Mechanical Vibrations (진동공학)
Ch.1 Introduction to Vibration and the Free Response
Prof. Gyuhae Park
1.1 Introduction: Modeling and Degrees of Freedom
• Recall from your study of statics and physics that a degree of freedom is
the independent parameter needed to describe the configuration of a
physical system
• So, a single degree of freedom system, which is where we start, is a
system whose position in time can be defined by one coordinate.
• In vibration, the coordinate typically used is displacement or position.
x
y
2/43
x
1-2
Degrees of Freedom: The minimum number of coordinates
to specify a configuration
◼ For a single particle confined to a line, one coordinate suffices,
so it has one degree of freedom
3/43
1-3
Degrees of Freedom: The minimum number of coordinates
to specify a configuration
◼ Examples of two degree-of-freedom systems:
© 2011 Mechanical Vibrations Fifth Edition in SI Units
4
1-4
Degrees of Freedom: The minimum number of coordinates
to specify a configuration
◼ Examples of three degree-of-freedom systems:
© 2011 Mechanical Vibrations Fifth Edition in SI Units
5
1-5
Stiffness and Mass
Vibration is cause by the interaction between two different forces
one related to position (stiffness) and one related to acceleration
(mass).
Proportional to displacement
Stiffness (k)
Displacement
fk = -kx(t)
Mass (m)
x
k
statics
fm = ma(t) = mx(t)
m
Mass
Spring
dynamics
Proportional to acceleration
6/43
1-6
Spring-Mass Model
◼ Single DOF Spring-Mass Model
Free Body Diagram
f k = kx + k
=
mg
k
Applying Newton's 2nd Law
 F = mx
mg − k − kx = mx
 mx + kx = 0
Kinetic energy
Inertia
force
Spring
force
(mg = k )
(1)
Potential energy
1-7
Spring-Mass Model – cont’d
Note
① Physics of Vibration = Interplay between Potential Energy and Kinetic Energy
②
 mx + kx = 0
③ Solution - Theory of Differential Equation
- Observation (i.e., experiment)
1-8
Spring-Mass Model – cont’d
◼ Solution to 2nd order ODEs
◆ Let’s assume a solution
x = Cet
x =  2Cet
(2)
◆ Substituting Eq. (2) into Eq. (1) gives :
k
)Cet = 0
m
• For nontrivial solution
( 2 +
 2 +
k
=0
m
1,2 =  −
: Characteristic Equation
k
k
=  − n2 ,  n2 =
m
m
1,2 = in
1,2 : eigenvalues, characteristic values
 n : natural frequency, rad/sec
1-9
Spring-Mass Model – cont’d
◆ General Solution is
x = C1eint + C2e −int
◆ The following relationship
eint = cos  nt + i sin  nt
e−int = cos  nt − i sin  nt
◆ Substituting back into the general solution
x = A1 cos  nt + B1 sin  nt
where
A1 = C1 + C2 ,
B1 = i (C1 − C2 )
◆ This equation becomes
x = A sin(nt +  ) or
x = A cos(nt +  )
where A = A12 + B12 ,
tan  =
A1
= cot 
B1
10
1-10
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 to
find the equations of
motion
å M = J a, J = m
0
0
2
0
11/43
1-11
The problem is one dimensional, hence a scalar equation
results
Here the over dots denote differentiation with respect to time t
This is a second order, nonlinear ordinary differential equation
We can linearize the equation by using the approximation
Requires knowledge of
sinq » q
q (0) and q (0)
the initial position and velocity.
12/43
1-12
Examples of Single-Degree-of-Freedom Systems
Pendulum
Shaft and Disk
l =length
θ
Gravity g
Torsional
Stiffness
k
Moment of
inertia J
m
θ
g
q (t) + q (t) = 0
l
Jq (t) + kq (t) = 0
1-13
Equivalent Solutions to 2nd order
Differential Equations (see Window 1.4)
All of the following homogeneous solutions are equivalent:
x(t) = Asin(w nt + f )
Called the magnitude and phase form
x(t) = A1 sin w nt + A2 cos w nt Sometimes called the Cartesian form
x(t) = a1e jw n t + a2 e- jw n t
Called the polar form
The relationships between A and ϕ, A1 and A2, and a1 and a2 can be
found in Window 1.4 of the text, page 31..
• Each represents the same information
• Each is useful in different situations
• Each solves the equation of motion
1-14
Solution of 2nd order DEs
x(t)
Lets assume a solution:
x(t) = Asin(w nt + f ) (1.3)
t
Differentiating twice gives:
x(t) = w n A cos(w nt + f )
( 1.4)
x(t) = -w n2 Asin(w nt + f ) = -w n2 x(t)
(1.5)
Substituting back into the equations of motion gives:
-mw n2 Asin(w nt + f ) + kAsin(w nt + f ) = 0
-mw + k = 0
2
n
or
wn =
k
m
Natural
frequency
rad/s
1-15
Summary of simple harmonic motion
x(t ) = A sin(nt +  )
x(t)
k
m
Period
T =
x0
n =
2
n
= 2
m
k
Amplitude A
(sec)
Slope here
is v0
t
f
fn =
Maximum
Velocity
wn
wn A
n rad/s
 cycles n
1 1 k
= n
=
Hz = =
2 rad/cycle
2 s
2
T T m
1-16
Initial Conditions
If 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 some kinetic energy) or
• Some combination of the two above cases
• From the solution
x(t ) = A sin( nt +  )
we know that:
x0 = x(0) = Asin(w n 0 + f ) = Asin(f )
v0 = x(0) = w n Acos(w n 0 + f ) = w n A cos(f )
1-17
Initial Conditions Determine the Constants of Integration
x(t ) = A sin( nt +  )
x(0) = x0 , x(0) = v0
Solving these two simultaneous equations for A and ϕ gives:
A=
Slope
here is
v0
x(t)
x0
1
wn
æ w n x0 ö
w x + v , f = tan ç
è v ÷ø
2 2
n 0
0
Amplitude
wn
Phase
1
t
f
-1
2
0
wn
w n2 x02 + v02
x0
f
v0
wn
1-18
Thus the solution for the spring mass system
becomes:
w n2 x02 + v02 æ
-1 w n x0 ö
x(t) =
sin ç w nt + tan
wn
v0 ÷ø
è
(1.10)
Called the solution to a simple harmonic oscillator
and describes oscillatory motion, or simple harmonic motion.
Physical Description: When an undamped system is subjected initial disturbances
(𝑥0 , 𝑣0 ), the system will vibrate at the frequency of 𝜔𝑛 , and its amplitude and phase
determined by 𝑥0 and 𝑣0 .
Equation (1-10) is referred to as Free response of an undamped system
Note (Example 1.1.2)
𝜔𝑛2 𝑥02 + 𝑣02
𝑥(0) =
𝜔𝑛
𝜔𝑛 𝑥0
= 𝑥0
𝜔𝑛2 𝑥02 + 𝑣02
as it should
1-19
Example 1.1.3 wheel, tire suspensio
n m = 30 kg, ωn = 10 hz, what is k?
 cylce 2 rad 
5
k = m = ( 30 kg ) 10
=
1.184

