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)=F0cosωt

• ω is the driving frequency

• F0 is the magnitude of the applied force

• We take c = 0 to start with

M

F=F0cosωt

Displacement

x

k

0

2

0

00

( ) ( ) cos( )

( ) ( ) cos( )

where ,

n

n

mx t kx t F t

x t x t f t

F kfm m

Equations of motion

( ) cos( )px t X t

• Solution is the sum of

homogenous and

particular solution

• The particular solution

assumes form of forcing

function (physically the

input wins):

Substitute particular solution into the equation of motion:

( ) cos( )px t X t

2

2 2

0

0

2 2

cos cos cos

solving yields:

p n px x

n

n

X t X t f t

fX

Thus the particular solution has the form:

0

2 2( ) cos( )p

n

fx t t

Add particular and homogeneous solutions to get general

solution:

01 2 2 2

homogeneous

1 2

( )

sin cos cos

and are constants of integration.

particular

n n

n

x t

fA t A t t

A A

(2.8)

Apply the initial conditions to evaluate the constants

0 01 2 2 02 2 2 2

02 0 2 2

01 2 1 02 2

01

0

(0) sin 0 cos0 cos0

(0) ( cos0 sin 0) sin 0

( )

n n

n

n n

n

n

f fx A A A x

fA x

fx A A A v

vA

vx t

0 00 2 2 2 2

sin cos cosn n

n n n

f ft x t t

(2.11)

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

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

0

Displacement (x)

Time (sec)

0

What happens when ω is near ωn?

0

2 2

2( ) sin sin (2.13)

2 2

n n

n

fx t t t

When the drive frequency and natural frequency

are close a beating phenomena occurs

0

2 2

2sin

2

n

n

ft

Larger

amplitude

Time (sec)

Displacement (x)

What happens when ω is ωn?

0

( ) sin( )

substitute into eq. and solve for

2

px t tX t

X

fX

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

2 2

2sin

2

n

n

ft

0 5 10

15

20

25

x(t) = A1 sinwt + A2 coswt +

f0

2wt sin(wt)

grows with out bound

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 =2ωn.

00

30

2 2 2 2 2 2

1000 N/m10 rad/s, 2 20 rad/s

10 kg

23 N 0.2 m/s2.3 N/kg, 0.02 m

10 kg 10 rad/s

2.3 N/kg7.9667 10 m

(10 20 ) rad / s

Equation (2.11) then yields:

( ) 0.02sin10 7.96

n n

n

n

k

m

vFf

m

f

x t t

367 10 (cos10 cos 20 )t t

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.

0

2 2

0

2 2

0.1 m

0

2 2 2

( ) cos 20 cos for zero initial conditions

2trig identity ( ) sin sin

2 2

2 2(20 / )0.1 0.1

2000 (20 )

0.405 kg

n

n

n n

n

n

fx t t t

fx t t t

f m

m

m

Example 2.1.3 Design a rectangular mount for a

security camera.

0.02 x 0.02 m

in cross section

Compute l > 0.5 m 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.

Solution: Modeling the mount and camera as a beam

with a tip mass, and the wind as harmonic, the equation

of motion becomes:

03

3( ) cos

EImx x t F t

From strength of materials:

3

12

bhI

Thus the frequency expression is: 3 3

2

3 3

3

12 14n

Ebh Ebh

m m

Here we are interested computing l that will make the

amplitude less then 0.01m:

2 20

2 2

0

2 22 20

2 2

2( ) 0.01 , for 0

20.01

2( ) 0.01, for 0

n

n

nn

n

fa

f

fb

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

32 2 20

0 02 2 3

33

2

0

20.01 2 0.01 0.01 0.01 2 0.01

4

0.01 0.02 0.272 m4 (0.01 2 )

n

n

f Ebhf f

m

Ebh

m f

Case (b):

32 2 20

0 02 2 3

33

2

0

20.01 2 0.01 0.01 2 0.01 0.01

4

0.01 0.012 0.229 m4 (2 0.01 )

n

n

f Ebhf f

m

Ebh

m f

3(2.7 10 )(0.22)(0.01)(0.01)

0.149 kg

m bh

Remembering the constraint that the length must be at

least 0.2 m, (a) and (b) yield

0.2 0.229, or choose = 0.22 m To check, note that

3

22 2

3

320 1609 0

12n

Ebh

m

A harmonic force may also be represented by sine or a

complex exponential. How does this

change the solution?

