional longitudinal vibrations of the fixed-fixed bar as excited by the distributed

axial force ( ),p x t . Our discovery in Sec. 4.3 that the general free response of the bar may be decomposed

into a linear combination of the normal modes is extended here to the case of the general forced response. With the general governing equation, repeated below from (4.1.11), we have

( ) ( ) ( ) ( ) ( )2

2

, ,,

u x t u x tEA x p x t A x

x x tρ

∂ ∂∂+ =

∂ ∂ ∂ (4.4.1)

As in the case of the free response, we again assume that the forced vibrations of this linear system may be decomposed into a linear combination of the normal modes. This approach is the assumed modes method, here applied to analyze the forced response of the system.

( ) ( ) ( )1

, n nn

u x t X x tξ∞

=

=∑ (4.4.2)

We substitute (4.4.2) into (4.4.1) to yield

( ) ( ) ( ) ( ) ( ) ( ) ( )1

,n n n nn

EA x X x t A x X x t p x tx

ξ ρ ξ∞

=

∂ ′− + = ∂ ∑ (4.4.3)

When the Young's modulus and cross-section area do not vary over the bar length, we have

( ) ( ) ( ) ( ) ( )1

,n n n nn

EAX x t AX x t p x tξ ρ ξ∞

=

′′− + = ∑ (4.4.4)

Then, (4.4.4) is multiplied by another normal mode ( )mX x and integrated over the bar length.

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )0 0

1,

L L

n m n n m n mn

EAX x X x t AX x X x t dx p x t X x dxξ ρ ξ∞

=

′′− + = ∑ ∫ ∫ (4.4.5)

Simplifying (4.4.5) according to the properties (4.3.4)-(4.3.6) of normal modes, we find

( ); 1,2,...n n n n nm k f t nξ ξ+ = =

(4.4.6)

where

( ) ( )0

L

n n nk EA X x X x dx′′= − ∫ (4.4.7)


70

( )2

0

L

n nm A X x dxρ= ∫ (4.4.8)

( ) ( )0

,L

n nf p x t X x dx= ∫ (4.4.9)

Then, the decoupled equations (4.4.6) may be individually solved for their respective values of the time-

dependent unknowns ( )n tξ , after which the total forced response may be reconstructed from (4.4.2).

From this derivation, it is apparent that the method established for lumped parameter systems in Sec. 2.4.2 extends with ease to the study of forced continuous structures via the occurrence of normal modes of vibration. The assumed modes method is therefore a valuable tool to analyze the free or forced response of continuous structures.

Example Consider the fixed-fixed bar with point axial force application at 2 / 3x L= , so that

( ) ( )0 02 / 3, cosp L t p t P tω= = and ( ), 0p x t = for all other [ ]0,x L∈ . The axial point force is harmonic in

time with angular frequency of oscillation ω . Determine the steady-state oscillation response of the bar induced by the point force load.

Answer The time-harmonic nature of the force encourages us to consider a complex exponential notation by

( )0 0j tt P e ω=p

Of course, because the actual force is 0 0 cosp P tω= , the complex exponential formulation inherently

assumed that the actual force is ( )0Re t p . Subsequently, solution to the EOM for the bar requires a final

consideration of the real part of the complex longitudinal deformation. If the actual force is 0 0 sinp P tω= ,

this indicates that the actual force is ( )0Im t p , and likewise we must take the imaginary part of the

complex longitudinal deformation as the final result.

We adopt a complex exponential form of time-harmonic dependence for the longitudinal vibrations of the

bar as modified by the spatial dependence of the normal modes ( )nX x

( ) ( )1

, j tn n

nx t e X xω

∞

=

= ∑u ξ

where the coefficients nξ are complex valued in the general case. Note that the coefficients nξ are no longer

time-dependent due to the time-harmonic assumption j te ω . In this way, substitution of such an assumed solution into (4.4.6) results in

( )2 ; 1,2,...j tn n n nm k e t nωω − + = = ξ f

where we have that


71

( ) ( )002 / 3

Lj tn ne P x L X x dxω= =∫f

Due to the point spatial extent of the harmonic axial force, we have that

( )0 2 / 3 j tn nP X L e ω=f

As before, the mass and stiffness terms associated with the thn normal mode are respectively

( )2

0

L

n nm A X x dxρ= ∫

( ) ( )0

L

n n nk EA X x X x dx′′= − ∫

The thn normal mode for the fixed-fixed bar is

( ) sinn nn xX x C

Lπ

=

Then, the evaluation of the mass and stiffness terms results in

; 1,2,3,...2nLm A nρ= =

2 2

; 1,2,3,...2n

nk EA nLπ

= =

Then, substitution of this time-harmonic force nf into the complex exponential form of the decoupled

governing equations results in

( )20 2 / 3 ; 1,2,...j t j t

n n n nm k e P X L e nω ωω − + = = ξ

As a result, we find that the thn complex amplitude of the longitudinal deformation ( )n tξ is computed from

0

2

2sin3

nn n

nP

k m

π

ω=

−ξ

Therefore, the complex valued vibrations of the bar are all together determined from the linear combination

( )0

21

2sin sin3, j t

n n n

n n xPLx t e

k mω

π π

ω

∞

=

=−∑u

whereas only the real part of this result corresponds to the actual, physical solution for the deformation response. In other words, the actual vibrations of the bar are


72

( ) ( )0

21

2sin sin3Re , , cos

n n n

n n xPLx t u x t t

k m

π π

ωω

∞

=

= = −∑u

It is clear that, similar to the harmonically forced lumped parameter systems investigated in Sec. 0, the bar may be driven in a state of resonance whereby the deformations of the bar are most greatly magnified around the respective normal mode natural frequency.

4.5 Longitudinal vibration of bars with continuously varying cross-section

Recall (4.1.9), which is repeated in (4.5.1) for convenience. If the bar is composed of a uniform material such that the Young's modulus E is independent of the spatial location along the bar, then a continuously varying bar cross-section will change the means by which we may solve the EOM.

( ) ( ) ( ) ( )2

2


x x tρ

∂ ∂∂=

∂ ∂ ∂ (4.5.1)

Rearranging terms, the EOM (4.5.1) is expressed

( ) ( ) ( ) ( )2

2

, ,u x t u x tA x A x

x x E tρ ∂ ∂∂

= ∂ ∂ ∂

(4.5.2)

Considering a separable form of the bar free longitudinal vibration

( ) ( ) [ ], sinu x t X x tω φ= + (4.5.3)

and substituting (4.5.3) into (4.5.2), one obtains

2 2

2

1 0d X dA dX Xdx A dx dx E

ρω + + =

(4.5.4)

When the cross-sectional area varies via

( ) 0nA x A x= (4.5.5)

then the (4.5.4) may be written

2

0nX X Xx E

ρω′′ ′+ + = (4.5.6)

Defining terms of

12

nv −= ;

22k

Eρω

= (4.5.7)


73

we have the EOM of

22 1 0vX X k Xx+′′ ′+ + = (4.5.8)

Applying the coordinate transform ( ) ( )X x x vU x= − results in an equation

( )2 2 2 2 0x U xU k x v U′′ ′+ + − = (4.5.9)

The (4.5.9) is the Bessel equation, whose solutions are combinations of Bessel functions of the first and second order, vJ and vY , respectively:

( ) ( )1 2v vU C J kx C Y kx= + (4.5.10)

The frequencies and mode shapes of the bar with continuously varying cross-section are obtained in the same way as for constant cross-section bars: by applying boundary conditions. The initial conditions likewise determine the amplitudes of the free vibration response. By substitution, the longitudinal deformation of the bar is ultimately found to be

( ) ( ) ( ) [ ]1 2, sinvv vu x t C J kx C Y kx x tω φ−= + + (4.5.11)

Exact solutions may be obtained for this complex problem in the case of ( ) 20A x A x= as described in [11].

Yet, the challenge of undertaking the math via this approach encourages alternative means of solution. In Sec. 5.4 we will learn how to handle sudden discontinuities in the solution method for the vibration of continuous structures. A bar with continuously varying cross-section may be considered as a collection of bar segments each with a sudden cross-section difference, such that in the limit of an infinite number of such discrete bar segments we then realize the continuously varying cross-section. Thus, the more straightforward analysis of Sec. 5.4 provides a useful means of approximately solving the problem of continuous variation of the continuous structure properties.

4.6 Summary

As shown in Sec. 3, variational methods may be utilized to derive governing equations of motion for continuous structures. Sec. 4.1 exemplifies that an equilibrium method based on Newton's laws of motion may likewise derive the same governing equations. Both methods also retrieve the associated boundary conditions. Once obtained, these EOMs are solved in the same ways for free or forced response via the solution approaches described in Secs. 4.3 and 4.4.

Consequently, in practice it is up to the investigator to select the method that is more beneficial to use for deriving the EOMs of a given continuous structure and application of interest. Experience in utilizing the methods is therefore an advantage in order to understand the strengths and challenges involved in their application.


74

To this point, we have established understanding on how to apply the variational and equilibrium methods for the case of the longitudinal vibrations of the bar to find the EOMs and then to scrutinize them for the free and forced responses. To enrich this expertise and familiarization with using the different methods to obtain the EOMs and subsequently to solve them, the following Secs. 5, 6, and 7 investigate the vibrations of three continuous systems also of widespread interest in engineering applications: Sec. 5 transverse vibrations of stretched strings, Sec. 6 rotational oscillations of shafts, and Sec. 7 transverse vibration beams.


75

5 Transverse vibrations of a stretched string

Vibrating strings proliferate in the engineered world. From electrical transmission lines, to musical instruments, to belts in automotive engines, to biological strings that include human organs and DNA: there are innumerable number of applications for which the investigation of the dynamics of these continuous structures are critical to understanding and decision-making.

This Sec. 5 will analyze the transverse vibrations of strings through the strategies described in the prior sections in the context of studying the longitudinal deformations of the one-dimensional bar. By applying these strategies to the investigation of string vibrations, we will uncover parallels between the oscillatory behaviors as well as contrasts that differentiate strings from other continuous structures.

5.1 Equations of motion and boundary conditions

5.1.1 Equilibrium approach

Consider the string in Figure 26 that is stretched along its length L . The mass per unit length of the string

is ρ . The string deflects in the transverse direction by an amount ( ),w x t . Such deflections are caused by

initial conditions and an applied force per unit length ( ),f x t . Additional reaction forces are generated at

the boundaries by end springs with stiffness constants 1k and 2k . We assume that transverse deflections of

the string ( ),w x t are sufficiently small with respect to the string length so as to induce only linear stretching

behaviors.

Figure 26. Stretched string with force application and spring boundary conditions. Free-body diagram of string differential element.

A free-body diagram of a differential element of the string is shown at right in Figure 26. The tension T is equal and opposite at the ends of the string by assuming a sufficiently small differential string length ds so that the tension difference is negligible. The angle made by the string above the horizontal according to the transverse deflection is θ . In other words, by the assumptions of linear string deflection from equilibrium, we find that /w xθ ≈ ∂ ∂ .

The difference in the vertical, z axis, tension force from one end of the string differential element to the other is


76

( ) ( )sin sinz x dx xdf T Tθ θ+

= − (5.1.1.1)

Taking a Taylor's series expansion of ( )sin x dxT θ+

, we find

( ) ( ) ( )sinsin sin x

x dx x

TT T dx

xθ

θ θ+

∂≈ +

∂ (5.1.1.2)

In this way, the net tension force is found to be

( )sin xz

Tdf dx

xθ∂

≈∂

(5.1.1.3)

Given the linearity of the string transverse displacement, /w xθ ≈ ∂ ∂ , we find

2

2zw wdf T dx T dx

x x x∂ ∂ ∂ = = ∂ ∂ ∂

(5.1.1.4)

The mass of the differential string element is dxρ . Consequently, by Newton's 2nd law the governing

equation of motion for the string is

( )2 2

2 2 ,w wdx T dx f x t dxt x

ρ ∂ ∂= +

∂ ∂ (5.1.1.5)

Simplifying, we obtain the EOM for the string

( )2 2

2 2 ,w wT f x tt x

ρ ∂ ∂− =

∂ ∂ (5.1.1.6)

When the string is unforced, we obtain the familiar equation

2 2

2 2

w wTx t

ρ∂ ∂=

∂ ∂ (5.1.1.7)

2 22

2 2

w wcx t

∂ ∂=

∂ ∂ (5.1.1.8)

where /c T ρ= is a constant. The (5.1.1.8) is an analogous equation to the EOM derived for the one-

dimensional longitudinal vibrations of the bar (4.2.2). Thus, while the deformation of the bar is along its axis, the deformation of the string is out-of-axis although the dynamics of both continuous structures are governed by the same underlying physics and equation. Also, the (5.1.1.8) is likewise a linear wave equation. Students interested in a course with emphasis on principles of wave propagation should consider enrolling in the OSU MAE courses on "Engineering Acoustics".

The boundary conditions for the string are obtained by force balances at the respective locations. At 0x =, the vertical force in the string must be balanced by the force of the spring 1k . Thus the force balance

expression is


77

( ) ( )1

0,0,

w tT k w t

x∂

=∂

(5.1.1.9)

while a similar relation is obtained for the boundary condition at x L= respecting the spring 2k

( ) ( )2

,,

w L tT k w L t

x∂

= −∂

(5.1.1.10)

The sign change on the RHS of (5.1.1.10) is because positive ( ),w x t near x L= results in negative slope

/w x∂ ∂ . As a result, since we have positive ( ),w x t near x L= , the negative LHS must be balanced by a

negative RHS to (5.1.1.10) presuming that the spring stiffness 2k >0.

5.1.2 Variational approach

To use the variational approach to derive the EOMs for the vibration of the stretched string, we must first identify the total kinetic and strain energies for the string.

Consider the differential length ds of the string element shown in Figure 26. This length is related to the horizontal and vertical change along the ds by

( )2 2

2 112

w wds dx dx dxx x

∂ ∂ = + ≈ + ∂ ∂ (5.1.2.1)

The approximation in (5.1.2.1) is made by a linearized binomial expansion.

The strain energy associated with the string transverse deflection is associated with the tension causing the change of string length given by ds dx− . In other words,

( ) ( ) ( )2

0 0

1,2

L L wT x ds x t dx T x dxx

∂ − = ∂ ∫ ∫ (5.1.2.2)

When the tension is presumed to be constant along the string length, the strain energy associated with the string stretching is

2

0

12

L wT dxx

∂ ∂ ∫ (5.1.2.3)

The strain energy associated with the end springs is

( ) ( )2 21 2

1 10, ,2 2

k w t k w L t+ (5.1.2.4)

Thus, the total strain energy is

( ) ( )2

2 21 20

1 1 10, ,2 2 2

L wU T dx k w t k w L tx

∂ = + + ∂ ∫ (5.1.2.5)


78

The kinetic energy of the string is

( )2

0

12

L wT x dxt

ρ ∂ = ∂ ∫ (5.1.2.6)

With constant string density ρ along the string length, the total kinetic energy is simplified to be

2

0

12

L wT dxt

ρ ∂ = ∂ ∫ (5.1.2.7)

Finally, the potential of the applied loads is

( ) ( )0

, ,L

W f x t w x t dx= ∫ (5.1.2.8)

Thus, the Lagrangian for the string is

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

2 22 2

1 20 0 0

2 22 2

1 20

1 1 1 10, , , ,2 2 2 2

1 1 1 1, , 0, ,2 2 2 2

L L L

L

L T U W

w wdx T dx k w t k w L t f x t w x t dxt x

w wT f x t w x t dx k w t k w L tt x

ρ

ρ

= − +

∂ ∂ = − − − + ∂ ∂ ∂ ∂ = − + − − ∂ ∂

∫ ∫ ∫

∫

(5.1.2.9)

Hamilton's principle then states

2

1

0t

tI Ldtδ δ= =∫ (5.1.2.10)

By computing (5.1.2.10), an EOM and associated boundary conditions will be obtained whose solutions

are the stationary values of the function ( ),w x t that minimize the functional (5.1.2.10).

