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Engineering Vibration3rd Edition
Prentice Hall© D. J. [email protected],
We make our living in dynamics, vibration and control
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Motivation The Millennium Bridgea Recent Vibration Problem (2000-1)
OPENED AND CLOSED WITHIN A VIEW HOURSBECAUSE OF UNDESIRABLE VIBRATION
(Show movies)
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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.
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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
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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.
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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
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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)
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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
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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
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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
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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
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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
= = + =
= = + =
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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
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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
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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.
+
+
-
_+
+
-
-
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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
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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:
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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
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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
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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
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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)
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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°
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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 )
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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
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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)
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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.
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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)
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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.
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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.
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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
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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
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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
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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
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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
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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
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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.
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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
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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
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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?
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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( )
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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
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Here is a plot of the displacement response versus time
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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⎛
⎝ ⎜ ⎞
⎠ ⎟