2

0 0( ) ( ) sin or ( ) ( ) sin (2.18)nmx t kx t F t x t x t f t

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 n

n n n n

v f fx t x t t t

Section 2.2 Harmonic Excitation of Damped Systems

Extending resonance and response calculation to damped

systems

0

2

0

now includes a phase shift

( ) ( ) ( ) cos

( ) 2 ( ) ( ) cos

( ) cos ( )

n n

p

mx t cx t kx t F t

x t x t x t f t

x t X t

2.2 Harmonic excitation of damped systems

(2.26)

(2.27)

M

x

k c

Displacement

F=F0cosωt

Let xp have the form:

2 2 1

2 2

( ) cos sin

, tan

sin cos

cos sin

p s s

ss s

s

p s s

p s s

x t A t B t

BX A B

A

x A t B t

x A t B t

Note that we are using the rectangular form, but we could use one of the

other forms of the solution.

Substitute into the equations of motion

2 2

0

2 2

2 2

0

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

Write as a matrix equation:

2 2

0

2 2

( ) 2

02 ( )

sn n

sn n

A f

B

Solving for As and Bs:

2 2

0

2 2 2 2

( )

( ) (2 )

ns

n n

fA

0

2 2 2 2

2

( ) (2 )

ns

n n

fB

Substitute the values of As and Bs into xp:

10

2 22 2 2 2

2( ) cos( tan )

( ) (2 )

np

nn n

X

fx t t

Add homogeneous and particular to get total solution:

particular or homogeneous or transient solutionsteady state solution

( ) sin( ) cos( )nt

dx t Ae t X t

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

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

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.

(0) 0.05

sin 0.133cos( 0.013)

(0) 0 0.01 sin

9.999 cos 0.665sin 0.013

x

A

v A

A

Example 2.2.3: 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 amplitude A and phase Φ for forced and unforced case:

Using the equations on slide 23:

0.1

0.133, 0.013

( ) sin(9.99 ) 0.133cos(5 0.013)t

X

x t Ae t t

Differentiating yields:

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

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

2

2 2 20 0

1

(1 ) (2 )

nXXk

F f r r

Proceeding with ignoring the transient

Magnitude:

0

2 2 2 2( ) (2 )n n

fX

(2.39)

Frequency ratio: n

r

Non dimensional

Form:

(2.40)

Phase: 12

2tan1

rr

2 2 2

1

(1 ) (2 )X

r r

0

0.5

1

1.5

2

Magnitude plot

• 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

12

2tan1

rr

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 2 0

0.5

1

1.5

2

2.5

3

3.5

r

Phase (rad)

=0.1

=0.3

=0.5

=1

Fig 2.7

2 2 20

2

peak

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)

Effect of Damping on Peak Value

• 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

0 0.2 0.4 0.6 0.8 0

5

10

15

20

25

30

ζ

0 0.2 0.4 0.6 0.8 0

0.2

0.4

0.6

0.8

1

rpeak

(dB)0F

Xk

Fig 2.9

ζ

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.

2.3.1 Geometric Approach

• Position, velocity and acceleration phase shifted each by

π/2

• Therefore write each as a vector

• Compute X in terms of F0 via vector addition

Re

Im

)cos( qw -tkX

A

B

C

D

E

kX

XcwXm 2w

0F

C

B A

Xmk )( 2w-

Xcw0F

Using vector addition on the diagram:

2 2 2 2 2 2

0

0

2 2 2

( ) ( )

( ) ( )

F k m X c X

FX

k m c

At resonance:

0, 2

FX

c

real part imaginary part

0

harmonic input

cos ( sin )

( ) ( ) ( )

j t

j t

Ae A t A t j

mx t cx t kx t F e

2.3.2 Complex response method

(2.47)

(2.48)

Real part of this complex solution corresponds to the physical

solution

2

0

002

2

the frequency response function

( )

( )

