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Chapter 2 Response to Harmonic Excitation
Introduces the important concept of resonance
2.1 Harmonic Excitation of Undamped Systems
• Consider the usual spring mass damper system with
applied force F(t)=F0cosωt
• ω is the driving frequency
• F0 is the magnitude of the applied force
• We take c = 0 to start with
M
F=F0cosωt
Displacement
x
k
0
2
0
00
( ) ( ) cos( )
( ) ( ) cos( )
where ,
n
n
mx t kx t F t
x t x t f t
F kfm m
Equations of motion
( ) cos( )px t X t
• Solution is the sum of
homogenous and
particular solution
• The particular solution
assumes form of forcing
function (physically the
input wins):
Substitute particular solution into the equation of motion:
( ) cos( )px t X t
2
2 2
0
0
2 2
cos cos cos
solving yields:
p n px x
n
n
X t X t f t
fX
Thus the particular solution has the form:
0
2 2( ) cos( )p
n
fx t t
Add particular and homogeneous solutions to get general
solution:
01 2 2 2
homogeneous
1 2
( )
sin cos cos
and are constants of integration.
particular
n n
n
x t
fA t A t t
A A
(2.8)
Apply the initial conditions to evaluate the constants
0 01 2 2 02 2 2 2
02 0 2 2
01 2 1 02 2
01
0
(0) sin 0 cos0 cos0
(0) ( cos0 sin 0) sin 0
( )
n n
n
n n
n
n
f fx A A A x
fA x
fx A A A v
vA
vx t
0 00 2 2 2 2
sin cos cosn n
n n n
f ft x t t
(2.11)
Comparison of free and forced response
• Sum of two harmonic terms of different frequency
• Free response has amplitude and phase effected by forcing
function
• Our solution is not defined for ωn = ω because it produces
division by 0.
• If forcing frequency is close to natural frequency the
amplitude of particular solution is very large
Response for m=100 kg, k=1000 N/m, F=100 N, ω = ωn +5
v0=0.1m/s and x0= -0.02 m.
Note the obvious presence of two harmonic signals
0
Displacement (x)
Time (sec)
0
What happens when ω is near ωn?
0
2 2
2( ) sin sin (2.13)
2 2
n n
n
fx t t t
When the drive frequency and natural frequency
are close a beating phenomena occurs
0
2 2
2sin
2
n
n
ft
Larger
amplitude
Time (sec)
Displacement (x)
What happens when ω is ωn?
0
( ) sin( )
substitute into eq. and solve for
2
px t tX t
X
fX
When the drive frequency and natural
frequency are the same the amplitude
of the vibration grows without bounds.
This is known as a resonance
condition.
The most important
concept in Chapter 2!
0
2 2
2sin
2
n
n
ft
0 5 10
15
20
25
x(t) = A1 sinwt + A2 coswt +
f0
2wt sin(wt)
grows with out bound
Example 2.1.1: Compute and plot the response for
m=10 kg, k=1000 N/m, x0=0,v0=0.2 m/s, F=23 N, ω =2ωn.
00
30
2 2 2 2 2 2
1000 N/m10 rad/s, 2 20 rad/s
10 kg
23 N 0.2 m/s2.3 N/kg, 0.02 m
10 kg 10 rad/s
2.3 N/kg7.9667 10 m
(10 20 ) rad / s
Equation (2.11) then yields:
( ) 0.02sin10 7.96
n n
n
n
k
m
vFf
m
f
x t t
367 10 (cos10 cos 20 )t t
Example 2.1.2 Given zero initial conditions a
harmonic input of 10 Hz with 20 N magnitude and k=
2000 N/m, and measured response amplitude of 0.1m,
compute the mass of the system.
0
2 2
0
2 2
0.1 m
0
2 2 2
( ) cos 20 cos for zero initial conditions
2trig identity ( ) sin sin
2 2
2 2(20 / )0.1 0.1
2000 (20 )
0.405 kg
n
n
n n
n
n
fx t t t
fx t t t
f m
m
m
Example 2.1.3 Design a rectangular mount for a
security camera.
0.02 x 0.02 m
in cross section
Compute l > 0.5 m so that the mount keeps the camera from
vibrating more then 0.01 m of maximum amplitude under a
wind load of 15 N at 10 Hz. The mass of the camera is 3 kg.