Following through on the first variation, we find

( ) ( ) ( ) ( )2

1

2 22 2

1 20

1 1 1 1, , 0, ,2 2 2 2

t L

t

w wI T f x t w x t dx k w t k w L t dtt x

δ δ ρ ∂ ∂ = − + − − ∂ ∂

∫ ∫ (5.1.2.11)

( ) ( ) ( )2

11 0 0 20

,t L

L Lt

w wI w T w f x t w dx k w w k w w dtt t x x

δ ρ δ δ δ δ δ ∂ ∂ ∂ ∂ = − + − − ∂ ∂ ∂ ∂

∫ ∫ (5.1.2.12)

Applying integration by parts to the first and second terms of the integrand in (5.1.2.12), we have

( )2

1

2 2

1 0 0 22 200

, 0L

t L

L Lt

w w wI T f x t wdx T w k w w k w w dtt x x

δ ρ δ δ δ δ ∂ ∂ ∂ = − + + − − − = ∂ ∂ ∂

∫ ∫ (5.1.2.13)

This results in the governing equation of motion (5.1.2.14) for the transverse vibrations of the stretched string


79

( )2 2

2 2 ,w wT f x tt x

ρ ∂ ∂− =

∂ ∂ (5.1.2.14)

The corresponding boundary conditions are

( ) ( )1

0,0,

w tT k w t

x∂

=∂

(5.1.2.15)

( ) ( )2

,,

w L tT k w L t

x∂

= −∂

(5.1.2.16)

As should be anticipated, the EOM (5.1.2.14) is the same as (5.1.1.6). The boundary conditions (5.1.2.15) and (5.1.2.16) are the same, respectively, as (5.1.1.9) and (5.1.1.10). This demonstrates the equivalence between the equilibrium and variational approaches to deriving the EOM for continuous systems.

5.2 Analysis of the free response

The EOM for the transverse vibrations of the stretched string, either (5.1.1.6) or (5.1.2.14), is the same as the EOM for the one-dimensional longitudinal deformations of the rod (4.2.2). Consequently, by the mathematical analogy, we may expect that the same assumed solution routines will work to predict the dynamic response of the string, whether by initial conditions or by force excitation. In this Sec. 5.2 and in Sec. 5.3 we confirm this intuition by examples.

Example A string stretched between two free ends is shown in Figure 27. The free boundary conditions are realized by massless rollers that slide in vertical passages. The string has uniform tension T and density per length

ρ along its length L . The initial transverse deflection of the string is given by ( ) ( )0 0, 2 1xw x w tL

= = −

while the initial velocity is zero ( ) ( )0 0, 0w x w t= = . Determine the transverse vibration of the string

( ),w x t for all time t .

Figure 27. Free-free stretched string between sliding rollers.

Answer The governing equation for the string is


80

2 22

2 2

w wcx t

∂ ∂=

∂ ∂

where /c T ρ= . The free-free boundary conditions indicate that

( )0

0,0 0

x

w twTx x=

∂∂= → =

∂ ∂

( ),0

w L tx

∂=

∂

We use the method of separation of variables and assume that the string responds in a way such that the

spatial variation of deflection ( )X x is independent of the time variation of deflection ( )T t .

( ) ( ) ( ),w x t X x T t=

Therefore, by substitution of this assumed solution into the EOM, we have

2 X TcX T′′=

Being independent of each other, both sides of the above equation must be equal to the same constant, 2ω−. We therefore obtain the following pair of equations

2

2 0X Xcω′′ + =

2 0T Tω+ =

Appropriate assumed solutions to these two, second-order ordinary differential equations are combinations of sinusoids, via


= +

( ) cos sinT t C t D tω ω= +

The assumed spatial response distribution ( )X x is substituted into the associated boundary conditions to

yield

0 sin 0 cos0 0A B Bc cω ω

= − + → =

0 sin nL ncA

c c Lω ω πω= − → =

Therefore, the normal modes are identified to be


81

( ) cos cosnn n n

x nxX x A Ac L

ω π= =

The natural frequencies associated with the normal modes are

; 1,2,...nnc nL

πω = =

As a result, the time response of the transverse vibrations of the free-free string associated with the thn mode are

( ), cos cos cos sinn n n n nnx nxw x t C t D tL L

π πω ω= +

so that the total response is a linear combinations of the all of the time histories associated with the normal modes:

( )1

, cos cos cos sinn n n nn

nx nxw x t C t D tL L

π πω ω∞

=

= +∑

Then, we apply the initial conditions to determine the coefficients nC and nD .

( ) ( )0 00 01 1

cos cos cos cosL L

n nn n

nx mx nx mxw x C w x dx C dxL L L L

π π π π∞ ∞

= =

= → =∑ ∑∫ ∫

( )01

0 cos 0n n nn

nxw x D DL

πω∞

=

= = → =∑

Due to the orthogonality of the modes, we find the nC to be

( )00

0

2 cos

2 2 1 cos

L

n

L

n

nxC w x dxL L

x nxC dxL L L

π

π

=

= −

∫

∫

The Fourier cosine series for nC is evaluated and found to be

2 2

8 ; 1,3,5,...

0; 2,4,6,...n

nC n

nπ

− == =

In consequence, the total time response of the stretched, free-free string with the initial condition

( ) ( )0,0w x w x= is given by

( )1

, cos cosn nn

nxw x t C tL

π ω∞

=

=∑ , with nncL

πω = and 2 2

8 ; 1,3,5,...

0; 2,4,6,...n

nC n

nπ

− == =


82

Representative results of reconstructing this initial condition with different numbers of the normal modes are given in Figure 28. Only the odd order modes are induced by the initial condition. This is because the initial condition is antisymmetric about the string center point. The even order modes are symmetric about the center point. Thus, the even order modes do not contribute to the free response of the string due to the antisymmetric initial condition.

Similar situations arise in the use of other boundary conditions and initial conditions. The consideration of symmetry in each case can be fruitful to quickly predicting whether even or odd order modes will contribute to the free response.

Figure 28. Examples of using different number of normal modes to reconstruct the initial condition of the free-free stretched

string, with 1[ ]L m= .


n=7; % number of normal modes in linear combination L=1; % string length x=linspace(0,L,201); w=zeros(size(x)); for iii=1:2:n, w=w+-8/iii^2/pi^2*cos(pi*iii*x/L); % fourier cosine coefficients, only for n odd end; figure(1); clf;


83

plot(x,w); xlabel('string length [m]'); ylabel('transverse string deflection w(x,0)'); title(['number of normal modes: ' num2str(n) ]);

5.3 Forced response

In the Example of Sec. 5.2, the excitation force per unit length ( ),f x t is absent, so that the string vibration

is due to initial conditions alone. Here, we examine the steady-state, forced response of the stretched string and find parallels to the forced response of the longitudinal deformations of the bar.

Example Consider the fixed-fixed stretched string shown in Figure 29. The string has uniform tension T and uniform density per length ρ along the string length L . The string is comparable to a musical instrument string.

Like certain stringed musical instruments the string is 'bowed' with an excitation distributed over a portion of the string length near a fixed end. The force per unit length is then expressed by

( ) 0

0; 2 / 3, cos ;2 / 3 5 / 6

0; 5 / 6

x Lf x t f t L x L

x Lω

<= ≤ ≤ >

Determine the steady-state transverse vibration response of the string due to the harmonic, distributed force.

Figure 29. String with distributed uniform excitation force per unit length over portion of string length.

Answer The EOM for the forced vibrations of the string is

( )2 2

2 2,w wT f x tx t

ρ∂ ∂+ =

∂ ∂

The boundary conditions for the fixed-fixed string are

( ) ( )0, , 0w t w L t= =

The initial conditions are not considered due to the attention given to steady-state string response.


84

To solve for the forced response of the string, we first determine the linear normal modes. Once determined, the normal modes will serve as our basis of applying the assumed modes method to analyze the forced response.

The normal modes are identified by solving the homogeneous form of the EOM.

2 2

2 2

w wTx t

ρ∂ ∂=

∂ ∂

By assuming a separable solution to the homogeneous EOM via ( ) ( ) ( ),w x t X x T t= , we obtain two

separable equations for both space and time

( )( )

22 2

2 0X x

c X XX x c

ωω′′

′′= − → + =

( )( )

2 2 0T t

T TT t

ω ω= − → + =

where /c T ρ= . The solution to the spatial equation is of the form

( ) sin cosx xX x A Bc cω ω

= +

Substituting the ( )X x into the boundary conditions results in

( ) ( )0, 0 0 0 0 sin 0 cos0 0w t X A B B= → = → = + → =

( ) ( ), 0 0 0 sin ; 1,2,3,...nL n cw L t X L A n

c Lω πω= → = → = → = =

So that the thn linear normal mode is

( ) sin sinnn n n

x n xX x A Ac L

ω π= =

Therefore, using the assumed modes method, we assume that the transverse vibration of the string is composed from a linear combination of the normal modes.

( )1

, sin sin sin cosn n n nn

n x n xw x t C t D tL Lπ πω ω

∞

=

= +∑

Such response is applicable whether for free or forced response.

For the homogeneous EOM, the unknown coefficients nC and nD are determined by applying the initial

conditions to the total vibration response expression.


85

For the inhomogeneous EOM with the forcing excitation, we employ a linear combination of the normal modes with frequencies equivalent to the frequency of the harmonic excitation. For mathematical convenience, we use the complex exponential notation, which reduces the assumed modes response expression.

( )1

, sin j tn

n

n xx t eL

ωπ∞

=

=∑w G

The nG are complex-valued amplitudes to be determined. In the same spirit, the excitation force is

expressed as a complex exponential using

( ) ( )0

0; 2 / 3, ;2 / 3 5 / 6

0; 5 / 6

j t j tx

x Lx t f e L x L F x e

x L

ω ω

< = ≤ ≤ = >

f

It is understood that only the real part of ( ),x tf is the actual excitation force, since Re cosj te tω ω = .

Similarly, only the real part of ( ),x tw is the actual transverse vibration response of the string.

By substituting these expressions into the inhomogeneous EOM, we obtain

( )2

2

1sinj t j t

x nn

n n xF x e T eL L

ω ωπ πω ρ∞

=

= −

∑ G

Multiplying both sides by a normal mode sin m xLπ and integrating over the string length enables us to

compute the complex coefficients nG .

( )2 02

2 1 sinL

n xn xF x dx

L LnTL

ππ ω ρ

= −

∫G

The expression for the complex-valued constants nG may be simplified by recalling the expression for the thn natural frequency

nn c

Lπω =

( )2 2 0

2 1 sinL

n xn

n xF x dxL L

πρ ω ω

=− ∫G

We also may note that since the spatial variation of the harmonic force ( )xF x is purely real, the coefficients

nG are likewise only real.


86

( )2 2 0

2 1 sinL

n xn

n xG F x dxL L

πρ ω ω

=− ∫

Evaluating the integral with the known form of the force distribution along the string length, we find

2 2

2 1 2 5cos cos3 6n

n

n nGLn

π πρ π ω ω

= − −

Putting it all back together, the forced, steady-state vibration of the string due to the distributed load is

( ) ( ) 2 21

2 1 2 5Re , , cos cos cos sin3 6n n

n n n xx t w x t tLn L

π π πωρ π ω ω

∞

=

= = − − ∑w

This solution result is exemplified in Figure 30 for the case of the vibration at location /x L π= [m] and using different number of normal modes in the expansion. In general, a few modes are sufficient to accurately determine the amplitude of the lowest order mode. An increasing number of modes used in the reconstruction leads to enhanced fidelity of the prediction and ability to predict the vibration response to higher frequencies.


87

Figure 30. Representative results of forced string vibration at location /x L π= [m] for different number of linear normal modes in assumed solution.


rho=900; % [kg/m] string linear density T=3000; % [N] tension c=sqrt(T/rho); % [m/s] phase speed f_0=1; % [N/m] force per unit length n=3; % number of normal modes to use L=1; % [m] string length xc=L/pi; % [m] length along string to plot string amplitude omega=linspace(1,80,2001); % [rad/s] range of excitation frequencies w=zeros(size(omega)); % preallocate transverse vibration response for ooo=1:length(omega) for iii=1:n w(ooo)=w(ooo)+2/rho/L/iii/pi/((iii*pi*c/L)^2-omega(ooo)^2)*(cos(2*iii*pi/3)-cos(5*iii*pi/6))*sin(iii*pi*xc/L); end end figure(1); clf; semilogy(omega/2/pi,abs(w),'-r'); xlabel('frequency [Hz]'); ylabel('w(t) amplitude of string vibration'); xlim([0 10]) ylim([1e-8 1e-2]) title(['string vibration amplitude evaluated at ' num2str(xc/L) ' along string length'],['number of modes in expansion: ' num2str(n) '']);


88

As seen from the Example, the investigation of forced vibration follows naturally from the investigation of free vibration. This is because the assumed modes methods requires one to first compute the normal modes before solving for the forced response. This assumed solution approach is valid under a vast range of circumstances in the study of continuous structures.

A primary conclusion from these results is that determining the normal modes is critical to the subsequent computation of either free or forced response of the continuous system.

5.4 Vibration of discontinuous strings

Discontinuities in continuous structures are common. A rib-stiffener on a beam system will create a local increase in stiffness similar to a point-attached spring and will cause the vibration of the beam to deviate from the behavior observed in the absence of the local stiffener. A construction truck parked on a bridge creates a mass discontinuity along the length of the bridge span that will change the bridge vibration from the normal mode behavior strictly associated with the bridge boundary conditions.

To investigate how discontinuities influence the solution procedure in predicting the vibration response of continuous systems, we examine a straightforward case of a mass discontinuity for the fixed-fixed string shown in Figure 31. The mass M is located one-quarter of the string length L from one of the edges. The string tension T and linear density ρ are unchanging along the string.

Figure 31. String with point mass creating a discontinuity along the string length.

Because of the discontinuity, the string vibration is governed by two separate equations respecting either side denoted with spatial positions 1x or 2x . These spatial locations are expressed as functions of the

respective string ends that are fixed, for sake of mathematical ease in manipulating later expressions. Thus, the two EOMs are

( ) ( )2 21 1 1 1

12 21

, ,; 0 / 4

w x t w x tT x L

x tρ

∂ ∂= < <

∂ ∂ (5.4.1)

( ) ( )2 22 2 2 2

22 22

, ,; 0 3 / 4

w x t w x tT x L

x tρ

∂ ∂= < <

∂ ∂ (5.4.2)

Separable solutions for 1w and 2w are assumed.


89

( ) ( ) ( )1 1 1 1 1,w x t X x T t= (5.4.3)

( ) ( ) ( )2 2 2 2 2,w x t X x T t= (5.4.4)

Considering the spatial components, we take the general expansions

( ) 1 1 1 11 1 1 1sin cosx xX x A B

c cω ω

= + (5.4.5)

( ) 2 2 2 22 2 2 2sin cosx xX x A B

c cω ω

= + (5.4.6)

where /c T ρ= is the phase speed, as before. The application of the boundary conditions at the respective

locations 1 0x = and 2 0x = locations results in

( )1 10, 0 0w t B= → = (5.4.7)

( )2 20, 0 0w t B= → = (5.4.8)

At the discontinuity itself, the displacement of both parts of the string must be equal. In other words,

( ) ( )1 2/ 4, 3 / 4,w L t w L t= (5.4.9)

Using the assumed solutions from above, and expanding the time components of the separable solutions, the equality (5.4.9) reads

( ) ( )1 21 1 1 1 1 2 2 2 2 2

3sin sin cos sin sin cos4 4

L LA C t D t A C t D tc c

ω ωω ω ω ω+ = + (5.4.10)

The (5.4.10) must be satisfied for all time. This indicates that 1 2ω ω ω= = . Then, the equality (5.4.10)

reduces further to

1 23sin sin

4 4L LA Ac c

ω ω= (5.4.11)

where the coefficients 1A and 2A have absorbed the respective iC and iD coefficients. In addition to the

displacement continuity that must be maintained at the junction, the forces must balance from one side to the next with respect to the mass that causes the discontinuity. A force balance at the junction then results in

( ) ( ) ( )21 2 1

21 2

/ 4, 3 / 4, / 4,w L t w L t w L tT T M

x x t∂ ∂ ∂

− − =∂ ∂ ∂

(5.4.12)

The second term on the LHS side of (5.4.12) may appear to have an incorrect sign, but it must be recalled that the direction of 2x is reversed that of the direction of 1x . The signs of the equality (5.4.12) are correct.