( )( ) ( )

1( )

( ) ( )

j t

p

j t j t

x t Xe

m cj k Xe F e

FX H j F

k m c j

H jk m c j

Choose complex exponential as a solution

(2.49)

(2.50)

(2.51)

(2.52)

Note: These are all complex functions

0

2 2 2

1

2

( )0

2 2 2

( ) ( )

tan

( )( ) ( )

j

j t

p

FX e

k m c

c

k m

Fx t e

k m c

Using complex arithmetic:

(2.53)

(2.54)

(2.55)

Has real part = to previous solution

Comments:

• Label x-axis Re(ejωt

) and y-axis Im(ejωt

) 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

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

2

0 0

2 2

0

0

2 2

( ) ( ) ( ) ( )

Let ( )

j t j t

n

j t

p

j t j t

n

n

mx t kx t F e x t x t f e

x t Xe

Xe f e

fX

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

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

0

( ) ( ( )) ( ) stX s x t x t e dt

L

Take the transform of the equation of motion:

0

2 0

2 2

cos

( ) ( )

mx cx kx F t

F sms cs k X s

s

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”

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

Transfer Function Method

With zero initial conditions:

(2.59)

(2.60)

The transfer

function

Example 2.3.2 Compute forced response of the

suspension system shown using the Laplace transform

Summing moments about the shaft:

0( ) ( ) sinJ t k t aF t

Taking the Laplace transform

2

0 2 2

0 2 2 2

( ) ( )

1( )

Js X s kX s aFs

X s a Fs Js k

Taking the inverse Laplace transform:

1

0 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

Notes on Phase for Homogeneous and Particular

Solutions

Equation (2.37) gives the full solution for a harmonically

driven underdamped SDOF oscillator to be

homogeneous solution particular solution

( ) sin cosnt

dx t Ae t X t

How do we interpret these phase angles?

Why is one added and the other subtracted?

sin dt

1tan o d

o n o

x

v x

0 00 0x v

Non-Zero initial conditions

Zero initial displacement

Zero initial velocity

Phase on Particular Solution

• Simple “atan” gives -π/2 < Φ < π/2

• Four-quadrant “atan2” gives 0 < Φ < π

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.

)( yxk )yxc(

FBD of SDOF Base Excitation

=- ( - )- ( - )=

+ + = + (2.61)

F k x y c x y mx

mx cx kx cy ky

System Sketch System FBD

M

M x(t)

k c

base

y(t)

SDOF Base Excitation (cont)

Assume: ( ) sin( ) and plug into Equation(2.61)

+ + = cos( ) + sin( ) (2.63)

harmonic forcing functions

y t Y t

mx cx kx c Y t kY t

For a car, 2 2 V

The steady-state solution is just the superposition of the two

individual particular solutions (system is linear).

00

2 2+2 + = 2 cos( ) + sin( ) (2.64)

sc ff

n n n nx x x Y t Y t

Particular Solution (sine term)

With a sine for the forcing function,

2

0

0

22 2 2

2 2

0

22 2 2

+2 + = sin

cos sin sin( )

where

2

( ) 2

( )

( ) 2

n n s

ps s s s s

n ss

n n

n ss

n n

x x x f t

x A t B t X t

fA

fB

Use rectangular form to

make it easier to add

the cos term

Particular Solution (cos term)

With a cosine for the forcing function, we showed

2

0

2 2

0

22 2 2

0

22 2 2

+2 + = cos

cos sin cos( )

where

( )

( ) 2

2

( ) 2

n n c

pc c c c c

n cc

n n

n cc

n n

x x x f t

x A t B t X t

fA

fB

Magnitude X/Y

Now add the sin and cos terms to get the magnitude of

the full particular solution

2 2 2 2

0 0

2 22 2 2 2 2 2

2

0 0

2

22 2

(2 )

( ) 2 ( ) 2

where 2 and

1 (2 )if we define this becomes (2.70)

(1 ) 2n

c s nn

n n n n

c n s n

f fX Y

f Y f Y

rr X Y

r r

2

22 2

1 (2 ) (2.71)