10
N/m

sec cylce 

2
n
There are of course more complex models of suspension systems
and these appear later in the course as our tools develop
1-20
1.2 Harmonic Motion
The period is the time elapsed to complete one complete cylce
2p rad 2p
T=
=
s
w n rad/s w n
(1.11)
The natural frequency in the commonly used units of hertz:
fn =
wn
w n rad/s
w cycles w n
=
= n
=
Hz
2p 2p rad/cycle
2p s
2p
For the pendulum:
wn =
g
rad/s, T = 2p
g
(1.12)
s
For the disk and shaft:
wn =
k
J
rad/s, T = 2p
s
J
k
1-21
Relationship between Displacement, Velocity and
Acceleration
A=1, ωn=12
1
x(t) = Asin(w nt + f )
x
Displacement
-1
0
20
v
Velocity
x(t) = w n Acos(w nt + f )
a
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0.1
0.2
0.3
0.4
0.5 0.6
Time (sec)
0.7
0.8
0.9
1
0
-20
0
200
Acceleration
x(t) = -w n2 Asin(w nt + f )
0
0
-200
0
Note how the relative magnitude of each increases for ωn>1
22/43
If
 n  1 rad / sec,
x  x  x
Acceleration is better to be used
If
 n  1 rad / sec,
x  x  x
Displacement is better to be used
1-22
Example 1.2.1Hardware store spring, bolt: m= 49.2x10 kg, k=
-3
857.8 N/m and x0 =10 mm. Compute ωn and the max amplitude of vibration.
k
857.8 N/m
wn =
=
= 132 rad/s
-3
m
49.2 ´10 kg
wn
fn =
= 21 Hz
2p
2p 1
1
T=
= =
wn fn 21 cyles
x(t ) max = A =
1
wn
Note:
common
Units are
Hertz
To avoid Costly errors use fn
when working in Hertz and
ωn when in rad/s
0.0476 s
sec
0
w x + v = x0 = 10 mm
2 2
n 0
2
0
23/43
Units depend
on system
1-23
Compute the solution and max velocity and acceleration
v(t )max = wn A = 1320 mm/s= 1.32 m/s
a(t ) max = w n A = 174.24 ´10 mm/s
2
3
2
= 174.24 m/s » 17.8g!
g = 9.8 m/s2
p
-1 æw n x0 ö
90°
f = tan ç
÷ = rad
è 0 ø 2
x(t ) = 10 sin(132t + p / 2) = 10 cos(132t ) mm
2
~0.4 in max
24/43
1-24
Example 1.2.2 Pendulums and measuring g
◼ A 2 m pendulum
T=
2p
= 2p
swings with a period
wn
g
of 2.893 s
2
2
4p
4p
◼ What is the
g= 2 =
2m
2
2
acceleration due to
T
2.893 s
gravity at that
2
Þ g = 9.796 m/s
location?
This is g in Denver, CO USA, at 1638m
and a latitude of 40°
25/43
1-25
Spring-Mass Model – cont’d
◼ Calculating RMS
◆ Peak value : A
◆ Average value T
1
x(t )dt = average value
T → T 
0
- Not very useful since for a
sine function the average
value is zero
x = lim
◆ Mean-square value
T
1
x 2 = lim  x 2 (t )dt = mean-square value
T → T
0
◆ Root Mean Square value (RMS)
xrms = x = root mean square value
2
◆ dB (desibel) scale
-Proportional to energy
- Commonly used in specifying
vibration
- Also useful when the vibration is
random
2
 x
 x
dB = 10 log10   = 20 log10  
 x0 
 x0 
Reference value
26
1-26
1.3 Viscous Damping
• Spring-mass model:
x(t ) = A sin( nt +  )
(Oscillates infinitely)
• All real systems dissipate energy when they vibrate. To account for this, we
must consider damping.
• The simplest 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) unit: ( N  s / m)
fc = -cv(t) = -cx(t)
x
fc
1-27
Viscous Damping
◼ Viscous damping force
f c = −cx ; c = damping coefficient ( N  s / m)
◆ Always opposite direction to the motion
◆ Mathematically simple and Linear
◆ Examples : oil-filled dashpot (shock absorber), block of rubber
◼ Responses of Damped and Un-damped Systems
Damped system
Undamped System
1-28
Viscous Damping
◼ Damped Single DOF System
Free body diagram
Displacement
x
k
M
c
◆ Equation of motion
 F = mx
••
− fk − fc = m x
cx
kx
mx + cx + kx = 0,
I.C.
x(0) = x0 , x(0) = v0
(1)
29/43
1-29
Viscous Damping
◼ Solution to DE with damping
◆ Assuming
t
x(t) = ae
◆ The velocity and acceleration can then be calculated as
𝜆𝑡
𝑥(𝑡)
ሶ
= 𝜆𝑎𝑒
𝑥(𝑡)
ሷ
= 𝜆2 𝑎𝑒 𝜆𝑡
◆ Substituting
back Into Eq. (1)
(m 2 + c + k )aet = 0
◆ Again,
aet  0
 m 2 + c + k = 0
•
Characteristic equation : Determine vibration characteristics
◆ Two solutions
1,2 = −
c
1