90

Also, we will obtain the same final result if we elect for the RHS of (5.4.12) to include the term

( )2 22 3 / 4, /w L t t∂ ∂ instead of the current use in (5.4.12) with ( )2 2

1 / 4, /w L t t∂ ∂ .

Prior to substituting the assumed solution forms into (5.4.12), one is encouraged to express the findings to this point, via

( ) [ ]11 1 1, sin sinxw x t A t

cω ω φ= + (5.4.13)

( ) [ ]22 2 2, sin sinxw x t A t

cω ω φ= + (5.4.14)

where the sine and cosine terms are collected together for ease in the ensuing math. Then, substituting (5.4.13) and (5.4.14) into (5.4.12) leads to

21 2 1

3cos cos sin4 4 4

L L LT A T A MAc c c c cω ω ω ω ωω− − = − (5.4.15)

Introducing terms /L cβ ω= and /M M Lρ= , we express (5.4.11) and (5.4.15) as a system of equations

1

2

3sin sin 04 43 0sin cos cos

4 4 4

AAM

β β

β β ββ

− = − −

(5.4.16)

For non-trivial solutions of 1A and 2A to exist, the determinant of the coefficient matrix must vanish.

3 3sin cos sin cos sin 04 4 4 4 4

Mβ β β β ββ − + − = (5.4.17)

The (5.4.17) is in the form of a transcendental characteristic equation for the frequencies nω , via

/n nL cβ ω= that cause the LHS of (5.4.17) to vanish. Simplifying, we find that

3cot cot4 4

Mβ β β+ = (5.4.18)

The characteristic equation (5.4.18) is solved numerically with results shown in Figure 32. The results show that the lowest order natural frequencies are

[ ]2.414 4.803 8.641 13.34 17.20 ...ncL

ω = (5.4.19)


91

Figure 32. Solving the transcendental equation for the natural frequencies of the discontinuous string.


L=1; % [m] string length rho=1; % [kg/m] linear density T=10; % [N] tension force c=sqrt(T/rho); % [m/s] phase speed M=.5; % [kg] mass discontinuity Mtilde=M/rho/L; % [dim] normalized mass beta=linspace(0.01,20,301); % define beta parameter range x1=cot(beta/4)+cot(3*beta/4); % LHS of characteristic equation x2=Mtilde*beta; % RHS of characteristic equation func=@(v)cot(v/4)+cot(3*v/4)-Mtilde*v; func2=@(v)cot(v/4)+cot(3*v/4); numerical_solve=nan(length(beta),1); % preallocate for iii=1:1:length(beta) numerical_solve(iii)=fzero(func,beta(iii)); end unique_out=unique(numerical_solve); figure(1); clf; plot(beta,x1,'--r',beta,x2,'-b',unique_out,func2(unique_out),'ok'); axis([0 max(beta) -10 10]) xlabel('beta'); ylabel('characteristic equation parts'); title('red dashed, cos(beta/4)+cos(3*beta/4). blue solid, Mtilde*beta');

Consequently, using the first equation of (5.4.16), the relation between the displacement amplitudes of both sides of the string is

1

2

sin4

3sin4

AA

β

β=

(5.4.20)

This indicates that the string response for the thn mode is

0 2 4 6 8 10 12 14 16 18 20-10

-8

-6

-4

-2

0

2

4

6

8

10

beta

char

acte

ristic

equ

atio

n pa

rtsred dashed, cos(beta/4)+cos(3*beta/4). blue solid, Mtilde*beta


92

( ) [ ]11 1 1, sin sinnn n n n

xw x t A tc

ωω φ= + (5.4.21)

( ) [ ]22 2 1

sin4, sin sin3sin4

n

nn n n n

n

Lxcw x t A tL c

c

ωω

ω φω

= + (5.4.22)

while a summation over an infinite number of the normal modes yields the total string vibration. As before,

the coefficients 1nA and nφ are determined by applying the initial conditions.

Plots of the spatial components of the four lowest order normal modes are shown in Figure 33. The discontinuity at the quarter length of the string is evident. Also, a qualitative similarity among the three lowest order modes of the discontinuous string is observed when compared to the three lowest order modes

of the fixed-fixed string, given by ( )sin /n x Lπ . A key difference between the fixed-fixed string modes

and the discontinuous string modes is the fourth mode, which is antisymmetric for the fixed-fixed string and is quasi-symmetric for the discontinuous string. We also note from Figure 33 that the displacements of the two portions of the string are continuous at the mass discontinuity.


93

Figure 33. Representative lower order modes of the discontinuous string.


L=1; % [m] string length rho=1; % [kg/m] linear density T=10; % [N] tension force c=sqrt(T/rho); % [m/s] phase speed x1=linspace(0,L/4,81); % [m] string length from left x2=linspace(0,3*L/4,121); % [m] string length from right omega_n=c/L*[2.414 4.803 8.641 13.34 17.20]; % [rad/s] lowest order natural frequencies mode_number=4; % select a mode number to plot mode_left=sin(omega_n(mode_number)*x1/c); % [dim] modal displacement from left of string mode_right=sin(omega_n(mode_number)*L/4/c)/sin(3*omega_n(mode_number)*L/4/c)*sin(omega_n(mode_number)*x2/c); % [dim] modal displacement from right of string figure(1); clf; plot(x1,mode_left,'--r',L-x2,mode_right,'-b'); xlabel('length along string [m]'); ylabel('modal displacement [dim]'); title(['mode number: ' num2str(mode_number) ]);


94

Similar procedures are undertaken in the solution to the EOMs for discontinuous structures, whether by lumped mass, damping, or stiffness discontinuities or by discrete changes in the structure density or other properties.

In Sec. 4.5 we examined how to investigate the problem of the longitudinal vibrations of a bar with continuously varying cross-section. The analytical results became associated with the Bessel equation for a subset of cross-section variations realized by polynomials. The math of the derivation was shown to be complex. In contrast, the math of this Sec. 5.4 is relatively straightforward to handle. We solve for the vibration response of each discrete component of the continuous system from one discontinuity to the next, making sure to have the appropriate boundary conditions between segments. If multiple discontinuities exist, the math will not be as straightforward as the example given above. In fact, the bar with continuously varying cross-section may be considered to be composed of a large number of discrete bar segments each with a sudden change in the bar cross-section from one neighbor to the next. As a result, the theory developed in this Sec. 5.4 may also be used to analyze the vibrations of structures with continuously varying material properties and geometries, so long as a large enough number of discrete transitions among the properties/geometries is allotted for in the modeling.


95

6 Torsional vibrations of a bar or shaft

In many engineering applications, bars are subjected to torque so as to induce dynamic motions that rotate about the longitudinal axis of the bar. We refer to bar that rotate about their longitudinal axes as shafts. Shafts are common throughout engineering applications, especially in power transmission systems, electric motors or generators, and positioning equipment. It is common that the shafts may rotate at constant angular speeds and will be subjected to variations of torque from one portion of the shaft length to another. These torque differences will cause torsional vibrations, which are the focus of this Sec. 6.



Consider the differential shaft element in Figure 34(a). The shaft has solid or hollow circular cross-section. Considerations of other cross-section geometries changes the derivation slightly.

The element is subjected to a moment difference from one side of the differential shaft length dx to the

other in addition to a distributed moment per unit length ( ),tm x t . This causes a difference of rotation from

one end of the differential element to the other, ( ) ( ), ,x dx t x tθ θ+ − . We assume that the center of mass

and center of rotation of the shaft coincide and that the rotations involved are very small so as to warrant linearization of series expansions. With the free-body diagram in mind from Figure 34(b), we balance the moments about the differential element to find

( ) ( ) ( ) ( ) ( )2

2

, ,, , ,t t t c

M x t x tM x t M x t dx m x t dx dI

x tθ ∂ ∂

− + + + = ∂ ∂ (6.1.1.1)

In (6.1.1.1), the moment at x dx+ is expanded via the linearized Taylor's series: ( ),M x dx t+

( ) ( ), , /M x t M x t x dx≈ + ∂ ∂ . The mass moment of inertia cdI is computed from

2 2c A A

dI r dm dx r dA Jdxρ ρ= = =∫ ∫ ∫ ∫ (6.1.1.2)

where r is the polar coordinate a point in the shaft cross-section, ρ is the mass per unit volume, and J is

the polar moment of area defined by

2

AJ r dydx= ∫ ∫ (6.1.1.3)

The twisting moment ( ),tM x t is related to the angle of twist ( ),x tθ through the shear modulus G and

the polar moment of area.

( ) ( ) ( ),,t

x tM x t GJ x

xθ∂

=∂

(6.1.1.4)


96

In general, the twisting stiffness ( )GJ x may vary along the shaft length. Substituting (6.1.1.2) and (6.1.1.4)

into (6.1.1.1) leads to the governing equation for the torsional motions of the shaft

( ) ( ) ( ) ( )2

2

, ,,t

x t x tGJ x m x t J

x x tθ θ

ρ ∂ ∂∂

+ = ∂ ∂ ∂ (6.1.1.5)

When the properties and geometry of the shaft are constant along the length and when considering the free, transient vibration response, the governing equation reduces to

2 2 2 22

2 2 2 2 ; /G c c Gx t x tθ θ θ θ ρ

ρ∂ ∂ ∂ ∂

= → = =∂ ∂ ∂ ∂

(6.1.1.6)

It is apparent that the EOM for the torsional vibrations of the shaft is equivalent to the EOM for the longitudinal vibrations of the bar (4.1.11) and equivalent to the EOM for the transverse vibrations of the string (5.1.1.5). As a result, we may anticipate that the fundamental dynamic behavior of torsional shaft vibration is analogous to the transverse vibrations of strings and longitudinal vibrations of bars. Moreover, we may anticipate that the mathematical techniques used to analyze the string and bar deflections are similarly relevant to investigate the torsional deformations of the shaft.

Figure 34. (a) Schematic of shaft element with torque variation from one end to the other. (b) Free-body diagram of the shaft differential element.

The boundary conditions of a shaft with a fixed end at 0x = and x L= are

( ) ( )0, , 0t L tθ θ= = (6.1.1.7)

If the shaft is loaded by an end moment of inertia dI at 0x = or at x L= , then the boundary conditions

read

2

20 0

dx x

GJ Ix tθ θ

= =

∂ ∂=

∂ ∂ (6.1.1.8)


97

2

2dx L x L

GJ Ix tθ θ

= =

∂ ∂= −

∂ ∂ (6.1.1.9)

Torsional springs with torsional stiffness tk that support the ends of a shaft at 0x = and x L= result in

boundary conditions of the form

( )0

0,tx

GJ k txθ θ

=

∂=

∂ (6.1.1.10)

( ),tx L

GJ k L txθ θ

=

∂= −

∂ (6.1.1.11)


The total kinetic energy of the shaft is computed from

2

0

12

LT J dx

tθρ ∂ = ∂ ∫ (6.1.2.1)

The total strain energy of the shaft that is supported at each end by torsional springs of stiffness tk is

( ) ( )2

2 2

0

1 1 10, ,2 2 2

L

t tU GJ dx k t k L txθ θ θ∂ = + + ∂ ∫ (6.1.2.2)

The potential of the applied moment ( ),tm x t per unit length is

( ) ( )0

, ,L

tW m x t x t dxθ= ∫ (6.1.2.3)

Applying Hamilton's principle, the first variation of the Lagrangian L T U W= − + is

2

1

0t

t

I

I Ldt

δ

δ δ

=

= ∫ (6.1.2.4)

( ) ( )2

10 00

t L

t t t L LtI J GJ m dx k k dt

t t x xθ θδ ρ δθ δθ δθ θ δθ θ δθ

∂ ∂ ∂ ∂ = − + − − ∂ ∂ ∂ ∂ ∫ ∫ (6.1.2.5)

Applying integration by parts, we find

2

1

2

0 0200

0L

t L

t t t L LtI J GJ m dx GJ k k dt

t x x xθ θ θδ ρ δθ δθ θ δθ θ δθ

δ

∂ ∂ ∂ ∂ = − + + − − − = ∂ ∂ ∂ ∫ ∫ (6.1.2.6)

By the same principles as utilized in early sections, we obtain from (6.1.2.6) the EOM for the torsional vibration of the shaft along with the natural boundary condition expressions at 0x = and x L= .


98

2

2tGJ m Jx x t

θ θρ∂ ∂ ∂ + = ∂ ∂ ∂ (6.1.2.7)

00

t xx

GJ kxθ θ

==

∂=

∂ (6.1.2.8)

t x Lx L

GJ kxθ θ

==

∂= −

∂ (6.1.2.9)

Of course, (6.1.2.7) is the same as (6.1.1.5) that was derived by the equilibrium approach. The boundary conditions associated with the torsional springs (6.1.2.8) and (6.1.2.9) are also the same as those identified by the force balance at the ends (6.1.1.10) and (6.1.1.11). As a result, we again see the equivalence of deriving the EOMs and determining the boundary conditions for continuous structures whether by utilizing the equilibrium or variational approach.


We investigate the free vibration response of the shaft by examples. For shafts, it is common to have inertial elements near terminating ends of a shaft length. Such inertial elements may be wheels, gears, or other mounted components that the shaft rotates for the component use. These inertial elements resist the torque provided by the remainder of the shaft and must be accelerated up to and around angular speed. Thus, the following example considers this realistic case of shaft vibration when the shaft is loaded by end-mounted disks.

Example Consider the shaft shown in Figure 35. End-mounted disks with inertia 1I and 2I are affixed to the ends of

the shaft. The shaft rotates with twist angle ( ),x tθ . The shaft density ρ , shear modulus G , and polar area

moment of inertia J are constant along the shaft length. The shaft is given an initial twist ( ) ( )0,0x xθ θ=

and initial twist rate ( ) ( )0,0x xθ θ= . Find the free response of the shaft for all time t .

Figure 35. Schematic of shaft with inertial disks at both ends.

Answer We assume that the linear shaft vibration is separable into components dependent only on space or time.


99

( ) ( ) ( ),x t X x T tθ =

Substituting this assumed solution into the governing equation (6.1.1.6), we find that combinations of sinusoids in space and time must satisfy the resulting two governing equations in space and time so that


= +


( ) ( ), cos sin cos sinx xx t A B C t D tc cω ωθ ω ω = + +

The boundary conditions for the shaft are

2

1 20 0x x

GJ Ix tθ θ

= =

∂ ∂=

∂ ∂

2

2 20 0x x

GJ Ix tθ θ

= =

∂ ∂= −

∂ ∂

Substitution of the total assumed separable solution into the boundary conditions, we find that the following expressions must hold for all time.

21GJ B I A

cω ω= −

22sin cos cos sinL L L LGJ A B I A B

c c c c cω ω ω ω ωω − + = +

Expressing the two equations in a matrix form, we have

21

2 22 2

00cos sin sin cos

I GJ AcL L L L BI GJ I GJ

c c c c c c

ωω

ω ω ω ω ω ωω ω

= + −

The determinant of the coefficient matrix must vanish for non-trivial A and B to be found. The determinant is

2 24 3 2

1 2 1 2 2sin cos cos sin 0L GJ L GJ L G J LI I I Ic c c c c c cω ω ω ωω ω ω − + − =

A transcendental equation may be obtained by rearranging the characteristic equation above for the thn frequency nω that satisfies the equation.


100

2

1 2 1 2

1 11 tannn n

αα α

β β β β

− = +

with

nn

Lc

ωα = , 1

1

JLIρβ = , and 2

2

JLIρβ =

The mode shape amplitude coefficients are related through

1

nn nB A α

β= −

As a result, the mode shapes are

( )1

cos sinn n nn n

x xX x AL Lα α α

β

= −

Then, the torsional vibration of the shaft is subsequently constructed from a weighted combination of the normal modes.

( )1 1 1

, cos sin cos cos sin sinn n n n n nn n n n

n

x x x xx t C t D tL L L Lα α α α α α

θ ω ωβ β

∞

=

= − + −

∑

The coefficients nC and nD have absorbed the coefficients nA . The free torsional vibration of the shaft is

therefore given above.

Applying the initial conditions and taking advantage of the orthogonality of the normal modes‡, we find the relation between the amplitude coefficients and initial conditions.

