Dynamic response to harmonic excitation
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Transcript of Dynamic response to harmonic excitation
© Eng. Vib, 3rd Ed.1/49 @ProfAdhikari, #EG260
Chapter 2 Response to Harmonic Excitation
Introduces the important concept of resonance
© Eng. Vib, 3rd Ed.2/49 College of Engineering
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
Mk
xDisplacement
F=F0cosωt
© Eng. Vib, 3rd Ed.3/49 College of Engineering
Equations of motion
• Solution is the sum of homogenous and particular solution
• The particular solution assumes form of forcing function (physically the input wins):
( ) cos( )px t X t
Figure 2.1
© Eng. Vib, 3rd Ed.4/49 College of Engineering
Substitute particular solution into the equation of motion:
Thus the particular solution has the form:
02 2
( ) cos( )pn
fx t t
( ) cospx t X t
© Eng. Vib, 3rd Ed.5/49 College of Engineering
Add particular and homogeneous solutions to get general solution:
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Apply the initial conditions to evaluate the constants
(2.11)
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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
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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
Go to code demo
0 2 4 6 8 10-0.05
0
0.05
Time (sec)
Dis
plac
emen
t (x)
© Eng. Vib, 3rd Ed.9/49 College of Engineering
What happens when ω is near ωn?
When the drive frequency and natural frequency are close a beating phenomena occurs
0 5 10 15 20 25 30-1
-0.5
0
0.5
1
Time (sec)
Dis
plac
emen
t (x)
02 2
2sin
2n
n
ft
Larger amplitude
0
2 2
2( ) sin sin (2.13)
2 2n n
n
fx t t t
Called beating
© Eng. Vib, 3rd Ed.10/49 College of Engineering
What happens when ω is ωn?
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 5 10 15 20 25 30
-5
0
5
Time (sec)
Dis
pla
cem
ent (
x)
0
( ) sin( )
substitute into eq. and solve for
2
px t tX t
X
fX
© Eng. Vib, 3rd Ed.11/49 College of Engineering
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 =2 ω n.
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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.
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Example 2.1.3 Design a rectangular mount for a security camera.
Compute l > 0.5 m so that the mount keeps the camera from vibrating more then 0.01 m of maximum amplitude under a windload of 15 N at 10 Hz. The mass of the camera is 3 kg.
0.02 x 0.02 m in cross section
© Eng. Vib, 3rd Ed.14/49 College of Engineering
Solution:Modeling the mount and camera as a beam with a tip mass, and the wind as harmonic, the equation of motion becomes:
From strength of materials:
Thus the frequency expression is:
Here we are interested computing l that will make the amplitude less then 0.01m:
2 202 2
02 2
2 202 2
2( ) 0.01 , for 0
20.01
2( ) 0.01, for 0
nn
nn
n
fa
f
fb
© Eng. Vib, 3rd Ed.15/49 College of Engineering
Case (a) (assume aluminum for the material):3
2 2 200 02 2 3
33
20
20.01 2 0.01 0.01 0.01 2 0.01
4
0.01 0.321 0.6848 m4 (0.01 2 )
nn
f Ebhf f
m
Ebh
m f
Case (b):3
2 2 200 02 2 3
33
20
20.01 2 0.01 0.01 2 0.01 0.01
4
0.01 0.191 0.576 m4 (2 0.01 )
nn
f Ebhf f
m
Ebh
m f
© Eng. Vib, 3rd Ed.16/49 College of Engineering
Remembering the constraint that the length must be at least 0.5 m, (a) and (b) yield
0.5 0.576, or > 0.6848 m
Less material is usually desired, so chose case
b, say l = 0.55 m. To check, note that
3
22 23
320 1742 0
12n
Ebh
m
Thus the case a condition is met.Next check the mass of the designed beam to insure it does not change the frequency. Note
it is less then m. 3(2.7 10 )(0.55)(0.02)(0.02)
0.594 kg
m bh
© Eng. Vib, 3rd Ed.17/49 College of Engineering
A harmonic force may also be represented by sine or a complex exponential. How does thischange the solution?
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 nn n n n
v f fx t x t t t
© Eng. Vib, 3rd Ed.18/49 College of Engineering
Section 2.2 Harmonic Excitation of Damped Systems
Extending resonance and response calculation to damped systems
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2.2 Harmonic excitation of damped systems
M
kx
Displacement
c
F=F0cosωt
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Let xp have the form:
Note that we are using the rectangular form, but we could use one of the other forms of the solution.
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2 2
0
2 2
2 20
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
Substitute into the equations of motion
© Eng. Vib, 3rd Ed.22/49 College of Engineering
Write as a matrix equation:
2 20
2 2
( ) 2
02 ( )sn n
sn n
A f
B
2 20
2 2 2 2
( )
( ) (2 )n
sn n
fA
02 2 2 2
2
( ) (2 )n
sn n
fB
Solving for As and Bs:
© Eng. Vib, 3rd Ed.23/49 College of Engineering
Substitute the values of As and Bs into xp:
Add homogeneous and particular to get total solution:
Note: that A and will not have the same values as in Ch 1,for the free response. Also as t gets large, transient dies out.
