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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 )= F 0 cos w t w is the driving frequency F 0 is the magnitude of the applied force

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chapter 2 response to harmonic excitation

Chapter 2 Response to Harmonic Excitation

Introduces the important concept of resonance

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2 1 harmonic excitation of undamped systems
2.1 Harmonic Excitation of Undamped Systems
  • Consider the usual spring mass damper system with applied force F(t)=F0coswt
  • w is the driving frequency
  • F0 is the magnitude of the applied force
  • We take c = 0 to start with

Displacement

x

F=F0cost

k

M

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equations of motion
Equations of motion

Figure 2.1

  • Solution is the sum of homogenous and particular solution
  • The particular solution assumes form of forcing function (physically the input wins):

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substitute particular solution into the equation of motion
Substitute particular solution into the equation of motion:

Thus the particular solution has the form:

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add particular and homogeneous solutions to get general solution
Add particular and homogeneous solutions to get general solution:

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slide6

Apply the initial conditions to evaluate the constants

(2.11)

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comparison of free and forced response
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 wn = w 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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slide8

Response for m=100 kg, k=1000 N/m, F=100 N, w = wn +5

v0=0.1m/s and x0= -0.02 m.

0.05

0

Displacement (x)

-0.05

0

2

4

6

8

10

Time (sec)

Note the obvious presence of two harmonic signals

Go to code demo

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what happens when w is near w n
What happens when w is near wn?

When the drive frequency and natural frequency are close a beating phenomena occurs

1

0.5

0

Displacement (x)

Larger amplitude

-0.5

-1

0

5

10

15

20

25

30

Time (sec)

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what happens when w is w n

5

Displacement (x)

0

-5

0

5

10

15

20

25

30

Time (sec)

What happens when w is wn?

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!

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example 2 1 1 compute and plot the response for m 10 kg k 1000 n m x 0 0 v 0 0 2 m s f 23 n w 2 w n
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, w=2wn.

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slide12

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
Example 2.1.3 Design a rectangular mount for a security camera.

Compute l 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.

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slide14
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:

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case a assume aluminum for the material
Case (a) (assume aluminum for the material):

Case (b):

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remembering the constraint that the length must be at least 0 5 m a and b yield
Remembering the constraint that the length must be at least 0.5 m, (a) and (b) yield

Less material is usually desired, so chose case

a, say l = 0.55 m.

To check, note that

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 much less then m.

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slide17
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:

Substitution of (2.19) into (2.18) yields:

Solving for the homogenous solution and evaluating the constants yields

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section 2 2 harmonic excitation of damped systems

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

Displacement

x

k

F=F0coswt

M

c

2.2 Harmonic excitation of damped systems

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slide20

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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slide21

Substitute into the equations of motion

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slide22

Write as a matrix equation:

Solving for As and Bs:

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slide23

Substitute the values of As and Bs into xp:

Add homogeneous and particular to get total solution:

Note: that A and f will not have the same values as in Ch 1,

for the free response. Also as t gets large, transient dies out.

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things to notice about damped forced response
Things to notice about damped forced response
  • If z = 0, undamped equations result
  • Steady state solution prevails for large t
  • Often we ignore the transient term (how large is z, 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.

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slide25

Example 2.2.1: wn = 10 rad/s, w = 5 rad/s, z = 0.01, F0= 1000 N, m = 100 kg, and the initial conditions x0 = 0.05 m and v0 = 0. Compare A and f for forced and unforced case:

Using the equations on slide 6:

Differentiating yields:

The numbers in ( ) are those obtained by incorrectly using the free

response values

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proceeding with ignoring the transient
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

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slide27

Proceeding with ignoring the transient

Magnitude:

(2.39)

Frequency ratio:

Non dimensional

Form:

(2.40)

Phase:

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magnitude plot

40

z

=0.01

z

=0.1

30

z

=0.3

z

=0.5

20

z

=1

X (dB)

10

0

-10

-20

0

0.5

1

1.5

2

r

Magnitude plot
  • Resonance is close to r = 1
  • For z = 0, r =1 defines resonance
  • As z grows resonance moves r <1, and X decreases
  • The exact value of r, can be found from differentiating the magnitude

Fig 2.7

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phase plot

3.5

z

=0.01

3

z

=0.1

z

=0.3

2.5

z

=0.5

z

=1

2

Phase (rad)

1.5

1

0.5

0

0

0.5

1

1.5

2

r

Phase plot
  • Resonance occurs at f = p/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 p radians

Fig 2.7

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slide30

Example 2.2.3 Compute max peak by differentiating:

(2.41)

(2.42)

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effect of damping on peak value
Effect of Damping on Peak Value

30

  • 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 z<0.707

25

20

15

10

5

0

0

0.2

0.4

0.6

0.8

z

1

0.8

0.6

rpeak

0.4

0.2

0

0

0.2

0.4

0.6

0.8

z

Fig 2.9

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section 2 3 alternative representations

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.

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2 3 1 geometric approach
2.3.1 Geometric Approach
  • Position, velocity and acceleration phase shifted each by p/2
  • Therefore write each as a vector
  • Compute X in terms of F0 via vector addition

Im

C

D

C

E

q

A

B

q

B

Re

A

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slide34

Using vector addition on the diagram:

At resonance:

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2 3 2 complex response method
2.3.2 Complex response method

(2.47)

(2.48)

Real part of this complex solution corresponds to the physical solution

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slide36

Choose complex exponential as a solution

(2.49)

(2.50)

(2.51)

(2.52)

Note: These are all complex functions

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slide37

Using complex arithmetic:

(2.53)

(2.54)

(2.55)

Has real part = to previous solution

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comments
Comments:
  • Label x-axis Re(ejwt) and y-axis Im(ejwt) 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

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slide39
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
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

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the laplace transform approach
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

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slide42

Take the transform of the equation of motion:

Now solve algebraic equation in s for X(s)

To get the time response this must be “inverse transformed”

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transfer function method
Transfer Function Method

With zero initial conditions:

The transfer

function

(2.59)

(2.60)

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example 2 3 2 compute forced response of the suspension system shown using the laplace transform
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:

Taking the inverse Laplace transform:

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notes on phase for homogeneous and particular solutions
Notes on Phase for Homogeneous and Particular Solutions
  • Equation (2.37) gives the full solution for a harmonically driven underdamped SDOF oscillator to beHow do we interpret these phase angles?Why is one added and the other subtracted?

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non zero initial conditions
Non-Zero initial conditions

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zero initial displacement
Zero initial displacement

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zero initial velocity
Zero initial velocity

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phase on particular solution
Phase on Particular Solution
  • Simple “atan” gives -π/2 < f < π/2
  • Four-quadrant “atan2” gives 0 < f < π

F0cos(wt)

f

Xcos(wt-f)

f

2zwnw

2zwnw

f

wt

wn2-w2

wn2-w2

w>wn

w<wn

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