Engineering Vibrations

1/43Virginia Tech© D. J. Inman

Engineering Vibration3rd Edition

Prentice Hall© D. J. [email protected],

We make our living in dynamics, vibration and control


Motivation The Millennium Bridgea Recent Vibration Problem (2000-1)

OPENED AND CLOSED WITHIN A VIEW HOURSBECAUSE OF UNDESIRABLE VIBRATION

(Show movies)


Modeling and Degrees of Freedom

Lack of consideration of dynamic loads and Vibration caused this to new bridge to vibrate wildly

The goal of this course is to understand such phenomenon and how to prevent it

The bridge has many degrees of freedom, we will start with oneand work towards many.


Example 1.1.1 The Pendulum

• Sketch the structure or part of interest

• Write down all the forces and make a “free body diagram”

• Use Newton’s Law and/or Euler’s Law to find the equations of motion

M0∑ = J0α, J0 = ml 2


The problem is one dimensional, hence a scalar equation results

20 ( ) sin ( ) ( ) sin ( ) 0

restoringforce

J t mgl t m t mg tα θ θ θ= − ⇒ + =

This is a second order, nonlinear ordinary differential equation

We can linearize the equation by using the approximation

Here the over dots denote differentiation with respect to time t

sinθ ≈ θ

2 ( ) ( ) 0 ( ) ( ) 0gm t mg t t tθ θ θ θ⇒ + = ⇒ + =

Requires knowledge of (0) and (0)θ θ

the initial position and velocity.


Next consider a spring mass system and perform a static experiment:• From strength of

materials recall:

A plot of force versus displacement:

FBD:

linear

nonlinear

experiment ⇒ fk = kx


Free-body diagram and equation of motion

0 0

( ) ( ) ( ) ( ) 0(0) , (0)

mx t kx t mx t kx tx x x v

= − ⇒ + == =

•Newton’s Law:

Again a 2nd order ordinary differential equation

(1.2)


Stiffness and MassVibration is cause by the interaction of two different forces one related to position (stiffness) and one related to acceleration (mass).

mk

xDisplacement

Mass Spring

Stiffness (k)

Mass (m)

fk = −kx(t)

( ) ( )mf ma t mx t= =

Proportional to displacement

Proportional to acceleration

statics

dynamics


Examples of Single-Degree-of-Freedom Systems

m

θ

l =length

Pendulum

θ(t) +

gl

θ(t) = 0

Gravity g

Shaft and Disk

θ

( ) ( ) 0J t k tθ θ+ =

Torsional Stiffness

kMoment of inertia

J


Solution to 2nd order DEs

x(t) = Asin(ωnt + φ)

Lets assume a solution:

Differentiating twice gives:

2 2

( ) cos( )

( ) sin( ) - ( )n n

n n n

x t A t

x t A t x t

ω ω φ

ω ω φ ω

= +

= − + =

Substituting back into the equations of motion gives:

−mωn2Asin(ω nt + φ) + kAsin(ω nt + φ) = 0

−mωn2 + k = 0 or ωn =

km

Natural frequency

t

x(t)

rad/s


Summary of simple harmonic motion

t

x(t)

x0Slope

here is v0

nωφ

Period

T =2πωn

Amplitude A

Maximum Velocity

Anω

fn =ωn rad/s

2π rad/cycle=

ωn cycles2π s

=ωn

2πHz


Initial ConditionsIf a system is vibrating then we must assume that something must have (in the past) transferred energy into to the system and caused it to move. For example the mass could have been:

•moved a distance x0 and then released at t = 0 (i.e. given Potential energy) or

•given an initial velocity v0 (i.e. given Kinetic energy) or

•Some combination of the two above cases

From our earlier solution we know that:

0

0

(0) sin( 0 ) sin( ) (0) cos( 0 ) cos( )

n

n n n

x x A Av x A A

ω φ φω ω φ ω φ

= = + =

= = + =


Initial ConditionsSolving these equation gives:

2 2 2 1 00 0

0

Amplitude Phase

1 , tan nn

n

xA x vv

ωω φω

− ⎛ ⎞= + = ⎜ ⎟

⎝ ⎠

φ

v0ωn

x0

1ωn

ωn2x0

2 + v02

t

x(t)

x0

Slope here is v0

nωφ


The total solution for the spring mass system is

x(t) =ωn

2x02 + v0

2

ωn

sin ωnt + tan−1 ωnx0

v0

⎛⎝⎜

⎞⎠⎟

(1.10)

Called the solution to a simple harmonic oscillatorand describes oscillatory motion, or simple harmonic motion.

Note (Example 1.1.2)

x(0) =ωn

2x02 + v0

2

ω n

ω nx0

ωn2x0

2 + v02

= x0

as it should


A note on arctangents• Note that calculating arctangent from a

calculator requires some attention. First, all machines work in radians.