Solution: Modeling the mount and camera as a beam
with a tip mass, and the wind as harmonic, the equation
of motion becomes:
03
3( ) cos
EImx x t F t
From strength of materials:
3
12
bhI
Thus the frequency expression is: 3 3
2
3 3
3
12 14n
Ebh Ebh
m m
Here we are interested computing l that will make the
amplitude less then 0.01m:
2 20
2 2
0
2 22 20
2 2
2( ) 0.01 , for 0
20.01
2( ) 0.01, for 0
n
n
nn
n
fa
f
fb
Case (a) (assume aluminum for the material):
32 2 20
0 02 2 3
33
2
0
20.01 2 0.01 0.01 0.01 2 0.01
4
0.01 0.02 0.272 m4 (0.01 2 )
n
n
f Ebhf f
m
Ebh
m f
Case (b):
32 2 20
0 02 2 3
33
2
0
20.01 2 0.01 0.01 2 0.01 0.01
4
0.01 0.012 0.229 m4 (2 0.01 )
n
n
f Ebhf f
m
Ebh
m f
3(2.7 10 )(0.22)(0.01)(0.01)
0.149 kg
m bh
Remembering the constraint that the length must be at
least 0.2 m, (a) and (b) yield
0.2 0.229, or choose = 0.22 m To check, note that
3
22 2
3
320 1609 0
12n
Ebh
m
A harmonic force may also be represented by sine or a
complex exponential. How does this
change the solution?
2
0 0( ) ( ) sin or ( ) ( ) sin (2.18)nmx t kx t F t x t x t f t
The particular solution then becomes a sine:
( ) sin (2.19)px t X t
Substitution of (2.19) into (2.18) yields:
0
2 2( ) sinp
n
fx t t
Solving for the homogenous solution and evaluating the constants yields
0 0 00 2 2 2 2
( ) cos sin sin (2.25)n n
n n n n
v f fx t x t t t
Section 2.2 Harmonic Excitation of Damped Systems
Extending resonance and response calculation to damped
systems
0
2
0
now includes a phase shift
( ) ( ) ( ) cos
( ) 2 ( ) ( ) cos
( ) cos ( )
n n
p
mx t cx t kx t F t
x t x t x t f t
x t X t
2.2 Harmonic excitation of damped systems
(2.26)
(2.27)
M
x
k c
Displacement
F=F0cosωt
Let xp have the form:
2 2 1
2 2
( ) cos sin
, tan
sin cos
cos sin
p s s
ss s
s
p s s
p s s
x t A t B t
BX A B
A
x A t B t
x A t B t
Note that we are using the rectangular form, but we could use one of the
other forms of the solution.
Substitute into the equations of motion
2 2
0
2 2
2 2
0
2 2
( 2 )cos
2 sin 0
for all time. Specifically for 0,2 /
( ) (2 )
( 2 ) ( ) 0
s n s n s
s n s n s
n s n s
n s n s
A B A f t
B A B t
t
A B f
A B
Write as a matrix equation:
2 2
0
2 2
( ) 2
02 ( )
sn n
sn n
A f
B
Solving for As and Bs:
2 2
0
2 2 2 2
( )
( ) (2 )
ns
n n
fA
0
2 2 2 2
2
( ) (2 )
ns
n n
fB
Substitute the values of As and Bs into xp:
10
2 22 2 2 2
2( ) cos( tan )
( ) (2 )
np
nn n
X
fx t t
Add homogeneous and particular to get total solution:
particular or homogeneous or transient solutionsteady state solution
( ) sin( ) cos( )nt
dx t Ae t X t
Note: that A and Φ will not have the same values as in Ch 1,
for the free response. Also as t gets large, transient dies out.
Things to notice about damped forced response
• If ζ = 0, undamped equations result
• Steady state solution prevails for large t
• Often we ignore the transient term (how large is ζ, how
long is t?)
• Coefficients of transient terms (constants of integration)
are effected by the initial conditions AND the forcing
function
• For underdamped systems at resonance the, amplitude
is finite.
(0) 0.05
sin 0.133cos( 0.013)
(0) 0 0.01 sin
9.999 cos 0.665sin 0.013
x
A
v A
A
Example 2.2.3: ωn = 10 rad/s, ω = 5 rad/s, ζ = 0.01, F0= 1000 N, m =
100 kg, and the initial conditions x0 = 0.05 m and v0 = 0.