c 2 − 4km
2m 2m
(3)
discriminant
1-30
Viscous Damping
◼ Standard forms of Single DOF system
mx + cx + kx = 0
(3 parameters m, c, k )
c
k
x + x + x=0
m
m
c
k
= 2n ,
= n 2
m
m
x + 2 n x +  n x = 0
2
(2 parameters
 , n )
31/43
1-31
Viscous Damping
◆ Defining the damping ratio
 =
c
c
c
=
=
ccr 2 mk 2mn
◆ The solutions become
 1,2 = −n  n  2 − 1
Characteristic values will be real or complex depending on (c2 − 4km )
or (  2 − 1)
◆ Discriminant
•
c 2 − 4km  0,   1 Overdamped Motion
No Oscillation
c 2 − 4km = 0,  = 1 Critically Damped motion
No Oscillation
c 2 − 4km  0,   1 Underdamped Motion
Oscillation
◆ Critical damping coefficient
ccr = 2 km = 2mn
32/43
1-32
Viscous Damping
◼ Underdamped motion
0  1
◆ The roots are complex conjugate pairs
1,2 = −n  i n 1 −  2
◆ The solution then takes the form
x(t ) = a1e1t + a2e2t
= e −nt (a1e i 1− nt + a2e −i 1− nt )
2
◆ Using Euler relations and
2
d = 1 −  2 n
x(t ) = Ae −nt sin (d t +  ) :
◆ Initial conditions
A=
x(0) = x0 , x0 = v0
(v0 + n x0 )2 + ( x0d )2
d 2
d = damped natural freq.
,  = tan −1 (
x0d
)
v0 + n x0
33/43
1-33
Viscous Damping
◼ The solution to underdamped motion
0  1
x(t ) = Ae −nt sin (d t +  ):
◆ Initial conditions
A=
(v0 + n x0 )2 + ( x0d )2
d 2
d = 1 −  2 n
d = damped natural freq.
x(0) = x0 , x0 = v0
,  = tan −1 (
x0d
)
v0 + n x0
Physical Description: When an underdamped system is subjected initial
disturbances (𝑥0 , 𝑣0 ), the system will vibrate at the frequency of
𝜔𝑑 (𝐼𝑛𝑠𝑡𝑒𝑑 𝑜𝑓 𝜔𝑛 𝑎𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑐𝑎𝑠𝑒 𝑜𝑓 𝑢𝑛𝑑𝑎𝑚𝑝𝑒𝑑 𝑠𝑦𝑠𝑡𝑒𝑚), and its amplitude
and phase determined by 𝑥0 and 𝑣0 .
This defines the Free response of an underdamped system
34/43
1-34
Viscous Damping
◼ Underdamped motion – cont’d
envelope
:  → determines the rate of decay
e−nt
Note
① In theory, the vibration continues and the amplitude never
goes to zero.
② If  =0, the response becomes the same as an undmped system.
35/43
1-35
Viscous Damping
◼ Critically Damped Motion
 =1
 c2 − 4mk = 0
k=225N/m m=100kg and
=1
0.6
◆ The roots are repeated
◆ The solution takes the form
 x(t ) = (a1 + a2t )e−n t
x(0) = x0 , x0 = v0
a1 = x0 , a2 = v0 + n x0
◆ Initial Conditions
 x(t ) =  x0 + (v0 + n x0 )t  e −n t
Displacement (mm)
1 = 2 = − n
x0=0.4mm v0=1mm/s
0.5
x0=0.4mm v0=0mm/s
x0=0.4mm v0=-1mm/s
0.4
0.3
0.2
0.1
0
-0.1
0
1
2
Time (sec)
3
4
Note
① The smallest value of damping ratio that yields aperiodic motion.
② The damping case which separates nonoscillation from oscillation.
36/43
1-36
Viscous Damping
◼ Overdamped motion
 1
◆ The roots are a pair of distinct reals
k=225N/m m=100kg andz=2
1,2 = −n  n  2 − 1
0.6
x0=0.4mm v0=1mm/s
x0=0.4mm v0=0mm/s
x0=0.4mm v0=-1mm/s
◆
The solution then becomes
x(t ) = a1e1t + a2e2t
=e
−nt
(a1e
◆ Initial conditions
a1 =
a2 =
−n  2 −1 t
+ a2e
n  2 −1 t
)
x(0) = x0 , x0 = v0
−v0 + (− +  2 − 1) n x0
2 n  2 − 1
+v0 + ( +  2 − 1) n x0
2 n  2 − 1
Displacement (mm)
0.5
0.4
0.3
0.2
0.1
0
-0.1
0
1
2
Time (sec)
3
4
Slower to respond than
critically damped case
37/43
1-37
Viscous Damping
Note
① The role of damping
Dissipate vibration energy
Slower to reach to the response
② The smaller damping ratio, the higher overshoot we have and
the slower to reach to the steady-state response.
The higher damping ratio, the overshoot may become smaller
and the faster to reach to the steady-state response.
38/43
1-38
Viscous Damping
Note
② With a fixed value of damping ratio, we may not able to reach
optimal performance
③ Vibration reduction, which will be studied in Chapter 5, will be
concerned about how to change value of damping ratio in a passive or
active manner
39/43
1-39
Torsion Vibration
◼ Torsion
◆ From the material mechanics, the torque T of a rod can be expressed
as
GI
T=
p
L