• The argument atan(-/+) is in a different quadrant then atan(+/-), and usual machine calculations will return an arctangent in between -π/2 and +π/2, reading only the atan(-) for both of the above two cases.

φ

φ

In MATLAB, use the atan2(x,y) function to getthe correct phase.

+

+

-

_+

+

-

-


Example 1.1.3 wheel, tire suspension m=30 kg, ωn= 10 hz, what is k?

( )2 5cylce 2 rad30 kg 10 1.184 10 N/msec cylcenk m πω

⎛ ⎞= = = ×⎜ ⎟

⎝ ⎠i

There are of course more complex models of suspension systemsand these appear latter in the course as our tools develope


Section 1.2 Harmonic Motion

T =2π rad

ωn rad/s=

2π ωn

s (1.11)

The natural frequency in the commonly used units of hertz:

The period is the time elapsed to complete one complete cylce

fn =ωn

2π=

ω n rad/s2π rad/cycle

=ω n cycles

2π s=

ωn

2π Hz (1.12)

For the pendulum:ωn =

gl

rad/s, T = 2πlg

s

ωn =kJ

rad/s, T = 2πJk

s

For the disk and shaft:


Relationship between Displacement, Velocity and Acceleration

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-1

0

1

x

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-20

0

20

v

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-200

0

200

Time (sec)

a

A=1, ωn=12

x(t) = Asin(ωnt + φ)

x(t) = ωnAcos(ωnt + φ)

x(t) = −ωn2Asin(ωnt + φ)

Note how the relative magnitude of each increases for ωn>1

Displacement

Velocity

Acceleration


Example 1.2.1Hardware store spring, bolt: m= 49.2x10-3

kg, k=857.8 N/m and x0 =10 mm. Compute ωn and the max amplitude of vibration.

ωn = km

= 857.8 N/m49.2 ×10-3 kg

=132 rad/s

fn =ωn

2π= 21 Hz

T =2πωn

=1fn

=1

21cylessec

0.0476 s

x(t)max = A =1

ωn

ωn2x0

2 + v02 = x0 =10 mm

0

Note: commonUnits are Hertz

To avoid Costly errors use fnwhen working in Hertz and ωnwhen in rad/s

Units depend on system


Compute the solution and max velocity and acceleration

v(t)max = ωnA =1320 mm/s = 1.32 m/s

a(t)max =ωn

2 A =174.24 ×103 mm/s2

=174.24 m/s2 ≈ 17.8g!

φ = tan−1 ωnx0

0⎛ ⎝ ⎜

⎞ ⎠ ⎟ =

π2

rad

x(t) =10 sin(132t + π / 2) =10 cos(132t) mm

g = 9.8 m/s2

2.92 mph

90°

~0.4 in max


Does gravity matter in spring problems?

Let Δ be the deflection caused byhanging a mass on a spring(Δ = x1-x0 in the figure)

Then from static equilibrium: mg = kΔNext sum the forces in the vertical for some point x > x1 measured from Δ:

( )0

( ) ( ) 0

mx k x mg kx mg k

mx t kx t=

= − + Δ + = − + − Δ

⇒ + =

So no, gravity does not have an effect on the vibration(note that this is not the case if the spring is nonlinear)


Example 1.2.2 Pendulums and measuring g

• A 2 m pendulum swings with a period of 2.893 s

• What is the acceleration due to gravity at that location?

g =4π 2

T 2 l =4π 2

2.8932 s2 2 m

⇒ g = 9.434 m/s2

This is g in Denver, CO USA, at 1638m and a latitude of 40°


Review of Complex Numbers and Complex Exponential (See Appendix A)

ℜ

ℑ

b

a

A

c = a + jb = Ae jθ

A complex number can be written with a real and imaginary part or as a complex exponential

Where

a = Acosθ,b = AsinθMultiplying two complex numbers:

c1c2 = A1A2ej (θ1 +θ2 )

Dividing two complex numbers:

c1

c2

=A1

A2

e j (θ1 −θ2 )


Equivalent Solutions to 2nd order Differential Equations (see Window 1.4)

x(t) = Asin(ωnt + φ)x(t) = A1 sinωnt + A2 cosωntx(t) = a1e

jωnt + a2e− jωnt

All of the following solutions are equivalent:

The relationships between A and φ, A1 and A2, and a1 and a2can be found in Window 1.4 of the course text, page 17.

Sometimes called the Cartesian form

Called the magnitude and phase form

Called the polar form

•Each is useful in different situations•Each represents the same information•Each solves the equation of motion


Derivation of the solution

2

2

1 2

1 2

Substitute ( ) into 000

( ) and ( )

( )

n n

n n

t

t t

n

jt jt

jt jt

x t ae mx kxm ae kae

m k

k k j jm m

x t a e x t a e

x t a e a e

λ

λ λ

ω ω

ω ω

λ

λ

λ ω

−

−

= + = ⇒

+ = ⇒

+ = ⇒

= ± − = ± = ± ⇒

= = ⇒

= +

This approach will be used again for more complicated problems

(1.18)


Is frequency always positive?