Compare amplitude A and phase Φ for forced and unforced case:
Using the equations on slide 23:
0.1
0.133, 0.013
( ) sin(9.99 ) 0.133cos(5 0.013)t
X
x t Ae t t
Differentiating yields:
0.1
0.1
( ) 0.01 sin(9.999 )
9.999 cos(9.999 )
0.665sin(5 0.013)
applying the intial conditions :
0.083 (0.05), 1.55 (1.561)
t
t
v t Ae t
Ae t
t
A
Proceeding with ignoring the transient
• Always check to make sure the transient is not significant
• For example, transients are very important in
earthquakes
• However, in many machine applications transients may be
ignored
2
2 2 20 0
1
(1 ) (2 )
nXXk
F f r r
Proceeding with ignoring the transient
Magnitude:
0
2 2 2 2( ) (2 )n n
fX
(2.39)
Frequency ratio: n
r
Non dimensional
Form:
(2.40)
Phase: 12
2tan1
rr
2 2 2
1
(1 ) (2 )X
r r
0
0.5
1
1.5
2
Magnitude plot
• Resonance is close to r = 1
• For ζ = 0, r =1 defines
resonance
• As ζ grows resonance moves
r <1, and X decreases
• The exact value of r, can be
found from differentiating the
magnitude
12
2tan1
rr
Phase plot
• Resonance occurs at Φ =
π/2
• The phase changes more
rapidly when the damping
is small
• From low to high values
of r the phase always
changes by 1800 or π
radians
0 0.5 1 1.5 2 0
0.5
1
1.5
2
2.5
3
3.5
r
Phase (rad)
=0.1
=0.3
=0.5
=1
Fig 2.7
2 2 20
2
peak
20 max
10
(1 ) (2 )
1 2 1 1/ 2
1
2 1
d Xk d
dr F dr r r
r
Xk
F
Example 2.2.3 Compute max peak by differentiating:
(2.41)
(2.42)
Effect of Damping on Peak Value
• The top plot shows how
the peak value becomes
very large when the
damping level is small
• The lower plot shows how
the frequency at which the
peak value occurs reduces
with increased damping
• Note that the peak value is
only defined for values
ζ<0.707
0 0.2 0.4 0.6 0.8 0
5
10
15
20
25
30
ζ
0 0.2 0.4 0.6 0.8 0
0.2
0.4
0.6
0.8
1
rpeak
(dB)0F
Xk
Fig 2.9
ζ
Section 2.3 Alternative Representations
• A variety methods for solving differential equations
• So far, we used the method of undetermined coefficients
• Now we look at 3 alternatives:
a geometric approach
a frequency response approach
a transform approach
• These also give us some insight and additional useful tools.
2.3.1 Geometric Approach
• Position, velocity and acceleration phase shifted each by
π/2
• Therefore write each as a vector
• Compute X in terms of F0 via vector addition
Re
Im
)cos( qw -tkX
A
B
C
D
E
kX
XcwXm 2w
0F
C
B A
Xmk )( 2w-
Xcw0F
Using vector addition on the diagram:
2 2 2 2 2 2
0
0
2 2 2
( ) ( )
( ) ( )
F k m X c X
FX
k m c
At resonance:
0, 2
FX
c
real part imaginary part
0
harmonic input
cos ( sin )
( ) ( ) ( )
j t
j t
Ae A t A t j
mx t cx t kx t F e
2.3.2 Complex response method
(2.47)
(2.48)
Real part of this complex solution corresponds to the physical
solution
2
0
002
2
the frequency response function
( )
( )
( )( ) ( )
1( )
( ) ( )
j t
p
j t j t
x t Xe
m cj k Xe F e
FX H j F
k m c j
H jk m c j
Choose complex exponential as a solution
(2.49)
(2.50)
(2.51)
(2.52)
Note: These are all complex functions
0
2 2 2
1
2
( )0
2 2 2
( ) ( )
tan
( )( ) ( )
j
j t
p
FX e
k m c
c
k m
Fx t e
k m c
Using complex arithmetic:
(2.53)
(2.54)
(2.55)
Has real part = to previous solution
Comments:
• Label x-axis Re(ejωt
) and y-axis Im(ejωt
) results in the
graphical approach
• It is the real part of this complex solution that is physical
• The approach is useful in more complicated problems
Example 2.3.1: Use the frequency response approach to
compute the particular solution of an undamped system
The equation of motion is written as
2
0 0
2 2
0
0
2 2
( ) ( ) ( ) ( )
Let ( )
j t j t
n
j t
p
j t j t
n
n
mx t kx t F e x t x t f e
x t Xe
