1Spring 05 AME 463

Review of Mechanical Vibrations

Ara

Arabyan

Week 12

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Vibration Analysis

Deformable structures vibrate (undergo oscillatory motion) under the action of impulsive or time-varying forcesVibration in a structure results from the transformation of potential energy into kinetic energy and vice versaPotential energy in the structure is stored in elastic elements and kinetic energy is stored in mass elements.

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Simple Harmonic Oscillator (SHO)

Consider the mass-spring system comprised of a block of mass m and a spring of stiffness k and unstretched length

The motion of the block is measured by its displacement u from the unstretched position of the spring

0l

mk

u0l

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SHO Equation of Motion

The free-body diagram of the SHO with no applied forces acting on the block is as shown below (forces shown only in direction of motion)

From Newton’s 2nd Law ormu ku= −&&

0mu ku+ =&&

u

mku

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Free Undamped

Vibrations of SHO

The solution of this equation is given by

where is known as the natural frequency

of the oscillator and A

and B

are constants determined by initial conditions

This solution can also be written as

where U

is known as the amplitude

and as the phase

of oscillation

( ) sin cosn nu t A t B tω ω= +

/n k mω =

( ) ( )0 00 0u u u u= =& &

( ) ( )sin nu t U tω φ= +

φ

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Frequency and Period

The amplitude and phase are given in terms of the undetermined constants as

When the frequency is computed from its units are in rad/s and is also known as circular frequency; this can also be expressed in Hz or cycles/s by

The period of the motion (in seconds) is given by

2 2 arctan BU A BA

φ ⎛ ⎞= + = ⎜ ⎟⎝ ⎠

/n k mω =

2nf ωπ

=

1 2nfπτ

ω= =

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Motion of Undamped

SHO

The motion of the oscillator and the quantities defined are displayed graphically in the figure below

The equation of motion of the oscillator can also be written as

0

-U

0

U

t

u(t)

−φτ 2 0nu uω+ =&&

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Effect of Damping

If a viscous damper is added to the SHO we obtain

The quantity c is known as the damping constant of the damperThe resulting equation of motion is

mk

0l u

c

kum

cu&

u

u&

0mu cu ku+ + =&& &

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Damped Vibrations of SHO

This equation can also be written as

Where is known as the damping ratio

For (underdamped) the solution of this equation is given by

where is known as the damped natural frequency

22 0n nu u uζω ω+ + =&& &

( )/ 2 nc mζ ω=

0 1ζ≤ <

( ) ( )sinntd du t Ue tζω ω φ−= +

21d nω ω ζ= −

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Motion of Damped SHOThe underdamped motion of the SHO is depicted graphically below

In the presence of damping the oscillator always comes to restThis motion is also known as transient vibration

0

-U

0

U

t

u(t)

ntUe ζω−

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Effect of Applied Forces

If an applied force acts on the block we obtain

The applied force F(t) is often known as an exciting or disturbingforceThe resulting equation of motion is

mk

0l u

c

( )F t kum

cu&

u

u&( )F t

( )mu cu ku F t+ + =&& &

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Effect of Harmonic Disturbances

Harmonic disturbance is one of the most common classes of excitation (e.g. rotating machinery, road surface, etc.)In such cases the disturbing force can be written as

Where is the excitation amplitude and is the excitation or forcing frequency

The quantity is known as the static deflection

( ) 0 sin fF t F tω=

fω0F

0 /st F kδ =

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Forced Harmonic Motion of SHO

The motion of an underdamped SHO under harmonic excitation is given by

where

( ) ( ) ( )sin sinntd d f fu t Ue t Uζω ω φ ω ψ−= + + +

( )0

1/222 2 2f

f f

FUk m cω ω

=⎡ ⎤− +⎢ ⎥⎣ ⎦

Transient responseForced or steady-

state response

2arctan f

f

ck m

ωψ

ω

⎛ ⎞= ⎜ ⎟⎜ ⎟−⎝ ⎠

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Forced Damped Vibration of SHO

The response of the SHO to harmonic excitation is depicted the rightAs is evident the motion follows the forcing function after the transient response dies outThe frequency at steady-state is the same as the forcing frequency

0

-Uf

0

Uf

t

u(t)

Steady-state Transient + steady-state

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Amplification of Response

Because of the effect of damping the transient response eventually dies out and the forced response (more commonly known as the steady-state response) remainsThe amplitude of the steady-state response is more generally written in terms of the static deflection as

where is known as the frequency ratio

( ) ( ) ( ) ( )0

2 22 22 2

/

1 2 1 2st

fF kU

r r r r

δ

ζ ζ= =

− + − +

/f nr ω ω=

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Resonance

When the frequency ratio r = 1 resonance is said to have occurredWhen resonance occurs the amplitude of the steady-state response becomes