From the preceding analysis, λ = + ωn then

x(t) = a1eωn jt + a2e

−ωn jt

Using the Euler relations for trigonometric functions, theabove solution can be written as (recall Window 1.4)

x(t) = Asin ωnt + φ( ) (1.19)It is in this form that we identify as the natural frequency ωnand this is positive, the + sign being used up in the transformationfrom exponentials to the sine function.


Calculating RMS

A = peak value

x = limT →∞

1T

x(t)dt = average value 0

T

∫

x 2 = limT →∞

1T

x2 (t)dt0

T

∫ = mean-square value

xrms = x 2 = root mean square valueProportional

to energy

Not very useful since for a sine function the

average value is zero

May need to be limited due to physical constraints

Also useful when the vibration is random

(1.21)


The Decibel or dB scaleIt is often useful to use a logarithmic scale to plot vibration levels (or noise levels). One such scale is called the decibel or dB scale. The dB scale is always relative to some reference value x0. It is define as:

dB = 10 log10xx0

⎛

⎝⎜⎞

⎠⎟

2

= 20 log10xx0

⎛

⎝⎜⎞

⎠⎟ (1.22)

For example: if an acceleration value was 19.6m/s2 then relative to 1g (or 9.8m/s2) the level would be 6dB,

10 log1019.69.8

⎛⎝⎜

⎞⎠⎟

2

= 20 log10 2( )= 6dB

Or for Example 1.2.1: The Acceleration Magnitudeis 20log10(17.8)=25dB relative to 1g.


1.3 Viscous DampingAll real systems dissipate energy when they vibrate. To account for this we must consider damping. The most simple form of damping (from a mathematical point of view) is called viscous damping. A viscous damper (or dashpot) produces a force that is proportional to velocity.

Damper (c)

( ) ( )cf cv t cx t= − = − x

fc

Mostly a mathematically motivated form, allowinga solution to the resulting equations of motion that predictsreasonable (observed) amounts of energy dissipation.


Differential Equation Including Damping

Mkx

Displacement

c

For this damped single degree of freedom system the force actingon the mass is due to the spring and the dashpot i.e. fm= - fk - fc.

( ) ( ) ( ) or

( ) ( ) ( ) 0 (1.25)

mx t kx t cx t

mx t cx t kx t

= − −

+ + =

To solve this for of the equation it is useful to assume a solution of the form:

x(t) = aeλt


Solution to DE with damping included (dates to 1743 by Euler)

ae(t)x ae(t)xt

t

λ

λ

λ

λ2=

=

The velocity and acceleration can then be calculated as:

If this is substituted into the equation of motion we get:

aeλt (mλ2 + cλ + k) = 0 (1.26)Divide equation by m, substitute for natural frequency and assume a non-trivial solution

aeλt ≠ 0 ⇒ (λ2 + cλm +ωn

2) = 0


Solution to DE with Damping Included

ζ =c

2 km (1.30)

For convenience we will define a term known as the damping ratio as:

The equation of motion then becomes:

(λ2 + 2ζωnλ +ωn2) = 0

Solving for λ then gives,

λ1,2 = −ζωn ± ωn ζ 2 − 1 (1.31)

Lower case Greek zeta


Possibility 1. Critically damped motion

Critical damping occurs when ζ=1. The damping coefficient c in this case is given by:

λ1,2 = −1ωn ±ωn 12 −1 = −ωn

definition of criticaldamping coefficient

=1 2 2cr nc c km mζ ω⇒ = = =

The solution then takes the form

x(t) = a1e−ωnt + a2te

−ωnt

Solving for λ then gives,

A repeated, real root

Needs two independent solutions, hence the tin the second term


Critically damped motiona1 and a2 can be calculated from initial conditions (t=0),

x = (a1 + a2t)e−ωnt

⇒ a1 = x0

v = (−ωna1 − ωna2t + a2 )e−ωnt v0 = −ωna1 + a2

⇒ a2 = v0 + ωnx0

0 1 2 3 4-0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

Time (sec)

Dis

plac

emen

t (m

m)

k=225N/m m=100kg and ζ=1

x0=0.4mm v0=1mm/s x0=0.4mm v0=0mm/s x0=0.4mm v0=-1mm/s

• No oscillation occurs• Useful in door

mechanisms, analog gauges


Possibility 2: Overdamped motionAn overdamped case occurs when ζ>1. Both of the roots of the equation are again real.