I p: polar moment of inertia of the rod
T=
G : Shear modulus of rigidity
L : Length of shaft
GJ p
l

◆ Letting the torsional stiffness
Kt =
GJ p
l
◆ Applying Newton’s 2nd law
 M  = J
− Kt = J
+
Jp
 T = Kt
Kt
 =0
J
F.B.D.
+
K t
J
Natural frequency
n =
Kt
J
40/43
1-40
Simple Pendulum
◆ Using Newton’s law
M = J 
0
0

𝐽0 𝜃ሷ 𝑡 = −𝑚𝑔𝑙 sin 𝜃 (𝑡)
ሷ
⇒ 𝑚ℓ2 𝜃(𝑡)
+ 𝑚𝑔ℓ sin 𝜃 (𝑡) = 0

m
J = m 2
𝑟𝑒𝑠𝑡𝑜𝑟𝑖𝑛𝑔
𝑓𝑜𝑟𝑐𝑒
mg
Using the approximation sin  
𝑚𝑔𝑙 sin 𝜃 (𝑡)
≈ 𝑚𝑔𝑙𝜃(𝑡)
ሷ
⇒ 𝑚ℓ2 𝜃(𝑡)
+ 𝑚𝑔ℓ𝜃(𝑡) = 0
• Finally, the equation of motion is
•
ሷ
𝜃(𝑡)
+
𝑔
𝜃(𝑡) = 0
ℓ
① Natural frequency
n =
g
l
②  n is independent of mass, only depends on length.
41/43
1-41
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
c
0.11 kg/s
z=
=
= 0.0085 Þ
ccr 12.993 kg/s
the motion is underdamped
and the bolt will oscillate
42/43
1-42
Example 1.3.2
The human leg has a measured natural frequency of around 20 Hz when in
its rigid (knee locked) position, in the longitudinal direction (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?
43/43
1-43
Solution:
From
x(t ) = Ae
wn =
−nt
sin (d t +  )
A=
(v0 + n x0 )2 + ( x0d )2
d 2
,  = tan −1 (
x0d
)
v0 + n x0
20 cycles 2p rad
= 125.66 rad/s
1
s cycles
w d = 125.66 1- (.224) = 122.467 rad/s
2
(0.6 + (0.224 )(125.66)( 0)) + (0)(122.467)2
2
A=
122.467
= 0.005 m
æ (0)(w d ) ö
÷ =0
f = tan ç
è v0 + zw n (0 )ø
-1
Þ x( t ) = 0.005e -2 8.148t sin(122.467t )
44/43
1-44
Use the undamped formula to get maximum acceleration:
x(t) = Asin(w nt + f )
2
æ
ö
v
A = x 02 + ç 0 ÷ , w n = 125.66, v 0 = 0.6, x 0 = 0
èwn ø
v0
0.6
A=
m=
m
wn
wn
æ 0.6 ö
max( x˙˙) = -w A = -w ç ÷ = (0.6)(125.66 m/s 2 ) = 75.396 m/s 2
è wn ø
2
n
2
n
2
75.396
m/s
maximum acceleration =
g = 7.68 g's
2
9.81 m/s
45/43
1-45
Here is a plot of the displacement response
versus time
46/43
1-46
1.4 Modeling and Energy Methods
◼ Two methods of Modeling
◆ Newton’s 2nd Law
 F = mx
x
M = I 
0
0
◆ Energy methods – for conservative system
An alternative way to determine the equation of motion and an
alternative way to calculate the natural frequency of a system
• Useful if the forces or torques acting on the object or mechanical
part are difficult to be determined
V1 + T1 = V2 + T2  V + T = const.
U1−2 = T2 − T1
•
Conservation of Energy
V1 − V2
①
d (T + V )
= 0 From this, we are able to derive an equation of motion of a system
dt
② If x1
If x2
static equilibrium position,
V1 = 0
maximum displacement,
T2 = 0
If V1  0 ,
(T ) max = (V ) max
T1 = V2
 Tmax = Vmax
⇒ extract natural frequency
47/43
1-47
Modeling and Energy Methods
◼ Potential Energy
◆ The potential energy of mechanical system
U is often stored in “springs”(remember that
F = kx for a spring )
x0
x0
x=0
1 2
Vspring =  F dx =  kx dx = kx0
2
0
0
◼ Kinetic Energy
◆ The kinetic energy of mechanical systems T
is due to the motion of the “mass” in the
system
1
1 2
2
𝑇𝑡𝑟𝑎𝑛𝑠 = 𝑚𝑥ሶ , 𝑇𝑟𝑜𝑡 = 𝐽𝜃ሶ
2
2
x0
k
M
Spring
Mass
48/43
1-48
Modeling and Energy Methods
◼ Deriving the equation of motion from the energy
𝑑
𝑑 1
1
(𝑇 + 𝑉) =
𝑚𝑥ሶ 2 + 𝑘𝑥 2
𝑑𝑡
𝑑𝑡 2
2
=0
x=0
x
k
⇒ 𝑥ሶ 𝑚𝑥ሷ + 𝑘𝑥 = 0
M
◆ Since 𝑥ሶ
cannot be zero for all time, then
∴ 𝑚𝑥ሷ + 𝑘𝑥 = 0
Spring
Mass
◼ Determining the Natural frequency directly from the energy
◆ Assuming the solution is
x = A sin( nt +  )
x =  n A cos( nt +  )
k
k = mn2  n =
1
1
m
V = kA2 T = m( A)2
max