Clearly if the damping ratio is very small then the amplitude ofthe forced response becomes very large which is generally undesirableResonance is a function of both the forcing frequency and the system natural frequency which are fixed by the physical parameters of the system

2st

fU δζ

=

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Amplification and Resonance

The ratio of the steady-state response amplitude and the static deflection is called the amplitude or amplification ratio

The amplification of the steady-state response as a function of the frequency ratio and the damping ratio is plotted on the right

0/

fU

F k

0 /fU

RF k

=Resonance

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Energy Considerations

If there is no damping the total mechanical energy of the SHO isconserved

where T

is the kinetic energy of the mass and Ω

is the potential energy stored in the spring at any time

Oscillatory motion results from the exchange of kinetic and potential energy between the spring and mass elementsThe energy is exchanged at the frequency of oscillation

0 0 0(0) (0)T T T E+ Ω = + Ω = + Ω =

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Two-DOF SystemsNow consider the two-mass, two-spring system below with no applied forces

The number of independent displacements determines the number of DOF (in this case )The displacements are measured from the unstretched positions of the springs

1m1k 2k

2m

1u 2u

1 2,u u

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Two-DOF System Equations of Motion

The free-body diagrams of the two masses with no applied forces acting on the block is as shown below (forces shown only in direction of motion)

From Newton’s 2nd Law

1 1k u ( )2 2 1k u u−1m 2m

1u 2u

( )( )

1 1 1 1 2 2 1

2 2 2 2 1

m u k u k u um u k u u

= − + −= − −

&&

&&

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Mass, Stiffness Matrices

These equations can also be written as

They can also be written in more concise form as

Note that the stiffness matrix is identical to the stiffness matrix obtained for two springs in series with one end constrained

1 1 1 2 2 1

2 2 2 2 2

0 00 0

m u k k k um u k k u

+ −⎡ ⎤ ⎧ ⎫ ⎡ ⎤ ⎧ ⎫ ⎧ ⎫+ =⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎩ ⎭ ⎣ ⎦ ⎩ ⎭ ⎩ ⎭

&&

&&

Mass matrix Stiffness matrix

+ =Mu Ku 0&&

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Solution of Two-DOF Systems

Assume that the two masses will undergo harmonic motion as in the single DOF case with some unknown frequency, amplitude, and phase

Substituting these in the equations of motion we obtain

( )( )

( )11

22

sinUu t

tUu t

ω φ⎧ ⎫ ⎧ ⎫

= +⎨ ⎬ ⎨ ⎬⎩ ⎭⎩ ⎭

( ) ( )1 1 1 2 2 12

2 2 2 2 2

0 0sin sin

0 0m U k k k U

t tm U k k U

ω ω φ ω φ+ −⎡ ⎤ ⎧ ⎫ ⎡ ⎤ ⎧ ⎫ ⎧ ⎫

− + + + =⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎩ ⎭ ⎣ ⎦ ⎩ ⎭ ⎩ ⎭

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Eigenvalue

ProblemThese equations can be written more concisely as

or

These equations can have a nontrivial solution for U for all time if and only if the determinant of the matrix in the parenthesis vanishes

This is equivalent to finding the eigenvalues of (in this case two values of which we will call and )

( ) ( )2 sin tω ω φ− + =K M U 0

( ) ( )1 2 sin tω ω φ− − + =M K I U 0

( )1 2det 0ω− − =M K I1−M K

2ω 21ω 2

2ω

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Natural Frequencies, Mode Shapes

The square roots of the eigenvalues of yield the two natural frequencies of the two-DOF system and ; the smaller of the two frequencies is labeled and is known as the fundamental frequencyThe assumed amplitudes of oscillation can now be determined from

where each vector is the ith

eigenvector of Recall that the components of cannot be determined explicitly; only the ratio between the two components can be determinedThe resulting are known as the mode shapes of vibration

1−M K1ω 2ω

( ) ( )1 2 1,2ii iω− − = =M K I U 0

1−M K( )iU( )iU

( )iU

1ω

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Multi-DOF Systems

When an oscillatory system comprised of discrete masses and springs has N DOF, then the mass and stiffness matrices of the system will be of order N

This system, in general, has N distinct natural frequencies and N distinct mode shapesThe natural frequencies of the system are the square roots of the eigenvalues of and the mode shapes are the eigenvectors of

1 1 1N N N N N N N× × × × ×+ =M u K u 0&&

1−M K1−M K

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Modal Analysis

The free motion of the masses is determined by taking a weighted superposition of the different mode shapes (as in the case of the two-DOF system)The determination of the free motion of the masses using a weighted superposition of the mode shapes is known as modal analysisLower frequency modes are more dominantConsequently the motion of the masses can be approximated by taking a weighted superposition of only the first few modes