( )001

2 cos sinL n n n

nx xC x dx

L L Lα α α

θβ

= −

∫

( )001

2 cos sinL n n n

nn

x xD x dxL L L

α α αθ

ω β

= −

∫

6.3 Forced response

In many engineering applications, shafts are driven by torques (moments) near the shaft ends. We examine a case of forced response of the shaft vibration in this spirit. Due to the equivalence of the EOM that governs

‡ Although not immediately obvious, one may check that the modes are orthogonal. For instance, numerically integrating a product of any modes with orders i and j such that i j≠ shows that the integral returns a null.


101

the angular vibrations of the shaft to the EOM of transverse vibration of the string and to the EOM of longitudinal deformation of the bar, we leverage the similar mathematical analysis techniques for the forced response prediction of the shaft vibration.

Example Consider the free-free shaft shown in Figure 36. The shaft density ρ , shear modulus G , and polar area

moment of inertia J are constant along the shaft length in the length coordinate x . A time-varying moment

is applied to the shaft ( ),tm x t that is concentrated at the end of the shaft x L= . The moment is time-

harmonic with angular frequency of oscillation ω and amplitude 0M .

( )0

0 0, ;

cost

x Lm x t

M t x Lω< <

= =

Determine the steady-state angular vibration response ( ),x tθ of the shaft.

Figure 36. Schematic of free-free shaft with end-acting moment.

Answer As undertaken in other examples, we assume that the forced response of the shaft is composed from a linear combination of the normal modes. To determine the normal modes, we must solve the homogeneous form of the EOM for the shaft vibrations.

2 22

2 2cx tθ θ∂ ∂=

∂ ∂

where the constant /c G ρ= is the phase speed. The boundary conditions for the free-free shaft are

0

0x x Lx x

θ θ

= =

∂ ∂= =

∂ ∂

Assuming a separable form of solution to the EOM via

( ) ( ) ( ),x t X x T tθ =

by substitution of the assumed solution into the EOM, we find that the spatial and temporal components must each be a combination of sinusoids.


102


= +


( ) ( ), cos sin cos sinx xx t A B C t D tc cω ωθ ω ω = + +

where ω is currently an unknown constant. The spatial component of the separable solution ( )X x must

satisfy the boundary conditions so that

0

0x x L

dX dXdx dx= =

= =

Substituting in the separable spatial component ( )X x of the solution, we find

0

0 00 sin cos 0x

dX A B Bdx c c c c

ω ω ω ω

=

= = − + → =

0 sin ; 1,2,3,...nx L

dX L n cA ndx c c L

ω ω πω=

= = − → = =

As a result, the thn normal mode is expressed by

( ) cos cos ; 1,2,3,...nn n n

x n xX x A A nc L

ω π= = =

The six lowest order modes of the free-free shaft angular vibration are shown in Figure 37. The natural frequencies associated with these modes are computed from /n n c Lω π= .

It is interesting to observe that there is no distinction between the natural frequencies of the longitudinal deformations of a free-free bar, the natural frequencies of transverse vibration of the free-free string, and those of the free-free shaft torsional vibration. Indeed, only the system phase speed (and thus material properties via its definition) and the system length play roles in determining the natural frequencies.

Similar simplifying statements may be made about fixed-fixed strings, bars, and shafts.

The similarity may be associated with the equivalence of the equations of motion. It suggests that a different form of EOM is required in order to have normal modes and natural frequencies that differ despite the presence of the same boundary conditions. Indeed, the beam vibrations studied in Sec. 7 are governed by a different form of EOM than the vibrations of the bar, string, and shaft. Thus, we will later find that the free-free beam normal modes and natural frequencies are not the same as the normal modes and natural frequencies for free-free bars, strings, and shafts.


103

Figure 37. Modes of torsional vibration for free-free shaft.

Having found the normal modes, we are in a position to return to the forced vibration problem represented by the inhomogeneous EOM.

( )2 2

2 2,tGJ m x t Jx tθ θρ∂ ∂+ =

∂ ∂

In the following undertaking, it is convenient to presume a general distributed torque ( ),tm x t . We then

use our knowledge of the normal modes to construct a complex exponential form of the assumed solution

( )1

, cos j tn

n

n xx t eL

ωπ∞

=

=∑θ A

where nA are in general complex coefficients to be determined. Similar to employing the technique to

study the forced vibration response of the bar or string, this is an application of the assumed modes method to study the rotary vibrations of the shaft.

By adopting the complex exponential form of response to investigate the time-harmonic behavior, we similarly presume

( ) ( ) 0

0 0, ;t j t j t

t

x Lx t

M x e M e x Lω ω

< <= = =

m

Substituting these forms of excitation and assumed torsional vibration into the inhomogeneous EOM, we have /n n c Lω π=

( )2

2

1cos j t j t

n tn

n n xJ GJ e M x eL L

ω ωπ πρ ω∞

=

− + =

∑ A

Given the definition of the natural frequencies and constant c , we may rearrange the equations above to find

0 0.2 0.4 0.6 0.8 1

-1

-0.5

0

0.5

1

normalized length along shaft [dim]

mod

al ro

tatio

n di

strib

utio

n [d

im]

red: mode 1. green: mode 2. blue: mode 3. cyan: mode 4. magenta: mode 5. black: mode 6.


104

( )

( )

22

1

2 2

1

1cos

1cos

j t j tn t

n

j t j tn n t

n

n c n x e M x eL L J

n x e M x eL J

ω ω

ω ω

π πωρ

πω ωρ

∞

=

∞

=

− + =

− + =

∑

∑

A

A

Multiplying the latter equation by another normal mode cos /m x Lπ and integrating over the shaft length

identifies the relation between the unknown complex coefficients nA and the distributed moment ( )tM x .

( ) ( )2 2 0

2 cosL

n tn

n xM x dxLJLπ

ρ ω ω=

− ∫A

Given the point-based application of the driving moment, the integral simplifies to be

( )0

2 2

2 cosn

n

M nJL

πρ ω ω

=−

A

In this way, the complex forced response of the torsional vibration of the shaft with applied end torque is

( ) ( )0

2 21

2 cos, cos j t

n n

M n n xx t eLJL

ωπ πρ ω ω

∞

=

=−

∑θ

The real part of ( ),x tθ is the actual angular displacement of the shaft

( ) ( )0

2 21

2 cos, cos cosn n

M n n xx t tLJL

π πθ ωρ ω ω

∞

=

=−

∑

Using 12 modes in the expansion, Figure 38 shows representative results of the amplitude of torsional vibration for the shaft at a location /x L π= along the shaft length. At any given frequency ω of the applied moment, an increase in the number of modes in the expansion will increase the accuracy of the predicted result.


105

Figure 38. Representative results of torsional oscillation at location along shaft.


M0=1; % [N.m] end-acting torque or moment L=1; % [m] length of shaft rho=1; % [kg/m^3] density of shaft material J=1; % [m^4] polar moment of area G=1; % [Pa] shear modulus c=sqrt(G/rho); % [m/s] phase speed x0=L/pi; % [m] evaluation point along shaft for torsional vibration n=12; % [m] number of modes to consider in superposition omega=linspace(0.1,30,601); % [rad/s] frequency range of acting moment theta=zeros(size(omega)); % preallocate for ooo=1:length(omega) for nnn=1:n theta(ooo)=theta(ooo)+2*M0*cos(nnn*pi)/rho/J/L/((nnn*pi*c/L)^2-omega(ooo)^2)*cos(nnn*pi*x0/L); end end figure(1); clf; semilogy(omega/2/pi,abs(theta),'-r'); axis([0 4 1e-4 1e2]); box on xlabel('frequency [Hz]'); ylabel('amplitude of torsional rotation [deg]'); title(['shaft torsional vibration amplitude evaluated at ' num2str(x0/L) ' along shaft length'],['number of modes in expansion: ' num2str(n) '']);

0 0.5 1 1.5 2 2.5 3 3.5 410

-4

10-3

10-2

10-1

100

101

102

frequency [Hz]

ampl

itude

of t

orsi

onal

rota

tion

[deg

]shaft torsional vibration amplitude evaluated at 0.31831 along shaft length

number of modes in expansion: 12


106

6.4 Forced torsional vibration of a damped shaft

Persistent free vibration resulting from initial conditions is rare in practice. In addition, as seen by the final torsional vibration expression from the Example above, the amplitude of the vibration grows without bound when the shaft is driven at resonance, nω ω= . In practice, such unbounded growth of vibration does not

occur. Damping phenomena suppress oscillations from either elapsing for all time (case of free vibration) or from becoming unbounded in amplitude (case of forced vibration).

Thus, almost all practical continuous structures are acted upon by various damping mechanisms that suppress vibration. The damping mechanisms are numerous and include external and internal phenomena. Often, the ways by which such damping phenomena act to dissipate the structural vibrations are known only empirically. Few theoretical models of energy dissipation are available since the origins of energy dissipation are often at length scales many orders of magnitude smaller than the structures under consideration (e.g. Joule heating of neighboring material particles rubbing together, opening and closing of microcracks, and deformation of nanoscale defects or voids in a material).

In the following two examples, we consider two common forms of damping that act on continuous structures: viscous damping and hysteretic damping.

Example Consider the free-free shaft shown in Figure 39. This is the same situation as seen in Figure 36 but now the shaft is immersed in an oil bath. As a result, viscous forces will work to suppress the rotation of the shaft as a function of the local shaft angular velocity. A constant dc is identified with units [N.s] that results in a

moment per unit length that resists the rotation of the shaft

( ) ( ),,v d

x tm x t c

tθ∂

=∂

As a result, the EOM for the harmonically forced shaft vibration is

( )2 2

2 2,t dGJ m x t J cx t tθ θ θρ∂ ∂ ∂+ = +

∂ ∂ ∂

Repeat the Example of Sec. 6.3 to determine the forced, damped response of the shaft shown in Figure 39.


107

Figure 39. Forced shaft immersed in an oil bath that exerts viscous damping forces against the rotational velocity of the shaft.

Answer We have already identified the normal modes and natural frequencies of the free-free shaft of Figure 39.

( ) cos cos ; 1,2,3,...nn n n

x n xX x A A nc L

ω π= = = with /n n c Lω π=

Then, assuming a complex exponential form of torsional vibration response

( )1

, cos j tn

n

n xx t eL

ωπ∞

=

=∑θ A

we substitute this torsional response into the EOM and simplify.

( )2

2

1cos j t j t

d n tn

n n xJ j c GJ e M x eL L

ω ωπ πρ ω ω∞

=

− + + =

∑ A

( )2 2

1

1cos j t j tdn n t

n

c n xj e M x eJ L J

ω ωπω ω ωρ ρ

∞

=

− + + =

∑ A

( )2 2

1

12 cos2

j t j tdn n n t

n n

c n xj e M x eJ L J

ω ωπω ωω ωω ρ ρ

∞

=

− + + =

∑ A

We then introduce , 2n cr nc Jω ρ= to be the critical damping factor for the thn normal mode. As a result,

we define the thn mode damping ratio as ,/ / 2n d n cr d nc c c Jζ ω ρ= = .

( )2 2

1

12 cos j t j tn n n n t

n

n xj e M x eL J

ω ωπω ωω ζ ωρ

∞

=

− + + = ∑ A


108

The remainder of the solution routine for this case of damped vibrations of the shaft is analogous to the Example of Sec. 6.3. We multiply both sides of the equation by an arbitrary normal mode and integrate over the shaft length. After simplification we find

( ) ( )2 2 0

2 cos2

L

n tn n n

n xM x dxLJL jπ

ρ ω ω ωω ζ=

− + ∫A

Given the end-point application of the moment, the torsional vibration of the shaft is ultimately found to be

( ) ( ) 02 2

1

2 cosRe , , Re cos2

j t

n n n n

M n n xx t x t eJL j L

ωπ πθρ ω ω ωω ζ

∞

=

= = − +

∑θ

This amplitude of torsional shaft vibration is plotted in Figure 40 considering the location /x L π= . The

increase in the torsional damping constant dc results in a reduction of the vibration amplitudes. The modal

damping ratios in the case of dc =0.1 [N.s], are [ ]0.0156 0.00778 0.00518 ...nζ = . Thus, viscous

damping that uniformly influences the continuous structure vibration does not uniformly influence the different normal modes.

Figure 40. Amplitude of torsional vibration at /x L π= for forced, viscously damped shaft.


M0=1; % [N.m] end-acting torque or moment

0 0.5 1 1.5 2 2.5 310

-4

10-3

10-2

10-1

100

101

102

frequency [Hz]

ampl

itude

of t

orsi

onal

rota

tion

[deg

]

shaft torsional vibration amplitude evaluated at 0.31831 along shaft lengthnumber of modes in expansion: 12

red, cd=0.001. green, cd=0.01. blue, cd=0.1. cyan, cd=1.0.


109

cd=0.001; % [N.s] damping constant L=1; % [m] length of shaft rho=1; % [kg/m^3] density of shaft material J=1; % [m^4] polar moment of area G=pi/3; % [Pa] shear modulus c=sqrt(G/rho); % [m/s] phase speed x0=L/pi; % [m] evaluation point along shaft for torsional vibration n=12; % [m] number of modes to consider in superposition omega=linspace(0.1,20,1601); % [rad/s] frequency range of acting moment theta=zeros(size(omega)); % preallocate for ooo=1:length(omega) for nnn=1:n theta(ooo)=theta(ooo)+2*M0*cos(nnn*pi)/rho/J/L/((nnn*pi*c/L)^2-omega(ooo)^2+j*2*omega(ooo)*(nnn*pi*c/L)*(cd/2/(nnn*pi*c/L)/rho/J))*cos(nnn*pi*x0/L); end end figure(1); clf; hold on semilogy(omega/2/pi,abs(theta),'-r'); axis([0 3 1e-4 1e2]); box on xlabel('frequency [Hz]'); ylabel('amplitude of torsional rotation [deg]'); title(['shaft torsional vibration amplitude evaluated at ' num2str(x0/L) ' along shaft length'],['number of modes in expansion: ' num2str(n) '']);

Example A common internal structural damping phenomena is hysteretic damping, also termed structural or material damping. This damping phenomenon is associated with inherent loss mechanisms manifest in delays between stress and strain states of material deformation [12]. Hysteretic damping is typically characterized using storage and loss moduli, where the storage modulus is the conventional shear (or Young's) modulus while the loss modulus is a considered to be a fraction of the shear (or Young's) modulus. Often, the total modulus is denoted via

*G jG+

where the loss component features the asterisk superscript and the imaginary number is 1j = − . The

alternative means to characterize a loss modulus is in terms of a loss factor η . The loss factor§ expresses a

proportion of the storage modulus as the lossy component via

( )1G jη+

Thus, * /G Gη = . The imaginary component of the total modulus delays states of stress or force to the states

of strain or displacement that work on the structure

Use a hysteretic damping model to determine the forced vibration of the free-free shaft shown in Figure 41.

§ The hysteretic loss factor η should not be confused with the loss factor 2η ζ= that is used in mechanical vibrations to denote a quantity that is twice the damping ratio ζ . These two loss factor representations have no comparability.


110

Figure 41. Forced shaft with hysteretic damping due to a loss shear modulus *G .

Answer In fact, we have already solved this problem via the Example of Sec. 6.3! The result was

( ) ( )0

2 21

2 cos, cos cosn n

M n n xx t tLJL

π πθ ωρ ω ω

∞

=

=−

∑

Considering this solution as a whole, the interpretation of the natural frequencies nω requires modification

in order to account for hysteretic damping. In other words,

/n n c Lω π=

where now we have that the phase speed is ( )1 /c G jη ρ= + .

Figure 42 shows representative results, using values taken from the prior Example for comparison to the case of viscous damping. Contrasting viscous damping, the influence of hysteretic damping by the loss factor η is such that higher frequency vibrations are more substantially attenuated. This is indeed a

common reality in experimentation: while lower order modes may be damped, it is often found that the higher order modes are much more damped than would be explained by a viscous damping phenomena. Respectively, even though viscous damping indeed influences each mode independently as was observed in the prior Example, the hysteretic damping exerts such mode dependent damping via a more dramatic frequency dependent influence on the vibration amplitude.


111

Figure 42. Amplitude of torsional vibration at /x L π= for forced, hysteretically damped shaft.