(1 ) 2

X r

Y r r

0

2 2

2sin

2

n

n

ft

W

h

e

n

t

h

e

d

r

i

v

e

f

r

e

q

0 0.5 1 1.5 2 2.5 3 -20

-10

0

2 2

2sin

2

n

n

ft

0

10

20

30

40

Frequency ratio r

X/Y (dB)

e is close to

r = 1

• For ζ = 0,

r =1 defines

resonance

• As ζ

grows

resonance

moves r <1,

and X

The relative magnitude plot

of X/Y versus frequency ratio: Called the Displacement Transmissibility

=0.01

=0.1

=0.3

=0.7

Figure 2.13

From the plot of relative Displacement Transmissibility

observe that:

• X/Y is called Displacement Transmissibility Ratio

• Potentially severe amplification at resonance

• Attenuation for r > sqrt(2) Isolation Zone

• If r < sqrt(2) transmissibility decreases with damping ratio

Amplification Zone

• If r >> 1 then transmissibility increases with damping

ratio Xp~2Yζ/r

Next examine the Force Transmitted to the mass as a function

of the frequency ratio

2

2 2

( ) ( )

At steady state, ( ) cos( ),

so =- cos( )

T

T

F k x y c x y mx

x t X t

x X t

F m X kr X

From FBD

m x(t)

FT

c

k

base y(t)

Plot of Force Transmissibility (in dB) versus frequency ratio

0 0.5 1 1.5 2 2.5 3 -20

-10

0

10

20

30

40 3

.5

Frequency ratio r

F/kY (dB)

=0.1

ζ =0.01

ζ =0.1

ζ =0.3 ζ =0.7

Figure 2.14

Figure 2.16 Comparison between force and displacement

transmissibility

Force

Transmissibility

Displacement

Transmissibility

Example 2.4.2: Effect of speed on the amplitude of

car vibration

Figure 2.17

Model the road as a sinusoidal input to base motion of

the car model

Approximation of road surface:

( ) (0.01 m)sin

1 hour 2 rad(km/hr) 0.2909 rad/s

0.006 km 3600 s cycle

(20km/hr) = 5.818 rad/s

b

b

b

y t t

v v

From the data give, determine the frequency and

damping ratio of the car suspension:

4

4

4 10 N/m6.303 rad/s ( 1 Hz)

1007 kg

2000 Ns/m= 0.158

2 2 4 10 N/m 1007 kg

n

k

m

c

km

From the input frequency, input amplitude, natural

frequency and damping ratio use equation (2.70) to

compute the amplitude of the response:

5.818

6.303

br

2

2 2 2

2

2 22

1 (2 )

(1 ) (2 )

1 2(0.158)(0.923)0.01 m 0.0319 m

1 0.923 2 0.158 0.923

rX Y

r r

What happens as the car goes faster? See Table 2.1.

9000.04

2 2 40,000 3000

c

km

Example 2.4.3: Compute the force transmitted to a

machine through base motion at resonance

1/22

2

2

1 (2 )1 4

(2 ) 2

TT

F kYF

kY

From measured excitation Y = 0.001 m:

2 2(40,000 N/m)(0.001 m)1 4 1 4(0.04) 501.6 N

2 2(0.04)T

kYF

From (2.77) at r =1:

From given m, c, and k:

rt

rt

2.5 Rotating

Unbalance

• Gyros

• Cryo-coolers

• Tires

• Washing machines

Machine of total mass m i.e. m0

included in m

e = eccentricity

mo = mass unbalance

=rotation frequency

m0

e

k c

texa

tex

rrrx

rr

sin

sin

2

rt

Rotating Unbalance (cont)

From sophomore dynamics,

tememamR

tememamR

rroroyy

rroroxx

coscos

sinsin

220

220

What force is imparted on the

structure? Note it rotates with x

component:

Rx

m0

e

Ry

2

0 sin( )r rm e t 2

2 2

sin (2.82)

or 2 sin

o r r

on n r r

mx cx kx m e t

mx x x e t

m

Rotating Unbalance (cont)

The problem is now just like any other SDOF system with a

harmonic excitation

Note the influences on the

forcing function (we are assuming that the

mass m is held in place in the y direction as indicated in Figure

2.19)

m

x(t)

k

c

Rotating Unbalance (cont)