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Example 1: Free Vibrations of a Two-DOF System

In the two-DOF system below

Determine the natural frequencies and free response of the system; verify your results with ANSYS

( ) ( ) ( ) ( )1 2 1 2

1 2 1 2

120 lb/in 80 lb/in 20 lbm0 1 in 0 0.5 in 0 0 0

k k m mu u u u

= = = == = = =& &

1m1k 2k

2m

1u 2u

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Example 1: Natural Frequencies

In this case the mass and stiffness matrices are

Thus

The eigenvalues of the matrix are (from Matlab)

Thus the natural frequencies of the system are

20 0 200 8010 20 80 80386.4

−⎡ ⎤ ⎡ ⎤= =⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦

M K

1 33.864 1.54610

1.546 1.546− −⎡ ⎤

= ×⎢ ⎥−⎣ ⎦M K

2 21 2772.8 4636.8ω ω= =

1 227.8 rad/s 4.42 Hz 68.1 rad/s 10.8 Hzω ω= = = =

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Example 1: Mode Shapes

The normalized eigenvectors (mode shapes) of are (from Matlab)

In vibration analysis this is often written as

Physically this means

( ) [ ] ( ) [ ]1 20.447 0.894 0.894 0.447T T= − − = −U U

( ) [ ] ( ) [ ]1 21.0 2.0 1.0 0.5T T= = −U U

1m

1.0

2m

2.0

1m

1.0

2m

0.5−

Mode 1 Mode 2

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Example 1: Time-domain Solution

Thus the free vibration response is given by

The four unknowns in these equations ( ) can be determined from the four initial conditions given (quite complicated) resulting in

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

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

1 21 1 1 1 1 2 2

1 21 1 1 2

1 22 2 1 1 2 2 2

1 21 1 1 2

sin sin

sin 27.8 sin 68.1

sin sin

2 sin 27.8 0.5 sin 68.1

u t U t U t

U t U t

u t U t U t

U t U t

ω φ ω φ

φ φ

ω φ ω φ

φ φ

= + + +

= + + +

= + + +

= + − +

( ) ( )1 21 1 1 2, , ,U U φ φ

( )

( )

1

2

0.4 sin 27.8 0.6 sin 68.12 2

0.8sin 27.8 0.3sin 68.12 2

in

in

u t t t

u t t t

π π

π π

= + + +

= + +

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎛ ⎞ ⎛ ⎞−⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

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Example 1: Time-domain Solution

The motion of the two masses are depicted in the figure on the rightNote that both modes are present in the motions of both massesThe motion of the masses is comprised of a weighted superposition of the two modes

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4-1.5

-1

-0.5

0

0.5

1

1.5

t (sec)

u1, u

2 (in

)

u1(t)

u2(t)

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Example 1: ANSYS Modal Analysis (1st

Mode)

The output of ANSYS modal analysis for this problem is shown on the rightThe first natural frequency is at top leftThe first mode shape is indicated by the position of the blue dots relative to the black dots

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Example 1: ANSYS Modal Analysis (2nd

Mode)

The second-mode solutions shown on the right also agree with hand-computed results

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Example 1: ANSYS Transient Analysis

A transient analysis performed on ANSYS using the initial conditions given yields the same results as those computed by hand

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Example 2: ANSYS Harmonic Analysis

Perform an ANSYS harmonic analysis on the two-DOF system shown below with Hz and following two forcemagnitude combinations

a)

b)

1m1k 2k

2m

1u 2u

1 sin fF tω

2 sin fF tω

0 20fω =

1 2100 lb 0F F= =

1 2100 lb 50 lbF F= =

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Example 2: Frequency Response (Case a)

The response of the system to the applied harmonic force is shown on the right As is evident resonance occurs at both natural frequencies but the response at the first natural frequency is more dominant

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Example 2: Frequency Response (Case b)

In this case the entire response of the system is the first mode because the two forcing functions are such that they excite only the first mode (amplitude of force on mass 2 is twice the amplitude of force on mass 1)

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Conclusions

The response of a structure to a pattern of harmonic excitationsdepends on the points of application and the amplitudes (as well as phases) of the forcing functionsIn general the strongest response will be in the first or fundamental mode (corresponding to lowest natural frequency) but depending on special circumstances higher modes may also be excited disproportionatelyIn all structural analysis first a modal analysis must be performed to determine natural frequencies and mode shapes; this must be followed by a harmonic analysis over a frequency range that brackets the first several natural frequencies to determine their response to expected loading conditions