Xe f e
fX
2.3.3 Transfer Function Method
The Laplace Transform
• Changes ODE into algebraic equation
• Solve algebraic equation then compute the inverse
transform
• Rule and table based in many cases
• Is used extensively in control analysis to examine the
response
• Related to the frequency response function
The Laplace Transform approach:
• See appendix B and section 3.4 for details
• Transforms the time variable into an algebraic, complex
variable
• Transforms differential equations into an algebraic
equation
• Related to the frequency response method
0
( ) ( ( )) ( ) stX s x t x t e dt
L
Take the transform of the equation of motion:
0
2 0
2 2
cos
( ) ( )
mx cx kx F t
F sms cs k X s
s
Now solve algebraic equation in s for X(s)
0
2 2 2( )
( )( )
F sX s
ms cs k s
To get the time response this must be “inverse
transformed”
2
2
2
( ) ( ) ( )
( ) 1( )
( )
1( )
frequency response function
ms cs k X s F s
X sH s
F s ms cs k
H jk m c j
Transfer Function Method
With zero initial conditions:
(2.59)
(2.60)
The transfer
function
Example 2.3.2 Compute forced response of the
suspension system shown using the Laplace transform
Summing moments about the shaft:
0( ) ( ) sinJ t k t aF t
Taking the Laplace transform
2
0 2 2
0 2 2 2
( ) ( )
1( )
Js X s kX s aFs
X s a Fs Js k
Taking the inverse Laplace transform:
1
0 2 2 2
2 2
1( )
1 1 1( ) sin sin , n n
n n
t a F Ls Js k
a F kt t t
J J
Notes on Phase for Homogeneous and Particular
Solutions
Equation (2.37) gives the full solution for a harmonically
driven underdamped SDOF oscillator to be
homogeneous solution particular solution
( ) sin cosnt
dx t Ae t X t
How do we interpret these phase angles?
Why is one added and the other subtracted?
sin dt
1tan o d
o n o
x
v x
0 00 0x v
Non-Zero initial conditions
Zero initial displacement
Zero initial velocity
Phase on Particular Solution
• Simple “atan” gives -π/2 < Φ < π/2
• Four-quadrant “atan2” gives 0 < Φ < π
2.4 Base Excitation
• Important class of vibration analysis
– Preventing excitations from passing from a vibrating
base through its mount into a structure
• Vibration isolation
– Vibrations in your car
– Satellite operation
– Disk drives, etc.
)( yxk )yxc(
FBD of SDOF Base Excitation
=- ( - )- ( - )=
+ + = + (2.61)
F k x y c x y mx
mx cx kx cy ky
System Sketch System FBD
M
M x(t)
k c
base
y(t)
SDOF Base Excitation (cont)
Assume: ( ) sin( ) and plug into Equation(2.61)
+ + = cos( ) + sin( ) (2.63)
harmonic forcing functions
y t Y t
mx cx kx c Y t kY t
For a car, 2 2 V
The steady-state solution is just the superposition of the two
individual particular solutions (system is linear).
00
2 2+2 + = 2 cos( ) + sin( ) (2.64)
sc ff
n n n nx x x Y t Y t
Particular Solution (sine term)
With a sine for the forcing function,
2
0
0
22 2 2
2 2
0
22 2 2
+2 + = sin
cos sin sin( )
where
2
( ) 2
( )
( ) 2
n n s
ps s s s s
n ss
n n
n ss
n n
x x x f t
x A t B t X t
fA
fB
Use rectangular form to
make it easier to add
the cos term
Particular Solution (cos term)
With a cosine for the forcing function, we showed
2
0
2 2
0
22 2 2
0
22 2 2
+2 + = cos
cos sin cos( )
where
( )
( ) 2
2
( ) 2
n n c
pc c c c c
n cc
n n
n cc
n n
x x x f t
x A t B t X t
fA
fB
Magnitude X/Y
Now add the sin and cos terms to get the magnitude of
the full particular solution
2 2 2 2
0 0
2 22 2 2 2 2 2
2
0 0
2
22 2
(2 )
( ) 2 ( ) 2
where 2 and
1 (2 )if we define this becomes (2.70)
(1 ) 2n
c s nn
n n n n
c n s n
f fX Y
f Y f Y
rr X Y
r r
2
22 2
1 (2 ) (2.71)
(1 ) 2
X r
Y r r
0
2 2
2sin
2
n
n
ft
W
h
e
n
t
h
e
d
r
i
v
e
f
r
e
q
0 0.5 1 1.5 2 2.5 3 -20
-10
0