λ1,2 = −ζωn ±ωn ζ 2 −1

x(t) = e−ζωn t(a1e−ωn t ζ 2 −1 + a2e

ωn t ζ 2 −1)

a1 and a2 can again be calculated from initial conditions (t=0),

a1 = −v0 + ( −ζ + ζ 2 −1)ωnx0

2ωn ζ 2 −1

a2 =v0 + (ζ + ζ 2 −1)ωnx0

2ωn ζ 2 −1

0 1 2 3 4-0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

Time (sec)

Dis

plac

emen

t (m

m)

k=225N/m m=100kg and z=2

x0=0.4mm v0=1mm/s x0=0.4mm v0=0mm/s x0=0.4mm v0=-1mm/s

Slower to respond than critically damped case


Possibility 3: Underdamped motionAn underdamped case occurs when ζ<1. The roots of the equation are complex conjugate pairs. This is the most common case and the only one that yields oscillation.

λ1,2 = −ζωn ±ωn j 1−ζ 2

x(t) = e−ζωn t(a1ejωn t 1−ζ 2

+ a2e− jωnt 1−ζ 2

) = Ae−ζωn t sin(ωdt +φ)

ωd = ωn 1 − ζ 2 (1.37)

The frequency of oscillation ωd is called the damped natural frequency is given by.


Underdamped motionA and φ can be calculated from initial conditions (t=0),

A = 1ωd

(v0 +ζωnx0)2 + (x0ωd)2

φ = tan−1 x0ωd

v0 +ζωnx0

⎛

⎝ ⎜

⎞

⎠ ⎟

0 1 2 3 4 5-1

-0.5

0

0.5

1

Time (sec)

Dis

plac

emen

t

• Gives an oscillating response with exponential decay

• Most natural systems vibrate with and underdamped response

• See Window 1.5 for details and other representations


Example 1.3.1: consider the spring of 1.2.1, if c = 0.11

kg/s, determine the damping ratio of the spring-bolt system.

m = 49.2 × 10−3 kg, k = 857.8 N/m

ccr = 2 km = 2 49.2 ×10−3 ×857.8 = 12.993 kg/s

ζ =cccr

=0.11 kg/s

12.993 kg/s= 0.0085 ⇒

the motion is underdamped and the bolt will oscillate


Example 1.3.2The human leg has a measured natural frequency of around 20 Hz when in its rigid (knee locked) position, in the longitudinaldirection (i.e., along the length of the bone) with a damping ratio of ζ = 0.224. Calculate the response of the tip if the leg bone to an initial velocity of v0 = 0.6 m/s and zero initial displacement (this would correspond to the vibration induced while landing on your feet, with your knees locked form a height of 18 mm) and plot the response. What is the maximum acceleration experienced by the leg assuming no damping?


Solution:

ωn = 201

cycless

2π radcycles

= 125.66 rad/s

ωd =125.66 1− .224( )2 = 122.467 rad/s

A =0.6 + 0.224( ) 125.66( ) 0( )( )2 + 0( ) 122.467( )2

122.467= 0.005 m

φ = tan-1 0( ) ωd( )v0 +ζω n 0( )

⎛

⎝ ⎜ ⎞

⎠ ⎟ = 0

⇒ x t( )= 0.005e−28.148t sin 122.467t( )


Use the undamped formula to get maximum acceleration:

( ) ( )( ) 2222

0

00

2

020

m/s 396.75m/s 66.1256.06.0max

m6.0m

0 ,6.0 ,66.125 ,

==⎟⎟⎠

⎞⎜⎜⎝

⎛−=−=

==

===⎟⎟⎠

⎞⎜⎜⎝

⎛+=

nnn

nn

nn

Ax

vA

xvvxA

ωωω

ωω

ωω

maximum acceleration = 75.396 m/s2

9.81 m/s2 g = 7.68g' s


Here is a plot of the displacement response versus time


Example 1.3.3 Compute the form of the response of an underdamped system using the Cartesian form of the solution given in Window 1.5.

sin(x + y) = sin x sin y + cos x cosy ⇒

x(t) = Ae−ζωnt sin(ω dt + φ) = e−ζωnt (A1 sinωdt + A2 cosωdt)

x(0) = x0 = e0( A1 sin(0) + A2 cos(0)) ⇒ A2 = x0

Ý x = −ζω ne−ζω nt(A1 sinωdt + A2 cosωdt)

+ ωde−ζωnt (A1 cosωdt − A2 sinωdt)

v0 = −ζωn (A1 sin 0 + x0 cos0) + ωd (A1 cos 0 − x0 sin 0)

⇒ A1 = v0 +ζωn x0

ω d

⇒

x(t) = e−ζω nt v0 +ζωn x0

ω d

sinωdt + x0 cosωdt⎛

⎝ ⎜ ⎞

⎠ ⎟

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