2
max
2
1 2 1
kA = m(n A)2
2
2
n
49/43
1-49
Modeling and Energy Methods
◼ Example 1.4.1 – Natural frequency
◆ Conservative system and no slipping
1
1
𝑇 = 𝐽𝜃ሶ 2 + 𝑚𝑥ሶ 2
2
2
1 2
𝑉 = 𝑘𝑥
2
1 𝑥ሶ 2 1
𝑥 = 𝑟𝜃 ⇒ 𝑥ሶ = 𝑟𝜃ሶ ⇒ 𝑇 =
•
1 (𝜔𝑛 𝐴)2 1
𝐽
+ 𝑚(𝜔𝑛 𝐴)2 =
2
𝑟2
2
From x = A sin( nt +  )
Vmax =
•
𝑟2
+ 𝑚𝑥ሶ 2
2
From x =  n A cos( nt +  ) the max value of T happens at
𝑇𝑚𝑎𝑥 =
•
2
𝐽
Hence,
1
𝐽
𝑚+ 2
2
𝑟
xmax = n A
(𝜔𝑛 𝐴)2
the max value of V happens at
xmax = A
1 2
kA
2
1
𝐽
1
𝑚 + 2 𝜔𝑛2 𝐴2 = 𝑘𝐴2
2
𝑟
2
⇒ 𝜔𝑛 =
𝑘
𝑚+
𝐽
𝑟2
50/43
1-50
Modeling and Energy Methods
◼ Example 1.4.2
◆ Using Energy method

1
1
2
ሶ
𝑇 = 𝐽0 𝜃 = 𝑚ℓ2 𝜃ሶ 2
2
2
J = m 2
V= 𝑚𝑔ℓ(1 − cos 𝜃), the change in elevation is ℓ(1 − cos 𝜃)
•

m
mg
Using the conservation of energy and differentiating
𝑑
𝑑 1
(𝑇 + 𝑉) =
𝑚ℓ2 𝜃ሶ 2 + 𝑚𝑔ℓ(1 − cos 𝜃) = 0
𝑑𝑡
𝑑𝑡 2
𝑚ℓ2 𝜃ሶ 𝜃ሷ + 𝑚𝑔ℓ(sin 𝜃)𝜃ሶ = 0 ⇒ 𝜃ሶ 𝑚ℓ2 𝜃ሷ + 𝑚𝑔ℓ(sin 𝜃) = 0
𝑔
2
ሷ
ሷ
⇒ 𝑚ℓ 𝜃 + 𝑚𝑔ℓ(sin 𝜃) = 0 ⇒ 𝜃(𝑡) + sin 𝜃 (𝑡) = 0
ℓ
•
Finally,
ሷ
⇒ 𝜃(𝑡)
+
𝑔
𝜃(𝑡) = 0
ℓ
⇒ 𝜔𝑛 =
𝑔
ℓ
51/43
1-51
Modeling and Energy Methods
◼ Example 1.4.4 – mass of springs
◆ Assumptions
mass of element dy :
ms
y +dy
ms, k
dy
velocity of element dy: vdy =
y
y