2 2
2sin
2
n
n
ft
0
10
20
30
40
Frequency ratio r
X/Y (dB)
e is close to
r = 1
• For ζ = 0,
r =1 defines
resonance
• As ζ
grows
resonance
moves r <1,
and X
The relative magnitude plot
of X/Y versus frequency ratio: Called the Displacement Transmissibility
=0.01
=0.1
=0.3
=0.7
Figure 2.13
From the plot of relative Displacement Transmissibility
observe that:
• X/Y is called Displacement Transmissibility Ratio
• Potentially severe amplification at resonance
• Attenuation for r > sqrt(2) Isolation Zone
• If r < sqrt(2) transmissibility decreases with damping ratio
Amplification Zone
• If r >> 1 then transmissibility increases with damping
ratio Xp~2Yζ/r
Next examine the Force Transmitted to the mass as a function
of the frequency ratio
2
2 2
( ) ( )
At steady state, ( ) cos( ),
so =- cos( )
T
T
F k x y c x y mx
x t X t
x X t
F m X kr X
From FBD
m x(t)
FT
c
k
base y(t)
Plot of Force Transmissibility (in dB) versus frequency ratio
0 0.5 1 1.5 2 2.5 3 -20
-10
0
10
20
30
40 3
.5
Frequency ratio r
F/kY (dB)
=0.1
ζ =0.01
ζ =0.1
ζ =0.3 ζ =0.7
Figure 2.14
Figure 2.16 Comparison between force and displacement
transmissibility
Force
Transmissibility
Displacement
Transmissibility
Example 2.4.2: Effect of speed on the amplitude of
car vibration
Figure 2.17
Model the road as a sinusoidal input to base motion of
the car model
Approximation of road surface:
( ) (0.01 m)sin
1 hour 2 rad(km/hr) 0.2909 rad/s
0.006 km 3600 s cycle
(20km/hr) = 5.818 rad/s
b
b
b
y t t
v v
From the data give, determine the frequency and
damping ratio of the car suspension:
4
4
4 10 N/m6.303 rad/s ( 1 Hz)
1007 kg
2000 Ns/m= 0.158
2 2 4 10 N/m 1007 kg
n
k
m
c
km
From the input frequency, input amplitude, natural
frequency and damping ratio use equation (2.70) to
compute the amplitude of the response:
5.818
6.303
br
2
2 2 2
2
2 22
1 (2 )
(1 ) (2 )
1 2(0.158)(0.923)0.01 m 0.0319 m
1 0.923 2 0.158 0.923
rX Y
r r
What happens as the car goes faster? See Table 2.1.
9000.04
2 2 40,000 3000
c
km
Example 2.4.3: Compute the force transmitted to a
machine through base motion at resonance
1/22
2
2
1 (2 )1 4
(2 ) 2
TT
F kYF
kY
From measured excitation Y = 0.001 m:
2 2(40,000 N/m)(0.001 m)1 4 1 4(0.04) 501.6 N
2 2(0.04)T
kYF
From (2.77) at r =1:
From given m, c, and k:
rt
rt
2.5 Rotating
Unbalance
• Gyros
• Cryo-coolers
• Tires
• Washing machines
Machine of total mass m i.e. m0
included in m
e = eccentricity
mo = mass unbalance
=rotation frequency
m0
e
k c
texa
tex
rrrx
rr
sin
sin
2
rt
Rotating Unbalance (cont)
From sophomore dynamics,
tememamR
tememamR
rroroyy
rroroxx
coscos
sinsin
220
220
What force is imparted on the
structure? Note it rotates with x
component:
Rx
m0
e
Ry
2
0 sin( )r rm e t 2
2 2
sin (2.82)
or 2 sin
o r r
on n r r
mx cx kx m e t
mx x x e t
m
Rotating Unbalance (cont)
The problem is now just like any other SDOF system with a
harmonic excitation
Note the influences on the
forcing function (we are assuming that the
mass m is held in place in the y direction as indicated in Figure
2.19)
m
x(t)
k
c
Rotating Unbalance (cont)
• Just another SDOF oscillator with a harmonic forcing
function
• Expressed in terms of frequency ratio r
2
22 2
1
2
( ) sin( ) (2.83)
(2.84)(1 ) 2
2tan (2.85)
1
p r
o
x t X t
m e rX
m r r
r
r
Figure 2.21: Displacement magnitude vs frequency
caused by rotating unbalance
0
1 1 0.1 m 110 10 0.1 m
2 2(0.05) 2
mXe
m e e
0
100.1 m
m X
m
Example 2.5.1:Given the deflection at resonance (0.1m),
ζ = 0.05 and a 10% out of balance, compute e and the amount of added
mass needed to reduce the maximum amplitude to 0.01 m.
