webinar_18_force_VRS

Report
Unit 18
Vibrationdata
Force Vibration Response Spectrum
1
Introduction

SDOF systems may be subjected to an applied force

Modal testing, impact or steady-state force

Wind, fluid, or gas pressure

Acoustic pressure field

Rotating or reciprocating parts
Vibrationdata
Rotating imbalance
Shaft misalignment
Bearings
Blade passing frequencies
Electromagnetic force, magnetostriction
2
Vibrationdata
SDOF System, Applied Force
m
= mass
c
= viscous damping coefficient
k
= stiffness
x
= displacement of the mass
f(t) = applied force
Governing equation of motion
m x  cx  kx  f (t )
3
Rayleigh Peak Response Formula
Vibrationdata
Consider a single-degree-of-freedom system with the index n.
The maximum response can be estimated by the following equations.
cn 
2 ln fn T 
Cn  cn 
Maximum Peak
fn
T
ln
n
0.5772
cn
 Cn n
is the natural frequency
is the duration
is the natural logarithm function
is the standard deviation of the oscillator response
4
Vibrationdata
Steady-State Response to Sine Force
The normalized displacement is
kx

F


1
2
2
1 
 2 2
where F is the applied force magnitude
The natural frequency fn is
fn 
1
2
k
m
  f / fn
f is the applied force frequency
fn is the natural frequency
5
Steady-State Response to Sine Force (cont)
Vibrationdata
The transmitted force to ground ratio is
Ft

F
1  2 2
1  2  2  2 2
,
  f / fn
where
Ft is the transmitted force magnitude
F
is the applied force magnitude
The transmitted force ratio is the same as that for the acceleration response to
base excitation.
6
SDOF STEADY-STATE RESPONSE TO APPLIED SINUSOIDAL FORCE
DISPLACEMENT MAGNITUDE [ k x / F ]
20
Q = 10
Q=2
Q=1
Vibrationdata
10
1
0.1
0.01
0.1
1
10
FREQUENCY ( f / fn )
Control by Frequency Domain
Low Freq
Resonance
High Freq
Stiffness
Damping
Mass
7
SDOF STEADY-STATE TRANSMITTED FORCE
20
Q = 10
Q=2
Q=1
10
TRANS FORCE MAG | Ft / F |
Vibrationdata
1
0.1
0.01
0.1
1
10
FREQUENCY ( f / fn )
8
Exercise
Vibrationdata
vibrationdata > Miscellaneous Functions >
SDOF Response: Steady-State Sine Force or Acceleration Input
Practice some sample calculations for applied force using your own parameters.
Try resonant excitation and then +/- one octave separation between the excitation
and natural frequencies.
9
SDOF Response to Force PSD, Miles Equation
Vibrationdata
The overall displacement x is
1/ 2
1/ 4 1 3 / 4
A
1
 
x RMS   
m
k 
 
 
8  
where
m
is the mass
k
is the stiffness

is viscous damping ratio
is the amplitude of the force PSD in dimensions of [force^2 / Hz]
at the natural frequency
A
Miles equation assumes that the PSD is white noise from 0 to infinity Hz.
10
Miles Equation, Velocity & Acceleration
The overall velocity is
Vibrationdata
x RMS n x RMS
•
An accelerance FRF curve is shown for a sample system in the next slide
•
The normalized accelerance converges to 1 as the excitation frequency becomes much
larger than the natural frequency
•
The acceleration response would be infinitely high for a white noise force excitation
which extended up to an infinitely high frequency
•
A Miles equation for the acceleration response to a white noise applied force cannot
be derived
11
Miles Equation, Acceleration
Vibrationdata
ACCELERANCE MAGNITUDE ( ACCELERATION / FORCE )
SDOF SYSTEM: mass= 1 kg fn = 100 Hz Damp = 0.05
10
2
ACCELERANCE ( m/sec / N )
100
1
0.1
0.01
0.001
1
10
100
1000
EXCITATION FREQUENCY (Hz)
12
SDOF Response to Force PSD, General Method
Vibrationdata
Displacement
1
x RMSf n , 
k
N



1
2
i 1 1  i 2  2  i  2
F PSD(fi ) fi
, i  f i / f n
Velocity
2
x RMSf n , 
k
N



fi 2
2
i 1 1  i 2  2  i  2
F PSD(f i ) f i
13
SDOF Response to Force PSD, General Method
Vibrationdata
Acceleration
x RMSf n ,  4
2
k
N



fi 4
2
i 1 1  i 2  2  i  2
F PSD(f i ) f i
, i  f i / f n
Transmitted Force
Ft
RMSf n ,  
N


1  2  i  2

2
i 1 1  i 2  2  i  2
F PSD(f i ) f i
14
Force PSD
Vibrationdata
Frequency
(Hz)
Force
(lbf^2/Hz)
10
0.1
1000
0.1
Duration = 60 sec
The same PSD was used for the time domain calculation in Webinar 17.
15
Vibrationdata
SDOF Example
Apply the Force PSD on the previous slide to the
SDOF system.
Duration = 60 seconds (but only affects peak value)
Mass = 20 lbm, Q=10,
Natural Frequency = independent variable
16
SDOF Response to Force PSD, Acceleration
Vibrationdata
Response at 400 Hz agrees
with time domain result in
previous webinar unit.
fn
(Hz)
Accel
(GRMS)
100
0.80
200
1.0
400
1.3
vibrationdata > Power Spectral Density > Force > SDOF Response to Force PSD
17
SDOF Response to Force PSD, Transmitted Force
Vibrationdata
18
Acceleration VRS
Vibrationdata
fn
(Hz)
Accel
(GRMS)
100
0.80
200
1.0
400
1.3
vibrationdata > Power Spectral Density > Force > Vibration Response Spectrum (VRS)
19
Velocity VRS
Vibrationdata
20
Displacement VRS
Vibrationdata
21
Transmitted Force VRS
Vibrationdata
22
Homework

Vibrationdata
Repeat the examples in the presentation using the Matlab scripts
23

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