M0=1; % [N.m] end-acting torque or moment L=1; % [m] length of shaft rho=1; % [kg/m^3] density of shaft material J=1; % [m^4] polar moment of area G=pi/3; % [Pa] shear modulus eta=.001; % [dim] loss factor c=sqrt(G*(1+j*eta)/rho); % [m/s] phase speed x0=L/pi; % [m] evaluation point along shaft for torsional vibration n=12; % [m] number of modes to consider in superposition omega=linspace(0.1,20,1601); % [rad/s] frequency range of acting moment theta=zeros(size(omega)); % preallocate for ooo=1:length(omega) for nnn=1:n theta(ooo)=theta(ooo)+2*M0*cos(nnn*pi)/rho/J/L/((nnn*pi*c/L)^2-omega(ooo)^2)*cos(nnn*pi*x0/L); end end figure(1); clf; % hold on semilogy(omega/2/pi,abs(theta),'-b'); axis([0 3 1e-4 1e2]); box on xlabel('frequency [Hz]'); ylabel('amplitude of torsional rotation [deg]'); title(['shaft torsional vibration amplitude evaluated at ' num2str(x0/L) ' along shaft length'],['number of modes in expansion: ' num2str(n) '']);

A side-by-side comparison of the forced, damped shaft torsional vibration results is given in Figure 43.

0 0.5 1 1.5 2 2.5 310

-4

10-3

10-2

10-1

100

101

102

frequency [Hz]

ampl

itude

of t

orsi

onal

rota

tion

[deg

]shaft torsional vibration amplitude evaluated at 0.31831 along shaft length

number of modes in expansion: 12red, η=0.001. green, η=0.01. blue, η=0.1


112

Figure 43. Comparison of viscously and hysteretically damped forced shaft vibration response. Generated using respective codes of Table 9 and Table 10.


113

7 Transverse vibrations of a beam

Beams are ubiquitous members of engineering structures. Beams are often joined and connected together to make up structural systems. Example include machinery and vehicle frames, building frames, bridge decks, robotic manipulators, furniture, gym equipment, and the list goes on. This Sec. 7 introduces the methods to examine the free and forced vibration response of beams.

Because the transverse deformation of beams is due to bending and shear actions, the dynamic behavior of beams is more intricate than that of bars, strings, or shafts. Thus, although the mathematics becomes more involved to analyze the behavior of beams, the theoretical foundations that we have laid down in earlier sections of this course will be utilized for their broad strengths to enable us to study beam vibration.



Consider the beam shown in Figure 44(a). In general, the Young's modulus E , shear modulus G , cross-sectional area A , and volumetric density ρ may vary along the beam length L . The transverse vibration

of the beam ( ),w x t from an equilibrium is caused by an applied force per unit length ( ),f x t , by reactions

at the beam ends from masses, torsional springs, and translational springs, and by initial conditions. The end masses at the beam ends 0x = and x L= are respectively 1m and 2m . Similarly, the translational

springs at the beam ends are 1k and 2k while the torsional springs are 1tk and 2tk .

Figure 44. Transverse vibrations of a beam with ends constrained by translational and torsional springs and loaded with end masses. (a) Full schematic and (b) free-body diagram of beam differential element.

A free-body diagram of a beam differential element is shown in Figure 44(b). First, we assume that the centerline OO′ is such that the undeformed beam cross-sections that are originally normal to the centerline remain normal after deformation. This is an assumption of negligible shearing stress through the beam cross-section. The centerline OO′ is deformed to a length ds when compared to the original centerline length dx . By linearized Taylor's series expansion, the moment M and shear V on the RHS face of the beam differential element are


114

( ) ( ) ( ),, ,

M x tM x dx t M x t ds

s∂

+ = +∂

(7.1.1.1)

( ) ( ) ( ),, ,

V x tV x dx t V x t ds

s∂

+ = +∂

(7.1.1.2)

For small slopes of the deformation due to the transverse vibration, the differential element deformation is approximately assessed according to / /s x∂ ∂ ≈ ∂ ∂ . As a result, a balance of vertical forces over the differential element leads to

( )2

2,V wV V ds f x t ds Adsx t

ρ∂ ∂ − + + = ∂ ∂ (7.1.1.3)

Simplifying, (7.1.1.3) becomes

( )2

2,V wf x t Ax t

ρ∂ ∂− + =∂ ∂

(7.1.1.4)

Similarly, a moment balance around the center of the differential element leads to

02 2 2 2

M ds V ds ds V dsM M ds V Vs s s

∂ ∂ ∂ − + + − − − + = ∂ ∂ ∂ (7.1.1.5)

The RHS of (7.1.1.5) omits the rotational inertia that will be included in Sec. 7.1.1.1. Simplifying (7.1.1.5), we have

M MVs x

∂ ∂= ≈

∂ ∂ (7.1.1.6)

Textbooks in mechanics of materials show that the moment in a beam is related to the beam curvature κ . For small slopes, this is an expression of

2

2

wM EI EIx

κ ∂= ≈

∂ (7.1.1.7)

where I is the second moment of cross-sectional area with respect to the neutral axis or centerline. Then, by substitution of (7.1.1.7) into (7.1.1.6) and then substitution into (7.1.1.4), we have the EOM for the transverse vibration of the beam

( )2 2 2

2 2 2 ,w wEI A f x tx x t

ρ ∂ ∂ ∂

+ = ∂ ∂ ∂ (7.1.1.8)

The (7.1.1.8) is the EOM for the beam vibration according to Euler-Bernoulli beam theory, which adopts the assumptions of negligible rotary inertia and shear stress. The (7.1.1.8) is applicable in the general case

for ( )E x , ( )I x , ( )A x , and ( )xρ that are functions of the spatial position x along the beam.


115

When the beam material and geometric properties do not vary along the beam length, the Euler-Bernoulli beam equation becomes

( )4 2

4 2 ,w wEI A f x tx t

ρ∂ ∂+ =

∂ ∂ (7.1.1.9)

The EOM is fourth order in space, so that four boundary conditions are required. At each end, the boundary conditions may be specified according to Euler-Bernoulli beam theory assumptions by

the transverse displacement w (7.1.1.10)

the displacement slope wx

∂∂

(7.1.1.11)

the moment 2

2

wEIx

∂∂

(7.1.1.12)

the shear force 2

2

wEIx x ∂ ∂ ∂ ∂

(7.1.1.13)

The associations of common beam boundary conditions respecting the expressions of (7.1.1.10)-(7.1.1.13) are as follows:

fixed (also "clamped"): 0w = and 0wx

∂=

∂ (7.1.1.14)

simply supported (also "pinned"): 0w = and 2

2 0wM EIx

∂= =

∂ (7.1.1.15)

free: 2

2 0wM EIx

∂= =

∂ and

2

2 0wV EIx x ∂ ∂

= = ∂ ∂ (7.1.1.16)

In fact, arbitrary combinations of these conditions are not possible, as will be shown in the derivation of the EOM by the variational approach in Sec. 7.1.2. In the case of the specific boundaries shown in Figure 44, we must balance the moment and shear force to the respective quantities induced by the end springs and masses. These expressions then read

2

1200

txx

w wEI kx x ==

∂ ∂=

∂ ∂ (7.1.1.17)

( )2 2

1 12 20 0

0,x x

w wEI k w t mx x t

= =

∂ ∂ ∂= − −

∂ ∂ ∂ (7.1.1.18)

2

22 tx Lx L

w wEI kx x ==

∂ ∂= −

∂ ∂ (7.1.1.19)


116

( )2 2

2 22 2,x L x L

w wEI k w L t mx x t

= =

∂ ∂ ∂= +

∂ ∂ ∂ (7.1.1.20)

The boundary conditions (7.1.1.17)-(7.1.1.20) are derived from force and moment balances based on an assumption of positive /w x∂ ∂ at both boundaries.

7.1.1.1 Accounting for axial force, rotary inertia, and shear flexibility

Consider that an axial force ( )P x is applied along the length of the beam centerline, Figure 45. The force

( )P x is considered positive as applied in the positive x direction of the beam axis. Mathematically, no

constraints are required to realize such a force. Yet, in practice, at least one end of the beam must have zero displacement 0w = while the other end must be allowed to axially move, as shown in Figure 45.

From an undeformed to a deformed state, the axial force follows the centerline of the beam. As a result, the moment balance in (7.1.1.5) is unaffected. On the other hand, the vertical force balance (7.1.1.4) is adjusted to be

( ) ( )2

2,w V wP x f x t Ax x x t

ρ∂ ∂ ∂ ∂ − + = ∂ ∂ ∂ ∂ (7.1.1.1.1)

Following through on the similar substitution steps of the derivation in Sec. 7.1.1, the EOM becomes

( ) ( )2 2 2

2 2 2 ,w w wEI A f x t P xx x t x x

ρ ∂ ∂ ∂ ∂ ∂ + = + ∂ ∂ ∂ ∂ ∂

(7.1.1.1.2)

Figure 45. Axial force applied along beam centerline.

When the beam material and geometric properties and the axial load are unchanging along the beam length, the EOM becomes

( )4 2 2

4 2 2 ,w w wEI P A f x tx x t

ρ∂ ∂ ∂− + =

∂ ∂ ∂ (7.1.1.1.3)

The (7.1.1.1.2) and (7.1.1.1.3) are applicable whether the applied load exerts tensile forces (positive values of P ) or compressive forces (negative values of P ). This will be exemplified in an Example of Sec. 7.2.


117

Shear flexibility and rotary inertia are introduced together due to a close correlation of their working mechanisms that are associated with the rotation of beam cross-sections during transverse vibration. To account for rotary inertia, the (7.1.1.5) is modified to be

2

2

MM M ds Vds Idss t

φρ∂ ∂ − + + − = ∂ ∂ (7.1.1.1.4)

where ( ),x tφ is the rotational angle of the cross-section and I is the area moment of inertia. Simplifying,

the (7.1.1.1.4) becomes

2

2

M V Ix t

φρ∂ ∂− =

∂ ∂ (7.1.1.1.5)

To include shear deformation effects, we consider the example of a cantilever beam subjected to an upward static force 0p at the free end, Figure 46. A uniform shearing force 0V p= − is thus induced along the

length of the beam. This results in a uniform shear strain / Lγ δ= where δ is the vertical deflection of the

cantilever beam free tip.

Figure 46. Cantilevered beam under uniform shear stress.

The shear force is the integral of the shear stress over the beam cross-section area

( )xyAV dA kG Aτ γ= = −∫ (7.1.1.1.6)

where kGγ− is the average shear stress. The k is a shear correction factor oftentimes tabulated in

textbooks [10] that accounts for the cross-section shape dependence on the inducement of shear stress.

By considering the rotation of the cross-section of the beam due to bending and shear, the total slope of the neutral axis is

wx

φ γ∂= +

∂ (7.1.1.1.7)

The moment-curvature relation is known from mechanics of materials texts as

M EIxφ∂

=∂

(7.1.1.1.8)

so that the shear force is


118

wV kAGx

φ∂ = − − ∂ (7.1.1.1.9)

Substitution of (7.1.1.1.9) into (7.1.1.4) results in

( )2 2

2 2,w wkAG f x t Ax x t

φ ρ ∂ ∂ ∂

− + = ∂ ∂ ∂ (7.1.1.1.10)

Then, substituting (7.1.1.1.8) and (7.1.1.1.9) into (7.1.1.1.5) gives

2

2

wEI kAG Ix x x t

φ φφ ρ∂ ∂ ∂ ∂ + − = ∂ ∂ ∂ ∂ (7.1.1.1.11)

Together, (7.1.1.1.10) and (7.1.1.1.11) are the EOMs for the transverse deflection ( ),w x t and cross-section

rotation ( ),x tφ of the beam that accounts for both rotary inertia and shear flexibility. This derivation is

associated with Stephen Timoshenko (1878-1972), a pioneer in mechanics and structural sciences. The equations (7.1.1.1.10) and (7.1.1.1.11) are referred to as Timoshenko beam theory, which introduce shear and rotary inertia corrections to Euler-Bernoulli beam theory (7.1.1.8).

When the beam material and geometric properties are uniform, the equations (7.1.1.1.10) and (7.1.1.1.11) may be simplified into one equation through substitutions. One finds that the EOM for the beam transverse

vibration ( ),w x t that inherently accounts for bending, shear, and rotary inertia is

( )4 2 4 2 4 2 2

4 2 2 2 4 2 2,w w EI w I w I f EI fEI A I f x tx t kG x t kG t kAG t kAG x

ρ ρ ρρ ρ∂ ∂ ∂ ∂ ∂ ∂ + − + + = + − ∂ ∂ ∂ ∂ ∂ ∂ ∂ (7.1.1.1.12)

By the substitution procedure, a fourth-order time derivative appears in (7.1.1.1.12) which is not intuitive in interpretation.

The (7.1.1.1.12) may be simplified. For instance, to neglect rotary inertia, one sets Iρ =0. Omitting shear

flexibility involves G →∞ . Of course, applying both such simplifications turns (7.1.1.1.12) into (7.1.1.9) (in the absence of applied load).

The boundary conditions in Timoshenko beam theory include

fixed or clamped: 0w = and 0φ = (7.1.1.1.13)

simply supported or pinned: 0w = and 0M EIxφ∂

= =∂

(7.1.1.1.14)

free: 0M EIxφ∂

= =∂

and 0wV kAGx

φ∂ = − = ∂ (7.1.1.1.15)


119


To utilize the variational approach to derive the governing equations of motion for the transverse vibrations of the beam, we identify the total kinetic and strain energies of the beam along with the potential of the applied loads.

Table 2 gives the strain and kinetic energies associated with beam deformation. The total kinetic energies associated with beam bending and shear deformation are given by (7.1.2.1) and (7.1.2.2), respectively.

2

0

12

L

bwT A dxt

ρ ∂ = ∂ ∫ (7.1.2.1)

2

0

12

L

sT I dxtφρ ∂ = ∂ ∫ (7.1.2.2)

The total strain energy associated with beam bending is

2

0

12

L

bU EI dxxφ∂ = ∂ ∫ (7.1.2.3)

The more general expression for the bending rotation φ is given in (7.1.2.3) in contrast to the use of the

simplification wx

φ ∂≈∂

that presumes the cross-sections originally normal to the beam axis remain normal

during deformation.

The total strain energy associated with the shearing deformation is

2

02L

swU kAG dxx

φ1 ∂ = − ∂ ∫ (7.1.2.4)

Finally, the potential of the transverse applied loads and axial applied loads are given in (7.1.2.5) and (7.1.2.6), respectively.

( ) ( )0

, ,L

tW f x t w x t dx= ∫ (7.1.2.5)

( ) 2

0

12

L

aW P x dxφ= − ∫ (7.1.2.6)

All together, the Lagrangian b s b s t aL T T U U W W= + − − + + is

( ) ( ) ( )2 2 2 2

2

0

1 1 1 1 1 , ,2 2 2 2 2

L w wL A I EI kAG P x f x t w x t dxt t x x

φ φρ ρ φ φ ∂ ∂ ∂ ∂ = + − − − − + ∂ ∂ ∂ ∂

∫

(7.1.2.7)

To (7.1.2.7) we may include the additional kinetic energies associated with the end-mounted masses and springs from Figure 44.