© Eng. Vib, 3rd Ed.24/49 College of Engineering
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.
© Eng. Vib, 3rd Ed.25/49 College of Engineering
Example 2.2.1: 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 A and for forced and unforced case:
0.1
0.133, 0.013
( ) sin(9.99 ) 0.133cos(5 0.013)t
X
x t Ae t t
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
Using the equations on slide 23:
Differentiating yields:
The numbers in ( ) are those obtained by
incorrectly using the freeresponse values
x(0)0.05
Asin 0.133cos( 0.013)
v(0)0 0.01Asin 9.999A cos 0.665sin 0.013
© Eng. Vib, 3rd Ed.26/49 College of Engineering
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
© Eng. Vib, 3rd Ed.27/49 College of Engineering
12
2tan1r
r
Magnitude:
Non dimensionalForm:
Phase:
Frequency ratio:
Proceeding with ignoring the transient
0
2 2 2 2( ) (2 )n n
fX
2
2 2 20 0
1
(1 ) (2 )nXXk
F f r r
n
r
(2.39)
(2.40)
© Eng. Vib, 3rd Ed.28/49 College of Engineering
Magnitude plot
0 0.5 1 1.5 2-20
-10
0
10
20
30
40
r
X (
dB
)
2 2 2
1
(1 ) (2 )X
r r• 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
Fig 2.7
© Eng. Vib, 3rd Ed.29/49 College of Engineering
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 20
0.5
1
1.5
2
2.5
3
3.5
r
Ph
ase
(rad
)
=0.01 =0.1 =0.3 =0.5 =1
12
2tan1r
r
Fig 2.7
© Eng. Vib, 3rd Ed.30/49 College of Engineering
2 2 20
2peak
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)
© Eng. Vib, 3rd Ed.31/49 College of Engineering
Effect of Damping on Peak Value
0 0.2 0.4 0.6 0.80
5
10
15
20
25
30
ζ
0 0.2 0.4 0.6 0.80
0.2
0.4
0.6
0.8
1
rpeak
(dB)0F
Xk
• 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
Fig 2.9ζ
© Eng. Vib, 3rd Ed.32/49 College of Engineering
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 2 alternatives:a frequency response approacha transform approach
• These also give us some insight and additional useful tools.
© Eng. Vib, 3rd Ed.35/49 College of Engineering
2.3.2 Complex response method
Real part of this complex solution corresponds to the physical solution
(2.47)
(2.48)
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Note: These are all complex functions
Choose complex exponential as a solution
(2.49)
(2.50)
(2.51)
(2.52)
© Eng. Vib, 3rd Ed.37/49 College of Engineering
0
2 2 2
12
( )0
2 2 2
( ) ( )
tan
( )( ) ( )
j
j tp
FX e
k m c
c
k m
Fx t e
k m c
Using complex arithmetic:
Has real part = to previous solution
(2.53)
(2.54)
(2.55)
© Eng. Vib, 3rd Ed.38/49 College of Engineering
Comments:
• It is the real part of this complex solution that is physical
• The approach is useful in more complicated problems
© Eng. Vib, 3rd Ed.39/49 College of Engineering
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
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2.3.3 Transfer Function Method
• 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
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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
X(s) L (x(t)) x(t)
0
e stdt
© Eng. Vib, 3rd Ed.42/49 College of Engineering
Take the transform of the equation of motion:
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”
© Eng. Vib, 3rd Ed.43/49 College of Engineering
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
The transferfunction
With zero initial conditions:
Transfer Function Method
(2.59)
(2.60)
© Eng. Vib, 3rd Ed.44/49 College of Engineering
Example 2.3.2 Compute forced response of the suspension system shown using the Laplace transform
Summing moments about the shaft:
Taking the Laplace transform:
20 2 2
0 2 2 2
( ) ( )
1( )
Js X s kX s aFs
X s a Fs Js k
Taking the inverse Laplace transform:
10 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
© Eng. Vib, 3rd Ed.45/49 College of Engineering
Notes on Phase for Homogeneous and Particular Solutions
• Equation (2.37) gives the full solution for a harmonically driven underdamped SDOF oscillator to be
How do we interpret these phase angles?Why is one added and the other subtracted?
© Eng. Vib, 3rd Ed.46/49 College of Engineering
sin dt tan 1 xod
vo nxo
0 00 0x v
Non-Zero initial conditions
Zero initial displacement
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Zero initial velocity
© Eng. Vib, 3rd Ed.48/49 College of Engineering
© Eng. Vib, 3rd Ed.49/49 College of Engineering
Phase on Particular Solution
• Simple “atan” gives -π/2 < Φ < π/2• Four-quadrant “atan2” gives 0 < Φ < π