• Just another SDOF oscillator with a harmonic forcing

function

• Expressed in terms of frequency ratio r

2

22 2

1

2

( ) sin( ) (2.83)

(2.84)(1 ) 2

2tan (2.85)

1

p r

o

x t X t

m e rX

m r r

r

r

Figure 2.21: Displacement magnitude vs frequency

caused by rotating unbalance

0

1 1 0.1 m 110 10 0.1 m

2 2(0.05) 2

mXe

m e e

0

100.1 m

m X

m

Example 2.5.1:Given the deflection at resonance (0.1m),

ζ = 0.05 and a 10% out of balance, compute e and the amount of added

mass needed to reduce the maximum amplitude to 0.01 m.

Now to compute the added mass, again at resonance;

0

0.01 m10 100 9

0.1 m (0.1)

m m m mm m

m m

At resonance r = 1 and

Use this to find Δm so that X is 0.01:

5

0

1 10 N/m

60 kg

20 kg

0.5 kg

= 0.01

tail

rot

k

m

m

m

Example 2.5.2 Helicopter rotor unbalance

Given Fig 2.22

Fig 2.23

Compute the deflection at

1500 rpm and find the rotor

speed at which the deflection is maximum

Example 2.5.2 Solution

The rotating mass is 20 + 0.5 or 20.5. The stiffness is provided

by the Tail section and the corresponding mass is that

determined in the example of a heavy beam. So the system

natural frequency is

510 N/m53.72rad/s

33 3320.5 60kg

140 140

n

tail

k

m m

The frequency of rotation is

rev min 2 rad1500 rpm = 1500 157 rad/s

min 60 s rev

157 rad/s 2.92

53.96 rad/s

r

r

Now compute the deflection at r = 2.91 and ζ =0.01 using

eq (2.84)

2

0

2 2 2

2

2 22

(1 ) (2 )

0.5 kg 0.15 m 2.92 0.002 m

34.64 kg1 (2.92) 2(0.01)(2.92)

m e rX

m r r

At around r = 1, the max deflection occurs:

rad rev 60 s1 53.72 rad/s = 53.72 515.1 rpm

s 2 rad minrr

At r = 1:

0

eq

0.5 kg 0.15 m1 10.108 m or 10.8 cm

2 34.34 kg 2(0.01)

m eX

m

= - ( - ) - ( - ) =

= - ( ) - ( )

(2.86) and (2.61)

F k x y c x y mx

mx c x y k x y

2.6 Measurement Devices

• A basic transducer

used in vibration

measurement is the

accelerometer.

• This device can be

modeled using the

base equations developed in the

previous section

Figure 2.24

Here, y(t) is the measured

response of the structure

2

2

2 2 2

1

2

Let ( ) ( ) ( ) (2.87) :

( ) ( ) cos (2.88)

(2.90)(1 ) (2 )

and

2 tan (2.91)

1

b b

z t x t y t

mz cz t kz t m Y t

Z r

Y r r

r

r

Base motion applied to measurement devices

Accelerometer

These equations should be familiar

from base motion.

Here they describe measurement!

Magnitude and sensitivity plots for accelerometers.

Effect of damping on

proportionality constant

Fig 2.27

Fig 2.28

Magnitude plot showing

Regions of measurement

In the accel region, output voltage is

nearly proportional to displacement

2.7 Other forms of damping

These various other forms of damping are all nonlinear.

They can be compared to linear damping by the method of

“equivalent viscous damping” discussed next. A numerical

treatment of the exact response is given in section 2.9.

The method of equivalent viscous damping: consists of

comparing the energy dissipated during one cycle of forced

response

Assume a stead state resulting from a harmonic

input and compute the energy dissipated per one cycle

sinssx X t

The energy per cycle for a viscously damped system is

2 / 2 /

2

0 0

(2.99)d

dxE F dx cx dt cx dt

dt

2 /

2 2

0

sin cos

cos

ssx X t x X t

E c X t dt c X

(2.101)

Next compute the energy dissipated per cycle for

Coulomb damping:

2 /

0

/2 3 /2 2

0 /2 3 /2

sgn( )

( cos cos cos ) 4

E mg x xdt mg

mgX udu udu udu mgX

Here we let u = ωt and du =ωdt and split up the integral according to the

sign changes in velocity.