120

( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

2 2 2 22

0

2 22 2 2 2

1 1 2 2 1 2

1 1 1 1 1 , ,2 2 2 2 2

0, ,1 1 1 1 1 10, 0, , ,2 2 2 2 2 2

L

t t

w wL A I EI kAG P x f x t w x t dxt t x x

w t w L tk w t k t k w L t k L t m m

t t

φ φρ ρ φ φ

φ φ

∂ ∂ ∂ ∂ = + − − − − + ∂ ∂ ∂ ∂

∂ ∂ − − − − + + ∂ ∂

∫

(7.1.2.8)

Using Hamilton's principle, we have that

2

1

0t

tI Ldtδ δ= =∫ (7.1.2.9)

Undertaking the operations in (7.1.2.9), we apply integration by parts on the needed terms. This results in two groups of terms either associated with the variation in transverse deflection wδ or variation in cross-section rotation δφ . We ultimately obtain the combination of EOMs

( )2

2 ,w wkAG A f x tx x t

φ ρ ∂ ∂ ∂ − − + = ∂ ∂ ∂ (7.1.2.10)

2

2 0wEI kAG P Ix x x t

φ φφ φ ρ∂ ∂ ∂ ∂ − − − − + = ∂ ∂ ∂ ∂ (7.1.2.11)

The boundary conditions are likewise found to be

( )10

0,tx

EI k txφ φ

=

∂=

∂ (7.1.2.12)

( )2

1 1 20 0

0,x x

wEI k w t mx x t

φ

= =

∂ ∂ ∂= − − ∂ ∂ ∂

(7.1.2.13)

( )2 ,tx L

EI k L txφ φ

=

∂=

∂ (7.1.2.14)

( )2

2 2 2,x L x L

wEI k w L t mx x t

φ

= =

∂ ∂ ∂= + ∂ ∂ ∂

(7.1.2.15)

When the material and geometric properties of the beam are presumed to be constant along the beam length, the EOMs (7.1.2.10) and (7.1.2.11) become

( )2 2

2 2 ,w wkAG A f x tx x t

φ ρ ∂ ∂ ∂

− − + = ∂ ∂ ∂ (7.1.2.16)

2 2

2 2 0wEI kAG P Ix x tφ φφ φ ρ∂ ∂ ∂ − − − − + = ∂ ∂ ∂

(7.1.2.17)


121

Of course, the (7.1.2.16) and (7.1.2.17) are the same as that derived by the equilibrium approach in

(7.1.1.1.10) and (7.1.1.1.11), with the addition of the influence of the axial force ( ),P x t .

To use the variational approach without accommodation of shear flexibility or rotary inertia, we may set

0Iρ = in our assessment of the kinetic energy for shear, and may substitute back wx

φ ∂≈∂

in other

expressions. Namely, we have that

0sT = ; 22

20

12

L

bwU EI dx

x ∂

= ∂ ∫ ;

2

00

2L

sw wU kAG dxx x

1 ∂ ∂ = − = ∂ ∂ ∫ ; ( )2

0

12

L

awW P x dxx

∂ = − ∂ ∫ (7.1.2.18)

The new Lagrangian b s b s t aL T T U U W W= + − − + + is therefore

( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

22 22

20

2 2 2 22 2

1 1 2 2 1 2

1 1 1 , ,2 2 2

0, , 0, ,1 1 1 1 1 10, ,2 2 2 2 2 2

L

t t

w w wL A EI P x f x t w x t dxt x x

w t w L t w t w L tk w t k k w L t k m m

x x t t

ρ ∂ ∂ ∂ = − − + ∂ ∂ ∂

∂ ∂ ∂ ∂ − − − − + + ∂ ∂ ∂ ∂

∫

(7.1.1.19)

By using Hamilton's principle and undertaking the appropriate integration by parts, we ultimately obtain the EOM

( ) ( )2 2 2

2 2 2 ,w w wEI A f x t P xx x t x x

ρ ∂ ∂ ∂ ∂ ∂ + = + ∂ ∂ ∂ ∂ ∂

(7.1.1.20)

and boundary conditions

2

1200

txx

w wEI kx x ==

∂ ∂=

∂ ∂ (7.1.2.21)

( )2 2

1 12 200 0

0,xx x

w w wEI P k w t mx x x t== =

∂ ∂ ∂ ∂− = − −

∂ ∂ ∂ ∂ (7.1.2.22)

2

22 tx Lx L

w wEI kx x ==

∂ ∂=

∂ ∂ (7.1.2.23)

( )2 2

2 22 2,x Lx L x L

w w wEI P k w L t mx x x t== =

∂ ∂ ∂ ∂− = +

∂ ∂ ∂ ∂ (7.1.2.24)

As expected, the EOM (7.1.1.20) is the same as that derived by the equilibrium approach in (7.1.1.1.2). The boundary conditions presented in (7.1.2.22) and (7.1.2.24) describe the requirement for the shear force to be included among the contributions to the transverse deflection force balance.


122

Of course, the boundary conditions in the derivation of this Sec. 7.1 take into account the general mounting situation schematically shown in Figure 44. The commonly considered boundary conditions such as free, pinned, or clamped beams result in boundary condition expressions of reduced complexity when compared to (7.1.2.12)-(7.1.2.15) or (7.1.2.21)-(7.1.2.24).


As shown through the earlier sections, the free response for the linear structural dynamics of continuous systems may be analyzed by applying the method of separation of variables and then taking into account the boundary and initial conditions. We employ such as technique in this Sec. 7.2 to study the free transverse vibration response of the beam.

Consider the clamped-free beam shown in Figure 47. Such a combination of boundary conditions is alternatively termed cantilevered. Beams that are clamped-free at the ends are consequently termed cantilevers. Along the length L , the beam has uniform Young's modulus E , second moment of area I , cross-sectional area A , and density ρ . The roles of shear flexibility and rotary inertia are presumed to be

negligible. The transverse displacement of the beam ( ),w x t is excited by initial conditions of displacement

( ) ( )0,0w x w x= and velocity ( ) ( )0,0w x w x= . We wish to predict this free, dynamic response associated

with the initial conditions.

Based on the formulation of the beam, we have the governing equation of motion (7.1.1.9) that omits the applied distributed force, repeated here for convenience

4 2

4 2 0w wEI Ax t

ρ∂ ∂+ =

∂ ∂ (7.2.1)

The associated boundary conditions for the beam are

( ) ( )0,0, 0

w tw t

x∂

= =∂

at 0x = (7.2.2)

( ) ( )2 3

2 3

, ,0

w L t w L tx x

∂ ∂= =

∂ ∂ at x L= (7.2.3)

The boundary conditions (7.2.2) and (7.2.3) are simplifications of the general Euler-Bernoulli boundary conditions given in (7.1.1.14)- (7.1.1.16).

We apply the method of separation of variables in our assumed solution. As such, we assume that the

transverse displacement of the beam ( ),w x t may be separated into spatially and time dependent

components

( ) ( ) ( ),w x t X x T t= (7.2.4)

Substituting (7.2.4) into (7.2.1), we have that


123

X A TX EI T

ρ′′′′= −

(7.2.5)

Figure 47. Cantilevered (clamped-free) beam.

Due to the independence of the LH and RH sides of (7.2.5), both sides must be equal to the same constant

that we here defined as 4α . Consequently, we have the pair of equations

4 0X Xα′′′′ − = (7.2.6)

2 0T Tω+ = (7.2.7)

where we introduce 4

2 EIAαωρ

= . As such, we have that 2

4 AEI

ρ ωα = .

For the equation (7.2.6) that governs the spatial variation of the transverse beam displacement, we see that it may be factored into the components

( )( )2 2 0X X X Xα α′′ ′′+ − = (7.2.8)

As such, both (7.2.9) and (7.2.10) must be satisfied for (7.2.6) to hold.

2 0X Xα′′ + = (7.2.9)

2 0X Xα′′ − = (7.2.10)

If the unknown constant 2α is positive, which turns out to be the only possibility [7], the general solution to (7.2.9) and (7.2.10) that collectively solves (7.2.6) is a combination of trigonometric sinusoids and hyperbolic sinusoids given in (7.2.11).

( ) 1 2 3 4sin cos sinh coshX x C x C x C x C xα α α α= + + + (7.2.11)

Substitution of (7.2.11) into the expressions for the cantilevered boundary conditions (7.2.2) and (7.2.3) results in

2 40 C C= + (7.2.12)

1 30 C C= + (7.2.13)

1 2 3 40 sin cos sinh coshC L C L C L C Lα α α α= − − + + (7.2.14)


124

1 2 3 40 cos sin cosh sinhC L C L C L C Lα α α α= − + + + (7.2.15)

The coefficient matrix associated with the system of equations is

1

2

3

4

0 1 0 1 01 0 1 0 0

sin cos sinh cosh 0cos sin cosh sinh 0

CCCL L L LCL L L L

α α α αα α α α

= − − −

(7.2.16)

For non-trivial values of nC , the determinant of the coefficient matrix in (7.2.16) must be zero. After

simplifications, the determinant is

1 cos cosh 0L Lα α+ = (7.2.17)

The equation (7.2.17) is the characteristic equation for the roots ( )nLα that satisfy the boundary conditions

for the thn mode. Subsequently, by computing the roots, one determines the unknown constants nα . This

procedure is typically a numerical computation, such as by the fzero command in the MATLAB software. Once the constants nα are determined through such numerical procedures, then the relation among the

constants nC may be determined with respect to one of the other constants by substituting a given nα into

three of the four boundary condition equations. In the case of the fixed-free beam, we are able to determine how 2C , 3C , and 4C relate to 1C .

2

1

1 cos cosh sin sinhsin cosh cos sinh

C L L L LC L L L L

α α α αα α α α

+ −=

+ (7.2.18)

3

1

1CC

= − (7.2.19)

4

1

cos sech sin tanhsin cos tanh

C L L L LC L L L

α α α αα α α+ +

=+

(7.2.20)

Because each ratio of (7.2.18), (7.2.19), and (7.2.20) is a function of α , which are infinite in number as computed from the characteristic equation (7.2.17), the ratios are more generally written as ( )1/i nC C for

the thn ratio between the thi constant and the 1st constant. As a result, we may write the thn normal mode for the cantilever beam in the form

( ) 32 41

1 1 1

sin cos sinh coshn n n n nn n n

CC CX x C x x x xC C C

α α α α

= + + + (7.2.21)

The four lowest order normal modes of the cantilever beam are shown in Figure 48.


125

Figure 48. Lowest order normal modes for the cantilever beam.


L=1; % [m] beam length beta=linspace(0,10,151); % create range for unknown alpha*L f1=1+1*cos(beta).*cosh(beta); % characteristic equation f2=@(x)1+cos(x).*cosh(x); % characteristic equation function for numerical solution figure(1); clf; plot(beta,f1,'-r'); box on xlabel('\alpha L'); ylabel('characteristic equation') ylim([-10 10]); temp=fzero(f2,1); % the initial guess should change based on interest of normal mode disp(['\alpha L = ' num2str(temp) ]); x=linspace(0,L,51); c2onc1=(1+cos(temp)*cosh(temp)-sin(temp)*sinh(temp))/(sin(temp)*cosh(temp)+cos(temp)*sinh(temp)); c3onc1=-1; c4onc1=(cos(temp)+sech(temp)+sin(temp)*tanh(temp))/(sin(temp)+cos(temp)*tanh(temp)); normal_mode=sin(temp/L.*x)+c2onc1*cos(temp/L.*x)+c3onc1*sinh(temp/L.*x)+c4onc1*cosh(temp/L.*x); figure(2); clf; plot(x/L,normal_mode/max(abs(normal_mode)),'-r'); box on ylim([-1.1 1.1]) xlabel('normalized length along beam [dim]'); ylabel('modal amplitude [dim]'); title(['cantilever beam normal mode for \alpha L=' num2str(temp) ', (\alpha L)^2=' num2str(temp^2) '']);

Having solved for the constants nα , we return to the time dependent equation (7.2.7). It is not clear that we

have an infinite number of equations since we have an infinite number of normal modes and associated constants nα .

2 0n n nT Tω+ = (7.2.22)


126

Of course, the solution to (7.2.22) is of the general form

( ) sin cosn n n n nT t A t B tω ω= + (7.2.23)

such that the total solution for the transverse beam displacement is

( ) ( )[ ]1

, sin cosn n n n nn

w x t X x A t B tω ω∞

=

= +∑ (7.2.24)

where the unknown amplitude constant 1C and the time-dependent constants nA and nB may be reduced

by way of their products.

Given the initial conditions ( )0w x and ( )0w x we have that

( ) ( )01

n nn

w x X x B∞

=

=∑ (7.2.25)

( ) ( )01

n n nn

w x X x Aω∞

=

=∑ (7.2.26)

Multiplying both sides of (7.2.25) and (7.2.26) by the normal mode ( )mX x and integrating over the beam

length, we find the relation between the initial conditions, normal modes, and unknown constants

( ) ( )

( )00

2

0

L

nn L

n

w x X x dxB

X x dx= ∫

∫ (7.2.27)

( ) ( )

( )00

2

0

L

nn L

n n

w x X x dxA

X x dxω= ∫

∫

(7.2.28)

Note that this procedure determines all of the needed constants. In other words, the constants nα are

determined numerically from the characteristic equation. This allows one to determine the ratios ( )1/i nC C

. Consequently, applying the initial conditions to compute nA and nB which are equivalent to 1 n nC A A→

and 1 n nC B B→ . As a result, one determines the total free transverse vibration response of the beam as given

in (7.2.24).

7.3 Forced response

Predicting the forced vibration response of beams is conducted in an analogous manner as that undertaken for other continuous systems. While the free response associated with initial conditions may be solved for via the homogeneous equation and the boundary and initial conditions, once the normal modes are identified the forced response may be predicted via the assumed modes method. When the steady-state behavior is of interest, we may neglect the free response solution to the homogeneous equation of motion. Therefore, in


127

the following Example, we give specific attention to the study of the steady-state vibration of the beam induced by a distributed force per unit length ( ),f x t .

Example Consider the simply-supported beam in Figure 49. Along its length L , the beam has uniform Young's modulus E , second moment of area I , cross-sectional area A , and density ρ , while the roles of shear

flexibility and rotary inertia are presumed to be negligible. A force per unit length ( ),f x t drives the beam

into steady-state vibration. The force occurs with spatially varying amplitude ( )0F x and harmonic

frequency ω such that ( ) ( )0, cosf x t F x tω= .

Determine the steady-state transverse vibration displacement ( ),w x t of the beam induced by the harmonic

force.

Figure 49. Simply-supported beam with distributed force per unit length excitation.

Answer The boundary conditions of the simply-supported beam are given in (7.1.1.15). Repeated here for convenience, the boundary conditions are

( ) ( )0, , 0w t w L t= =

( ) ( )2 2

2 2

0, ,0

w t w L tx x

∂ ∂= =

∂ ∂

Using the method of separation of variables with an assumed solution

( ) ( ) ( ),w x t X x T t=

we find that the response of the beam is governed by the combination of equations

4 0X Xα′′′′ − =

2 0T Tω+ =

where we introduce 4

2 EIAαωρ

= . The spatially dependent equation results in the equation pair


128

2 0X Xα′′ + =

2 0X Xα′′ − =

whose general solution is of the form

( ) 1 2 3 4sin cos sinh coshX x C x C x C x C xα α α α= + + +

Applying the general solution to the boundary conditions, we have

2 40 C C= +

1 2 3 40 sin cos sinh coshC L C L C L C Lα α α α= + + +

2 40 C C= − +

1 2 3 40 sin cos sinh coshC L C L C L C Lα α α α= − − + +

From these results, we see that 2 4 0C C= = . As a result, the characteristic equation is found by the

determinant of the coefficient matrix

1

3

sin sinh 0sin sinh 0

CL LCL L

α αα α

= −

sin sinh 0L Lα α =

Because sinh Lα will not take on a null value unless the root is zero, the roots of this characteristic equation are found from sin 0Lα = . This equality is realized when

nL nα π= , 1,2,3,...n =

Substituting these roots into one of the equations forming the coefficient matrix results in the finding that

3 0C = . Then, the normal modes are given by

( ) 1 sinn nX x C xα=

Assuming a complex exponential form of harmonic force excitation

( ) ( )0, j tx t F x e ω=f

we likewise tailor our assumed solution for the beam transverse displacement via

( )1

, sinj tn n

nx t e xω α

∞

=

=∑w A

where only the real part of ( ),x tw is recognized to be the actual or physical displacement response.

Substituting this assumed solution into the inhomogeneous governing equation of motion in complex exponential form


129

( )4 2

4 2 ,EI A x tx t

ρ∂ ∂+ =

∂ ∂w w f

we obtain

( )4 20sinj t j t

n n nEI A e x F x eω ωα ρ ω α − = A

( )4 20

1sinj t j tn n n

EI e x F x eA A

ω ωα ω αρ ρ

− =

A

Recalling the previous definition,

42 nn

EIAα

ωρ

=

we find that the equation determining the modal amplitude coefficients nA is

( )2 20

1sinn n n x F xA

ω ω αρ

− = A

Multiplying both sides of the equation by another normal mode sin m xα and integrating over the beam

length, we find that the modal amplitude coefficients are computed from

( ) ( )02 2 0

2 sinL

n nn

F x xdxAL

αρ ω ω

=− ∫A

In this case, because the harmonic force amplitude ( )0F x is purely real, the resulting coefficients are

likewise real

( ) ( )02 2 0

2 sinL

n nn

A F x xdxAL

αρ ω ω

=− ∫

so that the actual transverse displacement of the beam is given by

( )1

, sin cosn nn

w x t A x tα ω∞

=

=∑

Considering a specific example of the harmonic force amplitude distribution given by

( )0

0;0 / 62cos ; / 6 4 / 970;4 / 9

x LxF x L x L

LL x L

π< <

= ≤ ≤

< <

The amplitude coefficients are then determined by


130

( )4 /9

2 2 /6

2 2cos sin7

L

n Ln

x n xA dxL LALπ π

ρ ω ω=

− ∫

recalling again /n n Lα π= , 2 4 /n nEI Aω α ρ= , and 1,2,3,...n = .