x(t )
m
x(t)
◆ Kinetic energy & potential energy
2
1 m y 
Tspring =  s  x  dy
20
 
1m 
=  s  x2
2 3 
 n =
k
m+
ms
3
1  m  1 
1
Tmass = mx 2  Ttot =   s  + m  x 2
2
2 3  2 
m 
1
 Tmax =  m + s  n2 A2
2
3 
1
U max = kA2
2
52/43
1-52
1.5 More on Springs and Stiffness
◼ Rod – longitudinal motion
◆ The mass of a rod is ignored
◆ From the mechanics of materials
x=
Pl
EA
k =
EA
l
53/43
1-53
Stiffness
◼ Shaft – torsional motion
◆ The mass of a shaft is ignored
k=
GJ p
l
54/43
1-54
Stiffness
◼ Example 1.5.1
ሷ
෍ 𝑀 = 𝐽𝜃ሷ ⇒ 𝐽𝜃(𝑡)
+ 𝑘𝜃(𝑡) = 0
𝑘
ሷ
⇒ 𝜃(𝑡) + 𝜃(𝑡) = 0
𝐽
⇒ 𝜔𝑛 =
𝑘
=
𝐽
𝐺𝐽𝑝
,
ℓ𝐽
𝜋𝑑 4
𝐽𝑝 =
32
For a 2 m steel shaft, diameter of 0.5 cm ⇒
𝜔𝑛 =
𝐺𝐽𝑝
=
ℓ𝐽
(8 × 1010 N/m2 )[𝜋(0.5 × 10−2 m)4 /32]
(2 m)(0.5kg ⋅ m2 )
= 2.2 rad/s
55/43
1-55
Stiffness
◼ Helical Spring
d = diameter of wire
2R= diameter of turns
n = number of turns
x(t)= end deflection
G= shear modulus of spring material
Gd 4
k=
64nR 3
Allows the design of springs
to have specific stiffness
56/43
1-56
Stiffness
◼ Beam
◆ From the mechanics of materials
Pl 3
x=
3EI
k=
3EI
l3
With a mass at the tip: n =
3EI
m 3
57/43
1-57
Stiffness
◼ Example 1.5.2
◆ Model wing as transverse beam
◆ Model fuel as tip mass
◆ Ignore the mass of the wing and see
how the frequency of the system
changes as the fuel is used up
◆ Mass of pod 10 kg empty 1000 kg full
E, I
m
l = 5.2x10-5 m4, E =6.9x109 N/m, l = 2 m
 full =
 empty =
3EI
3(6.9  10 9 )(5.2  10 −5 )
=
ml 3
1000  2 3
= 11.6 rad/s=1.8 Hz
l
x(t)
3EI
3(6.9  10 9 )(5.2  10 −5 )
=
ml 3
10  2 3
= 115 rad/s=18.5 Hz
Hence the natural frequency changes by an order of magnitude
while it empties out fuel.
58/43
1-58
Stiffness
◼ Example 1.5.3 – Rolling motion of a ship
ሷ
𝐽𝜃(𝑡)
= −𝑊𝐺𝑍 = −𝑊ℎ sin 𝜃 (𝑡)
For small angles this becomes
ሷ
𝐽𝜃(𝑡)
+ 𝑊ℎ𝜃(𝑡) = 0
⇒ 𝜔𝑛 =
ℎ𝑊
𝐽
59/43
1-59
Stiffness
Solve for example 1.5.5, 1.5.6
Note
① Springs in series
x = x1 + x2
x1 =
x=
x
F
F
, x2 =
k1
k2
F F
1 1
+ = F( + )
k1 k2
k1 k2
1
1 1
= +
kac k1 k2
F
kac =
=
1
1/ k1+ 1/ k2
k1k2
k1 + k2
(kac  k1 ,
kac  k2)
② Springs in parallel
F = k1 x + k2 x = (k1 + k2 ) x
 kab = k1 + k2
xx
FF
60/43
1-60
1.6 Measurements
◼ Measurement
◆ The quantities required to analyze the vibrating motion of a system
all require measurements
◆ Verifying and improving analytical models
◆ Single DOF systems
mx + cx + kx = 0, I + ct + kt = 0
Mass : usually pretty easy to measure using a balance – static
experiment
• Stiffness : can be measured statically by using a simple
displacement measurement and knowing the applied force
• Damping : can only be measured dynamically
•
m, k ( kt )
: can be measured statically
I , c ( ct )
: must be measured by vibration experiments
61/43
1-61
Measurements
◼ Stiffness
◆ From Static deflection
Linear
Nonlinear
F = k x or  = E 
k=
F
x
Deflection or strain
◆ From Dynamic frequency
k
n =
 k = mn2
m
62/43
1-62
Measurements
Example 1.6.1 Use the beam stiffness equation to compute the modulus of
a material
Figure 1.24 l = 1 m, m = 6 kg, I = 10-9 m4 , and measured T = 0.62 s
m 3
T = 2p
= 0.62 s
3EI
4p 2 ( 6 kg ) (1 m )
4p 2 m 3
11
2
ÞE=
=
=
2.05
´
10
N/m
3T 2 I
3(0.62 s)2 10 -9 m 4
3
(
)
1-63
Measurements
◼ Damping Coefficient (underdamped case)
◆ Using the concept of logarithmic decrement
 = ln
x(t )
x(t + T )
x(t)
Ae−nt sin( d t +  )
= ln −n ( t +T )
Ae
sin( d t +  d T +  )
◆ Since
x(t+T)
 d T = 2 ,
t
t+T
Ae−nt sin(d t +  )
 = ln −n (t +T )
Ae
sin(d t + d T +  )
 = ln e T = nT = n
n
 =