Next compare this energy to that of a viscous system:

2 44eq eq

mgc X mgX c

X

This yields a linear viscous system dissipating the same amount of energy per cycle.

(2.105)

Using the equivalent viscous damping calculations, each

of the systems in Table 2.2 can be approximated by a

linear viscous system

In particular, ceq can be used to derive amplitude expressions.

However, as indicated in Section 2.8 and 2.9 the response can

be simulated numerically to provide more accurate

magnitude and response information.

Hysteresis: an important concept characterizing

damping

• A plot of displacement versus spring/damping force for viscous damping yields a loop

• At the bottom is a stress strain plot for a system with material damping of the hysteretic type

• The enclosed area is equal to the energy lost per cycle

The measured area yields the energy dissipated. For

some materials, called hysteretic this is 2 (2.120)E k X

Here the constant β, a measured quantity is called

the hysteretic damping constant, k is the stiffness

and X is the amplitude.

Comparing this to the viscous energy yields:

eq

kc

Hysteresis gives rise to the concept of complex stiffness

Substitution of the equivalent damping coefficient and using

the complex exponential to describe a harmonic input yields:

0

j tkmx x kx F e

0

complex stiffness

Assuming ( ) and ( )

yields

( ) (1 ) ( )

j t j t

j t

x t Xe x t Xj e

mx t k j x t F e

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

2 2

2 2

2

2

n n

n n

A

0

0

fx

Symbolic Manipulation

1

nA A x

Let

and

What is

This can be solved using Matlab, Mathcad or Mathematica

Symbolic Manipulation

Solve equations (2.34) using Mathcad symbolics :

Enter this .n2

2

...2 n

...2 n

n2

2

1

f0

0

.n2

2

n4

..2 n22

4

...4 2n22f0

....2 n

n4

..2 n22

4

...4 2n22f0

Choose evaluate

under symbolics

to get this

In MATLAB Command Window

>> syms z wn w f0

>> A=[wn^2-w^2 2*z*wn*w;-2*z*wn*w wn^2-w^2];

>> x=[f0 ;0];

>> An=inv(A)*x

An =

[ (wn^2-w^2)/(wn^4-2*wn^2*w^2+w^4+4*z^2*wn^2*w^2)*f0]

[ 2*z*wn*w/(wn^4-2*wn^2*w^2+w^4+4*z^2*wn^2*w^2)*f0]

>> 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 ]

Magnitude plots: Base Excitation

%m-file to plot base excitation to mass vibration

r=linspace(0,3,500); ze=[0.01;0.05;0.1;0.20;0.50]; X=sqrt( ((2*ze*r).^2+1) ./ ( (ones(size(ze))*(1-r.*r).^2) + (2*ze*r).^2) ); figure(1) plot(r,20*log10(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).

0.5 1 1.5 2 2.5 3 -20

-10

0

1

2

3

4

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

20

Frequency ratio r

X/Y (dB)

real part imaginary part

0

harmonic input

cos ( sin )

( ) ( ) ( )

j t

j t

Ae A t A t j

mx t cx t kx t F e

2

0

002

2

the frequency response function

( )

( )

( )( ) ( )

1( )

( ) ( )

j t

p

j t j t

x t Xe

m cj k Xe F e

FX H j F

k m c j

H jk m c j

0

2 2 2

1

2

( )0

2 2 2

( ) ( )

tan

( )( ) ( )

j

j t

p

FX e

k m c

c

k m

Fx t e

k m c

T

h

e

t

r

a

n

s

f

e

r

f

0 00 0x v

sin dt

System Ske

System FBD

=0.05

=0.1

=0.2

=0.5

Force Magnitude plots: Base Excitation

0.5

1

0. 1 1.5 2 2. 3 -20

-10

0

10

20

30

40

Frequency ratio

FT

/kY

=

0

.