Figure 50 presents results of the harmonically forced beam vibration evaluated at /x L π= by using a modal summation accounting for 16N = modes. Many resonant peaks appear. Examining the frequencies of these peaks, we find that they are located at / 2nω π . This indicates that nω are the natural frequencies

of the beam, and that for this case of harmonic excitation the beam is being driven at resonance when

nω ω= , thus explaining the large amplitude response.

Yet, in some cases the resonant driving does not yield so large amplitude of the beam displacement, for instance around 530 [Hz]. In this case, the evaluation of the vibration response of the beam at the location

/x L π= occurs near a location corresponding to a node of the third mode. This is exemplified in Figure 51 that compares the third mode shape to the evaluation location (data point circle). Consequently, a common issue that may be resolved in experimentation is an inadequate sensor placement that results in the ability to accurately measure modal response. This is typically associated with placing the sensor near to a node of the vibration mode in question, as shown by this Example.

Figure 50. Forced beam vibration evaluated at /x L π= .


rho=2100; % [kg/m^3] beam density E=72e9; % [N/m^2] beam young's modulus t=2e-3; % [m] beam thickness w=15e-3; % [m] beam width L=300e-3; % [m] beam length I=w*t^3/12; % [m^4] second moment of area A=w*t; % [m] beam cross-sectional area F0=1; % [N] force amplitude num_modes=16; % [dim] number of modes to take into summation x_eval=L/pi; % [m] length along beam to evaluate the displacement

101

102

103

104

10-6

10-4

10-2

100

frequency [Hz]

beam

tran

sver

se d

ispl

acem

ent a

mpl

itude

[mm

] number of modes in summation=16


131

omega=2*pi*logspace(1,4,501); % [rad/s] frequencies of harmonic force w=zeros(length(omega),1); % preallocate beam displacement vector for ooo=1:length(omega) for nnn=1:num_modes alpha_n=nnn*pi/L; % coefficient omega_n=sqrt(E*I*alpha_n^4/rho/A); % [rad/s] natural frequency weight=quad(@(x)cos(2*pi*x/7/L).*sin(nnn*pi*x/L),L/6,4*L/9); % numerically compute modal amplitude coefficient w(ooo)=w(ooo)+2/rho/A/L/(omega_n^2-omega(ooo)^2)*weight*sin(alpha_n*x_eval); end end figure(1); clf; loglog(omega/2/pi,1e3*abs(w),'r'); xlabel('frequency [Hz]') ylabel('beam transverse displacement amplitude [mm]'); box on ylim([1e-7 1e1]) title(['number of modes in summation=' num2str(num_modes) '']);

Figure 51. Third mode of simply-supported beam and evaluation location used to plot beam transverse displacement in Figure 50.

The transverse vibration of beams is of enduring interest throughout engineering applications. Yet, the presence of an ideal beam geometry configured in one of the ideal boundary conditions described in this Sec. 7 may not be the case of a particular practical application. As a result, the determination of the modes of vibration may be greatly challenged or convoluted.

This encourages one to obtain an understanding on how to utilize approximate techniques to solve the governing equations of motion for such more difficult and practical problems.

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

-1

-0.5

0

0.5

1

normalized length along beam [dim]

norm

aliz

ed m

odal

am

plitu

de [d

im]


132

8 Approximate methods of analyzing vibrations of deformable solids

Approximate methods of analyzing the vibrations of continuous structures are a common means employed to investigate the dynamic behavior of the systems. This is because structures do not often accommodate ideal geometries or ideal boundary conditions (i.e. free, simply-supported, or fixed) that permit reduction of the complexity of associated characteristic equation solution.

This Sec. 8 introduces the common approximate analytical methods utilized in the study of the vibrations of continuous systems. Final remarks conclude this Sec. 8 to underscore how these approaches are the foundation for the finite element method. It is emphasized that this Sec. 8 is an introduction to these approximate methods. Interested readers should pursue a follow-on course that details methods of approximately analyzing continuous structures.

8.1 The Ritz method

The concept of the Ritz method is to solve the governing equation of motion for the structure by a variational principle.

For problems in mechanics, this is the principle of minimum total potential energy. In Sec. 3.1.2 we denoted the total potential energy by Π . Consequently, the first variation of the total potential is zero in order for a solution to the governing equation, associated with the potential energy expression, to minimize the potential energy. The expression of the principle of minimum total potential energy is

0δΠ = (8.1.1)

For problems in dynamics, the principle employed by the Ritz method is Hamilton's principle, described in Sec. 3.1.5. The expression of Hamilton's principle is

0Iδ = (8.1.2)

where the functional I is given by

2

1

t

tI Ldt= ∫ (8.1.3)

and L is the Lagrangian of the system.

In our prior use of these principles (8.1.1) and (8.1.2), we made no assumptions about the relevant deformations of the structure. In fact, the act of taking (8.1.1) or (8.1.2) result in the corresponding governing equations of motion and associated boundary conditions. Thus, nothing is actually solved by the application of these principles.

We later employed the method of separation of variables (see Sec. 4.2 for a first overview) to solve the equations of motion. Although we investigated one example of a more realistic and discontinuous structure in Sec. 5.4, the math behind the strategy was observed to become more complex. Indeed, approximate


133

methods of analysis that may employ the principles represented by (8.1.1) and (8.1.2) may more easily obtain solutions for the mechanical configuration and dynamic behavior of non-ideal structural systems.

For instance, consider the segmented beam in Figure 52(a) bounded by a fixed end at 0x = and a simple support at x L= . This may be an ideal model of a test setup in the laboratory, for example of a robot manipulator connected between two working ends. We may also have reason to believe that the simple support effected in practice is not ideal. In other words, while permitting free rotation, the "pinned" support may actually provide compliance in the transverse direction. Thus, we would want to model the end support at x L= using a translational spring, Figure 52(b).

Needless to say, analyzing the system shown in Figure 52(b) using the method of separation of variables on each segment of the beam, and mating each segment according to neighboring boundary conditions, will be time-consuming. An approximate method to study the dynamic response is preferred.

Figure 52. (a) Segmented beam with distributed forcing. (b) Approximation of the beam boundary condition at x L= by translational springs.

For problems of structural vibrations, the Ritz method starts by assuming that the vibration response in

space and time is composed from a linear combinations of trial functions ( )n xφ multiplied by weighting

constants. The weights are provided by time-dependent terms ( )na t . In other words, we assume

( ) ( ) ( ) ( )01

,N

n nn

w x t x a t xφ φ=

= +∑ (8.1.4)

where N is the total number of functions assumed in the summation. The function ( )0 xφ is chosen such

that the natural or forced boundary conditions are satisfied. No unknown multiplier coefficient is required

for ( )0 xφ .

The trial functions ( )n xφ must satisfy the following three requirements.

1. The trial functions must be linearly independent. In other words, ( ) ( )1

0N

n nn

a t xφ=

=∑ iff ( )na t =0

for all n .


134

2. The trial functions must satisfy all essential boundary conditions. For the longitudinal vibration of bars, the transverse vibration of strings, and the torsional vibration of shafts, this requirement means that the displacement (or angular displacement) at the ends must be exactly met by the trial functions. For the transverse vibration of beams, this means that the displacement and slope at the ends must be exactly met by the trial functions.

3. The trial functions must form a complete set. In other words, in using the linear combination 1

N

n=∑, the 3rd trial function 3n = must be preceded by the use of the 1st and 2nd trial functions. As an

example, if considering polynomials ( ) nn x xφ = as trial functions, the polynomial 3x must be

preceded by x and 2x .

For the problems in dynamics, substitution of (8.1.4) into Hamilton's principle (8.1.2) results in a minimization expressed by

10

N

nn n

II aa

δ δ=

∂= =

∂∑ (8.1.5)

In problems of dynamics the ( )na t are termed generalized coordinates.

Because the variations in the generalized coordinates ( )na tδ are arbitrary, we have that

0n

Ia∂

=∂

for 1 n N≤ ≤ (8.1.6)

The (8.1.6) represents N equations whose simultaneous solution yields the unknown generalized coordinates na . Given the knowledge of the trial functions ( )n xφ and assumed response composition

(8.1.4), the dynamics of the system may be approximately reconstructed.

Undertaking a specific example helps to solidify this approximate method of analysis.

Consider the segmented beam system in Figure 52(b). The beam has constant Young's modulus E and density ρ over its whole length L . Due to the segmentation, the second moment of area I and the cross-

sectional area A vary in the different beam segments. For the thi segment, the Young's modulus and cross-sectional area are respectfully identified by iA and iI . We neglect shear flexibility and rotary inertia in this

analysis.

The total kinetic energy T of the system is

( ) 1 1 2

1 1 2

2 2 2 2

1 2 30 0

1 1 1 12 2 2 2

L L L L L

L L L

w w w wT A x dx A dx A dx A dxt t t t

ρ ρ ρ ρ+

+

∂ ∂ ∂ ∂ = = + + ∂ ∂ ∂ ∂ ∫ ∫ ∫ ∫ (8.1.7)

The total potential energy U of the system is


135

( )

( )1 1 2

1 1 2

222

20

2 2 22 2 22

1 2 32 2 20

1 1 ,2 2

1 1 1 1 ,2 2 2 2

L

L L L L

L L L

wU EI dx kw L tx

w w wEI dx EI dx EI dx kw L tx x x

+

+

∂= + ∂

∂ ∂ ∂= + + + ∂ ∂ ∂

∫

∫ ∫ ∫ (8.1.8)

The potential of the applied loads W is

( ) ( )0

, ,L

W f x t w x t dx= ∫ (8.1.9)

The Lagrangian in the generic form is thus

( ) ( ) ( )22 2

220

1 1 1, , ,2 2 2

L w wL A EI f x t w x t dx kw L tt x

ρ ∂ ∂ = − + − ∂ ∂

∫ (8.1.10)

Hamilton's principle then reads

( ) ( ) ( )2

1

22 22

20

1 1 1, , , 02 2 2

t L

t

w wI A EI f x t w x t dx kw L t dtt x

δ δ ρ ∂ ∂ = − + − = ∂ ∂

∫ ∫ (8.1.11)

( ) ( ) ( ) ( )2

1

2 2

2 20, , , , 0

t L

t

w w wA w EI f x t w x t dx kw L t w L t dtt t x x

ρ δ δ δ δ ∂ ∂ ∂ ∂ − + − = ∂ ∂ ∂ ∂

∫ ∫ (8.1.12)

In (8.1.12), we have selectively employed the characteristic of the variation operator that derivatives and variations are interchangeable. Performing integration by parts on the first term of (8.1.12), we find

( ) ( ) ( ) ( )2

1

2 2 2

2 2 20, , , , 0

t L

t

w w wA w EI f x t w x t dx kw L t w L t dtt x x

ρ δ δ δ δ ∂ ∂ ∂ − − + − = ∂ ∂ ∂

∫ ∫ (8.1.13)

At this point, we are able to substitute in our assumed solution composed in a form like (8.1.4).

To meet the essential boundary conditions ( ) ( )0, 0, 0w t w t′= = , we must select trial functions such that

( ) ( )0 0 0n nφ φ′= = .

Consequently, as a first guess we assume the trial functions ( )n xφ are of a form

( )1n

nxxL

φ+

=

for 1 n N≤ ≤ (8.1.14)

Note that if the superscript was n , instead of 1n + , then we would not meet the essential boundary conditions. Thus, (8.1.14) are a complete set of trial functions.

In this example of Figure 52(b), we do not need to meet explicit natural boundary conditions, so that

( )0 0xφ = (8.1.15)


136

Consequently, our assumed solution for the response of the segmented beam is

( ) ( ) ( ) ( )01

,N

n nn

w x t x a t xφ φ=

= +∑ (8.1.16)

Substituting this assumed solution into (8.1.13) results in

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( ) ( )

2

1

1 1

01 1

1

1 1

0,

N N

n n n nn n

N NL

n n n nt n n

Nt

n nn

N N

n n n nn n

A a t x a t x

EI a t x a t x dx

f x t a t x

k a t L a t L dt

ρ φ δ φ

φ δ φ

δ φ

φ δ φ

= =

= =

=

= =

−

′′ ′′−

= +

−

∑ ∑

∑ ∑∫∫

∑

∑ ∑

(8.1.17)

The variations are with respect to the unknown generalized coordinates na since the trial functions ( )n xφ

are known. In other words, ( ) ( ) ( )1

N

n n mn

a t x xδ φ φ=

=

∑ .

We can thus simplify (8.1.17) to be

( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) ( )2

1 0 01

, 0Nt L L

n n m n n m n n m mtn

A a t x x EI a t x x dx k a t L L f x t x dtρ φ φ φ φ φ φ φ=

′′ ′′− − − + = ∑∫ ∫ ∫

(8.1.18)

Because (8.1.18) must be valid for all time, the bracketed term of the time integral must independently be equal to zero.

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )0 0

1, 0

N L L

n n m n n m n n m mn

Aa t x x EIa t x x dx ka t L L f x t xρ φ φ φ φ φ φ φ=

′′ ′′− − − + =∑∫ ∫ (8.1.19)

We find that (8.1.19) may be written in the more compact form

( )M Ka a f t+ = (8.1.20)

The overbar indicates a vector and standard serif capital letters denote a matrix. The associated entries to (8.1.20) are

( ) ( ) ( ), 0M

L

n mn mA x x dxρ φ φ= ∫ (8.1.21)

( ) ( ) ( ) ( ) ( ), 0K

L

n m n mn mEI x x dx k L Lφ φ φ φ′′ ′′= +∫ (8.1.22)

( )( ) ( ) ( )0

,L

mmf t f x t x dxφ= ∫ (8.1.23)


137

Summarizing the analytical efforts so far, for problems in structural dynamics the Ritz method uses Hamilton's principle to culminate in equations of motion that are as numerous as the number of trial functions utilized in the response reconstruction (8.1.4). Each entry of this matrix system of equations is determined by a combination of the trial functions, such as those entries shown in (8.1.21) through (8.1.23). Solution to these equations yield the generalized coordinates ( )na t . With knowledge of the trial functions

( )n xφ according to the assumed response (8.1.4), the spatial and temporal dependence of the structural

dynamics may be approximately determined. In general, the accuracy of the approximation is strongly dependent on (a) the number N of trial functions used in the assumed solution (8.1.4), and (b) the quality of the trial functions in the sense of the similarity of the spatial variation of the trial functions when compared to the actual system response.

Continuing the derivation according to the example of the segmented beam shown in Figure 52(b), we put (8.1.21) to (8.1.23) into forms representative of the specific problem of interest.

( ) ( ) ( ) ( ) ( ) ( ) ( )1 1 2

1 1 21 2 3, 0

ML L L L

n m n m n mn m L L LA x x dx A x x dx A x x dxρ φ φ ρ φ φ ρ φ φ

+

+= + +∫ ∫ ∫ (8.1.24)

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )1 1 2

1 1 21 2 3, 0

KL L L L

n m n m n m n mn m L L LEI x x dx EI x x dx EI x x dx k L Lφ φ φ φ φ φ φ φ

+

+′′ ′′ ′′ ′′ ′′ ′′= + + +∫ ∫ ∫ (8.1.25)

( )( ) ( ) ( )0

,L

mmf t f x t x dxφ= ∫ (8.1.26)

Substituting in the specific trial functions (8.1.14) into (8.1.24), we find

( ) 1 1 2

1 1 2

1 1 1 1 1 1

1 2 3, 0M

n m n m n mL L L L

n m L L L

x x x x x xA dx A dx A dxL L L L L L

ρ ρ ρ+ + + + + +

+

+

= + + ∫ ∫ ∫ (8.1.27)

Whether by hand or by symbolic mathematical software like Mathematica or Maple, we evaluate the integrals in (8.1.27) to find

( )( )

( )

( )

( )

( ) ( ) ( )

3 33 31 1 2 1 1 21 2 1 1 23

11 2 32 2 2,

3 33 3 31 1 2 11 1 2 1 2 1 3 1 22

M3 3 3

3

m n m n m n m n

n m

m n m n m n

L L L L L LL L L L L LL L L L LA A AL m n L m n L m n

L L L L LA L A L L L A L L LL m n L L L

ρ ρ ρ

ρ

+ + + +

+ + +

+ + + − − + = + ++ + + + + +

+ = + + − + − + + +

1 2m nL

L

+ +

(8.1.28)

It is important to note that while remaining in a symbolic form, (8.1.28) evaluates to a number since all of the variables in the expression are known.

