Loading in 5 sec....

Chapter 2 Response to Harmonic ExcitationPowerPoint Presentation

Chapter 2 Response to Harmonic Excitation

- By
**gauri** - Follow User

- 230 Views
- Uploaded on

Download Presentation
## PowerPoint Slideshow about ' Chapter 2 Response to Harmonic Excitation' - gauri

**An Image/Link below is provided (as is) to download presentation**

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript

### Chapter 2 Response to Harmonic Excitation

### Section 2.2 Harmonic Excitation of Damped Systems complex exponential. How does this

### Section 2.3 Alternative Representations complex exponential. How does this

Introduces the important concept of resonance

Mechanical Engineering at Virginia Tech

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=F0cost

k

M

Mechanical Engineering at Virginia Tech

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):

Mechanical Engineering at Virginia Tech

Substitute particular solution into the equation of motion:

Thus the particular solution has the form:

Mechanical Engineering at Virginia Tech

Add particular and homogeneous solutions to get general solution:

Mechanical Engineering at Virginia Tech

Apply the initial conditions to evaluate the constants solution:

(2.11)

Mechanical Engineering at Virginia Tech

Comparison of free and forced response solution:

- 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

Mechanical Engineering at Virginia Tech

Response for solution: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

Mechanical Engineering at Virginia Tech

What happens when solution: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)

Mechanical Engineering at Virginia Tech

5 solution:

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!

Mechanical Engineering at Virginia Tech

Example 2.1.1: solution: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.

Mechanical Engineering at Virginia Tech

Example 2.1.2 solution: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.

Mechanical Engineering at Virginia Tech

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.

Mechanical Engineering at Virginia Tech

Solution: camera.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:

Mechanical Engineering at Virginia Tech

Case (a) camera.(assume aluminum for the material):

Case (b):

Mechanical Engineering at Virginia Tech

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.

Mechanical Engineering at Virginia Tech

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

Mechanical Engineering at Virginia Tech

Extending resonance and response calculation to damped systems

Mechanical Engineering at Virginia Tech

Displacement complex exponential. How does this

x

k

F=F0coswt

M

c

2.2 Harmonic excitation of damped systemsMechanical Engineering at Virginia Tech

Let complex exponential. How does thisxp have the form:

Note that we are using the rectangular form, but

we could use one of the other forms of the solution.

Mechanical Engineering at Virginia Tech

Substitute into the equations of motion complex exponential. How does this

Mechanical Engineering at Virginia Tech

Write as a matrix equation: complex exponential. How does this

Solving for As and Bs:

Mechanical Engineering at Virginia Tech

Substitute the values of complex exponential. How does thisAs 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.

Mechanical Engineering at Virginia Tech

Things to notice about damped forced response complex exponential. How does this

- 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.

Mechanical Engineering at Virginia Tech

Example 2.2.1: complex exponential. How does thiswn = 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

Mechanical Engineering at Virginia Tech

Proceeding with ignoring the transient complex exponential. How does this

- 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

Mechanical Engineering at Virginia Tech

Proceeding with ignoring the transient complex exponential. How does this

Magnitude:

(2.39)

Frequency ratio:

Non dimensional

Form:

(2.40)

Phase:

Mechanical Engineering at Virginia Tech

40 complex exponential. How does this

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

Mechanical Engineering at Virginia Tech

3.5 complex exponential. How does this

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

Mechanical Engineering at Virginia Tech

Example 2.2.3 complex exponential. How does this Compute max peak by differentiating:

(2.41)

(2.42)

Mechanical Engineering at Virginia Tech

Effect of Damping on Peak Value complex exponential. How does this

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

Mechanical Engineering at Virginia Tech

- 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.

Mechanical Engineering at Virginia Tech

2.3.1 Geometric Approach complex exponential. How does this

- 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

Mechanical Engineering at Virginia Tech

Using vector addition on the diagram: complex exponential. How does this

At resonance:

Mechanical Engineering at Virginia Tech

2.3.2 Complex response method complex exponential. How does this

(2.47)

(2.48)

Real part of this complex solution corresponds to the physical solution

Mechanical Engineering at Virginia Tech

Choose complex exponential as a solution complex exponential. How does this

(2.49)

(2.50)

(2.51)

(2.52)

Note: These are all complex functions

Mechanical Engineering at Virginia Tech

Using complex arithmetic: complex exponential. How does this

(2.53)

(2.54)

(2.55)

Has real part = to previous solution

Mechanical Engineering at Virginia Tech

Comments: complex exponential. How does this

- 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

Mechanical Engineering at Virginia Tech

Example 2.3.1: complex exponential. How does this Use the frequency response approach to compute the particular solution of an undamped system

The equation of motion is written as

Mechanical Engineering at Virginia Tech

2.3.3 Transfer Function Method complex exponential. How does this

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

Mechanical Engineering at Virginia Tech

The Laplace Transform approach: complex exponential. How does this

- 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

Mechanical Engineering at Virginia Tech

Take the transform of the equation of motion: complex exponential. How does this

Now solve algebraic equation in s for X(s)

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

Mechanical Engineering at Virginia Tech

Transfer Function Method complex exponential. How does this

With zero initial conditions:

The transfer

function

(2.59)

(2.60)

Mechanical Engineering at Virginia Tech

Example 2.3.2 complex exponential. How does thisCompute 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:

Mechanical Engineering at Virginia Tech

Notes on Phase for Homogeneous and Particular Solutions complex exponential. How does this

- 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?

Mechanical Engineering at Virginia Tech

Non-Zero initial conditions complex exponential. How does this

Mechanical Engineering at Virginia Tech

Zero initial displacement complex exponential. How does this

Mechanical Engineering at Virginia Tech

Zero initial velocity complex exponential. How does this

Mechanical Engineering at Virginia Tech

Phase on Particular Solution complex exponential. How does this

- 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

Mechanical Engineering at Virginia Tech

Download Presentation

Connecting to Server..