Now to compute the added mass, again at resonance;
0
0.01 m10 100 9
0.1 m (0.1)
m m m mm m
m m
At resonance r = 1 and
Use this to find Δm so that X is 0.01:
5
0
1 10 N/m
60 kg
20 kg
0.5 kg
= 0.01
tail
rot
k
m
m
m
Example 2.5.2 Helicopter rotor unbalance
Given Fig 2.22
Fig 2.23
Compute the deflection at
1500 rpm and find the rotor
speed at which the deflection is maximum
Example 2.5.2 Solution
The rotating mass is 20 + 0.5 or 20.5. The stiffness is provided
by the Tail section and the corresponding mass is that
determined in the example of a heavy beam. So the system
natural frequency is
510 N/m53.72rad/s
33 3320.5 60kg
140 140
n
tail
k
m m
The frequency of rotation is
rev min 2 rad1500 rpm = 1500 157 rad/s
min 60 s rev
157 rad/s 2.92
53.96 rad/s
r
r
Now compute the deflection at r = 2.91 and ζ =0.01 using
eq (2.84)
2
0
2 2 2
2
2 22
(1 ) (2 )
0.5 kg 0.15 m 2.92 0.002 m
34.64 kg1 (2.92) 2(0.01)(2.92)
m e rX
m r r
At around r = 1, the max deflection occurs:
rad rev 60 s1 53.72 rad/s = 53.72 515.1 rpm
s 2 rad minrr
At r = 1:
0
eq
0.5 kg 0.15 m1 10.108 m or 10.8 cm
2 34.34 kg 2(0.01)
m eX
m
= - ( - ) - ( - ) =
= - ( ) - ( )
(2.86) and (2.61)
F k x y c x y mx
mx c x y k x y
2.6 Measurement Devices
• A basic transducer
used in vibration
measurement is the
accelerometer.
• This device can be
modeled using the
base equations developed in the
previous section
Figure 2.24
Here, y(t) is the measured
response of the structure
2
2
2 2 2
1
2
Let ( ) ( ) ( ) (2.87) :
( ) ( ) cos (2.88)
(2.90)(1 ) (2 )
and
2 tan (2.91)
1
b b
z t x t y t
mz cz t kz t m Y t
Z r
Y r r
r
r
Base motion applied to measurement devices
Accelerometer
These equations should be familiar
from base motion.
Here they describe measurement!
Magnitude and sensitivity plots for accelerometers.
Effect of damping on
proportionality constant
Fig 2.27
Fig 2.28
Magnitude plot showing
Regions of measurement
In the accel region, output voltage is
nearly proportional to displacement
2.7 Other forms of damping
These various other forms of damping are all nonlinear.
They can be compared to linear damping by the method of
“equivalent viscous damping” discussed next. A numerical
treatment of the exact response is given in section 2.9.
The method of equivalent viscous damping: consists of
comparing the energy dissipated during one cycle of forced
response
Assume a stead state resulting from a harmonic
input and compute the energy dissipated per one cycle
sinssx X t
The energy per cycle for a viscously damped system is
2 / 2 /
2
0 0
(2.99)d
dxE F dx cx dt cx dt
dt
2 /
2 2
0
sin cos
cos
ssx X t x X t
E c X t dt c X
(2.101)
Next compute the energy dissipated per cycle for
Coulomb damping:
2 /
0
/2 3 /2 2
0 /2 3 /2
sgn( )
( cos cos cos ) 4
E mg x xdt mg
mgX udu udu udu mgX
Here we let u = ωt and du =ωdt and split up the integral according to the
sign changes in velocity.
Next compare this energy to that of a viscous system:
2 44eq eq
mgc X mgX c
X
This yields a linear viscous system dissipating the same amount of energy per cycle.
(2.105)
Using the equivalent viscous damping calculations, each
of the systems in Table 2.2 can be approximated by a
linear viscous system
In particular, ceq can be used to derive amplitude expressions.
However, as indicated in Section 2.8 and 2.9 the response can
be simulated numerically to provide more accurate
magnitude and response information.
