Feb 14, 2003

1
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.

2
FBD of SDOF Base Excitation
System Sketch
System FBD
x(t)
k
c
y(t)
base
3
SDOF Base Excitation (cont)
The steady-state solution is just the
superposition of the two individual particular
solutions (system is linear).
4
Particular Solution (sin term)
With a sine for the forcing function,
Use rectangular form to make it easier to add
the cos term
5
Particular Solution (cos term)
With a cosine for the forcing function, we showed
6
Magnitude X/Y
Now add the sin and cos terms to get the
magnitude of the full particular solution
7
The relative magnitude plot of X/Y versus
frequency ratio
8
From the plot of relative Displacement
Transmissibility observe that

• Potentially severe amplification at resonance
• Attenuation only for r gt sqrt(2)
• If rlt sqrt(2) transmissibility decreases with
  damping ratio
• If r gtgt 1 then transmissibility increases with
  damping ratio Xp2Yz/r

9
Next examine the relative Force Transmissibility
as function of frequency ratio
From FBD
x(t)
FT
k
c
y(t)
base
10
Plot of Force Transmissibility (in dB) versus
frequency ratio
11
Section 5.2 Isolation

• A major job of vibration engineers is to isolate
  systems from vibration disturbances or visa
  versa. Uses heavily material from Sections 2.4
  on Base Excitation

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Isolation a sdof concept

• Two types moving base and fixed base
• Three magnitude plots to consider TR used to
  denote the transmissibility ratio

Moving base, displacement TR

Moving base, force TR
Fixed base, force TR
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Example 5.2.1

• Design an isolator (chose k, c) to hold a 3 kg
  electronics module to less then 0.005 m
  deflection if the base is moving at
  y(t)(0.01)sin(35t)
• Calculate the force transmitted through the
  isolator

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Figure 5.6 Car, module and model
Detail sketch
System sketch
Free Body Diagram
15
TR Plot for moving base displacement
1.5
For TR 0.5
z
1
z
0.01
TR
z
0.05
z
0.1
0.5
z
0.2
z
0.5
z
1.2
0
0
0.5
1
1.5
2
2.5
3
Frequency ratio r
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Computing a design from formulas and the TR plot
17
Compute the Rattle Space
Choice of k and c must also be reasonable As must
force transmitted
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Transmitted force
Transmitted force, static deflection, damping
and stiffness values must all be reasonable and
obtainable for the application.
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2.5 Rotating Imbalance

• Gyros
• Cryo-coolers
• Tires
• Washing machines

Machine of total mass m i.e. m0 included in m
m0
e
??t
e eccentricitymo mass imbalancew?
rotation frequency
k
c
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Rotating Imbalance (cont)
What force is imparted on the structure? Note it
rotates at with x component
From sophomore dynamics,
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Rotating Imbalance (cont)
The problem is now just like any other SDOF
system with a harmonic excitation
m0ew?2sin(w?t)
x(t)
m
k
Note the influences on the forcing function (we
are assuming that the mass m is held in place in
the y direction)
c
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Rotating Imbalance (cont)

• Just another SDOF oscillator with a harmonic
  forcing function
• Expressed in terms of frequency ratio r

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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

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Symbolic Manipulation
Let
and
What is
This can be solved using Matlab, Mathcad or
Mathematica
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Symbolic Manipulation
Solve equations (2.26) using Mathcad symbolic s
Enter this
Choose evaluate under symbolics to get this
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In MATLAB Command Window

• gtgt syms z wn w f0
• gtgt Awn2-w2 2zwnw-2zwnw wn2-w2
• gtgt xf0 0
• gtgt Aninv(A)x
• An
• (wn2-w2)/(wn4-2wn2w2w44z2wn2w2)
  f0
• 2zwnw/(wn4-2wn2w2w44z2wn2w2)
  f0
• gtgt 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
  

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Magnitude plots Base Excitation
m-file to plot base excitation to mass
vibration rlinspace(0,3,500) ze0.010.050.10
.200.50 Xsqrt( ((2zer).21) ./ (
(ones(size(ze))(1-r.r).2) (2zer).2)
) figure(1) plot(r,20log10(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).
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Magnitude plots Base Excitation
m-file to plot base excitation to mass
vibration rlinspace(0,3,500) ze0.010.050.10
.200.50 Xsqrt( ((2zer).21) ./ (
(ones(size(ze))(1-r.r).2) (2zer).2)
) FX.(ones(length(ze),1)r).2 figure(1) plot(
r,20log10(F))
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Numerical Simulation
We can put the forced case
Into a state space form
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Numerical Integration
Zero initial conditions
Using the ODE45 function
gtgtTSPAN0 10 gtgtY000 gtgtt,y
ode45('num_for',TSPAN,Y0) gtgtplot(t,y(,1))
5
4
3
2
Including forcing
1
function Xdotnum_for(t,X) m100k1000c25 zec
/(2sqrt(km)) wnsqrt(k/m) w2.5F1000fF/m
f0 fcos(wt) A0 1-wnwn
-2zewn XdotAXf
0
Displacement (m)
-1
-2
-3
-4
-5
0
2
4
6
8
10
Time (sec)
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Example 2.8.2 Design damping for an electronics
model

• 100 kg mass, subject to 150cos(5t) N
• Stiffness k500 N/m, c 10kg/s
• Usually x00.01 m, v0 0.5 m/s
• Find a new c such that the max transient value is
  0.2 m.

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Response of the board is
transient exceeds design specification value
0.4
0.2
0
Displacement (m)
-0.2
-0.4
0
10
20
30
40
Time (sec)
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To run this use the following file
function Xdotnum_for(t,X) m100k500c10 zec/
(2sqrt(km)) wnsqrt(k/m) w5F150fF/m f0
fcos(wt) A0 1-wnwn -2zewn XdotAXf

Create function to model forcing
gtgtTSPAN0 40 gtgt Y00.010.5 gtgtt,y
ode45('num_for',TSPAN,Y0) gtgt plot(t,y(,1)) gtgt
xlabel('Time (sec)') gtgt ylabel('Displacement
(m)') gtgt grid
Matlab command window
Rerun this code, increasing c each time until a
response that satisfies the design limits results.
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Solution code it, plot it and change c until the
desired response bound is obtained.
0.3
Meets amplitude limit when c195kg/s
0.2
0.1
Displacement (m)
0
-0.1
0
10
20
30
40
Time (sec)
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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

36
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
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Numerical form
Vector of nonlinear dynamics
Input force vector
Euler equation is
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Cubic nonlinear spring (2.9.1)
2
1
0
Displacement (m)
-1
Non-linearity included
Linear system
-2
0
2
4
6
8
10
Time (sec)
Superharmonic resonance
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Cubic nonlinear spring near resonance
3
2
1
0
Displacement (m)
-1
-2
Non-linearity included
Linear system
-3
0
2
4
6
8
10
Time (sec)
Superharmonic resonance