0

1

ζ=0.01

ζ=0.05

ζ=0.1

ζ=0.2

ζ=0.5

Numerical Simulation

We can put the forced case:

0

2

0

( ) ( ) ( ) cos

( ) 2 ( ) ( ) cosn n

mx t cx t kx t F t

x t x t x t f t

Into a state space form

1 2

2

2 2 1 0

0

2 cos

0( ) ( ) ( ), ( )

cos

n n

x x

x x x f t

t A t t tf t

x x f f

Numerical Integration

1Euler: ( ) ( ) ( ) ( )i i i it t A t t t t x x x f

>>TSPAN=[0 10]; >>Y0=[0;0]; >>[t,y] =ode45('num_for',TSPAN,Y0); >>plot(t,y(:,1))

Including forcing

function Xdot=num_for(t,X) m=100;k=1000;c=25; ze=c/(2*sqrt(k*m)); wn=sqrt(k/m); w=2.5;F=1000;f=F/m; f=[0 ;f*cos(w*t)]; A=[0 1;-wn*wn -2*ze*wn]; Xdot=A*X+f;

2 4 6 8 10 -5

-4

-3

-2

-1

0

1

2

3

4

2

2 2 20 0

1

(1 ) (2 )

nXXk

F f r r

Displacement (m)

Example 2.8.2: Design damping for an electronics

model

• 100 kg mass, subject to 150cos(5t) N

• Stiffness k=500 N/m, c = 10kg/s

• Usually x0=0.01 m, v0 = 0.5 m/s

• Find a new c such that the max transient value is 0.2 m.

Response of the board is;

transient exceeds design specification value

0 10 20 30 40

-0.4

-0.2

0

0.2

0.4

Time (sec)

Displacement (m)

Figure 2.33

To run this use the following file:

Rerun this code, increasing c each time until a response that

satisfies the design limits results.

Create function to

model forcing

function Xdot=num_for(t,X) m=100;k=500;c=10; ze=c/(2*sqrt(k*m)); wn=sqrt(k/m); w=5;F=150;f=F/m; f=[0 ;f*cos(w*t)]; A=[0 1;-wn*wn -2*ze*wn]; Xdot=A*X+f;

Matlab command

window

>>TSPAN=[0 40]; >> Y0=[0.01;0.5]; >>[t,y] = ode45('num_for',TSPAN,Y0); >> plot(t,y(:,1)) >> xlabel('Time (sec)') >> ylabel('Displacement (m)') >> grid

Solution: code it, plot it and change c until the desired

response bound is obtained.

Meets amplitude limit when c=195kg/s

Machine of total mass

rt

k

rt

10 texa

tex

rrrx

rr

sin

sin

2

20

2

0 sin( )r rm e t

2

2 2

sin (2.82)

or 2 sin

o r r

on n r r

mx cx kx m e t

mx x x e t

m

30 40

-0.1

0

100.1 m

m X

m

0

5

0

1 10 N/m

60 kg

20 kg

0.5 kg

= 0.01

tail

rot

k

m

m

m

0.1

b

a

s

i

c

t

r

a

n

= - ( - ) - ( - ) =

= - ( ) - ( )

(2.86) and (2.61)

F k x y c x y mx

mx c x y k x y

0.2

0.3

M

a

g

n

i

t

u

d

e

p

l

o

t

s

I

n

t

h

e

a

c

c

e

l

r

e

g

Time (sec)

Displacement (m)

Figure 2.34

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

0

1 2

2 1 2 0

( ) ( , ) cos

( ) ( )

( ) ( , ) cos

( ) ( ) ( )

x t f x x f t

x t x t

x t f x x f t

t t

x F x f

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:

2

01 2

1

0( )( ) , ( )

cos( , )

( ) ( ) ( ( )) ( )i i i i

x tt

f tf x x

t t t t t t

F x f

x x F x f

Numerical form

Vector of nonlinear dynamics Input force vector

Euler equation is

964.2n

Cubic nonlinear spring (2.9.1)

tfxxxx nn cos2 032

2 4 6 8 10

-2

-1

0

1

2 2

0

Time (sec)

Displacement (m)

09.1n

Cubic nonlinear spring near resonance

tfxxxx nn cos2 032

09.1n

=0.05

-2

-1

0

1

2

Time (sec)

Displacement (m)