Hysteresis: an important concept characterizing
damping
• A plot of displacement versus spring/damping force for viscous damping yields a loop
• At the bottom is a stress strain plot for a system with material damping of the hysteretic type
• The enclosed area is equal to the energy lost per cycle
The measured area yields the energy dissipated. For
some materials, called hysteretic this is 2 (2.120)E k X
Here the constant β, a measured quantity is called
the hysteretic damping constant, k is the stiffness
and X is the amplitude.
Comparing this to the viscous energy yields:
eq
kc
Hysteresis gives rise to the concept of complex stiffness
Substitution of the equivalent damping coefficient and using
the complex exponential to describe a harmonic input yields:
0
j tkmx x kx F e
0
complex stiffness
Assuming ( ) and ( )
yields
( ) (1 ) ( )
j t j t
j t
x t Xe x t Xj e
mx t k j x t F e
2.8 Numerical Simulation and Design
• Four things we can do computationally to help solve,
understand and design vibration problems subject to
harmonic excitation
• Symbolic manipulation
• Plotting of the time response
• Solution and plotting of the time response
• Plotting magnitude and phase
2 2
2 2
2
2
n n
n n
A
0
0
fx
Symbolic Manipulation
1
nA A x
Let
and
What is
This can be solved using Matlab, Mathcad or Mathematica
Symbolic Manipulation
Solve equations (2.34) using Mathcad symbolics :
Enter this .n2
2
...2 n
...2 n
n2
2
1
f0
0
.n2
2
n4
..2 n22
4
...4 2n22f0
....2 n
n4
..2 n22
4
...4 2n22f0
Choose evaluate
under symbolics
to get this
In MATLAB Command Window
>> syms z wn w f0
>> A=[wn^2-w^2 2*z*wn*w;-2*z*wn*w wn^2-w^2];
>> x=[f0 ;0];
>> An=inv(A)*x
An =
[ (wn^2-w^2)/(wn^4-2*wn^2*w^2+w^4+4*z^2*wn^2*w^2)*f0]
[ 2*z*wn*w/(wn^4-2*wn^2*w^2+w^4+4*z^2*wn^2*w^2)*f0]
>> pretty(An)
[ 2 2 ]
[ (wn - w ) f0 ]
[ --------------------------------- ]
[ 4 2 2 4 2 2 2 ]
[ wn - 2 wn w + w + 4 z wn w ]
[ ]
[ z wn w f0 ]
[2 ---------------------------------]
[ 4 2 2 4 2 2 2]
[ wn - 2 wn w + w + 4 z wn w ]
Magnitude plots: Base Excitation
%m-file to plot base excitation to mass vibration
r=linspace(0,3,500); ze=[0.01;0.05;0.1;0.20;0.50]; X=sqrt( ((2*ze*r).^2+1) ./ ( (ones(size(ze))*(1-r.*r).^2) + (2*ze*r).^2) ); figure(1) plot(r,20*log10(X))
The values of z can
then be chosen directly
off of the plot.
For Example:
If the T.R. needs to be
less than 2 (or 6dB)
and r is close to 1
then z must be more
than 0.2 (probably
about 0.3).
0.5 1 1.5 2 2.5 3 -20
-10
0
1
2
3
4
value becomes very large when
the damping level is small
• The lower plot shows how the
frequency at which the peak value
occurs reduces with increased
damping
• Note that the peak value is only
defined for values ζ<0.707
20
Frequency ratio r
X/Y (dB)
real part imaginary part
0
harmonic input
cos ( sin )
( ) ( ) ( )
j t
j t
Ae A t A t j
mx t cx t kx t F e
2
0
002
2
the frequency response function
( )
( )
( )( ) ( )
1( )
( ) ( )
j t
p
j t j t
x t Xe
m cj k Xe F e
FX H j F
k m c j
H jk m c j
0
2 2 2
1
2
( )0
2 2 2
( ) ( )
tan
( )( ) ( )
j
j t
p
FX e
k m c
c
k m
Fx t e
k m c
T
h
e
t
r
a
n
s
f
e
r
f
0 00 0x v
sin dt
System Ske
System FBD
=0.05
=0.1
=0.2
=0.5
Force Magnitude plots: Base Excitation
0.5
1
0. 1 1.5 2 2. 3 -20
-10
0
10
20
30
40
Frequency ratio
FT
/kY
=
0
.