Continuing, we evaluate (8.1.25) and consider the case of 33 /k EI Lα= .


138

( ) ( ) ( ) 1 1 1 131 1 2 1 1 2

1 2 33 3,

1 1K 1

1

m n m n m n m n

n m

m m n n EIL L L L L LE I I IL m n L L L L L

α+ − + − + − + − + + + + = + − + − + + −

(8.1.29)

Likewise, we consider a relatively straightforward case of harmonic point force excitation at 7 / 9x L= , such that

( ) 0

0;0 7 / 9, cos ; 7 / 9

0;7 / 9

x Lf x t F t x L

L x Lω< <

= = < <

(8.1.30)

Evaluating the (8.1.26) gives

( )( )1

07 cos9

m

mf t F tω

+ =

(8.1.31)

For such harmonic excitation, we may use the complex exponential form of assumed solution for the time dependence of the generalized coordinates.

( )( )1

079

mj t

mt F e ω

+ =

f (8.1.32)

( ) j tt e ω=a A (8.1.33)

Note that the bold font and overbar indicate complex exponential time dependence and vectors, respectively. Substitution of (8.1.32) and (8.1.33) into (8.1.20) leads to

( )2- M+K tω = a f (8.1.34)

Thus, the complex valued amplitudes A of the generalized coordinates a of (8.1.34) may be determined

by multiplying the inversion of 2M Kω − + by both sides of the equation.

In one example, we consider the values: E =2 [GPa], ρ =1100 [kg/m3], b =12 [mm] as a width dimension,

1t =2 [mm], 2t =1 [mm], 3t =2.5 [mm], 1L =100 [mm], 2L =80 [mm], 3L =120 [mm], 0F =1 [mN], the range

of frequencies [ ]/ 2 1,100ω π = [Hz], and the selection of [ ]1 100 10000α = .

Once the generalized coordinates are obtained from (8.1.34), the beam displacement response may be reconstructed by the summation of (8.1.16). Then, the results may be shown [on a two-dimensional plot] as beam displacement evaluated at a spatial position x for all frequencies ω , or beam displacement

evaluated at a frequency ω across the whole spatial extent of the beam [ ]0,x L= .


139

Considering α =1 and N =6 modes, Figure 53 at left gives one example when the beam displacement is evaluated at 5 / 7x L= . The lowest resonant frequency occurs around 4.5 [Hz]. Using conventional Euler-Bernoulli beam theory for a cantilever beam having cross-sectional area 1A and second moment of area 1I

, the lowest order resonance is computed to be 4.84 [Hz]. Due to the similarity in these frequencies, it suggests that the small value of α =1 is insufficient to greatly deviate the dynamics of the beam away from conventional cantilever response. It also suggests that the segmentation of the beam does not greatly influence the spatial distribution of the vibration from conventional cantilever fundamental mode response. This is observed in the right in Figure 53. The low frequency vibration around 11 [Hz] closely parallels the lowest order mode of a cantilever in terms of spatial distribution of the displacement from the clamped end at 0x = .

Using the same layout and evaluation conditions, Figure 54 shows results for the case of α =100. More substantially restrained beam displacement is observed near x L= . This is expected according to the stiffer

0 10 20 30 40 50 60 70 80 90 10010

-8

10-7

10-6

10-5

10-4

10-3

10-2

10-1

frequency [Hz]

beam

dis

plac

emen

t am

plitu

de [m

]

number of modes used: 6spatial evaluation at 0.71429*L

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-9

-8

-7

-6

-5

-4

-3

-2

-1

0x 10

-5

normalized length along beam, x/L [dim]

beam

dis

plac

emen

t [m

]

number of modes used: 6frequency evaluation at 11 [Hz]

Figure 53. At left, beam displacement amplitude as function of harmonic excitation frequency. At right, beam displacement across spatial extent. =1

0 10 20 30 40 50 60 70 80 90 10010

-9

10-8

10-7

10-6

10-5

10-4

10-3

10-2

frequency [Hz]

beam

dis

plac

emen

t am

plitu

de [m

]

α=100. number of modes used: 6spatial evaluation at 0.71429*L

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.5

1

1.5

2

2.5x 10

-5


beam

dis

plac

emen

t [m

]

α=100. number of modes used: 6frequency evaluation at 11 [Hz]



140

end spring condition. The resonances in the left of Figure 54 are also increased in frequency respecting the case of α =1 shown in Figure 53.

Finally, Figure 55 shows the similar results in the case of the end translational spring α =10000. In this

case, due to the relationship between α and the beam translational stiffness by 33 /k EI Lα= , the end spring

stiffness k is considerably stiffer in translation than the beam. It is apparent that the actual simple support, shown in Figure 52(a) is effectively realized by increasing the stiffness of an end translational spring that acts on the beam at the same end. Using conventional Euler-Bernoulli beam theory for a beam with fixed-simple support boundary conditions and cross-sectional area 1A and second moment of area 1I , the lowest

order natural frequency occurs at 21.2 [Hz]. This value is in good agreement with the lowest order resonance shown at left in Figure 55 that is 21.4 [Hz]. This shows the strengths of the approximate analytical method to realize a wide range of structural modeling conditions without the need for new derivations and mode shape considerations.

We also investigate the capability of the Ritz method to characterize the segmentation aspect of the beam. We increase the number of modes to N =12 to use polynomials with greater capability to model the discreteness of the beam. It is worth noting that polynomials are not especially well suited for the Ritz method when compared to trigonometric functions. Polynomials often lead to an ill-conditioned matrix

2M Kω − + when the polynomial order increases. Despite this fact, we examine the variation of

[ ]2 0.1 0.5 1t = [mm] at frequency 5 [Hz] in Figure 56. From left to right, we see the influence of

increasing value of 2t . Clearly, for the smaller 2t at left, there is a sharper discontinuity of the beam

deflection than for the thickest 2t and case shown at right. This is intuitive because a thin segment of the

beam will act like a hinge about which a rotation will occur dominate when compared to bending.

0 10 20 30 40 50 60 70 80 90 10010

-8

10-7

10-6

10-5

10-4

10-3

10-2

frequency [Hz]

beam

dis

plac

emen

t am

plitu

de [m

]α=10000. number of modes used: 6

spatial evaluation at 0.71429*L

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8x 10

-5


beam

dis

plac

emen

t [m

]

α=10000. number of modes used: 6frequency evaluation at 11 [Hz]



141

It is evident through this detailed example that the Ritz method is amenable to predicting the dynamics of continuous structures having arbitrary composition, boundaries, and loads. Interested readers in the topic should consider an advanced course on the subject as well as reference the resources [13] [14] [15] [16].

Table 13. MATLAB code used to generate Figure 53 to Figure 56.

E=2e9; % [Pa] beam young's modulus rho=1100; % [kg/m^3] beam density b=12e-3; % [m] beam width t1=2e-3; % [m] beam thickness in segment 1 t2=1e-3; % [m] beam thickness in segment 2 t3=2.5e-3; % [m] beam thickness in segment 3 L1=100e-3; % [m] beam length in segment 1 L2=80e-3; % [m] beam length in segment 2 L3=120e-3; % [m] beam length in segment 3 L=L1+L2+L3; % [m] total beam length A1=b*t1; % [m^2] area of segment 1 A2=b*t2; % [m^2] area of segment 2 A3=b*t3; % [m^2] area of segment 3 I1=b*t1^3/12; % [m^4] second moment of area for segment 1 I2=b*t2^3/12; % [m^4] second moment of area for segment 2 I3=b*t3^3/12; % [m^4] second moment of area for segment 3 alpha=1e0; % [dim] ratio of translational end spring stiffness to translational stiffness of third beam segment F0=1e-3; % [N] harmonic force amplitude omega=2*pi*logspace(0,2,401);%(1e-2,250,401); % [Hz] frequencies of harmonic excitation num_modes=12; % [dim] number of modes taken in ritz expansion a=zeros(num_modes,length(omega)); % preallocate the generalized coordinates M=zeros(num_modes); % preallocate mass matrix K=zeros(num_modes); % preallocate stiffness matrix f=zeros(num_modes,1); % preallocate forcing vector for mmm=1:num_modes for nnn=1:num_modes M(mmm,nnn)=rho/L^2/(mmm+nnn+3)*(A1*L1^3*(L1/L)^(mmm+nnn)+A2*((L1+L2)^3*((L1+L2)/L)^(mmm+nnn)-L1^3*(L1/L)^(mmm+nnn))+A3*(L^3-(L1+L2)^3*((L1+L2)/L)^(mmm+nnn))); K(mmm,nnn)=E/L^3*mmm*(mmm+1)*nnn*(nnn+1)/(mmm+nnn-1)*(I1*(L1/L)^(mmm+nnn-1)+I2*(((L1+L2)/L)^(mmm+nnn-1)-(L1/L)^(mmm+nnn-1))+I3*(1-((L1+L2)/L)^(mmm+nnn-1)))+alpha*E*I3/L^3; end f(mmm)=F0*(7/9)^(mmm+1); end for ooo=1:length(omega) a(:,ooo)=(-omega(ooo)^2*M+K)\f; end x=linspace(0,L,51); % [m] make range of spatial location along beam w_freq=zeros(length(omega),1); % preallocate the beam transverse displacement evaluation w_space=zeros(length(x),1); % preallocate the beam transverse displacement in space space_eval=5*L/7; % [m] dimension along beam length to evaluate the beam displacement in frequency freq_eval=5; % [Hz] frequency of evaluation for beam displacement in space freq_index=max(find(omega/2/pi<freq_eval)); % [dim] find index of frequency evaluation for nnn=1:num_modes w_freq=w_freq+a(nnn,:).'*(space_eval/L)^(nnn+1);

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-14

-12

-10

-8

-6

-4

-2

0

2x 10

-4


beam

dis

plac

emen

t [m

]t2=0.1 [mm]. α=1. number of modes used: 12

frequency evaluation at 5 [Hz]

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-1.5

-1

-0.5

0x 10

-3


beam

dis

plac

emen

t [m

]

t2=0.5 [mm]. α=1. number of modes used: 12


0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-0.06

-0.05

-0.04

-0.03

-0.02

-0.01

0


beam

dis

plac

emen

t [m

]

t2=1 [mm]. α=1. number of modes used: 12


Figure 56. Beam displacement along length for cases of (left) =0.1 [mm], (middle) =0.5 [mm], (right) =1.0 [mm].


142

w_space=w_space+a(nnn,freq_index)*(x.'/L).^(nnn+1); end figure(1); clf; semilogy(omega/2/pi,abs(w_freq),'-r'); xlabel('frequency [Hz]'); ylabel('beam displacement amplitude [m]'); box on title(['\alpha=' num2str(alpha) '. number of modes used: ' num2str(num_modes) ''],['spatial evaluation at ' num2str(space_eval/L) '*L']); figure(2); clf plot(x/L,w_space,'-r'); xlabel('normalized length along beam, x/L [dim]'); ylabel('beam displacement [m]'); box on title(['t_2=' num2str(t2*1e3) ' [mm]. \alpha=' num2str(alpha) '. number of modes used: ' num2str(num_modes) ''],['frequency evaluation at ' num2str(freq_eval) ' [Hz]']);

8.2 Finite element method (FEM)

The Ritz method is a semi-analytical approach to approximately solving the governing equations of continuous structures. We note that the Ritz method operates on the functional whose minimization results in the governing equations. The functional is the principle of minimum potential energy for problems in mechanics and is Hamilton's principle for problems in structural dynamics. Consequently, the Ritz method is an energy-based method of analysis.

The finite element method (FEM) is the discrete counterpart of the Ritz method. The applicability of the FEM extends beyond problems in mechanics and dynamics, since the FEM is used in electromagnetics, acoustics, thermodynamics, and other fields. In the FEM, the problem domain, meaning the regions where variables are governed by various physics, is discretized into sub-domains. In these sub-domains, local properties, geometries, and loads are identified. This procedure has analogy to the segmented beam example considered in Sec. 8.1. On the other hand, the domain discretization is not limited to the points where the system properties or geometries specifically change. Such enhanced discretization results in improved accuracy of the response prediction, at the expense of increasing computational cost due to growing sizes of stiffness and mass matrices, like the Ritz method analogs in (8.1.28) and (8.1.29).

In the Ritz method of approximate structural analysis evaluated in Sec. 8.1, the size of the matrices is determined by the number of trial functions N used. The trial functions apply to the whole domain and must satisfy the essential boundary conditions of the whole domain.

In the FEM, the size of the matrices is associated with the number of nodes used in the domain discretization. At least two nodes are associated with each element, since start and terminating element ends are identified. The trial function for the element needs only to satisfy relevant essential boundary conditions for that local element. Thus, rather than use polynomial or trigonometric trial functions as required in the Ritz method for high fidelity of response prediction, in the FEM linear or quadratic functions are used to simplify the mathematical integration operations, often achieved by a quadrature rule.

Interested readers in the topic should consider an advanced course on the subject as well as reference the resource [17].


143

References

[1] A.W. Leissa, Vibration of Plates, Scientific and Technical Information Division, National Aeronautics and Space Administration, Washington D.C., 1969.

[2] A.W. Leissa, Vibration of Shells, Scientific and Technical Information Office, National Aeronautics and Space Administration, Washington D.C., 1973.

[3] L.E. Kinsler, A.R. Frey, A.B. Coppens, and J.V. Sanders, Fundamentals of Acoustics, John Wiley and Sons, New York, 2000.

[4] W.E. Boyce and R.C. DiPrima, Elementary Differential Equations and Boundary Value Problems, John Wiley and Sons, Inc., New York, 2010.

[5] K. Ogata, System Dynamics, Prentice Hall, Englewood Cliffs, New Jersey, 1992.

[6] W.J. Palm, System Dynamics, McGraw-Hill, New York, 2014.

[7] D.J. Inman, Engineering Vibration, Prentice Hall, Saddle River, New Jersey, 2001.

[8] D.J. Ewins, Modal Testing: Theory and Practice, John Wiley & Sons, Inc., Letchworth, 1984.

[9] P. Avitabile, Experimental modal analysis: a simple non-mathematical presentation. Sound and Vibration (2001) 1-11.

[10] J.M. Gere and S.P. Timoshenko, Mechanics of Materials, PWS Publishing Company, Boston, 1997.

[11] A.W. Leissa and M.S. Qatu, Vibrations of Continuous Systems, McGraw-Hill, New York, 2011.

[12] R. Lakes, Viscoelastic Materials, Cambridge University Press, Cambridge, 2009.

[13] J.N. Reddy, Energy and Variational Methods in Applied Mechanics, John Wiley & Sons, Inc., New York, 1984.

[14] L. Dozio, On the use of the trigonometric Ritz method for general vibration analysis of rectangular Kirchhoff plates. Thin-Walled Structures 49 (2011) 129-144.

[15] R.L. Harne and C.R. Fuller, Modeling of a passive distributed vibration control device using a superposition technique. Journal of Sound and Vibration 331 (2012) 1859-1869.

[16] O. Beslin and J. Nicolas, A hierarchical functions set for predicting very high order plate bending modes with any boundary conditions. Journal of Sound and Vibration 202 (1997) 633-655.

[17] M. Petyt, Introduction to Finite Element Vibration Analysis, Cambridge University Press, Cambridge, 1990.


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course notes for osu me 5134 introduction to … harne me 5134, intro. vib. deform. solids the ohio...

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