4 2 +  2
2
d
= n
2
n 1 −  2
=
2
1−  2
 c =  ccr , ccr = 2 km = 2mn
64/43
1-64
1.8 Stability
s in 2 . t
0.1. t .
0.1. t
◼ Stability is defined for the solution of free response case
1
◆ Stable (marginally stable)
x t
x(t)  M,  t  0
•
s in 2 t
y t
e
x tz t
0.1. t .
0.1. t
e
r t
e
z t x t
◆ Asymptotically Stable
x t
0
0
10
5
yt
yt
z5 t
0
0
10
2
5
1
r t
1
1
0
5
t
5 t
t
0
10
Divergent instability
5
t
10
4
4
2
2
r t
0
2
t
3
2
2
10
1
4
z t
0
3
10
2
yt
10
t
t
1
◆ Unstable : if it is not stable
or asymptotically
unstable
t
0
z t
1
1
1
1
3
z t .x t
r t
4
Example : for Single DOF system, m>0, k>0, c>0
1
5
e
1
lim 𝑥 (𝑡) = 0
5
x tz t
z t .x t
r t
1
x t 𝑡→∞
e
0
1
1
•
x t
t 0.1.system,
st in 2 . t ym>0,
t 0.1. k>0,
x tz t
Example : for. SinglexDOF
c=0
te
.
.
x t
y t
r t
0
5
2
10
4
t
Flutter instability
2
0
10
5
10
4
65/43
t
1-65
1.8 Stability
◼ Instability
◆ Divergent instability
• Growing without bound and not oscillating
• m-k system : k<0
◆ Flutter instability (flutter)
Growing without bound and oscillating
• m-c-k system : c<0
• For a fixed wing
•
mx + cx + kx = F ( x, x, t , va )
Aerodynamic force
= c p x + F1 ( x, t , va )
mx + (c − c p ) x + kx = F1
< 0 : Flutter occurs
66/43
1-66
Stability
Example: 1.8.1: For what values of the spring constant will the response be
stable?
Figure 1.37
æk 2
ö
k 2
2
m q +ç
sin q ÷ cos q - mg sin q = 0 Þ m q +
q - mg q = 0
2
è 2
ø
2
Þ 2m q + ( k - 2mg )q = 0
(for small q )
Þ k > 2mg for a stable response
67/43
1-67
1.9 Numerical Simulation
◼ Numerical Methods for Solving Ordinary Differential Equations
Euler method, Runge-Kutta Method, etc.
◆ Available in FORTRAN, C, Matlab, Mathematica, Maple, Python, Octave etc.
◆ Will use these to examine nonlinear vibration problems that do not have
analytical expressions for solutions
◆
◼ First Order Differential Equation
𝑥(𝑡)
ሶ
= 𝑎𝑥(𝑡),
𝑥(0) = 𝑥0
𝑥𝑖+1 − 𝑥𝑖
= 𝑎𝑥𝑖
Δ𝑡
◆ where
𝑥𝑖 = 𝑥(𝑡𝑖 ), Δ𝑡 = 𝑡𝑖+1 − 𝑡𝑖
◆ Algebraic equation
𝑥𝑖+1 = 𝑥𝑖 1 + 𝑎Δ𝑡
68/43
1-68
Numerical Solution
Example 1.9.1 solve dx/dt = -3x, x(0)=1
a = -3, take Dt = 0.5
x0 = 1
x1 = x0 + (0.5)(-3)(x0 ) = -0.5
x2 = x1 + (0.5)(-3)(x1 ) = 0.25
=
Numerical solution
Note that the numerical
Solution is different for
Each choice of Δt
x(t ) = Ae lt
x˙(t ) = -3x(t ) Þ -l Ae lt = -3Ae lt
Þ l = 3,
x(0) = x0 = 1 = Ae 0 Þ A = 1 so that
Analytical solution
x(t ) = e-3t
69/43
1-69
Numerical Solution
◼ Example 1.9.1
Large errors can be
caused by choosing the
time step to be too large
Small time steps require
more computation
1
Amplitude of x
With time step at 0.5
sec the numerical
solution oscillates about
the exact solution
Numerical solution  t=0.5sec
Numerical solution  t=0.05sec
Exact solution
0.5
0
-0.5
0
1
2
Time (sec)
3
4
70/43
1-70
Numerical Solution
◼ 2nd order Differential Equation of Vibration
Converting the 2nd order equation into two 1st order equation
◆ Defining two new variables
◆
𝑥1 = 𝑥,
◆
𝑥2 = 𝑥ሶ
Substituting these variables into the vibration equation
𝑚𝑥ሷ + 𝑐 𝑥ሶ + 𝑘𝑥 = 0
◆
Then 𝑥ሶ = 𝑥
1
2
𝑐
𝑘
𝑥ሶ 2 = − 𝑥2 − 𝑥1
𝑚
𝑚
◆
State-space equation
0
𝑥ሶ 1
𝑘
=
𝑥
ሶ
−
ถ
2
𝑚
ሶ
𝑋
𝐴
Using Euler method
𝑥 𝑡𝑖+1 = 𝑥 𝑡𝑖 + Δ𝑡𝐴𝑥 𝑡𝑖
or
𝑥𝑖+1 = 𝐼 + Δ𝑡𝐴 𝑥𝑖
1
𝑥1
𝑐
𝑥2
ถ
−
𝑚 𝑋
71/43
1-71
Numerical Simulation
◼ Use of Matlab (ode23, or ode45)
◆ Runge-Kutta Method:
More sophisticated than the Euler method but more accurate
◆ Working for nonlinear equation too
◆ Variable time step Δt
◆ Example of Matlab Program
•
Create Matlab function
function xdot=sdof(t,x)
k=2;c=1;m=3;
A=[0 1;-k/m -c/m];
xdot=A*x;
In the command window
» t0=0;tf=20;
» x0=[0 ; 0.25];
» [t,x]=ode45('sdof',[t0 tf],x0);
» plot(t,x)
Saved as sdof.m
72/43
1-72
Numerical Simulation
◼ Use of Matlab
3 x + x + 2 x = 0,
I.C.
x(0) = 0 , x(0) = 0.25, 0  t  20
0.3
Displacement
Velocity
Amplitude
0.2
0.1
0
-0.1
-0.2
0
5
10
Time (sec)
15
20
73/43
1-73
Numerical Solution
◼ Why use numerical simulation when we can compute the analytical
solution and plot it?
◆ To have a tool that we are confident with that will allow us to solve for
the response when an analytical solution cannot be found
◆ Nonlinear systems do not have analytical solutions, but can be
simulated numerically
74/43
1-74
1.10 Coulomb Frication
◼ Coulomb Friction
−𝜇𝑁
𝑓𝑐 (𝑥)
ሶ =൞ 0
𝜇𝑁
x0
x(t)
𝑥(𝑡)
ሶ
>0
𝑥(𝑡)
ሶ
=0
𝑥(𝑡)
ሶ
<0
k
m
◆ The force due to Coulomb friction oppose motion
N=mg
◆ Using the “sgn” function
𝑚𝑥ሷ + 𝜇𝑚𝑔 sgn( 𝑥)
ሶ + 𝑘𝑥 = 0
⇒ Nonlinear Vibration
75/43
1-75
Coulomb Friction
◼ Response
◆ Nonlinear
◆ Can be solved as piecewise
linear
◆ Can be solved numerically
◆ More than one equilibrium
position
◆ Linear decay rather than the
exponential
◆ Comes to rest when spring
cannot overcome friction at the
instant the velocity is zero
76/43
1-76
Coulomb Friction
◼ Analytical Solution – Piecewise linear
𝑚𝑥ሷ + 𝜇𝑚𝑔 sgn( 𝑥)
ሶ + 𝑘𝑥 = 0,
◆
𝑥 0 = 𝑥0 > 0,
𝑥ሶ 0 = 0
The velocity is negative
𝑚𝑥ሷ + 𝑘𝑥 = 𝜇𝑚𝑔
•
The form of its general solution is
𝑥 𝑡 = 𝐴1 cos 𝜔𝑛 𝑡 + 𝐵1 sin 𝜔𝑛 𝑡 +
•
𝜇𝑚𝑔
𝐴1 = 𝑥0 −
𝑘
•
Thus, the solution is
and B1 = 0 since
𝜇𝑚𝑔
𝑥 0 = 𝐴1 +
= 𝑥0
𝑘
𝑥ሶ 0 = 𝜔𝑛 𝐵1 = 0
𝑥 𝑡 = 𝑥0 −
•
𝜇𝑚𝑔
𝑘
𝜇𝑚𝑔
𝜇𝑚𝑔
cos 𝜔𝑛 𝑡 +
𝑘
𝑘
Since 𝑡1 = 𝜋/𝜔𝑛 , 𝑥ሶ 𝑡 = 0
the above solution can be applied
when
𝑡 < 𝑡1 = 𝜋/𝜔𝑛
77/43
1-77
Coulomb Friction
◼ Analytical Solution – cont’d
◆ The velocity is positive
𝑥 𝑡 = 𝐴2 cos 𝜔𝑛 𝑡 + 𝐵2 sin 𝜔𝑛 𝑡 −
•
𝜇𝑚𝑔
𝑘
From the previous solution
2𝜇𝑚𝑔
𝜇𝑚𝑔
− 𝑥0 = −𝐴2 −
𝑘
𝑘
= 0 = −𝜔𝑛 𝐵2
𝑥 𝜋/𝜔𝑛 =
𝑥ሶ 𝜋/𝜔𝑛
•
The solution is
𝑥 𝑡 = 𝑥0 −
3𝜇𝑚𝑔
𝜇𝑚𝑔
cos 𝜔𝑛 𝑡 −
𝑘
𝑘
𝜋
2𝜋
<𝑡<
𝜔𝑛
𝜔𝑛
78/43
1-78
Nonlinear Vibration
Example 1.10.2
◼ Equilibrium Positions of Nonlinear Systems
◆ Equations of motion
x + x −  2 x3 = 0
◆ State form
𝑥ሶ 1 = 𝑥2
𝑥ሶ 2 = 𝑥1 (𝛽 2 𝑥12 − 1)
◆ Equilibrium positions
x2 = 0
x1 ( 2 x12 − 1) = 0 
1
0   
xe =   ,  ,
0   0 
 