0
1
ζ=0.01
ζ=0.05
ζ=0.1
ζ=0.2
ζ=0.5
Numerical Simulation
We can put the forced case:
0
2
0
( ) ( ) ( ) cos
( ) 2 ( ) ( ) cosn n
mx t cx t kx t F t
x t x t x t f t
Into a state space form
1 2
2
2 2 1 0
0
2 cos
0( ) ( ) ( ), ( )
cos
n n
x x
x x x f t
t A t t tf t
x x f f
Numerical Integration
1Euler: ( ) ( ) ( ) ( )i i i it t A t t t t x x x f
>>TSPAN=[0 10]; >>Y0=[0;0]; >>[t,y] =ode45('num_for',TSPAN,Y0); >>plot(t,y(:,1))
Including forcing
function Xdot=num_for(t,X) m=100;k=1000;c=25; ze=c/(2*sqrt(k*m)); wn=sqrt(k/m); w=2.5;F=1000;f=F/m; f=[0 ;f*cos(w*t)]; A=[0 1;-wn*wn -2*ze*wn]; Xdot=A*X+f;
2 4 6 8 10 -5
-4
-3
-2
-1
0
1
2
3
4
2
2 2 20 0
1
(1 ) (2 )
nXXk
F f r r
Displacement (m)
Example 2.8.2: Design damping for an electronics
model
• 100 kg mass, subject to 150cos(5t) N
• Stiffness k=500 N/m, c = 10kg/s
• Usually x0=0.01 m, v0 = 0.5 m/s
• Find a new c such that the max transient value is 0.2 m.
Response of the board is;
transient exceeds design specification value
0 10 20 30 40
-0.4
-0.2
0
0.2
0.4
Time (sec)
Displacement (m)
Figure 2.33
To run this use the following file:
Rerun this code, increasing c each time until a response that
satisfies the design limits results.
Create function to
model forcing
function Xdot=num_for(t,X) m=100;k=500;c=10; ze=c/(2*sqrt(k*m)); wn=sqrt(k/m); w=5;F=150;f=F/m; f=[0 ;f*cos(w*t)]; A=[0 1;-wn*wn -2*ze*wn]; Xdot=A*X+f;
Matlab command
window
>>TSPAN=[0 40]; >> Y0=[0.01;0.5]; >>[t,y] = ode45('num_for',TSPAN,Y0); >> plot(t,y(:,1)) >> xlabel('Time (sec)') >> ylabel('Displacement (m)') >> grid
Solution: code it, plot it and change c until the desired
response bound is obtained.
Meets amplitude limit when c=195kg/s
Machine of total mass
rt
k
rt
10 texa
tex
rrrx
rr
sin
sin
2
20
2
0 sin( )r rm e t
2
2 2
sin (2.82)
or 2 sin
o r r
on n r r
mx cx kx m e t
mx x x e t
m
30 40
-0.1
0
100.1 m
m X
m
0
5
0
1 10 N/m
60 kg
20 kg
0.5 kg
= 0.01
tail
rot
k
m
m
m
0.1
b
a
s
i
c
t
r
a
n
= - ( - ) - ( - ) =
= - ( ) - ( )
(2.86) and (2.61)
F k x y c x y mx
mx c x y k x y
0.2
0.3
M
a
g
n
i
t
u
d
e
p
l
o
t
s
I
n
t
h
e
a
c
c
e
l
r
e
g
Time (sec)
Displacement (m)
Figure 2.34
2.9 Nonlinear Response Properties
• More than one equilibrium
• Steady state depends on initial conditions
• Period depends on I.C. and amplitude
• Sub and super harmonic resonance
• No superposition
• Harmonic input resulting in nonperiodic motion
• Jumps appear in response amplitude
0
1 2
2 1 2 0
( ) ( , ) cos
( ) ( )
( ) ( , ) cos
( ) ( ) ( )
x t f x x f t
x t x t
x t f x x f t
t t
x F x f
Computing the forced response of a non-linear system
A non-linear system has a equation of motion given by:
Put this expression into state-space form:
In vector form:
2
01 2
1
0( )( ) , ( )
cos( , )
( ) ( ) ( ( )) ( )i i i i
x tt
f tf x x
t t t t t t
F x f
x x F x f
Numerical form
Vector of nonlinear dynamics Input force vector
Euler equation is
964.2n
Cubic nonlinear spring (2.9.1)
tfxxxx nn cos2 032
2 4 6 8 10
-2
-1
0
1
2 2
0
Time (sec)
Displacement (m)
09.1n
Cubic nonlinear spring near resonance
tfxxxx nn cos2 032
09.1n
=0.05
-2
-1
0
1
2
Time (sec)
Displacement (m)
Recommended