 −1 

 
 0 
Multiple equilibrium
points possible
79/43
1-79
Nonlinear Vibration
Example 1.10.3
◼ Equilibrium points of pendulum
𝑔
ሷ𝜃(𝑡) + sin 𝜃 (𝑡) = 0
ℓ
𝑥1 = 𝜃,
𝑥2 = 𝜃ሶ
◆ State-space form
𝑥ሶ 1 = 𝑥2
𝑔
𝑥ሶ 2 = − sin 𝑥1
ℓ
◆ Equilibrium position
𝑥2
⇒ 𝐅 𝐱 = − 𝑔 sin 𝑥 = 0
1
ℓ
⇒
𝑥2 = 0
sin 𝑥1 = 0
⇒ 𝑥2 = 0 and 𝑥1 = 𝑛𝜋,
𝑛 = 0,1,2 ⋯
80/43
1-80
Nonlinear Vibration
Example 1.10.4
◼ Solution to the pendulum
◆ Can use numerical simulation to examine both linear and nonlinear
response
◆ Let (g/L)=(0.1)2 so that wn = 0.1
◆ a) use q(0)=0.3 rad & initial vel: 0.3 rad/s
◆ b) change the initial position to: q(0)= π rad which is near the
unstable equilibrium
(b)
(a)
1.5
30
Non-linear
Linear
1
Non-linear
Linear
20
Angle (q)
Angle (q)
0.5
0
-0.5
10
0
-1
-1.5
0
10
20
30
Time (sec)
40
50
-10
0
10
81/43
20
30
Time (sec)
40
50
1-81
Summary of Nonlinear Vibration
◼ Additional phenomena over linear case
◼ Multiple equilibrium
◼ Instabilities possible with positive coefficients
◼ Form of response dependent on initial conditions
◼ Closed form solutions usually not available
◼ Can simulate numerically
◼ Linear model has tremendous advantages
◼ Linear combination of inputs yields linear combination of outputs
◼ Linear ODE techniques very powerful
◼ But don’t make a design error by ignoring important nonlinear situations
◼ All systems have nonlinear ranges of operation
◼ Need to sort out when nonlinearity is important to consider and when to
ignore it
82/43
1-82